diff --git a/.github/FUNDING.yml b/.github/FUNDING.yml
new file mode 100644
index 000000000..2f943cf35
--- /dev/null
+++ b/.github/FUNDING.yml
@@ -0,0 +1,13 @@
+# These are supported funding model platforms
+
+github: [halfrost]
+patreon: # Replace with a single Patreon username
+open_collective: # Replace with a single Open Collective username
+ko_fi: # Replace with a single Ko-fi username
+tidelift: # Replace with a single Tidelift platform-name/package-name e.g., npm/babel
+community_bridge: # Replace with a single Community Bridge project-name e.g., cloud-foundry
+liberapay: # Replace with a single Liberapay username
+issuehunt: # Replace with a single IssueHunt username
+otechie: # Replace with a single Otechie username
+lfx_crowdfunding: # Replace with a single LFX Crowdfunding project-name e.g., cloud-foundry
+custom: # Replace with up to 4 custom sponsorship URLs e.g., ['link1', 'link2']
diff --git a/.github/workflows/deploy.yml b/.github/workflows/deploy.yml
new file mode 100644
index 000000000..42d1586b1
--- /dev/null
+++ b/.github/workflows/deploy.yml
@@ -0,0 +1,38 @@
+name: Deploy《LeetCode-Cookbook》
+
+on:
+  push:
+    branches:
+      - master  # 只在master上push触发部署
+      - add_hugo
+    paths-ignore:   # 下列文件的变更不触发部署，可以自行添加
+      - README.md
+      - LICENSE
+
+jobs:
+  deploy:
+
+    runs-on: ubuntu-latest   # 使用ubuntu系统镜像运行自动化脚本
+
+    steps:  # 自动化步骤
+    - uses: actions/checkout@v2   # 第一步，下载代码仓库
+
+    - name: Setup Hugo  # 第二步，安装 hugo
+      uses: peaceiris/actions-hugo@v2
+      with:
+        hugo-version: '0.74.3'
+
+    - name: Build  # 第三步，编译 hugo
+      run: |
+        cd ./website
+        hugo -D --minify
+
+    - name: Deploy to Server  # 第四步，rsync推文件
+      uses: appleboy/ssh-action@v0.1.3  # 使用别人包装好的步骤镜像
+      with:
+        host: ${{ secrets.SSH_HOST }}
+        username: ${{ secrets.SSH_USERNAME }}
+        key: ${{ secrets.DEPLOY_KEY }}
+        port: ${{ secrets.SERVER_PORT }}
+        script: |
+          rsync -avz --delete --exclude='*.pyc' ./website/public/ ${{ secrets.SSH_USERNAME }}@${{ secrets.SSH_HOST }}:${{ secrets.SERVER_DESTINATION }}
diff --git a/.gitignore b/.gitignore
new file mode 100644
index 000000000..e1e2e99aa
--- /dev/null
+++ b/.gitignore
@@ -0,0 +1,2 @@
+*.toml
+.idea
diff --git a/.travis.yml b/.travis.yml
new file mode 100644
index 000000000..35679106c
--- /dev/null
+++ b/.travis.yml
@@ -0,0 +1,17 @@
+language: go
+
+go:
+  - master
+
+# whitelist
+branches:
+  only:
+  - master
+
+script:
+  - go get -t -v ./...
+  - go vet ./...
+  - bash ./gotest.sh
+
+after_success:
+  - bash <(curl -s https://codecov.io/bash)
diff --git a/.vscode/launch.json b/.vscode/launch.json
new file mode 100644
index 000000000..3f8bab8b0
--- /dev/null
+++ b/.vscode/launch.json
@@ -0,0 +1,17 @@
+{
+    // 使用 IntelliSense 了解相关属性。 
+    // 悬停以查看现有属性的描述。
+    // 欲了解更多信息，请访问: https://go.microsoft.com/fwlink/?linkid=830387
+    "version": "0.2.0",
+    "configurations": [
+        {
+            "name": "Launch",
+            "type": "go",
+            "request": "launch",
+            "mode": "auto",
+            "program": "${fileDirname}",
+            "env": {},
+            "args": []
+        }
+    ]
+}
\ No newline at end of file
diff --git a/.vscode/settings.json b/.vscode/settings.json
new file mode 100644
index 000000000..eb946948d
--- /dev/null
+++ b/.vscode/settings.json
@@ -0,0 +1,14 @@
+{
+    "go.formatTool": "gofmt",
+    "go.formatFlags": [
+        "-s"
+    ],
+    "go.autocompleteUnimportedPackages": true,
+    "[go]": {
+        "editor.insertSpaces": false,
+        "editor.formatOnSave": true,
+        "editor.codeActionsOnSave": {
+            "source.organizeImports": true
+        }
+    }
+}
\ No newline at end of file
diff --git a/Algorithms/0001. Two Sum/1. Two Sum.go b/Algorithms/0001. Two Sum/1. Two Sum.go
deleted file mode 100644
index dda3d53fd..000000000
--- a/Algorithms/0001. Two Sum/1. Two Sum.go	
+++ /dev/null
@@ -1,13 +0,0 @@
-package leetcode
-
-func twoSum(nums []int, target int) []int {
-	m := make(map[int]int)
-	for i := 0; i < len(nums); i++ {
-		another := target - nums[i]
-		if _, ok := m[another]; ok {
-			return []int{m[another], i}
-		}
-		m[nums[i]] = i
-	}
-	return nil
-}
diff --git a/Algorithms/0001. Two Sum/1. Two Sum_test.go b/Algorithms/0001. Two Sum/1. Two Sum_test.go
deleted file mode 100644
index 3270cc750..000000000
--- a/Algorithms/0001. Two Sum/1. Two Sum_test.go	
+++ /dev/null
@@ -1,34 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-func TestTwoSum(t *testing.T) {
-	tests := [][]int{
-		[]int{3, 2, 4},
-		[]int{3, 2, 4},
-		[]int{0, 8, 7, 3, 3, 4, 2},
-		[]int{0, 1},
-	}
-	targets := []int{
-		6,
-		5,
-		11,
-		1,
-	}
-	results := [][]int{
-		[]int{1, 2},
-		[]int{0, 1},
-		[]int{1, 3},
-		[]int{0, 1},
-	}
-	fmt.Printf("------------------------Leetcode Problem 1------------------------\n")
-	for i := 0; i < len(targets); i++ {
-		fmt.Printf("nums = %v target = %v result = %v\n", tests[i], targets[i], twoSum(tests[i], targets[i]))
-		if ret := twoSum(tests[i], targets[i]); ret[0] != results[i][0] && ret[1] != results[i][1] {
-			t.Fatalf("case %d fails: %v\n", i, ret)
-		}
-	}
-}
diff --git a/Algorithms/0002. Add Two Numbers/2. Add Two Numbers.go b/Algorithms/0002. Add Two Numbers/2. Add Two Numbers.go
deleted file mode 100644
index 0b85531e4..000000000
--- a/Algorithms/0002. Add Two Numbers/2. Add Two Numbers.go	
+++ /dev/null
@@ -1,43 +0,0 @@
-package leetcode
-
-/**
- * Definition for singly-linked list.
- * type ListNode struct {
- *     Val int
- *     Next *ListNode
- * }
- */
-func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
-	if l1 == nil || l2 == nil {
-		return nil
-	}
-	head := &ListNode{Val: 0, Next: nil}
-	current := head
-	carry := 0
-	for l1 != nil || l2 != nil {
-		var x, y int
-		if l1 == nil {
-			x = 0
-		} else {
-			x = l1.Val
-		}
-		if l2 == nil {
-			y = 0
-		} else {
-			y = l2.Val
-		}
-		current.Next = &ListNode{Val: (x + y + carry) % 10, Next: nil}
-		current = current.Next
-		carry = (x + y + carry) / 10
-		if l1 != nil {
-			l1 = l1.Next
-		}
-		if l2 != nil {
-			l2 = l2.Next
-		}
-	}
-	if carry > 0 {
-		current.Next = &ListNode{Val: carry % 10, Next: nil}
-	}
-	return head.Next
-}
diff --git a/Algorithms/0003. Longest Substring Without Repeating Characters/3. Longest Substring Without Repeating Characters.go b/Algorithms/0003. Longest Substring Without Repeating Characters/3. Longest Substring Without Repeating Characters.go
deleted file mode 100644
index 635db849c..000000000
--- a/Algorithms/0003. Longest Substring Without Repeating Characters/3. Longest Substring Without Repeating Characters.go	
+++ /dev/null
@@ -1,22 +0,0 @@
-package leetcode
-
-func lengthOfLongestSubstring(s string) int {
-	if len(s) == 0 {
-		return 0
-	}
-	var freq [256]int
-	result := 0
-	left, right := 0, -1
-
-	for left < len(s) {
-		if right+1 < len(s) && freq[s[right+1]-'a'] == 0 {
-			freq[s[right+1]-'a']++
-			right++
-		} else {
-			freq[s[left]-'a']--
-			left++
-		}
-		result = max(result, right-left+1)
-	}
-	return result
-}
diff --git a/Algorithms/0003. Longest Substring Without Repeating Characters/README.md b/Algorithms/0003. Longest Substring Without Repeating Characters/README.md
deleted file mode 100644
index 02367f3ec..000000000
--- a/Algorithms/0003. Longest Substring Without Repeating Characters/README.md	
+++ /dev/null
@@ -1,56 +0,0 @@
-# [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/)
-
-## 题目
-
-Given a string, find the length of the longest substring without repeating characters.
-
-
-
-Example 1:
-
-```c
-Input: "abcabcbb"
-Output: 3 
-Explanation: The answer is "abc", with the length of 3. 
-```
-
-Example 2:
-
-```c
-Input: "bbbbb"
-Output: 1
-Explanation: The answer is "b", with the length of 1.
-```
-
-Example 3:
-
-```c
-Input: "pwwkew"
-Output: 3
-Explanation: The answer is "wke", with the length of 3. 
-             Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
-```
-
-## 题目大意
-
-
-在一个字符串重寻找没有重复字母的最长子串。
-
-## 解题思路
-
-这一题和第 438 题，第 3 题，第 76 题，第 567 题类似，用的思想都是"滑动窗口"。
-
-滑动窗口的右边界不断的右移，只要没有重复的字符，就不用的向右扩大窗口边界。一旦出现了重复字符，此时先计算一下滑动窗口的大小，记录下来。再需要缩小左边界。直到重复的字符移出了左边界。接着又可以开始移动滑动窗口的右边界。以此类推，不断的刷新记录的窗口大小。最终最大的值就是题目中的所求。
-
-
-
-
-
-
-
-
-
-
-
-
-
diff --git a/Algorithms/0007. Reverse Integer/README.md b/Algorithms/0007. Reverse Integer/README.md
deleted file mode 100755
index 8f89a895c..000000000
--- a/Algorithms/0007. Reverse Integer/README.md	
+++ /dev/null
@@ -1,35 +0,0 @@
-# [7. Reverse Integer](https://leetcode.com/problems/reverse-integer/)
-
-
-## 题目:
-
-Given a 32-bit signed integer, reverse digits of an integer.
-
-**Example 1:**
-
-    Input: 123
-    Output: 321
-
-**Example 2:**
-
-    Input: -123
-    Output: -321
-
-**Example 3:**
-
-    Input: 120
-    Output: 21
-
-**Note:**Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
-
-## 题目大意
-
-给出一个 32 位的有符号整数，你需要将这个整数中每位上的数字进行反转。注意:假设我们的环境只能存储得下 32 位的有符号整数，则其数值范围为 [−2^31,  2^31 − 1]。请根据这个假设，如果反转后整数溢出那么就返回 0。
-
-
-
-## 解题思路
-
-
-- 这一题是简单题，要求反转 10 进制数。类似的题目有第 190 题。
-- 这一题只需要注意一点，反转以后的数字要求在 [−2^31, 2^31 − 1]范围内，超过这个范围的数字都要输出 0 。
diff --git a/Algorithms/0015. 3Sum/15. 3Sum.go b/Algorithms/0015. 3Sum/15. 3Sum.go
deleted file mode 100644
index ea643c8c6..000000000
--- a/Algorithms/0015. 3Sum/15. 3Sum.go	
+++ /dev/null
@@ -1,38 +0,0 @@
-package leetcode
-
-import (
-	"sort"
-)
-
-func threeSum(nums []int) [][]int {
-	res := [][]int{}
-	counter := map[int]int{}
-	for _, value := range nums {
-		counter[value]++
-	}
-
-	uniqNums := []int{}
-	for key := range counter {
-		uniqNums = append(uniqNums, key)
-	}
-	sort.Ints(uniqNums)
-
-	for i := 0; i < len(uniqNums); i++ {
-		if (uniqNums[i]*3 == 0) && counter[uniqNums[i]] >= 3 {
-			res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i]})
-		}
-		for j := i + 1; j < len(uniqNums); j++ {
-			if (uniqNums[i]*2+uniqNums[j] == 0) && counter[uniqNums[i]] > 1 {
-				res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j]})
-			}
-			if (uniqNums[j]*2+uniqNums[i] == 0) && counter[uniqNums[j]] > 1 {
-				res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j]})
-			}
-			c := 0 - uniqNums[i] - uniqNums[j]
-			if c > uniqNums[j] && counter[c] > 0 {
-				res = append(res, []int{uniqNums[i], uniqNums[j], c})
-			}
-		}
-	}
-	return res
-}
diff --git a/Algorithms/0015. 3Sum/15. 3Sum_test.go b/Algorithms/0015. 3Sum/15. 3Sum_test.go
deleted file mode 100644
index d44134115..000000000
--- a/Algorithms/0015. 3Sum/15. 3Sum_test.go	
+++ /dev/null
@@ -1,59 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question15 struct {
-	para15
-	ans15
-}
-
-// para 是参数
-// one 代表第一个参数
-type para15 struct {
-	a []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans15 struct {
-	one [][]int
-}
-
-func Test_Problem15(t *testing.T) {
-
-	qs := []question15{
-
-		question15{
-			para15{[]int{0, 0, 0}},
-			ans15{[][]int{[]int{0, 0, 0}}},
-		},
-
-		question15{
-			para15{[]int{-1, 0, 1, 2, -1, -4}},
-			ans15{[][]int{[]int{-1, 0, 1}, []int{-1, -1, 2}}},
-		},
-
-		question15{
-			para15{[]int{-4, -2, -2, -2, 0, 1, 2, 2, 2, 3, 3, 4, 4, 6, 6}},
-			ans15{[][]int{[]int{-4, -2, 6}, []int{-4, 0, 4}, []int{-4, 1, 3}, []int{-4, 2, 2}, []int{-2, -2, 4}, []int{-2, 0, 2}}},
-		},
-
-		question15{
-			para15{[]int{5, -7, 3, -3, 5, -10, 4, 8, -3, -8, -3, -3, -1, -8, 6, 4, -4, 7, 2, -5, -2, -7, -3, 7, 2, 4, -6, 5}},
-			ans15{[][]int{[]int{-10, 2, 8}, []int{-10, 3, 7}, []int{-10, 4, 6}, []int{-10, 5, 5}, []int{-8, 2, 6}, []int{-8, 3, 5}, []int{-8, 4, 4}, []int{-7, -1, 8},
-				[]int{-7, 2, 5}, []int{-7, 3, 4}, []int{-6, -2, 8}, []int{-6, -1, 7}, []int{-6, 2, 4}, []int{-5, -3, 8}, []int{-5, -2, 7}, []int{-5, -1, 6}, []int{-5, 2, 3},
-				[]int{-4, -3, 7}, []int{-4, -2, 6}, []int{-4, -1, 5}, []int{-4, 2, 2}, []int{-3, -3, 6}, []int{-3, -2, 5}, []int{-3, -1, 4}, []int{-2, -1, 3}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 15------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans15, q.para15
-		fmt.Printf("【input】:%v       【output】:%v\n", p, threeSum(p.a))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0017. Letter Combinations of a Phone Number/17. Letter Combinations of a Phone Number.go b/Algorithms/0017. Letter Combinations of a Phone Number/17. Letter Combinations of a Phone Number.go
deleted file mode 100644
index 4015656fe..000000000
--- a/Algorithms/0017. Letter Combinations of a Phone Number/17. Letter Combinations of a Phone Number.go	
+++ /dev/null
@@ -1,39 +0,0 @@
-package leetcode
-
-var (
-	letterMap = []string{
-		" ",    //0
-		"",     //1
-		"abc",  //2
-		"def",  //3
-		"ghi",  //4
-		"jkl",  //5
-		"mno",  //6
-		"pqrs", //7
-		"tuv",  //8
-		"wxyz", //9
-	}
-	res = []string{}
-)
-
-func letterCombinations(digits string) []string {
-	if digits == "" {
-		return []string{}
-	}
-	res = []string{}
-	findCombination(&digits, 0, "")
-	return res
-}
-
-func findCombination(digits *string, index int, s string) {
-	if index == len(*digits) {
-		res = append(res, s)
-		return
-	}
-	num := (*digits)[index]
-	letter := letterMap[num-'0']
-	for i := 0; i < len(letter); i++ {
-		findCombination(digits, index+1, s+string(letter[i]))
-	}
-	return
-}
diff --git a/Algorithms/0017. Letter Combinations of a Phone Number/README.md b/Algorithms/0017. Letter Combinations of a Phone Number/README.md
deleted file mode 100755
index f60b400a2..000000000
--- a/Algorithms/0017. Letter Combinations of a Phone Number/README.md	
+++ /dev/null
@@ -1,30 +0,0 @@
-# [17. Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number/)
-
-
-## 题目:
-
-Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent.
-
-A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
-
-![](http://upload.wikimedia.org/wikipedia/commons/thumb/7/73/Telephone-keypad2.svg/200px-Telephone-keypad2.svg.png)
-
-**Example:**
-
-```c
-Input: "23"
-Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
-```
-
-**Note:**
-
-Although the above answer is in lexicographical order, your answer could be in any order you want.
-
-## 题目大意
-
-给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。
-
-
-## 解题思路
-
-- DFS 递归深搜即可
\ No newline at end of file
diff --git a/Algorithms/0018. 4Sum/18. 4Sum.go b/Algorithms/0018. 4Sum/18. 4Sum.go
deleted file mode 100644
index 38f8da3d5..000000000
--- a/Algorithms/0018. 4Sum/18. 4Sum.go	
+++ /dev/null
@@ -1,50 +0,0 @@
-package leetcode
-
-import "sort"
-
-func fourSum(nums []int, target int) [][]int {
-	res := [][]int{}
-	counter := map[int]int{}
-	for _, value := range nums {
-		counter[value]++
-	}
-
-	uniqNums := []int{}
-	for key := range counter {
-		uniqNums = append(uniqNums, key)
-	}
-	sort.Ints(uniqNums)
-
-	for i := 0; i < len(uniqNums); i++ {
-		if (uniqNums[i]*4 == target) && counter[uniqNums[i]] >= 4 {
-			res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i], uniqNums[i]})
-		}
-		for j := i + 1; j < len(uniqNums); j++ {
-			if (uniqNums[i]*3+uniqNums[j] == target) && counter[uniqNums[i]] > 2 {
-				res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i], uniqNums[j]})
-			}
-			if (uniqNums[j]*3+uniqNums[i] == target) && counter[uniqNums[j]] > 2 {
-				res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j], uniqNums[j]})
-			}
-			if (uniqNums[j]*2+uniqNums[i]*2 == target) && counter[uniqNums[j]] > 1 && counter[uniqNums[i]] > 1 {
-				res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j], uniqNums[j]})
-			}
-			for k := j + 1; k < len(uniqNums); k++ {
-				if (uniqNums[i]*2+uniqNums[j]+uniqNums[k] == target) && counter[uniqNums[i]] > 1 {
-					res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j], uniqNums[k]})
-				}
-				if (uniqNums[j]*2+uniqNums[i]+uniqNums[k] == target) && counter[uniqNums[j]] > 1 {
-					res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j], uniqNums[k]})
-				}
-				if (uniqNums[k]*2+uniqNums[i]+uniqNums[j] == target) && counter[uniqNums[k]] > 1 {
-					res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[k], uniqNums[k]})
-				}
-				c := target - uniqNums[i] - uniqNums[j] - uniqNums[k]
-				if c > uniqNums[k] && counter[c] > 0 {
-					res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[k], c})
-				}
-			}
-		}
-	}
-	return res
-}
diff --git a/Algorithms/0018. 4Sum/18. 4Sum_test.go b/Algorithms/0018. 4Sum/18. 4Sum_test.go
deleted file mode 100644
index c0d6373a3..000000000
--- a/Algorithms/0018. 4Sum/18. 4Sum_test.go	
+++ /dev/null
@@ -1,63 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question18 struct {
-	para18
-	ans18
-}
-
-// para 是参数
-// one 代表第一个参数
-type para18 struct {
-	a []int
-	t int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans18 struct {
-	one [][]int
-}
-
-func Test_Problem18(t *testing.T) {
-
-	qs := []question18{
-
-		question18{
-			para18{[]int{1, 1, 1, 1}, 4},
-			ans18{[][]int{[]int{1, 1, 1, 1}}},
-		},
-
-		question18{
-			para18{[]int{0, 1, 5, 0, 1, 5, 5, -4}, 11},
-			ans18{[][]int{[]int{-4, 5, 5, 5}, []int{0, 1, 5, 5}}},
-		},
-
-		question18{
-			para18{[]int{1, 0, -1, 0, -2, 2}, 0},
-			ans18{[][]int{[]int{-1, 0, 0, 1}, []int{-2, -1, 1, 2}, []int{-2, 0, 0, 2}}},
-		},
-
-		question18{
-			para18{[]int{1, 0, -1, 0, -2, 2, 0, 0, 0, 0}, 0},
-			ans18{[][]int{[]int{-1, 0, 0, 1}, []int{-2, -1, 1, 2}, []int{-2, 0, 0, 2}, []int{0, 0, 0, 0}}},
-		},
-
-		question18{
-			para18{[]int{1, 0, -1, 0, -2, 2, 0, 0, 0, 0}, 1},
-			ans18{[][]int{[]int{-1, 0, 0, 1}, []int{-2, -1, 1, 2}, []int{-2, 0, 0, 2}, []int{0, 0, 0, 0}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 18------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans18, q.para18
-		fmt.Printf("【input】:%v       【output】:%v\n", p, fourSum(p.a, p.t))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0019. Remove Nth Node From End of List/19. Remove Nth Node From End of List.go b/Algorithms/0019. Remove Nth Node From End of List/19. Remove Nth Node From End of List.go
deleted file mode 100644
index 63a4e285b..000000000
--- a/Algorithms/0019. Remove Nth Node From End of List/19. Remove Nth Node From End of List.go	
+++ /dev/null
@@ -1,67 +0,0 @@
-package leetcode
-
-/**
- * Definition for singly-linked list.
- * type ListNode struct {
- *     Val int
- *     Next *ListNode
- * }
- */
-
-// 解法一
-func removeNthFromEnd(head *ListNode, n int) *ListNode {
-	var fast, slow *ListNode
-	fast = head
-	slow = head
-	for i := 0; i < n; i++ {
-		fast = fast.Next
-	}
-	if fast == nil {
-		head = head.Next
-		return head
-	}
-	for fast.Next != nil {
-		fast = fast.Next
-		slow = slow.Next
-	}
-	slow.Next = slow.Next.Next
-	return head
-}
-
-// 解法二
-func removeNthFromEnd1(head *ListNode, n int) *ListNode {
-	if head == nil {
-		return nil
-	}
-	if n <= 0 {
-		return head
-	}
-	current := head
-	len := 0
-	for current != nil {
-		len++
-		current = current.Next
-	}
-	if n > len {
-		return head
-	}
-	if n == len {
-		current := head
-		head = head.Next
-		current.Next = nil
-		return head
-	}
-	current = head
-	i := 0
-	for current != nil {
-		if i == len-n-1 {
-			deleteNode := current.Next
-			current.Next = current.Next.Next
-			deleteNode.Next = nil
-			break
-		}
-		i++
-		current = current.Next
-	}
-	return head
-}
diff --git a/Algorithms/0019. Remove Nth Node From End of List/19. Remove Nth Node From End of List_test.go b/Algorithms/0019. Remove Nth Node From End of List/19. Remove Nth Node From End of List_test.go
deleted file mode 100644
index b8303f0f9..000000000
--- a/Algorithms/0019. Remove Nth Node From End of List/19. Remove Nth Node From End of List_test.go	
+++ /dev/null
@@ -1,72 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question19 struct {
-	para19
-	ans19
-}
-
-// para 是参数
-// one 代表第一个参数
-type para19 struct {
-	one []int
-	n   int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans19 struct {
-	one []int
-}
-
-func Test_Problem19(t *testing.T) {
-
-	qs := []question19{
-
-		question19{
-			para19{[]int{1, 2, 3, 4, 5}, 1},
-			ans19{[]int{1, 2, 3, 4}},
-		},
-
-		question19{
-			para19{[]int{1, 2, 3, 4, 5}, 2},
-			ans19{[]int{1, 2, 3, 5}},
-		},
-
-		question19{
-			para19{[]int{1, 2, 3, 4, 5}, 3},
-			ans19{[]int{1, 2, 4, 5}},
-		},
-		question19{
-			para19{[]int{1, 2, 3, 4, 5}, 4},
-			ans19{[]int{1, 3, 4, 5}},
-		},
-
-		question19{
-			para19{[]int{1, 2, 3, 4, 5}, 5},
-			ans19{[]int{2, 3, 4, 5}},
-		},
-
-		question19{
-			para19{[]int{1, 2, 3, 4, 5}, 0},
-			ans19{[]int{1, 2, 3, 4, 5}},
-		},
-
-		question19{
-			para19{[]int{1, 2, 3, 4, 5}, 10},
-			ans19{[]int{1, 2, 3, 4, 5}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 19------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans19, q.para19
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(removeNthFromEnd(S2l(p.one), p.n)))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0019. Remove Nth Node From End of List/README.md b/Algorithms/0019. Remove Nth Node From End of List/README.md
deleted file mode 100644
index 807a6cbc1..000000000
--- a/Algorithms/0019. Remove Nth Node From End of List/README.md	
+++ /dev/null
@@ -1,23 +0,0 @@
-# [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)
-
-## 题目
-
-Given a linked list, remove the n-th node from the end of list and return its head.
-
-Example:
-
-```
-Given linked list: 1->2->3->4->5, and n = 2.
-
-After removing the second node from the end, the linked list becomes 1->2->3->5.
-```
-
-## 题目大意
-
-删除链表中倒数第 n 个结点。
-
-## 解题思路
-
-这道题比较简单，先循环一次拿到链表的总长度，然后循环到要删除的结点的前一个结点开始删除操作。需要注意的一个特例是，有可能要删除头结点，要单独处理。
-
-这道题有一种特别简单的解法。设置 2 个指针，一个指针距离前一个指针 n 个距离。同时移动 2 个指针，2 个指针都移动相同的距离。当一个指针移动到了终点，那么前一个指针就是倒数第 n 个节点了。
diff --git a/Algorithms/0023. Merge k Sorted Lists/23. Merge k Sorted Lists_test.go b/Algorithms/0023. Merge k Sorted Lists/23. Merge k Sorted Lists_test.go
deleted file mode 100644
index 9f7d1ad19..000000000
--- a/Algorithms/0023. Merge k Sorted Lists/23. Merge k Sorted Lists_test.go	
+++ /dev/null
@@ -1,102 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question23 struct {
-	para23
-	ans23
-}
-
-// para 是参数
-// one 代表第一个参数
-type para23 struct {
-	one [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans23 struct {
-	one []int
-}
-
-func Test_Problem23(t *testing.T) {
-
-	qs := []question23{
-
-		question23{
-			para23{[][]int{}},
-			ans23{[]int{}},
-		},
-
-		question23{
-			para23{[][]int{
-				[]int{1},
-				[]int{1},
-			}},
-			ans23{[]int{1, 1}},
-		},
-
-		question23{
-			para23{[][]int{
-				[]int{1, 2, 3, 4},
-				[]int{1, 2, 3, 4},
-			}},
-			ans23{[]int{1, 1, 2, 2, 3, 3, 4, 4}},
-		},
-
-		question23{
-			para23{[][]int{
-				[]int{1, 2, 3, 4, 5},
-				[]int{1, 2, 3, 4, 5},
-			}},
-			ans23{[]int{1, 1, 2, 2, 3, 3, 4, 4, 5, 5}},
-		},
-
-		question23{
-			para23{[][]int{
-				[]int{1},
-				[]int{9, 9, 9, 9, 9},
-			}},
-			ans23{[]int{1, 9, 9, 9, 9, 9}},
-		},
-
-		question23{
-			para23{[][]int{
-				[]int{9, 9, 9, 9, 9},
-				[]int{1},
-			}},
-			ans23{[]int{1, 9, 9, 9, 9, 9}},
-		},
-
-		question23{
-			para23{[][]int{
-				[]int{2, 3, 4},
-				[]int{4, 5, 6},
-			}},
-			ans23{[]int{2, 3, 4, 4, 5, 6}},
-		},
-
-		question23{
-			para23{[][]int{
-				[]int{1, 3, 8},
-				[]int{1, 7},
-			}},
-			ans23{[]int{1, 1, 3, 7, 8}},
-		},
-		// 如需多个测试，可以复制上方元素。
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 23------------------------\n")
-
-	for _, q := range qs {
-		var ls []*ListNode
-		for _, qq := range q.para23.one {
-			ls = append(ls, S2l(qq))
-		}
-		fmt.Printf("【input】:%v       【output】:%v\n", q.para23.one, L2s(mergeKLists(ls)))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0024. Swap Nodes in Pairs/24. Swap Nodes in Pairs.go b/Algorithms/0024. Swap Nodes in Pairs/24. Swap Nodes in Pairs.go
deleted file mode 100644
index 467adb751..000000000
--- a/Algorithms/0024. Swap Nodes in Pairs/24. Swap Nodes in Pairs.go	
+++ /dev/null
@@ -1,38 +0,0 @@
-package leetcode
-
-/**
- * Definition for singly-linked list.
- * type ListNode struct {
- *     Val int
- *     Next *ListNode
- * }
- */
-
-func swapPairs(head *ListNode) *ListNode {
-	if head == nil || head.Next == nil {
-		return head
-	}
-	s := head.Next
-	var behind *ListNode
-	for head.Next != nil {
-		headNext := head.Next
-		if behind != nil && behind.Next != nil {
-			behind.Next = headNext
-		}
-		var next *ListNode
-		if head.Next.Next != nil {
-			next = head.Next.Next
-		}
-		if head.Next.Next != nil {
-			head.Next = next
-		} else {
-			head.Next = nil
-		}
-		headNext.Next = head
-		behind = head
-		if head.Next != nil {
-			head = next
-		}
-	}
-	return s
-}
diff --git a/Algorithms/0024. Swap Nodes in Pairs/24. Swap Nodes in Pairs_test.go b/Algorithms/0024. Swap Nodes in Pairs/24. Swap Nodes in Pairs_test.go
deleted file mode 100644
index 5cf86fcfa..000000000
--- a/Algorithms/0024. Swap Nodes in Pairs/24. Swap Nodes in Pairs_test.go	
+++ /dev/null
@@ -1,91 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question24 struct {
-	para24
-	ans24
-}
-
-// para 是参数
-// one 代表第一个参数
-type para24 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans24 struct {
-	one []int
-}
-
-func Test_Problem24(t *testing.T) {
-
-	qs := []question24{
-
-		question24{
-			para24{[]int{}},
-			ans24{[]int{}},
-		},
-
-		question24{
-			para24{[]int{1}},
-			ans24{[]int{1}},
-		},
-
-		question24{
-			para24{[]int{1, 2, 3, 4}},
-			ans24{[]int{2, 1, 4, 3}},
-		},
-
-		question24{
-			para24{[]int{1, 2, 3, 4, 5}},
-			ans24{[]int{2, 1, 4, 3, 5}},
-		},
-
-		// 如需多个测试，可以复制上方元素。
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 24------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans24, q.para24
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(swapPairs(S2l(p.one))))
-	}
-	fmt.Printf("\n\n\n")
-}
-
-// convert *ListNode to []int
-func L2s(head *ListNode) []int {
-	res := []int{}
-
-	for head != nil {
-		res = append(res, head.Val)
-		head = head.Next
-	}
-
-	return res
-}
-
-// convert []int to *ListNode
-func S2l(nums []int) *ListNode {
-	if len(nums) == 0 {
-		return nil
-	}
-
-	res := &ListNode{
-		Val: nums[0],
-	}
-	temp := res
-	for i := 1; i < len(nums); i++ {
-		temp.Next = &ListNode{
-			Val: nums[i],
-		}
-		temp = temp.Next
-	}
-
-	return res
-}
diff --git a/Algorithms/0028. Implement strStr()/28. Implement strStr()_test.go b/Algorithms/0028. Implement strStr()/28. Implement strStr()_test.go
deleted file mode 100644
index e0e928ffb..000000000
--- a/Algorithms/0028. Implement strStr()/28. Implement strStr()_test.go	
+++ /dev/null
@@ -1,68 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question28 struct {
-	para28
-	ans28
-}
-
-// para 是参数
-// one 代表第一个参数
-type para28 struct {
-	s string
-	p string
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans28 struct {
-	one int
-}
-
-func Test_Problem28(t *testing.T) {
-
-	qs := []question28{
-
-		question28{
-			para28{"abab", "ab"},
-			ans28{0},
-		},
-
-		question28{
-			para28{"hello", "ll"},
-			ans28{2},
-		},
-
-		question28{
-			para28{"", "abc"},
-			ans28{0},
-		},
-
-		question28{
-			para28{"abacbabc", "abc"},
-			ans28{5},
-		},
-
-		question28{
-			para28{"abacbabc", "abcd"},
-			ans28{-1},
-		},
-
-		question28{
-			para28{"abacbabc", ""},
-			ans28{0},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 28------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans28, q.para28
-		fmt.Printf("【input】:%v       【output】:%v\n", p, strStr(p.s, p.p))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0028. Implement strStr()/README.md b/Algorithms/0028. Implement strStr()/README.md
deleted file mode 100644
index 5fb575296..000000000
--- a/Algorithms/0028. Implement strStr()/README.md	
+++ /dev/null
@@ -1,50 +0,0 @@
-# [28. Implement strStr()](https://leetcode.com/problems/implement-strstr/)
-
-## 题目
-
-Implement strStr().
-
-Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
-
-
-Example 1:
-
-```c
-Input: haystack = "hello", needle = "ll"
-Output: 2
-```
-
-Example 2:
-
-```c
-Input: haystack = "aaaaa", needle = "bba"
-Output: -1
-```
-
-Clarification:  
-
-What should we return when needle is an empty string? This is a great question to ask during an interview.
-
-For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().
-
-## 题目大意
-
-
-实现一个查找 substring 的函数。如果在母串中找到了子串，返回子串在母串中出现的下标，如果没有找到，返回 -1，如果子串是空串，则返回 0 。
-
-## 解题思路
-
-这一题比较简单，直接写即可。
-
-
-
-
-
-
-
-
-
-
-
-
-
diff --git a/Algorithms/0029. Divide Two Integers/README.md b/Algorithms/0029. Divide Two Integers/README.md
deleted file mode 100755
index 5fee84b14..000000000
--- a/Algorithms/0029. Divide Two Integers/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# [29. Divide Two Integers](https://leetcode.com/problems/divide-two-integers/)
-
-
-## 题目:
-
-Given two integers `dividend` and `divisor`, divide two integers without using multiplication, division and mod operator.
-
-Return the quotient after dividing `dividend` by `divisor`.
-
-The integer division should truncate toward zero.
-
-**Example 1:**
-
-    Input: dividend = 10, divisor = 3
-    Output: 3
-
-**Example 2:**
-
-    Input: dividend = 7, divisor = -3
-    Output: -2
-
-**Note:**
-
-- Both dividend and divisor will be 32-bit signed integers.
-- The divisor will never be 0.
-- Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 2^31 − 1 when the division result overflows.
-
-
-## 题目大意
-
-给定两个整数，被除数 dividend 和除数 divisor。将两数相除，要求不使用乘法、除法和 mod 运算符。返回被除数 dividend 除以除数 divisor 得到的商。
-
-说明:
-
-- 被除数和除数均为 32 位有符号整数。
-- 除数不为 0。
-- 假设我们的环境只能存储 32 位有符号整数，其数值范围是 [−2^31,  2^31 − 1]。本题中，如果除法结果溢出，则返回 2^31 − 1。
-
-
-## 解题思路
-
-- 给出除数和被除数，要求计算除法运算以后的商。注意值的取值范围在 [−2^31, 2^31 − 1] 之中。超过范围的都按边界计算。
-- 这一题可以用二分搜索来做。要求除法运算之后的商，把商作为要搜索的目标。商的取值范围是 [0, dividend]，所以从 0 到被除数之间搜索。利用二分，找到(商 + 1 ) * 除数 > 被除数并且 商 * 除数 ≤ 被除数 或者 (商+1)* 除数 ≥ 被除数并且商 * 除数 < 被除数的时候，就算找到了商，其余情况继续二分即可。最后还要注意符号和题目规定的 Int32 取值范围。
-- 二分的写法常写错的 3 点：
-    1. low ≤ high (注意二分循环退出的条件是小于等于)
-    2. mid = low + (high-low)>>1 (防止溢出)
-    3. low = mid + 1 ; high = mid - 1 (注意更新 low 和 high 的值，如果更新不对就会死循环)
diff --git a/Algorithms/0033. Search in Rotated Sorted Array/README.md b/Algorithms/0033. Search in Rotated Sorted Array/README.md
deleted file mode 100755
index 6f69b7437..000000000
--- a/Algorithms/0033. Search in Rotated Sorted Array/README.md	
+++ /dev/null
@@ -1,36 +0,0 @@
-# [33. Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)
-
-## 题目:
-
-Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
-
-(i.e., `[0,1,2,4,5,6,7]` might become `[4,5,6,7,0,1,2]`).
-
-You are given a target value to search. If found in the array return its index, otherwise return `-1`.
-
-You may assume no duplicate exists in the array.
-
-Your algorithm's runtime complexity must be in the order of *O*(log *n*).
-
-**Example 1:**
-
-    Input: nums = [4,5,6,7,0,1,2], target = 0
-    Output: 4
-
-**Example 2:**
-
-    Input: nums = [4,5,6,7,0,1,2], target = 3
-    Output: -1
-
-
-## 题目大意
-
-假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。你可以假设数组中不存在重复的元素。
-
-你的算法时间复杂度必须是 O(log n) 级别。
-
-
-## 解题思路
-
-- 给出一个数组，数组中本来是从小到大排列的，并且数组中没有重复数字。但是现在把后面随机一段有序的放到数组前面，这样形成了前后两端有序的子序列。在这样的一个数组里面查找一个数，设计一个 O(log n) 的算法。如果找到就输出数组的小标，如果没有找到，就输出 -1 。
-- 由于数组基本有序，虽然中间有一个“断开点”，还是可以使用二分搜索的算法来实现。现在数组前面一段是数值比较大的数，后面一段是数值偏小的数。如果 mid 落在了前一段数值比较大的区间内了，那么一定有 `nums[mid] > nums[low]`，如果是落在后面一段数值比较小的区间内，`nums[mid] ≤ nums[low]` 。如果 mid 落在了后一段数值比较小的区间内了，那么一定有 `nums[mid] < nums[high]`，如果是落在前面一段数值比较大的区间内，`nums[mid] ≤ nums[high]` 。还有 `nums[low] == nums[mid]` 和 `nums[high] == nums[mid]` 的情况，单独处理即可。最后找到则输出 mid，没有找到则输出 -1 。
diff --git a/Algorithms/0034. Find First and Last Position of Element in Sorted Array/README.md b/Algorithms/0034. Find First and Last Position of Element in Sorted Array/README.md
deleted file mode 100755
index 831b0756b..000000000
--- a/Algorithms/0034. Find First and Last Position of Element in Sorted Array/README.md	
+++ /dev/null
@@ -1,36 +0,0 @@
-# [34. Find First and Last Position of Element in Sorted Array](https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/)
-
-
-## 题目:
-
-Given an array of integers `nums` sorted in ascending order, find the starting and ending position of a given `target` value.
-
-Your algorithm's runtime complexity must be in the order of *O*(log *n*).
-
-If the target is not found in the array, return `[-1, -1]`.
-
-**Example 1:**
-
-    Input: nums = [5,7,7,8,8,10], target = 8
-    Output: [3,4]
-
-**Example 2:**
-
-    Input: nums = [5,7,7,8,8,10], target = 6
-    Output: [-1,-1]
-
-## 题目大意
-
-给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。你的算法时间复杂度必须是 O(log n) 级别。如果数组中不存在目标值，返回 [-1, -1]。
-
-
-## 解题思路
-
-- 给出一个有序数组 `nums` 和一个数 `target`，要求在数组中找到第一个和这个元素相等的元素下标，最后一个和这个元素相等的元素下标。
-- 这一题是经典的二分搜索变种题。二分搜索有 4 大基础变种题：
-    1. 查找第一个值等于给定值的元素
-    2. 查找最后一个值等于给定值的元素
-    3. 查找第一个大于等于给定值的元素
-    4. 查找最后一个小于等于给定值的元素
-
-    这一题的解题思路可以分别利用变种 1 和变种 2 的解法就可以做出此题。或者用一次变种 1 的方法，然后循环往后找到最后一个与给定值相等的元素。不过后者这种方法可能会使时间复杂度下降到 O(n)，因为有可能数组中 n 个元素都和给定元素相同。(4 大基础变种的实现见代码)
diff --git a/Algorithms/0035. Search Insert Position/README.md b/Algorithms/0035. Search Insert Position/README.md
deleted file mode 100755
index cc592b663..000000000
--- a/Algorithms/0035. Search Insert Position/README.md	
+++ /dev/null
@@ -1,40 +0,0 @@
-# [35. Search Insert Position](https://leetcode.com/problems/search-insert-position/)
-
-
-## 题目:
-
-Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
-
-You may assume no duplicates in the array.
-
-**Example 1:**
-
-    Input: [1,3,5,6], 5
-    Output: 2
-
-**Example 2:**
-
-    Input: [1,3,5,6], 2
-    Output: 1
-
-**Example 3:**
-
-    Input: [1,3,5,6], 7
-    Output: 4
-
-**Example 4:**
-
-    Input: [1,3,5,6], 0
-    Output: 0
-
-
-## 题目大意
-
-给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。
-
-你可以假设数组中无重复元素。
-
-## 解题思路
-
-- 给出一个已经从小到大排序后的数组，要求在数组中找到插入 target 元素的位置。
-- 这一题是经典的二分搜索的变种题，在有序数组中找到最后一个比 target 小的元素。
diff --git a/Algorithms/0036. Valid Sudoku/36. Valid Sudoku_test.go b/Algorithms/0036. Valid Sudoku/36. Valid Sudoku_test.go
deleted file mode 100644
index 0d4ad83b6..000000000
--- a/Algorithms/0036. Valid Sudoku/36. Valid Sudoku_test.go	
+++ /dev/null
@@ -1,79 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question36 struct {
-	para36
-	ans36
-}
-
-// para 是参数
-// one 代表第一个参数
-type para36 struct {
-	s [][]byte
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans36 struct {
-	s bool
-}
-
-func Test_Problem36(t *testing.T) {
-
-	qs := []question36{
-
-		question36{
-			para36{[][]byte{
-				[]byte{'5', '3', '.', '.', '7', '.', '.', '.', '.'},
-				[]byte{'6', '.', '.', '1', '9', '5', '.', '.', '.'},
-				[]byte{'.', '9', '8', '.', '.', '.', '.', '6', '.'},
-				[]byte{'8', '.', '.', '.', '6', '.', '.', '.', '3'},
-				[]byte{'4', '.', '.', '8', '.', '3', '.', '.', '1'},
-				[]byte{'7', '.', '.', '.', '2', '.', '.', '.', '6'},
-				[]byte{'.', '6', '.', '.', '.', '.', '2', '8', '.'},
-				[]byte{'.', '.', '.', '4', '1', '9', '.', '.', '5'},
-				[]byte{'.', '.', '.', '.', '8', '.', '.', '7', '9'}}},
-			ans36{true},
-		},
-
-		question36{
-			para36{[][]byte{
-				[]byte{'8', '3', '.', '.', '7', '.', '.', '.', '.'},
-				[]byte{'6', '.', '.', '1', '9', '5', '.', '.', '.'},
-				[]byte{'.', '9', '8', '.', '.', '.', '.', '6', '.'},
-				[]byte{'8', '.', '.', '.', '6', '.', '.', '.', '3'},
-				[]byte{'4', '.', '.', '8', '.', '3', '.', '.', '1'},
-				[]byte{'7', '.', '.', '.', '2', '.', '.', '.', '6'},
-				[]byte{'.', '6', '.', '.', '.', '.', '2', '8', '.'},
-				[]byte{'.', '.', '.', '4', '1', '9', '.', '.', '5'},
-				[]byte{'.', '.', '.', '.', '8', '.', '.', '7', '9'}}},
-			ans36{false},
-		},
-
-		question36{
-			para36{[][]byte{
-				[]byte{'.', '8', '7', '6', '5', '4', '3', '2', '1'},
-				[]byte{'2', '.', '.', '.', '.', '.', '.', '.', '.'},
-				[]byte{'3', '.', '.', '.', '.', '.', '.', '.', '.'},
-				[]byte{'4', '.', '.', '.', '.', '.', '.', '.', '.'},
-				[]byte{'5', '.', '.', '.', '.', '.', '.', '.', '.'},
-				[]byte{'6', '.', '.', '.', '.', '.', '.', '.', '.'},
-				[]byte{'7', '.', '.', '.', '.', '.', '.', '.', '.'},
-				[]byte{'8', '.', '.', '.', '.', '.', '.', '.', '.'},
-				[]byte{'9', '.', '.', '.', '.', '.', '.', '.', '.'}}},
-			ans36{true},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 36------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans36, q.para36
-		fmt.Printf("【input】:%v       【output】:%v\n", p, isValidSudoku1(p.s))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0036. Valid Sudoku/README.md b/Algorithms/0036. Valid Sudoku/README.md
deleted file mode 100755
index f839190c3..000000000
--- a/Algorithms/0036. Valid Sudoku/README.md	
+++ /dev/null
@@ -1,75 +0,0 @@
-# [36. Valid Sudoku](https://leetcode.com/problems/valid-sudoku/)
-
-
-## 题目:
-
-Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
-
-1. Each row must contain the digits `1-9` without repetition.
-2. Each column must contain the digits `1-9` without repetition.
-3. Each of the 9 `3x3` sub-boxes of the grid must contain the digits `1-9` without repetition.
-
-![](https://upload.wikimedia.org/wikipedia/commons/thumb/f/ff/Sudoku-by-L2G-20050714.svg/250px-Sudoku-by-L2G-20050714.svg.png)
-
-A partially filled sudoku which is valid.
-
-The Sudoku board could be partially filled, where empty cells are filled with the character `'.'`.
-
-**Example 1:**
-
-
-    Input:
-    [
-      ["5","3",".",".","7",".",".",".","."],
-      ["6",".",".","1","9","5",".",".","."],
-      [".","9","8",".",".",".",".","6","."],
-      ["8",".",".",".","6",".",".",".","3"],
-      ["4",".",".","8",".","3",".",".","1"],
-      ["7",".",".",".","2",".",".",".","6"],
-      [".","6",".",".",".",".","2","8","."],
-      [".",".",".","4","1","9",".",".","5"],
-      [".",".",".",".","8",".",".","7","9"]
-    ]
-    Output: true
-
-
-**Example 2:**
-
-
-    Input:
-    [
-      ["8","3",".",".","7",".",".",".","."],
-      ["6",".",".","1","9","5",".",".","."],
-      [".","9","8",".",".",".",".","6","."],
-      ["8",".",".",".","6",".",".",".","3"],
-      ["4",".",".","8",".","3",".",".","1"],
-      ["7",".",".",".","2",".",".",".","6"],
-      [".","6",".",".",".",".","2","8","."],
-      [".",".",".","4","1","9",".",".","5"],
-      [".",".",".",".","8",".",".","7","9"]
-    ]
-    Output: false
-    Explanation: Same as Example 1, except with the 5 in the top left corner being 
-        modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.
-
-
-**Note:**
-
-- A Sudoku board (partially filled) could be valid but is not necessarily solvable.
-- Only the filled cells need to be validated according to the mentioned rules.
-- The given board contain only digits `1-9` and the character `'.'`.
-- The given board size is always `9x9`.
-
-## 题目大意
-
-判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。
-
-1. 数字 1-9 在每一行只能出现一次。
-2. 数字 1-9 在每一列只能出现一次。
-3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
-
-
-## 解题思路
-
-- 给出一个数独的棋盘，要求判断这个棋盘当前是否满足数独的要求：即行列是否都只包含 1-9，每个九宫格里面是否也只包含 1-9 。
-- 注意这题和第 37 题是不同的，这一题是判断当前棋盘状态是否满足数独的要求，而第 37 题是要求求解数独。本题中的棋盘有些是无解的，但是棋盘状态是满足题意的。
diff --git a/Algorithms/0037. Sudoku Solver/37. Sudoku Solver_test.go b/Algorithms/0037. Sudoku Solver/37. Sudoku Solver_test.go
deleted file mode 100644
index b6b71f81b..000000000
--- a/Algorithms/0037. Sudoku Solver/37. Sudoku Solver_test.go	
+++ /dev/null
@@ -1,62 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question37 struct {
-	para37
-	ans37
-}
-
-// para 是参数
-// one 代表第一个参数
-type para37 struct {
-	s [][]byte
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans37 struct {
-	s [][]byte
-}
-
-func Test_Problem37(t *testing.T) {
-
-	qs := []question37{
-
-		question37{
-			para37{[][]byte{
-				[]byte{'5', '3', '.', '.', '7', '.', '.', '.', '.'},
-				[]byte{'6', '.', '.', '1', '9', '5', '.', '.', '.'},
-				[]byte{'.', '9', '8', '.', '.', '.', '.', '6', '.'},
-				[]byte{'8', '.', '.', '.', '6', '.', '.', '.', '3'},
-				[]byte{'4', '.', '.', '8', '.', '3', '.', '.', '1'},
-				[]byte{'7', '.', '.', '.', '2', '.', '.', '.', '6'},
-				[]byte{'.', '6', '.', '.', '.', '.', '2', '8', '.'},
-				[]byte{'.', '.', '.', '4', '1', '9', '.', '.', '5'},
-				[]byte{'.', '.', '.', '.', '8', '.', '.', '7', '9'}}},
-			ans37{[][]byte{
-				[]byte{'5', '3', '4', '6', '7', '8', '9', '1', '2'},
-				[]byte{'6', '7', '2', '1', '9', '5', '3', '4', '8'},
-				[]byte{'1', '9', '8', '3', '4', '2', '5', '6', '7'},
-				[]byte{'8', '5', '9', '7', '6', '1', '4', '2', '3'},
-				[]byte{'4', '2', '6', '8', '5', '3', '7', '9', '1'},
-				[]byte{'7', '1', '3', '9', '2', '4', '8', '5', '6'},
-				[]byte{'9', '6', '1', '5', '3', '7', '2', '8', '4'},
-				[]byte{'2', '8', '7', '4', '1', '9', '6', '3', '5'},
-				[]byte{'3', '4', '5', '2', '8', '6', '1', '7', '9'}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 37------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans37, q.para37
-		fmt.Printf("【input】:%v \n\n", p)
-		solveSudoku(p.s)
-		fmt.Printf("【output】:%v \n\n", p)
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0037. Sudoku Solver/README.md b/Algorithms/0037. Sudoku Solver/README.md
deleted file mode 100755
index 6bc84c788..000000000
--- a/Algorithms/0037. Sudoku Solver/README.md	
+++ /dev/null
@@ -1,48 +0,0 @@
-# [37. Sudoku Solver](https://leetcode.com/problems/sudoku-solver/)
-
-
-
-## 题目:
-
-Write a program to solve a Sudoku puzzle by filling the empty cells.
-
-A sudoku solution must satisfy **all of the following rules**:
-
-1. Each of the digits `1-9` must occur exactly once in each row.
-2. Each of the digits `1-9` must occur exactly once in each column.
-3. Each of the the digits `1-9` must occur exactly once in each of the 9 `3x3` sub-boxes of the grid.
-
-Empty cells are indicated by the character `'.'`.
-
-![](https://upload.wikimedia.org/wikipedia/commons/thumb/f/ff/Sudoku-by-L2G-20050714.svg/250px-Sudoku-by-L2G-20050714.svg.png)
-
-A sudoku puzzle...
-
-![](https://upload.wikimedia.org/wikipedia/commons/thumb/3/31/Sudoku-by-L2G-20050714_solution.svg/250px-Sudoku-by-L2G-20050714_solution.svg.png)
-
-...and its solution numbers marked in red.
-
-**Note:**
-
-- The given board contain only digits `1-9` and the character `'.'`.
-- You may assume that the given Sudoku puzzle will have a single unique solution.
-- The given board size is always `9x9`.
-
-## 题目大意
-
-
-编写一个程序，通过已填充的空格来解决数独问题。一个数独的解法需遵循如下规则：
-
-1. 数字 1-9 在每一行只能出现一次。
-2. 数字 1-9 在每一列只能出现一次。
-3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
-
-空白格用 '.' 表示。
-
-
-## 解题思路
-
-- 给出一个数独谜题，要求解出这个数独
-- 解题思路 DFS 暴力回溯枚举。数独要求每横行，每竖行，每九宫格内，`1-9` 的数字不能重复，每次放下一个数字的时候，在这 3 个地方都需要判断一次。
-- 另外找到一组解以后就不需要再继续回溯了，直接返回即可。
-
diff --git a/Algorithms/0040. Combination Sum II/README.md b/Algorithms/0040. Combination Sum II/README.md
deleted file mode 100755
index 5375739bb..000000000
--- a/Algorithms/0040. Combination Sum II/README.md	
+++ /dev/null
@@ -1,48 +0,0 @@
-# [40. Combination Sum II](https://leetcode.com/problems/combination-sum-ii/)
-
-
-## 题目
-
-Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sums to `target`.
-
-Each number in `candidates` may only be used **once** in the combination.
-
-**Note:**
-
-- All numbers (including `target`) will be positive integers.
-- The solution set must not contain duplicate combinations.
-
-**Example 1:**
-
-
-    Input: candidates = [10,1,2,7,6,1,5], target = 8,
-    A solution set is:
-    [
-      [1, 7],
-      [1, 2, 5],
-      [2, 6],
-      [1, 1, 6]
-    ]
-
-
-**Example 2:**
-
-
-    Input: candidates = [2,5,2,1,2], target = 5,
-    A solution set is:
-    [
-      [1,2,2],
-      [5]
-    ]
-
-## 题目大意
-
-给定一个数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
-
-candidates 中的每个数字在每个组合中只能使用一次。
-
-
-## 解题思路
-
-- 题目要求出总和为 sum 的所有组合，组合需要去重。这一题是第 39 题的加强版，第 39 题中元素可以重复利用(重复元素可无限次使用)，这一题中元素只能有限次数的利用，因为存在重复元素，并且每个元素只能用一次(重复元素只能使用有限次)
-- 这一题和第 47 题类似，只不过元素可以反复使用。
diff --git a/Algorithms/0042. Trapping Rain Water/README.md b/Algorithms/0042. Trapping Rain Water/README.md
deleted file mode 100644
index 4b8fbf871..000000000
--- a/Algorithms/0042. Trapping Rain Water/README.md	
+++ /dev/null
@@ -1,29 +0,0 @@
-# [42. Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water/)
-
-## 题目
-
-Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
-
-![](https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png)
-
-
-The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!
-
-
-Example:
-
-```c
-Input: [0,1,0,2,1,0,1,3,2,1,2,1]
-Output: 6
-```
-
-## 题目大意
-
-从 x 轴开始，给出一个数组，数组里面的数字代表从 (0,0) 点开始，宽度为 1 个单位，高度为数组元素的值。如果下雨了，问这样一个容器能装多少单位的水？
-
-## 解题思路
-
-
-![](https://image.ibb.co/d6A2ZU/IMG-0139.jpg)
-
-每个数组里面的元素值可以想象成一个左右都有壁的圆柱筒。例如上图中左边的第二个元素 1，当前左边最大的元素是 2 ，所以 2 高度的水会装到 1 的上面(因为想象成了左右都有筒壁)。这道题的思路就是左指针从 0 开始往右扫，右指针从最右边开始往左扫。额外还需要 2 个变量分别记住左边最大的高度和右边最大高度。遍历扫数组元素的过程中，如果左指针的高度比右指针的高度小，就不断的移动左指针，否则移动右指针。循环的终止条件就是左右指针碰上以后就结束。只要数组中元素的高度比保存的局部最大高度小，就累加 res 的值，否则更新局部最大高度。最终解就是 res 的值。
\ No newline at end of file
diff --git a/Algorithms/0048. Rotate Image/48. Rotate Image.go b/Algorithms/0048. Rotate Image/48. Rotate Image.go
deleted file mode 100644
index a21d5fa80..000000000
--- a/Algorithms/0048. Rotate Image/48. Rotate Image.go	
+++ /dev/null
@@ -1,26 +0,0 @@
-package leetcode
-
-func rotate(matrix [][]int) {
-	row := len(matrix)
-	if row <= 0 {
-		return
-	}
-	column := len(matrix[0])
-	// rotate by diagonal 对角线变换
-	for i := 0; i < row; i++ {
-		for j := i + 1; j < column; j++ {
-			tmp := matrix[i][j]
-			matrix[i][j] = matrix[j][i]
-			matrix[j][i] = tmp
-		}
-	}
-	// rotate by vertical centerline 竖直轴对称翻转
-	halfColumn := column / 2
-	for i := 0; i < row; i++ {
-		for j := 0; j < halfColumn; j++ {
-			tmp := matrix[i][j]
-			matrix[i][j] = matrix[i][column-j-1]
-			matrix[i][column-j-1] = tmp
-		}
-	}
-}
diff --git a/Algorithms/0048. Rotate Image/48. Rotate Image_test.go b/Algorithms/0048. Rotate Image/48. Rotate Image_test.go
deleted file mode 100644
index 2f5278ae4..000000000
--- a/Algorithms/0048. Rotate Image/48. Rotate Image_test.go	
+++ /dev/null
@@ -1,50 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question48 struct {
-	para48
-	ans48
-}
-
-// para 是参数
-// one 代表第一个参数
-type para48 struct {
-	s [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans48 struct {
-	s [][]int
-}
-
-func Test_Problem48(t *testing.T) {
-
-	qs := []question48{
-
-		question48{
-			para48{[][]int{[]int{1, 2, 3}, []int{4, 5, 6}, []int{7, 8, 9}}},
-			ans48{[][]int{[]int{7, 4, 1}, []int{8, 5, 2}, []int{9, 6, 3}}},
-		},
-
-		question48{
-			para48{[][]int{[]int{5, 1, 9, 11}, []int{2, 4, 8, 10}, []int{13, 3, 6, 7}, []int{15, 14, 12, 16}}},
-			ans48{[][]int{[]int{15, 13, 2, 5}, []int{14, 3, 4, 1}, []int{12, 6, 8, 9}, []int{16, 7, 10, 11}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 48------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans48, q.para48
-		fmt.Printf("【input】:%v \n", p)
-		rotate(p.s)
-		fmt.Printf("【output】:%v\n", p)
-		fmt.Printf("\n")
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0048. Rotate Image/README.md b/Algorithms/0048. Rotate Image/README.md
deleted file mode 100755
index d05ee7d3e..000000000
--- a/Algorithms/0048. Rotate Image/README.md	
+++ /dev/null
@@ -1,91 +0,0 @@
-# [48. Rotate Image](https://leetcode.com/problems/rotate-image/)
-
-## 题目
-
-You are given an *n* x *n* 2D matrix representing an image.
-
-Rotate the image by 90 degrees (clockwise).
-
-**Note:**
-
-You have to rotate the image **[in-place](https://en.wikipedia.org/wiki/In-place_algorithm)**, which means you have to modify the input 2D matrix directly. **DO NOT** allocate another 2D matrix and do the rotation.
-
-**Example 1:**
-
-
-    Given input matrix = 
-    [
-      [1,2,3],
-      [4,5,6],
-      [7,8,9]
-    ],
-    
-    rotate the input matrix in-place such that it becomes:
-    [
-      [7,4,1],
-      [8,5,2],
-      [9,6,3]
-    ]
-
-
-**Example 2:**
-
-
-    Given input matrix =
-    [
-      [ 5, 1, 9,11],
-      [ 2, 4, 8,10],
-      [13, 3, 6, 7],
-      [15,14,12,16]
-    ], 
-    
-    rotate the input matrix in-place such that it becomes:
-    [
-      [15,13, 2, 5],
-      [14, 3, 4, 1],
-      [12, 6, 8, 9],
-      [16, 7,10,11]
-    ]
-
-
-## 题目大意
-
-给定一个 n × n 的二维矩阵表示一个图像。将图像顺时针旋转 90 度。说明：你必须在原地旋转图像，这意味着你需要直接修改输入的二维矩阵。请不要使用另一个矩阵来旋转图像。
-
-
-## 解题思路
-
-- 给出一个二维数组，要求顺时针旋转 90 度。
-- 这一题比较简单，按照题意做就可以。这里给出 2 种旋转方法的实现，顺时针旋转和逆时针旋转。
-
-```c
-    /*
-     * clockwise rotate 顺时针旋转
-     * first reverse up to down, then swap the symmetry 
-     * 1 2 3     7 8 9     7 4 1
-     * 4 5 6  => 4 5 6  => 8 5 2
-     * 7 8 9     1 2 3     9 6 3
-    */
-    void rotate(vector<vector<int> > &matrix) {
-        reverse(matrix.begin(), matrix.end());
-        for (int i = 0; i < matrix.size(); ++i) {
-            for (int j = i + 1; j < matrix[i].size(); ++j)
-                swap(matrix[i][j], matrix[j][i]);
-        }
-    }
-    
-    /*
-     * anticlockwise rotate 逆时针旋转
-     * first reverse left to right, then swap the symmetry
-     * 1 2 3     3 2 1     3 6 9
-     * 4 5 6  => 6 5 4  => 2 5 8
-     * 7 8 9     9 8 7     1 4 7
-    */
-    void anti_rotate(vector<vector<int> > &matrix) {
-        for (auto vi : matrix) reverse(vi.begin(), vi.end());
-        for (int i = 0; i < matrix.size(); ++i) {
-            for (int j = i + 1; j < matrix[i].size(); ++j)
-                swap(matrix[i][j], matrix[j][i]);
-        }
-    }
-```
\ No newline at end of file
diff --git a/Algorithms/0051. N-Queens/51. N-Queens.go b/Algorithms/0051. N-Queens/51. N-Queens.go
deleted file mode 100644
index de68f47bd..000000000
--- a/Algorithms/0051. N-Queens/51. N-Queens.go	
+++ /dev/null
@@ -1,85 +0,0 @@
-package leetcode
-
-// 解法一 DFS
-func solveNQueens(n int) [][]string {
-	col, dia1, dia2, row, res := make([]bool, n), make([]bool, 2*n-1), make([]bool, 2*n-1), []int{}, [][]string{}
-	putQueen(n, 0, &col, &dia1, &dia2, &row, &res)
-	return res
-}
-
-// 尝试在一个n皇后问题中, 摆放第index行的皇后位置
-func putQueen(n, index int, col, dia1, dia2 *[]bool, row *[]int, res *[][]string) {
-	if index == n {
-		*res = append(*res, generateBoard(n, row))
-		return
-	}
-	for i := 0; i < n; i++ {
-		// 尝试将第index行的皇后摆放在第i列
-		if !(*col)[i] && !(*dia1)[index+i] && !(*dia2)[index-i+n-1] {
-			*row = append(*row, i)
-			(*col)[i] = true
-			(*dia1)[index+i] = true
-			(*dia2)[index-i+n-1] = true
-			putQueen(n, index+1, col, dia1, dia2, row, res)
-			(*col)[i] = false
-			(*dia1)[index+i] = false
-			(*dia2)[index-i+n-1] = false
-			*row = (*row)[:len(*row)-1]
-		}
-	}
-	return
-}
-
-func generateBoard(n int, row *[]int) []string {
-	board := []string{}
-	res := ""
-	for i := 0; i < n; i++ {
-		res += "."
-	}
-	for i := 0; i < n; i++ {
-		board = append(board, res)
-	}
-	for i := 0; i < n; i++ {
-		tmp := []byte(board[i])
-		tmp[(*row)[i]] = 'Q'
-		board[i] = string(tmp)
-	}
-	return board
-}
-
-// 解法二 二进制操作法
-// class Solution
-// {
-//     int n;
-//     string getNq(int p)
-//     {
-//         string s(n, '.');
-//         s[p] = 'Q';
-//         return s;
-//     }
-//     void nQueens(int p, int l, int m, int r, vector<vector<string>> &res)
-//     {
-//         static vector<string> ans;
-//         if (p >= n)
-//         {
-//             res.push_back(ans);
-//             return ;
-//         }
-//         int mask = l | m | r;
-//         for (int i = 0, b = 1; i < n; ++ i, b <<= 1)
-//             if (!(mask & b))
-//             {
-//                 ans.push_back(getNq(i));
-//                 nQueens(p + 1, (l | b) >> 1, m | b, (r | b) << 1, res);
-//                 ans.pop_back();
-//             }
-//     }
-// public:
-//     vector<vector<string> > solveNQueens(int n)
-//     {
-//         this->n = n;
-//         vector<vector<string>> res;
-//         nQueens(0, 0, 0, 0, res);
-//         return res;
-//     }
-// };
diff --git a/Algorithms/0051. N-Queens/README.md b/Algorithms/0051. N-Queens/README.md
deleted file mode 100755
index c2be733ec..000000000
--- a/Algorithms/0051. N-Queens/README.md	
+++ /dev/null
@@ -1,43 +0,0 @@
-# [51. N-Queens](https://leetcode.com/problems/n-queens/)
-
-
-## 题目
-
-The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
-
-![](https://assets.leetcode.com/uploads/2018/10/12/8-queens.png)
-
-Given an integer *n*, return all distinct solutions to the *n*-queens puzzle.
-
-Each solution contains a distinct board configuration of the *n*-queens' placement, where `'Q'` and `'.'` both indicate a queen and an empty space respectively.
-
-**Example:**
-
-
-    Input: 4
-    Output: [
-     [".Q..",  // Solution 1
-      "...Q",
-      "Q...",
-      "..Q."],
-    
-     ["..Q.",  // Solution 2
-      "Q...",
-      "...Q",
-      ".Q.."]
-    ]
-    Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.
-
-
-## 题目大意
-
-给定一个整数 n，返回所有不同的 n 皇后问题的解决方案。每一种解法包含一个明确的 n 皇后问题的棋子放置方案，该方案中 'Q' 和 '.' 分别代表了皇后和空位。
-
-
-## 解题思路
-
-- 求解 n 皇后问题
-- 利用 col 数组记录列信息，col 有 `n` 列。用 dia1，dia2 记录从左下到右上的对角线，从左上到右下的对角线的信息，dia1 和 dia2 分别都有 `2*n-1` 个。
-- dia1 对角线的规律是 `i + j 是定值`，例如[0,0]，为 0；[1,0]、[0,1] 为 1；[2,0]、[1,1]、[0,2] 为 2；
-- dia2 对角线的规律是 `i - j 是定值`，例如[0,7]，为 -7；[0,6]、[1,7] 为 -6；[0,5]、[1,6]、[2,7] 为 -5；为了使他们从 0 开始，i - j + n - 1 偏移到 0 开始，所以 dia2 的规律是 `i - j + n - 1 为定值`。
-
diff --git a/Algorithms/0054. Spiral Matrix/54. Spiral Matrix.go b/Algorithms/0054. Spiral Matrix/54. Spiral Matrix.go
deleted file mode 100644
index 4829d39dc..000000000
--- a/Algorithms/0054. Spiral Matrix/54. Spiral Matrix.go	
+++ /dev/null
@@ -1,68 +0,0 @@
-package leetcode
-
-func spiralOrder(matrix [][]int) []int {
-	if len(matrix) == 0 {
-		return []int{}
-	}
-	res := []int{}
-	if len(matrix) == 1 {
-		for i := 0; i < len(matrix[0]); i++ {
-			res = append(res, matrix[0][i])
-		}
-		return res
-	}
-	if len(matrix[0]) == 1 {
-		for i := 0; i < len(matrix); i++ {
-			res = append(res, matrix[i][0])
-		}
-		return res
-	}
-	visit, m, n, round, x, y, spDir := make([][]int, len(matrix)), len(matrix), len(matrix[0]), 0, 0, 0, [][]int{
-		[]int{0, 1},  // 朝右
-		[]int{1, 0},  // 朝下
-		[]int{0, -1}, // 朝左
-		[]int{-1, 0}, // 朝上
-	}
-	for i := 0; i < m; i++ {
-		visit[i] = make([]int, n)
-	}
-	visit[x][y] = 1
-	res = append(res, matrix[x][y])
-	for i := 0; i < m*n; i++ {
-		x += spDir[round%4][0]
-		y += spDir[round%4][1]
-		if (x == 0 && y == n-1) || (x == m-1 && y == n-1) || (y == 0 && x == m-1) {
-			round++
-		}
-		if x > m-1 || y > n-1 || x < 0 || y < 0 {
-			return res
-		}
-		if visit[x][y] == 0 {
-			visit[x][y] = 1
-			res = append(res, matrix[x][y])
-		}
-		switch round % 4 {
-		case 0:
-			if y+1 <= n-1 && visit[x][y+1] == 1 {
-				round++
-				continue
-			}
-		case 1:
-			if x+1 <= m-1 && visit[x+1][y] == 1 {
-				round++
-				continue
-			}
-		case 2:
-			if y-1 >= 0 && visit[x][y-1] == 1 {
-				round++
-				continue
-			}
-		case 3:
-			if x-1 >= 0 && visit[x-1][y] == 1 {
-				round++
-				continue
-			}
-		}
-	}
-	return res
-}
diff --git a/Algorithms/0054. Spiral Matrix/54. Spiral Matrix_test.go b/Algorithms/0054. Spiral Matrix/54. Spiral Matrix_test.go
deleted file mode 100644
index e7af57f1c..000000000
--- a/Algorithms/0054. Spiral Matrix/54. Spiral Matrix_test.go	
+++ /dev/null
@@ -1,61 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question54 struct {
-	para54
-	ans54
-}
-
-// para 是参数
-// one 代表第一个参数
-type para54 struct {
-	one [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans54 struct {
-	one []int
-}
-
-func Test_Problem54(t *testing.T) {
-
-	qs := []question54{
-
-		question54{
-			para54{[][]int{[]int{3}, []int{2}}},
-			ans54{[]int{3, 2}},
-		},
-
-		question54{
-			para54{[][]int{[]int{2, 3}}},
-			ans54{[]int{2, 3}},
-		},
-
-		question54{
-			para54{[][]int{[]int{1}}},
-			ans54{[]int{1}},
-		},
-
-		question54{
-			para54{[][]int{[]int{1, 2, 3}, []int{4, 5, 6}, []int{7, 8, 9}}},
-			ans54{[]int{1, 2, 3, 6, 9, 8, 7, 4, 5}},
-		},
-		question54{
-			para54{[][]int{[]int{1, 2, 3, 4}, []int{5, 6, 7, 8}, []int{9, 10, 11, 12}}},
-			ans54{[]int{1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 54------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans54, q.para54
-		fmt.Printf("【input】:%v       【output】:%v\n", p, spiralOrder(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0054. Spiral Matrix/README.md b/Algorithms/0054. Spiral Matrix/README.md
deleted file mode 100755
index 50e663258..000000000
--- a/Algorithms/0054. Spiral Matrix/README.md	
+++ /dev/null
@@ -1,40 +0,0 @@
-# [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)
-
-
-## 题目
-
-Given a matrix of *m* x *n* elements (*m* rows, *n* columns), return all elements of the matrix in spiral order.
-
-**Example 1:**
-
-
-    Input:
-    [
-     [ 1, 2, 3 ],
-     [ 4, 5, 6 ],
-     [ 7, 8, 9 ]
-    ]
-    Output: [1,2,3,6,9,8,7,4,5]
-
-
-**Example 2:**
-
-
-    Input:
-    [
-      [1, 2, 3, 4],
-      [5, 6, 7, 8],
-      [9,10,11,12]
-    ]
-    Output: [1,2,3,4,8,12,11,10,9,5,6,7]
-
-
-## 题目大意
-
-给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。
-
-## 解题思路
-
-- 给出一个二维数组，按照螺旋的方式输出
-- 需要注意的是特殊情况，比如二维数组退化成一维或者一列或者一个元素。注意了这些情况，基本就可以一次通过了。
-
diff --git a/Algorithms/0056. Merge Intervals/56. Merge Intervals_test.go b/Algorithms/0056. Merge Intervals/56. Merge Intervals_test.go
deleted file mode 100644
index 30232c712..000000000
--- a/Algorithms/0056. Merge Intervals/56. Merge Intervals_test.go	
+++ /dev/null
@@ -1,72 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question56 struct {
-	para56
-	ans56
-}
-
-// para 是参数
-// one 代表第一个参数
-type para56 struct {
-	one []Interval
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans56 struct {
-	one []Interval
-}
-
-func Test_Problem56(t *testing.T) {
-
-	qs := []question56{
-
-		question56{
-			para56{[]Interval{}},
-			ans56{[]Interval{}},
-		},
-
-		question56{
-			para56{[]Interval{Interval{Start: 1, End: 3}, Interval{Start: 2, End: 6}, Interval{Start: 8, End: 10}, Interval{Start: 15, End: 18}}},
-			ans56{[]Interval{Interval{Start: 1, End: 6}, Interval{Start: 8, End: 10}, Interval{Start: 15, End: 18}}},
-		},
-
-		question56{
-			para56{[]Interval{Interval{Start: 1, End: 4}, Interval{Start: 4, End: 5}}},
-			ans56{[]Interval{Interval{Start: 1, End: 5}}},
-		},
-
-		question56{
-			para56{[]Interval{Interval{Start: 1, End: 3}, Interval{Start: 3, End: 6}, Interval{Start: 5, End: 10}, Interval{Start: 9, End: 18}}},
-			ans56{[]Interval{Interval{Start: 1, End: 18}}},
-		},
-
-		question56{
-			para56{[]Interval{Interval{Start: 15, End: 18}, Interval{Start: 8, End: 10}, Interval{Start: 2, End: 6}, Interval{Start: 1, End: 3}}},
-			ans56{[]Interval{Interval{Start: 1, End: 6}, Interval{Start: 8, End: 10}, Interval{Start: 15, End: 18}}},
-		},
-
-		question56{
-			para56{[]Interval{Interval{Start: 1, End: 3}, Interval{Start: 2, End: 6}, Interval{Start: 8, End: 10}, Interval{Start: 15, End: 18}}},
-			ans56{[]Interval{Interval{Start: 1, End: 6}, Interval{Start: 8, End: 10}, Interval{Start: 15, End: 18}}},
-		},
-
-		question56{
-			para56{[]Interval{Interval{Start: 1, End: 4}, Interval{Start: 1, End: 5}}},
-			ans56{[]Interval{Interval{Start: 1, End: 5}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 56------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans56, q.para56
-		fmt.Printf("【input】:%v       【output】:%v\n", p, merge56(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0057. Insert Interval/57. Insert Interval_test.go b/Algorithms/0057. Insert Interval/57. Insert Interval_test.go
deleted file mode 100644
index 9ce3a7b37..000000000
--- a/Algorithms/0057. Insert Interval/57. Insert Interval_test.go	
+++ /dev/null
@@ -1,68 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question57 struct {
-	para57
-	ans57
-}
-
-// para 是参数
-// one 代表第一个参数
-type para57 struct {
-	one []Interval
-	two Interval
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans57 struct {
-	one []Interval
-}
-
-func Test_Problem57(t *testing.T) {
-
-	qs := []question57{
-
-		question57{
-			para57{[]Interval{}, Interval{}},
-			ans57{[]Interval{}},
-		},
-
-		question57{
-			para57{[]Interval{Interval{Start: 1, End: 3}, Interval{Start: 6, End: 9}}, Interval{Start: 4, End: 8}},
-			ans57{[]Interval{Interval{Start: 1, End: 5}, Interval{Start: 6, End: 9}}},
-		},
-
-		question57{
-			para57{[]Interval{Interval{Start: 1, End: 3}, Interval{Start: 6, End: 9}}, Interval{Start: 2, End: 5}},
-			ans57{[]Interval{Interval{Start: 1, End: 5}, Interval{Start: 6, End: 9}}},
-		},
-
-		question57{
-			para57{[]Interval{Interval{Start: 1, End: 2}, Interval{Start: 3, End: 5}, Interval{Start: 6, End: 7}, Interval{Start: 8, End: 10}, Interval{Start: 12, End: 16}}, Interval{Start: 4, End: 8}},
-			ans57{[]Interval{Interval{Start: 1, End: 2}, Interval{Start: 3, End: 10}, Interval{Start: 12, End: 16}}},
-		},
-
-		question57{
-			para57{[]Interval{Interval{Start: 1, End: 5}}, Interval{Start: 5, End: 7}},
-			ans57{[]Interval{Interval{Start: 1, End: 7}}},
-		},
-
-		question57{
-			para57{[]Interval{Interval{Start: 1, End: 2}, Interval{Start: 3, End: 5}, Interval{Start: 6, End: 7}, Interval{Start: 8, End: 10}, Interval{Start: 12, End: 16}}, Interval{Start: 9, End: 12}},
-			ans57{[]Interval{Interval{Start: 1, End: 2}, Interval{Start: 3, End: 5}, Interval{Start: 6, End: 7}, Interval{Start: 8, End: 16}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 57------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans57, q.para57
-		fmt.Printf("【input】:%v       【output】:%v\n", p, insert(p.one, p.two))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0062. Unique Paths/62. Unique Paths.go b/Algorithms/0062. Unique Paths/62. Unique Paths.go
deleted file mode 100644
index f4d6b2859..000000000
--- a/Algorithms/0062. Unique Paths/62. Unique Paths.go	
+++ /dev/null
@@ -1,20 +0,0 @@
-package leetcode
-
-func uniquePaths(m int, n int) int {
-	dp := make([][]int, n)
-	for i := 0; i < n; i++ {
-		dp[i] = make([]int, m)
-	}
-	for i := 0; i < m; i++ {
-		dp[0][i] = 1
-	}
-	for i := 0; i < n; i++ {
-		dp[i][0] = 1
-	}
-	for i := 1; i < n; i++ {
-		for j := 1; j < m; j++ {
-			dp[i][j] = dp[i-1][j] + dp[i][j-1]
-		}
-	}
-	return dp[n-1][m-1]
-}
diff --git a/Algorithms/0066. Plus One/66. Plus One.go b/Algorithms/0066. Plus One/66. Plus One.go
deleted file mode 100644
index 1e6572e8d..000000000
--- a/Algorithms/0066. Plus One/66. Plus One.go	
+++ /dev/null
@@ -1,21 +0,0 @@
-package leetcode
-
-func plusOne(digits []int) []int {
-	if len(digits) == 0 {
-		return []int{}
-	}
-	carry := 1
-	for i := len(digits) - 1; i >= 0; i-- {
-		if digits[i]+carry > 9 {
-			digits[i] = 0
-			carry = 1
-		} else {
-			digits[i] += carry
-			carry = 0
-		}
-	}
-	if digits[0] == 0 && carry == 1 {
-		digits = append([]int{1}, digits...)
-	}
-	return digits
-}
diff --git a/Algorithms/0066. Plus One/README.md b/Algorithms/0066. Plus One/README.md
deleted file mode 100755
index 465bdcf1c..000000000
--- a/Algorithms/0066. Plus One/README.md	
+++ /dev/null
@@ -1,35 +0,0 @@
-# [66. Plus One](https://leetcode.com/problems/plus-one/)
-
-
-## 题目:
-
-Given a **non-empty** array of digits representing a non-negative integer, plus one to the integer.
-
-The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.
-
-You may assume the integer does not contain any leading zero, except the number 0 itself.
-
-**Example 1:**
-
-    Input: [1,2,3]
-    Output: [1,2,4]
-    Explanation: The array represents the integer 123.
-
-**Example 2:**
-
-    Input: [4,3,2,1]
-    Output: [4,3,2,2]
-    Explanation: The array represents the integer 4321.
-
-
-## 题目大意
-
-
-给定一个由整数组成的非空数组所表示的非负整数，在该数的基础上加一。最高位数字存放在数组的首位， 数组中每个元素只存储单个数字。你可以假设除了整数 0 之外，这个整数不会以零开头。
-
-
-
-## 解题思路
-
-- 给出一个数组，代表一个十进制数，数组的 0 下标是十进制数的高位。要求计算这个十进制数加一以后的结果。
-- 简单的模拟题。从数组尾部开始往前扫，逐位进位即可。最高位如果还有进位需要在数组里面第 0 位再插入一个 1 。
diff --git a/Algorithms/0074. Search a 2D Matrix/README.md b/Algorithms/0074. Search a 2D Matrix/README.md
deleted file mode 100755
index b81dec7d7..000000000
--- a/Algorithms/0074. Search a 2D Matrix/README.md	
+++ /dev/null
@@ -1,46 +0,0 @@
-# [74. Search a 2D Matrix](https://leetcode.com/problems/search-a-2d-matrix/)
-
-
-## 题目:
-
-Write an efficient algorithm that searches for a value in an *m* x *n* matrix. This matrix has the following properties:
-
-- Integers in each row are sorted from left to right.
-- The first integer of each row is greater than the last integer of the previous row.
-
-**Example 1:**
-
-    Input:
-    matrix = [
-      [1,   3,  5,  7],
-      [10, 11, 16, 20],
-      [23, 30, 34, 50]
-    ]
-    target = 3
-    Output: true
-
-**Example 2:**
-
-    Input:
-    matrix = [
-      [1,   3,  5,  7],
-      [10, 11, 16, 20],
-      [23, 30, 34, 50]
-    ]
-    target = 13
-    Output: false
-
-
-## 题目大意
-
-编写一个高效的算法来判断 m x n 矩阵中，是否存在一个目标值。该矩阵具有如下特性：
-
-- 每行中的整数从左到右按升序排列。
-- 每行的第一个整数大于前一行的最后一个整数。
-
-
-## 解题思路
-
-
-- 给出一个二维矩阵，矩阵的特点是随着矩阵的下标增大而增大。要求设计一个算法能在这个矩阵中高效的找到一个数，如果找到就输出 true，找不到就输出 false。
-- 虽然是一个二维矩阵，但是由于它特殊的有序性，所以完全可以按照下标把它看成一个一维矩阵，只不过需要行列坐标转换。最后利用二分搜索直接搜索即可。
diff --git a/Algorithms/0075. Sort Colors/75. Sort Colors.go b/Algorithms/0075. Sort Colors/75. Sort Colors.go
deleted file mode 100644
index 6440d550c..000000000
--- a/Algorithms/0075. Sort Colors/75. Sort Colors.go	
+++ /dev/null
@@ -1,28 +0,0 @@
-package leetcode
-
-func sortColors(nums []int) {
-	if len(nums) == 0 {
-		return
-	}
-
-	r := 0
-	w := 0
-	b := 0 // label the end of different colors;
-	for _, num := range nums {
-		if num == 0 {
-			nums[b] = 2
-			b++
-			nums[w] = 1
-			w++
-			nums[r] = 0
-			r++
-		} else if num == 1 {
-			nums[b] = 2
-			b++
-			nums[w] = 1
-			w++
-		} else if num == 2 {
-			b++
-		}
-	}
-}
diff --git a/Algorithms/0078. Subsets/78. Subsets.go b/Algorithms/0078. Subsets/78. Subsets.go
deleted file mode 100644
index f69f541fd..000000000
--- a/Algorithms/0078. Subsets/78. Subsets.go	
+++ /dev/null
@@ -1,43 +0,0 @@
-package leetcode
-
-import "sort"
-
-// 解法一
-func subsets(nums []int) [][]int {
-	c, res := []int{}, [][]int{}
-	for k := 0; k <= len(nums); k++ {
-		generateSubsets(nums, k, 0, c, &res)
-	}
-	return res
-}
-
-func generateSubsets(nums []int, k, start int, c []int, res *[][]int) {
-	if len(c) == k {
-		b := make([]int, len(c))
-		copy(b, c)
-		*res = append(*res, b)
-		return
-	}
-	// i will at most be n - (k - c.size()) + 1
-	for i := start; i < len(nums)-(k-len(c))+1; i++ {
-		c = append(c, nums[i])
-		generateSubsets(nums, k, i+1, c, res)
-		c = c[:len(c)-1]
-	}
-	return
-}
-
-// 解法二
-func subsets1(nums []int) [][]int {
-	res := make([][]int, 1)
-	sort.Ints(nums)
-	for i := range nums {
-		for _, org := range res {
-			clone := make([]int, len(org), len(org)+1)
-			copy(clone, org)
-			clone = append(clone, nums[i])
-			res = append(res, clone)
-		}
-	}
-	return res
-}
diff --git a/Algorithms/0079. Word Search/79. Word Search_test.go b/Algorithms/0079. Word Search/79. Word Search_test.go
deleted file mode 100644
index ae47e951e..000000000
--- a/Algorithms/0079. Word Search/79. Word Search_test.go	
+++ /dev/null
@@ -1,105 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question79 struct {
-	para79
-	ans79
-}
-
-// para 是参数
-// one 代表第一个参数
-type para79 struct {
-	b    [][]byte
-	word string
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans79 struct {
-	one bool
-}
-
-func Test_Problem79(t *testing.T) {
-
-	qs := []question79{
-
-		question79{
-			para79{[][]byte{
-				[]byte{'A', 'B', 'C', 'E'},
-				[]byte{'S', 'F', 'C', 'S'},
-				[]byte{'A', 'D', 'E', 'E'},
-			}, "ABCCED"},
-			ans79{true},
-		},
-
-		question79{
-			para79{[][]byte{
-				[]byte{'A', 'B', 'C', 'E'},
-				[]byte{'S', 'F', 'C', 'S'},
-				[]byte{'A', 'D', 'E', 'E'},
-			}, "SEE"},
-			ans79{true},
-		},
-
-		question79{
-			para79{[][]byte{
-				[]byte{'A', 'B', 'C', 'E'},
-				[]byte{'S', 'F', 'C', 'S'},
-				[]byte{'A', 'D', 'E', 'E'},
-			}, "ABCB"},
-			ans79{false},
-		},
-
-		question79{
-			para79{[][]byte{
-				[]byte{'o', 'a', 'a', 'n'},
-				[]byte{'e', 't', 'a', 'e'},
-				[]byte{'i', 'h', 'k', 'r'},
-				[]byte{'i', 'f', 'l', 'v'},
-			}, "oath"},
-			ans79{true},
-		},
-
-		question79{
-			para79{[][]byte{
-				[]byte{'o', 'a', 'a', 'n'},
-				[]byte{'e', 't', 'a', 'e'},
-				[]byte{'i', 'h', 'k', 'r'},
-				[]byte{'i', 'f', 'l', 'v'},
-			}, "pea"},
-			ans79{false},
-		},
-
-		question79{
-			para79{[][]byte{
-				[]byte{'o', 'a', 'a', 'n'},
-				[]byte{'e', 't', 'a', 'e'},
-				[]byte{'i', 'h', 'k', 'r'},
-				[]byte{'i', 'f', 'l', 'v'},
-			}, "eat"},
-			ans79{true},
-		},
-
-		question79{
-			para79{[][]byte{
-				[]byte{'o', 'a', 'a', 'n'},
-				[]byte{'e', 't', 'a', 'e'},
-				[]byte{'i', 'h', 'k', 'r'},
-				[]byte{'i', 'f', 'l', 'v'},
-			}, "rain"},
-			ans79{false},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 79------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans79, q.para79
-		fmt.Printf("【input】:%v       【output】:%v\n\n\n", p, exist(p.b, p.word))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0080. Remove Duplicates from Sorted Array II/80. Remove Duplicates from Sorted Array II.go b/Algorithms/0080. Remove Duplicates from Sorted Array II/80. Remove Duplicates from Sorted Array II.go
deleted file mode 100644
index 4b8227b0c..000000000
--- a/Algorithms/0080. Remove Duplicates from Sorted Array II/80. Remove Duplicates from Sorted Array II.go	
+++ /dev/null
@@ -1,35 +0,0 @@
-package leetcode
-
-func removeDuplicates80(nums []int) int {
-	if len(nums) == 0 {
-		return 0
-	}
-	last, finder, startFinder := 0, 0, -1
-	for last < len(nums)-1 {
-		startFinder = -1
-		for nums[finder] == nums[last] {
-			if startFinder == -1 || startFinder > finder {
-				startFinder = finder
-			}
-			if finder == len(nums)-1 {
-				break
-			}
-			finder++
-		}
-		if finder-startFinder >= 2 && nums[finder-1] == nums[last] && nums[finder] != nums[last] {
-			nums[last+1] = nums[finder-1]
-			nums[last+2] = nums[finder]
-			last += 2
-		} else {
-			nums[last+1] = nums[finder]
-			last++
-		}
-		if finder == len(nums)-1 {
-			if nums[finder] != nums[last-1] {
-				nums[last] = nums[finder]
-			}
-			return last + 1
-		}
-	}
-	return last + 1
-}
diff --git a/Algorithms/0080. Remove Duplicates from Sorted Array II/README.md b/Algorithms/0080. Remove Duplicates from Sorted Array II/README.md
deleted file mode 100644
index e6cbed428..000000000
--- a/Algorithms/0080. Remove Duplicates from Sorted Array II/README.md	
+++ /dev/null
@@ -1,56 +0,0 @@
-# [80. Remove Duplicates from Sorted Array II](https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/)
-
-## 题目
-
-Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.
-
-Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
-
-Example 1:
-
-```c
-Given nums = [1,1,1,2,2,3],
-
-Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
-
-It doesn't matter what you leave beyond the returned length.
-```
-
-Example 2:
-
-```c
-Given nums = [0,0,1,1,1,1,2,3,3],
-
-Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively.
-
-It doesn't matter what values are set beyond the returned length.
-```
-
-Clarification:
-
-Confused why the returned value is an integer but your answer is an array?
-
-Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
-
-Internally you can think of this:
-
-```c
-// nums is passed in by reference. (i.e., without making a copy)
-int len = removeElement(nums, val);
-
-// any modification to nums in your function would be known by the caller.
-// using the length returned by your function, it prints the first len elements.
-for (int i = 0; i < len; i++) {
-    print(nums[i]);
-}
-```
-
-## 题目大意
-
-给定一个有序数组 nums，对数组中的元素进行去重，使得原数组中的每个元素最多暴露 2 个。最后返回去重以后数组的长度值。
-
-## 解题思路
-
-这道题和第 26 题很像。是第 26 题的加强版。这道题和第 283 题，第 27 题基本一致，283 题是删除 0，27 题是删除指定元素，这一题是删除重复元素，实质是一样的。
-
-这里数组的删除并不是真的删除，只是将删除的元素移动到数组后面的空间内，然后返回数组实际剩余的元素个数，OJ 最终判断题目的时候会读取数组剩余个数的元素进行输出。
diff --git a/Algorithms/0081. Search in Rotated Sorted Array II/README.md b/Algorithms/0081. Search in Rotated Sorted Array II/README.md
deleted file mode 100755
index 133741cd1..000000000
--- a/Algorithms/0081. Search in Rotated Sorted Array II/README.md	
+++ /dev/null
@@ -1,44 +0,0 @@
-# [81. Search in Rotated Sorted Array II](https://leetcode.com/problems/search-in-rotated-sorted-array-ii/)
-
-
-## 题目:
-
-Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
-
-(i.e., `[0,0,1,2,2,5,6]` might become `[2,5,6,0,0,1,2]`).
-
-You are given a target value to search. If found in the array return `true`, otherwise return `false`.
-
-**Example 1:**
-
-    Input: nums = [2,5,6,0,0,1,2], target = 0
-    Output: true
-
-**Example 2:**
-
-    Input: nums = [2,5,6,0,0,1,2], target = 3
-    Output: false
-
-**Follow up:**
-
-- This is a follow up problem to [Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/description/), where `nums` may contain duplicates.
-- Would this affect the run-time complexity? How and why?
-
-
-## 题目大意
-
-假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,0,1,2,2,5,6] 可能变为 [2,5,6,0,0,1,2] )。
-
-编写一个函数来判断给定的目标值是否存在于数组中。若存在返回 true，否则返回 false。
-
-进阶:
-
-- 这是搜索旋转排序数组 的延伸题目，本题中的 nums  可能包含重复元素。
-- 这会影响到程序的时间复杂度吗？会有怎样的影响，为什么？
-
-
-## 解题思路
-
-
-- 给出一个数组，数组中本来是从小到大排列的，并且数组中有重复数字。但是现在把后面随机一段有序的放到数组前面，这样形成了前后两端有序的子序列。在这样的一个数组里面查找一个数，设计一个 O(log n) 的算法。如果找到就输出 `true`，如果没有找到，就输出 `false` 。
-- 这一题是第 33 题的加强版，实现代码完全一样，只不过输出变了。这一题输出 `true` 和 `false` 了。具体思路见第 33 题。
diff --git a/Algorithms/0082. Remove Duplicates from Sorted List II/82. Remove Duplicates from Sorted List II.go b/Algorithms/0082. Remove Duplicates from Sorted List II/82. Remove Duplicates from Sorted List II.go
deleted file mode 100644
index 60c2cd92e..000000000
--- a/Algorithms/0082. Remove Duplicates from Sorted List II/82. Remove Duplicates from Sorted List II.go	
+++ /dev/null
@@ -1,68 +0,0 @@
-package leetcode
-
-/**
- * Definition for singly-linked list.
- * type ListNode struct {
- *     Val int
- *     Next *ListNode
- * }
- */
-func deleteDuplicates1(head *ListNode) *ListNode {
-	if head == nil {
-		return nil
-	}
-	if head.Next == nil {
-		return head
-	}
-	newHead := &ListNode{Next: head, Val: -999999}
-	cur := newHead
-	last := newHead
-	front := head
-	for front.Next != nil {
-		if front.Val == cur.Val {
-			// fmt.Printf("相同节点front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
-			front = front.Next
-			continue
-		} else {
-			if cur.Next != front {
-				// fmt.Printf("删除重复节点front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
-				last.Next = front
-				if front.Next != nil && front.Next.Val != front.Val {
-					last = front
-				}
-				cur = front
-				front = front.Next
-			} else {
-				// fmt.Printf("常规循环前front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
-				last = cur
-				cur = cur.Next
-				front = front.Next
-				// fmt.Printf("常规循环后front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
-
-			}
-		}
-	}
-	if front.Val == cur.Val {
-		// fmt.Printf("相同节点front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
-		last.Next = nil
-	} else {
-		if cur.Next != front {
-			last.Next = front
-		}
-	}
-	return newHead.Next
-}
-
-func deleteDuplicates2(head *ListNode) *ListNode {
-	if head == nil {
-		return nil
-	}
-	if head.Next != nil && head.Val == head.Next.Val {
-		for head.Next != nil && head.Val == head.Next.Val {
-			head = head.Next
-		}
-		return deleteDuplicates(head.Next)
-	}
-	head.Next = deleteDuplicates(head.Next)
-	return head
-}
diff --git a/Algorithms/0084. Largest Rectangle in Histogram/84. Largest Rectangle in Histogram.go b/Algorithms/0084. Largest Rectangle in Histogram/84. Largest Rectangle in Histogram.go
deleted file mode 100644
index 3c56534d5..000000000
--- a/Algorithms/0084. Largest Rectangle in Histogram/84. Largest Rectangle in Histogram.go	
+++ /dev/null
@@ -1,32 +0,0 @@
-package leetcode
-
-import "fmt"
-
-func largestRectangleArea(heights []int) int {
-	maxArea, stack, height := 0, []int{}, 0
-	for i := 0; i <= len(heights); i++ {
-		if i == len(heights) {
-			height = 0
-		} else {
-			height = heights[i]
-		}
-		if len(stack) == 0 || height >= heights[stack[len(stack)-1]] {
-			stack = append(stack, i)
-		} else {
-			tmp := stack[len(stack)-1]
-			fmt.Printf("1. tmp = %v stack = %v\n", tmp, stack)
-			stack = stack[:len(stack)-1]
-			length := 0
-			if len(stack) == 0 {
-				length = i
-			} else {
-				length = i - 1 - stack[len(stack)-1]
-				fmt.Printf("2. length = %v stack = %v i = %v\n", length, stack, i)
-			}
-			maxArea = max(maxArea, heights[tmp]*length)
-			fmt.Printf("3. maxArea = %v heights[tmp]*length = %v\n", maxArea, heights[tmp]*length)
-			i--
-		}
-	}
-	return maxArea
-}
diff --git a/Algorithms/0088. Merge-Sorted-Array/88. Merge Sorted Array.go b/Algorithms/0088. Merge-Sorted-Array/88. Merge Sorted Array.go
deleted file mode 100644
index ec275970c..000000000
--- a/Algorithms/0088. Merge-Sorted-Array/88. Merge Sorted Array.go	
+++ /dev/null
@@ -1,29 +0,0 @@
-package leetcode
-
-func merge(nums1 []int, m int, nums2 []int, n int) {
-	if m == 0 {
-		copy(nums1, nums2)
-		return
-	}
-	// 这里不需要，因为测试数据考虑到了第一个数组的空间问题
-	// for index := 0; index < n; index++ {
-	// 	nums1 = append(nums1, nums2[index])
-	// }
-	i := m - 1
-	j := n - 1
-	k := m + n - 1
-	// 从后面往前放，只需要循环一次即可
-	for ; i >= 0 && j >= 0; k-- {
-		if nums1[i] > nums2[j] {
-			nums1[k] = nums1[i]
-			i--
-		} else {
-			nums1[k] = nums2[j]
-			j--
-		}
-	}
-	for ; j >= 0; k-- {
-		nums1[k] = nums2[j]
-		j--
-	}
-}
diff --git a/Algorithms/0088. Merge-Sorted-Array/88. Merge Sorted Array_test.go b/Algorithms/0088. Merge-Sorted-Array/88. Merge Sorted Array_test.go
deleted file mode 100644
index 3d078d120..000000000
--- a/Algorithms/0088. Merge-Sorted-Array/88. Merge Sorted Array_test.go	
+++ /dev/null
@@ -1,60 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question88 struct {
-	para88
-	ans88
-}
-
-// para 是参数
-// one 代表第一个参数
-type para88 struct {
-	one []int
-	m   int
-	two []int
-	n   int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans88 struct {
-	one []int
-}
-
-func Test_Problem88(t *testing.T) {
-
-	qs := []question88{
-
-		// question{
-		// 	para{[]int{0}, 0, []int{1}, 1},
-		// 	ans{[]int{1}},
-		// },
-		//
-		// question{
-		// 	para{[]int{1, 3, 5, 7}, 4, []int{2, 4}, 2},
-		// 	ans{[]int{1, 2, 3, 4}},
-		// },
-		//
-		// question{
-		// 	para{[]int{1, 3, 5, 7}, 4, []int{2, 2}, 2},
-		// 	ans{[]int{1, 2, 2, 3}},
-		// },
-
-		question88{
-			para88{[]int{1, 2, 3, 0, 0, 0}, 3, []int{2, 5, 6}, 3},
-			ans88{[]int{1, 2, 2, 3, 5, 6}},
-		},
-	}
-	fmt.Printf("------------------------Leetcode Problem 88------------------------\n")
-	for _, q := range qs {
-		_, p := q.ans88, q.para88
-		fmt.Printf("【intput】:%v,%v,%v,%v       ", p.one, p.m, p.two, p.n)
-		merge(p.one, p.m, p.two, p.n)
-		fmt.Printf("【output】:%v\n", p)
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0091. Decode Ways/91. Decode Ways.go b/Algorithms/0091. Decode Ways/91. Decode Ways.go
deleted file mode 100644
index 35a97845b..000000000
--- a/Algorithms/0091. Decode Ways/91. Decode Ways.go	
+++ /dev/null
@@ -1,29 +0,0 @@
-package leetcode
-
-import (
-	"strconv"
-)
-
-func numDecodings(s string) int {
-	if len(s) == 0 {
-		return 0
-	}
-	dp := make([]int, len(s)+1)
-	dp[0] = 1
-	if s[:1] == "0" {
-		dp[1] = 0
-	} else {
-		dp[1] = 1
-	}
-	for i := 2; i <= len(s); i++ {
-		lastNum, _ := strconv.Atoi(s[i-1 : i])
-		if lastNum >= 1 && lastNum <= 9 {
-			dp[i] += dp[i-1]
-		}
-		lastNum, _ = strconv.Atoi(s[i-2 : i])
-		if lastNum >= 10 && lastNum <= 26 {
-			dp[i] += dp[i-2]
-		}
-	}
-	return dp[len(s)]
-}
diff --git a/Algorithms/0095. Unique Binary Search Trees II/95. Unique Binary Search Trees II_test.go b/Algorithms/0095. Unique Binary Search Trees II/95. Unique Binary Search Trees II_test.go
deleted file mode 100644
index c6a6e0418..000000000
--- a/Algorithms/0095. Unique Binary Search Trees II/95. Unique Binary Search Trees II_test.go	
+++ /dev/null
@@ -1,56 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question95 struct {
-	para95
-	ans95
-}
-
-// para 是参数
-// one 代表第一个参数
-type para95 struct {
-	one int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans95 struct {
-	one []*TreeNode
-}
-
-func Test_Problem95(t *testing.T) {
-
-	qs := []question95{
-
-		question95{
-			para95{1},
-			ans95{[]*TreeNode{&TreeNode{Val: 1, Left: nil, Right: nil}}},
-		},
-
-		question95{
-			para95{3},
-			ans95{[]*TreeNode{
-				&TreeNode{Val: 1, Left: nil, Right: &TreeNode{Val: 3, Left: &TreeNode{Val: 2, Left: nil, Right: nil}, Right: nil}},
-				&TreeNode{Val: 1, Left: nil, Right: &TreeNode{Val: 2, Left: nil, Right: &TreeNode{Val: 3, Left: nil, Right: nil}}},
-				&TreeNode{Val: 3, Left: &TreeNode{Val: 2, Left: &TreeNode{Val: 1, Left: nil, Right: nil}, Right: nil}, Right: nil},
-				&TreeNode{Val: 3, Left: &TreeNode{Val: 1, Left: nil, Right: &TreeNode{Val: 2, Left: nil, Right: nil}}, Right: nil},
-				&TreeNode{Val: 2, Left: &TreeNode{Val: 1, Left: nil, Right: nil}, Right: &TreeNode{Val: 3, Left: nil, Right: nil}},
-			}}},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 95------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans95, q.para95
-		fmt.Printf("【input】:%v  \n", p)
-		trees := generateTrees(p.one)
-		for _, t := range trees {
-			fmt.Printf("【output】:%v\n", Tree2Preorder(t))
-		}
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0095. Unique Binary Search Trees II/README.md b/Algorithms/0095. Unique Binary Search Trees II/README.md
deleted file mode 100755
index 26a626382..000000000
--- a/Algorithms/0095. Unique Binary Search Trees II/README.md	
+++ /dev/null
@@ -1,35 +0,0 @@
-# [95. Unique Binary Search Trees II](https://leetcode.com/problems/unique-binary-search-trees-ii/)
-
-
-## 题目:
-
-Given an integer *n*, generate all structurally unique **BST's** (binary search trees) that store values 1 ... *n*.
-
-**Example:**
-
-    Input: 3
-    Output:
-    [
-      [1,null,3,2],
-      [3,2,null,1],
-      [3,1,null,null,2],
-      [2,1,3],
-      [1,null,2,null,3]
-    ]
-    Explanation:
-    The above output corresponds to the 5 unique BST's shown below:
-    
-       1         3     3      2      1
-        \       /     /      / \      \
-         3     2     1      1   3      2
-        /     /       \                 \
-       2     1         2                 3
-
-
-## 题目大意
-
-给定一个整数 n，生成所有由 1 ... n 为节点所组成的二叉搜索树。
-
-## 解题思路
-
-- 输出 1~n 元素组成的 BST 所有解。这一题递归求解即可。外层循环遍历 1~n 所有结点，作为根结点，内层双层递归分别求出左子树和右子树。
diff --git a/Algorithms/0098. Validate Binary Search Tree/98. Validate Binary Search Tree_test.go b/Algorithms/0098. Validate Binary Search Tree/98. Validate Binary Search Tree_test.go
deleted file mode 100644
index f0c54077d..000000000
--- a/Algorithms/0098. Validate Binary Search Tree/98. Validate Binary Search Tree_test.go	
+++ /dev/null
@@ -1,59 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question98 struct {
-	para98
-	ans98
-}
-
-// para 是参数
-// one 代表第一个参数
-type para98 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans98 struct {
-	one bool
-}
-
-func Test_Problem98(t *testing.T) {
-
-	qs := []question98{
-
-		question98{
-			para98{[]int{10, 5, 15, NULL, NULL, 6, 20}},
-			ans98{false},
-		},
-
-		question98{
-			para98{[]int{}},
-			ans98{true},
-		},
-
-		question98{
-			para98{[]int{2, 1, 3}},
-			ans98{true},
-		},
-
-		question98{
-			para98{[]int{5, 1, 4, NULL, NULL, 3, 6}},
-			ans98{false},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 98------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans98, q.para98
-		fmt.Printf("【input】:%v      ", p)
-		rootOne := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", isValidBST(rootOne))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0099. Recover Binary Search Tree/99. Recover Binary Search Tree_test.go b/Algorithms/0099. Recover Binary Search Tree/99. Recover Binary Search Tree_test.go
deleted file mode 100644
index 472d69e0e..000000000
--- a/Algorithms/0099. Recover Binary Search Tree/99. Recover Binary Search Tree_test.go	
+++ /dev/null
@@ -1,50 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question99 struct {
-	para99
-	ans99
-}
-
-// para 是参数
-// one 代表第一个参数
-type para99 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans99 struct {
-	one []int
-}
-
-func Test_Problem99(t *testing.T) {
-
-	qs := []question99{
-
-		question99{
-			para99{[]int{1, 3, NULL, NULL, 2}},
-			ans99{[]int{3, 1, NULL, NULL, 2}},
-		},
-
-		question99{
-			para99{[]int{3, 1, 4, NULL, NULL, 2}},
-			ans99{[]int{2, 1, 4, NULL, NULL, 3}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 99------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans99, q.para99
-		fmt.Printf("【input】:%v      ", p)
-		rootOne := Ints2TreeNode(p.one)
-		recoverTree(rootOne)
-		fmt.Printf("【output】:%v      \n", p)
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0101. Symmetric Tree/101. Symmetric Tree.go b/Algorithms/0101. Symmetric Tree/101. Symmetric Tree.go
deleted file mode 100644
index 8941b092b..000000000
--- a/Algorithms/0101. Symmetric Tree/101. Symmetric Tree.go	
+++ /dev/null
@@ -1,16 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-func isSymmetric(root *TreeNode) bool {
-	if root == nil {
-		return true
-	}
-	return isSameTree(invertTree(root.Left), root.Right)
-}
diff --git a/Algorithms/0101. Symmetric Tree/101. Symmetric Tree_test.go b/Algorithms/0101. Symmetric Tree/101. Symmetric Tree_test.go
deleted file mode 100644
index 99bcf0ba2..000000000
--- a/Algorithms/0101. Symmetric Tree/101. Symmetric Tree_test.go	
+++ /dev/null
@@ -1,74 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question101 struct {
-	para101
-	ans101
-}
-
-// para 是参数
-// one 代表第一个参数
-type para101 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans101 struct {
-	one bool
-}
-
-func Test_Problem101(t *testing.T) {
-
-	qs := []question101{
-
-		question101{
-			para101{[]int{3, 4, 4, 5, NULL, NULL, 5, 6, NULL, NULL, 6}},
-			ans101{true},
-		},
-
-		question101{
-			para101{[]int{1, 2, 2, NULL, 3, 3}},
-			ans101{true},
-		},
-
-		question101{
-			para101{[]int{}},
-			ans101{true},
-		},
-
-		question101{
-			para101{[]int{1}},
-			ans101{true},
-		},
-
-		question101{
-			para101{[]int{1, 2, 3}},
-			ans101{false},
-		},
-
-		question101{
-			para101{[]int{1, 2, 2, 3, 4, 4, 3}},
-			ans101{true},
-		},
-
-		question101{
-			para101{[]int{1, 2, 2, NULL, 3, NULL, 3}},
-			ans101{false},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 101------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans101, q.para101
-		fmt.Printf("【input】:%v      ", p)
-		rootOne := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", isSymmetric(rootOne))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0102. Binary Tree Level Order Traversal/102. Binary Tree Level Order Traversal.go b/Algorithms/0102. Binary Tree Level Order Traversal/102. Binary Tree Level Order Traversal.go
deleted file mode 100644
index 07ab3b1fb..000000000
--- a/Algorithms/0102. Binary Tree Level Order Traversal/102. Binary Tree Level Order Traversal.go	
+++ /dev/null
@@ -1,63 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-
-// 解法一 BFS
-func levelOrder(root *TreeNode) [][]int {
-	if root == nil {
-		return [][]int{}
-	}
-	queue := []*TreeNode{}
-	queue = append(queue, root)
-	curNum, nextLevelNum, res, tmp := 1, 0, [][]int{}, []int{}
-	for len(queue) != 0 {
-		if curNum > 0 {
-			node := queue[0]
-			if node.Left != nil {
-				queue = append(queue, node.Left)
-				nextLevelNum++
-			}
-			if node.Right != nil {
-				queue = append(queue, node.Right)
-				nextLevelNum++
-			}
-			curNum--
-			tmp = append(tmp, node.Val)
-			queue = queue[1:]
-		}
-		if curNum == 0 {
-			res = append(res, tmp)
-			curNum = nextLevelNum
-			nextLevelNum = 0
-			tmp = []int{}
-		}
-	}
-	return res
-}
-
-// 解法二 DFS
-func levelOrder1(root *TreeNode) [][]int {
-	levels := [][]int{}
-	dfsLevel(root, -1, &levels)
-	return levels
-}
-
-func dfsLevel(node *TreeNode, level int, res *[][]int) {
-	if node == nil {
-		return
-	}
-	currLevel := level + 1
-	for len(*res) <= currLevel {
-		*res = append(*res, []int{})
-	}
-	(*res)[currLevel] = append((*res)[currLevel], node.Val)
-	dfsLevel(node.Left, currLevel, res)
-	dfsLevel(node.Right, currLevel, res)
-}
diff --git a/Algorithms/0102. Binary Tree Level Order Traversal/102. Binary Tree Level Order Traversal_test.go b/Algorithms/0102. Binary Tree Level Order Traversal/102. Binary Tree Level Order Traversal_test.go
deleted file mode 100644
index 66679c918..000000000
--- a/Algorithms/0102. Binary Tree Level Order Traversal/102. Binary Tree Level Order Traversal_test.go	
+++ /dev/null
@@ -1,54 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question102 struct {
-	para102
-	ans102
-}
-
-// para 是参数
-// one 代表第一个参数
-type para102 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans102 struct {
-	one [][]int
-}
-
-func Test_Problem102(t *testing.T) {
-
-	qs := []question102{
-
-		question102{
-			para102{[]int{}},
-			ans102{[][]int{}},
-		},
-
-		question102{
-			para102{[]int{1}},
-			ans102{[][]int{[]int{1}}},
-		},
-
-		question102{
-			para102{[]int{3, 9, 20, NULL, NULL, 15, 7}},
-			ans102{[][]int{[]int{3}, []int{9, 20}, []int{15, 7}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 102------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans102, q.para102
-		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", levelOrder(root))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0103. Binary Tree Zigzag Level Order Traversal/103. Binary Tree Zigzag Level Order Traversal.go b/Algorithms/0103. Binary Tree Zigzag Level Order Traversal/103. Binary Tree Zigzag Level Order Traversal.go
deleted file mode 100644
index 8d5897d81..000000000
--- a/Algorithms/0103. Binary Tree Zigzag Level Order Traversal/103. Binary Tree Zigzag Level Order Traversal.go	
+++ /dev/null
@@ -1,51 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-func zigzagLevelOrder(root *TreeNode) [][]int {
-	if root == nil {
-		return [][]int{}
-	}
-	queue := []*TreeNode{}
-	queue = append(queue, root)
-	curNum, nextLevelNum, res, tmp, curDir := 1, 0, [][]int{}, []int{}, 0
-	for len(queue) != 0 {
-		if curNum > 0 {
-			node := queue[0]
-			if node.Left != nil {
-				queue = append(queue, node.Left)
-				nextLevelNum++
-			}
-			if node.Right != nil {
-				queue = append(queue, node.Right)
-				nextLevelNum++
-			}
-			curNum--
-			tmp = append(tmp, node.Val)
-			queue = queue[1:]
-		}
-		if curNum == 0 {
-			if curDir == 1 {
-				for i, j := 0, len(tmp)-1; i < j; i, j = i+1, j-1 {
-					tmp[i], tmp[j] = tmp[j], tmp[i]
-				}
-			}
-			res = append(res, tmp)
-			curNum = nextLevelNum
-			nextLevelNum = 0
-			tmp = []int{}
-			if curDir == 0 {
-				curDir = 1
-			} else {
-				curDir = 0
-			}
-		}
-	}
-	return res
-}
diff --git a/Algorithms/0103. Binary Tree Zigzag Level Order Traversal/103. Binary Tree Zigzag Level Order Traversal_test.go b/Algorithms/0103. Binary Tree Zigzag Level Order Traversal/103. Binary Tree Zigzag Level Order Traversal_test.go
deleted file mode 100644
index 7f7bf0591..000000000
--- a/Algorithms/0103. Binary Tree Zigzag Level Order Traversal/103. Binary Tree Zigzag Level Order Traversal_test.go	
+++ /dev/null
@@ -1,59 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question103 struct {
-	para103
-	ans103
-}
-
-// para 是参数
-// one 代表第一个参数
-type para103 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans103 struct {
-	one [][]int
-}
-
-func Test_Problem103(t *testing.T) {
-
-	qs := []question103{
-
-		question103{
-			para103{[]int{}},
-			ans103{[][]int{}},
-		},
-
-		question103{
-			para103{[]int{1}},
-			ans103{[][]int{[]int{1}}},
-		},
-
-		question103{
-			para103{[]int{3, 9, 20, NULL, NULL, 15, 7}},
-			ans103{[][]int{[]int{3}, []int{9, 20}, []int{15, 7}}},
-		},
-
-		question103{
-			para103{[]int{1, 2, 3, 4, NULL, NULL, 5}},
-			ans103{[][]int{[]int{1}, []int{3, 2}, []int{4, 5}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 103------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans103, q.para103
-		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", zigzagLevelOrder(root))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0104. Maximum Depth of Binary Tree/104. Maximum Depth of Binary Tree.go b/Algorithms/0104. Maximum Depth of Binary Tree/104. Maximum Depth of Binary Tree.go
deleted file mode 100644
index 73ca1a6e8..000000000
--- a/Algorithms/0104. Maximum Depth of Binary Tree/104. Maximum Depth of Binary Tree.go	
+++ /dev/null
@@ -1,16 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-func maxDepth(root *TreeNode) int {
-	if root == nil {
-		return 0
-	}
-	return max(maxDepth(root.Left), maxDepth(root.Right)) + 1
-}
diff --git a/Algorithms/0107. Binary Tree Level Order Traversal II/107. Binary Tree Level Order Traversal II.go b/Algorithms/0107. Binary Tree Level Order Traversal II/107. Binary Tree Level Order Traversal II.go
deleted file mode 100644
index 682dc1bb7..000000000
--- a/Algorithms/0107. Binary Tree Level Order Traversal II/107. Binary Tree Level Order Traversal II.go	
+++ /dev/null
@@ -1,18 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-func levelOrderBottom(root *TreeNode) [][]int {
-	tmp := levelOrder(root)
-	res := [][]int{}
-	for i := len(tmp) - 1; i >= 0; i-- {
-		res = append(res, tmp[i])
-	}
-	return res
-}
diff --git a/Algorithms/0107. Binary Tree Level Order Traversal II/107. Binary Tree Level Order Traversal II_test.go b/Algorithms/0107. Binary Tree Level Order Traversal II/107. Binary Tree Level Order Traversal II_test.go
deleted file mode 100644
index 211cc7f4e..000000000
--- a/Algorithms/0107. Binary Tree Level Order Traversal II/107. Binary Tree Level Order Traversal II_test.go	
+++ /dev/null
@@ -1,54 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question107 struct {
-	para107
-	ans107
-}
-
-// para 是参数
-// one 代表第一个参数
-type para107 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans107 struct {
-	one [][]int
-}
-
-func Test_Problem107(t *testing.T) {
-
-	qs := []question107{
-
-		question107{
-			para107{[]int{}},
-			ans107{[][]int{}},
-		},
-
-		question107{
-			para107{[]int{1}},
-			ans107{[][]int{[]int{1}}},
-		},
-
-		question107{
-			para107{[]int{3, 9, 20, NULL, NULL, 15, 7}},
-			ans107{[][]int{[]int{15, 7}, []int{9, 20}, []int{3}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 107------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans107, q.para107
-		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", levelOrderBottom(root))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0109. Convert Sorted List to Binary Search Tree/109. Convert Sorted List to Binary Search Tree_test.go b/Algorithms/0109. Convert Sorted List to Binary Search Tree/109. Convert Sorted List to Binary Search Tree_test.go
deleted file mode 100644
index 459c46afd..000000000
--- a/Algorithms/0109. Convert Sorted List to Binary Search Tree/109. Convert Sorted List to Binary Search Tree_test.go	
+++ /dev/null
@@ -1,83 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question109 struct {
-	para109
-	ans109
-}
-
-// para 是参数
-// one 代表第一个参数
-type para109 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans109 struct {
-	one []int
-}
-
-func Test_Problem109(t *testing.T) {
-
-	qs := []question109{
-
-		question109{
-			para109{[]int{-10, -3, 0, 5, 9}},
-			ans109{[]int{0, -3, 9, -10, -9999, 5}},
-		},
-
-		question109{
-			para109{[]int{-10}},
-			ans109{[]int{-10}},
-		},
-
-		question109{
-			para109{[]int{1, 2}},
-			ans109{[]int{2, 1}},
-		},
-
-		question109{
-			para109{[]int{1, 2, 3}},
-			ans109{[]int{2, 1, 3}},
-		},
-		// question109{
-		// 	para109{[]int{1, 2, 3, 4, 5}, 2, 2},
-		// 	ans109{[]int{1, 2, 3, 4, 5}},
-		// },
-
-		// question109{
-		// 	para109{[]int{1, 2, 3, 4, 5}, 1, 5},
-		// 	ans109{[]int{5, 4, 3, 2, 1}},
-		// },
-
-		// question109{
-		// 	para109{[]int{1, 2, 3, 4, 5, 6}, 3, 4},
-		// 	ans109{[]int{1, 2, 4, 3, 5, 6}},
-		// },
-
-		// question109{
-		// 	para109{[]int{3, 5}, 1, 2},
-		// 	ans109{[]int{5, 3}},
-		// },
-
-		// question109{
-		// 	para109{[]int{3}, 3, 5},
-		// 	ans109{[]int{3}},
-		// },
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 109------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans109, q.para109
-		arr := []int{}
-		T2s(sortedListToBST(S2l(p.one)), &arr)
-		fmt.Printf("【input】:%v       【output】:%v\n", p, arr)
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0109. Convert Sorted List to Binary Search Tree/README.md b/Algorithms/0109. Convert Sorted List to Binary Search Tree/README.md
deleted file mode 100644
index 0c0c91aef..000000000
--- a/Algorithms/0109. Convert Sorted List to Binary Search Tree/README.md	
+++ /dev/null
@@ -1,30 +0,0 @@
-# [109. Convert Sorted List to Binary Search Tree](https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/)
-
-## 题目
-
-Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
-
-For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
-
-Example:
-
-```
-Given the sorted linked list: [-10,-3,0,5,9],
-
-One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
-
-      0
-     / \
-   -3   9
-   /   /
- -10  5
-```
-
-
-## 题目大意
-
-将链表转化为高度平衡的二叉搜索树。高度平衡的定义：每个结点的 2 个子结点的深度不能相差超过 1 。
-
-## 解题思路
-
-思路比较简单，依次把链表的中间点作为根结点，类似二分的思想，递归排列所有结点即可。
\ No newline at end of file
diff --git a/Algorithms/0110. Balanced Binary Tree/110. Balanced Binary Tree.go b/Algorithms/0110. Balanced Binary Tree/110. Balanced Binary Tree.go
deleted file mode 100644
index d7018c329..000000000
--- a/Algorithms/0110. Balanced Binary Tree/110. Balanced Binary Tree.go	
+++ /dev/null
@@ -1,25 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-func isBalanced(root *TreeNode) bool {
-	if root == nil {
-		return true
-	}
-	leftHight := depth(root.Left)
-	rightHight := depth(root.Right)
-	return abs(leftHight-rightHight) <= 1 && isBalanced(root.Left) && isBalanced(root.Right)
-}
-
-func depth(root *TreeNode) int {
-	if root == nil {
-		return 0
-	}
-	return max(depth(root.Left), depth(root.Right)) + 1
-}
diff --git a/Algorithms/0110. Balanced Binary Tree/110. Balanced Binary Tree_test.go b/Algorithms/0110. Balanced Binary Tree/110. Balanced Binary Tree_test.go
deleted file mode 100644
index 9e8b4ae70..000000000
--- a/Algorithms/0110. Balanced Binary Tree/110. Balanced Binary Tree_test.go	
+++ /dev/null
@@ -1,74 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question110 struct {
-	para110
-	ans110
-}
-
-// para 是参数
-// one 代表第一个参数
-type para110 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans110 struct {
-	one bool
-}
-
-func Test_Problem110(t *testing.T) {
-
-	qs := []question110{
-
-		question110{
-			para110{[]int{3, 4, 4, 5, NULL, NULL, 5, 6, NULL, NULL, 6}},
-			ans110{false},
-		},
-
-		question110{
-			para110{[]int{1, 2, 2, NULL, 3, 3}},
-			ans110{true},
-		},
-
-		question110{
-			para110{[]int{}},
-			ans110{true},
-		},
-
-		question110{
-			para110{[]int{1}},
-			ans110{true},
-		},
-
-		question110{
-			para110{[]int{1, 2, 3}},
-			ans110{true},
-		},
-
-		question110{
-			para110{[]int{1, 2, 2, 3, 4, 4, 3}},
-			ans110{true},
-		},
-
-		question110{
-			para110{[]int{1, 2, 2, NULL, 3, NULL, 3}},
-			ans110{true},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 110------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans110, q.para110
-		fmt.Printf("【input】:%v      ", p)
-		rootOne := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", isBalanced(rootOne))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0111. Minimum Depth of Binary Tree/111. Minimum Depth of Binary Tree.go b/Algorithms/0111. Minimum Depth of Binary Tree/111. Minimum Depth of Binary Tree.go
deleted file mode 100644
index c3f19bc35..000000000
--- a/Algorithms/0111. Minimum Depth of Binary Tree/111. Minimum Depth of Binary Tree.go	
+++ /dev/null
@@ -1,22 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-func minDepth(root *TreeNode) int {
-	if root == nil {
-		return 0
-	}
-	if root.Left == nil {
-		return minDepth(root.Right) + 1
-	}
-	if root.Right == nil {
-		return minDepth(root.Left) + 1
-	}
-	return min(minDepth(root.Left), minDepth(root.Right)) + 1
-}
diff --git a/Algorithms/0113. Path Sum II/113. Path Sum II_test.go b/Algorithms/0113. Path Sum II/113. Path Sum II_test.go
deleted file mode 100644
index bce9f60fd..000000000
--- a/Algorithms/0113. Path Sum II/113. Path Sum II_test.go	
+++ /dev/null
@@ -1,54 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question113 struct {
-	para113
-	ans113
-}
-
-// para 是参数
-// one 代表第一个参数
-type para113 struct {
-	one []int
-	sum int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans113 struct {
-	one [][]int
-}
-
-func Test_Problem113(t *testing.T) {
-
-	qs := []question113{
-		question113{
-			para113{[]int{1, 0, 1, 1, 2, 0, -1, 0, 1, -1, 0, -1, 0, 1, 0}, 2},
-			ans113{[][]int{[]int{1, 0, 1, 0}, []int{1, 0, 2, -1}, []int{1, 1, 0, 0}, []int{1, 1, -1, 1}}},
-		},
-
-		question113{
-			para113{[]int{}, 0},
-			ans113{[][]int{[]int{}}},
-		},
-
-		question113{
-			para113{[]int{5, 4, 8, 11, NULL, 13, 4, 7, 2, NULL, NULL, 5, 1}, 22},
-			ans113{[][]int{[]int{5, 4, 11, 2}, []int{5, 8, 4, 5}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 113------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans113, q.para113
-		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", pathSum(root, p.sum))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0114. Flatten Binary Tree to Linked List/114. Flatten Binary Tree to Linked List.go b/Algorithms/0114. Flatten Binary Tree to Linked List/114. Flatten Binary Tree to Linked List.go
deleted file mode 100644
index 865988e5a..000000000
--- a/Algorithms/0114. Flatten Binary Tree to Linked List/114. Flatten Binary Tree to Linked List.go	
+++ /dev/null
@@ -1,58 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-
-// 解法一 非递归
-func flatten(root *TreeNode) {
-	list, cur := []int{}, &TreeNode{}
-	preorder(root, &list)
-	cur = root
-	for i := 1; i < len(list); i++ {
-		cur.Left = nil
-		cur.Right = &TreeNode{Val: list[i], Left: nil, Right: nil}
-		cur = cur.Right
-	}
-	return
-}
-
-// 解法二 递归
-func flatten1(root *TreeNode) {
-	if root == nil || (root.Left == nil && root.Right == nil) {
-		return
-	}
-	flatten(root.Left)
-	flatten(root.Right)
-	currRight := root.Right
-	root.Right = root.Left
-	root.Left = nil
-	for root.Right != nil {
-		root = root.Right
-	}
-	root.Right = currRight
-}
-
-// 解法三 递归
-func flatten2(root *TreeNode) {
-	if root == nil {
-		return
-	}
-	flatten(root.Right)
-	if root.Left == nil {
-		return
-	}
-	flatten(root.Left)
-	p := root.Left
-	for p.Right != nil {
-		p = p.Right
-	}
-	p.Right = root.Right
-	root.Right = root.Left
-	root.Left = nil
-}
diff --git a/Algorithms/0114. Flatten Binary Tree to Linked List/114. Flatten Binary Tree to Linked List_test.go b/Algorithms/0114. Flatten Binary Tree to Linked List/114. Flatten Binary Tree to Linked List_test.go
deleted file mode 100644
index 92f2ae6d3..000000000
--- a/Algorithms/0114. Flatten Binary Tree to Linked List/114. Flatten Binary Tree to Linked List_test.go	
+++ /dev/null
@@ -1,45 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question114 struct {
-	para114
-	ans114
-}
-
-// para 是参数
-// one 代表第一个参数
-type para114 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans114 struct {
-	one []int
-}
-
-func Test_Problem114(t *testing.T) {
-
-	qs := []question114{
-
-		question114{
-			para114{[]int{1, 2, 3, 4, 5, 6}},
-			ans114{[]int{1, 2, 3, 4, 5, 6}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 114------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans114, q.para114
-		fmt.Printf("【input】:%v       \n", p)
-		rootOne := Ints2TreeNode(p.one)
-		flatten(rootOne)
-		fmt.Printf("【output】:%v      \n", Tree2Preorder(rootOne))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0127. Word Ladder/README.md b/Algorithms/0127. Word Ladder/README.md
deleted file mode 100755
index 9a207836f..000000000
--- a/Algorithms/0127. Word Ladder/README.md	
+++ /dev/null
@@ -1,65 +0,0 @@
-# [127. Word Ladder](https://leetcode.com/problems/word-ladder/)
-
-
-## 题目
-
-Given two words (*beginWord* and *endWord*), and a dictionary's word list, find the length of shortest transformation sequence from *beginWord* to *endWord*, such that:
-
-1. Only one letter can be changed at a time.
-2. Each transformed word must exist in the word list. Note that *beginWord* is *not* a transformed word.
-
-**Note:**
-
-- Return 0 if there is no such transformation sequence.
-- All words have the same length.
-- All words contain only lowercase alphabetic characters.
-- You may assume no duplicates in the word list.
-- You may assume *beginWord* and *endWord* are non-empty and are not the same.
-
-**Example 1:**
-
-    Input:
-    beginWord = "hit",
-    endWord = "cog",
-    wordList = ["hot","dot","dog","lot","log","cog"]
-    
-    Output: 5
-    
-    Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
-    return its length 5.
-
-**Example 2:**
-
-    Input:
-    beginWord = "hit"
-    endWord = "cog"
-    wordList = ["hot","dot","dog","lot","log"]
-    
-    Output: 0
-    
-    Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
-
-
-## 题目大意
-
-给定两个单词（beginWord 和 endWord）和一个字典，找到从 beginWord 到 endWord 的最短转换序列的长度。转换需遵循如下规则：
-
-1. 每次转换只能改变一个字母。
-2. 转换过程中的中间单词必须是字典中的单词。
-
-说明:
-
-- 如果不存在这样的转换序列，返回 0。
-- 所有单词具有相同的长度。
-- 所有单词只由小写字母组成。
-- 字典中不存在重复的单词。
-- 你可以假设 beginWord 和 endWord 是非空的，且二者不相同。
-
-
-## 解题思路
-
-- 这一题要求输出从 `beginWord` 变换到 `endWord` 最短变换次数。可以用 BFS，从 `beginWord` 开始变换，把该单词的每个字母都用 `'a'~'z'` 变换一次，生成的数组到 `wordList` 中查找，这里用 Map 来记录查找。找得到就入队列，找不到就输出 0 。入队以后按照 BFS 的算法依次遍历完，当所有单词都 `len(queue)<=0` 出队以后，整个程序结束。
-- 这一题题目中虽然说了要求找到一条最短的路径，但是实际上最短的路径的寻找方法已经告诉你了：
-	1.  每次只变换一个字母
-	2. 每次变换都必须在 `wordList` 中  
-所以不需要单独考虑何种方式是最短的。 
diff --git a/Algorithms/0128. Longest Consecutive Sequence/README.md b/Algorithms/0128. Longest Consecutive Sequence/README.md
deleted file mode 100755
index 4809fe7ad..000000000
--- a/Algorithms/0128. Longest Consecutive Sequence/README.md	
+++ /dev/null
@@ -1,31 +0,0 @@
-# [128. Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/)
-
-
-## 题目:
-
-Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
-
-Your algorithm should run in O(*n*) complexity.
-
-**Example:**
-
-    Input: [100, 4, 200, 1, 3, 2]
-    Output: 4
-    Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.
-
-
-## 题目大意
-
-
-给定一个未排序的整数数组，找出最长连续序列的长度。要求算法的时间复杂度为 O(n)。
-
-
-
-
-## 解题思路
-
-
-- 给出一个数组，要求找出最长连续序列，输出这个最长的长度。要求时间复杂度为 `O(n)`。
-- 这一题可以先用暴力解决解决，代码见解法三。思路是把每个数都存在 `map` 中，先删去 `map` 中没有前一个数 `nums[i]-1` 也没有后一个数 `nums[i]+1` 的数 `nums[i]`，这种数前后都不连续。然后在 `map` 中找到前一个数 `nums[i]-1` 不存在，但是后一个数 `nums[i]+1` 存在的数，这种数是连续序列的起点，那么不断的往后搜，直到序列“断”了。最后输出最长序列的长度。
-- 这一题最优的解法是解法一，针对每一个 `map` 中不存在的数 `n`，插入进去都做 2 件事情。第一件事，先查看 `n - 1` 和 `n + 1` 是否都存在于 `map` 中，如果都存在，代表存在连续的序列，那么就更新 `left`，`right` 边界。那么 `n` 对应的这个小的子连续序列长度为 `sum = left + right + 1`。第二件事就是更新 `left` 和 `right` 左右边界对应的 `length = sum`。
-- 这一题还可以用并查集解决，见解法二。利用每个数在 `nums` 中的下标，把下标和下标进行 `union()`，具体做法是看前一个数 `nums[i]-1` 和后一个数 `nums[i]+1` 在 `map` 中是否存在，如果存在就 `union()`，最终输出整个并查集中包含最多元素的那个集合的元素总数。
\ No newline at end of file
diff --git a/Algorithms/0129. Sum Root to Leaf Numbers/129. Sum Root to Leaf Numbers.go b/Algorithms/0129. Sum Root to Leaf Numbers/129. Sum Root to Leaf Numbers.go
deleted file mode 100644
index 58dbd8710..000000000
--- a/Algorithms/0129. Sum Root to Leaf Numbers/129. Sum Root to Leaf Numbers.go	
+++ /dev/null
@@ -1,41 +0,0 @@
-package leetcode
-
-import (
-	"strconv"
-)
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-func sumNumbers(root *TreeNode) int {
-	res, nums := 0, binaryTreeNums(root)
-	for _, n := range nums {
-		num, _ := strconv.Atoi(n)
-		res += num
-	}
-	return res
-}
-
-func binaryTreeNums(root *TreeNode) []string {
-	if root == nil {
-		return []string{}
-	}
-	res := []string{}
-	if root.Left == nil && root.Right == nil {
-		return []string{strconv.Itoa(root.Val)}
-	}
-	tmpLeft := binaryTreeNums(root.Left)
-	for i := 0; i < len(tmpLeft); i++ {
-		res = append(res, strconv.Itoa(root.Val)+tmpLeft[i])
-	}
-	tmpRight := binaryTreeNums(root.Right)
-	for i := 0; i < len(tmpRight); i++ {
-		res = append(res, strconv.Itoa(root.Val)+tmpRight[i])
-	}
-	return res
-}
diff --git a/Algorithms/0129. Sum Root to Leaf Numbers/README.md b/Algorithms/0129. Sum Root to Leaf Numbers/README.md
deleted file mode 100755
index 3bb303520..000000000
--- a/Algorithms/0129. Sum Root to Leaf Numbers/README.md	
+++ /dev/null
@@ -1,48 +0,0 @@
-# [129. Sum Root to Leaf Numbers](https://leetcode.com/problems/sum-root-to-leaf-numbers/)
-
-
-## 题目
-
-Given a binary tree containing digits from `0-9` only, each root-to-leaf path could represent a number.
-
-An example is the root-to-leaf path `1->2->3` which represents the number `123`.
-
-Find the total sum of all root-to-leaf numbers.
-
-**Note:** A leaf is a node with no children.
-
-**Example:**
-
-    Input: [1,2,3]
-        1
-       / \
-      2   3
-    Output: 25
-    Explanation:
-    The root-to-leaf path 1->2 represents the number 12.
-    The root-to-leaf path 1->3 represents the number 13.
-    Therefore, sum = 12 + 13 = 25.
-
-**Example 2:**
-
-    Input: [4,9,0,5,1]
-        4
-       / \
-      9   0
-     / \
-    5   1
-    Output: 1026
-    Explanation:
-    The root-to-leaf path 4->9->5 represents the number 495.
-    The root-to-leaf path 4->9->1 represents the number 491.
-    The root-to-leaf path 4->0 represents the number 40.
-    Therefore, sum = 495 + 491 + 40 = 1026.
-
-## 题目大意
-
-给定一个二叉树，它的每个结点都存放一个 0-9 的数字，每条从根到叶子节点的路径都代表一个数字。例如，从根到叶子节点路径 1->2->3 代表数字 123。计算从根到叶子节点生成的所有数字之和。说明: 叶子节点是指没有子节点的节点。
-
-
-## 解题思路
-
-- 这一题是第 257 题的变形题，第 257 题要求输出每条从根节点到叶子节点的路径，这一题变成了把每一个从根节点到叶子节点的数字都串联起来，再累加每条路径，求出最后的总和。实际做题思路基本没变
diff --git a/Algorithms/0130. Surrounded Regions/130. Surrounded Regions_test.go b/Algorithms/0130. Surrounded Regions/130. Surrounded Regions_test.go
deleted file mode 100644
index a363b5f0e..000000000
--- a/Algorithms/0130. Surrounded Regions/130. Surrounded Regions_test.go	
+++ /dev/null
@@ -1,49 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question130 struct {
-	para130
-	ans130
-}
-
-// para 是参数
-// one 代表第一个参数
-type para130 struct {
-	one [][]byte
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans130 struct {
-	one [][]byte
-}
-
-func Test_Problem130(t *testing.T) {
-
-	qs := []question130{
-
-		question130{
-			para130{[][]byte{}},
-			ans130{[][]byte{}},
-		},
-
-		question130{
-			para130{[][]byte{[]byte{'X', 'X', 'X', 'X'}, []byte{'X', 'O', 'O', 'X'}, []byte{'X', 'X', 'O', 'X'}, []byte{'X', 'O', 'X', 'X'}}},
-			ans130{[][]byte{[]byte{'X', 'X', 'X', 'X'}, []byte{'X', 'X', 'X', 'X'}, []byte{'X', 'X', 'X', 'X'}, []byte{'X', 'O', 'X', 'X'}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 130------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans130, q.para130
-		fmt.Printf("【input】:%v      ", p)
-		solve1(p.one)
-		fmt.Printf("【output】:%v      \n", p)
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0130. Surrounded Regions/README.md b/Algorithms/0130. Surrounded Regions/README.md
deleted file mode 100755
index f8664a197..000000000
--- a/Algorithms/0130. Surrounded Regions/README.md	
+++ /dev/null
@@ -1,39 +0,0 @@
-# [130. Surrounded Regions](https://leetcode.com/problems/surrounded-regions/)
-
-
-
-## 题目:
-
-Given a 2D board containing `'X'` and `'O'` (**the letter O**), capture all regions surrounded by `'X'`.
-
-A region is captured by flipping all `'O'`s into `'X'`s in that surrounded region.
-
-**Example:**
-
-    X X X X
-    X O O X
-    X X O X
-    X O X X
-
-After running your function, the board should be:
-
-    X X X X
-    X X X X
-    X X X X
-    X O X X
-
-**Explanation:**
-
-Surrounded regions shouldn’t be on the border, which means that any `'O'` on the border of the board are not flipped to `'X'`. Any `'O'` that is not on the border and it is not connected to an `'O'` on the border will be flipped to `'X'`. Two cells are connected if they are adjacent cells connected horizontally or vertically.
-
-## 题目大意
-
-给定一个二维的矩阵，包含 'X' 和 'O'（字母 O）。找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O' 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
-
-
-## 解题思路
-
-
-- 给出一张二维地图，要求把地图上非边缘上的 'O' 都用 'X' 覆盖掉。
-- 这一题有多种解法。第一种解法是并查集。先将边缘上的 'O' 全部都和一个特殊的点进行 `union()` 。然后再把地图中间的 'O' 都进行 `union()`，最后把和特殊点不是同一个集合的点都标记成 'X'。第二种解法是 DFS 或者 BFS，可以先将边缘上的 'O' 先标记成另外一个字符，然后在递归遍历过程中，把剩下的 'O' 都标记成 'X'。
-
diff --git a/Algorithms/0131. Palindrome Partitioning/131. Palindrome Partitioning_test.go b/Algorithms/0131. Palindrome Partitioning/131. Palindrome Partitioning_test.go
deleted file mode 100644
index d482ce22b..000000000
--- a/Algorithms/0131. Palindrome Partitioning/131. Palindrome Partitioning_test.go	
+++ /dev/null
@@ -1,57 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question131 struct {
-	para131
-	ans131
-}
-
-// para 是参数
-// one 代表第一个参数
-type para131 struct {
-	s string
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans131 struct {
-	one [][]string
-}
-
-func Test_Problem131(t *testing.T) {
-
-	qs := []question131{
-
-		question131{
-			para131{"aab"},
-			ans131{[][]string{[]string{"aa", "b"}, []string{"a", "a", "b"}}},
-		},
-
-		question131{
-			para131{"bb"},
-			ans131{[][]string{[]string{"b", "b"}, []string{"bb"}}},
-		},
-
-		question131{
-			para131{"efe"},
-			ans131{[][]string{[]string{"e", "f", "e"}, []string{"efe"}}},
-		},
-
-		question131{
-			para131{"abbab"},
-			ans131{[][]string{[]string{"a", "b", "b", "a", "b"}, []string{"a", "b", "bab"}, []string{"a", "bb", "a", "b"}, []string{"abba", "b"}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 131------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans131, q.para131
-		fmt.Printf("【input】:%v       【output】:%v\n", p, partition131(p.s))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0136. Single Number/README.md b/Algorithms/0136. Single Number/README.md
deleted file mode 100755
index ba4920e8a..000000000
--- a/Algorithms/0136. Single Number/README.md	
+++ /dev/null
@@ -1,29 +0,0 @@
-# [136. Single Number](https://leetcode.com/problems/single-number/)
-
-## 题目:
-
-Given a **non-empty** array of integers, every element appears *twice* except for one. Find that single one.
-
-**Note:**
-
-Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
-
-**Example 1:**
-
-    Input: [2,2,1]
-    Output: 1
-
-**Example 2:**
-
-    Input: [4,1,2,1,2]
-    Output: 4
-
-## 题目大意
-
-给定一个非空整数数组，除了某个元素只出现一次以外，其余每个元素均出现两次。找出那个只出现了一次的元素。要求算法时间复杂度是线性的，并且不使用额外的辅助空间。
-
-
-## 解题思路
-
-- 题目要求不能使用辅助空间，并且时间复杂度只能是线性的。
-- 题目为什么要强调有一个数字出现一次，其他的出现两次？我们想到了异或运算的性质：任何一个数字异或它自己都等于0。也就是说，如果我们从头到尾依次异或数组中的每一个数字，那么最终的结果刚好是那个只出现依次的数字，因为那些出现两次的数字全部在异或中抵消掉了。于是最终做法是从头到尾依次异或数组中的每一个数字，那么最终得到的结果就是两个只出现一次的数字的异或结果。因为其他数字都出现了两次，在异或中全部抵消掉了。**利用的性质是 x^x = 0**。
diff --git a/Algorithms/0137. Single Number II/README.md b/Algorithms/0137. Single Number II/README.md
deleted file mode 100755
index 62ebda38c..000000000
--- a/Algorithms/0137. Single Number II/README.md	
+++ /dev/null
@@ -1,66 +0,0 @@
-# [137. Single Number II](https://leetcode.com/problems/single-number-ii/)
-
-
-## 题目:
-
-Given a **non-empty** array of integers, every element appears *three* times except for one, which appears exactly once. Find that single one.
-
-**Note:**
-
-Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
-
-**Example 1:**
-
-    Input: [2,2,3,2]
-    Output: 3
-
-**Example 2:**
-
-    Input: [0,1,0,1,0,1,99]
-    Output: 99
-
-
-## 题目大意
-
-给定一个非空整数数组，除了某个元素只出现一次以外，其余每个元素均出现了三次。找出那个只出现了一次的元素。要求算法时间复杂度是线性的，并且不使用额外的辅助空间。
-
-
-
-
-## 解题思路
-
-- 这一题是第 136 题的加强版。这类题也可以扩展，在数组中每个元素都出现 5 次，找出只出现 1 次的数。
-- 本题中要求找出只出现 1 次的数，出现 3 次的数都要被消除。第 136 题是消除出现 2 次的数。这一题也会相当相同的解法，出现 3 次的数也要被消除。定义状态，00、10、01，这 3 个状态。当一个数出现 3 次，那么它每个位置上的 1 出现的次数肯定是 3 的倍数，所以当 1 出现 3 次以后，就归零清除。如何能做到这点呢？仿造`三进制(00，10，01)` 就可以做到。
-- 变量  ones 中记录遍历中每个位上出现 1 的个数。将它与 A[i] 进行异或，目的是：
-    - 每位上两者都是 1 的，表示历史统计结果 ones 出现1次、A[i]中又出现1次，则是出现 2 次，需要进位到 twos 变量中。
-    - 每位上两者分别为 0、1 的，加入到 ones 统计结果中。
-    - 最后还要 & ^twos ，是为了能做到三进制，出现 3 次就清零。例如 ones = x，那么 twos = 0，当 twos = x，那么 ones = 0；
-- 变量 twos 中记录遍历中每个位上出现 1 ，2次 的个数。与 A[i] 进行异或的目的和上述描述相同，不再赘述。
-
-> 在 golang 中，&^ 表示 AND NOT 的意思。这里的 ^ 作为一元操作符，表示按位取反 (^0001 0100 = 1110 1011)，X &^ Y 的意思是将 X 中与 Y 相异的位保留，相同的位清零。
-
-> 在 golang 中没有 Java 中的 ~ 位操作运算符，Java 中的 ~ 运算符代表按位取反。这个操作就想当于 golang 中的 ^ 运算符当做一元运算符使用的效果。
-
-这一题还可以继续扩展，在数组中每个元素都出现 5 次，找出只出现 1 次的数。那该怎么做呢？思路还是一样的，模拟一个五进制，5 次就会消除。代码如下：
-
-    // 解法一
-    func singleNumberIII(nums []int) int {
-    	na, nb, nc := 0, 0, 0
-    	for i := 0; i < len(nums); i++ {
-    		nb = nb ^ (nums[i] & na)
-    		na = (na ^ nums[i]) & ^nc
-    		nc = nc ^ (nums[i] & ^na & ^nb)
-    	}
-    	return na & ^nb & ^nc
-    }
-    
-    // 解法二
-    func singleNumberIIII(nums []int) int {
-    	twos, threes, ones := 0xffffffff, 0xffffffff, 0
-    	for i := 0; i < len(nums); i++ {
-    		threes = threes ^ (nums[i] & twos)
-    		twos = (twos ^ nums[i]) & ^ones
-    		ones = ones ^ (nums[i] & ^twos & ^threes)
-    	}
-    	return ones
-    }
diff --git a/Algorithms/0141. Linked List Cycle/141. Linked List Cycle.go b/Algorithms/0141. Linked List Cycle/141. Linked List Cycle.go
deleted file mode 100644
index e04ae0484..000000000
--- a/Algorithms/0141. Linked List Cycle/141. Linked List Cycle.go	
+++ /dev/null
@@ -1,23 +0,0 @@
-package leetcode
-
-/**
- * Definition for singly-linked list.
- * struct ListNode {
- *     int val;
- *     ListNode *next;
- *     ListNode(int x) : val(x), next(NULL) {}
- * };
- */
-
-func hasCycle(head *ListNode) bool {
-	fast := head
-	slow := head
-	for slow != nil && fast != nil && fast.Next != nil {
-		fast = fast.Next.Next
-		slow = slow.Next
-		if fast == slow {
-			return true
-		}
-	}
-	return false
-}
diff --git a/Algorithms/0141. Linked List Cycle/141. Linked List Cycle_test.go b/Algorithms/0141. Linked List Cycle/141. Linked List Cycle_test.go
deleted file mode 100644
index 06d9b570a..000000000
--- a/Algorithms/0141. Linked List Cycle/141. Linked List Cycle_test.go	
+++ /dev/null
@@ -1,6 +0,0 @@
-package leetcode
-
-import "testing"
-
-func Test_Problem141(t *testing.T) {
-}
diff --git a/Algorithms/0143. Reorder List/README.md b/Algorithms/0143. Reorder List/README.md
deleted file mode 100644
index 8c1046f7d..000000000
--- a/Algorithms/0143. Reorder List/README.md	
+++ /dev/null
@@ -1,34 +0,0 @@
-# [143. Reorder List](https://leetcode.com/problems/reorder-list/)
-
-## 题目
-
-Given a singly linked list L: L0→L1→…→Ln-1→Ln,
-reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
-
-You may not modify the values in the list's nodes, only nodes itself may be changed.
-
-Example 1:
-
-```c
-Given 1->2->3->4, reorder it to 1->4->2->3.
-```
-
-Example 2:
-
-```c
-Given 1->2->3->4->5, reorder it to 1->5->2->4->3.
-```
-
-## 题目大意
-
-按照指定规则重新排序链表：第一个元素和最后一个元素排列在一起，接着第二个元素和倒数第二个元素排在一起，接着第三个元素和倒数第三个元素排在一起。
-
-
-## 解题思路
-
-
-最近简单的方法是先把链表存储到数组里，然后找到链表中间的结点，按照规则拼接即可。这样时间复杂度是 O(n)，空间复杂度是 O(n)。
-
-更好的做法是结合之前几道题的操作：链表逆序，找中间结点。
-
-先找到链表的中间结点，然后利用逆序区间的操作，如 [第 92 题](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0092.%20Reverse%20Linked%20List%20II) 里的 reverseBetween() 操作，只不过这里的反转区间是从中点一直到末尾。最后利用 2 个指针，一个指向头结点，一个指向中间结点，开始拼接最终的结果。这种做法的时间复杂度是 O(n)，空间复杂度是 O(1)。
\ No newline at end of file
diff --git a/Algorithms/0144. Binary Tree Preorder Traversal/144. Binary Tree Preorder Traversal_test.go b/Algorithms/0144. Binary Tree Preorder Traversal/144. Binary Tree Preorder Traversal_test.go
deleted file mode 100644
index 1ea66cb25..000000000
--- a/Algorithms/0144. Binary Tree Preorder Traversal/144. Binary Tree Preorder Traversal_test.go	
+++ /dev/null
@@ -1,213 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question144 struct {
-	para144
-	ans144
-}
-
-// para 是参数
-// one 代表第一个参数
-type para144 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans144 struct {
-	one []int
-}
-
-func Test_Problem144(t *testing.T) {
-
-	qs := []question144{
-
-		question144{
-			para144{[]int{}},
-			ans144{[]int{}},
-		},
-
-		question144{
-			para144{[]int{1}},
-			ans144{[]int{1}},
-		},
-
-		question144{
-			para144{[]int{1, NULL, 2, 3}},
-			ans144{[]int{1, 2, 3}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 144------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans144, q.para144
-		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", preorderTraversal(root))
-	}
-	fmt.Printf("\n\n\n")
-}
-
-var NULL = -1 << 63
-
-// Ints2TreeNode 利用 []int 生成 *TreeNode
-func Ints2TreeNode(ints []int) *TreeNode {
-	n := len(ints)
-	if n == 0 {
-		return nil
-	}
-	root := &TreeNode{
-		Val: ints[0],
-	}
-	queue := make([]*TreeNode, 1, n*2)
-	queue[0] = root
-	i := 1
-	for i < n {
-		node := queue[0]
-		queue = queue[1:]
-		if i < n && ints[i] != NULL {
-			node.Left = &TreeNode{Val: ints[i]}
-			queue = append(queue, node.Left)
-		}
-		i++
-		if i < n && ints[i] != NULL {
-			node.Right = &TreeNode{Val: ints[i]}
-			queue = append(queue, node.Right)
-		}
-		i++
-	}
-	return root
-}
-
-func indexOf(val int, nums []int) int {
-	for i, v := range nums {
-		if v == val {
-			return i
-		}
-	}
-
-	msg := fmt.Sprintf("%d 不存在于 %v 中", val, nums)
-	panic(msg)
-}
-
-// PreIn2Tree 把 preorder 和 inorder 切片转换成 二叉树
-func PreIn2Tree(pre, in []int) *TreeNode {
-	if len(pre) != len(in) {
-		panic("preIn2Tree 中两个切片的长度不相等")
-	}
-	if len(in) == 0 {
-		return nil
-	}
-	res := &TreeNode{
-		Val: pre[0],
-	}
-	if len(in) == 1 {
-		return res
-	}
-	idx := indexOf(res.Val, in)
-	res.Left = PreIn2Tree(pre[1:idx+1], in[:idx])
-	res.Right = PreIn2Tree(pre[idx+1:], in[idx+1:])
-	return res
-}
-
-// InPost2Tree 把 inorder 和 postorder 切片转换成 二叉树
-func InPost2Tree(in, post []int) *TreeNode {
-	if len(post) != len(in) {
-		panic("inPost2Tree 中两个切片的长度不相等")
-	}
-	if len(in) == 0 {
-		return nil
-	}
-	res := &TreeNode{
-		Val: post[len(post)-1],
-	}
-	if len(in) == 1 {
-		return res
-	}
-	idx := indexOf(res.Val, in)
-	res.Left = InPost2Tree(in[:idx], post[:idx])
-	res.Right = InPost2Tree(in[idx+1:], post[idx:len(post)-1])
-	return res
-}
-
-// Tree2Preorder 把 二叉树 转换成 preorder 的切片
-func Tree2Preorder(root *TreeNode) []int {
-	if root == nil {
-		return nil
-	}
-	if root.Left == nil && root.Right == nil {
-		return []int{root.Val}
-	}
-	res := []int{root.Val}
-	res = append(res, Tree2Preorder(root.Left)...)
-	res = append(res, Tree2Preorder(root.Right)...)
-	return res
-}
-
-// Tree2Inorder 把 二叉树转换成 inorder 的切片
-func Tree2Inorder(root *TreeNode) []int {
-	if root == nil {
-		return nil
-	}
-	if root.Left == nil && root.Right == nil {
-		return []int{root.Val}
-	}
-	res := Tree2Inorder(root.Left)
-	res = append(res, root.Val)
-	res = append(res, Tree2Inorder(root.Right)...)
-	return res
-}
-
-// Tree2Postorder 把 二叉树 转换成 postorder 的切片
-func Tree2Postorder(root *TreeNode) []int {
-	if root == nil {
-		return nil
-	}
-	if root.Left == nil && root.Right == nil {
-		return []int{root.Val}
-	}
-	res := Tree2Postorder(root.Left)
-	res = append(res, Tree2Postorder(root.Right)...)
-	res = append(res, root.Val)
-	return res
-}
-
-// Equal return ture if tn == a
-func (tn *TreeNode) Equal(a *TreeNode) bool {
-	if tn == nil && a == nil {
-		return true
-	}
-	if tn == nil || a == nil || tn.Val != a.Val {
-		return false
-	}
-	return tn.Left.Equal(a.Left) && tn.Right.Equal(a.Right)
-}
-
-// Tree2ints 把 *TreeNode 按照行还原成 []int
-func Tree2ints(tn *TreeNode) []int {
-	res := make([]int, 0, 1024)
-	queue := []*TreeNode{tn}
-	for len(queue) > 0 {
-		size := len(queue)
-		for i := 0; i < size; i++ {
-			nd := queue[i]
-			if nd == nil {
-				res = append(res, NULL)
-			} else {
-				res = append(res, nd.Val)
-				queue = append(queue, nd.Left, nd.Right)
-			}
-		}
-		queue = queue[size:]
-	}
-	i := len(res)
-	for i > 0 && res[i-1] == NULL {
-		i--
-	}
-	return res[:i]
-}
diff --git a/Algorithms/0148. Sort List/148. Sort List.go b/Algorithms/0148. Sort List/148. Sort List.go
deleted file mode 100644
index 049dbb58a..000000000
--- a/Algorithms/0148. Sort List/148. Sort List.go	
+++ /dev/null
@@ -1,57 +0,0 @@
-package leetcode
-
-/**
- * Definition for singly-linked list.
- * type ListNode struct {
- *     Val int
- *     Next *ListNode
- * }
- */
-func sortList(head *ListNode) *ListNode {
-	length := 0
-	cur := head
-	for cur != nil {
-		length++
-		cur = cur.Next
-	}
-	if length <= 1 {
-		return head
-	}
-
-	middleNode := middleNode1(head)
-	cur = middleNode.Next
-	middleNode.Next = nil
-	middleNode = cur
-
-	left := sortList(head)
-	right := sortList(middleNode)
-	return mergeTwoLists148(left, right)
-}
-
-func middleNode1(head *ListNode) *ListNode {
-	if head == nil || head.Next == nil {
-		return head
-	}
-	p1 := head
-	p2 := head
-	for p2.Next != nil && p2.Next.Next != nil {
-		p1 = p1.Next
-		p2 = p2.Next.Next
-	}
-	return p1
-}
-
-func mergeTwoLists148(l1 *ListNode, l2 *ListNode) *ListNode {
-	if l1 == nil {
-		return l2
-	}
-	if l2 == nil {
-		return l1
-	}
-	if l1.Val < l2.Val {
-		l1.Next = mergeTwoLists(l1.Next, l2)
-		return l1
-	}
-	l2.Next = mergeTwoLists(l1, l2.Next)
-	return l2
-}
diff --git a/Algorithms/0150. Evaluate Reverse Polish Notation/150. Evaluate Reverse Polish Notation.go b/Algorithms/0150. Evaluate Reverse Polish Notation/150. Evaluate Reverse Polish Notation.go
deleted file mode 100644
index eac594102..000000000
--- a/Algorithms/0150. Evaluate Reverse Polish Notation/150. Evaluate Reverse Polish Notation.go	
+++ /dev/null
@@ -1,52 +0,0 @@
-package leetcode
-
-import (
-	"strconv"
-)
-
-func evalRPN(tokens []string) int {
-	if len(tokens) == 1 {
-		i, _ := strconv.Atoi(tokens[0])
-		return i
-	}
-	stack, top := []int{}, 0
-	for _, v := range tokens {
-		switch v {
-		case "+":
-			{
-				sum := stack[top-2] + stack[top-1]
-				stack = stack[:top-2]
-				stack = append(stack, sum)
-				top--
-			}
-		case "-":
-			{
-				sub := stack[top-2] - stack[top-1]
-				stack = stack[:top-2]
-				stack = append(stack, sub)
-				top--
-			}
-		case "*":
-			{
-				mul := stack[top-2] * stack[top-1]
-				stack = stack[:top-2]
-				stack = append(stack, mul)
-				top--
-			}
-		case "/":
-			{
-				div := stack[top-2] / stack[top-1]
-				stack = stack[:top-2]
-				stack = append(stack, div)
-				top--
-			}
-		default:
-			{
-				i, _ := strconv.Atoi(v)
-				stack = append(stack, i)
-				top++
-			}
-		}
-	}
-	return stack[0]
-}
diff --git a/Algorithms/0151. Reverse Words in a String/README.md b/Algorithms/0151. Reverse Words in a String/README.md
deleted file mode 100755
index fa7ff4f2b..000000000
--- a/Algorithms/0151. Reverse Words in a String/README.md	
+++ /dev/null
@@ -1,57 +0,0 @@
-# [151. Reverse Words in a String](https://leetcode.com/problems/reverse-words-in-a-string/)
-
-
-
-## 题目:
-
-Given an input string, reverse the string word by word.
-
-**Example 1:**
-
-    Input: "the sky is blue"
-    Output: "blue is sky the"
-
-**Example 2:**
-
-    Input: "  hello world!  "
-    Output: "world! hello"
-    Explanation: Your reversed string should not contain leading or trailing spaces.
-
-**Example 3:**
-
-    Input: "a good   example"
-    Output: "example good a"
-    Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.
-
-**Note:**
-
-- A word is defined as a sequence of non-space characters.
-- Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
-- You need to reduce multiple spaces between two words to a single space in the reversed string.
-
-**Follow up:**
-
-For C programmers, try to solve it *in-place* in *O*(1) extra space.
-
-
-## 题目大意
-
-给定一个字符串，逐个翻转字符串中的每个单词。
-
-说明：
-
-- 无空格字符构成一个单词。
-- 输入字符串可以在前面或者后面包含多余的空格，但是反转后的字符不能包括。
-- 如果两个单词间有多余的空格，将反转后单词间的空格减少到只含一个。
- 
-
-进阶：
-
-- 请选用 C 语言的用户尝试使用 O(1) 额外空间复杂度的原地解法。
-
-
-## 解题思路
-
-
-- 给出一个中间有空格分隔的字符串，要求把这个字符串按照单词的维度前后翻转。
-- 依照题意，先把字符串按照空格分隔成每个小单词，然后把单词前后翻转，最后再把每个单词中间添加空格。
diff --git a/Algorithms/0152. Maximum Product Subarray/152. Maximum Product Subarray.go b/Algorithms/0152. Maximum Product Subarray/152. Maximum Product Subarray.go
deleted file mode 100644
index eb2c367ab..000000000
--- a/Algorithms/0152. Maximum Product Subarray/152. Maximum Product Subarray.go	
+++ /dev/null
@@ -1,14 +0,0 @@
-package leetcode
-
-func maxProduct(nums []int) int {
-	minimum, maximum, res := nums[0], nums[0], nums[0]
-	for i := 1; i < len(nums); i++ {
-		if nums[i] < 0 {
-			maximum, minimum = minimum, maximum
-		}
-		maximum = max(nums[i], maximum*nums[i])
-		minimum = min(nums[i], minimum*nums[i])
-		res = max(res, maximum)
-	}
-	return res
-}
diff --git a/Algorithms/0153. Find Minimum in Rotated Sorted Array/README.md b/Algorithms/0153. Find Minimum in Rotated Sorted Array/README.md
deleted file mode 100755
index 17cf0ab93..000000000
--- a/Algorithms/0153. Find Minimum in Rotated Sorted Array/README.md	
+++ /dev/null
@@ -1,35 +0,0 @@
-# [153. Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)
-
-
-## 题目:
-
-Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
-
-(i.e., `[0,1,2,4,5,6,7]` might become `[4,5,6,7,0,1,2]`).
-
-Find the minimum element.
-
-You may assume no duplicate exists in the array.
-
-**Example 1:**
-
-    Input: [3,4,5,1,2] 
-    Output: 1
-
-**Example 2:**
-
-    Input: [4,5,6,7,0,1,2]
-    Output: 0
-
-
-## 题目大意
-
-假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。请找出其中最小的元素。
-
-你可以假设数组中不存在重复元素。
-
-
-## 解题思路
-
-- 给出一个原本从小到大排序过的数组，但是在某一个分割点上，把数组切分后的两部分对调位置，数值偏大的放到了数组的前部。求这个数组中最小的元素。
-- 求数组最小的元素其实就是找分割点，前一个数比当前数大，后一个数比当前数也要大。可以用二分搜索查找，需要查找的两个有序区间。时间复杂度  O(log n)。这一题也可以用暴力解法，从头开始遍历，动态维护一个最小值即可，时间复杂度 O(n)。
diff --git a/Algorithms/0154. Find Minimum in Rotated Sorted Array II/README.md b/Algorithms/0154. Find Minimum in Rotated Sorted Array II/README.md
deleted file mode 100755
index 4cb627457..000000000
--- a/Algorithms/0154. Find Minimum in Rotated Sorted Array II/README.md	
+++ /dev/null
@@ -1,40 +0,0 @@
-# [154. Find Minimum in Rotated Sorted Array II](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/)
-
-
-## 题目:
-
-Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
-
-(i.e., `[0,1,2,4,5,6,7]` might become `[4,5,6,7,0,1,2]`).
-
-Find the minimum element.
-
-The array may contain duplicates.
-
-**Example 1:**
-
-    Input: [1,3,5]
-    Output: 1
-
-**Example 2:**
-
-    Input: [2,2,2,0,1]
-    Output: 0
-
-**Note:**
-
-- This is a follow up problem to [Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/).
-- Would allow duplicates affect the run-time complexity? How and why?
-
-
-## 题目大意
-
-假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。请找出其中最小的元素。
-
-注意数组中可能存在重复的元素。
-
-## 解题思路
-
-
-- 给出一个原本从小到大排序过的数组，注意数组中有重复的元素。但是在某一个分割点上，把数组切分后的两部分对调位置，数值偏大的放到了数组的前部。求这个数组中最小的元素。
-- 这一题是第 153 题的加强版，增加了重复元素的条件。但是实际做法还是没有变，还是用二分搜索，只不过在相等元素上多增加一个判断即可。时间复杂度 O(log n)。
diff --git a/Algorithms/0162. Find Peak Element/README.md b/Algorithms/0162. Find Peak Element/README.md
deleted file mode 100755
index 9304f0ab4..000000000
--- a/Algorithms/0162. Find Peak Element/README.md	
+++ /dev/null
@@ -1,43 +0,0 @@
-# [162. Find Peak Element](https://leetcode.com/problems/find-peak-element/)
-
-
-## 题目:
-
-A peak element is an element that is greater than its neighbors.
-
-Given an input array `nums`, where `nums[i] ≠ nums[i+1]`, find a peak element and return its index.
-
-The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
-
-You may imagine that `nums[-1] = nums[n] = -∞`.
-
-**Example 1:**
-
-    Input: nums = [1,2,3,1]
-    Output: 2
-    Explanation: 3 is a peak element and your function should return the index number 2.
-
-**Example 2:**
-
-    Input: nums = [1,2,1,3,5,6,4]
-    Output: 1 or 5 
-    Explanation: Your function can return either index number 1 where the peak element is 2, 
-                 or index number 5 where the peak element is 6.
-
-**Note:**
-
-Your solution should be in logarithmic complexity.
-
-## 题目大意
-
-峰值元素是指其值大于左右相邻值的元素。给定一个输入数组 nums，其中 nums[i] ≠ nums[i+1]，找到峰值元素并返回其索引。数组可能包含多个峰值，在这种情况下，返回任何一个峰值所在位置即可。你可以假设 nums[-1] = nums[n] = -∞。
-
-说明:
-
-- 你的解法应该是 O(logN) 时间复杂度的。
-
-
-## 解题思路
-
-- 给出一个数组，数组里面存在多个“山峰”，(山峰的定义是，下标 `i` 比 `i-1`、`i+1` 位置上的元素都要大)，找到这个“山峰”，并输出其中一个山峰的下标。
-- 这一题是第 852 题的伪加强版，第 852 题中只存在一个山峰，这一题存在多个山峰。但是实际上搜索的代码是一样的，因为此题只要求随便输出一个山峰的下标即可。思路同第 852 题。
diff --git a/Algorithms/0169. Majority Element/README.md b/Algorithms/0169. Majority Element/README.md
deleted file mode 100755
index b2e898acf..000000000
--- a/Algorithms/0169. Majority Element/README.md	
+++ /dev/null
@@ -1,29 +0,0 @@
-# [169. Majority Element](https://leetcode.com/problems/majority-element/)
-
-
-## 题目:
-
-Given an array of size n, find the majority element. The majority element is the element that appears **more than** `⌊ n/2 ⌋` times.
-
-You may assume that the array is non-empty and the majority element always exist in the array.
-
-**Example 1:**
-
-    Input: [3,2,3]
-    Output: 3
-
-**Example 2:**
-
-    Input: [2,2,1,1,1,2,2]
-    Output: 2
-
-## 题目大意
-
-
-给定一个大小为 n 的数组，找到其中的众数。众数是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素。你可以假设数组是非空的，并且给定的数组总是存在众数。
-
-
-## 解题思路
-
-- 题目要求找出数组中出现次数大于 `⌊ n/2 ⌋` 次的数。要求空间复杂度为 O(1)。简单题。
-- 这一题利用的算法是 Boyer-Moore Majority Vote Algorithm。[https://www.zhihu.com/question/49973163/answer/235921864](https://www.zhihu.com/question/49973163/answer/235921864)
\ No newline at end of file
diff --git a/Algorithms/0174. Dungeon Game/174. Dungeon Game_test.go b/Algorithms/0174. Dungeon Game/174. Dungeon Game_test.go
deleted file mode 100644
index baa06fcb5..000000000
--- a/Algorithms/0174. Dungeon Game/174. Dungeon Game_test.go	
+++ /dev/null
@@ -1,57 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question174 struct {
-	para174
-	ans174
-}
-
-// para 是参数
-// one 代表第一个参数
-type para174 struct {
-	s [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans174 struct {
-	one int
-}
-
-func Test_Problem174(t *testing.T) {
-
-	qs := []question174{
-
-		question174{
-			para174{[][]int{[]int{2, 1}, []int{1, -1}}},
-			ans174{1},
-		},
-
-		question174{
-			para174{[][]int{[]int{-3, 5}}},
-			ans174{4},
-		},
-
-		question174{
-			para174{[][]int{[]int{100}}},
-			ans174{1},
-		},
-
-		question174{
-			para174{[][]int{[]int{-2, -3, 3}, []int{-5, -10, 1}, []int{10, 30, -5}}},
-			ans174{7},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 174------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans174, q.para174
-		fmt.Printf("【input】:%v       【output】:%v\n", p, calculateMinimumHP1(p.s))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0174. Dungeon Game/README.md b/Algorithms/0174. Dungeon Game/README.md
deleted file mode 100755
index 756d39d48..000000000
--- a/Algorithms/0174. Dungeon Game/README.md	
+++ /dev/null
@@ -1,49 +0,0 @@
-# [174. Dungeon Game](https://leetcode.com/problems/dungeon-game/)
-
-## 题目:
-
-The demons had captured the princess (**P**) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (**K**) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
-
-The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
-
-Some of the rooms are guarded by demons, so the knight loses health (*negative* integers) upon entering these rooms; other rooms are either empty (*0's*) or contain magic orbs that increase the knight's health (*positive* integers).
-
-In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
-
-**Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.**
-
-For example, given the dungeon below, the initial health of the knight must be at least **7** if he follows the optimal path `RIGHT-> RIGHT -> DOWN -> DOWN`.
-
-
-![](https://img.halfrost.com/Leetcode/leetcode_174_0.png)
-
-**Note:**
-
-- The knight's health has no upper bound.
-- Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.
-
-
-## 题目大意
-
-一些恶魔抓住了公主（P）并将她关在了地下城的右下角。地下城是由 M x N 个房间组成的二维网格。我们英勇的骑士（K）最初被安置在左上角的房间里，他必须穿过地下城并通过对抗恶魔来拯救公主。
-
-骑士的初始健康点数为一个正整数。如果他的健康点数在某一时刻降至 0 或以下，他会立即死亡。
-
-有些房间由恶魔守卫，因此骑士在进入这些房间时会失去健康点数（若房间里的值为负整数，则表示骑士将损失健康点数）；其他房间要么是空的（房间里的值为 0），要么包含增加骑士健康点数的魔法球（若房间里的值为正整数，则表示骑士将增加健康点数）。
-
-为了尽快到达公主，骑士决定每次只向右或向下移动一步。编写一个函数来计算确保骑士能够拯救到公主所需的最低初始健康点数。
-
-说明:
-
-- 骑士的健康点数没有上限。
-- 任何房间都可能对骑士的健康点数造成威胁，也可能增加骑士的健康点数，包括骑士进入的左上角房间以及公主被监禁的右下角房间。
-
-## 解题思路
-
-- 在二维地图上给出每个格子扣血数，负数代表扣血，正数代表补血。左上角第一个格子是起点，右下角最后一个格子是终点。问骑士初始最少多少血才能走完迷宫，顺利营救位于终点的公主。需要注意的是，起点和终点都会对血量进行影响。每到一个格子，骑士的血都不能少于 1，一旦少于 1 点血，骑士就会死去。
-- 这一题首先想到的解题思路是动态规划。从终点逆推回起点。`dp[i][j]` 代表骑士进入坐标为 `(i,j)` 的格子之前最少的血量值。 那么 `dp[m-1][n-1]` 应该同时满足两个条件，`dp[m-1][n-1] + dungeon[m-1][n-1] ≥ 1` 并且 `dp[m-1][n-1] ≥ 1`，由于这两个不等式的方向是相同的，取交集以后，起决定作用的是数轴最右边的数，即 `max(1-dungeon[m-1][n-1] , 1)`。算出 `dp[m-1][n-1]` 以后，接着可以推出 `dp[m-1][i]` 这一行和 `dp[i][n-1]` 这一列的值。因为骑士只能往右走和往下走。往回推，即只能往上走和往左走。到这里，DP 的初始条件都准备好了。那么状态转移方程是什么呢？分析一般的情况，`dp[i][j]` 这个值应该是和 `dp[i+1][j]` 和 `dp[i][j+1]` 这两者有关系。即 `dp[i][j]` 经过自己本格子的扣血以后，要能至少满足下一行和右一列格子血量的最少要求。并且自己的血量也应该 `≥1`。即需要满足下面这两组不等式。   
-	 ![](https://img.halfrost.com/Leetcode/leetcode_174_1.png)  
-	 ![](https://img.halfrost.com/Leetcode/leetcode_174_2.png)    
-    上面不等式中第一组不等式是满足下一行格子的最低血量要求，第二组不等式是满足右一列格子的最低血量要求。第一个式子化简即 `dp[i][j] = max(1, dp[i+1][j]-dungeon[i][j])`，第二个式子化简即 `dp[i][j] = max(1, dp[i][j+1]-dungeon[i][j])`。求得了这两种走法的最低血量值，从这两个值里面取最小，即是当前格子所需的最低血量，所以状态转移方程为 `dp[i][j] = min(max(1, dp[i][j+1]-dungeon[i][j]), max(1, dp[i+1][j]-dungeon[i][j]))`。DP 完成以后，`dp[0][0]` 中记录的就是骑士初始最低血量值。时间复杂度 O(m\*n)，空间复杂度 O(m\*n)。
-
-- 这一题还可以用二分搜索来求解。骑士的血量取值范围一定是在 `[1，+∞)` 这个区间内。那么二分这个区间，每次二分的中间值，再用 dp 在地图中去判断是否能到达终点，如果能，就缩小搜索空间至 `[1，mid]`，否则搜索空间为 `[mid + 1，+∞)` 。时间复杂度 O(m\*n\* log math.MaxInt64)，空间复杂度 O(m\*n)。
diff --git a/Algorithms/0187. Repeated DNA Sequences/README.md b/Algorithms/0187. Repeated DNA Sequences/README.md
deleted file mode 100755
index b0c4b7fa4..000000000
--- a/Algorithms/0187. Repeated DNA Sequences/README.md	
+++ /dev/null
@@ -1,24 +0,0 @@
-# [187. Repeated DNA Sequences](https://leetcode.com/problems/repeated-dna-sequences/)
-
-
-## 题目:
-
-All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA.
-
-Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.
-
-**Example:**
-
-    Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
-    
-    Output: ["AAAAACCCCC", "CCCCCAAAAA"]
-
-
-## 题目大意
-
-所有 DNA 由一系列缩写为 A，C，G 和 T 的核苷酸组成，例如：“ACGAATTCCG”。在研究 DNA 时，识别 DNA 中的重复序列有时会对研究非常有帮助。编写一个函数来查找 DNA 分子中所有出现超多一次的10个字母长的序列（子串）。
-
-## 解题思路
-
-- 这一题不用位运算比较好做，维护一个长度为 10 的字符串，在 map 中出现次数 > 1 就输出。
-- 用位运算想做这一题，需要动态的维护长度为 10 的 hashkey，先计算开头长度为 9 的 hash，在往后面扫描的过程中，如果长度超过了 10 ，就移除 hash 开头的一个字符，加入后面一个字符。具体做法是先将 ATCG 变成 00，01，10，11 的编码，那么长度为 10 ，hashkey 就需要维护在 20 位。mask = 0xFFFFF 就是 20 位的。维护了 hashkey 以后，根据这个 hashkey 进行去重和统计频次。
diff --git a/Algorithms/0190. Reverse Bits/190. Reverse Bits_test.go b/Algorithms/0190. Reverse Bits/190. Reverse Bits_test.go
deleted file mode 100644
index 637d65e76..000000000
--- a/Algorithms/0190. Reverse Bits/190. Reverse Bits_test.go	
+++ /dev/null
@@ -1,47 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question190 struct {
-	para190
-	ans190
-}
-
-// para 是参数
-// one 代表第一个参数
-type para190 struct {
-	one uint32
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans190 struct {
-	one uint32
-}
-
-func Test_Problem190(t *testing.T) {
-
-	qs := []question190{
-
-		question190{
-			para190{43261596},
-			ans190{964176192},
-		},
-
-		question190{
-			para190{4294967293},
-			ans190{3221225471},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 190------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans190, q.para190
-		fmt.Printf("【input】:%v       【output】:%v\n", p, reverseBits(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0190. Reverse Bits/README.md b/Algorithms/0190. Reverse Bits/README.md
deleted file mode 100755
index b6fda04ce..000000000
--- a/Algorithms/0190. Reverse Bits/README.md	
+++ /dev/null
@@ -1,35 +0,0 @@
-# [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)
-
-
-## 题目:
-
-Reverse bits of a given 32 bits unsigned integer.
-
-**Example 1:**
-
-    Input: 00000010100101000001111010011100
-    Output: 00111001011110000010100101000000
-    Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
-
-**Example 2:**
-
-    Input: 11111111111111111111111111111101
-    Output: 10111111111111111111111111111111
-    Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10101111110010110010011101101001.
-
-**Note:**
-
-- Note that in some languages such as Java, there is no unsigned integer type. In this case, both input and output will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
-- In Java, the compiler represents the signed integers using [2's complement notation](https://en.wikipedia.org/wiki/Two%27s_complement). Therefore, in **Example 2** above the input represents the signed integer `-3` and the output represents the signed integer `-1073741825`.
-
-## 题目大意
-
-颠倒给定的 32 位无符号整数的二进制位。提示：
-
-- 请注意，在某些语言（如 Java）中，没有无符号整数类型。在这种情况下，输入和输出都将被指定为有符号整数类型，并且不应影响您的实现，因为无论整数是有符号的还是无符号的，其内部的二进制表示形式都是相同的。
-- 在 Java 中，编译器使用二进制补码记法来表示有符号整数。因此，在上面的 示例 2 中，输入表示有符号整数 -3，输出表示有符号整数 -1073741825。
-
-## 解题思路
-
-- 简单题，要求反转 32 位的二进制位。
-- 把 num 往右移动，不断的消灭右边最低位的 1，将这个 1 给 res，res 不断的左移即可实现反转二进制位的目的。
diff --git a/Algorithms/0191. Number of 1 Bits/191. Number of 1 Bits_test.go b/Algorithms/0191. Number of 1 Bits/191. Number of 1 Bits_test.go
deleted file mode 100644
index be7ff715f..000000000
--- a/Algorithms/0191. Number of 1 Bits/191. Number of 1 Bits_test.go	
+++ /dev/null
@@ -1,46 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question191 struct {
-	para191
-	ans191
-}
-
-// para 是参数
-// one 代表第一个参数
-type para191 struct {
-	one uint32
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans191 struct {
-	one int
-}
-
-func Test_Problem191(t *testing.T) {
-
-	qs := []question191{
-
-		question191{
-			para191{5},
-			ans191{1},
-		},
-		question191{
-			para191{13},
-			ans191{2},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 191------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans191, q.para191
-		fmt.Printf("【input】:%v       【output】:%v\n", p, hammingWeight(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0199. Binary Tree Right Side View/199. Binary Tree Right Side View.go b/Algorithms/0199. Binary Tree Right Side View/199. Binary Tree Right Side View.go
deleted file mode 100644
index 70454692b..000000000
--- a/Algorithms/0199. Binary Tree Right Side View/199. Binary Tree Right Side View.go	
+++ /dev/null
@@ -1,41 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-func rightSideView(root *TreeNode) []int {
-	if root == nil {
-		return []int{}
-	}
-	queue := []*TreeNode{}
-	queue = append(queue, root)
-	curNum, nextLevelNum, res, tmp := 1, 0, []int{}, []int{}
-	for len(queue) != 0 {
-		if curNum > 0 {
-			node := queue[0]
-			if node.Left != nil {
-				queue = append(queue, node.Left)
-				nextLevelNum++
-			}
-			if node.Right != nil {
-				queue = append(queue, node.Right)
-				nextLevelNum++
-			}
-			curNum--
-			tmp = append(tmp, node.Val)
-			queue = queue[1:]
-		}
-		if curNum == 0 {
-			res = append(res, tmp[len(tmp)-1])
-			curNum = nextLevelNum
-			nextLevelNum = 0
-			tmp = []int{}
-		}
-	}
-	return res
-}
diff --git a/Algorithms/0200. Number of Islands/200. Number of Islands_test.go b/Algorithms/0200. Number of Islands/200. Number of Islands_test.go
deleted file mode 100644
index 3f7805d8a..000000000
--- a/Algorithms/0200. Number of Islands/200. Number of Islands_test.go	
+++ /dev/null
@@ -1,57 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question200 struct {
-	para200
-	ans200
-}
-
-// para 是参数
-// one 代表第一个参数
-type para200 struct {
-	one [][]byte
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans200 struct {
-	one int
-}
-
-func Test_Problem200(t *testing.T) {
-
-	qs := []question200{
-
-		question200{
-			para200{[][]byte{
-				[]byte{'1', '1', '1', '1', '0'},
-				[]byte{'1', '1', '0', '1', '0'},
-				[]byte{'1', '1', '0', '0', '0'},
-				[]byte{'0', '0', '0', '0', '0'},
-			}},
-			ans200{1},
-		},
-
-		question200{
-			para200{[][]byte{
-				[]byte{'1', '1', '0', '0', '0'},
-				[]byte{'1', '1', '0', '0', '0'},
-				[]byte{'0', '0', '1', '0', '0'},
-				[]byte{'0', '0', '0', '1', '1'},
-			}},
-			ans200{3},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 200------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans200, q.para200
-		fmt.Printf("【input】:%v       【output】:%v\n", p, numIslands(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0201. Bitwise AND of Numbers Range/README.md b/Algorithms/0201. Bitwise AND of Numbers Range/README.md
deleted file mode 100755
index a6e34d0d8..000000000
--- a/Algorithms/0201. Bitwise AND of Numbers Range/README.md	
+++ /dev/null
@@ -1,35 +0,0 @@
-# [201. Bitwise AND of Numbers Range](https://leetcode.com/problems/bitwise-and-of-numbers-range/)
-
-
-## 题目:
-
-Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.
-
-**Example 1:**
-
-    Input: [5,7]
-    Output: 4
-
-**Example 2:**
-
-    Input: [0,1]
-    Output: 0
-
-## 题目大意
-
-给定范围 [m, n]，其中 0 <= m <= n <= 2147483647，返回此范围内所有数字的按位与（包含 m, n 两端点）。
-
-
-## 解题思路
-
-- 这一题要求输出 [m,n] 区间内所有数的 AND 与操作之后的结果。
-- 举个例子，假设区间是 [26,30]，那么这个区间内的数用二进制表示出来为：
-
-    11010  
-    11011  
-    11100   
-    11101   
-    11110  
-
-- 可以观察到，把这些数都 AND 起来，只要有 0 的位，最终结果都是 0，所以需要从右往前找到某一位上不为 0 的。不断的右移左边界和右边界，把右边的 0 都移走，直到它们俩相等，就找到了某一位上开始都不为 0 的了。在右移的过程中记录下右移了多少位，最后把 m 或者 n 的右边添上 0 即可。按照上面这个例子来看，11000 是最终的结果。
-- 这一题还有解法二，还是以 [26,30] 这个区间为例。这个区间内的数末尾 3 位不断的 0，1 变化着。那么如果把末尾的 1 都打掉，就是最终要求的结果了。当 n == m 或者 n < m 的时候就退出循环，说明后面不同的位数已经都被抹平了，1 都被打掉为 0 了。所以关键的操作为 `n &= (n - 1)` ，清除最低位的 1 。这个算法名叫 `Brian Kernighan` 算法。
diff --git a/Algorithms/0202. Happy Number/202. Happy Number.go b/Algorithms/0202. Happy Number/202. Happy Number.go
deleted file mode 100644
index 3764600ef..000000000
--- a/Algorithms/0202. Happy Number/202. Happy Number.go	
+++ /dev/null
@@ -1,27 +0,0 @@
-package leetcode
-
-func isHappy(n int) bool {
-	if n == 0 {
-		return false
-	}
-	res := 0
-	num := n
-	record := map[int]int{}
-	for {
-		for num != 0 {
-			res += (num % 10) * (num % 10)
-			num = num / 10
-		}
-		if _, ok := record[res]; !ok {
-			if res == 1 {
-				return true
-			}
-			record[res] = res
-			num = res
-			res = 0
-			continue
-		} else {
-			return false
-		}
-	}
-}
diff --git a/Algorithms/0204. Count Primes/README.md b/Algorithms/0204. Count Primes/README.md
deleted file mode 100755
index 94bdd565c..000000000
--- a/Algorithms/0204. Count Primes/README.md	
+++ /dev/null
@@ -1,22 +0,0 @@
-# [204. Count Primes](https://leetcode.com/problems/count-primes/)
-
-
-## 题目:
-
-Count the number of prime numbers less than a non-negative number, **n**.
-
-**Example:**
-
-    Input: 10
-    Output: 4
-    Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.
-
-
-## 题目大意
-
-统计所有小于非负整数 n 的质数的数量。
-
-
-## 解题思路
-
-- 给出一个数字 n，要求输出小于 n 的所有素数的个数总和。简单题。
diff --git a/Algorithms/0206. Reverse-Linked-List/206. Reverse Linked List_test.go b/Algorithms/0206. Reverse-Linked-List/206. Reverse Linked List_test.go
deleted file mode 100644
index 1ce806f6a..000000000
--- a/Algorithms/0206. Reverse-Linked-List/206. Reverse Linked List_test.go	
+++ /dev/null
@@ -1,73 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question206 struct {
-	para206
-	ans206
-}
-
-// para 是参数
-// one 代表第一个参数
-type para206 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans206 struct {
-	one []int
-}
-
-func Test_Problem206(t *testing.T) {
-
-	qs := []question206{
-
-		question206{
-			para206{[]int{1, 2, 3, 4, 5}},
-			ans206{[]int{5, 4, 3, 2, 1}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 206------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans206, q.para206
-		fmt.Printf("【input】:%v       【output】:%v\n", p, l2s(reverseList(s2l(p.one))))
-	}
-}
-
-// convert *ListNode to []int
-func l2s(head *ListNode) []int {
-	res := []int{}
-
-	for head != nil {
-		res = append(res, head.Val)
-		head = head.Next
-	}
-
-	return res
-}
-
-// convert []int to *ListNode
-func s2l(nums []int) *ListNode {
-	if len(nums) == 0 {
-		return nil
-	}
-
-	res := &ListNode{
-		Val: nums[0],
-	}
-	temp := res
-	for i := 1; i < len(nums); i++ {
-		temp.Next = &ListNode{
-			Val: nums[i],
-		}
-		temp = temp.Next
-	}
-
-	return res
-}
diff --git a/Algorithms/0209. Minimum Size Subarray Sum/209. Minimum Size Subarray Sum.go b/Algorithms/0209. Minimum Size Subarray Sum/209. Minimum Size Subarray Sum.go
deleted file mode 100644
index 277d1f6f9..000000000
--- a/Algorithms/0209. Minimum Size Subarray Sum/209. Minimum Size Subarray Sum.go	
+++ /dev/null
@@ -1,27 +0,0 @@
-package leetcode
-
-func minSubArrayLen(s int, nums []int) int {
-	n := len(nums)
-	if n == 0 {
-		return 0
-	}
-	l, r := 0, -1
-	res := n + 1
-	sum := 0
-	for l < n {
-		if (r+1) < n && sum < s {
-			r++
-			sum += nums[r]
-		} else {
-			sum -= nums[l]
-			l++
-		}
-		if sum >= s {
-			res = min(res, r-l+1)
-		}
-	}
-	if res == n+1 {
-		return 0
-	}
-	return res
-}
diff --git a/Algorithms/0210. Course Schedule II/210. Course Schedule II_test.go b/Algorithms/0210. Course Schedule II/210. Course Schedule II_test.go
deleted file mode 100644
index 6e75c853d..000000000
--- a/Algorithms/0210. Course Schedule II/210. Course Schedule II_test.go	
+++ /dev/null
@@ -1,58 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question210 struct {
-	para210
-	ans210
-}
-
-// para 是参数
-// one 代表第一个参数
-type para210 struct {
-	one int
-	pre [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans210 struct {
-	one []int
-}
-
-func Test_Problem210(t *testing.T) {
-
-	qs := []question210{
-
-		question210{
-			para210{2, [][]int{[]int{1, 0}}},
-			ans210{[]int{0, 1}},
-		},
-
-		question210{
-			para210{2, [][]int{[]int{1, 0}, []int{0, 1}}},
-			ans210{[]int{0, 1, 2, 3}},
-		},
-
-		question210{
-			para210{4, [][]int{[]int{1, 0}, []int{2, 0}, []int{3, 1}, []int{3, 2}}},
-			ans210{[]int{0, 1, 2, 3}},
-		},
-
-		question210{
-			para210{3, [][]int{[]int{1, 0}, []int{1, 2}, []int{0, 1}}},
-			ans210{[]int{}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 210------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans210, q.para210
-		fmt.Printf("【input】:%v       【output】:%v\n", p, findOrder(p.one, p.pre))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0211. Add and Search Word - Data structure design/README.md b/Algorithms/0211. Add and Search Word - Data structure design/README.md
deleted file mode 100755
index 57f8c3358..000000000
--- a/Algorithms/0211. Add and Search Word - Data structure design/README.md	
+++ /dev/null
@@ -1,34 +0,0 @@
-# [211. Add and Search Word - Data structure design](https://leetcode.com/problems/add-and-search-word-data-structure-design/)
-
-
-## 题目
-
-Design a data structure that supports the following two operations:
-
-    void addWord(word)
-    bool search(word)
-
-search(word) can search a literal word or a regular expression string containing only letters `a-z` or `.`. A `.` means it can represent any one letter.
-
-**Example:**
-
-    addWord("bad")
-    addWord("dad")
-    addWord("mad")
-    search("pad") -> false
-    search("bad") -> true
-    search(".ad") -> true
-    search("b..") -> true
-
-**Note:**You may assume that all words are consist of lowercase letters `a-z`.
-
-## 题目大意
-
-设计一个支持以下两种操作的数据结构：`void addWord(word)`、`bool search(word)`。`search(word)` 可以搜索文字或正则表达式字符串，字符串只包含字母 . 或 a-z 。 "." 可以表示任何一个字母。
-
-
-
-## 解题思路
-
-- 设计一个 `WordDictionary` 的数据结构，要求具有 `addWord(word)` 和 `search(word)` 的操作，并且具有模糊查找的功能。
-- 这一题是第 208 题的加强版，在第 208 题经典的 Trie 上加上了模糊查找的功能。其他实现一模一样。
diff --git a/Algorithms/0212. Word Search II/212. Word Search II.go b/Algorithms/0212. Word Search II/212. Word Search II.go
deleted file mode 100644
index 461e617f1..000000000
--- a/Algorithms/0212. Word Search II/212. Word Search II.go	
+++ /dev/null
@@ -1,11 +0,0 @@
-package leetcode
-
-func findWords(board [][]byte, words []string) []string {
-	res := []string{}
-	for _, v := range words {
-		if exist(board, v) {
-			res = append(res, v)
-		}
-	}
-	return res
-}
diff --git a/Algorithms/0215. Kth Largest Element in an Array/215. Kth Largest Element in an Array.go b/Algorithms/0215. Kth Largest Element in an Array/215. Kth Largest Element in an Array.go
deleted file mode 100644
index 3f7015ff4..000000000
--- a/Algorithms/0215. Kth Largest Element in an Array/215. Kth Largest Element in an Array.go	
+++ /dev/null
@@ -1,31 +0,0 @@
-package leetcode
-
-import "sort"
-
-// 解法一 排序，排序的方法反而速度是最快的
-func findKthLargest1(nums []int, k int) int {
-	sort.Ints(nums)
-	return nums[len(nums)-k]
-}
-
-// 解法二 这个方法的理论依据是 partition 得到的点的下标就是最终排序之后的下标，根据这个下标，我们可以判断第 K 大的数在哪里
-func findKthLargest(nums []int, k int) int {
-	if len(nums) == 0 {
-		return 0
-	}
-	return selection(nums, 0, len(nums)-1, len(nums)-k)
-}
-
-func selection(arr []int, l, r, k int) int {
-	if l == r {
-		return arr[l]
-	}
-	p := partition164(arr, l, r)
-	if k == p {
-		return arr[p]
-	} else if k < p {
-		return selection(arr, l, p-1, k)
-	} else {
-		return selection(arr, p+1, r, k)
-	}
-}
diff --git a/Algorithms/0215. Kth Largest Element in an Array/README.md b/Algorithms/0215. Kth Largest Element in an Array/README.md
deleted file mode 100644
index 63506d275..000000000
--- a/Algorithms/0215. Kth Largest Element in an Array/README.md	
+++ /dev/null
@@ -1,33 +0,0 @@
-# [215. Kth Largest Element in an Array](https://leetcode.com/problems/kth-largest-element-in-an-array/)
-
-## 题目
-
-Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
-
-Example 1:
-
-```c
-Input: [3,2,1,5,6,4] and k = 2
-Output: 5
-```
-
-Example 2:
-
-```c
-Input: [3,2,3,1,2,4,5,5,6] and k = 4
-Output: 4
-```
-
-Note:     
-
-You may assume k is always valid, 1 ≤ k ≤ array's length.
-
-
-## 题目大意
-
-找出数组中第 K 大的元素。这一题非常经典。可以用 O(n) 的时间复杂度实现。
-
-## 解题思路
-
-在快排的 partition 操作中，每次 partition 操作结束都会返回一个点，这个标定点的下标和最终排序之后有序数组中这个元素所在的下标是一致的。利用这个特性，我们可以不断的划分数组区间，最终找到第 K 大的元素。执行一次 partition 操作以后，如果这个元素的下标比 K 小，那么接着就在后边的区间继续执行 partition 操作；如果这个元素的下标比 K 大，那么就在左边的区间继续执行 partition 操作；如果相等就直接输出这个下标对应的数组元素即可。
-
diff --git a/Algorithms/0218. The Skyline Problem/218. The Skyline Problem.go b/Algorithms/0218. The Skyline Problem/218. The Skyline Problem.go
deleted file mode 100644
index 20178e5e6..000000000
--- a/Algorithms/0218. The Skyline Problem/218. The Skyline Problem.go	
+++ /dev/null
@@ -1,209 +0,0 @@
-package leetcode
-
-import (
-	"sort"
-
-	"github.com/halfrost/LeetCode-Go/template"
-)
-
-// 解法一 线段树 Segment Tree，时间复杂度 O(n log n)
-func getSkyline(buildings [][]int) [][]int {
-	st, ans, lastHeight, check := template.SegmentTree{}, [][]int{}, 0, false
-	posMap, pos := discretization218(buildings)
-	tmp := make([]int, len(posMap))
-	st.Init(tmp, func(i, j int) int {
-		return max(i, j)
-	})
-	for _, b := range buildings {
-		st.UpdateLazy(posMap[b[0]], posMap[b[1]-1], b[2])
-	}
-	for i := 0; i < len(pos); i++ {
-		h := st.QueryLazy(posMap[pos[i]], posMap[pos[i]])
-		if check == false && h != 0 {
-			ans = append(ans, []int{pos[i], h})
-			check = true
-		} else if i > 0 && h != lastHeight {
-			ans = append(ans, []int{pos[i], h})
-		}
-		lastHeight = h
-	}
-	return ans
-}
-
-func discretization218(positions [][]int) (map[int]int, []int) {
-	tmpMap, posArray, posMap := map[int]int{}, []int{}, map[int]int{}
-	for _, pos := range positions {
-		tmpMap[pos[0]]++
-		tmpMap[pos[1]-1]++
-		tmpMap[pos[1]]++
-	}
-	for k := range tmpMap {
-		posArray = append(posArray, k)
-	}
-	sort.Ints(posArray)
-	for i, pos := range posArray {
-		posMap[pos] = i
-	}
-	return posMap, posArray
-}
-
-// 解法二 扫描线 Sweep Line，时间复杂度 O(n log n)
-func getSkyline1(buildings [][]int) [][]int {
-	size := len(buildings)
-	es := make([]E, 0)
-	for i, b := range buildings {
-		l := b[0]
-		r := b[1]
-		h := b[2]
-		// 1-- enter
-		el := NewE(i, l, h, 0)
-		es = append(es, el)
-		// 0 -- leave
-		er := NewE(i, r, h, 1)
-		es = append(es, er)
-	}
-	skyline := make([][]int, 0)
-	sort.Slice(es, func(i, j int) bool {
-		if es[i].X == es[j].X {
-			if es[i].T == es[j].T {
-				if es[i].T == 0 {
-					return es[i].H > es[j].H
-				}
-				return es[i].H < es[j].H
-			}
-			return es[i].T < es[j].T
-		}
-		return es[i].X < es[j].X
-	})
-	pq := NewIndexMaxPQ(size)
-	for _, e := range es {
-		curH := pq.Front()
-		if e.T == 0 {
-			if e.H > curH {
-				skyline = append(skyline, []int{e.X, e.H})
-			}
-			pq.Enque(e.N, e.H)
-		} else {
-			pq.Remove(e.N)
-			h := pq.Front()
-			if curH > h {
-				skyline = append(skyline, []int{e.X, h})
-			}
-		}
-	}
-	return skyline
-}
-
-// 扫面线伪代码
-// events = {{x: L , height: H , type: entering},
-// 		  {x: R , height: H , type: leaving}}
-// event.SortByX()
-// ds = new DS()
-
-// for e in events:
-// 	if entering(e):
-// 		if e.height > ds.max(): ans += [e.height]
-// 		ds.add(e.height)
-// 	if leaving(e):
-// 		ds.remove(e.height)
-// 		if e.height > ds.max(): ans += [ds.max()]
-
-// E define
-type E struct { // 定义一个 event 事件
-	N int // number 编号
-	X int // x 坐标
-	H int // height 高度
-	T int // type  0-进入 1-离开
-}
-
-// NewE define
-func NewE(n, x, h, t int) E {
-	return E{
-		N: n,
-		X: x,
-		H: h,
-		T: t,
-	}
-}
-
-// IndexMaxPQ define
-type IndexMaxPQ struct {
-	items []int
-	pq    []int
-	qp    []int
-	total int
-}
-
-// NewIndexMaxPQ define
-func NewIndexMaxPQ(n int) IndexMaxPQ {
-	qp := make([]int, n)
-	for i := 0; i < n; i++ {
-		qp[i] = -1
-	}
-	return IndexMaxPQ{
-		items: make([]int, n),
-		pq:    make([]int, n+1),
-		qp:    qp,
-	}
-}
-
-// Enque define
-func (q *IndexMaxPQ) Enque(key, val int) {
-	q.total++
-	q.items[key] = val
-	q.pq[q.total] = key
-	q.qp[key] = q.total
-	q.swim(q.total)
-}
-
-// Front define
-func (q *IndexMaxPQ) Front() int {
-	if q.total < 1 {
-		return 0
-	}
-	return q.items[q.pq[1]]
-}
-
-// Remove define
-func (q *IndexMaxPQ) Remove(key int) {
-	rank := q.qp[key]
-	q.exch(rank, q.total)
-	q.total--
-	q.qp[key] = -1
-	q.sink(rank)
-}
-
-func (q *IndexMaxPQ) sink(n int) {
-	for 2*n <= q.total {
-		k := 2 * n
-		if k < q.total && q.less(k, k+1) {
-			k++
-		}
-		if q.less(k, n) {
-			break
-		}
-		q.exch(k, n)
-		n = k
-	}
-}
-
-func (q *IndexMaxPQ) swim(n int) {
-	for n > 1 {
-		k := n / 2
-		if q.less(n, k) {
-			break
-		}
-		q.exch(n, k)
-		n = k
-	}
-}
-
-func (q *IndexMaxPQ) exch(i, j int) {
-	q.pq[i], q.pq[j] = q.pq[j], q.pq[i]
-	q.qp[q.pq[i]] = i
-	q.qp[q.pq[j]] = j
-}
-
-func (q *IndexMaxPQ) less(i, j int) bool {
-	return q.items[q.pq[i]] < q.items[q.pq[j]]
-}
diff --git a/Algorithms/0218. The Skyline Problem/218. The Skyline Problem_test.go b/Algorithms/0218. The Skyline Problem/218. The Skyline Problem_test.go
deleted file mode 100644
index f51b33641..000000000
--- a/Algorithms/0218. The Skyline Problem/218. The Skyline Problem_test.go	
+++ /dev/null
@@ -1,42 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question218 struct {
-	para218
-	ans218
-}
-
-// para 是参数
-// one 代表第一个参数
-type para218 struct {
-	one [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans218 struct {
-	one [][]int
-}
-
-func Test_Problem218(t *testing.T) {
-
-	qs := []question218{
-
-		question218{
-			para218{[][]int{[]int{2, 9, 10}, []int{3, 7, 15}, []int{5, 12, 12}, []int{15, 20, 10}, []int{19, 24, 8}}},
-			ans218{[][]int{[]int{2, 10}, []int{3, 15}, []int{7, 12}, []int{12, 0}, []int{15, 10}, []int{20, 8}, []int{24, 0}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 218------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans218, q.para218
-		fmt.Printf("【input】:%v       【output】:%v\n", p, getSkyline(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0218. The Skyline Problem/README.md b/Algorithms/0218. The Skyline Problem/README.md
deleted file mode 100755
index 314d973c8..000000000
--- a/Algorithms/0218. The Skyline Problem/README.md	
+++ /dev/null
@@ -1,70 +0,0 @@
-# [218. The Skyline Problem](https://leetcode.com/problems/the-skyline-problem/)
-
-## 题目:
-
-A city's skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Now suppose you are **given the locations and height of all the buildings** as shown on a cityscape photo (Figure A), write a program to **output the skyline** formed by these buildings collectively (Figure B).
-
-![](https://img.halfrost.com/Leetcode/leetcode_218_0.png)
-
-![](https://img.halfrost.com/Leetcode/leetcode_218_1.png)
-
-The geometric information of each building is represented by a triplet of integers `[Li, Ri, Hi]`, where `Li` and `Ri` are the x coordinates of the left and right edge of the ith building, respectively, and `Hi` is its height. It is guaranteed that `0 ≤ Li, Ri ≤ INT_MAX`, `0 < Hi ≤ INT_MAX`, and `Ri - Li > 0`. You may assume all buildings are perfect rectangles grounded on an absolutely flat surface at height 0.
-
-For instance, the dimensions of all buildings in Figure A are recorded as: `[ [2 9 10], [3 7 15], [5 12 12], [15 20 10], [19 24 8] ]` .
-
-The output is a list of "**key points**" (red dots in Figure B) in the format of `[ [x1,y1], [x2, y2], [x3, y3], ... ]` that uniquely defines a skyline. **A key point is the left endpoint of a horizontal line segment**. Note that the last key point, where the rightmost building ends, is merely used to mark the termination of the skyline, and always has zero height. Also, the ground in between any two adjacent buildings should be considered part of the skyline contour.
-
-For instance, the skyline in Figure B should be represented as:`[ [2 10], [3 15], [7 12], [12 0], [15 10], [20 8], [24, 0] ]`.
-
-**Notes:**
-
-- The number of buildings in any input list is guaranteed to be in the range `[0, 10000]`.
-- The input list is already sorted in ascending order by the left x position `Li`.
-- The output list must be sorted by the x position.
-- There must be no consecutive horizontal lines of equal height in the output skyline. For instance, `[...[2 3], [4 5], [7 5], [11 5], [12 7]...]` is not acceptable; the three lines of height 5 should be merged into one in the final output as such: `[...[2 3], [4 5], [12 7], ...]`
-
-
-## 题目大意
-
-城市的天际线是从远处观看该城市中所有建筑物形成的轮廓的外部轮廓。现在，假设您获得了城市风光照片（图A）上显示的所有建筑物的位置和高度，请编写一个程序以输出由这些建筑物形成的天际线（图B）。
-
-每个建筑物的几何信息用三元组 [Li，Ri，Hi] 表示，其中 Li 和 Ri 分别是第 i 座建筑物左右边缘的 x 坐标，Hi 是其高度。可以保证 0 ≤ Li, Ri ≤ INT\_MAX, 0 < Hi ≤ INT\_MAX 和 Ri - Li > 0。您可以假设所有建筑物都是在绝对平坦且高度为 0 的表面上的完美矩形。
-
-例如，图 A 中所有建筑物的尺寸记录为：[ [2 9 10], [3 7 15], [5 12 12], [15 20 10], [19 24 8] ] 。
-
-输出是以 [ [x1,y1], [x2, y2], [x3, y3], ... ] 格式的“关键点”（图 B 中的红点）的列表，它们唯一地定义了天际线。关键点是水平线段的左端点。请注意，最右侧建筑物的最后一个关键点仅用于标记天际线的终点，并始终为零高度。此外，任何两个相邻建筑物之间的地面都应被视为天际线轮廓的一部分。
-
-例如，图 B 中的天际线应该表示为：[ [2 10], [3 15], [7 12], [12 0], [15 10], [20 8], [24, 0] ]。
-
-说明:
-
-- 任何输入列表中的建筑物数量保证在 [0, 10000] 范围内。
-- 输入列表已经按左 x 坐标 Li 进行升序排列。
-- 输出列表必须按 x 位排序。
-- 输出天际线中不得有连续的相同高度的水平线。例如 [...[2 3], [4 5], [7 5], [11 5], [12 7]...] 是不正确的答案；三条高度为 5 的线应该在最终输出中合并为一个：[...[2 3], [4 5], [12 7], ...]
-
-
-## 解题思路
-
-
-- 给出一个二维数组，每个子数组里面代表一个高楼的信息，一个高楼的信息包含 3 个信息，高楼起始坐标，高楼终止坐标，高楼高度。要求找到这些高楼的边际点，并输出这些边际点的高度信息。
-- 这一题可以用线段树来解。用线段树来解答，可以不用关心“楼挡住楼”的情况。由于楼的坐标是离散的，所以先把楼在 X 轴上两个坐标离散化。同第 699 题一样，楼的宽度是一个区间，但是离散的过程中，楼的宽度右边界需要减一，不然查询一个区间会包含两个点，导致错误的结果，例如，第一个楼是 [1,3)，楼高 10，第二个楼是 [3,6)，楼高 20 。第一个楼如果算上右边界 3，查询 [1,3] 的结果是 20，因为 [3,3] 这个点会查询到第二个楼上面去。所以每个楼的右边界应该减一。但是每个楼的右边界也要加入到 query 中，因为最终 query 的结果需要包含这些边界。将离散的数据排序以后，按照楼的信息，每个区间依次 update。最后统计的时候依次统计每个区间，如果当前区间的高度和前一个区间的高度一样，就算是等高的楼。当高度与前一个高度不相同的时候就算是天际线的边缘，就要添加到最后输出数组中。
-- 类似的线段树的题目有：第 715 题，第 732 题，第 699 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和第 699 题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
-- 这一题用线段树做时间复杂度有点高，可以用扫描线解题。扫描线的思路很简单，用一根根垂直于 X 轴的竖线，从最左边依次扫到最右边，扫描每一条大楼的边界，当进入大楼的左边界的时候，如果没有比这个左边界最高点更高的点，就记录下这个最高点 keyPoint，状态是进入状态。如果扫到一个大楼的左边界，有比它更高的高度，就不记录，因为它不是天际线，它被楼挡楼，挡在其他楼后面了。当扫到一个大楼的右边界的时候，如果是最高点，那么记录下它的状态是离开状态，此时还需要记录下第二高的点。在扫描线扫描的过程中，动态的维护大楼的高度，同时维护最高的高度和第二高的高度。其实只需要维护最高的高度这一个高度，因为当离开状态到来的时候，移除掉当前最高的，剩下的高度里面最高的就是第二高的高度。描述的伪代码如下：
-
-        // 扫描线伪代码
-        events = {{x: L , height: H , type: entering},
-        		  {x: R , height: H , type: leaving}}
-        event.SortByX()
-        ds = new DS()
-        
-        for e in events:
-        	if entering(e):
-        		if e.height > ds.max(): ans += [e.height]
-        		ds.add(e.height)
-        	if leaving(e):
-        		ds.remove(e.height)
-        		if e.height > ds.max(): ans += [ds.max()]
-
-- 动态插入，查找最大值可以选用的数据结构有，最大堆和二叉搜索树。最大堆找最大值 O(1)，插入 O(log n)，但是 remove_by_key 需要 O(n) 的时间复杂度，并且需要自己实现。二叉搜索树，查找 max，添加和删除元素都是 O(log n) 的时间复杂度。
-- 排序的时候也需要注意几个问题：如果大楼的边界相等，并且是进入状态，那么再按照高度从大到小进行排序；如果大楼的边界相等，并且是离开状态，那么高度按照从小到大进行排序。
diff --git a/Algorithms/0220. Contains Duplicate III/220. Contains Duplicate III.go b/Algorithms/0220. Contains Duplicate III/220. Contains Duplicate III.go
deleted file mode 100644
index cd0e9f0dd..000000000
--- a/Algorithms/0220. Contains Duplicate III/220. Contains Duplicate III.go	
+++ /dev/null
@@ -1,61 +0,0 @@
-package leetcode
-
-import "sort"
-
-// 解法一 排序 + 滑动窗口
-func containsNearbyAlmostDuplicate(nums []int, k int, t int) bool {
-	if len(nums) < 2 {
-		return false
-	}
-	elemList := make([]*elem, len(nums))
-	for i, num := range nums {
-		elemList[i] = &elem{num, i}
-	}
-	sort.SliceStable(elemList, func(i, j int) bool {
-		if elemList[i].val != elemList[j].val {
-			return elemList[i].val < elemList[j].val
-		}
-		return elemList[i].idx < elemList[j].idx
-	})
-	i, j := 0, 1
-	for j < len(elemList) {
-		if elemList[j].val-elemList[i].val <= t {
-			if abs(elemList[j].idx-elemList[i].idx) <= k {
-				return true
-			}
-			j++
-		} else {
-			i++
-			if j <= i {
-				j++
-			}
-		}
-	}
-	return false
-}
-
-type elem struct {
-	val int
-	idx int
-}
-
-// 解法二 滑动窗口 + 剪枝
-func containsNearbyAlmostDuplicate1(nums []int, k int, t int) bool {
-	if len(nums) <= 1 {
-		return false
-	}
-	if k <= 0 {
-		return false
-	}
-	n := len(nums)
-	for i := 0; i < n; i++ {
-		count := 0
-		for j := i + 1; j < n && count < k; j++ {
-			if abs(nums[i]-nums[j]) <= t {
-				return true
-			}
-			count++
-		}
-	}
-	return false
-}
diff --git a/Algorithms/0220. Contains Duplicate III/README.md b/Algorithms/0220. Contains Duplicate III/README.md
deleted file mode 100644
index 4087683e6..000000000
--- a/Algorithms/0220. Contains Duplicate III/README.md	
+++ /dev/null
@@ -1,34 +0,0 @@
-# [220. Contains Duplicate III](https://leetcode.com/problems/contains-duplicate-iii/)
-
-## 题目
-
-Given an array of integers, find out whether there are two distinct indices i and j in the array such that the **absolute** difference between **nums[i]** and **nums[j]** is at most t and the **absolute** difference between i and j is at most k.
-
-**Example 1:**
-
-    Input: nums = [1,2,3,1], k = 3, t = 0
-    Output: true
-
-**Example 2:**
-
-    Input: nums = [1,0,1,1], k = 1, t = 2
-    Output: true
-
-**Example 3:**
-
-    Input: nums = [1,5,9,1,5,9], k = 2, t = 3
-    Output: false
-
-
-
-
-## 题目大意
-
-给出一个数组 num，再给 K 和 t。问在 num 中能否找到一组 i 和 j，使得 num[i] 和 num[j] 的绝对差值最大为 t，并且 i 和 j 之前的绝对差值最大为 k。
-
-## 解题思路
-
-
-- 给出一个数组，要求在数组里面找到 2 个索引，`i` 和 `j`，使得 `| nums[i] - nums[j] | ≤ t` ，并且 `| i - j | ≤  k` 。
-- 这是一道滑动窗口的题目。第一想法就是用 `i` 和 `j` 两个指针，针对每个 `i` ，都从 `i + 1` 往后扫完整个数组，判断每个 `i` 和 `j` ，判断是否满足题意。`j` 在循环的过程中注意判断剪枝条件 `| i - j | ≤  k`。这个做法的时间复杂度是 O(n^2)。这个做法慢的原因在于滑动窗口的左边界和右边界在滑动过程中不是联动滑动的。
-- 于是考虑，如果数组是有序的呢？把数组按照元素值从小到大进行排序，如果元素值相等，就按照 index 从小到大进行排序。在这样有序的数组中找满足题意的 `i` 和 `j`，滑动窗口左边界和右边界就是联动的了。窗口的右边界滑到与左边界元素值的差值 ≤ t 的地方，满足了这个条件再判断 `| i - j | ≤  k`，如果右边界与左边界元素值的差值 > t 了，说明该把左边界往右移动了(能这样移动的原因就是因为我们将数组元素大小排序了，右移是增大元素的方向)。移动左边界的时候需要注意左边界不能超过右边界。这样滑动窗口一次滑过整个排序后的数组，就可以判断是否存在满足题意的 `i` 和 `j` 。这个做法的时间主要花在排序上了，时间复杂度是 O(n log n)。
diff --git a/Algorithms/0223. Rectangle Area/223. Rectangle Area.go b/Algorithms/0223. Rectangle Area/223. Rectangle Area.go
deleted file mode 100644
index c3d604ccd..000000000
--- a/Algorithms/0223. Rectangle Area/223. Rectangle Area.go	
+++ /dev/null
@@ -1,14 +0,0 @@
-package leetcode
-
-func computeArea(A int, B int, C int, D int, E int, F int, G int, H int) int {
-	X0, Y0, X1, Y1 := max(A, E), max(B, F), min(C, G), min(D, H)
-	return area(A, B, C, D) + area(E, F, G, H) - area(X0, Y0, X1, Y1)
-}
-
-func area(x0, y0, x1, y1 int) int {
-	l, h := x1-x0, y1-y0
-	if l <= 0 || h <= 0 {
-		return 0
-	}
-	return l * h
-}
diff --git a/Algorithms/0223. Rectangle Area/README.md b/Algorithms/0223. Rectangle Area/README.md
deleted file mode 100755
index 90968f8f3..000000000
--- a/Algorithms/0223. Rectangle Area/README.md	
+++ /dev/null
@@ -1,31 +0,0 @@
-# [223. Rectangle Area](https://leetcode.com/problems/rectangle-area/)
-
-
-## 题目:
-
-Find the total area covered by two **rectilinear** rectangles in a **2D** plane.
-
-Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/22/rectangle_area.png)
-
-**Example:**
-
-    Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2
-    Output: 45
-
-**Note:**
-
-Assume that the total area is never beyond the maximum possible value of **int**.
-
-
-
-## 题目大意
-
-在二维平面上计算出两个由直线构成的矩形重叠后形成的总面积。每个矩形由其左下顶点和右上顶点坐标表示，如图所示。说明: 假设矩形面积不会超出 int 的范围。
-
-## 解题思路
-
-
-- 给出两个矩形的坐标，求这两个矩形在坐标轴上覆盖的总面积。
-- 几何题，由于只有 2 个矩形，所以按照题意做即可。先分别求两个矩形的面积，加起来再减去两个矩形重叠的面积。
diff --git a/Algorithms/0226. Invert Binary Tree/226. Invert Binary Tree.go b/Algorithms/0226. Invert Binary Tree/226. Invert Binary Tree.go
deleted file mode 100644
index e43d8b2cc..000000000
--- a/Algorithms/0226. Invert Binary Tree/226. Invert Binary Tree.go	
+++ /dev/null
@@ -1,19 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-func invertTree(root *TreeNode) *TreeNode {
-	if root == nil {
-		return nil
-	}
-	invertTree(root.Left)
-	invertTree(root.Right)
-	root.Left, root.Right = root.Right, root.Left
-	return root
-}
diff --git a/Algorithms/0229. Majority Element II/README.md b/Algorithms/0229. Majority Element II/README.md
deleted file mode 100755
index e01c44e25..000000000
--- a/Algorithms/0229. Majority Element II/README.md	
+++ /dev/null
@@ -1,29 +0,0 @@
-# [229. Majority Element II](https://leetcode.com/problems/majority-element-ii/)
-
-
-## 题目:
-
-Given an integer array of size n, find all elements that appear more than `⌊ n/3 ⌋` times.
-
-**Note:** The algorithm should run in linear time and in O(1) space.
-
-**Example 1:**
-
-    Input: [3,2,3]
-    Output: [3]
-
-**Example 2:**
-
-    Input: [1,1,1,3,3,2,2,2]
-    Output: [1,2]
-
-
-## 题目大意
-
-给定一个大小为 n 的数组，找出其中所有出现超过 ⌊ n/3 ⌋ 次的元素。说明: 要求算法的时间复杂度为 O(n)，空间复杂度为 O(1)。
-
-## 解题思路
-
-- 这一题是第 169 题的加强版。Boyer-Moore Majority Vote algorithm 算法的扩展版。
-- 题目要求找出数组中出现次数大于 `⌊ n/3 ⌋` 次的数。要求空间复杂度为 O(1)。简单题。
-- 这篇文章写的不错，可参考：[https://gregable.com/2013/10/majority-vote-algorithm-find-majority.html](https://gregable.com/2013/10/majority-vote-algorithm-find-majority.html)
\ No newline at end of file
diff --git a/Algorithms/0235. Lowest Common Ancestor of a Binary Search Tree/235. Lowest Common Ancestor of a Binary Search Tree.go b/Algorithms/0235. Lowest Common Ancestor of a Binary Search Tree/235. Lowest Common Ancestor of a Binary Search Tree.go
deleted file mode 100644
index 97a4d8c11..000000000
--- a/Algorithms/0235. Lowest Common Ancestor of a Binary Search Tree/235. Lowest Common Ancestor of a Binary Search Tree.go	
+++ /dev/null
@@ -1,22 +0,0 @@
-package leetcode
-
-/**
- * Definition for TreeNode.
- * type TreeNode struct {
- *     Val int
- *     Left *ListNode
- *     Right *ListNode
- * }
- */
-func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
-	if p == nil || q == nil || root == nil {
-		return nil
-	}
-	if p.Val < root.Val && q.Val < root.Val {
-		return lowestCommonAncestor(root.Left, p, q)
-	}
-	if p.Val > root.Val && q.Val > root.Val {
-		return lowestCommonAncestor(root.Right, p, q)
-	}
-	return root
-}
diff --git a/Algorithms/0235. Lowest Common Ancestor of a Binary Search Tree/235. Lowest Common Ancestor of a Binary Search Tree_test.go b/Algorithms/0235. Lowest Common Ancestor of a Binary Search Tree/235. Lowest Common Ancestor of a Binary Search Tree_test.go
deleted file mode 100644
index cb1bc7496..000000000
--- a/Algorithms/0235. Lowest Common Ancestor of a Binary Search Tree/235. Lowest Common Ancestor of a Binary Search Tree_test.go	
+++ /dev/null
@@ -1,58 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question235 struct {
-	para235
-	ans235
-}
-
-// para 是参数
-// one 代表第一个参数
-type para235 struct {
-	one []int
-	two []int
-	thr []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans235 struct {
-	one []int
-}
-
-func Test_Problem235(t *testing.T) {
-
-	qs := []question235{
-
-		question235{
-			para235{[]int{}, []int{}, []int{}},
-			ans235{[]int{}},
-		},
-
-		question235{
-			para235{[]int{6, 2, 8, 0, 4, 7, 9, NULL, NULL, 3, 5}, []int{2}, []int{8}},
-			ans235{[]int{6}},
-		},
-
-		question235{
-			para235{[]int{6, 2, 8, 0, 4, 7, 9, NULL, NULL, 3, 5}, []int{2}, []int{4}},
-			ans235{[]int{2}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 235------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans235, q.para235
-		fmt.Printf("【input】:%v      ", p)
-		rootOne := Ints2TreeNode(p.one)
-		rootTwo := Ints2TreeNode(p.two)
-		rootThr := Ints2TreeNode(p.thr)
-		fmt.Printf("【output】:%v      \n", lowestCommonAncestor(rootOne, rootTwo, rootThr))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0236. Lowest Common Ancestor of a Binary Tree/236. Lowest Common Ancestor of a Binary Tree.go b/Algorithms/0236. Lowest Common Ancestor of a Binary Tree/236. Lowest Common Ancestor of a Binary Tree.go
deleted file mode 100644
index 6f5028d0e..000000000
--- a/Algorithms/0236. Lowest Common Ancestor of a Binary Tree/236. Lowest Common Ancestor of a Binary Tree.go	
+++ /dev/null
@@ -1,24 +0,0 @@
-package leetcode
-
-/**
- * Definition for TreeNode.
- * type TreeNode struct {
- *     Val int
- *     Left *ListNode
- *     Right *ListNode
- * }
- */
-func lowestCommonAncestor236(root, p, q *TreeNode) *TreeNode {
-	if root == nil || root == q || root == p {
-		return root
-	}
-	left := lowestCommonAncestor236(root.Left, p, q)
-	right := lowestCommonAncestor236(root.Right, p, q)
-	if left != nil {
-		if right != nil {
-			return root
-		}
-		return left
-	}
-	return right
-}
diff --git a/Algorithms/0236. Lowest Common Ancestor of a Binary Tree/236. Lowest Common Ancestor of a Binary Tree_test.go b/Algorithms/0236. Lowest Common Ancestor of a Binary Tree/236. Lowest Common Ancestor of a Binary Tree_test.go
deleted file mode 100644
index 5561ef4bf..000000000
--- a/Algorithms/0236. Lowest Common Ancestor of a Binary Tree/236. Lowest Common Ancestor of a Binary Tree_test.go	
+++ /dev/null
@@ -1,68 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question236 struct {
-	para236
-	ans236
-}
-
-// para 是参数
-// one 代表第一个参数
-type para236 struct {
-	one []int
-	two []int
-	thr []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans236 struct {
-	one []int
-}
-
-func Test_Problem236(t *testing.T) {
-
-	qs := []question236{
-
-		question236{
-			para236{[]int{}, []int{}, []int{}},
-			ans236{[]int{}},
-		},
-
-		question236{
-			para236{[]int{3, 5, 1, 6, 2, 0, 8, NULL, NULL, 7, 4}, []int{5}, []int{1}},
-			ans236{[]int{3}},
-		},
-
-		question236{
-			para236{[]int{3, 5, 1, 6, 2, 0, 8, NULL, NULL, 7, 4}, []int{5}, []int{4}},
-			ans236{[]int{5}},
-		},
-
-		question236{
-			para236{[]int{6, 2, 8, 0, 4, 7, 9, NULL, NULL, 3, 5}, []int{2}, []int{8}},
-			ans236{[]int{6}},
-		},
-
-		question236{
-			para236{[]int{6, 2, 8, 0, 4, 7, 9, NULL, NULL, 3, 5}, []int{2}, []int{4}},
-			ans236{[]int{2}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 236------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans236, q.para236
-		fmt.Printf("【input】:%v      ", p)
-		rootOne := Ints2TreeNode(p.one)
-		rootTwo := Ints2TreeNode(p.two)
-		rootThr := Ints2TreeNode(p.thr)
-		fmt.Printf("【output】:%v      \n", lowestCommonAncestor236(rootOne, rootTwo, rootThr))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0237. Delete Node in a Linked List/237. Delete Node in a Linked List.go b/Algorithms/0237. Delete Node in a Linked List/237. Delete Node in a Linked List.go
deleted file mode 100644
index ba5356b7f..000000000
--- a/Algorithms/0237. Delete Node in a Linked List/237. Delete Node in a Linked List.go	
+++ /dev/null
@@ -1,21 +0,0 @@
-package leetcode
-
-/**
- * Definition for singly-linked list.
- * type ListNode struct {
- *     Val int
- *     Next *ListNode
- * }
- */
-func deleteNode(node *ListNode) {
-	if node == nil {
-		return
-	}
-	cur := node
-	for cur.Next.Next != nil {
-		cur.Val = cur.Next.Val
-		cur = cur.Next
-	}
-	cur.Val = cur.Next.Val
-	cur.Next = nil
-}
diff --git a/Algorithms/0237. Delete Node in a Linked List/237. Delete Node in a Linked List_test.go b/Algorithms/0237. Delete Node in a Linked List/237. Delete Node in a Linked List_test.go
deleted file mode 100644
index 727011f03..000000000
--- a/Algorithms/0237. Delete Node in a Linked List/237. Delete Node in a Linked List_test.go	
+++ /dev/null
@@ -1,73 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question237 struct {
-	para237
-	ans237
-}
-
-// para 是参数
-// one 代表第一个参数
-type para237 struct {
-	one []int
-	n   int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans237 struct {
-	one []int
-}
-
-func Test_Problem237(t *testing.T) {
-
-	qs := []question237{
-
-		question237{
-			para237{[]int{1, 2, 3, 4, 5}, 1},
-			ans237{[]int{2, 3, 4, 5}},
-		},
-
-		question237{
-			para237{[]int{1, 2, 3, 4, 5}, 2},
-			ans237{[]int{1, 3, 4, 5}},
-		},
-
-		question237{
-			para237{[]int{1, 1, 1, 1, 1}, 1},
-			ans237{[]int{}},
-		},
-
-		question237{
-			para237{[]int{1, 2, 3, 4, 5}, 5},
-			ans237{[]int{1, 2, 3, 4}},
-		},
-
-		question237{
-			para237{[]int{}, 5},
-			ans237{[]int{}},
-		},
-
-		question237{
-			para237{[]int{1, 2, 3, 4, 5}, 10},
-			ans237{[]int{1, 2, 3, 4, 5}},
-		},
-
-		question237{
-			para237{[]int{1}, 1},
-			ans237{[]int{}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 237------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans237, q.para237
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(removeElements(S2l(p.one), p.n)))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0237. Delete Node in a Linked List/README.md b/Algorithms/0237. Delete Node in a Linked List/README.md
deleted file mode 100644
index 7c6e4954a..000000000
--- a/Algorithms/0237. Delete Node in a Linked List/README.md	
+++ /dev/null
@@ -1,40 +0,0 @@
-# [237. Delete Node in a Linked List](https://leetcode.com/problems/delete-node-in-a-linked-list/)
-
-## 题目
-
-Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.
-
-Given linked list -- head = [4,5,1,9], which looks like following:
-
-![](https://assets.leetcode.com/uploads/2018/12/28/237_example.png)
-
-Example 1:
-
-```c
-Input: head = [4,5,1,9], node = 5
-Output: [4,1,9]
-Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.
-```
-
-Example 2:
-
-```c
-Input: head = [4,5,1,9], node = 1
-Output: [4,5,9]
-Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.
-```
-
-Note:
-
-- The linked list will have at least two elements.
-- All of the nodes' values will be unique.
-- The given node will not be the tail and it will always be a valid node of the linked list.
-- Do not return anything from your function.
-
-## 题目大意
-
-删除给点结点。没有给链表的头结点。
-
-## 解题思路
-
-其实就是把后面的结点都覆盖上来即可。或者直接当前结点的值等于下一个结点，Next 指针指向下下个结点，这样做也可以，只不过中间有一个结点不被释放，内存消耗多一些。
\ No newline at end of file
diff --git a/Algorithms/0240. Search a 2D Matrix II/240. Search a 2D Matrix II_test.go b/Algorithms/0240. Search a 2D Matrix II/240. Search a 2D Matrix II_test.go
deleted file mode 100644
index 22e5efe66..000000000
--- a/Algorithms/0240. Search a 2D Matrix II/240. Search a 2D Matrix II_test.go	
+++ /dev/null
@@ -1,48 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question240 struct {
-	para240
-	ans240
-}
-
-// para 是参数
-// one 代表第一个参数
-type para240 struct {
-	matrix [][]int
-	target int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans240 struct {
-	one bool
-}
-
-func Test_Problem240(t *testing.T) {
-
-	qs := []question240{
-
-		question240{
-			para240{[][]int{[]int{1, 4, 7, 11, 15}, []int{2, 5, 8, 12, 19}, []int{3, 6, 9, 16, 22}, []int{10, 13, 14, 17, 24}, []int{18, 21, 23, 26, 30}}, 5},
-			ans240{true},
-		},
-
-		question240{
-			para240{[][]int{[]int{1, 4, 7, 11, 15}, []int{2, 5, 8, 12, 19}, []int{3, 6, 9, 16, 22}, []int{10, 13, 14, 17, 24}, []int{18, 21, 23, 26, 30}}, 20},
-			ans240{false},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 240------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans240, q.para240
-		fmt.Printf("【input】:%v       【output】:%v\n", p, searchMatrix240(p.matrix, p.target))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0240. Search a 2D Matrix II/README.md b/Algorithms/0240. Search a 2D Matrix II/README.md
deleted file mode 100755
index d9b3e859e..000000000
--- a/Algorithms/0240. Search a 2D Matrix II/README.md	
+++ /dev/null
@@ -1,41 +0,0 @@
-# [240. Search a 2D Matrix II](https://leetcode.com/problems/search-a-2d-matrix-ii/)
-
-
-## 题目:
-
-Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
-
-- Integers in each row are sorted in ascending from left to right.
-- Integers in each column are sorted in ascending from top to bottom.
-
-**Example:**
-
-Consider the following matrix:
-
-    [
-      [1,   4,  7, 11, 15],
-      [2,   5,  8, 12, 19],
-      [3,   6,  9, 16, 22],
-      [10, 13, 14, 17, 24],
-      [18, 21, 23, 26, 30]
-    ]
-
-Given target = `5`, return `true`.
-
-Given target = `20`, return `false`.
-
-
-## 题目大意
-
-编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target。该矩阵具有以下特性：
-
-- 每行的元素从左到右升序排列。
-- 每列的元素从上到下升序排列。
-
-
-
-## 解题思路
-
-- 给出一个二维矩阵，矩阵的特点是每一个行内，元素随着下标增大而增大，每一列内，元素也是随着下标增大而增大。但是相邻两行的元素并没有大小关系。例如第一行最后一个元素就比第二行第一个元素要大。要求设计一个算法能在这个矩阵中高效的找到一个数，如果找到就输出 true，找不到就输出 false。
-- 这一题是第 74 题的加强版。第 74 题中的二维矩阵完全是一个有序的一维矩阵，但是这一题如果把它拍扁成一维，并不是有序的。首先每一个行或者每一列是有序的 ，那么我们可以依次在每一行或者每一列中利用二分去搜索。这样做时间复杂度为  O(n log n)。
-- 还有一个模拟的解法。通过观察，我们发现了这个矩阵的一个特点，最右边一列的元素是本行中最大的元素，所以我们可以先从最右边一列开始找到第一个比 target 元素大的元素，这个元素所在的行，是我们接着要搜索的。在行中搜索是从最右边开始往左边搜索，时间复杂度是 O(n)，算上一开始在最右边一列中查找的时间复杂度是 O(m)，所以最终的时间复杂度为 O(m+n)。
diff --git a/Algorithms/0242. Valid Anagram/242. Valid Anagram.go b/Algorithms/0242. Valid Anagram/242. Valid Anagram.go
deleted file mode 100644
index 81920dd7f..000000000
--- a/Algorithms/0242. Valid Anagram/242. Valid Anagram.go	
+++ /dev/null
@@ -1,67 +0,0 @@
-package leetcode
-
-// 解法一
-func isAnagram(s string, t string) bool {
-	alphabet := make([]int, 26)
-	sBytes := []byte(s)
-	tBytes := []byte(t)
-	if len(sBytes) != len(tBytes) {
-		return false
-	}
-	for i := 0; i < len(sBytes); i++ {
-		alphabet[sBytes[i]-'a']++
-	}
-	for i := 0; i < len(tBytes); i++ {
-		alphabet[tBytes[i]-'a']--
-	}
-	for i := 0; i < 26; i++ {
-		if alphabet[i] != 0 {
-			return false
-		}
-	}
-	return true
-}
-
-// 解法二
-func isAnagram1(s string, t string) bool {
-	if s == "" && t == "" {
-		return true
-	}
-	if s == "" || t == "" {
-		return false
-	}
-	sBytes := []byte(s)
-	tBytes := []byte(t)
-	if len(sBytes) != len(tBytes) {
-		return false
-	}
-	quickSortByte(sBytes, 0, len(sBytes)-1)
-	quickSortByte(tBytes, 0, len(tBytes)-1)
-
-	for i := 0; i < len(sBytes); i++ {
-		if sBytes[i] != tBytes[i] {
-			return false
-		}
-	}
-	return true
-}
-func partitionByte(a []byte, lo, hi int) int {
-	pivot := a[hi]
-	i := lo - 1
-	for j := lo; j < hi; j++ {
-		if a[j] > pivot {
-			i++
-			a[j], a[i] = a[i], a[j]
-		}
-	}
-	a[i+1], a[hi] = a[hi], a[i+1]
-	return i + 1
-}
-func quickSortByte(a []byte, lo, hi int) {
-	if lo >= hi {
-		return
-	}
-	p := partitionByte(a, lo, hi)
-	quickSortByte(a, lo, p-1)
-	quickSortByte(a, p+1, hi)
-}
diff --git a/Algorithms/0260. Single Number III/README.md b/Algorithms/0260. Single Number III/README.md
deleted file mode 100755
index f7dccbf1a..000000000
--- a/Algorithms/0260. Single Number III/README.md	
+++ /dev/null
@@ -1,34 +0,0 @@
-# [260. Single Number III](https://leetcode.com/problems/single-number-iii/)
-
-
-## 题目:
-
-Given an array of numbers `nums`, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.
-
-**Example:**
-
-    Input:  [1,2,1,3,2,5]
-    Output: [3,5]
-
-**Note**:
-
-1. The order of the result is not important. So in the above example, `[5, 3]` is also correct.
-2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?
-
-
-## 题目大意
-
-给定一个整数数组 nums，其中恰好有两个元素只出现一次，其余所有元素均出现两次。 找出只出现一次的那两个元素。
-
-注意：  
-
-- 结果输出的顺序并不重要，对于上面的例子，[5, 3] 也是正确答案。
-- 要求算法时间复杂度是线性的，并且不使用额外的辅助空间。
-
-
-
-## 解题思路
-
-- 这一题是第 136 题的加强版。第 136 题里面只有一个数出现一次，其他数都出现 2 次。而这一次有 2 个数字出现一次，其他数出现 2 次。
-- 解题思路还是利用异或，把出现 2 次的数先消除。最后我们要找的 2 个数肯定也是不同的，所以最后 2 个数对一个数进行异或，答案肯定是不同的。那么我们找哪个数为参照物呢？可以随便取，不如就取 lsb 最低位为 1 的数吧
-- 于是整个数组会被分为 2 部分，异或 lsb 为 0 的和异或 lsb 为 1 的，在这 2 部分中，用异或操作把出现 2 次的数都消除，那么剩下的 2 个数分别就在这 2 部分中。
diff --git a/Algorithms/0263. Ugly Number/263. Ugly Number.go b/Algorithms/0263. Ugly Number/263. Ugly Number.go
deleted file mode 100644
index f155ab422..000000000
--- a/Algorithms/0263. Ugly Number/263. Ugly Number.go	
+++ /dev/null
@@ -1,10 +0,0 @@
-package leetcode
-
-func isUgly(num int) bool {
-	for i := 2; i < 6 && num > 0; i++ {
-		for num%i == 0 {
-			num /= i
-		}
-	}
-	return num == 1
-}
diff --git a/Algorithms/0268. Missing Number/README.md b/Algorithms/0268. Missing Number/README.md
deleted file mode 100755
index d0d104939..000000000
--- a/Algorithms/0268. Missing Number/README.md	
+++ /dev/null
@@ -1,30 +0,0 @@
-# [268. Missing Number](https://leetcode.com/problems/missing-number/)
-
-
-## 题目:
-
-Given an array containing n distinct numbers taken from `0, 1, 2, ..., n`, find the one that is missing from the array.
-
-**Example 1:**
-
-    Input: [3,0,1]
-    Output: 2
-
-**Example 2:**
-
-    Input: [9,6,4,2,3,5,7,0,1]
-    Output: 8
-
-**Note**:Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
-
-
-## 题目大意
-
-给定一个包含 0, 1, 2, ..., n 中 n 个数的序列，找出 0 .. n 中没有出现在序列中的那个数。算法应该具有线性时间复杂度。你能否仅使用额外常数空间来实现?
-
-
-
-## 解题思路
-
-
-- 要求找出 `0, 1, 2, ..., n` 中缺失的那个数。还是利用异或的性质，`X^X = 0`。这里我们需要构造一个 X，用数组下标就可以了。数字下标是从 `[0，n-1]`，数字是 `[0，n]`，依次把数组里面的数组进行异或，把结果和 n 再异或一次，中和掉出现的数字，剩下的那个数字就是之前没有出现过的，缺失的数字。
diff --git a/Algorithms/0274. H-Index/274. H-Index.go b/Algorithms/0274. H-Index/274. H-Index.go
deleted file mode 100644
index 9d59c9dbc..000000000
--- a/Algorithms/0274. H-Index/274. H-Index.go	
+++ /dev/null
@@ -1,36 +0,0 @@
-package leetcode
-
-// 解法一
-func hIndex(citations []int) int {
-	n := len(citations)
-	buckets := make([]int, n+1)
-	for _, c := range citations {
-		if c >= n {
-			buckets[n]++
-		} else {
-			buckets[c]++
-		}
-	}
-	count := 0
-	for i := n; i >= 0; i-- {
-		count += buckets[i]
-		if count >= i {
-			return i
-		}
-	}
-	return 0
-}
-
-// 解法二
-func hIndex1(citations []int) int {
-	quickSort164(citations, 0, len(citations)-1)
-	hIndex := 0
-	for i := len(citations) - 1; i >= 0; i-- {
-		if citations[i] >= len(citations)-i {
-			hIndex++
-		} else {
-			break
-		}
-	}
-	return hIndex
-}
diff --git a/Algorithms/0275. H-Index II/README.md b/Algorithms/0275. H-Index II/README.md
deleted file mode 100755
index 1cfe6ffd7..000000000
--- a/Algorithms/0275. H-Index II/README.md	
+++ /dev/null
@@ -1,49 +0,0 @@
-# [275. H-Index II](https://leetcode.com/problems/h-index-ii/)
-
-## 题目:
-
-Given an array of citations **sorted in ascending order** (each citation is a non-negative integer) of a researcher, write a function to compute the researcher's h-index.
-
-According to the [definition of h-index on Wikipedia](https://en.wikipedia.org/wiki/H-index): "A scientist has index h if h of his/her N papers have **at least** h citations each, and the other N − h papers have **no more than** h citations each."
-
-**Example:**
-
-    Input: citations = [0,1,3,5,6]
-    Output: 3 
-    Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had 
-                 received 0, 1, 3, 5, 6 citations respectively. 
-                 Since the researcher has 3 papers with at least 3 citations each and the remaining 
-                 two with no more than 3 citations each, her h-index is 3.
-
-**Note:**
-
-If there are several possible values for *h*, the maximum one is taken as the h-index.
-
-**Follow up:**
-
-- This is a follow up problem to [H-Index](https://leetcode.com/problems/h-index/description/), where `citations` is now guaranteed to be sorted in ascending order.
-- Could you solve it in logarithmic time complexity?
-
-
-
-## 题目大意
-
-
-给定一位研究者论文被引用次数的数组（被引用次数是非负整数），数组已经按照升序排列。编写一个方法，计算出研究者的 h 指数。
-
-h 指数的定义: “h 代表“高引用次数”（high citations），一名科研人员的 h 指数是指他（她）的 （N 篇论文中）至多有 h 篇论文分别被引用了至少 h 次。（其余的 N - h 篇论文每篇被引用次数不多于 h 次。）"
-
-说明:
-
-- 如果 h 有多有种可能的值 ，h 指数是其中最大的那个。
-
-进阶：
-
-- 这是 H 指数 的延伸题目，本题中的 citations 数组是保证有序的。
-你可以优化你的算法到对数时间复杂度吗？
-
-
-## 解题思路
-
-- 给出一个数组，代表该作者论文被引用次数，要求这个作者的 h 指数。h 指数定义："高引用次数”（`high citations`），一名科研人员的 h 指数是指他（她）的 （N 篇论文中）至多有 h 篇论文分别被引用了至少 h 次。（其余的 N - h 篇论文每篇被引用次数不多于 h 次。）
-- 这一题要找出 h 指数，即要找到一个边界，在这个边界上为最多的 h 指数。可以用二分搜索来解决这道题。当 `len(citations)-mid > citations[mid]` 时，说明 h 指数的边界一定在右边，因为最多 `len(citations)-mid` 篇数比引用数 `citations[mid]` 还要大。否则 h 指数的边界在左边界，缩小边界以后继续二分。找到边界以后，最终求的是 h 指数，用 `len(citations) - low` 即是结果。
diff --git a/Algorithms/0287. Find the Duplicate Number/README.md b/Algorithms/0287. Find the Duplicate Number/README.md
deleted file mode 100644
index aa72f7a6b..000000000
--- a/Algorithms/0287. Find the Duplicate Number/README.md	
+++ /dev/null
@@ -1,36 +0,0 @@
-# [287. Find the Duplicate Number](https://leetcode.com/problems/find-the-duplicate-number/)
-
-## 题目
-
-Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
-
-Example 1:
-
-```c
-Input: [1,3,4,2,2]
-Output: 2
-```
-
-Example 2:
-
-```c
-Input: [3,1,3,4,2]
-Output: 3
-```
-
-Note:  
-
-- You must not modify the array (assume the array is read only).
-- You must use only constant, O(1) extra space.
-- Your runtime complexity should be less than O(n^2).
-- There is only one duplicate number in the array, but it could be repeated more than once.
-
-## 题目大意
-
-给出 n + 1 个数，这些数是在 1-n 中取值的，同一个数字可以出现多次。要求找出这些数中重复的数字。时间复杂度最好低于 O(n^2)，空间复杂度为 O(1)。
-
-## 解题思路
-
-- 这道题比较巧的思路是，将这些数字想象成链表中的结点，数组中数字代表下一个结点的数组下标。找重复的数字就是找链表中成环的那个点。由于题目保证了一定会有重复的数字，所以一定会成环。所以用快慢指针的方法，快指针一次走 2 步，慢指针一次走 1 步，相交以后，快指针从头开始，每次走一步，再次遇见的时候就是成环的交点处，也即是重复数字所在的地方。
-- 这一题有多种做法。可以用快慢指针求解。还可以用二分搜索，题目中规定，给出 [1，n] 个数，如果每个数都不重复，那么就有 `n + 1` 个数。以 [1, 2, 2, 3, 4, 5, 6, 7] 为例，一共有 8 个数，每个数都在 1 和 7 之间。1 和 7 的中位数是 4，遍历整个数组，统计小于 4 的整数的个数，至多为 3 个，如果超过 3 个就说明重复的数存在于区间 `[1,4)` 中；否则，重复的数存在于区间 `[4,7]` 中。这里小于 4 的整数有 4 个（它们是 1, 2, 2, 3），因此砍掉右半区间，连中位数也去掉。如此反复，最后二分区间越来越小，直到变成 1 个整数，这个整数就是我们要找的重复的数。
-- 另外一个做法是，先将数组排序，依照下标是从 0 开始递增的特性，那么数组里面的数字与下标的差值应该是越来越大。如果出现了相同的数字，下标变大，差值应该比前一个数字小，出现了这个情况就说明找到了相同数字了。
\ No newline at end of file
diff --git a/Algorithms/0300. Longest Increasing Subsequence/README.md b/Algorithms/0300. Longest Increasing Subsequence/README.md
deleted file mode 100755
index afec2c534..000000000
--- a/Algorithms/0300. Longest Increasing Subsequence/README.md	
+++ /dev/null
@@ -1,37 +0,0 @@
-# [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/)
-
-
-## 题目
-
-Given an unsorted array of integers, find the length of longest increasing subsequence.
-
-**Example:**
-
-    Input: [10,9,2,5,3,7,101,18]
-    Output: 4 
-    Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
-
-**Note:**
-
-- There may be more than one LIS combination, it is only necessary for you to return the length.
-- Your algorithm should run in O(n^2) complexity.
-
-**Follow up:** Could you improve it to O(n log n) time complexity?
-
-## 题目大意
-
-给定一个无序的整数数组，找到其中最长上升子序列的长度。
-
-
-## 解题思路
-
-- 给定一个整数序列，求其中的最长上升子序列的长度。这一题就是经典的最长上升子序列的问题。
-- `dp[i]` 代表为第 i 个数字为结尾的最长上升子序列的长度。换种表述，dp[i] 代表 [0,i] 范围内，选择数字 nums[i] 可以获得的最长上升子序列的长度。状态转移方程为 `dp[i] = max( 1 + dp[j]) ，其中 j < i && nums[j] > nums[i]`，取所有满足条件的最大值。时间复杂度 O(n^2)
-- 这道题还有一种更快的解法。考虑这样一个问题，我们是否能用一个数组，记录上升子序列的最末尾的一个数字呢？如果这个数字越小，那么这个子序列往后面添加数字的几率就越大，那么就越可能成为最长的上升子序列。举个例子：nums = [4,5,6,3]，它的所有的上升子序列为
-
-    len = 1   :      [4], [5], [6], [3]   => tails[0] = 3
-    len = 2   :      [4, 5], [5, 6]       => tails[1] = 5
-    len = 3   :      [4, 5, 6]            => tails[2] = 6
-
-- 其中 `tails[i]` 中存储的是所有长度为 i + 1 的上升子序列中末尾最小的值。也很容易证明 `tails` 数组里面的值一定是递增的(因为我们用末尾的数字描述最长递增子序列)。既然 tails 是有序的，我们就可以用二分查找的方法去更新这个 tail 数组里面的值。更新策略如下：(1). 如果 x 比所有的 tails 元素都要大，那么就直接放在末尾，并且 tails 数组长度加一；(2). 如果 `tails[i-1] < x <= tails[i]`，则更新 tails[i]，因为 x 更小，更能获得最长上升子序列。最终 tails 数组的长度即为最长的上升子序列。这种做法的时间复杂度 O(n log n)。
-
diff --git a/Algorithms/0303. Range Sum Query - Immutable/README.md b/Algorithms/0303. Range Sum Query - Immutable/README.md
deleted file mode 100755
index 5b4542073..000000000
--- a/Algorithms/0303. Range Sum Query - Immutable/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# [303. Range Sum Query - Immutable](https://leetcode.com/problems/range-sum-query-immutable/)
-
-
-## 题目:
-
-Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
-
-**Example:**
-
-    Given nums = [-2, 0, 3, -5, 2, -1]
-    
-    sumRange(0, 2) -> 1
-    sumRange(2, 5) -> -1
-    sumRange(0, 5) -> -3
-
-**Note:**
-
-1. You may assume that the array does not change.
-2. There are many calls to sumRange function.
-
-
-## 题目大意
-
-给定一个整数数组  nums，求出数组从索引 i 到 j  (i ≤ j) 范围内元素的总和，包含 i,  j 两点。
-
-示例：
-
-```
-给定 nums = [-2, 0, 3, -5, 2, -1]，求和函数为 sumRange()
-
-sumRange(0, 2) -> 1
-sumRange(2, 5) -> -1
-sumRange(0, 5) -> -3
-
-```
-
-说明:
-
-- 你可以假设数组不可变。
-- 会多次调用 sumRange 方法。
-
-
-## 解题思路
-
-
-- 给出一个数组，数组里面的数都是`**不可变**`的，设计一个数据结构能够满足查询数组任意区间内元素的和。
-- 这一题由于数组里面的元素都是`**不可变**`的，所以可以用 2 种方式来解答，第一种解法是用 prefixSum，通过累计和相减的办法来计算区间内的元素和，初始化的时间复杂度是 O(n)，但是查询区间元素和的时间复杂度是 O(1)。第二种解法是利用线段树，构建一颗线段树，父结点内存的是两个子结点的和，初始化建树的时间复杂度是 O(log n)，查询区间元素和的时间复杂度是 O(log n)。
diff --git a/Algorithms/0307. Range Sum Query - Mutable/307. Range Sum Query - Mutable.go b/Algorithms/0307. Range Sum Query - Mutable/307. Range Sum Query - Mutable.go
deleted file mode 100644
index 5b84a304f..000000000
--- a/Algorithms/0307. Range Sum Query - Mutable/307. Range Sum Query - Mutable.go	
+++ /dev/null
@@ -1,63 +0,0 @@
-package leetcode
-
-import (
-	"github.com/halfrost/LeetCode-Go/template"
-)
-
-// Constructor307 define
-func Constructor307(nums []int) NumArray {
-	st := template.SegmentTree{}
-	st.Init(nums, func(i, j int) int {
-		return i + j
-	})
-	return NumArray{st: &st}
-}
-
-// Update define
-func (this *NumArray) Update(i int, val int) {
-	this.st.Update(i, val)
-}
-
-//解法二 prefixSum，sumRange 时间复杂度 O(1)
-
-// // NumArray define
-// type NumArray307 struct {
-// 	prefixSum []int
-// 	data   []int
-// }
-
-// // Constructor307 define
-// func Constructor307(nums []int) NumArray307 {
-// 	data := make([]int, len(nums))
-// 	for i := 0; i < len(nums); i++ {
-// 		data[i] = nums[i]
-// 	}
-// 	for i := 1; i < len(nums); i++ {
-// 		nums[i] += nums[i-1]
-// 	}
-// 	return NumArray307{prefixSum: nums, data: data}
-// }
-
-// // Update define
-// func (this *NumArray307) Update(i int, val int) {
-// 	this.data[i] = val
-// 	this.prefixSum[0] = this.data[0]
-// 	for i := 1; i < len(this.data); i++ {
-// 		this.prefixSum[i] = this.prefixSum[i-1] + this.data[i]
-// 	}
-// }
-
-// // SumRange define
-// func (this *NumArray307) SumRange(i int, j int) int {
-// 	if i > 0 {
-// 		return this.prefixSum[j] - this.prefixSum[i-1]
-// 	}
-// 	return this.prefixSum[j]
-// }
-
-/**
- * Your NumArray object will be instantiated and called as such:
- * obj := Constructor(nums);
- * obj.Update(i,val);
- * param_2 := obj.SumRange(i,j);
- */
diff --git a/Algorithms/0307. Range Sum Query - Mutable/307. Range Sum Query - Mutable_test.go b/Algorithms/0307. Range Sum Query - Mutable/307. Range Sum Query - Mutable_test.go
deleted file mode 100644
index 6ea6ccd1f..000000000
--- a/Algorithms/0307. Range Sum Query - Mutable/307. Range Sum Query - Mutable_test.go	
+++ /dev/null
@@ -1,15 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-func Test_Problem307(t *testing.T) {
-	obj := Constructor307([]int{1, 3, 5})
-	fmt.Printf("obj = %v\n", obj)
-	fmt.Printf("SumRange(0,2) = %v\n", obj.SumRange(0, 2))
-	obj.Update(1, 2)
-	fmt.Printf("obj = %v\n", obj)
-	fmt.Printf("SumRange(0,2) = %v\n", obj.SumRange(0, 2))
-}
diff --git a/Algorithms/0307. Range Sum Query - Mutable/README.md b/Algorithms/0307. Range Sum Query - Mutable/README.md
deleted file mode 100755
index 6cafb629f..000000000
--- a/Algorithms/0307. Range Sum Query - Mutable/README.md	
+++ /dev/null
@@ -1,50 +0,0 @@
-# [307. Range Sum Query - Mutable](https://leetcode.com/problems/range-sum-query-mutable/)
-
-
-## 题目:
-
-Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
-
-The update(i, val) function modifies nums by updating the element at index i to val.
-
-**Example:**
-
-    Given nums = [1, 3, 5]
-    
-    sumRange(0, 2) -> 9
-    update(1, 2)
-    sumRange(0, 2) -> 8
-
-**Note:**
-
-1. The array is only modifiable by the update function.
-2. You may assume the number of calls to update and sumRange function is distributed evenly.
-
-
-## 题目大意
-
-给定一个整数数组  nums，求出数组从索引 i 到 j  (i ≤ j) 范围内元素的总和，包含 i,  j 两点。
-
-update(i, val) 函数可以通过将下标为 i 的数值更新为 val，从而对数列进行修改。
-
-示例:
-
-```
-Given nums = [1, 3, 5]
-
-sumRange(0, 2) -> 9  
-update(1, 2)  
-sumRange(0, 2) -> 8  
-```
-
-说明:
-
-- 数组仅可以在 update 函数下进行修改。
-- 你可以假设 update 函数与 sumRange 函数的调用次数是均匀分布的。
-
-## 解题思路
-
-
-- 给出一个数组，数组里面的数都是`**可变**`的，设计一个数据结构能够满足查询数组任意区间内元素的和。
-- 对比第 303 题，这一题由于数组里面的元素都是**`可变`**的，所以第一个想到的解法就是线段树，构建一颗线段树，父结点内存的是两个子结点的和，初始化建树的时间复杂度是 O(log n)，查询区间元素和的时间复杂度是 O(log n)，更新元素值的时间复杂度是 O(log n)。
-- 如果此题还用 prefixSum 的思路解答呢？那每次 update 操作的时间复杂度都是 O(n)，因为每次更改一个值，最坏情况就是所有的 prefixSum 都要更新一次。prefixSum 的方法在这道题上面也可以 AC，只不过时间排名在 5%，非常差。
diff --git a/Algorithms/0315. Count of Smaller Numbers After Self/315. Count of Smaller Numbers After Self.go b/Algorithms/0315. Count of Smaller Numbers After Self/315. Count of Smaller Numbers After Self.go
deleted file mode 100644
index c6ab601b7..000000000
--- a/Algorithms/0315. Count of Smaller Numbers After Self/315. Count of Smaller Numbers After Self.go	
+++ /dev/null
@@ -1,37 +0,0 @@
-package leetcode
-
-import (
-	"sort"
-
-	"github.com/halfrost/LeetCode-Go/template"
-)
-
-func countSmaller(nums []int) []int {
-	if len(nums) == 0 {
-		return []int{}
-	}
-	st, minNum, numsMap, numsArray, res := template.SegmentCountTree{}, 0, make(map[int]int, 0), []int{}, make([]int, len(nums))
-	for i := 0; i < len(nums); i++ {
-		numsMap[nums[i]] = nums[i]
-	}
-	for _, v := range numsMap {
-		numsArray = append(numsArray, v)
-	}
-	// 排序是为了使得线段树中的区间 left <= right，如果此处不排序，线段树中的区间有很多不合法。
-	sort.Ints(numsArray)
-	minNum = numsArray[0]
-	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
-	st.Init(numsArray, func(i, j int) int {
-		return 0
-	})
-	for i := len(nums) - 1; i >= 0; i-- {
-		if nums[i] == minNum {
-			res[i] = 0
-			st.UpdateCount(nums[i])
-			continue
-		}
-		st.UpdateCount(nums[i])
-		res[i] = st.Query(minNum, nums[i]-1)
-	}
-	return res
-}
diff --git a/Algorithms/0315. Count of Smaller Numbers After Self/README.md b/Algorithms/0315. Count of Smaller Numbers After Self/README.md
deleted file mode 100755
index 6c106b2cb..000000000
--- a/Algorithms/0315. Count of Smaller Numbers After Self/README.md	
+++ /dev/null
@@ -1,41 +0,0 @@
-# [315. Count of Smaller Numbers After Self](https://leetcode.com/problems/count-of-smaller-numbers-after-self/)
-
-
-## 题目:
-
-You are given an integer array nums and you have to return a new counts array. The counts array has the property where `counts[i]` is the number of smaller elements to the right of `nums[i]`.
-
-**Example:**
-
-    Input: [5,2,6,1]
-    Output: [2,1,1,0] 
-    Explanation:
-    To the right of 5 there are 2 smaller elements (2 and 1).
-    To the right of 2 there is only 1 smaller element (1).
-    To the right of 6 there is 1 smaller element (1).
-    To the right of 1 there is 0 smaller element.
-
-
-## 题目大意
-
-
-给定一个整数数组 nums，按要求返回一个新数组 counts。数组 counts 有该性质： counts[i] 的值是  nums[i] 右侧小于 nums[i] 的元素的数量。
-
-示例:
-
-```
-输入: [5,2,6,1]
-输出: [2,1,1,0] 
-解释:
-5 的右侧有 2 个更小的元素 (2 和 1).
-2 的右侧仅有 1 个更小的元素 (1).
-6 的右侧有 1 个更小的元素 (1).
-1 的右侧有 0 个更小的元素.
-```
-
-
-## 解题思路
-
-
-- 给出一个数组，要求输出数组中每个元素相对于数组中的位置右边比它小的元素。
-- 这一题是第 327 题的缩水版。由于需要找数组位置右边比当前位置元素小的元素，所以从数组右边开始往左边扫。构造一颗线段树，线段树里面父节点存的是子节点出现的次数和。有可能给的数据会很大，所以构造线段树的时候先离散化。还需要注意的是数组里面可能有重复元素，所以构造线段树要先去重并排序。从右往左扫的过程中，依次添加数组中的元素，添加了一次就立即 query 一次。query 的区间是 [minNum, nums[i]-1]。如果是 minNum 则输出 0，并且也要记得插入这个最小值。这一题的思路和第 327 题大体类似，详解可见第 327 题。
diff --git a/Algorithms/0318. Maximum Product of Word Lengths/README.md b/Algorithms/0318. Maximum Product of Word Lengths/README.md
deleted file mode 100755
index 03d60e429..000000000
--- a/Algorithms/0318. Maximum Product of Word Lengths/README.md	
+++ /dev/null
@@ -1,46 +0,0 @@
-# [318. Maximum Product of Word Lengths](https://leetcode.com/problems/maximum-product-of-word-lengths/)
-
-
-## 题目:
-
-Given a string array `words`, find the maximum value of `length(word[i]) * length(word[j])` where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.
-
-**Example 1:**
-
-    Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
-    Output: 16 
-    Explanation: The two words can be "abcw", "xtfn".
-
-**Example 2:**
-
-    Input: ["a","ab","abc","d","cd","bcd","abcd"]
-    Output: 4 
-    Explanation: The two words can be "ab", "cd".
-
-**Example 3:**
-
-    Input: ["a","aa","aaa","aaaa"]
-    Output: 0 
-    Explanation: No such pair of words.
-
-
-
-## 题目大意
-
-给定一个字符串数组 words，找到 length(word[i]) * length(word[j]) 的最大值，并且这两个单词不含有公共字母。你可以认为每个单词只包含小写字母。如果不存在这样的两个单词，返回 0。
-
-
-## 解题思路
-
-- 在字符串数组中找到 2 个没有公共字符的字符串，并且这两个字符串的长度乘积要是最大的，求这个最大的乘积。
-- 这里需要利用位运算 `&` 运算的性质，如果 `X & Y = 0`，说明 X 和 Y 完全不相同。那么我们将字符串都编码成二进制数，进行 `&` 运算即可分出没有公共字符的字符串，最后动态维护长度乘积最大值即可。将字符串编码成二进制数的规则比较简单，每个字符相对于 'a' 的距离，根据这个距离将 1 左移多少位。
-
-```c
-    a 1->1  
-    b 2->10  
-    c 4->100  
-    ab 3->11  
-    ac 5->101  
-    abc 7->111  
-    az 33554433->10000000000000000000000001  
-```
diff --git a/Algorithms/0327. Count of Range Sum/327. Count of Range Sum.go b/Algorithms/0327. Count of Range Sum/327. Count of Range Sum.go
deleted file mode 100644
index 281594d3c..000000000
--- a/Algorithms/0327. Count of Range Sum/327. Count of Range Sum.go	
+++ /dev/null
@@ -1,60 +0,0 @@
-package leetcode
-
-import (
-	"sort"
-
-	"github.com/halfrost/LeetCode-Go/template"
-)
-
-// 解法一 线段树，时间复杂度 O(n log n)
-func countRangeSum(nums []int, lower int, upper int) int {
-	if len(nums) == 0 {
-		return 0
-	}
-	st, prefixSum, sumMap, sumArray, res := template.SegmentCountTree{}, make([]int, len(nums)), make(map[int]int, 0), []int{}, 0
-	prefixSum[0], sumMap[nums[0]] = nums[0], nums[0]
-	for i := 1; i < len(nums); i++ {
-		prefixSum[i] = prefixSum[i-1] + nums[i]
-		sumMap[prefixSum[i]] = prefixSum[i]
-	}
-	// sumArray 是 prefixSum 去重之后的版本，利用 sumMap 去重
-	for _, v := range sumMap {
-		sumArray = append(sumArray, v)
-	}
-	// 排序是为了使得线段树中的区间 left <= right，如果此处不排序，线段树中的区间有很多不合法。
-	sort.Ints(sumArray)
-	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
-	st.Init(sumArray, func(i, j int) int {
-		return 0
-	})
-	// 倒序是为了方便寻找 j，sum(i，j) 规定了 j >= i，所以倒序遍历，i 从大到小
-	for i := len(nums) - 1; i >= 0; i-- {
-		// 插入的 prefixSum[i] 即是 j
-		st.UpdateCount(prefixSum[i])
-		if i > 0 {
-			res += st.Query(lower+prefixSum[i-1], upper+prefixSum[i-1])
-		} else {
-			res += st.Query(lower, upper)
-		}
-	}
-	return res
-}
-
-// 解法二 暴力，时间复杂度 O(n^2)
-func countRangeSum1(nums []int, lower int, upper int) int {
-	res, n := 0, len(nums)
-	for i := 0; i < n; i++ {
-		tmp := 0
-		for j := i; j < n; j++ {
-			if i == j {
-				tmp = nums[i]
-			} else {
-				tmp += nums[j]
-			}
-			if tmp <= upper && tmp >= lower {
-				res++
-			}
-		}
-	}
-	return res
-}
diff --git a/Algorithms/0327. Count of Range Sum/README.md b/Algorithms/0327. Count of Range Sum/README.md
deleted file mode 100755
index d874dbc83..000000000
--- a/Algorithms/0327. Count of Range Sum/README.md	
+++ /dev/null
@@ -1,77 +0,0 @@
-# [327. Count of Range Sum](https://leetcode.com/problems/count-of-range-sum/)
-
-
-## 题目:
-
-Given an integer array `nums`, return the number of range sums that lie in `[lower, upper]` inclusive.Range sum `S(i, j)` is defined as the sum of the elements in `nums` between indices `i` and `j` (`i` ≤ `j`), inclusive.
-
-**Note:**A naive algorithm of O(n2) is trivial. You MUST do better than that.
-
-**Example:**
-
-    Input: nums = [-2,5,-1], lower = -2, upper = 2,
-    Output: 3 
-    Explanation: The three ranges are : [0,0], [2,2], [0,2] and their respective sums are: -2, -1, 2.
-
-
-## 题目大意
-
-
-给定一个整数数组 nums，返回区间和在 [lower, upper] 之间的个数，包含 lower 和 upper。区间和 S(i, j) 表示在 nums 中，位置从 i 到 j 的元素之和，包含 i 和 j (i ≤ j)。
-
-说明:   
-最直观的算法复杂度是 O(n^2) ，请在此基础上优化你的算法。
-
-
-## 解题思路
-
-- 给出一个数组，要求在这个数组中找出任意一段子区间的和，位于 [lower,upper] 之间。
-- 这一题可以用暴力解法，2 层循环，遍历所有子区间，求和并判断是否位于 [lower,upper] 之间，时间复杂度 O(n^2)。
-- 这一题当然还有更优的解法，用线段树或者树状数组，将时间复杂度降为 O(n log n)。题目中要求 `lower ≤ sum(i,j) ≤ upper`，`sum(i,j) = prefixSum(j) - prefixSum(i-1)`，那么 `lower + prefixSum(i-1) ≤ prefixSum(j) ≤ upper + prefixSum(i-1)`。所以利用前缀和将区间和转换成了前缀和在线段树中 `query` 的问题，只不过线段树中父节点中存的不是子节点的和，而应该是子节点出现的次数。第二个转换，由于前缀和会很大，所以需要离散化。例如 `prefixSum = [-3,-2,-1,0]`，用前缀和下标进行离散化，所以线段树中左右区间变成了 0-3 。
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_327_0.png)
-
-    利用 `prefixSum` 下标离散化：
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_327_1.png)
-
-- 还需要注意一些小细节，`prefixSum` 计算完以后需要去重，去重以后并排序，方便构造线段树的有效区间。如果不去重，线段树中可能出现非法区间(left > right)或者重叠区间。最后一步往线段树中倒序插入 `prefixSum` 的时候，用的是非去重的，插入 `prefixSum[j]` 代表 sum(i,j) 中的 j，例如往线段树中插入 `prefixSum[5]`，代表当前树中加入了 j = 5 的情况。query 操作实质是在做区间匹配，例如当前 i 循环到 i = 3，累计往线段树中插入了 `prefixSum[5]`，`prefixSum[4]`，`prefixSum[3]`，那么 query 操作实质是在判断：`lower ≤ sum(i=3,j=3) ≤ upper`，`lower ≤ sum(i=3,j=4) ≤ upper`，`lower ≤ sum(i=3,j=5) ≤ upper`，这 3 个等式是否成立，有几个成立就返回几个，即是最终要求得的结果的一部分。
-- 举个例子，`nums = [-3,1,2,-2,2,-1]`，`prefixSum = [-3,-2,0,-2,0,-1]`，去重以后并排序得到 `sum = [-3,-2,-1,0]`。离散化构造线段树，这里出于演示的方便，下图中就不画出离散后的线段树了，用非离散的线段树展示：
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_327_2_.png)
-
-    倒序插入 `len(prefixSum)-1 = prefixSum[5] = -1`：
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_327_3_.png)
-
-    这时候查找区间变为了 `[-3 + prefixSum[5-1], -1 + prefixSum[5-1]] = [-3,-1]`，即判断 `-3 ≤ sum(5,5) ≤ -1`，满足等式的有几种情况，这里明显只有一种情况，即 `j = 5`，也满足等式，所以这一步 `res = 1`。
-
-- 倒序插入 `len(prefixSum)-2 = prefixSum[4] = 0`：
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_327_4_.png)
-
-    这时候查找区间变为了 `[-3 + prefixSum[4-1], -1 + prefixSum[4-1]] = [-5,-3]`，即判断 `-5 ≤ sum(4,  4,5) ≤ -3`，满足等式的有几种情况，这里有两种情况，即 `j = 4` 或者 `j = 5`，都不满足等式，所以这一步 `res = 0`。
-
-- 倒序插入 `len(prefixSum)-3 = prefixSum[3] = -2`：
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_327_5_.png)
-
-    这时候查找区间变为了 `[-3 + prefixSum[3-1], -1 + prefixSum[3-1]] = [-3,-1]`，即判断 `-3 ≤ sum(3,  3,4,5) ≤ -1`，满足等式的有几种情况，这里有三种情况，即 `j = 3` 、`j = 4` 或者 `j = 5`，满足等式的有 `j = 3` 和 `j = 5`，即 `-3 ≤ sum(3, 3) ≤ -1` 和 `-3 ≤ sum(3, 5) ≤ -1`。所以这一步 `res = 2`。
-
-- 倒序插入 `len(prefixSum)-4 = prefixSum[2] = 0`：
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_327_6_.png)
-
-    这时候查找区间变为了 `[-3 + prefixSum[2-1], -1 + prefixSum[2-1]] = [-5,-3]`，即判断 `-5 ≤ sum(2,  2,3,4,5) ≤ -3`，满足等式的有几种情况，这里有四种情况，即 `j = 2`、 `j = 3` 、`j = 4` 或者 `j = 5`，都不满足等式。所以这一步 `res = 0`。
-
-- 倒序插入 `len(prefixSum)-5 = prefixSum[1] = -2`：
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_327_7_.png)
-
-    这时候查找区间变为了 `[-3 + prefixSum[1-1], -1 + prefixSum[1-1]] = [-6,-4]`，即判断 `-6 ≤ sum(1,  1,2,3,4,5) ≤ -4`，满足等式的有几种情况，这里有五种情况，即 `j = 1`、 `j = 2`、 `j = 3` 、`j = 4` 或者 `j = 5`，都不满足等式。所以这一步 `res = 0`。
-
-- 倒序插入 `len(prefixSum)-6 = prefixSum[0] = -3`：
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_327_8_.png)
-
-    这时候查找区间变为了 `[-3 + prefixSum[0-1], -1 + prefixSum[0-1]] = [-3,-1]`，注意 `prefixSum[-1] = 0`，即判断 `-3 ≤ sum(0,  0,1,2,3,4,5) ≤ -1`，满足等式的有几种情况，这里有六种情况，即 `j = 0`、`j = 1`、`j = 2`、 `j = 3` 、`j = 4` 或者 `j = 5`，满足等式的有 `j = 0`、`j = 1`、 `j = 3` 和 `j = 5`，即 `-3 ≤ sum(0, 0) ≤ -1` 、 `-3 ≤ sum(0, 1) ≤ -1`、`-3 ≤ sum(0, 3) ≤ -1` 和 `-3 ≤ sum(0, 5) ≤ -1`。所以这一步 `res = 4`。最后的答案就是把每一步的结果都累加，`res = 1 + 0 + 2 + 0 + 0 + 4 = 7`。
diff --git a/Algorithms/0328. Odd Even Linked List/README.md b/Algorithms/0328. Odd Even Linked List/README.md
deleted file mode 100644
index 760a9e596..000000000
--- a/Algorithms/0328. Odd Even Linked List/README.md	
+++ /dev/null
@@ -1,34 +0,0 @@
-# [328. Odd Even Linked List](https://leetcode.com/problems/odd-even-linked-list/)
-
-## 题目
-
-Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.
-
-You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.
-
-Example 1:
-
-```c
-Input: 1->2->3->4->5->NULL
-Output: 1->3->5->2->4->NULL
-```
-
-Example 2:
-
-```c
-Input: 2->1->3->5->6->4->7->NULL
-Output: 2->3->6->7->1->5->4->NULL
-```
-
-Note:
-
-- The relative order inside both the even and odd groups should remain as it was in the input.
-- The first node is considered odd, the second node even and so on ...
-
-## 题目大意
-
-这道题和第 86 题非常类型。第 86 题是把排在某个点前面的小值放在一个链表中，排在某个点后端的大值放在另外一个链表中，最后 2 个链表首尾拼接一下就是答案。
-
-## 解题思路
-
-这道题思路也是一样的，分别把奇数和偶数都放在 2 个链表中，最后首尾拼接就是答案。
\ No newline at end of file
diff --git a/Algorithms/0345. Reverse Vowels of a String/345. Reverse Vowels of a String.go b/Algorithms/0345. Reverse Vowels of a String/345. Reverse Vowels of a String.go
deleted file mode 100644
index 3bd3fbee3..000000000
--- a/Algorithms/0345. Reverse Vowels of a String/345. Reverse Vowels of a String.go	
+++ /dev/null
@@ -1,27 +0,0 @@
-package leetcode
-
-func reverseVowels(s string) string {
-	b := []byte(s)
-	for i, j := 0, len(b)-1; i < j; {
-		if isVowels(b[i]) && isVowels(b[j]) {
-			b[i], b[j] = b[j], b[i]
-			i++
-			j--
-		} else if isVowels(b[i]) && !isVowels(b[j]) {
-			j--
-		} else if !isVowels(b[i]) && isVowels(b[j]) {
-			i++
-		} else {
-			i++
-			j--
-		}
-	}
-	return string(b)
-}
-
-func isVowels(s byte) bool {
-	if s == 'a' || s == 'e' || s == 'i' || s == 'o' || s == 'u' || s == 'A' || s == 'E' || s == 'I' || s == 'O' || s == 'U' {
-		return true
-	}
-	return false
-}
diff --git a/Algorithms/0367. Valid Perfect Square/README.md b/Algorithms/0367. Valid Perfect Square/README.md
deleted file mode 100755
index 430bd1eb6..000000000
--- a/Algorithms/0367. Valid Perfect Square/README.md	
+++ /dev/null
@@ -1,33 +0,0 @@
-# [367. Valid Perfect Square](https://leetcode.com/problems/valid-perfect-square/)
-
-## 题目:
-
-Given a positive integer num, write a function which returns True if num is a perfect square else False.
-
-**Note:** **Do not** use any built-in library function such as `sqrt`.
-
-**Example 1:**
-
-    Input: 16
-    Output: true
-
-**Example 2:**
-
-    Input: 14
-    Output: false
-
-
-## 题目大意
-
-给定一个正整数 num，编写一个函数，如果 num 是一个完全平方数，则返回 True，否则返回 False。
-
-说明：不要使用任何内置的库函数，如 sqrt。
-
-
-
-
-## 解题思路
-
-
-- 给出一个数，要求判断这个数是不是完全平方数。
-- 可以用二分搜索来解答这道题。判断完全平方数，根据它的定义来，是否能被开根号，即找到一个数的平方是否可以等于待判断的数字。从 [1, n] 区间内进行二分，若能找到则返回 true，找不到就返回 false 。
diff --git a/Algorithms/0371. Sum of Two Integers/README.md b/Algorithms/0371. Sum of Two Integers/README.md
deleted file mode 100755
index 57887ffe8..000000000
--- a/Algorithms/0371. Sum of Two Integers/README.md	
+++ /dev/null
@@ -1,26 +0,0 @@
-# [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)
-
-
-## 题目:
-
-Calculate the sum of two integers a and b, but you are **not allowed** to use the operator `+` and `-`.
-
-**Example 1:**
-
-    Input: a = 1, b = 2
-    Output: 3
-
-**Example 2:**
-
-    Input: a = -2, b = 3
-    Output: 1
-
-
-## 题目大意
-
-不使用运算符 + 和 - ​​​​​​​，计算两整数 ​​​​​​​a 、b ​​​​​​​之和。
-
-## 解题思路
-
-- 要求不用加法和减法运算符计算 `a+b`。这一题需要用到 `^` 和 `&` 运算符的性质，两个数 ^ 可以实现两个数不带进位的二进制加法。这里需要实现加法，肯定需要进位。所以如何找到进位是本题的关键。
-- 在二进制中，只有 1 和 1 加在一起才会进位，0 和 0，0 和 1，1 和 0，这三种情况都不会进位，规律就是 `a & b` 为 0 的时候就不用进位，为 1 的时候代表需要进位。进位是往前进一位，所以还需要左移操作，所以加上的进位为 `(a&b)<<1`。
diff --git a/Algorithms/0372. Super Pow/README.md b/Algorithms/0372. Super Pow/README.md
deleted file mode 100755
index 563a28a86..000000000
--- a/Algorithms/0372. Super Pow/README.md	
+++ /dev/null
@@ -1,43 +0,0 @@
-# [372. Super Pow](https://leetcode.com/problems/super-pow/)
-
-
-## 题目:
-
-Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.
-
-**Example 1:**
-
-    Input: a = 2, b = [3]
-    Output: 8
-
-**Example 2:**
-
-    Input: a = 2, b = [1,0]
-    Output: 1024
-
-
-## 题目大意
-
-
-你的任务是计算 a^b 对 1337 取模，a 是一个正整数，b 是一个非常大的正整数且会以数组形式给出。
-
-## 解题思路
-
-- 求 a^b mod p 的结果，b 是大数。
-- 这一题可以用暴力解法尝试。需要用到 mod 计算的几个运算性质：
-
-        模运算性质一：(a + b) % p = (a % p + b % p) % p
-        模运算性质二：(a - b) % p = (a % p - b % p + p) % p
-        模运算性质三：(a * b) % p = (a % p * b % p) % p
-        模运算性质四：a ^ b % p = ((a % p)^b) % p
-        模运算性质五：ab % p = ((a % p) * ( b % p)) % p, 其中 ab 是一个数字，如:2874，98374 等等
-
-    这一题需要用到性质三、四、五。举个例子：
-
-        12345^678 % 1337 = (12345^670 * 12345^8) % 1337
-        				    = ((12345^670 % 1337) * (12345^8 % 1337)) % 1337  ---> 利用性质 三
-        				    = (((12345^67)^10 % 1337) * (12345^8 % 1337)) % 1337  ---> 乘方性质
-                    = ((12345^67 % 1337)^10) % 1337 * (12345^8 % 1337)) % 1337  ---> 利用性质 四
-        				    = (((12345^67 % 1337)^10) * (12345^8 % 1337)) % 1337  ---> 反向利用性质 三
-
-    经过上面这样的变换，把指数 678 的个位分离出来了，可以单独求解。继续经过上面的变换，可以把指数的 6 和 7 也分离出来。最终可以把大数 b 一位一位的分离出来。至于计算 a^b 就结果快速幂求解。
diff --git a/Algorithms/0385. Mini Parser/README.md b/Algorithms/0385. Mini Parser/README.md
deleted file mode 100755
index 20c810cad..000000000
--- a/Algorithms/0385. Mini Parser/README.md	
+++ /dev/null
@@ -1,50 +0,0 @@
-# [385. Mini Parser](https://leetcode.com/problems/mini-parser/)
-
-
-## 题目:
-
-Given a nested list of integers represented as a string, implement a parser to deserialize it.
-
-Each element is either an integer, or a list -- whose elements may also be integers or other lists.
-
-**Note:** You may assume that the string is well-formed:
-
-- String is non-empty.
-- String does not contain white spaces.
-- String contains only digits `0-9`, `[`, `-` `,`, `]`.
-
-**Example 1:**
-
-    Given s = "324",
-    
-    You should return a NestedInteger object which contains a single integer 324.
-
-**Example 2:**
-
-    Given s = "[123,[456,[789]]]",
-    
-    Return a NestedInteger object containing a nested list with 2 elements:
-    
-    1. An integer containing value 123.
-    2. A nested list containing two elements:
-        i.  An integer containing value 456.
-        ii. A nested list with one element:
-             a. An integer containing value 789.
-
-
-## 题目大意
-
-给定一个用字符串表示的整数的嵌套列表，实现一个解析它的语法分析器。列表中的每个元素只可能是整数或整数嵌套列表
-
-提示：你可以假定这些字符串都是格式良好的：
-
-- 字符串非空
-- 字符串不包含空格
-- 字符串只包含数字0-9, [, - ,, ]
-
-
-
-## 解题思路
-
-- 将一个嵌套的数据结构中的数字转换成 NestedInteger 数据结构。
-- 这一题用栈一层一层的处理就行。有一些比较坑的特殊的边界数据见测试文件。这一题正确率比很多 Hard 题还要低的原因应该是没有理解好题目和边界测试数据没有考虑到。NestedInteger 这个数据结构笔者实现了一遍，见代码。
diff --git a/Algorithms/0386. Lexicographical Numbers/README.md b/Algorithms/0386. Lexicographical Numbers/README.md
deleted file mode 100755
index f3e6b0a0a..000000000
--- a/Algorithms/0386. Lexicographical Numbers/README.md	
+++ /dev/null
@@ -1,25 +0,0 @@
-# [386. Lexicographical Numbers](https://leetcode.com/problems/lexicographical-numbers/)
-
-
-## 题目:
-
-Given an integer n, return 1 - n in lexicographical order.
-
-For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9].
-
-Please optimize your algorithm to use less time and space. The input size may be as large as 5,000,000.
-
-
-## 题目大意
-
-给定一个整数 n, 返回从 1 到 n 的字典顺序。例如，给定 n =13，返回 [1,10,11,12,13,2,3,4,5,6,7,8,9] 。
-
-请尽可能的优化算法的时间复杂度和空间复杂度。 输入的数据 n 小于等于 5,000,000。
-
-
-
-## 解题思路
-
-
-- 给出一个数字 n ，要求按照字典序对 1-n 这 n 个数排序。
-- DFS 暴力求解即可。
diff --git a/Algorithms/0387. First Unique Character in a String/387. First Unique Character in a String.go b/Algorithms/0387. First Unique Character in a String/387. First Unique Character in a String.go
deleted file mode 100644
index 591c37f80..000000000
--- a/Algorithms/0387. First Unique Character in a String/387. First Unique Character in a String.go	
+++ /dev/null
@@ -1,14 +0,0 @@
-package leetcode
-
-func firstUniqChar(s string) int {
-	result := make([]int, 26)
-	for i := 0; i < len(s); i++ {
-		result[s[i]-'a']++
-	}
-	for i := 0; i < len(s); i++ {
-		if result[s[i]-'a'] == 1 {
-			return i
-		}
-	}
-	return -1
-}
diff --git a/Algorithms/0387. First Unique Character in a String/README.md b/Algorithms/0387. First Unique Character in a String/README.md
deleted file mode 100755
index 2de711047..000000000
--- a/Algorithms/0387. First Unique Character in a String/README.md	
+++ /dev/null
@@ -1,26 +0,0 @@
-# [387. First Unique Character in a String](https://leetcode.com/problems/first-unique-character-in-a-string/)
-
-## 题目:
-
-Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.
-
-**Examples:**
-
-    s = "leetcode"
-    return 0.
-    
-    s = "loveleetcode",
-    return 2.
-
-**Note:** You may assume the string contain only lowercase letters.
-
-
-
-## 题目大意
-
-给定一个字符串，找到它的第一个不重复的字符，并返回它的索引。如果不存在，则返回 -1。
-
-
-## 解题思路
-
-- 简单题，要求输出第一个没有重复的字符。
diff --git a/Algorithms/0389. Find the Difference/README.md b/Algorithms/0389. Find the Difference/README.md
deleted file mode 100755
index ac93ced7e..000000000
--- a/Algorithms/0389. Find the Difference/README.md	
+++ /dev/null
@@ -1,30 +0,0 @@
-# [389. Find the Difference](https://leetcode.com/problems/find-the-difference/)
-
-## 题目:
-
-Given two strings **s** and **t** which consist of only lowercase letters.
-
-String **t** is generated by random shuffling string **s** and then add one more letter at a random position.
-
-Find the letter that was added in **t**.
-
-**Example:**
-
-    Input:
-    s = "abcd"
-    t = "abcde"
-    
-    Output:
-    e
-    
-    Explanation:
-    'e' is the letter that was added.
-
-## 题目大意
-
-给定两个字符串 s 和 t，它们只包含小写字母。字符串 t 由字符串 s 随机重排，然后在随机位置添加一个字母。请找出在 t 中被添加的字母。
-
-
-## 解题思路
-
-- 题目要求找出 t 字符串中比 s 字符串多出的一个字符。思路还是利用异或的性质，`X^X = 0`，将 s 和 t 依次异或，最终多出来的字符就是最后异或的结果。
diff --git a/Algorithms/0393. UTF-8 Validation/README.md b/Algorithms/0393. UTF-8 Validation/README.md
deleted file mode 100755
index 0120374ac..000000000
--- a/Algorithms/0393. UTF-8 Validation/README.md	
+++ /dev/null
@@ -1,71 +0,0 @@
-# [393. UTF-8 Validation](https://leetcode.com/problems/utf-8-validation/)
-
-
-
-## 题目:
-
-A character in UTF8 can be from **1 to 4 bytes** long, subjected to the following rules:
-
-1. For 1-byte character, the first bit is a 0, followed by its unicode code.
-2. For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
-
-This is how the UTF-8 encoding would work:
-
-       Char. number range  |        UTF-8 octet sequence
-          (hexadecimal)    |              (binary)
-       --------------------+---------------------------------------------
-       0000 0000-0000 007F | 0xxxxxxx
-       0000 0080-0000 07FF | 110xxxxx 10xxxxxx
-       0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
-       0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
-
-Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
-
-**Note:**The input is an array of integers. Only the **least significant 8 bits** of each integer is used to store the data. This means each integer represents only 1 byte of data.
-
-**Example 1:**
-
-    data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
-    
-    Return true.
-    It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
-
-**Example 2:**
-
-    data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
-    
-    Return false.
-    The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
-    The next byte is a continuation byte which starts with 10 and that's correct.
-    But the second continuation byte does not start with 10, so it is invalid.
-
-## 题目大意
-
-UTF-8 中的一个字符可能的长度为 1 到 4 字节，遵循以下的规则：
-
-对于 1 字节的字符，字节的第一位设为 0，后面 7 位为这个符号的 unicode 码。
-对于 n 字节的字符 (n > 1)，第一个字节的前 n 位都设为 1，第 n+1 位设为 0，后面字节的前两位一律设为 10。剩下的没有提及的二进制位，全部为这个符号的 unicode 码。
-这是 UTF-8 编码的工作方式：
-
-```c
-   Char. number range  |        UTF-8 octet sequence
-      (hexadecimal)    |              (binary)
-   --------------------+---------------------------------------------
-   0000 0000-0000 007F | 0xxxxxxx
-   0000 0080-0000 07FF | 110xxxxx 10xxxxxx
-   0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
-   0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
-
-```
-
-给定一个表示数据的整数数组，返回它是否为有效的 utf-8 编码。
-
-注意:
-
-输入是整数数组。只有每个整数的最低 8 个有效位用来存储数据。这意味着每个整数只表示 1 字节的数据。
-
-
-## 解题思路
-
-- 这一题看似很复杂，其实严格按照 UTF8 定义来模拟就可以了。
-
diff --git a/Algorithms/0397. Integer Replacement/README.md b/Algorithms/0397. Integer Replacement/README.md
deleted file mode 100755
index a94c45013..000000000
--- a/Algorithms/0397. Integer Replacement/README.md	
+++ /dev/null
@@ -1,62 +0,0 @@
-# [397. Integer Replacement](https://leetcode.com/problems/integer-replacement/)
-
-
-## 题目:
-
-Given a positive integer n and you can do operations as follow:
-
-1. If  n is even, replace  n with `n/2`.
-2. If  n is odd, you can replace n with either `n + 1` or `n - 1`.
-
-What is the minimum number of replacements needed for n to become 1?
-
-**Example 1:**
-
-    Input:
-    8
-    
-    Output:
-    3
-    
-    Explanation:
-    8 -> 4 -> 2 -> 1
-
-**Example 2:**
-
-    Input:
-    7
-    
-    Output:
-    4
-    
-    Explanation:
-    7 -> 8 -> 4 -> 2 -> 1
-    or
-    7 -> 6 -> 3 -> 2 -> 1
-
-
-## 题目大意
-
-给定一个正整数 n，你可以做如下操作：
-
-1. 如果 n 是偶数，则用 n / 2 替换 n。
-2. 如果 n 是奇数，则可以用 n + 1 或 n - 1 替换 n。
-
-问 n 变为 1 所需的最小替换次数是多少？
-
-
-## 解题思路
-
-
-- 题目给出一个整数 `n`，然后让我们通过变换将它为 1，如果 `n` 是偶数，可以直接变为 `n/2`，如果是奇数，可以先 `n+1` 或 `n-1`，问最终变为 1 的最少步骤。
-- 当 n 为奇数的时候，什么时候需要加 1 ，什么时候需要减 1 ，通过观察规律可以发现，除了 3 和 7 以外，所有加 1 就变成 4 的倍数的奇数，都适合先加 1 运算，比如 15:
-
-        15 -> 16 -> 8 -> 4 -> 2 -> 1
-        15 -> 14 -> 7 -> 6 -> 3 -> 2 -> 1
-        
-        111011 -> 111010 -> 11101 -> 11100 -> 1110 -> 111 -> 1000 -> 100 -> 10 -> 1
-        111011 -> 111100 -> 11110 -> 1111 -> 10000 -> 1000 -> 100 -> 10 -> 1
-
-- 对于 7 来说，加 1 和减 1 的结果相同，可以不用管，对于 3 来说，减 1 的步骤更少，所以需要先去掉这种特殊情况。
-- 最后如何判断某个数字加 1 后是 4 的倍数呢？这里有一个小技巧，由于之前判断了其是奇数了，那么最右边一位肯定是 1，如果其右边第二位也是 1 的话，那么进行加 1 运算，进位后右边肯定会出现两个 0，则一定是 4 的倍数。于是就可以判断出来了。剩下的情况就是偶数的情况，如果之前判定是偶数，那么直接除以 2 (右移一位)即可。
-
diff --git a/Algorithms/0399. Evaluate Division/README.md b/Algorithms/0399. Evaluate Division/README.md
deleted file mode 100755
index 4494c7ce6..000000000
--- a/Algorithms/0399. Evaluate Division/README.md	
+++ /dev/null
@@ -1,42 +0,0 @@
-# [399. Evaluate Division](https://leetcode.com/problems/evaluate-division/)
-
-
-## 题目:
-
-Equations are given in the format `A / B = k`, where `A` and `B` are variables represented as strings, and `k` is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return `-1.0`.
-
-**Example:**
-
-Given `a / b = 2.0, b / c = 3.0.`queries are: `a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .`return `[6.0, 0.5, -1.0, 1.0, -1.0 ].`
-
-The input is: `vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries` , where `equations.size() == values.size()`, and the values are positive. This represents the equations. Return `vector<double>`.
-
-According to the example above:
-
-    equations = [ ["a", "b"], ["b", "c"] ],
-    values = [2.0, 3.0],
-    queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ].
-
-The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.
-
-
-## 题目大意
-
-给出方程式 A / B = k, 其中 A 和 B 均为代表字符串的变量， k 是一个浮点型数字。根据已知方程式求解问题，并返回计算结果。如果结果不存在，则返回 -1.0。
-
-示例 :
-给定 a / b = 2.0, b / c = 3.0
-问题: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? 
-返回 [6.0, 0.5, -1.0, 1.0, -1.0 ]
-
-输入为: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries(方程式，方程式结果，问题方程式)， 其中 equations.size() == values.size()，即方程式的长度与方程式结果长度相等（程式与结果一一对应），并且结果值均为正数。以上为方程式的描述。 返回vector<double>类型。
-
-假设输入总是有效的。你可以假设除法运算中不会出现除数为0的情况，且不存在任何矛盾的结果。
-
-
-## 解题思路
-
-
-- 给出一些字母变量的倍数关系，问给出任意两个字母的倍数是多少。
-- 这一题可以用 DFS 或者并查集来解题。先来看看 DFS 的做法。先建图。每个字母或者字母组合可以看做成一个节点，给出的 `equations` 关系可以看成两个节点之间的有向边。每条有向边都有权值。那么问题可以转换成是否存在一条从起点节点到终点节点的路径，如果存在，输出这条路径上所有有向边权值的累乘结果。如果不存在这条路径，就返回 -1 。如果给的起点和终点不在给出的节点集里面，也输出 -1 。
-- 再来看看并查集的做法。先将每两个有倍数关系的节点做并查集 `union()` 操作。例如 A/B = 2，那么把 B 作为 `parent` 节点，`parents[A] = {B，2}`，`parents[B] = {B，1}`，B 指向自己是 1 。还有一个关系是 `B/C=3`，由于 B 已经在并查集中了，所以这个时候需要把这个关系反过来，处理成 `C/B = 1/3` ，即 `parents[C] = {B，1/3}`。这样把所有有关系的字母都 `union()` 起来。如何求任意两个字母的倍数关系呢？例如 `A/C = ?` 在并查集中查找，可以找到 `parents[C] == parents[A] == B`，那么就用 `parents[A]/parents[C] = 2/(1/3) = 6`。为什么可以这样做呢？因为 `A/B = 2`，`C/B = 1/3`，那么 `A/C = (A/B)/(C/B)` 即 `parents[A]/parents[C] = 2/(1/3) = 6`。
diff --git a/Algorithms/0401. Binary Watch/401. Binary Watch.go b/Algorithms/0401. Binary Watch/401. Binary Watch.go
deleted file mode 100644
index cf8c8488e..000000000
--- a/Algorithms/0401. Binary Watch/401. Binary Watch.go	
+++ /dev/null
@@ -1,90 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"strconv"
-)
-
-var (
-	hour    = []string{"1", "2", "4", "8"}
-	minute  = []string{"01", "02", "04", "08", "16", "32"}
-	hourMap = map[int][]string{
-		0: []string{"0"},
-		1: []string{"1", "2", "4", "8"},
-		2: []string{"3", "5", "9", "6", "10"},
-		3: []string{"7", "11"},
-	}
-	minuteMap = map[int][]string{
-		0: []string{"00"},
-		1: []string{"01", "02", "04", "08", "16", "32"},
-		2: []string{"03", "05", "09", "17", "33", "06", "10", "18", "34", "12", "20", "36", "24", "40", "48"},
-		3: []string{"07", "11", "19", "35", "13", "21", "37", "25", "41", "49", "14", "22", "38", "26", "42", "50", "28", "44", "52", "56"},
-		4: []string{"15", "23", "39", "27", "43", "51", "29", "45", "53", "57", "30", "46", "54", "58"},
-		5: []string{"31", "47", "55", "59"},
-	}
-)
-
-func readBinaryWatch(num int) []string {
-	if num > 8 {
-		return []string{}
-	}
-	res := []string{}
-	for i := 0; i <= num; i++ {
-		for j := 0; j < len(hourMap[i]); j++ {
-			for k := 0; k < len(minuteMap[num-i]); k++ {
-				res = append(res, hourMap[i][j]+":"+minuteMap[num-i][k])
-			}
-		}
-	}
-	return res
-}
-
-/// ---------------------------------------
-/// ---------------------------------------
-/// ---------------------------------------
-/// ---------------------------------------
-/// ---------------------------------------
-// 以下是打表用到的函数
-// 调用 findReadBinaryWatchMinute(num, 0, c, &res) 打表
-func findReadBinaryWatchMinute(target, index int, c []int, res *[]string) {
-	if target == 0 {
-		str, tmp := "", 0
-		for i := 0; i < len(c); i++ {
-			t, _ := strconv.Atoi(minute[c[i]])
-			tmp += t
-		}
-		if tmp < 10 {
-			str = "0" + strconv.Itoa(tmp)
-		} else {
-			str = strconv.Itoa(tmp)
-		}
-		// fmt.Printf("找到解了 c = %v str = %v\n", c, str)
-		fmt.Printf("\"%v\", ", str)
-		return
-	}
-	for i := index; i < 6; i++ {
-		c = append(c, i)
-		findReadBinaryWatchMinute(target-1, i+1, c, res)
-		c = c[:len(c)-1]
-	}
-}
-
-// 调用 findReadBinaryWatchHour(num, 0, c, &res) 打表
-func findReadBinaryWatchHour(target, index int, c []int, res *[]string) {
-	if target == 0 {
-		str, tmp := "", 0
-		for i := 0; i < len(c); i++ {
-			t, _ := strconv.Atoi(hour[c[i]])
-			tmp += t
-		}
-		str = strconv.Itoa(tmp)
-		//fmt.Printf("找到解了 c = %v str = %v\n", c, str)
-		fmt.Printf("\"%v\", ", str)
-		return
-	}
-	for i := index; i < 4; i++ {
-		c = append(c, i)
-		findReadBinaryWatchHour(target-1, i+1, c, res)
-		c = c[:len(c)-1]
-	}
-}
diff --git a/Algorithms/0405. Convert a Number to Hexadecimal/README.md b/Algorithms/0405. Convert a Number to Hexadecimal/README.md
deleted file mode 100755
index 764349dfe..000000000
--- a/Algorithms/0405. Convert a Number to Hexadecimal/README.md	
+++ /dev/null
@@ -1,48 +0,0 @@
-# [405. Convert a Number to Hexadecimal](https://leetcode.com/problems/convert-a-number-to-hexadecimal/)
-
-
-## 题目:
-
-Given an integer, write an algorithm to convert it to hexadecimal. For negative integer, [two’s complement](https://en.wikipedia.org/wiki/Two%27s_complement) method is used.
-
-**Note:**
-
-1. All letters in hexadecimal (`a-f`) must be in lowercase.
-2. The hexadecimal string must not contain extra leading `0`s. If the number is zero, it is represented by a single zero character `'0'`; otherwise, the first character in the hexadecimal string will not be the zero character.
-3. The given number is guaranteed to fit within the range of a 32-bit signed integer.
-4. You **must not use any method provided by the library** which converts/formats the number to hex directly.
-
-**Example 1:**
-
-    Input:
-    26
-    
-    Output:
-    "1a"
-
-**Example 2:**
-
-    Input:
-    -1
-    
-    Output:
-    "ffffffff"
-
-
-## 题目大意
-
-给定一个整数，编写一个算法将这个数转换为十六进制数。对于负整数，我们通常使用[补码运算](https://baike.baidu.com/item/%E8%A1%A5%E7%A0%81/6854613?fr=aladdin)方法。
-
-注意:
-
-1. 十六进制中所有字母(a-f)都必须是小写。
-2. 十六进制字符串中不能包含多余的前导零。如果要转化的数为 0，那么以单个字符 '0' 来表示；对于其他情况，十六进制字符串中的第一个字符将不会是 0 字符。 
-3. 给定的数确保在 32 位有符号整数范围内。
-4. 不能使用任何由库提供的将数字直接转换或格式化为十六进制的方法。
-
-
-
-## 解题思路
-
-- 这一题是水题，将十进制数转换成十六进制的数。需要额外注意 0 和负数的情况。
-
diff --git a/Algorithms/0409. Longest Palindrome/README.md b/Algorithms/0409. Longest Palindrome/README.md
deleted file mode 100755
index df76aec39..000000000
--- a/Algorithms/0409. Longest Palindrome/README.md	
+++ /dev/null
@@ -1,33 +0,0 @@
-# [409. Longest Palindrome](https://leetcode.com/problems/longest-palindrome/)
-
-
-## 题目:
-
-Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
-
-This is case sensitive, for example `"Aa"` is not considered a palindrome here.
-
-**Note:**Assume the length of given string will not exceed 1,010.
-
-**Example:**
-
-    Input:
-    "abccccdd"
-    
-    Output:
-    7
-    
-    Explanation:
-    One longest palindrome that can be built is "dccaccd", whose length is 7.
-
-
-## 题目大意
-
-给定一个包含大写字母和小写字母的字符串，找到通过这些字母构造成的最长的回文串。在构造过程中，请注意区分大小写。比如 "Aa" 不能当做一个回文字符串。注意:假设字符串的长度不会超过 1010。
-
-
-## 解题思路
-
-
-- 给出一个字符串，要求用这个字符串里面的字符组成一个回文串，问回文串最长可以组合成多长的？
-- 这也是一题水题，先统计每个字符的频次，然后每个字符能取 2 个的取 2 个，不足 2 个的并且当前构造中的回文串是偶数的情况下(即每 2 个都配对了)，可以取 1 个。最后组合出来的就是最长回文串。
diff --git a/Algorithms/0412. Fizz Buzz/412. Fizz Buzz.go b/Algorithms/0412. Fizz Buzz/412. Fizz Buzz.go
deleted file mode 100644
index bfac3f7f4..000000000
--- a/Algorithms/0412. Fizz Buzz/412. Fizz Buzz.go	
+++ /dev/null
@@ -1,22 +0,0 @@
-package leetcode
-
-import "strconv"
-
-func fizzBuzz(n int) []string {
-	if n < 0 {
-		return []string{}
-	}
-	solution := make([]string, n)
-	for i := 1; i <= n; i++ {
-		if i%3 == 0 && i%5 == 0 {
-			solution[i-1] = "FizzBuzz"
-		} else if i%3 == 0 {
-			solution[i-1] = "Fizz"
-		} else if i%5 == 0 {
-			solution[i-1] = "Buzz"
-		} else {
-			solution[i-1] = strconv.Itoa(i)
-		}
-	}
-	return solution
-}
diff --git a/Algorithms/0421. Maximum XOR of Two Numbers in an Array/README.md b/Algorithms/0421. Maximum XOR of Two Numbers in an Array/README.md
deleted file mode 100755
index b76b497f2..000000000
--- a/Algorithms/0421. Maximum XOR of Two Numbers in an Array/README.md	
+++ /dev/null
@@ -1,33 +0,0 @@
-# [421. Maximum XOR of Two Numbers in an Array](https://leetcode.com/problems/maximum-xor-of-two-numbers-in-an-array/)
-
-
-## 题目:
-
-Given a **non-empty** array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231.
-
-Find the maximum result of ai XOR aj, where 0 ≤ i, j < n.
-
-Could you do this in O(n) runtime?
-
-**Example:**
-
-    Input: [3, 10, 5, 25, 2, 8]
-    
-    Output: 28
-    
-    Explanation: The maximum result is 5 ^ 25 = 28.
-
-
-## 题目大意
-
-给定一个非空数组，数组中元素为 a0, a1, a2, … , an-1，其中 0 ≤ ai < 2^31 。找到 ai 和 aj 最大的异或 (XOR) 运算结果，其中0 ≤ i,  j < n 。你能在O(n)的时间解决这个问题吗？
-
-
-## 解题思路
-
-
-- 这一题最先考虑到的解法就是暴力解法，2 层循环，依次计算两两数之间的异或值，动态维护最大的值，遍历完成以后输出最大值即可。提交代码会发现超时。
-- 改进一点的做法就是一层循环。试想，求的最终结果是一个 32 位的二进制数，如果想要这个数最大，那么高位都填满 1 就是最大。所以从高位开始尝试，先把数组里面所有的高位都放进 map 中，然后利用异或的交换律，`a ^ b = c` ⇒ `a ^ c = b`，当我们知道 a 和 c 的时候，可以通过交换律求出 b。a 就是我们遍历的每个数，c 是我们想要尝试的高位最大值，例如，111…000，从高位逐渐往低位填 1 。如果我们求的 b 也在 map 中，那么就代表 c 是可以求出来的。如果 c 比当前的 max 值要大，就更新。按照这样的方式遍历往 32 位，每次也遍历完整个数组中的每个数，最终 max 里面就是需要求的最大值。
-- 还有更好的做法是利用 Trie 这个数据结构。构建一棵深度为 33 的二叉树。root 节点左孩子为 1，右孩子为 0 代表着所有数字的最高位，其次根据次高位继续往下。如果某一个节点左右子树都不为空，那么得到最终答案的两个数字肯定分别出自于左右子树且此位为 1；如果任意一个为空，那么最终答案该位为 0，依次迭代得到最终结果。具体做法见：[Java O(n) solution using Trie - LeetCode Discuss](https://discuss.leetcode.com/topic/63207/java-o-n-solution-using-trie)
-
-- 最后还有更“完美的做法”，利用 leetcode 网站判题的特性，我们可以测出比较弱的数据，绕过这组弱数据可以直接 AC。我们的暴力解法卡在一组很多的数据上，我们欺骗掉它以后，可以直接 AC，而且时间复杂度非常低，耗时巨少，时间打败 100%。
diff --git a/Algorithms/0424. Longest Repeating Character Replacement/424. Longest Repeating Character Replacement.go b/Algorithms/0424. Longest Repeating Character Replacement/424. Longest Repeating Character Replacement.go
deleted file mode 100644
index 874b00280..000000000
--- a/Algorithms/0424. Longest Repeating Character Replacement/424. Longest Repeating Character Replacement.go	
+++ /dev/null
@@ -1,15 +0,0 @@
-package leetcode
-
-func characterReplacement(s string, k int) int {
-	res, left, right, counter, freq := 0, 0, 1, 0, make([]int, 26)
-	for right = 0; right < len(s); right++ {
-		freq[s[right]-'A']++
-		counter = max(counter, freq[s[right]-'A'])
-		for right-left+1-counter > k {
-			freq[s[left]-'A']--
-			left++
-		}
-		res = max(res, right-left+1)
-	}
-	return res
-}
diff --git a/Algorithms/0435. Non-overlapping Intervals/435. Non-overlapping Intervals_test.go b/Algorithms/0435. Non-overlapping Intervals/435. Non-overlapping Intervals_test.go
deleted file mode 100644
index 3050bc8a1..000000000
--- a/Algorithms/0435. Non-overlapping Intervals/435. Non-overlapping Intervals_test.go	
+++ /dev/null
@@ -1,52 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question435 struct {
-	para435
-	ans435
-}
-
-// para 是参数
-// one 代表第一个参数
-type para435 struct {
-	one [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans435 struct {
-	one int
-}
-
-func Test_Problem435(t *testing.T) {
-
-	qs := []question435{
-
-		question435{
-			para435{[][]int{[]int{1, 2}, []int{2, 3}, []int{3, 4}, []int{1, 3}}},
-			ans435{1},
-		},
-
-		question435{
-			para435{[][]int{[]int{1, 2}, []int{1, 2}, []int{1, 2}}},
-			ans435{2},
-		},
-
-		question435{
-			para435{[][]int{[]int{1, 2}, []int{2, 3}}},
-			ans435{0},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 435------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans435, q.para435
-		fmt.Printf("【input】:%v       【output】:%v\n", p, eraseOverlapIntervals1(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0437. Path Sum III/437. Path Sum III.go b/Algorithms/0437. Path Sum III/437. Path Sum III.go
deleted file mode 100644
index 29adc8e7d..000000000
--- a/Algorithms/0437. Path Sum III/437. Path Sum III.go	
+++ /dev/null
@@ -1,33 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-func pathSumIII(root *TreeNode, sum int) int {
-	if root == nil {
-		return 0
-	}
-	res := findPath437(root, sum)
-	res += pathSumIII(root.Left, sum)
-	res += pathSumIII(root.Right, sum)
-	return res
-}
-
-// 寻找包含 root 这个结点，且和为 sum 的路径
-func findPath437(root *TreeNode, sum int) int {
-	if root == nil {
-		return 0
-	}
-	res := 0
-	if root.Val == sum {
-		res++
-	}
-	res += findPath437(root.Left, sum-root.Val)
-	res += findPath437(root.Right, sum-root.Val)
-	return res
-}
diff --git a/Algorithms/0437. Path Sum III/437. Path Sum III_test.go b/Algorithms/0437. Path Sum III/437. Path Sum III_test.go
deleted file mode 100644
index 71a727d70..000000000
--- a/Algorithms/0437. Path Sum III/437. Path Sum III_test.go	
+++ /dev/null
@@ -1,49 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question437 struct {
-	para437
-	ans437
-}
-
-// para 是参数
-// one 代表第一个参数
-type para437 struct {
-	one []int
-	sum int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans437 struct {
-	one int
-}
-
-func Test_Problem437(t *testing.T) {
-
-	qs := []question437{
-		question437{
-			para437{[]int{}, 0},
-			ans437{0},
-		},
-
-		question437{
-			para437{[]int{10, 5, -3, 3, 2, NULL, 11, 3, -2, NULL, 1}, 8},
-			ans437{3},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 437------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans437, q.para437
-		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", pathSumIII(root, p.sum))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0437. Path Sum III/README.md b/Algorithms/0437. Path Sum III/README.md
deleted file mode 100755
index cad475bdd..000000000
--- a/Algorithms/0437. Path Sum III/README.md	
+++ /dev/null
@@ -1,44 +0,0 @@
-# [437. Path Sum III](https://leetcode.com/problems/path-sum-iii/)
-
-
-## 题目
-
-You are given a binary tree in which each node contains an integer value.
-
-Find the number of paths that sum to a given value.
-
-The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
-
-The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
-
-**Example:**
-
-    root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
-    
-          10
-         /  \
-        5   -3
-       / \    \
-      3   2   11
-     / \   \
-    3  -2   1
-    
-    Return 3. The paths that sum to 8 are:
-    
-    1.  5 -> 3
-    2.  5 -> 2 -> 1
-    3. -3 -> 11
-
-
-## 题目大意
-
-给定一个二叉树，它的每个结点都存放着一个整数值。找出路径和等于给定数值的路径总数。路径不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。二叉树不超过1000个节点，且节点数值范围是 [-1000000,1000000] 的整数。
-
-
-## 解题思路
-
-
-- 这一题是第 112 题 Path Sum 和第 113 题 Path Sum II 的加强版，这一题要求求出任意一条路径的和为 sum，起点不一定是根节点，可以是任意节点开始。
-- 注意这一题可能出现负数的情况，节点和为 sum，并不一定是最终情况，有可能下面还有正数节点和负数节点相加正好为 0，那么这也是一种情况。一定要遍历到底。
-- 一个点是否为 sum 的起点，有 3 种情况，第一种情况路径包含该 root 节点，如果包含该结点，就在它的左子树和右子树中寻找和为 `sum-root.Val` 的情况。第二种情况路径不包含该 root 节点，那么就需要在它的左子树和右子树中分别寻找和为 sum 的结点。
-
diff --git a/Algorithms/0441. Arranging Coins/README.md b/Algorithms/0441. Arranging Coins/README.md
deleted file mode 100755
index 85df77bad..000000000
--- a/Algorithms/0441. Arranging Coins/README.md	
+++ /dev/null
@@ -1,45 +0,0 @@
-# [441. Arranging Coins](https://leetcode.com/problems/arranging-coins/)
-
-## 题目:
-
-You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins.
-
-Given n, find the total number of **full** staircase rows that can be formed.
-
-n is a non-negative integer and fits within the range of a 32-bit signed integer.
-
-**Example 1:**
-
-    n = 5
-    
-    The coins can form the following rows:
-    ¤
-    ¤ ¤
-    ¤ ¤
-    
-    Because the 3rd row is incomplete, we return 2.
-
-**Example 2:**
-
-    n = 8
-    
-    The coins can form the following rows:
-    ¤
-    ¤ ¤
-    ¤ ¤ ¤
-    ¤ ¤
-    
-    Because the 4th row is incomplete, we return 3.
-
-
-## 题目大意
-
-你总共有 n 枚硬币，你需要将它们摆成一个阶梯形状，第 k 行就必须正好有 k 枚硬币。给定一个数字 n，找出可形成完整阶梯行的总行数。n 是一个非负整数，并且在32位有符号整型的范围内。
-
-
-
-## 解题思路
-
-
-- n 个硬币，按照递增的方式排列搭楼梯，第一层一个，第二层二个，……第 n 层需要 n 个硬币。问硬币 n 能够搭建到第几层？
-- 这一题有 2 种解法，第一种解法就是解方程求出 X，`(1+x)x/2 = n`，即 `x = floor(sqrt(2*n+1/4) - 1/2)`，第二种解法是模拟。
diff --git a/Algorithms/0445. Add Two Numbers II/445. Add Two Numbers II.go b/Algorithms/0445. Add Two Numbers II/445. Add Two Numbers II.go
deleted file mode 100644
index 68822c1cd..000000000
--- a/Algorithms/0445. Add Two Numbers II/445. Add Two Numbers II.go	
+++ /dev/null
@@ -1,62 +0,0 @@
-package leetcode
-
-/**
- * Definition for singly-linked list.
- * type ListNode struct {
- *     Val int
- *     Next *ListNode
- * }
- */
-func addTwoNumbers445(l1 *ListNode, l2 *ListNode) *ListNode {
-	if l1 == nil {
-		return l2
-	}
-	if l2 == nil {
-		return l1
-	}
-	l1Length := getLength(l1)
-	l2Length := getLength(l2)
-	newHeader := &ListNode{Val: 1, Next: nil}
-	if l1Length < l2Length {
-		newHeader.Next = addNode(l2, l1, l2Length-l1Length)
-	} else {
-		newHeader.Next = addNode(l1, l2, l1Length-l2Length)
-	}
-	if newHeader.Next.Val > 9 {
-		newHeader.Next.Val = newHeader.Next.Val % 10
-		return newHeader
-	}
-	return newHeader.Next
-}
-
-func addNode(l1 *ListNode, l2 *ListNode, offset int) *ListNode {
-	if l1 == nil {
-		return nil
-	}
-	var (
-		res, node *ListNode
-	)
-	if offset == 0 {
-		res = &ListNode{Val: l1.Val + l2.Val, Next: nil}
-		node = addNode(l1.Next, l2.Next, 0)
-	} else {
-		res = &ListNode{Val: l1.Val, Next: nil}
-		node = addNode(l1.Next, l2, offset-1)
-	}
-	if node != nil && node.Val > 9 {
-		res.Val++
-		node.Val = node.Val % 10
-	}
-	res.Next = node
-	return res
-}
-
-func getLength(l *ListNode) int {
-	count := 0
-	cur := l
-	for cur != nil {
-		count++
-		cur = cur.Next
-	}
-	return count
-}
diff --git a/Algorithms/0457. Circular Array Loop/README.md b/Algorithms/0457. Circular Array Loop/README.md
deleted file mode 100755
index 72222ea19..000000000
--- a/Algorithms/0457. Circular Array Loop/README.md	
+++ /dev/null
@@ -1,59 +0,0 @@
-# [457. Circular Array Loop](https://leetcode.com/problems/circular-array-loop/)
-
-
-## 题目:
-
-You are given a **circular** array `nums` of positive and negative integers. If a number k at an index is positive, then move forward k steps. Conversely, if it's negative (-k), move backward k steps. Since the array is circular, you may assume that the last element's next element is the first element, and the first element's previous element is the last element.
-
-Determine if there is a loop (or a cycle) in `nums`. A cycle must start and end at the same index and the cycle's length > 1. Furthermore, movements in a cycle must all follow a single direction. In other words, a cycle must not consist of both forward and backward movements.
-
-**Example 1:**
-
-    Input: [2,-1,1,2,2]
-    Output: true
-    Explanation: There is a cycle, from index 0 -> 2 -> 3 -> 0. The cycle's length is 3.
-
-**Example 2:**
-
-    Input: [-1,2]
-    Output: false
-    Explanation: The movement from index 1 -> 1 -> 1 ... is not a cycle, because the cycle's length is 1. By definition the cycle's length must be greater than 1.
-
-**Example 3:**
-
-    Input: [-2,1,-1,-2,-2]
-    Output: false
-    Explanation: The movement from index 1 -> 2 -> 1 -> ... is not a cycle, because movement from index 1 -> 2 is a forward movement, but movement from index 2 -> 1 is a backward movement. All movements in a cycle must follow a single direction.
-
-**Note:**
-
-1. -1000 ≤ nums[i] ≤ 1000
-2. nums[i] ≠ 0
-3. 1 ≤ nums.length ≤ 5000
-
-**Follow up:**
-
-Could you solve it in **O(n)** time complexity and **O(1)** extra space complexity?
-
-
-## 题目大意
-
-给定一个含有正整数和负整数的环形数组 nums。 如果某个索引中的数 k 为正数，则向前移动 k 个索引。相反，如果是负数 (-k)，则向后移动 k 个索引。因为数组是环形的，所以可以假设最后一个元素的下一个元素是第一个元素，而第一个元素的前一个元素是最后一个元素。
-
-确定 nums 中是否存在循环（或周期）。循环必须在相同的索引处开始和结束并且循环长度 > 1。此外，一个循环中的所有运动都必须沿着同一方向进行。换句话说，一个循环中不能同时包括向前的运动和向后的运动。
-
-提示：
-
-- -1000 ≤ nums[i] ≤ 1000
-- nums[i] ≠ 0
-- 1 ≤ nums.length ≤ 5000
-
-进阶：
-
-- 你能写出时间时间复杂度为 O(n) 和额外空间复杂度为 O(1) 的算法吗？
-
-## 解题思路
-
-
-- 给出一个循环数组，数组的数字代表了前进和后退的步数，+ 代表往右(前进)，- 代表往左(后退)。问这个循环数组中是否存在一个循环，并且这个循环内不能只有一个元素，循环的方向都必须是同方向的。
-- 遇到循环就可以优先考虑用快慢指针的方法判断循环，这一题对循环增加了一个条件，循环不能只是单元素的循环，所以在快慢指针中加入这个判断条件。还有一个判断条件是循环的方向必须是同向的，这个简单，用 `num[i] * num[j] > 0` 就可以判断出是同向的(如果是反向的，那么两者的乘积必然是负数)，如果没有找到循环，可以将当前已经走过的路径上的 num[] 值都置为 0，标记已经访问过了。下次循环遇到访问过的元素，`num[i] * num[j] > 0` 就会是 0，提前退出找循环的过程。
diff --git a/Algorithms/0461. Hamming Distance/README.md b/Algorithms/0461. Hamming Distance/README.md
deleted file mode 100755
index 34a8a426c..000000000
--- a/Algorithms/0461. Hamming Distance/README.md	
+++ /dev/null
@@ -1,36 +0,0 @@
-# [461. Hamming Distance](https://leetcode.com/problems/hamming-distance/)
-
-## 题目:
-
-The [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) between two integers is the number of positions at which the corresponding bits are different.
-
-Given two integers `x` and `y`, calculate the Hamming distance.
-
-**Note:**0 ≤ `x`, `y` < 231.
-
-**Example:**
-
-    Input: x = 1, y = 4
-    
-    Output: 2
-    
-    Explanation:
-    1   (0 0 0 1)
-    4   (0 1 0 0)
-           ↑   ↑
-    
-    The above arrows point to positions where the corresponding bits are different.
-
-
-## 题目大意
-
-两个整数之间的汉明距离指的是这两个数字对应二进制位不同的位置的数目。给出两个整数 x 和 y，计算它们之间的汉明距离。
-
-注意：
-0 ≤ x, y < 231.
-
-
-
-## 解题思路
-
-- 求 2 个数的海明距离。海明距离的定义是两个数二进制位不同的总个数。这一题利用的位操作的是 X &= (X - 1) 不断的清除最低位的 1 。先将这两个数异或，异或以后清除低位的 1 就是最终答案。
diff --git a/Algorithms/0463. Island Perimeter/README.md b/Algorithms/0463. Island Perimeter/README.md
deleted file mode 100755
index c702c23f0..000000000
--- a/Algorithms/0463. Island Perimeter/README.md	
+++ /dev/null
@@ -1,39 +0,0 @@
-# [463. Island Perimeter](https://leetcode.com/problems/island-perimeter/)
-
-## 题目:
-
-You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water.
-
-Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells).
-
-The island doesn't have "lakes" (water inside that isn't connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don't exceed 100. Determine the perimeter of the island.
-
-**Example:**
-
-    Input:
-    [[0,1,0,0],
-     [1,1,1,0],
-     [0,1,0,0],
-     [1,1,0,0]]
-    
-    Output: 16
-    
-    Explanation: The perimeter is the 16 yellow stripes in the image below:
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/island.png)
-
-
-## 题目大意
-
-给定一个包含 0 和 1 的二维网格地图，其中 1 表示陆地 0 表示水域。
-
-网格中的格子水平和垂直方向相连（对角线方向不相连）。整个网格被水完全包围，但其中恰好有一个岛屿（或者说，一个或多个表示陆地的格子相连组成的岛屿）。
-
-岛屿中没有“湖”（“湖” 指水域在岛屿内部且不和岛屿周围的水相连）。格子是边长为 1 的正方形。网格为长方形，且宽度和高度均不超过 100 。计算这个岛屿的周长。
-
-
-
-## 解题思路
-
-- 给出一个二维数组，二维数组中有一些连在一起的 1 ，这是一个岛屿，求这个岛屿的周长。
-- 这是一道水题，判断四周边界的情况依次加一即可。
diff --git a/Algorithms/0470. Implement Rand10() Using Rand7()/README.md b/Algorithms/0470. Implement Rand10() Using Rand7()/README.md
deleted file mode 100755
index c88edd10f..000000000
--- a/Algorithms/0470. Implement Rand10() Using Rand7()/README.md	
+++ /dev/null
@@ -1,69 +0,0 @@
-# [470. Implement Rand10() Using Rand7()](https://leetcode.com/problems/implement-rand10-using-rand7/)
-
-
-## 题目:
-
-Given a function `rand7` which generates a uniform random integer in the range 1 to 7, write a function `rand10` which generates a uniform random integer in the range 1 to 10.
-
-Do NOT use system's `Math.random()`.
-
-**Example 1:**
-
-    Input: 1
-    Output: [7]
-
-**Example 2:**
-
-    Input: 2
-    Output: [8,4]
-
-**Example 3:**
-
-    Input: 3
-    Output: [8,1,10]
-
-**Note:**
-
-1. `rand7` is predefined.
-2. Each testcase has one argument: `n`, the number of times that `rand10` is called.
-
-**Follow up:**
-
-1. What is the [expected value](https://en.wikipedia.org/wiki/Expected_value) for the number of calls to `rand7()` function?
-2. Could you minimize the number of calls to `rand7()`?
-
-
-## 题目大意
-
-已有方法 rand7 可生成 1 到 7 范围内的均匀随机整数，试写一个方法 rand10 生成 1 到 10 范围内的均匀随机整数。不要使用系统的 Math.random() 方法。
-
-
-提示:
-
-- rand7 已定义。
-- 传入参数: n 表示 rand10 的调用次数。
- 
-
-进阶:
-
-- rand7()调用次数的 期望值 是多少 ?
-- 你能否尽量少调用 rand7() ?
-
-
-
-## 解题思路
-
-
-- 给出 `rand7()` 要求实现 `rand10()`。
-- `rand7()` 等概率地产生1，2，3，4，5，6，7。要想得到 `rand10()` 即等概率的生成 1-10 。解题思路是先构造一个 `randN()`，这个 N 必须是 10 的整数倍，然后 randN % 10 就可以得到 `rand10()` 了。所以可以从 `rand7()` 先构造出 `rand49()`，再把 `rand49()` 中大于等于 40 的都过滤掉，这样就得到了 `rand40()`，在对 10 取余即可。
-- 具体构造步骤，`rand7() --> rand49() --> rand40() --> rand10()`：
-    1. `rand7()` 等概率地产生 1,2,3,4,5,6,7.
-    2. `rand7() - 1` 等概率地产生 [0,6].
-    3. `(rand7() - 1) *7` 等概率地产生 0, 7, 14, 21, 28, 35, 42
-    4. `(rand7() - 1) * 7 + (rand7() - 1)` 等概率地产生 [0, 48] 这 49 个数字
-    5. 如果步骤 4 的结果大于等于 40，那么就重复步骤 4，直到产生的数小于 40
-    6. 把步骤 5 的结果 mod 10 再加 1， 就会等概率的随机生成 [1, 10]
-- 这道题可以推广到生成任意数的随机数问题。用 `randN()` 实现 `randM()`，`M>N`。步骤如下：
-    1. 用 `randN()` 先实现 `randX()`，其中 X ≥ M，X 是 M 的整数倍。如这题中的 49 > 10；
-    2. 再用 `randX()` 生成 `randM()`，如这题中的 49 —> 40 —> 10。
-- 举个例子，用 `rand3()` 生成 `rand11()`，可以先生成 `rand27()`，然后变成以 22 为界限，因为 22 是 11 的倍数。生成 `rand27()` 的方式：`3 * 3 * (rand3() - 1) + 3 * (rand3() - 1) + (rand3() - 1)`，最后生成了 `rand11()`；用 `rand7()` 生成 `rand9()`，可以先生成 `rand49()`，然后变成以 45 为界限，因为 45 是 9 的倍数。生成 `rand49()` 的方式：`(rand7() - 1) * 7 + (rand7() - 1)`，最后生成了 `rand9()`；用 `rand6()` 生成 `rand13()`，可以先生成 `rand36()`，然后变成以 26 为界限，因为 26 是 13 的倍数。生成 `rand36()` 的方式：`(rand6() - 1) * 6 + (rand6() - 1)`，最后生成了 `rand13()`；
diff --git a/Algorithms/0474. Ones and Zeroes/README.md b/Algorithms/0474. Ones and Zeroes/README.md
deleted file mode 100755
index 0b11e9967..000000000
--- a/Algorithms/0474. Ones and Zeroes/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# [474. Ones and Zeroes](https://leetcode.com/problems/ones-and-zeroes/)
-
-
-## 题目:
-
-In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.
-
-For now, suppose you are a dominator of **m** `0s` and **n** `1s` respectively. On the other hand, there is an array with strings consisting of only `0s` and `1s`.
-
-Now your task is to find the maximum number of strings that you can form with given **m**`0s` and **n** `1s`. Each `0` and `1` can be used at most **once**.
-
-**Note:**
-
-1. The given numbers of `0s` and `1s` will both not exceed `100`
-2. The size of given string array won't exceed `600`.
-
-**Example 1:**
-
-    Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3
-    Output: 4
-    
-    Explanation: This are totally 4 strings can be formed by the using of 5 0s and 3 1s, which are “10,”0001”,”1”,”0”
-
-**Example 2:**
-
-    Input: Array = {"10", "0", "1"}, m = 1, n = 1
-    Output: 2
-    
-    Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".
-
-
-## 题目大意
-
-在计算机界中，我们总是追求用有限的资源获取最大的收益。现在，假设你分别支配着 m 个 0 和 n 个 1。另外，还有一个仅包含 0 和 1 字符串的数组。你的任务是使用给定的 m 个 0 和 n 个 1 ，找到能拼出存在于数组中的字符串的最大数量。每个 0 和 1 至多被使用一次。
-
-注意:
-
-1. 给定 0 和 1 的数量都不会超过 100。
-2. 给定字符串数组的长度不会超过 600。
-
-
-
-## 解题思路
-
-- 给定一个字符串数组和 m，n，其中所有的字符都是由 0 和 1 组成的。问能否从数组中取出最多的字符串，使得这些取出的字符串中所有的 0 的个数 ≤ m，1 的个数 ≤ n。
-- 这一题是典型的 01 背包的题型。只不过是一个二维的背包问题。在 n 个物品中选出一定物品，**尽量完全填满** m 维和 n 维的背包。为什么是尽量填满？因为不一定能完全填满背包。
-- `dp[i][j]` 代表尽量填满容量为 `(i,j)` 的背包装下的物品总数，状态转移方程为 `dp[i][j] = max(dp[i][j], 1+dp[i-zero][j-one])`。其中 zero 代表的当前装入物品在 m 维上的体积，也即 0 的个数。one 代表的是当前装入物品在 n 维上的体积，也即 1 的个数。每添加一个物品，比较当前 (i,j) 的背包装下的物品总数和 (i-zero,j-one) 的背包装下的物品总数 + 1，比较这两者的大小，保存两者的最大值。每添加一个物品就刷新这个二维背包，直到所有物品都扫完一遍。dp[m][n] 中存储的就是最终的答案。时间复杂度 `O( n * M * N )`。
diff --git a/Algorithms/0475. Heaters/README.md b/Algorithms/0475. Heaters/README.md
deleted file mode 100755
index a0375170b..000000000
--- a/Algorithms/0475. Heaters/README.md	
+++ /dev/null
@@ -1,51 +0,0 @@
-# [475. Heaters](https://leetcode.com/problems/heaters/)
-
-## 题目:
-
-Winter is coming! Your first job during the contest is to design a standard heater with fixed warm radius to warm all the houses.
-
-Now, you are given positions of houses and heaters on a horizontal line, find out minimum radius of heaters so that all houses could be covered by those heaters.
-
-So, your input will be the positions of houses and heaters seperately, and your expected output will be the minimum radius standard of heaters.
-
-**Note:**
-
-1. Numbers of houses and heaters you are given are non-negative and will not exceed 25000.
-2. Positions of houses and heaters you are given are non-negative and will not exceed 10^9.
-3. As long as a house is in the heaters' warm radius range, it can be warmed.
-4. All the heaters follow your radius standard and the warm radius will the same.
-
-**Example 1:**
-
-    Input: [1,2,3],[2]
-    Output: 1
-    Explanation: The only heater was placed in the position 2, and if we use the radius 1 standard, then all the houses can be warmed.
-
-**Example 2:**
-
-    Input: [1,2,3,4],[1,4]
-    Output: 1
-    Explanation: The two heater was placed in the position 1 and 4. We need to use radius 1 standard, then all the houses can be warmed.
-
-
-
-## 题目大意
-
-
-冬季已经来临。 你的任务是设计一个有固定加热半径的供暖器向所有房屋供暖。现在，给出位于一条水平线上的房屋和供暖器的位置，找到可以覆盖所有房屋的最小加热半径。所以，你的输入将会是房屋和供暖器的位置。你将输出供暖器的最小加热半径。
-
-说明:
-
-- 给出的房屋和供暖器的数目是非负数且不会超过 25000。
-- 给出的房屋和供暖器的位置均是非负数且不会超过10^9。
-- 只要房屋位于供暖器的半径内(包括在边缘上)，它就可以得到供暖。
-- 所有供暖器都遵循你的半径标准，加热的半径也一样。
-
-
-
-## 解题思路
-
-
-- 给出一个房子坐标的数组，和一些供暖器坐标的数组，分别表示房子的坐标和供暖器的坐标。要求找到供暖器最小的半径能使得所有的房子都能享受到暖气。
-- 这一题可以用暴力的解法，暴力解法依次遍历每个房子的坐标，再遍历每个供暖器，找到距离房子最近的供暖器坐标。在所有这些距离的长度里面找到最大值，这个距离的最大值就是供暖器半径的最小值。时间复杂度 O(n^2)。
-- 这一题最优解是二分搜索。在查找距离房子最近的供暖器的时候，先将供暖器排序，然后用二分搜索的方法查找。其他的做法和暴力解法一致。时间复杂度 O(n log n)。
diff --git a/Algorithms/0476. Number Complement/README.md b/Algorithms/0476. Number Complement/README.md
deleted file mode 100755
index 5c48041d4..000000000
--- a/Algorithms/0476. Number Complement/README.md	
+++ /dev/null
@@ -1,41 +0,0 @@
-# [476. Number Complement](https://leetcode.com/problems/number-complement/)
-
-
-## 题目:
-
-Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
-
-**Note:**
-
-1. The given integer is guaranteed to fit within the range of a 32-bit signed integer.
-2. You could assume no leading zero bit in the integer’s binary representation.
-
-**Example 1:**
-
-    Input: 5
-    Output: 2
-    Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.
-
-**Example 2:**
-
-    Input: 1
-    Output: 0
-    Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.
-
-
-## 题目大意
-
-给定一个正整数，输出它的补数。补数是对该数的二进制表示取反。
-
-注意:
-
-给定的整数保证在32位带符号整数的范围内。
-你可以假定二进制数不包含前导零位。
-
-
-
-## 解题思路
-
-
-- 求一个正数的补数，补数的定义是对该数的二进制表示取反。当前不能改变符号位。按照题意构造相应的 mask 再取反即可。
-
diff --git a/Algorithms/0477. Total Hamming Distance/README.md b/Algorithms/0477. Total Hamming Distance/README.md
deleted file mode 100755
index bc25bb38c..000000000
--- a/Algorithms/0477. Total Hamming Distance/README.md	
+++ /dev/null
@@ -1,34 +0,0 @@
-# [477. Total Hamming Distance](https://leetcode.com/problems/total-hamming-distance/)
-
-
-## 题目:
-
-The [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) between two integers is the number of positions at which the corresponding bits are different.
-
-Now your job is to find the total Hamming distance between all pairs of the given numbers.
-
-**Example:**
-
-    Input: 4, 14, 2
-    
-    Output: 6
-    
-    Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
-    showing the four bits relevant in this case). So the answer will be:
-    HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
-
-**Note:**
-
-1. Elements of the given array are in the range of `0` to `10^9`
-2. Length of the array will not exceed `10^4`.
-
-
-## 题目大意
-
-两个整数的[汉明距离](https://baike.baidu.com/item/%E6%B1%89%E6%98%8E%E8%B7%9D%E7%A6%BB/475174?fr=aladdin)指的是这两个数字的二进制数对应位不同的数量。计算一个数组中，任意两个数之间汉明距离的总和。
-
-
-## 解题思路
-
-- 计算一个数组内两两元素的海明距离总和。海明距离的定义是两个数二进制位不同的总个数。那么可以把数组中的每个元素 32 位的二进制位依次扫一遍，当扫到某一位上的时候，有 k 个元素在这个位上的值是 1，n - k 个元素在这个位上的值是 0，那么在这一位上所有两两元素的海明距离是 k*(n-k) ，当把 32 位全部都扫完以后，累加出来的海明距离就是所有两两元素的海明距离。
-
diff --git a/Algorithms/0491. Increasing Subsequences/491. Increasing Subsequences_test.go b/Algorithms/0491. Increasing Subsequences/491. Increasing Subsequences_test.go
deleted file mode 100644
index 18f6e7422..000000000
--- a/Algorithms/0491. Increasing Subsequences/491. Increasing Subsequences_test.go	
+++ /dev/null
@@ -1,47 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question491 struct {
-	para491
-	ans491
-}
-
-// para 是参数
-// one 代表第一个参数
-type para491 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans491 struct {
-	one [][]int
-}
-
-func Test_Problem491(t *testing.T) {
-
-	qs := []question491{
-
-		question491{
-			para491{[]int{4, 3, 2, 1}},
-			ans491{[][]int{}},
-		},
-
-		question491{
-			para491{[]int{4, 6, 7, 7}},
-			ans491{[][]int{[]int{4, 6}, []int{4, 7}, []int{4, 6, 7}, []int{4, 6, 7, 7}, []int{6, 7}, []int{6, 7, 7}, []int{7, 7}, []int{4, 7, 7}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 491------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans491, q.para491
-		fmt.Printf("【input】:%v       【output】:%v\n", p, findSubsequences(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0491. Increasing Subsequences/README.md b/Algorithms/0491. Increasing Subsequences/README.md
deleted file mode 100755
index bb39e93de..000000000
--- a/Algorithms/0491. Increasing Subsequences/README.md	
+++ /dev/null
@@ -1,40 +0,0 @@
-# [491. Increasing Subsequences](https://leetcode.com/problems/increasing-subsequences/)
-
-
-## 题目:
-
-Given an integer array, your task is to find all the different possible increasing subsequences of the given array, and the length of an increasing subsequence should be at least 2.
-
-**Example:**
-
-    Input: [4, 6, 7, 7]
-    Output: [[4, 6], [4, 7], [4, 6, 7], [4, 6, 7, 7], [6, 7], [6, 7, 7], [7,7], [4,7,7]]
-
-**Note:**
-
-1. The length of the given array will not exceed 15.
-2. The range of integer in the given array is [-100,100].
-3. The given array may contain duplicates, and two equal integers should also be considered as a special case of increasing sequence.
-
-
-
-## 题目大意
-
-
-给定一个整型数组, 你的任务是找到所有该数组的递增子序列，递增子序列的长度至少是 2。
-
-说明:
-
-1. 给定数组的长度不会超过15。
-2. 数组中的整数范围是 [-100,100]。
-3. 给定数组中可能包含重复数字，相等的数字应该被视为递增的一种情况。
-
-
-
-
-## 解题思路
-
-
-- 给出一个数组，要求找出这个数组中所有长度大于 2 的非递减子序列。子序列顺序和原数组元素下标必须是顺序的，不能是逆序的。
-- 这一题和第 78 题和第 90 题是类似的题目。第 78 题和第 90 题是求所有子序列，这一题在这两题的基础上增加了非递减和长度大于 2 的条件。需要注意的两点是，原数组中元素可能会重复，最终结果输出的时候需要去重。最终结果输出的去重用 map 处理，数组中重复元素用 DFS 遍历搜索。在每次 DFS 中，用 map 记录遍历过的元素，保证本轮 DFS 中不出现重复的元素，递归到下一层还可以选择值相同，但是下标不同的另外一个元素。外层循环也要加一个  map，这个 map 是过滤每组解因为重复元素导致的重复解，经过过滤以后，起点不同了，最终的解也会不同。
-- 这一题和第 78 题，第 90 题类似，可以一起解答和复习。
diff --git a/Algorithms/0493. Reverse Pairs/493. Reverse Pairs.go b/Algorithms/0493. Reverse Pairs/493. Reverse Pairs.go
deleted file mode 100644
index b21a85a53..000000000
--- a/Algorithms/0493. Reverse Pairs/493. Reverse Pairs.go	
+++ /dev/null
@@ -1,76 +0,0 @@
-package leetcode
-
-import (
-	"sort"
-
-	"github.com/halfrost/LeetCode-Go/template"
-)
-
-// 解法一 线段树，时间复杂度 O(n log n)
-func reversePairs(nums []int) int {
-	if len(nums) < 2 {
-		return 0
-	}
-	st, numsMap, indexMap, numsArray, res := template.SegmentCountTree{}, make(map[int]int, 0), make(map[int]int, 0), []int{}, 0
-	numsMap[nums[0]] = nums[0]
-	for _, num := range nums {
-		numsMap[num] = num
-		numsMap[2*num+1] = 2*num + 1
-	}
-	// numsArray 是 prefixSum 去重之后的版本，利用 numsMap 去重
-	for _, v := range numsMap {
-		numsArray = append(numsArray, v)
-	}
-	// 排序是为了使得线段树中的区间 left <= right，如果此处不排序，线段树中的区间有很多不合法。
-	sort.Ints(numsArray)
-	// 离散化，构建映射关系
-	for i, n := range numsArray {
-		indexMap[n] = i
-	}
-	numsArray = []int{}
-	// 离散化，此题如果不离散化，MaxInt32 的数据会使得数字越界。
-	for i := 0; i < len(indexMap); i++ {
-		numsArray = append(numsArray, i)
-	}
-	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
-	st.Init(numsArray, func(i, j int) int {
-		return 0
-	})
-	for _, num := range nums {
-		res += st.Query(indexMap[num*2+1], len(indexMap)-1)
-		st.UpdateCount(indexMap[num])
-	}
-	return res
-}
-
-// 解法二 mergesort
-func reversePairs1(nums []int) int {
-	buf := make([]int, len(nums))
-	return mergesortCount(nums, buf)
-}
-
-func mergesortCount(nums, buf []int) int {
-	if len(nums) <= 1 {
-		return 0
-	}
-	mid := (len(nums) - 1) / 2
-	cnt := mergesortCount(nums[:mid+1], buf)
-	cnt += mergesortCount(nums[mid+1:], buf)
-	for i, j := 0, mid+1; i < mid+1; i++ { // Note!!! j is increasing.
-		for ; j < len(nums) && nums[i] <= 2*nums[j]; j++ {
-		}
-		cnt += len(nums) - j
-	}
-	copy(buf, nums)
-	for i, j, k := 0, mid+1, 0; k < len(nums); {
-		if j >= len(nums) || i < mid+1 && buf[i] > buf[j] {
-			nums[k] = buf[i]
-			i++
-		} else {
-			nums[k] = buf[j]
-			j++
-		}
-		k++
-	}
-	return cnt
-}
diff --git a/Algorithms/0493. Reverse Pairs/README.md b/Algorithms/0493. Reverse Pairs/README.md
deleted file mode 100755
index 82eb9119e..000000000
--- a/Algorithms/0493. Reverse Pairs/README.md	
+++ /dev/null
@@ -1,42 +0,0 @@
-# [493. Reverse Pairs](https://leetcode.com/problems/reverse-pairs/)
-
-
-## 题目:
-
-Given an array `nums`, we call `(i, j)` an **important reverse pair** if `i < j` and `nums[i] > 2*nums[j]`.
-
-You need to return the number of important reverse pairs in the given array.
-
-**Example1:**
-
-    Input: [1,3,2,3,1]
-    Output: 2
-
-**Example2:**
-
-    Input: [2,4,3,5,1]
-    Output: 3
-
-**Note:**
-
-1. The length of the given array will not exceed `50,000`.
-2. All the numbers in the input array are in the range of 32-bit integer.
-
-
-## 题目大意
-
-给定一个数组 nums ，如果 i < j 且 nums[i] > 2\*nums[j] 我们就将 (i, j) 称作一个重要翻转对。你需要返回给定数组中的重要翻转对的数量。
-
-注意:
-
-- 给定数组的长度不会超过 50000。
-- 输入数组中的所有数字都在 32 位整数的表示范围内。
-
-
-## 解题思路
-
-
-- 给出一个数组，要求找出满足条件的所有的“重要的反转对” (i,j)。重要的反转对的定义是：`i<j`，并且 `nums[i] > 2*nums[j]`。
-- 这一题是 327 题的变种题。首先将数组中所有的元素以及各自的 `2*nums[i] + 1` 都放在字典中去重。去重以后再做离散化处理。这一题的测试用例会卡离散化，如果不离散化，Math.MaxInt32 会导致数字溢出，见测试用例中 2147483647, -2147483647 这组测试用例。离散后，映射关系 保存在字典中。从左往右遍历数组，先 query ，再 update ，这个顺序和第 327 题是反的。先 query 查找 `[2*nums[i] + 1, len(indexMap)-1]` 这个区间内满足条件的值，这个区间内的值都是 `> 2*nums[j]` 的。这一题移动的是 `j`，`j` 不断的变化，往线段树中不断插入的是 `i`。每轮循环先 query 一次前一轮循环中累积插入线段树中的 `i`，这些累积在线段树中的代表的是所有在 `j` 前面的 `i`。query 查询的是本轮 `[2*nums[j] + 1, len(indexMap)-1]`，如果能找到，即找到了这样一个 `j`，能满足 `nums[i] > 2*nums[j`， 把整个数组都扫完，累加的 query 出来的 count 计数就是最终答案。
-- 类似的题目：第 327 题，第 315 题。
-- 这一题用线段树并不是最优解，用线段树解这一题是为了训练线段树这个数据结构。最优解是解法二中的 mergesort。
diff --git a/Algorithms/0498. Diagonal Traverse/498. Diagonal Traverse_test.go b/Algorithms/0498. Diagonal Traverse/498. Diagonal Traverse_test.go
deleted file mode 100644
index b78ae3ffe..000000000
--- a/Algorithms/0498. Diagonal Traverse/498. Diagonal Traverse_test.go	
+++ /dev/null
@@ -1,67 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question498 struct {
-	para498
-	ans498
-}
-
-// para 是参数
-// one 代表第一个参数
-type para498 struct {
-	one [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans498 struct {
-	one []int
-}
-
-func Test_Problem498(t *testing.T) {
-
-	qs := []question498{
-
-		question498{
-			para498{[][]int{[]int{3}, []int{2}, []int{9}}},
-			ans498{[]int{3, 2, 9}},
-		},
-
-		question498{
-			para498{[][]int{[]int{6, 9, 7}}},
-			ans498{[]int{6, 9, 7}},
-		},
-
-		question498{
-			para498{[][]int{[]int{3}, []int{2}}},
-			ans498{[]int{3, 2}},
-		},
-
-		question498{
-			para498{[][]int{[]int{1, 2, 3}, []int{4, 5, 6}, []int{7, 8, 9}}},
-			ans498{[]int{1, 2, 4, 7, 5, 3, 6, 8, 9}},
-		},
-
-		question498{
-			para498{[][]int{[]int{0}}},
-			ans498{[]int{0}},
-		},
-
-		question498{
-			para498{[][]int{[]int{}}},
-			ans498{[]int{}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 498------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans498, q.para498
-		fmt.Printf("【input】:%v       【output】:%v\n", p, findDiagonalOrder(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0498. Diagonal Traverse/README.md b/Algorithms/0498. Diagonal Traverse/README.md
deleted file mode 100755
index bc12d88b4..000000000
--- a/Algorithms/0498. Diagonal Traverse/README.md	
+++ /dev/null
@@ -1,39 +0,0 @@
-# [498. Diagonal Traverse](https://leetcode.com/problems/diagonal-traverse/)
-
-
-## 题目:
-
-Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image.
-
-**Example:**
-
-    Input:
-    [
-     [ 1, 2, 3 ],
-     [ 4, 5, 6 ],
-     [ 7, 8, 9 ]
-    ]
-    
-    Output:  [1,2,4,7,5,3,6,8,9]
-    
-    Explanation:
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/diagonal_traverse.png)
-
-**Note:**
-
-The total number of elements of the given matrix will not exceed 10,000.
-
-
-## 题目大意
-
-给定一个含有 M x N 个元素的矩阵（M 行，N 列），请以对角线遍历的顺序返回这个矩阵中的所有元素，对角线遍历如下图所示。
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/diagonal_traverse.png)
-
-说明: 给定矩阵中的元素总数不会超过 100000 。
-
-## 解题思路
-
-- 给出一个二维数组，要求按照如图的方式遍历整个数组。
-- 这一题用模拟的方式就可以解出来。需要注意的是边界条件：比如二维数组为空，二维数组退化为一行或者一列，退化为一个元素。具体例子见测试用例。
diff --git a/Algorithms/0500. Keyboard Row/README.md b/Algorithms/0500. Keyboard Row/README.md
deleted file mode 100755
index 74599011f..000000000
--- a/Algorithms/0500. Keyboard Row/README.md	
+++ /dev/null
@@ -1,27 +0,0 @@
-# [500. Keyboard Row](https://leetcode.com/problems/keyboard-row/)
-
-
-## 题目:
-
-Given a List of words, return the words that can be typed using letters of **alphabet** on only one row's of American keyboard like the image below.
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/keyboard.png)
-
-**Example:**
-
-    Input: ["Hello", "Alaska", "Dad", "Peace"]
-    Output: ["Alaska", "Dad"]
-
-**Note:**
-
-1. You may use one character in the keyboard more than once.
-2. You may assume the input string will only contain letters of alphabet.
-
-
-## 题目大意
-
-给定一个单词列表，只返回可以使用在键盘同一行的字母打印出来的单词。键盘如上图所示。
-
-## 解题思路
-
-- 给出一个字符串数组，要求依次判断数组中的每个字符串是否都位于键盘上的同一个行，如果是就输出。这也是一道水题。
diff --git a/Algorithms/0508. Most Frequent Subtree Sum/README.md b/Algorithms/0508. Most Frequent Subtree Sum/README.md
deleted file mode 100755
index b142324f0..000000000
--- a/Algorithms/0508. Most Frequent Subtree Sum/README.md	
+++ /dev/null
@@ -1,35 +0,0 @@
-# [508. Most Frequent Subtree Sum](https://leetcode.com/problems/most-frequent-subtree-sum/)
-
-
-## 题目:
-
-Given the root of a tree, you are asked to find the most frequent subtree sum. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.
-
-**Examples 1**Input:
-
-      5
-     /  \
-    2   -3
-
-return [2, -3, 4], since all the values happen only once, return all of them in any order.
-
-**Examples 2**Input:
-
-      5
-     /  \
-    2   -5
-
-return [2], since 2 happens twice, however -5 only occur once.
-
-**Note:** You may assume the sum of values in any subtree is in the range of 32-bit signed integer.
-
-
-## 题目大意
-
-给出二叉树的根，找出出现次数最多的子树元素和。一个结点的子树元素和定义为以该结点为根的二叉树上所有结点的元素之和（包括结点本身）。然后求出出现次数最多的子树元素和。如果有多个元素出现的次数相同，返回所有出现次数最多的元素（不限顺序）。提示： 假设任意子树元素和均可以用 32 位有符号整数表示。
-
-## 解题思路
-
-
-- 给出一个树，要求求出每个节点以自己为根节点的子树的所有节点值的和，最后按照这些和出现的频次，输出频次最多的和，如果频次出现次数最多的对应多个和，则全部输出。
-- 递归找出每个节点的累加和，用 map 记录频次，最后把频次最多的输出即可。
diff --git a/Algorithms/0509. Fibonacci Number/README.md b/Algorithms/0509. Fibonacci Number/README.md
deleted file mode 100755
index 4f37321a1..000000000
--- a/Algorithms/0509. Fibonacci Number/README.md	
+++ /dev/null
@@ -1,53 +0,0 @@
-# [509. Fibonacci Number](https://leetcode.com/problems/fibonacci-number/)
-
-
-## 题目:
-
-The **Fibonacci numbers**, commonly denoted `F(n)` form a sequence, called the **Fibonacci sequence**, such that each number is the sum of the two preceding ones, starting from `0` and `1`. That is,
-
-    F(0) = 0,   F(1) = 1
-    F(N) = F(N - 1) + F(N - 2), for N > 1.
-
-Given `N`, calculate `F(N)`.
-
-**Example 1:**
-
-    Input: 2
-    Output: 1
-    Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
-
-**Example 2:**
-
-    Input: 3
-    Output: 2
-    Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
-
-**Example 3:**
-
-    Input: 4
-    Output: 3
-    Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
-
-**Note:**
-
-0 ≤ `N` ≤ 30.
-
-
-## 题目大意
-
-斐波那契数，通常用 F(n) 表示，形成的序列称为斐波那契数列。该数列由 0 和 1 开始，后面的每一项数字都是前面两项数字的和。也就是：
-
-```
-F(0) = 0,   F(1) = 1
-F(N) = F(N - 1) + F(N - 2), 其中 N > 1.
-```
-
-给定 N，计算 F(N)。
-
-提示：0 ≤ N ≤ 30
-
-## 解题思路
-
-
-- 求斐波那契数列
-- 这一题解法很多，大的分类是四种，递归，记忆化搜索(dp)，矩阵快速幂，通项公式。其中记忆化搜索可以写 3 种方法，自底向上的，自顶向下的，优化空间复杂度版的。通项公式方法实质是求 a^b 这个还可以用快速幂优化时间复杂度到 O(log n) 。
diff --git a/Algorithms/0515. Find Largest Value in Each Tree Row/515. Find Largest Value in Each Tree Row.go b/Algorithms/0515. Find Largest Value in Each Tree Row/515. Find Largest Value in Each Tree Row.go
deleted file mode 100644
index 812adbe82..000000000
--- a/Algorithms/0515. Find Largest Value in Each Tree Row/515. Find Largest Value in Each Tree Row.go	
+++ /dev/null
@@ -1,54 +0,0 @@
-package leetcode
-
-import (
-	"math"
-	"sort"
-)
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-
-// 解法一 层序遍历二叉树，再将每层排序取出最大值
-func largestValues(root *TreeNode) []int {
-	tmp := levelOrder(root)
-	res := []int{}
-	for i := 0; i < len(tmp); i++ {
-		sort.Ints(tmp[i])
-		res = append(res, tmp[i][len(tmp[i])-1])
-	}
-	return res
-}
-
-// 解法二 层序遍历二叉树，遍历过程中不断更新最大值
-func largestValues1(root *TreeNode) []int {
-	if root == nil {
-		return []int{}
-	}
-	q := []*TreeNode{root}
-	var res []int
-	for len(q) > 0 {
-		qlen := len(q)
-		max := math.MinInt32
-		for i := 0; i < qlen; i++ {
-			node := q[0]
-			q = q[1:]
-			if node.Val > max {
-				max = node.Val
-			}
-			if node.Left != nil {
-				q = append(q, node.Left)
-			}
-			if node.Right != nil {
-				q = append(q, node.Right)
-			}
-		}
-		res = append(res, max)
-	}
-	return res
-}
diff --git a/Algorithms/0541. Reverse String II/541. Reverse String II.go b/Algorithms/0541. Reverse String II/541. Reverse String II.go
deleted file mode 100644
index fe300150f..000000000
--- a/Algorithms/0541. Reverse String II/541. Reverse String II.go	
+++ /dev/null
@@ -1,17 +0,0 @@
-package leetcode
-
-func reverseStr(s string, k int) string {
-	if k > len(s) {
-		k = len(s)
-	}
-	for i := 0; i < len(s); i = i + 2*k {
-		if len(s)-i >= k {
-			ss := revers(s[i : i+k])
-			s = s[:i] + ss + s[i+k:]
-		} else {
-			ss := revers(s[i:])
-			s = s[:i] + ss
-		}
-	}
-	return s
-}
diff --git a/Algorithms/0541. Reverse String II/README.md b/Algorithms/0541. Reverse String II/README.md
deleted file mode 100755
index f368ba5f8..000000000
--- a/Algorithms/0541. Reverse String II/README.md	
+++ /dev/null
@@ -1,31 +0,0 @@
-# [541. Reverse String II](https://leetcode.com/problems/reverse-string-ii/)
-
-
-## 题目:
-
-Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.
-
-**Example:**
-
-    Input: s = "abcdefg", k = 2
-    Output: "bacdfeg"
-
-**Restrictions:**
-
-1. The string consists of lower English letters only.
-2. Length of the given string and k will in the range [1, 10000]
-
-## 题目大意
-
-给定一个字符串和一个整数 k，你需要对从字符串开头算起的每个 2k 个字符的前k个字符进行反转。如果剩余少于 k 个字符，则将剩余的所有全部反转。如果有小于 2k 但大于或等于 k 个字符，则反转前 k 个字符，并将剩余的字符保持原样。
-
-要求:
-
-- 该字符串只包含小写的英文字母。
-- 给定字符串的长度和 k 在[1, 10000]范围内。
-
-
-## 解题思路
-
-- 要求按照一定规则反转字符串：每 `2 * K` 长度的字符串，反转前 `K` 个字符，后 `K` 个字符串保持不变；对于末尾不够 `2 * K` 的字符串，如果长度大于 `K`，那么反转前 `K` 个字符串，剩下的保持不变。如果长度小于 `K`，则把小于 `K` 的这部分字符串全部反转。
-- 这一题是简单题，按照题意反转字符串即可。
diff --git a/Algorithms/0542. 01 Matrix/542. 01 Matrix_test.go b/Algorithms/0542. 01 Matrix/542. 01 Matrix_test.go
deleted file mode 100644
index efebe43bf..000000000
--- a/Algorithms/0542. 01 Matrix/542. 01 Matrix_test.go	
+++ /dev/null
@@ -1,52 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question542 struct {
-	para542
-	ans542
-}
-
-// para 是参数
-// one 代表第一个参数
-type para542 struct {
-	one [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans542 struct {
-	one [][]int
-}
-
-func Test_Problem542(t *testing.T) {
-
-	qs := []question542{
-
-		question542{
-			para542{[][]int{[]int{0, 0, 0}, []int{0, 1, 0}, []int{0, 0, 0}}},
-			ans542{[][]int{[]int{0, 0, 0}, []int{0, 1, 0}, []int{0, 0, 0}}},
-		},
-
-		question542{
-			para542{[][]int{[]int{0, 0, 0}, []int{0, 1, 0}, []int{1, 1, 1}}},
-			ans542{[][]int{[]int{0, 0, 0}, []int{0, 1, 0}, []int{1, 2, 1}}},
-		},
-
-		question542{
-			para542{[][]int{[]int{1, 1, 1, 1, 1, 1}, []int{1, 1, 1, 1, 1, 1}, []int{1, 1, 0, 0, 1, 1}, []int{1, 1, 0, 0, 1, 1}, []int{1, 1, 1, 1, 1, 1}, []int{1, 1, 1, 1, 1, 1}}},
-			ans542{[][]int{[]int{4, 3, 2, 2, 3, 4}, []int{3, 2, 1, 1, 2, 3}, []int{2, 1, 0, 0, 1, 2}, []int{2, 1, 0, 0, 1, 2}, []int{3, 2, 1, 1, 2, 3}, []int{4, 3, 2, 2, 3, 4}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 542------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans542, q.para542
-		fmt.Printf("【input】:%v       【output】:%v\n", p, updateMatrixDP(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0547. Friend Circles/547. Friend Circles_test.go b/Algorithms/0547. Friend Circles/547. Friend Circles_test.go
deleted file mode 100644
index d481f1a03..000000000
--- a/Algorithms/0547. Friend Circles/547. Friend Circles_test.go	
+++ /dev/null
@@ -1,52 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question547 struct {
-	para547
-	ans547
-}
-
-// para 是参数
-// one 代表第一个参数
-type para547 struct {
-	one [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans547 struct {
-	one int
-}
-
-func Test_Problem547(t *testing.T) {
-
-	qs := []question547{
-
-		question547{
-			para547{[][]int{[]int{0, 0, 0}, []int{0, 1, 0}, []int{0, 0, 0}}},
-			ans547{3},
-		},
-
-		question547{
-			para547{[][]int{[]int{1, 1, 0}, []int{1, 1, 0}, []int{0, 0, 1}}},
-			ans547{2},
-		},
-
-		question547{
-			para547{[][]int{[]int{1, 1, 0}, []int{1, 1, 1}, []int{0, 1, 1}}},
-			ans547{1},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 547------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans547, q.para547
-		fmt.Printf("【input】:%v       【output】:%v\n", p, findCircleNum(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0547. Friend Circles/README.md b/Algorithms/0547. Friend Circles/README.md
deleted file mode 100755
index 92b187a91..000000000
--- a/Algorithms/0547. Friend Circles/README.md	
+++ /dev/null
@@ -1,54 +0,0 @@
-# [547. Friend Circles](https://leetcode.com/problems/friend-circles/)
-
-## 题目:
-
-There are **N** students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a **direct** friend of B, and B is a **direct**friend of C, then A is an **indirect** friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.
-
-Given a **N*N** matrix **M** representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are **direct** friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.
-
-**Example 1:**
-
-    Input: 
-    [[1,1,0],
-     [1,1,0],
-     [0,0,1]]
-    Output: 2
-    Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. 
-    The 2nd student himself is in a friend circle. So return 2.
-
-**Example 2:**
-
-    Input: 
-    [[1,1,0],
-     [1,1,1],
-     [0,1,1]]
-    Output: 1
-    Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, 
-    so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.
-
-**Note:**
-
-1. N is in range [1,200].
-2. M[i][i] = 1 for all students.
-3. If M[i][j] = 1, then M[j][i] = 1.
-
-
-## 题目大意
-
-班上有 N 名学生。其中有些人是朋友，有些则不是。他们的友谊具有是传递性。如果已知 A 是 B 的朋友，B 是 C 的朋友，那么我们可以认为 A 也是 C 的朋友。所谓的朋友圈，是指所有朋友的集合。
-
-给定一个 N * N 的矩阵 M，表示班级中学生之间的朋友关系。如果 M[i][j] = 1，表示已知第 i 个和 j 个学生互为朋友关系，否则为不知道。你必须输出所有学生中的已知的朋友圈总数。
-
-
-注意：
-
-- N 在[1,200]的范围内。
-- 对于所有学生，有M[i][i] = 1。
-- 如果有 M[i][j] = 1，则有 M[j][i] = 1。
-
-
-## 解题思路
-
-
-- 给出一个二维矩阵，矩阵中的行列表示的是两个人之间是否是朋友关系，如果是 1，代表两个人是朋友关系。由于自己和自肯定朋友关系，所以对角线上都是 1，并且矩阵也是关于从左往右下的这条对角线对称。
-- 这题有 2 种解法，第一种解法是并查集，依次扫描矩阵，如果两个人认识，并且 root 并不相等就执行 union 操作。扫完所有矩阵，最后数一下还有几个不同的 root 就是最终答案。第二种解法是 DFS 或者 BFS。利用 FloodFill 的想法去染色，每次染色一次，计数器加一。最终扫完整个矩阵，计数器的结果就是最终结果。
diff --git a/Algorithms/0557. Reverse Words in a String III/README.md b/Algorithms/0557. Reverse Words in a String III/README.md
deleted file mode 100755
index 5ce638e73..000000000
--- a/Algorithms/0557. Reverse Words in a String III/README.md	
+++ /dev/null
@@ -1,25 +0,0 @@
-# [557. Reverse Words in a String III](https://leetcode.com/problems/reverse-words-in-a-string-iii/)
-
-
-## 题目:
-
-Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.
-
-**Example 1:**
-
-    Input: "Let's take LeetCode contest"
-    Output: "s'teL ekat edoCteeL tsetnoc"
-
-**Note:** In the string, each word is separated by single space and there will not be any extra space in the string.
-
-
-## 题目大意
-
-给定一个字符串，你需要反转字符串中每个单词的字符顺序，同时仍保留空格和单词的初始顺序。注意：在字符串中，每个单词由单个空格分隔，并且字符串中不会有任何额外的空格。
-
-
-## 解题思路
-
-
-- 反转字符串，要求按照空格隔开的小字符串为单位反转。
-- 这是一道简单题。按照题意反转每个空格隔开的单词即可。
diff --git a/Algorithms/0566. Reshape the Matrix/566. Reshape the Matrix_test.go b/Algorithms/0566. Reshape the Matrix/566. Reshape the Matrix_test.go
deleted file mode 100644
index 918ba9ec5..000000000
--- a/Algorithms/0566. Reshape the Matrix/566. Reshape the Matrix_test.go	
+++ /dev/null
@@ -1,49 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question566 struct {
-	para566
-	ans566
-}
-
-// para 是参数
-// one 代表第一个参数
-type para566 struct {
-	nums [][]int
-	r    int
-	c    int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans566 struct {
-	one [][]int
-}
-
-func Test_Problem566(t *testing.T) {
-
-	qs := []question566{
-
-		question566{
-			para566{[][]int{[]int{1, 2}, []int{3, 4}}, 1, 4},
-			ans566{[][]int{[]int{1, 2, 3, 4}}},
-		},
-
-		question566{
-			para566{[][]int{[]int{1, 2}, []int{3, 4}}, 2, 4},
-			ans566{[][]int{[]int{1, 2}, []int{3, 4}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 566------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans566, q.para566
-		fmt.Printf("【input】:%v       【output】:%v\n", p, matrixReshape(p.nums, p.r, p.c))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0567. Permutation in String/README.md b/Algorithms/0567. Permutation in String/README.md
deleted file mode 100644
index bcbccf293..000000000
--- a/Algorithms/0567. Permutation in String/README.md	
+++ /dev/null
@@ -1,44 +0,0 @@
-# [567. Permutation in String](https://leetcode.com/problems/permutation-in-string/)
-
-## 题目
-
-Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.
-
-
-Example 1:
-
-```c
-Input:s1 = "ab" s2 = "eidbaooo"
-Output:True
-Explanation: s2 contains one permutation of s1 ("ba").
-```
-
-Example 2:
-
-```c
-Input:s1= "ab" s2 = "eidboaoo"
-Output: False
-```
-
-Note:
-
-1. The input strings only contain lower case letters.
-2. The length of both given strings is in range [1, 10,000].
-
-## 题目大意
-
-
-在一个字符串重寻找子串出现的位置。子串可以是 Anagrams 形式存在的。Anagrams 是一个字符串任意字符的全排列组合。
-
-## 解题思路
-
-这一题和第 438 题，第 3 题，第 76 题，第 567 题类似，用的思想都是"滑动窗口"。
-
-
-这道题只需要判断是否存在，而不需要输出子串所在的下标起始位置。所以这道题是第 438 题的缩水版。具体解题思路见第 438 题。
-
-
-
-
-
-
diff --git a/Algorithms/0572. Subtree of Another Tree/572. Subtree of Another Tree.go b/Algorithms/0572. Subtree of Another Tree/572. Subtree of Another Tree.go
deleted file mode 100644
index de74977ba..000000000
--- a/Algorithms/0572. Subtree of Another Tree/572. Subtree of Another Tree.go	
+++ /dev/null
@@ -1,22 +0,0 @@
-package leetcode
-
-/**
- * Definition for a binary tree node.
- * type TreeNode struct {
- *     Val int
- *     Left *TreeNode
- *     Right *TreeNode
- * }
- */
-func isSubtree(s *TreeNode, t *TreeNode) bool {
-	if isSameTree(s, t) {
-		return true
-	}
-	if s == nil {
-		return false
-	}
-	if isSubtree(s.Left, t) || isSubtree(s.Right, t) {
-		return true
-	}
-	return false
-}
diff --git a/Algorithms/0572. Subtree of Another Tree/572. Subtree of Another Tree_test.go b/Algorithms/0572. Subtree of Another Tree/572. Subtree of Another Tree_test.go
deleted file mode 100644
index b65303702..000000000
--- a/Algorithms/0572. Subtree of Another Tree/572. Subtree of Another Tree_test.go	
+++ /dev/null
@@ -1,61 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question572 struct {
-	para572
-	ans572
-}
-
-// para 是参数
-// one 代表第一个参数
-type para572 struct {
-	s []int
-	t []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans572 struct {
-	one bool
-}
-
-func Test_Problem572(t *testing.T) {
-
-	qs := []question572{
-
-		question572{
-			para572{[]int{}, []int{}},
-			ans572{false},
-		},
-
-		question572{
-			para572{[]int{3, 4, 5, 1, 2}, []int{4, 1, 2}},
-			ans572{true},
-		},
-
-		question572{
-			para572{[]int{1, 1}, []int{1}},
-			ans572{true},
-		},
-
-		question572{
-			para572{[]int{1, NULL, 1, NULL, 1, NULL, 1, NULL, 1, NULL, 1, NULL, 1, NULL, 1, NULL, 1, NULL, 1, NULL, 1, 2}, []int{1, NULL, 1, NULL, 1, NULL, 1, NULL, 1, NULL, 1, 2}},
-			ans572{true},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 572------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans572, q.para572
-		fmt.Printf("【input】:%v      ", p)
-		roots := Ints2TreeNode(p.s)
-		roott := Ints2TreeNode(p.t)
-		fmt.Printf("【output】:%v      \n", isSubtree(roots, roott))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0572. Subtree of Another Tree/README.md b/Algorithms/0572. Subtree of Another Tree/README.md
deleted file mode 100755
index 479869a38..000000000
--- a/Algorithms/0572. Subtree of Another Tree/README.md	
+++ /dev/null
@@ -1,53 +0,0 @@
-# [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)
-
-
-## 题目:
-
-Given two non-empty binary trees **s** and **t**, check whether tree **t** has exactly the same structure and node values with a subtree of **s**. A subtree of **s** is a tree consists of a node in **s** and all of this node's descendants. The tree **s** could also be considered as a subtree of itself.
-
-**Example 1:** Given tree s:
-
-		 3
-        / \
-       4   5
-      / \
-     1   2
-
-Given tree t:
-
-       4 
-      / \
-     1   2
-
-Return **true**, because t has the same structure and node values with a subtree of s.
-
-**Example 2:**Given tree s:
-
-         3
-        / \
-       4   5
-      / \
-     1   2
-        /
-       0
-
-Given tree t:
-
-       4
-      / \
-     1   2
-
-Return **false**.
-
-
-## 题目大意
-
-给定两个非空二叉树 s 和 t，检验 s 中是否包含和 t 具有相同结构和节点值的子树。s 的一个子树包括 s 的一个节点和这个节点的所有子孙。s 也可以看做它自身的一棵子树。
-
-
-## 解题思路
-
-
-- 给出 2 棵树 `s` 和 `t`，要求判断 `t` 是否是 `s` 的子树🌲。
-- 这一题比较简单，针对 3 种情况依次递归判断，第一种情况 `s` 和 `t` 是完全一样的两棵树，第二种情况 `t` 是 `s` 左子树中的子树，第三种情况 `t` 是 `s` 右子树中的子树。第一种情况判断两棵数是否完全一致是第 100 题的原题。
-
diff --git a/Algorithms/0575. Distribute Candies/README.md b/Algorithms/0575. Distribute Candies/README.md
deleted file mode 100755
index 4fe69a71b..000000000
--- a/Algorithms/0575. Distribute Candies/README.md	
+++ /dev/null
@@ -1,38 +0,0 @@
-# [575. Distribute Candies](https://leetcode.com/problems/distribute-candies/)
-
-
-## 题目:
-
-Given an integer array with **even** length, where different numbers in this array represent different **kinds** of candies. Each number means one candy of the corresponding kind. You need to distribute these candies **equally** in number to brother and sister. Return the maximum number of **kinds** of candies the sister could gain.
-
-**Example 1:**
-
-    Input: candies = [1,1,2,2,3,3]
-    Output: 3
-    Explanation:
-    There are three different kinds of candies (1, 2 and 3), and two candies for each kind.
-    Optimal distribution: The sister has candies [1,2,3] and the brother has candies [1,2,3], too. 
-    The sister has three different kinds of candies.
-
-**Example 2:**
-
-    Input: candies = [1,1,2,3]
-    Output: 2
-    Explanation: For example, the sister has candies [2,3] and the brother has candies [1,1]. 
-    The sister has two different kinds of candies, the brother has only one kind of candies.
-
-**Note:**
-
-1. The length of the given array is in range [2, 10,000], and will be even.
-2. The number in given array is in range [-100,000, 100,000].
-
-
-## 题目大意
-
-给定一个偶数长度的数组，其中不同的数字代表着不同种类的糖果，每一个数字代表一个糖果。你需要把这些糖果平均分给一个弟弟和一个妹妹。返回妹妹可以获得的最大糖果的种类数。
-
-
-## 解题思路
-
-
-- 给出一个糖果数组，里面每个元素代表糖果的种类，相同数字代表相同种类。把这些糖果分给兄弟姐妹，问姐妹最多可以分到多少种糖果。这一题比较简单，用 map 统计每个糖果的出现频次，如果总数比 `n/2` 小，那么就返回 `len(map)`，否则返回 `n/2` (即一半都分给姐妹)。
diff --git a/Algorithms/0594. Longest Harmonious Subsequence/README.md b/Algorithms/0594. Longest Harmonious Subsequence/README.md
deleted file mode 100755
index ce5484c88..000000000
--- a/Algorithms/0594. Longest Harmonious Subsequence/README.md	
+++ /dev/null
@@ -1,25 +0,0 @@
-# [594. Longest Harmonious Subsequence](https://leetcode.com/problems/longest-harmonious-subsequence/)
-
-
-## 题目:
-
-We define a harmounious array as an array where the difference between its maximum value and its minimum value is **exactly** 1.
-
-Now, given an integer array, you need to find the length of its longest harmonious subsequence among all its possible [subsequences](https://en.wikipedia.org/wiki/Subsequence).
-
-**Example 1:**
-
-    Input: [1,3,2,2,5,2,3,7]
-    Output: 5
-    Explanation: The longest harmonious subsequence is [3,2,2,2,3].
-
-**Note:** The length of the input array will not exceed 20,000.
-
-
-## 题目大意
-
-和谐数组是指一个数组里元素的最大值和最小值之间的差别正好是1。现在，给定一个整数数组，你需要在所有可能的子序列中找到最长的和谐子序列的长度。说明: 输入的数组长度最大不超过20,000.
-
-## 解题思路
-
-- 在给出的数组里面找到这样一个子数组：要求子数组中的最大值和最小值相差 1 。这一题是简单题。先统计每个数字出现的频次，然后在 map 找相差 1 的 2 个数组的频次和，动态的维护两个数的频次和就是最后要求的子数组的最大长度。
diff --git a/Algorithms/0599. Minimum Index Sum of Two Lists/README.md b/Algorithms/0599. Minimum Index Sum of Two Lists/README.md
deleted file mode 100755
index 9c3630e36..000000000
--- a/Algorithms/0599. Minimum Index Sum of Two Lists/README.md	
+++ /dev/null
@@ -1,50 +0,0 @@
-# [599. Minimum Index Sum of Two Lists](https://leetcode.com/problems/minimum-index-sum-of-two-lists/)
-
-## 题目:
-
-Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings.
-
-You need to help them find out their **common interest** with the **least list index sum**. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.
-
-**Example 1:**
-
-    Input:
-    ["Shogun", "Tapioca Express", "Burger King", "KFC"]
-    ["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
-    Output: ["Shogun"]
-    Explanation: The only restaurant they both like is "Shogun".
-
-**Example 2:**
-
-    Input:
-    ["Shogun", "Tapioca Express", "Burger King", "KFC"]
-    ["KFC", "Shogun", "Burger King"]
-    Output: ["Shogun"]
-    Explanation: The restaurant they both like and have the least index sum is "Shogun" with index sum 1 (0+1).
-
-**Note:**
-
-1. The length of both lists will be in the range of [1, 1000].
-2. The length of strings in both lists will be in the range of [1, 30].
-3. The index is starting from 0 to the list length minus 1.
-4. No duplicates in both lists.
-
-
-## 题目大意
-
-假设 Andy 和 Doris 想在晚餐时选择一家餐厅，并且他们都有一个表示最喜爱餐厅的列表，每个餐厅的名字用字符串表示。你需要帮助他们用最少的索引和找出他们共同喜爱的餐厅。 如果答案不止一个，则输出所有答案并且不考虑顺序。 你可以假设总是存在一个答案。
-
-
-提示:  
-
-- 两个列表的长度范围都在 [1, 1000] 内。
-- 两个列表中的字符串的长度将在 [1，30] 的范围内。
-- 下标从 0 开始，到列表的长度减 1。
-- 两个列表都没有重复的元素。
-
-
-
-## 解题思路
-
-
-- 在 Andy 和 Doris 两人分别有各自的餐厅喜欢列表，要求找出两人公共喜欢的一家餐厅，如果共同喜欢的次数相同，都输出。这一题是简单题，用 map 统计频次，输出频次最多的餐厅。
diff --git a/Algorithms/0628. Maximum Product of Three Numbers/628. Maximum Product of Three Numbers.go b/Algorithms/0628. Maximum Product of Three Numbers/628. Maximum Product of Three Numbers.go
deleted file mode 100644
index 41b7a78ac..000000000
--- a/Algorithms/0628. Maximum Product of Three Numbers/628. Maximum Product of Three Numbers.go	
+++ /dev/null
@@ -1,52 +0,0 @@
-package leetcode
-
-import (
-	"sort"
-)
-
-// 解法一 排序，时间复杂度 O(n log n)
-func maximumProduct(nums []int) int {
-	if len(nums) == 0 {
-		return 0
-	}
-	res := 1
-	if len(nums) <= 3 {
-		for i := 0; i < len(nums); i++ {
-			res = res * nums[i]
-		}
-		return res
-	}
-	sort.Ints(nums)
-	if nums[len(nums)-1] <= 0 {
-		return 0
-	}
-	return max(nums[0]*nums[1]*nums[len(nums)-1], nums[len(nums)-1]*nums[len(nums)-2]*nums[len(nums)-3])
-}
-
-// 解法二 模拟，时间复杂度 O(n)
-func maximumProduct1(nums []int) int {
-	n1, n2, n3 := -1<<63, -1<<63, -1<<63
-	n4, n5 := 0, 0
-	for _, v := range nums {
-		if v > n1 {
-			n3 = n2
-			n2 = n1
-			n1 = v
-		} else if v > n2 {
-			n3 = n2
-			n2 = v
-		} else if v > n3 {
-			n3 = v
-		}
-		if v < n4 {
-			n5 = n4
-			n4 = v
-		} else if v < n5 {
-			n5 = v
-		}
-	}
-	if n2*n3 > n4*n5 {
-		return n1 * n2 * n3
-	}
-	return n1 * n4 * n5
-}
diff --git a/Algorithms/0632. Smallest Range Covering Elements from K Lists/README.md b/Algorithms/0632. Smallest Range Covering Elements from K Lists/README.md
deleted file mode 100755
index 391259c0c..000000000
--- a/Algorithms/0632. Smallest Range Covering Elements from K Lists/README.md	
+++ /dev/null
@@ -1,46 +0,0 @@
-# [632. Smallest Range Covering Elements from K Lists](https://leetcode.com/problems/smallest-range-covering-elements-from-k-lists/)
-
-
-## 题目:
-
-You have `k` lists of sorted integers in ascending order. Find the **smallest** range that includes at least one number from each of the `k` lists.
-
-We define the range [a,b] is smaller than range [c,d] if `b-a < d-c` or `a < c` if `b-a == d-c`.
-
-**Example 1:**
-
-    Input: [[4,10,15,24,26], [0,9,12,20], [5,18,22,30]]
-    Output: [20,24]
-    Explanation: 
-    List 1: [4, 10, 15, 24,26], 24 is in range [20,24].
-    List 2: [0, 9, 12, 20], 20 is in range [20,24].
-    List 3: [5, 18, 22, 30], 22 is in range [20,24].
-
-**Note:**
-
-1. The given list may contain duplicates, so ascending order means >= here.
-2. 1 <= `k` <= 3500
-3. -10^5 <= `value of elements` <= 10^5.
-
-
-## 题目大意
-
-你有 k 个升序排列的整数数组。找到一个最小区间，使得 k 个列表中的每个列表至少有一个数包含在其中。
-
-我们定义如果 b-a < d-c 或者在 b-a == d-c 时 a < c，则区间 [a,b] 比 [c,d] 小。
-
-注意:
-
-- 给定的列表可能包含重复元素，所以在这里升序表示 >= 。
-- 1 <= k <= 3500
-- -105 <= 元素的值 <= 105
-- 对于使用Java的用户，请注意传入类型已修改为List<List<Integer>>。重置代码模板后可以看到这项改动。
-
-
-
-## 解题思路
-
-
-- 给出 K 个数组，要求在这 K 个数组中找到一个区间，至少能包含这 K 个数组中每个数组中的一个元素。
-- 这一题是第 76 题的变种版。第 76 题是用滑动窗口来解答的，它要求在母字符串 S 中找到最小的子串能包含 T 串的所有字母。这一题类似的，可以把母字符串看成 K 个数组合并起来的大数组，那么 T 串是由 K 个数组中每个数组中抽一个元素出来组成的。求的区间相同，都是能包含 T 的最小区间。另外一个区别在于，第 76 题里面都是字符串，这一题都是数字，在最终拼接成 T 串的时候需要保证 K 个数组中每个都有一个元素，所以理所当然的想到需要维护每个元素所在数组编号。经过上述的转换，可以把这道题转换成第 76 题的解法了。
-- 在具体解题过程中，用 map 来维护窗口内 K 个数组出现的频次。时间复杂度 O(n*log n)，空间复杂度是O(n)。
\ No newline at end of file
diff --git a/Algorithms/0633. Sum of Square Numbers/README.md b/Algorithms/0633. Sum of Square Numbers/README.md
deleted file mode 100755
index b581b1ab3..000000000
--- a/Algorithms/0633. Sum of Square Numbers/README.md	
+++ /dev/null
@@ -1,28 +0,0 @@
-# [633. Sum of Square Numbers](https://leetcode.com/problems/sum-of-square-numbers/)
-
-
-## 题目:
-
-Given a non-negative integer `c`, your task is to decide whether there're two integers `a` and `b` such that a^2 + b^2 = c.
-
-**Example 1:**
-
-    Input: 5
-    Output: True
-    Explanation: 1 * 1 + 2 * 2 = 5
-
-**Example 2:**
-
-    Input: 3
-    Output: False
-
-
-## 题目大意
-
-给定一个非负整数 c ，你要判断是否存在两个整数 a 和 b，使得 a^2 + b^2 = c。
-
-
-## 解题思路
-
-- 给出一个数，要求判断这个数能否由由 2 个完全平方数组成。能则输出 true，不能则输出 false。
-- 可以用二分搜索来解答这道题。判断题意，依次计算 `low * low + high * high`  和 c 是否相等。从 [1, sqrt(n)] 区间内进行二分，若能找到则返回 true，找不到就返回 false 。
diff --git a/Algorithms/0637. Average of Levels in Binary Tree/637. Average of Levels in Binary Tree_test.go b/Algorithms/0637. Average of Levels in Binary Tree/637. Average of Levels in Binary Tree_test.go
deleted file mode 100644
index 50b94f35e..000000000
--- a/Algorithms/0637. Average of Levels in Binary Tree/637. Average of Levels in Binary Tree_test.go	
+++ /dev/null
@@ -1,54 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question637 struct {
-	para637
-	ans637
-}
-
-// para 是参数
-// one 代表第一个参数
-type para637 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans637 struct {
-	one [][]int
-}
-
-func Test_Problem637(t *testing.T) {
-
-	qs := []question637{
-
-		question637{
-			para637{[]int{}},
-			ans637{[][]int{}},
-		},
-
-		question637{
-			para637{[]int{1}},
-			ans637{[][]int{[]int{1}}},
-		},
-
-		question637{
-			para637{[]int{3, 9, 20, NULL, NULL, 15, 7}},
-			ans637{[][]int{[]int{3}, []int{9, 20}, []int{15, 7}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 637------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans637, q.para637
-		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", averageOfLevels(root))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0645. Set Mismatch/README.md b/Algorithms/0645. Set Mismatch/README.md
deleted file mode 100755
index 09649d613..000000000
--- a/Algorithms/0645. Set Mismatch/README.md	
+++ /dev/null
@@ -1,35 +0,0 @@
-# [645. Set Mismatch](https://leetcode.com/problems/set-mismatch/)
-
-
-## 题目:
-
-The set `S` originally contains numbers from 1 to `n`. But unfortunately, due to the data error, one of the numbers in the set got duplicated to **another** number in the set, which results in repetition of one number and loss of another number.
-
-Given an array `nums` representing the data status of this set after the error. Your task is to firstly find the number occurs twice and then find the number that is missing. Return them in the form of an array.
-
-**Example 1:**
-
-    Input: nums = [1,2,2,4]
-    Output: [2,3]
-
-**Note:**
-
-1. The given array size will in the range [2, 10000].
-2. The given array's numbers won't have any order.
-
-
-## 题目大意
-
-
-集合 S 包含从1到 n 的整数。不幸的是，因为数据错误，导致集合里面某一个元素复制了成了集合里面的另外一个元素的值，导致集合丢失了一个整数并且有一个元素重复。给定一个数组 nums 代表了集合 S 发生错误后的结果。你的任务是首先寻找到重复出现的整数，再找到丢失的整数，将它们以数组的形式返回。
-
-注意:  
-
-- 给定数组的长度范围是 [2, 10000]。
-- 给定的数组是无序的。
-
-
-## 解题思路
-
-
-- 给出一个数组，数组里面装的是 1-n 的数字，由于错误导致有一个数字变成了另外一个数字，要求找出重复的一个数字和正确的数字。这一题是简单题，根据下标比对就可以找到哪个数字重复了，哪个数字缺少了。
diff --git a/Algorithms/0648. Replace Words/648. Replace Words.go b/Algorithms/0648. Replace Words/648. Replace Words.go
deleted file mode 100644
index 037452083..000000000
--- a/Algorithms/0648. Replace Words/648. Replace Words.go	
+++ /dev/null
@@ -1,64 +0,0 @@
-package leetcode
-
-import "strings"
-
-// 解法一 哈希表
-func replaceWords(dict []string, sentence string) string {
-	roots := make(map[byte][]string)
-	for _, root := range dict {
-		b := root[0]
-		roots[b] = append(roots[b], root)
-	}
-	words := strings.Split(sentence, " ")
-	for i, word := range words {
-		b := []byte(word)
-		for j := 1; j < len(b) && j <= 100; j++ {
-			if findWord(roots, b[0:j]) {
-				words[i] = string(b[0:j])
-				break
-			}
-		}
-	}
-	return strings.Join(words, " ")
-}
-
-func findWord(roots map[byte][]string, word []byte) bool {
-	if roots[word[0]] == nil {
-		return false
-	}
-	for _, root := range roots[word[0]] {
-		if root == string(word) {
-			return true
-		}
-	}
-	return false
-}
-
-//解法二 Trie
-func replaceWords1(dict []string, sentence string) string {
-	trie := Constructor208()
-	for _, v := range dict {
-		trie.Insert(v)
-	}
-	words := strings.Split(sentence, " ")
-	var result []string
-	word := ""
-	i := 0
-	for _, value := range words {
-		word = ""
-		for i = 1; i < len(value); i++ {
-			if trie.Search(value[:i]) {
-				word = value[:i]
-				break
-			}
-		}
-
-		if len(word) == 0 {
-			result = append(result, value)
-		} else {
-			result = append(result, word)
-		}
-
-	}
-	return strings.Join(result, " ")
-}
diff --git a/Algorithms/0662. Maximum Width of Binary Tree/662. Maximum Width of Binary Tree_test.go b/Algorithms/0662. Maximum Width of Binary Tree/662. Maximum Width of Binary Tree_test.go
deleted file mode 100644
index 8f4e3d617..000000000
--- a/Algorithms/0662. Maximum Width of Binary Tree/662. Maximum Width of Binary Tree_test.go	
+++ /dev/null
@@ -1,64 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question662 struct {
-	para662
-	ans662
-}
-
-// para 是参数
-// one 代表第一个参数
-type para662 struct {
-	one []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans662 struct {
-	one int
-}
-
-func Test_Problem662(t *testing.T) {
-
-	qs := []question662{
-
-		question662{
-			para662{[]int{1, 1, 1, 1, 1, 1, 1, NULL, NULL, NULL, 1, NULL, NULL, NULL, NULL, 2, 2, 2, 2, 2, 2, 2, NULL, 2, NULL, NULL, 2, NULL, 2}},
-			ans662{8},
-		},
-
-		question662{
-			para662{[]int{1, 1, 1, 1, NULL, NULL, 1, 1, NULL, NULL, 1}},
-			ans662{8},
-		},
-
-		question662{
-			para662{[]int{}},
-			ans662{0},
-		},
-
-		question662{
-			para662{[]int{1}},
-			ans662{1},
-		},
-
-		question662{
-			para662{[]int{1, 3, 2, 5, 3, NULL, 9}},
-			ans662{4},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 662------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans662, q.para662
-		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", widthOfBinaryTree(root))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0662. Maximum Width of Binary Tree/README.md b/Algorithms/0662. Maximum Width of Binary Tree/README.md
deleted file mode 100755
index a158b5ac3..000000000
--- a/Algorithms/0662. Maximum Width of Binary Tree/README.md	
+++ /dev/null
@@ -1,82 +0,0 @@
-# [662. Maximum Width of Binary Tree](https://leetcode.com/problems/maximum-width-of-binary-tree/)
-
-
-## 题目:
-
-Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a **full binary tree**, but some nodes are null.
-
-The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the `null` nodes between the end-nodes are also counted into the length calculation.
-
-**Example 1:**
-
-    Input: 
-    
-               1
-             /   \
-            3     2
-           / \     \  
-          5   3     9 
-    
-    Output: 4
-    Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).
-
-**Example 2:**
-
-    Input: 
-    
-              1
-             /  
-            3    
-           / \       
-          5   3     
-    
-    Output: 2
-    Explanation: The maximum width existing in the third level with the length 2 (5,3).
-
-**Example 3:**
-
-    Input: 
-    
-              1
-             / \
-            3   2 
-           /        
-          5      
-    
-    Output: 2
-    Explanation: The maximum width existing in the second level with the length 2 (3,2).
-
-**Example 4:**
-
-    Input: 
-    
-              1
-             / \
-            3   2
-           /     \  
-          5       9 
-         /         \
-        6           7
-    Output: 8
-    Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).
-
-**Note:** Answer will in the range of 32-bit signed integer.
-
-
-## 题目大意
-
-给定一个二叉树，编写一个函数来获取这个树的最大宽度。树的宽度是所有层中的最大宽度。这个二叉树与满二叉树（full binary tree）结构相同，但一些节点为空。
-
-每一层的宽度被定义为两个端点（该层最左和最右的非空节点，两端点间的null节点也计入长度）之间的长度。
-
-注意: 答案在32位有符号整数的表示范围内。
-
-
-
-## 解题思路
-
-
-- 给出一个二叉树，求这棵树最宽的部分。
-- 这一题可以用 BFS 也可以用 DFS，但是用 BFS 比较方便。按照层序遍历，依次算出每层最左边不为 `null` 的节点和最右边不为 `null` 的节点。这两个节点之间都是算宽度的。最终输出最大的宽度即可。此题的关键在于如何有效的找到每一层的左右边界。
-- 这一题可能有人会想着先补全满二叉树，然后每层分别找左右边界。这种方法提交以后会卡在 `104 / 108` 这组测试用例上，这组测试用例会使得最后某几层填充出现的满二叉树节点特别多，最终导致 `Memory Limit Exceeded` 了。
-- 由于此题要找每层的左右边界，实际上每个节点的 `Val` 值是我们不关心的，那么可以把这个值用来标号，标记成该节点在每层中的序号。父亲节点在上一层中的序号是 x，那么它的左孩子在下一层满二叉树中的序号是 `2*x`，它的右孩子在下一层满二叉树中的序号是 `2*x + 1`。将所有节点都标上号，用 BFS 层序遍历每一层，每一层都找到左右边界，相减拿到宽度，动态维护最大宽度，就是本题的最终答案。
\ No newline at end of file
diff --git a/Algorithms/0684. Redundant Connection/README.md b/Algorithms/0684. Redundant Connection/README.md
deleted file mode 100755
index 798d9e7c0..000000000
--- a/Algorithms/0684. Redundant Connection/README.md	
+++ /dev/null
@@ -1,55 +0,0 @@
-# [684. Redundant Connection](https://leetcode.com/problems/redundant-connection/)
-
-
-## 题目:
-
-In this problem, a tree is an **undirected** graph that is connected and has no cycles.
-
-The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, ..., N), with one additional edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.
-
-The resulting graph is given as a 2D-array of `edges`. Each element of `edges` is a pair `[u, v]` with `u < v`, that represents an **undirected** edge connecting nodes `u` and `v`.
-
-Return an edge that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge `[u, v]` should be in the same format, with `u < v`.
-
-**Example 1:**
-
-    Input: [[1,2], [1,3], [2,3]]
-    Output: [2,3]
-    Explanation: The given undirected graph will be like this:
-      1
-     / \
-    2 - 3
-
-**Example 2:**
-
-    Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
-    Output: [1,4]
-    Explanation: The given undirected graph will be like this:
-    5 - 1 - 2
-        |   |
-        4 - 3
-
-**Note:**
-
-- The size of the input 2D-array will be between 3 and 1000.
-- Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.
-
-**Update (2017-09-26):**We have overhauled the problem description + test cases and specified clearly the graph is an **undirected** graph. For the **directed** graph follow up please see **[Redundant Connection II](https://leetcode.com/problems/redundant-connection-ii/description/)**). We apologize for any inconvenience caused.
-
-
-## 题目大意
-
-在本问题中, 树指的是一个连通且无环的无向图。输入一个图，该图由一个有着N个节点 (节点值不重复1, 2, ..., N) 的树及一条附加的边构成。附加的边的两个顶点包含在1到N中间，这条附加的边不属于树中已存在的边。结果图是一个以边组成的二维数组。每一个边的元素是一对[u, v] ，满足 u < v，表示连接顶点u 和v的无向图的边。
-
-返回一条可以删去的边，使得结果图是一个有着N个节点的树。如果有多个答案，则返回二维数组中最后出现的边。答案边 [u, v] 应满足相同的格式 u < v。
-
-注意:
-
-- 输入的二维数组大小在 3 到 1000。
-- 二维数组中的整数在 1 到 N 之间，其中 N 是输入数组的大小。
-
-
-## 解题思路
-
-- 给出一个连通无环无向图和一些连通的边，要求在这些边中删除一条边以后，图中的 N 个节点依旧是连通的。如果有多条边，输出最后一条。
-- 这一题可以用并查集直接秒杀。依次扫描所有的边，把边的两端点都合并 `union()` 到一起。如果遇到一条边的两端点已经在一个集合里面了，就说明是多余边，删除。最后输出这些边即可。
diff --git a/Algorithms/0685. Redundant Connection II/685. Redundant Connection II_test.go b/Algorithms/0685. Redundant Connection II/685. Redundant Connection II_test.go
deleted file mode 100644
index 96a4ff22b..000000000
--- a/Algorithms/0685. Redundant Connection II/685. Redundant Connection II_test.go	
+++ /dev/null
@@ -1,62 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question685 struct {
-	para685
-	ans685
-}
-
-// para 是参数
-// one 代表第一个参数
-type para685 struct {
-	one [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans685 struct {
-	one []int
-}
-
-func Test_Problem685(t *testing.T) {
-
-	qs := []question685{
-
-		question685{
-			para685{[][]int{[]int{3, 5}, []int{1, 3}, []int{2, 1}, []int{5, 4}, []int{2, 3}}},
-			ans685{[]int{2, 3}},
-		},
-
-		question685{
-			para685{[][]int{[]int{4, 2}, []int{1, 5}, []int{5, 2}, []int{5, 3}, []int{2, 4}}},
-			ans685{[]int{4, 2}},
-		},
-
-		question685{
-			para685{[][]int{[]int{1, 2}, []int{1, 3}, []int{2, 3}}},
-			ans685{[]int{2, 3}},
-		},
-
-		question685{
-			para685{[][]int{[]int{1, 2}, []int{2, 3}, []int{3, 4}, []int{1, 4}, []int{1, 5}}},
-			ans685{[]int{1, 4}},
-		},
-
-		question685{
-			para685{[][]int{[]int{2, 1}, []int{3, 1}, []int{4, 2}, []int{1, 4}}},
-			ans685{[]int{2, 1}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 685------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans685, q.para685
-		fmt.Printf("【input】:%v       【output】:%v\n", p, findRedundantDirectedConnection(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0685. Redundant Connection II/README.md b/Algorithms/0685. Redundant Connection II/README.md
deleted file mode 100755
index cb0831046..000000000
--- a/Algorithms/0685. Redundant Connection II/README.md	
+++ /dev/null
@@ -1,59 +0,0 @@
-# [685. Redundant Connection II](https://leetcode.com/problems/redundant-connection-ii/)
-
-
-## 题目:
-
-In this problem, a rooted tree is a **directed** graph such that, there is exactly one node (the root) for which all other nodes are descendants of this node, plus every node has exactly one parent, except for the root node which has no parents.
-
-The given input is a directed graph that started as a rooted tree with N nodes (with distinct values 1, 2, ..., N), with one additional directed edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.
-
-The resulting graph is given as a 2D-array of `edges`. Each element of `edges` is a pair `[u, v]` that represents a **directed** edge connecting nodes `u` and `v`, where `u` is a parent of child `v`.
-
-Return an edge that can be removed so that the resulting graph is a rooted tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array.
-
-**Example 1:**
-
-    Input: [[1,2], [1,3], [2,3]]
-    Output: [2,3]
-    Explanation: The given directed graph will be like this:
-      1
-     / \
-    v   v
-    2-->3
-
-**Example 2:**
-
-    Input: [[1,2], [2,3], [3,4], [4,1], [1,5]]
-    Output: [4,1]
-    Explanation: The given directed graph will be like this:
-    5 <- 1 -> 2
-         ^    |
-         |    v
-         4 <- 3
-
-**Note:**
-
-- The size of the input 2D-array will be between 3 and 1000.
-- Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.
-
-
-## 题目大意
-
-
-在本问题中，有根树指满足以下条件的有向图。该树只有一个根节点，所有其他节点都是该根节点的后继。每一个节点只有一个父节点，除了根节点没有父节点。输入一个有向图，该图由一个有着 N 个节点 (节点值不重复1, 2, ..., N) 的树及一条附加的边构成。附加的边的两个顶点包含在1到N中间，这条附加的边不属于树中已存在的边。结果图是一个以边组成的二维数组。 每一个边的元素是一对 [u, v]，用以表示有向图中连接顶点 u and v 和顶点的边，其中父节点 u 是子节点 v 的一个父节点。返回一条能删除的边，使得剩下的图是有 N 个节点的有根树。若有多个答案，返回最后出现在给定二维数组的答案。
-
-注意:
-
-- 二维数组大小的在 3 到 1000 范围内。
-- 二维数组中的每个整数在 1 到 N 之间，其中 N 是二维数组的大小。
-
-
-## 解题思路
-
-- 这一题是第 684 题的加强版。第 684 题中的图是无向图，这一题中的图是有向图。
-- 这一题的解法也是用并查集，不过需要灵活一点，不要用模板，因为在模板中，存在路径压缩和 `rank()` 优化，这些优化会改变有向边原始的方向。所以并查集只需要记录 `parent()` 就够用了。
-
-![](https://img.halfrost.com/Leetcode/leetcode_685.png)
-
-- 经过分析，可以得到上面这 3 种情况，红色的边是我们实际应该删除的。先来看情况 2 和情况 3 。当不断 `union()` 时，加入一条边以后，会使一个节点的入度变成 2，那么记录下这两条边为 `candidate1` 和 `candidate2`。将后加入的 `candidate2` 这条边先放在一边，继续往下 `union()`。如果 `candidate2` 是红色的边，那么合并到最后，也不会出现任何异常，那么 `candidate2` 就是红色的边，即找到了要删除的边了。如果合并到最后出现了环的问题了，那说明 `candidate2` 是黑色的边，`candidate1` 才是红色的边，那么 `candidate1` 是要删除的边。
-- 再来看看情况 1。如果一路合并到结束也没有发现出现入度为 2 的情况，那么说明遇到了情况 1 。情况 1 会出现环的情况。题目中说如果要删除边，就删除最后出现的那条边。**具体实现见代码注释**。
\ No newline at end of file
diff --git a/Algorithms/0693. Binary Number with Alternating Bits/README.md b/Algorithms/0693. Binary Number with Alternating Bits/README.md
deleted file mode 100755
index 71090ebb0..000000000
--- a/Algorithms/0693. Binary Number with Alternating Bits/README.md	
+++ /dev/null
@@ -1,43 +0,0 @@
-# [693. Binary Number with Alternating Bits](https://leetcode.com/problems/binary-number-with-alternating-bits/)
-
-## 题目:
-
-Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.
-
-**Example 1:**
-
-    Input: 5
-    Output: True
-    Explanation:
-    The binary representation of 5 is: 101
-
-**Example 2:**
-
-    Input: 7
-    Output: False
-    Explanation:
-    The binary representation of 7 is: 111.
-
-**Example 3:**
-
-    Input: 11
-    Output: False
-    Explanation:
-    The binary representation of 11 is: 1011.
-
-**Example 4:**
-
-    Input: 10
-    Output: True
-    Explanation:
-    The binary representation of 10 is: 1010.
-
-
-## 题目大意
-
-给定一个正整数，检查他是否为交替位二进制数：换句话说，就是他的二进制数相邻的两个位数永不相等。
-
-## 解题思路
-
-
-- 判断一个数的二进制位相邻两个数是不相等的，即 `0101` 交叉间隔的，如果是，输出 true。这一题有多种做法，最简单的方法就是直接模拟。比较巧妙的方法是通过位运算，合理构造特殊数据进行位运算到达目的。`010101` 构造出 `101010` 两者相互 `&` 位运算以后就为 0，因为都“插空”了。
diff --git a/Algorithms/0699. Falling Squares/699. Falling Squares_test.go b/Algorithms/0699. Falling Squares/699. Falling Squares_test.go
deleted file mode 100644
index 7089d7e83..000000000
--- a/Algorithms/0699. Falling Squares/699. Falling Squares_test.go	
+++ /dev/null
@@ -1,57 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question699 struct {
-	para699
-	ans699
-}
-
-// para 是参数
-// one 代表第一个参数
-type para699 struct {
-	one [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans699 struct {
-	one []int
-}
-
-func Test_Problem699(t *testing.T) {
-
-	qs := []question699{
-
-		question699{
-			para699{[][]int{[]int{6, 1}, []int{9, 2}, []int{2, 4}}},
-			ans699{[]int{1, 2, 4}},
-		},
-
-		question699{
-			para699{[][]int{[]int{1, 2}, []int{1, 3}}},
-			ans699{[]int{2, 5}},
-		},
-
-		question699{
-			para699{[][]int{[]int{1, 2}, []int{2, 3}, []int{6, 1}}},
-			ans699{[]int{2, 5, 5}},
-		},
-
-		question699{
-			para699{[][]int{[]int{100, 100}, []int{200, 100}}},
-			ans699{[]int{100, 100}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 699------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans699, q.para699
-		fmt.Printf("【input】:%v       【output】:%v\n", p, fallingSquares(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0699. Falling Squares/README.md b/Algorithms/0699. Falling Squares/README.md
deleted file mode 100755
index ea5fd2088..000000000
--- a/Algorithms/0699. Falling Squares/README.md	
+++ /dev/null
@@ -1,99 +0,0 @@
-# [699. Falling Squares](https://leetcode.com/problems/falling-squares/)
-
-
-## 题目:
-
-On an infinite number line (x-axis), we drop given squares in the order they are given.
-
-The `i`-th square dropped (`positions[i] = (left, side_length)`) is a square with the left-most point being `positions[i][0]` and sidelength `positions[i][1]`.
-
-The square is dropped with the bottom edge parallel to the number line, and from a higher height than all currently landed squares. We wait for each square to stick before dropping the next.
-
-The squares are infinitely sticky on their bottom edge, and will remain fixed to any positive length surface they touch (either the number line or another square). Squares dropped adjacent to each other will not stick together prematurely.
-
-Return a list `ans` of heights. Each height `ans[i]` represents the current highest height of any square we have dropped, after dropping squares represented by `positions[0], positions[1], ..., positions[i]`.
-
-**Example 1:**
-
-    Input: [[1, 2], [2, 3], [6, 1]]
-    Output: [2, 5, 5]
-    Explanation:
-
-After the first drop of `positions[0] = [1, 2]: _aa _aa -------` The maximum height of any square is 2.
-
-After the second drop of `positions[1] = [2, 3]: __aaa __aaa __aaa _aa__ _aa__ --------------` The maximum height of any square is 5. The larger square stays on top of the smaller square despite where its center of gravity is, because squares are infinitely sticky on their bottom edge.
-
-After the third drop of `positions[1] = [6, 1]: __aaa __aaa __aaa _aa _aa___a --------------` The maximum height of any square is still 5. Thus, we return an answer of `[2, 5, 5]`.
-
-**Example 2:**
-
-    Input: [[100, 100], [200, 100]]
-    Output: [100, 100]
-    Explanation: Adjacent squares don't get stuck prematurely - only their bottom edge can stick to surfaces.
-
-**Note:**
-
-- `1 <= positions.length <= 1000`.
-- `1 <= positions[i][0] <= 10^8`.
-- `1 <= positions[i][1] <= 10^6`.
-
-
-## 题目大意
-
-在无限长的数轴（即 x 轴）上，我们根据给定的顺序放置对应的正方形方块。第 i 个掉落的方块（positions[i] = (left, side\_length)）是正方形，其中 left 表示该方块最左边的点位置(positions[i][0])，side\_length 表示该方块的边长(positions[i][1])。
-
-每个方块的底部边缘平行于数轴（即 x 轴），并且从一个比目前所有的落地方块更高的高度掉落而下。在上一个方块结束掉落，并保持静止后，才开始掉落新方块。方块的底边具有非常大的粘性，并将保持固定在它们所接触的任何长度表面上（无论是数轴还是其他方块）。邻接掉落的边不会过早地粘合在一起，因为只有底边才具有粘性。
-
-返回一个堆叠高度列表 ans 。每一个堆叠高度 ans[i] 表示在通过 positions[0], positions[1], ..., positions[i] 表示的方块掉落结束后，目前所有已经落稳的方块堆叠的最高高度。
-
-示例 1:
-
-```c
-输入: [[1, 2], [2, 3], [6, 1]]
-输出: [2, 5, 5]
-解释:
-
-第一个方块 positions[0] = [1, 2] 掉落：
-_aa
-_aa
--------
-方块最大高度为 2 。
-
-第二个方块 positions[1] = [2, 3] 掉落：
-__aaa
-__aaa
-__aaa
-_aa__
-_aa__
---------------
-方块最大高度为5。
-大的方块保持在较小的方块的顶部，不论它的重心在哪里，因为方块的底部边缘有非常大的粘性。
-
-第三个方块 positions[1] = [6, 1] 掉落：
-__aaa
-__aaa
-__aaa
-_aa
-_aa___a
--------------- 
-方块最大高度为5。
-
-因此，我们返回结果[2, 5, 5]。
-
-```
-
-注意:
-
-- 1 <= positions.length <= 1000.
-- 1 <= positions[i][0] <= 10^8.
-- 1 <= positions[i][1] <= 10^6.
- 
-
-
-## 解题思路
-
-- 给出一个二维数组，每个一维数组中只有 2 个值，分别代表的是正方形砖块所在 x 轴的坐标起始点，和边长。要求输出每次砖块落下以后，当前最大的高度。正方形砖块落下如同俄罗斯方块，落下的过程中如果遇到了砖块会落在砖块的上面。如果砖块摞起来了以后，下方有空间，是不可能再把砖块挪进去的，因为此题砖块只会垂直落下，不会水平移动(这一点和俄罗斯方块不同)。
-- 这一题可以用线段树解答。由于方块在 x 轴上的坐标范围特别大，如果不离散化，这一题就会 MTE。所以首先去重 - 排序 - 离散化。先把每个砖块所在区间都算出来，每个正方形的方块所在区间是 `[pos[0] , pos[0]+pos[1]-1]` ，为什么右边界要减一呢？因为每个方块占据的区间其实应该是左闭右开的，即 `[pos[0] , pos[0]+pos[1])`，如果右边是开的，那么这个边界会被 2 个区间查询共用，从而导致错误结果。例如 [2,3]，[3,4]，这两个区间的砖块实际是不会摞在一起的。但是如果右边都是闭区间，用线段树 query 查询的时候，会都找到 [3，3]，从而导致这两个区间都会判断 3 这一点的情况。正确的做法应该是 [2,3)，[3,4）这样就避免了上述可能导致错误的情况了。离散化以后，所有的坐标区间都在 0~n 之间了。
-- 遍历每个砖块所在区间，先查询这个区间内的值，再加上当前砖块的高度，即为这个区间的最新高度。并更新该区间的值。更新区间的值用到懒惰更新。然后和动态维护的当前最大高度进行比较，将最大值放入最终输出的数组中。
-- 类似的题目有：第 715 题，第 218 题，第 732 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和本题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
-- leetcode  上也有线段树的讲解：[Get Solutions to Interview Questions](https://leetcode.com/articles/a-recursive-approach-to-segment-trees-range-sum-queries-lazy-propagation/)
diff --git a/Algorithms/0705. Design HashSet/README.md b/Algorithms/0705. Design HashSet/README.md
deleted file mode 100755
index d4bb59c88..000000000
--- a/Algorithms/0705. Design HashSet/README.md	
+++ /dev/null
@@ -1,53 +0,0 @@
-# [705. Design HashSet](https://leetcode.com/problems/design-hashset/)
-
-
-## 题目:
-
-Design a HashSet without using any built-in hash table libraries.
-
-To be specific, your design should include these functions:
-
-- `add(value)`: Insert a value into the HashSet.
-- `contains(value)` : Return whether the value exists in the HashSet or not.
-- `remove(value)`: Remove a value in the HashSet. If the value does not exist in the HashSet, do nothing.
-
-**Example:**
-
-    MyHashSet hashSet = new MyHashSet();
-    hashSet.add(1);         
-    hashSet.add(2);         
-    hashSet.contains(1);    // returns true
-    hashSet.contains(3);    // returns false (not found)
-    hashSet.add(2);          
-    hashSet.contains(2);    // returns true
-    hashSet.remove(2);          
-    hashSet.contains(2);    // returns false (already removed)
-
-**Note:**
-
-- All values will be in the range of `[0, 1000000]`.
-- The number of operations will be in the range of `[1, 10000]`.
-- Please do not use the built-in HashSet library.
-
-
-## 题目大意
-
-不使用任何内建的哈希表库设计一个哈希集合具体地说，你的设计应该包含以下的功能：
-
-- add(value)：向哈希集合中插入一个值。
-- contains(value) ：返回哈希集合中是否存在这个值。
-- remove(value)：将给定值从哈希集合中删除。如果哈希集合中没有这个值，什么也不做。
-
-
-注意：
-
-- 所有的值都在 [1, 1000000] 的范围内。
-- 操作的总数目在 [1, 10000] 范围内。
-- 不要使用内建的哈希集合库。
-
-
-
-## 解题思路
-
-
-- 简单题，设计一个 hashset 的数据结构，要求有 `add(value)`，`contains(value)`，`remove(value)`，这 3 个方法。
diff --git a/Algorithms/0706. Design HashMap/README.md b/Algorithms/0706. Design HashMap/README.md
deleted file mode 100755
index e60ccc517..000000000
--- a/Algorithms/0706. Design HashMap/README.md	
+++ /dev/null
@@ -1,51 +0,0 @@
-# [706. Design HashMap](https://leetcode.com/problems/design-hashmap/)
-
-
-## 题目:
-
-Design a HashMap without using any built-in hash table libraries.
-
-To be specific, your design should include these functions:
-
-- `put(key, value)` : Insert a (key, value) pair into the HashMap. If the value already exists in the HashMap, update the value.
-- `get(key)`: Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
-- `remove(key)` : Remove the mapping for the value key if this map contains the mapping for the key.
-
-**Example:**
-
-    MyHashMap hashMap = new MyHashMap();
-    hashMap.put(1, 1);          
-    hashMap.put(2, 2);         
-    hashMap.get(1);            // returns 1
-    hashMap.get(3);            // returns -1 (not found)
-    hashMap.put(2, 1);          // update the existing value
-    hashMap.get(2);            // returns 1 
-    hashMap.remove(2);          // remove the mapping for 2
-    hashMap.get(2);            // returns -1 (not found)
-
-**Note:**
-
-- All keys and values will be in the range of `[0, 1000000]`.
-- The number of operations will be in the range of `[1, 10000]`.
-- Please do not use the built-in HashMap library.
-
-
-## 题目大意
-
-不使用任何内建的哈希表库设计一个哈希映射具体地说，你的设计应该包含以下的功能：
-
-- put(key, value)：向哈希映射中插入(键,值)的数值对。如果键对应的值已经存在，更新这个值。
-- get(key)：返回给定的键所对应的值，如果映射中不包含这个键，返回 -1。
-- remove(key)：如果映射中存在这个键，删除这个数值对。
-
-注意：
-
-- 所有的值都在 [1, 1000000] 的范围内。
-- 操作的总数目在 [1, 10000] 范围内。
-- 不要使用内建的哈希库。
-
-
-## 解题思路
-
-
-- 简单题，设计一个 hashmap 的数据结构，要求有 `put(key, value)`，`get(key)`，`remove(key)`，这 3 个方法。设计一个 map 主要需要处理哈希冲突，一般都是链表法解决冲突。
diff --git a/Algorithms/0707. Design Linked List/707. Design Linked List.go b/Algorithms/0707. Design Linked List/707. Design Linked List.go
deleted file mode 100644
index 788b9930d..000000000
--- a/Algorithms/0707. Design Linked List/707. Design Linked List.go	
+++ /dev/null
@@ -1,91 +0,0 @@
-package leetcode
-
-type MyLinkedList struct {
-	Val  int
-	Next *MyLinkedList
-}
-
-/** Initialize your data structure here. */
-func Constructor() MyLinkedList {
-	return MyLinkedList{Val: -999, Next: nil}
-}
-
-/** Get the value of the index-th node in the linked list. If the index is invalid, return -1. */
-func (this *MyLinkedList) Get(index int) int {
-	cur := this
-	for i := 0; cur != nil; i++ {
-		if i == index {
-			if cur.Val == -999 {
-				return -1
-			} else {
-				return cur.Val
-			}
-		}
-		cur = cur.Next
-	}
-	return -1
-}
-
-/** Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list. */
-func (this *MyLinkedList) AddAtHead(val int) {
-	if this.Val == -999 {
-		this.Val = val
-	} else {
-		tmp := &MyLinkedList{Val: this.Val, Next: this.Next}
-		this.Val = val
-		this.Next = tmp
-	}
-}
-
-/** Append a node of value val to the last element of the linked list. */
-func (this *MyLinkedList) AddAtTail(val int) {
-	cur := this
-	for cur.Next != nil {
-		cur = cur.Next
-	}
-	tmp := &MyLinkedList{Val: val, Next: nil}
-	cur.Next = tmp
-}
-
-/** Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted. */
-func (this *MyLinkedList) AddAtIndex(index int, val int) {
-	cur := this
-	if index == 0 {
-		this.AddAtHead(val)
-		return
-	}
-	for i := 0; cur != nil; i++ {
-		if i == index-1 {
-			break
-		}
-		cur = cur.Next
-	}
-	if cur != nil && cur.Val != -999 {
-		tmp := &MyLinkedList{Val: val, Next: cur.Next}
-		cur.Next = tmp
-	}
-}
-
-/** Delete the index-th node in the linked list, if the index is valid. */
-func (this *MyLinkedList) DeleteAtIndex(index int) {
-	cur := this
-	for i := 0; cur != nil; i++ {
-		if i == index-1 {
-			break
-		}
-		cur = cur.Next
-	}
-	if cur != nil && cur.Next != nil {
-		cur.Next = cur.Next.Next
-	}
-}
-
-/**
- * Your MyLinkedList object will be instantiated and called as such:
- * obj := Constructor();
- * param_1 := obj.Get(index);
- * obj.AddAtHead(val);
- * obj.AddAtTail(val);
- * obj.AddAtIndex(index,val);
- * obj.DeleteAtIndex(index);
- */
diff --git a/Algorithms/0707. Design Linked List/707. Design Linked List_test.go b/Algorithms/0707. Design Linked List/707. Design Linked List_test.go
deleted file mode 100644
index 790da6eab..000000000
--- a/Algorithms/0707. Design Linked List/707. Design Linked List_test.go	
+++ /dev/null
@@ -1,64 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-func Test_Problem707(t *testing.T) {
-	obj := Constructor()
-	fmt.Printf("obj = %v\n", ML2s(&obj))
-	param1 := obj.Get(1)
-	fmt.Printf("param_1 = %v obj = %v\n", param1, ML2s(&obj))
-	obj.AddAtHead(38)
-	fmt.Printf("obj = %v\n", ML2s(&obj))
-	obj.AddAtHead(45)
-	fmt.Printf("obj = %v\n", ML2s(&obj))
-	obj.DeleteAtIndex(2)
-	fmt.Printf("obj = %v\n", ML2s(&obj))
-	obj.AddAtIndex(1, 24)
-	fmt.Printf("obj = %v\n", ML2s(&obj))
-	obj.AddAtTail(36)
-	fmt.Printf("obj = %v\n", ML2s(&obj))
-	obj.AddAtIndex(3, 72)
-	obj.AddAtTail(76)
-	fmt.Printf("obj = %v\n", ML2s(&obj))
-	obj.AddAtHead(7)
-	fmt.Printf("obj = %v\n", ML2s(&obj))
-	obj.AddAtHead(36)
-	fmt.Printf("obj = %v\n", ML2s(&obj))
-	obj.AddAtHead(34)
-	fmt.Printf("obj = %v\n", ML2s(&obj))
-	obj.AddAtTail(91)
-	fmt.Printf("obj = %v\n", ML2s(&obj))
-
-	fmt.Printf("\n\n\n")
-
-	obj1 := Constructor()
-	fmt.Printf("obj1 = %v\n", ML2s(&obj1))
-	param2 := obj1.Get(0)
-	fmt.Printf("param_2 = %v obj1 = %v\n", param2, ML2s(&obj1))
-	obj1.AddAtIndex(1, 2)
-	fmt.Printf("obj1 = %v\n", ML2s(&obj1))
-	param2 = obj1.Get(0)
-	fmt.Printf("param_2 = %v obj1 = %v\n", param2, ML2s(&obj1))
-	param2 = obj1.Get(1)
-	fmt.Printf("param_2 = %v obj1 = %v\n", param2, ML2s(&obj1))
-	obj1.AddAtIndex(0, 1)
-	fmt.Printf("obj1 = %v\n", ML2s(&obj1))
-	param2 = obj1.Get(0)
-	fmt.Printf("param_1 = %v obj1 = %v\n", param2, ML2s(&obj1))
-	param2 = obj1.Get(1)
-	fmt.Printf("param_2 = %v obj1 = %v\n", param2, ML2s(&obj1))
-}
-
-func ML2s(head *MyLinkedList) []int {
-	res := []int{}
-
-	for head != nil {
-		res = append(res, head.Val)
-		head = head.Next
-	}
-
-	return res
-}
diff --git a/Algorithms/0715. Range Module/README.md b/Algorithms/0715. Range Module/README.md
deleted file mode 100755
index 501a16786..000000000
--- a/Algorithms/0715. Range Module/README.md	
+++ /dev/null
@@ -1,60 +0,0 @@
-# [715. Range Module](https://leetcode.com/problems/range-module/)
-
-
-## 题目:
-
-A Range Module is a module that tracks ranges of numbers. Your task is to design and implement the following interfaces in an efficient manner.
-
-- `addRange(int left, int right)` Adds the half-open interval `[left, right)`, tracking every real number in that interval. Adding an interval that partially overlaps with currently tracked numbers should add any numbers in the interval `[left, right)` that are not already tracked.
-- `queryRange(int left, int right)` Returns true if and only if every real number in the interval `[left, right)` is currently being tracked.
-- `removeRange(int left, int right)` Stops tracking every real number currently being tracked in the interval `[left, right)`.
-
-**Example 1:**
-
-    addRange(10, 20): null
-    removeRange(14, 16): null
-    queryRange(10, 14): true (Every number in [10, 14) is being tracked)
-    queryRange(13, 15): false (Numbers like 14, 14.03, 14.17 in [13, 15) are not being tracked)
-    queryRange(16, 17): true (The number 16 in [16, 17) is still being tracked, despite the remove operation)
-
-**Note:**
-
-- A half open interval `[left, right)` denotes all real numbers `left <= x < right`.
-- `0 < left < right < 10^9` in all calls to `addRange, queryRange, removeRange`.
-- The total number of calls to `addRange` in a single test case is at most `1000`.
-- The total number of calls to `queryRange` in a single test case is at most `5000`.
-- The total number of calls to `removeRange` in a single test case is at most `1000`.
-
-## 题目大意
-
-Range 模块是跟踪数字范围的模块。你的任务是以一种有效的方式设计和实现以下接口。
-
-- addRange(int left, int right) 添加半开区间 [left, right)，跟踪该区间中的每个实数。添加与当前跟踪的数字部分重叠的区间时，应当添加在区间 [left, right) 中尚未跟踪的任何数字到该区间中。
-- queryRange(int left, int right) 只有在当前正在跟踪区间 [left, right) 中的每一个实数时，才返回 true。
-- removeRange(int left, int right) 停止跟踪区间 [left, right) 中当前正在跟踪的每个实数。
- 
-
-示例：
-
-```
-addRange(10, 20): null
-removeRange(14, 16): null
-queryRange(10, 14): true （区间 [10, 14) 中的每个数都正在被跟踪）
-queryRange(13, 15): false （未跟踪区间 [13, 15) 中像 14, 14.03, 14.17 这样的数字）
-queryRange(16, 17): true （尽管执行了删除操作，区间 [16, 17) 中的数字 16 仍然会被跟踪）
-```
-
-提示：
-
-- 半开区间 [left, right) 表示所有满足 left <= x < right 的实数。
-- 对 addRange, queryRange, removeRange 的所有调用中 0 < left < right < 10^9。
-- 在单个测试用例中，对 addRange 的调用总数不超过 1000 次。
-- 在单个测试用例中，对  queryRange 的调用总数不超过 5000 次。
-- 在单个测试用例中，对 removeRange 的调用总数不超过 1000 次。
-
-
-## 解题思路
-
-- 设计一个数据结构，能完成添加区间 `addRange`，查询区间 `queryRange`，移除区间 `removeRange` 三种操作。查询区间的操作需要更加高效一点。
-- 这一题可以用线段树来解答，但是时间复杂度不高，最优解是用二叉排序树 BST 来解答。先来看线段树。这一题是更新区间内的值，所以需要用到懒惰更新。添加区间可以把区间内的值都赋值为 1 。由于题目中未预先确定区间范围，选用树的形式实现线段树比数组实现更加节约空间(当然用数组也可以，区间最大是 1000，点至多有 2000 个)。移除区间的时候就是把区间内的值都赋值标记为 0 。
-- 类似的题目有：第 699 题，第 218 题，第 732 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和第 699 题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
\ No newline at end of file
diff --git a/Algorithms/0717. 1-bit and 2-bit Characters/README.md b/Algorithms/0717. 1-bit and 2-bit Characters/README.md
deleted file mode 100755
index 44224df0f..000000000
--- a/Algorithms/0717. 1-bit and 2-bit Characters/README.md	
+++ /dev/null
@@ -1,50 +0,0 @@
-# 717. 1-bit and 2-bit Characters
-
-Created Time: Jul 25, 2019 10:23 AM
-Difficulty: Easy
-Link: https://leetcode.com/problems/1-bit-and-2-bit-characters/
-Tags: Array
-
-# 题目:
-
-We have two special characters. The first character can be represented by one bit `0`. The second character can be represented by two bits (`10` or `11`).
-
-Now given a string represented by several bits. Return whether the last character must be a one-bit character or not. The given string will always end with a zero.
-
-**Example 1:**
-
-    Input: 
-    bits = [1, 0, 0]
-    Output: True
-    Explanation: 
-    The only way to decode it is two-bit character and one-bit character. So the last character is one-bit character.
-
-**Example 2:**
-
-    Input: 
-    bits = [1, 1, 1, 0]
-    Output: False
-    Explanation: 
-    The only way to decode it is two-bit character and two-bit character. So the last character is NOT one-bit character.
-
-**Note:**
-
-- `1 <= len(bits) <= 1000`.
-- `bits[i]` is always `0` or `1`.
-
-## 题目大意
-
-有两种特殊字符。第一种字符可以用一比特0来表示。第二种字符可以用两比特(10 或 11)来表示。
-
-现给一个由若干比特组成的字符串。问最后一个字符是否必定为一个一比特字符。给定的字符串总是由0结束。
-
-注意:
-
-- 1 <= len(bits) <= 1000.
-- bits[i] 总是0 或 1.
-
-
-## 解题思路
-
-- 给出一个数组，数组里面的元素只有 0 和 1，并且数组的最后一个元素一定是 0。有 2 种特殊的字符，第一类字符是 "0"，第二类字符是 "11" 和 "10"，请判断这个数组最后一个元素是否一定是属于第一类字符？
-- 依题意， 0 的来源有 2 处，可以是第一类字符，也可以是第二类字符，1 的来源只有 1 处，一定出自第二类字符。最后一个 0 当前仅当为第一类字符的情况有 2 种，第一种情况，前面出现有 0，但是 0 和 1 配对形成了第二类字符。第二种情况，前面没有出现 0 。这两种情况的共同点是除去最后一个元素，数组中前面所有的1 都“结对子”。所以利用第二类字符的特征，"1X"，遍历整个数组，如果遇到 "1"，就跳 2 步，因为 1 后面出现什么数字( 0 或者 1 )并不需要关心。如果 `i` 能在 `len(bits) - 1` 的地方`(数组最后一个元素)`停下，那么对应的是情况一或者情况二，前面的 0 都和 1 匹配上了，最后一个 0 一定是第一类字符。如果 `i` 在 `len(bit)` 的位置`(超出数组下标)`停下，说明 `bits[len(bits) - 1] == 1`，这个时候最后一个 0 一定属于第二类字符。
diff --git a/Algorithms/0721. Accounts Merge/721. Accounts Merge_test.go b/Algorithms/0721. Accounts Merge/721. Accounts Merge_test.go
deleted file mode 100644
index d0dd8bcd2..000000000
--- a/Algorithms/0721. Accounts Merge/721. Accounts Merge_test.go	
+++ /dev/null
@@ -1,73 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question721 struct {
-	para721
-	ans721
-}
-
-// para 是参数
-// one 代表第一个参数
-type para721 struct {
-	w [][]string
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans721 struct {
-	one [][]string
-}
-
-func Test_Problem721(t *testing.T) {
-
-	qs := []question721{
-
-		question721{
-			para721{[][]string{[]string{"John", "johnsmith@mail.com", "john00@mail.com"},
-				[]string{"John", "johnnybravo@mail.com"},
-				[]string{"John", "johnsmith@mail.com", "john_newyork@mail.com"},
-				[]string{"Mary", "mary@mail.com"}}},
-			ans721{[][]string{[]string{"John", "john00@mail.com", "john_newyork@mail.com", "johnsmith@mail.com"},
-				[]string{"John", "johnnybravo@mail.com"},
-				[]string{"Mary", "mary@mail.com"}}},
-		},
-
-		question721{
-			para721{[][]string{[]string{"Alex", "Alex5@m.co", "Alex4@m.co", "Alex0@m.co"},
-				[]string{"Ethan", "Ethan3@m.co", "Ethan3@m.co", "Ethan0@m.co"},
-				[]string{"Kevin", "Kevin4@m.co", "Kevin2@m.co", "Kevin2@m.co"},
-				[]string{"Gabe", "Gabe0@m.co", "Gabe3@m.co", "Gabe2@m.co"},
-				[]string{"Gabe", "Gabe3@m.co", "Gabe4@m.co", "Gabe2@m.co"}}},
-			ans721{[][]string{[]string{"Alex", "Alex0@m.co", "Alex4@m.co", "Alex5@m.co"},
-				[]string{"Ethan", "Ethan0@m.co", "Ethan3@m.co"},
-				[]string{"Gabe", "Gabe0@m.co", "Gabe2@m.co", "Gabe3@m.co", "Gabe4@m.co"},
-				[]string{"Kevin", "Kevin2@m.co", "Kevin4@m.co"}}},
-		},
-
-		question721{
-			para721{[][]string{[]string{"David", "David0@m.co", "David4@m.co", "David3@m.co"},
-				[]string{"David", "David5@m.co", "David5@m.co", "David0@m.co"},
-				[]string{"David", "David1@m.co", "David4@m.co", "David0@m.co"},
-				[]string{"David", "David0@m.co", "David1@m.co", "David3@m.co"},
-				[]string{"David", "David4@m.co", "David1@m.co", "David3@m.co"}}},
-			ans721{[][]string{[]string{"David", "David0@m.co", "David1@m.co", "David3@m.co", "David4@m.co", "David5@m.co"}}},
-		},
-
-		question721{
-			para721{[][]string{}},
-			ans721{[][]string{}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 721------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans721, q.para721
-		fmt.Printf("【input】:%v       【output】:%v\n", p, accountsMerge(p.w))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0721. Accounts Merge/README.md b/Algorithms/0721. Accounts Merge/README.md
deleted file mode 100755
index cb8f3bd2d..000000000
--- a/Algorithms/0721. Accounts Merge/README.md	
+++ /dev/null
@@ -1,49 +0,0 @@
-# [721. Accounts Merge](https://leetcode.com/problems/accounts-merge/)
-
-
-## 题目:
-
-Given a list `accounts`, each element `accounts[i]` is a list of strings, where the first element `accounts[i][0]` is a name, and the rest of the elements are emailsrepresenting emails of the account.
-
-Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some email that is common to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.
-
-After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails **in sorted order**. The accounts themselves can be returned in any order.
-
-**Example 1:**
-
-    Input: 
-    accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
-    Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'],  ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
-    Explanation: 
-    The first and third John's are the same person as they have the common email "johnsmith@mail.com".
-    The second John and Mary are different people as none of their email addresses are used by other accounts.
-    We could return these lists in any order, for example the answer [['Mary', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], 
-    ['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.
-
-**Note:**
-
-- The length of `accounts` will be in the range `[1, 1000]`.
-- The length of `accounts[i]` will be in the range `[1, 10]`.
-- The length of `accounts[i][j]` will be in the range `[1, 30]`.
-
-
-## 题目大意
-
-
-给定一个列表 accounts，每个元素 accounts[i] 是一个字符串列表，其中第一个元素 accounts[i][0] 是 名称 (name)，其余元素是 emails 表示该帐户的邮箱地址。现在，我们想合并这些帐户。如果两个帐户都有一些共同的邮件地址，则两个帐户必定属于同一个人。请注意，即使两个帐户具有相同的名称，它们也可能属于不同的人，因为人们可能具有相同的名称。一个人最初可以拥有任意数量的帐户，但其所有帐户都具有相同的名称。合并帐户后，按以下格式返回帐户：每个帐户的第一个元素是名称，其余元素是按顺序排列的邮箱地址。accounts 本身可以以任意顺序返回。
-
-
-注意：
-
-- accounts 的长度将在 [1，1000] 的范围内。
-- accounts[i] 的长度将在 [1，10] 的范围内。
-- accounts[i][j] 的长度将在 [1，30] 的范围内。
-
-
-
-## 解题思路
-
-
-- 给出一堆账户和对应的邮箱。要求合并同一个人的多个邮箱账户。如果判断是同一个人呢？如果这个人名和所属的其中之一的邮箱是相同的，就判定这是同一个人的邮箱，那么就合并这些邮箱。
-- 这题的解题思路是并查集。不过如果用暴力合并的方法，时间复杂度非常差。优化方法是先把每组数据都进行编号，人编号，每个邮箱都进行编号。这个映射关系用 `map` 记录起来。如果利用并查集的 `union()` 操作，把这些编号都进行合并。最后把人的编号和对应邮箱的编号拼接起来。
-- 这一题有 2 处比较“坑”的是，不需要合并的用户的邮箱列表也是需要排序和去重的，同一个人的所有邮箱集合都要合并到一起。具体见测试用例。不过题目中也提到了这些点，也不能算题目坑，只能归自己没注意这些边界情况。
diff --git a/Algorithms/0725. Split Linked List in Parts/README.md b/Algorithms/0725. Split Linked List in Parts/README.md
deleted file mode 100644
index 74112a288..000000000
--- a/Algorithms/0725. Split Linked List in Parts/README.md	
+++ /dev/null
@@ -1,57 +0,0 @@
-# [725. Split Linked List in Parts](https://leetcode.com/problems/split-linked-list-in-parts/)
-
-## 题目
-
-Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts".
-
-The length of each part should be as equal as possible: no two parts should have a size differing by more than 1. This may lead to some parts being null.
-
-The parts should be in order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal parts occurring later.
-
-Return a List of ListNode's representing the linked list parts that are formed.
-
-Examples 1->2->3->4, k = 5 // 5 equal parts [ [1], [2], [3], [4], null ]
-
-Example 1:
-
-```c
-Input: 
-root = [1, 2, 3], k = 5
-Output: [[1],[2],[3],[],[]]
-Explanation:
-The input and each element of the output are ListNodes, not arrays.
-For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null.
-The first element output[0] has output[0].val = 1, output[0].next = null.
-The last element output[4] is null, but it's string representation as a ListNode is [].
-```
-
-Example 2:
-
-```c
-Input: 
-root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
-Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
-Explanation:
-The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.
-```
-
-Note:
-
-- The length of root will be in the range [0, 1000].
-- Each value of a node in the input will be an integer in the range [0, 999].
-- k will be an integer in the range [1, 50].
-
-
-
-## 题目大意
-
-把链表分成 K 个部分，要求这 K 个部分尽量两两长度相差不超过 1，并且长度尽量相同。
-
-## 解题思路
-
-把链表长度对 K 进行除法，结果就是最终每组的长度 n。把链表长度对 K 进行取余操作，得到的结果 m，代表前 m 组链表长度为 n + 1 。相当于把多出来的部分都分摊到前面 m 组链表中了。最终链表是前 m 组长度为 n + 1，后 K - m 组链表长度是 n。
-
-注意长度不足 K 的时候要用 nil 进行填充。
-
-
-
diff --git a/Algorithms/0729. My Calendar I/README.md b/Algorithms/0729. My Calendar I/README.md
deleted file mode 100755
index 63ec80aad..000000000
--- a/Algorithms/0729. My Calendar I/README.md	
+++ /dev/null
@@ -1,60 +0,0 @@
-# [729. My Calendar I](https://leetcode.com/problems/my-calendar-i/)
-
-
-## 题目:
-
-Implement a `MyCalendar` class to store your events. A new event can be added if adding the event will not cause a double booking.
-
-Your class will have the method, `book(int start, int end)`. Formally, this represents a booking on the half open interval `[start, end)`, the range of real numbers `x` such that `start <= x < end`.
-
-A double booking happens when two events have some non-empty intersection (ie., there is some time that is common to both events.)
-
-For each call to the method `MyCalendar.book`, return `true` if the event can be added to the calendar successfully without causing a double booking. Otherwise, return `false` and do not add the event to the calendar.
-
-Your class will be called like this:
-
-`MyCalendar cal = new MyCalendar();`
-
-`MyCalendar.book(start, end)`
-
-**Example 1:**
-
-    MyCalendar();
-    MyCalendar.book(10, 20); // returns true
-    MyCalendar.book(15, 25); // returns false
-    MyCalendar.book(20, 30); // returns true
-    Explanation: 
-    The first event can be booked.  The second can't because time 15 is already booked by another event.
-    The third event can be booked, as the first event takes every time less than 20, but not including 20.
-
-**Note:**
-
-- The number of calls to `MyCalendar.book` per test case will be at most `1000`.
-- In calls to `MyCalendar.book(start, end)`, `start` and `end` are integers in the range `[0, 10^9]`.
-
-
-
-## 题目大意
-
-实现一个 MyCalendar 类来存放你的日程安排。如果要添加的时间内没有其他安排，则可以存储这个新的日程安排。
-
-MyCalendar 有一个 book(int start, int end) 方法。它意味着在 start 到 end 时间内增加一个日程安排，注意，这里的时间是半开区间，即 [start, end), 实数 x 的范围为，  start <= x < end。
-
-当两个日程安排有一些时间上的交叉时（例如两个日程安排都在同一时间内），就会产生重复预订。
-
-每次调用 MyCalendar.book 方法时，如果可以将日程安排成功添加到日历中而不会导致重复预订，返回 true。否则，返回 false 并且不要将该日程安排添加到日历中。
-
-请按照以下步骤调用 MyCalendar 类: MyCalendar cal = new MyCalendar(); MyCalendar.book(start, end)
-
-说明:
-
-- 每个测试用例，调用 MyCalendar.book 函数最多不超过 100次。
-- 调用函数 MyCalendar.book(start, end) 时， start 和 end 的取值范围为 [0, 10^9]。
-
-
-## 解题思路
-
-
-- 要求实现一个日程安排的功能，如果有日程安排冲突了，就返回 false，如果不冲突则返回 ture
-- 这一题有多种解法，第一种解法可以用类似第 34 题的解法。先排序每个区间，然后再这个集合中用二分搜索找到最后一个区间的左值比当前要比较的区间左值小的，如果找到，再判断能否插入进去(判断右区间是否比下一个区间的左区间小)，此方法时间复杂度 O(n log n)
-- 第二种解法是用生成一个 BST 树。在插入树中先排除不能插入的情况，例如区间有重合。然后以区间左值为依据，递归插入，每次插入依次会继续判断区间是否重合。直到不能插入，则返回 fasle。整个查找的时间复杂度是 O(log n)。
\ No newline at end of file
diff --git a/Algorithms/0732. My Calendar III/README.md b/Algorithms/0732. My Calendar III/README.md
deleted file mode 100755
index 0f050dd1d..000000000
--- a/Algorithms/0732. My Calendar III/README.md	
+++ /dev/null
@@ -1,63 +0,0 @@
-# [732. My Calendar III](https://leetcode.com/problems/my-calendar-iii/)
-
-
-## 题目:
-
-Implement a `MyCalendarThree` class to store your events. A new event can **always** be added.
-
-Your class will have one method, `book(int start, int end)`. Formally, this represents a booking on the half open interval `[start, end)`, the range of real numbers `x` such that `start <= x < end`.
-
-A K-booking happens when **K** events have some non-empty intersection (ie., there is some time that is common to all K events.)
-
-For each call to the method `MyCalendar.book`, return an integer `K` representing the largest integer such that there exists a `K`-booking in the calendar.
-
-Your class will be called like this:
-
-`MyCalendarThree cal = new MyCalendarThree();`
-
-`MyCalendarThree.book(start, end)`
-
-**Example 1:**
-
-    MyCalendarThree();
-    MyCalendarThree.book(10, 20); // returns 1
-    MyCalendarThree.book(50, 60); // returns 1
-    MyCalendarThree.book(10, 40); // returns 2
-    MyCalendarThree.book(5, 15); // returns 3
-    MyCalendarThree.book(5, 10); // returns 3
-    MyCalendarThree.book(25, 55); // returns 3
-    Explanation: 
-    The first two events can be booked and are disjoint, so the maximum K-booking is a 1-booking.
-    The third event [10, 40) intersects the first event, and the maximum K-booking is a 2-booking.
-    The remaining events cause the maximum K-booking to be only a 3-booking.
-    Note that the last event locally causes a 2-booking, but the answer is still 3 because
-    eg. [10, 20), [10, 40), and [5, 15) are still triple booked.
-
-**Note:**
-
-- The number of calls to `MyCalendarThree.book` per test case will be at most `400`.
-- In calls to `MyCalendarThree.book(start, end)`, `start` and `end` are integers in the range `[0, 10^9]`.
-
-
-## 题目大意
-
-实现一个 MyCalendar 类来存放你的日程安排，你可以一直添加新的日程安排。
-
-MyCalendar 有一个 book(int start, int end)方法。它意味着在start到end时间内增加一个日程安排，注意，这里的时间是半开区间，即 [start, end), 实数 x 的范围为，  start <= x < end。当 K 个日程安排有一些时间上的交叉时（例如K个日程安排都在同一时间内），就会产生 K 次预订。每次调用 MyCalendar.book方法时，返回一个整数 K ，表示最大的 K 次预订。
-
-请按照以下步骤调用MyCalendar 类: MyCalendar cal = new MyCalendar(); MyCalendar.book(start, end)
-
-说明:
-
-- 每个测试用例，调用 MyCalendar.book 函数最多不超过 400 次。
-- 调用函数 MyCalendar.book(start, end)时， start 和 end 的取值范围为 [0, 10^9]。
-
-
-
-
-## 解题思路
-
-- 设计一个日程类，每添加一个日程，实时显示出当前排期中累计日程最多的个数，例如在一段时间内，排了 3 个日程，其他时间内都只有 0，1，2 个日程，则输出 3 。
-- 拿到这个题目以后会立即想到线段树。由于题目中只有增加日程，所以这一题难度不大。这一题和第 699 题也类似，但是有区别，第 699 题中，俄罗斯方块会依次摞起来，而这一题中，俄罗斯方块也就摞起来，但是方块下面如果是空挡，方块会断掉。举个例子：依次增加区间 [10,20]，[10,40]，[5,15]，[5,10]，如果是第 699 题的规则，这 [5,10] 的这块砖块会落在 [5,15] 上，从而使得高度为 4，但是这一题是日程，日程不一样，[5,15] 这个区间内有 3 个日程，但是其他部分都没有 3 个日程，所以第三块砖块 [5,15] 中的 [5,10] 会“断裂”，掉下去，第四块砖块还是 [5,10]，落在第三块砖块断落下去的位置，它们俩落在一起的高度是 2 。
-- 构造一颗线段树，这里用树来构造，如果用数组需要开辟很大的空间。当区间左右边界和查询边界完全相同的时候再累加技术，否则不加，继续划分区间。以区间的左边界作为划分区间的标准，因为区间左边界是开区间，右边是闭区间。一个区间的计数值以区间左边界的计数为准。还是上面的例子，[5,10) 计数以 5 为标准，count = 2，[10,15) 计数以 10 为标准，count = 3 。还需要再动态维护一个最大值。这个线段树的实现比较简单。
-- 类似的题目有：第 715 题，第 218 题，第 699 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和第 699 题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
diff --git a/Algorithms/0733. Flood Fill/README.md b/Algorithms/0733. Flood Fill/README.md
deleted file mode 100755
index 740e7faaf..000000000
--- a/Algorithms/0733. Flood Fill/README.md	
+++ /dev/null
@@ -1,50 +0,0 @@
-# [733. Flood Fill](https://leetcode.com/problems/flood-fill/)
-
-
-## 题目:
-
-An `image` is represented by a 2-D array of integers, each integer representing the pixel value of the image (from 0 to 65535).
-
-Given a coordinate `(sr, sc)` representing the starting pixel (row and column) of the flood fill, and a pixel value `newColor`, "flood fill" the image.
-
-To perform a "flood fill", consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color as the starting pixel), and so on. Replace the color of all of the aforementioned pixels with the newColor.
-
-At the end, return the modified image.
-
-**Example 1:**
-
-    Input: 
-    image = [[1,1,1],[1,1,0],[1,0,1]]
-    sr = 1, sc = 1, newColor = 2
-    Output: [[2,2,2],[2,2,0],[2,0,1]]
-    Explanation: 
-    From the center of the image (with position (sr, sc) = (1, 1)), all pixels connected 
-    by a path of the same color as the starting pixel are colored with the new color.
-    Note the bottom corner is not colored 2, because it is not 4-directionally connected
-    to the starting pixel.
-
-**Note:**
-
-- The length of `image` and `image[0]` will be in the range `[1, 50]`.
-- The given starting pixel will satisfy `0 <= sr < image.length` and `0 <= sc < image[0].length`.
-- The value of each color in `image[i][j]` and `newColor` will be an integer in `[0, 65535]`.
-
-
-## 题目大意
-
-有一幅以二维整数数组表示的图画，每一个整数表示该图画的像素值大小，数值在 0 到 65535 之间。给你一个坐标 (sr, sc) 表示图像渲染开始的像素值（行 ，列）和一个新的颜色值 newColor，让你重新上色这幅图像。
-
-为了完成上色工作，从初始坐标开始，记录初始坐标的上下左右四个方向上像素值与初始坐标相同的相连像素点，接着再记录这四个方向上符合条件的像素点与他们对应四个方向上像素值与初始坐标相同的相连像素点，……，重复该过程。将所有有记录的像素点的颜色值改为新的颜色值。最后返回经过上色渲染后的图像。
-
-注意:
-
-- image 和 image[0] 的长度在范围 [1, 50] 内。
-- 给出的初始点将满足 0 <= sr < image.length 和 0 <= sc < image[0].length。
-- image[i][j] 和 newColor 表示的颜色值在范围 [0, 65535]内。
-
-
-## 解题思路
-
-
-- 给出一个二维的图片点阵，每个点阵都有一个数字。给出一个起点坐标，要求从这个起点坐标开始，把所有与这个起点连通的点都染色成 newColor。
-- 这一题是标准的 Flood Fill 算法。可以用 DFS 也可以用 BFS 。
diff --git a/Algorithms/0748. Shortest Completing Word/README.md b/Algorithms/0748. Shortest Completing Word/README.md
deleted file mode 100755
index 94d227c2b..000000000
--- a/Algorithms/0748. Shortest Completing Word/README.md	
+++ /dev/null
@@ -1,55 +0,0 @@
-# [748. Shortest Completing Word](https://leetcode.com/problems/shortest-completing-word/)
-
-
-## 题目:
-
-Find the minimum length word from a given dictionary `words`, which has all the letters from the string `licensePlate`. Such a word is said to complete the given string `licensePlate`
-
-Here, for letters we ignore case. For example, `"P"` on the `licensePlate` still matches `"p"` on the word.
-
-It is guaranteed an answer exists. If there are multiple answers, return the one that occurs first in the array.
-
-The license plate might have the same letter occurring multiple times. For example, given a `licensePlate` of `"PP"`, the word `"pair"` does not complete the `licensePlate`, but the word `"supper"` does.
-
-**Example 1:**
-
-    Input: licensePlate = "1s3 PSt", words = ["step", "steps", "stripe", "stepple"]
-    Output: "steps"
-    Explanation: The smallest length word that contains the letters "S", "P", "S", and "T".
-    Note that the answer is not "step", because the letter "s" must occur in the word twice.
-    Also note that we ignored case for the purposes of comparing whether a letter exists in the word.
-
-**Example 2:**
-
-    Input: licensePlate = "1s3 456", words = ["looks", "pest", "stew", "show"]
-    Output: "pest"
-    Explanation: There are 3 smallest length words that contains the letters "s".
-    We return the one that occurred first.
-
-**Note:**
-
-1. `licensePlate` will be a string with length in range `[1, 7]`.
-2. `licensePlate` will contain digits, spaces, or letters (uppercase or lowercase).
-3. `words` will have a length in the range `[10, 1000]`.
-4. Every `words[i]` will consist of lowercase letters, and have length in range `[1, 15]`.
-
-
-## 题目大意
-
-如果单词列表（words）中的一个单词包含牌照（licensePlate）中所有的字母，那么我们称之为完整词。在所有完整词中，最短的单词我们称之为最短完整词。
-
-单词在匹配牌照中的字母时不区分大小写，比如牌照中的 "P" 依然可以匹配单词中的 "p" 字母。我们保证一定存在一个最短完整词。当有多个单词都符合最短完整词的匹配条件时取单词列表中最靠前的一个。牌照中可能包含多个相同的字符，比如说：对于牌照 "PP"，单词 "pair" 无法匹配，但是 "supper" 可以匹配。
-
-注意:
-
-- 牌照（licensePlate）的长度在区域[1, 7]中。
-- 牌照（licensePlate）将会包含数字、空格、或者字母（大写和小写）。
-- 单词列表（words）长度在区间 [10, 1000] 中。
-- 每一个单词 words[i] 都是小写，并且长度在区间 [1, 15] 中。
-
-
-
-## 解题思路
-
-
-- 给出一个数组，要求找出能包含 `licensePlate` 字符串中所有字符的最短长度的字符串。如果最短长度的字符串有多个，输出 word 下标小的那个。这一题也是简单题，不过有 2 个需要注意的点，第一点，`licensePlate` 中可能包含 `Unicode` 任意的字符，所以要先把字母的字符筛选出来，第二点是题目中保证了一定存在一个最短的单词能满足题意，并且忽略大小写。具体做法按照题意模拟即可。
diff --git a/Algorithms/0756. Pyramid Transition Matrix/README.md b/Algorithms/0756. Pyramid Transition Matrix/README.md
deleted file mode 100755
index 445ab0e6f..000000000
--- a/Algorithms/0756. Pyramid Transition Matrix/README.md	
+++ /dev/null
@@ -1,57 +0,0 @@
-# [756. Pyramid Transition Matrix](https://leetcode.com/problems/pyramid-transition-matrix/)
-
-
-## 题目:
-
-We are stacking blocks to form a pyramid. Each block has a color which is a one letter string.
-
-We are allowed to place any color block `C` on top of two adjacent blocks of colors `A` and `B`, if and only if `ABC` is an allowed triple.
-
-We start with a bottom row of `bottom`, represented as a single string. We also start with a list of allowed triples `allowed`. Each allowed triple is represented as a string of length 3.
-
-Return true if we can build the pyramid all the way to the top, otherwise false.
-
-**Example 1:**
-
-    Input: bottom = "BCD", allowed = ["BCG", "CDE", "GEA", "FFF"]
-    Output: true
-    Explanation:
-    We can stack the pyramid like this:
-        A
-       / \
-      G   E
-     / \ / \
-    B   C   D
-    
-    We are allowed to place G on top of B and C because BCG is an allowed triple.  Similarly, we can place E on top of C and D, then A on top of G and E.
-
-**Example 2:**
-
-    Input: bottom = "AABA", allowed = ["AAA", "AAB", "ABA", "ABB", "BAC"]
-    Output: false
-    Explanation:
-    We can't stack the pyramid to the top.
-    Note that there could be allowed triples (A, B, C) and (A, B, D) with C != D.
-
-**Note:**
-
-1. `bottom` will be a string with length in range `[2, 8]`.
-2. `allowed` will have length in range `[0, 200]`.
-3. Letters in all strings will be chosen from the set `{'A', 'B', 'C', 'D', 'E', 'F', 'G'}`.
-
-
-## 题目大意
-
-现在，我们用一些方块来堆砌一个金字塔。 每个方块用仅包含一个字母的字符串表示，例如 “Z”。使用三元组表示金字塔的堆砌规则如下：
-
-(A, B, C) 表示，“C” 为顶层方块，方块 “A”、“B” 分别作为方块 “C” 下一层的的左、右子块。当且仅当(A, B, C)是被允许的三元组，我们才可以将其堆砌上。
-
-初始时，给定金字塔的基层 bottom，用一个字符串表示。一个允许的三元组列表 allowed，每个三元组用一个长度为 3 的字符串表示。如果可以由基层一直堆到塔尖返回 true，否则返回 false。
-
-
-
-## 解题思路
-
-- 这一题是一道 DFS 的题目。题目给出金字塔的底座字符串。然后还会给一个字符串数组，字符串数组里面代表的字符串的砖块。砖块是 3 个字符串组成的。前两个字符代表的是砖块的底边，后一个字符代表的是砖块的顶部。问给出的字符能拼成一个金字塔么？金字塔的特点是顶端就一个字符。  
-
-- 这一题用 DFS 深搜每个砖块，从底层砖块开始逐渐往上层码。每递归一层，新一层底部的砖块都会变。当递归到了一层底部只有 2 个字符，顶部只有一个字符的时候，就到金字塔顶端了，就算是完成了。这一题为了挑选合适的砖块，需要把每个砖块底部的 2 个字符作为 key 放进 map 中，加速查找。题目中也给出了特殊情况，相同底部可能存在多种砖块，所以一个 key 可能对应多个 value 的情况，即可能存在多个顶部砖块的情况。这种情况在递归遍历中需要考虑。
diff --git a/Algorithms/0762. Prime Number of Set Bits in Binary Representation/README.md b/Algorithms/0762. Prime Number of Set Bits in Binary Representation/README.md
deleted file mode 100755
index 1c81d5cd4..000000000
--- a/Algorithms/0762. Prime Number of Set Bits in Binary Representation/README.md	
+++ /dev/null
@@ -1,53 +0,0 @@
-# [762. Prime Number of Set Bits in Binary Representation](https://leetcode.com/problems/prime-number-of-set-bits-in-binary-representation/)
-
-
-## 题目:
-
-Given two integers `L` and `R`, find the count of numbers in the range `[L, R]` (inclusive) having a prime number of set bits in their binary representation.
-
-(Recall that the number of set bits an integer has is the number of `1`s present when written in binary. For example, `21` written in binary is `10101` which has 3 set bits. Also, 1 is not a prime.)
-
-**Example 1:**
-
-    Input: L = 6, R = 10
-    Output: 4
-    Explanation:
-    6 -> 110 (2 set bits, 2 is prime)
-    7 -> 111 (3 set bits, 3 is prime)
-    9 -> 1001 (2 set bits , 2 is prime)
-    10->1010 (2 set bits , 2 is prime)
-
-**Example 2:**
-
-    Input: L = 10, R = 15
-    Output: 5
-    Explanation:
-    10 -> 1010 (2 set bits, 2 is prime)
-    11 -> 1011 (3 set bits, 3 is prime)
-    12 -> 1100 (2 set bits, 2 is prime)
-    13 -> 1101 (3 set bits, 3 is prime)
-    14 -> 1110 (3 set bits, 3 is prime)
-    15 -> 1111 (4 set bits, 4 is not prime)
-
-**Note:**
-
-1. `L, R` will be integers `L <= R` in the range `[1, 10^6]`.
-2. `R - L` will be at most 10000.
-
-
-## 题目大意
-
-给定两个整数 L 和 R ，找到闭区间 [L, R] 范围内，计算置位位数为质数的整数个数。（注意，计算置位代表二进制表示中1的个数。例如 21 的二进制表示 10101 有 3 个计算置位。还有，1 不是质数。）
-
-
-注意:
-
-- L, R 是 L <= R 且在 [1, 10^6] 中的整数。
-- R - L 的最大值为 10000。
-
-
-
-## 解题思路
-
-
-- 题目给出 `[L, R]` 区间，在这个区间内的每个整数的二进制表示中 1 的个数如果是素数，那么最终结果就加一，问最终结果是多少？这一题是一个组合题，判断一个数的二进制位有多少位 1，是第 191 题。题目中限定了区间最大不超过 10^6 ，所以 1 的位数最大是 19 位，也就是说素数最大就是 19 。那么素数可以有限枚举出来。最后按照题目的意思累积结果就可以了。
diff --git a/Algorithms/0763. Partition Labels/README.md b/Algorithms/0763. Partition Labels/README.md
deleted file mode 100644
index 68ecff8a0..000000000
--- a/Algorithms/0763. Partition Labels/README.md	
+++ /dev/null
@@ -1,37 +0,0 @@
-# [763. Partition Labels](https://leetcode.com/problems/partition-labels/)
-
-## 题目
-
-A string S of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.
-
-
-
-Example 1:
-
-
-```c
-Input: S = "ababcbacadefegdehijhklij"
-Output: [9,7,8]
-Explanation:
-The partition is "ababcbaca", "defegde", "hijhklij".
-This is a partition so that each letter appears in at most one part.
-A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.
-```
-
-Note:
-
-- S will have length in range [1, 500].
-- S will consist of lowercase letters ('a' to 'z') only.
-
-
-## 题目大意
-
-这道题考察的是滑动窗口的问题。
-
-给出一个字符串，要求输出满足条件窗口的长度，条件是在这个窗口内，字母中出现在这一个窗口内，不出现在其他窗口内。
-
-## 解题思路
-
-这一题有 2 种思路，第一种思路是先记录下每个字母的出现次数，然后对滑动窗口中的每个字母判断次数是否用尽为 0，如果这个窗口内的所有字母次数都为 0，这个窗口就是符合条件的窗口。时间复杂度为 O(n^2)
-
-另外一种思路是记录下每个字符最后一次出现的下标，这样就不用记录次数。在每个滑动窗口中，依次判断每个字母最后一次出现的位置，如果在一个下标内，所有字母的最后一次出现的位置都包含进来了，那么这个下标就是这个满足条件的窗口大小。时间复杂度为 O(n^2)
diff --git a/Algorithms/0765. Couples Holding Hands/README.md b/Algorithms/0765. Couples Holding Hands/README.md
deleted file mode 100755
index e833ead32..000000000
--- a/Algorithms/0765. Couples Holding Hands/README.md	
+++ /dev/null
@@ -1,63 +0,0 @@
-# [765. Couples Holding Hands](https://leetcode.com/problems/couples-holding-hands/)
-
-
-## 题目:
-
-N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum number of swaps so that every couple is sitting side by side. A swap consists of choosing **any** two people, then they stand up and switch seats.
-
-The people and seats are represented by an integer from `0` to `2N-1`, the couples are numbered in order, the first couple being `(0, 1)`, the second couple being `(2, 3)`, and so on with the last couple being `(2N-2, 2N-1)`.
-
-The couples' initial seating is given by `row[i]` being the value of the person who is initially sitting in the i-th seat.
-
-**Example 1:**
-
-    Input: row = [0, 2, 1, 3]
-    Output: 1
-    Explanation: We only need to swap the second (row[1]) and third (row[2]) person.
-
-**Example 2:**
-
-    Input: row = [3, 2, 0, 1]
-    Output: 0
-    Explanation: All couples are already seated side by side.
-
-**Note:**
-
-1. `len(row)` is even and in the range of `[4, 60]`.
-2. `row` is guaranteed to be a permutation of `0...len(row)-1`.
-
-
-## 题目大意
-
-N 对情侣坐在连续排列的 2N 个座位上，想要牵到对方的手。 计算最少交换座位的次数，以便每对情侣可以并肩坐在一起。 一次交换可选择任意两人，让他们站起来交换座位。人和座位用 0 到 2N-1 的整数表示，情侣们按顺序编号，第一对是 (0, 1)，第二对是 (2, 3)，以此类推，最后一对是 (2N-2, 2N-1)。这些情侣的初始座位 row[i] 是由最初始坐在第 i 个座位上的人决定的。
-
-说明:
-
-1. len(row) 是偶数且数值在 [4, 60]范围内。  
-2. 可以保证 row 是序列 0...len(row)-1 的一个全排列。  
-
-
-## 解题思路
-
-- 给出一个数组，数组里面两两相邻的元素代表一对情侣。情侣编号是从 0 开始的：0 和 1 是情侣，2 和 3 是情侣……这些情侣坐在一排，但是并非成对坐着一起的，问如何用最小的次数交换座位以后，情侣能两两坐在一起。
-- 这道题的突破口是如何找到最小的交换次数。乍一想可能没有思路。直觉告诉我们，这种难题，很可能最后推出来的结论，或者公式是一个很简单的式子。(事实此题确实是这种情况)先不考虑最小交换次数，用正常的方法来处理这道题。举个例子：【3 1 4 0 2 5】，从数组 0 下标开始往后扫。
-
-        初始状态
-        
-        集合 0：0，1
-        集合 1：2，3
-        集合 2：4，5
-
-    3 和 1 不是情侣，将 3 和 1 所在集合 `union()` 起来。3 所在集合是 1 ，1 所在集合是 0，将 0 和 1 号集合 `union()` 起来。因为情侣 0 和情侣 1 是集合 0 ，情侣 2 和情侣 3 是集合 1，以此类推。
-
-        集合 0 和 1：0，1，2，3
-        集合 2：4，5
-
-- 继续往后扫，4 和 0 不在同一个集合，4 在集合 3，0 在集合 0，那么把它们 `union()` 起来。
-
-        集合 0 和 1 和 2：0，1，2，3，4，5
-
-    在上面集合合并的过程中，合并了 2 次。那么就代表最少需要交换 2 次。也可以通过 `len(row)/2 - uf.count` 来计算。`len(row)/2` 是初始集合总数，`uf.count` 是最后剩下的集合数，两者相减就是中间交换的次数。
-
-- 最后实现的代码非常简单。并查集先相邻的两两元素 `union()` 在一起。然后扫原数组，每次扫相邻的两个，通过这两个元素值所在集合，进行 `union()`。扫完以后就可以得到最后的答案。
-- 回过头来看这道题，为什么我们从数组开头往后依次调整每一对情侣，这样交换的次数是最少的呢？其实这个方法的思想是贪心思想。从头开始往后一对一对的调整，就是可以最终做到次数最少。(具体证明笔者不会)交换到最后，最后一对情侣一定是正确的，无须交换。(因为前面每一对都调整完了，最后一对一定是正确的)
diff --git a/Algorithms/0771. Jewels and Stones/README.md b/Algorithms/0771. Jewels and Stones/README.md
deleted file mode 100755
index 9b7ac381b..000000000
--- a/Algorithms/0771. Jewels and Stones/README.md	
+++ /dev/null
@@ -1,38 +0,0 @@
-# [771. Jewels and Stones](https://leetcode.com/problems/jewels-and-stones/)
-
-
-
-## 题目:
-
-You're given strings `J` representing the types of stones that are jewels, and `S` representing the stones you have. Each character in `S` is a type of stone you have. You want to know how many of the stones you have are also jewels.
-
-The letters in `J` are guaranteed distinct, and all characters in `J` and `S` are letters. Letters are case sensitive, so `"a"` is considered a different type of stone from `"A"`.
-
-**Example 1:**
-
-    Input: J = "aA", S = "aAAbbbb"
-    Output: 3
-
-**Example 2:**
-
-    Input: J = "z", S = "ZZ"
-    Output: 0
-
-**Note:**
-
-- `S` and `J` will consist of letters and have length at most 50.
-- The characters in `J` are distinct.
-
-
-## 题目大意
-
-给定字符串 J 代表石头中宝石的类型，和字符串 S 代表你拥有的石头。S 中每个字符代表了一种你拥有的石头的类型，你想知道你拥有的石头中有多少是宝石。
-
-J 中的字母不重复，J 和 S 中的所有字符都是字母。字母区分大小写，因此 "a" 和 "A" 是不同类型的石头。
-
-
-
-## 解题思路
-
-
-- 给出 2 个字符串，要求在 S 字符串中找出在 J 字符串里面出现的字符个数。这是一道简单题。
diff --git a/Algorithms/0778. Swim in Rising Water/778. Swim in Rising Water_test.go b/Algorithms/0778. Swim in Rising Water/778. Swim in Rising Water_test.go
deleted file mode 100644
index d78aeff25..000000000
--- a/Algorithms/0778. Swim in Rising Water/778. Swim in Rising Water_test.go	
+++ /dev/null
@@ -1,46 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question778 struct {
-	para778
-	ans778
-}
-
-// para 是参数
-// one 代表第一个参数
-type para778 struct {
-	grid [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans778 struct {
-	one int
-}
-
-func Test_Problem778(t *testing.T) {
-
-	qs := []question778{
-		question778{
-			para778{[][]int{[]int{0, 2}, []int{1, 3}}},
-			ans778{3},
-		},
-
-		question778{
-			para778{[][]int{[]int{0, 1, 2, 3, 4}, []int{24, 23, 22, 21, 5}, []int{12, 13, 14, 15, 16}, []int{11, 17, 18, 19, 20}, []int{10, 9, 8, 7, 6}}},
-			ans778{16},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 778------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans778, q.para778
-		fmt.Printf("【input】:%v       【output】:%v\n", p, swimInWater(p.grid))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0778. Swim in Rising Water/README.md b/Algorithms/0778. Swim in Rising Water/README.md
deleted file mode 100755
index 237e932e7..000000000
--- a/Algorithms/0778. Swim in Rising Water/README.md	
+++ /dev/null
@@ -1,59 +0,0 @@
-# [778. Swim in Rising Water](https://leetcode.com/problems/swim-in-rising-water/)
-
-
-## 题目:
-
-On an N x N `grid`, each square `grid[i][j]` represents the elevation at that point `(i,j)`.
-
-Now rain starts to fall. At time `t`, the depth of the water everywhere is `t`. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most `t`. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.
-
-You start at the top left square `(0, 0)`. What is the least time until you can reach the bottom right square `(N-1, N-1)`?
-
-**Example 1:**
-
-    Input: [[0,2],[1,3]]
-    Output: 3
-    Explanation:
-    At time 0, you are in grid location (0, 0).
-    You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
-    
-    You cannot reach point (1, 1) until time 3.
-    When the depth of water is 3, we can swim anywhere inside the grid.
-
-**Example 2:**
-
-    Input: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
-    Output: 16
-    Explanation:
-     0  1  2  3  4
-    24 23 22 21  5
-    12 13 14 15 16
-    11 17 18 19 20
-    10  9  8  7  6
-    
-    The final route is marked in bold.
-    We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
-
-**Note:**
-
-1. `2 <= N <= 50`.
-2. grid[i][j] is a permutation of [0, ..., N*N - 1].
-
-## 题目大意
-
-
-在一个 N x N 的坐标方格 grid 中，每一个方格的值 grid[i][j] 表示在位置 (i,j) 的平台高度。现在开始下雨了。当时间为 t 时，此时雨水导致水池中任意位置的水位为 t 。你可以从一个平台游向四周相邻的任意一个平台，但是前提是此时水位必须同时淹没这两个平台。假定你可以瞬间移动无限距离，也就是默认在方格内部游动是不耗时的。当然，在你游泳的时候你必须待在坐标方格里面。
-
-你从坐标方格的左上平台 (0，0) 出发。最少耗时多久你才能到达坐标方格的右下平台 (N-1, N-1)？
-
-提示:
-
-- 2 <= N <= 50.
-- grid[i][j] 位于区间 [0, ..., N*N - 1] 内。
-
-
-## 解题思路
-
-- 给出一个 grid[i][j] 方格，每个格子里面表示游泳池里面平台的高度。t 时刻，游泳池中的水的高度是 t。只有水的高度到达了平台的高度以后才能游过去。问从 (0,0) 开始，最短多长时间能到达 (N-1, N-1) 。
-- 这一题有多种解法。第一种解题思路是利用 DFS + 二分。DFS 是用来遍历是否可达。利用时间(即当前水淹过的高度)来判断是否能到达终点 (N-1, N-1) 点。二分用来搜索最终结果的时间。为什么会考虑用二分加速呢？原因是：时间从 0 - max 依次递增。max 是游泳池最高的平台高度。当时间从 0 增加到 max 以后，肯定能到达终点 (N-1, N-1) 点，因为水比所有平台都要高了。想快速找到一个时间 t 能使得 (0,0) 点和 (N-1, N-1) 点之间连通，那么就想到用二分加速了。判断是否取中值的条件是 (0,0) 点和 (N-1, N-1) 点之间是否连通。
-- 第二种解题思路是并查集。只要是 (0,0) 点和 (N-1, N-1) 点没有连通，即不能游到终点，那么就开始 `union()` 操作，由于起点是 (0,0)，所以向右边 `i + 1` 和向下边 `j + 1` 开始尝试。每尝试完一轮，时间会加 1 秒，即高度会加一。直到 (0,0) 点和 (N-1, N-1) 点刚好连通，那么这个时间点就是最终要求的。
\ No newline at end of file
diff --git a/Algorithms/0781. Rabbits in Forest/README.md b/Algorithms/0781. Rabbits in Forest/README.md
deleted file mode 100755
index c6c5576b0..000000000
--- a/Algorithms/0781. Rabbits in Forest/README.md	
+++ /dev/null
@@ -1,46 +0,0 @@
-# [781. Rabbits in Forest](https://leetcode.com/problems/rabbits-in-forest/)
-
-
-## 题目:
-
-In a forest, each rabbit has some color. Some subset of rabbits (possibly all of them) tell you how many other rabbits have the same color as them. Those `answers` are placed in an array.
-
-Return the minimum number of rabbits that could be in the forest.
-
-    Examples:
-    Input: answers = [1, 1, 2]
-    Output: 5
-    Explanation:
-    The two rabbits that answered "1" could both be the same color, say red.
-    The rabbit than answered "2" can't be red or the answers would be inconsistent.
-    Say the rabbit that answered "2" was blue.
-    Then there should be 2 other blue rabbits in the forest that didn't answer into the array.
-    The smallest possible number of rabbits in the forest is therefore 5: 3 that answered plus 2 that didn't.
-    
-    Input: answers = [10, 10, 10]
-    Output: 11
-    
-    Input: answers = []
-    Output: 0
-
-**Note:**
-
-1. `answers` will have length at most `1000`.
-2. Each `answers[i]` will be an integer in the range `[0, 999]`.
-
-
-## 题目大意
-
-森林中，每个兔子都有颜色。其中一些兔子（可能是全部）告诉你还有多少其他的兔子和自己有相同的颜色。我们将这些回答放在 answers 数组里。返回森林中兔子的最少数量。
-
-说明:
-
-- answers 的长度最大为1000。
-- answers[i] 是在 [0, 999] 范围内的整数。
-
-
-## 解题思路
-
-
-- 给出一个数组，数组里面代表的是每个兔子说自己同类还有多少个。要求输出总共有多少只兔子。数字中可能兔子汇报的人数小于总兔子数。
-- 这一题关键在于如何划分不同种类的兔子，有可能相同种类的兔子的个数是一样的，比如 `[2,2,2,2,2,2]`，这其实是 3 个种类，总共 6 只兔子。用 map 去重相同种类的兔子，不断的减少，当有种类的兔子为 0 以后，还有该种类的兔子报数，需要当做另外一个种类的兔子来看待。
diff --git a/Algorithms/0803. Bricks Falling When Hit/803. Bricks Falling When Hit_test.go b/Algorithms/0803. Bricks Falling When Hit/803. Bricks Falling When Hit_test.go
deleted file mode 100644
index 35bfa3de7..000000000
--- a/Algorithms/0803. Bricks Falling When Hit/803. Bricks Falling When Hit_test.go	
+++ /dev/null
@@ -1,58 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question803 struct {
-	para803
-	ans803
-}
-
-// para 是参数
-// one 代表第一个参数
-type para803 struct {
-	grid [][]int
-	hits [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans803 struct {
-	one []int
-}
-
-func Test_Problem803(t *testing.T) {
-
-	qs := []question803{
-
-		question803{
-			para803{[][]int{[]int{1, 1, 1, 0, 1, 1, 1, 1}, []int{1, 0, 0, 0, 0, 1, 1, 1}, []int{1, 1, 1, 0, 0, 0, 1, 1}, []int{1, 1, 0, 0, 0, 0, 0, 0}, []int{1, 0, 0, 0, 0, 0, 0, 0}, []int{1, 0, 0, 0, 0, 0, 0, 0}},
-				[][]int{[]int{4, 6}, []int{3, 0}, []int{2, 3}, []int{2, 6}, []int{4, 1}, []int{5, 2}, []int{2, 1}}},
-			ans803{[]int{0, 2, 0, 0, 0, 0, 2}},
-		},
-
-		question803{
-			para803{[][]int{[]int{1, 0, 1}, []int{1, 1, 1}}, [][]int{[]int{0, 0}, []int{0, 2}, []int{1, 1}}},
-			ans803{[]int{0, 3, 0}},
-		},
-		question803{
-			para803{[][]int{[]int{1, 0, 0, 0}, []int{1, 1, 1, 0}}, [][]int{[]int{1, 0}}},
-			ans803{[]int{2}},
-		},
-
-		question803{
-			para803{[][]int{[]int{1, 0, 0, 0}, []int{1, 1, 0, 0}}, [][]int{[]int{1, 1}, []int{1, 0}}},
-			ans803{[]int{0, 0}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 803------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans803, q.para803
-		fmt.Printf("【input】:%v       【output】:%v\n", p, hitBricks(p.grid, p.hits))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0803. Bricks Falling When Hit/README.md b/Algorithms/0803. Bricks Falling When Hit/README.md
deleted file mode 100755
index 744d86e75..000000000
--- a/Algorithms/0803. Bricks Falling When Hit/README.md	
+++ /dev/null
@@ -1,56 +0,0 @@
-# [803. Bricks Falling When Hit](https://leetcode.com/problems/bricks-falling-when-hit/)
-
-
-## 题目:
-
-We have a grid of 1s and 0s; the 1s in a cell represent bricks. A brick will not drop if and only if it is directly connected to the top of the grid, or at least one of its (4-way) adjacent bricks will not drop.
-
-We will do some erasures sequentially. Each time we want to do the erasure at the location (i, j), the brick (if it exists) on that location will disappear, and then some other bricks may drop because of that erasure.
-
-Return an array representing the number of bricks that will drop after each erasure in sequence.
-
-    Example 1:
-    Input: 
-    grid = [[1,0,0,0],[1,1,1,0]]
-    hits = [[1,0]]
-    Output: [2]
-    Explanation: 
-    If we erase the brick at (1, 0), the brick at (1, 1) and (1, 2) will drop. So we should return 2.
-
-    Example 2:
-    Input: 
-    grid = [[1,0,0,0],[1,1,0,0]]
-    hits = [[1,1],[1,0]]
-    Output: [0,0]
-    Explanation: 
-    When we erase the brick at (1, 0), the brick at (1, 1) has already disappeared due to the last move. So each erasure will cause no bricks dropping.  Note that the erased brick (1, 0) will not be counted as a dropped brick.
-
-**Note:**
-
-- The number of rows and columns in the grid will be in the range [1, 200].
-- The number of erasures will not exceed the area of the grid.
-- It is guaranteed that each erasure will be different from any other erasure, and located inside the grid.
-- An erasure may refer to a location with no brick - if it does, no bricks drop.
-
-
-## 题目大意
-
-我们有一组包含1和0的网格；其中1表示砖块。 当且仅当一块砖直接连接到网格的顶部，或者它至少有一块相邻（4 个方向之一）砖块不会掉落时，它才不会落下。我们会依次消除一些砖块。每当我们消除 (i, j) 位置时， 对应位置的砖块（若存在）会消失，然后其他的砖块可能因为这个消除而落下。返回一个数组表示每次消除操作对应落下的砖块数目。
-
-
-注意:
-
-- 网格的行数和列数的范围是[1, 200]。
-- 消除的数字不会超过网格的区域。
-- 可以保证每次的消除都不相同，并且位于网格的内部。
-- 一个消除的位置可能没有砖块，如果这样的话，就不会有砖块落下。
-
-
-
-## 解题思路
-
-
-- 有一些砖块连接在天花板上，问，如果打掉某个砖块，会掉落几块砖块？打掉的每个砖块不参与计数。
-- 这一题可以用并查集和 DFS 求解。不过尝试用 DFS 的同学就会知道，这一题卡时间卡的很紧。用 DFS 虽然能 AC，但是耗时非常长。用并查集也必须进行秩压缩，不然耗时也非常长。另外，如果用了并查集，每个集合的总数单独统计，不随着 union() 操作，也会导致超时，笔者在这里被 LTE 了多次，最后只能重写 UnionFind 并查集类，将统计操作和 union() 操作写在一起，这一题才 faster than 100.00% AC。
-- 拿到题以后，首先尝试暴力解法，按照顺序打掉砖块，每次打掉砖块以后，都重建并查集。题目要求每次掉落几块砖块，实际上比较每次和天花板连通的砖块个数变化了多少块就可以了。那么解法就出来了，先把和天花板连通的砖块都 union() 起来，记录这个集合中砖块的个数 `count`，然后每次打掉一个砖块以后，重建并查集，计算与天花板连通的砖块的个数 `newCount`，`newCount - count -1` 就是最终答案(打掉的那块砖块不计算其中)，提交代码以后，发现 TLE。
-- 出现 TLE 以后一般思路都是对的，只是时间复杂度过高，需要优化。很明显，需要优化的地方是每次都重建了新的并查集，有没有办法能在上一次状态上进行变更，不用重建并查集呢？如果正向的打掉砖块，那么每次还需要以这个砖块为起点进行 DFS，时间复杂度还是很高。如果反向考虑呢？先把所有要打掉的砖块都打掉，构建打掉这些砖块以后剩下与天花板连通的并查集。然后反向添加打掉的砖块，每次添加一块就刷新一次它周围的 4 个砖块，不用 DFS，这样时间复杂度优化了很多。最后在按照 `newCount - count -1` 方式计算最终答案。注意每次还原一个砖块的时候需要染色回原有砖块的颜色 `1` 。优化成这样的做法，基本不会 TLE 了，如果计算 count 是单独计算的，还是会 TLE。如果没有进行秩压缩，时间会超过 1500 ms，所以这一题想拿到 100%，每步优化都要做好。最终 100% 的答案见代码。
diff --git a/Algorithms/0811. Subdomain Visit Count/README.md b/Algorithms/0811. Subdomain Visit Count/README.md
deleted file mode 100755
index e49339428..000000000
--- a/Algorithms/0811. Subdomain Visit Count/README.md	
+++ /dev/null
@@ -1,49 +0,0 @@
-# [811. Subdomain Visit Count](https://leetcode.com/problems/subdomain-visit-count/)
-
-
-## 题目:
-
-A website domain like "discuss.leetcode.com" consists of various subdomains. At the top level, we have "com", at the next level, we have "leetcode.com", and at the lowest level, "discuss.leetcode.com". When we visit a domain like "discuss.leetcode.com", we will also visit the parent domains "leetcode.com" and "com" implicitly.
-
-Now, call a "count-paired domain" to be a count (representing the number of visits this domain received), followed by a space, followed by the address. An example of a count-paired domain might be "9001 discuss.leetcode.com".
-
-We are given a list `cpdomains` of count-paired domains. We would like a list of count-paired domains, (in the same format as the input, and in any order), that explicitly counts the number of visits to each subdomain.
-
-    Example 1:
-    Input: 
-    ["9001 discuss.leetcode.com"]
-    Output: 
-    ["9001 discuss.leetcode.com", "9001 leetcode.com", "9001 com"]
-    Explanation: 
-    We only have one website domain: "discuss.leetcode.com". As discussed above, the subdomain "leetcode.com" and "com" will also be visited. So they will all be visited 9001 times.
-
-    Example 2:
-    Input: 
-    ["900 google.mail.com", "50 yahoo.com", "1 intel.mail.com", "5 wiki.org"]
-    Output: 
-    ["901 mail.com","50 yahoo.com","900 google.mail.com","5 wiki.org","5 org","1 intel.mail.com","951 com"]
-    Explanation: 
-    We will visit "google.mail.com" 900 times, "yahoo.com" 50 times, "intel.mail.com" once and "wiki.org" 5 times. For the subdomains, we will visit "mail.com" 900 + 1 = 901 times, "com" 900 + 50 + 1 = 951 times, and "org" 5 times.
-
-**Notes:**
-
-- The length of `cpdomains` will not exceed `100`.
-- The length of each domain name will not exceed `100`.
-- Each address will have either 1 or 2 "." characters.
-- The input count in any count-paired domain will not exceed `10000`.
-- The answer output can be returned in any order.
-
-
-## 题目大意
-
-
-一个网站域名，如 "discuss.leetcode.com"，包含了多个子域名。作为顶级域名，常用的有 "com"，下一级则有 "leetcode.com"，最低的一级为 "discuss.leetcode.com"。当我们访问域名 "discuss.leetcode.com" 时，也同时访问了其父域名 "leetcode.com" 以及顶级域名 "com"。给定一个带访问次数和域名的组合，要求分别计算每个域名被访问的次数。其格式为访问次数+空格+地址，例如："9001 discuss.leetcode.com"。
-
-接下来会给出一组访问次数和域名组合的列表 cpdomains 。要求解析出所有域名的访问次数，输出格式和输入格式相同，不限定先后顺序。
-
-
-
-## 解题思路
-
-
-- 这一题是简单题，统计每个 domain 的出现频次。每个域名根据层级，一级一级的累加频次，比如 `discuss.leetcode.com`、`discuss.leetcode.com` 这个域名频次为 1，`leetcode.com` 这个域名频次为 1，`com` 这个域名频次为 1。用 map 依次统计每个 domain 出现的频次，按照格式要求输出。
diff --git a/Algorithms/0815. Bus Routes/README.md b/Algorithms/0815. Bus Routes/README.md
deleted file mode 100755
index 59c8b1b90..000000000
--- a/Algorithms/0815. Bus Routes/README.md	
+++ /dev/null
@@ -1,41 +0,0 @@
-# [815. Bus Routes](https://leetcode.com/problems/bus-routes/)
-
-
-## 题目:
-
-We have a list of bus routes. Each `routes[i]` is a bus route that the i-th bus repeats forever. For example if `routes[0] = [1, 5, 7]`, this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->... forever.
-
-We start at bus stop `S` (initially not on a bus), and we want to go to bus stop `T`. Travelling by buses only, what is the least number of buses we must take to reach our destination? Return -1 if it is not possible.
-
-    Example:
-    Input: 
-    routes = [[1, 2, 7], [3, 6, 7]]
-    S = 1
-    T = 6
-    Output: 2
-    Explanation: 
-    The best strategy is take the first bus to the bus stop 7, then take the second bus to the bus stop 6.
-
-**Note:**
-
-- `1 <= routes.length <= 500`.
-- `1 <= routes[i].length <= 500`.
-- `0 <= routes[i][j] < 10 ^ 6`.
-
-
-## 题目大意
-
-我们有一系列公交路线。每一条路线 routes[i] 上都有一辆公交车在上面循环行驶。例如，有一条路线 routes[0] = [1, 5, 7]，表示第一辆 (下标为0) 公交车会一直按照 1->5->7->1->5->7->1->... 的车站路线行驶。假设我们从 S 车站开始（初始时不在公交车上），要去往 T 站。 期间仅可乘坐公交车，求出最少乘坐的公交车数量。返回 -1 表示不可能到达终点车站。
-
-
-说明:
-
-- 1 <= routes.length <= 500.
-- 1 <= routes[i].length <= 500.
-- 0 <= routes[i][j] < 10 ^ 6.
-
-
-## 解题思路
-
-- 给出一些公交路线，公交路径代表经过的哪些站。现在给出起点和终点站，问最少需要换多少辆公交车才能从起点到终点？
-- 这一题可以转换成图论的问题，将每个站台看成顶点，公交路径看成每个顶点的边。同一个公交的边染色相同。题目即可转化为从顶点 S 到顶点 T 需要经过最少多少条不同的染色边。用 BFS 即可轻松解决。从起点 S 开始，不断的扩展它能到达的站点。用 visited 数组防止放入已经可达的站点引起的环。用 map 存储站点和公交车的映射关系(即某个站点可以由哪些公交车到达)，BFS 的过程中可以用这个映射关系，拿到公交车的其他站点信息，从而扩张队列里面的可达站点。一旦扩展出现了终点 T，就可以返回结果了。
diff --git a/Algorithms/0819. Most Common Word/README.md b/Algorithms/0819. Most Common Word/README.md
deleted file mode 100755
index ec3713736..000000000
--- a/Algorithms/0819. Most Common Word/README.md	
+++ /dev/null
@@ -1,44 +0,0 @@
-# [819. Most Common Word](https://leetcode.com/problems/most-common-word/)
-
-
-## 题目:
-
-Given a paragraph and a list of banned words, return the most frequent word that is not in the list of banned words. It is guaranteed there is at least one word that isn't banned, and that the answer is unique.
-
-Words in the list of banned words are given in lowercase, and free of punctuation. Words in the paragraph are not case sensitive. The answer is in lowercase.
-
-**Example:**
-
-    Input: 
-    paragraph = "Bob hit a ball, the hit BALL flew far after it was hit."
-    banned = ["hit"]
-    Output: "ball"
-    Explanation: 
-    "hit" occurs 3 times, but it is a banned word.
-    "ball" occurs twice (and no other word does), so it is the most frequent non-banned word in the paragraph. 
-    Note that words in the paragraph are not case sensitive,
-    that punctuation is ignored (even if adjacent to words, such as "ball,"), 
-    and that "hit" isn't the answer even though it occurs more because it is banned.
-
-**Note:**
-
-- `1 <= paragraph.length <= 1000`.
-- `0 <= banned.length <= 100`.
-- `1 <= banned[i].length <= 10`.
-- The answer is unique, and written in lowercase (even if its occurrences in `paragraph` may have uppercase symbols, and even if it is a proper noun.)
-- `paragraph` only consists of letters, spaces, or the punctuation symbols `!?',;.`
-- There are no hyphens or hyphenated words.
-- Words only consist of letters, never apostrophes or other punctuation symbols.
-
-
-## 题目大意
-
-
-给定一个段落 (paragraph) 和一个禁用单词列表 (banned)。返回出现次数最多，同时不在禁用列表中的单词。题目保证至少有一个词不在禁用列表中，而且答案唯一。
-
-禁用列表中的单词用小写字母表示，不含标点符号。段落中的单词不区分大小写。答案都是小写字母。
-
-
-## 解题思路
-
-- 给出一段话和一个 banned 的字符串，要求输出这段话中出现频次最高的并且不出现在 banned 数组里面的字符串，答案唯一。这题是简单题，依次统计每个单词的频次，然后从 map 中删除 banned 里面的单词，取出剩下频次最高的单词即可。
diff --git a/Algorithms/0828. Unique Letter String/828. Unique Letter String.go b/Algorithms/0828. Unique Letter String/828. Unique Letter String.go
deleted file mode 100644
index 92dd5f760..000000000
--- a/Algorithms/0828. Unique Letter String/828. Unique Letter String.go	
+++ /dev/null
@@ -1,44 +0,0 @@
-package leetcode
-
-func uniqueLetterString(S string) int {
-	res, left, right := 0, 0, 0
-	for i := 0; i < len(S); i++ {
-		left = i - 1
-		for left >= 0 && S[left] != S[i] {
-			left--
-		}
-		right = i + 1
-		for right < len(S) && S[right] != S[i] {
-			right++
-		}
-		res += (i - left) * (right - i)
-	}
-	return res % 1000000007
-}
-
-// 暴力解法，超时！时间复杂度 O(n^2)
-func uniqueLetterString1(S string) int {
-	if len(S) == 0 {
-		return 0
-	}
-	letterMap := map[byte]int{}
-	res, mod := 0, 1000000007
-	for i := 0; i < len(S); i++ {
-		letterMap = map[byte]int{}
-		for j := i; j < len(S); j++ {
-			letterMap[S[j]]++
-			tmp := 0
-			for _, v := range letterMap {
-				if v > 1 {
-					tmp++
-				}
-			}
-			if tmp == len(letterMap) {
-				continue
-			} else {
-				res += len(letterMap) - tmp
-			}
-		}
-	}
-	return res % mod
-}
diff --git a/Algorithms/0828. Unique Letter String/828. Unique Letter String_test.go b/Algorithms/0828. Unique Letter String/828. Unique Letter String_test.go
deleted file mode 100644
index 15a21e407..000000000
--- a/Algorithms/0828. Unique Letter String/828. Unique Letter String_test.go	
+++ /dev/null
@@ -1,57 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question828 struct {
-	para828
-	ans828
-}
-
-// para 是参数
-// one 代表第一个参数
-type para828 struct {
-	one string
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans828 struct {
-	one int
-}
-
-func Test_Problem828(t *testing.T) {
-
-	qs := []question828{
-
-		question828{
-			para828{"BABABBABAA"},
-			ans828{35},
-		},
-
-		question828{
-			para828{"ABC"},
-			ans828{10},
-		},
-
-		question828{
-			para828{"ABA"},
-			ans828{8},
-		},
-
-		question828{
-			para828{"ABAB"},
-			ans828{12},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 828------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans828, q.para828
-		fmt.Printf("【input】:%v       【output】:%v\n", p, uniqueLetterString(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0828. Unique Letter String/README.md b/Algorithms/0828. Unique Letter String/README.md
deleted file mode 100644
index 8f7914e56..000000000
--- a/Algorithms/0828. Unique Letter String/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# [828. Unique Letter String](https://leetcode.com/problems/unique-letter-string/)
-
-## 题目
-
-A character is unique in string `S` if it occurs exactly once in it.
-
-For example, in string `S = "LETTER"`, the only unique characters are `"L"` and `"R"`.
-
-Let's define `UNIQ(S)` as the number of unique characters in string `S`.
-
-For example, `UNIQ("LETTER") = 2`.
-
-Given a string `S` with only uppercases, calculate the sum of `UNIQ(substring)` over all non-empty substrings of `S`.
-
-If there are two or more equal substrings at different positions in `S`, we consider them different.
-
-Since the answer can be very large, return the answer modulo `10 ^ 9 + 7`.
-
-**Example 1:**
-
-    Input: "ABC"
-    Output: 10
-    Explanation: All possible substrings are: "A","B","C","AB","BC" and "ABC".
-    Evey substring is composed with only unique letters.
-    Sum of lengths of all substring is 1 + 1 + 1 + 2 + 2 + 3 = 10
-
-**Example 2:**
-
-    Input: "ABA"
-    Output: 8
-    Explanation: The same as example 1, except uni("ABA") = 1.
-
-
-## 题目大意
-
-如果一个字符在字符串 S 中有且仅有出现一次，那么我们称其为独特字符。例如，在字符串 S = "LETTER" 中，"L" 和 "R" 可以被称为独特字符。我们再定义 UNIQ(S) 作为字符串 S 中独特字符的个数。那么，在 S = "LETTER" 中， UNIQ("LETTER") =  2。
-
-对于给定字符串 S，计算其所有非空子串的独特字符的个数（即 UNIQ(substring)）之和。如果在 S 的不同位置上出现两个甚至多个相同的子串，那么我们认为这些子串是不同的。考虑到答案可能会非常大，规定返回格式为：结果 mod 10 ^ 9 + 7。
-
-
-## 解题思路
-
-- 这一题可以先用暴力解法尝试解题，不过提交以后会发现判题结果是超时。出错的一组数据是一个有 10000 个字符的字符串。暴力解法中间由于遍历了太多的子区间，导致了超时。
-- 这道题换一个角度思考问题。当子字符串中字符 X 出现了 2 次以上，那么它就对最终结果没有任何影响，所以只有当某个字符只出现一次的时候才会影响最终结果。再者，一个子字符串中不重复的字符的总个数，也就是这个子字符串 UNIQ 值。例如，“ABC”，这个子字符串的 UNIQ 值是 3，可以这样计算，它属于 A 的独特的字符串，也属于 B 的独特的字符串，也属于 C 的独特的字符串，那么计算这个子字符串的问题可以分解成计算 A 有多少个独特的子字符串，B 有多少个独特的子字符串，C 有多少个独特的子字符串的问题。在计算 A 有多少个子字符串的问题的时候，里面肯定会包含 "ABC" 这个子字符串的。所以原问题就转换成了分别计算给出的字符串中每个字符出现在独特字符串中的总数之和。
-- 假设原字符串是 BAABBABBBAAABA，这个字符串中出现了很多 A 和很多 B，假设我们当前计算到了第 3 个 A 的位置了(index = 5)，即标红色的那个 A。如何计算这个 A 在哪些子字符串中是独特的呢？由于子字符串题目中要求必须是连续的区间，所以这个问题很简单。找到这个 A 前一个 A 的下标位置(index = 2)，再找到这个 A 后一个 A 的下标位置(index = 9)，即 BAABBABBBAAABA，第一个 A 和当前计算的 A 中间区间有 2 个字符，第三个 A 和当前计算的 A 中间有 3 个字符。那么当前计算的 A 出现在 `(2 + 1) * (3 + 1) = 12` 个子字符串中是独特的，这 12 个字符串是：`A`，`BA`，`BBA`，`AB`，`ABB`，`ABBB`，`BAB`，`BABB`，`BABBB`，`BBAB`，`BBABB`，`BBABBB`。计算方法，假设当前待计算的字符的下标是 i ，找到当前字符前一次出现的下标位置 left，再找到当前字符后一次出现的下标位置 right，那么左边区间 (left,i) 的**开区间**内包含的字符数是 i - left - 1，右边区间 (i,right) 的**开区间**内包含的字符数是 right - i - 1。左右两边都还需要考虑空字符串的情况，即左右两边都可以不取任何字符，那么对应的就是只有中间这个待计算的字符 `A`。所以左右两边都还需要再加上空串的情况，左边 i - left - 1 + 1 = i - left，右边 right - i - 1 + 1 = right - i。左右两边的情况进行排列组合，即 (i - left) * (right - i)。针对字符串的每个字符都计算这样的值，最后累积的总和就是题目中要求的总 UNIQ 值。
-
-
diff --git a/Algorithms/0834. Sum of Distances in Tree/README.md b/Algorithms/0834. Sum of Distances in Tree/README.md
deleted file mode 100755
index c29d9c442..000000000
--- a/Algorithms/0834. Sum of Distances in Tree/README.md	
+++ /dev/null
@@ -1,41 +0,0 @@
-# [834. Sum of Distances in Tree](https://leetcode.com/problems/sum-of-distances-in-tree/)
-
-
-## 题目:
-
-An undirected, connected tree with `N` nodes labelled `0...N-1` and `N-1edges` are given.
-
-The `i`th edge connects nodes `edges[i][0]` and `edges[i][1]` together.
-
-Return a list `ans`, where `ans[i]` is the sum of the distances between node `i`and all other nodes.
-
-**Example 1:**
-
-    Input: N = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
-    Output: [8,12,6,10,10,10]
-    Explanation: 
-    Here is a diagram of the given tree:
-      0
-     / \
-    1   2
-       /|\
-      3 4 5
-    We can see that dist(0,1) + dist(0,2) + dist(0,3) + dist(0,4) + dist(0,5)
-    equals 1 + 1 + 2 + 2 + 2 = 8.  Hence, answer[0] = 8, and so on.
-
-Note: `1 <= N <= 10000`
-
-## 题目大意
-
-给定一个无向、连通的树。树中有 N 个标记为 0...N-1 的节点以及 N-1 条边。第 i 条边连接节点 edges[i][0] 和 edges[i][1] 。返回一个表示节点 i 与其他所有节点距离之和的列表 ans。
-
-说明: 1 <= N <= 10000
-
-
-
-## 解题思路
-
-- 给出 N 个节点和这些节点之间的一些边的关系。要求求出分别以 x 为根节点到所有节点路径和。
-- 这一题虽说描述的是求树的路径，但是完全可以当做图来做，因为并不是二叉树，是多叉树。这一题的解题思路是先一次 DFS 求出以 0 为根节点到各个节点的路径和(不以 0 为节点也可以，可以取任意节点作为开始)。第二次 DFS 求出从 0 根节点转换到其他各个节点的路径和。由于第一次计算出来以 0 为节点的路径和是正确的，所以计算其他节点为根节点的路径和只需要转换一下就可以得到正确结果。经过 2 次 DFS 之后就可以得到所有节点以自己为根节点到所有节点的路径和了。
-- 如何从以 0 为根节点到其他所有节点的路径和转换到以其他节点为根节点到所有节点的路径和呢？从 0 节点换成 x 节点，只需要在 0 到所有节点的路径和基础上增增减减就可以了。增加的是 x 节点到除去以 x 为根节点所有子树以外的节点的路径，有多少个节点就增加多少条路径。减少的是 0 到以 x 为根节点所有子树节点的路径和，包含 0 到 x 根节点，有多少节点就减少多少条路径。所以在第一次 DFS 中需要计算好每个节点以自己为根节点的子树总数和(包含自己在内)，这样在第二次 DFS 中可以直接拿来做转换。具体细节的实现见代码。
-
diff --git a/Algorithms/0836. Rectangle Overlap/README.md b/Algorithms/0836. Rectangle Overlap/README.md
deleted file mode 100755
index fa9e4c778..000000000
--- a/Algorithms/0836. Rectangle Overlap/README.md	
+++ /dev/null
@@ -1,41 +0,0 @@
-# [836. Rectangle Overlap](https://leetcode.com/problems/rectangle-overlap/)
-
-
-## 题目:
-
-A rectangle is represented as a list `[x1, y1, x2, y2]`, where `(x1, y1)` are the coordinates of its bottom-left corner, and `(x2, y2)` are the coordinates of its top-right corner.
-
-Two rectangles overlap if the area of their intersection is positive. To be clear, two rectangles that only touch at the corner or edges do not overlap.
-
-Given two (axis-aligned) rectangles, return whether they overlap.
-
-**Example 1:**
-
-    Input: rec1 = [0,0,2,2], rec2 = [1,1,3,3]
-    Output: true
-
-**Example 2:**
-
-    Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]
-    Output: false
-
-**Notes:**
-
-1. Both rectangles `rec1` and `rec2` are lists of 4 integers.
-2. All coordinates in rectangles will be between `-10^9` and `10^9`.
-
-
-## 题目大意
-
-矩形以列表 [x1, y1, x2, y2] 的形式表示，其中 (x1, y1) 为左下角的坐标，(x2, y2) 是右上角的坐标。如果相交的面积为正，则称两矩形重叠。需要明确的是，只在角或边接触的两个矩形不构成重叠。给出两个矩形，判断它们是否重叠并返回结果。
-
-说明：
-
-1. 两个矩形 rec1 和 rec2 都以含有四个整数的列表的形式给出。
-2. 矩形中的所有坐标都处于 -10^9 和 10^9 之间。
-
-
-## 解题思路
-
-- 给出两个矩形的坐标，判断两个矩形是否重叠。
-- 几何题，按照几何方法判断即可。
diff --git a/Algorithms/0839. Similar String Groups/README.md b/Algorithms/0839. Similar String Groups/README.md
deleted file mode 100755
index 4d20c6017..000000000
--- a/Algorithms/0839. Similar String Groups/README.md	
+++ /dev/null
@@ -1,59 +0,0 @@
-# [839. Similar String Groups](https://leetcode.com/problems/similar-string-groups/)
-
-
-## 题目:
-
-Two strings `X` and `Y` are similar if we can swap two letters (in different positions) of `X`, so that it equals `Y`.
-
-For example, `"tars"` and `"rats"` are similar (swapping at positions `0` and `2`), and `"rats"` and `"arts"` are similar, but `"star"` is not similar to `"tars"`, `"rats"`, or `"arts"`.
-
-Together, these form two connected groups by similarity: `{"tars", "rats", "arts"}` and `{"star"}`. Notice that `"tars"` and `"arts"` are in the same group even though they are not similar. Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.
-
-We are given a list `A` of strings. Every string in `A` is an anagram of every other string in `A`. How many groups are there?
-
-**Example 1:**
-
-    Input: ["tars","rats","arts","star"]
-    Output: 2
-
-**Note:**
-
-1. `A.length <= 2000`
-2. `A[i].length <= 1000`
-3. `A.length * A[i].length <= 20000`
-4. All words in `A` consist of lowercase letters only.
-5. All words in `A` have the same length and are anagrams of each other.
-6. The judging time limit has been increased for this question.
-
-
-## 题目大意
-
-如果我们交换字符串 X 中的两个不同位置的字母，使得它和字符串 Y 相等，那么称 X 和 Y 两个字符串相似。
-
-例如，"tars" 和 "rats" 是相似的 (交换 0 与 2 的位置)； "rats" 和 "arts" 也是相似的，但是 "star" 不与 "tars"，"rats"，或 "arts" 相似。
-
-总之，它们通过相似性形成了两个关联组：{"tars", "rats", "arts"} 和 {"star"}。注意，"tars" 和 "arts" 是在同一组中，即使它们并不相似。形式上，对每个组而言，要确定一个单词在组中，只需要这个词和该组中至少一个单词相似。我们给出了一个不包含重复的字符串列表 A。列表中的每个字符串都是 A 中其它所有字符串的一个字母异位词。请问 A 中有多少个相似字符串组？
-
-
-提示：
-
-- A.length <= 2000
-- A[i].length <= 1000
-- A.length * A[i].length <= 20000
-- A 中的所有单词都只包含小写字母。
-- A 中的所有单词都具有相同的长度，且是彼此的字母异位词。
-- 此问题的判断限制时间已经延长。
-
-
-备注：  
-
-- 字母异位词[anagram]，一种把某个字符串的字母的位置（顺序）加以改换所形成的新词。
-
-
-
-
-## 解题思路
-
-
-- 给出一个字符串数组，要求找出这个数组中，"不相似"的字符串有多少种。相似的字符串的定义是：如果 A 和 B 字符串只需要交换一次字母的位置就能变成两个相等的字符串，那么 A 和 B 是相似的。
-- 这一题的解题思路是用并查集。先将题目中的“相似”的定义转换一下，A 和 B 相似的意思是，A 和 B 中只有 2 个字符不相等，其他字符都相等，这样交换一次才能完全相等。有没有可能这两个字符交换了也不相等呢？这种情况不用考虑，因为题目中提到了给的字符串都是 `anagram` 的(`anagram` 的意思是，字符串的任意排列组合)。那么这题就比较简单了，只需要判断每两个字符串是否“相似”，如果相似就 `union()`，最后看并查集中有几个集合即可。
diff --git a/Algorithms/0850. Rectangle Area II/850. Rectangle Area II_test.go b/Algorithms/0850. Rectangle Area II/850. Rectangle Area II_test.go
deleted file mode 100644
index fa7498156..000000000
--- a/Algorithms/0850. Rectangle Area II/850. Rectangle Area II_test.go	
+++ /dev/null
@@ -1,57 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question850 struct {
-	para850
-	ans850
-}
-
-// para 是参数
-// one 代表第一个参数
-type para850 struct {
-	one [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans850 struct {
-	one int
-}
-
-func Test_Problem850(t *testing.T) {
-
-	qs := []question850{
-
-		question850{
-			para850{[][]int{[]int{0, 0, 3, 3}, []int{2, 0, 5, 3}, []int{1, 1, 4, 4}}},
-			ans850{18},
-		},
-
-		question850{
-			para850{[][]int{[]int{0, 0, 1, 1}, []int{2, 2, 3, 3}}},
-			ans850{2},
-		},
-
-		question850{
-			para850{[][]int{[]int{0, 0, 2, 2}, []int{1, 0, 2, 3}, []int{1, 0, 3, 1}}},
-			ans850{6},
-		},
-
-		question850{
-			para850{[][]int{[]int{0, 0, 1000000000, 1000000000}}},
-			ans850{49},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 850------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans850, q.para850
-		fmt.Printf("【input】:%v       【output】:%v\n", p, rectangleArea(p.one))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0850. Rectangle Area II/README.md b/Algorithms/0850. Rectangle Area II/README.md
deleted file mode 100755
index 4951f3dc9..000000000
--- a/Algorithms/0850. Rectangle Area II/README.md	
+++ /dev/null
@@ -1,99 +0,0 @@
-# [850. Rectangle Area II](https://leetcode.com/problems/rectangle-area-ii/)
-
-
-## 题目:
-
-We are given a list of (axis-aligned) `rectangles`. Each `rectangle[i] = [x1, y1, x2, y2]` , where (x1, y1) are the coordinates of the bottom-left corner, and (x2, y2) are the coordinates of the top-right corner of the `i`th rectangle.
-
-Find the total area covered by all `rectangles` in the plane. Since the answer may be too large, **return it modulo 10^9 + 7**.
-
-![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/06/rectangle_area_ii_pic.png)
-
-**Example 1:**
-
-    Input: [[0,0,2,2],[1,0,2,3],[1,0,3,1]]
-    Output: 6
-    Explanation: As illustrated in the picture.
-
-**Example 2:**
-
-    Input: [[0,0,1000000000,1000000000]]
-    Output: 49
-    Explanation: The answer is 10^18 modulo (10^9 + 7), which is (10^9)^2 = (-7)^2 = 49.
-
-**Note:**
-
-- `1 <= rectangles.length <= 200`
-- `rectanges[i].length = 4`
-- `0 <= rectangles[i][j] <= 10^9`
-- The total area covered by all rectangles will never exceed `2^63 - 1` and thus will fit in a 64-bit signed integer.
-
-
-## 题目大意
-
-我们给出了一个（轴对齐的）矩形列表 rectangles。 对于 rectangle[i] = [x1, y1, x2, y2]，其中（x1，y1）是矩形 i 左下角的坐标，（x2，y2）是该矩形右上角的坐标。找出平面中所有矩形叠加覆盖后的总面积。由于答案可能太大，请返回它对 10 ^ 9 + 7 取模的结果。
-
-提示：
-
-- 1 <= rectangles.length <= 200
-- rectanges[i].length = 4
-- 0 <= rectangles[i][j] <= 10^9
-- 矩形叠加覆盖后的总面积不会超越 2^63 - 1 ，这意味着可以用一个 64 位有符号整数来保存面积结果。
-
-
-## 解题思路
-
-
-- 在二维坐标系中给出一些矩形，要求这些矩形合并之后的面积。由于矩形有重叠，所以需要考虑合并以后的面积。矩形的坐标值也会很大。
-- 这一题给人的感觉很像第 218 题，求天际线的过程也是有楼挡楼，重叠的情况。不过那一题只用求天际线的拐点，所以我们可以对区间做“右边界减一”的处理，防止两个相邻区间因为共点，而导致结果错误。但是这一题如果还是用相同的做法，就会出错，因为“右边界减一”以后，面积会少一部分，最终得到的结果也是偏小的。所以这一题要将线段树改造一下。
-- 思路是先讲 Y 轴上的坐标离线化，转换成线段树。将矩形的 2 条边变成扫描线，左边是入边，右边是出边。
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_850_8.png)
-
-- 再从左往右遍历每条扫描线，并对 Y 轴上的线段树进行 update。X 轴上的每个坐标区间 * query 线段树总高度的结果 = 区间面积。最后将 X 轴对应的每个区间面积加起来，就是最终矩形合并以后的面积。如下图中间的图。
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_850_9.png)
-
-    需要注意的一点是，**每次 query 的结果并不一定是连续线段**。如上图最右边的图，中间有一段是可能出现镂空的。这种情况看似复杂，其实很简单，因为每段线段树的线段代表的权值高度是不同的，每次 query 最大高度得到的结果已经考虑了中间可能有镂空的情况了。
-
-- 具体做法，先把各个矩形在 Y 轴方向上离散化，这里的**线段树叶子节点不再是一个点了，而是一个区间长度为 1 的区间段。**
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_850_0.png)
-
-    每个叶子节点也不再是存储一个 int 值，而是存 2 个值，一个是 count 值，用来记录这条区间被覆盖的次数，另一个值是 val 值，用来反映射该线段长度是多少，因为 Y 轴被离散化了，区间坐标间隔都是 1，但是实际 Y 轴的高度并不是 1 ，所以用 val 来反映射原来的高度。
-
-- 初始化线段树，叶子节点的 count = 0，val 根据题目给的 Y 坐标进行计算。
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_850_1.png)
-
-- 从左往右遍历每个扫描线。每条扫面线都把对应 update 更新到叶子节点。pushUp 的时候需要合并每个区间段的高度 val 值。如果有区间没有被覆盖，那么这个区间高度 val 为 0，这也就处理了可能“中间镂空”的情况。
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_850_2.png)
-
-        func (sat *SegmentAreaTree) pushUp(treeIndex, leftTreeIndex, rightTreeIndex int) {
-        	newCount, newValue := sat.merge(sat.tree[leftTreeIndex].count, sat.tree[rightTreeIndex].count), 0
-        	if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count > 0 {
-        		newValue = sat.merge(sat.tree[leftTreeIndex].val, sat.tree[rightTreeIndex].val)
-        	} else if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count == 0 {
-        		newValue = sat.tree[leftTreeIndex].val
-        	} else if sat.tree[leftTreeIndex].count == 0 && sat.tree[rightTreeIndex].count > 0 {
-        		newValue = sat.tree[rightTreeIndex].val
-        	}
-        	sat.tree[treeIndex] = SegmentItem{count: newCount, val: newValue}
-        }
-
-- 扫描每一个扫描线，先 pushDown 到叶子节点，再 pushUp 到根节点。
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_850_3.png)
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_850_4.png)
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_850_5.png)
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_850_6.png)
-
-- 遍历到倒数第 2 根扫描线的时候就能得到结果了。因为最后一根扫描线 update 以后，整个线段树全部都归为初始化状态了。
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_850_7.png)
-
-- 这一题是线段树扫面线解法的经典题。
diff --git a/Algorithms/0852. Peak Index in a Mountain Array/README.md b/Algorithms/0852. Peak Index in a Mountain Array/README.md
deleted file mode 100755
index 3d1ed48d8..000000000
--- a/Algorithms/0852. Peak Index in a Mountain Array/README.md	
+++ /dev/null
@@ -1,49 +0,0 @@
-# [852. Peak Index in a Mountain Array](https://leetcode.com/problems/peak-index-in-a-mountain-array/)
-
-
-## 题目:
-
-Let's call an array `A` a *mountain* if the following properties hold:
-
-- `A.length >= 3`
-- There exists some `0 < i < A.length - 1` such that `A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]`
-
-Given an array that is definitely a mountain, return any `i` such that `A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]`.
-
-**Example 1:**
-
-    Input: [0,1,0]
-    Output: 1
-
-**Example 2:**
-
-    Input: [0,2,1,0]
-    Output: 1
-
-**Note:**
-
-1. `3 <= A.length <= 10000`
-2. `0 <= A[i] <= 10^6`
-3. A is a mountain, as defined above.
-
-
-## 题目大意
-
-我们把符合下列属性的数组 A 称作山脉：
-
-- A.length >= 3
-- 存在 0 < i < A.length - 1 使得A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]
-给定一个确定为山脉的数组，返回任何满足 A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1] 的 i 的值。
-
-提示：
-
-- 3 <= A.length <= 10000
-- 0 <= A[i] <= 10^6
-- A 是如上定义的山脉
- 
- 
-
-## 解题思路
-
-- 给出一个数组，数组里面存在有且仅有一个“山峰”，(山峰的定义是，下标 `i` 比 `i-1`、`i+1` 位置上的元素都要大)，找到这个“山峰”，并输出其中一个山峰的下标。
-- 这一题直接用二分搜索即可，数组中的元素算基本有序。判断是否为山峰的条件为比较左右两个数，如果当前的数比左右两个数都大，即找到了山峰。其他的情况都在山坡上。这一题有两种写法，第一种写法是标准的二分写法，第二种写法是变形的二分写法。
diff --git a/Algorithms/0864. Shortest Path to Get All Keys/README.md b/Algorithms/0864. Shortest Path to Get All Keys/README.md
deleted file mode 100755
index 3b217024c..000000000
--- a/Algorithms/0864. Shortest Path to Get All Keys/README.md	
+++ /dev/null
@@ -1,55 +0,0 @@
-# [864. Shortest Path to Get All Keys](https://leetcode.com/problems/shortest-path-to-get-all-keys/)
-
-
-## 题目:
-
-We are given a 2-dimensional `grid`. `"."` is an empty cell, `"#"` is a wall, `"@"` is the starting point, (`"a"`, `"b"`, ...) are keys, and (`"A"`, `"B"`, ...) are locks.
-
-We start at the starting point, and one move consists of walking one space in one of the 4 cardinal directions. We cannot walk outside the grid, or walk into a wall. If we walk over a key, we pick it up. We can't walk over a lock unless we have the corresponding key.
-
-For some 1 <= K <= 6, there is exactly one lowercase and one uppercase letter of the first `K` letters of the English alphabet in the grid. This means that there is exactly one key for each lock, and one lock for each key; and also that the letters used to represent the keys and locks were chosen in the same order as the English alphabet.
-
-Return the lowest number of moves to acquire all keys. If it's impossible, return `-1`.
-
-**Example 1:**
-
-    Input: ["@.a.#","###.#","b.A.B"]
-    Output: 8
-
-**Example 2:**
-
-    Input: ["@..aA","..B#.","....b"]
-    Output: 6
-
-**Note:**
-
-1. `1 <= grid.length <= 30`
-2. `1 <= grid[0].length <= 30`
-3. `grid[i][j]` contains only `'.'`, `'#'`, `'@'`, `'a'-'f'` and `'A'-'F'`
-4. The number of keys is in `[1, 6]`. Each key has a different letter and opens exactly one lock.
-
-
-## 题目大意
-
-给定一个二维网格 grid。 "." 代表一个空房间， "#" 代表一堵墙， "@" 是起点，（"a", "b", ...）代表钥匙，（"A", "B", ...）代表锁。
-
-我们从起点开始出发，一次移动是指向四个基本方向之一行走一个单位空间。我们不能在网格外面行走，也无法穿过一堵墙。如果途经一个钥匙，我们就把它捡起来。除非我们手里有对应的钥匙，否则无法通过锁。
-
-假设 K 为钥匙/锁的个数，且满足 1 <= K <= 6，字母表中的前 K 个字母在网格中都有自己对应的一个小写和一个大写字母。换言之，每个锁有唯一对应的钥匙，每个钥匙也有唯一对应的锁。另外，代表钥匙和锁的字母互为大小写并按字母顺序排列。
-
-返回获取所有钥匙所需要的移动的最少次数。如果无法获取所有钥匙，返回 -1 。
-
-提示：
-
-1. 1 <= grid.length <= 30
-2. 1 <= grid[0].length <= 30
-3. grid[i][j] 只含有 '.', '#', '@', 'a'-'f' 以及 'A'-'F'
-4. 钥匙的数目范围是 [1, 6]，每个钥匙都对应一个不同的字母，正好打开一个对应的锁。
-
-
-## 解题思路
-
-
-- 给出一个地图，在图中有钥匙和锁，当锁在没有钥匙的时候不能通行，问从起点 @ 开始，到最终获得所有钥匙，最短需要走多少步。
-- 这一题可以用 BFS 来解答。由于钥匙的种类比较多，所以 visited 数组需要 3 个维度，一个是 x 坐标，一个是 y 坐标，最后一个是当前获取钥匙的状态。每把钥匙都有获取了和没有获取两种状态，题目中说最多有 6 把钥匙，那么排列组合最多是 2^6 = 64 种状态。用一个十进制数的二进制位来压缩这些状态，二进制位分别来表示这些钥匙是否已经获取了。既然钥匙的状态可以压缩，其实 x 和 y 的坐标也可以一并压缩到这个数中。BFS 中存的数字是坐标 + 钥匙状态的状态。在 BFS 遍历的过程中，用 visited 数组来过滤遍历过的情况，来保证走的路是最短的。其他的情况无非是判断锁的状态，是否能通过，判断钥匙获取状态。
-- 这一题不知道是否能用 DFS 来解答。我实现了一版，但是在 18 / 35 这组 case 上超时了，具体 case 见测试文件第一个 case。
diff --git a/Algorithms/0867. Transpose Matrix/867. Transpose Matrix_test.go b/Algorithms/0867. Transpose Matrix/867. Transpose Matrix_test.go
deleted file mode 100644
index e204553fa..000000000
--- a/Algorithms/0867. Transpose Matrix/867. Transpose Matrix_test.go	
+++ /dev/null
@@ -1,47 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question867 struct {
-	para867
-	ans867
-}
-
-// para 是参数
-// one 代表第一个参数
-type para867 struct {
-	A [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans867 struct {
-	B [][]int
-}
-
-func Test_Problem867(t *testing.T) {
-
-	qs := []question867{
-
-		question867{
-			para867{[][]int{[]int{1, 2, 3}, []int{4, 5, 6}, []int{7, 8, 9}}},
-			ans867{[][]int{[]int{1, 4, 7}, []int{2, 5, 8}, []int{3, 6, 9}}},
-		},
-
-		question867{
-			para867{[][]int{[]int{1, 2, 3}, []int{4, 5, 6}}},
-			ans867{[][]int{[]int{1, 4}, []int{2, 5}, []int{3, 6}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 867------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans867, q.para867
-		fmt.Printf("【input】:%v       【output】:%v\n", p, transpose(p.A))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0884. Uncommon Words from Two Sentences/README.md b/Algorithms/0884. Uncommon Words from Two Sentences/README.md
deleted file mode 100755
index 98be131c1..000000000
--- a/Algorithms/0884. Uncommon Words from Two Sentences/README.md	
+++ /dev/null
@@ -1,40 +0,0 @@
-# [884. Uncommon Words from Two Sentences](https://leetcode.com/problems/uncommon-words-from-two-sentences/)
-
-
-## 题目:
-
-We are given two sentences `A` and `B`. (A *sentence* is a string of space separated words. Each *word* consists only of lowercase letters.)
-
-A word is *uncommon* if it appears exactly once in one of the sentences, and does not appear in the other sentence.
-
-Return a list of all uncommon words.
-
-You may return the list in any order.
-
-**Example 1:**
-
-    Input: A = "this apple is sweet", B = "this apple is sour"
-    Output: ["sweet","sour"]
-
-**Example 2:**
-
-    Input: A = "apple apple", B = "banana"
-    Output: ["banana"]
-
-**Note:**
-
-1. `0 <= A.length <= 200`
-2. `0 <= B.length <= 200`
-3. `A` and `B` both contain only spaces and lowercase letters.
-
-
-## 题目大意
-
-给定两个句子 A 和 B 。（句子是一串由空格分隔的单词。每个单词仅由小写字母组成。）
-
-如果一个单词在其中一个句子中只出现一次，在另一个句子中却没有出现，那么这个单词就是不常见的。返回所有不常用单词的列表。您可以按任何顺序返回列表。
-
-
-## 解题思路
-
-- 找出 2 个句子中不同的单词，将它们俩都打印出来。简单题，先将 2 个句子的单词都拆开放入 map 中进行词频统计，不同的两个单词的词频肯定都为 1，输出它们即可。
diff --git a/Algorithms/0885. Spiral Matrix III/885. Spiral Matrix III_test.go b/Algorithms/0885. Spiral Matrix III/885. Spiral Matrix III_test.go
deleted file mode 100644
index fa45c821e..000000000
--- a/Algorithms/0885. Spiral Matrix III/885. Spiral Matrix III_test.go	
+++ /dev/null
@@ -1,52 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question885 struct {
-	para885
-	ans885
-}
-
-// para 是参数
-// one 代表第一个参数
-type para885 struct {
-	R  int
-	C  int
-	r0 int
-	c0 int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans885 struct {
-	one [][]int
-}
-
-func Test_Problem885(t *testing.T) {
-
-	qs := []question885{
-
-		question885{
-			para885{1, 4, 0, 0},
-			ans885{[][]int{[]int{0, 0}, []int{0, 1}, []int{0, 2}, []int{0, 3}}},
-		},
-
-		question885{
-			para885{5, 6, 1, 4},
-			ans885{[][]int{[]int{1, 4}, []int{1, 5}, []int{2, 5}, []int{2, 4}, []int{2, 3}, []int{1, 3}, []int{0, 3}, []int{0, 4}, []int{0, 5}, []int{3, 5}, []int{3, 4},
-				[]int{3, 3}, []int{3, 2}, []int{2, 2}, []int{1, 2}, []int{0, 2}, []int{4, 5}, []int{4, 4}, []int{4, 3}, []int{4, 2}, []int{4, 1}, []int{3, 1}, []int{2, 1},
-				[]int{1, 1}, []int{0, 1}, []int{4, 0}, []int{3, 0}, []int{2, 0}, []int{1, 0}, []int{0, 0}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 885------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans885, q.para885
-		fmt.Printf("【input】:%v       【output】:%v\n", p, spiralMatrixIII(p.R, p.C, p.r0, p.c0))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0898. Bitwise ORs of Subarrays/README.md b/Algorithms/0898. Bitwise ORs of Subarrays/README.md
deleted file mode 100755
index 23df36dd0..000000000
--- a/Algorithms/0898. Bitwise ORs of Subarrays/README.md	
+++ /dev/null
@@ -1,82 +0,0 @@
-# [898. Bitwise ORs of Subarrays](https://leetcode.com/problems/bitwise-ors-of-subarrays/)
-
-
-## 题目:
-
-We have an array `A` of non-negative integers.
-
-For every (contiguous) subarray `B = [A[i], A[i+1], ..., A[j]]` (with `i <= j`), we take the bitwise OR of all the elements in `B`, obtaining a result `A[i] | A[i+1] | ... | A[j]`.
-
-Return the number of possible results. (Results that occur more than once are only counted once in the final answer.)
-
-**Example 1:**
-
-    Input: [0]
-    Output: 1
-    Explanation: 
-    There is only one possible result: 0.
-
-**Example 2:**
-
-    Input: [1,1,2]
-    Output: 3
-    Explanation: 
-    The possible subarrays are [1], [1], [2], [1, 1], [1, 2], [1, 1, 2].
-    These yield the results 1, 1, 2, 1, 3, 3.
-    There are 3 unique values, so the answer is 3.
-
-**Example 3:**
-
-    Input: [1,2,4]
-    Output: 6
-    Explanation: 
-    The possible results are 1, 2, 3, 4, 6, and 7.
-
-**Note:**
-
-1. `1 <= A.length <= 50000`
-2. `0 <= A[i] <= 10^9`
-
-
-## 题目大意
-
-我们有一个非负整数数组 A。对于每个（连续的）子数组 B = [A[i], A[i+1], ..., A[j]] （ i <= j），我们对 B 中的每个元素进行按位或操作，获得结果 A[i] | A[i+1] | ... | A[j]。返回可能结果的数量。（多次出现的结果在最终答案中仅计算一次。）
-
-
-
-## 解题思路
-
-- 给出一个数组，要求求出这个数组所有的子数组中，每个集合内所有数字取 `|` 运算以后，不同结果的种类数。
-- 这道题可以这样考虑，第一步，先考虑所有的子数组如何得到，以 `[001, 011, 100, 110, 101]` 为例，所有的子数组集合如下：
-
-```c
-    [001]  
-    [001 011] [011]  
-    [001 011 100] [011 100] [100]  
-    [001 011 100 110] [011 100 110] [100 110] [110]  
-    [001 011 100 110 101] [011 100 110 101] [100 110 101] [110 101] [101]  
-```
-
-可以发现，从左往右遍历原数组，每次新来的一个元素，依次加入到之前已经生成过的集合中，再以自己为单独集合。这样就可以生成原数组的所有子集。
-
-- 第二步，将每一行的子集内的所有元素都进行 `|` 运算，得到：
-
-```c
-    001  
-    011 011  
-    111 111 100  
-    111 111 110 110  
-    111 111 111 111 101  
-```
-
-- 第三步，去重：
-
-```c
-    001  
-    011  
-    111 100  
-    111 110  
-    111 101  
-```
-
-由于二进制位不超过 32 位，所以这里每一行最多不会超过 32 个数。所以最终时间复杂度不会超过 O(32 N)，即 O(K * N)。最后将这每一行的数字都放入最终的 map 中去重即可。
diff --git a/Algorithms/0918. Maximum Sum Circular Subarray/918. Maximum Sum Circular Subarray.go b/Algorithms/0918. Maximum Sum Circular Subarray/918. Maximum Sum Circular Subarray.go
deleted file mode 100644
index be73a4f8b..000000000
--- a/Algorithms/0918. Maximum Sum Circular Subarray/918. Maximum Sum Circular Subarray.go	
+++ /dev/null
@@ -1,28 +0,0 @@
-package leetcode
-
-import "math"
-
-func maxSubarraySumCircular(A []int) int {
-	n, sum := len(A), 0
-	for _, v := range A {
-		sum += v
-	}
-	kad := kadane(A)
-	for i := 0; i < n; i++ {
-		A[i] = -A[i]
-	}
-	negativeMax := kadane(A)
-	if sum+negativeMax <= 0 {
-		return kad
-	}
-	return max(kad, sum+negativeMax)
-}
-
-func kadane(a []int) int {
-	n, MaxEndingHere, maxSoFar := len(a), a[0], math.MinInt32
-	for i := 1; i < n; i++ {
-		MaxEndingHere = max(a[i], MaxEndingHere+a[i])
-		maxSoFar = max(MaxEndingHere, maxSoFar)
-	}
-	return maxSoFar
-}
diff --git a/Algorithms/0918. Maximum Sum Circular Subarray/README.md b/Algorithms/0918. Maximum Sum Circular Subarray/README.md
deleted file mode 100755
index 080d27d3e..000000000
--- a/Algorithms/0918. Maximum Sum Circular Subarray/README.md	
+++ /dev/null
@@ -1,65 +0,0 @@
-# [918. Maximum Sum Circular Subarray](https://leetcode.com/problems/maximum-sum-circular-subarray/)
-
-
-## 题目:
-
-Given a **circular array** **C** of integers represented by `A`, find the maximum possible sum of a non-empty subarray of **C**.
-
-Here, a *circular array* means the end of the array connects to the beginning of the array. (Formally, `C[i] = A[i]` when `0 <= i < A.length`, and `C[i+A.length] = C[i]` when `i >= 0`.)
-
-Also, a subarray may only include each element of the fixed buffer `A` at most once. (Formally, for a subarray `C[i], C[i+1], ..., C[j]`, there does not exist `i <= k1, k2 <= j` with `k1 % A.length = k2 % A.length`.)
-
-**Example 1:**
-
-    Input: [1,-2,3,-2]
-    Output: 3
-    Explanation: Subarray [3] has maximum sum 3
-
-**Example 2:**
-
-    Input: [5,-3,5]
-    Output: 10
-    Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10
-
-**Example 3:**
-
-    Input: [3,-1,2,-1]
-    Output: 4
-    Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4
-
-**Example 4:**
-
-    Input: [3,-2,2,-3]
-    Output: 3
-    Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3
-
-**Example 5:**
-
-    Input: [-2,-3,-1]
-    Output: -1
-    Explanation: Subarray [-1] has maximum sum -1
-
-**Note:**
-
-1. `-30000 <= A[i] <= 30000`
-2. `1 <= A.length <= 30000`
-
-
-## 题目大意
-
-给定一个由整数数组 A 表示的环形数组 C，求 C 的非空子数组的最大可能和。在此处，环形数组意味着数组的末端将会与开头相连呈环状。（形式上，当0 <= i < A.length 时 C[i] = A[i]，而当 i >= 0 时 C[i+A.length] = C[i]）
-
-此外，子数组最多只能包含固定缓冲区 A 中的每个元素一次。（形式上，对于子数组 C[i], C[i+1], ..., C[j]，不存在 i <= k1, k2 <= j 其中 k1 % A.length = k2 % A.length）
-
-提示：
-
-- -30000 <= A[i] <= 30000
-- 1 <= A.length <= 30000
-
-
-## 解题思路
-
-
-- 给出一个环形数组，要求出这个环形数组中的连续子数组的最大和。
-- 拿到这题最先想到的思路是把这个数组再拼接一个，在这两个数组中查找连续子数组的最大和。这种做法是错误的，例如在 `[5,-3,5]` 这个数组中会得出 `7` 的结果，但是实际结果是 `10` 。那么这题怎么做呢？仔细分析可以得到，环形数组的最大连续子段和有两种情况，第一种情况是这个连续子段就出现在数组中， 不存在循环衔接的情况。针对这种情况就比较简单，用 `kadane` 算法(也是动态规划的思想)，`O(n)` 的时间复杂度就可以求出结果。第二种情况是这个连续的子段出现在跨数组的情况，即会出现首尾相连的情况。要想找到这样一个连续子段，可以反向考虑。想找到跨段的连续子段，那么这个数组剩下的这一段就是不跨段的连续子段。想要跨段的子段和最大，那么剩下的这段连续子段和最小。如果能找到这个数组的每个元素取相反数组成的数组中的最大连续子段和，那么反过来就能找到原数组的连续子段和最小。举个例子：`[1，2，-3，-4，5]` ，取它的每个元素的相反数 `[-1，-2，3，4，-5]`，构造的数组中最大连续子段和是 `3 + 4 = 7`，由于取了相反数，所以可以得到原数组中最小连续子段和是 `-7` 。所以跨段的最大连续子段和就是剩下的那段 `[1,2,5]`。
-- 还有一些边界的情况，例如，`[1，2，-2，-3，5，5，-4，6]` 和 `[1，2，-2，-3，5，5，-4，8]`，所以还需要比较一下情况一和情况二的值，它们两者最大值才是最终环形数组的连续子数组的最大和。
diff --git a/Algorithms/0920. Number of Music Playlists/README.md b/Algorithms/0920. Number of Music Playlists/README.md
deleted file mode 100755
index a328fde00..000000000
--- a/Algorithms/0920. Number of Music Playlists/README.md	
+++ /dev/null
@@ -1,59 +0,0 @@
-# [920. Number of Music Playlists](https://leetcode.com/problems/number-of-music-playlists/)
-
-
-## 题目:
-
-Your music player contains `N` different songs and she wants to listen to `L` ****(not necessarily different) songs during your trip. You create a playlist so that:
-
-- Every song is played at least once
-- A song can only be played again only if `K` other songs have been played
-
-Return the number of possible playlists. **As the answer can be very large, return it modulo `10^9 + 7`**.
-
-**Example 1:**
-
-    Input: N = 3, L = 3, K = 1
-    Output: 6
-    Explanation: There are 6 possible playlists. [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].
-
-**Example 2:**
-
-    Input: N = 2, L = 3, K = 0
-    Output: 6
-    Explanation: There are 6 possible playlists. [1, 1, 2], [1, 2, 1], [2, 1, 1], [2, 2, 1], [2, 1, 2], [1, 2, 2]
-
-**Example 3:**
-
-    Input: N = 2, L = 3, K = 1
-    Output: 2
-    Explanation: There are 2 possible playlists. [1, 2, 1], [2, 1, 2]
-
-**Note:**
-
-1. `0 <= K < N <= L <= 100`
-
-## 题目大意
-
-你的音乐播放器里有 N 首不同的歌，在旅途中，你的旅伴想要听 L 首歌（不一定不同，即，允许歌曲重复）。请你为她按如下规则创建一个播放列表：
-
-- 每首歌至少播放一次。
-- 一首歌只有在其他 K 首歌播放完之后才能再次播放。
-
-返回可以满足要求的播放列表的数量。由于答案可能非常大，请返回它模 10^9 + 7 的结果。
-
-提示：
-
-- 0 <= K < N <= L <= 100
-
-
-
-
-## 解题思路
-
-- 简化抽象一下题意，给 N 个数，要求从这 N 个数里面组成一个长度为 L 的序列，并且相同元素的间隔不能小于 K 个数。问总共有多少组组成方法。
-- 一拿到题，会觉得这一题是三维 DP，因为存在 3 个变量，但是实际考虑一下，可以降一维。我们先不考虑 K 的限制，只考虑 N 和 L。定义 `dp[i][j]` 代表播放列表里面有 `i` 首歌，其中包含 `j` 首不同的歌曲，那么题目要求的最终解存在 `dp[L][N]` 中。考虑 `dp[i][j]` 的递归公式，音乐列表当前需要组成 `i` 首歌，有 2 种方式可以得到，由 `i - 1` 首歌的列表中添加一首列表中**不存在**的新歌曲，或者由 `i - 1` 首歌的列表中添加一首列表中**已经存在**的歌曲。即，`dp[i][j]` 可以由 `dp[i - 1][j - 1]` 得到，也可以由 `dp[i - 1][j]` 得到。如果是第一种情况，添加一首新歌，那么新歌有 N - ( j - 1 ) 首，如果是第二种情况，添加一首已经存在的歌，歌有 j 首，所以状态转移方程是 `dp[i][j] = dp[i - 1][j - 1] * ( N - ( j - 1 ) ) + dp[i - 1][j] * j` 。但是这个方程是在不考虑 K 的限制条件下得到的，距离满足题意还差一步。接下来需要考虑加入 K 这个限制条件以后，状态转移方程该如何推导。
-- 如果是添加一首新歌，是不受 K 限制的，所以 `dp[i - 1][j - 1] * ( N - ( j - 1 ) )` 这里不需要变化。如果是添加一首存在的歌曲，这个时候就会受到 K 的限制了。如果当前播放列表里面的歌曲有 `j` 首，并且 `j > K`，那么选择歌曲只能从 `j - K` 里面选，因为不能选择 `j - 1` 到 `j - k` 的这些歌，选择了就不满足重复的歌之间间隔不能小于 `K` 的限制条件了。那 j ≤ K 呢？这个时候一首歌都不能选，因为歌曲数都没有超过 K，当然不能再选择重复的歌曲。(选择了就再次不满足重复的歌之间间隔不能小于 `K` 的限制条件了)。经过上述分析，可以得到最终的状态转移方程：
-
-![](https://img.halfrost.com/Leetcode/leetcode_920.gif)
-
-- 上面的式子可以合并简化成下面这个式子：`dp[i][j] = dp[i - 1][j - 1]*(N - (j - 1)) + dp[i-1][j]*max(j-K, 0)`，递归初始值 `dp[0][0] = 1`。
diff --git a/Algorithms/0924. Minimize Malware Spread/924. Minimize Malware Spread.go b/Algorithms/0924. Minimize Malware Spread/924. Minimize Malware Spread.go
deleted file mode 100644
index b868b1570..000000000
--- a/Algorithms/0924. Minimize Malware Spread/924. Minimize Malware Spread.go	
+++ /dev/null
@@ -1,39 +0,0 @@
-package leetcode
-
-import (
-	"math"
-
-	"github.com/halfrost/LeetCode-Go/template"
-)
-
-func minMalwareSpread(graph [][]int, initial []int) int {
-	if len(initial) == 0 {
-		return 0
-	}
-	uf, minIndex, count, countMap := template.UnionFind{}, 0, math.MinInt64, map[int]int{}
-	uf.Init(len(graph))
-	for i := range graph {
-		for j := range graph[i] {
-			if i == j {
-				break
-			}
-			if graph[i][j] == 1 {
-				uf.Union(i, j)
-			}
-		}
-	}
-	for i := 0; i < len(graph); i++ {
-		countMap[uf.Find(i)]++
-	}
-	for _, v := range initial {
-		tmp := countMap[uf.Find(v)]
-		if count == tmp && minIndex > v {
-			minIndex = v
-		}
-		if count < tmp {
-			minIndex = v
-			count = tmp
-		}
-	}
-	return minIndex
-}
diff --git a/Algorithms/0924. Minimize Malware Spread/924. Minimize Malware Spread_test.go b/Algorithms/0924. Minimize Malware Spread/924. Minimize Malware Spread_test.go
deleted file mode 100644
index 6973f0d80..000000000
--- a/Algorithms/0924. Minimize Malware Spread/924. Minimize Malware Spread_test.go	
+++ /dev/null
@@ -1,63 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question924 struct {
-	para924
-	ans924
-}
-
-// para 是参数
-// one 代表第一个参数
-type para924 struct {
-	graph   [][]int
-	initial []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans924 struct {
-	one int
-}
-
-func Test_Problem924(t *testing.T) {
-
-	qs := []question924{
-
-		question924{
-			para924{[][]int{[]int{1, 0, 0, 0}, []int{0, 1, 0, 0}, []int{0, 0, 1, 1}, []int{0, 0, 1, 1}}, []int{3, 1}},
-			ans924{3},
-		},
-
-		question924{
-			para924{[][]int{[]int{1, 0, 0, 0, 0, 0}, []int{0, 1, 0, 0, 0, 0}, []int{0, 0, 1, 0, 0, 0}, []int{0, 0, 0, 1, 1, 0}, []int{0, 0, 0, 1, 1, 0}, []int{0, 0, 0, 0, 0, 1}}, []int{5, 0}},
-			ans924{0},
-		},
-
-		question924{
-			para924{[][]int{[]int{1, 1, 1}, []int{1, 1, 1}, []int{1, 1, 1}}, []int{1, 2}},
-			ans924{1},
-		},
-
-		question924{
-			para924{[][]int{[]int{1, 1, 0}, []int{1, 1, 0}, []int{0, 0, 1}}, []int{0, 1}},
-			ans924{0},
-		},
-
-		question924{
-			para924{[][]int{[]int{1, 0, 0}, []int{0, 1, 0}, []int{0, 0, 1}}, []int{0, 2}},
-			ans924{0},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 924------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans924, q.para924
-		fmt.Printf("【input】:%v       【output】:%v\n", p, minMalwareSpread(p.graph, p.initial))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0924. Minimize Malware Spread/README.md b/Algorithms/0924. Minimize Malware Spread/README.md
deleted file mode 100755
index b293461f3..000000000
--- a/Algorithms/0924. Minimize Malware Spread/README.md	
+++ /dev/null
@@ -1,58 +0,0 @@
-# [924. Minimize Malware Spread](https://leetcode.com/problems/minimize-malware-spread/)
-
-
-## 题目:
-
-In a network of nodes, each node `i` is directly connected to another node `j` if and only if `graph[i][j] = 1`.
-
-Some nodes `initial` are initially infected by malware. Whenever two nodes are directly connected and at least one of those two nodes is infected by malware, both nodes will be infected by malware. This spread of malware will continue until no more nodes can be infected in this manner.
-
-Suppose `M(initial)` is the final number of nodes infected with malware in the entire network, after the spread of malware stops.
-
-We will remove one node from the initial list. Return the node that if removed, would minimize `M(initial)`. If multiple nodes could be removed to minimize `M(initial)`, return such a node with the smallest index.
-
-Note that if a node was removed from the `initial` list of infected nodes, it may still be infected later as a result of the malware spread.
-
-**Example 1:**
-
-    Input: graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]
-    Output: 0
-
-**Example 2:**
-
-    Input: graph = [[1,0,0],[0,1,0],[0,0,1]], initial = [0,2]
-    Output: 0
-
-**Example 3:**
-
-    Input: graph = [[1,1,1],[1,1,1],[1,1,1]], initial = [1,2]
-    Output: 1
-
-**Note:**
-
-1. `1 < graph.length = graph[0].length <= 300`
-2. `0 <= graph[i][j] == graph[j][i] <= 1`
-3. `graph[i][i] = 1`
-4. `1 <= initial.length < graph.length`
-5. `0 <= initial[i] < graph.length`
-
-
-## 题目大意
-
-在节点网络中，只有当 graph[i][j] = 1 时，每个节点 i 能够直接连接到另一个节点 j。一些节点 initial 最初被恶意软件感染。只要两个节点直接连接，且其中至少一个节点受到恶意软件的感染，那么两个节点都将被恶意软件感染。这种恶意软件的传播将继续，直到没有更多的节点可以被这种方式感染。假设 M(initial) 是在恶意软件停止传播之后，整个网络中感染恶意软件的最终节点数。我们可以从初始列表中删除一个节点。如果移除这一节点将最小化 M(initial)， 则返回该节点。如果有多个节点满足条件，就返回索引最小的节点。请注意，如果某个节点已从受感染节点的列表 initial 中删除，它以后可能仍然因恶意软件传播而受到感染。
-
-
-提示：
-
-- 1 < graph.length = graph[0].length <= 300
-- 0 <= graph[i][j] == graph[j][i] <= 1
-- graph[i][i] = 1
-- 1 <= initial.length < graph.length
-- 0 <= initial[i] < graph.length
-
-
-## 解题思路
-
-
-- 给出一个节点之间的关系图，如果两个节点是连通的，那么病毒软件就会感染到连通的所有节点。现在如果想移除一个病毒节点，能最大减少感染，请问移除哪个节点？如果多个节点都能减少感染量，优先移除序号偏小的那个节点。
-- 这一题一看就是考察的并查集。利用节点的连通关系，把题目中给的所有节点都 `union()` 起来，然后依次统计每个集合内有多少个点。最后扫描一次 initial 数组，选出这个数组中节点小的并且所在集合节点多，这个节点就是最终答案。
diff --git a/Algorithms/0925. Long Pressed Name/925. Long Pressed Name.go b/Algorithms/0925. Long Pressed Name/925. Long Pressed Name.go
deleted file mode 100644
index 69b27ea54..000000000
--- a/Algorithms/0925. Long Pressed Name/925. Long Pressed Name.go	
+++ /dev/null
@@ -1,26 +0,0 @@
-package leetcode
-
-func isLongPressedName(name string, typed string) bool {
-	if len(name) == 0 && len(typed) == 0 {
-		return true
-	}
-	if (len(name) == 0 && len(typed) != 0) || (len(name) != 0 && len(typed) == 0) {
-		return false
-	}
-
-	j := 0
-	for i := 0; i < len(name); i++ {
-		if j < len(typed) && name[i] == typed[j] {
-			j++
-			continue
-		} else {
-			if i > 0 && j < len(typed) && name[i-1] == typed[j] {
-				j++
-				i--
-			} else {
-				return false
-			}
-		}
-	}
-	return true
-}
diff --git a/Algorithms/0925. Long Pressed Name/README.md b/Algorithms/0925. Long Pressed Name/README.md
deleted file mode 100644
index 95dfbaf51..000000000
--- a/Algorithms/0925. Long Pressed Name/README.md	
+++ /dev/null
@@ -1,67 +0,0 @@
-# [925. Long Pressed Name](https://leetcode.com/problems/long-pressed-name/)
-
-## 题目
-
-Your friend is typing his name into a keyboard.  Sometimes, when typing a character c, the key might get long pressed, and the character will be typed 1 or more times.
-
-You examine the typed characters of the keyboard.  Return True if it is possible that it was your friends name, with some characters (possibly none) being long pressed.
-
-
-
-Example 1:
-
-```c
-Input: name = "alex", typed = "aaleex"
-Output: true
-Explanation: 'a' and 'e' in 'alex' were long pressed.
-```
-
-Example 2:
-
-```c
-Input: name = "saeed", typed = "ssaaedd"
-Output: false
-Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.
-```
-
-Example 3:
-
-```c
-Input: name = "leelee", typed = "lleeelee"
-Output: true
-```
-
-Example 4:
-
-```c
-Input: name = "laiden", typed = "laiden"
-Output: true
-Explanation: It's not necessary to long press any character.
-```
-
-
-Note:  
-
-1. name.length <= 1000
-2. typed.length <= 1000
-3. The characters of name and typed are lowercase letters.
-
-## 题目大意
-
-
-给定 2 个字符串，后者的字符串中包含前者的字符串。比如在打字的过程中，某个字符会多按了几下。判断后者字符串是不是比前者字符串存在这样的“长按”键盘的情况。
-
-## 解题思路
-
-这一题也可以借助滑动窗口的思想。2 个字符串一起比较，如果遇到有相同的字符串，窗口继续往后滑动。直到遇到了第一个不同的字符，如果遇到两个字符串不相等的情况，可以直接返回 false。具体实现见代码。
-
-
-
-
-
-
-
-
-
-
-
diff --git a/Algorithms/0928. Minimize Malware Spread II/928. Minimize Malware Spread II_test.go b/Algorithms/0928. Minimize Malware Spread II/928. Minimize Malware Spread II_test.go
deleted file mode 100644
index 7d8552c29..000000000
--- a/Algorithms/0928. Minimize Malware Spread II/928. Minimize Malware Spread II_test.go	
+++ /dev/null
@@ -1,72 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question928 struct {
-	para928
-	ans928
-}
-
-// para 是参数
-// one 代表第一个参数
-type para928 struct {
-	graph   [][]int
-	initial []int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans928 struct {
-	one int
-}
-
-func Test_Problem928(t *testing.T) {
-
-	qs := []question928{
-
-		question928{
-			para928{[][]int{[]int{1, 0, 0, 0}, []int{0, 1, 1, 1}, []int{0, 1, 1, 0}, []int{0, 1, 0, 1}}, []int{1, 3}},
-			ans928{1},
-		},
-		question928{
-			para928{[][]int{[]int{1, 1, 1, 0}, []int{1, 1, 0, 0}, []int{1, 0, 1, 0}, []int{0, 0, 0, 1}}, []int{3, 2}},
-			ans928{2},
-		},
-
-		question928{
-			para928{[][]int{[]int{1, 1, 0}, []int{1, 1, 1}, []int{0, 1, 1}}, []int{0, 1}},
-			ans928{1},
-		},
-
-		question928{
-			para928{[][]int{[]int{1, 1, 0}, []int{1, 1, 0}, []int{0, 0, 1}}, []int{0, 1}},
-			ans928{0},
-		},
-
-		question928{
-			para928{[][]int{[]int{1, 1, 0, 0}, []int{1, 1, 1, 0}, []int{0, 1, 1, 1}, []int{0, 0, 1, 1}}, []int{0, 1}},
-			ans928{1},
-		},
-
-		question928{
-			para928{[][]int{[]int{1, 0, 0, 0}, []int{0, 1, 0, 0}, []int{0, 0, 1, 1}, []int{0, 0, 1, 1}}, []int{3, 1}},
-			ans928{3},
-		},
-
-		question928{
-			para928{[][]int{[]int{1, 0, 0, 0, 0, 0}, []int{0, 1, 0, 0, 0, 0}, []int{0, 0, 1, 0, 0, 0}, []int{0, 0, 0, 1, 1, 0}, []int{0, 0, 0, 1, 1, 0}, []int{0, 0, 0, 0, 0, 1}}, []int{5, 0}},
-			ans928{0},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 928------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans928, q.para928
-		fmt.Printf("【input】:%v       【output】:%v\n", p, minMalwareSpread2(p.graph, p.initial))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0928. Minimize Malware Spread II/README.md b/Algorithms/0928. Minimize Malware Spread II/README.md
deleted file mode 100755
index 3e39df6a1..000000000
--- a/Algorithms/0928. Minimize Malware Spread II/README.md	
+++ /dev/null
@@ -1,61 +0,0 @@
-# [928. Minimize Malware Spread II](https://leetcode.com/problems/minimize-malware-spread-ii/)
-
-
-## 题目:
-
-(This problem is the same as *Minimize Malware Spread*, with the differences bolded.)
-
-In a network of nodes, each node `i` is directly connected to another node `j` if and only if `graph[i][j] = 1`.
-
-Some nodes `initial` are initially infected by malware. Whenever two nodes are directly connected and at least one of those two nodes is infected by malware, both nodes will be infected by malware. This spread of malware will continue until no more nodes can be infected in this manner.
-
-Suppose `M(initial)` is the final number of nodes infected with malware in the entire network, after the spread of malware stops.
-
-We will remove one node from the initial list, **completely removing it and any connections from this node to any other node**. Return the node that if removed, would minimize `M(initial)`. If multiple nodes could be removed to minimize `M(initial)`, return such a node with the smallest index.
-
-**Example 1:**
-
-    Input: graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]
-    Output: 0
-
-**Example 2:**
-
-    Input: graph = [[1,1,0],[1,1,1],[0,1,1]], initial = [0,1]
-    Output: 1
-
-**Example 3:**
-
-    Input: graph = [[1,1,0,0],[1,1,1,0],[0,1,1,1],[0,0,1,1]], initial = [0,1]
-    Output: 1
-
-**Note:**
-
-1. `1 < graph.length = graph[0].length <= 300`
-2. `0 <= graph[i][j] == graph[j][i] <= 1`
-3. `graph[i][i] = 1`
-4. `1 <= initial.length < graph.length`
-5. `0 <= initial[i] < graph.length`
-
-
-## 题目大意
-
-(这个问题与 尽量减少恶意软件的传播 是一样的，不同之处用粗体表示。)在节点网络中，只有当 graph[i][j] = 1 时，每个节点 i 能够直接连接到另一个节点 j。一些节点 initial 最初被恶意软件感染。只要两个节点直接连接，且其中至少一个节点受到恶意软件的感染，那么两个节点都将被恶意软件感染。这种恶意软件的传播将继续，直到没有更多的节点可以被这种方式感染。假设 M(initial) 是在恶意软件停止传播之后，整个网络中感染恶意软件的最终节点数。我们可以从初始列表中删除一个节点，并完全移除该节点以及从该节点到任何其他节点的任何连接。如果移除这一节点将最小化 M(initial)， 则返回该节点。如果有多个节点满足条件，就返回索引最小的节点。
-
-提示：
-
-- 1 < graph.length = graph[0].length <= 300
-- 0 <= graph[i][j] == graph[j][i] <= 1
-- graph[i][i] = 1
-- 1 <= initial.length < graph.length
-- 0 <= initial[i] < graph.length
-
-
-## 解题思路
-
-
-- 这一题是第 924 题的加强版。给出一个节点之间的关系图，如果两个节点是连通的，那么病毒软件就会感染到连通的所有节点。现在如果想**完全彻底**移除一个病毒节点，能最大减少感染，请问移除哪个节点？如果多个节点都能减少感染量，优先移除序号偏小的那个节点。这一题的输入输出要求和第 924 题是完全一样的，区别在于第 924 题实际上是要求把一个病毒节点变成非病毒节点，而这道题是完全删除一个病毒节点以及它连接的所有边。
-- 这一题考察的是并查集。当然用 DFS 也可以解答这一题。并查集的做法如下，首先先将所有的病毒节点去掉，然后将所有连通块合并成一个节点。因为一个连通集合中的节点，要么全部被感染，要么全部不被感染，所以可以把每个集合整体考虑。然后统计所有集合直接相邻的病毒节点的个数。对于一个集合来说：
-    1. 如果直接相邻的病毒节点的个数为 0，则一定不会被感染，忽略这种情况；
-    2. 如果直接相邻的病毒节点的个数为 1，则将该病毒节点删除后，整个连通块就可以避免被感染，这种情况是我们寻找的答案；
-    3. 如果直接相邻的病毒节点的个数大于等于2，则不管删除哪个病毒节点，该连通块都仍会被感染，忽略这种情况；
-- 所以只需在所有第二种连通块（直接相邻的病毒节点的个数为 1 的连通块）中，找出节点个数最多的连通块，与它相邻的病毒节点就是我们要删除的节点；如果有多个连通块节点个数相同，再找出与之对应的编号最小的病毒节点即可。
diff --git a/Algorithms/0946. Validate Stack Sequences/946. Validate Stack Sequences.go b/Algorithms/0946. Validate Stack Sequences/946. Validate Stack Sequences.go
deleted file mode 100644
index a98ce9435..000000000
--- a/Algorithms/0946. Validate Stack Sequences/946. Validate Stack Sequences.go	
+++ /dev/null
@@ -1,17 +0,0 @@
-package leetcode
-
-import "fmt"
-
-func validateStackSequences(pushed []int, popped []int) bool {
-	stack, j, N := []int{}, 0, len(pushed)
-	for _, x := range pushed {
-		stack = append(stack, x)
-		fmt.Printf("stack = %v j = %v\n", stack, j)
-		for len(stack) != 0 && j < N && stack[len(stack)-1] == popped[j] {
-			stack = stack[0 : len(stack)-1]
-			j++
-		}
-		fmt.Printf("*****stack = %v j = %v\n", stack, j)
-	}
-	return j == N
-}
diff --git a/Algorithms/0947. Most Stones Removed with Same Row or Column/947. Most Stones Removed with Same Row or Column_test.go b/Algorithms/0947. Most Stones Removed with Same Row or Column/947. Most Stones Removed with Same Row or Column_test.go
deleted file mode 100644
index 54328dced..000000000
--- a/Algorithms/0947. Most Stones Removed with Same Row or Column/947. Most Stones Removed with Same Row or Column_test.go	
+++ /dev/null
@@ -1,51 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question947 struct {
-	para947
-	ans947
-}
-
-// para 是参数
-// one 代表第一个参数
-type para947 struct {
-	stones [][]int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans947 struct {
-	one int
-}
-
-func Test_Problem947(t *testing.T) {
-
-	qs := []question947{
-		question947{
-			para947{[][]int{[]int{0, 0}, []int{0, 1}, []int{1, 0}, []int{1, 2}, []int{2, 1}, []int{2, 2}}},
-			ans947{5},
-		},
-
-		question947{
-			para947{[][]int{[]int{0, 0}, []int{0, 2}, []int{1, 1}, []int{2, 0}, []int{2, 2}}},
-			ans947{3},
-		},
-
-		question947{
-			para947{[][]int{[]int{0, 0}}},
-			ans947{0},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 947------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans947, q.para947
-		fmt.Printf("【input】:%v       【output】:%v\n", p, removeStones(p.stones))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0947. Most Stones Removed with Same Row or Column/README.md b/Algorithms/0947. Most Stones Removed with Same Row or Column/README.md
deleted file mode 100755
index 1021516fb..000000000
--- a/Algorithms/0947. Most Stones Removed with Same Row or Column/README.md	
+++ /dev/null
@@ -1,48 +0,0 @@
-# [947. Most Stones Removed with Same Row or Column](https://leetcode.com/problems/most-stones-removed-with-same-row-or-column/)
-
-
-## 题目:
-
-On a 2D plane, we place stones at some integer coordinate points. Each coordinate point may have at most one stone.
-
-Now, a *move* consists of removing a stone that shares a column or row with another stone on the grid.
-
-What is the largest possible number of moves we can make?
-
-**Example 1:**
-
-    Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
-    Output: 5
-
-**Example 2:**
-
-    Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
-    Output: 3
-
-**Example 3:**
-
-    Input: stones = [[0,0]]
-    Output: 0
-
-**Note:**
-
-1. `1 <= stones.length <= 1000`
-2. `0 <= stones[i][j] < 10000`
-
-
-## 题目大意
-
-在二维平面上，我们将石头放置在一些整数坐标点上。每个坐标点上最多只能有一块石头。现在，move 操作将会移除与网格上的某一块石头共享一列或一行的一块石头。我们最多能执行多少次 move 操作？
-
-提示：
-
-- 1 <= stones.length <= 1000
-- 0 <= stones[i][j] < 10000
-
-
-## 解题思路
-
-
-- 给出一个数组，数组中的元素是一系列的坐标点。现在可以移除一些坐标点，移除必须满足：移除的这个点，在相同的行或者列上有一个点。问最终可以移除多少个点。移除到最后必然有些点独占一行，那么这些点都不能被移除。
-- 这一题的解题思路是并查集。把所有共行或者共列的点都 `union()` 起来。不同集合之间是不能相互移除的。反证法：如果能移除，代表存在共行或者共列的情况，那么肯定是同一个集合了，这样就不满足不同集合了。最终剩下的点就是集合的个数，每个集合只会留下一个点。所以移除的点就是点的总数减去集合的个数 `len(stones) - uf.totalCount()`。
-- 如果暴力合并集合，时间复杂度也非常差，可以由优化的地方。再遍历所有点的过程中，可以把遍历过的行和列存起来。这里可以用 map 来记录，key 为行号，value 为上一次遍历过的点的序号。同样，列也可以用 map 存起来，key 为列号，value 为上一次遍历过的点的序号。经过这样的优化以后，时间复杂度会提高不少。
diff --git a/Algorithms/0952. Largest Component Size by Common Factor/README.md b/Algorithms/0952. Largest Component Size by Common Factor/README.md
deleted file mode 100755
index 52074d799..000000000
--- a/Algorithms/0952. Largest Component Size by Common Factor/README.md	
+++ /dev/null
@@ -1,56 +0,0 @@
-# [952. Largest Component Size by Common Factor](https://leetcode.com/problems/largest-component-size-by-common-factor/)
-
-
-## 题目:
-
-Given a non-empty array of unique positive integers `A`, consider the following graph:
-
-- There are `A.length` nodes, labelled `A[0]` to `A[A.length - 1];`
-- There is an edge between `A[i]` and `A[j]` if and only if `A[i]`and `A[j]` share a common factor greater than 1.
-
-Return the size of the largest connected component in the graph.
-
-**Example 1:**
-
-    Input: [4,6,15,35]
-    Output: 4
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/01/ex1.png)
-
-**Example 2:**
-
-    Input: [20,50,9,63]
-    Output: 2
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/01/ex2.png)
-
-**Example 3:**
-
-    Input: [2,3,6,7,4,12,21,39]
-    Output: 8
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/01/ex3.png)
-
-**Note:**
-
-1. `1 <= A.length <= 20000`
-2. `1 <= A[i] <= 100000`
-
-
-## 题目大意
-
-给定一个由不同正整数的组成的非空数组 A，考虑下面的图：
-
-有 A.length 个节点，按从 A[0] 到 A[A.length - 1] 标记；
-只有当 A[i] 和 A[j] 共用一个大于 1 的公因数时，A[i] 和 A[j] 之间才有一条边。
-返回图中最大连通组件的大小。
-
-提示：
-
-1. 1 <= A.length <= 20000
-2. 1 <= A[i] <= 100000
-
-## 解题思路
-
-- 给出一个数组，数组中的元素如果每两个元素有公约数，那么这两个元素可以算有关系。所有有关系的数可以放在一个集合里，问这个数组里面有关系的元素组成的集合里面最多有多少个元素。
-- 这一题读完题直觉就是用并查集来解题。首先可以用暴力的解法尝试。用 2 层循环，两两比较有没有公约数，如果有公约数就 `union()` 到一起。提交以后出现 TLE，其实看一下数据规模就知道会超时，`1 <= A.length <= 20000`。注意到 `1 <= A[i] <= 100000`，开根号以后最后才 316.66666，这个规模的数不大。所以把每个数小于根号自己的因子都找出来，例如 `6 = 2 * 3`，`15 = 3 * 5`，那么把 6 和 2，6 和 3 都 `union()`，15 和 3，15 和 5 都 `union()`，最终遍历所有的集合，找到最多元素的集合，输出它包含的元素值。
diff --git a/Algorithms/0953. Verifying an Alien Dictionary/README.md b/Algorithms/0953. Verifying an Alien Dictionary/README.md
deleted file mode 100755
index 3a1cadacc..000000000
--- a/Algorithms/0953. Verifying an Alien Dictionary/README.md	
+++ /dev/null
@@ -1,45 +0,0 @@
-# [953. Verifying an Alien Dictionary](https://leetcode.com/problems/verifying-an-alien-dictionary/)
-
-
-## 题目:
-
-In an alien language, surprisingly they also use english lowercase letters, but possibly in a different `order`. The `order`of the alphabet is some permutation of lowercase letters.
-
-Given a sequence of `words` written in the alien language, and the `order` of the alphabet, return `true` if and only if the given `words` are sorted lexicographicaly in this alien language.
-
-**Example 1:**
-
-    Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
-    Output: true
-    Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.
-
-**Example 2:**
-
-    Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
-    Output: false
-    Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.
-
-**Example 3:**
-
-    Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
-    Output: false
-    Explanation: The first three characters "app" match, and the second string is shorter (in size.) According to lexicographical rules "apple" > "app", because 'l' > '∅', where '∅' is defined as the blank character which is less than any other character (More info).
-
-**Note:**
-
-1. `1 <= words.length <= 100`
-2. `1 <= words[i].length <= 20`
-3. `order.length == 26`
-4. All characters in `words[i]` and `order` are english lowercase letters.
-
-
-## 题目大意
-
-某种外星语也使用英文小写字母，但可能顺序 order 不同。字母表的顺序（order）是一些小写字母的排列。给定一组用外星语书写的单词 words，以及其字母表的顺序 order，只有当给定的单词在这种外星语中按字典序排列时，返回 true；否则，返回 false。
-
-
-
-## 解题思路
-
-
-- 这一题是简单题。给出一个字符串数组，判断把字符串数组里面字符串是否是按照 order 的排序排列的。order 是给出个一个字符串排序。这道题的解法是把 26 个字母的顺序先存在 map 中，然后依次遍历判断字符串数组里面字符串的大小。
diff --git a/Algorithms/0959. Regions Cut By Slashes/README.md b/Algorithms/0959. Regions Cut By Slashes/README.md
deleted file mode 100755
index 918e05310..000000000
--- a/Algorithms/0959. Regions Cut By Slashes/README.md	
+++ /dev/null
@@ -1,98 +0,0 @@
-# [959. Regions Cut By Slashes](https://leetcode.com/problems/regions-cut-by-slashes/)
-
-
-## 题目:
-
-In a N x N `grid` composed of 1 x 1 squares, each 1 x 1 square consists of a `/`, `\`, or blank space. These characters divide the square into contiguous regions.
-
-(Note that backslash characters are escaped, so a `\` is represented as `"\\"`.)
-
-Return the number of regions.
-
-**Example 1:**
-
-    Input:
-    [
-      " /",
-      "/ "
-    ]
-    Output: 2
-    Explanation: The 2x2 grid is as follows:
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/1.png)
-
-**Example 2:**
-
-    Input:
-    [
-      " /",
-      "  "
-    ]
-    Output: 1
-    Explanation: The 2x2 grid is as follows:
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/2.png)
-
-**Example 3:**
-
-    Input:
-    [
-      "\\/",
-      "/\\"
-    ]
-    Output: 4
-    Explanation: (Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.)
-    The 2x2 grid is as follows:
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/3.png)
-
-**Example 4:**
-
-    Input:
-    [
-      "/\\",
-      "\\/"
-    ]
-    Output: 5
-    Explanation: (Recall that because \ characters are escaped, "/\\" refers to /\, and "\\/" refers to \/.)
-    The 2x2 grid is as follows:
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/4.png)
-
-**Example 5:**
-
-    Input:
-    [
-      "//",
-      "/ "
-    ]
-    Output: 3
-    Explanation: The 2x2 grid is as follows:
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/5.png)
-
-**Note:**
-
-1. `1 <= grid.length == grid[0].length <= 30`
-2. `grid[i][j]` is either `'/'`, `'\'`, or `' '`.
-
-
-## 题目大意
-
-在由 1 x 1 方格组成的 N x N 网格 grid 中，每个 1 x 1 方块由 /、\ 或空格构成。这些字符会将方块划分为一些共边的区域。(请注意，反斜杠字符是转义的，因此 \ 用 "\\" 表示)返回区域的数目。
-
-
-提示：
-
-- 1 <= grid.length == grid[0].length <= 30
-- grid[i][j] 是 '/'、'\'、或 ' '。
-
-## 解题思路
-
-
-- 给出一个字符串，代表的是 `N x N` 正方形中切分的情况，有 2 种切分的情况 `'\'` 和 `'/'` ，即从左上往右下切和从右上往左下切。问按照给出的切分方法，能把 `N x N` 正方形切成几部分？
-- 这一题解题思路是并查集。先将每个 `1*1` 的正方形切分成下图的样子。分成 4 小块。然后按照题目给的切分图来合并各个小块。
-
-![](https://img.halfrost.com/Leetcode/leetcode_959.png)
-
-- 遇到 `'\\'`，就把第 0 块和第 1 块 `union()` 起来，第 2 块和第 3 块 `union()` 起来；遇到 `'/'`，就把第 0 块和第 3 块 `union()` 起来，第 2 块和第 1 块 `union()` 起来；遇到 `' '`，就把第 0 块和第 1 块 `union()` 起来，第 2 块和第 1 块 `union()` 起来，第 2 块和第 3 块 `union()` 起来，即 4 块都 `union()` 起来；最后还需要记得上一行和下一行还需要 `union()`，即本行的第 2 块和下一行的第 0 块 `union()` 起来；左边一列和右边一列也需要 `union()`。即本列的第 1 块和右边一列的第 3 块 `union()` 起来。最后计算出集合总个数就是最终答案了。
diff --git a/Algorithms/0961. N-Repeated Element in Size 2N Array/README.md b/Algorithms/0961. N-Repeated Element in Size 2N Array/README.md
deleted file mode 100755
index 4ba8d3905..000000000
--- a/Algorithms/0961. N-Repeated Element in Size 2N Array/README.md	
+++ /dev/null
@@ -1,40 +0,0 @@
-# [961. N-Repeated Element in Size 2N Array](https://leetcode.com/problems/n-repeated-element-in-size-2n-array/)
-
-
-## 题目:
-
-In a array `A` of size `2N`, there are `N+1` unique elements, and exactly one of these elements is repeated N times.
-
-Return the element repeated `N` times.
-
-**Example 1:**
-
-    Input: [1,2,3,3]
-    Output: 3
-
-**Example 2:**
-
-    Input: [2,1,2,5,3,2]
-    Output: 2
-
-**Example 3:**
-
-    Input: [5,1,5,2,5,3,5,4]
-    Output: 5
-
-**Note:**
-
-1. `4 <= A.length <= 10000`
-2. `0 <= A[i] < 10000`
-3. `A.length` is even
-
-
-## 题目大意
-
-在大小为 2N 的数组 A 中有 N+1 个不同的元素，其中有一个元素重复了 N 次。返回重复了 N 次的那个元素。
-
-
-## 解题思路
-
-
-- 简单题。数组中有 2N 个数，有 N + 1 个数是不重复的，这之中有一个数重复了 N 次，请找出这个数。解法非常简单，把所有数都存入 map 中，如果遇到存在的 key 就返回这个数。
diff --git a/Algorithms/0970. Powerful Integers/README.md b/Algorithms/0970. Powerful Integers/README.md
deleted file mode 100755
index 7f49306b9..000000000
--- a/Algorithms/0970. Powerful Integers/README.md	
+++ /dev/null
@@ -1,45 +0,0 @@
-# [970. Powerful Integers](https://leetcode.com/problems/powerful-integers/)
-
-
-## 题目:
-
-Given two positive integers `x` and `y`, an integer is *powerful* if it is equal to `x^i + y^j` for some integers `i >= 0` and `j >= 0`.
-
-Return a list of all *powerful* integers that have value less than or equal to `bound`.
-
-You may return the answer in any order. In your answer, each value should occur at most once.
-
-**Example 1:**
-
-    Input: x = 2, y = 3, bound = 10
-    Output: [2,3,4,5,7,9,10]
-    Explanation: 
-    2 = 2^0 + 3^0
-    3 = 2^1 + 3^0
-    4 = 2^0 + 3^1
-    5 = 2^1 + 3^1
-    7 = 2^2 + 3^1
-    9 = 2^3 + 3^0
-    10 = 2^0 + 3^2
-
-**Example 2:**
-
-    Input: x = 3, y = 5, bound = 15
-    Output: [2,4,6,8,10,14]
-
-**Note:**
-
-- `1 <= x <= 100`
-- `1 <= y <= 100`
-- `0 <= bound <= 10^6`
-
-
-## 题目大意
-
-给定两个正整数 x 和 y，如果某一整数等于 x^i + y^j，其中整数 i >= 0 且 j >= 0，那么我们认为该整数是一个强整数。返回值小于或等于 bound 的所有强整数组成的列表。你可以按任何顺序返回答案。在你的回答中，每个值最多出现一次。
-
-
-## 解题思路
-
-
-- 简答题，题目要求找出满足 `x^i + y^j ≤ bound` 条件的所有解。题目要求输出中不能重复，所以用 map 来去重。剩下的就是 `n^2` 暴力循环枚举所有解。
diff --git a/Algorithms/0973. K Closest Points to Origin/973. K Closest Points to Origin_test.go b/Algorithms/0973. K Closest Points to Origin/973. K Closest Points to Origin_test.go
deleted file mode 100644
index ea223ffa6..000000000
--- a/Algorithms/0973. K Closest Points to Origin/973. K Closest Points to Origin_test.go	
+++ /dev/null
@@ -1,63 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question973 struct {
-	para973
-	ans973
-}
-
-// para 是参数
-// one 代表第一个参数
-type para973 struct {
-	one [][]int
-	two int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans973 struct {
-	one [][]int
-}
-
-func Test_Problem973(t *testing.T) {
-
-	qs := []question973{
-
-		question973{
-			para973{[][]int{[]int{1, 3}, []int{-2, 2}}, 1},
-			ans973{[][]int{[]int{-2, 2}}},
-		},
-
-		question973{
-			para973{[][]int{[]int{}, []int{}}, 1},
-			ans973{[][]int{[]int{}}},
-		},
-
-		question973{
-			para973{[][]int{[]int{1, 3}, []int{-2, 2}}, 0},
-			ans973{[][]int{[]int{}}},
-		},
-
-		question973{
-			para973{[][]int{[]int{3, 3}, []int{5, -1}, []int{-2, 4}}, 2},
-			ans973{[][]int{[]int{3, 3}, []int{-2, 4}}},
-		},
-
-		question973{
-			para973{[][]int{[]int{0, 1}, []int{1, 0}}, 2},
-			ans973{[][]int{[]int{1, 0}, []int{0, 1}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 973------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans973, q.para973
-		fmt.Printf("【input】:%v       【output】:%v\n", p, KClosest(p.one, p.two))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0976. Largest Perimeter Triangle/976. Largest Perimeter Triangle.go b/Algorithms/0976. Largest Perimeter Triangle/976. Largest Perimeter Triangle.go
deleted file mode 100644
index 03f426cb7..000000000
--- a/Algorithms/0976. Largest Perimeter Triangle/976. Largest Perimeter Triangle.go	
+++ /dev/null
@@ -1,14 +0,0 @@
-package leetcode
-
-func largestPerimeter(A []int) int {
-	if len(A) < 3 {
-		return 0
-	}
-	quickSort164(A, 0, len(A)-1)
-	for i := len(A) - 1; i >= 2; i-- {
-		if (A[i]+A[i-1] > A[i-2]) && (A[i]+A[i-2] > A[i-1]) && (A[i-2]+A[i-1] > A[i]) {
-			return A[i] + A[i-1] + A[i-2]
-		}
-	}
-	return 0
-}
diff --git a/Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray.go b/Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray.go
deleted file mode 100644
index 6c04a15cb..000000000
--- a/Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray.go	
+++ /dev/null
@@ -1,43 +0,0 @@
-package leetcode
-
-// 解法一 模拟法
-func maxTurbulenceSize(A []int) int {
-	inc, dec := 1, 1
-	maxLen := min(1, len(A))
-	for i := 1; i < len(A); i++ {
-		if A[i-1] < A[i] {
-			inc = dec + 1
-			dec = 1
-		} else if A[i-1] > A[i] {
-			dec = inc + 1
-			inc = 1
-		} else {
-			inc = 1
-			dec = 1
-		}
-		maxLen = max(maxLen, max(inc, dec))
-	}
-	return maxLen
-}
-
-// 解法二 滑动窗口
-func maxTurbulenceSize1(A []int) int {
-	if len(A) == 1 {
-		return 1
-	}
-	// flag > 0 代表下一个数要大于前一个数，flag < 0 代表下一个数要小于前一个数
-	res, left, right, flag, lastNum := 0, 0, 0, A[1]-A[0], A[0]
-	for left < len(A) {
-		if right < len(A)-1 && ((A[right+1] > lastNum && flag > 0) || (A[right+1] < lastNum && flag < 0) || (right == left)) {
-			right++
-			flag = lastNum - A[right]
-			lastNum = A[right]
-		} else {
-			if right != left && flag != 0 {
-				res = max(res, right-left+1)
-			}
-			left++
-		}
-	}
-	return max(res, 1)
-}
diff --git a/Algorithms/0978. Longest Turbulent Subarray/README.md b/Algorithms/0978. Longest Turbulent Subarray/README.md
deleted file mode 100755
index 4b0bbd490..000000000
--- a/Algorithms/0978. Longest Turbulent Subarray/README.md	
+++ /dev/null
@@ -1,58 +0,0 @@
-# [978. Longest Turbulent Subarray](https://leetcode.com/problems/longest-turbulent-subarray/)
-
-## 题目:
-
-A subarray `A[i], A[i+1], ..., A[j]` of `A` is said to be *turbulent* if and only if:
-
-- For `i <= k < j`, `A[k] > A[k+1]` when `k` is odd, and `A[k] < A[k+1]` when `k` is even;
-- **OR**, for `i <= k < j`, `A[k] > A[k+1]` when `k` is even, and `A[k] < A[k+1]` when `k` is odd.
-
-That is, the subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray.
-
-Return the **length** of a maximum size turbulent subarray of A.
-
-**Example 1:**
-
-    Input: [9,4,2,10,7,8,8,1,9]
-    Output: 5
-    Explanation: (A[1] > A[2] < A[3] > A[4] < A[5])
-
-**Example 2:**
-
-    Input: [4,8,12,16]
-    Output: 2
-
-**Example 3:**
-
-    Input: [100]
-    Output: 1
-
-**Note:**
-
-1. `1 <= A.length <= 40000`
-2. `0 <= A[i] <= 10^9`
-
-
-## 题目大意
-
-
-当 A 的子数组 A[i], A[i+1], ..., A[j] 满足下列条件时，我们称其为湍流子数组：
-
-若 i <= k < j，当 k 为奇数时， A[k] > A[k+1]，且当 k 为偶数时，A[k] < A[k+1]；
-或 若 i <= k < j，当 k 为偶数时，A[k] > A[k+1] ，且当 k 为奇数时， A[k] < A[k+1]。
-也就是说，如果比较符号在子数组中的每个相邻元素对之间翻转，则该子数组是湍流子数组。
-
-返回 A 的最大湍流子数组的长度。
-
-提示：
-
-- 1 <= A.length <= 40000
-- 0 <= A[i] <= 10^9
-
-
-
-## 解题思路
-
-
-- 给出一个数组，要求找出“摆动数组”的最大长度。所谓“摆动数组”的意思是，元素一大一小间隔的。
-- 这一题可以用滑动窗口来解答。用一个变量记住下次出现的元素需要大于还是需要小于前一个元素。也可以用模拟的方法，用两个变量分别记录上升和下降数字的长度。一旦元素相等了，上升和下降数字长度都置为 1，其他时候按照上升和下降的关系增加队列长度即可，最后输出动态维护的最长长度。
diff --git a/Algorithms/0984. String Without AAA or BBB/README.md b/Algorithms/0984. String Without AAA or BBB/README.md
deleted file mode 100755
index 9b7a0406e..000000000
--- a/Algorithms/0984. String Without AAA or BBB/README.md	
+++ /dev/null
@@ -1,49 +0,0 @@
-# [984. String Without AAA or BBB](https://leetcode.com/problems/string-without-aaa-or-bbb/)
-
-
-## 题目:
-
-Given two integers `A` and `B`, return **any** string `S` such that:
-
-- `S` has length `A + B` and contains exactly `A` `'a'` letters, and exactly `B` `'b'`letters;
-- The substring `'aaa'` does not occur in `S`;
-- The substring `'bbb'` does not occur in `S`.
-
-**Example 1:**
-
-    Input: A = 1, B = 2
-    Output: "abb"
-    Explanation: "abb", "bab" and "bba" are all correct answers.
-
-**Example 2:**
-
-    Input: A = 4, B = 1
-    Output: "aabaa"
-
-**Note:**
-
-1. `0 <= A <= 100`
-2. `0 <= B <= 100`
-3. It is guaranteed such an `S` exists for the given `A` and `B`.
-
-
-## 题目大意
-
-给定两个整数 A 和 B，返回任意字符串 S，要求满足：
-
-- S 的长度为 A + B，且正好包含 A 个 'a' 字母与 B 个 'b' 字母；
-- 子串 'aaa' 没有出现在 S 中；
-- 子串 'bbb' 没有出现在 S 中。
-
-
-提示：
-
-- 0 <= A <= 100
-- 0 <= B <= 100
-- 对于给定的 A 和 B，保证存在满足要求的 S。
-
-
-## 解题思路
-
-
-- 给出 A 和 B 的个数，要求组合出字符串，不能出现 3 个连续的 A 和 3 个连续的 B。这题由于只可能有 4 种情况，暴力枚举就可以了。假设 B 的个数比 A 多(如果 A 多，就交换一下 A 和 B)，最终可能的情况只可能是这 4 种情况中的一种： `ba`，`bbabb`，`bbabbabb`，`bbabbabbabbabababa`。
diff --git a/Algorithms/0986. Interval List Intersections/986. Interval List Intersections.go b/Algorithms/0986. Interval List Intersections/986. Interval List Intersections.go
deleted file mode 100644
index 0edc4b504..000000000
--- a/Algorithms/0986. Interval List Intersections/986. Interval List Intersections.go	
+++ /dev/null
@@ -1,25 +0,0 @@
-package leetcode
-
-/**
- * Definition for an interval.
- * type Interval struct {
- *	   Start int
- *	   End   int
- * }
- */
-func intervalIntersection(A []Interval, B []Interval) []Interval {
-	res := []Interval{}
-	for i, j := 0, 0; i < len(A) && j < len(B); {
-		start := max(A[i].Start, B[j].Start)
-		end := min(A[i].End, B[j].End)
-		if start <= end {
-			res = append(res, Interval{Start: start, End: end})
-		}
-		if A[i].End <= B[j].End {
-			i++
-		} else {
-			j++
-		}
-	}
-	return res
-}
diff --git a/Algorithms/0986. Interval List Intersections/986. Interval List Intersections_test.go b/Algorithms/0986. Interval List Intersections/986. Interval List Intersections_test.go
deleted file mode 100644
index a39c4a9ec..000000000
--- a/Algorithms/0986. Interval List Intersections/986. Interval List Intersections_test.go	
+++ /dev/null
@@ -1,45 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question986 struct {
-	para986
-	ans986
-}
-
-// para 是参数
-// one 代表第一个参数
-type para986 struct {
-	one []Interval
-	two []Interval
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans986 struct {
-	one []Interval
-}
-
-func Test_Problem986(t *testing.T) {
-
-	qs := []question986{
-
-		question986{
-			para986{[]Interval{Interval{Start: 0, End: 2}, Interval{Start: 5, End: 10}, Interval{Start: 13, End: 23}, Interval{Start: 24, End: 25}},
-				[]Interval{Interval{Start: 1, End: 5}, Interval{Start: 8, End: 12}, Interval{Start: 15, End: 24}, Interval{Start: 25, End: 26}}},
-			ans986{[]Interval{Interval{Start: 1, End: 2}, Interval{Start: 5, End: 5}, Interval{Start: 8, End: 10},
-				Interval{Start: 15, End: 23}, Interval{Start: 24, End: 24}, Interval{Start: 25, End: 25}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 986------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans986, q.para986
-		fmt.Printf("【input】:%v       【output】:%v\n", p, intervalIntersection(p.one, p.two))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/0990. Satisfiability of Equality Equations/README.md b/Algorithms/0990. Satisfiability of Equality Equations/README.md
deleted file mode 100755
index 2d32532b7..000000000
--- a/Algorithms/0990. Satisfiability of Equality Equations/README.md	
+++ /dev/null
@@ -1,65 +0,0 @@
-# [990. Satisfiability of Equality Equations](https://leetcode.com/problems/satisfiability-of-equality-equations/)
-
-
-## 题目:
-
-Given an array equations of strings that represent relationships between variables, each string `equations[i]` has length `4` and takes one of two different forms: `"a==b"` or `"a!=b"`. Here, `a` and `b` are lowercase letters (not necessarily different) that represent one-letter variable names.
-
-Return `true` if and only if it is possible to assign integers to variable names so as to satisfy all the given equations.
-
-**Example 1:**
-
-    Input: ["a==b","b!=a"]
-    Output: false
-    Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second.  There is no way to assign the variables to satisfy both equations.
-
-**Example 2:**
-
-    Input: ["b==a","a==b"]
-    Output: true
-    Explanation: We could assign a = 1 and b = 1 to satisfy both equations.
-
-**Example 3:**
-
-    Input: ["a==b","b==c","a==c"]
-    Output: true
-
-**Example 4:**
-
-    Input: ["a==b","b!=c","c==a"]
-    Output: false
-
-**Example 5:**
-
-    Input: ["c==c","b==d","x!=z"]
-    Output: true
-
-**Note:**
-
-1. `1 <= equations.length <= 500`
-2. `equations[i].length == 4`
-3. `equations[i][0]` and `equations[i][3]` are lowercase letters
-4. `equations[i][1]` is either `'='` or `'!'`
-5. `equations[i][2]` is `'='`
-
-
-
-## 题目大意
-
-给定一个由表示变量之间关系的字符串方程组成的数组，每个字符串方程 equations[i] 的长度为 4，并采用两种不同的形式之一："a==b" 或 "a!=b"。在这里，a 和 b 是小写字母（不一定不同），表示单字母变量名。只有当可以将整数分配给变量名，以便满足所有给定的方程时才返回 true，否则返回 false。 
-
-提示：
-
-1. 1 <= equations.length <= 500
-2. equations[i].length == 4
-3. equations[i][0] 和 equations[i][3] 是小写字母
-4. equations[i][1] 要么是 '='，要么是 '!'
-5. equations[i][2] 是 '='
-
-
-
-## 解题思路
-
-
-- 给出一个字符串数组，数组里面给出的是一些字母的关系，只有 `'=='` 和 `'! ='` 两种关系。问给出的这些关系中是否存在悖论？
-- 这一题是简单的并查集的问题。先将所有 `'=='` 关系的字母 `union()` 起来，然后再一一查看 `'! ='` 关系中是否有 `'=='` 关系的组合，如果有，就返回 `false`，如果遍历完都没有找到，则返回 `true`。
diff --git a/Algorithms/0995. Minimum Number of K Consecutive Bit Flips/README.md b/Algorithms/0995. Minimum Number of K Consecutive Bit Flips/README.md
deleted file mode 100755
index 8a47847d2..000000000
--- a/Algorithms/0995. Minimum Number of K Consecutive Bit Flips/README.md	
+++ /dev/null
@@ -1,67 +0,0 @@
-# [995. Minimum Number of K Consecutive Bit Flips](https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips/)
-
-
-## 题目:
-
-In an array `A` containing only 0s and 1s, a `K`-bit flip consists of choosing a (contiguous) subarray of length `K` and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0.
-
-Return the minimum number of `K`-bit flips required so that there is no 0 in the array. If it is not possible, return `-1`.
-
-**Example 1:**
-
-    Input: A = [0,1,0], K = 1
-    Output: 2
-    Explanation: Flip A[0], then flip A[2].
-
-**Example 2:**
-
-    Input: A = [1,1,0], K = 2
-    Output: -1
-    Explanation: No matter how we flip subarrays of size 2, we can't make the array become [1,1,1].
-
-**Example 3:**
-
-    Input: A = [0,0,0,1,0,1,1,0], K = 3
-    Output: 3
-    Explanation:
-    Flip A[0],A[1],A[2]: A becomes [1,1,1,1,0,1,1,0]
-    Flip A[4],A[5],A[6]: A becomes [1,1,1,1,1,0,0,0]
-    Flip A[5],A[6],A[7]: A becomes [1,1,1,1,1,1,1,1]
-
-**Note:**
-
-1. `1 <= A.length <= 30000`
-2. `1 <= K <= A.length`
-
-
-## 题目大意
-
-在仅包含 0 和 1 的数组 A 中，一次 K 位翻转包括选择一个长度为 K 的（连续）子数组，同时将子数组中的每个 0 更改为 1，而每个 1 更改为 0。返回所需的 K 位翻转的次数，以便数组没有值为 0 的元素。如果不可能，返回 -1。
-
-提示：
-
-1. 1 <= A.length <= 30000
-2. 1 <= K <= A.length
-
-
-## 解题思路
-
-
-- 给出一个数组，数组里面的元素只有 0 和 1。给一个长度为 K 的窗口，在这个窗口内的所有元素都会 0-1 翻转。问最后需要翻转几次，使得整个数组都为 1 。如果不能翻转使得最后数组元素都为 1，则输出 -1。
-- 拿到这一题首先想到的是贪心算法。例如第 765 题，这类题的描述都是这样的：在一个数组中或者环形数组中通过交换位置，或者翻转变换，达到最终结果，要求找到最少步数。贪心能保证是最小步数(证明略)。按照贪心的思想，这一题也这样做，从数组 0 下标开始往后扫，依次翻转每个 K 大小的窗口内元素。
-- 由于窗口大小限制了，所以这题滑动窗口只需要一个边界坐标，用左边界就可以判断了。每一个 `A[i]` 是否需要翻转，是由于 `[ i-k+1，i ]`、`[ i-k+2，i+1 ]`、`[ i-k+3，i+2 ]`……`[ i-1，i+k ]` 这一系列的窗口翻转`累积影响`的。那如何之前这些窗口`累积`到 `A[i]` 上翻转的次数呢？可以动态的维护一个翻转次数，当 `i` 摆脱了上一次翻转窗口 `K` 的时候，翻转次数就 `-1` 。举个例子：
-
-        A = [0 0 0 1 0 1 1 0] K = 3
-        
-        A = [2 0 0 1 0 1 1 0] i = 0 flippedTime = 1
-        A = [2 0 0 1 0 1 1 0] i = 1 flippedTime = 1
-        A = [2 0 0 1 0 1 1 0] i = 2 flippedTime = 1
-        A = [2 0 0 1 0 1 1 0] i = 3 flippedTime = 0
-        A = [2 0 0 1 2 1 1 0] i = 4 flippedTime = 1
-        A = [2 0 0 1 2 2 1 0] i = 5 flippedTime = 2
-        A = [2 0 0 1 2 2 1 0] i = 6 flippedTime = 2
-        A = [2 0 0 1 2 2 1 0] i = 7 flippedTime = 1
-
-    当判断 `A[i]` 是否需要翻转的时候，只需要留意每个宽度为 `K` 窗口的左边界。会影响到 A[i] 的窗口的左边界分别是 `i-k+1`、`i-k+2`、`i-k+3`、…… `i-1`，只需要分别看这些窗口有没有翻转就行。这里可以用特殊标记来记录这些窗口的左边界是否被翻转了。如果翻转过，则把窗口左边界的那个数字标记为 2 (为什么要标记为 2 呢？其实设置成什么都可以，只要不是 0 和 1 ，和原有的数字区分开就行)。当 `i≥k` 的时候，代表 `i` 已经脱离了 `i-k` 的这个窗口，因为能影响 `A[i]` 的窗口是从 `i-k+1` 开始的，如果 `A[i-k] == 2` 代表 `i-k` 窗口已经翻转过了，现在既然脱离了它的窗口影响，那么就要把累积的 `flippedTime - 1` 。这样就维护了累积 `flippedTime` 和滑动窗口中累积影响的关系。
-
-- 接下来还需要处理的是 `flippedTime` 与当前 `A[i]` 是否翻转的问题。如果 `flippedTime` 是偶数次，原来的 0 还是 0，就需要再次翻转，如果 `flippedTime` 是奇数次，原来的 0 变成了 1 就不需要翻转了。总结成一条结论就是 `A[i]` 与 `flippedTime` 同奇偶性的时候就要翻转。当 `i + K` 比 `len(A)` 大的时候，代表剩下的这些元素肯定不能在一个窗口里面翻转，则输出 -1 。
diff --git a/Algorithms/1002. Find Common Characters/README.md b/Algorithms/1002. Find Common Characters/README.md
deleted file mode 100755
index 612bb2149..000000000
--- a/Algorithms/1002. Find Common Characters/README.md	
+++ /dev/null
@@ -1,33 +0,0 @@
-# [1002. Find Common Characters](https://leetcode.com/problems/find-common-characters/)
-
-
-## 题目:
-
-Given an array `A` of strings made only from lowercase letters, return a list of all characters that show up in all strings within the list **(including duplicates)**. For example, if a character occurs 3 times in all strings but not 4 times, you need to include that character three times in the final answer.
-
-You may return the answer in any order.
-
-**Example 1:**
-
-    Input: ["bella","label","roller"]
-    Output: ["e","l","l"]
-
-**Example 2:**
-
-    Input: ["cool","lock","cook"]
-    Output: ["c","o"]
-
-**Note:**
-
-1. `1 <= A.length <= 100`
-2. `1 <= A[i].length <= 100`
-3. `A[i][j]` is a lowercase letter
-
-## 题目大意
-
-给定仅有小写字母组成的字符串数组 A，返回列表中的每个字符串中都显示的全部字符（包括重复字符）组成的列表。例如，如果一个字符在每个字符串中出现 3 次，但不是 4 次，则需要在最终答案中包含该字符 3 次。你可以按任意顺序返回答案。
-
-
-## 解题思路
-
-- 简单题。给出一个字符串数组 A，要求找出这个数组中每个字符串都包含字符，如果字符出现多次，在最终结果中也需要出现多次。这一题可以用 map 来统计每个字符串的频次，但是如果用数组统计会更快。题目中说了只有小写字母，那么用 2 个 26 位长度的数组就可以统计出来了。遍历字符串数组的过程中，不过的缩小每个字符在每个字符串中出现的频次(因为需要找所有字符串公共的字符，公共的频次肯定就是最小的频次)，得到了最终公共字符的频次数组以后，按顺序输出就可以了。
diff --git a/Algorithms/1011. Capacity To Ship Packages Within D Days/1011. Capacity To Ship Packages Within D Days.go b/Algorithms/1011. Capacity To Ship Packages Within D Days/1011. Capacity To Ship Packages Within D Days.go
deleted file mode 100644
index 34dd4ccb9..000000000
--- a/Algorithms/1011. Capacity To Ship Packages Within D Days/1011. Capacity To Ship Packages Within D Days.go	
+++ /dev/null
@@ -1,24 +0,0 @@
-package leetcode
-
-func shipWithinDays(weights []int, D int) int {
-	maxNum, sum := 0, 0
-	for _, num := range weights {
-		sum += num
-		if num > maxNum {
-			maxNum = num
-		}
-	}
-	if D == 1 {
-		return sum
-	}
-	low, high := maxNum, sum
-	for low < high {
-		mid := low + (high-low)>>1
-		if calSum(mid, D, weights) {
-			high = mid
-		} else {
-			low = mid + 1
-		}
-	}
-	return low
-}
diff --git a/Algorithms/1011. Capacity To Ship Packages Within D Days/README.md b/Algorithms/1011. Capacity To Ship Packages Within D Days/README.md
deleted file mode 100755
index 2da8d0d6b..000000000
--- a/Algorithms/1011. Capacity To Ship Packages Within D Days/README.md	
+++ /dev/null
@@ -1,69 +0,0 @@
-# [1011. Capacity To Ship Packages Within D Days](https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/)
-
-
-## 题目:
-
-A conveyor belt has packages that must be shipped from one port to another within `D` days.
-
-The `i`-th package on the conveyor belt has a weight of `weights[i]`. Each day, we load the ship with packages on the conveyor belt (in the order given by `weights`). We may not load more weight than the maximum weight capacity of the ship.
-
-Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within `D` days.
-
-**Example 1:**
-
-    Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5
-    Output: 15
-    Explanation: 
-    A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
-    1st day: 1, 2, 3, 4, 5
-    2nd day: 6, 7
-    3rd day: 8
-    4th day: 9
-    5th day: 10
-    
-    Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed.
-
-**Example 2:**
-
-    Input: weights = [3,2,2,4,1,4], D = 3
-    Output: 6
-    Explanation: 
-    A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
-    1st day: 3, 2
-    2nd day: 2, 4
-    3rd day: 1, 4
-
-**Example 3:**
-
-    Input: weights = [1,2,3,1,1], D = 4
-    Output: 3
-    Explanation: 
-    1st day: 1
-    2nd day: 2
-    3rd day: 3
-    4th day: 1, 1
-
-**Note:**
-
-1. `1 <= D <= weights.length <= 50000`
-2. `1 <= weights[i] <= 500`
-
-
-## 题目大意
-
-传送带上的包裹必须在 D 天内从一个港口运送到另一个港口。
-
-传送带上的第 i 个包裹的重量为 weights[i]。每一天，我们都会按给出重量的顺序往传送带上装载包裹。我们装载的重量不会超过船的最大运载重量。
-
-返回能在 D 天内将传送带上的所有包裹送达的船的最低运载能力。
-
-提示：
-
-- 1 <= D <= weights.length <= 50000
-- 1 <= weights[i] <= 500
-
-
-## 解题思路
-
-- 给出一个数组和天数 D，要求正好在 D 天把数组中的货物都运完。求传输带上能承受的最小货物重量是多少。
-- 这一题和第 410 题完全一样，只不过换了一个题面。代码完全不变。思路解析见第 410 题。
diff --git a/Algorithms/1017. Convert to Base -2/README.md b/Algorithms/1017. Convert to Base -2/README.md
deleted file mode 100755
index 7c4bbfdd0..000000000
--- a/Algorithms/1017. Convert to Base -2/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# [1017. Convert to Base -2](https://leetcode.com/problems/convert-to-base-2/)
-
-
-## 题目:
-
-Given a number `N`, return a string consisting of `"0"`s and `"1"`s that represents its value in base **`-2`** (negative two).
-
-The returned string must have no leading zeroes, unless the string is `"0"`.
-
-**Example 1:**
-
-    Input: 2
-    Output: "110"
-    Explantion: (-2) ^ 2 + (-2) ^ 1 = 2
-
-**Example 2:**
-
-    Input: 3
-    Output: "111"
-    Explantion: (-2) ^ 2 + (-2) ^ 1 + (-2) ^ 0 = 3
-
-**Example 3:**
-
-    Input: 4
-    Output: "100"
-    Explantion: (-2) ^ 2 = 4
-
-**Note:**
-
-1. `0 <= N <= 10^9`
-
-
-## 题目大意
-
-给出数字 N，返回由若干 "0" 和 "1"组成的字符串，该字符串为 N 的负二进制（base -2）表示。除非字符串就是 "0"，否则返回的字符串中不能含有前导零。
-
-提示：
-
-- 0 <= N <= 10^9
-
-
-
-## 解题思路
-
-- 给出一个十进制的数，要求转换成 -2 进制的数
-- 这一题仿造十进制转二进制的思路，短除法即可。
-
diff --git a/Algorithms/1019. Next Greater Node In Linked List/1019. Next Greater Node In Linked List.go b/Algorithms/1019. Next Greater Node In Linked List/1019. Next Greater Node In Linked List.go
deleted file mode 100644
index 52bd3a596..000000000
--- a/Algorithms/1019. Next Greater Node In Linked List/1019. Next Greater Node In Linked List.go	
+++ /dev/null
@@ -1,31 +0,0 @@
-package leetcode
-
-/**
- * Definition for singly-linked list.
- * type ListNode struct {
- *     Val int
- *     Next *ListNode
- * }
- */
-// 解法一 单调栈
-func nextLargerNodes(head *ListNode) []int {
-	res, indexes, nums := make([]int, 0), make([]int, 0), make([]int, 0)
-	p := head
-	for p != nil {
-		nums = append(nums, p.Val)
-		p = p.Next
-	}
-	for i := 0; i < len(nums); i++ {
-		res = append(res, 0)
-	}
-	for i := 0; i < len(nums); i++ {
-		num := nums[i]
-		for len(indexes) > 0 && nums[indexes[len(indexes)-1]] < num {
-			index := indexes[len(indexes)-1]
-			res[index] = num
-			indexes = indexes[:len(indexes)-1]
-		}
-		indexes = append(indexes, i)
-	}
-	return res
-}
diff --git a/Algorithms/1030. Matrix Cells in Distance Order/1030. Matrix Cells in Distance Order_test.go b/Algorithms/1030. Matrix Cells in Distance Order/1030. Matrix Cells in Distance Order_test.go
deleted file mode 100644
index 91d8597f6..000000000
--- a/Algorithms/1030. Matrix Cells in Distance Order/1030. Matrix Cells in Distance Order_test.go	
+++ /dev/null
@@ -1,55 +0,0 @@
-package leetcode
-
-import (
-	"fmt"
-	"testing"
-)
-
-type question1030 struct {
-	para1030
-	ans1030
-}
-
-// para 是参数
-// one 代表第一个参数
-type para1030 struct {
-	R  int
-	C  int
-	r0 int
-	c0 int
-}
-
-// ans 是答案
-// one 代表第一个答案
-type ans1030 struct {
-	one [][]int
-}
-
-func Test_Problem1030(t *testing.T) {
-
-	qs := []question1030{
-
-		question1030{
-			para1030{1, 2, 0, 0},
-			ans1030{[][]int{[]int{0, 0}, []int{0, 1}}},
-		},
-
-		question1030{
-			para1030{2, 2, 0, 1},
-			ans1030{[][]int{[]int{0, 1}, []int{0, 0}, []int{1, 1}, []int{1, 0}}},
-		},
-
-		question1030{
-			para1030{2, 3, 1, 2},
-			ans1030{[][]int{[]int{1, 2}, []int{0, 2}, []int{1, 1}, []int{0, 1}, []int{1, 0}, []int{0, 0}}},
-		},
-	}
-
-	fmt.Printf("------------------------Leetcode Problem 1030------------------------\n")
-
-	for _, q := range qs {
-		_, p := q.ans1030, q.para1030
-		fmt.Printf("【input】:%v       【output】:%v\n", p, allCellsDistOrder(p.R, p.C, p.r0, p.c0))
-	}
-	fmt.Printf("\n\n\n")
-}
diff --git a/Algorithms/1040. Moving Stones Until Consecutive II/README.md b/Algorithms/1040. Moving Stones Until Consecutive II/README.md
deleted file mode 100755
index 8867a5352..000000000
--- a/Algorithms/1040. Moving Stones Until Consecutive II/README.md	
+++ /dev/null
@@ -1,79 +0,0 @@
-# [1040. Moving Stones Until Consecutive II](https://leetcode.com/problems/moving-stones-until-consecutive-ii/)
-
-
-## 题目:
-
-On an **infinite** number line, the position of the i-th stone is given by `stones[i]`. Call a stone an *endpoint stone* if it has the smallest or largest position.
-
-Each turn, you pick up an endpoint stone and move it to an unoccupied position so that it is no longer an endpoint stone.
-
-In particular, if the stones are at say, `stones = [1,2,5]`, you **cannot** move the endpoint stone at position 5, since moving it to any position (such as 0, or 3) will still keep that stone as an endpoint stone.
-
-The game ends when you cannot make any more moves, ie. the stones are in consecutive positions.
-
-When the game ends, what is the minimum and maximum number of moves that you could have made? Return the answer as an length 2 array: `answer = [minimum_moves, maximum_moves]`
-
-**Example 1:**
-
-    Input: [7,4,9]
-    Output: [1,2]
-    Explanation: 
-    We can move 4 -> 8 for one move to finish the game.
-    Or, we can move 9 -> 5, 4 -> 6 for two moves to finish the game.
-
-**Example 2:**
-
-    Input: [6,5,4,3,10]
-    Output: [2,3]
-    We can move 3 -> 8 then 10 -> 7 to finish the game.
-    Or, we can move 3 -> 7, 4 -> 8, 5 -> 9 to finish the game.
-    Notice we cannot move 10 -> 2 to finish the game, because that would be an illegal move.
-
-**Example 3:**
-
-    Input: [100,101,104,102,103]
-    Output: [0,0]
-
-**Note:**
-
-1. `3 <= stones.length <= 10^4`
-2. `1 <= stones[i] <= 10^9`
-3. `stones[i]` have distinct values.
-
-
-## 题目大意
-
-在一个长度无限的数轴上，第 i 颗石子的位置为 stones[i]。如果一颗石子的位置最小/最大，那么该石子被称作端点石子。每个回合，你可以将一颗端点石子拿起并移动到一个未占用的位置，使得该石子不再是一颗端点石子。值得注意的是，如果石子像 stones = [1,2,5] 这样，你将无法移动位于位置 5 的端点石子，因为无论将它移动到任何位置（例如 0 或 3），该石子都仍然会是端点石子。当你无法进行任何移动时，即，这些石子的位置连续时，游戏结束。
-
-要使游戏结束，你可以执行的最小和最大移动次数分别是多少？ 以长度为 2 的数组形式返回答案：answer = [minimum\_moves, maximum\_moves] 。
-
-提示：
-
-1. 3 <= stones.length <= 10^4
-2. 1 <= stones[i] <= 10^9
-3. stones[i] 的值各不相同。
-
-
-## 解题思路
-
-
-- 给出一个数组，数组里面代表的是石头的坐标。要求移动石头，最终使得这些石头的坐标是一个连续的自然数列。但是规定，当一个石头是端点的时候，是不能移动的，例如 [1,2,5]，5 是端点，不能把 5 移到 3 或者 0 的位置，因为移动之后，这个石头仍然是端点。最终输出将所有石头排成连续的自然数列所需的最小步数和最大步数。
-- 这道题的关键就是如何保证端点石头不能再次移动到端点的限制。例如，[5,6,8,9,20]，20 是端点，但是 20 就可以移动到 7 的位置，最终形成 [5,6,7,8,9] 的连续序列。但是 [5,6,7,8,20]，这种情况 20 就不能移动到 9 了，只能让 8 移动到 9，20 再移动到 8 的位置，最终还是形成了 [5,6,7,8,9]，但是步数需要 2 步。经过上述分析，可以得到，端点石头只能往中间空挡的位置移动，如果中间没有空挡，那么需要借助一个石头先制造一个空挡，然后端点石头再插入到中间，这样最少是需要 2 步。
-- 再来考虑极值的情况。先看最大步数，最大步数肯定慢慢移动，一次移动一格，并且移动的格数最多。这里有两个极端情况，把数组里面的数全部都移动到最左端点，把数组里面的数全部都移动到最右端点。每次只移动一格。例如，全部都移到最右端点：
-
-        [3,4,5,6,10] // 初始状态，连续的情况
-        [4,5,6,7,10] // 第一步，把 3 挪到右边第一个可以插入的位置，即 7
-        [5,6,7,8,10] // 第二步，把 4 挪到右边第一个可以插入的位置，即 8
-        [6,7,8,9,10] // 第三步，把 5 挪到右边第一个可以插入的位置，即 9
-        
-        
-        [1,3,5,7,10] // 初始状态，不连续的情况
-        [3,4,5,7,10] // 第一步，把 1 挪到右边第一个可以插入的位置，即 4
-        [4,5,6,7,10] // 第二步，把 3 挪到右边第一个可以插入的位置，即 6
-        [5,6,7,8,10] // 第三步，把 4 挪到右边第一个可以插入的位置，即 8
-        [6,7,8,9,10] // 第四步，把 5 挪到右边第一个可以插入的位置，即 9
-
-    挪动的过程类似翻滚，最左边的石头挪到右边第一个可以放下的地方。然后不断的往右翻滚。把数组中的数全部都移动到最左边也同理。对比这两种情况的最大值，即是移动的最大步数。
-
-- 再看最小步数。这里就涉及到了滑动窗口了。由于最终是要形成连续的自然数列，所以滑动窗口的大小已经固定成 n 了，从数组的 0 下标可以往右滑动窗口，这个窗口中能包含的数字越多，代表窗口外的数字越少，那么把这些数字放进窗口内的步数也最小。于是可以求得最小步数。这里有一个比较坑的地方就是题目中的那个`“端点不能移动以后还是端点”`的限制。针对这种情况，需要额外的判断。如果当前窗口内有 n-1 个元素了，即只有一个端点在窗口外，并且窗口右边界的值减去左边界的值也等于 n-1，代表这个窗口内已经都是连续数字了。这种情况端点想融合到这个连续数列中，最少需要 2 步(上文已经分析过了)。
-- 注意一些边界情况。如果窗口从左往右滑动，窗口右边界滑到最右边了，但是窗口右边界的数字减去左边界的数字还是小于窗口大小 n，代表已经滑到头了，可以直接 break 出去。为什么滑到头了呢？由于数组经过从小到大排序以后，数字越往右边越大，当前数字是小值，已经满足了 `stones[right]-stones[left] < n`，左边界继续往右移动只会使得 `stones[left]` 更大，就更加小于 n 了。而我们需要寻找的是 `stones[right]-stones[left] >= n` 的边界点，肯定再也找不到了。
diff --git a/Algorithms/1049. Last Stone Weight II/README.md b/Algorithms/1049. Last Stone Weight II/README.md
deleted file mode 100755
index 865456fab..000000000
--- a/Algorithms/1049. Last Stone Weight II/README.md	
+++ /dev/null
@@ -1,49 +0,0 @@
-# [1049. Last Stone Weight II](https://leetcode.com/problems/last-stone-weight-ii/)
-
-## 题目:
-
-We have a collection of rocks, each rock has a positive integer weight.
-
-Each turn, we choose **any two rocks** and smash them together. Suppose the stones have weights `x` and `y` with `x <= y`. The result of this smash is:
-
-- If `x == y`, both stones are totally destroyed;
-- If `x != y`, the stone of weight `x` is totally destroyed, and the stone of weight `y`has new weight `y-x`.
-
-At the end, there is at most 1 stone left. Return the **smallest possible** weight of this stone (the weight is 0 if there are no stones left.)
-
-**Example 1:**
-
-    Input: [2,7,4,1,8,1]
-    Output: 1
-    Explanation: 
-    We can combine 2 and 4 to get 2 so the array converts to [2,7,1,8,1] then,
-    we can combine 7 and 8 to get 1 so the array converts to [2,1,1,1] then,
-    we can combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
-    we can combine 1 and 1 to get 0 so the array converts to [1] then that's the optimal value.
-
-**Note:**
-
-1. `1 <= stones.length <= 30`
-2. `1 <= stones[i] <= 100`
-
-
-
-## 题目大意
-
-有一堆石头，每块石头的重量都是正整数。每一回合，从中选出任意两块石头，然后将它们一起粉碎。假设石头的重量分别为 x 和 y，且 x <= y。那么粉碎的可能结果如下：
-
-如果 x == y，那么两块石头都会被完全粉碎；
-如果 x != y，那么重量为 x 的石头将会完全粉碎，而重量为 y 的石头新重量为 y-x。
-最后，最多只会剩下一块石头。返回此石头最小的可能重量。如果没有石头剩下，就返回 0。
-
-提示：
-
-1. 1 <= stones.length <= 30
-2. 1 <= stones[i] <= 1000
-
-
-## 解题思路
-
-
-- 给出一个数组，数组里面的元素代表的是石头的重量。现在要求两个石头对碰，如果重量相同，两个石头都消失，如果一个重一个轻，剩下的石头是两者的差值。问经过这样的多次碰撞以后，能剩下的石头的重量最轻是多少？
-- 由于两两石头要发生碰撞，所以可以将整个数组可以分为两部分，如果这两部分的石头重量总和相差不大，那么经过若干次碰撞以后，剩下的石头重量一定是最小的。现在就需要找到这样两堆总重量差不多的两堆石头。这个问题就可以转化为 01 背包问题。从数组中找到 `sum/2` 重量的石头集合，如果一半能尽量达到 `sum/2`，那么另外一半和 `sum/2` 的差是最小的，最好的情况就是两堆石头的重量都是 `sum/2`，那么两两石头对碰以后最后都能消失。01 背包的经典模板可以参考第 416 题。
diff --git a/Algorithms/1052. Grumpy Bookstore Owner/README.md b/Algorithms/1052. Grumpy Bookstore Owner/README.md
deleted file mode 100755
index 01d3f866d..000000000
--- a/Algorithms/1052. Grumpy Bookstore Owner/README.md	
+++ /dev/null
@@ -1,45 +0,0 @@
-# [1052. Grumpy Bookstore Owner](https://leetcode.com/problems/grumpy-bookstore-owner/)
-
-
-## 题目:
-
-Today, the bookstore owner has a store open for `customers.length`minutes. Every minute, some number of customers (`customers[i]`) enter the store, and all those customers leave after the end of that minute.
-
-On some minutes, the bookstore owner is grumpy. If the bookstore owner is grumpy on the i-th minute, `grumpy[i] = 1`, otherwise `grumpy[i] = 0`. When the bookstore owner is grumpy, the customers of that minute are not satisfied, otherwise they are satisfied.
-
-The bookstore owner knows a secret technique to keep themselves not grumpy for `X` minutes straight, but can only use it once.
-
-Return the maximum number of customers that can be satisfied throughout the day.
-
-**Example 1:**
-
-    Input: customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], X = 3
-    Output: 16
-    Explanation: The bookstore owner keeps themselves not grumpy for the last 3 minutes. 
-    The maximum number of customers that can be satisfied = 1 + 1 + 1 + 1 + 7 + 5 = 16.
-
-**Note:**
-
-- `1 <= X <= customers.length == grumpy.length <= 20000`
-- `0 <= customers[i] <= 1000`
-- `0 <= grumpy[i] <= 1`
-
-
-## 题目大意
-
-今天，书店老板有一家店打算试营业 customers.length 分钟。每分钟都有一些顾客（customers[i]）会进入书店，所有这些顾客都会在那一分钟结束后离开。在某些时候，书店老板会生气。 如果书店老板在第 i 分钟生气，那么 grumpy[i] = 1，否则 grumpy[i] = 0。 当书店老板生气时，那一分钟的顾客就会不满意，不生气则他们是满意的。书店老板知道一个秘密技巧，能抑制自己的情绪，可以让自己连续 X 分钟不生气，但却只能使用一次。请你返回这一天营业下来，最多有多少客户能够感到满意的数量。
-
-提示：
-
-1. 1 <= X <= customers.length == grumpy.length <= 20000
-2. 0 <= customers[i] <= 1000
-3. 0 <= grumpy[i] <= 1
-
-
-
-## 解题思路
-
-
-- 给出一个顾客入店时间表和书店老板发脾气的时间表。两个数组的时间是一一对应的，即相同下标对应的相同的时间。书店老板可以控制自己在 X 分钟内不发火，但是只能控制一次。问有多少顾客能在书店老板不发火的时候在书店里看书。抽象一下，给出一个价值数组和一个装着 0 和 1 的数组，当价值数组的下标对应另外一个数组相同下标的值是 0 的时候，那么这个价值可以累加，当对应是 1 的时候，就不能加上这个价值。现在可以让装着 0 和 1 的数组中连续 X 个数都变成 0，问最终价值最大是多少？
-- 这道题是典型的滑动窗口的题目。最暴力的解法是滑动窗口右边界，当与左边界的距离等于 X 的时候，计算此刻对应的数组的总价值。当整个宽度为 X 的窗口滑过整个数组以后，输出维护的最大值即可。这个方法耗时比较长。因为每次计算数组总价值的时候都要遍历整个数组。这里是可以优化的地方。
-- 每次计算数组总价值的时候，其实目的是为了找到宽度为 X 的窗口对应里面为 1 的数累加和最大，因为如果把这个窗口里面的 1 都变成 0 以后，那么对最终价值的影响也最大。所以用一个变量 `customer0` 专门记录脾气数组中为 0 的对应的价值，累加起来。因为不管怎么改变，为 0 的永远为 0，唯一变化的是 1 变成 0 。用 `customer1` 专门记录脾气数组中为 1 的对应的价值。在窗口滑动过程中找到 `customer1` 的最大值。最终要求的最大值就是 `customer0 + maxCustomer1`。
diff --git a/Algorithms/1073. Adding Two Negabinary Numbers/README.md b/Algorithms/1073. Adding Two Negabinary Numbers/README.md
deleted file mode 100755
index 120958bad..000000000
--- a/Algorithms/1073. Adding Two Negabinary Numbers/README.md	
+++ /dev/null
@@ -1,75 +0,0 @@
-# [1073. Adding Two Negabinary Numbers](https://leetcode.com/problems/adding-two-negabinary-numbers/)
-
-
-## 题目:
-
-Given two numbers `arr1` and `arr2` in base **-2**, return the result of adding them together.
-
-Each number is given in *array format*: as an array of 0s and 1s, from most significant bit to least significant bit. For example, `arr = [1,1,0,1]`represents the number `(-2)^3 + (-2)^2 + (-2)^0 = -3`. A number `arr` in *array format* is also guaranteed to have no leading zeros: either `arr == [0]` or `arr[0] == 1`.
-
-Return the result of adding `arr1` and `arr2` in the same format: as an array of 0s and 1s with no leading zeros.
-
-**Example 1:**
-
-    Input: arr1 = [1,1,1,1,1], arr2 = [1,0,1]
-    Output: [1,0,0,0,0]
-    Explanation: arr1 represents 11, arr2 represents 5, the output represents 16.
-
-**Note:**
-
-1. `1 <= arr1.length <= 1000`
-2. `1 <= arr2.length <= 1000`
-3. `arr1` and `arr2` have no leading zeros
-4. `arr1[i]` is `0` or `1`
-5. `arr2[i]` is `0` or `1`
-
-
-## 题目大意
-
-给出基数为 -2 的两个数 arr1 和 arr2，返回两数相加的结果。数字以 数组形式 给出：数组由若干 0 和 1 组成，按最高有效位到最低有效位的顺序排列。例如，arr = [1,1,0,1] 表示数字 (-2)^3 + (-2)^2 + (-2)^0 = -3。数组形式 的数字也同样不含前导零：以 arr 为例，这意味着要么 arr == [0]，要么 arr[0] == 1。
-
-返回相同表示形式的 arr1 和 arr2 相加的结果。两数的表示形式为：不含前导零、由若干 0 和 1 组成的数组。
-
-提示：
-
-- 1 <= arr1.length <= 1000
-- 1 <= arr2.length <= 1000
-- arr1 和 arr2 都不含前导零
-- arr1[i] 为 0 或 1
-- arr2[i] 为 0 或 1
-
-
-
-## 解题思路
-
-- 给出两个 -2 进制的数，要求计算出这两个数的和，最终表示形式还是 -2 进制。
-- 这一题最先想到的思路是先把两个 -2 进制的数转成 10 进制以后做加法，然后把结果表示成 -2 进制。这个思路可行，但是在提交以后会发现数据溢出 int64 了。在第 257 / 267 组测试数据会出现 WA。测试数据见 test 文件。另外换成 big.Add 也不是很方便。所以考虑换一个思路。
-- 这道题实际上就是求两个 -2 进制数的加法，为什么还要先转到 10 进制再换回 -2 进制呢？为何不直接进行 -2 进制的加法。所以开始尝试直接进行加法运算。加法是从低位到高位依次累加，遇到进位要从低往高位进位。所以从两个数组的末尾往前扫，模拟低位相加的过程。关键的是进位问题。进位分 3 种情况，依次来讨论：
-
-1. 进位到高位 k ，高位 k 上的两个数数字分别是 0 和 0 。这种情况最终 k 位为 1 。
-```c
-        证明：由于进位是由 k - 1 位进过来的，所以 k - 1 位是 2 个 1 。现在 k 位是 2 个 0，
-        所以加起来的和是 2 * (-2)^(k - 1)。
-        当 k 为奇数的时候，2 * (-2)^(k - 1) = (-1)^(k - 1)* 2 * 2^(k - 1) = 2^k
-        当 k 为偶数的时候，2 * (-2)^(k - 1) = (-1)^(k - 1)* 2 * 2^(k - 1) = -2^k
-        综合起来就是 (-2)^k，所以最终 k 位上有一个 1
-```
-2. 进位到高位 k ，高位 k 上的两个数数字分别是 0 和 1 。这种情况最终 k 位为 0 。
-```c
-        证明：由于进位是由 k - 1 位进过来的，所以 k - 1 位是 2 个 1。现在 k 位是 1 个 0 和 1 个 1,
-        所以加起来的和是 (-2)^k + 2 * (-2)^(k - 1)。
-        当 k 为奇数的时候，(-2)^k + 2 * (-2)^(k - 1) = -2^k + 2^k = 0
-        当 k 为偶数的时候，(-2)^k + 2 * (-2)^(k - 1) = 2^k - 2^k = 0
-        综合起来就是 0，所以最终 k 位上有一个 0
-```
-3. 进位到高位 k ，高位 k 上的两个数数字分别是 1 和 1 。这种情况最终 k 位为 1 。
-```c
-        证明：由于进位是由 k - 1 位进过来的，所以 k - 1 位是 2 个 1 。现在 k 位是 2 个 1，
-        所以加起来的和是 2 * (-2)^k + 2 * (-2)^(k - 1)。
-        当 k 为奇数的时候，2 * (-2)^k + 2 * (-2)^(k - 1) = -2^(k + 1) + 2^k = 2^k*(1 - 2) = -2^k
-        当 k 为偶数的时候，2 * (-2)^k + 2 * (-2)^(k - 1) = 2^(k + 1) - 2^k = 2^k*(2 - 1) = 2^k
-        综合起来就是 (-2)^k，所以最终 k 位上有一个 1
-```
-
-- 所以综上所属，-2 进制的进位和 2 进制的进位原理是完全一致的，只不过 -2 进制的进位是 -1，而 2 进制的进位是 1 。由于进位可能在 -2 进制上出现前导 0 ，所以最终结果需要再去除前导 0 。
-
diff --git a/Algorithms/1074. Number of Submatrices That Sum to Target/README.md b/Algorithms/1074. Number of Submatrices That Sum to Target/README.md
deleted file mode 100755
index 7866daa21..000000000
--- a/Algorithms/1074. Number of Submatrices That Sum to Target/README.md	
+++ /dev/null
@@ -1,61 +0,0 @@
-# [1074. Number of Submatrices That Sum to Target](https://leetcode.com/problems/number-of-submatrices-that-sum-to-target/)
-
-
-## 题目:
-
-Given a `matrix`, and a `target`, return the number of non-empty submatrices that sum to target.
-
-A submatrix `x1, y1, x2, y2` is the set of all cells `matrix[y]` with `x1 <= x <= x2` and `y1 <= y <= y2`.
-
-Two submatrices `(x1, y1, x2, y2)` and `(x1', y1', x2', y2')` are different if they have some coordinate that is different: for example, if `x1 != x1'`.
-
-**Example 1:**
-
-    Input: matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
-    Output: 4
-    Explanation: The four 1x1 submatrices that only contain 0.
-
-**Example 2:**
-
-    Input: matrix = [[1,-1],[-1,1]], target = 0
-    Output: 5
-    Explanation: The two 1x2 submatrices, plus the two 2x1 submatrices, plus the 2x2 submatrix.
-
-**Note:**
-
-1. `1 <= matrix.length <= 300`
-2. `1 <= matrix[0].length <= 300`
-3. `-1000 <= matrix[i] <= 1000`
-4. `-10^8 <= target <= 10^8`
-
-
-## 题目大意
-
-给出矩阵 matrix 和目标值 target，返回元素总和等于目标值的非空子矩阵的数量。
-
-子矩阵 x1, y1, x2, y2 是满足 x1 <= x <= x2 且 y1 <= y <= y2 的所有单元 matrix[x][y] 的集合。
-
-如果 (x1, y1, x2, y2) 和 (x1', y1', x2', y2') 两个子矩阵中部分坐标不同（如：x1 != x1'），那么这两个子矩阵也不同。
-
-
-提示：
-
-1. 1 <= matrix.length <= 300
-2. 1 <= matrix[0].length <= 300
-3. -1000 <= matrix[i] <= 1000
-4. -10^8 <= target <= 10^8
-
-
-
-
-## 解题思路
-
-- 给出一个矩阵，要求在这个矩阵中找出子矩阵的和等于 target 的矩阵个数。
-- 这一题读完题感觉是滑动窗口的二维版本。如果把它拍扁，在一维数组中，求连续的子数组和为 target，这样就很好做。如果这题不降维，纯暴力解是 O(n^6)。如何优化降低时间复杂度呢？
-- 联想到第 1 题 Two Sum 问题，可以把 2 个数求和的问题优化到 O(n)。这里也用类似的思想，用一个 map 来保存行方向上曾经出现过的累加和，相减就可以得到本行的和。这里可能读者会有疑惑，为什么不能每一行都单独保存呢？为什么一定要用累加和相减的方式来获取每一行的和呢？因为这一题要求子矩阵所有解，如果只单独保存每一行的和，只能求得小的子矩阵，子矩阵和子矩阵组成的大矩阵的情况会漏掉(当然再循环一遍，把子矩阵累加起来也可以，但是这样就多了一层循环了)，例如子矩阵是 1*4 的，但是 2 个这样的子矩阵摞在一起形成 2 * 4 也能满足条件，如果不用累加和的办法，只单独存每一行的和，最终还要有组合的步骤。经过这样的优化，可以从 O(n^6) 优化到 O(n^4)，能 AC 这道题，但是时间复杂度太高了。如何优化？
-- 首先，子矩阵需要上下左右 4 个边界，4 个变量控制循环就需要 O(n^4)，行和列的区间累加还需要 O(n^2)。行和列的区间累加可以通过 preSum 来解决。例如 `sum[i,j] = sum[j] - sum[i - 1]`，其中 sum[k] 中存的是从 0 到 K 的累加和： ![](https://img.halfrost.com/Leetcode/leetcode_1074.gif)
-
-    那么一个区间内的累加和可以由这个区间的右边界减去区间左边界左边的那个累加和得到(由于是闭区间，所需要取左边界左边的和)。经过这样的处理，列方向的维度就被我们拍扁了。
-
-- 再来看看行方向的和，现在每一列的和都可以通过区间相减的方法得到。那么这道题就变成了第 1 题 Two Sum 的问题了。Two Sum 问题只需要 O(n) 的时间复杂度求解，这一题由于是二维的，所以两个列的边界还需要循环，所以最终优化下来的时间复杂度是 O(n^3)。计算 presum 可以直接用原数组，所以空间复杂度只有一个 O(n) 的字典。
-- 类似思路的题目有第 560 题，第 304 题。
diff --git a/Algorithms/1078. Occurrences After Bigram/README.md b/Algorithms/1078. Occurrences After Bigram/README.md
deleted file mode 100755
index 73de067bb..000000000
--- a/Algorithms/1078. Occurrences After Bigram/README.md	
+++ /dev/null
@@ -1,39 +0,0 @@
-# [1078. Occurrences After Bigram](https://leetcode.com/problems/occurrences-after-bigram/)
-
-
-## 题目:
-
-Given words `first` and `second`, consider occurrences in some `text` of the form "`first second third`", where `second` comes immediately after `first`, and `third`comes immediately after `second`.
-
-For each such occurrence, add "`third`" to the answer, and return the answer.
-
-**Example 1:**
-
-    Input: text = "alice is a good girl she is a good student", first = "a", second = "good"
-    Output: ["girl","student"]
-
-**Example 2:**
-
-    Input: text = "we will we will rock you", first = "we", second = "will"
-    Output: ["we","rock"]
-
-**Note:**
-
-1. `1 <= text.length <= 1000`
-2. `text` consists of space separated words, where each word consists of lowercase English letters.
-3. `1 <= first.length, second.length <= 10`
-4. `first` and `second` consist of lowercase English letters.
-
-
-## 题目大意
-
-
-给出第一个词 first 和第二个词 second，考虑在某些文本 text 中可能以 "first second third" 形式出现的情况，其中 second 紧随 first 出现，third 紧随 second 出现。对于每种这样的情况，将第三个词 "third" 添加到答案中，并返回答案。
-
-
-
-
-## 解题思路
-
-
-- 简单题。给出一个 text，要求找出紧接在 first 和 second 后面的那个字符串，有多个就输出多个。解法很简单，先分解出 words 每个字符串，然后依次遍历进行字符串匹配。匹配到 first 和 second 以后，输出之后的那个字符串。
diff --git a/Algorithms/1079. Letter Tile Possibilities/README.md b/Algorithms/1079. Letter Tile Possibilities/README.md
deleted file mode 100755
index e8e12a5e5..000000000
--- a/Algorithms/1079. Letter Tile Possibilities/README.md	
+++ /dev/null
@@ -1,36 +0,0 @@
-# [1079. Letter Tile Possibilities](https://leetcode.com/problems/letter-tile-possibilities/)
-
-
-## 题目
-
-You have a set of `tiles`, where each tile has one letter `tiles[i]` printed on it. Return the number of possible non-empty sequences of letters you can make.
-
-**Example 1:**
-
-    Input: "AAB"
-    Output: 8
-    Explanation: The possible sequences are "A", "B", "AA", "AB", "BA", "AAB", "ABA", "BAA".
-
-**Example 2:**
-
-    Input: "AAABBC"
-    Output: 188
-
-**Note:**
-
-1. `1 <= tiles.length <= 7`
-2. `tiles` consists of uppercase English letters.
-
-## 题目大意
-
-你有一套活字字模 tiles，其中每个字模上都刻有一个字母 tiles[i]。返回你可以印出的非空字母序列的数目。提示：  
-
-- 1 <= tiles.length <= 7
-- tiles 由大写英文字母组成
-
-## 解题思路
-
-- 题目要求输出所有非空字母序列的数目。这一题是排列和组合的结合题目。组合是可以选择一个字母，二个字母，…… n 个字母。每个组合内是排列问题。比如选择 2 个字母，字母之间相互排序不同是影响最终结果的，不同的排列顺序是不同的解。
-- 这道题目由于不需要输出所有解，所以解法可以优化，例如我们在递归计算解的时候，不需要真的遍历原字符串，只需要累加一些字母的频次就可以。当然如果要输出所有解，就需要真实遍历原字符串了(见解法二)。简单的做法是每次递归按照频次累加。因为每次增加一个字母一定是 26 个大写字母中的一个。这里需要注意的是，增加的只能是 26 个字母里面还能取出“机会”的字母，例如递归到到第 3 轮了，A 用完了，这个时候只能取频次还不为 0 的字母拼上去。
-
-![](https://pic.cdn.lizenghai.com/uploads/2019/06/15604249050330956_414598-48d6698970379275.png?x-oss-process=style%2Ffull)
\ No newline at end of file
diff --git a/Algorithms/1093. Statistics from a Large Sample/README.md b/Algorithms/1093. Statistics from a Large Sample/README.md
deleted file mode 100755
index c0d634a3b..000000000
--- a/Algorithms/1093. Statistics from a Large Sample/README.md	
+++ /dev/null
@@ -1,51 +0,0 @@
-# [1093. Statistics from a Large Sample](https://leetcode.com/problems/statistics-from-a-large-sample/)
-
-
-## 题目
-
-We sampled integers between `0` and `255`, and stored the results in an array `count`: `count[k]` is the number of integers we sampled equal to `k`.
-
-Return the minimum, maximum, mean, median, and mode of the sample respectively, as an array of **floating point numbers**. The mode is guaranteed to be unique.
-
-*(Recall that the median of a sample is:*
-
-- *The middle element, if the elements of the sample were sorted and the number of elements is odd;*
-- *The average of the middle two elements, if the elements of the sample were sorted and the number of elements is even.)*
-
-**Example 1:**
-
-    Input: count = [0,1,3,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
-    Output: [1.00000,3.00000,2.37500,2.50000,3.00000]
-
-**Example 2:**
-
-    Input: count = [0,4,3,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
-    Output: [1.00000,4.00000,2.18182,2.00000,1.00000]
-
-**Constraints:**
-
-1. `count.length == 256`
-2. `1 <= sum(count) <= 10^9`
-3. The mode of the sample that count represents is unique.
-4. Answers within `10^-5` of the true value will be accepted as correct.
-
-
-## 题目大意
-
-我们对 0 到 255 之间的整数进行采样，并将结果存储在数组 count 中：count[k] 就是整数 k 的采样个数。
-
-我们以 浮点数 数组的形式，分别返回样本的最小值、最大值、平均值、中位数和众数。其中，众数是保证唯一的。我们先来回顾一下中位数的知识：
-
-- 如果样本中的元素有序，并且元素数量为奇数时，中位数为最中间的那个元素；
-- 如果样本中的元素有序，并且元素数量为偶数时，中位数为中间的两个元素的平均值。
-
-
-
-## 解题思路
-
-
-- 这个问题的关键需要理解题目的意思，什么是采样？`count[k]` 就是整数 `k` 的采样个数。
-- 题目要求返回样本的最小值、最大值、平均值、中位数和众数。最大值和最小值就很好处理，只需要遍历 count 判断最小的非 0 的 index 就是最小值，最大的非 0 的 index 就是最大值。平均值也非常好处理，对于所有非 0 的 count，我们通过累加 count[k] * index 得到所有数的和，然后除上所有非 0 的 count 的和。![](https://latex.codecogs.com/gif.latex?%24%24%5Csum_%7Bn%3D0%7D%5E%7B256%7Dcount%5Bn%5D%28while%5C%20%5C%20count%5Bn%5D%21%3D0%29%24%24)
-
-- 众数也非常容易，只需统计 count 值最大时的 index 即可。
-- 中位数的处理相对麻烦一些，因为需要分非 0 的 count 之和是奇数还是偶数两种情况。先假设非 0 的 count 和为 cnt，那么如果 cnt 是奇数的话，只需要找到 cnt/2 的位置即可，通过不断累加 count 的值，直到累加和超过 ≥ cnt/2。如果 cnt 是偶数的话，需要找到 cnt/2 + 1 和 cnt/2 的位置，找法和奇数情况相同，不过需要找两次(可以放到一个循环中做两次判断)。
\ No newline at end of file
diff --git a/Algorithms/1105. Filling Bookcase Shelves/README.md b/Algorithms/1105. Filling Bookcase Shelves/README.md
deleted file mode 100755
index 5dc7bc7c6..000000000
--- a/Algorithms/1105. Filling Bookcase Shelves/README.md	
+++ /dev/null
@@ -1,45 +0,0 @@
-# [1105. Filling Bookcase Shelves](https://leetcode.com/problems/filling-bookcase-shelves/)
-
-
-## 题目:
-
-We have a sequence of `books`: the `i`-th book has thickness `books[i][0]`and height `books[i][1]`.
-
-We want to place these books **in order** onto bookcase shelves that have total width `shelf_width`.
-
-We choose some of the books to place on this shelf (such that the sum of their thickness is `<= shelf_width`), then build another level of shelf of the bookcase so that the total height of the bookcase has increased by the maximum height of the books we just put down. We repeat this process until there are no more books to place.
-
-Note again that at each step of the above process, the order of the books we place is the same order as the given sequence of books. For example, if we have an ordered list of 5 books, we might place the first and second book onto the first shelf, the third book on the second shelf, and the fourth and fifth book on the last shelf.
-
-Return the minimum possible height that the total bookshelf can be after placing shelves in this manner.
-
-**Example 1:**
-
-![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/06/28/shelves.png)
-
-    Input: books = [[1,1],[2,3],[2,3],[1,1],[1,1],[1,1],[1,2]], shelf_width = 4
-    Output: 6
-    Explanation:
-    The sum of the heights of the 3 shelves are 1 + 3 + 2 = 6.
-    Notice that book number 2 does not have to be on the first shelf.
-
-**Constraints:**
-
-- `1 <= books.length <= 1000`
-- `1 <= books[i][0] <= shelf_width <= 1000`
-- `1 <= books[i][1] <= 1000`
-
-## 题目大意
-
-附近的家居城促销，你买回了一直心仪的可调节书架，打算把自己的书都整理到新的书架上。你把要摆放的书 books 都整理好，叠成一摞：从上往下，第 i 本书的厚度为 books[i][0]，高度为 books[i][1]。按顺序将这些书摆放到总宽度为 shelf\_width 的书架上。
-
-先选几本书放在书架上（它们的厚度之和小于等于书架的宽度 shelf_width），然后再建一层书架。重复这个过程，直到把所有的书都放在书架上。
-
-需要注意的是，在上述过程的每个步骤中，摆放书的顺序与你整理好的顺序相同。 例如，如果这里有 5 本书，那么可能的一种摆放情况是：第一和第二本书放在第一层书架上，第三本书放在第二层书架上，第四和第五本书放在最后一层书架上。每一层所摆放的书的最大高度就是这一层书架的层高，书架整体的高度为各层高之和。以这种方式布置书架，返回书架整体可能的最小高度。
-
-
-
-## 解题思路
-
-- 给出一个数组，数组里面每个元素代表的是一个本书的宽度和高度。要求按照书籍的顺序，把书摆到宽度为 `shelf_width` 的书架上。问最终放下所有书籍以后，书架的最小高度。
-- 这一题的解题思路是动态规划。`dp[i]` 代表放置前 `i` 本书所需要的书架最小高度。初始值 dp[0] = 0，其他为最大值 1000*1000。遍历每一本书，把当前这本书作为书架最后一层的最后一本书，将这本书之前的书向后调整，看看是否可以减少之前的书架高度。状态转移方程为 `dp[i] = min(dp[i] , dp[j - 1] + h)`，其中 `j` 表示最后一层所能容下书籍的索引，`h` 表示最后一层最大高度。`j` 调整完一遍以后就能找出书架最小高度值了。时间复杂度 O(n^2)。
\ No newline at end of file
diff --git a/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/README.md b/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/README.md
deleted file mode 100755
index 56bdf8d67..000000000
--- a/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/README.md	
+++ /dev/null
@@ -1,75 +0,0 @@
-# [1111. Maximum Nesting Depth of Two Valid Parentheses Strings](https://leetcode.com/problems/maximum-nesting-depth-of-two-valid-parentheses-strings/)
-
-
-## 题目:
-
-A string is a *valid parentheses string* (denoted VPS) if and only if it consists of `"("` and `")"` characters only, and:
-
-- It is the empty string, or
-- It can be written as `AB` (`A` concatenated with `B`), where `A` and `B` are VPS's, or
-- It can be written as `(A)`, where `A` is a VPS.
-
-We can similarly define the *nesting depth* `depth(S)` of any VPS `S` as follows:
-
-- `depth("") = 0`
-- `depth(A + B) = max(depth(A), depth(B))`, where `A` and `B` are VPS's
-- `depth("(" + A + ")") = 1 + depth(A)`, where `A` is a VPS.
-
-For example, `""`, `"()()"`, and `"()(()())"` are VPS's (with nesting depths 0, 1, and 2), and `")("` and `"(()"` are not VPS's.
-
-Given a VPS seq, split it into two disjoint subsequences `A` and `B`, such that `A` and `B` are VPS's (and `A.length + B.length = seq.length`).
-
-Now choose **any** such `A` and `B` such that `max(depth(A), depth(B))` is the minimum possible value.
-
-Return an `answer` array (of length `seq.length`) that encodes such a choice of `A` and `B`: `answer[i] = 0` if `seq[i]` is part of `A`, else `answer[i] = 1`. Note that even though multiple answers may exist, you may return any of them.
-
-**Example 1:**
-
-    Input: seq = "(()())"
-    Output: [0,1,1,1,1,0]
-
-**Example 2:**
-
-    Input: seq = "()(())()"
-    Output: [0,0,0,1,1,0,1,1]
-
-**Constraints:**
-
-- `1 <= seq.size <= 10000`
-
-
-## 题目大意
-
-
-有效括号字符串 仅由 "(" 和 ")" 构成，并符合下述几个条件之一：
-
-- 空字符串
-- 连接，可以记作 AB（A 与 B 连接），其中 A 和 B 都是有效括号字符串
-- 嵌套，可以记作 (A)，其中 A 是有效括号字符串
-
-类似地，我们可以定义任意有效括号字符串 s 的 嵌套深度 depth(S)：
-
-- s 为空时，depth("") = 0
-- s 为 A 与 B 连接时，depth(A + B) = max(depth(A), depth(B))，其中 A 和 B 都是有效括号字符串
-- s 为嵌套情况，depth("(" + A + ")") = 1 + depth(A)，其中 A 是有效括号字符串
-
-
-例如：""，"()()"，和 "()(()())" 都是有效括号字符串，嵌套深度分别为 0，1，2，而 ")(" 和 "(()" 都不是有效括号字符串。
-
- 
-
-给你一个有效括号字符串 seq，将其分成两个不相交的子序列 A 和 B，且 A 和 B 满足有效括号字符串的定义（注意：A.length + B.length = seq.length）。
-
-现在，你需要从中选出 任意 一组有效括号字符串 A 和 B，使 max(depth(A), depth(B)) 的可能取值最小。
-
-返回长度为 seq.length 答案数组 answer ，选择 A 还是 B 的编码规则是：如果 seq[i] 是 A 的一部分，那么 answer[i] = 0。否则，answer[i] = 1。即便有多个满足要求的答案存在，你也只需返回 一个。
-
-
-
-## 解题思路
-
-- 给出一个括号字符串。选出 A 部分和 B 部分，使得 `max(depth(A), depth(B))` 值最小。在最终的数组中输出 0 和 1，0 标识是 A 部分，1 标识是 B 部分。
-- 这一题想要 `max(depth(A), depth(B))` 值最小，可以使用贪心思想。如果 A 部分和 B 部分都尽快括号匹配，不深层次嵌套，那么总的层次就会变小。只要让嵌套的括号中属于 A 的和属于 B 的间隔排列即可。例如：“`(((())))`”，上面的字符串的嵌套深度是 4，按照上述的贪心思想，则标记为 0101 1010。
-- 这一题也可以用二分的思想来解答。把深度平分给 A 部分和 B 部分。
-    - 第一次遍历，先计算最大深度
-    - 第二次遍历，把深度小于等于最大深度一半的括号标记为 0(给 A 部分)，否则标记为 1(给 B 部分)
diff --git a/Algorithms/1157. Online Majority Element In Subarray/README.md b/Algorithms/1157. Online Majority Element In Subarray/README.md
deleted file mode 100755
index 5c7e8a13e..000000000
--- a/Algorithms/1157. Online Majority Element In Subarray/README.md	
+++ /dev/null
@@ -1,78 +0,0 @@
-# [1157. Online Majority Element In Subarray](https://leetcode.com/problems/online-majority-element-in-subarray/)
-
-
-## 题目:
-
-Implementing the class `MajorityChecker`, which has the following API:
-
-- `MajorityChecker(int[] arr)` constructs an instance of MajorityChecker with the given array `arr`;
-- `int query(int left, int right, int threshold)` has arguments such that:
-    - `0 <= left <= right < arr.length` representing a subarray of `arr`;
-    - `2 * threshold > right - left + 1`, ie. the threshold is always a strict majority of the length of the subarray
-
-Each `query(...)` returns the element in `arr[left], arr[left+1], ..., arr[right]` that occurs at least `threshold` times, or `-1` if no such element exists.
-
-**Example:**
-
-    MajorityChecker majorityChecker = new MajorityChecker([1,1,2,2,1,1]);
-    majorityChecker.query(0,5,4); // returns 1
-    majorityChecker.query(0,3,3); // returns -1
-    majorityChecker.query(2,3,2); // returns 2
-
-**Constraints:**
-
-- `1 <= arr.length <= 20000`
-- `1 <= arr[i] <= 20000`
-- For each query, `0 <= left <= right < len(arr)`
-- For each query, `2 * threshold > right - left + 1`
-- The number of queries is at most `10000`
-
-
-## 题目大意
-
-实现一个 MajorityChecker 的类，它应该具有下述几个 API：
-
-- MajorityChecker(int[] arr) 会用给定的数组 arr 来构造一个 MajorityChecker 的实例。
-- int query(int left, int right, int threshold) 有这么几个参数：
-- 0 <= left <= right < arr.length 表示数组 arr 的子数组的长度。
-- 2 * threshold > right - left + 1，也就是说阈值 threshold 始终比子序列长度的一半还要大。
-
-每次查询 query(...) 会返回在 arr[left], arr[left+1], ..., arr[right] 中至少出现阈值次数 threshold 的元素，如果不存在这样的元素，就返回 -1。
-
-提示：
-
-- 1 <= arr.length <= 20000
-- 1 <= arr[i] <= 20000
-- 对于每次查询，0 <= left <= right < len(arr)
-- 对于每次查询，2 * threshold > right - left + 1
-- 查询次数最多为 10000
-
-
-
-
-## 解题思路
-
-
-- 设计一个数据结构，能在任意的一个区间内，查找是否存在众数，众数的定义是：该数字出现的次数大于区间的一半。如果存在众数，一定唯一。如果在给定的区间内找不到众数，则输出 -1 。
-- 这一题有一个很显眼的“暗示”，`2 * threshold > right - left + 1`，这个条件就是摩尔投票算法的前提条件。摩尔投票的思想可以见第 169 题。这一题又要在区间内查询，所以选用线段树这个数据结构来实现。经过分析，可以确定此题的解题思路，摩尔投票 + 线段树。
-- 摩尔投票的思想是用两个变量，candidate 和 count，用来记录待被投票投出去的元素，和候选人累积没被投出去的轮数。如果候选人累积没有被投出去的轮数越多，那么最终成为众数的可能越大。从左往右扫描整个数组，先去第一个元素为 candidate，如果遇到相同的元素就累加轮数，如果遇到不同的元素，就把 candidate 和不同的元素一起投出去。当轮数变成 0 了，再选下一个元素作为 candidate。从左扫到右，就能找到众数了。那怎么和线段树结合起来呢？
-- 线段树是把一个大的区间拆分成很多个小区间，那么考虑这样一个问题。每个小区间内使用摩尔投票，最终把所有小区间合并起来再用一次摩尔投票，得到的结果和对整个区间使用一次摩尔投票的结果是一样的么？答案是一样的。可以这样想，众数总会在一个区间内被选出来，那么其他区间的摩尔投票都是起“中和”作用的，即两两元素一起出局。这个问题想通以后，说明摩尔投票具有可加的性质。既然满足可加，就可以和线段树结合，因为线段树每个线段就是加起来，最终合并成大区间的。
-- 举个例子，arr = [1,1,2,2,1,1]，先构造线段树，如下左图。
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_1157_0.png)
-
-    现在每个线段树的节点不是只存一个 int 数字了，而是存 candidate 和 count。每个节点的 candidate 和 count 分别代表的是该区间内摩尔投票的结果。初始化的时候，先把每个叶子都填满，candidate 是自己，count = 1 。即右图绿色节点。然后在 pushUp 的时候，进行摩尔投票：
-
-        mc.merge = func(i, j segmentItem) segmentItem {
-        		if i.candidate == j.candidate {
-        			return segmentItem{candidate: i.candidate, count: i.count + j.count}
-        		}
-        		if i.count > j.count {
-        			return segmentItem{candidate: i.candidate, count: i.count - j.count}
-        		}
-        		return segmentItem{candidate: j.candidate, count: j.count - i.count}
-        	}
-
-    直到根节点的 candidate 和 count 都填满。**注意，这里的 count 并不是元素出现的总次数，而是摩尔投票中坚持没有被投出去的轮数。**当线段树构建完成以后，就可以开始查询任意区间内的众数了，candidate 即为众数。接下来还要确定众数是否满足 `threshold` 的条件。
-
-- 用一个字典记录每个元素在数组中出现位置的下标，例如上述这个例子，用 map 记录下标：count = map[1:[0 1 4 5] 2:[2 3]]。由于下标在记录过程中是递增的，所以满足二分查找的条件。利用这个字典就可以查出在任意区间内，指定元素出现的次数。例如这里要查找 1 在 [0,5] 区间内出现的个数，那么利用 2 次二分查找，分别找到 `lowerBound` 和 `upperBound`，在 [lowerBound，upperBound) 区间内，都是元素 1 ，那么区间长度即是该元素重复出现的次数，和 `threshold` 比较，如果 ≥ `threshold` 说明找到了答案，否则没有找到就输出 -1 。
diff --git a/Algorithms/1184. Distance Between Bus Stops/README.md b/Algorithms/1184. Distance Between Bus Stops/README.md
deleted file mode 100755
index b77bce660..000000000
--- a/Algorithms/1184. Distance Between Bus Stops/README.md	
+++ /dev/null
@@ -1,59 +0,0 @@
-# [1184. Distance Between Bus Stops](https://leetcode.com/problems/distance-between-bus-stops/)
-
-## 题目:
-
-A bus has `n` stops numbered from `0` to `n - 1` that form a circle. We know the distance between all pairs of neighboring stops where `distance[i]` is the distance between the stops number `i` and `(i + 1) % n`.
-
-The bus goes along both directions i.e. clockwise and counterclockwise.
-
-Return the shortest distance between the given `start` and `destination` stops.
-
-**Example 1:**
-
-![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1.jpg)
-
-    Input: distance = [1,2,3,4], start = 0, destination = 1
-    Output: 1
-    Explanation: Distance between 0 and 1 is 1 or 9, minimum is 1.
-
-**Example 2:**
-
-![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1-1.jpg)
-
-    Input: distance = [1,2,3,4], start = 0, destination = 2
-    Output: 3
-    Explanation: Distance between 0 and 2 is 3 or 7, minimum is 3.
-
-**Example 3:**
-
-![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1-2.jpg)
-
-    Input: distance = [1,2,3,4], start = 0, destination = 3
-    Output: 4
-    Explanation: Distance between 0 and 3 is 6 or 4, minimum is 4.
-
-**Constraints:**
-
-- `1 <= n <= 10^4`
-- `distance.length == n`
-- `0 <= start, destination < n`
-- `0 <= distance[i] <= 10^4`
-
-
-## 题目大意
-
-环形公交路线上有 n 个站，按次序从 0 到 n - 1 进行编号。我们已知每一对相邻公交站之间的距离，distance[i] 表示编号为 i 的车站和编号为 (i + 1) % n 的车站之间的距离。环线上的公交车都可以按顺时针和逆时针的方向行驶。返回乘客从出发点 start 到目的地 destination 之间的最短距离。
-
-提示：
-
-- 1 <= n <= 10^4
-- distance.length == n
-- 0 <= start, destination < n
-- 0 <= distance[i] <= 10^4
-
-
-## 解题思路
-
-
-- 给出一个数组，代表的是公交车站每站直接的距离。距离是按照数组下标的顺序给出的，公交车可以按照顺时针行驶，也可以按照逆时针行驶。问行驶的最短距离是多少。
-- 按照题意，分别算出顺时针和逆时针的行驶距离，比较两者距离，取出小值就是结果。
diff --git a/Algorithms/1185. Day of the Week/README.md b/Algorithms/1185. Day of the Week/README.md
deleted file mode 100755
index c8389b70f..000000000
--- a/Algorithms/1185. Day of the Week/README.md	
+++ /dev/null
@@ -1,46 +0,0 @@
-# [1185. Day of the Week](https://leetcode.com/problems/day-of-the-week/)
-
-
-## 题目:
-
-Given a date, return the corresponding day of the week for that date.
-
-The input is given as three integers representing the `day`, `month` and `year` respectively.
-
-Return the answer as one of the following values `{"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}`.
-
-**Example 1:**
-
-    Input: day = 31, month = 8, year = 2019
-    Output: "Saturday"
-
-**Example 2:**
-
-    Input: day = 18, month = 7, year = 1999
-    Output: "Sunday"
-
-**Example 3:**
-
-    Input: day = 15, month = 8, year = 1993
-    Output: "Sunday"
-
-**Constraints:**
-
-- The given dates are valid dates between the years `1971` and `2100`.
-
-
-## 题目大意
-
-给你一个日期，请你设计一个算法来判断它是对应一周中的哪一天。输入为三个整数：day、month 和 year，分别表示日、月、年。
-
-您返回的结果必须是这几个值中的一个 {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}。
-
-提示：
-
-- 给出的日期一定是在 1971 到 2100 年之间的有效日期。
-
-## 解题思路
-
-
-- 给出一个日期，要求算出这一天是星期几。
-- 简单题，按照常识计算即可。
diff --git a/Algorithms/1189. Maximum Number of Balloons/README.md b/Algorithms/1189. Maximum Number of Balloons/README.md
deleted file mode 100755
index b55129e82..000000000
--- a/Algorithms/1189. Maximum Number of Balloons/README.md	
+++ /dev/null
@@ -1,48 +0,0 @@
-# [1189. Maximum Number of Balloons](https://leetcode.com/problems/maximum-number-of-balloons/)
-
-
-## 题目:
-
-Given a string `text`, you want to use the characters of `text` to form as many instances of the word **"balloon"** as possible.
-
-You can use each character in `text` **at most once**. Return the maximum number of instances that can be formed.
-
-**Example 1:**
-
-![](https://assets.leetcode.com/uploads/2019/09/05/1536_ex1_upd.JPG)
-
-    Input: text = "nlaebolko"
-    Output: 1
-
-**Example 2:**
-
-![](https://assets.leetcode.com/uploads/2019/09/05/1536_ex2_upd.JPG)
-
-    Input: text = "loonbalxballpoon"
-    Output: 2
-
-**Example 3:**
-
-    Input: text = "leetcode"
-    Output: 0
-
-**Constraints:**
-
-- `1 <= text.length <= 10^4`
-- `text` consists of lower case English letters only.
-
-
-## 题目大意
-
-给你一个字符串 text，你需要使用 text 中的字母来拼凑尽可能多的单词 "balloon"（气球）。字符串 text 中的每个字母最多只能被使用一次。请你返回最多可以拼凑出多少个单词 "balloon"。
-
-提示：
-
-- 1 <= text.length <= 10^4
-- text 全部由小写英文字母组成
-
-## 解题思路
-
-
-- 给出一个字符串，问这个字符串里面的数组能组成多少个 **balloon** 这个单词。
-- 简单题，先统计 26 个字母每个字母的频次，然后取出 balloon 这 5 个字母出现频次最小的值就是结果。
diff --git a/Algorithms/1200. Minimum Absolute Difference/README.md b/Algorithms/1200. Minimum Absolute Difference/README.md
deleted file mode 100755
index f028d2f6c..000000000
--- a/Algorithms/1200. Minimum Absolute Difference/README.md	
+++ /dev/null
@@ -1,44 +0,0 @@
-# [1200. Minimum Absolute Difference](https://leetcode.com/problems/minimum-absolute-difference/)
-
-
-## 题目:
-
-Given an array of **distinct** integers `arr`, find all pairs of elements with the minimum absolute difference of any two elements.
-
-Return a list of pairs in ascending order(with respect to pairs), each pair `[a, b]` follows
-
-- `a, b` are from `arr`
-- `a < b`
-- `b - a` equals to the minimum absolute difference of any two elements in `arr`
-
-**Example 1:**
-
-    Input: arr = [4,2,1,3]
-    Output: [[1,2],[2,3],[3,4]]
-    Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order.
-
-**Example 2:**
-
-    Input: arr = [1,3,6,10,15]
-    Output: [[1,3]]
-
-**Example 3:**
-
-    Input: arr = [3,8,-10,23,19,-4,-14,27]
-    Output: [[-14,-10],[19,23],[23,27]]
-
-**Constraints:**
-
-- `2 <= arr.length <= 10^5`
-- `-10^6 <= arr[i] <= 10^6`
-
-
-## 题目大意
-
-给出一个数组，要求找出所有满足条件的数值对 [a,b]：`a<b` 并且 `b-a` 的差值是数组中所有两个元素差值的最小值。
-
-## 解题思路
-
-
-- 给出一个数组，要求找出所有满足条件的数值对 [a,b]：`a<b` 并且 `b-a` 的差值是数组中所有两个元素差值的最小值。
-- 简单题，按照题意先排序，然后依次求出两个相邻元素的差值，求出最小的差值。最后遍历一遍数组，把所有等于最小差值的数值对都输出。
diff --git a/Algorithms/1201. Ugly Number III/README.md b/Algorithms/1201. Ugly Number III/README.md
deleted file mode 100755
index f33a936fc..000000000
--- a/Algorithms/1201. Ugly Number III/README.md	
+++ /dev/null
@@ -1,57 +0,0 @@
-# [1201. Ugly Number III](https://leetcode.com/problems/ugly-number-iii/)
-
-
-## 题目:
-
-Write a program to find the `n`-th ugly number.
-
-Ugly numbers are **positive integers** which are divisible by `a` **or** `b` **or** `c`.
-
-**Example 1:**
-
-    Input: n = 3, a = 2, b = 3, c = 5
-    Output: 4
-    Explanation: The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3rd is 4.
-
-**Example 2:**
-
-    Input: n = 4, a = 2, b = 3, c = 4
-    Output: 6
-    Explanation: The ugly numbers are 2, 3, 4, 6, 8, 9, 10, 12... The 4th is 6.
-
-**Example 3:**
-
-    Input: n = 5, a = 2, b = 11, c = 13
-    Output: 10
-    Explanation: The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5th is 10.
-
-**Example 4:**
-
-    Input: n = 1000000000, a = 2, b = 217983653, c = 336916467
-    Output: 1999999984
-
-**Constraints:**
-
-- `1 <= n, a, b, c <= 10^9`
-- `1 <= a * b * c <= 10^18`
-- It's guaranteed that the result will be in range `[1, 2 * 10^9]`
-
-
-## 题目大意
-
-
-请你帮忙设计一个程序，用来找出第 n 个丑数。丑数是可以被 a 或 b 或 c 整除的 正整数。
-
-
-提示：
-
-- 1 <= n, a, b, c <= 10^9
-- 1 <= a * b * c <= 10^18
-- 本题结果在 [1, 2 * 10^9] 的范围内
-
-## 解题思路
-
-
-- 给出 4 个数字，a，b，c，n。要求输出可以整除 a 或者整除 b 或者整除 c 的第 n 个数。
-- 这一题限定了解的范围， `[1, 2 * 10^9]`，所以直接二分搜索来求解。逐步二分逼近 low 值，直到找到能满足条件的 low 的最小值，即为最终答案。
-- 这一题的关键在如何判断一个数是第几个数。一个数能整除 a，能整除 b，能整除 c，那么它应该是第 `num/a + num/b + num/c - num/ab - num/bc - num/ac + num/abc` 个数。这个就是韦恩图。需要注意的是，求 `ab`、`bc`、`ac`、`abc` 的时候需要再除以各自的最大公约数 `gcd()`。
diff --git a/Algorithms/1207. Unique Number of Occurrences/README.md b/Algorithms/1207. Unique Number of Occurrences/README.md
deleted file mode 100755
index 9d0287115..000000000
--- a/Algorithms/1207. Unique Number of Occurrences/README.md	
+++ /dev/null
@@ -1,44 +0,0 @@
-# [1207. Unique Number of Occurrences](https://leetcode.com/problems/unique-number-of-occurrences/)
-
-
-## 题目:
-
-Given an array of integers `arr`, write a function that returns `true` if and only if the number of occurrences of each value in the array is unique.
-
-**Example 1:**
-
-    Input: arr = [1,2,2,1,1,3]
-    Output: true
-    Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences.
-
-**Example 2:**
-
-    Input: arr = [1,2]
-    Output: false
-
-**Example 3:**
-
-    Input: arr = [-3,0,1,-3,1,1,1,-3,10,0]
-    Output: true
-
-**Constraints:**
-
-- `1 <= arr.length <= 1000`
-- `-1000 <= arr[i] <= 1000`
-
-
-
-## 题目大意
-
-给你一个整数数组 arr，请你帮忙统计数组中每个数的出现次数。如果每个数的出现次数都是独一无二的，就返回 true；否则返回 false。
-
-提示：
-
-- 1 <= arr.length <= 1000
-- -1000 <= arr[i] <= 1000
-
-## 解题思路
-
-
-- 给出一个数组，先统计每个数字出现的频次，判断在这个数组中是否存在相同的频次。
-- 简单题，先统计数组中每个数字的频次，然后用一个 map 判断频次是否重复。
diff --git a/Algorithms/378. Kth Smallest Element in a Sorted Matrix/README.md b/Algorithms/378. Kth Smallest Element in a Sorted Matrix/README.md
deleted file mode 100755
index a14e3380c..000000000
--- a/Algorithms/378. Kth Smallest Element in a Sorted Matrix/README.md	
+++ /dev/null
@@ -1,40 +0,0 @@
-# [378. Kth Smallest Element in a Sorted Matrix](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/)
-
-
-## 题目:
-
-Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.
-
-Note that it is the kth smallest element in the sorted order, not the kth distinct element.
-
-**Example:**
-
-    matrix = [
-       [ 1,  5,  9],
-       [10, 11, 13],
-       [12, 13, 15]
-    ],
-    k = 8,
-    
-    return 13.
-
-**Note:**You may assume k is always valid, 1 ≤ k ≤ n2.
-
-
-## 题目大意
-
-给定一个 n x n 矩阵，其中每行和每列元素均按升序排序，找到矩阵中第 k 小的元素。请注意，它是排序后的第 k 小元素，而不是第 k 个元素。
-
-
-说明:
-你可以假设 k 的值永远是有效的, 1 ≤ k ≤ n2 。
-
-
-## 解题思路
-
-
-- 给出一个行有序，列有序的矩阵(并非是按照下标有序的)，要求找出这个矩阵中第 K 小的元素。注意找的第 K 小元素指的不是 k 个不同的元素，可能存在相同的元素。
-- 最容易想到的就解法是优先队列。依次把矩阵中的元素推入到优先队列中。维护一个最小堆，一旦优先队列里面的元素有 k 个了，就算找到结果了。
-- 这一题最优解法是二分搜索。那搜索的空间是什么呢？根据题意，可以知道，矩阵左上角的那个元素是最小的，右下角的元素是最大的。即矩阵第一个元素确定了下界，矩阵的最后一个元素确定了上界。在这个解空间里面二分搜索所有值，找到第 K  小的元素。判断是否找到的条件是，在矩阵中比 mid 小的元素个数等于 K。不断的逼近 low，使得 low == high 的时候，就是找到了第 K 小的元素了。(因为题目中说了，一定会存在第 K 小元素，所以二分搜索到一个元素的时候，一定会得出结果)。
-
-![](https://img.halfrost.com/Leetcode/leetcode_378.png)
diff --git a/Algorithms/4. Median of Two Sorted Arrays/README.md b/Algorithms/4. Median of Two Sorted Arrays/README.md
deleted file mode 100755
index 43f84e640..000000000
--- a/Algorithms/4. Median of Two Sorted Arrays/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# [4. Median of Two Sorted Arrays](https://leetcode.com/problems/median-of-two-sorted-arrays/)
-
-
-## 题目:
-
-There are two sorted arrays **nums1** and **nums2** of size m and n respectively.
-
-Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
-
-You may assume **nums1** and **nums2** cannot be both empty.
-
-**Example 1:**
-
-    nums1 = [1, 3]
-    nums2 = [2]
-    
-    The median is 2.0
-
-**Example 2:**
-
-    nums1 = [1, 2]
-    nums2 = [3, 4]
-    
-    The median is (2 + 3)/2 = 2.5
-
-
-## 题目大意
-
-
-给定两个大小为 m 和 n 的有序数组 nums1 和 nums2。
-
-请你找出这两个有序数组的中位数，并且要求算法的时间复杂度为 O(log(m + n))。
-
-你可以假设 nums1 和 nums2 不会同时为空。
-
-
-
-## 解题思路
-
-
-- 给出两个有序数组，要求找出这两个数组合并以后的有序数组中的中位数。要求时间复杂度为 O(log (m+n))。
-- 这一题最容易想到的办法是把两个数组合并，然后取出中位数。但是合并有序数组的操作是 `O(max(n,m))` 的，不符合题意。看到题目给的 `log` 的时间复杂度，很容易联想到二分搜索。
-- 由于要找到最终合并以后数组的中位数，两个数组的总大小也知道，所以中间这个位置也是知道的。只需要二分搜索一个数组中切分的位置，另一个数组中切分的位置也能得到。为了使得时间复杂度最小，所以二分搜索两个数组中长度较小的那个数组。
-- 关键的问题是如何切分数组 1 和数组 2 。其实就是如何切分数组 1 。先随便二分产生一个 `midA`，切分的线何时算满足了中位数的条件呢？即，线左边的数都小于右边的数，即，`nums1[midA-1] ≤ nums2[midB] && nums2[midB-1] ≤ nums1[midA]` 。如果这些条件都不满足，切分线就需要调整。如果 `nums1[midA] < nums2[midB-1]`，说明 `midA` 这条线划分出来左边的数小了，切分线应该右移；如果 `nums1[midA-1] > nums2[midB]`，说明 midA 这条线划分出来左边的数大了，切分线应该左移。经过多次调整以后，切分线总能找到满足条件的解。
-- 假设现在找到了切分的两条线了，`数组 1` 在切分线两边的下标分别是 `midA - 1` 和 `midA`。`数组 2` 在切分线两边的下标分别是 `midB - 1` 和 `midB`。最终合并成最终数组，如果数组长度是奇数，那么中位数就是 `max(nums1[midA-1], nums2[midB-1])`。如果数组长度是偶数，那么中间位置的两个数依次是：`max(nums1[midA-1], nums2[midB-1])` 和 `min(nums1[midA], nums2[midB])`，那么中位数就是 `(max(nums1[midA-1], nums2[midB-1]) + min(nums1[midA], nums2[midB])) / 2`。图示见下图：
-
-    ![](https://img.halfrost.com/Leetcode/leetcode_4.png)
diff --git a/README.md b/README.md
old mode 100644
new mode 100755
index 03c31dff1..ebb30e7d1
--- a/README.md
+++ b/README.md
@@ -1,1399 +1,2616 @@
 
-# LeetCode by Go
-[LeetCode Online Judge] (https://leetcode.com/) is a website containing many **algorithm questions**. Most of them are real interview questions of **Google, Facebook, LinkedIn, Apple**, etc. This repo shows my solutions by Go with the code style strictly follows the [Google Golang Style Guide](https://github.com/golang/go/wiki/CodeReviewComments). Please feel free to reference and **STAR** to support this repo, thank you!
+# LeetCode in Go
+[LeetCode Online Judge](https://leetcode.com/) is a website containing many **algorithm questions**. Most of them are real interview questions of **Google, Facebook, LinkedIn, Apple**, etc. and it always help to sharp our algorithm Skills. Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. This repo shows my solutions in Go with the code style strictly follows the [Google Golang Style Guide](https://github.com/golang/go/wiki/CodeReviewComments). Please feel free to reference and **STAR** to support this repo, thank you!
 
 
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-<a href="https://www.zhihu.com/people/halfrost/activities"><img src="https://img.shields.io/badge/%E7%9F%A5%E4%B9%8E-@halfrost-50E3C2.svg?style=flat&colorA=0083ea"></a>
-<img src="https://img.shields.io/badge/PR-welcome%20!-brightgreen.svg?colorA=a0cd34">
-<a href=""><img src="https://img.shields.io/badge/license-CC-000000.svg"></a>
+<a href="https://twitter.com/halffrost"><img src="https://img.shields.io/badge/twitter-@halffrost-F8E81C.svg?style=flat&colorA=009df2"></a>
+<a href="https://www.zhihu.com/people/halfrost/activities"><img src="https://img.shields.io/badge/%E7%9F%A5%E4%B9%8E-@halfrost-fd6f32.svg?style=flat&colorA=0083ea"></a>
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+<a href="https://github.com/halfrost/LeetCode-Go/pulls"><img src="https://img.shields.io/badge/PR-Welcome-brightgreen.svg"></a>
 </p>
 
+支持 Progressive Web Apps 和 Dark Mode 的题解电子书《LeetCode Cookbook》 <a href="https://books.halfrost.com/leetcode/" rel="nofollow">Online Reading</a>
+
+<p align='center'>
+<a href="https://books.halfrost.com/leetcode/"><img src="https://img.halfrost.com/Leetcode/Cookbook_Safari_0.png"></a>
+<a href="https://books.halfrost.com/leetcode/"><img src="https://img.halfrost.com/Leetcode/Cookbook_Chrome_PWA.png"></a>
+</p>
+
+离线版本的电子书《LeetCode Cookbook》PDF <a href="https://github.com/halfrost/LeetCode-Go/releases/" rel="nofollow">Download here</a>
+
+<p align='center'>
+<a href="https://github.com/halfrost/LeetCode-Go/releases/"><img src="https://img.halfrost.com/Leetcode/Cookbook.png"></a>
+</p>
+
+通过 iOS / Android 浏览器安装 PWA 版《LeetCode Cookbook》至设备桌面随时学习
+
+<p align='center'>
+<a href="https://books.halfrost.com/leetcode/"><img src="https://img.halfrost.com/Leetcode/Cookbook_PWA_iPad.png"></a>
+<a href="https://books.halfrost.com/leetcode/"><img src="https://img.halfrost.com/Leetcode/Cookbook_PWA_iPad_example1__.png"></a>
+<a href="https://books.halfrost.com/leetcode/"><img src="https://img.halfrost.com/Leetcode/Cookbook_PWA_iPad_example2__.png"></a>
+</p>
+
+
 ## Data Structures
 
 > 标识了 ✅ 的专题是完成所有题目了的，没有标识的是还没有做完所有题目的
 
+<a href="https://books.halfrost.com/leetcode/"><img src="./website/static/logo.png" alt="logo" height="550" align="right" /></a>
+
 * [Array](#array)
 * [String](#string)
-* [Two Pointers ✅](#two-pointers)
-* [Linked List ✅](#linked-list)
-* [Stack ✅](#stack)
+* [✅ Two Pointers](#two-pointers)
+* [✅ Linked List](#linked-list)
+* [✅ Stack](#stack)
 * [Tree](#tree)
 * [Dynamic programming](#dynamic-programming)
-* [Backtracking ✅](#backtracking)
-* [Depth-first search](#depth-first-search)
-* [Breadth-first Search](#breadth-first-search)
+* [✅ Backtracking](#backtracking)
+* [Depth First Search](#depth-first-search)
+* [Breadth First Search](#breadth-first-search)
 * [Binary Search](#binary-search)
 * [Math](#math)
 * [Hash Table](#hash-table)
-* [Sort ✅](#sort)
-* [Bit Manipulation ✅](#bit-manipulation)
-* [Union Find ✅](#union-find)
-* [Sliding Window ✅](#sliding-window)
-* [Segment Tree ✅](#segment-tree)
-* [Binary Indexed Tree ✅](#binary-indexed-tree)
-
-
-| 数据结构 | 变种 | 相关题目 |
-|:-------:|:-------|:------|
-|顺序线性表：向量|||
-|单链表|1.双链表<br>2.静态链表<br>3.对称矩阵<br>4.稀疏矩阵||
-|栈|广义栈||
-|队列|1.链表实现<br>2.循环数组实现<br>3.双端队列||
-|字符串|1.KMP算法<br>2.有限状态自动机<br>3.模式匹配有限状态自动机<br>4.BM模式匹配算法<br>5.BM-KMP算法||
-|树|1.二叉树<br>2.并查集<br>3.Huffman数||
-|数组实现的堆|1.极大堆和极小堆<br>2.极大极小堆<br>3.双端堆<br>4.d叉堆||
-|树实现的堆|1.左堆<br>2.扁堆<br>3.二项式堆<br>4.斐波那契堆<br>5.配对堆||
-|查找|1.哈希表<br>2.跳跃表<br>3.排序二叉树<br>4.AVL树<br>5.B树<br>6.AA树<br>7.红黑树<br>8.排序二叉堆<br>9.Splay树<br>10.双链树<br>11.Trie树||
-
-
-
-## 算法
-
-## 一. 目录
-
-| #    |  Title                                                       |  Solution                                                                                                                                           |  Acceptance |  Difficulty |  Frequency |
+* [✅ Sort](#sort)
+* [✅ Bit Manipulation](#bit-manipulation)
+* [✅ Union Find](#union-find)
+* [✅ Sliding Window](#sliding-window)
+* [✅ Segment Tree](#segment-tree)
+* [✅ Binary Indexed Tree](#binary-indexed-tree)
+
+<br>
+<br>
+
+| 数据结构 | 变种 | 相关题目 | 讲解文章 | 
+|:-------:|:-------|:------|:------|
+|顺序线性表：向量||||
+|单链表|1. 双向链表<br>2. 静态链表<br>3. 对称矩阵<br>4. 稀疏矩阵|||
+|哈希表|1. 散列函数<br>2. 解决碰撞/填充因子<br>|||
+|栈和队列|1. 广义栈<br>2. 双端队列<br>|||
+|队列|1. 链表实现<br>2. 循环数组实现<br>3. 双端队列|||
+|字符串|1. KMP算法<br>2. 有限状态自动机<br>3. 模式匹配有限状态自动机<br>4. BM 模式匹配算法<br>5. BM-KMP 算法<br>6. BF 算法|||
+|树|1. 二叉树<br>2. 并查集<br>3. Huffman 树|||
+|数组实现的堆|1. 极大堆和极小堆<br>2. 极大极小堆<br>3. 双端堆<br>4. d 叉堆|||
+|树实现的堆|1. 左堆<br>2. 扁堆<br>3. 二项式堆<br>4. 斐波那契堆<br>5. 配对堆|||
+|查找|1. 哈希表<br>2. 跳跃表<br>3. 排序二叉树<br>4. AVL 树<br>5. B 树 / B+ 树 / B* 树<br>6. AA 树<br>7. 红黑树<br>8. 排序二叉堆<br>9. Splay 树<br>10. 双链树<br>11. Trie 树<br>12. R 树|||
+|--------------------------------------------|--------------------------------------------------------------------------------------------|---------------------------|-----------------------------------|
+
+
+## Algorithm
+
+
+| 算法 | 具体类型 | 相关题目 | 讲解文章 | 
+|:-------:|:-------|:------|:------|
+|排序算法|1. 冒泡排序<br>2. 插入排序<br>3. 选择排序<br>4. 希尔 Shell 排序<br>5. 快速排序<br>6. 归并排序<br>7. 堆排序<br>8. 线性排序算法<br>9. 自省排序<br>10. 间接排序<br>11. 计数排序<br>12. 基数排序<br>13. 桶排序<br>14. 外部排序 - k 路归并败者树<br>15. 外部排序 - 最佳归并树|||
+|递归与分治||1. 二分搜索/查找<br>2. 大整数的乘法<br>3. Strassen 矩阵乘法<br>4. 棋盘覆盖<br>5. 合并排序<br>6. 快速排序<br>7. 线性时间选择<br>8. 最接近点对问题<br>9. 循环赛日程表<br>||
+|动态规划||1. 矩阵连乘问题<br>2. 最长公共子序列<br>3. 最大子段和<br>4. 凸多边形最优三角剖分<br>5. 多边形游戏<br>6. 图像压缩<br>7. 电路布线<br>8. 流水作业调度<br>9. 0-1 背包问题/背包九讲<br>10. 最优二叉搜索树<br>11. 动态规划加速原理<br>12. 树型 DP<br>||
+|贪心||1. 活动安排问题<br>2. 最优装载<br>3. 哈夫曼编码<br>4. 单源最短路径<br>5. 最小生成树<br>6. 多机调度问题<br>||
+|回溯法||1. 装载问题<br>2. 批处理作业调度<br>3. 符号三角形问题<br>4. n 后问题<br>5. 0-1 背包问题<br>6. 最大团问题<br>7. 图的 m 着色问题<br>8. 旅行售货员问题<br>9. 圆排列问题<br>10. 电路板排列问题<br>11. 连续邮资问题<br>||
+|搜索|1. 枚举<br>2. DFS<br>3. BFS<br>4. 启发式搜索<br>|||
+|随机化|1. 随机数<br>2. 数值随机化算法<br>3. Sherwood 舍伍德算法<br>4. Las Vegas 拉斯维加斯算法<br>5. Monte Carlo 蒙特卡罗算法<br>|1. 计算 π 值<br>2. 计算定积分<br>3. 解非线性方程组<br>4. 线性时间选择算法<br>5. 跳跃表<br>6. n 后问题<br>7. 整数因子分解<br>8. 主元素问题<br>9. 素数测试<br>||
+|图论|1. 遍历 DFS / BFS<br>2. AOV / AOE 网络<br>3. Kruskal 算法(最小生成树)<br>4. Prim 算法(最小生成树)<br>5. Boruvka 算法(最小生成树)<br>6. Dijkstra 算法(单源最短路径)<br>7. Bellman-Ford 算法(单源最短路径)<br>8. SPFA 算法(单源最短路径)<br>9. Floyd 算法(多源最短路径)<br>10. Johnson 算法(多源最短路径)<br>11. Fleury 算法(欧拉回路)<br>12. Ford-Fulkerson 算法(最大网络流增广路)<br>13. Edmonds-Karp 算法(最大网络流)<br>14. Dinic 算法(最大网络流)<br>15. 一般预流推进算法<br>16. 最高标号预流推进 HLPP 算法<br>17. Primal-Dual 原始对偶算法(最小费用流)18. Kosaraju 算法(有向图强连通分量)<br>19. Tarjan 算法(有向图强连通分量)<br>20. Gabow 算法(有向图强连通分量)<br>21. 匈牙利算法(二分图匹配)<br>22. Hopcroft－Karp 算法(二分图匹配)<br>23. kuhn munkras 算法(二分图最佳匹配)<br>24. Edmonds’ Blossom-Contraction 算法(一般图匹配)<br>|1. 图遍历<br>2. 有向图和无向图的强弱连通性<br>3. 割点/割边<br>3. AOV 网络和拓扑排序<br>4. AOE 网络和关键路径<br>5. 最小代价生成树/次小生成树<br>6. 最短路径问题/第 K 短路问题<br>7. 最大网络流问题<br>8. 最小费用流问题<br>9. 图着色问题<br>10. 差分约束系统<br>11. 欧拉回路<br>12. 中国邮递员问题<br>13. 汉密尔顿回路<br>14. 最佳边割集/最佳点割集/最小边割集/最小点割集/最小路径覆盖/最小点集覆盖 <br>15. 边覆盖集<br>16. 二分图完美匹配和最大匹配问题<br>17. 仙人掌图<br>18. 弦图<br>19. 稳定婚姻问题<br>20. 最大团问题<br>||
+|数论||1. 最大公约数<br> 2. 最小公倍数<br>3. 分解质因数<br>4. 素数判定<br>5. 进制转换<br>6. 高精度计算<br>7. 整除问题<br>8. 同余问题<br>9. 欧拉函数<br>10. 扩展欧几里得<br>11. 置换群<br>12. 母函数<br>13. 离散变换<br>14. 康托展开<br>15. 矩阵<br>16. 向量<br>17. 线性方程组<br>18. 线性规划<br> ||
+|几何||1. 凸包 - Gift wrapping<br>2. 凸包 - Graham scan<br>3. 线段问题<br> 4. 多边形和多面体相关问题<br>||
+|NP 完全|1. 计算模型<br>2. P 类与 NP 类问题<br>3. NP 完全问题<br>4. NP 完全问题的近似算法<br>|1. 随机存取机 RAM<br>2. 随机存取存储程序机 RASP<br>3. 图灵机<br>4. 非确定性图灵机<br>5. P 类与 NP 类语言<br>6. 多项式时间验证<br>7. 多项式时间变换<br>8. Cook定理<br>9. 合取范式的可满足性问题 CNF-SAT<br>10. 3 元合取范式的可满足性问题 3-SAT<br>11. 团问题 CLIQUE<br>12. 顶点覆盖问题 VERTEX-COVER<br>13. 子集和问题 SUBSET-SUM<br>14. 哈密顿回路问题 HAM-CYCLE<br>15. 旅行售货员问题 TSP<br>16. 顶点覆盖问题的近似算法<br>17. 旅行售货员问题近似算法<br>18. 具有三角不等式性质的旅行售货员问题<br>19. 一般的旅行售货员问题<br>20. 集合覆盖问题的近似算法<br>21. 子集和问题的近似算法<br>22. 子集和问题的指数时间算法<br>23. 子集和问题的多项式时间近似格式<br>||
+|------------|------------------------------------------------------------------|-----------------------------------------------------------------|--------------------|
+
+
+## LeetCode Problems
+
+## 一. 个人数据
+
+|    |  Easy  |  Medium  |  Hard |  Total |
+|:--------:|:--------:|:--------:|:--------:|:--------:|
+|Optimizing|31|78|43|152|
+|Accepted|**287**|**484**|**142**|**913**|
+|Total|600|1305|539|2444|
+|Perfection Rate|89.2%|83.9%|69.7%|83.4%|
+|Completion Rate|47.8%|37.1%|26.3%|37.4%|
+|------------|----------------------------|----------------------------|----------------------------|----------------------------|
+
+## 二. 目录
+
+以下已经收录了 787 道题的题解，还有 11 道题在尝试优化到 beats 100%
+
+| No.    |  Title  |  Solution  |  Acceptance |  Difficulty |  Frequency |
 |:--------:|:--------------------------------------------------------------|:--------:|:--------:|:--------:|:--------:|
-| 0001 |  Two Sum                                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0001.%20Two%20Sum)                                                              | 44.30%      |  Easy       |            |
-| 0002 |  Add Two Numbers                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0002.%20Add%20Two%20Numbers)                                                         | 31.30%      |  Medium     |            |
-| 0003 |  Longest Substring Without Repeating Characters              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0003.%20Longest%20Substring%20Without%20Repeating%20Characters)                 | 28.50%      |  Medium     |            |
-| 0004 |  Median of Two Sorted Arrays                                 |                                                                                                                                               | 26.60%      |  Hard       |            |
-| 0005 |  Longest Palindromic Substring                               |                                                                                                                                               | 27.50%      |  Medium     |            |
-| 0006 |  ZigZag Conversion                                           |                                                                                                                                               | 32.20%      |  Medium     |            |
-| 0007 |  Reverse Integer                                             |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0007.%20Reverse%20Integer)                                                                                                                                               | 25.40%      |  Easy       |            |
-| 0008 |  String to Integer (atoi)                                    |                                                                                                                                               | 14.70%      |  Medium     |            |
-| 0009 |  Palindrome Number                                           |                                                                                                                                               | 43.70%      |  Easy       |            |
-| 0010 |  Regular Expression Matching                                 |                                                                                                                                               | 25.40%      |  Hard       |            |
-| 0011 |  Container With Most Water                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0011.%20Container%20With%20Most%20Water)                                        | 45.10%      |  Medium     |            |
-| 0012 |  Integer to Roman                                            |                                                                                                                                               | 51.30%      |  Medium     |            |
-| 0013 |  Roman to Integer                                            |                                                                                                                                               | 52.60%      |  Easy       |            |
-| 0014 |  Longest Common Prefix                                       |                                                                                                                                               | 33.70%      |  Easy       |            |
-| 0015 |  3Sum                                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0015.%203Sum)                                                                   | 24.20%      |  Medium     |            |
-| 0016 |  3Sum Closest                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0016.%203Sum%20Closest)                                                         | 45.80%      |  Medium     |            |
-| 0017 |  Letter Combinations of a Phone Number                       | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0017.%20Letter%20Combinations%20of%20a%20Phone%20Number)                        | 41.90%      |  Medium     |            |
-| 0018 |  4Sum                                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0018.%204Sum)                                                                   | 30.80%      |  Medium     |            |
-| 0019 |  Remove Nth Node From End of List                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0019.%20Remove%20Nth%20Node%20From%20End%20of%20List)                           | 34.30%      |  Medium     |            |
-| 0020 |  Valid Parentheses                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0020.%20Valid-Parentheses)                                                      | 36.70%      |  Easy       |            |
-| 0021 |  Merge Two Sorted Lists                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0021.%20Merge%20Two%20Sorted%20Lists)                                           | 47.70%      |  Easy       |            |
-| 0022 |  Generate Parentheses                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0022.%20Generate%20Parentheses)                                                 | 55.50%      |  Medium     |            |
-| 0023 |  Merge k Sorted Lists                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0023.%20Merge%20k%20Sorted%20Lists)                                             | 34.90%      |  Hard       |            |
-| 0024 |  Swap Nodes in Pairs                                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0024.%20Swap%20Nodes%20in%20Pairs)                                              | 45.10%      |  Medium     |            |
-| 0025 |  Reverse Nodes in k-Group                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0025.%20Reverse%20Nodes%20in%20k%20Group)                                       | 36.80%      |  Hard       |            |
-| 0026 |  Remove Duplicates from Sorted Array                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0026.%20Remove%20Duplicates%20from%20Sorted%20Array)                            | 41.00%      |  Easy       |            |
-| 0027 |  Remove Element                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0027.%20Remove%20Element)                                                       | 44.80%      |  Easy       |            |
-| 0028 |  Implement strStr()                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0028.%20Implement%20strStr())                                                   | 32.20%      |  Easy       |            |
-| 0029 |  Divide Two Integers                                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0029.%20Divide%20Two%20Integers)                                                                                                                                              | 16.10%      |  Medium     |            |
-| 0030 |  Substring with Concatenation of All Words                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0030.%20Substring%20with%20Concatenation%20of%20All%20Words)(是否还有更优解)                    | 23.70%      |  Hard       |            |
-| 0031 |  Next Permutation                                            |                                                                                                                                               | 30.60%      |  Medium     |            |
-| 0032 |  Longest Valid Parentheses                                   |                                                                                                                                               | 25.70%      |  Hard       |            |
-| 0033 |  Search in Rotated Sorted Array                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0033.%20Search%20in%20Rotated%20Sorted%20Array)                                                                                                                                              | 33.00%      |  Medium     |            |
-| 0034 |  Find First and Last Position of Element in Sorted Array     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0034.%20Find%20First%20and%20Last%20Position%20of%20Element%20in%20Sorted%20Array)                                                                                                                                              | 33.70%      |  Medium     |            |
-| 0035 |  Search Insert Position                                      |                                                                                                                                               | 40.90%      |  Easy       |            |
-| 0036 |  Valid Sudoku                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0036.%20Valid%20Sudoku)                                                         | 43.50%      |  Medium     |            |
-| 0037 |  Sudoku Solver                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0037.%20Sudoku%20Solver)                                                        | 37.40%      |  Hard       |            |
-| 0038 |  Count and Say                                               |                                                                                                                                               | 40.80%      |  Easy       |            |
-| 0039 |  Combination Sum                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0039.%20Combination%20Sum)                                                      | 49.10%      |  Medium     |            |
-| 0040 |  Combination Sum II                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0040.%20Combination%20Sum%20II)                                                 | 42.10%      |  Medium     |            |
-| 0041 |  First Missing Positive                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0041.%20First-Missing-Positive)                                                 | 29.10%      |  Hard       |            |
-| 0042 |  Trapping Rain Water                                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0042.%20Trapping%20Rain%20Water)                                                | 43.50%      |  Hard       |            |
-| 0043 |  Multiply Strings                                            |                                                                                                                                               | 30.90%      |  Medium     |            |
-| 0044 |  Wildcard Matching                                           |                                                                                                                                               | 23.00%      |  Hard       |            |
-| 0045 |  Jump Game II                                                |                                                                                                                                               | 28.20%      |  Hard       |            |
-| 0046 |  Permutations                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0046.%20Permutations)                                                           | 55.70%      |  Medium     |            |
-| 0047 |  Permutations II                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0047.%20Permutations%20II)                                                      | 40.90%      |  Medium     |            |
-| 0048 |  Rotate Image                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0048.%20Rotate%20Image)                                                         | 49.00%      |  Medium     |            |
-| 0049 |  Group Anagrams                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0049.%20Group%20Anagrams)                                                       | 47.40%      |  Medium     |            |
-| 0050 |  Pow(x, n)                                                       | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0050.%20Pow(x%2C%20n))                                                          | 28.10%      |  Medium    |
-| 0051 |  N-Queens                                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0051.%20N-Queens)                                                               | 39.80%      |  Hard       |            |
-| 0052 |  N-Queens II                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0052.%20N-Queens%20II)                                                          | 52.40%      |  Hard       |            |
-| 0053 |  Maximum Subarray                                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0053.%20Maximum%20Subarray)                                                     | 43.80%      |  Easy       |            |
-| 0054 |  Spiral Matrix                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0054.%20Spiral%20Matrix)                                                        | 30.70%      |  Medium     |            |
-| 0055 |  Jump Game                                                   |                                                                                                                                               | 32.10%      |  Medium     |            |
-| 0056 |  Merge Intervals                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0056.%20Merge%20Intervals)                                                      | 35.90%      |  Medium     |            |
-| 0057 |  Insert Interval                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0057.%20Insert%20Interval)                                                      | 31.40%      |  Hard       |            |
-| 0058 |  Length of Last Word                                         |                                                                                                                                               | 32.30%      |  Easy       |            |
-| 0059 |  Spiral Matrix II                                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0059.%20Spiral%20Matrix%20II)                                                   | 47.00%      |  Medium     |            |
-| 0060 |  Permutation Sequence                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0060.%20Permutation%20Sequence)                                                 | 33.40%      |  Medium     |            |
-| 0061 |  Rotate List                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0061.%20Rotate%20List)                                                          | 27.30%      |  Medium     |            |
-| 0062 |  Unique Paths                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0062.%20Unique%20Paths)                                                         | 48.00%      |  Medium     |            |
-| 0063 |  Unique Paths II                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0063.%20Unique%20Paths%20II)                                                    | 33.50%      |  Medium     |            |
-| 0064 |  Minimum Path Sum                                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0064.%20Minimum%20Path%20Sum)                                                   | 47.30%      |  Medium     |            |
-| 0065 |  Valid Number                                                |                                                                                                                                               | 14.10%      |  Hard       |            |
-| 0066 |  Plus One                                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0066.%20Plus%20One)                                                                                                                                              | 41.40%      |  Easy       |            |
-| 0067 |  Add Binary                                                  |                                                                                                                                               | 39.50%      |  Easy       |            |
-| 0068 |  Text Justification                                          |                                                                                                                                               | 23.50%      |  Hard       |            |
-| 0069 |  Sqrt(x)                                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0069.%20Sqrt(x))                                                                | 31.50%      |  Easy       |            |
-| 0070 |  Climbing Stairs                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0070.%20Climbing%20Stairs)                                                      | 44.40%      |  Easy       |            |
-| 0071 |  Simplify Path                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0071.%20Simplify%20Path)                                                        | 29.00%      |  Medium     |            |
-| 0072 |  Edit Distance                                               |                                                                                                                                               | 38.20%      |  Hard       |            |
-| 0073 |  Set Matrix Zeroes                                           |                                                                                                                                               | 40.10%      |  Medium     |            |
-| 0074 |  Search a 2D Matrix                                          |                                                                                                                                               | 35.00%      |  Medium     |            |
-| 0075 |  Sort Colors                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0075.%20Sort%20Colors)                                                          | 42.40%      |  Medium     |            |
-| 0076 |  Minimum Window Substring                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0076.%20Minimum%20Window%20Substring)                                           | 31.10%      |  Hard       |            |
-| 0077 |  Combinations                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0077.%20Combinations)                                                           | 48.20%      |  Medium     |            |
-| 0078 |  Subsets                                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0078.%20Subsets)                                                                | 53.40%      |  Medium     |            |
-| 0079 |  Word Search                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0079.%20Word%20Search)                                                          | 31.50%      |  Medium     |            |
-| 0080 |  Remove Duplicates from Sorted Array II                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0080.%20Remove%20Duplicates%20from%20Sorted%20Array%20II)                       | 40.60%      |  Medium     |            |
-| 0081 |  Search in Rotated Sorted Array II                           |                                                                                                                                               | 32.70%      |  Medium     |            |
-| 0082 |  Remove Duplicates from Sorted List II                       | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0082.%20Remove%20Duplicates%20from%20Sorted%20List%20II)                        | 33.30%      |  Medium     |            |
-| 0083 |  Remove Duplicates from Sorted List                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0083.%20Remove%20Duplicates%20from%20Sorted%20List)                             | 42.80%      |  Easy       |            |
-| 0084 |  Largest Rectangle in Histogram                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0084.%20Largest%20Rectangle%20in%20Histogram)                                   | 31.40%      |  Hard       |            |
-| 0085 |  Maximal Rectangle                                           |                                                                                                                                               | 33.60%      |  Hard       |            |
-| 0086 |  Partition List                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0086.%20Partition%20List)                                                       | 37.60%      |  Medium     |            |
-| 0087 |  Scramble String                                             |                                                                                                                                               | 31.70%      |  Hard       |            |
-| 0088 |  Merge Sorted Array                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0088.%20Merge-Sorted-Array)                                                     | 36.00%      |  Easy       |            |
-| 0089 |  Gray Code                                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0089.%20Gray%20Code)                                                            | 46.00%      |  Medium     |            |
-| 0090 |  Subsets II                                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0090.%20Subsets%20II)                                                           | 42.70%      |  Medium     |            |
-| 0091 |  Decode Ways                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0091.%20Decode%20Ways)                                                          | 22.50%      |  Medium     |            |
-| 0092 |  Reverse Linked List II                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0092.%20Reverse%20Linked%20List%20II)                                           | 35.20%      |  Medium     |            |
-| 0093 |  Restore IP Addresses                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0093.%20Restore%20IP%20Addresses)                                               | 31.70%      |  Medium     |            |
-| 0094 |  Binary Tree Inorder Traversal                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0094.%20Binary%20Tree%20Inorder%20Traversal)                                    | 57.10%      |  Medium     |            |
-| 0095 |  Unique Binary Search Trees II                               |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0095.%20Unique%20Binary%20Search%20Trees%20II)                                                                                                                                               | 36.00%      |  Medium     |            |
-| 0096 |  Unique Binary Search Trees                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0096.%20Unique%20Binary%20Search%20Trees)                                       | 46.60%      |  Medium     |            |
-| 0097 |  Interleaving String                                         |                                                                                                                                               | 28.20%      |  Hard       |            |
-| 0098 |  Validate Binary Search Tree                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0098.%20Validate%20Binary%20Search%20Tree)                                      | 25.90%      |  Medium     |            |
-| 0099 |  Recover Binary Search Tree                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0099.%20Recover%20Binary%20Search%20Tree)                                       | 34.90%      |  Hard       |            |
-| 0100 |  Same Tree                                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0100.%20Same%20Tree)                                                            | 50.20%      |  Easy       |            |
-| 0101 |  Symmetric Tree                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0101.%20Symmetric%20Tree)                                                       | 43.70%      |  Easy       |            |
-| 0102 |  Binary Tree Level Order Traversal                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0102.%20Binary%20Tree%20Level%20Order%20Traversal)                              | 48.90%      |  Medium     |            |
-| 0103 |  Binary Tree Zigzag Level Order Traversal                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0103.%20Binary%20Tree%20Zigzag%20Level%20Order%20Traversal)                     | 42.10%      |  Medium     |            |
-| 0104 |  Maximum Depth of Binary Tree                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0104.%20Maximum%20Depth%20of%20Binary%20Tree)                                   | 60.90%      |  Easy       |            |
-| 0105 |  Construct Binary Tree from Preorder and Inorder Traversal   |                                                                                                                                               | 41.60%      |  Medium     |            |
-| 0106 |  Construct Binary Tree from Inorder and Postorder Traversal  |                                                                                                                                               | 39.70%      |  Medium     |            |
-| 0107 |  Binary Tree Level Order Traversal II                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0107.%20Binary%20Tree%20Level%20Order%20Traversal%20II)                         | 47.20%      |  Easy       |            |
-| 0108 |  Convert Sorted Array to Binary Search Tree                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0108.%20Convert%20Sorted%20Array%20to%20Binary%20Search%20Tree)                 | 51.20%      |  Easy       |            |
-| 0109 |  Convert Sorted List to Binary Search Tree                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0109.%20Convert%20Sorted%20List%20to%20Binary%20Search%20Tree)                  | 41.20%      |  Medium     |            |
-| 0110 |  Balanced Binary Tree                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0110.%20Balanced%20Binary%20Tree)                                               | 41.20%      |  Easy       |            |
-| 0111 |  Minimum Depth of Binary Tree                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0111.%20Minimum%20Depth%20of%20Binary%20Tree)                                   | 35.40%      |  Easy       |            |
-| 0112 |  Path Sum                                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0112.%20Path%20Sum)                                                             | 38.00%      |  Easy       |            |
-| 0113 |  Path Sum II                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0113.%20Path%20Sum%20II)                                                        | 41.00%      |  Medium     |            |
-| 0114 |  Flatten Binary Tree to Linked List                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0114.%20Flatten%20Binary%20Tree%20to%20Linked%20List)                           | 42.80%      |  Medium     |            |
-| 0115 |  Distinct Subsequences                                       |                                                                                                                                               | 35.20%      |  Hard       |            |
-| 0116 |  Populating Next Right Pointers in Each Node                 |                                                                                                                                               | 38.20%      |  Medium     |            |
-| 0117 |  Populating Next Right Pointers in Each Node II              |                                                                                                                                               | 34.70%      |  Medium     |            |
-| 0118 |  Pascal's Triangle                                           |                                                                                                                                               | 46.50%      |  Easy       |            |
-| 0119 |  Pascal's Triangle II                                        |                                                                                                                                               | 44.00%      |  Easy       |            |
-| 0120 |  Triangle                                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0120.%20Triangle)                                                               | 39.70%      |  Medium     |            |
-| 0121 |  Best Time to Buy and Sell Stock                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock)                            | 47.50%      |  Easy       |            |
-| 0122 |  Best Time to Buy and Sell Stock II                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0122.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20II)                       | 52.20%      |  Easy       |            |
-| 0123 |  Best Time to Buy and Sell Stock III                         |                                                                                                                                               | 33.90%      |  Hard       |            |
-| 0124 |  Binary Tree Maximum Path Sum                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0124.%20Binary%20Tree%20Maximum%20Path%20Sum)                                   | 30.20%      |  Hard       |            |
-| 0125 |  Valid Palindrome                                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0125.%20Valid-Palindrome)                                                       | 31.40%      |  Easy       |            |
-| 0126 |  Word Ladder II                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0126.%20Word%20Ladder%20II)                                                     | 18.00%      |  Hard       |            |
-| 0127 |  Word Ladder                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0127.%20Word%20Ladder)                                                          | 24.40%      |  Medium     |            |
-| 0128 |  Longest Consecutive Sequence                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0128.%20Longest%20Consecutive%20Sequence)                                                                                                                                              | 41.90%      |  Hard       |            |
-| 0129 |  Sum Root to Leaf Numbers                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0129.%20Sum%20Root%20to%20Leaf%20Numbers)                                       | 42.80%      |  Medium     |            |
-| 0130 |  Surrounded Regions                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0130.%20Surrounded%20Regions)                                                                                                                                               | 23.10%      |  Medium     |            |
-| 0131 |  Palindrome Partitioning                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0131.%20Palindrome%20Partitioning)                                              | 41.30%      |  Medium     |            |
-| 0132 |  Palindrome Partitioning II                                  |                                                                                                                                               | 27.60%      |  Hard       |            |
-| 0133 |  Clone Graph                                                 |                                                                                                                                               | 27.00%      |  Medium     |            |
-| 0134 |  Gas Station                                                 |                                                                                                                                               | 34.10%      |  Medium     |            |
-| 0135 |  Candy                                                       |                                                                                                                                               | 28.60%      |  Hard       |            |
-| 0136 |  Single Number                                               |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0136.%20Single%20Number)                                                                                                                                               | 60.50%      |  Easy       |            |
-| 0137 |  Single Number II                                            |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0137.%20Single%20Number%20II)                                                                                                                                               | 46.20%      |  Medium     |            |
-| 0138 |  Copy List with Random Pointer                               |                                                                                                                                               | 27.50%      |  Medium     |            |
-| 0139 |  Word Break                                                  |                                                                                                                                               | 35.60%      |  Medium     |            |
-| 0140 |  Word Break II                                               |                                                                                                                                               | 27.50%      |  Hard       |            |
-| 0141 |  Linked List Cycle                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0141.%20Linked%20List%20Cycle)                                                  | 37.20%      |  Easy       |            |
-| 0142 |  Linked List Cycle II                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0142.%20Linked%20List%20Cycle%20II)                                             | 32.40%      |  Medium     |            |
-| 0143 |  Reorder List                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0143.%20Reorder%20List)                                                         | 31.20%      |  Medium     |            |
-| 0144 |  Binary Tree Preorder Traversal                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0144.%20Binary%20Tree%20Preorder%20Traversal)                                   | 51.70%      |  Medium     |            |
-| 0145 |  Binary Tree Postorder Traversal                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0145.%20Binary%20Tree%20Postorder%20Traversal)                                  | 48.90%      |  Hard       |            |
-| 0146 |  LRU Cache                                                   |                                                                                                                                               | 25.90%      |  Medium     |            |
-| 0147 |  Insertion Sort List                                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0147.%20Insertion%20Sort%20List)                                                | 37.60%      |  Medium     |            |
-| 0148 |  Sort List                                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0148.%20Sort%20List)                                                            | 35.80%      |  Medium     |            |
-| 0149 |  Max Points on a Line                                        |                                                                                                                                               | 15.80%      |  Hard       |            |
-| 0150 |  Evaluate Reverse Polish Notation                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0150.%20Evaluate%20Reverse%20Polish%20Notation)                                 | 32.50%      |  Medium     |            |
-| 0151 |  Reverse Words in a String                                   |                                                                                                                                               | 17.00%      |  Medium     |            |
-| 0152 |  Maximum Product Subarray                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0152.%20Maximum%20Product%20Subarray)                                           | 29.40%      |  Medium     |            |
-| 0153 |  Find Minimum in Rotated Sorted Array                        |                                                                                                                                               | 43.10%      |  Medium     |            |
-| 0154 |  Find Minimum in Rotated Sorted Array II                     |                                                                                                                                               | 39.40%      |  Hard       |            |
-| 0155 |  Min Stack                                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0155.%20Min%20Stack)                                                            | 37.50%      |  Easy       |            |
-| 0156 |  Binary Tree Upside Down                                     |                                                                                                                                               | 51.10%      |  Medium     |            |
-| 0157 |  Read N Characters Given Read4                               |                                                                                                                                               | 29.70%      |  Easy       |            |
-| 0158 |  Read N Characters Given Read4 II - Call multiple times      |                                                                                                                                               | 26.60%      |  Hard       |            |
-| 0159 |  Longest Substring with At Most Two Distinct Characters      |                                                                                                                                               | 47.20%      |  Hard       |            |
-| 0160 |  Intersection of Two Linked Lists                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0160.%20Intersection%20of%20Two%20Linked%20Lists)                               | 34.30%      |  Easy       |            |
-| 0161 |  One Edit Distance                                           |                                                                                                                                               | 31.70%      |  Medium     |            |
-| 0162 |  Find Peak Element                                           |                                                                                                                                               | 41.40%      |  Medium     |            |
-| 0163 |  Missing Ranges                                              |                                                                                                                                               | 23.30%      |  Medium     |            |
-| 0164 |  Maximum Gap                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0164.%20Maximum%20Gap)                                                          | 32.70%      |  Hard       |            |
-| 0165 |  Compare Version Numbers                                     |                                                                                                                                               | 23.80%      |  Medium     |            |
-| 0166 |  Fraction to Recurring Decimal                               |                                                                                                                                               | 19.60%      |  Medium     |            |
-| 0167 |  Two Sum II - Input array is sorted                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0167.%20Two%20Sum%20II%20-%20Input%20array%20is%20sorted)                       | 50.60%      |  Easy       |            |
-| 0168 |  Excel Sheet Column Title                                    |                                                                                                                                               | 29.10%      |  Easy       |            |
-| 0169 |  Majority Element                                            |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0169.%20Majority%20Element)                                                                                                                                               | 53.00%      |  Easy       |            |
-| 0170 |  Two Sum III - Data structure design                         |                                                                                                                                               | 30.60%      |  Easy       |            |
-| 0171 |  Excel Sheet Column Number                                   |                                                                                                                                               | 51.70%      |  Easy       |            |
-| 0172 |  Factorial Trailing Zeroes                                   |                                                                                                                                               | 37.40%      |  Easy       |            |
-| 0173 |  Binary Search Tree Iterator                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0173.%20Binary%20Search%20Tree%20Iterator)                                      | 49.00%      |  Medium     |            |
-| 0174 |  Dungeon Game                                                |                                                                                                                                               | 27.30%      |  Hard       |            |
-| 0175 |  Combine Two Tables                                          |                                                                                                                                               | 52.40%      |  Easy       |            |
-| 0176 |  Second Highest Salary                                       |                                                                                                                                               | 27.50%      |  Easy       |            |
-| 0177 |  Nth Highest Salary                                          |                                                                                                                                               | 26.60%      |  Medium     |            |
-| 0178 |  Rank Scores                                                 |                                                                                                                                               | 37.10%      |  Medium     |            |
-| 0179 |  Largest Number                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0179.%20Largest%20Number)                                                       | 26.00%      |  Medium     |            |
-| 0180 |  Consecutive Numbers                                         |                                                                                                                                               | 34.00%      |  Medium     |            |
-| 0181 |  Employees Earning More Than Their Managers                  |                                                                                                                                               | 48.90%      |  Easy       |            |
-| 0182 |  Duplicate Emails                                            |                                                                                                                                               | 55.20%      |  Easy       |            |
-| 0183 |  Customers Who Never Order                                   |                                                                                                                                               | 45.40%      |  Easy       |            |
-| 0184 |  Department Highest Salary                                   |                                                                                                                                               | 29.40%      |  Medium     |            |
-| 0185 |  Department Top Three Salaries                               |                                                                                                                                               | 26.50%      |  Hard       |            |
-| 0186 |  Reverse Words in a String II                                |                                                                                                                                               | 38.10%      |  Medium     |            |
-| 0187 |  Repeated DNA Sequences                                      |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0187.%20Repeated%20DNA%20Sequences)                                                                                                                                               | 36.30%      |  Medium     |            |
-| 0188 |  Best Time to Buy and Sell Stock IV                          |                                                                                                                                               | 26.40%      |  Hard       |            |
-| 0189 |  Rotate Array                                                |                                                                                                                                               | 30.50%      |  Easy       |            |
-| 0190 |  Reverse Bits                                                |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0190.%20Reverse%20Bits)                                                                                                                                               | 31.70%      |  Easy       |            |
-| 0191 |  Number of 1 Bits                                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0191.%20Number%20of%201%20Bits)                                                 | 43.60%      |  Easy       |            |
-| 0192 |  Word Frequency                                              |                                                                                                                                               | 26.80%      |  Medium     |            |
-| 0193 |  Valid Phone Numbers                                         |                                                                                                                                               | 25.20%      |  Easy       |            |
-| 0194 |  Transpose File                                              |                                                                                                                                               | 22.90%      |  Medium     |            |
-| 0195 |  Tenth Line                                                  |                                                                                                                                               | 33.90%      |  Easy       |            |
-| 0196 |  Delete Duplicate Emails                                     |                                                                                                                                               | 33.40%      |  Easy       |            |
-| 0197 |  Rising Temperature                                          |                                                                                                                                               | 35.10%      |  Easy       |            |
-| 0198 |  House Robber                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0198.%20House%20Robber)                                                         | 41.10%      |  Easy       |            |
-| 0199 |  Binary Tree Right Side View                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0199.%20Binary%20Tree%20Right%20Side%20View)                                    | 48.30%      |  Medium     |            |
-| 0200 |  Number of Islands                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0200.%20Number%20of%20Islands)                                                  | 41.90%      |  Medium     |            |
-| 0201 |  Bitwise AND of Numbers Range                                |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0201.%20Bitwise%20AND%20of%20Numbers%20Range)                                                                                                                                               | 36.10%      |  Medium     |            |
-| 0202 |  Happy Number                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0202.%20Happy%20Number)                                                         | 45.60%      |  Easy       |            |
-| 0203 |  Remove Linked List Elements                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0203.%20Remove%20Linked%20List%20Elements)                                      | 35.90%      |  Easy       |            |
-| 0204 |  Count Primes                                                |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0204.%20Count%20Primes)                                                                                                                                               | 29.20%      |  Easy       |            |
-| 0205 |  Isomorphic Strings                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0205.%20Isomorphic%20Strings)                                                   | 37.60%      |  Easy       |            |
-| 0206 |  Reverse Linked List                                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0206.%20Reverse-Linked-List)                                                    | 55.30%      |  Easy       |            |
-| 0207 |  Course Schedule                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0207.%20Course%20Schedule)                                                      | 38.20%      |  Medium     |            |
-| 0208 |  Implement Trie (Prefix Tree)                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0208.%20Implement%20Trie%20(Prefix%20Tree))                                     | 39.00%      |  Medium     |            |
-| 0209 |  Minimum Size Subarray Sum                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0209.%20Minimum%20Size%20Subarray%20Sum)                                        | 35.10%      |  Medium     |            |
-| 0210 |  Course Schedule II                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0210.%20Course%20Schedule%20II)                                                 | 35.20%      |  Medium     |            |
-| 0211 |  Add and Search Word - Data structure design                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0211.%20Add%20and%20Search%20Word%20-%20Data%20structure%20design)              | 30.70%      |  Medium     |            |
-| 0212 |  Word Search II                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0212.%20Word%20Search%20II)                                                     | 29.00%      |  Hard       |            |
-| 0213 |  House Robber II                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0213.%20House%20Robber%20II)                                                    | 35.40%      |  Medium     |            |
-| 0214 |  Shortest Palindrome                                         |                                                                                                                                               | 27.70%      |  Hard       |            |
-| 0215 |  Kth Largest Element in an Array                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0215.%20Kth%20Largest%20Element%20in%20an%20Array)                              | 48.30%      |  Medium     |            |
-| 0216 |  Combination Sum III                                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0216.%20Combination%20Sum%20III)                                                | 51.90%      |  Medium     |            |
-| 0217 |  Contains Duplicate                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0217.%20Contains%20Duplicate)                                                   | 52.30%      |  Easy       |            |
-| 0218 |  The Skyline Problem                                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0218.%20The%20Skyline%20Problem)                                                                                                                                              | 31.80%      |  Hard       |            |
-| 0219 |  Contains Duplicate II                                       | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0219.%20Contains%20Duplicate%20II)                                              | 35.50%      |  Easy       |            |
-| 0220 |  Contains Duplicate III                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0220.%20Contains%20Duplicate%20III)                                             | 19.80%      |  Medium     |            |
-| 0221 |  Maximal Square                                              |                                                                                                                                               | 33.30%      |  Medium     |            |
-| 0222 |  Count Complete Tree Nodes                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0222.%20Count%20Complete%20Tree%20Nodes)                                        | 34.40%      |  Medium     |            |
-| 0223 |  Rectangle Area                                              |                                                                                                                                               | 36.00%      |  Medium     |            |
-| 0224 |  Basic Calculator                                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0224.%20Basic%20Calculator)                                                     | 33.00%      |  Hard       |            |
-| 0225 |  Implement Stack using Queues                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0225.%20Implement%20Stack%20using%20Queues)                                     | 39.60%      |  Easy       |            |
-| 0226 |  Invert Binary Tree                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0226.%20Invert%20Binary%20Tree)                                                 | 58.50%      |  Easy       |            |
-| 0227 |  Basic Calculator II                                         |                                                                                                                                               | 33.70%      |  Medium     |            |
-| 0228 |  Summary Ranges                                              |                                                                                                                                               | 36.20%      |  Medium     |            |
-| 0229 |  Majority Element II                                         |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0229.%20Majority%20Element%20II)                                                                                                                                               | 32.30%      |  Medium     |            |
-| 0230 |  Kth Smallest Element in a BST                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0230.%20Kth%20Smallest%20Element%20in%20a%20BST)                                | 51.90%      |  Medium     |            |
-| 0231 |  Power of Two                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0231.%20Power%20of%20Two)                                                       | 42.10%      |  Easy       |            |
-| 0232 |  Implement Queue using Stacks                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0232.%20Implement%20Queue%20using%20Stacks)                                     | 43.70%      |  Easy       |            |
-| 0233 |  Number of Digit One                                         |                                                                                                                                               | 30.30%      |  Hard       |            |
-| 0234 |  Palindrome Linked List                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0234.%20Palindrome%20Linked%20List)                                             | 36.30%      |  Easy       |            |
-| 0235 |  Lowest Common Ancestor of a Binary Search Tree              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree)           | 45.10%      |  Easy       |            |
-| 0236 |  Lowest Common Ancestor of a Binary Tree                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0236.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree)                    | 37.90%      |  Medium     |            |
-| 0237 |  Delete Node in a Linked List                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0237.%20Delete%20Node%20in%20a%20Linked%20List)                                 | 54.00%      |  Easy       |            |
-| 0238 |  Product of Array Except Self                                |                                                                                                                                               | 55.40%      |  Medium     |            |
-| 0239 |  Sliding Window Maximum                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0239.%20Sliding%20Window%20Maximum)                                             | 38.40%      |  Hard       |            |
-| 0240 |  Search a 2D Matrix II                                       |                                                                                                                                               | 41.00%      |  Medium     |            |
-| 0241 |  Different Ways to Add Parentheses                           |                                                                                                                                               | 50.30%      |  Medium     |            |
-| 0242 |  Valid Anagram                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0242.%20Valid%20Anagram)                                                        | 52.50%      |  Easy       |            |
-| 0243 |  Shortest Word Distance                                      |                                                                                                                                               | 57.60%      |  Easy       |            |
-| 0244 |  Shortest Word Distance II                                   |                                                                                                                                               | 47.60%      |  Medium     |            |
-| 0245 |  Shortest Word Distance III                                  |                                                                                                                                               | 53.60%      |  Medium     |            |
-| 0246 |  Strobogrammatic Number                                      |                                                                                                                                               | 42.40%      |  Easy       |            |
-| 0247 |  Strobogrammatic Number II                                   |                                                                                                                                               | 44.60%      |  Medium     |            |
-| 0248 |  Strobogrammatic Number III                                  |                                                                                                                                               | 36.70%      |  Hard       |            |
-| 0249 |  Group Shifted Strings                                       |                                                                                                                                               | 49.10%      |  Medium     |            |
-| 0250 |  Count Univalue Subtrees                                     |                                                                                                                                               | 49.10%      |  Medium     |            |
-| 0251 |  Flatten 2D Vector                                           |                                                                                                                                               | 43.90%      |  Medium     |            |
-| 0252 |  Meeting Rooms                                               |                                                                                                                                               | 52.30%      |  Easy       |            |
-| 0253 |  Meeting Rooms II                                            |                                                                                                                                               | 43.10%      |  Medium     |            |
-| 0254 |  Factor Combinations                                         |                                                                                                                                               | 44.50%      |  Medium     |            |
-| 0255 |  Verify Preorder Sequence in Binary Search Tree              |                                                                                                                                               | 43.70%      |  Medium     |            |
-| 0256 |  Paint House                                                 |                                                                                                                                               | 49.10%      |  Easy       |            |
-| 0257 |  Binary Tree Paths                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0257.%20Binary%20Tree%20Paths)                                                  | 46.30%      |  Easy       |            |
-| 0258 |  Add Digits                                                  |                                                                                                                                               | 54.30%      |  Easy       |            |
-| 0259 |  3Sum Smaller                                                |                                                                                                                                               | 45.10%      |  Medium     |            |
-| 0260 |  Single Number III                                           |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0260.%20Single%20Number%20III)                                                                                                                                               | 57.30%      |  Medium     |            |
-| 0261 |  Graph Valid Tree                                            |                                                                                                                                               | 40.00%      |  Medium     |            |
-| 0262 |  Trips and Users                                             |                                                                                                                                               | 25.40%      |  Hard       |            |
-| 0263 |  Ugly Number                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0263.%20Ugly%20Number)                                                          | 40.70%      |  Easy       |            |
-| 0264 |  Ugly Number II                                              |                                                                                                                                               | 36.60%      |  Medium     |            |
-| 0265 |  Paint House II                                              |                                                                                                                                               | 41.70%      |  Hard       |            |
-| 0266 |  Palindrome Permutation                                      |                                                                                                                                               | 60.20%      |  Easy       |            |
-| 0267 |  Palindrome Permutation II                                   |                                                                                                                                               | 33.90%      |  Medium     |            |
-| 0268 |  Missing Number                                              |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0268.%20Missing%20Number)                                                                                                                                               | 48.60%      |  Easy       |            |
-| 0269 |  Alien Dictionary                                            |                                                                                                                                               | 31.30%      |  Hard       |            |
-| 0270 |  Closest Binary Search Tree Value                            |                                                                                                                                               | 44.00%      |  Easy       |            |
-| 0271 |  Encode and Decode Strings                                   |                                                                                                                                               | 27.00%      |  Medium     |            |
-| 0272 |  Closest Binary Search Tree Value II                         |                                                                                                                                               | 45.40%      |  Hard       |            |
-| 0273 |  Integer to English Words                                    |                                                                                                                                               | 24.50%      |  Hard       |            |
-| 0274 |  H-Index                                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0274.%20H-Index)                                                                | 34.70%      |  Medium     |            |
-| 0275 |  H-Index II                                                  |                                                                                                                                               | 35.50%      |  Medium     |            |
-| 0276 |  Paint Fence                                                 |                                                                                                                                               | 36.60%      |  Easy       |            |
-| 0277 |  Find the Celebrity                                          |                                                                                                                                               | 37.10%      |  Medium     |            |
-| 0278 |  First Bad Version                                           |                                                                                                                                               | 30.30%      |  Easy       |            |
-| 0279 |  Perfect Squares                                             |                                                                                                                                               | 42.30%      |  Medium     |            |
-| 0280 |  Wiggle Sort                                                 |                                                                                                                                               | 61.20%      |  Medium     |            |
-| 0281 |  Zigzag Iterator                                             |                                                                                                                                               | 56.10%      |  Medium     |            |
-| 0282 |  Expression Add Operators                                    |                                                                                                                                               | 32.90%      |  Hard       |            |
-| 0283 |  Move Zeroes                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0283.%20Move%20Zeroes)                                                          | 54.50%      |  Easy       |            |
-| 0284 |  Peeking Iterator                                            |                                                                                                                                               | 40.90%      |  Medium     |            |
-| 0285 |  Inorder Successor in BST                                    |                                                                                                                                               | 35.10%      |  Medium     |            |
-| 0286 |  Walls and Gates                                             |                                                                                                                                               | 49.80%      |  Medium     |            |
-| 0287 |  Find the Duplicate Number                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0287.%20Find%20the%20Duplicate%20Number)                                        | 50.00%      |  Medium     |            |
-| 0288 |  Unique Word Abbreviation                                    |                                                                                                                                               | 20.10%      |  Medium     |            |
-| 0289 |  Game of Life                                                |                                                                                                                                               | 45.90%      |  Medium     |            |
-| 0290 |  Word Pattern                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0290.%20Word%20Pattern)                                                         | 35.20%      |  Easy       |            |
-| 0291 |  Word Pattern II                                             |                                                                                                                                               | 41.10%      |  Hard       |            |
-| 0292 |  Nim Game                                                    |                                                                                                                                               | 55.70%      |  Easy       |            |
-| 0293 |  Flip Game                                                   |                                                                                                                                               | 59.00%      |  Easy       |            |
-| 0294 |  Flip Game II                                                |                                                                                                                                               | 48.40%      |  Medium     |            |
-| 0295 |  Find Median from Data Stream                                |                                                                                                                                               | 37.10%      |  Hard       |            |
-| 0296 |  Best Meeting Point                                          |                                                                                                                                               | 55.00%      |  Hard       |            |
-| 0297 |  Serialize and Deserialize Binary Tree                       |                                                                                                                                               | 41.30%      |  Hard       |            |
-| 0298 |  Binary Tree Longest Consecutive Sequence                    |                                                                                                                                               | 44.20%      |  Medium     |            |
-| 0299 |  Bulls and Cows                                              |                                                                                                                                               | 39.70%      |  Medium     |            |
-| 0300 |  Longest Increasing Subsequence                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0300.%20Longest%20Increasing%20Subsequence)                                     | 41.00%      |  Medium     |            |
-| 0301 |  Remove Invalid Parentheses                                  |                                                                                                                                               | 39.50%      |  Hard       |            |
-| 0302 |  Smallest Rectangle Enclosing Black Pixels                   |                                                                                                                                               | 49.40%      |  Hard       |            |
-| 0303 |  Range Sum Query - Immutable                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0303.%20Range%20Sum%20Query%20-%20Immutable)                                                                                                                                              | 38.40%      |  Easy       |            |
-| 0304 |  Range Sum Query 2D - Immutable                              |                                                                                                                                               | 32.70%      |  Medium     |            |
-| 0305 |  Number of Islands II                                        |                                                                                                                                               | 41.70%      |  Hard       |            |
-| 0306 |  Additive Number                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0306.%20Additive%20Number)                                                      | 28.40%      |  Medium     |            |
-| 0307 |  Range Sum Query - Mutable                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0307.%20Range%20Sum%20Query%20-%20Mutable)                                                                                                                                              | 29.00%      |  Medium     |            |
-| 0308 |  Range Sum Query 2D - Mutable                                |                                                                                                                                               | 32.20%      |  Hard       |            |
-| 0309 |  Best Time to Buy and Sell Stock with Cooldown               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0309.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Cooldown)          | 44.20%      |  Medium     |            |
-| 0310 |  Minimum Height Trees                                        |                                                                                                                                               | 30.30%      |  Medium     |            |
-| 0311 |  Sparse Matrix Multiplication                                |                                                                                                                                               | 56.70%      |  Medium     |            |
-| 0312 |  Burst Balloons                                              |                                                                                                                                               | 47.50%      |  Hard       |            |
-| 0313 |  Super Ugly Number                                           |                                                                                                                                               | 41.70%      |  Medium     |            |
-| 0314 |  Binary Tree Vertical Order Traversal                        |                                                                                                                                               | 41.20%      |  Medium     |            |
-| 0315 |  Count of Smaller Numbers After Self                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0315.%20Count%20of%20Smaller%20Numbers%20After%20Self)                                                                                                                                              | 38.40%      |  Hard       |            |
-| 0316 |  Remove Duplicate Letters                                    |                                                                                                                                               | 32.80%      |  Hard       |            |
-| 0317 |  Shortest Distance from All Buildings                        |                                                                                                                                               | 38.10%      |  Hard       |            |
-| 0318 |  Maximum Product of Word Lengths                             |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0318.%20Maximum%20Product%20of%20Word%20Lengths)                                                                                                                                               | 48.70%      |  Medium     |            |
-| 0319 |  Bulb Switcher                                               |                                                                                                                                               | 44.00%      |  Medium     |            |
-| 0320 |  Generalized Abbreviation                                    |                                                                                                                                               | 48.90%      |  Medium     |            |
-| 0321 |  Create Maximum Number                                       |                                                                                                                                               | 25.50%      |  Hard       |            |
-| 0322 |  Coin Change                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0322.%20Coin%20Change)                                                          | 30.80%      |  Medium     |            |
-| 0323 |  Number of Connected Components in an Undirected Graph       |                                                                                                                                               | 52.30%      |  Medium     |            |
-| 0324 |  Wiggle Sort II                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0324.%20Wiggle%20Sort%20II)                                                     | 28.10%      |  Medium     |            |
-| 0325 |  Maximum Size Subarray Sum Equals k                          |                                                                                                                                               | 44.80%      |  Medium     |            |
-| 0326 |  Power of Three                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0326.%20Power%20of%20Three)                                                     | 41.70%      |  Easy       |            |
-| 0327 |  Count of Range Sum                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0327.%20Count%20of%20Range%20Sum)                                                                                                                                              | 32.90%      |  Hard       |            |
-| 0328 |  Odd Even Linked List                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0328.%20Odd%20Even%20Linked%20List)                                             | 49.50%      |  Medium     |            |
-| 0329 |  Longest Increasing Path in a Matrix                         |                                                                                                                                               | 40.30%      |  Hard       |            |
-| 0330 |  Patching Array                                              |                                                                                                                                               | 33.40%      |  Hard       |            |
-| 0331 |  Verify Preorder Serialization of a Binary Tree              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0331.%20Verify%20Preorder%20Serialization%20of%20a%20Binary%20Tree)             | 38.80%      |  Medium     |            |
-| 0332 |  Reconstruct Itinerary                                       |                                                                                                                                               | 31.70%      |  Medium     |            |
-| 0333 |  Largest BST Subtree                                         |                                                                                                                                               | 33.20%      |  Medium     |            |
-| 0334 |  Increasing Triplet Subsequence                              |                                                                                                                                               | 39.60%      |  Medium     |            |
-| 0335 |  Self Crossing                                               |                                                                                                                                               | 27.00%      |  Hard       |            |
-| 0336 |  Palindrome Pairs                                            |                                                                                                                                               | 31.10%      |  Hard       |            |
-| 0337 |  House Robber III                                            |                                                                                                                                               | 48.30%      |  Medium     |            |
-| 0338 |  Counting Bits                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0338.%20Counting%20Bits)                                                        | 64.90%      |  Medium     |            |
-| 0339 |  Nested List Weight Sum                                      |                                                                                                                                               | 68.50%      |  Easy       |            |
-| 0340 |  Longest Substring with At Most K Distinct Characters        |                                                                                                                                               | 40.00%      |  Hard       |            |
-| 0341 |  Flatten Nested List Iterator                                |                                                                                                                                               | 48.20%      |  Medium     |            |
-| 0342 |  Power of Four                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0342.%20Power%20of%20Four)                                                      | 40.40%      |  Easy       |            |
-| 0343 |  Integer Break                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0343.%20Integer%20Break)                                                        | 48.00%      |  Medium     |            |
-| 0344 |  Reverse String                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0344.%20Reverse%20String)                                                       | 63.50%      |  Easy       |            |
-| 0345 |  Reverse Vowels of a String                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0345.%20Reverse%20Vowels%20of%20a%20String)                                     | 41.70%      |  Easy       |            |
-| 0346 |  Moving Average from Data Stream                             |                                                                                                                                               | 66.50%      |  Easy       |            |
-| 0347 |  Top K Frequent Elements                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0347.%20Top%20K%20Frequent%20Elements)                                          | 55.30%      |  Medium     |            |
-| 0348 |  Design Tic-Tac-Toe                                          |                                                                                                                                               | 50.00%      |  Medium     |            |
-| 0349 |  Intersection of Two Arrays                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0349.%20Intersection%20of%20Two%20Arrays)                                       | 55.10%      |  Easy       |            |
-| 0350 |  Intersection of Two Arrays II                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0350.%20Intersection%20of%20Two%20Arrays%20II)                                  | 48.10%      |  Easy       |            |
-| 0351 |  Android Unlock Patterns                                     |                                                                                                                                               | 46.00%      |  Medium     |            |
-| 0352 |  Data Stream as Disjoint Intervals                           |                                                                                                                                               | 43.70%      |  Hard       |            |
-| 0353 |  Design Snake Game                                           |                                                                                                                                               | 30.80%      |  Medium     |            |
-| 0354 |  Russian Doll Envelopes                                      |                                                                                                                                               | 34.10%      |  Hard       |            |
-| 0355 |  Design Twitter                                              |                                                                                                                                               | 27.50%      |  Medium     |            |
-| 0356 |  Line Reflection                                             |                                                                                                                                               | 30.90%      |  Medium     |            |
-| 0357 |  Count Numbers with Unique Digits                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0357.%20Count%20Numbers%20with%20Unique%20Digits)                               | 47.10%      |  Medium     |            |
-| 0358 |  Rearrange String k Distance Apart                           |                                                                                                                                               | 32.90%      |  Hard       |            |
-| 0359 |  Logger Rate Limiter                                         |                                                                                                                                               | 65.40%      |  Easy       |            |
-| 0360 |  Sort Transformed Array                                      |                                                                                                                                               | 46.90%      |  Medium     |            |
-| 0361 |  Bomb Enemy                                                  |                                                                                                                                               | 43.60%      |  Medium     |            |
-| 0362 |  Design Hit Counter                                          |                                                                                                                                               | 59.20%      |  Medium     |            |
-| 0363 |  Max Sum of Rectangle No Larger Than K                       |                                                                                                                                               | 35.30%      |  Hard       |            |
-| 0364 |  Nested List Weight Sum II                                   |                                                                                                                                               | 58.00%      |  Medium     |            |
-| 0365 |  Water and Jug Problem                                       |                                                                                                                                               | 29.00%      |  Medium     |            |
-| 0366 |  Find Leaves of Binary Tree                                  |                                                                                                                                               | 66.00%      |  Medium     |            |
-| 0367 |  Valid Perfect Square                                        |                                                                                                                                               | 40.00%      |  Easy       |            |
-| 0368 |  Largest Divisible Subset                                    |                                                                                                                                               | 34.90%      |  Medium     |            |
-| 0369 |  Plus One Linked List                                        |                                                                                                                                               | 56.40%      |  Medium     |            |
-| 0370 |  Range Addition                                              |                                                                                                                                               | 60.60%      |  Medium     |            |
-| 0371 |  Sum of Two Integers                                         |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0371.%20Sum%20of%20Two%20Integers)                                                                                                                                               | 50.90%      |  Easy       |            |
-| 0372 |  Super Pow                                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0372.%20Super%20Pow)                                                                                                                                              | 35.70%      |  Medium     |            |
-| 0373 |  Find K Pairs with Smallest Sums                             |                                                                                                                                               | 34.00%      |  Medium     |            |
-| 0374 |  Guess Number Higher or Lower                                |                                                                                                                                               | 39.60%      |  Easy       |            |
-| 0375 |  Guess Number Higher or Lower II                             |                                                                                                                                               | 37.80%      |  Medium     |            |
-| 0376 |  Wiggle Subsequence                                          |                                                                                                                                               | 37.60%      |  Medium     |            |
-| 0377 |  Combination Sum IV                                          |                                                                                                                                               | 43.70%      |  Medium     |            |
-| 0378 |  Kth Smallest Element in a Sorted Matrix                     |                                                                                                                                               | 49.70%      |  Medium     |            |
-| 0379 |  Design Phone Directory                                      |                                                                                                                                               | 41.70%      |  Medium     |            |
-| 0380 |  Insert Delete GetRandom O(1)                                |                                                                                                                                               | 43.00%      |  Medium     |            |
-| 0381 |  Insert Delete GetRandom O(1) - Duplicates allowed           |                                                                                                                                               | 32.10%      |  Hard       |            |
-| 0382 |  Linked List Random Node                                     |                                                                                                                                               | 49.40%      |  Medium     |            |
-| 0383 |  Ransom Note                                                 |                                                                                                                                               | 50.10%      |  Easy       |            |
-| 0384 |  Shuffle an Array                                            |                                                                                                                                               | 50.30%      |  Medium     |            |
-| 0385 |  Mini Parser                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0385.%20Mini%20Parser)                                                                                                                                              | 31.90%      |  Medium     |            |
-| 0386 |  Lexicographical Numbers                                     |                                                                                                                                               | 46.50%      |  Medium     |            |
-| 0387 |  First Unique Character in a String                          |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0387.%20First%20Unique%20Character%20in%20a%20String)                                                                                                                                               | 50.20%      |  Easy       |            |
-| 0388 |  Longest Absolute File Path                                  |                                                                                                                                               | 39.30%      |  Medium     |            |
-| 0389 |  Find the Difference                                         |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0389.%20Find%20the%20Difference)                                                                                                                                               | 53.20%      |  Easy       |            |
-| 0390 |  Elimination Game                                            |                                                                                                                                               | 43.40%      |  Medium     |            |
-| 0391 |  Perfect Rectangle                                           |                                                                                                                                               | 28.30%      |  Hard       |            |
-| 0392 |  Is Subsequence                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0392.%20Is%20Subsequence)                                                       | 46.90%      |  Medium     |            |
-| 0393 |  UTF-8 Validation                                            |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0393.%20UTF-8%20Validation)                                                                                                                                               | 36.00%      |  Medium     |            |
-| 0394 |  Decode String                                               |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0394.%20Decode%20String)                                                        | 45.20%      |  Medium     |            |
-| 0395 |  Longest Substring with At Least K Repeating Characters      |                                                                                                                                               | 38.80%      |  Medium     |            |
-| 0396 |  Rotate Function                                             |                                                                                                                                               | 35.20%      |  Medium     |            |
-| 0397 |  Integer Replacement                                         |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0397.%20Integer%20Replacement)                                                                                                                                               | 31.50%      |  Medium     |            |
-| 0398 |  Random Pick Index                                           |                                                                                                                                               | 50.20%      |  Medium     |            |
-| 0399 |  Evaluate Division                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0399.%20Evaluate%20Division)                                                                                                                                              | 47.80%      |  Medium     |            |
-| 0400 |  Nth Digit                                                   |                                                                                                                                               | 30.40%      |  Easy       |            |
-| 0401 |  Binary Watch                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0401.%20Binary%20Watch)                                                         | 45.40%      |  Easy       |            |
-| 0402 |  Remove K Digits                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0402.%20Remove%20K%20Digits)                                                    | 26.70%      |  Medium     |            |
-| 0403 |  Frog Jump                                                   |                                                                                                                                               | 36.40%      |  Hard       |            |
-| 0404 |  Sum of Left Leaves                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0404.%20Sum%20of%20Left%20Leaves)                                               | 49.20%      |  Easy       |            |
-| 0405 |  Convert a Number to Hexadecimal                             |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0405.%20Convert%20a%20Number%20to%20Hexadecimal)                                                                                                                                               | 42.00%      |  Easy       |            |
-| 0406 |  Queue Reconstruction by Height                              |                                                                                                                                               | 60.10%      |  Medium     |            |
-| 0407 |  Trapping Rain Water II                                      |                                                                                                                                               | 39.40%      |  Hard       |            |
-| 0408 |  Valid Word Abbreviation                                     |                                                                                                                                               | 29.60%      |  Easy       |            |
-| 0409 |  Longest Palindrome                                          |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0409.%20Longest%20Palindrome)                                                                                                                                               | 48.20%      |  Easy       |            |
-| 0410 |  Split Array Largest Sum                                     |                                                                                                                                               | 42.60%      |  Hard       |            |
-| 0411 |  Minimum Unique Word Abbreviation                            |                                                                                                                                               | 35.10%      |  Hard       |            |
-| 0412 |  Fizz Buzz                                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0412.%20Fizz%20Buzz)                                                              | 59.70%      |  Easy       |            |
-| 0413 |  Arithmetic Slices                                           |                                                                                                                                               | 56.00%      |  Medium     |            |
-| 0414 |  Third Maximum Number                                        |
-[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0414.%20Third%20Maximum%20Number)                                                                                                                                               | 29.10%      |  Easy       |            |
-| 0415 |  Add Strings                                                 |                                                                                                                                               | 43.90%      |  Easy       |            |
-| 0416 |  Partition Equal Subset Sum                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0416.%20Partition%20Equal%20Subset%20Sum)                                       | 40.90%      |  Medium     |            |
-| 0417 |  Pacific Atlantic Water Flow                                 |                                                                                                                                               | 37.60%      |  Medium     |            |
-| 0418 |  Sentence Screen Fitting                                     |                                                                                                                                               | 31.20%      |  Medium     |            |
-| 0419 |  Battleships in a Board                                      |                                                                                                                                               | 65.90%      |  Medium     |            |
-| 0420 |  Strong Password Checker                                     |                                                                                                                                               | 17.20%      |  Hard       |            |
-| 0421 |  Maximum XOR of Two Numbers in an Array                      |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0421.%20Maximum%20XOR%20of%20Two%20Numbers%20in%20an%20Array)                                                                                                                                               | 51.20%      |  Medium     |            |
-| 0422 |  Valid Word Square                                           |                                                                                                                                               | 36.60%      |  Easy       |            |
-| 0423 |  Reconstruct Original Digits from English                    |                                                                                                                                               | 45.70%      |  Medium     |            |
-| 0424 |  Longest Repeating Character Replacement                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0424.%20Longest%20Repeating%20Character%20Replacement)                          | 44.30%      |  Medium     |            |
-| 0425 |  Word Squares                                                |                                                                                                                                               | 44.50%      |  Hard       |            |
-| 0426 |  Convert Binary Search Tree to Sorted Doubly Linked List     |                                                                                                                                               | 52.50%      |  Medium     |            |
-| 0427 |  Construct Quad Tree                                         |                                                                                                                                               | 57.20%      |  Medium     |            |
-| 0428 |  Serialize and Deserialize N-ary Tree                        |                                                                                                                                               | 54.20%      |  Hard       |            |
-| 0429 |  N-ary Tree Level Order Traversal                            |                                                                                                                                               | 59.80%      |  Easy       |            |
-| 0430 |  Flatten a Multilevel Doubly Linked List                     |                                                                                                                                               | 42.60%      |  Medium     |            |
-| 0431 |  Encode N-ary Tree to Binary Tree                            |                                                                                                                                               | 64.20%      |  Hard       |            |
-| 0432 |  All O`one Data Structure                                    |                                                                                                                                               | 29.70%      |  Hard       |            |
-| 0433 |  Minimum Genetic Mutation                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0433.%20Minimum%20Genetic%20Mutation)                                           | 38.40%      |  Medium     |            |
-| 0434 |  Number of Segments in a String                              |                                                                                                                                               | 37.00%      |  Easy       |            |
-| 0435 |  Non-overlapping Intervals                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0435.%20Non-overlapping%20Intervals)                                            | 41.60%      |  Medium     |            |
-| 0436 |  Find Right Interval                                         |                                                                                                                                               | 42.90%      |  Medium     |            |
-| 0437 |  Path Sum III                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0437.%20Path%20Sum%20III)                                                       | 43.00%      |  Easy       |            |
-| 0438 |  Find All Anagrams in a String                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0438.%20Find%20All%20Anagrams%20in%20a%20String)                                | 37.50%      |  Easy       |            |
-| 0439 |  Ternary Expression Parser                                   |                                                                                                                                               | 53.70%      |  Medium     |            |
-| 0440 |  K-th Smallest in Lexicographical Order                      |                                                                                                                                               | 26.60%      |  Hard       |            |
-| 0441 |  Arranging Coins                                             |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0441.%20Arranging%20Coins)                                                                                                                                               | 38.10%      |  Easy       |            |
-| 0442 |  Find All Duplicates in an Array                             |                                                                                                                                               | 61.30%      |  Medium     |            |
-| 0443 |  String Compression                                          |                                                                                                                                               | 37.90%      |  Easy       |            |
-| 0444 |  Sequence Reconstruction                                     |                                                                                                                                               | 20.50%      |  Medium     |            |
-| 0445 |  Add Two Numbers II                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0445.%20Add%20Two%20Numbers%20II)                                               | 50.40%      |  Medium     |            |
-| 0446 |  Arithmetic Slices II - Subsequence                          |                                                                                                                                               | 30.30%      |  Hard       |            |
-| 0447 |  Number of Boomerangs                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0447.%20Number%20of%20Boomerangs)                                               | 50.00%      |  Easy       |            |
-| 0448 |  Find All Numbers Disappeared in an Array                    |                                                                                                                                               | 53.60%      |  Easy       |            |
-| 0449 |  Serialize and Deserialize BST                               |                                                                                                                                               | 47.50%      |  Medium     |            |
-| 0450 |  Delete Node in a BST                                        |                                                                                                                                               | 40.20%      |  Medium     |            |
-| 0451 |  Sort Characters By Frequency                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0451.%20Sort%20Characters%20By%20Frequency)                                     | 56.40%      |  Medium     |            |
-| 0452 |  Minimum Number of Arrows to Burst Balloons                  |                                                                                                                                               | 46.60%      |  Medium     |            |
-| 0453 |  Minimum Moves to Equal Array Elements                       |                                                                                                                                               | 49.30%      |  Easy       |            |
-| 0454 |  4Sum II                                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0454.%204Sum%20II)                                                              | 50.80%      |  Medium     |            |
-| 0455 |  Assign Cookies                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0455.%20Assign%20Cookies)                                                       | 48.50%      |  Easy       |            |
-| 0456 |  132 Pattern                                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0456.%20132%20Pattern)                                                          | 27.40%      |  Medium     |            |
-| 0457 |  Circular Array Loop                                         |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0457.%20Circular%20Array%20Loop)                                                                                                                                               | 27.70%      |  Medium     |            |
-| 0458 |  Poor Pigs                                                   |                                                                                                                                               | 45.50%      |  Hard       |            |
-| 0459 |  Repeated Substring Pattern                                  |                                                                                                                                               | 40.00%      |  Easy       |            |
-| 0460 |  LFU Cache                                                   |                                                                                                                                               | 29.00%      |  Hard       |            |
-| 0461 |  Hamming Distance                                            |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0461.%20Hamming%20Distance)                                                                                                                                               | 70.40%      |  Easy       |            |
-| 0462 |  Minimum Moves to Equal Array Elements II                    |                                                                                                                                               | 52.50%      |  Medium     |            |
-| 0463 |  Island Perimeter                                            |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0463.%20Island%20Perimeter)                                                                                                                                               | 61.20%      |  Easy       |            |
-| 0464 |  Can I Win                                                   |                                                                                                                                               | 27.30%      |  Medium     |            |
-| 0465 |  Optimal Account Balancing                                   |                                                                                                                                               | 42.90%      |  Hard       |            |
-| 0466 |  Count The Repetitions                                       |                                                                                                                                               | 27.30%      |  Hard       |            |
-| 0467 |  Unique Substrings in Wraparound String                      |                                                                                                                                               | 34.10%      |  Medium     |            |
-| 0468 |  Validate IP Address                                         |                                                                                                                                               | 21.40%      |  Medium     |            |
-| 0469 |  Convex Polygon                                              |                                                                                                                                               | 35.50%      |  Medium     |            |
-| 0470 |  Implement Rand10() Using Rand7()                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0470.%20Implement%20Rand10()%20Using%20Rand7())                                                                                                                                               | 45.10%      |  Medium     |            |
-| 0471 |  Encode String with Shortest Length                          |                                                                                                                                               | 45.20%      |  Hard       |            |
-| 0472 |  Concatenated Words                                          |                                                                                                                                               | 35.10%      |  Hard       |            |
-| 0473 |  Matchsticks to Square                                       |                                                                                                                                               | 36.00%      |  Medium     |            |
-| 0474 |  Ones and Zeroes                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0474.%20Ones%20and%20Zeroes)                                                                                                                                              | 39.80%      |  Medium     |            |
-| 0475 |  Heaters                                                     |                                                                                                                                               | 31.90%      |  Easy       |            |
-| 0476 |  Number Complement                                           |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0476.%20Number%20Complement)                                                                                                                                               | 62.50%      |  Easy       |            |
-| 0477 |  Total Hamming Distance                                      |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0477.%20Total%20Hamming%20Distance)                                                                                                                                               | 49.00%      |  Medium     |            |
-| 0478 |  Generate Random Point in a Circle                           |                                                                                                                                               | 36.90%      |  Medium     |            |
-| 0479 |  Largest Palindrome Product                                  |                                                                                                                                               | 27.50%      |  Hard       |            |
-| 0480 |  Sliding Window Median                                       | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0480.%20Sliding%20Window%20Median)                                              | 32.80%      |  Hard       |            |
-| 0481 |  Magical String                                              |                                                                                                                                               | 46.30%      |  Medium     |            |
-| 0482 |  License Key Formatting                                      |                                                                                                                                               | 41.10%      |  Easy       |            |
-| 0483 |  Smallest Good Base                                          |                                                                                                                                               | 34.20%      |  Hard       |            |
-| 0484 |  Find Permutation                                            |                                                                                                                                               | 57.80%      |  Medium     |            |
-| 0485 |  Max Consecutive Ones                                        |                                                                                                                                               | 55.10%      |  Easy       |            |
-| 0486 |  Predict the Winner                                          |                                                                                                                                               | 46.80%      |  Medium     |            |
-| 0487 |  Max Consecutive Ones II                                     |                                                                                                                                               | 46.70%      |  Medium     |            |
-| 0488 |  Zuma Game                                                   |                                                                                                                                               | 39.10%      |  Hard       |            |
-| 0489 |  Robot Room Cleaner                                          |                                                                                                                                               | 64.70%      |  Hard       |            |
-| 0490 |  The Maze                                                    |                                                                                                                                               | 47.60%      |  Medium     |            |
-| 0491 |  Increasing Subsequences                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0491.%20Increasing%20Subsequences)                                                                                                                                              | 42.20%      |  Medium     |            |
-| 0492 |  Construct the Rectangle                                     |                                                                                                                                               | 48.80%      |  Easy       |            |
-| 0493 |  Reverse Pairs                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0493.%20Reverse%20Pairs)                                                                                                                                              | 23.30%      |  Hard       |            |
-| 0494 |  Target Sum                                                  |                                                                                                                                               | 45.40%      |  Medium     |            |
-| 0495 |  Teemo Attacking                                             |                                                                                                                                               | 52.30%      |  Medium     |            |
-| 0496 |  Next Greater Element I                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0496.%20Next%20Greater%20Element%20I)                                           | 59.80%      |  Easy       |            |
-| 0497 |  Random Point in Non-overlapping Rectangles                  |                                                                                                                                               | 35.90%      |  Medium     |            |
-| 0498 |  Diagonal Traverse                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0498.%20Diagonal%20Traverse)                                                                                                                                              | 45.40%      |  Medium     |            |
-| 0499 |  The Maze III                                                |                                                                                                                                               | 37.40%      |  Hard       |            |
-| 0500 |  Keyboard Row                                                |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0500.%20Keyboard%20Row)                                                                                                                                               | 62.40%      |  Easy       |            |
-| 0501 |  Find Mode in Binary Search Tree                             |                                                                                                                                               | 39.60%      |  Easy       |            |
-| 0502 |  IPO                                                         |                                                                                                                                               | 38.00%      |  Hard       |            |
-| 0503 |  Next Greater Element II                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0503.%20Next%20Greater%20Element%20II)                                          | 51.30%      |  Medium     |            |
-| 0504 |  Base 7                                                      |                                                                                                                                               | 45.00%      |  Easy       |            |
-| 0505 |  The Maze II                                                 |                                                                                                                                               | 44.10%      |  Medium     |            |
-| 0506 |  Relative Ranks                                              |                                                                                                                                               | 48.40%      |  Easy       |            |
-| 0507 |  Perfect Number                                              |                                                                                                                                               | 34.40%      |  Easy       |            |
-| 0508 |  Most Frequent Subtree Sum                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0508.%20Most%20Frequent%20Subtree%20Sum)                                                                                                                                              | 54.80%      |  Medium     |            |
-| 0509 |  Fibonacci Number                                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0509.%20Fibonacci%20Number)                                                                                                                                              | 66.70%      |  Easy       |            |
-| 0510 |  Inorder Successor in BST II                                 |                                                                                                                                               | 53.30%      |  Medium     |            |
-| 0511 |  Game Play Analysis I                                        |                                                                                                                                               | 74.40%      |  Easy       |            |
-| 0512 |  Game Play Analysis II                                       |                                                                                                                                               | 56.80%      |  Easy       |            |
-| 0513 |  Find Bottom Left Tree Value                                 |                                                                                                                                               | 58.60%      |  Medium     |            |
-| 0514 |  Freedom Trail                                               |                                                                                                                                               | 40.70%      |  Hard       |            |
-| 0515 |  Find Largest Value in Each Tree Row                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0515.%20Find%20Largest%20Value%20in%20Each%20Tree%20Row)                        | 58.00%      |  Medium     |            |
-| 0516 |  Longest Palindromic Subsequence                             |                                                                                                                                               | 47.10%      |  Medium     |            |
-| 0517 |  Super Washing Machines                                      |                                                                                                                                               | 37.00%      |  Hard       |            |
-| 0518 |  Coin Change 2                                               |                                                                                                                                               | 43.10%      |  Medium     |            |
-| 0519 |  Random Flip Matrix                                          |                                                                                                                                               | 33.00%      |  Medium     |            |
-| 0520 |  Detect Capital                                              |                                                                                                                                               | 52.60%      |  Easy       |            |
-| 0521 |  Longest Uncommon Subsequence I                              |                                                                                                                                               | 56.50%      |  Easy       |            |
-| 0522 |  Longest Uncommon Subsequence II                             |                                                                                                                                               | 32.90%      |  Medium     |            |
-| 0523 |  Continuous Subarray Sum                                     |                                                                                                                                               | 24.20%      |  Medium     |            |
-| 0524 |  Longest Word in Dictionary through Deleting                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0524.%20Longest%20Word%20in%20Dictionary%20through%20Deleting)                  | 46.00%      |  Medium     |            |
-| 0525 |  Contiguous Array                                            |                                                                                                                                               | 42.80%      |  Medium     |            |
-| 0526 |  Beautiful Arrangement                                       | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0526.%20Beautiful%20Arrangement)                                                | 54.80%      |  Medium     |            |
-| 0527 |  Word Abbreviation                                           |                                                                                                                                               | 50.20%      |  Hard       |            |
-| 0528 |  Random Pick with Weight                                     |                                                                                                                                               | 42.80%      |  Medium     |            |
-| 0529 |  Minesweeper                                                 |                                                                                                                                               | 53.10%      |  Medium     |            |
-| 0530 |  Minimum Absolute Difference in BST                          |                                                                                                                                               | 50.70%      |  Easy       |            |
-| 0531 |  Lonely Pixel I                                              |                                                                                                                                               | 57.60%      |  Medium     |            |
-| 0532 |  K-diff Pairs in an Array                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0532.%20K-diff%20Pairs%20in%20an%20Array)                                       | 30.00%      |  Easy       |            |
-| 0533 |  Lonely Pixel II                                             |                                                                                                                                               | 46.20%      |  Medium     |            |
-| 0534 |  Game Play Analysis III                                      |                                                                                                                                               | 64.60%      |  Medium     |            |
-| 0535 |  Encode and Decode TinyURL                                   |                                                                                                                                               | 76.90%      |  Medium     |            |
-| 0536 |  Construct Binary Tree from String                           |                                                                                                                                               | 45.20%      |  Medium     |            |
-| 0537 |  Complex Number Multiplication                               |                                                                                                                                               | 65.70%      |  Medium     |            |
-| 0538 |  Convert BST to Greater Tree                                 |                                                                                                                                               | 51.30%      |  Easy       |            |
-| 0539 |  Minimum Time Difference                                     |                                                                                                                                               | 48.10%      |  Medium     |            |
-| 0540 |  Single Element in a Sorted Array                            |                                                                                                                                               | 57.40%      |  Medium     |            |
-| 0541 |  Reverse String II                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0541.%20Reverse%20String%20II)                                                                                                                                              | 45.70%      |  Easy       |            |
-| 0542 |  01 Matrix                                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0542.%2001%20Matrix)                                                            | 36.00%      |  Medium     |            |
-| 0543 |  Diameter of Binary Tree                                     |                                                                                                                                               | 46.90%      |  Easy       |            |
-| 0544 |  Output Contest Matches                                      |                                                                                                                                               | 73.50%      |  Medium     |            |
-| 0545 |  Boundary of Binary Tree                                     |                                                                                                                                               | 35.30%      |  Medium     |            |
-| 0546 |  Remove Boxes                                                |                                                                                                                                               | 38.50%      |  Hard       |            |
-| 0547 |  Friend Circles                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0547.%20Friend%20Circles)                                                                                                                                               | 53.90%      |  Medium     |            |
-| 0548 |  Split Array with Equal Sum                                  |                                                                                                                                               | 43.50%      |  Medium     |            |
-| 0549 |  Binary Tree Longest Consecutive Sequence II                 |                                                                                                                                               | 44.50%      |  Medium     |            |
-| 0550 |  Game Play Analysis IV                                       |                                                                                                                                               | 42.20%      |  Medium     |            |
-| 0551 |  Student Attendance Record I                                 |                                                                                                                                               | 45.30%      |  Easy       |            |
-| 0552 |  Student Attendance Record II                                |                                                                                                                                               | 33.30%      |  Hard       |            |
-| 0553 |  Optimal Division                                            |                                                                                                                                               | 55.50%      |  Medium     |            |
-| 0554 |  Brick Wall                                                  |                                                                                                                                               | 47.80%      |  Medium     |            |
-| 0555 |  Split Concatenated Strings                                  |                                                                                                                                               | 40.00%      |  Medium     |            |
-| 0556 |  Next Greater Element III                                    |                                                                                                                                               | 30.00%      |  Medium     |            |
-| 0557 |  Reverse Words in a String III                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0557.%20Reverse%20Words%20in%20a%20String%20III)                                                                                                                                              | 64.40%      |  Easy       |            |
-| 0558 |  Quad Tree Intersection                                      |                                                                                                                                               | 41.10%      |  Easy       |            |
-| 0559 |  Maximum Depth of N-ary Tree                                 |                                                                                                                                               | 65.60%      |  Easy       |            |
-| 0560 |  Subarray Sum Equals K                                       |                                                                                                                                               | 42.30%      |  Medium     |            |
-| 0561 |  Array Partition I                                           |                                                                                                                                               | 69.30%      |  Easy       |            |
-| 0562 |  Longest Line of Consecutive One in Matrix                   |                                                                                                                                               | 43.80%      |  Medium     |            |
-| 0563 |  Binary Tree Tilt                                            |                                                                                                                                               | 47.10%      |  Easy       |            |
-| 0564 |  Find the Closest Palindrome                                 |                                                                                                                                               | 18.90%      |  Hard       |            |
-| 0565 |  Array Nesting                                               |                                                                                                                                               | 52.80%      |  Medium     |            |
-| 0566 |  Reshape the Matrix                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0566.%20Reshape%20the%20Matrix)                                                 | 59.00%      |  Easy       |            |
-| 0567 |  Permutation in String                                       | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0567.%20Permutation%20in%20String)                                              | 38.60%      |  Medium     |            |
-| 0568 |  Maximum Vacation Days                                       |                                                                                                                                               | 38.30%      |  Hard       |            |
-| 0569 |  Median Employee Salary                                      |                                                                                                                                               | 47.60%      |  Hard       |            |
-| 0570 |  Managers with at Least 5 Direct Reports                     |                                                                                                                                               | 62.90%      |  Medium     |            |
-| 0571 |  Find Median Given Frequency of Numbers                      |                                                                                                                                               | 46.80%      |  Hard       |            |
-| 0572 |  Subtree of Another Tree                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0572.%20Subtree%20of%20Another%20Tree)                                                                                                                                              | 41.80%      |  Easy       |            |
-| 0573 |  Squirrel Simulation                                         |                                                                                                                                               | 53.80%      |  Medium     |            |
-| 0574 |  Winning Candidate                                           |                                                                                                                                               | 36.80%      |  Medium     |            |
-| 0575 |  Distribute Candies                                          |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0575.%20Distribute%20Candies)                                                                                                                                               | 59.80%      |  Easy       |            |
-| 0576 |  Out of Boundary Paths                                       |                                                                                                                                               | 32.30%      |  Medium     |            |
-| 0577 |  Employee Bonus                                              |                                                                                                                                               | 59.90%      |  Easy       |            |
-| 0578 |  Get Highest Answer Rate Question                            |                                                                                                                                               | 35.70%      |  Medium     |            |
-| 0579 |  Find Cumulative Salary of an Employee                       |                                                                                                                                               | 34.30%      |  Hard       |            |
-| 0580 |  Count Student Number in Departments                         |                                                                                                                                               | 42.90%      |  Medium     |            |
-| 0581 |  Shortest Unsorted Continuous Subarray                       |                                                                                                                                               | 30.20%      |  Easy       |            |
-| 0582 |  Kill Process                                                |                                                                                                                                               | 56.50%      |  Medium     |            |
-| 0583 |  Delete Operation for Two Strings                            |                                                                                                                                               | 45.10%      |  Medium     |            |
-| 0584 |  Find Customer Referee                                       |                                                                                                                                               | 67.90%      |  Easy       |            |
-| 0585 |  Investments in 2016                                         |                                                                                                                                               | 47.70%      |  Medium     |            |
-| 0586 |  Customer Placing the Largest Number of Orders               |                                                                                                                                               | 66.90%      |  Easy       |            |
-| 0587 |  Erect the Fence                                             |                                                                                                                                               | 34.40%      |  Hard       |            |
-| 0588 |  Design In-Memory File System                                |                                                                                                                                               | 40.00%      |  Hard       |            |
-| 0589 |  N-ary Tree Preorder Traversal                               |                                                                                                                                               | 67.80%      |  Easy       |            |
-| 0590 |  N-ary Tree Postorder Traversal                              |                                                                                                                                               | 67.90%      |  Easy       |            |
-| 0591 |  Tag Validator                                               |                                                                                                                                               | 32.90%      |  Hard       |            |
-| 0592 |  Fraction Addition and Subtraction                           |                                                                                                                                               | 47.20%      |  Medium     |            |
-| 0593 |  Valid Square                                                |                                                                                                                                               | 40.60%      |  Medium     |            |
-| 0594 |  Longest Harmonious Subsequence                              |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0594.%20Longest%20Harmonious%20Subsequence)                                                                                                                                               | 44.00%      |  Easy       |            |
-| 0595 |  Big Countries                                               |                                                                                                                                               | 74.00%      |  Easy       |            |
-| 0596 |  Classes More Than 5 Students                                |                                                                                                                                               | 35.80%      |  Easy       |            |
-| 0597 |  Friend Requests I: Overall Acceptance Rate                  |                                                                                                                                               | 39.80%      |  Easy       |            |
-| 0598 |  Range Addition II                                           |                                                                                                                                               | 48.70%      |  Easy       |            |
-| 0599 |  Minimum Index Sum of Two Lists                              |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0599.%20Minimum%20Index%20Sum%20of%20Two%20Lists)                                                                                                                                               | 48.20%      |  Easy       |            |
-| 0600 |  Non-negative Integers without Consecutive Ones              |                                                                                                                                               | 32.80%      |  Hard       |            |
-| 0601 |  Human Traffic of Stadium                                    |                                                                                                                                               | 36.80%      |  Hard       |            |
-| 0602 |  Friend Requests II: Who Has the Most Friends                |                                                                                                                                               | 45.40%      |  Medium     |            |
-| 0603 |  Consecutive Available Seats                                 |                                                                                                                                               | 58.60%      |  Easy       |            |
-| 0604 |  Design Compressed String Iterator                           |                                                                                                                                               | 34.90%      |  Easy       |            |
-| 0605 |  Can Place Flowers                                           |                                                                                                                                               | 31.00%      |  Easy       |            |
-| 0606 |  Construct String from Binary Tree                           |                                                                                                                                               | 51.70%      |  Easy       |            |
-| 0607 |  Sales Person                                                |                                                                                                                                               | 57.30%      |  Easy       |            |
-| 0608 |  Tree Node                                                   |                                                                                                                                               | 59.10%      |  Medium     |            |
-| 0609 |  Find Duplicate File in System                               |                                                                                                                                               | 55.50%      |  Medium     |            |
-| 0610 |  Triangle Judgement                                          |                                                                                                                                               | 61.60%      |  Easy       |            |
-| 0611 |  Valid Triangle Number                                       |                                                                                                                                               | 45.40%      |  Medium     |            |
-| 0612 |  Shortest Distance in a Plane                                |                                                                                                                                               | 54.00%      |  Medium     |            |
-| 0613 |  Shortest Distance in a Line                                 |                                                                                                                                               | 73.20%      |  Easy       |            |
-| 0614 |  Second Degree Follower                                      |                                                                                                                                               | 22.60%      |  Medium     |            |
-| 0615 |  Average Salary: Departments VS Company                      |                                                                                                                                               | 38.00%      |  Hard       |            |
-| 0616 |  Add Bold Tag in String                                      |                                                                                                                                               | 39.40%      |  Medium     |            |
-| 0617 |  Merge Two Binary Trees                                      |                                                                                                                                               | 70.40%      |  Easy       |            |
-| 0618 |  Students Report By Geography                                |                                                                                                                                               | 43.10%      |  Hard       |            |
-| 0619 |  Biggest Single Number                                       |                                                                                                                                               | 39.10%      |  Easy       |            |
-| 0620 |  Not Boring Movies                                           |                                                                                                                                               | 62.50%      |  Easy       |            |
-| 0621 |  Task Scheduler                                              |                                                                                                                                               | 45.70%      |  Medium     |            |
-| 0622 |  Design Circular Queue                                       |                                                                                                                                               | 39.70%      |  Medium     |            |
-| 0623 |  Add One Row to Tree                                         |                                                                                                                                               | 47.40%      |  Medium     |            |
-| 0624 |  Maximum Distance in Arrays                                  |                                                                                                                                               | 37.40%      |  Easy       |            |
-| 0625 |  Minimum Factorization                                       |                                                                                                                                               | 32.00%      |  Medium     |            |
-| 0626 |  Exchange Seats                                              |                                                                                                                                               | 54.80%      |  Medium     |            |
-| 0627 |  Swap Salary                                                 |                                                                                                                                               | 69.20%      |  Easy       |            |
-| 0628 |  Maximum Product of Three Numbers                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0628.%20Maximum%20Product%20of%20Three%20Numbers)                               | 46.20%      |  Easy       |            |
-| 0629 |  K Inverse Pairs Array                                       |                                                                                                                                               | 29.30%      |  Hard       |            |
-| 0630 |  Course Schedule III                                         |                                                                                                                                               | 31.90%      |  Hard       |            |
-| 0631 |  Design Excel Sum Formula                                    |                                                                                                                                               | 29.30%      |  Hard       |            |
-| 0632 |  Smallest Range Covering Elements from K Lists                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0632.%20Smallest%20Range%20Covering%20Elements%20from%20K%20Lists)                                                                                                                                              | 47.90%      |  Hard       |            |
-| 0633 |  Sum of Square Numbers                                       |                                                                                                                                               | 32.70%      |  Easy       |            |
-| 0634 |  Find the Derangement of An Array                            |                                                                                                                                               | 37.80%      |  Medium     |            |
-| 0635 |  Design Log Storage System                                   |                                                                                                                                               | 54.40%      |  Medium     |            |
-| 0636 |  Exclusive Time of Functions                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0636.%20Exclusive%20Time%20of%20Functions)                                      | 48.80%      |  Medium     |            |
-| 0637 |  Average of Levels in Binary Tree                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0637.%20Average%20of%20Levels%20in%20Binary%20Tree)                             | 59.00%      |  Easy       |            |
-| 0638 |  Shopping Offers                                             |                                                                                                                                               | 48.90%      |  Medium     |            |
-| 0639 |  Decode Ways II                                              |                                                                                                                                               | 25.30%      |  Hard       |            |
-| 0640 |  Solve the Equation                                          |                                                                                                                                               | 40.40%      |  Medium     |            |
-| 0641 |  Design Circular Deque                                       |                                                                                                                                               | 49.80%      |  Medium     |            |
-| 0642 |  Design Search Autocomplete System                           |                                                                                                                                               | 38.20%      |  Hard       |            |
-| 0643 |  Maximum Average Subarray I                                  |                                                                                                                                               | 39.70%      |  Easy       |            |
-| 0644 |  Maximum Average Subarray II                                 |                                                                                                                                               | 28.70%      |  Hard       |            |
-| 0645 |  Set Mismatch                                                |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0645.%20Set%20Mismatch)                                                                                                                                               | 40.80%      |  Easy       |            |
-| 0646 |  Maximum Length of Pair Chain                                |                                                                                                                                               | 49.00%      |  Medium     |            |
-| 0647 |  Palindromic Substrings                                      |                                                                                                                                               | 57.20%      |  Medium     |            |
-| 0648 |  Replace Words                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0648.%20Replace%20Words)                                                        | 52.20%      |  Medium     |            |
-| 0649 |  Dota2 Senate                                                |                                                                                                                                               | 37.80%      |  Medium     |            |
-| 0650 |  2 Keys Keyboard                                             |                                                                                                                                               | 46.80%      |  Medium     |            |
-| 0651 |  4 Keys Keyboard                                             |                                                                                                                                               | 50.80%      |  Medium     |            |
-| 0652 |  Find Duplicate Subtrees                                     |                                                                                                                                               | 45.90%      |  Medium     |            |
-| 0653 |  Two Sum IV - Input is a BST                                 |                                                                                                                                               | 52.60%      |  Easy       |            |
-| 0654 |  Maximum Binary Tree                                         |                                                                                                                                               | 76.40%      |  Medium     |            |
-| 0655 |  Print Binary Tree                                           |                                                                                                                                               | 52.10%      |  Medium     |            |
-| 0656 |  Coin Path                                                   |                                                                                                                                               | 27.00%      |  Hard       |            |
-| 0657 |  Robot Return to Origin                                      |                                                                                                                                               | 71.70%      |  Easy       |            |
-| 0658 |  Find K Closest Elements                                     |                                                                                                                                               | 38.20%      |  Medium     |            |
-| 0659 |  Split Array into Consecutive Subsequences                   |                                                                                                                                               | 40.80%      |  Medium     |            |
-| 0660 |  Remove 9                                                    |                                                                                                                                               | 51.60%      |  Hard       |            |
-| 0661 |  Image Smoother                                              |                                                                                                                                               | 49.00%      |  Easy       |            |
-| 0662 |  Maximum Width of Binary Tree                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0662.%20Maximum%20Width%20of%20Binary%20Tree)                                                                                                                                              | 39.60%      |  Medium     |            |
-| 0663 |  Equal Tree Partition                                        |                                                                                                                                               | 38.10%      |  Medium     |            |
-| 0664 |  Strange Printer                                             |                                                                                                                                               | 36.90%      |  Hard       |            |
-| 0665 |  Non-decreasing Array                                        |                                                                                                                                               | 19.50%      |  Easy       |            |
-| 0666 |  Path Sum IV                                                 |                                                                                                                                               | 52.50%      |  Medium     |            |
-| 0667 |  Beautiful Arrangement II                                    |                                                                                                                                               | 52.10%      |  Medium     |            |
-| 0668 |  Kth Smallest Number in Multiplication Table                 |                                                                                                                                               | 42.20%      |  Hard       |            |
-| 0669 |  Trim a Binary Search Tree                                   |                                                                                                                                               | 60.50%      |  Easy       |            |
-| 0670 |  Maximum Swap                                                |                                                                                                                                               | 39.80%      |  Medium     |            |
-| 0671 |  Second Minimum Node In a Binary Tree                        |                                                                                                                                               | 43.30%      |  Easy       |            |
-| 0672 |  Bulb Switcher II                                            |                                                                                                                                               | 50.10%      |  Medium     |            |
-| 0673 |  Number of Longest Increasing Subsequence                    |                                                                                                                                               | 33.70%      |  Medium     |            |
-| 0674 |  Longest Continuous Increasing Subsequence                   |                                                                                                                                               | 44.40%      |  Easy       |            |
-| 0675 |  Cut Off Trees for Golf Event                                |                                                                                                                                               | 30.80%      |  Hard       |            |
-| 0676 |  Implement Magic Dictionary                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0676.%20Implement%20Magic%20Dictionary)                                         | 51.70%      |  Medium     |            |
-| 0677 |  Map Sum Pairs                                               |                                                                                                                                               | 51.80%      |  Medium     |            |
-| 0678 |  Valid Parenthesis String                                    |                                                                                                                                               | 32.90%      |  Medium     |            |
-| 0679 |  24 Game                                                     |                                                                                                                                               | 43.00%      |  Hard       |            |
-| 0680 |  Valid Palindrome II                                         |                                                                                                                                               | 34.30%      |  Easy       |            |
-| 0681 |  Next Closest Time                                           |                                                                                                                                               | 42.70%      |  Medium     |            |
-| 0682 |  Baseball Game                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0682.%20Baseball%20Game)                                                        | 61.20%      |  Easy       |            |
-| 0683 |  K Empty Slots                                               |                                                                                                                                               | 34.20%      |  Hard       |            |
-| 0684 |  Redundant Connection                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0684.%20Redundant%20Connection)                                                                                                                                              | 52.10%      |  Medium     |            |
-| 0685 |  Redundant Connection II                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0685.%20Redundant%20Connection%20II)                                                                                                                                              | 30.80%      |  Hard       |            |
-| 0686 |  Repeated String Match                                       |                                                                                                                                               | 31.50%      |  Easy       |            |
-| 0687 |  Longest Univalue Path                                       |                                                                                                                                               | 33.90%      |  Easy       |            |
-| 0688 |  Knight Probability in Chessboard                            |                                                                                                                                               | 44.60%      |  Medium     |            |
-| 0689 |  Maximum Sum of 3 Non-Overlapping Subarrays                  |                                                                                                                                               | 44.20%      |  Hard       |            |
-| 0690 |  Employee Importance                                         |                                                                                                                                               | 54.40%      |  Easy       |            |
-| 0691 |  Stickers to Spell Word                                      |                                                                                                                                               | 38.50%      |  Hard       |            |
-| 0692 |  Top K Frequent Words                                        |                                                                                                                                               | 46.00%      |  Medium     |            |
-| 0693 |  Binary Number with Alternating Bits                         |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0693.%20Binary%20Number%20with%20Alternating%20Bits)                                                                                                                                               | 58.00%      |  Easy       |            |
-| 0694 |  Number of Distinct Islands                                  |                                                                                                                                               | 51.40%      |  Medium     |            |
-| 0695 |  Max Area of Island                                          |                                                                                                                                               | 57.60%      |  Medium     |            |
-| 0696 |  Count Binary Substrings                                     |                                                                                                                                               | 53.60%      |  Easy       |            |
-| 0697 |  Degree of an Array                                          |                                                                                                                                               | 50.60%      |  Easy       |            |
-| 0698 |  Partition to K Equal Sum Subsets                            |                                                                                                                                               | 42.50%      |  Medium     |            |
-| 0699 |  Falling Squares                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0699.%20Falling%20Squares)                                                                                                                                              | 40.10%      |  Hard       |            |
-| 0700 |  Search in a Binary Search Tree                              |                                                                                                                                               | 68.60%      |  Easy       |            |
-| 0701 |  Insert into a Binary Search Tree                            |                                                                                                                                               | 76.10%      |  Medium     |            |
-| 0702 |  Search in a Sorted Array of Unknown Size                    |                                                                                                                                               | 59.30%      |  Medium     |            |
-| 0703 |  Kth Largest Element in a Stream                             |                                                                                                                                               | 46.60%      |  Easy       |            |
-| 0704 |  Binary Search                                               |                                                                                                                                               | 48.00%      |  Easy       |            |
-| 0705 |  Design HashSet                                              |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0705.%20Design%20HashSet)                                                                                                                                               | 53.90%      |  Easy       |            |
-| 0706 |  Design HashMap                                              |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0706.%20Design%20HashMap)                                                                                                                                               | 56.30%      |  Easy       |            |
-| 0707 |  Design Linked List                                          |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0707.%20Design%20Linked%20List)                                                                                                                                               | 21.60%      |  Easy       |            |
-| 0708 |  Insert into a Cyclic Sorted List                            |                                                                                                                                               | 29.20%      |  Medium     |            |
-| 0709 |  To Lower Case                                               |                                                                                                                                               | 77.00%      |  Easy       |            |
-| 0710 |  Random Pick with Blacklist                                  |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0710.%20Random%20Pick%20with%20Blacklist)                                                                                                                                                | 31.30%      |  Hard       |            |
-| 0711 |  Number of Distinct Islands II                               |                                                                                                                                               | 46.20%      |  Hard       |            |
-| 0712 |  Minimum ASCII Delete Sum for Two Strings                    |                                                                                                                                               | 54.80%      |  Medium     |            |
-| 0713 |  Subarray Product Less Than K                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0713.%20Subarray%20Product%20Less%20Than%20K)                                   | 36.80%      |  Medium     |            |
-| 0714 |  Best Time to Buy and Sell Stock with Transaction Fee        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0714.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Transaction%20Fee) | 50.60%      |  Medium     |            |
-| 0715 |  Range Module                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0715.%20Range%20Module)                                                                                                                                              | 35.70%      |  Hard       |            |
-| 0716 |  Max Stack                                                   |                                                                                                                                               | 40.20%      |  Easy       |            |
-| 0717 |  1-bit and 2-bit Characters                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0717.%201-bit%20and%202-bit%20Characters)                                                                                                                                               | 49.20%      |  Easy       |            |
-| 0718 |  Maximum Length of Repeated Subarray                         |                                                                                                                                               | 46.30%      |  Medium     |            |
-| 0719 |  Find K-th Smallest Pair Distance                            |                                                                                                                                               | 29.30%      |  Hard       |            |
-| 0720 |  Longest Word in Dictionary                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0720.%20Longest%20Word%20in%20Dictionary)                                       | 45.00%      |  Easy       |            |
-| 0721 |  Accounts Merge                                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0721.%20Accounts%20Merge)                                                                                                                                              | 40.80%      |  Medium     |            |
-| 0722 |  Remove Comments                                             |                                                                                                                                               | 31.40%      |  Medium     |            |
-| 0723 |  Candy Crush                                                 |                                                                                                                                               | 63.40%      |  Medium     |            |
-| 0724 |  Find Pivot Index                                            |                                                                                                                                               | 41.30%      |  Easy       |            |
-| 0725 |  Split Linked List in Parts                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0725.%20Split%20Linked%20List%20in%20Parts)                                     | 49.10%      |  Medium     |            |
-| 0726 |  Number of Atoms                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0726.%20Number%20of%20Atoms)                                                    | 44.90%      |  Hard       |            |
-| 0727 |  Minimum Window Subsequence                                  |                                                                                                                                               | 37.50%      |  Hard       |            |
-| 0728 |  Self Dividing Numbers                                       |                                                                                                                                               | 70.50%      |  Easy       |            |
-| 0729 |  My Calendar I                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0729.%20My%20Calendar%20I)                                                                                                                                              | 47.60%      |  Medium     |            |
-| 0730 |  Count Different Palindromic Subsequences                    |                                                                                                                                               | 39.20%      |  Hard       |            |
-| 0731 |  My Calendar II                                              |                                                                                                                                               | 44.80%      |  Medium     |            |
-| 0732 |  My Calendar III                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0732.%20My%20Calendar%20III)                                                                                                                                              | 55.20%      |  Hard       |            |
-| 0733 |  Flood Fill                                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0733.%20Flood%20Fill)                                                                                                                                              | 51.10%      |  Easy       |            |
-| 0734 |  Sentence Similarity                                         |                                                                                                                                               | 40.80%      |  Easy       |            |
-| 0735 |  Asteroid Collision                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0735.%20Asteroid%20Collision)                                                   | 38.60%      |  Medium     |            |
-| 0736 |  Parse Lisp Expression                                       |                                                                                                                                               | 44.30%      |  Hard       |            |
-| 0737 |  Sentence Similarity II                                      |                                                                                                                                               | 43.60%      |  Medium     |            |
-| 0738 |  Monotone Increasing Digits                                  |                                                                                                                                               | 42.00%      |  Medium     |            |
-| 0739 |  Daily Temperatures                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0739.%20Daily%20Temperatures)                                                   | 60.10%      |  Medium     |            |
-| 0740 |  Delete and Earn                                             |                                                                                                                                               | 46.10%      |  Medium     |            |
-| 0741 |  Cherry Pickup                                               |                                                                                                                                               | 30.10%      |  Hard       |            |
-| 0742 |  Closest Leaf in a Binary Tree                               |                                                                                                                                               | 39.40%      |  Medium     |            |
-| 0743 |  Network Delay Time                                          |                                                                                                                                               | 42.40%      |  Medium     |            |
-| 0744 |  Find Smallest Letter Greater Than Target                    |                                                                                                                                               | 44.10%      |  Easy       |            |
-| 0745 |  Prefix and Suffix Search                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0745.%20Prefix%20and%20Suffix%20Search)                                         | 30.90%      |  Hard       |            |
-| 0746 |  Min Cost Climbing Stairs                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0746.%20Min%20Cost%20Climbing%20Stairs)                                         | 47.50%      |  Easy       |            |
-| 0747 |  Largest Number At Least Twice of Others                     |                                                                                                                                               | 40.70%      |  Easy       |            |
-| 0748 |  Shortest Completing Word                                    |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0748.%20Shortest%20Completing%20Word)                                                                                                                                               | 54.50%      |  Easy       |            |
-| 0749 |  Contain Virus                                               |                                                                                                                                               | 41.40%      |  Hard       |            |
-| 0750 |  Number Of Corner Rectangles                                 |                                                                                                                                               | 64.80%      |  Medium     |            |
-| 0751 |  IP to CIDR                                                  |                                                                                                                                               | 59.70%      |  Easy       |            |
-| 0752 |  Open the Lock                                               |                                                                                                                                               | 46.30%      |  Medium     |            |
-| 0753 |  Cracking the Safe                                           |                                                                                                                                               | 46.60%      |  Hard       |            |
-| 0754 |  Reach a Number                                              |                                                                                                                                               | 32.80%      |  Easy       |            |
-| 0755 |  Pour Water                                                  |                                                                                                                                               | 40.80%      |  Medium     |            |
-| 0756 |  Pyramid Transition Matrix                                   |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0756.%20Pyramid%20Transition%20Matrix)                                                                                                                                               | 51.80%      |  Medium     |            |
-| 0757 |  Set Intersection Size At Least Two                          |                                                                                                                                               | 37.00%      |  Hard       |            |
-| 0758 |  Bold Words in String                                        |                                                                                                                                               | 42.70%      |  Easy       |            |
-| 0759 |  Employee Free Time                                          |                                                                                                                                               | 61.50%      |  Hard       |            |
-| 0760 |  Find Anagram Mappings                                       |                                                                                                                                               | 79.40%      |  Easy       |            |
-| 0761 |  Special Binary String                                       |                                                                                                                                               | 52.00%      |  Hard       |            |
-| 0762 |  Prime Number of Set Bits in Binary Representation           |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0762.%20Prime%20Number%20of%20Set%20Bits%20in%20Binary%20Representation)                                                                                                                                               | 59.60%      |  Easy       |            |
-| 0763 |  Partition Labels                                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0763.%20Partition%20Labels)                                                     | 70.80%      |  Medium     |            |
-| 0764 |  Largest Plus Sign                                           |                                                                                                                                               | 43.50%      |  Medium     |            |
-| 0765 |  Couples Holding Hands                                       | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0765.%20Couples%20Holding%20Hands)                                                                                                                                              | 51.70%      |  Hard       |            |
-| 0766 |  Toeplitz Matrix                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0766.%20Toeplitz%20Matrix)                                                      | 61.90%      |  Easy       |            |
-| 0767 |  Reorganize String                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0767.%20Reorganize%20String)                                                    | 42.70%      |  Medium     |            |
-| 0768 |  Max Chunks To Make Sorted II                                |                                                                                                                                               | 46.10%      |  Hard       |            |
-| 0769 |  Max Chunks To Make Sorted                                   |                                                                                                                                               | 51.90%      |  Medium     |            |
-| 0770 |  Basic Calculator IV                                         |                                                                                                                                               | 45.70%      |  Hard       |            |
-| 0771 |  Jewels and Stones                                           |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0771.%20Jewels%20and%20Stones)                                                                                                                                               | 83.20%      |  Easy       |            |
-| 0772 |  Basic Calculator III                                        |                                                                                                                                               | 43.40%      |  Hard       |            |
-| 0773 |  Sliding Puzzle                                              |                                                                                                                                               | 52.90%      |  Hard       |            |
-| 0774 |  Minimize Max Distance to Gas Station                        |                                                                                                                                               | 42.00%      |  Hard       |            |
-| 0775 |  Global and Local Inversions                                 |                                                                                                                                               | 39.10%      |  Medium     |            |
-| 0776 |  Split BST                                                   |                                                                                                                                               | 52.70%      |  Medium     |            |
-| 0777 |  Swap Adjacent in LR String                                  |                                                                                                                                               | 33.40%      |  Medium     |            |
-| 0778 |  Swim in Rising Water                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0778.%20Swim%20in%20Rising%20Water)                                                                                                                                              | 47.80%      |  Hard       |            |
-| 0779 |  K-th Symbol in Grammar                                      |                                                                                                                                               | 37.50%      |  Medium     |            |
-| 0780 |  Reaching Points                                             |                                                                                                                                               | 27.60%      |  Hard       |            |
-| 0781 |  Rabbits in Forest                                           |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0781.%20Rabbits%20in%20Forest)                                                                                                                                               | 51.80%      |  Medium     |            |
-| 0782 |  Transform to Chessboard                                     |                                                                                                                                               | 39.80%      |  Hard       |            |
-| 0783 |  Minimum Distance Between BST Nodes                          |                                                                                                                                               | 50.60%      |  Easy       |            |
-| 0784 |  Letter Case Permutation                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0784.%20Letter%20Case%20Permutation)                                            | 57.40%      |  Easy       |            |
-| 0785 |  Is Graph Bipartite?                                         |                                                                                                                                               | 43.50%      |  Medium     |            |
-| 0786 |  K-th Smallest Prime Fraction                                |                                                                                                                                               | 39.90%      |  Hard       |            |
-| 0787 |  Cheapest Flights Within K Stops                             |                                                                                                                                               | 35.20%      |  Medium     |            |
-| 0788 |  Rotated Digits                                              |                                                                                                                                               | 54.40%      |  Easy       |            |
-| 0789 |  Escape The Ghosts                                           |                                                                                                                                               | 55.40%      |  Medium     |            |
-| 0790 |  Domino and Tromino Tiling                                   |                                                                                                                                               | 36.00%      |  Medium     |            |
-| 0791 |  Custom Sort String                                          |                                                                                                                                               | 62.30%      |  Medium     |            |
-| 0792 |  Number of Matching Subsequences                             |                                                                                                                                               | 43.40%      |  Medium     |            |
-| 0793 |  Preimage Size of Factorial Zeroes Function                  |                                                                                                                                               | 39.10%      |  Hard       |            |
-| 0794 |  Valid Tic-Tac-Toe State                                     |                                                                                                                                               | 29.80%      |  Medium     |            |
-| 0795 |  Number of Subarrays with Bounded Maximum                    |                                                                                                                                               | 43.40%      |  Medium     |            |
-| 0796 |  Rotate String                                               |                                                                                                                                               | 49.10%      |  Easy       |            |
-| 0797 |  All Paths From Source to Target                             |                                                                                                                                               | 70.70%      |  Medium     |            |
-| 0798 |  Smallest Rotation with Highest Score                        |                                                                                                                                               | 40.40%      |  Hard       |            |
-| 0799 |  Champagne Tower                                             |                                                                                                                                               | 34.20%      |  Medium     |            |
-| 0800 |  Similar RGB Color                                           |                                                                                                                                               | 59.80%      |  Easy       |            |
-| 0801 |  Minimum Swaps To Make Sequences Increasing                  |                                                                                                                                               | 34.90%      |  Medium     |            |
-| 0802 |  Find Eventual Safe States                                   |                                                                                                                                               | 44.10%      |  Medium     |            |
-| 0803 |  Bricks Falling When Hit                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0803.%20Bricks%20Falling%20When%20Hit)                                                                                                                                              | 28.80%      |  Hard       |            |
-| 0804 |  Unique Morse Code Words                                     |                                                                                                                                               | 74.90%      |  Easy       |            |
-| 0805 |  Split Array With Same Average                               |                                                                                                                                               | 24.50%      |  Hard       |            |
-| 0806 |  Number of Lines To Write String                             |                                                                                                                                               | 63.20%      |  Easy       |            |
-| 0807 |  Max Increase to Keep City Skyline                           |                                                                                                                                               | 81.60%      |  Medium     |            |
-| 0808 |  Soup Servings                                               |                                                                                                                                               | 37.50%      |  Medium     |            |
-| 0809 |  Expressive Words                                            |                                                                                                                                               | 43.70%      |  Medium     |            |
-| 0810 |  Chalkboard XOR Game                                         |                                                                                                                                               | 45.10%      |  Hard       |            |
-| 0811 |  Subdomain Visit Count                                       |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0811.%20Subdomain%20Visit%20Count)                                                                                                                                               | 65.60%      |  Easy       |            |
-| 0812 |  Largest Triangle Area                                       |                                                                                                                                               | 56.30%      |  Easy       |            |
-| 0813 |  Largest Sum of Averages                                     |                                                                                                                                               | 45.50%      |  Medium     |            |
-| 0814 |  Binary Tree Pruning                                         |                                                                                                                                               | 71.20%      |  Medium     |            |
-| 0815 |  Bus Routes                                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0815.%20Bus%20Routes)                                                                                                                                              | 40.30%      |  Hard       |            |
-| 0816 |  Ambiguous Coordinates                                       |                                                                                                                                               | 44.20%      |  Medium     |            |
-| 0817 |  Linked List Components                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0817.%20Linked%20List%20Components)                                             | 54.80%      |  Medium     |            |
-| 0818 |  Race Car                                                    |                                                                                                                                               | 35.20%      |  Hard       |            |
-| 0819 |  Most Common Word                                            |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0819.%20Most%20Common%20Word)                                                                                                                                               | 42.20%      |  Easy       |            |
-| 0820 |  Short Encoding of Words                                     |                                                                                                                                               | 47.40%      |  Medium     |            |
-| 0821 |  Shortest Distance to a Character                            |                                                                                                                                               | 63.60%      |  Easy       |            |
-| 0822 |  Card Flipping Game                                          |                                                                                                                                               | 40.50%      |  Medium     |            |
-| 0823 |  Binary Trees With Factors                                   |                                                                                                                                               | 32.60%      |  Medium     |            |
-| 0824 |  Goat Latin                                                  |                                                                                                                                               | 57.90%      |  Easy       |            |
-| 0825 |  Friends Of Appropriate Ages                                 |                                                                                                                                               | 36.50%      |  Medium     |            |
-| 0826 |  Most Profit Assigning Work                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0826.%20Most%20Profit%20Assigning%20Work)                                       | 35.90%      |  Medium     |            |
-| 0827 |  Making A Large Island                                       |                                                                                                                                               | 43.40%      |  Hard       |            |
-| 0828 |  Unique Letter String                                        |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0828.%20Unique%20Letter%20String)                                                                                                                                                | 39.80%      |  Hard       |            |
-| 0829 |  Consecutive Numbers Sum                                     |                                                                                                                                               | 33.50%      |  Hard       |            |
-| 0830 |  Positions of Large Groups                                   |                                                                                                                                               | 47.90%      |  Easy       |            |
-| 0831 |  Masking Personal Information                                |                                                                                                                                               | 42.20%      |  Medium     |            |
-| 0832 |  Flipping an Image                                           |                                                                                                                                               | 72.90%      |  Easy       |            |
-| 0833 |  Find And Replace in String                                  |                                                                                                                                               | 46.50%      |  Medium     |            |
-| 0834 |  Sum of Distances in Tree                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0834.%20Sum%20of%20Distances%20in%20Tree)                                                                                                                                               | 39.70%      |  Hard       |            |
-| 0835 |  Image Overlap                                               |                                                                                                                                               | 52.90%      |  Medium     |            |
-| 0836 |  Rectangle Overlap                                           |                                                                                                                                               | 46.60%      |  Easy       |            |
-| 0837 |  New 21 Game                                                 |                                                                                                                                               | 31.80%      |  Medium     |            |
-| 0838 |  Push Dominoes                                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0838.%20Push%20Dominoes)                                                        | 44.00%      |  Medium     |            |
-| 0839 |  Similar String Groups                                       | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0839.%20Similar%20String%20Groups)                                                                                                                                              | 34.70%      |  Hard       |            |
-| 0840 |  Magic Squares In Grid                                       |                                                                                                                                               | 35.60%      |  Easy       |            |
-| 0841 |  Keys and Rooms                                              |                                                                                                                                               | 60.50%      |  Medium     |            |
-| 0842 |  Split Array into Fibonacci Sequence                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0842.%20Split%20Array%20into%20Fibonacci%20Sequence)                            | 34.90%      |  Medium     |            |
-| 0843 |  Guess the Word                                              |                                                                                                                                               | 43.60%      |  Hard       |            |
-| 0844 |  Backspace String Compare                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0844.%20Backspace%20String%20Compare)                                           | 46.10%      |  Easy       |            |
-| 0845 |  Longest Mountain in Array                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0845.%20Longest%20Mountain%20in%20Array)                                        | 34.40%      |  Medium     |            |
-| 0846 |  Hand of Straights                                           |                                                                                                                                               | 49.50%      |  Medium     |            |
-| 0847 |  Shortest Path Visiting All Nodes                            |                                                                                                                                               | 47.10%      |  Hard       |            |
-| 0848 |  Shifting Letters                                            |                                                                                                                                               | 41.00%      |  Medium     |            |
-| 0849 |  Maximize Distance to Closest Person                         |                                                                                                                                               | 41.00%      |  Easy       |            |
-| 0850 |  Rectangle Area II                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0850.%20Rectangle%20Area%20II)                                                                                                                                              | 45.00%      |  Hard       |            |
-| 0851 |  Loud and Rich                                               |                                                                                                                                               | 47.80%      |  Medium     |            |
-| 0852 |  Peak Index in a Mountain Array                              |                                                                                                                                               | 69.70%      |  Easy       |            |
-| 0853 |  Car Fleet                                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0853.%20Car%20Fleet)                                                            | 40.00%      |  Medium     |            |
-| 0854 |  K-Similar Strings                                           |                                                                                                                                               | 33.80%      |  Hard       |            |
-| 0855 |  Exam Room                                                   |                                                                                                                                               | 38.60%      |  Medium     |            |
-| 0856 |  Score of Parentheses                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0856.%20Score%20of%20Parentheses)                                               | 56.50%      |  Medium     |            |
-| 0857 |  Minimum Cost to Hire K Workers                              |                                                                                                                                               | 47.60%      |  Hard       |            |
-| 0858 |  Mirror Reflection                                           |                                                                                                                                               | 52.00%      |  Medium     |            |
-| 0859 |  Buddy Strings                                               |                                                                                                                                               | 27.60%      |  Easy       |            |
-| 0860 |  Lemonade Change                                             |                                                                                                                                               | 50.60%      |  Easy       |            |
-| 0861 |  Score After Flipping Matrix                                 |                                                                                                                                               | 69.80%      |  Medium     |            |
-| 0862 |  Shortest Subarray with Sum at Least K                       |                                                                                                                                               | 22.20%      |  Hard       |            |
-| 0863 |  All Nodes Distance K in Binary Tree                         |                                                                                                                                               | 48.00%      |  Medium     |            |
-| 0864 |  Shortest Path to Get All Keys                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0864.%20Shortest%20Path%20to%20Get%20All%20Keys)                                                                                                                                              | 36.20%      |  Hard       |            |
-| 0865 |  Smallest Subtree with all the Deepest Nodes                 |                                                                                                                                               | 56.00%      |  Medium     |            |
-| 0866 |  Prime Palindrome                                            |                                                                                                                                               | 20.30%      |  Medium     |            |
-| 0867 |  Transpose Matrix                                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0867.%20Transpose%20Matrix)                                                     | 63.90%      |  Easy       |            |
-| 0868 |  Binary Gap                                                  |                                                                                                                                               | 59.50%      |  Easy       |            |
-| 0869 |  Reordered Power of 2                                        |                                                                                                                                               | 51.10%      |  Medium     |            |
-| 0870 |  Advantage Shuffle                                           |                                                                                                                                               | 42.60%      |  Medium     |            |
-| 0871 |  Minimum Number of Refueling Stops                           |                                                                                                                                               | 29.10%      |  Hard       |            |
-| 0872 |  Leaf-Similar Trees                                          |                                                                                                                                               | 63.50%      |  Easy       |            |
-| 0873 |  Length of Longest Fibonacci Subsequence                     |                                                                                                                                               | 46.40%      |  Medium     |            |
-| 0874 |  Walking Robot Simulation                                    |                                                                                                                                               | 32.30%      |  Easy       |            |
-| 0875 |  Koko Eating Bananas                                         |                                                                                                                                               | 46.20%      |  Medium     |            |
-| 0876 |  Middle of the Linked List                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0876.%20Middle%20of%20the%20Linked%20List)                                      | 64.40%      |  Easy       |            |
-| 0877 |  Stone Game                                                  |                                                                                                                                               | 61.50%      |  Medium     |            |
-| 0878 |  Nth Magical Number                                          |                                                                                                                                               | 25.60%      |  Hard       |            |
-| 0879 |  Profitable Schemes                                          |                                                                                                                                               | 36.80%      |  Hard       |            |
-| 0880 |  Decoded String at Index                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0880.%20Decoded%20String%20at%20Index)                                          | 23.00%      |  Medium     |            |
-| 0881 |  Boats to Save People                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0881.%20Boats%20to%20Save%20People)                                             | 44.00%      |  Medium     |            |
-| 0882 |  Reachable Nodes In Subdivided Graph                         |                                                                                                                                               | 38.30%      |  Hard       |            |
-| 0883 |  Projection Area of 3D Shapes                                |                                                                                                                                               | 65.80%      |  Easy       |            |
-| 0884 |  Uncommon Words from Two Sentences                           |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0884.%20Uncommon%20Words%20from%20Two%20Sentences)                                                                                                                                               | 60.90%      |  Easy       |            |
-| 0885 |  Spiral Matrix III                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0885.%20Spiral%20Matrix%20III)                                                  | 64.40%      |  Medium     |            |
-| 0886 |  Possible Bipartition                                        |                                                                                                                                               | 40.90%      |  Medium     |            |
-| 0887 |  Super Egg Drop                                              |                                                                                                                                               | 24.80%      |  Hard       |            |
-| 0888 |  Fair Candy Swap                                             |                                                                                                                                               | 56.80%      |  Easy       |            |
-| 0889 |  Construct Binary Tree from Preorder and Postorder Traversal |                                                                                                                                               | 60.30%      |  Medium     |            |
-| 0890 |  Find and Replace Pattern                                    |                                                                                                                                               | 71.30%      |  Medium     |            |
-| 0891 |  Sum of Subsequence Widths                                   |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0891.%20Sum%20of%20Subsequence%20Widths)                                                                                                                                               | 29.10%      |  Hard       |            |
-| 0892 |  Surface Area of 3D Shapes                                   |                                                                                                                                               | 56.10%      |  Easy       |            |
-| 0893 |  Groups of Special-Equivalent Strings                        |                                                                                                                                               | 62.80%      |  Easy       |            |
-| 0894 |  All Possible Full Binary Trees                              |                                                                                                                                               | 71.00%      |  Medium     |            |
-| 0895 |  Maximum Frequency Stack                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0895.%20Maximum%20Frequency%20Stack)                                            | 56.40%      |  Hard       |            |
-| 0896 |  Monotonic Array                                             |                                                                                                                                               | 55.40%      |  Easy       |            |
-| 0897 |  Increasing Order Search Tree                                |                                                                                                                                               | 64.90%      |  Easy       |            |
-| 0898 |  Bitwise ORs of Subarrays                                    |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0898.%20Bitwise%20ORs%20of%20Subarrays)                                                                                                                                               | 34.40%      |  Medium     |            |
-| 0899 |  Orderly Queue                                               |                                                                                                                                               | 47.50%      |  Hard       |            |
-| 0900 |  RLE Iterator                                                |                                                                                                                                               | 50.30%      |  Medium     |            |
-| 0901 |  Online Stock Span                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0901.%20Online%20Stock%20Span)                                                  | 49.40%      |  Medium     |            |
-| 0902 |  Numbers At Most N Given Digit Set                           |                                                                                                                                               | 28.70%      |  Hard       |            |
-| 0903 |  Valid Permutations for DI Sequence                          |                                                                                                                                               | 44.70%      |  Hard       |            |
-| 0904 |  Fruit Into Baskets                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0904.%20Fruit%20Into%20Baskets)                                                 | 41.60%      |  Medium     |            |
-| 0905 |  Sort Array By Parity                                        |                                                                                                                                               | 72.60%      |  Easy       |            |
-| 0906 |  Super Palindromes                                           |                                                                                                                                               | 30.40%      |  Hard       |            |
-| 0907 |  Sum of Subarray Minimums                                    |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0907.%20Sum%20of%20Subarray%20Minimums)                                                                                                                                                | 27.60%      |  Medium     |            |
-| 0908 |  Smallest Range I                                            |                                                                                                                                               | 64.60%      |  Easy       |            |
-| 0909 |  Snakes and Ladders                                          |                                                                                                                                               | 33.80%      |  Medium     |            |
-| 0910 |  Smallest Range II                                           |                                                                                                                                               | 23.70%      |  Medium     |            |
-| 0911 |  Online Election                                             |                                                                                                                                               | 47.10%      |  Medium     |            |
-| 0912 |  Sort an Array                                               |                                                                                                                                               | 63.60%      |  Medium     |            |
-| 0913 |  Cat and Mouse                                               |                                                                                                                                               | 28.80%      |  Hard       |            |
-| 0914 |  X of a Kind in a Deck of Cards                              |                                                                                                                                               | 34.10%      |  Easy       |            |
-| 0915 |  Partition Array into Disjoint Intervals                     |                                                                                                                                               | 43.50%      |  Medium     |            |
-| 0916 |  Word Subsets                                                |                                                                                                                                               | 45.40%      |  Medium     |            |
-| 0917 |  Reverse Only Letters                                        |                                                                                                                                               | 55.90%      |  Easy       |            |
-| 0918 |  Maximum Sum Circular Subarray                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0918.%20Maximum%20Sum%20Circular%20Subarray)                                                                                                                                              | 32.00%      |  Medium     |            |
-| 0919 |  Complete Binary Tree Inserter                               |                                                                                                                                               | 55.20%      |  Medium     |            |
-| 0920 |  Number of Music Playlists                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0920.%20Number%20of%20Music%20Playlists)                                                                                                                                               | 43.90%      |  Hard       |            |
-| 0921 |  Minimum Add to Make Parentheses Valid                       | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0921.%20Minimum%20Add%20to%20Make%20Parentheses%20Valid)                        | 70.30%      |  Medium     |            |
-| 0922 |  Sort Array By Parity II                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0922.%20Sort%20Array%20By%20Parity%20II)                                        | 67.10%      |  Easy       |            |
-| 0923 |  3Sum With Multiplicity                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0923.%203Sum%20With%20Multiplicity)                                             | 33.90%      |  Medium     |            |
-| 0924 |  Minimize Malware Spread                                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0924.%20Minimize%20Malware%20Spread)                                                                                                                                              | 40.10%      |  Hard       |            |
-| 0925 |  Long Pressed Name                                           | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0925.%20Long%20Pressed%20Name)                                                  | 44.50%      |  Easy       |            |
-| 0926 |  Flip String to Monotone Increasing                          |                                                                                                                                               | 49.80%      |  Medium     |            |
-| 0927 |  Three Equal Parts                                           |                                                                                                                                               | 30.40%      |  Hard       |            |
-| 0928 |  Minimize Malware Spread II                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0928.%20Minimize%20Malware%20Spread%20II)                                                                                                                                              | 39.50%      |  Hard       |            |
-| 0929 |  Unique Email Addresses                                      |                                                                                                                                               | 70.50%      |  Easy       |            |
-| 0930 |  Binary Subarrays With Sum                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0930.%20Binary%20Subarrays%20With%20Sum)                                        | 38.20%      |  Medium     |            |
-| 0931 |  Minimum Falling Path Sum                                    |                                                                                                                                               | 59.00%      |  Medium     |            |
-| 0932 |  Beautiful Array                                             |                                                                                                                                               | 53.60%      |  Medium     |            |
-| 0933 |  Number of Recent Calls                                      |                                                                                                                                               | 69.40%      |  Easy       |            |
-| 0934 |  Shortest Bridge                                             |                                                                                                                                               | 44.20%      |  Medium     |            |
-| 0935 |  Knight Dialer                                               |                                                                                                                                               | 41.10%      |  Medium     |            |
-| 0936 |  Stamping The Sequence                                       |                                                                                                                                               | 36.20%      |  Hard       |            |
-| 0937 |  Reorder Log Files                                           |                                                                                                                                               | 56.90%      |  Easy       |            |
-| 0938 |  Range Sum of BST                                            |                                                                                                                                               | 78.80%      |  Easy       |            |
-| 0939 |  Minimum Area Rectangle                                      |                                                                                                                                               | 50.50%      |  Medium     |            |
-| 0940 |  Distinct Subsequences II                                    |                                                                                                                                               | 39.80%      |  Hard       |            |
-| 0941 |  Valid Mountain Array                                        |                                                                                                                                               | 35.50%      |  Easy       |            |
-| 0942 |  DI String Match                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0942.%20DI%20String%20Match)                                                    | 70.00%      |  Easy       |            |
-| 0943 |  Find the Shortest Superstring                               |                                                                                                                                               | 38.30%      |  Hard       |            |
-| 0944 |  Delete Columns to Make Sorted                               |                                                                                                                                               | 69.40%      |  Easy       |            |
-| 0945 |  Minimum Increment to Make Array Unique                      |                                                                                                                                               | 42.90%      |  Medium     |            |
-| 0946 |  Validate Stack Sequences                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0946.%20Validate%20Stack%20Sequences)                                           | 57.60%      |  Medium     |            |
-| 0947 |  Most Stones Removed with Same Row or Column                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0947.%20Most%20Stones%20Removed%20with%20Same%20Row%20or%20Column)                                                                                                                                              | 54.20%      |  Medium     |            |
-| 0948 |  Bag of Tokens                                               |                                                                                                                                               | 39.30%      |  Medium     |            |
-| 0949 |  Largest Time for Given Digits                               |                                                                                                                                               | 34.20%      |  Easy       |            |
-| 0950 |  Reveal Cards In Increasing Order                            |                                                                                                                                               | 72.20%      |  Medium     |            |
-| 0951 |  Flip Equivalent Binary Trees                                |                                                                                                                                               | 65.20%      |  Medium     |            |
-| 0952 |  Largest Component Size by Common Factor                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0952.%20Largest%20Component%20Size%20by%20Common%20Factor)                                                                                                                                              | 26.60%      |  Hard       |            |
-| 0953 |  Verifying an Alien Dictionary                               |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0953.%20Verifying%20an%20Alien%20Dictionary)                                                                                                                                               | 55.60%      |  Easy       |            |
-| 0954 |  Array of Doubled Pairs                                      |                                                                                                                                               | 34.40%      |  Medium     |            |
-| 0955 |  Delete Columns to Make Sorted II                            |                                                                                                                                               | 31.60%      |  Medium     |            |
-| 0956 |  Tallest Billboard                                           |                                                                                                                                               | 38.40%      |  Hard       |            |
-| 0957 |  Prison Cells After N Days                                   |                                                                                                                                               | 38.20%      |  Medium     |            |
-| 0958 |  Check Completeness of a Binary Tree                         |                                                                                                                                               | 47.60%      |  Medium     |            |
-| 0959 |  Regions Cut By Slashes                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0959.%20Regions%20Cut%20By%20Slashes)                                                                                                                                              | 62.50%      |  Medium     |            |
-| 0960 |  Delete Columns to Make Sorted III                           |                                                                                                                                               | 52.80%      |  Hard       |            |
-| 0961 |  N-Repeated Element in Size 2N Array                         |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0961.%20N-Repeated%20Element%20in%20Size%202N%20Array)                                                                                                                                               | 72.40%      |  Easy       |            |
-| 0962 |  Maximum Width Ramp                                          |                                                                                                                                               | 41.80%      |  Medium     |            |
-| 0963 |  Minimum Area Rectangle II                                   |                                                                                                                                               | 44.50%      |  Medium     |            |
-| 0964 |  Least Operators to Express Number                           |                                                                                                                                               | 40.90%      |  Hard       |            |
-| 0965 |  Univalued Binary Tree                                       |                                                                                                                                               | 66.90%      |  Easy       |            |
-| 0966 |  Vowel Spellchecker                                          |                                                                                                                                               | 41.90%      |  Medium     |            |
-| 0967 |  Numbers With Same Consecutive Differences                   |                                                                                                                                               | 37.10%      |  Medium     |            |
-| 0968 |  Binary Tree Cameras                                         |                                                                                                                                               | 35.40%      |  Hard       |            |
-| 0969 |  Pancake Sorting                                             | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0969.%20Pancake%20Sorting)                                                      | 62.40%      |  Medium     |            |
-| 0970 |  Powerful Integers                                           |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0970.%20Powerful%20Integers)                                                                                                                                               | 39.30%      |  Easy       |            |
-| 0971 |  Flip Binary Tree To Match Preorder Traversal                |                                                                                                                                               | 43.00%      |  Medium     |            |
-| 0972 |  Equal Rational Numbers                                      |                                                                                                                                               | 40.20%      |  Hard       |            |
-| 0973 |  K Closest Points to Origin                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0973.%20K%20Closest%20Points%20to%20Origin)                                     | 62.20%      |  Medium     |            |
-| 0974 |  Subarray Sums Divisible by K                                |                                                                                                                                               | 45.20%      |  Medium     |            |
-| 0975 |  Odd Even Jump                                               |                                                                                                                                               | 47.20%      |  Hard       |            |
-| 0976 |  Largest Perimeter Triangle                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0976.%20Largest%20Perimeter%20Triangle)                                         | 57.20%      |  Easy       |            |
-| 0977 |  Squares of a Sorted Array                                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0977.%20Squares%20of%20a%20Sorted%20Array)                                      | 72.00%      |  Easy       |            |
-| 0978 |  Longest Turbulent Subarray                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0978.%20Longest%20Turbulent%20Subarray)                                                                                                                                              | 45.60%      |  Medium     |            |
-| 0979 |  Distribute Coins in Binary Tree                             |                                                                                                                                               | 67.20%      |  Medium     |            |
-| 0980 |  Unique Paths III                                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0980.%20Unique%20Paths%20III)                                                   | 71.20%      |  Hard       |            |
-| 0981 |  Time Based Key-Value Store                                  |                                                                                                                                               | 50.80%      |  Medium     |            |
-| 0982 |  Triples with Bitwise AND Equal To Zero                      |                                                                                                                                               | 54.00%      |  Hard       |            |
-| 0983 |  Minimum Cost For Tickets                                    |                                                                                                                                               | 57.30%      |  Medium     |            |
-| 0984 |  String Without AAA or BBB                                   |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0984.%20String%20Without%20AAA%20or%20BBB)                                                                                                                                               | 33.90%      |  Medium     |            |
-| 0985 |  Sum of Even Numbers After Queries                           |                                                                                                                                               | 63.40%      |  Easy       |            |
-| 0986 |  Interval List Intersections                                 | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0986.%20Interval%20List%20Intersections)                                        | 63.30%      |  Medium     |            |
-| 0987 |  Vertical Order Traversal of a Binary Tree                   |                                                                                                                                               | 31.80%      |  Medium     |            |
-| 0988 |  Smallest String Starting From Leaf                          |                                                                                                                                               | 46.00%      |  Medium     |            |
-| 0989 |  Add to Array-Form of Integer                                |                                                                                                                                               | 44.70%      |  Easy       |            |
-| 0990 |  Satisfiability of Equality Equations                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0990.%20Satisfiability%20of%20Equality%20Equations)                                                                                                                                              | 40.10%      |  Medium     |            |
-| 0991 |  Broken Calculator                                           |                                                                                                                                               | 40.40%      |  Medium     |            |
-| 0992 |  Subarrays with K Different Integers                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0992.%20Subarrays%20with%20K%20Different%20Integers)                            | 45.20%      |  Hard       |            |
-| 0993 |  Cousins in Binary Tree                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0993.%20Cousins%20in%20Binary%20Tree)                                           | 52.40%      |  Easy       |            |
-| 0994 |  Rotting Oranges                                             |                                                                                                                                               | 46.40%      |  Easy       |            |
-| 0995 |  Minimum Number of K Consecutive Bit Flips                   | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0995.%20Minimum%20Number%20of%20K%20Consecutive%20Bit%20Flips)                                                                                                                                              | 48.40%      |  Hard       |            |
-| 0996 |  Number of Squareful Arrays                                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0996.%20Number%20of%20Squareful%20Arrays)                                       | 47.80%      |  Hard       |            |
-| 0997 |  Find the Town Judge                                         |                                                                                                                                               | 49.10%      |  Easy       |            |
-| 0998 |  Maximum Binary Tree II                                      |                                                                                                                                               | 61.60%      |  Medium     |            |
-| 0999 |  Available Captures for Rook                                 |                                                                                                                                               | 66.10%      |  Easy       |            |
-| 1000 |  Minimum Cost to Merge Stones                                |                                                                                                                                               | 32.10%      |  Hard       |            |
-| 1001 |  Grid Illumination                                           |                                                                                                                                               | 34.40%      |  Hard       |            |
-| 1002 |  Find Common Characters                                      |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1002.%20Find%20Common%20Characters)                                                                                                                                               | 65.80%      |  Easy       |            |
-| 1003 |  Check If Word Is Valid After Substitutions                  | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1003.%20Check%20If%20Word%20Is%20Valid%20After%20Substitutions)                 | 51.80%      |  Medium     |            |
-| 1004 |  Max Consecutive Ones III                                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1004.%20Max%20Consecutive%20Ones%20III)                                         | 53.40%      |  Medium     |            |
-| 1005 |  Maximize Sum Of Array After K Negations                     | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1005.%20Maximize%20Sum%20Of%20Array%20After%20K%20Negations)                    | 49.70%      |  Easy       |            |
-| 1006 |  Clumsy Factorial                                            |                                                                                                                                               | 53.70%      |  Medium     |            |
-| 1007 |  Minimum Domino Rotations For Equal Row                      |                                                                                                                                               | 47.90%      |  Medium     |            |
-| 1008 |  Construct Binary Search Tree from Preorder Traversal        |                                                                                                                                               | 73.10%      |  Medium     |            |
-| 1009 |  Complement of Base 10 Integer                               |                                                                                                                                               | 58.90%      |  Easy       |            |
-| 1010 |  Pairs of Songs With Total Durations Divisible by 60         |                                                                                                                                               | 45.60%      |  Easy       |            |
-| 1011 |  Capacity To Ship Packages Within D Days                     |                                                                                                                                               | 53.30%      |  Medium     |            |
-| 1012 |  Numbers With Repeated Digits                                |                                                                                                                                               | 34.80%      |  Hard       |            |
-| 1013 |  Partition Array Into Three Parts With Equal Sum             |                                                                                                                                               | 55.50%      |  Easy       |            |
-| 1014 |  Best Sightseeing Pair                                       |                                                                                                                                               | 49.30%      |  Medium     |            |
-| 1015 |  Smallest Integer Divisible by K                             |                                                                                                                                               | 28.40%      |  Medium     |            |
-| 1016 |  Binary String With Substrings Representing 1 To N           |                                                                                                                                               | 60.70%      |  Medium     |            |
-| 1017 |  Convert to Base -2                                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1017.%20Convert%20to%20Base%20-2)                                                                                                                                              | 56.70%      |  Medium     |            |
-| 1018 |  Binary Prefix Divisible By 5                                |                                                                                                                                               | 46.60%      |  Easy       |            |
-| 1019 |  Next Greater Node In Linked List                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1019.%20Next%20Greater%20Node%20In%20Linked%20List)                             | 56.50%      |  Medium     |            |
-| 1020 |  Number of Enclaves                                          |                                                                                                                                               | 54.70%      |  Medium     |            |
-| 1021 |  Remove Outermost Parentheses                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1021.%20Remove%20Outermost%20Parentheses)                                       | 75.60%      |  Easy       |            |
-| 1022 |  Sum of Root To Leaf Binary Numbers                          |                                                                                                                                               | 55.90%      |  Easy       |            |
-| 1023 |  Camelcase Matching                                          |                                                                                                                                               | 56.50%      |  Medium     |            |
-| 1024 |  Video Stitching                                             |                                                                                                                                               | 47.20%      |  Medium     |            |
-| 1025 |  Divisor Game                                                | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1025.%20Divisor%20Game)                                                         | 63.30%      |  Easy       |            |
-| 1026 |  Maximum Difference Between Node and Ancestor                |                                                                                                                                               | 59.60%      |  Medium     |            |
-| 1027 |  Longest Arithmetic Sequence                                 |                                                                                                                                               | 47.90%      |  Medium     |            |
-| 1028 |  Recover a Tree From Preorder Traversal                      |                                                                                                                                               | 70.20%      |  Hard       |            |
-| 1029 |  Two City Scheduling                                         |                                                                                                                                               | 53.90%      |  Easy       |            |
-| 1030 |  Matrix Cells in Distance Order                              | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1030.%20Matrix%20Cells%20in%20Distance%20Order)                                 | 65.20%      |  Easy       |            |
-| 1031 |  Maximum Sum of Two Non-Overlapping Subarrays                |                                                                                                                                               | 55.30%      |  Medium     |            |
-| 1032 |  Stream of Characters                                        |                                                                                                                                               | 41.30%      |  Hard       |            |
-| 1033 |  Moving Stones Until Consecutive                             |                                                                                                                                               | 35.50%      |  Easy       |            |
-| 1034 |  Coloring A Border                                           |                                                                                                                                               | 42.50%      |  Medium     |            |
-| 1035 |  Uncrossed Lines                                             |                                                                                                                                               | 51.80%      |  Medium     |            |
-| 1036 |  Escape a Large Maze                                         |                                                                                                                                               | 36.40%      |  Hard       |            |
-| 1037 |  Valid Boomerang                                             |                                                                                                                                               | 37.60%      |  Easy       |            |
-| 1038 |  Binary Search Tree to Greater Sum Tree                      |                                                                                                                                               | 79.70%      |  Medium     |            |
-| 1039 |  Minimum Score Triangulation of Polygon                      |                                                                                                                                               | 41.50%      |  Medium     |            |
-| 1040 |  Moving Stones Until Consecutive II                          | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1040.%20Moving%20Stones%20Until%20Consecutive%20II)                                                                                                                                              | 46.20%      |  Medium     |            |
-| 1041 |  Robot Bounded In Circle                                     |                                                                                                                                               | 43.40%      |  Easy       |            |
-| 1042 |  Flower Planting With No Adjacent                            |                                                                                                                                               | 48.40%      |  Easy       |            |
-| 1043 |  Partition Array for Maximum Sum                             |                                                                                                                                               | 62.40%      |  Medium     |            |
-| 1044 |  Longest Duplicate Substring                                 |                                                                                                                                               | 22.80%      |  Hard       |            |
-| 1045 |  Customers Who Bought All Products                           |                                                                                                                                               | 64.10%      |  Medium     |            |
-| 1046 |  Last Stone Weight                                           |                                                                                                                                               | 62.40%      |  Easy       |            |
-| 1047 |  Remove All Adjacent Duplicates In String                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1047.%20Remove%20All%20Adjacent%20Duplicates%20In%20String)                     | 63.60%      |  Easy       |            |
-| 1048 |  Longest String Chain                                        |                                                                                                                                               | 47.30%      |  Medium     |            |
-| 1049 |  Last Stone Weight II                                        | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1049.%20Last%20Stone%20Weight%20II)                                                                                                                                              | 39.90%      |  Medium     |            |
-| 1050 |  Actors and Directors Who Cooperated At Least Three Times    |                                                                                                                                               | 75.40%      |  Easy       |            |
-| 1051 |  Height Checker                                              |                                                                                                                                               | 69.40%      |  Easy       |            |
-| 1052 |  Grumpy Bookstore Owner                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1052.%20Grumpy%20Bookstore%20Owner)                                                                                                                                              | 52.00%      |  Medium     |            |
-| 1053 |  Previous Permutation With One Swap                          |                                                                                                                                               | 47.10%      |  Medium     |            |
-| 1054 |  Distant Barcodes                                            | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1054.%20Distant%20Barcodes)                                                     | 38.60%      |  Medium     |            |
-| 1055 |  Shortest Way to Form String                                 |                                                                                                                                               | 58.20%      |  Medium     |            |
-| 1056 |  Confusing Number                                            |                                                                                                                                               | 52.10%      |  Easy       |            |
-| 1057 |  Campus Bikes                                                |                                                                                                                                               | 58.90%      |  Medium     |            |
-| 1058 |  Minimize Rounding Error to Meet Target                      |                                                                                                                                               | 41.30%      |  Medium     |            |
-| 1059 |  All Paths from Source Lead to Destination                   |                                                                                                                                               | 45.60%      |  Medium     |            |
-| 1060 |  Missing Element in Sorted Array                             |                                                                                                                                               | 54.20%      |  Medium     |            |
-| 1061 |  Lexicographically Smallest Equivalent String                |                                                                                                                                               | 59.30%      |  Medium     |            |
-| 1062 |  Longest Repeating Substring                                 |                                                                                                                                               | 50.80%      |  Medium     |            |
-| 1063 |  Number of Valid Subarrays                                   |                                                                                                                                               | 74.30%      |  Hard       |            |
-| 1064 |  Fixed Point                                                 |                                                                                                                                               | 79.90%      |  Easy       |            |
-| 1065 |  Index Pairs of a String                                     |                                                                                                                                               | 59.30%      |  Easy       |            |
-| 1066 |  Campus Bikes II                                             |                                                                                                                                               | 43.30%      |  Medium     |            |
-| 1067 |  Digit Count in Range                                        |                                                                                                                                               | 34.40%      |  Hard       |            |
-| 1068 |  Product Sales Analysis I                                    |                                                                                                                                               | 77.50%      |  Easy       |            |
-| 1069 |  Product Sales Analysis II                                   |                                                                                                                                               | 79.90%      |  Easy       |            |
-| 1070 |  Product Sales Analysis III                                  |                                                                                                                                               | 42.70%      |  Medium     |            |
-| 1071 |  Greatest Common Divisor of Strings                          |                                                                                                                                               | 53.60%      |  Easy       |            |
-| 1072 |  Flip Columns For Maximum Number of Equal Rows               |                                                                                                                                               | 56.20%      |  Medium     |            |
-| 1073 |  Adding Two Negabinary Numbers                               | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1073.%20Adding%20Two%20Negabinary%20Numbers)                                                                                                                                              | 31.60%      |  Medium     |            |
-| 1074 |  Number of Submatrices That Sum to Target                    | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1074.%20Number%20of%20Submatrices%20That%20Sum%20to%20Target)                                                                                                                                              | 58.50%      |  Hard       |            |
-| 1075 |  Project Employees I                                         |                                                                                                                                               | 58.70%      |  Easy       |            |
-| 1076 |  Project Employees II                                        |                                                                                                                                               | 52.80%      |  Easy       |            |
-| 1077 |  Project Employees III                                       |                                                                                                                                               | 76.90%      |  Medium     |            |
-| 1078 |  Occurrences After Bigram                                    |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1078.%20Occurrences%20After%20Bigram)                                                                                                                                               | 67.40%      |  Easy       |            |
-| 1079 |  Letter Tile Possibilities                                   |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1079.%20Letter%20Tile%20Possibilities)                                                                                                                                                | 76.50%      |  Medium     |            |
-| 1080 |  Insufficient Nodes in Root to Leaf Paths                    |                                                                                                                                               | 41.10%      |  Medium     |            |
-| 1081 |  Smallest Subsequence of Distinct Characters                 |                                                                                                                                               | 42.80%      |  Medium     |            |
-| 1082 |  Sales Analysis I                                            |                                                                                                                                               | 74.20%      |  Easy       |            |
-| 1083 |  Sales Analysis II                                           |                                                                                                                                               | 55.00%      |  Easy       |            |
-| 1084 |  Sales Analysis III                                          |                                                                                                                                               | 57.80%      |  Easy       |            |
-| 1085 |  Sum of Digits in the Minimum Number                         |                                                                                                                                               | 75.30%      |  Easy       |            |
-| 1086 |  High Five                                                   |                                                                                                                                               | 76.90%      |  Easy       |            |
-| 1087 |  Brace Expansion                                             |                                                                                                                                               | 57.50%      |  Medium     |            |
-| 1088 |  Confusing Number II                                         |                                                                                                                                               | 34.10%      |  Hard       |            |
-| 1089 |  Duplicate Zeros                                             |                                                                                                                                               | 58.80%      |  Easy       |            |
-| 1090 |  Largest Values From Labels                                  |                                                                                                                                               | 56.00%      |  Medium     |            |
-| 1091 |  Shortest Path in Binary Matrix                              |                                                                                                                                               | 35.90%      |  Medium     |            |
-| 1092 |  Shortest Common Supersequence                               |                                                                                                                                               | 47.20%      |  Hard       |            |
-| 1093 |  Statistics from a Large Sample                              |[Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1093.%20Statistics%20from%20a%20Large%20Sample)                                                                                                                                                | 41.10%      |  Medium     |            |
-| 1094 |  Car Pooling                                                 |                                                                                                                                               | 57.10%      |  Medium     |            |
-| 1095 |  Find in Mountain Array                                      |                                                                                                                                               | 32.50%      |  Hard       |            |
-| 1096 |  Brace Expansion II                                          |                                                                                                                                               | 52.00%      |  Hard       |            |
-| 1097 |  Game Play Analysis V                                        |                                                                                                                                               | 51.10%      |  Hard       |            |
-| 1098 |  Unpopular Books                                             |                                                                                                                                               | 43.20%      |  Medium     |            |
-| 1099 | Two Sum Less Than K                                        |  | 62.80% | Easy    |            |
-| 1100 | Find K-Length Substrings With No Repeated Characters       |  | 72.40% | Medium  |            |
-| 1101 | The Earliest Moment When Everyone Become Friends           |  | 62.20% | Medium  |            |
-| 1102 | Path With Maximum Minimum Value                            |  | 42.50% | Medium  |            |
-| 1103 | Distribute Candies to People                               |  | 62.00% | Easy    |            |
-| 1104 | Path In Zigzag Labelled Binary Tree                        |  | 71.10% | Medium  |            |
-| 1105 | Filling Bookcase Shelves                                      | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1105.%20Filling%20Bookcase%20Shelves) | 53.80% | Medium  |            |
-| 1106 | Parsing A Boolean Expression                               |  | 59.00% | Hard    |            |
-| 1107 | New Users Daily Count                                      |  | 46.30% | Medium  |            |
-| 1108 | Defanging an IP Address                                    |  | 87.50% | Easy    |            |
-| 1109 | Corporate Flight Bookings                                  |  | 47.30% | Medium  |            |
-| 1110 | Delete Nodes And Return Forest                             |  | 61.90% | Medium  |            |
-| 1111 | Maximum Nesting Depth of Two Valid Parentheses Strings     |  | 63.30% | Medium  |            |
-| 1112 | Highest Grade For Each Student                             |  | 60.80% | Medium  |            |
-| 1113 | Reported Posts                                             |  | 68.20% | Easy    |            |
-| 1114 | Print in Order                                             |  | 59.20% | Easy    |            |
-| 1115 | Print FooBar Alternately                                   |  | 59.40% | Medium  |            |
-| 1116 | Print Zero Even Odd                                        |  | 55.40% | Medium  |            |
-| 1117 | Building H2O                                               |  | 50.40% | Hard    |            |
-| 1118 | Number of Days in a Month                                  |  | 57.50% | Easy    |            |
-| 1119 | Remove Vowels from a String                                |  | 91.10% | Easy    |            |
-| 1120 | Maximum Average Subtree                                    |  | 60.90% | Medium  |            |
-| 1121 | Divide Array Into Increasing Sequences                     |  | 51.20% | Hard    |            |
-| 1122 | Relative Sort Array                                        |  | 68.50% | Easy    |            |
-| 1123 | Lowest Common Ancestor of Deepest Leaves                   |  | 64.60% | Medium  |            |
-| 1124 | Longest Well-Performing Interval                           |  | 27.40% | Medium  |            |
-| 1125 | Smallest Sufficient Team                                   |  | 43.50% | Hard    |            |
-| 1126 | Active Businesses                                          |  | 70.80% | Medium  |            |
-| 1127 | User Purchase Platform                                     |  | 64.50% | Hard    |            |
-| 1128 | Number of Equivalent Domino Pairs                          |  | 40.30% | Easy    |            |
-| 1129 | Shortest Path with Alternating Colors                      |  | 34.20% | Medium  |            |
-| 1130 | Minimum Cost Tree From Leaf Values                          |                                                                                                                                                 | 58.80% | Medium |            |
-| 1131 | Maximum of Absolute Value Expression                        |                                                                                                                                                 | 50.30% | Medium |            |
-| 1132 | Reported Posts II                                           |                                                                                                                                                 | 35.50% | Medium |            |
-| 1133 | Largest Unique Number                                       |                                                                                                                                                 | 70.40% | Easy   |            |
-| 1134 | Armstrong Number                                            |                                                                                                                                                 | 79.90% | Easy   |            |
-| 1135 | Connecting Cities With Minimum Cost                         |                                                                                                                                                 | 49.40% | Medium |            |
-| 1136 | Parallel Courses                                            |                                                                                                                                                 | 58.80% | Hard   |            |
-| 1137 | N-th Tribonacci Number                                      |                                                                                                                                                 | 61.10% | Easy   |            |
-| 1138 | Alphabet Board Path                                         |                                                                                                                                                 | 42.40% | Medium |            |
-| 1139 | Largest 1-Bordered Square                                   |                                                                                                                                                 | 42.80% | Medium |            |
-| 1140 | Stone Game II                                               |                                                                                                                                                 | 60.60% | Medium |            |
-| 1141 | User Activity for the Past 30 Days I                        |                                                                                                                                                 | 55.30% | Easy   |            |
-| 1142 | User Activity for the Past 30 Days II                       |                                                                                                                                                 | 38.10% | Easy   |            |
-| 1143 | Longest Common Subsequence                                  |                                                                                                                                                 | 58.80% | Medium |            |
-| 1144 | Decrease Elements To Make Array Zigzag                      |                                                                                                                                                 | 40.50% | Medium |            |
-| 1145 | Binary Tree Coloring Game                                   |                                                                                                                                                 | 45.10% | Medium |            |
-| 1146 | Snapshot Array                                              |                                                                                                                                                 | 29.80% | Medium |            |
-| 1147 | Longest Chunked Palindrome Decomposition                    |                                                                                                                                                 | 58.40% | Hard   |            |
-| 1148 | Article Views I                                             |                                                                                                                                                 | 68.30% | Easy   |            |
-| 1149 | Article Views II                                            |                                                                                                                                                 | 50.10% | Medium |            |
-| 1150 | Check If a Number Is Majority Element in a Sorted Array     |                                                                                                                                                 | 72.40% | Easy   |            |
-| 1151 | Minimum Swaps to Group All 1's Together                     |                                                                                                                                                 | 51.60% | Medium |            |
-| 1152 | Analyze User Website Visit Pattern                          |                                                                                                                                                 | 33.50% | Medium |            |
-| 1153 | String Transforms Into Another String                       |                                                                                                                                                 | 23.60% | Hard   |            |
-| 1154 | Day of the Year                                             |                                                                                                                                                 | 52.10% | Easy   |            |
-| 1155 | Number of Dice Rolls With Target Sum                        |                                                                                                                                                 | 48.60% | Medium |            |
-| 1156 | Swap For Longest Repeated Character Substring               |                                                                                                                                                 | 41.10% | Medium |            |
-| 1157 | Online Majority Element In Subarray                         | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1157.%20Online%20Majority%20Element%20In%20Subarray)                                                                                                                                                | 20.50% | Hard   |            |
-| 1158 | Market Analysis I                                               |          | 50.5%      | Medium     |             |
-| 1159 | Market Analysis II                                              |          | 46.8%      | Hard       |             |
-| 1160 | Find Words That Can Be Formed by Characters                     |          | 67.6%      | Easy       |             |
-| 1161 | Maximum Level Sum of a Binary Tree                              |          | 71.4%      | Medium     |             |
-| 1162 | As Far from Land as Possible                                    |          | 39.5%      | Medium     |             |
-| 1163 | Last Substring in Lexicographical Order                         |          | 30.2%      | Hard       |             |
-| 1164 | Product Price at a Given Date                                   |          | 59.1%      | Medium     |             |
-| 1165 | Single-Row Keyboard                                             |          | 84.9%      | Easy       |             |
-| 1166 | Design File System                                              |          | 60.0%      | Medium     |             |
-| 1167 | Minimum Cost to Connect Sticks                                  |          | 60.1%      | Medium     |             |
-| 1168 | Optimize Water Distribution in a Village                        |          | 52.6%      | Hard       |             |
-| 1169 | Invalid Transactions                                            |          | 27.4%      | Medium     |             |
-| 1170 | Compare Strings by Frequency of the Smallest Character          |          | 60.5%      | Easy       |             |
-| 1171 | Remove Zero Sum Consecutive Nodes from Linked List              |          | 42.3%      | Medium     |             |
-| 1172 | Dinner Plate Stacks                                             |          | 41.6%      | Hard       |             |
-| 1173 | Immediate Food Delivery I                                       |          | 72.5%      | Easy       |             |
-| 1174 | Immediate Food Delivery II                                      |          | 52.3%      | Medium     |             |
-| 1175 | Prime Arrangements                                              |          | 50.3%      | Easy       |             |
-| 1176 | Diet Plan Performance                                           |          | 48.6%      | Easy       |             |
-| 1177 | Can Make Palindrome from Substring                              |          | 32.0%      | Medium     |             |
-| 1178 | Number of Valid Words for Each Puzzle                           |          | 34.7%      | Hard       |             |
-| 1179 | Reformat Department Table                                       |          | 67.1%      | Easy       |             |
-| 1180 | Count Substrings with Only One Distinct Letter                  |          | 78.3%      | Easy       |             |
-| 1181 | Before and After Puzzle                                         |          | 41.9%      | Medium     |             |
-| 1182 | Shortest Distance to Target Color                               |          | 50.7%      | Medium     |             |
-| 1183 | Maximum Number of Ones                                          |          | 45.7%      | Hard       |             |
-| 1184 | Distance Between Bus Stops                                      |          | 56.7%      | Easy       |             |
-| 1185 | Day of the Week                                                 |          | 65.8%      | Easy       |             |
-| 1186 | Maximum Subarray Sum with One Deletion                          |          | 33.2%      | Medium     |             |
-| 1187 | Make Array Strictly Increasing                                  |          | 39.6%      | Hard       |             |
-| 1188 | Design Bounded Blocking Queue                                   |          | 65.0%      | Medium     |             |
-| 1189 | Maximum Number of Balloons                                      |          | 62.7%      | Easy       |             |
-| 1190 | Reverse Substrings Between Each Pair of Parentheses             |          | 57.8%      | Medium     |             |
-| 1191 | K-Concatenation Maximum Sum                                     |          | 24.0%      | Medium     |             |
-| 1192 | Critical Connections in a Network                               |          | 45.2%      | Hard       |             |
-| 1193 | Monthly Transactions I                                          |          | 58.8%      | Medium     |             |
-| 1194 | Tournament Winners                                              |          | 52.2%      | Hard       |             |
-| 1195 | Fizz Buzz Multithreaded                                         |          | 67.7%      | Medium     |             |
-| 1196 | How Many Apples Can You Put into the Basket                     |          | 68.7%      | Easy       |             |
-| 1197 | Minimum Knight Moves                                            |          | 30.1%      | Medium     |             |
-| 1198 | Find Smallest Common Element in All Rows                        |          | 71.8%      | Medium     |             |
-| 1199 | Minimum Time to Build Blocks                                    |          | 30.1%      | Hard       |             |
-| 1200 | Minimum Absolute Difference                                     |          | 66.3%      | Easy       |             |
-| 1201 | Ugly Number III                                                 |          | 23.9%      | Medium     |             |
-| 1202 | Smallest String With Swaps                                      |          | 39.2%      | Medium     |             |
-| 1203 | Sort Items by Groups Respecting Dependencies                    |          | 46.9%      | Hard       |             |
-| 1204 | Last Person to Fit in the Elevator    New                       |          | 65.8%      | Medium     |             |
-| 1205 | Monthly Transactions II    New                                  |          | 49.3%      | Medium     |             |
-| 1206 | Design Skiplist    New                                          |          | 61.8%      | Hard       |             |
-| 1207 | Unique Number of Occurrences                                    |          | 77.4%      | Easy       |             |
-| 1208 | Get Equal Substrings Within Budget                              |          | 33.3%      | Medium     |             |
-| 1209 | Remove All Adjacent Duplicates in String II                     |          | 59.6%      | Medium     |             |
-| 1210 | Minimum Moves to Reach Target with Rotations                    |          | 40.1%      | Hard       |             |
+|0001|Two Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0001.Two-Sum)|49.1%|Easy||
+|0002|Add Two Numbers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0002.Add-Two-Numbers)|39.7%|Medium||
+|0003|Longest Substring Without Repeating Characters|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0003.Longest-Substring-Without-Repeating-Characters)|33.8%|Medium||
+|0004|Median of Two Sorted Arrays|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0004.Median-of-Two-Sorted-Arrays)|35.1%|Hard||
+|0005|Longest Palindromic Substring|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0005.Longest-Palindromic-Substring)|32.4%|Medium||
+|0006|Zigzag Conversion|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0006.Zigzag-Conversion)|43.0%|Medium||
+|0007|Reverse Integer|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0007.Reverse-Integer)|27.2%|Medium||
+|0008|String to Integer (atoi)|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0008.String-to-Integer-atoi)|16.6%|Medium||
+|0009|Palindrome Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0009.Palindrome-Number)|52.8%|Easy||
+|0010|Regular Expression Matching||28.3%|Hard||
+|0011|Container With Most Water|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0011.Container-With-Most-Water)|54.3%|Medium||
+|0012|Integer to Roman|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0012.Integer-to-Roman)|60.5%|Medium||
+|0013|Roman to Integer|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0013.Roman-to-Integer)|58.2%|Easy||
+|0014|Longest Common Prefix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0014.Longest-Common-Prefix)|40.7%|Easy||
+|0015|3Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0015.3Sum)|32.2%|Medium||
+|0016|3Sum Closest|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0016.3Sum-Closest)|46.2%|Medium||
+|0017|Letter Combinations of a Phone Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0017.Letter-Combinations-of-a-Phone-Number)|55.5%|Medium||
+|0018|4Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0018.4Sum)|36.5%|Medium||
+|0019|Remove Nth Node From End of List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0019.Remove-Nth-Node-From-End-of-List)|39.9%|Medium||
+|0020|Valid Parentheses|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0020.Valid-Parentheses)|40.7%|Easy||
+|0021|Merge Two Sorted Lists|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0021.Merge-Two-Sorted-Lists)|61.8%|Easy||
+|0022|Generate Parentheses|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0022.Generate-Parentheses)|71.7%|Medium||
+|0023|Merge k Sorted Lists|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0023.Merge-k-Sorted-Lists)|48.3%|Hard||
+|0024|Swap Nodes in Pairs|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0024.Swap-Nodes-in-Pairs)|60.3%|Medium||
+|0025|Reverse Nodes in k-Group|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0025.Reverse-Nodes-in-k-Group)|53.4%|Hard||
+|0026|Remove Duplicates from Sorted Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0026.Remove-Duplicates-from-Sorted-Array)|50.3%|Easy||
+|0027|Remove Element|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0027.Remove-Element)|52.0%|Easy||
+|0028|Find the Index of the First Occurrence in a String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0028.Find-the-Index-of-the-First-Occurrence-in-a-String)|37.4%|Medium||
+|0029|Divide Two Integers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0029.Divide-Two-Integers)|17.4%|Medium||
+|0030|Substring with Concatenation of All Words|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0030.Substring-with-Concatenation-of-All-Words)|30.9%|Hard||
+|0031|Next Permutation|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0031.Next-Permutation)|37.1%|Medium||
+|0032|Longest Valid Parentheses|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0032.Longest-Valid-Parentheses)|32.7%|Hard||
+|0033|Search in Rotated Sorted Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0033.Search-in-Rotated-Sorted-Array)|38.6%|Medium||
+|0034|Find First and Last Position of Element in Sorted Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array)|41.5%|Medium||
+|0035|Search Insert Position|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0035.Search-Insert-Position)|42.0%|Easy||
+|0036|Valid Sudoku|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0036.Valid-Sudoku)|56.7%|Medium||
+|0037|Sudoku Solver|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0037.Sudoku-Solver)|56.6%|Hard||
+|0038|Count and Say||51.1%|Medium||
+|0039|Combination Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0039.Combination-Sum)|67.5%|Medium||
+|0040|Combination Sum II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0040.Combination-Sum-II)|53.3%|Medium||
+|0041|First Missing Positive|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0041.First-Missing-Positive)|36.5%|Hard||
+|0042|Trapping Rain Water|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0042.Trapping-Rain-Water)|58.7%|Hard||
+|0043|Multiply Strings|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0043.Multiply-Strings)|38.7%|Medium||
+|0044|Wildcard Matching||26.8%|Hard||
+|0045|Jump Game II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0045.Jump-Game-II)|38.5%|Medium||
+|0046|Permutations|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0046.Permutations)|74.6%|Medium||
+|0047|Permutations II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0047.Permutations-II)|56.6%|Medium||
+|0048|Rotate Image|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0048.Rotate-Image)|69.8%|Medium||
+|0049|Group Anagrams|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0049.Group-Anagrams)|65.9%|Medium||
+|0050|Pow(x, n)|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0050.Powx-n)|32.8%|Medium||
+|0051|N-Queens|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0051.N-Queens)|62.8%|Hard||
+|0052|N-Queens II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0052.N-Queens-II)|70.8%|Hard||
+|0053|Maximum Subarray|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0053.Maximum-Subarray)|50.0%|Medium||
+|0054|Spiral Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0054.Spiral-Matrix)|43.6%|Medium||
+|0055|Jump Game|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0055.Jump-Game)|38.4%|Medium||
+|0056|Merge Intervals|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0056.Merge-Intervals)|45.9%|Medium||
+|0057|Insert Interval|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0057.Insert-Interval)|37.9%|Medium||
+|0058|Length of Last Word|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0058.Length-of-Last-Word)|40.3%|Easy||
+|0059|Spiral Matrix II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0059.Spiral-Matrix-II)|66.5%|Medium||
+|0060|Permutation Sequence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0060.Permutation-Sequence)|43.7%|Hard||
+|0061|Rotate List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0061.Rotate-List)|35.7%|Medium||
+|0062|Unique Paths|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0062.Unique-Paths)|62.2%|Medium||
+|0063|Unique Paths II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0063.Unique-Paths-II)|39.1%|Medium||
+|0064|Minimum Path Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0064.Minimum-Path-Sum)|60.6%|Medium||
+|0065|Valid Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0065.Valid-Number)|18.6%|Hard||
+|0066|Plus One|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0066.Plus-One)|43.3%|Easy||
+|0067|Add Binary|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0067.Add-Binary)|51.3%|Easy||
+|0068|Text Justification||36.6%|Hard||
+|0069|Sqrt(x)|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0069.Sqrtx)|37.0%|Easy||
+|0070|Climbing Stairs|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0070.Climbing-Stairs)|51.7%|Easy||
+|0071|Simplify Path|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0071.Simplify-Path)|39.2%|Medium||
+|0072|Edit Distance||52.6%|Hard||
+|0073|Set Matrix Zeroes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0073.Set-Matrix-Zeroes)|49.9%|Medium||
+|0074|Search a 2D Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0074.Search-a-2D-Matrix)|46.7%|Medium||
+|0075|Sort Colors|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0075.Sort-Colors)|57.1%|Medium||
+|0076|Minimum Window Substring|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0076.Minimum-Window-Substring)|40.0%|Hard||
+|0077|Combinations|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0077.Combinations)|66.0%|Medium||
+|0078|Subsets|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0078.Subsets)|73.7%|Medium||
+|0079|Word Search|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0079.Word-Search)|39.8%|Medium||
+|0080|Remove Duplicates from Sorted Array II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0080.Remove-Duplicates-from-Sorted-Array-II)|51.5%|Medium||
+|0081|Search in Rotated Sorted Array II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0081.Search-in-Rotated-Sorted-Array-II)|35.7%|Medium||
+|0082|Remove Duplicates from Sorted List II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0082.Remove-Duplicates-from-Sorted-List-II)|45.4%|Medium||
+|0083|Remove Duplicates from Sorted List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0083.Remove-Duplicates-from-Sorted-List)|49.8%|Easy||
+|0084|Largest Rectangle in Histogram|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0084.Largest-Rectangle-in-Histogram)|42.1%|Hard||
+|0085|Maximal Rectangle||44.1%|Hard||
+|0086|Partition List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0086.Partition-List)|51.3%|Medium||
+|0087|Scramble String||36.1%|Hard||
+|0088|Merge Sorted Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0088.Merge-Sorted-Array)|45.7%|Easy||
+|0089|Gray Code|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0089.Gray-Code)|56.4%|Medium||
+|0090|Subsets II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0090.Subsets-II)|55.2%|Medium||
+|0091|Decode Ways|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0091.Decode-Ways)|32.2%|Medium||
+|0092|Reverse Linked List II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0092.Reverse-Linked-List-II)|45.3%|Medium||
+|0093|Restore IP Addresses|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0093.Restore-IP-Addresses)|43.3%|Medium||
+|0094|Binary Tree Inorder Traversal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0094.Binary-Tree-Inorder-Traversal)|72.9%|Easy||
+|0095|Unique Binary Search Trees II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0095.Unique-Binary-Search-Trees-II)|51.4%|Medium||
+|0096|Unique Binary Search Trees|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0096.Unique-Binary-Search-Trees)|59.2%|Medium||
+|0097|Interleaving String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0097.Interleaving-String)|37.1%|Medium||
+|0098|Validate Binary Search Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0098.Validate-Binary-Search-Tree)|31.7%|Medium||
+|0099|Recover Binary Search Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0099.Recover-Binary-Search-Tree)|50.2%|Medium||
+|0100|Same Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0100.Same-Tree)|56.3%|Easy||
+|0101|Symmetric Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0101.Symmetric-Tree)|52.8%|Easy||
+|0102|Binary Tree Level Order Traversal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0102.Binary-Tree-Level-Order-Traversal)|63.2%|Medium||
+|0103|Binary Tree Zigzag Level Order Traversal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0103.Binary-Tree-Zigzag-Level-Order-Traversal)|55.0%|Medium||
+|0104|Maximum Depth of Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0104.Maximum-Depth-of-Binary-Tree)|73.0%|Easy||
+|0105|Construct Binary Tree from Preorder and Inorder Traversal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal)|60.6%|Medium||
+|0106|Construct Binary Tree from Inorder and Postorder Traversal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal)|57.3%|Medium||
+|0107|Binary Tree Level Order Traversal II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0107.Binary-Tree-Level-Order-Traversal-II)|60.2%|Medium||
+|0108|Convert Sorted Array to Binary Search Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0108.Convert-Sorted-Array-to-Binary-Search-Tree)|68.9%|Easy||
+|0109|Convert Sorted List to Binary Search Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0109.Convert-Sorted-List-to-Binary-Search-Tree)|57.2%|Medium||
+|0110|Balanced Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0110.Balanced-Binary-Tree)|48.1%|Easy||
+|0111|Minimum Depth of Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0111.Minimum-Depth-of-Binary-Tree)|43.5%|Easy||
+|0112|Path Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0112.Path-Sum)|47.6%|Easy||
+|0113|Path Sum II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0113.Path-Sum-II)|56.5%|Medium||
+|0114|Flatten Binary Tree to Linked List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0114.Flatten-Binary-Tree-to-Linked-List)|61.0%|Medium||
+|0115|Distinct Subsequences|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0115.Distinct-Subsequences)|43.7%|Hard||
+|0116|Populating Next Right Pointers in Each Node|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0116.Populating-Next-Right-Pointers-in-Each-Node)|59.3%|Medium||
+|0117|Populating Next Right Pointers in Each Node II||49.6%|Medium||
+|0118|Pascal's Triangle|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0118.Pascals-Triangle)|68.9%|Easy||
+|0119|Pascal's Triangle II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0119.Pascals-Triangle-II)|59.5%|Easy||
+|0120|Triangle|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0120.Triangle)|53.8%|Medium||
+|0121|Best Time to Buy and Sell Stock|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0121.Best-Time-to-Buy-and-Sell-Stock)|54.5%|Easy||
+|0122|Best Time to Buy and Sell Stock II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0122.Best-Time-to-Buy-and-Sell-Stock-II)|63.2%|Medium||
+|0123|Best Time to Buy and Sell Stock III||44.8%|Hard||
+|0124|Binary Tree Maximum Path Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0124.Binary-Tree-Maximum-Path-Sum)|38.4%|Hard||
+|0125|Valid Palindrome|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0125.Valid-Palindrome)|43.5%|Easy||
+|0126|Word Ladder II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0126.Word-Ladder-II)|27.6%|Hard||
+|0127|Word Ladder|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0127.Word-Ladder)|36.6%|Hard||
+|0128|Longest Consecutive Sequence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0128.Longest-Consecutive-Sequence)|49.0%|Medium||
+|0129|Sum Root to Leaf Numbers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0129.Sum-Root-to-Leaf-Numbers)|58.6%|Medium||
+|0130|Surrounded Regions|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0130.Surrounded-Regions)|35.8%|Medium||
+|0131|Palindrome Partitioning|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0131.Palindrome-Partitioning)|62.3%|Medium||
+|0132|Palindrome Partitioning II||33.6%|Hard||
+|0133|Clone Graph||50.5%|Medium||
+|0134|Gas Station||45.0%|Medium||
+|0135|Candy|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0135.Candy)|40.7%|Hard||
+|0136|Single Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0136.Single-Number)|70.0%|Easy||
+|0137|Single Number II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0137.Single-Number-II)|57.7%|Medium||
+|0138|Copy List with Random Pointer|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0138.Copy-List-with-Random-Pointer)|50.4%|Medium||
+|0139|Word Break||45.4%|Medium||
+|0140|Word Break II||44.3%|Hard||
+|0141|Linked List Cycle|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0141.Linked-List-Cycle)|46.8%|Easy||
+|0142|Linked List Cycle II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0142.Linked-List-Cycle-II)|46.2%|Medium||
+|0143|Reorder List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0143.Reorder-List)|50.9%|Medium||
+|0144|Binary Tree Preorder Traversal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0144.Binary-Tree-Preorder-Traversal)|64.5%|Easy||
+|0145|Binary Tree Postorder Traversal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0145.Binary-Tree-Postorder-Traversal)|66.5%|Easy||
+|0146|LRU Cache|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0146.LRU-Cache)|40.5%|Medium||
+|0147|Insertion Sort List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0147.Insertion-Sort-List)|50.1%|Medium||
+|0148|Sort List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0148.Sort-List)|54.0%|Medium||
+|0149|Max Points on a Line||21.7%|Hard||
+|0150|Evaluate Reverse Polish Notation|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0150.Evaluate-Reverse-Polish-Notation)|44.1%|Medium||
+|0151|Reverse Words in a String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0151.Reverse-Words-in-a-String)|30.2%|Medium||
+|0152|Maximum Product Subarray|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0152.Maximum-Product-Subarray)|34.9%|Medium||
+|0153|Find Minimum in Rotated Sorted Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0153.Find-Minimum-in-Rotated-Sorted-Array)|48.4%|Medium||
+|0154|Find Minimum in Rotated Sorted Array II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0154.Find-Minimum-in-Rotated-Sorted-Array-II)|43.4%|Hard||
+|0155|Min Stack|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0155.Min-Stack)|51.7%|Medium||
+|0156|Binary Tree Upside Down||61.5%|Medium||
+|0157|Read N Characters Given Read4||40.7%|Easy||
+|0158|Read N Characters Given read4 II - Call Multiple Times||41.4%|Hard||
+|0159|Longest Substring with At Most Two Distinct Characters||53.5%|Medium||
+|0160|Intersection of Two Linked Lists|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0160.Intersection-of-Two-Linked-Lists)|53.1%|Easy||
+|0161|One Edit Distance||34.1%|Medium||
+|0162|Find Peak Element|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0162.Find-Peak-Element)|46.2%|Medium||
+|0163|Missing Ranges||31.9%|Easy||
+|0164|Maximum Gap|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0164.Maximum-Gap)|42.6%|Hard||
+|0165|Compare Version Numbers||35.3%|Medium||
+|0166|Fraction to Recurring Decimal||24.0%|Medium||
+|0167|Two Sum II - Input Array Is Sorted|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0167.Two-Sum-II-Input-Array-Is-Sorted)|60.0%|Medium||
+|0168|Excel Sheet Column Title|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0168.Excel-Sheet-Column-Title)|34.7%|Easy||
+|0169|Majority Element|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0169.Majority-Element)|63.8%|Easy||
+|0170|Two Sum III - Data structure design||37.3%|Easy||
+|0171|Excel Sheet Column Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0171.Excel-Sheet-Column-Number)|61.3%|Easy||
+|0172|Factorial Trailing Zeroes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0172.Factorial-Trailing-Zeroes)|41.6%|Medium||
+|0173|Binary Search Tree Iterator|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0173.Binary-Search-Tree-Iterator)|69.0%|Medium||
+|0174|Dungeon Game|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0174.Dungeon-Game)|37.2%|Hard||
+|0175|Combine Two Tables||72.8%|Easy||
+|0176|Second Highest Salary||36.4%|Medium||
+|0177|Nth Highest Salary||37.3%|Medium||
+|0178|Rank Scores||59.7%|Medium||
+|0179|Largest Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0179.Largest-Number)|33.9%|Medium||
+|0180|Consecutive Numbers||46.7%|Medium||
+|0181|Employees Earning More Than Their Managers||68.4%|Easy||
+|0182|Duplicate Emails||70.5%|Easy||
+|0183|Customers Who Never Order||67.6%|Easy||
+|0184|Department Highest Salary||49.5%|Medium||
+|0185|Department Top Three Salaries||49.9%|Hard||
+|0186|Reverse Words in a String II||52.3%|Medium||
+|0187|Repeated DNA Sequences|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0187.Repeated-DNA-Sequences)|46.1%|Medium||
+|0188|Best Time to Buy and Sell Stock IV||37.8%|Hard||
+|0189|Rotate Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0189.Rotate-Array)|39.2%|Medium||
+|0190|Reverse Bits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0190.Reverse-Bits)|51.9%|Easy||
+|0191|Number of 1 Bits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0191.Number-of-1-Bits)|64.5%|Easy||
+|0192|Word Frequency||25.6%|Medium||
+|0193|Valid Phone Numbers||26.0%|Easy||
+|0194|Transpose File||25.3%|Medium||
+|0195|Tenth Line||32.9%|Easy||
+|0196|Delete Duplicate Emails||58.3%|Easy||
+|0197|Rising Temperature||44.4%|Easy||
+|0198|House Robber|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0198.House-Robber)|48.6%|Medium||
+|0199|Binary Tree Right Side View|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0199.Binary-Tree-Right-Side-View)|61.1%|Medium||
+|0200|Number of Islands|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0200.Number-of-Islands)|56.1%|Medium||
+|0201|Bitwise AND of Numbers Range|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0201.Bitwise-AND-of-Numbers-Range)|42.2%|Medium||
+|0202|Happy Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0202.Happy-Number)|54.3%|Easy||
+|0203|Remove Linked List Elements|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0203.Remove-Linked-List-Elements)|44.7%|Easy||
+|0204|Count Primes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0204.Count-Primes)|33.1%|Medium||
+|0205|Isomorphic Strings|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0205.Isomorphic-Strings)|42.5%|Easy||
+|0206|Reverse Linked List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0206.Reverse-Linked-List)|72.3%|Easy||
+|0207|Course Schedule|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0207.Course-Schedule)|45.3%|Medium||
+|0208|Implement Trie (Prefix Tree)|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0208.Implement-Trie-Prefix-Tree)|60.7%|Medium||
+|0209|Minimum Size Subarray Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0209.Minimum-Size-Subarray-Sum)|44.4%|Medium||
+|0210|Course Schedule II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0210.Course-Schedule-II)|47.9%|Medium||
+|0211|Design Add and Search Words Data Structure|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0211.Design-Add-and-Search-Words-Data-Structure)|43.2%|Medium||
+|0212|Word Search II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0212.Word-Search-II)|37.0%|Hard||
+|0213|House Robber II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0213.House-Robber-II)|40.6%|Medium||
+|0214|Shortest Palindrome||32.2%|Hard||
+|0215|Kth Largest Element in an Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0215.Kth-Largest-Element-in-an-Array)|65.7%|Medium||
+|0216|Combination Sum III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0216.Combination-Sum-III)|67.0%|Medium||
+|0217|Contains Duplicate|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0217.Contains-Duplicate)|61.2%|Easy||
+|0218|The Skyline Problem|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0218.The-Skyline-Problem)|41.5%|Hard||
+|0219|Contains Duplicate II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0219.Contains-Duplicate-II)|40.9%|Easy||
+|0220|Contains Duplicate III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0220.Contains-Duplicate-III)|21.8%|Hard||
+|0221|Maximal Square||44.4%|Medium||
+|0222|Count Complete Tree Nodes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0222.Count-Complete-Tree-Nodes)|57.4%|Medium||
+|0223|Rectangle Area|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0223.Rectangle-Area)|40.8%|Medium||
+|0224|Basic Calculator|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0224.Basic-Calculator)|41.1%|Hard||
+|0225|Implement Stack using Queues|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0225.Implement-Stack-using-Queues)|57.4%|Easy||
+|0226|Invert Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0226.Invert-Binary-Tree)|73.1%|Easy||
+|0227|Basic Calculator II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0227.Basic-Calculator-II)|42.2%|Medium||
+|0228|Summary Ranges|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0228.Summary-Ranges)|46.8%|Easy||
+|0229|Majority Element II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0229.Majority-Element-II)|44.0%|Medium||
+|0230|Kth Smallest Element in a BST|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0230.Kth-Smallest-Element-in-a-BST)|69.2%|Medium||
+|0231|Power of Two|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0231.Power-of-Two)|45.6%|Easy||
+|0232|Implement Queue using Stacks|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0232.Implement-Queue-using-Stacks)|61.0%|Easy||
+|0233|Number of Digit One||34.2%|Hard||
+|0234|Palindrome Linked List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0234.Palindrome-Linked-List)|49.4%|Easy||
+|0235|Lowest Common Ancestor of a Binary Search Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree)|60.1%|Medium||
+|0236|Lowest Common Ancestor of a Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0236.Lowest-Common-Ancestor-of-a-Binary-Tree)|57.9%|Medium||
+|0237|Delete Node in a Linked List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0237.Delete-Node-in-a-Linked-List)|75.1%|Medium||
+|0238|Product of Array Except Self||64.7%|Medium||
+|0239|Sliding Window Maximum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0239.Sliding-Window-Maximum)|46.6%|Hard||
+|0240|Search a 2D Matrix II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0240.Search-a-2D-Matrix-II)|50.5%|Medium||
+|0241|Different Ways to Add Parentheses||63.1%|Medium||
+|0242|Valid Anagram|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0242.Valid-Anagram)|62.7%|Easy||
+|0243|Shortest Word Distance||64.9%|Easy||
+|0244|Shortest Word Distance II||60.7%|Medium||
+|0245|Shortest Word Distance III||57.5%|Medium||
+|0246|Strobogrammatic Number||47.6%|Easy||
+|0247|Strobogrammatic Number II||51.4%|Medium||
+|0248|Strobogrammatic Number III||41.7%|Hard||
+|0249|Group Shifted Strings||64.2%|Medium||
+|0250|Count Univalue Subtrees||55.2%|Medium||
+|0251|Flatten 2D Vector||48.9%|Medium||
+|0252|Meeting Rooms||57.0%|Easy||
+|0253|Meeting Rooms II||50.4%|Medium||
+|0254|Factor Combinations||48.8%|Medium||
+|0255|Verify Preorder Sequence in Binary Search Tree||48.0%|Medium||
+|0256|Paint House||60.5%|Medium||
+|0257|Binary Tree Paths|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0257.Binary-Tree-Paths)|60.5%|Easy||
+|0258|Add Digits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0258.Add-Digits)|63.3%|Easy||
+|0259|3Sum Smaller||50.7%|Medium||
+|0260|Single Number III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0260.Single-Number-III)|67.4%|Medium||
+|0261|Graph Valid Tree||46.8%|Medium||
+|0262|Trips and Users||38.4%|Hard||
+|0263|Ugly Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0263.Ugly-Number)|41.7%|Easy||
+|0264|Ugly Number II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0264.Ugly-Number-II)|46.1%|Medium||
+|0265|Paint House II||52.2%|Hard||
+|0266|Palindrome Permutation||65.8%|Easy||
+|0267|Palindrome Permutation II||40.4%|Medium||
+|0268|Missing Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0268.Missing-Number)|61.5%|Easy||
+|0269|Alien Dictionary||35.2%|Hard||
+|0270|Closest Binary Search Tree Value||54.5%|Easy||
+|0271|Encode and Decode Strings||41.3%|Medium||
+|0272|Closest Binary Search Tree Value II||58.2%|Hard||
+|0273|Integer to English Words||29.9%|Hard||
+|0274|H-Index|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0274.H-Index)|38.1%|Medium||
+|0275|H-Index II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0275.H-Index-II)|37.4%|Medium||
+|0276|Paint Fence||44.0%|Medium||
+|0277|Find the Celebrity||46.7%|Medium||
+|0278|First Bad Version|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0278.First-Bad-Version)|42.9%|Easy||
+|0279|Perfect Squares|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0279.Perfect-Squares)|52.2%|Medium||
+|0280|Wiggle Sort||66.3%|Medium||
+|0281|Zigzag Iterator||62.3%|Medium||
+|0282|Expression Add Operators||39.2%|Hard||
+|0283|Move Zeroes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0283.Move-Zeroes)|61.3%|Easy||
+|0284|Peeking Iterator|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0284.Peeking-Iterator)|58.2%|Medium||
+|0285|Inorder Successor in BST||48.3%|Medium||
+|0286|Walls and Gates||60.2%|Medium||
+|0287|Find the Duplicate Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0287.Find-the-Duplicate-Number)|59.1%|Medium||
+|0288|Unique Word Abbreviation||25.2%|Medium||
+|0289|Game of Life||66.6%|Medium||
+|0290|Word Pattern|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0290.Word-Pattern)|40.4%|Easy||
+|0291|Word Pattern II||46.9%|Medium||
+|0292|Nim Game||55.8%|Easy||
+|0293|Flip Game||63.0%|Easy||
+|0294|Flip Game II||51.7%|Medium||
+|0295|Find Median from Data Stream||51.1%|Hard||
+|0296|Best Meeting Point||59.6%|Hard||
+|0297|Serialize and Deserialize Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0297.Serialize-and-Deserialize-Binary-Tree)|54.9%|Hard||
+|0298|Binary Tree Longest Consecutive Sequence||52.5%|Medium||
+|0299|Bulls and Cows|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0299.Bulls-and-Cows)|48.5%|Medium||
+|0300|Longest Increasing Subsequence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0300.Longest-Increasing-Subsequence)|51.5%|Medium||
+|0301|Remove Invalid Parentheses|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0301.Remove-Invalid-Parentheses)|47.1%|Hard||
+|0302|Smallest Rectangle Enclosing Black Pixels||58.2%|Hard||
+|0303|Range Sum Query - Immutable|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0303.Range-Sum-Query-Immutable)|57.9%|Easy||
+|0304|Range Sum Query 2D - Immutable|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0304.Range-Sum-Query-2D-Immutable)|52.1%|Medium||
+|0305|Number of Islands II||39.5%|Hard||
+|0306|Additive Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0306.Additive-Number)|30.9%|Medium||
+|0307|Range Sum Query - Mutable|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0307.Range-Sum-Query-Mutable)|40.7%|Medium||
+|0308|Range Sum Query 2D - Mutable||42.2%|Hard||
+|0309|Best Time to Buy and Sell Stock with Cooldown|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown)|54.4%|Medium||
+|0310|Minimum Height Trees||38.5%|Medium||
+|0311|Sparse Matrix Multiplication||67.1%|Medium||
+|0312|Burst Balloons||56.9%|Hard||
+|0313|Super Ugly Number||45.8%|Medium||
+|0314|Binary Tree Vertical Order Traversal||52.0%|Medium||
+|0315|Count of Smaller Numbers After Self|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0315.Count-of-Smaller-Numbers-After-Self)|42.8%|Hard||
+|0316|Remove Duplicate Letters||44.5%|Medium||
+|0317|Shortest Distance from All Buildings||43.0%|Hard||
+|0318|Maximum Product of Word Lengths|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0318.Maximum-Product-of-Word-Lengths)|60.1%|Medium||
+|0319|Bulb Switcher|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0319.Bulb-Switcher)|48.0%|Medium||
+|0320|Generalized Abbreviation||57.3%|Medium||
+|0321|Create Maximum Number||28.8%|Hard||
+|0322|Coin Change|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0322.Coin-Change)|41.5%|Medium||
+|0323|Number of Connected Components in an Undirected Graph||62.0%|Medium||
+|0324|Wiggle Sort II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0324.Wiggle-Sort-II)|32.9%|Medium||
+|0325|Maximum Size Subarray Sum Equals k||49.3%|Medium||
+|0326|Power of Three|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0326.Power-of-Three)|45.2%|Easy||
+|0327|Count of Range Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0327.Count-of-Range-Sum)|36.0%|Hard||
+|0328|Odd Even Linked List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0328.Odd-Even-Linked-List)|60.2%|Medium||
+|0329|Longest Increasing Path in a Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0329.Longest-Increasing-Path-in-a-Matrix)|52.1%|Hard||
+|0330|Patching Array||40.0%|Hard||
+|0331|Verify Preorder Serialization of a Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0331.Verify-Preorder-Serialization-of-a-Binary-Tree)|44.2%|Medium||
+|0332|Reconstruct Itinerary||40.9%|Hard||
+|0333|Largest BST Subtree||42.3%|Medium||
+|0334|Increasing Triplet Subsequence||42.7%|Medium||
+|0335|Self Crossing||29.3%|Hard||
+|0336|Palindrome Pairs||35.2%|Hard||
+|0337|House Robber III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0337.House-Robber-III)|53.8%|Medium||
+|0338|Counting Bits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0338.Counting-Bits)|75.2%|Easy||
+|0339|Nested List Weight Sum||82.0%|Medium||
+|0340|Longest Substring with At Most K Distinct Characters||47.8%|Medium||
+|0341|Flatten Nested List Iterator|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0341.Flatten-Nested-List-Iterator)|61.5%|Medium||
+|0342|Power of Four|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0342.Power-of-Four)|45.6%|Easy||
+|0343|Integer Break|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0343.Integer-Break)|55.3%|Medium||
+|0344|Reverse String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0344.Reverse-String)|76.1%|Easy||
+|0345|Reverse Vowels of a String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0345.Reverse-Vowels-of-a-String)|47.7%|Easy||
+|0346|Moving Average from Data Stream||77.0%|Easy||
+|0347|Top K Frequent Elements|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0347.Top-K-Frequent-Elements)|64.8%|Medium||
+|0348|Design Tic-Tac-Toe||57.6%|Medium||
+|0349|Intersection of Two Arrays|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0349.Intersection-of-Two-Arrays)|70.2%|Easy||
+|0350|Intersection of Two Arrays II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0350.Intersection-of-Two-Arrays-II)|55.5%|Easy||
+|0351|Android Unlock Patterns||51.3%|Medium||
+|0352|Data Stream as Disjoint Intervals|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0352.Data-Stream-as-Disjoint-Intervals)|51.5%|Hard||
+|0353|Design Snake Game||39.0%|Medium||
+|0354|Russian Doll Envelopes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0354.Russian-Doll-Envelopes)|38.3%|Hard||
+|0355|Design Twitter||36.5%|Medium||
+|0356|Line Reflection||34.6%|Medium||
+|0357|Count Numbers with Unique Digits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0357.Count-Numbers-with-Unique-Digits)|51.5%|Medium||
+|0358|Rearrange String k Distance Apart||37.5%|Hard||
+|0359|Logger Rate Limiter||75.5%|Easy||
+|0360|Sort Transformed Array||54.6%|Medium||
+|0361|Bomb Enemy||50.9%|Medium||
+|0362|Design Hit Counter||68.3%|Medium||
+|0363|Max Sum of Rectangle No Larger Than K||44.1%|Hard||
+|0364|Nested List Weight Sum II||67.3%|Medium||
+|0365|Water and Jug Problem||36.5%|Medium||
+|0366|Find Leaves of Binary Tree||80.1%|Medium||
+|0367|Valid Perfect Square|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0367.Valid-Perfect-Square)|43.3%|Easy||
+|0368|Largest Divisible Subset|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0368.Largest-Divisible-Subset)|41.2%|Medium||
+|0369|Plus One Linked List||60.9%|Medium||
+|0370|Range Addition||70.8%|Medium||
+|0371|Sum of Two Integers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0371.Sum-of-Two-Integers)|50.7%|Medium||
+|0372|Super Pow|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0372.Super-Pow)|37.2%|Medium||
+|0373|Find K Pairs with Smallest Sums|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0373.Find-K-Pairs-with-Smallest-Sums)|38.4%|Medium||
+|0374|Guess Number Higher or Lower|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0374.Guess-Number-Higher-or-Lower)|50.4%|Easy||
+|0375|Guess Number Higher or Lower II||46.3%|Medium||
+|0376|Wiggle Subsequence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0376.Wiggle-Subsequence)|48.2%|Medium||
+|0377|Combination Sum IV|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0377.Combination-Sum-IV)|52.1%|Medium||
+|0378|Kth Smallest Element in a Sorted Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0378.Kth-Smallest-Element-in-a-Sorted-Matrix)|61.6%|Medium||
+|0379|Design Phone Directory||50.9%|Medium||
+|0380|Insert Delete GetRandom O(1)||52.0%|Medium||
+|0381|Insert Delete GetRandom O(1) - Duplicates allowed||35.7%|Hard||
+|0382|Linked List Random Node|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0382.Linked-List-Random-Node)|59.5%|Medium||
+|0383|Ransom Note|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0383.Ransom-Note)|57.6%|Easy||
+|0384|Shuffle an Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0384.Shuffle-an-Array)|57.6%|Medium||
+|0385|Mini Parser|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0385.Mini-Parser)|36.5%|Medium||
+|0386|Lexicographical Numbers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0386.Lexicographical-Numbers)|60.5%|Medium||
+|0387|First Unique Character in a String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0387.First-Unique-Character-in-a-String)|58.8%|Easy||
+|0388|Longest Absolute File Path||46.5%|Medium||
+|0389|Find the Difference|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0389.Find-the-Difference)|60.4%|Easy||
+|0390|Elimination Game|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0390.Elimination-Game)|46.6%|Medium||
+|0391|Perfect Rectangle|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0391.Perfect-Rectangle)|32.5%|Hard||
+|0392|Is Subsequence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0392.Is-Subsequence)|49.3%|Easy||
+|0393|UTF-8 Validation|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0393.UTF-8-Validation)|45.2%|Medium||
+|0394|Decode String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0394.Decode-String)|57.5%|Medium||
+|0395|Longest Substring with At Least K Repeating Characters|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters)|44.8%|Medium||
+|0396|Rotate Function|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0396.Rotate-Function)|40.3%|Medium||
+|0397|Integer Replacement|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0397.Integer-Replacement)|35.1%|Medium||
+|0398|Random Pick Index||62.9%|Medium||
+|0399|Evaluate Division|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0399.Evaluate-Division)|59.5%|Medium||
+|0400|Nth Digit|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0400.Nth-Digit)|34.0%|Medium||
+|0401|Binary Watch|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0401.Binary-Watch)|51.4%|Easy||
+|0402|Remove K Digits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0402.Remove-K-Digits)|30.5%|Medium||
+|0403|Frog Jump||43.1%|Hard||
+|0404|Sum of Left Leaves|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0404.Sum-of-Left-Leaves)|56.2%|Easy||
+|0405|Convert a Number to Hexadecimal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0405.Convert-a-Number-to-Hexadecimal)|46.1%|Easy||
+|0406|Queue Reconstruction by Height||72.8%|Medium||
+|0407|Trapping Rain Water II||47.4%|Hard||
+|0408|Valid Word Abbreviation||34.8%|Easy||
+|0409|Longest Palindrome|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0409.Longest-Palindrome)|54.6%|Easy||
+|0410|Split Array Largest Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0410.Split-Array-Largest-Sum)|53.2%|Hard||
+|0411|Minimum Unique Word Abbreviation||39.2%|Hard||
+|0412|Fizz Buzz|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0412.Fizz-Buzz)|68.6%|Easy||
+|0413|Arithmetic Slices|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0413.Arithmetic-Slices)|65.0%|Medium||
+|0414|Third Maximum Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0414.Third-Maximum-Number)|32.5%|Easy||
+|0415|Add Strings||52.6%|Easy||
+|0416|Partition Equal Subset Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0416.Partition-Equal-Subset-Sum)|46.7%|Medium||
+|0417|Pacific Atlantic Water Flow|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0417.Pacific-Atlantic-Water-Flow)|53.9%|Medium||
+|0418|Sentence Screen Fitting||35.6%|Medium||
+|0419|Battleships in a Board|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0419.Battleships-in-a-Board)|74.6%|Medium||
+|0420|Strong Password Checker||14.3%|Hard||
+|0421|Maximum XOR of Two Numbers in an Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0421.Maximum-XOR-of-Two-Numbers-in-an-Array)|54.6%|Medium||
+|0422|Valid Word Square||38.8%|Easy||
+|0423|Reconstruct Original Digits from English|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0423.Reconstruct-Original-Digits-from-English)|51.3%|Medium||
+|0424|Longest Repeating Character Replacement|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0424.Longest-Repeating-Character-Replacement)|51.4%|Medium||
+|0425|Word Squares||52.6%|Hard||
+|0426|Convert Binary Search Tree to Sorted Doubly Linked List||64.6%|Medium||
+|0427|Construct Quad Tree||66.3%|Medium||
+|0428|Serialize and Deserialize N-ary Tree||65.3%|Hard||
+|0429|N-ary Tree Level Order Traversal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0429.N-ary-Tree-Level-Order-Traversal)|70.6%|Medium||
+|0430|Flatten a Multilevel Doubly Linked List||59.4%|Medium||
+|0431|Encode N-ary Tree to Binary Tree||78.6%|Hard||
+|0432|All O`one Data Structure||36.6%|Hard||
+|0433|Minimum Genetic Mutation|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0433.Minimum-Genetic-Mutation)|48.1%|Medium||
+|0434|Number of Segments in a String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0434.Number-of-Segments-in-a-String)|37.7%|Easy||
+|0435|Non-overlapping Intervals|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0435.Non-overlapping-Intervals)|49.8%|Medium||
+|0436|Find Right Interval|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0436.Find-Right-Interval)|50.3%|Medium||
+|0437|Path Sum III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0437.Path-Sum-III)|48.8%|Medium||
+|0438|Find All Anagrams in a String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0438.Find-All-Anagrams-in-a-String)|48.9%|Medium||
+|0439|Ternary Expression Parser||58.2%|Medium||
+|0440|K-th Smallest in Lexicographical Order||30.7%|Hard||
+|0441|Arranging Coins|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0441.Arranging-Coins)|46.0%|Easy||
+|0442|Find All Duplicates in an Array||73.3%|Medium||
+|0443|String Compression||48.7%|Medium||
+|0444|Sequence Reconstruction||26.3%|Medium||
+|0445|Add Two Numbers II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0445.Add-Two-Numbers-II)|59.4%|Medium||
+|0446|Arithmetic Slices II - Subsequence||39.8%|Hard||
+|0447|Number of Boomerangs|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0447.Number-of-Boomerangs)|54.6%|Medium||
+|0448|Find All Numbers Disappeared in an Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array)|59.6%|Easy||
+|0449|Serialize and Deserialize BST||56.7%|Medium||
+|0450|Delete Node in a BST||49.9%|Medium||
+|0451|Sort Characters By Frequency|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0451.Sort-Characters-By-Frequency)|68.5%|Medium||
+|0452|Minimum Number of Arrows to Burst Balloons||53.2%|Medium||
+|0453|Minimum Moves to Equal Array Elements|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements)|55.6%|Medium||
+|0454|4Sum II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0454.4Sum-II)|57.3%|Medium||
+|0455|Assign Cookies|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0455.Assign-Cookies)|50.6%|Easy||
+|0456|132 Pattern|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0456.132-Pattern)|32.4%|Medium||
+|0457|Circular Array Loop|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0457.Circular-Array-Loop)|32.1%|Medium||
+|0458|Poor Pigs|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0458.Poor-Pigs)|65.0%|Hard||
+|0459|Repeated Substring Pattern||43.7%|Easy||
+|0460|LFU Cache|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0460.LFU-Cache)|40.3%|Hard||
+|0461|Hamming Distance|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0461.Hamming-Distance)|74.8%|Easy||
+|0462|Minimum Moves to Equal Array Elements II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0462.Minimum-Moves-to-Equal-Array-Elements-II)|60.1%|Medium||
+|0463|Island Perimeter|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0463.Island-Perimeter)|69.4%|Easy||
+|0464|Can I Win||29.8%|Medium||
+|0465|Optimal Account Balancing||49.3%|Hard||
+|0466|Count The Repetitions||29.2%|Hard||
+|0467|Unique Substrings in Wraparound String||38.2%|Medium||
+|0468|Validate IP Address||26.5%|Medium||
+|0469|Convex Polygon||38.5%|Medium||
+|0470|Implement Rand10() Using Rand7()|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0470.Implement-Rand10-Using-Rand7)|46.7%|Medium||
+|0471|Encode String with Shortest Length||50.7%|Hard||
+|0472|Concatenated Words||45.5%|Hard||
+|0473|Matchsticks to Square|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0473.Matchsticks-to-Square)|40.5%|Medium||
+|0474|Ones and Zeroes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0474.Ones-and-Zeroes)|46.7%|Medium||
+|0475|Heaters|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0475.Heaters)|36.0%|Medium||
+|0476|Number Complement|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0476.Number-Complement)|67.1%|Easy||
+|0477|Total Hamming Distance|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0477.Total-Hamming-Distance)|52.2%|Medium||
+|0478|Generate Random Point in a Circle|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0478.Generate-Random-Point-in-a-Circle)|39.6%|Medium||
+|0479|Largest Palindrome Product||31.6%|Hard||
+|0480|Sliding Window Median|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0480.Sliding-Window-Median)|41.4%|Hard||
+|0481|Magical String||50.4%|Medium||
+|0482|License Key Formatting||43.2%|Easy||
+|0483|Smallest Good Base|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0483.Smallest-Good-Base)|38.4%|Hard||
+|0484|Find Permutation||67.0%|Medium||
+|0485|Max Consecutive Ones|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0485.Max-Consecutive-Ones)|56.0%|Easy||
+|0486|Predict the Winner||50.8%|Medium||
+|0487|Max Consecutive Ones II||49.2%|Medium||
+|0488|Zuma Game|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0488.Zuma-Game)|34.9%|Hard||
+|0489|Robot Room Cleaner||76.5%|Hard||
+|0490|The Maze||55.4%|Medium||
+|0491|Increasing Subsequences|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0491.Increasing-Subsequences)|52.0%|Medium||
+|0492|Construct the Rectangle|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0492.Construct-the-Rectangle)|53.6%|Easy||
+|0493|Reverse Pairs|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0493.Reverse-Pairs)|30.8%|Hard||
+|0494|Target Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0494.Target-Sum)|45.6%|Medium||
+|0495|Teemo Attacking|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0495.Teemo-Attacking)|57.0%|Easy||
+|0496|Next Greater Element I|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0496.Next-Greater-Element-I)|71.3%|Easy||
+|0497|Random Point in Non-overlapping Rectangles|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0497.Random-Point-in-Non-overlapping-Rectangles)|39.3%|Medium||
+|0498|Diagonal Traverse|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0498.Diagonal-Traverse)|58.0%|Medium||
+|0499|The Maze III||46.9%|Hard||
+|0500|Keyboard Row|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0500.Keyboard-Row)|69.0%|Easy||
+|0501|Find Mode in Binary Search Tree||48.5%|Easy||
+|0502|IPO||44.9%|Hard||
+|0503|Next Greater Element II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0503.Next-Greater-Element-II)|63.0%|Medium||
+|0504|Base 7|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0504.Base-7)|47.9%|Easy||
+|0505|The Maze II||52.4%|Medium||
+|0506|Relative Ranks|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0506.Relative-Ranks)|58.8%|Easy||
+|0507|Perfect Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0507.Perfect-Number)|37.7%|Easy||
+|0508|Most Frequent Subtree Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0508.Most-Frequent-Subtree-Sum)|64.2%|Medium||
+|0509|Fibonacci Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0509.Fibonacci-Number)|69.2%|Easy||
+|0510|Inorder Successor in BST II||61.1%|Medium||
+|0511|Game Play Analysis I||78.8%|Easy||
+|0512|Game Play Analysis II||54.1%|Easy||
+|0513|Find Bottom Left Tree Value|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0513.Find-Bottom-Left-Tree-Value)|66.3%|Medium||
+|0514|Freedom Trail||46.8%|Hard||
+|0515|Find Largest Value in Each Tree Row|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0515.Find-Largest-Value-in-Each-Tree-Row)|64.6%|Medium||
+|0516|Longest Palindromic Subsequence||60.6%|Medium||
+|0517|Super Washing Machines||39.7%|Hard||
+|0518|Coin Change II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0518.Coin-Change-II)|59.7%|Medium||
+|0519|Random Flip Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0519.Random-Flip-Matrix)|39.6%|Medium||
+|0520|Detect Capital|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0520.Detect-Capital)|55.6%|Easy||
+|0521|Longest Uncommon Subsequence I||60.3%|Easy||
+|0522|Longest Uncommon Subsequence II||40.4%|Medium||
+|0523|Continuous Subarray Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0523.Continuous-Subarray-Sum)|27.7%|Medium||
+|0524|Longest Word in Dictionary through Deleting|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0524.Longest-Word-in-Dictionary-through-Deleting)|51.2%|Medium||
+|0525|Contiguous Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0525.Contiguous-Array)|46.8%|Medium||
+|0526|Beautiful Arrangement|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0526.Beautiful-Arrangement)|64.6%|Medium||
+|0527|Word Abbreviation||60.3%|Hard||
+|0528|Random Pick with Weight|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0528.Random-Pick-with-Weight)|46.1%|Medium||
+|0529|Minesweeper|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0529.Minesweeper)|65.5%|Medium||
+|0530|Minimum Absolute Difference in BST|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0530.Minimum-Absolute-Difference-in-BST)|56.7%|Easy||
+|0531|Lonely Pixel I||62.0%|Medium||
+|0532|K-diff Pairs in an Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0532.K-diff-Pairs-in-an-Array)|40.7%|Medium||
+|0533|Lonely Pixel II||48.4%|Medium||
+|0534|Game Play Analysis III||82.5%|Medium||
+|0535|Encode and Decode TinyURL|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0535.Encode-and-Decode-TinyURL)|85.7%|Medium||
+|0536|Construct Binary Tree from String||56.1%|Medium||
+|0537|Complex Number Multiplication|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0537.Complex-Number-Multiplication)|71.3%|Medium||
+|0538|Convert BST to Greater Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0538.Convert-BST-to-Greater-Tree)|67.3%|Medium||
+|0539|Minimum Time Difference||56.3%|Medium||
+|0540|Single Element in a Sorted Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0540.Single-Element-in-a-Sorted-Array)|58.5%|Medium||
+|0541|Reverse String II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0541.Reverse-String-II)|50.5%|Easy||
+|0542|01 Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0542.01-Matrix)|44.1%|Medium||
+|0543|Diameter of Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0543.Diameter-of-Binary-Tree)|55.9%|Easy||
+|0544|Output Contest Matches||76.7%|Medium||
+|0545|Boundary of Binary Tree||44.2%|Medium||
+|0546|Remove Boxes||47.6%|Hard||
+|0547|Number of Provinces|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0547.Number-of-Provinces)|63.2%|Medium||
+|0548|Split Array with Equal Sum||50.1%|Hard||
+|0549|Binary Tree Longest Consecutive Sequence II||49.5%|Medium||
+|0550|Game Play Analysis IV||44.1%|Medium||
+|0551|Student Attendance Record I|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0551.Student-Attendance-Record-I)|48.0%|Easy||
+|0552|Student Attendance Record II||41.1%|Hard||
+|0553|Optimal Division||59.7%|Medium||
+|0554|Brick Wall|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0554.Brick-Wall)|53.1%|Medium||
+|0555|Split Concatenated Strings||43.5%|Medium||
+|0556|Next Greater Element III||34.1%|Medium||
+|0557|Reverse Words in a String III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0557.Reverse-Words-in-a-String-III)|81.5%|Easy||
+|0558|Logical OR of Two Binary Grids Represented as Quad-Trees||48.2%|Medium||
+|0559|Maximum Depth of N-ary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0559.Maximum-Depth-of-N-ary-Tree)|71.6%|Easy||
+|0560|Subarray Sum Equals K|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0560.Subarray-Sum-Equals-K)|44.0%|Medium||
+|0561|Array Partition|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0561.Array-Partition)|76.5%|Easy||
+|0562|Longest Line of Consecutive One in Matrix||50.0%|Medium||
+|0563|Binary Tree Tilt|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0563.Binary-Tree-Tilt)|59.3%|Easy||
+|0564|Find the Closest Palindrome||21.9%|Hard||
+|0565|Array Nesting||56.5%|Medium||
+|0566|Reshape the Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0566.Reshape-the-Matrix)|62.7%|Easy||
+|0567|Permutation in String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0567.Permutation-in-String)|43.8%|Medium||
+|0568|Maximum Vacation Days||44.8%|Hard||
+|0569|Median Employee Salary||68.3%|Hard||
+|0570|Managers with at Least 5 Direct Reports||67.1%|Medium||
+|0571|Find Median Given Frequency of Numbers||44.7%|Hard||
+|0572|Subtree of Another Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0572.Subtree-of-Another-Tree)|45.9%|Easy||
+|0573|Squirrel Simulation||55.0%|Medium||
+|0574|Winning Candidate||59.7%|Medium||
+|0575|Distribute Candies|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0575.Distribute-Candies)|66.1%|Easy||
+|0576|Out of Boundary Paths|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0576.Out-of-Boundary-Paths)|44.3%|Medium||
+|0577|Employee Bonus||75.3%|Easy||
+|0578|Get Highest Answer Rate Question||41.7%|Medium||
+|0579|Find Cumulative Salary of an Employee||45.0%|Hard||
+|0580|Count Student Number in Departments||58.2%|Medium||
+|0581|Shortest Unsorted Continuous Subarray|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0581.Shortest-Unsorted-Continuous-Subarray)|36.2%|Medium||
+|0582|Kill Process||68.4%|Medium||
+|0583|Delete Operation for Two Strings|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0583.Delete-Operation-for-Two-Strings)|59.2%|Medium||
+|0584|Find Customer Referee||76.0%|Easy||
+|0585|Investments in 2016||53.5%|Medium||
+|0586|Customer Placing the Largest Number of Orders||72.5%|Easy||
+|0587|Erect the Fence||43.2%|Hard||
+|0588|Design In-Memory File System||48.8%|Hard||
+|0589|N-ary Tree Preorder Traversal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0589.N-ary-Tree-Preorder-Traversal)|76.3%|Easy||
+|0590|N-ary Tree Postorder Traversal||77.1%|Easy||
+|0591|Tag Validator||37.0%|Hard||
+|0592|Fraction Addition and Subtraction||52.1%|Medium||
+|0593|Valid Square||44.1%|Medium||
+|0594|Longest Harmonious Subsequence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0594.Longest-Harmonious-Subsequence)|53.1%|Easy||
+|0595|Big Countries||73.8%|Easy||
+|0596|Classes More Than 5 Students||46.5%|Easy||
+|0597|Friend Requests I: Overall Acceptance Rate||42.9%|Easy||
+|0598|Range Addition II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0598.Range-Addition-II)|55.0%|Easy||
+|0599|Minimum Index Sum of Two Lists|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0599.Minimum-Index-Sum-of-Two-Lists)|52.7%|Easy||
+|0600|Non-negative Integers without Consecutive Ones||39.0%|Hard||
+|0601|Human Traffic of Stadium||50.6%|Hard||
+|0602|Friend Requests II: Who Has the Most Friends||61.3%|Medium||
+|0603|Consecutive Available Seats||68.2%|Easy||
+|0604|Design Compressed String Iterator||39.4%|Easy||
+|0605|Can Place Flowers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0605.Can-Place-Flowers)|33.0%|Easy||
+|0606|Construct String from Binary Tree||63.5%|Easy||
+|0607|Sales Person||72.1%|Easy||
+|0608|Tree Node||74.9%|Medium||
+|0609|Find Duplicate File in System|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0609.Find-Duplicate-File-in-System)|67.8%|Medium||
+|0610|Triangle Judgement||71.1%|Easy||
+|0611|Valid Triangle Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0611.Valid-Triangle-Number)|50.3%|Medium||
+|0612|Shortest Distance in a Plane||63.5%|Medium||
+|0613|Shortest Distance in a Line||81.5%|Easy||
+|0614|Second Degree Follower||36.9%|Medium||
+|0615|Average Salary: Departments VS Company||57.3%|Hard||
+|0616|Add Bold Tag in String||48.6%|Medium||
+|0617|Merge Two Binary Trees|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0617.Merge-Two-Binary-Trees)|78.5%|Easy||
+|0618|Students Report By Geography||64.2%|Hard||
+|0619|Biggest Single Number||48.7%|Easy||
+|0620|Not Boring Movies||73.3%|Easy||
+|0621|Task Scheduler||55.6%|Medium||
+|0622|Design Circular Queue|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0622.Design-Circular-Queue)|51.8%|Medium||
+|0623|Add One Row to Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0623.Add-One-Row-to-Tree)|59.4%|Medium||
+|0624|Maximum Distance in Arrays||41.7%|Medium||
+|0625|Minimum Factorization||33.4%|Medium||
+|0626|Exchange Seats||70.6%|Medium||
+|0627|Swap Salary||82.8%|Easy||
+|0628|Maximum Product of Three Numbers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0628.Maximum-Product-of-Three-Numbers)|46.4%|Easy||
+|0629|K Inverse Pairs Array||43.0%|Hard||
+|0630|Course Schedule III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0630.Course-Schedule-III)|40.2%|Hard||
+|0631|Design Excel Sum Formula||43.3%|Hard||
+|0632|Smallest Range Covering Elements from K Lists|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0632.Smallest-Range-Covering-Elements-from-K-Lists)|60.4%|Hard||
+|0633|Sum of Square Numbers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0633.Sum-of-Square-Numbers)|34.7%|Medium||
+|0634|Find the Derangement of An Array||41.8%|Medium||
+|0635|Design Log Storage System||62.7%|Medium||
+|0636|Exclusive Time of Functions|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0636.Exclusive-Time-of-Functions)|61.0%|Medium||
+|0637|Average of Levels in Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0637.Average-of-Levels-in-Binary-Tree)|71.6%|Easy||
+|0638|Shopping Offers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0638.Shopping-Offers)|54.3%|Medium||
+|0639|Decode Ways II||30.4%|Hard||
+|0640|Solve the Equation||43.3%|Medium||
+|0641|Design Circular Deque||57.6%|Medium||
+|0642|Design Search Autocomplete System||48.7%|Hard||
+|0643|Maximum Average Subarray I|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0643.Maximum-Average-Subarray-I)|43.8%|Easy||
+|0644|Maximum Average Subarray II||35.7%|Hard||
+|0645|Set Mismatch|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0645.Set-Mismatch)|41.4%|Easy||
+|0646|Maximum Length of Pair Chain||56.4%|Medium||
+|0647|Palindromic Substrings|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0647.Palindromic-Substrings)|66.3%|Medium||
+|0648|Replace Words|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0648.Replace-Words)|62.7%|Medium||
+|0649|Dota2 Senate||40.4%|Medium||
+|0650|2 Keys Keyboard||53.1%|Medium||
+|0651|4 Keys Keyboard||54.5%|Medium||
+|0652|Find Duplicate Subtrees||56.5%|Medium||
+|0653|Two Sum IV - Input is a BST|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0653.Two-Sum-IV-Input-is-a-BST)|60.9%|Easy||
+|0654|Maximum Binary Tree||84.4%|Medium||
+|0655|Print Binary Tree||61.3%|Medium||
+|0656|Coin Path||31.6%|Hard||
+|0657|Robot Return to Origin||75.3%|Easy||
+|0658|Find K Closest Elements|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0658.Find-K-Closest-Elements)|46.7%|Medium||
+|0659|Split Array into Consecutive Subsequences||50.6%|Medium||
+|0660|Remove 9||56.8%|Hard||
+|0661|Image Smoother|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0661.Image-Smoother)|55.0%|Easy||
+|0662|Maximum Width of Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0662.Maximum-Width-of-Binary-Tree)|40.6%|Medium||
+|0663|Equal Tree Partition||41.4%|Medium||
+|0664|Strange Printer||46.8%|Hard||
+|0665|Non-decreasing Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0665.Non-decreasing-Array)|24.1%|Medium||
+|0666|Path Sum IV||59.1%|Medium||
+|0667|Beautiful Arrangement II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0667.Beautiful-Arrangement-II)|59.7%|Medium||
+|0668|Kth Smallest Number in Multiplication Table|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0668.Kth-Smallest-Number-in-Multiplication-Table)|51.5%|Hard||
+|0669|Trim a Binary Search Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0669.Trim-a-Binary-Search-Tree)|66.3%|Medium||
+|0670|Maximum Swap||47.8%|Medium||
+|0671|Second Minimum Node In a Binary Tree||44.0%|Easy||
+|0672|Bulb Switcher II||50.9%|Medium||
+|0673|Number of Longest Increasing Subsequence||42.1%|Medium||
+|0674|Longest Continuous Increasing Subsequence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0674.Longest-Continuous-Increasing-Subsequence)|49.0%|Easy||
+|0675|Cut Off Trees for Golf Event||34.3%|Hard||
+|0676|Implement Magic Dictionary|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0676.Implement-Magic-Dictionary)|56.9%|Medium||
+|0677|Map Sum Pairs|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0677.Map-Sum-Pairs)|57.0%|Medium||
+|0678|Valid Parenthesis String||33.9%|Medium||
+|0679|24 Game||49.1%|Hard||
+|0680|Valid Palindrome II||39.3%|Easy||
+|0681|Next Closest Time||46.4%|Medium||
+|0682|Baseball Game|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0682.Baseball-Game)|73.6%|Easy||
+|0683|K Empty Slots||36.9%|Hard||
+|0684|Redundant Connection|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0684.Redundant-Connection)|62.0%|Medium||
+|0685|Redundant Connection II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0685.Redundant-Connection-II)|34.1%|Hard||
+|0686|Repeated String Match||34.0%|Medium||
+|0687|Longest Univalue Path||40.1%|Medium||
+|0688|Knight Probability in Chessboard||52.0%|Medium||
+|0689|Maximum Sum of 3 Non-Overlapping Subarrays||48.8%|Hard||
+|0690|Employee Importance|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0690.Employee-Importance)|65.1%|Medium||
+|0691|Stickers to Spell Word||46.3%|Hard||
+|0692|Top K Frequent Words|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0692.Top-K-Frequent-Words)|55.2%|Medium||
+|0693|Binary Number with Alternating Bits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0693.Binary-Number-with-Alternating-Bits)|61.2%|Easy||
+|0694|Number of Distinct Islands||60.6%|Medium||
+|0695|Max Area of Island|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0695.Max-Area-of-Island)|71.6%|Medium||
+|0696|Count Binary Substrings|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0696.Count-Binary-Substrings)|65.5%|Easy||
+|0697|Degree of an Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0697.Degree-of-an-Array)|55.8%|Easy||
+|0698|Partition to K Equal Sum Subsets||41.1%|Medium||
+|0699|Falling Squares|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0699.Falling-Squares)|44.4%|Hard||
+|0700|Search in a Binary Search Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0700.Search-in-a-Binary-Search-Tree)|77.1%|Easy||
+|0701|Insert into a Binary Search Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0701.Insert-into-a-Binary-Search-Tree)|74.6%|Medium||
+|0702|Search in a Sorted Array of Unknown Size||71.4%|Medium||
+|0703|Kth Largest Element in a Stream|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0703.Kth-Largest-Element-in-a-Stream)|55.5%|Easy||
+|0704|Binary Search|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0704.Binary-Search)|55.1%|Easy||
+|0705|Design HashSet|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0705.Design-HashSet)|66.0%|Easy||
+|0706|Design HashMap|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0706.Design-HashMap)|65.1%|Easy||
+|0707|Design Linked List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0707.Design-Linked-List)|27.5%|Medium||
+|0708|Insert into a Sorted Circular Linked List||34.5%|Medium||
+|0709|To Lower Case|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0709.To-Lower-Case)|81.9%|Easy||
+|0710|Random Pick with Blacklist|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0710.Random-Pick-with-Blacklist)|33.6%|Hard||
+|0711|Number of Distinct Islands II||51.8%|Hard||
+|0712|Minimum ASCII Delete Sum for Two Strings||62.2%|Medium||
+|0713|Subarray Product Less Than K|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0713.Subarray-Product-Less-Than-K)|45.0%|Medium||
+|0714|Best Time to Buy and Sell Stock with Transaction Fee|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee)|64.2%|Medium||
+|0715|Range Module|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0715.Range-Module)|44.6%|Hard||
+|0716|Max Stack||45.3%|Hard||
+|0717|1-bit and 2-bit Characters|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0717.1-bit-and-2-bit-Characters)|46.0%|Easy||
+|0718|Maximum Length of Repeated Subarray|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0718.Maximum-Length-of-Repeated-Subarray)|51.6%|Medium||
+|0719|Find K-th Smallest Pair Distance|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0719.Find-K-th-Smallest-Pair-Distance)|36.3%|Hard||
+|0720|Longest Word in Dictionary|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0720.Longest-Word-in-Dictionary)|51.8%|Medium||
+|0721|Accounts Merge|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0721.Accounts-Merge)|56.3%|Medium||
+|0722|Remove Comments||38.0%|Medium||
+|0723|Candy Crush||76.3%|Medium||
+|0724|Find Pivot Index|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0724.Find-Pivot-Index)|53.3%|Easy||
+|0725|Split Linked List in Parts|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0725.Split-Linked-List-in-Parts)|57.2%|Medium||
+|0726|Number of Atoms|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0726.Number-of-Atoms)|52.2%|Hard||
+|0727|Minimum Window Subsequence||42.8%|Hard||
+|0728|Self Dividing Numbers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0728.Self-Dividing-Numbers)|77.5%|Easy||
+|0729|My Calendar I|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0729.My-Calendar-I)|57.2%|Medium||
+|0730|Count Different Palindromic Subsequences||44.4%|Hard||
+|0731|My Calendar II||54.7%|Medium||
+|0732|My Calendar III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0732.My-Calendar-III)|71.6%|Hard||
+|0733|Flood Fill|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0733.Flood-Fill)|60.3%|Easy||
+|0734|Sentence Similarity||43.1%|Easy||
+|0735|Asteroid Collision|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0735.Asteroid-Collision)|44.4%|Medium||
+|0736|Parse Lisp Expression||51.6%|Hard||
+|0737|Sentence Similarity II||48.7%|Medium||
+|0738|Monotone Increasing Digits||47.0%|Medium||
+|0739|Daily Temperatures|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0739.Daily-Temperatures)|66.5%|Medium||
+|0740|Delete and Earn||57.4%|Medium||
+|0741|Cherry Pickup||36.3%|Hard||
+|0742|Closest Leaf in a Binary Tree||45.8%|Medium||
+|0743|Network Delay Time||51.4%|Medium||
+|0744|Find Smallest Letter Greater Than Target|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0744.Find-Smallest-Letter-Greater-Than-Target)|44.6%|Easy||
+|0745|Prefix and Suffix Search|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0745.Prefix-and-Suffix-Search)|41.3%|Hard||
+|0746|Min Cost Climbing Stairs|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0746.Min-Cost-Climbing-Stairs)|62.3%|Easy||
+|0747|Largest Number At Least Twice of Others|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0747.Largest-Number-At-Least-Twice-of-Others)|46.3%|Easy||
+|0748|Shortest Completing Word|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0748.Shortest-Completing-Word)|59.1%|Easy||
+|0749|Contain Virus||50.8%|Hard||
+|0750|Number Of Corner Rectangles||67.5%|Medium||
+|0751|IP to CIDR||54.7%|Medium||
+|0752|Open the Lock|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0752.Open-the-Lock)|55.5%|Medium||
+|0753|Cracking the Safe|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0753.Cracking-the-Safe)|55.4%|Hard||
+|0754|Reach a Number||42.5%|Medium||
+|0755|Pour Water||46.1%|Medium||
+|0756|Pyramid Transition Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0756.Pyramid-Transition-Matrix)|53.3%|Medium||
+|0757|Set Intersection Size At Least Two||43.8%|Hard||
+|0758|Bold Words in String||50.6%|Medium||
+|0759|Employee Free Time||71.7%|Hard||
+|0760|Find Anagram Mappings||82.8%|Easy||
+|0761|Special Binary String||60.3%|Hard||
+|0762|Prime Number of Set Bits in Binary Representation|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0762.Prime-Number-of-Set-Bits-in-Binary-Representation)|67.6%|Easy||
+|0763|Partition Labels|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0763.Partition-Labels)|79.8%|Medium||
+|0764|Largest Plus Sign||48.4%|Medium||
+|0765|Couples Holding Hands|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0765.Couples-Holding-Hands)|56.9%|Hard||
+|0766|Toeplitz Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0766.Toeplitz-Matrix)|68.1%|Easy||
+|0767|Reorganize String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0767.Reorganize-String)|52.7%|Medium||
+|0768|Max Chunks To Make Sorted II||52.7%|Hard||
+|0769|Max Chunks To Make Sorted||58.2%|Medium||
+|0770|Basic Calculator IV||56.0%|Hard||
+|0771|Jewels and Stones|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0771.Jewels-and-Stones)|88.0%|Easy||
+|0772|Basic Calculator III||48.4%|Hard||
+|0773|Sliding Puzzle||63.7%|Hard||
+|0774|Minimize Max Distance to Gas Station||51.4%|Hard||
+|0775|Global and Local Inversions|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0775.Global-and-Local-Inversions)|43.8%|Medium||
+|0776|Split BST||58.7%|Medium||
+|0777|Swap Adjacent in LR String||37.1%|Medium||
+|0778|Swim in Rising Water|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0778.Swim-in-Rising-Water)|59.6%|Hard||
+|0779|K-th Symbol in Grammar||40.7%|Medium||
+|0780|Reaching Points||32.3%|Hard||
+|0781|Rabbits in Forest|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0781.Rabbits-in-Forest)|55.3%|Medium||
+|0782|Transform to Chessboard||51.9%|Hard||
+|0783|Minimum Distance Between BST Nodes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0783.Minimum-Distance-Between-BST-Nodes)|56.8%|Easy||
+|0784|Letter Case Permutation|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0784.Letter-Case-Permutation)|73.4%|Medium||
+|0785|Is Graph Bipartite?|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0785.Is-Graph-Bipartite)|52.6%|Medium||
+|0786|K-th Smallest Prime Fraction|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0786.K-th-Smallest-Prime-Fraction)|50.6%|Medium||
+|0787|Cheapest Flights Within K Stops||35.9%|Medium||
+|0788|Rotated Digits||56.9%|Medium||
+|0789|Escape The Ghosts||60.6%|Medium||
+|0790|Domino and Tromino Tiling||48.4%|Medium||
+|0791|Custom Sort String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0791.Custom-Sort-String)|69.4%|Medium||
+|0792|Number of Matching Subsequences|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0792.Number-of-Matching-Subsequences)|51.9%|Medium||
+|0793|Preimage Size of Factorial Zeroes Function|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0793.Preimage-Size-of-Factorial-Zeroes-Function)|42.7%|Hard||
+|0794|Valid Tic-Tac-Toe State|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0794.Valid-Tic-Tac-Toe-State)|35.2%|Medium||
+|0795|Number of Subarrays with Bounded Maximum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum)|52.7%|Medium||
+|0796|Rotate String||54.0%|Easy||
+|0797|All Paths From Source to Target||81.4%|Medium||
+|0798|Smallest Rotation with Highest Score||49.6%|Hard||
+|0799|Champagne Tower||51.2%|Medium||
+|0800|Similar RGB Color||66.4%|Easy||
+|0801|Minimum Swaps To Make Sequences Increasing||39.2%|Hard||
+|0802|Find Eventual Safe States|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0802.Find-Eventual-Safe-States)|54.9%|Medium||
+|0803|Bricks Falling When Hit|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0803.Bricks-Falling-When-Hit)|34.3%|Hard||
+|0804|Unique Morse Code Words||82.6%|Easy||
+|0805|Split Array With Same Average||25.9%|Hard||
+|0806|Number of Lines To Write String||66.1%|Easy||
+|0807|Max Increase to Keep City Skyline|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0807.Max-Increase-to-Keep-City-Skyline)|85.9%|Medium||
+|0808|Soup Servings||43.1%|Medium||
+|0809|Expressive Words||46.3%|Medium||
+|0810|Chalkboard XOR Game|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0810.Chalkboard-XOR-Game)|55.1%|Hard||
+|0811|Subdomain Visit Count|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0811.Subdomain-Visit-Count)|75.1%|Medium||
+|0812|Largest Triangle Area|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0812.Largest-Triangle-Area)|60.1%|Easy||
+|0813|Largest Sum of Averages||52.9%|Medium||
+|0814|Binary Tree Pruning||72.6%|Medium||
+|0815|Bus Routes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0815.Bus-Routes)|45.7%|Hard||
+|0816|Ambiguous Coordinates|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0816.Ambiguous-Coordinates)|56.0%|Medium||
+|0817|Linked List Components|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0817.Linked-List-Components)|58.1%|Medium||
+|0818|Race Car||43.6%|Hard||
+|0819|Most Common Word|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0819.Most-Common-Word)|45.0%|Easy||
+|0820|Short Encoding of Words|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0820.Short-Encoding-of-Words)|60.7%|Medium||
+|0821|Shortest Distance to a Character|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0821.Shortest-Distance-to-a-Character)|71.3%|Easy||
+|0822|Card Flipping Game||45.4%|Medium||
+|0823|Binary Trees With Factors|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0823.Binary-Trees-With-Factors)|50.0%|Medium||
+|0824|Goat Latin||67.8%|Easy||
+|0825|Friends Of Appropriate Ages|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0825.Friends-Of-Appropriate-Ages)|46.3%|Medium||
+|0826|Most Profit Assigning Work|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0826.Most-Profit-Assigning-Work)|44.3%|Medium||
+|0827|Making A Large Island||44.7%|Hard||
+|0828|Count Unique Characters of All Substrings of a Given String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String)|51.5%|Hard||
+|0829|Consecutive Numbers Sum||41.5%|Hard||
+|0830|Positions of Large Groups|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0830.Positions-of-Large-Groups)|51.7%|Easy||
+|0831|Masking Personal Information||46.8%|Medium||
+|0832|Flipping an Image|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0832.Flipping-an-Image)|80.3%|Easy||
+|0833|Find And Replace in String||54.1%|Medium||
+|0834|Sum of Distances in Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0834.Sum-of-Distances-in-Tree)|54.1%|Hard||
+|0835|Image Overlap||61.0%|Medium||
+|0836|Rectangle Overlap|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0836.Rectangle-Overlap)|43.3%|Easy||
+|0837|New 21 Game||36.2%|Medium||
+|0838|Push Dominoes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0838.Push-Dominoes)|57.0%|Medium||
+|0839|Similar String Groups|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0839.Similar-String-Groups)|47.5%|Hard||
+|0840|Magic Squares In Grid||38.5%|Medium||
+|0841|Keys and Rooms|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0841.Keys-and-Rooms)|70.1%|Medium||
+|0842|Split Array into Fibonacci Sequence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0842.Split-Array-into-Fibonacci-Sequence)|38.2%|Medium||
+|0843|Guess the Word||42.0%|Hard||
+|0844|Backspace String Compare|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0844.Backspace-String-Compare)|48.0%|Easy||
+|0845|Longest Mountain in Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0845.Longest-Mountain-in-Array)|40.1%|Medium||
+|0846|Hand of Straights|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0846.Hand-of-Straights)|56.5%|Medium||
+|0847|Shortest Path Visiting All Nodes||61.3%|Hard||
+|0848|Shifting Letters||45.4%|Medium||
+|0849|Maximize Distance to Closest Person||47.6%|Medium||
+|0850|Rectangle Area II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0850.Rectangle-Area-II)|53.7%|Hard||
+|0851|Loud and Rich|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0851.Loud-and-Rich)|58.1%|Medium||
+|0852|Peak Index in a Mountain Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0852.Peak-Index-in-a-Mountain-Array)|69.5%|Medium||
+|0853|Car Fleet|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0853.Car-Fleet)|50.0%|Medium||
+|0854|K-Similar Strings||40.0%|Hard||
+|0855|Exam Room||43.5%|Medium||
+|0856|Score of Parentheses|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0856.Score-of-Parentheses)|65.1%|Medium||
+|0857|Minimum Cost to Hire K Workers||52.0%|Hard||
+|0858|Mirror Reflection||63.3%|Medium||
+|0859|Buddy Strings|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0859.Buddy-Strings)|29.0%|Easy||
+|0860|Lemonade Change||52.8%|Easy||
+|0861|Score After Flipping Matrix||75.1%|Medium||
+|0862|Shortest Subarray with Sum at Least K|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0862.Shortest-Subarray-with-Sum-at-Least-K)|26.1%|Hard||
+|0863|All Nodes Distance K in Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0863.All-Nodes-Distance-K-in-Binary-Tree)|62.1%|Medium||
+|0864|Shortest Path to Get All Keys|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0864.Shortest-Path-to-Get-All-Keys)|45.3%|Hard||
+|0865|Smallest Subtree with all the Deepest Nodes||68.5%|Medium||
+|0866|Prime Palindrome||25.8%|Medium||
+|0867|Transpose Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0867.Transpose-Matrix)|63.4%|Easy||
+|0868|Binary Gap||61.9%|Easy||
+|0869|Reordered Power of 2|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0869.Reordered-Power-of-2)|64.1%|Medium||
+|0870|Advantage Shuffle|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0870.Advantage-Shuffle)|51.6%|Medium||
+|0871|Minimum Number of Refueling Stops||39.9%|Hard||
+|0872|Leaf-Similar Trees|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0872.Leaf-Similar-Trees)|65.2%|Easy||
+|0873|Length of Longest Fibonacci Subsequence||48.6%|Medium||
+|0874|Walking Robot Simulation|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0874.Walking-Robot-Simulation)|38.3%|Medium||
+|0875|Koko Eating Bananas|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0875.Koko-Eating-Bananas)|52.4%|Medium||
+|0876|Middle of the Linked List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0876.Middle-of-the-Linked-List)|73.7%|Easy||
+|0877|Stone Game|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0877.Stone-Game)|69.7%|Medium||
+|0878|Nth Magical Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0878.Nth-Magical-Number)|35.7%|Hard||
+|0879|Profitable Schemes||40.5%|Hard||
+|0880|Decoded String at Index|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0880.Decoded-String-at-Index)|28.3%|Medium||
+|0881|Boats to Save People|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0881.Boats-to-Save-People)|52.7%|Medium||
+|0882|Reachable Nodes In Subdivided Graph||50.3%|Hard||
+|0883|Projection Area of 3D Shapes||70.7%|Easy||
+|0884|Uncommon Words from Two Sentences|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0884.Uncommon-Words-from-Two-Sentences)|65.9%|Easy||
+|0885|Spiral Matrix III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0885.Spiral-Matrix-III)|73.0%|Medium||
+|0886|Possible Bipartition||48.4%|Medium||
+|0887|Super Egg Drop|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0887.Super-Egg-Drop)|27.2%|Hard||
+|0888|Fair Candy Swap|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0888.Fair-Candy-Swap)|60.5%|Easy||
+|0889|Construct Binary Tree from Preorder and Postorder Traversal||70.8%|Medium||
+|0890|Find and Replace Pattern|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0890.Find-and-Replace-Pattern)|77.9%|Medium||
+|0891|Sum of Subsequence Widths|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0891.Sum-of-Subsequence-Widths)|36.4%|Hard||
+|0892|Surface Area of 3D Shapes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0892.Surface-Area-of-3D-Shapes)|63.0%|Easy||
+|0893|Groups of Special-Equivalent Strings||70.8%|Medium||
+|0894|All Possible Full Binary Trees||80.0%|Medium||
+|0895|Maximum Frequency Stack|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0895.Maximum-Frequency-Stack)|66.8%|Hard||
+|0896|Monotonic Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0896.Monotonic-Array)|58.2%|Easy||
+|0897|Increasing Order Search Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0897.Increasing-Order-Search-Tree)|78.4%|Easy||
+|0898|Bitwise ORs of Subarrays|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0898.Bitwise-ORs-of-Subarrays)|36.8%|Medium||
+|0899|Orderly Queue||59.0%|Hard||
+|0900|RLE Iterator||59.5%|Medium||
+|0901|Online Stock Span|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0901.Online-Stock-Span)|63.9%|Medium||
+|0902|Numbers At Most N Given Digit Set||41.4%|Hard||
+|0903|Valid Permutations for DI Sequence||57.7%|Hard||
+|0904|Fruit Into Baskets|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0904.Fruit-Into-Baskets)|42.6%|Medium||
+|0905|Sort Array By Parity||75.7%|Easy||
+|0906|Super Palindromes||39.2%|Hard||
+|0907|Sum of Subarray Minimums|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0907.Sum-of-Subarray-Minimums)|34.3%|Medium||
+|0908|Smallest Range I||67.6%|Easy||
+|0909|Snakes and Ladders|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0909.Snakes-and-Ladders)|40.8%|Medium||
+|0910|Smallest Range II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0910.Smallest-Range-II)|34.4%|Medium||
+|0911|Online Election|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0911.Online-Election)|52.1%|Medium||
+|0912|Sort an Array||60.0%|Medium||
+|0913|Cat and Mouse||35.3%|Hard||
+|0914|X of a Kind in a Deck of Cards|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0914.X-of-a-Kind-in-a-Deck-of-Cards)|32.2%|Easy||
+|0915|Partition Array into Disjoint Intervals||48.6%|Medium||
+|0916|Word Subsets|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0916.Word-Subsets)|54.1%|Medium||
+|0917|Reverse Only Letters||61.4%|Easy||
+|0918|Maximum Sum Circular Subarray|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0918.Maximum-Sum-Circular-Subarray)|38.1%|Medium||
+|0919|Complete Binary Tree Inserter||64.9%|Medium||
+|0920|Number of Music Playlists|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0920.Number-of-Music-Playlists)|50.6%|Hard||
+|0921|Minimum Add to Make Parentheses Valid|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0921.Minimum-Add-to-Make-Parentheses-Valid)|76.4%|Medium||
+|0922|Sort Array By Parity II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0922.Sort-Array-By-Parity-II)|70.7%|Easy||
+|0923|3Sum With Multiplicity|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0923.3Sum-With-Multiplicity)|45.4%|Medium||
+|0924|Minimize Malware Spread|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0924.Minimize-Malware-Spread)|42.1%|Hard||
+|0925|Long Pressed Name|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0925.Long-Pressed-Name)|33.8%|Easy||
+|0926|Flip String to Monotone Increasing||59.6%|Medium||
+|0927|Three Equal Parts|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0927.Three-Equal-Parts)|39.6%|Hard||
+|0928|Minimize Malware Spread II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0928.Minimize-Malware-Spread-II)|42.6%|Hard||
+|0929|Unique Email Addresses||67.2%|Easy||
+|0930|Binary Subarrays With Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0930.Binary-Subarrays-With-Sum)|50.9%|Medium||
+|0931|Minimum Falling Path Sum||68.4%|Medium||
+|0932|Beautiful Array||65.1%|Medium||
+|0933|Number of Recent Calls|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0933.Number-of-Recent-Calls)|73.2%|Easy||
+|0934|Shortest Bridge||53.9%|Medium||
+|0935|Knight Dialer||49.9%|Medium||
+|0936|Stamping The Sequence||63.3%|Hard||
+|0937|Reorder Data in Log Files||56.4%|Medium||
+|0938|Range Sum of BST|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0938.Range-Sum-of-BST)|85.3%|Easy||
+|0939|Minimum Area Rectangle||53.2%|Medium||
+|0940|Distinct Subsequences II||44.4%|Hard||
+|0941|Valid Mountain Array||33.5%|Easy||
+|0942|DI String Match|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0942.DI-String-Match)|76.6%|Easy||
+|0943|Find the Shortest Superstring||45.0%|Hard||
+|0944|Delete Columns to Make Sorted||69.7%|Easy||
+|0945|Minimum Increment to Make Array Unique||49.8%|Medium||
+|0946|Validate Stack Sequences|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0946.Validate-Stack-Sequences)|67.6%|Medium||
+|0947|Most Stones Removed with Same Row or Column|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0947.Most-Stones-Removed-with-Same-Row-or-Column)|57.0%|Medium||
+|0948|Bag of Tokens||52.0%|Medium||
+|0949|Largest Time for Given Digits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0949.Largest-Time-for-Given-Digits)|35.2%|Medium||
+|0950|Reveal Cards In Increasing Order||77.6%|Medium||
+|0951|Flip Equivalent Binary Trees||66.8%|Medium||
+|0952|Largest Component Size by Common Factor|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0952.Largest-Component-Size-by-Common-Factor)|40.3%|Hard||
+|0953|Verifying an Alien Dictionary|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0953.Verifying-an-Alien-Dictionary)|52.7%|Easy||
+|0954|Array of Doubled Pairs||39.1%|Medium||
+|0955|Delete Columns to Make Sorted II||34.6%|Medium||
+|0956|Tallest Billboard||40.0%|Hard||
+|0957|Prison Cells After N Days||39.2%|Medium||
+|0958|Check Completeness of a Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0958.Check-Completeness-of-a-Binary-Tree)|53.8%|Medium||
+|0959|Regions Cut By Slashes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0959.Regions-Cut-By-Slashes)|69.1%|Medium||
+|0960|Delete Columns to Make Sorted III||57.1%|Hard||
+|0961|N-Repeated Element in Size 2N Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0961.N-Repeated-Element-in-Size-2N-Array)|75.8%|Easy||
+|0962|Maximum Width Ramp||48.9%|Medium||
+|0963|Minimum Area Rectangle II||54.7%|Medium||
+|0964|Least Operators to Express Number||47.8%|Hard||
+|0965|Univalued Binary Tree||69.2%|Easy||
+|0966|Vowel Spellchecker|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0966.Vowel-Spellchecker)|51.5%|Medium||
+|0967|Numbers With Same Consecutive Differences||57.0%|Medium||
+|0968|Binary Tree Cameras|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0968.Binary-Tree-Cameras)|46.8%|Hard||
+|0969|Pancake Sorting|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0969.Pancake-Sorting)|70.0%|Medium||
+|0970|Powerful Integers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0970.Powerful-Integers)|43.6%|Medium||
+|0971|Flip Binary Tree To Match Preorder Traversal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal)|49.9%|Medium||
+|0972|Equal Rational Numbers||43.0%|Hard||
+|0973|K Closest Points to Origin|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0973.K-Closest-Points-to-Origin)|65.9%|Medium||
+|0974|Subarray Sums Divisible by K||53.6%|Medium||
+|0975|Odd Even Jump||38.9%|Hard||
+|0976|Largest Perimeter Triangle|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0976.Largest-Perimeter-Triangle)|54.8%|Easy||
+|0977|Squares of a Sorted Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0977.Squares-of-a-Sorted-Array)|71.9%|Easy||
+|0978|Longest Turbulent Subarray|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0978.Longest-Turbulent-Subarray)|47.4%|Medium||
+|0979|Distribute Coins in Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0979.Distribute-Coins-in-Binary-Tree)|72.0%|Medium||
+|0980|Unique Paths III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0980.Unique-Paths-III)|79.6%|Hard||
+|0981|Time Based Key-Value Store|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0981.Time-Based-Key-Value-Store)|53.6%|Medium||
+|0982|Triples with Bitwise AND Equal To Zero||57.6%|Hard||
+|0983|Minimum Cost For Tickets||64.4%|Medium||
+|0984|String Without AAA or BBB|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0984.String-Without-AAA-or-BBB)|42.9%|Medium||
+|0985|Sum of Even Numbers After Queries|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0985.Sum-of-Even-Numbers-After-Queries)|68.3%|Medium||
+|0986|Interval List Intersections|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0986.Interval-List-Intersections)|71.4%|Medium||
+|0987|Vertical Order Traversal of a Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0987.Vertical-Order-Traversal-of-a-Binary-Tree)|44.6%|Hard||
+|0988|Smallest String Starting From Leaf||49.6%|Medium||
+|0989|Add to Array-Form of Integer|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0989.Add-to-Array-Form-of-Integer)|45.5%|Easy||
+|0990|Satisfiability of Equality Equations|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0990.Satisfiability-of-Equality-Equations)|50.7%|Medium||
+|0991|Broken Calculator|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0991.Broken-Calculator)|54.1%|Medium||
+|0992|Subarrays with K Different Integers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0992.Subarrays-with-K-Different-Integers)|54.3%|Hard||
+|0993|Cousins in Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0993.Cousins-in-Binary-Tree)|54.1%|Easy||
+|0994|Rotting Oranges||52.4%|Medium||
+|0995|Minimum Number of K Consecutive Bit Flips|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0995.Minimum-Number-of-K-Consecutive-Bit-Flips)|51.1%|Hard||
+|0996|Number of Squareful Arrays|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0996.Number-of-Squareful-Arrays)|49.2%|Hard||
+|0997|Find the Town Judge|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0997.Find-the-Town-Judge)|49.4%|Easy||
+|0998|Maximum Binary Tree II||66.3%|Medium||
+|0999|Available Captures for Rook|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0999.Available-Captures-for-Rook)|67.8%|Easy||
+|1000|Minimum Cost to Merge Stones||42.3%|Hard||
+|1001|Grid Illumination||36.2%|Hard||
+|1002|Find Common Characters|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1002.Find-Common-Characters)|68.3%|Easy||
+|1003|Check If Word Is Valid After Substitutions|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1003.Check-If-Word-Is-Valid-After-Substitutions)|58.2%|Medium||
+|1004|Max Consecutive Ones III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1004.Max-Consecutive-Ones-III)|63.5%|Medium||
+|1005|Maximize Sum Of Array After K Negations|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1005.Maximize-Sum-Of-Array-After-K-Negations)|51.1%|Easy||
+|1006|Clumsy Factorial|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1006.Clumsy-Factorial)|54.7%|Medium||
+|1007|Minimum Domino Rotations For Equal Row||52.4%|Medium||
+|1008|Construct Binary Search Tree from Preorder Traversal||80.9%|Medium||
+|1009|Complement of Base 10 Integer|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1009.Complement-of-Base-10-Integer)|62.0%|Easy||
+|1010|Pairs of Songs With Total Durations Divisible by 60|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60)|53.0%|Medium||
+|1011|Capacity To Ship Packages Within D Days|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1011.Capacity-To-Ship-Packages-Within-D-Days)|64.4%|Medium||
+|1012|Numbers With Repeated Digits||40.4%|Hard||
+|1013|Partition Array Into Three Parts With Equal Sum||43.5%|Easy||
+|1014|Best Sightseeing Pair||59.5%|Medium||
+|1015|Smallest Integer Divisible by K||47.1%|Medium||
+|1016|Binary String With Substrings Representing 1 To N||57.6%|Medium||
+|1017|Convert to Base -2|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1017.Convert-to-Base-2)|60.9%|Medium||
+|1018|Binary Prefix Divisible By 5|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1018.Binary-Prefix-Divisible-By-5)|47.4%|Easy||
+|1019|Next Greater Node In Linked List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1019.Next-Greater-Node-In-Linked-List)|59.8%|Medium||
+|1020|Number of Enclaves|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1020.Number-of-Enclaves)|64.8%|Medium||
+|1021|Remove Outermost Parentheses|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1021.Remove-Outermost-Parentheses)|80.1%|Easy||
+|1022|Sum of Root To Leaf Binary Numbers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1022.Sum-of-Root-To-Leaf-Binary-Numbers)|73.8%|Easy||
+|1023|Camelcase Matching||60.1%|Medium||
+|1024|Video Stitching||50.5%|Medium||
+|1025|Divisor Game|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1025.Divisor-Game)|67.1%|Easy||
+|1026|Maximum Difference Between Node and Ancestor|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1026.Maximum-Difference-Between-Node-and-Ancestor)|73.4%|Medium||
+|1027|Longest Arithmetic Subsequence||47.2%|Medium||
+|1028|Recover a Tree From Preorder Traversal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1028.Recover-a-Tree-From-Preorder-Traversal)|72.8%|Hard||
+|1029|Two City Scheduling||64.5%|Medium||
+|1030|Matrix Cells in Distance Order|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1030.Matrix-Cells-in-Distance-Order)|69.3%|Easy||
+|1031|Maximum Sum of Two Non-Overlapping Subarrays||59.4%|Medium||
+|1032|Stream of Characters||51.5%|Hard||
+|1033|Moving Stones Until Consecutive||45.6%|Medium||
+|1034|Coloring A Border|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1034.Coloring-A-Border)|48.9%|Medium||
+|1035|Uncrossed Lines||58.7%|Medium||
+|1036|Escape a Large Maze||34.2%|Hard||
+|1037|Valid Boomerang|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1037.Valid-Boomerang)|37.5%|Easy||
+|1038|Binary Search Tree to Greater Sum Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1038.Binary-Search-Tree-to-Greater-Sum-Tree)|85.3%|Medium||
+|1039|Minimum Score Triangulation of Polygon||54.5%|Medium||
+|1040|Moving Stones Until Consecutive II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1040.Moving-Stones-Until-Consecutive-II)|55.6%|Medium||
+|1041|Robot Bounded In Circle||55.3%|Medium||
+|1042|Flower Planting With No Adjacent||50.4%|Medium||
+|1043|Partition Array for Maximum Sum||71.2%|Medium||
+|1044|Longest Duplicate Substring||30.7%|Hard||
+|1045|Customers Who Bought All Products||67.6%|Medium||
+|1046|Last Stone Weight||64.7%|Easy||
+|1047|Remove All Adjacent Duplicates In String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1047.Remove-All-Adjacent-Duplicates-In-String)|70.5%|Easy||
+|1048|Longest String Chain|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1048.Longest-String-Chain)|59.2%|Medium||
+|1049|Last Stone Weight II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1049.Last-Stone-Weight-II)|52.4%|Medium||
+|1050|Actors and Directors Who Cooperated At Least Three Times||72.4%|Easy||
+|1051|Height Checker|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1051.Height-Checker)|75.0%|Easy||
+|1052|Grumpy Bookstore Owner|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1052.Grumpy-Bookstore-Owner)|57.0%|Medium||
+|1053|Previous Permutation With One Swap||50.8%|Medium||
+|1054|Distant Barcodes|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1054.Distant-Barcodes)|45.7%|Medium||
+|1055|Shortest Way to Form String||59.1%|Medium||
+|1056|Confusing Number||46.1%|Easy||
+|1057|Campus Bikes||57.7%|Medium||
+|1058|Minimize Rounding Error to Meet Target||44.9%|Medium||
+|1059|All Paths from Source Lead to Destination||40.5%|Medium||
+|1060|Missing Element in Sorted Array||54.6%|Medium||
+|1061|Lexicographically Smallest Equivalent String||70.4%|Medium||
+|1062|Longest Repeating Substring||59.1%|Medium||
+|1063|Number of Valid Subarrays||74.1%|Hard||
+|1064|Fixed Point||63.6%|Easy||
+|1065|Index Pairs of a String||63.0%|Easy||
+|1066|Campus Bikes II||54.5%|Medium||
+|1067|Digit Count in Range||44.6%|Hard||
+|1068|Product Sales Analysis I||80.4%|Easy||
+|1069|Product Sales Analysis II||82.0%|Easy||
+|1070|Product Sales Analysis III||49.2%|Medium||
+|1071|Greatest Common Divisor of Strings||51.0%|Easy||
+|1072|Flip Columns For Maximum Number of Equal Rows||63.1%|Medium||
+|1073|Adding Two Negabinary Numbers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1073.Adding-Two-Negabinary-Numbers)|36.4%|Medium||
+|1074|Number of Submatrices That Sum to Target|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1074.Number-of-Submatrices-That-Sum-to-Target)|69.8%|Hard||
+|1075|Project Employees I||67.2%|Easy||
+|1076|Project Employees II||51.0%|Easy||
+|1077|Project Employees III||78.7%|Medium||
+|1078|Occurrences After Bigram|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1078.Occurrences-After-Bigram)|63.9%|Easy||
+|1079|Letter Tile Possibilities|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1079.Letter-Tile-Possibilities)|76.1%|Medium||
+|1080|Insufficient Nodes in Root to Leaf Paths||52.8%|Medium||
+|1081|Smallest Subsequence of Distinct Characters||57.6%|Medium||
+|1082|Sales Analysis I||75.4%|Easy||
+|1083|Sales Analysis II||50.5%|Easy||
+|1084|Sales Analysis III||52.2%|Easy||
+|1085|Sum of Digits in the Minimum Number||75.9%|Easy||
+|1086|High Five||75.2%|Easy||
+|1087|Brace Expansion||66.1%|Medium||
+|1088|Confusing Number II||46.5%|Hard||
+|1089|Duplicate Zeros|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1089.Duplicate-Zeros)|51.5%|Easy||
+|1090|Largest Values From Labels||60.9%|Medium||
+|1091|Shortest Path in Binary Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1091.Shortest-Path-in-Binary-Matrix)|44.5%|Medium||
+|1092|Shortest Common Supersequence||57.8%|Hard||
+|1093|Statistics from a Large Sample|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1093.Statistics-from-a-Large-Sample)|44.7%|Medium||
+|1094|Car Pooling||57.4%|Medium||
+|1095|Find in Mountain Array||35.8%|Hard||
+|1096|Brace Expansion II||63.5%|Hard||
+|1097|Game Play Analysis V||55.2%|Hard||
+|1098|Unpopular Books||45.2%|Medium||
+|1099|Two Sum Less Than K||60.4%|Easy||
+|1100|Find K-Length Substrings With No Repeated Characters||74.7%|Medium||
+|1101|The Earliest Moment When Everyone Become Friends||65.0%|Medium||
+|1102|Path With Maximum Minimum Value||53.3%|Medium||
+|1103|Distribute Candies to People||63.9%|Easy||
+|1104|Path In Zigzag Labelled Binary Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1104.Path-In-Zigzag-Labelled-Binary-Tree)|74.9%|Medium||
+|1105|Filling Bookcase Shelves|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1105.Filling-Bookcase-Shelves)|58.9%|Medium||
+|1106|Parsing A Boolean Expression||58.5%|Hard||
+|1107|New Users Daily Count||45.9%|Medium||
+|1108|Defanging an IP Address|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1108.Defanging-an-IP-Address)|89.3%|Easy||
+|1109|Corporate Flight Bookings||60.3%|Medium||
+|1110|Delete Nodes And Return Forest|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1110.Delete-Nodes-And-Return-Forest)|69.4%|Medium||
+|1111|Maximum Nesting Depth of Two Valid Parentheses Strings|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings)|73.3%|Medium||
+|1112|Highest Grade For Each Student||73.8%|Medium||
+|1113|Reported Posts||66.1%|Easy||
+|1114|Print in Order||68.1%|Easy||
+|1115|Print FooBar Alternately||61.7%|Medium||
+|1116|Print Zero Even Odd||60.1%|Medium||
+|1117|Building H2O||55.7%|Medium||
+|1118|Number of Days in a Month||56.7%|Easy||
+|1119|Remove Vowels from a String||90.7%|Easy||
+|1120|Maximum Average Subtree||65.5%|Medium||
+|1121|Divide Array Into Increasing Sequences||60.0%|Hard||
+|1122|Relative Sort Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1122.Relative-Sort-Array)|68.4%|Easy||
+|1123|Lowest Common Ancestor of Deepest Leaves|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1123.Lowest-Common-Ancestor-of-Deepest-Leaves)|70.6%|Medium||
+|1124|Longest Well-Performing Interval||34.6%|Medium||
+|1125|Smallest Sufficient Team||47.1%|Hard||
+|1126|Active Businesses||67.9%|Medium||
+|1127|User Purchase Platform||51.2%|Hard||
+|1128|Number of Equivalent Domino Pairs|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1128.Number-of-Equivalent-Domino-Pairs)|46.9%|Easy||
+|1129|Shortest Path with Alternating Colors||42.9%|Medium||
+|1130|Minimum Cost Tree From Leaf Values||68.5%|Medium||
+|1131|Maximum of Absolute Value Expression||49.5%|Medium||
+|1132|Reported Posts II||33.6%|Medium||
+|1133|Largest Unique Number||67.5%|Easy||
+|1134|Armstrong Number||78.1%|Easy||
+|1135|Connecting Cities With Minimum Cost||61.1%|Medium||
+|1136|Parallel Courses||62.0%|Medium||
+|1137|N-th Tribonacci Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1137.N-th-Tribonacci-Number)|63.3%|Easy||
+|1138|Alphabet Board Path||52.3%|Medium||
+|1139|Largest 1-Bordered Square||50.0%|Medium||
+|1140|Stone Game II||64.9%|Medium||
+|1141|User Activity for the Past 30 Days I||50.2%|Easy||
+|1142|User Activity for the Past 30 Days II||36.0%|Easy||
+|1143|Longest Common Subsequence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1143.Longest-Common-Subsequence)|58.8%|Medium||
+|1144|Decrease Elements To Make Array Zigzag||47.0%|Medium||
+|1145|Binary Tree Coloring Game|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1145.Binary-Tree-Coloring-Game)|51.4%|Medium||
+|1146|Snapshot Array||37.3%|Medium||
+|1147|Longest Chunked Palindrome Decomposition||60.0%|Hard||
+|1148|Article Views I||77.0%|Easy||
+|1149|Article Views II||47.6%|Medium||
+|1150|Check If a Number Is Majority Element in a Sorted Array||56.8%|Easy||
+|1151|Minimum Swaps to Group All 1's Together||60.9%|Medium||
+|1152|Analyze User Website Visit Pattern||43.4%|Medium||
+|1153|String Transforms Into Another String||35.3%|Hard||
+|1154|Day of the Year|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1154.Day-of-the-Year)|50.3%|Easy||
+|1155|Number of Dice Rolls With Target Sum||53.6%|Medium||
+|1156|Swap For Longest Repeated Character Substring||45.5%|Medium||
+|1157|Online Majority Element In Subarray|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1157.Online-Majority-Element-In-Subarray)|42.0%|Hard||
+|1158|Market Analysis I||63.9%|Medium||
+|1159|Market Analysis II||58.7%|Hard||
+|1160|Find Words That Can Be Formed by Characters|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1160.Find-Words-That-Can-Be-Formed-by-Characters)|67.8%|Easy||
+|1161|Maximum Level Sum of a Binary Tree||66.2%|Medium||
+|1162|As Far from Land as Possible||48.5%|Medium||
+|1163|Last Substring in Lexicographical Order||35.1%|Hard||
+|1164|Product Price at a Given Date||68.4%|Medium||
+|1165|Single-Row Keyboard||85.7%|Easy||
+|1166|Design File System||61.8%|Medium||
+|1167|Minimum Cost to Connect Sticks||67.7%|Medium||
+|1168|Optimize Water Distribution in a Village||64.4%|Hard||
+|1169|Invalid Transactions||31.0%|Medium||
+|1170|Compare Strings by Frequency of the Smallest Character|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character)|61.3%|Medium||
+|1171|Remove Zero Sum Consecutive Nodes from Linked List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List)|42.9%|Medium||
+|1172|Dinner Plate Stacks||33.7%|Hard||
+|1173|Immediate Food Delivery I||83.4%|Easy||
+|1174|Immediate Food Delivery II||63.9%|Medium||
+|1175|Prime Arrangements|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1175.Prime-Arrangements)|53.5%|Easy||
+|1176|Diet Plan Performance||52.4%|Easy||
+|1177|Can Make Palindrome from Substring||37.8%|Medium||
+|1178|Number of Valid Words for Each Puzzle|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1178.Number-of-Valid-Words-for-Each-Puzzle)|46.6%|Hard||
+|1179|Reformat Department Table||82.7%|Easy||
+|1180|Count Substrings with Only One Distinct Letter||78.9%|Easy||
+|1181|Before and After Puzzle||45.1%|Medium||
+|1182|Shortest Distance to Target Color||55.5%|Medium||
+|1183|Maximum Number of Ones||60.9%|Hard||
+|1184|Distance Between Bus Stops|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1184.Distance-Between-Bus-Stops)|54.1%|Easy||
+|1185|Day of the Week|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1185.Day-of-the-Week)|57.8%|Easy||
+|1186|Maximum Subarray Sum with One Deletion||41.2%|Medium||
+|1187|Make Array Strictly Increasing||45.3%|Hard||
+|1188|Design Bounded Blocking Queue||72.9%|Medium||
+|1189|Maximum Number of Balloons|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1189.Maximum-Number-of-Balloons)|62.2%|Easy||
+|1190|Reverse Substrings Between Each Pair of Parentheses|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses)|65.9%|Medium||
+|1191|K-Concatenation Maximum Sum||24.0%|Medium||
+|1192|Critical Connections in a Network||54.5%|Hard||
+|1193|Monthly Transactions I||67.2%|Medium||
+|1194|Tournament Winners||51.7%|Hard||
+|1195|Fizz Buzz Multithreaded||72.5%|Medium||
+|1196|How Many Apples Can You Put into the Basket||67.0%|Easy||
+|1197|Minimum Knight Moves||39.8%|Medium||
+|1198|Find Smallest Common Element in All Rows||76.0%|Medium||
+|1199|Minimum Time to Build Blocks||40.7%|Hard||
+|1200|Minimum Absolute Difference|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1200.Minimum-Absolute-Difference)|69.8%|Easy||
+|1201|Ugly Number III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1201.Ugly-Number-III)|28.5%|Medium||
+|1202|Smallest String With Swaps|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1202.Smallest-String-With-Swaps)|57.5%|Medium||
+|1203|Sort Items by Groups Respecting Dependencies|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1203.Sort-Items-by-Groups-Respecting-Dependencies)|50.6%|Hard||
+|1204|Last Person to Fit in the Bus||74.1%|Medium||
+|1205|Monthly Transactions II||43.8%|Medium||
+|1206|Design Skiplist||60.5%|Hard||
+|1207|Unique Number of Occurrences|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1207.Unique-Number-of-Occurrences)|70.9%|Easy||
+|1208|Get Equal Substrings Within Budget|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1208.Get-Equal-Substrings-Within-Budget)|47.6%|Medium||
+|1209|Remove All Adjacent Duplicates in String II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1209.Remove-All-Adjacent-Duplicates-in-String-II)|56.0%|Medium||
+|1210|Minimum Moves to Reach Target with Rotations||48.9%|Hard||
+|1211|Queries Quality and Percentage||71.9%|Easy||
+|1212|Team Scores in Football Tournament||57.6%|Medium||
+|1213|Intersection of Three Sorted Arrays||80.0%|Easy||
+|1214|Two Sum BSTs||66.2%|Medium||
+|1215|Stepping Numbers||45.8%|Medium||
+|1216|Valid Palindrome III||50.1%|Hard||
+|1217|Minimum Cost to Move Chips to The Same Position|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position)|72.2%|Easy||
+|1218|Longest Arithmetic Subsequence of Given Difference||51.8%|Medium||
+|1219|Path with Maximum Gold||64.4%|Medium||
+|1220|Count Vowels Permutation||60.7%|Hard||
+|1221|Split a String in Balanced Strings|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1221.Split-a-String-in-Balanced-Strings)|84.7%|Easy||
+|1222|Queens That Can Attack the King||71.7%|Medium||
+|1223|Dice Roll Simulation||48.4%|Hard||
+|1224|Maximum Equal Frequency||36.8%|Hard||
+|1225|Report Contiguous Dates||63.4%|Hard||
+|1226|The Dining Philosophers||56.6%|Medium||
+|1227|Airplane Seat Assignment Probability||64.9%|Medium||
+|1228|Missing Number In Arithmetic Progression||51.4%|Easy||
+|1229|Meeting Scheduler||55.4%|Medium||
+|1230|Toss Strange Coins||53.4%|Medium||
+|1231|Divide Chocolate||56.9%|Hard||
+|1232|Check If It Is a Straight Line|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1232.Check-If-It-Is-a-Straight-Line)|41.2%|Easy||
+|1233|Remove Sub-Folders from the Filesystem||65.4%|Medium||
+|1234|Replace the Substring for Balanced String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1234.Replace-the-Substring-for-Balanced-String)|36.8%|Medium||
+|1235|Maximum Profit in Job Scheduling|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1235.Maximum-Profit-in-Job-Scheduling)|51.1%|Hard||
+|1236|Web Crawler||66.3%|Medium||
+|1237|Find Positive Integer Solution for a Given Equation||69.3%|Medium||
+|1238|Circular Permutation in Binary Representation||68.8%|Medium||
+|1239|Maximum Length of a Concatenated String with Unique Characters|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters)|50.6%|Medium||
+|1240|Tiling a Rectangle with the Fewest Squares||54.0%|Hard||
+|1241|Number of Comments per Post||68.0%|Easy||
+|1242|Web Crawler Multithreaded||49.0%|Medium||
+|1243|Array Transformation||50.7%|Easy||
+|1244|Design A Leaderboard||68.8%|Medium||
+|1245|Tree Diameter||61.7%|Medium||
+|1246|Palindrome Removal||45.9%|Hard||
+|1247|Minimum Swaps to Make Strings Equal||63.7%|Medium||
+|1248|Count Number of Nice Subarrays||59.5%|Medium||
+|1249|Minimum Remove to Make Valid Parentheses|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1249.Minimum-Remove-to-Make-Valid-Parentheses)|65.6%|Medium||
+|1250|Check If It Is a Good Array||58.7%|Hard||
+|1251|Average Selling Price||83.2%|Easy||
+|1252|Cells with Odd Values in a Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1252.Cells-with-Odd-Values-in-a-Matrix)|78.6%|Easy||
+|1253|Reconstruct a 2-Row Binary Matrix||43.8%|Medium||
+|1254|Number of Closed Islands|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1254.Number-of-Closed-Islands)|64.2%|Medium||
+|1255|Maximum Score Words Formed by Letters||72.8%|Hard||
+|1256|Encode Number||69.9%|Medium||
+|1257|Smallest Common Region||64.1%|Medium||
+|1258|Synonymous Sentences||56.4%|Medium||
+|1259|Handshakes That Don't Cross||56.2%|Hard||
+|1260|Shift 2D Grid|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1260.Shift-2D-Grid)|68.0%|Easy||
+|1261|Find Elements in a Contaminated Binary Tree||76.1%|Medium||
+|1262|Greatest Sum Divisible by Three||50.9%|Medium||
+|1263|Minimum Moves to Move a Box to Their Target Location||49.0%|Hard||
+|1264|Page Recommendations||67.8%|Medium||
+|1265|Print Immutable Linked List in Reverse||94.3%|Medium||
+|1266|Minimum Time Visiting All Points|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1266.Minimum-Time-Visiting-All-Points)|79.1%|Easy||
+|1267|Count Servers that Communicate||59.1%|Medium||
+|1268|Search Suggestions System|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1268.Search-Suggestions-System)|66.7%|Medium||
+|1269|Number of Ways to Stay in the Same Place After Some Steps||43.6%|Hard||
+|1270|All People Report to the Given Manager||88.0%|Medium||
+|1271|Hexspeak||57.0%|Easy||
+|1272|Remove Interval||63.3%|Medium||
+|1273|Delete Tree Nodes||61.2%|Medium||
+|1274|Number of Ships in a Rectangle||69.3%|Hard||
+|1275|Find Winner on a Tic Tac Toe Game|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1275.Find-Winner-on-a-Tic-Tac-Toe-Game)|54.3%|Easy||
+|1276|Number of Burgers with No Waste of Ingredients||50.6%|Medium||
+|1277|Count Square Submatrices with All Ones||74.4%|Medium||
+|1278|Palindrome Partitioning III||60.8%|Hard||
+|1279|Traffic Light Controlled Intersection||74.5%|Easy||
+|1280|Students and Examinations||74.5%|Easy||
+|1281|Subtract the Product and Sum of Digits of an Integer|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer)|86.7%|Easy||
+|1282|Group the People Given the Group Size They Belong To||85.7%|Medium||
+|1283|Find the Smallest Divisor Given a Threshold|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1283.Find-the-Smallest-Divisor-Given-a-Threshold)|55.3%|Medium||
+|1284|Minimum Number of Flips to Convert Binary Matrix to Zero Matrix||72.1%|Hard||
+|1285|Find the Start and End Number of Continuous Ranges||88.2%|Medium||
+|1286|Iterator for Combination||73.5%|Medium||
+|1287|Element Appearing More Than 25% In Sorted Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1287.Element-Appearing-More-Than-25-In-Sorted-Array)|59.5%|Easy||
+|1288|Remove Covered Intervals||57.3%|Medium||
+|1289|Minimum Falling Path Sum II||59.6%|Hard||
+|1290|Convert Binary Number in a Linked List to Integer|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer)|82.6%|Easy||
+|1291|Sequential Digits||61.2%|Medium||
+|1292|Maximum Side Length of a Square with Sum Less than or Equal to Threshold||53.1%|Medium||
+|1293|Shortest Path in a Grid with Obstacles Elimination|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination)|43.6%|Hard||
+|1294|Weather Type in Each Country||67.9%|Easy||
+|1295|Find Numbers with Even Number of Digits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1295.Find-Numbers-with-Even-Number-of-Digits)|76.9%|Easy||
+|1296|Divide Array in Sets of K Consecutive Numbers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers)|56.6%|Medium||
+|1297|Maximum Number of Occurrences of a Substring||52.1%|Medium||
+|1298|Maximum Candies You Can Get from Boxes||61.1%|Hard||
+|1299|Replace Elements with Greatest Element on Right Side|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1299.Replace-Elements-with-Greatest-Element-on-Right-Side)|74.7%|Easy||
+|1300|Sum of Mutated Array Closest to Target|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1300.Sum-of-Mutated-Array-Closest-to-Target)|43.1%|Medium||
+|1301|Number of Paths with Max Score||38.7%|Hard||
+|1302|Deepest Leaves Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1302.Deepest-Leaves-Sum)|87.0%|Medium||
+|1303|Find the Team Size||90.8%|Easy||
+|1304|Find N Unique Integers Sum up to Zero|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1304.Find-N-Unique-Integers-Sum-up-to-Zero)|77.1%|Easy||
+|1305|All Elements in Two Binary Search Trees|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1305.All-Elements-in-Two-Binary-Search-Trees)|79.8%|Medium||
+|1306|Jump Game III|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1306.Jump-Game-III)|63.0%|Medium||
+|1307|Verbal Arithmetic Puzzle||34.5%|Hard||
+|1308|Running Total for Different Genders||88.3%|Medium||
+|1309|Decrypt String from Alphabet to Integer Mapping||79.4%|Easy||
+|1310|XOR Queries of a Subarray|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1310.XOR-Queries-of-a-Subarray)|72.1%|Medium||
+|1311|Get Watched Videos by Your Friends||45.9%|Medium||
+|1312|Minimum Insertion Steps to Make a String Palindrome||65.5%|Hard||
+|1313|Decompress Run-Length Encoded List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1313.Decompress-Run-Length-Encoded-List)|85.9%|Easy||
+|1314|Matrix Block Sum||75.4%|Medium||
+|1315|Sum of Nodes with Even-Valued Grandparent||85.6%|Medium||
+|1316|Distinct Echo Substrings||49.8%|Hard||
+|1317|Convert Integer to the Sum of Two No-Zero Integers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers)|56.2%|Easy||
+|1318|Minimum Flips to Make a OR b Equal to c||66.0%|Medium||
+|1319|Number of Operations to Make Network Connected|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1319.Number-of-Operations-to-Make-Network-Connected)|58.3%|Medium||
+|1320|Minimum Distance to Type a Word Using Two Fingers||59.8%|Hard||
+|1321|Restaurant Growth||72.2%|Medium||
+|1322|Ads Performance||60.9%|Easy||
+|1323|Maximum 69 Number||79.1%|Easy||
+|1324|Print Words Vertically||60.0%|Medium||
+|1325|Delete Leaves With a Given Value||74.8%|Medium||
+|1326|Minimum Number of Taps to Open to Water a Garden||47.8%|Hard||
+|1327|List the Products Ordered in a Period||77.1%|Easy||
+|1328|Break a Palindrome||53.4%|Medium||
+|1329|Sort the Matrix Diagonally|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1329.Sort-the-Matrix-Diagonally)|83.6%|Medium||
+|1330|Reverse Subarray To Maximize Array Value||39.9%|Hard||
+|1331|Rank Transform of an Array||59.0%|Easy||
+|1332|Remove Palindromic Subsequences|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1332.Remove-Palindromic-Subsequences)|76.1%|Easy||
+|1333|Filter Restaurants by Vegan-Friendly, Price and Distance||59.4%|Medium||
+|1334|Find the City With the Smallest Number of Neighbors at a Threshold Distance||52.9%|Medium||
+|1335|Minimum Difficulty of a Job Schedule||58.8%|Hard||
+|1336|Number of Transactions per Visit||51.4%|Hard||
+|1337|The K Weakest Rows in a Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1337.The-K-Weakest-Rows-in-a-Matrix)|73.1%|Easy||
+|1338|Reduce Array Size to The Half||69.8%|Medium||
+|1339|Maximum Product of Splitted Binary Tree||43.3%|Medium||
+|1340|Jump Game V||62.6%|Hard||
+|1341|Movie Rating||58.6%|Medium||
+|1342|Number of Steps to Reduce a Number to Zero||85.6%|Easy||
+|1343|Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold||67.7%|Medium||
+|1344|Angle Between Hands of a Clock||63.4%|Medium||
+|1345|Jump Game IV||44.1%|Hard||
+|1346|Check If N and Its Double Exist||36.1%|Easy||
+|1347|Minimum Number of Steps to Make Two Strings Anagram||77.2%|Medium||
+|1348|Tweet Counts Per Frequency||43.5%|Medium||
+|1349|Maximum Students Taking Exam||48.0%|Hard||
+|1350|Students With Invalid Departments||90.5%|Easy||
+|1351|Count Negative Numbers in a Sorted Matrix||75.3%|Easy||
+|1352|Product of the Last K Numbers||49.4%|Medium||
+|1353|Maximum Number of Events That Can Be Attended|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended)|32.9%|Medium||
+|1354|Construct Target Array With Multiple Sums||36.3%|Hard||
+|1355|Activity Participants||74.4%|Medium||
+|1356|Sort Integers by The Number of 1 Bits||72.0%|Easy||
+|1357|Apply Discount Every n Orders||69.9%|Medium||
+|1358|Number of Substrings Containing All Three Characters||62.9%|Medium||
+|1359|Count All Valid Pickup and Delivery Options||63.0%|Hard||
+|1360|Number of Days Between Two Dates||47.5%|Easy||
+|1361|Validate Binary Tree Nodes||40.4%|Medium||
+|1362|Closest Divisors||59.8%|Medium||
+|1363|Largest Multiple of Three||33.5%|Hard||
+|1364|Number of Trusted Contacts of a Customer||79.0%|Medium||
+|1365|How Many Numbers Are Smaller Than the Current Number||86.6%|Easy||
+|1366|Rank Teams by Votes||58.6%|Medium||
+|1367|Linked List in Binary Tree||43.4%|Medium||
+|1368|Minimum Cost to Make at Least One Valid Path in a Grid||61.3%|Hard||
+|1369|Get the Second Most Recent Activity||69.6%|Hard||
+|1370|Increasing Decreasing String||77.5%|Easy||
+|1371|Find the Longest Substring Containing Vowels in Even Counts||63.0%|Medium||
+|1372|Longest ZigZag Path in a Binary Tree||59.6%|Medium||
+|1373|Maximum Sum BST in Binary Tree||39.2%|Hard||
+|1374|Generate a String With Characters That Have Odd Counts||77.5%|Easy||
+|1375|Number of Times Binary String Is Prefix-Aligned||65.9%|Medium||
+|1376|Time Needed to Inform All Employees||58.4%|Medium||
+|1377|Frog Position After T Seconds||36.1%|Hard||
+|1378|Replace Employee ID With The Unique Identifier||91.4%|Easy||
+|1379|Find a Corresponding Node of a Binary Tree in a Clone of That Tree||87.1%|Easy||
+|1380|Lucky Numbers in a Matrix|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1380.Lucky-Numbers-in-a-Matrix)|70.5%|Easy||
+|1381|Design a Stack With Increment Operation||77.4%|Medium||
+|1382|Balance a Binary Search Tree||80.7%|Medium||
+|1383|Maximum Performance of a Team|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1383.Maximum-Performance-of-a-Team)|49.0%|Hard||
+|1384|Total Sales Amount by Year||67.6%|Hard||
+|1385|Find the Distance Value Between Two Arrays|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1385.Find-the-Distance-Value-Between-Two-Arrays)|65.3%|Easy||
+|1386|Cinema Seat Allocation||40.9%|Medium||
+|1387|Sort Integers by The Power Value||69.9%|Medium||
+|1388|Pizza With 3n Slices||50.1%|Hard||
+|1389|Create Target Array in the Given Order|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1389.Create-Target-Array-in-the-Given-Order)|85.8%|Easy||
+|1390|Four Divisors||41.2%|Medium||
+|1391|Check if There is a Valid Path in a Grid||47.0%|Medium||
+|1392|Longest Happy Prefix||45.0%|Hard||
+|1393|Capital Gain/Loss||91.1%|Medium||
+|1394|Find Lucky Integer in an Array||63.5%|Easy||
+|1395|Count Number of Teams||68.3%|Medium||
+|1396|Design Underground System|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1396.Design-Underground-System)|73.7%|Medium||
+|1397|Find All Good Strings||42.1%|Hard||
+|1398|Customers Who Bought Products A and B but Not C||77.7%|Medium||
+|1399|Count Largest Group||67.2%|Easy||
+|1400|Construct K Palindrome Strings||63.3%|Medium||
+|1401|Circle and Rectangle Overlapping||44.2%|Medium||
+|1402|Reducing Dishes||72.0%|Hard||
+|1403|Minimum Subsequence in Non-Increasing Order||72.2%|Easy||
+|1404|Number of Steps to Reduce a Number in Binary Representation to One||52.0%|Medium||
+|1405|Longest Happy String||57.4%|Medium||
+|1406|Stone Game III||59.7%|Hard||
+|1407|Top Travellers||68.7%|Easy||
+|1408|String Matching in an Array||63.9%|Easy||
+|1409|Queries on a Permutation With Key||83.3%|Medium||
+|1410|HTML Entity Parser||52.1%|Medium||
+|1411|Number of Ways to Paint N × 3 Grid||62.3%|Hard||
+|1412|Find the Quiet Students in All Exams||63.0%|Hard||
+|1413|Minimum Value to Get Positive Step by Step Sum||68.4%|Easy||
+|1414|Find the Minimum Number of Fibonacci Numbers Whose Sum Is K||65.4%|Medium||
+|1415|The k-th Lexicographical String of All Happy Strings of Length n||72.0%|Medium||
+|1416|Restore The Array||38.4%|Hard||
+|1417|Reformat The String||55.8%|Easy||
+|1418|Display Table of Food Orders in a Restaurant||73.7%|Medium||
+|1419|Minimum Number of Frogs Croaking||49.9%|Medium||
+|1420|Build Array Where You Can Find The Maximum Exactly K Comparisons||63.6%|Hard||
+|1421|NPV Queries||84.1%|Easy||
+|1422|Maximum Score After Splitting a String||57.8%|Easy||
+|1423|Maximum Points You Can Obtain from Cards|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1423.Maximum-Points-You-Can-Obtain-from-Cards)|52.3%|Medium||
+|1424|Diagonal Traverse II||50.3%|Medium||
+|1425|Constrained Subsequence Sum||47.4%|Hard||
+|1426|Counting Elements||59.5%|Easy||
+|1427|Perform String Shifts||54.2%|Easy||
+|1428|Leftmost Column with at Least a One||53.0%|Medium||
+|1429|First Unique Number||52.7%|Medium||
+|1430|Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree||46.1%|Medium||
+|1431|Kids With the Greatest Number of Candies||87.5%|Easy||
+|1432|Max Difference You Can Get From Changing an Integer||43.0%|Medium||
+|1433|Check If a String Can Break Another String||68.8%|Medium||
+|1434|Number of Ways to Wear Different Hats to Each Other||42.8%|Hard||
+|1435|Create a Session Bar Chart||78.3%|Easy||
+|1436|Destination City||77.6%|Easy||
+|1437|Check If All 1's Are at Least Length K Places Away|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away)|59.2%|Easy||
+|1438|Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit)|48.0%|Medium||
+|1439|Find the Kth Smallest Sum of a Matrix With Sorted Rows|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows)|61.4%|Hard||
+|1440|Evaluate Boolean Expression||76.5%|Medium||
+|1441|Build an Array With Stack Operations||71.3%|Medium||
+|1442|Count Triplets That Can Form Two Arrays of Equal XOR|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR)|75.6%|Medium||
+|1443|Minimum Time to Collect All Apples in a Tree||56.0%|Medium||
+|1444|Number of Ways of Cutting a Pizza||58.6%|Hard||
+|1445|Apples & Oranges||91.3%|Medium||
+|1446|Consecutive Characters|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1446.Consecutive-Characters)|61.7%|Easy||
+|1447|Simplified Fractions||64.6%|Medium||
+|1448|Count Good Nodes in Binary Tree||74.6%|Medium||
+|1449|Form Largest Integer With Digits That Add up to Target||47.1%|Hard||
+|1450|Number of Students Doing Homework at a Given Time||75.9%|Easy||
+|1451|Rearrange Words in a Sentence||62.5%|Medium||
+|1452|People Whose List of Favorite Companies Is Not a Subset of Another List||56.8%|Medium||
+|1453|Maximum Number of Darts Inside of a Circular Dartboard||36.8%|Hard||
+|1454|Active Users||38.4%|Medium||
+|1455|Check If a Word Occurs As a Prefix of Any Word in a Sentence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence)|64.2%|Easy||
+|1456|Maximum Number of Vowels in a Substring of Given Length||57.9%|Medium||
+|1457|Pseudo-Palindromic Paths in a Binary Tree||68.1%|Medium||
+|1458|Max Dot Product of Two Subsequences||46.2%|Hard||
+|1459|Rectangles Area||70.0%|Medium||
+|1460|Make Two Arrays Equal by Reversing Subarrays||72.2%|Easy||
+|1461|Check If a String Contains All Binary Codes of Size K|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K)|56.8%|Medium||
+|1462|Course Schedule IV||48.9%|Medium||
+|1463|Cherry Pickup II|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1463.Cherry-Pickup-II)|70.2%|Hard||
+|1464|Maximum Product of Two Elements in an Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array)|79.2%|Easy||
+|1465|Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts)|40.8%|Medium||
+|1466|Reorder Routes to Make All Paths Lead to the City Zero||61.8%|Medium||
+|1467|Probability of a Two Boxes Having The Same Number of Distinct Balls||60.8%|Hard||
+|1468|Calculate Salaries||81.9%|Medium||
+|1469|Find All The Lonely Nodes||81.7%|Easy||
+|1470|Shuffle the Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1470.Shuffle-the-Array)|88.5%|Easy||
+|1471|The k Strongest Values in an Array||60.1%|Medium||
+|1472|Design Browser History||76.1%|Medium||
+|1473|Paint House III||62.0%|Hard||
+|1474|Delete N Nodes After M Nodes of a Linked List||73.8%|Easy||
+|1475|Final Prices With a Special Discount in a Shop||75.4%|Easy||
+|1476|Subrectangle Queries||88.5%|Medium||
+|1477|Find Two Non-overlapping Sub-arrays Each With Target Sum||37.0%|Medium||
+|1478|Allocate Mailboxes||55.3%|Hard||
+|1479|Sales by Day of the Week||82.4%|Hard||
+|1480|Running Sum of 1d Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1480.Running-Sum-of-1d-Array)|89.8%|Easy||
+|1481|Least Number of Unique Integers after K Removals||57.2%|Medium||
+|1482|Minimum Number of Days to Make m Bouquets|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1482.Minimum-Number-of-Days-to-Make-m-Bouquets)|56.7%|Medium||
+|1483|Kth Ancestor of a Tree Node||33.8%|Hard||
+|1484|Group Sold Products By The Date||84.1%|Easy||
+|1485|Clone Binary Tree With Random Pointer||79.5%|Medium||
+|1486|XOR Operation in an Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1486.XOR-Operation-in-an-Array)|84.2%|Easy||
+|1487|Making File Names Unique||35.6%|Medium||
+|1488|Avoid Flood in The City||26.1%|Medium||
+|1489|Find Critical and Pseudo-Critical Edges in Minimum Spanning Tree||53.4%|Hard||
+|1490|Clone N-ary Tree||83.6%|Medium||
+|1491|Average Salary Excluding the Minimum and Maximum Salary||63.1%|Easy||
+|1492|The kth Factor of n||62.3%|Medium||
+|1493|Longest Subarray of 1's After Deleting One Element||60.1%|Medium||
+|1494|Parallel Courses II||31.1%|Hard||
+|1495|Friendly Movies Streamed Last Month||49.8%|Easy||
+|1496|Path Crossing||55.8%|Easy||
+|1497|Check If Array Pairs Are Divisible by k||39.8%|Medium||
+|1498|Number of Subsequences That Satisfy the Given Sum Condition||38.4%|Medium||
+|1499|Max Value of Equation||46.5%|Hard||
+|1500|Design a File Sharing System||44.9%|Medium||
+|1501|Countries You Can Safely Invest In||58.3%|Medium||
+|1502|Can Make Arithmetic Progression From Sequence||68.3%|Easy||
+|1503|Last Moment Before All Ants Fall Out of a Plank||55.2%|Medium||
+|1504|Count Submatrices With All Ones||57.9%|Medium||
+|1505|Minimum Possible Integer After at Most K Adjacent Swaps On Digits||38.1%|Hard||
+|1506|Find Root of N-Ary Tree||78.1%|Medium||
+|1507|Reformat Date||62.4%|Easy||
+|1508|Range Sum of Sorted Subarray Sums||59.4%|Medium||
+|1509|Minimum Difference Between Largest and Smallest Value in Three Moves||54.6%|Medium||
+|1510|Stone Game IV||60.6%|Hard||
+|1511|Customer Order Frequency||73.1%|Easy||
+|1512|Number of Good Pairs|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1512.Number-of-Good-Pairs)|88.2%|Easy||
+|1513|Number of Substrings With Only 1s||45.2%|Medium||
+|1514|Path with Maximum Probability||48.3%|Medium||
+|1515|Best Position for a Service Centre||38.4%|Hard||
+|1516|Move Sub-Tree of N-Ary Tree||63.9%|Hard||
+|1517|Find Users With Valid E-Mails||57.1%|Easy||
+|1518|Water Bottles|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1518.Water-Bottles)|60.3%|Easy||
+|1519|Number of Nodes in the Sub-Tree With the Same Label||40.9%|Medium||
+|1520|Maximum Number of Non-Overlapping Substrings||37.9%|Hard||
+|1521|Find a Value of a Mysterious Function Closest to Target||43.6%|Hard||
+|1522|Diameter of N-Ary Tree||73.4%|Medium||
+|1523|Count Odd Numbers in an Interval Range||46.4%|Easy||
+|1524|Number of Sub-arrays With Odd Sum||43.6%|Medium||
+|1525|Number of Good Ways to Split a String||69.5%|Medium||
+|1526|Minimum Number of Increments on Subarrays to Form a Target Array||68.6%|Hard||
+|1527|Patients With a Condition||42.9%|Easy||
+|1528|Shuffle String||85.7%|Easy||
+|1529|Minimum Suffix Flips||72.5%|Medium||
+|1530|Number of Good Leaf Nodes Pairs||60.5%|Medium||
+|1531|String Compression II||50.2%|Hard||
+|1532|The Most Recent Three Orders||71.1%|Medium||
+|1533|Find the Index of the Large Integer||50.7%|Medium||
+|1534|Count Good Triplets||80.8%|Easy||
+|1535|Find the Winner of an Array Game||48.8%|Medium||
+|1536|Minimum Swaps to Arrange a Binary Grid||46.4%|Medium||
+|1537|Get the Maximum Score||39.3%|Hard||
+|1538|Guess the Majority in a Hidden Array||63.0%|Medium||
+|1539|Kth Missing Positive Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1539.Kth-Missing-Positive-Number)|55.9%|Easy||
+|1540|Can Convert String in K Moves||33.1%|Medium||
+|1541|Minimum Insertions to Balance a Parentheses String||49.9%|Medium||
+|1542|Find Longest Awesome Substring||41.6%|Hard||
+|1543|Fix Product Name Format||62.4%|Easy||
+|1544|Make The String Great||57.0%|Easy||
+|1545|Find Kth Bit in Nth Binary String||58.2%|Medium||
+|1546|Maximum Number of Non-Overlapping Subarrays With Sum Equals Target||47.2%|Medium||
+|1547|Minimum Cost to Cut a Stick||56.8%|Hard||
+|1548|The Most Similar Path in a Graph||56.9%|Hard||
+|1549|The Most Recent Orders for Each Product||67.9%|Medium||
+|1550|Three Consecutive Odds||63.7%|Easy||
+|1551|Minimum Operations to Make Array Equal|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1551.Minimum-Operations-to-Make-Array-Equal)|81.1%|Medium||
+|1552|Magnetic Force Between Two Balls||56.8%|Medium||
+|1553|Minimum Number of Days to Eat N Oranges||34.4%|Hard||
+|1554|Strings Differ by One Character||43.6%|Medium||
+|1555|Bank Account Summary||53.0%|Medium||
+|1556|Thousand Separator||55.1%|Easy||
+|1557|Minimum Number of Vertices to Reach All Nodes||79.6%|Medium||
+|1558|Minimum Numbers of Function Calls to Make Target Array||64.3%|Medium||
+|1559|Detect Cycles in 2D Grid||48.1%|Medium||
+|1560|Most Visited Sector in  a Circular Track||58.4%|Easy||
+|1561|Maximum Number of Coins You Can Get||78.6%|Medium||
+|1562|Find Latest Group of Size M||42.4%|Medium||
+|1563|Stone Game V||40.6%|Hard||
+|1564|Put Boxes Into the Warehouse I||67.1%|Medium||
+|1565|Unique Orders and Customers Per Month||83.6%|Easy||
+|1566|Detect Pattern of Length M Repeated K or More Times||43.6%|Easy||
+|1567|Maximum Length of Subarray With Positive Product||43.7%|Medium||
+|1568|Minimum Number of Days to Disconnect Island||47.1%|Hard||
+|1569|Number of Ways to Reorder Array to Get Same BST||48.2%|Hard||
+|1570|Dot Product of Two Sparse Vectors||90.4%|Medium||
+|1571|Warehouse Manager||90.1%|Easy||
+|1572|Matrix Diagonal Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1572.Matrix-Diagonal-Sum)|79.7%|Easy||
+|1573|Number of Ways to Split a String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1573.Number-of-Ways-to-Split-a-String)|32.4%|Medium||
+|1574|Shortest Subarray to be Removed to Make Array Sorted||36.5%|Medium||
+|1575|Count All Possible Routes||56.9%|Hard||
+|1576|Replace All ?'s to Avoid Consecutive Repeating Characters|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters)|49.2%|Easy||
+|1577|Number of Ways Where Square of Number Is Equal to Product of Two Numbers||39.9%|Medium||
+|1578|Minimum Time to Make Rope Colorful||63.7%|Medium||
+|1579|Remove Max Number of Edges to Keep Graph Fully Traversable|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable)|52.7%|Hard||
+|1580|Put Boxes Into the Warehouse II||63.5%|Medium||
+|1581|Customer Who Visited but Did Not Make Any Transactions||89.0%|Easy||
+|1582|Special Positions in a Binary Matrix||65.3%|Easy||
+|1583|Count Unhappy Friends||59.8%|Medium||
+|1584|Min Cost to Connect All Points||64.3%|Medium||
+|1585|Check If String Is Transformable With Substring Sort Operations||48.4%|Hard||
+|1586|Binary Search Tree Iterator II||70.8%|Medium||
+|1587|Bank Account Summary II||90.2%|Easy||
+|1588|Sum of All Odd Length Subarrays||83.6%|Easy||
+|1589|Maximum Sum Obtained of Any Permutation||37.0%|Medium||
+|1590|Make Sum Divisible by P||28.1%|Medium||
+|1591|Strange Printer II||58.4%|Hard||
+|1592|Rearrange Spaces Between Words||43.8%|Easy||
+|1593|Split a String Into the Max Number of Unique Substrings||55.1%|Medium||
+|1594|Maximum Non Negative Product in a Matrix||33.1%|Medium||
+|1595|Minimum Cost to Connect Two Groups of Points||46.3%|Hard||
+|1596|The Most Frequently Ordered Products for Each Customer||85.1%|Medium||
+|1597|Build Binary Expression Tree From Infix Expression||62.2%|Hard||
+|1598|Crawler Log Folder||64.3%|Easy||
+|1599|Maximum Profit of Operating a Centennial Wheel||43.7%|Medium||
+|1600|Throne Inheritance|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1600.Throne-Inheritance)|63.6%|Medium||
+|1601|Maximum Number of Achievable Transfer Requests||51.3%|Hard||
+|1602|Find Nearest Right Node in Binary Tree||75.3%|Medium||
+|1603|Design Parking System|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1603.Design-Parking-System)|88.1%|Easy||
+|1604|Alert Using Same Key-Card Three or More Times in a One Hour Period||47.4%|Medium||
+|1605|Find Valid Matrix Given Row and Column Sums||78.1%|Medium||
+|1606|Find Servers That Handled Most Number of Requests||42.8%|Hard||
+|1607|Sellers With No Sales||55.3%|Easy||
+|1608|Special Array With X Elements Greater Than or Equal X|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X)|60.0%|Easy||
+|1609|Even Odd Tree|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1609.Even-Odd-Tree)|53.6%|Medium||
+|1610|Maximum Number of Visible Points||37.4%|Hard||
+|1611|Minimum One Bit Operations to Make Integers Zero||63.2%|Hard||
+|1612|Check If Two Expression Trees are Equivalent||69.9%|Medium||
+|1613|Find the Missing IDs||75.9%|Medium||
+|1614|Maximum Nesting Depth of the Parentheses|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1614.Maximum-Nesting-Depth-of-the-Parentheses)|82.7%|Easy||
+|1615|Maximal Network Rank||58.1%|Medium||
+|1616|Split Two Strings to Make Palindrome||31.3%|Medium||
+|1617|Count Subtrees With Max Distance Between Cities||65.9%|Hard||
+|1618|Maximum Font to Fit a Sentence in a Screen||59.1%|Medium||
+|1619|Mean of Array After Removing Some Elements|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1619.Mean-of-Array-After-Removing-Some-Elements)|64.6%|Easy||
+|1620|Coordinate With Maximum Network Quality||37.3%|Medium||
+|1621|Number of Sets of K Non-Overlapping Line Segments||42.3%|Medium||
+|1622|Fancy Sequence||16.1%|Hard||
+|1623|All Valid Triplets That Can Represent a Country||88.0%|Easy||
+|1624|Largest Substring Between Two Equal Characters|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1624.Largest-Substring-Between-Two-Equal-Characters)|59.0%|Easy||
+|1625|Lexicographically Smallest String After Applying Operations||66.0%|Medium||
+|1626|Best Team With No Conflicts||41.1%|Medium||
+|1627|Graph Connectivity With Threshold||45.5%|Hard||
+|1628|Design an Expression Tree With Evaluate Function||82.7%|Medium||
+|1629|Slowest Key|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1629.Slowest-Key)|59.4%|Easy||
+|1630|Arithmetic Subarrays||79.9%|Medium||
+|1631|Path With Minimum Effort|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1631.Path-With-Minimum-Effort)|55.3%|Medium||
+|1632|Rank Transform of a Matrix||41.0%|Hard||
+|1633|Percentage of Users Attended a Contest||68.8%|Easy||
+|1634|Add Two Polynomials Represented as Linked Lists||53.4%|Medium||
+|1635|Hopper Company Queries I||53.0%|Hard||
+|1636|Sort Array by Increasing Frequency|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1636.Sort-Array-by-Increasing-Frequency)|68.6%|Easy||
+|1637|Widest Vertical Area Between Two Points Containing No Points||84.2%|Medium||
+|1638|Count Substrings That Differ by One Character||71.5%|Medium||
+|1639|Number of Ways to Form a Target String Given a Dictionary||42.9%|Hard||
+|1640|Check Array Formation Through Concatenation|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1640.Check-Array-Formation-Through-Concatenation)|56.1%|Easy||
+|1641|Count Sorted Vowel Strings|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1641.Count-Sorted-Vowel-Strings)|77.3%|Medium||
+|1642|Furthest Building You Can Reach|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1642.Furthest-Building-You-Can-Reach)|48.3%|Medium||
+|1643|Kth Smallest Instructions||46.3%|Hard||
+|1644|Lowest Common Ancestor of a Binary Tree II||59.3%|Medium||
+|1645|Hopper Company Queries II||38.8%|Hard||
+|1646|Get Maximum in Generated Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1646.Get-Maximum-in-Generated-Array)|50.2%|Easy||
+|1647|Minimum Deletions to Make Character Frequencies Unique|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique)|59.2%|Medium||
+|1648|Sell Diminishing-Valued Colored Balls|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1648.Sell-Diminishing-Valued-Colored-Balls)|30.6%|Medium||
+|1649|Create Sorted Array through Instructions|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1649.Create-Sorted-Array-through-Instructions)|37.2%|Hard||
+|1650|Lowest Common Ancestor of a Binary Tree III||77.3%|Medium||
+|1651|Hopper Company Queries III||68.0%|Hard||
+|1652|Defuse the Bomb|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1652.Defuse-the-Bomb)|61.1%|Easy||
+|1653|Minimum Deletions to Make String Balanced|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced)|58.7%|Medium||
+|1654|Minimum Jumps to Reach Home|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1654.Minimum-Jumps-to-Reach-Home)|28.6%|Medium||
+|1655|Distribute Repeating Integers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1655.Distribute-Repeating-Integers)|39.7%|Hard||
+|1656|Design an Ordered Stream|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1656.Design-an-Ordered-Stream)|85.3%|Easy||
+|1657|Determine if Two Strings Are Close|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1657.Determine-if-Two-Strings-Are-Close)|54.2%|Medium||
+|1658|Minimum Operations to Reduce X to Zero|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero)|37.6%|Medium||
+|1659|Maximize Grid Happiness|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1659.Maximize-Grid-Happiness)|38.4%|Hard||
+|1660|Correct a Binary Tree||72.5%|Medium||
+|1661|Average Time of Process per Machine||79.4%|Easy||
+|1662|Check If Two String Arrays are Equivalent|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1662.Check-If-Two-String-Arrays-are-Equivalent)|81.8%|Easy||
+|1663|Smallest String With A Given Numeric Value|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1663.Smallest-String-With-A-Given-Numeric-Value)|67.0%|Medium||
+|1664|Ways to Make a Fair Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1664.Ways-to-Make-a-Fair-Array)|63.6%|Medium||
+|1665|Minimum Initial Energy to Finish Tasks|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1665.Minimum-Initial-Energy-to-Finish-Tasks)|56.2%|Hard||
+|1666|Change the Root of a Binary Tree||69.4%|Medium||
+|1667|Fix Names in a Table||66.8%|Easy||
+|1668|Maximum Repeating Substring|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1668.Maximum-Repeating-Substring)|39.6%|Easy||
+|1669|Merge In Between Linked Lists|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1669.Merge-In-Between-Linked-Lists)|74.5%|Medium||
+|1670|Design Front Middle Back Queue|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1670.Design-Front-Middle-Back-Queue)|56.4%|Medium||
+|1671|Minimum Number of Removals to Make Mountain Array||42.6%|Hard||
+|1672|Richest Customer Wealth|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1672.Richest-Customer-Wealth)|88.4%|Easy||
+|1673|Find the Most Competitive Subsequence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1673.Find-the-Most-Competitive-Subsequence)|49.3%|Medium||
+|1674|Minimum Moves to Make Array Complementary|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1674.Minimum-Moves-to-Make-Array-Complementary)|38.5%|Medium||
+|1675|Minimize Deviation in Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1675.Minimize-Deviation-in-Array)|52.1%|Hard||
+|1676|Lowest Common Ancestor of a Binary Tree IV||79.3%|Medium||
+|1677|Product's Worth Over Invoices||39.4%|Easy||
+|1678|Goal Parser Interpretation|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1678.Goal-Parser-Interpretation)|86.0%|Easy||
+|1679|Max Number of K-Sum Pairs|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1679.Max-Number-of-K-Sum-Pairs)|57.4%|Medium||
+|1680|Concatenation of Consecutive Binary Numbers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1680.Concatenation-of-Consecutive-Binary-Numbers)|57.0%|Medium||
+|1681|Minimum Incompatibility|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1681.Minimum-Incompatibility)|37.3%|Hard||
+|1682|Longest Palindromic Subsequence II||49.7%|Medium||
+|1683|Invalid Tweets||91.0%|Easy||
+|1684|Count the Number of Consistent Strings|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1684.Count-the-Number-of-Consistent-Strings)|81.8%|Easy||
+|1685|Sum of Absolute Differences in a Sorted Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array)|65.3%|Medium||
+|1686|Stone Game VI||54.3%|Medium||
+|1687|Delivering Boxes from Storage to Ports||38.2%|Hard||
+|1688|Count of Matches in Tournament|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1688.Count-of-Matches-in-Tournament)|83.1%|Easy||
+|1689|Partitioning Into Minimum Number Of Deci-Binary Numbers|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers)|89.7%|Medium||
+|1690|Stone Game VII|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1690.Stone-Game-VII)|58.7%|Medium||
+|1691|Maximum Height by Stacking Cuboids|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1691.Maximum-Height-by-Stacking-Cuboids)|54.2%|Hard||
+|1692|Count Ways to Distribute Candies||62.0%|Hard||
+|1693|Daily Leads and Partners||90.4%|Easy||
+|1694|Reformat Phone Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1694.Reformat-Phone-Number)|64.7%|Easy||
+|1695|Maximum Erasure Value|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1695.Maximum-Erasure-Value)|57.7%|Medium||
+|1696|Jump Game VI|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1696.Jump-Game-VI)|46.3%|Medium||
+|1697|Checking Existence of Edge Length Limited Paths||50.5%|Hard||
+|1698|Number of Distinct Substrings in a String||63.1%|Medium||
+|1699|Number of Calls Between Two Persons||85.8%|Medium||
+|1700|Number of Students Unable to Eat Lunch|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1700.Number-of-Students-Unable-to-Eat-Lunch)|67.7%|Easy||
+|1701|Average Waiting Time||62.3%|Medium||
+|1702|Maximum Binary String After Change||46.1%|Medium||
+|1703|Minimum Adjacent Swaps for K Consecutive Ones||41.3%|Hard||
+|1704|Determine if String Halves Are Alike|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1704.Determine-if-String-Halves-Are-Alike)|77.4%|Easy||
+|1705|Maximum Number of Eaten Apples|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1705.Maximum-Number-of-Eaten-Apples)|37.9%|Medium||
+|1706|Where Will the Ball Fall||65.8%|Medium||
+|1707|Maximum XOR With an Element From Array||44.3%|Hard||
+|1708|Largest Subarray Length K||63.5%|Easy||
+|1709|Biggest Window Between Visits||77.8%|Medium||
+|1710|Maximum Units on a Truck|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1710.Maximum-Units-on-a-Truck)|74.0%|Easy||
+|1711|Count Good Meals||29.0%|Medium||
+|1712|Ways to Split Array Into Three Subarrays||32.4%|Medium||
+|1713|Minimum Operations to Make a Subsequence||49.2%|Hard||
+|1714|Sum Of Special Evenly-Spaced Elements In Array||49.8%|Hard||
+|1715|Count Apples and Oranges||77.8%|Medium||
+|1716|Calculate Money in Leetcode Bank|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1716.Calculate-Money-in-Leetcode-Bank)|65.0%|Easy||
+|1717|Maximum Score From Removing Substrings||46.0%|Medium||
+|1718|Construct the Lexicographically Largest Valid Sequence||51.7%|Medium||
+|1719|Number Of Ways To Reconstruct A Tree||42.5%|Hard||
+|1720|Decode XORed Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1720.Decode-XORed-Array)|86.0%|Easy||
+|1721|Swapping Nodes in a Linked List|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1721.Swapping-Nodes-in-a-Linked-List)|67.8%|Medium||
+|1722|Minimize Hamming Distance After Swap Operations||48.6%|Medium||
+|1723|Find Minimum Time to Finish All Jobs||42.7%|Hard||
+|1724|Checking Existence of Edge Length Limited Paths II||51.3%|Hard||
+|1725|Number Of Rectangles That Can Form The Largest Square|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square)|78.7%|Easy||
+|1726|Tuple with Same Product||60.7%|Medium||
+|1727|Largest Submatrix With Rearrangements||60.8%|Medium||
+|1728|Cat and Mouse II||40.5%|Hard||
+|1729|Find Followers Count||71.6%|Easy||
+|1730|Shortest Path to Get Food||54.0%|Medium||
+|1731|The Number of Employees Which Report to Each Employee||51.1%|Easy||
+|1732|Find the Highest Altitude|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1732.Find-the-Highest-Altitude)|78.7%|Easy||
+|1733|Minimum Number of People to Teach||41.6%|Medium||
+|1734|Decode XORed Permutation|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1734.Decode-XORed-Permutation)|62.2%|Medium||
+|1735|Count Ways to Make Array With Product||49.2%|Hard||
+|1736|Latest Time by Replacing Hidden Digits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1736.Latest-Time-by-Replacing-Hidden-Digits)|42.2%|Easy||
+|1737|Change Minimum Characters to Satisfy One of Three Conditions||35.2%|Medium||
+|1738|Find Kth Largest XOR Coordinate Value|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1738.Find-Kth-Largest-XOR-Coordinate-Value)|61.6%|Medium||
+|1739|Building Boxes||51.7%|Hard||
+|1740|Find Distance in a Binary Tree||68.7%|Medium||
+|1741|Find Total Time Spent by Each Employee||92.0%|Easy||
+|1742|Maximum Number of Balls in a Box|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1742.Maximum-Number-of-Balls-in-a-Box)|73.9%|Easy||
+|1743|Restore the Array From Adjacent Pairs||68.7%|Medium||
+|1744|Can You Eat Your Favorite Candy on Your Favorite Day?|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day)|32.8%|Medium||
+|1745|Palindrome Partitioning IV||46.3%|Hard||
+|1746|Maximum Subarray Sum After One Operation||62.2%|Medium||
+|1747|Leetflex Banned Accounts||68.2%|Medium||
+|1748|Sum of Unique Elements|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1748.Sum-of-Unique-Elements)|75.6%|Easy||
+|1749|Maximum Absolute Sum of Any Subarray||58.2%|Medium||
+|1750|Minimum Length of String After Deleting Similar Ends||43.6%|Medium||
+|1751|Maximum Number of Events That Can Be Attended II||55.8%|Hard||
+|1752|Check if Array Is Sorted and Rotated|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1752.Check-if-Array-Is-Sorted-and-Rotated)|49.2%|Easy||
+|1753|Maximum Score From Removing Stones||66.1%|Medium||
+|1754|Largest Merge Of Two Strings||45.0%|Medium||
+|1755|Closest Subsequence Sum||36.6%|Hard||
+|1756|Design Most Recently Used Queue||79.0%|Medium||
+|1757|Recyclable and Low Fat Products||93.9%|Easy||
+|1758|Minimum Changes To Make Alternating Binary String|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String)|58.3%|Easy||
+|1759|Count Number of Homogenous Substrings||47.8%|Medium||
+|1760|Minimum Limit of Balls in a Bag||60.3%|Medium||
+|1761|Minimum Degree of a Connected Trio in a Graph||41.7%|Hard||
+|1762|Buildings With an Ocean View||79.2%|Medium||
+|1763|Longest Nice Substring|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1763.Longest-Nice-Substring)|61.7%|Easy||
+|1764|Form Array by Concatenating Subarrays of Another Array||52.7%|Medium||
+|1765|Map of Highest Peak||60.3%|Medium||
+|1766|Tree of Coprimes||38.5%|Hard||
+|1767|Find the Subtasks That Did Not Execute||84.7%|Hard||
+|1768|Merge Strings Alternately||75.9%|Easy||
+|1769|Minimum Number of Operations to Move All Balls to Each Box||85.4%|Medium||
+|1770|Maximum Score from Performing Multiplication Operations||36.3%|Hard||
+|1771|Maximize Palindrome Length From Subsequences||35.2%|Hard||
+|1772|Sort Features by Popularity||65.1%|Medium||
+|1773|Count Items Matching a Rule||84.3%|Easy||
+|1774|Closest Dessert Cost||46.7%|Medium||
+|1775|Equal Sum Arrays With Minimum Number of Operations||52.6%|Medium||
+|1776|Car Fleet II||53.3%|Hard||
+|1777|Product's Price for Each Store||85.4%|Easy||
+|1778|Shortest Path in a Hidden Grid||39.7%|Medium||
+|1779|Find Nearest Point That Has the Same X or Y Coordinate||67.2%|Easy||
+|1780|Check if Number is a Sum of Powers of Three||65.3%|Medium||
+|1781|Sum of Beauty of All Substrings||60.4%|Medium||
+|1782|Count Pairs Of Nodes||37.9%|Hard||
+|1783|Grand Slam Titles||88.9%|Medium||
+|1784|Check if Binary String Has at Most One Segment of Ones||40.5%|Easy||
+|1785|Minimum Elements to Add to Form a Given Sum||42.4%|Medium||
+|1786|Number of Restricted Paths From First to Last Node||39.2%|Medium||
+|1787|Make the XOR of All Segments Equal to Zero||39.5%|Hard||
+|1788|Maximize the Beauty of the Garden||66.4%|Hard||
+|1789|Primary Department for Each Employee||80.0%|Easy||
+|1790|Check if One String Swap Can Make Strings Equal||45.6%|Easy||
+|1791|Find Center of Star Graph|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1791.Find-Center-of-Star-Graph)|83.6%|Easy||
+|1792|Maximum Average Pass Ratio||51.9%|Medium||
+|1793|Maximum Score of a Good Subarray||53.2%|Hard||
+|1794|Count Pairs of Equal Substrings With Minimum Difference||65.2%|Medium||
+|1795|Rearrange Products Table||90.4%|Easy||
+|1796|Second Largest Digit in a String||49.0%|Easy||
+|1797|Design Authentication Manager||56.1%|Medium||
+|1798|Maximum Number of Consecutive Values You Can Make||54.1%|Medium||
+|1799|Maximize Score After N Operations||45.7%|Hard||
+|1800|Maximum Ascending Subarray Sum||63.8%|Easy||
+|1801|Number of Orders in the Backlog||47.2%|Medium||
+|1802|Maximum Value at a Given Index in a Bounded Array||31.8%|Medium||
+|1803|Count Pairs With XOR in a Range||46.8%|Hard||
+|1804|Implement Trie II (Prefix Tree)||59.8%|Medium||
+|1805|Number of Different Integers in a String||36.1%|Easy||
+|1806|Minimum Number of Operations to Reinitialize a Permutation||71.3%|Medium||
+|1807|Evaluate the Bracket Pairs of a String||66.7%|Medium||
+|1808|Maximize Number of Nice Divisors||31.2%|Hard||
+|1809|Ad-Free Sessions||60.0%|Easy||
+|1810|Minimum Path Cost in a Hidden Grid||53.8%|Medium||
+|1811|Find Interview Candidates||65.3%|Medium||
+|1812|Determine Color of a Chessboard Square||77.4%|Easy||
+|1813|Sentence Similarity III||33.0%|Medium||
+|1814|Count Nice Pairs in an Array||41.9%|Medium||
+|1815|Maximum Number of Groups Getting Fresh Donuts||40.1%|Hard||
+|1816|Truncate Sentence|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1816.Truncate-Sentence)|81.9%|Easy||
+|1817|Finding the Users Active Minutes||80.7%|Medium||
+|1818|Minimum Absolute Sum Difference|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1818.Minimum-Absolute-Sum-Difference)|30.1%|Medium||
+|1819|Number of Different Subsequences GCDs||37.9%|Hard||
+|1820|Maximum Number of Accepted Invitations||49.7%|Medium||
+|1821|Find Customers With Positive Revenue this Year||89.4%|Easy||
+|1822|Sign of the Product of an Array||66.3%|Easy||
+|1823|Find the Winner of the Circular Game||77.7%|Medium||
+|1824|Minimum Sideway Jumps||49.8%|Medium||
+|1825|Finding MK Average||35.1%|Hard||
+|1826|Faulty Sensor||49.5%|Easy||
+|1827|Minimum Operations to Make the Array Increasing||78.2%|Easy||
+|1828|Queries on Number of Points Inside a Circle||86.5%|Medium||
+|1829|Maximum XOR for Each Query||77.0%|Medium||
+|1830|Minimum Number of Operations to Make String Sorted||49.2%|Hard||
+|1831|Maximum Transaction Each Day||84.3%|Medium||
+|1832|Check if the Sentence Is Pangram||83.9%|Easy||
+|1833|Maximum Ice Cream Bars||65.5%|Medium||
+|1834|Single-Threaded CPU||41.9%|Medium||
+|1835|Find XOR Sum of All Pairs Bitwise AND||60.0%|Hard||
+|1836|Remove Duplicates From an Unsorted Linked List||69.3%|Medium||
+|1837|Sum of Digits in Base K||76.8%|Easy||
+|1838|Frequency of the Most Frequent Element||38.3%|Medium||
+|1839|Longest Substring Of All Vowels in Order||48.5%|Medium||
+|1840|Maximum Building Height||35.3%|Hard||
+|1841|League Statistics||57.4%|Medium||
+|1842|Next Palindrome Using Same Digits||53.6%|Hard||
+|1843|Suspicious Bank Accounts||48.0%|Medium||
+|1844|Replace All Digits with Characters||79.7%|Easy||
+|1845|Seat Reservation Manager||64.2%|Medium||
+|1846|Maximum Element After Decreasing and Rearranging|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1846.Maximum-Element-After-Decreasing-and-Rearranging)|59.1%|Medium||
+|1847|Closest Room||35.1%|Hard||
+|1848|Minimum Distance to the Target Element||58.6%|Easy||
+|1849|Splitting a String Into Descending Consecutive Values||32.1%|Medium||
+|1850|Minimum Adjacent Swaps to Reach the Kth Smallest Number||72.0%|Medium||
+|1851|Minimum Interval to Include Each Query||47.7%|Hard||
+|1852|Distinct Numbers in Each Subarray||71.5%|Medium||
+|1853|Convert Date Format||88.0%|Easy||
+|1854|Maximum Population Year||59.8%|Easy||
+|1855|Maximum Distance Between a Pair of Values||52.6%|Medium||
+|1856|Maximum Subarray Min-Product||37.7%|Medium||
+|1857|Largest Color Value in a Directed Graph||40.6%|Hard||
+|1858|Longest Word With All Prefixes||66.3%|Medium||
+|1859|Sorting the Sentence||84.4%|Easy||
+|1860|Incremental Memory Leak||71.6%|Medium||
+|1861|Rotating the Box||64.7%|Medium||
+|1862|Sum of Floored Pairs||28.3%|Hard||
+|1863|Sum of All Subset XOR Totals||79.0%|Easy||
+|1864|Minimum Number of Swaps to Make the Binary String Alternating||42.0%|Medium||
+|1865|Finding Pairs With a Certain Sum||50.3%|Medium||
+|1866|Number of Ways to Rearrange Sticks With K Sticks Visible||55.4%|Hard||
+|1867|Orders With Maximum Quantity Above Average||75.9%|Medium||
+|1868|Product of Two Run-Length Encoded Arrays||57.8%|Medium||
+|1869|Longer Contiguous Segments of Ones than Zeros||60.2%|Easy||
+|1870|Minimum Speed to Arrive on Time||37.3%|Medium||
+|1871|Jump Game VII||25.1%|Medium||
+|1872|Stone Game VIII||52.4%|Hard||
+|1873|Calculate Special Bonus||64.7%|Easy||
+|1874|Minimize Product Sum of Two Arrays||90.4%|Medium||
+|1875|Group Employees of the Same Salary||75.6%|Medium||
+|1876|Substrings of Size Three with Distinct Characters||70.2%|Easy||
+|1877|Minimize Maximum Pair Sum in Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1877.Minimize-Maximum-Pair-Sum-in-Array)|80.4%|Medium||
+|1878|Get Biggest Three Rhombus Sums in a Grid||46.4%|Medium||
+|1879|Minimum XOR Sum of Two Arrays||44.6%|Hard||
+|1880|Check if Word Equals Summation of Two Words||73.6%|Easy||
+|1881|Maximum Value after Insertion||36.5%|Medium||
+|1882|Process Tasks Using Servers||39.5%|Medium||
+|1883|Minimum Skips to Arrive at Meeting On Time||38.5%|Hard||
+|1884|Egg Drop With 2 Eggs and N Floors||70.3%|Medium||
+|1885|Count Pairs in Two Arrays||59.2%|Medium||
+|1886|Determine Whether Matrix Can Be Obtained By Rotation||55.3%|Easy||
+|1887|Reduction Operations to Make the Array Elements Equal||62.3%|Medium||
+|1888|Minimum Number of Flips to Make the Binary String Alternating||37.8%|Medium||
+|1889|Minimum Space Wasted From Packaging||30.6%|Hard||
+|1890|The Latest Login in 2020||81.8%|Easy||
+|1891|Cutting Ribbons||48.1%|Medium||
+|1892|Page Recommendations II||44.8%|Hard||
+|1893|Check if All the Integers in a Range Are Covered||50.8%|Easy||
+|1894|Find the Student that Will Replace the Chalk||43.6%|Medium||
+|1895|Largest Magic Square||51.9%|Medium||
+|1896|Minimum Cost to Change the Final Value of Expression||54.9%|Hard||
+|1897|Redistribute Characters to Make All Strings Equal||60.0%|Easy||
+|1898|Maximum Number of Removable Characters||39.2%|Medium||
+|1899|Merge Triplets to Form Target Triplet||64.3%|Medium||
+|1900|The Earliest and Latest Rounds Where Players Compete||51.8%|Hard||
+|1901|Find a Peak Element II||53.3%|Medium||
+|1902|Depth of BST Given Insertion Order||45.0%|Medium||
+|1903|Largest Odd Number in String||55.7%|Easy||
+|1904|The Number of Full Rounds You Have Played||45.7%|Medium||
+|1905|Count Sub Islands||67.9%|Medium||
+|1906|Minimum Absolute Difference Queries||43.9%|Medium||
+|1907|Count Salary Categories||64.6%|Medium||
+|1908|Game of Nim||57.3%|Medium||
+|1909|Remove One Element to Make the Array Strictly Increasing||26.0%|Easy||
+|1910|Remove All Occurrences of a Substring||74.0%|Medium||
+|1911|Maximum Alternating Subsequence Sum||59.2%|Medium||
+|1912|Design Movie Rental System||41.2%|Hard||
+|1913|Maximum Product Difference Between Two Pairs||81.4%|Easy||
+|1914|Cyclically Rotating a Grid||48.0%|Medium||
+|1915|Number of Wonderful Substrings||44.8%|Medium||
+|1916|Count Ways to Build Rooms in an Ant Colony||48.7%|Hard||
+|1917|Leetcodify Friends Recommendations||28.9%|Hard||
+|1918|Kth Smallest Subarray Sum||52.9%|Medium||
+|1919|Leetcodify Similar Friends||43.2%|Hard||
+|1920|Build Array from Permutation||91.5%|Easy||
+|1921|Eliminate Maximum Number of Monsters||37.8%|Medium||
+|1922|Count Good Numbers||38.5%|Medium||
+|1923|Longest Common Subpath||27.6%|Hard||
+|1924|Erect the Fence II||53.9%|Hard||
+|1925|Count Square Sum Triples||67.9%|Easy||
+|1926|Nearest Exit from Entrance in Maze||43.1%|Medium||
+|1927|Sum Game||47.0%|Medium||
+|1928|Minimum Cost to Reach Destination in Time||37.6%|Hard||
+|1929|Concatenation of Array||91.5%|Easy||
+|1930|Unique Length-3 Palindromic Subsequences||51.8%|Medium||
+|1931|Painting a Grid With Three Different Colors||57.3%|Hard||
+|1932|Merge BSTs to Create Single BST||35.3%|Hard||
+|1933|Check if String Is Decomposable Into Value-Equal Substrings||50.4%|Easy||
+|1934|Confirmation Rate||77.8%|Medium||
+|1935|Maximum Number of Words You Can Type||71.0%|Easy||
+|1936|Add Minimum Number of Rungs||42.8%|Medium||
+|1937|Maximum Number of Points with Cost||36.2%|Medium||
+|1938|Maximum Genetic Difference Query||39.5%|Hard||
+|1939|Users That Actively Request Confirmation Messages||61.7%|Easy||
+|1940|Longest Common Subsequence Between Sorted Arrays||79.4%|Medium||
+|1941|Check if All Characters Have Equal Number of Occurrences||76.9%|Easy||
+|1942|The Number of the Smallest Unoccupied Chair||40.4%|Medium||
+|1943|Describe the Painting||47.8%|Medium||
+|1944|Number of Visible People in a Queue||69.9%|Hard||
+|1945|Sum of Digits of String After Convert||61.1%|Easy||
+|1946|Largest Number After Mutating Substring||34.5%|Medium||
+|1947|Maximum Compatibility Score Sum||60.9%|Medium||
+|1948|Delete Duplicate Folders in System||58.0%|Hard||
+|1949|Strong Friendship||58.7%|Medium||
+|1950|Maximum of Minimum Values in All Subarrays||50.0%|Medium||
+|1951|All the Pairs With the Maximum Number of Common Followers||72.9%|Medium||
+|1952|Three Divisors||57.0%|Easy||
+|1953|Maximum Number of Weeks for Which You Can Work||39.0%|Medium||
+|1954|Minimum Garden Perimeter to Collect Enough Apples||53.0%|Medium||
+|1955|Count Number of Special Subsequences||51.3%|Hard||
+|1956|Minimum Time For K Virus Variants to Spread||46.5%|Hard||
+|1957|Delete Characters to Make Fancy String||56.6%|Easy||
+|1958|Check if Move is Legal||44.2%|Medium||
+|1959|Minimum Total Space Wasted With K Resizing Operations||41.9%|Medium||
+|1960|Maximum Product of the Length of Two Palindromic Substrings||29.6%|Hard||
+|1961|Check If String Is a Prefix of Array||54.2%|Easy||
+|1962|Remove Stones to Minimize the Total||58.8%|Medium||
+|1963|Minimum Number of Swaps to Make the String Balanced||68.3%|Medium||
+|1964|Find the Longest Valid Obstacle Course at Each Position||46.9%|Hard||
+|1965|Employees With Missing Information||81.1%|Easy||
+|1966|Binary Searchable Numbers in an Unsorted Array||65.5%|Medium||
+|1967|Number of Strings That Appear as Substrings in Word||79.9%|Easy||
+|1968|Array With Elements Not Equal to Average of Neighbors||49.4%|Medium||
+|1969|Minimum Non-Zero Product of the Array Elements||33.7%|Medium||
+|1970|Last Day Where You Can Still Cross||49.5%|Hard||
+|1971|Find if Path Exists in Graph||50.5%|Easy||
+|1972|First and Last Call On the Same Day||54.4%|Hard||
+|1973|Count Nodes Equal to Sum of Descendants||75.3%|Medium||
+|1974|Minimum Time to Type Word Using Special Typewriter||71.5%|Easy||
+|1975|Maximum Matrix Sum||45.6%|Medium||
+|1976|Number of Ways to Arrive at Destination||32.4%|Medium||
+|1977|Number of Ways to Separate Numbers||21.2%|Hard||
+|1978|Employees Whose Manager Left the Company||50.5%|Easy||
+|1979|Find Greatest Common Divisor of Array||76.8%|Easy||
+|1980|Find Unique Binary String||64.2%|Medium||
+|1981|Minimize the Difference Between Target and Chosen Elements||32.4%|Medium||
+|1982|Find Array Given Subset Sums||48.7%|Hard||
+|1983|Widest Pair of Indices With Equal Range Sum||53.1%|Medium||
+|1984|Minimum Difference Between Highest and Lowest of K Scores|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores)|53.5%|Easy||
+|1985|Find the Kth Largest Integer in the Array||44.6%|Medium||
+|1986|Minimum Number of Work Sessions to Finish the Tasks||33.1%|Medium||
+|1987|Number of Unique Good Subsequences||52.2%|Hard||
+|1988|Find Cutoff Score for Each School||70.5%|Medium||
+|1989|Maximum Number of People That Can Be Caught in Tag||53.8%|Medium||
+|1990|Count the Number of Experiments||51.7%|Medium||
+|1991|Find the Middle Index in Array||67.3%|Easy||
+|1992|Find All Groups of Farmland||68.5%|Medium||
+|1993|Operations on Tree||43.5%|Medium||
+|1994|The Number of Good Subsets||34.1%|Hard||
+|1995|Count Special Quadruplets||59.0%|Easy||
+|1996|The Number of Weak Characters in the Game||44.0%|Medium||
+|1997|First Day Where You Have Been in All the Rooms||36.0%|Medium||
+|1998|GCD Sort of an Array||45.5%|Hard||
+|1999|Smallest Greater Multiple Made of Two Digits||50.0%|Medium||
+|2000|Reverse Prefix of Word||77.7%|Easy||
+|2001|Number of Pairs of Interchangeable Rectangles||44.6%|Medium||
+|2002|Maximum Product of the Length of Two Palindromic Subsequences||53.4%|Medium||
+|2003|Smallest Missing Genetic Value in Each Subtree||44.1%|Hard||
+|2004|The Number of Seniors and Juniors to Join the Company||38.8%|Hard||
+|2005|Subtree Removal Game with Fibonacci Tree||62.8%|Hard||
+|2006|Count Number of Pairs With Absolute Difference K||82.2%|Easy||
+|2007|Find Original Array From Doubled Array||41.0%|Medium||
+|2008|Maximum Earnings From Taxi||43.0%|Medium||
+|2009|Minimum Number of Operations to Make Array Continuous||45.5%|Hard||
+|2010|The Number of Seniors and Juniors to Join the Company II||57.8%|Hard||
+|2011|Final Value of Variable After Performing Operations||88.8%|Easy||
+|2012|Sum of Beauty in the Array||46.7%|Medium||
+|2013|Detect Squares||50.1%|Medium||
+|2014|Longest Subsequence Repeated k Times||55.7%|Hard||
+|2015|Average Height of Buildings in Each Segment||58.6%|Medium||
+|2016|Maximum Difference Between Increasing Elements||53.4%|Easy||
+|2017|Grid Game||42.7%|Medium||
+|2018|Check if Word Can Be Placed In Crossword||49.3%|Medium||
+|2019|The Score of Students Solving Math Expression||33.2%|Hard||
+|2020|Number of Accounts That Did Not Stream||73.2%|Medium||
+|2021|Brightest Position on Street|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2021.Brightest-Position-on-Street)|62.9%|Medium||
+|2022|Convert 1D Array Into 2D Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2022.Convert-1D-Array-Into-2D-Array)|58.3%|Easy||
+|2023|Number of Pairs of Strings With Concatenation Equal to Target||72.8%|Medium||
+|2024|Maximize the Confusion of an Exam||59.4%|Medium||
+|2025|Maximum Number of Ways to Partition an Array||32.0%|Hard||
+|2026|Low-Quality Problems||85.5%|Easy||
+|2027|Minimum Moves to Convert String||53.8%|Easy||
+|2028|Find Missing Observations||43.5%|Medium||
+|2029|Stone Game IX||26.1%|Medium||
+|2030|Smallest K-Length Subsequence With Occurrences of a Letter||38.9%|Hard||
+|2031|Count Subarrays With More Ones Than Zeros||53.2%|Medium||
+|2032|Two Out of Three||72.6%|Easy||
+|2033|Minimum Operations to Make a Uni-Value Grid||52.3%|Medium||
+|2034|Stock Price Fluctuation||49.3%|Medium||
+|2035|Partition Array Into Two Arrays to Minimize Sum Difference||17.9%|Hard||
+|2036|Maximum Alternating Subarray Sum||41.2%|Medium||
+|2037|Minimum Number of Moves to Seat Everyone|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2037.Minimum-Number-of-Moves-to-Seat-Everyone)|82.2%|Easy||
+|2038|Remove Colored Pieces if Both Neighbors are the Same Color|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color)|58.1%|Medium||
+|2039|The Time When the Network Becomes Idle||50.6%|Medium||
+|2040|Kth Smallest Product of Two Sorted Arrays||29.1%|Hard||
+|2041|Accepted Candidates From the Interviews||79.4%|Medium||
+|2042|Check if Numbers Are Ascending in a Sentence||65.9%|Easy||
+|2043|Simple Bank System|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2043.Simple-Bank-System)|65.9%|Medium||
+|2044|Count Number of Maximum Bitwise-OR Subsets||74.8%|Medium||
+|2045|Second Minimum Time to Reach Destination||38.5%|Hard||
+|2046|Sort Linked List Already Sorted Using Absolute Values||68.6%|Medium||
+|2047|Number of Valid Words in a Sentence||29.5%|Easy||
+|2048|Next Greater Numerically Balanced Number||47.1%|Medium||
+|2049|Count Nodes With the Highest Score||47.1%|Medium||
+|2050|Parallel Courses III||59.4%|Hard||
+|2051|The Category of Each Member in the Store||73.5%|Medium||
+|2052|Minimum Cost to Separate Sentence Into Rows||51.0%|Medium||
+|2053|Kth Distinct String in an Array||71.9%|Easy||
+|2054|Two Best Non-Overlapping Events||44.8%|Medium||
+|2055|Plates Between Candles||44.7%|Medium||
+|2056|Number of Valid Move Combinations On Chessboard||59.1%|Hard||
+|2057|Smallest Index With Equal Value||71.2%|Easy||
+|2058|Find the Minimum and Maximum Number of Nodes Between Critical Points||57.1%|Medium||
+|2059|Minimum Operations to Convert Number||47.2%|Medium||
+|2060|Check if an Original String Exists Given Two Encoded Strings||41.0%|Hard||
+|2061|Number of Spaces Cleaning Robot Cleaned||54.9%|Medium||
+|2062|Count Vowel Substrings of a String||65.9%|Easy||
+|2063|Vowels of All Substrings||55.0%|Medium||
+|2064|Minimized Maximum of Products Distributed to Any Store||50.0%|Medium||
+|2065|Maximum Path Quality of a Graph||57.7%|Hard||
+|2066|Account Balance||85.5%|Medium||
+|2067|Number of Equal Count Substrings||49.6%|Medium||
+|2068|Check Whether Two Strings are Almost Equivalent||64.7%|Easy||
+|2069|Walking Robot Simulation II||23.0%|Medium||
+|2070|Most Beautiful Item for Each Query||49.5%|Medium||
+|2071|Maximum Number of Tasks You Can Assign||34.8%|Hard||
+|2072|The Winner University||72.4%|Easy||
+|2073|Time Needed to Buy Tickets||62.0%|Easy||
+|2074|Reverse Nodes in Even Length Groups||51.5%|Medium||
+|2075|Decode the Slanted Ciphertext||50.2%|Medium||
+|2076|Process Restricted Friend Requests||53.4%|Hard||
+|2077|Paths in Maze That Lead to Same Room||56.8%|Medium||
+|2078|Two Furthest Houses With Different Colors||67.3%|Easy||
+|2079|Watering Plants||80.2%|Medium||
+|2080|Range Frequency Queries||38.1%|Medium||
+|2081|Sum of k-Mirror Numbers||42.1%|Hard||
+|2082|The Number of Rich Customers||80.7%|Easy||
+|2083|Substrings That Begin and End With the Same Letter||68.0%|Medium||
+|2084|Drop Type 1 Orders for Customers With Type 0 Orders||91.2%|Medium||
+|2085|Count Common Words With One Occurrence||69.6%|Easy||
+|2086|Minimum Number of Buckets Required to Collect Rainwater from Houses||45.0%|Medium||
+|2087|Minimum Cost Homecoming of a Robot in a Grid||51.4%|Medium||
+|2088|Count Fertile Pyramids in a Land||63.3%|Hard||
+|2089|Find Target Indices After Sorting Array||76.9%|Easy||
+|2090|K Radius Subarray Averages||42.4%|Medium||
+|2091|Removing Minimum and Maximum From Array||56.8%|Medium||
+|2092|Find All People With Secret||34.2%|Hard||
+|2093|Minimum Cost to Reach City With Discounts||56.3%|Medium||
+|2094|Finding 3-Digit Even Numbers||57.4%|Easy||
+|2095|Delete the Middle Node of a Linked List||60.7%|Medium||
+|2096|Step-By-Step Directions From a Binary Tree Node to Another|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another)|48.8%|Medium||
+|2097|Valid Arrangement of Pairs||40.9%|Hard||
+|2098|Subsequence of Size K With the Largest Even Sum||38.7%|Medium||
+|2099|Find Subsequence of Length K With the Largest Sum||42.5%|Easy||
+|2100|Find Good Days to Rob the Bank||49.1%|Medium||
+|2101|Detonate the Maximum Bombs||40.9%|Medium||
+|2102|Sequentially Ordinal Rank Tracker||66.1%|Hard||
+|2103|Rings and Rods||81.5%|Easy||
+|2104|Sum of Subarray Ranges||60.2%|Medium||
+|2105|Watering Plants II||50.2%|Medium||
+|2106|Maximum Fruits Harvested After at Most K Steps||35.1%|Hard||
+|2107|Number of Unique Flavors After Sharing K Candies||57.2%|Medium||
+|2108|Find First Palindromic String in the Array||78.6%|Easy||
+|2109|Adding Spaces to a String||56.2%|Medium||
+|2110|Number of Smooth Descent Periods of a Stock||57.4%|Medium||
+|2111|Minimum Operations to Make the Array K-Increasing||37.7%|Hard||
+|2112|The Airport With the Most Traffic||71.2%|Medium||
+|2113|Elements in Array After Removing and Replacing Elements||73.4%|Medium||
+|2114|Maximum Number of Words Found in Sentences||88.2%|Easy||
+|2115|Find All Possible Recipes from Given Supplies||48.2%|Medium||
+|2116|Check if a Parentheses String Can Be Valid||31.4%|Medium||
+|2117|Abbreviating the Product of a Range||28.1%|Hard||
+|2118|Build the Equation||57.8%|Hard||
+|2119|A Number After a Double Reversal||75.7%|Easy||
+|2120|Execution of All Suffix Instructions Staying in a Grid||83.5%|Medium||
+|2121|Intervals Between Identical Elements||43.0%|Medium||
+|2122|Recover the Original Array||38.1%|Hard||
+|2123|Minimum Operations to Remove Adjacent Ones in Matrix||41.1%|Hard||
+|2124|Check if All A's Appears Before All B's||71.5%|Easy||
+|2125|Number of Laser Beams in a Bank||82.7%|Medium||
+|2126|Destroying Asteroids||49.4%|Medium||
+|2127|Maximum Employees to Be Invited to a Meeting||32.9%|Hard||
+|2128|Remove All Ones With Row and Column Flips||76.4%|Medium||
+|2129|Capitalize the Title||60.1%|Easy||
+|2130|Maximum Twin Sum of a Linked List||81.6%|Medium||
+|2131|Longest Palindrome by Concatenating Two Letter Words||41.3%|Medium||
+|2132|Stamping the Grid||30.6%|Hard||
+|2133|Check if Every Row and Column Contains All Numbers||52.8%|Easy||
+|2134|Minimum Swaps to Group All 1's Together II||50.3%|Medium||
+|2135|Count Words Obtained After Adding a Letter||42.8%|Medium||
+|2136|Earliest Possible Day of Full Bloom||68.4%|Hard||
+|2137|Pour Water Between Buckets to Make Water Levels Equal||67.1%|Medium||
+|2138|Divide a String Into Groups of Size k||65.1%|Easy||
+|2139|Minimum Moves to Reach Target Score||48.4%|Medium||
+|2140|Solving Questions With Brainpower||45.8%|Medium||
+|2141|Maximum Running Time of N Computers||38.7%|Hard||
+|2142|The Number of Passengers in Each Bus I||51.1%|Medium||
+|2143|Choose Numbers From Two Arrays in Range||51.8%|Hard||
+|2144|Minimum Cost of Buying Candies With Discount||60.8%|Easy||
+|2145|Count the Hidden Sequences||36.3%|Medium||
+|2146|K Highest Ranked Items Within a Price Range||41.2%|Medium||
+|2147|Number of Ways to Divide a Long Corridor||39.9%|Hard||
+|2148|Count Elements With Strictly Smaller and Greater Elements||60.0%|Easy||
+|2149|Rearrange Array Elements by Sign||81.0%|Medium||
+|2150|Find All Lonely Numbers in the Array||60.7%|Medium||
+|2151|Maximum Good People Based on Statements||48.4%|Hard||
+|2152|Minimum Number of Lines to Cover Points||46.7%|Medium||
+|2153|The Number of Passengers in Each Bus II||50.4%|Hard||
+|2154|Keep Multiplying Found Values by Two||73.3%|Easy||
+|2155|All Divisions With the Highest Score of a Binary Array||63.3%|Medium||
+|2156|Find Substring With Given Hash Value||21.9%|Hard||
+|2157|Groups of Strings||25.2%|Hard||
+|2158|Amount of New Area Painted Each Day||55.7%|Hard||
+|2159|Order Two Columns Independently||63.5%|Medium||
+|2160|Minimum Sum of Four Digit Number After Splitting Digits||88.2%|Easy||
+|2161|Partition Array According to Given Pivot||84.3%|Medium||
+|2162|Minimum Cost to Set Cooking Time||39.2%|Medium||
+|2163|Minimum Difference in Sums After Removal of Elements||46.5%|Hard||
+|2164|Sort Even and Odd Indices Independently|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2164.Sort-Even-and-Odd-Indices-Independently)|66.6%|Easy||
+|2165|Smallest Value of the Rearranged Number|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2165.Smallest-Value-of-the-Rearranged-Number)|51.1%|Medium||
+|2166|Design Bitset|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2166.Design-Bitset)|31.3%|Medium||
+|2167|Minimum Time to Remove All Cars Containing Illegal Goods|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods)|40.1%|Hard||
+|2168|Unique Substrings With Equal Digit Frequency||60.2%|Medium||
+|2169|Count Operations to Obtain Zero|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2169.Count-Operations-to-Obtain-Zero)|75.6%|Easy||
+|2170|Minimum Operations to Make the Array Alternating|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2170.Minimum-Operations-to-Make-the-Array-Alternating)|33.2%|Medium||
+|2171|Removing Minimum Number of Magic Beans|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2171.Removing-Minimum-Number-of-Magic-Beans)|42.0%|Medium||
+|2172|Maximum AND Sum of Array||46.8%|Hard||
+|2173|Longest Winning Streak||59.7%|Hard||
+|2174|Remove All Ones With Row and Column Flips II||68.8%|Medium||
+|2175|The Change in Global Rankings||68.5%|Medium||
+|2176|Count Equal and Divisible Pairs in an Array||80.3%|Easy||
+|2177|Find Three Consecutive Integers That Sum to a Given Number||63.5%|Medium||
+|2178|Maximum Split of Positive Even Integers||59.1%|Medium||
+|2179|Count Good Triplets in an Array||36.7%|Hard||
+|2180|Count Integers With Even Digit Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2180.Count-Integers-With-Even-Digit-Sum)|64.5%|Easy||
+|2181|Merge Nodes in Between Zeros|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2181.Merge-Nodes-in-Between-Zeros)|86.9%|Medium||
+|2182|Construct String With Repeat Limit|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2182.Construct-String-With-Repeat-Limit)|51.8%|Medium||
+|2183|Count Array Pairs Divisible by K|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/2183.Count-Array-Pairs-Divisible-by-K)|28.3%|Hard||
+|2184|Number of Ways to Build Sturdy Brick Wall||52.1%|Medium||
+|2185|Counting Words With a Given Prefix||77.1%|Easy||
+|2186|Minimum Number of Steps to Make Two Strings Anagram II||71.9%|Medium||
+|2187|Minimum Time to Complete Trips||31.8%|Medium||
+|2188|Minimum Time to Finish the Race||41.9%|Hard||
+|2189|Number of Ways to Build House of Cards||63.2%|Medium||
+|2190|Most Frequent Number Following Key In an Array||60.4%|Easy||
+|2191|Sort the Jumbled Numbers||45.1%|Medium||
+|2192|All Ancestors of a Node in a Directed Acyclic Graph||50.5%|Medium||
+|2193|Minimum Number of Moves to Make Palindrome||50.8%|Hard||
+|2194|Cells in a Range on an Excel Sheet||85.7%|Easy||
+|2195|Append K Integers With Minimal Sum||24.9%|Medium||
+|2196|Create Binary Tree From Descriptions||72.0%|Medium||
+|2197|Replace Non-Coprime Numbers in Array||38.3%|Hard||
+|2198|Number of Single Divisor Triplets||56.2%|Medium||
+|2199|Finding the Topic of Each Post||49.8%|Hard||
+|2200|Find All K-Distant Indices in an Array||64.4%|Easy||
+|2201|Count Artifacts That Can Be Extracted||54.9%|Medium||
+|2202|Maximize the Topmost Element After K Moves||22.7%|Medium||
+|2203|Minimum Weighted Subgraph With the Required Paths||35.6%|Hard||
+|2204|Distance to a Cycle in Undirected Graph||71.4%|Hard||
+|2205|The Number of Users That Are Eligible for Discount||50.4%|Easy||
+|2206|Divide Array Into Equal Pairs||74.8%|Easy||
+|2207|Maximize Number of Subsequences in a String||32.7%|Medium||
+|2208|Minimum Operations to Halve Array Sum||45.1%|Medium||
+|2209|Minimum White Tiles After Covering With Carpets||33.7%|Hard||
+|2210|Count Hills and Valleys in an Array||57.7%|Easy||
+|2211|Count Collisions on a Road||41.7%|Medium||
+|2212|Maximum Points in an Archery Competition||48.7%|Medium||
+|2213|Longest Substring of One Repeating Character||31.1%|Hard||
+|2214|Minimum Health to Beat Game||57.6%|Medium||
+|2215|Find the Difference of Two Arrays||69.1%|Easy||
+|2216|Minimum Deletions to Make Array Beautiful||46.1%|Medium||
+|2217|Find Palindrome With Fixed Length||34.4%|Medium||
+|2218|Maximum Value of K Coins From Piles||48.5%|Hard||
+|2219|Maximum Sum Score of Array||60.6%|Medium||
+|2220|Minimum Bit Flips to Convert Number||82.3%|Easy||
+|2221|Find Triangular Sum of an Array||79.1%|Medium||
+|2222|Number of Ways to Select Buildings||50.7%|Medium||
+|2223|Sum of Scores of Built Strings||36.5%|Hard||
+|2224|Minimum Number of Operations to Convert Time||65.2%|Easy||
+|2225|Find Players With Zero or One Losses||69.3%|Medium||
+|2226|Maximum Candies Allocated to K Children||35.9%|Medium||
+|2227|Encrypt and Decrypt Strings||39.0%|Hard||
+|2228|Users With Two Purchases Within Seven Days||44.8%|Medium||
+|2229|Check if an Array Is Consecutive||61.7%|Easy||
+|2230|The Users That Are Eligible for Discount||51.1%|Easy||
+|2231|Largest Number After Digit Swaps by Parity||60.2%|Easy||
+|2232|Minimize Result by Adding Parentheses to Expression||64.7%|Medium||
+|2233|Maximum Product After K Increments||41.3%|Medium||
+|2234|Maximum Total Beauty of the Gardens||28.1%|Hard||
+|2235|Add Two Integers||90.0%|Easy||
+|2236|Root Equals Sum of Children||87.6%|Easy||
+|2237|Count Positions on Street With Required Brightness||68.6%|Medium||
+|2238|Number of Times a Driver Was a Passenger||75.3%|Medium||
+|2239|Find Closest Number to Zero||45.8%|Easy||
+|2240|Number of Ways to Buy Pens and Pencils||56.8%|Medium||
+|2241|Design an ATM Machine||38.6%|Medium||
+|2242|Maximum Score of a Node Sequence||37.4%|Hard||
+|2243|Calculate Digit Sum of a String||66.9%|Easy||
+|2244|Minimum Rounds to Complete All Tasks||57.0%|Medium||
+|2245|Maximum Trailing Zeros in a Cornered Path||35.2%|Medium||
+|2246|Longest Path With Different Adjacent Characters||44.9%|Hard||
+|2247|Maximum Cost of Trip With K Highways||50.5%|Hard||
+|2248|Intersection of Multiple Arrays||69.5%|Easy||
+|2249|Count Lattice Points Inside a Circle||50.4%|Medium||
+|2250|Count Number of Rectangles Containing Each Point||33.9%|Medium||
+|2251|Number of Flowers in Full Bloom||51.9%|Hard||
+|2252|Dynamic Pivoting of a Table||55.5%|Hard||
+|2253|Dynamic Unpivoting of a Table||67.4%|Hard||
+|2254|Design Video Sharing Platform||65.8%|Hard||
+|2255|Count Prefixes of a Given String||73.3%|Easy||
+|2256|Minimum Average Difference||35.7%|Medium||
+|2257|Count Unguarded Cells in the Grid||52.2%|Medium||
+|2258|Escape the Spreading Fire||34.7%|Hard||
+|2259|Remove Digit From Number to Maximize Result||47.0%|Easy||
+|2260|Minimum Consecutive Cards to Pick Up||51.9%|Medium||
+|2261|K Divisible Elements Subarrays||47.4%|Medium||
+|2262|Total Appeal of A String||58.1%|Hard||
+|2263|Make Array Non-decreasing or Non-increasing||68.0%|Hard||
+|2264|Largest 3-Same-Digit Number in String||58.9%|Easy||
+|2265|Count Nodes Equal to Average of Subtree||85.6%|Medium||
+|2266|Count Number of Texts||47.3%|Medium||
+|2267|Check if There Is a Valid Parentheses String Path||37.9%|Hard||
+|2268|Minimum Number of Keypresses||74.4%|Medium||
+|2269|Find the K-Beauty of a Number||57.2%|Easy||
+|2270|Number of Ways to Split Array||44.2%|Medium||
+|2271|Maximum White Tiles Covered by a Carpet||32.4%|Medium||
+|2272|Substring With Largest Variance||37.1%|Hard||
+|2273|Find Resultant Array After Removing Anagrams||58.0%|Easy||
+|2274|Maximum Consecutive Floors Without Special Floors||52.2%|Medium||
+|2275|Largest Combination With Bitwise AND Greater Than Zero||72.3%|Medium||
+|2276|Count Integers in Intervals||33.9%|Hard||
+|2277|Closest Node to Path in Tree||64.1%|Hard||
+|2278|Percentage of Letter in String||74.0%|Easy||
+|2279|Maximum Bags With Full Capacity of Rocks||62.6%|Medium||
+|2280|Minimum Lines to Represent a Line Chart||23.7%|Medium||
+|2281|Sum of Total Strength of Wizards||27.9%|Hard||
+|2282|Number of People That Can Be Seen in a Grid||50.4%|Medium||
+|2283|Check if Number Has Equal Digit Count and Digit Value||73.6%|Easy||
+|2284|Sender With Largest Word Count||55.8%|Medium||
+|2285|Maximum Total Importance of Roads||60.7%|Medium||
+|2286|Booking Concert Tickets in Groups||15.7%|Hard||
+|2287|Rearrange Characters to Make Target String||57.7%|Easy||
+|2288|Apply Discount to Prices||27.3%|Medium||
+|2289|Steps to Make Array Non-decreasing||21.4%|Medium||
+|2290|Minimum Obstacle Removal to Reach Corner||49.2%|Hard||
+|2291|Maximum Profit From Trading Stocks||46.4%|Medium||
+|2292|Products With Three or More Orders in Two Consecutive Years||39.9%|Medium||
+|2293|Min Max Game||64.2%|Easy||
+|2294|Partition Array Such That Maximum Difference Is K||72.4%|Medium||
+|2295|Replace Elements in an Array||57.6%|Medium||
+|2296|Design a Text Editor||39.5%|Hard||
+|2297|Jump Game VIII||57.0%|Medium||
+|2298|Tasks Count in the Weekend||87.7%|Medium||
+|2299|Strong Password Checker II||56.8%|Easy||
+|2300|Successful Pairs of Spells and Potions||31.5%|Medium||
+|2301|Match Substring After Replacement||39.2%|Hard||
+|2302|Count Subarrays With Score Less Than K||51.9%|Hard||
+|2303|Calculate Amount Paid in Taxes||63.0%|Easy||
+|2304|Minimum Path Cost in a Grid||65.4%|Medium||
+|2305|Fair Distribution of Cookies||62.9%|Medium||
+|2306|Naming a Company||34.4%|Hard||
+|2307|Check for Contradictions in Equations||40.9%|Hard||
+|2308|Arrange Table by Gender||79.5%|Medium||
+|2309|Greatest English Letter in Upper and Lower Case||68.5%|Easy||
+|2310|Sum of Numbers With Units Digit K||25.4%|Medium||
+|2311|Longest Binary Subsequence Less Than or Equal to K||36.1%|Medium||
+|2312|Selling Pieces of Wood||48.0%|Hard||
+|2313|Minimum Flips in Binary Tree to Get Result||67.5%|Hard||
+|2314|The First Day of the Maximum Recorded Degree in Each City||76.9%|Medium||
+|2315|Count Asterisks||82.2%|Easy||
+|2316|Count Unreachable Pairs of Nodes in an Undirected Graph||38.6%|Medium||
+|2317|Maximum XOR After Operations||78.2%|Medium||
+|2318|Number of Distinct Roll Sequences||56.3%|Hard||
+|2319|Check if Matrix Is X-Matrix||67.3%|Easy||
+|2320|Count Number of Ways to Place Houses||39.9%|Medium||
+|2321|Maximum Score Of Spliced Array||55.1%|Hard||
+|2322|Minimum Score After Removals on a Tree||50.3%|Hard||
+|2323|Find Minimum Time to Finish All Jobs II||75.4%|Medium||
+|2324|Product Sales Analysis IV||84.5%|Medium||
+|2325|Decode the Message||84.7%|Easy||
+|2326|Spiral Matrix IV||74.6%|Medium||
+|2327|Number of People Aware of a Secret||44.4%|Medium||
+|2328|Number of Increasing Paths in a Grid||47.7%|Hard||
+|2329|Product Sales Analysis V||70.2%|Easy||
+|2330|Valid Palindrome IV||76.2%|Medium||
+|2331|Evaluate Boolean Binary Tree||79.5%|Easy||
+|2332|The Latest Time to Catch a Bus||22.7%|Medium||
+|2333|Minimum Sum of Squared Difference||24.9%|Medium||
+|2334|Subarray With Elements Greater Than Varying Threshold||40.1%|Hard||
+|2335|Minimum Amount of Time to Fill Cups||55.3%|Easy||
+|2336|Smallest Number in Infinite Set||71.8%|Medium||
+|2337|Move Pieces to Obtain a String||48.0%|Medium||
+|2338|Count the Number of Ideal Arrays||25.3%|Hard||
+|2339|All the Matches of the League||88.8%|Easy||
+|2340|Minimum Adjacent Swaps to Make a Valid Array||76.6%|Medium||
+|2341|Maximum Number of Pairs in Array||76.6%|Easy||
+|2342|Max Sum of a Pair With Equal Sum of Digits||52.9%|Medium||
+|2343|Query Kth Smallest Trimmed Number||40.7%|Medium||
+|2344|Minimum Deletions to Make Array Divisible||56.9%|Hard||
+|2345|Finding the Number of Visible Mountains||44.5%|Medium||
+|2346|Compute the Rank as a Percentage||32.9%|Medium||
+|2347|Best Poker Hand||60.7%|Easy||
+|2348|Number of Zero-Filled Subarrays||56.8%|Medium||
+|2349|Design a Number Container System||46.8%|Medium||
+|2350|Shortest Impossible Sequence of Rolls||68.2%|Hard||
+|2351|First Letter to Appear Twice||76.6%|Easy||
+|2352|Equal Row and Column Pairs||71.1%|Medium||
+|2353|Design a Food Rating System||34.4%|Medium||
+|2354|Number of Excellent Pairs||45.8%|Hard||
+|2355|Maximum Number of Books You Can Take||47.0%|Hard||
+|2356|Number of Unique Subjects Taught by Each Teacher||91.5%|Easy||
+|2357|Make Array Zero by Subtracting Equal Amounts||72.5%|Easy||
+|2358|Maximum Number of Groups Entering a Competition||67.3%|Medium||
+|2359|Find Closest Node to Given Two Nodes||33.8%|Medium||
+|2360|Longest Cycle in a Graph||38.5%|Hard||
+|2361|Minimum Costs Using the Train Line||77.1%|Hard||
+|2362|Generate the Invoice||89.1%|Hard||
+|2363|Merge Similar Items||75.1%|Easy||
+|2364|Count Number of Bad Pairs||40.5%|Medium||
+|2365|Task Scheduler II||46.0%|Medium||
+|2366|Minimum Replacements to Sort the Array||39.5%|Hard||
+|2367|Number of Arithmetic Triplets||83.7%|Easy||
+|2368|Reachable Nodes With Restrictions||57.2%|Medium||
+|2369|Check if There is a Valid Partition For The Array||40.0%|Medium||
+|2370|Longest Ideal Subsequence||37.8%|Medium||
+|2371|Minimize Maximum Value in a Grid||69.5%|Hard||
+|2372|Calculate the Influence of Each Salesperson||88.6%|Medium||
+|2373|Largest Local Values in a Matrix||84.1%|Easy||
+|2374|Node With Highest Edge Score||45.9%|Medium||
+|2375|Construct Smallest Number From DI String||73.7%|Medium||
+|2376|Count Special Integers||36.0%|Hard||
+|2377|Sort the Olympic Table||80.2%|Easy||
+|2378|Choose Edges to Maximize Score in a Tree||61.9%|Medium||
+|2379|Minimum Recolors to Get K Consecutive Black Blocks||56.6%|Easy||
+|2380|Time Needed to Rearrange a Binary String||47.7%|Medium||
+|2381|Shifting Letters II||33.9%|Medium||
+|2382|Maximum Segment Sum After Removals||47.7%|Hard||
+|2383|Minimum Hours of Training to Win a Competition||40.8%|Easy||
+|2384|Largest Palindromic Number||30.1%|Medium||
+|2385|Amount of Time for Binary Tree to Be Infected||56.0%|Medium||
+|2386|Find the K-Sum of an Array||36.2%|Hard||
+|2387|Median of a Row Wise Sorted Matrix||67.7%|Medium||
+|2388|Change Null Values in a Table to the Previous Value||79.7%|Medium||
+|2389|Longest Subsequence With Limited Sum||64.4%|Easy||
+|2390|Removing Stars From a String||62.9%|Medium||
+|2391|Minimum Amount of Time to Collect Garbage||85.4%|Medium||
+|2392|Build a Matrix With Conditions||59.1%|Hard||
+|2393|Count Strictly Increasing Subarrays||76.4%|Medium||
+|2394|Employees With Deductions||50.9%|Medium||
+|2395|Find Subarrays With Equal Sum||63.6%|Easy||
+|2396|Strictly Palindromic Number||87.7%|Medium||
+|2397|Maximum Rows Covered by Columns||52.2%|Medium||
+|2398|Maximum Number of Robots Within Budget||31.6%|Hard||
+|2399|Check Distances Between Same Letters||70.5%|Easy||
+|2400|Number of Ways to Reach a Position After Exactly k Steps||31.8%|Medium||
+|2401|Longest Nice Subarray||47.7%|Medium||
+|2402|Meeting Rooms III||32.7%|Hard||
+|2403|Minimum Time to Kill All Monsters||52.3%|Hard||
+|2404|Most Frequent Even Element||51.7%|Easy||
+|2405|Optimal Partition of String||74.1%|Medium||
+|2406|Divide Intervals Into Minimum Number of Groups||45.0%|Medium||
+|2407|Longest Increasing Subsequence II||20.6%|Hard||
+|2408|Design SQL||87.3%|Medium||
+|2409|Count Days Spent Together||42.3%|Easy||
+|2410|Maximum Matching of Players With Trainers||59.3%|Medium||
+|2411|Smallest Subarrays With Maximum Bitwise OR||40.0%|Medium||
+|2412|Minimum Money Required Before Transactions||38.8%|Hard||
+|2413|Smallest Even Multiple||88.2%|Easy||
+|2414|Length of the Longest Alphabetical Continuous Substring||55.5%|Medium||
+|2415|Reverse Odd Levels of Binary Tree||75.6%|Medium||
+|2416|Sum of Prefix Scores of Strings||42.3%|Hard||
+|2417|Closest Fair Integer||48.5%|Medium||
+|2418|Sort the People||82.7%|Easy||
+|2419|Longest Subarray With Maximum Bitwise AND||47.3%|Medium||
+|2420|Find All Good Indices||36.8%|Medium||
+|2421|Number of Good Paths||37.5%|Hard||
+|2422|Merge Operations to Turn Array Into a Palindrome||74.8%|Medium||
+|2423|Remove Letter To Equalize Frequency||19.4%|Easy||
+|2424|Longest Uploaded Prefix||53.2%|Medium||
+|2425|Bitwise XOR of All Pairings||58.3%|Medium||
+|2426|Number of Pairs Satisfying Inequality||41.1%|Hard||
+|2427|Number of Common Factors||80.3%|Easy||
+|2428|Maximum Sum of an Hourglass||73.6%|Medium||
+|2429|Minimize XOR||41.4%|Medium||
+|2430|Maximum Deletions on a String||32.8%|Hard||
+|2431|Maximize Total Tastiness of Purchased Fruits||76.6%|Medium||
+|2432|The Employee That Worked on the Longest Task||48.5%|Easy||
+|2433|Find The Original Array of Prefix Xor||85.1%|Medium||
+|2434|Using a Robot to Print the Lexicographically Smallest String||37.5%|Medium||
+|2435|Paths in Matrix Whose Sum Is Divisible by K||40.8%|Hard||
+|2436|Minimum Split Into Subarrays With GCD Greater Than One||85.0%|Medium||
+|2437|Number of Valid Clock Times||40.3%|Easy||
+|2438|Range Product Queries of Powers||36.6%|Medium||
+|2439|Minimize Maximum of Array||30.1%|Medium||
+|2440|Create Components With Same Value||53.4%|Hard||
+|2441|Largest Positive Integer That Exists With Its Negative||68.7%|Easy||
+|2442|Count Number of Distinct Integers After Reverse Operations||78.8%|Medium||
+|2443|Sum of Number and Its Reverse||40.6%|Medium||
+|2444|Count Subarrays With Fixed Bounds||38.4%|Hard||
 |------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|
 
-
 ------------------------------------------------------------------
 
 下面这些是免费的算法题，但是暂时还不能使用 Go 解答的：
 
-- [116.Populating Next Right Pointers in Each Node](https://leetcode.com/problems/populating-next-right-pointers-in-each-node/)
-- [117.Populating Next Right Pointers in Each Node II](https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/)
-- [133.Clone Graph](https://leetcode.com/problems/clone-graph/)
-- [138.Copy List with Random Pointer](https://leetcode.com/problems/copy-list-with-random-pointer/)
-- [278.First Bad Version](https://leetcode.com/problems/first-bad-version/)
-- [284.Peeking Iterator](https://leetcode.com/problems/peeking-iterator/)
-- [297.Serialize and Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)
-- [341.Flatten Nested List Iterator](https://leetcode.com/problems/flatten-nested-list-iterator/)
-- [374.Guess Number Higher or Lower](https://leetcode.com/problems/guess-number-higher-or-lower/)
-- [449.Serialize and Deserialize BST](https://leetcode.com/problems/serialize-and-deserialize-bst/)
-- [535.Encode and Decode TinyURL](https://leetcode.com/problems/encode-and-decode-tinyurl/)
-- [690.Employee Importance](https://leetcode.com/problems/employee-importance/)
-
+暂无
 
 ------------------------------------------------------------------
 
 
-## 二.分类
+## 三.分类
 
 ## Array
 
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[1. Two Sum](https://leetcode.com/problems/two-sum/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0001.%20Two%20Sum)| Easy | O(n)| O(n)||
-|[11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0011.%20Container%20With%20Most%20Water)| Medium | O(n)| O(1)||
-|[15. 3Sum](https://leetcode.com/problems/3sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0015.%203Sum)| Medium | O(n^2)| O(n)|❤️|
-|[16. 3Sum Closest](https://leetcode.com/problems/3sum-closest)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0016.%203Sum%20Closest)| Medium | O(n^2)| O(1)|❤️|
-|[18. 4Sum](https://leetcode.com/problems/4sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0018.%204Sum)| Medium | O(n^3)| O(n^2)|❤️|
-|[26. Remove Duplicates from Sorted Array](https://leetcode.com/problems/remove-duplicates-from-sorted-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0026.%20Remove%20Duplicates%20from%20Sorted%20Array)| Easy | O(n)| O(1)||
-|[27. Remove Element](https://leetcode.com/problems/remove-element)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0027.%20Remove%20Element)| Easy | O(n)| O(1)||
-|[39. Combination Sum](https://leetcode.com/problems/combination-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0039.%20Combination%20Sum)| Medium | O(n log n)| O(n)||
-|[40. Combination Sum II](https://leetcode.com/problems/combination-sum-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0040.%20Combination%20Sum%20II)| Medium | O(n log n)| O(n)||
-|[41. First Missing Positive](https://leetcode.com/problems/first-missing-positive)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0041.%20First-Missing-Positive)| Hard | O(n)| O(n)||
-|[42. Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0042.%20Trapping%20Rain%20Water)| Hard | O(n)| O(1)|❤️|
-|[48. Rotate Image](https://leetcode.com/problems/rotate-image)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0048.%20Rotate%20Image)| Medium | O(n)| O(1)||
-|[53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0053.%20Maximum%20Subarray)| Easy | O(n)| O(n)||
-|[54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0054.%20Spiral%20Matrix)| Medium | O(n)| O(n^2)||
-|[56. Merge Intervals](https://leetcode.com/problems/merge-intervals)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0056.%20Merge%20Intervals)| Medium | O(n log n)| O(1)||
-|[57. Insert Interval](https://leetcode.com/problems/insert-interval)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0057.%20Insert%20Interval)| Hard | O(n)| O(1)||
-|[59. Spiral Matrix II](https://leetcode.com/problems/spiral-matrix-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0059.%20Spiral%20Matrix%20II)| Medium | O(n)| O(n^2)||
-|[62. Unique Paths](https://leetcode.com/problems/unique-paths)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0062.%20Unique%20Paths)| Medium | O(n^2)| O(n^2)||
-|[63. Unique Paths II](https://leetcode.com/problems/unique-paths-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0063.%20Unique%20Paths%20II)| Medium | O(n^2)| O(n^2)||
-|[64. Minimum Path Sum](https://leetcode.com/problems/minimum-path-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0064.%20Minimum%20Path%20Sum)| Medium | O(n^2)| O(n^2)||
-|[75. Sort Colors](https://leetcode.com/problems/sort-colors/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0075.%20Sort%20Colors)| Medium| O(n)| O(1)|❤️|
-|[78. Subsets](https://leetcode.com/problems/subsets)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0078.%20Subsets)| Medium | O(n^2)| O(n)|❤️|
-|[79. Word Search](https://leetcode.com/problems/word-search)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0079.%20Word%20Search)| Medium | O(n^2)| O(n^2)|❤️|
-|[80. Remove Duplicates from Sorted Array II](https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0080.%20Remove%20Duplicates%20from%20Sorted%20Array%20II)| Medium | O(n)| O(1||
-|[84. Largest Rectangle in Histogram](https://leetcode.com/problems/largest-rectangle-in-histogram)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0084.%20Largest%20Rectangle%20in%20Histogram)| Medium | O(n)| O(n)|❤️|
-|[88. Merge Sorted Array](https://leetcode.com/problems/merge-sorted-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0088.%20Merge-Sorted-Array)| Easy | O(n)| O(1)|❤️|
-|[90. Subsets II](https://leetcode.com/problems/subsets-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0090.%20Subsets%20II)| Medium | O(n^2)| O(n)|❤️|
-|[120. Triangle](https://leetcode.com/problems/triangle)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0120.%20Triangle)| Medium | O(n^2)| O(n)||
-|[121. Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock)| Easy | O(n)| O(1)||
-|[122. Best Time to Buy and Sell Stock II](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0122.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20II)| Easy | O(n)| O(1)||
-|[126. Word Ladder II](https://leetcode.com/problems/word-ladder-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0126.%20Word%20Ladder%20II)| Hard | O(n)| O(n^2)|❤️|
-|[152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0152.%20Maximum%20Product%20Subarray)| Medium | O(n)| O(1)||
-|[167. Two Sum II - Input array is sorted](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0167.%20Two%20Sum%20II%20-%20Input%20array%20is%20sorted)| Easy | O(n)| O(1)||
-|[209. Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0209.%20Minimum%20Size%20Subarray%20Sum)| Medium | O(n)| O(1)||
-|[216. Combination Sum III](https://leetcode.com/problems/combination-sum-iii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0216.%20Combination%20Sum%20III)| Medium | O(n)| O(1)|❤️|
-|[217. Contains Duplicate](https://leetcode.com/problems/contains-duplicate)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0217.%20Contains%20Duplicate)| Easy | O(n)| O(n)||
-|[219. Contains Duplicate II](https://leetcode.com/problems/contains-duplicate-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0219.%20Contains%20Duplicate%20II)| Easy | O(n)| O(n)||
-|[283. Move Zeroes](https://leetcode.com/problems/move-zeroes)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0283.%20Move%20Zeroes)| Easy | O(n)| O(1)||
-|[287. Find the Duplicate Number](https://leetcode.com/problems/find-the-duplicate-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0287.%20Find%20the%20Duplicate%20Number)| Easy | O(n)| O(1)|❤️|
-|[532. K-diff Pairs in an Array](https://leetcode.com/problems/k-diff-pairs-in-an-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0532.%20K-diff%20Pairs%20in%20an%20Array)| Easy | O(n)| O(n)||
-|[566. Reshape the Matrix](https://leetcode.com/problems/reshape-the-matrix)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0566.%20Reshape%20the%20Matrix)| Easy | O(n^2)| O(n^2)||
-|[628. Maximum Product of Three Numbers](https://leetcode.com/problems/maximum-product-of-three-numbers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0628.%20Maximum%20Product%20of%20Three%20Numbers)| Easy | O(n)| O(1)||
-|[713. Subarray Product Less Than K](https://leetcode.com/problems/subarray-product-less-than-k)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0713.%20Subarray%20Product%20Less%20Than%20K)| Medium | O(n)| O(1)||
-|[714. Best Time to Buy and Sell Stock with Transaction Fee](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0714.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Transaction%20Fee)| Medium | O(n)| O(1)||
-|[746. Min Cost Climbing Stairs](https://leetcode.com/problems/min-cost-climbing-stairs)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0746.%20Min%20Cost%20Climbing%20Stairs)| Easy | O(n)| O(1)||
-|[766. Toeplitz Matrix](https://leetcode.com/problems/toeplitz-matrix)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0766.%20Toeplitz%20Matrix)| Easy | O(n)| O(1)||
-|[867. Transpose Matrix](https://leetcode.com/problems/transpose-matrix)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0867.%20Transpose%20Matrix)| Easy | O(n)| O(1)||
-|[891. Sum of Subsequence Widths](https://leetcode.com/problems/sum-of-subsequence-widths)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0891.%20Sum%20of%20Subsequence%20Widths)| Hard | O(n log n)| O(1)||
-|[907. Sum of Subarray Minimums](https://leetcode.com/problems/sum-of-subarray-minimums)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0907.%20Sum%20of%20Subarray%20Minimums)| Medium | O(n)| O(n)|❤️|
-|[922. Sort Array By Parity II](https://leetcode.com/problems/sum-of-subarray-minimums)| [Go]()| Medium | O(n)| O(1)||
-|[969. Pancake Sorting](https://leetcode.com/problems/pancake-sorting)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0969.%20Pancake%20Sorting)| Medium | O(n)| O(1)|❤️|
-|[977. Squares of a Sorted Array](https://leetcode.com/problems/squares-of-a-sorted-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0977.%20Squares%20of%20a%20Sorted%20Array)| Easy | O(n)| O(1)||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Array/)
 
 
 
 ## String
 
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0003.%20Longest%20Substring%20Without%20Repeating%20Characters)| Medium | O(n)| O(1)|❤️|
-|[17. Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0017.%20Letter%20Combinations%20of%20a%20Phone%20Number)| Medium | O(log n)| O(1)||
-|[20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0020.%20Valid-Parentheses)| Easy | O(log n)| O(1)||
-|[22. Generate Parentheses](https://leetcode.com/problems/generate-parentheses)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0022.%20Generate%20Parentheses)| Medium | O(log n)| O(1)||
-|[28. Implement strStr()](https://leetcode.com/problems/implement-strstr)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0028.%20Implement%20strStr())| Easy | O(n)| O(1)||
-|[30. Substring with Concatenation of All Words](https://leetcode.com/problems/substring-with-concatenation-of-all-words)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0030.%20Substring%20with%20Concatenation%20of%20All%20Words)| Hard | O(n)| O(n)|❤️|
-|[49. Group Anagrams](https://leetcode.com/problems/group-anagrams)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0049.%20Group%20Anagrams)| Medium | O(n log n)| O(n)||
-|[71. Simplify Path](https://leetcode.com/problems/simplify-path)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0071.%20Simplify%20Path)| Medium | O(n)| O(n)||
-|[76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0076.%20Minimum%20Window%20Substring)| Hard | O(n)| O(n)|❤️|
-|[91. Decode Ways](https://leetcode.com/problems/decode-ways)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0091.%20Decode%20Ways)| Medium | O(n)| O(n)||
-|[93. Restore IP Addresses](https://leetcode.com/problems/restore-ip-addresses)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0093.%20Restore%20IP%20Addresses)| Medium | O(n)| O(n)|❤️|
-|[125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0125.%20Valid-Palindrome)| Easy | O(n)| O(1)||
-|[126. Word Ladder II](https://leetcode.com/problems/word-ladder-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0126.%20Word%20Ladder%20II)| Hard | O(n)| O(n^2)|❤️|
-|[344. Reverse String](https://leetcode.com/problems/reverse-string)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0344.%20Reverse%20String)| Easy | O(n)| O(1)||
-|[345. Reverse Vowels of a String](https://leetcode.com/problems/reverse-vowels-of-a-string)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0345.%20Reverse%20Vowels%20of%20a%20String)| Easy | O(n)| O(1)||
-|[767. Reorganize String](https://leetcode.com/problems/reorganize-string/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0767.%20Reorganize%20String)| Medium | O(n log n)| O(log n)  |❤️|
-|[842. Split Array into Fibonacci Sequence](https://leetcode.com/problems/split-array-into-fibonacci-sequence)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0842.%20Split%20Array%20into%20Fibonacci%20Sequence)| Medium | O(n^2)| O(1)|❤️|
-|[856. Score of Parentheses](https://leetcode.com/problems/score-of-parentheses)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0856.%20Score%20of%20Parentheses)| Medium | O(n)| O(n)||
-|[925. Long Pressed Name](https://leetcode.com/problems/long-pressed-name)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0925.%20Long%20Pressed%20Name)| Easy | O(n)| O(1)||
-|[1003. Check If Word Is Valid After Substitutions](https://leetcode.com/problems/check-if-word-is-valid-after-substitutions)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1003.%20Check%20If%20Word%20Is%20Valid%20After%20Substitutions)| Medium | O(n)| O(1)||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/String/)
 
 
 ## Two Pointers
 
-![](./Topics/Two_pointers.png)
+![](./topic/Two_pointers.png)
 
 - 双指针滑动窗口的经典写法。右指针不断往右移，移动到不能往右移动为止(具体条件根据题目而定)。当右指针到最右边以后，开始挪动左指针，释放窗口左边界。第 3 题，第 76 题，第 209 题，第 424 题，第 438 题，第 567 题，第 713 题，第 763 题，第 845 题，第 881 题，第 904 题，第 978 题，第 992 题，第 1004 题，第 1040 题，第 1052 题。
 
@@ -1416,64 +2633,13 @@
 - 替换字母以后，相同字母能出现连续最长的长度。第 424 题。
 - SUM 问题集。第 1 题，第 15 题，第 16 题，第 18 题，第 167 题，第 923 题，第 1074 题。
 
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0003.%20Longest%20Substring%20Without%20Repeating%20Characters)| Medium | O(n)| O(1)|❤️|
-|[11. Container With Most Water](https://leetcode.com/problems/container-with-most-water)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0011.%20Container%20With%20Most%20Water)| Medium | O(n)| O(1)||
-|[15. 3Sum](https://leetcode.com/problems/3sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0015.%203Sum)| Medium | O(n^2)| O(n)|❤️|
-|[16. 3Sum Closest](https://leetcode.com/problems/3sum-closest)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0016.%203Sum%20Closest)| Medium | O(n^2)| O(1)|❤️|
-|[18. 4Sum](https://leetcode.com/problems/4sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0018.%204Sum)| Medium | O(n^3)| O(n^2)|❤️|
-|[19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0019.%20Remove%20Nth%20Node%20From%20End%20of%20List)| Medium | O(n)| O(1)||
-|[26. Remove Duplicates from Sorted Array](https://leetcode.com/problems/remove-duplicates-from-sorted-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0026.%20Remove%20Duplicates%20from%20Sorted%20Array)| Easy | O(n)| O(1)||
-|[27. Remove Element](https://leetcode.com/problems/remove-element)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0027.%20Remove%20Element)| Easy | O(n)| O(1)||
-|[28. Implement strStr()](https://leetcode.com/problems/implement-strstr)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0028.%20Implement%20strStr())| Easy | O(n)| O(1)||
-|[30. Substring with Concatenation of All Words](https://leetcode.com/problems/substring-with-concatenation-of-all-words)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0030.%20Substring%20with%20Concatenation%20of%20All%20Words)| Hard | O(n)| O(n)|❤️|
-|[42. Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0042.%20Trapping%20Rain%20Water)| Hard | O(n)| O(1)|❤️|
-|[61. Rotate List](https://leetcode.com/problems/rotate-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0061.%20Rotate%20List)| Medium | O(n)| O(1)||
-|[75. Sort Colors](https://leetcode.com/problems/sort-colors/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0075.%20Sort%20Colors)| Medium| O(n)| O(1)|❤️|
-|[76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0076.%20Minimum%20Window%20Substring)| Hard | O(n)| O(n)|❤️|
-|[80. Remove Duplicates from Sorted Array II](https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0080.%20Remove%20Duplicates%20from%20Sorted%20Array%20II)| Medium | O(n)| O(1||
-|[86. Partition List](https://leetcode.com/problems/partition-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0086.%20Partition%20List)| Medium | O(n)| O(1)|❤️|
-|[88. Merge Sorted Array](https://leetcode.com/problems/merge-sorted-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0088.%20Merge-Sorted-Array)| Easy | O(n)| O(1)|❤️|
-|[125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0125.%20Valid-Palindrome)| Easy | O(n)| O(1)||
-|[141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0141.%20Linked%20List%20Cycle)| Easy | O(n)| O(1)|❤️|
-|[142. Linked List Cycle II](https://leetcode.com/problems/linked-list-cycle-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0142.%20Linked%20List%20Cycle%20II)| Medium | O(n)| O(1)|❤️|
-|[167. Two Sum II - Input array is sorted](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0167.%20Two%20Sum%20II%20-%20Input%20array%20is%20sorted)| Easy | O(n)| O(1)||
-|[209. Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0209.%20Minimum%20Size%20Subarray%20Sum)| Medium | O(n)| O(1)||
-|[234. Palindrome Linked List](https://leetcode.com/problems/palindrome-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0234.%20Palindrome%20Linked%20List)| Easy | O(n)| O(1)||
-|[283. Move Zeroes](https://leetcode.com/problems/move-zeroes)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0283.%20Move%20Zeroes)| Easy | O(n)| O(1)||
-|[287. Find the Duplicate Number](https://leetcode.com/problems/find-the-duplicate-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0287.%20Find%20the%20Duplicate%20Number)| Easy | O(n)| O(1)|❤️|
-|[344. Reverse String](https://leetcode.com/problems/reverse-string)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0344.%20Reverse%20String)| Easy | O(n)| O(1)||
-|[345. Reverse Vowels of a String](https://leetcode.com/problems/reverse-vowels-of-a-string)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0345.%20Reverse%20Vowels%20of%20a%20String)| Easy | O(n)| O(1)||
-|[349. Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0349.%20Intersection%20of%20Two%20Arrays)| Easy | O(n)| O(n) ||
-|[350. Intersection of Two Arrays II](https://leetcode.com/problems/intersection-of-two-arrays-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0350.%20Intersection%20of%20Two%20Arrays%20II)| Easy | O(n)| O(n) ||
-|[424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0424.%20Longest%20Repeating%20Character%20Replacement)| Medium | O(n)| O(1) ||
-|[524. Longest Word in Dictionary through Deleting](https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0524.%20Longest%20Word%20in%20Dictionary%20through%20Deleting)| Medium | O(n)| O(1) ||
-|[532. K-diff Pairs in an Array](https://leetcode.com/problems/k-diff-pairs-in-an-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0532.%20K-diff%20Pairs%20in%20an%20Array)| Easy | O(n)| O(n)||
-|[567. Permutation in String](https://leetcode.com/problems/permutation-in-string)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0567.%20Permutation%20in%20String)| Medium | O(n)| O(1)|❤️|
-|[713. Subarray Product Less Than K](https://leetcode.com/problems/subarray-product-less-than-k)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0713.%20Subarray%20Product%20Less%20Than%20K)| Medium | O(n)| O(1)||
-|[763. Partition Labels](https://leetcode.com/problems/partition-labels)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0763.%20Partition%20Labels)| Medium | O(n)| O(1)|❤️|
-|[826. Most Profit Assigning Work](https://leetcode.com/problems/most-profit-assigning-work)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0826.%20Most%20Profit%20Assigning%20Work)| Medium | O(n log n)| O(n)||
-|[828. Unique Letter String](https://leetcode.com/problems/unique-letter-string)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0828.%20Unique%20Letter%20String)| Hard | O(n)| O(1)|❤️|
-|[838. Push Dominoes](https://leetcode.com/problems/push-dominoes)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0838.%20Push%20Dominoes)| Medium | O(n)| O(n)||
-|[844. Backspace String Compare](https://leetcode.com/problems/backspace-string-compare)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0844.%20Backspace%20String%20Compare)| Easy | O(n)| O(n) ||
-|[845. Longest Mountain in Array](https://leetcode.com/problems/longest-mountain-in-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0845.%20Longest%20Mountain%20in%20Array)| Medium | O(n)| O(1) ||
-|[881. Boats to Save People](https://leetcode.com/problems/boats-to-save-people)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0881.%20Boats%20to%20Save%20People)| Medium | O(n log n)| O(1) ||
-|[904. Fruit Into Baskets](https://leetcode.com/problems/fruit-into-baskets)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0904.%20Fruit%20Into%20Baskets)| Medium | O(n log n)| O(1) ||
-|[923. 3Sum With Multiplicity](https://leetcode.com/problems/3sum-with-multiplicity)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0923.%203Sum%20With%20Multiplicity)| Medium | O(n^2)| O(n) ||
-|[925. Long Pressed Name](https://leetcode.com/problems/long-pressed-name)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0925.%20Long%20Pressed%20Name)| Easy | O(n)| O(1)||
-|[930. Binary Subarrays With Sum](https://leetcode.com/problems/binary-subarrays-with-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0930.%20Binary%20Subarrays%20With%20Sum)| Medium | O(n)| O(n)  |❤️|
-|[977. Squares of a Sorted Array](https://leetcode.com/problems/squares-of-a-sorted-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0977.%20Squares%20of%20a%20Sorted%20Array)| Easy | O(n)| O(1)||
-|[986. Interval List Intersections](https://leetcode.com/problems/interval-list-intersections)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0986.%20Interval%20List%20Intersections)| Medium | O(n)| O(1)||
-|[992. Subarrays with K Different Integers](https://leetcode.com/problems/subarrays-with-k-different-integers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0992.%20Subarrays%20with%20K%20Different%20Integers)| Hard | O(n)| O(n)|❤️|
-|[1004. Max Consecutive Ones III](https://leetcode.com/problems/max-consecutive-ones-iii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1004.%20Max%20Consecutive%20Ones%20III)| Medium | O(n)| O(1) ||
-|[1093. Statistics from a Large Sample](https://leetcode.com/problems/statistics-from-a-large-sample)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1093.%20Statistics%20from%20a%20Large%20Sample)| Medium | O(n)| O(1) ||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Two_Pointers/)
 
 
 ## Linked List
 
-![](./Topics/Linked_List.png)
+![](./topic/Linked_List.png)
 
 
 - 巧妙的构造虚拟头结点。可以使遍历处理逻辑更加统一。
@@ -1486,133 +2652,28 @@
 - 判断链表是否存在环，如果有环，输出环的交叉点的下标；判断 2 个链表是否有交叉点，如果有交叉点，输出交叉点。第 141 题，第 142 题，第 160 题。
 
 
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Linked_List/)
 
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[2. Add Two Numbers](https://leetcode.com/problems/add-two-numbers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0002.%20Add%20Two%20Numbers)| Medium | O(n)| O(1)||
-|[19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0019.%20Remove%20Nth%20Node%20From%20End%20of%20List)| Medium | O(n)| O(1)||
-|[21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0021.%20Merge%20Two%20Sorted%20Lists)| Easy | O(log n)| O(1)||
-|[23. Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0023.%20Merge%20k%20Sorted%20Lists)| Hard | O(log n)| O(1)|❤️|
-|[24. Swap Nodes in Pairs](https://leetcode.com/problems/swap-nodes-in-pairs/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0024.%20Swap%20Nodes%20in%20Pairs)| Medium | O(n)| O(1)||
-|[25. Reverse Nodes in k-Group](https://leetcode.com/problems/reverse-nodes-in-k-group/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0025.%20Reverse%20Nodes%20in%20k%20Group)| Hard | O(log n)| O(1)|❤️|
-|[61. Rotate List](https://leetcode.com/problems/rotate-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0061.%20Rotate%20List)| Medium | O(n)| O(1)||
-|[82. Remove Duplicates from Sorted List II](https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0082.%20Remove%20Duplicates%20from%20Sorted%20List%20II)| Medium | O(n)| O(1)||
-|[83. Remove Duplicates from Sorted List](https://leetcode.com/problems/remove-duplicates-from-sorted-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0083.%20Remove%20Duplicates%20from%20Sorted%20List)| Easy | O(n)| O(1)||
-|[86. Partition List](https://leetcode.com/problems/partition-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0086.%20Partition%20List)| Medium | O(n)| O(1)|❤️|
-|[92. Reverse Linked List II](https://leetcode.com/problems/reverse-linked-list-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0092.%20Reverse%20Linked%20List%20II)| Medium | O(n)| O(1)|❤️|
-|[109. Convert Sorted List to Binary Search Tree](https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0109.%20Convert%20Sorted%20List%20to%20Binary%20Search%20Tree)| Medium | O(log n)| O(n)||
-|[141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0141.%20Linked%20List%20Cycle)| Easy | O(n)| O(1)|❤️|
-|[142. Linked List Cycle II](https://leetcode.com/problems/linked-list-cycle-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0142.%20Linked%20List%20Cycle%20II)| Medium | O(n)| O(1)|❤️|
-|[143. Reorder List](https://leetcode.com/problems/reorder-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0143.%20Reorder%20List)| Medium | O(n)| O(1)|❤️|
-|[147. Insertion Sort List](https://leetcode.com/problems/insertion-sort-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0147.%20Insertion%20Sort%20List)| Medium | O(n)| O(1)|❤️|
-|[148. Sort List](https://leetcode.com/problems/sort-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0148.%20Sort%20List)| Medium | O(n log n)| O(n)|❤️|
-|[160. Intersection of Two Linked Lists](https://leetcode.com/problems/intersection-of-two-linked-lists/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0160.%20Intersection%20of%20Two%20Linked%20Lists)| Easy | O(n)| O(1)|❤️|
-|[203. Remove Linked List Elements](https://leetcode.com/problems/remove-linked-list-elements/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0203.%20Remove%20Linked%20List%20Elements)| Easy | O(n)| O(1)||
-|[206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0206.%20Reverse-Linked-List)| Easy | O(n)| O(1)||
-|[234. Palindrome Linked List](https://leetcode.com/problems/palindrome-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0234.%20Palindrome%20Linked%20List)| Easy | O(n)| O(1)||
-|[237. Delete Node in a Linked List](https://leetcode.com/problems/delete-node-in-a-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0237.%20Delete%20Node%20in%20a%20Linked%20List)| Easy | O(n)| O(1)||
-|[328. Odd Even Linked List](https://leetcode.com/problems/odd-even-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0328.%20Odd%20Even%20Linked%20List)| Medium | O(n)| O(1)||
-|[445. Add Two Numbers II](https://leetcode.com/problems/add-two-numbers-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0445.%20Add%20Two%20Numbers%20II)| Medium | O(n)| O(n)||
-|[725. Split Linked List in Parts](https://leetcode.com/problems/split-linked-list-in-parts/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0725.%20Split%20Linked%20List%20in%20Parts)| Medium | O(n)| O(1)||
-|[817. Linked List Components](https://leetcode.com/problems/linked-list-components/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0817.%20Linked%20List%20Components)| Medium | O(n)| O(1)||
-|[707. Design Linked List](https://leetcode.com/problems/design-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0707.%20Design%20Linked%20List)| Easy | O(n)| O(1)||
-|[876. Middle of the Linked List](https://leetcode.com/problems/middle-of-the-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0876.%20Middle%20of%20the%20Linked%20List)| Easy | O(n)| O(1)|❤️|
-|[1019. Next Greater Node In Linked List](https://leetcode.com/problems/next-greater-node-in-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1019.%20Next%20Greater%20Node%20In%20Linked%20List)| Medium | O(n)| O(1)||
-|----------------------------------------------------------------------------|-------------|-------------| -------------| -------------|-------------|
 
 
 
 ## Stack
 
-![](./Topics/Stack.png)
+![](./topic/Stack.png)
 
 - 括号匹配问题及类似问题。第 20 题，第 921 题，第 1021 题。
 - 栈的基本 pop 和 push 操作。第 71 题，第 150 题，第 155 题，第 224 题，第 225 题，第 232 题，第 946 题，第 1047 题。
 - 利用栈进行编码问题。第 394 题，第 682 题，第 856 题，第 880 题。
 - **单调栈**。**利用栈维护一个单调递增或者递减的下标数组**。第 84 题，第 456 题，第 496 题，第 503 题，第 739 题，第 901 题，第 907 题，第 1019 题。
 
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0020.%20Valid-Parentheses)| Easy | O(log n)| O(1)||
-|[42. Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0042.%20Trapping%20Rain%20Water)| Hard | O(n)| O(1)|❤️|
-|[71. Simplify Path](https://leetcode.com/problems/simplify-path)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0071.%20Simplify%20Path)| Medium | O(n)| O(n)|❤️|
-|[84. Largest Rectangle in Histogram](https://leetcode.com/problems/largest-rectangle-in-histogram)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0084.%20Largest%20Rectangle%20in%20Histogram)| Medium | O(n)| O(n)|❤️|
-|[94. Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0094.%20Binary%20Tree%20Inorder%20Traversal)| Medium | O(n)| O(1)||
-|[103. Binary Tree Zigzag Level Order Traversal](https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0103.%20Binary%20Tree%20Zigzag%20Level%20Order%20Traversal)| Medium | O(n)| O(n)||
-|[144. Binary Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0144.%20Binary%20Tree%20Preorder%20Traversal)| Medium | O(n)| O(1)||
-|[145. Binary Tree Postorder Traversal](https://leetcode.com/problems/binary-tree-postorder-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0145.%20Binary%20Tree%20Postorder%20Traversal)| Hard | O(n)| O(1)||
-|[150. Evaluate Reverse Polish Notation](https://leetcode.com/problems/evaluate-reverse-polish-notation)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0150.%20Evaluate%20Reverse%20Polish%20Notation)| Medium | O(n)| O(1)||
-|[155. Min Stack](https://leetcode.com/problems/min-stack)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0155.%20Min%20Stack)| Easy | O(n)| O(n)||
-|[173. Binary Search Tree Iterator](https://leetcode.com/problems/binary-search-tree-iterator)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0173.%20Binary%20Search%20Tree%20Iterator)| Medium | O(n)| O(1)||
-|[224. Basic Calculator](https://leetcode.com/problems/basic-calculator)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0224.%20Basic%20Calculator)| Hard | O(n)| O(n)||
-|[225. Implement Stack using Queues](https://leetcode.com/problems/implement-stack-using-queues)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0225.%20Implement%20Stack%20using%20Queues)| Easy | O(n)| O(n)||
-|[232. Implement Queue using Stacks](https://leetcode.com/problems/implement-queue-using-stacks)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0232.%20Implement%20Queue%20using%20Stacks)| Easy | O(n)| O(n)||
-|[331. Verify Preorder Serialization of a Binary Tree](https://leetcode.com/problems/verify-preorder-serialization-of-a-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0331.%20Verify%20Preorder%20Serialization%20of%20a%20Binary%20Tree)| Medium | O(n)| O(1)||
-|[394. Decode String](https://leetcode.com/problems/decode-string)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0394.%20Decode%20String)| Medium | O(n)| O(n)||
-|[402. Remove K Digits](https://leetcode.com/problems/remove-k-digits)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0402.%20Remove%20K%20Digits)| Medium | O(n)| O(1)||
-|[456. 132 Pattern](https://leetcode.com/problems/132-pattern)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0456.%20132%20Pattern)| Medium | O(n)| O(n)||
-|[496. Next Greater Element I](https://leetcode.com/problems/next-greater-element-i)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0496.%20Next%20Greater%20Element%20I)| Easy | O(n)| O(n)||
-|[503. Next Greater Element II](https://leetcode.com/problems/next-greater-element-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0503.%20Next%20Greater%20Element%20II)| Medium | O(n)| O(n)||
-|[636. Exclusive Time of Functions](https://leetcode.com/problems/exclusive-time-of-functions)| [Go](https://leetcode.com/problems/exclusive-time-of-functions)| Medium | O(n)| O(n)||
-|[682. Baseball Game](https://leetcode.com/problems/baseball-game)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0682.%20Baseball%20Game)| Easy | O(n)| O(n)||
-|[726. Number of Atoms](https://leetcode.com/problems/number-of-atoms)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0726.%20Number%20of%20Atoms)| Hard | O(n)| O(n) |❤️|
-|[735. Asteroid Collision](https://leetcode.com/problems/asteroid-collision)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0735.%20Asteroid%20Collision)| Medium | O(n)| O(n) ||
-|[739. Daily Temperatures](https://leetcode.com/problems/daily-temperatures)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0739.%20Daily%20Temperatures)| Medium | O(n)| O(n) ||
-|[844. Backspace String Compare](https://leetcode.com/problems/backspace-string-compare)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0844.%20Backspace%20String%20Compare)| Easy | O(n)| O(n) ||
-|[856. Score of Parentheses](https://leetcode.com/problems/score-of-parentheses)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0856.%20Score%20of%20Parentheses)| Medium | O(n)| O(n)||
-|[880. Decoded String at Index](https://leetcode.com/problems/decoded-string-at-index)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0880.%20Decoded%20String%20at%20Index)| Medium | O(n)| O(n)||
-|[895. Maximum Frequency Stack](https://leetcode.com/problems/maximum-frequency-stack)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0895.%20Maximum%20Frequency%20Stack)| Hard | O(n)| O(n)  ||
-|[901. Online Stock Span](https://leetcode.com/problems/online-stock-span)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0901.%20Online%20Stock%20Span)| Medium | O(n)| O(n)  ||
-|[907. Sum of Subarray Minimums](https://leetcode.com/problems/sum-of-subarray-minimums)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0907.%20Sum%20of%20Subarray%20Minimums)| Medium | O(n)| O(n)|❤️|
-|[921. Minimum Add to Make Parentheses Valid](https://leetcode.com/problems/minimum-add-to-make-parentheses-valid)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0921.%20Minimum%20Add%20to%20Make%20Parentheses%20Valid)| Medium | O(n)| O(n)||
-|[946. Validate Stack Sequences](https://leetcode.com/problems/validate-stack-sequences)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0946.%20Validate%20Stack%20Sequences)| Medium | O(n)| O(n)||
-|[1003. Check If Word Is Valid After Substitutions](https://leetcode.com/problems/check-if-word-is-valid-after-substitutions)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1003.%20Check%20If%20Word%20Is%20Valid%20After%20Substitutions)| Medium | O(n)| O(1)||
-|[1019. Next Greater Node In Linked List](https://leetcode.com/problems/next-greater-node-in-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1019.%20Next%20Greater%20Node%20In%20Linked%20List)| Medium | O(n)| O(1)||
-|[1021. Remove Outermost Parentheses](https://leetcode.com/problems/remove-outermost-parentheses)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1021.%20Remove%20Outermost%20Parentheses)| Medium | O(n)| O(1)||
-|[1047. Remove All Adjacent Duplicates In String](https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1047.%20Remove%20All%20Adjacent%20Duplicates%20In%20String)| Medium | O(n)| O(1)||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Stack/)
 
 
 
 ## Tree
 
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[94. Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0094.%20Binary%20Tree%20Inorder%20Traversal)| Medium | O(n)| O(1)||
-|[96. Unique Binary Search Trees](https://leetcode.com/problems/unique-binary-search-trees)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0096.%20Unique%20Binary%20Search%20Trees)| Medium | O(n^2)| O(n)||
-|[98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0098.%20Validate%20Binary%20Search%20Tree)| Medium | O(n)| O(1)||
-|[99. Recover Binary Search Tree](https://leetcode.com/problems/recover-binary-search-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0099.%20Recover%20Binary%20Search%20Tree)| Hard | O(n)| O(1)||
-|[100. Same Tree](https://leetcode.com/problems/same-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0100.%20Same%20Tree)| Easy | O(n)| O(1)||
-|[101. Symmetric Tree](https://leetcode.com/problems/symmetric-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0101.%20Symmetric%20Tree)| Easy | O(n)| O(1)||
-|[102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0102.%20Binary%20Tree%20Level%20Order%20Traversal)| Medium | O(n)| O(1)||
-|[103. Binary Tree Zigzag Level Order Traversal](https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0103.%20Binary%20Tree%20Zigzag%20Level%20Order%20Traversal)| Medium | O(n)| O(n)||
-|[104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0104.%20Maximum%20Depth%20of%20Binary%20Tree)| Easy | O(n)| O(1)||
-|[107. Binary Tree Level Order Traversal II](https://leetcode.com/problems/binary-tree-level-order-traversal-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0107.%20Binary%20Tree%20Level%20Order%20Traversal%20II)| Easy | O(n)| O(1)||
-|[108. Convert Sorted Array to Binary Search Tree](https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0108.%20Convert%20Sorted%20Array%20to%20Binary%20Search%20Tree)| Easy | O(n)| O(1)||
-|[110. Balanced Binary Tree](https://leetcode.com/problems/balanced-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0110.%20Balanced%20Binary%20Tree)| Easy | O(n)| O(1)||
-|[111. Minimum Depth of Binary Tree](https://leetcode.com/problems/minimum-depth-of-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0111.%20Minimum%20Depth%20of%20Binary%20Tree)| Easy | O(n)| O(1)||
-|[112. Path Sum](https://leetcode.com/problems/path-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0112.%20Path%20Sum)| Easy | O(n)| O(1)||
-|[113. Path Sum II](https://leetcode.com/problems/path-sum-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0113.%20Path%20Sum%20II)| Medium | O(n)| O(1)||
-|[114. Flatten Binary Tree to Linked List](https://leetcode.com/problems/flatten-binary-tree-to-linked-list)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0114.%20Flatten%20Binary%20Tree%20to%20Linked%20List)| Medium | O(n)| O(1)||
-|[124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0124.%20Binary%20Tree%20Maximum%20Path%20Sum)| Hard | O(n)| O(1)||
-|[129. Sum Root to Leaf Numbers](https://leetcode.com/problems/sum-root-to-leaf-numbers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0129.%20Sum%20Root%20to%20Leaf%20Numbers)| Medium | O(n)| O(1)||
-|[144. Binary Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0144.%20Binary%20Tree%20Preorder%20Traversal)| Medium | O(n)| O(1)||
-|[145. Binary Tree Postorder Traversal](https://leetcode.com/problems/binary-tree-postorder-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0145.%20Binary%20Tree%20Postorder%20Traversal)| Hard | O(n)| O(1)||
-|[173. Binary Search Tree Iterator](https://leetcode.com/problems/binary-search-tree-iterator)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0173.%20Binary%20Search%20Tree%20Iterator)| Medium | O(n)| O(1)||
-|[199. Binary Tree Right Side View](https://leetcode.com/problems/binary-tree-right-side-view)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0199.%20Binary%20Tree%20Right%20Side%20Views)| Medium | O(n)| O(1)||
-|[222. Count Complete Tree Nodes](https://leetcode.com/problems/count-complete-tree-nodes)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0222.%20Count%20Complete%20Tree%20Nodes)| Medium | O(n)| O(1)||
-|[226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0226.%20Invert%20Binary%20Tree)| Easy | O(n)| O(1)||
-|[230. Kth Smallest Element in a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0230.%20Kth%20Smallest%20Element%20in%20a%20BST)| Medium | O(n)| O(1)||
-|[235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree)| Easy | O(n)| O(1)||
-|[236. Lowest Common Ancestor of a Binary Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0236.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree)| Medium | O(n)| O(1)||
-|[257. Binary Tree Paths](https://leetcode.com/problems/binary-tree-paths)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0257.%20Binary%20Tree%20Paths)| Easy | O(n)| O(1)||
-|[404. Sum of Left Leaves](https://leetcode.com/problems/sum-of-left-leaves)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0404.%20Sum%20of%20Left%20Leaves)| Easy | O(n)| O(1)||
-|[437. Path Sum III](https://leetcode.com/problems/path-sum-iii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0437.%20Path%20Sum%20III)| Easy | O(n)| O(1)||
-|[515. Find Largest Value in Each Tree Row](https://leetcode.com/problems/find-largest-value-in-each-tree-row)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0515.%20Find%20Largest%20Value%20in%20Each%20Tree%20Row)| Medium | O(n)| O(n)||
-|[637. Average of Levels in Binary Tree](https://leetcode.com/problems/average-of-levels-in-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0637.%20Average%20of%20Levels%20in%20Binary%20Tree)| Easy | O(n)| O(n)||
-|[993. Cousins in Binary Tree](https://leetcode.com/problems/cousins-in-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0993.%20Cousins%20in%20Binary%20Tree)| Easy | O(n)| O(1)||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
 
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Tree/)
 
 
 
@@ -1620,40 +2681,13 @@
 
 ## Dynamic Programming
 
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0053.%20Maximum%20Subarray)| Easy | O(n)| O(n)||
-|[62. Unique Paths](https://leetcode.com/problems/unique-paths)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0062.%20Unique%20Paths)| Medium | O(n^2)| O(n^2)||
-|[63. Unique Paths II](https://leetcode.com/problems/unique-paths-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0063.%20Unique%20Paths%20II)| Medium | O(n^2)| O(n^2)||
-|[64. Minimum Path Sum](https://leetcode.com/problems/minimum-path-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0064.%20Minimum%20Path%20Sum)| Medium | O(n^2)| O(n^2)||
-|[70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0070.%20Climbing%20Stairs)| Easy | O(n)| O(n)||
-|[91. Decode Ways](https://leetcode.com/problems/decode-ways)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0091.%20Decode%20Ways)| Medium | O(n)| O(n)||
-|[96. Unique Binary Search Trees](https://leetcode.com/problems/unique-binary-search-trees)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0096.%20Unique%20Binary%20Search%20Trees)| Medium | O(n)| O(n)||
-|[120. Triangle](https://leetcode.com/problems/triangle)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0120.%20Triangle)| Medium | O(n^2)| O(n)||
-|[121. Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock)| Easy | O(n)| O(1)||
-|[152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0152.%20Maximum%20Product%20Subarray)| Medium | O(n)| O(1)||
-|[198. House Robber](https://leetcode.com/problems/house-robber)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0198.%20House%20Robber)| Easy | O(n)| O(n)||
-|[213. House Robber II](https://leetcode.com/problems/house-robber-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0213.%20House%20Robber%20II)| Medium | O(n)| O(n)||
-|[300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0300.%20Longest%20Increasing%20Subsequence)| Medium | O(n log n)| O(n)||
-|[309. Best Time to Buy and Sell Stock with Cooldown](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0309.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Cooldown)| Medium | O(n)| O(n)||
-|[322. Coin Change](https://leetcode.com/problems/coin-change)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0322.%20Coin%20Change)| Medium | O(n)| O(n)||
-|[338. Counting Bits](https://leetcode.com/problems/counting-bits)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0338.%20Counting%20Bits)| Medium | O(n)| O(n)||
-|[343. Integer Break](https://leetcode.com/problems/integer-break)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0343.%20Integer%20Break)| Medium | O(n^2)| O(n)||
-|[357. Count Numbers with Unique Digits](https://leetcode.com/problems/count-numbers-with-unique-digits)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0357.%20Count%20Numbers%20with%20Unique%20Digits)| Medium | O(1)| O(1)||
-|[392. Is Subsequence](https://leetcode.com/problems/is-subsequence)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0392.%20Is%20Subsequence)| Medium | O(n)| O(1)||
-|[416. Partition Equal Subset Sum](https://leetcode.com/problems/partition-equal-subset-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0416.%20Partition%20Equal%20Subset%20Sum)| Medium | O(n^2)| O(n)||
-|[714. Best Time to Buy and Sell Stock with Transaction Fee](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0714.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Transaction%20Fee)| Medium | O(n)| O(1)||
-|[746. Min Cost Climbing Stairs](https://leetcode.com/problems/min-cost-climbing-stairs)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0746.%20Min%20Cost%20Climbing%20Stairs)| Easy | O(n)| O(1)||
-|[838. Push Dominoes](https://leetcode.com/problems/push-dominoes)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0838.%20Push%20Dominoes)| Medium | O(n)| O(n)||
-|[1025. Divisor Game](https://leetcode.com/problems/divisor-game)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1025.%20Divisor%20Game)| Easy | O(1)| O(1)||
-|[891. Sum of Subsequence Widths](https://leetcode.com/problems/sum-of-subsequence-widths)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0891.%20Sum%20of%20Subsequence%20Widths)| Hard | O(n log n)| O(1)||
-|[942. DI String Match](https://leetcode.com/problems/di-string-match)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0942.%20DI%20String%20Match)| Easy | O(n)| O(1)||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Dynamic_Programming/)
+
 
 
 ## Backtracking
 
-![](./Topics/Backtracking.png)
+![](./topic/Backtracking.png)
 
 - 排列问题 Permutations。第 46 题，第 47 题。第 60 题，第 526 题，第 996 题。
 - 组合问题 Combination。第 39 题，第 40 题，第 77 题，第 216 题。
@@ -1745,97 +2779,23 @@ func updateMatrix_BFS(matrix [][]int) [][]int {
 }
 ```
 
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[17. Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0017.%20Letter%20Combinations%20of%20a%20Phone%20Number)| Medium | O(log n)| O(1)||
-|[22. Generate Parentheses](https://leetcode.com/problems/generate-parentheses)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0022.%20Generate%20Parentheses)| Medium | O(log n)| O(1)||
-|[37. Sudoku Solver](https://leetcode.com/problems/sudoku-solver)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0037.%20Sudoku%20Solver)| Hard | O(n^2)| O(n^2)|❤️|
-|[39. Combination Sum](https://leetcode.com/problems/combination-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0039.%20Combination%20Sum)| Medium | O(n log n)| O(n)||
-|[40. Combination Sum II](https://leetcode.com/problems/combination-sum-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0040.%20Combination%20Sum%20II)| Medium | O(n log n)| O(n)||
-|[46. Permutations](https://leetcode.com/problems/permutations)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0046.%20Permutations)| Medium | O(n)| O(n)|❤️|
-|[47. Permutations II](https://leetcode.com/problems/permutations-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0047.%20Permutations%20II)| Medium | O(n^2)| O(n)|❤️|
-|[51. N-Queens](https://leetcode.com/problems/n-queens)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0051.%20N-Queens)| Hard | O(n^2)| O(n)|❤️|
-|[52. N-Queens II](https://leetcode.com/problems/n-queens-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0052.%20N-Queens%20II)| Hard | O(n^2)| O(n)|❤️|
-|[60. Permutation Sequence](https://leetcode.com/problems/permutation-sequence)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0060.%20Permutation%20Sequence)| Medium | O(n log n)| O(1)||
-|[77. Combinations](https://leetcode.com/problems/combinations)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0077.%20Combinations)| Medium | O(n)| O(n)|❤️|
-|[78. Subsets](https://leetcode.com/problems/subsets)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0078.%20Subsets)| Medium | O(n^2)| O(n)|❤️|
-|[79. Word Search](https://leetcode.com/problems/word-search)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0079.%20Word%20Search)| Medium | O(n^2)| O(n^2)|❤️|
-|[89. Gray Codes](https://leetcode.com/problems/gray-code)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0089.%20Gray%20Code)| Medium | O(n)| O(1)||
-|[90. Subsets II](https://leetcode.com/problems/subsets-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0090.%20Subsets%20II)| Medium | O(n^2)| O(n)|❤️|
-|[93. Restore IP Addresses](https://leetcode.com/problems/restore-ip-addresses)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0093.%20Restore%20IP%20Addresses)| Medium | O(n)| O(n)|❤️|
-|[126. Word Ladder II](https://leetcode.com/problems/word-ladder-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0126.%20Word%20Ladder%20II)| Hard | O(n)| O(n^2)|❤️|
-|[131. Palindrome Partitioning](https://leetcode.com/problems/palindrome-partitioning)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0131.%20Palindrome%20Partitioning)| Medium | O(n)| O(n^2)|❤️|
-|[211. Add and Search Word - Data structure design](https://leetcode.com/problems/add-and-search-word-data-structure-design)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0211.%20Add%20and%20Search%20Word%20-%20Data%20structure%20design)| Medium | O(n)| O(n)|❤️|
-|[212. Word Search II](https://leetcode.com/problems/word-search-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0212.%20Word%20Search%20II)| Hard | O(n^2)| O(n^2)|❤️|
-|[216. Combination Sum III](https://leetcode.com/problems/combination-sum-iii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0216.%20Combination%20Sum%20III)| Medium | O(n)| O(1)|❤️|
-|[306. Additive Number](https://leetcode.com/problems/additive-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0306.%20Additive%20Number)| Medium | O(n^2)| O(1)|❤️|
-|[357. Count Numbers with Unique Digits](https://leetcode.com/problems/count-numbers-with-unique-digits)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0357.%20Count%20Numbers%20with%20Unique%20Digits)| Medium | O(1)| O(1)||
-|[401. Binary Watch](https://leetcode.com/problems/binary-watch)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0401.%20Binary%20Watch)| Easy | O(1)| O(1)||
-|[526. Beautiful Arrangement](https://leetcode.com/problems/beautiful-arrangement)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0526.%20Beautiful%20Arrangement)| Medium | O(n^2)| O(1)|❤️|
-|[784. Letter Case Permutation](https://leetcode.com/problems/letter-case-permutation)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0784.%20Letter%20Case%20Permutation)| Easy | O(n)| O(n)||
-|[842. Split Array into Fibonacci Sequence](https://leetcode.com/problems/split-array-into-fibonacci-sequence)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0842.%20Split%20Array%20into%20Fibonacci%20Sequence)| Medium | O(n^2)| O(1)|❤️|
-|[980. Unique Paths III](https://leetcode.com/problems/unique-paths-iii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0980.%20Unique%20Paths%20III)| Hard | O(n log n)| O(n)||
-|[996. Number of Squareful Arrays](https://leetcode.com/problems/number-of-squareful-arrays)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0996.%20Number%20of%20Squareful%20Arrays)| Hard | O(n log n)| O(n) ||
-|[1079. Letter Tile Possibilities](https://leetcode.com/problems/letter-tile-possibilities)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1079.%20Letter%20Tile%20Possibilities)| Medium | O(n^2)| O(1)|❤️|
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
-
-
-## Depth-first Search
-
-
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0098.%20Validate%20Binary%20Search%20Tree)| Medium | O(n)| O(1)||
-|[99. Recover Binary Search Tree](https://leetcode.com/problems/recover-binary-search-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0099.%20Recover%20Binary%20Search%20Tree)| Hard | O(n)| O(1)||
-|[100. Same Tree](https://leetcode.com/problems/same-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0100.%20Same%20Tree)| Easy | O(n)| O(1)||
-|[101. Symmetric Tree](https://leetcode.com/problems/symmetric-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0101.%20Symmetric%20Tree)| Easy | O(n)| O(1)||
-|[104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0104.%20Maximum%20Depth%20of%20Binary%20Tree)| Easy | O(n)| O(1)||
-|[108. Convert Sorted Array to Binary Search Tree](https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0108.%20Convert%20Sorted%20Array%20to%20Binary%20Search%20Tree)| Easy | O(n)| O(1)||
-|[109. Convert Sorted List to Binary Search Tree](https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0109.%20Convert%20Sorted%20List%20to%20Binary%20Search%20Tree)| Medium | O(log n)| O(n)||
-|[110. Balanced Binary Tree](https://leetcode.com/problems/balanced-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0110.%20Balanced%20Binary%20Tree)| Easy | O(n)| O(1)||
-|[111. Minimum Depth of Binary Tree](https://leetcode.com/problems/minimum-depth-of-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0111.%20Minimum%20Depth%20of%20Binary%20Tree)| Easy | O(n)| O(1)||
-|[112. Path Sum](https://leetcode.com/problems/path-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0112.%20Path%20Sum)| Easy | O(n)| O(1)||
-|[113. Path Sum II](https://leetcode.com/problems/path-sum-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0113.%20Path%20Sum%20II)| Medium | O(n)| O(1)||
-|[114. Flatten Binary Tree to Linked List](https://leetcode.com/problems/flatten-binary-tree-to-linked-list)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0114.%20Flatten%20Binary%20Tree%20to%20Linked%20List)| Medium | O(n)| O(1)||
-|[124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0124.%20Binary%20Tree%20Maximum%20Path%20Sum)| Hard | O(n)| O(1)||
-|[129. Sum Root to Leaf Numbers](https://leetcode.com/problems/sum-root-to-leaf-numbers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0129.%20Sum%20Root%20to%20Leaf%20Numbers)| Medium | O(n)| O(1)||
-|[199. Binary Tree Right Side View](https://leetcode.com/problems/binary-tree-right-side-view)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0199.%20Binary%20Tree%20Right%20Side%20Views)| Medium | O(n)| O(1)||
-|[200. Number of Islands](https://leetcode.com/problems/number-of-islands)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0200.%20Number%20of%20Islands)| Medium | O(n^2)| O(n^2)||
-|[207. Course Schedule](https://leetcode.com/problems/course-schedule)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0207.%20Course%20Schedule)| Medium | O(n^2)| O(n^2)||
-|[210. Course Schedule II](https://leetcode.com/problems/course-schedule-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0210.%20Course%20Schedule%20II)| Medium | O(n^2)| O(n^2)||
-|[257. Binary Tree Paths](https://leetcode.com/problems/binary-tree-paths)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0257.%20Binary%20Tree%20Paths)| Easy | O(n)| O(1)||
-|[394. Decode String](https://leetcode.com/problems/decode-string)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0394.%20Decode%20String)| Medium | O(n)| O(n)||
-|[515. Find Largest Value in Each Tree Row](https://leetcode.com/problems/find-largest-value-in-each-tree-row)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0515.%20Find%20Largest%20Value%20in%20Each%20Tree%20Row)| Medium | O(n)| O(n)||
-|[542. 01 Matrix](https://leetcode.com/problems/01-matrix)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0542.%2001%20Matrix)| Medium | O(n)| O(1)||
-|[980. Unique Paths III](https://leetcode.com/problems/unique-paths-iii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0980.%20Unique%20Paths%20III)| Hard | O(n log n)| O(n)||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
-
-
-
-
-
-
-## Breadth-first Search
-
-
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[101. Symmetric Tree](https://leetcode.com/problems/symmetric-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0101.%20Symmetric%20Tree)| Easy | O(n)| O(1)||
-|[102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0102.%20Binary%20Tree%20Level%20Order%20Traversal)| Medium | O(n)| O(1)||
-|[103. Binary Tree Zigzag Level Order Traversal](https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0103.%20Binary%20Tree%20Zigzag%20Level%20Order%20Traversal)| Medium | O(n)| O(n)||
-|[107. Binary Tree Level Order Traversal II](https://leetcode.com/problems/binary-tree-level-order-traversal-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0107.%20Binary%20Tree%20Level%20Order%20Traversal%20II)| Easy | O(n)| O(1)||
-|[111. Minimum Depth of Binary Tree](https://leetcode.com/problems/minimum-depth-of-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0111.%20Minimum%20Depth%20of%20Binary%20Tree)| Easy | O(n)| O(1)||
-|[126. Word Ladder II](https://leetcode.com/problems/word-ladder-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0126.%20Word%20Ladder%20II)| Hard | O(n)| O(n^2)|❤️|
-|[127. Word Ladder](https://leetcode.com/problems/word-ladder)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0127.%20Word%20Ladder)| Medium | O(n)| O(n)||
-|[199. Binary Tree Right Side View](https://leetcode.com/problems/binary-tree-right-side-view)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0199.%20Binary%20Tree%20Right%20Side%20Views)| Medium | O(n)| O(1)||
-|[200. Number of Islands](https://leetcode.com/problems/number-of-islands)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0200.%20Number%20of%20Islands)| Medium | O(n^2)| O(n^2)||
-|[207. Course Schedule](https://leetcode.com/problems/course-schedule)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0207.%20Course%20Schedule)| Medium | O(n^2)| O(n^2)||
-|[210. Course Schedule II](https://leetcode.com/problems/course-schedule-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0210.%20Course%20Schedule%20II)| Medium | O(n^2)| O(n^2)||
-|[515. Find Largest Value in Each Tree Row](https://leetcode.com/problems/find-largest-value-in-each-tree-row)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0515.%20Find%20Largest%20Value%20in%20Each%20Tree%20Row)| Medium | O(n)| O(n)||
-|[542. 01 Matrix](https://leetcode.com/problems/01-matrix)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0542.%2001%20Matrix)| Medium | O(n)| O(1)||
-|[993. Cousins in Binary Tree](https://leetcode.com/problems/cousins-in-binary-tree)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0993.%20Cousins%20in%20Binary%20Tree)| Easy | O(n)| O(1)||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
 
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Backtracking/)
+
+
+## Depth First Search
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Depth_First_Search/)
+
+
+
+
+## Breadth First Search
+
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Breadth_First_Search/)
 
 
 
@@ -1965,109 +2925,30 @@ func peakIndexInMountainArray(A []int) int {
 }
 ```
 
-- max-min 最大值最小化问题。求在最小满足条件的情况下的最大值。第 410 题，第 1011 题。
-
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[50. Pow(x, n)](https://leetcode.com/problems/powx-n)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0050.%20Pow(x%2C%20n))| Medium | O(log n)| O(1)||
-|[69. Sqrt(x)](https://leetcode.com/problems/sqrtx)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0069.%20Sqrt(x))| Easy | O(log n)| O(1)||
-|[167. Two Sum II - Input array is sorted](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0167.%20Two%20Sum%20II%20-%20Input%20array%20is%20sorted)| Easy | O(n)| O(1)||
-|[209. Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0209.%20Minimum%20Size%20Subarray%20Sum)| Medium | O(n)| O(1)||
-|[222. Count Complete Tree Nodes](https://leetcode.com/problems/count-complete-tree-nodes)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0222.%20Count%20Complete%20Tree%20Nodes)| Medium | O(n)| O(1)||
-|[230. Kth Smallest Element in a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0230.%20Kth%20Smallest%20Element%20in%20a%20BST)| Medium | O(n)| O(1)||
-|[287. Find the Duplicate Number](https://leetcode.com/problems/find-the-duplicate-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0287.%20Find%20the%20Duplicate%20Number)| Easy | O(n)| O(1)|❤️|
-|[300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0300.%20Longest%20Increasing%20Subsequence)| Medium | O(n log n)| O(n)||
-|[349. Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0349.%20Intersection%20of%20Two%20Arrays)| Easy | O(n)| O(n) ||
-|[350. Intersection of Two Arrays II](https://leetcode.com/problems/intersection-of-two-arrays-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0350.%20Intersection%20of%20Two%20Arrays%20II)| Easy | O(n)| O(n) ||
-|[392. Is Subsequence](https://leetcode.com/problems/is-subsequence)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0392.%20Is%20Subsequence)| Medium | O(n)| O(1)||
-|[454. 4Sum II](https://leetcode.com/problems/4sum-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0454.%204Sum%20II)| Medium | O(n^2)| O(n) ||
-|[710. Random Pick with Blacklist](https://leetcode.com/problems/random-pick-with-blacklist/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0710.%20Random%20Pick%20with%20Blacklist)| Hard | O(n)| O(n)  ||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
-
-
-
-
-
-
-
-
-
+- max-min 最大值最小化问题。求在最小满足条件的情况下的最大值。第 410 题，第 875 题，第 1011 题，第 1283 题。
 
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Binary_Search/)
 
 
 
 ## Math
 
 
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[2. Add Two Numbers](https://leetcode.com/problems/add-two-numbers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0002.%20Add%20Two%20Numbers)| Medium | O(n)| O(1)||
-|[50. Pow(x, n)](https://leetcode.com/problems/powx-n)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0050.%20Pow(x%2C%20n))| Medium | O(log n)| O(1)||
-|[60. Permutation Sequence](https://leetcode.com/problems/permutation-sequence)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0060.%20Permutation%20Sequence)| Medium | O(n log n)| O(1)||
-|[69. Sqrt(x)](https://leetcode.com/problems/sqrtx)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0069.%20Sqrt(x))| Easy | O(log n)| O(1)||
-|[202. Happy Number](https://leetcode.com/problems/happy-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0202.%20Happy%20Number)| Easy | O(log n)| O(1)||
-|[224. Basic Calculator](https://leetcode.com/problems/basic-calculator)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0224.%20Basic%20Calculator)| Hard | O(n)| O(n)||
-|[231. Power of Two](https://leetcode.com/problems/power-of-twor)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0231.%20Power%20of%20Two)| Easy | O(1)| O(1)||
-|[263. Ugly Number](https://leetcode.com/problems/ugly-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0263.%20Ugly%20Number)| Easy | O(log n)| O(1)||
-|[326. Power of Three](https://leetcode.com/problems/power-of-three)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0326.%20Power%20of%20Three)| Easy | O(1)| O(1)||
-|[343. Integer Break](https://leetcode.com/problems/integer-break)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0343.%20Integer%20Break)| Medium | O(n^2)| O(n)||
-|[357. Count Numbers with Unique Digits](https://leetcode.com/problems/count-numbers-with-unique-digits)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0357.%20Count%20Numbers%20with%20Unique%20Digits)| Medium | O(1)| O(1)||
-|[628. Maximum Product of Three Numbers](https://leetcode.com/problems/maximum-product-of-three-numbers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0628.%20Maximum%20Product%20of%20Three%20Numbers)| Easy | O(n)| O(1)||
-|[885. Spiral Matrix III](https://leetcode.com/problems/spiral-matrix-iii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0885.%20Spiral%20Matrix%20III)| Medium | O(n^2)| O(1)||
-|[891. Sum of Subsequence Widths](https://leetcode.com/problems/sum-of-subsequence-widths)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0891.%20Sum%20of%20Subsequence%20Widths)| Hard | O(n log n)| O(1)||
-|[942. DI String Match](https://leetcode.com/problems/di-string-match)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0942.%20DI%20String%20Match)| Easy | O(n)| O(1)||
-|[976. Largest Perimeter Triangle](https://leetcode.com/problems/largest-perimeter-triangle/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0976.%20Largest%20Perimeter%20Triangle)| Easy | O(n log n)| O(log n) ||
-|[996. Number of Squareful Arrays](https://leetcode.com/problems/number-of-squareful-arrays)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0996.%20Number%20of%20Squareful%20Arrays)| Hard | O(n log n)| O(n) ||
-|[1025. Divisor Game](https://leetcode.com/problems/divisor-game)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1025.%20Divisor%20Game)| Easy | O(1)| O(1)||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Math/)
+
 
 
 
 ## Hash Table
 
 
-| Title | Solution | Difficulty | Time | Space |收藏| 
-| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
-|[1. Two Sum](https://leetcode.com/problems/two-sum/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0001.%20Two%20Sum)| Easy | O(n)| O(n)||
-|[3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0003.%20Longest%20Substring%20Without%20Repeating%20Characters)| Medium | O(n)| O(1)|❤️|
-|[18. 4Sum](https://leetcode.com/problems/4sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0018.%204Sum)| Medium | O(n^3)| O(n^2)|❤️|
-|[30. Substring with Concatenation of All Words](https://leetcode.com/problems/substring-with-concatenation-of-all-words)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0030.%20Substring%20with%20Concatenation%20of%20All%20Words)| Hard | O(n)| O(n)|❤️|
-|[36. Valid Sudoku](https://leetcode.com/problems/valid-sudoku)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0036.%20Valid%20Sudoku)| Medium | O(n^2)| O(n^2)||
-|[37. Sudoku Solver](https://leetcode.com/problems/sudoku-solver)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0037.%20Sudoku%20Solver)| Hard | O(n^2)| O(n^2)|❤️|
-|[49. Group Anagrams](https://leetcode.com/problems/group-anagrams)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0049.%20Group%20Anagrams)| Medium | O(n log n)| O(n)||
-|[76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0076.%20Minimum%20Window%20Substring)| Hard | O(n)| O(n)|❤️|
-|[94. Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0094.%20Binary%20Tree%20Inorder%20Traversal)| Medium | O(n)| O(1)||
-|[138. Copy List with Random Pointer](https://leetcode.com/problems/copy-list-with-random-pointer)| [Go]()| Medium | O(n)| O(1)||
-|[202. Happy Number](https://leetcode.com/problems/happy-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0202.%20Happy%20Number)| Easy | O(log n)| O(1)||
-|[205. Isomorphic Strings](https://leetcode.com/problems/isomorphic-strings)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0205.%20Isomorphic%20Strings)| Easy | O(log n)| O(n)||
-|[217. Contains Duplicate](https://leetcode.com/problems/contains-duplicate)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0217.%20Contains%20Duplicate)| Easy | O(n)| O(n)||
-|[219. Contains Duplicate II](https://leetcode.com/problems/contains-duplicate-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0219.%20Contains%20Duplicate%20II)| Easy | O(n)| O(n)||
-|[242. Valid Anagram](https://leetcode.com/problems/valid-anagram/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0242.%20Valid%20Anagram)| Easy | O(n)| O(n) ||
-|[274. H-Index](https://leetcode.com/problems/h-index/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0274.%20H-Index)| Medium |  O(n)| O(n) ||
-|[290. Word Pattern](https://leetcode.com/problems/word-pattern)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0290.%20Word%20Pattern)| Easy |  O(n)| O(n) ||
-|[347. Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0347.%20Top%20K%20Frequent%20Elements)| Medium |  O(n)| O(n) ||
-|[349. Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0349.%20Intersection%20of%20Two%20Arrays)| Easy | O(n)| O(n) ||
-|[350. Intersection of Two Arrays II](https://leetcode.com/problems/intersection-of-two-arrays-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0350.%20Intersection%20of%20Two%20Arrays%20II)| Easy | O(n)| O(n) ||
-|[438. Find All Anagrams in a String](https://leetcode.com/problems/find-all-anagrams-in-a-string)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0438.%20Find%20All%20Anagrams%20in%20a%20String)| Easy | O(n)| O(1) ||
-|[447. Number of Boomerangs](https://leetcode.com/problems/number-of-boomerangs)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0447.%20Number%20of%20Boomerangs)| Easy | O(n)| O(1) ||
-|[451. Sort Characters By Frequency](https://leetcode.com/problems/sort-characters-by-frequency)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0451.%20Sort%20Characters%20By%20Frequency)| Medium | O(n log n)| O(1) ||
-|[454. 4Sum II](https://leetcode.com/problems/4sum-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0454.%204Sum%20II)| Medium | O(n^2)| O(n) ||
-|[648. Replace Words](https://leetcode.com/problems/replace-words)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0648.%20Replace%20Words)| Medium | O(n)| O(n) ||
-|[676. Implement Magic Dictionary](https://leetcode.com/problems/implement-magic-dictionary)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0676.%20Implement%20Magic%20Dictionary)| Medium | O(n)| O(n) ||
-|[720. Longest Word in Dictionary](https://leetcode.com/problems/longest-word-in-dictionary)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0720.%20Longest%20Word%20in%20Dictionary)| Easy | O(n)| O(n) ||
-|[726. Number of Atoms](https://leetcode.com/problems/number-of-atoms)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0726.%20Number%20of%20Atoms)| Hard | O(n)| O(n) |❤️|
-|[739. Daily Temperatures](https://leetcode.com/problems/daily-temperatures)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0739.%20Daily%20Temperatures)| Medium | O(n)| O(n) ||
-|[710. Random Pick with Blacklist](https://leetcode.com/problems/random-pick-with-blacklist/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0710.%20Random%20Pick%20with%20Blacklist)| Hard | O(n)| O(n)  ||
-|[895. Maximum Frequency Stack](https://leetcode.com/problems/maximum-frequency-stack)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0895.%20Maximum%20Frequency%20Stack)| Hard | O(n)| O(n)  ||
-|[930. Binary Subarrays With Sum](https://leetcode.com/problems/binary-subarrays-with-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0930.%20Binary%20Subarrays%20With%20Sum)| Medium | O(n)| O(n)  |❤️|
-|[992. Subarrays with K Different Integers](https://leetcode.com/problems/subarrays-with-k-different-integers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0992.%20Subarrays%20with%20K%20Different%20Integers)| Hard | O(n)| O(n)  |❤️|
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Hash_Table/)
 
 
 
 ## Sort
 
-![](./Topics/Sort.png)
+![](./topic/Sort.png)
 
 - 深刻的理解多路快排。第 75 题。
 - 链表的排序，插入排序(第 147 题)和归并排序(第 148 题)
@@ -2076,37 +2957,12 @@ func peakIndexInMountainArray(A []int) int {
 - 两两不相邻的排序。第 767 题，第 1054 题。
 - "饼子排序"。第 969 题。
 
-| Title | Solution | Difficulty | Time | Space | 收藏 |
-| ----- | :--------: | :----------: | :----: | :-----: |:-----: |
-|[56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0056.%20Merge%20Intervals)| Medium | O(n log n)| O(log n)||
-|[57. Insert Interval](https://leetcode.com/problems/insert-interval/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0057.%20Insert%20Interval)| Hard | O(n)| O(1)||
-|[75. Sort Colors](https://leetcode.com/problems/sort-colors/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0075.%20Sort%20Colors)| Medium| O(n)| O(1)|❤️|
-|[147. Insertion Sort List](https://leetcode.com/problems/insertion-sort-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0147.%20Insertion%20Sort%20List)| Medium | O(n)| O(1) |❤️|
-|[148. Sort List](https://leetcode.com/problems/sort-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0148.%20Sort%20List)| Medium |O(n log n)| O(log n)|❤️|
-|[164. Maximum Gap](https://leetcode.com/problems/maximum-gap/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0164.%20Maximum%20Gap)| Hard | O(n log n)| O(log n) |❤️|
-|[179. Largest Number](https://leetcode.com/problems/largest-number/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0179.%20Largest%20Number)| Medium | O(n log n)| O(log n) |❤️|
-|[220. Contains Duplicate III](https://leetcode.com/problems/contains-duplicate-iii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0220.%20Contains%20Duplicate%20III)| Medium | O(n log n)| O(1) |❤️|
-|[242. Valid Anagram](https://leetcode.com/problems/valid-anagram/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0242.%20Valid%20Anagram)| Easy | O(n)| O(n) ||
-|[274. H-Index](https://leetcode.com/problems/h-index/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0274.%20H-Index)| Medium |  O(n)| O(n) ||
-|[324. Wiggle Sort II](https://leetcode.com/problems/wiggle-sort-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0324.%20Wiggle%20Sort%20II)| Medium| O(n)| O(n)|❤️|
-|[349. Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0349.%20Intersection%20of%20Two%20Arrays)| Easy | O(n)| O(n) ||
-|[350. Intersection of Two Arrays II](https://leetcode.com/problems/intersection-of-two-arrays-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0350.%20Intersection%20of%20Two%20Arrays%20II)| Easy | O(n)| O(n) ||
-|[524. Longest Word in Dictionary through Deleting](https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0524.%20Longest%20Word%20in%20Dictionary%20through%20Deleting)| Medium | O(n)| O(1) ||
-|[767. Reorganize String](https://leetcode.com/problems/reorganize-string/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0767.%20Reorganize%20String)| Medium | O(n log n)| O(log n)  |❤️|
-|[853. Car Fleet](https://leetcode.com/problems/car-fleet/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0853.%20Car%20Fleet)| Medium | O(n log n)| O(log n)  ||
-|[710. Random Pick with Blacklist](https://leetcode.com/problems/random-pick-with-blacklist/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0710.%20Random%20Pick%20with%20Blacklist)| Hard | O(n)| O(n)  ||
-|[922. Sort Array By Parity II](https://leetcode.com/problems/sort-array-by-parity-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0922.%20Sort%20Array%20By%20Parity%20II)| Easy | O(n)| O(1) ||
-|[969. Pancake Sorting](https://leetcode.com/problems/pancake-sorting/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0969.%20Pancake%20Sorting)| Medium | O(n log n)| O(log n) |❤️|
-|[973. K Closest Points to Origin](https://leetcode.com/problems/k-closest-points-to-origin/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0973.%20K%20Closest%20Points%20to%20Origin)| Medium | O(n log n)| O(log n) ||
-|[976. Largest Perimeter Triangle](https://leetcode.com/problems/largest-perimeter-triangle/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0976.%20Largest%20Perimeter%20Triangle)| Easy | O(n log n)| O(log n) ||
-|[1030. Matrix Cells in Distance Order](https://leetcode.com/problems/matrix-cells-in-distance-order/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1030.%20Matrix%20Cells%20in%20Distance%20Order)| Easy | O(n^2)| O(1) ||
-|[1054. Distant Barcodes](https://leetcode.com/problems/distant-barcodes/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1054.%20Distant%20Barcodes)| Medium | O(n log n)| O(log n) |❤️|
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Sort/)
 
 
 ## Bit Manipulation
 
-![](./Topics/Bit_Manipulation.png)
+![](./topic/Bit_Manipulation.png)
 
 - 异或的特性。第 136 题，第 268 题，第 389 题，第 421 题，
 
@@ -2140,45 +2996,12 @@ X & -X 得到最低位(LSB)的 1
 X & ~X = 0
 ```
 
-
-| Title | Solution | Difficulty | Time | Space | 收藏 |
-| ----- | :--------: | :----------: | :----: | :-----: |:-----: |
-|[78. Subsets](https://leetcode.com/problems/subsets)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0078.%20Subsets)| Medium | O(n^2)| O(n)|❤️|
-|[136. Single Number](https://leetcode.com/problems/single-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0136.%20Single%20Number)| Easy | O(n)| O(1)||
-|[137. Single Number II](https://leetcode.com/problems/single-number-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0137.%20Single%20Number%20II)| Medium | O(n)| O(1)|❤️|
-|[169. Majority Element](https://leetcode.com/problems/majority-element)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0169.%20Majority%20Element)| Easy | O(n)| O(1)|❤️|
-|[187. Repeated DNA Sequences](https://leetcode.com/problems/repeated-dna-sequences)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0187.%20Repeated%20DNA%20Sequences)| Medium | O(n)| O(1)||
-|[190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0190.%20Reverse%20Bits)| Easy | O(n)| O(1)|❤️|
-|[191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0191.%20Number%20of%201%20Bits)| Easy | O(n)| O(1)||
-|[201. Bitwise AND of Numbers Range](https://leetcode.com/problems/bitwise-and-of-numbers-range)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0201.%20Bitwise%20AND%20of%20Numbers%20Range)| Medium | O(n)| O(1)|❤️|
-|[231. Power of Two](https://leetcode.com/problems/power-of-twor)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0231.%20Power%20of%20Two)| Easy | O(1)| O(1)||
-|[260. Single Number III](https://leetcode.com/problems/single-number-iii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0260.%20Single%20Number%20III)| Medium | O(n)| O(1)|❤️|
-|[268. Missing Number](https://leetcode.com/problems/missing-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0268.%20Missing%20Number)| Easy | O(n)| O(1)||
-|[318. Maximum Product of Word Lengths](https://leetcode.com/problems/maximum-product-of-word-lengths)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0318.%20Maximum%20Product%20of%20Word%20Lengths)| Medium | O(n)| O(1)||
-|[338. Counting Bits](https://leetcode.com/problems/counting-bits)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0338.%20Counting%20Bits)| Medium | O(n)| O(n)||
-|[342. Power of Four](https://leetcode.com/problems/power-of-four)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0342.%20Power%20of%20Four)| Easy | O(n)| O(1)||
-|[371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0371.%20Sum%20of%20Two%20Integers)| Easy | O(n)| O(1)||
-|[389. Find the Difference](https://leetcode.com/problems/find-the-difference)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0389.%20Find%20the%20Difference)| Easy | O(n)| O(1)||
-|[393. UTF-8 Validation](https://leetcode.com/problems/utf-8-validation)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0393.%20UTF-8%20Validation)| Medium | O(n)| O(1)||
-|[397. Integer Replacement](https://leetcode.com/problems/integer-replacement)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0397.%20Integer%20Replacement)| Medium | O(n)| O(1)||
-|[401. Binary Watch](https://leetcode.com/problems/binary-watch)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0401.%20Binary%20Watch)| Easy | O(1)| O(1)||
-|[405. Convert a Number to Hexadecimal](https://leetcode.com/problems/convert-a-number-to-hexadecimal)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0405.%20Convert%20a%20Number%20to%20Hexadecimal)| Easy | O(n)| O(1)||
-|[421. Maximum XOR of Two Numbers in an Array](https://leetcode.com/problems/maximum-xor-of-two-numbers-in-an-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0421.%20Maximum%20XOR%20of%20Two%20Numbers%20in%20an%20Array)| Medium | O(n)| O(1)|❤️|
-|[461. Hamming Distance](https://leetcode.com/problems/hamming-distance)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0461.%20Hamming%20Distance)| Easy | O(n)| O(1)||
-|[476. Number Complement](https://leetcode.com/problems/number-complement)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0476.%20Number%20Complement)| Easy | O(n)| O(1)||
-|[477. Total Hamming Distance](https://leetcode.com/problems/total-hamming-distance)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0477.%20Total%20Hamming%20Distance)| Medium | O(n)| O(1)||
-|[693. Binary Number with Alternating Bits](https://leetcode.com/problems/binary-number-with-alternating-bits)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0693.%20Binary%20Number%20with%20Alternating%20Bits)| Easy | O(n)| O(1)|❤️|
-|[756. Pyramid Transition Matrix](https://leetcode.com/problems/pyramid-transition-matrix)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0756.%20Pyramid%20Transition%20Matrix)| Medium | O(n log n)| O(n)||
-|[762. Prime Number of Set Bits in Binary Representation](https://leetcode.com/problems/prime-number-of-set-bits-in-binary-representation)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0762.%20Prime%20Number%20of%20Set%20Bits%20in%20Binary%20Representation)| Easy | O(n)| O(1)||
-|[784. Letter Case Permutation](https://leetcode.com/problems/letter-case-permutation)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0784.%20Letter%20Case%20Permutation)| Easy | O(n)| O(1)||
-|[898. Bitwise ORs of Subarrays](https://leetcode.com/problems/letter-case-permutation)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0898.%20Bitwise%20ORs%20of%20Subarrays)| Medium | O(n)| O(1)||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
-
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Bit_Manipulation/)
 
 
 ## Union Find
 
-![](./Topics/Union_Find.png)
+![](./topic/Union_Find.png)
 
 - 灵活使用并查集的思想，熟练掌握并查集的[模板](https://github.com/halfrost/LeetCode-Go/blob/master/template/UnionFind.go)，模板中有两种并查集的实现方式，一种是路径压缩 + 秩优化的版本，另外一种是计算每个集合中元素的个数 + 最大集合元素个数的版本，这两种版本都有各自使用的地方。能使用第一类并查集模板的题目有：第 128 题，第 130 题，第 547 题，第 684 题，第 721 题，第 765 题，第 778 题，第 839 题，第 924 题，第 928 题，第 947 题，第 952 题，第 959 题，第 990 题。能使用第二类并查集模板的题目有：第 803 题，第 952 题。第 803 题秩优化和统计集合个数这些地方会卡时间，如果不优化，会 TLE。
 - 并查集是一种思想，有些题需要灵活使用这种思想，而不是死套模板，如第 399 题，这一题是 stringUnionFind，利用并查集思想实现的。这里每个节点是基于字符串和 map 的，而不是单纯的用 int 节点编号实现的。
@@ -2188,35 +3011,13 @@ X & ~X = 0
 - 能用并查集的题目，一般也可以用 DFS 和 BFS 解答，只不过时间复杂度会高一点。
 
 
-| Title | Solution | Difficulty | Time | Space | 收藏 |
-| ----- | :--------: | :----------: | :----: | :-----: |:-----: |
-|[128. Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0128.%20Longest%20Consecutive%20Sequence)| Hard | O(n)| O(n)|❤️|
-|[130. Surrounded Regions](https://leetcode.com/problems/surrounded-regions)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0130.%20Surrounded%20Regions)| Medium | O(m\*n)| O(m\*n)||
-|[200. Number of Islands](https://leetcode.com/problems/number-of-islands)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0200.%20Number%20of%20Islands)| Medium | O(m\*n)| O(m\*n)||
-|[399. Evaluate Division](https://leetcode.com/problems/evaluate-division)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0399.%20Evaluate%20Division)| Medium | O(n)| O(n)||
-|[547. Friend Circles](https://leetcode.com/problems/friend-circles)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0547.%20Friend%20Circles)| Medium | O(n^2)| O(n)||
-|[684. Redundant Connection](https://leetcode.com/problems/redundant-connections)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0684.%20Redundant%20Connection)| Medium | O(n)| O(n)||
-|[685. Redundant Connection II](https://leetcode.com/problems/redundant-connection-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0685.%20Redundant%20Connection%20II)| Hard | O(n)| O(n)||
-|[721. Accounts Merge](https://leetcode.com/problems/accounts-merge)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0721.%20Accounts%20Merge)| Medium | O(n)| O(n)|❤️|
-|[765. Couples Holding Hands](https://leetcode.com/problems/couples-holding-hands)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0765.%20Couples%20Holding%20Hands)| Hard | O(n)| O(n)|❤️|
-|[778. Swim in Rising Water](https://leetcode.com/problems/swim-in-rising-water)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0778.%20Swim%20in%20Rising%20Water)| Hard | O(n^2)| O(n)|❤️|
-|[803. Bricks Falling When Hit](https://leetcode.com/problems/bricks-falling-when-hit)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0803.%20Bricks%20Falling%20When%20Hit)| Hard | O(n^2)| O(n)|❤️|
-|[839. Similar String Groups](https://leetcode.com/problems/similar-string-groups)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0839.%20Similar%20String%20Groups)| Hard | O(n^2)| O(n)||
-|[924. Minimize Malware Spread](https://leetcode.com/problems/minimize-malware-spread)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0924.%20Minimize%20Malware%20Spread)| Hard | O(m\*n)| O(n)||
-|[928. Minimize Malware Spread II](https://leetcode.com/problems/minimize-malware-spread-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0928.%20Minimize%20Malware%20Spread%20II)| Hard | O(m\*n)| O(n)|❤️|
-|[947. Most Stones Removed with Same Row or Column](https://leetcode.com/problems/most-stones-removed-with-same-row-or-column)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0947.%20Most%20Stones%20Removed%20with%20Same%20Row%20or%20Column)| Medium | O(n)| O(n)||
-|[952. Largest Component Size by Common Factor](https://leetcode.com/problems/largest-component-size-by-common-factor)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0952.%20Largest%20Component%20Size%20by%20Common%20Factor)| Hard | O(n)| O(n)|❤️|
-|[959. Regions Cut By Slashes](https://leetcode.com/problems/regions-cut-by-slashes)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0959.%20Regions%20Cut%20By%20Slashes)| Medium | O(n^2)| O(n^2)|❤️|
-|[990. Satisfiability of Equality Equations](https://leetcode.com/problems/satisfiability-of-equality-equations)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0990.%20Satisfiability%20of%20Equality%20Equations)| Medium | O(n)| O(n)||
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
-
-
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Union_Find/)
 
 
 
 ## Sliding Window
 
-![](./Topics/Sliding_Window.png)
+![](./topic/Sliding_Window.png)
 
 - 双指针滑动窗口的经典写法。右指针不断往右移，移动到不能往右移动为止(具体条件根据题目而定)。当右指针到最右边以后，开始挪动左指针，释放窗口左边界。第 3 题，第 76 题，第 209 题，第 424 题，第 438 题，第 567 题，第 713 题，第 763 题，第 845 题，第 881 题，第 904 题，第 978 题，第 992 题，第 1004 题，第 1040 题，第 1052 题。
 
@@ -2236,29 +3037,14 @@ X & ~X = 0
 ```
 - 滑动窗口经典题。第 239 题，第 480 题。
 
-| Title | Solution | Difficulty | Time | Space | 收藏 |
-| ----- | :--------: | :----------: | :----: | :-----: |:-----: |
-|[3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0003.%20Longest%20Substring%20Without%20Repeating%20Characters)| Medium | O(n)| O(1)|❤️|
-|[76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0076.%20Minimum%20Window%20Substring)| Hard | O(n)| O(n)|❤️|
-|[239. Sliding Window Maximum](https://leetcode.com/problems/sliding-window-maximum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0239.%20Sliding%20Window%20Maximum)| Hard | O(n * k)| O(n)|❤️|
-|[424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0424.%20Longest%20Repeating%20Character%20Replacement)| Medium | O(n)| O(1) ||
-|[480. Sliding Window Median](https://leetcode.com/problems/sliding-window-median)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0480.%20Sliding%20Window%20Median)| Hard | O(n * log k)| O(k)|❤️|
-|[567. Permutation in String](https://leetcode.com/problems/permutation-in-string)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0567.%20Permutation%20in%20String)| Medium | O(n)| O(1)|❤️|
-|[978. Longest Turbulent Subarray](https://leetcode.com/problems/longest-turbulent-subarray)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0978.%20Longest%20Turbulent%20Subarray)| Medium | O(n)| O(1)|❤️|
-|[992. Subarrays with K Different Integers](https://leetcode.com/problems/subarrays-with-k-different-integers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0992.%20Subarrays%20with%20K%20Different%20Integers)| Hard | O(n)| O(n)|❤️|
-|[995. Minimum Number of K Consecutive Bit Flips](https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0995.%20Minimum%20Number%20of%20K%20Consecutive%20Bit%20Flips)| Hard | O(n)| O(1)|❤️|
-|[1004. Max Consecutive Ones III](https://leetcode.com/problems/max-consecutive-ones-iii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1004.%20Max%20Consecutive%20Ones%20III)| Medium | O(n)| O(1) ||
-|[1040. Moving Stones Until Consecutive II](https://leetcode.com/problems/moving-stones-until-consecutive-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1040.%20Moving%20Stones%20Until%20Consecutive%20II)| Medium | O(n log n)| O(1) |❤️|
-|[1052. Grumpy Bookstore Owner](https://leetcode.com/problems/grumpy-bookstore-owner)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1052.%20Grumpy%20Bookstore%20Owner)| Medium | O(n log n)| O(1) ||
-|[1074. Number of Submatrices That Sum to Target](https://leetcode.com/problems/number-of-submatrices-that-sum-to-target)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1074.%20Number%20of%20Submatrices%20That%20Sum%20to%20Target)| Hard | O(n^3)| O(n) |❤️|
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Sliding_Window/)
 
 
 ## Segment Tree
 
-![](./Topics/Segment_Tree.png)
+![](./topic/Segment_Tree.png)
 
-- 线段数的经典数组实现写法。将合并两个节点 pushUp 逻辑抽象出来了，可以实现任意操作(常见的操作有：加法，取 max，min 等等)。第 218 题，第 303 题，第 307 题，第 699 题。
+- 线段树的经典数组实现写法。将合并两个节点 pushUp 逻辑抽象出来了，可以实现任意操作(常见的操作有：加法，取 max，min 等等)。第 218 题，第 303 题，第 307 题，第 699 题。
 - 计数线段树的经典写法。第 315 题，第 327 题，第 493 题。
 - 线段树的树的实现写法。第 715 题，第 732 题。
 - 区间懒惰更新。第 218 题，第 699 题。
@@ -2284,23 +3070,29 @@ X & ~X = 0
 	[HDU 1542 Atlantis](http://acm.hdu.edu.cn/showproblem.php?pid=1542) update:区间增减 query:直接取根节点的值  
 	[HDU 1828 Picture](http://acm.hdu.edu.cn/showproblem.php?pid=1828) update:区间增减 query:直接取根节点的值
 
-| Title | Solution | Difficulty | Time | Space | 收藏 |
-| ----- | :--------: | :----------: | :----: | :-----: |:-----: |
-|[218. The Skyline Problem](https://leetcode.com/problems/the-skyline-problem)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0218.%20The%20Skyline%20Problem)| Hard | O(n log n)| O(n)|❤️|
-|[307. Range Sum Query - Mutable](https://leetcode.com/problems/range-sum-query-mutable)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0307.%20Range%20Sum%20Query%20-%20Mutable)| Hard | O(1)| O(n)||
-|[315. Count of Smaller Numbers After Self](https://leetcode.com/problems/count-of-smaller-numbers-after-self)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0315.%20Count%20of%20Smaller%20Numbers%20After%20Self)| Hard | O(n log n)| O(n)||
-|[327. Count of Range Sum](https://leetcode.com/problems/count-of-range-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0327.%20Count%20of%20Range%20Sum)| Hard | O(n log n)| O(n)|❤️|
-|[493. Reverse Pairs](https://leetcode.com/problems/reverse-pairs)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0493.%20Reverse%20Pairs)| Hard | O(n log n)| O(n)||
-|[699. Falling Squares](https://leetcode.com/problems/falling-squares)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0699.%20Falling%20Squares)| Hard | O(n log n)| O(n)|❤️|
-|[715. Range Module](https://leetcode.com/problems/range-module)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0715.%20Range%20Module)| Hard | O(log n)| O(n)|❤️|
-|[732. My Calendar III](https://leetcode.com/problems/my-calendar-iii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0732.%20My%20Calendar%20III)| Hard | O(log n)| O(n)|❤️|
-|[850. Rectangle Area II](https://leetcode.com/problems/rectangle-area-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0850.%20Rectangle%20Area%20II)| Hard | O(n log n)| O(n)|❤️|
-|[1157. Online Majority Element In Subarray](https://leetcode.com/problems/online-majority-element-in-subarray)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1157.%20Online%20Majority%20Element%20In%20Subarray)| Hard | O(log n)| O(n)|❤️|
-|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Segment_Tree/)
 
 
 ## Binary Indexed Tree
 
-![](./Topics/Binary_Indexed_Tree.png)
+![](./topic/Binary_Indexed_Tree.png)
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Binary_Indexed_Tree/)
+
 
 ----------------------------------------------------------------------------------------
+
+<p align='center'>
+<a href="https://github.com/halfrost/LeetCode-Go/releases/tag/Special"><img src="https://img.halfrost.com/Leetcode/ACM-ICPC_Algorithm_Template.png"></a>
+</p>
+
+Thank you for reading here. This is bonus. You can download my [《ACM-ICPC Algorithm Template》](https://github.com/halfrost/LeetCode-Go/releases/tag/Special/)
+
+
+
+## ♥️ Thanks
+
+Thanks for your Star！
+
+[![Stargazers over time](https://starchart.cc/halfrost/LeetCode-Go.svg)](https://starchart.cc/halfrost/LeetCode-Go)
+
diff --git a/README_old.md b/README_old.md
new file mode 100644
index 000000000..bc1892c44
--- /dev/null
+++ b/README_old.md
@@ -0,0 +1,1173 @@
+
+# LeetCode by Go
+[LeetCode Online Judge] (https://leetcode.com/) is a website containing many **algorithm questions**. Most of them are real interview questions of **Google, Facebook, LinkedIn, Apple**, etc. This repo shows my solutions by Go with the code style strictly follows the [Google Golang Style Guide](https://github.com/golang/go/wiki/CodeReviewComments). Please feel free to reference and **STAR** to support this repo, thank you!
+
+
+<p align='center'>
+<img src='./logo.png'>
+</p>
+
+
+
+
+## 数据结构
+
+| 数据结构 | 变种 | 相关题目 |
+|:-------:|:-------|:------|
+|顺序线性表：向量|||
+|单链表|1.双链表<br>2.静态链表<br>3.对称矩阵<br>4.稀疏矩阵||
+|栈|广义栈||
+|队列|1.链表实现<br>2.循环数组实现<br>3.双端队列||
+|字符串|1.KMP算法<br>2.有限状态自动机<br>3.模式匹配有限状态自动机<br>4.BM模式匹配算法<br>5.BM-KMP算法||
+|树|1.二叉树<br>2.并查集<br>3.Huffman数||
+|数组实现的堆|1.极大堆和极小堆<br>2.极大极小堆<br>3.双端堆<br>4.d叉堆||
+|树实现的堆|1.左堆<br>2.扁堆<br>3.二项式堆<br>4.斐波那契堆<br>5.配对堆||
+|查找|1.哈希表<br>2.跳跃表<br>3.排序二叉树<br>4.AVL树<br>5.B树<br>6.AA树<br>7.红黑树<br>8.排序二叉堆<br>9.Splay树<br>10.双链树<br>11.Trie树||
+
+
+## Data Structures
+
+> 标识了 (已全部做完) 的专题是全部完成了的，没有标识的是还没有做完的
+
+* [Array](#array)
+* [String](#string)
+* [Linked List](#linked-list)
+* [Stack](#stack)
+* [Tree](#tree)
+* [Dynamic programming](#dynamic-programming)
+* [Depth-first search](#depth-first-search)
+* [Math](#math)
+* [Search](#search)
+* [Sort](#sort)
+* [Union Find](#union-find)
+
+## Companies
+* [Google](#google)
+* [Facebook](#facebook)
+* [Snapchat](#snapchat)
+* [Uber](#uber)
+* [Airbnb](#airbnb)
+* [LinkedIn](#linkedin)
+* [Amazon](#amazon)
+* [Microsoft](#microsoft)
+
+
+## 算法
+
+
+## 一. 目录
+
+|题号|题目|通过率|难度|收藏|  
+|:--:|:--|:--: | :--: | :--: |  
+|1|[Two Sum](./Algorithms/1.two-sum)|38%|Easy||  
+|2|[Add Two Numbers](./Algorithms/2.add-two-numbers)|28%|Medium||
+|3|[Longest Substring Without Repeating Characters](./Algorithms/3.longest-substring-without-repeating-characters)|24%|Medium||
+|4|[Median of Two Sorted Arrays](./Algorithms/4.median-of-two-sorted-arrays)|23%|Hard||
+|5|[Longest Palindromic Substring](./Algorithms/5.longest-palindromic-substring)|25%|Medium||
+|6|[ZigZag Conversion](./Algorithms/6.zigzag-conversion)|27%|Medium||
+|7|[Reverse Integer](./Algorithms/7.reverse-integer)|24%|Easy||
+|8|[String to Integer (atoi)](./Algorithms/8.string-to-integer-atoi)|14%|Medium||
+|9|[Palindrome Number](./Algorithms/9.palindrome-number)|35%|Easy||
+|10|[Regular Expression Matching](./Algorithms/10.regular-expression-matching)|24%|Hard|❤️|
+|11|[Container With Most Water](./Algorithms/11.container-with-most-water)|37%|Medium||
+|12|[Integer to Roman](./Algorithms/12.integer-to-roman)|46%|Medium||
+|13|[Roman to Integer](./Algorithms/13.roman-to-integer)|48%|Easy||
+|14|[Longest Common Prefix](./Algorithms/14.longest-common-prefix)|31%|Easy||
+|15|[3Sum](./Algorithms/15.3sum)|21%|Medium||
+|16|[3Sum Closest](./Algorithms/16.3sum-closest)|31%|Medium||
+|17|[Letter Combinations of a Phone Number](./Algorithms/17.letter-combinations-of-a-phone-number)|36%|Medium||
+|18|[4Sum](./Algorithms/18.4sum)|27%|Medium||
+|19|[Remove Nth Node From End of List](./Algorithms/19.remove-nth-node-from-end-of-list)|33%|Medium||
+|20|[Valid Parentheses](./Algorithms/20.Valid-Parentheses)|33%|Easy||
+|21|[Merge Two Sorted Lists](./Algorithms/21.merge-two-sorted-lists)|41%|Easy||
+|22|[Generate Parentheses](./Algorithms/22.generate-parentheses)|48%|Medium|❤️|
+|23|[Merge k Sorted Lists](./Algorithms/23.merge-k-sorted-lists)|28%|Hard||
+|24|[Swap Nodes in Pairs](./Algorithms/24.Swap-Nodes-In-Pairs)|39%|Medium|❤️|
+|25|[Reverse Nodes in k-Group](./Algorithms/25.Reverse-Nodes-In-k-Group)|31%|Hard|❤️|
+|26|[Remove Duplicates from Sorted Array](./Algorithms/26.remove-duplicates-from-sorted-array)|36%|Easy|❤️|
+|27|[Remove Element](./Algorithms/27.remove-element)|40%|Easy||
+|28|[Implement strStr()](./Algorithms/28.implement-strstr)|28%|Easy||
+|29|[Divide Two Integers](./Algorithms/29.divide-two-integers)|15%|Medium||
+|30|[Substring with Concatenation of All Words](./Algorithms/30.substring-with-concatenation-of-all-words)|22%|Hard|❤️|
+|31|[Next Permutation](./Algorithms/31.next-permutation)|29%|Medium|❤️|
+|32|[Longest Valid Parentheses](./Algorithms/32.longest-valid-parentheses)|23%|Hard||
+|33|[Search in Rotated Sorted Array](./Algorithms/33.search-in-rotated-sorted-array)|31%|Medium|❤️|
+|34|[Search for a Range](./Algorithms/34.search-for-a-range)|31%|Medium||
+|35|[Search Insert Position](./Algorithms/35.search-insert-position)|40%|Easy||
+|36|[Valid Sudoku](./Algorithms/36.valid-sudoku)|37%|Medium|❤️|
+|37|[Sudoku Solver](./Algorithms/37.sudoku-solver)|32%|Hard|❤️|
+|38|[Count and Say](./Algorithms/38.count-and-say)|36%|Easy||
+|39|[Combination Sum](./Algorithms/39.combination-sum)|41%|Medium||
+|40|[Combination Sum II](./Algorithms/40.combination-sum-ii)|35%|Medium||
+|41|[First Missing Positive](./Algorithms/41.First-Missing-Positive)|25%|Hard||
+|42|[Trapping Rain Water](./Algorithms/42.trapping-rain-water)|37%|Hard||
+|43|[Multiply Strings](./Algorithms/43.multiply-strings)|28%|Medium||
+|44|[Wildcard Matching](./Algorithms/44.wildcard-matching)|21%|Hard|❤️|
+|45|[Jump Game II](./Algorithms/45.jump-game-ii)|26%|Hard||
+|46|[Permutations](./Algorithms/46.permutations)|47%|Medium||
+|47|[Permutations II](./Algorithms/47.permutations-ii)|35%|Medium||
+|48|[Rotate Image](./Algorithms/48.rotate-image)|41%|Medium||
+|49|[Group Anagrams](./Algorithms/49.group-anagrams)|38%|Medium|❤️|
+|50|[Pow(x, n)](./Algorithms/50.powx-n)|26%|Medium||
+|51|[N-Queens](./Algorithms/51.n-queens)|33%|Hard||
+|52|[N-Queens II](./Algorithms/52.n-queens-ii)|46%|Hard||
+|53|[Maximum Subarray](./Algorithms/53.maximum-subarray)|40%|Easy|❤️|
+|54|[Spiral Matrix](./Algorithms/54.spiral-matrix)|27%|Medium|❤️|
+|55|[Jump Game](./Algorithms/55.jump-game)|29%|Medium||
+|56|[Merge Intervals](./Algorithms/56.merge-intervals)|31%|Medium|❤️|
+|57|[Insert Interval](./Algorithms/57.insert-interval)|28%|Hard||
+|58|[Length of Last Word](./Algorithms/58.length-of-last-word)|32%|Easy||
+|59|[Spiral Matrix II](./Algorithms/59.spiral-matrix-ii)|41%|Medium|❤️|
+|60|[Permutation Sequence](./Algorithms/60.permutation-sequence)|29%|Medium||
+|61|[Rotate List](./Algorithms/61.rotate-list)|24%|Medium|❤️|
+|62|[Unique Paths](./Algorithms/62.unique-paths)|42%|Medium|❤️|
+|63|[Unique Paths II](./Algorithms/63.unique-paths-ii)|32%|Medium||
+|64|[Minimum Path Sum](./Algorithms/64.minimum-path-sum)|40%|Medium||
+|65|[Valid Number](./Algorithms/65.valid-number)|12%|Hard||
+|66|[Plus One](./Algorithms/66.plus-one)|39%|Easy||
+|67|[Add Binary](./Algorithms/67.add-binary)|34%|Easy||
+|68|[Text Justification](./Algorithms/68.text-justification)|20%|Hard||
+|69|[Sqrt(x)](./Algorithms/69.sqrtx)|28%|Easy||
+|70|[Climbing Stairs](./Algorithms/70.climbing-stairs)|41%|Easy||
+|71|[Simplify Path](./Algorithms/71.simplify-path)|26%|Medium||
+|72|[Edit Distance](./Algorithms/72.edit-distance)|32%|Hard|❤️|
+|73|[Set Matrix Zeroes](./Algorithms/73.set-matrix-zeroes)|36%|Medium|❤️|
+|74|[Search a 2D Matrix](./Algorithms/74.search-a-2d-matrix)|34%|Medium||
+|75|[Sort Colors](./Algorithms/75.sort-colors)|38%|Medium|❤️|
+|76|[Minimum Window Substring](./Algorithms/76.minimum-window-substring)|26%|Hard|❤️|
+|77|[Combinations](./Algorithms/77.combinations)|41%|Medium||
+|78|[Subsets](./Algorithms/78.subsets)|44%|Medium|❤️|
+|79|[Word Search](./Algorithms/79.word-search)|28%|Medium||
+|80|[Remove Duplicates from Sorted Array II](./Algorithms/80.remove-duplicates-from-sorted-array-ii)|36%|Medium|❤️|
+|81|[Search in Rotated Sorted Array II](./Algorithms/81.search-in-rotated-sorted-array-ii)|32%|Medium||
+|82|[Remove Duplicates from Sorted List II](./Algorithms/82.remove-duplicates-from-sorted-list-ii)|29%|Medium|❤️|
+|83|[Remove Duplicates from Sorted List](./Algorithms/83.remove-duplicates-from-sorted-list)|40%|Easy||
+|84|[Largest Rectangle in Histogram](./Algorithms/84.largest-rectangle-in-histogram)|27%|Hard|❤️|
+|85|[Maximal Rectangle](./Algorithms/85.maximal-rectangle)|29%|Hard|❤️|
+|86|[Partition List](./Algorithms/86.partition-list)|33%|Medium||
+|87|[Scramble String](./Algorithms/87.scramble-string)|29%|Hard|❤️|
+|88|[Merge Sorted Array](./Algorithms/88.Merge-Sorted-Array)|32%|Easy||
+|89|[Gray Code](./Algorithms/89.gray-code)|42%|Medium||
+|90|[Subsets II](./Algorithms/90.subsets-ii)|38%|Medium|❤️|
+|91|[Decode Ways](./Algorithms/91.decode-ways)|20%|Medium|❤️|
+|92|[Reverse Linked List II](./Algorithms/92.reverse-linked-list-ii)|31%|Medium||
+|93|[Restore IP Addresses](./Algorithms/93.restore-ip-addresses)|28%|Medium|❤️|
+|94|[Binary Tree Inorder Traversal](./Algorithms/94.binary-tree-inorder-traversal)|49%|Medium||
+|95|[Unique Binary Search Trees II](./Algorithms/95.unique-binary-search-trees-ii)|32%|Medium|❤️|
+|96|[Unique Binary Search Trees](./Algorithms/96.unique-binary-search-trees)|41%|Medium||
+|97|[Interleaving String](./Algorithms/97.interleaving-string)|25%|Hard|❤️|
+|98|[Validate Binary Search Tree](./Algorithms/98.validate-binary-search-tree)|24%|Medium|❤️|
+|99|[Recover Binary Search Tree](./Algorithms/99.recover-binary-search-tree)|31%|Hard|❤️|
+|100|[Same Tree](./Algorithms/100.same-tree)|47%|Easy||
+|101|[Symmetric Tree](./Algorithms/101.symmetric-tree)|40%|Easy|❤️|
+|102|[Binary Tree Level Order Traversal](./Algorithms/102.binary-tree-level-order-traversal)|42%|Medium||
+|103|[Binary Tree Zigzag Level Order Traversal](./Algorithms/103.binary-tree-zigzag-level-order-traversal)|36%|Medium||
+|104|[Maximum Depth of Binary Tree](./Algorithms/104.maximum-depth-of-binary-tree)|54%|Easy||
+|105|[Construct Binary Tree from Preorder and Inorder Traversal](./Algorithms/105.construct-binary-tree-from-preorder-and-inorder-traversal)|34%|Medium|❤️|
+|106|[Construct Binary Tree from Inorder and Postorder Traversal](./Algorithms/106.construct-binary-tree-from-inorder-and-postorder-traversal)|33%|Medium|❤️|
+|107|[Binary Tree Level Order Traversal II](./Algorithms/107.binary-tree-level-order-traversal-ii)|42%|Easy||
+|108|[Convert Sorted Array to Binary Search Tree](./Algorithms/108.convert-sorted-array-to-binary-search-tree)|44%|Easy||
+|109|[Convert Sorted List to Binary Search Tree](./Algorithms/109.convert-sorted-list-to-binary-search-tree)|35%|Medium||
+|110|[Balanced Binary Tree](./Algorithms/110.balanced-binary-tree)|38%|Easy||
+|111|[Minimum Depth of Binary Tree](./Algorithms/111.minimum-depth-of-binary-tree)|33%|Easy||
+|112|[Path Sum](./Algorithms/112.path-sum)|34%|Easy||
+|113|[Path Sum II](./Algorithms/113.path-sum-ii)|35%|Medium||
+|114|[Flatten Binary Tree to Linked List](./Algorithms/114.flatten-binary-tree-to-linked-list)|36%|Medium|❤️|
+|115|[Distinct Subsequences](./Algorithms/115.distinct-subsequences)|32%|Hard|❤️|
+|118|[Pascal's Triangle](./Algorithms/118.pascals-triangle)|40%|Easy||
+|119|[Pascal's Triangle II](./Algorithms/119.pascals-triangle-ii)|38%|Easy||
+|120|[Triangle](./Algorithms/120.triangle)|34%|Medium|❤️|
+|121|[Best Time to Buy and Sell Stock](./Algorithms/121.best-time-to-buy-and-sell-stock)|42%|Easy||
+|122|[Best Time to Buy and Sell Stock II](./Algorithms/122.best-time-to-buy-and-sell-stock-ii)|47%|Easy||
+|123|[Best Time to Buy and Sell Stock III](./Algorithms/123.best-time-to-buy-and-sell-stock-iii)|30%|Hard||
+|124|[Binary Tree Maximum Path Sum](./Algorithms/124.binary-tree-maximum-path-sum)|27%|Hard|❤️|
+|125|[Valid Palindrome](./Algorithms/125.Valid-Palindrome)|27.0%|Easy||
+|126|[Word Ladder II](./Algorithms/126.word-ladder-ii)|14%|Hard|❤️|
+|127|[Word Ladder](./Algorithms/127.word-ladder)|20%|Medium|❤️|
+|128|[Longest Consecutive Sequence](./Algorithms/128.longest-consecutive-sequence)|38%|Hard||
+|129|[Sum Root to Leaf Numbers](./Algorithms/129.sum-root-to-leaf-numbers)|37%|Medium||
+|130|[Surrounded Regions](./Algorithms/130.surrounded-regions)|19%|Medium|❤️|
+|131|[Palindrome Partitioning](./Algorithms/131.palindrome-partitioning)|35%|Medium|❤️|
+|132|[Palindrome Partitioning II](./Algorithms/132.palindrome-partitioning-ii)|24%|Hard|❤️|
+|134|[Gas Station](./Algorithms/134.gas-station)|29%|Medium|❤️|
+|135|[Candy](./Algorithms/135.candy)|25%|Hard||
+|136|[Single Number](./Algorithms/136.single-number)|55%|Easy||
+|137|[Single Number II](./Algorithms/137.single-number-ii)|42%|Medium|❤️|
+|139|[Word Break](./Algorithms/139.word-break)|31%|Medium|❤️|
+|140|[Word Break II](./Algorithms/140.word-break-ii)|24%|Hard|❤️|
+|141|[Linked List Cycle](./Algorithms/140.Linked-List-Cycle)|24%|Hard|❤️|
+|142|[Linked List Cycle II](./Algorithms/140.Linked-List-Cycle-II)|24%|Hard|❤️|
+|143|[Reorder List](./Algorithms/143.reorder-list)|26%|Medium|❤️|
+|144|[Binary Tree Preorder Traversal](./Algorithms/144.binary-tree-preorder-traversal)|46%|Medium|❤️|
+|145|[Binary Tree Postorder Traversal](./Algorithms/145.binary-tree-postorder-traversal)|42%|Hard||
+|146|[LRU Cache](./Algorithms/146.lru-cache)|19%|Hard|❤️|
+|147|[Insertion Sort List](./Algorithms/147.insertion-sort-list)|33%|Medium|❤️|
+|148|[Sort List](./Algorithms/148.sort-list)|29%|Medium|❤️|
+|149|[Max Points on a Line](./Algorithms/149.max-points-on-a-line)|15%|Hard|❤️|
+|150|[Evaluate Reverse Polish Notation](./Algorithms/150.evaluate-reverse-polish-notation)|28%|Medium||
+|152|[Maximum Product Subarray](./Algorithms/152.maximum-product-subarray)|26%|Medium|❤️|
+|153|[Find Minimum in Rotated Sorted Array](./Algorithms/153.find-minimum-in-rotated-sorted-array)|40%|Medium||
+|154|[Find Minimum in Rotated Sorted Array II](./Algorithms/154.find-minimum-in-rotated-sorted-array-ii)|37%|Hard||
+|155|[Min Stack](./Algorithms/155.min-stack)|31%|Easy||
+|162|[Find Peak Element](./Algorithms/162.find-peak-element)|38%|Medium||
+|164|[Maximum Gap](./Algorithms/164.maximum-gap)|30%|Hard||
+|165|[Compare Version Numbers](./Algorithms/165.compare-version-numbers)|20%|Medium||
+|166|[Fraction to Recurring Decimal](./Algorithms/166.fraction-to-recurring-decimal)|18%|Medium|❤️|
+|167|[Two Sum II - Input array is sorted](./Algorithms/167.two-sum-ii-input-array-is-sorted)|47%|Easy||
+|168|[Excel Sheet Column Title](./Algorithms/168.excel-sheet-column-title)|27%|Easy|❤️|
+|169|[Majority Element](./Algorithms/169.majority-element)|48%|Easy|❤️|
+|171|[Excel Sheet Column Number](./Algorithms/171.excel-sheet-column-number)|48%|Easy||
+|172|[Factorial Trailing Zeroes](./Algorithms/172.factorial-trailing-zeroes)|36%|Easy||
+|174|[Dungeon Game](./Algorithms/174.dungeon-game)|24%|Hard|❤️|
+|179|[Largest Number](./Algorithms/179.largest-number)|23%|Medium|❤️|
+|187|[Repeated DNA Sequences](./Algorithms/187.repeated-dna-sequences)|33%|Medium||
+|188|[Best Time to Buy and Sell Stock IV](./Algorithms/188.best-time-to-buy-and-sell-stock-iv)|24%|Hard|❤️|
+|189|[Rotate Array](./Algorithms/189.rotate-array)|25%|Easy||
+|198|[House Robber](./Algorithms/198.house-robber)|39%|Easy|❤️|
+|199|[Binary Tree Right Side View](./Algorithms/199.binary-tree-right-side-view)|42%|Medium||
+|200|[Number of Islands](./Algorithms/200.number-of-islands)|36%|Medium||
+|201|[Bitwise AND of Numbers Range](./Algorithms/201.bitwise-and-of-numbers-range)|34%|Medium|❤️|
+|202|[Happy Number](./Algorithms/202.happy-number)|41%|Easy||
+|203|[Remove Linked List Elements](./Algorithms/203.remove-linked-list-elements)|33%|Easy||
+|204|[Count Primes](./Algorithms/204.count-primes)|26%|Easy|❤️|
+|205|[Isomorphic Strings](./Algorithms/205.isomorphic-strings)|34%|Easy|❤️|
+|206|[Reverse Linked List](./Algorithms/206.Reverse-Linked-List)|46%|Easy||
+|207|[Course Schedule](./Algorithms/207.course-schedule)|33%|Medium|❤️|
+|208|[Implement Trie (Prefix Tree)](./Algorithms/208.implement-trie-prefix-tree)|30%|Medium|❤️|
+|209|[Minimum Size Subarray Sum](./Algorithms/209.minimum-size-subarray-sum)|32%|Medium||
+|210|[Course Schedule II](./Algorithms/210.course-schedule-ii)|30%|Medium||
+|211|[Add and Search Word - Data structure design](./Algorithms/211.add-and-search-word-data-structure-design)|25%|Medium|❤️|
+|212|[Word Search II](./Algorithms/212.word-search-ii)|24%|Hard|❤️|
+|213|[House Robber II](./Algorithms/213.house-robber-ii)|34%|Medium||
+|214|[Shortest Palindrome](./Algorithms/214.shortest-palindrome)|25%|Hard|❤️|
+|215|[Kth Largest Element in an Array](./Algorithms/215.kth-largest-element-in-an-array)|40%|Medium|❤️|
+|216|[Combination Sum III](./Algorithms/216.combination-sum-iii)|47%|Medium||
+|217|[Contains Duplicate](./Algorithms/217.contains-duplicate)|47%|Easy||
+|218|[The Skyline Problem](./Algorithms/218.the-skyline-problem)|29%|Hard|❤️|
+|219|[Contains Duplicate II](./Algorithms/219.contains-duplicate-ii)|32%|Easy||
+|220|[Contains Duplicate III](./Algorithms/220.contains-duplicate-iii)|18%|Medium|❤️|
+|221|[Maximal Square](./Algorithms/221.maximal-square)|30%|Medium|❤️|
+|223|[Rectangle Area](./Algorithms/223.rectangle-area)|33%|Medium||
+|224|[Basic Calculator](./Algorithms/224.basic-calculator)|28%|Hard||
+|225|[Implement Stack using Queues](./Algorithms/225.implement-stack-using-queues)|34%|Easy||
+|226|[Invert Binary Tree](./Algorithms/226.invert-binary-tree)|53%|Easy||
+|227|[Basic Calculator II](./Algorithms/227.basic-calculator-ii)|30%|Medium||
+|228|[Summary Ranges](./Algorithms/228.summary-ranges)|32%|Medium||
+|229|[Majority Element II](./Algorithms/229.majority-element-ii)|29%|Medium|❤️|
+|230|[Kth Smallest Element in a BST](./Algorithms/230.kth-smallest-element-in-a-bst)|45%|Medium||
+|231|[Power of Two](./Algorithms/231.power-of-two)|40%|Easy||
+|232|[Implement Queue using Stacks](./Algorithms/232.implement-queue-using-stacks)|38%|Easy||
+|233|[Number of Digit One](./Algorithms/233.number-of-digit-one)|29%|Hard|❤️|
+|234|[Palindrome Linked List](./Algorithms/234.palindrome-linked-list)|33%|Easy||
+|238|[Product of Array Except Self](./Algorithms/238.product-of-array-except-self)|50%|Medium||
+|239|[Sliding Window Maximum](./Algorithms/239.sliding-window-maximum)|34%|Hard|❤️|
+|240|[Search a 2D Matrix II](./Algorithms/240.search-a-2d-matrix-ii)|39%|Medium|❤️|
+|241|[Different Ways to Add Parentheses](./Algorithms/241.different-ways-to-add-parentheses)|46%|Medium||
+|242|[Valid Anagram](./Algorithms/242.valid-anagram)|47%|Easy||
+|257|[Binary Tree Paths](./Algorithms/257.binary-tree-paths)|41%|Easy||
+|258|[Add Digits](./Algorithms/258.add-digits)|51%|Easy||
+|260|[Single Number III](./Algorithms/260.single-number-iii)|53%|Medium|❤️|
+|263|[Ugly Number](./Algorithms/263.ugly-number)|39%|Easy||
+|264|[Ugly Number II](./Algorithms/264.ugly-number-ii)|33%|Medium|❤️|
+|268|[Missing Number](./Algorithms/268.missing-number)|45%|Easy||
+|273|[Integer to English Words](./Algorithms/273.integer-to-english-words)|22%|Hard|❤️|
+|274|[H-Index](./Algorithms/274.h-index)|33%|Medium||
+|275|[H-Index II](./Algorithms/275.h-index-ii)|34%|Medium|❤️|
+|279|[Perfect Squares](./Algorithms/279.perfect-squares)|37%|Medium|❤️|
+|282|[Expression Add Operators](./Algorithms/282.expression-add-operators)|30%|Hard|❤️|
+|283|[Move Zeroes](./Algorithms/283.move-zeroes)|51%|Easy||
+|287|[Find the Duplicate Number](./Algorithms/287.find-the-duplicate-number)|44%|Medium|❤️|
+|289|[Game of Life](./Algorithms/289.game-of-life)|37%|Medium|❤️|
+|290|[Word Pattern](./Algorithms/290.word-pattern)|33%|Easy||
+|292|[Nim Game](./Algorithms/292.nim-game)|55%|Easy|❤️|
+|295|[Find Median from Data Stream](./Algorithms/295.find-median-from-data-stream)|29%|Hard|❤️|
+|299|[Bulls and Cows](./Algorithms/299.bulls-and-cows)|35%|Medium||
+|300|[Longest Increasing Subsequence](./Algorithms/300.longest-increasing-subsequence)|38%|Medium|❤️|
+|301|[Remove Invalid Parentheses](./Algorithms/301.remove-invalid-parentheses)|35%|Hard|❤️|
+|303|[Range Sum Query - Immutable](./Algorithms/303.range-sum-query-immutable)|32%|Easy||
+|304|[Range Sum Query 2D - Immutable](./Algorithms/304.range-sum-query-2d-immutable)|27%|Medium||
+|306|[Additive Number](./Algorithms/306.additive-number)|27%|Medium||
+|307|[Range Sum Query - Mutable](./Algorithms/307.range-sum-query-mutable)|22%|Medium||
+|309|[Best Time to Buy and Sell Stock with Cooldown](./Algorithms/309.best-time-to-buy-and-sell-stock-with-cooldown)|41%|Medium|❤️|
+|310|[Minimum Height Trees](./Algorithms/310.minimum-height-trees)|28%|Medium||
+|312|[Burst Balloons](./Algorithms/312.burst-balloons)|43%|Hard|❤️|
+|313|[Super Ugly Number](./Algorithms/313.super-ugly-number)|38%|Medium|❤️|
+|315|[Count of Smaller Numbers After Self](./Algorithms/315.count-of-smaller-numbers-after-self)|34%|Hard|❤️|
+|316|[Remove Duplicate Letters](./Algorithms/316.remove-duplicate-letters)|30%|Hard|❤️|
+|318|[Maximum Product of Word Lengths](./Algorithms/318.maximum-product-of-word-lengths)|45%|Medium|❤️|
+|319|[Bulb Switcher](./Algorithms/319.bulb-switcher)|42%|Medium|❤️|
+|321|[Create Maximum Number](./Algorithms/321.create-maximum-number)|24%|Hard|❤️|
+|322|[Coin Change](./Algorithms/322.coin-change)|26%|Medium|❤️|
+|324|[Wiggle Sort II](./Algorithms/324.wiggle-sort-ii)|26%|Medium||
+|326|[Power of Three](./Algorithms/326.power-of-three)|40%|Easy|❤️|
+|327|[Count of Range Sum](./Algorithms/327.count-of-range-sum)|30%|Hard|❤️|
+|328|[Odd Even Linked List](./Algorithms/328.odd-even-linked-list)|44%|Medium|❤️|
+|329|[Longest Increasing Path in a Matrix](./Algorithms/329.longest-increasing-path-in-a-matrix)|37%|Hard|❤️|
+|330|[Patching Array](./Algorithms/330.patching-array)|32%|Hard||
+|331|[Verify Preorder Serialization of a Binary Tree](./Algorithms/331.verify-preorder-serialization-of-a-binary-tree)|37%|Medium|❤️|
+|332|[Reconstruct Itinerary](./Algorithms/332.reconstruct-itinerary)|29%|Medium|❤️|
+|334|[Increasing Triplet Subsequence](./Algorithms/334.increasing-triplet-subsequence)|39%|Medium|❤️|
+|335|[Self Crossing](./Algorithms/335.self-crossing)|26%|Hard||
+|336|[Palindrome Pairs](./Algorithms/336.palindrome-pairs)|27%|Hard|❤️|
+|337|[House Robber III](./Algorithms/337.house-robber-iii)|44%|Medium|❤️|
+|338|[Counting Bits](./Algorithms/338.counting-bits)|62%|Medium||
+|342|[Power of Four](./Algorithms/342.power-of-four)|39%|Easy||
+|343|[Integer Break](./Algorithms/343.integer-break)|46%|Medium||
+|344|[Reverse String](./Algorithms/344.reverse-string)|60%|Easy||
+|345|[Reverse Vowels of a String](./Algorithms/345.reverse-vowels-of-a-string)|39%|Easy||
+|347|[Top K Frequent Elements](./Algorithms/347.top-k-frequent-elements)|49%|Medium||
+|349|[Intersection of Two Arrays](./Algorithms/349.intersection-of-two-arrays)|48%|Easy||
+|350|[Intersection of Two Arrays II](./Algorithms/350.intersection-of-two-arrays-ii)|44%|Easy|❤️|
+|352|[Data Stream as Disjoint Intervals](./Algorithms/352.data-stream-as-disjoint-intervals)|41%|Hard||
+|354|[Russian Doll Envelopes](./Algorithms/354.russian-doll-envelopes)|32%|Hard|❤️|
+|355|[Design Twitter](./Algorithms/355.design-twitter)|25%|Medium|❤️|
+|357|[Count Numbers with Unique Digits](./Algorithms/357.count-numbers-with-unique-digits)|46%|Medium|❤️|
+|363|[Max Sum of Rectangle No Larger Than K](./Algorithms/363.max-sum-of-rectangle-no-larger-than-k)|33%|Hard|❤️|
+|365|[Water and Jug Problem](./Algorithms/365.water-and-jug-problem)|28%|Medium|❤️|
+|367|[Valid Perfect Square](./Algorithms/367.valid-perfect-square)|38%|Easy||
+|368|[Largest Divisible Subset](./Algorithms/368.largest-divisible-subset)|33%|Medium|❤️|
+|371|[Sum of Two Integers](./Algorithms/371.sum-of-two-integers)|50%|Easy|❤️|
+|372|[Super Pow](./Algorithms/372.super-pow)|34%|Medium||
+|373|[Find K Pairs with Smallest Sums](./Algorithms/373.find-k-pairs-with-smallest-sums)|31%|Medium|❤️|
+|375|[Guess Number Higher or Lower II](./Algorithms/375.guess-number-higher-or-lower-ii)|36%|Medium||
+|376|[Wiggle Subsequence](./Algorithms/376.wiggle-subsequence)|36%|Medium|❤️|
+|377|[Combination Sum IV](./Algorithms/377.combination-sum-iv)|42%|Medium||
+|378|[Kth Smallest Element in a Sorted Matrix](./Algorithms/378.kth-smallest-element-in-a-sorted-matrix)|45%|Medium|❤️|
+|380|[Insert Delete GetRandom O(1)](./Algorithms/380.insert-delete-getrandom-o1)|39%|Medium|❤️|
+|381|[Insert Delete GetRandom O(1) - Duplicates allowed](./Algorithms/381.insert-delete-getrandom-o1-duplicates-allowed)|29%|Hard|❤️|
+|382|[Linked List Random Node](./Algorithms/382.linked-list-random-node)|47%|Medium||
+|383|[Ransom Note](./Algorithms/383.ransom-note)|47%|Easy||
+|384|[Shuffle an Array](./Algorithms/384.shuffle-an-array)|47%|Medium|❤️|
+|385|[Mini Parser](./Algorithms/385.mini-parser)|30%|Medium|❤️|
+|387|[First Unique Character in a String](./Algorithms/387.first-unique-character-in-a-string)|47%|Easy||
+|388|[Longest Absolute File Path](./Algorithms/388.longest-absolute-file-path)|37%|Medium||
+|389|[Find the Difference](./Algorithms/389.find-the-difference)|51%|Easy||
+|390|[Elimination Game](./Algorithms/390.elimination-game)|42%|Medium|❤️|
+|391|[Perfect Rectangle](./Algorithms/391.perfect-rectangle)|27%|Hard||
+|392|[Is Subsequence](./Algorithms/392.is-subsequence)|44%|Medium|❤️|
+|393|[UTF-8 Validation](./Algorithms/393.utf-8-validation)|34%|Medium|❤️|
+|394|[Decode String](./Algorithms/394.decode-string)|42%|Medium||
+|395|[Longest Substring with At Least K Repeating Characters](./Algorithms/395.longest-substring-with-at-least-k-repeating-characters)|35%|Medium||
+|396|[Rotate Function](./Algorithms/396.rotate-function)|33%|Medium|❤️|
+|397|[Integer Replacement](./Algorithms/397.integer-replacement)|30%|Medium|❤️|
+|398|[Random Pick Index](./Algorithms/398.random-pick-index)|44%|Medium||
+|399|[Evaluate Division](./Algorithms/399.evaluate-division)|42%|Medium|❤️|
+|400|[Nth Digit](./Algorithms/400.nth-digit)|30%|Easy|❤️|
+|401|[Binary Watch](./Algorithms/401.binary-watch)|44%|Easy||
+|402|[Remove K Digits](./Algorithms/402.remove-k-digits)|25%|Medium|❤️|
+|403|[Frog Jump](./Algorithms/403.frog-jump)|32%|Hard|❤️|
+|404|[Sum of Left Leaves](./Algorithms/404.sum-of-left-leaves)|47%|Easy||
+|405|[Convert a Number to Hexadecimal](./Algorithms/405.convert-a-number-to-hexadecimal)|41%|Easy||
+|406|[Queue Reconstruction by Height](./Algorithms/406.queue-reconstruction-by-height)|56%|Medium|❤️|
+|407|[Trapping Rain Water II](./Algorithms/407.trapping-rain-water-ii)|37%|Hard||
+|409|[Longest Palindrome](./Algorithms/409.longest-palindrome)|45%|Easy||
+|410|[Split Array Largest Sum](./Algorithms/410.split-array-largest-sum)|39%|Hard||
+|412|[Fizz Buzz](./Algorithms/412.Fizz-Buzz)|58%|Easy||
+|413|[Arithmetic Slices](./Algorithms/413.arithmetic-slices)|54%|Medium||
+|414|[Third Maximum Number](./Algorithms/414.third-maximum-number)|28%|Easy||
+|415|[Add Strings](./Algorithms/415.add-strings)|41%|Easy||
+|416|[Partition Equal Subset Sum](./Algorithms/416.partition-equal-subset-sum)|38%|Medium|❤️|
+|417|[Pacific Atlantic Water Flow](./Algorithms/417.pacific-atlantic-water-flow)|34%|Medium||
+|419|[Battleships in a Board](./Algorithms/419.battleships-in-a-board)|63%|Medium||
+|420|[Strong Password Checker](./Algorithms/420.strong-password-checker)|19%|Hard|❤️|
+|421|[Maximum XOR of Two Numbers in an Array](./Algorithms/421.maximum-xor-of-two-numbers-in-an-array)|47%|Medium|❤️|
+|423|[Reconstruct Original Digits from English](./Algorithms/423.reconstruct-original-digits-from-english)|44%|Medium||
+|424|[Longest Repeating Character Replacement](./Algorithms/424.longest-repeating-character-replacement)|42%|Medium|❤️|
+|432|[All O`one Data Structure](./Algorithms/432.all-oone-data-structure)|27%|Hard|❤️|
+|434|[Number of Segments in a String](./Algorithms/434.number-of-segments-in-a-string)|36%|Easy||
+|435|[Non-overlapping Intervals](./Algorithms/435.non-overlapping-intervals)|41%|Medium|❤️|
+|436|[Find Right Interval](./Algorithms/436.find-right-interval)|41%|Medium||
+|437|[Path Sum III](./Algorithms/437.path-sum-iii)|40%|Easy|❤️|
+|438|[Find All Anagrams in a String](./Algorithms/438.find-all-anagrams-in-a-string)|33%|Easy||
+|440|[K-th Smallest in Lexicographical Order](./Algorithms/440.k-th-smallest-in-lexicographical-order)|25%|Hard|❤️|
+|441|[Arranging Coins](./Algorithms/441.arranging-coins)|36%|Easy||
+|442|[Find All Duplicates in an Array](./Algorithms/442.find-all-duplicates-in-an-array)|56%|Medium||
+|443|[String Compression](./Algorithms/443.string-compression)|35%|Easy|❤️|
+|445|[Add Two Numbers II](./Algorithms/445.add-two-numbers-ii)|46%|Medium|❤️|
+|446|[Arithmetic Slices II - Subsequence](./Algorithms/446.arithmetic-slices-ii-subsequence)|28%|Hard|❤️|
+|447|[Number of Boomerangs](./Algorithms/447.number-of-boomerangs)|46%|Easy||
+|448|[Find All Numbers Disappeared in an Array](./Algorithms/448.find-all-numbers-disappeared-in-an-array)|51%|Easy||
+|450|[Delete Node in a BST](./Algorithms/450.delete-node-in-a-bst)|37%|Medium|❤️|
+|451|[Sort Characters By Frequency](./Algorithms/451.sort-characters-by-frequency)|51%|Medium||
+|452|[Minimum Number of Arrows to Burst Balloons](./Algorithms/452.minimum-number-of-arrows-to-burst-balloons)|44%|Medium||
+|453|[Minimum Moves to Equal Array Elements](./Algorithms/453.minimum-moves-to-equal-array-elements)|48%|Easy||
+|454|[4Sum II](./Algorithms/454.4sum-ii)|47%|Medium||
+|455|[Assign Cookies](./Algorithms/455.assign-cookies)|47%|Easy||
+|456|[132 Pattern](./Algorithms/456.132-pattern)|27%|Medium|❤️|
+|459|[Repeated Substring Pattern](./Algorithms/459.repeated-substring-pattern)|38%|Easy|❤️|
+|460|[LFU Cache](./Algorithms/460.lfu-cache)|25%|Hard||
+|461|[Hamming Distance](./Algorithms/461.hamming-distance)|69%|Easy||
+|462|[Minimum Moves to Equal Array Elements II](./Algorithms/462.minimum-moves-to-equal-array-elements-ii)|51%|Medium||
+|463|[Island Perimeter](./Algorithms/463.island-perimeter)|57%|Easy||
+|464|[Can I Win](./Algorithms/464.can-i-win)|25%|Medium|❤️|
+|466|[Count The Repetitions](./Algorithms/466.count-the-repetitions)|27%|Hard|❤️|
+|467|[Unique Substrings in Wraparound String](./Algorithms/467.unique-substrings-in-wraparound-string)|33%|Medium|❤️|
+|468|[Validate IP Address](./Algorithms/468.validate-ip-address)|20%|Medium||
+|472|[Concatenated Words](./Algorithms/472.concatenated-words)|30%|Hard||
+|473|[Matchsticks to Square](./Algorithms/473.matchsticks-to-square)|35%|Medium|❤️|
+|474|[Ones and Zeroes](./Algorithms/474.ones-and-zeroes)|38%|Medium||
+|475|[Heaters](./Algorithms/475.heaters)|29%|Easy||
+|476|[Number Complement](./Algorithms/476.number-complement)|61%|Easy||
+|477|[Total Hamming Distance](./Algorithms/477.total-hamming-distance)|47%|Medium|❤️|
+|479|[Largest Palindrome Product](./Algorithms/479.largest-palindrome-product)|25%|Easy||
+|480|[Sliding Window Median](./Algorithms/480.sliding-window-median)|30%|Hard|❤️|
+|481|[Magical String](./Algorithms/481.magical-string)|45%|Medium||
+|482|[License Key Formatting](./Algorithms/482.license-key-formatting)|39%|Easy||
+|483|[Smallest Good Base](./Algorithms/483.smallest-good-base)|33%|Hard|❤️|
+|485|[Max Consecutive Ones](./Algorithms/485.max-consecutive-ones)|53%|Easy||
+|486|[Predict the Winner](./Algorithms/486.predict-the-winner)|45%|Medium||
+|488|[Zuma Game](./Algorithms/488.zuma-game)|37%|Hard|❤️|
+|491|[Increasing Subsequences](./Algorithms/491.increasing-subsequences)|38%|Medium|❤️|
+|492|[Construct the Rectangle](./Algorithms/492.construct-the-rectangle)|48%|Easy|❤️|
+|493|[Reverse Pairs](./Algorithms/493.reverse-pairs)|20%|Hard||
+|494|[Target Sum](./Algorithms/494.target-sum)|43%|Medium|❤️|
+|495|[Teemo Attacking](./Algorithms/495.teemo-attacking)|51%|Medium||
+|496|[Next Greater Element I](./Algorithms/496.next-greater-element-i)|56%|Easy||
+|498|[Diagonal Traverse](./Algorithms/498.diagonal-traverse)|46%|Medium||
+|500|[Keyboard Row](./Algorithms/500.keyboard-row)|59%|Easy||
+|501|[Find Mode in Binary Search Tree](./Algorithms/501.find-mode-in-binary-search-tree)|37%|Easy|❤️|
+|502|[IPO](./Algorithms/502.ipo)|36%|Hard||
+|503|[Next Greater Element II](./Algorithms/503.next-greater-element-ii)|48%|Medium|❤️|
+|504|[Base 7](./Algorithms/504.base-7)|43%|Easy||
+|506|[Relative Ranks](./Algorithms/506.relative-ranks)|46%|Easy||
+|507|[Perfect Number](./Algorithms/507.perfect-number)|32%|Easy||
+|508|[Most Frequent Subtree Sum](./Algorithms/508.most-frequent-subtree-sum)|52%|Medium||
+|513|[Find Bottom Left Tree Value](./Algorithms/513.find-bottom-left-tree-value)|56%|Medium||
+|514|[Freedom Trail](./Algorithms/514.freedom-trail)|39%|Hard||
+|515|[Find Largest Value in Each Tree Row](./Algorithms/515.find-largest-value-in-each-tree-row)|55%|Medium||
+|516|[Longest Palindromic Subsequence](./Algorithms/516.longest-palindromic-subsequence)|42%|Medium|❤️|
+|517|[Super Washing Machines](./Algorithms/517.super-washing-machines)|36%|Hard|❤️|
+|520|[Detect Capital](./Algorithms/520.detect-capital)|51%|Easy||
+|521|[Longest Uncommon Subsequence I](./Algorithms/521.longest-uncommon-subsequence-i)|55%|Easy||
+|522|[Longest Uncommon Subsequence II](./Algorithms/522.longest-uncommon-subsequence-ii)|31%|Medium||
+|523|[Continuous Subarray Sum](./Algorithms/523.continuous-subarray-sum)|23%|Medium|❤️|
+|524|[Longest Word in Dictionary through Deleting](./Algorithms/524.longest-word-in-dictionary-through-deleting)|43%|Medium||
+|525|[Contiguous Array](./Algorithms/525.contiguous-array)|41%|Medium|❤️|
+|526|[Beautiful Arrangement](./Algorithms/526.beautiful-arrangement)|53%|Medium|❤️|
+|529|[Minesweeper](./Algorithms/529.minesweeper)|49%|Medium||
+|530|[Minimum Absolute Difference in BST](./Algorithms/530.minimum-absolute-difference-in-bst)|47%|Easy|❤️|
+|532|[K-diff Pairs in an Array](./Algorithms/532.k-diff-pairs-in-an-array)|28%|Easy||
+|537|[Complex Number Multiplication](./Algorithms/537.complex-number-multiplication)|63%|Medium||
+|538|[Convert BST to Greater Tree](./Algorithms/538.convert-bst-to-greater-tree)|48%|Easy||
+|539|[Minimum Time Difference](./Algorithms/539.minimum-time-difference)|46%|Medium||
+|540|[Single Element in a Sorted Array](./Algorithms/540.single-element-in-a-sorted-array)|55%|Medium||
+|541|[Reverse String II](./Algorithms/541.reverse-string-ii)|43%|Easy||
+|542|[01 Matrix](./Algorithms/542.01-matrix)|33%|Medium|❤️|
+|543|[Diameter of Binary Tree](./Algorithms/543.diameter-of-binary-tree)|44%|Easy||
+|546|[Remove Boxes](./Algorithms/546.remove-boxes)|35%|Hard|❤️|
+|547|[Friend Circles](./Algorithms/547.friend-circles)|49%|Medium|❤️|
+|551|[Student Attendance Record I](./Algorithms/551.student-attendance-record-i)|44%|Easy||
+|552|[Student Attendance Record II](./Algorithms/552.student-attendance-record-ii)|31%|Hard|❤️|
+|553|[Optimal Division](./Algorithms/553.optimal-division)|55%|Medium||
+|554|[Brick Wall](./Algorithms/554.brick-wall)|46%|Medium||
+|556|[Next Greater Element III](./Algorithms/556.next-greater-element-iii)|28%|Medium||
+|557|[Reverse Words in a String III](./Algorithms/557.reverse-words-in-a-string-iii)|60%|Easy||
+|560|[Subarray Sum Equals K](./Algorithms/560.subarray-sum-equals-k)|39%|Medium|❤️|
+|561|[Array Partition I](./Algorithms/561.array-partition-i)|66%|Easy||
+|563|[Binary Tree Tilt](./Algorithms/563.binary-tree-tilt)|47%|Easy||
+|564|[Find the Closest Palindrome](./Algorithms/564.find-the-closest-palindrome)|17%|Hard||
+|565|[Array Nesting](./Algorithms/565.array-nesting)|49%|Medium||
+|566|[Reshape the Matrix](./Algorithms/566.reshape-the-matrix)|57%|Easy||
+|567|[Permutation in String](./Algorithms/567.permutation-in-string)|36%|Medium||
+|572|[Subtree of Another Tree](./Algorithms/572.subtree-of-another-tree)|40%|Easy|❤️|
+|575|[Distribute Candies](./Algorithms/575.distribute-candies)|57%|Easy||
+|576|[Out of Boundary Paths](./Algorithms/576.out-of-boundary-paths)|30%|Medium|❤️|
+|581|[Shortest Unsorted Continuous Subarray](./Algorithms/581.shortest-unsorted-continuous-subarray)|29%|Easy|❤️|
+|583|[Delete Operation for Two Strings](./Algorithms/583.delete-operation-for-two-strings)|44%|Medium|❤️|
+|587|[Erect the Fence](./Algorithms/587.erect-the-fence)|33%|Hard|❤️|
+|591|[Tag Validator](./Algorithms/591.tag-validator)|31%|Hard|❤️|
+|592|[Fraction Addition and Subtraction](./Algorithms/592.fraction-addition-and-subtraction)|46%|Medium||
+|593|[Valid Square](./Algorithms/593.valid-square)|39%|Medium||
+|594|[Longest Harmonious Subsequence](./Algorithms/594.longest-harmonious-subsequence)|40%|Easy||
+|598|[Range Addition II](./Algorithms/598.range-addition-ii)|48%|Easy||
+|599|[Minimum Index Sum of Two Lists](./Algorithms/599.minimum-index-sum-of-two-lists)|46%|Easy||
+|600|[Non-negative Integers without Consecutive Ones](./Algorithms/600.non-negative-integers-without-consecutive-ones)|31%|Hard|❤️|
+|605|[Can Place Flowers](./Algorithms/605.can-place-flowers)|30%|Easy|❤️|
+|606|[Construct String from Binary Tree](./Algorithms/606.construct-string-from-binary-tree)|49%|Easy||
+|609|[Find Duplicate File in System](./Algorithms/609.find-duplicate-file-in-system)|52%|Medium||
+|611|[Valid Triangle Number](./Algorithms/611.valid-triangle-number)|42%|Medium|❤️|
+|617|[Merge Two Binary Trees](./Algorithms/617.merge-two-binary-trees)|67%|Easy||
+|621|[Task Scheduler](./Algorithms/621.task-scheduler)|42%|Medium|❤️|
+|623|[Add One Row to Tree](./Algorithms/623.add-one-row-to-tree)|46%|Medium||
+|628|[Maximum Product of Three Numbers](./Algorithms/628.maximum-product-of-three-numbers)|44%|Easy|❤️|
+|629|[K Inverse Pairs Array](./Algorithms/629.k-inverse-pairs-array)|27%|Hard||
+|630|[Course Schedule III](./Algorithms/630.course-schedule-iii)|29%|Hard|❤️|
+|632|[Smallest Range](./Algorithms/632.smallest-range)|41%|Hard||
+|633|[Sum of Square Numbers](./Algorithms/633.sum-of-square-numbers)|32%|Easy||
+|636|[Exclusive Time of Functions](./Algorithms/636.exclusive-time-of-functions)|44%|Medium|❤️|
+|637|[Average of Levels in Binary Tree](./Algorithms/637.average-of-levels-in-binary-tree)|55%|Easy||
+|638|[Shopping Offers](./Algorithms/638.shopping-offers)|45%|Medium|❤️|
+|639|[Decode Ways II](./Algorithms/639.decode-ways-ii)|24%|Hard||
+|640|[Solve the Equation](./Algorithms/640.solve-the-equation)|38%|Medium||
+|643|[Maximum Average Subarray I](./Algorithms/643.maximum-average-subarray-i)|37%|Easy||
+|645|[Set Mismatch](./Algorithms/645.set-mismatch)|39%|Easy|❤️|
+|646|[Maximum Length of Pair Chain](./Algorithms/646.maximum-length-of-pair-chain)|47%|Medium||
+|647|[Palindromic Substrings](./Algorithms/647.palindromic-substrings)|54%|Medium|❤️|
+|648|[Replace Words](./Algorithms/648.replace-words)|47%|Medium||
+|649|[Dota2 Senate](./Algorithms/649.dota2-senate)|36%|Medium|❤️|
+|650|[2 Keys Keyboard](./Algorithms/650.2-keys-keyboard)|44%|Medium||
+|652|[Find Duplicate Subtrees](./Algorithms/652.find-duplicate-subtrees)|36%|Medium||
+|653|[Two Sum IV - Input is a BST](./Algorithms/653.two-sum-iv-input-is-a-bst)|49%|Easy||
+|654|[Maximum Binary Tree](./Algorithms/654.maximum-binary-tree)|69%|Medium|❤️|
+|655|[Print Binary Tree](./Algorithms/655.print-binary-tree)|49%|Medium||
+|657|[Judge Route Circle](./Algorithms/657.judge-route-circle)|68%|Easy||
+|658|[Find K Closest Elements](./Algorithms/658.find-k-closest-elements)|34%|Medium|❤️|
+|659|[Split Array into Consecutive Subsequences](./Algorithms/659.split-array-into-consecutive-subsequences)|36%|Medium|❤️|
+|661|[Image Smoother](./Algorithms/661.image-smoother)|46%|Easy||
+|662|[Maximum Width of Binary Tree](./Algorithms/662.maximum-width-of-binary-tree)|37%|Medium||
+|664|[Strange Printer](./Algorithms/664.strange-printer)|34%|Hard|❤️|
+|665|[Non-decreasing Array](./Algorithms/665.non-decreasing-array)|20%|Easy||
+|667|[Beautiful Arrangement II](./Algorithms/667.beautiful-arrangement-ii)|51%|Medium||
+|668|[Kth Smallest Number in Multiplication Table](./Algorithms/668.kth-smallest-number-in-multiplication-table)|40%|Hard||
+|669|[Trim a Binary Search Tree](./Algorithms/669.trim-a-binary-search-tree)|58%|Easy||
+|670|[Maximum Swap](./Algorithms/670.maximum-swap)|38%|Medium|❤️|
+|671|[Second Minimum Node In a Binary Tree](./Algorithms/671.second-minimum-node-in-a-binary-tree)|41%|Easy||
+|672|[Bulb Switcher II](./Algorithms/672.bulb-switcher-ii)|49%|Medium||
+|673|[Number of Longest Increasing Subsequence](./Algorithms/673.number-of-longest-increasing-subsequence)|31%|Medium|❤️|
+|674|[Longest Continuous Increasing Subsequence](./Algorithms/674.longest-continuous-increasing-subsequence)|42%|Easy||
+|675|[Cut Off Trees for Golf Event](./Algorithms/675.cut-off-trees-for-golf-event)|27%|Hard|❤️|
+|676|[Implement Magic Dictionary](./Algorithms/676.implement-magic-dictionary)|49%|Medium||
+|677|[Map Sum Pairs](./Algorithms/677.map-sum-pairs)|51%|Medium|❤️|
+|678|[Valid Parenthesis String](./Algorithms/678.valid-parenthesis-string)|29%|Medium|❤️|
+|679|[24 Game](./Algorithms/679.24-game)|38%|Hard|❤️|
+|680|[Valid Palindrome II](./Algorithms/680.valid-palindrome-ii)|32%|Easy||
+|682|[Baseball Game](./Algorithms/682.baseball-game)|58%|Easy||
+|684|[Redundant Connection](./Algorithms/684.redundant-connection)|43%|Medium||
+|685|[Redundant Connection II](./Algorithms/685.redundant-connection-ii)|27%|Hard|❤️|
+|686|[Repeated String Match](./Algorithms/686.repeated-string-match)|32%|Easy|❤️|
+|687|[Longest Univalue Path](./Algorithms/687.longest-univalue-path)|32%|Easy||
+|688|[Knight Probability in Chessboard](./Algorithms/688.knight-probability-in-chessboard)|39%|Medium||
+|689|[Maximum Sum of 3 Non-Overlapping Subarrays](./Algorithms/689.maximum-sum-of-3-non-overlapping-subarrays)|41%|Hard||
+|691|[Stickers to Spell Word](./Algorithms/691.stickers-to-spell-word)|34%|Hard|❤️|
+|692|[Top K Frequent Words](./Algorithms/692.top-k-frequent-words)|41%|Medium||
+|693|[Binary Number with Alternating Bits](./Algorithms/693.binary-number-with-alternating-bits)|55%|Easy||
+|695|[Max Area of Island](./Algorithms/695.max-area-of-island)|51%|Easy|❤️|
+|696|[Count Binary Substrings](./Algorithms/696.count-binary-substrings)|50%|Easy||
+|697|[Degree of an Array](./Algorithms/697.degree-of-an-array)|46%|Easy||
+|698|[Partition to K Equal Sum Subsets](./Algorithms/698.partition-to-k-equal-sum-subsets)|37%|Medium|❤️|
+|699|[Falling Squares](./Algorithms/699.falling-squares)|37%|Hard||
+|712|[Minimum ASCII Delete Sum for Two Strings](./Algorithms/712.minimum-ascii-delete-sum-for-two-strings)|51%|Medium|❤️|
+|713|[Subarray Product Less Than K](./Algorithms/713.subarray-product-less-than-k)|33%|Medium|❤️|
+|714|[Best Time to Buy and Sell Stock with Transaction Fee](./Algorithms/714.best-time-to-buy-and-sell-stock-with-transaction-fee)|45%|Medium|❤️|
+|715|[Range Module](./Algorithms/715.range-module)|31%|Hard||
+|717|[1-bit and 2-bit Characters](./Algorithms/717.1-bit-and-2-bit-characters)|49%|Easy||
+|718|[Maximum Length of Repeated Subarray](./Algorithms/718.maximum-length-of-repeated-subarray)|41%|Medium|❤️|
+|719|[Find K-th Smallest Pair Distance](./Algorithms/719.find-k-th-smallest-pair-distance)|27%|Hard|❤️|
+|720|[Longest Word in Dictionary](./Algorithms/720.longest-word-in-dictionary)|41%|Easy||
+|721|[Accounts Merge](./Algorithms/721.accounts-merge)|32%|Medium|❤️|
+|722|[Remove Comments](./Algorithms/722.remove-comments)|27%|Medium|❤️|
+|724|[Find Pivot Index](./Algorithms/724.find-pivot-index)|39%|Easy||
+|725|[Split Linked List in Parts](./Algorithms/725.split-linked-list-in-parts)|47%|Medium||
+|726|[Number of Atoms](./Algorithms/726.number-of-atoms)|43%|Hard|❤️|
+|728|[Self Dividing Numbers](./Algorithms/728.self-dividing-numbers)|66%|Easy||
+|729|[My Calendar I](./Algorithms/729.my-calendar-i)|42%|Medium||
+|730|[Count Different Palindromic Subsequences](./Algorithms/730.count-different-palindromic-subsequences)|34%|Hard|❤️|
+|731|[My Calendar II](./Algorithms/731.my-calendar-ii)|37%|Medium|❤️|
+|732|[My Calendar III](./Algorithms/732.my-calendar-iii)|50%|Hard||
+|733|[Flood Fill](./Algorithms/733.flood-fill)|47%|Easy||
+|735|[Asteroid Collision](./Algorithms/735.asteroid-collision)|37%|Medium||
+|736|[Parse Lisp Expression](./Algorithms/736.parse-lisp-expression)|42%|Hard|❤️|
+|738|[Monotone Increasing Digits](./Algorithms/738.monotone-increasing-digits)|40%|Medium||
+|739|[Daily Temperatures](./Algorithms/739.daily-temperatures)|52%|Medium|❤️|
+|740|[Delete and Earn](./Algorithms/740.delete-and-earn)|43%|Medium|❤️|
+|741|[Cherry Pickup](./Algorithms/741.cherry-pickup)|23%|Hard||
+|743|[Network Delay Time](./Algorithms/743.network-delay-time)|34%|Medium||
+|744|[Find Smallest Letter Greater Than Target](./Algorithms/744.find-smallest-letter-greater-than-target)|44%|Easy||
+|745|[Prefix and Suffix Search](./Algorithms/745.prefix-and-suffix-search)|25%|Hard||
+|746|[Min Cost Climbing Stairs](./Algorithms/746.min-cost-climbing-stairs)|43%|Easy||
+|747|[Largest Number At Least Twice of Others](./Algorithms/747.largest-number-at-least-twice-of-others)|41%|Easy||
+|748|[Shortest Completing Word](./Algorithms/748.shortest-completing-word)|51%|Medium||
+|749|[Contain Virus](./Algorithms/749.contain-virus)|39%|Hard||
+|752|[Open the Lock](./Algorithms/752.open-the-lock)|38%|Medium|❤️|
+|753|[Cracking the Safe](./Algorithms/753.cracking-the-safe)|39%|Hard|❤️|
+|754|[Reach a Number](./Algorithms/754.reach-a-number)|26%|Medium|❤️|
+|756|[Pyramid Transition Matrix](./Algorithms/756.pyramid-transition-matrix)|45%|Medium|❤️|
+|757|[Set Intersection Size At Least Two](./Algorithms/757.set-intersection-size-at-least-two)|34%|Hard||
+|761|[Special Binary String](./Algorithms/761.special-binary-string)|41%|Hard||
+|762|[Prime Number of Set Bits in Binary Representation](./Algorithms/762.prime-number-of-set-bits-in-binary-representation)|55%|Easy||
+|763|[Partition Labels](./Algorithms/763.partition-labels)|64%|Medium||
+|764|[Largest Plus Sign](./Algorithms/764.largest-plus-sign)|37%|Medium||
+|765|[Couples Holding Hands](./Algorithms/765.couples-holding-hands)|48%|Hard||
+|766|[Toeplitz Matrix](./Algorithms/766.toeplitz-matrix)|57%|Easy||
+|767|[Reorganize String](./Algorithms/767.reorganize-string)|35%|Medium||
+|768|[Max Chunks To Make Sorted II](./Algorithms/768.max-chunks-to-make-sorted-ii)|42%|Hard||
+|769|[Max Chunks To Make Sorted](./Algorithms/769.max-chunks-to-make-sorted)|47%|Medium||
+|770|[Basic Calculator IV](./Algorithms/770.basic-calculator-iv)|43%|Hard|❤️|
+|771|[Jewels and Stones](./Algorithms/771.jewels-and-stones)|82%|Easy||
+|773|[Sliding Puzzle](./Algorithms/773.sliding-puzzle)|47%|Hard||
+|775|[Global and Local Inversions](./Algorithms/775.global-and-local-inversions)|31%|Medium||
+|777|[Swap Adjacent in LR String](./Algorithms/777.swap-adjacent-in-lr-string)|27%|Medium||
+|778|[Swim in Rising Water](./Algorithms/778.swim-in-rising-water)|44%|Hard||
+|779|[K-th Symbol in Grammar](./Algorithms/779.k-th-symbol-in-grammar)|35%|Medium||
+|780|[Reaching Points](./Algorithms/780.reaching-points)|21%|Hard||
+|781|[Rabbits in Forest](./Algorithms/781.rabbits-in-forest)|49%|Medium||
+|782|[Transform to Chessboard](./Algorithms/782.transform-to-chessboard)|35%|Hard||
+|783|[Minimum Distance Between BST Nodes](./Algorithms/783.minimum-distance-between-bst-nodes)|47%|Easy||
+|784|[Letter Case Permutation](./Algorithms/784.letter-case-permutation)|52%|Easy|❤️|
+|785|[Is Graph Bipartite?](./Algorithms/785.is-graph-bipartite)|38%|Medium|❤️|
+|786|[K-th Smallest Prime Fraction](./Algorithms/786.k-th-smallest-prime-fraction)|29%|Hard|❤️|
+|787|[Cheapest Flights Within K Stops](./Algorithms/787.cheapest-flights-within-k-stops)|30%|Medium|❤️|
+|788|[Rotated Digits](./Algorithms/788.rotated-digits)|50%|Easy||
+|789|[Escape The Ghosts](./Algorithms/789.escape-the-ghosts)|47%|Medium||
+|790|[Domino and Tromino Tiling](./Algorithms/790.domino-and-tromino-tiling)|32%|Medium|❤️|
+|791|[Custom Sort String](./Algorithms/791.custom-sort-string)|60%|Medium||
+|792|[Number of Matching Subsequences](./Algorithms/792.number-of-matching-subsequences)|35%|Medium|❤️|
+|793| * Preimage Size of Factorial Zeroes Function|45%|Hard||
+|794| * Valid Tic-Tac-Toe State|27%|Medium||
+|795| * Number of Subarrays with Bounded Maximum|40%|Medium||
+|796| * Rotate String|54%|Easy||
+|797| * All Paths From Source to Target|69%|Medium||
+|798| * Smallest Rotation with Highest Score|31%|Hard||
+|799| * Champagne Tower|28%|Medium||
+|801| * Minimum Swaps To Make Sequences Increasing|23%|Medium||
+|802| * Find Eventual Safe States|35%|Medium||
+|803| * Bricks Falling When Hit|20%|Hard||
+|804| * Unique Morse Code Words|76%|Easy||
+|805| * Split Array With Same Average|19%|Hard||
+|806| * Number of Lines To Write String|64%|Easy||
+|807| * Max Increase to Keep City Skyline|82%|Medium||
+|808| * Soup Servings|30%|Medium||
+|809| * Expressive Words|34%|Medium||
+|810| * Chalkboard XOR Game|35%|Hard||
+|811| * Subdomain Visit Count|65%|Easy||
+|812| * Largest Triangle Area :new: |51%|Easy||
+|813| * Largest Sum of Averages :new: |37%|Medium||
+|814| * Binary Tree Pruning :new: |73%|Medium||
+|815| * Bus Routes :new: |30%|Hard||
+
+
+------------------------------------------------------------------
+
+以下免费的算法题，暂时不能使用 Go 解答
+
+- [116.Populating Next Right Pointers in Each Node](https://leetcode.com/problems/populating-next-right-pointers-in-each-node/)
+- [117.Populating Next Right Pointers in Each Node II](https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/)
+- [133.Clone Graph](https://leetcode.com/problems/clone-graph/)
+- [138.Copy List with Random Pointer](https://leetcode.com/problems/copy-list-with-random-pointer/)
+- [141.Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)
+- [142.Linked List Cycle II](https://leetcode.com/problems/linked-list-cycle-ii/)
+- [151.Reverse Words in a String](https://leetcode.com/problems/reverse-words-in-a-string/)
+- [160.Intersection of Two Linked Lists](https://leetcode.com/problems/intersection-of-two-linked-lists/)
+- [173.Binary Search Tree Iterator](https://leetcode.com/problems/binary-search-tree-iterator/)
+- [190.Reverse Bits](https://leetcode.com/problems/reverse-bits/)
+- [191.Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)
+- [222.Count Complete Tree Nodes](https://leetcode.com/problems/count-complete-tree-nodes/)
+- [235.Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)
+- [236.Lowest Common Ancestor of a Binary Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/)
+- [237.Delete Node in a Linked List](https://leetcode.com/problems/delete-node-in-a-linked-list/)
+- [278.First Bad Version](https://leetcode.com/problems/first-bad-version/)
+- [284.Peeking Iterator](https://leetcode.com/problems/peeking-iterator/)
+- [297.Serialize and Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)
+- [341.Flatten Nested List Iterator](https://leetcode.com/problems/flatten-nested-list-iterator/)
+- [374.Guess Number Higher or Lower](https://leetcode.com/problems/guess-number-higher-or-lower/)
+- [386.Lexicographical Numbers](https://leetcode.com/problems/lexicographical-numbers/)
+- [449.Serialize and Deserialize BST](https://leetcode.com/problems/serialize-and-deserialize-bst/)
+- [535.Encode and Decode TinyURL](https://leetcode.com/problems/encode-and-decode-tinyurl/)
+- [690.Employee Importance](https://leetcode.com/problems/employee-importance/)
+
+------------------------------------------------------------------
+
+## 二.分类
+
+## Array
+
+| Title | Solution | Difficulty | Time | Space |收藏| 
+| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
+|[1. Two Sum](https://leetcode.com/problems/two-sum/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0001.%20Two%20Sum)| Easy | O(n)| O(n)||
+|[11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0011.%20Container%20With%20Most%20Water)| Medium | O(n)| O(1)||
+|[15. 3Sum](https://leetcode.com/problems/3sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0015.%203Sum)| Medium | O(n^2)| O(n)||
+|[16. 3Sum Closest](https://leetcode.com/problems/3sum-closest)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0016.%203Sum%20Closest)| Medium | O(n^2)| O(1)||
+|[18. 4Sum](https://leetcode.com/problems/4sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0018.%204Sum)| Medium | O(n^3)| O(n^2)||
+|[26. Remove Duplicates from Sorted Array](https://leetcode.com/problems/remove-duplicates-from-sorted-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0026.%20Remove%20Duplicates%20from%20Sorted%20Array)| Easy | O(n)| O(1)||
+|[27. Remove Element](https://leetcode.com/problems/remove-element)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0027.%20Remove%20Element)| Easy | O(n)| O(1)||
+|[39. Combination Sum](https://leetcode.com/problems/combination-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0039.%20Combination%20Sum)| Medium | O(n log n)| O(n)||
+|[40. Combination Sum II](https://leetcode.com/problems/combination-sum-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0040.%20Combination%20Sum%20II)| Medium | O(n log n)| O(n)||
+|[41. First Missing Positive](https://leetcode.com/problems/first-missing-positive)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0041.%20First-Missing-Positive)| Hard | O(n)| O(n)||
+|[42. Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0042.%20Trapping%20Rain%20Water)| Hard | O(n)| O(1)||
+|[48. Rotate Image](https://leetcode.com/problems/rotate-image)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0048.%20Rotate%20Image)| Medium | O(n)| O(1)||
+|[53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0053.%20Maximum%20Subarray)| Easy | O(n)| O(n)||
+|[54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0054.%20Spiral%20Matrix)| Medium | O(n)| O(n^2)||
+|[56. Merge Intervals](https://leetcode.com/problems/merge-intervals)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0056.%20Merge%20Intervals)| Medium | O(n log n)| O(1)||
+|[57. Insert Interval](https://leetcode.com/problems/insert-interval)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0057.%20Insert%20Interval)| Hard | O(n)| O(1)||
+|[59. Spiral Matrix II](https://leetcode.com/problems/spiral-matrix-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0059.%20Spiral%20Matrix%20II)| Medium | O(n)| O(n^2)||
+|[62. Unique Paths](https://leetcode.com/problems/unique-paths)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0062.%20Unique%20Paths)| Medium | O(n^2)| O(n^2)||
+|[63. Unique Paths II](https://leetcode.com/problems/unique-paths-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0063.%20Unique%20Paths%20II)| Medium | O(n^2)| O(n^2)||
+|[64. Minimum Path Sum](https://leetcode.com/problems/minimum-path-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0064.%20Minimum%20Path%20Sum)| Medium | O(n^2)| O(n^2)||
+|[75. Sort Colors](https://leetcode.com/problems/sort-colors)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0075.%20Sort%20Colors)| Medium | O(n)| O(1)||
+|[78. Subsets](https://leetcode.com/problems/subsets)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0078.%20Subsets)| Medium | O(n^2)| O(n)||
+|[79. Word Search](https://leetcode.com/problems/word-search)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0079.%20Word%20Search)| Medium | O(n^2)| O(n^2)||
+|[80. Remove Duplicates from Sorted Array II](https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0080.%20Remove%20Duplicates%20from%20Sorted%20Array%20II)| Medium | O(n^2)| O(n^2)||
+|[84. Largest Rectangle in Histogram](https://leetcode.com/problems/largest-rectangle-in-histogram)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0084.%20Largest%20Rectangle%20in%20Histogram)| Medium | O(n)| O(n)||
+|[88. Merge Sorted Array](https://leetcode.com/problems/merge-sorted-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0088.%20Merge-Sorted-Array)| Easy | O(n)| O(1)||
+|[90. Subsets II](https://leetcode.com/problems/subsets-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0090.%20Subsets%20II)| Medium | O(n^2)| O(n)||
+|[120. Triangle](https://leetcode.com/problems/triangle)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0120.%20Triangle)| Medium | O(n^2)| O(n)||
+|[121. Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock)| Easy | O(n)| O(1)||
+|[122. Best Time to Buy and Sell Stock II](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0122.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20II)| Easy | O(n)| O(1)||
+|[126. Word Ladder II](https://leetcode.com/problems/word-ladder-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0126.%20Word%20Ladder%20II)| Hard | O(n)| O(n^2)||
+|[152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0152.%20Maximum%20Product%20Subarray)| Medium | O(n)| O(1)||
+|[167. Two Sum II - Input array is sorted](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0167.%20Two%20Sum%20II%20-%20Input%20array%20is%20sorted)| Easy | O(n)| O(1)||
+|[209. Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0209.%20Minimum%20Size%20Subarray%20Sum)| Medium | O(n)| O(1)||
+|[216. Combination Sum III](https://leetcode.com/problems/combination-sum-iii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0216.%20Combination%20Sum%20III)| Medium | O(n)| O(1)||
+|[217. Contains Duplicate](https://leetcode.com/problems/contains-duplicate)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0217.%20Contains%20Duplicate)| Easy | O(n)| O(n)||
+|[219. Contains Duplicate II](https://leetcode.com/problems/contains-duplicate-ii)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0219.%20Contains%20Duplicate%20II)| Easy | O(n)| O(n)||
+|[283. Move Zeroes](https://leetcode.com/problems/move-zeroes)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0283.%20Move%20Zeroes)| Easy | O(n)| O(1)||
+|[287. Find the Duplicate Number](https://leetcode.com/problems/find-the-duplicate-number)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0287.%20Find%20the%20Duplicate%20Number)| Easy | O(n)| O(1)||
+|[532. K-diff Pairs in an Array](https://leetcode.com/problems/k-diff-pairs-in-an-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0532.%20K-diff%20Pairs%20in%20an%20Array)| Easy | O(n)| O(n)||
+|[566. Reshape the Matrix](https://leetcode.com/problems/reshape-the-matrix)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0566.%20Reshape%20the%20Matrix)| Easy | O(n^2)| O(n^2)||
+|[628. Maximum Product of Three Numbers](https://leetcode.com/problems/maximum-product-of-three-numbers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0628.%20Maximum%20Product%20of%20Three%20Numbers)| Easy | O(n)| O(1)||
+|[713. Subarray Product Less Than K](https://leetcode.com/problems/subarray-product-less-than-k)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0713.%20Subarray%20Product%20Less%20Than%20K)| Medium | O(n)| O(1)||
+|[714. Best Time to Buy and Sell Stock with Transaction Fee](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0714.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Transaction%20Fee)| Medium | O(n)| O(1)||
+|[746. Min Cost Climbing Stairs](https://leetcode.com/problems/min-cost-climbing-stairs)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0746.%20Min%20Cost%20Climbing%20Stairs)| Easy | O(n)| O(1)||
+|[766. Toeplitz Matrix](https://leetcode.com/problems/toeplitz-matrix)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0766.%20Toeplitz%20Matrix)| Easy | O(n)| O(1)||
+|[867. Transpose Matrix](https://leetcode.com/problems/transpose-matrix)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0867.%20Transpose%20Matrix)| Easy | O(n)| O(1)||
+|[891. Sum of Subsequence Widths](https://leetcode.com/problems/sum-of-subsequence-widths)| [Go]()| Hard | O(n)| O(1)||
+|[907. Sum of Subarray Minimums](https://leetcode.com/problems/sum-of-subarray-minimums)| [Go]()| Medium | O(n)| O(1)||
+|[922. Sort Array By Parity II](https://leetcode.com/problems/sum-of-subarray-minimums)| [Go]()| Medium | O(n)| O(1)||
+|[969. Pancake Sorting](https://leetcode.com/problems/pancake-sorting)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0969.%20Pancake%20Sorting)| Medium | O(n)| O(1)||
+|[977. Squares of a Sorted Array](https://leetcode.com/problems/squares-of-a-sorted-array)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0977.%20Squares%20of%20a%20Sorted%20Array)| Easy | O(n)| O(1)||
+
+
+
+
+| Title | Solution | Difficulty | Time | Space |
+| ----- | -------- | ---------- | ---- | ----- |
+[Max Consecutive Ones](https://leetcode.com/problems/max-consecutive-ones/)| [Go](./Array/MaxConsecutiveOnes.Go)| Easy| O(n)| O(1)|
+[Heaters](https://leetcode.com/problems/heaters/)| [Go](./Array/Heaters.Go)| Easy| O(nlogn)| O(1)|
+[Number of Boomerangs](https://leetcode.com/problems/number-of-boomerangs/)| [Go](./Array/NumberBoomerangs.Go)| Easy| O(n ^ 2)| O(n)|
+[Island Perimeter](https://leetcode.com/problems/island-perimeter/)| [Go](./Array/IslandPerimeter.Go)| Easy| O(nm)| O(1)|
+[Majority Element](https://leetcode.com/problems/majority-element/)| [Go](./Array/MajorityElement.Go)| Easy| O(n)| O(1)|
+[Majority Element II](https://leetcode.com/problems/majority-element-ii/)| [Go](./Array/MajorityElementII.Go)| Medium| O(n)| O(1)|
+[Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays/)| [Go](./Array/IntersectionTwoArrays.Go)| Easy| O(n)| O(n)|
+[Intersection of Two Arrays II](https://leetcode.com/problems/intersection-of-two-arrays-ii/)| [Go](./Array/IntersectionTwoArraysII.Go)| Easy| O(n)| O(n)|
+[Contains Duplicate](https://leetcode.com/problems/contains-duplicate/)| [Go](./Array/ContainsDuplicate.Go)| Easy| O(n)| O(n)|
+[Contains Duplicate II](https://leetcode.com/problems/contains-duplicate-ii/)| [Go](./Array/ContainsDuplicateII.Go)| Easy| O(n)| O(n)|
+[Remove Duplicates from Sorted Array](https://leetcode.com/problems/remove-duplicates-from-sorted-array/)| [Go](./Array/RemoveDuplicatesFromSortedArray.Go)| Easy| O(n)| O(1)|
+[Remove Duplicates from Sorted Array II](https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/)| [Go](./Array/RemoveDuplicatesFromSortedArrayII.Go)| Medium| O(n)| O(1)|
+[Move Zeroes](https://leetcode.com/problems/move-zeroes/)| [Go](./Array/MoveZeroes.Go)| Easy| O(n)| O(1)|
+[Remove Element](https://leetcode.com/problems/remove-element/)| [Go](./Array/RemoveElement.Go)| Easy| O(n)| O(1)|
+[Two Sum](https://leetcode.com/problems/two-sum/)| [Go](./Array/TwoSum.Go)| Easy| O(n)| O(n)|
+[3Sum](https://leetcode.com/problems/3sum/)| [Go](./Array/ThreeSum.Go)| Medium| O(n^2)| O(nC3)|
+[3Sum Closest](https://leetcode.com/problems/3sum-closest/)| [Go](./Array/ThreeSum.Go)| Medium| O(n^2)| O(nC3)|
+[4Sum](https://leetcode.com/problems/4sum/)| [Go](./Array/FourSum.Go)| Medium| O(n^3)| O(nC4)|
+[Summary Ranges](https://leetcode.com/problems/summary-ranges/)| [Go](./Array/SummaryRanges.Go)| Medium| O(n)| O(n)|
+[Shortest Word Distance](https://leetcode.com/problems/shortest-word-distance/)| [Go](./Array/ShortestWordDistance.Go)| Easy| O(n)| O(1)|
+[Shortest Word Distance III](https://leetcode.com/problems/shortest-word-distance-iii/)| [Go](./Array/ShortestWordDistanceIII.Go)| Medium| O(n)| O(1)|
+[Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum/)| [Go](./Array/MinimumSizeSubarraySum.Go)| Medium| O(n)| O(1)|
+[Maximum Size Subarray Sum Equals k](https://leetcode.com/problems/maximum-size-subarray-sum-equals-k/)| [Go](./Array/MaximumSizeSubarraySumEqualsK.Go)| Medium| O(n)| O(n)|
+[Smallest Range](https://leetcode.com/problems/smallest-range/)| [Go](./Array/SmallestRange.Go)| Hard | O(nm)| O(nm)|
+[Product of Array Except Self](https://leetcode.com/problems/product-of-array-except-self/)| [Go](./Array/ProductExceptSelf.Go)| Medium| O(n)| O(n)|
+[Rotate Array](https://leetcode.com/problems/rotate-array/)| [Go](./Array/RotateArray.Go)| Easy| O(n)| O(1)|
+[Rotate Image](https://leetcode.com/problems/rotate-image/)| [Go](./Array/RotateImage.Go)| Medium| O(n^2)| O(1)|
+[Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)| [Go](./Array/SpiralMatrix.Go)| Medium| O(n^2)| O(1)|
+[Spiral Matrix II](https://leetcode.com/problems/spiral-matrix/)| [Go](./Array/SpiralMatrixII.Go)| Medium| O(n^2)| O(1)|
+[Valid Sudoku](https://leetcode.com/problems/valid-sudoku/)| [Go](./Array/ValidSudoku.Go)| Easy| O(n^2)| O(n)|
+[Set Matrix Zero](https://leetcode.com/problems/set-matrix-zeroes/)| [Go](./Array/SetMatrixZero.Go)| Medium| O(n^2)| O(1)|
+[Next Permutation](https://leetcode.com/problems/next-permutation/)| [Go](./Array/NextPermutation.Go)| Medium| O(n)| O(1)|
+[Gas Station](https://leetcode.com/problems/gas-station/)| [Go](./Array/GasStation.Go)| Medium| O(n)| O(1)|
+[Game of Life](https://leetcode.com/problems/game-of-life/)| [Go](./Array/GameLife.Go)| Medium| O(n)| O(1)|
+[Task Scheduler](https://leetcode.com/problems/task-scheduler/)| [Go](./Array/TaskScheduler.Go)| Medium| O(nlogn)| O(n)|
+[Sliding Window Maximum ](https://leetcode.com/problems/sliding-window-maximum/)| [Go](./Array/SlidingWindowMaximum.Go)| Hard| O(n)| O(n)|
+[Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/)| [Go](./Array/LongestConsecutiveSequence.Go)| Hard| O(n)| O(n)|
+
+
+## String
+| Title | Solution | Difficulty | Time | Space |
+| ----- | -------- | ---------- | ---- | ----- |
+[Fizz Buzz](https://leetcode.com/problems/fizz-buzz/)| [Go](./String/FizzBuzz.Go)| Easy| O(n)| O(1)|
+[First Unique Character in a String](https://leetcode.com/problems/first-unique-character-in-a-string/)| [Go](./String/FirstUniqueCharacterInString.Go)| Easy| O(n)| O(1)|
+[Keyboard Row](https://leetcode.com/problems/keyboard-row/)| [Go](./String/KeyboardRow.Go)| Easy| O(nm)| O(n)|
+[Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)| [Go](./String/ValidPalindrome.Go)| Easy| O(n)| O(n)|
+[Valid Palindrome II](https://leetcode.com/problems/valid-palindrome-ii/)| [Go](./String/ValidPalindromeII.Go)| Easy| O(n)| O(n)|
+[Detect Capital](https://leetcode.com/problems/detect-capital/)| [Go](./String/DetectCapital.Go)| Easy| O(n)| O(1)|
+[Count and Say](https://leetcode.com/problems/count-and-say/)| [Go](./String/CountAndSay.Go)| Easy| O(n^2)| O(n)|
+[Flip Game](https://leetcode.com/problems/flip-game/)| [Go](./String/FlipGame.Go)| Easy| O(n)| O(n)|
+[Implement strStr()](https://leetcode.com/problems/implement-strstr/)| [Go](./String/StrStr.Go)| Easy| O(nm)| O(n)|
+[Isomorphic Strings](https://leetcode.com/problems/isomorphic-strings/)| [Go](./String/IsomorphicStrings.Go)| Easy| O(n)| O(n)|
+[Reverse String](https://leetcode.com/problems/reverse-string/)| [Go](./String/ReverseString.Go)| Easy| O(n)| O(n)|
+[Reverse String II](https://leetcode.com/problems/reverse-string-ii/)| [Go](./String/ReverseStringII.Go)| Easy| O(n)| O(n)|
+[Reverse Vowels of a String](https://leetcode.com/problems/reverse-vowels-of-a-string/)| [Go](./String/ReverseVowelsOfAString.Go)| Easy| O(n)| O(n)|
+[Length of Last Word](https://leetcode.com/problems/length-of-last-word/)| [Go](./String/AddStrings.Go)| Easy| O(n)| O(n)|
+[Add Strings](https://leetcode.com/problems/add-strings/)| [Go](./String/LengthLastWord.Go)| Easy| O(n)| O(1)|
+[Multiply Strings](https://leetcode.com/problems/multiply-strings/)| [Go](./String/MultiplyStrings.Go)| Medium| O(n)| O(1)|
+[Palindrome Permutation](https://leetcode.com/problems/palindrome-permutation/)| [Go](./String/PalindromePermutation.Go)| Easy| O(n)| O(n)|
+[Valid Anagram](https://leetcode.com/problems/valid-anagram/)| [Go](./String/ValidAnagram.Go)| Easy| O(nlogn)| O(1)|
+[Ransom Note](https://leetcode.com/problems/ransom-note/)| [Go](./String/RansomNote.Go)| Easy| O(n)| O(n)|
+[Group Anagrams](https://leetcode.com/problems/anagrams/)| [Go](./String/GroupAnagrams.Go)| Medium| O(nmlogm + nlogn)| O(n)
+[Longest Common Prefix](https://leetcode.com/problems/longest-common-prefix/)| [Go](./String/LongestCommonPrefix.Go)| Easy| O(nm)| O(m)|
+[Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/)| [Go](./String/LongestSubstringWithoutRepeatingCharacters.Go)| Medium| O(n)| O(n)|
+[One Edit Distance](https://leetcode.com/problems/one-edit-distance/)| [Go](./String/OneEditDistance.Go)| Medium| O(n)| O(n)|
+[Word Pattern](https://leetcode.com/problems/word-pattern/)| [Go](./String/WordPattern.Go)| Easy| O(n)| O(n)|
+[Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)| [Go](./Array/MinimumWindowSubstring.Go)| Hard| O(n^2)| O(n)|
+[Text Justification](https://leetcode.com/problems/text-justification/)| [Go](./String/TextJustification.Go)| Hard| O(n)| O(n)|
+
+## Linked List (已全部做完)
+
+- 巧妙的构造虚拟头结点。可以使遍历处理逻辑更加统一。
+- 灵活使用递归。构造递归条件，使用递归可以巧妙的解题。不过需要注意有些题目不能使用递归，因为递归深度太深会导致超时和栈溢出。
+- 链表区间逆序。第 92 题。
+- 链表寻找中间节点。第 876 题。链表寻找倒数第 n 个节点。第 19 题。只需要一次遍历就可以得到答案。
+- 合并 K 个有序链表。第 21 题，第 23 题。
+- 链表归类。第 86 题，第 328 题。
+- 链表排序，时间复杂度要求 O(n * log n)，空间复杂度 O(1)。只有一种做法，归并排序，至顶向下归并。第 148 题。
+- 判断链表是否存在环，如果有环，输出环的交叉点的下标；判断 2 个链表是否有交叉点，如果有交叉点，输出交叉点。第 141 题，第 142 题，第 160 题。
+
+
+
+| Title | Solution | Difficulty | Time | Space |收藏| 
+| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
+|[2. Add Two Numbers](https://leetcode.com/problems/add-two-numbers)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0002.%20Add-Two-Number)| Medium | O(n)| O(1)||
+|[19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0019.%20Remove%20Nth%20Node%20From%20End%20of%20List)| Medium | O(n)| O(1)||
+|[21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0021.%20Merge%20Two%20Sorted%20Lists)| Easy | O(log n)| O(1)||
+|[23. Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0023.%20Merge%20k%20Sorted%20Lists)| Hard | O(log n)| O(1)|❤️|
+|[24. Swap Nodes in Pairs](https://leetcode.com/problems/swap-nodes-in-pairs/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0024.%20Swap%20Nodes%20in%20Pairs)| Medium | O(n)| O(1)||
+|[25. Reverse Nodes in k-Group](https://leetcode.com/problems/reverse-nodes-in-k-group/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0025.%20Reverse%20Nodes%20in%20k%20Group)| Hard | O(log n)| O(1)|❤️|
+|[61. Rotate List](https://leetcode.com/problems/rotate-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0061.%20Rotate%20List)| Medium | O(n)| O(1)||
+|[82. Remove Duplicates from Sorted List II](https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0082.%20Remove%20Duplicates%20from%20Sorted%20List%20II)| Medium | O(n)| O(1)||
+|[83. Remove Duplicates from Sorted List](https://leetcode.com/problems/remove-duplicates-from-sorted-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0083.%20Remove%20Duplicates%20from%20Sorted%20List)| Easy | O(n)| O(1)||
+|[86. Partition List](https://leetcode.com/problems/partition-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0086.%20Partition%20List)| Medium | O(n)| O(1)|❤️|
+|[92. Reverse Linked List II](https://leetcode.com/problems/reverse-linked-list-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0092.%20Reverse%20Linked%20List%20II)| Medium | O(n)| O(1)|❤️|
+|[109. Convert Sorted List to Binary Search Tree](https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0109.%20Convert%20Sorted%20List%20to%20Binary%20Search%20Tree)| Medium | O(log n)| O(n)||
+|[141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0141.%20Linked%20List%20Cycle)| Easy | O(n)| O(1)|❤️|
+|[142. Linked List Cycle II](https://leetcode.com/problems/linked-list-cycle-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0142.%20Linked%20List%20Cycle%20II)| Medium | O(n)| O(1)|❤️|
+|[143. Reorder List](https://leetcode.com/problems/reorder-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0143.%20Reorder%20List)| Medium | O(n)| O(1)|❤️|
+|[147. Insertion Sort List](https://leetcode.com/problems/insertion-sort-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0147.%20Insertion%20Sort%20List)| Medium | O(n)| O(1)||
+|[148. Sort List](https://leetcode.com/problems/sort-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0148.%20Sort%20List)| Medium | O(log n)| O(n)|❤️|
+|[160. Intersection of Two Linked Lists](https://leetcode.com/problems/intersection-of-two-linked-lists/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0160.%20Intersection%20of%20Two%20Linked%20Lists)| Easy | O(n)| O(1)|❤️|
+|[203. Remove Linked List Elements](https://leetcode.com/problems/remove-linked-list-elements/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0203.%20Remove%20Linked%20List%20Elements)| Easy | O(n)| O(1)||
+|[206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0206.%20Reverse-Linked-List)| Easy | O(n)| O(1)||
+|[234. Palindrome Linked List](https://leetcode.com/problems/palindrome-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0234.%20Palindrome%20Linked%20List)| Easy | O(n)| O(1)||
+|[237. Delete Node in a Linked List](https://leetcode.com/problems/delete-node-in-a-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0237.%20Delete%20Node%20in%20a%20Linked%20List)| Easy | O(n)| O(1)||
+|[328. Odd Even Linked List](https://leetcode.com/problems/odd-even-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0328.%20Odd%20Even%20Linked%20List)| Medium | O(n)| O(1)||
+|[445. Add Two Numbers II](https://leetcode.com/problems/add-two-numbers-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0445.%20Add%20Two%20Numbers%20II)| Medium | O(n)| O(n)||
+|[725. Split Linked List in Parts](https://leetcode.com/problems/split-linked-list-in-parts/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0725.%20Split%20Linked%20List%20in%20Parts)| Medium | O(n)| O(1)||
+|[817. Linked List Components](https://leetcode.com/problems/linked-list-components/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0817.%20Linked%20List%20Components)| Medium | O(n)| O(1)||
+|[707. Design Linked List](https://leetcode.com/problems/design-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0707.%20Design%20Linked%20List)| Easy | O(n)| O(1)||
+|[876. Middle of the Linked List](https://leetcode.com/problems/middle-of-the-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0876.%20Middle%20of%20the%20Linked%20List)| Easy | O(n)| O(1)|❤️|
+|[1019. Next Greater Node In Linked List](https://leetcode.com/problems/next-greater-node-in-linked-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1019.%20Next%20Greater%20Node%20In%20Linked%20List)| Medium | O(n)| O(1)||
+|----------------------------------------------------------------------------|-------------|-------------| -------------| -------------|-------------|
+
+## Stack (已全部做完)
+
+| Title | Solution | Difficulty | Time | Space |
+| ----- | -------- | ---------- | ---- | ----- |
+[Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)| [Go](./Stack/ValidParentheses.Go)| Easy| O(n)| O(n)|
+[Longest Valid Parentheses](https://leetcode.com/problems/longest-valid-parentheses/)| [Go](./Stack/LongestValidParentheses.Go)| Hard| O(n)| O(n)|
+[Evaluate Reverse Polish Notation](https://leetcode.com/problems/evaluate-reverse-polish-notation/)| [Go](./Stack/EvaluateReversePolishNotation.Go)| Medium| O(n)| O(n)|
+[Simplify Path](https://leetcode.com/problems/simplify-path/)| [Go](./Stack/SimplifyPath.Go)| Medium| O(n)| O(n)|
+[Remove K Digits](https://leetcode.com/problems/remove-k-digits/)| [Go](./Stack/RemoveKDigits.Go)| Medium| O(n)| O(n)|
+[Ternary Expression Parser](https://leetcode.com/problems/ternary-expression-parser/)| [Go](./Stack/TernaryExpressionParser.Go)| Medium| O(n)| O(n)|
+[Binary Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal/)| [Go](./Stack/PreorderTraversal.Go)| Medium| O(n)| O(n)|
+[Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal/)| [Go](./Stack/InorderTraversal.Go)| Medium| O(n)| O(n)|
+[Binary Tree Postorder Traversal](https://leetcode.com/problems/binary-tree-postorder-traversal/)| [Go](./Stack/PostorderTraversal.Go)| Hard| O(n)| O(n)|
+
+
+## Tree
+| Title | Solution | Difficulty | Time | Space |
+| ----- | -------- | ---------- | ---- | ----- |
+[Same Tree](https://oj.leetcode.com/problems/same-tree/)| [Go](./Tree/SameTree.Go)| Easy| O(n)| O(n)|
+[Symmetric Tree](https://leetcode.com/problems/symmetric-tree/)| [Go](./Tree/SymmetricTree.Go)| Easy| O(n)| O(n)|
+[Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)| [Go](./Tree/InvertBinaryTree)| Easy| O(n)| O(n)|
+[Binary Tree Upside Down](https://leetcode.com/problems/binary-tree-upside-down/)| [Go](./Tree/BinaryTreeUpsideDown)| Medium| O(n)| O(1)|
+[Minimum Depth of Binary Tree](https://leetcode.com/problems/minimum-depth-of-binary-tree/)| [Go](./Tree/MinimumDepthOfBinaryTree.Go)| Easy| O(n)| O(1)|
+[Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)| [Go](./Tree/MaximumDepthOfBinaryTree.Go)| Easy| O(n)| O(1)|
+[Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/)| [Go](./Tree/DiameterBinaryTree.Go)| Easy| O(n)| O(1)|
+[Balanced Binary Tree](https://leetcode.com/problems/balanced-binary-tree/)| [Go](./Tree/BalancedBinaryTree.Go)| Easy|  O(n)| O(n)|
+[Sum of Left Leaves](https://leetcode.com/problems/sum-of-left-leaves/)| [Go](./Tree/SumLeftLeaves.Go)| Easy|  O(n)| O(1)|
+[Flatten Binary Tree to Linked List](https://leetcode.com/problems/flatten-binary-tree-to-linked-list/)| [Go](./Tree/FlattenBinaryTreeLinkedList.Go)| Medium| O(n)| O(1)|
+[Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)| [Go](./Tree/ValidateBinarySearchTree.Go)| Medium| O(n)| O(n)|
+[Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/)| [Go](./Tree/BinaryTreeLevelOrderTraversal.Go)| Easy| O(n)| O(n)|
+[Binary Tree Level Order Traversal II](https://leetcode.com/problems/binary-tree-level-order-traversal-ii/)| [Go](./Tree/BinaryTreeLevelOrderTraversalII.Go)| Easy| O(n)| O(n)|
+[Binary Tree Zigzag Level Order Traversal](https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/)| [Go](./Tree/BinaryTreeZigzagLevelOrderTraversal.Go)| Medium| O(n)| O(n)|
+[Binary Tree Vertical Order Traversal](https://leetcode.com/problems/binary-tree-vertical-order-traversal/)| [Go](./Tree/BinaryTreeVerticalOrderTraversal.Go)| Medium| O(n)| O(n)|
+[Binary Tree Right Side View](https://leetcode.com/problems/binary-tree-right-side-view/)| [Go](./Tree/BinaryTreeRightSideView.Go)| Medium| O(n)| O(n)|
+[Construct Binary Tree from Preorder and Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)| [Go](./Tree/ConstructBinaryTreePreorderInorder.Go)| Medium| O(n)| O(n)|
+[Construct Binary Tree from Inorder and Postorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/)| [Go](./Tree/ConstructBinaryTreeInorderPostorder.Go)| Medium| O(n)| O(n)|
+[Path Sum](https://leetcode.com/problems/path-sum/)| [Go](./Tree/PathSum.Go)| Easy| O(n)| O(n)|
+[Path Sum II](https://leetcode.com/problems/path-sum-ii/)| [Go](./Tree/PathSumII.Go)| Medium| O(n)| O(n)|
+[Path Sum III](https://leetcode.com/problems/path-sum-iiI/)| [Go](./Tree/PathSumIII.Go)| Easy| O(n^2)| O(1)|
+[Bnary Tree Paths](https://leetcode.com/problems/binary-tree-paths/)| [Go](./Tree/BnaryTreePaths.Go)| Easy| O(n)| O(n)|
+[Unique Binary Search Trees](https://leetcode.com/problems/unique-binary-search-trees/)| [Go](./Tree/UniqueBinarySearchTrees.Go)| Medium| O(n^2)| O(n)|
+[Recover Binary Search Tree](https://leetcode.com/problems/recover-binary-search-tree/)| [Go](./Tree/RecoverBinarySearchTree.Go)| Hard| O(n)| O(1)|
+[Merge Two Binary Trees](https://leetcode.com/problems/merge-two-binary-trees/description/) | [Go](./Tree/MergeTwoBinaryTrees.Go) | Easy | O(n) | O(n) |
+
+## Dynamic programming
+| Title | Solution | Difficulty | Time | Space |
+| ----- | -------- | ---------- | ---- | ----- |
+[Nested List Weight Sum](https://leetcode.com/problems/nested-list-weight-sum/)| [Go](./DP/NestedListWeightSum.Go)| Easy| O(n)| O(1)|
+[Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)| [Go](./DP/ClimbingStairs.Go)| Easy| O(n)| O(1)|
+[Min Cost Climbing Stairs](https://leetcode.com/problems/min-cost-climbing-stairs/)| [Go](./DP/MinCostClimbingStairs.Go)| Easy| O(n)| O(n)|
+[Unique Paths](https://leetcode.com/problems/unique-paths/)| [Go](./DP/UniquePaths.Go)| Medium| O(mn)| O(mn)|
+[Unique Paths II](https://leetcode.com/problems/unique-paths-ii/)| [Go](./DP/UniquePathsII.Go)| Medium| O(mn)| O(mn)|
+[Decode Ways](https://leetcode.com/problems/decode-ways/)| [Go](./DP/DecodeWays.Go) | O(n)|O(n)|
+[Minimum Path Sum](https://leetcode.com/problems/minimum-path-sum/)| [Go](./DP/MinimumPathSum.Go)| Medium| O(mn)| O(mn)|
+[Generate Parentheses](https://leetcode.com/problems/generate-parentheses/)| [Go](./DP/GenerateParentheses.Go)| Medium| O(2^n)| O(n)|
+[Different Ways to Add Parentheses](https://leetcode.com/problems/different-ways-to-add-parentheses/)| [Go](./DP/DifferentWaysAddParentheses.Go)| Medium| O(n^n)| O(n)|
+[Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)| [Go](./DP/BestTimeBuySellStock.Go)| Easy| O(n)| O(1)|
+[Best Time to Buy and Sell Stock II](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/)| [Go](./DP/BestTimeBuySellStockII.Go)| Medium| O(n)| O(1)|
+[Best Time to Buy and Sell Stock III](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/)| [Go](./DP/BestTimeBuySellStockIII.Go)| Hard| O(n)| O(n)|
+[Best Time to Buy and Sell Stock IV](https://leetcode.com/problems/https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/)| [Go](./DP/BestTimeBuySellStockIV.Go)| Hard| O(n^2)| O(n)|
+[Best Time to Buy and Sell Stock with Cooldown](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/)| [Go](./DP/BestTimeBuySellStockCooldown.Go)| Medium| O(n^2)| O(n)|
+[Coin Change](https://leetcode.com/problems/coin-change/)| [Go](./DP/CoinChange.Go)| Medium| O(n^2)| O(n)|
+[Coin Change II](https://leetcode.com/problems/coin-change-ii/)| [Go](./DP/CoinChangeII.Go)| Medium| O(n^2)| O(n)|
+[Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/)| [Go](./DP/LongestIncreasingSubsequence.Go)| Medium| O(n^2)| O(n)|
+[Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)| [Go](./DP/LongestPalindromicSubstring.Go)| Medium| O(n^2)| O(n^2)|
+[Perfect Squares](https://leetcode.com/problems/perfect-squares/)| [Go](./DP/PerfectSquares.Go)| Medium| O(n^2)| O(n)|
+[House Robber](https://leetcode.com/problems/house-robber/)| [Go](./DP/HouseRobber.Go)| Easy| O(n)| O(1)|
+[House Robber II](https://leetcode.com/problems/house-robber-ii/)| [Go](./DP/HouseRobberII.Go)| Medium| O(n)| O(1)|
+[Paint Fence](https://leetcode.com/problems/paint-fence/)| [Go](./DP/PaintFence.Go)| Easy| O(n)| O(n)|
+[Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)| [Go](./DP/MaximumSubarray.Go)| Medium| O(n)| O(1)|
+[Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)| [Go](./DP/MaximumProductSubarray.Go)| Medium| O(n)| O(1)|
+[Maximal Square](https://leetcode.com/problems/maximal-square/)| [Go](./DP/MaximalSquare.Go)| Medium| O(mn)| O(mn)|
+[Edit Distance](https://leetcode.com/problems/edit-distance/)| [Go](./DP/EditDistance.Go)| Hard| O(mn)| O(mn)|
+[Combination Sum IV](https://leetcode.com/problems/combination-sum-iv/)| [Go](./DP/CombinationSumIV.Go)| Medium| O(2^n)| O(n)|
+[Triangle](https://leetcode.com/problems/triangle/)| [Go](./DP/Triangle.Go)| Medium| O(2^n - 1)| O(m)|
+[Guess Number Higher or Lower II](https://leetcode.com/problems/guess-number-higher-or-lower-ii/)| [Go](./DP/GuessNumberHigherOrLowerII.Go)| Medium| O(nlogn)| O(n^2)|
+[Burst Ballons](https://leetcode.com/problems/burst-balloons/)| [Go](./DP/BurstBalloons.Go)| Hard| O(n^3)| O(n)|
+[Frog Jump](https://leetcode.com/problems/frog-jump/)| [Go](./DP/FrogJump.Go)| Hard| O(n^2)| O(n)|
+
+## Depth-first search
+| Title | Solution | Difficulty | Time | Space |
+| ----- | -------- | ---------- | ---- | ----- |
+[Permutations](https://leetcode.com/problems/permutations/)| [Go](./DFS/Permutations.Go)| Medium| O(n!)| O(n)|
+[Permutations II](https://leetcode.com/problems/permutations-ii/)| [Go](./DFS/PermutationsII.Go)| Medium| O(n!)| O(n)|
+[Subsets](https://leetcode.com/problems/subsets/)| [Go](./DFS/Subsets.Go)| Medium| O(n!)| O(n)|
+[Subsets II](https://leetcode.com/problems/subsets-ii/)| [Go](./DFS/SubsetsII.Go)| Medium| O(n!)| O(n)|
+[Combinations](https://leetcode.com/problems/combinations/)| [Go](./DFS/Combinations.Go)| Medium| O(n!)| O(n)|
+[Combination Sum](https://leetcode.com/problems/combination-sum/)| [Go](./DFS/CombinationSum.Go)| Medium| O(n^n)| O(2^n - 1)|
+[Combination Sum II](https://leetcode.com/problems/combination-sum-ii/)| [Go](./DFS/CombinationSumII.Go)| Medium| O(n!)| O(2^n - 2)|
+[Combination Sum III](https://leetcode.com/problems/combination-sum-iii/)| [Go](./DFS/CombinationSumIII.Go)| Medium| O(n!)| O(nCk)|
+[Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number/)| [Go](./DFS/LetterCombinationsPhoneNumber.Go)| Medium| O(mn)| O(n)|
+[Factor Combinations](https://leetcode.com/problems/factor-combinations/)| [Go](./DFS/FactorCombinations.Go)| Medium| O(n^n))| O(2^n - 1)|
+[Generalized Abbreviation](https://leetcode.com/problems/generalized-abbreviation/)| [Go](./DFS/GeneralizedAbbreviation.Go)| Medium| O(n!)| O(2^n)|
+[Palindrome Partitioning](https://leetcode.com/problems/palindrome-partitioning/)| [Go](./DFS/PalindromePartitioning.Go)| Medium| O(n!)| O(n)|
+[Number of Islands](https://leetcode.com/problems/number-of-islands/)| [Go](./DFS/NumberofIslands.Go)| Medium| O((mn)^2)| O(1)|
+[Walls and Gates](https://leetcode.com/problems/walls-and-gates/)| [Go](./DFS/WallsGates.Go)| Medium| O(n!)| O(2^n)|
+[Word Search](https://leetcode.com/problems/word-search/)| [Go](./DFS/WordSearch.Go)| Medium| O((n^2)!)| O(n^2)|
+[Word Search II](https://leetcode.com/problems/word-search-ii/)| [Go](./DFS/WordSearchII.Go)| Hard| O(((mn)^2))| O(n^2)|
+[Add and Search Word - Data structure design](https://leetcode.com/problems/add-and-search-word-data-structure-design/)| [Go](./DFS/WordDictionary.Go)| Medium| O(n)| O(n)|
+[N-Queens](https://leetcode.com/problems/n-queens/)| [Go](./DFS/NQueens.Go)| Hard| O((n^4))| O(n^2)|
+[N-Queens II](https://leetcode.com/problems/n-queens-ii/)| [Go](./DFS/NQueensII.Go)| Hard| O((n^3))| O(n)|
+[Sudoku Solver](https://leetcode.com/problems/sudoku-solver/)| [Go](./DFS/SudokuSolver.Go)| Hard| O(n^4)| O(1)|
+[Remove Invalid Parentheses](https://leetcode.com/problems/remove-invalid-parentheses/)| [Go](./DFS/RemoveInvalidParentheses.Go)| Hard| O(n!)| O(n)|
+[Expression Add Operators](https://leetcode.com/problems/expression-add-operators/)| [Go](./DFS/ExpressionAddOperators.Go)| Hard| O(n!)| O(n)|
+
+## Math
+
+| Title | Solution | Difficulty | Time | Space |
+| ----- | -------- | ---------- | ---- | ----- |
+[Add Binary](https://leetcode.com/problems/add-binary/)| [Go](./Math/AddBinary.Go)| Easy| O(n)| O(n)|
+[Add Two Numbers](https://leetcode.com/problems/add-two-numbers/)| [Go](./Math/AddTwoNumbers.Go)| Medium| O(n)| O(1)|
+[Add Digits](https://leetcode.com/problems/add-digits/)| [Go](./Math/AddDigits.Go)| Easy| O(1)| O(1)|
+[Plus One](https://leetcode.com/problems/plus-one/)| [Go](./Math/PlusOne.Go)| Easy| O(n)| O(1)|
+[Divide Two Integers](https://leetcode.com/problems/divide-two-integers/)| [Go](./Math/DivideTwoIntegers.Go)| Medium| O(logn)| O(1)|
+[Number Complement](https://leetcode.com/problems/number-complement/)| [Go](./Math/NumberComplement.Go)| Easy| O(n)| O(1)|
+[Hamming Distance](https://leetcode.com/problems/hamming-distance/)| [Go](./Math/HammingDistance.Go)| Easy| O(n)| O(1)|
+[Integer Break](https://leetcode.com/problems/integer-break/)| [Go](./Math/IntegerBreak.Go)| Medium| O(logn)| O(1)|
+[Happy Number](https://leetcode.com/problems/happy-number/)| [Go](./Math/HappyNumber.Go)| Easy| O(n)| O(n)|
+[Single Number](https://leetcode.com/problems/single-number/)| [Go](./Math/SingleNumber.Go)| Medium| O(n)| O(1)|
+[Ugly Number](https://leetcode.com/problems/ugly-number/)| [Go](./Math/UglyNumber.Go)| Easy| O(logn)| O(1)|
+[Ugly Number II](https://leetcode.com/problems/ugly-number-ii/)| [Go](./Math/UglyNumberII.Go)| Medium| O(n)| O(n)|
+[Super Ugly Number](https://leetcode.com/problems/super-ugly-number/)| [Go](./Math/SuperUglyNumber.Go)| Medium| O(n^2)| O(n)|
+[Count Primes](https://leetcode.com/problems/count-primes/)| [Go](./Math/CountPrimes.Go)| Easy| O(n)| O(n)|
+[String to Integer (atoi)](https://leetcode.com/problems/string-to-integer-atoi/)| [Go](./Math/Atoi.Go)| Easy| O(n)| O(1)|
+[Pow(x, n)](https://leetcode.com/problems/isomorphic-strings/)| [Go](./Math/Pow.Go)| Medium| O(logn)| O(1)|
+[Power of Two](https://leetcode.com/problems/power-of-two/)| [Go](./Math/PowerTwo.Go)| Easy| O(1)| O(1)|
+[Power of Three](https://leetcode.com/problems/power-of-three/)| [Go](./Math/PowerThree.Go)| Easy| O(1)| O(1)|
+[Super Power](https://leetcode.com/problems/super-pow/)| [Go](./Math/SuperPow.Go)| Medium| O(n)| O(1)|
+[Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)| [Go](./Math/SumTwoIntegers.Go)| Easy| O(n)| O(1)|
+[Reverse Integer](https://leetcode.com/problems/reverse-integer/)| [Go](./Math/ReverseInteger.Go)| Easy| O(n)| O(1)|
+[Excel Sheet Column Number](https://leetcode.com/problems/excel-sheet-column-number/)| [Go](./Math/ExcelSheetColumnNumber.Go)| Easy| O(n)| O(1)|
+[Integer to Roman](https://leetcode.com/problems/integer-to-roman/)| [Go](./Math/IntegerToRoman.Go)| Medium| O(n)| O(1)|
+[Roman to Integer](https://leetcode.com/problems/roman-to-integer/)| [Go](./Math/RomanToInteger.Go)| Easy| O(n)| O(n)|
+[Integer to English Words](https://leetcode.com/problems/integer-to-english-words/)| [Go](./Math/IntegerEnglishWords.Go)| Hard| O(n)| O(1)|
+[Rectangle Area](https://leetcode.com/problems/rectangle-area/)| [Go](./Math/RectangleArea.Go)| Easy| O(1)| O(1)|
+[Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water/)| [Go](./Math/TrappingRainWater.Go)| Hard| O(n)| O(n)|
+[Container With Most Water](https://leetcode.com/problems/container-with-most-water/)| [Go](./Math/ContainerMostWater.Go)| Medium| O(n)| O(1)|
+[Counting Bits](https://leetcode.com/problems/counting-bits/)| [Go](./Math/CountingBits.Go)| Medium| O(n)| O(n)|
+[K-th Smallest in Lexicographical Order](https://leetcode.com/problems/k-th-smallest-in-lexicographical-order/)| [Go](./Math/KthSmallestLexicographicalOrder.Go)| Hard| O(n)| O(1)|
+
+## Search
+
+| Title | Solution | Difficulty | Time | Space |
+| ----- | -------- | ---------- | ---- | ----- |
+[Closest Binary Search Tree Value](https://leetcode.com/problems/closest-binary-search-tree-value/)| [Go](./Search/ClosestBinarySearchTreeValue.Go)| Easy| O(logn)| O(1)|
+[Closest Binary Search Tree Value II](https://leetcode.com/problems/closest-binary-search-tree-value-ii/)| [Go](./Search/ClosestBinarySearchTreeValueII.Go)| Hard| O(n)| O(n)|
+[Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)| [Go](./Search/SearchInRotatedSortedArray.Go)| Hard| O(logn)| O(1)|
+[Search in Rotated Sorted Array II](https://leetcode.com/problems/search-in-rotated-sorted-array-ii/)| [Go](./Search/SearchInRotatedSortedArrayII.Go)| Medium| O(logn)| O(1)|
+[Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)| [Go](./Search/FindMinimumRotatedSortedArray.Go)| Medium| O(logn)| O(1)|
+[Find Minimum in Rotated Sorted Array II](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/)| [Go](./Search/FindMinimumRotatedSortedArrayII.Go)| Hard| O(logn)| O(1)|
+[Search a 2D Matrix](https://leetcode.com/problems/search-a-2d-matrix/)| [Go](./Search/Search2DMatrix.Go)| Medium| O(log(m + n))| O(1)|
+[Search a 2D Matrix II](https://leetcode.com/problems/search-a-2d-matrix-ii/)| [Go](./Search/Search2DMatrixII.Go)| Medium| O(m + n)| O(1)|
+[Search for a Range](https://leetcode.com/problems/search-for-a-range/)| [Go](./Search/SearchForARange.Go)| Medium| O(logn)| O(1)|
+[Search Insert Position](https://leetcode.com/problems/search-insert-position/)| [Go](./Search/SearchForARange.Go)| Medium| O(logn)| O(1)|
+[Find Peak Element](https://leetcode.com/problems/find-peak-element/)| [Go](./Search/FindPeakElement.Go)| Medium| O(logn)| O(1)|
+[Sqrt(x)](https://leetcode.com/problems/sqrtx/)| [Go](./Search/Sqrtx.Go)| Medium| O(logn)| O(1)|
+[Median of Two Sorted Arrays](https://leetcode.com/problems/median-of-two-sorted-arrays/)| [Go](./Search/MedianTwoSortedArrays.Go)| Hard| O(log(m + n))| O(1)|
+
+
+## Sort (已全部做完)
+
+- 深刻的理解多路快排。第 75 题。
+- 链表的排序，插入排序(第 147 题)和归并排序(第 148 题)
+- 桶排序和基数排序。第 164 题。
+- "摆动排序"。第 324 题。
+- 两两不相邻的排序。第 767 题，第 1054 题。
+- "饼子排序"。第 969 题。
+
+| Title | Solution | Difficulty | Time | Space | 收藏 |
+| ----- | :--------: | :----------: | :----: | :-----: |:-----: |
+|[56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0056.%20Merge%20Intervals)| Medium | O(n log n)| O(log n)||
+|[57. Insert Interval](https://leetcode.com/problems/insert-interval/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0057.%20Insert%20Interval)| Hard | O(n)| O(1)||
+|[75. Sort Colors](https://leetcode.com/problems/sort-colors/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0075.%20Sort%20Colors)| Medium| O(n)| O(1)|❤️|
+|[147. Insertion Sort List](https://leetcode.com/problems/insertion-sort-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0147.%20Insertion%20Sort%20List)| Medium | O(n)| O(1) |❤️|
+|[148. Sort List](https://leetcode.com/problems/sort-list/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0148.%20Sort%20List)| Medium |O(n log n)| O(log n)|❤️|
+|[164. Maximum Gap](https://leetcode.com/problems/maximum-gap/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0164.%20Maximum%20Gap)| Hard | O(n log n)| O(log n) |❤️|
+|[179. Largest Number](https://leetcode.com/problems/largest-number/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0179.%20Largest%20Number)| Medium | O(n log n)| O(log n) |❤️|
+|[220. Contains Duplicate III](https://leetcode.com/problems/contains-duplicate-iii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0220.%20Contains%20Duplicate%20III)| Medium | O(n^2)| O(1) ||
+|[242. Valid Anagram](https://leetcode.com/problems/valid-anagram/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0242.%20Valid%20Anagram)| Easy | O(n)| O(n) ||
+|[274. H-Index](https://leetcode.com/problems/h-index/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0274.%20H-Index)| Medium |  O(n)| O(n) ||
+|[324. Wiggle Sort II](https://leetcode.com/problems/wiggle-sort-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0324.%20Wiggle%20Sort%20II)| Medium| O(n)| O(n)|❤️|
+|[349. Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0349.%20Intersection%20of%20Two%20Arrays)| Easy | O(n)| O(n) ||
+|[350. Intersection of Two Arrays II](https://leetcode.com/problems/intersection-of-two-arrays-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0350.%20Intersection%20of%20Two%20Arrays%20II)| Easy | O(n)| O(n) ||
+|[524. Longest Word in Dictionary through Deleting](https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0524.%20Longest%20Word%20in%20Dictionary%20through%20Deleting)| Medium | O(n)| O(1) ||
+|[767. Reorganize String](https://leetcode.com/problems/reorganize-string/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0767.%20Reorganize%20String)| Medium | O(n log n)| O(log n)  |❤️|
+|[853. Car Fleet](https://leetcode.com/problems/car-fleet/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0853.%20Car%20Fleet)| Medium | O(n log n)| O(log n)  ||
+|[710. Random Pick with Blacklist](https://leetcode.com/problems/random-pick-with-blacklist/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0710.%20Random%20Pick%20with%20Blacklist)| Hard | O(n)| O(n)  ||
+|[922. Sort Array By Parity II](https://leetcode.com/problems/sort-array-by-parity-ii/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0922.%20Sort%20Array%20By%20Parity%20II)| Easy | O(n)| O(1) ||
+|[969. Pancake Sorting](https://leetcode.com/problems/pancake-sorting/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0969.%20Pancake%20Sorting)| Medium | O(n log n)| O(log n) |❤️|
+|[973. K Closest Points to Origin](https://leetcode.com/problems/k-closest-points-to-origin/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0973.%20K%20Closest%20Points%20to%20Origin)| Medium | O(n log n)| O(log n) ||
+|[976. Largest Perimeter Triangle](https://leetcode.com/problems/largest-perimeter-triangle/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0976.%20Largest%20Perimeter%20Triangle)| Easy | O(n log n)| O(log n) ||
+|[1030. Matrix Cells in Distance Order](https://leetcode.com/problems/matrix-cells-in-distance-order/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1030.%20Matrix%20Cells%20in%20Distance%20Order)| Easy | O(n^2)| O(1) ||
+|[1054. Distant Barcodes](https://leetcode.com/problems/distant-barcodes/)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/1054.%20Distant%20Barcodes)| Medium | O(n log n)| O(log n) ||
+|-----------------------------------------------------------------|-------------|-------------| --------------------------| --------------------------|-------------|
+
+
+## Union Find
+| Title | Solution | Difficulty | Time | Space |
+| ----- | -------- | ---------- | ---- | ----- |
+[Number of Connected Components in an Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/)| [Go](./UnionFind/NumberConnectedComponentsUndirectedGraph.Go)| Medium| O(nlogn)| O(n)|
+[Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)| [Go](./UnionFind/GraphValidTree.Go)| Medium| O(nlogn)| O(n)|
+
+## Google
+| Title | Solution | Difficulty | Frequency |
+| ----- | -------- | ---------- | --------- |
+[Plus One](https://leetcode.com/problems/plus-one/)| [Go](./Math/PlusOne.Go)| Easy| ★★★★★★|
+[Number of Islands](https://leetcode.com/problems/number-of-islands/)| [Go](./DFS/NumberofIslands.Go)| Medium| ★★★★|
+[Summary Ranges](https://leetcode.com/problems/summary-ranges/)| [Go](./Array/SummaryRanges.Go)| Medium| ★★★★|
+[Perfect Squares](https://leetcode.com/problems/perfect-squares/)| [Go](./DP/PerfectSquares.Go)| Medium| ★★★★|
+[Merge Intervals](https://leetcode.com/problems/merge-intervals/)| [Go](./Sort/MergeIntervals.Go)| Hard| ★★★|
+[Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)| [Go](./Stack/ValidParentheses.Go)| Easy| ★★★|
+[Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water/)| [Go](./Math/TrappingRainWater.Go)| Hard| ★★|
+[Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)| [Go](./LinkedList/MergeKSortedLists.Go)| Hard| ★★|
+[Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/)| [Go](./Array/LongestConsecutiveSequence.Go)| Hard| ★★|
+[Find Peak Element](https://leetcode.com/problems/find-peak-element/)| [Go](./Search/FindPeakElement.Go)| Medium| ★★|
+[Power of Two](https://leetcode.com/problems/power-of-two/)| [Go](./Math/PowerTwo.Go)| Easy| ★★|
+[Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)| [Go](./Array/SpiralMatrix.Go)| Medium| ★★|
+[Sliding Window Maximum ](https://leetcode.com/problems/sliding-window-maximum/)| [Go](./Array/SlidingWindowMaximum.Go)| Hard| ★★|
+[Pow(x, n)](https://leetcode.com/problems/isomorphic-strings/)| [Go](./Math/Pow.Go)| Medium| ★★|
+[Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number/)| [Go](./DFS/LetterCombinationsPhoneNumber.Go)| Medium| ★★|
+[Heaters](https://leetcode.com/problems/heaters/)| [Go](./Array/Heaters.Go)| Easy| ★|
+
+## Facebook
+| Title | Solution | Difficulty | Frequency |
+| ----- | -------- | ---------- | --------- |
+[3Sum](https://leetcode.com/problems/3sum/)| [Go](./Array/ThreeSum.Go)| Medium| ★★★★★★|
+[Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)| [Go](./String/ValidPalindrome.Go)| Easy| ★★★★★★|
+[Valid Palindrome II](https://leetcode.com/problems/valid-palindrome-ii/)| [Go](./String/ValidPalindromeII.Go)| Easy| ★★★★★★|
+[Move Zeroes](https://leetcode.com/problems/move-zeroes/)| [Go](./Array/MoveZeroes.Go)| Easy| ★★★★★★|
+[Remove Invalid Parentheses](https://leetcode.com/problems/remove-invalid-parentheses/)| [Go](./DFS/RemoveInvalidParentheses.Go)| Hard| ★★★★★★|
+[Add Binary](https://leetcode.com/problems/add-binary/)| [Go](./Math/AddBinary.Go)| Easy| ★★★★★|
+[Two Sum](https://leetcode.com/problems/two-sum/)| [Go](./Array/TwoSum.Go)| Easy| ★★★★★|
+[Bnary Tree Paths](https://leetcode.com/problems/binary-tree-paths/)| [Go](./Tree/BnaryTreePaths.Go)| Easy| ★★★★|
+[Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number/)| [Go](./DFS/LetterCombinationsPhoneNumber.Go)| Medium| ★★★★|
+[Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)| [Go](./LinkedList/MergeKSortedLists.Go)| Hard| ★★★★|
+[Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)| [Go](./LinkedList/ReverseLinkedList.Go)| Easy| ★★★|
+[Merge Intervals](https://leetcode.com/problems/merge-intervals/)| [Go](./Sort/MergeIntervals.Go)| Hard| ★★★|
+[Number of Islands](https://leetcode.com/problems/number-of-islands/)| [Go](./DFS/NumberofIslands.Go)| Medium| ★★★|
+[Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)| [Go](./LinkedList/ReverseLinkedList.Go)| Easy| ★★★|
+[Expression Add Operators](https://leetcode.com/problems/expression-add-operators/)| [Go](./DFS/ExpressionAddOperators.Go)| Hard| ★★★|
+[Subsets](https://leetcode.com/problems/subsets/)| [Go](./DFS/Subsets.Go)| Medium| ★★★|
+[Sort Colors](https://leetcode.com/problems/sort-colors/)| [Go](./Sort/SortColors.Go)| Medium| ★★|
+
+## Snapchat
+| Title | Solution | Difficulty | Frequency |
+| ----- | -------- | ---------- | --------- |
+[Game of Life](https://leetcode.com/problems/game-of-life/)	|	[Go](./Array/GameLife.Go)| Medium| ★★★★★★|
+[Meeting Rooms II](https://leetcode.com/problems/meeting-rooms-ii/)| [Go](./Sort/MeetingRoomsII.Go)| Medium| ★★★★★★|
+[Valid Sudoku](https://leetcode.com/problems/valid-sudoku/)| [Go](./Array/ValidSudoku.Go)| Easy| ★★★★★|
+[Binary Tree Vertical Order Traversal](https://leetcode.com/problems/binary-tree-vertical-order-traversal/)| [Go](./Tree/BinaryTreeVerticalOrderTraversal.Go)| Medium| ★★★★|
+[Alien Dictionary](https://leetcode.com/problems/alien-dictionary/)| [Go](./Sort/AlienDictionary.Go)| Hard| ★★★★|
+[One Edit Distance](https://leetcode.com/problems/one-edit-distance/)| [Go](./String/OneEditDistance.Go)| Medium| ★★★|
+[Sudoku Solver](https://leetcode.com/problems/sudoku-solver/)| [Go](./Math/SudokuSolver.Go)| Hard| ★★★|
+[Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)| [Go](./LinkedList/ReverseLinkedList.Go)| Easy| ★★|
+[Unique Binary Search Trees](https://leetcode.com/problems/unique-binary-search-trees/)| [Go](./Tree/UniqueBinarySearchTrees.Go)| Medium| ★★|
+[Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)| [Go](./Array/MinimumWindowSubstring.Go)| Hard| ★★|
+[Remove K Digits](https://leetcode.com/problems/remove-k-digits/)| [Go](./Stack/RemoveKDigits.Go)| Medium| ★|
+[Ternary Expression Parser](https://leetcode.com/problems/ternary-expression-parser/)| [Go](./Stack/TernaryExpressionParser.Go)| Medium| ★|
+
+## Uber
+| Title | Solution | Difficulty | Frequency |
+| ----- | -------- | ---------- | --------- |
+[Valid Sudoku](https://leetcode.com/problems/valid-sudoku/)| [Go](./Array/ValidSudoku.Go)| Easy| ★★★★|
+[Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)| [Go](./Array/SpiralMatrix.Go)| Medium| ★★★★|
+[Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number/)| [Go](./DFS/LetterCombinationsPhoneNumber.Go)| Medium| ★★★★|
+[Group Anagrams](https://leetcode.com/problems/anagrams/)| [Go](./String/GroupAnagrams.Go)| Medium| ★★★★|
+[Word Pattern](https://leetcode.com/problems/word-pattern/)| [Go](./String/WordPattern.Go)| Easy| ★★★|
+[Roman to Integer](https://leetcode.com/problems/roman-to-integer/)| [Go](./Math/RomanToInteger.Go)| Easy| ★★★|
+[Combination Sum](https://leetcode.com/problems/combination-sum/)| [Go](./DFS/CombinationSum.Go)| Medium| ★★|
+
+## Airbnb
+| Title | Solution | Difficulty | Frequency |
+| ----- | -------- | ---------- | --------- |
+[Two Sum](https://leetcode.com/problems/two-sum/)| [Go](./Array/TwoSum.Go)| Easy| ★★★★★|
+[Text Justification](https://leetcode.com/problems/text-justification/)| [Go](./String/TextJustification.Go)| Hard| ★★★★|
+[House Robber](https://leetcode.com/problems/house-robber/)| [Go](./DP/HouseRobber.Go)| Easy| ★★|
+[Single Number](https://leetcode.com/problems/single-number/)| [Go](./Math/SingleNumber.Go)| Medium| ★★|
+[Word Search II](https://leetcode.com/problems/word-search-ii/)| [Go](./DFS/WordSearchII.Go)| Hard| ★★|
+[Add Two Numbers](https://leetcode.com/problems/add-two-numbers/)| [Go](./Math/AddTwoNumbers.Go)| Medium| ★★|
+
+## LinkedIn
+| Title | Solution | Difficulty | Frequency |
+| ----- | -------- | ---------- | --------- |
+[Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)| [Go](./DP/MaximumSubarray.Go)| Medium| ★★★★★★|
+[Pow(x, n)](https://leetcode.com/problems/isomorphic-strings/)| [Go](./Math/Pow.Go)| Medium| ★★★★★★|
+[Merge Intervals](https://leetcode.com/problems/merge-intervals/)| [Go](./Sort/MergeIntervals.Go)| Hard| ★★★★★★|
+[Isomorphic Strings](https://leetcode.com/problems/isomorphic-strings/)| [Go](./String/IsomorphicStrings.Go)| Easy| ★★★★★★|
+[Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)| [Go](./Search/SearchInRotatedSortedArray.Go)| Hard| ★★★★★|
+[Search for a Range](https://leetcode.com/problems/search-for-a-range/)| [Go](./Search/SearchForARange.Go)| Medium| ★★★★★|
+[Two Sum](https://leetcode.com/problems/two-sum/)| [Go](./Array/TwoSum.Go)| Easy| ★★★★|
+[Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/)| [Go](./Tree/BinaryTreeLevelOrderTraversal.Go)| Easy| ★★★★|
+[Evaluate Reverse Polish Notation](https://leetcode.com/problems/evaluate-reverse-polish-notation/)| [Go](./Stack/EvaluateReversePolishNotation.Go)| Medium| ★★★|
+[Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)| [Go](./DP/MaximumProductSubarray.Go)| Medium| ★★★|
+[Product of Array Except Self](https://leetcode.com/problems/product-of-array-except-self/)| [Go](./Array/ProductExceptSelf.Go)| Medium| ★★★|
+[Symmetric Tree](https://leetcode.com/problems/symmetric-tree/)| [Go](./Tree/SymmetricTree.Go)| Easy| ★★|
+
+## Amazon
+| Title | Solution | Difficulty | Frequency |
+| ----- | -------- | ---------- | --------- |
+[Two Sum](https://leetcode.com/problems/two-sum/)| [Go](./Array/TwoSum.Go)| Easy| ★★★★★★|
+[Min Cost Climbing Stairs](https://leetcode.com/problems/min-cost-climbing-stairs/)| [Go](./DP/MinCostClimbingStairs.Go)| Easy| ★★★★|
+[Number of Islands](https://leetcode.com/problems/number-of-islands/)| [Go](./DFS/NumberofIslands.Go)| Medium| ★★|
+[Add Two Numbers](https://leetcode.com/problems/add-two-numbers/)| [Go](./Math/AddTwoNumbers.Go)| Medium| ★★|
+[Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)| [Go](./LinkedList/ReverseLinkedList.Go)| Easy| ★★|
+[Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)| [Go](./Stack/ValidParentheses.Go)| Easy| ★★|
+[Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)| [Go](./DP/LongestPalindromicSubstring.Go)| Medium| ★★|
+[Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water/)| [Go](./Math/TrappingRainWater.Go)| Hard| ★★|
+[Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/)| [Go](./String/LongestSubstringWithoutRepeatingCharacters.Go)| Medium| ★★|
+[Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number/)| [Go](./DFS/LetterCombinationsPhoneNumber.Go)| Medium| ★★|
+[Valid Anagram](https://leetcode.com/problems/valid-anagram/)| [Go](./String/ValidAnagram.Go)| Easy| ★★|
+[Rotate Image](https://leetcode.com/problems/rotate-image/)| [Go](./Array/RotateImage.Go)| Medium| ★★|
+[Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)| [Go](./DP/BestTimeBuySellStock.Go)| Easy| ★★|
+[3Sum](https://leetcode.com/problems/3sum/)| [Go](./Array/ThreeSum.Go)| Medium| ★★|
+[Sliding Window Maximum ](https://leetcode.com/problems/sliding-window-maximum/)| [Go](./Array/SlidingWindowMaximum.Go)| Hard| ★★|
+
+## Microsoft
+| Title | Solution | Difficulty | Frequency |
+| ----- | -------- | ---------- | --------- |
+[Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)| [Go](./LinkedList/ReverseLinkedList.Go)| Easy| ★★★★★★|
+[Two Sum](https://leetcode.com/problems/two-sum/)| [Go](./Array/TwoSum.Go)| Easy| ★★★★★|
+[String to Integer (atoi)](https://leetcode.com/problems/string-to-integer-atoi/)| [Go](./Math/Atoi.Go)| Easy| ★★★★|
+[Add Two Numbers](https://leetcode.com/problems/add-two-numbers/)| [Go](./Math/AddTwoNumbers.Go)| Medium| ★★★★|
+[Excel Sheet Column Number](https://leetcode.com/problems/excel-sheet-column-number/)| [Go](./Math/ExcelSheetColumnNumber.Go)| Easy| ★★★★|
+[Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)| [Go](./Tree/ValidateBinarySearchTree.Go)| Medium| ★★★|
+[Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)| [Go](./LinkedList/MergeTwoSortedLists.Go)| Easy| ★★★|
\ No newline at end of file
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diff --git a/Topics/Binary_Indexed_Tree.png b/Topics/Binary_Indexed_Tree.png
deleted file mode 100644
index 446594e56..000000000
Binary files a/Topics/Binary_Indexed_Tree.png and /dev/null differ
diff --git a/Topics/Bit_Manipulation.png b/Topics/Bit_Manipulation.png
deleted file mode 100644
index 81b0eca99..000000000
Binary files a/Topics/Bit_Manipulation.png and /dev/null differ
diff --git a/Topics/Linked_List.png b/Topics/Linked_List.png
deleted file mode 100644
index d7abbeedb..000000000
Binary files a/Topics/Linked_List.png and /dev/null differ
diff --git a/Topics/Sliding_Window.png b/Topics/Sliding_Window.png
deleted file mode 100644
index 24b992a83..000000000
Binary files a/Topics/Sliding_Window.png and /dev/null differ
diff --git a/Topics/Sort.png b/Topics/Sort.png
deleted file mode 100644
index 83c787da4..000000000
Binary files a/Topics/Sort.png and /dev/null differ
diff --git a/Topics/Stack.png b/Topics/Stack.png
deleted file mode 100644
index 9dfa95195..000000000
Binary files a/Topics/Stack.png and /dev/null differ
diff --git a/Topics/Two_pointers.png b/Topics/Two_pointers.png
deleted file mode 100644
index 54d03e5df..000000000
Binary files a/Topics/Two_pointers.png and /dev/null differ
diff --git a/Topics/Union_Find.png b/Topics/Union_Find.png
deleted file mode 100644
index 0a3fef998..000000000
Binary files a/Topics/Union_Find.png and /dev/null differ
diff --git a/ctl/README.md b/ctl/README.md
new file mode 100644
index 000000000..cbbc8327f
--- /dev/null
+++ b/ctl/README.md
@@ -0,0 +1,20 @@
+# LeetCode-Go ctl
+
+## 配置方法
+
+1. 在`.gitignore`中，添加一行`*.toml`
+2. 在`LeetCode-Go`目录下，添加文本文件`config.toml`。
+3. 把以下内容复制到`config.toml`中。
+4. 把`config.toml`中的`test`分别修改为你的 leetcode `用户名`和`密码`。
+5. 去 leetcode 登录后，把网站 cookie 复制 (需要复制 csrftoken 和 LEETCODE_SESSION) 并替换 `config.toml`中的`Cookie`。
+
+```toml
+Username="test"
+Password="test"
+Cookie="csrftoken=XXXXXXXXX; LEETCODE_SESSION=YYYYYYYY;"
+CSRFtoken="ZZZZZZZZ"
+```
+
+## PDF 生成
+
+用 `leetcode-go pdf` 命令先生成书籍内容的合并版本 pdf.md，再用 vscode 或者其他支持 toc 目录生成的工具，生成 toc。最后用 Typora 把 md 文件转换成 pdf。就可以发布 release 新版本了。
\ No newline at end of file
diff --git a/ctl/command.go b/ctl/command.go
new file mode 100644
index 000000000..a440a5710
--- /dev/null
+++ b/ctl/command.go
@@ -0,0 +1,30 @@
+package main
+
+import (
+	"fmt"
+	"os"
+
+	"github.com/spf13/cobra"
+)
+
+var rootCmd = &cobra.Command{
+	Use:   "leetcode-go",
+	Short: "A simple command line client for leetcode-go.",
+}
+
+func execute() {
+	if err := rootCmd.Execute(); err != nil {
+		fmt.Println(err)
+		os.Exit(-1)
+	}
+}
+
+func init() {
+	rootCmd.AddCommand(
+		versionCmd,
+		newBuildCommand(),
+		newLabelCommand(),
+		newPDFCommand(),
+		newRefresh(),
+	)
+}
diff --git a/ctl/config.go b/ctl/config.go
new file mode 100644
index 000000000..4288c3d25
--- /dev/null
+++ b/ctl/config.go
@@ -0,0 +1,32 @@
+package main
+
+import (
+	"fmt"
+	"log"
+
+	"github.com/BurntSushi/toml"
+)
+
+const (
+	configTOML = "config.toml"
+)
+
+type config struct {
+	Username  string
+	Password  string
+	Cookie    string
+	CSRFtoken string
+}
+
+func (c config) String() string {
+	return fmt.Sprintf("Username: %s, Password: %s", c.Username, c.Password)
+}
+
+func getConfig() *config {
+	cfg := new(config)
+	if _, err := toml.DecodeFile(configTOML, &cfg); err != nil {
+		log.Panicf(err.Error())
+	}
+	// log.Printf("get config: %s", cfg)
+	return cfg
+}
diff --git a/ctl/error.go b/ctl/error.go
new file mode 100644
index 000000000..903460a83
--- /dev/null
+++ b/ctl/error.go
@@ -0,0 +1,21 @@
+package main
+
+import (
+	"fmt"
+	"os"
+)
+
+const (
+	// ExitSuccess define
+	ExitSuccess = iota
+	// ExitError define
+	ExitError
+	// ExitBadArgs define
+	ExitBadArgs
+)
+
+// ExitWithError define
+func ExitWithError(code int, err error) {
+	fmt.Fprintln(os.Stderr, "Error: ", err)
+	os.Exit(code)
+}
diff --git a/ctl/label.go b/ctl/label.go
new file mode 100644
index 000000000..db94bf720
--- /dev/null
+++ b/ctl/label.go
@@ -0,0 +1,348 @@
+package main
+
+import (
+	"bufio"
+	"errors"
+	"fmt"
+	"io"
+	"io/ioutil"
+	"os"
+	"regexp"
+	"strings"
+
+	"github.com/halfrost/LeetCode-Go/ctl/util"
+	"github.com/spf13/cobra"
+)
+
+var (
+	chapterOneFileOrder = []string{"_index", "Data_Structure", "Algorithm", "Time_Complexity"}
+	chapterOneMenuOrder = []string{"_index", "#关于作者", "Data_Structure", "Algorithm", "Time_Complexity"}
+	chapterTwoFileOrder = []string{"_index", "Array", "String", "Two_Pointers", "Linked_List", "Stack", "Tree", "Dynamic_Programming", "Backtracking", "Depth_First_Search", "Breadth_First_Search",
+		"Binary_Search", "Math", "Hash_Table", "Sorting", "Bit_Manipulation", "Union_Find", "Sliding_Window", "Segment_Tree", "Binary_Indexed_Tree"}
+	chapterThreeFileOrder = []string{"_index", "Segment_Tree", "UnionFind", "LRUCache", "LFUCache"}
+	preNextHeader         = "----------------------------------------------\n<div style=\"display: flex;justify-content: space-between;align-items: center;\">\n"
+	preNextFotter         = "</div>"
+	delLine               = "----------------------------------------------\n"
+	delHeader             = "<div style=\"display: flex;justify-content: space-between;align-items: center;\">"
+	delLabel              = "<[a-zA-Z]+.*?>([\\s\\S]*?)</[a-zA-Z]*?>"
+	delFooter             = "</div>"
+
+	//ErrNoFilename is thrown when the path the the file to tail was not given
+	ErrNoFilename = errors.New("You must provide the path to a file in the \"-file\" flag.")
+
+	//ErrInvalidLineCount is thrown when the user provided 0 (zero) as the value for number of lines to tail
+	ErrInvalidLineCount = errors.New("You cannot tail zero lines.")
+
+	chapterMap = map[string]map[string]string{
+		"ChapterOne": {
+			"_index":          "第一章 序章",
+			"Data_Structure":  "1.1 数据结构知识",
+			"Algorithm":       "1.2 算法知识",
+			"Time_Complexity": "1.3 时间复杂度",
+		},
+		"ChapterTwo": {
+			"_index":               "第二章 算法专题",
+			"Array":                "2.01 Array",
+			"String":               "2.02 String",
+			"Two_Pointers":         "2.03 ✅ Two Pointers",
+			"Linked_List":          "2.04 ✅ Linked List",
+			"Stack":                "2.05 ✅ Stack",
+			"Tree":                 "2.06 Tree",
+			"Dynamic_Programming":  "2.07 Dynamic Programming",
+			"Backtracking":         "2.08 ✅ Backtracking",
+			"Depth_First_Search":   "2.09 Depth First Search",
+			"Breadth_First_Search": "2.10 Breadth First Search",
+			"Binary_Search":        "2.11 Binary Search",
+			"Math":                 "2.12 Math",
+			"Hash_Table":           "2.13 Hash Table",
+			"Sorting":              "2.14 ✅ Sorting",
+			"Bit_Manipulation":     "2.15 ✅ Bit Manipulation",
+			"Union_Find":           "2.16 ✅ Union Find",
+			"Sliding_Window":       "2.17 ✅ Sliding Window",
+			"Segment_Tree":         "2.18 ✅ Segment Tree",
+			"Binary_Indexed_Tree":  "2.19 ✅ Binary Indexed Tree",
+		},
+		"ChapterThree": {
+			"_index":       "第三章 一些模板",
+			"Segment_Tree": "3.1 Segment Tree",
+			"UnionFind":    "3.2 UnionFind",
+			"LRUCache":     "3.3 LRUCache",
+			"LFUCache":     "3.4 LFUCache",
+		},
+		"ChapterFour": {
+			"_index": "第四章 Leetcode 题解",
+		},
+	}
+)
+
+func getChapterFourFileOrder() ([]string, []int) {
+	solutions, solutionIds := util.LoadChapterFourDir()
+	chapterFourFileOrder := []string{"_index"}
+	chapterFourFileOrder = append(chapterFourFileOrder, solutions...)
+	fmt.Printf("ChapterFour 中包括 _index 有 %v 个文件, len(id) = %v\n", len(chapterFourFileOrder), len(solutionIds))
+	return chapterFourFileOrder, solutionIds
+}
+
+func newLabelCommand() *cobra.Command {
+	mc := &cobra.Command{
+		Use:   "label <subcommand>",
+		Short: "Label related commands",
+	}
+	//mc.PersistentFlags().StringVar(&logicEndpoint, "endpoint", "localhost:5880", "endpoint of logic service")
+	mc.AddCommand(
+		newAddPreNext(),
+		newDeletePreNext(),
+	)
+	return mc
+}
+
+func newAddPreNext() *cobra.Command {
+	cmd := &cobra.Command{
+		Use:   "add-pre-next",
+		Short: "Add pre-next label",
+		Run: func(cmd *cobra.Command, args []string) {
+			addPreNext()
+		},
+	}
+	// cmd.Flags().StringVar(&alias, "alias", "", "alias")
+	// cmd.Flags().StringVar(&appId, "appid", "", "appid")
+	return cmd
+}
+
+func newDeletePreNext() *cobra.Command {
+	cmd := &cobra.Command{
+		Use:   "del-pre-next",
+		Short: "Delete pre-next label",
+		Run: func(cmd *cobra.Command, args []string) {
+			delPreNext()
+		},
+	}
+	// cmd.Flags().StringVar(&alias, "alias", "", "alias")
+	// cmd.Flags().StringVar(&appId, "appid", "", "appid")
+	return cmd
+}
+
+func addPreNext() {
+	// Chpater one add pre-next
+	addPreNextLabel(chapterOneFileOrder, []string{}, []int{}, "", "ChapterOne", "ChapterTwo")
+	// Chpater two add pre-next
+	addPreNextLabel(chapterTwoFileOrder, chapterOneFileOrder, []int{}, "ChapterOne", "ChapterTwo", "ChapterThree")
+	// Chpater three add pre-next
+	addPreNextLabel(chapterThreeFileOrder, chapterTwoFileOrder, []int{}, "ChapterTwo", "ChapterThree", "ChapterFour")
+	// Chpater four add pre-next
+	//fmt.Printf("%v\n", getChapterFourFileOrder())
+	chapterFourFileOrder, solutionIds := getChapterFourFileOrder()
+	addPreNextLabel(chapterFourFileOrder, chapterThreeFileOrder, solutionIds, "ChapterThree", "ChapterFour", "")
+}
+
+func addPreNextLabel(order, preOrder []string, chapterFourIds []int, preChapter, chapter, nextChapter string) {
+	var (
+		exist bool
+		err   error
+		res   []byte
+		count int
+	)
+	for index, path := range order {
+		tmp := ""
+		if index == 0 {
+			if chapter == "ChapterOne" {
+				// 第一页不需要“上一章”
+				tmp = "\n\n" + delLine + fmt.Sprintf("<p align = \"right\"><a href=\"https://books.halfrost.com/leetcode/%v/%v/\">下一页➡️</a></p>\n", chapter, order[index+1])
+			} else {
+				if chapter == "ChapterFour" {
+					tmp = "\n\n" + preNextHeader + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/\">⬅️上一章</a></p>\n", preChapter, preOrder[len(preOrder)-1]) + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/%v/\">下一页➡️</a></p>\n", chapter, util.GetChpaterFourFileNum(chapterFourIds[(index-1)+1]), order[index+1]) + preNextFotter
+				} else {
+					tmp = "\n\n" + preNextHeader + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/\">⬅️上一章</a></p>\n", preChapter, preOrder[len(preOrder)-1]) + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/\">下一页➡️</a></p>\n", chapter, order[index+1]) + preNextFotter
+				}
+			}
+		} else if index == len(order)-1 {
+			if chapter == "ChapterFour" {
+				// 最后一页不需要“下一页”
+				tmp = "\n\n" + delLine + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/%v/\">⬅️上一页</a></p>\n", chapter, util.GetChpaterFourFileNum(chapterFourIds[(index-1)-1]), order[index-1])
+			} else {
+				tmp = "\n\n" + preNextHeader + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/\">⬅️上一页</a></p>\n", chapter, order[index-1]) + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/\">下一章➡️</a></p>\n", nextChapter) + preNextFotter
+			}
+		} else if index == 1 {
+			if chapter == "ChapterFour" {
+				tmp = "\n\n" + preNextHeader + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/\">⬅️上一页</a></p>\n", chapter) + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/%v/\">下一页➡️</a></p>\n", chapter, util.GetChpaterFourFileNum(chapterFourIds[(index-1)+1]), order[index+1]) + preNextFotter
+			} else {
+				tmp = "\n\n" + preNextHeader + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/\">⬅️上一页</a></p>\n", chapter) + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/\">下一页➡️</a></p>\n", chapter, order[index+1]) + preNextFotter
+			}
+		} else {
+			if chapter == "ChapterFour" {
+				tmp = "\n\n" + preNextHeader + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/%v/\">⬅️上一页</a></p>\n", chapter, util.GetChpaterFourFileNum(chapterFourIds[(index-1)-1]), order[index-1]) + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/%v/\">下一页➡️</a></p>\n", chapter, util.GetChpaterFourFileNum(chapterFourIds[(index-1)+1]), order[index+1]) + preNextFotter
+			} else {
+				tmp = "\n\n" + preNextHeader + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/\">⬅️上一页</a></p>\n", chapter, order[index-1]) + fmt.Sprintf("<p><a href=\"https://books.halfrost.com/leetcode/%v/%v/\">下一页➡️</a></p>\n", chapter, order[index+1]) + preNextFotter
+			}
+		}
+
+		if chapter == "ChapterFour" && index > 0 {
+			path = fmt.Sprintf("%v/%v", util.GetChpaterFourFileNum(chapterFourIds[(index-1)]), path)
+		}
+
+		exist, err = needAdd(fmt.Sprintf("../website/content/%v/%v.md", chapter, path))
+		if err != nil {
+			fmt.Println(err)
+			return
+		}
+		// 当前没有上一页和下一页，才添加
+		if !exist && err == nil {
+			res, err = eofAdd(fmt.Sprintf("../website/content/%v/%v.md", chapter, path), tmp)
+			if err != nil {
+				fmt.Println(err)
+				return
+			}
+			util.WriteFile(fmt.Sprintf("../website/content/%v/%v.md", chapter, path), res)
+			count++
+		}
+	}
+	fmt.Printf("添加了 %v 个文件的 pre-next\n", count)
+}
+
+func eofAdd(filePath string, labelString string) ([]byte, error) {
+	f, err := os.OpenFile(filePath, os.O_RDONLY, 0644)
+	if err != nil {
+		return nil, err
+	}
+	defer f.Close()
+	reader, output := bufio.NewReader(f), []byte{}
+
+	for {
+		line, _, err := reader.ReadLine()
+		if err != nil {
+			if err == io.EOF {
+				output = append(output, []byte(labelString)...)
+				output = append(output, []byte("\n")...)
+				return output, nil
+			}
+			return nil, err
+		}
+		output = append(output, line...)
+		output = append(output, []byte("\n")...)
+	}
+}
+
+func delPreNext() {
+	// Chpater one del pre-next
+	delPreNextLabel(chapterOneFileOrder, []int{}, "ChapterOne")
+	// Chpater two del pre-next
+	delPreNextLabel(chapterTwoFileOrder, []int{}, "ChapterTwo")
+	// Chpater three del pre-next
+	delPreNextLabel(chapterThreeFileOrder, []int{}, "ChapterThree")
+	// Chpater four del pre-next
+	chapterFourFileOrder, solutionIds := getChapterFourFileOrder()
+	delPreNextLabel(chapterFourFileOrder, solutionIds, "ChapterFour")
+}
+
+func delPreNextLabel(order []string, chapterFourIds []int, chapter string) {
+	count := 0
+	for index, path := range order {
+		lineNum := 5
+		if index == 0 && chapter == "ChapterOne" || index == len(order)-1 && chapter == "ChapterFour" {
+			lineNum = 3
+		}
+		if chapter == "ChapterFour" && index > 0 {
+			path = fmt.Sprintf("%v/%v", util.GetChpaterFourFileNum(chapterFourIds[(index-1)]), path)
+		}
+
+		exist, err := needAdd(fmt.Sprintf("../website/content/%v/%v.md", chapter, path))
+		if err != nil {
+			fmt.Println(err)
+			return
+		}
+		// 存在才删除
+		if exist && err == nil {
+			removeLine(fmt.Sprintf("../website/content/%v/%v.md", chapter, path), lineNum+1)
+			count++
+		}
+	}
+	fmt.Printf("删除了 %v 个文件的 pre-next\n", count)
+	// 另外一种删除方法
+	// res, err := eofDel(fmt.Sprintf("../website/content/ChapterOne/%v.md", v))
+	// if err != nil {
+	// 	fmt.Println(err)
+	// 	return
+	// }
+	// util.WriteFile(fmt.Sprintf("../website/content/ChapterOne/%v.md", v), res)
+}
+
+func needAdd(filePath string) (bool, error) {
+	f, err := os.OpenFile(filePath, os.O_RDONLY, 0644)
+	if err != nil {
+		return false, err
+	}
+	defer f.Close()
+	reader, output := bufio.NewReader(f), []byte{}
+	for {
+		line, _, err := reader.ReadLine()
+		if err != nil {
+			if err == io.EOF {
+				return false, nil
+			}
+			return false, err
+		}
+		if ok, _ := regexp.Match(delHeader, line); ok {
+			return true, nil
+		} else if ok, _ := regexp.Match(delLabel, line); ok {
+			return true, nil
+		} else {
+			output = append(output, line...)
+			output = append(output, []byte("\n")...)
+		}
+	}
+}
+
+func eofDel(filePath string) ([]byte, error) {
+	f, err := os.OpenFile(filePath, os.O_RDONLY, 0644)
+	if err != nil {
+		return nil, err
+	}
+	defer f.Close()
+	reader, output := bufio.NewReader(f), []byte{}
+	for {
+		line, _, err := reader.ReadLine()
+		if err != nil {
+			if err == io.EOF {
+				return output, nil
+			}
+			return nil, err
+		}
+		if ok, _ := regexp.Match(delLine, line); ok {
+			reg := regexp.MustCompile(delLine)
+			newByte := reg.ReplaceAll(line, []byte(""))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else if ok, _ := regexp.Match(delHeader, line); ok {
+			reg := regexp.MustCompile(delHeader)
+			newByte := reg.ReplaceAll(line, []byte(""))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else if ok, _ := regexp.Match(delLabel, line); ok {
+			reg := regexp.MustCompile(delLabel)
+			newByte := reg.ReplaceAll(line, []byte(""))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else if ok, _ := regexp.Match(delFooter, line); ok {
+			reg := regexp.MustCompile(delFooter)
+			newByte := reg.ReplaceAll(line, []byte(""))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else {
+			output = append(output, line...)
+			output = append(output, []byte("\n")...)
+		}
+	}
+}
+
+func removeLine(path string, lineNumber int) {
+	file, err := ioutil.ReadFile(path)
+	if err != nil {
+		panic(err)
+	}
+	info, _ := os.Stat(path)
+	mode := info.Mode()
+	array := strings.Split(string(file), "\n")
+	array = array[:len(array)-lineNumber-1]
+	ioutil.WriteFile(path, []byte(strings.Join(array, "\n")), mode)
+	//fmt.Println("remove line successful")
+}
diff --git a/ctl/main.go b/ctl/main.go
new file mode 100644
index 000000000..f7d07b015
--- /dev/null
+++ b/ctl/main.go
@@ -0,0 +1,5 @@
+package main
+
+func main() {
+	execute()
+}
diff --git a/ctl/meta/Array b/ctl/meta/Array
new file mode 100644
index 000000000..6611e480c
--- /dev/null
+++ b/ctl/meta/Array
@@ -0,0 +1,52 @@
+|1. Two Sum| [Go]({{< relref "/ChapterFour/0001.Two-Sum.md" >}})| Easy | O(n)| O(n)||
+|11. Container With Most Water| [Go]({{< relref "/ChapterFour/0011.Container-With-Most-Water.md" >}})| Medium | O(n)| O(1)||
+|15. 3Sum | [Go]({{< relref "/ChapterFour/0015.3Sum.md" >}})| Medium | O(n^2)| O(n)|❤️|
+|16. 3Sum Closest | [Go]({{< relref "/ChapterFour/0016.3Sum-Closest.md" >}})| Medium | O(n^2)| O(1)|❤️|
+|18. 4Sum | [Go]({{< relref "/ChapterFour/0018.4Sum.md" >}})| Medium | O(n^3)| O(n^2)|❤️|
+|26. Remove Duplicates from Sorted Array | [Go]({{< relref "/ChapterFour/0026.Remove-Duplicates-from-Sorted-Array.md" >}})| Easy | O(n)| O(1)||
+|27. Remove Element   | [Go]({{< relref "/ChapterFour/0027.Remove-Element.md" >}})| Easy | O(n)| O(1)||
+|39. Combination Sum    | [Go]({{< relref "/ChapterFour/0039.Combination-Sum.md" >}})| Medium | O(n log n)| O(n)||
+|40. Combination Sum II  | [Go]({{< relref "/ChapterFour/0040.Combination-Sum-II.md" >}})| Medium | O(n log n)| O(n)||
+|41. First Missing Positive | [Go]({{< relref "/ChapterFour/0041.First-Missing-Positive.md" >}})| Hard | O(n)| O(n)||
+|42. Trapping Rain Water | [Go]({{< relref "/ChapterFour/0042.Trapping-Rain-Water.md" >}})| Hard | O(n)| O(1)|❤️|
+|48. Rotate Image | [Go]({{< relref "/ChapterFour/0048.Rotate-Image.md" >}})| Medium | O(n)| O(1)||
+|53. Maximum Subarray| [Go]({{< relref "/ChapterFour/0053.Maximum-Subarray.md" >}})| Easy | O(n)| O(n)||
+|54. Spiral Matrix| [Go]({{< relref "/ChapterFour/0054.Spiral-Matrix.md" >}})| Medium | O(n)| O(n^2)||
+|56. Merge Intervals | [Go]({{< relref "/ChapterFour/0056.Merge-Intervals.md" >}})| Medium | O(n log n)| O(1)||
+|57. Insert Interval | [Go]({{< relref "/ChapterFour/0057.Insert-Interval.md" >}})| Hard | O(n)| O(1)||
+|59. Spiral Matrix II | [Go]({{< relref "/ChapterFour/0059.Spiral-Matrix-II.md" >}})| Medium | O(n)| O(n^2)||
+|62. Unique Paths | [Go]({{< relref "/ChapterFour/0062.Unique-Paths.md" >}})| Medium | O(n^2)| O(n^2)||
+|63. Unique Paths II  | [Go]({{< relref "/ChapterFour/0063.Unique-Paths-II.md" >}})| Medium | O(n^2)| O(n^2)||
+|64. Minimum Path Sum | [Go]({{< relref "/ChapterFour/0064.Minimum-Path-Sum.md" >}})| Medium | O(n^2)| O(n^2)||
+|75. Sort Colors | [Go]({{< relref "/ChapterFour/0075.Sort-Colors.md" >}})| Medium| O(n)| O(1)|❤️|
+|78. Subsets| [Go]({{< relref "/ChapterFour/0078.Subsets.md" >}})| Medium | O(n^2)| O(n)|❤️|
+|79. Word Search  | [Go]({{< relref "/ChapterFour/0079.Word-Search.md" >}})| Medium | O(n^2)| O(n^2)|❤️|
+|80. Remove Duplicates from Sorted Array II| [Go]({{< relref "/ChapterFour/0080.Remove-Duplicates-from-Sorted-Array-II.md" >}})| Medium | O(n)| O(1||
+|84. Largest Rectangle in Histogram | [Go]({{< relref "/ChapterFour/0084.Largest-Rectangle-in-Histogram.md" >}})| Medium | O(n)| O(n)|❤️|
+|88. Merge Sorted Array | [Go]({{< relref "/ChapterFour/0088.Merge-Sorted-Array.md" >}})| Easy | O(n)| O(1)|❤️|
+|90. Subsets II | [Go]({{< relref "/ChapterFour/0090.Subsets-II.md" >}})| Medium | O(n^2)| O(n)|❤️|
+|120. Triangle | [Go]({{< relref "/ChapterFour/0120.Triangle.md" >}})| Medium | O(n^2)| O(n)||
+|121. Best Time to Buy and Sell Stock | [Go]({{< relref "/ChapterFour/0121.Best-Time-to-Buy-and-Sell-Stock.md" >}})| Easy | O(n)| O(1)||
+|122. Best Time to Buy and Sell Stock II | [Go]({{< relref "/ChapterFour/0122.Best-Time-to-Buy-and-Sell-Stock-II.md" >}})| Easy | O(n)| O(1)||
+|126. Word Ladder II  | [Go]({{< relref "/ChapterFour/0126.Word-Ladder-II.md" >}})| Hard | O(n)| O(n^2)|❤️|
+|152. Maximum Product Subarray  | [Go]({{< relref "/ChapterFour/0152.Maximum-Product-Subarray.md" >}})| Medium | O(n)| O(1)||
+|167. Two Sum II - Input array is sorted  | [Go]({{< relref "/ChapterFour/0167.Two-Sum-II---Input-array-is-sorted.md" >}})| Easy | O(n)| O(1)||
+|209. Minimum Size Subarray Sum | [Go]({{< relref "/ChapterFour/0209.Minimum-Size-Subarray-Sum.md" >}})| Medium | O(n)| O(1)||
+|216. Combination Sum III  | [Go]({{< relref "/ChapterFour/0216.Combination-Sum-III.md" >}})| Medium | O(n)| O(1)|❤️|
+|217. Contains Duplicate  | [Go]({{< relref "/ChapterFour/0217.Contains-Duplicate.md" >}})| Easy | O(n)| O(n)||
+|219. Contains Duplicate II  | [Go]({{< relref "/ChapterFour/0219.Contains-Duplicate-II.md" >}})| Easy | O(n)| O(n)||
+|283. Move Zeroes  | [Go]({{< relref "/ChapterFour/0283.Move-Zeroes.md" >}})| Easy | O(n)| O(1)||
+|287. Find the Duplicate Number | [Go]({{< relref "/ChapterFour/0287.Find-the-Duplicate-Number.md" >}})| Easy | O(n)| O(1)|❤️|
+|532. K-diff Pairs in an Array  | [Go]({{< relref "/ChapterFour/0532.K-diff-Pairs-in-an-Array.md" >}})| Easy | O(n)| O(n)||
+|566. Reshape the Matrix  | [Go]({{< relref "/ChapterFour/0566.Reshape-the-Matrix.md" >}})| Easy | O(n^2)| O(n^2)||
+|628. Maximum Product of Three Numbers | [Go]({{< relref "/ChapterFour/0628.Maximum-Product-of-Three-Numbers.md" >}})| Easy | O(n)| O(1)||
+|713. Subarray Product Less Than K | [Go]({{< relref "/ChapterFour/0713.Subarray-Product-Less-Than-K.md" >}})| Medium | O(n)| O(1)||
+|714. Best Time to Buy and Sell Stock with Transaction Fee| [Go]({{< relref "/ChapterFour/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee.md" >}})| Medium | O(n)| O(1)||
+|746. Min Cost Climbing Stairs | [Go]({{< relref "/ChapterFour/0746.Min-Cost-Climbing-Stairs.md" >}})| Easy | O(n)| O(1)||
+|766. Toeplitz Matrix | [Go]({{< relref "/ChapterFour/0766.Toeplitz-Matrix.md" >}})| Easy | O(n)| O(1)||
+|867. Transpose Matrix | [Go]({{< relref "/ChapterFour/0867.Transpose-Matrix.md" >}})| Easy | O(n)| O(1)||
+|891. Sum of Subsequence Widths  | [Go]({{< relref "/ChapterFour/0891.Sum-of-Subsequence-Widths.md" >}})| Hard | O(n log n)| O(1)||
+|907. Sum of Subarray Minimums | [Go]({{< relref "/ChapterFour/0907.Sum-of-Subarray-Minimums.md" >}})| Medium | O(n)| O(n)|❤️|
+|922. Sort Array By Parity II  | [Go]({{< relref "/ChapterFour/0922.Sort-Array-By-Parity-II.md" >}})| Medium | O(n)| O(1)||
+|969. Pancake Sorting  | [Go]({{< relref "/ChapterFour/0969.Pancake-Sorting.md" >}})| Medium | O(n)| O(1)|❤️|
+|977. Squares of a Sorted Array  | [Go]({{< relref "/ChapterFour/0977.Squares-of-a-Sorted-Array.md" >}})| Easy | O(n)| O(1)||
\ No newline at end of file
diff --git a/ctl/meta/Backtracking b/ctl/meta/Backtracking
new file mode 100644
index 000000000..9c140d2dd
--- /dev/null
+++ b/ctl/meta/Backtracking
@@ -0,0 +1,30 @@
+|17. Letter Combinations of a Phone Number | [Go]({{< relref "/ChapterFour/0017.Letter-Combinations-of-a-Phone-Number.md" >}})| Medium | O(log n)| O(1)||
+|22. Generate Parentheses| [Go]({{< relref "/ChapterFour/0022.Generate-Parentheses.md" >}})| Medium | O(log n)| O(1)||
+|37. Sudoku Solver  | [Go]({{< relref "/ChapterFour/0037.Sudoku-Solver.md" >}})| Hard | O(n^2)| O(n^2)|❤️|
+|39. Combination Sum | [Go]({{< relref "/ChapterFour/0039.Combination-Sum.md" >}})| Medium | O(n log n)| O(n)||
+|40. Combination Sum II | [Go]({{< relref "/ChapterFour/0040.Combination-Sum-II.md" >}})| Medium | O(n log n)| O(n)||
+|46. Permutations | [Go]({{< relref "/ChapterFour/0046.Permutations.md" >}})| Medium | O(n)| O(n)|❤️|
+|47. Permutations II | [Go]({{< relref "/ChapterFour/0047.Permutations-II.md" >}})| Medium | O(n^2)| O(n)|❤️|
+|51. N-Queens | [Go]({{< relref "/ChapterFour/0051.N-Queens.md" >}})| Hard | O(n!)| O(n)|❤️|
+|52. N-Queens II  | [Go]({{< relref "/ChapterFour/0052.N-Queens-II.md" >}})| Hard | O(n!)| O(n)|❤️|
+|60. Permutation Sequence  | [Go]({{< relref "/ChapterFour/0060.Permutation-Sequence.md" >}})| Medium | O(n log n)| O(1)||
+|77. Combinations  | [Go]({{< relref "/ChapterFour/0077.Combinations.md" >}})| Medium | O(n)| O(n)|❤️|
+|78. Subsets   | [Go]({{< relref "/ChapterFour/0078.Subsets.md" >}})| Medium | O(n^2)| O(n)|❤️|
+|79. Word Search  | [Go]({{< relref "/ChapterFour/0079.Word-Search.md" >}})| Medium | O(n^2)| O(n^2)|❤️|
+|89. Gray Codes | [Go]({{< relref "/ChapterFour/0089.Gray-Code.md" >}})| Medium | O(n)| O(1)||
+|90. Subsets II  | [Go]({{< relref "/ChapterFour/0090.Subsets-II.md" >}})| Medium | O(n^2)| O(n)|❤️|
+|93. Restore IP Addresses | [Go]({{< relref "/ChapterFour/0093.Restore-IP-Addresses.md" >}})| Medium | O(n)| O(n)|❤️|
+|126. Word Ladder II | [Go]({{< relref "/ChapterFour/0126.Word-Ladder-II.md" >}})| Hard | O(n)| O(n^2)|❤️|
+|131. Palindrome Partitioning | [Go]({{< relref "/ChapterFour/0131.Palindrome-Partitioning.md" >}})| Medium | O(n)| O(n^2)|❤️|
+|211. Add and Search Word - Data structure design | [Go]({{< relref "/ChapterFour/0211.Add-and-Search-Word---Data-structure-design.md" >}})| Medium | O(n)| O(n)|❤️|
+|212. Word Search II | [Go]({{< relref "/ChapterFour/0212.Word-Search-II.md" >}})| Hard | O(n^2)| O(n^2)|❤️|
+|216. Combination Sum III | [Go]({{< relref "/ChapterFour/0216.Combination-Sum-III.md" >}})| Medium | O(n)| O(1)|❤️|
+|306. Additive Number  | [Go]({{< relref "/ChapterFour/0306.Additive-Number.md" >}})| Medium | O(n^2)| O(1)|❤️|
+|357. Count Numbers with Unique Digits  | [Go]({{< relref "/ChapterFour/0357.Count-Numbers-with-Unique-Digits.md" >}})| Medium | O(1)| O(1)||
+|401. Binary Watch | [Go]({{< relref "/ChapterFour/0401.Binary-Watch.md" >}})| Easy | O(1)| O(1)||
+|526. Beautiful Arrangement | [Go]({{< relref "/ChapterFour/0526.Beautiful-Arrangement.md" >}})| Medium | O(n^2)| O(1)|❤️|
+|784. Letter Case Permutation | [Go]({{< relref "/ChapterFour/0784.Letter-Case-Permutation.md" >}})| Easy | O(n)| O(n)||
+|842. Split Array into Fibonacci Sequence  | [Go]({{< relref "/ChapterFour/0842.Split-Array-into-Fibonacci-Sequence.md" >}})| Medium | O(n^2)| O(1)|❤️|
+|980. Unique Paths III | [Go]({{< relref "/ChapterFour/0980.Unique-Paths-III.md" >}})| Hard | O(n log n)| O(n)||
+|996. Number of Squareful Arrays | [Go]({{< relref "/ChapterFour/0996.Number-of-Squareful-Arrays.md" >}})| Hard | O(n log n)| O(n) ||
+|1079. Letter Tile Possibilities | [Go]({{< relref "/ChapterFour/1079.Letter-Tile-Possibilities.md" >}})| Medium | O(n^2)| O(1)|❤️|
\ No newline at end of file
diff --git a/ctl/meta/Binary_Indexed_Tree b/ctl/meta/Binary_Indexed_Tree
new file mode 100644
index 000000000..e69de29bb
diff --git a/ctl/meta/Binary_Search b/ctl/meta/Binary_Search
new file mode 100644
index 000000000..427924180
--- /dev/null
+++ b/ctl/meta/Binary_Search
@@ -0,0 +1,13 @@
+|50. Pow(x, n) | [Go]({{< relref "/ChapterFour/0050.Powx-n.md" >}})| Medium | O(log n)| O(1)||
+|69. Sqrt(x)  | [Go]({{< relref "/ChapterFour/0069.Sqrtx.md" >}})| Easy | O(log n)| O(1)||
+|167. Two Sum II - Input array is sorted | [Go]({{< relref "/ChapterFour/0167.Two-Sum-II---Input-array-is-sorted.md" >}})| Easy | O(n)| O(1)||
+|209. Minimum Size Subarray Sum  | [Go]({{< relref "/ChapterFour/0209.Minimum-Size-Subarray-Sum.md" >}})| Medium | O(n)| O(1)||
+|222. Count Complete Tree Nodes  | [Go]({{< relref "/ChapterFour/0222.Count-Complete-Tree-Nodes.md" >}})| Medium | O(n)| O(1)||
+|230. Kth Smallest Element in a BST  | [Go]({{< relref "/ChapterFour/0230.Kth-Smallest-Element-in-a-BST.md" >}})| Medium | O(n)| O(1)||
+|287. Find the Duplicate Number   | [Go]({{< relref "/ChapterFour/0287.Find-the-Duplicate-Number.md" >}})| Easy | O(n)| O(1)|❤️|
+|300. Longest Increasing Subsequence  | [Go]({{< relref "/ChapterFour/0300.Longest-Increasing-Subsequence.md" >}})| Medium | O(n log n)| O(n)||
+|349. Intersection of Two Arrays  | [Go]({{< relref "/ChapterFour/0349.Intersection-of-Two-Arrays.md" >}})| Easy | O(n)| O(n) ||
+|350. Intersection of Two Arrays II  | [Go]({{< relref "/ChapterFour/0350.Intersection-of-Two-Arrays-II.md" >}})| Easy | O(n)| O(n) ||
+|392. Is Subsequence  | [Go]({{< relref "/ChapterFour/0392.Is-Subsequence.md" >}})| Medium | O(n)| O(1)||
+|454. 4Sum II | [Go]({{< relref "/ChapterFour/0454.4Sum-II.md" >}})| Medium | O(n^2)| O(n) ||
+|710. Random Pick with Blacklist | [Go]({{< relref "/ChapterFour/0710.Random-Pick-with-Blacklist.md" >}})| Hard | O(n)| O(n)  ||
\ No newline at end of file
diff --git a/ctl/meta/Bit_Manipulation b/ctl/meta/Bit_Manipulation
new file mode 100644
index 000000000..5043dbf04
--- /dev/null
+++ b/ctl/meta/Bit_Manipulation
@@ -0,0 +1,29 @@
+|78. Subsets | [Go]({{< relref "/ChapterFour/0078.Subsets.md" >}})| Medium | O(n^2)| O(n)|❤️|
+|136. Single Number | [Go]({{< relref "/ChapterFour/0136.Single-Number.md" >}})| Easy | O(n)| O(1)||
+|137. Single Number II  | [Go]({{< relref "/ChapterFour/0137.Single-Number-II.md" >}})| Medium | O(n)| O(1)|❤️|
+|169. Majority Element | [Go]({{< relref "/ChapterFour/0169.Majority-Element.md" >}})| Easy | O(n)| O(1)|❤️|
+|187. Repeated DNA Sequences  | [Go]({{< relref "/ChapterFour/0187.Repeated-DNA-Sequences.md" >}})| Medium | O(n)| O(1)||
+|190. Reverse Bits | [Go]({{< relref "/ChapterFour/0190.Reverse-Bits.md" >}})| Easy | O(n)| O(1)|❤️|
+|191. Number of 1 Bits  | [Go]({{< relref "/ChapterFour/0191.Number-of-1-Bits.md" >}})| Easy | O(n)| O(1)||
+|201. Bitwise AND of Numbers Range  | [Go]({{< relref "/ChapterFour/0201.Bitwise-AND-of-Numbers-Range.md" >}})| Medium | O(n)| O(1)|❤️|
+|231. Power of Two  | [Go]({{< relref "/ChapterFour/0231.Power-of-Two.md" >}})| Easy | O(1)| O(1)||
+|260. Single Number III  | [Go]({{< relref "/ChapterFour/0260.Single-Number-III.md" >}})| Medium | O(n)| O(1)|❤️|
+|268. Missing Number  | [Go]({{< relref "/ChapterFour/0268.Missing-Number.md" >}})| Easy | O(n)| O(1)||
+|318. Maximum Product of Word Lengths | [Go]({{< relref "/ChapterFour/0318.Maximum-Product-of-Word-Lengths.md" >}})| Medium | O(n)| O(1)||
+|338. Counting Bits   | [Go]({{< relref "/ChapterFour/0338.Counting-Bits.md" >}})| Medium | O(n)| O(n)||
+|342. Power of Four  | [Go]({{< relref "/ChapterFour/0342.Power-of-Four.md" >}})| Easy | O(n)| O(1)||
+|371. Sum of Two Integers  | [Go]({{< relref "/ChapterFour/0371.Sum-of-Two-Integers.md" >}})| Easy | O(n)| O(1)||
+|389. Find the Difference  | [Go]({{< relref "/ChapterFour/0389.Find-the-Difference.md" >}})| Easy | O(n)| O(1)||
+|393. UTF-8 Validation   | [Go]({{< relref "/ChapterFour/0393.UTF-8-Validation.md" >}})| Medium | O(n)| O(1)||
+|397. Integer Replacement   | [Go]({{< relref "/ChapterFour/0397.Integer-Replacement.md" >}})| Medium | O(n)| O(1)||
+|401. Binary Watch  | [Go]({{< relref "/ChapterFour/0401.Binary-Watch.md" >}})| Easy | O(1)| O(1)||
+|405. Convert a Number to Hexadecimal | [Go]({{< relref "/ChapterFour/0405.Convert-a-Number-to-Hexadecimal.md" >}})| Easy | O(n)| O(1)||
+|421. Maximum XOR of Two Numbers in an Array  | [Go]({{< relref "/ChapterFour/0421.Maximum-XOR-of-Two-Numbers-in-an-Array.md" >}})| Medium | O(n)| O(1)|❤️|
+|461. Hamming Distance  | [Go]({{< relref "/ChapterFour/0461.Hamming-Distance.md" >}})| Easy | O(n)| O(1)||
+|476. Number Complement  | [Go]({{< relref "/ChapterFour/0476.Number-Complement.md" >}})| Easy | O(n)| O(1)||
+|477. Total Hamming Distance  | [Go]({{< relref "/ChapterFour/0477.Total-Hamming-Distance.md" >}})| Medium | O(n)| O(1)||
+|693. Binary Number with Alternating Bits | [Go]({{< relref "/ChapterFour/0693.Binary-Number-with-Alternating-Bits.md" >}})| Easy | O(n)| O(1)|❤️|
+|756. Pyramid Transition Matrix  | [Go]({{< relref "/ChapterFour/0756.Pyramid-Transition-Matrix.md" >}})| Medium | O(n log n)| O(n)||
+|762. Prime Number of Set Bits in Binary Representation | [Go]({{< relref "/ChapterFour/0762.Prime-Number-of-Set-Bits-in-Binary-Representation.md" >}})| Easy | O(n)| O(1)||
+|784. Letter Case Permutation | [Go]({{< relref "/ChapterFour/0784.Letter-Case-Permutation.md" >}})| Easy | O(n)| O(1)||
+|898. Bitwise ORs of Subarrays | [Go]({{< relref "/ChapterFour/0898.Bitwise-ORs-of-Subarrays.md" >}})| Medium | O(n)| O(1)||
\ No newline at end of file
diff --git a/ctl/meta/Breadth_First_Search b/ctl/meta/Breadth_First_Search
new file mode 100644
index 000000000..4eb99cfe7
--- /dev/null
+++ b/ctl/meta/Breadth_First_Search
@@ -0,0 +1,14 @@
+|101. Symmetric Tree | [Go]({{< relref "/ChapterFour/0101.Symmetric-Tree.md" >}})| Easy | O(n)| O(1)||
+|102. Binary Tree Level Order Traversal | [Go]({{< relref "/ChapterFour/0102.Binary-Tree-Level-Order-Traversal.md" >}})| Medium | O(n)| O(1)||
+|103. Binary Tree Zigzag Level Order Traversal | [Go]({{< relref "/ChapterFour/0103.Binary-Tree-Zigzag-Level-Order-Traversal.md" >}})| Medium | O(n)| O(n)||
+|107. Binary Tree Level Order Traversal II | [Go]({{< relref "/ChapterFour/0107.Binary-Tree-Level-Order-Traversal-II.md" >}})| Easy | O(n)| O(1)||
+|111. Minimum Depth of Binary Tree | [Go]({{< relref "/ChapterFour/0111.Minimum-Depth-of-Binary-Tree.md" >}})| Easy | O(n)| O(1)||
+|126. Word Ladder II | [Go]({{< relref "/ChapterFour/0126.Word-Ladder-II.md" >}})| Hard | O(n)| O(n^2)|❤️|
+|127. Word Ladder  | [Go]({{< relref "/ChapterFour/0127.Word-Ladder.md" >}})| Medium | O(n)| O(n)||
+|199. Binary Tree Right Side View | [Go]({{< relref "/ChapterFour/0199.Binary-Tree-Right-Side-View.md" >}})| Medium | O(n)| O(1)||
+|200. Number of Islands | [Go]({{< relref "/ChapterFour/0200.Number-of-Islands.md" >}})| Medium | O(n^2)| O(n^2)||
+|207. Course Schedule  | [Go]({{< relref "/ChapterFour/0207.Course-Schedule.md" >}})| Medium | O(n^2)| O(n^2)||
+|210. Course Schedule II  | [Go]({{< relref "/ChapterFour/0210.Course-Schedule-II.md" >}})| Medium | O(n^2)| O(n^2)||
+|515. Find Largest Value in Each Tree Row | [Go]({{< relref "/ChapterFour/0515.Find-Largest-Value-in-Each-Tree-Row.md" >}})| Medium | O(n)| O(n)||
+|542. 01 Matrix  | [Go]({{< relref "/ChapterFour/0542.01-Matrix.md" >}})| Medium | O(n)| O(1)||
+|993. Cousins in Binary Tree | [Go]({{< relref "/ChapterFour/0993.Cousins-in-Binary-Tree.md" >}})| Easy | O(n)| O(1)||
\ No newline at end of file
diff --git a/ctl/meta/Depth_First_Search b/ctl/meta/Depth_First_Search
new file mode 100644
index 000000000..9bf247b18
--- /dev/null
+++ b/ctl/meta/Depth_First_Search
@@ -0,0 +1,23 @@
+|98. Validate Binary Search Tree | [Go]({{< relref "/ChapterFour/0098.Validate-Binary-Search-Tree.md" >}})| Medium | O(n)| O(1)||
+|99. Recover Binary Search Tree | [Go]({{< relref "/ChapterFour/0099.Recover-Binary-Search-Tree.md" >}})| Hard | O(n)| O(1)||
+|100. Same Tree | [Go]({{< relref "/ChapterFour/0100.Same-Tree.md" >}})| Easy | O(n)| O(1)||
+|101. Symmetric Tree | [Go]({{< relref "/ChapterFour/0101.Symmetric-Tree.md" >}})| Easy | O(n)| O(1)||
+|104. Maximum Depth of Binary Tree  | [Go]({{< relref "/ChapterFour/0104.Maximum-Depth-of-Binary-Tree.md" >}})| Easy | O(n)| O(1)||
+|108. Convert Sorted Array to Binary Search Tree | [Go]({{< relref "/ChapterFour/0108.Convert-Sorted-Array-to-Binary-Search-Tree.md" >}})| Easy | O(n)| O(1)||
+|109. Convert Sorted List to Binary Search Tree | [Go]({{< relref "/ChapterFour/0109.Convert-Sorted-List-to-Binary-Search-Tree.md" >}})| Medium | O(log n)| O(n)||
+|110. Balanced Binary Tree | [Go]({{< relref "/ChapterFour/0110.Balanced-Binary-Tree.md" >}})| Easy | O(n)| O(1)||
+|111. Minimum Depth of Binary Tree | [Go]({{< relref "/ChapterFour/0111.Minimum-Depth-of-Binary-Tree.md" >}})| Easy | O(n)| O(1)||
+|112. Path Sum | [Go]({{< relref "/ChapterFour/0112.Path-Sum.md" >}})| Easy | O(n)| O(1)||
+|113. Path Sum II | [Go]({{< relref "/ChapterFour/0113.Path-Sum-II.md" >}})| Medium | O(n)| O(1)||
+|114. Flatten Binary Tree to Linked List | [Go]({{< relref "/ChapterFour/0114.Flatten-Binary-Tree-to-Linked-List.md" >}})| Medium | O(n)| O(1)||
+|124. Binary Tree Maximum Path Sum | [Go]({{< relref "/ChapterFour/0124.Binary-Tree-Maximum-Path-Sum.md" >}})| Hard | O(n)| O(1)||
+|129. Sum Root to Leaf Numbers | [Go]({{< relref "/ChapterFour/0129.Sum-Root-to-Leaf-Numbers.md" >}})| Medium | O(n)| O(1)||
+|199. Binary Tree Right Side View | [Go]({{< relref "/ChapterFour/0199.Binary-Tree-Right-Side-View.md" >}})| Medium | O(n)| O(1)||
+|200. Number of Islands | [Go]({{< relref "/ChapterFour/0200.Number-of-Islands.md" >}})| Medium | O(n^2)| O(n^2)||
+|207. Course Schedule | [Go]({{< relref "/ChapterFour/0207.Course-Schedule.md" >}})| Medium | O(n^2)| O(n^2)||
+|210. Course Schedule II | [Go]({{< relref "/ChapterFour/0210.Course-Schedule-II.md" >}})| Medium | O(n^2)| O(n^2)||
+|257. Binary Tree Paths | [Go]({{< relref "/ChapterFour/0257.Binary-Tree-Paths.md" >}})| Easy | O(n)| O(1)||
+|394. Decode String | [Go]({{< relref "/ChapterFour/0394.Decode-String.md" >}})| Medium | O(n)| O(n)||
+|515. Find Largest Value in Each Tree Row | [Go]({{< relref "/ChapterFour/0515.Find-Largest-Value-in-Each-Tree-Row.md" >}})| Medium | O(n)| O(n)||
+|542. 01 Matrix | [Go]({{< relref "/ChapterFour/0542.01-Matrix.md" >}})| Medium | O(n)| O(1)||
+|980. Unique Paths III | [Go]({{< relref "/ChapterFour/0980.Unique-Paths-III.md" >}})| Hard | O(n log n)| O(n)||
\ No newline at end of file
diff --git a/ctl/meta/Dynamic_Programming b/ctl/meta/Dynamic_Programming
new file mode 100644
index 000000000..0f937f916
--- /dev/null
+++ b/ctl/meta/Dynamic_Programming
@@ -0,0 +1,26 @@
+|53. Maximum Subarray| [Go]({{< relref "/ChapterFour/0053.Maximum-Subarray.md" >}})| Easy | O(n)| O(n)||
+|62. Unique Paths | [Go]({{< relref "/ChapterFour/0062.Unique-Paths.md" >}})| Medium | O(n^2)| O(n^2)||
+|63. Unique Paths II | [Go]({{< relref "/ChapterFour/0063.Unique-Paths-II.md" >}})| Medium | O(n^2)| O(n^2)||
+|64. Minimum Path Sum | [Go]({{< relref "/ChapterFour/0064.Minimum-Path-Sum.md" >}})| Medium | O(n^2)| O(n^2)||
+|70. Climbing Stairs | [Go]({{< relref "/ChapterFour/0070.Climbing-Stairs.md" >}})| Easy | O(n)| O(n)||
+|91. Decode Ways | [Go]({{< relref "/ChapterFour/0091.Decode-Ways.md" >}})| Medium | O(n)| O(n)||
+|96. Unique Binary Search Trees | [Go]({{< relref "/ChapterFour/0096.Unique-Binary-Search-Trees.md" >}})| Medium | O(n)| O(n)||
+|120. Triangle | [Go]({{< relref "/ChapterFour/0120.Triangle.md" >}})| Medium | O(n^2)| O(n)||
+|121. Best Time to Buy and Sell Stock | [Go]({{< relref "/ChapterFour/0121.Best-Time-to-Buy-and-Sell-Stock.md" >}})| Easy | O(n)| O(1)||
+|152. Maximum Product Subarray | [Go]({{< relref "/ChapterFour/0152.Maximum-Product-Subarray.md" >}})| Medium | O(n)| O(1)||
+|198. House Robber | [Go]({{< relref "/ChapterFour/0198.House-Robber.md" >}})| Easy | O(n)| O(n)||
+|213. House Robber II | [Go]({{< relref "/ChapterFour/0213.House-Robber-II.md" >}})| Medium | O(n)| O(n)||
+|300. Longest Increasing Subsequence | [Go]({{< relref "/ChapterFour/0300.Longest-Increasing-Subsequence.md" >}})| Medium | O(n log n)| O(n)||
+|309. Best Time to Buy and Sell Stock with Cooldown | [Go]({{< relref "/ChapterFour/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown.md" >}})| Medium | O(n)| O(n)||
+|322. Coin Change | [Go]({{< relref "/ChapterFour/0322.Coin-Change.md" >}})| Medium | O(n)| O(n)||
+|338. Counting Bits | [Go]({{< relref "/ChapterFour/0338.Counting-Bits.md" >}})| Medium | O(n)| O(n)||
+|343. Integer Break | [Go]({{< relref "/ChapterFour/0343.Integer-Break.md" >}})| Medium | O(n^2)| O(n)||
+|357. Count Numbers with Unique Digits | [Go]({{< relref "/ChapterFour/0357.Count-Numbers-with-Unique-Digits.md" >}})| Medium | O(1)| O(1)||
+|392. Is Subsequence  | [Go]({{< relref "/ChapterFour/0392.Is-Subsequence.md" >}})| Medium | O(n)| O(1)||
+|416. Partition Equal Subset Sum  | [Go]({{< relref "/ChapterFour/0416.Partition-Equal-Subset-Sum.md" >}})| Medium | O(n^2)| O(n)||
+|714. Best Time to Buy and Sell Stock with Transaction Fee | [Go]({{< relref "/ChapterFour/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee.md" >}})| Medium | O(n)| O(1)||
+|746. Min Cost Climbing Stairs | [Go]({{< relref "/ChapterFour/0746.Min-Cost-Climbing-Stairs.md" >}})| Easy | O(n)| O(1)||
+|838. Push Dominoes | [Go]({{< relref "/ChapterFour/0838.Push-Dominoes.md" >}})| Medium | O(n)| O(n)||
+|1025. Divisor Game | [Go]({{< relref "/ChapterFour/1025.Divisor-Game.md" >}})| Easy | O(1)| O(1)||
+|891. Sum of Subsequence Widths | [Go]({{< relref "/ChapterFour/0891.Sum-of-Subsequence-Widths.md" >}})| Hard | O(n log n)| O(1)||
+|942. DI String Match | [Go]({{< relref "/ChapterFour/0942.DI-String-Match.md" >}})| Easy | O(n)| O(1)||
\ No newline at end of file
diff --git a/ctl/meta/Hash_Table b/ctl/meta/Hash_Table
new file mode 100644
index 000000000..56a6bde09
--- /dev/null
+++ b/ctl/meta/Hash_Table
@@ -0,0 +1,33 @@
+|1. Two Sum | [Go]({{< relref "/ChapterFour/0001.Two-Sum.md" >}})| Easy | O(n)| O(n)||
+|3. Longest Substring Without Repeating Characters  | [Go]({{< relref "/ChapterFour/0003.Longest-Substring-Without-Repeating-Characters.md" >}})| Medium | O(n)| O(1)|❤️|
+|18. 4Sum  | [Go]({{< relref "/ChapterFour/0018.4Sum.md" >}})| Medium | O(n^3)| O(n^2)|❤️|
+|30. Substring with Concatenation of All Words | [Go]({{< relref "/ChapterFour/0030.Substring-with-Concatenation-of-All-Words.md" >}})| Hard | O(n)| O(n)|❤️|
+|36. Valid Sudoku | [Go]({{< relref "/ChapterFour/0036.Valid-Sudoku.md" >}})| Medium | O(n^2)| O(n^2)||
+|37. Sudoku Solver  | [Go]({{< relref "/ChapterFour/0037.Sudoku-Solver.md" >}})| Hard | O(n^2)| O(n^2)|❤️|
+|49. Group Anagrams | [Go]({{< relref "/ChapterFour/0049.Group-Anagrams.md" >}})| Medium | O(n log n)| O(n)||
+|76. Minimum Window Substring | [Go]({{< relref "/ChapterFour/0076.Minimum-Window-Substring.md" >}})| Hard | O(n)| O(n)|❤️|
+|94. Binary Tree Inorder Traversal | [Go]({{< relref "/ChapterFour/0094.Binary-Tree-Inorder-Traversal.md" >}})| Medium | O(n)| O(1)||
+|138. Copy List with Random Pointer | [Go]()| Medium | O(n)| O(1)||
+|202. Happy Number | [Go]({{< relref "/ChapterFour/0202.Happy-Number.md" >}})| Easy | O(log n)| O(1)||
+|205. Isomorphic Strings  | [Go]({{< relref "/ChapterFour/0205.Isomorphic-Strings.md" >}})| Easy | O(log n)| O(n)||
+|217. Contains Duplicate  | [Go]({{< relref "/ChapterFour/0217.Contains-Duplicate.md" >}})| Easy | O(n)| O(n)||
+|219. Contains Duplicate II  | [Go]({{< relref "/ChapterFour/0219.Contains-Duplicate-II.md" >}})| Easy | O(n)| O(n)||
+|242. Valid Anagram  | [Go]({{< relref "/ChapterFour/0242.Valid-Anagram.md" >}})| Easy | O(n)| O(n) ||
+|274. H-Index  | [Go]({{< relref "/ChapterFour/0274.H-Index.md" >}})| Medium |  O(n)| O(n) ||
+|290. Word Pattern | [Go]({{< relref "/ChapterFour/0290.Word-Pattern.md" >}})| Easy |  O(n)| O(n) ||
+|347. Top K Frequent Elements  | [Go]({{< relref "/ChapterFour/0347.Top-K-Frequent-Elements.md" >}})| Medium |  O(n)| O(n) ||
+|349. Intersection of Two Arrays  | [Go]({{< relref "/ChapterFour/0349.Intersection-of-Two-Arrays.md" >}})| Easy | O(n)| O(n) ||
+|350. Intersection of Two Arrays II | [Go]({{< relref "/ChapterFour/0350.Intersection-of-Two-Arrays-II.md" >}})| Easy | O(n)| O(n) ||
+|438. Find All Anagrams in a String | [Go]({{< relref "/ChapterFour/0438.Find-All-Anagrams-in-a-String.md" >}})| Easy | O(n)| O(1) ||
+|447. Number of Boomerangs | [Go]({{< relref "/ChapterFour/0447.Number-of-Boomerangs.md" >}})| Easy | O(n)| O(1) ||
+|451. Sort Characters By Frequency  | [Go]({{< relref "/ChapterFour/0451.Sort-Characters-By-Frequency.md" >}})| Medium | O(n log n)| O(1) ||
+|454. 4Sum II | [Go]({{< relref "/ChapterFour/0454.4Sum-II.md" >}})| Medium | O(n^2)| O(n) ||
+|648. Replace Words  | [Go]({{< relref "/ChapterFour/0648.Replace-Words.md" >}})| Medium | O(n)| O(n) ||
+|676. Implement Magic Dictionary | [Go]({{< relref "/ChapterFour/0676.Implement-Magic-Dictionary.md" >}})| Medium | O(n)| O(n) ||
+|720. Longest Word in Dictionary | [Go]({{< relref "/ChapterFour/0720.Longest-Word-in-Dictionary.md" >}})| Easy | O(n)| O(n) ||
+|726. Number of Atoms  | [Go]({{< relref "/ChapterFour/0726.Number-of-Atoms.md" >}})| Hard | O(n)| O(n) |❤️|
+|739. Daily Temperatures | [Go]({{< relref "/ChapterFour/0739.Daily-Temperatures.md" >}})| Medium | O(n)| O(n) ||
+|710. Random Pick with Blacklist | [Go]({{< relref "/ChapterFour/0710.Random-Pick-with-Blacklist.md" >}})| Hard | O(n)| O(n)  ||
+|895. Maximum Frequency Stack | [Go]({{< relref "/ChapterFour/0895.Maximum-Frequency-Stack.md" >}})| Hard | O(n)| O(n)  ||
+|930. Binary Subarrays With Sum | [Go]({{< relref "/ChapterFour/0930.Binary-Subarrays-With-Sum.md" >}})| Medium | O(n)| O(n)  |❤️|
+|992. Subarrays with K Different Integers | [Go]({{< relref "/ChapterFour/0992.Subarrays-with-K-Different-Integers.md" >}})| Hard | O(n)| O(n)  |❤️|
\ No newline at end of file
diff --git a/ctl/meta/Linked_List b/ctl/meta/Linked_List
new file mode 100644
index 000000000..326edc579
--- /dev/null
+++ b/ctl/meta/Linked_List
@@ -0,0 +1,29 @@
+|2. Add Two Numbers | [Go]({{< relref "/ChapterFour/0002.Add-Two-Numbers.md" >}})| Medium | O(n)| O(1)||
+|19. Remove Nth Node From End of List | [Go]({{< relref "/ChapterFour/0019.Remove-Nth-Node-From-End-of-List.md" >}})| Medium | O(n)| O(1)||
+|21. Merge Two Sorted Lists | [Go]({{< relref "/ChapterFour/0021.Merge-Two-Sorted-Lists.md" >}})| Easy | O(log n)| O(1)||
+|23. Merge k Sorted Lists| [Go]({{< relref "/ChapterFour/0023.Merge-k-Sorted-Lists.md" >}})| Hard | O(log n)| O(1)|❤️|
+|24. Swap Nodes in Pairs | [Go]({{< relref "/ChapterFour/0024.Swap-Nodes-in-Pairs.md" >}})| Medium | O(n)| O(1)||
+|25. Reverse Nodes in k-Group | [Go]({{< relref "/ChapterFour/0025.Reverse-Nodes-in-k-Group.md" >}})| Hard | O(log n)| O(1)|❤️|
+|61. Rotate List | [Go]({{< relref "/ChapterFour/0061.Rotate-List.md" >}})| Medium | O(n)| O(1)||
+|82. Remove Duplicates from Sorted List II | [Go]({{< relref "/ChapterFour/0082.Remove-Duplicates-from-Sorted-List-II.md" >}})| Medium | O(n)| O(1)||
+|83. Remove Duplicates from Sorted List  | [Go]({{< relref "/ChapterFour/0083.Remove-Duplicates-from-Sorted-List.md" >}})| Easy | O(n)| O(1)||
+|86. Partition List  | [Go]({{< relref "/ChapterFour/0086.Partition-List.md" >}})| Medium | O(n)| O(1)|❤️|
+|92. Reverse Linked List II  | [Go]({{< relref "/ChapterFour/0092.Reverse-Linked-List-II.md" >}})| Medium | O(n)| O(1)|❤️|
+|109. Convert Sorted List to Binary Search Tree | [Go]({{< relref "/ChapterFour/0109.Convert-Sorted-List-to-Binary-Search-Tree.md" >}})| Medium | O(log n)| O(n)||
+|141. Linked List Cycle | [Go]({{< relref "/ChapterFour/0141.Linked-List-Cycle.md" >}})| Easy | O(n)| O(1)|❤️|
+|142. Linked List Cycle II | [Go]({{< relref "/ChapterFour/0142.Linked-List-Cycle-II.md" >}})| Medium | O(n)| O(1)|❤️|
+|143. Reorder List | [Go]({{< relref "/ChapterFour/0143.Reorder-List.md" >}})| Medium | O(n)| O(1)|❤️|
+|147. Insertion Sort List | [Go]({{< relref "/ChapterFour/0147.Insertion-Sort-List.md" >}})| Medium | O(n)| O(1)|❤️|
+|148. Sort List | [Go]({{< relref "/ChapterFour/0148.Sort-List.md" >}})| Medium | O(n log n)| O(n)|❤️|
+|160. Intersection of Two Linked Lists | [Go]({{< relref "/ChapterFour/0160.Intersection-of-Two-Linked-Lists.md" >}})| Easy | O(n)| O(1)|❤️|
+|203. Remove Linked List Elements | [Go]({{< relref "/ChapterFour/0203.Remove-Linked-List-Elements.md" >}})| Easy | O(n)| O(1)||
+|206. Reverse Linked List  | [Go]({{< relref "/ChapterFour/0206.Reverse-Linked-List.md" >}})| Easy | O(n)| O(1)||
+|234. Palindrome Linked List | [Go]({{< relref "/ChapterFour/0234.Palindrome-Linked-List.md" >}})| Easy | O(n)| O(1)||
+|237. Delete Node in a Linked List  | [Go]({{< relref "/ChapterFour/0237.Delete-Node-in-a-Linked-List.md" >}})| Easy | O(n)| O(1)||
+|328. Odd Even Linked List | [Go]({{< relref "/ChapterFour/0328.Odd-Even-Linked-List.md" >}})| Medium | O(n)| O(1)||
+|445. Add Two Numbers II | [Go]({{< relref "/ChapterFour/0445.Add-Two-Numbers-II.md" >}})| Medium | O(n)| O(n)||
+|725. Split Linked List in Parts  | [Go]({{< relref "/ChapterFour/0725.Split-Linked-List-in-Parts.md" >}})| Medium | O(n)| O(1)||
+|817. Linked List Components | [Go]({{< relref "/ChapterFour/0817.Linked-List-Components.md" >}})| Medium | O(n)| O(1)||
+|707. Design Linked List | [Go]({{< relref "/ChapterFour/0707.Design-Linked-List.md" >}})| Easy | O(n)| O(1)||
+|876. Middle of the Linked List | [Go]({{< relref "/ChapterFour/0876.Middle-of-the-Linked-List.md" >}})| Easy | O(n)| O(1)|❤️|
+|1019. Next Greater Node In Linked List | [Go]({{< relref "/ChapterFour/1019.Next-Greater-Node-In-Linked-List.md" >}})| Medium | O(n)| O(1)||
\ No newline at end of file
diff --git a/ctl/meta/Math b/ctl/meta/Math
new file mode 100644
index 000000000..e78509551
--- /dev/null
+++ b/ctl/meta/Math
@@ -0,0 +1,18 @@
+|2. Add Two Numbers  | [Go]({{< relref "/ChapterFour/0002.Add-Two-Numbers.md" >}})| Medium | O(n)| O(1)||
+|50. Pow(x, n)  | [Go]({{< relref "/ChapterFour/0050.Powx-n.md" >}})| Medium | O(log n)| O(1)||
+|60. Permutation Sequence   | [Go]({{< relref "/ChapterFour/0060.Permutation-Sequence.md" >}})| Medium | O(n log n)| O(1)||
+|69. Sqrt(x)  | [Go]({{< relref "/ChapterFour/0069.Sqrtx.md" >}})| Easy | O(log n)| O(1)||
+|202. Happy Number  | [Go]({{< relref "/ChapterFour/0202.Happy-Number.md" >}})| Easy | O(log n)| O(1)||
+|224. Basic Calculator  | [Go]({{< relref "/ChapterFour/0224.Basic-Calculator.md" >}})| Hard | O(n)| O(n)||
+|231. Power of Two | [Go]({{< relref "/ChapterFour/0231.Power-of-Two.md" >}})| Easy | O(1)| O(1)||
+|263. Ugly Number  | [Go]({{< relref "/ChapterFour/0263.Ugly-Number.md" >}})| Easy | O(log n)| O(1)||
+|326. Power of Three  | [Go]({{< relref "/ChapterFour/0326.Power-of-Three.md" >}})| Easy | O(1)| O(1)||
+|343. Integer Break | [Go]({{< relref "/ChapterFour/0343.Integer-Break.md" >}})| Medium | O(n^2)| O(n)||
+|357. Count Numbers with Unique Digits | [Go]({{< relref "/ChapterFour/0357.Count-Numbers-with-Unique-Digits.md" >}})| Medium | O(1)| O(1)||
+|628. Maximum Product of Three Numbers  | [Go]({{< relref "/ChapterFour/0628.Maximum-Product-of-Three-Numbers.md" >}})| Easy | O(n)| O(1)||
+|885. Spiral Matrix III | [Go]({{< relref "/ChapterFour/0885.Spiral-Matrix-III.md" >}})| Medium | O(n^2)| O(1)||
+|891. Sum of Subsequence Widths | [Go]({{< relref "/ChapterFour/0891.Sum-of-Subsequence-Widths.md" >}})| Hard | O(n log n)| O(1)||
+|942. DI String Match  | [Go]({{< relref "/ChapterFour/0942.DI-String-Match.md" >}})| Easy | O(n)| O(1)||
+|976. Largest Perimeter Triangle | [Go]({{< relref "/ChapterFour/0976.Largest-Perimeter-Triangle.md" >}})| Easy | O(n log n)| O(log n) ||
+|996. Number of Squareful Arrays | [Go]({{< relref "/ChapterFour/0996.Number-of-Squareful-Arrays.md" >}})| Hard | O(n log n)| O(n) ||
+|1025. Divisor Game | [Go]({{< relref "/ChapterFour/1025.Divisor-Game.md" >}})| Easy | O(1)| O(1)||
\ No newline at end of file
diff --git a/ctl/meta/PDFPreface b/ctl/meta/PDFPreface
new file mode 100644
index 000000000..465400d8a
--- /dev/null
+++ b/ctl/meta/PDFPreface
@@ -0,0 +1,12 @@
+<img src="https://books.halfrost.com/leetcode/logo.png" alt="logo" heigth="1300px" align="center"/>
+
+
+# 说明
+
+此版本是 https://books.halfrost.com/leetcode 网页的离线版，由于网页版实时会更新，所以此 PDF 版难免会有一些排版或者错别字。如果读者遇到了，可以到网页版相应页面，点击页面 edit 按钮，提交 pr 进行更改。此 PDF 版本号是 V1.5.20。PDF 永久更新地址是 https://github.com/halfrost/leetcode-go/releases/，以版本号区分不同版本。笔者还是强烈推荐看在线版，有任何错误都会立即更新。如果觉得此书对刷题有一点点帮助，可以给此书点一个 star，鼓励一下笔者早点更新更多题解。
+
+> 版本号说明，V1.5.20，1 是大版本号，5 代表当前题解中有几百题，目前是 520 题，所以第二个版本号是 5，20 代表当前题解中有几十题，目前是 520 题，所以第三个版本号是 20 。
+
+# 目录
+
+[toc]
\ No newline at end of file
diff --git a/ctl/meta/Segment_Tree b/ctl/meta/Segment_Tree
new file mode 100644
index 000000000..0e7f9c743
--- /dev/null
+++ b/ctl/meta/Segment_Tree
@@ -0,0 +1,10 @@
+|218. The Skyline Problem | [Go]({{< relref "/ChapterFour/0218.The-Skyline-Problem.md" >}})| Hard | O(n log n)| O(n)|❤️|
+|307. Range Sum Query - Mutable | [Go]({{< relref "/ChapterFour/0307.Range-Sum-Query---Mutable.md" >}})| Hard | O(1)| O(n)||
+|315. Count of Smaller Numbers After Self | [Go]({{< relref "/ChapterFour/0315.Count-of-Smaller-Numbers-After-Self.md" >}})| Hard | O(n log n)| O(n)||
+|327. Count of Range Sum | [Go]({{< relref "/ChapterFour/0327.Count-of-Range-Sum.md" >}})| Hard | O(n log n)| O(n)|❤️|
+|493. Reverse Pairs | [Go]({{< relref "/ChapterFour/0493.Reverse-Pairs.md" >}})| Hard | O(n log n)| O(n)||
+|699. Falling Squares | [Go]({{< relref "/ChapterFour/0699.Falling-Squares.md" >}})| Hard | O(n log n)| O(n)|❤️|
+|715. Range Module | [Go]({{< relref "/ChapterFour/0715.Range-Module.md" >}})| Hard | O(log n)| O(n)|❤️|
+|732. My Calendar III  | [Go]({{< relref "/ChapterFour/0732.My-Calendar-III.md" >}})| Hard | O(log n)| O(n)|❤️|
+|850. Rectangle Area II  | [Go]({{< relref "/ChapterFour/0850.Rectangle-Area-II.md" >}})| Hard | O(n log n)| O(n)|❤️|
+|1157. Online Majority Element In Subarray | [Go]({{< relref "/ChapterFour/1157.Online-Majority-Element-In-Subarray.md" >}})| Hard | O(log n)| O(n)|❤️|
\ No newline at end of file
diff --git a/ctl/meta/Sliding_Window b/ctl/meta/Sliding_Window
new file mode 100644
index 000000000..e3ec8c4a2
--- /dev/null
+++ b/ctl/meta/Sliding_Window
@@ -0,0 +1,13 @@
+|3. Longest Substring Without Repeating Characters | [Go]({{< relref "/ChapterFour/0003.Longest-Substring-Without-Repeating-Characters.md" >}})| Medium | O(n)| O(1)|❤️|
+|76. Minimum Window Substring | [Go]({{< relref "/ChapterFour/0076.Minimum-Window-Substring.md" >}})| Hard | O(n)| O(n)|❤️|
+|239. Sliding Window Maximum  | [Go]({{< relref "/ChapterFour/0239.Sliding-Window-Maximum.md" >}})| Hard | O(n * k)| O(n)|❤️|
+|424. Longest Repeating Character Replacement | [Go]({{< relref "/ChapterFour/0424.Longest-Repeating-Character-Replacement.md" >}})| Medium | O(n)| O(1) ||
+|480. Sliding Window Median | [Go]({{< relref "/ChapterFour/0480.Sliding-Window-Median.md" >}})| Hard | O(n * log k)| O(k)|❤️|
+|567. Permutation in String | [Go]({{< relref "/ChapterFour/0567.Permutation-in-String.md" >}})| Medium | O(n)| O(1)|❤️|
+|978. Longest Turbulent Subarray | [Go]({{< relref "/ChapterFour/0978.Longest-Turbulent-Subarray.md" >}})| Medium | O(n)| O(1)|❤️|
+|992. Subarrays with K Different Integers | [Go]({{< relref "/ChapterFour/0992.Subarrays-with-K-Different-Integers.md" >}})| Hard | O(n)| O(n)|❤️|
+|995. Minimum Number of K Consecutive Bit Flips | [Go]({{< relref "/ChapterFour/0995.Minimum-Number-of-K-Consecutive-Bit-Flips.md" >}})| Hard | O(n)| O(1)|❤️|
+|1004. Max Consecutive Ones III | [Go]({{< relref "/ChapterFour/1004.Max-Consecutive-Ones-III.md" >}})| Medium | O(n)| O(1) ||
+|1040. Moving Stones Until Consecutive II | [Go]({{< relref "/ChapterFour/1040.Moving-Stones-Until-Consecutive-II.md" >}})| Medium | O(n log n)| O(1) |❤️|
+|1052. Grumpy Bookstore Owner | [Go]({{< relref "/ChapterFour/1052.Grumpy-Bookstore-Owner.md" >}})| Medium | O(n log n)| O(1) ||
+|1074. Number of Submatrices That Sum to Target | [Go]({{< relref "/ChapterFour/1074.Number-of-Submatrices-That-Sum-to-Target.md" >}})| Hard | O(n^3)| O(n) |❤️|
\ No newline at end of file
diff --git a/ctl/meta/Sorting b/ctl/meta/Sorting
new file mode 100644
index 000000000..abe19d2ad
--- /dev/null
+++ b/ctl/meta/Sorting
@@ -0,0 +1,23 @@
+|56. Merge Intervals | [Go]({{< relref "/ChapterFour/0056.Merge-Intervals.md" >}})| Medium | O(n log n)| O(log n)||
+|57. Insert Interval  | [Go]({{< relref "/ChapterFour/0057.Insert-Interval.md" >}})| Hard | O(n)| O(1)||
+|75. Sort Colors  | [Go]({{< relref "/ChapterFour/0075.Sort-Colors.md" >}})| Medium| O(n)| O(1)|❤️|
+|147. Insertion Sort List | [Go]({{< relref "/ChapterFour/0147.Insertion-Sort-List.md" >}})| Medium | O(n^2)| O(1) |❤️|
+|148. Sort List  | [Go]({{< relref "/ChapterFour/0148.Sort-List.md" >}})| Medium |O(n log n)| O(log n)|❤️|
+|164. Maximum Gap  | [Go]({{< relref "/ChapterFour/0164.Maximum-Gap.md" >}})| Hard | O(n log n)| O(log n) |❤️|
+|179. Largest Number  | [Go]({{< relref "/ChapterFour/0179.Largest-Number.md" >}})| Medium | O(n log n)| O(log n) |❤️|
+|220. Contains Duplicate III  | [Go]({{< relref "/ChapterFour/0220.Contains-Duplicate-III.md" >}})| Medium | O(n log n)| O(1) |❤️|
+|242. Valid Anagram   | [Go]({{< relref "/ChapterFour/0242.Valid-Anagram.md" >}})| Easy | O(n)| O(n) ||
+|274. H-Index  | [Go]({{< relref "/ChapterFour/0274.H-Index.md" >}})| Medium |  O(n)| O(n) ||
+|324. Wiggle Sort II  | [Go]({{< relref "/ChapterFour/0324.Wiggle-Sort-II.md" >}})| Medium| O(n)| O(n)|❤️|
+|349. Intersection of Two Arrays | [Go]({{< relref "/ChapterFour/0349.Intersection-of-Two-Arrays.md" >}})| Easy | O(n)| O(n) ||
+|350. Intersection of Two Arrays II  | [Go]({{< relref "/ChapterFour/0350.Intersection-of-Two-Arrays-II.md" >}})| Easy | O(n)| O(n) ||
+|524. Longest Word in Dictionary through Deleting  | [Go]({{< relref "/ChapterFour/0524.Longest-Word-in-Dictionary-through-Deleting.md" >}})| Medium | O(n)| O(1) ||
+|767. Reorganize String | [Go]({{< relref "/ChapterFour/0767.Reorganize-String.md" >}})| Medium | O(n log n)| O(log n)  |❤️|
+|853. Car Fleet | [Go]({{< relref "/ChapterFour/0853.Car-Fleet.md" >}})| Medium | O(n log n)| O(log n)  ||
+|710. Random Pick with Blacklist | [Go]({{< relref "/ChapterFour/0710.Random-Pick-with-Blacklist.md" >}})| Hard | O(n)| O(n)  ||
+|922. Sort Array By Parity II | [Go]({{< relref "/ChapterFour/0922.Sort-Array-By-Parity-II.md" >}})| Easy | O(n)| O(1) ||
+|969. Pancake Sorting | [Go]({{< relref "/ChapterFour/0969.Pancake-Sorting.md" >}})| Medium | O(n log n)| O(log n) |❤️|
+|973. K Closest Points to Origin | [Go]({{< relref "/ChapterFour/0973.K-Closest-Points-to-Origin.md" >}})| Medium | O(n log n)| O(log n) ||
+|976. Largest Perimeter Triangle  | [Go]({{< relref "/ChapterFour/0976.Largest-Perimeter-Triangle.md" >}})| Easy | O(n log n)| O(log n) ||
+|1030. Matrix Cells in Distance Order  | [Go]({{< relref "/ChapterFour/1030.Matrix-Cells-in-Distance-Order.md" >}})| Easy | O(n^2)| O(1) ||
+|1054. Distant Barcodes | [Go]({{< relref "/ChapterFour/1054.Distant-Barcodes.md" >}})| Medium | O(n log n)| O(log n) |❤️|
\ No newline at end of file
diff --git a/ctl/meta/Stack b/ctl/meta/Stack
new file mode 100644
index 000000000..a145368c2
--- /dev/null
+++ b/ctl/meta/Stack
@@ -0,0 +1,37 @@
+|20. Valid Parentheses | [Go]({{< relref "/ChapterFour/0020.Valid-Parentheses.md" >}})| Easy | O(log n)| O(1)||
+|42. Trapping Rain Water | [Go]({{< relref "/ChapterFour/0042.Trapping-Rain-Water.md" >}})| Hard | O(n)| O(1)|❤️|
+|71. Simplify Path | [Go]({{< relref "/ChapterFour/0071.Simplify-Path.md" >}})| Medium | O(n)| O(n)|❤️|
+|84. Largest Rectangle in Histogram | [Go]({{< relref "/ChapterFour/0084.Largest-Rectangle-in-Histogram.md" >}})| Medium | O(n)| O(n)|❤️|
+|94. Binary Tree Inorder Traversal | [Go]({{< relref "/ChapterFour/0094.Binary-Tree-Inorder-Traversal.md" >}})| Medium | O(n)| O(1)||
+|103. Binary Tree Zigzag Level Order Traversal | [Go]({{< relref "/ChapterFour/0103.Binary-Tree-Zigzag-Level-Order-Traversal.md" >}})| Medium | O(n)| O(n)||
+|144. Binary Tree Preorder Traversal | [Go]({{< relref "/ChapterFour/0144.Binary-Tree-Preorder-Traversal.md" >}})| Medium | O(n)| O(1)||
+|145. Binary Tree Postorder Traversal | [Go]({{< relref "/ChapterFour/0145.Binary-Tree-Postorder-Traversal.md" >}})| Hard | O(n)| O(1)||
+|150. Evaluate Reverse Polish Notation | [Go]({{< relref "/ChapterFour/0150.Evaluate-Reverse-Polish-Notation.md" >}})| Medium | O(n)| O(1)||
+|155. Min Stack | [Go]({{< relref "/ChapterFour/0155.Min-Stack.md" >}})| Easy | O(n)| O(n)||
+|173. Binary Search Tree Iterator | [Go]({{< relref "/ChapterFour/0173.Binary-Search-Tree-Iterator.md" >}})| Medium | O(n)| O(1)||
+|224. Basic Calculator | [Go]({{< relref "/ChapterFour/0224.Basic-Calculator.md" >}})| Hard | O(n)| O(n)||
+|225. Implement Stack using Queues | [Go]({{< relref "/ChapterFour/0225.Implement-Stack-using-Queues.md" >}})| Easy | O(n)| O(n)||
+|232. Implement Queue using Stacks | [Go]({{< relref "/ChapterFour/0232.Implement-Queue-using-Stacks.md" >}})| Easy | O(n)| O(n)||
+|331. Verify Preorder Serialization of a Binary Tree | [Go]({{< relref "/ChapterFour/0331.Verify-Preorder-Serialization-of-a-Binary-Tree.md" >}})| Medium | O(n)| O(1)||
+|394. Decode String  | [Go]({{< relref "/ChapterFour/0394.Decode-String.md" >}})| Medium | O(n)| O(n)||
+|402. Remove K Digits | [Go]({{< relref "/ChapterFour/0402.Remove-K-Digits.md" >}})| Medium | O(n)| O(1)||
+|456. 132 Pattern  | [Go]({{< relref "/ChapterFour/0456.132-Pattern.md" >}})| Medium | O(n)| O(n)||
+|496. Next Greater Element I | [Go]({{< relref "/ChapterFour/0496.Next-Greater-Element-I.md" >}})| Easy | O(n)| O(n)||
+|503. Next Greater Element II  | [Go]({{< relref "/ChapterFour/0503.Next-Greater-Element-II.md" >}})| Medium | O(n)| O(n)||
+|636. Exclusive Time of Functions | [Go]({{< relref "/ChapterFour/0636.Exclusive-Time-of-Functions.md" >}})| Medium | O(n)| O(n)||
+|682. Baseball Game  | [Go]({{< relref "/ChapterFour/0682.Baseball-Game.md" >}})| Easy | O(n)| O(n)||
+|726. Number of Atoms | [Go]({{< relref "/ChapterFour/0726.Number-of-Atoms.md" >}})| Hard | O(n)| O(n) |❤️|
+|735. Asteroid Collision | [Go]({{< relref "/ChapterFour/0735.Asteroid-Collision.md" >}})| Medium | O(n)| O(n) ||
+|739. Daily Temperatures | [Go]({{< relref "/ChapterFour/0739.Daily-Temperatures.md" >}})| Medium | O(n)| O(n) ||
+|844. Backspace String Compare | [Go]({{< relref "/ChapterFour/0844.Backspace-String-Compare.md" >}})| Easy | O(n)| O(n) ||
+|856. Score of Parentheses  | [Go]({{< relref "/ChapterFour/0856.Score-of-Parentheses.md" >}})| Medium | O(n)| O(n)||
+|880. Decoded String at Index | [Go]({{< relref "/ChapterFour/0880.Decoded-String-at-Index.md" >}})| Medium | O(n)| O(n)||
+|895. Maximum Frequency Stack | [Go]({{< relref "/ChapterFour/0895.Maximum-Frequency-Stack.md" >}})| Hard | O(n)| O(n)  ||
+|901. Online Stock Span | [Go]({{< relref "/ChapterFour/0901.Online-Stock-Span.md" >}})| Medium | O(n)| O(n)  ||
+|907. Sum of Subarray Minimums | [Go]({{< relref "/ChapterFour/0907.Sum-of-Subarray-Minimums.md" >}})| Medium | O(n)| O(n)|❤️|
+|921. Minimum Add to Make Parentheses Valid | [Go]({{< relref "/ChapterFour/0921.Minimum-Add-to-Make-Parentheses-Valid.md" >}})| Medium | O(n)| O(n)||
+|946. Validate Stack Sequences | [Go]({{< relref "/ChapterFour/0946.Validate-Stack-Sequences.md" >}})| Medium | O(n)| O(n)||
+|1003. Check If Word Is Valid After Substitutions | [Go]({{< relref "/ChapterFour/1003.Check-If-Word-Is-Valid-After-Substitutions.md" >}})| Medium | O(n)| O(1)||
+|1019. Next Greater Node In Linked List | [Go]({{< relref "/ChapterFour/1019.Next-Greater-Node-In-Linked-List.md" >}})| Medium | O(n)| O(1)||
+|1021. Remove Outermost Parentheses | [Go]({{< relref "/ChapterFour/1021.Remove-Outermost-Parentheses.md" >}})| Medium | O(n)| O(1)||
+|1047. Remove All Adjacent Duplicates In String | [Go]({{< relref "/ChapterFour/1047.Remove-All-Adjacent-Duplicates-In-String.md" >}})| Medium | O(n)| O(1)||
\ No newline at end of file
diff --git a/ctl/meta/String b/ctl/meta/String
new file mode 100644
index 000000000..e2d905f71
--- /dev/null
+++ b/ctl/meta/String
@@ -0,0 +1,20 @@
+|3. Longest Substring Without Repeating Characters | [Go]({{< relref "/ChapterFour/0003.Longest-Substring-Without-Repeating-Characters.md" >}})| Medium | O(n)| O(1)|❤️|
+|17. Letter Combinations of a Phone Number | [Go]({{< relref "/ChapterFour/0017.Letter-Combinations-of-a-Phone-Number.md" >}})| Medium | O(log n)| O(1)||
+|20. Valid Parentheses | [Go]({{< relref "/ChapterFour/0020.Valid-Parentheses.md" >}})| Easy | O(log n)| O(1)||
+|22. Generate Parentheses | [Go]({{< relref "/ChapterFour/0022.Generate-Parentheses.md" >}})| Medium | O(log n)| O(1)||
+|28. Implement strStr()   | [Go]({{< relref "/ChapterFour/0028.Implement-strStr.md" >}})| Easy | O(n)| O(1)||
+|30. Substring with Concatenation of All Words | [Go]({{< relref "/ChapterFour/0030.Substring-with-Concatenation-of-All-Words.md" >}})| Hard | O(n)| O(n)|❤️|
+|49. Group Anagrams | [Go]({{< relref "/ChapterFour/0049.Group-Anagrams.md" >}})| Medium | O(n log n)| O(n)||
+|71. Simplify Path | [Go]({{< relref "/ChapterFour/0071.Simplify-Path.md" >}})| Medium | O(n)| O(n)||
+|76. Minimum Window Substring | [Go]({{< relref "/ChapterFour/0076.Minimum-Window-Substring.md" >}})| Hard | O(n)| O(n)|❤️|
+|91. Decode Ways  | [Go]({{< relref "/ChapterFour/0091.Decode-Ways.md" >}})| Medium | O(n)| O(n)||
+|93. Restore IP Addresses | [Go]({{< relref "/ChapterFour/0093.Restore-IP-Addresses.md" >}})| Medium | O(n)| O(n)|❤️|
+|125. Valid Palindrome | [Go]({{< relref "/ChapterFour/0125.Valid-Palindrome.md" >}})| Easy | O(n)| O(1)||
+|126. Word Ladder II | [Go]({{< relref "/ChapterFour/0126.Word-Ladder-II.md" >}})| Hard | O(n)| O(n^2)|❤️|
+|344. Reverse String | [Go]({{< relref "/ChapterFour/0344.Reverse-String.md" >}})| Easy | O(n)| O(1)||
+|345. Reverse Vowels of a String | [Go]({{< relref "/ChapterFour/0345.Reverse-Vowels-of-a-String.md" >}})| Easy | O(n)| O(1)||
+|767. Reorganize String  | [Go]({{< relref "/ChapterFour/0767.Reorganize-String.md" >}})| Medium | O(n log n)| O(log n)  |❤️|
+|842. Split Array into Fibonacci Sequence | [Go]({{< relref "/ChapterFour/0842.Split-Array-into-Fibonacci-Sequence.md" >}})| Medium | O(n^2)| O(1)|❤️|
+|856. Score of Parentheses | [Go]({{< relref "/ChapterFour/0856.Score-of-Parentheses.md" >}})| Medium | O(n)| O(n)||
+|925. Long Pressed Name | [Go]({{< relref "/ChapterFour/0925.Long-Pressed-Name.md" >}})| Easy | O(n)| O(1)||
+|1003. Check If Word Is Valid After Substitutions | [Go]({{< relref "/ChapterFour/1003.Check-If-Word-Is-Valid-After-Substitutions.md" >}})| Medium | O(n)| O(1)||
\ No newline at end of file
diff --git a/ctl/meta/Tree b/ctl/meta/Tree
new file mode 100644
index 000000000..ce2a837bb
--- /dev/null
+++ b/ctl/meta/Tree
@@ -0,0 +1,33 @@
+|94. Binary Tree Inorder Traversal | [Go]({{< relref "/ChapterFour/0094.Binary-Tree-Inorder-Traversal.md" >}})| Medium | O(n)| O(1)||
+|96. Unique Binary Search Trees | [Go]({{< relref "/ChapterFour/0096.Unique-Binary-Search-Trees.md" >}})| Medium | O(n^2)| O(n)||
+|98. Validate Binary Search Tree | [Go]({{< relref "/ChapterFour/0098.Validate-Binary-Search-Tree.md" >}})| Medium | O(n)| O(1)||
+|99. Recover Binary Search Tree  | [Go]({{< relref "/ChapterFour/0099.Recover-Binary-Search-Tree.md" >}})| Hard | O(n)| O(1)||
+|100. Same Tree  | [Go]({{< relref "/ChapterFour/0100.Same-Tree.md" >}})| Easy | O(n)| O(1)||
+|101. Symmetric Tree  | [Go]({{< relref "/ChapterFour/0101.Symmetric-Tree.md" >}})| Easy | O(n)| O(1)||
+|102. Binary Tree Level Order Traversal  | [Go]({{< relref "/ChapterFour/0102.Binary-Tree-Level-Order-Traversal.md" >}})| Medium | O(n)| O(1)||
+|103. Binary Tree Zigzag Level Order Traversal | [Go]({{< relref "/ChapterFour/0103.Binary-Tree-Zigzag-Level-Order-Traversal.md" >}})| Medium | O(n)| O(n)||
+|104. Maximum Depth of Binary Tree | [Go]({{< relref "/ChapterFour/0104.Maximum-Depth-of-Binary-Tree.md" >}})| Easy | O(n)| O(1)||
+|107. Binary Tree Level Order Traversal II | [Go]({{< relref "/ChapterFour/0107.Binary-Tree-Level-Order-Traversal-II.md" >}})| Easy | O(n)| O(1)||
+|108. Convert Sorted Array to Binary Search Tree | [Go]({{< relref "/ChapterFour/0108.Convert-Sorted-Array-to-Binary-Search-Tree.md" >}})| Easy | O(n)| O(1)||
+|110. Balanced Binary Tree | [Go]({{< relref "/ChapterFour/0110.Balanced-Binary-Tree.md" >}})| Easy | O(n)| O(1)||
+|111. Minimum Depth of Binary Tree | [Go]({{< relref "/ChapterFour/0111.Minimum-Depth-of-Binary-Tree.md" >}})| Easy | O(n)| O(1)||
+|112. Path Sum | [Go]({{< relref "/ChapterFour/0112.Path-Sum.md" >}})| Easy | O(n)| O(1)||
+|113. Path Sum II | [Go]({{< relref "/ChapterFour/0113.Path-Sum-II.md" >}})| Medium | O(n)| O(1)||
+|114. Flatten Binary Tree to Linked List | [Go]({{< relref "/ChapterFour/0114.Flatten-Binary-Tree-to-Linked-List.md" >}})| Medium | O(n)| O(1)||
+|124. Binary Tree Maximum Path Sum | [Go]({{< relref "/ChapterFour/0124.Binary-Tree-Maximum-Path-Sum.md" >}})| Hard | O(n)| O(1)||
+|129. Sum Root to Leaf Numbers | [Go]({{< relref "/ChapterFour/0129.Sum-Root-to-Leaf-Numbers.md" >}})| Medium | O(n)| O(1)||
+|144. Binary Tree Preorder Traversal  | [Go]({{< relref "/ChapterFour/0144.Binary-Tree-Preorder-Traversal.md" >}})| Medium | O(n)| O(1)||
+|145. Binary Tree Postorder Traversal | [Go]({{< relref "/ChapterFour/0145.Binary-Tree-Postorder-Traversal.md" >}})| Hard | O(n)| O(1)||
+|173. Binary Search Tree Iterator | [Go]({{< relref "/ChapterFour/0173.Binary-Search-Tree-Iterator.md" >}})| Medium | O(n)| O(1)||
+|199. Binary Tree Right Side View | [Go]({{< relref "/ChapterFour/0199.Binary-Tree-Right-Side-View.md" >}})| Medium | O(n)| O(1)||
+|222. Count Complete Tree Nodes | [Go]({{< relref "/ChapterFour/0222.Count-Complete-Tree-Nodes.md" >}})| Medium | O(n)| O(1)||
+|226. Invert Binary Tree | [Go]({{< relref "/ChapterFour/0226.Invert-Binary-Tree.md" >}})| Easy | O(n)| O(1)||
+|230. Kth Smallest Element in a BST | [Go]({{< relref "/ChapterFour/0230.Kth-Smallest-Element-in-a-BST.md" >}})| Medium | O(n)| O(1)||
+|235. Lowest Common Ancestor of a Binary Search Tree | [Go]({{< relref "/ChapterFour/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree.md" >}})| Easy | O(n)| O(1)||
+|236. Lowest Common Ancestor of a Binary Tree | [Go]({{< relref "/ChapterFour/0236.Lowest-Common-Ancestor-of-a-Binary-Tree.md" >}})| Medium | O(n)| O(1)||
+|257. Binary Tree Paths | [Go]({{< relref "/ChapterFour/0257.Binary-Tree-Paths.md" >}})| Easy | O(n)| O(1)||
+|404. Sum of Left Leaves | [Go]({{< relref "/ChapterFour/0404.Sum-of-Left-Leaves.md" >}})| Easy | O(n)| O(1)||
+|437. Path Sum III  | [Go]({{< relref "/ChapterFour/0437.Path-Sum-III.md" >}})| Easy | O(n)| O(1)||
+|515. Find Largest Value in Each Tree Row | [Go]({{< relref "/ChapterFour/0515.Find-Largest-Value-in-Each-Tree-Row.md" >}})| Medium | O(n)| O(n)||
+|637. Average of Levels in Binary Tree | [Go]({{< relref "/ChapterFour/0637.Average-of-Levels-in-Binary-Tree.md" >}})| Easy | O(n)| O(n)||
+|993. Cousins in Binary Tree | [Go]({{< relref "/ChapterFour/0993.Cousins-in-Binary-Tree.md" >}})| Easy | O(n)| O(1)||
\ No newline at end of file
diff --git a/ctl/meta/Two_Pointers b/ctl/meta/Two_Pointers
new file mode 100644
index 000000000..2c238556f
--- /dev/null
+++ b/ctl/meta/Two_Pointers
@@ -0,0 +1,50 @@
+|3. Longest Substring Without Repeating Characters | [Go]({{< relref "/ChapterFour/0003.Longest-Substring-Without-Repeating-Characters.md" >}})| Medium | O(n)| O(1)|❤️|
+|11. Container With Most Water  | [Go]({{< relref "/ChapterFour/0011.Container-With-Most-Water.md" >}})| Medium | O(n)| O(1)||
+|15. 3Sum  | [Go]({{< relref "/ChapterFour/0015.3Sum.md" >}})| Medium | O(n^2)| O(n)|❤️|
+|16. 3Sum Closest  | [Go]({{< relref "/ChapterFour/0016.3Sum-Closest.md" >}})| Medium | O(n^2)| O(1)|❤️|
+|18. 4Sum  | [Go]({{< relref "/ChapterFour/0018.4Sum.md" >}})| Medium | O(n^3)| O(n^2)|❤️|
+|19. Remove Nth Node From End of List | [Go]({{< relref "/ChapterFour/0019.Remove-Nth-Node-From-End-of-List.md" >}})| Medium | O(n)| O(1)||
+|26. Remove Duplicates from Sorted Array | [Go]({{< relref "/ChapterFour/0026.Remove-Duplicates-from-Sorted-Array.md" >}})| Easy | O(n)| O(1)||
+|27. Remove Element | [Go]({{< relref "/ChapterFour/0027.Remove-Element.md" >}})| Easy | O(n)| O(1)||
+|28. Implement strStr() | [Go]({{< relref "/ChapterFour/0028.Implement-strStr.md" >}})| Easy | O(n)| O(1)||
+|30. Substring with Concatenation of All Words | [Go]({{< relref "/ChapterFour/0030.Substring-with-Concatenation-of-All-Words.md" >}})| Hard | O(n)| O(n)|❤️|
+|42. Trapping Rain Water | [Go]({{< relref "/ChapterFour/0042.Trapping-Rain-Water.md" >}})| Hard | O(n)| O(1)|❤️|
+|61. Rotate List  | [Go]({{< relref "/ChapterFour/0061.Rotate-List.md" >}})| Medium | O(n)| O(1)||
+|75. Sort Colors  | [Go]({{< relref "/ChapterFour/0075.Sort-Colors.md" >}})| Medium| O(n)| O(1)|❤️|
+|76. Minimum Window Substring  | [Go]({{< relref "/ChapterFour/0076.Minimum-Window-Substring.md" >}})| Hard | O(n)| O(n)|❤️|
+|80. Remove Duplicates from Sorted Array II  | [Go]({{< relref "/ChapterFour/0080.Remove-Duplicates-from-Sorted-Array-II.md" >}})| Medium | O(n)| O(1||
+|86. Partition List | [Go]({{< relref "/ChapterFour/0086.Partition-List.md" >}})| Medium | O(n)| O(1)|❤️|
+|88. Merge Sorted Array | [Go]({{< relref "/ChapterFour/0088.Merge-Sorted-Array.md" >}})| Easy | O(n)| O(1)|❤️|
+|125. Valid Palindrome  | [Go]({{< relref "/ChapterFour/0125.Valid-Palindrome.md" >}})| Easy | O(n)| O(1)||
+|141. Linked List Cycle | [Go]({{< relref "/ChapterFour/0141.Linked-List-Cycle.md" >}})| Easy | O(n)| O(1)|❤️|
+|142. Linked List Cycle II | [Go]({{< relref "/ChapterFour/0142.Linked-List-Cycle-II.md" >}})| Medium | O(n)| O(1)|❤️|
+|167. Two Sum II - Input array is sorted | [Go]({{< relref "/ChapterFour/0167.Two-Sum-II---Input-array-is-sorted.md" >}})| Easy | O(n)| O(1)||
+|209. Minimum Size Subarray Sum | [Go]({{< relref "/ChapterFour/0209.Minimum-Size-Subarray-Sum.md" >}})| Medium | O(n)| O(1)||
+|234. Palindrome Linked List | [Go]({{< relref "/ChapterFour/0234.Palindrome-Linked-List.md" >}})| Easy | O(n)| O(1)||
+|283. Move Zeroes | [Go]({{< relref "/ChapterFour/0283.Move-Zeroes.md" >}})| Easy | O(n)| O(1)||
+|287. Find the Duplicate Number  | [Go]({{< relref "/ChapterFour/0287.Find-the-Duplicate-Number.md" >}})| Easy | O(n)| O(1)|❤️|
+|344. Reverse String | [Go]({{< relref "/ChapterFour/0344.Reverse-String.md" >}})| Easy | O(n)| O(1)||
+|345. Reverse Vowels of a String | [Go]({{< relref "/ChapterFour/0345.Reverse-Vowels-of-a-String.md" >}})| Easy | O(n)| O(1)||
+|349. Intersection of Two Arrays | [Go]({{< relref "/ChapterFour/0349.Intersection-of-Two-Arrays.md" >}})| Easy | O(n)| O(n) ||
+|350. Intersection of Two Arrays II | [Go]({{< relref "/ChapterFour/0350.Intersection-of-Two-Arrays-II.md" >}})| Easy | O(n)| O(n) ||
+|424. Longest Repeating Character Replacement | [Go]({{< relref "/ChapterFour/0424.Longest-Repeating-Character-Replacement.md" >}})| Medium | O(n)| O(1) ||
+|524. Longest Word in Dictionary through Deleting | [Go]({{< relref "/ChapterFour/0524.Longest-Word-in-Dictionary-through-Deleting.md" >}})| Medium | O(n)| O(1) ||
+|532. K-diff Pairs in an Array | [Go]({{< relref "/ChapterFour/0532.K-diff-Pairs-in-an-Array.md" >}})| Easy | O(n)| O(n)||
+|567. Permutation in String | [Go]({{< relref "/ChapterFour/0567.Permutation-in-String.md" >}})| Medium | O(n)| O(1)|❤️|
+|713. Subarray Product Less Than K  | [Go]({{< relref "/ChapterFour/0713.Subarray-Product-Less-Than-K.md" >}})| Medium | O(n)| O(1)||
+|763. Partition Labels | [Go]({{< relref "/ChapterFour/0763.Partition-Labels.md" >}})| Medium | O(n)| O(1)|❤️|
+|826. Most Profit Assigning Work | [Go]({{< relref "/ChapterFour/0826.Most-Profit-Assigning-Work.md" >}})| Medium | O(n log n)| O(n)||
+|828. Count Unique Characters of All Substrings of a Given String | [Go]({{< relref "/ChapterFour/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String.md" >}})| Hard | O(n)| O(1)|❤️|
+|838. Push Dominoes  | [Go]({{< relref "/ChapterFour/0838.Push-Dominoes.md" >}})| Medium | O(n)| O(n)||
+|844. Backspace String Compare  | [Go]({{< relref "/ChapterFour/0844.Backspace-String-Compare.md" >}})| Easy | O(n)| O(n) ||
+|845. Longest Mountain in Array | [Go]({{< relref "/ChapterFour/0845.Longest-Mountain-in-Array.md" >}})| Medium | O(n)| O(1) ||
+|881. Boats to Save People | [Go]({{< relref "/ChapterFour/0881.Boats-to-Save-People.md" >}})| Medium | O(n log n)| O(1) ||
+|904. Fruit Into Baskets | [Go]({{< relref "/ChapterFour/0904.Fruit-Into-Baskets.md" >}})| Medium | O(n log n)| O(1) ||
+|923. 3Sum With Multiplicity | [Go]({{< relref "/ChapterFour/0923.3Sum-With-Multiplicity.md" >}})| Medium | O(n^2)| O(n) ||
+|925. Long Pressed Name  | [Go]({{< relref "/ChapterFour/0925.Long-Pressed-Name.md" >}})| Easy | O(n)| O(1)||
+|930. Binary Subarrays With Sum  | [Go]({{< relref "/ChapterFour/0930.Binary-Subarrays-With-Sum.md" >}})| Medium | O(n)| O(n)  |❤️|
+|977. Squares of a Sorted Array | [Go]({{< relref "/ChapterFour/0977.Squares-of-a-Sorted-Array.md" >}})| Easy | O(n)| O(1)||
+|986. Interval List Intersections | [Go]({{< relref "/ChapterFour/0986.Interval-List-Intersections.md" >}})| Medium | O(n)| O(1)||
+|992. Subarrays with K Different Integers | [Go]({{< relref "/ChapterFour/0992.Subarrays-with-K-Different-Integers.md" >}})| Hard | O(n)| O(n)|❤️|
+|1004. Max Consecutive Ones III | [Go]({{< relref "/ChapterFour/1004.Max-Consecutive-Ones-III.md" >}})| Medium | O(n)| O(1) ||
+|1093. Statistics from a Large Sample | [Go]({{< relref "/ChapterFour/1093.Statistics-from-a-Large-Sample.md" >}})| Medium | O(n)| O(1) ||
\ No newline at end of file
diff --git a/ctl/meta/Union_Find b/ctl/meta/Union_Find
new file mode 100644
index 000000000..66424cf4a
--- /dev/null
+++ b/ctl/meta/Union_Find
@@ -0,0 +1,18 @@
+|128. Longest Consecutive Sequence | [Go]({{< relref "/ChapterFour/0128.Longest-Consecutive-Sequence.md" >}})| Hard | O(n)| O(n)|❤️|
+|130. Surrounded Regions | [Go]({{< relref "/ChapterFour/0130.Surrounded-Regions.md" >}})| Medium | O(m\*n)| O(m\*n)||
+|200. Number of Islands | [Go]({{< relref "/ChapterFour/0200.Number-of-Islands.md" >}})| Medium | O(m\*n)| O(m\*n)||
+|399. Evaluate Division | [Go]({{< relref "/ChapterFour/0399.Evaluate-Division.md" >}})| Medium | O(n)| O(n)||
+|547. Number of Provinces  | [Go]({{< relref "/ChapterFour/0547.Number-of-Provinces.md" >}})| Medium | O(n^2)| O(n)||
+|684. Redundant Connection  | [Go]({{< relref "/ChapterFour/0684.Redundant-Connection.md" >}})| Medium | O(n)| O(n)||
+|685. Redundant Connection II | [Go]({{< relref "/ChapterFour/0685.Redundant-Connection-II.md" >}})| Hard | O(n)| O(n)||
+|721. Accounts Merge  | [Go]({{< relref "/ChapterFour/0721.Accounts-Merge.md" >}})| Medium | O(n)| O(n)|❤️|
+|765. Couples Holding Hands | [Go]({{< relref "/ChapterFour/0765.Couples-Holding-Hands.md" >}})| Hard | O(n)| O(n)|❤️|
+|778. Swim in Rising Water  | [Go]({{< relref "/ChapterFour/0778.Swim-in-Rising-Water.md" >}})| Hard | O(n^2)| O(n)|❤️|
+|803. Bricks Falling When Hit  | [Go]({{< relref "/ChapterFour/0803.Bricks-Falling-When-Hit.md" >}})| Hard | O(n^2)| O(n)|❤️|
+|839. Similar String Groups  | [Go]({{< relref "/ChapterFour/0839.Similar-String-Groups.md" >}})| Hard | O(n^2)| O(n)||
+|924. Minimize Malware Spread  | [Go]({{< relref "/ChapterFour/0924.Minimize-Malware-Spread.md" >}})| Hard | O(m\*n)| O(n)||
+|928. Minimize Malware Spread II  | [Go]({{< relref "/ChapterFour/0928.Minimize-Malware-Spread-II.md" >}})| Hard | O(m\*n)| O(n)|❤️|
+|947. Most Stones Removed with Same Row or Column  | [Go]({{< relref "/ChapterFour/0947.Most-Stones-Removed-with-Same-Row-or-Column.md" >}})| Medium | O(n)| O(n)||
+|952. Largest Component Size by Common Factor | [Go]({{< relref "/ChapterFour/0952.Largest-Component-Size-by-Common-Factor.md" >}})| Hard | O(n)| O(n)|❤️|
+|959. Regions Cut By Slashes | [Go]({{< relref "/ChapterFour/0959.Regions-Cut-By-Slashes.md" >}})| Medium | O(n^2)| O(n^2)|❤️|
+|990. Satisfiability of Equality Equations | [Go]({{< relref "/ChapterFour/0990.Satisfiability-of-Equality-Equations.md" >}})| Medium | O(n)| O(n)||
\ No newline at end of file
diff --git a/ctl/meta/meta b/ctl/meta/meta
new file mode 100644
index 000000000..a28e1a67e
--- /dev/null
+++ b/ctl/meta/meta
@@ -0,0 +1,2 @@
+| Title | Solution | Difficulty | Time | Space |收藏| 
+| ----- | :--------: | :----------: | :----: | :-----: | :-----: |
\ No newline at end of file
diff --git a/ctl/models/go.mod b/ctl/models/go.mod
new file mode 100644
index 000000000..2b1389a7f
--- /dev/null
+++ b/ctl/models/go.mod
@@ -0,0 +1,7 @@
+module github.com/halfrost/LeetCode-Go/ctl/models
+
+go 1.19
+
+replace github.com/halfrost/LeetCode-Go/ctl/models => ../util
+
+require github.com/halfrost/LeetCode-Go/ctl/util v0.0.0-20220910225043-e3bb5aff34d0
diff --git a/ctl/models/go.sum b/ctl/models/go.sum
new file mode 100644
index 000000000..2c3eec0c9
--- /dev/null
+++ b/ctl/models/go.sum
@@ -0,0 +1,2 @@
+github.com/halfrost/LeetCode-Go/ctl/util v0.0.0-20220910225043-e3bb5aff34d0 h1:WAOAj59szR52uAnEQljAt7ucpbGGOsy0xgR/NeP4Xbc=
+github.com/halfrost/LeetCode-Go/ctl/util v0.0.0-20220910225043-e3bb5aff34d0/go.mod h1:+cA8KYcbGxP2Itd3NG+QJVGL/MEZISKlei0tvgDeEag=
diff --git a/ctl/models/lcproblems.go b/ctl/models/lcproblems.go
new file mode 100644
index 000000000..e73537d64
--- /dev/null
+++ b/ctl/models/lcproblems.go
@@ -0,0 +1,104 @@
+package models
+
+import (
+	"fmt"
+	"strings"
+)
+
+// LeetCodeProblemAll define
+type LeetCodeProblemAll struct {
+	UserName        string            `json:"user_name"`
+	NumSolved       int32             `json:"num_solved"`
+	NumTotal        int32             `json:"num_total"`
+	AcEasy          int32             `json:"ac_easy"`
+	AcMedium        int32             `json:"ac_medium"`
+	AcHard          int32             `json:"ac_hard"`
+	StatStatusPairs []StatStatusPairs `json:"stat_status_pairs"`
+	FrequencyHigh   float64           `json:"frequency_high"`
+	FrequencyMid    float64           `json:"frequency_mid"`
+	CategorySlug    string            `json:"category_slug"`
+	AcEasyTotal     int32
+	AcMediumTotal   int32
+	AcHardTotal     int32
+}
+
+// ConvertUserInfoModel define
+func ConvertUserInfoModel(lpa LeetCodeProblemAll) UserInfo {
+	info := UserInfo{}
+	info.UserName = lpa.UserName
+	info.NumSolved = lpa.NumSolved
+	info.NumTotal = lpa.NumTotal
+	info.AcEasy = lpa.AcEasy
+	info.AcMedium = lpa.AcMedium
+	info.AcHard = lpa.AcHard
+	info.FrequencyHigh = lpa.FrequencyHigh
+	info.FrequencyMid = lpa.FrequencyMid
+	info.CategorySlug = lpa.CategorySlug
+	return info
+}
+
+// StatStatusPairs define
+type StatStatusPairs struct {
+	Stat       Stat       `json:"stat"`
+	Status     string     `json:"status"`
+	Difficulty Difficulty `json:"difficulty"`
+	PaidOnly   bool       `json:"paid_only"`
+	IsFavor    bool       `json:"is_favor"`
+	Frequency  float64    `json:"frequency"`
+	Progress   float64    `json:"progress"`
+}
+
+// ConvertMdModelFromSsp define
+func ConvertMdModelFromSsp(problems []StatStatusPairs) []Mdrow {
+	mdrows := []Mdrow{}
+	for _, problem := range problems {
+		res := Mdrow{}
+		res.FrontendQuestionID = problem.Stat.FrontendQuestionID
+		res.QuestionTitle = strings.TrimSpace(problem.Stat.QuestionTitle)
+		res.QuestionTitleSlug = strings.TrimSpace(problem.Stat.QuestionTitleSlug)
+		res.Acceptance = fmt.Sprintf("%.1f%%", (problem.Stat.TotalAcs/problem.Stat.TotalSubmitted)*100)
+		res.Difficulty = DifficultyMap[problem.Difficulty.Level]
+		res.Frequency = fmt.Sprintf("%f", problem.Frequency)
+		mdrows = append(mdrows, res)
+	}
+	return mdrows
+}
+
+// ConvertMdModelFromIds define
+func ConvertMdModelFromIds(problemsMap map[int]StatStatusPairs, ids []int) []Mdrow {
+	mdrows := []Mdrow{}
+	for _, v := range ids {
+		res, problem := Mdrow{}, problemsMap[v]
+		res.FrontendQuestionID = problem.Stat.FrontendQuestionID
+		res.QuestionTitle = strings.TrimSpace(problem.Stat.QuestionTitle)
+		res.QuestionTitleSlug = strings.TrimSpace(problem.Stat.QuestionTitleSlug)
+		res.Acceptance = fmt.Sprintf("%.1f%%", (problem.Stat.TotalAcs/problem.Stat.TotalSubmitted)*100)
+		res.Difficulty = DifficultyMap[problem.Difficulty.Level]
+		res.Frequency = fmt.Sprintf("%f", problem.Frequency)
+		mdrows = append(mdrows, res)
+	}
+	return mdrows
+}
+
+// Stat define
+type Stat struct {
+	QuestionTitle      string  `json:"question__title"`
+	QuestionTitleSlug  string  `json:"question__title_slug"`
+	TotalAcs           float64 `json:"total_acs"`
+	TotalSubmitted     float64 `json:"total_submitted"`
+	Acceptance         string
+	Difficulty         string
+	FrontendQuestionID int32 `json:"frontend_question_id"`
+}
+
+// Difficulty define
+type Difficulty struct {
+	Level int32 `json:"level"`
+}
+
+// DifficultyMap define
+var DifficultyMap = map[int32]string{
+	1: "Easy",
+	2: "Medium",
+	3: "Hard",
+}
diff --git a/ctl/models/mdrow.go b/ctl/models/mdrow.go
new file mode 100644
index 000000000..5ad90ec0e
--- /dev/null
+++ b/ctl/models/mdrow.go
@@ -0,0 +1,90 @@
+package models
+
+import (
+	"fmt"
+	"strings"
+)
+
+// Mdrow define
+type Mdrow struct {
+	FrontendQuestionID int32  `json:"question_id"`
+	QuestionTitle      string `json:"question__title"`
+	QuestionTitleSlug  string `json:"question__title_slug"`
+	SolutionPath       string `json:"solution_path"`
+	Acceptance         string `json:"acceptance"`
+	Difficulty         string `json:"difficulty"`
+	Frequency          string `json:"frequency"`
+}
+
+// GenerateMdRows define
+func GenerateMdRows(solutionIds []int, mdrows []Mdrow) {
+	mdMap := map[int]Mdrow{}
+	for _, row := range mdrows {
+		mdMap[int(row.FrontendQuestionID)] = row
+	}
+	for i := 0; i < len(solutionIds); i++ {
+		if row, ok := mdMap[solutionIds[i]]; ok {
+			s7 := standardizedTitle(row.QuestionTitle, row.FrontendQuestionID)
+			mdMap[solutionIds[i]] = Mdrow{
+				FrontendQuestionID: row.FrontendQuestionID,
+				QuestionTitle:      strings.TrimSpace(row.QuestionTitle),
+				QuestionTitleSlug:  row.QuestionTitleSlug,
+				SolutionPath:       fmt.Sprintf("[Go](https://github.com/halfrost/leetcode-go/tree/master/leetcode/%v)", fmt.Sprintf("%04d.%v", solutionIds[i], s7)),
+				Acceptance:         row.Acceptance,
+				Difficulty:         row.Difficulty,
+				Frequency:          row.Frequency,
+			}
+		} else {
+			fmt.Printf("序号不存在 len(solutionIds) = %v len(mdrows) = %v len(solutionIds) = %v solutionIds[i] = %v QuestionTitle = %v\n", len(solutionIds), len(mdrows), len(solutionIds), solutionIds[i], mdrows[solutionIds[i]-1].QuestionTitle)
+		}
+	}
+	for i := range mdrows {
+		mdrows[i] = Mdrow{
+			FrontendQuestionID: mdrows[i].FrontendQuestionID,
+			QuestionTitle:      strings.TrimSpace(mdrows[i].QuestionTitle),
+			QuestionTitleSlug:  mdrows[i].QuestionTitleSlug,
+			SolutionPath:       mdMap[int(mdrows[i].FrontendQuestionID)].SolutionPath,
+			Acceptance:         mdrows[i].Acceptance,
+			Difficulty:         mdrows[i].Difficulty,
+			Frequency:          mdrows[i].Frequency,
+		}
+	}
+	// fmt.Printf("mdrows = %v\n\n", mdrows)
+}
+
+// | 0001 | Two Sum  | [Go](https://github.com/halfrost/leetcode-go/tree/master/leetcode/0001.Two-Sum)| 45.6%  | Easy | |
+func (m Mdrow) tableLine() string {
+	return fmt.Sprintf("|%04d|%v|%v|%v|%v||\n", m.FrontendQuestionID, m.QuestionTitle, m.SolutionPath, m.Acceptance, m.Difficulty)
+}
+
+// SortByQuestionID define
+type SortByQuestionID []Mdrow
+
+func (a SortByQuestionID) Len() int      { return len(a) }
+func (a SortByQuestionID) Swap(i, j int) { a[i], a[j] = a[j], a[i] }
+func (a SortByQuestionID) Less(i, j int) bool {
+	return a[i].FrontendQuestionID < a[j].FrontendQuestionID
+}
+
+// Mdrows define
+type Mdrows struct {
+	Mdrows []Mdrow
+}
+
+// | No.    |  Title  |  Solution  |  Acceptance |  Difficulty |  Frequency |
+// |:--------:|:--------------------------------------------------------------|:--------:|:--------:|:--------:|:--------:|
+func (mds Mdrows) table() string {
+	res := "| No.    |  Title  |  Solution  |  Acceptance |  Difficulty |  Frequency |\n"
+	res += "|:--------:|:--------------------------------------------------------------|:--------:|:--------:|:--------:|:--------:|\n"
+	for _, p := range mds.Mdrows {
+		res += p.tableLine()
+	}
+	// 加这一行是为了撑开整个表格
+	res += "|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|"
+	return res
+}
+
+// AvailableTable define
+func (mds Mdrows) AvailableTable() string {
+	return mds.table()
+}
diff --git a/ctl/models/tagproblem.go b/ctl/models/tagproblem.go
new file mode 100644
index 000000000..5bc04ab8c
--- /dev/null
+++ b/ctl/models/tagproblem.go
@@ -0,0 +1,284 @@
+package models
+
+import (
+	"encoding/json"
+	"fmt"
+	"strconv"
+	"strings"
+
+	"github.com/halfrost/LeetCode-Go/ctl/util"
+)
+
+// Graphql define
+type Graphql struct {
+	OperationName string `json:"operationName"`
+	Variables     struct {
+		TitleSlug string `json:"titleSlug"`
+	} `json:"variables"`
+	Query string `json:"query"`
+}
+
+// GraphQLResp define
+type GraphQLResp struct {
+	Data struct {
+		TopicTag       TopicTag       `json:"topicTag"`
+		FavoritesLists FavoritesLists `json:"favoritesLists"`
+	} `json:"data"`
+}
+
+// TopicTag define
+type TopicTag struct {
+	Name           string     `json:"name"`
+	TranslatedName string     `json:"translatedName"`
+	Slug           string     `json:"slug"`
+	Questions      []Question `json:"questions"`
+	Frequencies    string     `json:"frequencies"`
+	Typename       string     `json:"__typename"`
+}
+
+// Question define
+type Question struct {
+	Status             string      `json:"status"`
+	QuestionID         string      `json:"questionId"`
+	QuestionFrontendID string      `json:"questionFrontendId"`
+	Title              string      `json:"title"`
+	TitleSlug          string      `json:"titleSlug"`
+	TranslatedTitle    string      `json:"translatedTitle"`
+	Stats              string      `json:"stats"`
+	Difficulty         string      `json:"difficulty"`
+	TopicTags          []TopicTags `json:"topicTags"`
+	CompanyTags        interface{} `json:"companyTags"`
+	Typename           string      `json:"__typename"`
+}
+
+// TopicTags define
+type TopicTags struct {
+	Name           string `json:"name"`
+	TranslatedName string `json:"translatedName"`
+	Slug           string `json:"slug"`
+	Typename       string `json:"__typename"`
+}
+
+func (q Question) generateTagStatus() (TagStatus, error) {
+	var ts TagStatus
+	err := json.Unmarshal([]byte(q.Stats), &ts)
+	if err != nil {
+		fmt.Println(err)
+		return ts, err
+	}
+	return ts, nil
+}
+
+// TagStatus define
+type TagStatus struct {
+	TotalAccepted      string `json:"totalAccepted"`
+	TotalSubmission    string `json:"totalSubmission"`
+	TotalAcceptedRaw   int32  `json:"totalAcceptedRaw"`
+	TotalSubmissionRaw int32  `json:"totalSubmissionRaw"`
+	AcRate             string `json:"acRate"`
+}
+
+// ConvertMdModelFromQuestions define
+func ConvertMdModelFromQuestions(questions []Question) []Mdrow {
+	mdrows := []Mdrow{}
+	for _, question := range questions {
+		res := Mdrow{}
+		v, _ := strconv.Atoi(question.QuestionFrontendID)
+		res.FrontendQuestionID = int32(v)
+		res.QuestionTitle = strings.TrimSpace(question.Title)
+		res.QuestionTitleSlug = strings.TrimSpace(question.TitleSlug)
+		q, err := question.generateTagStatus()
+		if err != nil {
+			fmt.Println(err)
+		}
+		res.Acceptance = q.AcRate
+		res.Difficulty = question.Difficulty
+		mdrows = append(mdrows, res)
+	}
+	return mdrows
+}
+
+// TagList define
+type TagList struct {
+	FrontendQuestionID int32  `json:"question_id"`
+	QuestionTitle      string `json:"question__title"`
+	SolutionPath       string `json:"solution_path"`
+	Acceptance         string `json:"acceptance"`
+	Difficulty         string `json:"difficulty"`
+	TimeComplexity     string `json:"time_complexity"`
+	SpaceComplexity    string `json:"space_complexity"`
+	Favorite           string `json:"favorite"`
+}
+
+// | 0001 | Two Sum  | [Go]({{< relref "/ChapterFour/0001.Two-Sum.md" >}})| Easy | O(n)| O(n)|❤️|50%|
+func (t TagList) tableLine() string {
+	return fmt.Sprintf("|%04d|%v|%v|%v|%v|%v|%v|%v|\n", t.FrontendQuestionID, t.QuestionTitle, t.SolutionPath, t.Difficulty, t.TimeComplexity, t.SpaceComplexity, t.Favorite, t.Acceptance)
+}
+
+func standardizedTitle(orig string, frontendQuestionID int32) string {
+	s0 := strings.TrimSpace(orig)
+	s1 := strings.Replace(s0, " ", "-", -1)
+	s2 := strings.Replace(s1, "'", "", -1)
+	s3 := strings.Replace(s2, "%", "", -1)
+	s4 := strings.Replace(s3, "(", "", -1)
+	s5 := strings.Replace(s4, ")", "", -1)
+	s6 := strings.Replace(s5, ",", "", -1)
+	s7 := strings.Replace(s6, "?", "", -1)
+	count := 0
+	// 去掉 --- 这种情况，这种情况是由于题目标题中包含 - ，左右有空格，左右一填充，造成了 ---，3 个 -
+	for i := 0; i < len(s7)-2; i++ {
+		if s7[i] == '-' && s7[i+1] == '-' && s7[i+2] == '-' {
+			fmt.Printf("【需要修正 --- 的标题是 %v】\n", fmt.Sprintf("%04d.%v", int(frontendQuestionID), s7))
+			s7 = s7[:i+1] + s7[i+3:]
+			count++
+		}
+	}
+	if count > 0 {
+		fmt.Printf("总共修正了 %v 个标题\n", count)
+	}
+	// 去掉 -- 这种情况，这种情况是由于题目标题中包含负号 -
+	for i := 0; i < len(s7)-2; i++ {
+		if s7[i] == '-' && s7[i+1] == '-' {
+			fmt.Printf("【需要修正 -- 的标题是 %v】\n", fmt.Sprintf("%04d.%v", int(frontendQuestionID), s7))
+			s7 = s7[:i+1] + s7[i+2:]
+			count++
+		}
+	}
+	if count > 0 {
+		fmt.Printf("总共修正了 %v 个标题\n", count)
+	}
+	return s7
+}
+
+// GenerateTagMdRows define
+func GenerateTagMdRows(solutionIds []int, metaMap map[int]TagList, mdrows []Mdrow, internal bool) []TagList {
+	tl := []TagList{}
+	for _, row := range mdrows {
+		if util.BinarySearch(solutionIds, int(row.FrontendQuestionID)) != -1 {
+			tmp := TagList{}
+			tmp.FrontendQuestionID = row.FrontendQuestionID
+			tmp.QuestionTitle = strings.TrimSpace(row.QuestionTitle)
+			s7 := standardizedTitle(row.QuestionTitle, row.FrontendQuestionID)
+			if internal {
+				tmp.SolutionPath = fmt.Sprintf("[Go]({{< relref \"/ChapterFour/%v/%v.md\" >}})", util.GetChpaterFourFileNum(int(row.FrontendQuestionID)), fmt.Sprintf("%04d.%v", int(row.FrontendQuestionID), s7))
+			} else {
+				tmp.SolutionPath = fmt.Sprintf("[Go](https://books.halfrost.com/leetcode/ChapterFour/%v/%v)", util.GetChpaterFourFileNum(int(row.FrontendQuestionID)), fmt.Sprintf("%04d.%v", int(row.FrontendQuestionID), s7))
+			}
+			tmp.Acceptance = row.Acceptance
+			tmp.Difficulty = row.Difficulty
+			tmp.TimeComplexity = metaMap[int(row.FrontendQuestionID)].TimeComplexity
+			tmp.SpaceComplexity = metaMap[int(row.FrontendQuestionID)].SpaceComplexity
+			tmp.Favorite = metaMap[int(row.FrontendQuestionID)].Favorite
+			tl = append(tl, tmp)
+		}
+	}
+	return tl
+}
+
+// TagLists define
+type TagLists struct {
+	TagLists []TagList
+}
+
+// | No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+// |:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+func (tls TagLists) table() string {
+	res := "| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |\n"
+	res += "|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |\n"
+	for _, p := range tls.TagLists {
+		res += p.tableLine()
+	}
+	// 加这一行是为了撑开整个表格
+	res += "|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|"
+	return res
+}
+
+// AvailableTagTable define
+func (tls TagLists) AvailableTagTable() string {
+	return tls.table()
+}
+
+// FavoritesLists define
+type FavoritesLists struct {
+	PublicFavorites  []int `json:"publicFavorites"`
+	PrivateFavorites []struct {
+		IDHash           string `json:"idHash"`
+		ID               string `json:"id"`
+		Name             string `json:"name"`
+		IsPublicFavorite bool   `json:"isPublicFavorite"`
+		ViewCount        int    `json:"viewCount"`
+		Creator          string `json:"creator"`
+		IsWatched        bool   `json:"isWatched"`
+		Questions        []struct {
+			QuestionID string `json:"questionId"`
+			Title      string `json:"title"`
+			TitleSlug  string `json:"titleSlug"`
+			Typename   string `json:"__typename"`
+		} `json:"questions"`
+		Typename string `json:"__typename"`
+	} `json:"privateFavorites"`
+	Typename string `json:"__typename"`
+}
+
+// Gproblem define
+type Gproblem struct {
+	QuestionID            string        `json:"questionId"`
+	QuestionFrontendID    string        `json:"questionFrontendId"`
+	BoundTopicID          int           `json:"boundTopicId"`
+	Title                 string        `json:"title"`
+	TitleSlug             string        `json:"titleSlug"`
+	Content               string        `json:"content"`
+	TranslatedTitle       string        `json:"translatedTitle"`
+	TranslatedContent     string        `json:"translatedContent"`
+	IsPaidOnly            bool          `json:"isPaidOnly"`
+	Difficulty            string        `json:"difficulty"`
+	Likes                 int           `json:"likes"`
+	Dislikes              int           `json:"dislikes"`
+	IsLiked               interface{}   `json:"isLiked"`
+	SimilarQuestions      string        `json:"similarQuestions"`
+	Contributors          []interface{} `json:"contributors"`
+	LangToValidPlayground string        `json:"langToValidPlayground"`
+	TopicTags             []struct {
+		Name           string `json:"name"`
+		Slug           string `json:"slug"`
+		TranslatedName string `json:"translatedName"`
+		Typename       string `json:"__typename"`
+	} `json:"topicTags"`
+	CompanyTagStats interface{}    `json:"companyTagStats"`
+	CodeSnippets    []GcodeSnippet `json:"codeSnippets"`
+	Stats           string         `json:"stats"`
+	Hints           []interface{}  `json:"hints"`
+	Solution        interface{}    `json:"solution"`
+	Status          interface{}    `json:"status"`
+	SampleTestCase  string         `json:"sampleTestCase"`
+	MetaData        string         `json:"metaData"`
+	JudgerAvailable bool           `json:"judgerAvailable"`
+	JudgeType       string         `json:"judgeType"`
+	MysqlSchemas    []interface{}  `json:"mysqlSchemas"`
+	EnableRunCode   bool           `json:"enableRunCode"`
+	EnableTestMode  bool           `json:"enableTestMode"`
+	EnvInfo         string         `json:"envInfo"`
+	Typename        string         `json:"__typename"`
+}
+
+// Gstat define
+type Gstat struct {
+	TotalAcs            int    `json:"total_acs"`
+	QuestionTitle       string `json:"question__title"`
+	IsNewQuestion       bool   `json:"is_new_question"`
+	QuestionArticleSlug string `json:"question__article__slug"`
+	TotalSubmitted      int    `json:"total_submitted"`
+	FrontendQuestionID  int    `json:"frontend_question_id"`
+	QuestionTitleSlug   string `json:"question__title_slug"`
+	QuestionArticleLive bool   `json:"question__article__live"`
+	QuestionHide        bool   `json:"question__hide"`
+	QuestionID          int    `json:"question_id"`
+}
+
+// GcodeSnippet define
+type GcodeSnippet struct {
+	Lang     string `json:"lang"`
+	LangSlug string `json:"langSlug"`
+	Code     string `json:"code"`
+	Typename string `json:"__typename"`
+}
diff --git a/ctl/models/user.go b/ctl/models/user.go
new file mode 100644
index 000000000..d79dc6d5f
--- /dev/null
+++ b/ctl/models/user.go
@@ -0,0 +1,44 @@
+package models
+
+import (
+	"fmt"
+)
+
+// UserInfo define
+type UserInfo struct {
+	UserName         string `json:"user_name"`
+	NumSolved        int32  `json:"num_solved"`
+	NumTotal         int32  `json:"num_total"`
+	AcEasy           int32  `json:"ac_easy"`
+	AcMedium         int32  `json:"ac_medium"`
+	AcHard           int32  `json:"ac_hard"`
+	EasyTotal        int32
+	MediumTotal      int32
+	HardTotal        int32
+	OptimizingEasy   int32
+	OptimizingMedium int32
+	OptimizingHard   int32
+	FrequencyHigh    float64 `json:"frequency_high"`
+	FrequencyMid     float64 `json:"frequency_mid"`
+	CategorySlug     string  `json:"category_slug"`
+}
+
+// |    |  Easy  |  Medium  |  Hard |  Total |  optimizing |
+// |:--------:|:--------------------------------------------------------------|:--------:|:--------:|:--------:|:--------:|
+func (ui UserInfo) table() string {
+	res := "|    |  Easy  |  Medium  |  Hard |  Total |\n"
+	res += "|:--------:|:--------:|:--------:|:--------:|:--------:|\n"
+	res += fmt.Sprintf("|Optimizing|%v|%v|%v|%v|\n", ui.OptimizingEasy, ui.OptimizingMedium, ui.OptimizingHard, ui.OptimizingEasy+ui.OptimizingMedium+ui.OptimizingHard)
+	res += fmt.Sprintf("|Accepted|**%v**|**%v**|**%v**|**%v**|\n", ui.AcEasy, ui.AcMedium, ui.AcHard, ui.AcEasy+ui.AcMedium+ui.AcHard)
+	res += fmt.Sprintf("|Total|%v|%v|%v|%v|\n", ui.EasyTotal, ui.MediumTotal, ui.HardTotal, ui.EasyTotal+ui.MediumTotal+ui.HardTotal)
+	res += fmt.Sprintf("|Perfection Rate|%.1f%%|%.1f%%|%.1f%%|%.1f%%|\n", (1-float64(ui.OptimizingEasy)/float64(ui.AcEasy))*100, (1-float64(ui.OptimizingMedium)/float64(ui.AcMedium))*100, (1-float64(ui.OptimizingHard)/float64(ui.AcHard))*100, (1-float64(ui.OptimizingEasy+ui.OptimizingMedium+ui.OptimizingHard)/float64(ui.AcEasy+ui.AcMedium+ui.AcHard))*100)
+	res += fmt.Sprintf("|Completion Rate|%.1f%%|%.1f%%|%.1f%%|%.1f%%|\n", float64(ui.AcEasy)/float64(ui.EasyTotal)*100, float64(ui.AcMedium)/float64(ui.MediumTotal)*100, float64(ui.AcHard)/float64(ui.HardTotal)*100, float64(ui.AcEasy+ui.AcMedium+ui.AcHard)/float64(ui.EasyTotal+ui.MediumTotal+ui.HardTotal)*100)
+	// 加这一行是为了撑开整个表格
+	res += "|------------|----------------------------|----------------------------|----------------------------|----------------------------|"
+	return res
+}
+
+// PersonalData define
+func (ui UserInfo) PersonalData() string {
+	return ui.table()
+}
diff --git a/ctl/pdf.go b/ctl/pdf.go
new file mode 100644
index 000000000..d481df4c5
--- /dev/null
+++ b/ctl/pdf.go
@@ -0,0 +1,224 @@
+package main
+
+import (
+	"bufio"
+	"fmt"
+	"io"
+	"io/ioutil"
+	"os"
+	"regexp"
+	"strconv"
+	"strings"
+
+	"github.com/halfrost/LeetCode-Go/ctl/util"
+	"github.com/spf13/cobra"
+)
+
+var (
+	cleanString1 = "{{< columns >}}"
+	cleanString2 = "<--->"
+	cleanString3 = "{{< /columns >}}"
+	cleanString4 = "<img src=\"https://books.halfrost.com/leetcode/logo.png\" alt=\"logo\" height=\"600\" align=\"right\" style=\"padding-left: 30px;\"/>"
+	pdfPreface   = `<img src="https://books.halfrost.com/leetcode/logo.png" alt="logo" heigth="1300px" align="center"/>
+
+
+# 说明
+	
+此版本是 https://books.halfrost.com/leetcode 网页的离线版，由于网页版实时会更新，所以此 PDF 版难免会有一些排版或者错别字。如果读者遇到了，可以到网页版相应页面，点击页面 edit 按钮，提交 pr 进行更改。此 PDF 版本号是 V%v.%v.%v。PDF 永久更新地址是 https://github.com/halfrost/leetcode-go/releases/，以版本号区分不同版本。笔者还是强烈推荐看在线版，有任何错误都会立即更新。如果觉得此书对刷题有一点点帮助，可以给此书点一个 star，鼓励一下笔者早点更新更多题解。
+	
+> 版本号说明，V%v.%v.%v，%v 是大版本号，%v 代表当前题解中有几百题，目前是 %v 题，所以第二个版本号是 %v，%v 代表当前题解中有几十题，目前是 %v 题，所以第三个版本号是 %v 。
+	
+# 目录
+	
+[toc]
+
+`
+
+	majorVersion   = 1
+	midVersion     = 0
+	lastVersion    = 0
+	totalSolutions = 0
+)
+
+func newPDFCommand() *cobra.Command {
+	cmd := &cobra.Command{
+		Use:   "pdf",
+		Short: "PDF related commands",
+		Run: func(cmd *cobra.Command, args []string) {
+			generatePDF()
+		},
+	}
+	// cmd.Flags().StringVar(&alias, "alias", "", "alias")
+	// cmd.Flags().StringVar(&appId, "appid", "", "appid")
+	return cmd
+}
+
+func generatePDF() {
+	var (
+		pdf, tmp []byte
+		err      error
+	)
+	// 先删除 pre-next
+	delPreNext()
+
+	chapterFourFileOrder, _ := util.LoadChapterFourDir()
+	totalSolutions = len(chapterFourFileOrder)
+	midVersion = totalSolutions / 100
+	lastVersion = totalSolutions % 100
+	fmt.Printf("[当前的版本号是 V%v.%v.%v]\n", majorVersion, midVersion, lastVersion)
+	// 删除原始文档中的头部，并创建临时文件夹
+	prepare(fmt.Sprintf("../PDF v%v.%v.%v.md", majorVersion, midVersion, lastVersion))
+	// PDF 前言
+	pdf = append(pdf, []byte(fmt.Sprintf(pdfPreface, majorVersion, midVersion, lastVersion, majorVersion, midVersion, lastVersion, majorVersion, midVersion, totalSolutions, midVersion, lastVersion, totalSolutions, lastVersion))...)
+	// PDF 第一章
+	tmp, err = loadChapter(chapterOneFileOrder, "./pdftemp", "ChapterOne")
+	pdf = append(pdf, tmp...)
+	// PDF 第二章
+	tmp, err = loadChapter(chapterTwoFileOrder, "./pdftemp", "ChapterTwo")
+	pdf = append(pdf, tmp...)
+	// PDF 第三章
+	tmp, err = loadChapter(chapterThreeFileOrder, "./pdftemp", "ChapterThree")
+	pdf = append(pdf, tmp...)
+	// PDF 第四章
+	tmp, err = util.LoadFile("./pdftemp/ChapterFour/_index.md")
+	pdf = append(pdf, tmp...)
+	tmp, err = loadChapter(chapterFourFileOrder, "../website/content", "ChapterFour")
+	pdf = append(pdf, tmp...)
+	if err != nil {
+		fmt.Println(err)
+	}
+	// 生成 PDF
+	util.WriteFile(fmt.Sprintf("../PDF v%v.%v.%v.md", majorVersion, midVersion, lastVersion), pdf)
+	// 还原现场
+	addPreNext()
+	util.DestoryDir("./pdftemp")
+}
+
+func loadChapter(order []string, path, chapter string) ([]byte, error) {
+	var (
+		res, tmp []byte
+		err      error
+	)
+	for index, v := range order {
+		if chapter == "ChapterOne" && index == 0 {
+			// 清理不支持的特殊 MarkDown 语法
+			tmp, err = clean(fmt.Sprintf("%v/%v/%v.md", path, chapter, v))
+		} else {
+			if chapter == "ChapterFour" {
+				if v[4] == '.' {
+					num, err := strconv.Atoi(v[:4])
+					if err != nil {
+						fmt.Println(err)
+					}
+					tmp, err = util.LoadFile(fmt.Sprintf("%v/%v/%v/%v.md", path, chapter, util.GetChpaterFourFileNum(num), v))
+				}
+			} else {
+				tmp, err = util.LoadFile(fmt.Sprintf("%v/%v/%v.md", path, chapter, v))
+			}
+		}
+		if err != nil {
+			fmt.Println(err)
+			return []byte{}, err
+		}
+		res = append(res, tmp...)
+	}
+	return res, err
+}
+
+func prepare(path string) {
+	err := os.Remove(path)
+	if err != nil {
+		fmt.Println("pdf 还没有创建")
+		fmt.Println(err)
+	}
+	fmt.Println("pdf 删除成功，开始构建全新版本")
+
+	err = os.MkdirAll("./pdftemp/ChapterOne", os.ModePerm)
+	if err != nil {
+		fmt.Println(err)
+	}
+	for _, v := range chapterOneFileOrder {
+		removeHeader(fmt.Sprintf("../website/content/ChapterOne/%v.md", v), fmt.Sprintf("./pdftemp/ChapterOne/%v.md", v), 5)
+	}
+
+	err = os.MkdirAll("./pdftemp/ChapterTwo", os.ModePerm)
+	if err != nil {
+		fmt.Println(err)
+	}
+	// 生成外部链接的 ChapterTwo
+	buildChapterTwo(false)
+	util.CopyFile("./pdftemp/ChapterTwo/_index.md", "../website/content/ChapterTwo/_index.md")
+
+	for _, v := range chapterTwoFileOrder {
+		removeHeader(fmt.Sprintf("./pdftemp/ChapterTwo/%v.md", v), fmt.Sprintf("./pdftemp/ChapterTwo/%v.md", v), 5)
+	}
+
+	err = os.MkdirAll("./pdftemp/ChapterThree", os.ModePerm)
+	if err != nil {
+		fmt.Println(err)
+	}
+	for _, v := range chapterThreeFileOrder {
+		removeHeader(fmt.Sprintf("../website/content/ChapterThree/%v.md", v), fmt.Sprintf("./pdftemp/ChapterThree/%v.md", v), 5)
+	}
+
+	err = os.MkdirAll("./pdftemp/ChapterFour", os.ModePerm)
+	if err != nil {
+		fmt.Println(err)
+	}
+	removeHeader(fmt.Sprintf("../website/content/ChapterFour/_index.md"), fmt.Sprintf("./pdftemp/ChapterFour/_index.md"), 5)
+}
+
+func clean(filePath string) ([]byte, error) {
+	f, err := os.OpenFile(filePath, os.O_RDONLY, 0644)
+	if err != nil {
+		return nil, err
+	}
+	defer f.Close()
+	reader, output := bufio.NewReader(f), []byte{}
+	for {
+		line, _, err := reader.ReadLine()
+		if err != nil {
+			if err == io.EOF {
+				return output, nil
+			}
+			return nil, err
+		}
+		if ok, _ := regexp.Match(cleanString1, line); ok {
+			reg := regexp.MustCompile(cleanString1)
+			newByte := reg.ReplaceAll(line, []byte(""))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else if ok, _ := regexp.Match(cleanString2, line); ok {
+			reg := regexp.MustCompile(cleanString2)
+			newByte := reg.ReplaceAll(line, []byte(""))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else if ok, _ := regexp.Match(cleanString3, line); ok {
+			reg := regexp.MustCompile(cleanString3)
+			newByte := reg.ReplaceAll(line, []byte(""))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else if ok, _ := regexp.Match(cleanString4, line); ok {
+			reg := regexp.MustCompile(cleanString4)
+			newByte := reg.ReplaceAll(line, []byte(""))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else {
+			output = append(output, line...)
+			output = append(output, []byte("\n")...)
+		}
+	}
+}
+
+func removeHeader(path, newPath string, lineNumber int) {
+	file, err := ioutil.ReadFile(path)
+	if err != nil {
+		panic(err)
+	}
+	info, _ := os.Stat(path)
+	mode := info.Mode()
+	array := strings.Split(string(file), "\n")
+	array = array[lineNumber:]
+	ioutil.WriteFile(newPath, []byte(strings.Join(array, "\n")), mode)
+	//fmt.Println("remove line successful")
+}
diff --git a/ctl/rangking.go b/ctl/rangking.go
new file mode 100644
index 000000000..3e8196913
--- /dev/null
+++ b/ctl/rangking.go
@@ -0,0 +1,36 @@
+package main
+
+import (
+	"fmt"
+	"strconv"
+	"strings"
+)
+
+// getRanking 让这个方法优雅一点
+func getRanking() int {
+	// 获取网页数据
+	URL := fmt.Sprintf("https://leetcode.com/%s/", getConfig().Username)
+	data := getRaw(URL)
+	str := string(data)
+	// 通过不断裁剪 str 获取排名信息
+	fmt.Println(str)
+	i := strings.Index(str, "ng-init")
+	j := i + strings.Index(str[i:], "ng-cloak")
+	str = str[i:j]
+	i = strings.Index(str, "(")
+	j = strings.Index(str, ")")
+	str = str[i:j]
+	//	fmt.Println("2\n", str)
+	strs := strings.Split(str, ",")
+	str = strs[6]
+	//	fmt.Println("1\n", str)
+	i = strings.Index(str, "'")
+	j = 2 + strings.Index(str[2:], "'")
+	//	fmt.Println("0\n", str)
+	str = str[i+1 : j]
+	r, err := strconv.Atoi(str)
+	if err != nil {
+		fmt.Printf("无法把 %s 转换成数字Ranking", str)
+	}
+	return r
+}
diff --git a/ctl/refresh.go b/ctl/refresh.go
new file mode 100644
index 000000000..501031a6a
--- /dev/null
+++ b/ctl/refresh.go
@@ -0,0 +1,27 @@
+package main
+
+import (
+	"github.com/spf13/cobra"
+)
+
+func newRefresh() *cobra.Command {
+	cmd := &cobra.Command{
+		Use:   "refresh",
+		Short: "Refresh all document",
+		Run: func(cmd *cobra.Command, args []string) {
+			refresh()
+		},
+	}
+	// cmd.Flags().StringVar(&alias, "alias", "", "alias")
+	// cmd.Flags().StringVar(&appId, "appid", "", "appid")
+	return cmd
+}
+
+func refresh() {
+	//buildBookMenu()
+	copyLackFile()
+	delPreNext()
+	buildREADME()
+	buildChapterTwo(true)
+	addPreNext()
+}
diff --git a/ctl/render.go b/ctl/render.go
new file mode 100644
index 000000000..3e62a36eb
--- /dev/null
+++ b/ctl/render.go
@@ -0,0 +1,367 @@
+package main
+
+import (
+	"bufio"
+	"encoding/json"
+	"fmt"
+	"io"
+	"os"
+	"regexp"
+	"sort"
+	"strconv"
+	"strings"
+
+	m "github.com/halfrost/LeetCode-Go/ctl/models"
+	"github.com/halfrost/LeetCode-Go/ctl/util"
+	"github.com/spf13/cobra"
+)
+
+var (
+	chapterTwoList = []string{"Array", "String", "Two Pointers", "Linked List", "Stack", "Tree", "Dynamic Programming", "Backtracking", "Depth First Search", "Breadth First Search",
+		"Binary Search", "Math", "Hash Table", "Sorting", "Bit Manipulation", "Union Find", "Sliding Window", "Segment Tree", "Binary Indexed Tree"}
+	chapterTwoFileName = []string{"Array", "String", "Two_Pointers", "Linked_List", "Stack", "Tree", "Dynamic_Programming", "Backtracking", "Depth_First_Search", "Breadth_First_Search",
+		"Binary_Search", "Math", "Hash_Table", "Sorting", "Bit_Manipulation", "Union_Find", "Sliding_Window", "Segment_Tree", "Binary_Indexed_Tree"}
+	chapterTwoSlug = []string{"array", "string", "two-pointers", "linked-list", "stack", "tree", "dynamic-programming", "backtracking", "depth-first-search", "breadth-first-search",
+		"binary-search", "math", "hash-table", "sorting", "bit-manipulation", "union-find", "sliding-window", "segment-tree", "binary-indexed-tree"}
+)
+
+func newBuildCommand() *cobra.Command {
+	mc := &cobra.Command{
+		Use:   "build <subcommand>",
+		Short: "Build doc related commands",
+	}
+	//mc.PersistentFlags().StringVar(&logicEndpoint, "endpoint", "localhost:5880", "endpoint of logic service")
+	mc.AddCommand(
+		newBuildREADME(),
+		newBuildChapterTwo(),
+		// newBuildMenu(),
+	)
+	return mc
+}
+
+func newBuildREADME() *cobra.Command {
+	cmd := &cobra.Command{
+		Use:   "readme",
+		Short: "Build readme.md commands",
+		Run: func(cmd *cobra.Command, args []string) {
+			buildREADME()
+		},
+	}
+	// cmd.Flags().StringVar(&alias, "alias", "", "alias")
+	// cmd.Flags().StringVar(&appId, "appid", "", "appid")
+	return cmd
+}
+
+func newBuildChapterTwo() *cobra.Command {
+	cmd := &cobra.Command{
+		Use:   "chapter-two",
+		Short: "Build Chapter Two commands",
+		Run: func(cmd *cobra.Command, args []string) {
+			buildChapterTwo(true)
+		},
+	}
+	// cmd.Flags().StringVar(&alias, "alias", "", "alias")
+	// cmd.Flags().StringVar(&appId, "appid", "", "appid")
+	return cmd
+}
+
+func newBuildMenu() *cobra.Command {
+	cmd := &cobra.Command{
+		Use:   "menu",
+		Short: "Build Menu commands",
+		Run: func(cmd *cobra.Command, args []string) {
+			buildBookMenu()
+		},
+	}
+	// cmd.Flags().StringVar(&alias, "alias", "", "alias")
+	// cmd.Flags().StringVar(&appId, "appid", "", "appid")
+	return cmd
+}
+
+func buildREADME() {
+	var (
+		problems []m.StatStatusPairs
+		lpa      m.LeetCodeProblemAll
+		info     m.UserInfo
+	)
+	// 请求所有题目信息
+	body := getProblemAllList()
+	problemsMap, optimizingIds := map[int]m.StatStatusPairs{}, []int{}
+	err := json.Unmarshal(body, &lpa)
+	if err != nil {
+		fmt.Println(err)
+		return
+	}
+	//writeFile("leetcode_problem.json", body)
+
+	// 拼凑 README 需要渲染的数据
+	problems = lpa.StatStatusPairs
+	info = m.ConvertUserInfoModel(lpa)
+	for _, v := range problems {
+		problemsMap[int(v.Stat.FrontendQuestionID)] = v
+	}
+	mdrows := m.ConvertMdModelFromSsp(problems)
+	sort.Sort(m.SortByQuestionID(mdrows))
+	solutionIds, _, try := util.LoadSolutionsDir()
+	m.GenerateMdRows(solutionIds, mdrows)
+	info.EasyTotal, info.MediumTotal, info.HardTotal, info.OptimizingEasy, info.OptimizingMedium, info.OptimizingHard, optimizingIds = statisticalData(problemsMap, solutionIds)
+	omdrows := m.ConvertMdModelFromIds(problemsMap, optimizingIds)
+	sort.Sort(m.SortByQuestionID(omdrows))
+
+	// 按照模板渲染 README
+	res, err := renderReadme("./template/template.markdown", len(solutionIds), try, m.Mdrows{Mdrows: mdrows}, m.Mdrows{Mdrows: omdrows}, info)
+	if err != nil {
+		fmt.Println(err)
+		return
+	}
+	util.WriteFile("../README.md", res)
+	fmt.Println("write file successful")
+	//makeReadmeFile(mds)
+}
+
+func renderReadme(filePath string, total, try int, mdrows, omdrows m.Mdrows, user m.UserInfo) ([]byte, error) {
+	f, err := os.OpenFile(filePath, os.O_RDONLY, 0644)
+	if err != nil {
+		return nil, err
+	}
+	defer f.Close()
+	reader, output := bufio.NewReader(f), []byte{}
+
+	for {
+		line, _, err := reader.ReadLine()
+		if err != nil {
+			if err == io.EOF {
+				return output, nil
+			}
+			return nil, err
+		}
+		if ok, _ := regexp.Match("{{.AvailableTable}}", line); ok {
+			reg := regexp.MustCompile("{{.AvailableTable}}")
+			newByte := reg.ReplaceAll(line, []byte(mdrows.AvailableTable()))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else if ok, _ := regexp.Match("{{.TotalNum}}", line); ok {
+			reg := regexp.MustCompile("{{.TotalNum}}")
+			newByte := reg.ReplaceAll(line, []byte(fmt.Sprintf("以下已经收录了 %v 道题的题解，还有 %v 道题在尝试优化到 beats 100%%", total, try)))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else if ok, _ := regexp.Match("{{.PersonalData}}", line); ok {
+			reg := regexp.MustCompile("{{.PersonalData}}")
+			newByte := reg.ReplaceAll(line, []byte(user.PersonalData()))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else if ok, _ := regexp.Match("{{.OptimizingTable}}", line); ok {
+			reg := regexp.MustCompile("{{.OptimizingTable}}")
+			newByte := reg.ReplaceAll(line, []byte(fmt.Sprintf("以下 %v 道题还需要优化到 100%% 的题目列表\n\n%v", (user.OptimizingEasy+user.OptimizingMedium+user.OptimizingHard), omdrows.AvailableTable())))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else {
+			output = append(output, line...)
+			output = append(output, []byte("\n")...)
+		}
+	}
+}
+
+// internal: true  渲染的链接都是 hugo 内部链接，用户生成 hugo web
+//
+//	false 渲染的链接是外部 HTTPS 链接，用于生成 PDF
+func buildChapterTwo(internal bool) {
+	var (
+		gr        m.GraphQLResp
+		questions []m.Question
+		count     int
+	)
+	for index, tag := range chapterTwoSlug {
+		body := getTagProblemList(tag)
+		// fmt.Printf("%v\n", string(body))
+		err := json.Unmarshal(body, &gr)
+		if err != nil {
+			fmt.Println(err)
+			return
+		}
+		questions = gr.Data.TopicTag.Questions
+		mdrows := m.ConvertMdModelFromQuestions(questions)
+		sort.Sort(m.SortByQuestionID(mdrows))
+		solutionIds, _, _ := util.LoadSolutionsDir()
+		tl, err := loadMetaData(fmt.Sprintf("./meta/%v", chapterTwoFileName[index]))
+		if err != nil {
+			fmt.Printf("err = %v\n", err)
+		}
+		tls := m.GenerateTagMdRows(solutionIds, tl, mdrows, internal)
+		//fmt.Printf("tls = %v\n", tls)
+		//  按照模板渲染 README
+		res, err := renderChapterTwo(fmt.Sprintf("./template/%v.md", chapterTwoFileName[index]), m.TagLists{TagLists: tls})
+		if err != nil {
+			fmt.Println(err)
+			return
+		}
+		if internal {
+			util.WriteFile(fmt.Sprintf("../website/content/ChapterTwo/%v.md", chapterTwoFileName[index]), res)
+		} else {
+			util.WriteFile(fmt.Sprintf("./pdftemp/ChapterTwo/%v.md", chapterTwoFileName[index]), res)
+		}
+
+		count++
+	}
+	fmt.Printf("write %v files successful", count)
+}
+
+func loadMetaData(filePath string) (map[int]m.TagList, error) {
+	f, err := os.OpenFile(filePath, os.O_RDONLY, 0644)
+	if err != nil {
+		return nil, err
+	}
+	defer f.Close()
+	reader, metaMap := bufio.NewReader(f), map[int]m.TagList{}
+
+	for {
+		line, _, err := reader.ReadLine()
+		if err != nil {
+			if err == io.EOF {
+				return metaMap, nil
+			}
+			return nil, err
+		}
+		s := strings.Split(string(line), "|")
+		v, _ := strconv.Atoi(strings.Split(s[1], ".")[0])
+		// v[0] 是题号，s[4] time, s[5] space, s[6] favorite
+		metaMap[v] = m.TagList{
+			FrontendQuestionID: int32(v),
+			Acceptance:         "",
+			Difficulty:         "",
+			TimeComplexity:     s[4],
+			SpaceComplexity:    s[5],
+			Favorite:           s[6],
+		}
+	}
+}
+
+func renderChapterTwo(filePath string, tls m.TagLists) ([]byte, error) {
+	f, err := os.OpenFile(filePath, os.O_RDONLY, 0644)
+	if err != nil {
+		return nil, err
+	}
+	defer f.Close()
+	reader, output := bufio.NewReader(f), []byte{}
+
+	for {
+		line, _, err := reader.ReadLine()
+		if err != nil {
+			if err == io.EOF {
+				return output, nil
+			}
+			return nil, err
+		}
+		if ok, _ := regexp.Match("{{.AvailableTagTable}}", line); ok {
+			reg := regexp.MustCompile("{{.AvailableTagTable}}")
+			newByte := reg.ReplaceAll(line, []byte(tls.AvailableTagTable()))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else {
+			output = append(output, line...)
+			output = append(output, []byte("\n")...)
+		}
+	}
+}
+
+func buildBookMenu() {
+	copyLackFile()
+	// 按照模板重新渲染 Menu
+	res, err := renderBookMenu("./template/menu.md")
+	if err != nil {
+		fmt.Println(err)
+		return
+	}
+	util.WriteFile("../website/content/menu/index.md", res)
+	fmt.Println("generate Menu successful")
+}
+
+// 拷贝 leetcode 目录下的题解 README 文件至第四章对应文件夹中
+func copyLackFile() {
+	solutionIds, soName, _ := util.LoadSolutionsDir()
+	_, ch4Ids := util.LoadChapterFourDir()
+
+	needCopy := []string{}
+	for i := 0; i < len(solutionIds); i++ {
+		if util.BinarySearch(ch4Ids, solutionIds[i]) == -1 {
+			needCopy = append(needCopy, soName[i])
+		}
+	}
+	if len(needCopy) > 0 {
+		fmt.Printf("有 %v 道题需要拷贝到第四章中\n", len(needCopy))
+		for i := 0; i < len(needCopy); i++ {
+			if needCopy[i][4] == '.' {
+				tmp, err := strconv.Atoi(needCopy[i][:4])
+				if err != nil {
+					fmt.Println(err)
+				}
+				err = os.MkdirAll(fmt.Sprintf("../website/content/ChapterFour/%v", util.GetChpaterFourFileNum(tmp)), os.ModePerm)
+				if err != nil {
+					fmt.Println(err)
+				}
+				util.CopyFile(fmt.Sprintf("../website/content/ChapterFour/%v/%v.md", util.GetChpaterFourFileNum(tmp), needCopy[i]), fmt.Sprintf("../leetcode/%v/README.md", needCopy[i]))
+				util.CopyFile(fmt.Sprintf("../website/content/ChapterFour/%v/_index.md", util.GetChpaterFourFileNum(tmp)), "./template/collapseSection.md")
+			}
+		}
+	} else {
+		fmt.Printf("【第四章没有需要添加的题解，已经完整了】\n")
+	}
+}
+
+func generateMenu() string {
+	res := ""
+	res += menuLine(chapterOneMenuOrder, "ChapterOne")
+	res += menuLine(chapterTwoFileOrder, "ChapterTwo")
+	res += menuLine(chapterThreeFileOrder, "ChapterThree")
+	chapterFourFileOrder, _ := getChapterFourFileOrder()
+	res += menuLine(chapterFourFileOrder, "ChapterFour")
+	return res
+}
+
+func menuLine(order []string, chapter string) string {
+	res := ""
+	for i := 0; i < len(order); i++ {
+		if i == 1 && chapter == "ChapterOne" {
+			res += fmt.Sprintf("  - [%v]({{< relref \"/%v/%v\" >}})\n", chapterMap[chapter][order[i]], chapter, order[i])
+			continue
+		}
+		if i == 0 {
+			res += fmt.Sprintf("- [%v]({{< relref \"/%v/%v.md\" >}})\n", chapterMap[chapter][order[i]], chapter, order[i])
+		} else {
+			if chapter == "ChapterFour" {
+				res += fmt.Sprintf("    - [%v]({{< relref \"/%v/%v.md\" >}})\n", order[i], chapter, order[i])
+			} else {
+				res += fmt.Sprintf("  - [%v]({{< relref \"/%v/%v.md\" >}})\n", chapterMap[chapter][order[i]], chapter, order[i])
+			}
+		}
+	}
+	return res
+}
+
+func renderBookMenu(filePath string) ([]byte, error) {
+	f, err := os.OpenFile(filePath, os.O_RDONLY, 0644)
+	if err != nil {
+		return nil, err
+	}
+	defer f.Close()
+	reader, output := bufio.NewReader(f), []byte{}
+
+	for {
+		line, _, err := reader.ReadLine()
+		if err != nil {
+			if err == io.EOF {
+				return output, nil
+			}
+			return nil, err
+		}
+		if ok, _ := regexp.Match("{{.BookMenu}}", line); ok {
+			reg := regexp.MustCompile("{{.BookMenu}}")
+			newByte := reg.ReplaceAll(line, []byte(generateMenu()))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
+		} else {
+			output = append(output, line...)
+			output = append(output, []byte("\n")...)
+		}
+	}
+}
diff --git a/ctl/request.go b/ctl/request.go
new file mode 100644
index 000000000..8222a1fc0
--- /dev/null
+++ b/ctl/request.go
@@ -0,0 +1,110 @@
+package main
+
+import (
+	"bytes"
+	"fmt"
+	"io/ioutil"
+	"net/http"
+
+	"github.com/mozillazg/request"
+)
+
+const (
+	// AllProblemURL define
+	AllProblemURL = "https://leetcode.com/api/problems/all/"
+	// QraphqlURL define
+	QraphqlURL = "https://leetcode.com/graphql"
+	// LoginPageURL define
+	LoginPageURL = "https://leetcode.com/accounts/login/"
+	// AlgorithmsURL define
+	AlgorithmsURL = "https://leetcode.com/api/problems/Algorithms/"
+
+	// ArrayProblemURL define
+	ArrayProblemURL = "https://leetcode.com/tag/array/"
+)
+
+var req *request.Request
+
+func newReq() *request.Request {
+	if req == nil {
+		req = signin()
+	}
+	return req
+}
+
+func signin() *request.Request {
+	cfg := getConfig()
+	req := request.NewRequest(new(http.Client))
+	req.Headers = map[string]string{
+		"Content-Type":    "application/json",
+		"Accept-Encoding": "",
+		"cookie":          cfg.Cookie,
+		"x-csrftoken":     cfg.CSRFtoken,
+		"Referer":         "https://leetcode.com/accounts/login/",
+		"origin":          "https://leetcode.com",
+	}
+	return req
+}
+
+func getRaw(URL string) []byte {
+	req := newReq()
+	resp, err := req.Get(URL)
+	if err != nil {
+		fmt.Printf("getRaw: Get Error: " + err.Error())
+		return []byte{}
+	}
+	defer resp.Body.Close()
+
+	body, err := ioutil.ReadAll(resp.Body)
+	if err != nil {
+		fmt.Printf("getRaw: Read Error: " + err.Error())
+		return []byte{}
+	}
+	if resp.StatusCode == 200 {
+		fmt.Println("Get problem Success!")
+	}
+	return body
+}
+
+func getProblemAllList() []byte {
+	return getRaw(AllProblemURL)
+}
+
+// Variables define
+type Variables struct {
+	slug string
+}
+
+func getQraphql(payload string) []byte {
+	req := newReq()
+	resp, err := req.PostForm(QraphqlURL, bytes.NewBuffer([]byte(payload)))
+	if err != nil {
+		fmt.Printf("getRaw: Get Error: " + err.Error())
+		return []byte{}
+	}
+	defer resp.Body.Close()
+
+	body, err := ioutil.ReadAll(resp.Body)
+	if err != nil {
+		fmt.Printf("getRaw: Read Error: " + err.Error())
+		return []byte{}
+	}
+	if resp.StatusCode == 200 {
+		fmt.Println("Get problem Success!")
+	}
+	return body
+}
+
+func getTopicTag(variable string) string {
+	return fmt.Sprintf(`{
+		"operationName": "getTopicTag",
+		"variables": {
+			"slug": "%s"
+		},
+		"query": "query getTopicTag($slug: String!) {  topicTag(slug: $slug) {    name    translatedName    slug    questions {      status      questionId      questionFrontendId      title      titleSlug      translatedTitle      stats      difficulty      isPaidOnly      topicTags {        name        translatedName        slug        __typename      }      companyTags {        name        translatedName        slug        __typename      }      __typename    }    frequencies    __typename  }  favoritesLists {    publicFavorites {      ...favoriteFields      __typename    }    privateFavorites {      ...favoriteFields      __typename    }    __typename  }}fragment favoriteFields on FavoriteNode {  idHash  id  name  isPublicFavorite  viewCount  creator  isWatched  questions {    questionId    title    titleSlug    __typename  }  __typename}"
+		}`, variable)
+}
+
+func getTagProblemList(tag string) []byte {
+	return getQraphql(getTopicTag(tag))
+}
diff --git a/ctl/statistic.go b/ctl/statistic.go
new file mode 100644
index 000000000..20b84ccb9
--- /dev/null
+++ b/ctl/statistic.go
@@ -0,0 +1,42 @@
+package main
+
+import (
+	"sort"
+
+	m "github.com/halfrost/LeetCode-Go/ctl/models"
+	"github.com/halfrost/LeetCode-Go/ctl/util"
+)
+
+func statisticalData(problemsMap map[int]m.StatStatusPairs, solutionIds []int) (easyTotal, mediumTotal, hardTotal, optimizingEasy, optimizingMedium, optimizingHard int32, optimizingIds []int) {
+	easyTotal, mediumTotal, hardTotal, optimizingEasy, optimizingMedium, optimizingHard, optimizingIds = 0, 0, 0, 0, 0, 0, []int{}
+	for _, v := range problemsMap {
+		switch m.DifficultyMap[v.Difficulty.Level] {
+		case "Easy":
+			{
+				easyTotal++
+				if v.Status == "ac" && util.BinarySearch(solutionIds, int(v.Stat.FrontendQuestionID)) == -1 {
+					optimizingEasy++
+					optimizingIds = append(optimizingIds, int(v.Stat.FrontendQuestionID))
+				}
+			}
+		case "Medium":
+			{
+				mediumTotal++
+				if v.Status == "ac" && util.BinarySearch(solutionIds, int(v.Stat.FrontendQuestionID)) == -1 {
+					optimizingMedium++
+					optimizingIds = append(optimizingIds, int(v.Stat.FrontendQuestionID))
+				}
+			}
+		case "Hard":
+			{
+				hardTotal++
+				if v.Status == "ac" && util.BinarySearch(solutionIds, int(v.Stat.FrontendQuestionID)) == -1 {
+					optimizingHard++
+					optimizingIds = append(optimizingIds, int(v.Stat.FrontendQuestionID))
+				}
+			}
+		}
+	}
+	sort.Ints(optimizingIds)
+	return easyTotal, mediumTotal, hardTotal, optimizingEasy, optimizingMedium, optimizingHard, optimizingIds
+}
diff --git a/ctl/template/Array.md b/ctl/template/Array.md
new file mode 100644
index 000000000..257fd3625
--- /dev/null
+++ b/ctl/template/Array.md
@@ -0,0 +1,9 @@
+---
+title: 2.01 Array
+type: docs
+weight: 1
+---
+
+# Array
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Backtracking.md b/ctl/template/Backtracking.md
new file mode 100644
index 000000000..fe1a54e6b
--- /dev/null
+++ b/ctl/template/Backtracking.md
@@ -0,0 +1,101 @@
+---
+title: 2.08 ✅ Backtracking
+type: docs
+weight: 8
+---
+
+# Backtracking
+
+![](https://img.halfrost.com/Leetcode/Backtracking.png)
+
+- 排列问题 Permutations。第 46 题，第 47 题。第 60 题，第 526 题，第 996 题。
+- 组合问题 Combination。第 39 题，第 40 题，第 77 题，第 216 题。
+- 排列和组合杂交问题。第 1079 题。
+- N 皇后终极解法(二进制解法)。第 51 题，第 52 题。
+- 数独问题。第 37 题。
+- 四个方向搜索。第 79 题，第 212 题，第 980 题。
+- 子集合问题。第 78 题，第 90 题。
+- Trie。第 208 题，第 211 题。
+- BFS 优化。第 126 题，第 127 题。
+- DFS 模板。(只是一个例子，不对应任何题)
+
+```go
+func combinationSum2(candidates []int, target int) [][]int {
+	if len(candidates) == 0 {
+		return [][]int{}
+	}
+	c, res := []int{}, [][]int{}
+	sort.Ints(candidates)
+	findcombinationSum2(candidates, target, 0, c, &res)
+	return res
+}
+
+func findcombinationSum2(nums []int, target, index int, c []int, res *[][]int) {
+	if target == 0 {
+		b := make([]int, len(c))
+		copy(b, c)
+		*res = append(*res, b)
+		return
+	}
+	for i := index; i < len(nums); i++ {
+		if i > index && nums[i] == nums[i-1] { // 这里是去重的关键逻辑
+			continue
+		}
+		if target >= nums[i] {
+			c = append(c, nums[i])
+			findcombinationSum2(nums, target-nums[i], i+1, c, res)
+			c = c[:len(c)-1]
+		}
+	}
+}
+```
+- BFS 模板。(只是一个例子，不对应任何题)
+
+```go
+func updateMatrix_BFS(matrix [][]int) [][]int {
+	res := make([][]int, len(matrix))
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return res
+	}
+	queue := make([][]int, 0)
+	for i, _ := range matrix {
+		res[i] = make([]int, len(matrix[0]))
+		for j, _ := range res[i] {
+			if matrix[i][j] == 0 {
+				res[i][j] = -1
+				queue = append(queue, []int{i, j})
+			}
+		}
+	}
+	level := 1
+	for len(queue) > 0 {
+		size := len(queue)
+		for size > 0 {
+			size -= 1
+			node := queue[0]
+			queue = queue[1:]
+			i, j := node[0], node[1]
+			for _, direction := range [][]int{{-1, 0}, {1, 0}, {0, 1}, {0, -1}} {
+				x := i + direction[0]
+				y := j + direction[1]
+				if x < 0 || x >= len(matrix) || y < 0 || y >= len(matrix[0]) || res[x][y] < 0 || res[x][y] > 0 {
+					continue
+				}
+				res[x][y] = level
+				queue = append(queue, []int{x, y})
+			}
+		}
+		level++
+	}
+	for i, row := range res {
+		for j, cell := range row {
+			if cell == -1 {
+				res[i][j] = 0
+			}
+		}
+	}
+	return res
+}
+```
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Binary_Indexed_Tree.md b/ctl/template/Binary_Indexed_Tree.md
new file mode 100644
index 000000000..039b49ccb
--- /dev/null
+++ b/ctl/template/Binary_Indexed_Tree.md
@@ -0,0 +1,11 @@
+---
+title: 2.19 ✅ Binary Indexed Tree
+type: docs
+weight: 19
+---
+
+# Binary Indexed Tree
+
+![](https://img.halfrost.com/Leetcode/Binary_Indexed_Tree.png)
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Binary_Search.md b/ctl/template/Binary_Search.md
new file mode 100644
index 000000000..e7a5de3de
--- /dev/null
+++ b/ctl/template/Binary_Search.md
@@ -0,0 +1,132 @@
+---
+title: 2.11 Binary Search
+type: docs
+weight: 11
+---
+
+# Binary Search
+
+- 二分搜索的经典写法。需要注意的三点：
+	1. 循环退出条件，注意是 low <= high，而不是 low < high。
+	2. mid 的取值，mid := low + (high-low)>>1
+	3. low 和 high 的更新。low = mid + 1，high = mid - 1。
+
+```go
+func binarySearchMatrix(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[mid] == target {
+			return mid
+		} else if nums[mid] > target {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+```
+
+- 二分搜索的变种写法。有 4 个基本变种:
+	1. 查找第一个与 target 相等的元素，时间复杂度 O(logn)
+	2. 查找最后一个与 target 相等的元素，时间复杂度 O(logn)
+	3. 查找第一个大于等于 target 的元素，时间复杂度 O(logn)
+	4. 查找最后一个小于等于 target 的元素，时间复杂度 O(logn)
+
+```go
+// 二分查找第一个与 target 相等的元素，时间复杂度 O(logn)
+func searchFirstEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == 0) || (nums[mid-1] != target) { // 找到第一个与 target 相等的元素
+				return mid
+			}
+			high = mid - 1
+		}
+	}
+	return -1
+}
+
+// 二分查找最后一个与 target 相等的元素，时间复杂度 O(logn)
+func searchLastEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == len(nums)-1) || (nums[mid+1] != target) { // 找到最后一个与 target 相等的元素
+				return mid
+			}
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// 二分查找第一个大于等于 target 的元素，时间复杂度 O(logn)
+func searchFirstGreaterElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] >= target {
+			if (mid == 0) || (nums[mid-1] < target) { // 找到第一个大于等于 target 的元素
+				return mid
+			}
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// 二分查找最后一个小于等于 target 的元素，时间复杂度 O(logn)
+func searchLastLessElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] <= target {
+			if (mid == len(nums)-1) || (nums[mid+1] > target) { // 找到最后一个小于等于 target 的元素
+				return mid
+			}
+			low = mid + 1
+		} else {
+			high = mid - 1
+		}
+	}
+	return -1
+}
+```
+
+- 在基本有序的数组中用二分搜索。经典解法可以解，变种写法也可以写，常见的题型，在山峰数组中找山峰，在旋转有序数组中找分界点。第 33 题，第 81 题，第 153 题，第 154 题，第 162 题，第 852 题
+
+```go
+func peakIndexInMountainArray(A []int) int {
+	low, high := 0, len(A)-1
+	for low < high {
+		mid := low + (high-low)>>1
+		// 如果 mid 较大，则左侧存在峰值，high = m，如果 mid + 1 较大，则右侧存在峰值，low = mid + 1
+		if A[mid] > A[mid+1] {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+```
+
+- max-min 最大值最小化问题。求在最小满足条件的情况下的最大值。第 410 题，第 875 题，第 1011 题，第 1283 题。
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Bit_Manipulation.md b/ctl/template/Bit_Manipulation.md
new file mode 100644
index 000000000..6c643994a
--- /dev/null
+++ b/ctl/template/Bit_Manipulation.md
@@ -0,0 +1,45 @@
+---
+title: 2.15 ✅ Bit Manipulation
+type: docs
+weight: 15
+---
+
+# Bit Manipulation
+
+![](https://img.halfrost.com/Leetcode/Bit_Manipulation.png)
+
+- 异或的特性。第 136 题，第 268 题，第 389 题，第 421 题，
+
+```go
+x ^ 0 = x
+x ^ 11111……1111 = ~x
+x ^ (~x) = 11111……1111
+x ^ x = 0
+a ^ b = c  => a ^ c = b  => b ^ c = a (交换律)
+a ^ b ^ c = a ^ (b ^ c) = (a ^ b）^ c (结合律)
+```
+
+- 构造特殊 Mask，将特殊位置放 0 或 1。
+
+```go
+将 x 最右边的 n 位清零， x & ( ~0 << n )
+获取 x 的第 n 位值(0 或者 1)，(x >> n) & 1
+获取 x 的第 n 位的幂值，x & (1 << (n - 1))
+仅将第 n 位置为 1，x | (1 << n)
+仅将第 n 位置为 0，x & (~(1 << n))
+将 x 最高位至第 n 位(含)清零，x & ((1 << n) - 1)
+将第 n 位至第 0 位(含)清零，x & (~((1 << (n + 1)) - 1)）
+```
+
+- 有特殊意义的 & 位操作运算。第 260 题，第 201 题，第 318 题，第 371 题，第 397 题，第 461 题，第 693 题，
+
+```go
+X & 1 == 1 判断是否是奇数(偶数)
+X & = (X - 1) 将最低位(LSB)的 1 清零
+X & -X 得到最低位(LSB)的 1
+X & ~X = 0
+```
+
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Breadth_First_Search.md b/ctl/template/Breadth_First_Search.md
new file mode 100644
index 000000000..64e56acba
--- /dev/null
+++ b/ctl/template/Breadth_First_Search.md
@@ -0,0 +1,10 @@
+---
+title: 2.10 Breadth First Search
+type: docs
+weight: 10
+---
+
+# Breadth First Search
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Depth_First_Search.md b/ctl/template/Depth_First_Search.md
new file mode 100644
index 000000000..e8fb44153
--- /dev/null
+++ b/ctl/template/Depth_First_Search.md
@@ -0,0 +1,10 @@
+---
+title: 2.09 Depth First Search
+type: docs
+weight: 9
+---
+
+# Depth First Search
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Dynamic_Programming.md b/ctl/template/Dynamic_Programming.md
new file mode 100644
index 000000000..f18015f77
--- /dev/null
+++ b/ctl/template/Dynamic_Programming.md
@@ -0,0 +1,10 @@
+---
+title: 2.07 Dynamic Programming
+type: docs
+weight: 7
+---
+
+# Dynamic Programming
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Hash_Table.md b/ctl/template/Hash_Table.md
new file mode 100644
index 000000000..03fb07d62
--- /dev/null
+++ b/ctl/template/Hash_Table.md
@@ -0,0 +1,10 @@
+---
+title: 2.13 Hash Table
+type: docs
+weight: 13
+---
+
+# Hash Table
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Linked_List.md b/ctl/template/Linked_List.md
new file mode 100644
index 000000000..b8673cc39
--- /dev/null
+++ b/ctl/template/Linked_List.md
@@ -0,0 +1,24 @@
+---
+title: 2.04 ✅ Linked List
+type: docs
+weight: 4
+---
+
+# Linked List
+
+![](https://img.halfrost.com/Leetcode/Linked_List.png)
+
+
+- 巧妙的构造虚拟头结点。可以使遍历处理逻辑更加统一。
+- 灵活使用递归。构造递归条件，使用递归可以巧妙的解题。不过需要注意有些题目不能使用递归，因为递归深度太深会导致超时和栈溢出。
+- 链表区间逆序。第 92 题。
+- 链表寻找中间节点。第 876 题。链表寻找倒数第 n 个节点。第 19 题。只需要一次遍历就可以得到答案。
+- 合并 K 个有序链表。第 21 题，第 23 题。
+- 链表归类。第 86 题，第 328 题。
+- 链表排序，时间复杂度要求 O(n * log n)，空间复杂度 O(1)。只有一种做法，归并排序，至顶向下归并。第 148 题。
+- 判断链表是否存在环，如果有环，输出环的交叉点的下标；判断 2 个链表是否有交叉点，如果有交叉点，输出交叉点。第 141 题，第 142 题，第 160 题。
+
+
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Math.md b/ctl/template/Math.md
new file mode 100644
index 000000000..dacbb1ff7
--- /dev/null
+++ b/ctl/template/Math.md
@@ -0,0 +1,10 @@
+---
+title: 2.12 Math
+type: docs
+weight: 12
+---
+
+# Math
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Segment_Tree.md b/ctl/template/Segment_Tree.md
new file mode 100644
index 000000000..220ac679e
--- /dev/null
+++ b/ctl/template/Segment_Tree.md
@@ -0,0 +1,38 @@
+---
+title: 2.18 ✅ Segment Tree
+type: docs
+weight: 18
+---
+
+# Segment Tree
+
+![](https://img.halfrost.com/Leetcode/Segment_Tree.png)
+
+- 线段树的经典数组实现写法。将合并两个节点 pushUp 逻辑抽象出来了，可以实现任意操作(常见的操作有：加法，取 max，min 等等)。第 218 题，第 303 题，第 307 题，第 699 题。
+- 计数线段树的经典写法。第 315 题，第 327 题，第 493 题。
+- 线段树的树的实现写法。第 715 题，第 732 题。
+- 区间懒惰更新。第 218 题，第 699 题。
+- 离散化。离散化需要注意一个特殊情况：假如三个区间为 [1,10] [1,4] [6,10]，离散化后 x[1]=1,x[2]=4,x[3]=6,x[4]=10。第一个区间为 [1,4]，第二个区间为 [1,2]，第三个区间为 [3,4]，这样一来，区间一 = 区间二 + 区间三，这和离散前的模型不符，离散前，很明显，区间一 > 区间二 + 区间三。正确的做法是：在相差大于 1 的数间加一个数，例如在上面 1 4 6 10 中间加 5，即可 x[1]=1,x[2]=4,x[3]=5,x[4]=6,x[5]=10。这样处理之后，区间一是 1-5 ，区间二是 1-2 ，区间三是 4-5 。
+- 灵活构建线段树。线段树节点可以存储多条信息，合并两个节点的 pushUp 操作也可以是多样的。第 850 题，第 1157 题。
+
+
+线段树[题型](https://blog.csdn.net/xuechelingxiao/article/details/38313105)从简单到困难:
+
+1. 单点更新:  
+	[HDU 1166 敌兵布阵](http://acm.hdu.edu.cn/showproblem.php?pid=1166) update:单点增减 query:区间求和  
+	[HDU 1754 I Hate It](http://acm.hdu.edu.cn/showproblem.php?pid=1754) update:单点替换 query:区间最值  
+	[HDU 1394 Minimum Inversion Number](http://acm.hdu.edu.cn/showproblem.php?pid=1394) update:单点增减 query:区间求和  
+	[HDU 2795 Billboard](http://acm.hdu.edu.cn/showproblem.php?pid=2795) query:区间求最大值的位子(直接把update的操作在query里做了)
+2. 区间更新:  
+	[HDU 1698 Just a Hook](http://acm.hdu.edu.cn/showproblem.php?pid=1698) update:成段替换 (由于只query一次总区间,所以可以直接输出 1 结点的信息)  
+	[POJ 3468 A Simple Problem with Integers](http://poj.org/problem?id=3468) update:成段增减 query:区间求和  
+	[POJ 2528 Mayor’s posters](http://poj.org/problem?id=2528) 离散化 + update:成段替换 query:简单hash  
+	[POJ 3225 Help with Intervals](http://poj.org/problem?id=3225) update:成段替换,区间异或 query:简单hash
+3. 区间合并(这类题目会询问区间中满足条件的连续最长区间,所以PushUp的时候需要对左右儿子的区间进行合并):  
+	[POJ 3667 Hotel](http://poj.org/problem?id=3667) update:区间替换 query:询问满足条件的最左端点
+4. 扫描线(这类题目需要将一些操作排序,然后从左到右用一根扫描线扫过去最典型的就是矩形面积并,周长并等题):  
+	[HDU 1542 Atlantis](http://acm.hdu.edu.cn/showproblem.php?pid=1542) update:区间增减 query:直接取根节点的值  
+	[HDU 1828 Picture](http://acm.hdu.edu.cn/showproblem.php?pid=1828) update:区间增减 query:直接取根节点的值
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Sliding_Window.md b/ctl/template/Sliding_Window.md
new file mode 100644
index 000000000..117e50223
--- /dev/null
+++ b/ctl/template/Sliding_Window.md
@@ -0,0 +1,30 @@
+---
+title: 2.17 ✅ Sliding Window
+type: docs
+weight: 17
+---
+
+# Sliding Window
+
+![](https://img.halfrost.com/Leetcode/Sliding_Window.png)
+
+- 双指针滑动窗口的经典写法。右指针不断往右移，移动到不能往右移动为止(具体条件根据题目而定)。当右指针到最右边以后，开始挪动左指针，释放窗口左边界。第 3 题，第 76 题，第 209 题，第 424 题，第 438 题，第 567 题，第 713 题，第 763 题，第 845 题，第 881 题，第 904 题，第 978 题，第 992 题，第 1004 题，第 1040 题，第 1052 题。
+
+```c
+	left, right := 0, -1
+
+	for left < len(s) {
+		if right+1 < len(s) && freq[s[right+1]-'a'] == 0 {
+			freq[s[right+1]-'a']++
+			right++
+		} else {
+			freq[s[left]-'a']--
+			left++
+		}
+		result = max(result, right-left+1)
+	}
+```
+- 滑动窗口经典题。第 239 题，第 480 题。
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Sorting.md b/ctl/template/Sorting.md
new file mode 100644
index 000000000..5038c03cf
--- /dev/null
+++ b/ctl/template/Sorting.md
@@ -0,0 +1,19 @@
+---
+title: 2.14 ✅ Sorting
+type: docs
+weight: 14
+---
+
+# Sorting
+
+![](https://img.halfrost.com/Leetcode/Sort.png)
+
+- 深刻的理解多路快排。第 75 题。
+- 链表的排序，插入排序(第 147 题)和归并排序(第 148 题)
+- 桶排序和基数排序。第 164 题。
+- "摆动排序"。第 324 题。
+- 两两不相邻的排序。第 767 题，第 1054 题。
+- "饼子排序"。第 969 题。
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Stack.md b/ctl/template/Stack.md
new file mode 100644
index 000000000..0cc1f39b6
--- /dev/null
+++ b/ctl/template/Stack.md
@@ -0,0 +1,18 @@
+---
+title: 2.05 ✅ Stack
+type: docs
+weight: 5
+---
+
+# Stack
+
+![](https://img.halfrost.com/Leetcode/Stack.png)
+
+- 括号匹配问题及类似问题。第 20 题，第 921 题，第 1021 题。
+- 栈的基本 pop 和 push 操作。第 71 题，第 150 题，第 155 题，第 224 题，第 225 题，第 232 题，第 946 题，第 1047 题。
+- 利用栈进行编码问题。第 394 题，第 682 题，第 856 题，第 880 题。
+- **单调栈**。**利用栈维护一个单调递增或者递减的下标数组**。第 84 题，第 456 题，第 496 题，第 503 题，第 739 题，第 901 题，第 907 题，第 1019 题。
+
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/String.md b/ctl/template/String.md
new file mode 100644
index 000000000..e0505e337
--- /dev/null
+++ b/ctl/template/String.md
@@ -0,0 +1,10 @@
+---
+title: 2.02 String
+type: docs
+weight: 2
+---
+
+# String
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Tree.md b/ctl/template/Tree.md
new file mode 100644
index 000000000..19bdee6ba
--- /dev/null
+++ b/ctl/template/Tree.md
@@ -0,0 +1,10 @@
+---
+title: 2.06 Tree
+type: docs
+weight: 6
+---
+
+# Tree
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Two_Pointers.md b/ctl/template/Two_Pointers.md
new file mode 100644
index 000000000..e25406e0c
--- /dev/null
+++ b/ctl/template/Two_Pointers.md
@@ -0,0 +1,33 @@
+---
+title: 2.03 ✅ Two Pointers
+type: docs
+weight: 3
+---
+
+# Two Pointers
+
+![](https://img.halfrost.com/Leetcode/Two_pointers.png)
+
+- 双指针滑动窗口的经典写法。右指针不断往右移，移动到不能往右移动为止(具体条件根据题目而定)。当右指针到最右边以后，开始挪动左指针，释放窗口左边界。第 3 题，第 76 题，第 209 题，第 424 题，第 438 题，第 567 题，第 713 题，第 763 题，第 845 题，第 881 题，第 904 题，第 978 题，第 992 题，第 1004 题，第 1040 题，第 1052 题。
+
+```c
+	left, right := 0, -1
+
+	for left < len(s) {
+		if right+1 < len(s) && freq[s[right+1]-'a'] == 0 {
+			freq[s[right+1]-'a']++
+			right++
+		} else {
+			freq[s[left]-'a']--
+			left++
+		}
+		result = max(result, right-left+1)
+	}
+```
+
+- 快慢指针可以查找重复数字，时间复杂度 O(n)，第 287 题。
+- 替换字母以后，相同字母能出现连续最长的长度。第 424 题。
+- SUM 问题集。第 1 题，第 15 题，第 16 题，第 18 题，第 167 题，第 923 题，第 1074 题。
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/Union_Find.md b/ctl/template/Union_Find.md
new file mode 100644
index 000000000..0b23116ff
--- /dev/null
+++ b/ctl/template/Union_Find.md
@@ -0,0 +1,20 @@
+---
+title: 2.16 ✅ Union Find
+type: docs
+weight: 16
+---
+
+# Union Find
+
+![](https://img.halfrost.com/Leetcode/Union_Find.png)
+
+- 灵活使用并查集的思想，熟练掌握并查集的[模板]({{< relref "/ChapterThree/UnionFind.md" >}})，模板中有两种并查集的实现方式，一种是路径压缩 + 秩优化的版本，另外一种是计算每个集合中元素的个数 + 最大集合元素个数的版本，这两种版本都有各自使用的地方。能使用第一类并查集模板的题目有：第 128 题，第 130 题，第 547 题，第 684 题，第 721 题，第 765 题，第 778 题，第 839 题，第 924 题，第 928 题，第 947 题，第 952 题，第 959 题，第 990 题。能使用第二类并查集模板的题目有：第 803 题，第 952 题。第 803 题秩优化和统计集合个数这些地方会卡时间，如果不优化，会 TLE。
+- 并查集是一种思想，有些题需要灵活使用这种思想，而不是死套模板，如第 399 题，这一题是 stringUnionFind，利用并查集思想实现的。这里每个节点是基于字符串和 map 的，而不是单纯的用 int 节点编号实现的。
+- 有些题死套模板反而做不出来，比如第 685 题，这一题不能路径压缩和秩优化，因为题目中涉及到有向图，需要知道节点的前驱节点，如果路径压缩了，这一题就没法做了。这一题不需要路径压缩和秩优化。
+- 灵活的抽象题目给的信息，将给定的信息合理的编号，使用并查集解题，并用 map 降低时间复杂度，如第 721 题，第 959 题。
+- 关于地图，砖块，网格的题目，可以新建一个特殊节点，将四周边缘的砖块或者网格都 union() 到这个特殊节点上。第 130 题，第 803 题。
+- 能用并查集的题目，一般也可以用 DFS 和 BFS 解答，只不过时间复杂度会高一点。
+
+
+
+{{.AvailableTagTable}}
\ No newline at end of file
diff --git a/ctl/template/collapseSection.md b/ctl/template/collapseSection.md
new file mode 100644
index 000000000..e954f0877
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+++ b/ctl/template/collapseSection.md
@@ -0,0 +1,4 @@
+---
+bookCollapseSection: true
+weight: 20
+---
diff --git a/ctl/template/menu.md b/ctl/template/menu.md
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--- /dev/null
+++ b/ctl/template/menu.md
@@ -0,0 +1,9 @@
+---
+headless: true
+---
+
+<hr>
+
+{{.BookMenu}}
+
+<br />
diff --git a/ctl/template/template.markdown b/ctl/template/template.markdown
new file mode 100644
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+++ b/ctl/template/template.markdown
@@ -0,0 +1,645 @@
+
+# LeetCode in Go
+[LeetCode Online Judge](https://leetcode.com/) is a website containing many **algorithm questions**. Most of them are real interview questions of **Google, Facebook, LinkedIn, Apple**, etc. and it always help to sharp our algorithm Skills. Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. This repo shows my solutions in Go with the code style strictly follows the [Google Golang Style Guide](https://github.com/golang/go/wiki/CodeReviewComments). Please feel free to reference and **STAR** to support this repo, thank you!
+
+
+<p align='center'>
+<img src='./logo.png'>
+</p>
+
+![](./website/static/wechat-qr-code.png)
+
+<p align='center'>
+<a href="https://github.com/halfrost/leetcode-go/releases/" rel="nofollow"><img alt="GitHub All Releases" src="https://img.shields.io/github/downloads/halfrost/LeetCode-Go/total?label=PDF%20downloads"></a>
+<img src="https://img.shields.io/badge/Total%20Word%20Count-738884-success">
+<a href="https://github.com/halfrost/leetcode-go/actions" rel="nofollow"><img src="https://github.com/halfrost/leetcode-go/workflows/Deploy%20leetcode-cookbook/badge.svg?branch=master"></a>
+<a href="https://travis-ci.org/github/halfrost/LeetCode-Go" rel="nofollow"><img src="https://travis-ci.org/halfrost/LeetCode-Go.svg?branch=master"></a>
+<a href="https://goreportcard.com/report/github.com/halfrost/LeetCode-Go" rel="nofollow"><img src="https://goreportcard.com/badge/github.com/halfrost/LeetCode-Go"></a>
+<img src="https://img.shields.io/badge/runtime%20beats-100%25-success">
+<a href="https://codecov.io/gh/halfrost/LeetCode-Go"><img src="https://codecov.io/gh/halfrost/LeetCode-Go/branch/master/graph/badge.svg" /></a>
+<!--<img alt="GitHub go.mod Go version" src="https://img.shields.io/github/go-mod/go-version/halfrost/LeetCode-Go?color=26C2F0">-->
+<img alt="Support Go version" src="https://img.shields.io/badge/Go-v1.15-26C2F0">
+<img src="https://visitor-badge.laobi.icu/badge?page_id=halfrost.LeetCode-Go">
+</p>
+
+<p align='center'>
+<a href="https://github.com/halfrost/leetcode-go/blob/master/LICENSE"><img alt="GitHub" src="https://img.shields.io/github/license/halfrost/LeetCode-Go?label=License"></a>
+<img src="https://img.shields.io/badge/License-CC-000000.svg">
+<a href="https://leetcode.com/halfrost/"><img src="https://img.shields.io/badge/@halfrost-8751-yellow.svg">
+<img src="https://img.shields.io/badge/language-Golang-26C2F0.svg">
+<a href="https://halfrost.com"><img src="https://img.shields.io/badge/Blog-Halfrost--Field-80d4f9.svg?style=flat"></a>
+<a href="http://weibo.com/halfrost"><img src="https://img.shields.io/badge/weibo-@halfrost-f974ce.svg?style=flat&colorA=f4292e"></a>
+<a href="https://twitter.com/halffrost"><img src="https://img.shields.io/badge/twitter-@halffrost-F8E81C.svg?style=flat&colorA=009df2"></a>
+<a href="https://www.zhihu.com/people/halfrost/activities"><img src="https://img.shields.io/badge/%E7%9F%A5%E4%B9%8E-@halfrost-fd6f32.svg?style=flat&colorA=0083ea"></a>
+<img src="https://img.shields.io/badge/made%20with-=1-blue.svg">
+<a href="https://github.com/halfrost/leetcode-go/pulls"><img src="https://img.shields.io/badge/PR-Welcome-brightgreen.svg"></a>
+</p>
+
+支持 Progressive Web Apps 和 Dark Mode 的题解电子书《LeetCode Cookbook》 <a href="https://books.halfrost.com/leetcode/" rel="nofollow">Online Reading</a>
+
+<p align='center'>
+<a href="https://books.halfrost.com/leetcode/"><img src="https://img.halfrost.com/Leetcode/Cookbook_Safari_0.png"></a>
+<a href="https://books.halfrost.com/leetcode/"><img src="https://img.halfrost.com/Leetcode/Cookbook_Chrome_PWA.png"></a>
+</p>
+
+离线版本的电子书《LeetCode Cookbook》PDF <a href="https://github.com/halfrost/leetcode-go/releases/" rel="nofollow">Download here</a>
+
+<p align='center'>
+<a href="https://github.com/halfrost/leetcode-go/releases/"><img src="https://img.halfrost.com/Leetcode/Cookbook.png"></a>
+</p>
+
+通过 iOS / Android 浏览器安装 PWA 版《LeetCode Cookbook》至设备桌面随时学习
+
+<p align='center'>
+<a href="https://books.halfrost.com/leetcode/"><img src="https://img.halfrost.com/Leetcode/Cookbook_PWA_iPad.png"></a>
+<a href="https://books.halfrost.com/leetcode/"><img src="https://img.halfrost.com/Leetcode/Cookbook_PWA_iPad_example1__.png"></a>
+<a href="https://books.halfrost.com/leetcode/"><img src="https://img.halfrost.com/Leetcode/Cookbook_PWA_iPad_example2__.png"></a>
+</p>
+
+
+## Data Structures
+
+> 标识了 ✅ 的专题是完成所有题目了的，没有标识的是还没有做完所有题目的
+
+<a href="https://books.halfrost.com/leetcode/"><img src="./website/static/logo.png" alt="logo" height="550" align="right" /></a>
+
+* [Array](#array)
+* [String](#string)
+* [✅ Two Pointers](#two-pointers)
+* [✅ Linked List](#linked-list)
+* [✅ Stack](#stack)
+* [Tree](#tree)
+* [Dynamic programming](#dynamic-programming)
+* [✅ Backtracking](#backtracking)
+* [Depth First Search](#depth-first-search)
+* [Breadth First Search](#breadth-first-search)
+* [Binary Search](#binary-search)
+* [Math](#math)
+* [Hash Table](#hash-table)
+* [✅ Sort](#sort)
+* [✅ Bit Manipulation](#bit-manipulation)
+* [✅ Union Find](#union-find)
+* [✅ Sliding Window](#sliding-window)
+* [✅ Segment Tree](#segment-tree)
+* [✅ Binary Indexed Tree](#binary-indexed-tree)
+
+<br>
+<br>
+
+| 数据结构 | 变种 | 相关题目 | 讲解文章 | 
+|:-------:|:-------|:------|:------|
+|顺序线性表：向量||||
+|单链表|1. 双向链表<br>2. 静态链表<br>3. 对称矩阵<br>4. 稀疏矩阵|||
+|哈希表|1. 散列函数<br>2. 解决碰撞/填充因子<br>|||
+|栈和队列|1. 广义栈<br>2. 双端队列<br>|||
+|队列|1. 链表实现<br>2. 循环数组实现<br>3. 双端队列|||
+|字符串|1. KMP算法<br>2. 有限状态自动机<br>3. 模式匹配有限状态自动机<br>4. BM 模式匹配算法<br>5. BM-KMP 算法<br>6. BF 算法|||
+|树|1. 二叉树<br>2. 并查集<br>3. Huffman 树|||
+|数组实现的堆|1. 极大堆和极小堆<br>2. 极大极小堆<br>3. 双端堆<br>4. d 叉堆|||
+|树实现的堆|1. 左堆<br>2. 扁堆<br>3. 二项式堆<br>4. 斐波那契堆<br>5. 配对堆|||
+|查找|1. 哈希表<br>2. 跳跃表<br>3. 排序二叉树<br>4. AVL 树<br>5. B 树 / B+ 树 / B* 树<br>6. AA 树<br>7. 红黑树<br>8. 排序二叉堆<br>9. Splay 树<br>10. 双链树<br>11. Trie 树<br>12. R 树|||
+|--------------------------------------------|--------------------------------------------------------------------------------------------|---------------------------|-----------------------------------|
+
+
+## Algorithm
+
+
+| 算法 | 具体类型 | 相关题目 | 讲解文章 | 
+|:-------:|:-------|:------|:------|
+|排序算法|1. 冒泡排序<br>2. 插入排序<br>3. 选择排序<br>4. 希尔 Shell 排序<br>5. 快速排序<br>6. 归并排序<br>7. 堆排序<br>8. 线性排序算法<br>9. 自省排序<br>10. 间接排序<br>11. 计数排序<br>12. 基数排序<br>13. 桶排序<br>14. 外部排序 - k 路归并败者树<br>15. 外部排序 - 最佳归并树|||
+|递归与分治||1. 二分搜索/查找<br>2. 大整数的乘法<br>3. Strassen 矩阵乘法<br>4. 棋盘覆盖<br>5. 合并排序<br>6. 快速排序<br>7. 线性时间选择<br>8. 最接近点对问题<br>9. 循环赛日程表<br>||
+|动态规划||1. 矩阵连乘问题<br>2. 最长公共子序列<br>3. 最大子段和<br>4. 凸多边形最优三角剖分<br>5. 多边形游戏<br>6. 图像压缩<br>7. 电路布线<br>8. 流水作业调度<br>9. 0-1 背包问题/背包九讲<br>10. 最优二叉搜索树<br>11. 动态规划加速原理<br>12. 树型 DP<br>||
+|贪心||1. 活动安排问题<br>2. 最优装载<br>3. 哈夫曼编码<br>4. 单源最短路径<br>5. 最小生成树<br>6. 多机调度问题<br>||
+|回溯法||1. 装载问题<br>2. 批处理作业调度<br>3. 符号三角形问题<br>4. n 后问题<br>5. 0-1 背包问题<br>6. 最大团问题<br>7. 图的 m 着色问题<br>8. 旅行售货员问题<br>9. 圆排列问题<br>10. 电路板排列问题<br>11. 连续邮资问题<br>||
+|搜索|1. 枚举<br>2. DFS<br>3. BFS<br>4. 启发式搜索<br>|||
+|随机化|1. 随机数<br>2. 数值随机化算法<br>3. Sherwood 舍伍德算法<br>4. Las Vegas 拉斯维加斯算法<br>5. Monte Carlo 蒙特卡罗算法<br>|1. 计算 π 值<br>2. 计算定积分<br>3. 解非线性方程组<br>4. 线性时间选择算法<br>5. 跳跃表<br>6. n 后问题<br>7. 整数因子分解<br>8. 主元素问题<br>9. 素数测试<br>||
+|图论|1. 遍历 DFS / BFS<br>2. AOV / AOE 网络<br>3. Kruskal 算法(最小生成树)<br>4. Prim 算法(最小生成树)<br>5. Boruvka 算法(最小生成树)<br>6. Dijkstra 算法(单源最短路径)<br>7. Bellman-Ford 算法(单源最短路径)<br>8. SPFA 算法(单源最短路径)<br>9. Floyd 算法(多源最短路径)<br>10. Johnson 算法(多源最短路径)<br>11. Fleury 算法(欧拉回路)<br>12. Ford-Fulkerson 算法(最大网络流增广路)<br>13. Edmonds-Karp 算法(最大网络流)<br>14. Dinic 算法(最大网络流)<br>15. 一般预流推进算法<br>16. 最高标号预流推进 HLPP 算法<br>17. Primal-Dual 原始对偶算法(最小费用流)18. Kosaraju 算法(有向图强连通分量)<br>19. Tarjan 算法(有向图强连通分量)<br>20. Gabow 算法(有向图强连通分量)<br>21. 匈牙利算法(二分图匹配)<br>22. Hopcroft－Karp 算法(二分图匹配)<br>23. kuhn munkras 算法(二分图最佳匹配)<br>24. Edmonds’ Blossom-Contraction 算法(一般图匹配)<br>|1. 图遍历<br>2. 有向图和无向图的强弱连通性<br>3. 割点/割边<br>3. AOV 网络和拓扑排序<br>4. AOE 网络和关键路径<br>5. 最小代价生成树/次小生成树<br>6. 最短路径问题/第 K 短路问题<br>7. 最大网络流问题<br>8. 最小费用流问题<br>9. 图着色问题<br>10. 差分约束系统<br>11. 欧拉回路<br>12. 中国邮递员问题<br>13. 汉密尔顿回路<br>14. 最佳边割集/最佳点割集/最小边割集/最小点割集/最小路径覆盖/最小点集覆盖 <br>15. 边覆盖集<br>16. 二分图完美匹配和最大匹配问题<br>17. 仙人掌图<br>18. 弦图<br>19. 稳定婚姻问题<br>20. 最大团问题<br>||
+|数论||1. 最大公约数<br> 2. 最小公倍数<br>3. 分解质因数<br>4. 素数判定<br>5. 进制转换<br>6. 高精度计算<br>7. 整除问题<br>8. 同余问题<br>9. 欧拉函数<br>10. 扩展欧几里得<br>11. 置换群<br>12. 母函数<br>13. 离散变换<br>14. 康托展开<br>15. 矩阵<br>16. 向量<br>17. 线性方程组<br>18. 线性规划<br> ||
+|几何||1. 凸包 - Gift wrapping<br>2. 凸包 - Graham scan<br>3. 线段问题<br> 4. 多边形和多面体相关问题<br>||
+|NP 完全|1. 计算模型<br>2. P 类与 NP 类问题<br>3. NP 完全问题<br>4. NP 完全问题的近似算法<br>|1. 随机存取机 RAM<br>2. 随机存取存储程序机 RASP<br>3. 图灵机<br>4. 非确定性图灵机<br>5. P 类与 NP 类语言<br>6. 多项式时间验证<br>7. 多项式时间变换<br>8. Cook定理<br>9. 合取范式的可满足性问题 CNF-SAT<br>10. 3 元合取范式的可满足性问题 3-SAT<br>11. 团问题 CLIQUE<br>12. 顶点覆盖问题 VERTEX-COVER<br>13. 子集和问题 SUBSET-SUM<br>14. 哈密顿回路问题 HAM-CYCLE<br>15. 旅行售货员问题 TSP<br>16. 顶点覆盖问题的近似算法<br>17. 旅行售货员问题近似算法<br>18. 具有三角不等式性质的旅行售货员问题<br>19. 一般的旅行售货员问题<br>20. 集合覆盖问题的近似算法<br>21. 子集和问题的近似算法<br>22. 子集和问题的指数时间算法<br>23. 子集和问题的多项式时间近似格式<br>||
+|------------|------------------------------------------------------------------|-----------------------------------------------------------------|--------------------|
+
+
+## LeetCode Problems
+
+## 一. 个人数据
+
+{{.PersonalData}}
+
+## 二. 目录
+
+{{.TotalNum}}
+
+{{.AvailableTable}}
+
+------------------------------------------------------------------
+
+下面这些是免费的算法题，但是暂时还不能使用 Go 解答的：
+
+暂无
+
+------------------------------------------------------------------
+
+
+## 三.分类
+
+## Array
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Array/)
+
+
+
+## String
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/String/)
+
+
+## Two Pointers
+
+![](./topic/Two_pointers.png)
+
+- 双指针滑动窗口的经典写法。右指针不断往右移，移动到不能往右移动为止(具体条件根据题目而定)。当右指针到最右边以后，开始挪动左指针，释放窗口左边界。第 3 题，第 76 题，第 209 题，第 424 题，第 438 题，第 567 题，第 713 题，第 763 题，第 845 题，第 881 题，第 904 题，第 978 题，第 992 题，第 1004 题，第 1040 题，第 1052 题。
+
+```c
+	left, right := 0, -1
+
+	for left < len(s) {
+		if right+1 < len(s) && freq[s[right+1]-'a'] == 0 {
+			freq[s[right+1]-'a']++
+			right++
+		} else {
+			freq[s[left]-'a']--
+			left++
+		}
+		result = max(result, right-left+1)
+	}
+```
+
+- 快慢指针可以查找重复数字，时间复杂度 O(n)，第 287 题。
+- 替换字母以后，相同字母能出现连续最长的长度。第 424 题。
+- SUM 问题集。第 1 题，第 15 题，第 16 题，第 18 题，第 167 题，第 923 题，第 1074 题。
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Two_Pointers/)
+
+
+## Linked List
+
+![](./topic/Linked_List.png)
+
+
+- 巧妙的构造虚拟头结点。可以使遍历处理逻辑更加统一。
+- 灵活使用递归。构造递归条件，使用递归可以巧妙的解题。不过需要注意有些题目不能使用递归，因为递归深度太深会导致超时和栈溢出。
+- 链表区间逆序。第 92 题。
+- 链表寻找中间节点。第 876 题。链表寻找倒数第 n 个节点。第 19 题。只需要一次遍历就可以得到答案。
+- 合并 K 个有序链表。第 21 题，第 23 题。
+- 链表归类。第 86 题，第 328 题。
+- 链表排序，时间复杂度要求 O(n * log n)，空间复杂度 O(1)。只有一种做法，归并排序，至顶向下归并。第 148 题。
+- 判断链表是否存在环，如果有环，输出环的交叉点的下标；判断 2 个链表是否有交叉点，如果有交叉点，输出交叉点。第 141 题，第 142 题，第 160 题。
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Linked_List/)
+
+
+
+
+## Stack
+
+![](./topic/Stack.png)
+
+- 括号匹配问题及类似问题。第 20 题，第 921 题，第 1021 题。
+- 栈的基本 pop 和 push 操作。第 71 题，第 150 题，第 155 题，第 224 题，第 225 题，第 232 题，第 946 题，第 1047 题。
+- 利用栈进行编码问题。第 394 题，第 682 题，第 856 题，第 880 题。
+- **单调栈**。**利用栈维护一个单调递增或者递减的下标数组**。第 84 题，第 456 题，第 496 题，第 503 题，第 739 题，第 901 题，第 907 题，第 1019 题。
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Stack/)
+
+
+
+## Tree
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Tree/)
+
+
+
+
+
+## Dynamic Programming
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Dynamic_Programming/)
+
+
+
+## Backtracking
+
+![](./topic/Backtracking.png)
+
+- 排列问题 Permutations。第 46 题，第 47 题。第 60 题，第 526 题，第 996 题。
+- 组合问题 Combination。第 39 题，第 40 题，第 77 题，第 216 题。
+- 排列和组合杂交问题。第 1079 题。
+- N 皇后终极解法(二进制解法)。第 51 题，第 52 题。
+- 数独问题。第 37 题。
+- 四个方向搜索。第 79 题，第 212 题，第 980 题。
+- 子集合问题。第 78 题，第 90 题。
+- Trie。第 208 题，第 211 题。
+- BFS 优化。第 126 题，第 127 题。
+- DFS 模板。(只是一个例子，不对应任何题)
+
+```go
+func combinationSum2(candidates []int, target int) [][]int {
+	if len(candidates) == 0 {
+		return [][]int{}
+	}
+	c, res := []int{}, [][]int{}
+	sort.Ints(candidates)
+	findcombinationSum2(candidates, target, 0, c, &res)
+	return res
+}
+
+func findcombinationSum2(nums []int, target, index int, c []int, res *[][]int) {
+	if target == 0 {
+		b := make([]int, len(c))
+		copy(b, c)
+		*res = append(*res, b)
+		return
+	}
+	for i := index; i < len(nums); i++ {
+		if i > index && nums[i] == nums[i-1] { // 这里是去重的关键逻辑
+			continue
+		}
+		if target >= nums[i] {
+			c = append(c, nums[i])
+			findcombinationSum2(nums, target-nums[i], i+1, c, res)
+			c = c[:len(c)-1]
+		}
+	}
+}
+```
+- BFS 模板。(只是一个例子，不对应任何题)
+
+```go
+func updateMatrix_BFS(matrix [][]int) [][]int {
+	res := make([][]int, len(matrix))
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return res
+	}
+	queue := make([][]int, 0)
+	for i, _ := range matrix {
+		res[i] = make([]int, len(matrix[0]))
+		for j, _ := range res[i] {
+			if matrix[i][j] == 0 {
+				res[i][j] = -1
+				queue = append(queue, []int{i, j})
+			}
+		}
+	}
+	level := 1
+	for len(queue) > 0 {
+		size := len(queue)
+		for size > 0 {
+			size -= 1
+			node := queue[0]
+			queue = queue[1:]
+			i, j := node[0], node[1]
+			for _, direction := range [][]int{{-1, 0}, {1, 0}, {0, 1}, {0, -1}} {
+				x := i + direction[0]
+				y := j + direction[1]
+				if x < 0 || x >= len(matrix) || y < 0 || y >= len(matrix[0]) || res[x][y] < 0 || res[x][y] > 0 {
+					continue
+				}
+				res[x][y] = level
+				queue = append(queue, []int{x, y})
+			}
+		}
+		level++
+	}
+	for i, row := range res {
+		for j, cell := range row {
+			if cell == -1 {
+				res[i][j] = 0
+			}
+		}
+	}
+	return res
+}
+```
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Backtracking/)
+
+
+## Depth First Search
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Depth_First_Search/)
+
+
+
+
+## Breadth First Search
+
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Breadth_First_Search/)
+
+
+
+
+
+## Binary Search
+
+![]()
+
+- 二分搜索的经典写法。需要注意的三点：
+	1. 循环退出条件，注意是 low <= high，而不是 low < high。
+	2. mid 的取值，mid := low + (high-low)>>1
+	3. low 和 high 的更新。low = mid + 1，high = mid - 1。
+
+```go
+func binarySearchMatrix(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[mid] == target {
+			return mid
+		} else if nums[mid] > target {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+```
+
+- 二分搜索的变种写法。有 4 个基本变种:
+	1. 查找第一个与 target 相等的元素，时间复杂度 O(logn)
+	2. 查找最后一个与 target 相等的元素，时间复杂度 O(logn)
+	3. 查找第一个大于等于 target 的元素，时间复杂度 O(logn)
+	4. 查找最后一个小于等于 target 的元素，时间复杂度 O(logn)
+
+```go
+// 二分查找第一个与 target 相等的元素，时间复杂度 O(logn)
+func searchFirstEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == 0) || (nums[mid-1] != target) { // 找到第一个与 target 相等的元素
+				return mid
+			}
+			high = mid - 1
+		}
+	}
+	return -1
+}
+
+// 二分查找最后一个与 target 相等的元素，时间复杂度 O(logn)
+func searchLastEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == len(nums)-1) || (nums[mid+1] != target) { // 找到最后一个与 target 相等的元素
+				return mid
+			}
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// 二分查找第一个大于等于 target 的元素，时间复杂度 O(logn)
+func searchFirstGreaterElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] >= target {
+			if (mid == 0) || (nums[mid-1] < target) { // 找到第一个大于等于 target 的元素
+				return mid
+			}
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// 二分查找最后一个小于等于 target 的元素，时间复杂度 O(logn)
+func searchLastLessElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] <= target {
+			if (mid == len(nums)-1) || (nums[mid+1] > target) { // 找到最后一个小于等于 target 的元素
+				return mid
+			}
+			low = mid + 1
+		} else {
+			high = mid - 1
+		}
+	}
+	return -1
+}
+```
+
+- 在基本有序的数组中用二分搜索。经典解法可以解，变种写法也可以写，常见的题型，在山峰数组中找山峰，在旋转有序数组中找分界点。第 33 题，第 81 题，第 153 题，第 154 题，第 162 题，第 852 题
+
+```go
+func peakIndexInMountainArray(A []int) int {
+	low, high := 0, len(A)-1
+	for low < high {
+		mid := low + (high-low)>>1
+		// 如果 mid 较大，则左侧存在峰值，high = m，如果 mid + 1 较大，则右侧存在峰值，low = mid + 1
+		if A[mid] > A[mid+1] {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+```
+
+- max-min 最大值最小化问题。求在最小满足条件的情况下的最大值。第 410 题，第 875 题，第 1011 题，第 1283 题。
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Binary_Search/)
+
+
+
+## Math
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Math/)
+
+
+
+
+## Hash Table
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Hash_Table/)
+
+
+
+## Sort
+
+![](./topic/Sort.png)
+
+- 深刻的理解多路快排。第 75 题。
+- 链表的排序，插入排序(第 147 题)和归并排序(第 148 题)
+- 桶排序和基数排序。第 164 题。
+- "摆动排序"。第 324 题。
+- 两两不相邻的排序。第 767 题，第 1054 题。
+- "饼子排序"。第 969 题。
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Sort/)
+
+
+## Bit Manipulation
+
+![](./topic/Bit_Manipulation.png)
+
+- 异或的特性。第 136 题，第 268 题，第 389 题，第 421 题，
+
+```go
+x ^ 0 = x
+x ^ 11111……1111 = ~x
+x ^ (~x) = 11111……1111
+x ^ x = 0
+a ^ b = c  => a ^ c = b  => b ^ c = a (交换律)
+a ^ b ^ c = a ^ (b ^ c) = (a ^ b）^ c (结合律)
+```
+
+- 构造特殊 Mask，将特殊位置放 0 或 1。
+
+```go
+将 x 最右边的 n 位清零， x & ( ~0 << n )
+获取 x 的第 n 位值(0 或者 1)，(x >> n) & 1
+获取 x 的第 n 位的幂值，x & (1 << (n - 1))
+仅将第 n 位置为 1，x | (1 << n)
+仅将第 n 位置为 0，x & (~(1 << n))
+将 x 最高位至第 n 位(含)清零，x & ((1 << n) - 1)
+将第 n 位至第 0 位(含)清零，x & (~((1 << (n + 1)) - 1)）
+```
+
+- 有特殊意义的 & 位操作运算。第 260 题，第 201 题，第 318 题，第 371 题，第 397 题，第 461 题，第 693 题，
+
+```go
+X & 1 == 1 判断是否是奇数(偶数)
+X & = (X - 1) 将最低位(LSB)的 1 清零
+X & -X 得到最低位(LSB)的 1
+X & ~X = 0
+```
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Bit_Manipulation/)
+
+
+## Union Find
+
+![](./topic/Union_Find.png)
+
+- 灵活使用并查集的思想，熟练掌握并查集的[模板](https://github.com/halfrost/leetcode-go/blob/master/template/UnionFind.go)，模板中有两种并查集的实现方式，一种是路径压缩 + 秩优化的版本，另外一种是计算每个集合中元素的个数 + 最大集合元素个数的版本，这两种版本都有各自使用的地方。能使用第一类并查集模板的题目有：第 128 题，第 130 题，第 547 题，第 684 题，第 721 题，第 765 题，第 778 题，第 839 题，第 924 题，第 928 题，第 947 题，第 952 题，第 959 题，第 990 题。能使用第二类并查集模板的题目有：第 803 题，第 952 题。第 803 题秩优化和统计集合个数这些地方会卡时间，如果不优化，会 TLE。
+- 并查集是一种思想，有些题需要灵活使用这种思想，而不是死套模板，如第 399 题，这一题是 stringUnionFind，利用并查集思想实现的。这里每个节点是基于字符串和 map 的，而不是单纯的用 int 节点编号实现的。
+- 有些题死套模板反而做不出来，比如第 685 题，这一题不能路径压缩和秩优化，因为题目中涉及到有向图，需要知道节点的前驱节点，如果路径压缩了，这一题就没法做了。这一题不需要路径压缩和秩优化。
+- 灵活的抽象题目给的信息，将给定的信息合理的编号，使用并查集解题，并用 map 降低时间复杂度，如第 721 题，第 959 题。
+- 关于地图，砖块，网格的题目，可以新建一个特殊节点，将四周边缘的砖块或者网格都 union() 到这个特殊节点上。第 130 题，第 803 题。
+- 能用并查集的题目，一般也可以用 DFS 和 BFS 解答，只不过时间复杂度会高一点。
+
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Union_Find/)
+
+
+
+## Sliding Window
+
+![](./topic/Sliding_Window.png)
+
+- 双指针滑动窗口的经典写法。右指针不断往右移，移动到不能往右移动为止(具体条件根据题目而定)。当右指针到最右边以后，开始挪动左指针，释放窗口左边界。第 3 题，第 76 题，第 209 题，第 424 题，第 438 题，第 567 题，第 713 题，第 763 题，第 845 题，第 881 题，第 904 题，第 978 题，第 992 题，第 1004 题，第 1040 题，第 1052 题。
+
+```c
+	left, right := 0, -1
+
+	for left < len(s) {
+		if right+1 < len(s) && freq[s[right+1]-'a'] == 0 {
+			freq[s[right+1]-'a']++
+			right++
+		} else {
+			freq[s[left]-'a']--
+			left++
+		}
+		result = max(result, right-left+1)
+	}
+```
+- 滑动窗口经典题。第 239 题，第 480 题。
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Sliding_Window/)
+
+
+## Segment Tree
+
+![](./topic/Segment_Tree.png)
+
+- 线段树的经典数组实现写法。将合并两个节点 pushUp 逻辑抽象出来了，可以实现任意操作(常见的操作有：加法，取 max，min 等等)。第 218 题，第 303 题，第 307 题，第 699 题。
+- 计数线段树的经典写法。第 315 题，第 327 题，第 493 题。
+- 线段树的树的实现写法。第 715 题，第 732 题。
+- 区间懒惰更新。第 218 题，第 699 题。
+- 离散化。离散化需要注意一个特殊情况：假如三个区间为 [1,10] [1,4] [6,10]，离散化后 x[1]=1,x[2]=4,x[3]=6,x[4]=10。第一个区间为 [1,4]，第二个区间为 [1,2]，第三个区间为 [3,4]，这样一来，区间一 = 区间二 + 区间三，这和离散前的模型不符，离散前，很明显，区间一 > 区间二 + 区间三。正确的做法是：在相差大于 1 的数间加一个数，例如在上面 1 4 6 10 中间加 5，即可 x[1]=1,x[2]=4,x[3]=5,x[4]=6,x[5]=10。这样处理之后，区间一是 1-5 ，区间二是 1-2 ，区间三是 4-5 。
+- 灵活构建线段树。线段树节点可以存储多条信息，合并两个节点的 pushUp 操作也可以是多样的。第 850 题，第 1157 题。
+
+
+线段树[题型](https://blog.csdn.net/xuechelingxiao/article/details/38313105)从简单到困难:
+
+1. 单点更新:  
+	[HDU 1166 敌兵布阵](http://acm.hdu.edu.cn/showproblem.php?pid=1166) update:单点增减 query:区间求和  
+	[HDU 1754 I Hate It](http://acm.hdu.edu.cn/showproblem.php?pid=1754) update:单点替换 query:区间最值  
+	[HDU 1394 Minimum Inversion Number](http://acm.hdu.edu.cn/showproblem.php?pid=1394) update:单点增减 query:区间求和  
+	[HDU 2795 Billboard](http://acm.hdu.edu.cn/showproblem.php?pid=2795) query:区间求最大值的位子(直接把update的操作在query里做了)
+2. 区间更新:  
+	[HDU 1698 Just a Hook](http://acm.hdu.edu.cn/showproblem.php?pid=1698) update:成段替换 (由于只query一次总区间,所以可以直接输出 1 结点的信息)  
+	[POJ 3468 A Simple Problem with Integers](http://poj.org/problem?id=3468) update:成段增减 query:区间求和  
+	[POJ 2528 Mayor’s posters](http://poj.org/problem?id=2528) 离散化 + update:成段替换 query:简单hash  
+	[POJ 3225 Help with Intervals](http://poj.org/problem?id=3225) update:成段替换,区间异或 query:简单hash
+3. 区间合并(这类题目会询问区间中满足条件的连续最长区间,所以PushUp的时候需要对左右儿子的区间进行合并):  
+	[POJ 3667 Hotel](http://poj.org/problem?id=3667) update:区间替换 query:询问满足条件的最左端点
+4. 扫描线(这类题目需要将一些操作排序,然后从左到右用一根扫描线扫过去最典型的就是矩形面积并,周长并等题):  
+	[HDU 1542 Atlantis](http://acm.hdu.edu.cn/showproblem.php?pid=1542) update:区间增减 query:直接取根节点的值  
+	[HDU 1828 Picture](http://acm.hdu.edu.cn/showproblem.php?pid=1828) update:区间增减 query:直接取根节点的值
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Segment_Tree/)
+
+
+## Binary Indexed Tree
+
+![](./topic/Binary_Indexed_Tree.png)
+
+Problems List in [there](https://books.halfrost.com/leetcode/ChapterTwo/Binary_Indexed_Tree/)
+
+
+----------------------------------------------------------------------------------------
+
+<p align='center'>
+<a href="https://github.com/halfrost/leetcode-go/releases/tag/Special"><img src="https://img.halfrost.com/Leetcode/ACM-ICPC_Algorithm_Template.png"></a>
+</p>
+
+Thank you for reading here. This is bonus. You can download my [《ACM-ICPC Algorithm Template》](https://github.com/halfrost/leetcode-go/releases/tag/Special/)
+
+
+
+## ♥️ Thanks
+
+Thanks for your Star！
+
+[![Stargazers over time](https://starchart.cc/halfrost/LeetCode-Go.svg)](https://starchart.cc/halfrost/LeetCode-Go)
+
diff --git a/ctl/template_render.go b/ctl/template_render.go
new file mode 100644
index 000000000..01c65f28f
--- /dev/null
+++ b/ctl/template_render.go
@@ -0,0 +1,39 @@
+package main
+
+import (
+	"bytes"
+	"fmt"
+	"html/template"
+	"io/ioutil"
+	"os"
+
+	m "github.com/halfrost/LeetCode-Go/ctl/models"
+	"github.com/halfrost/LeetCode-Go/ctl/util"
+)
+
+func makeReadmeFile(mdrows m.Mdrows) {
+	file := "./README.md"
+	os.Remove(file)
+	var b bytes.Buffer
+	tmpl := template.Must(template.New("readme").Parse(readTMPL("template.markdown")))
+	err := tmpl.Execute(&b, mdrows)
+	if err != nil {
+		fmt.Println(err)
+	}
+	// 保存 README.md 文件
+	util.WriteFile(file, b.Bytes())
+}
+
+func readTMPL(path string) string {
+	file, err := os.Open(path)
+	if err != nil {
+		fmt.Println(err)
+	}
+	defer file.Close()
+
+	data, err := ioutil.ReadAll(file)
+	if err != nil {
+		fmt.Println(err)
+	}
+	return string(data)
+}
diff --git a/ctl/util/go.mod b/ctl/util/go.mod
new file mode 100644
index 000000000..6127aeea8
--- /dev/null
+++ b/ctl/util/go.mod
@@ -0,0 +1,3 @@
+module github.com/halfrost/LeetCode-Go/ctl/util
+
+go 1.19
diff --git a/ctl/util/util.go b/ctl/util/util.go
new file mode 100644
index 000000000..4fcf5b025
--- /dev/null
+++ b/ctl/util/util.go
@@ -0,0 +1,185 @@
+package util
+
+import (
+	"bufio"
+	"fmt"
+	"io"
+	"io/ioutil"
+	"os"
+	"path/filepath"
+	"sort"
+	"strconv"
+	"strings"
+)
+
+// LoadSolutionsDir define
+func LoadSolutionsDir() ([]int, []string, int) {
+	solutionIds, soNames, total := loadFile("../leetcode/")
+	fmt.Printf("读取了 %v 道题的题解，当前目录下有 %v 个文件(可能包含 .DS_Store)，目录中有 %v 道题在尝试中\n", len(solutionIds), total, total-len(solutionIds))
+	return solutionIds, soNames, total - len(solutionIds)
+}
+
+func loadFile(path string) ([]int, []string, int) {
+	files, err := ioutil.ReadDir(path)
+	if err != nil {
+		fmt.Println(err)
+	}
+	solutionIds, soNames, solutionsMap := []int{}, []string{}, map[int]string{}
+	for _, f := range files {
+		if f.Name()[4] == '.' {
+			tmp, err := strconv.Atoi(f.Name()[:4])
+			if err != nil {
+				fmt.Println(err)
+			}
+			solutionIds = append(solutionIds, tmp)
+			solutionsMap[tmp] = f.Name()
+		}
+	}
+	sort.Ints(solutionIds)
+	for _, v := range solutionIds {
+		if name, ok := solutionsMap[v]; ok {
+			soNames = append(soNames, name)
+		}
+	}
+	return solutionIds, soNames, len(files)
+}
+
+// GetAllFile define
+func GetAllFile(pathname string, fileList *[]string) ([]string, error) {
+	rd, err := ioutil.ReadDir(pathname)
+	for _, fi := range rd {
+		if fi.IsDir() {
+			//fmt.Printf("[%s]\n", pathname+"\\"+fi.Name())
+			GetAllFile(pathname+fi.Name()+"/", fileList)
+		} else {
+			//fmt.Println(fi.Name())
+			*fileList = append(*fileList, fi.Name())
+		}
+	}
+	return *fileList, err
+}
+
+// LoadChapterFourDir define
+func LoadChapterFourDir() ([]string, []int) {
+	files, err := GetAllFile("../website/content/ChapterFour/", &[]string{})
+	// files, err := ioutil.ReadDir("../website/content/ChapterFour/")
+	if err != nil {
+		fmt.Println(err)
+	}
+	solutions, solutionIds, solutionsMap := []string{}, []int{}, map[int]string{}
+	for _, f := range files {
+		if f[4] == '.' {
+			tmp, err := strconv.Atoi(f[:4])
+			if err != nil {
+				fmt.Println(err)
+			}
+			solutionIds = append(solutionIds, tmp)
+			// len(f.Name())-3 = 文件名去掉 .md 后缀
+			solutionsMap[tmp] = f[:len(f)-3]
+		}
+	}
+	sort.Ints(solutionIds)
+	fmt.Printf("读取了第四章的 %v 道题的题解\n", len(solutionIds))
+	for _, v := range solutionIds {
+		if name, ok := solutionsMap[v]; ok {
+			solutions = append(solutions, name)
+		}
+	}
+	return solutions, solutionIds
+}
+
+// WriteFile define
+func WriteFile(fileName string, content []byte) {
+	file, err := os.OpenFile(fileName, os.O_RDWR|os.O_CREATE, 0777)
+	if err != nil {
+		fmt.Println(err)
+	}
+	defer file.Close()
+
+	_, err = file.Write(content)
+	if err != nil {
+		fmt.Println(err)
+	}
+	//fmt.Println("write file successful")
+}
+
+// LoadFile define
+func LoadFile(filePath string) ([]byte, error) {
+	f, err := os.OpenFile(filePath, os.O_RDONLY, 0644)
+	if err != nil {
+		return nil, err
+	}
+	defer f.Close()
+	reader, output := bufio.NewReader(f), []byte{}
+	for {
+		line, _, err := reader.ReadLine()
+		if err != nil {
+			if err == io.EOF {
+				return output, nil
+			}
+			return nil, err
+		}
+		output = append(output, line...)
+		output = append(output, []byte("\n")...)
+	}
+}
+
+// DestoryDir define
+func DestoryDir(path string) {
+	filepath.Walk(path, func(path string, fi os.FileInfo, err error) error {
+		if nil == fi {
+			return err
+		}
+		if !fi.IsDir() {
+			return nil
+		}
+		name := fi.Name()
+		if strings.Contains(name, "temp") {
+			fmt.Println("temp file name:", path)
+			err := os.RemoveAll(path)
+			if err != nil {
+				fmt.Println("delet dir error:", err)
+			}
+		}
+		return nil
+	})
+}
+
+// CopyFile define
+func CopyFile(dstName, srcName string) (written int64, err error) {
+	src, err := os.Open(srcName)
+	if err != nil {
+		return
+	}
+	defer src.Close()
+	dst, err := os.OpenFile(dstName, os.O_WRONLY|os.O_CREATE, 0644)
+	if err != nil {
+		return
+	}
+	defer dst.Close()
+	return io.Copy(dst, src)
+}
+
+// BinarySearch define
+func BinarySearch(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[mid] == target {
+			return mid
+		} else if nums[mid] > target {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// GetChpaterFourFileNum define
+func GetChpaterFourFileNum(num int) string {
+	if num < 100 {
+		return fmt.Sprintf("%04d~%04d", (num/100)*100+1, (num/100)*100+99)
+	}
+	return fmt.Sprintf("%04d~%04d", (num/100)*100, (num/100)*100+99)
+}
diff --git a/ctl/version.go b/ctl/version.go
new file mode 100644
index 000000000..c996a651a
--- /dev/null
+++ b/ctl/version.go
@@ -0,0 +1,18 @@
+package main
+
+import (
+	"fmt"
+
+	"github.com/spf13/cobra"
+)
+
+var (
+	version    = "v1.0"
+	versionCmd = &cobra.Command{
+		Use:   "version",
+		Short: "Prints the version of tacoctl",
+		Run: func(cmd *cobra.Command, args []string) {
+			fmt.Println("tacoctl version:", version)
+		},
+	}
+)
diff --git a/go.mod b/go.mod
new file mode 100644
index 000000000..3b163f0e1
--- /dev/null
+++ b/go.mod
@@ -0,0 +1,30 @@
+module github.com/halfrost/LeetCode-Go
+
+go 1.19
+
+replace github.com/halfrost/LeetCode-Go/structures => ./structures
+
+replace github.com/halfrost/LeetCode-Go/template => ./template
+
+replace github.com/halfrost/LeetCode-Go/ctl/util => ./ctl/util
+
+replace github.com/halfrost/LeetCode-Go/ctl/models => ./ctl/models
+
+require (
+	github.com/BurntSushi/toml v1.2.0
+	github.com/halfrost/LeetCode-Go/ctl/models v0.0.0-20220910225043-e3bb5aff34d0
+	github.com/halfrost/LeetCode-Go/ctl/util v0.0.0-20220910225043-e3bb5aff34d0
+	github.com/halfrost/LeetCode-Go/structures v0.0.0-20220910233101-aa0e2c897b18
+	github.com/halfrost/LeetCode-Go/template v0.0.0-20220910233504-e2a72e6212ce
+	github.com/mozillazg/request v0.8.0
+	github.com/spf13/cobra v1.5.0
+)
+
+require (
+	github.com/bitly/go-simplejson v0.5.0 // indirect
+	github.com/bmizerany/assert v0.0.0-20160611221934-b7ed37b82869 // indirect
+	github.com/inconshreveable/mousetrap v1.0.0 // indirect
+	github.com/kr/pretty v0.3.0 // indirect
+	github.com/spf13/pflag v1.0.5 // indirect
+	golang.org/x/net v0.7.0 // indirect
+)
diff --git a/go.sum b/go.sum
new file mode 100644
index 000000000..d1e5e64b1
--- /dev/null
+++ b/go.sum
@@ -0,0 +1,36 @@
+github.com/BurntSushi/toml v1.2.0 h1:Rt8g24XnyGTyglgET/PRUNlrUeu9F5L+7FilkXfZgs0=
+github.com/BurntSushi/toml v1.2.0/go.mod h1:CxXYINrC8qIiEnFrOxCa7Jy5BFHlXnUU2pbicEuybxQ=
+github.com/bitly/go-simplejson v0.5.0 h1:6IH+V8/tVMab511d5bn4M7EwGXZf9Hj6i2xSwkNEM+Y=
+github.com/bitly/go-simplejson v0.5.0/go.mod h1:cXHtHw4XUPsvGaxgjIAn8PhEWG9NfngEKAMDJEczWVA=
+github.com/bmizerany/assert v0.0.0-20160611221934-b7ed37b82869 h1:DDGfHa7BWjL4YnC6+E63dPcxHo2sUxDIu8g3QgEJdRY=
+github.com/bmizerany/assert v0.0.0-20160611221934-b7ed37b82869/go.mod h1:Ekp36dRnpXw/yCqJaO+ZrUyxD+3VXMFFr56k5XYrpB4=
+github.com/cpuguy83/go-md2man/v2 v2.0.2/go.mod h1:tgQtvFlXSQOSOSIRvRPT7W67SCa46tRHOmNcaadrF8o=
+github.com/creack/pty v1.1.9/go.mod h1:oKZEueFk5CKHvIhNR5MUki03XCEU+Q6VDXinZuGJ33E=
+github.com/davecgh/go-spew v1.1.1 h1:vj9j/u1bqnvCEfJOwUhtlOARqs3+rkHYY13jYWTU97c=
+github.com/inconshreveable/mousetrap v1.0.0 h1:Z8tu5sraLXCXIcARxBp/8cbvlwVa7Z1NHg9XEKhtSvM=
+github.com/inconshreveable/mousetrap v1.0.0/go.mod h1:PxqpIevigyE2G7u3NXJIT2ANytuPF1OarO4DADm73n8=
+github.com/kr/pretty v0.1.0/go.mod h1:dAy3ld7l9f0ibDNOQOHHMYYIIbhfbHSm3C4ZsoJORNo=
+github.com/kr/pretty v0.3.0 h1:WgNl7dwNpEZ6jJ9k1snq4pZsg7DOEN8hP9Xw0Tsjwk0=
+github.com/kr/pretty v0.3.0/go.mod h1:640gp4NfQd8pI5XOwp5fnNeVWj67G7CFk/SaSQn7NBk=
+github.com/kr/pty v1.1.1/go.mod h1:pFQYn66WHrOpPYNljwOMqo10TkYh1fy3cYio2l3bCsQ=
+github.com/kr/text v0.1.0/go.mod h1:4Jbv+DJW3UT/LiOwJeYQe1efqtUx/iVham/4vfdArNI=
+github.com/kr/text v0.2.0 h1:5Nx0Ya0ZqY2ygV366QzturHI13Jq95ApcVaJBhpS+AY=
+github.com/kr/text v0.2.0/go.mod h1:eLer722TekiGuMkidMxC/pM04lWEeraHUUmBw8l2grE=
+github.com/mozillazg/request v0.8.0 h1:TbXeQUdBWr1J1df5Z+lQczDFzX9JD71kTCl7Zu/9rNM=
+github.com/mozillazg/request v0.8.0/go.mod h1:weoQ/mVFNbWgRBtivCGF1tUT9lwneFesues+CleXMWc=
+github.com/pmezard/go-difflib v1.0.0 h1:4DBwDE0NGyQoBHbLQYPwSUPoCMWR5BEzIk/f1lZbAQM=
+github.com/rogpeppe/go-internal v1.6.1 h1:/FiVV8dS/e+YqF2JvO3yXRFbBLTIuSDkuC7aBOAvL+k=
+github.com/rogpeppe/go-internal v1.6.1/go.mod h1:xXDCJY+GAPziupqXw64V24skbSoqbTEfhy4qGm1nDQc=
+github.com/russross/blackfriday/v2 v2.1.0/go.mod h1:+Rmxgy9KzJVeS9/2gXHxylqXiyQDYRxCVz55jmeOWTM=
+github.com/spf13/cobra v1.5.0 h1:X+jTBEBqF0bHN+9cSMgmfuvv2VHJ9ezmFNf9Y/XstYU=
+github.com/spf13/cobra v1.5.0/go.mod h1:dWXEIy2H428czQCjInthrTRUg7yKbok+2Qi/yBIJoUM=
+github.com/spf13/pflag v1.0.5 h1:iy+VFUOCP1a+8yFto/drg2CJ5u0yRoB7fZw3DKv/JXA=
+github.com/spf13/pflag v1.0.5/go.mod h1:McXfInJRrz4CZXVZOBLb0bTZqETkiAhM9Iw0y3An2Bg=
+github.com/stretchr/testify v1.8.0 h1:pSgiaMZlXftHpm5L7V1+rVB+AZJydKsMxsQBIJw4PKk=
+golang.org/x/net v0.7.0 h1:rJrUqqhjsgNp7KqAIc25s9pZnjU7TUcSY7HcVZjdn1g=
+golang.org/x/net v0.7.0/go.mod h1:2Tu9+aMcznHK/AK1HMvgo6xiTLG5rD5rZLDS+rp2Bjs=
+gopkg.in/check.v1 v0.0.0-20161208181325-20d25e280405/go.mod h1:Co6ibVJAznAaIkqp8huTwlJQCZ016jof/cbN4VW5Yz0=
+gopkg.in/check.v1 v1.0.0-20180628173108-788fd7840127/go.mod h1:Co6ibVJAznAaIkqp8huTwlJQCZ016jof/cbN4VW5Yz0=
+gopkg.in/errgo.v2 v2.1.0/go.mod h1:hNsd1EY+bozCKY1Ytp96fpM3vjJbqLJn88ws8XvfDNI=
+gopkg.in/yaml.v2 v2.4.0/go.mod h1:RDklbk79AGWmwhnvt/jBztapEOGDOx6ZbXqjP6csGnQ=
+gopkg.in/yaml.v3 v3.0.1 h1:fxVm/GzAzEWqLHuvctI91KS9hhNmmWOoWu0XTYJS7CA=
diff --git a/gotest.sh b/gotest.sh
new file mode 100644
index 000000000..26a68f384
--- /dev/null
+++ b/gotest.sh
@@ -0,0 +1,13 @@
+#!/usr/bin/env bash
+
+set -e
+echo "" > coverage.txt
+
+for d in $(go list ./leetcode/... | grep -v vendor); do
+    echo $d
+    go test -coverprofile=profile.out -covermode=atomic $d
+    if [ -f profile.out ]; then
+        cat profile.out >> coverage.txt
+        rm profile.out
+    fi
+done
diff --git a/leetcode/0001.Two-Sum/1. Two Sum.go b/leetcode/0001.Two-Sum/1. Two Sum.go
new file mode 100644
index 000000000..26aefd128
--- /dev/null
+++ b/leetcode/0001.Two-Sum/1. Two Sum.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+func twoSum(nums []int, target int) []int {
+	m := make(map[int]int)
+	for k, v := range nums {
+		if idx, ok := m[target-v]; ok {
+			return []int{idx, k}
+		}
+		m[v] = k
+	}
+	return nil
+}
diff --git a/leetcode/0001.Two-Sum/1. Two Sum_test.go b/leetcode/0001.Two-Sum/1. Two Sum_test.go
new file mode 100644
index 000000000..e6378be08
--- /dev/null
+++ b/leetcode/0001.Two-Sum/1. Two Sum_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1 struct {
+	para1
+	ans1
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1 struct {
+	nums   []int
+	target int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1 struct {
+	one []int
+}
+
+func Test_Problem1(t *testing.T) {
+
+	qs := []question1{
+		{
+			para1{[]int{3, 2, 4}, 6},
+			ans1{[]int{1, 2}},
+		},
+
+		{
+			para1{[]int{3, 2, 4}, 5},
+			ans1{[]int{0, 1}},
+		},
+
+		{
+			para1{[]int{0, 8, 7, 3, 3, 4, 2}, 11},
+			ans1{[]int{1, 3}},
+		},
+
+		{
+			para1{[]int{0, 1}, 1},
+			ans1{[]int{0, 1}},
+		},
+
+		{
+			para1{[]int{0, 3}, 5},
+			ans1{[]int{}},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1, q.para1
+		fmt.Printf("【input】:%v       【output】:%v\n", p, twoSum(p.nums, p.target))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0001. Two Sum/README.md b/leetcode/0001.Two-Sum/README.md
similarity index 100%
rename from Algorithms/0001. Two Sum/README.md
rename to leetcode/0001.Two-Sum/README.md
diff --git a/leetcode/0002.Add-Two-Numbers/2. Add Two Numbers.go b/leetcode/0002.Add-Two-Numbers/2. Add Two Numbers.go
new file mode 100644
index 000000000..398fb7e27
--- /dev/null
+++ b/leetcode/0002.Add-Two-Numbers/2. Add Two Numbers.go	
@@ -0,0 +1,39 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
+	head := &ListNode{Val: 0}
+	n1, n2, carry, current := 0, 0, 0, head
+	for l1 != nil || l2 != nil || carry != 0 {
+		if l1 == nil {
+			n1 = 0
+		} else {
+			n1 = l1.Val
+			l1 = l1.Next
+		}
+		if l2 == nil {
+			n2 = 0
+		} else {
+			n2 = l2.Val
+			l2 = l2.Next
+		}
+		current.Next = &ListNode{Val: (n1 + n2 + carry) % 10}
+		current = current.Next
+		carry = (n1 + n2 + carry) / 10
+	}
+	return head.Next
+}
diff --git a/Algorithms/0002. Add Two Numbers/2. Add Two Numbers_test.go b/leetcode/0002.Add-Two-Numbers/2. Add Two Numbers_test.go
similarity index 82%
rename from Algorithms/0002. Add Two Numbers/2. Add Two Numbers_test.go
rename to leetcode/0002.Add-Two-Numbers/2. Add Two Numbers_test.go
index 304ffa763..d615065b5 100644
--- a/Algorithms/0002. Add Two Numbers/2. Add Two Numbers_test.go	
+++ b/leetcode/0002.Add-Two-Numbers/2. Add Two Numbers_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question2 struct {
@@ -27,42 +29,42 @@ func Test_Problem2(t *testing.T) {
 
 	qs := []question2{
 
-		question2{
+		{
 			para2{[]int{}, []int{}},
 			ans2{[]int{}},
 		},
 
-		question2{
+		{
 			para2{[]int{1}, []int{1}},
 			ans2{[]int{2}},
 		},
 
-		question2{
+		{
 			para2{[]int{1, 2, 3, 4}, []int{1, 2, 3, 4}},
 			ans2{[]int{2, 4, 6, 8}},
 		},
 
-		question2{
+		{
 			para2{[]int{1, 2, 3, 4, 5}, []int{1, 2, 3, 4, 5}},
 			ans2{[]int{2, 4, 6, 8, 0, 1}},
 		},
 
-		question2{
+		{
 			para2{[]int{1}, []int{9, 9, 9, 9, 9}},
 			ans2{[]int{0, 0, 0, 0, 0, 1}},
 		},
 
-		question2{
+		{
 			para2{[]int{9, 9, 9, 9, 9}, []int{1}},
 			ans2{[]int{0, 0, 0, 0, 0, 1}},
 		},
 
-		question2{
+		{
 			para2{[]int{2, 4, 3}, []int{5, 6, 4}},
 			ans2{[]int{7, 0, 8}},
 		},
 
-		question2{
+		{
 			para2{[]int{1, 8, 3}, []int{7, 1}},
 			ans2{[]int{8, 9, 3}},
 		},
@@ -73,7 +75,7 @@ func Test_Problem2(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans2, q.para2
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(addTwoNumbers(S2l(p.one), S2l(p.another))))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(addTwoNumbers(structures.Ints2List(p.one), structures.Ints2List(p.another))))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0002. Add Two Numbers/README.md b/leetcode/0002.Add-Two-Numbers/README.md
similarity index 100%
rename from Algorithms/0002. Add Two Numbers/README.md
rename to leetcode/0002.Add-Two-Numbers/README.md
diff --git a/leetcode/0003.Longest-Substring-Without-Repeating-Characters/3. Longest Substring Without Repeating Characters.go b/leetcode/0003.Longest-Substring-Without-Repeating-Characters/3. Longest Substring Without Repeating Characters.go
new file mode 100644
index 000000000..541f868de
--- /dev/null
+++ b/leetcode/0003.Longest-Substring-Without-Repeating-Characters/3. Longest Substring Without Repeating Characters.go	
@@ -0,0 +1,71 @@
+package leetcode
+
+// 解法一 位图
+func lengthOfLongestSubstring(s string) int {
+	if len(s) == 0 {
+		return 0
+	}
+	var bitSet [256]bool
+	result, left, right := 0, 0, 0
+	for left < len(s) {
+		// 右侧字符对应的 bitSet 被标记 true，说明此字符在 X 位置重复，需要左侧向前移动，直到将 X 标记为 false
+		if bitSet[s[right]] {
+			bitSet[s[left]] = false
+			left++
+		} else {
+			bitSet[s[right]] = true
+			right++
+		}
+		if result < right-left {
+			result = right - left
+		}
+		if left+result >= len(s) || right >= len(s) {
+			break
+		}
+	}
+	return result
+}
+
+// 解法二 滑动窗口
+func lengthOfLongestSubstring1(s string) int {
+	if len(s) == 0 {
+		return 0
+	}
+	var freq [127]int
+	result, left, right := 0, 0, -1
+
+	for left < len(s) {
+		if right+1 < len(s) && freq[s[right+1]] == 0 {
+			freq[s[right+1]]++
+			right++
+
+		} else {
+			freq[s[left]]--
+			left++
+		}
+		result = max(result, right-left+1)
+	}
+	return result
+}
+
+// 解法三 滑动窗口-哈希桶
+func lengthOfLongestSubstring2(s string) int {
+	right, left, res := 0, 0, 0
+	indexes := make(map[byte]int, len(s))
+	for left < len(s) {
+		if idx, ok := indexes[s[left]]; ok && idx >= right {
+			right = idx + 1
+		}
+		indexes[s[left]] = left
+		left++
+		res = max(res, left-right)
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0003. Longest Substring Without Repeating Characters/3. Longest Substring Without Repeating Characters_test.go b/leetcode/0003.Longest-Substring-Without-Repeating-Characters/3. Longest Substring Without Repeating Characters_test.go
similarity index 93%
rename from Algorithms/0003. Longest Substring Without Repeating Characters/3. Longest Substring Without Repeating Characters_test.go
rename to leetcode/0003.Longest-Substring-Without-Repeating-Characters/3. Longest Substring Without Repeating Characters_test.go
index f49e1f38a..7eebdbfa8 100644
--- a/Algorithms/0003. Longest Substring Without Repeating Characters/3. Longest Substring Without Repeating Characters_test.go	
+++ b/leetcode/0003.Longest-Substring-Without-Repeating-Characters/3. Longest Substring Without Repeating Characters_test.go	
@@ -26,22 +26,22 @@ func Test_Problem3(t *testing.T) {
 
 	qs := []question3{
 
-		question3{
+		{
 			para3{"abcabcbb"},
 			ans3{3},
 		},
 
-		question3{
+		{
 			para3{"bbbbb"},
 			ans3{1},
 		},
 
-		question3{
+		{
 			para3{"pwwkew"},
 			ans3{3},
 		},
 
-		question3{
+		{
 			para3{""},
 			ans3{0},
 		},
diff --git a/leetcode/0003.Longest-Substring-Without-Repeating-Characters/README.md b/leetcode/0003.Longest-Substring-Without-Repeating-Characters/README.md
new file mode 100644
index 000000000..9f80ab4a5
--- /dev/null
+++ b/leetcode/0003.Longest-Substring-Without-Repeating-Characters/README.md
@@ -0,0 +1,56 @@
+# [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/)
+
+## 题目
+
+Given a string, find the length of the longest substring without repeating characters.
+
+
+
+Example 1:
+
+```c
+Input: "abcabcbb"
+Output: 3 
+Explanation: The answer is "abc", with the length of 3. 
+```
+
+Example 2:
+
+```c
+Input: "bbbbb"
+Output: 1
+Explanation: The answer is "b", with the length of 1.
+```
+
+Example 3:
+
+```c
+Input: "pwwkew"
+Output: 3
+Explanation: The answer is "wke", with the length of 3. 
+             Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
+```
+
+## 题目大意
+
+
+在一个字符串中寻找没有重复字母的最长子串。
+
+## 解题思路
+
+这一题和第 438 题，第 3 题，第 76 题，第 567 题类似，用的思想都是"滑动窗口"。
+
+滑动窗口的右边界不断的右移，只要没有重复的字符，就持续向右扩大窗口边界。一旦出现了重复字符，就需要缩小左边界，直到重复的字符移出了左边界，然后继续移动滑动窗口的右边界。以此类推，每次移动需要计算当前长度，并判断是否需要更新最大长度，最终最大的值就是题目中的所求。
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/Algorithms/4. Median of Two Sorted Arrays/4. Median of Two Sorted Arrays.go b/leetcode/0004.Median-of-Two-Sorted-Arrays/4. Median of Two Sorted Arrays.go
similarity index 91%
rename from Algorithms/4. Median of Two Sorted Arrays/4. Median of Two Sorted Arrays.go
rename to leetcode/0004.Median-of-Two-Sorted-Arrays/4. Median of Two Sorted Arrays.go
index 02a3abe67..098e34721 100644
--- a/Algorithms/4. Median of Two Sorted Arrays/4. Median of Two Sorted Arrays.go	
+++ b/leetcode/0004.Median-of-Two-Sorted-Arrays/4. Median of Two Sorted Arrays.go	
@@ -41,3 +41,17 @@ func findMedianSortedArrays(nums1 []int, nums2 []int) float64 {
 	}
 	return float64(midLeft+midRight) / 2
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/Algorithms/4. Median of Two Sorted Arrays/4. Median of Two Sorted Arrays_test.go b/leetcode/0004.Median-of-Two-Sorted-Arrays/4. Median of Two Sorted Arrays_test.go
similarity index 96%
rename from Algorithms/4. Median of Two Sorted Arrays/4. Median of Two Sorted Arrays_test.go
rename to leetcode/0004.Median-of-Two-Sorted-Arrays/4. Median of Two Sorted Arrays_test.go
index c24c8e531..14d21d0e0 100644
--- a/Algorithms/4. Median of Two Sorted Arrays/4. Median of Two Sorted Arrays_test.go	
+++ b/leetcode/0004.Median-of-Two-Sorted-Arrays/4. Median of Two Sorted Arrays_test.go	
@@ -27,12 +27,12 @@ func Test_Problem4(t *testing.T) {
 
 	qs := []question4{
 
-		question4{
+		{
 			para4{[]int{1, 3}, []int{2}},
 			ans4{2.0},
 		},
 
-		question4{
+		{
 			para4{[]int{1, 2}, []int{3, 4}},
 			ans4{2.5},
 		},
diff --git a/leetcode/0004.Median-of-Two-Sorted-Arrays/README.md b/leetcode/0004.Median-of-Two-Sorted-Arrays/README.md
new file mode 100755
index 000000000..1840f2833
--- /dev/null
+++ b/leetcode/0004.Median-of-Two-Sorted-Arrays/README.md
@@ -0,0 +1,47 @@
+# [4. Median of Two Sorted Arrays](https://leetcode.com/problems/median-of-two-sorted-arrays/)
+
+
+## 题目
+
+There are two sorted arrays **nums1** and **nums2** of size m and n respectively.
+
+Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
+
+You may assume **nums1** and **nums2** cannot be both empty.
+
+**Example 1:**
+
+    nums1 = [1, 3]
+    nums2 = [2]
+    
+    The median is 2.0
+
+**Example 2:**
+
+    nums1 = [1, 2]
+    nums2 = [3, 4]
+    
+    The median is (2 + 3)/2 = 2.5
+
+
+## 题目大意
+
+
+给定两个大小为 m 和 n 的有序数组 nums1 和 nums2。
+
+请你找出这两个有序数组的中位数，并且要求算法的时间复杂度为 O(log(m + n))。
+
+你可以假设 nums1 和 nums2 不会同时为空。
+
+
+
+## 解题思路
+
+
+- 给出两个有序数组，要求找出这两个数组合并以后的有序数组中的中位数。要求时间复杂度为 O(log (m+n))。
+- 这一题最容易想到的办法是把两个数组合并，然后取出中位数。但是合并有序数组的操作是 `O(m+n)` 的，不符合题意。看到题目给的 `log` 的时间复杂度，很容易联想到二分搜索。
+- 由于要找到最终合并以后数组的中位数，两个数组的总大小也知道，所以中间这个位置也是知道的。只需要二分搜索一个数组中切分的位置，另一个数组中切分的位置也能得到。为了使得时间复杂度最小，所以二分搜索两个数组中长度较小的那个数组。
+- 关键的问题是如何切分数组 1 和数组 2 。其实就是如何切分数组 1 。先随便二分产生一个 `midA`，切分的线何时算满足了中位数的条件呢？即，线左边的数都小于右边的数，即，`nums1[midA-1] ≤ nums2[midB] && nums2[midB-1] ≤ nums1[midA]` 。如果这些条件都不满足，切分线就需要调整。如果 `nums1[midA] < nums2[midB-1]`，说明 `midA` 这条线划分出来左边的数小了，切分线应该右移；如果 `nums1[midA-1] > nums2[midB]`，说明 midA 这条线划分出来左边的数大了，切分线应该左移。经过多次调整以后，切分线总能找到满足条件的解。
+- 假设现在找到了切分的两条线了，`数组 1` 在切分线两边的下标分别是 `midA - 1` 和 `midA`。`数组 2` 在切分线两边的下标分别是 `midB - 1` 和 `midB`。最终合并成最终数组，如果数组长度是奇数，那么中位数就是 `max(nums1[midA-1], nums2[midB-1])`。如果数组长度是偶数，那么中间位置的两个数依次是：`max(nums1[midA-1], nums2[midB-1])` 和 `min(nums1[midA], nums2[midB])`，那么中位数就是 `(max(nums1[midA-1], nums2[midB-1]) + min(nums1[midA], nums2[midB])) / 2`。图示见下图：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_4.png)
diff --git a/leetcode/0005.Longest-Palindromic-Substring/5. Longest Palindromic Substring.go b/leetcode/0005.Longest-Palindromic-Substring/5. Longest Palindromic Substring.go
new file mode 100644
index 000000000..4ede2f1a5
--- /dev/null
+++ b/leetcode/0005.Longest-Palindromic-Substring/5. Longest Palindromic Substring.go	
@@ -0,0 +1,117 @@
+package leetcode
+
+// 解法一 Manacher's algorithm，时间复杂度 O(n)，空间复杂度 O(n)
+func longestPalindrome(s string) string {
+	if len(s) < 2 {
+		return s
+	}
+	newS := make([]rune, 0)
+	newS = append(newS, '#')
+	for _, c := range s {
+		newS = append(newS, c)
+		newS = append(newS, '#')
+	}
+	// dp[i]:    以预处理字符串下标 i 为中心的回文半径(奇数长度时不包括中心)
+	// maxRight: 通过中心扩散的方式能够扩散的最右边的下标
+	// center:   与 maxRight 对应的中心字符的下标
+	// maxLen:   记录最长回文串的半径
+	// begin:    记录最长回文串在起始串 s 中的起始下标
+	dp, maxRight, center, maxLen, begin := make([]int, len(newS)), 0, 0, 1, 0
+	for i := 0; i < len(newS); i++ {
+		if i < maxRight {
+			// 这一行代码是 Manacher 算法的关键所在
+			dp[i] = min(maxRight-i, dp[2*center-i])
+		}
+		// 中心扩散法更新 dp[i]
+		left, right := i-(1+dp[i]), i+(1+dp[i])
+		for left >= 0 && right < len(newS) && newS[left] == newS[right] {
+			dp[i]++
+			left--
+			right++
+		}
+		// 更新 maxRight，它是遍历过的 i 的 i + dp[i] 的最大者
+		if i+dp[i] > maxRight {
+			maxRight = i + dp[i]
+			center = i
+		}
+		// 记录最长回文子串的长度和相应它在原始字符串中的起点
+		if dp[i] > maxLen {
+			maxLen = dp[i]
+			begin = (i - maxLen) / 2 // 这里要除以 2 因为有我们插入的辅助字符 #
+		}
+	}
+	return s[begin : begin+maxLen]
+}
+
+func min(x, y int) int {
+	if x < y {
+		return x
+	}
+	return y
+}
+
+// 解法二 滑动窗口，时间复杂度 O(n^2)，空间复杂度 O(1)
+func longestPalindrome1(s string) string {
+	if len(s) == 0 {
+		return ""
+	}
+	left, right, pl, pr := 0, -1, 0, 0
+	for left < len(s) {
+		// 移动到相同字母的最右边（如果有相同字母）
+		for right+1 < len(s) && s[left] == s[right+1] {
+			right++
+		}
+		// 找到回文的边界
+		for left-1 >= 0 && right+1 < len(s) && s[left-1] == s[right+1] {
+			left--
+			right++
+		}
+		if right-left > pr-pl {
+			pl, pr = left, right
+		}
+		// 重置到下一次寻找回文的中心
+		left = (left+right)/2 + 1
+		right = left
+	}
+	return s[pl : pr+1]
+}
+
+// 解法三 中心扩散法，时间复杂度 O(n^2)，空间复杂度 O(1)
+func longestPalindrome2(s string) string {
+	res := ""
+	for i := 0; i < len(s); i++ {
+		res = maxPalindrome(s, i, i, res)
+		res = maxPalindrome(s, i, i+1, res)
+	}
+	return res
+}
+
+func maxPalindrome(s string, i, j int, res string) string {
+	sub := ""
+	for i >= 0 && j < len(s) && s[i] == s[j] {
+		sub = s[i : j+1]
+		i--
+		j++
+	}
+	if len(res) < len(sub) {
+		return sub
+	}
+	return res
+}
+
+// 解法四 DP，时间复杂度 O(n^2)，空间复杂度 O(n^2)
+func longestPalindrome3(s string) string {
+	res, dp := "", make([][]bool, len(s))
+	for i := 0; i < len(s); i++ {
+		dp[i] = make([]bool, len(s))
+	}
+	for i := len(s) - 1; i >= 0; i-- {
+		for j := i; j < len(s); j++ {
+			dp[i][j] = (s[i] == s[j]) && ((j-i < 3) || dp[i+1][j-1])
+			if dp[i][j] && (res == "" || j-i+1 > len(res)) {
+				res = s[i : j+1]
+			}
+		}
+	}
+	return res
+}
diff --git a/leetcode/0005.Longest-Palindromic-Substring/5. Longest Palindromic Substring_test.go b/leetcode/0005.Longest-Palindromic-Substring/5. Longest Palindromic Substring_test.go
new file mode 100644
index 000000000..de64571bd
--- /dev/null
+++ b/leetcode/0005.Longest-Palindromic-Substring/5. Longest Palindromic Substring_test.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question5 struct {
+	para5
+	ans5
+}
+
+// para 是参数
+// one 代表第一个参数
+type para5 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans5 struct {
+	one string
+}
+
+func Test_Problem5(t *testing.T) {
+
+	qs := []question5{
+
+		{
+			para5{"babad"},
+			ans5{"bab"},
+		},
+
+		{
+			para5{"cbbd"},
+			ans5{"bb"},
+		},
+
+		{
+			para5{"a"},
+			ans5{"a"},
+		},
+
+		{
+			para5{"ac"},
+			ans5{"a"},
+		},
+
+		{
+			para5{"aa"},
+			ans5{"aa"},
+		},
+
+		{
+			para5{"ajgiljtperkvubjmdsefcylksrxtftqrehoitdgdtttswwttmfuvwgwrruuqmxttzsbmuhgfaoueumvbhajqsaxkkihjwevzzedizmrsmpxqavyryklbotwzngxscvyuqjkkaotitddlhhnutmotupwuwyltebtsdfssbwayuxrbgihmtphshdslktvsjadaykyjivbzhwujcdvzdxxfiixnzrmusqvwujjmxhbqbdpauacnzojnzxxgrkmupadfcsujkcwajsgintahwgbjnvjqubcxajdyyapposrkpqtpqfjcvbhlmwfutgognqxgaukpmdyaxghgoqkqnigcllachmwzrazwhpppmsodvxilrccfqgpkmdqhoorxpyjsrtbeeidsinpeyxxpsjnymxkouskyhenzgieybwkgzrhhrzgkwbyeigznehyksuokxmynjspxxyepnisdieebtrsjypxroohqdmkpgqfccrlixvdosmppphwzarzwmhcallcginqkqoghgxaydmpkuagxqngogtufwmlhbvcjfqptqpkrsoppayydjaxcbuqjvnjbgwhatnigsjawckjuscfdapumkrgxxznjozncauapdbqbhxmjjuwvqsumrznxiifxxdzvdcjuwhzbvijykyadajsvtklsdhshptmhigbrxuyawbssfdstbetlywuwputomtunhhlddtitoakkjquyvcsxgnzwtoblkyryvaqxpmsrmzidezzvewjhikkxasqjahbvmueuoafghumbszttxmquurrwgwvufmttwwstttdgdtioherqtftxrsklycfesdmjbuvkreptjligja"},
+			ans5{"ajgiljtperkvubjmdsefcylksrxtftqrehoitdgdtttswwttmfuvwgwrruuqmxttzsbmuhgfaoueumvbhajqsaxkkihjwevzzedizmrsmpxqavyryklbotwzngxscvyuqjkkaotitddlhhnutmotupwuwyltebtsdfssbwayuxrbgihmtphshdslktvsjadaykyjivbzhwujcdvzdxxfiixnzrmusqvwujjmxhbqbdpauacnzojnzxxgrkmupadfcsujkcwajsgintahwgbjnvjqubcxajdyyapposrkpqtpqfjcvbhlmwfutgognqxgaukpmdyaxghgoqkqnigcllachmwzrazwhpppmsodvxilrccfqgpkmdqhoorxpyjsrtbeeidsinpeyxxpsjnymxkouskyhenzgieybwkgzrhhrzgkwbyeigznehyksuokxmynjspxxyepnisdieebtrsjypxroohqdmkpgqfccrlixvdosmppphwzarzwmhcallcginqkqoghgxaydmpkuagxqngogtufwmlhbvcjfqptqpkrsoppayydjaxcbuqjvnjbgwhatnigsjawckjuscfdapumkrgxxznjozncauapdbqbhxmjjuwvqsumrznxiifxxdzvdcjuwhzbvijykyadajsvtklsdhshptmhigbrxuyawbssfdstbetlywuwputomtunhhlddtitoakkjquyvcsxgnzwtoblkyryvaqxpmsrmzidezzvewjhikkxasqjahbvmueuoafghumbszttxmquurrwgwvufmttwwstttdgdtioherqtftxrsklycfesdmjbuvkreptjligja"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 5------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans5, q.para5
+		fmt.Printf("【input】:%v       【output】:%v\n", p, longestPalindrome(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0005.Longest-Palindromic-Substring/README.md b/leetcode/0005.Longest-Palindromic-Substring/README.md
new file mode 100644
index 000000000..e02c14561
--- /dev/null
+++ b/leetcode/0005.Longest-Palindromic-Substring/README.md
@@ -0,0 +1,186 @@
+# [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)
+
+
+## 题目
+
+Given a string `s`, return *the longest palindromic substring* in `s`.
+
+**Example 1:**
+
+```
+Input: s = "babad"
+Output: "bab"
+Note: "aba" is also a valid answer.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "cbbd"
+Output: "bb"
+
+```
+
+**Example 3:**
+
+```
+Input: s = "a"
+Output: "a"
+
+```
+
+**Example 4:**
+
+```
+Input: s = "ac"
+Output: "a"
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 1000`
+- `s` consist of only digits and English letters (lower-case and/or upper-case),
+
+## 题目大意
+
+给你一个字符串 `s`，找到 `s` 中最长的回文子串。
+
+## 解题思路
+
+- 此题非常经典，并且有多种解法。
+- 解法一，动态规划。定义 `dp[i][j]` 表示从字符串第 `i` 个字符到第 `j` 个字符这一段子串是否是回文串。由回文串的性质可以得知，回文串去掉一头一尾相同的字符以后，剩下的还是回文串。所以状态转移方程是 `dp[i][j] = (s[i] == s[j]) && ((j-i < 3) || dp[i+1][j-1])`，注意特殊的情况，`j - i == 1` 的时候，即只有 2 个字符的情况，只需要判断这 2 个字符是否相同即可。`j - i == 2` 的时候，即只有 3 个字符的情况，只需要判断除去中心以外对称的 2 个字符是否相等。每次循环动态维护保存最长回文串即可。时间复杂度 O(n^2)，空间复杂度 O(n^2)。
+- 解法二，中心扩散法。动态规划的方法中，我们将任意起始，终止范围内的字符串都判断了一遍。其实没有这个必要，如果不是最长回文串，无需判断并保存结果。所以动态规划的方法在空间复杂度上还有优化空间。判断回文有一个核心问题是找到“轴心”。如果长度是偶数，那么轴心是中心虚拟的，如果长度是奇数，那么轴心正好是正中心的那个字母。中心扩散法的思想是枚举每个轴心的位置。然后做两次假设，假设最长回文串是偶数，那么以虚拟中心往 2 边扩散；假设最长回文串是奇数，那么以正中心的字符往 2 边扩散。扩散的过程就是对称判断两边字符是否相等的过程。这个方法时间复杂度和动态规划是一样的，但是空间复杂度降低了。时间复杂度 O(n^2)，空间复杂度 O(1)。
+- 解法三，滑动窗口。这个写法其实就是中心扩散法变了一个写法。中心扩散是依次枚举每一个轴心。滑动窗口的方法稍微优化了一点，有些轴心两边字符不相等，下次就不会枚举这些不可能形成回文子串的轴心了。不过这点优化并没有优化时间复杂度，时间复杂度 O(n^2)，空间复杂度 O(1)。
+- 解法四，马拉车算法。这个算法是本题的最优解，也是最复杂的解法。时间复杂度 O(n)，空间复杂度 O(n)。中心扩散法有 2 处有重复判断，第一处是每次都往两边扩散，不同中心扩散多次，实际上有很多重复判断的字符，能否不重复判断？第二处，中心能否跳跃选择，不是每次都枚举，是否可以利用前一次的信息，跳跃选择下一次的中心？马拉车算法针对重复判断的问题做了优化，增加了一个辅助数组，将时间复杂度从 O(n^2) 优化到了 O(n)，空间换了时间，空间复杂度增加到 O(n)。
+
+    ![https://img.halfrost.com/Leetcode/leetcode_5_1.png](https://img.halfrost.com/Leetcode/leetcode_5_1.png)
+
+- 首先是预处理，向字符串的头尾以及每两个字符中间添加一个特殊字符 `#`，比如字符串 `aaba` 处理后会变成 `#a#a#b#a#`。那么原先长度为偶数的回文字符串 `aa` 会变成长度为奇数的回文字符串 `#a#a#`，而长度为奇数的回文字符串 `aba` 会变成长度仍然为奇数的回文字符串 `#a#b#a#`，经过预处理以后，都会变成长度为奇数的字符串。**注意这里的特殊字符不需要是没有出现过的字母，也可以使用任何一个字符来作为这个特殊字符。**这是因为，当我们只考虑长度为奇数的回文字符串时，每次我们比较的两个字符奇偶性一定是相同的，所以原来字符串中的字符不会与插入的特殊字符互相比较，不会因此产生问题。**预处理以后，以某个中心扩散的步数和实际字符串长度是相等的。**因为半径里面包含了插入的特殊字符，又由于左右对称的性质，所以扩散半径就等于原来回文子串的长度。
+
+    ![https://img.halfrost.com/Leetcode/leetcode_5_2.png](https://img.halfrost.com/Leetcode/leetcode_5_2.png)
+
+- 核心部分是如何通过左边已经扫描过的数据推出右边下一次要扩散的中心。这里定义下一次要扩散的中心下标是 `i`。如果 `i` 比 `maxRight` 要小，只能继续中心扩散。如果 `i` 比 `maxRight` 大，这是又分为 3 种情况。三种情况见上图。将上述 3 种情况总结起来，就是 ：`dp[i] = min(maxRight-i, dp[2*center-i])`，其中，`mirror` 相对于 `center` 是和 `i` 中心对称的，所以它的下标可以计算出来是 `2*center-i`。更新完 `dp[i]` 以后，就要进行中心扩散了。中心扩散以后动态维护最长回文串并相应的更新 `center`，`maxRight`，并且记录下原始字符串的起始位置 `begin` 和 `maxLen`。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 Manacher's algorithm，时间复杂度 O(n)，空间复杂度 O(n)
+func longestPalindrome(s string) string {
+	if len(s) < 2 {
+		return s
+	}
+	newS := make([]rune, 0)
+	newS = append(newS, '#')
+	for _, c := range s {
+		newS = append(newS, c)
+		newS = append(newS, '#')
+	}
+	// dp[i]:    以预处理字符串下标 i 为中心的回文半径(奇数长度时不包括中心)
+	// maxRight: 通过中心扩散的方式能够扩散的最右边的下标
+	// center:   与 maxRight 对应的中心字符的下标
+	// maxLen:   记录最长回文串的半径
+	// begin:    记录最长回文串在起始串 s 中的起始下标
+	dp, maxRight, center, maxLen, begin := make([]int, len(newS)), 0, 0, 1, 0
+	for i := 0; i < len(newS); i++ {
+		if i < maxRight {
+			// 这一行代码是 Manacher 算法的关键所在
+			dp[i] = min(maxRight-i, dp[2*center-i])
+		}
+		// 中心扩散法更新 dp[i]
+		left, right := i-(1+dp[i]), i+(1+dp[i])
+		for left >= 0 && right < len(newS) && newS[left] == newS[right] {
+			dp[i]++
+			left--
+			right++
+		}
+		// 更新 maxRight，它是遍历过的 i 的 i + dp[i] 的最大者
+		if i+dp[i] > maxRight {
+			maxRight = i + dp[i]
+			center = i
+		}
+		// 记录最长回文子串的长度和相应它在原始字符串中的起点
+		if dp[i] > maxLen {
+			maxLen = dp[i]
+			begin = (i - maxLen) / 2 // 这里要除以 2 因为有我们插入的辅助字符 #
+		}
+	}
+	return s[begin : begin+maxLen]
+}
+
+func min(x, y int) int {
+	if x < y {
+		return x
+	}
+	return y
+}
+
+// 解法二 滑动窗口，时间复杂度 O(n^2)，空间复杂度 O(1)
+func longestPalindrome1(s string) string {
+	if len(s) == 0 {
+		return ""
+	}
+	left, right, pl, pr := 0, -1, 0, 0
+	for left < len(s) {
+		// 移动到相同字母的最右边（如果有相同字母）
+		for right+1 < len(s) && s[left] == s[right+1] {
+			right++
+		}
+		// 找到回文的边界
+		for left-1 >= 0 && right+1 < len(s) && s[left-1] == s[right+1] {
+			left--
+			right++
+		}
+		if right-left > pr-pl {
+			pl, pr = left, right
+		}
+		// 重置到下一次寻找回文的中心
+		left = (left+right)/2 + 1
+		right = left
+	}
+	return s[pl : pr+1]
+}
+
+// 解法三 中心扩散法，时间复杂度 O(n^2)，空间复杂度 O(1)
+func longestPalindrome2(s string) string {
+	res := ""
+	for i := 0; i < len(s); i++ {
+		res = maxPalindrome(s, i, i, res)
+		res = maxPalindrome(s, i, i+1, res)
+	}
+	return res
+}
+
+func maxPalindrome(s string, i, j int, res string) string {
+	sub := ""
+	for i >= 0 && j < len(s) && s[i] == s[j] {
+		sub = s[i : j+1]
+		i--
+		j++
+	}
+	if len(res) < len(sub) {
+		return sub
+	}
+	return res
+}
+
+// 解法四 DP，时间复杂度 O(n^2)，空间复杂度 O(n^2)
+func longestPalindrome3(s string) string {
+	res, dp := "", make([][]bool, len(s))
+	for i := 0; i < len(s); i++ {
+		dp[i] = make([]bool, len(s))
+	}
+	for i := len(s) - 1; i >= 0; i-- {
+		for j := i; j < len(s); j++ {
+			dp[i][j] = (s[i] == s[j]) && ((j-i < 3) || dp[i+1][j-1])
+			if dp[i][j] && (res == "" || j-i+1 > len(res)) {
+				res = s[i : j+1]
+			}
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion.go b/leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion.go
new file mode 100644
index 000000000..e35454ab2
--- /dev/null
+++ b/leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion.go	
@@ -0,0 +1,26 @@
+package leetcode
+
+func convert(s string, numRows int) string {
+	matrix, down, up := make([][]byte, numRows, numRows), 0, numRows-2
+	for i := 0; i != len(s); {
+		if down != numRows {
+			matrix[down] = append(matrix[down], byte(s[i]))
+			down++
+			i++
+		} else if up > 0 {
+			matrix[up] = append(matrix[up], byte(s[i]))
+			up--
+			i++
+		} else {
+			up = numRows - 2
+			down = 0
+		}
+	}
+	solution := make([]byte, 0, len(s))
+	for _, row := range matrix {
+		for _, item := range row {
+			solution = append(solution, item)
+		}
+	}
+	return string(solution)
+}
diff --git a/leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion_test.go b/leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion_test.go
new file mode 100644
index 000000000..c6bbd8242
--- /dev/null
+++ b/leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question6 struct {
+	para6
+	ans6
+}
+
+// para 是参数
+// one 代表第一个参数
+type para6 struct {
+	s       string
+	numRows int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans6 struct {
+	one string
+}
+
+func Test_Problem6(t *testing.T) {
+
+	qs := []question6{
+
+		{
+			para6{"PAYPALISHIRING", 3},
+			ans6{"PAHNAPLSIIGYIR"},
+		},
+
+		{
+			para6{"PAYPALISHIRING", 4},
+			ans6{"PINALSIGYAHRPI"},
+		},
+
+		{
+			para6{"A", 1},
+			ans6{"A"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 6------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans6, q.para6
+		fmt.Printf("【input】:%v       【output】:%v\n", p, convert(p.s, p.numRows))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0006.ZigZag-Conversion/README.md b/leetcode/0006.ZigZag-Conversion/README.md
new file mode 100644
index 000000000..b281e084c
--- /dev/null
+++ b/leetcode/0006.ZigZag-Conversion/README.md
@@ -0,0 +1,107 @@
+# [6. ZigZag Conversion](https://leetcode.com/problems/zigzag-conversion/)
+
+
+## 题目
+
+The string `"PAYPALISHIRING"` is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
+
+```
+P   A   H   N
+A P L S I I G
+Y   I   R
+```
+
+And then read line by line: `"PAHNAPLSIIGYIR"`
+
+Write the code that will take a string and make this conversion given a number of rows:
+
+```
+string convert(string s, int numRows);
+```
+
+**Example 1:**
+
+```
+Input: s = "PAYPALISHIRING", numRows = 3
+Output: "PAHNAPLSIIGYIR"
+```
+
+**Example 2:**
+
+```
+Input: s = "PAYPALISHIRING", numRows = 4
+Output: "PINALSIGYAHRPI"
+Explanation:
+P     I    N
+A   L S  I G
+Y A   H R
+P     I
+```
+
+**Example 3:**
+
+```
+Input: s = "A", numRows = 1
+Output: "A"
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 1000`
+- `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
+- `1 <= numRows <= 1000`
+
+## 题目大意
+
+将一个给定字符串 `s` 根据给定的行数 `numRows` ，以从上往下、从左到右进行 Z 字形排列。
+
+比如输入字符串为 `"PAYPALISHIRING"` 行数为 3 时，排列如下：
+
+```go
+P   A   H   N
+A P L S I I G
+Y   I   R
+```
+
+之后，你的输出需要从左往右逐行读取，产生出一个新的字符串，比如：`"PAHNAPLSIIGYIR"`。
+
+请你实现这个将字符串进行指定行数变换的函数：
+
+```go
+string convert(string s, int numRows);
+```
+
+## 解题思路
+
+- 这一题没有什么算法思想，考察的是对程序控制的能力。用 2 个变量保存方向，当垂直输出的行数达到了规定的目标行数以后，需要从下往上转折到第一行，循环中控制好方向ji
+
+## 代码
+
+```go
+package leetcode
+
+func convert(s string, numRows int) string {
+	matrix, down, up := make([][]byte, numRows, numRows), 0, numRows-2
+	for i := 0; i != len(s); {
+		if down != numRows {
+			matrix[down] = append(matrix[down], byte(s[i]))
+			down++
+			i++
+		} else if up > 0 {
+			matrix[up] = append(matrix[up], byte(s[i]))
+			up--
+			i++
+		} else {
+			up = numRows - 2
+			down = 0
+		}
+	}
+	solution := make([]byte, 0, len(s))
+	for _, row := range matrix {
+		for _, item := range row {
+			solution = append(solution, item)
+		}
+	}
+	return string(solution)
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0007. Reverse Integer/7. Reverse Integer.go b/leetcode/0007.Reverse-Integer/7. Reverse Integer.go
similarity index 100%
rename from Algorithms/0007. Reverse Integer/7. Reverse Integer.go
rename to leetcode/0007.Reverse-Integer/7. Reverse Integer.go
diff --git a/Algorithms/0007. Reverse Integer/7. Reverse Integer_test.go b/leetcode/0007.Reverse-Integer/7. Reverse Integer_test.go
similarity index 92%
rename from Algorithms/0007. Reverse Integer/7. Reverse Integer_test.go
rename to leetcode/0007.Reverse-Integer/7. Reverse Integer_test.go
index 8fb8e5942..4154fb9e9 100644
--- a/Algorithms/0007. Reverse Integer/7. Reverse Integer_test.go	
+++ b/leetcode/0007.Reverse-Integer/7. Reverse Integer_test.go	
@@ -26,22 +26,22 @@ func Test_Problem7(t *testing.T) {
 
 	qs := []question7{
 
-		question7{
+		{
 			para7{321},
 			ans7{123},
 		},
 
-		question7{
+		{
 			para7{-123},
 			ans7{-321},
 		},
 
-		question7{
+		{
 			para7{120},
 			ans7{21},
 		},
 
-		question7{
+		{
 			para7{1534236469},
 			ans7{0},
 		},
diff --git a/leetcode/0007.Reverse-Integer/README.md b/leetcode/0007.Reverse-Integer/README.md
new file mode 100755
index 000000000..84504f047
--- /dev/null
+++ b/leetcode/0007.Reverse-Integer/README.md
@@ -0,0 +1,35 @@
+# [7. Reverse Integer](https://leetcode.com/problems/reverse-integer/)
+
+
+## 题目
+
+Given a 32-bit signed integer, reverse digits of an integer.
+
+**Example 1:**
+
+    Input: 123
+    Output: 321
+
+**Example 2:**
+
+    Input: -123
+    Output: -321
+
+**Example 3:**
+
+    Input: 120
+    Output: 21
+
+**Note:**Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
+
+## 题目大意
+
+给出一个 32 位的有符号整数，你需要将这个整数中每位上的数字进行反转。注意:假设我们的环境只能存储得下 32 位的有符号整数，则其数值范围为 [−2^31,  2^31 − 1]。请根据这个假设，如果反转后整数溢出那么就返回 0。
+
+
+
+## 解题思路
+
+
+- 这一题是简单题，要求反转 10 进制数。类似的题目有第 190 题。
+- 这一题只需要注意一点，反转以后的数字要求在 [−2^31, 2^31 − 1]范围内，超过这个范围的数字都要输出 0 。
diff --git a/leetcode/0008.String-to-Integer-atoi/8. String to Integer atoi.go b/leetcode/0008.String-to-Integer-atoi/8. String to Integer atoi.go
new file mode 100644
index 000000000..24a04676d
--- /dev/null
+++ b/leetcode/0008.String-to-Integer-atoi/8. String to Integer atoi.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+func myAtoi(s string) int {
+	maxInt, signAllowed, whitespaceAllowed, sign, digits := int64(2<<30), true, true, 1, []int{}
+	for _, c := range s {
+		if c == ' ' && whitespaceAllowed {
+			continue
+		}
+		if signAllowed {
+			if c == '+' {
+				signAllowed = false
+				whitespaceAllowed = false
+				continue
+			} else if c == '-' {
+				sign = -1
+				signAllowed = false
+				whitespaceAllowed = false
+				continue
+			}
+		}
+		if c < '0' || c > '9' {
+			break
+		}
+		whitespaceAllowed, signAllowed = false, false
+		digits = append(digits, int(c-48))
+	}
+	var num, place int64
+	place, num = 1, 0
+	lastLeading0Index := -1
+	for i, d := range digits {
+		if d == 0 {
+			lastLeading0Index = i
+		} else {
+			break
+		}
+	}
+	if lastLeading0Index > -1 {
+		digits = digits[lastLeading0Index+1:]
+	}
+	var rtnMax int64
+	if sign > 0 {
+		rtnMax = maxInt - 1
+	} else {
+		rtnMax = maxInt
+	}
+	digitsCount := len(digits)
+	for i := digitsCount - 1; i >= 0; i-- {
+		num += int64(digits[i]) * place
+		place *= 10
+		if digitsCount-i > 10 || num > rtnMax {
+			return int(int64(sign) * rtnMax)
+		}
+	}
+	num *= int64(sign)
+	return int(num)
+}
diff --git a/leetcode/0008.String-to-Integer-atoi/8. String to Integer atoi_test.go b/leetcode/0008.String-to-Integer-atoi/8. String to Integer atoi_test.go
new file mode 100644
index 000000000..92b769ed4
--- /dev/null
+++ b/leetcode/0008.String-to-Integer-atoi/8. String to Integer atoi_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question8 struct {
+	para8
+	ans8
+}
+
+// para 是参数
+// one 代表第一个参数
+type para8 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans8 struct {
+	one int
+}
+
+func Test_Problem8(t *testing.T) {
+
+	qs := []question8{
+
+		{
+			para8{"42"},
+			ans8{42},
+		},
+
+		{
+			para8{"   -42"},
+			ans8{-42},
+		},
+
+		{
+			para8{"4193 with words"},
+			ans8{4193},
+		},
+
+		{
+			para8{"words and 987"},
+			ans8{0},
+		},
+
+		{
+			para8{"-91283472332"},
+			ans8{-2147483648},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 8------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans8, q.para8
+		fmt.Printf("【input】:%v    【output】:%v\n", p.one, myAtoi(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0008.String-to-Integer-atoi/README.md b/leetcode/0008.String-to-Integer-atoi/README.md
new file mode 100644
index 000000000..677ed3340
--- /dev/null
+++ b/leetcode/0008.String-to-Integer-atoi/README.md
@@ -0,0 +1,192 @@
+# [8. String to Integer (atoi)](https://leetcode.com/problems/string-to-integer-atoi/)
+
+
+## 题目
+
+Implement the `myAtoi(string s)` function, which converts a string to a 32-bit signed integer (similar to C/C++'s `atoi` function).
+
+The algorithm for `myAtoi(string s)` is as follows:
+
+1. Read in and ignore any leading whitespace.
+2. Check if the next character (if not already at the end of the string) is `'-'` or `'+'`. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
+3. Read in next the characters until the next non-digit charcter or the end of the input is reached. The rest of the string is ignored.
+4. Convert these digits into an integer (i.e. `"123" -> 123`, `"0032" -> 32`). If no digits were read, then the integer is `0`. Change the sign as necessary (from step 2).
+5. If the integer is out of the 32-bit signed integer range `[-231, 231 - 1]`, then clamp the integer so that it remains in the range. Specifically, integers less than `231` should be clamped to `231`, and integers greater than `231 - 1` should be clamped to `231 - 1`.
+6. Return the integer as the final result.
+
+**Note:**
+
+- Only the space character `' '` is considered a whitespace character.
+- **Do not ignore** any characters other than the leading whitespace or the rest of the string after the digits.
+
+**Example 1:**
+
+```
+Input: s = "42"
+Output: 42
+Explanation: The underlined characters are what is read in, the caret is the current reader position.
+Step 1: "42" (no characters read because there is no leading whitespace)
+         ^
+Step 2: "42" (no characters read because there is neither a '-' nor '+')
+         ^
+Step 3: "42" ("42" is read in)
+           ^
+The parsed integer is 42.
+Since 42 is in the range [-231, 231 - 1], the final result is 42.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "   -42"
+Output: -42
+Explanation:
+Step 1: "   -42" (leading whitespace is read and ignored)
+            ^
+Step 2: "   -42" ('-' is read, so the result should be negative)
+             ^
+Step 3: "   -42" ("42" is read in)
+               ^
+The parsed integer is -42.
+Since -42 is in the range [-231, 231 - 1], the final result is -42.
+
+```
+
+**Example 3:**
+
+```
+Input: s = "4193 with words"
+Output: 4193
+Explanation:
+Step 1: "4193 with words" (no characters read because there is no leading whitespace)
+         ^
+Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
+         ^
+Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
+             ^
+The parsed integer is 4193.
+Since 4193 is in the range [-231, 231 - 1], the final result is 4193.
+
+```
+
+**Example 4:**
+
+```
+Input: s = "words and 987"
+Output: 0
+Explanation:
+Step 1: "words and 987" (no characters read because there is no leading whitespace)
+         ^
+Step 2: "words and 987" (no characters read because there is neither a '-' nor '+')
+         ^
+Step 3: "words and 987" (reading stops immediately because there is a non-digit 'w')
+         ^
+The parsed integer is 0 because no digits were read.
+Since 0 is in the range [-231, 231 - 1], the final result is 0.
+
+```
+
+**Example 5:**
+
+```
+Input: s = "-91283472332"
+Output: -2147483648
+Explanation:
+Step 1: "-91283472332" (no characters read because there is no leading whitespace)
+         ^
+Step 2: "-91283472332" ('-' is read, so the result should be negative)
+          ^
+Step 3: "-91283472332" ("91283472332" is read in)
+                     ^
+The parsed integer is -91283472332.
+Since -91283472332 is less than the lower bound of the range [-231, 231 - 1], the final result is clamped to -231 = -2147483648.
+```
+
+**Constraints:**
+
+- `0 <= s.length <= 200`
+- `s` consists of English letters (lower-case and upper-case), digits (`0-9`), `' '`, `'+'`
+
+## 题目大意
+
+请你来实现一个 myAtoi(string s) 函数，使其能将字符串转换成一个 32 位有符号整数（类似 C/C++ 中的 atoi 函数）。
+
+函数 myAtoi(string s) 的算法如下：
+
+- 读入字符串并丢弃无用的前导空格
+- 检查下一个字符（假设还未到字符末尾）为正还是负号，读取该字符（如果有）。 确定最终结果是负数还是正数。 如果两者都不存在，则假定结果为正。
+- 读入下一个字符，直到到达下一个非数字字符或到达输入的结尾。字符串的其余部分将被忽略。
+- 将前面步骤读入的这些数字转换为整数（即，"123" -> 123， "0032" -> 32）。如果没有读入数字，则整数为 0 。必要时更改符号（从步骤 2 开始）。
+- 如果整数数超过 32 位有符号整数范围 [−231,  231 − 1] ，需要截断这个整数，使其保持在这个范围内。具体来说，小于 −231 的整数应该被固定为 −231 ，大于 231 − 1 的整数应该被固定为 231 − 1 。
+- 返回整数作为最终结果。
+
+注意：
+
+- 本题中的空白字符只包括空格字符 ' ' 。
+- 除前导空格或数字后的其余字符串外，请勿忽略 任何其他字符。
+
+## 解题思路
+
+- 这题是简单题。题目要求实现类似 `C++` 中 `atoi` 函数的功能。这个函数功能是将字符串类型的数字转成 `int` 类型数字。先去除字符串中的前导空格，并判断记录数字的符号。数字需要去掉前导 `0` 。最后将数字转换成数字类型，判断是否超过 `int` 类型的上限 `[-2^31, 2^31 - 1]`，如果超过上限，需要输出边界，即 `-2^31`，或者 `2^31 - 1`。
+
+## 代码
+
+```go
+package leetcode
+
+func myAtoi(s string) int {
+	maxInt, signAllowed, whitespaceAllowed, sign, digits := int64(2<<30), true, true, 1, []int{}
+	for _, c := range s {
+		if c == ' ' && whitespaceAllowed {
+			continue
+		}
+		if signAllowed {
+			if c == '+' {
+				signAllowed = false
+				whitespaceAllowed = false
+				continue
+			} else if c == '-' {
+				sign = -1
+				signAllowed = false
+				whitespaceAllowed = false
+				continue
+			}
+		}
+		if c < '0' || c > '9' {
+			break
+		}
+		whitespaceAllowed, signAllowed = false, false
+		digits = append(digits, int(c-48))
+	}
+	var num, place int64
+	place, num = 1, 0
+	lastLeading0Index := -1
+	for i, d := range digits {
+		if d == 0 {
+			lastLeading0Index = i
+		} else {
+			break
+		}
+	}
+	if lastLeading0Index > -1 {
+		digits = digits[lastLeading0Index+1:]
+	}
+	var rtnMax int64
+	if sign > 0 {
+		rtnMax = maxInt - 1
+	} else {
+		rtnMax = maxInt
+	}
+	digitsCount := len(digits)
+	for i := digitsCount - 1; i >= 0; i-- {
+		num += int64(digits[i]) * place
+		place *= 10
+		if digitsCount-i > 10 || num > rtnMax {
+			return int(int64(sign) * rtnMax)
+		}
+	}
+	num *= int64(sign)
+	return int(num)
+}
+```
\ No newline at end of file
diff --git a/leetcode/0009.Palindrome-Number/9. Palindrome Number.go b/leetcode/0009.Palindrome-Number/9. Palindrome Number.go
new file mode 100644
index 000000000..f5c284eea
--- /dev/null
+++ b/leetcode/0009.Palindrome-Number/9. Palindrome Number.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import "strconv"
+
+// 解法一
+func isPalindrome(x int) bool {
+	if x < 0 {
+		return false
+	}
+	if x == 0 {
+		return true
+	}
+	if x%10 == 0 {
+		return false
+	}
+	arr := make([]int, 0, 32)
+	for x > 0 {
+		arr = append(arr, x%10)
+		x = x / 10
+	}
+	sz := len(arr)
+	for i, j := 0, sz-1; i <= j; i, j = i+1, j-1 {
+		if arr[i] != arr[j] {
+			return false
+		}
+	}
+	return true
+}
+
+// 解法二 数字转字符串
+func isPalindrome1(x int) bool {
+	if x < 0 {
+		return false
+	}
+	if x < 10 {
+		return true
+	}
+	s := strconv.Itoa(x)
+	length := len(s)
+	for i := 0; i <= length/2; i++ {
+		if s[i] != s[length-1-i] {
+			return false
+		}
+	}
+	return true
+}
diff --git a/leetcode/0009.Palindrome-Number/9. Palindrome Number_test.go b/leetcode/0009.Palindrome-Number/9. Palindrome Number_test.go
new file mode 100644
index 000000000..57f465408
--- /dev/null
+++ b/leetcode/0009.Palindrome-Number/9. Palindrome Number_test.go	
@@ -0,0 +1,72 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question9 struct {
+	para9
+	ans9
+}
+
+// para 是参数
+// one 代表第一个参数
+type para9 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans9 struct {
+	one bool
+}
+
+func Test_Problem9(t *testing.T) {
+
+	qs := []question9{
+
+		{
+			para9{121},
+			ans9{true},
+		},
+
+		{
+			para9{-121},
+			ans9{false},
+		},
+
+		{
+			para9{10},
+			ans9{false},
+		},
+
+		{
+			para9{321},
+			ans9{false},
+		},
+
+		{
+			para9{-123},
+			ans9{false},
+		},
+
+		{
+			para9{120},
+			ans9{false},
+		},
+
+		{
+			para9{1534236469},
+			ans9{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 9------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans9, q.para9
+		fmt.Printf("【input】:%v    【output】:%v\n", p.one, isPalindrome(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0009.Palindrome-Number/README.md b/leetcode/0009.Palindrome-Number/README.md
new file mode 100644
index 000000000..168404fd2
--- /dev/null
+++ b/leetcode/0009.Palindrome-Number/README.md
@@ -0,0 +1,96 @@
+# [9. Palindrome Number](https://leetcode.com/problems/palindrome-number/)
+
+
+## 题目
+
+Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
+
+**Example 1**:
+
+```
+Input: 121
+Output: true
+```
+
+**Example 2**:
+
+```
+Input: -121
+Output: false
+Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
+```
+
+**Example 3**:
+
+```
+Input: 10
+Output: false
+Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
+```
+
+**Follow up**:
+
+Coud you solve it without converting the integer to a string?
+
+## 题目大意
+
+判断一个整数是否是回文数。回文数是指正序（从左向右）和倒序（从右向左）读都是一样的整数。
+
+## 解题思路
+
+- 判断一个整数是不是回文数。
+- 简单题。注意会有负数的情况，负数，个位数，10 都不是回文数。其他的整数再按照回文的规则判断。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strconv"
+
+// 解法一
+func isPalindrome(x int) bool {
+	if x < 0 {
+		return false
+	}
+	if x == 0 {
+		return true
+	}
+	if x%10 == 0 {
+		return false
+	}
+	arr := make([]int, 0, 32)
+	for x > 0 {
+		arr = append(arr, x%10)
+		x = x / 10
+	}
+	sz := len(arr)
+	for i, j := 0, sz-1; i <= j; i, j = i+1, j-1 {
+		if arr[i] != arr[j] {
+			return false
+		}
+	}
+	return true
+}
+
+// 解法二 数字转字符串
+func isPalindrome1(x int) bool {
+	if x < 0 {
+		return false
+	}
+	if x < 10 {
+		return true
+	}
+	s := strconv.Itoa(x)
+	length := len(s)
+	for i := 0; i <= length/2; i++ {
+		if s[i] != s[length-1-i] {
+			return false
+		}
+	}
+	return true
+}
+
+
+```
\ No newline at end of file
diff --git a/Algorithms/0011. Container With Most Water/11. Container With Most Water.go b/leetcode/0011.Container-With-Most-Water/11. Container With Most Water.go
similarity index 100%
rename from Algorithms/0011. Container With Most Water/11. Container With Most Water.go
rename to leetcode/0011.Container-With-Most-Water/11. Container With Most Water.go
diff --git a/Algorithms/0011. Container With Most Water/11. Container With Most Water_test.go b/leetcode/0011.Container-With-Most-Water/11. Container With Most Water_test.go
similarity index 95%
rename from Algorithms/0011. Container With Most Water/11. Container With Most Water_test.go
rename to leetcode/0011.Container-With-Most-Water/11. Container With Most Water_test.go
index f48a305e5..b9185a28c 100644
--- a/Algorithms/0011. Container With Most Water/11. Container With Most Water_test.go	
+++ b/leetcode/0011.Container-With-Most-Water/11. Container With Most Water_test.go	
@@ -26,12 +26,12 @@ func Test_Problem11(t *testing.T) {
 
 	qs := []question11{
 
-		question11{
+		{
 			para11{[]int{1, 8, 6, 2, 5, 4, 8, 3, 7}},
 			ans11{49},
 		},
 
-		question11{
+		{
 			para11{[]int{1, 1}},
 			ans11{1},
 		},
diff --git a/Algorithms/0011. Container With Most Water/README.md b/leetcode/0011.Container-With-Most-Water/README.md
similarity index 100%
rename from Algorithms/0011. Container With Most Water/README.md
rename to leetcode/0011.Container-With-Most-Water/README.md
diff --git a/leetcode/0012.Integer-to-Roman/12. Integer to Roman.go b/leetcode/0012.Integer-to-Roman/12. Integer to Roman.go
new file mode 100644
index 000000000..525e8aef1
--- /dev/null
+++ b/leetcode/0012.Integer-to-Roman/12. Integer to Roman.go	
@@ -0,0 +1,15 @@
+package leetcode
+
+func intToRoman(num int) string {
+	values := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
+	symbols := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
+	res, i := "", 0
+	for num != 0 {
+		for values[i] > num {
+			i++
+		}
+		num -= values[i]
+		res += symbols[i]
+	}
+	return res
+}
diff --git a/leetcode/0012.Integer-to-Roman/12. Integer to Roman_test.go b/leetcode/0012.Integer-to-Roman/12. Integer to Roman_test.go
new file mode 100644
index 000000000..c5fff602a
--- /dev/null
+++ b/leetcode/0012.Integer-to-Roman/12. Integer to Roman_test.go	
@@ -0,0 +1,71 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question12 struct {
+	para12
+	ans12
+}
+
+// para 是参数
+// one 代表第一个参数
+type para12 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans12 struct {
+	one string
+}
+
+func Test_Problem12(t *testing.T) {
+
+	qs := []question12{
+
+		{
+			para12{3},
+			ans12{"III"},
+		},
+
+		{
+			para12{4},
+			ans12{"IV"},
+		},
+
+		{
+			para12{9},
+			ans12{"IX"},
+		},
+
+		{
+			para12{58},
+			ans12{"LVIII"},
+		},
+
+		{
+			para12{1994},
+			ans12{"MCMXCIV"},
+		},
+		{
+			para12{123},
+			ans12{"CXXIII"},
+		},
+
+		{
+			para12{120},
+			ans12{"CXX"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 12------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans12, q.para12
+		fmt.Printf("【input】:%v    【output】:%v\n", p.one, intToRoman(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0012.Integer-to-Roman/README.md b/leetcode/0012.Integer-to-Roman/README.md
new file mode 100644
index 000000000..5a2882d6a
--- /dev/null
+++ b/leetcode/0012.Integer-to-Roman/README.md
@@ -0,0 +1,102 @@
+# [12. Integer to Roman](https://leetcode.com/problems/integer-to-roman/)
+
+
+## 题目
+
+Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
+
+```
+Symbol       Value
+I             1
+V             5
+X             10
+L             50
+C             100
+D             500
+M             1000
+```
+
+For example, `2` is written as `II` in Roman numeral, just two one's added together. `12` is written as `XII`, which is simply `X + II`. The number `27` is written as `XXVII`, which is `XX + V + II`.
+
+Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
+
+- `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
+- `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
+- `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
+
+Given an integer, convert it to a roman numeral.
+
+**Example 1:**
+
+```
+Input: num = 3
+Output: "III"
+```
+
+**Example 2:**
+
+```
+Input: num = 4
+Output: "IV"
+```
+
+**Example 3:**
+
+```
+Input: num = 9
+Output: "IX"
+```
+
+**Example 4:**
+
+```
+Input: num = 58
+Output: "LVIII"
+Explanation: L = 50, V = 5, III = 3.
+```
+
+**Example 5:**
+
+```
+Input: num = 1994
+Output: "MCMXCIV"
+Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
+```
+
+**Constraints:**
+
+- `1 <= num <= 3999`
+
+## 题目大意
+
+通常情况下，罗马数字中小的数字在大的数字的右边。但也存在特例，例如 4 不写做 IIII，而是 IV。数字 1 在数字 5 的左边，所表示的数等于大数 5 减小数 1 得到的数值 4 。同样地，数字 9 表示为 IX。这个特殊的规则只适用于以下六种情况：
+
+- I 可以放在 V (5) 和 X (10) 的左边，来表示 4 和 9。
+- X 可以放在 L (50) 和 C (100) 的左边，来表示 40 和 90。
+- C 可以放在 D (500) 和 M (1000) 的左边，来表示 400 和 900。
+
+给定一个整数，将其转为罗马数字。输入确保在 1 到 3999 的范围内。
+
+## 解题思路
+
+- 依照题意，优先选择大的数字，解题思路采用贪心算法。将 1-3999 范围内的罗马数字从大到小放在数组中，从头选择到尾，即可把整数转成罗马数字。
+
+## 代码
+
+```go
+package leetcode
+
+func intToRoman(num int) string {
+	values := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
+	symbols := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
+	res, i := "", 0
+	for num != 0 {
+		for values[i] > num {
+			i++
+		}
+		num -= values[i]
+		res += symbols[i]
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0013.Roman-to-Integer/13. Roman to Integer.go b/leetcode/0013.Roman-to-Integer/13. Roman to Integer.go
new file mode 100644
index 000000000..51fd47828
--- /dev/null
+++ b/leetcode/0013.Roman-to-Integer/13. Roman to Integer.go	
@@ -0,0 +1,29 @@
+package leetcode
+
+var roman = map[string]int{
+	"I": 1,
+	"V": 5,
+	"X": 10,
+	"L": 50,
+	"C": 100,
+	"D": 500,
+	"M": 1000,
+}
+
+func romanToInt(s string) int {
+	if s == "" {
+		return 0
+	}
+	num, lastint, total := 0, 0, 0
+	for i := 0; i < len(s); i++ {
+		char := s[len(s)-(i+1) : len(s)-i]
+		num = roman[char]
+		if num < lastint {
+			total = total - num
+		} else {
+			total = total + num
+		}
+		lastint = num
+	}
+	return total
+}
diff --git a/leetcode/0013.Roman-to-Integer/13. Roman to Integer_test.go b/leetcode/0013.Roman-to-Integer/13. Roman to Integer_test.go
new file mode 100644
index 000000000..b611319e3
--- /dev/null
+++ b/leetcode/0013.Roman-to-Integer/13. Roman to Integer_test.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question13 struct {
+	para13
+	ans13
+}
+
+// para 是参数
+// one 代表第一个参数
+type para13 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans13 struct {
+	one int
+}
+
+func Test_Problem13(t *testing.T) {
+
+	qs := []question13{
+
+		{
+			para13{"III"},
+			ans13{3},
+		},
+
+		{
+			para13{"IV"},
+			ans13{4},
+		},
+
+		{
+			para13{"IX"},
+			ans13{9},
+		},
+
+		{
+			para13{"LVIII"},
+			ans13{58},
+		},
+
+		{
+			para13{"MCMXCIV"},
+			ans13{1994},
+		},
+
+		{
+			para13{"MCMXICIVI"},
+			ans13{2014},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 13------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans13, q.para13
+		fmt.Printf("【input】:%v    【output】:%v\n", p.one, romanToInt(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0013.Roman-to-Integer/README.md b/leetcode/0013.Roman-to-Integer/README.md
new file mode 100644
index 000000000..44d758ebb
--- /dev/null
+++ b/leetcode/0013.Roman-to-Integer/README.md
@@ -0,0 +1,132 @@
+# [13. Roman to Integer](https://leetcode.com/problems/roman-to-integer/)
+
+
+## 题目
+
+Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
+
+```
+Symbol       Value
+I             1
+V             5
+X             10
+L             50
+C             100
+D             500
+M             1000
+```
+
+For example, two is written as `II` in Roman numeral, just two one's added together. Twelve is written as, `XII`, which is simply `X` + `II`. The number twenty seven is written as `XXVII`, which is `XX` + `V` + `II`.
+
+Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
+
+- `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
+- `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
+- `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
+
+Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.
+
+**Example 1**:
+
+```
+Input: "III"
+Output: 3
+```
+
+**Example 2**:
+
+```
+Input: "IV"
+Output: 4
+```
+
+**Example 3**:
+
+```
+Input: "IX"
+Output: 9
+```
+
+**Example 4**:
+
+```
+Input: "LVIII"
+Output: 58
+Explanation: L = 50, V= 5, III = 3.
+```
+
+**Example 5**:
+
+```
+Input: "MCMXCIV"
+Output: 1994
+Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
+```
+
+## 题目大意
+
+罗马数字包含以下七种字符: I， V， X， L，C，D 和 M。
+
+```go
+
+字符          数值
+I             1
+V             5
+X             10
+L             50
+C             100
+D             500
+M             1000
+
+```
+
+例如， 罗马数字 2 写做 II ，即为两个并列的 1。12 写做 XII ，即为 X + II 。 27 写做  XXVII, 即为 XX + V + II 。
+
+通常情况下，罗马数字中小的数字在大的数字的右边。但也存在特例，例如 4 不写做 IIII，而是 IV。数字 1 在数字 5 的左边，所表示的数等于大数 5 减小数 1 得到的数值 4 。同样地，数字 9 表示为 IX。这个特殊的规则只适用于以下六种情况：
+
+- I 可以放在 V (5) 和 X (10) 的左边，来表示 4 和 9。
+- X 可以放在 L (50) 和 C (100) 的左边，来表示 40 和 90。 
+- C 可以放在 D (500) 和 M (1000) 的左边，来表示 400 和 900。
+
+给定一个罗马数字，将其转换成整数。输入确保在 1 到 3999 的范围内。
+
+## 解题思路
+
+- 给定一个罗马数字，将其转换成整数。输入确保在 1 到 3999 的范围内。
+- 简单题。按照题目中罗马数字的字符数值，计算出对应罗马数字的十进制数即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+var roman = map[string]int{
+	"I": 1,
+	"V": 5,
+	"X": 10,
+	"L": 50,
+	"C": 100,
+	"D": 500,
+	"M": 1000,
+}
+
+func romanToInt(s string) int {
+	if s == "" {
+		return 0
+	}
+	num, lastint, total := 0, 0, 0
+	for i := 0; i < len(s); i++ {
+		char := s[len(s)-(i+1) : len(s)-i]
+		num = roman[char]
+		if num < lastint {
+			total = total - num
+		} else {
+			total = total + num
+		}
+		lastint = num
+	}
+	return total
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/0014.Longest-Common-Prefix/14.Longest Common Prefix.go b/leetcode/0014.Longest-Common-Prefix/14.Longest Common Prefix.go
new file mode 100644
index 000000000..259c5c835
--- /dev/null
+++ b/leetcode/0014.Longest-Common-Prefix/14.Longest Common Prefix.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func longestCommonPrefix(strs []string) string {
+	prefix := strs[0]
+
+	for i := 1; i < len(strs); i++ {
+		for j := 0; j < len(prefix); j++ {
+			if len(strs[i]) <= j || strs[i][j] != prefix[j] {
+				prefix = prefix[0:j]
+				break
+			}
+		}
+	}
+
+	return prefix
+}
diff --git a/leetcode/0014.Longest-Common-Prefix/14.Longest Common Prefix_test.go b/leetcode/0014.Longest-Common-Prefix/14.Longest Common Prefix_test.go
new file mode 100644
index 000000000..e7635d52f
--- /dev/null
+++ b/leetcode/0014.Longest-Common-Prefix/14.Longest Common Prefix_test.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question14 struct {
+	para14
+	ans14
+}
+
+// para 是参数
+type para14 struct {
+	strs []string
+}
+
+// ans 是答案
+type ans14 struct {
+	ans string
+}
+
+func Test_Problem14(t *testing.T) {
+
+	qs := []question14{
+
+		{
+			para14{[]string{"flower", "flow", "flight"}},
+			ans14{"fl"},
+		},
+
+		{
+			para14{[]string{"dog", "racecar", "car"}},
+			ans14{""},
+		},
+
+		{
+			para14{[]string{"ab", "a"}},
+			ans14{"a"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 14------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans14, q.para14
+		fmt.Printf("【input】:%v    【output】:%v\n", p.strs, longestCommonPrefix(p.strs))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0014.Longest-Common-Prefix/README.md b/leetcode/0014.Longest-Common-Prefix/README.md
new file mode 100644
index 000000000..26b9e03bf
--- /dev/null
+++ b/leetcode/0014.Longest-Common-Prefix/README.md
@@ -0,0 +1,57 @@
+# [14. Longest Common Prefix](https://leetcode.com/problems/longest-common-prefix/)
+
+## 题目
+
+Write a function to find the longest common prefix string amongst an array of strings.
+
+If there is no common prefix, return an empty string "".
+
+**Example 1**:
+
+    Input: strs = ["flower","flow","flight"]
+    Output: "fl"
+
+**Example 2**:
+
+    Input: strs = ["dog","racecar","car"]
+    Output: ""
+    Explanation: There is no common prefix among the input strings.
+
+**Constraints:**
+
+- 1 <= strs.length <= 200
+- 0 <= strs[i].length <= 200
+- strs[i] consists of only lower-case English letters.
+
+## 题目大意
+
+编写一个函数来查找字符串数组中的最长公共前缀。
+
+如果不存在公共前缀，返回空字符串 ""。
+
+## 解题思路
+
+- 对 strs 按照字符串长度进行升序排序，求出 strs 中长度最小字符串的长度 minLen
+- 逐个比较长度最小字符串与其它字符串中的字符，如果不相等就返回 commonPrefix,否则就把该字符加入 commonPrefix
+
+## 代码
+
+```go
+
+package leetcode
+
+func longestCommonPrefix(strs []string) string {
+	prefix := strs[0]
+
+	for i := 1; i < len(strs); i++ {
+		for j := 0; j < len(prefix); j++ {
+			if len(strs[i]) <= j || strs[i][j] != prefix[j] {
+				prefix = prefix[0:j]
+				break
+			}
+		}
+	}
+
+	return prefix
+}
+```
diff --git a/leetcode/0015.3Sum/15. 3Sum.go b/leetcode/0015.3Sum/15. 3Sum.go
new file mode 100644
index 000000000..a8be09e8e
--- /dev/null
+++ b/leetcode/0015.3Sum/15. 3Sum.go	
@@ -0,0 +1,72 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+// 解法一 最优解，双指针 + 排序
+func threeSum(nums []int) [][]int {
+	sort.Ints(nums)
+	result, start, end, index, addNum, length := make([][]int, 0), 0, 0, 0, 0, len(nums)
+	for index = 1; index < length-1; index++ {
+		start, end = 0, length-1
+		if index > 1 && nums[index] == nums[index-1] {
+			start = index - 1
+		}
+		for start < index && end > index {
+			if start > 0 && nums[start] == nums[start-1] {
+				start++
+				continue
+			}
+			if end < length-1 && nums[end] == nums[end+1] {
+				end--
+				continue
+			}
+			addNum = nums[start] + nums[end] + nums[index]
+			if addNum == 0 {
+				result = append(result, []int{nums[start], nums[index], nums[end]})
+				start++
+				end--
+			} else if addNum > 0 {
+				end--
+			} else {
+				start++
+			}
+		}
+	}
+	return result
+}
+
+// 解法二
+func threeSum1(nums []int) [][]int {
+	res := [][]int{}
+	counter := map[int]int{}
+	for _, value := range nums {
+		counter[value]++
+	}
+
+	uniqNums := []int{}
+	for key := range counter {
+		uniqNums = append(uniqNums, key)
+	}
+	sort.Ints(uniqNums)
+
+	for i := 0; i < len(uniqNums); i++ {
+		if (uniqNums[i]*3 == 0) && counter[uniqNums[i]] >= 3 {
+			res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i]})
+		}
+		for j := i + 1; j < len(uniqNums); j++ {
+			if (uniqNums[i]*2+uniqNums[j] == 0) && counter[uniqNums[i]] > 1 {
+				res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j]})
+			}
+			if (uniqNums[j]*2+uniqNums[i] == 0) && counter[uniqNums[j]] > 1 {
+				res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j]})
+			}
+			c := 0 - uniqNums[i] - uniqNums[j]
+			if c > uniqNums[j] && counter[c] > 0 {
+				res = append(res, []int{uniqNums[i], uniqNums[j], c})
+			}
+		}
+	}
+	return res
+}
diff --git a/leetcode/0015.3Sum/15. 3Sum_test.go b/leetcode/0015.3Sum/15. 3Sum_test.go
new file mode 100644
index 000000000..e48b7d146
--- /dev/null
+++ b/leetcode/0015.3Sum/15. 3Sum_test.go	
@@ -0,0 +1,59 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question15 struct {
+	para15
+	ans15
+}
+
+// para 是参数
+// one 代表第一个参数
+type para15 struct {
+	a []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans15 struct {
+	one [][]int
+}
+
+func Test_Problem15(t *testing.T) {
+
+	qs := []question15{
+
+		{
+			para15{[]int{0, 0, 0}},
+			ans15{[][]int{{0, 0, 0}}},
+		},
+
+		{
+			para15{[]int{-1, 0, 1, 2, -1, -4}},
+			ans15{[][]int{{-1, 0, 1}, {-1, -1, 2}}},
+		},
+
+		{
+			para15{[]int{-4, -2, -2, -2, 0, 1, 2, 2, 2, 3, 3, 4, 4, 6, 6}},
+			ans15{[][]int{{-4, -2, 6}, {-4, 0, 4}, {-4, 1, 3}, {-4, 2, 2}, {-2, -2, 4}, {-2, 0, 2}}},
+		},
+
+		{
+			para15{[]int{5, -7, 3, -3, 5, -10, 4, 8, -3, -8, -3, -3, -1, -8, 6, 4, -4, 7, 2, -5, -2, -7, -3, 7, 2, 4, -6, 5}},
+			ans15{[][]int{{-10, 2, 8}, {-10, 3, 7}, {-10, 4, 6}, {-10, 5, 5}, {-8, 2, 6}, {-8, 3, 5}, {-8, 4, 4}, {-7, -1, 8},
+				{-7, 2, 5}, {-7, 3, 4}, {-6, -2, 8}, {-6, -1, 7}, {-6, 2, 4}, {-5, -3, 8}, {-5, -2, 7}, {-5, -1, 6}, {-5, 2, 3},
+				{-4, -3, 7}, {-4, -2, 6}, {-4, -1, 5}, {-4, 2, 2}, {-3, -3, 6}, {-3, -2, 5}, {-3, -1, 4}, {-2, -1, 3}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 15------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans15, q.para15
+		fmt.Printf("【input】:%v       【output】:%v\n", p, threeSum(p.a))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0015. 3Sum/README.md b/leetcode/0015.3Sum/README.md
similarity index 100%
rename from Algorithms/0015. 3Sum/README.md
rename to leetcode/0015.3Sum/README.md
diff --git a/Algorithms/0016. 3Sum Closest/16. 3Sum Closest.go b/leetcode/0016.3Sum-Closest/16. 3Sum Closest.go
similarity index 94%
rename from Algorithms/0016. 3Sum Closest/16. 3Sum Closest.go
rename to leetcode/0016.3Sum-Closest/16. 3Sum Closest.go
index 0beaac707..ed4465d17 100644
--- a/Algorithms/0016. 3Sum Closest/16. 3Sum Closest.go	
+++ b/leetcode/0016.3Sum-Closest/16. 3Sum Closest.go	
@@ -11,6 +11,9 @@ func threeSumClosest(nums []int, target int) int {
 	if n > 2 {
 		sort.Ints(nums)
 		for i := 0; i < n-2; i++ {
+			if i > 0 && nums[i] == nums[i-1] {
+				continue
+			}
 			for j, k := i+1, n-1; j < k; {
 				sum := nums[i] + nums[j] + nums[k]
 				if abs(sum-target) < diff {
diff --git a/Algorithms/0016. 3Sum Closest/16. 3Sum Closest_test.go b/leetcode/0016.3Sum-Closest/16. 3Sum Closest_test.go
similarity index 92%
rename from Algorithms/0016. 3Sum Closest/16. 3Sum Closest_test.go
rename to leetcode/0016.3Sum-Closest/16. 3Sum Closest_test.go
index 0db2570e8..0fd4d392b 100644
--- a/Algorithms/0016. 3Sum Closest/16. 3Sum Closest_test.go	
+++ b/leetcode/0016.3Sum-Closest/16. 3Sum Closest_test.go	
@@ -27,27 +27,27 @@ func Test_Problem16(t *testing.T) {
 
 	qs := []question16{
 
-		question16{
+		{
 			para16{[]int{-1, 0, 1, 1, 55}, 3},
 			ans16{2},
 		},
 
-		question16{
+		{
 			para16{[]int{0, 0, 0}, 1},
 			ans16{0},
 		},
 
-		question16{
+		{
 			para16{[]int{-1, 2, 1, -4}, 1},
 			ans16{2},
 		},
 
-		question16{
+		{
 			para16{[]int{1, 1, -1}, 0},
 			ans16{1},
 		},
 
-		question16{
+		{
 			para16{[]int{-1, 2, 1, -4}, 1},
 			ans16{2},
 		},
diff --git a/Algorithms/0016. 3Sum Closest/README.md b/leetcode/0016.3Sum-Closest/README.md
similarity index 100%
rename from Algorithms/0016. 3Sum Closest/README.md
rename to leetcode/0016.3Sum-Closest/README.md
diff --git a/leetcode/0017.Letter-Combinations-of-a-Phone-Number/17. Letter Combinations of a Phone Number.go b/leetcode/0017.Letter-Combinations-of-a-Phone-Number/17. Letter Combinations of a Phone Number.go
new file mode 100644
index 000000000..4efc7aaf7
--- /dev/null
+++ b/leetcode/0017.Letter-Combinations-of-a-Phone-Number/17. Letter Combinations of a Phone Number.go	
@@ -0,0 +1,104 @@
+package leetcode
+
+var (
+	letterMap = []string{
+		" ",    //0
+		"",     //1
+		"abc",  //2
+		"def",  //3
+		"ghi",  //4
+		"jkl",  //5
+		"mno",  //6
+		"pqrs", //7
+		"tuv",  //8
+		"wxyz", //9
+	}
+	res   = []string{}
+	final = 0
+)
+
+// 解法一 DFS
+func letterCombinations(digits string) []string {
+	if digits == "" {
+		return []string{}
+	}
+	res = []string{}
+	findCombination(&digits, 0, "")
+	return res
+}
+
+func findCombination(digits *string, index int, s string) {
+	if index == len(*digits) {
+		res = append(res, s)
+		return
+	}
+	num := (*digits)[index]
+	letter := letterMap[num-'0']
+	for i := 0; i < len(letter); i++ {
+		findCombination(digits, index+1, s+string(letter[i]))
+	}
+	return
+}
+
+// 解法二 非递归
+func letterCombinations_(digits string) []string {
+	if digits == "" {
+		return []string{}
+	}
+	index := digits[0] - '0'
+	letter := letterMap[index]
+	tmp := []string{}
+	for i := 0; i < len(letter); i++ {
+		if len(res) == 0 {
+			res = append(res, "")
+		}
+		for j := 0; j < len(res); j++ {
+			tmp = append(tmp, res[j]+string(letter[i]))
+		}
+	}
+	res = tmp
+	final++
+	letterCombinations(digits[1:])
+	final--
+	if final == 0 {
+		tmp = res
+		res = []string{}
+	}
+	return tmp
+}
+
+// 解法三 回溯（参考回溯模板，类似DFS）
+var result []string
+var dict = map[string][]string{
+	"2": {"a", "b", "c"},
+	"3": {"d", "e", "f"},
+	"4": {"g", "h", "i"},
+	"5": {"j", "k", "l"},
+	"6": {"m", "n", "o"},
+	"7": {"p", "q", "r", "s"},
+	"8": {"t", "u", "v"},
+	"9": {"w", "x", "y", "z"},
+}
+
+func letterCombinationsBT(digits string) []string {
+	result = []string{}
+	if digits == "" {
+		return result
+	}
+	letterFunc("", digits)
+	return result
+}
+
+func letterFunc(res string, digits string) {
+	if digits == "" {
+		result = append(result, res)
+		return
+	}
+	k := digits[0:1]
+	digits = digits[1:]
+	for i := 0; i < len(dict[k]); i++ {
+		res += dict[k][i]
+		letterFunc(res, digits)
+		res = res[0 : len(res)-1]
+	}
+}
diff --git a/Algorithms/0017. Letter Combinations of a Phone Number/17. Letter Combinations of a Phone Number_test.go b/leetcode/0017.Letter-Combinations-of-a-Phone-Number/17. Letter Combinations of a Phone Number_test.go
similarity index 97%
rename from Algorithms/0017. Letter Combinations of a Phone Number/17. Letter Combinations of a Phone Number_test.go
rename to leetcode/0017.Letter-Combinations-of-a-Phone-Number/17. Letter Combinations of a Phone Number_test.go
index fffeeb44c..f384ff467 100644
--- a/Algorithms/0017. Letter Combinations of a Phone Number/17. Letter Combinations of a Phone Number_test.go	
+++ b/leetcode/0017.Letter-Combinations-of-a-Phone-Number/17. Letter Combinations of a Phone Number_test.go	
@@ -26,7 +26,7 @@ func Test_Problem17(t *testing.T) {
 
 	qs := []question17{
 
-		question17{
+		{
 			para17{"23"},
 			ans17{[]string{"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"}},
 		},
diff --git a/leetcode/0017.Letter-Combinations-of-a-Phone-Number/README.md b/leetcode/0017.Letter-Combinations-of-a-Phone-Number/README.md
new file mode 100755
index 000000000..46f759a83
--- /dev/null
+++ b/leetcode/0017.Letter-Combinations-of-a-Phone-Number/README.md
@@ -0,0 +1,30 @@
+# [17. Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number/)
+
+
+## 题目
+
+Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent.
+
+A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
+
+![](http://upload.wikimedia.org/wikipedia/commons/thumb/7/73/Telephone-keypad2.svg/200px-Telephone-keypad2.svg.png)
+
+**Example:**
+
+```c
+Input: "23"
+Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
+```
+
+**Note:**
+
+Although the above answer is in lexicographical order, your answer could be in any order you want.
+
+## 题目大意
+
+给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。
+
+
+## 解题思路
+
+- DFS 递归深搜即可
\ No newline at end of file
diff --git a/leetcode/0018.4Sum/18. 4Sum.go b/leetcode/0018.4Sum/18. 4Sum.go
new file mode 100644
index 000000000..be44e5c7f
--- /dev/null
+++ b/leetcode/0018.4Sum/18. 4Sum.go	
@@ -0,0 +1,135 @@
+package leetcode
+
+import "sort"
+
+// 解法一 双指针
+func fourSum(nums []int, target int) (quadruplets [][]int) {
+	sort.Ints(nums)
+	n := len(nums)
+	for i := 0; i < n-3 && nums[i]+nums[i+1]+nums[i+2]+nums[i+3] <= target; i++ {
+		if i > 0 && nums[i] == nums[i-1] || nums[i]+nums[n-3]+nums[n-2]+nums[n-1] < target {
+			continue
+		}
+		for j := i + 1; j < n-2 && nums[i]+nums[j]+nums[j+1]+nums[j+2] <= target; j++ {
+			if j > i+1 && nums[j] == nums[j-1] || nums[i]+nums[j]+nums[n-2]+nums[n-1] < target {
+				continue
+			}
+			for left, right := j+1, n-1; left < right; {
+				if sum := nums[i] + nums[j] + nums[left] + nums[right]; sum == target {
+					quadruplets = append(quadruplets, []int{nums[i], nums[j], nums[left], nums[right]})
+					for left++; left < right && nums[left] == nums[left-1]; left++ {
+					}
+					for right--; left < right && nums[right] == nums[right+1]; right-- {
+					}
+				} else if sum < target {
+					left++
+				} else {
+					right--
+				}
+			}
+		}
+	}
+	return
+}
+
+// 解法二 kSum
+func fourSum1(nums []int, target int) [][]int {
+	res, cur := make([][]int, 0), make([]int, 0)
+	sort.Ints(nums)
+	kSum(nums, 0, len(nums)-1, target, 4, cur, &res)
+	return res
+}
+
+func kSum(nums []int, left, right int, target int, k int, cur []int, res *[][]int) {
+	if right-left+1 < k || k < 2 || target < nums[left]*k || target > nums[right]*k {
+		return
+	}
+	if k == 2 {
+		// 2 sum
+		twoSum(nums, left, right, target, cur, res)
+	} else {
+		for i := left; i < len(nums); i++ {
+			if i == left || (i > left && nums[i-1] != nums[i]) {
+				next := make([]int, len(cur))
+				copy(next, cur)
+				next = append(next, nums[i])
+				kSum(nums, i+1, len(nums)-1, target-nums[i], k-1, next, res)
+			}
+		}
+	}
+
+}
+
+func twoSum(nums []int, left, right int, target int, cur []int, res *[][]int) {
+	for left < right {
+		sum := nums[left] + nums[right]
+		if sum == target {
+			cur = append(cur, nums[left], nums[right])
+			temp := make([]int, len(cur))
+			copy(temp, cur)
+			*res = append(*res, temp)
+			// reset cur to previous state
+			cur = cur[:len(cur)-2]
+			left++
+			right--
+			for left < right && nums[left] == nums[left-1] {
+				left++
+			}
+			for left < right && nums[right] == nums[right+1] {
+				right--
+			}
+		} else if sum < target {
+			left++
+		} else {
+			right--
+		}
+	}
+}
+
+// 解法三
+func fourSum2(nums []int, target int) [][]int {
+	res := [][]int{}
+	counter := map[int]int{}
+	for _, value := range nums {
+		counter[value]++
+	}
+
+	uniqNums := []int{}
+	for key := range counter {
+		uniqNums = append(uniqNums, key)
+	}
+	sort.Ints(uniqNums)
+
+	for i := 0; i < len(uniqNums); i++ {
+		if (uniqNums[i]*4 == target) && counter[uniqNums[i]] >= 4 {
+			res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i], uniqNums[i]})
+		}
+		for j := i + 1; j < len(uniqNums); j++ {
+			if (uniqNums[i]*3+uniqNums[j] == target) && counter[uniqNums[i]] > 2 {
+				res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i], uniqNums[j]})
+			}
+			if (uniqNums[j]*3+uniqNums[i] == target) && counter[uniqNums[j]] > 2 {
+				res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j], uniqNums[j]})
+			}
+			if (uniqNums[j]*2+uniqNums[i]*2 == target) && counter[uniqNums[j]] > 1 && counter[uniqNums[i]] > 1 {
+				res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j], uniqNums[j]})
+			}
+			for k := j + 1; k < len(uniqNums); k++ {
+				if (uniqNums[i]*2+uniqNums[j]+uniqNums[k] == target) && counter[uniqNums[i]] > 1 {
+					res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j], uniqNums[k]})
+				}
+				if (uniqNums[j]*2+uniqNums[i]+uniqNums[k] == target) && counter[uniqNums[j]] > 1 {
+					res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j], uniqNums[k]})
+				}
+				if (uniqNums[k]*2+uniqNums[i]+uniqNums[j] == target) && counter[uniqNums[k]] > 1 {
+					res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[k], uniqNums[k]})
+				}
+				c := target - uniqNums[i] - uniqNums[j] - uniqNums[k]
+				if c > uniqNums[k] && counter[c] > 0 {
+					res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[k], c})
+				}
+			}
+		}
+	}
+	return res
+}
diff --git a/leetcode/0018.4Sum/18. 4Sum_test.go b/leetcode/0018.4Sum/18. 4Sum_test.go
new file mode 100644
index 000000000..bc08f2fbf
--- /dev/null
+++ b/leetcode/0018.4Sum/18. 4Sum_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question18 struct {
+	para18
+	ans18
+}
+
+// para 是参数
+// one 代表第一个参数
+type para18 struct {
+	a []int
+	t int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans18 struct {
+	one [][]int
+}
+
+func Test_Problem18(t *testing.T) {
+
+	qs := []question18{
+
+		{
+			para18{[]int{1, 1, 1, 1}, 4},
+			ans18{[][]int{{1, 1, 1, 1}}},
+		},
+
+		{
+			para18{[]int{0, 1, 5, 0, 1, 5, 5, -4}, 11},
+			ans18{[][]int{{-4, 5, 5, 5}, {0, 1, 5, 5}}},
+		},
+
+		{
+			para18{[]int{1, 0, -1, 0, -2, 2}, 0},
+			ans18{[][]int{{-1, 0, 0, 1}, {-2, -1, 1, 2}, {-2, 0, 0, 2}}},
+		},
+
+		{
+			para18{[]int{1, 0, -1, 0, -2, 2, 0, 0, 0, 0}, 0},
+			ans18{[][]int{{-1, 0, 0, 1}, {-2, -1, 1, 2}, {-2, 0, 0, 2}, {0, 0, 0, 0}}},
+		},
+
+		{
+			para18{[]int{1, 0, -1, 0, -2, 2, 0, 0, 0, 0}, 1},
+			ans18{[][]int{{-1, 0, 0, 2}, {-2, 0, 1, 2}, {0, 0, 0, 1}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 18------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans18, q.para18
+		fmt.Printf("【input】:%v       【output】:%v\n", p, fourSum(p.a, p.t))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0018. 4Sum/README.md b/leetcode/0018.4Sum/README.md
similarity index 100%
rename from Algorithms/0018. 4Sum/README.md
rename to leetcode/0018.4Sum/README.md
diff --git a/leetcode/0019.Remove-Nth-Node-From-End-of-List/19. Remove Nth Node From End of List.go b/leetcode/0019.Remove-Nth-Node-From-End-of-List/19. Remove Nth Node From End of List.go
new file mode 100644
index 000000000..7c2855bbb
--- /dev/null
+++ b/leetcode/0019.Remove-Nth-Node-From-End-of-List/19. Remove Nth Node From End of List.go	
@@ -0,0 +1,70 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// 解法一
+func removeNthFromEnd(head *ListNode, n int) *ListNode {
+	dummyHead := &ListNode{Next: head}
+	preSlow, slow, fast := dummyHead, head, head
+	for fast != nil {
+		if n <= 0 {
+			preSlow = slow
+			slow = slow.Next
+		}
+		n--
+		fast = fast.Next
+	}
+	preSlow.Next = slow.Next
+	return dummyHead.Next
+}
+
+// 解法二
+func removeNthFromEnd1(head *ListNode, n int) *ListNode {
+	if head == nil {
+		return nil
+	}
+	if n <= 0 {
+		return head
+	}
+	current := head
+	len := 0
+	for current != nil {
+		len++
+		current = current.Next
+	}
+	if n > len {
+		return head
+	}
+	if n == len {
+		current := head
+		head = head.Next
+		current.Next = nil
+		return head
+	}
+	current = head
+	i := 0
+	for current != nil {
+		if i == len-n-1 {
+			deleteNode := current.Next
+			current.Next = current.Next.Next
+			deleteNode.Next = nil
+			break
+		}
+		i++
+		current = current.Next
+	}
+	return head
+}
diff --git a/leetcode/0019.Remove-Nth-Node-From-End-of-List/19. Remove Nth Node From End of List_test.go b/leetcode/0019.Remove-Nth-Node-From-End-of-List/19. Remove Nth Node From End of List_test.go
new file mode 100644
index 000000000..f112b55d3
--- /dev/null
+++ b/leetcode/0019.Remove-Nth-Node-From-End-of-List/19. Remove Nth Node From End of List_test.go	
@@ -0,0 +1,84 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question19 struct {
+	para19
+	ans19
+}
+
+// para 是参数
+// one 代表第一个参数
+type para19 struct {
+	one []int
+	n   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans19 struct {
+	one []int
+}
+
+func Test_Problem19(t *testing.T) {
+
+	qs := []question19{
+
+		{
+			para19{[]int{1}, 3},
+			ans19{[]int{1}},
+		},
+
+		{
+			para19{[]int{1, 2}, 2},
+			ans19{[]int{2}},
+		},
+
+		{
+			para19{[]int{1}, 1},
+			ans19{[]int{}},
+		},
+
+		{
+			para19{[]int{1, 2, 3, 4, 5}, 10},
+			ans19{[]int{1, 2, 3, 4, 5}},
+		},
+
+		{
+			para19{[]int{1, 2, 3, 4, 5}, 1},
+			ans19{[]int{1, 2, 3, 4}},
+		},
+
+		{
+			para19{[]int{1, 2, 3, 4, 5}, 2},
+			ans19{[]int{1, 2, 3, 5}},
+		},
+
+		{
+			para19{[]int{1, 2, 3, 4, 5}, 3},
+			ans19{[]int{1, 2, 4, 5}},
+		},
+		{
+			para19{[]int{1, 2, 3, 4, 5}, 4},
+			ans19{[]int{1, 3, 4, 5}},
+		},
+
+		{
+			para19{[]int{1, 2, 3, 4, 5}, 5},
+			ans19{[]int{2, 3, 4, 5}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 19------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans19, q.para19
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(removeNthFromEnd(structures.Ints2List(p.one), p.n)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0019.Remove-Nth-Node-From-End-of-List/README.md b/leetcode/0019.Remove-Nth-Node-From-End-of-List/README.md
new file mode 100644
index 000000000..090e74efd
--- /dev/null
+++ b/leetcode/0019.Remove-Nth-Node-From-End-of-List/README.md
@@ -0,0 +1,51 @@
+# [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)
+
+## 题目
+
+Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
+
+**Follow up:** Could you do this in one pass?
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.jpg)
+
+```
+Input: head = [1,2,3,4,5], n = 2
+Output: [1,2,3,5]
+
+```
+
+**Example 2:**
+
+```
+Input: head = [1], n = 1
+Output: []
+
+```
+
+**Example 3:**
+
+```
+Input: head = [1,2], n = 1
+Output: [1]
+
+```
+
+**Constraints:**
+
+- The number of nodes in the list is `sz`.
+- `1 <= sz <= 30`
+- `0 <= Node.val <= 100`
+- `1 <= n <= sz`
+
+
+## 题目大意
+
+删除链表中倒数第 n 个结点。
+
+## 解题思路
+
+这道题比较简单，先循环一次拿到链表的总长度，然后循环到要删除的结点的前一个结点开始删除操作。需要注意的一个特例是，有可能要删除头结点，要单独处理。
+
+这道题有一种特别简单的解法。设置 2 个指针，一个指针距离前一个指针 n 个距离。同时移动 2 个指针，2 个指针都移动相同的距离。当一个指针移动到了终点，那么前一个指针就是倒数第 n 个节点了。
diff --git a/Algorithms/0020. Valid-Parentheses/20. Valid Parentheses.go b/leetcode/0020.Valid-Parentheses/20. Valid Parentheses.go
similarity index 100%
rename from Algorithms/0020. Valid-Parentheses/20. Valid Parentheses.go
rename to leetcode/0020.Valid-Parentheses/20. Valid Parentheses.go
diff --git a/Algorithms/0020. Valid-Parentheses/20. Valid Parentheses_test.go b/leetcode/0020.Valid-Parentheses/20. Valid Parentheses_test.go
similarity index 88%
rename from Algorithms/0020. Valid-Parentheses/20. Valid Parentheses_test.go
rename to leetcode/0020.Valid-Parentheses/20. Valid Parentheses_test.go
index 76ce31467..482e54cb4 100644
--- a/Algorithms/0020. Valid-Parentheses/20. Valid Parentheses_test.go	
+++ b/leetcode/0020.Valid-Parentheses/20. Valid Parentheses_test.go	
@@ -26,35 +26,35 @@ func Test_Problem20(t *testing.T) {
 
 	qs := []question20{
 
-		question20{
+		{
 			para20{"()[]{}"},
 			ans20{true},
 		},
-		question20{
+		{
 			para20{"(]"},
 			ans20{false},
 		},
-		question20{
+		{
 			para20{"({[]})"},
 			ans20{true},
 		},
-		question20{
+		{
 			para20{"(){[({[]})]}"},
 			ans20{true},
 		},
-		question20{
+		{
 			para20{"((([[[{{{"},
 			ans20{false},
 		},
-		question20{
+		{
 			para20{"(())]]"},
 			ans20{false},
 		},
-		question20{
+		{
 			para20{""},
 			ans20{true},
 		},
-		question20{
+		{
 			para20{"["},
 			ans20{false},
 		},
diff --git a/Algorithms/0020. Valid-Parentheses/README.md b/leetcode/0020.Valid-Parentheses/README.md
similarity index 100%
rename from Algorithms/0020. Valid-Parentheses/README.md
rename to leetcode/0020.Valid-Parentheses/README.md
diff --git a/Algorithms/0021. Merge Two Sorted Lists/21. Merge Two Sorted Lists.go b/leetcode/0021.Merge-Two-Sorted-Lists/21. Merge Two Sorted Lists.go
similarity index 76%
rename from Algorithms/0021. Merge Two Sorted Lists/21. Merge Two Sorted Lists.go
rename to leetcode/0021.Merge-Two-Sorted-Lists/21. Merge Two Sorted Lists.go
index e3a71d9ce..53531b2b4 100644
--- a/Algorithms/0021. Merge Two Sorted Lists/21. Merge Two Sorted Lists.go	
+++ b/leetcode/0021.Merge-Two-Sorted-Lists/21. Merge Two Sorted Lists.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0021. Merge Two Sorted Lists/21. Merge Two Sorted Lists_test.go b/leetcode/0021.Merge-Two-Sorted-Lists/21. Merge Two Sorted Lists_test.go
similarity index 83%
rename from Algorithms/0021. Merge Two Sorted Lists/21. Merge Two Sorted Lists_test.go
rename to leetcode/0021.Merge-Two-Sorted-Lists/21. Merge Two Sorted Lists_test.go
index 499b1b5b5..f79777c0d 100644
--- a/Algorithms/0021. Merge Two Sorted Lists/21. Merge Two Sorted Lists_test.go	
+++ b/leetcode/0021.Merge-Two-Sorted-Lists/21. Merge Two Sorted Lists_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question21 struct {
@@ -27,42 +29,42 @@ func Test_Problem21(t *testing.T) {
 
 	qs := []question21{
 
-		question21{
+		{
 			para21{[]int{}, []int{}},
 			ans21{[]int{}},
 		},
 
-		question21{
+		{
 			para21{[]int{1}, []int{1}},
 			ans21{[]int{1, 1}},
 		},
 
-		question21{
+		{
 			para21{[]int{1, 2, 3, 4}, []int{1, 2, 3, 4}},
 			ans21{[]int{1, 1, 2, 2, 3, 3, 4, 4}},
 		},
 
-		question21{
+		{
 			para21{[]int{1, 2, 3, 4, 5}, []int{1, 2, 3, 4, 5}},
 			ans21{[]int{1, 1, 2, 2, 3, 3, 4, 4, 5, 5}},
 		},
 
-		question21{
+		{
 			para21{[]int{1}, []int{9, 9, 9, 9, 9}},
 			ans21{[]int{1, 9, 9, 9, 9, 9}},
 		},
 
-		question21{
+		{
 			para21{[]int{9, 9, 9, 9, 9}, []int{1}},
 			ans21{[]int{1, 9, 9, 9, 9, 9}},
 		},
 
-		question21{
+		{
 			para21{[]int{2, 3, 4}, []int{4, 5, 6}},
 			ans21{[]int{2, 3, 4, 4, 5, 6}},
 		},
 
-		question21{
+		{
 			para21{[]int{1, 3, 8}, []int{1, 7}},
 			ans21{[]int{1, 1, 3, 7, 8}},
 		},
@@ -73,7 +75,7 @@ func Test_Problem21(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans21, q.para21
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(mergeTwoLists(S2l(p.one), S2l(p.another))))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(mergeTwoLists(structures.Ints2List(p.one), structures.Ints2List(p.another))))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0021. Merge Two Sorted Lists/README.md b/leetcode/0021.Merge-Two-Sorted-Lists/README.md
similarity index 100%
rename from Algorithms/0021. Merge Two Sorted Lists/README.md
rename to leetcode/0021.Merge-Two-Sorted-Lists/README.md
diff --git a/Algorithms/0022. Generate Parentheses/22. Generate Parentheses.go b/leetcode/0022.Generate-Parentheses/22. Generate Parentheses.go
similarity index 100%
rename from Algorithms/0022. Generate Parentheses/22. Generate Parentheses.go
rename to leetcode/0022.Generate-Parentheses/22. Generate Parentheses.go
diff --git a/Algorithms/0022. Generate Parentheses/22. Generate Parentheses_test.go b/leetcode/0022.Generate-Parentheses/22. Generate Parentheses_test.go
similarity index 98%
rename from Algorithms/0022. Generate Parentheses/22. Generate Parentheses_test.go
rename to leetcode/0022.Generate-Parentheses/22. Generate Parentheses_test.go
index 5e21a3dac..7f41625e9 100644
--- a/Algorithms/0022. Generate Parentheses/22. Generate Parentheses_test.go	
+++ b/leetcode/0022.Generate-Parentheses/22. Generate Parentheses_test.go	
@@ -26,7 +26,7 @@ func Test_Problem22(t *testing.T) {
 
 	qs := []question22{
 
-		question22{
+		{
 			para22{3},
 			ans22{[]string{
 				"((()))",
diff --git a/Algorithms/0022. Generate Parentheses/README.md b/leetcode/0022.Generate-Parentheses/README.md
similarity index 100%
rename from Algorithms/0022. Generate Parentheses/README.md
rename to leetcode/0022.Generate-Parentheses/README.md
diff --git a/Algorithms/0023. Merge k Sorted Lists/23. Merge k Sorted Lists.go b/leetcode/0023.Merge-k-Sorted-Lists/23. Merge k Sorted Lists.go
similarity index 85%
rename from Algorithms/0023. Merge k Sorted Lists/23. Merge k Sorted Lists.go
rename to leetcode/0023.Merge-k-Sorted-Lists/23. Merge k Sorted Lists.go
index 5ae46b400..c25317b2f 100644
--- a/Algorithms/0023. Merge k Sorted Lists/23. Merge k Sorted Lists.go	
+++ b/leetcode/0023.Merge-k-Sorted-Lists/23. Merge k Sorted Lists.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/leetcode/0023.Merge-k-Sorted-Lists/23. Merge k Sorted Lists_test.go b/leetcode/0023.Merge-k-Sorted-Lists/23. Merge k Sorted Lists_test.go
new file mode 100644
index 000000000..309899032
--- /dev/null
+++ b/leetcode/0023.Merge-k-Sorted-Lists/23. Merge k Sorted Lists_test.go	
@@ -0,0 +1,104 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question23 struct {
+	para23
+	ans23
+}
+
+// para 是参数
+// one 代表第一个参数
+type para23 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans23 struct {
+	one []int
+}
+
+func Test_Problem23(t *testing.T) {
+
+	qs := []question23{
+
+		{
+			para23{[][]int{}},
+			ans23{[]int{}},
+		},
+
+		{
+			para23{[][]int{
+				{1},
+				{1},
+			}},
+			ans23{[]int{1, 1}},
+		},
+
+		{
+			para23{[][]int{
+				{1, 2, 3, 4},
+				{1, 2, 3, 4},
+			}},
+			ans23{[]int{1, 1, 2, 2, 3, 3, 4, 4}},
+		},
+
+		{
+			para23{[][]int{
+				{1, 2, 3, 4, 5},
+				{1, 2, 3, 4, 5},
+			}},
+			ans23{[]int{1, 1, 2, 2, 3, 3, 4, 4, 5, 5}},
+		},
+
+		{
+			para23{[][]int{
+				{1},
+				{9, 9, 9, 9, 9},
+			}},
+			ans23{[]int{1, 9, 9, 9, 9, 9}},
+		},
+
+		{
+			para23{[][]int{
+				{9, 9, 9, 9, 9},
+				{1},
+			}},
+			ans23{[]int{1, 9, 9, 9, 9, 9}},
+		},
+
+		{
+			para23{[][]int{
+				{2, 3, 4},
+				{4, 5, 6},
+			}},
+			ans23{[]int{2, 3, 4, 4, 5, 6}},
+		},
+
+		{
+			para23{[][]int{
+				{1, 3, 8},
+				{1, 7},
+			}},
+			ans23{[]int{1, 1, 3, 7, 8}},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 23------------------------\n")
+
+	for _, q := range qs {
+		var ls []*ListNode
+		for _, qq := range q.para23.one {
+			ls = append(ls, structures.Ints2List(qq))
+		}
+		fmt.Printf("【input】:%v       【output】:%v\n", q.para23.one, structures.List2Ints(mergeKLists(ls)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0023. Merge k Sorted Lists/README.md b/leetcode/0023.Merge-k-Sorted-Lists/README.md
similarity index 100%
rename from Algorithms/0023. Merge k Sorted Lists/README.md
rename to leetcode/0023.Merge-k-Sorted-Lists/README.md
diff --git a/leetcode/0024.Swap-Nodes-in-Pairs/24. Swap Nodes in Pairs.go b/leetcode/0024.Swap-Nodes-in-Pairs/24. Swap Nodes in Pairs.go
new file mode 100644
index 000000000..ac1e7e09f
--- /dev/null
+++ b/leetcode/0024.Swap-Nodes-in-Pairs/24. Swap Nodes in Pairs.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func swapPairs(head *ListNode) *ListNode {
+	dummy := &ListNode{Next: head}
+	for pt := dummy; pt != nil && pt.Next != nil && pt.Next.Next != nil; {
+		pt, pt.Next, pt.Next.Next, pt.Next.Next.Next = pt.Next, pt.Next.Next, pt.Next.Next.Next, pt.Next
+	}
+	return dummy.Next
+}
diff --git a/leetcode/0024.Swap-Nodes-in-Pairs/24. Swap Nodes in Pairs_test.go b/leetcode/0024.Swap-Nodes-in-Pairs/24. Swap Nodes in Pairs_test.go
new file mode 100644
index 000000000..9997800f5
--- /dev/null
+++ b/leetcode/0024.Swap-Nodes-in-Pairs/24. Swap Nodes in Pairs_test.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question24 struct {
+	para24
+	ans24
+}
+
+// para 是参数
+// one 代表第一个参数
+type para24 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans24 struct {
+	one []int
+}
+
+func Test_Problem24(t *testing.T) {
+
+	qs := []question24{
+
+		{
+			para24{[]int{}},
+			ans24{[]int{}},
+		},
+
+		{
+			para24{[]int{1}},
+			ans24{[]int{1}},
+		},
+
+		{
+			para24{[]int{1, 2, 3, 4}},
+			ans24{[]int{2, 1, 4, 3}},
+		},
+
+		{
+			para24{[]int{1, 2, 3, 4, 5}},
+			ans24{[]int{2, 1, 4, 3, 5}},
+		},
+
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 24------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans24, q.para24
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(swapPairs(structures.Ints2List(p.one))))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0024. Swap Nodes in Pairs/README.md b/leetcode/0024.Swap-Nodes-in-Pairs/README.md
similarity index 100%
rename from Algorithms/0024. Swap Nodes in Pairs/README.md
rename to leetcode/0024.Swap-Nodes-in-Pairs/README.md
diff --git a/Algorithms/0025. Reverse Nodes in k Group/25. Reverse Nodes in k Group.go b/leetcode/0025.Reverse-Nodes-in-k-Group/25. Reverse Nodes in k Group.go
similarity index 82%
rename from Algorithms/0025. Reverse Nodes in k Group/25. Reverse Nodes in k Group.go
rename to leetcode/0025.Reverse-Nodes-in-k-Group/25. Reverse Nodes in k Group.go
index 41ed9237d..d206f0761 100644
--- a/Algorithms/0025. Reverse Nodes in k Group/25. Reverse Nodes in k Group.go	
+++ b/leetcode/0025.Reverse-Nodes-in-k-Group/25. Reverse Nodes in k Group.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0025. Reverse Nodes in k Group/25. Reverse Nodes in k Group_test.go b/leetcode/0025.Reverse-Nodes-in-k-Group/25. Reverse Nodes in k Group_test.go
similarity index 78%
rename from Algorithms/0025. Reverse Nodes in k Group/25. Reverse Nodes in k Group_test.go
rename to leetcode/0025.Reverse-Nodes-in-k-Group/25. Reverse Nodes in k Group_test.go
index 9654a24d1..52221c072 100644
--- a/Algorithms/0025. Reverse Nodes in k Group/25. Reverse Nodes in k Group_test.go	
+++ b/leetcode/0025.Reverse-Nodes-in-k-Group/25. Reverse Nodes in k Group_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question25 struct {
@@ -27,7 +29,7 @@ func Test_Problem25(t *testing.T) {
 
 	qs := []question25{
 
-		question25{
+		{
 			para25{
 				[]int{1, 2, 3, 4, 5},
 				3,
@@ -35,7 +37,7 @@ func Test_Problem25(t *testing.T) {
 			ans25{[]int{3, 2, 1, 4, 5}},
 		},
 
-		question25{
+		{
 			para25{
 				[]int{1, 2, 3, 4, 5},
 				1,
@@ -48,7 +50,7 @@ func Test_Problem25(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans25, q.para25
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(reverseKGroup(S2l(p.one), p.two)))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(reverseKGroup(structures.Ints2List(p.one), p.two)))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0025. Reverse Nodes in k Group/README.md b/leetcode/0025.Reverse-Nodes-in-k-Group/README.md
similarity index 100%
rename from Algorithms/0025. Reverse Nodes in k Group/README.md
rename to leetcode/0025.Reverse-Nodes-in-k-Group/README.md
diff --git a/Algorithms/0026. Remove Duplicates from Sorted Array/26. Remove Duplicates from Sorted Array.go b/leetcode/0026.Remove-Duplicates-from-Sorted-Array/26. Remove Duplicates from Sorted Array.go
similarity index 100%
rename from Algorithms/0026. Remove Duplicates from Sorted Array/26. Remove Duplicates from Sorted Array.go
rename to leetcode/0026.Remove-Duplicates-from-Sorted-Array/26. Remove Duplicates from Sorted Array.go
diff --git a/Algorithms/0026. Remove Duplicates from Sorted Array/26. Remove Duplicates from Sorted Array_test.go b/leetcode/0026.Remove-Duplicates-from-Sorted-Array/26. Remove Duplicates from Sorted Array_test.go
similarity index 93%
rename from Algorithms/0026. Remove Duplicates from Sorted Array/26. Remove Duplicates from Sorted Array_test.go
rename to leetcode/0026.Remove-Duplicates-from-Sorted-Array/26. Remove Duplicates from Sorted Array_test.go
index 10ee62e1e..74cc3ca0b 100644
--- a/Algorithms/0026. Remove Duplicates from Sorted Array/26. Remove Duplicates from Sorted Array_test.go	
+++ b/leetcode/0026.Remove-Duplicates-from-Sorted-Array/26. Remove Duplicates from Sorted Array_test.go	
@@ -26,22 +26,22 @@ func Test_Problem26(t *testing.T) {
 
 	qs := []question26{
 
-		question26{
+		{
 			para26{[]int{1, 1, 2}},
 			ans26{2},
 		},
 
-		question26{
+		{
 			para26{[]int{0, 0, 1, 1, 1, 1, 2, 3, 4, 4}},
 			ans26{5},
 		},
 
-		question26{
+		{
 			para26{[]int{0, 0, 0, 0, 0}},
 			ans26{1},
 		},
 
-		question26{
+		{
 			para26{[]int{1}},
 			ans26{1},
 		},
diff --git a/Algorithms/0026. Remove Duplicates from Sorted Array/README.md b/leetcode/0026.Remove-Duplicates-from-Sorted-Array/README.md
similarity index 100%
rename from Algorithms/0026. Remove Duplicates from Sorted Array/README.md
rename to leetcode/0026.Remove-Duplicates-from-Sorted-Array/README.md
diff --git a/Algorithms/0027. Remove Element/27. Remove Element.go b/leetcode/0027.Remove-Element/27. Remove Element.go
similarity index 89%
rename from Algorithms/0027. Remove Element/27. Remove Element.go
rename to leetcode/0027.Remove-Element/27. Remove Element.go
index f0233b6ad..1395d1086 100644
--- a/Algorithms/0027. Remove Element/27. Remove Element.go	
+++ b/leetcode/0027.Remove-Element/27. Remove Element.go	
@@ -9,10 +9,8 @@ func removeElement(nums []int, val int) int {
 		if nums[i] != val {
 			if i != j {
 				nums[i], nums[j] = nums[j], nums[i]
-				j++
-			} else {
-				j++
 			}
+			j++
 		}
 	}
 	return j
diff --git a/Algorithms/0027. Remove Element/27. Remove Element_test.go b/leetcode/0027.Remove-Element/27. Remove Element_test.go
similarity index 91%
rename from Algorithms/0027. Remove Element/27. Remove Element_test.go
rename to leetcode/0027.Remove-Element/27. Remove Element_test.go
index fa3a1653a..719bf7dce 100644
--- a/Algorithms/0027. Remove Element/27. Remove Element_test.go	
+++ b/leetcode/0027.Remove-Element/27. Remove Element_test.go	
@@ -27,32 +27,32 @@ func Test_Problem27(t *testing.T) {
 
 	qs := []question27{
 
-		question27{
+		{
 			para27{[]int{1, 0, 1}, 1},
 			ans27{1},
 		},
 
-		question27{
+		{
 			para27{[]int{0, 1, 0, 3, 0, 12}, 0},
 			ans27{3},
 		},
 
-		question27{
+		{
 			para27{[]int{0, 1, 0, 3, 0, 0, 0, 0, 1, 12}, 0},
 			ans27{4},
 		},
 
-		question27{
+		{
 			para27{[]int{0, 0, 0, 0, 0}, 0},
 			ans27{0},
 		},
 
-		question27{
+		{
 			para27{[]int{1}, 1},
 			ans27{0},
 		},
 
-		question27{
+		{
 			para27{[]int{0, 1, 2, 2, 3, 0, 4, 2}, 2},
 			ans27{5},
 		},
diff --git a/Algorithms/0027. Remove Element/README.md b/leetcode/0027.Remove-Element/README.md
similarity index 100%
rename from Algorithms/0027. Remove Element/README.md
rename to leetcode/0027.Remove-Element/README.md
diff --git a/Algorithms/0028. Implement strStr()/28. Implement strStr().go b/leetcode/0028.Find-the-Index-of-the-First-Occurrence-in-a-String/28. Find the Index of the First Occurrence in a String.go
similarity index 100%
rename from Algorithms/0028. Implement strStr()/28. Implement strStr().go
rename to leetcode/0028.Find-the-Index-of-the-First-Occurrence-in-a-String/28. Find the Index of the First Occurrence in a String.go
diff --git a/leetcode/0028.Find-the-Index-of-the-First-Occurrence-in-a-String/28. Find the Index of the First Occurrence in a String_test.go b/leetcode/0028.Find-the-Index-of-the-First-Occurrence-in-a-String/28. Find the Index of the First Occurrence in a String_test.go
new file mode 100644
index 000000000..d06f2bf79
--- /dev/null
+++ b/leetcode/0028.Find-the-Index-of-the-First-Occurrence-in-a-String/28. Find the Index of the First Occurrence in a String_test.go	
@@ -0,0 +1,68 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question28 struct {
+	para28
+	ans28
+}
+
+// para 是参数
+// one 代表第一个参数
+type para28 struct {
+	s string
+	p string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans28 struct {
+	one int
+}
+
+func Test_Problem28(t *testing.T) {
+
+	qs := []question28{
+
+		{
+			para28{"abab", "ab"},
+			ans28{0},
+		},
+
+		{
+			para28{"hello", "ll"},
+			ans28{2},
+		},
+
+		{
+			para28{"", "abc"},
+			ans28{0},
+		},
+
+		{
+			para28{"abacbabc", "abc"},
+			ans28{5},
+		},
+
+		{
+			para28{"abacbabc", "abcd"},
+			ans28{-1},
+		},
+
+		{
+			para28{"abacbabc", ""},
+			ans28{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 28------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans28, q.para28
+		fmt.Printf("【input】:%v       【output】:%v\n", p, strStr(p.s, p.p))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0028.Find-the-Index-of-the-First-Occurrence-in-a-String/README.md b/leetcode/0028.Find-the-Index-of-the-First-Occurrence-in-a-String/README.md
new file mode 100644
index 000000000..d1e643bd9
--- /dev/null
+++ b/leetcode/0028.Find-the-Index-of-the-First-Occurrence-in-a-String/README.md
@@ -0,0 +1,50 @@
+# [28. Find the Index of the First Occurrence in a String](https://leetcode.com/problems/find-the-index-of-the-first-occurrence-in-a-string/)
+
+## 题目
+
+Implement strStr().
+
+Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
+
+
+Example 1:
+
+```c
+Input: haystack = "hello", needle = "ll"
+Output: 2
+```
+
+Example 2:
+
+```c
+Input: haystack = "aaaaa", needle = "bba"
+Output: -1
+```
+
+Clarification:  
+
+What should we return when needle is an empty string? This is a great question to ask during an interview.
+
+For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().
+
+## 题目大意
+
+
+实现一个查找 substring 的函数。如果在母串中找到了子串，返回子串在母串中出现的下标，如果没有找到，返回 -1，如果子串是空串，则返回 0 。
+
+## 解题思路
+
+这一题比较简单，直接写即可。
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/Algorithms/0029. Divide Two Integers/29. Divide Two Integers.go b/leetcode/0029.Divide-Two-Integers/29. Divide Two Integers.go
similarity index 97%
rename from Algorithms/0029. Divide Two Integers/29. Divide Two Integers.go
rename to leetcode/0029.Divide-Two-Integers/29. Divide Two Integers.go
index faeed5cdf..21fb06960 100644
--- a/Algorithms/0029. Divide Two Integers/29. Divide Two Integers.go	
+++ b/leetcode/0029.Divide-Two-Integers/29. Divide Two Integers.go	
@@ -91,3 +91,10 @@ func divide1(divided int, divisor int) int {
 	}
 	return sign * result
 }
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
diff --git a/Algorithms/0029. Divide Two Integers/29. Divide Two Integers_test.go b/leetcode/0029.Divide-Two-Integers/29. Divide Two Integers_test.go
similarity index 91%
rename from Algorithms/0029. Divide Two Integers/29. Divide Two Integers_test.go
rename to leetcode/0029.Divide-Two-Integers/29. Divide Two Integers_test.go
index 670ee6de7..9be6c89f9 100644
--- a/Algorithms/0029. Divide Two Integers/29. Divide Two Integers_test.go	
+++ b/leetcode/0029.Divide-Two-Integers/29. Divide Two Integers_test.go	
@@ -27,27 +27,27 @@ func Test_Problem29(t *testing.T) {
 
 	qs := []question29{
 
-		question29{
+		{
 			para29{10, 3},
 			ans29{3},
 		},
 
-		question29{
+		{
 			para29{7, -3},
 			ans29{-2},
 		},
 
-		question29{
+		{
 			para29{-1, 1},
 			ans29{-1},
 		},
 
-		question29{
+		{
 			para29{1, -1},
 			ans29{-1},
 		},
 
-		question29{
+		{
 			para29{2147483647, 3},
 			ans29{715827882},
 		},
diff --git a/leetcode/0029.Divide-Two-Integers/README.md b/leetcode/0029.Divide-Two-Integers/README.md
new file mode 100755
index 000000000..761e41252
--- /dev/null
+++ b/leetcode/0029.Divide-Two-Integers/README.md
@@ -0,0 +1,47 @@
+# [29. Divide Two Integers](https://leetcode.com/problems/divide-two-integers/)
+
+
+## 题目
+
+Given two integers `dividend` and `divisor`, divide two integers without using multiplication, division and mod operator.
+
+Return the quotient after dividing `dividend` by `divisor`.
+
+The integer division should truncate toward zero.
+
+**Example 1:**
+
+    Input: dividend = 10, divisor = 3
+    Output: 3
+
+**Example 2:**
+
+    Input: dividend = 7, divisor = -3
+    Output: -2
+
+**Note:**
+
+- Both dividend and divisor will be 32-bit signed integers.
+- The divisor will never be 0.
+- Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 2^31 − 1 when the division result overflows.
+
+
+## 题目大意
+
+给定两个整数，被除数 dividend 和除数 divisor。将两数相除，要求不使用乘法、除法和 mod 运算符。返回被除数 dividend 除以除数 divisor 得到的商。
+
+说明:
+
+- 被除数和除数均为 32 位有符号整数。
+- 除数不为 0。
+- 假设我们的环境只能存储 32 位有符号整数，其数值范围是 [−2^31,  2^31 − 1]。本题中，如果除法结果溢出，则返回 2^31 − 1。
+
+
+## 解题思路
+
+- 给出除数和被除数，要求计算除法运算以后的商。注意值的取值范围在 [−2^31, 2^31 − 1] 之中。超过范围的都按边界计算。
+- 这一题可以用二分搜索来做。要求除法运算之后的商，把商作为要搜索的目标。商的取值范围是 [0, dividend]，所以从 0 到被除数之间搜索。利用二分，找到(商 + 1 ) * 除数 > 被除数并且 商 * 除数 ≤ 被除数 或者 (商+1)* 除数 ≥ 被除数并且商 * 除数 < 被除数的时候，就算找到了商，其余情况继续二分即可。最后还要注意符号和题目规定的 Int32 取值范围。
+- 二分的写法常写错的 3 点：
+    1. low ≤ high (注意二分循环退出的条件是小于等于)
+    2. mid = low + (high-low)>>1 (防止溢出)
+    3. low = mid + 1 ; high = mid - 1 (注意更新 low 和 high 的值，如果更新不对就会死循环)
diff --git a/Algorithms/0030. Substring with Concatenation of All Words/30. Substring with Concatenation of All Words.go b/leetcode/0030.Substring-with-Concatenation-of-All-Words/30. Substring with Concatenation of All Words.go
similarity index 100%
rename from Algorithms/0030. Substring with Concatenation of All Words/30. Substring with Concatenation of All Words.go
rename to leetcode/0030.Substring-with-Concatenation-of-All-Words/30. Substring with Concatenation of All Words.go
diff --git a/Algorithms/0030. Substring with Concatenation of All Words/30. Substring with Concatenation of All Words_test.go b/leetcode/0030.Substring-with-Concatenation-of-All-Words/30. Substring with Concatenation of All Words_test.go
similarity index 90%
rename from Algorithms/0030. Substring with Concatenation of All Words/30. Substring with Concatenation of All Words_test.go
rename to leetcode/0030.Substring-with-Concatenation-of-All-Words/30. Substring with Concatenation of All Words_test.go
index 7ee2c50cd..c6ed5fc3c 100644
--- a/Algorithms/0030. Substring with Concatenation of All Words/30. Substring with Concatenation of All Words_test.go	
+++ b/leetcode/0030.Substring-with-Concatenation-of-All-Words/30. Substring with Concatenation of All Words_test.go	
@@ -27,57 +27,57 @@ func Test_Problem30(t *testing.T) {
 
 	qs := []question30{
 
-		question30{
+		{
 			para30{"aaaaaaaa", []string{"aa", "aa", "aa"}},
 			ans30{[]int{0, 1, 2}},
 		},
 
-		question30{
+		{
 			para30{"barfoothefoobarman", []string{"foo", "bar"}},
 			ans30{[]int{0, 9}},
 		},
 
-		question30{
+		{
 			para30{"wordgoodgoodgoodbestword", []string{"word", "good", "best", "word"}},
 			ans30{[]int{}},
 		},
 
-		question30{
+		{
 			para30{"goodgoodgoodgoodgood", []string{"good"}},
 			ans30{[]int{0, 4, 8, 12, 16}},
 		},
 
-		question30{
+		{
 			para30{"barofoothefoolbarman", []string{"foo", "bar"}},
 			ans30{[]int{}},
 		},
 
-		question30{
+		{
 			para30{"bbarffoothefoobarman", []string{"foo", "bar"}},
 			ans30{[]int{}},
 		},
 
-		question30{
+		{
 			para30{"ooroodoofoodtoo", []string{"foo", "doo", "roo", "tee", "oo"}},
 			ans30{[]int{}},
 		},
 
-		question30{
+		{
 			para30{"abc", []string{"a", "b", "c"}},
 			ans30{[]int{0}},
 		},
 
-		question30{
+		{
 			para30{"a", []string{"b"}},
 			ans30{[]int{}},
 		},
 
-		question30{
+		{
 			para30{"ab", []string{"ba"}},
 			ans30{[]int{}},
 		},
 
-		question30{
+		{
 			para30{"n", []string{}},
 			ans30{[]int{}},
 		},
diff --git a/Algorithms/0030. Substring with Concatenation of All Words/README.md b/leetcode/0030.Substring-with-Concatenation-of-All-Words/README.md
similarity index 100%
rename from Algorithms/0030. Substring with Concatenation of All Words/README.md
rename to leetcode/0030.Substring-with-Concatenation-of-All-Words/README.md
diff --git a/leetcode/0031.Next-Permutation/31. Next Permutation.go b/leetcode/0031.Next-Permutation/31. Next Permutation.go
new file mode 100644
index 000000000..04e07485e
--- /dev/null
+++ b/leetcode/0031.Next-Permutation/31. Next Permutation.go	
@@ -0,0 +1,72 @@
+package leetcode
+
+// 解法一
+func nextPermutation(nums []int) {
+	i, j := 0, 0
+	for i = len(nums) - 2; i >= 0; i-- {
+		if nums[i] < nums[i+1] {
+			break
+		}
+	}
+	if i >= 0 {
+		for j = len(nums) - 1; j > i; j-- {
+			if nums[j] > nums[i] {
+				break
+			}
+		}
+		swap(&nums, i, j)
+	}
+	reverse(&nums, i+1, len(nums)-1)
+}
+
+func reverse(nums *[]int, i, j int) {
+	for i < j {
+		swap(nums, i, j)
+		i++
+		j--
+	}
+}
+
+func swap(nums *[]int, i, j int) {
+	(*nums)[i], (*nums)[j] = (*nums)[j], (*nums)[i]
+}
+
+// 解法二
+// [2,(3),6,5,4,1] -> 2,(4),6,5,(3),1 -> 2,4, 1,3,5,6
+func nextPermutation1(nums []int) {
+	var n = len(nums)
+	var pIdx = checkPermutationPossibility(nums)
+	if pIdx == -1 {
+		reverse(&nums, 0, n-1)
+		return
+	}
+
+	var rp = len(nums) - 1
+	// start from right most to leftward,find the first number which is larger than PIVOT
+	for rp > 0 {
+		if nums[rp] > nums[pIdx] {
+			swap(&nums, pIdx, rp)
+			break
+		} else {
+			rp--
+		}
+	}
+	// Finally, Reverse all elements which are right from pivot
+	reverse(&nums, pIdx+1, n-1)
+}
+
+// checkPermutationPossibility returns 1st occurrence Index where
+// value is in decreasing order(from right to left)
+// returns -1 if not found(it's already in its last permutation)
+func checkPermutationPossibility(nums []int) (idx int) {
+	// search right to left for 1st number(from right) that is not in increasing order
+	var rp = len(nums) - 1
+	for rp > 0 {
+		if nums[rp-1] < nums[rp] {
+			idx = rp - 1
+			return idx
+		}
+		rp--
+	}
+	return -1
+}
diff --git a/leetcode/0031.Next-Permutation/31. Next Permutation_test.go b/leetcode/0031.Next-Permutation/31. Next Permutation_test.go
new file mode 100644
index 000000000..f42ffab45
--- /dev/null
+++ b/leetcode/0031.Next-Permutation/31. Next Permutation_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question31 struct {
+	para31
+	ans31
+}
+
+// para 是参数
+// one 代表第一个参数
+type para31 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans31 struct {
+	one []int
+}
+
+func Test_Problem31(t *testing.T) {
+
+	qs := []question31{
+
+		{
+			para31{[]int{1, 2, 3}},
+			ans31{[]int{1, 3, 2}},
+		},
+
+		{
+			para31{[]int{3, 2, 1}},
+			ans31{[]int{1, 2, 3}},
+		},
+
+		{
+			para31{[]int{1, 1, 5}},
+			ans31{[]int{1, 5, 1}},
+		},
+
+		{
+			para31{[]int{1}},
+			ans31{[]int{1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 31------------------------\n")
+	for _, q := range qs {
+		_, p := q.ans31, q.para31
+		fmt.Printf("【input】:%v       ", p)
+		nextPermutation(p.nums)
+		fmt.Printf("【output】:%v\n", p.nums)
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0031.Next-Permutation/README.md b/leetcode/0031.Next-Permutation/README.md
new file mode 100644
index 000000000..6ae4169a2
--- /dev/null
+++ b/leetcode/0031.Next-Permutation/README.md
@@ -0,0 +1,89 @@
+# [31. Next Permutation](https://leetcode.com/problems/next-permutation/)
+
+
+## 题目
+
+Implement **next permutation**, which rearranges numbers into the lexicographically next greater permutation of numbers.
+
+If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).
+
+The replacement must be **[in place](http://en.wikipedia.org/wiki/In-place_algorithm)** and use only constant extra memory.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3]
+Output: [1,3,2]
+```
+
+**Example 2:**
+
+```
+Input: nums = [3,2,1]
+Output: [1,2,3]
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,1,5]
+Output: [1,5,1]
+```
+
+**Example 4:**
+
+```
+Input: nums = [1]
+Output: [1]
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `0 <= nums[i] <= 100`
+
+## 题目大意
+
+实现获取 下一个排列 的函数，算法需要将给定数字序列重新排列成字典序中下一个更大的排列。如果不存在下一个更大的排列，则将数字重新排列成最小的排列（即升序排列）。必须 原地 修改，只允许使用额外常数空间。
+
+## 解题思路
+
+- 题目有 3 个问题需要解决。如何找到下一个排列。不存在下一个排列的时候如何生成最小的排列。如何原地修改。先解决第一个问题，如何找到下一个排列。下一个排列是找到一个大于当前排序的字典序，且变大的幅度最小。那么只能将较小的数与较大数做一次原地交换。并且较小数的下标要尽量靠右，较大数也要尽可能小。原地交换以后，还需要将较大数右边的数按照升序重新排列。这样交换以后，才能生成下一个排列。以排列 [8,9,6,10,7,2] 为例：能找到的符合条件的一对「较小数」与「较大数」的组合为 6 与 7，满足「较小数」尽量靠右，而「较大数」尽可能小。当完成交换后排列变为 [8,9,7,10,6,2]，此时我们可以重排「较小数」右边的序列，序列变为 [8,9,7,2,6,10]。
+- 第一步：在 `nums[i]` 中找到 `i` 使得 `nums[i] < nums[i+1]`，此时较小数为 `nums[i]`，并且 `[i+1, n)` 一定为下降区间。第二步：如果找到了这样的 `i` ，则在下降区间 `[i+1, n)` 中从后往前找到第一个 `j` ，使得 `nums[i] < nums[j]` ，此时较大数为 `nums[j]`。第三步，交换 `nums[i]` 和 `nums[j]`，此时区间 `[i+1, n)` 一定为降序区间。最后原地交换 `[i+1, n)` 区间内的元素，使其变为升序，无需对该区间进行排序。
+- 如果第一步找不到符合条件的下标 `i`，说明当前序列已经是一个最大的排列。那么应该直接执行第三步，生成最小的排列。
+
+## 代码
+
+```go
+package leetcode
+
+func nextPermutation(nums []int) {
+	i, j := 0, 0
+	for i = len(nums) - 2; i >= 0; i-- {
+		if nums[i] < nums[i+1] {
+			break
+		}
+	}
+	if i >= 0 {
+		for j = len(nums) - 1; j > i; j-- {
+			if nums[j] > nums[i] {
+				break
+			}
+		}
+		swap(&nums, i, j)
+	}
+	reverse(&nums, i+1, len(nums)-1)
+}
+
+func reverse(nums *[]int, i, j int) {
+	for i < j {
+		swap(nums, i, j)
+		i++
+		j--
+	}
+}
+
+func swap(nums *[]int, i, j int) {
+	(*nums)[i], (*nums)[j] = (*nums)[j], (*nums)[i]
+}
+```
\ No newline at end of file
diff --git a/leetcode/0032.Longest-Valid-Parentheses/32. Longest Valid Parentheses.go b/leetcode/0032.Longest-Valid-Parentheses/32. Longest Valid Parentheses.go
new file mode 100644
index 000000000..ae59da5ed
--- /dev/null
+++ b/leetcode/0032.Longest-Valid-Parentheses/32. Longest Valid Parentheses.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+// 解法一 栈
+func longestValidParentheses(s string) int {
+	stack, res := []int{}, 0
+	stack = append(stack, -1)
+	for i := 0; i < len(s); i++ {
+		if s[i] == '(' {
+			stack = append(stack, i)
+		} else {
+			stack = stack[:len(stack)-1]
+			if len(stack) == 0 {
+				stack = append(stack, i)
+			} else {
+				res = max(res, i-stack[len(stack)-1])
+			}
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法二 双指针
+func longestValidParentheses1(s string) int {
+	left, right, maxLength := 0, 0, 0
+	for i := 0; i < len(s); i++ {
+		if s[i] == '(' {
+			left++
+		} else {
+			right++
+		}
+		if left == right {
+			maxLength = max(maxLength, 2*right)
+		} else if right > left {
+			left, right = 0, 0
+		}
+	}
+	left, right = 0, 0
+	for i := len(s) - 1; i >= 0; i-- {
+		if s[i] == '(' {
+			left++
+		} else {
+			right++
+		}
+		if left == right {
+			maxLength = max(maxLength, 2*left)
+		} else if left > right {
+			left, right = 0, 0
+		}
+	}
+	return maxLength
+}
diff --git a/leetcode/0032.Longest-Valid-Parentheses/32. Longest Valid Parentheses_test.go b/leetcode/0032.Longest-Valid-Parentheses/32. Longest Valid Parentheses_test.go
new file mode 100644
index 000000000..4bdf8de2b
--- /dev/null
+++ b/leetcode/0032.Longest-Valid-Parentheses/32. Longest Valid Parentheses_test.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question32 struct {
+	para32
+	ans32
+}
+
+// para 是参数
+// one 代表第一个参数
+type para32 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans32 struct {
+	one int
+}
+
+func Test_Problem32(t *testing.T) {
+
+	qs := []question32{
+
+		{
+			para32{"(()"},
+			ans32{2},
+		},
+
+		{
+			para32{")()())"},
+			ans32{4},
+		},
+
+		{
+			para32{"()(())"},
+			ans32{6},
+		},
+
+		{
+			para32{"()(())))"},
+			ans32{6},
+		},
+
+		{
+			para32{"()(()"},
+			ans32{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 32------------------------\n")
+	for _, q := range qs {
+		_, p := q.ans32, q.para32
+		fmt.Printf("【input】:%v    【output】:%v\n", p, longestValidParentheses(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0032.Longest-Valid-Parentheses/README.md b/leetcode/0032.Longest-Valid-Parentheses/README.md
new file mode 100644
index 000000000..50f6ec116
--- /dev/null
+++ b/leetcode/0032.Longest-Valid-Parentheses/README.md
@@ -0,0 +1,108 @@
+# [32. Longest Valid Parentheses](https://leetcode.com/problems/longest-valid-parentheses/)
+
+
+## 题目
+
+Given a string containing just the characters `'('` and `')'`, find the length of the longest valid (well-formed) parentheses substring.
+
+**Example 1:**
+
+```
+Input: s = "(()"
+Output: 2
+Explanation: The longest valid parentheses substring is "()".
+```
+
+**Example 2:**
+
+```
+Input: s = ")()())"
+Output: 4
+Explanation: The longest valid parentheses substring is "()()".
+```
+
+**Example 3:**
+
+```
+Input: s = ""
+Output: 0
+```
+
+**Constraints:**
+
+- `0 <= s.length <= 3 * 104`
+- `s[i]` is `'('`, or `')'`.
+
+## 题目大意
+
+给你一个只包含 '(' 和 ')' 的字符串，找出最长有效（格式正确且连续）括号子串的长度。
+
+## 解题思路
+
+- 提到括号匹配，第一时间能让人想到的就是利用栈。这里需要计算嵌套括号的总长度，所以栈里面不能单纯的存左括号，而应该存左括号在原字符串的下标，这样通过下标相减可以获取长度。那么栈如果是非空，栈底永远存的是当前遍历过的字符串中**上一个没有被匹配的右括号的下标**。**上一个没有被匹配的右括号的下标**可以理解为每段括号匹配之间的“隔板”。例如，`())((()))`，第三个右括号，即为左右 2 段正确的括号匹配中间的“隔板”。“隔板”的存在影响计算最长括号长度。如果不存在“隔板”，前后 2 段正确的括号匹配应该“融合”在一起，最长长度为 `2 + 6 = 8`，但是这里存在了“隔板”，所以最长长度仅为 `6`。
+- 具体算法实现，遇到每个 `'('` ，将它的下标压入栈中。对于遇到的每个 `')'`，先弹出栈顶元素表示匹配了当前右括号。如果栈为空，说明当前的右括号为没有被匹配的右括号，于是将其下标放入栈中来更新**上一个没有被匹配的右括号的下标**。如果栈不为空，当前右括号的下标减去栈顶元素即为以该右括号为结尾的最长有效括号的长度。需要注意初始化时，不存在**上一个没有被匹配的右括号的下标**，那么将 `-1` 放入栈中，充当下标为 `0` 的“隔板”。时间复杂度 O(n)，空间复杂度 O(n)。
+- 在栈的方法中，实际用到的元素仅仅是栈底的**上一个没有被匹配的右括号的下标**。那么考虑能否把这个值存在一个变量中，这样可以省去栈 O(n) 的时间复杂度。利用两个计数器 left 和 right 。首先，从左到右遍历字符串，每当遇到 `'('`，增加 left 计数器，每当遇到 `')'` ，增加 right 计数器。每当 left 计数器与 right 计数器相等时，计算当前有效字符串的长度，并且记录目前为止找到的最长子字符串。当 right 计数器比 left 计数器大时，说明括号不匹配，于是将 left 和 right 计数器同时变回 0。这样的做法利用了贪心的思想，考虑了以当前字符下标结尾的有效括号长度，每次当右括号数量多于左括号数量的时候之前的字符就扔掉不再考虑，重新从下一个字符开始计算。
+- 但上面的做法会漏掉一种情况，就是遍历的时候左括号的数量始终大于右括号的数量，即 `(()` ，这种时候最长有效括号是求不出来的。解决办法是反向再计算一遍，如果从右往左计算，`(()` 先计算匹配的括号，最后只剩下 `'('`，这样依旧可以算出最长匹配的括号长度。反过来计算的方法和上述从左往右计算的方法一致：当 left 计数器比 right 计数器大时，将 left 和 right 计数器同时变回 0；当 left 计数器与 right 计数器相等时，计算当前有效字符串的长度，并且记录目前为止找到的最长子字符串。这种方法的时间复杂度是 O(n)，空间复杂度 O(1)。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 栈
+func longestValidParentheses(s string) int {
+	stack, res := []int{}, 0
+	stack = append(stack, -1)
+	for i := 0; i < len(s); i++ {
+		if s[i] == '(' {
+			stack = append(stack, i)
+		} else {
+			stack = stack[:len(stack)-1]
+			if len(stack) == 0 {
+				stack = append(stack, i)
+			} else {
+				res = max(res, i-stack[len(stack)-1])
+			}
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法二 双指针
+func longestValidParentheses1(s string) int {
+	left, right, maxLength := 0, 0, 0
+	for i := 0; i < len(s); i++ {
+		if s[i] == '(' {
+			left++
+		} else {
+			right++
+		}
+		if left == right {
+			maxLength = max(maxLength, 2*right)
+		} else if right > left {
+			left, right = 0, 0
+		}
+	}
+	left, right = 0, 0
+	for i := len(s) - 1; i >= 0; i-- {
+		if s[i] == '(' {
+			left++
+		} else {
+			right++
+		}
+		if left == right {
+			maxLength = max(maxLength, 2*left)
+		} else if left > right {
+			left, right = 0, 0
+		}
+	}
+	return maxLength
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0033. Search in Rotated Sorted Array/33. Search in Rotated Sorted Array.go b/leetcode/0033.Search-in-Rotated-Sorted-Array/33. Search in Rotated Sorted Array.go
similarity index 100%
rename from Algorithms/0033. Search in Rotated Sorted Array/33. Search in Rotated Sorted Array.go
rename to leetcode/0033.Search-in-Rotated-Sorted-Array/33. Search in Rotated Sorted Array.go
diff --git a/Algorithms/0033. Search in Rotated Sorted Array/33. Search in Rotated Sorted Array_test.go b/leetcode/0033.Search-in-Rotated-Sorted-Array/33. Search in Rotated Sorted Array_test.go
similarity index 94%
rename from Algorithms/0033. Search in Rotated Sorted Array/33. Search in Rotated Sorted Array_test.go
rename to leetcode/0033.Search-in-Rotated-Sorted-Array/33. Search in Rotated Sorted Array_test.go
index 3fcfbb5ad..b31eca3bc 100644
--- a/Algorithms/0033. Search in Rotated Sorted Array/33. Search in Rotated Sorted Array_test.go	
+++ b/leetcode/0033.Search-in-Rotated-Sorted-Array/33. Search in Rotated Sorted Array_test.go	
@@ -27,17 +27,17 @@ func Test_Problem33(t *testing.T) {
 
 	qs := []question33{
 
-		question33{
+		{
 			para33{[]int{3, 1}, 1},
 			ans33{1},
 		},
 
-		question33{
+		{
 			para33{[]int{4, 5, 6, 7, 0, 1, 2}, 0},
 			ans33{4},
 		},
 
-		question33{
+		{
 			para33{[]int{4, 5, 6, 7, 0, 1, 2}, 3},
 			ans33{-1},
 		},
diff --git a/leetcode/0033.Search-in-Rotated-Sorted-Array/README.md b/leetcode/0033.Search-in-Rotated-Sorted-Array/README.md
new file mode 100755
index 000000000..d7b7284e1
--- /dev/null
+++ b/leetcode/0033.Search-in-Rotated-Sorted-Array/README.md
@@ -0,0 +1,36 @@
+# [33. Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)
+
+## 题目
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., `[0,1,2,4,5,6,7]` might become `[4,5,6,7,0,1,2]`).
+
+You are given a target value to search. If found in the array return its index, otherwise return `-1`.
+
+You may assume no duplicate exists in the array.
+
+Your algorithm's runtime complexity must be in the order of *O*(log *n*).
+
+**Example 1:**
+
+    Input: nums = [4,5,6,7,0,1,2], target = 0
+    Output: 4
+
+**Example 2:**
+
+    Input: nums = [4,5,6,7,0,1,2], target = 3
+    Output: -1
+
+
+## 题目大意
+
+假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。你可以假设数组中不存在重复的元素。
+
+你的算法时间复杂度必须是 O(log n) 级别。
+
+
+## 解题思路
+
+- 给出一个数组，数组中本来是从小到大排列的，并且数组中没有重复数字。但是现在把后面随机一段有序的放到数组前面，这样形成了前后两端有序的子序列。在这样的一个数组里面查找一个数，设计一个 O(log n) 的算法。如果找到就输出数组的小标，如果没有找到，就输出 -1 。
+- 由于数组基本有序，虽然中间有一个“断开点”，还是可以使用二分搜索的算法来实现。现在数组前面一段是数值比较大的数，后面一段是数值偏小的数。如果 mid 落在了前一段数值比较大的区间内了，那么一定有 `nums[mid] > nums[low]`，如果是落在后面一段数值比较小的区间内，`nums[mid] ≤ nums[low]` 。如果 mid 落在了后一段数值比较小的区间内了，那么一定有 `nums[mid] < nums[high]`，如果是落在前面一段数值比较大的区间内，`nums[mid] ≤ nums[high]` 。还有 `nums[low] == nums[mid]` 和 `nums[high] == nums[mid]` 的情况，单独处理即可。最后找到则输出 mid，没有找到则输出 -1 。
diff --git a/Algorithms/0034. Find First and Last Position of Element in Sorted Array/34. Find First and Last Position of Element in Sorted Array.go b/leetcode/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array/34. Find First and Last Position of Element in Sorted Array.go
similarity index 100%
rename from Algorithms/0034. Find First and Last Position of Element in Sorted Array/34. Find First and Last Position of Element in Sorted Array.go
rename to leetcode/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array/34. Find First and Last Position of Element in Sorted Array.go
diff --git a/Algorithms/0034. Find First and Last Position of Element in Sorted Array/34. Find First and Last Position of Element in Sorted Array_test.go b/leetcode/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array/34. Find First and Last Position of Element in Sorted Array_test.go
similarity index 96%
rename from Algorithms/0034. Find First and Last Position of Element in Sorted Array/34. Find First and Last Position of Element in Sorted Array_test.go
rename to leetcode/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array/34. Find First and Last Position of Element in Sorted Array_test.go
index 52d408aa8..c8bd3306b 100644
--- a/Algorithms/0034. Find First and Last Position of Element in Sorted Array/34. Find First and Last Position of Element in Sorted Array_test.go	
+++ b/leetcode/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array/34. Find First and Last Position of Element in Sorted Array_test.go	
@@ -27,12 +27,12 @@ func Test_Problem34(t *testing.T) {
 
 	qs := []question34{
 
-		question34{
+		{
 			para34{[]int{5, 7, 7, 8, 8, 10}, 8},
 			ans34{[]int{3, 4}},
 		},
 
-		question34{
+		{
 			para34{[]int{5, 7, 7, 8, 8, 10}, 6},
 			ans34{[]int{-1, -1}},
 		},
diff --git a/leetcode/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array/README.md b/leetcode/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array/README.md
new file mode 100755
index 000000000..cdd5e99e3
--- /dev/null
+++ b/leetcode/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array/README.md
@@ -0,0 +1,36 @@
+# [34. Find First and Last Position of Element in Sorted Array](https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/)
+
+
+## 题目
+
+Given an array of integers `nums` sorted in ascending order, find the starting and ending position of a given `target` value.
+
+Your algorithm's runtime complexity must be in the order of *O*(log *n*).
+
+If the target is not found in the array, return `[-1, -1]`.
+
+**Example 1:**
+
+    Input: nums = [5,7,7,8,8,10], target = 8
+    Output: [3,4]
+
+**Example 2:**
+
+    Input: nums = [5,7,7,8,8,10], target = 6
+    Output: [-1,-1]
+
+## 题目大意
+
+给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。你的算法时间复杂度必须是 O(log n) 级别。如果数组中不存在目标值，返回 [-1, -1]。
+
+
+## 解题思路
+
+- 给出一个有序数组 `nums` 和一个数 `target`，要求在数组中找到第一个和这个元素相等的元素下标，最后一个和这个元素相等的元素下标。
+- 这一题是经典的二分搜索变种题。二分搜索有 4 大基础变种题：
+    1. 查找第一个值等于给定值的元素
+    2. 查找最后一个值等于给定值的元素
+    3. 查找第一个大于等于给定值的元素
+    4. 查找最后一个小于等于给定值的元素
+
+    这一题的解题思路可以分别利用变种 1 和变种 2 的解法就可以做出此题。或者用一次变种 1 的方法，然后循环往后找到最后一个与给定值相等的元素。不过后者这种方法可能会使时间复杂度下降到 O(n)，因为有可能数组中 n 个元素都和给定元素相同。(4 大基础变种的实现见代码)
diff --git a/Algorithms/0035. Search Insert Position/35. Search Insert Position.go b/leetcode/0035.Search-Insert-Position/35. Search Insert Position.go
similarity index 100%
rename from Algorithms/0035. Search Insert Position/35. Search Insert Position.go
rename to leetcode/0035.Search-Insert-Position/35. Search Insert Position.go
diff --git a/Algorithms/0035. Search Insert Position/35. Search Insert Position_test.go b/leetcode/0035.Search-Insert-Position/35. Search Insert Position_test.go
similarity index 93%
rename from Algorithms/0035. Search Insert Position/35. Search Insert Position_test.go
rename to leetcode/0035.Search-Insert-Position/35. Search Insert Position_test.go
index 6947076c2..1b9a30bce 100644
--- a/Algorithms/0035. Search Insert Position/35. Search Insert Position_test.go	
+++ b/leetcode/0035.Search-Insert-Position/35. Search Insert Position_test.go	
@@ -27,22 +27,22 @@ func Test_Problem35(t *testing.T) {
 
 	qs := []question35{
 
-		question35{
+		{
 			para35{[]int{1, 3, 5, 6}, 5},
 			ans35{2},
 		},
 
-		question35{
+		{
 			para35{[]int{1, 3, 5, 6}, 2},
 			ans35{1},
 		},
 
-		question35{
+		{
 			para35{[]int{1, 3, 5, 6}, 7},
 			ans35{4},
 		},
 
-		question35{
+		{
 			para35{[]int{1, 3, 5, 6}, 0},
 			ans35{0},
 		},
diff --git a/leetcode/0035.Search-Insert-Position/README.md b/leetcode/0035.Search-Insert-Position/README.md
new file mode 100755
index 000000000..9b8f67c8f
--- /dev/null
+++ b/leetcode/0035.Search-Insert-Position/README.md
@@ -0,0 +1,40 @@
+# [35. Search Insert Position](https://leetcode.com/problems/search-insert-position/)
+
+
+## 题目
+
+Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
+
+You may assume no duplicates in the array.
+
+**Example 1:**
+
+    Input: [1,3,5,6], 5
+    Output: 2
+
+**Example 2:**
+
+    Input: [1,3,5,6], 2
+    Output: 1
+
+**Example 3:**
+
+    Input: [1,3,5,6], 7
+    Output: 4
+
+**Example 4:**
+
+    Input: [1,3,5,6], 0
+    Output: 0
+
+
+## 题目大意
+
+给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。
+
+你可以假设数组中无重复元素。
+
+## 解题思路
+
+- 给出一个已经从小到大排序后的数组，要求在数组中找到插入 target 元素的位置。
+- 这一题是经典的二分搜索的变种题，在有序数组中找到最后一个比 target 小的元素。
diff --git a/Algorithms/0036. Valid Sudoku/36. Valid Sudoku.go b/leetcode/0036.Valid-Sudoku/36. Valid Sudoku.go
similarity index 100%
rename from Algorithms/0036. Valid Sudoku/36. Valid Sudoku.go
rename to leetcode/0036.Valid-Sudoku/36. Valid Sudoku.go
diff --git a/leetcode/0036.Valid-Sudoku/36. Valid Sudoku_test.go b/leetcode/0036.Valid-Sudoku/36. Valid Sudoku_test.go
new file mode 100644
index 000000000..0477efaa7
--- /dev/null
+++ b/leetcode/0036.Valid-Sudoku/36. Valid Sudoku_test.go	
@@ -0,0 +1,79 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question36 struct {
+	para36
+	ans36
+}
+
+// para 是参数
+// one 代表第一个参数
+type para36 struct {
+	s [][]byte
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans36 struct {
+	s bool
+}
+
+func Test_Problem36(t *testing.T) {
+
+	qs := []question36{
+
+		{
+			para36{[][]byte{
+				{'5', '3', '.', '.', '7', '.', '.', '.', '.'},
+				{'6', '.', '.', '1', '9', '5', '.', '.', '.'},
+				{'.', '9', '8', '.', '.', '.', '.', '6', '.'},
+				{'8', '.', '.', '.', '6', '.', '.', '.', '3'},
+				{'4', '.', '.', '8', '.', '3', '.', '.', '1'},
+				{'7', '.', '.', '.', '2', '.', '.', '.', '6'},
+				{'.', '6', '.', '.', '.', '.', '2', '8', '.'},
+				{'.', '.', '.', '4', '1', '9', '.', '.', '5'},
+				{'.', '.', '.', '.', '8', '.', '.', '7', '9'}}},
+			ans36{true},
+		},
+
+		{
+			para36{[][]byte{
+				{'8', '3', '.', '.', '7', '.', '.', '.', '.'},
+				{'6', '.', '.', '1', '9', '5', '.', '.', '.'},
+				{'.', '9', '8', '.', '.', '.', '.', '6', '.'},
+				{'8', '.', '.', '.', '6', '.', '.', '.', '3'},
+				{'4', '.', '.', '8', '.', '3', '.', '.', '1'},
+				{'7', '.', '.', '.', '2', '.', '.', '.', '6'},
+				{'.', '6', '.', '.', '.', '.', '2', '8', '.'},
+				{'.', '.', '.', '4', '1', '9', '.', '.', '5'},
+				{'.', '.', '.', '.', '8', '.', '.', '7', '9'}}},
+			ans36{false},
+		},
+
+		{
+			para36{[][]byte{
+				{'.', '8', '7', '6', '5', '4', '3', '2', '1'},
+				{'2', '.', '.', '.', '.', '.', '.', '.', '.'},
+				{'3', '.', '.', '.', '.', '.', '.', '.', '.'},
+				{'4', '.', '.', '.', '.', '.', '.', '.', '.'},
+				{'5', '.', '.', '.', '.', '.', '.', '.', '.'},
+				{'6', '.', '.', '.', '.', '.', '.', '.', '.'},
+				{'7', '.', '.', '.', '.', '.', '.', '.', '.'},
+				{'8', '.', '.', '.', '.', '.', '.', '.', '.'},
+				{'9', '.', '.', '.', '.', '.', '.', '.', '.'}}},
+			ans36{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 36------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans36, q.para36
+		fmt.Printf("【input】:%v       【output】:%v\n", p, isValidSudoku1(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0036.Valid-Sudoku/README.md b/leetcode/0036.Valid-Sudoku/README.md
new file mode 100755
index 000000000..cfede681e
--- /dev/null
+++ b/leetcode/0036.Valid-Sudoku/README.md
@@ -0,0 +1,75 @@
+# [36. Valid Sudoku](https://leetcode.com/problems/valid-sudoku/)
+
+
+## 题目
+
+Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
+
+1. Each row must contain the digits `1-9` without repetition.
+2. Each column must contain the digits `1-9` without repetition.
+3. Each of the 9 `3x3` sub-boxes of the grid must contain the digits `1-9` without repetition.
+
+![](https://upload.wikimedia.org/wikipedia/commons/thumb/f/ff/Sudoku-by-L2G-20050714.svg/250px-Sudoku-by-L2G-20050714.svg.png)
+
+A partially filled sudoku which is valid.
+
+The Sudoku board could be partially filled, where empty cells are filled with the character `'.'`.
+
+**Example 1:**
+
+
+    Input:
+    [
+      ["5","3",".",".","7",".",".",".","."],
+      ["6",".",".","1","9","5",".",".","."],
+      [".","9","8",".",".",".",".","6","."],
+      ["8",".",".",".","6",".",".",".","3"],
+      ["4",".",".","8",".","3",".",".","1"],
+      ["7",".",".",".","2",".",".",".","6"],
+      [".","6",".",".",".",".","2","8","."],
+      [".",".",".","4","1","9",".",".","5"],
+      [".",".",".",".","8",".",".","7","9"]
+    ]
+    Output: true
+
+
+**Example 2:**
+
+
+    Input:
+    [
+      ["8","3",".",".","7",".",".",".","."],
+      ["6",".",".","1","9","5",".",".","."],
+      [".","9","8",".",".",".",".","6","."],
+      ["8",".",".",".","6",".",".",".","3"],
+      ["4",".",".","8",".","3",".",".","1"],
+      ["7",".",".",".","2",".",".",".","6"],
+      [".","6",".",".",".",".","2","8","."],
+      [".",".",".","4","1","9",".",".","5"],
+      [".",".",".",".","8",".",".","7","9"]
+    ]
+    Output: false
+    Explanation: Same as Example 1, except with the 5 in the top left corner being 
+        modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.
+
+
+**Note:**
+
+- A Sudoku board (partially filled) could be valid but is not necessarily solvable.
+- Only the filled cells need to be validated according to the mentioned rules.
+- The given board contain only digits `1-9` and the character `'.'`.
+- The given board size is always `9x9`.
+
+## 题目大意
+
+判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。
+
+1. 数字 1-9 在每一行只能出现一次。
+2. 数字 1-9 在每一列只能出现一次。
+3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
+
+
+## 解题思路
+
+- 给出一个数独的棋盘，要求判断这个棋盘当前是否满足数独的要求：即行列是否都只包含 1-9，每个九宫格里面是否也只包含 1-9 。
+- 注意这题和第 37 题是不同的，这一题是判断当前棋盘状态是否满足数独的要求，而第 37 题是要求求解数独。本题中的棋盘有些是无解的，但是棋盘状态是满足题意的。
diff --git a/Algorithms/0037. Sudoku Solver/37. Sudoku Solver.go b/leetcode/0037.Sudoku-Solver/37. Sudoku Solver.go
similarity index 100%
rename from Algorithms/0037. Sudoku Solver/37. Sudoku Solver.go
rename to leetcode/0037.Sudoku-Solver/37. Sudoku Solver.go
diff --git a/leetcode/0037.Sudoku-Solver/37. Sudoku Solver_test.go b/leetcode/0037.Sudoku-Solver/37. Sudoku Solver_test.go
new file mode 100644
index 000000000..d56bd362c
--- /dev/null
+++ b/leetcode/0037.Sudoku-Solver/37. Sudoku Solver_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question37 struct {
+	para37
+	ans37
+}
+
+// para 是参数
+// one 代表第一个参数
+type para37 struct {
+	s [][]byte
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans37 struct {
+	s [][]byte
+}
+
+func Test_Problem37(t *testing.T) {
+
+	qs := []question37{
+
+		{
+			para37{[][]byte{
+				{'5', '3', '.', '.', '7', '.', '.', '.', '.'},
+				{'6', '.', '.', '1', '9', '5', '.', '.', '.'},
+				{'.', '9', '8', '.', '.', '.', '.', '6', '.'},
+				{'8', '.', '.', '.', '6', '.', '.', '.', '3'},
+				{'4', '.', '.', '8', '.', '3', '.', '.', '1'},
+				{'7', '.', '.', '.', '2', '.', '.', '.', '6'},
+				{'.', '6', '.', '.', '.', '.', '2', '8', '.'},
+				{'.', '.', '.', '4', '1', '9', '.', '.', '5'},
+				{'.', '.', '.', '.', '8', '.', '.', '7', '9'}}},
+			ans37{[][]byte{
+				{'5', '3', '4', '6', '7', '8', '9', '1', '2'},
+				{'6', '7', '2', '1', '9', '5', '3', '4', '8'},
+				{'1', '9', '8', '3', '4', '2', '5', '6', '7'},
+				{'8', '5', '9', '7', '6', '1', '4', '2', '3'},
+				{'4', '2', '6', '8', '5', '3', '7', '9', '1'},
+				{'7', '1', '3', '9', '2', '4', '8', '5', '6'},
+				{'9', '6', '1', '5', '3', '7', '2', '8', '4'},
+				{'2', '8', '7', '4', '1', '9', '6', '3', '5'},
+				{'3', '4', '5', '2', '8', '6', '1', '7', '9'}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 37------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans37, q.para37
+		fmt.Printf("【input】:%v \n\n", p)
+		solveSudoku(p.s)
+		fmt.Printf("【output】:%v \n\n", p)
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0037.Sudoku-Solver/README.md b/leetcode/0037.Sudoku-Solver/README.md
new file mode 100755
index 000000000..9a2a86ecf
--- /dev/null
+++ b/leetcode/0037.Sudoku-Solver/README.md
@@ -0,0 +1,48 @@
+# [37. Sudoku Solver](https://leetcode.com/problems/sudoku-solver/)
+
+
+
+## 题目
+
+Write a program to solve a Sudoku puzzle by filling the empty cells.
+
+A sudoku solution must satisfy **all of the following rules**:
+
+1. Each of the digits `1-9` must occur exactly once in each row.
+2. Each of the digits `1-9` must occur exactly once in each column.
+3. Each of the the digits `1-9` must occur exactly once in each of the 9 `3x3` sub-boxes of the grid.
+
+Empty cells are indicated by the character `'.'`.
+
+![](https://upload.wikimedia.org/wikipedia/commons/thumb/f/ff/Sudoku-by-L2G-20050714.svg/250px-Sudoku-by-L2G-20050714.svg.png)
+
+A sudoku puzzle...
+
+![](https://upload.wikimedia.org/wikipedia/commons/thumb/3/31/Sudoku-by-L2G-20050714_solution.svg/250px-Sudoku-by-L2G-20050714_solution.svg.png)
+
+...and its solution numbers marked in red.
+
+**Note:**
+
+- The given board contain only digits `1-9` and the character `'.'`.
+- You may assume that the given Sudoku puzzle will have a single unique solution.
+- The given board size is always `9x9`.
+
+## 题目大意
+
+
+编写一个程序，通过已填充的空格来解决数独问题。一个数独的解法需遵循如下规则：
+
+1. 数字 1-9 在每一行只能出现一次。
+2. 数字 1-9 在每一列只能出现一次。
+3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
+
+空白格用 '.' 表示。
+
+
+## 解题思路
+
+- 给出一个数独谜题，要求解出这个数独
+- 解题思路 DFS 暴力回溯枚举。数独要求每横行，每竖行，每九宫格内，`1-9` 的数字不能重复，每次放下一个数字的时候，在这 3 个地方都需要判断一次。
+- 另外找到一组解以后就不需要再继续回溯了，直接返回即可。
+
diff --git a/Algorithms/0039. Combination Sum/39. Combination Sum.go b/leetcode/0039.Combination-Sum/39. Combination Sum.go
similarity index 91%
rename from Algorithms/0039. Combination Sum/39. Combination Sum.go
rename to leetcode/0039.Combination-Sum/39. Combination Sum.go
index 7ce37162d..feefddceb 100644
--- a/Algorithms/0039. Combination Sum/39. Combination Sum.go	
+++ b/leetcode/0039.Combination-Sum/39. Combination Sum.go	
@@ -22,6 +22,9 @@ func findcombinationSum(nums []int, target, index int, c []int, res *[][]int) {
 		return
 	}
 	for i := index; i < len(nums); i++ {
+		if nums[i] > target { // 这里可以剪枝优化
+			break
+		}
 		c = append(c, nums[i])
 		findcombinationSum(nums, target-nums[i], i, c, res) // 注意这里迭代的时候 index 依旧不变，因为一个元素可以取多次
 		c = c[:len(c)-1]
diff --git a/Algorithms/0039. Combination Sum/39. Combination Sum_test.go b/leetcode/0039.Combination-Sum/39. Combination Sum_test.go
similarity index 82%
rename from Algorithms/0039. Combination Sum/39. Combination Sum_test.go
rename to leetcode/0039.Combination-Sum/39. Combination Sum_test.go
index b6df4147f..8b3a3568b 100644
--- a/Algorithms/0039. Combination Sum/39. Combination Sum_test.go	
+++ b/leetcode/0039.Combination-Sum/39. Combination Sum_test.go	
@@ -27,14 +27,14 @@ func Test_Problem39(t *testing.T) {
 
 	qs := []question39{
 
-		question39{
+		{
 			para39{[]int{2, 3, 6, 7}, 7},
-			ans39{[][]int{[]int{7}, []int{2, 2, 3}}},
+			ans39{[][]int{{7}, {2, 2, 3}}},
 		},
 
-		question39{
+		{
 			para39{[]int{2, 3, 5}, 8},
-			ans39{[][]int{[]int{2, 2, 2, 2}, []int{2, 3, 3}, []int{3, 5}}},
+			ans39{[][]int{{2, 2, 2, 2}, {2, 3, 3}, {3, 5}}},
 		},
 	}
 
diff --git a/Algorithms/0039. Combination Sum/README.md b/leetcode/0039.Combination-Sum/README.md
similarity index 100%
rename from Algorithms/0039. Combination Sum/README.md
rename to leetcode/0039.Combination-Sum/README.md
diff --git a/Algorithms/0040. Combination Sum II/40. Combination Sum II.go b/leetcode/0040.Combination-Sum-II/40. Combination Sum II.go
similarity index 100%
rename from Algorithms/0040. Combination Sum II/40. Combination Sum II.go
rename to leetcode/0040.Combination-Sum-II/40. Combination Sum II.go
diff --git a/Algorithms/0040. Combination Sum II/40. Combination Sum II_test.go b/leetcode/0040.Combination-Sum-II/40. Combination Sum II_test.go
similarity index 81%
rename from Algorithms/0040. Combination Sum II/40. Combination Sum II_test.go
rename to leetcode/0040.Combination-Sum-II/40. Combination Sum II_test.go
index ef9716237..8c46b3f70 100644
--- a/Algorithms/0040. Combination Sum II/40. Combination Sum II_test.go	
+++ b/leetcode/0040.Combination-Sum-II/40. Combination Sum II_test.go	
@@ -27,14 +27,14 @@ func Test_Problem40(t *testing.T) {
 
 	qs := []question40{
 
-		question40{
+		{
 			para40{[]int{10, 1, 2, 7, 6, 1, 5}, 8},
-			ans40{[][]int{[]int{1, 7}, []int{1, 2, 5}, []int{2, 6}, []int{1, 1, 6}}},
+			ans40{[][]int{{1, 7}, {1, 2, 5}, {2, 6}, {1, 1, 6}}},
 		},
 
-		question40{
+		{
 			para40{[]int{2, 5, 2, 1, 2}, 5},
-			ans40{[][]int{[]int{1, 2, 2}, []int{5}}},
+			ans40{[][]int{{1, 2, 2}, {5}}},
 		},
 	}
 
diff --git a/leetcode/0040.Combination-Sum-II/README.md b/leetcode/0040.Combination-Sum-II/README.md
new file mode 100755
index 000000000..b53f67dcd
--- /dev/null
+++ b/leetcode/0040.Combination-Sum-II/README.md
@@ -0,0 +1,48 @@
+# [40. Combination Sum II](https://leetcode.com/problems/combination-sum-ii/)
+
+
+## 题目
+
+Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sums to `target`.
+
+Each number in `candidates` may only be used **once** in the combination.
+
+**Note:**
+
+- All numbers (including `target`) will be positive integers.
+- The solution set must not contain duplicate combinations.
+
+**Example 1:**
+
+
+    Input: candidates = [10,1,2,7,6,1,5], target = 8,
+    A solution set is:
+    [
+      [1, 7],
+      [1, 2, 5],
+      [2, 6],
+      [1, 1, 6]
+    ]
+
+
+**Example 2:**
+
+
+    Input: candidates = [2,5,2,1,2], target = 5,
+    A solution set is:
+    [
+      [1,2,2],
+      [5]
+    ]
+
+## 题目大意
+
+给定一个数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
+
+candidates 中的每个数字在每个组合中只能使用一次。
+
+
+## 解题思路
+
+- 题目要求出总和为 sum 的所有组合，组合需要去重。这一题是第 39 题的加强版，第 39 题中元素可以重复利用(重复元素可无限次使用)，这一题中元素只能有限次数的利用，因为存在重复元素，并且每个元素只能用一次(重复元素只能使用有限次)
+- 这一题和第 47 题类似。
diff --git a/Algorithms/0041. First-Missing-Positive/41. First Missing Positive.go b/leetcode/0041.First-Missing-Positive/41. First Missing Positive.go
similarity index 100%
rename from Algorithms/0041. First-Missing-Positive/41. First Missing Positive.go
rename to leetcode/0041.First-Missing-Positive/41. First Missing Positive.go
diff --git a/Algorithms/0041. First-Missing-Positive/41. First Missing Positive_test.go b/leetcode/0041.First-Missing-Positive/41. First Missing Positive_test.go
similarity index 90%
rename from Algorithms/0041. First-Missing-Positive/41. First Missing Positive_test.go
rename to leetcode/0041.First-Missing-Positive/41. First Missing Positive_test.go
index e1d0ba59e..0edd5c0df 100644
--- a/Algorithms/0041. First-Missing-Positive/41. First Missing Positive_test.go	
+++ b/leetcode/0041.First-Missing-Positive/41. First Missing Positive_test.go	
@@ -26,37 +26,37 @@ func Test_Problem41(t *testing.T) {
 
 	qs := []question41{
 
-		question41{
+		{
 			para41{[]int{10, -1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, -3}},
 			ans41{6},
 		},
 
-		question41{
+		{
 			para41{[]int{10, -1, 8, 6, 7, 3, -2, 5, 4, 2, 1, -3}},
 			ans41{9},
 		},
 
-		question41{
+		{
 			para41{[]int{1}},
 			ans41{2},
 		},
 
-		question41{
+		{
 			para41{[]int{0, 2, 2, 1, 1}},
 			ans41{3},
 		},
 
-		question41{
+		{
 			para41{[]int{}},
 			ans41{1},
 		},
 
-		question41{
+		{
 			para41{[]int{1, 2, 0}},
 			ans41{3},
 		},
 
-		question41{
+		{
 			para41{[]int{3, 4, -1, 1}},
 			ans41{2},
 		},
diff --git a/Algorithms/0041. First-Missing-Positive/README.md b/leetcode/0041.First-Missing-Positive/README.md
similarity index 100%
rename from Algorithms/0041. First-Missing-Positive/README.md
rename to leetcode/0041.First-Missing-Positive/README.md
diff --git a/Algorithms/0042. Trapping Rain Water/42. Trapping Rain Water.go b/leetcode/0042.Trapping-Rain-Water/42. Trapping Rain Water.go
similarity index 100%
rename from Algorithms/0042. Trapping Rain Water/42. Trapping Rain Water.go
rename to leetcode/0042.Trapping-Rain-Water/42. Trapping Rain Water.go
diff --git a/Algorithms/0042. Trapping Rain Water/42. Trapping Rain Water_test.go b/leetcode/0042.Trapping-Rain-Water/42. Trapping Rain Water_test.go
similarity index 97%
rename from Algorithms/0042. Trapping Rain Water/42. Trapping Rain Water_test.go
rename to leetcode/0042.Trapping-Rain-Water/42. Trapping Rain Water_test.go
index a6dc5684d..918f5ebb2 100644
--- a/Algorithms/0042. Trapping Rain Water/42. Trapping Rain Water_test.go	
+++ b/leetcode/0042.Trapping-Rain-Water/42. Trapping Rain Water_test.go	
@@ -26,7 +26,7 @@ func Test_Problem42(t *testing.T) {
 
 	qs := []question42{
 
-		question42{
+		{
 			para42{[]int{0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1}},
 			ans42{6},
 		},
diff --git a/leetcode/0042.Trapping-Rain-Water/README.md b/leetcode/0042.Trapping-Rain-Water/README.md
new file mode 100644
index 000000000..9bafaa2d2
--- /dev/null
+++ b/leetcode/0042.Trapping-Rain-Water/README.md
@@ -0,0 +1,31 @@
+# [42. Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water/)
+
+## 题目
+
+Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
+
+![](https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png)
+
+
+The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!
+
+
+**Example**:
+
+```go
+Input: [0,1,0,2,1,0,1,3,2,1,2,1]
+Output: 6
+```
+
+## 题目大意
+
+从 x 轴开始，给出一个数组，数组里面的数字代表从 (0,0) 点开始，宽度为 1 个单位，高度为数组元素的值。如果下雨了，问这样一个容器能装多少单位的水？
+
+## 解题思路
+
+- 每个数组里面的元素值可以想象成一个左右都有壁的圆柱筒。例如下图中左边的第二个元素 1，当前左边最大的元素是 2 ，所以 2 高度的水会装到 1 的上面(因为想象成了左右都有筒壁)。这道题的思路就是左指针从 0 开始往右扫，右指针从最右边开始往左扫。额外还需要 2 个变量分别记住左边最大的高度和右边最大高度。遍历扫数组元素的过程中，如果左指针的高度比右指针的高度小，就不断的移动左指针，否则移动右指针。循环的终止条件就是左右指针碰上以后就结束。只要数组中元素的高度比保存的局部最大高度小，就累加 res 的值，否则更新局部最大高度。最终解就是 res 的值。
+    ![](https://image.ibb.co/d6A2ZU/IMG-0139.jpg)  
+- 抽象一下，本题是想求针对每个 i，找到它左边最大值 leftMax，右边的最大值 rightMax，然后 min(leftMax，rightMax) 为能够接到水的高度。left 和 right 指针是两边往中间移动的游标指针。最傻的解题思路是针对每个下标 i，往左循环找到第一个最大值，往右循环找到第一个最大值，然后把这两个最大值取出最小者，即为当前雨水的高度。这样做时间复杂度高，浪费了很多循环。i 在从左往右的过程中，是可以动态维护最大值的。右边的最大值用右边的游标指针来维护。从左往右扫一遍下标，和，从两边往中间遍历一遍下标，是相同的结果，每个下标都遍历了一次。    
+    ![](https://img.halfrost.com/Leetcode/leetcode_42_1.png)  
+- 每个 i 的宽度固定为 1，所以每个“坑”只需要求出高度，即当前这个“坑”能积攒的雨水。最后依次将每个“坑”中的雨水相加即是能接到的雨水数。    
+    ![](https://img.halfrost.com/Leetcode/leetcode_42_0.png)  
\ No newline at end of file
diff --git a/leetcode/0043.Multiply-Strings/43. Multiply Strings.go b/leetcode/0043.Multiply-Strings/43. Multiply Strings.go
new file mode 100644
index 000000000..88ab4b4c7
--- /dev/null
+++ b/leetcode/0043.Multiply-Strings/43. Multiply Strings.go	
@@ -0,0 +1,25 @@
+package leetcode
+
+func multiply(num1 string, num2 string) string {
+	if num1 == "0" || num2 == "0" {
+		return "0"
+	}
+	b1, b2, tmp := []byte(num1), []byte(num2), make([]int, len(num1)+len(num2))
+	for i := 0; i < len(b1); i++ {
+		for j := 0; j < len(b2); j++ {
+			tmp[i+j+1] += int(b1[i]-'0') * int(b2[j]-'0')
+		}
+	}
+	for i := len(tmp) - 1; i > 0; i-- {
+		tmp[i-1] += tmp[i] / 10
+		tmp[i] = tmp[i] % 10
+	}
+	if tmp[0] == 0 {
+		tmp = tmp[1:]
+	}
+	res := make([]byte, len(tmp))
+	for i := 0; i < len(tmp); i++ {
+		res[i] = '0' + byte(tmp[i])
+	}
+	return string(res)
+}
diff --git a/leetcode/0043.Multiply-Strings/43. Multiply Strings_test.go b/leetcode/0043.Multiply-Strings/43. Multiply Strings_test.go
new file mode 100644
index 000000000..6017786e0
--- /dev/null
+++ b/leetcode/0043.Multiply-Strings/43. Multiply Strings_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question43 struct {
+	para43
+	ans43
+}
+
+// para 是参数
+// one 代表第一个参数
+type para43 struct {
+	num1 string
+	num2 string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans43 struct {
+	one string
+}
+
+func Test_Problem43(t *testing.T) {
+
+	qs := []question43{
+
+		{
+			para43{"2", "3"},
+			ans43{"6"},
+		},
+
+		{
+			para43{"123", "456"},
+			ans43{"56088"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 43------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans43, q.para43
+		fmt.Printf("【input】:%v       【output】:%v\n", p, multiply(p.num1, p.num2))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0043.Multiply-Strings/README.md b/leetcode/0043.Multiply-Strings/README.md
new file mode 100644
index 000000000..46410bd38
--- /dev/null
+++ b/leetcode/0043.Multiply-Strings/README.md
@@ -0,0 +1,66 @@
+# [43. Multiply Strings](https://leetcode.com/problems/multiply-strings/)
+
+
+## 题目
+
+Given two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.
+
+**Note:** You must not use any built-in BigInteger library or convert the inputs to integer directly.
+
+**Example 1:**
+
+```
+Input: num1 = "2", num2 = "3"
+Output: "6"
+```
+
+**Example 2:**
+
+```
+Input: num1 = "123", num2 = "456"
+Output: "56088"
+```
+
+**Constraints:**
+
+- `1 <= num1.length, num2.length <= 200`
+- `num1` and `num2` consist of digits only.
+- Both `num1` and `num2` do not contain any leading zero, except the number `0` itself.
+
+## 题目大意
+
+给定两个以字符串形式表示的非负整数 num1 和 num2，返回 num1 和 num2 的乘积，它们的乘积也表示为字符串形式。
+
+## 解题思路
+
+- 用数组模拟乘法。创建一个数组长度为 `len(num1) + len(num2)` 的数组用于存储乘积。对于任意 `0 ≤ i < len(num1)`，`0 ≤ j < len(num2)`，`num1[i] * num2[j]` 的结果位于 `tmp[i+j+1]`，如果 `tmp[i+j+1]≥10`，则将进位部分加到 `tmp[i+j]`。最后，将数组 `tmp` 转成字符串，如果最高位是 0 则舍弃最高位。
+
+## 代码
+
+```go
+package leetcode
+
+func multiply(num1 string, num2 string) string {
+	if num1 == "0" || num2 == "0" {
+		return "0"
+	}
+	b1, b2, tmp := []byte(num1), []byte(num2), make([]int, len(num1)+len(num2))
+	for i := 0; i < len(b1); i++ {
+		for j := 0; j < len(b2); j++ {
+			tmp[i+j+1] += int(b1[i]-'0') * int(b2[j]-'0')
+		}
+	}
+	for i := len(tmp) - 1; i > 0; i-- {
+		tmp[i-1] += tmp[i] / 10
+		tmp[i] = tmp[i] % 10
+	}
+	if tmp[0] == 0 {
+		tmp = tmp[1:]
+	}
+	res := make([]byte, len(tmp))
+	for i := 0; i < len(tmp); i++ {
+		res[i] = '0' + byte(tmp[i])
+	}
+	return string(res)
+}
+```
\ No newline at end of file
diff --git a/leetcode/0045.Jump-Game-II/45. Jump Game II.go b/leetcode/0045.Jump-Game-II/45. Jump Game II.go
new file mode 100644
index 000000000..2707e5dbd
--- /dev/null
+++ b/leetcode/0045.Jump-Game-II/45. Jump Game II.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+func jump(nums []int) int {
+	if len(nums) == 1 {
+		return 0
+	}
+	needChoose, canReach, step := 0, 0, 0
+	for i, x := range nums {
+		if i+x > canReach {
+			canReach = i + x
+			if canReach >= len(nums)-1 {
+				return step + 1
+			}
+		}
+		if i == needChoose {
+			needChoose = canReach
+			step++
+		}
+	}
+	return step
+}
diff --git a/leetcode/0045.Jump-Game-II/45. Jump Game II_test.go b/leetcode/0045.Jump-Game-II/45. Jump Game II_test.go
new file mode 100644
index 000000000..0e58e91ae
--- /dev/null
+++ b/leetcode/0045.Jump-Game-II/45. Jump Game II_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question45 struct {
+	para45
+	ans45
+}
+
+// para 是参数
+// one 代表第一个参数
+type para45 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans45 struct {
+	one int
+}
+
+func Test_Problem45(t *testing.T) {
+
+	qs := []question45{
+
+		{
+			para45{[]int{2, 3, 1, 1, 4}},
+			ans45{2},
+		},
+
+		{
+			para45{[]int{2, 3, 0, 1, 4}},
+			ans45{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 45------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans45, q.para45
+		fmt.Printf("【input】:%v       【output】:%v\n", p, jump(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0045.Jump-Game-II/README.md b/leetcode/0045.Jump-Game-II/README.md
new file mode 100644
index 000000000..e5394aa22
--- /dev/null
+++ b/leetcode/0045.Jump-Game-II/README.md
@@ -0,0 +1,67 @@
+# [45. Jump Game II](https://leetcode.com/problems/jump-game-ii/)
+
+
+## 题目
+
+Given an array of non-negative integers `nums`, you are initially positioned at the first index of the array.
+
+Each element in the array represents your maximum jump length at that position.
+
+Your goal is to reach the last index in the minimum number of jumps.
+
+You can assume that you can always reach the last index.
+
+**Example 1:**
+
+```
+Input: nums = [2,3,1,1,4]
+Output: 2
+Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,3,0,1,4]
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 1000`
+- `0 <= nums[i] <= 10^5`
+
+## 题目大意
+
+给定一个非负整数数组，你最初位于数组的第一个位置。数组中的每个元素代表你在该位置可以跳跃的最大长度。你的目标是使用最少的跳跃次数到达数组的最后一个位置。
+
+## 解题思路
+
+- 要求找到最少跳跃次数，顺理成章的会想到用贪心算法解题。扫描步数数组，维护当前能够到达最大下标的位置，记为能到达的最远边界，如果扫描过程中到达了最远边界，更新边界并将跳跃次数 + 1。
+- 扫描数组的时候，其实不需要扫描最后一个元素，因为在跳到最后一个元素之前，能到达的最远边界一定大于等于最后一个元素的位置，不然就跳不到最后一个元素，到达不了终点了；如果遍历到最后一个元素，说明边界正好为最后一个位置，最终跳跃次数直接 + 1 即可，也不需要访问最后一个元素。
+
+## 代码
+
+```go
+package leetcode
+
+func jump(nums []int) int {
+	if len(nums) == 1 {
+		return 0
+	}
+	needChoose, canReach, step := 0, 0, 0
+	for i, x := range nums {
+		if i+x > canReach {
+			canReach = i + x
+			if canReach >= len(nums)-1 {
+				return step + 1
+			}
+		}
+		if i == needChoose {
+			needChoose = canReach
+			step++
+		}
+	}
+	return step
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0046. Permutations/46. Permutations.go b/leetcode/0046.Permutations/46. Permutations.go
similarity index 100%
rename from Algorithms/0046. Permutations/46. Permutations.go
rename to leetcode/0046.Permutations/46. Permutations.go
diff --git a/Algorithms/0046. Permutations/46. Permutations_test.go b/leetcode/0046.Permutations/46. Permutations_test.go
similarity index 81%
rename from Algorithms/0046. Permutations/46. Permutations_test.go
rename to leetcode/0046.Permutations/46. Permutations_test.go
index 74938e151..d90a4e9c3 100644
--- a/Algorithms/0046. Permutations/46. Permutations_test.go	
+++ b/leetcode/0046.Permutations/46. Permutations_test.go	
@@ -26,9 +26,9 @@ func Test_Problem46(t *testing.T) {
 
 	qs := []question46{
 
-		question46{
+		{
 			para46{[]int{1, 2, 3}},
-			ans46{[][]int{[]int{1, 2, 3}, []int{1, 3, 2}, []int{2, 1, 3}, []int{2, 3, 1}, []int{3, 1, 2}, []int{3, 2, 1}}},
+			ans46{[][]int{{1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1}}},
 		},
 	}
 
diff --git a/Algorithms/0046. Permutations/README.md b/leetcode/0046.Permutations/README.md
similarity index 100%
rename from Algorithms/0046. Permutations/README.md
rename to leetcode/0046.Permutations/README.md
diff --git a/Algorithms/0047. Permutations II/47. Permutations II.go b/leetcode/0047.Permutations-II/47. Permutations II.go
similarity index 100%
rename from Algorithms/0047. Permutations II/47. Permutations II.go
rename to leetcode/0047.Permutations-II/47. Permutations II.go
diff --git a/Algorithms/0047. Permutations II/47. Permutations II_test.go b/leetcode/0047.Permutations-II/47. Permutations II_test.go
similarity index 75%
rename from Algorithms/0047. Permutations II/47. Permutations II_test.go
rename to leetcode/0047.Permutations-II/47. Permutations II_test.go
index 233930756..f997b9a9d 100644
--- a/Algorithms/0047. Permutations II/47. Permutations II_test.go	
+++ b/leetcode/0047.Permutations-II/47. Permutations II_test.go	
@@ -26,19 +26,19 @@ func Test_Problem47(t *testing.T) {
 
 	qs := []question47{
 
-		question47{
+		{
 			para47{[]int{1, 1, 2}},
-			ans47{[][]int{[]int{1, 1, 2}, []int{1, 2, 1}, []int{2, 1, 1}}},
+			ans47{[][]int{{1, 1, 2}, {1, 2, 1}, {2, 1, 1}}},
 		},
 
-		question47{
+		{
 			para47{[]int{1, 2, 2}},
-			ans47{[][]int{[]int{1, 2, 2}, []int{2, 2, 1}, []int{2, 1, 2}}},
+			ans47{[][]int{{1, 2, 2}, {2, 2, 1}, {2, 1, 2}}},
 		},
 
-		question47{
+		{
 			para47{[]int{2, 2, 2}},
-			ans47{[][]int{[]int{2, 2, 2}}},
+			ans47{[][]int{{2, 2, 2}}},
 		},
 	}
 
diff --git a/Algorithms/0047. Permutations II/README.md b/leetcode/0047.Permutations-II/README.md
similarity index 100%
rename from Algorithms/0047. Permutations II/README.md
rename to leetcode/0047.Permutations-II/README.md
diff --git a/leetcode/0048.Rotate-Image/48. Rotate Image.go b/leetcode/0048.Rotate-Image/48. Rotate Image.go
new file mode 100644
index 000000000..f27e91e35
--- /dev/null
+++ b/leetcode/0048.Rotate-Image/48. Rotate Image.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+// 解法一
+func rotate(matrix [][]int) {
+	length := len(matrix)
+	// rotate by diagonal 对角线变换
+	for i := 0; i < length; i++ {
+		for j := i + 1; j < length; j++ {
+			matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
+		}
+	}
+	// rotate by vertical centerline 竖直轴对称翻转
+	for i := 0; i < length; i++ {
+		for j := 0; j < length/2; j++ {
+			matrix[i][j], matrix[i][length-j-1] = matrix[i][length-j-1], matrix[i][j]
+		}
+	}
+}
+
+// 解法二
+func rotate1(matrix [][]int) {
+	n := len(matrix)
+	if n == 1 {
+		return
+	}
+	/* rotate clock-wise = 1. transpose matrix => 2. reverse(matrix[i])
+
+	1   2  3  4      1   5  9  13        13  9  5  1
+	5   6  7  8  =>  2   6  10 14  =>    14  10 6  2
+	9  10 11 12      3   7  11 15        15  11 7  3
+	13 14 15 16      4   8  12 16        16  12 8  4
+
+	*/
+
+	for i := 0; i < n; i++ {
+		// transpose, i=rows, j=columns
+		// j = i+1, coz diagonal elements didn't change in a square matrix
+		for j := i + 1; j < n; j++ {
+			swap(matrix, i, j)
+		}
+		// reverse each row of the image
+		matrix[i] = reverse(matrix[i])
+	}
+}
+
+// swap changes original slice's i,j position
+func swap(nums [][]int, i, j int) {
+	nums[i][j], nums[j][i] = nums[j][i], nums[i][j]
+}
+
+// reverses a row of image, matrix[i]
+func reverse(nums []int) []int {
+	var lp, rp = 0, len(nums) - 1
+
+	for lp < rp {
+		nums[lp], nums[rp] = nums[rp], nums[lp]
+		lp++
+		rp--
+	}
+	return nums
+}
diff --git a/leetcode/0048.Rotate-Image/48. Rotate Image_test.go b/leetcode/0048.Rotate-Image/48. Rotate Image_test.go
new file mode 100644
index 000000000..76f8f704a
--- /dev/null
+++ b/leetcode/0048.Rotate-Image/48. Rotate Image_test.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question48 struct {
+	para48
+	ans48
+}
+
+// para 是参数
+// one 代表第一个参数
+type para48 struct {
+	s [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans48 struct {
+	s [][]int
+}
+
+func Test_Problem48(t *testing.T) {
+
+	qs := []question48{
+
+		{
+			para48{[][]int{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}},
+			ans48{[][]int{{7, 4, 1}, {8, 5, 2}, {9, 6, 3}}},
+		},
+
+		{
+			para48{[][]int{{5, 1, 9, 11}, {2, 4, 8, 10}, {13, 3, 6, 7}, {15, 14, 12, 16}}},
+			ans48{[][]int{{15, 13, 2, 5}, {14, 3, 4, 1}, {12, 6, 8, 9}, {16, 7, 10, 11}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 48------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans48, q.para48
+		fmt.Printf("【input】:%v \n", p)
+		rotate(p.s)
+		fmt.Printf("【output】:%v\n", p)
+		fmt.Printf("\n")
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0048.Rotate-Image/README.md b/leetcode/0048.Rotate-Image/README.md
new file mode 100755
index 000000000..1a07a97d9
--- /dev/null
+++ b/leetcode/0048.Rotate-Image/README.md
@@ -0,0 +1,93 @@
+# [48. Rotate Image](https://leetcode.com/problems/rotate-image/)
+
+## 题目
+
+You are given an *n* x *n* 2D matrix representing an image.
+
+Rotate the image by 90 degrees (clockwise).
+
+**Note:**
+
+You have to rotate the image **[in-place](https://en.wikipedia.org/wiki/In-place_algorithm)**, which means you have to modify the input 2D matrix directly. **DO NOT** allocate another 2D matrix and do the rotation.
+
+**Example 1**:
+
+![](https://assets.leetcode.com/uploads/2020/08/28/mat1.jpg)
+
+    Given input matrix = 
+    [
+      [1,2,3],
+      [4,5,6],
+      [7,8,9]
+    ],
+    
+    rotate the input matrix in-place such that it becomes:
+    [
+      [7,4,1],
+      [8,5,2],
+      [9,6,3]
+    ]
+
+
+**Example 2**:
+
+![](https://assets.leetcode.com/uploads/2020/08/28/mat2.jpg)
+
+    Given input matrix =
+    [
+      [ 5, 1, 9,11],
+      [ 2, 4, 8,10],
+      [13, 3, 6, 7],
+      [15,14,12,16]
+    ], 
+    
+    rotate the input matrix in-place such that it becomes:
+    [
+      [15,13, 2, 5],
+      [14, 3, 4, 1],
+      [12, 6, 8, 9],
+      [16, 7,10,11]
+    ]
+
+
+## 题目大意
+
+给定一个 n × n 的二维矩阵表示一个图像。将图像顺时针旋转 90 度。说明：你必须在原地旋转图像，这意味着你需要直接修改输入的二维矩阵。请不要使用另一个矩阵来旋转图像。
+
+
+## 解题思路
+
+- 给出一个二维数组，要求顺时针旋转 90 度。
+- 这一题比较简单，按照题意做就可以。这里给出 2 种旋转方法的实现，顺时针旋转和逆时针旋转。
+
+```c
+    /*
+     * clockwise rotate 顺时针旋转
+     * first reverse up to down, then swap the symmetry 
+     * 1 2 3     7 8 9     7 4 1
+     * 4 5 6  => 4 5 6  => 8 5 2
+     * 7 8 9     1 2 3     9 6 3
+    */
+    void rotate(vector<vector<int> > &matrix) {
+        reverse(matrix.begin(), matrix.end());
+        for (int i = 0; i < matrix.size(); ++i) {
+            for (int j = i + 1; j < matrix[i].size(); ++j)
+                swap(matrix[i][j], matrix[j][i]);
+        }
+    }
+    
+    /*
+     * anticlockwise rotate 逆时针旋转
+     * first reverse left to right, then swap the symmetry
+     * 1 2 3     3 2 1     3 6 9
+     * 4 5 6  => 6 5 4  => 2 5 8
+     * 7 8 9     9 8 7     1 4 7
+    */
+    void anti_rotate(vector<vector<int> > &matrix) {
+        for (auto vi : matrix) reverse(vi.begin(), vi.end());
+        for (int i = 0; i < matrix.size(); ++i) {
+            for (int j = i + 1; j < matrix[i].size(); ++j)
+                swap(matrix[i][j], matrix[j][i]);
+        }
+    }
+```
\ No newline at end of file
diff --git a/Algorithms/0049. Group Anagrams/49. Group Anagrams.go b/leetcode/0049.Group-Anagrams/49. Group Anagrams.go
similarity index 100%
rename from Algorithms/0049. Group Anagrams/49. Group Anagrams.go
rename to leetcode/0049.Group-Anagrams/49. Group Anagrams.go
diff --git a/Algorithms/0049. Group Anagrams/49. Group Anagrams_test.go b/leetcode/0049.Group-Anagrams/49. Group Anagrams_test.go
similarity index 85%
rename from Algorithms/0049. Group Anagrams/49. Group Anagrams_test.go
rename to leetcode/0049.Group-Anagrams/49. Group Anagrams_test.go
index f7c537cc3..bee3e303e 100644
--- a/Algorithms/0049. Group Anagrams/49. Group Anagrams_test.go	
+++ b/leetcode/0049.Group-Anagrams/49. Group Anagrams_test.go	
@@ -26,9 +26,9 @@ func Test_Problem49(t *testing.T) {
 
 	qs := []question49{
 
-		question49{
+		{
 			para49{[]string{"eat", "tea", "tan", "ate", "nat", "bat"}},
-			ans49{[][]string{[]string{"ate", "eat", "tea"}, []string{"nat", "tan"}, []string{"bat"}}},
+			ans49{[][]string{{"ate", "eat", "tea"}, {"nat", "tan"}, {"bat"}}},
 		},
 	}
 
diff --git a/Algorithms/0049. Group Anagrams/README.md b/leetcode/0049.Group-Anagrams/README.md
similarity index 100%
rename from Algorithms/0049. Group Anagrams/README.md
rename to leetcode/0049.Group-Anagrams/README.md
diff --git a/Algorithms/0050. Pow(x, n)/50. Pow(x, n).go b/leetcode/0050.Powx-n/50. Pow(x, n).go
similarity index 100%
rename from Algorithms/0050. Pow(x, n)/50. Pow(x, n).go
rename to leetcode/0050.Powx-n/50. Pow(x, n).go
diff --git a/Algorithms/0050. Pow(x, n)/50. Pow(x, n)_test.go b/leetcode/0050.Powx-n/50. Pow(x, n)_test.go
similarity index 94%
rename from Algorithms/0050. Pow(x, n)/50. Pow(x, n)_test.go
rename to leetcode/0050.Powx-n/50. Pow(x, n)_test.go
index b542b4469..2558a1b66 100644
--- a/Algorithms/0050. Pow(x, n)/50. Pow(x, n)_test.go	
+++ b/leetcode/0050.Powx-n/50. Pow(x, n)_test.go	
@@ -27,17 +27,17 @@ func Test_Problem50(t *testing.T) {
 
 	qs := []question50{
 
-		question50{
+		{
 			para50{2.00000, 10},
 			ans50{1024.00000},
 		},
 
-		question50{
+		{
 			para50{2.10000, 3},
 			ans50{9.26100},
 		},
 
-		question50{
+		{
 			para50{2.00000, -2},
 			ans50{0.25000},
 		},
diff --git a/Algorithms/0050. Pow(x, n)/README.md b/leetcode/0050.Powx-n/README.md
similarity index 100%
rename from Algorithms/0050. Pow(x, n)/README.md
rename to leetcode/0050.Powx-n/README.md
diff --git a/leetcode/0051.N-Queens/51. N-Queens.go b/leetcode/0051.N-Queens/51. N-Queens.go
new file mode 100644
index 000000000..ba63a9100
--- /dev/null
+++ b/leetcode/0051.N-Queens/51. N-Queens.go	
@@ -0,0 +1,91 @@
+package leetcode
+
+// 解法一 DFS
+func solveNQueens(n int) [][]string {
+	col, dia1, dia2, row, res := make([]bool, n), make([]bool, 2*n-1), make([]bool, 2*n-1), []int{}, [][]string{}
+	putQueen(n, 0, &col, &dia1, &dia2, &row, &res)
+	return res
+}
+
+// 尝试在一个n皇后问题中, 摆放第index行的皇后位置
+func putQueen(n, index int, col, dia1, dia2 *[]bool, row *[]int, res *[][]string) {
+	if index == n {
+		*res = append(*res, generateBoard(n, row))
+		return
+	}
+	for i := 0; i < n; i++ {
+		// 尝试将第index行的皇后摆放在第i列
+		if !(*col)[i] && !(*dia1)[index+i] && !(*dia2)[index-i+n-1] {
+			*row = append(*row, i)
+			(*col)[i] = true
+			(*dia1)[index+i] = true
+			(*dia2)[index-i+n-1] = true
+			putQueen(n, index+1, col, dia1, dia2, row, res)
+			(*col)[i] = false
+			(*dia1)[index+i] = false
+			(*dia2)[index-i+n-1] = false
+			*row = (*row)[:len(*row)-1]
+		}
+	}
+	return
+}
+
+func generateBoard(n int, row *[]int) []string {
+	board := []string{}
+	res := ""
+	for i := 0; i < n; i++ {
+		res += "."
+	}
+	for i := 0; i < n; i++ {
+		board = append(board, res)
+	}
+	for i := 0; i < n; i++ {
+		tmp := []byte(board[i])
+		tmp[(*row)[i]] = 'Q'
+		board[i] = string(tmp)
+	}
+	return board
+}
+
+// 解法二 二进制操作法 Signed-off-by: Hanlin Shi shihanlin9@gmail.com
+func solveNQueens2(n int) (res [][]string) {
+	placements := make([]string, n)
+	for i := range placements {
+		buf := make([]byte, n)
+		for j := range placements {
+			if i == j {
+				buf[j] = 'Q'
+			} else {
+				buf[j] = '.'
+			}
+		}
+		placements[i] = string(buf)
+	}
+	var construct func(prev []int)
+	construct = func(prev []int) {
+		if len(prev) == n {
+			plan := make([]string, n)
+			for i := 0; i < n; i++ {
+				plan[i] = placements[prev[i]]
+			}
+			res = append(res, plan)
+			return
+		}
+		occupied := 0
+		for i := range prev {
+			dist := len(prev) - i
+			bit := 1 << prev[i]
+			occupied |= bit | bit<<dist | bit>>dist
+		}
+		prev = append(prev, -1)
+		for i := 0; i < n; i++ {
+			if (occupied>>i)&1 != 0 {
+				continue
+			}
+			prev[len(prev)-1] = i
+			construct(prev)
+		}
+	}
+	construct(make([]int, 0, n))
+	return
+}
diff --git a/Algorithms/0051. N-Queens/51. N-Queens_test.go b/leetcode/0051.N-Queens/51. N-Queens_test.go
similarity index 92%
rename from Algorithms/0051. N-Queens/51. N-Queens_test.go
rename to leetcode/0051.N-Queens/51. N-Queens_test.go
index ad6653283..2f6cfd172 100644
--- a/Algorithms/0051. N-Queens/51. N-Queens_test.go	
+++ b/leetcode/0051.N-Queens/51. N-Queens_test.go	
@@ -26,14 +26,14 @@ func Test_Problem51(t *testing.T) {
 
 	qs := []question51{
 
-		question51{
+		{
 			para51{4},
 			ans51{[][]string{
-				[]string{".Q..",
+				{".Q..",
 					"...Q",
 					"Q...",
 					"..Q."},
-				[]string{"..Q.",
+				{"..Q.",
 					"Q...",
 					"...Q",
 					".Q.."},
diff --git a/leetcode/0051.N-Queens/README.md b/leetcode/0051.N-Queens/README.md
new file mode 100755
index 000000000..6fcabdee5
--- /dev/null
+++ b/leetcode/0051.N-Queens/README.md
@@ -0,0 +1,44 @@
+# [51. N-Queens](https://leetcode.com/problems/n-queens/)
+
+
+## 题目
+
+The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
+
+![](https://assets.leetcode.com/uploads/2018/10/12/8-queens.png)
+
+Given an integer *n*, return all distinct solutions to the *n*-queens puzzle.
+
+Each solution contains a distinct board configuration of the *n*-queens' placement, where `'Q'` and `'.'` both indicate a queen and an empty space respectively.
+
+**Example:**
+
+
+    Input: 4
+    Output: [
+     [".Q..",  // Solution 1
+      "...Q",
+      "Q...",
+      "..Q."],
+    
+     ["..Q.",  // Solution 2
+      "Q...",
+      "...Q",
+      ".Q.."]
+    ]
+    Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.
+
+
+## 题目大意
+
+给定一个整数 n，返回所有不同的 n 皇后问题的解决方案。每一种解法包含一个明确的 n 皇后问题的棋子放置方案，该方案中 'Q' 和 '.' 分别代表了皇后和空位。
+
+
+## 解题思路
+
+- 求解 n 皇后问题
+- 利用 col 数组记录列信息，col 有 `n` 列。用 dia1，dia2 记录从左下到右上的对角线，从左上到右下的对角线的信息，dia1 和 dia2 分别都有 `2*n-1` 个。
+- dia1 对角线的规律是 `i + j 是定值`，例如[0,0]，为 0；[1,0]、[0,1] 为 1；[2,0]、[1,1]、[0,2] 为 2；
+- dia2 对角线的规律是 `i - j 是定值`，例如[0,7]，为 -7；[0,6]、[1,7] 为 -6；[0,5]、[1,6]、[2,7] 为 -5；为了使他们从 0 开始，i - j + n - 1 偏移到 0 开始，所以 dia2 的规律是 `i - j + n - 1 为定值`。
+- 还有一个位运算的方法，每行只能选一个位置放皇后，那么对每行遍历可能放皇后的位置。如何高效判断哪些点不能放皇后呢？这里的做法毕竟巧妙，把所有之前选过的点按照顺序存下来，然后根据之前选的点到当前行的距离，就可以快速判断是不是会有冲突。举个例子: 假如在 4 皇后问题中，如果第一二行已经选择了位置 [1, 3]，那么在第三行选择时，首先不能再选 1, 3 列了，而对于第三行， 1 距离长度为2，所以它会影响到 -1, 3 两个列。同理，3 在第二行，距离第三行为 1，所以 3 会影响到列 2, 4。由上面的结果，我们知道 -1, 4 超出边界了不用去管，别的不能选的点是 1, 2, 3，所以第三行就只能选 0。在代码实现中，可以在每次遍历前根据之前选择的情况生成一个 occupied 用来记录当前这一行，已经被选了的和由于之前皇后攻击范围所以不能选的位置，然后只选择合法的位置进入到下一层递归。另外就是预处理了一个皇后放不同位置的字符串，这样这些字符串在返回结果的时候是可以在内存中复用的，省一点内存。
+
diff --git a/Algorithms/0052. N-Queens II/52. N-Queens II.go b/leetcode/0052.N-Queens-II/52. N-Queens II.go
similarity index 100%
rename from Algorithms/0052. N-Queens II/52. N-Queens II.go
rename to leetcode/0052.N-Queens-II/52. N-Queens II.go
diff --git a/Algorithms/0052. N-Queens II/52. N-Queens II_test.go b/leetcode/0052.N-Queens-II/52. N-Queens II_test.go
similarity index 86%
rename from Algorithms/0052. N-Queens II/52. N-Queens II_test.go
rename to leetcode/0052.N-Queens-II/52. N-Queens II_test.go
index 672af899a..a906c2a64 100644
--- a/Algorithms/0052. N-Queens II/52. N-Queens II_test.go	
+++ b/leetcode/0052.N-Queens-II/52. N-Queens II_test.go	
@@ -26,51 +26,51 @@ func Test_Problem52(t *testing.T) {
 
 	qs := []question52{
 
-		question52{
+		{
 			para52{1},
 			ans52{1},
 		},
 
-		question52{
+		{
 			para52{2},
 			ans52{0},
 		},
-		question52{
+		{
 			para52{3},
 			ans52{0},
 		},
 
-		question52{
+		{
 			para52{4},
 			ans52{2},
 		},
 
-		question52{
+		{
 			para52{5},
 			ans52{10},
 		},
 
-		question52{
+		{
 			para52{6},
 			ans52{4},
 		},
 
-		question52{
+		{
 			para52{7},
 			ans52{40},
 		},
 
-		question52{
+		{
 			para52{8},
 			ans52{92},
 		},
 
-		question52{
+		{
 			para52{9},
 			ans52{352},
 		},
 
-		question52{
+		{
 			para52{10},
 			ans52{724},
 		},
diff --git a/Algorithms/0052. N-Queens II/README.md b/leetcode/0052.N-Queens-II/README.md
similarity index 100%
rename from Algorithms/0052. N-Queens II/README.md
rename to leetcode/0052.N-Queens-II/README.md
diff --git a/Algorithms/0053. Maximum Subarray/53. Maximum Subarray.go b/leetcode/0053.Maximum-Subarray/53. Maximum Subarray.go
similarity index 90%
rename from Algorithms/0053. Maximum Subarray/53. Maximum Subarray.go
rename to leetcode/0053.Maximum-Subarray/53. Maximum Subarray.go
index 03e3e2270..8db251a2d 100644
--- a/Algorithms/0053. Maximum Subarray/53. Maximum Subarray.go	
+++ b/leetcode/0053.Maximum-Subarray/53. Maximum Subarray.go	
@@ -39,3 +39,10 @@ func maxSubArray1(nums []int) int {
 	}
 	return maxSum
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0053. Maximum Subarray/53. Maximum Subarray_test.go b/leetcode/0053.Maximum-Subarray/53. Maximum Subarray_test.go
similarity index 93%
rename from Algorithms/0053. Maximum Subarray/53. Maximum Subarray_test.go
rename to leetcode/0053.Maximum-Subarray/53. Maximum Subarray_test.go
index 04e07b05e..0a99d22fc 100644
--- a/Algorithms/0053. Maximum Subarray/53. Maximum Subarray_test.go	
+++ b/leetcode/0053.Maximum-Subarray/53. Maximum Subarray_test.go	
@@ -26,21 +26,21 @@ func Test_Problem53(t *testing.T) {
 
 	qs := []question53{
 
-		question53{
+		{
 			para53{[]int{-2, 1, -3, 4, -1, 2, 1, -5, 4}},
 			ans53{6},
 		},
-		question53{
+		{
 			para53{[]int{2, 7, 9, 3, 1}},
 			ans53{22},
 		},
 
-		question53{
+		{
 			para53{[]int{2}},
 			ans53{2},
 		},
 
-		question53{
+		{
 			para53{[]int{-1, -2}},
 			ans53{-1},
 		},
diff --git a/Algorithms/0053. Maximum Subarray/README.md b/leetcode/0053.Maximum-Subarray/README.md
similarity index 100%
rename from Algorithms/0053. Maximum Subarray/README.md
rename to leetcode/0053.Maximum-Subarray/README.md
diff --git a/leetcode/0054.Spiral-Matrix/54. Spiral Matrix.go b/leetcode/0054.Spiral-Matrix/54. Spiral Matrix.go
new file mode 100644
index 000000000..93bb35646
--- /dev/null
+++ b/leetcode/0054.Spiral-Matrix/54. Spiral Matrix.go	
@@ -0,0 +1,118 @@
+package leetcode
+
+// 解法 1
+func spiralOrder(matrix [][]int) []int {
+	if len(matrix) == 0 {
+		return []int{}
+	}
+	res := []int{}
+	if len(matrix) == 1 {
+		for i := 0; i < len(matrix[0]); i++ {
+			res = append(res, matrix[0][i])
+		}
+		return res
+	}
+	if len(matrix[0]) == 1 {
+		for i := 0; i < len(matrix); i++ {
+			res = append(res, matrix[i][0])
+		}
+		return res
+	}
+	visit, m, n, round, x, y, spDir := make([][]int, len(matrix)), len(matrix), len(matrix[0]), 0, 0, 0, [][]int{
+		{0, 1},  // 朝右
+		{1, 0},  // 朝下
+		{0, -1}, // 朝左
+		{-1, 0}, // 朝上
+	}
+	for i := 0; i < m; i++ {
+		visit[i] = make([]int, n)
+	}
+	visit[x][y] = 1
+	res = append(res, matrix[x][y])
+	for i := 0; i < m*n; i++ {
+		x += spDir[round%4][0]
+		y += spDir[round%4][1]
+		if (x == 0 && y == n-1) || (x == m-1 && y == n-1) || (y == 0 && x == m-1) {
+			round++
+		}
+		if x > m-1 || y > n-1 || x < 0 || y < 0 {
+			return res
+		}
+		if visit[x][y] == 0 {
+			visit[x][y] = 1
+			res = append(res, matrix[x][y])
+		}
+		switch round % 4 {
+		case 0:
+			if y+1 <= n-1 && visit[x][y+1] == 1 {
+				round++
+				continue
+			}
+		case 1:
+			if x+1 <= m-1 && visit[x+1][y] == 1 {
+				round++
+				continue
+			}
+		case 2:
+			if y-1 >= 0 && visit[x][y-1] == 1 {
+				round++
+				continue
+			}
+		case 3:
+			if x-1 >= 0 && visit[x-1][y] == 1 {
+				round++
+				continue
+			}
+		}
+	}
+	return res
+}
+
+// 解法 2
+func spiralOrder2(matrix [][]int) []int {
+	m := len(matrix)
+	if m == 0 {
+		return nil
+	}
+
+	n := len(matrix[0])
+	if n == 0 {
+		return nil
+	}
+
+	// top、left、right、bottom 分别是剩余区域的上、左、右、下的下标
+	top, left, bottom, right := 0, 0, m-1, n-1
+	count, sum := 0, m*n
+	res := []int{}
+
+	// 外层循环每次遍历一圈
+	for count < sum {
+		i, j := top, left
+		for j <= right && count < sum {
+			res = append(res, matrix[i][j])
+			count++
+			j++
+		}
+		i, j = top+1, right
+		for i <= bottom && count < sum {
+			res = append(res, matrix[i][j])
+			count++
+			i++
+		}
+		i, j = bottom, right-1
+		for j >= left && count < sum {
+			res = append(res, matrix[i][j])
+			count++
+			j--
+		}
+		i, j = bottom-1, left
+		for i > top && count < sum {
+			res = append(res, matrix[i][j])
+			count++
+			i--
+		}
+		// 进入到下一层
+		top, left, bottom, right = top+1, left+1, bottom-1, right-1
+	}
+	return res
+}
diff --git a/leetcode/0054.Spiral-Matrix/54. Spiral Matrix_test.go b/leetcode/0054.Spiral-Matrix/54. Spiral Matrix_test.go
new file mode 100644
index 000000000..39e0d2ecb
--- /dev/null
+++ b/leetcode/0054.Spiral-Matrix/54. Spiral Matrix_test.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question54 struct {
+	para54
+	ans54
+}
+
+// para 是参数
+// one 代表第一个参数
+type para54 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans54 struct {
+	one []int
+}
+
+func Test_Problem54(t *testing.T) {
+
+	qs := []question54{
+
+		{
+			para54{[][]int{{3}, {2}}},
+			ans54{[]int{3, 2}},
+		},
+
+		{
+			para54{[][]int{{2, 3}}},
+			ans54{[]int{2, 3}},
+		},
+
+		{
+			para54{[][]int{{1}}},
+			ans54{[]int{1}},
+		},
+
+		{
+			para54{[][]int{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}},
+			ans54{[]int{1, 2, 3, 6, 9, 8, 7, 4, 5}},
+		},
+		{
+			para54{[][]int{{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}}},
+			ans54{[]int{1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 54------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans54, q.para54
+		fmt.Printf("【input】:%v       【output】:%v\n", p, spiralOrder(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0054.Spiral-Matrix/README.md b/leetcode/0054.Spiral-Matrix/README.md
new file mode 100755
index 000000000..7a6b8f1ed
--- /dev/null
+++ b/leetcode/0054.Spiral-Matrix/README.md
@@ -0,0 +1,40 @@
+# [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)
+
+
+## 题目
+
+Given a matrix of *m* x *n* elements (*m* rows, *n* columns), return all elements of the matrix in spiral order.
+
+**Example 1:**
+
+
+    Input:
+    [
+     [ 1, 2, 3 ],
+     [ 4, 5, 6 ],
+     [ 7, 8, 9 ]
+    ]
+    Output: [1,2,3,6,9,8,7,4,5]
+
+
+**Example 2:**
+
+
+    Input:
+    [
+      [1, 2, 3, 4],
+      [5, 6, 7, 8],
+      [9,10,11,12]
+    ]
+    Output: [1,2,3,4,8,12,11,10,9,5,6,7]
+
+
+## 题目大意
+
+给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。
+
+## 解题思路
+
+- 给出一个二维数组，按照螺旋的方式输出
+- 解法一：需要注意的是特殊情况，比如二维数组退化成一维或者一列或者一个元素。注意了这些情况，基本就可以一次通过了。
+- 解法二：提前算出一共多少个元素，一圈一圈地遍历矩阵，停止条件就是遍历了所有元素（count == sum）
diff --git a/leetcode/0055.Jump-Game/55. Jump Game.go b/leetcode/0055.Jump-Game/55. Jump Game.go
new file mode 100644
index 000000000..62ed18492
--- /dev/null
+++ b/leetcode/0055.Jump-Game/55. Jump Game.go	
@@ -0,0 +1,26 @@
+package leetcode
+
+func canJump(nums []int) bool {
+	n := len(nums)
+	if n == 0 {
+		return false
+	}
+	if n == 1 {
+		return true
+	}
+	maxJump := 0
+	for i, v := range nums {
+		if i > maxJump {
+			return false
+		}
+		maxJump = max(maxJump, i+v)
+	}
+	return true
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0055.Jump-Game/55. Jump Game_test.go b/leetcode/0055.Jump-Game/55. Jump Game_test.go
new file mode 100644
index 000000000..e4e47e31a
--- /dev/null
+++ b/leetcode/0055.Jump-Game/55. Jump Game_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question55 struct {
+	para55
+	ans55
+}
+
+// para 是参数
+// one 代表第一个参数
+type para55 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans55 struct {
+	one bool
+}
+
+func Test_Problem55(t *testing.T) {
+
+	qs := []question55{
+
+		{
+			para55{[]int{2, 3, 1, 1, 4}},
+			ans55{true},
+		},
+		{
+			para55{[]int{3, 2, 1, 0, 4}},
+			ans55{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 55------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans55, q.para55
+		fmt.Printf("【input】:%v       【output】:%v\n", p, canJump(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0055.Jump-Game/README.md b/leetcode/0055.Jump-Game/README.md
new file mode 100644
index 000000000..84b7ddaf1
--- /dev/null
+++ b/leetcode/0055.Jump-Game/README.md
@@ -0,0 +1,58 @@
+# [55. Jump Game](https://leetcode.com/problems/jump-game/)
+
+
+## 题目
+
+Given an array of non-negative integers, you are initially positioned at the first index of the array.
+
+Each element in the array represents your maximum jump length at that position.
+
+Determine if you are able to reach the last index.
+
+**Example 1**:
+
+```
+Input: [2,3,1,1,4]
+Output: true
+Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
+```
+
+**Example 2**:
+
+```
+Input: [3,2,1,0,4]
+Output: false
+Explanation: You will always arrive at index 3 no matter what. Its maximum
+             jump length is 0, which makes it impossible to reach the last index.
+```
+
+## 题目大意
+
+给定一个非负整数数组，最初位于数组的第一个位置。数组中的每个元素代表在该位置可以跳跃的最大长度。判断是否能够到达最后一个位置。
+
+## 解题思路
+
+- 给出一个非负数组，要求判断从数组 0 下标开始，能否到达数组最后一个位置。
+- 这一题比较简单。如果某一个作为 `起跳点` 的格子可以跳跃的距离是 `n`，那么表示后面 `n` 个格子都可以作为 `起跳点`。可以对每一个能作为 `起跳点` 的格子都尝试跳一次，把 `能跳到最远的距离maxJump` 不断更新。如果可以一直跳到最后，就成功了。如果中间有一个点比 `maxJump` 还要大，说明在这个点和 maxJump 中间连不上了，有些点不能到达最后一个位置。
+
+## 代码
+
+```go
+func canJump(nums []int) bool {
+	n := len(nums)
+	if n == 0 {
+		return false
+	}
+	if n == 1 {
+		return true
+	}
+	maxJump := 0
+	for i, v := range nums {
+		if i > maxJump {
+			return false
+		}
+		maxJump = max(maxJump, i+v)
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0056. Merge Intervals/56. Merge Intervals.go b/leetcode/0056.Merge-Intervals/56. Merge Intervals.go
similarity index 92%
rename from Algorithms/0056. Merge Intervals/56. Merge Intervals.go
rename to leetcode/0056.Merge-Intervals/56. Merge Intervals.go
index 4ce74167d..e529aa3b3 100644
--- a/Algorithms/0056. Merge Intervals/56. Merge Intervals.go	
+++ b/leetcode/0056.Merge-Intervals/56. Merge Intervals.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// Interval define
+type Interval = structures.Interval
+
 /**
  * Definition for an interval.
  * type Interval struct {
@@ -8,12 +15,6 @@ package leetcode
  * }
  */
 
-// Interval define
-type Interval struct {
-	Start int
-	End   int
-}
-
 func merge56(intervals []Interval) []Interval {
 	if len(intervals) == 0 {
 		return intervals
diff --git a/leetcode/0056.Merge-Intervals/56. Merge Intervals_test.go b/leetcode/0056.Merge-Intervals/56. Merge Intervals_test.go
new file mode 100644
index 000000000..25cd44bab
--- /dev/null
+++ b/leetcode/0056.Merge-Intervals/56. Merge Intervals_test.go	
@@ -0,0 +1,72 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question56 struct {
+	para56
+	ans56
+}
+
+// para 是参数
+// one 代表第一个参数
+type para56 struct {
+	one []Interval
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans56 struct {
+	one []Interval
+}
+
+func Test_Problem56(t *testing.T) {
+
+	qs := []question56{
+
+		{
+			para56{[]Interval{}},
+			ans56{[]Interval{}},
+		},
+
+		{
+			para56{[]Interval{{Start: 1, End: 3}, {Start: 2, End: 6}, {Start: 8, End: 10}, {Start: 15, End: 18}}},
+			ans56{[]Interval{{Start: 1, End: 6}, {Start: 8, End: 10}, {Start: 15, End: 18}}},
+		},
+
+		{
+			para56{[]Interval{{Start: 1, End: 4}, {Start: 4, End: 5}}},
+			ans56{[]Interval{{Start: 1, End: 5}}},
+		},
+
+		{
+			para56{[]Interval{{Start: 1, End: 3}, {Start: 3, End: 6}, {Start: 5, End: 10}, {Start: 9, End: 18}}},
+			ans56{[]Interval{{Start: 1, End: 18}}},
+		},
+
+		{
+			para56{[]Interval{{Start: 15, End: 18}, {Start: 8, End: 10}, {Start: 2, End: 6}, {Start: 1, End: 3}}},
+			ans56{[]Interval{{Start: 1, End: 6}, {Start: 8, End: 10}, {Start: 15, End: 18}}},
+		},
+
+		{
+			para56{[]Interval{{Start: 1, End: 3}, {Start: 2, End: 6}, {Start: 8, End: 10}, {Start: 15, End: 18}}},
+			ans56{[]Interval{{Start: 1, End: 6}, {Start: 8, End: 10}, {Start: 15, End: 18}}},
+		},
+
+		{
+			para56{[]Interval{{Start: 1, End: 4}, {Start: 1, End: 5}}},
+			ans56{[]Interval{{Start: 1, End: 5}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 56------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans56, q.para56
+		fmt.Printf("【input】:%v       【output】:%v\n", p, merge56(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0056. Merge Intervals/README.md b/leetcode/0056.Merge-Intervals/README.md
similarity index 100%
rename from Algorithms/0056. Merge Intervals/README.md
rename to leetcode/0056.Merge-Intervals/README.md
diff --git a/Algorithms/0057. Insert Interval/57. Insert Interval.go b/leetcode/0057.Insert-Interval/57. Insert Interval.go
similarity index 76%
rename from Algorithms/0057. Insert Interval/57. Insert Interval.go
rename to leetcode/0057.Insert-Interval/57. Insert Interval.go
index fa5422e36..591ee3dfd 100644
--- a/Algorithms/0057. Insert Interval/57. Insert Interval.go	
+++ b/leetcode/0057.Insert-Interval/57. Insert Interval.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// Interval define
+type Interval = structures.Interval
+
 /**
  * Definition for an interval.
  * type Interval struct {
@@ -32,3 +39,17 @@ func insert(intervals []Interval, newInterval Interval) []Interval {
 	}
 	return res
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/0057.Insert-Interval/57. Insert Interval_test.go b/leetcode/0057.Insert-Interval/57. Insert Interval_test.go
new file mode 100644
index 000000000..8923829c7
--- /dev/null
+++ b/leetcode/0057.Insert-Interval/57. Insert Interval_test.go	
@@ -0,0 +1,68 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question57 struct {
+	para57
+	ans57
+}
+
+// para 是参数
+// one 代表第一个参数
+type para57 struct {
+	one []Interval
+	two Interval
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans57 struct {
+	one []Interval
+}
+
+func Test_Problem57(t *testing.T) {
+
+	qs := []question57{
+
+		{
+			para57{[]Interval{}, Interval{}},
+			ans57{[]Interval{}},
+		},
+
+		{
+			para57{[]Interval{{Start: 1, End: 3}, {Start: 6, End: 9}}, Interval{Start: 4, End: 8}},
+			ans57{[]Interval{{Start: 1, End: 5}, {Start: 6, End: 9}}},
+		},
+
+		{
+			para57{[]Interval{{Start: 1, End: 3}, {Start: 6, End: 9}}, Interval{Start: 2, End: 5}},
+			ans57{[]Interval{{Start: 1, End: 5}, {Start: 6, End: 9}}},
+		},
+
+		{
+			para57{[]Interval{{Start: 1, End: 2}, {Start: 3, End: 5}, {Start: 6, End: 7}, {Start: 8, End: 10}, {Start: 12, End: 16}}, Interval{Start: 4, End: 8}},
+			ans57{[]Interval{{Start: 1, End: 2}, {Start: 3, End: 10}, {Start: 12, End: 16}}},
+		},
+
+		{
+			para57{[]Interval{{Start: 1, End: 5}}, Interval{Start: 5, End: 7}},
+			ans57{[]Interval{{Start: 1, End: 7}}},
+		},
+
+		{
+			para57{[]Interval{{Start: 1, End: 2}, {Start: 3, End: 5}, {Start: 6, End: 7}, {Start: 8, End: 10}, {Start: 12, End: 16}}, Interval{Start: 9, End: 12}},
+			ans57{[]Interval{{Start: 1, End: 2}, {Start: 3, End: 5}, {Start: 6, End: 7}, {Start: 8, End: 16}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 57------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans57, q.para57
+		fmt.Printf("【input】:%v       【output】:%v\n", p, insert(p.one, p.two))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0057. Insert Interval/README.md b/leetcode/0057.Insert-Interval/README.md
similarity index 100%
rename from Algorithms/0057. Insert Interval/README.md
rename to leetcode/0057.Insert-Interval/README.md
diff --git a/leetcode/0058.Length-of-Last-Word/58.Length of Last Word.go b/leetcode/0058.Length-of-Last-Word/58.Length of Last Word.go
new file mode 100644
index 000000000..5fccdb64a
--- /dev/null
+++ b/leetcode/0058.Length-of-Last-Word/58.Length of Last Word.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func lengthOfLastWord(s string) int {
+	last := len(s) - 1
+	for last >= 0 && s[last] == ' ' {
+		last--
+	}
+	if last < 0 {
+		return 0
+	}
+	first := last
+	for first >= 0 && s[first] != ' ' {
+		first--
+	}
+	return last - first
+}
diff --git a/leetcode/0058.Length-of-Last-Word/58.Length of Last Word_test.go b/leetcode/0058.Length-of-Last-Word/58.Length of Last Word_test.go
new file mode 100644
index 000000000..b47de4b7a
--- /dev/null
+++ b/leetcode/0058.Length-of-Last-Word/58.Length of Last Word_test.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question58 struct {
+	para58
+	ans58
+}
+
+// para 是参数
+type para58 struct {
+	s string
+}
+
+// ans 是答案
+type ans58 struct {
+	ans int
+}
+
+func Test_Problem58(t *testing.T) {
+
+	qs := []question58{
+
+		{
+			para58{"Hello World"},
+			ans58{5},
+		},
+
+		{
+			para58{"   fly me   to   the moon  "},
+			ans58{4},
+		},
+
+		{
+			para58{"luffy is still joyboy"},
+			ans58{6},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 58------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans58, q.para58
+		fmt.Printf("【input】:%v       【output】:%v\n", p, lengthOfLastWord(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0058.Length-of-Last-Word/README.md b/leetcode/0058.Length-of-Last-Word/README.md
new file mode 100644
index 000000000..39df720a6
--- /dev/null
+++ b/leetcode/0058.Length-of-Last-Word/README.md
@@ -0,0 +1,70 @@
+# [58. Length of Last Word](https://leetcode.com/problems/length-of-last-word/)
+
+
+## 题目
+
+Given a string `s` consisting of some words separated by some number of spaces, return *the length of the **last** word in the string.*
+
+A **word** is a maximal substring consisting of non-space characters only.
+
+**Example 1:**
+
+```
+Input: s = "Hello World"
+Output: 5
+Explanation: The last word is "World" with length 5.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "   fly me   to   the moon  "
+Output: 4
+Explanation: The last word is "moon" with length 4.
+
+```
+
+**Example 3:**
+
+```
+Input: s = "luffy is still joyboy"
+Output: 6
+Explanation: The last word is "joyboy" with length 6.
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 104`
+- `s` consists of only English letters and spaces `' '`.
+- There will be at least one word in `s`.
+
+## 题目大意
+
+给你一个字符串 `s`，由若干单词组成，单词前后用一些空格字符隔开。返回字符串中最后一个单词的长度。**单词** 是指仅由字母组成、不包含任何空格字符的最大子字符串。
+
+## 解题思路
+
+- 先从后过滤掉空格找到单词尾部，再从尾部向前遍历，找到单词头部，最后两者相减，即为单词的长度。
+
+## 代码
+
+```go
+package leetcode
+
+func lengthOfLastWord(s string) int {
+	last := len(s) - 1
+	for last >= 0 && s[last] == ' ' {
+		last--
+	}
+	if last < 0 {
+		return 0
+	}
+	first := last
+	for first >= 0 && s[first] != ' ' {
+		first--
+	}
+	return last - first
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0059. Spiral Matrix II/59. Spiral Matrix II.go b/leetcode/0059.Spiral-Matrix-II/59. Spiral Matrix II.go
similarity index 88%
rename from Algorithms/0059. Spiral Matrix II/59. Spiral Matrix II.go
rename to leetcode/0059.Spiral-Matrix-II/59. Spiral Matrix II.go
index 8fc1def45..aa1657f02 100644
--- a/Algorithms/0059. Spiral Matrix II/59. Spiral Matrix II.go	
+++ b/leetcode/0059.Spiral-Matrix-II/59. Spiral Matrix II.go	
@@ -5,13 +5,13 @@ func generateMatrix(n int) [][]int {
 		return [][]int{}
 	}
 	if n == 1 {
-		return [][]int{[]int{1}}
+		return [][]int{{1}}
 	}
 	res, visit, round, x, y, spDir := make([][]int, n), make([][]int, n), 0, 0, 0, [][]int{
-		[]int{0, 1},  // 朝右
-		[]int{1, 0},  // 朝下
-		[]int{0, -1}, // 朝左
-		[]int{-1, 0}, // 朝上
+		{0, 1},  // 朝右
+		{1, 0},  // 朝下
+		{0, -1}, // 朝左
+		{-1, 0}, // 朝上
 	}
 	for i := 0; i < n; i++ {
 		visit[i] = make([]int, n)
diff --git a/Algorithms/0059. Spiral Matrix II/59. Spiral Matrix II_test.go b/leetcode/0059.Spiral-Matrix-II/59. Spiral Matrix II_test.go
similarity index 75%
rename from Algorithms/0059. Spiral Matrix II/59. Spiral Matrix II_test.go
rename to leetcode/0059.Spiral-Matrix-II/59. Spiral Matrix II_test.go
index 8448ced6e..be43c350d 100644
--- a/Algorithms/0059. Spiral Matrix II/59. Spiral Matrix II_test.go	
+++ b/leetcode/0059.Spiral-Matrix-II/59. Spiral Matrix II_test.go	
@@ -26,14 +26,14 @@ func Test_Problem59(t *testing.T) {
 
 	qs := []question59{
 
-		question59{
+		{
 			para59{3},
-			ans59{[][]int{[]int{1, 2, 3}, []int{8, 9, 4}, []int{7, 6, 5}}},
+			ans59{[][]int{{1, 2, 3}, {8, 9, 4}, {7, 6, 5}}},
 		},
 
-		question59{
+		{
 			para59{4},
-			ans59{[][]int{[]int{1, 2, 3, 4}, []int{12, 13, 14, 5}, []int{11, 16, 15, 6}, []int{10, 9, 8, 7}}},
+			ans59{[][]int{{1, 2, 3, 4}, {12, 13, 14, 5}, {11, 16, 15, 6}, {10, 9, 8, 7}}},
 		},
 	}
 
diff --git a/Algorithms/0059. Spiral Matrix II/README.md b/leetcode/0059.Spiral-Matrix-II/README.md
similarity index 100%
rename from Algorithms/0059. Spiral Matrix II/README.md
rename to leetcode/0059.Spiral-Matrix-II/README.md
diff --git a/Algorithms/0060. Permutation Sequence/60. Permutation Sequence.go b/leetcode/0060.Permutation-Sequence/60. Permutation Sequence.go
similarity index 100%
rename from Algorithms/0060. Permutation Sequence/60. Permutation Sequence.go
rename to leetcode/0060.Permutation-Sequence/60. Permutation Sequence.go
diff --git a/Algorithms/0060. Permutation Sequence/60. Permutation Sequence_test.go b/leetcode/0060.Permutation-Sequence/60. Permutation Sequence_test.go
similarity index 95%
rename from Algorithms/0060. Permutation Sequence/60. Permutation Sequence_test.go
rename to leetcode/0060.Permutation-Sequence/60. Permutation Sequence_test.go
index 7e14fb4e4..b9805ed25 100644
--- a/Algorithms/0060. Permutation Sequence/60. Permutation Sequence_test.go	
+++ b/leetcode/0060.Permutation-Sequence/60. Permutation Sequence_test.go	
@@ -27,12 +27,12 @@ func Test_Problem60(t *testing.T) {
 
 	qs := []question60{
 
-		question60{
+		{
 			para60{3, 3},
 			ans60{"213"},
 		},
 
-		question60{
+		{
 			para60{4, 9},
 			ans60{"2314"},
 		},
diff --git a/Algorithms/0060. Permutation Sequence/README.md b/leetcode/0060.Permutation-Sequence/README.md
similarity index 100%
rename from Algorithms/0060. Permutation Sequence/README.md
rename to leetcode/0060.Permutation-Sequence/README.md
diff --git a/Algorithms/0061. Rotate List/61. Rotate List.go b/leetcode/0061.Rotate-List/61. Rotate List.go
similarity index 83%
rename from Algorithms/0061. Rotate List/61. Rotate List.go
rename to leetcode/0061.Rotate-List/61. Rotate List.go
index 2fa6e24d4..a2f584ede 100644
--- a/Algorithms/0061. Rotate List/61. Rotate List.go	
+++ b/leetcode/0061.Rotate-List/61. Rotate List.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0061. Rotate List/61. Rotate List_test.go b/leetcode/0061.Rotate-List/61. Rotate List_test.go
similarity index 81%
rename from Algorithms/0061. Rotate List/61. Rotate List_test.go
rename to leetcode/0061.Rotate-List/61. Rotate List_test.go
index ca0f7cc92..7df9e550d 100644
--- a/Algorithms/0061. Rotate List/61. Rotate List_test.go	
+++ b/leetcode/0061.Rotate-List/61. Rotate List_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question61 struct {
@@ -27,32 +29,32 @@ func Test_Problem61(t *testing.T) {
 
 	qs := []question61{
 
-		question61{
+		{
 			para61{[]int{1, 2, 3, 4, 5}, 2},
 			ans61{[]int{4, 5, 1, 2, 3}},
 		},
 
-		question61{
+		{
 			para61{[]int{1, 2, 3, 4, 5}, 3},
 			ans61{[]int{4, 5, 1, 2, 3}},
 		},
 
-		question61{
+		{
 			para61{[]int{0, 1, 2}, 4},
 			ans61{[]int{2, 0, 1}},
 		},
 
-		question61{
+		{
 			para61{[]int{1, 1, 1, 2}, 3},
 			ans61{[]int{1, 1, 2, 1}},
 		},
 
-		question61{
+		{
 			para61{[]int{1}, 10},
 			ans61{[]int{1}},
 		},
 
-		question61{
+		{
 			para61{[]int{}, 100},
 			ans61{[]int{}},
 		},
@@ -62,7 +64,7 @@ func Test_Problem61(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans61, q.para61
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(rotateRight(S2l(p.one), p.k)))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(rotateRight(structures.Ints2List(p.one), p.k)))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0061. Rotate List/README.md b/leetcode/0061.Rotate-List/README.md
similarity index 100%
rename from Algorithms/0061. Rotate List/README.md
rename to leetcode/0061.Rotate-List/README.md
diff --git a/leetcode/0062.Unique-Paths/62. Unique Paths.go b/leetcode/0062.Unique-Paths/62. Unique Paths.go
new file mode 100644
index 000000000..aeb84056a
--- /dev/null
+++ b/leetcode/0062.Unique-Paths/62. Unique Paths.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func uniquePaths(m int, n int) int {
+	dp := make([][]int, n)
+	for i := 0; i < n; i++ {
+		dp[i] = make([]int, m)
+	}
+	for i := 0; i < n; i++ {
+		for j := 0; j < m; j++ {
+			if i == 0 || j == 0 {
+				dp[i][j] = 1
+				continue
+			}
+			dp[i][j] = dp[i-1][j] + dp[i][j-1]
+		}
+	}
+	return dp[n-1][m-1]
+}
diff --git a/Algorithms/0062. Unique Paths/62. Unique Paths_test.go b/leetcode/0062.Unique-Paths/62. Unique Paths_test.go
similarity index 94%
rename from Algorithms/0062. Unique Paths/62. Unique Paths_test.go
rename to leetcode/0062.Unique-Paths/62. Unique Paths_test.go
index 00fad12e3..e5e7f6575 100644
--- a/Algorithms/0062. Unique Paths/62. Unique Paths_test.go	
+++ b/leetcode/0062.Unique-Paths/62. Unique Paths_test.go	
@@ -27,17 +27,17 @@ func Test_Problem62(t *testing.T) {
 
 	qs := []question62{
 
-		question62{
+		{
 			para62{3, 2},
 			ans62{3},
 		},
 
-		question62{
+		{
 			para62{7, 3},
 			ans62{28},
 		},
 
-		question62{
+		{
 			para62{1, 2},
 			ans62{1},
 		},
diff --git a/Algorithms/0062. Unique Paths/README.md b/leetcode/0062.Unique-Paths/README.md
similarity index 100%
rename from Algorithms/0062. Unique Paths/README.md
rename to leetcode/0062.Unique-Paths/README.md
diff --git a/Algorithms/0063. Unique Paths II/63. Unique Paths II.go b/leetcode/0063.Unique-Paths-II/63. Unique Paths II.go
similarity index 100%
rename from Algorithms/0063. Unique Paths II/63. Unique Paths II.go
rename to leetcode/0063.Unique-Paths-II/63. Unique Paths II.go
diff --git a/Algorithms/0063. Unique Paths II/63. Unique Paths II_test.go b/leetcode/0063.Unique-Paths-II/63. Unique Paths II_test.go
similarity index 76%
rename from Algorithms/0063. Unique Paths II/63. Unique Paths II_test.go
rename to leetcode/0063.Unique-Paths-II/63. Unique Paths II_test.go
index 03536295d..e8bf80281 100644
--- a/Algorithms/0063. Unique Paths II/63. Unique Paths II_test.go	
+++ b/leetcode/0063.Unique-Paths-II/63. Unique Paths II_test.go	
@@ -26,29 +26,29 @@ func Test_Problem63(t *testing.T) {
 
 	qs := []question63{
 
-		question63{
+		{
 			para63{[][]int{
-				[]int{0, 0, 0},
-				[]int{0, 1, 0},
-				[]int{0, 0, 0},
+				{0, 0, 0},
+				{0, 1, 0},
+				{0, 0, 0},
 			}},
 			ans63{2},
 		},
 
-		question63{
+		{
 			para63{[][]int{
-				[]int{0, 0},
-				[]int{1, 1},
-				[]int{0, 0},
+				{0, 0},
+				{1, 1},
+				{0, 0},
 			}},
 			ans63{0},
 		},
 
-		question63{
+		{
 			para63{[][]int{
-				[]int{0, 1, 0, 0},
-				[]int{1, 0, 0, 0},
-				[]int{0, 0, 0, 0},
+				{0, 1, 0, 0},
+				{1, 0, 0, 0},
+				{0, 0, 0, 0},
 			}},
 			ans63{0},
 		},
diff --git a/Algorithms/0063. Unique Paths II/README.md b/leetcode/0063.Unique-Paths-II/README.md
similarity index 100%
rename from Algorithms/0063. Unique Paths II/README.md
rename to leetcode/0063.Unique-Paths-II/README.md
diff --git a/Algorithms/0064. Minimum Path Sum/64. Minimum Path Sum.go b/leetcode/0064.Minimum-Path-Sum/64. Minimum Path Sum.go
similarity index 94%
rename from Algorithms/0064. Minimum Path Sum/64. Minimum Path Sum.go
rename to leetcode/0064.Minimum-Path-Sum/64. Minimum Path Sum.go
index ab706adf4..ea2bcc04d 100644
--- a/Algorithms/0064. Minimum Path Sum/64. Minimum Path Sum.go	
+++ b/leetcode/0064.Minimum-Path-Sum/64. Minimum Path Sum.go	
@@ -55,3 +55,10 @@ func minPathSum1(grid [][]int) int {
 	}
 	return dp[m-1][n-1]
 }
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/Algorithms/0064. Minimum Path Sum/64. Minimum Path Sum_test.go b/leetcode/0064.Minimum-Path-Sum/64. Minimum Path Sum_test.go
similarity index 89%
rename from Algorithms/0064. Minimum Path Sum/64. Minimum Path Sum_test.go
rename to leetcode/0064.Minimum-Path-Sum/64. Minimum Path Sum_test.go
index ab5605308..afbfcacd5 100644
--- a/Algorithms/0064. Minimum Path Sum/64. Minimum Path Sum_test.go	
+++ b/leetcode/0064.Minimum-Path-Sum/64. Minimum Path Sum_test.go	
@@ -26,11 +26,11 @@ func Test_Problem64(t *testing.T) {
 
 	qs := []question64{
 
-		question64{
+		{
 			para64{[][]int{
-				[]int{1, 3, 1},
-				[]int{1, 5, 1},
-				[]int{4, 2, 1},
+				{1, 3, 1},
+				{1, 5, 1},
+				{4, 2, 1},
 			}},
 			ans64{7},
 		},
diff --git a/Algorithms/0064. Minimum Path Sum/README.md b/leetcode/0064.Minimum-Path-Sum/README.md
similarity index 100%
rename from Algorithms/0064. Minimum Path Sum/README.md
rename to leetcode/0064.Minimum-Path-Sum/README.md
diff --git a/leetcode/0065.Valid-Number/65. Valid Number.go b/leetcode/0065.Valid-Number/65. Valid Number.go
new file mode 100644
index 000000000..c1cd1fe7c
--- /dev/null
+++ b/leetcode/0065.Valid-Number/65. Valid Number.go	
@@ -0,0 +1,22 @@
+package leetcode
+
+func isNumber(s string) bool {
+	numFlag, dotFlag, eFlag := false, false, false
+	for i := 0; i < len(s); i++ {
+		if '0' <= s[i] && s[i] <= '9' {
+			numFlag = true
+		} else if s[i] == '.' && !dotFlag && !eFlag {
+			dotFlag = true
+		} else if (s[i] == 'e' || s[i] == 'E') && !eFlag && numFlag {
+			eFlag = true
+			numFlag = false // reJudge integer after 'e' or 'E'
+		} else if (s[i] == '+' || s[i] == '-') && (i == 0 || s[i-1] == 'e' || s[i-1] == 'E') {
+			continue
+		} else {
+			return false
+		}
+	}
+	// avoid case: s == '.' or 'e/E' or '+/-' and etc...
+	// string s must have num
+	return numFlag
+}
diff --git a/leetcode/0065.Valid-Number/65. Valid Number_test.go b/leetcode/0065.Valid-Number/65. Valid Number_test.go
new file mode 100644
index 000000000..184e1d4ab
--- /dev/null
+++ b/leetcode/0065.Valid-Number/65. Valid Number_test.go	
@@ -0,0 +1,41 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem65(t *testing.T) {
+
+	tcs := []struct {
+		s   string
+		ans bool
+	}{
+
+		{
+			"0",
+			true,
+		},
+
+		{
+			"e",
+			false,
+		},
+
+		{
+			".",
+			false,
+		},
+
+		{
+			".1",
+			true,
+		},
+	}
+	fmt.Printf("------------------------Leetcode Problem 65------------------------\n")
+
+	for _, tc := range tcs {
+		fmt.Printf("【input】:%v       【output】:%v\n", tc, isNumber(tc.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0065.Valid-Number/README.md b/leetcode/0065.Valid-Number/README.md
new file mode 100644
index 000000000..f64df4566
--- /dev/null
+++ b/leetcode/0065.Valid-Number/README.md
@@ -0,0 +1,78 @@
+# [65. Valid Number](https://leetcode.com/problems/valid-number/)
+
+
+## 题目
+
+A **valid number** can be split up into these components (in order):
+
+   1. A **decimal number** or an integer. 
+   2. (Optional) An 'e' or 'E', followed by an **integer.**
+   
+A **decimal number** can be split up into these components (in order):
+
+   1. (Optional) A sign character (either '+' or '-').
+   2. One of the following formats:
+        1. One or more digits, followed by a dot '.'.
+        2. One or more digits, followed by a dot '.', followed by one or more digits.
+        3. A dot '.', followed by one or more digits.
+        
+An **integer** can be split up into these components (in order):
+
+   1. (Optional) A sign character (either '+' or '-').
+   2. One or more digits.
+   
+For example, all the following are valid numbers: `["2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789"]`, while the following are not valid numbers: `["abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53"].`
+
+Given a string s, return true if s is a **valid number.**
+
+**Example:**
+
+    Input: s = "0"
+    Output: true
+    
+    Input: s = "e"
+    Output: false
+
+## 题目大意
+
+给定一个字符串S，请根据以上的规则判断该字符串是否是一个有效的数字字符串。
+
+
+## 解题思路
+
+- 用三个变量分别标记是否出现过数字、是否出现过'.'和 是否出现过 'e/E'
+- 从左到右依次遍历字符串中的每一个元素
+    - 如果是数字，则标记数字出现过
+    - 如果是 '.', 则需要 '.'没有出现过，并且 'e/E' 没有出现过，才会进行标记
+    - 如果是 'e/E', 则需要 'e/E'没有出现过，并且前面出现过数字，才会进行标记
+    - 如果是 '+/-', 则需要是第一个字符，或者前一个字符是 'e/E'，才会进行标记，并重置数字出现的标识
+    - 最后返回时，需要字符串中至少出现过数字，避免下列case: s == '.' or 'e/E' or '+/e' and etc...
+
+## 代码
+
+```go
+
+package leetcode
+
+func isNumber(s string) bool {
+	numFlag, dotFlag, eFlag := false, false, false
+	for i := 0; i < len(s); i++ {
+		if '0' <= s[i] && s[i] <= '9' {
+			numFlag = true
+		} else if s[i] == '.' && !dotFlag && !eFlag {
+			dotFlag = true
+		} else if (s[i] == 'e' || s[i] == 'E') && !eFlag && numFlag {
+			eFlag = true
+			numFlag = false // reJudge integer after 'e' or 'E'
+		} else if (s[i] == '+' || s[i] == '-') && (i == 0 || s[i-1] == 'e' || s[i-1] == 'E') {
+			continue
+		} else {
+			return false
+		}
+	}
+	// avoid case: s == '.' or 'e/E' or '+/-' and etc...
+	// string s must have num
+	return numFlag
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/0066.Plus-One/66. Plus One.go b/leetcode/0066.Plus-One/66. Plus One.go
new file mode 100644
index 000000000..9a58ea973
--- /dev/null
+++ b/leetcode/0066.Plus-One/66. Plus One.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+func plusOne(digits []int) []int {
+	for i := len(digits) - 1; i >= 0; i-- {
+		if digits[i] != 9 {
+			digits[i]++
+			return digits
+		}
+		digits[i] = 0
+	}
+	return append([]int{1}, digits...)
+}
diff --git a/Algorithms/0066. Plus One/66. Plus One_test.go b/leetcode/0066.Plus-One/66. Plus One_test.go
similarity index 93%
rename from Algorithms/0066. Plus One/66. Plus One_test.go
rename to leetcode/0066.Plus-One/66. Plus One_test.go
index 0a72dea8f..cc112f4dc 100644
--- a/Algorithms/0066. Plus One/66. Plus One_test.go	
+++ b/leetcode/0066.Plus-One/66. Plus One_test.go	
@@ -26,22 +26,22 @@ func Test_Problem66(t *testing.T) {
 
 	qs := []question66{
 
-		question66{
+		{
 			para66{[]int{1, 2, 3}},
 			ans66{[]int{1, 2, 4}},
 		},
 
-		question66{
+		{
 			para66{[]int{4, 3, 2, 1}},
 			ans66{[]int{4, 3, 2, 2}},
 		},
 
-		question66{
+		{
 			para66{[]int{9, 9}},
 			ans66{[]int{1, 0, 0}},
 		},
 
-		question66{
+		{
 			para66{[]int{0}},
 			ans66{[]int{0}},
 		},
diff --git a/leetcode/0066.Plus-One/README.md b/leetcode/0066.Plus-One/README.md
new file mode 100755
index 000000000..6422b0fd6
--- /dev/null
+++ b/leetcode/0066.Plus-One/README.md
@@ -0,0 +1,35 @@
+# [66. Plus One](https://leetcode.com/problems/plus-one/)
+
+
+## 题目
+
+Given a **non-empty** array of digits representing a non-negative integer, plus one to the integer.
+
+The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.
+
+You may assume the integer does not contain any leading zero, except the number 0 itself.
+
+**Example 1:**
+
+    Input: [1,2,3]
+    Output: [1,2,4]
+    Explanation: The array represents the integer 123.
+
+**Example 2:**
+
+    Input: [4,3,2,1]
+    Output: [4,3,2,2]
+    Explanation: The array represents the integer 4321.
+
+
+## 题目大意
+
+
+给定一个由整数组成的非空数组所表示的非负整数，在该数的基础上加一。最高位数字存放在数组的首位， 数组中每个元素只存储单个数字。你可以假设除了整数 0 之外，这个整数不会以零开头。
+
+
+
+## 解题思路
+
+- 给出一个数组，代表一个十进制数，数组的 0 下标是十进制数的高位。要求计算这个十进制数加一以后的结果。
+- 简单的模拟题。从数组尾部开始往前扫，逐位进位即可。最高位如果还有进位需要在数组里面第 0 位再插入一个 1 。
diff --git a/leetcode/0067.Add-Binary/67. Add Binary.go b/leetcode/0067.Add-Binary/67. Add Binary.go
new file mode 100644
index 000000000..8d22cf04b
--- /dev/null
+++ b/leetcode/0067.Add-Binary/67. Add Binary.go	
@@ -0,0 +1,38 @@
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+)
+
+func addBinary(a string, b string) string {
+	if len(b) > len(a) {
+		a, b = b, a
+	}
+
+	res := make([]string, len(a)+1)
+	i, j, k, c := len(a)-1, len(b)-1, len(a), 0
+	for i >= 0 && j >= 0 {
+		ai, _ := strconv.Atoi(string(a[i]))
+		bj, _ := strconv.Atoi(string(b[j]))
+		res[k] = strconv.Itoa((ai + bj + c) % 2)
+		c = (ai + bj + c) / 2
+		i--
+		j--
+		k--
+	}
+
+	for i >= 0 {
+		ai, _ := strconv.Atoi(string(a[i]))
+		res[k] = strconv.Itoa((ai + c) % 2)
+		c = (ai + c) / 2
+		i--
+		k--
+	}
+
+	if c > 0 {
+		res[k] = strconv.Itoa(c)
+	}
+
+	return strings.Join(res, "")
+}
diff --git a/leetcode/0067.Add-Binary/67. Add Binary_test.go b/leetcode/0067.Add-Binary/67. Add Binary_test.go
new file mode 100644
index 000000000..29adc0cd9
--- /dev/null
+++ b/leetcode/0067.Add-Binary/67. Add Binary_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question67 struct {
+	para67
+	ans67
+}
+
+// para 是参数
+// one 代表第一个参数
+type para67 struct {
+	a string
+	b string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans67 struct {
+	one string
+}
+
+func Test_Problem67(t *testing.T) {
+
+	qs := []question67{
+
+		{
+			para67{"11", "1"},
+			ans67{"100"},
+		},
+
+		{
+			para67{"1010", "1011"},
+			ans67{"10101"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 67------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans67, q.para67
+		fmt.Printf("【input】:%v       【output】:%v\n", p, addBinary(p.a, p.b))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0067.Add-Binary/README.md b/leetcode/0067.Add-Binary/README.md
new file mode 100644
index 000000000..91d977e58
--- /dev/null
+++ b/leetcode/0067.Add-Binary/README.md
@@ -0,0 +1,76 @@
+# [67. Add Binary](https://leetcode.com/problems/add-binary/)
+
+
+## 题目
+
+Given two binary strings, return their sum (also a binary string).
+
+The input strings are both **non-empty** and contains only characters `1` or `0`.
+
+**Example 1**:
+
+```
+Input: a = "11", b = "1"
+Output: "100"
+```
+
+**Example 2**:
+
+```
+Input: a = "1010", b = "1011"
+Output: "10101"
+```
+
+## 题目大意
+
+给你两个二进制字符串，返回它们的和（用二进制表示）。输入为 非空 字符串且只包含数字 1 和 0。
+
+## 解题思路
+
+- 要求输出 2 个二进制数的和，结果也用二进制表示。
+- 简单题。按照二进制的加法规则做加法即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+)
+
+func addBinary(a string, b string) string {
+	if len(b) > len(a) {
+		a, b = b, a
+	}
+
+	res := make([]string, len(a)+1)
+	i, j, k, c := len(a)-1, len(b)-1, len(a), 0
+	for i >= 0 && j >= 0 {
+		ai, _ := strconv.Atoi(string(a[i]))
+		bj, _ := strconv.Atoi(string(b[j]))
+		res[k] = strconv.Itoa((ai + bj + c) % 2)
+		c = (ai + bj + c) / 2
+		i--
+		j--
+		k--
+	}
+
+	for i >= 0 {
+		ai, _ := strconv.Atoi(string(a[i]))
+		res[k] = strconv.Itoa((ai + c) % 2)
+		c = (ai + c) / 2
+		i--
+		k--
+	}
+
+	if c > 0 {
+		res[k] = strconv.Itoa(c)
+	}
+
+	return strings.Join(res, "")
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0069. Sqrt(x)/69. Sqrt(x).go b/leetcode/0069.Sqrtx/69. Sqrt(x).go
similarity index 77%
rename from Algorithms/0069. Sqrt(x)/69. Sqrt(x).go
rename to leetcode/0069.Sqrtx/69. Sqrt(x).go
index c228c82d0..33d21aee2 100644
--- a/Algorithms/0069. Sqrt(x)/69. Sqrt(x).go	
+++ b/leetcode/0069.Sqrtx/69. Sqrt(x).go	
@@ -1,23 +1,17 @@
 package leetcode
 
-// 解法一 二分
+// 解法一 二分, 找到最后一个满足 n^2 <= x 的整数n
 func mySqrt(x int) int {
-	if x == 0 {
-		return 0
-	}
-	left, right, res := 1, x, 0
-	for left <= right {
-		mid := left + ((right - left) >> 1)
-		if mid < x/mid {
-			left = mid + 1
-			res = mid
-		} else if mid == x/mid {
-			return mid
+	l, r := 0, x
+	for l < r {
+		mid := (l + r + 1) / 2
+		if mid*mid > x {
+			r = mid - 1
 		} else {
-			right = mid - 1
+			l = mid
 		}
 	}
-	return res
+	return l
 }
 
 // 解法二 牛顿迭代法 https://en.wikipedia.org/wiki/Integer_square_root
diff --git a/Algorithms/0069. Sqrt(x)/69. Sqrt(x)_test.go b/leetcode/0069.Sqrtx/69. Sqrt(x)_test.go
similarity index 95%
rename from Algorithms/0069. Sqrt(x)/69. Sqrt(x)_test.go
rename to leetcode/0069.Sqrtx/69. Sqrt(x)_test.go
index 69e59cfcf..b31883482 100644
--- a/Algorithms/0069. Sqrt(x)/69. Sqrt(x)_test.go	
+++ b/leetcode/0069.Sqrtx/69. Sqrt(x)_test.go	
@@ -26,12 +26,12 @@ func Test_Problem69(t *testing.T) {
 
 	qs := []question69{
 
-		question69{
+		{
 			para69{4},
 			ans69{2},
 		},
 
-		question69{
+		{
 			para69{8},
 			ans69{2},
 		},
diff --git a/Algorithms/0069. Sqrt(x)/README.md b/leetcode/0069.Sqrtx/README.md
similarity index 100%
rename from Algorithms/0069. Sqrt(x)/README.md
rename to leetcode/0069.Sqrtx/README.md
diff --git a/Algorithms/0070. Climbing Stairs/70. Climbing Stairs.go b/leetcode/0070.Climbing-Stairs/70. Climbing Stairs.go
similarity index 100%
rename from Algorithms/0070. Climbing Stairs/70. Climbing Stairs.go
rename to leetcode/0070.Climbing-Stairs/70. Climbing Stairs.go
diff --git a/Algorithms/0070. Climbing Stairs/70. Climbing Stairs_test.go b/leetcode/0070.Climbing-Stairs/70. Climbing Stairs_test.go
similarity index 95%
rename from Algorithms/0070. Climbing Stairs/70. Climbing Stairs_test.go
rename to leetcode/0070.Climbing-Stairs/70. Climbing Stairs_test.go
index b574248d7..02634ec3a 100644
--- a/Algorithms/0070. Climbing Stairs/70. Climbing Stairs_test.go	
+++ b/leetcode/0070.Climbing-Stairs/70. Climbing Stairs_test.go	
@@ -26,12 +26,12 @@ func Test_Problem70(t *testing.T) {
 
 	qs := []question70{
 
-		question70{
+		{
 			para70{2},
 			ans70{2},
 		},
 
-		question70{
+		{
 			para70{3},
 			ans70{3},
 		},
diff --git a/Algorithms/0070. Climbing Stairs/README.md b/leetcode/0070.Climbing-Stairs/README.md
similarity index 100%
rename from Algorithms/0070. Climbing Stairs/README.md
rename to leetcode/0070.Climbing-Stairs/README.md
diff --git a/Algorithms/0071. Simplify Path/71. Simplify Path.go b/leetcode/0071.Simplify-Path/71. Simplify Path.go
similarity index 100%
rename from Algorithms/0071. Simplify Path/71. Simplify Path.go
rename to leetcode/0071.Simplify-Path/71. Simplify Path.go
diff --git a/Algorithms/0071. Simplify Path/71. Simplify Path_test.go b/leetcode/0071.Simplify-Path/71. Simplify Path_test.go
similarity index 89%
rename from Algorithms/0071. Simplify Path/71. Simplify Path_test.go
rename to leetcode/0071.Simplify-Path/71. Simplify Path_test.go
index 65dd575a7..583769210 100644
--- a/Algorithms/0071. Simplify Path/71. Simplify Path_test.go	
+++ b/leetcode/0071.Simplify-Path/71. Simplify Path_test.go	
@@ -26,42 +26,42 @@ func Test_Problem71(t *testing.T) {
 
 	qs := []question71{
 
-		question71{
+		{
 			para71{"/.hidden"},
 			ans71{"/.hidden"},
 		},
 
-		question71{
+		{
 			para71{"/..hidden"},
 			ans71{"/..hidden"},
 		},
 
-		question71{
+		{
 			para71{"/abc/..."},
 			ans71{"/abc/..."},
 		},
 
-		question71{
+		{
 			para71{"/home/"},
 			ans71{"/home"},
 		},
 
-		question71{
+		{
 			para71{"/..."},
 			ans71{"/..."},
 		},
 
-		question71{
+		{
 			para71{"/../"},
 			ans71{"/"},
 		},
 
-		question71{
+		{
 			para71{"/home//foo/"},
 			ans71{"/home/foo"},
 		},
 
-		question71{
+		{
 			para71{"/a/./b/../../c/"},
 			ans71{"/c"},
 		},
diff --git a/Algorithms/0071. Simplify Path/README.md b/leetcode/0071.Simplify-Path/README.md
similarity index 100%
rename from Algorithms/0071. Simplify Path/README.md
rename to leetcode/0071.Simplify-Path/README.md
diff --git a/leetcode/0073.Set-Matrix-Zeroes/73. Set Matrix Zeroes.go b/leetcode/0073.Set-Matrix-Zeroes/73. Set Matrix Zeroes.go
new file mode 100644
index 000000000..09882130f
--- /dev/null
+++ b/leetcode/0073.Set-Matrix-Zeroes/73. Set Matrix Zeroes.go	
@@ -0,0 +1,54 @@
+package leetcode
+
+func setZeroes(matrix [][]int) {
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return
+	}
+	isFirstRowExistZero, isFirstColExistZero := false, false
+	for i := 0; i < len(matrix); i++ {
+		if matrix[i][0] == 0 {
+			isFirstColExistZero = true
+			break
+		}
+	}
+	for j := 0; j < len(matrix[0]); j++ {
+		if matrix[0][j] == 0 {
+			isFirstRowExistZero = true
+			break
+		}
+	}
+	for i := 1; i < len(matrix); i++ {
+		for j := 1; j < len(matrix[0]); j++ {
+			if matrix[i][j] == 0 {
+				matrix[i][0] = 0
+				matrix[0][j] = 0
+			}
+		}
+	}
+	// 处理[1:]行全部置 0
+	for i := 1; i < len(matrix); i++ {
+		if matrix[i][0] == 0 {
+			for j := 1; j < len(matrix[0]); j++ {
+				matrix[i][j] = 0
+			}
+		}
+	}
+	// 处理[1:]列全部置 0
+	for j := 1; j < len(matrix[0]); j++ {
+		if matrix[0][j] == 0 {
+			for i := 1; i < len(matrix); i++ {
+				matrix[i][j] = 0
+			}
+		}
+	}
+	if isFirstRowExistZero {
+		for j := 0; j < len(matrix[0]); j++ {
+			matrix[0][j] = 0
+		}
+	}
+	if isFirstColExistZero {
+		for i := 0; i < len(matrix); i++ {
+			matrix[i][0] = 0
+		}
+	}
+}
diff --git a/leetcode/0073.Set-Matrix-Zeroes/73. Set Matrix Zeroes_test.go b/leetcode/0073.Set-Matrix-Zeroes/73. Set Matrix Zeroes_test.go
new file mode 100644
index 000000000..e4b23425d
--- /dev/null
+++ b/leetcode/0073.Set-Matrix-Zeroes/73. Set Matrix Zeroes_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question73 struct {
+	para73
+	ans73
+}
+
+// para 是参数
+// one 代表第一个参数
+type para73 struct {
+	matrix [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans73 struct {
+}
+
+func Test_Problem73(t *testing.T) {
+
+	qs := []question73{
+
+		{
+			para73{[][]int{
+				{0, 1, 2, 0},
+				{3, 4, 5, 2},
+				{1, 3, 1, 5},
+			}},
+			ans73{},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 73------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans73, q.para73
+		fmt.Printf("【input】:%v       ", p)
+		setZeroes(p.matrix)
+		fmt.Printf("【output】:%v\n", p)
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0073.Set-Matrix-Zeroes/README.md b/leetcode/0073.Set-Matrix-Zeroes/README.md
new file mode 100644
index 000000000..e3f7a2b83
--- /dev/null
+++ b/leetcode/0073.Set-Matrix-Zeroes/README.md
@@ -0,0 +1,104 @@
+# [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)
+
+
+## 题目
+
+Given an *`m* x *n*` matrix. If an element is **0**, set its entire row and column to **0**. Do it **[in-place](https://en.wikipedia.org/wiki/In-place_algorithm)**.
+
+**Follow up:**
+
+- A straight forward solution using O(*mn*) space is probably a bad idea.
+- A simple improvement uses O(*m* + *n*) space, but still not the best solution.
+- Could you devise a constant space solution?
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/08/17/mat1.jpg](https://assets.leetcode.com/uploads/2020/08/17/mat1.jpg)
+
+```
+Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
+Output: [[1,0,1],[0,0,0],[1,0,1]]
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/08/17/mat2.jpg](https://assets.leetcode.com/uploads/2020/08/17/mat2.jpg)
+
+```
+Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
+Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
+```
+
+**Constraints:**
+
+- `m == matrix.length`
+- `n == matrix[0].length`
+- `1 <= m, n <= 200`
+- `2^31 <= matrix[i][j] <= 2^31 - 1`
+
+## 题目大意
+
+给定一个 `m x n` 的矩阵，如果一个元素为 0，则将其所在行和列的所有元素都设为 0。请使用原地算法。
+
+## 解题思路
+
+- 此题考查对程序的控制能力，无算法思想。题目要求采用原地的算法，所有修改即在原二维数组上进行。在二维数组中有 2 个特殊位置，一个是第一行，一个是第一列。它们的特殊性在于，它们之间只要有一个 0，它们都会变为全 0 。先用 2 个变量记录这一行和这一列中是否有 0，防止之后的修改覆盖了这 2 个地方。然后除去这一行和这一列以外的部分判断是否有 0，如果有 0，将它们所在的行第一个元素标记为 0，所在列的第一个元素标记为 0 。最后通过标记，将对应的行列置 0 即可。
+
+## 代码
+
+```go
+package leetcode
+
+func setZeroes(matrix [][]int) {
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return
+	}
+	isFirstRowExistZero, isFirstColExistZero := false, false
+	for i := 0; i < len(matrix); i++ {
+		if matrix[i][0] == 0 {
+			isFirstColExistZero = true
+			break
+		}
+	}
+	for j := 0; j < len(matrix[0]); j++ {
+		if matrix[0][j] == 0 {
+			isFirstRowExistZero = true
+			break
+		}
+	}
+	for i := 1; i < len(matrix); i++ {
+		for j := 1; j < len(matrix[0]); j++ {
+			if matrix[i][j] == 0 {
+				matrix[i][0] = 0
+				matrix[0][j] = 0
+			}
+		}
+	}
+	// 处理[1:]行全部置 0
+	for i := 1; i < len(matrix); i++ {
+		if matrix[i][0] == 0 {
+			for j := 1; j < len(matrix[0]); j++ {
+				matrix[i][j] = 0
+			}
+		}
+	}
+	// 处理[1:]列全部置 0
+	for j := 1; j < len(matrix[0]); j++ {
+		if matrix[0][j] == 0 {
+			for i := 1; i < len(matrix); i++ {
+				matrix[i][j] = 0
+			}
+		}
+	}
+	if isFirstRowExistZero {
+		for j := 0; j < len(matrix[0]); j++ {
+			matrix[0][j] = 0
+		}
+	}
+	if isFirstColExistZero {
+		for i := 0; i < len(matrix); i++ {
+			matrix[i][0] = 0
+		}
+	}
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0074. Search a 2D Matrix/74. Search a 2D Matrix.go b/leetcode/0074.Search-a-2D-Matrix/74. Search a 2D Matrix.go
similarity index 100%
rename from Algorithms/0074. Search a 2D Matrix/74. Search a 2D Matrix.go
rename to leetcode/0074.Search-a-2D-Matrix/74. Search a 2D Matrix.go
diff --git a/Algorithms/0074. Search a 2D Matrix/74. Search a 2D Matrix_test.go b/leetcode/0074.Search-a-2D-Matrix/74. Search a 2D Matrix_test.go
similarity index 75%
rename from Algorithms/0074. Search a 2D Matrix/74. Search a 2D Matrix_test.go
rename to leetcode/0074.Search-a-2D-Matrix/74. Search a 2D Matrix_test.go
index 8890b2895..9bca5b95a 100644
--- a/Algorithms/0074. Search a 2D Matrix/74. Search a 2D Matrix_test.go	
+++ b/leetcode/0074.Search-a-2D-Matrix/74. Search a 2D Matrix_test.go	
@@ -27,13 +27,13 @@ func Test_Problem74(t *testing.T) {
 
 	qs := []question74{
 
-		question74{
-			para74{[][]int{[]int{1, 3, 5, 7}, []int{10, 11, 16, 20}, []int{23, 30, 34, 50}}, 3},
+		{
+			para74{[][]int{{1, 3, 5, 7}, {10, 11, 16, 20}, {23, 30, 34, 50}}, 3},
 			ans74{true},
 		},
 
-		question74{
-			para74{[][]int{[]int{1, 3, 5, 7}, []int{10, 11, 16, 20}, []int{23, 30, 34, 50}}, 13},
+		{
+			para74{[][]int{{1, 3, 5, 7}, {10, 11, 16, 20}, {23, 30, 34, 50}}, 13},
 			ans74{false},
 		},
 	}
diff --git a/leetcode/0074.Search-a-2D-Matrix/README.md b/leetcode/0074.Search-a-2D-Matrix/README.md
new file mode 100755
index 000000000..cd3f1ad8e
--- /dev/null
+++ b/leetcode/0074.Search-a-2D-Matrix/README.md
@@ -0,0 +1,46 @@
+# [74. Search a 2D Matrix](https://leetcode.com/problems/search-a-2d-matrix/)
+
+
+## 题目
+
+Write an efficient algorithm that searches for a value in an *m* x *n* matrix. This matrix has the following properties:
+
+- Integers in each row are sorted from left to right.
+- The first integer of each row is greater than the last integer of the previous row.
+
+**Example 1:**
+
+    Input:
+    matrix = [
+      [1,   3,  5,  7],
+      [10, 11, 16, 20],
+      [23, 30, 34, 50]
+    ]
+    target = 3
+    Output: true
+
+**Example 2:**
+
+    Input:
+    matrix = [
+      [1,   3,  5,  7],
+      [10, 11, 16, 20],
+      [23, 30, 34, 50]
+    ]
+    target = 13
+    Output: false
+
+
+## 题目大意
+
+编写一个高效的算法来判断 m x n 矩阵中，是否存在一个目标值。该矩阵具有如下特性：
+
+- 每行中的整数从左到右按升序排列。
+- 每行的第一个整数大于前一行的最后一个整数。
+
+
+## 解题思路
+
+
+- 给出一个二维矩阵，矩阵的特点是随着矩阵的下标增大而增大。要求设计一个算法能在这个矩阵中高效的找到一个数，如果找到就输出 true，找不到就输出 false。
+- 虽然是一个二维矩阵，但是由于它特殊的有序性，所以完全可以按照下标把它看成一个一维矩阵，只不过需要行列坐标转换。最后利用二分搜索直接搜索即可。
diff --git a/leetcode/0075.Sort-Colors/75. Sort Colors.go b/leetcode/0075.Sort-Colors/75. Sort Colors.go
new file mode 100644
index 000000000..573ef53fe
--- /dev/null
+++ b/leetcode/0075.Sort-Colors/75. Sort Colors.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func sortColors(nums []int) {
+	zero, one := 0, 0
+	for i, n := range nums {
+		nums[i] = 2
+		if n <= 1 {
+			nums[one] = 1
+			one++
+		}
+		if n == 0 {
+			nums[zero] = 0
+			zero++
+		}
+	}
+}
diff --git a/Algorithms/0075. Sort Colors/75. Sort Colors_test.go b/leetcode/0075.Sort-Colors/75. Sort Colors_test.go
similarity index 94%
rename from Algorithms/0075. Sort Colors/75. Sort Colors_test.go
rename to leetcode/0075.Sort-Colors/75. Sort Colors_test.go
index 26ad20036..6feecf3a6 100644
--- a/Algorithms/0075. Sort Colors/75. Sort Colors_test.go	
+++ b/leetcode/0075.Sort-Colors/75. Sort Colors_test.go	
@@ -26,22 +26,22 @@ func Test_Problem75(t *testing.T) {
 
 	qs := []question75{
 
-		question75{
+		{
 			para75{[]int{}},
 			ans75{[]int{}},
 		},
 
-		question75{
+		{
 			para75{[]int{1}},
 			ans75{[]int{1}},
 		},
 
-		question75{
+		{
 			para75{[]int{2, 0, 2, 1, 1, 0}},
 			ans75{[]int{0, 0, 1, 1, 2, 2}},
 		},
 
-		question75{
+		{
 			para75{[]int{2, 0, 1, 1, 2, 0, 2, 1, 2, 0, 0, 0, 1, 2, 2, 2, 0, 1, 1}},
 			ans75{[]int{0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2}},
 		},
diff --git a/Algorithms/0075. Sort Colors/README.md b/leetcode/0075.Sort-Colors/README.md
similarity index 100%
rename from Algorithms/0075. Sort Colors/README.md
rename to leetcode/0075.Sort-Colors/README.md
diff --git a/Algorithms/0076. Minimum Window Substring/76. Minimum Window Substring.go b/leetcode/0076.Minimum-Window-Substring/76. Minimum Window Substring.go
similarity index 90%
rename from Algorithms/0076. Minimum Window Substring/76. Minimum Window Substring.go
rename to leetcode/0076.Minimum-Window-Substring/76. Minimum Window Substring.go
index ece9380ca..8d8808f1d 100644
--- a/Algorithms/0076. Minimum Window Substring/76. Minimum Window Substring.go	
+++ b/leetcode/0076.Minimum-Window-Substring/76. Minimum Window Substring.go	
@@ -32,9 +32,7 @@ func minWindow(s string, t string) string {
 		}
 	}
 	if finalLeft != -1 {
-		for i := finalLeft; i < finalRight+1; i++ {
-			result += string(s[i])
-		}
+		result = string(s[finalLeft : finalRight+1])
 	}
 	return result
 }
diff --git a/Algorithms/0076. Minimum Window Substring/76. Minimum Window Substring_test.go b/leetcode/0076.Minimum-Window-Substring/76. Minimum Window Substring_test.go
similarity index 94%
rename from Algorithms/0076. Minimum Window Substring/76. Minimum Window Substring_test.go
rename to leetcode/0076.Minimum-Window-Substring/76. Minimum Window Substring_test.go
index 73338f495..95076c7a6 100644
--- a/Algorithms/0076. Minimum Window Substring/76. Minimum Window Substring_test.go	
+++ b/leetcode/0076.Minimum-Window-Substring/76. Minimum Window Substring_test.go	
@@ -27,17 +27,17 @@ func Test_Problem76(t *testing.T) {
 
 	qs := []question76{
 
-		question76{
+		{
 			para76{"ADOBECODEBANC", "ABC"},
 			ans76{"BANC"},
 		},
 
-		question76{
+		{
 			para76{"a", "aa"},
 			ans76{""},
 		},
 
-		question76{
+		{
 			para76{"a", "a"},
 			ans76{"a"},
 		},
diff --git a/Algorithms/0076. Minimum Window Substring/README.md b/leetcode/0076.Minimum-Window-Substring/README.md
similarity index 100%
rename from Algorithms/0076. Minimum Window Substring/README.md
rename to leetcode/0076.Minimum-Window-Substring/README.md
diff --git a/Algorithms/0077. Combinations/77. Combinations.go b/leetcode/0077.Combinations/77. Combinations.go
similarity index 100%
rename from Algorithms/0077. Combinations/77. Combinations.go
rename to leetcode/0077.Combinations/77. Combinations.go
diff --git a/Algorithms/0077. Combinations/77. Combinations_test.go b/leetcode/0077.Combinations/77. Combinations_test.go
similarity index 84%
rename from Algorithms/0077. Combinations/77. Combinations_test.go
rename to leetcode/0077.Combinations/77. Combinations_test.go
index a5da8b934..4d1a14f38 100644
--- a/Algorithms/0077. Combinations/77. Combinations_test.go	
+++ b/leetcode/0077.Combinations/77. Combinations_test.go	
@@ -27,9 +27,9 @@ func Test_Problem77(t *testing.T) {
 
 	qs := []question77{
 
-		question77{
+		{
 			para77{4, 2},
-			ans77{[][]int{[]int{2, 4}, []int{3, 4}, []int{2, 3}, []int{1, 2}, []int{1, 3}, []int{1, 4}}},
+			ans77{[][]int{{2, 4}, {3, 4}, {2, 3}, {1, 2}, {1, 3}, {1, 4}}},
 		},
 	}
 
diff --git a/Algorithms/0077. Combinations/README.md b/leetcode/0077.Combinations/README.md
similarity index 100%
rename from Algorithms/0077. Combinations/README.md
rename to leetcode/0077.Combinations/README.md
diff --git a/leetcode/0078.Subsets/78. Subsets.go b/leetcode/0078.Subsets/78. Subsets.go
new file mode 100644
index 000000000..1cca87952
--- /dev/null
+++ b/leetcode/0078.Subsets/78. Subsets.go	
@@ -0,0 +1,64 @@
+package leetcode
+
+import "sort"
+
+// 解法一
+func subsets(nums []int) [][]int {
+	c, res := []int{}, [][]int{}
+	for k := 0; k <= len(nums); k++ {
+		generateSubsets(nums, k, 0, c, &res)
+	}
+	return res
+}
+
+func generateSubsets(nums []int, k, start int, c []int, res *[][]int) {
+	if len(c) == k {
+		b := make([]int, len(c))
+		copy(b, c)
+		*res = append(*res, b)
+		return
+	}
+	// i will at most be n - (k - c.size()) + 1
+	for i := start; i < len(nums)-(k-len(c))+1; i++ {
+		c = append(c, nums[i])
+		generateSubsets(nums, k, i+1, c, res)
+		c = c[:len(c)-1]
+	}
+	return
+}
+
+// 解法二
+func subsets1(nums []int) [][]int {
+	res := make([][]int, 1)
+	sort.Ints(nums)
+	for i := range nums {
+		for _, org := range res {
+			clone := make([]int, len(org), len(org)+1)
+			copy(clone, org)
+			clone = append(clone, nums[i])
+			res = append(res, clone)
+		}
+	}
+	return res
+}
+
+// 解法三：位运算的方法
+func subsets2(nums []int) [][]int {
+	if len(nums) == 0 {
+		return nil
+	}
+	res := [][]int{}
+	sum := 1 << uint(len(nums))
+	for i := 0; i < sum; i++ {
+		stack := []int{}
+		tmp := i                              // i 从 000...000 到 111...111
+		for j := len(nums) - 1; j >= 0; j-- { // 遍历 i 的每一位
+			if tmp&1 == 1 {
+				stack = append([]int{nums[j]}, stack...)
+			}
+			tmp >>= 1
+		}
+		res = append(res, stack)
+	}
+	return res
+}
diff --git a/Algorithms/0078. Subsets/78. Subsets_test.go b/leetcode/0078.Subsets/78. Subsets_test.go
similarity index 78%
rename from Algorithms/0078. Subsets/78. Subsets_test.go
rename to leetcode/0078.Subsets/78. Subsets_test.go
index 462ea24d8..434f61cb0 100644
--- a/Algorithms/0078. Subsets/78. Subsets_test.go	
+++ b/leetcode/0078.Subsets/78. Subsets_test.go	
@@ -26,14 +26,14 @@ func Test_Problem78(t *testing.T) {
 
 	qs := []question78{
 
-		question78{
+		{
 			para78{[]int{}},
-			ans78{[][]int{[]int{}}},
+			ans78{[][]int{{}}},
 		},
 
-		question78{
+		{
 			para78{[]int{1, 2, 3}},
-			ans78{[][]int{[]int{}, []int{1}, []int{2}, []int{3}, []int{1, 2}, []int{2, 3}, []int{1, 3}, []int{1, 2, 3}}},
+			ans78{[][]int{{}, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}},
 		},
 	}
 
diff --git a/Algorithms/0078. Subsets/README.md b/leetcode/0078.Subsets/README.md
similarity index 100%
rename from Algorithms/0078. Subsets/README.md
rename to leetcode/0078.Subsets/README.md
diff --git a/Algorithms/0079. Word Search/79. Word Search.go b/leetcode/0079.Word-Search/79. Word Search.go
similarity index 94%
rename from Algorithms/0079. Word Search/79. Word Search.go
rename to leetcode/0079.Word-Search/79. Word Search.go
index 6a6bbca54..a2bd2207b 100644
--- a/Algorithms/0079. Word Search/79. Word Search.go	
+++ b/leetcode/0079.Word-Search/79. Word Search.go	
@@ -1,10 +1,10 @@
 package leetcode
 
 var dir = [][]int{
-	[]int{-1, 0},
-	[]int{0, 1},
-	[]int{1, 0},
-	[]int{0, -1},
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
 }
 
 func exist(board [][]byte, word string) bool {
diff --git a/leetcode/0079.Word-Search/79. Word Search_test.go b/leetcode/0079.Word-Search/79. Word Search_test.go
new file mode 100644
index 000000000..85d992c3b
--- /dev/null
+++ b/leetcode/0079.Word-Search/79. Word Search_test.go	
@@ -0,0 +1,105 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question79 struct {
+	para79
+	ans79
+}
+
+// para 是参数
+// one 代表第一个参数
+type para79 struct {
+	b    [][]byte
+	word string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans79 struct {
+	one bool
+}
+
+func Test_Problem79(t *testing.T) {
+
+	qs := []question79{
+
+		{
+			para79{[][]byte{
+				{'A', 'B', 'C', 'E'},
+				{'S', 'F', 'C', 'S'},
+				{'A', 'D', 'E', 'E'},
+			}, "ABCCED"},
+			ans79{true},
+		},
+
+		{
+			para79{[][]byte{
+				{'A', 'B', 'C', 'E'},
+				{'S', 'F', 'C', 'S'},
+				{'A', 'D', 'E', 'E'},
+			}, "SEE"},
+			ans79{true},
+		},
+
+		{
+			para79{[][]byte{
+				{'A', 'B', 'C', 'E'},
+				{'S', 'F', 'C', 'S'},
+				{'A', 'D', 'E', 'E'},
+			}, "ABCB"},
+			ans79{false},
+		},
+
+		{
+			para79{[][]byte{
+				{'o', 'a', 'a', 'n'},
+				{'e', 't', 'a', 'e'},
+				{'i', 'h', 'k', 'r'},
+				{'i', 'f', 'l', 'v'},
+			}, "oath"},
+			ans79{true},
+		},
+
+		{
+			para79{[][]byte{
+				{'o', 'a', 'a', 'n'},
+				{'e', 't', 'a', 'e'},
+				{'i', 'h', 'k', 'r'},
+				{'i', 'f', 'l', 'v'},
+			}, "pea"},
+			ans79{false},
+		},
+
+		{
+			para79{[][]byte{
+				{'o', 'a', 'a', 'n'},
+				{'e', 't', 'a', 'e'},
+				{'i', 'h', 'k', 'r'},
+				{'i', 'f', 'l', 'v'},
+			}, "eat"},
+			ans79{true},
+		},
+
+		{
+			para79{[][]byte{
+				{'o', 'a', 'a', 'n'},
+				{'e', 't', 'a', 'e'},
+				{'i', 'h', 'k', 'r'},
+				{'i', 'f', 'l', 'v'},
+			}, "rain"},
+			ans79{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 79------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans79, q.para79
+		fmt.Printf("【input】:%v       【output】:%v\n\n\n", p, exist(p.b, p.word))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0079. Word Search/README.md b/leetcode/0079.Word-Search/README.md
similarity index 100%
rename from Algorithms/0079. Word Search/README.md
rename to leetcode/0079.Word-Search/README.md
diff --git a/leetcode/0080.Remove-Duplicates-from-Sorted-Array-II/80. Remove Duplicates from Sorted Array II.go b/leetcode/0080.Remove-Duplicates-from-Sorted-Array-II/80. Remove Duplicates from Sorted Array II.go
new file mode 100644
index 000000000..dfb67fd53
--- /dev/null
+++ b/leetcode/0080.Remove-Duplicates-from-Sorted-Array-II/80. Remove Duplicates from Sorted Array II.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+func removeDuplicates(nums []int) int {
+	slow := 0
+	for fast, v := range nums {
+		if fast < 2 || nums[slow-2] != v {
+			nums[slow] = v
+			slow++
+		}
+	}
+	return slow
+}
diff --git a/Algorithms/0080. Remove Duplicates from Sorted Array II/80. Remove Duplicates from Sorted Array II_test.go b/leetcode/0080.Remove-Duplicates-from-Sorted-Array-II/80. Remove Duplicates from Sorted Array II_test.go
similarity index 89%
rename from Algorithms/0080. Remove Duplicates from Sorted Array II/80. Remove Duplicates from Sorted Array II_test.go
rename to leetcode/0080.Remove-Duplicates-from-Sorted-Array-II/80. Remove Duplicates from Sorted Array II_test.go
index 3e2e2dcec..76939aab3 100644
--- a/Algorithms/0080. Remove Duplicates from Sorted Array II/80. Remove Duplicates from Sorted Array II_test.go	
+++ b/leetcode/0080.Remove-Duplicates-from-Sorted-Array-II/80. Remove Duplicates from Sorted Array II_test.go	
@@ -26,32 +26,32 @@ func Test_Problem80(t *testing.T) {
 
 	qs := []question80{
 
-		question80{
+		{
 			para80{[]int{1, 1, 2}},
 			ans80{3},
 		},
 
-		question80{
+		{
 			para80{[]int{0, 0, 1, 1, 1, 1, 2, 3, 4, 4}},
 			ans80{8},
 		},
 
-		question80{
+		{
 			para80{[]int{0, 0, 0, 0, 0}},
 			ans80{2},
 		},
 
-		question80{
+		{
 			para80{[]int{1}},
 			ans80{1},
 		},
 
-		question80{
+		{
 			para80{[]int{0, 0, 1, 1, 1, 1, 2, 3, 3}},
 			ans80{7},
 		},
 
-		question80{
+		{
 			para80{[]int{1, 1, 1, 1, 2, 2, 3}},
 			ans80{5},
 		},
@@ -61,7 +61,7 @@ func Test_Problem80(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans80, q.para80
-		fmt.Printf("【input】:%v    【output】:%v\n", p.one, removeDuplicates80(p.one))
+		fmt.Printf("【input】:%v    【output】:%v\n", p.one, removeDuplicates(p.one))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/leetcode/0080.Remove-Duplicates-from-Sorted-Array-II/README.md b/leetcode/0080.Remove-Duplicates-from-Sorted-Array-II/README.md
new file mode 100644
index 000000000..1c1b25e39
--- /dev/null
+++ b/leetcode/0080.Remove-Duplicates-from-Sorted-Array-II/README.md
@@ -0,0 +1,56 @@
+# [80. Remove Duplicates from Sorted Array II](https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/)
+
+## 题目
+
+Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.
+
+Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
+
+Example 1:
+
+```c
+Given nums = [1,1,1,2,2,3],
+
+Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
+
+It doesn't matter what you leave beyond the returned length.
+```
+
+Example 2:
+
+```c
+Given nums = [0,0,1,1,1,1,2,3,3],
+
+Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively.
+
+It doesn't matter what values are set beyond the returned length.
+```
+
+Clarification:
+
+Confused why the returned value is an integer but your answer is an array?
+
+Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
+
+Internally you can think of this:
+
+```c
+// nums is passed in by reference. (i.e., without making a copy)
+int len = removeElement(nums, val);
+
+// any modification to nums in your function would be known by the caller.
+// using the length returned by your function, it prints the first len elements.
+for (int i = 0; i < len; i++) {
+    print(nums[i]);
+}
+```
+
+## 题目大意
+
+给定一个有序数组 nums，对数组中的元素进行去重，使得原数组中的每个元素最多暴露 2 个。最后返回去重以后数组的长度值。
+
+## 解题思路
+
+- 问题提示有序数组，一般最容易想到使用双指针的解法，双指针的关键点：移动两个指针的条件。
+- 在该题中移动的条件：快指针从头遍历数组，慢指针指向修改后的数组的末端，当慢指针指向倒数第二个数与快指针指向的数不相等时，才移动慢指针，同时赋值慢指针。
+- 处理边界条件：当数组小于两个元素时，不做处理。
diff --git a/Algorithms/0081. Search in Rotated Sorted Array II/81. Search in Rotated Sorted Array II.go b/leetcode/0081.Search-in-Rotated-Sorted-Array-II/81. Search in Rotated Sorted Array II.go
similarity index 100%
rename from Algorithms/0081. Search in Rotated Sorted Array II/81. Search in Rotated Sorted Array II.go
rename to leetcode/0081.Search-in-Rotated-Sorted-Array-II/81. Search in Rotated Sorted Array II.go
diff --git a/Algorithms/0081. Search in Rotated Sorted Array II/81. Search in Rotated Sorted Array II_test.go b/leetcode/0081.Search-in-Rotated-Sorted-Array-II/81. Search in Rotated Sorted Array II_test.go
similarity index 96%
rename from Algorithms/0081. Search in Rotated Sorted Array II/81. Search in Rotated Sorted Array II_test.go
rename to leetcode/0081.Search-in-Rotated-Sorted-Array-II/81. Search in Rotated Sorted Array II_test.go
index 9e90576e9..ba089f42d 100644
--- a/Algorithms/0081. Search in Rotated Sorted Array II/81. Search in Rotated Sorted Array II_test.go	
+++ b/leetcode/0081.Search-in-Rotated-Sorted-Array-II/81. Search in Rotated Sorted Array II_test.go	
@@ -27,12 +27,12 @@ func Test_Problem81(t *testing.T) {
 
 	qs := []question81{
 
-		question81{
+		{
 			para81{[]int{2, 5, 6, 0, 0, 1, 2}, 0},
 			ans81{true},
 		},
 
-		question81{
+		{
 			para81{[]int{2, 5, 6, 0, 0, 1, 2}, 3},
 			ans81{false},
 		},
diff --git a/leetcode/0081.Search-in-Rotated-Sorted-Array-II/README.md b/leetcode/0081.Search-in-Rotated-Sorted-Array-II/README.md
new file mode 100755
index 000000000..8ed8f89d5
--- /dev/null
+++ b/leetcode/0081.Search-in-Rotated-Sorted-Array-II/README.md
@@ -0,0 +1,44 @@
+# [81. Search in Rotated Sorted Array II](https://leetcode.com/problems/search-in-rotated-sorted-array-ii/)
+
+
+## 题目
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., `[0,0,1,2,2,5,6]` might become `[2,5,6,0,0,1,2]`).
+
+You are given a target value to search. If found in the array return `true`, otherwise return `false`.
+
+**Example 1:**
+
+    Input: nums = [2,5,6,0,0,1,2], target = 0
+    Output: true
+
+**Example 2:**
+
+    Input: nums = [2,5,6,0,0,1,2], target = 3
+    Output: false
+
+**Follow up:**
+
+- This is a follow up problem to [Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/description/), where `nums` may contain duplicates.
+- Would this affect the run-time complexity? How and why?
+
+
+## 题目大意
+
+假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,0,1,2,2,5,6] 可能变为 [2,5,6,0,0,1,2] )。
+
+编写一个函数来判断给定的目标值是否存在于数组中。若存在返回 true，否则返回 false。
+
+进阶:
+
+- 这是搜索旋转排序数组 的延伸题目，本题中的 nums  可能包含重复元素。
+- 这会影响到程序的时间复杂度吗？会有怎样的影响，为什么？
+
+
+## 解题思路
+
+
+- 给出一个数组，数组中本来是从小到大排列的，并且数组中有重复数字。但是现在把后面随机一段有序的放到数组前面，这样形成了前后两端有序的子序列。在这样的一个数组里面查找一个数，设计一个 O(log n) 的算法。如果找到就输出 `true`，如果没有找到，就输出 `false` 。
+- 这一题是第 33 题的加强版，实现代码完全一样，只不过输出变了。这一题输出 `true` 和 `false` 了。具体思路见第 33 题。
diff --git a/leetcode/0082.Remove-Duplicates-from-Sorted-List-II/82. Remove Duplicates from Sorted List II.go b/leetcode/0082.Remove-Duplicates-from-Sorted-List-II/82. Remove Duplicates from Sorted List II.go
new file mode 100644
index 000000000..aa08a6ce2
--- /dev/null
+++ b/leetcode/0082.Remove-Duplicates-from-Sorted-List-II/82. Remove Duplicates from Sorted List II.go	
@@ -0,0 +1,156 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func deleteDuplicates1(head *ListNode) *ListNode {
+	if head == nil {
+		return nil
+	}
+	if head.Next == nil {
+		return head
+	}
+	newHead := &ListNode{Next: head, Val: -999999}
+	cur := newHead
+	last := newHead
+	front := head
+	for front.Next != nil {
+		if front.Val == cur.Val {
+			// fmt.Printf("相同节点front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
+			front = front.Next
+			continue
+		} else {
+			if cur.Next != front {
+				// fmt.Printf("删除重复节点front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
+				last.Next = front
+				if front.Next != nil && front.Next.Val != front.Val {
+					last = front
+				}
+				cur = front
+				front = front.Next
+			} else {
+				// fmt.Printf("常规循环前front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
+				last = cur
+				cur = cur.Next
+				front = front.Next
+				// fmt.Printf("常规循环后front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
+
+			}
+		}
+	}
+	if front.Val == cur.Val {
+		// fmt.Printf("相同节点front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
+		last.Next = nil
+	} else {
+		if cur.Next != front {
+			last.Next = front
+		}
+	}
+	return newHead.Next
+}
+
+func deleteDuplicates2(head *ListNode) *ListNode {
+	if head == nil {
+		return nil
+	}
+	if head.Next != nil && head.Val == head.Next.Val {
+		for head.Next != nil && head.Val == head.Next.Val {
+			head = head.Next
+		}
+		return deleteDuplicates(head.Next)
+	}
+	head.Next = deleteDuplicates(head.Next)
+	return head
+}
+
+func deleteDuplicates(head *ListNode) *ListNode {
+	cur := head
+	if head == nil {
+		return nil
+	}
+	if head.Next == nil {
+		return head
+	}
+	for cur.Next != nil {
+		if cur.Next.Val == cur.Val {
+			cur.Next = cur.Next.Next
+		} else {
+			cur = cur.Next
+		}
+	}
+	return head
+}
+
+// 双循环简单解法 O(n*m)
+func deleteDuplicates3(head *ListNode) *ListNode {
+	if head == nil {
+		return head
+	}
+
+	nilNode := &ListNode{Val: 0, Next: head}
+	head = nilNode
+
+	lastVal := 0
+	for head.Next != nil && head.Next.Next != nil {
+		if head.Next.Val == head.Next.Next.Val {
+			lastVal = head.Next.Val
+			for head.Next != nil && lastVal == head.Next.Val {
+				head.Next = head.Next.Next
+			}
+		} else {
+			head = head.Next
+		}
+	}
+	return nilNode.Next
+}
+
+// 双指针+删除标志位，单循环解法 O(n)
+func deleteDuplicates4(head *ListNode) *ListNode {
+	if head == nil || head.Next == nil {
+		return head
+	}
+
+	nilNode := &ListNode{Val: 0, Next: head}
+	// 上次遍历有删除操作的标志位
+	lastIsDel := false
+	// 虚拟空结点
+	head = nilNode
+	// 前后指针用于判断
+	pre, back := head.Next, head.Next.Next
+	// 每次只删除前面的一个重复的元素，留一个用于下次遍历判重
+	// pre, back 指针的更新位置和值比较重要和巧妙
+	for head.Next != nil && head.Next.Next != nil {
+		if pre.Val != back.Val && lastIsDel {
+			head.Next = head.Next.Next
+			pre, back = head.Next, head.Next.Next
+			lastIsDel = false
+			continue
+		}
+
+		if pre.Val == back.Val {
+			head.Next = head.Next.Next
+			pre, back = head.Next, head.Next.Next
+			lastIsDel = true
+		} else {
+			head = head.Next
+			pre, back = head.Next, head.Next.Next
+			lastIsDel = false
+		}
+	}
+	// 处理 [1,1] 这种删除还剩一个的情况
+	if lastIsDel && head.Next != nil {
+		head.Next = nil
+	}
+	return nilNode.Next
+}
diff --git a/Algorithms/0082. Remove Duplicates from Sorted List II/82. Remove Duplicates from Sorted List II_test.go b/leetcode/0082.Remove-Duplicates-from-Sorted-List-II/82. Remove Duplicates from Sorted List II_test.go
similarity index 82%
rename from Algorithms/0082. Remove Duplicates from Sorted List II/82. Remove Duplicates from Sorted List II_test.go
rename to leetcode/0082.Remove-Duplicates-from-Sorted-List-II/82. Remove Duplicates from Sorted List II_test.go
index ebda8a62d..5bfd3ec28 100644
--- a/Algorithms/0082. Remove Duplicates from Sorted List II/82. Remove Duplicates from Sorted List II_test.go	
+++ b/leetcode/0082.Remove-Duplicates-from-Sorted-List-II/82. Remove Duplicates from Sorted List II_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question82 struct {
@@ -26,47 +28,47 @@ func Test_Problem82(t *testing.T) {
 
 	qs := []question82{
 
-		question82{
+		{
 			para82{[]int{1, 1, 2, 2, 3, 4, 4, 4}},
 			ans82{[]int{3}},
 		},
 
-		question82{
+		{
 			para82{[]int{1, 1, 1, 1, 1, 1}},
 			ans82{[]int{}},
 		},
 
-		question82{
+		{
 			para82{[]int{1, 1, 1, 2, 3}},
 			ans82{[]int{2, 3}},
 		},
 
-		question82{
+		{
 			para82{[]int{1}},
 			ans82{[]int{1}},
 		},
 
-		question82{
+		{
 			para82{[]int{}},
 			ans82{[]int{}},
 		},
 
-		question82{
+		{
 			para82{[]int{1, 2, 2, 2, 2}},
 			ans82{[]int{1}},
 		},
 
-		question82{
+		{
 			para82{[]int{1, 1, 2, 3, 3, 4, 5, 5, 6}},
 			ans82{[]int{2, 4, 6}},
 		},
 
-		question82{
+		{
 			para82{[]int{1, 1, 2, 3, 3, 4, 5, 6}},
 			ans82{[]int{2, 4, 5, 6}},
 		},
 
-		question82{
+		{
 			para82{[]int{0, 1, 2, 2, 3, 4}},
 			ans82{[]int{0, 1, 2, 2, 3, 4}},
 		},
@@ -76,7 +78,7 @@ func Test_Problem82(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans82, q.para82
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(deleteDuplicates1(S2l(p.one))))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(deleteDuplicates1(structures.Ints2List(p.one))))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0082. Remove Duplicates from Sorted List II/README.md b/leetcode/0082.Remove-Duplicates-from-Sorted-List-II/README.md
similarity index 100%
rename from Algorithms/0082. Remove Duplicates from Sorted List II/README.md
rename to leetcode/0082.Remove-Duplicates-from-Sorted-List-II/README.md
diff --git a/Algorithms/0083. Remove Duplicates from Sorted List/83. Remove Duplicates from Sorted List.go b/leetcode/0083.Remove-Duplicates-from-Sorted-List/83. Remove Duplicates from Sorted List.go
similarity index 77%
rename from Algorithms/0083. Remove Duplicates from Sorted List/83. Remove Duplicates from Sorted List.go
rename to leetcode/0083.Remove-Duplicates-from-Sorted-List/83. Remove Duplicates from Sorted List.go
index 5abf64311..4c9e711cc 100644
--- a/Algorithms/0083. Remove Duplicates from Sorted List/83. Remove Duplicates from Sorted List.go	
+++ b/leetcode/0083.Remove-Duplicates-from-Sorted-List/83. Remove Duplicates from Sorted List.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0083. Remove Duplicates from Sorted List/83. Remove Duplicates from Sorted List_test.go b/leetcode/0083.Remove-Duplicates-from-Sorted-List/83. Remove Duplicates from Sorted List_test.go
similarity index 78%
rename from Algorithms/0083. Remove Duplicates from Sorted List/83. Remove Duplicates from Sorted List_test.go
rename to leetcode/0083.Remove-Duplicates-from-Sorted-List/83. Remove Duplicates from Sorted List_test.go
index 54e9507a1..2feb9fee6 100644
--- a/Algorithms/0083. Remove Duplicates from Sorted List/83. Remove Duplicates from Sorted List_test.go	
+++ b/leetcode/0083.Remove-Duplicates-from-Sorted-List/83. Remove Duplicates from Sorted List_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question83 struct {
@@ -26,17 +28,17 @@ func Test_Problem83(t *testing.T) {
 
 	qs := []question83{
 
-		question83{
+		{
 			para83{[]int{1, 1, 2}},
 			ans83{[]int{1, 2}},
 		},
 
-		question83{
+		{
 			para83{[]int{1, 1, 2, 2, 3, 3, 3}},
 			ans83{[]int{1, 2, 3}},
 		},
 
-		question83{
+		{
 			para83{[]int{1, 1, 1, 1, 1, 1, 1, 1}},
 			ans83{[]int{1}},
 		},
@@ -46,7 +48,7 @@ func Test_Problem83(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans83, q.para83
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(deleteDuplicates(S2l(p.one))))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(deleteDuplicates(structures.Ints2List(p.one))))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0083. Remove Duplicates from Sorted List/README.md b/leetcode/0083.Remove-Duplicates-from-Sorted-List/README.md
similarity index 100%
rename from Algorithms/0083. Remove Duplicates from Sorted List/README.md
rename to leetcode/0083.Remove-Duplicates-from-Sorted-List/README.md
diff --git a/leetcode/0084.Largest-Rectangle-in-Histogram/84. Largest Rectangle in Histogram.go b/leetcode/0084.Largest-Rectangle-in-Histogram/84. Largest Rectangle in Histogram.go
new file mode 100644
index 000000000..d01401b05
--- /dev/null
+++ b/leetcode/0084.Largest-Rectangle-in-Histogram/84. Largest Rectangle in Histogram.go	
@@ -0,0 +1,32 @@
+package leetcode
+
+func largestRectangleArea(heights []int) int {
+	maxArea := 0
+	n := len(heights) + 2
+	// Add a sentry at the beginning and the end
+	getHeight := func(i int) int {
+		if i == 0 || n-1 == i {
+			return 0
+		}
+		return heights[i-1]
+	}
+	st := make([]int, 0, n/2)
+	for i := 0; i < n; i++ {
+		for len(st) > 0 && getHeight(st[len(st)-1]) > getHeight(i) {
+			// pop stack
+			idx := st[len(st)-1]
+			st = st[:len(st)-1]
+			maxArea = max(maxArea, getHeight(idx)*(i-st[len(st)-1]-1))
+		}
+		// push stack
+		st = append(st, i)
+	}
+	return maxArea
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0084. Largest Rectangle in Histogram/84. Largest Rectangle in Histogram_test.go b/leetcode/0084.Largest-Rectangle-in-Histogram/84. Largest Rectangle in Histogram_test.go
similarity index 92%
rename from Algorithms/0084. Largest Rectangle in Histogram/84. Largest Rectangle in Histogram_test.go
rename to leetcode/0084.Largest-Rectangle-in-Histogram/84. Largest Rectangle in Histogram_test.go
index 65f0621eb..0465a1eb6 100644
--- a/Algorithms/0084. Largest Rectangle in Histogram/84. Largest Rectangle in Histogram_test.go	
+++ b/leetcode/0084.Largest-Rectangle-in-Histogram/84. Largest Rectangle in Histogram_test.go	
@@ -26,20 +26,24 @@ func Test_Problem84(t *testing.T) {
 
 	qs := []question84{
 
-		question84{
+		{
 			para84{[]int{2, 1, 5, 6, 2, 3}},
 			ans84{10},
 		},
 
-		question84{
+		{
 			para84{[]int{1}},
 			ans84{1},
 		},
 
-		question84{
+		{
 			para84{[]int{1, 1}},
 			ans84{2},
 		},
+		{
+			para84{[]int{2, 1, 2}},
+			ans84{3},
+		},
 	}
 
 	fmt.Printf("------------------------Leetcode Problem 84------------------------\n")
diff --git a/Algorithms/0084. Largest Rectangle in Histogram/README.md b/leetcode/0084.Largest-Rectangle-in-Histogram/README.md
similarity index 100%
rename from Algorithms/0084. Largest Rectangle in Histogram/README.md
rename to leetcode/0084.Largest-Rectangle-in-Histogram/README.md
diff --git a/Algorithms/0086. Partition List/86. Partition List.go b/leetcode/0086.Partition-List/86. Partition List.go
similarity index 95%
rename from Algorithms/0086. Partition List/86. Partition List.go
rename to leetcode/0086.Partition-List/86. Partition List.go
index 4d4b40c32..d031e884c 100644
--- a/Algorithms/0086. Partition List/86. Partition List.go	
+++ b/leetcode/0086.Partition-List/86. Partition List.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0086. Partition List/86. Partition List_test.go b/leetcode/0086.Partition-List/86. Partition List_test.go
similarity index 83%
rename from Algorithms/0086. Partition List/86. Partition List_test.go
rename to leetcode/0086.Partition-List/86. Partition List_test.go
index 31a23cfb1..4fd4e5f74 100644
--- a/Algorithms/0086. Partition List/86. Partition List_test.go	
+++ b/leetcode/0086.Partition-List/86. Partition List_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question86 struct {
@@ -27,37 +29,37 @@ func Test_Problem86(t *testing.T) {
 
 	qs := []question86{
 
-		question86{
+		{
 			para86{[]int{1, 4, 3, 2, 5, 2}, 3},
 			ans86{[]int{1, 2, 2, 4, 3, 5}},
 		},
 
-		question86{
+		{
 			para86{[]int{1, 1, 2, 2, 3, 3, 3}, 2},
 			ans86{[]int{1, 1, 2, 2, 3, 3, 3}},
 		},
 
-		question86{
+		{
 			para86{[]int{1, 4, 3, 2, 5, 2}, 0},
 			ans86{[]int{1, 4, 3, 2, 5, 2}},
 		},
 
-		question86{
+		{
 			para86{[]int{4, 3, 2, 5, 2}, 3},
 			ans86{[]int{2, 2, 4, 3, 5}},
 		},
 
-		question86{
+		{
 			para86{[]int{1, 1, 1, 1, 1, 1}, 1},
 			ans86{[]int{1, 1, 1, 1, 1, 1}},
 		},
 
-		question86{
+		{
 			para86{[]int{3, 1}, 2},
 			ans86{[]int{1, 3}},
 		},
 
-		question86{
+		{
 			para86{[]int{1, 2}, 3},
 			ans86{[]int{1, 2}},
 		},
@@ -67,7 +69,7 @@ func Test_Problem86(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans86, q.para86
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(partition(S2l(p.one), p.x)))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(partition(structures.Ints2List(p.one), p.x)))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0086. Partition List/README.md b/leetcode/0086.Partition-List/README.md
similarity index 100%
rename from Algorithms/0086. Partition List/README.md
rename to leetcode/0086.Partition-List/README.md
diff --git a/leetcode/0088.Merge-Sorted-Array/88. Merge Sorted Array.go b/leetcode/0088.Merge-Sorted-Array/88. Merge Sorted Array.go
new file mode 100644
index 000000000..6b30f8e76
--- /dev/null
+++ b/leetcode/0088.Merge-Sorted-Array/88. Merge Sorted Array.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func merge(nums1 []int, m int, nums2 []int, n int) {
+	for p := m + n; m > 0 && n > 0; p-- {
+		if nums1[m-1] <= nums2[n-1] {
+			nums1[p-1] = nums2[n-1]
+			n--
+		} else {
+			nums1[p-1] = nums1[m-1]
+			m--
+		}
+	}
+	for ; n > 0; n-- {
+		nums1[n-1] = nums2[n-1]
+	}
+}
diff --git a/leetcode/0088.Merge-Sorted-Array/88. Merge Sorted Array_test.go b/leetcode/0088.Merge-Sorted-Array/88. Merge Sorted Array_test.go
new file mode 100644
index 000000000..3c4ebb1da
--- /dev/null
+++ b/leetcode/0088.Merge-Sorted-Array/88. Merge Sorted Array_test.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question88 struct {
+	para88
+	ans88
+}
+
+// para 是参数
+// one 代表第一个参数
+type para88 struct {
+	one []int
+	m   int
+	two []int
+	n   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans88 struct {
+	one []int
+}
+
+func Test_Problem88(t *testing.T) {
+
+	qs := []question88{
+
+		{
+			para88{[]int{0}, 0, []int{1}, 1},
+			ans88{[]int{1}},
+		},
+
+		{
+			para88{[]int{1, 2, 3, 0, 0, 0}, 3, []int{2, 5, 6}, 3},
+			ans88{[]int{1, 2, 2, 3, 5, 6}},
+		},
+	}
+	fmt.Printf("------------------------Leetcode Problem 88------------------------\n")
+	for _, q := range qs {
+		_, p := q.ans88, q.para88
+		fmt.Printf("【intput】:%v,%v,%v,%v       ", p.one, p.m, p.two, p.n)
+		merge(p.one, p.m, p.two, p.n)
+		fmt.Printf("【output】:%v\n", p)
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0088. Merge-Sorted-Array/README.md b/leetcode/0088.Merge-Sorted-Array/README.md
similarity index 100%
rename from Algorithms/0088. Merge-Sorted-Array/README.md
rename to leetcode/0088.Merge-Sorted-Array/README.md
diff --git a/Algorithms/0089. Gray Code/89. Gray Code.go b/leetcode/0089.Gray-Code/89. Gray Code.go
similarity index 100%
rename from Algorithms/0089. Gray Code/89. Gray Code.go
rename to leetcode/0089.Gray-Code/89. Gray Code.go
diff --git a/Algorithms/0089. Gray Code/89. Gray Code_test.go b/leetcode/0089.Gray-Code/89. Gray Code_test.go
similarity index 94%
rename from Algorithms/0089. Gray Code/89. Gray Code_test.go
rename to leetcode/0089.Gray-Code/89. Gray Code_test.go
index 80da647f4..5d27d3dc0 100644
--- a/Algorithms/0089. Gray Code/89. Gray Code_test.go	
+++ b/leetcode/0089.Gray-Code/89. Gray Code_test.go	
@@ -26,17 +26,17 @@ func Test_Problem89(t *testing.T) {
 
 	qs := []question89{
 
-		question89{
+		{
 			para89{2},
 			ans89{[]int{0, 1, 3, 2}},
 		},
 
-		question89{
+		{
 			para89{0},
 			ans89{[]int{0}},
 		},
 
-		question89{
+		{
 			para89{3},
 			ans89{[]int{0, 1, 3, 2, 6, 7, 5, 4}},
 		},
diff --git a/Algorithms/0089. Gray Code/README.md b/leetcode/0089.Gray-Code/README.md
similarity index 100%
rename from Algorithms/0089. Gray Code/README.md
rename to leetcode/0089.Gray-Code/README.md
diff --git a/Algorithms/0090. Subsets II/90. Subsets II.go b/leetcode/0090.Subsets-II/90. Subsets II.go
similarity index 100%
rename from Algorithms/0090. Subsets II/90. Subsets II.go
rename to leetcode/0090.Subsets-II/90. Subsets II.go
diff --git a/Algorithms/0090. Subsets II/90. Subsets II_test.go b/leetcode/0090.Subsets-II/90. Subsets II_test.go
similarity index 80%
rename from Algorithms/0090. Subsets II/90. Subsets II_test.go
rename to leetcode/0090.Subsets-II/90. Subsets II_test.go
index 8a5c1b9b1..6b099e76d 100644
--- a/Algorithms/0090. Subsets II/90. Subsets II_test.go	
+++ b/leetcode/0090.Subsets-II/90. Subsets II_test.go	
@@ -26,14 +26,14 @@ func Test_Problem90(t *testing.T) {
 
 	qs := []question90{
 
-		question90{
+		{
 			para90{[]int{}},
-			ans90{[][]int{[]int{}}},
+			ans90{[][]int{{}}},
 		},
 
-		question90{
+		{
 			para90{[]int{1, 2, 2}},
-			ans90{[][]int{[]int{}, []int{1}, []int{2}, []int{1, 2}, []int{2, 2}, []int{1, 2, 2}}},
+			ans90{[][]int{{}, {1}, {2}, {1, 2}, {2, 2}, {1, 2, 2}}},
 		},
 	}
 
diff --git a/Algorithms/0090. Subsets II/README.md b/leetcode/0090.Subsets-II/README.md
similarity index 100%
rename from Algorithms/0090. Subsets II/README.md
rename to leetcode/0090.Subsets-II/README.md
diff --git a/leetcode/0091.Decode-Ways/91. Decode Ways.go b/leetcode/0091.Decode-Ways/91. Decode Ways.go
new file mode 100644
index 000000000..6746cdc47
--- /dev/null
+++ b/leetcode/0091.Decode-Ways/91. Decode Ways.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func numDecodings(s string) int {
+	n := len(s)
+	dp := make([]int, n+1)
+	dp[0] = 1
+	for i := 1; i <= n; i++ {
+		if s[i-1] != '0' {
+			dp[i] += dp[i-1]
+		}
+		if i > 1 && s[i-2] != '0' && (s[i-2]-'0')*10+(s[i-1]-'0') <= 26 {
+			dp[i] += dp[i-2]
+		}
+	}
+	return dp[n]
+}
diff --git a/Algorithms/0091. Decode Ways/91. Decode Ways_test.go b/leetcode/0091.Decode-Ways/91. Decode Ways_test.go
similarity index 95%
rename from Algorithms/0091. Decode Ways/91. Decode Ways_test.go
rename to leetcode/0091.Decode-Ways/91. Decode Ways_test.go
index 584cfcb93..2e2740697 100644
--- a/Algorithms/0091. Decode Ways/91. Decode Ways_test.go	
+++ b/leetcode/0091.Decode-Ways/91. Decode Ways_test.go	
@@ -26,12 +26,12 @@ func Test_Problem91(t *testing.T) {
 
 	qs := []question91{
 
-		question91{
+		{
 			para91{"12"},
 			ans91{2},
 		},
 
-		question91{
+		{
 			para91{"226"},
 			ans91{3},
 		},
diff --git a/Algorithms/0091. Decode Ways/README.md b/leetcode/0091.Decode-Ways/README.md
similarity index 100%
rename from Algorithms/0091. Decode Ways/README.md
rename to leetcode/0091.Decode-Ways/README.md
diff --git a/Algorithms/0092. Reverse Linked List II/92. Reverse Linked List II.go b/leetcode/0092.Reverse-Linked-List-II/92. Reverse Linked List II.go
similarity index 83%
rename from Algorithms/0092. Reverse Linked List II/92. Reverse Linked List II.go
rename to leetcode/0092.Reverse-Linked-List-II/92. Reverse Linked List II.go
index 5b5fcba91..47ada37b1 100644
--- a/Algorithms/0092. Reverse Linked List II/92. Reverse Linked List II.go	
+++ b/leetcode/0092.Reverse-Linked-List-II/92. Reverse Linked List II.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0092. Reverse Linked List II/92. Reverse Linked List II_test.go b/leetcode/0092.Reverse-Linked-List-II/92. Reverse Linked List II_test.go
similarity index 81%
rename from Algorithms/0092. Reverse Linked List II/92. Reverse Linked List II_test.go
rename to leetcode/0092.Reverse-Linked-List-II/92. Reverse Linked List II_test.go
index 52b844a0f..a209bb26a 100644
--- a/Algorithms/0092. Reverse Linked List II/92. Reverse Linked List II_test.go	
+++ b/leetcode/0092.Reverse-Linked-List-II/92. Reverse Linked List II_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question92 struct {
@@ -27,32 +29,32 @@ func Test_Problem92(t *testing.T) {
 
 	qs := []question92{
 
-		question92{
+		{
 			para92{[]int{1, 2, 3, 4, 5}, 2, 4},
 			ans92{[]int{1, 4, 3, 2, 5}},
 		},
 
-		question92{
+		{
 			para92{[]int{1, 2, 3, 4, 5}, 2, 2},
 			ans92{[]int{1, 2, 3, 4, 5}},
 		},
 
-		question92{
+		{
 			para92{[]int{1, 2, 3, 4, 5}, 1, 5},
 			ans92{[]int{5, 4, 3, 2, 1}},
 		},
 
-		question92{
+		{
 			para92{[]int{1, 2, 3, 4, 5, 6}, 3, 4},
 			ans92{[]int{1, 2, 4, 3, 5, 6}},
 		},
 
-		question92{
+		{
 			para92{[]int{3, 5}, 1, 2},
 			ans92{[]int{5, 3}},
 		},
 
-		question92{
+		{
 			para92{[]int{3}, 3, 5},
 			ans92{[]int{3}},
 		},
@@ -62,7 +64,7 @@ func Test_Problem92(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans92, q.para92
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(reverseBetween(S2l(p.one), p.m, p.n)))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(reverseBetween(structures.Ints2List(p.one), p.m, p.n)))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0092. Reverse Linked List II/README.md b/leetcode/0092.Reverse-Linked-List-II/README.md
similarity index 100%
rename from Algorithms/0092. Reverse Linked List II/README.md
rename to leetcode/0092.Reverse-Linked-List-II/README.md
diff --git a/Algorithms/0093. Restore IP Addresses/93. Restore IP Addresses.go b/leetcode/0093.Restore-IP-Addresses/93. Restore IP Addresses.go
similarity index 100%
rename from Algorithms/0093. Restore IP Addresses/93. Restore IP Addresses.go
rename to leetcode/0093.Restore-IP-Addresses/93. Restore IP Addresses.go
diff --git a/Algorithms/0093. Restore IP Addresses/93. Restore IP Addresses_test.go b/leetcode/0093.Restore-IP-Addresses/93. Restore IP Addresses_test.go
similarity index 94%
rename from Algorithms/0093. Restore IP Addresses/93. Restore IP Addresses_test.go
rename to leetcode/0093.Restore-IP-Addresses/93. Restore IP Addresses_test.go
index f528a746e..143b9892e 100644
--- a/Algorithms/0093. Restore IP Addresses/93. Restore IP Addresses_test.go	
+++ b/leetcode/0093.Restore-IP-Addresses/93. Restore IP Addresses_test.go	
@@ -26,17 +26,17 @@ func Test_Problem93(t *testing.T) {
 
 	qs := []question93{
 
-		question93{
+		{
 			para93{"25525511135"},
 			ans93{[]string{"255.255.11.135", "255.255.111.35"}},
 		},
 
-		question93{
+		{
 			para93{"0000"},
 			ans93{[]string{"0.0.0.0"}},
 		},
 
-		question93{
+		{
 			para93{"010010"},
 			ans93{[]string{"0.10.0.10", "0.100.1.0"}},
 		},
diff --git a/Algorithms/0093. Restore IP Addresses/README.md b/leetcode/0093.Restore-IP-Addresses/README.md
similarity index 100%
rename from Algorithms/0093. Restore IP Addresses/README.md
rename to leetcode/0093.Restore-IP-Addresses/README.md
diff --git a/Algorithms/0094. Binary Tree Inorder Traversal/94. Binary Tree Inorder Traversal.go b/leetcode/0094.Binary-Tree-Inorder-Traversal/94. Binary Tree Inorder Traversal.go
similarity index 78%
rename from Algorithms/0094. Binary Tree Inorder Traversal/94. Binary Tree Inorder Traversal.go
rename to leetcode/0094.Binary-Tree-Inorder-Traversal/94. Binary Tree Inorder Traversal.go
index 6965b7ef9..95c6a8a3f 100644
--- a/Algorithms/0094. Binary Tree Inorder Traversal/94. Binary Tree Inorder Traversal.go	
+++ b/leetcode/0094.Binary-Tree-Inorder-Traversal/94. Binary Tree Inorder Traversal.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func inorderTraversal(root *TreeNode) []int {
 	var result []int
 	inorder(root, &result)
diff --git a/Algorithms/0094. Binary Tree Inorder Traversal/94. Binary Tree Inorder Traversal_test.go b/leetcode/0094.Binary-Tree-Inorder-Traversal/94. Binary Tree Inorder Traversal_test.go
similarity index 83%
rename from Algorithms/0094. Binary Tree Inorder Traversal/94. Binary Tree Inorder Traversal_test.go
rename to leetcode/0094.Binary-Tree-Inorder-Traversal/94. Binary Tree Inorder Traversal_test.go
index 5d5b44eb0..cb71b7293 100644
--- a/Algorithms/0094. Binary Tree Inorder Traversal/94. Binary Tree Inorder Traversal_test.go	
+++ b/leetcode/0094.Binary-Tree-Inorder-Traversal/94. Binary Tree Inorder Traversal_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question94 struct {
@@ -26,18 +28,18 @@ func Test_Problem94(t *testing.T) {
 
 	qs := []question94{
 
-		question94{
+		{
 			para94{[]int{}},
 			ans94{[]int{}},
 		},
 
-		question94{
+		{
 			para94{[]int{1}},
 			ans94{[]int{1}},
 		},
 
-		question94{
-			para94{[]int{1, NULL, 2, 3}},
+		{
+			para94{[]int{1, structures.NULL, 2, 3}},
 			ans94{[]int{1, 2, 3}},
 		},
 	}
@@ -47,7 +49,7 @@ func Test_Problem94(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans94, q.para94
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", inorderTraversal(root))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0094. Binary Tree Inorder Traversal/README.md b/leetcode/0094.Binary-Tree-Inorder-Traversal/README.md
similarity index 100%
rename from Algorithms/0094. Binary Tree Inorder Traversal/README.md
rename to leetcode/0094.Binary-Tree-Inorder-Traversal/README.md
diff --git a/Algorithms/0095. Unique Binary Search Trees II/95. Unique Binary Search Trees II.go b/leetcode/0095.Unique-Binary-Search-Trees-II/95. Unique Binary Search Trees II.go
similarity index 76%
rename from Algorithms/0095. Unique Binary Search Trees II/95. Unique Binary Search Trees II.go
rename to leetcode/0095.Unique-Binary-Search-Trees-II/95. Unique Binary Search Trees II.go
index 3d180a62f..27f4cf17d 100644
--- a/Algorithms/0095. Unique Binary Search Trees II/95. Unique Binary Search Trees II.go	
+++ b/leetcode/0095.Unique-Binary-Search-Trees-II/95. Unique Binary Search Trees II.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func generateTrees(n int) []*TreeNode {
 	if n == 0 {
 		return []*TreeNode{}
@@ -21,10 +29,9 @@ func generateBSTree(start, end int) []*TreeNode {
 		tree = append(tree, nil)
 		return tree
 	}
-	left, right := []*TreeNode{}, []*TreeNode{}
 	for i := start; i <= end; i++ {
-		left = generateBSTree(start, i-1)
-		right = generateBSTree(i+1, end)
+		left := generateBSTree(start, i-1)
+		right := generateBSTree(i+1, end)
 		for _, l := range left {
 			for _, r := range right {
 				root := &TreeNode{Val: i, Left: l, Right: r}
diff --git a/leetcode/0095.Unique-Binary-Search-Trees-II/95. Unique Binary Search Trees II_test.go b/leetcode/0095.Unique-Binary-Search-Trees-II/95. Unique Binary Search Trees II_test.go
new file mode 100644
index 000000000..3cac732d5
--- /dev/null
+++ b/leetcode/0095.Unique-Binary-Search-Trees-II/95. Unique Binary Search Trees II_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question95 struct {
+	para95
+	ans95
+}
+
+// para 是参数
+// one 代表第一个参数
+type para95 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans95 struct {
+	one []*TreeNode
+}
+
+func Test_Problem95(t *testing.T) {
+
+	qs := []question95{
+
+		{
+			para95{1},
+			ans95{[]*TreeNode{{Val: 1, Left: nil, Right: nil}}},
+		},
+
+		{
+			para95{3},
+			ans95{[]*TreeNode{
+				{Val: 1, Left: nil, Right: &TreeNode{Val: 3, Left: &TreeNode{Val: 2, Left: nil, Right: nil}, Right: nil}},
+				{Val: 1, Left: nil, Right: &TreeNode{Val: 2, Left: nil, Right: &TreeNode{Val: 3, Left: nil, Right: nil}}},
+				{Val: 3, Left: &TreeNode{Val: 2, Left: &TreeNode{Val: 1, Left: nil, Right: nil}, Right: nil}, Right: nil},
+				{Val: 3, Left: &TreeNode{Val: 1, Left: nil, Right: &TreeNode{Val: 2, Left: nil, Right: nil}}, Right: nil},
+				{Val: 2, Left: &TreeNode{Val: 1, Left: nil, Right: nil}, Right: &TreeNode{Val: 3, Left: nil, Right: nil}},
+			}}},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 95------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans95, q.para95
+		fmt.Printf("【input】:%v  \n", p)
+		trees := generateTrees(p.one)
+		for _, t := range trees {
+			fmt.Printf("【output】:%v\n", structures.Tree2Preorder(t))
+		}
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0095.Unique-Binary-Search-Trees-II/README.md b/leetcode/0095.Unique-Binary-Search-Trees-II/README.md
new file mode 100755
index 000000000..bda2dd0ef
--- /dev/null
+++ b/leetcode/0095.Unique-Binary-Search-Trees-II/README.md
@@ -0,0 +1,35 @@
+# [95. Unique Binary Search Trees II](https://leetcode.com/problems/unique-binary-search-trees-ii/)
+
+
+## 题目
+
+Given an integer *n*, generate all structurally unique **BST's** (binary search trees) that store values 1 ... *n*.
+
+**Example:**
+
+    Input: 3
+    Output:
+    [
+      [1,null,3,2],
+      [3,2,null,1],
+      [3,1,null,null,2],
+      [2,1,3],
+      [1,null,2,null,3]
+    ]
+    Explanation:
+    The above output corresponds to the 5 unique BST's shown below:
+    
+       1         3     3      2      1
+        \       /     /      / \      \
+         3     2     1      1   3      2
+        /     /       \                 \
+       2     1         2                 3
+
+
+## 题目大意
+
+给定一个整数 n，生成所有由 1 ... n 为节点所组成的二叉搜索树。
+
+## 解题思路
+
+- 输出 1~n 元素组成的 BST 所有解。这一题递归求解即可。外层循环遍历 1~n 所有结点，作为根结点，内层双层递归分别求出左子树和右子树。
diff --git a/Algorithms/0096. Unique Binary Search Trees/96. Unique Binary Search Trees.go b/leetcode/0096.Unique-Binary-Search-Trees/96. Unique Binary Search Trees.go
similarity index 100%
rename from Algorithms/0096. Unique Binary Search Trees/96. Unique Binary Search Trees.go
rename to leetcode/0096.Unique-Binary-Search-Trees/96. Unique Binary Search Trees.go
diff --git a/Algorithms/0096. Unique Binary Search Trees/96. Unique Binary Search Trees_test.go b/leetcode/0096.Unique-Binary-Search-Trees/96. Unique Binary Search Trees_test.go
similarity index 87%
rename from Algorithms/0096. Unique Binary Search Trees/96. Unique Binary Search Trees_test.go
rename to leetcode/0096.Unique-Binary-Search-Trees/96. Unique Binary Search Trees_test.go
index e573c1964..dcf364584 100644
--- a/Algorithms/0096. Unique Binary Search Trees/96. Unique Binary Search Trees_test.go	
+++ b/leetcode/0096.Unique-Binary-Search-Trees/96. Unique Binary Search Trees_test.go	
@@ -26,46 +26,46 @@ func Test_Problem96(t *testing.T) {
 
 	qs := []question96{
 
-		question96{
+		{
 			para96{1},
 			ans96{1},
 		},
 
-		question96{
+		{
 			para96{3},
 			ans96{5},
 		},
 
-		question96{
+		{
 			para96{4},
 			ans96{14},
 		},
 
-		question96{
+		{
 			para96{5},
 			ans96{42},
 		},
 
-		question96{
+		{
 			para96{6},
 			ans96{132},
 		},
 
-		question96{
+		{
 			para96{7},
 			ans96{429},
 		},
 
-		question96{
+		{
 			para96{8},
 			ans96{1430},
 		},
 
-		question96{
+		{
 			para96{9},
 			ans96{4862},
 		},
-		question96{
+		{
 			para96{10},
 			ans96{16796},
 		},
diff --git a/Algorithms/0096. Unique Binary Search Trees/README.md b/leetcode/0096.Unique-Binary-Search-Trees/README.md
similarity index 100%
rename from Algorithms/0096. Unique Binary Search Trees/README.md
rename to leetcode/0096.Unique-Binary-Search-Trees/README.md
diff --git a/leetcode/0097.Interleaving-String/97. Interleaving String.go b/leetcode/0097.Interleaving-String/97. Interleaving String.go
new file mode 100644
index 000000000..20d62f152
--- /dev/null
+++ b/leetcode/0097.Interleaving-String/97. Interleaving String.go	
@@ -0,0 +1,35 @@
+package leetcode
+
+func isInterleave(s1 string, s2 string, s3 string) bool {
+	if len(s1)+len(s2) != len(s3) {
+		return false
+	}
+	visited := make(map[int]bool)
+	return dfs(s1, s2, s3, 0, 0, visited)
+}
+
+func dfs(s1, s2, s3 string, p1, p2 int, visited map[int]bool) bool {
+	if p1+p2 == len(s3) {
+		return true
+	}
+	if _, ok := visited[(p1*len(s3))+p2]; ok {
+		return false
+	}
+	visited[(p1*len(s3))+p2] = true
+	var match1, match2 bool
+	if p1 < len(s1) && s3[p1+p2] == s1[p1] {
+		match1 = true
+	}
+	if p2 < len(s2) && s3[p1+p2] == s2[p2] {
+		match2 = true
+	}
+	if match1 && match2 {
+		return dfs(s1, s2, s3, p1+1, p2, visited) || dfs(s1, s2, s3, p1, p2+1, visited)
+	} else if match1 {
+		return dfs(s1, s2, s3, p1+1, p2, visited)
+	} else if match2 {
+		return dfs(s1, s2, s3, p1, p2+1, visited)
+	} else {
+		return false
+	}
+}
diff --git a/leetcode/0097.Interleaving-String/97. Interleaving String_test.go b/leetcode/0097.Interleaving-String/97. Interleaving String_test.go
new file mode 100644
index 000000000..20ee4e78d
--- /dev/null
+++ b/leetcode/0097.Interleaving-String/97. Interleaving String_test.go	
@@ -0,0 +1,54 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question97 struct {
+	para97
+	ans97
+}
+
+// para 是参数
+// one 代表第一个参数
+type para97 struct {
+	s1 string
+	s2 string
+	s3 string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans97 struct {
+	one bool
+}
+
+func Test_Problem97(t *testing.T) {
+
+	qs := []question97{
+
+		{
+			para97{"aabcc", "dbbca", "aadbbcbcac"},
+			ans97{true},
+		},
+
+		{
+			para97{"aabcc", "dbbca", "aadbbbaccc"},
+			ans97{false},
+		},
+
+		{
+			para97{"", "", ""},
+			ans97{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 97------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans97, q.para97
+		fmt.Printf("【input】:%v       【output】:%v\n", p, isInterleave(p.s1, p.s2, p.s3))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0097.Interleaving-String/README.md b/leetcode/0097.Interleaving-String/README.md
new file mode 100644
index 000000000..2a720371a
--- /dev/null
+++ b/leetcode/0097.Interleaving-String/README.md
@@ -0,0 +1,104 @@
+# [97. Interleaving String](https://leetcode.com/problems/interleaving-string/)
+
+
+## 题目
+
+Given strings `s1`, `s2`, and `s3`, find whether `s3` is formed by an **interleaving** of `s1` and `s2`.
+
+An **interleaving** of two strings `s` and `t` is a configuration where they are divided into **non-empty** substrings such that:
+
+- `s = s1 + s2 + ... + sn`
+- `t = t1 + t2 + ... + tm`
+- `|n - m| <= 1`
+- The **interleaving** is `s1 + t1 + s2 + t2 + s3 + t3 + ...` or `t1 + s1 + t2 + s2 + t3 + s3 + ...`
+
+**Note:** `a + b` is the concatenation of strings `a` and `b`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/09/02/interleave.jpg](https://assets.leetcode.com/uploads/2020/09/02/interleave.jpg)
+
+```
+Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
+Output: true
+
+```
+
+**Example 2:**
+
+```
+Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
+Output: false
+
+```
+
+**Example 3:**
+
+```
+Input: s1 = "", s2 = "", s3 = ""
+Output: true
+
+```
+
+**Constraints:**
+
+- `0 <= s1.length, s2.length <= 100`
+- `0 <= s3.length <= 200`
+- `s1`, `s2`, and `s3` consist of lowercase English letters.
+
+**Follow up:** Could you solve it using only `O(s2.length)` additional memory space?
+
+## 题目大意
+
+给定三个字符串 s1、s2、s3，请你帮忙验证 s3 是否是由 s1 和 s2 交错 组成的。两个字符串 s 和 t 交错 的定义与过程如下，其中每个字符串都会被分割成若干 非空 子字符串：
+
+- s = s1 + s2 + ... + sn
+- t = t1 + t2 + ... + tm
+- |n - m| <= 1
+- 交错 是 s1 + t1 + s2 + t2 + s3 + t3 + ... 或者 t1 + s1 + t2 + s2 + t3 + s3 + ...
+
+提示：a + b 意味着字符串 a 和 b 连接。
+
+## 解题思路
+
+- 深搜或者广搜暴力解题。笔者用深搜实现的。记录 s1 和 s2 串当前比较的位置 p1 和 p2。如果 s3[p1+p2] 的位置上等于 s1[p1] 或者 s2[p2] 代表能匹配上，那么继续往后移动 p1 和 p2 相应的位置。因为是交错字符串，所以判断匹配的位置是 s3[p1+p2] 的位置。如果仅仅这么写，会超时，s1 和 s2 两个字符串重复交叉判断的位置太多了。需要加上记忆化搜索。可以用 visited[i][j] 这样的二维数组来记录是否搜索过了。笔者为了压缩空间，将 i 和 j 编码压缩到一维数组了。i * len(s3) + j 是唯一下标，所以可以用这种方式存储是否搜索过。具体代码见下面的实现。
+
+## 代码
+
+```go
+package leetcode
+
+func isInterleave(s1 string, s2 string, s3 string) bool {
+	if len(s1)+len(s2) != len(s3) {
+		return false
+	}
+	visited := make(map[int]bool)
+	return dfs(s1, s2, s3, 0, 0, visited)
+}
+
+func dfs(s1, s2, s3 string, p1, p2 int, visited map[int]bool) bool {
+	if p1+p2 == len(s3) {
+		return true
+	}
+	if _, ok := visited[(p1*len(s3))+p2]; ok {
+		return false
+	}
+	visited[(p1*len(s3))+p2] = true
+	var match1, match2 bool
+	if p1 < len(s1) && s3[p1+p2] == s1[p1] {
+		match1 = true
+	}
+	if p2 < len(s2) && s3[p1+p2] == s2[p2] {
+		match2 = true
+	}
+	if match1 && match2 {
+		return dfs(s1, s2, s3, p1+1, p2, visited) || dfs(s1, s2, s3, p1, p2+1, visited)
+	} else if match1 {
+		return dfs(s1, s2, s3, p1+1, p2, visited)
+	} else if match2 {
+		return dfs(s1, s2, s3, p1, p2+1, visited)
+	} else {
+		return false
+	}
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0098. Validate Binary Search Tree/98. Validate Binary Search Tree.go b/leetcode/0098.Validate-Binary-Search-Tree/98. Validate Binary Search Tree.go
similarity index 89%
rename from Algorithms/0098. Validate Binary Search Tree/98. Validate Binary Search Tree.go
rename to leetcode/0098.Validate-Binary-Search-Tree/98. Validate Binary Search Tree.go
index 8e626a68a..1a4d530b9 100644
--- a/Algorithms/0098. Validate Binary Search Tree/98. Validate Binary Search Tree.go	
+++ b/leetcode/0098.Validate-Binary-Search-Tree/98. Validate Binary Search Tree.go	
@@ -1,6 +1,13 @@
 package leetcode
 
-import "math"
+import (
+	"math"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
 
 /**
  * Definition for a binary tree node.
diff --git a/leetcode/0098.Validate-Binary-Search-Tree/98. Validate Binary Search Tree_test.go b/leetcode/0098.Validate-Binary-Search-Tree/98. Validate Binary Search Tree_test.go
new file mode 100644
index 000000000..e9063d347
--- /dev/null
+++ b/leetcode/0098.Validate-Binary-Search-Tree/98. Validate Binary Search Tree_test.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question98 struct {
+	para98
+	ans98
+}
+
+// para 是参数
+// one 代表第一个参数
+type para98 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans98 struct {
+	one bool
+}
+
+func Test_Problem98(t *testing.T) {
+
+	qs := []question98{
+
+		{
+			para98{[]int{10, 5, 15, structures.NULL, structures.NULL, 6, 20}},
+			ans98{false},
+		},
+
+		{
+			para98{[]int{}},
+			ans98{true},
+		},
+
+		{
+			para98{[]int{2, 1, 3}},
+			ans98{true},
+		},
+
+		{
+			para98{[]int{5, 1, 4, structures.NULL, structures.NULL, 3, 6}},
+			ans98{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 98------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans98, q.para98
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", isValidBST(rootOne))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0098. Validate Binary Search Tree/README.md b/leetcode/0098.Validate-Binary-Search-Tree/README.md
similarity index 100%
rename from Algorithms/0098. Validate Binary Search Tree/README.md
rename to leetcode/0098.Validate-Binary-Search-Tree/README.md
diff --git a/Algorithms/0099. Recover Binary Search Tree/99. Recover Binary Search Tree.go b/leetcode/0099.Recover-Binary-Search-Tree/99. Recover Binary Search Tree.go
similarity index 88%
rename from Algorithms/0099. Recover Binary Search Tree/99. Recover Binary Search Tree.go
rename to leetcode/0099.Recover-Binary-Search-Tree/99. Recover Binary Search Tree.go
index 4b5427b07..57fb92958 100644
--- a/Algorithms/0099. Recover Binary Search Tree/99. Recover Binary Search Tree.go	
+++ b/leetcode/0099.Recover-Binary-Search-Tree/99. Recover Binary Search Tree.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func recoverTree(root *TreeNode) {
 	var prev, target1, target2 *TreeNode
 	_, target1, target2 = inOrderTraverse(root, prev, target1, target2)
diff --git a/leetcode/0099.Recover-Binary-Search-Tree/99. Recover Binary Search Tree_test.go b/leetcode/0099.Recover-Binary-Search-Tree/99. Recover Binary Search Tree_test.go
new file mode 100644
index 000000000..958c23417
--- /dev/null
+++ b/leetcode/0099.Recover-Binary-Search-Tree/99. Recover Binary Search Tree_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question99 struct {
+	para99
+	ans99
+}
+
+// para 是参数
+// one 代表第一个参数
+type para99 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans99 struct {
+	one []int
+}
+
+func Test_Problem99(t *testing.T) {
+
+	qs := []question99{
+
+		{
+			para99{[]int{1, 3, structures.NULL, structures.NULL, 2}},
+			ans99{[]int{3, 1, structures.NULL, structures.NULL, 2}},
+		},
+
+		{
+			para99{[]int{3, 1, 4, structures.NULL, structures.NULL, 2}},
+			ans99{[]int{2, 1, 4, structures.NULL, structures.NULL, 3}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 99------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans99, q.para99
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.one)
+		recoverTree(rootOne)
+		fmt.Printf("【output】:%v      \n", p)
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0099. Recover Binary Search Tree/README.md b/leetcode/0099.Recover-Binary-Search-Tree/README.md
similarity index 100%
rename from Algorithms/0099. Recover Binary Search Tree/README.md
rename to leetcode/0099.Recover-Binary-Search-Tree/README.md
diff --git a/Algorithms/0100. Same Tree/100. Same Tree.go b/leetcode/0100.Same-Tree/100. Same Tree.go
similarity index 78%
rename from Algorithms/0100. Same Tree/100. Same Tree.go
rename to leetcode/0100.Same-Tree/100. Same Tree.go
index ae4056bdb..ab65859b6 100644
--- a/Algorithms/0100. Same Tree/100. Same Tree.go	
+++ b/leetcode/0100.Same-Tree/100. Same Tree.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func isSameTree(p *TreeNode, q *TreeNode) bool {
 	if p == nil && q == nil {
 		return true
diff --git a/Algorithms/0100. Same Tree/100. Same Tree_test.go b/leetcode/0100.Same-Tree/100. Same Tree_test.go
similarity index 81%
rename from Algorithms/0100. Same Tree/100. Same Tree_test.go
rename to leetcode/0100.Same-Tree/100. Same Tree_test.go
index 7da838471..e448f8308 100644
--- a/Algorithms/0100. Same Tree/100. Same Tree_test.go	
+++ b/leetcode/0100.Same-Tree/100. Same Tree_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question100 struct {
@@ -27,32 +29,32 @@ func Test_Problem100(t *testing.T) {
 
 	qs := []question100{
 
-		question100{
+		{
 			para100{[]int{}, []int{}},
 			ans100{true},
 		},
 
-		question100{
+		{
 			para100{[]int{}, []int{1}},
 			ans100{false},
 		},
 
-		question100{
+		{
 			para100{[]int{1}, []int{1}},
 			ans100{true},
 		},
 
-		question100{
+		{
 			para100{[]int{1, 2, 3}, []int{1, 2, 3}},
 			ans100{true},
 		},
 
-		question100{
-			para100{[]int{1, 2}, []int{1, NULL, 2}},
+		{
+			para100{[]int{1, 2}, []int{1, structures.NULL, 2}},
 			ans100{false},
 		},
 
-		question100{
+		{
 			para100{[]int{1, 2, 1}, []int{1, 1, 2}},
 			ans100{false},
 		},
@@ -63,8 +65,8 @@ func Test_Problem100(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans100, q.para100
 		fmt.Printf("【input】:%v      ", p)
-		rootOne := Ints2TreeNode(p.one)
-		rootTwo := Ints2TreeNode(p.two)
+		rootOne := structures.Ints2TreeNode(p.one)
+		rootTwo := structures.Ints2TreeNode(p.two)
 		fmt.Printf("【output】:%v      \n", isSameTree(rootOne, rootTwo))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0100. Same Tree/README.md b/leetcode/0100.Same-Tree/README.md
similarity index 100%
rename from Algorithms/0100. Same Tree/README.md
rename to leetcode/0100.Same-Tree/README.md
diff --git a/leetcode/0101.Symmetric-Tree/101. Symmetric Tree.go b/leetcode/0101.Symmetric-Tree/101. Symmetric Tree.go
new file mode 100644
index 000000000..b8c26b9a1
--- /dev/null
+++ b/leetcode/0101.Symmetric-Tree/101. Symmetric Tree.go	
@@ -0,0 +1,66 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 dfs
+func isSymmetric(root *TreeNode) bool {
+	if root == nil {
+		return true
+	}
+	return isMirror(root.Left, root.Right)
+}
+
+func isMirror(left *TreeNode, right *TreeNode) bool {
+	if left == nil && right == nil {
+		return true
+	}
+	if left == nil || right == nil {
+		return false
+	}
+	return (left.Val == right.Val) && isMirror(left.Left, right.Right) && isMirror(left.Right, right.Left)
+}
+
+// 解法二
+func isSymmetric1(root *TreeNode) bool {
+	if root == nil {
+		return true
+	}
+	return isSameTree(invertTree(root.Left), root.Right)
+}
+
+func isSameTree(p *TreeNode, q *TreeNode) bool {
+	if p == nil && q == nil {
+		return true
+	} else if p != nil && q != nil {
+		if p.Val != q.Val {
+			return false
+		}
+		return isSameTree(p.Left, q.Left) && isSameTree(p.Right, q.Right)
+	} else {
+		return false
+	}
+}
+
+func invertTree(root *TreeNode) *TreeNode {
+	if root == nil {
+		return nil
+	}
+	invertTree(root.Left)
+	invertTree(root.Right)
+	root.Left, root.Right = root.Right, root.Left
+	return root
+}
diff --git a/leetcode/0101.Symmetric-Tree/101. Symmetric Tree_test.go b/leetcode/0101.Symmetric-Tree/101. Symmetric Tree_test.go
new file mode 100644
index 000000000..fbf8293ff
--- /dev/null
+++ b/leetcode/0101.Symmetric-Tree/101. Symmetric Tree_test.go	
@@ -0,0 +1,76 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question101 struct {
+	para101
+	ans101
+}
+
+// para 是参数
+// one 代表第一个参数
+type para101 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans101 struct {
+	one bool
+}
+
+func Test_Problem101(t *testing.T) {
+
+	qs := []question101{
+
+		{
+			para101{[]int{3, 4, 4, 5, structures.NULL, structures.NULL, 5, 6, structures.NULL, structures.NULL, 6}},
+			ans101{true},
+		},
+
+		{
+			para101{[]int{1, 2, 2, structures.NULL, 3, 3}},
+			ans101{true},
+		},
+
+		{
+			para101{[]int{}},
+			ans101{true},
+		},
+
+		{
+			para101{[]int{1}},
+			ans101{true},
+		},
+
+		{
+			para101{[]int{1, 2, 3}},
+			ans101{false},
+		},
+
+		{
+			para101{[]int{1, 2, 2, 3, 4, 4, 3}},
+			ans101{true},
+		},
+
+		{
+			para101{[]int{1, 2, 2, structures.NULL, 3, structures.NULL, 3}},
+			ans101{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 101------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans101, q.para101
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", isSymmetric(rootOne))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0101. Symmetric Tree/README.md b/leetcode/0101.Symmetric-Tree/README.md
similarity index 100%
rename from Algorithms/0101. Symmetric Tree/README.md
rename to leetcode/0101.Symmetric-Tree/README.md
diff --git a/leetcode/0102.Binary-Tree-Level-Order-Traversal/102. Binary Tree Level Order Traversal.go b/leetcode/0102.Binary-Tree-Level-Order-Traversal/102. Binary Tree Level Order Traversal.go
new file mode 100644
index 000000000..5df975c47
--- /dev/null
+++ b/leetcode/0102.Binary-Tree-Level-Order-Traversal/102. Binary Tree Level Order Traversal.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 BFS
+func levelOrder(root *TreeNode) [][]int {
+	if root == nil {
+		return [][]int{}
+	}
+	queue := []*TreeNode{root}
+	res := make([][]int, 0)
+	for len(queue) > 0 {
+		l := len(queue)
+		tmp := make([]int, 0, l)
+		for i := 0; i < l; i++ {
+			if queue[i].Left != nil {
+				queue = append(queue, queue[i].Left)
+			}
+			if queue[i].Right != nil {
+				queue = append(queue, queue[i].Right)
+			}
+			tmp = append(tmp, queue[i].Val)
+		}
+		queue = queue[l:]
+		res = append(res, tmp)
+	}
+	return res
+}
+
+// 解法二 DFS
+func levelOrder1(root *TreeNode) [][]int {
+	var res [][]int
+	var dfsLevel func(node *TreeNode, level int)
+	dfsLevel = func(node *TreeNode, level int) {
+		if node == nil {
+			return
+		}
+		if len(res) == level {
+			res = append(res, []int{node.Val})
+		} else {
+			res[level] = append(res[level], node.Val)
+		}
+		dfsLevel(node.Left, level+1)
+		dfsLevel(node.Right, level+1)
+	}
+	dfsLevel(root, 0)
+	return res
+}
diff --git a/leetcode/0102.Binary-Tree-Level-Order-Traversal/102. Binary Tree Level Order Traversal_test.go b/leetcode/0102.Binary-Tree-Level-Order-Traversal/102. Binary Tree Level Order Traversal_test.go
new file mode 100644
index 000000000..190db881a
--- /dev/null
+++ b/leetcode/0102.Binary-Tree-Level-Order-Traversal/102. Binary Tree Level Order Traversal_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question102 struct {
+	para102
+	ans102
+}
+
+// para 是参数
+// one 代表第一个参数
+type para102 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans102 struct {
+	one [][]int
+}
+
+func Test_Problem102(t *testing.T) {
+
+	qs := []question102{
+
+		{
+			para102{[]int{}},
+			ans102{[][]int{}},
+		},
+
+		{
+			para102{[]int{1}},
+			ans102{[][]int{{1}}},
+		},
+
+		{
+			para102{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
+			ans102{[][]int{{3}, {9, 20}, {15, 7}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 102------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans102, q.para102
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", levelOrder(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0102. Binary Tree Level Order Traversal/README.md b/leetcode/0102.Binary-Tree-Level-Order-Traversal/README.md
similarity index 100%
rename from Algorithms/0102. Binary Tree Level Order Traversal/README.md
rename to leetcode/0102.Binary-Tree-Level-Order-Traversal/README.md
diff --git a/leetcode/0103.Binary-Tree-Zigzag-Level-Order-Traversal/103. Binary Tree Zigzag Level Order Traversal.go b/leetcode/0103.Binary-Tree-Zigzag-Level-Order-Traversal/103. Binary Tree Zigzag Level Order Traversal.go
new file mode 100644
index 000000000..587ad0f43
--- /dev/null
+++ b/leetcode/0103.Binary-Tree-Zigzag-Level-Order-Traversal/103. Binary Tree Zigzag Level Order Traversal.go	
@@ -0,0 +1,120 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一
+func zigzagLevelOrder(root *TreeNode) [][]int {
+	if root == nil {
+		return [][]int{}
+	}
+	queue := []*TreeNode{}
+	queue = append(queue, root)
+	curNum, nextLevelNum, res, tmp, curDir := 1, 0, [][]int{}, []int{}, 0
+	for len(queue) != 0 {
+		if curNum > 0 {
+			node := queue[0]
+			if node.Left != nil {
+				queue = append(queue, node.Left)
+				nextLevelNum++
+			}
+			if node.Right != nil {
+				queue = append(queue, node.Right)
+				nextLevelNum++
+			}
+			curNum--
+			tmp = append(tmp, node.Val)
+			queue = queue[1:]
+		}
+		if curNum == 0 {
+			if curDir == 1 {
+				for i, j := 0, len(tmp)-1; i < j; i, j = i+1, j-1 {
+					tmp[i], tmp[j] = tmp[j], tmp[i]
+				}
+			}
+			res = append(res, tmp)
+			curNum = nextLevelNum
+			nextLevelNum = 0
+			tmp = []int{}
+			if curDir == 0 {
+				curDir = 1
+			} else {
+				curDir = 0
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 递归
+func zigzagLevelOrder0(root *TreeNode) [][]int {
+	var res [][]int
+	search(root, 0, &res)
+	return res
+}
+
+func search(root *TreeNode, depth int, res *[][]int) {
+	if root == nil {
+		return
+	}
+	for len(*res) < depth+1 {
+		*res = append(*res, []int{})
+	}
+	if depth%2 == 0 {
+		(*res)[depth] = append((*res)[depth], root.Val)
+	} else {
+		(*res)[depth] = append([]int{root.Val}, (*res)[depth]...)
+	}
+	search(root.Left, depth+1, res)
+	search(root.Right, depth+1, res)
+}
+
+// 解法三 BFS
+func zigzagLevelOrder1(root *TreeNode) [][]int {
+	res := [][]int{}
+	if root == nil {
+		return res
+	}
+	q := []*TreeNode{root}
+	size, i, j, lay, tmp, flag := 0, 0, 0, []int{}, []*TreeNode{}, false
+	for len(q) > 0 {
+		size = len(q)
+		tmp = []*TreeNode{}
+		lay = make([]int, size)
+		j = size - 1
+		for i = 0; i < size; i++ {
+			root = q[0]
+			q = q[1:]
+			if !flag {
+				lay[i] = root.Val
+			} else {
+				lay[j] = root.Val
+				j--
+			}
+			if root.Left != nil {
+				tmp = append(tmp, root.Left)
+			}
+			if root.Right != nil {
+				tmp = append(tmp, root.Right)
+			}
+
+		}
+		res = append(res, lay)
+		flag = !flag
+		q = tmp
+	}
+	return res
+}
diff --git a/leetcode/0103.Binary-Tree-Zigzag-Level-Order-Traversal/103. Binary Tree Zigzag Level Order Traversal_test.go b/leetcode/0103.Binary-Tree-Zigzag-Level-Order-Traversal/103. Binary Tree Zigzag Level Order Traversal_test.go
new file mode 100644
index 000000000..1b5506988
--- /dev/null
+++ b/leetcode/0103.Binary-Tree-Zigzag-Level-Order-Traversal/103. Binary Tree Zigzag Level Order Traversal_test.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question103 struct {
+	para103
+	ans103
+}
+
+// para 是参数
+// one 代表第一个参数
+type para103 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans103 struct {
+	one [][]int
+}
+
+func Test_Problem103(t *testing.T) {
+
+	qs := []question103{
+
+		{
+			para103{[]int{}},
+			ans103{[][]int{}},
+		},
+
+		{
+			para103{[]int{1}},
+			ans103{[][]int{{1}}},
+		},
+
+		{
+			para103{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
+			ans103{[][]int{{3}, {9, 20}, {15, 7}}},
+		},
+
+		{
+			para103{[]int{1, 2, 3, 4, structures.NULL, structures.NULL, 5}},
+			ans103{[][]int{{1}, {3, 2}, {4, 5}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 103------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans103, q.para103
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", zigzagLevelOrder(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0103. Binary Tree Zigzag Level Order Traversal/README.md b/leetcode/0103.Binary-Tree-Zigzag-Level-Order-Traversal/README.md
similarity index 100%
rename from Algorithms/0103. Binary Tree Zigzag Level Order Traversal/README.md
rename to leetcode/0103.Binary-Tree-Zigzag-Level-Order-Traversal/README.md
diff --git a/leetcode/0104.Maximum-Depth-of-Binary-Tree/104. Maximum Depth of Binary Tree.go b/leetcode/0104.Maximum-Depth-of-Binary-Tree/104. Maximum Depth of Binary Tree.go
new file mode 100644
index 000000000..7294bc3cb
--- /dev/null
+++ b/leetcode/0104.Maximum-Depth-of-Binary-Tree/104. Maximum Depth of Binary Tree.go	
@@ -0,0 +1,31 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func maxDepth(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	return max(maxDepth(root.Left), maxDepth(root.Right)) + 1
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0104. Maximum Depth of Binary Tree/104. Maximum Depth of Binary Tree_test.go b/leetcode/0104.Maximum-Depth-of-Binary-Tree/104. Maximum Depth of Binary Tree_test.go
similarity index 79%
rename from Algorithms/0104. Maximum Depth of Binary Tree/104. Maximum Depth of Binary Tree_test.go
rename to leetcode/0104.Maximum-Depth-of-Binary-Tree/104. Maximum Depth of Binary Tree_test.go
index 3a79f4006..a3ef5e1ea 100644
--- a/Algorithms/0104. Maximum Depth of Binary Tree/104. Maximum Depth of Binary Tree_test.go	
+++ b/leetcode/0104.Maximum-Depth-of-Binary-Tree/104. Maximum Depth of Binary Tree_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question104 struct {
@@ -26,13 +28,13 @@ func Test_Problem104(t *testing.T) {
 
 	qs := []question104{
 
-		question104{
+		{
 			para104{[]int{}},
 			ans104{0},
 		},
 
-		question104{
-			para104{[]int{3, 9, 20, NULL, NULL, 15, 7}},
+		{
+			para104{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
 			ans104{3},
 		},
 	}
@@ -42,7 +44,7 @@ func Test_Problem104(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans104, q.para104
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", maxDepth(root))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0104. Maximum Depth of Binary Tree/README.md b/leetcode/0104.Maximum-Depth-of-Binary-Tree/README.md
similarity index 100%
rename from Algorithms/0104. Maximum Depth of Binary Tree/README.md
rename to leetcode/0104.Maximum-Depth-of-Binary-Tree/README.md
diff --git a/leetcode/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal/105. Construct Binary Tree from Preorder and Inorder Traversal.go b/leetcode/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal/105. Construct Binary Tree from Preorder and Inorder Traversal.go
new file mode 100644
index 000000000..2654e7381
--- /dev/null
+++ b/leetcode/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal/105. Construct Binary Tree from Preorder and Inorder Traversal.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一, 直接传入需要的 slice 范围作为输入, 可以避免申请对应 inorder 索引的内存, 内存使用(leetcode test case) 4.7MB -> 4.3MB.
+func buildTree(preorder []int, inorder []int) *TreeNode {
+	if len(preorder) == 0 {
+		return nil
+	}
+	root := &TreeNode{Val: preorder[0]}
+	for pos, node := range inorder {
+		if node == root.Val {
+			root.Left = buildTree(preorder[1:pos+1], inorder[:pos])
+			root.Right = buildTree(preorder[pos+1:], inorder[pos+1:])
+		}
+	}
+	return root
+}
+
+// 解法二
+func buildTree1(preorder []int, inorder []int) *TreeNode {
+	inPos := make(map[int]int)
+	for i := 0; i < len(inorder); i++ {
+		inPos[inorder[i]] = i
+	}
+	return buildPreIn2TreeDFS(preorder, 0, len(preorder)-1, 0, inPos)
+}
+
+func buildPreIn2TreeDFS(pre []int, preStart int, preEnd int, inStart int, inPos map[int]int) *TreeNode {
+	if preStart > preEnd {
+		return nil
+	}
+	root := &TreeNode{Val: pre[preStart]}
+	rootIdx := inPos[pre[preStart]]
+	leftLen := rootIdx - inStart
+	root.Left = buildPreIn2TreeDFS(pre, preStart+1, preStart+leftLen, inStart, inPos)
+	root.Right = buildPreIn2TreeDFS(pre, preStart+leftLen+1, preEnd, rootIdx+1, inPos)
+	return root
+}
diff --git a/leetcode/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal/105. Construct Binary Tree from Preorder and Inorder Traversal_test.go b/leetcode/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal/105. Construct Binary Tree from Preorder and Inorder Traversal_test.go
new file mode 100644
index 000000000..1bef00178
--- /dev/null
+++ b/leetcode/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal/105. Construct Binary Tree from Preorder and Inorder Traversal_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question105 struct {
+	para105
+	ans105
+}
+
+// para 是参数
+// one 代表第一个参数
+type para105 struct {
+	preorder []int
+	inorder  []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans105 struct {
+	one []int
+}
+
+func Test_Problem105(t *testing.T) {
+
+	qs := []question105{
+
+		{
+			para105{[]int{3, 9, 20, 15, 7}, []int{9, 3, 15, 20, 7}},
+			ans105{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 105------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans105, q.para105
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", structures.Tree2ints(buildTree(p.preorder, p.inorder)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal/README.md b/leetcode/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal/README.md
new file mode 100755
index 000000000..f3ee8e511
--- /dev/null
+++ b/leetcode/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal/README.md
@@ -0,0 +1,36 @@
+# [105. Construct Binary Tree from Preorder and Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)
+
+
+## 题目
+
+Given preorder and inorder traversal of a tree, construct the binary tree.
+
+**Note:**You may assume that duplicates do not exist in the tree.
+
+For example, given
+
+    preorder = [3,9,20,15,7]
+    inorder = [9,3,15,20,7]
+
+Return the following binary tree:
+
+    	3
+       / \
+      9  20
+        /  \
+       15   7
+
+
+
+## 题目大意
+
+根据一棵树的前序遍历与中序遍历构造二叉树。
+
+注意:
+你可以假设树中没有重复的元素。
+
+
+## 解题思路
+
+- 给出 2 个数组，根据 preorder 和 inorder 数组构造一颗树。
+- 利用递归思想，从 preorder 可以得到根节点，从 inorder 中得到左子树和右子树。只剩一个节点的时候即为根节点。不断的递归直到所有的树都生成完成。
diff --git a/leetcode/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal/106. Construct Binary Tree from Inorder and Postorder Traversal.go b/leetcode/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal/106. Construct Binary Tree from Inorder and Postorder Traversal.go
new file mode 100644
index 000000000..1c7a86b66
--- /dev/null
+++ b/leetcode/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal/106. Construct Binary Tree from Inorder and Postorder Traversal.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一, 直接传入需要的 slice 范围作为输入, 可以避免申请对应 inorder 索引的内存, 内存使用(leetcode test case) 4.7MB -> 4.3MB.
+func buildTree(inorder []int, postorder []int) *TreeNode {
+	postorderLen := len(postorder)
+	if len(inorder) == 0 {
+		return nil
+	}
+	root := &TreeNode{Val: postorder[postorderLen-1]}
+	postorder = postorder[:postorderLen-1]
+	for pos, node := range inorder {
+		if node == root.Val {
+			root.Left = buildTree(inorder[:pos], postorder[:len(inorder[:pos])])
+			root.Right = buildTree(inorder[pos+1:], postorder[len(inorder[:pos]):])
+		}
+	}
+	return root
+}
+
+// 解法二
+func buildTree1(inorder []int, postorder []int) *TreeNode {
+	inPos := make(map[int]int)
+	for i := 0; i < len(inorder); i++ {
+		inPos[inorder[i]] = i
+	}
+	return buildInPos2TreeDFS(postorder, 0, len(postorder)-1, 0, inPos)
+}
+
+func buildInPos2TreeDFS(post []int, postStart int, postEnd int, inStart int, inPos map[int]int) *TreeNode {
+	if postStart > postEnd {
+		return nil
+	}
+	root := &TreeNode{Val: post[postEnd]}
+	rootIdx := inPos[post[postEnd]]
+	leftLen := rootIdx - inStart
+	root.Left = buildInPos2TreeDFS(post, postStart, postStart+leftLen-1, inStart, inPos)
+	root.Right = buildInPos2TreeDFS(post, postStart+leftLen, postEnd-1, rootIdx+1, inPos)
+	return root
+}
diff --git a/leetcode/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal/106. Construct Binary Tree from Inorder and Postorder Traversal_test.go b/leetcode/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal/106. Construct Binary Tree from Inorder and Postorder Traversal_test.go
new file mode 100644
index 000000000..eaa86c378
--- /dev/null
+++ b/leetcode/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal/106. Construct Binary Tree from Inorder and Postorder Traversal_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question106 struct {
+	para106
+	ans106
+}
+
+// para 是参数
+// one 代表第一个参数
+type para106 struct {
+	inorder   []int
+	postorder []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans106 struct {
+	one []int
+}
+
+func Test_Problem106(t *testing.T) {
+
+	qs := []question106{
+
+		{
+			para106{[]int{9, 3, 15, 20, 7}, []int{9, 15, 7, 20, 3}},
+			ans106{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 106------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans106, q.para106
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", structures.Tree2ints(buildTree(p.inorder, p.postorder)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal/README.md b/leetcode/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal/README.md
new file mode 100755
index 000000000..5abc5adfd
--- /dev/null
+++ b/leetcode/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal/README.md
@@ -0,0 +1,34 @@
+# [106. Construct Binary Tree from Inorder and Postorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/)
+
+## 题目
+
+Given inorder and postorder traversal of a tree, construct the binary tree.
+
+**Note:**You may assume that duplicates do not exist in the tree.
+
+For example, given
+
+    inorder = [9,3,15,20,7]
+    postorder = [9,15,7,20,3]
+
+Return the following binary tree:
+
+    	3
+       / \
+      9  20
+        /  \
+       15   7
+
+
+## 题目大意
+
+根据一棵树的中序遍历与后序遍历构造二叉树。
+
+注意:
+你可以假设树中没有重复的元素。
+
+
+## 解题思路
+
+- 给出 2 个数组，根据 inorder 和 postorder 数组构造一颗树。
+- 利用递归思想，从 postorder 可以得到根节点，从 inorder 中得到左子树和右子树。只剩一个节点的时候即为根节点。不断的递归直到所有的树都生成完成。
diff --git a/leetcode/0107.Binary-Tree-Level-Order-Traversal-II/107. Binary Tree Level Order Traversal II.go b/leetcode/0107.Binary-Tree-Level-Order-Traversal-II/107. Binary Tree Level Order Traversal II.go
new file mode 100644
index 000000000..84a88a7b1
--- /dev/null
+++ b/leetcode/0107.Binary-Tree-Level-Order-Traversal-II/107. Binary Tree Level Order Traversal II.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func levelOrderBottom(root *TreeNode) [][]int {
+	tmp := levelOrder(root)
+	res := [][]int{}
+	for i := len(tmp) - 1; i >= 0; i-- {
+		res = append(res, tmp[i])
+	}
+	return res
+}
+
+func levelOrder(root *TreeNode) [][]int {
+	if root == nil {
+		return [][]int{}
+	}
+	queue := []*TreeNode{}
+	queue = append(queue, root)
+	curNum, nextLevelNum, res, tmp := 1, 0, [][]int{}, []int{}
+	for len(queue) != 0 {
+		if curNum > 0 {
+			node := queue[0]
+			if node.Left != nil {
+				queue = append(queue, node.Left)
+				nextLevelNum++
+			}
+			if node.Right != nil {
+				queue = append(queue, node.Right)
+				nextLevelNum++
+			}
+			curNum--
+			tmp = append(tmp, node.Val)
+			queue = queue[1:]
+		}
+		if curNum == 0 {
+			res = append(res, tmp)
+			curNum = nextLevelNum
+			nextLevelNum = 0
+			tmp = []int{}
+		}
+	}
+	return res
+}
diff --git a/leetcode/0107.Binary-Tree-Level-Order-Traversal-II/107. Binary Tree Level Order Traversal II_test.go b/leetcode/0107.Binary-Tree-Level-Order-Traversal-II/107. Binary Tree Level Order Traversal II_test.go
new file mode 100644
index 000000000..8978135da
--- /dev/null
+++ b/leetcode/0107.Binary-Tree-Level-Order-Traversal-II/107. Binary Tree Level Order Traversal II_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question107 struct {
+	para107
+	ans107
+}
+
+// para 是参数
+// one 代表第一个参数
+type para107 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans107 struct {
+	one [][]int
+}
+
+func Test_Problem107(t *testing.T) {
+
+	qs := []question107{
+
+		{
+			para107{[]int{}},
+			ans107{[][]int{}},
+		},
+
+		{
+			para107{[]int{1}},
+			ans107{[][]int{{1}}},
+		},
+
+		{
+			para107{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
+			ans107{[][]int{{15, 7}, {9, 20}, {3}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 107------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans107, q.para107
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", levelOrderBottom(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0107. Binary Tree Level Order Traversal II/README.md b/leetcode/0107.Binary-Tree-Level-Order-Traversal-II/README.md
similarity index 100%
rename from Algorithms/0107. Binary Tree Level Order Traversal II/README.md
rename to leetcode/0107.Binary-Tree-Level-Order-Traversal-II/README.md
diff --git a/Algorithms/0108. Convert Sorted Array to Binary Search Tree/108. Convert Sorted Array to Binary Search Tree.go b/leetcode/0108.Convert-Sorted-Array-to-Binary-Search-Tree/108. Convert Sorted Array to Binary Search Tree.go
similarity index 76%
rename from Algorithms/0108. Convert Sorted Array to Binary Search Tree/108. Convert Sorted Array to Binary Search Tree.go
rename to leetcode/0108.Convert-Sorted-Array-to-Binary-Search-Tree/108. Convert Sorted Array to Binary Search Tree.go
index b5169630d..57e60d7d3 100644
--- a/Algorithms/0108. Convert Sorted Array to Binary Search Tree/108. Convert Sorted Array to Binary Search Tree.go	
+++ b/leetcode/0108.Convert-Sorted-Array-to-Binary-Search-Tree/108. Convert Sorted Array to Binary Search Tree.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func sortedArrayToBST(nums []int) *TreeNode {
 	if len(nums) == 0 {
 		return nil
diff --git a/Algorithms/0108. Convert Sorted Array to Binary Search Tree/108. Convert Sorted Array to Binary Search Tree_test.go b/leetcode/0108.Convert-Sorted-Array-to-Binary-Search-Tree/108. Convert Sorted Array to Binary Search Tree_test.go
similarity index 80%
rename from Algorithms/0108. Convert Sorted Array to Binary Search Tree/108. Convert Sorted Array to Binary Search Tree_test.go
rename to leetcode/0108.Convert-Sorted-Array-to-Binary-Search-Tree/108. Convert Sorted Array to Binary Search Tree_test.go
index ed2ca1dd4..44f4ae41e 100644
--- a/Algorithms/0108. Convert Sorted Array to Binary Search Tree/108. Convert Sorted Array to Binary Search Tree_test.go	
+++ b/leetcode/0108.Convert-Sorted-Array-to-Binary-Search-Tree/108. Convert Sorted Array to Binary Search Tree_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question108 struct {
@@ -26,9 +28,9 @@ func Test_Problem108(t *testing.T) {
 
 	qs := []question108{
 
-		question108{
+		{
 			para108{[]int{-10, -3, 0, 5, 9}},
-			ans108{[]int{0, -3, 9, -10, NULL, 5}},
+			ans108{[]int{0, -3, 9, -10, structures.NULL, 5}},
 		},
 	}
 
@@ -37,7 +39,7 @@ func Test_Problem108(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans108, q.para108
 		arr := []int{}
-		T2s(sortedArrayToBST(p.one), &arr)
+		structures.T2s(sortedArrayToBST(p.one), &arr)
 		fmt.Printf("【input】:%v       【output】:%v\n", p, arr)
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0108. Convert Sorted Array to Binary Search Tree/README.md b/leetcode/0108.Convert-Sorted-Array-to-Binary-Search-Tree/README.md
similarity index 100%
rename from Algorithms/0108. Convert Sorted Array to Binary Search Tree/README.md
rename to leetcode/0108.Convert-Sorted-Array-to-Binary-Search-Tree/README.md
diff --git a/Algorithms/0109. Convert Sorted List to Binary Search Tree/109. Convert Sorted List to Binary Search Tree.go b/leetcode/0109.Convert-Sorted-List-to-Binary-Search-Tree/109. Convert Sorted List to Binary Search Tree.go
similarity index 87%
rename from Algorithms/0109. Convert Sorted List to Binary Search Tree/109. Convert Sorted List to Binary Search Tree.go
rename to leetcode/0109.Convert-Sorted-List-to-Binary-Search-Tree/109. Convert Sorted List to Binary Search Tree.go
index a5b405738..8edf22873 100644
--- a/Algorithms/0109. Convert Sorted List to Binary Search Tree/109. Convert Sorted List to Binary Search Tree.go	
+++ b/leetcode/0109.Convert-Sorted-List-to-Binary-Search-Tree/109. Convert Sorted List to Binary Search Tree.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
@@ -7,6 +14,10 @@ package leetcode
  *     Next *ListNode
  * }
  */
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -16,13 +27,6 @@ package leetcode
  * }
  */
 
-// TreeNode define
-type TreeNode struct {
-	Val   int
-	Left  *TreeNode
-	Right *TreeNode
-}
-
 func sortedListToBST(head *ListNode) *TreeNode {
 	if head == nil {
 		return nil
diff --git a/leetcode/0109.Convert-Sorted-List-to-Binary-Search-Tree/109. Convert Sorted List to Binary Search Tree_test.go b/leetcode/0109.Convert-Sorted-List-to-Binary-Search-Tree/109. Convert Sorted List to Binary Search Tree_test.go
new file mode 100644
index 000000000..06f7f323b
--- /dev/null
+++ b/leetcode/0109.Convert-Sorted-List-to-Binary-Search-Tree/109. Convert Sorted List to Binary Search Tree_test.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question109 struct {
+	para109
+	ans109
+}
+
+// para 是参数
+// one 代表第一个参数
+type para109 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans109 struct {
+	one []int
+}
+
+func Test_Problem109(t *testing.T) {
+
+	qs := []question109{
+
+		{
+			para109{[]int{-10, -3, 0, 5, 9}},
+			ans109{[]int{0, -10, 5, structures.NULL, -3, structures.NULL, 9}},
+		},
+
+		{
+			para109{[]int{-10}},
+			ans109{[]int{-10}},
+		},
+
+		{
+			para109{[]int{1, 2}},
+			ans109{[]int{1, 2}},
+		},
+
+		{
+			para109{[]int{1, 2, 3}},
+			ans109{[]int{2, 1, 3}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 109------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans109, q.para109
+		arr := []int{}
+		structures.T2s(sortedListToBST(structures.Ints2List(p.one)), &arr)
+		fmt.Printf("【input】:%v       【output】:%v\n", p, arr)
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0109.Convert-Sorted-List-to-Binary-Search-Tree/README.md b/leetcode/0109.Convert-Sorted-List-to-Binary-Search-Tree/README.md
new file mode 100644
index 000000000..692a49ca3
--- /dev/null
+++ b/leetcode/0109.Convert-Sorted-List-to-Binary-Search-Tree/README.md
@@ -0,0 +1,31 @@
+# [109. Convert Sorted List to Binary Search Tree](https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/)
+
+## 题目
+
+Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
+
+For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
+
+Example:
+
+```
+Given the sorted linked list: [-10,-3,0,5,9],
+
+One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
+
+      0
+     / \
+   -3   9
+   /   /
+ -10  5
+```
+
+
+## 题目大意
+
+将链表转化为高度平衡的二叉搜索树。高度平衡的定义：每个结点的 2 个子结点的深度不能相差超过 1 。
+
+## 解题思路
+
+思路比较简单，依次把链表的中间点作为根结点，类似二分的思想，递归排列所有结点即可。
+
diff --git a/leetcode/0110.Balanced-Binary-Tree/110. Balanced Binary Tree.go b/leetcode/0110.Balanced-Binary-Tree/110. Balanced Binary Tree.go
new file mode 100644
index 000000000..57b2c7876
--- /dev/null
+++ b/leetcode/0110.Balanced-Binary-Tree/110. Balanced Binary Tree.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func isBalanced(root *TreeNode) bool {
+	if root == nil {
+		return true
+	}
+	leftHight := depth(root.Left)
+	rightHight := depth(root.Right)
+	return abs(leftHight-rightHight) <= 1 && isBalanced(root.Left) && isBalanced(root.Right)
+}
+
+func depth(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	return max(depth(root.Left), depth(root.Right)) + 1
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0110.Balanced-Binary-Tree/110. Balanced Binary Tree_test.go b/leetcode/0110.Balanced-Binary-Tree/110. Balanced Binary Tree_test.go
new file mode 100644
index 000000000..2868cfccd
--- /dev/null
+++ b/leetcode/0110.Balanced-Binary-Tree/110. Balanced Binary Tree_test.go	
@@ -0,0 +1,76 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question110 struct {
+	para110
+	ans110
+}
+
+// para 是参数
+// one 代表第一个参数
+type para110 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans110 struct {
+	one bool
+}
+
+func Test_Problem110(t *testing.T) {
+
+	qs := []question110{
+
+		{
+			para110{[]int{3, 4, 4, 5, structures.NULL, structures.NULL, 5, 6, structures.NULL, structures.NULL, 6}},
+			ans110{false},
+		},
+
+		{
+			para110{[]int{1, 2, 2, structures.NULL, 3, 3}},
+			ans110{true},
+		},
+
+		{
+			para110{[]int{}},
+			ans110{true},
+		},
+
+		{
+			para110{[]int{1}},
+			ans110{true},
+		},
+
+		{
+			para110{[]int{1, 2, 3}},
+			ans110{true},
+		},
+
+		{
+			para110{[]int{1, 2, 2, 3, 4, 4, 3}},
+			ans110{true},
+		},
+
+		{
+			para110{[]int{1, 2, 2, structures.NULL, 3, structures.NULL, 3}},
+			ans110{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 110------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans110, q.para110
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", isBalanced(rootOne))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0110. Balanced Binary Tree/README.md b/leetcode/0110.Balanced-Binary-Tree/README.md
similarity index 100%
rename from Algorithms/0110. Balanced Binary Tree/README.md
rename to leetcode/0110.Balanced-Binary-Tree/README.md
diff --git a/leetcode/0111.Minimum-Depth-of-Binary-Tree/111. Minimum Depth of Binary Tree.go b/leetcode/0111.Minimum-Depth-of-Binary-Tree/111. Minimum Depth of Binary Tree.go
new file mode 100644
index 000000000..c577ad71c
--- /dev/null
+++ b/leetcode/0111.Minimum-Depth-of-Binary-Tree/111. Minimum Depth of Binary Tree.go	
@@ -0,0 +1,37 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func minDepth(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	if root.Left == nil {
+		return minDepth(root.Right) + 1
+	}
+	if root.Right == nil {
+		return minDepth(root.Left) + 1
+	}
+	return min(minDepth(root.Left), minDepth(root.Right)) + 1
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/Algorithms/0111. Minimum Depth of Binary Tree/111. Minimum Depth of Binary Tree_test.go b/leetcode/0111.Minimum-Depth-of-Binary-Tree/111. Minimum Depth of Binary Tree_test.go
similarity index 80%
rename from Algorithms/0111. Minimum Depth of Binary Tree/111. Minimum Depth of Binary Tree_test.go
rename to leetcode/0111.Minimum-Depth-of-Binary-Tree/111. Minimum Depth of Binary Tree_test.go
index 8d648dc64..a2a3753d9 100644
--- a/Algorithms/0111. Minimum Depth of Binary Tree/111. Minimum Depth of Binary Tree_test.go	
+++ b/leetcode/0111.Minimum-Depth-of-Binary-Tree/111. Minimum Depth of Binary Tree_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question111 struct {
@@ -26,22 +28,22 @@ func Test_Problem111(t *testing.T) {
 
 	qs := []question111{
 
-		question111{
+		{
 			para111{[]int{}},
 			ans111{0},
 		},
 
-		question111{
+		{
 			para111{[]int{1}},
 			ans111{1},
 		},
 
-		question111{
-			para111{[]int{3, 9, 20, NULL, NULL, 15, 7}},
+		{
+			para111{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
 			ans111{2},
 		},
 
-		question111{
+		{
 			para111{[]int{1, 2}},
 			ans111{2},
 		},
@@ -52,7 +54,7 @@ func Test_Problem111(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans111, q.para111
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", minDepth(root))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0111. Minimum Depth of Binary Tree/README.md b/leetcode/0111.Minimum-Depth-of-Binary-Tree/README.md
similarity index 100%
rename from Algorithms/0111. Minimum Depth of Binary Tree/README.md
rename to leetcode/0111.Minimum-Depth-of-Binary-Tree/README.md
diff --git a/Algorithms/0112. Path Sum/112. Path Sum.go b/leetcode/0112.Path-Sum/112. Path Sum.go
similarity index 77%
rename from Algorithms/0112. Path Sum/112. Path Sum.go
rename to leetcode/0112.Path-Sum/112. Path Sum.go
index 78b783c47..5ced14fdf 100644
--- a/Algorithms/0112. Path Sum/112. Path Sum.go	
+++ b/leetcode/0112.Path-Sum/112. Path Sum.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func hasPathSum(root *TreeNode, sum int) bool {
 	if root == nil {
 		return false
diff --git a/Algorithms/0112. Path Sum/112. Path Sum_test.go b/leetcode/0112.Path-Sum/112. Path Sum_test.go
similarity index 75%
rename from Algorithms/0112. Path Sum/112. Path Sum_test.go
rename to leetcode/0112.Path-Sum/112. Path Sum_test.go
index ea6618309..70168357a 100644
--- a/Algorithms/0112. Path Sum/112. Path Sum_test.go	
+++ b/leetcode/0112.Path-Sum/112. Path Sum_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question112 struct {
@@ -27,13 +29,13 @@ func Test_Problem112(t *testing.T) {
 
 	qs := []question112{
 
-		question112{
+		{
 			para112{[]int{}, 0},
 			ans112{false},
 		},
 
-		question112{
-			para112{[]int{5, 4, 8, 11, NULL, 13, 4, 7, 2, NULL, NULL, NULL, 1}, 22},
+		{
+			para112{[]int{5, 4, 8, 11, structures.NULL, 13, 4, 7, 2, structures.NULL, structures.NULL, structures.NULL, 1}, 22},
 			ans112{true},
 		},
 	}
@@ -43,7 +45,7 @@ func Test_Problem112(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans112, q.para112
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", hasPathSum(root, p.sum))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0112. Path Sum/README.md b/leetcode/0112.Path-Sum/README.md
similarity index 100%
rename from Algorithms/0112. Path Sum/README.md
rename to leetcode/0112.Path-Sum/README.md
diff --git a/Algorithms/0113. Path Sum II/113. Path Sum II.go b/leetcode/0113.Path-Sum-II/113. Path Sum II.go
similarity index 90%
rename from Algorithms/0113. Path Sum II/113. Path Sum II.go
rename to leetcode/0113.Path-Sum-II/113. Path Sum II.go
index f1846f8d7..6bca79a44 100644
--- a/Algorithms/0113. Path Sum II/113. Path Sum II.go	
+++ b/leetcode/0113.Path-Sum-II/113. Path Sum II.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -38,7 +45,7 @@ func pathSum1(root *TreeNode, sum int) [][]int {
 	}
 	if root.Left == nil && root.Right == nil {
 		if sum == root.Val {
-			return [][]int{[]int{root.Val}}
+			return [][]int{{root.Val}}
 		}
 	}
 	path, res := []int{}, [][]int{}
diff --git a/leetcode/0113.Path-Sum-II/113. Path Sum II_test.go b/leetcode/0113.Path-Sum-II/113. Path Sum II_test.go
new file mode 100644
index 000000000..e7520d115
--- /dev/null
+++ b/leetcode/0113.Path-Sum-II/113. Path Sum II_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question113 struct {
+	para113
+	ans113
+}
+
+// para 是参数
+// one 代表第一个参数
+type para113 struct {
+	one []int
+	sum int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans113 struct {
+	one [][]int
+}
+
+func Test_Problem113(t *testing.T) {
+
+	qs := []question113{
+		{
+			para113{[]int{1, 0, 1, 1, 2, 0, -1, 0, 1, -1, 0, -1, 0, 1, 0}, 2},
+			ans113{[][]int{{1, 0, 1, 0}, {1, 0, 2, -1}, {1, 1, 0, 0}, {1, 1, -1, 1}}},
+		},
+
+		{
+			para113{[]int{}, 0},
+			ans113{[][]int{{}}},
+		},
+
+		{
+			para113{[]int{5, 4, 8, 11, structures.NULL, 13, 4, 7, 2, structures.NULL, structures.NULL, 5, 1}, 22},
+			ans113{[][]int{{5, 4, 11, 2}, {5, 8, 4, 5}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 113------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans113, q.para113
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", pathSum(root, p.sum))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0113. Path Sum II/README.md b/leetcode/0113.Path-Sum-II/README.md
similarity index 100%
rename from Algorithms/0113. Path Sum II/README.md
rename to leetcode/0113.Path-Sum-II/README.md
diff --git a/leetcode/0114.Flatten-Binary-Tree-to-Linked-List/114. Flatten Binary Tree to Linked List.go b/leetcode/0114.Flatten-Binary-Tree-to-Linked-List/114. Flatten Binary Tree to Linked List.go
new file mode 100644
index 000000000..74860eae8
--- /dev/null
+++ b/leetcode/0114.Flatten-Binary-Tree-to-Linked-List/114. Flatten Binary Tree to Linked List.go	
@@ -0,0 +1,71 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 非递归
+func flatten(root *TreeNode) {
+	list := preorder(root)
+	for i := 1; i < len(list); i++ {
+		prev, cur := list[i-1], list[i]
+		prev.Left, prev.Right = nil, cur
+	}
+	return
+}
+
+func preorder(root *TreeNode) (ans []*TreeNode) {
+	if root != nil {
+		ans = append(ans, root)
+		ans = append(ans, preorder(root.Left)...)
+		ans = append(ans, preorder(root.Right)...)
+	}
+	return
+}
+
+// 解法二 递归
+func flatten1(root *TreeNode) {
+	if root == nil || (root.Left == nil && root.Right == nil) {
+		return
+	}
+	flatten(root.Left)
+	flatten(root.Right)
+	currRight := root.Right
+	root.Right = root.Left
+	root.Left = nil
+	for root.Right != nil {
+		root = root.Right
+	}
+	root.Right = currRight
+}
+
+// 解法三 递归
+func flatten2(root *TreeNode) {
+	if root == nil {
+		return
+	}
+	flatten(root.Right)
+	if root.Left == nil {
+		return
+	}
+	flatten(root.Left)
+	p := root.Left
+	for p.Right != nil {
+		p = p.Right
+	}
+	p.Right = root.Right
+	root.Right = root.Left
+	root.Left = nil
+}
diff --git a/leetcode/0114.Flatten-Binary-Tree-to-Linked-List/114. Flatten Binary Tree to Linked List_test.go b/leetcode/0114.Flatten-Binary-Tree-to-Linked-List/114. Flatten Binary Tree to Linked List_test.go
new file mode 100644
index 000000000..aa20ab605
--- /dev/null
+++ b/leetcode/0114.Flatten-Binary-Tree-to-Linked-List/114. Flatten Binary Tree to Linked List_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question114 struct {
+	para114
+	ans114
+}
+
+// para 是参数
+// one 代表第一个参数
+type para114 struct {
+	one []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans114 struct {
+	one []string
+}
+
+func Test_Problem114(t *testing.T) {
+
+	qs := []question114{
+
+		{
+			para114{[]string{"1", "2", "5", "3", "4", "null", "6"}},
+			ans114{[]string{"1", "null", "2", "null", "3", "null", "4", "null", "5", "null", "6"}},
+		},
+
+		{
+			para114{[]string{"0"}},
+			ans114{[]string{"0"}},
+		},
+
+		{
+			para114{[]string{"1", "2", "3", "4", "5", "6"}},
+			ans114{[]string{"1", "2", "4", "5", "3", "6", "null"}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 114------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans114, q.para114
+		fmt.Printf("【input】:%v       \n", p)
+		rootOne := structures.Strings2TreeNode(p.one)
+		flatten(rootOne)
+		fmt.Printf("【levelorder output】:%v       \n", structures.Tree2LevelOrderStrings(rootOne))
+		fmt.Printf("【preorder output】:%v      \n", structures.Tree2PreOrderStrings(rootOne))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0114. Flatten Binary Tree to Linked List/README.md b/leetcode/0114.Flatten-Binary-Tree-to-Linked-List/README.md
similarity index 100%
rename from Algorithms/0114. Flatten Binary Tree to Linked List/README.md
rename to leetcode/0114.Flatten-Binary-Tree-to-Linked-List/README.md
diff --git a/leetcode/0115.Distinct-Subsequences/115. Distinct Subsequences.go b/leetcode/0115.Distinct-Subsequences/115. Distinct Subsequences.go
new file mode 100644
index 000000000..f8f1f2339
--- /dev/null
+++ b/leetcode/0115.Distinct-Subsequences/115. Distinct Subsequences.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+// 解法一 压缩版 DP
+func numDistinct(s string, t string) int {
+	dp := make([]int, len(s)+1)
+	for i, curT := range t {
+		pre := 0
+		for j, curS := range s {
+			if i == 0 {
+				pre = 1
+			}
+			newDP := dp[j+1]
+			if curT == curS {
+				dp[j+1] = dp[j] + pre
+			} else {
+				dp[j+1] = dp[j]
+			}
+			pre = newDP
+		}
+	}
+	return dp[len(s)]
+}
+
+// 解法二 普通 DP
+func numDistinct1(s, t string) int {
+	m, n := len(s), len(t)
+	if m < n {
+		return 0
+	}
+	dp := make([][]int, m+1)
+	for i := range dp {
+		dp[i] = make([]int, n+1)
+		dp[i][n] = 1
+	}
+	for i := m - 1; i >= 0; i-- {
+		for j := n - 1; j >= 0; j-- {
+			if s[i] == t[j] {
+				dp[i][j] = dp[i+1][j+1] + dp[i+1][j]
+			} else {
+				dp[i][j] = dp[i+1][j]
+			}
+		}
+	}
+	return dp[0][0]
+}
diff --git a/leetcode/0115.Distinct-Subsequences/115. Distinct Subsequences_test.go b/leetcode/0115.Distinct-Subsequences/115. Distinct Subsequences_test.go
new file mode 100644
index 000000000..d192137e2
--- /dev/null
+++ b/leetcode/0115.Distinct-Subsequences/115. Distinct Subsequences_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question115 struct {
+	para115
+	ans115
+}
+
+// para 是参数
+// one 代表第一个参数
+type para115 struct {
+	s string
+	t string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans115 struct {
+	one int
+}
+
+func Test_Problem115(t *testing.T) {
+
+	qs := []question115{
+
+		{
+			para115{"rabbbit", "rabbit"},
+			ans115{3},
+		},
+
+		{
+			para115{"babgbag", "bag"},
+			ans115{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 115------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans115, q.para115
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numDistinct(p.s, p.t))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0115.Distinct-Subsequences/README.md b/leetcode/0115.Distinct-Subsequences/README.md
new file mode 100644
index 000000000..244152a68
--- /dev/null
+++ b/leetcode/0115.Distinct-Subsequences/README.md
@@ -0,0 +1,99 @@
+# [115. Distinct Subsequences](https://leetcode.com/problems/distinct-subsequences/)
+
+
+## 题目
+
+Given two strings `s` and `t`, return *the number of distinct subsequences of `s` which equals `t`*.
+
+A string's **subsequence** is a new string formed from the original string by deleting some (can be none) of the characters without disturbing the remaining characters' relative positions. (i.e., `"ACE"` is a subsequence of `"ABCDE"` while `"AEC"` is not).
+
+It is guaranteed the answer fits on a 32-bit signed integer.
+
+**Example 1:**
+
+```
+Input: s = "rabbbit", t = "rabbit"
+Output: 3
+Explanation:
+As shown below, there are 3 ways you can generate "rabbit" from S.
+rabbbitrabbbitrabbbit
+```
+
+**Example 2:**
+
+```
+Input: s = "babgbag", t = "bag"
+Output: 5
+Explanation:
+As shown below, there are 5 ways you can generate "bag" from S.
+babgbagbabgbagbabgbagbabgbagbabgbag
+```
+
+**Constraints:**
+
+- `0 <= s.length, t.length <= 1000`
+- `s` and `t` consist of English letters.
+
+## 题目大意
+
+给定一个字符串 s 和一个字符串 t ，计算在 s 的子序列中 t 出现的个数。字符串的一个 子序列 是指，通过删除一些（也可以不删除）字符且不干扰剩余字符相对位置所组成的新字符串。（例如，"ACE" 是 "ABCDE" 的一个子序列，而 "AEC" 不是）题目数据保证答案符合 32 位带符号整数范围。
+
+## 解题思路
+
+- 在字符串 `s` 中最多包含多少个字符串 `t`。这里面包含很多重叠子问题，所以尝试用动态规划解决这个问题。定义 `dp[i][j]` 代表 `s[i:]` 的子序列中 `t[j:]` 出现的个数。初始化先判断边界条件。当 `i = len(s)` 且 `0≤ j < len(t)` 的时候，`s[i:]` 为空字符串，`t[j:]` 不为空，所以 `dp[len(s)][j] = 0`。当 `j = len(t)` 且 `0 ≤ i < len(s)` 的时候，`t[j:]` 不为空字符串，空字符串是任何字符串的子序列。所以 `dp[i][n] = 1`。
+- 当 `i < len(s)` 且 `j < len(t)` 的时候，如果 `s[i] == t[j]`，有 2 种匹配方式，第一种将 `s[i]` 与 `t[j]` 匹配，那么 `t[j+1:]` 匹配 `s[i+1:]` 的子序列，子序列数为 `dp[i+1][j+1]`；第二种将 `s[i]` 不与 `t[j]` 匹配，`t[j:]` 作为 `s[i+1:]` 的子序列，子序列数为 `dp[i+1][j]`。综合 2 种情况，当 `s[i] == t[j]` 时，`dp[i][j] = dp[i+1][j+1] + dp[i+1][j]`。
+- 如果 `s[i] != t[j]`，此时 `t[j:]` 只能作为 `s[i+1:]` 的子序列，子序列数为 `dp[i+1][j]`。所以当 `s[i] != t[j]` 时，`dp[i][j] = dp[i+1][j]`。综上分析得：
+
+	$$dp[i][j] = \left\{\begin{matrix}dp[i+1][j+1]+dp[i+1][j]&,s[i]=t[j]\\ dp[i+1][j]&,s[i]!=t[j]\end{matrix}\right.$$
+
+- 最后是优化版本。写出上述代码以后，可以发现填表的过程是从右下角一直填到左上角。填表顺序是 从下往上一行一行的填。行内从右往左填。于是可以将这个二维数据压缩到一维。因为填充当前行只需要用到它的下一行信息即可，更进一步，用到的是下一行中右边元素的信息。于是可以每次更新该行时，先将旧的值存起来，计算更新该行的时候从右往左更新。这样做即可减少一维空间，将原来的二维数组压缩到一维数组。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 压缩版 DP
+func numDistinct(s string, t string) int {
+	dp := make([]int, len(s)+1)
+	for i, curT := range t {
+		pre := 0
+		for j, curS := range s {
+			if i == 0 {
+				pre = 1
+			}
+			newDP := dp[j+1]
+			if curT == curS {
+				dp[j+1] = dp[j] + pre
+			} else {
+				dp[j+1] = dp[j]
+			}
+			pre = newDP
+		}
+	}
+	return dp[len(s)]
+}
+
+// 解法二 普通 DP
+func numDistinct1(s, t string) int {
+	m, n := len(s), len(t)
+	if m < n {
+		return 0
+	}
+	dp := make([][]int, m+1)
+	for i := range dp {
+		dp[i] = make([]int, n+1)
+		dp[i][n] = 1
+	}
+	for i := m - 1; i >= 0; i-- {
+		for j := n - 1; j >= 0; j-- {
+			if s[i] == t[j] {
+				dp[i][j] = dp[i+1][j+1] + dp[i+1][j]
+			} else {
+				dp[i][j] = dp[i+1][j]
+			}
+		}
+	}
+	return dp[0][0]
+}
+```
\ No newline at end of file
diff --git a/leetcode/0116.Populating-Next-Right-Pointers-in-Each-Node/116.Populating Next Right Pointers in Each Node.go b/leetcode/0116.Populating-Next-Right-Pointers-in-Each-Node/116.Populating Next Right Pointers in Each Node.go
new file mode 100644
index 000000000..4587c39d5
--- /dev/null
+++ b/leetcode/0116.Populating-Next-Right-Pointers-in-Each-Node/116.Populating Next Right Pointers in Each Node.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+type Node struct {
+	Val   int
+	Left  *Node
+	Right *Node
+	Next  *Node
+}
+
+// 解法一：迭代
+func connect(root *Node) *Node {
+	if root == nil {
+		return root
+	}
+	q := []*Node{root}
+	for len(q) > 0 {
+		var p []*Node
+		// 遍历这一层的所有节点
+		for i, node := range q {
+			if i+1 < len(q) {
+				node.Next = q[i+1]
+			}
+			if node.Left != nil {
+				p = append(p, node.Left)
+			}
+			if node.Right != nil {
+				p = append(p, node.Right)
+			}
+		}
+		q = p
+	}
+	return root
+}
+
+// 解法二 递归
+func connect2(root *Node) *Node {
+	if root == nil {
+		return nil
+	}
+	connectTwoNode(root.Left, root.Right)
+	return root
+}
+
+func connectTwoNode(node1, node2 *Node) {
+	if node1 == nil || node2 == nil {
+		return
+	}
+	node1.Next = node2
+	connectTwoNode(node1.Left, node1.Right)
+	connectTwoNode(node2.Left, node2.Right)
+	connectTwoNode(node1.Right, node2.Left)
+}
diff --git a/leetcode/0116.Populating-Next-Right-Pointers-in-Each-Node/116.Populating Next Right Pointers in Each Node_test.go b/leetcode/0116.Populating-Next-Right-Pointers-in-Each-Node/116.Populating Next Right Pointers in Each Node_test.go
new file mode 100644
index 000000000..6146f1579
--- /dev/null
+++ b/leetcode/0116.Populating-Next-Right-Pointers-in-Each-Node/116.Populating Next Right Pointers in Each Node_test.go	
@@ -0,0 +1,117 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question116 struct {
+	para116
+	ans116
+}
+
+// para 是参数
+// one 代表第一个参数
+type para116 struct {
+	one *Node
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans116 struct {
+	one *Node
+}
+
+func newQuestionNode() *Node {
+	node7 := &Node{}
+	node7.Val = 7
+
+	node6 := &Node{}
+	node6.Val = 6
+
+	node5 := &Node{}
+	node5.Val = 5
+
+	node4 := &Node{}
+	node4.Val = 4
+
+	node3 := &Node{}
+	node3.Val = 3
+
+	node2 := &Node{}
+	node2.Val = 2
+
+	node1 := &Node{}
+	node1.Val = 1
+
+	node1.Left = node2
+	node1.Right = node3
+
+	node2.Left = node4
+	node2.Right = node5
+
+	node3.Left = node6
+	node3.Right = node7
+	return node1
+}
+
+func newResultNode() *Node {
+	node7 := &Node{}
+	node7.Val = 7
+
+	node6 := &Node{}
+	node6.Val = 6
+
+	node5 := &Node{}
+	node5.Val = 5
+
+	node4 := &Node{}
+	node4.Val = 4
+
+	node3 := &Node{}
+	node3.Val = 3
+
+	node2 := &Node{}
+	node2.Val = 2
+
+	node1 := &Node{}
+	node1.Val = 1
+
+	node1.Left = node2
+	node1.Right = node3
+
+	node2.Left = node4
+	node2.Right = node5
+
+	node3.Left = node6
+	node3.Right = node7
+
+	node1.Next = nil
+	node2.Next = node3
+	node3.Next = nil
+	node4.Next = node5
+	node5.Next = node6
+	node6.Next = node7
+	node7.Next = nil
+
+	return node1
+}
+
+func Test_Problem116(t *testing.T) {
+
+	qs := []question116{
+		{
+			para116{newQuestionNode()},
+			ans116{newResultNode()},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 116------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans116, q.para116
+		fmt.Printf("【input】:%v      ", p.one)
+		fmt.Printf("【output】:%v      \n", connect(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0116.Populating-Next-Right-Pointers-in-Each-Node/README.md b/leetcode/0116.Populating-Next-Right-Pointers-in-Each-Node/README.md
new file mode 100644
index 000000000..f25539f42
--- /dev/null
+++ b/leetcode/0116.Populating-Next-Right-Pointers-in-Each-Node/README.md
@@ -0,0 +1,118 @@
+# [116. Populating Next Right Pointers in Each Node](https://leetcode.com/problems/populating-next-right-pointers-in-each-node/)
+
+
+## 题目
+
+You are given a **perfect binary tree** where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:
+
+```
+struct Node {
+  int val;
+  Node *left;
+  Node *right;
+  Node *next;
+}
+
+```
+
+Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to `NULL`.
+
+Initially, all next pointers are set to `NULL`.
+
+**Follow up:**
+
+- You may only use constant extra space.
+- Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/02/14/116_sample.png](https://assets.leetcode.com/uploads/2019/02/14/116_sample.png)
+
+```
+Input: root = [1,2,3,4,5,6,7]
+Output: [1,#,2,3,#,4,5,6,7,#]
+Explanation:Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the given tree is less than `4096`.
+- `1000 <= node.val <= 1000`
+
+## 题目大意
+
+给定一个 完美二叉树 ，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：
+
+```jsx
+struct Node {
+  int val;
+  Node *left;
+  Node *right;
+  Node *next;
+}
+
+```
+
+填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。初始状态下，所有 next 指针都被设置为 NULL。
+
+## 解题思路
+
+- 本质上是二叉树的层序遍历，基于广度优先搜索，将每层的节点放入队列，并遍历队列进行连接。
+
+## 代码
+
+```go
+package leetcode
+
+type Node struct {
+	Val   int
+	Left  *Node
+	Right *Node
+	Next  *Node
+}
+
+//解法一：迭代
+func connect(root *Node) *Node {
+	if root == nil {
+		return root
+	}
+	q := []*Node{root}
+	for len(q) > 0 {
+		var p []*Node
+		// 遍历这一层的所有节点
+		for i, node := range q {
+			if i+1 < len(q) {
+				node.Next = q[i+1]
+			}
+			if node.Left != nil {
+				p = append(p, node.Left)
+			}
+			if node.Right != nil {
+				p = append(p, node.Right)
+			}
+		}
+		q = p
+	}
+	return root
+}
+
+// 解法二 递归
+func connect2(root *Node) *Node {
+	if root == nil {
+		return nil
+	}
+	connectTwoNode(root.Left, root.Right)
+	return root
+}
+
+func connectTwoNode(node1, node2 *Node) {
+	if node1 == nil || node2 == nil {
+		return
+	}
+	node1.Next = node2
+	connectTwoNode(node1.Left, node1.Right)
+	connectTwoNode(node2.Left, node2.Right)
+	connectTwoNode(node1.Right, node2.Left)
+}
+```
\ No newline at end of file
diff --git a/leetcode/0118.Pascals-Triangle/118. Pascals Triangle.go b/leetcode/0118.Pascals-Triangle/118. Pascals Triangle.go
new file mode 100644
index 000000000..33b1a6e18
--- /dev/null
+++ b/leetcode/0118.Pascals-Triangle/118. Pascals Triangle.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+func generate(numRows int) [][]int {
+	result := [][]int{}
+	for i := 0; i < numRows; i++ {
+		row := []int{}
+		for j := 0; j < i+1; j++ {
+			if j == 0 || j == i {
+				row = append(row, 1)
+			} else if i > 1 {
+				row = append(row, result[i-1][j-1]+result[i-1][j])
+			}
+		}
+		result = append(result, row)
+	}
+	return result
+}
diff --git a/leetcode/0118.Pascals-Triangle/118. Pascals Triangle_test.go b/leetcode/0118.Pascals-Triangle/118. Pascals Triangle_test.go
new file mode 100644
index 000000000..5b8e87a7e
--- /dev/null
+++ b/leetcode/0118.Pascals-Triangle/118. Pascals Triangle_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question118 struct {
+	para118
+	ans118
+}
+
+// para 是参数
+// one 代表第一个参数
+type para118 struct {
+	numRows int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans118 struct {
+	one [][]int
+}
+
+func Test_Problem118(t *testing.T) {
+
+	qs := []question118{
+
+		{
+			para118{2},
+			ans118{[][]int{{1}, {1, 1}}},
+		},
+
+		{
+			para118{5},
+			ans118{[][]int{{1}, {1, 1}, {1, 2, 1}, {1, 3, 3, 1}, {1, 4, 6, 4, 1}}},
+		},
+
+		{
+			para118{10},
+			ans118{[][]int{{1}, {1, 1}, {1, 2, 1}, {1, 3, 3, 1}, {1, 4, 6, 4, 1}, {1, 5, 10, 10, 5, 1}, {1, 6, 15, 20, 15, 6, 1}, {1, 7, 21, 35, 35, 21, 7, 1}, {1, 8, 28, 56, 70, 56, 28, 8, 1}, {1, 9, 36, 84, 126, 126, 84, 36, 9, 1}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 118------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans118, q.para118
+		fmt.Printf("【input】:%v    【output】:%v\n", p, generate(p.numRows))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0118.Pascals-Triangle/README.md b/leetcode/0118.Pascals-Triangle/README.md
new file mode 100644
index 000000000..bcb79df33
--- /dev/null
+++ b/leetcode/0118.Pascals-Triangle/README.md
@@ -0,0 +1,60 @@
+# [118. Pascal's Triangle](https://leetcode.com/problems/pascals-triangle/)
+
+
+## 题目
+
+Given a non-negative integer numRows, generate the first numRows of Pascal's triangle.
+
+![](https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif)
+
+**Note:** In Pascal's triangle, each number is the sum of the two numbers directly above it.
+
+**Example:**
+
+```
+Input: 5
+Output:
+[
+     [1],
+    [1,1],
+   [1,2,1],
+  [1,3,3,1],
+ [1,4,6,4,1]
+]
+```
+
+## 题目大意
+
+给定一个非负整数 numRows，生成杨辉三角的前 numRows 行。在杨辉三角中，每个数是它左上方和右上方的数的和。
+
+
+## 解题思路
+
+- 给定一个 n，要求打印杨辉三角的前 n 行。
+- 简单题。按照杨辉三角的生成规则循环打印即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func generate(numRows int) [][]int {
+	result := [][]int{}
+	for i := 0; i < numRows; i++ {
+		row := []int{}
+		for j := 0; j < i+1; j++ {
+			if j == 0 || j == i {
+				row = append(row, 1)
+			} else if i > 1 {
+				row = append(row, result[i-1][j-1]+result[i-1][j])
+			}
+		}
+		result = append(result, row)
+	}
+	return result
+}
+
+```
+
diff --git a/leetcode/0119.Pascals-Triangle-II/119. Pascals Triangle II.go b/leetcode/0119.Pascals-Triangle-II/119. Pascals Triangle II.go
new file mode 100644
index 000000000..34e4401ea
--- /dev/null
+++ b/leetcode/0119.Pascals-Triangle-II/119. Pascals Triangle II.go	
@@ -0,0 +1,10 @@
+package leetcode
+
+func getRow(rowIndex int) []int {
+	row := make([]int, rowIndex+1)
+	row[0] = 1
+	for i := 1; i <= rowIndex; i++ {
+		row[i] = row[i-1] * (rowIndex - i + 1) / i
+	}
+	return row
+}
diff --git a/leetcode/0119.Pascals-Triangle-II/119. Pascals Triangle II_test.go b/leetcode/0119.Pascals-Triangle-II/119. Pascals Triangle II_test.go
new file mode 100644
index 000000000..85771cba4
--- /dev/null
+++ b/leetcode/0119.Pascals-Triangle-II/119. Pascals Triangle II_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question119 struct {
+	para119
+	ans119
+}
+
+// para 是参数
+// one 代表第一个参数
+type para119 struct {
+	rowIndex int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans119 struct {
+	one []int
+}
+
+func Test_Problem119(t *testing.T) {
+
+	qs := []question119{
+
+		{
+			para119{3},
+			ans119{[]int{1, 3, 3, 1}},
+		},
+
+		{
+			para119{0},
+			ans119{[]int{1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 119------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans119, q.para119
+		fmt.Printf("【input】:%v    【output】:%v\n", p, getRow(p.rowIndex))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0119.Pascals-Triangle-II/README.md b/leetcode/0119.Pascals-Triangle-II/README.md
new file mode 100644
index 000000000..c233fe4ee
--- /dev/null
+++ b/leetcode/0119.Pascals-Triangle-II/README.md
@@ -0,0 +1,72 @@
+# [119. Pascal's Triangle II](https://leetcode.com/problems/pascals-triangle-ii/)
+
+
+## 题目
+
+Given an integer `rowIndex`, return the `rowIndexth` row of the Pascal's triangle.
+
+Notice that the row index starts from **0**.
+
+![https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif](https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif)
+
+In Pascal's triangle, each number is the sum of the two numbers directly above it.
+
+**Follow up:**
+
+Could you optimize your algorithm to use only *O*(*k*) extra space?
+
+**Example 1:**
+
+```
+Input: rowIndex = 3
+Output: [1,3,3,1]
+```
+
+**Example 2:**
+
+```
+Input: rowIndex = 0
+Output: [1]
+```
+
+**Example 3:**
+
+```
+Input: rowIndex = 1
+Output: [1,1]
+```
+
+**Constraints:**
+
+- `0 <= rowIndex <= 33`
+
+## 题目大意
+
+给定一个非负索引 k，其中 k ≤ 33，返回杨辉三角的第 k 行。
+
+## 解题思路
+
+- 题目中的三角是杨辉三角，每个数字是 `(a+b)^n` 二项式展开的系数。题目要求我们只能使用 O(k) 的空间。那么需要找到两两项直接的递推关系。由组合知识得知：
+
+    $$\begin{aligned}C_{n}^{m} &= \frac{n!}{m!(n-m)!} \\C_{n}^{m-1} &= \frac{n!}{(m-1)!(n-m+1)!}\end{aligned}$$
+
+    于是得到递推公式：
+
+    $$C_{n}^{m} = C_{n}^{m-1} \times \frac{n-m+1}{m}$$
+
+    利用这个递推公式即可以把空间复杂度优化到 O(k)
+
+## 代码
+
+```go
+package leetcode
+
+func getRow(rowIndex int) []int {
+	row := make([]int, rowIndex+1)
+	row[0] = 1
+	for i := 1; i <= rowIndex; i++ {
+		row[i] = row[i-1] * (rowIndex - i + 1) / i
+	}
+	return row
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0120. Triangle/120. Triangle.go b/leetcode/0120.Triangle/120. Triangle.go
similarity index 94%
rename from Algorithms/0120. Triangle/120. Triangle.go
rename to leetcode/0120.Triangle/120. Triangle.go
index 83a1c47e7..f97b88acd 100644
--- a/Algorithms/0120. Triangle/120. Triangle.go	
+++ b/leetcode/0120.Triangle/120. Triangle.go	
@@ -17,6 +17,13 @@ func minimumTotal(triangle [][]int) int {
 	return triangle[0][0]
 }
 
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
 // 解法二 正常 DP，空间复杂度 O(n)
 func minimumTotal1(triangle [][]int) int {
 	if len(triangle) == 0 {
diff --git a/Algorithms/0120. Triangle/120. Triangle_test.go b/leetcode/0120.Triangle/120. Triangle_test.go
similarity index 86%
rename from Algorithms/0120. Triangle/120. Triangle_test.go
rename to leetcode/0120.Triangle/120. Triangle_test.go
index a79301033..2398cafe0 100644
--- a/Algorithms/0120. Triangle/120. Triangle_test.go	
+++ b/leetcode/0120.Triangle/120. Triangle_test.go	
@@ -25,8 +25,8 @@ type ans120 struct {
 func Test_Problem120(t *testing.T) {
 
 	qs := []question120{
-		question120{
-			para120{[][]int{[]int{2}, []int{3, 4}, []int{6, 5, 7}, []int{4, 1, 8, 3}}},
+		{
+			para120{[][]int{{2}, {3, 4}, {6, 5, 7}, {4, 1, 8, 3}}},
 			ans120{11},
 		},
 	}
diff --git a/Algorithms/0120. Triangle/README.md b/leetcode/0120.Triangle/README.md
similarity index 100%
rename from Algorithms/0120. Triangle/README.md
rename to leetcode/0120.Triangle/README.md
diff --git a/Algorithms/0121. Best Time to Buy and Sell Stock/121. Best Time to Buy and Sell Stock.go b/leetcode/0121.Best-Time-to-Buy-and-Sell-Stock/121. Best Time to Buy and Sell Stock.go
similarity index 92%
rename from Algorithms/0121. Best Time to Buy and Sell Stock/121. Best Time to Buy and Sell Stock.go
rename to leetcode/0121.Best-Time-to-Buy-and-Sell-Stock/121. Best Time to Buy and Sell Stock.go
index d0b2ca17e..012187fd7 100644
--- a/Algorithms/0121. Best Time to Buy and Sell Stock/121. Best Time to Buy and Sell Stock.go	
+++ b/leetcode/0121.Best-Time-to-Buy-and-Sell-Stock/121. Best Time to Buy and Sell Stock.go	
@@ -40,3 +40,10 @@ func maxProfit1(prices []int) int {
 	}
 	return res
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0121. Best Time to Buy and Sell Stock/121. Best Time to Buy and Sell Stock_test.go b/leetcode/0121.Best-Time-to-Buy-and-Sell-Stock/121. Best Time to Buy and Sell Stock_test.go
similarity index 92%
rename from Algorithms/0121. Best Time to Buy and Sell Stock/121. Best Time to Buy and Sell Stock_test.go
rename to leetcode/0121.Best-Time-to-Buy-and-Sell-Stock/121. Best Time to Buy and Sell Stock_test.go
index 10e031e09..681161ed2 100644
--- a/Algorithms/0121. Best Time to Buy and Sell Stock/121. Best Time to Buy and Sell Stock_test.go	
+++ b/leetcode/0121.Best-Time-to-Buy-and-Sell-Stock/121. Best Time to Buy and Sell Stock_test.go	
@@ -26,22 +26,22 @@ func Test_Problem121(t *testing.T) {
 
 	qs := []question121{
 
-		question121{
+		{
 			para121{[]int{}},
 			ans121{0},
 		},
 
-		question121{
+		{
 			para121{[]int{7, 1, 5, 3, 6, 4}},
 			ans121{5},
 		},
 
-		question121{
+		{
 			para121{[]int{7, 6, 4, 3, 1}},
 			ans121{0},
 		},
 
-		question121{
+		{
 			para121{[]int{1, 3, 2, 8, 4, 9}},
 			ans121{8},
 		},
diff --git a/Algorithms/0121. Best Time to Buy and Sell Stock/README.md b/leetcode/0121.Best-Time-to-Buy-and-Sell-Stock/README.md
similarity index 100%
rename from Algorithms/0121. Best Time to Buy and Sell Stock/README.md
rename to leetcode/0121.Best-Time-to-Buy-and-Sell-Stock/README.md
diff --git a/Algorithms/0122. Best Time to Buy and Sell Stock II/122. Best Time to Buy and Sell Stock II.go b/leetcode/0122.Best-Time-to-Buy-and-Sell-Stock-II/122. Best Time to Buy and Sell Stock II.go
similarity index 100%
rename from Algorithms/0122. Best Time to Buy and Sell Stock II/122. Best Time to Buy and Sell Stock II.go
rename to leetcode/0122.Best-Time-to-Buy-and-Sell-Stock-II/122. Best Time to Buy and Sell Stock II.go
diff --git a/Algorithms/0122. Best Time to Buy and Sell Stock II/122. Best Time to Buy and Sell Stock II_test.go b/leetcode/0122.Best-Time-to-Buy-and-Sell-Stock-II/122. Best Time to Buy and Sell Stock II_test.go
similarity index 91%
rename from Algorithms/0122. Best Time to Buy and Sell Stock II/122. Best Time to Buy and Sell Stock II_test.go
rename to leetcode/0122.Best-Time-to-Buy-and-Sell-Stock-II/122. Best Time to Buy and Sell Stock II_test.go
index 009f44c4c..dd58050fe 100644
--- a/Algorithms/0122. Best Time to Buy and Sell Stock II/122. Best Time to Buy and Sell Stock II_test.go	
+++ b/leetcode/0122.Best-Time-to-Buy-and-Sell-Stock-II/122. Best Time to Buy and Sell Stock II_test.go	
@@ -26,27 +26,27 @@ func Test_Problem122(t *testing.T) {
 
 	qs := []question122{
 
-		question122{
+		{
 			para122{[]int{}},
 			ans122{0},
 		},
 
-		question122{
+		{
 			para122{[]int{7, 1, 5, 3, 6, 4}},
 			ans122{7},
 		},
 
-		question122{
+		{
 			para122{[]int{7, 6, 4, 3, 1}},
 			ans122{0},
 		},
 
-		question122{
+		{
 			para122{[]int{1, 2, 3, 4, 5}},
 			ans122{4},
 		},
 
-		question122{
+		{
 			para122{[]int{1, 2, 10, 11, 12, 15}},
 			ans122{14},
 		},
diff --git a/Algorithms/0122. Best Time to Buy and Sell Stock II/README.md b/leetcode/0122.Best-Time-to-Buy-and-Sell-Stock-II/README.md
similarity index 100%
rename from Algorithms/0122. Best Time to Buy and Sell Stock II/README.md
rename to leetcode/0122.Best-Time-to-Buy-and-Sell-Stock-II/README.md
diff --git a/Algorithms/0124. Binary Tree Maximum Path Sum/124. Binary Tree Maximum Path Sum.go b/leetcode/0124.Binary-Tree-Maximum-Path-Sum/124. Binary Tree Maximum Path Sum.go
similarity index 75%
rename from Algorithms/0124. Binary Tree Maximum Path Sum/124. Binary Tree Maximum Path Sum.go
rename to leetcode/0124.Binary-Tree-Maximum-Path-Sum/124. Binary Tree Maximum Path Sum.go
index 186550d90..058193b7b 100644
--- a/Algorithms/0124. Binary Tree Maximum Path Sum/124. Binary Tree Maximum Path Sum.go	
+++ b/leetcode/0124.Binary-Tree-Maximum-Path-Sum/124. Binary Tree Maximum Path Sum.go	
@@ -1,6 +1,13 @@
 package leetcode
 
-import "math"
+import (
+	"math"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
 
 /**
  * Definition for a binary tree node.
@@ -10,6 +17,7 @@ import "math"
  *     Right *TreeNode
  * }
  */
+
 func maxPathSum(root *TreeNode) int {
 	if root == nil {
 		return 0
@@ -30,3 +38,10 @@ func getPathSum(root *TreeNode, maxSum *int) int {
 	*maxSum = max(*maxSum, max(currMax, left+right+root.Val))
 	return currMax
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0124. Binary Tree Maximum Path Sum/124. Binary Tree Maximum Path Sum_test.go b/leetcode/0124.Binary-Tree-Maximum-Path-Sum/124. Binary Tree Maximum Path Sum_test.go
similarity index 80%
rename from Algorithms/0124. Binary Tree Maximum Path Sum/124. Binary Tree Maximum Path Sum_test.go
rename to leetcode/0124.Binary-Tree-Maximum-Path-Sum/124. Binary Tree Maximum Path Sum_test.go
index f196df122..7e57778c6 100644
--- a/Algorithms/0124. Binary Tree Maximum Path Sum/124. Binary Tree Maximum Path Sum_test.go	
+++ b/leetcode/0124.Binary-Tree-Maximum-Path-Sum/124. Binary Tree Maximum Path Sum_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question124 struct {
@@ -26,23 +28,23 @@ func Test_Problem124(t *testing.T) {
 
 	qs := []question124{
 
-		question124{
+		{
 			para124{[]int{}},
 			ans124{0},
 		},
 
-		question124{
+		{
 			para124{[]int{1}},
 			ans124{1},
 		},
 
-		question124{
+		{
 			para124{[]int{1, 2, 3}},
 			ans124{6},
 		},
 
-		question124{
-			para124{[]int{-10, 9, 20, NULL, NULL, 15, 7}},
+		{
+			para124{[]int{-10, 9, 20, structures.NULL, structures.NULL, 15, 7}},
 			ans124{42},
 		},
 	}
@@ -52,7 +54,7 @@ func Test_Problem124(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans124, q.para124
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", maxPathSum(root))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0124. Binary Tree Maximum Path Sum/README.md b/leetcode/0124.Binary-Tree-Maximum-Path-Sum/README.md
similarity index 100%
rename from Algorithms/0124. Binary Tree Maximum Path Sum/README.md
rename to leetcode/0124.Binary-Tree-Maximum-Path-Sum/README.md
diff --git a/Algorithms/0125. Valid-Palindrome/125. Valid Palindrome.go b/leetcode/0125.Valid-Palindrome/125. Valid Palindrome.go
similarity index 91%
rename from Algorithms/0125. Valid-Palindrome/125. Valid Palindrome.go
rename to leetcode/0125.Valid-Palindrome/125. Valid Palindrome.go
index dad5d16e6..6ccd92002 100644
--- a/Algorithms/0125. Valid-Palindrome/125. Valid Palindrome.go	
+++ b/leetcode/0125.Valid-Palindrome/125. Valid Palindrome.go	
@@ -1,13 +1,11 @@
 package leetcode
 
 import (
-  "strings"
+	"strings"
 )
 
 func isPalindrome(s string) bool {
-
-  s = strings.ToLower(s)
-
+	s = strings.ToLower(s)
 	i, j := 0, len(s)-1
 	for i < j {
 		for i < j && !isChar(s[i]) {
@@ -22,7 +20,6 @@ func isPalindrome(s string) bool {
 		i++
 		j--
 	}
-
 	return true
 }
 
diff --git a/Algorithms/0125. Valid-Palindrome/125. Valid Palindrome_test.go b/leetcode/0125.Valid-Palindrome/125. Valid Palindrome_test.go
similarity index 100%
rename from Algorithms/0125. Valid-Palindrome/125. Valid Palindrome_test.go
rename to leetcode/0125.Valid-Palindrome/125. Valid Palindrome_test.go
diff --git a/Algorithms/0125. Valid-Palindrome/README.md b/leetcode/0125.Valid-Palindrome/README.md
similarity index 100%
rename from Algorithms/0125. Valid-Palindrome/README.md
rename to leetcode/0125.Valid-Palindrome/README.md
diff --git a/Algorithms/0126. Word Ladder II/126. Word Ladder II.go b/leetcode/0126.Word-Ladder-II/126. Word Ladder II.go
similarity index 100%
rename from Algorithms/0126. Word Ladder II/126. Word Ladder II.go
rename to leetcode/0126.Word-Ladder-II/126. Word Ladder II.go
diff --git a/Algorithms/0126. Word Ladder II/126. Word Ladder II_test.go b/leetcode/0126.Word-Ladder-II/126. Word Ladder II_test.go
similarity index 84%
rename from Algorithms/0126. Word Ladder II/126. Word Ladder II_test.go
rename to leetcode/0126.Word-Ladder-II/126. Word Ladder II_test.go
index 9745cabc0..7859a7b36 100644
--- a/Algorithms/0126. Word Ladder II/126. Word Ladder II_test.go	
+++ b/leetcode/0126.Word-Ladder-II/126. Word Ladder II_test.go	
@@ -27,12 +27,12 @@ type ans126 struct {
 func Test_Problem126(t *testing.T) {
 
 	qs := []question126{
-		question126{
+		{
 			para126{"hit", "cog", []string{"hot", "dot", "dog", "lot", "log", "cog"}},
-			ans126{[][]string{[]string{"hit", "hot", "dot", "dog", "cog"}, []string{"hit", "hot", "lot", "log", "cog"}}},
+			ans126{[][]string{{"hit", "hot", "dot", "dog", "cog"}, {"hit", "hot", "lot", "log", "cog"}}},
 		},
 
-		question126{
+		{
 			para126{"hit", "cog", []string{"hot", "dot", "dog", "lot", "log"}},
 			ans126{[][]string{}},
 		},
diff --git a/Algorithms/0126. Word Ladder II/README.md b/leetcode/0126.Word-Ladder-II/README.md
similarity index 100%
rename from Algorithms/0126. Word Ladder II/README.md
rename to leetcode/0126.Word-Ladder-II/README.md
diff --git a/Algorithms/0127. Word Ladder/127. Word Ladder.go b/leetcode/0127.Word-Ladder/127. Word Ladder.go
similarity index 93%
rename from Algorithms/0127. Word Ladder/127. Word Ladder.go
rename to leetcode/0127.Word-Ladder/127. Word Ladder.go
index a983bc525..ea4b9b2cb 100644
--- a/Algorithms/0127. Word Ladder/127. Word Ladder.go	
+++ b/leetcode/0127.Word-Ladder/127. Word Ladder.go	
@@ -40,7 +40,7 @@ func getCandidates(word string) []string {
 	for i := 0; i < 26; i++ {
 		for j := 0; j < len(word); j++ {
 			if word[j] != byte(int('a')+i) {
-				res = append(res, word[:j]+string(int('a')+i)+word[j+1:])
+				res = append(res, word[:j]+string(rune(int('a')+i))+word[j+1:])
 			}
 		}
 	}
diff --git a/Algorithms/0127. Word Ladder/127. Word Ladder_test.go b/leetcode/0127.Word-Ladder/127. Word Ladder_test.go
similarity index 96%
rename from Algorithms/0127. Word Ladder/127. Word Ladder_test.go
rename to leetcode/0127.Word-Ladder/127. Word Ladder_test.go
index 59591fe08..64c5f13f2 100644
--- a/Algorithms/0127. Word Ladder/127. Word Ladder_test.go	
+++ b/leetcode/0127.Word-Ladder/127. Word Ladder_test.go	
@@ -27,12 +27,12 @@ type ans127 struct {
 func Test_Problem127(t *testing.T) {
 
 	qs := []question127{
-		question127{
+		{
 			para127{"hit", "cog", []string{"hot", "dot", "dog", "lot", "log", "cog"}},
 			ans127{5},
 		},
 
-		question127{
+		{
 			para127{"hit", "cog", []string{"hot", "dot", "dog", "lot", "log"}},
 			ans127{0},
 		},
diff --git a/leetcode/0127.Word-Ladder/README.md b/leetcode/0127.Word-Ladder/README.md
new file mode 100755
index 000000000..6575eab24
--- /dev/null
+++ b/leetcode/0127.Word-Ladder/README.md
@@ -0,0 +1,65 @@
+# [127. Word Ladder](https://leetcode.com/problems/word-ladder/)
+
+
+## 题目
+
+Given two words (*beginWord* and *endWord*), and a dictionary's word list, find the length of shortest transformation sequence from *beginWord* to *endWord*, such that:
+
+1. Only one letter can be changed at a time.
+2. Each transformed word must exist in the word list. Note that *beginWord* is *not* a transformed word.
+
+**Note:**
+
+- Return 0 if there is no such transformation sequence.
+- All words have the same length.
+- All words contain only lowercase alphabetic characters.
+- You may assume no duplicates in the word list.
+- You may assume *beginWord* and *endWord* are non-empty and are not the same.
+
+**Example 1:**
+
+    Input:
+    beginWord = "hit",
+    endWord = "cog",
+    wordList = ["hot","dot","dog","lot","log","cog"]
+    
+    Output: 5
+    
+    Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
+    return its length 5.
+
+**Example 2:**
+
+    Input:
+    beginWord = "hit"
+    endWord = "cog"
+    wordList = ["hot","dot","dog","lot","log"]
+    
+    Output: 0
+    
+    Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
+
+
+## 题目大意
+
+给定两个单词（beginWord 和 endWord）和一个字典，找到从 beginWord 到 endWord 的最短转换序列的长度。转换需遵循如下规则：
+
+1. 每次转换只能改变一个字母。
+2. 转换过程中的中间单词必须是字典中的单词。
+
+说明:
+
+- 如果不存在这样的转换序列，返回 0。
+- 所有单词具有相同的长度。
+- 所有单词只由小写字母组成。
+- 字典中不存在重复的单词。
+- 你可以假设 beginWord 和 endWord 是非空的，且二者不相同。
+
+
+## 解题思路
+
+- 这一题要求输出从 `beginWord` 变换到 `endWord` 最短变换次数。可以用 BFS，从 `beginWord` 开始变换，把该单词的每个字母都用 `'a'~'z'` 变换一次，生成的数组到 `wordList` 中查找，这里用 Map 来记录查找。找得到就入队列，找不到就输出 0 。入队以后按照 BFS 的算法依次遍历完，当所有单词都 `len(queue)<=0` 出队以后，整个程序结束。
+- 这一题题目中虽然说了要求找到一条最短的路径，但是实际上最短的路径的寻找方法已经告诉你了：
+	1. 每次只变换一个字母
+	2. 每次变换都必须在 `wordList` 中  
+所以不需要单独考虑何种方式是最短的。 
diff --git a/Algorithms/0128. Longest Consecutive Sequence/128. Longest Consecutive Sequence.go b/leetcode/0128.Longest-Consecutive-Sequence/128. Longest Consecutive Sequence.go
similarity index 96%
rename from Algorithms/0128. Longest Consecutive Sequence/128. Longest Consecutive Sequence.go
rename to leetcode/0128.Longest-Consecutive-Sequence/128. Longest Consecutive Sequence.go
index 34e39771f..a43dc96d6 100644
--- a/Algorithms/0128. Longest Consecutive Sequence/128. Longest Consecutive Sequence.go	
+++ b/leetcode/0128.Longest-Consecutive-Sequence/128. Longest Consecutive Sequence.go	
@@ -36,6 +36,13 @@ func longestConsecutive(nums []int) int {
 	return res
 }
 
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
 // 解法二 并查集
 func longestConsecutive1(nums []int) int {
 	if len(nums) == 0 {
diff --git a/Algorithms/0128. Longest Consecutive Sequence/128. Longest Consecutive Sequence_test.go b/leetcode/0128.Longest-Consecutive-Sequence/128. Longest Consecutive Sequence_test.go
similarity index 92%
rename from Algorithms/0128. Longest Consecutive Sequence/128. Longest Consecutive Sequence_test.go
rename to leetcode/0128.Longest-Consecutive-Sequence/128. Longest Consecutive Sequence_test.go
index 0b7692414..1e2ee31db 100644
--- a/Algorithms/0128. Longest Consecutive Sequence/128. Longest Consecutive Sequence_test.go	
+++ b/leetcode/0128.Longest-Consecutive-Sequence/128. Longest Consecutive Sequence_test.go	
@@ -26,27 +26,27 @@ func Test_Problem128(t *testing.T) {
 
 	qs := []question128{
 
-		question128{
+		{
 			para128{[]int{}},
 			ans128{0},
 		},
 
-		question128{
+		{
 			para128{[]int{0}},
 			ans128{1},
 		},
 
-		question128{
+		{
 			para128{[]int{9, 1, 4, 7, 3, -1, 0, 5, 8, -1, 6}},
 			ans128{7},
 		},
 
-		question128{
+		{
 			para128{[]int{2147483646, -2147483647, 0, 2, 2147483644, -2147483645, 2147483645}},
 			ans128{3},
 		},
 
-		question128{
+		{
 			para128{[]int{100, 4, 200, 1, 3, 2}},
 			ans128{4},
 		},
diff --git a/leetcode/0128.Longest-Consecutive-Sequence/README.md b/leetcode/0128.Longest-Consecutive-Sequence/README.md
new file mode 100755
index 000000000..82c8bacea
--- /dev/null
+++ b/leetcode/0128.Longest-Consecutive-Sequence/README.md
@@ -0,0 +1,31 @@
+# [128. Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/)
+
+
+## 题目
+
+Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
+
+Your algorithm should run in O(*n*) complexity.
+
+**Example:**
+
+    Input: [100, 4, 200, 1, 3, 2]
+    Output: 4
+    Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.
+
+
+## 题目大意
+
+
+给定一个未排序的整数数组，找出最长连续序列的长度。要求算法的时间复杂度为 O(n)。
+
+
+
+
+## 解题思路
+
+
+- 给出一个数组，要求找出最长连续序列，输出这个最长的长度。要求时间复杂度为 `O(n)`。
+- 这一题可以先用暴力解决解决，代码见解法三。思路是把每个数都存在 `map` 中，先删去 `map` 中没有前一个数 `nums[i]-1` 也没有后一个数 `nums[i]+1` 的数 `nums[i]`，这种数前后都不连续。然后在 `map` 中找到前一个数 `nums[i]-1` 不存在，但是后一个数 `nums[i]+1` 存在的数，这种数是连续序列的起点，那么不断的往后搜，直到序列“断”了。最后输出最长序列的长度。
+- 这一题最优的解法是解法一，针对每一个 `map` 中不存在的数 `n`，插入进去都做 2 件事情。第一件事，先查看 `n - 1` 和 `n + 1` 是否都存在于 `map` 中，如果都存在，代表存在连续的序列，那么就更新 `left`，`right` 边界。那么 `n` 对应的这个小的子连续序列长度为 `sum = left + right + 1`。第二件事就是更新 `left` 和 `right` 左右边界对应的 `length = sum`。
+- 这一题还可以用并查集解决，见解法二。利用每个数在 `nums` 中的下标，把下标和下标进行 `union()`，具体做法是看前一个数 `nums[i]-1` 和后一个数 `nums[i]+1` 在 `map` 中是否存在，如果存在就 `union()`，最终输出整个并查集中包含最多元素的那个集合的元素总数。
\ No newline at end of file
diff --git a/leetcode/0129.Sum-Root-to-Leaf-Numbers/129. Sum Root to Leaf Numbers.go b/leetcode/0129.Sum-Root-to-Leaf-Numbers/129. Sum Root to Leaf Numbers.go
new file mode 100644
index 000000000..fefdd64b9
--- /dev/null
+++ b/leetcode/0129.Sum-Root-to-Leaf-Numbers/129. Sum Root to Leaf Numbers.go	
@@ -0,0 +1,36 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func sumNumbers(root *TreeNode) int {
+	res := 0
+	dfs(root, 0, &res)
+	return res
+}
+
+func dfs(root *TreeNode, sum int, res *int) {
+	if root == nil {
+		return
+	}
+	sum = sum*10 + root.Val
+	if root.Left == nil && root.Right == nil {
+		*res += sum
+		return
+	}
+	dfs(root.Left, sum, res)
+	dfs(root.Right, sum, res)
+}
diff --git a/Algorithms/0129. Sum Root to Leaf Numbers/129. Sum Root to Leaf Numbers_test.go b/leetcode/0129.Sum-Root-to-Leaf-Numbers/129. Sum Root to Leaf Numbers_test.go
similarity index 87%
rename from Algorithms/0129. Sum Root to Leaf Numbers/129. Sum Root to Leaf Numbers_test.go
rename to leetcode/0129.Sum-Root-to-Leaf-Numbers/129. Sum Root to Leaf Numbers_test.go
index 9d68c0f64..0b50a6620 100644
--- a/Algorithms/0129. Sum Root to Leaf Numbers/129. Sum Root to Leaf Numbers_test.go	
+++ b/leetcode/0129.Sum-Root-to-Leaf-Numbers/129. Sum Root to Leaf Numbers_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question129 struct {
@@ -26,17 +28,17 @@ func Test_Problem129(t *testing.T) {
 
 	qs := []question129{
 
-		question129{
+		{
 			para129{[]int{}},
 			ans129{0},
 		},
 
-		question129{
+		{
 			para129{[]int{1, 2, 3}},
 			ans129{25},
 		},
 
-		question129{
+		{
 			para129{[]int{4, 9, 0, 5, 1}},
 			ans129{1026},
 		},
@@ -47,7 +49,7 @@ func Test_Problem129(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans129, q.para129
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", sumNumbers(root))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/leetcode/0129.Sum-Root-to-Leaf-Numbers/README.md b/leetcode/0129.Sum-Root-to-Leaf-Numbers/README.md
new file mode 100755
index 000000000..7e7dababa
--- /dev/null
+++ b/leetcode/0129.Sum-Root-to-Leaf-Numbers/README.md
@@ -0,0 +1,48 @@
+# [129. Sum Root to Leaf Numbers](https://leetcode.com/problems/sum-root-to-leaf-numbers/)
+
+
+## 题目
+
+Given a binary tree containing digits from `0-9` only, each root-to-leaf path could represent a number.
+
+An example is the root-to-leaf path `1->2->3` which represents the number `123`.
+
+Find the total sum of all root-to-leaf numbers.
+
+**Note:** A leaf is a node with no children.
+
+**Example:**
+
+    Input: [1,2,3]
+        1
+       / \
+      2   3
+    Output: 25
+    Explanation:
+    The root-to-leaf path 1->2 represents the number 12.
+    The root-to-leaf path 1->3 represents the number 13.
+    Therefore, sum = 12 + 13 = 25.
+
+**Example 2:**
+
+    Input: [4,9,0,5,1]
+        4
+       / \
+      9   0
+     / \
+    5   1
+    Output: 1026
+    Explanation:
+    The root-to-leaf path 4->9->5 represents the number 495.
+    The root-to-leaf path 4->9->1 represents the number 491.
+    The root-to-leaf path 4->0 represents the number 40.
+    Therefore, sum = 495 + 491 + 40 = 1026.
+
+## 题目大意
+
+给定一个二叉树，它的每个结点都存放一个 0-9 的数字，每条从根到叶子节点的路径都代表一个数字。例如，从根到叶子节点路径 1->2->3 代表数字 123。计算从根到叶子节点生成的所有数字之和。说明: 叶子节点是指没有子节点的节点。
+
+
+## 解题思路
+
+- 运用前序遍历的思想，当从根节点出发一直加到叶子节点，每个叶子节点汇总一次。
\ No newline at end of file
diff --git a/Algorithms/0130. Surrounded Regions/130. Surrounded Regions.go b/leetcode/0130.Surrounded-Regions/130. Surrounded Regions.go
similarity index 96%
rename from Algorithms/0130. Surrounded Regions/130. Surrounded Regions.go
rename to leetcode/0130.Surrounded-Regions/130. Surrounded Regions.go
index 8621a0aee..9d904e36e 100644
--- a/Algorithms/0130. Surrounded Regions/130. Surrounded Regions.go	
+++ b/leetcode/0130.Surrounded-Regions/130. Surrounded Regions.go	
@@ -4,6 +4,13 @@ import (
 	"github.com/halfrost/LeetCode-Go/template"
 )
 
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
 // 解法一 并查集
 func solve(board [][]byte) {
 	if len(board) == 0 {
diff --git a/leetcode/0130.Surrounded-Regions/130. Surrounded Regions_test.go b/leetcode/0130.Surrounded-Regions/130. Surrounded Regions_test.go
new file mode 100644
index 000000000..1dec2cedc
--- /dev/null
+++ b/leetcode/0130.Surrounded-Regions/130. Surrounded Regions_test.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question130 struct {
+	para130
+	ans130
+}
+
+// para 是参数
+// one 代表第一个参数
+type para130 struct {
+	one [][]byte
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans130 struct {
+	one [][]byte
+}
+
+func Test_Problem130(t *testing.T) {
+
+	qs := []question130{
+
+		{
+			para130{[][]byte{}},
+			ans130{[][]byte{}},
+		},
+
+		{
+			para130{[][]byte{{'X', 'X', 'X', 'X'}, {'X', 'O', 'O', 'X'}, {'X', 'X', 'O', 'X'}, {'X', 'O', 'X', 'X'}}},
+			ans130{[][]byte{{'X', 'X', 'X', 'X'}, {'X', 'X', 'X', 'X'}, {'X', 'X', 'X', 'X'}, {'X', 'O', 'X', 'X'}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 130------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans130, q.para130
+		fmt.Printf("【input】:%v      ", p)
+		solve1(p.one)
+		fmt.Printf("【output】:%v      \n", p)
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0130.Surrounded-Regions/README.md b/leetcode/0130.Surrounded-Regions/README.md
new file mode 100755
index 000000000..2c161a8ea
--- /dev/null
+++ b/leetcode/0130.Surrounded-Regions/README.md
@@ -0,0 +1,39 @@
+# [130. Surrounded Regions](https://leetcode.com/problems/surrounded-regions/)
+
+
+
+## 题目
+
+Given a 2D board containing `'X'` and `'O'` (**the letter O**), capture all regions surrounded by `'X'`.
+
+A region is captured by flipping all `'O'`s into `'X'`s in that surrounded region.
+
+**Example:**
+
+    X X X X
+    X O O X
+    X X O X
+    X O X X
+
+After running your function, the board should be:
+
+    X X X X
+    X X X X
+    X X X X
+    X O X X
+
+**Explanation:**
+
+Surrounded regions shouldn’t be on the border, which means that any `'O'` on the border of the board are not flipped to `'X'`. Any `'O'` that is not on the border and it is not connected to an `'O'` on the border will be flipped to `'X'`. Two cells are connected if they are adjacent cells connected horizontally or vertically.
+
+## 题目大意
+
+给定一个二维的矩阵，包含 'X' 和 'O'（字母 O）。找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O' 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
+
+
+## 解题思路
+
+
+- 给出一张二维地图，要求把地图上非边缘上的 'O' 都用 'X' 覆盖掉。
+- 这一题有多种解法。第一种解法是并查集。先将边缘上的 'O' 全部都和一个特殊的点进行 `union()` 。然后再把地图中间的 'O' 都进行 `union()`，最后把和特殊点不是同一个集合的点都标记成 'X'。第二种解法是 DFS 或者 BFS，可以先将边缘上的 'O' 先标记成另外一个字符，然后在递归遍历过程中，把剩下的 'O' 都标记成 'X'。
+
diff --git a/Algorithms/0131. Palindrome Partitioning/131. Palindrome Partitioning.go b/leetcode/0131.Palindrome-Partitioning/131. Palindrome Partitioning.go
similarity index 100%
rename from Algorithms/0131. Palindrome Partitioning/131. Palindrome Partitioning.go
rename to leetcode/0131.Palindrome-Partitioning/131. Palindrome Partitioning.go
diff --git a/leetcode/0131.Palindrome-Partitioning/131. Palindrome Partitioning_test.go b/leetcode/0131.Palindrome-Partitioning/131. Palindrome Partitioning_test.go
new file mode 100644
index 000000000..7008f7e2b
--- /dev/null
+++ b/leetcode/0131.Palindrome-Partitioning/131. Palindrome Partitioning_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question131 struct {
+	para131
+	ans131
+}
+
+// para 是参数
+// one 代表第一个参数
+type para131 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans131 struct {
+	one [][]string
+}
+
+func Test_Problem131(t *testing.T) {
+
+	qs := []question131{
+
+		{
+			para131{"aab"},
+			ans131{[][]string{{"aa", "b"}, {"a", "a", "b"}}},
+		},
+
+		{
+			para131{"bb"},
+			ans131{[][]string{{"b", "b"}, {"bb"}}},
+		},
+
+		{
+			para131{"efe"},
+			ans131{[][]string{{"e", "f", "e"}, {"efe"}}},
+		},
+
+		{
+			para131{"abbab"},
+			ans131{[][]string{{"a", "b", "b", "a", "b"}, {"a", "b", "bab"}, {"a", "bb", "a", "b"}, {"abba", "b"}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 131------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans131, q.para131
+		fmt.Printf("【input】:%v       【output】:%v\n", p, partition131(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0131. Palindrome Partitioning/README.md b/leetcode/0131.Palindrome-Partitioning/README.md
similarity index 100%
rename from Algorithms/0131. Palindrome Partitioning/README.md
rename to leetcode/0131.Palindrome-Partitioning/README.md
diff --git a/leetcode/0134.Gas-Station/Gas.go b/leetcode/0134.Gas-Station/Gas.go
new file mode 100644
index 000000000..4277a791c
--- /dev/null
+++ b/leetcode/0134.Gas-Station/Gas.go
@@ -0,0 +1,26 @@
+package leetcode
+
+func canCompleteCircuit(gas []int, cost []int) int {
+	totalGas := 0
+	totalCost := 0
+	currGas := 0
+	start := 0
+
+	for i := 0; i < len(gas); i++ {
+		totalGas += gas[i]
+		totalCost += cost[i]
+		currGas += gas[i] - cost[i]
+
+		if currGas < 0 {
+			start = i + 1
+			currGas = 0
+		}
+	}
+
+	if totalGas < totalCost {
+		return -1
+	}
+
+	return start
+
+}
diff --git a/leetcode/0134.Gas-Station/Gas_test.go b/leetcode/0134.Gas-Station/Gas_test.go
new file mode 100644
index 000000000..f6c2a7893
--- /dev/null
+++ b/leetcode/0134.Gas-Station/Gas_test.go
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question134 struct {
+	para134
+	ans134
+}
+
+// para 是参数
+// one 代表第一个参数
+type para134 struct {
+	one []int
+	two []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans134 struct {
+	one int
+}
+
+func Test_Problem134(t *testing.T) {
+	qs := []question134{
+
+		{
+			para134{[]int{1, 2, 3, 4, 5}, []int{3, 4, 5, 1, 2}},
+			ans134{3},
+		},
+
+		{
+			para134{[]int{2, 3, 4}, []int{3, 4, 3}},
+			ans134{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 134------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans134, q.para134
+		fmt.Printf("【input】:%v， %v       【output】:%v\n", p.one, p.two, canCompleteCircuit(p.one, p.two))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0134.Gas-Station/README.md b/leetcode/0134.Gas-Station/README.md
new file mode 100644
index 000000000..e4cb07821
--- /dev/null
+++ b/leetcode/0134.Gas-Station/README.md
@@ -0,0 +1,88 @@
+# [134. Gas Stations](https://leetcode.com/problems/gas-station/)
+
+
+## 题目
+
+You are given a list of gas stations on a circular route, where the amount of gas at station i is gas[i], and the cost to travel from station i to its next station (i+1) is cost[i]. You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.
+
+Write a function that returns the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique.
+
+**Example 1:**
+
+```
+Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
+Output: 3
+Explanation: Start at station 3 (index 3), you have 4 liters of gas. Now you have a total of =0+4=4 liters of gas.
+Travel to station 4. Now you have 4 - 1 + 5 = 8 liters of gas.
+Travel to station 0. Now you have 8 - 2 + 1 = 7 liters of gas.
+Travel to station 1. Now you have 7 - 3 + 2 = 6 liters of gas.
+Travel to station 2. Now you have 6 - 4 + 3 = 5 liters of gas.
+Travel to station 3 again. This time you will have consumed 5 liters of gas, which is exactly enough for you to return to station 3.
+Therefore, 3 can be the starting index.
+```
+
+**Example 2:**
+
+```
+Input: gas = [2,3,4], cost = [3,4,3]
+Output: -1
+Explanation: You cannot start at station 0 or 1 because there is not enough gas to get you to the next station.
+We start at station 2, we get 4 liters of gas. Now you have a total of =0+4=4 liters of gas.
+Travel to station 0. Now you have 4 - 3 + 2 = 3 liters of gas.
+Travel to station 1. Now you have 3 - 3 + 3 = 3 liters of gas.
+You cannot return to station 2 because it requires 4 liters of gas but you only have 3 liters in your tank.
+Therefore, no matter what, you cannot travel around the ring.
+```
+
+**Constraints:**
+
+- `gas.length == n`
+- `cost.length == n`
+- `1 <= n <= 10^5`
+- `0 <= gas[i], cost[i] <= 10^4`
+
+## 题目大意
+
+在一条环路上有 n 个加油站，其中第 i 个加油站有汽油 gas[i] 升。
+
+你有一辆油箱容量无限的的汽车，从第 i 个加油站开往第 i+1 个加油站需要消耗汽油 cost[i] 升。你从其中的一个加油站出发，开始时油箱为空。
+
+给定两个整数数组 gas 和 cost ，如果你可以按顺序绕环路行驶一周，则返回出发时加油站的编号，否则返回 -1 。如果存在解，则 保证 它是 唯一 的。
+
+## 解题思路
+
+- 初始化总油量（totalGas）和总消耗（totalCost）为0，起始点（start）为0，当前油量（currGas）为0。
+  遍历每个加油站，累加油量和消耗，计算当前油量。
+  如果当前油量小于0，说明从上一个加油站到当前加油站的油量不足以到达下一个加油站，因此将起始点设置为当前加油站的下一个位置，并将当前油量重置为0。
+  遍历结束后，如果总油量小于总消耗，说明无法完成整个旅程，返回-1；否则返回起始点。
+
+## 代码
+
+```go
+package leetcode
+
+func canCompleteCircuit(gas []int, cost []int) int {
+	totalGas := 0
+	totalCost := 0
+	currGas := 0
+	start := 0
+
+	for i := 0; i < len(gas); i++ {
+		totalGas += gas[i]
+		totalCost += cost[i]
+		currGas += gas[i] - cost[i]
+
+		if currGas < 0 {
+			start = i + 1
+			currGas = 0
+		}
+	}
+
+	if totalGas < totalCost {
+		return -1
+	}
+
+	return start
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/0135.Candy/135. Candy.go b/leetcode/0135.Candy/135. Candy.go
new file mode 100644
index 000000000..0cf961e0b
--- /dev/null
+++ b/leetcode/0135.Candy/135. Candy.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+func candy(ratings []int) int {
+	candies := make([]int, len(ratings))
+	for i := 1; i < len(ratings); i++ {
+		if ratings[i] > ratings[i-1] {
+			candies[i] += candies[i-1] + 1
+		}
+	}
+	for i := len(ratings) - 2; i >= 0; i-- {
+		if ratings[i] > ratings[i+1] && candies[i] <= candies[i+1] {
+			candies[i] = candies[i+1] + 1
+		}
+	}
+	total := 0
+	for _, candy := range candies {
+		total += candy + 1
+	}
+	return total
+}
diff --git a/leetcode/0135.Candy/135. Candy_test.go b/leetcode/0135.Candy/135. Candy_test.go
new file mode 100644
index 000000000..edb15ad35
--- /dev/null
+++ b/leetcode/0135.Candy/135. Candy_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question135 struct {
+	para135
+	ans135
+}
+
+// para 是参数
+// one 代表第一个参数
+type para135 struct {
+	ratings []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans135 struct {
+	one int
+}
+
+func Test_Problem135(t *testing.T) {
+
+	qs := []question135{
+
+		{
+			para135{[]int{1, 0, 2}},
+			ans135{5},
+		},
+
+		{
+			para135{[]int{1, 2, 2}},
+			ans135{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 135------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans135, q.para135
+		fmt.Printf("【input】:%v       【output】:%v\n", p, candy(p.ratings))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0135.Candy/README.md b/leetcode/0135.Candy/README.md
new file mode 100644
index 000000000..9f6dc16ad
--- /dev/null
+++ b/leetcode/0135.Candy/README.md
@@ -0,0 +1,74 @@
+# [135. Candy](https://leetcode.com/problems/candy/)
+
+
+## 题目
+
+There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
+
+You are giving candies to these children subjected to the following requirements:
+
+- Each child must have at least one candy.
+- Children with a higher rating get more candies than their neighbors.
+
+Return *the minimum number of candies you need to have to distribute the candies to the children*.
+
+**Example 1:**
+
+```
+Input: ratings = [1,0,2]
+Output: 5
+Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
+```
+
+**Example 2:**
+
+```
+Input: ratings = [1,2,2]
+Output: 4
+Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
+The third child gets 1 candy because it satisfies the above two conditions.
+```
+
+**Constraints:**
+
+- `n == ratings.length`
+- `1 <= n <= 2 * 10^4`
+- `0 <= ratings[i] <= 2 * 10^4`
+
+## 题目大意
+
+老师想给孩子们分发糖果，有 N 个孩子站成了一条直线，老师会根据每个孩子的表现，预先给他们评分。你需要按照以下要求，帮助老师给这些孩子分发糖果：
+
+- 每个孩子至少分配到 1 个糖果。
+- 评分更高的孩子必须比他两侧的邻位孩子获得更多的糖果。
+
+那么这样下来，老师至少需要准备多少颗糖果呢？
+
+## 解题思路
+
+- 本题的突破口在于，评分更高的孩子必须比他两侧的邻位孩子获得更多的糖果，这句话。这个规则可以理解为 2 条规则，想象成按身高排队，站在下标为 0 的地方往后“看”，评分高即为个子高的，应该比前面个子矮(评分低)的分到糖果多；站在下标为 n - 1 的地方往后“看”，评分高即为个子高的，同样应该比前面个子矮(评分低)的分到糖果多。你可能会有疑问，规则都是一样的，为什么会出现至少需要多少糖果呢？因为可能出现评分一样高的同学。扫描数组两次，处理出每一个学生分别满足左规则或右规则时，最少需要被分得的糖果数量。每个人最终分得的糖果数量即为这两个数量的最大值。两次遍历结束，将所有糖果累加起来即为至少需要准备的糖果数。由于每个人至少分配到 1 个糖果，所以每个人糖果数再加一。
+
+## 代码
+
+```go
+package leetcode
+
+func candy(ratings []int) int {
+	candies := make([]int, len(ratings))
+	for i := 1; i < len(ratings); i++ {
+		if ratings[i] > ratings[i-1] {
+			candies[i] += candies[i-1] + 1
+		}
+	}
+	for i := len(ratings) - 2; i >= 0; i-- {
+		if ratings[i] > ratings[i+1] && candies[i] <= candies[i+1] {
+			candies[i] = candies[i+1] + 1
+		}
+	}
+	total := 0
+	for _, candy := range candies {
+		total += candy + 1
+	}
+	return total
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0136. Single Number/136. Single Number.go b/leetcode/0136.Single-Number/136. Single Number.go
similarity index 100%
rename from Algorithms/0136. Single Number/136. Single Number.go
rename to leetcode/0136.Single-Number/136. Single Number.go
diff --git a/Algorithms/0136. Single Number/136. Single Number_test.go b/leetcode/0136.Single-Number/136. Single Number_test.go
similarity index 95%
rename from Algorithms/0136. Single Number/136. Single Number_test.go
rename to leetcode/0136.Single-Number/136. Single Number_test.go
index fb2be390c..2360a740a 100644
--- a/Algorithms/0136. Single Number/136. Single Number_test.go	
+++ b/leetcode/0136.Single-Number/136. Single Number_test.go	
@@ -26,12 +26,12 @@ func Test_Problem136(t *testing.T) {
 
 	qs := []question136{
 
-		question136{
+		{
 			para136{[]int{2, 2, 1}},
 			ans136{1},
 		},
 
-		question136{
+		{
 			para136{[]int{4, 1, 2, 1, 2}},
 			ans136{4},
 		},
diff --git a/leetcode/0136.Single-Number/README.md b/leetcode/0136.Single-Number/README.md
new file mode 100755
index 000000000..fdf307eac
--- /dev/null
+++ b/leetcode/0136.Single-Number/README.md
@@ -0,0 +1,29 @@
+# [136. Single Number](https://leetcode.com/problems/single-number/)
+
+## 题目
+
+Given a **non-empty** array of integers, every element appears *twice* except for one. Find that single one.
+
+**Note:**
+
+Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
+
+**Example 1:**
+
+    Input: [2,2,1]
+    Output: 1
+
+**Example 2:**
+
+    Input: [4,1,2,1,2]
+    Output: 4
+
+## 题目大意
+
+给定一个非空整数数组，除了某个元素只出现一次以外，其余每个元素均出现两次。找出那个只出现了一次的元素。要求算法时间复杂度是线性的，并且不使用额外的辅助空间。
+
+
+## 解题思路
+
+- 题目要求不能使用辅助空间，并且时间复杂度只能是线性的。
+- 题目为什么要强调有一个数字出现一次，其他的出现两次？我们想到了异或运算的性质：任何一个数字异或它自己都等于0。也就是说，如果我们从头到尾依次异或数组中的每一个数字，那么最终的结果刚好是那个只出现一次的数字，因为那些出现两次的数字全部在异或中抵消掉了。于是最终做法是从头到尾依次异或数组中的每一个数字，那么最终得到的结果就是两个只出现一次的数字的异或结果。因为其他数字都出现了两次，在异或中全部抵消掉了。**利用的性质是 x^x = 0**。
diff --git a/Algorithms/0137. Single Number II/137. Single Number II.go b/leetcode/0137.Single-Number-II/137. Single Number II.go
similarity index 100%
rename from Algorithms/0137. Single Number II/137. Single Number II.go
rename to leetcode/0137.Single-Number-II/137. Single Number II.go
diff --git a/Algorithms/0137. Single Number II/137. Single Number II_test.go b/leetcode/0137.Single-Number-II/137. Single Number II_test.go
similarity index 95%
rename from Algorithms/0137. Single Number II/137. Single Number II_test.go
rename to leetcode/0137.Single-Number-II/137. Single Number II_test.go
index f637f0bbd..55c618034 100644
--- a/Algorithms/0137. Single Number II/137. Single Number II_test.go	
+++ b/leetcode/0137.Single-Number-II/137. Single Number II_test.go	
@@ -26,12 +26,12 @@ func Test_Problem137(t *testing.T) {
 
 	qs := []question137{
 
-		question137{
+		{
 			para137{[]int{2, 2, 3, 2}},
 			ans137{3},
 		},
 
-		question137{
+		{
 			para137{[]int{0, 1, 0, 1, 0, 1, 99}},
 			ans137{99},
 		},
diff --git a/leetcode/0137.Single-Number-II/README.md b/leetcode/0137.Single-Number-II/README.md
new file mode 100755
index 000000000..9733ff3c3
--- /dev/null
+++ b/leetcode/0137.Single-Number-II/README.md
@@ -0,0 +1,91 @@
+# [137. Single Number II](https://leetcode.com/problems/single-number-ii/)
+
+
+## 题目
+
+Given a **non-empty** array of integers, every element appears *three* times except for one, which appears exactly once. Find that single one.
+
+**Note:**
+
+Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
+
+**Example 1:**
+
+    Input: [2,2,3,2]
+    Output: 3
+
+**Example 2:**
+
+    Input: [0,1,0,1,0,1,99]
+    Output: 99
+
+
+## 题目大意
+
+给定一个非空整数数组，除了某个元素只出现一次以外，其余每个元素均出现了三次。找出那个只出现了一次的元素。要求算法时间复杂度是线性的，并且不使用额外的辅助空间。
+
+
+
+
+## 解题思路
+
+- 这一题是第 136 题的加强版。这类题也可以扩展，在数组中每个元素都出现 5 次，找出只出现 1 次的数。
+- 本题中要求找出只出现 1 次的数，出现 3 次的数都要被消除。第 136 题是消除出现 2 次的数。这一题也会相当相同的解法，出现 3 次的数也要被消除。定义状态，00、10、01，这 3 个状态。当一个数出现 3 次，那么它每个位置上的 1 出现的次数肯定是 3 的倍数，所以当 1 出现 3 次以后，就归零清除。如何能做到这点呢？仿造`三进制(00，01，10)` 就可以做到。
+- 变量  ones 中记录遍历中每个位上出现 1 的个数。将它与 A[i] 进行异或，目的是：
+    - 每位上两者都是 1 的，表示历史统计结果 ones 出现1次、A[i]中又出现 1 次，则是出现 2 次，需要进位到 twos 变量中。
+    - 每位上两者分别为 0、1 的，加入到 ones 统计结果中。
+    - 最后还要 & ^twos ，是为了能做到三进制，出现 3 次就清零。例如 ones = x，那么 twos = 0，当 twos = x，那么 ones = 0；
+- 变量 twos 中记录遍历中每个位上出现 1 ，2次 的个数。与 A[i] 进行异或的目的和上述描述相同，不再赘述。
+
+> 在 golang 中，&^ 表示 AND NOT 的意思。这里的 ^ 作为一元操作符，表示按位取反 (^0001 0100 = 1110 1011)，X &^ Y 的意思是将 X 中与 Y 相异的位保留，相同的位清零。
+
+> 在 golang 中没有 Java 中的 ~ 位操作运算符，Java 中的 ~ 运算符代表按位取反。这个操作就想当于 golang 中的 ^ 运算符当做一元运算符使用的效果。
+
+|(twos,ones)|xi|(twos'',ones')|ones'|
+|:----:|:----:|:----:|:----:|
+|00|0|00|0|
+|00|1|01|1|
+|01|0|01|1|
+|01|1|10|0|
+|10|0|10|0|
+|10|1|00|0|
+
+- 第一步，先将 ones -> ones'。通过观察可以看出 ones = (ones ^ nums[i]) & ^twos
+
+|(twos,ones')|xi|twos'|
+|:----:|:----:|:----:|
+|00|0|0|
+|01|1|0|
+|01|0|0|
+|00|1|1|
+|10|0|1|
+|10|1|0|
+
+
+- 第二步，再将 twos -> twos'。这一步需要用到前一步的 ones。通过观察可以看出 twos = (twos ^ nums[i]) & ^ones。
+
+--------------------------
+
+这一题还可以继续扩展，在数组中每个元素都出现 5 次，找出只出现 1 次的数。那该怎么做呢？思路还是一样的，模拟一个五进制，5 次就会消除。代码如下：
+
+    // 解法一
+    func singleNumberIII(nums []int) int {
+    	na, nb, nc := 0, 0, 0
+    	for i := 0; i < len(nums); i++ {
+    		nb = nb ^ (nums[i] & na)
+    		na = (na ^ nums[i]) & ^nc
+    		nc = nc ^ (nums[i] & ^na & ^nb)
+    	}
+    	return na & ^nb & ^nc
+    }
+    
+    // 解法二
+    func singleNumberIIII(nums []int) int {
+    	twos, threes, ones := 0xffffffff, 0xffffffff, 0
+    	for i := 0; i < len(nums); i++ {
+    		threes = threes ^ (nums[i] & twos)
+    		twos = (twos ^ nums[i]) & ^ones
+    		ones = ones ^ (nums[i] & ^twos & ^threes)
+    	}
+    	return ones
+    }
diff --git a/leetcode/0138.Copy-List-With-Random-Pointer/138. Copy List With Random Pointer.go b/leetcode/0138.Copy-List-With-Random-Pointer/138. Copy List With Random Pointer.go
new file mode 100644
index 000000000..357740638
--- /dev/null
+++ b/leetcode/0138.Copy-List-With-Random-Pointer/138. Copy List With Random Pointer.go	
@@ -0,0 +1,44 @@
+package leetcode
+
+// Node define
+type Node struct {
+	Val    int
+	Next   *Node
+	Random *Node
+}
+
+func copyRandomList(head *Node) *Node {
+	if head == nil {
+		return nil
+	}
+	tempHead := copyNodeToLinkedList(head)
+	return splitLinkedList(tempHead)
+}
+
+func splitLinkedList(head *Node) *Node {
+	cur := head
+	head = head.Next
+	for cur != nil && cur.Next != nil {
+		cur.Next, cur = cur.Next.Next, cur.Next
+	}
+	return head
+}
+
+func copyNodeToLinkedList(head *Node) *Node {
+	cur := head
+	for cur != nil {
+		node := &Node{
+			Val:  cur.Val,
+			Next: cur.Next,
+		}
+		cur.Next, cur = node, cur.Next
+	}
+	cur = head
+	for cur != nil {
+		if cur.Random != nil {
+			cur.Next.Random = cur.Random.Next
+		}
+		cur = cur.Next.Next
+	}
+	return head
+}
diff --git a/leetcode/0138.Copy-List-With-Random-Pointer/138. Copy List With Random Pointer_test.go b/leetcode/0138.Copy-List-With-Random-Pointer/138. Copy List With Random Pointer_test.go
new file mode 100644
index 000000000..2bd540920
--- /dev/null
+++ b/leetcode/0138.Copy-List-With-Random-Pointer/138. Copy List With Random Pointer_test.go	
@@ -0,0 +1,115 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question138 struct {
+	para138
+	ans138
+}
+
+// para 是参数
+// one 代表第一个参数
+type para138 struct {
+	head [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans138 struct {
+	one [][]int
+}
+
+func Test_Problem138(t *testing.T) {
+
+	qs := []question138{
+
+		{
+			para138{[][]int{{7, structures.NULL}, {13, 0}, {11, 4}, {10, 2}, {1, 0}}},
+			ans138{[][]int{{7, structures.NULL}, {13, 0}, {11, 4}, {10, 2}, {1, 0}}},
+		},
+
+		{
+			para138{[][]int{{1, 1}, {2, 1}}},
+			ans138{[][]int{{1, 1}, {2, 1}}},
+		},
+
+		{
+			para138{[][]int{{3, structures.NULL}, {3, 0}, {3, structures.NULL}}},
+			ans138{[][]int{{3, structures.NULL}, {3, 0}, {3, structures.NULL}}},
+		},
+
+		{
+			para138{[][]int{}},
+			ans138{[][]int{}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 138------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans138, q.para138
+		fmt.Printf("【input】:%v       【output】:%v\n", p, node2Ints(copyRandomList(ints2Node(p.head))))
+	}
+	fmt.Printf("\n\n\n")
+}
+
+func ints2Node(one [][]int) *Node {
+	nodesMap := map[int]*Node{}
+	for index, nodes := range one {
+		tmp := &Node{Val: nodes[0]}
+		nodesMap[index] = tmp
+	}
+	for index, node := range nodesMap {
+		if index != len(one)-1 {
+			node.Next = nodesMap[index+1]
+		} else {
+			node.Next = nil
+		}
+	}
+	for index, nodes := range one {
+		if nodes[1] == structures.NULL {
+			nodesMap[index].Random = nil
+		} else {
+			nodesMap[index].Random = nodesMap[nodes[1]]
+		}
+	}
+	return nodesMap[0]
+}
+
+func node2Ints(head *Node) (res [][]int) {
+	nodesMap, count, cur := map[*Node]int{}, 0, head
+	if head == nil {
+		return [][]int{}
+	}
+	for cur.Next != nil {
+		nodesMap[cur] = count
+		count++
+		cur = cur.Next
+	}
+	nodesMap[cur] = count
+	tmp := []int{}
+	cur = head
+	for cur.Next != nil {
+		tmp := append(tmp, cur.Val)
+		if cur.Random == nil {
+			tmp = append(tmp, structures.NULL)
+		} else {
+			tmp = append(tmp, nodesMap[cur.Random])
+		}
+		res = append(res, tmp)
+		cur = cur.Next
+	}
+	tmp = append(tmp, cur.Val)
+	if cur.Random == nil {
+		tmp = append(tmp, structures.NULL)
+	} else {
+		tmp = append(tmp, nodesMap[cur.Random])
+	}
+	res = append(res, tmp)
+	return res
+}
diff --git a/leetcode/0138.Copy-List-With-Random-Pointer/README.md b/leetcode/0138.Copy-List-With-Random-Pointer/README.md
new file mode 100644
index 000000000..a4062ec39
--- /dev/null
+++ b/leetcode/0138.Copy-List-With-Random-Pointer/README.md
@@ -0,0 +1,130 @@
+# [138. Copy List with Random Pointer](https://leetcode.com/problems/copy-list-with-random-pointer/)
+
+## 题目
+
+A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.
+
+Return a **[deep copy](https://en.wikipedia.org/wiki/Object_copying#Deep_copy)** of the list.
+
+The Linked List is represented in the input/output as a list of `n` nodes. Each node is represented as a pair of `[val, random_index]` where:
+
+- `val`: an integer representing `Node.val`
+- `random_index`: the index of the node (range from `0` to `n-1`) where random pointer points to, or `null` if it does not point to any node.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2019/12/18/e1.png)
+
+```
+Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
+Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]
+
+```
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2019/12/18/e2.png)
+
+```
+Input: head = [[1,1],[2,1]]
+Output: [[1,1],[2,1]]
+
+```
+
+**Example 3:**
+
+![](https://assets.leetcode.com/uploads/2019/12/18/e3.png)
+
+```
+Input: head = [[3,null],[3,0],[3,null]]
+Output: [[3,null],[3,0],[3,null]]
+
+```
+
+**Example 4:**
+
+```
+Input: head = []
+Output: []
+Explanation: Given linked list is empty (null pointer), so return null.
+
+```
+
+**Constraints:**
+
+- `10000 <= Node.val <= 10000`
+- `Node.random` is null or pointing to a node in the linked list.
+- The number of nodes will not exceed 1000.
+
+## 题目大意
+
+给定一个链表，每个节点包含一个额外增加的随机指针，该指针可以指向链表中的任何节点或空节点。要求返回这个链表的 深拷贝。
+
+我们用一个由 n 个节点组成的链表来表示输入/输出中的链表。每个节点用一个 [val, random_index] 表示：
+
+- val：一个表示 Node.val 的整数。
+- random_index：随机指针指向的节点索引（范围从 0 到 n-1）；如果不指向任何节点，则为  null 。
+
+## 解题思路
+
+- 这道题严格意义上是数据结构题，根据给定的数据结构，对它进行深拷贝。
+- 先将每个节点都复制一份，放在它的 next 节点中。如此穿插的复制一份链表。
+
+    ![https://img.halfrost.com/Leetcode/leetcode_138_1_0.png](https://img.halfrost.com/Leetcode/leetcode_138_1_0.png)
+
+    再将穿插版的链表的 random 指针指向正确的位置。
+
+    ![https://img.halfrost.com/Leetcode/leetcode_138_2.png](https://img.halfrost.com/Leetcode/leetcode_138_2.png)
+
+    再将穿插版的链表的 next 指针指向正确的位置。最后分开这交织在一起的两个链表的头节点，即可分开 2 个链表。
+
+    ![https://img.halfrost.com/Leetcode/leetcode_138_3.png](https://img.halfrost.com/Leetcode/leetcode_138_3.png)
+
+## 代码
+
+```go
+package leetcode
+
+// Node define
+type Node struct {
+	Val    int
+	Next   *Node
+	Random *Node
+}
+
+func copyRandomList(head *Node) *Node {
+	if head == nil {
+		return nil
+	}
+	tempHead := copyNodeToLinkedList(head)
+	return splitLinkedList(tempHead)
+}
+
+func splitLinkedList(head *Node) *Node {
+	cur := head
+	head = head.Next
+	for cur != nil && cur.Next != nil {
+		cur.Next, cur = cur.Next.Next, cur.Next
+	}
+	return head
+}
+
+func copyNodeToLinkedList(head *Node) *Node {
+	cur := head
+	for cur != nil {
+		node := &Node{
+			Val:  cur.Val,
+			Next: cur.Next,
+		}
+		cur.Next, cur = node, cur.Next
+	}
+	cur = head
+	for cur != nil {
+		if cur.Random != nil {
+			cur.Next.Random = cur.Random.Next
+		}
+		cur = cur.Next.Next
+	}
+	return head
+}
+```
\ No newline at end of file
diff --git a/leetcode/0141.Linked-List-Cycle/141. Linked List Cycle.go b/leetcode/0141.Linked-List-Cycle/141. Linked List Cycle.go
new file mode 100644
index 000000000..947c10ce9
--- /dev/null
+++ b/leetcode/0141.Linked-List-Cycle/141. Linked List Cycle.go	
@@ -0,0 +1,29 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func hasCycle(head *ListNode) bool {
+	fast := head
+	slow := head
+	for fast != nil && fast.Next != nil {
+		fast = fast.Next.Next
+		slow = slow.Next
+		if fast == slow {
+			return true
+		}
+	}
+	return false
+}
diff --git a/leetcode/0141.Linked-List-Cycle/141. Linked List Cycle_test.go b/leetcode/0141.Linked-List-Cycle/141. Linked List Cycle_test.go
new file mode 100644
index 000000000..06bf230fd
--- /dev/null
+++ b/leetcode/0141.Linked-List-Cycle/141. Linked List Cycle_test.go	
@@ -0,0 +1,54 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question141 struct {
+	para141
+	ans141
+}
+
+// para 是参数
+// one 代表第一个参数
+type para141 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans141 struct {
+	one bool
+}
+
+func Test_Problem141(t *testing.T) {
+
+	qs := []question141{
+
+		{
+			para141{[]int{3, 2, 0, -4}},
+			ans141{false},
+		},
+
+		{
+			para141{[]int{1, 2}},
+			ans141{false},
+		},
+
+		{
+			para141{[]int{1}},
+			ans141{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 141------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans141, q.para141
+		fmt.Printf("【input】:%v       【output】:%v\n", p, hasCycle(structures.Ints2List(p.one)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0141. Linked List Cycle/README.md b/leetcode/0141.Linked-List-Cycle/README.md
similarity index 100%
rename from Algorithms/0141. Linked List Cycle/README.md
rename to leetcode/0141.Linked-List-Cycle/README.md
diff --git a/Algorithms/0142. Linked List Cycle II/142. Linked List Cycle II.go b/leetcode/0142.Linked-List-Cycle-II/142. Linked List Cycle II.go
similarity index 84%
rename from Algorithms/0142. Linked List Cycle II/142. Linked List Cycle II.go
rename to leetcode/0142.Linked-List-Cycle-II/142. Linked List Cycle II.go
index 9d6130302..341d4a0a8 100644
--- a/Algorithms/0142. Linked List Cycle II/142. Linked List Cycle II.go	
+++ b/leetcode/0142.Linked-List-Cycle-II/142. Linked List Cycle II.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0142. Linked List Cycle II/142. Linked List Cycle II_test.go b/leetcode/0142.Linked-List-Cycle-II/142. Linked List Cycle II_test.go
similarity index 100%
rename from Algorithms/0142. Linked List Cycle II/142. Linked List Cycle II_test.go
rename to leetcode/0142.Linked-List-Cycle-II/142. Linked List Cycle II_test.go
diff --git a/Algorithms/0142. Linked List Cycle II/README.md b/leetcode/0142.Linked-List-Cycle-II/README.md
similarity index 100%
rename from Algorithms/0142. Linked List Cycle II/README.md
rename to leetcode/0142.Linked-List-Cycle-II/README.md
diff --git a/Algorithms/0143. Reorder List/143. Reorder List.go b/leetcode/0143.Reorder-List/143. Reorder List.go
similarity index 92%
rename from Algorithms/0143. Reorder List/143. Reorder List.go
rename to leetcode/0143.Reorder-List/143. Reorder List.go
index d48732c9d..d607a8cab 100644
--- a/Algorithms/0143. Reorder List/143. Reorder List.go	
+++ b/leetcode/0143.Reorder-List/143. Reorder List.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0143. Reorder List/143. Reorder List_test.go b/leetcode/0143.Reorder-List/143. Reorder List_test.go
similarity index 79%
rename from Algorithms/0143. Reorder List/143. Reorder List_test.go
rename to leetcode/0143.Reorder-List/143. Reorder List_test.go
index d9a8fb9ac..d8363de59 100644
--- a/Algorithms/0143. Reorder List/143. Reorder List_test.go	
+++ b/leetcode/0143.Reorder-List/143. Reorder List_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question143 struct {
@@ -26,21 +28,21 @@ func Test_Problem143(t *testing.T) {
 
 	qs := []question143{
 
-		question143{
+		{
 			para143{[]int{1, 2, 3, 4, 5}},
 			ans143{[]int{1, 5, 2, 4, 3}},
 		},
-		question143{
+		{
 			para143{[]int{1, 2, 3, 4}},
 			ans143{[]int{1, 4, 2, 3}},
 		},
 
-		question143{
+		{
 			para143{[]int{1}},
 			ans143{[]int{1}},
 		},
 
-		question143{
+		{
 			para143{[]int{}},
 			ans143{[]int{}},
 		},
@@ -50,7 +52,7 @@ func Test_Problem143(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans143, q.para143
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(reorderList(S2l(p.one))))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(reorderList(structures.Ints2List(p.one))))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/leetcode/0143.Reorder-List/README.md b/leetcode/0143.Reorder-List/README.md
new file mode 100644
index 000000000..0e0915d02
--- /dev/null
+++ b/leetcode/0143.Reorder-List/README.md
@@ -0,0 +1,34 @@
+# [143. Reorder List](https://leetcode.com/problems/reorder-list/)
+
+## 题目
+
+Given a singly linked list L: L0→L1→…→Ln-1→Ln,
+reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
+
+You may not modify the values in the list's nodes, only nodes itself may be changed.
+
+Example 1:
+
+```c
+Given 1->2->3->4, reorder it to 1->4->2->3.
+```
+
+Example 2:
+
+```c
+Given 1->2->3->4->5, reorder it to 1->5->2->4->3.
+```
+
+## 题目大意
+
+按照指定规则重新排序链表：第一个元素和最后一个元素排列在一起，接着第二个元素和倒数第二个元素排在一起，接着第三个元素和倒数第三个元素排在一起。
+
+
+## 解题思路
+
+
+最近简单的方法是先把链表存储到数组里，然后找到链表中间的结点，按照规则拼接即可。这样时间复杂度是 O(n)，空间复杂度是 O(n)。
+
+更好的做法是结合之前几道题的操作：链表逆序，找中间结点。
+
+先找到链表的中间结点，然后利用逆序区间的操作，如 [第 92 题](https://github.com/halfrost/leetcode-go/tree/master/leetcode/0092.Reverse-Linked-List-II) 里的 reverseBetween() 操作，只不过这里的反转区间是从中点一直到末尾。最后利用 2 个指针，一个指向头结点，一个指向中间结点，开始拼接最终的结果。这种做法的时间复杂度是 O(n)，空间复杂度是 O(1)。
\ No newline at end of file
diff --git a/Algorithms/0144. Binary Tree Preorder Traversal/144. Binary Tree Preorder Traversal.go b/leetcode/0144.Binary-Tree-Preorder-Traversal/144. Binary Tree Preorder Traversal.go
similarity index 91%
rename from Algorithms/0144. Binary Tree Preorder Traversal/144. Binary Tree Preorder Traversal.go
rename to leetcode/0144.Binary-Tree-Preorder-Traversal/144. Binary Tree Preorder Traversal.go
index fafa40c4b..816eb1779 100644
--- a/Algorithms/0144. Binary Tree Preorder Traversal/144. Binary Tree Preorder Traversal.go	
+++ b/leetcode/0144.Binary-Tree-Preorder-Traversal/144. Binary Tree Preorder Traversal.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
diff --git a/leetcode/0144.Binary-Tree-Preorder-Traversal/144. Binary Tree Preorder Traversal_test.go b/leetcode/0144.Binary-Tree-Preorder-Traversal/144. Binary Tree Preorder Traversal_test.go
new file mode 100644
index 000000000..c66a5474e
--- /dev/null
+++ b/leetcode/0144.Binary-Tree-Preorder-Traversal/144. Binary Tree Preorder Traversal_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question144 struct {
+	para144
+	ans144
+}
+
+// para 是参数
+// one 代表第一个参数
+type para144 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans144 struct {
+	one []int
+}
+
+func Test_Problem144(t *testing.T) {
+
+	qs := []question144{
+
+		{
+			para144{[]int{}},
+			ans144{[]int{}},
+		},
+
+		{
+			para144{[]int{1}},
+			ans144{[]int{1}},
+		},
+
+		{
+			para144{[]int{1, structures.NULL, 2, 3}},
+			ans144{[]int{1, 2, 3}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 144------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans144, q.para144
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", preorderTraversal(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0144. Binary Tree Preorder Traversal/README.md b/leetcode/0144.Binary-Tree-Preorder-Traversal/README.md
similarity index 100%
rename from Algorithms/0144. Binary Tree Preorder Traversal/README.md
rename to leetcode/0144.Binary-Tree-Preorder-Traversal/README.md
diff --git a/Algorithms/0145. Binary Tree Postorder Traversal/145. Binary Tree Postorder Traversal.go b/leetcode/0145.Binary-Tree-Postorder-Traversal/145. Binary Tree Postorder Traversal.go
similarity index 79%
rename from Algorithms/0145. Binary Tree Postorder Traversal/145. Binary Tree Postorder Traversal.go
rename to leetcode/0145.Binary-Tree-Postorder-Traversal/145. Binary Tree Postorder Traversal.go
index bb1844ed5..0923511d5 100644
--- a/Algorithms/0145. Binary Tree Postorder Traversal/145. Binary Tree Postorder Traversal.go	
+++ b/leetcode/0145.Binary-Tree-Postorder-Traversal/145. Binary Tree Postorder Traversal.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func postorderTraversal(root *TreeNode) []int {
 	var result []int
 	postorder(root, &result)
diff --git a/Algorithms/0145. Binary Tree Postorder Traversal/145. Binary Tree Postorder Traversal_test.go b/leetcode/0145.Binary-Tree-Postorder-Traversal/145. Binary Tree Postorder Traversal_test.go
similarity index 83%
rename from Algorithms/0145. Binary Tree Postorder Traversal/145. Binary Tree Postorder Traversal_test.go
rename to leetcode/0145.Binary-Tree-Postorder-Traversal/145. Binary Tree Postorder Traversal_test.go
index 6e786b6d5..37a1ef3ca 100644
--- a/Algorithms/0145. Binary Tree Postorder Traversal/145. Binary Tree Postorder Traversal_test.go	
+++ b/leetcode/0145.Binary-Tree-Postorder-Traversal/145. Binary Tree Postorder Traversal_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question145 struct {
@@ -26,18 +28,18 @@ func Test_Problem145(t *testing.T) {
 
 	qs := []question145{
 
-		question145{
+		{
 			para145{[]int{}},
 			ans145{[]int{}},
 		},
 
-		question145{
+		{
 			para145{[]int{1}},
 			ans145{[]int{1}},
 		},
 
-		question145{
-			para145{[]int{1, NULL, 2, 3}},
+		{
+			para145{[]int{1, structures.NULL, 2, 3}},
 			ans145{[]int{1, 2, 3}},
 		},
 	}
@@ -47,7 +49,7 @@ func Test_Problem145(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans145, q.para145
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", postorderTraversal(root))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0145. Binary Tree Postorder Traversal/README.md b/leetcode/0145.Binary-Tree-Postorder-Traversal/README.md
similarity index 100%
rename from Algorithms/0145. Binary Tree Postorder Traversal/README.md
rename to leetcode/0145.Binary-Tree-Postorder-Traversal/README.md
diff --git a/leetcode/0146.LRU-Cache/146. LRU Cache.go b/leetcode/0146.LRU-Cache/146. LRU Cache.go
new file mode 100644
index 000000000..f7a1ae051
--- /dev/null
+++ b/leetcode/0146.LRU-Cache/146. LRU Cache.go	
@@ -0,0 +1,71 @@
+package leetcode
+
+type LRUCache struct {
+	head, tail *Node
+	Keys       map[int]*Node
+	Cap        int
+}
+
+type Node struct {
+	Key, Val   int
+	Prev, Next *Node
+}
+
+func Constructor(capacity int) LRUCache {
+	return LRUCache{Keys: make(map[int]*Node), Cap: capacity}
+}
+
+func (this *LRUCache) Get(key int) int {
+	if node, ok := this.Keys[key]; ok {
+		this.Remove(node)
+		this.Add(node)
+		return node.Val
+	}
+	return -1
+}
+
+func (this *LRUCache) Put(key int, value int) {
+	if node, ok := this.Keys[key]; ok {
+		node.Val = value
+		this.Remove(node)
+		this.Add(node)
+		return
+	} else {
+		node = &Node{Key: key, Val: value}
+		this.Keys[key] = node
+		this.Add(node)
+	}
+	if len(this.Keys) > this.Cap {
+		delete(this.Keys, this.tail.Key)
+		this.Remove(this.tail)
+	}
+}
+
+func (this *LRUCache) Add(node *Node) {
+	node.Prev = nil
+	node.Next = this.head
+	if this.head != nil {
+		this.head.Prev = node
+	}
+	this.head = node
+	if this.tail == nil {
+		this.tail = node
+		this.tail.Next = nil
+	}
+}
+
+func (this *LRUCache) Remove(node *Node) {
+	if node == this.head {
+		this.head = node.Next
+		node.Next = nil
+		return
+	}
+	if node == this.tail {
+		this.tail = node.Prev
+		node.Prev.Next = nil
+		node.Prev = nil
+		return
+	}
+	node.Prev.Next = node.Next
+	node.Next.Prev = node.Prev
+}
diff --git a/leetcode/0146.LRU-Cache/146. LRU Cache_test.go b/leetcode/0146.LRU-Cache/146. LRU Cache_test.go
new file mode 100644
index 000000000..9d8d16047
--- /dev/null
+++ b/leetcode/0146.LRU-Cache/146. LRU Cache_test.go	
@@ -0,0 +1,40 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem146(t *testing.T) {
+	obj := Constructor(2)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.Put(1, 1)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.Put(2, 2)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	param1 := obj.Get(1)
+	fmt.Printf("param_1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	obj.Put(3, 3)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	param1 = obj.Get(2)
+	fmt.Printf("param_1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	obj.Put(4, 4)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	param1 = obj.Get(1)
+	fmt.Printf("param_1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.Get(3)
+	fmt.Printf("param_1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.Get(4)
+	fmt.Printf("param_1 = %v obj = %v\n", param1, MList2Ints(&obj))
+}
+
+func MList2Ints(lru *LRUCache) [][]int {
+	res := [][]int{}
+	head := lru.head
+	for head != nil {
+		tmp := []int{head.Key, head.Val}
+		res = append(res, tmp)
+		head = head.Next
+	}
+	return res
+}
diff --git a/leetcode/0146.LRU-Cache/README.md b/leetcode/0146.LRU-Cache/README.md
new file mode 100644
index 000000000..3e5ac858a
--- /dev/null
+++ b/leetcode/0146.LRU-Cache/README.md
@@ -0,0 +1,134 @@
+# [146. LRU Cache](https://leetcode.com/problems/lru-cache/)
+
+## 题目
+
+Design a data structure that follows the constraints of a **[Least Recently Used (LRU) cache](https://en.wikipedia.org/wiki/Cache_replacement_policies#LRU)**.
+
+Implement the `LRUCache` class:
+
+- `LRUCache(int capacity)` Initialize the LRU cache with **positive** size `capacity`.
+- `int get(int key)` Return the value of the `key` if the key exists, otherwise return `1`.
+- `void put(int key, int value)` Update the value of the `key` if the `key` exists. Otherwise, add the `key-value` pair to the cache. If the number of keys exceeds the `capacity` from this operation, **evict** the least recently used key.
+
+**Follow up:**Could you do `get` and `put` in `O(1)` time complexity?
+
+**Example 1:**
+
+```
+Input
+["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
+[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
+Output
+[null, null, null, 1, null, -1, null, -1, 3, 4]
+
+Explanation
+LRUCache lRUCache = new LRUCache(2);
+lRUCache.put(1, 1); // cache is {1=1}
+lRUCache.put(2, 2); // cache is {1=1, 2=2}
+lRUCache.get(1);    // return 1
+lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
+lRUCache.get(2);    // returns -1 (not found)
+lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
+lRUCache.get(1);    // return -1 (not found)
+lRUCache.get(3);    // return 3
+lRUCache.get(4);    // return 4
+
+```
+
+**Constraints:**
+
+- `1 <= capacity <= 3000`
+- `0 <= key <= 3000`
+- `0 <= value <= 104`
+- At most `3 * 104` calls will be made to `get` and `put`.
+
+## 题目大意
+
+运用你所掌握的数据结构，设计和实现一个  LRU (最近最少使用) 缓存机制 。
+实现 LRUCache 类：
+
+- LRUCache(int capacity) 以正整数作为容量 capacity 初始化 LRU 缓存
+- int get(int key) 如果关键字 key 存在于缓存中，则返回关键字的值，否则返回 -1 。
+- void put(int key, int value) 如果关键字已经存在，则变更其数据值；如果关键字不存在，则插入该组「关键字-值」。当缓存容量达到上限时，它应该在写入新数据之前删除最久未使用的数据值，从而为新的数据值留出空间。
+
+进阶：你是否可以在 O(1) 时间复杂度内完成这两种操作？
+
+## 解题思路
+
+- 这一题是 LRU 经典面试题，详细解释见[第三章 LRUCache 模板](https://books.halfrost.com/leetcode/ChapterThree/LRUCache/)。
+
+## 代码
+
+```go
+package leetcode
+
+type LRUCache struct {
+	head, tail *Node
+	Keys       map[int]*Node
+	Cap        int
+}
+
+type Node struct {
+	Key, Val   int
+	Prev, Next *Node
+}
+
+func Constructor(capacity int) LRUCache {
+	return LRUCache{Keys: make(map[int]*Node), Cap: capacity}
+}
+
+func (this *LRUCache) Get(key int) int {
+	if node, ok := this.Keys[key]; ok {
+		this.Remove(node)
+		this.Add(node)
+		return node.Val
+	}
+	return -1
+}
+
+func (this *LRUCache) Put(key int, value int) {
+	if node, ok := this.Keys[key]; ok {
+		node.Val = value
+		this.Remove(node)
+		this.Add(node)
+		return
+	} else {
+		node = &Node{Key: key, Val: value}
+		this.Keys[key] = node
+		this.Add(node)
+	}
+	if len(this.Keys) > this.Cap {
+		delete(this.Keys, this.tail.Key)
+		this.Remove(this.tail)
+	}
+}
+
+func (this *LRUCache) Add(node *Node) {
+	node.Prev = nil
+	node.Next = this.head
+	if this.head != nil {
+		this.head.Prev = node
+	}
+	this.head = node
+	if this.tail == nil {
+		this.tail = node
+		this.tail.Next = nil
+	}
+}
+
+func (this *LRUCache) Remove(node *Node) {
+	if node == this.head {
+		this.head = node.Next
+		node.Next = nil
+		return
+	}
+	if node == this.tail {
+		this.tail = node.Prev
+		node.Prev.Next = nil
+		node.Prev = nil
+		return
+	}
+	node.Prev.Next = node.Next
+	node.Next.Prev = node.Prev
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0147. Insertion Sort List/147. Insertion Sort List.go b/leetcode/0147.Insertion-Sort-List/147. Insertion Sort List.go
similarity index 77%
rename from Algorithms/0147. Insertion Sort List/147. Insertion Sort List.go
rename to leetcode/0147.Insertion-Sort-List/147. Insertion Sort List.go
index 4e98efa46..89ea6d7ef 100644
--- a/Algorithms/0147. Insertion Sort List/147. Insertion Sort List.go	
+++ b/leetcode/0147.Insertion-Sort-List/147. Insertion Sort List.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
@@ -12,11 +19,9 @@ func insertionSortList(head *ListNode) *ListNode {
 		return head
 	}
 	newHead := &ListNode{Val: 0, Next: nil} // 这里初始化不要直接指向 head，为了下面循环可以统一处理
-	cur := head
-	pre := newHead
-	next := &ListNode{Val: 0, Next: nil}
+	cur, pre := head, newHead
 	for cur != nil {
-		next = cur.Next
+		next := cur.Next
 		for pre.Next != nil && pre.Next.Val < cur.Val {
 			pre = pre.Next
 		}
diff --git a/Algorithms/0147. Insertion Sort List/147. Insertion Sort List_test.go b/leetcode/0147.Insertion-Sort-List/147. Insertion Sort List_test.go
similarity index 79%
rename from Algorithms/0147. Insertion Sort List/147. Insertion Sort List_test.go
rename to leetcode/0147.Insertion-Sort-List/147. Insertion Sort List_test.go
index bddea69a3..d1b0d3955 100644
--- a/Algorithms/0147. Insertion Sort List/147. Insertion Sort List_test.go	
+++ b/leetcode/0147.Insertion-Sort-List/147. Insertion Sort List_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question147 struct {
@@ -26,21 +28,21 @@ func Test_Problem147(t *testing.T) {
 
 	qs := []question147{
 
-		question147{
+		{
 			para147{[]int{4, 2, 1, 3}},
 			ans147{[]int{1, 2, 3, 4}},
 		},
-		question147{
+		{
 			para147{[]int{1}},
 			ans147{[]int{1}},
 		},
 
-		question147{
+		{
 			para147{[]int{-1, 5, 3, 4, 0}},
 			ans147{[]int{-1, 0, 3, 4, 5}},
 		},
 
-		question147{
+		{
 			para147{[]int{}},
 			ans147{[]int{}},
 		},
@@ -50,7 +52,7 @@ func Test_Problem147(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans147, q.para147
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(insertionSortList(S2l(p.one))))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(insertionSortList(structures.Ints2List(p.one))))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0147. Insertion Sort List/README.md b/leetcode/0147.Insertion-Sort-List/README.md
similarity index 100%
rename from Algorithms/0147. Insertion Sort List/README.md
rename to leetcode/0147.Insertion-Sort-List/README.md
diff --git a/leetcode/0148.Sort-List/148. Sort List.go b/leetcode/0148.Sort-List/148. Sort List.go
new file mode 100644
index 000000000..2461013f4
--- /dev/null
+++ b/leetcode/0148.Sort-List/148. Sort List.go	
@@ -0,0 +1,64 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func sortList(head *ListNode) *ListNode {
+	length := 0
+	cur := head
+	for cur != nil {
+		length++
+		cur = cur.Next
+	}
+	if length <= 1 {
+		return head
+	}
+
+	middleNode := middleNode(head)
+	cur = middleNode.Next
+	middleNode.Next = nil
+	middleNode = cur
+
+	left := sortList(head)
+	right := sortList(middleNode)
+	return mergeTwoLists(left, right)
+}
+
+func middleNode(head *ListNode) *ListNode {
+	if head == nil || head.Next == nil {
+		return head
+	}
+	p1 := head
+	p2 := head
+	for p2.Next != nil && p2.Next.Next != nil {
+		p1 = p1.Next
+		p2 = p2.Next.Next
+	}
+	return p1
+}
+
+func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
+	if l1 == nil {
+		return l2
+	}
+	if l2 == nil {
+		return l1
+	}
+	if l1.Val < l2.Val {
+		l1.Next = mergeTwoLists(l1.Next, l2)
+		return l1
+	}
+	l2.Next = mergeTwoLists(l1, l2.Next)
+	return l2
+}
diff --git a/Algorithms/0148. Sort List/148. Sort List_test.go b/leetcode/0148.Sort-List/148. Sort List_test.go
similarity index 80%
rename from Algorithms/0148. Sort List/148. Sort List_test.go
rename to leetcode/0148.Sort-List/148. Sort List_test.go
index 871011cb3..11258c23e 100644
--- a/Algorithms/0148. Sort List/148. Sort List_test.go	
+++ b/leetcode/0148.Sort-List/148. Sort List_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question148 struct {
@@ -26,21 +28,21 @@ func Test_Problem148(t *testing.T) {
 
 	qs := []question148{
 
-		question148{
+		{
 			para148{[]int{1, 2, 3, 4, 5}},
 			ans148{[]int{1, 2, 3, 4, 5}},
 		},
-		question148{
+		{
 			para148{[]int{1, 1, 2, 5, 5, 4, 10, 0}},
 			ans148{[]int{0, 1, 1, 2, 4, 5, 5}},
 		},
 
-		question148{
+		{
 			para148{[]int{1}},
 			ans148{[]int{1}},
 		},
 
-		question148{
+		{
 			para148{[]int{}},
 			ans148{[]int{}},
 		},
@@ -50,7 +52,7 @@ func Test_Problem148(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans148, q.para148
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(sortList(S2l(p.one))))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(sortList(structures.Ints2List(p.one))))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0148. Sort List/README.md b/leetcode/0148.Sort-List/README.md
similarity index 100%
rename from Algorithms/0148. Sort List/README.md
rename to leetcode/0148.Sort-List/README.md
diff --git a/leetcode/0150.Evaluate-Reverse-Polish-Notation/150. Evaluate Reverse Polish Notation.go b/leetcode/0150.Evaluate-Reverse-Polish-Notation/150. Evaluate Reverse Polish Notation.go
new file mode 100644
index 000000000..e91375491
--- /dev/null
+++ b/leetcode/0150.Evaluate-Reverse-Polish-Notation/150. Evaluate Reverse Polish Notation.go	
@@ -0,0 +1,29 @@
+package leetcode
+
+import (
+	"strconv"
+)
+
+func evalRPN(tokens []string) int {
+	stack := make([]int, 0, len(tokens))
+	for _, token := range tokens {
+		v, err := strconv.Atoi(token)
+		if err == nil {
+			stack = append(stack, v)
+		} else {
+			num1, num2 := stack[len(stack)-2], stack[len(stack)-1]
+			stack = stack[:len(stack)-2]
+			switch token {
+			case "+":
+				stack = append(stack, num1+num2)
+			case "-":
+				stack = append(stack, num1-num2)
+			case "*":
+				stack = append(stack, num1*num2)
+			case "/":
+				stack = append(stack, num1/num2)
+			}
+		}
+	}
+	return stack[0]
+}
diff --git a/Algorithms/0150. Evaluate Reverse Polish Notation/150. Evaluate Reverse Polish Notation_test.go b/leetcode/0150.Evaluate-Reverse-Polish-Notation/150. Evaluate Reverse Polish Notation_test.go
similarity index 93%
rename from Algorithms/0150. Evaluate Reverse Polish Notation/150. Evaluate Reverse Polish Notation_test.go
rename to leetcode/0150.Evaluate-Reverse-Polish-Notation/150. Evaluate Reverse Polish Notation_test.go
index 13339ae26..55fb44761 100644
--- a/Algorithms/0150. Evaluate Reverse Polish Notation/150. Evaluate Reverse Polish Notation_test.go	
+++ b/leetcode/0150.Evaluate-Reverse-Polish-Notation/150. Evaluate Reverse Polish Notation_test.go	
@@ -26,21 +26,21 @@ func Test_Problem150(t *testing.T) {
 
 	qs := []question150{
 
-		question150{
+		{
 			para150{[]string{"18"}},
 			ans150{18},
 		},
 
-		question150{
+		{
 			para150{[]string{"2", "1", "+", "3", "*"}},
 			ans150{9},
 		},
-		question150{
+		{
 			para150{[]string{"4", "13", "5", "/", "+"}},
 			ans150{6},
 		},
 
-		question150{
+		{
 			para150{[]string{"10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"}},
 			ans150{22},
 		},
diff --git a/Algorithms/0150. Evaluate Reverse Polish Notation/README.md b/leetcode/0150.Evaluate-Reverse-Polish-Notation/README.md
similarity index 100%
rename from Algorithms/0150. Evaluate Reverse Polish Notation/README.md
rename to leetcode/0150.Evaluate-Reverse-Polish-Notation/README.md
diff --git a/Algorithms/0151. Reverse Words in a String/151. Reverse Words in a String.go b/leetcode/0151.Reverse-Words-in-a-String/151. Reverse Words in a String.go
similarity index 100%
rename from Algorithms/0151. Reverse Words in a String/151. Reverse Words in a String.go
rename to leetcode/0151.Reverse-Words-in-a-String/151. Reverse Words in a String.go
diff --git a/Algorithms/0151. Reverse Words in a String/151. Reverse Words in a String_test.go b/leetcode/0151.Reverse-Words-in-a-String/151. Reverse Words in a String_test.go
similarity index 94%
rename from Algorithms/0151. Reverse Words in a String/151. Reverse Words in a String_test.go
rename to leetcode/0151.Reverse-Words-in-a-String/151. Reverse Words in a String_test.go
index aa5f48e05..195ac83df 100644
--- a/Algorithms/0151. Reverse Words in a String/151. Reverse Words in a String_test.go	
+++ b/leetcode/0151.Reverse-Words-in-a-String/151. Reverse Words in a String_test.go	
@@ -26,16 +26,16 @@ func Test_Problem151(t *testing.T) {
 
 	qs := []question151{
 
-		question151{
+		{
 			para151{"the sky is blue"},
 			ans151{"blue is sky the"},
 		},
 
-		question151{
+		{
 			para151{"  hello world!  "},
 			ans151{"world! hello"},
 		},
-		question151{
+		{
 			para151{"a good   example"},
 			ans151{"example good a"},
 		},
diff --git a/leetcode/0151.Reverse-Words-in-a-String/README.md b/leetcode/0151.Reverse-Words-in-a-String/README.md
new file mode 100755
index 000000000..e9c8cb314
--- /dev/null
+++ b/leetcode/0151.Reverse-Words-in-a-String/README.md
@@ -0,0 +1,57 @@
+# [151. Reverse Words in a String](https://leetcode.com/problems/reverse-words-in-a-string/)
+
+
+
+## 题目
+
+Given an input string, reverse the string word by word.
+
+**Example 1:**
+
+    Input: "the sky is blue"
+    Output: "blue is sky the"
+
+**Example 2:**
+
+    Input: "  hello world!  "
+    Output: "world! hello"
+    Explanation: Your reversed string should not contain leading or trailing spaces.
+
+**Example 3:**
+
+    Input: "a good   example"
+    Output: "example good a"
+    Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.
+
+**Note:**
+
+- A word is defined as a sequence of non-space characters.
+- Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
+- You need to reduce multiple spaces between two words to a single space in the reversed string.
+
+**Follow up:**
+
+For C programmers, try to solve it *in-place* in *O*(1) extra space.
+
+
+## 题目大意
+
+给定一个字符串，逐个翻转字符串中的每个单词。
+
+说明：
+
+- 无空格字符构成一个单词。
+- 输入字符串可以在前面或者后面包含多余的空格，但是反转后的字符不能包括。
+- 如果两个单词间有多余的空格，将反转后单词间的空格减少到只含一个。
+ 
+
+进阶：
+
+- 请选用 C 语言的用户尝试使用 O(1) 额外空间复杂度的原地解法。
+
+
+## 解题思路
+
+
+- 给出一个中间有空格分隔的字符串，要求把这个字符串按照单词的维度前后翻转。
+- 依照题意，先把字符串按照空格分隔成每个小单词，然后把单词前后翻转，最后再把每个单词中间添加空格。
diff --git a/leetcode/0152.Maximum-Product-Subarray/152. Maximum Product Subarray.go b/leetcode/0152.Maximum-Product-Subarray/152. Maximum Product Subarray.go
new file mode 100644
index 000000000..4acde4c87
--- /dev/null
+++ b/leetcode/0152.Maximum-Product-Subarray/152. Maximum Product Subarray.go	
@@ -0,0 +1,28 @@
+package leetcode
+
+func maxProduct(nums []int) int {
+	minimum, maximum, res := nums[0], nums[0], nums[0]
+	for i := 1; i < len(nums); i++ {
+		if nums[i] < 0 {
+			maximum, minimum = minimum, maximum
+		}
+		maximum = max(nums[i], maximum*nums[i])
+		minimum = min(nums[i], minimum*nums[i])
+		res = max(res, maximum)
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/Algorithms/0152. Maximum Product Subarray/152. Maximum Product Subarray_test.go b/leetcode/0152.Maximum-Product-Subarray/152. Maximum Product Subarray_test.go
similarity index 91%
rename from Algorithms/0152. Maximum Product Subarray/152. Maximum Product Subarray_test.go
rename to leetcode/0152.Maximum-Product-Subarray/152. Maximum Product Subarray_test.go
index e37728d7c..12a77add9 100644
--- a/Algorithms/0152. Maximum Product Subarray/152. Maximum Product Subarray_test.go	
+++ b/leetcode/0152.Maximum-Product-Subarray/152. Maximum Product Subarray_test.go	
@@ -26,27 +26,27 @@ func Test_Problem152(t *testing.T) {
 
 	qs := []question152{
 
-		question152{
+		{
 			para152{[]int{-2}},
 			ans152{-2},
 		},
 
-		question152{
+		{
 			para152{[]int{3, -1, 4}},
 			ans152{4},
 		},
 
-		question152{
+		{
 			para152{[]int{0}},
 			ans152{0},
 		},
 
-		question152{
+		{
 			para152{[]int{2, 3, -2, 4}},
 			ans152{6},
 		},
 
-		question152{
+		{
 			para152{[]int{-2, 0, -1}},
 			ans152{0},
 		},
diff --git a/Algorithms/0152. Maximum Product Subarray/README.md b/leetcode/0152.Maximum-Product-Subarray/README.md
similarity index 100%
rename from Algorithms/0152. Maximum Product Subarray/README.md
rename to leetcode/0152.Maximum-Product-Subarray/README.md
diff --git a/Algorithms/0153. Find Minimum in Rotated Sorted Array/153. Find Minimum in Rotated Sorted Array.go b/leetcode/0153.Find-Minimum-in-Rotated-Sorted-Array/153. Find Minimum in Rotated Sorted Array.go
similarity index 100%
rename from Algorithms/0153. Find Minimum in Rotated Sorted Array/153. Find Minimum in Rotated Sorted Array.go
rename to leetcode/0153.Find-Minimum-in-Rotated-Sorted-Array/153. Find Minimum in Rotated Sorted Array.go
diff --git a/Algorithms/0153. Find Minimum in Rotated Sorted Array/153. Find Minimum in Rotated Sorted Array_test.go b/leetcode/0153.Find-Minimum-in-Rotated-Sorted-Array/153. Find Minimum in Rotated Sorted Array_test.go
similarity index 88%
rename from Algorithms/0153. Find Minimum in Rotated Sorted Array/153. Find Minimum in Rotated Sorted Array_test.go
rename to leetcode/0153.Find-Minimum-in-Rotated-Sorted-Array/153. Find Minimum in Rotated Sorted Array_test.go
index 43d3cde8a..b6dce5e4c 100644
--- a/Algorithms/0153. Find Minimum in Rotated Sorted Array/153. Find Minimum in Rotated Sorted Array_test.go	
+++ b/leetcode/0153.Find-Minimum-in-Rotated-Sorted-Array/153. Find Minimum in Rotated Sorted Array_test.go	
@@ -26,42 +26,42 @@ func Test_Problem153(t *testing.T) {
 
 	qs := []question153{
 
-		question153{
+		{
 			para153{[]int{5, 1, 2, 3, 4}},
 			ans153{1},
 		},
 
-		question153{
+		{
 			para153{[]int{1}},
 			ans153{1},
 		},
 
-		question153{
+		{
 			para153{[]int{1, 2}},
 			ans153{1},
 		},
 
-		question153{
+		{
 			para153{[]int{2, 1}},
 			ans153{1},
 		},
 
-		question153{
+		{
 			para153{[]int{2, 3, 1}},
 			ans153{1},
 		},
 
-		question153{
+		{
 			para153{[]int{1, 2, 3}},
 			ans153{1},
 		},
 
-		question153{
+		{
 			para153{[]int{3, 4, 5, 1, 2}},
 			ans153{1},
 		},
 
-		question153{
+		{
 			para153{[]int{4, 5, 6, 7, 0, 1, 2}},
 			ans153{0},
 		},
diff --git a/leetcode/0153.Find-Minimum-in-Rotated-Sorted-Array/README.md b/leetcode/0153.Find-Minimum-in-Rotated-Sorted-Array/README.md
new file mode 100755
index 000000000..ed5994435
--- /dev/null
+++ b/leetcode/0153.Find-Minimum-in-Rotated-Sorted-Array/README.md
@@ -0,0 +1,35 @@
+# [153. Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)
+
+
+## 题目
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., `[0,1,2,4,5,6,7]` might become `[4,5,6,7,0,1,2]`).
+
+Find the minimum element.
+
+You may assume no duplicate exists in the array.
+
+**Example 1:**
+
+    Input: [3,4,5,1,2] 
+    Output: 1
+
+**Example 2:**
+
+    Input: [4,5,6,7,0,1,2]
+    Output: 0
+
+
+## 题目大意
+
+假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。请找出其中最小的元素。
+
+你可以假设数组中不存在重复元素。
+
+
+## 解题思路
+
+- 给出一个原本从小到大排序过的数组，但是在某一个分割点上，把数组切分后的两部分对调位置，数值偏大的放到了数组的前部。求这个数组中最小的元素。
+- 求数组最小的元素其实就是找分割点，前一个数比当前数大，后一个数比当前数也要大。可以用二分搜索查找，需要查找的两个有序区间。时间复杂度  O(log n)。这一题也可以用暴力解法，从头开始遍历，动态维护一个最小值即可，时间复杂度 O(n)。
diff --git a/Algorithms/0154. Find Minimum in Rotated Sorted Array II/154. Find Minimum in Rotated Sorted Array II.go b/leetcode/0154.Find-Minimum-in-Rotated-Sorted-Array-II/154. Find Minimum in Rotated Sorted Array II.go
similarity index 100%
rename from Algorithms/0154. Find Minimum in Rotated Sorted Array II/154. Find Minimum in Rotated Sorted Array II.go
rename to leetcode/0154.Find-Minimum-in-Rotated-Sorted-Array-II/154. Find Minimum in Rotated Sorted Array II.go
diff --git a/Algorithms/0154. Find Minimum in Rotated Sorted Array II/154. Find Minimum in Rotated Sorted Array II_test.go b/leetcode/0154.Find-Minimum-in-Rotated-Sorted-Array-II/154. Find Minimum in Rotated Sorted Array II_test.go
similarity index 87%
rename from Algorithms/0154. Find Minimum in Rotated Sorted Array II/154. Find Minimum in Rotated Sorted Array II_test.go
rename to leetcode/0154.Find-Minimum-in-Rotated-Sorted-Array-II/154. Find Minimum in Rotated Sorted Array II_test.go
index 7cfada0ee..8e329a8a4 100644
--- a/Algorithms/0154. Find Minimum in Rotated Sorted Array II/154. Find Minimum in Rotated Sorted Array II_test.go	
+++ b/leetcode/0154.Find-Minimum-in-Rotated-Sorted-Array-II/154. Find Minimum in Rotated Sorted Array II_test.go	
@@ -26,57 +26,57 @@ func Test_Problem154(t *testing.T) {
 
 	qs := []question154{
 
-		question154{
+		{
 			para154{[]int{10, 2, 10, 10, 10, 10}},
 			ans154{2},
 		},
 
-		question154{
+		{
 			para154{[]int{1, 1}},
 			ans154{1},
 		},
 
-		question154{
+		{
 			para154{[]int{2, 2, 2, 0, 1}},
 			ans154{0},
 		},
 
-		question154{
+		{
 			para154{[]int{5, 1, 2, 3, 4}},
 			ans154{1},
 		},
 
-		question154{
+		{
 			para154{[]int{1}},
 			ans154{1},
 		},
 
-		question154{
+		{
 			para154{[]int{1, 2}},
 			ans154{1},
 		},
 
-		question154{
+		{
 			para154{[]int{2, 1}},
 			ans154{1},
 		},
 
-		question154{
+		{
 			para154{[]int{2, 3, 1}},
 			ans154{1},
 		},
 
-		question154{
+		{
 			para154{[]int{1, 2, 3}},
 			ans154{1},
 		},
 
-		question154{
+		{
 			para154{[]int{3, 4, 5, 1, 2}},
 			ans154{1},
 		},
 
-		question154{
+		{
 			para154{[]int{4, 5, 6, 7, 0, 1, 2}},
 			ans154{0},
 		},
diff --git a/leetcode/0154.Find-Minimum-in-Rotated-Sorted-Array-II/README.md b/leetcode/0154.Find-Minimum-in-Rotated-Sorted-Array-II/README.md
new file mode 100755
index 000000000..ba635eb30
--- /dev/null
+++ b/leetcode/0154.Find-Minimum-in-Rotated-Sorted-Array-II/README.md
@@ -0,0 +1,40 @@
+# [154. Find Minimum in Rotated Sorted Array II](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/)
+
+
+## 题目
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., `[0,1,2,4,5,6,7]` might become `[4,5,6,7,0,1,2]`).
+
+Find the minimum element.
+
+The array may contain duplicates.
+
+**Example 1:**
+
+    Input: [1,3,5]
+    Output: 1
+
+**Example 2:**
+
+    Input: [2,2,2,0,1]
+    Output: 0
+
+**Note:**
+
+- This is a follow up problem to [Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/).
+- Would allow duplicates affect the run-time complexity? How and why?
+
+
+## 题目大意
+
+假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。请找出其中最小的元素。
+
+注意数组中可能存在重复的元素。
+
+## 解题思路
+
+
+- 给出一个原本从小到大排序过的数组，注意数组中有重复的元素。但是在某一个分割点上，把数组切分后的两部分对调位置，数值偏大的放到了数组的前部。求这个数组中最小的元素。
+- 这一题是第 153 题的加强版，增加了重复元素的条件。但是实际做法还是没有变，还是用二分搜索，只不过在相等元素上多增加一个判断即可。时间复杂度 O(log n)。
diff --git a/Algorithms/0155. Min Stack/155. Min Stack.go b/leetcode/0155.Min-Stack/155. Min Stack.go
similarity index 100%
rename from Algorithms/0155. Min Stack/155. Min Stack.go
rename to leetcode/0155.Min-Stack/155. Min Stack.go
diff --git a/Algorithms/0155. Min Stack/155. Min Stack_test.go b/leetcode/0155.Min-Stack/155. Min Stack_test.go
similarity index 100%
rename from Algorithms/0155. Min Stack/155. Min Stack_test.go
rename to leetcode/0155.Min-Stack/155. Min Stack_test.go
diff --git a/Algorithms/0155. Min Stack/README.md b/leetcode/0155.Min-Stack/README.md
similarity index 100%
rename from Algorithms/0155. Min Stack/README.md
rename to leetcode/0155.Min-Stack/README.md
diff --git a/Algorithms/0160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists.go b/leetcode/0160.Intersection-of-Two-Linked-Lists/160. Intersection of Two Linked Lists.go
similarity index 84%
rename from Algorithms/0160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists.go
rename to leetcode/0160.Intersection-of-Two-Linked-Lists/160. Intersection of Two Linked Lists.go
index ac27e70e7..b7b2ee342 100644
--- a/Algorithms/0160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists.go	
+++ b/leetcode/0160.Intersection-of-Two-Linked-Lists/160. Intersection of Two Linked Lists.go	
@@ -1,6 +1,13 @@
 package leetcode
 
-import "fmt"
+import (
+	"fmt"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
 
 /**
  * Definition for singly-linked list.
diff --git a/Algorithms/0160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists_test.go b/leetcode/0160.Intersection-of-Two-Linked-Lists/160. Intersection of Two Linked Lists_test.go
similarity index 80%
rename from Algorithms/0160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists_test.go
rename to leetcode/0160.Intersection-of-Two-Linked-Lists/160. Intersection of Two Linked Lists_test.go
index 95d9a6ecd..cc78f05a8 100644
--- a/Algorithms/0160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists_test.go	
+++ b/leetcode/0160.Intersection-of-Two-Linked-Lists/160. Intersection of Two Linked Lists_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question160 struct {
@@ -27,32 +29,32 @@ func Test_Problem160(t *testing.T) {
 
 	qs := []question160{
 
-		question160{
+		{
 			para160{[]int{}, []int{}},
 			ans160{[]int{}},
 		},
 
-		question160{
+		{
 			para160{[]int{3}, []int{1, 2, 3}},
 			ans160{[]int{3}},
 		},
 
-		question160{
+		{
 			para160{[]int{1, 2, 3, 4}, []int{1, 2, 3, 4}},
 			ans160{[]int{1, 2, 3, 4}},
 		},
 
-		question160{
+		{
 			para160{[]int{4, 1, 8, 4, 5}, []int{5, 0, 1, 8, 4, 5}},
 			ans160{[]int{8, 4, 5}},
 		},
 
-		question160{
+		{
 			para160{[]int{1}, []int{9, 9, 9, 9, 9}},
 			ans160{[]int{}},
 		},
 
-		question160{
+		{
 			para160{[]int{0, 9, 1, 2, 4}, []int{3, 2, 4}},
 			ans160{[]int{2, 4}},
 		},
@@ -62,7 +64,7 @@ func Test_Problem160(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans160, q.para160
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(getIntersectionNode(S2l(p.one), S2l(p.another))))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(getIntersectionNode(structures.Ints2List(p.one), structures.Ints2List(p.another))))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0160. Intersection of Two Linked Lists/README.md b/leetcode/0160.Intersection-of-Two-Linked-Lists/README.md
similarity index 100%
rename from Algorithms/0160. Intersection of Two Linked Lists/README.md
rename to leetcode/0160.Intersection-of-Two-Linked-Lists/README.md
diff --git a/Algorithms/0162. Find Peak Element/162. Find Peak Element.go b/leetcode/0162.Find-Peak-Element/162. Find Peak Element.go
similarity index 100%
rename from Algorithms/0162. Find Peak Element/162. Find Peak Element.go
rename to leetcode/0162.Find-Peak-Element/162. Find Peak Element.go
diff --git a/Algorithms/0162. Find Peak Element/162. Find Peak Element_test.go b/leetcode/0162.Find-Peak-Element/162. Find Peak Element_test.go
similarity index 89%
rename from Algorithms/0162. Find Peak Element/162. Find Peak Element_test.go
rename to leetcode/0162.Find-Peak-Element/162. Find Peak Element_test.go
index f5ef72bb8..d446b33dd 100644
--- a/Algorithms/0162. Find Peak Element/162. Find Peak Element_test.go	
+++ b/leetcode/0162.Find-Peak-Element/162. Find Peak Element_test.go	
@@ -26,37 +26,37 @@ func Test_Problem162(t *testing.T) {
 
 	qs := []question162{
 
-		question162{
+		{
 			para162{[]int{2, 1, 2}},
 			ans162{0},
 		},
 
-		question162{
+		{
 			para162{[]int{3, 2, 1}},
 			ans162{0},
 		},
 
-		question162{
+		{
 			para162{[]int{1, 2}},
 			ans162{1},
 		},
 
-		question162{
+		{
 			para162{[]int{2, 1}},
 			ans162{0},
 		},
 
-		question162{
+		{
 			para162{[]int{1}},
 			ans162{0},
 		},
 
-		question162{
+		{
 			para162{[]int{1, 2, 3, 1}},
 			ans162{2},
 		},
 
-		question162{
+		{
 			para162{[]int{1, 2, 1, 3, 5, 6, 4}},
 			ans162{5},
 		},
diff --git a/leetcode/0162.Find-Peak-Element/README.md b/leetcode/0162.Find-Peak-Element/README.md
new file mode 100755
index 000000000..9bc7c2200
--- /dev/null
+++ b/leetcode/0162.Find-Peak-Element/README.md
@@ -0,0 +1,43 @@
+# [162. Find Peak Element](https://leetcode.com/problems/find-peak-element/)
+
+
+## 题目
+
+A peak element is an element that is greater than its neighbors.
+
+Given an input array `nums`, where `nums[i] ≠ nums[i+1]`, find a peak element and return its index.
+
+The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
+
+You may imagine that `nums[-1] = nums[n] = -∞`.
+
+**Example 1:**
+
+    Input: nums = [1,2,3,1]
+    Output: 2
+    Explanation: 3 is a peak element and your function should return the index number 2.
+
+**Example 2:**
+
+    Input: nums = [1,2,1,3,5,6,4]
+    Output: 1 or 5 
+    Explanation: Your function can return either index number 1 where the peak element is 2, 
+                 or index number 5 where the peak element is 6.
+
+**Note:**
+
+Your solution should be in logarithmic complexity.
+
+## 题目大意
+
+峰值元素是指其值大于左右相邻值的元素。给定一个输入数组 nums，其中 nums[i] ≠ nums[i+1]，找到峰值元素并返回其索引。数组可能包含多个峰值，在这种情况下，返回任何一个峰值所在位置即可。你可以假设 nums[-1] = nums[n] = -∞。
+
+说明:
+
+- 你的解法应该是 O(logN) 时间复杂度的。
+
+
+## 解题思路
+
+- 给出一个数组，数组里面存在多个“山峰”，(山峰的定义是，下标 `i` 比 `i-1`、`i+1` 位置上的元素都要大)，找到这个“山峰”，并输出其中一个山峰的下标。
+- 这一题是第 852 题的伪加强版，第 852 题中只存在一个山峰，这一题存在多个山峰。但是实际上搜索的代码是一样的，因为此题只要求随便输出一个山峰的下标即可。思路同第 852 题。
diff --git a/Algorithms/0164. Maximum Gap/164. Maximum Gap.go b/leetcode/0164.Maximum-Gap/164. Maximum Gap.go
similarity index 90%
rename from Algorithms/0164. Maximum Gap/164. Maximum Gap.go
rename to leetcode/0164.Maximum-Gap/164. Maximum Gap.go
index 5462ddb83..120334ffd 100644
--- a/Algorithms/0164. Maximum Gap/164. Maximum Gap.go	
+++ b/leetcode/0164.Maximum-Gap/164. Maximum Gap.go	
@@ -1,12 +1,11 @@
 package leetcode
 
-// 解法一
+// 解法一 快排
 func maximumGap(nums []int) int {
 	if len(nums) < 2 {
 		return 0
 	}
 	quickSort164(nums, 0, len(nums)-1)
-
 	res := 0
 	for i := 0; i < len(nums)-1; i++ {
 		if (nums[i+1] - nums[i]) > res {
@@ -37,51 +36,48 @@ func quickSort164(a []int, lo, hi int) {
 	quickSort164(a, p+1, hi)
 }
 
-// 解法二
+// 解法二 基数排序
 func maximumGap1(nums []int) int {
-
 	if nums == nil || len(nums) < 2 {
 		return 0
 	}
-
 	// m is the maximal number in nums
 	m := nums[0]
 	for i := 1; i < len(nums); i++ {
 		m = max(m, nums[i])
 	}
-
 	exp := 1 // 1, 10, 100, 1000 ...
 	R := 10  // 10 digits
-
 	aux := make([]int, len(nums))
-
 	for (m / exp) > 0 { // Go through all digits from LSB to MSB
 		count := make([]int, R)
-
 		for i := 0; i < len(nums); i++ {
 			count[(nums[i]/exp)%10]++
 		}
-
 		for i := 1; i < len(count); i++ {
 			count[i] += count[i-1]
 		}
-
 		for i := len(nums) - 1; i >= 0; i-- {
 			tmp := count[(nums[i]/exp)%10]
 			tmp--
 			aux[tmp] = nums[i]
+			count[(nums[i]/exp)%10] = tmp
 		}
-
 		for i := 0; i < len(nums); i++ {
 			nums[i] = aux[i]
 		}
 		exp *= 10
 	}
-
 	maxValue := 0
 	for i := 1; i < len(aux); i++ {
 		maxValue = max(maxValue, aux[i]-aux[i-1])
 	}
-
 	return maxValue
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0164. Maximum Gap/164. Maximum Gap_test.go b/leetcode/0164.Maximum-Gap/164. Maximum Gap_test.go
similarity index 91%
rename from Algorithms/0164. Maximum Gap/164. Maximum Gap_test.go
rename to leetcode/0164.Maximum-Gap/164. Maximum Gap_test.go
index cb203d876..db38fb388 100644
--- a/Algorithms/0164. Maximum Gap/164. Maximum Gap_test.go	
+++ b/leetcode/0164.Maximum-Gap/164. Maximum Gap_test.go	
@@ -26,29 +26,34 @@ func Test_Problem164(t *testing.T) {
 
 	qs := []question164{
 
-		question164{
+		{
 			para164{[]int{3, 6, 9, 1}},
 			ans164{3},
 		},
-		question164{
+		{
 			para164{[]int{1}},
 			ans164{0},
 		},
 
-		question164{
+		{
 			para164{[]int{}},
 			ans164{0},
 		},
 
-		question164{
+		{
 			para164{[]int{2, 10}},
 			ans164{8},
 		},
 
-		question164{
+		{
 			para164{[]int{2, 435, 214, 64321, 643, 7234, 7, 436523, 7856, 8}},
 			ans164{372202},
 		},
+
+		{
+			para164{[]int{1, 10000000}},
+			ans164{9999999},
+		},
 	}
 
 	fmt.Printf("------------------------Leetcode Problem 164------------------------\n")
diff --git a/Algorithms/0164. Maximum Gap/README.md b/leetcode/0164.Maximum-Gap/README.md
similarity index 100%
rename from Algorithms/0164. Maximum Gap/README.md
rename to leetcode/0164.Maximum-Gap/README.md
diff --git a/Algorithms/0167. Two Sum II - Input array is sorted/167. Two Sum II - Input array is sorted.go b/leetcode/0167.Two-Sum-II-Input-array-is-sorted/167. Two Sum II - Input array is sorted.go
similarity index 79%
rename from Algorithms/0167. Two Sum II - Input array is sorted/167. Two Sum II - Input array is sorted.go
rename to leetcode/0167.Two-Sum-II-Input-array-is-sorted/167. Two Sum II - Input array is sorted.go
index b73a32e10..e77da7c0d 100644
--- a/Algorithms/0167. Two Sum II - Input array is sorted/167. Two Sum II - Input array is sorted.go	
+++ b/leetcode/0167.Two-Sum-II-Input-array-is-sorted/167. Two Sum II - Input array is sorted.go	
@@ -6,13 +6,14 @@ func twoSum167(numbers []int, target int) []int {
 	for i < j {
 		if numbers[i]+numbers[j] == target {
 			return []int{i + 1, j + 1}
-		} else if numbers[i]+numbers[j] < target {
+		}
+		if numbers[i]+numbers[j] < target {
 			i++
 		} else {
 			j--
 		}
 	}
-	return []int{-1, -1}
+	return nil
 }
 
 // 解法二 不管数组是否有序，空间复杂度比上一种解法要多 O(n)
@@ -20,8 +21,8 @@ func twoSum167_1(numbers []int, target int) []int {
 	m := make(map[int]int)
 	for i := 0; i < len(numbers); i++ {
 		another := target - numbers[i]
-		if _, ok := m[another]; ok {
-			return []int{m[another] + 1, i + 1}
+		if idx, ok := m[another]; ok {
+			return []int{idx + 1, i + 1}
 		}
 		m[numbers[i]] = i
 	}
diff --git a/Algorithms/0167. Two Sum II - Input array is sorted/167. Two Sum II - Input array is sorted_test.go b/leetcode/0167.Two-Sum-II-Input-array-is-sorted/167. Two Sum II - Input array is sorted_test.go
similarity index 97%
rename from Algorithms/0167. Two Sum II - Input array is sorted/167. Two Sum II - Input array is sorted_test.go
rename to leetcode/0167.Two-Sum-II-Input-array-is-sorted/167. Two Sum II - Input array is sorted_test.go
index e7336806f..c172641a2 100644
--- a/Algorithms/0167. Two Sum II - Input array is sorted/167. Two Sum II - Input array is sorted_test.go	
+++ b/leetcode/0167.Two-Sum-II-Input-array-is-sorted/167. Two Sum II - Input array is sorted_test.go	
@@ -27,7 +27,7 @@ func Test_Problem167(t *testing.T) {
 
 	qs := []question167{
 
-		question167{
+		{
 			para167{[]int{2, 7, 11, 15}, 9},
 			ans167{[]int{1, 2}},
 		},
diff --git a/Algorithms/0167. Two Sum II - Input array is sorted/README.md b/leetcode/0167.Two-Sum-II-Input-array-is-sorted/README.md
similarity index 100%
rename from Algorithms/0167. Two Sum II - Input array is sorted/README.md
rename to leetcode/0167.Two-Sum-II-Input-array-is-sorted/README.md
diff --git a/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title.go b/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title.go
new file mode 100644
index 000000000..e9db5f7c9
--- /dev/null
+++ b/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+func convertToTitle(n int) string {
+	result := []byte{}
+	for n > 0 {
+		result = append(result, 'A'+byte((n-1)%26))
+		n = (n - 1) / 26
+	}
+	for i, j := 0, len(result)-1; i < j; i, j = i+1, j-1 {
+		result[i], result[j] = result[j], result[i]
+	}
+	return string(result)
+}
diff --git a/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title_test.go b/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title_test.go
new file mode 100644
index 000000000..b0a6b4d8b
--- /dev/null
+++ b/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title_test.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question168 struct {
+	para168
+	ans168
+}
+
+// para 是参数
+// one 代表第一个参数
+type para168 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans168 struct {
+	one string
+}
+
+func Test_Problem168(t *testing.T) {
+
+	qs := []question168{
+
+		{
+			para168{1},
+			ans168{"A"},
+		},
+
+		{
+			para168{28},
+			ans168{"AB"},
+		},
+
+		{
+			para168{701},
+			ans168{"ZY"},
+		},
+
+		{
+			para168{10011},
+			ans168{"NUA"},
+		},
+
+		{
+			para168{999},
+			ans168{"ALK"},
+		},
+
+		{
+			para168{681},
+			ans168{"ZE"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 168------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans168, q.para168
+		fmt.Printf("【input】:%v    【output】:%v\n", p, convertToTitle(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0168.Excel-Sheet-Column-Title/README.md b/leetcode/0168.Excel-Sheet-Column-Title/README.md
new file mode 100644
index 000000000..4602e86b2
--- /dev/null
+++ b/leetcode/0168.Excel-Sheet-Column-Title/README.md
@@ -0,0 +1,80 @@
+# [168. Excel Sheet Column Title](https://leetcode.com/problems/excel-sheet-column-title/)
+
+## 题目
+
+Given a positive integer, return its corresponding column title as appear in an Excel sheet.
+
+For example:
+
+```
+    1 -> A
+    2 -> B
+    3 -> C
+    ...
+    26 -> Z
+    27 -> AA
+    28 -> AB 
+    ...
+```
+
+**Example 1**:
+
+```
+Input: 1
+Output: "A"
+```
+
+**Example 2**:
+
+```
+Input: 28
+Output: "AB"
+```
+
+**Example 3**:
+
+```
+Input: 701
+Output: "ZY"
+```
+
+## 题目大意
+
+给定一个正整数，返回它在 Excel 表中相对应的列名称。
+
+例如，
+
+    1 -> A
+    2 -> B
+    3 -> C
+    ...
+    26 -> Z
+    27 -> AA
+    28 -> AB 
+    ...
+
+
+## 解题思路
+
+- 给定一个正整数，返回它在 Excel 表中的对应的列名称
+- 简单题。这一题就类似短除法的计算过程。以 26 进制的字母编码。按照短除法先除，然后余数逆序输出即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func convertToTitle(n int) string {
+	result := []byte{}
+	for n > 0 {
+		result = append(result, 'A'+byte((n-1)%26))
+		n = (n - 1) / 26
+	}
+	for i, j := 0, len(result)-1; i < j; i, j = i+1, j-1 {
+		result[i], result[j] = result[j], result[i]
+	}
+	return string(result)
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0169. Majority Element/169. Majority Element.go b/leetcode/0169.Majority-Element/169. Majority Element.go
similarity index 100%
rename from Algorithms/0169. Majority Element/169. Majority Element.go
rename to leetcode/0169.Majority-Element/169. Majority Element.go
diff --git a/Algorithms/0169. Majority Element/169. Majority Element_test.go b/leetcode/0169.Majority-Element/169. Majority Element_test.go
similarity index 92%
rename from Algorithms/0169. Majority Element/169. Majority Element_test.go
rename to leetcode/0169.Majority-Element/169. Majority Element_test.go
index 08c9bb897..a6aa6dc42 100644
--- a/Algorithms/0169. Majority Element/169. Majority Element_test.go	
+++ b/leetcode/0169.Majority-Element/169. Majority Element_test.go	
@@ -26,22 +26,22 @@ func Test_Problem169(t *testing.T) {
 
 	qs := []question169{
 
-		question169{
+		{
 			para169{[]int{2, 2, 1}},
 			ans169{2},
 		},
 
-		question169{
+		{
 			para169{[]int{3, 2, 3}},
 			ans169{3},
 		},
 
-		question169{
+		{
 			para169{[]int{2, 2, 1, 1, 1, 2, 2}},
 			ans169{2},
 		},
 
-		question169{
+		{
 			para169{[]int{-2147483648}},
 			ans169{-2147483648},
 		},
diff --git a/leetcode/0169.Majority-Element/README.md b/leetcode/0169.Majority-Element/README.md
new file mode 100755
index 000000000..a8919533b
--- /dev/null
+++ b/leetcode/0169.Majority-Element/README.md
@@ -0,0 +1,29 @@
+# [169. Majority Element](https://leetcode.com/problems/majority-element/)
+
+
+## 题目
+
+Given an array of size n, find the majority element. The majority element is the element that appears **more than** `⌊ n/2 ⌋` times.
+
+You may assume that the array is non-empty and the majority element always exist in the array.
+
+**Example 1:**
+
+    Input: [3,2,3]
+    Output: 3
+
+**Example 2:**
+
+    Input: [2,2,1,1,1,2,2]
+    Output: 2
+
+## 题目大意
+
+
+给定一个大小为 n 的数组，找到其中的众数。众数是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素。你可以假设数组是非空的，并且给定的数组总是存在众数。
+
+
+## 解题思路
+
+- 题目要求找出数组中出现次数大于 `⌊ n/2 ⌋` 次的数。要求空间复杂度为 O(1)。简单题。
+- 这一题利用的算法是 Boyer-Moore Majority Vote Algorithm。[https://www.zhihu.com/question/49973163/answer/235921864](https://www.zhihu.com/question/49973163/answer/235921864)
\ No newline at end of file
diff --git a/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number.go b/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number.go
new file mode 100644
index 000000000..5bcfed11c
--- /dev/null
+++ b/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number.go	
@@ -0,0 +1,10 @@
+package leetcode
+
+func titleToNumber(s string) int {
+	val, res := 0, 0
+	for i := 0; i < len(s); i++ {
+		val = int(s[i] - 'A' + 1)
+		res = res*26 + val
+	}
+	return res
+}
diff --git a/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number_test.go b/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number_test.go
new file mode 100644
index 000000000..1b75d1adc
--- /dev/null
+++ b/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question171 struct {
+	para171
+	ans171
+}
+
+// para 是参数
+// one 代表第一个参数
+type para171 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans171 struct {
+	one int
+}
+
+func Test_Problem171(t *testing.T) {
+
+	qs := []question171{
+
+		{
+			para171{"A"},
+			ans171{1},
+		},
+
+		{
+			para171{"AB"},
+			ans171{28},
+		},
+
+		{
+			para171{"ZY"},
+			ans171{701},
+		},
+
+		{
+			para171{"ABC"},
+			ans171{731},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 171------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans171, q.para171
+		fmt.Printf("【input】:%v       【output】:%v\n", p, titleToNumber(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0171.Excel-Sheet-Column-Number/README.md b/leetcode/0171.Excel-Sheet-Column-Number/README.md
new file mode 100644
index 000000000..8c9f6651d
--- /dev/null
+++ b/leetcode/0171.Excel-Sheet-Column-Number/README.md
@@ -0,0 +1,67 @@
+# [171. Excel Sheet Column Number](https://leetcode.com/problems/excel-sheet-column-number/)
+
+
+## 题目
+
+Given a column title as appear in an Excel sheet, return its corresponding column number.
+
+For example:
+
+```
+    A -> 1
+    B -> 2
+    C -> 3
+    ...
+    Z -> 26
+    AA -> 27
+    AB -> 28 
+    ...
+```
+
+**Example 1**:
+
+```
+Input: "A"
+Output: 1
+```
+
+**Example 2**:
+
+```
+Input: "AB"
+Output: 28
+```
+
+**Example 3**:
+
+```
+Input: "ZY"
+Output: 701
+```
+
+## 题目大意
+
+给定一个 Excel 表格中的列名称，返回其相应的列序号。
+
+
+## 解题思路
+
+- 给出 Excel 中列的名称，输出其对应的列序号。
+- 简单题。这一题是第 168 题的逆序题。按照 26 进制还原成十进制即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func titleToNumber(s string) int {
+	val, res := 0, 0
+	for i := 0; i < len(s); i++ {
+		val = int(s[i] - 'A' + 1)
+		res = res*26 + val
+	}
+	return res
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/0172.Factorial-Trailing-Zeroes/172. Factorial Trailing Zeroes.go b/leetcode/0172.Factorial-Trailing-Zeroes/172. Factorial Trailing Zeroes.go
new file mode 100644
index 000000000..74f3081e5
--- /dev/null
+++ b/leetcode/0172.Factorial-Trailing-Zeroes/172. Factorial Trailing Zeroes.go	
@@ -0,0 +1,8 @@
+package leetcode
+
+func trailingZeroes(n int) int {
+	if n/5 == 0 {
+		return 0
+	}
+	return n/5 + trailingZeroes(n/5)
+}
diff --git a/leetcode/0172.Factorial-Trailing-Zeroes/172. Factorial Trailing Zeroes_test.go b/leetcode/0172.Factorial-Trailing-Zeroes/172. Factorial Trailing Zeroes_test.go
new file mode 100644
index 000000000..c29815fcc
--- /dev/null
+++ b/leetcode/0172.Factorial-Trailing-Zeroes/172. Factorial Trailing Zeroes_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question172 struct {
+	para172
+	ans172
+}
+
+// para 是参数
+// one 代表第一个参数
+type para172 struct {
+	s int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans172 struct {
+	one int
+}
+
+func Test_Problem172(t *testing.T) {
+
+	qs := []question172{
+
+		{
+			para172{3},
+			ans172{0},
+		},
+
+		{
+			para172{5},
+			ans172{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 172------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans172, q.para172
+		fmt.Printf("【input】:%v       【output】:%v\n", p, trailingZeroes(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0172.Factorial-Trailing-Zeroes/README.md b/leetcode/0172.Factorial-Trailing-Zeroes/README.md
new file mode 100755
index 000000000..90f37798a
--- /dev/null
+++ b/leetcode/0172.Factorial-Trailing-Zeroes/README.md
@@ -0,0 +1,35 @@
+# [172. Factorial Trailing Zeroes](https://leetcode.com/problems/factorial-trailing-zeroes/)
+
+
+## 题目
+
+Given an integer n, return the number of trailing zeroes in n!.
+
+**Example 1:**
+
+    Input: 3
+    Output: 0
+    Explanation: 3! = 6, no trailing zero.
+
+**Example 2:**
+
+    Input: 5
+    Output: 1
+    Explanation: 5! = 120, one trailing zero.
+
+**Note:** Your solution should be in logarithmic time complexity.
+
+
+## 题目大意
+
+
+给定一个整数 n，返回 n! 结果尾数中零的数量。说明: 你算法的时间复杂度应为 O(log n) 。
+
+
+
+
+## 解题思路
+
+- 给出一个数 n，要求 n！末尾 0 的个数。
+- 这是一道数学题。计算 N 的阶乘有多少个后缀 0，即计算 N! 里有多少个 10，也是计算 N! 里有多少个 2 和 5（分解质因数），最后结果即 2 的个数和 5 的个数取较小值。每两个数字就会多一个质因数 2，而每五个数字才多一个质因数 5。每 5 个数字就会多一个质因数 5。0~4 的阶乘里没有质因数 5，5~9 的阶乘里有 1 个质因数 5，10~14 的阶乘里有 2 个质因数 5，依此类推。所以 0 的个数即为 `min(阶乘中 5 的个数和 2 的个数)`。
+- N! 有多少个后缀 0，即 N! 有多少个质因数 5。N! 有多少个质因数 5，即 N 可以划分成多少组 5个数字一组，加上划分成多少组 25 个数字一组，加上划分多少组成 125 个数字一组，等等。即 `res = N/5 + N/(5^2) + N/(5^3) + ... = ((N / 5) / 5) / 5 /...` 。最终算法复杂度为 O(logN)。
diff --git a/Algorithms/0173. Binary Search Tree Iterator/173. Binary Search Tree Iterator.go b/leetcode/0173.Binary-Search-Tree-Iterator/173. Binary Search Tree Iterator.go
similarity index 81%
rename from Algorithms/0173. Binary Search Tree Iterator/173. Binary Search Tree Iterator.go
rename to leetcode/0173.Binary-Search-Tree-Iterator/173. Binary Search Tree Iterator.go
index c7986e35e..c7e8f2b2f 100644
--- a/Algorithms/0173. Binary Search Tree Iterator/173. Binary Search Tree Iterator.go	
+++ b/leetcode/0173.Binary-Search-Tree-Iterator/173. Binary Search Tree Iterator.go	
@@ -1,6 +1,13 @@
 package leetcode
 
-import "container/heap"
+import (
+	"container/heap"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
 
 /**
  * Definition for a binary tree node.
@@ -28,6 +35,14 @@ func Constructor173(root *TreeNode) BSTIterator {
 	return bs
 }
 
+func postorder(root *TreeNode, output *[]int) {
+	if root != nil {
+		postorder(root.Left, output)
+		postorder(root.Right, output)
+		*output = append(*output, root.Val)
+	}
+}
+
 /** @return the next smallest number */
 func (this *BSTIterator) Next() int {
 	this.count--
diff --git a/Algorithms/0173. Binary Search Tree Iterator/173. Binary Search Tree Iterator_test.go b/leetcode/0173.Binary-Search-Tree-Iterator/173. Binary Search Tree Iterator_test.go
similarity index 80%
rename from Algorithms/0173. Binary Search Tree Iterator/173. Binary Search Tree Iterator_test.go
rename to leetcode/0173.Binary-Search-Tree-Iterator/173. Binary Search Tree Iterator_test.go
index 6de0e627a..b9d80ee14 100644
--- a/Algorithms/0173. Binary Search Tree Iterator/173. Binary Search Tree Iterator_test.go	
+++ b/leetcode/0173.Binary-Search-Tree-Iterator/173. Binary Search Tree Iterator_test.go	
@@ -3,11 +3,13 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 func Test_Problem173(t *testing.T) {
 
-	root := Ints2TreeNode([]int{9, 7, 15, 3, NULL, NULL, 20})
+	root := structures.Ints2TreeNode([]int{9, 7, 15, 3, structures.NULL, structures.NULL, 20})
 	obj := Constructor173(root)
 	fmt.Printf("obj = %v\n", obj)
 	param1 := obj.Next()
diff --git a/Algorithms/0173. Binary Search Tree Iterator/README.md b/leetcode/0173.Binary-Search-Tree-Iterator/README.md
similarity index 100%
rename from Algorithms/0173. Binary Search Tree Iterator/README.md
rename to leetcode/0173.Binary-Search-Tree-Iterator/README.md
diff --git a/Algorithms/0174. Dungeon Game/174. Dungeon Game.go b/leetcode/0174.Dungeon-Game/174. Dungeon Game.go
similarity index 91%
rename from Algorithms/0174. Dungeon Game/174. Dungeon Game.go
rename to leetcode/0174.Dungeon-Game/174. Dungeon Game.go
index 0eba7dbc7..db9f0ebe2 100644
--- a/Algorithms/0174. Dungeon Game/174. Dungeon Game.go	
+++ b/leetcode/0174.Dungeon-Game/174. Dungeon Game.go	
@@ -27,6 +27,20 @@ func calculateMinimumHP(dungeon [][]int) int {
 	return dp[0][0]
 }
 
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
 // 解法二 二分搜索
 func calculateMinimumHP1(dungeon [][]int) int {
 	low, high := 1, math.MaxInt64
diff --git a/leetcode/0174.Dungeon-Game/174. Dungeon Game_test.go b/leetcode/0174.Dungeon-Game/174. Dungeon Game_test.go
new file mode 100644
index 000000000..fb2515c6b
--- /dev/null
+++ b/leetcode/0174.Dungeon-Game/174. Dungeon Game_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question174 struct {
+	para174
+	ans174
+}
+
+// para 是参数
+// one 代表第一个参数
+type para174 struct {
+	s [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans174 struct {
+	one int
+}
+
+func Test_Problem174(t *testing.T) {
+
+	qs := []question174{
+
+		{
+			para174{[][]int{{2, 1}, {1, -1}}},
+			ans174{1},
+		},
+
+		{
+			para174{[][]int{{-3, 5}}},
+			ans174{4},
+		},
+
+		{
+			para174{[][]int{{100}}},
+			ans174{1},
+		},
+
+		{
+			para174{[][]int{{-2, -3, 3}, {-5, -10, 1}, {10, 30, -5}}},
+			ans174{7},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 174------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans174, q.para174
+		fmt.Printf("【input】:%v       【output】:%v\n", p, calculateMinimumHP1(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0174.Dungeon-Game/README.md b/leetcode/0174.Dungeon-Game/README.md
new file mode 100755
index 000000000..886923cb2
--- /dev/null
+++ b/leetcode/0174.Dungeon-Game/README.md
@@ -0,0 +1,49 @@
+# [174. Dungeon Game](https://leetcode.com/problems/dungeon-game/)
+
+## 题目
+
+The demons had captured the princess (**P**) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (**K**) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
+
+The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
+
+Some of the rooms are guarded by demons, so the knight loses health (*negative* integers) upon entering these rooms; other rooms are either empty (*0's*) or contain magic orbs that increase the knight's health (*positive* integers).
+
+In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
+
+**Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.**
+
+For example, given the dungeon below, the initial health of the knight must be at least **7** if he follows the optimal path `RIGHT-> RIGHT -> DOWN -> DOWN`.
+
+
+![](https://img.halfrost.com/Leetcode/leetcode_174_0.png)
+
+**Note:**
+
+- The knight's health has no upper bound.
+- Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.
+
+
+## 题目大意
+
+一些恶魔抓住了公主（P）并将她关在了地下城的右下角。地下城是由 M x N 个房间组成的二维网格。我们英勇的骑士（K）最初被安置在左上角的房间里，他必须穿过地下城并通过对抗恶魔来拯救公主。
+
+骑士的初始健康点数为一个正整数。如果他的健康点数在某一时刻降至 0 或以下，他会立即死亡。
+
+有些房间由恶魔守卫，因此骑士在进入这些房间时会失去健康点数（若房间里的值为负整数，则表示骑士将损失健康点数）；其他房间要么是空的（房间里的值为 0），要么包含增加骑士健康点数的魔法球（若房间里的值为正整数，则表示骑士将增加健康点数）。
+
+为了尽快到达公主，骑士决定每次只向右或向下移动一步。编写一个函数来计算确保骑士能够拯救到公主所需的最低初始健康点数。
+
+说明:
+
+- 骑士的健康点数没有上限。
+- 任何房间都可能对骑士的健康点数造成威胁，也可能增加骑士的健康点数，包括骑士进入的左上角房间以及公主被监禁的右下角房间。
+
+## 解题思路
+
+- 在二维地图上给出每个格子扣血数，负数代表扣血，正数代表补血。左上角第一个格子是起点，右下角最后一个格子是终点。问骑士初始最少多少血才能走完迷宫，顺利营救位于终点的公主。需要注意的是，起点和终点都会对血量进行影响。每到一个格子，骑士的血都不能少于 1，一旦少于 1 点血，骑士就会死去。
+- 这一题首先想到的解题思路是动态规划。从终点逆推回起点。`dp[i][j]` 代表骑士进入坐标为 `(i,j)` 的格子之前最少的血量值。 那么 `dp[m-1][n-1]` 应该同时满足两个条件，`dp[m-1][n-1] + dungeon[m-1][n-1] ≥ 1` 并且 `dp[m-1][n-1] ≥ 1`，由于这两个不等式的方向是相同的，取交集以后，起决定作用的是数轴最右边的数，即 `max(1-dungeon[m-1][n-1] , 1)`。算出 `dp[m-1][n-1]` 以后，接着可以推出 `dp[m-1][i]` 这一行和 `dp[i][n-1]` 这一列的值。因为骑士只能往右走和往下走。往回推，即只能往上走和往左走。到这里，DP 的初始条件都准备好了。那么状态转移方程是什么呢？分析一般的情况，`dp[i][j]` 这个值应该是和 `dp[i+1][j]` 和 `dp[i][j+1]` 这两者有关系。即 `dp[i][j]` 经过自己本格子的扣血以后，要能至少满足下一行和右一列格子血量的最少要求。并且自己的血量也应该 `≥1`。即需要满足下面这两组不等式。   
+	 ![](https://img.halfrost.com/Leetcode/leetcode_174_1.png)  
+	 ![](https://img.halfrost.com/Leetcode/leetcode_174_2.png)    
+    上面不等式中第一组不等式是满足下一行格子的最低血量要求，第二组不等式是满足右一列格子的最低血量要求。第一个式子化简即 `dp[i][j] = max(1, dp[i+1][j]-dungeon[i][j])`，第二个式子化简即 `dp[i][j] = max(1, dp[i][j+1]-dungeon[i][j])`。求得了这两种走法的最低血量值，从这两个值里面取最小，即是当前格子所需的最低血量，所以状态转移方程为 `dp[i][j] = min(max(1, dp[i][j+1]-dungeon[i][j]), max(1, dp[i+1][j]-dungeon[i][j]))`。DP 完成以后，`dp[0][0]` 中记录的就是骑士初始最低血量值。时间复杂度 O(m\*n)，空间复杂度 O(m\*n)。
+
+- 这一题还可以用二分搜索来求解。骑士的血量取值范围一定是在 `[1，+∞)` 这个区间内。那么二分这个区间，每次二分的中间值，再用 dp 在地图中去判断是否能到达终点，如果能，就缩小搜索空间至 `[1，mid]`，否则搜索空间为 `[mid + 1，+∞)` 。时间复杂度 O(m\*n\* log math.MaxInt64)，空间复杂度 O(m\*n)。
diff --git a/Algorithms/0179. Largest Number/179. Largest Number.go b/leetcode/0179.Largest-Number/179. Largest Number.go
similarity index 100%
rename from Algorithms/0179. Largest Number/179. Largest Number.go
rename to leetcode/0179.Largest-Number/179. Largest Number.go
diff --git a/Algorithms/0179. Largest Number/179. Largest Number_test.go b/leetcode/0179.Largest-Number/179. Largest Number_test.go
similarity index 89%
rename from Algorithms/0179. Largest Number/179. Largest Number_test.go
rename to leetcode/0179.Largest-Number/179. Largest Number_test.go
index 3b40b3126..20d98c3dc 100644
--- a/Algorithms/0179. Largest Number/179. Largest Number_test.go	
+++ b/leetcode/0179.Largest-Number/179. Largest Number_test.go	
@@ -26,46 +26,46 @@ func Test_Problem179(t *testing.T) {
 
 	qs := []question179{
 
-		question179{
+		{
 			para179{[]int{3, 6, 9, 1}},
 			ans179{"9631"},
 		},
-		question179{
+		{
 			para179{[]int{1}},
 			ans179{"1"},
 		},
 
-		question179{
+		{
 			para179{[]int{}},
 			ans179{""},
 		},
 
-		question179{
+		{
 			para179{[]int{2, 10}},
 			ans179{"210"},
 		},
 
-		question179{
+		{
 			para179{[]int{3, 30, 34, 5, 9}},
 			ans179{"9534330"},
 		},
 
-		question179{
+		{
 			para179{[]int{12, 128}},
 			ans179{"12812"},
 		},
 
-		question179{
+		{
 			para179{[]int{12, 121}},
 			ans179{"12121"},
 		},
 
-		question179{
+		{
 			para179{[]int{0, 0}},
 			ans179{"0"},
 		},
 
-		question179{
+		{
 			para179{[]int{1440, 7548, 4240, 6616, 733, 4712, 883, 8, 9576}},
 			ans179{"9576888375487336616471242401440"},
 		},
diff --git a/Algorithms/0179. Largest Number/README.md b/leetcode/0179.Largest-Number/README.md
similarity index 100%
rename from Algorithms/0179. Largest Number/README.md
rename to leetcode/0179.Largest-Number/README.md
diff --git a/Algorithms/0187. Repeated DNA Sequences/187. Repeated DNA Sequences.go b/leetcode/0187.Repeated-DNA-Sequences/187. Repeated DNA Sequences.go
similarity index 100%
rename from Algorithms/0187. Repeated DNA Sequences/187. Repeated DNA Sequences.go
rename to leetcode/0187.Repeated-DNA-Sequences/187. Repeated DNA Sequences.go
diff --git a/Algorithms/0187. Repeated DNA Sequences/187. Repeated DNA Sequences_test.go b/leetcode/0187.Repeated-DNA-Sequences/187. Repeated DNA Sequences_test.go
similarity index 97%
rename from Algorithms/0187. Repeated DNA Sequences/187. Repeated DNA Sequences_test.go
rename to leetcode/0187.Repeated-DNA-Sequences/187. Repeated DNA Sequences_test.go
index 7d78d0e79..ae0e40c9f 100644
--- a/Algorithms/0187. Repeated DNA Sequences/187. Repeated DNA Sequences_test.go	
+++ b/leetcode/0187.Repeated-DNA-Sequences/187. Repeated DNA Sequences_test.go	
@@ -26,7 +26,7 @@ func Test_Problem187(t *testing.T) {
 
 	qs := []question187{
 
-		question187{
+		{
 			para187{"AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"},
 			ans187{[]string{"AAAAACCCCC", "CCCCCAAAAA"}},
 		},
diff --git a/leetcode/0187.Repeated-DNA-Sequences/README.md b/leetcode/0187.Repeated-DNA-Sequences/README.md
new file mode 100755
index 000000000..5532d5633
--- /dev/null
+++ b/leetcode/0187.Repeated-DNA-Sequences/README.md
@@ -0,0 +1,24 @@
+# [187. Repeated DNA Sequences](https://leetcode.com/problems/repeated-dna-sequences/)
+
+
+## 题目
+
+All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA.
+
+Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.
+
+**Example:**
+
+    Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
+    
+    Output: ["AAAAACCCCC", "CCCCCAAAAA"]
+
+
+## 题目大意
+
+所有 DNA 由一系列缩写为 A，C，G 和 T 的核苷酸组成，例如：“ACGAATTCCG”。在研究 DNA 时，识别 DNA 中的重复序列有时会对研究非常有帮助。编写一个函数来查找 DNA 分子中所有出现超多一次的10个字母长的序列（子串）。
+
+## 解题思路
+
+- 这一题不用位运算比较好做，维护一个长度为 10 的字符串，在 map 中出现次数 > 1 就输出。
+- 用位运算想做这一题，需要动态的维护长度为 10 的 hashkey，先计算开头长度为 9 的 hash，在往后面扫描的过程中，如果长度超过了 10 ，就移除 hash 开头的一个字符，加入后面一个字符。具体做法是先将 ATCG 变成 00，01，10，11 的编码，那么长度为 10 ，hashkey 就需要维护在 20 位。mask = 0xFFFFF 就是 20 位的。维护了 hashkey 以后，根据这个 hashkey 进行去重和统计频次。
diff --git a/leetcode/0189.Rotate-Array/189. Rotate Array.go b/leetcode/0189.Rotate-Array/189. Rotate Array.go
new file mode 100644
index 000000000..004b55e6f
--- /dev/null
+++ b/leetcode/0189.Rotate-Array/189. Rotate Array.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+// 解法一 时间复杂度 O(n)，空间复杂度 O(1)
+func rotate(nums []int, k int) {
+	k %= len(nums)
+	reverse(nums)
+	reverse(nums[:k])
+	reverse(nums[k:])
+}
+
+func reverse(a []int) {
+	for i, n := 0, len(a); i < n/2; i++ {
+		a[i], a[n-1-i] = a[n-1-i], a[i]
+	}
+}
+
+// 解法二 时间复杂度 O(n)，空间复杂度 O(n)
+func rotate1(nums []int, k int) {
+	newNums := make([]int, len(nums))
+	for i, v := range nums {
+		newNums[(i+k)%len(nums)] = v
+	}
+	copy(nums, newNums)
+}
diff --git a/leetcode/0189.Rotate-Array/189. Rotate Array_test.go b/leetcode/0189.Rotate-Array/189. Rotate Array_test.go
new file mode 100644
index 000000000..a3267c9a4
--- /dev/null
+++ b/leetcode/0189.Rotate-Array/189. Rotate Array_test.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question189 struct {
+	para189
+	ans189
+}
+
+// para 是参数
+// one 代表第一个参数
+type para189 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans189 struct {
+	one []int
+}
+
+func Test_Problem189(t *testing.T) {
+
+	qs := []question189{
+
+		{
+			para189{[]int{1, 2, 3, 4, 5, 6, 7}, 3},
+			ans189{[]int{5, 6, 7, 1, 2, 3, 4}},
+		},
+
+		{
+			para189{[]int{-1, -100, 3, 99}, 2},
+			ans189{[]int{3, 99, -1, -100}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 189------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans189, q.para189
+		fmt.Printf("【input】:%v       ", p)
+		rotate(p.nums, p.k)
+		fmt.Printf("【output】:%v\n", p.nums)
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0189.Rotate-Array/README.md b/leetcode/0189.Rotate-Array/README.md
new file mode 100644
index 000000000..a83b4ab60
--- /dev/null
+++ b/leetcode/0189.Rotate-Array/README.md
@@ -0,0 +1,75 @@
+# [189. Rotate Array](https://leetcode.com/problems/rotate-array/)
+
+## 题目
+
+Given an array, rotate the array to the right by *k* steps, where *k* is non-negative.
+
+**Follow up:**
+
+- Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
+- Could you do it in-place with O(1) extra space?
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3,4,5,6,7], k = 3
+Output: [5,6,7,1,2,3,4]
+Explanation:
+rotate 1 steps to the right: [7,1,2,3,4,5,6]
+rotate 2 steps to the right: [6,7,1,2,3,4,5]
+rotate 3 steps to the right: [5,6,7,1,2,3,4]
+```
+
+**Example 2:**
+
+```
+Input: nums = [-1,-100,3,99], k = 2
+Output: [3,99,-1,-100]
+Explanation: 
+rotate 1 steps to the right: [99,-1,-100,3]
+rotate 2 steps to the right: [3,99,-1,-100]
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 2 * 10^4`
+- `-2^31 <= nums[i] <= 2^31 - 1`
+- `0 <= k <= 10^5`
+
+## 题目大意
+
+给定一个数组，将数组中的元素向右移动 k 个位置，其中 k 是非负数。
+
+## 解题思路
+
+- 解法二，使用一个额外的数组，先将原数组下标为 i 的元素移动到 `(i+k) mod n` 的位置，再将剩下的元素拷贝回来即可。
+- 解法一，由于题目要求不能使用额外的空间，所以本题最佳解法不是解法二。翻转最终态是，末尾 `k mod n` 个元素移动至了数组头部，剩下的元素右移 `k mod n` 个位置至最尾部。确定了最终态以后再变换就很容易。先将数组中所有元素从头到尾翻转一次，尾部的所有元素都到了头部，然后再将 `[0,(k mod n) − 1]` 区间内的元素翻转一次，最后再将 `[k mod n, n − 1]` 区间内的元素翻转一次，即可满足题目要求。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 时间复杂度 O(n)，空间复杂度 O(1)
+func rotate(nums []int, k int) {
+	k %= len(nums)
+	reverse(nums)
+	reverse(nums[:k])
+	reverse(nums[k:])
+}
+
+func reverse(a []int) {
+	for i, n := 0, len(a); i < n/2; i++ {
+		a[i], a[n-1-i] = a[n-1-i], a[i]
+	}
+}
+
+// 解法二 时间复杂度 O(n)，空间复杂度 O(n)
+func rotate1(nums []int, k int) {
+	newNums := make([]int, len(nums))
+	for i, v := range nums {
+		newNums[(i+k)%len(nums)] = v
+	}
+	copy(nums, newNums)
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0190. Reverse Bits/190. Reverse Bits.go b/leetcode/0190.Reverse-Bits/190. Reverse Bits.go
similarity index 100%
rename from Algorithms/0190. Reverse Bits/190. Reverse Bits.go
rename to leetcode/0190.Reverse-Bits/190. Reverse Bits.go
diff --git a/leetcode/0190.Reverse-Bits/190. Reverse Bits_test.go b/leetcode/0190.Reverse-Bits/190. Reverse Bits_test.go
new file mode 100644
index 000000000..12c5c23d5
--- /dev/null
+++ b/leetcode/0190.Reverse-Bits/190. Reverse Bits_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"strconv"
+	"testing"
+)
+
+type question190 struct {
+	para190
+	ans190
+}
+
+// para 是参数
+// one 代表第一个参数
+type para190 struct {
+	one uint32
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans190 struct {
+	one uint32
+}
+
+func Test_Problem190(t *testing.T) {
+	qs := []question190{
+
+		{
+			para190{43261596},
+			ans190{964176192},
+		},
+
+		{
+			para190{4294967293},
+			ans190{3221225471},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 190------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans190, q.para190
+		input := strconv.FormatUint(uint64(p.one), 2) // 32位无符号整数转换为二进制字符串
+		input = fmt.Sprintf("%0*v", 32, input)        // 格式化输出32位,保留前置0
+		output := reverseBits(p.one)
+		outputBin := strconv.FormatUint(uint64(output), 2) // 32位无符号整数转换为二进制字符串
+		outputBin = fmt.Sprintf("%0*v", 32, outputBin)     // 格式化输出32位,保留前置0
+		fmt.Printf("【input】:%v       【output】:%v (%v)\n", input, output, outputBin)
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0190.Reverse-Bits/README.md b/leetcode/0190.Reverse-Bits/README.md
new file mode 100755
index 000000000..b04b5c29d
--- /dev/null
+++ b/leetcode/0190.Reverse-Bits/README.md
@@ -0,0 +1,35 @@
+# [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)
+
+
+## 题目
+
+Reverse bits of a given 32 bits unsigned integer.
+
+**Example 1:**
+
+    Input: 00000010100101000001111010011100
+    Output: 00111001011110000010100101000000
+    Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
+
+**Example 2:**
+
+    Input: 11111111111111111111111111111101
+    Output: 10111111111111111111111111111111
+    Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10101111110010110010011101101001.
+
+**Note:**
+
+- Note that in some languages such as Java, there is no unsigned integer type. In this case, both input and output will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
+- In Java, the compiler represents the signed integers using [2's complement notation](https://en.wikipedia.org/wiki/Two%27s_complement). Therefore, in **Example 2** above the input represents the signed integer `-3` and the output represents the signed integer `-1073741825`.
+
+## 题目大意
+
+颠倒给定的 32 位无符号整数的二进制位。提示：
+
+- 请注意，在某些语言（如 Java）中，没有无符号整数类型。在这种情况下，输入和输出都将被指定为有符号整数类型，并且不应影响您的实现，因为无论整数是有符号的还是无符号的，其内部的二进制表示形式都是相同的。
+- 在 Java 中，编译器使用二进制补码记法来表示有符号整数。因此，在上面的 示例 2 中，输入表示有符号整数 -3，输出表示有符号整数 -1073741825。
+
+## 解题思路
+
+- 简单题，要求反转 32 位的二进制位。
+- 把 num 往右移动，不断的消灭右边最低位的 1，将这个 1 给 res，res 不断的左移即可实现反转二进制位的目的。
diff --git a/Algorithms/0191. Number of 1 Bits/191. Number of 1 Bits.go b/leetcode/0191.Number-of-1-Bits/191. Number of 1 Bits.go
similarity index 100%
rename from Algorithms/0191. Number of 1 Bits/191. Number of 1 Bits.go
rename to leetcode/0191.Number-of-1-Bits/191. Number of 1 Bits.go
diff --git a/leetcode/0191.Number-of-1-Bits/191. Number of 1 Bits_test.go b/leetcode/0191.Number-of-1-Bits/191. Number of 1 Bits_test.go
new file mode 100644
index 000000000..ac789e86e
--- /dev/null
+++ b/leetcode/0191.Number-of-1-Bits/191. Number of 1 Bits_test.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import (
+	"fmt"
+	"strconv"
+	"testing"
+)
+
+type question191 struct {
+	para191
+	ans191
+}
+
+// para 是参数
+// one 代表第一个参数
+type para191 struct {
+	one uint32
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans191 struct {
+	one int
+}
+
+func Test_Problem191(t *testing.T) {
+
+	qs := []question191{
+
+		{
+			para191{5},
+			ans191{1},
+		},
+		{
+			para191{13},
+			ans191{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 191------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans191, q.para191
+		input := strconv.FormatUint(uint64(p.one), 2) // 32位无符号整数转换为二进制字符串
+		input = fmt.Sprintf("%0*v", 32, input)        // 格式化输出32位,保留前置0
+		fmt.Printf("【input】:%v       【output】:%v\n", input, hammingWeight(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0191. Number of 1 Bits/README.md b/leetcode/0191.Number-of-1-Bits/README.md
similarity index 100%
rename from Algorithms/0191. Number of 1 Bits/README.md
rename to leetcode/0191.Number-of-1-Bits/README.md
diff --git a/Algorithms/0198. House Robber/198. House Robber.go b/leetcode/0198.House-Robber/198. House Robber.go
similarity index 93%
rename from Algorithms/0198. House Robber/198. House Robber.go
rename to leetcode/0198.House-Robber/198. House Robber.go
index b8a82fca4..9d7b05be6 100644
--- a/Algorithms/0198. House Robber/198. House Robber.go	
+++ b/leetcode/0198.House-Robber/198. House Robber.go	
@@ -18,6 +18,13 @@ func rob198(nums []int) int {
 	return dp[n-1]
 }
 
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
 // 解法二 DP 优化辅助空间，把迭代的值保存在 2 个变量中
 func rob198_1(nums []int) int {
 	n := len(nums)
diff --git a/Algorithms/0198. House Robber/198. House Robber_test.go b/leetcode/0198.House-Robber/198. House Robber_test.go
similarity index 94%
rename from Algorithms/0198. House Robber/198. House Robber_test.go
rename to leetcode/0198.House-Robber/198. House Robber_test.go
index e2f9be992..95eb40ffe 100644
--- a/Algorithms/0198. House Robber/198. House Robber_test.go	
+++ b/leetcode/0198.House-Robber/198. House Robber_test.go	
@@ -26,16 +26,16 @@ func Test_Problem198(t *testing.T) {
 
 	qs := []question198{
 
-		question198{
+		{
 			para198{[]int{1, 2}},
 			ans198{2},
 		},
 
-		question198{
+		{
 			para198{[]int{1, 2, 3, 1}},
 			ans198{4},
 		},
-		question198{
+		{
 			para198{[]int{2, 7, 9, 3, 1}},
 			ans198{12},
 		},
diff --git a/Algorithms/0198. House Robber/README.md b/leetcode/0198.House-Robber/README.md
similarity index 100%
rename from Algorithms/0198. House Robber/README.md
rename to leetcode/0198.House-Robber/README.md
diff --git a/leetcode/0199.Binary-Tree-Right-Side-View/199. Binary Tree Right Side View.go b/leetcode/0199.Binary-Tree-Right-Side-View/199. Binary Tree Right Side View.go
new file mode 100644
index 000000000..de53951bb
--- /dev/null
+++ b/leetcode/0199.Binary-Tree-Right-Side-View/199. Binary Tree Right Side View.go	
@@ -0,0 +1,39 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func rightSideView(root *TreeNode) []int {
+	res := []int{}
+	if root == nil {
+		return res
+	}
+	queue := []*TreeNode{root}
+	for len(queue) > 0 {
+		n := len(queue)
+		for i := 0; i < n; i++ {
+			if queue[i].Left != nil {
+				queue = append(queue, queue[i].Left)
+			}
+			if queue[i].Right != nil {
+				queue = append(queue, queue[i].Right)
+			}
+		}
+		res = append(res, queue[n-1].Val)
+		queue = queue[n:]
+	}
+	return res
+}
diff --git a/Algorithms/0199. Binary Tree Right Side View/199. Binary Tree Right Side View_test.go b/leetcode/0199.Binary-Tree-Right-Side-View/199. Binary Tree Right Side View_test.go
similarity index 75%
rename from Algorithms/0199. Binary Tree Right Side View/199. Binary Tree Right Side View_test.go
rename to leetcode/0199.Binary-Tree-Right-Side-View/199. Binary Tree Right Side View_test.go
index dcb8b2cb1..58265fef4 100644
--- a/Algorithms/0199. Binary Tree Right Side View/199. Binary Tree Right Side View_test.go	
+++ b/leetcode/0199.Binary-Tree-Right-Side-View/199. Binary Tree Right Side View_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question199 struct {
@@ -26,23 +28,23 @@ func Test_Problem199(t *testing.T) {
 
 	qs := []question199{
 
-		question199{
+		{
 			para199{[]int{}},
 			ans199{[]int{}},
 		},
 
-		question199{
+		{
 			para199{[]int{1}},
 			ans199{[]int{1}},
 		},
 
-		question199{
-			para199{[]int{3, 9, 20, NULL, NULL, 15, 7}},
+		{
+			para199{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
 			ans199{[]int{3, 20, 7}},
 		},
 
-		question199{
-			para199{[]int{1, 2, 3, 4, NULL, NULL, 5}},
+		{
+			para199{[]int{1, 2, 3, 4, structures.NULL, structures.NULL, 5}},
 			ans199{[]int{1, 3, 5}},
 		},
 	}
@@ -52,7 +54,7 @@ func Test_Problem199(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans199, q.para199
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", rightSideView(root))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0199. Binary Tree Right Side View/README.md b/leetcode/0199.Binary-Tree-Right-Side-View/README.md
similarity index 100%
rename from Algorithms/0199. Binary Tree Right Side View/README.md
rename to leetcode/0199.Binary-Tree-Right-Side-View/README.md
diff --git a/Algorithms/0200. Number of Islands/200. Number of Islands.go b/leetcode/0200.Number-of-Islands/200. Number of Islands.go
similarity index 80%
rename from Algorithms/0200. Number of Islands/200. Number of Islands.go
rename to leetcode/0200.Number-of-Islands/200. Number of Islands.go
index 94ad94990..79829e074 100644
--- a/Algorithms/0200. Number of Islands/200. Number of Islands.go	
+++ b/leetcode/0200.Number-of-Islands/200. Number of Islands.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
 func numIslands(grid [][]byte) int {
 	m := len(grid)
 	if m == 0 {
@@ -34,3 +41,7 @@ func searchIslands(grid [][]byte, visited *[][]bool, x, y int) {
 		}
 	}
 }
+
+func isInBoard(board [][]byte, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
diff --git a/leetcode/0200.Number-of-Islands/200. Number of Islands_test.go b/leetcode/0200.Number-of-Islands/200. Number of Islands_test.go
new file mode 100644
index 000000000..8c16ac76b
--- /dev/null
+++ b/leetcode/0200.Number-of-Islands/200. Number of Islands_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question200 struct {
+	para200
+	ans200
+}
+
+// para 是参数
+// one 代表第一个参数
+type para200 struct {
+	one [][]byte
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans200 struct {
+	one int
+}
+
+func Test_Problem200(t *testing.T) {
+
+	qs := []question200{
+
+		{
+			para200{[][]byte{
+				{'1', '1', '1', '1', '0'},
+				{'1', '1', '0', '1', '0'},
+				{'1', '1', '0', '0', '0'},
+				{'0', '0', '0', '0', '0'},
+			}},
+			ans200{1},
+		},
+
+		{
+			para200{[][]byte{
+				{'1', '1', '0', '0', '0'},
+				{'1', '1', '0', '0', '0'},
+				{'0', '0', '1', '0', '0'},
+				{'0', '0', '0', '1', '1'},
+			}},
+			ans200{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 200------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans200, q.para200
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numIslands(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0200. Number of Islands/README.md b/leetcode/0200.Number-of-Islands/README.md
similarity index 100%
rename from Algorithms/0200. Number of Islands/README.md
rename to leetcode/0200.Number-of-Islands/README.md
diff --git a/Algorithms/0201. Bitwise AND of Numbers Range/201. Bitwise AND of Numbers Range.go b/leetcode/0201.Bitwise-AND-of-Numbers-Range/201. Bitwise AND of Numbers Range.go
similarity index 100%
rename from Algorithms/0201. Bitwise AND of Numbers Range/201. Bitwise AND of Numbers Range.go
rename to leetcode/0201.Bitwise-AND-of-Numbers-Range/201. Bitwise AND of Numbers Range.go
diff --git a/Algorithms/0201. Bitwise AND of Numbers Range/201. Bitwise AND of Numbers Range_test.go b/leetcode/0201.Bitwise-AND-of-Numbers-Range/201. Bitwise AND of Numbers Range_test.go
similarity index 95%
rename from Algorithms/0201. Bitwise AND of Numbers Range/201. Bitwise AND of Numbers Range_test.go
rename to leetcode/0201.Bitwise-AND-of-Numbers-Range/201. Bitwise AND of Numbers Range_test.go
index d68684cdd..047214a02 100644
--- a/Algorithms/0201. Bitwise AND of Numbers Range/201. Bitwise AND of Numbers Range_test.go	
+++ b/leetcode/0201.Bitwise-AND-of-Numbers-Range/201. Bitwise AND of Numbers Range_test.go	
@@ -27,12 +27,12 @@ func Test_Problem201(t *testing.T) {
 
 	qs := []question201{
 
-		question201{
+		{
 			para201{5, 7},
 			ans201{4},
 		},
 
-		question201{
+		{
 			para201{0, 1},
 			ans201{0},
 		},
diff --git a/leetcode/0201.Bitwise-AND-of-Numbers-Range/README.md b/leetcode/0201.Bitwise-AND-of-Numbers-Range/README.md
new file mode 100755
index 000000000..7f6eb4440
--- /dev/null
+++ b/leetcode/0201.Bitwise-AND-of-Numbers-Range/README.md
@@ -0,0 +1,35 @@
+# [201. Bitwise AND of Numbers Range](https://leetcode.com/problems/bitwise-and-of-numbers-range/)
+
+
+## 题目
+
+Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.
+
+**Example 1:**
+
+    Input: [5,7]
+    Output: 4
+
+**Example 2:**
+
+    Input: [0,1]
+    Output: 0
+
+## 题目大意
+
+给定范围 [m, n]，其中 0 <= m <= n <= 2147483647，返回此范围内所有数字的按位与（包含 m, n 两端点）。
+
+
+## 解题思路
+
+- 这一题要求输出 [m,n] 区间内所有数的 AND 与操作之后的结果。
+- 举个例子，假设区间是 [26,30]，那么这个区间内的数用二进制表示出来为：
+
+    11010  
+    11011  
+    11100   
+    11101   
+    11110  
+
+- 可以观察到，把这些数都 AND 起来，只要有 0 的位，最终结果都是 0，所以需要从右往前找到某一位上不为 0 的。不断的右移左边界和右边界，把右边的 0 都移走，直到它们俩相等，就找到了某一位上开始都不为 0 的了。在右移的过程中记录下右移了多少位，最后把 m 或者 n 的右边添上 0 即可。按照上面这个例子来看，11000 是最终的结果。
+- 这一题还有解法二，还是以 [26,30] 这个区间为例。这个区间内的数末尾 3 位不断的 0，1 变化着。那么如果把末尾的 1 都打掉，就是最终要求的结果了。当 n == m 或者 n < m 的时候就退出循环，说明后面不同的位数已经都被抹平了，1 都被打掉为 0 了。所以关键的操作为 `n &= (n - 1)` ，清除最低位的 1 。这个算法名叫 `Brian Kernighan` 算法。
diff --git a/leetcode/0202.Happy-Number/202. Happy Number.go b/leetcode/0202.Happy-Number/202. Happy Number.go
new file mode 100644
index 000000000..bd2b1b2ce
--- /dev/null
+++ b/leetcode/0202.Happy-Number/202. Happy Number.go	
@@ -0,0 +1,26 @@
+package leetcode
+
+func isHappy(n int) bool {
+	record := map[int]int{}
+	for n != 1 {
+		record[n] = n
+		n = getSquareOfDigits(n)
+		for _, previous := range record {
+			if n == previous {
+				return false
+			}
+		}
+	}
+	return true
+}
+
+func getSquareOfDigits(n int) int {
+	squareOfDigits := 0
+	temporary := n
+	for temporary != 0 {
+		remainder := temporary % 10
+		squareOfDigits += remainder * remainder
+		temporary /= 10
+	}
+	return squareOfDigits
+}
diff --git a/Algorithms/0202. Happy Number/202. Happy Number_test.go b/leetcode/0202.Happy-Number/202. Happy Number_test.go
similarity index 87%
rename from Algorithms/0202. Happy Number/202. Happy Number_test.go
rename to leetcode/0202.Happy-Number/202. Happy Number_test.go
index f4963e984..a80ff83cf 100644
--- a/Algorithms/0202. Happy Number/202. Happy Number_test.go	
+++ b/leetcode/0202.Happy-Number/202. Happy Number_test.go	
@@ -26,15 +26,25 @@ func Test_Problem202(t *testing.T) {
 
 	qs := []question202{
 
-		question202{
+		{
 			para202{202},
 			ans202{false},
 		},
 
-		question202{
+		{
 			para202{19},
 			ans202{true},
 		},
+
+		{
+			para202{2},
+			ans202{false},
+		},
+
+		{
+			para202{3},
+			ans202{false},
+		},
 	}
 
 	fmt.Printf("------------------------Leetcode Problem 202------------------------\n")
diff --git a/Algorithms/0202. Happy Number/README.md b/leetcode/0202.Happy-Number/README.md
similarity index 100%
rename from Algorithms/0202. Happy Number/README.md
rename to leetcode/0202.Happy-Number/README.md
diff --git a/Algorithms/0203. Remove Linked List Elements/203. Remove Linked List Elements.go b/leetcode/0203.Remove-Linked-List-Elements/203. Remove Linked List Elements.go
similarity index 79%
rename from Algorithms/0203. Remove Linked List Elements/203. Remove Linked List Elements.go
rename to leetcode/0203.Remove-Linked-List-Elements/203. Remove Linked List Elements.go
index 5b4e74b12..abb2bc9c3 100644
--- a/Algorithms/0203. Remove Linked List Elements/203. Remove Linked List Elements.go	
+++ b/leetcode/0203.Remove-Linked-List-Elements/203. Remove Linked List Elements.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0203. Remove Linked List Elements/203. Remove Linked List Elements_test.go b/leetcode/0203.Remove-Linked-List-Elements/203. Remove Linked List Elements_test.go
similarity index 82%
rename from Algorithms/0203. Remove Linked List Elements/203. Remove Linked List Elements_test.go
rename to leetcode/0203.Remove-Linked-List-Elements/203. Remove Linked List Elements_test.go
index 1cdb08961..f673d2761 100644
--- a/Algorithms/0203. Remove Linked List Elements/203. Remove Linked List Elements_test.go	
+++ b/leetcode/0203.Remove-Linked-List-Elements/203. Remove Linked List Elements_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question203 struct {
@@ -27,42 +29,42 @@ func Test_Problem203(t *testing.T) {
 
 	qs := []question203{
 
-		question203{
+		{
 			para203{[]int{1, 2, 3, 4, 5}, 1},
 			ans203{[]int{2, 3, 4, 5}},
 		},
 
-		question203{
+		{
 			para203{[]int{1, 2, 3, 4, 5}, 2},
 			ans203{[]int{1, 3, 4, 5}},
 		},
 
-		question203{
+		{
 			para203{[]int{1, 1, 1, 1, 1}, 1},
 			ans203{[]int{}},
 		},
 
-		question203{
+		{
 			para203{[]int{1, 2, 3, 2, 3, 2, 3, 2}, 2},
 			ans203{[]int{1, 3, 3, 3}},
 		},
 
-		question203{
+		{
 			para203{[]int{1, 2, 3, 4, 5}, 5},
 			ans203{[]int{1, 2, 3, 4}},
 		},
 
-		question203{
+		{
 			para203{[]int{}, 5},
 			ans203{[]int{}},
 		},
 
-		question203{
+		{
 			para203{[]int{1, 2, 3, 4, 5}, 10},
 			ans203{[]int{1, 2, 3, 4, 5}},
 		},
 
-		question203{
+		{
 			para203{[]int{1}, 1},
 			ans203{[]int{}},
 		},
@@ -72,7 +74,7 @@ func Test_Problem203(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans203, q.para203
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(removeElements(S2l(p.one), p.n)))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(removeElements(structures.Ints2List(p.one), p.n)))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0203. Remove Linked List Elements/README.md b/leetcode/0203.Remove-Linked-List-Elements/README.md
similarity index 100%
rename from Algorithms/0203. Remove Linked List Elements/README.md
rename to leetcode/0203.Remove-Linked-List-Elements/README.md
diff --git a/Algorithms/0204. Count Primes/204. Count Primes.go b/leetcode/0204.Count-Primes/204. Count Primes.go
similarity index 100%
rename from Algorithms/0204. Count Primes/204. Count Primes.go
rename to leetcode/0204.Count-Primes/204. Count Primes.go
diff --git a/Algorithms/0204. Count Primes/204. Count Primes_test.go b/leetcode/0204.Count-Primes/204. Count Primes_test.go
similarity index 93%
rename from Algorithms/0204. Count Primes/204. Count Primes_test.go
rename to leetcode/0204.Count-Primes/204. Count Primes_test.go
index b5c2cc891..884e9db48 100644
--- a/Algorithms/0204. Count Primes/204. Count Primes_test.go	
+++ b/leetcode/0204.Count-Primes/204. Count Primes_test.go	
@@ -26,17 +26,17 @@ func Test_Problem204(t *testing.T) {
 
 	qs := []question204{
 
-		question204{
+		{
 			para204{10},
 			ans204{4},
 		},
 
-		question204{
+		{
 			para204{100},
 			ans204{25},
 		},
 
-		question204{
+		{
 			para204{1000},
 			ans204{168},
 		},
diff --git a/leetcode/0204.Count-Primes/README.md b/leetcode/0204.Count-Primes/README.md
new file mode 100755
index 000000000..90b9b4d97
--- /dev/null
+++ b/leetcode/0204.Count-Primes/README.md
@@ -0,0 +1,22 @@
+# [204. Count Primes](https://leetcode.com/problems/count-primes/)
+
+
+## 题目
+
+Count the number of prime numbers less than a non-negative number, **n**.
+
+**Example:**
+
+    Input: 10
+    Output: 4
+    Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.
+
+
+## 题目大意
+
+统计所有小于非负整数 n 的质数的数量。
+
+
+## 解题思路
+
+- 给出一个数字 n，要求输出小于 n 的所有素数的个数总和。简单题。
diff --git a/Algorithms/0205. Isomorphic Strings/205. Isomorphic Strings.go b/leetcode/0205.Isomorphic-Strings/205. Isomorphic Strings.go
similarity index 100%
rename from Algorithms/0205. Isomorphic Strings/205. Isomorphic Strings.go
rename to leetcode/0205.Isomorphic-Strings/205. Isomorphic Strings.go
diff --git a/Algorithms/0205. Isomorphic Strings/205. Isomorphic Strings_test.go b/leetcode/0205.Isomorphic-Strings/205. Isomorphic Strings_test.go
similarity index 92%
rename from Algorithms/0205. Isomorphic Strings/205. Isomorphic Strings_test.go
rename to leetcode/0205.Isomorphic-Strings/205. Isomorphic Strings_test.go
index a8913622c..d18c1f43d 100644
--- a/Algorithms/0205. Isomorphic Strings/205. Isomorphic Strings_test.go	
+++ b/leetcode/0205.Isomorphic-Strings/205. Isomorphic Strings_test.go	
@@ -27,22 +27,22 @@ func Test_Problem205(t *testing.T) {
 
 	qs := []question205{
 
-		question205{
+		{
 			para205{"egg", "add"},
 			ans205{true},
 		},
 
-		question205{
+		{
 			para205{"foo", "bar"},
 			ans205{false},
 		},
 
-		question205{
+		{
 			para205{"paper", "title"},
 			ans205{true},
 		},
 
-		question205{
+		{
 			para205{"", ""},
 			ans205{true},
 		},
diff --git a/Algorithms/0205. Isomorphic Strings/README.md b/leetcode/0205.Isomorphic-Strings/README.md
similarity index 100%
rename from Algorithms/0205. Isomorphic Strings/README.md
rename to leetcode/0205.Isomorphic-Strings/README.md
diff --git a/Algorithms/0206. Reverse-Linked-List/206. Reverse Linked List.go b/leetcode/0206.Reverse-Linked-List/206. Reverse Linked List.go
similarity index 77%
rename from Algorithms/0206. Reverse-Linked-List/206. Reverse Linked List.go
rename to leetcode/0206.Reverse-Linked-List/206. Reverse Linked List.go
index 3ec9ed2a7..00d2a5682 100644
--- a/Algorithms/0206. Reverse-Linked-List/206. Reverse Linked List.go	
+++ b/leetcode/0206.Reverse-Linked-List/206. Reverse Linked List.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
@@ -8,12 +15,6 @@ package leetcode
  * }
  */
 
-// ListNode define
-type ListNode struct {
-	Val  int
-	Next *ListNode
-}
-
 func reverseList(head *ListNode) *ListNode {
 	var behind *ListNode
 	for head != nil {
diff --git a/leetcode/0206.Reverse-Linked-List/206. Reverse Linked List_test.go b/leetcode/0206.Reverse-Linked-List/206. Reverse Linked List_test.go
new file mode 100644
index 000000000..08e78a469
--- /dev/null
+++ b/leetcode/0206.Reverse-Linked-List/206. Reverse Linked List_test.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question206 struct {
+	para206
+	ans206
+}
+
+// para 是参数
+// one 代表第一个参数
+type para206 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans206 struct {
+	one []int
+}
+
+func Test_Problem206(t *testing.T) {
+
+	qs := []question206{
+
+		{
+			para206{[]int{1, 2, 3, 4, 5}},
+			ans206{[]int{5, 4, 3, 2, 1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 206------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans206, q.para206
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(reverseList(structures.Ints2List(p.one))))
+	}
+}
diff --git a/Algorithms/0206. Reverse-Linked-List/README.md b/leetcode/0206.Reverse-Linked-List/README.md
similarity index 100%
rename from Algorithms/0206. Reverse-Linked-List/README.md
rename to leetcode/0206.Reverse-Linked-List/README.md
diff --git a/Algorithms/0207. Course Schedule/207. Course Schedule.go b/leetcode/0207.Course-Schedule/207. Course Schedule.go
similarity index 100%
rename from Algorithms/0207. Course Schedule/207. Course Schedule.go
rename to leetcode/0207.Course-Schedule/207. Course Schedule.go
diff --git a/Algorithms/0207. Course Schedule/207. Course Schedule_test.go b/leetcode/0207.Course-Schedule/207. Course Schedule_test.go
similarity index 84%
rename from Algorithms/0207. Course Schedule/207. Course Schedule_test.go
rename to leetcode/0207.Course-Schedule/207. Course Schedule_test.go
index e1d79a0cd..0f5395dd2 100644
--- a/Algorithms/0207. Course Schedule/207. Course Schedule_test.go	
+++ b/leetcode/0207.Course-Schedule/207. Course Schedule_test.go	
@@ -27,13 +27,13 @@ func Test_Problem207(t *testing.T) {
 
 	qs := []question207{
 
-		question207{
-			para207{2, [][]int{[]int{1, 0}}},
+		{
+			para207{2, [][]int{{1, 0}}},
 			ans207{true},
 		},
 
-		question207{
-			para207{2, [][]int{[]int{1, 0}, []int{0, 1}}},
+		{
+			para207{2, [][]int{{1, 0}, {0, 1}}},
 			ans207{false},
 		},
 	}
diff --git a/Algorithms/0207. Course Schedule/README.md b/leetcode/0207.Course-Schedule/README.md
similarity index 100%
rename from Algorithms/0207. Course Schedule/README.md
rename to leetcode/0207.Course-Schedule/README.md
diff --git a/Algorithms/0208. Implement Trie (Prefix Tree)/208. Implement Trie (Prefix Tree).go b/leetcode/0208.Implement-Trie-Prefix-Tree/208. Implement Trie (Prefix Tree).go
similarity index 100%
rename from Algorithms/0208. Implement Trie (Prefix Tree)/208. Implement Trie (Prefix Tree).go
rename to leetcode/0208.Implement-Trie-Prefix-Tree/208. Implement Trie (Prefix Tree).go
diff --git a/Algorithms/0208. Implement Trie (Prefix Tree)/208. Implement Trie (Prefix Tree)_test.go b/leetcode/0208.Implement-Trie-Prefix-Tree/208. Implement Trie (Prefix Tree)_test.go
similarity index 100%
rename from Algorithms/0208. Implement Trie (Prefix Tree)/208. Implement Trie (Prefix Tree)_test.go
rename to leetcode/0208.Implement-Trie-Prefix-Tree/208. Implement Trie (Prefix Tree)_test.go
diff --git a/Algorithms/0208. Implement Trie (Prefix Tree)/README.md b/leetcode/0208.Implement-Trie-Prefix-Tree/README.md
similarity index 100%
rename from Algorithms/0208. Implement Trie (Prefix Tree)/README.md
rename to leetcode/0208.Implement-Trie-Prefix-Tree/README.md
diff --git a/leetcode/0209.Minimum-Size-Subarray-Sum/209. Minimum Size Subarray Sum.go b/leetcode/0209.Minimum-Size-Subarray-Sum/209. Minimum Size Subarray Sum.go
new file mode 100644
index 000000000..acedf56c6
--- /dev/null
+++ b/leetcode/0209.Minimum-Size-Subarray-Sum/209. Minimum Size Subarray Sum.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+func minSubArrayLen(target int, nums []int) int {
+	left, sum, res := 0, 0, len(nums)+1
+	for right, v := range nums {
+		sum += v
+		for sum >= target {
+			res = min(res, right-left+1)
+			sum -= nums[left]
+			left++
+		}
+	}
+	if res == len(nums)+1 {
+		return 0
+	}
+	return res
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/Algorithms/0209. Minimum Size Subarray Sum/209. Minimum Size Subarray Sum_test.go b/leetcode/0209.Minimum-Size-Subarray-Sum/209. Minimum Size Subarray Sum_test.go
similarity index 97%
rename from Algorithms/0209. Minimum Size Subarray Sum/209. Minimum Size Subarray Sum_test.go
rename to leetcode/0209.Minimum-Size-Subarray-Sum/209. Minimum Size Subarray Sum_test.go
index 0bc0537ec..446b943d8 100644
--- a/Algorithms/0209. Minimum Size Subarray Sum/209. Minimum Size Subarray Sum_test.go	
+++ b/leetcode/0209.Minimum-Size-Subarray-Sum/209. Minimum Size Subarray Sum_test.go	
@@ -27,7 +27,7 @@ func Test_Problem209(t *testing.T) {
 
 	qs := []question209{
 
-		question209{
+		{
 			para209{7, []int{2, 3, 1, 2, 4, 3}},
 			ans209{2},
 		},
diff --git a/Algorithms/0209. Minimum Size Subarray Sum/README.md b/leetcode/0209.Minimum-Size-Subarray-Sum/README.md
similarity index 100%
rename from Algorithms/0209. Minimum Size Subarray Sum/README.md
rename to leetcode/0209.Minimum-Size-Subarray-Sum/README.md
diff --git a/Algorithms/0210. Course Schedule II/210. Course Schedule II.go b/leetcode/0210.Course-Schedule-II/210. Course Schedule II.go
similarity index 100%
rename from Algorithms/0210. Course Schedule II/210. Course Schedule II.go
rename to leetcode/0210.Course-Schedule-II/210. Course Schedule II.go
diff --git a/leetcode/0210.Course-Schedule-II/210. Course Schedule II_test.go b/leetcode/0210.Course-Schedule-II/210. Course Schedule II_test.go
new file mode 100644
index 000000000..871bfc6e1
--- /dev/null
+++ b/leetcode/0210.Course-Schedule-II/210. Course Schedule II_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question210 struct {
+	para210
+	ans210
+}
+
+// para 是参数
+// one 代表第一个参数
+type para210 struct {
+	one int
+	pre [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans210 struct {
+	one []int
+}
+
+func Test_Problem210(t *testing.T) {
+
+	qs := []question210{
+
+		{
+			para210{2, [][]int{{1, 0}}},
+			ans210{[]int{0, 1}},
+		},
+
+		{
+			para210{2, [][]int{{1, 0}, {0, 1}}},
+			ans210{[]int{0, 1, 2, 3}},
+		},
+
+		{
+			para210{4, [][]int{{1, 0}, {2, 0}, {3, 1}, {3, 2}}},
+			ans210{[]int{0, 1, 2, 3}},
+		},
+
+		{
+			para210{3, [][]int{{1, 0}, {1, 2}, {0, 1}}},
+			ans210{[]int{}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 210------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans210, q.para210
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findOrder(p.one, p.pre))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0210. Course Schedule II/README.md b/leetcode/0210.Course-Schedule-II/README.md
similarity index 100%
rename from Algorithms/0210. Course Schedule II/README.md
rename to leetcode/0210.Course-Schedule-II/README.md
diff --git a/Algorithms/0211. Add and Search Word - Data structure design/211. Add and Search Word - Data structure design.go b/leetcode/0211.Design-Add-and-Search-Words-Data-Structure/211. Design Add and Search Words Data Structure.go
similarity index 100%
rename from Algorithms/0211. Add and Search Word - Data structure design/211. Add and Search Word - Data structure design.go
rename to leetcode/0211.Design-Add-and-Search-Words-Data-Structure/211. Design Add and Search Words Data Structure.go
diff --git a/Algorithms/0211. Add and Search Word - Data structure design/211. Add and Search Word - Data structure design_test.go b/leetcode/0211.Design-Add-and-Search-Words-Data-Structure/211. Design Add and Search Words Data Structure_test.go
similarity index 100%
rename from Algorithms/0211. Add and Search Word - Data structure design/211. Add and Search Word - Data structure design_test.go
rename to leetcode/0211.Design-Add-and-Search-Words-Data-Structure/211. Design Add and Search Words Data Structure_test.go
diff --git a/leetcode/0211.Design-Add-and-Search-Words-Data-Structure/README.md b/leetcode/0211.Design-Add-and-Search-Words-Data-Structure/README.md
new file mode 100755
index 000000000..cc646b06d
--- /dev/null
+++ b/leetcode/0211.Design-Add-and-Search-Words-Data-Structure/README.md
@@ -0,0 +1,34 @@
+# [211. Design Add and Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/)
+
+
+## 题目
+
+Design a data structure that supports the following two operations:
+
+    void addWord(word)
+    bool search(word)
+
+search(word) can search a literal word or a regular expression string containing only letters `a-z` or `.`. A `.` means it can represent any one letter.
+
+**Example:**
+
+    addWord("bad")
+    addWord("dad")
+    addWord("mad")
+    search("pad") -> false
+    search("bad") -> true
+    search(".ad") -> true
+    search("b..") -> true
+
+**Note:**You may assume that all words are consist of lowercase letters `a-z`.
+
+## 题目大意
+
+设计一个支持以下两种操作的数据结构：`void addWord(word)`、`bool search(word)`。`search(word)` 可以搜索文字或正则表达式字符串，字符串只包含字母 . 或 a-z 。 "." 可以表示任何一个字母。
+
+
+
+## 解题思路
+
+- 设计一个 `WordDictionary` 的数据结构，要求具有 `addWord(word)` 和 `search(word)` 的操作，并且具有模糊查找的功能。
+- 这一题是第 208 题的加强版，在第 208 题经典的 Trie 上加上了模糊查找的功能。其他实现一模一样。
diff --git a/leetcode/0212.Word-Search-II/212. Word Search II.go b/leetcode/0212.Word-Search-II/212. Word Search II.go
new file mode 100644
index 000000000..bea7241aa
--- /dev/null
+++ b/leetcode/0212.Word-Search-II/212. Word Search II.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+func findWords(board [][]byte, words []string) []string {
+	res := []string{}
+	for _, v := range words {
+		if exist(board, v) {
+			res = append(res, v)
+		}
+	}
+	return res
+}
+
+// these is 79 solution
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func exist(board [][]byte, word string) bool {
+	visited := make([][]bool, len(board))
+	for i := 0; i < len(visited); i++ {
+		visited[i] = make([]bool, len(board[0]))
+	}
+	for i, v := range board {
+		for j := range v {
+			if searchWord(board, visited, word, 0, i, j) {
+				return true
+			}
+		}
+	}
+	return false
+}
+
+func isInBoard(board [][]byte, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+
+func searchWord(board [][]byte, visited [][]bool, word string, index, x, y int) bool {
+	if index == len(word)-1 {
+		return board[x][y] == word[index]
+	}
+	if board[x][y] == word[index] {
+		visited[x][y] = true
+		for i := 0; i < 4; i++ {
+			nx := x + dir[i][0]
+			ny := y + dir[i][1]
+			if isInBoard(board, nx, ny) && !visited[nx][ny] && searchWord(board, visited, word, index+1, nx, ny) {
+				return true
+			}
+		}
+		visited[x][y] = false
+	}
+	return false
+}
diff --git a/Algorithms/0212. Word Search II/212. Word Search II_test.go b/leetcode/0212.Word-Search-II/212. Word Search II_test.go
similarity index 76%
rename from Algorithms/0212. Word Search II/212. Word Search II_test.go
rename to leetcode/0212.Word-Search-II/212. Word Search II_test.go
index 9bcf342b6..36780401e 100644
--- a/Algorithms/0212. Word Search II/212. Word Search II_test.go	
+++ b/leetcode/0212.Word-Search-II/212. Word Search II_test.go	
@@ -27,21 +27,21 @@ func Test_Problem212(t *testing.T) {
 
 	qs := []question212{
 
-		question212{
+		{
 			para212{[][]byte{
-				[]byte{'A', 'B', 'C', 'E'},
-				[]byte{'S', 'F', 'C', 'S'},
-				[]byte{'A', 'D', 'E', 'E'},
+				{'A', 'B', 'C', 'E'},
+				{'S', 'F', 'C', 'S'},
+				{'A', 'D', 'E', 'E'},
 			}, []string{"ABCCED", "SEE", "ABCB"}},
 			ans212{[]string{"ABCCED", "SEE"}},
 		},
 
-		question212{
+		{
 			para212{[][]byte{
-				[]byte{'o', 'a', 'a', 'n'},
-				[]byte{'e', 't', 'a', 'e'},
-				[]byte{'i', 'h', 'k', 'r'},
-				[]byte{'i', 'f', 'l', 'v'},
+				{'o', 'a', 'a', 'n'},
+				{'e', 't', 'a', 'e'},
+				{'i', 'h', 'k', 'r'},
+				{'i', 'f', 'l', 'v'},
 			}, []string{"oath", "pea", "eat", "rain"}},
 			ans212{[]string{"oath", "eat"}},
 		},
diff --git a/Algorithms/0212. Word Search II/README.md b/leetcode/0212.Word-Search-II/README.md
similarity index 100%
rename from Algorithms/0212. Word Search II/README.md
rename to leetcode/0212.Word-Search-II/README.md
diff --git a/Algorithms/0213. House Robber II/213. House Robber II.go b/leetcode/0213.House-Robber-II/213. House Robber II.go
similarity index 89%
rename from Algorithms/0213. House Robber II/213. House Robber II.go
rename to leetcode/0213.House-Robber-II/213. House Robber II.go
index fe1bf32d6..a20a7e105 100644
--- a/Algorithms/0213. House Robber II/213. House Robber II.go	
+++ b/leetcode/0213.House-Robber-II/213. House Robber II.go	
@@ -25,3 +25,10 @@ func rob213_1(nums []int, start, end int) int {
 	}
 	return curMax
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0213. House Robber II/213. House Robber II_test.go b/leetcode/0213.House-Robber-II/213. House Robber II_test.go
similarity index 93%
rename from Algorithms/0213. House Robber II/213. House Robber II_test.go
rename to leetcode/0213.House-Robber-II/213. House Robber II_test.go
index c2faf9277..9c09e9e4f 100644
--- a/Algorithms/0213. House Robber II/213. House Robber II_test.go	
+++ b/leetcode/0213.House-Robber-II/213. House Robber II_test.go	
@@ -26,16 +26,16 @@ func Test_Problem213(t *testing.T) {
 
 	qs := []question213{
 
-		question213{
+		{
 			para213{[]int{0, 0}},
 			ans213{0},
 		},
 
-		question213{
+		{
 			para213{[]int{2, 3, 2}},
 			ans213{3},
 		},
-		question213{
+		{
 			para213{[]int{1, 2, 3, 1}},
 			ans213{4},
 		},
diff --git a/Algorithms/0213. House Robber II/README.md b/leetcode/0213.House-Robber-II/README.md
similarity index 100%
rename from Algorithms/0213. House Robber II/README.md
rename to leetcode/0213.House-Robber-II/README.md
diff --git a/leetcode/0215.Kth-Largest-Element-in-an-Array/215. Kth Largest Element in an Array.go b/leetcode/0215.Kth-Largest-Element-in-an-Array/215. Kth Largest Element in an Array.go
new file mode 100644
index 000000000..c8e3a553b
--- /dev/null
+++ b/leetcode/0215.Kth-Largest-Element-in-an-Array/215. Kth Largest Element in an Array.go	
@@ -0,0 +1,73 @@
+package leetcode
+
+import (
+	"math/rand"
+	"sort"
+)
+
+// 解法一 排序，排序的方法反而速度是最快的
+func findKthLargest1(nums []int, k int) int {
+	sort.Ints(nums)
+	return nums[len(nums)-k]
+}
+
+// 解法二 这个方法的理论依据是 partition 得到的点的下标就是最终排序之后的下标，根据这个下标，我们可以判断第 K 大的数在哪里
+// 时间复杂度 O(n)，空间复杂度 O(log n)，最坏时间复杂度为 O(n^2)，空间复杂度 O(n)
+func findKthLargest(nums []int, k int) int {
+	m := len(nums) - k + 1 // mth smallest, from 1..len(nums)
+	return selectSmallest(nums, 0, len(nums)-1, m)
+}
+
+func selectSmallest(nums []int, l, r, i int) int {
+	if l >= r {
+		return nums[l]
+	}
+	q := partition(nums, l, r)
+	k := q - l + 1
+	if k == i {
+		return nums[q]
+	}
+	if i < k {
+		return selectSmallest(nums, l, q-1, i)
+	} else {
+		return selectSmallest(nums, q+1, r, i-k)
+	}
+}
+
+func partition(nums []int, l, r int) int {
+	k := l + rand.Intn(r-l+1) // 此处为优化，使得时间复杂度期望降为 O(n)，最坏时间复杂度为 O(n^2)
+	nums[k], nums[r] = nums[r], nums[k]
+	i := l - 1
+	// nums[l..i] <= nums[r]
+	// nums[i+1..j-1] > nums[r]
+	for j := l; j < r; j++ {
+		if nums[j] <= nums[r] {
+			i++
+			nums[i], nums[j] = nums[j], nums[i]
+		}
+	}
+	nums[i+1], nums[r] = nums[r], nums[i+1]
+	return i + 1
+}
+
+// 扩展题 剑指 Offer 40. 最小的 k 个数
+func getLeastNumbers(arr []int, k int) []int {
+	return selectSmallest1(arr, 0, len(arr)-1, k)[:k]
+}
+
+// 和 selectSmallest 实现完全一致，只是返回值不用再截取了，直接返回 nums 即可
+func selectSmallest1(nums []int, l, r, i int) []int {
+	if l >= r {
+		return nums
+	}
+	q := partition(nums, l, r)
+	k := q - l + 1
+	if k == i {
+		return nums
+	}
+	if i < k {
+		return selectSmallest1(nums, l, q-1, i)
+	} else {
+		return selectSmallest1(nums, q+1, r, i-k)
+	}
+}
diff --git a/Algorithms/0215. Kth Largest Element in an Array/215. Kth Largest Element in an Array_test.go b/leetcode/0215.Kth-Largest-Element-in-an-Array/215. Kth Largest Element in an Array_test.go
similarity index 92%
rename from Algorithms/0215. Kth Largest Element in an Array/215. Kth Largest Element in an Array_test.go
rename to leetcode/0215.Kth-Largest-Element-in-an-Array/215. Kth Largest Element in an Array_test.go
index e66bb9537..54a070a43 100644
--- a/Algorithms/0215. Kth Largest Element in an Array/215. Kth Largest Element in an Array_test.go	
+++ b/leetcode/0215.Kth-Largest-Element-in-an-Array/215. Kth Largest Element in an Array_test.go	
@@ -27,27 +27,32 @@ func Test_Problem215(t *testing.T) {
 
 	qs := []question215{
 
-		question215{
+		{
+			para215{[]int{3, 2, 1}, 2},
+			ans215{2},
+		},
+
+		{
 			para215{[]int{3, 2, 1, 5, 6, 4}, 2},
 			ans215{5},
 		},
 
-		question215{
+		{
 			para215{[]int{3, 2, 3, 1, 2, 4, 5, 5, 6}, 4},
 			ans215{4},
 		},
 
-		question215{
+		{
 			para215{[]int{0, 0, 0, 0, 0}, 2},
 			ans215{0},
 		},
 
-		question215{
+		{
 			para215{[]int{1}, 1},
 			ans215{1},
 		},
 
-		question215{
+		{
 			para215{[]int{3, 2, 3, 1, 2, 4, 5, 5, 6, 7, 7, 8, 2, 3, 1, 1, 1, 10, 11, 5, 6, 2, 4, 7, 8, 5, 6}, 20},
 			ans215{2},
 		},
diff --git a/leetcode/0215.Kth-Largest-Element-in-an-Array/README.md b/leetcode/0215.Kth-Largest-Element-in-an-Array/README.md
new file mode 100644
index 000000000..b3a2d21f3
--- /dev/null
+++ b/leetcode/0215.Kth-Largest-Element-in-an-Array/README.md
@@ -0,0 +1,33 @@
+# [215. Kth Largest Element in an Array](https://leetcode.com/problems/kth-largest-element-in-an-array/)
+
+## 题目
+
+Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
+
+Example 1:
+
+```c
+Input: [3,2,1,5,6,4] and k = 2
+Output: 5
+```
+
+Example 2:
+
+```c
+Input: [3,2,3,1,2,4,5,5,6] and k = 4
+Output: 4
+```
+
+Note:     
+
+You may assume k is always valid, 1 ≤ k ≤ array's length.
+
+
+## 题目大意
+
+找出数组中第 K 大的元素。这一题非常经典。可以用 O(n) 的时间复杂度实现。
+
+## 解题思路
+
+在快速选择 quickselect 的 partition 操作中，每次 partition 操作结束都会返回一个点，这个标定点的下标和最终排序之后有序数组中这个元素所在的下标是一致的。利用这个特性，我们可以不断的划分数组区间，最终找到第 K 大的元素。执行一次 partition 操作以后，如果这个元素的下标比 K 小，那么接着就在后边的区间继续执行 partition 操作；如果这个元素的下标比 K 大，那么就在左边的区间继续执行 partition 操作；如果相等就直接输出这个下标对应的数组元素即可。
+
diff --git a/Algorithms/0216. Combination Sum III/216. Combination Sum III.go b/leetcode/0216.Combination-Sum-III/216. Combination Sum III.go
similarity index 100%
rename from Algorithms/0216. Combination Sum III/216. Combination Sum III.go
rename to leetcode/0216.Combination-Sum-III/216. Combination Sum III.go
diff --git a/Algorithms/0216. Combination Sum III/216. Combination Sum III_test.go b/leetcode/0216.Combination-Sum-III/216. Combination Sum III_test.go
similarity index 82%
rename from Algorithms/0216. Combination Sum III/216. Combination Sum III_test.go
rename to leetcode/0216.Combination-Sum-III/216. Combination Sum III_test.go
index b58915974..c4bcaad6a 100644
--- a/Algorithms/0216. Combination Sum III/216. Combination Sum III_test.go	
+++ b/leetcode/0216.Combination-Sum-III/216. Combination Sum III_test.go	
@@ -27,13 +27,13 @@ func Test_Problem216(t *testing.T) {
 
 	qs := []question216{
 
-		question216{
+		{
 			para216{3, 7},
-			ans216{[][]int{[]int{1, 2, 4}}},
+			ans216{[][]int{{1, 2, 4}}},
 		},
-		question216{
+		{
 			para216{3, 9},
-			ans216{[][]int{[]int{1, 2, 6}, []int{1, 3, 5}, []int{2, 3, 4}}},
+			ans216{[][]int{{1, 2, 6}, {1, 3, 5}, {2, 3, 4}}},
 		},
 	}
 
diff --git a/Algorithms/0216. Combination Sum III/README.md b/leetcode/0216.Combination-Sum-III/README.md
similarity index 100%
rename from Algorithms/0216. Combination Sum III/README.md
rename to leetcode/0216.Combination-Sum-III/README.md
diff --git a/Algorithms/0217. Contains Duplicate/217. Contains Duplicate.go b/leetcode/0217.Contains-Duplicate/217. Contains Duplicate.go
similarity index 100%
rename from Algorithms/0217. Contains Duplicate/217. Contains Duplicate.go
rename to leetcode/0217.Contains-Duplicate/217. Contains Duplicate.go
diff --git a/Algorithms/0217. Contains Duplicate/217. Contains Duplicate_test.go b/leetcode/0217.Contains-Duplicate/217. Contains Duplicate_test.go
similarity index 94%
rename from Algorithms/0217. Contains Duplicate/217. Contains Duplicate_test.go
rename to leetcode/0217.Contains-Duplicate/217. Contains Duplicate_test.go
index 270d00411..32542d576 100644
--- a/Algorithms/0217. Contains Duplicate/217. Contains Duplicate_test.go	
+++ b/leetcode/0217.Contains-Duplicate/217. Contains Duplicate_test.go	
@@ -26,17 +26,17 @@ func Test_Problem217(t *testing.T) {
 
 	qs := []question217{
 
-		question217{
+		{
 			para217{[]int{1, 2, 3, 1}},
 			ans217{true},
 		},
 
-		question217{
+		{
 			para217{[]int{1, 2, 3, 4}},
 			ans217{false},
 		},
 
-		question217{
+		{
 			para217{[]int{1, 1, 1, 3, 3, 4, 3, 2, 4, 2}},
 			ans217{true},
 		},
diff --git a/Algorithms/0217. Contains Duplicate/README.md b/leetcode/0217.Contains-Duplicate/README.md
similarity index 100%
rename from Algorithms/0217. Contains Duplicate/README.md
rename to leetcode/0217.Contains-Duplicate/README.md
diff --git a/leetcode/0218.The-Skyline-Problem/218. The Skyline Problem.go b/leetcode/0218.The-Skyline-Problem/218. The Skyline Problem.go
new file mode 100644
index 000000000..ad13dc245
--- /dev/null
+++ b/leetcode/0218.The-Skyline-Problem/218. The Skyline Problem.go	
@@ -0,0 +1,291 @@
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+// 解法一 树状数组，时间复杂度 O(n log n)
+const LEFTSIDE = 1
+const RIGHTSIDE = 2
+
+type Point struct {
+	xAxis int
+	side  int
+	index int
+}
+
+func getSkyline(buildings [][]int) [][]int {
+	res := [][]int{}
+	if len(buildings) == 0 {
+		return res
+	}
+	allPoints, bit := make([]Point, 0), BinaryIndexedTree{}
+	// [x-axis (value), [1 (left) | 2 (right)], index (building number)]
+	for i, b := range buildings {
+		allPoints = append(allPoints, Point{xAxis: b[0], side: LEFTSIDE, index: i})
+		allPoints = append(allPoints, Point{xAxis: b[1], side: RIGHTSIDE, index: i})
+	}
+	sort.Slice(allPoints, func(i, j int) bool {
+		if allPoints[i].xAxis == allPoints[j].xAxis {
+			return allPoints[i].side < allPoints[j].side
+		}
+		return allPoints[i].xAxis < allPoints[j].xAxis
+	})
+	bit.Init(len(allPoints))
+	kth := make(map[Point]int)
+	for i := 0; i < len(allPoints); i++ {
+		kth[allPoints[i]] = i
+	}
+	for i := 0; i < len(allPoints); i++ {
+		pt := allPoints[i]
+		if pt.side == LEFTSIDE {
+			bit.Add(kth[Point{xAxis: buildings[pt.index][1], side: RIGHTSIDE, index: pt.index}], buildings[pt.index][2])
+		}
+		currHeight := bit.Query(kth[pt] + 1)
+		if len(res) == 0 || res[len(res)-1][1] != currHeight {
+			if len(res) > 0 && res[len(res)-1][0] == pt.xAxis {
+				res[len(res)-1][1] = currHeight
+			} else {
+				res = append(res, []int{pt.xAxis, currHeight})
+			}
+		}
+	}
+	return res
+}
+
+type BinaryIndexedTree struct {
+	tree     []int
+	capacity int
+}
+
+// Init define
+func (bit *BinaryIndexedTree) Init(capacity int) {
+	bit.tree, bit.capacity = make([]int, capacity+1), capacity
+}
+
+// Add define
+func (bit *BinaryIndexedTree) Add(index int, val int) {
+	for ; index > 0; index -= index & -index {
+		bit.tree[index] = max(bit.tree[index], val)
+	}
+}
+
+// Query define
+func (bit *BinaryIndexedTree) Query(index int) int {
+	sum := 0
+	for ; index <= bit.capacity; index += index & -index {
+		sum = max(sum, bit.tree[index])
+	}
+	return sum
+}
+
+// 解法二 线段树 Segment Tree，时间复杂度 O(n log n)
+func getSkyline1(buildings [][]int) [][]int {
+	st, ans, lastHeight, check := template.SegmentTree{}, [][]int{}, 0, false
+	posMap, pos := discretization218(buildings)
+	tmp := make([]int, len(posMap))
+	st.Init(tmp, func(i, j int) int {
+		return max(i, j)
+	})
+	for _, b := range buildings {
+		st.UpdateLazy(posMap[b[0]], posMap[b[1]-1], b[2])
+	}
+	for i := 0; i < len(pos); i++ {
+		h := st.QueryLazy(posMap[pos[i]], posMap[pos[i]])
+		if check == false && h != 0 {
+			ans = append(ans, []int{pos[i], h})
+			check = true
+		} else if i > 0 && h != lastHeight {
+			ans = append(ans, []int{pos[i], h})
+		}
+		lastHeight = h
+	}
+	return ans
+}
+
+func discretization218(positions [][]int) (map[int]int, []int) {
+	tmpMap, posArray, posMap := map[int]int{}, []int{}, map[int]int{}
+	for _, pos := range positions {
+		tmpMap[pos[0]]++
+		tmpMap[pos[1]-1]++
+		tmpMap[pos[1]]++
+	}
+	for k := range tmpMap {
+		posArray = append(posArray, k)
+	}
+	sort.Ints(posArray)
+	for i, pos := range posArray {
+		posMap[pos] = i
+	}
+	return posMap, posArray
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法三 扫描线 Sweep Line，时间复杂度 O(n log n)
+func getSkyline2(buildings [][]int) [][]int {
+	size := len(buildings)
+	es := make([]E, 0)
+	for i, b := range buildings {
+		l := b[0]
+		r := b[1]
+		h := b[2]
+		// 1-- enter
+		el := NewE(i, l, h, 0)
+		es = append(es, el)
+		// 0 -- leave
+		er := NewE(i, r, h, 1)
+		es = append(es, er)
+	}
+	skyline := make([][]int, 0)
+	sort.Slice(es, func(i, j int) bool {
+		if es[i].X == es[j].X {
+			if es[i].T == es[j].T {
+				if es[i].T == 0 {
+					return es[i].H > es[j].H
+				}
+				return es[i].H < es[j].H
+			}
+			return es[i].T < es[j].T
+		}
+		return es[i].X < es[j].X
+	})
+	pq := NewIndexMaxPQ(size)
+	for _, e := range es {
+		curH := pq.Front()
+		if e.T == 0 {
+			if e.H > curH {
+				skyline = append(skyline, []int{e.X, e.H})
+			}
+			pq.Enque(e.N, e.H)
+		} else {
+			pq.Remove(e.N)
+			h := pq.Front()
+			if curH > h {
+				skyline = append(skyline, []int{e.X, h})
+			}
+		}
+	}
+	return skyline
+}
+
+// 扫面线伪代码
+// events = {{x: L , height: H , type: entering},
+// 		  {x: R , height: H , type: leaving}}
+// event.SortByX()
+// ds = new DS()
+
+// for e in events:
+// 	if entering(e):
+// 		if e.height > ds.max(): ans += [e.height]
+// 		ds.add(e.height)
+// 	if leaving(e):
+// 		ds.remove(e.height)
+// 		if e.height > ds.max(): ans += [ds.max()]
+
+// E define
+type E struct { // 定义一个 event 事件
+	N int // number 编号
+	X int // x 坐标
+	H int // height 高度
+	T int // type  0-进入 1-离开
+}
+
+// NewE define
+func NewE(n, x, h, t int) E {
+	return E{
+		N: n,
+		X: x,
+		H: h,
+		T: t,
+	}
+}
+
+// IndexMaxPQ define
+type IndexMaxPQ struct {
+	items []int
+	pq    []int
+	qp    []int
+	total int
+}
+
+// NewIndexMaxPQ define
+func NewIndexMaxPQ(n int) IndexMaxPQ {
+	qp := make([]int, n)
+	for i := 0; i < n; i++ {
+		qp[i] = -1
+	}
+	return IndexMaxPQ{
+		items: make([]int, n),
+		pq:    make([]int, n+1),
+		qp:    qp,
+	}
+}
+
+// Enque define
+func (q *IndexMaxPQ) Enque(key, val int) {
+	q.total++
+	q.items[key] = val
+	q.pq[q.total] = key
+	q.qp[key] = q.total
+	q.swim(q.total)
+}
+
+// Front define
+func (q *IndexMaxPQ) Front() int {
+	if q.total < 1 {
+		return 0
+	}
+	return q.items[q.pq[1]]
+}
+
+// Remove define
+func (q *IndexMaxPQ) Remove(key int) {
+	rank := q.qp[key]
+	q.exch(rank, q.total)
+	q.total--
+	q.qp[key] = -1
+	q.sink(rank)
+}
+
+func (q *IndexMaxPQ) sink(n int) {
+	for 2*n <= q.total {
+		k := 2 * n
+		if k < q.total && q.less(k, k+1) {
+			k++
+		}
+		if q.less(k, n) {
+			break
+		}
+		q.exch(k, n)
+		n = k
+	}
+}
+
+func (q *IndexMaxPQ) swim(n int) {
+	for n > 1 {
+		k := n / 2
+		if q.less(n, k) {
+			break
+		}
+		q.exch(n, k)
+		n = k
+	}
+}
+
+func (q *IndexMaxPQ) exch(i, j int) {
+	q.pq[i], q.pq[j] = q.pq[j], q.pq[i]
+	q.qp[q.pq[i]] = i
+	q.qp[q.pq[j]] = j
+}
+
+func (q *IndexMaxPQ) less(i, j int) bool {
+	return q.items[q.pq[i]] < q.items[q.pq[j]]
+}
diff --git a/leetcode/0218.The-Skyline-Problem/218. The Skyline Problem_test.go b/leetcode/0218.The-Skyline-Problem/218. The Skyline Problem_test.go
new file mode 100644
index 000000000..62e09dfab
--- /dev/null
+++ b/leetcode/0218.The-Skyline-Problem/218. The Skyline Problem_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question218 struct {
+	para218
+	ans218
+}
+
+// para 是参数
+// one 代表第一个参数
+type para218 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans218 struct {
+	one [][]int
+}
+
+func Test_Problem218(t *testing.T) {
+
+	qs := []question218{
+
+		{
+			para218{[][]int{{2, 9, 10}, {3, 7, 15}, {5, 12, 12}, {15, 20, 10}, {19, 24, 8}}},
+			ans218{[][]int{{2, 10}, {3, 15}, {7, 12}, {12, 0}, {15, 10}, {20, 8}, {24, 0}}},
+		},
+
+		{
+			para218{[][]int{{1, 2, 1}, {1, 2, 2}, {1, 2, 3}, {2, 3, 1}, {2, 3, 2}, {2, 3, 3}}},
+			ans218{[][]int{{1, 3}, {3, 0}}},
+		},
+
+		{
+			para218{[][]int{{4, 9, 10}, {4, 9, 15}, {4, 9, 12}, {10, 12, 10}, {10, 12, 8}}},
+			ans218{[][]int{{4, 15}, {9, 0}, {10, 10}, {12, 0}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 218------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans218, q.para218
+		fmt.Printf("【input】:%v       【output】:%v\n", p, getSkyline(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0218.The-Skyline-Problem/README.md b/leetcode/0218.The-Skyline-Problem/README.md
new file mode 100755
index 000000000..f4b2e5ba3
--- /dev/null
+++ b/leetcode/0218.The-Skyline-Problem/README.md
@@ -0,0 +1,70 @@
+# [218. The Skyline Problem](https://leetcode.com/problems/the-skyline-problem/)
+
+## 题目
+
+A city's skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Now suppose you are **given the locations and height of all the buildings** as shown on a cityscape photo (Figure A), write a program to **output the skyline** formed by these buildings collectively (Figure B).
+
+![](https://img.halfrost.com/Leetcode/leetcode_218_0.png)
+
+![](https://img.halfrost.com/Leetcode/leetcode_218_1.png)
+
+The geometric information of each building is represented by a triplet of integers `[Li, Ri, Hi]`, where `Li` and `Ri` are the x coordinates of the left and right edge of the ith building, respectively, and `Hi` is its height. It is guaranteed that `0 ≤ Li, Ri ≤ INT_MAX`, `0 < Hi ≤ INT_MAX`, and `Ri - Li > 0`. You may assume all buildings are perfect rectangles grounded on an absolutely flat surface at height 0.
+
+For instance, the dimensions of all buildings in Figure A are recorded as: `[ [2 9 10], [3 7 15], [5 12 12], [15 20 10], [19 24 8] ]` .
+
+The output is a list of "**key points**" (red dots in Figure B) in the format of `[ [x1,y1], [x2, y2], [x3, y3], ... ]` that uniquely defines a skyline. **A key point is the left endpoint of a horizontal line segment**. Note that the last key point, where the rightmost building ends, is merely used to mark the termination of the skyline, and always has zero height. Also, the ground in between any two adjacent buildings should be considered part of the skyline contour.
+
+For instance, the skyline in Figure B should be represented as:`[ [2 10], [3 15], [7 12], [12 0], [15 10], [20 8], [24, 0] ]`.
+
+**Notes:**
+
+- The number of buildings in any input list is guaranteed to be in the range `[0, 10000]`.
+- The input list is already sorted in ascending order by the left x position `Li`.
+- The output list must be sorted by the x position.
+- There must be no consecutive horizontal lines of equal height in the output skyline. For instance, `[...[2 3], [4 5], [7 5], [11 5], [12 7]...]` is not acceptable; the three lines of height 5 should be merged into one in the final output as such: `[...[2 3], [4 5], [12 7], ...]`
+
+
+## 题目大意
+
+城市的天际线是从远处观看该城市中所有建筑物形成的轮廓的外部轮廓。现在，假设您获得了城市风光照片（图A）上显示的所有建筑物的位置和高度，请编写一个程序以输出由这些建筑物形成的天际线（图B）。
+
+每个建筑物的几何信息用三元组 [Li，Ri，Hi] 表示，其中 Li 和 Ri 分别是第 i 座建筑物左右边缘的 x 坐标，Hi 是其高度。可以保证 0 ≤ Li, Ri ≤ INT\_MAX, 0 < Hi ≤ INT\_MAX 和 Ri - Li > 0。您可以假设所有建筑物都是在绝对平坦且高度为 0 的表面上的完美矩形。
+
+例如，图 A 中所有建筑物的尺寸记录为：[ [2 9 10], [3 7 15], [5 12 12], [15 20 10], [19 24 8] ] 。
+
+输出是以 [ [x1,y1], [x2, y2], [x3, y3], ... ] 格式的“关键点”（图 B 中的红点）的列表，它们唯一地定义了天际线。关键点是水平线段的左端点。请注意，最右侧建筑物的最后一个关键点仅用于标记天际线的终点，并始终为零高度。此外，任何两个相邻建筑物之间的地面都应被视为天际线轮廓的一部分。
+
+例如，图 B 中的天际线应该表示为：[ [2 10], [3 15], [7 12], [12 0], [15 10], [20 8], [24, 0] ]。
+
+说明:
+
+- 任何输入列表中的建筑物数量保证在 [0, 10000] 范围内。
+- 输入列表已经按左 x 坐标 Li 进行升序排列。
+- 输出列表必须按 x 位排序。
+- 输出天际线中不得有连续的相同高度的水平线。例如 [...[2 3], [4 5], [7 5], [11 5], [12 7]...] 是不正确的答案；三条高度为 5 的线应该在最终输出中合并为一个：[...[2 3], [4 5], [12 7], ...]
+
+
+## 解题思路
+
+
+- 给出一个二维数组，每个子数组里面代表一个高楼的信息，一个高楼的信息包含 3 个信息，高楼起始坐标，高楼终止坐标，高楼高度。要求找到这些高楼的边际点，并输出这些边际点的高度信息。
+- 这一题可以用线段树来解。用线段树来解答，可以不用关心“楼挡住楼”的情况。由于楼的坐标是离散的，所以先把楼在 X 轴上两个坐标离散化。同第 699 题一样，楼的宽度是一个区间，但是离散的过程中，楼的宽度右边界需要减一，不然查询一个区间会包含两个点，导致错误的结果，例如，第一个楼是 [1,3)，楼高 10，第二个楼是 [3,6)，楼高 20 。第一个楼如果算上右边界 3，查询 [1,3] 的结果是 20，因为 [3,3] 这个点会查询到第二个楼上面去。所以每个楼的右边界应该减一。但是每个楼的右边界也要加入到 query 中，因为最终 query 的结果需要包含这些边界。将离散的数据排序以后，按照楼的信息，每个区间依次 update。最后统计的时候依次统计每个区间，如果当前区间的高度和前一个区间的高度一样，就算是等高的楼。当高度与前一个高度不相同的时候就算是天际线的边缘，就要添加到最后输出数组中。
+- 类似的线段树的题目有：第 715 题，第 732 题，第 699 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和第 699 题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
+- 这一题用线段树做时间复杂度有点高，可以用扫描线解题。扫描线的思路很简单，用一根根垂直于 X 轴的竖线，从最左边依次扫到最右边，扫描每一条大楼的边界，当进入大楼的左边界的时候，如果没有比这个左边界最高点更高的点，就记录下这个最高点 keyPoint，状态是进入状态。如果扫到一个大楼的左边界，有比它更高的高度，就不记录，因为它不是天际线，它被楼挡楼，挡在其他楼后面了。当扫到一个大楼的右边界的时候，如果是最高点，那么记录下它的状态是离开状态，此时还需要记录下第二高的点。在扫描线扫描的过程中，动态的维护大楼的高度，同时维护最高的高度和第二高的高度。其实只需要维护最高的高度这一个高度，因为当离开状态到来的时候，移除掉当前最高的，剩下的高度里面最高的就是第二高的高度。描述的伪代码如下：
+
+        // 扫描线伪代码
+        events = {{x: L , height: H , type: entering},
+        		  {x: R , height: H , type: leaving}}
+        event.SortByX()
+        ds = new DS()
+        
+        for e in events:
+        	if entering(e):
+        		if e.height > ds.max(): ans += [e.height]
+        		ds.add(e.height)
+        	if leaving(e):
+        		ds.remove(e.height)
+        		if e.height > ds.max(): ans += [ds.max()]
+
+- 动态插入，查找最大值可以选用的数据结构有，最大堆和二叉搜索树。最大堆找最大值 O(1)，插入 O(log n)，但是 remove_by_key 需要 O(n) 的时间复杂度，并且需要自己实现。二叉搜索树，查找 max，添加和删除元素都是 O(log n) 的时间复杂度。
+- 排序的时候也需要注意几个问题：如果大楼的边界相等，并且是进入状态，那么再按照高度从大到小进行排序；如果大楼的边界相等，并且是离开状态，那么高度按照从小到大进行排序。
diff --git a/Algorithms/0219. Contains Duplicate II/219. Contains Duplicate II.go b/leetcode/0219.Contains-Duplicate-II/219. Contains Duplicate II.go
similarity index 100%
rename from Algorithms/0219. Contains Duplicate II/219. Contains Duplicate II.go
rename to leetcode/0219.Contains-Duplicate-II/219. Contains Duplicate II.go
diff --git a/Algorithms/0219. Contains Duplicate II/219. Contains Duplicate II_test.go b/leetcode/0219.Contains-Duplicate-II/219. Contains Duplicate II_test.go
similarity index 94%
rename from Algorithms/0219. Contains Duplicate II/219. Contains Duplicate II_test.go
rename to leetcode/0219.Contains-Duplicate-II/219. Contains Duplicate II_test.go
index e666345bf..39bb6a6c3 100644
--- a/Algorithms/0219. Contains Duplicate II/219. Contains Duplicate II_test.go	
+++ b/leetcode/0219.Contains-Duplicate-II/219. Contains Duplicate II_test.go	
@@ -27,17 +27,17 @@ func Test_Problem219(t *testing.T) {
 
 	qs := []question219{
 
-		question219{
+		{
 			para219{[]int{1, 2, 3, 1}, 3},
 			ans219{true},
 		},
 
-		question219{
+		{
 			para219{[]int{1, 0, 0, 1}, 1},
 			ans219{true},
 		},
 
-		question219{
+		{
 			para219{[]int{1, 2, 3, 1, 2, 3}, 2},
 			ans219{false},
 		},
diff --git a/Algorithms/0219. Contains Duplicate II/README.md b/leetcode/0219.Contains-Duplicate-II/README.md
similarity index 100%
rename from Algorithms/0219. Contains Duplicate II/README.md
rename to leetcode/0219.Contains-Duplicate-II/README.md
diff --git a/leetcode/0220.Contains-Duplicate-III/220. Contains Duplicate III.go b/leetcode/0220.Contains-Duplicate-III/220. Contains Duplicate III.go
new file mode 100644
index 000000000..3f635fafa
--- /dev/null
+++ b/leetcode/0220.Contains-Duplicate-III/220. Contains Duplicate III.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+// 解法一 桶排序
+func containsNearbyAlmostDuplicate(nums []int, k int, t int) bool {
+	if k <= 0 || t < 0 || len(nums) < 2 {
+		return false
+	}
+	buckets := map[int]int{}
+	for i := 0; i < len(nums); i++ {
+		// Get the ID of the bucket from element value nums[i] and bucket width t + 1
+		key := nums[i] / (t + 1)
+		// -7/9 = 0, but need -7/9 = -1
+		if nums[i] < 0 {
+			key--
+		}
+		if _, ok := buckets[key]; ok {
+			return true
+		}
+		// check the lower bucket, and have to check the value too
+		if v, ok := buckets[key-1]; ok && nums[i]-v <= t {
+			return true
+		}
+		// check the upper bucket, and have to check the value too
+		if v, ok := buckets[key+1]; ok && v-nums[i] <= t {
+			return true
+		}
+		// maintain k size of window
+		if len(buckets) >= k {
+			delete(buckets, nums[i-k]/(t+1))
+		}
+		buckets[key] = nums[i]
+	}
+	return false
+}
+
+// 解法二 滑动窗口 + 剪枝
+func containsNearbyAlmostDuplicate1(nums []int, k int, t int) bool {
+	if len(nums) <= 1 {
+		return false
+	}
+	if k <= 0 {
+		return false
+	}
+	n := len(nums)
+	for i := 0; i < n; i++ {
+		count := 0
+		for j := i + 1; j < n && count < k; j++ {
+			if abs(nums[i]-nums[j]) <= t {
+				return true
+			}
+			count++
+		}
+	}
+	return false
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
diff --git a/Algorithms/0220. Contains Duplicate III/220. Contains Duplicate III_test.go b/leetcode/0220.Contains-Duplicate-III/220. Contains Duplicate III_test.go
similarity index 91%
rename from Algorithms/0220. Contains Duplicate III/220. Contains Duplicate III_test.go
rename to leetcode/0220.Contains-Duplicate-III/220. Contains Duplicate III_test.go
index 18761e0d7..3f0aedd0a 100644
--- a/Algorithms/0220. Contains Duplicate III/220. Contains Duplicate III_test.go	
+++ b/leetcode/0220.Contains-Duplicate-III/220. Contains Duplicate III_test.go	
@@ -28,30 +28,35 @@ func Test_Problem220(t *testing.T) {
 
 	qs := []question220{
 
-		question220{
+		{
 			para220{[]int{7, 1, 3}, 2, 3},
 			ans220{true},
 		},
 
-		question220{
+		{
 			para220{[]int{-1, -1}, 1, -1},
 			ans220{false},
 		},
 
-		question220{
+		{
 			para220{[]int{1, 2, 3, 1}, 3, 0},
 			ans220{true},
 		},
 
-		question220{
+		{
 			para220{[]int{1, 0, 1, 1}, 1, 2},
 			ans220{true},
 		},
 
-		question220{
+		{
 			para220{[]int{1, 5, 9, 1, 5, 9}, 2, 3},
 			ans220{false},
 		},
+
+		{
+			para220{[]int{1, 2, 1, 1}, 1, 0},
+			ans220{true},
+		},
 	}
 
 	fmt.Printf("------------------------Leetcode Problem 220------------------------\n")
diff --git a/leetcode/0220.Contains-Duplicate-III/README.md b/leetcode/0220.Contains-Duplicate-III/README.md
new file mode 100644
index 000000000..591330e91
--- /dev/null
+++ b/leetcode/0220.Contains-Duplicate-III/README.md
@@ -0,0 +1,35 @@
+# [220. Contains Duplicate III](https://leetcode.com/problems/contains-duplicate-iii/)
+
+## 题目
+
+Given an array of integers, find out whether there are two distinct indices i and j in the array such that the **absolute** difference between **nums[i]** and **nums[j]** is at most t and the **absolute** difference between i and j is at most k.
+
+**Example 1:**
+
+    Input: nums = [1,2,3,1], k = 3, t = 0
+    Output: true
+
+**Example 2:**
+
+    Input: nums = [1,0,1,1], k = 1, t = 2
+    Output: true
+
+**Example 3:**
+
+    Input: nums = [1,5,9,1,5,9], k = 2, t = 3
+    Output: false
+
+
+
+
+## 题目大意
+
+给出一个数组 num，再给 K 和 t。问在 num 中能否找到一组 i 和 j，使得 num[i] 和 num[j] 的绝对差值最大为 t，并且 i 和 j 之前的绝对差值最大为 k。
+
+## 解题思路
+
+
+- 给出一个数组，要求在数组里面找到 2 个索引，`i` 和 `j`，使得 `| nums[i] - nums[j] | ≤ t` ，并且 `| i - j | ≤  k` 。
+- 这是一道滑动窗口的题目。第一想法就是用 `i` 和 `j` 两个指针，针对每个 `i` ，都从 `i + 1` 往后扫完整个数组，判断每个 `i` 和 `j` ，判断是否满足题意。`j` 在循环的过程中注意判断剪枝条件 `| i - j | ≤  k`。这个做法的时间复杂度是 O(n^2)。这个做法慢的原因在于滑动窗口的左边界和右边界在滑动过程中不是联动滑动的。
+- 于是考虑，如果数组是有序的呢？把数组按照元素值从小到大进行排序，如果元素值相等，就按照 index 从小到大进行排序。在这样有序的数组中找满足题意的 `i` 和 `j`，滑动窗口左边界和右边界就是联动的了。窗口的右边界滑到与左边界元素值的差值 ≤ t 的地方，满足了这个条件再判断 `| i - j | ≤  k`，如果右边界与左边界元素值的差值 > t 了，说明该把左边界往右移动了(能这样移动的原因就是因为我们将数组元素大小排序了，右移是增大元素的方向)。移动左边界的时候需要注意左边界不能超过右边界。这样滑动窗口一次滑过整个排序后的数组，就可以判断是否存在满足题意的 `i` 和 `j` 。这个做法的时间主要花在排序上了，时间复杂度是 O(n log n)。
+- 本题最优解是利用桶排序的思想。`| i - j | ≤  k` 这个条件利用一个窗口大小为 k 来维护。重点在 `| nums[i] - nums[j] | ≤ t` 这个条件如何满足。利用桶排序的思想，将 `nums[i]` 所有元素分为 ...,`[0,t]`,`[t+1,2t+1]`,...。每个区间的大小为 `t + 1`。每个元素现在都对应一个桶编号。进行 3 次查找即可确定能否找到满足这个 `| nums[i] - nums[j] | ≤ t` 条件的数对。如果在相同的桶中找到了元素，那么说明能找到这样的 i 和 j。还有 2 种可能对应桶边界的情况。如果存在前一个桶中的元素能使得相差的值也 `≤ t`，这样的数对同样满足题意。最后一种情况是，如果存在后一个桶中的元素能使得相差的值也 `≤ t`，这样的数对同样满足题意。查询 3 次，如果都不存在，说明当前的 i 找不到满足题意的 j。继续循环寻找。循环一遍都找不到满足题意的数对，输出 false。
\ No newline at end of file
diff --git a/Algorithms/0222. Count Complete Tree Nodes/222. Count Complete Tree Nodes.go b/leetcode/0222.Count-Complete-Tree-Nodes/222. Count Complete Tree Nodes.go
similarity index 85%
rename from Algorithms/0222. Count Complete Tree Nodes/222. Count Complete Tree Nodes.go
rename to leetcode/0222.Count-Complete-Tree-Nodes/222. Count Complete Tree Nodes.go
index 9eee619f4..32c4d4e14 100644
--- a/Algorithms/0222. Count Complete Tree Nodes/222. Count Complete Tree Nodes.go	
+++ b/leetcode/0222.Count-Complete-Tree-Nodes/222. Count Complete Tree Nodes.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func countNodes(root *TreeNode) int {
 	if root == nil {
 		return 0
diff --git a/Algorithms/0222. Count Complete Tree Nodes/222. Count Complete Tree Nodes_test.go b/leetcode/0222.Count-Complete-Tree-Nodes/222. Count Complete Tree Nodes_test.go
similarity index 87%
rename from Algorithms/0222. Count Complete Tree Nodes/222. Count Complete Tree Nodes_test.go
rename to leetcode/0222.Count-Complete-Tree-Nodes/222. Count Complete Tree Nodes_test.go
index 6ce347572..2c7b8b8e1 100644
--- a/Algorithms/0222. Count Complete Tree Nodes/222. Count Complete Tree Nodes_test.go	
+++ b/leetcode/0222.Count-Complete-Tree-Nodes/222. Count Complete Tree Nodes_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question222 struct {
@@ -26,22 +28,22 @@ func Test_Problem222(t *testing.T) {
 
 	qs := []question222{
 
-		question222{
+		{
 			para222{[]int{}},
 			ans222{0},
 		},
 
-		question222{
+		{
 			para222{[]int{1}},
 			ans222{1},
 		},
 
-		question222{
+		{
 			para222{[]int{1, 2, 3}},
 			ans222{3},
 		},
 
-		question222{
+		{
 			para222{[]int{1, 2, 3, 4, 5, 6}},
 			ans222{6},
 		},
@@ -52,7 +54,7 @@ func Test_Problem222(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans222, q.para222
 		fmt.Printf("【input】:%v      ", p)
-		rootOne := Ints2TreeNode(p.one)
+		rootOne := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", countNodes(rootOne))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0222. Count Complete Tree Nodes/README.md b/leetcode/0222.Count-Complete-Tree-Nodes/README.md
similarity index 100%
rename from Algorithms/0222. Count Complete Tree Nodes/README.md
rename to leetcode/0222.Count-Complete-Tree-Nodes/README.md
diff --git a/leetcode/0223.Rectangle-Area/223. Rectangle Area.go b/leetcode/0223.Rectangle-Area/223. Rectangle Area.go
new file mode 100644
index 000000000..8fd9dbdd8
--- /dev/null
+++ b/leetcode/0223.Rectangle-Area/223. Rectangle Area.go	
@@ -0,0 +1,28 @@
+package leetcode
+
+func computeArea(A int, B int, C int, D int, E int, F int, G int, H int) int {
+	X0, Y0, X1, Y1 := max(A, E), max(B, F), min(C, G), min(D, H)
+	return area(A, B, C, D) + area(E, F, G, H) - area(X0, Y0, X1, Y1)
+}
+
+func area(x0, y0, x1, y1 int) int {
+	l, h := x1-x0, y1-y0
+	if l <= 0 || h <= 0 {
+		return 0
+	}
+	return l * h
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/Algorithms/0223. Rectangle Area/223. Rectangle Area_test.go b/leetcode/0223.Rectangle-Area/223. Rectangle Area_test.go
similarity index 97%
rename from Algorithms/0223. Rectangle Area/223. Rectangle Area_test.go
rename to leetcode/0223.Rectangle-Area/223. Rectangle Area_test.go
index c7f8c7b9f..5b092d784 100644
--- a/Algorithms/0223. Rectangle Area/223. Rectangle Area_test.go	
+++ b/leetcode/0223.Rectangle-Area/223. Rectangle Area_test.go	
@@ -33,7 +33,7 @@ func Test_Problem223(t *testing.T) {
 
 	qs := []question223{
 
-		question223{
+		{
 			para223{-3, 0, 3, 4, 0, -1, 9, 2},
 			ans223{45},
 		},
diff --git a/leetcode/0223.Rectangle-Area/README.md b/leetcode/0223.Rectangle-Area/README.md
new file mode 100755
index 000000000..bbb4b061c
--- /dev/null
+++ b/leetcode/0223.Rectangle-Area/README.md
@@ -0,0 +1,31 @@
+# [223. Rectangle Area](https://leetcode.com/problems/rectangle-area/)
+
+
+## 题目
+
+Find the total area covered by two **rectilinear** rectangles in a **2D** plane.
+
+Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/22/rectangle_area.png)
+
+**Example:**
+
+    Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2
+    Output: 45
+
+**Note:**
+
+Assume that the total area is never beyond the maximum possible value of **int**.
+
+
+
+## 题目大意
+
+在二维平面上计算出两个由直线构成的矩形重叠后形成的总面积。每个矩形由其左下顶点和右上顶点坐标表示，如图所示。说明: 假设矩形面积不会超出 int 的范围。
+
+## 解题思路
+
+
+- 给出两个矩形的坐标，求这两个矩形在坐标轴上覆盖的总面积。
+- 几何题，由于只有 2 个矩形，所以按照题意做即可。先分别求两个矩形的面积，加起来再减去两个矩形重叠的面积。
diff --git a/Algorithms/0224. Basic Calculator/224. Basic Calculator.go b/leetcode/0224.Basic-Calculator/224. Basic Calculator.go
similarity index 95%
rename from Algorithms/0224. Basic Calculator/224. Basic Calculator.go
rename to leetcode/0224.Basic-Calculator/224. Basic Calculator.go
index 5c1f8d6f8..29400c92c 100644
--- a/Algorithms/0224. Basic Calculator/224. Basic Calculator.go	
+++ b/leetcode/0224.Basic-Calculator/224. Basic Calculator.go	
@@ -53,7 +53,7 @@ func calculate1(s string) int {
 					break
 				}
 			}
-			tmp = string(stack[index+1 : len(stack)])
+			tmp = string(stack[index+1:])
 			stack = stack[:index]
 			res := strconv.Itoa(calculateStr(tmp))
 			for j := 0; j < len(res); j++ {
@@ -98,7 +98,6 @@ func calculateStr(str string) int {
 	if tmpStr != "" {
 		num, _ := strconv.Atoi(tmpStr)
 		nums = append(nums, num)
-		tmpStr = ""
 	}
 	res = nums[0]
 	for i := 0; i < len(s); i++ {
@@ -111,3 +110,10 @@ func calculateStr(str string) int {
 	fmt.Printf("s = %v nums = %v res = %v\n", string(s), nums, res)
 	return res
 }
+
+func isDigital(v byte) bool {
+	if v >= '0' && v <= '9' {
+		return true
+	}
+	return false
+}
diff --git a/Algorithms/0224. Basic Calculator/224. Basic Calculator_test.go b/leetcode/0224.Basic-Calculator/224. Basic Calculator_test.go
similarity index 92%
rename from Algorithms/0224. Basic Calculator/224. Basic Calculator_test.go
rename to leetcode/0224.Basic-Calculator/224. Basic Calculator_test.go
index 35e7b0678..47331a695 100644
--- a/Algorithms/0224. Basic Calculator/224. Basic Calculator_test.go	
+++ b/leetcode/0224.Basic-Calculator/224. Basic Calculator_test.go	
@@ -26,21 +26,21 @@ func Test_Problem224(t *testing.T) {
 
 	qs := []question224{
 
-		question224{
+		{
 			para224{"1 + 1"},
 			ans224{2},
 		},
-		question224{
+		{
 			para224{" 2-1 + 2 "},
 			ans224{3},
 		},
 
-		question224{
+		{
 			para224{"(1+(4+5+2)-3)+(6+8)"},
 			ans224{23},
 		},
 
-		question224{
+		{
 			para224{"2-(5-6)"},
 			ans224{3},
 		},
diff --git a/Algorithms/0224. Basic Calculator/README.md b/leetcode/0224.Basic-Calculator/README.md
similarity index 100%
rename from Algorithms/0224. Basic Calculator/README.md
rename to leetcode/0224.Basic-Calculator/README.md
diff --git a/Algorithms/0225. Implement Stack using Queues/225. Implement Stack using Queues.go b/leetcode/0225.Implement-Stack-using-Queues/225. Implement Stack using Queues.go
similarity index 100%
rename from Algorithms/0225. Implement Stack using Queues/225. Implement Stack using Queues.go
rename to leetcode/0225.Implement-Stack-using-Queues/225. Implement Stack using Queues.go
diff --git a/Algorithms/0225. Implement Stack using Queues/225. Implement Stack using Queues_test.go b/leetcode/0225.Implement-Stack-using-Queues/225. Implement Stack using Queues_test.go
similarity index 100%
rename from Algorithms/0225. Implement Stack using Queues/225. Implement Stack using Queues_test.go
rename to leetcode/0225.Implement-Stack-using-Queues/225. Implement Stack using Queues_test.go
diff --git a/Algorithms/0225. Implement Stack using Queues/README.md b/leetcode/0225.Implement-Stack-using-Queues/README.md
similarity index 100%
rename from Algorithms/0225. Implement Stack using Queues/README.md
rename to leetcode/0225.Implement-Stack-using-Queues/README.md
diff --git a/leetcode/0226.Invert-Binary-Tree/226. Invert Binary Tree.go b/leetcode/0226.Invert-Binary-Tree/226. Invert Binary Tree.go
new file mode 100644
index 000000000..bce4e6286
--- /dev/null
+++ b/leetcode/0226.Invert-Binary-Tree/226. Invert Binary Tree.go	
@@ -0,0 +1,27 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func invertTree(root *TreeNode) *TreeNode {
+	if root == nil {
+		return nil
+	}
+	invertTree(root.Left)
+	invertTree(root.Right)
+	root.Left, root.Right = root.Right, root.Left
+	return root
+}
diff --git a/Algorithms/0226. Invert Binary Tree/226. Invert Binary Tree_test.go b/leetcode/0226.Invert-Binary-Tree/226. Invert Binary Tree_test.go
similarity index 79%
rename from Algorithms/0226. Invert Binary Tree/226. Invert Binary Tree_test.go
rename to leetcode/0226.Invert-Binary-Tree/226. Invert Binary Tree_test.go
index 266c5f35d..fecd35364 100644
--- a/Algorithms/0226. Invert Binary Tree/226. Invert Binary Tree_test.go	
+++ b/leetcode/0226.Invert-Binary-Tree/226. Invert Binary Tree_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question226 struct {
@@ -26,17 +28,17 @@ func Test_Problem226(t *testing.T) {
 
 	qs := []question226{
 
-		question226{
+		{
 			para226{[]int{}},
 			ans226{[]int{}},
 		},
 
-		question226{
+		{
 			para226{[]int{1}},
 			ans226{[]int{1}},
 		},
 
-		question226{
+		{
 			para226{[]int{4, 2, 7, 1, 3, 6, 9}},
 			ans226{[]int{4, 7, 2, 9, 6, 3, 1}},
 		},
@@ -47,8 +49,8 @@ func Test_Problem226(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans226, q.para226
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
-		fmt.Printf("【output】:%v      \n", Tree2Preorder(invertTree(root)))
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", structures.Tree2ints(invertTree(root)))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0226. Invert Binary Tree/README.md b/leetcode/0226.Invert-Binary-Tree/README.md
similarity index 100%
rename from Algorithms/0226. Invert Binary Tree/README.md
rename to leetcode/0226.Invert-Binary-Tree/README.md
diff --git a/leetcode/0227.Basic-Calculator-II/227. Basic Calculator II.go b/leetcode/0227.Basic-Calculator-II/227. Basic Calculator II.go
new file mode 100644
index 000000000..a20168b9d
--- /dev/null
+++ b/leetcode/0227.Basic-Calculator-II/227. Basic Calculator II.go	
@@ -0,0 +1,29 @@
+package leetcode
+
+func calculate(s string) int {
+	stack, preSign, num, res := []int{}, '+', 0, 0
+	for i, ch := range s {
+		isDigit := '0' <= ch && ch <= '9'
+		if isDigit {
+			num = num*10 + int(ch-'0')
+		}
+		if !isDigit && ch != ' ' || i == len(s)-1 {
+			switch preSign {
+			case '+':
+				stack = append(stack, num)
+			case '-':
+				stack = append(stack, -num)
+			case '*':
+				stack[len(stack)-1] *= num
+			default:
+				stack[len(stack)-1] /= num
+			}
+			preSign = ch
+			num = 0
+		}
+	}
+	for _, v := range stack {
+		res += v
+	}
+	return res
+}
diff --git a/leetcode/0227.Basic-Calculator-II/227. Basic Calculator II_test.go b/leetcode/0227.Basic-Calculator-II/227. Basic Calculator II_test.go
new file mode 100644
index 000000000..131d97baf
--- /dev/null
+++ b/leetcode/0227.Basic-Calculator-II/227. Basic Calculator II_test.go	
@@ -0,0 +1,66 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question227 struct {
+	para227
+	ans227
+}
+
+// para 是参数
+// one 代表第一个参数
+type para227 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans227 struct {
+	one int
+}
+
+func Test_Problem227(t *testing.T) {
+
+	qs := []question227{
+
+		{
+			para227{"3+2*2"},
+			ans227{7},
+		},
+
+		{
+			para227{"3/2"},
+			ans227{1},
+		},
+
+		{
+			para227{" 3+5 / 2 "},
+			ans227{5},
+		},
+
+		{
+			para227{"1 + 1"},
+			ans227{2},
+		},
+		{
+			para227{" 2-1 + 2 "},
+			ans227{3},
+		},
+
+		{
+			para227{"2-5/6"},
+			ans227{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 227------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans227, q.para227
+		fmt.Printf("【input】:%v       【output】:%v\n", p, calculate(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0227.Basic-Calculator-II/README.md b/leetcode/0227.Basic-Calculator-II/README.md
new file mode 100644
index 000000000..470a58af5
--- /dev/null
+++ b/leetcode/0227.Basic-Calculator-II/README.md
@@ -0,0 +1,80 @@
+# [227. Basic Calculator II](https://leetcode.com/problems/basic-calculator-ii/)
+
+
+## 题目
+
+Given a string `s` which represents an expression, *evaluate this expression and return its value*.
+
+The integer division should truncate toward zero.
+
+**Example 1:**
+
+```
+Input: s = "3+2*2"
+Output: 7
+```
+
+**Example 2:**
+
+```
+Input: s = " 3/2 "
+Output: 1
+```
+
+**Example 3:**
+
+```
+Input: s = " 3+5 / 2 "
+Output: 5
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 3 * 10^5`
+- `s` consists of integers and operators `('+', '-', '*', '/')` separated by some number of spaces.
+- `s` represents **a valid expression**.
+- All the integers in the expression are non-negative integers in the range `[0, 2^31 - 1]`.
+- The answer is **guaranteed** to fit in a **32-bit integer**.
+
+## 题目大意
+
+给你一个字符串表达式 `s` ，请你实现一个基本计算器来计算并返回它的值。整数除法仅保留整数部分。
+
+## 解题思路
+
+- 这道题是第 224 题的加强版。第 224 题中只有加减运算和括号，这一题增加了乘除运算。由于乘除运算的优先级高于加减。所以先计算乘除运算，将算出来的结果再替换回原来的算式中。最后只剩下加减运算，于是题目降级成了第 224 题。
+- 把加减运算符号后面的数字压入栈中，遇到乘除运算，直接将它与栈顶的元素计算，并将计算后的结果放入栈顶。若读到一个运算符，或者遍历到字符串末尾，即认为是遍历到了数字末尾。处理完该数字后，更新 `preSign` 为当前遍历的字符。遍历完字符串 `s` 后，将栈中元素累加，即为该字符串表达式的值。时间复杂度 O(n)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+func calculate(s string) int {
+	stack, preSign, num, res := []int{}, '+', 0, 0
+	for i, ch := range s {
+		isDigit := '0' <= ch && ch <= '9'
+		if isDigit {
+			num = num*10 + int(ch-'0')
+		}
+		if !isDigit && ch != ' ' || i == len(s)-1 {
+			switch preSign {
+			case '+':
+				stack = append(stack, num)
+			case '-':
+				stack = append(stack, -num)
+			case '*':
+				stack[len(stack)-1] *= num
+			default:
+				stack[len(stack)-1] /= num
+			}
+			preSign = ch
+			num = 0
+		}
+	}
+	for _, v := range stack {
+		res += v
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0228.Summary-Ranges/228. Summary Ranges.go b/leetcode/0228.Summary-Ranges/228. Summary Ranges.go
new file mode 100644
index 000000000..b19c0e862
--- /dev/null
+++ b/leetcode/0228.Summary-Ranges/228. Summary Ranges.go	
@@ -0,0 +1,19 @@
+package leetcode
+
+import (
+	"strconv"
+)
+
+func summaryRanges(nums []int) (ans []string) {
+	for i, n := 0, len(nums); i < n; {
+		left := i
+		for i++; i < n && nums[i-1]+1 == nums[i]; i++ {
+		}
+		s := strconv.Itoa(nums[left])
+		if left != i-1 {
+			s += "->" + strconv.Itoa(nums[i-1])
+		}
+		ans = append(ans, s)
+	}
+	return
+}
diff --git a/leetcode/0228.Summary-Ranges/228. Summary Ranges_test.go b/leetcode/0228.Summary-Ranges/228. Summary Ranges_test.go
new file mode 100644
index 000000000..dd18d007e
--- /dev/null
+++ b/leetcode/0228.Summary-Ranges/228. Summary Ranges_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question228 struct {
+	para228
+	ans228
+}
+
+// para 是参数
+// one 代表第一个参数
+type para228 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans228 struct {
+	ans []string
+}
+
+func Test_Problem228(t *testing.T) {
+
+	qs := []question228{
+
+		{
+			para228{[]int{0, 1, 2, 4, 5, 7}},
+			ans228{[]string{"0->2", "4->5", "7"}},
+		},
+
+		{
+			para228{[]int{0, 2, 3, 4, 6, 8, 9}},
+			ans228{[]string{"0", "2->4", "6", "8->9"}},
+		},
+
+		{
+			para228{[]int{}},
+			ans228{[]string{}},
+		},
+
+		{
+			para228{[]int{-1}},
+			ans228{[]string{"-1"}},
+		},
+
+		{
+			para228{[]int{0}},
+			ans228{[]string{"0"}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 228------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans228, q.para228
+		fmt.Printf("【input】:%v       【output】:%v\n", p, summaryRanges(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0228.Summary-Ranges/README.md b/leetcode/0228.Summary-Ranges/README.md
new file mode 100644
index 000000000..85530b5aa
--- /dev/null
+++ b/leetcode/0228.Summary-Ranges/README.md
@@ -0,0 +1,107 @@
+# [228. Summary Ranges](https://leetcode.com/problems/summary-ranges/)
+
+## 题目
+
+You are given a **sorted unique** integer array `nums`.
+
+Return *the **smallest sorted** list of ranges that **cover all the numbers in the array exactly***. That is, each element of `nums` is covered by exactly one of the ranges, and there is no integer `x` such that `x` is in one of the ranges but not in `nums`.
+
+Each range `[a,b]` in the list should be output as:
+
+- `"a->b"` if `a != b`
+- `"a"` if `a == b`
+
+**Example 1:**
+
+```
+Input: nums = [0,1,2,4,5,7]
+Output: ["0->2","4->5","7"]
+Explanation: The ranges are:
+[0,2] --> "0->2"
+[4,5] --> "4->5"
+[7,7] --> "7"
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [0,2,3,4,6,8,9]
+Output: ["0","2->4","6","8->9"]
+Explanation: The ranges are:
+[0,0] --> "0"
+[2,4] --> "2->4"
+[6,6] --> "6"
+[8,9] --> "8->9"
+
+```
+
+**Example 3:**
+
+```
+Input: nums = []
+Output: []
+
+```
+
+**Example 4:**
+
+```
+Input: nums = [-1]
+Output: ["-1"]
+
+```
+
+**Example 5:**
+
+```
+Input: nums = [0]
+Output: ["0"]
+
+```
+
+**Constraints:**
+
+- `0 <= nums.length <= 20`
+- `231 <= nums[i] <= 231 - 1`
+- All the values of `nums` are **unique**.
+- `nums` is sorted in ascending order.
+
+## 题目大意
+
+给定一个无重复元素的有序整数数组 nums 。
+
+返回 恰好覆盖数组中所有数字 的 最小有序 区间范围列表。也就是说，nums 的每个元素都恰好被某个区间范围所覆盖，并且不存在属于某个范围但不属于 nums 的数字 x 。
+
+列表中的每个区间范围 [a,b] 应该按如下格式输出：
+
+- "a->b" ，如果 a != b
+- "a" ，如果 a == b
+
+## 解题思路
+
+- 简单题。按照题意，用一个游标变量累加寻找连续的区间。一旦出现了中断，就按照题意格式输出。输出的规则有多种，带箭头的区间，单个元素区间，空区间。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"strconv"
+)
+
+func summaryRanges(nums []int) (ans []string) {
+	for i, n := 0, len(nums); i < n; {
+		left := i
+		for i++; i < n && nums[i-1]+1 == nums[i]; i++ {
+		}
+		s := strconv.Itoa(nums[left])
+		if left != i-1 {
+			s += "->" + strconv.Itoa(nums[i-1])
+		}
+		ans = append(ans, s)
+	}
+	return
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0229. Majority Element II/229. Majority Element II.go b/leetcode/0229.Majority-Element-II/229. Majority Element II.go
similarity index 100%
rename from Algorithms/0229. Majority Element II/229. Majority Element II.go
rename to leetcode/0229.Majority-Element-II/229. Majority Element II.go
diff --git a/Algorithms/0229. Majority Element II/229. Majority Element II_test.go b/leetcode/0229.Majority-Element-II/229. Majority Element II_test.go
similarity index 95%
rename from Algorithms/0229. Majority Element II/229. Majority Element II_test.go
rename to leetcode/0229.Majority-Element-II/229. Majority Element II_test.go
index 2eb99c513..2450ae509 100644
--- a/Algorithms/0229. Majority Element II/229. Majority Element II_test.go	
+++ b/leetcode/0229.Majority-Element-II/229. Majority Element II_test.go	
@@ -26,12 +26,12 @@ func Test_Problem229(t *testing.T) {
 
 	qs := []question229{
 
-		question229{
+		{
 			para229{[]int{3, 2, 3}},
 			ans229{[]int{3}},
 		},
 
-		question229{
+		{
 			para229{[]int{1, 1, 1, 3, 3, 2, 2, 2}},
 			ans229{[]int{1, 2}},
 		},
diff --git a/leetcode/0229.Majority-Element-II/README.md b/leetcode/0229.Majority-Element-II/README.md
new file mode 100755
index 000000000..cf5fbea18
--- /dev/null
+++ b/leetcode/0229.Majority-Element-II/README.md
@@ -0,0 +1,29 @@
+# [229. Majority Element II](https://leetcode.com/problems/majority-element-ii/)
+
+
+## 题目
+
+Given an integer array of size n, find all elements that appear more than `⌊ n/3 ⌋` times.
+
+**Note:** The algorithm should run in linear time and in O(1) space.
+
+**Example 1:**
+
+    Input: [3,2,3]
+    Output: [3]
+
+**Example 2:**
+
+    Input: [1,1,1,3,3,2,2,2]
+    Output: [1,2]
+
+
+## 题目大意
+
+给定一个大小为 n 的数组，找出其中所有出现超过 ⌊ n/3 ⌋ 次的元素。说明: 要求算法的时间复杂度为 O(n)，空间复杂度为 O(1)。
+
+## 解题思路
+
+- 这一题是第 169 题的加强版。Boyer-Moore Majority Vote algorithm 算法的扩展版。
+- 题目要求找出数组中出现次数大于 `⌊ n/3 ⌋` 次的数。要求空间复杂度为 O(1)。简单题。
+- 这篇文章写的不错，可参考：[https://gregable.com/2013/10/majority-vote-algorithm-find-majority.html](https://gregable.com/2013/10/majority-vote-algorithm-find-majority.html)
\ No newline at end of file
diff --git a/Algorithms/0230. Kth Smallest Element in a BST/230. Kth Smallest Element in a BST.go b/leetcode/0230.Kth-Smallest-Element-in-a-BST/230. Kth Smallest Element in a BST.go
similarity index 81%
rename from Algorithms/0230. Kth Smallest Element in a BST/230. Kth Smallest Element in a BST.go
rename to leetcode/0230.Kth-Smallest-Element-in-a-BST/230. Kth Smallest Element in a BST.go
index f50f8845e..1cef23158 100644
--- a/Algorithms/0230. Kth Smallest Element in a BST/230. Kth Smallest Element in a BST.go	
+++ b/leetcode/0230.Kth-Smallest-Element-in-a-BST/230. Kth Smallest Element in a BST.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func kthSmallest(root *TreeNode, k int) int {
 	res, count := 0, 0
 	inorder230(root, k, &count, &res)
diff --git a/Algorithms/0230. Kth Smallest Element in a BST/230. Kth Smallest Element in a BST_test.go b/leetcode/0230.Kth-Smallest-Element-in-a-BST/230. Kth Smallest Element in a BST_test.go
similarity index 75%
rename from Algorithms/0230. Kth Smallest Element in a BST/230. Kth Smallest Element in a BST_test.go
rename to leetcode/0230.Kth-Smallest-Element-in-a-BST/230. Kth Smallest Element in a BST_test.go
index cceea1012..4eb806fff 100644
--- a/Algorithms/0230. Kth Smallest Element in a BST/230. Kth Smallest Element in a BST_test.go	
+++ b/leetcode/0230.Kth-Smallest-Element-in-a-BST/230. Kth Smallest Element in a BST_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question230 struct {
@@ -27,18 +29,18 @@ func Test_Problem230(t *testing.T) {
 
 	qs := []question230{
 
-		question230{
+		{
 			para230{[]int{}, 0},
 			ans230{0},
 		},
 
-		question230{
-			para230{[]int{3, 1, 4, NULL, 2}, 1},
+		{
+			para230{[]int{3, 1, 4, structures.NULL, 2}, 1},
 			ans230{1},
 		},
 
-		question230{
-			para230{[]int{5, 3, 6, 2, 4, NULL, NULL, 1}, 3},
+		{
+			para230{[]int{5, 3, 6, 2, 4, structures.NULL, structures.NULL, 1}, 3},
 			ans230{3},
 		},
 	}
@@ -48,7 +50,7 @@ func Test_Problem230(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans230, q.para230
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", kthSmallest(root, p.k))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0230. Kth Smallest Element in a BST/README.md b/leetcode/0230.Kth-Smallest-Element-in-a-BST/README.md
similarity index 100%
rename from Algorithms/0230. Kth Smallest Element in a BST/README.md
rename to leetcode/0230.Kth-Smallest-Element-in-a-BST/README.md
diff --git a/Algorithms/0231. Power of Two/231. Power of Two.go b/leetcode/0231.Power-of-Two/231. Power of Two.go
similarity index 100%
rename from Algorithms/0231. Power of Two/231. Power of Two.go
rename to leetcode/0231.Power-of-Two/231. Power of Two.go
diff --git a/Algorithms/0231. Power of Two/231. Power of Two_test.go b/leetcode/0231.Power-of-Two/231. Power of Two_test.go
similarity index 93%
rename from Algorithms/0231. Power of Two/231. Power of Two_test.go
rename to leetcode/0231.Power-of-Two/231. Power of Two_test.go
index 5566b017a..15b9681ff 100644
--- a/Algorithms/0231. Power of Two/231. Power of Two_test.go	
+++ b/leetcode/0231.Power-of-Two/231. Power of Two_test.go	
@@ -26,17 +26,17 @@ func Test_Problem231(t *testing.T) {
 
 	qs := []question231{
 
-		question231{
+		{
 			para231{1},
 			ans231{true},
 		},
 
-		question231{
+		{
 			para231{16},
 			ans231{true},
 		},
 
-		question231{
+		{
 			para231{218},
 			ans231{false},
 		},
diff --git a/Algorithms/0231. Power of Two/README.md b/leetcode/0231.Power-of-Two/README.md
similarity index 100%
rename from Algorithms/0231. Power of Two/README.md
rename to leetcode/0231.Power-of-Two/README.md
diff --git a/Algorithms/0232. Implement Queue using Stacks/232. Implement Queue using Stacks.go b/leetcode/0232.Implement-Queue-using-Stacks/232. Implement Queue using Stacks.go
similarity index 100%
rename from Algorithms/0232. Implement Queue using Stacks/232. Implement Queue using Stacks.go
rename to leetcode/0232.Implement-Queue-using-Stacks/232. Implement Queue using Stacks.go
diff --git a/Algorithms/0232. Implement Queue using Stacks/232. Implement Queue using Stacks_test.go b/leetcode/0232.Implement-Queue-using-Stacks/232. Implement Queue using Stacks_test.go
similarity index 100%
rename from Algorithms/0232. Implement Queue using Stacks/232. Implement Queue using Stacks_test.go
rename to leetcode/0232.Implement-Queue-using-Stacks/232. Implement Queue using Stacks_test.go
diff --git a/Algorithms/0232. Implement Queue using Stacks/README.md b/leetcode/0232.Implement-Queue-using-Stacks/README.md
similarity index 100%
rename from Algorithms/0232. Implement Queue using Stacks/README.md
rename to leetcode/0232.Implement-Queue-using-Stacks/README.md
diff --git a/Algorithms/0234. Palindrome Linked List/234. Palindrome Linked List.go b/leetcode/0234.Palindrome-Linked-List/234. Palindrome Linked List.go
similarity index 76%
rename from Algorithms/0234. Palindrome Linked List/234. Palindrome Linked List.go
rename to leetcode/0234.Palindrome-Linked-List/234. Palindrome Linked List.go
index b0d06e2de..4a7f05c66 100644
--- a/Algorithms/0234. Palindrome Linked List/234. Palindrome Linked List.go	
+++ b/leetcode/0234.Palindrome-Linked-List/234. Palindrome Linked List.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
@@ -8,8 +15,26 @@ package leetcode
  * }
  */
 
+// 解法一
+func isPalindrome(head *ListNode) bool {
+	slice := []int{}
+	for head != nil {
+		slice = append(slice, head.Val)
+		head = head.Next
+	}
+	for i, j := 0, len(slice)-1; i < j; {
+		if slice[i] != slice[j] {
+			return false
+		}
+		i++
+		j--
+	}
+	return true
+}
+
+// 解法二
 // 此题和 143 题 Reorder List 思路基本一致
-func isPalindrome234(head *ListNode) bool {
+func isPalindrome1(head *ListNode) bool {
 	if head == nil || head.Next == nil {
 		return true
 	}
@@ -21,7 +46,6 @@ func isPalindrome234(head *ListNode) bool {
 		p1 = p1.Next
 		p2 = p2.Next.Next
 	}
-
 	// 反转链表后半部分  1->2->3->4->5->6 to 1->2->3->6->5->4
 	preMiddle := p1
 	preCurrent := p1.Next
@@ -31,7 +55,6 @@ func isPalindrome234(head *ListNode) bool {
 		current.Next = preMiddle.Next
 		preMiddle.Next = current
 	}
-
 	// 扫描表，判断是否是回文
 	p1 = head
 	p2 = preMiddle.Next
@@ -52,18 +75,15 @@ func isPalindrome234(head *ListNode) bool {
 			return false
 		}
 	}
-
 	return res
 }
 
 // L2ss define
 func L2ss(head *ListNode) []int {
 	res := []int{}
-
 	for head != nil {
 		res = append(res, head.Val)
 		head = head.Next
 	}
-
 	return res
 }
diff --git a/Algorithms/0234. Palindrome Linked List/234. Palindrome Linked List_test.go b/leetcode/0234.Palindrome-Linked-List/234. Palindrome Linked List_test.go
similarity index 86%
rename from Algorithms/0234. Palindrome Linked List/234. Palindrome Linked List_test.go
rename to leetcode/0234.Palindrome-Linked-List/234. Palindrome Linked List_test.go
index 7995be3bb..57cdb5dc5 100644
--- a/Algorithms/0234. Palindrome Linked List/234. Palindrome Linked List_test.go	
+++ b/leetcode/0234.Palindrome-Linked-List/234. Palindrome Linked List_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question234 struct {
@@ -26,52 +28,52 @@ func Test_Problem234(t *testing.T) {
 
 	qs := []question234{
 
-		question234{
+		{
 			para234{[]int{1, 1, 2, 2, 3, 4, 4, 4}},
 			ans234{false},
 		},
 
-		question234{
+		{
 			para234{[]int{1, 1, 1, 1, 1, 1}},
 			ans234{true},
 		},
 
-		question234{
+		{
 			para234{[]int{1, 2, 2, 1, 3}},
 			ans234{false},
 		},
 
-		question234{
+		{
 			para234{[]int{1}},
 			ans234{true},
 		},
 
-		question234{
+		{
 			para234{[]int{}},
 			ans234{true},
 		},
 
-		question234{
+		{
 			para234{[]int{1, 2, 2, 2, 2, 1}},
 			ans234{true},
 		},
 
-		question234{
+		{
 			para234{[]int{1, 2, 2, 3, 3, 3, 3, 2, 2, 1}},
 			ans234{true},
 		},
 
-		question234{
+		{
 			para234{[]int{1, 2}},
 			ans234{false},
 		},
 
-		question234{
+		{
 			para234{[]int{1, 0, 1}},
 			ans234{true},
 		},
 
-		question234{
+		{
 			para234{[]int{1, 1, 2, 1}},
 			ans234{false},
 		},
@@ -81,7 +83,7 @@ func Test_Problem234(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans234, q.para234
-		fmt.Printf("【input】:%v       【output】:%v\n", p, isPalindrome234(S2l(p.one)))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, isPalindrome(structures.Ints2List(p.one)))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0234. Palindrome Linked List/README.md b/leetcode/0234.Palindrome-Linked-List/README.md
similarity index 100%
rename from Algorithms/0234. Palindrome Linked List/README.md
rename to leetcode/0234.Palindrome-Linked-List/README.md
diff --git a/leetcode/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree/235. Lowest Common Ancestor of a Binary Search Tree.go b/leetcode/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree/235. Lowest Common Ancestor of a Binary Search Tree.go
new file mode 100644
index 000000000..688acd213
--- /dev/null
+++ b/leetcode/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree/235. Lowest Common Ancestor of a Binary Search Tree.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
+	if p == nil || q == nil || root == nil {
+		return nil
+	}
+	if p.Val < root.Val && q.Val < root.Val {
+		return lowestCommonAncestor(root.Left, p, q)
+	}
+	if p.Val > root.Val && q.Val > root.Val {
+		return lowestCommonAncestor(root.Right, p, q)
+	}
+	return root
+}
diff --git a/leetcode/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree/235. Lowest Common Ancestor of a Binary Search Tree_test.go b/leetcode/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree/235. Lowest Common Ancestor of a Binary Search Tree_test.go
new file mode 100644
index 000000000..675dddc13
--- /dev/null
+++ b/leetcode/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree/235. Lowest Common Ancestor of a Binary Search Tree_test.go	
@@ -0,0 +1,60 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question235 struct {
+	para235
+	ans235
+}
+
+// para 是参数
+// one 代表第一个参数
+type para235 struct {
+	one []int
+	two []int
+	thr []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans235 struct {
+	one []int
+}
+
+func Test_Problem235(t *testing.T) {
+
+	qs := []question235{
+
+		{
+			para235{[]int{}, []int{}, []int{}},
+			ans235{[]int{}},
+		},
+
+		{
+			para235{[]int{6, 2, 8, 0, 4, 7, 9, structures.NULL, structures.NULL, 3, 5}, []int{2}, []int{8}},
+			ans235{[]int{6}},
+		},
+
+		{
+			para235{[]int{6, 2, 8, 0, 4, 7, 9, structures.NULL, structures.NULL, 3, 5}, []int{2}, []int{4}},
+			ans235{[]int{2}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 235------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans235, q.para235
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.one)
+		rootTwo := structures.Ints2TreeNode(p.two)
+		rootThr := structures.Ints2TreeNode(p.thr)
+		fmt.Printf("【output】:%v      \n", lowestCommonAncestor(rootOne, rootTwo, rootThr))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0235. Lowest Common Ancestor of a Binary Search Tree/README.md b/leetcode/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree/README.md
similarity index 100%
rename from Algorithms/0235. Lowest Common Ancestor of a Binary Search Tree/README.md
rename to leetcode/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree/README.md
diff --git a/leetcode/0236.Lowest-Common-Ancestor-of-a-Binary-Tree/236. Lowest Common Ancestor of a Binary Tree.go b/leetcode/0236.Lowest-Common-Ancestor-of-a-Binary-Tree/236. Lowest Common Ancestor of a Binary Tree.go
new file mode 100644
index 000000000..f2c0e8451
--- /dev/null
+++ b/leetcode/0236.Lowest-Common-Ancestor-of-a-Binary-Tree/236. Lowest Common Ancestor of a Binary Tree.go	
@@ -0,0 +1,32 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func lowestCommonAncestor236(root, p, q *TreeNode) *TreeNode {
+	if root == nil || root == q || root == p {
+		return root
+	}
+	left := lowestCommonAncestor236(root.Left, p, q)
+	right := lowestCommonAncestor236(root.Right, p, q)
+	if left != nil {
+		if right != nil {
+			return root
+		}
+		return left
+	}
+	return right
+}
diff --git a/leetcode/0236.Lowest-Common-Ancestor-of-a-Binary-Tree/236. Lowest Common Ancestor of a Binary Tree_test.go b/leetcode/0236.Lowest-Common-Ancestor-of-a-Binary-Tree/236. Lowest Common Ancestor of a Binary Tree_test.go
new file mode 100644
index 000000000..48a471f42
--- /dev/null
+++ b/leetcode/0236.Lowest-Common-Ancestor-of-a-Binary-Tree/236. Lowest Common Ancestor of a Binary Tree_test.go	
@@ -0,0 +1,70 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question236 struct {
+	para236
+	ans236
+}
+
+// para 是参数
+// one 代表第一个参数
+type para236 struct {
+	one []int
+	two []int
+	thr []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans236 struct {
+	one []int
+}
+
+func Test_Problem236(t *testing.T) {
+
+	qs := []question236{
+
+		{
+			para236{[]int{}, []int{}, []int{}},
+			ans236{[]int{}},
+		},
+
+		{
+			para236{[]int{3, 5, 1, 6, 2, 0, 8, structures.NULL, structures.NULL, 7, 4}, []int{5}, []int{1}},
+			ans236{[]int{3}},
+		},
+
+		{
+			para236{[]int{3, 5, 1, 6, 2, 0, 8, structures.NULL, structures.NULL, 7, 4}, []int{5}, []int{4}},
+			ans236{[]int{5}},
+		},
+
+		{
+			para236{[]int{6, 2, 8, 0, 4, 7, 9, structures.NULL, structures.NULL, 3, 5}, []int{2}, []int{8}},
+			ans236{[]int{6}},
+		},
+
+		{
+			para236{[]int{6, 2, 8, 0, 4, 7, 9, structures.NULL, structures.NULL, 3, 5}, []int{2}, []int{4}},
+			ans236{[]int{2}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 236------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans236, q.para236
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.one)
+		rootTwo := structures.Ints2TreeNode(p.two)
+		rootThr := structures.Ints2TreeNode(p.thr)
+		fmt.Printf("【output】:%v      \n", lowestCommonAncestor236(rootOne, rootTwo, rootThr))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0236. Lowest Common Ancestor of a Binary Tree/README.md b/leetcode/0236.Lowest-Common-Ancestor-of-a-Binary-Tree/README.md
similarity index 100%
rename from Algorithms/0236. Lowest Common Ancestor of a Binary Tree/README.md
rename to leetcode/0236.Lowest-Common-Ancestor-of-a-Binary-Tree/README.md
diff --git a/leetcode/0237.Delete-Node-in-a-Linked-List/237. Delete Node in a Linked List.go b/leetcode/0237.Delete-Node-in-a-Linked-List/237. Delete Node in a Linked List.go
new file mode 100644
index 000000000..bb33b98ad
--- /dev/null
+++ b/leetcode/0237.Delete-Node-in-a-Linked-List/237. Delete Node in a Linked List.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func deleteNode(node *ListNode) {
+	node.Val = node.Next.Val
+	node.Next = node.Next.Next
+}
diff --git a/leetcode/0237.Delete-Node-in-a-Linked-List/237. Delete Node in a Linked List_test.go b/leetcode/0237.Delete-Node-in-a-Linked-List/237. Delete Node in a Linked List_test.go
new file mode 100644
index 000000000..29e43c343
--- /dev/null
+++ b/leetcode/0237.Delete-Node-in-a-Linked-List/237. Delete Node in a Linked List_test.go	
@@ -0,0 +1,93 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question237 struct {
+	para237
+	ans237
+}
+
+// para 是参数
+// one 代表第一个参数
+type para237 struct {
+	one []int
+	n   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans237 struct {
+	one []int
+}
+
+func Test_Problem237(t *testing.T) {
+
+	qs := []question237{
+
+		{
+			para237{[]int{1, 2, 3, 4, 5}, 1},
+			ans237{[]int{2, 3, 4, 5}},
+		},
+
+		{
+			para237{[]int{1, 2, 3, 4, 5}, 2},
+			ans237{[]int{1, 3, 4, 5}},
+		},
+
+		{
+			para237{[]int{1, 1, 1, 1, 1}, 1},
+			ans237{[]int{}},
+		},
+
+		{
+			para237{[]int{1, 2, 3, 4, 5}, 5},
+			ans237{[]int{1, 2, 3, 4}},
+		},
+
+		{
+			para237{[]int{}, 5},
+			ans237{[]int{}},
+		},
+
+		{
+			para237{[]int{1, 2, 3, 4, 5}, 10},
+			ans237{[]int{1, 2, 3, 4, 5}},
+		},
+
+		{
+			para237{[]int{1}, 1},
+			ans237{[]int{}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 237------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans237, q.para237
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(removeElements(structures.Ints2List(p.one), p.n)))
+	}
+	fmt.Printf("\n\n\n")
+}
+
+func removeElements(head *ListNode, val int) *ListNode {
+	if head == nil {
+		return head
+	}
+	newHead := &ListNode{Val: 0, Next: head}
+	pre := newHead
+	cur := head
+	for cur != nil {
+		if cur.Val == val {
+			pre.Next = cur.Next
+		} else {
+			pre = cur
+		}
+		cur = cur.Next
+	}
+	return newHead.Next
+}
diff --git a/leetcode/0237.Delete-Node-in-a-Linked-List/README.md b/leetcode/0237.Delete-Node-in-a-Linked-List/README.md
new file mode 100644
index 000000000..fa135fa7d
--- /dev/null
+++ b/leetcode/0237.Delete-Node-in-a-Linked-List/README.md
@@ -0,0 +1,94 @@
+# [237. Delete Node in a Linked List](https://leetcode.com/problems/delete-node-in-a-linked-list/)
+
+
+## 题目
+
+Write a function to **delete a node** in a singly-linked list. You will **not** be given access to the `head` of the list, instead you will be given access to **the node to be deleted** directly.
+
+It is **guaranteed** that the node to be deleted is **not a tail node** in the list.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/09/01/node1.jpg](https://assets.leetcode.com/uploads/2020/09/01/node1.jpg)
+
+```
+Input: head = [4,5,1,9], node = 5
+Output: [4,1,9]
+Explanation:You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/09/01/node2.jpg](https://assets.leetcode.com/uploads/2020/09/01/node2.jpg)
+
+```
+Input: head = [4,5,1,9], node = 1
+Output: [4,5,9]
+Explanation:You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.
+
+```
+
+**Example 3:**
+
+```
+Input: head = [1,2,3,4], node = 3
+Output: [1,2,4]
+
+```
+
+**Example 4:**
+
+```
+Input: head = [0,1], node = 0
+Output: [1]
+
+```
+
+**Example 5:**
+
+```
+Input: head = [-3,5,-99], node = -3
+Output: [5,-99]
+
+```
+
+**Constraints:**
+
+- The number of the nodes in the given list is in the range `[2, 1000]`.
+- `1000 <= Node.val <= 1000`
+- The value of each node in the list is **unique**.
+- The `node` to be deleted is **in the list** and is **not a tail** node
+
+## 题目大意
+
+删除给点结点。没有给链表的头结点。
+
+## 解题思路
+
+其实就是把后面的结点都覆盖上来即可。或者直接当前结点的值等于下一个结点，Next 指针指向下下个结点，这样做也可以，只不过中间有一个结点不被释放，内存消耗多一些。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func deleteNode(node *ListNode) {
+	node.Val = node.Next.Val
+	node.Next = node.Next.Next
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0239. Sliding Window Maximum/239. Sliding Window Maximum.go b/leetcode/0239.Sliding-Window-Maximum/239. Sliding Window Maximum.go
similarity index 96%
rename from Algorithms/0239. Sliding Window Maximum/239. Sliding Window Maximum.go
rename to leetcode/0239.Sliding-Window-Maximum/239. Sliding Window Maximum.go
index 5b0425eb0..805ea08b4 100644
--- a/Algorithms/0239. Sliding Window Maximum/239. Sliding Window Maximum.go	
+++ b/leetcode/0239.Sliding-Window-Maximum/239. Sliding Window Maximum.go	
@@ -29,7 +29,7 @@ func maxSlidingWindow(nums []int, k int) []int {
 	result := make([]int, 0, len(nums)-k+1)
 	for i, v := range nums { // if the left-most index is out of window, remove it
 		if i >= k && window[0] <= i-k {
-			window = window[1:len(window)]
+			window = window[1:]
 		}
 		for len(window) > 0 && nums[window[len(window)-1]] < v { // maintain window
 			window = window[0 : len(window)-1]
diff --git a/Algorithms/0239. Sliding Window Maximum/239. Sliding Window Maximum_test.go b/leetcode/0239.Sliding-Window-Maximum/239. Sliding Window Maximum_test.go
similarity index 97%
rename from Algorithms/0239. Sliding Window Maximum/239. Sliding Window Maximum_test.go
rename to leetcode/0239.Sliding-Window-Maximum/239. Sliding Window Maximum_test.go
index ce348a429..a8e542983 100644
--- a/Algorithms/0239. Sliding Window Maximum/239. Sliding Window Maximum_test.go	
+++ b/leetcode/0239.Sliding-Window-Maximum/239. Sliding Window Maximum_test.go	
@@ -27,7 +27,7 @@ func Test_Problem239(t *testing.T) {
 
 	qs := []question239{
 
-		question239{
+		{
 			para239{[]int{1, 3, -1, -3, 5, 3, 6, 7}, 3},
 			ans239{[]int{3, 3, 5, 5, 6, 7}},
 		},
diff --git a/Algorithms/0239. Sliding Window Maximum/README.md b/leetcode/0239.Sliding-Window-Maximum/README.md
similarity index 100%
rename from Algorithms/0239. Sliding Window Maximum/README.md
rename to leetcode/0239.Sliding-Window-Maximum/README.md
diff --git a/Algorithms/0240. Search a 2D Matrix II/240. Search a 2D Matrix II.go b/leetcode/0240.Search-a-2D-Matrix-II/240. Search a 2D Matrix II.go
similarity index 100%
rename from Algorithms/0240. Search a 2D Matrix II/240. Search a 2D Matrix II.go
rename to leetcode/0240.Search-a-2D-Matrix-II/240. Search a 2D Matrix II.go
diff --git a/leetcode/0240.Search-a-2D-Matrix-II/240. Search a 2D Matrix II_test.go b/leetcode/0240.Search-a-2D-Matrix-II/240. Search a 2D Matrix II_test.go
new file mode 100644
index 000000000..e8a3d915a
--- /dev/null
+++ b/leetcode/0240.Search-a-2D-Matrix-II/240. Search a 2D Matrix II_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question240 struct {
+	para240
+	ans240
+}
+
+// para 是参数
+// one 代表第一个参数
+type para240 struct {
+	matrix [][]int
+	target int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans240 struct {
+	one bool
+}
+
+func Test_Problem240(t *testing.T) {
+
+	qs := []question240{
+
+		{
+			para240{[][]int{{1, 4, 7, 11, 15}, {2, 5, 8, 12, 19}, {3, 6, 9, 16, 22}, {10, 13, 14, 17, 24}, {18, 21, 23, 26, 30}}, 5},
+			ans240{true},
+		},
+
+		{
+			para240{[][]int{{1, 4, 7, 11, 15}, {2, 5, 8, 12, 19}, {3, 6, 9, 16, 22}, {10, 13, 14, 17, 24}, {18, 21, 23, 26, 30}}, 20},
+			ans240{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 240------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans240, q.para240
+		fmt.Printf("【input】:%v       【output】:%v\n", p, searchMatrix240(p.matrix, p.target))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0240.Search-a-2D-Matrix-II/README.md b/leetcode/0240.Search-a-2D-Matrix-II/README.md
new file mode 100755
index 000000000..97cf4fc72
--- /dev/null
+++ b/leetcode/0240.Search-a-2D-Matrix-II/README.md
@@ -0,0 +1,41 @@
+# [240. Search a 2D Matrix II](https://leetcode.com/problems/search-a-2d-matrix-ii/)
+
+
+## 题目
+
+Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
+
+- Integers in each row are sorted in ascending from left to right.
+- Integers in each column are sorted in ascending from top to bottom.
+
+**Example:**
+
+Consider the following matrix:
+
+    [
+      [1,   4,  7, 11, 15],
+      [2,   5,  8, 12, 19],
+      [3,   6,  9, 16, 22],
+      [10, 13, 14, 17, 24],
+      [18, 21, 23, 26, 30]
+    ]
+
+Given target = `5`, return `true`.
+
+Given target = `20`, return `false`.
+
+
+## 题目大意
+
+编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target。该矩阵具有以下特性：
+
+- 每行的元素从左到右升序排列。
+- 每列的元素从上到下升序排列。
+
+
+
+## 解题思路
+
+- 给出一个二维矩阵，矩阵的特点是每一个行内，元素随着下标增大而增大，每一列内，元素也是随着下标增大而增大。但是相邻两行的元素并没有大小关系。例如第一行最后一个元素就比第二行第一个元素要大。要求设计一个算法能在这个矩阵中高效的找到一个数，如果找到就输出 true，找不到就输出 false。
+- 这一题是第 74 题的加强版。第 74 题中的二维矩阵完全是一个有序的一维矩阵，但是这一题如果把它拍扁成一维，并不是有序的。首先每一个行或者每一列是有序的 ，那么我们可以依次在每一行或者每一列中利用二分去搜索。这样做时间复杂度为  O(n log n)。
+- 还有一个模拟的解法。通过观察，我们发现了这个矩阵的一个特点，最右边一列的元素是本行中最大的元素，所以我们可以先从最右边一列开始找到第一个比 target 元素大的元素，这个元素所在的行，是我们接着要搜索的。在行中搜索是从最右边开始往左边搜索，时间复杂度是 O(n)，算上一开始在最右边一列中查找的时间复杂度是 O(m)，所以最终的时间复杂度为 O(m+n)。
diff --git a/leetcode/0242.Valid-Anagram/242. Valid Anagram.go b/leetcode/0242.Valid-Anagram/242. Valid Anagram.go
new file mode 100644
index 000000000..ae3a7287c
--- /dev/null
+++ b/leetcode/0242.Valid-Anagram/242. Valid Anagram.go	
@@ -0,0 +1,40 @@
+package leetcode
+
+// 解法一
+func isAnagram(s string, t string) bool {
+	alphabet := make([]int, 26)
+	sBytes := []byte(s)
+	tBytes := []byte(t)
+	if len(sBytes) != len(tBytes) {
+		return false
+	}
+	for i := 0; i < len(sBytes); i++ {
+		alphabet[sBytes[i]-'a']++
+	}
+	for i := 0; i < len(tBytes); i++ {
+		alphabet[tBytes[i]-'a']--
+	}
+	for i := 0; i < 26; i++ {
+		if alphabet[i] != 0 {
+			return false
+		}
+	}
+	return true
+}
+
+// 解法二
+func isAnagram1(s string, t string) bool {
+	hash := map[rune]int{}
+	for _, value := range s {
+		hash[value]++
+	}
+	for _, value := range t {
+		hash[value]--
+	}
+	for _, value := range hash {
+		if value != 0 {
+			return false
+		}
+	}
+	return true
+}
diff --git a/Algorithms/0242. Valid Anagram/242. Valid Anagram_test.go b/leetcode/0242.Valid-Anagram/242. Valid Anagram_test.go
similarity index 82%
rename from Algorithms/0242. Valid Anagram/242. Valid Anagram_test.go
rename to leetcode/0242.Valid-Anagram/242. Valid Anagram_test.go
index e9b363fad..8ece1c25a 100644
--- a/Algorithms/0242. Valid Anagram/242. Valid Anagram_test.go	
+++ b/leetcode/0242.Valid-Anagram/242. Valid Anagram_test.go	
@@ -27,24 +27,39 @@ func Test_Problem242(t *testing.T) {
 
 	qs := []question242{
 
-		question242{
+		{
 			para242{"", ""},
 			ans242{true},
 		},
-		question242{
+		{
 			para242{"", "1"},
 			ans242{false},
 		},
 
-		question242{
+		{
 			para242{"anagram", "nagaram"},
 			ans242{true},
 		},
 
-		question242{
+		{
 			para242{"rat", "car"},
 			ans242{false},
 		},
+
+		{
+			para242{"a", "ab"},
+			ans242{false},
+		},
+
+		{
+			para242{"ab", "a"},
+			ans242{false},
+		},
+
+		{
+			para242{"aa", "bb"},
+			ans242{false},
+		},
 	}
 
 	fmt.Printf("------------------------Leetcode Problem 242------------------------\n")
diff --git a/Algorithms/0242. Valid Anagram/README.md b/leetcode/0242.Valid-Anagram/README.md
similarity index 100%
rename from Algorithms/0242. Valid Anagram/README.md
rename to leetcode/0242.Valid-Anagram/README.md
diff --git a/Algorithms/0257. Binary Tree Paths/257. Binary Tree Paths.go b/leetcode/0257.Binary-Tree-Paths/257. Binary Tree Paths.go
similarity index 86%
rename from Algorithms/0257. Binary Tree Paths/257. Binary Tree Paths.go
rename to leetcode/0257.Binary-Tree-Paths/257. Binary Tree Paths.go
index 0ea32c52f..0759ad1cc 100644
--- a/Algorithms/0257. Binary Tree Paths/257. Binary Tree Paths.go	
+++ b/leetcode/0257.Binary-Tree-Paths/257. Binary Tree Paths.go	
@@ -2,8 +2,13 @@ package leetcode
 
 import (
 	"strconv"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -12,6 +17,7 @@ import (
  *     Right *TreeNode
  * }
  */
+
 func binaryTreePaths(root *TreeNode) []string {
 	if root == nil {
 		return []string{}
diff --git a/Algorithms/0257. Binary Tree Paths/257. Binary Tree Paths_test.go b/leetcode/0257.Binary-Tree-Paths/257. Binary Tree Paths_test.go
similarity index 81%
rename from Algorithms/0257. Binary Tree Paths/257. Binary Tree Paths_test.go
rename to leetcode/0257.Binary-Tree-Paths/257. Binary Tree Paths_test.go
index 194d1427e..dcea1685f 100644
--- a/Algorithms/0257. Binary Tree Paths/257. Binary Tree Paths_test.go	
+++ b/leetcode/0257.Binary-Tree-Paths/257. Binary Tree Paths_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question257 struct {
@@ -26,17 +28,17 @@ func Test_Problem257(t *testing.T) {
 
 	qs := []question257{
 
-		question257{
+		{
 			para257{[]int{}},
 			ans257{[]string{""}},
 		},
 
-		question257{
-			para257{[]int{1, 2, 3, NULL, 5, NULL, NULL}},
+		{
+			para257{[]int{1, 2, 3, structures.NULL, 5, structures.NULL, structures.NULL}},
 			ans257{[]string{"1->2->5", "1->3"}},
 		},
 
-		question257{
+		{
 			para257{[]int{1, 2, 3, 4, 5, 6, 7}},
 			ans257{[]string{"1->2->4", "1->2->5", "1->2->4", "1->2->5", "1->3->6", "1->3->7"}},
 		},
@@ -47,7 +49,7 @@ func Test_Problem257(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans257, q.para257
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", binaryTreePaths(root))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0257. Binary Tree Paths/README.md b/leetcode/0257.Binary-Tree-Paths/README.md
similarity index 100%
rename from Algorithms/0257. Binary Tree Paths/README.md
rename to leetcode/0257.Binary-Tree-Paths/README.md
diff --git a/leetcode/0258.Add-Digits/258. Add Digits.go b/leetcode/0258.Add-Digits/258. Add Digits.go
new file mode 100644
index 000000000..b64893b85
--- /dev/null
+++ b/leetcode/0258.Add-Digits/258. Add Digits.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+func addDigits(num int) int {
+	for num > 9 {
+		cur := 0
+		for num != 0 {
+			cur += num % 10
+			num /= 10
+		}
+		num = cur
+	}
+	return num
+}
diff --git a/leetcode/0258.Add-Digits/258. Add Digits_test.go b/leetcode/0258.Add-Digits/258. Add Digits_test.go
new file mode 100644
index 000000000..3aed67777
--- /dev/null
+++ b/leetcode/0258.Add-Digits/258. Add Digits_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question258 struct {
+	para258
+	ans258
+}
+
+// para 是参数
+// one 代表第一个参数
+type para258 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans258 struct {
+	one int
+}
+
+func Test_Problem258(t *testing.T) {
+
+	qs := []question258{
+
+		{
+			para258{38},
+			ans258{2},
+		},
+
+		{
+			para258{88},
+			ans258{7},
+		},
+
+		{
+			para258{96},
+			ans258{6},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 258------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans258, q.para258
+		fmt.Printf("【input】:%v       【output】:%v\n", p, addDigits(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0258.Add-Digits/README.md b/leetcode/0258.Add-Digits/README.md
new file mode 100644
index 000000000..8c8119043
--- /dev/null
+++ b/leetcode/0258.Add-Digits/README.md
@@ -0,0 +1,47 @@
+# [258. Add Digits](https://leetcode.com/problems/add-digits/)
+
+
+## 题目
+
+Given a non-negative integer `num`, repeatedly add all its digits until the result has only one digit.
+
+**Example**:
+
+```
+Input: 38
+Output: 2 
+Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2. 
+             Since 2 has only one digit, return it.
+```
+
+**Follow up**: Could you do it without any loop/recursion in O(1) runtime?
+
+## 题目大意
+
+给定一个非负整数 num，反复将各个位上的数字相加，直到结果为一位数。
+
+
+## 解题思路
+
+- 给定一个非负整数，反复加各个位上的数，直到结果为一位数为止，最后输出这一位数。
+- 简单题。按照题意循环累加即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func addDigits(num int) int {
+	for num > 9 {
+		cur := 0
+		for num != 0 {
+			cur += num % 10
+			num /= 10
+		}
+		num = cur
+	}
+	return num
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0260. Single Number III/260. Single Number III.go b/leetcode/0260.Single-Number-III/260. Single Number III.go
similarity index 100%
rename from Algorithms/0260. Single Number III/260. Single Number III.go
rename to leetcode/0260.Single-Number-III/260. Single Number III.go
diff --git a/Algorithms/0260. Single Number III/260. Single Number III_test.go b/leetcode/0260.Single-Number-III/260. Single Number III_test.go
similarity index 97%
rename from Algorithms/0260. Single Number III/260. Single Number III_test.go
rename to leetcode/0260.Single-Number-III/260. Single Number III_test.go
index 860605b45..720f4c30e 100644
--- a/Algorithms/0260. Single Number III/260. Single Number III_test.go	
+++ b/leetcode/0260.Single-Number-III/260. Single Number III_test.go	
@@ -26,7 +26,7 @@ func Test_Problem260(t *testing.T) {
 
 	qs := []question260{
 
-		question260{
+		{
 			para260{[]int{1, 2, 1, 3, 2, 5}},
 			ans260{[]int{3, 5}},
 		},
diff --git a/leetcode/0260.Single-Number-III/README.md b/leetcode/0260.Single-Number-III/README.md
new file mode 100755
index 000000000..ae445a5eb
--- /dev/null
+++ b/leetcode/0260.Single-Number-III/README.md
@@ -0,0 +1,34 @@
+# [260. Single Number III](https://leetcode.com/problems/single-number-iii/)
+
+
+## 题目
+
+Given an array of numbers `nums`, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.
+
+**Example:**
+
+    Input:  [1,2,1,3,2,5]
+    Output: [3,5]
+
+**Note**:
+
+1. The order of the result is not important. So in the above example, `[5, 3]` is also correct.
+2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?
+
+
+## 题目大意
+
+给定一个整数数组 nums，其中恰好有两个元素只出现一次，其余所有元素均出现两次。 找出只出现一次的那两个元素。
+
+注意：  
+
+- 结果输出的顺序并不重要，对于上面的例子，[5, 3] 也是正确答案。
+- 要求算法时间复杂度是线性的，并且不使用额外的辅助空间。
+
+
+
+## 解题思路
+
+- 这一题是第 136 题的加强版。第 136 题里面只有一个数出现一次，其他数都出现 2 次。而这一次有 2 个数字出现一次，其他数出现 2 次。
+- 解题思路还是利用异或，把出现 2 次的数先消除。最后我们要找的 2 个数肯定也是不同的，所以最后 2 个数对一个数进行异或，答案肯定是不同的。那么我们找哪个数为参照物呢？可以随便取，不如就取 lsb 最低位为 1 的数吧
+- 于是整个数组会被分为 2 部分，异或 lsb 为 0 的和异或 lsb 为 1 的，在这 2 部分中，用异或操作把出现 2 次的数都消除，那么剩下的 2 个数分别就在这 2 部分中。
diff --git a/leetcode/0263.Ugly-Number/263. Ugly Number.go b/leetcode/0263.Ugly-Number/263. Ugly Number.go
new file mode 100644
index 000000000..42e66c217
--- /dev/null
+++ b/leetcode/0263.Ugly-Number/263. Ugly Number.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+func isUgly(num int) bool {
+	if num > 0 {
+		for _, i := range []int{2, 3, 5} {
+			for num%i == 0 {
+				num /= i
+			}
+		}
+	}
+	return num == 1
+}
diff --git a/Algorithms/0263. Ugly Number/263. Ugly Number_test.go b/leetcode/0263.Ugly-Number/263. Ugly Number_test.go
similarity index 93%
rename from Algorithms/0263. Ugly Number/263. Ugly Number_test.go
rename to leetcode/0263.Ugly-Number/263. Ugly Number_test.go
index 83a96e003..72ba1385a 100644
--- a/Algorithms/0263. Ugly Number/263. Ugly Number_test.go	
+++ b/leetcode/0263.Ugly-Number/263. Ugly Number_test.go	
@@ -26,17 +26,17 @@ func Test_Problem263(t *testing.T) {
 
 	qs := []question263{
 
-		question263{
+		{
 			para263{6},
 			ans263{true},
 		},
 
-		question263{
+		{
 			para263{8},
 			ans263{true},
 		},
 
-		question263{
+		{
 			para263{14},
 			ans263{false},
 		},
diff --git a/Algorithms/0263. Ugly Number/README.md b/leetcode/0263.Ugly-Number/README.md
similarity index 100%
rename from Algorithms/0263. Ugly Number/README.md
rename to leetcode/0263.Ugly-Number/README.md
diff --git a/leetcode/0264.Ugly-Number-II/264. Ugly Number II.go b/leetcode/0264.Ugly-Number-II/264. Ugly Number II.go
new file mode 100644
index 000000000..a34eb0b91
--- /dev/null
+++ b/leetcode/0264.Ugly-Number-II/264. Ugly Number II.go	
@@ -0,0 +1,27 @@
+package leetcode
+
+func nthUglyNumber(n int) int {
+	dp, p2, p3, p5 := make([]int, n+1), 1, 1, 1
+	dp[0], dp[1] = 0, 1
+	for i := 2; i <= n; i++ {
+		x2, x3, x5 := dp[p2]*2, dp[p3]*3, dp[p5]*5
+		dp[i] = min(min(x2, x3), x5)
+		if dp[i] == x2 {
+			p2++
+		}
+		if dp[i] == x3 {
+			p3++
+		}
+		if dp[i] == x5 {
+			p5++
+		}
+	}
+	return dp[n]
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0264.Ugly-Number-II/264. Ugly Number II_test.go b/leetcode/0264.Ugly-Number-II/264. Ugly Number II_test.go
new file mode 100644
index 000000000..b5c732140
--- /dev/null
+++ b/leetcode/0264.Ugly-Number-II/264. Ugly Number II_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question264 struct {
+	para264
+	ans264
+}
+
+// para 是参数
+// one 代表第一个参数
+type para264 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans264 struct {
+	one int
+}
+
+func Test_Problem264(t *testing.T) {
+
+	qs := []question264{
+
+		{
+			para264{10},
+			ans264{12},
+		},
+
+		{
+			para264{1},
+			ans264{1},
+		},
+
+		{
+			para264{6},
+			ans264{6},
+		},
+
+		{
+			para264{8},
+			ans264{9},
+		},
+
+		{
+			para264{14},
+			ans264{20},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 264------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans264, q.para264
+		fmt.Printf("【input】:%v       【output】:%v\n", p, nthUglyNumber(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0264.Ugly-Number-II/README.md b/leetcode/0264.Ugly-Number-II/README.md
new file mode 100644
index 000000000..431eadfa3
--- /dev/null
+++ b/leetcode/0264.Ugly-Number-II/README.md
@@ -0,0 +1,69 @@
+# [264. Ugly Number II](https://leetcode.com/problems/ugly-number-ii/)
+
+
+## 题目
+
+Given an integer `n`, return *the* `nth` ***ugly number***.
+
+**Ugly number** is a positive number whose prime factors only include `2`, `3`, and/or `5`.
+
+**Example 1:**
+
+```
+Input: n = 10
+Output: 12
+Explanation: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12] is the sequence of the first 10 ugly numbers.
+```
+
+**Example 2:**
+
+```
+Input: n = 1
+Output: 1
+Explanation: 1 is typically treated as an ugly number.
+```
+
+**Constraints:**
+
+- `1 <= n <= 1690`
+
+## 题目大意
+
+给你一个整数 `n` ，请你找出并返回第 `n` 个 **丑数** 。**丑数** 就是只包含质因数 `2`、`3` 和/或 `5` 的正整数。
+
+## 解题思路
+
+- 解法一，生成丑数的方法：先用最小质因数 1，分别和 2，3，5 相乘，得到的数是丑数，不断的将这些数分别和 2，3，5 相乘，得到的数去重以后，从小到大排列，第 n 个数即为所求。排序可用最小堆实现，去重用 map 去重。时间复杂度 O(n log n)，空间复杂度 O(n)
+- 上面的解法耗时在排序中，需要排序的根源是小的丑数乘以 5 大于了大的丑数乘以 2 。如何保证每次乘积以后，找出有序的丑数，是去掉排序，提升时间复杂度的关键。举个例子很容易想通：初始状态丑数只有 {1}，乘以 2，3，5 以后，将最小的结果存入集合中 {1,2}。下一轮再相乘，由于上一轮 1 已经和 2 相乘过了，1 不要再和 2 相乘了，所以这一轮 1 和 3，5 相乘。2 和 2，3，5 相乘。将最小的结果存入集合中 {1,2,3}，按照这样的策略往下比较，每轮选出的丑数是有序且不重复的。具体实现利用 3 个指针和一个数组即可实现。时间复杂度 O(n)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+func nthUglyNumber(n int) int {
+	dp, p2, p3, p5 := make([]int, n+1), 1, 1, 1
+	dp[0], dp[1] = 0, 1
+	for i := 2; i <= n; i++ {
+		x2, x3, x5 := dp[p2]*2, dp[p3]*3, dp[p5]*5
+		dp[i] = min(min(x2, x3), x5)
+		if dp[i] == x2 {
+			p2++
+		}
+		if dp[i] == x3 {
+			p3++
+		}
+		if dp[i] == x5 {
+			p5++
+		}
+	}
+	return dp[n]
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0268. Missing Number/268. Missing Number.go b/leetcode/0268.Missing-Number/268. Missing Number.go
similarity index 100%
rename from Algorithms/0268. Missing Number/268. Missing Number.go
rename to leetcode/0268.Missing-Number/268. Missing Number.go
diff --git a/Algorithms/0268. Missing Number/268. Missing Number_test.go b/leetcode/0268.Missing-Number/268. Missing Number_test.go
similarity index 95%
rename from Algorithms/0268. Missing Number/268. Missing Number_test.go
rename to leetcode/0268.Missing-Number/268. Missing Number_test.go
index 5997ec6ed..a3a7246c0 100644
--- a/Algorithms/0268. Missing Number/268. Missing Number_test.go	
+++ b/leetcode/0268.Missing-Number/268. Missing Number_test.go	
@@ -26,12 +26,12 @@ func Test_Problem268(t *testing.T) {
 
 	qs := []question268{
 
-		question268{
+		{
 			para268{[]int{3, 0, 1}},
 			ans268{2},
 		},
 
-		question268{
+		{
 			para268{[]int{9, 6, 4, 2, 3, 5, 7, 0, 1}},
 			ans268{8},
 		},
diff --git a/leetcode/0268.Missing-Number/README.md b/leetcode/0268.Missing-Number/README.md
new file mode 100755
index 000000000..97895940d
--- /dev/null
+++ b/leetcode/0268.Missing-Number/README.md
@@ -0,0 +1,30 @@
+# [268. Missing Number](https://leetcode.com/problems/missing-number/)
+
+
+## 题目
+
+Given an array containing n distinct numbers taken from `0, 1, 2, ..., n`, find the one that is missing from the array.
+
+**Example 1:**
+
+    Input: [3,0,1]
+    Output: 2
+
+**Example 2:**
+
+    Input: [9,6,4,2,3,5,7,0,1]
+    Output: 8
+
+**Note**:Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
+
+
+## 题目大意
+
+给定一个包含 0, 1, 2, ..., n 中 n 个数的序列，找出 0 .. n 中没有出现在序列中的那个数。算法应该具有线性时间复杂度。你能否仅使用额外常数空间来实现?
+
+
+
+## 解题思路
+
+
+- 要求找出 `0, 1, 2, ..., n` 中缺失的那个数。还是利用异或的性质，`X^X = 0`。这里我们需要构造一个 X，用数组下标就可以了。数字下标是从 `[0，n-1]`，数字是 `[0，n]`，依次把数组里面的数组进行异或，把结果和 n 再异或一次，中和掉出现的数字，剩下的那个数字就是之前没有出现过的，缺失的数字。
diff --git a/leetcode/0274.H-Index/274. H-Index.go b/leetcode/0274.H-Index/274. H-Index.go
new file mode 100644
index 000000000..b26d4d823
--- /dev/null
+++ b/leetcode/0274.H-Index/274. H-Index.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+// 解法一
+func hIndex(citations []int) int {
+	n := len(citations)
+	buckets := make([]int, n+1)
+	for _, c := range citations {
+		if c >= n {
+			buckets[n]++
+		} else {
+			buckets[c]++
+		}
+	}
+	count := 0
+	for i := n; i >= 0; i-- {
+		count += buckets[i]
+		if count >= i {
+			return i
+		}
+	}
+	return 0
+}
+
+// 解法二
+func hIndex1(citations []int) int {
+	quickSort164(citations, 0, len(citations)-1)
+	hIndex := 0
+	for i := len(citations) - 1; i >= 0; i-- {
+		if citations[i] >= len(citations)-i {
+			hIndex++
+		} else {
+			break
+		}
+	}
+	return hIndex
+}
+
+func quickSort164(a []int, lo, hi int) {
+	if lo >= hi {
+		return
+	}
+	p := partition164(a, lo, hi)
+	quickSort164(a, lo, p-1)
+	quickSort164(a, p+1, hi)
+}
+
+func partition164(a []int, lo, hi int) int {
+	pivot := a[hi]
+	i := lo - 1
+	for j := lo; j < hi; j++ {
+		if a[j] < pivot {
+			i++
+			a[j], a[i] = a[i], a[j]
+		}
+	}
+	a[i+1], a[hi] = a[hi], a[i+1]
+	return i + 1
+}
diff --git a/Algorithms/0274. H-Index/274. H-Index_test.go b/leetcode/0274.H-Index/274. H-Index_test.go
similarity index 92%
rename from Algorithms/0274. H-Index/274. H-Index_test.go
rename to leetcode/0274.H-Index/274. H-Index_test.go
index b302ab8eb..35e178da6 100644
--- a/Algorithms/0274. H-Index/274. H-Index_test.go	
+++ b/leetcode/0274.H-Index/274. H-Index_test.go	
@@ -26,21 +26,21 @@ func Test_Problem274(t *testing.T) {
 
 	qs := []question274{
 
-		question274{
+		{
 			para274{[]int{3, 6, 9, 1}},
 			ans274{3},
 		},
-		question274{
+		{
 			para274{[]int{1}},
 			ans274{1},
 		},
 
-		question274{
+		{
 			para274{[]int{}},
 			ans274{0},
 		},
 
-		question274{
+		{
 			para274{[]int{3, 0, 6, 1, 5}},
 			ans274{3},
 		},
diff --git a/Algorithms/0274. H-Index/README.md b/leetcode/0274.H-Index/README.md
similarity index 100%
rename from Algorithms/0274. H-Index/README.md
rename to leetcode/0274.H-Index/README.md
diff --git a/Algorithms/0275. H-Index II/275. H-Index II.go b/leetcode/0275.H-Index-II/275. H-Index II.go
similarity index 100%
rename from Algorithms/0275. H-Index II/275. H-Index II.go
rename to leetcode/0275.H-Index-II/275. H-Index II.go
diff --git a/Algorithms/0275. H-Index II/275. H-Index II_test.go b/leetcode/0275.H-Index-II/275. H-Index II_test.go
similarity index 91%
rename from Algorithms/0275. H-Index II/275. H-Index II_test.go
rename to leetcode/0275.H-Index-II/275. H-Index II_test.go
index 87b8a55b1..c0ac4e36b 100644
--- a/Algorithms/0275. H-Index II/275. H-Index II_test.go	
+++ b/leetcode/0275.H-Index-II/275. H-Index II_test.go	
@@ -26,26 +26,26 @@ func Test_Problem275(t *testing.T) {
 
 	qs := []question275{
 
-		question275{
+		{
 			para275{[]int{3, 6, 9, 1}},
 			ans275{3},
 		},
-		question275{
+		{
 			para275{[]int{1}},
 			ans275{1},
 		},
 
-		question275{
+		{
 			para275{[]int{}},
 			ans275{0},
 		},
 
-		question275{
+		{
 			para275{[]int{3, 0, 6, 1, 5}},
 			ans275{3},
 		},
 
-		question275{
+		{
 			para275{[]int{0, 1, 3, 5, 6}},
 			ans275{3},
 		},
diff --git a/leetcode/0275.H-Index-II/README.md b/leetcode/0275.H-Index-II/README.md
new file mode 100755
index 000000000..cc94f3343
--- /dev/null
+++ b/leetcode/0275.H-Index-II/README.md
@@ -0,0 +1,49 @@
+# [275. H-Index II](https://leetcode.com/problems/h-index-ii/)
+
+## 题目
+
+Given an array of citations **sorted in ascending order** (each citation is a non-negative integer) of a researcher, write a function to compute the researcher's h-index.
+
+According to the [definition of h-index on Wikipedia](https://en.wikipedia.org/wiki/H-index): "A scientist has index h if h of his/her N papers have **at least** h citations each, and the other N − h papers have **no more than** h citations each."
+
+**Example:**
+
+    Input: citations = [0,1,3,5,6]
+    Output: 3 
+    Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had 
+                 received 0, 1, 3, 5, 6 citations respectively. 
+                 Since the researcher has 3 papers with at least 3 citations each and the remaining 
+                 two with no more than 3 citations each, her h-index is 3.
+
+**Note:**
+
+If there are several possible values for *h*, the maximum one is taken as the h-index.
+
+**Follow up:**
+
+- This is a follow up problem to [H-Index](https://leetcode.com/problems/h-index/description/), where `citations` is now guaranteed to be sorted in ascending order.
+- Could you solve it in logarithmic time complexity?
+
+
+
+## 题目大意
+
+
+给定一位研究者论文被引用次数的数组（被引用次数是非负整数），数组已经按照升序排列。编写一个方法，计算出研究者的 h 指数。
+
+h 指数的定义: “h 代表“高引用次数”（high citations），一名科研人员的 h 指数是指他（她）的 （N 篇论文中）至多有 h 篇论文分别被引用了至少 h 次。（其余的 N - h 篇论文每篇被引用次数不多于 h 次。）"
+
+说明:
+
+- 如果 h 有多有种可能的值 ，h 指数是其中最大的那个。
+
+进阶：
+
+- 这是 H 指数 的延伸题目，本题中的 citations 数组是保证有序的。
+你可以优化你的算法到对数时间复杂度吗？
+
+
+## 解题思路
+
+- 给出一个数组，代表该作者论文被引用次数，要求这个作者的 h 指数。h 指数定义："高引用次数”（`high citations`），一名科研人员的 h 指数是指他（她）的 （N 篇论文中）至多有 h 篇论文分别被引用了至少 h 次。（其余的 N - h 篇论文每篇被引用次数不多于 h 次。）
+- 这一题要找出 h 指数，即要找到一个边界，在这个边界上为最多的 h 指数。可以用二分搜索来解决这道题。当 `len(citations)-mid > citations[mid]` 时，说明 h 指数的边界一定在右边，因为最多 `len(citations)-mid` 篇数比引用数 `citations[mid]` 还要大。否则 h 指数的边界在左边界，缩小边界以后继续二分。找到边界以后，最终求的是 h 指数，用 `len(citations) - low` 即是结果。
diff --git a/leetcode/0278.First-Bad-Version/278. First Bad Version.go b/leetcode/0278.First-Bad-Version/278. First Bad Version.go
new file mode 100644
index 000000000..d4517c087
--- /dev/null
+++ b/leetcode/0278.First-Bad-Version/278. First Bad Version.go	
@@ -0,0 +1,19 @@
+package leetcode
+
+import "sort"
+
+/**
+ * Forward declaration of isBadVersion API.
+ * @param   version   your guess about first bad version
+ * @return 	 	      true if current version is bad
+ *			          false if current version is good
+ * func isBadVersion(version int) bool;
+ */
+
+func firstBadVersion(n int) int {
+	return sort.Search(n, func(version int) bool { return isBadVersion(version) })
+}
+
+func isBadVersion(version int) bool {
+	return true
+}
diff --git a/leetcode/0278.First-Bad-Version/278. First Bad Version_test.go b/leetcode/0278.First-Bad-Version/278. First Bad Version_test.go
new file mode 100644
index 000000000..5792af66c
--- /dev/null
+++ b/leetcode/0278.First-Bad-Version/278. First Bad Version_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question278 struct {
+	para278
+	ans278
+}
+
+// para 是参数
+// one 代表第一个参数
+type para278 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans278 struct {
+	one int
+}
+
+func Test_Problem278(t *testing.T) {
+
+	qs := []question278{
+
+		{
+			para278{5},
+			ans278{4},
+		},
+		{
+			para278{1},
+			ans278{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 278------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans278, q.para278
+		fmt.Printf("【input】:%v       【output】:%v\n", p, firstBadVersion(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0278.First-Bad-Version/README.md b/leetcode/0278.First-Bad-Version/README.md
new file mode 100644
index 000000000..cb9bfe5ab
--- /dev/null
+++ b/leetcode/0278.First-Bad-Version/README.md
@@ -0,0 +1,62 @@
+# [278. First Bad Version](https://leetcode.com/problems/first-bad-version/)
+
+## 题目
+
+You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.
+
+Suppose you have `n` versions `[1, 2, ..., n]` and you want to find out the first bad one, which causes all the following ones to be bad.
+
+You are given an API `bool isBadVersion(version)` which returns whether `version` is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
+
+**Example 1:**
+
+```
+Input: n = 5, bad = 4
+Output: 4
+Explanation:
+call isBadVersion(3) -> false
+call isBadVersion(5) -> true
+call isBadVersion(4) -> true
+Then 4 is the first bad version.
+
+```
+
+**Example 2:**
+
+```
+Input: n = 1, bad = 1
+Output: 1
+
+```
+
+**Constraints:**
+
+- `1 <= bad <= n <= 231 - 1`
+
+## 题目大意
+
+你是产品经理，目前正在带领一个团队开发新的产品。不幸的是，你的产品的最新版本没有通过质量检测。由于每个版本都是基于之前的版本开发的，所以错误的版本之后的所有版本都是错的。假设你有 n 个版本 [1, 2, ..., n]，你想找出导致之后所有版本出错的第一个错误的版本。你可以通过调用 bool isBadVersion(version) 接口来判断版本号 version 是否在单元测试中出错。实现一个函数来查找第一个错误的版本。你应该尽量减少对调用 API 的次数。
+
+## 解题思路
+
+- 我们知道开发产品迭代的版本，如果当一个版本为正确版本，则该版本之前的所有版本均为正确版本；当一个版本为错误版本，则该版本之后的所有版本均为错误版本。利用这个性质就可以进行二分查找。利用二分搜索，也可以满足减少对调用 API 的次数的要求。时间复杂度：O(logn)，其中 n 是给定版本的数量。空间复杂度：O(1)。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+/**
+ * Forward declaration of isBadVersion API.
+ * @param   version   your guess about first bad version
+ * @return 	 	      true if current version is bad
+ *			          false if current version is good
+ * func isBadVersion(version int) bool;
+ */
+
+func firstBadVersion(n int) int {
+	return sort.Search(n, func(version int) bool { return isBadVersion(version) })
+}
+```
\ No newline at end of file
diff --git a/leetcode/0279.Perfect-Squares/279. Perfect Squares.go b/leetcode/0279.Perfect-Squares/279. Perfect Squares.go
new file mode 100644
index 000000000..99771598c
--- /dev/null
+++ b/leetcode/0279.Perfect-Squares/279. Perfect Squares.go	
@@ -0,0 +1,33 @@
+package leetcode
+
+import "math"
+
+func numSquares(n int) int {
+	if isPerfectSquare(n) {
+		return 1
+	}
+	if checkAnswer4(n) {
+		return 4
+	}
+	for i := 1; i*i <= n; i++ {
+		j := n - i*i
+		if isPerfectSquare(j) {
+			return 2
+		}
+	}
+	return 3
+}
+
+// 判断是否为完全平方数
+func isPerfectSquare(n int) bool {
+	sq := int(math.Floor(math.Sqrt(float64(n))))
+	return sq*sq == n
+}
+
+// 判断是否能表示为 4^k*(8m+7)
+func checkAnswer4(x int) bool {
+	for x%4 == 0 {
+		x /= 4
+	}
+	return x%8 == 7
+}
diff --git a/leetcode/0279.Perfect-Squares/279. Perfect Squares_test.go b/leetcode/0279.Perfect-Squares/279. Perfect Squares_test.go
new file mode 100644
index 000000000..6c8a13ae1
--- /dev/null
+++ b/leetcode/0279.Perfect-Squares/279. Perfect Squares_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question279 struct {
+	para279
+	ans279
+}
+
+// para 是参数
+// one 代表第一个参数
+type para279 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans279 struct {
+	one int
+}
+
+func Test_Problem279(t *testing.T) {
+
+	qs := []question279{
+
+		{
+			para279{13},
+			ans279{2},
+		},
+		{
+			para279{12},
+			ans279{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 279------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans279, q.para279
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numSquares(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0279.Perfect-Squares/README.md b/leetcode/0279.Perfect-Squares/README.md
new file mode 100644
index 000000000..5e8e466df
--- /dev/null
+++ b/leetcode/0279.Perfect-Squares/README.md
@@ -0,0 +1,78 @@
+# [279. Perfect Squares](https://leetcode.com/problems/perfect-squares/)
+
+
+## 题目
+
+Given an integer `n`, return *the least number of perfect square numbers that sum to* `n`.
+
+A **perfect square** is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, `1`, `4`, `9`, and `16` are perfect squares while `3` and `11` are not.
+
+**Example 1:**
+
+```
+Input: n = 12
+Output: 3
+Explanation: 12 = 4 + 4 + 4.
+```
+
+**Example 2:**
+
+```
+Input: n = 13
+Output: 2
+Explanation: 13 = 4 + 9.
+```
+
+**Constraints:**
+
+- `1 <= n <= 104`
+
+## 题目大意
+
+给定正整数 n，找到若干个完全平方数（比如 1, 4, 9, 16, ...）使得它们的和等于 n。你需要让组成和的完全平方数的个数最少。给你一个整数 n ，返回和为 n 的完全平方数的 最少数量 。
+
+完全平方数 是一个整数，其值等于另一个整数的平方；换句话说，其值等于一个整数自乘的积。例如，1、4、9 和 16 都是完全平方数，而 3 和 11 不是。
+
+## 解题思路
+
+- 由拉格朗日的四平方定理可得，每个自然数都可以表示为四个整数平方之和。 其中四个数字是整数。四平方和定理证明了任意一个正整数都可以被表示为至多四个正整数的平方和。这给出了本题的答案的上界。
+- 四平方和定理可以推出三平方和推论：当且仅当 {{< katex >}} n \neq 4^{k} \times (8*m + 7){{< /katex >}} 时，n 可以被表示为至多三个正整数的平方和。所以当 {{< katex >}} n = 4^{k} * (8*m + 7){{< /katex >}}  时，n 只能被表示为四个正整数的平方和。此时我们可以直接返回 4。
+- 当 {{< katex >}} n \neq 4^{k} \times (8*m + 7){{< /katex >}} 时，需要判断 n 到底可以分解成几个完全平方数之和。答案肯定是 1，2，3 中的一个。题目要求我们求最小的，所以从 1 开始一个个判断是否满足。如果答案为 1，代表 n 为完全平方数，这很好判断。如果答案为 2，代表 {{< katex >}} n = a^{2} + b^{2} {{< /katex >}}，枚举 {{< katex >}} 1 \leqslant a \leqslant \sqrt{n} {{< /katex >}}，判断  {{< katex >}} n - a^{2} {{< /katex >}} 是否为完全平方数。当 1 和 2 都排除了，剩下的答案只能为 3 了。
+
+## 代码
+
+```go
+package leetcode
+
+import "math"
+
+func numSquares(n int) int {
+	if isPerfectSquare(n) {
+		return 1
+	}
+	if checkAnswer4(n) {
+		return 4
+	}
+	for i := 1; i*i <= n; i++ {
+		j := n - i*i
+		if isPerfectSquare(j) {
+			return 2
+		}
+	}
+	return 3
+}
+
+// 判断是否为完全平方数
+func isPerfectSquare(n int) bool {
+	sq := int(math.Floor(math.Sqrt(float64(n))))
+	return sq*sq == n
+}
+
+// 判断是否能表示为 4^k*(8m+7)
+func checkAnswer4(x int) bool {
+	for x%4 == 0 {
+		x /= 4
+	}
+	return x%8 == 7
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0283. Move Zeroes/283. Move Zeroes.go b/leetcode/0283.Move-Zeroes/283. Move Zeroes.go
similarity index 88%
rename from Algorithms/0283. Move Zeroes/283. Move Zeroes.go
rename to leetcode/0283.Move-Zeroes/283. Move Zeroes.go
index a342e66ee..f3a1f4c12 100644
--- a/Algorithms/0283. Move Zeroes/283. Move Zeroes.go	
+++ b/leetcode/0283.Move-Zeroes/283. Move Zeroes.go	
@@ -9,10 +9,8 @@ func moveZeroes(nums []int) {
 		if nums[i] != 0 {
 			if i != j {
 				nums[i], nums[j] = nums[j], nums[i]
-				j++
-			} else {
-				j++
 			}
+			j++
 		}
 	}
 }
diff --git a/Algorithms/0283. Move Zeroes/283. Move Zeroes_test.go b/leetcode/0283.Move-Zeroes/283. Move Zeroes_test.go
similarity index 92%
rename from Algorithms/0283. Move Zeroes/283. Move Zeroes_test.go
rename to leetcode/0283.Move-Zeroes/283. Move Zeroes_test.go
index 68bc1042a..1cc7e531f 100644
--- a/Algorithms/0283. Move Zeroes/283. Move Zeroes_test.go	
+++ b/leetcode/0283.Move-Zeroes/283. Move Zeroes_test.go	
@@ -26,32 +26,32 @@ func Test_Problem283(t *testing.T) {
 
 	qs := []question283{
 
-		question283{
+		{
 			para283{[]int{1, 0, 1}},
 			ans283{[]int{1, 1, 0}},
 		},
 
-		question283{
+		{
 			para283{[]int{0, 1, 0, 3, 0, 12}},
 			ans283{[]int{1, 3, 12, 0, 0, 0}},
 		},
 
-		question283{
+		{
 			para283{[]int{0, 1, 0, 3, 0, 0, 0, 0, 1, 12}},
 			ans283{[]int{1, 3, 1, 12, 0, 0, 0, 0, 0}},
 		},
 
-		question283{
+		{
 			para283{[]int{0, 0, 0, 0, 0, 0, 0, 0, 12, 1}},
 			ans283{[]int{12, 1, 0, 0, 0, 0, 0, 0, 0, 0}},
 		},
 
-		question283{
+		{
 			para283{[]int{0, 0, 0, 0, 0}},
 			ans283{[]int{0, 0, 0, 0, 0}},
 		},
 
-		question283{
+		{
 			para283{[]int{1}},
 			ans283{[]int{1}},
 		},
diff --git a/Algorithms/0283. Move Zeroes/README.md b/leetcode/0283.Move-Zeroes/README.md
similarity index 100%
rename from Algorithms/0283. Move Zeroes/README.md
rename to leetcode/0283.Move-Zeroes/README.md
diff --git a/leetcode/0284.Peeking-Iterator/284. Peeking Iterator.go b/leetcode/0284.Peeking-Iterator/284. Peeking Iterator.go
new file mode 100644
index 000000000..4b5b578be
--- /dev/null
+++ b/leetcode/0284.Peeking-Iterator/284. Peeking Iterator.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+//Below is the interface for Iterator, which is already defined for you.
+
+type Iterator struct {
+}
+
+func (this *Iterator) hasNext() bool {
+	// Returns true if the iteration has more elements.
+	return true
+}
+
+func (this *Iterator) next() int {
+	// Returns the next element in the iteration.
+	return 0
+}
+
+type PeekingIterator struct {
+	nextEl int
+	hasEl  bool
+	iter   *Iterator
+}
+
+func Constructor(iter *Iterator) *PeekingIterator {
+	return &PeekingIterator{
+		iter: iter,
+	}
+}
+
+func (this *PeekingIterator) hasNext() bool {
+	if this.hasEl {
+		return true
+	}
+	return this.iter.hasNext()
+}
+
+func (this *PeekingIterator) next() int {
+	if this.hasEl {
+		this.hasEl = false
+		return this.nextEl
+	} else {
+		return this.iter.next()
+	}
+}
+
+func (this *PeekingIterator) peek() int {
+	if this.hasEl {
+		return this.nextEl
+	}
+	this.hasEl = true
+	this.nextEl = this.iter.next()
+	return this.nextEl
+}
diff --git a/leetcode/0284.Peeking-Iterator/284. Peeking Iterator_test.go b/leetcode/0284.Peeking-Iterator/284. Peeking Iterator_test.go
new file mode 100644
index 000000000..5d0d958fd
--- /dev/null
+++ b/leetcode/0284.Peeking-Iterator/284. Peeking Iterator_test.go	
@@ -0,0 +1,9 @@
+package leetcode
+
+import (
+	"testing"
+)
+
+func Test_Problem284(t *testing.T) {
+
+}
diff --git a/leetcode/0284.Peeking-Iterator/README.md b/leetcode/0284.Peeking-Iterator/README.md
new file mode 100644
index 000000000..f2562d3fa
--- /dev/null
+++ b/leetcode/0284.Peeking-Iterator/README.md
@@ -0,0 +1,87 @@
+# [284. Peeking Iterator](https://leetcode.com/problems/peeking-iterator/)
+
+## 题目
+
+Given an Iterator class interface with methods: `next()` and `hasNext()`, design and implement a PeekingIterator that support the `peek()` operation -- it essentially peek() at the element that will be returned by the next call to next().
+
+**Example:**
+
+```
+Assume that the iterator is initialized to the beginning of the list: [1,2,3].
+
+Call next() gets you 1, the first element in the list.
+Now you call peek() and it returns 2, the next element. Calling next() after that still return 2. 
+You call next() the final time and it returns 3, the last element. 
+Calling hasNext() after that should return false.
+```
+
+**Follow up**: How would you extend your design to be generic and work with all types, not just integer?
+
+## 题目大意
+
+给定一个迭代器类的接口，接口包含两个方法： next() 和 hasNext()。设计并实现一个支持 peek() 操作的顶端迭代器 -- 其本质就是把原本应由 next() 方法返回的元素 peek() 出来。
+
+> peek() 是偷看的意思。偷偷看一看下一个元素是什么，但是并不是 next() 访问。
+
+## 解题思路
+
+- 简单题。在 PeekingIterator 内部保存 2 个变量，一个是下一个元素值，另一个是是否有下一个元素。在 next() 操作和 hasNext() 操作时，访问保存的这 2 个变量。peek() 操作也比较简单，判断是否有下一个元素，如果有，即返回该元素值。这里实现了迭代指针不移动的功能。如果没有保存下一个元素值，即没有 peek() 偷看，next() 操作继续往后移动指针，读取后一位元素。
+- 这里复用了是否有下一个元素值，来判断 hasNext() 和 peek() 操作中不移动指针的逻辑。
+
+## 代码
+
+```go
+package leetcode
+
+//Below is the interface for Iterator, which is already defined for you.
+
+type Iterator struct {
+}
+
+func (this *Iterator) hasNext() bool {
+	// Returns true if the iteration has more elements.
+	return true
+}
+
+func (this *Iterator) next() int {
+	// Returns the next element in the iteration.
+	return 0
+}
+
+type PeekingIterator struct {
+	nextEl int
+	hasEl  bool
+	iter   *Iterator
+}
+
+func Constructor(iter *Iterator) *PeekingIterator {
+	return &PeekingIterator{
+		iter: iter,
+	}
+}
+
+func (this *PeekingIterator) hasNext() bool {
+	if this.hasEl {
+		return true
+	}
+	return this.iter.hasNext()
+}
+
+func (this *PeekingIterator) next() int {
+	if this.hasEl {
+		this.hasEl = false
+		return this.nextEl
+	} else {
+		return this.iter.next()
+	}
+}
+
+func (this *PeekingIterator) peek() int {
+	if this.hasEl {
+		return this.nextEl
+	}
+	this.hasEl = true
+	this.nextEl = this.iter.next()
+	return this.nextEl
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0287. Find the Duplicate Number/287. Find the Duplicate Number.go b/leetcode/0287.Find-the-Duplicate-Number/287. Find the Duplicate Number.go
similarity index 100%
rename from Algorithms/0287. Find the Duplicate Number/287. Find the Duplicate Number.go
rename to leetcode/0287.Find-the-Duplicate-Number/287. Find the Duplicate Number.go
diff --git a/Algorithms/0287. Find the Duplicate Number/287. Find the Duplicate Number_test.go b/leetcode/0287.Find-the-Duplicate-Number/287. Find the Duplicate Number_test.go
similarity index 87%
rename from Algorithms/0287. Find the Duplicate Number/287. Find the Duplicate Number_test.go
rename to leetcode/0287.Find-the-Duplicate-Number/287. Find the Duplicate Number_test.go
index 68b126aa7..6e4aa1b0b 100644
--- a/Algorithms/0287. Find the Duplicate Number/287. Find the Duplicate Number_test.go	
+++ b/leetcode/0287.Find-the-Duplicate-Number/287. Find the Duplicate Number_test.go	
@@ -26,22 +26,17 @@ func Test_Problem287(t *testing.T) {
 
 	qs := []question287{
 
-		question287{
+		{
 			para287{[]int{1, 3, 4, 2, 2}},
 			ans287{2},
 		},
 
-		question287{
+		{
 			para287{[]int{3, 1, 3, 4, 2}},
 			ans287{3},
 		},
 
-		question287{
-			para287{[]int{}},
-			ans287{0},
-		},
-
-		question287{
+		{
 			para287{[]int{2, 2, 2, 2, 2}},
 			ans287{2},
 		},
diff --git a/leetcode/0287.Find-the-Duplicate-Number/README.md b/leetcode/0287.Find-the-Duplicate-Number/README.md
new file mode 100644
index 000000000..72ceb6fe0
--- /dev/null
+++ b/leetcode/0287.Find-the-Duplicate-Number/README.md
@@ -0,0 +1,41 @@
+# [287. Find the Duplicate Number](https://leetcode.com/problems/find-the-duplicate-number/)
+
+## 题目
+
+Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
+
+Example 1:
+
+```c
+Input: [1,3,4,2,2]
+Output: 2
+```
+
+Example 2:
+
+```c
+Input: [3,1,3,4,2]
+Output: 3
+```
+
+Note:  
+
+- You must not modify the array (assume the array is read only).
+- You must use only constant, O(1) extra space.
+- Your runtime complexity should be less than O(n^2).
+- There is only one duplicate number in the array, but it could be repeated more than once.
+
+## 题目大意
+
+给出 n + 1 个数，这些数是在 1-n 中取值的，同一个数字可以出现多次。要求找出这些数中重复的数字。时间复杂度最好低于 O(n^2)，空间复杂度为 O(1)。
+
+## 解题思路
+
+- 这道题比较巧的思路是，将这些数字想象成链表中的结点，数组中数字代表下一个结点的数组下标。找重复的数字就是找链表中成环的那个点。由于题目保证了一定会有重复的数字，所以一定会成环。所以用快慢指针的方法，快指针一次走 2 步，慢指针一次走 1 步，相交以后，快指针从头开始，每次走一步，再次遇见的时候就是成环的交点处，也即是重复数字所在的地方。
+- 这一题有多种做法。可以用快慢指针求解。还可以用二分搜索：(这里的题解感谢 [@imageslr](https://github.com/imageslr) 指出错误）：
+	1. 假设有 n+1 个数，则可能重复的数位于区间 [1, n] 中。记该区间最小值、最大值和中间值为 low、high、mid
+	2. 遍历整个数组，统计小于等于 mid 的整数的个数，至多为 mid 个
+	3. 如果超过 mid 个就说明重复的数存在于区间 [low,mid] （闭区间）中；否则，重复的数存在于区间 (mid, high] （左开右闭）中
+	4. 缩小区间，继续重复步骤 2、3，直到区间变成 1 个整数，即 low == high
+	5. 整数 low 就是要找的重复的数
+- 另外一个做法是，先将数组排序，依照下标是从 0 开始递增的特性，那么数组里面的数字与下标的差值应该是越来越大。如果出现了相同的数字，下标变大，差值应该比前一个数字小，出现了这个情况就说明找到了相同数字了。
\ No newline at end of file
diff --git a/Algorithms/0290. Word Pattern/290. Word Pattern.go b/leetcode/0290.Word-Pattern/290. Word Pattern.go
similarity index 100%
rename from Algorithms/0290. Word Pattern/290. Word Pattern.go
rename to leetcode/0290.Word-Pattern/290. Word Pattern.go
diff --git a/Algorithms/0290. Word Pattern/290. Word Pattern_test.go b/leetcode/0290.Word-Pattern/290. Word Pattern_test.go
similarity index 93%
rename from Algorithms/0290. Word Pattern/290. Word Pattern_test.go
rename to leetcode/0290.Word-Pattern/290. Word Pattern_test.go
index c9356501a..4997c0ed7 100644
--- a/Algorithms/0290. Word Pattern/290. Word Pattern_test.go	
+++ b/leetcode/0290.Word-Pattern/290. Word Pattern_test.go	
@@ -27,22 +27,22 @@ func Test_Problem290(t *testing.T) {
 
 	qs := []question290{
 
-		question290{
+		{
 			para290{"abba", "dog cat cat dog"},
 			ans290{true},
 		},
 
-		question290{
+		{
 			para290{"abba", "dog cat cat fish"},
 			ans290{false},
 		},
 
-		question290{
+		{
 			para290{"aaaa", "dog cat cat dog"},
 			ans290{false},
 		},
 
-		question290{
+		{
 			para290{"abba", "dog dog dog dog"},
 			ans290{false},
 		},
diff --git a/Algorithms/0290. Word Pattern/README.md b/leetcode/0290.Word-Pattern/README.md
similarity index 100%
rename from Algorithms/0290. Word Pattern/README.md
rename to leetcode/0290.Word-Pattern/README.md
diff --git a/leetcode/0297.Serialize-and-Deserialize-Binary-Tree/297.Serialize and Deserialize Binary Tree.go b/leetcode/0297.Serialize-and-Deserialize-Binary-Tree/297.Serialize and Deserialize Binary Tree.go
new file mode 100644
index 000000000..33a631370
--- /dev/null
+++ b/leetcode/0297.Serialize-and-Deserialize-Binary-Tree/297.Serialize and Deserialize Binary Tree.go	
@@ -0,0 +1,54 @@
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type TreeNode = structures.TreeNode
+
+type Codec struct {
+	builder strings.Builder
+	input   []string
+}
+
+func Constructor() Codec {
+	return Codec{}
+}
+
+// Serializes a tree to a single string.
+func (this *Codec) serialize(root *TreeNode) string {
+	if root == nil {
+		this.builder.WriteString("#,")
+		return ""
+	}
+	this.builder.WriteString(strconv.Itoa(root.Val) + ",")
+	this.serialize(root.Left)
+	this.serialize(root.Right)
+	return this.builder.String()
+}
+
+// Deserializes your encoded data to tree.
+func (this *Codec) deserialize(data string) *TreeNode {
+	if len(data) == 0 {
+		return nil
+	}
+	this.input = strings.Split(data, ",")
+	return this.deserializeHelper()
+}
+
+func (this *Codec) deserializeHelper() *TreeNode {
+	if this.input[0] == "#" {
+		this.input = this.input[1:]
+		return nil
+	}
+	val, _ := strconv.Atoi(this.input[0])
+	this.input = this.input[1:]
+	return &TreeNode{
+		Val:   val,
+		Left:  this.deserializeHelper(),
+		Right: this.deserializeHelper(),
+	}
+}
diff --git a/leetcode/0297.Serialize-and-Deserialize-Binary-Tree/297.Serialize and Deserialize Binary Tree_test.go b/leetcode/0297.Serialize-and-Deserialize-Binary-Tree/297.Serialize and Deserialize Binary Tree_test.go
new file mode 100644
index 000000000..cf7031c4a
--- /dev/null
+++ b/leetcode/0297.Serialize-and-Deserialize-Binary-Tree/297.Serialize and Deserialize Binary Tree_test.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question297 struct {
+	para297
+	ans297
+}
+
+// para 是参数
+// one 代表第一个参数
+type para297 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans297 struct {
+	one []int
+}
+
+func Test_Problem297(t *testing.T) {
+	qs := []question297{
+		{
+			para297{[]int{}},
+			ans297{[]int{}},
+		},
+		{
+			para297{[]int{1, 2, 3, -1, -1, 4, 5}},
+			ans297{[]int{1, 2, 3, -1, -1, 4, 5}},
+		},
+		{
+			para297{[]int{1, 2}},
+			ans297{[]int{1, 2}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 297------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans297, q.para297
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+
+		tree297 := Constructor()
+		serialized := tree297.serialize(root)
+		fmt.Printf("【output】:%v      \n", structures.Tree2Preorder(tree297.deserialize(serialized)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0297.Serialize-and-Deserialize-Binary-Tree/README.md b/leetcode/0297.Serialize-and-Deserialize-Binary-Tree/README.md
new file mode 100644
index 000000000..15f4afa5f
--- /dev/null
+++ b/leetcode/0297.Serialize-and-Deserialize-Binary-Tree/README.md
@@ -0,0 +1,113 @@
+# [297. Serialize and Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)
+
+
+## 题目
+
+Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
+
+Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.
+
+**Clarification:** The input/output format is the same as [how LeetCode serializes a binary tree](https://leetcode.com/faq/#binary-tree). You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/09/15/serdeser.jpg](https://assets.leetcode.com/uploads/2020/09/15/serdeser.jpg)
+
+```
+Input: root = [1,2,3,null,null,4,5]
+Output: [1,2,3,null,null,4,5]
+```
+
+**Example 2:**
+
+```
+Input: root = []
+Output: []
+```
+
+**Example 3:**
+
+```
+Input: root = [1]
+Output: [1]
+```
+
+**Example 4:**
+
+```
+Input: root = [1,2]
+Output: [1,2]
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[0, 104]`.
+- `1000 <= Node.val <= 1000`
+
+## 题目大意
+
+设计一个算法，来序列化和反序列化二叉树。并不限制如何进行序列化和反序列化，但是你需要保证二叉树可以序列化为字符串，并且这个字符串可以被反序列化成原有的二叉树。
+
+## 解题思路
+
+- 1. 将给定的二叉树想象成一颗满二叉树(不存在的结点用 null 填充)。
+- 2. 通过前序遍历，可以得到一个第一个结点为根的序列，然后递归进行序列化/反序列化即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type TreeNode = structures.TreeNode
+
+type Codec struct {
+	builder strings.Builder
+	input   []string
+}
+
+func Constructor() Codec {
+	return Codec{}
+}
+
+// Serializes a tree to a single string.
+func (this *Codec) serialize(root *TreeNode) string {
+	if root == nil {
+		this.builder.WriteString("#,")
+		return ""
+	}
+	this.builder.WriteString(strconv.Itoa(root.Val) + ",")
+	this.serialize(root.Left)
+	this.serialize(root.Right)
+	return this.builder.String()
+}
+
+// Deserializes your encoded data to tree.
+func (this *Codec) deserialize(data string) *TreeNode {
+	if len(data) == 0 {
+		return nil
+	}
+	this.input = strings.Split(data, ",")
+	return this.deserializeHelper()
+}
+
+func (this *Codec) deserializeHelper() *TreeNode {
+	if this.input[0] == "#" {
+		this.input = this.input[1:]
+		return nil
+	}
+	val, _ := strconv.Atoi(this.input[0])
+	this.input = this.input[1:]
+	return &TreeNode{
+		Val:   val,
+		Left:  this.deserializeHelper(),
+		Right: this.deserializeHelper(),
+	}
+}
+```
\ No newline at end of file
diff --git a/leetcode/0299.Bulls-and-Cows/299.Bulls and Cows.go b/leetcode/0299.Bulls-and-Cows/299.Bulls and Cows.go
new file mode 100644
index 000000000..920827f07
--- /dev/null
+++ b/leetcode/0299.Bulls-and-Cows/299.Bulls and Cows.go	
@@ -0,0 +1,31 @@
+package leetcode
+
+import "strconv"
+
+func getHint(secret string, guess string) string {
+	cntA, cntB := 0, 0
+	mpS := make(map[byte]int)
+	var strG []byte
+	n := len(secret)
+	var ans string
+	for i := 0; i < n; i++ {
+		if secret[i] == guess[i] {
+			cntA++
+		} else {
+			mpS[secret[i]] += 1
+			strG = append(strG, guess[i])
+		}
+	}
+	for _, v := range strG {
+		if _, ok := mpS[v]; ok {
+			if mpS[v] > 1 {
+				mpS[v] -= 1
+			} else {
+				delete(mpS, v)
+			}
+			cntB++
+		}
+	}
+	ans += strconv.Itoa(cntA) + "A" + strconv.Itoa(cntB) + "B"
+	return ans
+}
diff --git a/leetcode/0299.Bulls-and-Cows/299.Bulls and Cows_test.go b/leetcode/0299.Bulls-and-Cows/299.Bulls and Cows_test.go
new file mode 100644
index 000000000..72028efb6
--- /dev/null
+++ b/leetcode/0299.Bulls-and-Cows/299.Bulls and Cows_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question299 struct {
+	para299
+	ans299
+}
+
+// para 是参数
+type para299 struct {
+	secret string
+	guess  string
+}
+
+// ans 是答案
+type ans299 struct {
+	ans string
+}
+
+func Test_Problem299(t *testing.T) {
+
+	qs := []question299{
+
+		{
+			para299{"1807", "7810"},
+			ans299{"1A3B"},
+		},
+
+		{
+			para299{"1123", "0111"},
+			ans299{"1A1B"},
+		},
+
+		{
+			para299{"1", "0"},
+			ans299{"0A0B"},
+		},
+
+		{
+			para299{"1", "1"},
+			ans299{"1A0B"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 299------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans299, q.para299
+		fmt.Printf("【input】:%v       【output】:%v\n", p, getHint(p.secret, p.guess))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0299.Bulls-and-Cows/README.md b/leetcode/0299.Bulls-and-Cows/README.md
new file mode 100644
index 000000000..61f7b164b
--- /dev/null
+++ b/leetcode/0299.Bulls-and-Cows/README.md
@@ -0,0 +1,114 @@
+# [299. Bulls and Cows](https://leetcode.com/problems/bulls-and-cows/)
+
+
+## 题目
+
+You are playing the Bulls and Cows game with your friend.
+
+You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
+
+The number of "bulls", which are digits in the guess that are in the correct position.
+The number of "cows", which are digits in the guess that are in your secret number but are located in the wrong position. Specifically, the non-bull digits in the guess that could be rearranged such that they become bulls.
+Given the secret number secret and your friend's guess guess, return the hint for your friend's guess.
+
+The hint should be formatted as "xAyB", where x is the number of bulls and y is the number of cows. Note that both secret and guess may contain duplicate digits.
+
+**Example 1:**
+
+```
+Input: secret = "1807", guess = "7810"
+Output: "1A3B"
+Explanation: Bulls are connected with a '|' and cows are underlined:
+"1807"
+  |
+"7810"
+```
+
+**Example 2:**
+
+```
+Input: secret = "1123", guess = "0111"
+Output: "1A1B"
+Explanation: Bulls are connected with a '|' and cows are underlined:
+"1123"        "1123"
+  |      or     |
+"0111"        "0111"
+Note that only one of the two unmatched 1s is counted as a cow since the non-bull digits can only be rearranged to allow one 1 to be a bull.
+```
+
+**Example 3:**
+
+```
+Input: secret = "1", guess = "0"
+Output: "0A0B"
+```
+
+**Example 4:**
+
+```
+Input: secret = "1", guess = "1"
+Output: "1A0B"
+```
+
+**Constraints:**
+
+- 1 <= secret.length, guess.length <= 1000
+- secret.length == guess.length
+- secret and guess consist of digits only.
+
+## 题目大意
+
+你在和朋友一起玩 猜数字（Bulls and Cows）游戏，该游戏规则如下：
+
+写出一个秘密数字，并请朋友猜这个数字是多少。朋友每猜测一次，你就会给他一个包含下述信息的提示：
+
+猜测数字中有多少位属于数字和确切位置都猜对了（称为 "Bulls", 公牛），
+有多少位属于数字猜对了但是位置不对（称为 "Cows", 奶牛）。也就是说，这次猜测中有多少位非公牛数字可以通过重新排列转换成公牛数字。
+给你一个秘密数字secret 和朋友猜测的数字guess ，请你返回对朋友这次猜测的提示。
+
+提示的格式为 "xAyB" ，x 是公牛个数， y 是奶牛个数，A 表示公牛，B表示奶牛。
+
+请注意秘密数字和朋友猜测的数字都可能含有重复数字。
+
+## 解题思路
+
+- 计算下标一致并且对应下标的元素一致的个数，即 x
+- secret 和 guess 分别去除 x 个公牛的元素,剩下 secret 和 guess 求共同的元素个数就是 y
+- 把 x， y 转换成字符串，分别与 "A" 和 "B" 进行拼接返回结果
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strconv"
+
+func getHint(secret string, guess string) string {
+	cntA, cntB := 0, 0
+	mpS := make(map[byte]int)
+	var strG []byte
+	n := len(secret)
+	var ans string
+	for i := 0; i < n; i++ {
+		if secret[i] == guess[i] {
+			cntA++
+		} else {
+			mpS[secret[i]] += 1
+			strG = append(strG, guess[i])
+		}
+	}
+	for _, v := range strG {
+		if _, ok := mpS[v]; ok {
+			if mpS[v] > 1 {
+				mpS[v] -= 1
+			} else {
+				delete(mpS, v)
+			}
+			cntB++
+		}
+	}
+	ans += strconv.Itoa(cntA) + "A" + strconv.Itoa(cntB) + "B"
+	return ans
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0300. Longest Increasing Subsequence/300. Longest Increasing Subsequence.go b/leetcode/0300.Longest-Increasing-Subsequence/300. Longest Increasing Subsequence.go
similarity index 89%
rename from Algorithms/0300. Longest Increasing Subsequence/300. Longest Increasing Subsequence.go
rename to leetcode/0300.Longest-Increasing-Subsequence/300. Longest Increasing Subsequence.go
index a0e1eb32b..a658bab8b 100644
--- a/Algorithms/0300. Longest Increasing Subsequence/300. Longest Increasing Subsequence.go	
+++ b/leetcode/0300.Longest-Increasing-Subsequence/300. Longest Increasing Subsequence.go	
@@ -18,6 +18,13 @@ func lengthOfLIS(nums []int) int {
 	return res
 }
 
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
 // 解法二 O(n log n) DP
 func lengthOfLIS1(nums []int) int {
 	dp := []int{}
diff --git a/Algorithms/0300. Longest Increasing Subsequence/300. Longest Increasing Subsequence_test.go b/leetcode/0300.Longest-Increasing-Subsequence/300. Longest Increasing Subsequence_test.go
similarity index 97%
rename from Algorithms/0300. Longest Increasing Subsequence/300. Longest Increasing Subsequence_test.go
rename to leetcode/0300.Longest-Increasing-Subsequence/300. Longest Increasing Subsequence_test.go
index 17d5a36b7..074c712ea 100644
--- a/Algorithms/0300. Longest Increasing Subsequence/300. Longest Increasing Subsequence_test.go	
+++ b/leetcode/0300.Longest-Increasing-Subsequence/300. Longest Increasing Subsequence_test.go	
@@ -26,7 +26,7 @@ func Test_Problem300(t *testing.T) {
 
 	qs := []question300{
 
-		question300{
+		{
 			para300{[]int{10, 9, 2, 5, 3, 7, 101, 18}},
 			ans300{4},
 		},
diff --git a/leetcode/0300.Longest-Increasing-Subsequence/README.md b/leetcode/0300.Longest-Increasing-Subsequence/README.md
new file mode 100755
index 000000000..fc967cdbd
--- /dev/null
+++ b/leetcode/0300.Longest-Increasing-Subsequence/README.md
@@ -0,0 +1,38 @@
+# [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/)
+
+
+## 题目
+
+Given an unsorted array of integers, find the length of longest increasing subsequence.
+
+**Example:**
+
+    Input: [10,9,2,5,3,7,101,18]
+    Output: 4 
+    Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
+
+**Note:**
+
+- There may be more than one LIS combination, it is only necessary for you to return the length.
+- Your algorithm should run in O(n^2) complexity.
+
+**Follow up:** Could you improve it to O(n log n) time complexity?
+
+## 题目大意
+
+给定一个无序的整数数组，找到其中最长上升子序列的长度。
+
+
+## 解题思路
+
+- 给定一个整数序列，求其中的最长上升子序列的长度。这一题就是经典的最长上升子序列的问题。
+- `dp[i]` 代表为第 i 个数字为结尾的最长上升子序列的长度。换种表述，dp[i] 代表 [0,i] 范围内，选择数字 nums[i] 可以获得的最长上升子序列的长度。状态转移方程为 `dp[i] = max( 1 + dp[j]) ，其中 j < i && nums[j] > nums[i]`，取所有满足条件的最大值。时间复杂度 O(n^2)
+- 这道题还有一种更快的解法。考虑这样一个问题，我们是否能用一个数组，记录上升子序列的最末尾的一个数字呢？如果这个数字越小，那么这个子序列往后面添加数字的几率就越大，那么就越可能成为最长的上升子序列。举个例子：nums = [4,5,6,3]，它的所有的上升子序列为
+
+```
+    len = 1   :      [4], [5], [6], [3]   => tails[0] = 3
+    len = 2   :      [4, 5], [5, 6]       => tails[1] = 5
+    len = 3   :      [4, 5, 6]            => tails[2] = 6
+```
+- 其中 `tails[i]` 中存储的是所有长度为 i + 1 的上升子序列中末尾最小的值。也很容易证明 `tails` 数组里面的值一定是递增的(因为我们用末尾的数字描述最长递增子序列)。既然 tails 是有序的，我们就可以用二分查找的方法去更新这个 tail 数组里面的值。更新策略如下：(1). 如果 x 比所有的 tails 元素都要大，那么就直接放在末尾，并且 tails 数组长度加一；(2). 如果 `tails[i-1] < x <= tails[i]`，则更新 tails[i]，因为 x 更小，更能获得最长上升子序列。最终 tails 数组的长度即为最长的上升子序列。这种做法的时间复杂度 O(n log n)。
+
diff --git a/leetcode/0301.Remove-Invalid-Parentheses/301.Remove Invalid Parentheses.go b/leetcode/0301.Remove-Invalid-Parentheses/301.Remove Invalid Parentheses.go
new file mode 100644
index 000000000..0e53f6bdd
--- /dev/null
+++ b/leetcode/0301.Remove-Invalid-Parentheses/301.Remove Invalid Parentheses.go	
@@ -0,0 +1,69 @@
+package leetcode
+
+var (
+	res      []string
+	mp       map[string]int
+	n        int
+	length   int
+	maxScore int
+	str      string
+)
+
+func removeInvalidParentheses(s string) []string {
+	lmoves, rmoves, lcnt, rcnt := 0, 0, 0, 0
+	for _, v := range s {
+		if v == '(' {
+			lmoves++
+			lcnt++
+		} else if v == ')' {
+			if lmoves != 0 {
+				lmoves--
+			} else {
+				rmoves++
+			}
+			rcnt++
+		}
+	}
+	n = len(s)
+	length = n - lmoves - rmoves
+	res = []string{}
+	mp = make(map[string]int)
+	maxScore = min(lcnt, rcnt)
+	str = s
+	backtrace(0, "", lmoves, rmoves, 0)
+	return res
+}
+
+func backtrace(i int, cur string, lmoves int, rmoves int, score int) {
+	if lmoves < 0 || rmoves < 0 || score < 0 || score > maxScore {
+		return
+	}
+	if lmoves == 0 && rmoves == 0 {
+		if len(cur) == length {
+			if _, ok := mp[cur]; !ok {
+				res = append(res, cur)
+				mp[cur] = 1
+			}
+			return
+		}
+	}
+	if i == n {
+		return
+	}
+	if str[i] == '(' {
+		backtrace(i+1, cur+string('('), lmoves, rmoves, score+1)
+		backtrace(i+1, cur, lmoves-1, rmoves, score)
+	} else if str[i] == ')' {
+		backtrace(i+1, cur+string(')'), lmoves, rmoves, score-1)
+		backtrace(i+1, cur, lmoves, rmoves-1, score)
+	} else {
+		backtrace(i+1, cur+string(str[i]), lmoves, rmoves, score)
+	}
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0301.Remove-Invalid-Parentheses/301.Remove Invalid Parentheses_test.go b/leetcode/0301.Remove-Invalid-Parentheses/301.Remove Invalid Parentheses_test.go
new file mode 100644
index 000000000..c49d72df7
--- /dev/null
+++ b/leetcode/0301.Remove-Invalid-Parentheses/301.Remove Invalid Parentheses_test.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question301 struct {
+	para301
+	ans301
+}
+
+// s 是参数
+type para301 struct {
+	s string
+}
+
+// ans 是答案
+type ans301 struct {
+	ans []string
+}
+
+func Test_Problem301(t *testing.T) {
+
+	qs := []question301{
+
+		{
+			para301{"()())()"},
+			ans301{[]string{"(())()", "()()()"}},
+		},
+
+		{
+			para301{"(a)())()"},
+			ans301{[]string{"(a())()", "(a)()()"}},
+		},
+
+		{
+			para301{")("},
+			ans301{[]string{""}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 301------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans301, q.para301
+		fmt.Printf("【input】:%v       【output】:%v\n", p, removeInvalidParentheses(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0301.Remove-Invalid-Parentheses/README.md b/leetcode/0301.Remove-Invalid-Parentheses/README.md
new file mode 100644
index 000000000..cd1d80b81
--- /dev/null
+++ b/leetcode/0301.Remove-Invalid-Parentheses/README.md
@@ -0,0 +1,131 @@
+# [301. Remove Invalid Parentheses](https://leetcode.com/problems/remove-invalid-parentheses/)
+
+
+## 题目
+
+Given a string s that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
+
+Return all the possible results. You may return the answer in any order.
+
+**Example 1:**
+
+    Input: s = "()())()"
+    Output: ["(())()","()()()"]
+
+**Example 2:**
+
+    Input: s = "(a)())()"
+    Output: ["(a())()","(a)()()"]
+
+**Example 3:**
+
+    Input: s = ")("
+    Output: [""]
+
+**Constraints:**
+
+- 1 <= s.length <= 25
+- s consists of lowercase English letters and parentheses '(' and ')'.
+- There will be at most 20 parentheses in s.
+
+## 题目大意
+
+给你一个由若干括号和字母组成的字符串 s ，删除最小数量的无效括号，使得输入的字符串有效。
+
+返回所有可能的结果。答案可以按 任意顺序 返回。
+
+说明:
+
+- 1 <= s.length <= 25
+- s 由小写英文字母以及括号 '(' 和 ')' 组成
+- s 中至多含 20 个括号
+
+
+## 解题思路
+
+回溯和剪枝
+- 计算最大得分数maxScore，合法字符串的长度length，左括号和右括号的移除次数lmoves,rmoves
+- 加一个左括号的得分加1；加一个右括号的得分减1
+- 对于一个合法的字符串，左括号等于右括号，得分最终为0；
+- 搜索过程中出现以下任何一种情况都直接返回
+    - 得分值为负数
+    - 得分大于最大得分数
+    - 得分小于0
+    - lmoves小于0
+    - rmoves小于0
+
+## 代码
+
+```go
+
+package leetcode
+
+var (
+	res      []string
+	mp       map[string]int
+	n        int
+	length   int
+	maxScore int
+	str      string
+)
+
+func removeInvalidParentheses(s string) []string {
+	lmoves, rmoves, lcnt, rcnt := 0, 0, 0, 0
+	for _, v := range s {
+		if v == '(' {
+			lmoves++
+			lcnt++
+		} else if v == ')' {
+			if lmoves != 0 {
+				lmoves--
+			} else {
+				rmoves++
+			}
+			rcnt++
+		}
+	}
+	n = len(s)
+	length = n - lmoves - rmoves
+	res = []string{}
+	mp = make(map[string]int)
+	maxScore = min(lcnt, rcnt)
+	str = s
+	backtrace(0, "", lmoves, rmoves, 0)
+	return res
+}
+
+func backtrace(i int, cur string, lmoves int, rmoves int, score int) {
+	if lmoves < 0 || rmoves < 0 || score < 0 || score > maxScore {
+		return
+	}
+	if lmoves == 0 && rmoves == 0 {
+		if len(cur) == length {
+			if _, ok := mp[cur]; !ok {
+				res = append(res, cur)
+				mp[cur] = 1
+			}
+			return
+		}
+	}
+	if i == n {
+		return
+	}
+	if str[i] == '(' {
+		backtrace(i+1, cur+string('('), lmoves, rmoves, score+1)
+		backtrace(i+1, cur, lmoves-1, rmoves, score)
+	} else if str[i] == ')' {
+		backtrace(i+1, cur+string(')'), lmoves, rmoves, score-1)
+		backtrace(i+1, cur, lmoves, rmoves-1, score)
+	} else {
+		backtrace(i+1, cur+string(str[i]), lmoves, rmoves, score)
+	}
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0303. Range Sum Query - Immutable/303. Range Sum Query - Immutable.go b/leetcode/0303.Range-Sum-Query-Immutable/303. Range Sum Query - Immutable.go
similarity index 100%
rename from Algorithms/0303. Range Sum Query - Immutable/303. Range Sum Query - Immutable.go
rename to leetcode/0303.Range-Sum-Query-Immutable/303. Range Sum Query - Immutable.go
diff --git a/Algorithms/0303. Range Sum Query - Immutable/303. Range Sum Query - Immutable_test.go b/leetcode/0303.Range-Sum-Query-Immutable/303. Range Sum Query - Immutable_test.go
similarity index 100%
rename from Algorithms/0303. Range Sum Query - Immutable/303. Range Sum Query - Immutable_test.go
rename to leetcode/0303.Range-Sum-Query-Immutable/303. Range Sum Query - Immutable_test.go
diff --git a/leetcode/0303.Range-Sum-Query-Immutable/README.md b/leetcode/0303.Range-Sum-Query-Immutable/README.md
new file mode 100755
index 000000000..95996ad0f
--- /dev/null
+++ b/leetcode/0303.Range-Sum-Query-Immutable/README.md
@@ -0,0 +1,47 @@
+# [303. Range Sum Query - Immutable](https://leetcode.com/problems/range-sum-query-immutable/)
+
+
+## 题目
+
+Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
+
+**Example:**
+
+    Given nums = [-2, 0, 3, -5, 2, -1]
+    
+    sumRange(0, 2) -> 1
+    sumRange(2, 5) -> -1
+    sumRange(0, 5) -> -3
+
+**Note:**
+
+1. You may assume that the array does not change.
+2. There are many calls to sumRange function.
+
+
+## 题目大意
+
+给定一个整数数组  nums，求出数组从索引 i 到 j  (i ≤ j) 范围内元素的总和，包含 i,  j 两点。
+
+示例：
+
+```
+给定 nums = [-2, 0, 3, -5, 2, -1]，求和函数为 sumRange()
+
+sumRange(0, 2) -> 1
+sumRange(2, 5) -> -1
+sumRange(0, 5) -> -3
+
+```
+
+说明:
+
+- 你可以假设数组不可变。
+- 会多次调用 sumRange 方法。
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面的数都是`**不可变**`的，设计一个数据结构能够满足查询数组任意区间内元素的和。
+- 这一题由于数组里面的元素都是`**不可变**`的，所以可以用 2 种方式来解答，第一种解法是用 prefixSum，通过累计和相减的办法来计算区间内的元素和，初始化的时间复杂度是 O(n)，但是查询区间元素和的时间复杂度是 O(1)。第二种解法是利用线段树，构建一颗线段树，父结点内存的是两个子结点的和，初始化建树的时间复杂度是 O(log n)，查询区间元素和的时间复杂度是 O(log n)。
diff --git a/leetcode/0304.Range-Sum-Query-2D-Immutable/304. Range Sum Query 2D - Immutable.go b/leetcode/0304.Range-Sum-Query-2D-Immutable/304. Range Sum Query 2D - Immutable.go
new file mode 100644
index 000000000..ab28dd0f8
--- /dev/null
+++ b/leetcode/0304.Range-Sum-Query-2D-Immutable/304. Range Sum Query 2D - Immutable.go	
@@ -0,0 +1,31 @@
+package leetcode
+
+type NumMatrix struct {
+	cumsum [][]int
+}
+
+func Constructor(matrix [][]int) NumMatrix {
+	if len(matrix) == 0 {
+		return NumMatrix{nil}
+	}
+	cumsum := make([][]int, len(matrix)+1)
+	cumsum[0] = make([]int, len(matrix[0])+1)
+	for i := range matrix {
+		cumsum[i+1] = make([]int, len(matrix[i])+1)
+		for j := range matrix[i] {
+			cumsum[i+1][j+1] = matrix[i][j] + cumsum[i][j+1] + cumsum[i+1][j] - cumsum[i][j]
+		}
+	}
+	return NumMatrix{cumsum}
+}
+
+func (this *NumMatrix) SumRegion(row1 int, col1 int, row2 int, col2 int) int {
+	cumsum := this.cumsum
+	return cumsum[row2+1][col2+1] - cumsum[row1][col2+1] - cumsum[row2+1][col1] + cumsum[row1][col1]
+}
+
+/**
+ * Your NumMatrix object will be instantiated and called as such:
+ * obj := Constructor(matrix);
+ * param_1 := obj.SumRegion(row1,col1,row2,col2);
+ */
diff --git a/leetcode/0304.Range-Sum-Query-2D-Immutable/304. Range Sum Query 2D - Immutable_test.go b/leetcode/0304.Range-Sum-Query-2D-Immutable/304. Range Sum Query 2D - Immutable_test.go
new file mode 100644
index 000000000..10e56d406
--- /dev/null
+++ b/leetcode/0304.Range-Sum-Query-2D-Immutable/304. Range Sum Query 2D - Immutable_test.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem304(t *testing.T) {
+	obj := Constructor(
+		[][]int{
+			{3, 0, 1, 4, 2},
+			{5, 6, 3, 2, 1},
+			{1, 2, 0, 1, 5},
+			{4, 1, 0, 1, 7},
+			{1, 0, 3, 0, 5},
+		},
+	)
+	fmt.Printf("obj = %v\n", obj.SumRegion(2, 1, 4, 3))
+	fmt.Printf("obj = %v\n", obj.SumRegion(1, 1, 2, 2))
+	fmt.Printf("obj = %v\n", obj.SumRegion(1, 2, 2, 4))
+}
diff --git a/leetcode/0304.Range-Sum-Query-2D-Immutable/README.md b/leetcode/0304.Range-Sum-Query-2D-Immutable/README.md
new file mode 100644
index 000000000..3ed2a65d2
--- /dev/null
+++ b/leetcode/0304.Range-Sum-Query-2D-Immutable/README.md
@@ -0,0 +1,89 @@
+# [304. Range Sum Query 2D - Immutable](https://leetcode.com/problems/range-sum-query-2d-immutable/)
+
+
+## 题目
+
+Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).
+
+![https://leetcode.com/static/images/courses/range_sum_query_2d.png](https://assets.leetcode.com/uploads/2021/03/14/sum-grid.jpg)
+
+The above rectangle (with the red border) is defined by (row1, col1) = **(2, 1)** and (row2, col2) = **(4, 3)**, which contains sum = **8**.
+
+**Example:**
+
+```
+Given matrix = [
+  [3, 0, 1, 4, 2],
+  [5, 6, 3, 2, 1],
+  [1, 2, 0, 1, 5],
+  [4, 1, 0, 1, 7],
+  [1, 0, 3, 0, 5]
+]
+
+sumRegion(2, 1, 4, 3) -> 8
+sumRegion(1, 1, 2, 2) -> 11
+sumRegion(1, 2, 2, 4) -> 12
+
+```
+
+**Note:**
+
+1. You may assume that the matrix does not change.
+2. There are many calls to sumRegion function.
+3. You may assume that row1 ≤ row2 and col1 ≤ col2.
+
+## 题目大意
+
+给定一个二维矩阵，计算其子矩形范围内元素的总和，该子矩阵的左上角为 (row1, col1) ，右下角为 (row2, col2) 。
+
+## 解题思路
+
+- 这一题是一维数组前缀和的进阶版本。定义 f(x,y) 代表矩形左上角 (0,0)，右下角 (x,y) 内的元素和。{{< katex display >}} f(i,j) = \sum_{x=0}^{i}\sum_{y=0}^{j} Matrix[x][y]{{< /katex >}}
+
+	{{< katex display >}}
+    \begin{aligned}f(i,j) &= \sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + \sum_{x=0}^{i-1} Matrix[x][j] + \sum_{y=0}^{j-1} Matrix[i][y] + Matrix[i][j]\\&= (\sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + \sum_{x=0}^{i-1} Matrix[x][j]) + (\sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + \sum_{y=0}^{j-1} Matrix[i][y]) - \sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + Matrix[i][j]\\&= \sum_{x=0}^{i-1}\sum_{y=0}^{j} Matrix[x][y] + \sum_{x=0}^{i}\sum_{y=0}^{j-1} Matrix[x][y] - \sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + Matrix[i][j]\\&= f(i-1,j) + f(i,j-1) - f(i-1,j-1) + Matrix[i][j]\end{aligned}
+	{{< /katex >}}
+
+- 于是得到递推的关系式：`f(i, j) = f(i-1, j) + f(i, j-1) - f(i-1, j-1) + matrix[i][j]`，写代码为了方便，新建一个 `m+1 * n+1` 的矩阵，这样就不需要对 `row = 0` 和 `col = 0` 做单独处理了。上述推导公式如果画成图也很好理解：
+
+    ![https://img.halfrost.com/Leetcode/leetcode_304.png](https://img.halfrost.com/Leetcode/leetcode_304.png)
+
+    左图中大的矩形由粉红色的矩形 + 绿色矩形 - 粉红色和绿色重叠部分 + 黄色部分。这就对应的是上面推导出来的递推公式。左图是矩形左上角为 (0，0) 的情况，更加一般的情况是右图，左上角是任意的坐标，公式不变。
+
+- 时间复杂度：初始化 O(mn)，查询 O(1)。空间复杂度 O(mn)
+
+## 代码
+
+```go
+package leetcode
+
+type NumMatrix struct {
+	cumsum [][]int
+}
+
+func Constructor(matrix [][]int) NumMatrix {
+	if len(matrix) == 0 {
+		return NumMatrix{nil}
+	}
+	cumsum := make([][]int, len(matrix)+1)
+	cumsum[0] = make([]int, len(matrix[0])+1)
+	for i := range matrix {
+		cumsum[i+1] = make([]int, len(matrix[i])+1)
+		for j := range matrix[i] {
+			cumsum[i+1][j+1] = matrix[i][j] + cumsum[i][j+1] + cumsum[i+1][j] - cumsum[i][j]
+		}
+	}
+	return NumMatrix{cumsum}
+}
+
+func (this *NumMatrix) SumRegion(row1 int, col1 int, row2 int, col2 int) int {
+	cumsum := this.cumsum
+	return cumsum[row2+1][col2+1] - cumsum[row1][col2+1] - cumsum[row2+1][col1] + cumsum[row1][col1]
+}
+
+/**
+ * Your NumMatrix object will be instantiated and called as such:
+ * obj := Constructor(matrix);
+ * param_1 := obj.SumRegion(row1,col1,row2,col2);
+ */
+```
\ No newline at end of file
diff --git a/Algorithms/0306. Additive Number/306. Additive Number.go b/leetcode/0306.Additive-Number/306. Additive Number.go
similarity index 90%
rename from Algorithms/0306. Additive Number/306. Additive Number.go
rename to leetcode/0306.Additive-Number/306. Additive Number.go
index 9aac8ee56..bd0429a16 100644
--- a/Algorithms/0306. Additive Number/306. Additive Number.go	
+++ b/leetcode/0306.Additive-Number/306. Additive Number.go	
@@ -28,7 +28,14 @@ func isAdditiveNumber(num string) bool {
 	return false
 }
 
-//Propagate for rest of the string
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// Propagate for rest of the string
 func recursiveCheck(num string, x1 int, x2 int, left int) bool {
 	if left == len(num) {
 		return true
diff --git a/Algorithms/0306. Additive Number/306. Additive Number_test.go b/leetcode/0306.Additive-Number/306. Additive Number_test.go
similarity index 95%
rename from Algorithms/0306. Additive Number/306. Additive Number_test.go
rename to leetcode/0306.Additive-Number/306. Additive Number_test.go
index 80e1478ef..8191d720c 100644
--- a/Algorithms/0306. Additive Number/306. Additive Number_test.go	
+++ b/leetcode/0306.Additive-Number/306. Additive Number_test.go	
@@ -26,12 +26,12 @@ func Test_Problem306(t *testing.T) {
 
 	qs := []question306{
 
-		question306{
+		{
 			para306{"112358"},
 			ans306{true},
 		},
 
-		question306{
+		{
 			para306{"199100199"},
 			ans306{true},
 		},
diff --git a/Algorithms/0306. Additive Number/README.md b/leetcode/0306.Additive-Number/README.md
similarity index 100%
rename from Algorithms/0306. Additive Number/README.md
rename to leetcode/0306.Additive-Number/README.md
diff --git a/leetcode/0307.Range-Sum-Query-Mutable/307. Range Sum Query - Mutable.go b/leetcode/0307.Range-Sum-Query-Mutable/307. Range Sum Query - Mutable.go
new file mode 100644
index 000000000..e16c7c3f6
--- /dev/null
+++ b/leetcode/0307.Range-Sum-Query-Mutable/307. Range Sum Query - Mutable.go	
@@ -0,0 +1,95 @@
+package leetcode
+
+import "github.com/halfrost/LeetCode-Go/template"
+
+// NumArray define
+type NumArray struct {
+	st *template.SegmentTree
+}
+
+// Constructor307 define
+func Constructor307(nums []int) NumArray {
+	st := template.SegmentTree{}
+	st.Init(nums, func(i, j int) int {
+		return i + j
+	})
+	return NumArray{st: &st}
+}
+
+// Update define
+func (this *NumArray) Update(i int, val int) {
+	this.st.Update(i, val)
+}
+
+// SumRange define
+func (this *NumArray) SumRange(i int, j int) int {
+	return this.st.Query(i, j)
+}
+
+//解法二 prefixSum，sumRange 时间复杂度 O(1)
+
+// // NumArray define
+// type NumArray307 struct {
+// 	prefixSum []int
+// 	data   []int
+// }
+
+// // Constructor307 define
+// func Constructor307(nums []int) NumArray307 {
+// 	data := make([]int, len(nums))
+// 	for i := 0; i < len(nums); i++ {
+// 		data[i] = nums[i]
+// 	}
+// 	for i := 1; i < len(nums); i++ {
+// 		nums[i] += nums[i-1]
+// 	}
+// 	return NumArray307{prefixSum: nums, data: data}
+// }
+
+// // Update define
+// func (this *NumArray307) Update(i int, val int) {
+// 	this.data[i] = val
+// 	this.prefixSum[0] = this.data[0]
+// 	for i := 1; i < len(this.data); i++ {
+// 		this.prefixSum[i] = this.prefixSum[i-1] + this.data[i]
+// 	}
+// }
+
+// // SumRange define
+// func (this *NumArray307) SumRange(i int, j int) int {
+// 	if i > 0 {
+// 		return this.prefixSum[j] - this.prefixSum[i-1]
+// 	}
+// 	return this.prefixSum[j]
+// }
+
+// 解法三 树状数组
+// type NumArray struct {
+// 	bit  template.BinaryIndexedTree
+// 	data []int
+// }
+
+// // Constructor define
+// func Constructor307(nums []int) NumArray {
+// 	bit := template.BinaryIndexedTree{}
+// 	bit.InitWithNums(nums)
+// 	return NumArray{bit: bit, data: nums}
+// }
+
+// // Update define
+// func (this *NumArray) Update(i int, val int) {
+// 	this.bit.Add(i+1, val-this.data[i])
+// 	this.data[i] = val
+// }
+
+// // SumRange define
+// func (this *NumArray) SumRange(i int, j int) int {
+// 	return this.bit.Query(j+1) - this.bit.Query(i)
+// }
+
+/**
+ * Your NumArray object will be instantiated and called as such:
+ * obj := Constructor(nums);
+ * obj.Update(i,val);
+ * param_2 := obj.SumRange(i,j);
+ */
diff --git a/leetcode/0307.Range-Sum-Query-Mutable/307. Range Sum Query - Mutable_test.go b/leetcode/0307.Range-Sum-Query-Mutable/307. Range Sum Query - Mutable_test.go
new file mode 100644
index 000000000..49a548142
--- /dev/null
+++ b/leetcode/0307.Range-Sum-Query-Mutable/307. Range Sum Query - Mutable_test.go	
@@ -0,0 +1,36 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem307(t *testing.T) {
+	obj := Constructor307([]int{1, 3, 5})
+	fmt.Printf("obj = %v\n", obj)
+	fmt.Printf("SumRange(0,2) = %v\n", obj.SumRange(0, 2))
+	obj.Update(1, 2)
+	fmt.Printf("obj = %v\n", obj)
+	fmt.Printf("SumRange(0,2) = %v\n", obj.SumRange(0, 2))
+
+	obj = Constructor307([]int{-1})
+	fmt.Printf("obj = %v\n", obj)
+	fmt.Printf("SumRange(0,2) = %v\n", obj.SumRange(0, 0))
+	obj.Update(0, 1)
+	fmt.Printf("obj = %v\n", obj)
+	fmt.Printf("SumRange(0,2) = %v\n", obj.SumRange(0, 0))
+
+	obj = Constructor307([]int{7, 2, 7, 2, 0})
+	fmt.Printf("obj = %v\n", obj)
+	obj.Update(4, 6)
+	obj.Update(0, 2)
+	obj.Update(0, 9)
+	fmt.Printf("SumRange(0,2) = %v\n", obj.SumRange(4, 4))
+	obj.Update(3, 8)
+	fmt.Printf("obj = %v\n", obj)
+	fmt.Printf("SumRange(0,2) = %v\n", obj.SumRange(0, 4))
+	obj.Update(4, 1)
+	fmt.Printf("SumRange(0,2) = %v\n", obj.SumRange(0, 3))
+	fmt.Printf("SumRange(0,2) = %v\n", obj.SumRange(0, 4))
+	obj.Update(0, 4)
+}
diff --git a/leetcode/0307.Range-Sum-Query-Mutable/README.md b/leetcode/0307.Range-Sum-Query-Mutable/README.md
new file mode 100755
index 000000000..854501e0d
--- /dev/null
+++ b/leetcode/0307.Range-Sum-Query-Mutable/README.md
@@ -0,0 +1,51 @@
+# [307. Range Sum Query - Mutable](https://leetcode.com/problems/range-sum-query-mutable/)
+
+
+## 题目
+
+Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
+
+The update(i, val) function modifies nums by updating the element at index i to val.
+
+**Example:**
+
+    Given nums = [1, 3, 5]
+    
+    sumRange(0, 2) -> 9
+    update(1, 2)
+    sumRange(0, 2) -> 8
+
+**Note:**
+
+1. The array is only modifiable by the update function.
+2. You may assume the number of calls to update and sumRange function is distributed evenly.
+
+
+## 题目大意
+
+给定一个整数数组  nums，求出数组从索引 i 到 j  (i ≤ j) 范围内元素的总和，包含 i,  j 两点。
+
+update(i, val) 函数可以通过将下标为 i 的数值更新为 val，从而对数列进行修改。
+
+示例:
+
+```
+Given nums = [1, 3, 5]
+
+sumRange(0, 2) -> 9  
+update(1, 2)  
+sumRange(0, 2) -> 8  
+```
+
+说明:
+
+- 数组仅可以在 update 函数下进行修改。
+- 你可以假设 update 函数与 sumRange 函数的调用次数是均匀分布的。
+
+## 解题思路
+
+
+- 给出一个数组，数组里面的数都是`**可变**`的，设计一个数据结构能够满足查询数组任意区间内元素的和。
+- 对比第 303 题，这一题由于数组里面的元素都是**`可变`**的，所以第一个想到的解法就是线段树，构建一颗线段树，父结点内存的是两个子结点的和，初始化建树的时间复杂度是 O(log n)，查询区间元素和的时间复杂度是 O(log n)，更新元素值的时间复杂度是 O(log n)。
+- 如果此题还用 prefixSum 的思路解答呢？那每次 update 操作的时间复杂度都是 O(n)，因为每次更改一个值，最坏情况就是所有的 prefixSum 都要更新一次。prefixSum 的方法在这道题上面也可以 AC，只不过时间排名在 5%，非常差。
+- 此题也可以用树状数组解决。代码很直白，区间查询即是两个区间前缀和相减。最简单的树状数组应用。
diff --git a/Algorithms/0309. Best Time to Buy and Sell Stock with Cooldown/309. Best Time to Buy and Sell Stock with Cooldown.go b/leetcode/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown/309. Best Time to Buy and Sell Stock with Cooldown.go
similarity index 93%
rename from Algorithms/0309. Best Time to Buy and Sell Stock with Cooldown/309. Best Time to Buy and Sell Stock with Cooldown.go
rename to leetcode/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown/309. Best Time to Buy and Sell Stock with Cooldown.go
index 5e2c02f4d..f6bd86d00 100644
--- a/Algorithms/0309. Best Time to Buy and Sell Stock with Cooldown/309. Best Time to Buy and Sell Stock with Cooldown.go	
+++ b/leetcode/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown/309. Best Time to Buy and Sell Stock with Cooldown.go	
@@ -23,6 +23,13 @@ func maxProfit309(prices []int) int {
 	return sell[len(sell)-1]
 }
 
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
 // 解法二 优化辅助空间的 DP
 func maxProfit309_1(prices []int) int {
 	if len(prices) <= 1 {
diff --git a/Algorithms/0309. Best Time to Buy and Sell Stock with Cooldown/309. Best Time to Buy and Sell Stock with Cooldown_test.go b/leetcode/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown/309. Best Time to Buy and Sell Stock with Cooldown_test.go
similarity index 90%
rename from Algorithms/0309. Best Time to Buy and Sell Stock with Cooldown/309. Best Time to Buy and Sell Stock with Cooldown_test.go
rename to leetcode/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown/309. Best Time to Buy and Sell Stock with Cooldown_test.go
index ed8a04188..1204bef6e 100644
--- a/Algorithms/0309. Best Time to Buy and Sell Stock with Cooldown/309. Best Time to Buy and Sell Stock with Cooldown_test.go	
+++ b/leetcode/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown/309. Best Time to Buy and Sell Stock with Cooldown_test.go	
@@ -26,32 +26,32 @@ func Test_Problem309(t *testing.T) {
 
 	qs := []question309{
 
-		question309{
+		{
 			para309{[]int{}},
 			ans309{0},
 		},
 
-		question309{
+		{
 			para309{[]int{2, 1, 4, 5, 2, 9, 7}},
 			ans309{10},
 		},
 
-		question309{
+		{
 			para309{[]int{6, 1, 3, 2, 4, 7}},
 			ans309{6},
 		},
 
-		question309{
+		{
 			para309{[]int{1, 2, 3, 0, 2}},
 			ans309{3},
 		},
 
-		question309{
+		{
 			para309{[]int{7, 1, 5, 3, 6, 4}},
 			ans309{5},
 		},
 
-		question309{
+		{
 			para309{[]int{7, 6, 4, 3, 1}},
 			ans309{0},
 		},
diff --git a/Algorithms/0309. Best Time to Buy and Sell Stock with Cooldown/README.md b/leetcode/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown/README.md
similarity index 100%
rename from Algorithms/0309. Best Time to Buy and Sell Stock with Cooldown/README.md
rename to leetcode/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown/README.md
diff --git a/leetcode/0315.Count-of-Smaller-Numbers-After-Self/315. Count of Smaller Numbers After Self.go b/leetcode/0315.Count-of-Smaller-Numbers-After-Self/315. Count of Smaller Numbers After Self.go
new file mode 100644
index 000000000..d5908ce7a
--- /dev/null
+++ b/leetcode/0315.Count-of-Smaller-Numbers-After-Self/315. Count of Smaller Numbers After Self.go	
@@ -0,0 +1,66 @@
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+// 解法一 线段树
+func countSmaller(nums []int) []int {
+	if len(nums) == 0 {
+		return []int{}
+	}
+	st, minNum, numsMap, numsArray, res := template.SegmentCountTree{}, 0, make(map[int]int, 0), []int{}, make([]int, len(nums))
+	for i := 0; i < len(nums); i++ {
+		numsMap[nums[i]] = nums[i]
+	}
+	for _, v := range numsMap {
+		numsArray = append(numsArray, v)
+	}
+	// 排序是为了使得线段树中的区间 left <= right，如果此处不排序，线段树中的区间有很多不合法。
+	sort.Ints(numsArray)
+	minNum = numsArray[0]
+	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
+	st.Init(numsArray, func(i, j int) int {
+		return 0
+	})
+	for i := len(nums) - 1; i >= 0; i-- {
+		if nums[i] == minNum {
+			res[i] = 0
+			st.UpdateCount(nums[i])
+			continue
+		}
+		st.UpdateCount(nums[i])
+		res[i] = st.Query(minNum, nums[i]-1)
+	}
+	return res
+}
+
+// 解法二 树状数组
+func countSmaller1(nums []int) []int {
+	// copy 一份原数组至所有数字 allNums 数组中
+	allNums, res := make([]int, len(nums)), []int{}
+	copy(allNums, nums)
+	// 将 allNums 离散化
+	sort.Ints(allNums)
+	k := 1
+	kth := map[int]int{allNums[0]: k}
+	for i := 1; i < len(allNums); i++ {
+		if allNums[i] != allNums[i-1] {
+			k++
+			kth[allNums[i]] = k
+		}
+	}
+	// 树状数组 Query
+	bit := template.BinaryIndexedTree{}
+	bit.Init(k)
+	for i := len(nums) - 1; i >= 0; i-- {
+		res = append(res, bit.Query(kth[nums[i]]-1))
+		bit.Add(kth[nums[i]], 1)
+	}
+	for i := 0; i < len(res)/2; i++ {
+		res[i], res[len(res)-1-i] = res[len(res)-1-i], res[i]
+	}
+	return res
+}
diff --git a/Algorithms/0315. Count of Smaller Numbers After Self/315. Count of Smaller Numbers After Self_test.go b/leetcode/0315.Count-of-Smaller-Numbers-After-Self/315. Count of Smaller Numbers After Self_test.go
similarity index 95%
rename from Algorithms/0315. Count of Smaller Numbers After Self/315. Count of Smaller Numbers After Self_test.go
rename to leetcode/0315.Count-of-Smaller-Numbers-After-Self/315. Count of Smaller Numbers After Self_test.go
index 82b31ad68..35579fcf0 100644
--- a/Algorithms/0315. Count of Smaller Numbers After Self/315. Count of Smaller Numbers After Self_test.go	
+++ b/leetcode/0315.Count-of-Smaller-Numbers-After-Self/315. Count of Smaller Numbers After Self_test.go	
@@ -26,17 +26,17 @@ func Test_Problem315(t *testing.T) {
 
 	qs := []question315{
 
-		question315{
+		{
 			para315{[]int{5, 2, 6, 1}},
 			ans315{[]int{2, 1, 1, 0}},
 		},
 
-		question315{
+		{
 			para315{[]int{-1, -1}},
 			ans315{[]int{0, 0}},
 		},
 
-		question315{
+		{
 			para315{[]int{26, 78, 27, 100, 33, 67, 90, 23, 66, 5, 38, 7, 35, 23, 52, 22, 83, 51, 98, 69, 81, 32, 78, 28, 94, 13, 2, 97, 3, 76, 99, 51, 9, 21, 84, 66, 65, 36, 100, 41}},
 			ans315{[]int{10, 27, 10, 35, 12, 22, 28, 8, 19, 2, 12, 2, 9, 6, 12, 5, 17, 9, 19, 12, 14, 6, 12, 5, 12, 3, 0, 10, 0, 7, 8, 4, 0, 0, 4, 3, 2, 0, 1, 0}},
 		},
diff --git a/leetcode/0315.Count-of-Smaller-Numbers-After-Self/README.md b/leetcode/0315.Count-of-Smaller-Numbers-After-Self/README.md
new file mode 100755
index 000000000..680fef3be
--- /dev/null
+++ b/leetcode/0315.Count-of-Smaller-Numbers-After-Self/README.md
@@ -0,0 +1,42 @@
+# [315. Count of Smaller Numbers After Self](https://leetcode.com/problems/count-of-smaller-numbers-after-self/)
+
+
+## 题目
+
+You are given an integer array nums and you have to return a new counts array. The counts array has the property where `counts[i]` is the number of smaller elements to the right of `nums[i]`.
+
+**Example:**
+
+    Input: [5,2,6,1]
+    Output: [2,1,1,0] 
+    Explanation:
+    To the right of 5 there are 2 smaller elements (2 and 1).
+    To the right of 2 there is only 1 smaller element (1).
+    To the right of 6 there is 1 smaller element (1).
+    To the right of 1 there is 0 smaller element.
+
+
+## 题目大意
+
+
+给定一个整数数组 nums，按要求返回一个新数组 counts。数组 counts 有该性质： counts[i] 的值是  nums[i] 右侧小于 nums[i] 的元素的数量。
+
+示例:
+
+```
+输入: [5,2,6,1]
+输出: [2,1,1,0] 
+解释:
+5 的右侧有 2 个更小的元素 (2 和 1).
+2 的右侧仅有 1 个更小的元素 (1).
+6 的右侧有 1 个更小的元素 (1).
+1 的右侧有 0 个更小的元素.
+```
+
+
+## 解题思路
+
+
+- 给出一个数组，要求输出数组中每个元素相对于数组中的位置右边比它小的元素。
+- 这一题是第 327 题的缩水版。由于需要找数组位置右边比当前位置元素小的元素，所以从数组右边开始往左边扫。构造一颗线段树，线段树里面父节点存的是子节点出现的次数和。有可能给的数据会很大，所以构造线段树的时候先离散化。还需要注意的是数组里面可能有重复元素，所以构造线段树要先去重并排序。从右往左扫的过程中，依次添加数组中的元素，添加了一次就立即 query 一次。query 的区间是 [minNum, nums[i]-1]。如果是 minNum 则输出 0，并且也要记得插入这个最小值。这一题的思路和第 327 题大体类似，详解可见第 327 题。
+- 这一题同样可以用树状数组来解答。相比 327 题简单很多。第一步还是把所有用到的元素放入 allNums 数组中，第二步排序 + 离散化。由于题目要求输出右侧更小的元素，所以第三步倒序插入构造树状数组，Query 查询 `[1,i-1]` 区间内元素总数即为右侧更小元素个数。注意最终输出是顺序输出，计算是逆序计算的，最终数组里面的答案还需要逆序一遍。相同的套路题有，第 327 题，第 493 题。
\ No newline at end of file
diff --git a/Algorithms/0318. Maximum Product of Word Lengths/318. Maximum Product of Word Lengths.go b/leetcode/0318.Maximum-Product-of-Word-Lengths/318. Maximum Product of Word Lengths.go
similarity index 100%
rename from Algorithms/0318. Maximum Product of Word Lengths/318. Maximum Product of Word Lengths.go
rename to leetcode/0318.Maximum-Product-of-Word-Lengths/318. Maximum Product of Word Lengths.go
diff --git a/Algorithms/0318. Maximum Product of Word Lengths/318. Maximum Product of Word Lengths_test.go b/leetcode/0318.Maximum-Product-of-Word-Lengths/318. Maximum Product of Word Lengths_test.go
similarity index 94%
rename from Algorithms/0318. Maximum Product of Word Lengths/318. Maximum Product of Word Lengths_test.go
rename to leetcode/0318.Maximum-Product-of-Word-Lengths/318. Maximum Product of Word Lengths_test.go
index 84078b207..3c7a10fd2 100644
--- a/Algorithms/0318. Maximum Product of Word Lengths/318. Maximum Product of Word Lengths_test.go	
+++ b/leetcode/0318.Maximum-Product-of-Word-Lengths/318. Maximum Product of Word Lengths_test.go	
@@ -26,17 +26,17 @@ func Test_Problem318(t *testing.T) {
 
 	qs := []question318{
 
-		question318{
+		{
 			para318{[]string{"abcw", "baz", "foo", "bar", "xtfn", "abcdef"}},
 			ans318{16},
 		},
 
-		question318{
+		{
 			para318{[]string{"a", "ab", "abc", "d", "cd", "bcd", "abcd"}},
 			ans318{4},
 		},
 
-		question318{
+		{
 			para318{[]string{"a", "aa", "aaa", "aaaa"}},
 			ans318{0},
 		},
diff --git a/leetcode/0318.Maximum-Product-of-Word-Lengths/README.md b/leetcode/0318.Maximum-Product-of-Word-Lengths/README.md
new file mode 100755
index 000000000..b41899281
--- /dev/null
+++ b/leetcode/0318.Maximum-Product-of-Word-Lengths/README.md
@@ -0,0 +1,46 @@
+# [318. Maximum Product of Word Lengths](https://leetcode.com/problems/maximum-product-of-word-lengths/)
+
+
+## 题目
+
+Given a string array `words`, find the maximum value of `length(word[i]) * length(word[j])` where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.
+
+**Example 1:**
+
+    Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
+    Output: 16 
+    Explanation: The two words can be "abcw", "xtfn".
+
+**Example 2:**
+
+    Input: ["a","ab","abc","d","cd","bcd","abcd"]
+    Output: 4 
+    Explanation: The two words can be "ab", "cd".
+
+**Example 3:**
+
+    Input: ["a","aa","aaa","aaaa"]
+    Output: 0 
+    Explanation: No such pair of words.
+
+
+
+## 题目大意
+
+给定一个字符串数组 words，找到 length(word[i]) * length(word[j]) 的最大值，并且这两个单词不含有公共字母。你可以认为每个单词只包含小写字母。如果不存在这样的两个单词，返回 0。
+
+
+## 解题思路
+
+- 在字符串数组中找到 2 个没有公共字符的字符串，并且这两个字符串的长度乘积要是最大的，求这个最大的乘积。
+- 这里需要利用位运算 `&` 运算的性质，如果 `X & Y = 0`，说明 X 和 Y 完全不相同。那么我们将字符串都编码成二进制数，进行 `&` 运算即可分出没有公共字符的字符串，最后动态维护长度乘积最大值即可。将字符串编码成二进制数的规则比较简单，每个字符相对于 'a' 的距离，根据这个距离将 1 左移多少位。
+
+```c
+    a 1->1  
+    b 2->10  
+    c 4->100  
+    ab 3->11  
+    ac 5->101  
+    abc 7->111  
+    az 33554433->10000000000000000000000001  
+```
diff --git a/leetcode/0319.Bulb-Switcher/319.Bulb Switcher.go b/leetcode/0319.Bulb-Switcher/319.Bulb Switcher.go
new file mode 100644
index 000000000..fba8ea41e
--- /dev/null
+++ b/leetcode/0319.Bulb-Switcher/319.Bulb Switcher.go	
@@ -0,0 +1,7 @@
+package leetcode
+
+import "math"
+
+func bulbSwitch(n int) int {
+	return int(math.Sqrt(float64(n)))
+}
diff --git a/leetcode/0319.Bulb-Switcher/319.Bulb Switcher_test.go b/leetcode/0319.Bulb-Switcher/319.Bulb Switcher_test.go
new file mode 100644
index 000000000..81f40a62e
--- /dev/null
+++ b/leetcode/0319.Bulb-Switcher/319.Bulb Switcher_test.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question319 struct {
+	para319
+	ans319
+}
+
+// para 是参数
+type para319 struct {
+	n int
+}
+
+// ans 是答案
+type ans319 struct {
+	ans int
+}
+
+func Test_Problem319(t *testing.T) {
+
+	qs := []question319{
+
+		{
+			para319{3},
+			ans319{1},
+		},
+
+		{
+			para319{0},
+			ans319{0},
+		},
+
+		{
+			para319{1},
+			ans319{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 319------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans319, q.para319
+		fmt.Printf("【input】:%v       【output】:%v\n", p, bulbSwitch(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0319.Bulb-Switcher/README.md b/leetcode/0319.Bulb-Switcher/README.md
new file mode 100644
index 000000000..830228df8
--- /dev/null
+++ b/leetcode/0319.Bulb-Switcher/README.md
@@ -0,0 +1,58 @@
+# [319. Bulb Switcher](https://leetcode.com/problems/bulb-switcher/)
+
+
+## 题目
+
+There are n bulbs that are initially off. You first turn on all the bulbs, then you turn off every second bulb.
+
+On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb.
+
+Return the number of bulbs that are on after n rounds.
+
+**Example 1:**
+
+    Input: n = 3
+    Output: 1
+    Explanation: At first, the three bulbs are [off, off, off].
+    After the first round, the three bulbs are [on, on, on].
+    After the second round, the three bulbs are [on, off, on].
+    After the third round, the three bulbs are [on, off, off].
+    So you should return 1 because there is only one bulb is on.
+
+**Example 2:**
+
+    Input: n = 0
+    Output: 0
+
+**Example 3:**
+
+    Input: n = 1
+    Output: 1
+
+## 题目大意
+
+初始时有 n 个灯泡处于关闭状态。第一轮，你将会打开所有灯泡。接下来的第二轮，你将会每两个灯泡关闭一个。
+
+第三轮，你每三个灯泡就切换一个灯泡的开关（即，打开变关闭，关闭变打开）。第 i 轮，你每 i 个灯泡就切换一个灯泡的开关。直到第 n 轮，你只需要切换最后一个灯泡的开关。
+
+找出并返回 n 轮后有多少个亮着的灯泡。
+
+## 解题思路
+
+- 计算 1 到 n 中有奇数个约数的个数
+- 1 到 n 中的某个数 x 有奇数个约数,也即 x 是完全平方数
+- 计算 1 到 n 中完全平方数的个数 sqrt(n)
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func bulbSwitch(n int) int {
+	return int(math.Sqrt(float64(n)))
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0322. Coin Change/322. Coin Change.go b/leetcode/0322.Coin-Change/322. Coin Change.go
similarity index 84%
rename from Algorithms/0322. Coin Change/322. Coin Change.go
rename to leetcode/0322.Coin-Change/322. Coin Change.go
index c9c7e1b89..a8fdcb994 100644
--- a/Algorithms/0322. Coin Change/322. Coin Change.go	
+++ b/leetcode/0322.Coin-Change/322. Coin Change.go	
@@ -18,3 +18,10 @@ func coinChange(coins []int, amount int) int {
 	}
 	return dp[amount]
 }
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/Algorithms/0322. Coin Change/322. Coin Change_test.go b/leetcode/0322.Coin-Change/322. Coin Change_test.go
similarity index 91%
rename from Algorithms/0322. Coin Change/322. Coin Change_test.go
rename to leetcode/0322.Coin-Change/322. Coin Change_test.go
index 903634032..8f7fe2f5a 100644
--- a/Algorithms/0322. Coin Change/322. Coin Change_test.go	
+++ b/leetcode/0322.Coin-Change/322. Coin Change_test.go	
@@ -27,26 +27,26 @@ func Test_Problem322(t *testing.T) {
 
 	qs := []question322{
 
-		question322{
+		{
 			para322{[]int{186, 419, 83, 408}, 6249},
 			ans322{20},
 		},
 
-		question322{
+		{
 			para322{[]int{1, 2147483647}, 2},
 			ans322{2},
 		},
 
-		question322{
+		{
 			para322{[]int{1, 2, 5}, 11},
 			ans322{3},
 		},
-		question322{
+		{
 			para322{[]int{2}, 3},
 			ans322{-1},
 		},
 
-		question322{
+		{
 			para322{[]int{1}, 0},
 			ans322{0},
 		},
diff --git a/Algorithms/0322. Coin Change/README.md b/leetcode/0322.Coin-Change/README.md
similarity index 100%
rename from Algorithms/0322. Coin Change/README.md
rename to leetcode/0322.Coin-Change/README.md
diff --git a/Algorithms/0324. Wiggle Sort II/324. Wiggle Sort II.go b/leetcode/0324.Wiggle-Sort-II/324. Wiggle Sort II.go
similarity index 100%
rename from Algorithms/0324. Wiggle Sort II/324. Wiggle Sort II.go
rename to leetcode/0324.Wiggle-Sort-II/324. Wiggle Sort II.go
diff --git a/Algorithms/0324. Wiggle Sort II/324. Wiggle Sort II_test.go b/leetcode/0324.Wiggle-Sort-II/324. Wiggle Sort II_test.go
similarity index 92%
rename from Algorithms/0324. Wiggle Sort II/324. Wiggle Sort II_test.go
rename to leetcode/0324.Wiggle-Sort-II/324. Wiggle Sort II_test.go
index a9c4cba52..9885673bb 100644
--- a/Algorithms/0324. Wiggle Sort II/324. Wiggle Sort II_test.go	
+++ b/leetcode/0324.Wiggle-Sort-II/324. Wiggle Sort II_test.go	
@@ -26,27 +26,27 @@ func Test_Problem324(t *testing.T) {
 
 	qs := []question324{
 
-		question324{
+		{
 			para324{[]int{}},
 			ans324{[]int{}},
 		},
 
-		question324{
+		{
 			para324{[]int{1}},
 			ans324{[]int{1}},
 		},
 
-		question324{
+		{
 			para324{[]int{1, 5, 1, 1, 6, 4}},
 			ans324{[]int{1, 4, 1, 5, 1, 6}},
 		},
 
-		question324{
+		{
 			para324{[]int{1, 3, 2, 2, 3, 1}},
 			ans324{[]int{2, 3, 1, 3, 1, 2}},
 		},
 
-		question324{
+		{
 			para324{[]int{1, 1, 2, 1, 2, 2, 1}},
 			ans324{[]int{1, 2, 1, 2, 1, 2, 1}},
 		},
diff --git a/Algorithms/0324. Wiggle Sort II/README.md b/leetcode/0324.Wiggle-Sort-II/README.md
similarity index 100%
rename from Algorithms/0324. Wiggle Sort II/README.md
rename to leetcode/0324.Wiggle-Sort-II/README.md
diff --git a/Algorithms/0326. Power of Three/326. Power of Three.go b/leetcode/0326.Power-of-Three/326. Power of Three.go
similarity index 100%
rename from Algorithms/0326. Power of Three/326. Power of Three.go
rename to leetcode/0326.Power-of-Three/326. Power of Three.go
diff --git a/Algorithms/0326. Power of Three/326. Power of Three_test.go b/leetcode/0326.Power-of-Three/326. Power of Three_test.go
similarity index 92%
rename from Algorithms/0326. Power of Three/326. Power of Three_test.go
rename to leetcode/0326.Power-of-Three/326. Power of Three_test.go
index b338d8722..48ba0a6fc 100644
--- a/Algorithms/0326. Power of Three/326. Power of Three_test.go	
+++ b/leetcode/0326.Power-of-Three/326. Power of Three_test.go	
@@ -26,22 +26,22 @@ func Test_Problem326(t *testing.T) {
 
 	qs := []question326{
 
-		question326{
+		{
 			para326{27},
 			ans326{true},
 		},
 
-		question326{
+		{
 			para326{0},
 			ans326{false},
 		},
 
-		question326{
+		{
 			para326{9},
 			ans326{true},
 		},
 
-		question326{
+		{
 			para326{45},
 			ans326{false},
 		},
diff --git a/Algorithms/0326. Power of Three/README.md b/leetcode/0326.Power-of-Three/README.md
similarity index 100%
rename from Algorithms/0326. Power of Three/README.md
rename to leetcode/0326.Power-of-Three/README.md
diff --git a/leetcode/0327.Count-of-Range-Sum/327. Count of Range Sum.go b/leetcode/0327.Count-of-Range-Sum/327. Count of Range Sum.go
new file mode 100644
index 000000000..68c7b4665
--- /dev/null
+++ b/leetcode/0327.Count-of-Range-Sum/327. Count of Range Sum.go	
@@ -0,0 +1,91 @@
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+// 解法一 线段树，时间复杂度 O(n log n)
+func countRangeSum(nums []int, lower int, upper int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	st, prefixSum, sumMap, sumArray, res := template.SegmentCountTree{}, make([]int, len(nums)), make(map[int]int, 0), []int{}, 0
+	prefixSum[0], sumMap[nums[0]] = nums[0], nums[0]
+	for i := 1; i < len(nums); i++ {
+		prefixSum[i] = prefixSum[i-1] + nums[i]
+		sumMap[prefixSum[i]] = prefixSum[i]
+	}
+	// sumArray 是 prefixSum 去重之后的版本，利用 sumMap 去重
+	for _, v := range sumMap {
+		sumArray = append(sumArray, v)
+	}
+	// 排序是为了使得线段树中的区间 left <= right，如果此处不排序，线段树中的区间有很多不合法。
+	sort.Ints(sumArray)
+	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
+	st.Init(sumArray, func(i, j int) int {
+		return 0
+	})
+	// 倒序是为了方便寻找 j，sum(i，j) 规定了 j >= i，所以倒序遍历，i 从大到小
+	for i := len(nums) - 1; i >= 0; i-- {
+		// 插入的 prefixSum[i] 即是 j
+		st.UpdateCount(prefixSum[i])
+		if i > 0 {
+			res += st.Query(lower+prefixSum[i-1], upper+prefixSum[i-1])
+		} else {
+			res += st.Query(lower, upper)
+		}
+	}
+	return res
+}
+
+// 解法二 树状数组，时间复杂度 O(n log n)
+func countRangeSum1(nums []int, lower int, upper int) int {
+	n := len(nums)
+	// 计算前缀和 preSum，以及后面统计时会用到的所有数字 allNums
+	allNums, preSum, res := make([]int, 1, 3*n+1), make([]int, n+1), 0
+	for i, v := range nums {
+		preSum[i+1] = preSum[i] + v
+		allNums = append(allNums, preSum[i+1], preSum[i+1]-lower, preSum[i+1]-upper)
+	}
+	// 将 allNums 离散化
+	sort.Ints(allNums)
+	k := 1
+	kth := map[int]int{allNums[0]: k}
+	for i := 1; i <= 3*n; i++ {
+		if allNums[i] != allNums[i-1] {
+			k++
+			kth[allNums[i]] = k
+		}
+	}
+	// 遍历 preSum，利用树状数组计算每个前缀和对应的合法区间数
+	bit := template.BinaryIndexedTree{}
+	bit.Init(k)
+	bit.Add(kth[0], 1)
+	for _, sum := range preSum[1:] {
+		left, right := kth[sum-upper], kth[sum-lower]
+		res += bit.Query(right) - bit.Query(left-1)
+		bit.Add(kth[sum], 1)
+	}
+	return res
+}
+
+// 解法三 暴力，时间复杂度 O(n^2)
+func countRangeSum2(nums []int, lower int, upper int) int {
+	res, n := 0, len(nums)
+	for i := 0; i < n; i++ {
+		tmp := 0
+		for j := i; j < n; j++ {
+			if i == j {
+				tmp = nums[i]
+			} else {
+				tmp += nums[j]
+			}
+			if tmp <= upper && tmp >= lower {
+				res++
+			}
+		}
+	}
+	return res
+}
diff --git a/Algorithms/0327. Count of Range Sum/327. Count of Range Sum_test.go b/leetcode/0327.Count-of-Range-Sum/327. Count of Range Sum_test.go
similarity index 94%
rename from Algorithms/0327. Count of Range Sum/327. Count of Range Sum_test.go
rename to leetcode/0327.Count-of-Range-Sum/327. Count of Range Sum_test.go
index e0512ed74..555a63e43 100644
--- a/Algorithms/0327. Count of Range Sum/327. Count of Range Sum_test.go	
+++ b/leetcode/0327.Count-of-Range-Sum/327. Count of Range Sum_test.go	
@@ -28,17 +28,17 @@ func Test_Problem327(t *testing.T) {
 
 	qs := []question327{
 
-		question327{
+		{
 			para327{[]int{-2, 5, -1}, -2, 2},
 			ans327{3},
 		},
 
-		question327{
+		{
 			para327{[]int{0, -3, -3, 1, 1, 2}, 3, 5},
 			ans327{2},
 		},
 
-		question327{
+		{
 			para327{[]int{-3, 1, 2, -2, 2, -1}, -3, -1},
 			ans327{7},
 		},
diff --git a/leetcode/0327.Count-of-Range-Sum/README.md b/leetcode/0327.Count-of-Range-Sum/README.md
new file mode 100755
index 000000000..be08237ff
--- /dev/null
+++ b/leetcode/0327.Count-of-Range-Sum/README.md
@@ -0,0 +1,78 @@
+# [327. Count of Range Sum](https://leetcode.com/problems/count-of-range-sum/)
+
+
+## 题目
+
+Given an integer array `nums`, return the number of range sums that lie in `[lower, upper]` inclusive.Range sum `S(i, j)` is defined as the sum of the elements in `nums` between indices `i` and `j` (`i` ≤ `j`), inclusive.
+
+**Note:**A naive algorithm of O(n2) is trivial. You MUST do better than that.
+
+**Example:**
+
+    Input: nums = [-2,5,-1], lower = -2, upper = 2,
+    Output: 3 
+    Explanation: The three ranges are : [0,0], [2,2], [0,2] and their respective sums are: -2, -1, 2.
+
+
+## 题目大意
+
+
+给定一个整数数组 nums 。区间和 S(i, j) 表示在 nums 中，位置从 i 到 j 的元素之和，包含 i 和 j (i ≤ j)。请你以下标 i （0 <= i <= nums.length ）为起点，元素个数逐次递增，计算子数组内的元素和。当元素和落在范围 [lower, upper] （包含 lower 和 upper）之内时，记录子数组当前最末元素下标 j ，记作 有效 区间和 S(i, j) 。求数组中，值位于范围 [lower, upper] （包含 lower 和 upper）之内的 有效 区间和的个数。
+
+注意：
+最直观的算法复杂度是 O(n2) ，请在此基础上优化你的算法。
+
+## 解题思路
+
+- 给出一个数组，要求在这个数组中找出任意一段子区间的和，位于 [lower,upper] 之间。
+- 这一题可以用暴力解法，2 层循环，遍历所有子区间，求和并判断是否位于 [lower,upper] 之间，时间复杂度 O(n^2)。
+- 这一题当然还有更优的解法，用线段树或者树状数组，将时间复杂度降为 O(n log n)。题目中要求 `lower ≤ sum(i,j) ≤ upper`，`sum(i,j) = prefixSum(j) - prefixSum(i-1)`，那么 `lower + prefixSum(i-1) ≤ prefixSum(j) ≤ upper + prefixSum(i-1)`。所以利用前缀和将区间和转换成了前缀和在线段树中 `query` 的问题，只不过线段树中父节点中存的不是子节点的和，而应该是子节点出现的次数。第二个转换，由于前缀和会很大，所以需要离散化。例如 `prefixSum = [-3,-2,-1,0]`，用前缀和下标进行离散化，所以线段树中左右区间变成了 0-3 。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_0.png)
+
+    利用 `prefixSum` 下标离散化：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_1.png)
+
+- 还需要注意一些小细节，`prefixSum` 计算完以后需要去重，去重以后并排序，方便构造线段树的有效区间。如果不去重，线段树中可能出现非法区间(left > right)或者重叠区间。最后一步往线段树中倒序插入 `prefixSum` 的时候，用的是非去重的，插入 `prefixSum[j]` 代表 sum(i,j) 中的 j，例如往线段树中插入 `prefixSum[5]`，代表当前树中加入了 j = 5 的情况。query 操作实质是在做区间匹配，例如当前 i 循环到 i = 3，累计往线段树中插入了 `prefixSum[5]`，`prefixSum[4]`，`prefixSum[3]`，那么 query 操作实质是在判断：`lower ≤ sum(i=3,j=3) ≤ upper`，`lower ≤ sum(i=3,j=4) ≤ upper`，`lower ≤ sum(i=3,j=5) ≤ upper`，这 3 个等式是否成立，有几个成立就返回几个，即是最终要求得的结果的一部分。
+- 举个例子，`nums = [-3,1,2,-2,2,-1]`，`prefixSum = [-3,-2,0,-2,0,-1]`，去重以后并排序得到 `sum = [-3,-2,-1,0]`。离散化构造线段树，这里出于演示的方便，下图中就不画出离散后的线段树了，用非离散的线段树展示：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_2_.png)
+
+    倒序插入 `len(prefixSum)-1 = prefixSum[5] = -1`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_3_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[5-1], -1 + prefixSum[5-1]] = [-3,-1]`，即判断 `-3 ≤ sum(5,5) ≤ -1`，满足等式的有几种情况，这里明显只有一种情况，即 `j = 5`，也满足等式，所以这一步 `res = 1`。
+
+- 倒序插入 `len(prefixSum)-2 = prefixSum[4] = 0`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_4_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[4-1], -1 + prefixSum[4-1]] = [-5,-3]`，即判断 `-5 ≤ sum(4,  4,5) ≤ -3`，满足等式的有几种情况，这里有两种情况，即 `j = 4` 或者 `j = 5`，都不满足等式，所以这一步 `res = 0`。
+
+- 倒序插入 `len(prefixSum)-3 = prefixSum[3] = -2`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_5_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[3-1], -1 + prefixSum[3-1]] = [-3,-1]`，即判断 `-3 ≤ sum(3,  3,4,5) ≤ -1`，满足等式的有几种情况，这里有三种情况，即 `j = 3` 、`j = 4` 或者 `j = 5`，满足等式的有 `j = 3` 和 `j = 5`，即 `-3 ≤ sum(3, 3) ≤ -1` 和 `-3 ≤ sum(3, 5) ≤ -1`。所以这一步 `res = 2`。
+
+- 倒序插入 `len(prefixSum)-4 = prefixSum[2] = 0`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_6_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[2-1], -1 + prefixSum[2-1]] = [-5,-3]`，即判断 `-5 ≤ sum(2,  2,3,4,5) ≤ -3`，满足等式的有几种情况，这里有四种情况，即 `j = 2`、 `j = 3` 、`j = 4` 或者 `j = 5`，都不满足等式。所以这一步 `res = 0`。
+
+- 倒序插入 `len(prefixSum)-5 = prefixSum[1] = -2`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_7_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[1-1], -1 + prefixSum[1-1]] = [-6,-4]`，即判断 `-6 ≤ sum(1,  1,2,3,4,5) ≤ -4`，满足等式的有几种情况，这里有五种情况，即 `j = 1`、 `j = 2`、 `j = 3` 、`j = 4` 或者 `j = 5`，都不满足等式。所以这一步 `res = 0`。
+
+- 倒序插入 `len(prefixSum)-6 = prefixSum[0] = -3`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_8_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[0-1], -1 + prefixSum[0-1]] = [-3,-1]`，注意 `prefixSum[-1] = 0`，即判断 `-3 ≤ sum(0,  0,1,2,3,4,5) ≤ -1`，满足等式的有几种情况，这里有六种情况，即 `j = 0`、`j = 1`、`j = 2`、 `j = 3` 、`j = 4` 或者 `j = 5`，满足等式的有 `j = 0`、`j = 1`、 `j = 3` 和 `j = 5`，即 `-3 ≤ sum(0, 0) ≤ -1` 、 `-3 ≤ sum(0, 1) ≤ -1`、`-3 ≤ sum(0, 3) ≤ -1` 和 `-3 ≤ sum(0, 5) ≤ -1`。所以这一步 `res = 4`。最后的答案就是把每一步的结果都累加，`res = 1 + 0 + 2 + 0 + 0 + 4 = 7`。
+
+- 此题同样可以用树状数组来解答。同样把问题先转化成区间 Query 的模型，`lower ≤ prefixSum(j) - prefixSum(i-1) ≤ upper` 等价于 `prefixSum(j) - upper ≤ prefixSum(i-1) ≤ prefixSum(j) - lower`，`i` 的取值在 `[0,j-1]` 区间内。所以题目可以转化为 `i` 在 `[0,j-1]` 区间内取值，问数组 `prefixSum[0...j-1]` 中的所有取值，位于区间 `[prefixSum(j) - upper, prefixSum(j) - lower]` 内的次数。在树状数组中，区间内的前缀和可以转化为 2 个区间的前缀和相减，即 `Query([i,j]) = Query(j) - Query(i-1)`。所以这道题枚举数组 `prefixSum[0...j-1]` 中每个值是否出现在指定区间内出现次数即可。第一步先将所有的前缀和 `prefixSum(j)` 以及 `[prefixSum(j) - upper, prefixSum(j) - lower]` 计算出来。第二步排序和离散化，离散化以后的点区间为 `[1,n]`。最后根据数组 `prefixSum(j)` 的值在指定区间内查询出现次数即可。相同的套路题有，第 315 题，第 493 题。
\ No newline at end of file
diff --git a/Algorithms/0328. Odd Even Linked List/328. Odd Even Linked List.go b/leetcode/0328.Odd-Even-Linked-List/328. Odd Even Linked List.go
similarity index 82%
rename from Algorithms/0328. Odd Even Linked List/328. Odd Even Linked List.go
rename to leetcode/0328.Odd-Even-Linked-List/328. Odd Even Linked List.go
index 4f771cc6b..5281610c0 100644
--- a/Algorithms/0328. Odd Even Linked List/328. Odd Even Linked List.go	
+++ b/leetcode/0328.Odd-Even-Linked-List/328. Odd Even Linked List.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0328. Odd Even Linked List/328. Odd Even Linked List_test.go b/leetcode/0328.Odd-Even-Linked-List/328. Odd Even Linked List_test.go
similarity index 83%
rename from Algorithms/0328. Odd Even Linked List/328. Odd Even Linked List_test.go
rename to leetcode/0328.Odd-Even-Linked-List/328. Odd Even Linked List_test.go
index d66cedcfa..6488dd54d 100644
--- a/Algorithms/0328. Odd Even Linked List/328. Odd Even Linked List_test.go	
+++ b/leetcode/0328.Odd-Even-Linked-List/328. Odd Even Linked List_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question328 struct {
@@ -26,37 +28,37 @@ func Test_Problem328(t *testing.T) {
 
 	qs := []question328{
 
-		question328{
+		{
 			para328{[]int{1, 4, 3, 2, 5, 2}},
 			ans328{[]int{1, 3, 5, 4, 2, 2}},
 		},
 
-		question328{
+		{
 			para328{[]int{1, 1, 2, 2, 3, 3, 3}},
 			ans328{[]int{1, 2, 3, 3, 1, 2, 3}},
 		},
 
-		question328{
+		{
 			para328{[]int{4, 3, 2, 5, 2}},
 			ans328{[]int{4, 2, 2, 3, 5}},
 		},
 
-		question328{
+		{
 			para328{[]int{1, 1, 1, 1, 1, 1}},
 			ans328{[]int{1, 1, 1, 1, 1, 1}},
 		},
 
-		question328{
+		{
 			para328{[]int{3, 1}},
 			ans328{[]int{3, 1}},
 		},
 
-		question328{
+		{
 			para328{[]int{1, 2, 3, 4, 5}},
 			ans328{[]int{1, 3, 5, 2, 4}},
 		},
 
-		question328{
+		{
 			para328{[]int{}},
 			ans328{[]int{}},
 		},
@@ -66,7 +68,7 @@ func Test_Problem328(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans328, q.para328
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(oddEvenList(S2l(p.one))))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(oddEvenList(structures.Ints2List(p.one))))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/leetcode/0328.Odd-Even-Linked-List/README.md b/leetcode/0328.Odd-Even-Linked-List/README.md
new file mode 100644
index 000000000..4caa26c68
--- /dev/null
+++ b/leetcode/0328.Odd-Even-Linked-List/README.md
@@ -0,0 +1,34 @@
+# [328. Odd Even Linked List](https://leetcode.com/problems/odd-even-linked-list/)
+
+## 题目
+
+Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.
+
+You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.
+
+Example 1:
+
+```c
+Input: 1->2->3->4->5->NULL
+Output: 1->3->5->2->4->NULL
+```
+
+Example 2:
+
+```c
+Input: 2->1->3->5->6->4->7->NULL
+Output: 2->3->6->7->1->5->4->NULL
+```
+
+Note:
+
+- The relative order inside both the even and odd groups should remain as it was in the input.
+- The first node is considered odd, the second node even and so on ...
+
+## 题目大意
+
+这道题和第 86 题非常类型。第 86 题是把排在某个点前面的小值放在一个链表中，排在某个点后端的大值放在另外一个链表中，最后 2 个链表首尾拼接一下就是答案。
+
+## 解题思路
+
+这道题思路也是一样的，分别把奇数节点和偶数节点都放在 2 个链表中，最后首尾拼接就是答案。
diff --git a/leetcode/0329.Longest-Increasing-Path-in-a-Matrix/329. Longest Increasing Path in a Matrix.go b/leetcode/0329.Longest-Increasing-Path-in-a-Matrix/329. Longest Increasing Path in a Matrix.go
new file mode 100644
index 000000000..74287a05c
--- /dev/null
+++ b/leetcode/0329.Longest-Increasing-Path-in-a-Matrix/329. Longest Increasing Path in a Matrix.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"math"
+)
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func longestIncreasingPath(matrix [][]int) int {
+	cache, res := make([][]int, len(matrix)), 0
+	for i := 0; i < len(cache); i++ {
+		cache[i] = make([]int, len(matrix[0]))
+	}
+	for i, v := range matrix {
+		for j := range v {
+			searchPath(matrix, cache, math.MinInt64, i, j)
+			res = max(res, cache[i][j])
+		}
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func isInIntBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+
+func searchPath(board, cache [][]int, lastNum, x, y int) int {
+	if board[x][y] <= lastNum {
+		return 0
+	}
+	if cache[x][y] > 0 {
+		return cache[x][y]
+	}
+	count := 1
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		if isInIntBoard(board, nx, ny) {
+			count = max(count, searchPath(board, cache, board[x][y], nx, ny)+1)
+		}
+	}
+	cache[x][y] = count
+	return count
+}
diff --git a/leetcode/0329.Longest-Increasing-Path-in-a-Matrix/329. Longest Increasing Path in a Matrix_test.go b/leetcode/0329.Longest-Increasing-Path-in-a-Matrix/329. Longest Increasing Path in a Matrix_test.go
new file mode 100644
index 000000000..cda6bfe2d
--- /dev/null
+++ b/leetcode/0329.Longest-Increasing-Path-in-a-Matrix/329. Longest Increasing Path in a Matrix_test.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question329 struct {
+	para329
+	ans329
+}
+
+// para 是参数
+// one 代表第一个参数
+type para329 struct {
+	matrix [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans329 struct {
+	one int
+}
+
+func Test_Problem329(t *testing.T) {
+
+	qs := []question329{
+
+		{
+			para329{[][]int{{1}}},
+			ans329{1},
+		},
+
+		{
+			para329{[][]int{{}}},
+			ans329{0},
+		},
+
+		{
+			para329{[][]int{{9, 9, 4}, {6, 6, 8}, {2, 1, 1}}},
+			ans329{4},
+		},
+
+		{
+			para329{[][]int{{3, 4, 5}, {3, 2, 6}, {2, 2, 1}}},
+			ans329{4},
+		},
+
+		{
+			para329{[][]int{{1, 5, 9}, {10, 11, 13}, {12, 13, 15}}},
+			ans329{5},
+		},
+
+		{
+			para329{[][]int{{1, 5, 7}, {11, 12, 13}, {12, 13, 15}}},
+			ans329{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 329------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans329, q.para329
+		fmt.Printf("【input】:%v       【output】:%v\n", p, longestIncreasingPath(p.matrix))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0329.Longest-Increasing-Path-in-a-Matrix/README.md b/leetcode/0329.Longest-Increasing-Path-in-a-Matrix/README.md
new file mode 100755
index 000000000..2bd4e8142
--- /dev/null
+++ b/leetcode/0329.Longest-Increasing-Path-in-a-Matrix/README.md
@@ -0,0 +1,42 @@
+# [329. Longest Increasing Path in a Matrix](https://leetcode.com/problems/longest-increasing-path-in-a-matrix/)
+
+
+## 题目
+
+Given an integer matrix, find the length of the longest increasing path.
+
+From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).
+
+**Example 1:**
+
+    Input: nums = 
+    [
+      [9,9,4],
+      [6,6,8],
+      [2,1,1]
+    ] 
+    Output: 4 
+    Explanation: The longest increasing path is [1, 2, 6, 9].
+
+**Example 2:**
+
+    Input: nums = 
+    [
+      [3,4,5],
+      [3,2,6],
+      [2,2,1]
+    ] 
+    Output: 4 
+    Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.
+
+
+## 题目大意
+
+给定一个整数矩阵，找出最长递增路径的长度。对于每个单元格，你可以往上，下，左，右四个方向移动。 你不能在对角线方向上移动或移动到边界外（即不允许环绕）。
+
+
+## 解题思路
+
+
+- 给出一个矩阵，要求在这个矩阵中找到一个最长递增的路径。路径有上下左右 4 个方向。
+- 这一题解题思路很明显，用 DFS 即可。在提交完第一版以后会发现 TLE，因为题目给出了一个非常大的矩阵，搜索次数太多。所以需要用到记忆化，把曾经搜索过的最大长度缓存起来，增加了记忆化以后再次提交 AC。
diff --git a/Algorithms/0331. Verify Preorder Serialization of a Binary Tree/331. Verify Preorder Serialization of a Binary Tree.go b/leetcode/0331.Verify-Preorder-Serialization-of-a-Binary-Tree/331. Verify Preorder Serialization of a Binary Tree.go
similarity index 100%
rename from Algorithms/0331. Verify Preorder Serialization of a Binary Tree/331. Verify Preorder Serialization of a Binary Tree.go
rename to leetcode/0331.Verify-Preorder-Serialization-of-a-Binary-Tree/331. Verify Preorder Serialization of a Binary Tree.go
diff --git a/Algorithms/0331. Verify Preorder Serialization of a Binary Tree/331. Verify Preorder Serialization of a Binary Tree_test.go b/leetcode/0331.Verify-Preorder-Serialization-of-a-Binary-Tree/331. Verify Preorder Serialization of a Binary Tree_test.go
similarity index 94%
rename from Algorithms/0331. Verify Preorder Serialization of a Binary Tree/331. Verify Preorder Serialization of a Binary Tree_test.go
rename to leetcode/0331.Verify-Preorder-Serialization-of-a-Binary-Tree/331. Verify Preorder Serialization of a Binary Tree_test.go
index 026f13d23..977626fcd 100644
--- a/Algorithms/0331. Verify Preorder Serialization of a Binary Tree/331. Verify Preorder Serialization of a Binary Tree_test.go	
+++ b/leetcode/0331.Verify-Preorder-Serialization-of-a-Binary-Tree/331. Verify Preorder Serialization of a Binary Tree_test.go	
@@ -26,17 +26,17 @@ func Test_Problem331(t *testing.T) {
 
 	qs := []question331{
 
-		question331{
+		{
 			para331{"9,3,4,#,#,1,#,#,2,#,6,#,#"},
 			ans331{true},
 		},
 
-		question331{
+		{
 			para331{"1,#"},
 			ans331{false},
 		},
 
-		question331{
+		{
 			para331{"9,#,#,1"},
 			ans331{false},
 		},
diff --git a/Algorithms/0331. Verify Preorder Serialization of a Binary Tree/README.md b/leetcode/0331.Verify-Preorder-Serialization-of-a-Binary-Tree/README.md
similarity index 100%
rename from Algorithms/0331. Verify Preorder Serialization of a Binary Tree/README.md
rename to leetcode/0331.Verify-Preorder-Serialization-of-a-Binary-Tree/README.md
diff --git a/leetcode/0337.House-Robber-III/337. House Robber III.go b/leetcode/0337.House-Robber-III/337. House Robber III.go
new file mode 100644
index 000000000..4be71f546
--- /dev/null
+++ b/leetcode/0337.House-Robber-III/337. House Robber III.go	
@@ -0,0 +1,41 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func rob337(root *TreeNode) int {
+	a, b := dfsTreeRob(root)
+	return max(a, b)
+}
+
+func dfsTreeRob(root *TreeNode) (a, b int) {
+	if root == nil {
+		return 0, 0
+	}
+	l0, l1 := dfsTreeRob(root.Left)
+	r0, r1 := dfsTreeRob(root.Right)
+	// 当前节点没有被打劫
+	tmp0 := max(l0, l1) + max(r0, r1)
+	// 当前节点被打劫
+	tmp1 := root.Val + l0 + r0
+	return tmp0, tmp1
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0337.House-Robber-III/337. House Robber III_test.go b/leetcode/0337.House-Robber-III/337. House Robber III_test.go
new file mode 100644
index 000000000..a05c7edba
--- /dev/null
+++ b/leetcode/0337.House-Robber-III/337. House Robber III_test.go	
@@ -0,0 +1,54 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question337 struct {
+	para337
+	ans337
+}
+
+// para 是参数
+// one 代表第一个参数
+type para337 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans337 struct {
+	one int
+}
+
+func Test_Problem337(t *testing.T) {
+
+	qs := []question337{
+
+		{
+			para337{[]int{3, 2, 3, structures.NULL, 3, structures.NULL, 1}},
+			ans337{7},
+		},
+
+		{
+			para337{[]int{}},
+			ans337{0},
+		},
+
+		{
+			para337{[]int{3, 4, 5, 1, 3, structures.NULL, 1}},
+			ans337{9},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 337------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans337, q.para337
+		fmt.Printf("【input】:%v       【output】:%v\n", p, rob337(structures.Ints2TreeNode(p.one)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0337.House-Robber-III/README.md b/leetcode/0337.House-Robber-III/README.md
new file mode 100644
index 000000000..e84518812
--- /dev/null
+++ b/leetcode/0337.House-Robber-III/README.md
@@ -0,0 +1,73 @@
+# [337. House Robber III](https://leetcode.com/problems/house-robber-iii/)
+
+
+
+## 题目
+
+The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.
+
+Determine the maximum amount of money the thief can rob tonight without alerting the police.
+
+**Example 1**:
+
+```
+Input: [3,2,3,null,3,null,1]
+
+     3
+    / \
+   2   3
+    \   \ 
+     3   1
+
+Output: 7 
+Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
+```
+
+**Example 2**:
+
+```
+Input: [3,4,5,1,3,null,1]
+
+     3
+    / \
+   4   5
+  / \   \ 
+ 1   3   1
+
+Output: 9
+Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.
+```
+
+## 题目大意
+
+一个新的可行窃的地区只有一个入口，称之为“根”。 除了“根”之外，每栋房子有且只有一个“父“房子与之相连。一番侦察之后，聪明的小偷意识到“这个地方的所有房屋的排列类似于一棵二叉树”。 如果两个直接相连的房子在同一天晚上被打劫，房屋将自动报警。计算在不触动警报的情况下，小偷一晚能够盗取的最高金额。
+
+
+## 解题思路
+
+- 这一题是打家劫舍的第 3 题。这一题需要偷的房子是树状的。报警的条件还是相邻的房子如果都被偷了，就会触发报警。只不过这里相邻的房子是树上的。问小偷在不触发报警的条件下最终能偷的最高金额。
+- 解题思路是 DFS。当前节点是否被打劫，会产生 2 种结果。如果当前节点被打劫，那么它的孩子节点可以被打劫；如果当前节点没有被打劫，那么它的孩子节点不能被打劫。按照这个逻辑递归，最终递归到根节点，取最大值输出即可。
+
+## 代码
+
+```go
+
+func rob337(root *TreeNode) int {
+	a, b := dfsTreeRob(root)
+	return max(a, b)
+}
+
+func dfsTreeRob(root *TreeNode) (a, b int) {
+	if root == nil {
+		return 0, 0
+	}
+	l0, l1 := dfsTreeRob(root.Left)
+	r0, r1 := dfsTreeRob(root.Right)
+	// 当前节点没有被打劫
+	tmp0 := max(l0, l1) + max(r0, r1)
+	// 当前节点被打劫
+	tmp1 := root.Val + l0 + r0
+	return tmp0, tmp1
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0338. Counting Bits/338. Counting Bits.go b/leetcode/0338.Counting-Bits/338. Counting Bits.go
similarity index 100%
rename from Algorithms/0338. Counting Bits/338. Counting Bits.go
rename to leetcode/0338.Counting-Bits/338. Counting Bits.go
diff --git a/Algorithms/0338. Counting Bits/338. Counting Bits_test.go b/leetcode/0338.Counting-Bits/338. Counting Bits_test.go
similarity index 95%
rename from Algorithms/0338. Counting Bits/338. Counting Bits_test.go
rename to leetcode/0338.Counting-Bits/338. Counting Bits_test.go
index b69f10e13..2ae0f7b5c 100644
--- a/Algorithms/0338. Counting Bits/338. Counting Bits_test.go	
+++ b/leetcode/0338.Counting-Bits/338. Counting Bits_test.go	
@@ -26,12 +26,12 @@ func Test_Problem338(t *testing.T) {
 
 	qs := []question338{
 
-		question338{
+		{
 			para338{2},
 			ans338{[]int{0, 1, 1}},
 		},
 
-		question338{
+		{
 			para338{5},
 			ans338{[]int{0, 1, 1, 2, 1, 2}},
 		},
diff --git a/Algorithms/0338. Counting Bits/README.md b/leetcode/0338.Counting-Bits/README.md
similarity index 100%
rename from Algorithms/0338. Counting Bits/README.md
rename to leetcode/0338.Counting-Bits/README.md
diff --git a/leetcode/0341.Flatten-Nested-List-Iterator/341. Flatten Nested List Iterator.go b/leetcode/0341.Flatten-Nested-List-Iterator/341. Flatten Nested List Iterator.go
new file mode 100644
index 000000000..02a1d3e31
--- /dev/null
+++ b/leetcode/0341.Flatten-Nested-List-Iterator/341. Flatten Nested List Iterator.go	
@@ -0,0 +1,72 @@
+package leetcode
+
+import "container/list"
+
+// This is the interface that allows for creating nested lists.
+// You should not implement it, or speculate about its implementation
+type NestedInteger struct {
+}
+
+// Return true if this NestedInteger holds a single integer, rather than a nested list.
+func (this NestedInteger) IsInteger() bool { return true }
+
+// Return the single integer that this NestedInteger holds, if it holds a single integer
+// The result is undefined if this NestedInteger holds a nested list
+// So before calling this method, you should have a check
+func (this NestedInteger) GetInteger() int { return 0 }
+
+// Set this NestedInteger to hold a single integer.
+func (n *NestedInteger) SetInteger(value int) {}
+
+// Set this NestedInteger to hold a nested list and adds a nested integer to it.
+func (this *NestedInteger) Add(elem NestedInteger) {}
+
+// Return the nested list that this NestedInteger holds, if it holds a nested list
+// The list length is zero if this NestedInteger holds a single integer
+// You can access NestedInteger's List element directly if you want to modify it
+func (this NestedInteger) GetList() []*NestedInteger { return nil }
+
+type NestedIterator struct {
+	stack *list.List
+}
+
+type listIndex struct {
+	nestedList []*NestedInteger
+	index      int
+}
+
+func Constructor(nestedList []*NestedInteger) *NestedIterator {
+	stack := list.New()
+	stack.PushBack(&listIndex{nestedList, 0})
+	return &NestedIterator{stack}
+}
+
+func (this *NestedIterator) Next() int {
+	if !this.HasNext() {
+		return -1
+	}
+	last := this.stack.Back().Value.(*listIndex)
+	nestedList, i := last.nestedList, last.index
+	val := nestedList[i].GetInteger()
+	last.index++
+	return val
+}
+
+func (this *NestedIterator) HasNext() bool {
+	stack := this.stack
+	for stack.Len() > 0 {
+		last := stack.Back().Value.(*listIndex)
+		nestedList, i := last.nestedList, last.index
+		if i >= len(nestedList) {
+			stack.Remove(stack.Back())
+		} else {
+			val := nestedList[i]
+			if val.IsInteger() {
+				return true
+			}
+			last.index++
+			stack.PushBack(&listIndex{val.GetList(), 0})
+		}
+	}
+	return false
+}
diff --git a/leetcode/0341.Flatten-Nested-List-Iterator/341. Flatten Nested List Iterator_test.go b/leetcode/0341.Flatten-Nested-List-Iterator/341. Flatten Nested List Iterator_test.go
new file mode 100644
index 000000000..b81c22c4a
--- /dev/null
+++ b/leetcode/0341.Flatten-Nested-List-Iterator/341. Flatten Nested List Iterator_test.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem338(t *testing.T) {
+	obj := Constructor([]*NestedInteger{})
+	fmt.Printf("obj = %v\n", obj)
+	fmt.Printf("obj = %v\n", obj.Next())
+	fmt.Printf("obj = %v\n", obj.HasNext())
+}
diff --git a/leetcode/0341.Flatten-Nested-List-Iterator/README.md b/leetcode/0341.Flatten-Nested-List-Iterator/README.md
new file mode 100644
index 000000000..5d04e5e02
--- /dev/null
+++ b/leetcode/0341.Flatten-Nested-List-Iterator/README.md
@@ -0,0 +1,110 @@
+# [341. Flatten Nested List Iterator](https://leetcode.com/problems/flatten-nested-list-iterator/)
+
+
+## 题目
+
+Given a nested list of integers, implement an iterator to flatten it.
+
+Each element is either an integer, or a list -- whose elements may also be integers or other lists.
+
+**Example 1:**
+
+```
+Input:[[1,1],2,[1,1]]
+Output:[1,1,2,1,1]
+Explanation:By callingnext repeatedly untilhasNext returns false,
+             the order of elements returned bynext should be:[1,1,2,1,1].
+```
+
+**Example 2:**
+
+```
+Input:[1,[4,[6]]]
+Output:[1,4,6]
+Explanation:By callingnext repeatedly untilhasNext returns false,
+             the order of elements returned bynext should be:[1,4,6].
+
+```
+
+## 题目大意
+
+给你一个嵌套的整型列表。请你设计一个迭代器，使其能够遍历这个整型列表中的所有整数。列表中的每一项或者为一个整数，或者是另一个列表。其中列表的元素也可能是整数或是其他列表。
+
+## 解题思路
+
+- 题目要求实现一个嵌套版的数组。可以用 `[]int` 实现，也可以用链表实现。笔者此处用链表实现。外层构造一个一维数组，一维数组内部每个元素是一个链表。额外还需要记录这个嵌套链表在原数组中的 `index` 索引。`Next()` 实现比较简单，取出对应的嵌套节点。`HasNext()` 方法则感觉嵌套节点里面的 `index` 信息判断是否还有 `next` 元素。
+
+## 代码
+
+```go
+/**
+ * // This is the interface that allows for creating nested lists.
+ * // You should not implement it, or speculate about its implementation
+ * type NestedInteger struct {
+ * }
+ *
+ * // Return true if this NestedInteger holds a single integer, rather than a nested list.
+ * func (this NestedInteger) IsInteger() bool {}
+ *
+ * // Return the single integer that this NestedInteger holds, if it holds a single integer
+ * // The result is undefined if this NestedInteger holds a nested list
+ * // So before calling this method, you should have a check
+ * func (this NestedInteger) GetInteger() int {}
+ *
+ * // Set this NestedInteger to hold a single integer.
+ * func (n *NestedInteger) SetInteger(value int) {}
+ *
+ * // Set this NestedInteger to hold a nested list and adds a nested integer to it.
+ * func (this *NestedInteger) Add(elem NestedInteger) {}
+ *
+ * // Return the nested list that this NestedInteger holds, if it holds a nested list
+ * // The list length is zero if this NestedInteger holds a single integer
+ * // You can access NestedInteger's List element directly if you want to modify it
+ * func (this NestedInteger) GetList() []*NestedInteger {}
+ */
+
+type NestedIterator struct {
+    stack *list.List
+}
+
+type listIndex struct {
+    nestedList []*NestedInteger
+    index int
+}
+
+func Constructor(nestedList []*NestedInteger) *NestedIterator {
+    stack := list.New()
+    stack.PushBack(&listIndex{nestedList, 0})
+    return &NestedIterator{stack}
+}
+
+func (this *NestedIterator) Next() int {
+    if !this.HasNext() {
+        return -1
+    }
+    last := this.stack.Back().Value.(*listIndex)
+    nestedList, i := last.nestedList, last.index
+    val := nestedList[i].GetInteger()
+    last.index++
+    return val
+}
+
+func (this *NestedIterator) HasNext() bool {
+    stack := this.stack
+    for stack.Len() > 0 {
+        last := stack.Back().Value.(*listIndex)
+        nestedList, i := last.nestedList, last.index
+        if i >= len(nestedList) {
+            stack.Remove(stack.Back())
+        } else {
+            val := nestedList[i]
+            if val.IsInteger() {
+                return true
+            }
+            last.index++
+            stack.PushBack(&listIndex{val.GetList(), 0})
+        }
+    }
+    return false
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0342. Power of Four/342. Power of Four.go b/leetcode/0342.Power-of-Four/342. Power of Four.go
similarity index 100%
rename from Algorithms/0342. Power of Four/342. Power of Four.go
rename to leetcode/0342.Power-of-Four/342. Power of Four.go
diff --git a/Algorithms/0342. Power of Four/342. Power of Four_test.go b/leetcode/0342.Power-of-Four/342. Power of Four_test.go
similarity index 95%
rename from Algorithms/0342. Power of Four/342. Power of Four_test.go
rename to leetcode/0342.Power-of-Four/342. Power of Four_test.go
index 7c5c4cfc7..637f3b981 100644
--- a/Algorithms/0342. Power of Four/342. Power of Four_test.go	
+++ b/leetcode/0342.Power-of-Four/342. Power of Four_test.go	
@@ -26,12 +26,12 @@ func Test_Problem342(t *testing.T) {
 
 	qs := []question342{
 
-		question342{
+		{
 			para342{16},
 			ans342{true},
 		},
 
-		question342{
+		{
 			para342{5},
 			ans342{false},
 		},
diff --git a/Algorithms/0342. Power of Four/README.md b/leetcode/0342.Power-of-Four/README.md
similarity index 100%
rename from Algorithms/0342. Power of Four/README.md
rename to leetcode/0342.Power-of-Four/README.md
diff --git a/Algorithms/0343. Integer Break/343. Integer Break.go b/leetcode/0343.Integer-Break/343. Integer Break.go
similarity index 79%
rename from Algorithms/0343. Integer Break/343. Integer Break.go
rename to leetcode/0343.Integer-Break/343. Integer Break.go
index 0a30e3438..fea86ccfe 100644
--- a/Algorithms/0343. Integer Break/343. Integer Break.go	
+++ b/leetcode/0343.Integer-Break/343. Integer Break.go	
@@ -11,3 +11,10 @@ func integerBreak(n int) int {
 	}
 	return dp[n]
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0343. Integer Break/343. Integer Break_test.go b/leetcode/0343.Integer-Break/343. Integer Break_test.go
similarity index 95%
rename from Algorithms/0343. Integer Break/343. Integer Break_test.go
rename to leetcode/0343.Integer-Break/343. Integer Break_test.go
index e82a448a4..b6228af4a 100644
--- a/Algorithms/0343. Integer Break/343. Integer Break_test.go	
+++ b/leetcode/0343.Integer-Break/343. Integer Break_test.go	
@@ -26,12 +26,12 @@ func Test_Problem343(t *testing.T) {
 
 	qs := []question343{
 
-		question343{
+		{
 			para343{2},
 			ans343{1},
 		},
 
-		question343{
+		{
 			para343{10},
 			ans343{36},
 		},
diff --git a/Algorithms/0343. Integer Break/README.md b/leetcode/0343.Integer-Break/README.md
similarity index 100%
rename from Algorithms/0343. Integer Break/README.md
rename to leetcode/0343.Integer-Break/README.md
diff --git a/Algorithms/0344. Reverse String/344. Reverse String.go b/leetcode/0344.Reverse-String/344. Reverse String.go
similarity index 100%
rename from Algorithms/0344. Reverse String/344. Reverse String.go
rename to leetcode/0344.Reverse-String/344. Reverse String.go
diff --git a/Algorithms/0344. Reverse String/344. Reverse String_test.go b/leetcode/0344.Reverse-String/344. Reverse String_test.go
similarity index 96%
rename from Algorithms/0344. Reverse String/344. Reverse String_test.go
rename to leetcode/0344.Reverse-String/344. Reverse String_test.go
index 71f4f5544..68aa77b96 100644
--- a/Algorithms/0344. Reverse String/344. Reverse String_test.go	
+++ b/leetcode/0344.Reverse-String/344. Reverse String_test.go	
@@ -26,12 +26,12 @@ func Test_Problem344(t *testing.T) {
 
 	qs := []question344{
 
-		question344{
+		{
 			para344{[]byte{'h', 'e', 'l', 'l', 'o'}},
 			ans344{[]byte{'o', 'l', 'l', 'e', 'h'}},
 		},
 
-		question344{
+		{
 			para344{[]byte{'H', 'a', 'n', 'n', 'a', 'h'}},
 			ans344{[]byte{'h', 'a', 'n', 'n', 'a', 'H'}},
 		},
diff --git a/Algorithms/0344. Reverse String/README.md b/leetcode/0344.Reverse-String/README.md
similarity index 100%
rename from Algorithms/0344. Reverse String/README.md
rename to leetcode/0344.Reverse-String/README.md
diff --git a/leetcode/0345.Reverse-Vowels-of-a-String/345. Reverse Vowels of a String.go b/leetcode/0345.Reverse-Vowels-of-a-String/345. Reverse Vowels of a String.go
new file mode 100644
index 000000000..ff1e1be04
--- /dev/null
+++ b/leetcode/0345.Reverse-Vowels-of-a-String/345. Reverse Vowels of a String.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+func reverseVowels(s string) string {
+	b := []byte(s)
+	for i, j := 0, len(b)-1; i < j; {
+		if !isVowel(b[i]) {
+			i++
+			continue
+		}
+		if !isVowel(b[j]) {
+			j--
+			continue
+		}
+		b[i], b[j] = b[j], b[i]
+		i++
+		j--
+	}
+	return string(b)
+}
+
+func isVowel(s byte) bool {
+	return s == 'a' || s == 'e' || s == 'i' || s == 'o' || s == 'u' || s == 'A' ||
+		s == 'E' || s == 'I' || s == 'O' || s == 'U'
+}
diff --git a/Algorithms/0345. Reverse Vowels of a String/345. Reverse Vowels of a String_test.go b/leetcode/0345.Reverse-Vowels-of-a-String/345. Reverse Vowels of a String_test.go
similarity index 94%
rename from Algorithms/0345. Reverse Vowels of a String/345. Reverse Vowels of a String_test.go
rename to leetcode/0345.Reverse-Vowels-of-a-String/345. Reverse Vowels of a String_test.go
index c47420a16..74585a121 100644
--- a/Algorithms/0345. Reverse Vowels of a String/345. Reverse Vowels of a String_test.go	
+++ b/leetcode/0345.Reverse-Vowels-of-a-String/345. Reverse Vowels of a String_test.go	
@@ -26,17 +26,17 @@ func Test_Problem345(t *testing.T) {
 
 	qs := []question345{
 
-		question345{
+		{
 			para345{"hello"},
 			ans345{"holle"},
 		},
 
-		question345{
+		{
 			para345{"leetcode"},
 			ans345{"leotcede"},
 		},
 
-		question345{
+		{
 			para345{"aA"},
 			ans345{"Aa"},
 		},
diff --git a/Algorithms/0345. Reverse Vowels of a String/README.md b/leetcode/0345.Reverse-Vowels-of-a-String/README.md
similarity index 100%
rename from Algorithms/0345. Reverse Vowels of a String/README.md
rename to leetcode/0345.Reverse-Vowels-of-a-String/README.md
diff --git a/Algorithms/0347. Top K Frequent Elements/347. Top K Frequent Elements.go b/leetcode/0347.Top-K-Frequent-Elements/347. Top K Frequent Elements.go
similarity index 100%
rename from Algorithms/0347. Top K Frequent Elements/347. Top K Frequent Elements.go
rename to leetcode/0347.Top-K-Frequent-Elements/347. Top K Frequent Elements.go
diff --git a/Algorithms/0347. Top K Frequent Elements/347. Top K Frequent Elements_test.go b/leetcode/0347.Top-K-Frequent-Elements/347. Top K Frequent Elements_test.go
similarity index 95%
rename from Algorithms/0347. Top K Frequent Elements/347. Top K Frequent Elements_test.go
rename to leetcode/0347.Top-K-Frequent-Elements/347. Top K Frequent Elements_test.go
index 8d28f9adb..0e6f050f7 100644
--- a/Algorithms/0347. Top K Frequent Elements/347. Top K Frequent Elements_test.go	
+++ b/leetcode/0347.Top-K-Frequent-Elements/347. Top K Frequent Elements_test.go	
@@ -27,12 +27,12 @@ func Test_Problem347(t *testing.T) {
 
 	qs := []question347{
 
-		question347{
+		{
 			para347{[]int{1, 1, 1, 2, 2, 3}, 2},
 			ans347{[]int{1, 2}},
 		},
 
-		question347{
+		{
 			para347{[]int{1}, 1},
 			ans347{[]int{1}},
 		},
diff --git a/Algorithms/0347. Top K Frequent Elements/README.md b/leetcode/0347.Top-K-Frequent-Elements/README.md
similarity index 100%
rename from Algorithms/0347. Top K Frequent Elements/README.md
rename to leetcode/0347.Top-K-Frequent-Elements/README.md
diff --git a/Algorithms/0349. Intersection of Two Arrays/349. Intersection of Two Arrays.go b/leetcode/0349.Intersection-of-Two-Arrays/349. Intersection of Two Arrays.go
similarity index 100%
rename from Algorithms/0349. Intersection of Two Arrays/349. Intersection of Two Arrays.go
rename to leetcode/0349.Intersection-of-Two-Arrays/349. Intersection of Two Arrays.go
diff --git a/Algorithms/0349. Intersection of Two Arrays/349. Intersection of Two Arrays_test.go b/leetcode/0349.Intersection-of-Two-Arrays/349. Intersection of Two Arrays_test.go
similarity index 92%
rename from Algorithms/0349. Intersection of Two Arrays/349. Intersection of Two Arrays_test.go
rename to leetcode/0349.Intersection-of-Two-Arrays/349. Intersection of Two Arrays_test.go
index b17d4b5b4..f4cc33194 100644
--- a/Algorithms/0349. Intersection of Two Arrays/349. Intersection of Two Arrays_test.go	
+++ b/leetcode/0349.Intersection-of-Two-Arrays/349. Intersection of Two Arrays_test.go	
@@ -27,32 +27,32 @@ func Test_Problem349(t *testing.T) {
 
 	qs := []question349{
 
-		question349{
+		{
 			para349{[]int{}, []int{}},
 			ans349{[]int{}},
 		},
 
-		question349{
+		{
 			para349{[]int{1}, []int{1}},
 			ans349{[]int{1}},
 		},
 
-		question349{
+		{
 			para349{[]int{1, 2, 3, 4}, []int{1, 2, 3, 4}},
 			ans349{[]int{1, 2, 3, 4}},
 		},
 
-		question349{
+		{
 			para349{[]int{1, 2, 2, 1}, []int{2, 2}},
 			ans349{[]int{2}},
 		},
 
-		question349{
+		{
 			para349{[]int{1}, []int{9, 9, 9, 9, 9}},
 			ans349{[]int{}},
 		},
 
-		question349{
+		{
 			para349{[]int{4, 9, 5}, []int{9, 4, 9, 8, 4}},
 			ans349{[]int{9, 4}},
 		},
diff --git a/Algorithms/0349. Intersection of Two Arrays/README.md b/leetcode/0349.Intersection-of-Two-Arrays/README.md
similarity index 100%
rename from Algorithms/0349. Intersection of Two Arrays/README.md
rename to leetcode/0349.Intersection-of-Two-Arrays/README.md
diff --git a/Algorithms/0350. Intersection of Two Arrays II/350. Intersection of Two Arrays II.go b/leetcode/0350.Intersection-of-Two-Arrays-II/350. Intersection of Two Arrays II.go
similarity index 100%
rename from Algorithms/0350. Intersection of Two Arrays II/350. Intersection of Two Arrays II.go
rename to leetcode/0350.Intersection-of-Two-Arrays-II/350. Intersection of Two Arrays II.go
diff --git a/Algorithms/0350. Intersection of Two Arrays II/350. Intersection of Two Arrays II_test.go b/leetcode/0350.Intersection-of-Two-Arrays-II/350. Intersection of Two Arrays II_test.go
similarity index 91%
rename from Algorithms/0350. Intersection of Two Arrays II/350. Intersection of Two Arrays II_test.go
rename to leetcode/0350.Intersection-of-Two-Arrays-II/350. Intersection of Two Arrays II_test.go
index 5ab6f411c..cdfd4f76c 100644
--- a/Algorithms/0350. Intersection of Two Arrays II/350. Intersection of Two Arrays II_test.go	
+++ b/leetcode/0350.Intersection-of-Two-Arrays-II/350. Intersection of Two Arrays II_test.go	
@@ -27,37 +27,37 @@ func Test_Problem350(t *testing.T) {
 
 	qs := []question350{
 
-		question350{
+		{
 			para350{[]int{}, []int{}},
 			ans350{[]int{}},
 		},
 
-		question350{
+		{
 			para350{[]int{1}, []int{1}},
 			ans350{[]int{1}},
 		},
 
-		question350{
+		{
 			para350{[]int{1, 2, 3, 4}, []int{1, 2, 3, 4}},
 			ans350{[]int{1, 2, 3, 4}},
 		},
 
-		question350{
+		{
 			para350{[]int{1, 2, 2, 1}, []int{2, 2}},
 			ans350{[]int{2, 2}},
 		},
 
-		question350{
+		{
 			para350{[]int{1}, []int{9, 9, 9, 9, 9}},
 			ans350{[]int{}},
 		},
 
-		question350{
+		{
 			para350{[]int{4, 9, 5}, []int{9, 4, 9, 8, 4}},
 			ans350{[]int{9, 4}},
 		},
 
-		question350{
+		{
 			para350{[]int{4, 9, 5, 9, 4}, []int{9, 4, 9, 8, 4, 7, 9, 4, 4, 9}},
 			ans350{[]int{9, 4, 9, 4}},
 		},
diff --git a/Algorithms/0350. Intersection of Two Arrays II/README.md b/leetcode/0350.Intersection-of-Two-Arrays-II/README.md
similarity index 100%
rename from Algorithms/0350. Intersection of Two Arrays II/README.md
rename to leetcode/0350.Intersection-of-Two-Arrays-II/README.md
diff --git a/leetcode/0352.Data-Stream-as-Disjoint-Intervals/352.Data Stream as Disjoint Intervals.go b/leetcode/0352.Data-Stream-as-Disjoint-Intervals/352.Data Stream as Disjoint Intervals.go
new file mode 100644
index 000000000..84d382d12
--- /dev/null
+++ b/leetcode/0352.Data-Stream-as-Disjoint-Intervals/352.Data Stream as Disjoint Intervals.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import "sort"
+
+type SummaryRanges struct {
+	nums []int
+	mp   map[int]int
+}
+
+func Constructor() SummaryRanges {
+	return SummaryRanges{
+		nums: []int{},
+		mp:   map[int]int{},
+	}
+}
+
+func (this *SummaryRanges) AddNum(val int) {
+	if _, ok := this.mp[val]; !ok {
+		this.mp[val] = 1
+		this.nums = append(this.nums, val)
+	}
+	sort.Ints(this.nums)
+}
+
+func (this *SummaryRanges) GetIntervals() [][]int {
+	n := len(this.nums)
+	var ans [][]int
+	if n == 0 {
+		return ans
+	}
+	if n == 1 {
+		ans = append(ans, []int{this.nums[0], this.nums[0]})
+		return ans
+	}
+	start, end := this.nums[0], this.nums[0]
+	ans = append(ans, []int{start, end})
+	index := 0
+	for i := 1; i < n; i++ {
+		if this.nums[i] == end+1 {
+			end = this.nums[i]
+			ans[index][1] = end
+		} else {
+			start, end = this.nums[i], this.nums[i]
+			ans = append(ans, []int{start, end})
+			index++
+		}
+	}
+	return ans
+}
diff --git a/leetcode/0352.Data-Stream-as-Disjoint-Intervals/352.Data Stream as Disjoint Intervals_test.go b/leetcode/0352.Data-Stream-as-Disjoint-Intervals/352.Data Stream as Disjoint Intervals_test.go
new file mode 100644
index 000000000..b19897391
--- /dev/null
+++ b/leetcode/0352.Data-Stream-as-Disjoint-Intervals/352.Data Stream as Disjoint Intervals_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question352 struct {
+	para352
+	ans352
+}
+
+// para 是参数
+type para352 struct {
+	para string
+	num  int
+}
+
+// ans 是答案
+type ans352 struct {
+	ans [][]int
+}
+
+func Test_Problem352(t *testing.T) {
+
+	qs := []question352{
+
+		{
+			para352{"addNum", 1},
+			ans352{[][]int{{1, 1}}},
+		},
+
+		{
+			para352{"addNum", 3},
+			ans352{[][]int{{1, 1}, {3, 3}}},
+		},
+
+		{
+			para352{"addNum", 7},
+			ans352{[][]int{{1, 1}, {3, 3}, {7, 7}}},
+		},
+
+		{
+			para352{"addNum", 2},
+			ans352{[][]int{{1, 3}, {7, 7}}},
+		},
+
+		{
+			para352{"addNum", 6},
+			ans352{[][]int{{1, 3}, {6, 7}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 352------------------------\n")
+
+	obj := Constructor()
+	for _, q := range qs {
+		_, p := q.ans352, q.para352
+		obj.AddNum(p.num)
+		fmt.Printf("【input】:%v       【output】:%v\n", p, obj.GetIntervals())
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0352.Data-Stream-as-Disjoint-Intervals/README.md b/leetcode/0352.Data-Stream-as-Disjoint-Intervals/README.md
new file mode 100644
index 000000000..a3f7abd75
--- /dev/null
+++ b/leetcode/0352.Data-Stream-as-Disjoint-Intervals/README.md
@@ -0,0 +1,108 @@
+# [352. Data Stream as Disjoint Intervals](https://leetcode.com/problems/data-stream-as-disjoint-intervals/)
+
+
+## 题目
+
+Given a data stream input of non-negative integers a1, a2, ..., an, summarize the numbers seen so far as a list of disjoint intervals.
+
+Implement the SummaryRanges class:
+
+   - SummaryRanges() Initializes the object with an empty stream.
+   - void addNum(int val) Adds the integer val to the stream. 
+   - int[][] getIntervals() Returns a summary of the integers in the stream currently as a list of disjoint intervals [starti, endi].
+
+**Example 1:**
+
+    Input
+    ["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"]
+    [[], [1], [], [3], [], [7], [], [2], [], [6], []]
+    Output
+    [null, null, [[1, 1]], null, [[1, 1], [3, 3]], null, [[1, 1], [3, 3], [7, 7]], null, [[1, 3], [7, 7]], null, [[1, 3], [6, 7]]]
+
+    Explanation
+    SummaryRanges summaryRanges = new SummaryRanges();
+    summaryRanges.addNum(1);      // arr = [1]
+    summaryRanges.getIntervals(); // return [[1, 1]]
+    summaryRanges.addNum(3);      // arr = [1, 3]
+    summaryRanges.getIntervals(); // return [[1, 1], [3, 3]]
+    summaryRanges.addNum(7);      // arr = [1, 3, 7]
+    summaryRanges.getIntervals(); // return [[1, 1], [3, 3], [7, 7]]
+    summaryRanges.addNum(2);      // arr = [1, 2, 3, 7]
+    summaryRanges.getIntervals(); // return [[1, 3], [7, 7]]
+    summaryRanges.addNum(6);      // arr = [1, 2, 3, 6, 7]
+    summaryRanges.getIntervals(); // return [[1, 3], [6, 7]]
+
+**Constraints**
+
+   - 0 <= val <= 10000
+   - At most 3 * 10000 calls will be made to addNum and getIntervals.
+
+## 题目大意
+
+给你一个由非负整数a1, a2, ..., an 组成的数据流输入，请你将到目前为止看到的数字总结为不相交的区间列表。
+
+实现 SummaryRanges 类：
+
+   - SummaryRanges() 使用一个空数据流初始化对象。
+   - void addNum(int val) 向数据流中加入整数 val 。
+   - int[][] getIntervals() 以不相交区间[starti, endi] 的列表形式返回对数据流中整数的总结
+
+## 解题思路
+
+- 使用字典过滤掉重复的数字
+- 把过滤后的数字放到nums中,并进行排序 
+- 使用nums构建不重复的区间
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+type SummaryRanges struct {
+	nums []int
+	mp map[int]int
+}
+
+func Constructor() SummaryRanges {
+	return SummaryRanges{
+		nums: []int{},
+		mp : map[int]int{},
+	}
+}
+
+func (this *SummaryRanges) AddNum(val int) {
+	if _, ok := this.mp[val]; !ok {
+		this.mp[val] = 1
+		this.nums = append(this.nums, val)
+	}
+	sort.Ints(this.nums)
+}
+
+func (this *SummaryRanges) GetIntervals() [][]int {
+	n := len(this.nums)
+	var ans [][]int
+	if n == 0 {
+		return ans
+	}
+	if n == 1 {
+		ans = append(ans, []int{this.nums[0], this.nums[0]})
+		return ans
+	}
+	start, end := this.nums[0], this.nums[0]
+	ans = append(ans, []int{start, end})
+	index := 0
+	for i := 1; i < n; i++ {
+		if this.nums[i] == end + 1 {
+			end = this.nums[i]
+			ans[index][1] = end
+		} else {
+			start, end = this.nums[i], this.nums[i]
+			ans = append(ans, []int{start, end})
+			index++
+		}
+	}
+	return ans
+}
+```
\ No newline at end of file
diff --git a/leetcode/0354.Russian-Doll-Envelopes/354. Russian Doll Envelopes.go b/leetcode/0354.Russian-Doll-Envelopes/354. Russian Doll Envelopes.go
new file mode 100644
index 000000000..51ce4c7ca
--- /dev/null
+++ b/leetcode/0354.Russian-Doll-Envelopes/354. Russian Doll Envelopes.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+type sortEnvelopes [][]int
+
+func (s sortEnvelopes) Len() int {
+	return len(s)
+}
+func (s sortEnvelopes) Less(i, j int) bool {
+	if s[i][0] == s[j][0] {
+		return s[i][1] > s[j][1]
+	}
+	return s[i][0] < s[j][0]
+}
+func (s sortEnvelopes) Swap(i, j int) {
+	s[i], s[j] = s[j], s[i]
+}
+
+func maxEnvelopes(envelopes [][]int) int {
+	sort.Sort(sortEnvelopes(envelopes))
+	dp := []int{}
+	for _, e := range envelopes {
+		low, high := 0, len(dp)
+		for low < high {
+			mid := low + (high-low)>>1
+			if dp[mid] >= e[1] {
+				high = mid
+			} else {
+				low = mid + 1
+			}
+		}
+		if low == len(dp) {
+			dp = append(dp, e[1])
+		} else {
+			dp[low] = e[1]
+		}
+	}
+	return len(dp)
+}
diff --git a/leetcode/0354.Russian-Doll-Envelopes/354. Russian Doll Envelopes_test.go b/leetcode/0354.Russian-Doll-Envelopes/354. Russian Doll Envelopes_test.go
new file mode 100644
index 000000000..a477e0199
--- /dev/null
+++ b/leetcode/0354.Russian-Doll-Envelopes/354. Russian Doll Envelopes_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question354 struct {
+	para354
+	ans354
+}
+
+// para 是参数
+// one 代表第一个参数
+type para354 struct {
+	envelopes [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans354 struct {
+	one int
+}
+
+func Test_Problem354(t *testing.T) {
+
+	qs := []question354{
+
+		{
+			para354{[][]int{{5, 4}, {6, 4}, {6, 7}, {2, 3}}},
+			ans354{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 354------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans354, q.para354
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxEnvelopes(p.envelopes))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0354.Russian-Doll-Envelopes/README.md b/leetcode/0354.Russian-Doll-Envelopes/README.md
new file mode 100755
index 000000000..826264a89
--- /dev/null
+++ b/leetcode/0354.Russian-Doll-Envelopes/README.md
@@ -0,0 +1,31 @@
+# [354. Russian Doll Envelopes](https://leetcode.com/problems/russian-doll-envelopes/)
+
+
+## 题目
+
+You have a number of envelopes with widths and heights given as a pair of integers `(w, h)`. One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.
+
+What is the maximum number of envelopes can you Russian doll? (put one inside other)
+
+**Note:**Rotation is not allowed.
+
+**Example:**
+
+    **Input:** [[5,4],[6,4],[6,7],[2,3]]
+    **Output:** 3
+    **Explanation: T**he maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).
+
+
+## 题目大意
+
+给定一些标记了宽度和高度的信封，宽度和高度以整数对形式 (w, h) 出现。当另一个信封的宽度和高度都比这个信封大的时候，这个信封就可以放进另一个信封里，如同俄罗斯套娃一样。
+
+请计算最多能有多少个信封能组成一组“俄罗斯套娃”信封（即可以把一个信封放到另一个信封里面）。
+
+说明:
+- 不允许旋转信封。
+
+## 解题思路
+
+- 给出一组信封的宽度和高度，如果组成俄罗斯套娃，问最多能套几层。只有当一个信封的宽度和高度都比另外一个信封大的时候，才能套在小信封上面。
+- 这一题的实质是第 300 题 Longest Increasing Subsequence 的加强版。能组成俄罗斯套娃的条件就是能找到一个最长上升子序列。但是这题的条件是二维的，要求能找到在二维上都能满足条件的最长上升子序列。先降维，把宽度排序。然后在高度上寻找最长上升子序列。这里用到的方法和第 300 题的方法一致。解题思路详解见第 300 题。
diff --git a/Algorithms/0357. Count Numbers with Unique Digits/357. Count Numbers with Unique Digits.go b/leetcode/0357.Count-Numbers-with-Unique-Digits/357. Count Numbers with Unique Digits.go
similarity index 100%
rename from Algorithms/0357. Count Numbers with Unique Digits/357. Count Numbers with Unique Digits.go
rename to leetcode/0357.Count-Numbers-with-Unique-Digits/357. Count Numbers with Unique Digits.go
diff --git a/Algorithms/0357. Count Numbers with Unique Digits/357. Count Numbers with Unique Digits_test.go b/leetcode/0357.Count-Numbers-with-Unique-Digits/357. Count Numbers with Unique Digits_test.go
similarity index 90%
rename from Algorithms/0357. Count Numbers with Unique Digits/357. Count Numbers with Unique Digits_test.go
rename to leetcode/0357.Count-Numbers-with-Unique-Digits/357. Count Numbers with Unique Digits_test.go
index 60b252d92..a2ea89555 100644
--- a/Algorithms/0357. Count Numbers with Unique Digits/357. Count Numbers with Unique Digits_test.go	
+++ b/leetcode/0357.Count-Numbers-with-Unique-Digits/357. Count Numbers with Unique Digits_test.go	
@@ -26,32 +26,32 @@ func Test_Problem357(t *testing.T) {
 
 	qs := []question357{
 
-		question357{
+		{
 			para357{1},
 			ans357{10},
 		},
 
-		question357{
+		{
 			para357{2},
 			ans357{91},
 		},
 
-		question357{
+		{
 			para357{3},
 			ans357{739},
 		},
 
-		question357{
+		{
 			para357{4},
 			ans357{5275},
 		},
 
-		question357{
+		{
 			para357{5},
 			ans357{32491},
 		},
 
-		question357{
+		{
 			para357{6},
 			ans357{168571},
 		},
diff --git a/Algorithms/0357. Count Numbers with Unique Digits/README.md b/leetcode/0357.Count-Numbers-with-Unique-Digits/README.md
similarity index 100%
rename from Algorithms/0357. Count Numbers with Unique Digits/README.md
rename to leetcode/0357.Count-Numbers-with-Unique-Digits/README.md
diff --git a/Algorithms/0367. Valid Perfect Square/367. Valid Perfect Square.go b/leetcode/0367.Valid-Perfect-Square/367. Valid Perfect Square.go
similarity index 100%
rename from Algorithms/0367. Valid Perfect Square/367. Valid Perfect Square.go
rename to leetcode/0367.Valid-Perfect-Square/367. Valid Perfect Square.go
diff --git a/Algorithms/0367. Valid Perfect Square/367. Valid Perfect Square_test.go b/leetcode/0367.Valid-Perfect-Square/367. Valid Perfect Square_test.go
similarity index 89%
rename from Algorithms/0367. Valid Perfect Square/367. Valid Perfect Square_test.go
rename to leetcode/0367.Valid-Perfect-Square/367. Valid Perfect Square_test.go
index 2877dc4cd..e7a1e6eb6 100644
--- a/Algorithms/0367. Valid Perfect Square/367. Valid Perfect Square_test.go	
+++ b/leetcode/0367.Valid-Perfect-Square/367. Valid Perfect Square_test.go	
@@ -26,37 +26,37 @@ func Test_Problem367(t *testing.T) {
 
 	qs := []question367{
 
-		question367{
+		{
 			para367{1},
 			ans367{true},
 		},
 
-		question367{
+		{
 			para367{2},
 			ans367{false},
 		},
 
-		question367{
+		{
 			para367{3},
 			ans367{false},
 		},
 
-		question367{
+		{
 			para367{4},
 			ans367{true},
 		},
 
-		question367{
+		{
 			para367{5},
 			ans367{false},
 		},
 
-		question367{
+		{
 			para367{6},
 			ans367{false},
 		},
 
-		question367{
+		{
 			para367{104976},
 			ans367{true},
 		},
diff --git a/leetcode/0367.Valid-Perfect-Square/README.md b/leetcode/0367.Valid-Perfect-Square/README.md
new file mode 100755
index 000000000..7d8a80528
--- /dev/null
+++ b/leetcode/0367.Valid-Perfect-Square/README.md
@@ -0,0 +1,33 @@
+# [367. Valid Perfect Square](https://leetcode.com/problems/valid-perfect-square/)
+
+## 题目
+
+Given a positive integer num, write a function which returns True if num is a perfect square else False.
+
+**Note:** **Do not** use any built-in library function such as `sqrt`.
+
+**Example 1:**
+
+    Input: 16
+    Output: true
+
+**Example 2:**
+
+    Input: 14
+    Output: false
+
+
+## 题目大意
+
+给定一个正整数 num，编写一个函数，如果 num 是一个完全平方数，则返回 True，否则返回 False。
+
+说明：不要使用任何内置的库函数，如 sqrt。
+
+
+
+
+## 解题思路
+
+
+- 给出一个数，要求判断这个数是不是完全平方数。
+- 可以用二分搜索来解答这道题。判断完全平方数，根据它的定义来，是否能被开根号，即找到一个数的平方是否可以等于待判断的数字。从 [1, n] 区间内进行二分，若能找到则返回 true，找不到就返回 false 。
diff --git a/leetcode/0368.Largest-Divisible-Subset/368. Largest Divisible Subset.go b/leetcode/0368.Largest-Divisible-Subset/368. Largest Divisible Subset.go
new file mode 100644
index 000000000..3888ff768
--- /dev/null
+++ b/leetcode/0368.Largest-Divisible-Subset/368. Largest Divisible Subset.go	
@@ -0,0 +1,33 @@
+package leetcode
+
+import "sort"
+
+func largestDivisibleSubset(nums []int) []int {
+	sort.Ints(nums)
+	dp, res := make([]int, len(nums)), []int{}
+	for i := range dp {
+		dp[i] = 1
+	}
+	maxSize, maxVal := 1, 1
+	for i := 1; i < len(nums); i++ {
+		for j, v := range nums[:i] {
+			if nums[i]%v == 0 && dp[j]+1 > dp[i] {
+				dp[i] = dp[j] + 1
+			}
+		}
+		if dp[i] > maxSize {
+			maxSize, maxVal = dp[i], nums[i]
+		}
+	}
+	if maxSize == 1 {
+		return []int{nums[0]}
+	}
+	for i := len(nums) - 1; i >= 0 && maxSize > 0; i-- {
+		if dp[i] == maxSize && maxVal%nums[i] == 0 {
+			res = append(res, nums[i])
+			maxVal = nums[i]
+			maxSize--
+		}
+	}
+	return res
+}
diff --git a/leetcode/0368.Largest-Divisible-Subset/368. Largest Divisible Subset_test.go b/leetcode/0368.Largest-Divisible-Subset/368. Largest Divisible Subset_test.go
new file mode 100644
index 000000000..558e75a04
--- /dev/null
+++ b/leetcode/0368.Largest-Divisible-Subset/368. Largest Divisible Subset_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question368 struct {
+	para368
+	ans368
+}
+
+// para 是参数
+// one 代表第一个参数
+type para368 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans368 struct {
+	one []int
+}
+
+func Test_Problem368(t *testing.T) {
+
+	qs := []question368{
+
+		{
+			para368{[]int{1, 2, 3}},
+			ans368{[]int{1, 2}},
+		},
+
+		{
+			para368{[]int{1, 2, 4, 8}},
+			ans368{[]int{1, 2, 4, 8}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 368------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans368, q.para368
+		fmt.Printf("【input】:%v       【output】:%v\n", p, largestDivisibleSubset(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0368.Largest-Divisible-Subset/README.md b/leetcode/0368.Largest-Divisible-Subset/README.md
new file mode 100644
index 000000000..53eb0fa92
--- /dev/null
+++ b/leetcode/0368.Largest-Divisible-Subset/README.md
@@ -0,0 +1,86 @@
+# [368. Largest Divisible Subset](https://leetcode.com/problems/largest-divisible-subset/)
+
+
+## 题目
+
+Given a set of **distinct** positive integers `nums`, return the largest subset `answer` such that every pair `(answer[i], answer[j])` of elements in this subset satisfies:
+
+- `answer[i] % answer[j] == 0`, or
+- `answer[j] % answer[i] == 0`
+
+If there are multiple solutions, return any of them.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3]
+Output: [1,2]
+Explanation: [1,3] is also accepted.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,4,8]
+Output: [1,2,4,8]
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 1000`
+- `1 <= nums[i] <= 2 * 109`
+- All the integers in `nums` are **unique**.
+
+## 题目大意
+
+给你一个由 无重复 正整数组成的集合 nums ，请你找出并返回其中最大的整除子集 answer ，子集中每一元素对 (answer[i], answer[j]) 都应当满足：
+
+- answer[i] % answer[j] == 0 ，或
+- answer[j] % answer[i] == 0
+
+如果存在多个有效解子集，返回其中任何一个均可。
+
+## 解题思路
+
+- 根据题目数据规模 1000，可以估计此题大概率是动态规划，并且时间复杂度是 O(n^2)。先将集合排序，以某一个小的数作为基准，不断的选择能整除的数加入集合。按照这个思路考虑，此题和第 300 题经典的 LIS 解题思路一致。只不过 LIS 每次选择更大的数，此题除了选择更大的数，只不过多了一个判断，这个更大的数能否整除当前集合里面的所有元素。按照此法一定可以找出最大的集合。
+- 剩下的问题是如何输出最大集合。这道题的集合具有重叠子集的性质，例如 [2,4,8,16] 这个集合，长度是 4，它一定包含长度为 3 的子集，因为从它里面随便取 3 个数形成的子集也满足元素相互能整除的条件。同理，它也一定包含长度为 2，长度为 1 的子集。由于有这个性质，可以利用 dp 数组里面的数据，输出最大集合。例如，[2,4,6,8,9,13,16,40]，由动态规划可以找到最大集合是 [2,4,8,16]。长度为 4 的找到了，再找长度为 3 的，[2,4,8]，[2,4,40]。在最大集合中，最大元素是 16，所以 [2,4,40] 这个集合排除，它的最大元素大于 16。选定 [2,4,8] 这个集合，此时最大元素是 8 。以此类推，筛选到最后，便可以输出 [16,8,4,2] 这个组最大集合的答案了。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func largestDivisibleSubset(nums []int) []int {
+	sort.Ints(nums)
+	dp, res := make([]int, len(nums)), []int{}
+	for i := range dp {
+		dp[i] = 1
+	}
+	maxSize, maxVal := 1, 1
+	for i := 1; i < len(nums); i++ {
+		for j, v := range nums[:i] {
+			if nums[i]%v == 0 && dp[j]+1 > dp[i] {
+				dp[i] = dp[j] + 1
+			}
+		}
+		if dp[i] > maxSize {
+			maxSize, maxVal = dp[i], nums[i]
+		}
+	}
+	if maxSize == 1 {
+		return []int{nums[0]}
+	}
+	for i := len(nums) - 1; i >= 0 && maxSize > 0; i-- {
+		if dp[i] == maxSize && maxVal%nums[i] == 0 {
+			res = append(res, nums[i])
+			maxVal = nums[i]
+			maxSize--
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0371. Sum of Two Integers/371. Sum of Two Integers.go b/leetcode/0371.Sum-of-Two-Integers/371. Sum of Two Integers.go
similarity index 100%
rename from Algorithms/0371. Sum of Two Integers/371. Sum of Two Integers.go
rename to leetcode/0371.Sum-of-Two-Integers/371. Sum of Two Integers.go
diff --git a/Algorithms/0371. Sum of Two Integers/371. Sum of Two Integers_test.go b/leetcode/0371.Sum-of-Two-Integers/371. Sum of Two Integers_test.go
similarity index 95%
rename from Algorithms/0371. Sum of Two Integers/371. Sum of Two Integers_test.go
rename to leetcode/0371.Sum-of-Two-Integers/371. Sum of Two Integers_test.go
index fa065c933..8a76c8dc5 100644
--- a/Algorithms/0371. Sum of Two Integers/371. Sum of Two Integers_test.go	
+++ b/leetcode/0371.Sum-of-Two-Integers/371. Sum of Two Integers_test.go	
@@ -27,12 +27,12 @@ func Test_Problem371(t *testing.T) {
 
 	qs := []question371{
 
-		question371{
+		{
 			para371{1, 2},
 			ans371{3},
 		},
 
-		question371{
+		{
 			para371{-2, 3},
 			ans371{1},
 		},
diff --git a/leetcode/0371.Sum-of-Two-Integers/README.md b/leetcode/0371.Sum-of-Two-Integers/README.md
new file mode 100755
index 000000000..187fbc8f1
--- /dev/null
+++ b/leetcode/0371.Sum-of-Two-Integers/README.md
@@ -0,0 +1,26 @@
+# [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)
+
+
+## 题目
+
+Calculate the sum of two integers a and b, but you are **not allowed** to use the operator `+` and `-`.
+
+**Example 1:**
+
+    Input: a = 1, b = 2
+    Output: 3
+
+**Example 2:**
+
+    Input: a = -2, b = 3
+    Output: 1
+
+
+## 题目大意
+
+不使用运算符 + 和 - ​​​​​​​，计算两整数 ​​​​​​​a 、b ​​​​​​​之和。
+
+## 解题思路
+
+- 要求不用加法和减法运算符计算 `a+b`。这一题需要用到 `^` 和 `&` 运算符的性质，两个数 ^ 可以实现两个数不带进位的二进制加法。这里需要实现加法，肯定需要进位。所以如何找到进位是本题的关键。
+- 在二进制中，只有 1 和 1 加在一起才会进位，0 和 0，0 和 1，1 和 0，这三种情况都不会进位，规律就是 `a & b` 为 0 的时候就不用进位，为 1 的时候代表需要进位。进位是往前进一位，所以还需要左移操作，所以加上的进位为 `(a&b)<<1`。
diff --git a/Algorithms/0372. Super Pow/372. Super Pow.go b/leetcode/0372.Super-Pow/372. Super Pow.go
similarity index 84%
rename from Algorithms/0372. Super Pow/372. Super Pow.go
rename to leetcode/0372.Super-Pow/372. Super Pow.go
index aacc6c7ae..3f4082c7a 100644
--- a/Algorithms/0372. Super Pow/372. Super Pow.go	
+++ b/leetcode/0372.Super-Pow/372. Super Pow.go	
@@ -8,10 +8,11 @@ package leetcode
 // 模运算性质五：ab % p = ((a % p) * ( b % p)) % p, 其中 ab 是一个数字，如:2874，98374 等等
 // 举个例子
 // 12345^678 % 1337 = (12345^670 * 12345^8) % 1337
-//				    = ((12345^670 % 1337) * (12345^8 % 1337)) % 1337  ---> 利用性质 三
+//
+//	                = ((12345^670 % 1337) * (12345^8 % 1337)) % 1337  ---> 利用性质 三
 //				    = (((12345^67)^10 % 1337) * (12345^8 % 1337)) % 1337  ---> 乘方性质
-//                  = ((12345^67 % 1337)^10) % 1337 * (12345^8 % 1337)) % 1337  ---> 利用性质 四
-//				    = (((12345^67 % 1337)^10) * (12345^8 % 1337)) % 1337  ---> 反向利用性质 三
+//		            = ((12345^67 % 1337)^10) % 1337 * (12345^8 % 1337)) % 1337  ---> 利用性质 四
+//					= (((12345^67 % 1337)^10) * (12345^8 % 1337)) % 1337  ---> 反向利用性质 三
 func superPow(a int, b []int) int {
 	res := 1
 	for i := 0; i < len(b); i++ {
diff --git a/Algorithms/0372. Super Pow/372. Super Pow_test.go b/leetcode/0372.Super-Pow/372. Super Pow_test.go
similarity index 96%
rename from Algorithms/0372. Super Pow/372. Super Pow_test.go
rename to leetcode/0372.Super-Pow/372. Super Pow_test.go
index c862d96c6..75347ad4d 100644
--- a/Algorithms/0372. Super Pow/372. Super Pow_test.go	
+++ b/leetcode/0372.Super-Pow/372. Super Pow_test.go	
@@ -27,12 +27,12 @@ func Test_Problem372(t *testing.T) {
 
 	qs := []question372{
 
-		question372{
+		{
 			para372{2, []int{3}},
 			ans372{8},
 		},
 
-		question372{
+		{
 			para372{2, []int{1, 0}},
 			ans372{1024},
 		},
diff --git a/leetcode/0372.Super-Pow/README.md b/leetcode/0372.Super-Pow/README.md
new file mode 100755
index 000000000..8d7bcc7b7
--- /dev/null
+++ b/leetcode/0372.Super-Pow/README.md
@@ -0,0 +1,42 @@
+# [372. Super Pow](https://leetcode.com/problems/super-pow/)
+
+
+## 题目
+
+Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.
+
+**Example 1:**
+
+    Input: a = 2, b = [3]
+    Output: 8
+
+**Example 2:**
+
+    Input: a = 2, b = [1,0]
+    Output: 1024
+
+
+## 题目大意
+
+
+你的任务是计算 a^b 对 1337 取模，a 是一个正整数，b 是一个非常大的正整数且会以数组形式给出。
+
+## 解题思路
+
+- 求 a^b mod p 的结果，b 是大数。
+- 这一题可以用暴力解法尝试。需要用到 mod 计算的几个运算性质：
+
+        模运算性质一：(a + b) % p = (a % p + b % p) % p
+        模运算性质二：(a - b) % p = (a % p - b % p + p) % p
+        模运算性质三：(a * b) % p = (a % p * b % p) % p
+        模运算性质四：a ^ b % p = ((a % p)^b) % p
+
+    这一题需要用到性质三、四。举个例子：
+
+        12345^678 % 1337 = (12345^670 * 12345^8) % 1337
+        				 = ((12345^670 % 1337) * (12345^8 % 1337)) % 1337  ---> 利用性质 三
+        			     = (((12345^67)^10 % 1337) * (12345^8 % 1337)) % 1337  ---> 乘方性质
+                         = ((12345^67 % 1337)^10) % 1337 * (12345^8 % 1337)) % 1337  ---> 利用性质 四
+        				 = (((12345^67 % 1337)^10) * (12345^8 % 1337)) % 1337  ---> 反向利用性质 三
+
+    经过上面这样的变换，把指数 678 的个位分离出来了，可以单独求解。继续经过上面的变换，可以把指数的 6 和 7 也分离出来。最终可以把大数 b 一位一位的分离出来。至于计算 a^b 就结果快速幂求解。
diff --git a/leetcode/0373.Find-K-Pairs-with-Smallest-Sums/373. Find K Pairs with Smallest Sums.go b/leetcode/0373.Find-K-Pairs-with-Smallest-Sums/373. Find K Pairs with Smallest Sums.go
new file mode 100644
index 000000000..2e78cd688
--- /dev/null
+++ b/leetcode/0373.Find-K-Pairs-with-Smallest-Sums/373. Find K Pairs with Smallest Sums.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+// 解法一 优先队列
+func kSmallestPairs(nums1 []int, nums2 []int, k int) [][]int {
+	result, h := [][]int{}, &minHeap{}
+	if len(nums1) == 0 || len(nums2) == 0 || k == 0 {
+		return result
+	}
+	if len(nums1)*len(nums2) < k {
+		k = len(nums1) * len(nums2)
+	}
+	heap.Init(h)
+	for _, num := range nums1 {
+		heap.Push(h, []int{num, nums2[0], 0})
+	}
+	for len(result) < k {
+		min := heap.Pop(h).([]int)
+		result = append(result, min[:2])
+		if min[2] < len(nums2)-1 {
+			heap.Push(h, []int{min[0], nums2[min[2]+1], min[2] + 1})
+		}
+	}
+	return result
+}
+
+type minHeap [][]int
+
+func (h minHeap) Len() int           { return len(h) }
+func (h minHeap) Less(i, j int) bool { return h[i][0]+h[i][1] < h[j][0]+h[j][1] }
+func (h minHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
+
+func (h *minHeap) Push(x interface{}) {
+	*h = append(*h, x.([]int))
+}
+
+func (h *minHeap) Pop() interface{} {
+	old := *h
+	n := len(old)
+	x := old[n-1]
+	*h = old[0 : n-1]
+	return x
+}
+
+// 解法二 暴力解法
+func kSmallestPairs1(nums1 []int, nums2 []int, k int) [][]int {
+	size1, size2, res := len(nums1), len(nums2), [][]int{}
+	if size1 == 0 || size2 == 0 || k < 0 {
+		return nil
+	}
+	for i := 0; i < size1; i++ {
+		for j := 0; j < size2; j++ {
+			res = append(res, []int{nums1[i], nums2[j]})
+		}
+	}
+	sort.Slice(res, func(i, j int) bool {
+		return res[i][0]+res[i][1] < res[j][0]+res[j][1]
+	})
+	if len(res) >= k {
+		return res[:k]
+	}
+	return res
+}
diff --git a/leetcode/0373.Find-K-Pairs-with-Smallest-Sums/373. Find K Pairs with Smallest Sums_test.go b/leetcode/0373.Find-K-Pairs-with-Smallest-Sums/373. Find K Pairs with Smallest Sums_test.go
new file mode 100644
index 000000000..214ace275
--- /dev/null
+++ b/leetcode/0373.Find-K-Pairs-with-Smallest-Sums/373. Find K Pairs with Smallest Sums_test.go	
@@ -0,0 +1,74 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question373 struct {
+	para373
+	ans373
+}
+
+// para 是参数
+// one 代表第一个参数
+type para373 struct {
+	nums1 []int
+	nums2 []int
+	k     int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans373 struct {
+	one [][]int
+}
+
+func Test_Problem373(t *testing.T) {
+
+	qs := []question373{
+
+		{
+			para373{[]int{1, 7, 11}, []int{2, 4, 6}, 3},
+			ans373{[][]int{{1, 2}, {1, 4}, {1, 6}}},
+		},
+
+		{
+			para373{[]int{1, 1, 2}, []int{1, 2, 3}, 2},
+			ans373{[][]int{{1, 1}, {1, 1}}},
+		},
+
+		{
+			para373{[]int{1, 2}, []int{3}, 3},
+			ans373{[][]int{{1, 3}, {2, 3}}},
+		},
+
+		{
+			para373{[]int{1, 2, 4, 5, 6}, []int{3, 5, 7, 9}, 3},
+			ans373{[][]int{{1, 3}, {2, 3}, {1, 5}}},
+		},
+
+		{
+			para373{[]int{1, 1, 2}, []int{1, 2, 3}, 10},
+			ans373{[][]int{{1, 1}, {1, 1}, {2, 1}, {1, 2}, {1, 2}, {2, 2}, {1, 3}, {1, 3}, {2, 3}}},
+		},
+
+		{
+			para373{[]int{-10, -4, 0, 0, 6}, []int{3, 5, 6, 7, 8, 100}, 10},
+			ans373{[][]int{{-10, 3}, {-10, 5}, {-10, 6}, {-10, 7}, {-10, 8}, {-4, 3}, {-4, 5}, {-4, 6}, {0, 3}, {0, 3}}},
+		},
+
+		{
+			para373{[]int{0, 0, 0, 0, 0}, []int{-3, 22, 35, 56, 76}, 22},
+			ans373{[][]int{{0, -3}, {0, -3}, {0, -3}, {0, -3}, {0, -3}, {0, 22}, {0, 22}, {0, 22}, {0, 22}, {0, 22}, {0, 35}, {0, 35}, {0, 35}, {0, 35}, {0, 35}, {0, 56}, {0, 56}, {0, 56}, {0, 56}, {0, 56}, {0, 76}, {0, 76}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 373------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans373, q.para373
+		fmt.Printf("【input】:%v       【output】:%v\n", p, kSmallestPairs(p.nums1, p.nums2, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0373.Find-K-Pairs-with-Smallest-Sums/README.md b/leetcode/0373.Find-K-Pairs-with-Smallest-Sums/README.md
new file mode 100755
index 000000000..68d5336c3
--- /dev/null
+++ b/leetcode/0373.Find-K-Pairs-with-Smallest-Sums/README.md
@@ -0,0 +1,51 @@
+# [373. Find K Pairs with Smallest Sums](https://leetcode.com/problems/find-k-pairs-with-smallest-sums/)
+
+
+## 题目
+
+You are given two integer arrays **nums1** and **nums2** sorted in ascending order and an integer **k**.
+
+Define a pair **(u,v)** which consists of one element from the first array and one element from the second array.
+
+Find the k pairs **(u1,v1),(u2,v2) ...(uk,vk)** with the smallest sums.
+
+**Example 1:**
+
+    Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
+    Output: [[1,2],[1,4],[1,6]] 
+    Explanation: The first 3 pairs are returned from the sequence: 
+                 [1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]
+
+**Example 2:**
+
+    Input: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
+    Output: [1,1],[1,1]
+    Explanation: The first 2 pairs are returned from the sequence: 
+                 [1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]
+
+**Example 3:**
+
+    Input: nums1 = [1,2], nums2 = [3], k = 3
+    Output: [1,3],[2,3]
+    Explanation: All possible pairs are returned from the sequence: [1,3],[2,3]
+
+
+## 题目大意
+
+
+给定两个以升序排列的整形数组 nums1 和 nums2, 以及一个整数 k。
+
+定义一对值 (u,v)，其中第一个元素来自 nums1，第二个元素来自 nums2。
+
+找到和最小的 k 对数字 (u1,v1), (u2,v2) ... (uk,vk)。
+
+
+
+## 解题思路
+
+
+- 给出 2 个数组，和数字 k，要求找到 k 个数值对，数值对两个数的和最小。
+- 这一题咋一看可以用二分搜索，两个数组两个组合有 `m * n` 个数值对。然后找到最小的和，最大的和，在这个范围内进行二分搜索，每分出一个 `mid`，再去找比 `mid` 小的数值对有多少个，如果个数小于 `k` 个，那么在右区间上继续二分，如果个数大于 `k` 个，那么在左区间上继续二分。到目前为止，这个思路看似可行。但是每次搜索的数值对是无序的。这会导致最终出现错误的结果。例如 `mid = 10` 的时候，小于 10 的和有 22 个，而 `k = 25` 。这说明 `mid` 偏小，`mid` 增大，`mid = 11` 的时候，小于 11 的和有 30 个，而 `k = 25` 。这时候应该从这 30 个和中取前 25 个。但是我们遍历数值对的时候，和并不是从小到大排序的。这时候还需要额外对这 30 个候选值进行排序。这样时间复杂度又增大了。
+- 可以先用暴力解法解答。将所有的和都遍历出来，排序以后，取前 k 个。这个暴力方法可以 AC。
+- 本题最优解应该是优先队列。维护一个最小堆。把数值对的和放在这个最小堆中，不断的 pop 出 k 个最小值到数组中，即为答案。
+- 在已排序的矩阵中寻找最 K 小的元素这一系列的题目有：第 373 题，第 378 题，第 668 题，第 719 题，第 786 题。
diff --git a/leetcode/0374.Guess-Number-Higher-or-Lower/374. Guess Number Higher or Lower.go b/leetcode/0374.Guess-Number-Higher-or-Lower/374. Guess Number Higher or Lower.go
new file mode 100644
index 000000000..fafc4a8c4
--- /dev/null
+++ b/leetcode/0374.Guess-Number-Higher-or-Lower/374. Guess Number Higher or Lower.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+import "sort"
+
+/**
+ * Forward declaration of guess API.
+ * @param  num   your guess
+ * @return 	     -1 if num is lower than the guess number
+ *			      1 if num is higher than the guess number
+ *               otherwise return 0
+ * func guess(num int) int;
+ */
+
+func guessNumber(n int) int {
+	return sort.Search(n, func(x int) bool { return guess(x) <= 0 })
+}
+
+func guess(num int) int {
+	return 0
+}
diff --git a/leetcode/0374.Guess-Number-Higher-or-Lower/374. Guess Number Higher or Lower_test.go b/leetcode/0374.Guess-Number-Higher-or-Lower/374. Guess Number Higher or Lower_test.go
new file mode 100644
index 000000000..9c66680cc
--- /dev/null
+++ b/leetcode/0374.Guess-Number-Higher-or-Lower/374. Guess Number Higher or Lower_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question374 struct {
+	para374
+	ans374
+}
+
+// para 是参数
+// one 代表第一个参数
+type para374 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans374 struct {
+	one int
+}
+
+func Test_Problem374(t *testing.T) {
+
+	qs := []question374{
+
+		{
+			para374{10},
+			ans374{6},
+		},
+
+		{
+			para374{1},
+			ans374{1},
+		},
+
+		{
+			para374{2},
+			ans374{1},
+		},
+
+		{
+			para374{2},
+			ans374{2},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 374------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans374, q.para374
+		fmt.Printf("【input】:%v       【output】:%v\n", p, guessNumber(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0374.Guess-Number-Higher-or-Lower/README.md b/leetcode/0374.Guess-Number-Higher-or-Lower/README.md
new file mode 100644
index 000000000..795efe6f3
--- /dev/null
+++ b/leetcode/0374.Guess-Number-Higher-or-Lower/README.md
@@ -0,0 +1,94 @@
+# [374. Guess Number Higher or Lower](https://leetcode.com/problems/guess-number-higher-or-lower/)
+
+## 题目
+
+We are playing the Guess Game. The game is as follows:
+
+I pick a number from `1` to `n`. You have to guess which number I picked.
+
+Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
+
+You call a pre-defined API `int guess(int num)`, which returns 3 possible results:
+
+- `1`: The number I picked is lower than your guess (i.e. `pick < num`).
+- `1`: The number I picked is higher than your guess (i.e. `pick > num`).
+- `0`: The number I picked is equal to your guess (i.e. `pick == num`).
+
+Return *the number that I picked*.
+
+**Example 1:**
+
+```
+Input: n = 10, pick = 6
+Output: 6
+```
+
+**Example 2:**
+
+```
+Input: n = 1, pick = 1
+Output: 1
+```
+
+**Example 3:**
+
+```
+Input: n = 2, pick = 1
+Output: 1
+```
+
+**Example 4:**
+
+```
+Input: n = 2, pick = 2
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= n <= 231 - 1`
+- `1 <= pick <= n`
+
+## 题目大意
+
+猜数字游戏的规则如下：
+
+- 每轮游戏，我都会从 1 到 n 随机选择一个数字。 请你猜选出的是哪个数字。
+- 如果你猜错了，我会告诉你，你猜测的数字比我选出的数字是大了还是小了。
+
+你可以通过调用一个预先定义好的接口 int guess(int num) 来获取猜测结果，返回值一共有 3 种可能的情况（-1，1 或 0）：
+
+- 1：我选出的数字比你猜的数字小 pick < num
+- 1：我选出的数字比你猜的数字大 pick > num
+- 0：我选出的数字和你猜的数字一样。恭喜！你猜对了！pick == num
+
+返回我选出的数字。
+
+## 解题思路
+
+- 这一题是简单题，和小时候玩的猜大猜小的游戏一样。思路很简单，二分查找即可。这一题和第 278 题类似。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+/**
+ * Forward declaration of guess API.
+ * @param  num   your guess
+ * @return 	     -1 if num is lower than the guess number
+ *			      1 if num is higher than the guess number
+ *               otherwise return 0
+ * func guess(num int) int;
+ */
+
+func guessNumber(n int) int {
+	return sort.Search(n, func(x int) bool { return guess(x) <= 0 })
+}
+
+func guess(num int) int {
+	return 0
+}
+```
\ No newline at end of file
diff --git a/leetcode/0376.Wiggle-Subsequence/376. Wiggle Subsequence.go b/leetcode/0376.Wiggle-Subsequence/376. Wiggle Subsequence.go
new file mode 100644
index 000000000..2d01c7227
--- /dev/null
+++ b/leetcode/0376.Wiggle-Subsequence/376. Wiggle Subsequence.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+func wiggleMaxLength(nums []int) int {
+	if len(nums) < 2 {
+		return len(nums)
+	}
+	res := 1
+	prevDiff := nums[1] - nums[0]
+	if prevDiff != 0 {
+		res = 2
+	}
+	for i := 2; i < len(nums); i++ {
+		diff := nums[i] - nums[i-1]
+		if diff > 0 && prevDiff <= 0 || diff < 0 && prevDiff >= 0 {
+			res++
+			prevDiff = diff
+		}
+	}
+	return res
+}
diff --git a/leetcode/0376.Wiggle-Subsequence/376. Wiggle Subsequence_test.go b/leetcode/0376.Wiggle-Subsequence/376. Wiggle Subsequence_test.go
new file mode 100644
index 000000000..7abf91c46
--- /dev/null
+++ b/leetcode/0376.Wiggle-Subsequence/376. Wiggle Subsequence_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question376 struct {
+	para376
+	ans376
+}
+
+// para 是参数
+// one 代表第一个参数
+type para376 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans376 struct {
+	one int
+}
+
+func Test_Problem376(t *testing.T) {
+
+	qs := []question376{
+
+		{
+			para376{[]int{1, 7, 4, 9, 2, 5}},
+			ans376{6},
+		},
+
+		{
+			para376{[]int{1, 17, 5, 10, 13, 15, 10, 5, 16, 8}},
+			ans376{7},
+		},
+
+		{
+			para376{[]int{1, 2, 3, 4, 5, 6, 7, 8, 9}},
+			ans376{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 376------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans376, q.para376
+		fmt.Printf("【input】:%v       【output】:%v\n", p, wiggleMaxLength(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0376.Wiggle-Subsequence/README.md b/leetcode/0376.Wiggle-Subsequence/README.md
new file mode 100644
index 000000000..d66ed01f7
--- /dev/null
+++ b/leetcode/0376.Wiggle-Subsequence/README.md
@@ -0,0 +1,76 @@
+# [376. Wiggle Subsequence](https://leetcode.com/problems/wiggle-subsequence/)
+
+
+## 题目
+
+Given an integer array `nums`, return *the length of the longest **wiggle sequence***.
+
+A **wiggle sequence** is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two elements is trivially a wiggle sequence.
+
+- For example, `[1, 7, 4, 9, 2, 5]` is a **wiggle sequence** because the differences `(6, -3, 5, -7, 3)` are alternately positive and negative.
+- In contrast, `[1, 4, 7, 2, 5]` and `[1, 7, 4, 5, 5]` are not wiggle sequences, the first because its first two differences are positive and the second because its last difference is zero.
+
+A **subsequence** is obtained by deleting some elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.
+
+**Example 1:**
+
+```
+Input: nums = [1,7,4,9,2,5]
+Output: 6
+Explanation: The entire sequence is a wiggle sequence.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,17,5,10,13,15,10,5,16,8]
+Output: 7
+Explanation: There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,2,3,4,5,6,7,8,9]
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 1000`
+- `0 <= nums[i] <= 1000`
+
+**Follow up:** Could you solve this in `O(n)` time?
+
+## 题目大意
+
+如果连续数字之间的差严格地在正数和负数之间交替，则数字序列称为摆动序列。第一个差（如果存在的话）可能是正数或负数。少于两个元素的序列也是摆动序列。例如， [1,7,4,9,2,5] 是一个摆动序列，因为差值 (6,-3,5,-7,3) 是正负交替出现的。相反, [1,4,7,2,5] 和 [1,7,4,5,5] 不是摆动序列，第一个序列是因为它的前两个差值都是正数，第二个序列是因为它的最后一个差值为零。给定一个整数序列，返回作为摆动序列的最长子序列的长度。 通过从原始序列中删除一些（也可以不删除）元素来获得子序列，剩下的元素保持其原始顺序。
+
+## 解题思路
+
+- 题目要求找到摆动序列最长的子序列。本题可以用贪心的思路，记录当前序列的上升和下降的趋势。扫描数组过程中，每扫描一个元素都判断是“峰”还是“谷”，根据前一个是“峰”还是“谷”做出对应的决定。利用贪心的思想找到最长的摆动子序列。
+
+## 代码
+
+```go
+package leetcode
+
+func wiggleMaxLength(nums []int) int {
+	if len(nums) < 2 {
+		return len(nums)
+	}
+	res := 1
+	prevDiff := nums[1] - nums[0]
+	if prevDiff != 0 {
+		res = 2
+	}
+	for i := 2; i < len(nums); i++ {
+		diff := nums[i] - nums[i-1]
+		if diff > 0 && prevDiff <= 0 || diff < 0 && prevDiff >= 0 {
+			res++
+			prevDiff = diff
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0377.Combination-Sum-IV/377. Combination Sum IV.go b/leetcode/0377.Combination-Sum-IV/377. Combination Sum IV.go
new file mode 100644
index 000000000..6a6d8ada1
--- /dev/null
+++ b/leetcode/0377.Combination-Sum-IV/377. Combination Sum IV.go	
@@ -0,0 +1,38 @@
+package leetcode
+
+func combinationSum4(nums []int, target int) int {
+	dp := make([]int, target+1)
+	dp[0] = 1
+	for i := 1; i <= target; i++ {
+		for _, num := range nums {
+			if i-num >= 0 {
+				dp[i] += dp[i-num]
+			}
+		}
+	}
+	return dp[target]
+}
+
+// 暴力解法超时
+func combinationSum41(nums []int, target int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	c, res := []int{}, 0
+	findcombinationSum4(nums, target, 0, c, &res)
+	return res
+}
+
+func findcombinationSum4(nums []int, target, index int, c []int, res *int) {
+	if target <= 0 {
+		if target == 0 {
+			*res++
+		}
+		return
+	}
+	for i := 0; i < len(nums); i++ {
+		c = append(c, nums[i])
+		findcombinationSum4(nums, target-nums[i], i, c, res)
+		c = c[:len(c)-1]
+	}
+}
diff --git a/leetcode/0377.Combination-Sum-IV/377. Combination Sum IV_test.go b/leetcode/0377.Combination-Sum-IV/377. Combination Sum IV_test.go
new file mode 100644
index 000000000..297e31c95
--- /dev/null
+++ b/leetcode/0377.Combination-Sum-IV/377. Combination Sum IV_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question377 struct {
+	para377
+	ans377
+}
+
+// para 是参数
+// one 代表第一个参数
+type para377 struct {
+	n []int
+	k int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans377 struct {
+	one int
+}
+
+func Test_Problem377(t *testing.T) {
+
+	qs := []question377{
+
+		{
+			para377{[]int{1, 2, 3}, 4},
+			ans377{7},
+		},
+
+		{
+			para377{[]int{1, 2, 3}, 32},
+			ans377{181997601},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 377------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans377, q.para377
+		fmt.Printf("【input】:%v       【output】:%v\n", p, combinationSum4(p.n, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0377.Combination-Sum-IV/README.md b/leetcode/0377.Combination-Sum-IV/README.md
new file mode 100644
index 000000000..d0920bf62
--- /dev/null
+++ b/leetcode/0377.Combination-Sum-IV/README.md
@@ -0,0 +1,98 @@
+# [377. Combination Sum IV](https://leetcode.com/problems/combination-sum-iv/)
+
+
+## 题目
+
+Given an array of **distinct** integers `nums` and a target integer `target`, return *the number of possible combinations that add up to* `target`.
+
+The answer is **guaranteed** to fit in a **32-bit** integer.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3], target = 4
+Output: 7
+Explanation:
+The possible combination ways are:
+(1, 1, 1, 1)
+(1, 1, 2)
+(1, 2, 1)
+(1, 3)
+(2, 1, 1)
+(2, 2)
+(3, 1)
+Note that different sequences are counted as different combinations.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [9], target = 3
+Output: 0
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 200`
+- `1 <= nums[i] <= 1000`
+- All the elements of `nums` are **unique**.
+- `1 <= target <= 1000`
+
+**Follow up:** What if negative numbers are allowed in the given array? How does it change the problem? What limitation we need to add to the question to allow negative numbers?
+
+## 题目大意
+
+给你一个由 不同 整数组成的数组 nums ，和一个目标整数 target 。请你从 nums 中找出并返回总和为 target 的元素组合的个数。题目数据保证答案符合 32 位整数范围。
+
+## 解题思路
+
+- Combination Sum 这是系列问题。拿到题目，笔者先用暴力解法 dfs 尝试了一版，包含的重叠子问题特别多，剪枝条件也没有写好，果然超时。元素只有 [1,2,3] 这三种，target = 32，这组数据居然有 181997601 这么多种情况。仔细看了题目数据规模 1000，基本可以断定此题是动态规划，并且时间复杂度是 O(n^2)。
+- 本题和完全背包有点像，但是还是有区别。完全背包的取法内部不区分顺序。例如 5 = 1 + 2 + 2。但是本题是 3 种答案 (1，2，2)，(2，1，2)，(2，2，1)。定义 dp[i] 为总和为 target = i 的组合总数。最终答案存在 dp[target] 中。状态转移方程为：
+
+    $$dp[i] =\left\{\begin{matrix}1,i=0\\ \sum dp[i-j],i\neq 0\end{matrix}\right.$$
+
+- 这道题最后有一个进阶问题。如果给定的数组中含有负数，则会导致出现无限长度的排列。例如，假设数组 nums 中含有正整数 a 和负整数 −b（其中 a>0,b>0,-b<0），则有 a×b+(−b)×a=0，对于任意一个元素之和等于 target 的排列，在该排列的后面添加 b 个 a 和 a 个 −b 之后，得到的新排列的元素之和仍然等于 target，而且还可以在新排列的后面继续 b 个 a 和 a 个 −b。因此只要存在元素之和等于 target 的排列，就能构造出无限长度的排列。如果允许负数出现，则必须限制排列的最大长度，不然会出现无限长度的排列。
+
+## 代码
+
+```go
+package leetcode
+
+func combinationSum4(nums []int, target int) int {
+	dp := make([]int, target+1)
+	dp[0] = 1
+	for i := 1; i <= target; i++ {
+		for _, num := range nums {
+			if i-num >= 0 {
+				dp[i] += dp[i-num]
+			}
+		}
+	}
+	return dp[target]
+}
+
+// 暴力解法超时
+func combinationSum41(nums []int, target int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	c, res := []int{}, 0
+	findcombinationSum4(nums, target, 0, c, &res)
+	return res
+}
+
+func findcombinationSum4(nums []int, target, index int, c []int, res *int) {
+	if target <= 0 {
+		if target == 0 {
+			*res++
+		}
+		return
+	}
+	for i := 0; i < len(nums); i++ {
+		c = append(c, nums[i])
+		findcombinationSum4(nums, target-nums[i], i, c, res)
+		c = c[:len(c)-1]
+	}
+}
+```
\ No newline at end of file
diff --git a/Algorithms/378. Kth Smallest Element in a Sorted Matrix/378. Kth Smallest Element in a Sorted Matrix.go b/leetcode/0378.Kth-Smallest-Element-in-a-Sorted-Matrix/378. Kth Smallest Element in a Sorted Matrix.go
similarity index 100%
rename from Algorithms/378. Kth Smallest Element in a Sorted Matrix/378. Kth Smallest Element in a Sorted Matrix.go
rename to leetcode/0378.Kth-Smallest-Element-in-a-Sorted-Matrix/378. Kth Smallest Element in a Sorted Matrix.go
diff --git a/Algorithms/378. Kth Smallest Element in a Sorted Matrix/378. Kth Smallest Element in a Sorted Matrix_test.go b/leetcode/0378.Kth-Smallest-Element-in-a-Sorted-Matrix/378. Kth Smallest Element in a Sorted Matrix_test.go
similarity index 77%
rename from Algorithms/378. Kth Smallest Element in a Sorted Matrix/378. Kth Smallest Element in a Sorted Matrix_test.go
rename to leetcode/0378.Kth-Smallest-Element-in-a-Sorted-Matrix/378. Kth Smallest Element in a Sorted Matrix_test.go
index 8019a0211..021828aa0 100644
--- a/Algorithms/378. Kth Smallest Element in a Sorted Matrix/378. Kth Smallest Element in a Sorted Matrix_test.go	
+++ b/leetcode/0378.Kth-Smallest-Element-in-a-Sorted-Matrix/378. Kth Smallest Element in a Sorted Matrix_test.go	
@@ -27,13 +27,13 @@ func Test_Problem378(t *testing.T) {
 
 	qs := []question378{
 
-		question378{
-			para378{[][]int{[]int{1, 5, 9}, []int{10, 11, 13}, []int{12, 13, 15}}, 8},
+		{
+			para378{[][]int{{1, 5, 9}, {10, 11, 13}, {12, 13, 15}}, 8},
 			ans378{13},
 		},
 
-		question378{
-			para378{[][]int{[]int{1, 5, 7}, []int{11, 12, 13}, []int{12, 13, 15}}, 3},
+		{
+			para378{[][]int{{1, 5, 7}, {11, 12, 13}, {12, 13, 15}}, 3},
 			ans378{9},
 		},
 	}
diff --git a/leetcode/0378.Kth-Smallest-Element-in-a-Sorted-Matrix/README.md b/leetcode/0378.Kth-Smallest-Element-in-a-Sorted-Matrix/README.md
new file mode 100755
index 000000000..0eab93620
--- /dev/null
+++ b/leetcode/0378.Kth-Smallest-Element-in-a-Sorted-Matrix/README.md
@@ -0,0 +1,40 @@
+# [378. Kth Smallest Element in a Sorted Matrix](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/)
+
+
+## 题目
+
+Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.
+
+Note that it is the kth smallest element in the sorted order, not the kth distinct element.
+
+**Example:**
+
+    matrix = [
+       [ 1,  5,  9],
+       [10, 11, 13],
+       [12, 13, 15]
+    ],
+    k = 8,
+    
+    return 13.
+
+**Note:**You may assume k is always valid, 1 ≤ k ≤ n2.
+
+
+## 题目大意
+
+给定一个 n x n 矩阵，其中每行和每列元素均按升序排序，找到矩阵中第 k 小的元素。请注意，它是排序后的第 k 小元素，而不是第 k 个元素。
+
+
+说明:
+你可以假设 k 的值永远是有效的, 1 ≤ k ≤ n2 。
+
+
+## 解题思路
+
+
+- 给出一个行有序，列有序的矩阵(并非是按照下标有序的)，要求找出这个矩阵中第 K 小的元素。注意找的第 K 小元素指的不是 k 个不同的元素，可能存在相同的元素。
+- 最容易想到的就解法是优先队列。依次把矩阵中的元素推入到优先队列中。维护一个最小堆，一旦优先队列里面的元素有 k 个了，就算找到结果了。
+- 这一题最优解法是二分搜索。那搜索的空间是什么呢？根据题意，可以知道，矩阵左上角的那个元素是最小的，右下角的元素是最大的。即矩阵第一个元素确定了下界，矩阵的最后一个元素确定了上界。在这个解空间里面二分搜索所有值，找到第 K  小的元素。判断是否找到的条件是，在矩阵中比 mid 小的元素个数等于 K。不断的逼近 low，使得 low == high 的时候，就是找到了第 K 小的元素了。(因为题目中说了，一定会存在第 K 小元素，所以二分搜索到一个元素的时候，一定会得出结果)。
+
+![](https://img.halfrost.com/Leetcode/leetcode_378.png)
diff --git a/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node.go b/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node.go
new file mode 100644
index 000000000..892ca7075
--- /dev/null
+++ b/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"math/rand"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+type Solution struct {
+	head *ListNode
+}
+
+/*
+  - @param head The linked list's head.
+    Note that the head is guaranteed to be not null, so it contains at least one node.
+*/
+func Constructor(head *ListNode) Solution {
+	return Solution{head: head}
+}
+
+/** Returns a random node's value. */
+func (this *Solution) GetRandom() int {
+	scope, selectPoint, curr := 1, 0, this.head
+	for curr != nil {
+		if rand.Float64() < 1.0/float64(scope) {
+			selectPoint = curr.Val
+		}
+		scope += 1
+		curr = curr.Next
+	}
+	return selectPoint
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(head);
+ * param_1 := obj.GetRandom();
+ */
diff --git a/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node_test.go b/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node_test.go
new file mode 100644
index 000000000..f33713446
--- /dev/null
+++ b/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node_test.go	
@@ -0,0 +1,28 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+func Test_Problem382(t *testing.T) {
+	header := structures.Ints2List([]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 0})
+	obj := Constructor(header)
+	fmt.Printf("obj = %v\n", structures.List2Ints(header))
+	param1 := obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+	param1 = obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+	param1 = obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+	param1 = obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+	param1 = obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+	param1 = obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0382.Linked-List-Random-Node/README.md b/leetcode/0382.Linked-List-Random-Node/README.md
new file mode 100644
index 000000000..db5ecc06c
--- /dev/null
+++ b/leetcode/0382.Linked-List-Random-Node/README.md
@@ -0,0 +1,105 @@
+# [382. Linked List Random Node](https://leetcode.com/problems/linked-list-random-node/)
+
+
+## 题目
+
+Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
+
+Implement the `Solution` class:
+
+- `Solution(ListNode head)` Initializes the object with the integer array nums.
+- `int getRandom()` Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be choosen.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/03/16/getrand-linked-list.jpg](https://assets.leetcode.com/uploads/2021/03/16/getrand-linked-list.jpg)
+
+```
+Input
+["Solution", "getRandom", "getRandom", "getRandom", "getRandom", "getRandom"]
+[[[1, 2, 3]], [], [], [], [], []]
+Output
+[null, 1, 3, 2, 2, 3]
+
+Explanation
+Solution solution = new Solution([1, 2, 3]);
+solution.getRandom(); // return 1
+solution.getRandom(); // return 3
+solution.getRandom(); // return 2
+solution.getRandom(); // return 2
+solution.getRandom(); // return 3
+// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the linked list will be in the range `[1, 104]`.
+- `-10^4 <= Node.val <= 10^4`
+- At most `10^4` calls will be made to `getRandom`.
+
+**Follow up:**
+
+- What if the linked list is extremely large and its length is unknown to you?
+- Could you solve this efficiently without using extra space?
+
+## 题目大意
+
+给定一个单链表，随机选择链表的一个节点，并返回相应的节点值。保证每个节点被选的概率一样。
+
+进阶: 如果链表十分大且长度未知，如何解决这个问题？你能否使用常数级空间复杂度实现？
+
+## 解题思路
+
+- rand.Float64() 可以返回 [0.0,1.0) 之间的随机数。利用这个函数完成我们的随机化取节点的过程。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math/rand"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+type Solution struct {
+	head *ListNode
+}
+
+/** @param head The linked list's head.
+  Note that the head is guaranteed to be not null, so it contains at least one node. */
+func Constructor(head *ListNode) Solution {
+	return Solution{head: head}
+}
+
+/** Returns a random node's value. */
+func (this *Solution) GetRandom() int {
+	scope, selectPoint, curr := 1, 0, this.head
+	for curr != nil {
+		if rand.Float64() < 1.0/float64(scope) {
+			selectPoint = curr.Val
+		}
+		scope += 1
+		curr = curr.Next
+	}
+	return selectPoint
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(head);
+ * param_1 := obj.GetRandom();
+ */
+```
\ No newline at end of file
diff --git a/leetcode/0383.Ransom-Note/383.Ransom Note.go b/leetcode/0383.Ransom-Note/383.Ransom Note.go
new file mode 100644
index 000000000..a20162c61
--- /dev/null
+++ b/leetcode/0383.Ransom-Note/383.Ransom Note.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func canConstruct(ransomNote string, magazine string) bool {
+	if len(ransomNote) > len(magazine) {
+		return false
+	}
+	var cnt [26]int
+	for _, v := range magazine {
+		cnt[v-'a']++
+	}
+	for _, v := range ransomNote {
+		cnt[v-'a']--
+		if cnt[v-'a'] < 0 {
+			return false
+		}
+	}
+	return true
+}
diff --git a/leetcode/0383.Ransom-Note/383.Ransom Note_test.go b/leetcode/0383.Ransom-Note/383.Ransom Note_test.go
new file mode 100644
index 000000000..b17147686
--- /dev/null
+++ b/leetcode/0383.Ransom-Note/383.Ransom Note_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question383 struct {
+	para383
+	ans383
+}
+
+// para 是参数
+type para383 struct {
+	ransomNote string
+	magazine   string
+}
+
+// ans 是答案
+type ans383 struct {
+	ans bool
+}
+
+func Test_Problem383(t *testing.T) {
+
+	qs := []question383{
+
+		{
+			para383{"a", "b"},
+			ans383{false},
+		},
+
+		{
+			para383{"aa", "ab"},
+			ans383{false},
+		},
+
+		{
+			para383{"aa", "aab"},
+			ans383{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 383------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans383, q.para383
+		fmt.Printf("【input】:%v    【output】:%v\n", p, canConstruct(p.ransomNote, p.magazine))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0383.Ransom-Note/README.md b/leetcode/0383.Ransom-Note/README.md
new file mode 100644
index 000000000..9fe9941c5
--- /dev/null
+++ b/leetcode/0383.Ransom-Note/README.md
@@ -0,0 +1,64 @@
+# [383. Ransom Note](https://leetcode.com/problems/ransom-note/)
+
+## 题目
+
+Given two stings ransomNote and magazine, return true if ransomNote can be constructed from magazine and false otherwise.
+
+Each letter in magazine can only be used once in ransomNote.
+
+**Example 1**:
+
+    Input: ransomNote = "a", magazine = "b"
+    Output: false
+
+**Example 2**:
+
+    Input: ransomNote = "aa", magazine = "ab"
+    Output: false
+
+**Example 3**:
+
+    Input: ransomNote = "aa", magazine = "aab"
+    Output: true
+
+**Constraints:**
+
+- 1 <= ransomNote.length, magazine.length <= 100000
+- ransomNote and magazine consist of lowercase English letters.
+
+## 题目大意
+
+为了不在赎金信中暴露字迹，从杂志上搜索各个需要的字母，组成单词来表达意思。
+
+给你一个赎金信 (ransomNote) 字符串和一个杂志(magazine)字符串，判断 ransomNote 能不能由 magazines 里面的字符构成。
+
+如果可以构成，返回 true ；否则返回 false 。
+
+magazine 中的每个字符只能在 ransomNote 中使用一次。
+
+## 解题思路
+
+- ransomNote 和 magazine 都是由小写字母组成，所以用数组进行简单的字符统计
+
+## 代码
+
+````go
+package leetcode
+
+func canConstruct(ransomNote string, magazine string) bool {
+	if len(ransomNote) > len(magazine) {
+		return false
+	}
+	var cnt [26]int
+	for _, v := range magazine {
+		cnt[v-'a']++
+	}
+	for _, v := range ransomNote {
+		cnt[v-'a']--
+		if cnt[v-'a'] < 0 {
+			return false
+		}
+	}
+	return true
+}
+````
\ No newline at end of file
diff --git a/leetcode/0384.Shuffle-an-Array/384.Shuffle an Array.go b/leetcode/0384.Shuffle-an-Array/384.Shuffle an Array.go
new file mode 100644
index 000000000..541908424
--- /dev/null
+++ b/leetcode/0384.Shuffle-an-Array/384.Shuffle an Array.go	
@@ -0,0 +1,28 @@
+package leetcode
+
+import "math/rand"
+
+type Solution struct {
+	nums []int
+}
+
+func Constructor(nums []int) Solution {
+	return Solution{
+		nums: nums,
+	}
+}
+
+/** Resets the array to its original configuration and return it. */
+func (this *Solution) Reset() []int {
+	return this.nums
+}
+
+/** Returns a random shuffling of the array. */
+func (this *Solution) Shuffle() []int {
+	arr := make([]int, len(this.nums))
+	copy(arr, this.nums)
+	rand.Shuffle(len(arr), func(i, j int) {
+		arr[i], arr[j] = arr[j], arr[i]
+	})
+	return arr
+}
diff --git a/leetcode/0384.Shuffle-an-Array/384.Shuffle an Array_test.go b/leetcode/0384.Shuffle-an-Array/384.Shuffle an Array_test.go
new file mode 100644
index 000000000..9ecebf0f7
--- /dev/null
+++ b/leetcode/0384.Shuffle-an-Array/384.Shuffle an Array_test.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question384 struct {
+	para384
+	ans384
+}
+
+// para 是参数
+type para384 struct {
+	ops   []string
+	value [][]int
+}
+
+// ans 是答案
+type ans384 struct {
+	ans [][]int
+}
+
+func Test_Problem384(t *testing.T) {
+
+	qs := []question384{
+
+		{
+			para384{ops: []string{"Solution", "shuffle", "reset", "shuffle"}, value: [][]int{{1, 2, 3}, {}, {}, {}}},
+			ans384{[][]int{nil, {3, 1, 2}, {1, 2, 3}, {1, 3, 2}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 384------------------------\n")
+
+	for _, q := range qs {
+		sol := Constructor(nil)
+		_, p := q.ans384, q.para384
+		for _, op := range p.ops {
+			if op == "Solution" {
+				sol = Constructor(q.value[0])
+			} else if op == "reset" {
+				fmt.Printf("【input】:%v    【output】:%v\n", op, sol.Reset())
+			} else {
+				fmt.Printf("【input】:%v    【output】:%v\n", op, sol.Shuffle())
+			}
+		}
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0384.Shuffle-an-Array/README.md b/leetcode/0384.Shuffle-an-Array/README.md
new file mode 100644
index 000000000..df22193a8
--- /dev/null
+++ b/leetcode/0384.Shuffle-an-Array/README.md
@@ -0,0 +1,82 @@
+# [384.Shuffle an Array](https://leetcode.com/problems/shuffle-an-array/)
+
+## 题目
+
+Given an integer array nums, design an algorithm to randomly shuffle the array. All permutations of the array should be equally likely as a result of the shuffling.
+
+Implement the Solution class:
+
+- Solution(int[] nums) Initializes the object with the integer array nums.
+- int[] reset() Resets the array to its original configuration and returns it.
+- int[] shuffle() Returns a random shuffling of the array.
+
+**Example 1**:
+
+    Input
+    ["Solution", "shuffle", "reset", "shuffle"]
+    [[[1, 2, 3]], [], [], []]
+    Output
+    [null, [3, 1, 2], [1, 2, 3], [1, 3, 2]]
+
+    Explanation
+    Solution solution = new Solution([1, 2, 3]);
+    solution.shuffle();    // Shuffle the array [1,2,3] and return its result.
+    // Any permutation of [1,2,3] must be equally likely to be returned.
+    // Example: return [3, 1, 2]
+    solution.reset();      // Resets the array back to its original configuration [1,2,3]. Return [1, 2, 3]
+    solution.shuffle();    // Returns the random shuffling of array [1,2,3]. Example: return [1, 3, 2]
+
+**Constraints:**
+
+- 1 <= nums.length <= 200
+- -1000000 <= nums[i] <= 1000000
+- All the elements of nums are unique.
+- At most 5 * 10000 calls in total will be made to reset and shuffle.
+
+## 题目大意
+
+给你一个整数数组 nums ，设计算法来打乱一个没有重复元素的数组。
+
+实现 Solution class:
+
+- Solution(int[] nums) 使用整数数组 nums 初始化对象
+- int[] reset() 重设数组到它的初始状态并返回
+- int[] shuffle() 返回数组随机打乱后的结果
+
+## 解题思路
+
+- 使用 rand.Shuffle 进行数组随机打乱
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math/rand"
+
+type Solution struct {
+	nums []int
+}
+
+func Constructor(nums []int) Solution {
+	return Solution{
+		nums: nums,
+	}
+}
+
+/** Resets the array to its original configuration and return it. */
+func (this *Solution) Reset() []int {
+	return this.nums
+}
+
+/** Returns a random shuffling of the array. */
+func (this *Solution) Shuffle() []int {
+	arr := make([]int, len(this.nums))
+	copy(arr, this.nums)
+	rand.Shuffle(len(arr), func(i, j int) {
+		arr[i], arr[j] = arr[j], arr[i]
+	})
+	return arr
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0385. Mini Parser/385. Mini Parser.go b/leetcode/0385.Mini-Parser/385. Mini Parser.go
similarity index 96%
rename from Algorithms/0385. Mini Parser/385. Mini Parser.go
rename to leetcode/0385.Mini-Parser/385. Mini Parser.go
index ffbd1818d..fee25200b 100644
--- a/Algorithms/0385. Mini Parser/385. Mini Parser.go	
+++ b/leetcode/0385.Mini-Parser/385. Mini Parser.go	
@@ -115,3 +115,10 @@ func deserialize(s string) *NestedInteger {
 	}
 	return cur
 }
+
+func isDigital(v byte) bool {
+	if v >= '0' && v <= '9' {
+		return true
+	}
+	return false
+}
diff --git a/Algorithms/0385. Mini Parser/385. Mini Parser_test.go b/leetcode/0385.Mini-Parser/385. Mini Parser_test.go
similarity index 92%
rename from Algorithms/0385. Mini Parser/385. Mini Parser_test.go
rename to leetcode/0385.Mini-Parser/385. Mini Parser_test.go
index 8ca47fb5e..f3efe64d3 100644
--- a/Algorithms/0385. Mini Parser/385. Mini Parser_test.go	
+++ b/leetcode/0385.Mini-Parser/385. Mini Parser_test.go	
@@ -26,27 +26,27 @@ func Test_Problem385(t *testing.T) {
 
 	qs := []question385{
 
-		question385{
+		{
 			para385{"[[]]"},
 			ans385{[]int{}},
 		},
 
-		question385{
+		{
 			para385{"[]"},
 			ans385{[]int{}},
 		},
 
-		question385{
+		{
 			para385{"[-1]"},
 			ans385{[]int{-1}},
 		},
 
-		question385{
+		{
 			para385{"[123,[456,[789]]]"},
 			ans385{[]int{123, 456, 789}},
 		},
 
-		question385{
+		{
 			para385{"324"},
 			ans385{[]int{324}},
 		},
diff --git a/leetcode/0385.Mini-Parser/README.md b/leetcode/0385.Mini-Parser/README.md
new file mode 100755
index 000000000..a7705d4e7
--- /dev/null
+++ b/leetcode/0385.Mini-Parser/README.md
@@ -0,0 +1,50 @@
+# [385. Mini Parser](https://leetcode.com/problems/mini-parser/)
+
+
+## 题目
+
+Given a nested list of integers represented as a string, implement a parser to deserialize it.
+
+Each element is either an integer, or a list -- whose elements may also be integers or other lists.
+
+**Note:** You may assume that the string is well-formed:
+
+- String is non-empty.
+- String does not contain white spaces.
+- String contains only digits `0-9`, `[`, `-` `,`, `]`.
+
+**Example 1:**
+
+    Given s = "324",
+    
+    You should return a NestedInteger object which contains a single integer 324.
+
+**Example 2:**
+
+    Given s = "[123,[456,[789]]]",
+    
+    Return a NestedInteger object containing a nested list with 2 elements:
+    
+    1. An integer containing value 123.
+    2. A nested list containing two elements:
+        i.  An integer containing value 456.
+        ii. A nested list with one element:
+             a. An integer containing value 789.
+
+
+## 题目大意
+
+给定一个用字符串表示的整数的嵌套列表，实现一个解析它的语法分析器。列表中的每个元素只可能是整数或整数嵌套列表
+
+提示：你可以假定这些字符串都是格式良好的：
+
+- 字符串非空
+- 字符串不包含空格
+- 字符串只包含数字0-9, [, - ,, ]
+
+
+
+## 解题思路
+
+- 将一个嵌套的数据结构中的数字转换成 NestedInteger 数据结构。
+- 这一题用栈一层一层的处理就行。有一些比较坑的特殊的边界数据见测试文件。这一题正确率比很多 Hard 题还要低的原因应该是没有理解好题目和边界测试数据没有考虑到。NestedInteger 这个数据结构笔者实现了一遍，见代码。
diff --git a/Algorithms/0386. Lexicographical Numbers/386. Lexicographical Numbers.go b/leetcode/0386.Lexicographical-Numbers/386. Lexicographical Numbers.go
similarity index 100%
rename from Algorithms/0386. Lexicographical Numbers/386. Lexicographical Numbers.go
rename to leetcode/0386.Lexicographical-Numbers/386. Lexicographical Numbers.go
diff --git a/Algorithms/0386. Lexicographical Numbers/386. Lexicographical Numbers_test.go b/leetcode/0386.Lexicographical-Numbers/386. Lexicographical Numbers_test.go
similarity index 97%
rename from Algorithms/0386. Lexicographical Numbers/386. Lexicographical Numbers_test.go
rename to leetcode/0386.Lexicographical-Numbers/386. Lexicographical Numbers_test.go
index 4a72edef9..eb0d60fb1 100644
--- a/Algorithms/0386. Lexicographical Numbers/386. Lexicographical Numbers_test.go	
+++ b/leetcode/0386.Lexicographical-Numbers/386. Lexicographical Numbers_test.go	
@@ -26,7 +26,7 @@ func Test_Problem386(t *testing.T) {
 
 	qs := []question386{
 
-		question386{
+		{
 			para386{13},
 			ans386{[]int{1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9}},
 		},
diff --git a/leetcode/0386.Lexicographical-Numbers/README.md b/leetcode/0386.Lexicographical-Numbers/README.md
new file mode 100755
index 000000000..ff39b8bd5
--- /dev/null
+++ b/leetcode/0386.Lexicographical-Numbers/README.md
@@ -0,0 +1,25 @@
+# [386. Lexicographical Numbers](https://leetcode.com/problems/lexicographical-numbers/)
+
+
+## 题目
+
+Given an integer n, return 1 - n in lexicographical order.
+
+For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9].
+
+Please optimize your algorithm to use less time and space. The input size may be as large as 5,000,000.
+
+
+## 题目大意
+
+给定一个整数 n, 返回从 1 到 n 的字典顺序。例如，给定 n =13，返回 [1,10,11,12,13,2,3,4,5,6,7,8,9] 。
+
+请尽可能的优化算法的时间复杂度和空间复杂度。 输入的数据 n 小于等于 5,000,000。
+
+
+
+## 解题思路
+
+
+- 给出一个数字 n ，要求按照字典序对 1-n 这 n 个数排序。
+- DFS 暴力求解即可。
diff --git a/leetcode/0387.First-Unique-Character-in-a-String/387. First Unique Character in a String.go b/leetcode/0387.First-Unique-Character-in-a-String/387. First Unique Character in a String.go
new file mode 100644
index 000000000..1ce1fe898
--- /dev/null
+++ b/leetcode/0387.First-Unique-Character-in-a-String/387. First Unique Character in a String.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+// 解法一
+func firstUniqChar(s string) int {
+	result := make([]int, 26)
+	for i := 0; i < len(s); i++ {
+		result[s[i]-'a']++
+	}
+	for i := 0; i < len(s); i++ {
+		if result[s[i]-'a'] == 1 {
+			return i
+		}
+	}
+	return -1
+}
+
+// 解法二
+// 执行用时: 8 ms
+// 内存消耗: 5.2 MB
+func firstUniqChar1(s string) int {
+	charMap := make([][2]int, 26)
+	for i := 0; i < 26; i++ {
+		charMap[i][0] = -1
+		charMap[i][1] = -1
+	}
+	for i := 0; i < len(s); i++ {
+		if charMap[s[i]-'a'][0] == -1 {
+			charMap[s[i]-'a'][0] = i
+		} else { //已经出现过
+			charMap[s[i]-'a'][1] = i
+		}
+	}
+	res := len(s)
+	for i := 0; i < 26; i++ {
+		//只出现了一次
+		if charMap[i][0] >= 0 && charMap[i][1] == -1 {
+			if charMap[i][0] < res {
+				res = charMap[i][0]
+			}
+		}
+	}
+	if res == len(s) {
+		return -1
+	}
+	return res
+}
diff --git a/Algorithms/0387. First Unique Character in a String/387. First Unique Character in a String_test.go b/leetcode/0387.First-Unique-Character-in-a-String/387. First Unique Character in a String_test.go
similarity index 95%
rename from Algorithms/0387. First Unique Character in a String/387. First Unique Character in a String_test.go
rename to leetcode/0387.First-Unique-Character-in-a-String/387. First Unique Character in a String_test.go
index a400f081a..fd0b6d1f8 100644
--- a/Algorithms/0387. First Unique Character in a String/387. First Unique Character in a String_test.go	
+++ b/leetcode/0387.First-Unique-Character-in-a-String/387. First Unique Character in a String_test.go	
@@ -26,12 +26,12 @@ func Test_Problem387(t *testing.T) {
 
 	qs := []question387{
 
-		question387{
+		{
 			para387{"leetcode"},
 			ans387{0},
 		},
 
-		question387{
+		{
 			para387{"loveleetcode"},
 			ans387{2},
 		},
diff --git a/leetcode/0387.First-Unique-Character-in-a-String/README.md b/leetcode/0387.First-Unique-Character-in-a-String/README.md
new file mode 100755
index 000000000..f54515311
--- /dev/null
+++ b/leetcode/0387.First-Unique-Character-in-a-String/README.md
@@ -0,0 +1,26 @@
+# [387. First Unique Character in a String](https://leetcode.com/problems/first-unique-character-in-a-string/)
+
+## 题目
+
+Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.
+
+**Examples:**
+
+    s = "leetcode"
+    return 0.
+    
+    s = "loveleetcode",
+    return 2.
+
+**Note:** You may assume the string contain only lowercase letters.
+
+
+
+## 题目大意
+
+给定一个字符串，找到它的第一个不重复的字符，并返回它的索引。如果不存在，则返回 -1。
+
+
+## 解题思路
+
+- 简单题，要求输出第一个没有重复的字符。
diff --git a/Algorithms/0389. Find the Difference/389. Find the Difference.go b/leetcode/0389.Find-the-Difference/389. Find the Difference.go
similarity index 100%
rename from Algorithms/0389. Find the Difference/389. Find the Difference.go
rename to leetcode/0389.Find-the-Difference/389. Find the Difference.go
diff --git a/Algorithms/0389. Find the Difference/389. Find the Difference_test.go b/leetcode/0389.Find-the-Difference/389. Find the Difference_test.go
similarity index 97%
rename from Algorithms/0389. Find the Difference/389. Find the Difference_test.go
rename to leetcode/0389.Find-the-Difference/389. Find the Difference_test.go
index a42b722e9..6f31d7adb 100644
--- a/Algorithms/0389. Find the Difference/389. Find the Difference_test.go	
+++ b/leetcode/0389.Find-the-Difference/389. Find the Difference_test.go	
@@ -27,7 +27,7 @@ func Test_Problem389(t *testing.T) {
 
 	qs := []question389{
 
-		question389{
+		{
 			para389{"abcd", "abcde"},
 			ans389{'e'},
 		},
diff --git a/leetcode/0389.Find-the-Difference/README.md b/leetcode/0389.Find-the-Difference/README.md
new file mode 100755
index 000000000..2c5cee2a7
--- /dev/null
+++ b/leetcode/0389.Find-the-Difference/README.md
@@ -0,0 +1,30 @@
+# [389. Find the Difference](https://leetcode.com/problems/find-the-difference/)
+
+## 题目
+
+Given two strings **s** and **t** which consist of only lowercase letters.
+
+String **t** is generated by random shuffling string **s** and then add one more letter at a random position.
+
+Find the letter that was added in **t**.
+
+**Example:**
+
+    Input:
+    s = "abcd"
+    t = "abcde"
+    
+    Output:
+    e
+    
+    Explanation:
+    'e' is the letter that was added.
+
+## 题目大意
+
+给定两个字符串 s 和 t，它们只包含小写字母。字符串 t 由字符串 s 随机重排，然后在随机位置添加一个字母。请找出在 t 中被添加的字母。
+
+
+## 解题思路
+
+- 题目要求找出 t 字符串中比 s 字符串多出的一个字符。思路还是利用异或的性质，`X^X = 0`，将 s 和 t 依次异或，最终多出来的字符就是最后异或的结果。
diff --git a/leetcode/0390.Elimination-Game/390. Elimination Game.go b/leetcode/0390.Elimination-Game/390. Elimination Game.go
new file mode 100644
index 000000000..d741e387f
--- /dev/null
+++ b/leetcode/0390.Elimination-Game/390. Elimination Game.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func lastRemaining(n int) int {
+	start, dir, step := 1, true, 1
+	for n > 1 {
+		if dir { // 正向
+			start += step
+		} else { // 反向
+			if n%2 == 1 {
+				start += step
+			}
+		}
+		dir = !dir
+		n >>= 1
+		step <<= 1
+	}
+	return start
+}
diff --git a/leetcode/0390.Elimination-Game/390. Elimination Game_test.go b/leetcode/0390.Elimination-Game/390. Elimination Game_test.go
new file mode 100644
index 000000000..d7f0c4efd
--- /dev/null
+++ b/leetcode/0390.Elimination-Game/390. Elimination Game_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question390 struct {
+	para390
+	ans390
+}
+
+// para 是参数
+// one 代表第一个参数
+type para390 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans390 struct {
+	one int
+}
+
+func Test_Problem390(t *testing.T) {
+
+	qs := []question390{
+
+		{
+			para390{9},
+			ans390{6},
+		},
+
+		{
+			para390{1},
+			ans390{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 390------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans390, q.para390
+		fmt.Printf("【input】:%v       【output】:%v\n", p, lastRemaining(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0390.Elimination-Game/README.md b/leetcode/0390.Elimination-Game/README.md
new file mode 100644
index 000000000..91eea4eea
--- /dev/null
+++ b/leetcode/0390.Elimination-Game/README.md
@@ -0,0 +1,74 @@
+# [390. Elimination Game](https://leetcode.com/problems/elimination-game/)
+
+
+## 题目
+
+You have a list `arr` of all integers in the range `[1, n]` sorted in a strictly increasing order. Apply the following algorithm on `arr`:
+
+- Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.
+- Repeat the previous step again, but this time from right to left, remove the rightmost number and every other number from the remaining numbers.
+- Keep repeating the steps again, alternating left to right and right to left, until a single number remains.
+
+Given the integer `n`, return *the last number that remains in* `arr`.
+
+**Example 1:**
+
+```
+Input: n = 9
+Output: 6
+Explanation:
+arr = [1, 2,3, 4,5, 6,7, 8,9]
+arr = [2,4, 6,8]
+arr = [2, 6]
+arr = [6]
+
+```
+
+**Example 2:**
+
+```
+Input: n = 1
+Output: 1
+
+```
+
+**Constraints:**
+
+- `1 <= n <= 109`
+
+## 题目大意
+
+列表 arr 由在范围 [1, n] 中的所有整数组成，并按严格递增排序。请你对 arr 应用下述算法：
+
+- 从左到右，删除第一个数字，然后每隔一个数字删除一个，直到到达列表末尾。
+- 重复上面的步骤，但这次是从右到左。也就是，删除最右侧的数字，然后剩下的数字每隔一个删除一个。
+- 不断重复这两步，从左到右和从右到左交替进行，直到只剩下一个数字。
+
+给你整数 n ，返回 arr 最后剩下的数字。
+
+## 解题思路
+
+- 模拟题。按照题意，第一轮从左往右删除数字，第二轮从右往左删除数字。题目要求最后剩下的数字，模拟过程中不需要真的删除元素。只需要标记起始元素，该轮步长和方向即可。最后总元素只剩下一个即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+func lastRemaining(n int) int {
+	start, dir, step := 1, true, 1
+	for n > 1 {
+		if dir { // 正向
+			start += step
+		} else { // 反向
+			if n%2 == 1 {
+				start += step
+			}
+		}
+		dir = !dir
+		n >>= 1
+		step <<= 1
+	}
+	return start
+}
+```
\ No newline at end of file
diff --git a/leetcode/0391.Perfect-Rectangle/391.Perfect Rectangle.go b/leetcode/0391.Perfect-Rectangle/391.Perfect Rectangle.go
new file mode 100644
index 000000000..0d30790e0
--- /dev/null
+++ b/leetcode/0391.Perfect-Rectangle/391.Perfect Rectangle.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+type point struct {
+	x int
+	y int
+}
+
+func isRectangleCover(rectangles [][]int) bool {
+	minX, minY, maxA, maxB := rectangles[0][0], rectangles[0][1], rectangles[0][2], rectangles[0][3]
+	area := 0
+	cnt := make(map[point]int)
+	for _, v := range rectangles {
+		x, y, a, b := v[0], v[1], v[2], v[3]
+		area += (a - x) * (b - y)
+		minX = min(minX, x)
+		minY = min(minY, y)
+		maxA = max(maxA, a)
+		maxB = max(maxB, b)
+		cnt[point{x, y}]++
+		cnt[point{a, b}]++
+		cnt[point{x, b}]++
+		cnt[point{a, y}]++
+	}
+	if area != (maxA-minX)*(maxB-minY) ||
+		cnt[point{minX, minY}] != 1 || cnt[point{maxA, maxB}] != 1 ||
+		cnt[point{minX, maxB}] != 1 || cnt[point{maxA, minY}] != 1 {
+		return false
+	}
+	delete(cnt, point{minX, minY})
+	delete(cnt, point{maxA, maxB})
+	delete(cnt, point{minX, maxB})
+	delete(cnt, point{maxA, minY})
+	for _, v := range cnt {
+		if v != 2 && v != 4 {
+			return false
+		}
+	}
+	return true
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0391.Perfect-Rectangle/391.Perfect Rectangle_test.go b/leetcode/0391.Perfect-Rectangle/391.Perfect Rectangle_test.go
new file mode 100644
index 000000000..0efa7efa7
--- /dev/null
+++ b/leetcode/0391.Perfect-Rectangle/391.Perfect Rectangle_test.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question391 struct {
+	para391
+	ans391
+}
+
+// para 是参数
+type para391 struct {
+	rectangles [][]int
+}
+
+// ans 是答案
+type ans391 struct {
+	ans bool
+}
+
+func Test_Problem391(t *testing.T) {
+
+	qs := []question391{
+
+		{
+			para391{[][]int{{1, 1, 3, 3}, {3, 1, 4, 2}, {3, 2, 4, 4}, {1, 3, 2, 4}, {2, 3, 3, 4}}},
+			ans391{true},
+		},
+
+		{
+			para391{[][]int{{1, 1, 2, 3}, {1, 3, 2, 4}, {3, 1, 4, 2}, {3, 2, 4, 4}}},
+			ans391{false},
+		},
+
+		{
+			para391{[][]int{{1, 1, 3, 3}, {3, 1, 4, 2}, {1, 3, 2, 4}, {3, 2, 4, 4}}},
+			ans391{false},
+		},
+
+		{
+			para391{[][]int{{1, 1, 3, 3}, {3, 1, 4, 2}, {1, 3, 2, 4}, {2, 2, 4, 4}}},
+			ans391{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 391------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans391, q.para391
+		fmt.Printf("【input】:%v       【output】:%v\n", p, isRectangleCover(p.rectangles))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0391.Perfect-Rectangle/README.md b/leetcode/0391.Perfect-Rectangle/README.md
new file mode 100644
index 000000000..9a66b3b22
--- /dev/null
+++ b/leetcode/0391.Perfect-Rectangle/README.md
@@ -0,0 +1,114 @@
+# [391. Perfect Rectangle](https://leetcode.com/problems/perfect-rectangle/)
+
+## 题目
+
+Given an array rectangles where rectangles[i] = [xi, yi, ai, bi] represents an axis-aligned rectangle. The bottom-left point of the rectangle is (xi, yi) and the top-right point of it is (ai, bi).
+
+Return true if all the rectangles together form an exact cover of a rectangular region.
+
+**Example1:**
+
+![https://assets.leetcode.com/uploads/2021/03/27/perectrec1-plane.jpg](https://assets.leetcode.com/uploads/2021/03/27/perectrec1-plane.jpg)
+
+    Input: rectangles = [[1,1,3,3],[3,1,4,2],[3,2,4,4],[1,3,2,4],[2,3,3,4]]
+    Output: true
+    Explanation: All 5 rectangles together form an exact cover of a rectangular region.
+
+**Example2:**
+
+![https://assets.leetcode.com/uploads/2021/03/27/perfectrec2-plane.jpg](https://assets.leetcode.com/uploads/2021/03/27/perfectrec2-plane.jpg)
+
+    Input: rectangles = [[1,1,2,3],[1,3,2,4],[3,1,4,2],[3,2,4,4]]
+    Output: false
+    Explanation: Because there is a gap between the two rectangular regions.
+
+**Example3:**
+
+![https://assets.leetcode.com/uploads/2021/03/27/perfectrec3-plane.jpg](https://assets.leetcode.com/uploads/2021/03/27/perfectrec3-plane.jpg)
+
+    Input: rectangles = [[1,1,3,3],[3,1,4,2],[1,3,2,4],[3,2,4,4]]
+    Output: false
+    Explanation: Because there is a gap in the top center.
+
+**Example4:**
+
+![https://assets.leetcode.com/uploads/2021/03/27/perfecrrec4-plane.jpg](https://assets.leetcode.com/uploads/2021/03/27/perfecrrec4-plane.jpg)
+
+    Input: rectangles = [[1,1,3,3],[3,1,4,2],[1,3,2,4],[2,2,4,4]]
+    Output: false
+    Explanation: Because two of the rectangles overlap with each other.
+
+**Constraints:**
+
+- 1 <= rectangles.length <= 2 * 10000
+- rectangles[i].length == 4
+- -100000 <= xi, yi, ai, bi <= 100000
+
+## 题目大意
+
+给你一个数组 rectangles ，其中 rectangles[i] = [xi, yi, ai, bi] 表示一个坐标轴平行的矩形。这个矩形的左下顶点是 (xi, yi) ，右上顶点是 (ai, bi) 。
+
+如果所有矩形一起精确覆盖了某个矩形区域，则返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 矩形区域的面积等于所有矩形的面积之和并且满足矩形区域四角的顶点只能出现一次，且其余顶点的出现次数只能是两次或四次则返回 true,否则返回 false
+
+## 代码
+
+```go
+
+package leetcode
+
+type point struct {
+	x int
+	y int
+}
+
+func isRectangleCover(rectangles [][]int) bool {
+	minX, minY, maxA, maxB := rectangles[0][0], rectangles[0][1], rectangles[0][2], rectangles[0][3]
+	area := 0
+	cnt := make(map[point]int)
+	for _, v := range rectangles {
+		x, y, a, b := v[0], v[1], v[2], v[3]
+		area += (a - x) * (b - y)
+		minX = min(minX, x)
+		minY = min(minY, y)
+		maxA = max(maxA, a)
+		maxB = max(maxB, b)
+		cnt[point{x, y}]++
+		cnt[point{a, b}]++
+		cnt[point{x, b}]++
+		cnt[point{a, y}]++
+	}
+	if area != (maxA - minX) * (maxB - minY) ||
+		cnt[point{minX, minY}] != 1 || cnt[point{maxA, maxB}] != 1 ||
+		cnt[point{minX, maxB}] != 1 || cnt[point{maxA, minY}] != 1 {
+		return false
+	}
+	delete(cnt, point{minX, minY})
+	delete(cnt, point{maxA, maxB})
+	delete(cnt, point{minX, maxB})
+	delete(cnt, point{maxA, minY})
+	for _, v := range cnt {
+		if v != 2 && v != 4 {
+			return false
+		}
+	}
+	return true
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0392. Is Subsequence/392. Is Subsequence.go b/leetcode/0392.Is-Subsequence/392. Is Subsequence.go
similarity index 92%
rename from Algorithms/0392. Is Subsequence/392. Is Subsequence.go
rename to leetcode/0392.Is-Subsequence/392. Is Subsequence.go
index 0c9027b92..2098cb93b 100644
--- a/Algorithms/0392. Is Subsequence/392. Is Subsequence.go	
+++ b/leetcode/0392.Is-Subsequence/392. Is Subsequence.go	
@@ -2,9 +2,9 @@ package leetcode
 
 // 解法一 O(n^2)
 func isSubsequence(s string, t string) bool {
-	index, flag := 0, false
+	index := 0
 	for i := 0; i < len(s); i++ {
-		flag = false
+		flag := false
 		for ; index < len(t); index++ {
 			if s[i] == t[index] {
 				flag = true
diff --git a/Algorithms/0392. Is Subsequence/392. Is Subsequence_test.go b/leetcode/0392.Is-Subsequence/392. Is Subsequence_test.go
similarity index 99%
rename from Algorithms/0392. Is Subsequence/392. Is Subsequence_test.go
rename to leetcode/0392.Is-Subsequence/392. Is Subsequence_test.go
index 8e2c023bf..8c565e661 100644
--- a/Algorithms/0392. Is Subsequence/392. Is Subsequence_test.go	
+++ b/leetcode/0392.Is-Subsequence/392. Is Subsequence_test.go	
@@ -27,22 +27,22 @@ func Test_Problem392(t *testing.T) {
 
 	qs := []question392{
 
-		question392{
+		{
 			para392{"abc", "ahbgdc"},
 			ans392{true},
 		},
 
-		question392{
+		{
 			para392{"axc", "ahbgdc"},
 			ans392{false},
 		},
 
-		question392{
+		{
 			para392{"acb", "ahbgdc"},
 			ans392{false},
 		},
 
-		question392{
+		{
 			para392{"leeeeetcode", "yyyyylyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyeyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyeyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyeyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy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y"},
 			ans392{false},
 		},
diff --git a/Algorithms/0392. Is Subsequence/README.md b/leetcode/0392.Is-Subsequence/README.md
similarity index 100%
rename from Algorithms/0392. Is Subsequence/README.md
rename to leetcode/0392.Is-Subsequence/README.md
diff --git a/Algorithms/0393. UTF-8 Validation/393. UTF-8 Validation.go b/leetcode/0393.UTF-8-Validation/393. UTF-8 Validation.go
similarity index 100%
rename from Algorithms/0393. UTF-8 Validation/393. UTF-8 Validation.go
rename to leetcode/0393.UTF-8-Validation/393. UTF-8 Validation.go
diff --git a/Algorithms/0393. UTF-8 Validation/393. UTF-8 Validation_test.go b/leetcode/0393.UTF-8-Validation/393. UTF-8 Validation_test.go
similarity index 95%
rename from Algorithms/0393. UTF-8 Validation/393. UTF-8 Validation_test.go
rename to leetcode/0393.UTF-8-Validation/393. UTF-8 Validation_test.go
index 9700becc4..25ba67e63 100644
--- a/Algorithms/0393. UTF-8 Validation/393. UTF-8 Validation_test.go	
+++ b/leetcode/0393.UTF-8-Validation/393. UTF-8 Validation_test.go	
@@ -26,12 +26,12 @@ func Test_Problem393(t *testing.T) {
 
 	qs := []question393{
 
-		question393{
+		{
 			para393{[]int{197, 130, 1}},
 			ans393{true},
 		},
 
-		question393{
+		{
 			para393{[]int{235, 140, 4}},
 			ans393{false},
 		},
diff --git a/leetcode/0393.UTF-8-Validation/README.md b/leetcode/0393.UTF-8-Validation/README.md
new file mode 100755
index 000000000..9f859f285
--- /dev/null
+++ b/leetcode/0393.UTF-8-Validation/README.md
@@ -0,0 +1,71 @@
+# [393. UTF-8 Validation](https://leetcode.com/problems/utf-8-validation/)
+
+
+
+## 题目
+
+A character in UTF8 can be from **1 to 4 bytes** long, subjected to the following rules:
+
+1. For 1-byte character, the first bit is a 0, followed by its unicode code.
+2. For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
+
+This is how the UTF-8 encoding would work:
+
+       Char. number range  |        UTF-8 octet sequence
+          (hexadecimal)    |              (binary)
+       --------------------+---------------------------------------------
+       0000 0000-0000 007F | 0xxxxxxx
+       0000 0080-0000 07FF | 110xxxxx 10xxxxxx
+       0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
+       0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
+
+Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
+
+**Note:**The input is an array of integers. Only the **least significant 8 bits** of each integer is used to store the data. This means each integer represents only 1 byte of data.
+
+**Example 1:**
+
+    data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
+    
+    Return true.
+    It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
+
+**Example 2:**
+
+    data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
+    
+    Return false.
+    The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
+    The next byte is a continuation byte which starts with 10 and that's correct.
+    But the second continuation byte does not start with 10, so it is invalid.
+
+## 题目大意
+
+UTF-8 中的一个字符可能的长度为 1 到 4 字节，遵循以下的规则：
+
+对于 1 字节的字符，字节的第一位设为 0，后面 7 位为这个符号的 unicode 码。
+对于 n 字节的字符 (n > 1)，第一个字节的前 n 位都设为 1，第 n+1 位设为 0，后面字节的前两位一律设为 10。剩下的没有提及的二进制位，全部为这个符号的 unicode 码。
+这是 UTF-8 编码的工作方式：
+
+```c
+   Char. number range  |        UTF-8 octet sequence
+      (hexadecimal)    |              (binary)
+   --------------------+---------------------------------------------
+   0000 0000-0000 007F | 0xxxxxxx
+   0000 0080-0000 07FF | 110xxxxx 10xxxxxx
+   0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
+   0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
+
+```
+
+给定一个表示数据的整数数组，返回它是否为有效的 utf-8 编码。
+
+注意:
+
+输入是整数数组。只有每个整数的最低 8 个有效位用来存储数据。这意味着每个整数只表示 1 字节的数据。
+
+
+## 解题思路
+
+- 这一题看似很复杂，其实严格按照 UTF8 定义来模拟就可以了。
+
diff --git a/Algorithms/0394. Decode String/394. Decode String.go b/leetcode/0394.Decode-String/394. Decode String.go
similarity index 100%
rename from Algorithms/0394. Decode String/394. Decode String.go
rename to leetcode/0394.Decode-String/394. Decode String.go
diff --git a/Algorithms/0394. Decode String/394. Decode String_test.go b/leetcode/0394.Decode-String/394. Decode String_test.go
similarity index 92%
rename from Algorithms/0394. Decode String/394. Decode String_test.go
rename to leetcode/0394.Decode-String/394. Decode String_test.go
index 50846ccc6..b7434e462 100644
--- a/Algorithms/0394. Decode String/394. Decode String_test.go	
+++ b/leetcode/0394.Decode-String/394. Decode String_test.go	
@@ -26,22 +26,22 @@ func Test_Problem394(t *testing.T) {
 
 	qs := []question394{
 
-		question394{
+		{
 			para394{"10[a]"},
 			ans394{"aaaaaaaaaa"},
 		},
 
-		question394{
+		{
 			para394{"3[a]2[bc]"},
 			ans394{"aaabcbc"},
 		},
 
-		question394{
+		{
 			para394{"3[a2[c]]"},
 			ans394{"accaccacc"},
 		},
 
-		question394{
+		{
 			para394{"2[abc]3[cd]ef"},
 			ans394{"abcabccdcdcdef"},
 		},
diff --git a/Algorithms/0394. Decode String/README.md b/leetcode/0394.Decode-String/README.md
similarity index 100%
rename from Algorithms/0394. Decode String/README.md
rename to leetcode/0394.Decode-String/README.md
diff --git a/leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/395. Longest Substring with At Least K Repeating Characters.go b/leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/395. Longest Substring with At Least K Repeating Characters.go
new file mode 100644
index 000000000..079a6bcf2
--- /dev/null
+++ b/leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/395. Longest Substring with At Least K Repeating Characters.go	
@@ -0,0 +1,70 @@
+package leetcode
+
+import "strings"
+
+// 解法一 滑动窗口
+func longestSubstring(s string, k int) int {
+	res := 0
+	for t := 1; t <= 26; t++ {
+		freq, total, lessK, left, right := [26]int{}, 0, 0, 0, -1
+		for left < len(s) {
+			if right+1 < len(s) && total <= t {
+				if freq[s[right+1]-'a'] == 0 {
+					total++
+					lessK++
+				}
+				freq[s[right+1]-'a']++
+				if freq[s[right+1]-'a'] == k {
+					lessK--
+				}
+				right++
+			} else {
+				if freq[s[left]-'a'] == k {
+					lessK++
+				}
+				freq[s[left]-'a']--
+				if freq[s[left]-'a'] == 0 {
+					total--
+					lessK--
+				}
+				left++
+			}
+			if lessK == 0 {
+				res = max(res, right-left+1)
+			}
+
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法二 递归分治
+func longestSubstring1(s string, k int) int {
+	if s == "" {
+		return 0
+	}
+	freq, split, res := [26]int{}, byte(0), 0
+	for _, ch := range s {
+		freq[ch-'a']++
+	}
+	for i, c := range freq[:] {
+		if 0 < c && c < k {
+			split = 'a' + byte(i)
+			break
+		}
+	}
+	if split == 0 {
+		return len(s)
+	}
+	for _, subStr := range strings.Split(s, string(split)) {
+		res = max(res, longestSubstring1(subStr, k))
+	}
+	return res
+}
diff --git a/leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/395. Longest Substring with At Least K Repeating Characters_test.go b/leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/395. Longest Substring with At Least K Repeating Characters_test.go
new file mode 100644
index 000000000..3d5179e3a
--- /dev/null
+++ b/leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/395. Longest Substring with At Least K Repeating Characters_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question395 struct {
+	para395
+	ans395
+}
+
+// para 是参数
+// one 代表第一个参数
+type para395 struct {
+	s string
+	k int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans395 struct {
+	one int
+}
+
+func Test_Problem395(t *testing.T) {
+
+	qs := []question395{
+
+		{
+			para395{"aaabb", 3},
+			ans395{3},
+		},
+
+		{
+			para395{"ababbc", 2},
+			ans395{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 395------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans395, q.para395
+		fmt.Printf("【input】:%v       【output】:%v\n", p, longestSubstring(p.s, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/README.md b/leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/README.md
new file mode 100644
index 000000000..cd84eba52
--- /dev/null
+++ b/leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/README.md
@@ -0,0 +1,112 @@
+# [395. Longest Substring with At Least K Repeating Characters](https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/)
+
+
+## 题目
+
+Given a string `s` and an integer `k`, return *the length of the longest substring of* `s` *such that the frequency of each character in this substring is greater than or equal to* `k`.
+
+**Example 1:**
+
+```
+Input: s = "aaabb", k = 3
+Output: 3
+Explanation: The longest substring is "aaa", as 'a' is repeated 3 times.
+```
+
+**Example 2:**
+
+```
+Input: s = "ababbc", k = 2
+Output: 5
+Explanation: The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 10^4`
+- `s` consists of only lowercase English letters.
+- `1 <= k <= 10^5`
+
+## 题目大意
+
+给你一个字符串 s 和一个整数 k ，请你找出 s 中的最长子串， 要求该子串中的每一字符出现次数都不少于 k 。返回这一子串的长度。
+
+## 解题思路
+
+- 最容易想到的思路是递归。如果某个字符出现次数大于 0 小于 k，那么包含这个字符的子串都不满足要求。所以按照这个字符来切分整个字符串，满足题意的最长子串一定不包含切分的字符。切分完取出最长子串即可。时间复杂度 O(26*n)，空间复杂度 O(26^2)
+- 此题另外一个思路是滑动窗口。有一个需要解决的问题是右窗口移动的条件。此题要求最长字符串，那么这个最终的字符串内包含的字符种类最多是 26 种。字符种类就是右窗口移动的条件。依次枚举字符种类，如果当前窗口内的字符种类小于当前枚举的字符种类，那么窗口右移，否则左移。窗口移动中需要动态维护 freq 频次数组。可以每次都循环一遍这个数组，计算出出现次数大于 k 的字符。虽然这种做法只最多循环 26 次，但是还是不高效。更高效的做法是维护 1 个值，一个用来记录当前出现次数小于 k 次的字符种类数 `less`。如果 freq 为 0 ，说明小于 k 次的字符种类数要发生变化，如果是右窗口移动，那么 `less++`，如果是左窗口移动，那么`less--`。同理，如果 freq 为 k ，说明小于 k 次的字符种类数要发生变化，如果是右窗口移动，那么 `less--`，如果是左窗口移动，那么`less++`。在枚举 26 个字符种类中，动态维护记录出最长字符串。枚举完成，最长字符串长度也就求出来了。时间复杂度 O(26*n)，空间复杂度 O(26)
+
+## 代码
+
+```go
+package leetcode
+
+import "strings"
+
+// 解法一 滑动窗口
+func longestSubstring(s string, k int) int {
+	res := 0
+	for t := 1; t <= 26; t++ {
+		freq, total, lessK, left, right := [26]int{}, 0, 0, 0, -1
+		for left < len(s) {
+			if right+1 < len(s) && total <= t {
+				if freq[s[right+1]-'a'] == 0 {
+					total++
+					lessK++
+				}
+				freq[s[right+1]-'a']++
+				if freq[s[right+1]-'a'] == k {
+					lessK--
+				}
+				right++
+			} else {
+				if freq[s[left]-'a'] == k {
+					lessK++
+				}
+				freq[s[left]-'a']--
+				if freq[s[left]-'a'] == 0 {
+					total--
+					lessK--
+				}
+				left++
+			}
+			if lessK == 0 {
+				res = max(res, right-left+1)
+			}
+
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法二 递归分治
+func longestSubstring1(s string, k int) int {
+	if s == "" {
+		return 0
+	}
+	freq, split, res := [26]int{}, byte(0), 0
+	for _, ch := range s {
+		freq[ch-'a']++
+	}
+	for i, c := range freq[:] {
+		if 0 < c && c < k {
+			split = 'a' + byte(i)
+			break
+		}
+	}
+	if split == 0 {
+		return len(s)
+	}
+	for _, subStr := range strings.Split(s, string(split)) {
+		res = max(res, longestSubstring1(subStr, k))
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0396.Rotate-Function/396. Rotate Function.go b/leetcode/0396.Rotate-Function/396. Rotate Function.go
new file mode 100644
index 000000000..aecc5ae1c
--- /dev/null
+++ b/leetcode/0396.Rotate-Function/396. Rotate Function.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func maxRotateFunction(nums []int) int {
+	n := len(nums)
+	var sum, f int
+	for i, num := range nums {
+		sum += num
+		f += i * num // F(0)
+	}
+	ans := f
+	for i := 1; i < n; i++ {
+		f += sum - n*nums[n-i] // F(i) = F(i-1) + sum - n*nums[n-i]
+		if f > ans {
+			ans = f
+		}
+	}
+	return ans
+}
diff --git a/leetcode/0396.Rotate-Function/396. Rotate Function_test.go b/leetcode/0396.Rotate-Function/396. Rotate Function_test.go
new file mode 100644
index 000000000..855323a15
--- /dev/null
+++ b/leetcode/0396.Rotate-Function/396. Rotate Function_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question396 struct {
+	para396
+	ans396
+}
+
+// para 是参数
+// one 代表第一个参数
+type para396 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans396 struct {
+	one int
+}
+
+func Test_Problem396(t *testing.T) {
+
+	qs := []question396{
+		{
+			para396{[]int{4, 3, 2, 6}},
+			ans396{26},
+		},
+
+		{
+			para396{[]int{100}},
+			ans396{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 396------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans396, q.para396
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxRotateFunction(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0396.Rotate-Function/README.md b/leetcode/0396.Rotate-Function/README.md
new file mode 100644
index 000000000..ed1ab9c5a
--- /dev/null
+++ b/leetcode/0396.Rotate-Function/README.md
@@ -0,0 +1,110 @@
+# [396. Rotate Function](https://leetcode.com/problems/rotate-function/)
+
+## 题目
+
+You are given an integer array `nums` of length `n`.
+
+Assume `arrk` to be an array obtained by rotating `nums` by `k` positions clock-wise. We define the **rotation function** `F` on `nums` as follow:
+
+- `F(k) = 0 * arrk[0] + 1 * arrk[1] + ... + (n - 1) * arrk[n - 1]`.
+
+Return the maximum value of `F(0), F(1), ..., F(n-1)`.
+
+The test cases are generated so that the answer fits in a **32-bit** integer.
+
+**Example 1:**
+
+```c
+Input: nums = [4,3,2,6]
+Output: 26
+Explanation:
+F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
+F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
+F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
+F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
+So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.
+```
+
+**Example 2:**
+
+```c
+Input: nums = [100]
+Output: 0
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `1 <= n <= 105`
+- `-100 <= nums[i] <= 100`
+
+## 题目大意
+
+给定一个长度为`n`的整数数组`nums`，设`arrk`是数组`nums`顺时针旋转`k`个位置后的数组。
+
+定义`nums`的旋转函数`F`为：
+
+- `F(k) = 0 * arrk[0] + 1 * arrk[1] + ... + (n - 1) * arrk[n - 1]`
+
+返回`F(0), F(1), ..., F(n-1)`中的最大值。
+
+## 解题思路
+
+**抽象化观察：**
+
+```c
+nums = [A0, A1, A2, A3]
+
+sum  = A0 + A1 + A2+ A3
+F(0) = 0*A0 +0*A0 + 1*A1 + 2*A2 + 3*A3
+    
+F(1) = 0*A3 + 1*A0 + 2*A1 + 3*A2 
+     = F(0) + (A0 + A1 + A2) - 3*A3 
+     = F(0) + (sum-A3) - 3*A3 
+     = F(0) + sum - 4*A3
+
+F(2) = 0*A2 + 1*A3 + 2*A0 + 3*A1 
+     = F(1) + A3 + A0 + A1 - 3*A2 
+     = F(1) + sum - 4*A2
+
+F(3) = 0*A1 + 1*A2 + 2*A3 + 3*A0 
+     = F(2) + A2 + A3 + A0 - 3*A1 
+     = F(2) + sum - 4*A1
+    
+// 记sum为nums数组中所有元素和
+// 可以猜测当0 ≤ i < n时存在公式:
+F(i) = F(i-1) + sum - n * A(n-i)
+```
+
+**数学归纳法证明迭代公式：**
+
+根据题目中给定的旋转函数公式可得已知条件：
+
+- `F(0) = 0×nums[0] + 1×nums[1] + ... + (n−1)×nums[n−1]`；
+
+- `F(1) = 1×nums[0] + 2×nums[1] + ... + 0×nums[n-1]`。
+
+令数组`nums`中所有元素和为`sum`，用数学归纳法验证：当`1 ≤ k < n`时，`F(k) = F(k-1) + sum - n×nums[n-k]`成立。
+
+**归纳奠基**：证明`k=1`时命题成立。
+
+```c
+F(1) = 1×nums[0] + 2×nums[1] + ... + 0×nums[n-1]
+     = F(0) + sum - n×nums[n-1]
+```
+
+**归纳假设**：假设`F(k) = F(k-1) + sum - n×nums[n-k]`成立。
+
+**归纳递推**：由归纳假设推出`F(k+1) = F(k) + sum - n×nums[n-(k+1)]`成立，则假设的递推公式成立。
+
+```c
+F(k+1) = (k+1)×nums[0] + k×nums[1] + ... + 0×nums[n-1]
+       = F(k) + sum - n×nums[n-(k+1)] 
+```
+
+因此可以得到递推公式：
+
+- 当`n = 0`时，`F(0) = 0×nums[0] + 1×nums[1] + ... + (n−1)×nums[n−1]`
+- 当`1 ≤ k < n`时，`F(k) = F(k-1) + sum - n×nums[n-k]`成立。
+
+循环遍历`0 ≤ k < n`，计算出不同的`F(k)`并不断更新最大值，就能求出`F(0), F(1), ..., F(n-1)`中的最大值。
diff --git a/Algorithms/0397. Integer Replacement/397. Integer Replacement.go b/leetcode/0397.Integer-Replacement/397. Integer Replacement.go
similarity index 100%
rename from Algorithms/0397. Integer Replacement/397. Integer Replacement.go
rename to leetcode/0397.Integer-Replacement/397. Integer Replacement.go
diff --git a/Algorithms/0397. Integer Replacement/397. Integer Replacement_test.go b/leetcode/0397.Integer-Replacement/397. Integer Replacement_test.go
similarity index 95%
rename from Algorithms/0397. Integer Replacement/397. Integer Replacement_test.go
rename to leetcode/0397.Integer-Replacement/397. Integer Replacement_test.go
index ffdab849e..dc9ab7d3d 100644
--- a/Algorithms/0397. Integer Replacement/397. Integer Replacement_test.go	
+++ b/leetcode/0397.Integer-Replacement/397. Integer Replacement_test.go	
@@ -26,12 +26,12 @@ func Test_Problem397(t *testing.T) {
 
 	qs := []question397{
 
-		question397{
+		{
 			para397{8},
 			ans397{3},
 		},
 
-		question397{
+		{
 			para397{7},
 			ans397{4},
 		},
diff --git a/leetcode/0397.Integer-Replacement/README.md b/leetcode/0397.Integer-Replacement/README.md
new file mode 100755
index 000000000..953cbfc81
--- /dev/null
+++ b/leetcode/0397.Integer-Replacement/README.md
@@ -0,0 +1,62 @@
+# [397. Integer Replacement](https://leetcode.com/problems/integer-replacement/)
+
+
+## 题目
+
+Given a positive integer n and you can do operations as follow:
+
+1. If  n is even, replace  n with `n/2`.
+2. If  n is odd, you can replace n with either `n + 1` or `n - 1`.
+
+What is the minimum number of replacements needed for n to become 1?
+
+**Example 1:**
+
+    Input:
+    8
+    
+    Output:
+    3
+    
+    Explanation:
+    8 -> 4 -> 2 -> 1
+
+**Example 2:**
+
+    Input:
+    7
+    
+    Output:
+    4
+    
+    Explanation:
+    7 -> 8 -> 4 -> 2 -> 1
+    or
+    7 -> 6 -> 3 -> 2 -> 1
+
+
+## 题目大意
+
+给定一个正整数 n，你可以做如下操作：
+
+1. 如果 n 是偶数，则用 n / 2 替换 n。
+2. 如果 n 是奇数，则可以用 n + 1 或 n - 1 替换 n。
+
+问 n 变为 1 所需的最小替换次数是多少？
+
+
+## 解题思路
+
+
+- 题目给出一个整数 `n`，然后让我们通过变换将它为 1，如果 `n` 是偶数，可以直接变为 `n/2`，如果是奇数，可以先 `n+1` 或 `n-1`，问最终变为 1 的最少步骤。
+- 当 n 为奇数的时候，什么时候需要加 1 ，什么时候需要减 1 ，通过观察规律可以发现，除了 3 和 7 以外，所有加 1 就变成 4 的倍数的奇数，都适合先加 1 运算，比如 15:
+
+        15 -> 16 -> 8 -> 4 -> 2 -> 1
+        15 -> 14 -> 7 -> 6 -> 3 -> 2 -> 1
+        
+        111011 -> 111010 -> 11101 -> 11100 -> 1110 -> 111 -> 1000 -> 100 -> 10 -> 1
+        111011 -> 111100 -> 11110 -> 1111 -> 10000 -> 1000 -> 100 -> 10 -> 1
+
+- 对于 7 来说，加 1 和减 1 的结果相同，可以不用管，对于 3 来说，减 1 的步骤更少，所以需要先去掉这种特殊情况。
+- 最后如何判断某个数字加 1 后是 4 的倍数呢？这里有一个小技巧，由于之前判断了其是奇数了，那么最右边一位肯定是 1，如果其右边第二位也是 1 的话，那么进行加 1 运算，进位后右边肯定会出现两个 0，则一定是 4 的倍数。于是就可以判断出来了。剩下的情况就是偶数的情况，如果之前判定是偶数，那么直接除以 2 (右移一位)即可。
+
diff --git a/Algorithms/0399. Evaluate Division/399. Evaluate Division.go b/leetcode/0399.Evaluate-Division/399. Evaluate Division.go
similarity index 100%
rename from Algorithms/0399. Evaluate Division/399. Evaluate Division.go
rename to leetcode/0399.Evaluate-Division/399. Evaluate Division.go
diff --git a/Algorithms/0399. Evaluate Division/399. Evaluate Division_test.go b/leetcode/0399.Evaluate-Division/399. Evaluate Division_test.go
similarity index 75%
rename from Algorithms/0399. Evaluate Division/399. Evaluate Division_test.go
rename to leetcode/0399.Evaluate-Division/399. Evaluate Division_test.go
index 3959ce144..abbcf0afa 100644
--- a/Algorithms/0399. Evaluate Division/399. Evaluate Division_test.go	
+++ b/leetcode/0399.Evaluate-Division/399. Evaluate Division_test.go	
@@ -28,8 +28,8 @@ func Test_Problem399(t *testing.T) {
 
 	qs := []question399{
 
-		question399{
-			para399{[][]string{[]string{"a", "b"}, []string{"b", "c"}}, []float64{2.0, 3.0}, [][]string{[]string{"a", "c"}, []string{"b", "a"}, []string{"a", "e"}, []string{"a", "a"}, []string{"x", "x"}}},
+		{
+			para399{[][]string{{"a", "b"}, {"b", "c"}}, []float64{2.0, 3.0}, [][]string{{"a", "c"}, {"b", "a"}, {"a", "e"}, {"a", "a"}, {"x", "x"}}},
 			ans399{[]float64{6.0, 0.5, -1.0, 1.0, -1.0}},
 		},
 	}
diff --git a/leetcode/0399.Evaluate-Division/README.md b/leetcode/0399.Evaluate-Division/README.md
new file mode 100755
index 000000000..69be00eb5
--- /dev/null
+++ b/leetcode/0399.Evaluate-Division/README.md
@@ -0,0 +1,42 @@
+# [399. Evaluate Division](https://leetcode.com/problems/evaluate-division/)
+
+
+## 题目
+
+Equations are given in the format `A / B = k`, where `A` and `B` are variables represented as strings, and `k` is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return `-1.0`.
+
+**Example:**
+
+Given `a / b = 2.0, b / c = 3.0.`queries are: `a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .`return `[6.0, 0.5, -1.0, 1.0, -1.0 ].`
+
+The input is: `vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries` , where `equations.size() == values.size()`, and the values are positive. This represents the equations. Return `vector<double>`.
+
+According to the example above:
+
+    equations = [ ["a", "b"], ["b", "c"] ],
+    values = [2.0, 3.0],
+    queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ].
+
+The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.
+
+
+## 题目大意
+
+给出方程式 A / B = k, 其中 A 和 B 均为代表字符串的变量， k 是一个浮点型数字。根据已知方程式求解问题，并返回计算结果。如果结果不存在，则返回 -1.0。
+
+示例 :
+给定 a / b = 2.0, b / c = 3.0
+问题: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? 
+返回 [6.0, 0.5, -1.0, 1.0, -1.0 ]
+
+输入为: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries(方程式，方程式结果，问题方程式)， 其中 equations.size() == values.size()，即方程式的长度与方程式结果长度相等（程式与结果一一对应），并且结果值均为正数。以上为方程式的描述。 返回vector<double>类型。
+
+假设输入总是有效的。你可以假设除法运算中不会出现除数为0的情况，且不存在任何矛盾的结果。
+
+
+## 解题思路
+
+
+- 给出一些字母变量的倍数关系，问给出任意两个字母的倍数是多少。
+- 这一题可以用 DFS 或者并查集来解题。先来看看 DFS 的做法。先建图。每个字母或者字母组合可以看做成一个节点，给出的 `equations` 关系可以看成两个节点之间的有向边。每条有向边都有权值。那么问题可以转换成是否存在一条从起点节点到终点节点的路径，如果存在，输出这条路径上所有有向边权值的累乘结果。如果不存在这条路径，就返回 -1 。如果给的起点和终点不在给出的节点集里面，也输出 -1 。
+- 再来看看并查集的做法。先将每两个有倍数关系的节点做并查集 `union()` 操作。例如 A/B = 2，那么把 B 作为 `parent` 节点，`parents[A] = {B，2}`，`parents[B] = {B，1}`，B 指向自己是 1 。还有一个关系是 `B/C=3`，由于 B 已经在并查集中了，所以这个时候需要把这个关系反过来，处理成 `C/B = 1/3` ，即 `parents[C] = {B，1/3}`。这样把所有有关系的字母都 `union()` 起来。如何求任意两个字母的倍数关系呢？例如 `A/C = ?` 在并查集中查找，可以找到 `parents[C] == parents[A] == B`，那么就用 `parents[A]/parents[C] = 2/(1/3) = 6`。为什么可以这样做呢？因为 `A/B = 2`，`C/B = 1/3`，那么 `A/C = (A/B)/(C/B)` 即 `parents[A]/parents[C] = 2/(1/3) = 6`。
diff --git a/leetcode/0400.Nth-Digit/400.Nth Digit.go b/leetcode/0400.Nth-Digit/400.Nth Digit.go
new file mode 100644
index 000000000..26d518223
--- /dev/null
+++ b/leetcode/0400.Nth-Digit/400.Nth Digit.go	
@@ -0,0 +1,19 @@
+package leetcode
+
+import "math"
+
+func findNthDigit(n int) int {
+	if n <= 9 {
+		return n
+	}
+	bits := 1
+	for n > 9*int(math.Pow10(bits-1))*bits {
+		n -= 9 * int(math.Pow10(bits-1)) * bits
+		bits++
+	}
+	idx := n - 1
+	start := int(math.Pow10(bits - 1))
+	num := start + idx/bits
+	digitIdx := idx % bits
+	return num / int(math.Pow10(bits-digitIdx-1)) % 10
+}
diff --git a/leetcode/0400.Nth-Digit/400.Nth Digit_test.go b/leetcode/0400.Nth-Digit/400.Nth Digit_test.go
new file mode 100644
index 000000000..b7ecfeaf7
--- /dev/null
+++ b/leetcode/0400.Nth-Digit/400.Nth Digit_test.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question400 struct {
+	para400
+	ans400
+}
+
+// para 是参数
+type para400 struct {
+	n int
+}
+
+// ans 是答案
+type ans400 struct {
+	ans int
+}
+
+func Test_Problem400(t *testing.T) {
+
+	qs := []question400{
+
+		{
+			para400{3},
+			ans400{3},
+		},
+
+		{
+			para400{11},
+			ans400{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 400------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans400, q.para400
+		fmt.Printf("【input】:%v    【output】:%v\n", p.n, findNthDigit(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0400.Nth-Digit/README.md b/leetcode/0400.Nth-Digit/README.md
new file mode 100644
index 000000000..fefe14588
--- /dev/null
+++ b/leetcode/0400.Nth-Digit/README.md
@@ -0,0 +1,67 @@
+# [400. Nth Digit](https://leetcode.com/problems/nth-digit/)
+
+## 题目
+
+Given an integer n, return the nth digit of the infinite integer sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...].
+
+**Example 1**:
+
+    Input: n = 3
+    Output: 3
+
+**Example 2**:
+
+    Input: n = 11
+    Output: 0
+    Explanation: The 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10.
+
+**Constraints:**
+
+- 1 <= n <= int(math.Pow(2, 31)) - 1
+
+## 题目大意
+
+给你一个整数 n ，请你在无限的整数序列 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...] 中找出并返回第 n 位数字。
+
+## 解题思路
+
+- bits = 1 的时候有 1,2,3,4,5,6,7,8,9 这 9 个数; 9 = math.Pow10(bits - 1) * bits
+- bits = 2 的时候有 10-99 这 90 个数; 90 = math.Pow10(bits - 1) * bits
+- n 不断减去 bits 从 1 开始的数字总数,求出 n 所在的数字是几位数即 bits
+- 计算 n 所在的数字 num，等于初始值加上 (n - 1) / bits
+- 计算 n 所在这个数字的第几位 digitIdx 等于 (n - 1) % bits
+- 计算出 digitIdx 位的数字
+  
+  ### 以11 为例:
+  11 - 9 = 2 
+  
+  (2 - 1) / 2 = 0
+  
+  (2 - 1) % 2 = 1
+  
+  也就是说第 11 位数字是位数是 2 的第一个数字的第二位，即是 0
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func findNthDigit(n int) int {
+	if n <= 9 {
+		return n
+	}
+	bits := 1
+	for n > 9*int(math.Pow10(bits-1))*bits {
+		n -= 9 * int(math.Pow10(bits-1)) * bits
+		bits++
+	}
+	idx := n - 1
+	start := int(math.Pow10(bits - 1))
+	num := start + idx/bits
+	digitIdx := idx % bits
+	return num / int(math.Pow10(bits-digitIdx-1)) % 10
+}
+```
\ No newline at end of file
diff --git a/leetcode/0401.Binary-Watch/401. Binary Watch.go b/leetcode/0401.Binary-Watch/401. Binary Watch.go
new file mode 100644
index 000000000..39cbf8393
--- /dev/null
+++ b/leetcode/0401.Binary-Watch/401. Binary Watch.go	
@@ -0,0 +1,127 @@
+package leetcode
+
+import (
+	"fmt"
+	"strconv"
+)
+
+// 解法一
+func readBinaryWatch(num int) []string {
+	memo := make([]int, 60)
+	// count the number of 1 in a binary number
+	count := func(n int) int {
+		if memo[n] != 0 {
+			return memo[n]
+		}
+		originN, res := n, 0
+		for n != 0 {
+			n = n & (n - 1)
+			res++
+		}
+		memo[originN] = res
+		return res
+	}
+	// fmtMinute format minute 0:1 -> 0:01
+	fmtMinute := func(m int) string {
+		if m < 10 {
+			return "0" + strconv.Itoa(m)
+		}
+		return strconv.Itoa(m)
+	}
+
+	var res []string
+	// traverse 0:00 -> 12:00
+	for i := 0; i < 12; i++ {
+		for j := 0; j < 60; j++ {
+			if count(i)+count(j) == num {
+				res = append(res, strconv.Itoa(i)+":"+fmtMinute(j))
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 打表
+var (
+	hour    = []string{"1", "2", "4", "8"}
+	minute  = []string{"01", "02", "04", "08", "16", "32"}
+	hourMap = map[int][]string{
+		0: {"0"},
+		1: {"1", "2", "4", "8"},
+		2: {"3", "5", "9", "6", "10"},
+		3: {"7", "11"},
+	}
+	minuteMap = map[int][]string{
+		0: {"00"},
+		1: {"01", "02", "04", "08", "16", "32"},
+		2: {"03", "05", "09", "17", "33", "06", "10", "18", "34", "12", "20", "36", "24", "40", "48"},
+		3: {"07", "11", "19", "35", "13", "21", "37", "25", "41", "49", "14", "22", "38", "26", "42", "50", "28", "44", "52", "56"},
+		4: {"15", "23", "39", "27", "43", "51", "29", "45", "53", "57", "30", "46", "54", "58"},
+		5: {"31", "47", "55", "59"},
+	}
+)
+
+func readBinaryWatch1(num int) []string {
+	var res []string
+	if num > 8 {
+		return res
+	}
+	for i := 0; i <= num; i++ {
+		for j := 0; j < len(hourMap[i]); j++ {
+			for k := 0; k < len(minuteMap[num-i]); k++ {
+				res = append(res, hourMap[i][j]+":"+minuteMap[num-i][k])
+			}
+		}
+	}
+	return res
+}
+
+// / ---------------------------------------
+// / ---------------------------------------
+// / ---------------------------------------
+// / ---------------------------------------
+// / ---------------------------------------
+// 以下是打表用到的函数
+// 调用 findReadBinaryWatchMinute(num, 0, c, &res) 打表
+func findReadBinaryWatchMinute(target, index int, c []int, res *[]string) {
+	if target == 0 {
+		str, tmp := "", 0
+		for i := 0; i < len(c); i++ {
+			t, _ := strconv.Atoi(minute[c[i]])
+			tmp += t
+		}
+		if tmp < 10 {
+			str = "0" + strconv.Itoa(tmp)
+		} else {
+			str = strconv.Itoa(tmp)
+		}
+		// fmt.Printf("找到解了 c = %v str = %v\n", c, str)
+		fmt.Printf("\"%v\", ", str)
+		return
+	}
+	for i := index; i < 6; i++ {
+		c = append(c, i)
+		findReadBinaryWatchMinute(target-1, i+1, c, res)
+		c = c[:len(c)-1]
+	}
+}
+
+// 调用 findReadBinaryWatchHour(num, 0, c, &res) 打表
+func findReadBinaryWatchHour(target, index int, c []int, res *[]string) {
+	if target == 0 {
+		str, tmp := "", 0
+		for i := 0; i < len(c); i++ {
+			t, _ := strconv.Atoi(hour[c[i]])
+			tmp += t
+		}
+		str = strconv.Itoa(tmp)
+		//fmt.Printf("找到解了 c = %v str = %v\n", c, str)
+		fmt.Printf("\"%v\", ", str)
+		return
+	}
+	for i := index; i < 4; i++ {
+		c = append(c, i)
+		findReadBinaryWatchHour(target-1, i+1, c, res)
+		c = c[:len(c)-1]
+	}
+}
diff --git a/Algorithms/0401. Binary Watch/401. Binary Watch_test.go b/leetcode/0401.Binary-Watch/401. Binary Watch_test.go
similarity index 97%
rename from Algorithms/0401. Binary Watch/401. Binary Watch_test.go
rename to leetcode/0401.Binary-Watch/401. Binary Watch_test.go
index 5eb18e8e7..745cd4bbb 100644
--- a/Algorithms/0401. Binary Watch/401. Binary Watch_test.go	
+++ b/leetcode/0401.Binary-Watch/401. Binary Watch_test.go	
@@ -26,7 +26,7 @@ func Test_Problem401(t *testing.T) {
 
 	qs := []question401{
 
-		question401{
+		{
 			para401{1},
 			ans401{[]string{"1:00", "2:00", "4:00", "8:00", "0:01", "0:02", "0:04", "0:08", "0:16", "0:32"}},
 		},
diff --git a/Algorithms/0401. Binary Watch/README.md b/leetcode/0401.Binary-Watch/README.md
similarity index 100%
rename from Algorithms/0401. Binary Watch/README.md
rename to leetcode/0401.Binary-Watch/README.md
diff --git a/Algorithms/0402. Remove K Digits/402. Remove K Digits.go b/leetcode/0402.Remove-K-Digits/402. Remove K Digits.go
similarity index 100%
rename from Algorithms/0402. Remove K Digits/402. Remove K Digits.go
rename to leetcode/0402.Remove-K-Digits/402. Remove K Digits.go
diff --git a/Algorithms/0402. Remove K Digits/402. Remove K Digits_test.go b/leetcode/0402.Remove-K-Digits/402. Remove K Digits_test.go
similarity index 88%
rename from Algorithms/0402. Remove K Digits/402. Remove K Digits_test.go
rename to leetcode/0402.Remove-K-Digits/402. Remove K Digits_test.go
index 5b0b502d5..f9b23418b 100644
--- a/Algorithms/0402. Remove K Digits/402. Remove K Digits_test.go	
+++ b/leetcode/0402.Remove-K-Digits/402. Remove K Digits_test.go	
@@ -26,42 +26,42 @@ type ans402 struct {
 func Test_Problem402(t *testing.T) {
 
 	qs := []question402{
-		question402{
+		{
 			para402{"10", 1},
 			ans402{"0"},
 		},
 
-		question402{
+		{
 			para402{"1111111", 3},
 			ans402{"1111"},
 		},
 
-		question402{
+		{
 			para402{"5337", 2},
 			ans402{"33"},
 		},
 
-		question402{
+		{
 			para402{"112", 1},
 			ans402{"11"},
 		},
 
-		question402{
+		{
 			para402{"1432219", 3},
 			ans402{"1219"},
 		},
 
-		question402{
+		{
 			para402{"10200", 1},
 			ans402{"200"},
 		},
 
-		question402{
+		{
 			para402{"10", 2},
 			ans402{"0"},
 		},
 
-		question402{
+		{
 			para402{"19", 2},
 			ans402{"0"},
 		},
diff --git a/Algorithms/0402. Remove K Digits/README.md b/leetcode/0402.Remove-K-Digits/README.md
similarity index 100%
rename from Algorithms/0402. Remove K Digits/README.md
rename to leetcode/0402.Remove-K-Digits/README.md
diff --git a/Algorithms/0404. Sum of Left Leaves/404. Sum of Left Leaves.go b/leetcode/0404.Sum-of-Left-Leaves/404. Sum of Left Leaves.go
similarity index 78%
rename from Algorithms/0404. Sum of Left Leaves/404. Sum of Left Leaves.go
rename to leetcode/0404.Sum-of-Left-Leaves/404. Sum of Left Leaves.go
index a089b1417..6e0cdf5eb 100644
--- a/Algorithms/0404. Sum of Left Leaves/404. Sum of Left Leaves.go	
+++ b/leetcode/0404.Sum-of-Left-Leaves/404. Sum of Left Leaves.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func sumOfLeftLeaves(root *TreeNode) int {
 	if root == nil {
 		return 0
diff --git a/Algorithms/0404. Sum of Left Leaves/404. Sum of Left Leaves_test.go b/leetcode/0404.Sum-of-Left-Leaves/404. Sum of Left Leaves_test.go
similarity index 79%
rename from Algorithms/0404. Sum of Left Leaves/404. Sum of Left Leaves_test.go
rename to leetcode/0404.Sum-of-Left-Leaves/404. Sum of Left Leaves_test.go
index c593abe7b..09a665211 100644
--- a/Algorithms/0404. Sum of Left Leaves/404. Sum of Left Leaves_test.go	
+++ b/leetcode/0404.Sum-of-Left-Leaves/404. Sum of Left Leaves_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question404 struct {
@@ -26,13 +28,13 @@ func Test_Problem404(t *testing.T) {
 
 	qs := []question404{
 
-		question404{
+		{
 			para404{[]int{}},
 			ans404{0},
 		},
 
-		question404{
-			para404{[]int{3, 9, 20, NULL, NULL, 15, 7}},
+		{
+			para404{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
 			ans404{24},
 		},
 	}
@@ -42,7 +44,7 @@ func Test_Problem404(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans404, q.para404
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", sumOfLeftLeaves(root))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0404. Sum of Left Leaves/README.md b/leetcode/0404.Sum-of-Left-Leaves/README.md
similarity index 100%
rename from Algorithms/0404. Sum of Left Leaves/README.md
rename to leetcode/0404.Sum-of-Left-Leaves/README.md
diff --git a/Algorithms/0405. Convert a Number to Hexadecimal/405. Convert a Number to Hexadecimal.go b/leetcode/0405.Convert-a-Number-to-Hexadecimal/405. Convert a Number to Hexadecimal.go
similarity index 100%
rename from Algorithms/0405. Convert a Number to Hexadecimal/405. Convert a Number to Hexadecimal.go
rename to leetcode/0405.Convert-a-Number-to-Hexadecimal/405. Convert a Number to Hexadecimal.go
diff --git a/Algorithms/0405. Convert a Number to Hexadecimal/405. Convert a Number to Hexadecimal_test.go b/leetcode/0405.Convert-a-Number-to-Hexadecimal/405. Convert a Number to Hexadecimal_test.go
similarity index 95%
rename from Algorithms/0405. Convert a Number to Hexadecimal/405. Convert a Number to Hexadecimal_test.go
rename to leetcode/0405.Convert-a-Number-to-Hexadecimal/405. Convert a Number to Hexadecimal_test.go
index ef7c7fa5a..e4aafd853 100644
--- a/Algorithms/0405. Convert a Number to Hexadecimal/405. Convert a Number to Hexadecimal_test.go	
+++ b/leetcode/0405.Convert-a-Number-to-Hexadecimal/405. Convert a Number to Hexadecimal_test.go	
@@ -26,12 +26,12 @@ func Test_Problem405(t *testing.T) {
 
 	qs := []question405{
 
-		question405{
+		{
 			para405{26},
 			ans405{"1a"},
 		},
 
-		question405{
+		{
 			para405{-1},
 			ans405{"ffffffff"},
 		},
diff --git a/leetcode/0405.Convert-a-Number-to-Hexadecimal/README.md b/leetcode/0405.Convert-a-Number-to-Hexadecimal/README.md
new file mode 100755
index 000000000..0026e4bb8
--- /dev/null
+++ b/leetcode/0405.Convert-a-Number-to-Hexadecimal/README.md
@@ -0,0 +1,48 @@
+# [405. Convert a Number to Hexadecimal](https://leetcode.com/problems/convert-a-number-to-hexadecimal/)
+
+
+## 题目
+
+Given an integer, write an algorithm to convert it to hexadecimal. For negative integer, [two’s complement](https://en.wikipedia.org/wiki/Two%27s_complement) method is used.
+
+**Note:**
+
+1. All letters in hexadecimal (`a-f`) must be in lowercase.
+2. The hexadecimal string must not contain extra leading `0`s. If the number is zero, it is represented by a single zero character `'0'`; otherwise, the first character in the hexadecimal string will not be the zero character.
+3. The given number is guaranteed to fit within the range of a 32-bit signed integer.
+4. You **must not use any method provided by the library** which converts/formats the number to hex directly.
+
+**Example 1:**
+
+    Input:
+    26
+    
+    Output:
+    "1a"
+
+**Example 2:**
+
+    Input:
+    -1
+    
+    Output:
+    "ffffffff"
+
+
+## 题目大意
+
+给定一个整数，编写一个算法将这个数转换为十六进制数。对于负整数，我们通常使用[补码运算](https://baike.baidu.com/item/%E8%A1%A5%E7%A0%81/6854613?fr=aladdin)方法。
+
+注意:
+
+1. 十六进制中所有字母(a-f)都必须是小写。
+2. 十六进制字符串中不能包含多余的前导零。如果要转化的数为 0，那么以单个字符 '0' 来表示；对于其他情况，十六进制字符串中的第一个字符将不会是 0 字符。 
+3. 给定的数确保在 32 位有符号整数范围内。
+4. 不能使用任何由库提供的将数字直接转换或格式化为十六进制的方法。
+
+
+
+## 解题思路
+
+- 这一题是水题，将十进制数转换成十六进制的数。需要额外注意 0 和负数的情况。
+
diff --git a/Algorithms/0409. Longest Palindrome/409. Longest Palindrome.go b/leetcode/0409.Longest-Palindrome/409. Longest Palindrome.go
similarity index 100%
rename from Algorithms/0409. Longest Palindrome/409. Longest Palindrome.go
rename to leetcode/0409.Longest-Palindrome/409. Longest Palindrome.go
diff --git a/Algorithms/0409. Longest Palindrome/409. Longest Palindrome_test.go b/leetcode/0409.Longest-Palindrome/409. Longest Palindrome_test.go
similarity index 97%
rename from Algorithms/0409. Longest Palindrome/409. Longest Palindrome_test.go
rename to leetcode/0409.Longest-Palindrome/409. Longest Palindrome_test.go
index c6f49e731..ee023dc85 100644
--- a/Algorithms/0409. Longest Palindrome/409. Longest Palindrome_test.go	
+++ b/leetcode/0409.Longest-Palindrome/409. Longest Palindrome_test.go	
@@ -26,7 +26,7 @@ func Test_Problem409(t *testing.T) {
 
 	qs := []question409{
 
-		question409{
+		{
 			para409{"abccccdd"},
 			ans409{7},
 		},
diff --git a/leetcode/0409.Longest-Palindrome/README.md b/leetcode/0409.Longest-Palindrome/README.md
new file mode 100755
index 000000000..eb515067b
--- /dev/null
+++ b/leetcode/0409.Longest-Palindrome/README.md
@@ -0,0 +1,33 @@
+# [409. Longest Palindrome](https://leetcode.com/problems/longest-palindrome/)
+
+
+## 题目
+
+Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
+
+This is case sensitive, for example `"Aa"` is not considered a palindrome here.
+
+**Note:**Assume the length of given string will not exceed 1,010.
+
+**Example:**
+
+    Input:
+    "abccccdd"
+    
+    Output:
+    7
+    
+    Explanation:
+    One longest palindrome that can be built is "dccaccd", whose length is 7.
+
+
+## 题目大意
+
+给定一个包含大写字母和小写字母的字符串，找到通过这些字母构造成的最长的回文串。在构造过程中，请注意区分大小写。比如 "Aa" 不能当做一个回文字符串。注意:假设字符串的长度不会超过 1010。
+
+
+## 解题思路
+
+
+- 给出一个字符串，要求用这个字符串里面的字符组成一个回文串，问回文串最长可以组合成多长的？
+- 这也是一题水题，先统计每个字符的频次，然后每个字符能取 2 个的取 2 个，不足 2 个的并且当前构造中的回文串是偶数的情况下(即每 2 个都配对了)，可以取 1 个。最后组合出来的就是最长回文串。
diff --git a/leetcode/0410.Split-Array-Largest-Sum/410. Split Array Largest Sum.go b/leetcode/0410.Split-Array-Largest-Sum/410. Split Array Largest Sum.go
new file mode 100644
index 000000000..3152248e2
--- /dev/null
+++ b/leetcode/0410.Split-Array-Largest-Sum/410. Split Array Largest Sum.go	
@@ -0,0 +1,40 @@
+package leetcode
+
+func splitArray(nums []int, m int) int {
+	maxNum, sum := 0, 0
+	for _, num := range nums {
+		sum += num
+		if num > maxNum {
+			maxNum = num
+		}
+	}
+	if m == 1 {
+		return sum
+	}
+	low, high := maxNum, sum
+	for low < high {
+		mid := low + (high-low)>>1
+		if calSum(mid, m, nums) {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+func calSum(mid, m int, nums []int) bool {
+	sum, count := 0, 0
+	for _, v := range nums {
+		sum += v
+		if sum > mid {
+			sum = v
+			count++
+			// 分成 m 块，只需要插桩 m -1 个
+			if count > m-1 {
+				return false
+			}
+		}
+	}
+	return true
+}
diff --git a/leetcode/0410.Split-Array-Largest-Sum/410. Split Array Largest Sum_test.go b/leetcode/0410.Split-Array-Largest-Sum/410. Split Array Largest Sum_test.go
new file mode 100644
index 000000000..31e5fd058
--- /dev/null
+++ b/leetcode/0410.Split-Array-Largest-Sum/410. Split Array Largest Sum_test.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question410 struct {
+	para410
+	ans410
+}
+
+// para 是参数
+// one 代表第一个参数
+type para410 struct {
+	nums []int
+	m    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans410 struct {
+	one int
+}
+
+func Test_Problem410(t *testing.T) {
+
+	qs := []question410{
+
+		{
+			para410{[]int{7, 2, 5, 10, 8}, 2},
+			ans410{18},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 410------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans410, q.para410
+		fmt.Printf("【input】:%v       【output】:%v\n", p, splitArray(p.nums, p.m))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0410.Split-Array-Largest-Sum/README.md b/leetcode/0410.Split-Array-Largest-Sum/README.md
new file mode 100755
index 000000000..45fef4ccf
--- /dev/null
+++ b/leetcode/0410.Split-Array-Largest-Sum/README.md
@@ -0,0 +1,44 @@
+# [410. Split Array Largest Sum](https://leetcode.com/problems/split-array-largest-sum/)
+
+
+## 题目
+
+Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays.
+
+**Note:**If n is the length of array, assume the following constraints are satisfied:
+
+- 1 ≤ n ≤ 1000
+- 1 ≤ m ≤ min(50, n)
+
+**Examples:**
+
+    Input:
+    nums = [7,2,5,10,8]
+    m = 2
+    
+    Output:
+    18
+    
+    Explanation:
+    There are four ways to split nums into two subarrays.
+    The best way is to split it into [7,2,5] and [10,8],
+    where the largest sum among the two subarrays is only 18.
+
+
+## 题目大意
+
+
+给定一个非负整数数组和一个整数 m，你需要将这个数组分成 m 个非空的连续子数组。设计一个算法使得这 m 个子数组各自和的最大值最小。
+
+注意:
+数组长度 n 满足以下条件:
+
+- 1 ≤ n ≤ 1000
+- 1 ≤ m ≤ min(50, n)
+
+
+
+## 解题思路
+
+- 给出一个数组和分割的个数 M。要求把数组分成 M 个子数组，输出子数组和的最大值。
+- 这一题可以用动态规划 DP 解答，也可以用二分搜索来解答。这一题是二分搜索里面的 max-min 最大最小值问题。题目可以转化为在 `M` 次划分中，求一个 `x`，使得 `x` 满足：对任意的`S(i)`，都满足 `S(i) ≤ x`。这个条件保证了 `x` 是所有 `S(i)` 中的最大值。要求的是满足该条件的最小的 `x`。`x` 的搜索范围在 `[max, sum]` 中。逐步二分逼近 low 值，直到找到能满足条件的 low 的最小值，即为最终答案。
diff --git a/leetcode/0412.Fizz-Buzz/412. Fizz Buzz.go b/leetcode/0412.Fizz-Buzz/412. Fizz Buzz.go
new file mode 100644
index 000000000..cd3fbcd2f
--- /dev/null
+++ b/leetcode/0412.Fizz-Buzz/412. Fizz Buzz.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+import "strconv"
+
+func fizzBuzz(n int) []string {
+	solution := make([]string, n)
+	for i := 1; i <= n; i++ {
+		solution[i-1] = ""
+		if i%3 == 0 {
+			solution[i-1] += "Fizz"
+		}
+		if i%5 == 0 {
+			solution[i-1] += "Buzz"
+		}
+		if solution[i-1] == "" {
+			solution[i-1] = strconv.Itoa(i)
+		}
+	}
+	return solution
+}
diff --git a/Algorithms/0412. Fizz Buzz/412. Fizz Buzz_test.go b/leetcode/0412.Fizz-Buzz/412. Fizz Buzz_test.go
similarity index 100%
rename from Algorithms/0412. Fizz Buzz/412. Fizz Buzz_test.go
rename to leetcode/0412.Fizz-Buzz/412. Fizz Buzz_test.go
diff --git a/Algorithms/0412. Fizz Buzz/README.md b/leetcode/0412.Fizz-Buzz/README.md
similarity index 100%
rename from Algorithms/0412. Fizz Buzz/README.md
rename to leetcode/0412.Fizz-Buzz/README.md
diff --git a/leetcode/0413.Arithmetic-Slices/413. Arithmetic Slices.go b/leetcode/0413.Arithmetic-Slices/413. Arithmetic Slices.go
new file mode 100644
index 000000000..289caa828
--- /dev/null
+++ b/leetcode/0413.Arithmetic-Slices/413. Arithmetic Slices.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+func numberOfArithmeticSlices(A []int) int {
+	if len(A) < 3 {
+		return 0
+	}
+	res, dp := 0, 0
+	for i := 1; i < len(A)-1; i++ {
+		if A[i+1]-A[i] == A[i]-A[i-1] {
+			dp++
+			res += dp
+		} else {
+			dp = 0
+		}
+	}
+	return res
+}
diff --git a/leetcode/0413.Arithmetic-Slices/413. Arithmetic Slices_test.go b/leetcode/0413.Arithmetic-Slices/413. Arithmetic Slices_test.go
new file mode 100644
index 000000000..40fa2245c
--- /dev/null
+++ b/leetcode/0413.Arithmetic-Slices/413. Arithmetic Slices_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question413 struct {
+	para413
+	ans413
+}
+
+// para 是参数
+// one 代表第一个参数
+type para413 struct {
+	A []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans413 struct {
+	one int
+}
+
+func Test_Problem413(t *testing.T) {
+
+	qs := []question413{
+
+		{
+			para413{[]int{1, 2, 3, 4}},
+			ans413{3},
+		},
+
+		{
+			para413{[]int{1, 2, 3, 4, 9}},
+			ans413{3},
+		},
+
+		{
+			para413{[]int{1, 2, 3, 4, 5, 6, 7}},
+			ans413{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 413------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans413, q.para413
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numberOfArithmeticSlices(p.A))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0413.Arithmetic-Slices/README.md b/leetcode/0413.Arithmetic-Slices/README.md
new file mode 100644
index 000000000..3a8575ba1
--- /dev/null
+++ b/leetcode/0413.Arithmetic-Slices/README.md
@@ -0,0 +1,64 @@
+# [413. Arithmetic Slices](https://leetcode.com/problems/arithmetic-slices/)
+
+
+## 题目
+
+A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.
+
+For example, these are arithmetic sequences:
+
+```
+1, 3, 5, 7, 9
+7, 7, 7, 7
+3, -1, -5, -9
+```
+
+The following sequence is not arithmetic.
+
+```
+1, 1, 2, 5, 7
+```
+
+A zero-indexed array A consisting of N numbers is given. A slice of that array is any pair of integers (P, Q) such that 0 <= P < Q < N.
+
+A slice (P, Q) of the array A is called arithmetic if the sequence:A[P], A[P + 1], ..., A[Q - 1], A[Q] is arithmetic. In particular, this means that P + 1 < Q.
+
+The function should return the number of arithmetic slices in the array A.
+
+**Example:**
+
+```
+A = [1, 2, 3, 4]
+
+return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.
+```
+
+## 题目大意
+
+数组 A 包含 N 个数，且索引从0开始。数组 A 的一个子数组划分为数组 (P, Q)，P 与 Q 是整数且满足 0<=P<Q<N 。如果满足以下条件，则称子数组(P, Q)为等差数组：元素 A[P], A[p + 1], ..., A[Q - 1], A[Q] 是等差的。并且 P + 1 < Q 。函数要返回数组 A 中所有为等差数组的子数组个数。
+
+## 解题思路
+
+- 由题目给出的定义，至少 3 个数字以上的等差数列才满足题意。连续 k 个连续等差的元素，包含的子等差数列是底层的，1，2，3…… k。所以每判断一组 3 个连续的数列，只需要用一个变量累加前面已经有多少个满足题意的连续元素，只要满足题意的等差数列就加上这个累加值。一旦不满足等差的条件，累加值置 0。如此循环一次即可找到题目要求的答案。
+
+## 代码
+
+```go
+package leetcode
+
+func numberOfArithmeticSlices(A []int) int {
+	if len(A) < 3 {
+		return 0
+	}
+	res, dp := 0, 0
+	for i := 1; i < len(A)-1; i++ {
+		if A[i+1]-A[i] == A[i]-A[i-1] {
+			dp++
+			res += dp
+		} else {
+			dp = 0
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0414. Third Maximum Number/414. Third Maximum Number.go b/leetcode/0414.Third-Maximum-Number/414. Third Maximum Number.go
similarity index 100%
rename from Algorithms/0414. Third Maximum Number/414. Third Maximum Number.go
rename to leetcode/0414.Third-Maximum-Number/414. Third Maximum Number.go
diff --git a/Algorithms/0414. Third Maximum Number/414. Third Maximum Number_test.go b/leetcode/0414.Third-Maximum-Number/414. Third Maximum Number_test.go
similarity index 92%
rename from Algorithms/0414. Third Maximum Number/414. Third Maximum Number_test.go
rename to leetcode/0414.Third-Maximum-Number/414. Third Maximum Number_test.go
index 062cc829f..522784dde 100644
--- a/Algorithms/0414. Third Maximum Number/414. Third Maximum Number_test.go	
+++ b/leetcode/0414.Third-Maximum-Number/414. Third Maximum Number_test.go	
@@ -26,22 +26,22 @@ func Test_Problem414(t *testing.T) {
 
 	qs := []question414{
 
-		question414{
+		{
 			para414{[]int{1, 1, 2}},
 			ans414{2},
 		},
 
-		question414{
+		{
 			para414{[]int{3, 2, 1}},
 			ans414{1},
 		},
 
-		question414{
+		{
 			para414{[]int{1, 2}},
 			ans414{2},
 		},
 
-		question414{
+		{
 			para414{[]int{2, 2, 3, 1}},
 			ans414{1},
 		},
diff --git a/Algorithms/0414. Third Maximum Number/README.md b/leetcode/0414.Third-Maximum-Number/README.md
similarity index 100%
rename from Algorithms/0414. Third Maximum Number/README.md
rename to leetcode/0414.Third-Maximum-Number/README.md
diff --git a/Algorithms/0416. Partition Equal Subset Sum/416. Partition Equal Subset Sum.go b/leetcode/0416.Partition-Equal-Subset-Sum/416. Partition Equal Subset Sum.go
similarity index 100%
rename from Algorithms/0416. Partition Equal Subset Sum/416. Partition Equal Subset Sum.go
rename to leetcode/0416.Partition-Equal-Subset-Sum/416. Partition Equal Subset Sum.go
diff --git a/Algorithms/0416. Partition Equal Subset Sum/416. Partition Equal Subset Sum_test.go b/leetcode/0416.Partition-Equal-Subset-Sum/416. Partition Equal Subset Sum_test.go
similarity index 94%
rename from Algorithms/0416. Partition Equal Subset Sum/416. Partition Equal Subset Sum_test.go
rename to leetcode/0416.Partition-Equal-Subset-Sum/416. Partition Equal Subset Sum_test.go
index 4e26f6327..fdceb0b09 100644
--- a/Algorithms/0416. Partition Equal Subset Sum/416. Partition Equal Subset Sum_test.go	
+++ b/leetcode/0416.Partition-Equal-Subset-Sum/416. Partition Equal Subset Sum_test.go	
@@ -26,17 +26,17 @@ func Test_Problem416(t *testing.T) {
 
 	qs := []question416{
 
-		question416{
+		{
 			para416{[]int{1, 5, 11, 5}},
 			ans416{true},
 		},
 
-		question416{
+		{
 			para416{[]int{1, 2, 3, 5}},
 			ans416{false},
 		},
 
-		question416{
+		{
 			para416{[]int{1, 2, 5}},
 			ans416{false},
 		},
diff --git a/Algorithms/0416. Partition Equal Subset Sum/README.md b/leetcode/0416.Partition-Equal-Subset-Sum/README.md
similarity index 100%
rename from Algorithms/0416. Partition Equal Subset Sum/README.md
rename to leetcode/0416.Partition-Equal-Subset-Sum/README.md
diff --git a/leetcode/0417.Pacific-Atlantic-Water-Flow/417. Pacific Atlantic Water Flow.go b/leetcode/0417.Pacific-Atlantic-Water-Flow/417. Pacific Atlantic Water Flow.go
new file mode 100644
index 000000000..cfe62561a
--- /dev/null
+++ b/leetcode/0417.Pacific-Atlantic-Water-Flow/417. Pacific Atlantic Water Flow.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+import "math"
+
+func pacificAtlantic(matrix [][]int) [][]int {
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return nil
+	}
+	row, col, res := len(matrix), len(matrix[0]), make([][]int, 0)
+	pacific, atlantic := make([][]bool, row), make([][]bool, row)
+	for i := 0; i < row; i++ {
+		pacific[i] = make([]bool, col)
+		atlantic[i] = make([]bool, col)
+	}
+	for i := 0; i < row; i++ {
+		dfs(matrix, i, 0, &pacific, math.MinInt32)
+		dfs(matrix, i, col-1, &atlantic, math.MinInt32)
+	}
+	for j := 0; j < col; j++ {
+		dfs(matrix, 0, j, &pacific, math.MinInt32)
+		dfs(matrix, row-1, j, &atlantic, math.MinInt32)
+	}
+	for i := 0; i < row; i++ {
+		for j := 0; j < col; j++ {
+			if atlantic[i][j] && pacific[i][j] {
+				res = append(res, []int{i, j})
+			}
+		}
+	}
+	return res
+}
+
+func dfs(matrix [][]int, row, col int, visited *[][]bool, height int) {
+	if row < 0 || row >= len(matrix) || col < 0 || col >= len(matrix[0]) {
+		return
+	}
+	if (*visited)[row][col] || matrix[row][col] < height {
+		return
+	}
+	(*visited)[row][col] = true
+	dfs(matrix, row+1, col, visited, matrix[row][col])
+	dfs(matrix, row-1, col, visited, matrix[row][col])
+	dfs(matrix, row, col+1, visited, matrix[row][col])
+	dfs(matrix, row, col-1, visited, matrix[row][col])
+}
diff --git a/leetcode/0417.Pacific-Atlantic-Water-Flow/417. Pacific Atlantic Water Flow_test.go b/leetcode/0417.Pacific-Atlantic-Water-Flow/417. Pacific Atlantic Water Flow_test.go
new file mode 100644
index 000000000..a7278808f
--- /dev/null
+++ b/leetcode/0417.Pacific-Atlantic-Water-Flow/417. Pacific Atlantic Water Flow_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question417 struct {
+	para417
+	ans417
+}
+
+// para 是参数
+// one 代表第一个参数
+type para417 struct {
+	matrix [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans417 struct {
+	one [][]int
+}
+
+func Test_Problem417(t *testing.T) {
+
+	qs := []question417{
+
+		{
+			para417{[][]int{
+				{1, 2, 2, 3, 5},
+				{3, 2, 3, 4, 4},
+				{2, 4, 5, 3, 1},
+				{6, 7, 1, 4, 5},
+				{5, 1, 1, 2, 4},
+			}},
+			ans417{[][]int{
+				{0, 4},
+				{1, 3},
+				{1, 4},
+				{2, 2},
+				{3, 0},
+				{3, 1},
+				{4, 0},
+			}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 417------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans417, q.para417
+		fmt.Printf("【input】:%v       【output】:%v\n", p, pacificAtlantic(p.matrix))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0417.Pacific-Atlantic-Water-Flow/README.md b/leetcode/0417.Pacific-Atlantic-Water-Flow/README.md
new file mode 100644
index 000000000..124ce1fa5
--- /dev/null
+++ b/leetcode/0417.Pacific-Atlantic-Water-Flow/README.md
@@ -0,0 +1,92 @@
+# [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)
+
+
+## 题目
+
+Given an `m x n` matrix of non-negative integers representing the height of each unit cell in a continent, the "Pacific ocean" touches the left and top edges of the matrix and the "Atlantic ocean" touches the right and bottom edges.
+
+Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower.
+
+Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean.
+
+**Note:**
+
+1. The order of returned grid coordinates does not matter.
+2. Both m and n are less than 150.
+
+**Example:**
+
+```
+Given the following 5x5 matrix:
+
+  Pacific ~   ~   ~   ~   ~
+       ~  1   2   2   3  (5) *
+       ~  3   2   3  (4) (4) *
+       ~  2   4  (5)  3   1  *
+       ~ (6) (7)  1   4   5  *
+       ~ (5)  1   1   2   4  *
+          *   *   *   *   * Atlantic
+
+Return:
+
+[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).
+
+```
+
+## 题目大意
+
+给定一个 m x n 的非负整数矩阵来表示一片大陆上各个单元格的高度。“太平洋”处于大陆的左边界和上边界，而“大西洋”处于大陆的右边界和下边界。规定水流只能按照上、下、左、右四个方向流动，且只能从高到低或者在同等高度上流动。请找出那些水流既可以流动到“太平洋”，又能流动到“大西洋”的陆地单元的坐标。
+
+## 解题思路
+
+- 暴力解法，利用 DFS 把二维数据按照行优先搜索一遍，分别标记出太平洋和大西洋水流能到达的位置。再按照列优先搜索一遍，标记出太平洋和大西洋水流能到达的位置。最后两者都能到达的坐标即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+import "math"
+
+func pacificAtlantic(matrix [][]int) [][]int {
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return nil
+	}
+	row, col, res := len(matrix), len(matrix[0]), make([][]int, 0)
+	pacific, atlantic := make([][]bool, row), make([][]bool, row)
+	for i := 0; i < row; i++ {
+		pacific[i] = make([]bool, col)
+		atlantic[i] = make([]bool, col)
+	}
+	for i := 0; i < row; i++ {
+		dfs(matrix, i, 0, &pacific, math.MinInt32)
+		dfs(matrix, i, col-1, &atlantic, math.MinInt32)
+	}
+	for j := 0; j < col; j++ {
+		dfs(matrix, 0, j, &pacific, math.MinInt32)
+		dfs(matrix, row-1, j, &atlantic, math.MinInt32)
+	}
+	for i := 0; i < row; i++ {
+		for j := 0; j < col; j++ {
+			if atlantic[i][j] && pacific[i][j] {
+				res = append(res, []int{i, j})
+			}
+		}
+	}
+	return res
+}
+
+func dfs(matrix [][]int, row, col int, visited *[][]bool, height int) {
+	if row < 0 || row >= len(matrix) || col < 0 || col >= len(matrix[0]) {
+		return
+	}
+	if (*visited)[row][col] || matrix[row][col] < height {
+		return
+	}
+	(*visited)[row][col] = true
+	dfs(matrix, row+1, col, visited, matrix[row][col])
+	dfs(matrix, row-1, col, visited, matrix[row][col])
+	dfs(matrix, row, col+1, visited, matrix[row][col])
+	dfs(matrix, row, col-1, visited, matrix[row][col])
+}
+```
\ No newline at end of file
diff --git a/leetcode/0419.Battleships-in-a-Board/419. Battleships in a Board.go b/leetcode/0419.Battleships-in-a-Board/419. Battleships in a Board.go
new file mode 100644
index 000000000..6e519b858
--- /dev/null
+++ b/leetcode/0419.Battleships-in-a-Board/419. Battleships in a Board.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+func countBattleships(board [][]byte) (ans int) {
+	if len(board) == 0 || len(board[0]) == 0 {
+		return 0
+	}
+	for i := range board {
+		for j := range board[i] {
+			if board[i][j] == 'X' {
+				if i > 0 && board[i-1][j] == 'X' {
+					continue
+				}
+				if j > 0 && board[i][j-1] == 'X' {
+					continue
+				}
+				ans++
+			}
+		}
+	}
+	return
+}
diff --git a/leetcode/0419.Battleships-in-a-Board/419. Battleships in a Board_test.go b/leetcode/0419.Battleships-in-a-Board/419. Battleships in a Board_test.go
new file mode 100644
index 000000000..3669aeec3
--- /dev/null
+++ b/leetcode/0419.Battleships-in-a-Board/419. Battleships in a Board_test.go	
@@ -0,0 +1,59 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+	"unsafe"
+)
+
+type question419 struct {
+	para419
+	ans419
+}
+
+// para 是参数
+// one 代表第一个参数
+type para419 struct {
+	one [][]byte
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans419 struct {
+	one int
+}
+
+func Test_Problem419(t *testing.T) {
+
+	qs := []question419{
+
+		{
+			para419{[][]byte{{'X', '.', '.', 'X'}, {'.', '.', '.', 'X'}, {'.', '.', '.', 'X'}}},
+			ans419{2},
+		},
+
+		{
+			para419{[][]byte{{'.'}}},
+			ans419{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 419------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans419, q.para419
+		fmt.Printf("【input】:%v       【output】:%v\n", bytesArrayToStringArray(p.one), countBattleships(p.one))
+	}
+	fmt.Printf("\n\n\n")
+
+}
+
+// 在运行go test时 为了更直观地显示[][]byte中的字符而非ASCII码数值
+// bytesArrayToStringArray converts [][]byte to []string
+func bytesArrayToStringArray(b [][]byte) []string {
+	s := make([]string, len(b))
+	for i := range b {
+		s[i] = fmt.Sprintf("[%v]", *(*string)(unsafe.Pointer(&b[i])))
+	}
+	return s
+}
diff --git a/leetcode/0419.Battleships-in-a-Board/README.md b/leetcode/0419.Battleships-in-a-Board/README.md
new file mode 100644
index 000000000..8754bfb91
--- /dev/null
+++ b/leetcode/0419.Battleships-in-a-Board/README.md
@@ -0,0 +1,51 @@
+# [419. Battleships in a Board](https://leetcode.com/problems/battleships-in-a-board/)
+
+## 题目
+
+Given an `m x n` matrix `board` where each cell is a battleship `'X'` or empty `'.'`, return the number of the **battleships** on `board`.
+
+**Battleships** can only be placed horizontally or vertically on `board`. In other words, they can only be made of the shape `1 x k` (`1` row, `k` columns) or `k x 1` (`k` rows, `1` column), where `k` can be of any size. At least one horizontal or vertical cell separates between two battleships (i.e., there are no adjacent battleships).
+
+**Example 1:**
+
+![img](https://assets.leetcode.com/uploads/2021/04/10/battelship-grid.jpg)
+
+```c
+Input: board = [["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]]
+Output: 2
+```
+
+**Example 2:**
+
+```c
+Input: board = [["."]]
+Output: 0
+```
+
+**Constraints:**
+
+- `m == board.length`
+- `n == board[i].length`
+- `1 <= m, n <= 200`
+- `board[i][j] is either '.' or 'X'`.
+
+**Follow up:** Could you do it in one-pass, using only `O(1)` extra memory and without modifying the values `board`?
+
+## 题目大意
+
+给定一个大小为`m × n`的矩阵 称之为甲板，矩阵单元格中的`'X'`表示战舰，`'.'`表示空位。
+
+战舰只能水平或竖直摆放在甲板上（换句话说，可以理解为联通的同一行`'X'`或同一列`'X'`只算作一个“战舰群”），任意俩个“战舰群”间都是不相邻的。返回甲板上“战舰群”的数量。
+
+## 解题思路
+
+题目进阶要求一次扫描算法，空间复杂度为`O(1)`，且不能修改矩阵中的值。
+
+因为题目中给定的两个“战舰群”间至少有一个水平或垂直的空位分隔，所以可以通过枚举每个战舰的左上顶点即可统计“战舰群”的个数。
+
+假设当前遍历到矩阵中`'X'`的位置为`(i, j)`，即 `board[i][j]='X'`。如果当前战舰属于一个新的“战舰群”，则需要满足以下条件：
+
+- 当前位置的上方位为空，即 `board[i-1][j]='.'`；
+- 当前位置的左方位为空，即 `board[i][j-1]='.'`；
+
+统计出所有左方位和上方位为空的战舰个数，即可得到“战舰群”的数量。
\ No newline at end of file
diff --git a/Algorithms/0421. Maximum XOR of Two Numbers in an Array/421. Maximum XOR of Two Numbers in an Array.go b/leetcode/0421.Maximum-XOR-of-Two-Numbers-in-an-Array/421. Maximum XOR of Two Numbers in an Array.go
similarity index 100%
rename from Algorithms/0421. Maximum XOR of Two Numbers in an Array/421. Maximum XOR of Two Numbers in an Array.go
rename to leetcode/0421.Maximum-XOR-of-Two-Numbers-in-an-Array/421. Maximum XOR of Two Numbers in an Array.go
diff --git a/Algorithms/0421. Maximum XOR of Two Numbers in an Array/421. Maximum XOR of Two Numbers in an Array_test.go b/leetcode/0421.Maximum-XOR-of-Two-Numbers-in-an-Array/421. Maximum XOR of Two Numbers in an Array_test.go
similarity index 97%
rename from Algorithms/0421. Maximum XOR of Two Numbers in an Array/421. Maximum XOR of Two Numbers in an Array_test.go
rename to leetcode/0421.Maximum-XOR-of-Two-Numbers-in-an-Array/421. Maximum XOR of Two Numbers in an Array_test.go
index c7225f6f3..7369f8458 100644
--- a/Algorithms/0421. Maximum XOR of Two Numbers in an Array/421. Maximum XOR of Two Numbers in an Array_test.go	
+++ b/leetcode/0421.Maximum-XOR-of-Two-Numbers-in-an-Array/421. Maximum XOR of Two Numbers in an Array_test.go	
@@ -26,7 +26,7 @@ func Test_Problem421(t *testing.T) {
 
 	qs := []question421{
 
-		question421{
+		{
 			para421{[]int{3, 10, 5, 25, 2, 8}},
 			ans421{28},
 		},
diff --git a/leetcode/0421.Maximum-XOR-of-Two-Numbers-in-an-Array/README.md b/leetcode/0421.Maximum-XOR-of-Two-Numbers-in-an-Array/README.md
new file mode 100755
index 000000000..54cf2f6e9
--- /dev/null
+++ b/leetcode/0421.Maximum-XOR-of-Two-Numbers-in-an-Array/README.md
@@ -0,0 +1,33 @@
+# [421. Maximum XOR of Two Numbers in an Array](https://leetcode.com/problems/maximum-xor-of-two-numbers-in-an-array/)
+
+
+## 题目
+
+Given a **non-empty** array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231.
+
+Find the maximum result of ai XOR aj, where 0 ≤ i, j < n.
+
+Could you do this in O(n) runtime?
+
+**Example:**
+
+    Input: [3, 10, 5, 25, 2, 8]
+    
+    Output: 28
+    
+    Explanation: The maximum result is 5 ^ 25 = 28.
+
+
+## 题目大意
+
+给定一个非空数组，数组中元素为 a0, a1, a2, … , an-1，其中 0 ≤ ai < 2^31 。找到 ai 和 aj 最大的异或 (XOR) 运算结果，其中0 ≤ i,  j < n 。你能在O(n)的时间解决这个问题吗？
+
+
+## 解题思路
+
+
+- 这一题最先考虑到的解法就是暴力解法，2 层循环，依次计算两两数之间的异或值，动态维护最大的值，遍历完成以后输出最大值即可。提交代码会发现超时。
+- 改进一点的做法就是一层循环。试想，求的最终结果是一个 32 位的二进制数，如果想要这个数最大，那么高位都填满 1 就是最大。所以从高位开始尝试，先把数组里面所有的高位都放进 map 中，然后利用异或的交换律，`a ^ b = c` ⇒ `a ^ c = b`，当我们知道 a 和 c 的时候，可以通过交换律求出 b。a 就是我们遍历的每个数，c 是我们想要尝试的高位最大值，例如，111…000，从高位逐渐往低位填 1 。如果我们求的 b 也在 map 中，那么就代表 c 是可以求出来的。如果 c 比当前的 max 值要大，就更新。按照这样的方式遍历往 32 位，每次也遍历完整个数组中的每个数，最终 max 里面就是需要求的最大值。
+- 还有更好的做法是利用 Trie 这个数据结构。构建一棵深度为 33 的二叉树。root 节点左孩子为 1，右孩子为 0 代表着所有数字的最高位，其次根据次高位继续往下。如果某一个节点左右子树都不为空，那么得到最终答案的两个数字肯定分别出自于左右子树且此位为 1；如果任意一个为空，那么最终答案该位为 0，依次迭代得到最终结果。具体做法见：[Java O(n) solution using Trie - LeetCode Discuss](https://discuss.leetcode.com/topic/63207/java-o-n-solution-using-trie)
+
+- 最后还有更“完美的做法”，利用 leetcode 网站判题的特性，我们可以测出比较弱的数据，绕过这组弱数据可以直接 AC。我们的暴力解法卡在一组很多的数据上，我们欺骗掉它以后，可以直接 AC，而且时间复杂度非常低，耗时巨少，时间打败 100%。
diff --git a/leetcode/0423.Reconstruct-Original-Digits-from-English/423. Reconstruct Original Digits from English.go b/leetcode/0423.Reconstruct-Original-Digits-from-English/423. Reconstruct Original Digits from English.go
new file mode 100644
index 000000000..77577c712
--- /dev/null
+++ b/leetcode/0423.Reconstruct-Original-Digits-from-English/423. Reconstruct Original Digits from English.go	
@@ -0,0 +1,32 @@
+package leetcode
+
+import (
+	"strings"
+)
+
+func originalDigits(s string) string {
+	digits := make([]int, 26)
+	for i := 0; i < len(s); i++ {
+		digits[int(s[i]-'a')]++
+	}
+	res := make([]string, 10)
+	res[0] = convert('z', digits, "zero", "0")
+	res[6] = convert('x', digits, "six", "6")
+	res[2] = convert('w', digits, "two", "2")
+	res[4] = convert('u', digits, "four", "4")
+	res[5] = convert('f', digits, "five", "5")
+	res[1] = convert('o', digits, "one", "1")
+	res[7] = convert('s', digits, "seven", "7")
+	res[3] = convert('r', digits, "three", "3")
+	res[8] = convert('t', digits, "eight", "8")
+	res[9] = convert('i', digits, "nine", "9")
+	return strings.Join(res, "")
+}
+
+func convert(b byte, digits []int, s string, num string) string {
+	v := digits[int(b-'a')]
+	for i := 0; i < len(s); i++ {
+		digits[int(s[i]-'a')] -= v
+	}
+	return strings.Repeat(num, v)
+}
diff --git a/leetcode/0423.Reconstruct-Original-Digits-from-English/423. Reconstruct Original Digits from English_test.go b/leetcode/0423.Reconstruct-Original-Digits-from-English/423. Reconstruct Original Digits from English_test.go
new file mode 100644
index 000000000..758aa950f
--- /dev/null
+++ b/leetcode/0423.Reconstruct-Original-Digits-from-English/423. Reconstruct Original Digits from English_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question423 struct {
+	para423
+	ans423
+}
+
+// para 是参数
+// one 代表第一个参数
+type para423 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans423 struct {
+	one int
+}
+
+func Test_Problem423(t *testing.T) {
+
+	qs := []question423{
+
+		{
+			para423{"owoztneoer"},
+			ans423{012},
+		},
+
+		{
+			para423{"fviefuro"},
+			ans423{45},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 423------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans423, q.para423
+		fmt.Printf("【input】:%v       【output】:%v\n", p, originalDigits(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0423.Reconstruct-Original-Digits-from-English/README.md b/leetcode/0423.Reconstruct-Original-Digits-from-English/README.md
new file mode 100644
index 000000000..ab95e6698
--- /dev/null
+++ b/leetcode/0423.Reconstruct-Original-Digits-from-English/README.md
@@ -0,0 +1,103 @@
+# [423. Reconstruct Original Digits from English](https://leetcode.com/problems/reconstruct-original-digits-from-english/)
+
+
+## 题目
+
+Given a **non-empty** string containing an out-of-order English representation of digits `0-9`, output the digits in ascending order.
+
+**Note:**
+
+1. Input contains only lowercase English letters.
+2. Input is guaranteed to be valid and can be transformed to its original digits. That means invalid inputs such as "abc" or "zerone" are not permitted.
+3. Input length is less than 50,000.
+
+**Example 1:**
+
+```
+Input: "owoztneoer"
+Output: "012"
+```
+
+**Example 2:**
+
+```
+Input: "fviefuro"
+Output: "45"
+```
+
+## 题目大意
+
+给定一个非空字符串，其中包含字母顺序打乱的英文单词表示的数字0-9。按升序输出原始的数字。
+
+注意:
+
+- 输入只包含小写英文字母。
+- 输入保证合法并可以转换为原始的数字，这意味着像 "abc" 或 "zerone" 的输入是不允许的。
+- 输入字符串的长度小于 50,000。
+
+## 解题思路
+
+- 这道题是一道找规律的题目。首先观察 0-9 对应的英文单词，找到特殊规律：所有的偶数都包含一个独特的字母：
+
+    `z` 只在 `zero` 中出现。
+
+    `w` 只在 `two` 中出现。
+
+    `u` 只在 `four` 中出现。
+
+    `x` 只在 `six` 中出现。
+
+    `g` 只在 `eight` 中出现。
+
+- 所以先排除掉这些偶数。然后在看剩下来几个数字对应的英文字母，这也是计算 3，5 和 7 的关键，因为有些单词只在一个奇数和一个偶数中出现（而且偶数已经被计算过了）：
+
+    `h` 只在 `three` 和 `eight` 中出现。
+
+    `f` 只在 `five` 和 `four` 中出现。
+
+    `s` 只在 `seven` 和 `six` 中出现。
+
+- 接下来只需要处理 9 和 0，思路依然相同。
+
+    `i` 在 `nine`，`five`，`six` 和 `eight` 中出现。
+
+    `n` 在 `one`，`seven` 和 `nine` 中出现。
+
+- 最后按照上述的优先级，依次消耗对应的英文字母，生成最终的原始数字。注意按照优先级换算数字的时候，注意有多个重复数字的情况，比如多个 `1`，多个 `5` 等等。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"strings"
+)
+
+func originalDigits(s string) string {
+	digits := make([]int, 26)
+	for i := 0; i < len(s); i++ {
+		digits[int(s[i]-'a')]++
+	}
+	res := make([]string, 10)
+	res[0] = convert('z', digits, "zero", "0")
+	res[6] = convert('x', digits, "six", "6")
+	res[2] = convert('w', digits, "two", "2")
+	res[4] = convert('u', digits, "four", "4")
+	res[5] = convert('f', digits, "five", "5")
+	res[1] = convert('o', digits, "one", "1")
+	res[7] = convert('s', digits, "seven", "7")
+	res[3] = convert('r', digits, "three", "3")
+	res[8] = convert('t', digits, "eight", "8")
+	res[9] = convert('i', digits, "nine", "9")
+	return strings.Join(res, "")
+}
+
+func convert(b byte, digits []int, s string, num string) string {
+	v := digits[int(b-'a')]
+	for i := 0; i < len(s); i++ {
+		digits[int(s[i]-'a')] -= v
+	}
+	return strings.Repeat(num, v)
+}
+```
\ No newline at end of file
diff --git a/leetcode/0424.Longest-Repeating-Character-Replacement/424. Longest Repeating Character Replacement.go b/leetcode/0424.Longest-Repeating-Character-Replacement/424. Longest Repeating Character Replacement.go
new file mode 100644
index 000000000..8ccd0c082
--- /dev/null
+++ b/leetcode/0424.Longest-Repeating-Character-Replacement/424. Longest Repeating Character Replacement.go	
@@ -0,0 +1,22 @@
+package leetcode
+
+func characterReplacement(s string, k int) int {
+	res, left, counter, freq := 0, 0, 0, make([]int, 26)
+	for right := 0; right < len(s); right++ {
+		freq[s[right]-'A']++
+		counter = max(counter, freq[s[right]-'A'])
+		for right-left+1-counter > k {
+			freq[s[left]-'A']--
+			left++
+		}
+		res = max(res, right-left+1)
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0424. Longest Repeating Character Replacement/424. Longest Repeating Character Replacement_test.go b/leetcode/0424.Longest-Repeating-Character-Replacement/424. Longest Repeating Character Replacement_test.go
similarity index 91%
rename from Algorithms/0424. Longest Repeating Character Replacement/424. Longest Repeating Character Replacement_test.go
rename to leetcode/0424.Longest-Repeating-Character-Replacement/424. Longest Repeating Character Replacement_test.go
index 2c3d5bd39..3ec26313d 100644
--- a/Algorithms/0424. Longest Repeating Character Replacement/424. Longest Repeating Character Replacement_test.go	
+++ b/leetcode/0424.Longest-Repeating-Character-Replacement/424. Longest Repeating Character Replacement_test.go	
@@ -27,27 +27,27 @@ func Test_Problem424(t *testing.T) {
 
 	qs := []question424{
 
-		question424{
+		{
 			para424{"AABABBA", 1},
 			ans424{4},
 		},
 
-		question424{
+		{
 			para424{"ABBB", 2},
 			ans424{4},
 		},
 
-		question424{
+		{
 			para424{"BAAA", 0},
 			ans424{3},
 		},
 
-		question424{
+		{
 			para424{"ABCDE", 1},
 			ans424{2},
 		},
 
-		question424{
+		{
 			para424{"BAAAB", 2},
 			ans424{5},
 		},
diff --git a/Algorithms/0424. Longest Repeating Character Replacement/README.md b/leetcode/0424.Longest-Repeating-Character-Replacement/README.md
similarity index 100%
rename from Algorithms/0424. Longest Repeating Character Replacement/README.md
rename to leetcode/0424.Longest-Repeating-Character-Replacement/README.md
diff --git a/leetcode/0429.N-ary-Tree-Level-Order-Traversal/429. N-ary Tree Level Order Traversal.go b/leetcode/0429.N-ary-Tree-Level-Order-Traversal/429. N-ary Tree Level Order Traversal.go
new file mode 100644
index 000000000..dfcf57a1e
--- /dev/null
+++ b/leetcode/0429.N-ary-Tree-Level-Order-Traversal/429. N-ary Tree Level Order Traversal.go	
@@ -0,0 +1,39 @@
+package leetcode
+
+/**
+ * Definition for a Node.
+ * type Node struct {
+ *     Val int
+ *     Children []*Node
+ * }
+ */
+
+type Node struct {
+	Val      int
+	Children []*Node
+}
+
+func levelOrder(root *Node) [][]int {
+	var res [][]int
+	var temp []int
+	if root == nil {
+		return res
+	}
+	queue := []*Node{root, nil}
+	for len(queue) > 1 {
+		node := queue[0]
+		queue = queue[1:]
+		if node == nil {
+			queue = append(queue, nil)
+			res = append(res, temp)
+			temp = []int{}
+		} else {
+			temp = append(temp, node.Val)
+			if len(node.Children) > 0 {
+				queue = append(queue, node.Children...)
+			}
+		}
+	}
+	res = append(res, temp)
+	return res
+}
diff --git a/leetcode/0429.N-ary-Tree-Level-Order-Traversal/429. N-ary Tree Level Order Traversal_test.go b/leetcode/0429.N-ary-Tree-Level-Order-Traversal/429. N-ary Tree Level Order Traversal_test.go
new file mode 100644
index 000000000..fe56775a4
--- /dev/null
+++ b/leetcode/0429.N-ary-Tree-Level-Order-Traversal/429. N-ary Tree Level Order Traversal_test.go	
@@ -0,0 +1,10 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem429(t *testing.T) {
+	fmt.Printf("success\n")
+}
diff --git a/leetcode/0429.N-ary-Tree-Level-Order-Traversal/README.md b/leetcode/0429.N-ary-Tree-Level-Order-Traversal/README.md
new file mode 100644
index 000000000..79d4912d9
--- /dev/null
+++ b/leetcode/0429.N-ary-Tree-Level-Order-Traversal/README.md
@@ -0,0 +1,85 @@
+# [429. N-ary Tree Level Order Traversal](https://leetcode.com/problems/n-ary-tree-level-order-traversal/)
+
+
+## 题目
+
+Given an n-ary tree, return the *level order* traversal of its nodes' values.
+
+*Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png](https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png)
+
+```
+Input: root = [1,null,3,2,4,null,5,6]
+Output: [[1],[3,2,4],[5,6]]
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png](https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png)
+
+```
+Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
+Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]
+
+```
+
+**Constraints:**
+
+- The height of the n-ary tree is less than or equal to `1000`
+- The total number of nodes is between `[0, 104]`
+
+## 题目大意
+
+给定一个 N 叉树，返回其节点值的层序遍历。（即从左到右，逐层遍历）。树的序列化输入是用层序遍历，每组子节点都由 null 值分隔（参见示例）。
+
+## 解题思路
+
+- 这是 n 叉树的系列题，第 589 题也是这一系列的题目。这一题思路不难，既然是层序遍历，用 BFS 解答。
+
+## 代码
+
+```go
+package leetcode
+
+/**
+ * Definition for a Node.
+ * type Node struct {
+ *     Val int
+ *     Children []*Node
+ * }
+ */
+
+type Node struct {
+	Val      int
+	Children []*Node
+}
+
+func levelOrder(root *Node) [][]int {
+	var res [][]int
+	var temp []int
+	if root == nil {
+		return res
+	}
+	queue := []*Node{root, nil}
+	for len(queue) > 1 {
+		node := queue[0]
+		queue = queue[1:]
+		if node == nil {
+			queue = append(queue, nil)
+			res = append(res, temp)
+			temp = []int{}
+		} else {
+			temp = append(temp, node.Val)
+			if len(node.Children) > 0 {
+				queue = append(queue, node.Children...)
+			}
+		}
+	}
+	res = append(res, temp)
+	return res
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0433. Minimum Genetic Mutation/433. Minimum Genetic Mutation.go b/leetcode/0433.Minimum-Genetic-Mutation/433. Minimum Genetic Mutation.go
similarity index 85%
rename from Algorithms/0433. Minimum Genetic Mutation/433. Minimum Genetic Mutation.go
rename to leetcode/0433.Minimum-Genetic-Mutation/433. Minimum Genetic Mutation.go
index e4e9cf4f9..d273b843e 100644
--- a/Algorithms/0433. Minimum Genetic Mutation/433. Minimum Genetic Mutation.go	
+++ b/leetcode/0433.Minimum-Genetic-Mutation/433. Minimum Genetic Mutation.go	
@@ -24,12 +24,24 @@ func minMutation(start string, end string, bank []string) int {
 	return -1
 }
 
+func getWordMap(wordList []string, beginWord string) map[string]int {
+	wordMap := make(map[string]int)
+	for i, word := range wordList {
+		if _, ok := wordMap[word]; !ok {
+			if word != beginWord {
+				wordMap[word] = i
+			}
+		}
+	}
+	return wordMap
+}
+
 func getCandidates433(word string) []string {
 	var res []string
 	for i := 0; i < 26; i++ {
 		for j := 0; j < len(word); j++ {
 			if word[j] != byte(int('A')+i) {
-				res = append(res, word[:j]+string(int('A')+i)+word[j+1:])
+				res = append(res, word[:j]+string(rune(int('A')+i))+word[j+1:])
 			}
 		}
 	}
diff --git a/Algorithms/0433. Minimum Genetic Mutation/433. Minimum Genetic Mutation_test.go b/leetcode/0433.Minimum-Genetic-Mutation/433. Minimum Genetic Mutation_test.go
similarity index 94%
rename from Algorithms/0433. Minimum Genetic Mutation/433. Minimum Genetic Mutation_test.go
rename to leetcode/0433.Minimum-Genetic-Mutation/433. Minimum Genetic Mutation_test.go
index 0bec4713d..57e7a2cb8 100644
--- a/Algorithms/0433. Minimum Genetic Mutation/433. Minimum Genetic Mutation_test.go	
+++ b/leetcode/0433.Minimum-Genetic-Mutation/433. Minimum Genetic Mutation_test.go	
@@ -27,22 +27,22 @@ type ans433 struct {
 func Test_Problem433(t *testing.T) {
 
 	qs := []question433{
-		question433{
+		{
 			para433{"AACCGGTT", "AACCGGTA", []string{"AACCGGTA"}},
 			ans433{1},
 		},
 
-		question433{
+		{
 			para433{"AACCGGTT", "AAACGGTA", []string{"AACCGGTA", "AACCGCTA", "AAACGGTA"}},
 			ans433{2},
 		},
 
-		question433{
+		{
 			para433{"AAAAACCC", "AACCCCCC", []string{"AAAACCCC", "AAACCCCC", "AACCCCCC"}},
 			ans433{3},
 		},
 
-		question433{
+		{
 			para433{"AACCGGTT", "AACCGGTA", []string{}},
 			ans433{-1},
 		},
diff --git a/Algorithms/0433. Minimum Genetic Mutation/README.md b/leetcode/0433.Minimum-Genetic-Mutation/README.md
similarity index 100%
rename from Algorithms/0433. Minimum Genetic Mutation/README.md
rename to leetcode/0433.Minimum-Genetic-Mutation/README.md
diff --git a/leetcode/0434.Number-of-Segments-in-a-String/434.Number of Segments in a String.go b/leetcode/0434.Number-of-Segments-in-a-String/434.Number of Segments in a String.go
new file mode 100644
index 000000000..626caebe8
--- /dev/null
+++ b/leetcode/0434.Number-of-Segments-in-a-String/434.Number of Segments in a String.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func countSegments(s string) int {
+	segments := false
+	cnt := 0
+	for _, v := range s {
+		if v == ' ' && segments {
+			segments = false
+			cnt += 1
+		} else if v != ' ' {
+			segments = true
+		}
+	}
+	if segments {
+		cnt++
+	}
+	return cnt
+}
diff --git a/leetcode/0434.Number-of-Segments-in-a-String/434.Number of Segments in a String_test.go b/leetcode/0434.Number-of-Segments-in-a-String/434.Number of Segments in a String_test.go
new file mode 100644
index 000000000..7c634601e
--- /dev/null
+++ b/leetcode/0434.Number-of-Segments-in-a-String/434.Number of Segments in a String_test.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question434 struct {
+	para434
+	ans434
+}
+
+// s 是参数
+type para434 struct {
+	s string
+}
+
+// ans 是答案
+type ans434 struct {
+	ans int
+}
+
+func Test_Problem434(t *testing.T) {
+
+	qs := []question434{
+
+		{
+			para434{"Hello, my name is John"},
+			ans434{5},
+		},
+
+		{
+			para434{"Hello"},
+			ans434{1},
+		},
+
+		{
+			para434{"love live! mu'sic forever"},
+			ans434{4},
+		},
+
+		{
+			para434{""},
+			ans434{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 434------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans434, q.para434
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countSegments(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0434.Number-of-Segments-in-a-String/README.md b/leetcode/0434.Number-of-Segments-in-a-String/README.md
new file mode 100644
index 000000000..48fd4b8da
--- /dev/null
+++ b/leetcode/0434.Number-of-Segments-in-a-String/README.md
@@ -0,0 +1,70 @@
+# [434. Number of Segments in a String](https://leetcode.com/problems/number-of-segments-in-a-string/)
+
+
+## 题目
+
+You are given a string s, return the number of segments in the string.
+
+A segment is defined to be a contiguous sequence of non-space characters.
+
+**Example 1:**
+
+    Input: s = "Hello, my name is John"
+    Output: 5
+    Explanation: The five segments are ["Hello,", "my", "name", "is", "John"]
+
+**Example 2:**
+
+    Input: s = "Hello"
+    Output: 1
+
+**Example 3:**
+
+    Input: s = "love live! mu'sic forever"
+    Output: 4
+
+**Example 4:**
+
+    Input: s = ""
+    Output: 0
+
+**Constraints**
+
+ - 0 <= s.length <= 300
+ - s consists of lower-case and upper-case English letters, digits or one of the following characters "!@#$%^&*()_+-=',.:".
+ - The only space character in s is ' '.
+
+## 题目大意
+
+统计字符串中的单词个数，这里的单词指的是连续的不是空格的字符。
+
+请注意，你可以假定字符串里不包括任何不可打印的字符。
+
+## 解题思路
+
+- 以空格为分割计算元素个数
+
+## 代码
+
+```go
+
+package leetcode
+
+func countSegments(s string) int {
+	segments := false
+	cnt := 0
+	for _, v := range s {
+		if v == ' ' && segments {
+			segments = false
+			cnt += 1
+		} else if v != ' ' {
+			segments = true
+		}
+	}
+	if segments {
+		cnt++
+	}
+	return cnt
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0435. Non-overlapping Intervals/435. Non-overlapping Intervals.go b/leetcode/0435.Non-overlapping-Intervals/435. Non-overlapping Intervals.go
similarity index 94%
rename from Algorithms/0435. Non-overlapping Intervals/435. Non-overlapping Intervals.go
rename to leetcode/0435.Non-overlapping-Intervals/435. Non-overlapping Intervals.go
index 6a3fcc4dd..5abdb20d9 100644
--- a/Algorithms/0435. Non-overlapping Intervals/435. Non-overlapping Intervals.go	
+++ b/leetcode/0435.Non-overlapping-Intervals/435. Non-overlapping Intervals.go	
@@ -27,6 +27,13 @@ func eraseOverlapIntervals(intervals [][]int) int {
 	return len(intervals) - res
 }
 
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
 // Intervals define
 type Intervals [][]int
 
diff --git a/leetcode/0435.Non-overlapping-Intervals/435. Non-overlapping Intervals_test.go b/leetcode/0435.Non-overlapping-Intervals/435. Non-overlapping Intervals_test.go
new file mode 100644
index 000000000..059a118bb
--- /dev/null
+++ b/leetcode/0435.Non-overlapping-Intervals/435. Non-overlapping Intervals_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question435 struct {
+	para435
+	ans435
+}
+
+// para 是参数
+// one 代表第一个参数
+type para435 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans435 struct {
+	one int
+}
+
+func Test_Problem435(t *testing.T) {
+
+	qs := []question435{
+
+		{
+			para435{[][]int{{1, 2}, {2, 3}, {3, 4}, {1, 3}}},
+			ans435{1},
+		},
+
+		{
+			para435{[][]int{{1, 2}, {1, 2}, {1, 2}}},
+			ans435{2},
+		},
+
+		{
+			para435{[][]int{{1, 2}, {2, 3}}},
+			ans435{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 435------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans435, q.para435
+		fmt.Printf("【input】:%v       【output】:%v\n", p, eraseOverlapIntervals1(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0435. Non-overlapping Intervals/README.md b/leetcode/0435.Non-overlapping-Intervals/README.md
similarity index 100%
rename from Algorithms/0435. Non-overlapping Intervals/README.md
rename to leetcode/0435.Non-overlapping-Intervals/README.md
diff --git a/leetcode/0436.Find-Right-Interval/436. Find Right Interval.go b/leetcode/0436.Find-Right-Interval/436. Find Right Interval.go
new file mode 100644
index 000000000..101730533
--- /dev/null
+++ b/leetcode/0436.Find-Right-Interval/436. Find Right Interval.go	
@@ -0,0 +1,80 @@
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// Interval define
+type Interval = structures.Interval
+
+// 解法一 利用系统函数 sort + 二分搜索
+func findRightInterval(intervals [][]int) []int {
+	intervalList := make(intervalList, len(intervals))
+	// 转换成 interval 类型
+	for i, v := range intervals {
+		intervalList[i] = interval{interval: v, index: i}
+	}
+	sort.Sort(intervalList)
+	out := make([]int, len(intervalList))
+	for i := 0; i < len(intervalList); i++ {
+		index := sort.Search(len(intervalList), func(p int) bool { return intervalList[p].interval[0] >= intervalList[i].interval[1] })
+		if index == len(intervalList) {
+			out[intervalList[i].index] = -1
+		} else {
+			out[intervalList[i].index] = intervalList[index].index
+		}
+	}
+	return out
+}
+
+type interval struct {
+	interval []int
+	index    int
+}
+
+type intervalList []interval
+
+func (in intervalList) Len() int { return len(in) }
+func (in intervalList) Less(i, j int) bool {
+	return in[i].interval[0] <= in[j].interval[0]
+}
+func (in intervalList) Swap(i, j int) { in[i], in[j] = in[j], in[i] }
+
+// 解法二 手撸 sort + 二分搜索
+func findRightInterval1(intervals [][]int) []int {
+	if len(intervals) == 0 {
+		return []int{}
+	}
+	intervalsList, res, intervalMap := []Interval{}, []int{}, map[Interval]int{}
+	for k, v := range intervals {
+		intervalsList = append(intervalsList, Interval{Start: v[0], End: v[1]})
+		intervalMap[Interval{Start: v[0], End: v[1]}] = k
+	}
+	structures.QuickSort(intervalsList, 0, len(intervalsList)-1)
+	for _, v := range intervals {
+		tmp := searchFirstGreaterInterval(intervalsList, v[1])
+		if tmp > 0 {
+			tmp = intervalMap[intervalsList[tmp]]
+		}
+		res = append(res, tmp)
+	}
+	return res
+}
+
+func searchFirstGreaterInterval(nums []Interval, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid].Start >= target {
+			if (mid == 0) || (nums[mid-1].Start < target) { // 找到第一个大于等于 target 的元素
+				return mid
+			}
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
diff --git a/leetcode/0436.Find-Right-Interval/436. Find Right Interval_test.go b/leetcode/0436.Find-Right-Interval/436. Find Right Interval_test.go
new file mode 100644
index 000000000..d3d209dfa
--- /dev/null
+++ b/leetcode/0436.Find-Right-Interval/436. Find Right Interval_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question436 struct {
+	para436
+	ans436
+}
+
+// para 是参数
+// one 代表第一个参数
+type para436 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans436 struct {
+	one []int
+}
+
+func Test_Problem436(t *testing.T) {
+
+	qs := []question436{
+
+		{
+			para436{[][]int{{3, 4}, {2, 3}, {1, 2}}},
+			ans436{[]int{-1, 0, 1}},
+		},
+
+		{
+			para436{[][]int{{1, 4}, {2, 3}, {3, 4}}},
+			ans436{[]int{-1, 2, -1}},
+		},
+
+		{
+			para436{[][]int{{1, 2}}},
+			ans436{[]int{-1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 436------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans436, q.para436
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findRightInterval(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0436.Find-Right-Interval/README.md b/leetcode/0436.Find-Right-Interval/README.md
new file mode 100755
index 000000000..056222d2b
--- /dev/null
+++ b/leetcode/0436.Find-Right-Interval/README.md
@@ -0,0 +1,61 @@
+# [436. Find Right Interval](https://leetcode.com/problems/find-right-interval/)
+
+
+## 题目
+
+Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.
+
+For any interval i, you need to store the minimum interval j's index, which means that the interval j has the minimum start point to build the "right" relationship for interval i. If the interval j doesn't exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.
+
+**Note:**
+
+1. You may assume the interval's end point is always bigger than its start point.
+2. You may assume none of these intervals have the same start point.
+
+**Example 1:**
+
+    Input: [ [1,2] ]
+    
+    Output: [-1]
+    
+    Explanation: There is only one interval in the collection, so it outputs -1.
+
+**Example 2:**
+
+    Input: [ [3,4], [2,3], [1,2] ]
+    
+    Output: [-1, 0, 1]
+    
+    Explanation: There is no satisfied "right" interval for [3,4].
+    For [2,3], the interval [3,4] has minimum-"right" start point;
+    For [1,2], the interval [2,3] has minimum-"right" start point.
+
+**Example 3:**
+
+    Input: [ [1,4], [2,3], [3,4] ]
+    
+    Output: [-1, 2, -1]
+    
+    Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
+    For [2,3], the interval [3,4] has minimum-"right" start point.
+
+**NOTE:** input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
+
+
+## 题目大意
+
+给定一组区间，对于每一个区间 i，检查是否存在一个区间 j，它的起始点大于或等于区间 i 的终点，这可以称为 j 在 i 的“右侧”。
+
+对于任何区间，你需要存储的满足条件的区间 j 的最小索引，这意味着区间 j 有最小的起始点可以使其成为“右侧”区间。如果区间 j 不存在，则将区间 i 存储为 -1。最后，你需要输出一个值为存储的区间值的数组。
+
+注意:
+
+- 你可以假设区间的终点总是大于它的起始点。
+- 你可以假定这些区间都不具有相同的起始点。
+
+
+## 解题思路
+
+
+- 给出一个 `interval` 的 数组，要求找到每个 `interval` 在它右边第一个 `interval` 的下标。A 区间在 B 区间的右边：A 区间的左边界的值大于等于 B 区间的右边界。
+- 这一题很明显可以用二分搜索来解答。先将 `interval` 数组排序，然后针对每个 `interval`，用二分搜索搜索大于等于 `interval` 右边界值的 `interval`。如果找到就把下标存入最终数组中，如果没有找到，把 `-1` 存入最终数组中。
diff --git a/leetcode/0437.Path-Sum-III/437. Path Sum III.go b/leetcode/0437.Path-Sum-III/437. Path Sum III.go
new file mode 100644
index 000000000..cdbe52bf6
--- /dev/null
+++ b/leetcode/0437.Path-Sum-III/437. Path Sum III.go	
@@ -0,0 +1,65 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 带缓存 dfs
+func pathSum(root *TreeNode, targetSum int) int {
+	prefixSum := make(map[int]int)
+	prefixSum[0] = 1
+	return dfs(root, prefixSum, 0, targetSum)
+}
+
+func dfs(root *TreeNode, prefixSum map[int]int, cur, sum int) int {
+	if root == nil {
+		return 0
+	}
+	cur += root.Val
+	cnt := 0
+	if v, ok := prefixSum[cur-sum]; ok {
+		cnt = v
+	}
+	prefixSum[cur]++
+	cnt += dfs(root.Left, prefixSum, cur, sum)
+	cnt += dfs(root.Right, prefixSum, cur, sum)
+	prefixSum[cur]--
+	return cnt
+}
+
+// 解法二
+func pathSumIII(root *TreeNode, sum int) int {
+	if root == nil {
+		return 0
+	}
+	res := findPath437(root, sum)
+	res += pathSumIII(root.Left, sum)
+	res += pathSumIII(root.Right, sum)
+	return res
+}
+
+// 寻找包含 root 这个结点，且和为 sum 的路径
+func findPath437(root *TreeNode, sum int) int {
+	if root == nil {
+		return 0
+	}
+	res := 0
+	if root.Val == sum {
+		res++
+	}
+	res += findPath437(root.Left, sum-root.Val)
+	res += findPath437(root.Right, sum-root.Val)
+	return res
+}
diff --git a/leetcode/0437.Path-Sum-III/437. Path Sum III_test.go b/leetcode/0437.Path-Sum-III/437. Path Sum III_test.go
new file mode 100644
index 000000000..089bef78a
--- /dev/null
+++ b/leetcode/0437.Path-Sum-III/437. Path Sum III_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question437 struct {
+	para437
+	ans437
+}
+
+// para 是参数
+// one 代表第一个参数
+type para437 struct {
+	one []int
+	sum int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans437 struct {
+	one int
+}
+
+func Test_Problem437(t *testing.T) {
+
+	qs := []question437{
+		{
+			para437{[]int{}, 0},
+			ans437{0},
+		},
+
+		{
+			para437{[]int{10, 5, -3, 3, 2, structures.NULL, 11, 3, -2, structures.NULL, 1}, 8},
+			ans437{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 437------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans437, q.para437
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", pathSum(root, p.sum))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0437.Path-Sum-III/README.md b/leetcode/0437.Path-Sum-III/README.md
new file mode 100644
index 000000000..755a4e263
--- /dev/null
+++ b/leetcode/0437.Path-Sum-III/README.md
@@ -0,0 +1,119 @@
+# [437. Path Sum III](https://leetcode.com/problems/path-sum-iii/)
+
+
+## 题目
+
+Given the `root` of a binary tree and an integer `targetSum`, return *the number of paths where the sum of the values along the path equals* `targetSum`.
+
+The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/04/09/pathsum3-1-tree.jpg](https://assets.leetcode.com/uploads/2021/04/09/pathsum3-1-tree.jpg)
+
+```
+Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
+Output: 3
+Explanation: The paths that sum to 8 are shown.
+
+```
+
+**Example 2:**
+
+```
+Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
+Output: 3
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[0, 1000]`.
+- `109 <= Node.val <= 109`
+- `1000 <= targetSum <= 1000`
+
+## 题目大意
+
+给定一个二叉树，它的每个结点都存放着一个整数值。找出路径和等于给定数值的路径总数。路径不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。二叉树不超过1000个节点，且节点数值范围是 [-1000000,1000000] 的整数。
+
+
+## 解题思路
+
+
+- 这一题是第 112 题 Path Sum 和第 113 题 Path Sum II 的加强版，这一题要求求出任意一条路径的和为 sum，起点不一定是根节点，可以是任意节点开始。
+- 注意这一题可能出现负数的情况，节点和为 sum，并不一定是最终情况，有可能下面还有正数节点和负数节点相加正好为 0，那么这也是一种情况。一定要遍历到底。
+- 一个点是否为 sum 的起点，有 3 种情况，第一种情况路径包含该 root 节点，如果包含该结点，就在它的左子树和右子树中寻找和为 `sum-root.Val` 的情况。第二种情况路径不包含该 root 节点，那么就需要在它的左子树和右子树中分别寻找和为 sum 的结点。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 带缓存 dfs
+func pathSum(root *TreeNode, targetSum int) int {
+	prefixSum := make(map[int]int)
+	prefixSum[0] = 1
+	return dfs(root, prefixSum, 0, targetSum)
+}
+
+func dfs(root *TreeNode, prefixSum map[int]int, cur, sum int) int {
+	if root == nil {
+		return 0
+	}
+	cur += root.Val
+	cnt := 0
+	if v, ok := prefixSum[cur-sum]; ok {
+		cnt = v
+	}
+	prefixSum[cur]++
+	cnt += dfs(root.Left, prefixSum, cur, sum)
+	cnt += dfs(root.Right, prefixSum, cur, sum)
+	prefixSum[cur]--
+	return cnt
+}
+
+// 解法二
+func pathSumIII(root *TreeNode, sum int) int {
+	if root == nil {
+		return 0
+	}
+	res := findPath437(root, sum)
+	res += pathSumIII(root.Left, sum)
+	res += pathSumIII(root.Right, sum)
+	return res
+}
+
+// 寻找包含 root 这个结点，且和为 sum 的路径
+func findPath437(root *TreeNode, sum int) int {
+	if root == nil {
+		return 0
+	}
+	res := 0
+	if root.Val == sum {
+		res++
+	}
+	res += findPath437(root.Left, sum-root.Val)
+	res += findPath437(root.Right, sum-root.Val)
+	return res
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0438. Find All Anagrams in a String/438. Find All Anagrams in a String.go b/leetcode/0438.Find-All-Anagrams-in-a-String/438. Find All Anagrams in a String.go
similarity index 100%
rename from Algorithms/0438. Find All Anagrams in a String/438. Find All Anagrams in a String.go
rename to leetcode/0438.Find-All-Anagrams-in-a-String/438. Find All Anagrams in a String.go
diff --git a/Algorithms/0438. Find All Anagrams in a String/438. Find All Anagrams in a String_test.go b/leetcode/0438.Find-All-Anagrams-in-a-String/438. Find All Anagrams in a String_test.go
similarity index 93%
rename from Algorithms/0438. Find All Anagrams in a String/438. Find All Anagrams in a String_test.go
rename to leetcode/0438.Find-All-Anagrams-in-a-String/438. Find All Anagrams in a String_test.go
index 1fcd08cd6..8d457e4e6 100644
--- a/Algorithms/0438. Find All Anagrams in a String/438. Find All Anagrams in a String_test.go	
+++ b/leetcode/0438.Find-All-Anagrams-in-a-String/438. Find All Anagrams in a String_test.go	
@@ -27,22 +27,22 @@ func Test_Problem438(t *testing.T) {
 
 	qs := []question438{
 
-		question438{
+		{
 			para438{"abab", "ab"},
 			ans438{[]int{0, 1, 2}},
 		},
 
-		question438{
+		{
 			para438{"cbaebabacd", "abc"},
 			ans438{[]int{0, 6}},
 		},
 
-		question438{
+		{
 			para438{"", "abc"},
 			ans438{[]int{}},
 		},
 
-		question438{
+		{
 			para438{"abacbabc", "abc"},
 			ans438{[]int{1, 2, 3, 5}},
 		},
diff --git a/Algorithms/0438. Find All Anagrams in a String/README.md b/leetcode/0438.Find-All-Anagrams-in-a-String/README.md
similarity index 100%
rename from Algorithms/0438. Find All Anagrams in a String/README.md
rename to leetcode/0438.Find-All-Anagrams-in-a-String/README.md
diff --git a/Algorithms/0441. Arranging Coins/441. Arranging Coins.go b/leetcode/0441.Arranging-Coins/441. Arranging Coins.go
similarity index 100%
rename from Algorithms/0441. Arranging Coins/441. Arranging Coins.go
rename to leetcode/0441.Arranging-Coins/441. Arranging Coins.go
diff --git a/Algorithms/0441. Arranging Coins/441. Arranging Coins_test.go b/leetcode/0441.Arranging-Coins/441. Arranging Coins_test.go
similarity index 95%
rename from Algorithms/0441. Arranging Coins/441. Arranging Coins_test.go
rename to leetcode/0441.Arranging-Coins/441. Arranging Coins_test.go
index a5ac60cd8..51f8403cf 100644
--- a/Algorithms/0441. Arranging Coins/441. Arranging Coins_test.go	
+++ b/leetcode/0441.Arranging-Coins/441. Arranging Coins_test.go	
@@ -26,12 +26,12 @@ func Test_Problem441(t *testing.T) {
 
 	qs := []question441{
 
-		question441{
+		{
 			para441{5},
 			ans441{2},
 		},
 
-		question441{
+		{
 			para441{8},
 			ans441{3},
 		},
diff --git a/leetcode/0441.Arranging-Coins/README.md b/leetcode/0441.Arranging-Coins/README.md
new file mode 100755
index 000000000..83cc7c1de
--- /dev/null
+++ b/leetcode/0441.Arranging-Coins/README.md
@@ -0,0 +1,45 @@
+# [441. Arranging Coins](https://leetcode.com/problems/arranging-coins/)
+
+## 题目
+
+You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins.
+
+Given n, find the total number of **full** staircase rows that can be formed.
+
+n is a non-negative integer and fits within the range of a 32-bit signed integer.
+
+**Example 1:**
+
+    n = 5
+    
+    The coins can form the following rows:
+    ¤
+    ¤ ¤
+    ¤ ¤
+    
+    Because the 3rd row is incomplete, we return 2.
+
+**Example 2:**
+
+    n = 8
+    
+    The coins can form the following rows:
+    ¤
+    ¤ ¤
+    ¤ ¤ ¤
+    ¤ ¤
+    
+    Because the 4th row is incomplete, we return 3.
+
+
+## 题目大意
+
+你总共有 n 枚硬币，你需要将它们摆成一个阶梯形状，第 k 行就必须正好有 k 枚硬币。给定一个数字 n，找出可形成完整阶梯行的总行数。n 是一个非负整数，并且在32位有符号整型的范围内。
+
+
+
+## 解题思路
+
+
+- n 个硬币，按照递增的方式排列搭楼梯，第一层一个，第二层二个，……第 n 层需要 n 个硬币。问硬币 n 能够搭建到第几层？
+- 这一题有 2 种解法，第一种解法就是解方程求出 X，`(1+x)x/2 = n`，即 `x = floor(sqrt(2*n+1/4) - 1/2)`，第二种解法是模拟。
diff --git a/leetcode/0445.Add-Two-Numbers-II/445. Add Two Numbers II.go b/leetcode/0445.Add-Two-Numbers-II/445. Add Two Numbers II.go
new file mode 100644
index 000000000..e4ddee058
--- /dev/null
+++ b/leetcode/0445.Add-Two-Numbers-II/445. Add Two Numbers II.go	
@@ -0,0 +1,138 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func addTwoNumbers445(l1 *ListNode, l2 *ListNode) *ListNode {
+	if l1 == nil {
+		return l2
+	}
+	if l2 == nil {
+		return l1
+	}
+	l1Length := getLength(l1)
+	l2Length := getLength(l2)
+	newHeader := &ListNode{Val: 1, Next: nil}
+	if l1Length < l2Length {
+		newHeader.Next = addNode(l2, l1, l2Length-l1Length)
+	} else {
+		newHeader.Next = addNode(l1, l2, l1Length-l2Length)
+	}
+	if newHeader.Next.Val > 9 {
+		newHeader.Next.Val = newHeader.Next.Val % 10
+		return newHeader
+	}
+	return newHeader.Next
+}
+
+func addNode(l1 *ListNode, l2 *ListNode, offset int) *ListNode {
+	if l1 == nil {
+		return nil
+	}
+	var (
+		res, node *ListNode
+	)
+	if offset == 0 {
+		res = &ListNode{Val: l1.Val + l2.Val, Next: nil}
+		node = addNode(l1.Next, l2.Next, 0)
+	} else {
+		res = &ListNode{Val: l1.Val, Next: nil}
+		node = addNode(l1.Next, l2, offset-1)
+	}
+	if node != nil && node.Val > 9 {
+		res.Val++
+		node.Val = node.Val % 10
+	}
+	res.Next = node
+	return res
+}
+
+func getLength(l *ListNode) int {
+	count := 0
+	cur := l
+	for cur != nil {
+		count++
+		cur = cur.Next
+	}
+	return count
+}
+
+func addTwoNumbers1(l1 *ListNode, l2 *ListNode) *ListNode {
+	reservedL1 := reverseList(l1)
+	reservedL2 := reverseList(l2)
+
+	dummyHead := &ListNode{}
+	head := dummyHead
+	carry := 0
+	for reservedL1 != nil || reservedL2 != nil || carry > 0 {
+		val := carry
+		if reservedL1 != nil {
+			val = reservedL1.Val + val
+			reservedL1 = reservedL1.Next
+		}
+		if reservedL2 != nil {
+			val = reservedL2.Val + val
+			reservedL2 = reservedL2.Next
+		}
+		carry = val / 10
+		head.Next = &ListNode{Val: val % 10}
+		head = head.Next
+	}
+	return reverseList(dummyHead.Next)
+}
+
+func reverseList(head *ListNode) *ListNode {
+	var prev *ListNode
+	for head != nil {
+		tmp := head.Next
+		head.Next = prev
+
+		prev = head
+		head = tmp
+	}
+	return prev
+}
+
+func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
+	stack1 := pushStack(l1)
+	stack2 := pushStack(l2)
+
+	dummyHead := &ListNode{}
+	head := dummyHead
+	carry := 0
+	for len(stack1) > 0 || len(stack2) > 0 || carry > 0 {
+		val := carry
+		if len(stack1) > 0 {
+			val = val + stack1[len(stack1)-1]
+			stack1 = stack1[:len(stack1)-1]
+		}
+		if len(stack2) > 0 {
+			val = val + stack2[len(stack2)-1]
+			stack2 = stack2[:len(stack2)-1]
+		}
+		carry = val / 10
+		tmp := head.Next
+		head.Next = &ListNode{Val: val % 10, Next: tmp}
+	}
+	return dummyHead.Next
+}
+
+func pushStack(l *ListNode) []int {
+	var stack []int
+	for l != nil {
+		stack = append(stack, l.Val)
+		l = l.Next
+	}
+	return stack
+}
diff --git a/Algorithms/0445. Add Two Numbers II/445. Add Two Numbers II_test.go b/leetcode/0445.Add-Two-Numbers-II/445. Add Two Numbers II_test.go
similarity index 83%
rename from Algorithms/0445. Add Two Numbers II/445. Add Two Numbers II_test.go
rename to leetcode/0445.Add-Two-Numbers-II/445. Add Two Numbers II_test.go
index 2c1bd3526..fbdd48dd7 100644
--- a/Algorithms/0445. Add Two Numbers II/445. Add Two Numbers II_test.go	
+++ b/leetcode/0445.Add-Two-Numbers-II/445. Add Two Numbers II_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question445 struct {
@@ -27,42 +29,42 @@ func Test_Problem445(t *testing.T) {
 
 	qs := []question445{
 
-		question445{
+		{
 			para445{[]int{}, []int{}},
 			ans445{[]int{}},
 		},
 
-		question445{
+		{
 			para445{[]int{1}, []int{1}},
 			ans445{[]int{2}},
 		},
 
-		question445{
+		{
 			para445{[]int{1, 2, 3, 4}, []int{1, 2, 3, 4}},
 			ans445{[]int{2, 4, 6, 8}},
 		},
 
-		question445{
+		{
 			para445{[]int{1, 2, 3, 4, 5}, []int{1, 2, 3, 4, 5}},
 			ans445{[]int{2, 4, 6, 9, 0}},
 		},
 
-		question445{
+		{
 			para445{[]int{1}, []int{9, 9, 9, 9, 9}},
 			ans445{[]int{1, 0, 0, 0, 0, 0}},
 		},
 
-		question445{
+		{
 			para445{[]int{9, 9, 9, 9, 9}, []int{1}},
 			ans445{[]int{1, 0, 0, 0, 0, 0}},
 		},
 
-		question445{
+		{
 			para445{[]int{2, 4, 3}, []int{5, 6, 4}},
 			ans445{[]int{8, 0, 7}},
 		},
 
-		question445{
+		{
 			para445{[]int{1, 8, 3}, []int{7, 1}},
 			ans445{[]int{2, 5, 4}},
 		},
@@ -73,7 +75,7 @@ func Test_Problem445(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans445, q.para445
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(addTwoNumbers445(S2l(p.one), S2l(p.another))))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(addTwoNumbers445(structures.Ints2List(p.one), structures.Ints2List(p.another))))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0445. Add Two Numbers II/README.md b/leetcode/0445.Add-Two-Numbers-II/README.md
similarity index 100%
rename from Algorithms/0445. Add Two Numbers II/README.md
rename to leetcode/0445.Add-Two-Numbers-II/README.md
diff --git a/Algorithms/0447. Number of Boomerangs/447. Number of Boomerangs.go b/leetcode/0447.Number-of-Boomerangs/447. Number of Boomerangs.go
similarity index 100%
rename from Algorithms/0447. Number of Boomerangs/447. Number of Boomerangs.go
rename to leetcode/0447.Number-of-Boomerangs/447. Number of Boomerangs.go
diff --git a/Algorithms/0447. Number of Boomerangs/447. Number of Boomerangs_test.go b/leetcode/0447.Number-of-Boomerangs/447. Number of Boomerangs_test.go
similarity index 89%
rename from Algorithms/0447. Number of Boomerangs/447. Number of Boomerangs_test.go
rename to leetcode/0447.Number-of-Boomerangs/447. Number of Boomerangs_test.go
index efa0c8987..19e84708a 100644
--- a/Algorithms/0447. Number of Boomerangs/447. Number of Boomerangs_test.go	
+++ b/leetcode/0447.Number-of-Boomerangs/447. Number of Boomerangs_test.go	
@@ -26,8 +26,8 @@ func Test_Problem447(t *testing.T) {
 
 	qs := []question447{
 
-		question447{
-			para447{[][]int{[]int{0, 0}, []int{1, 0}, []int{2, 0}}},
+		{
+			para447{[][]int{{0, 0}, {1, 0}, {2, 0}}},
 			ans447{2},
 		},
 
diff --git a/Algorithms/0447. Number of Boomerangs/README.md b/leetcode/0447.Number-of-Boomerangs/README.md
similarity index 100%
rename from Algorithms/0447. Number of Boomerangs/README.md
rename to leetcode/0447.Number-of-Boomerangs/README.md
diff --git a/leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/448. Find All Numbers Disappeared in an Array.go b/leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/448. Find All Numbers Disappeared in an Array.go
new file mode 100644
index 000000000..931133a37
--- /dev/null
+++ b/leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/448. Find All Numbers Disappeared in an Array.go	
@@ -0,0 +1,19 @@
+package leetcode
+
+func findDisappearedNumbers(nums []int) []int {
+	res := []int{}
+	for _, v := range nums {
+		if v < 0 {
+			v = -v
+		}
+		if nums[v-1] > 0 {
+			nums[v-1] = -nums[v-1]
+		}
+	}
+	for i, v := range nums {
+		if v > 0 {
+			res = append(res, i+1)
+		}
+	}
+	return res
+}
diff --git a/leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/448. Find All Numbers Disappeared in an Array_test.go b/leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/448. Find All Numbers Disappeared in an Array_test.go
new file mode 100644
index 000000000..232e56bfb
--- /dev/null
+++ b/leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/448. Find All Numbers Disappeared in an Array_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question448 struct {
+	para448
+	ans448
+}
+
+// para 是参数
+// one 代表第一个参数
+type para448 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans448 struct {
+	one []int
+}
+
+func Test_Problem448(t *testing.T) {
+
+	qs := []question448{
+
+		{
+			para448{[]int{4, 3, 2, 7, 8, 2, 3, 1}},
+			ans448{[]int{5, 6}},
+		},
+
+		{
+			para448{[]int{4, 3, 2, 10, 9, 2, 3, 1, 1, 1, 1}},
+			ans448{[]int{5, 6, 7, 8, 11}},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 448------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans448, q.para448
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findDisappearedNumbers(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/README.md b/leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/README.md
new file mode 100644
index 000000000..e1c6e556a
--- /dev/null
+++ b/leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/README.md
@@ -0,0 +1,53 @@
+# [448. Find All Numbers Disappeared in an Array](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/)
+
+
+## 题目
+
+Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
+
+Find all the elements of [1, n] inclusive that do not appear in this array.
+
+Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.
+
+**Example**:
+
+```
+Input:
+[4,3,2,7,8,2,3,1]
+
+Output:
+[5,6]
+```
+
+## 题目大意
+
+给定一个范围在  1 ≤ a[i] ≤ n ( n = 数组大小 ) 的 整型数组，数组中的元素一些出现了两次，另一些只出现一次。找到所有在 [1, n] 范围之间没有出现在数组中的数字。你能在不使用额外空间且时间复杂度为 O(n) 的情况下完成这个任务吗? 你可以假定返回的数组不算在额外空间内。
+
+
+
+## 解题思路
+
+- 找出 [1,n] 范围内没有出现在数组中的数字。要求不使用额外空间，并且时间复杂度为 O(n)。
+- 要求不能使用额外的空间，那么只能想办法在原有数组上进行修改，并且这个修改是可还原的。时间复杂度也只能允许我们一层循环。只要循环一次能标记出已经出现过的数字，这道题就可以按要求解答出来。这里笔者的标记方法是把 |nums[i]|-1 索引位置的元素标记为负数。即 nums[| nums[i] |- 1] * -1。这里需要注意的是，nums[i] 需要加绝对值，因为它可能被之前的数置为负数了，需要还原一下。最后再遍历一次数组，若当前数组元素 nums[i] 为负数，说明我们在数组中存在数字 i+1。把结果输出到最终数组里即可。
+
+## 代码
+
+```go
+func findDisappearedNumbers(nums []int) []int {
+	res := []int{}
+	for _, v := range nums {
+		if v < 0 {
+			v = -v
+		}
+		if nums[v-1] > 0 {
+			nums[v-1] = -nums[v-1]
+		}
+	}
+	for i, v := range nums {
+		if v > 0 {
+			res = append(res, i+1)
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0451. Sort Characters By Frequency/451. Sort Characters By Frequency.go b/leetcode/0451.Sort-Characters-By-Frequency/451. Sort Characters By Frequency.go
similarity index 100%
rename from Algorithms/0451. Sort Characters By Frequency/451. Sort Characters By Frequency.go
rename to leetcode/0451.Sort-Characters-By-Frequency/451. Sort Characters By Frequency.go
diff --git a/Algorithms/0451. Sort Characters By Frequency/451. Sort Characters By Frequency_test.go b/leetcode/0451.Sort-Characters-By-Frequency/451. Sort Characters By Frequency_test.go
similarity index 94%
rename from Algorithms/0451. Sort Characters By Frequency/451. Sort Characters By Frequency_test.go
rename to leetcode/0451.Sort-Characters-By-Frequency/451. Sort Characters By Frequency_test.go
index e578af168..a46b9acd0 100644
--- a/Algorithms/0451. Sort Characters By Frequency/451. Sort Characters By Frequency_test.go	
+++ b/leetcode/0451.Sort-Characters-By-Frequency/451. Sort Characters By Frequency_test.go	
@@ -25,17 +25,17 @@ type ans451 struct {
 func Test_Problem451(t *testing.T) {
 
 	qs := []question451{
-		question451{
+		{
 			para451{"tree"},
 			ans451{"eert"},
 		},
 
-		question451{
+		{
 			para451{"cccaaa"},
 			ans451{"cccaaa"},
 		},
 
-		question451{
+		{
 			para451{"Aabb"},
 			ans451{"bbAa"},
 		},
diff --git a/Algorithms/0451. Sort Characters By Frequency/README.md b/leetcode/0451.Sort-Characters-By-Frequency/README.md
similarity index 100%
rename from Algorithms/0451. Sort Characters By Frequency/README.md
rename to leetcode/0451.Sort-Characters-By-Frequency/README.md
diff --git a/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements.go b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements.go
new file mode 100644
index 000000000..fbdfeb299
--- /dev/null
+++ b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements.go	
@@ -0,0 +1,14 @@
+package leetcode
+
+import "math"
+
+func minMoves(nums []int) int {
+	sum, min, l := 0, math.MaxInt32, len(nums)
+	for _, v := range nums {
+		sum += v
+		if min > v {
+			min = v
+		}
+	}
+	return sum - min*l
+}
diff --git a/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements_test.go b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements_test.go
new file mode 100644
index 000000000..93e378825
--- /dev/null
+++ b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question453 struct {
+	para453
+	ans453
+}
+
+// para 是参数
+// one 代表第一个参数
+type para453 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans453 struct {
+	one int
+}
+
+func Test_Problem453(t *testing.T) {
+
+	qs := []question453{
+
+		{
+			para453{[]int{4, 3, 2, 7, 8, 2, 3, 1}},
+			ans453{22},
+		},
+
+		{
+			para453{[]int{1, 2, 3}},
+			ans453{3},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 453------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans453, q.para453
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minMoves(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/README.md b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/README.md
new file mode 100644
index 000000000..280f22a25
--- /dev/null
+++ b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/README.md
@@ -0,0 +1,51 @@
+# [453. Minimum Moves to Equal Array Elements](https://leetcode.com/problems/minimum-moves-to-equal-array-elements/)
+
+
+## 题目
+
+Given a **non-empty** integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1.
+
+**Example**:
+
+```
+Input:
+[1,2,3]
+
+Output:
+3
+
+Explanation:
+Only three moves are needed (remember each move increments two elements):
+
+[1,2,3]  =>  [2,3,3]  =>  [3,4,3]  =>  [4,4,4]
+```
+
+## 题目大意
+
+给定一个长度为 n 的非空整数数组，找到让数组所有元素相等的最小移动次数。每次移动将会使 n - 1 个元素增加 1。
+
+## 解题思路
+
+- 给定一个数组，要求输出让所有元素都相等的最小步数。每移动一步都会使得 n - 1 个元素 + 1 。
+- 数学题。这道题正着思考会考虑到排序或者暴力的方法上去。反过来思考一下，使得每个元素都相同，意思让所有元素的差异变为 0 。每次移动的过程中，都有 n - 1 个元素 + 1，那么没有 + 1 的那个元素和其他 n - 1 个元素相对差异就缩小了。所以这道题让所有元素都变为相等的最少步数，即等于让所有元素相对差异减少到最小的那个数。想到这里，此题就可以优雅的解出来了。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func minMoves(nums []int) int {
+	sum, min, l := 0, math.MaxInt32, len(nums)
+	for _, v := range nums {
+		sum += v
+		if min > v {
+			min = v
+		}
+	}
+	return sum - min*l
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0454. 4Sum II/454. 4Sum II.go b/leetcode/0454.4Sum-II/454. 4Sum II.go
similarity index 100%
rename from Algorithms/0454. 4Sum II/454. 4Sum II.go
rename to leetcode/0454.4Sum-II/454. 4Sum II.go
diff --git a/Algorithms/0454. 4Sum II/454. 4Sum II_test.go b/leetcode/0454.4Sum-II/454. 4Sum II_test.go
similarity index 97%
rename from Algorithms/0454. 4Sum II/454. 4Sum II_test.go
rename to leetcode/0454.4Sum-II/454. 4Sum II_test.go
index 54fc5b425..dc5677837 100644
--- a/Algorithms/0454. 4Sum II/454. 4Sum II_test.go	
+++ b/leetcode/0454.4Sum-II/454. 4Sum II_test.go	
@@ -29,7 +29,7 @@ func Test_Problem454(t *testing.T) {
 
 	qs := []question454{
 
-		question454{
+		{
 			para454{[]int{1, 2}, []int{-2, -1}, []int{-1, 2}, []int{0, 2}},
 			ans454{2},
 		},
diff --git a/Algorithms/0454. 4Sum II/README.md b/leetcode/0454.4Sum-II/README.md
similarity index 100%
rename from Algorithms/0454. 4Sum II/README.md
rename to leetcode/0454.4Sum-II/README.md
diff --git a/Algorithms/0455. Assign Cookies/455. Assign Cookies.go b/leetcode/0455.Assign-Cookies/455. Assign Cookies.go
similarity index 100%
rename from Algorithms/0455. Assign Cookies/455. Assign Cookies.go
rename to leetcode/0455.Assign-Cookies/455. Assign Cookies.go
diff --git a/Algorithms/0455. Assign Cookies/455. Assign Cookies_test.go b/leetcode/0455.Assign-Cookies/455. Assign Cookies_test.go
similarity index 95%
rename from Algorithms/0455. Assign Cookies/455. Assign Cookies_test.go
rename to leetcode/0455.Assign-Cookies/455. Assign Cookies_test.go
index e337930df..14fd09d36 100644
--- a/Algorithms/0455. Assign Cookies/455. Assign Cookies_test.go	
+++ b/leetcode/0455.Assign-Cookies/455. Assign Cookies_test.go	
@@ -27,12 +27,12 @@ func Test_Problem455(t *testing.T) {
 
 	qs := []question455{
 
-		question455{
+		{
 			para455{[]int{1, 2, 3}, []int{1, 1}},
 			ans455{1},
 		},
 
-		question455{
+		{
 			para455{[]int{1, 2}, []int{1, 2, 3}},
 			ans455{2},
 		},
diff --git a/Algorithms/0455. Assign Cookies/README.md b/leetcode/0455.Assign-Cookies/README.md
similarity index 100%
rename from Algorithms/0455. Assign Cookies/README.md
rename to leetcode/0455.Assign-Cookies/README.md
diff --git a/Algorithms/0456. 132 Pattern/456. 132 Pattern.go b/leetcode/0456.132-Pattern/456. 132 Pattern.go
similarity index 95%
rename from Algorithms/0456. 132 Pattern/456. 132 Pattern.go
rename to leetcode/0456.132-Pattern/456. 132 Pattern.go
index 89a75e6b1..af7db50b2 100644
--- a/Algorithms/0456. 132 Pattern/456. 132 Pattern.go	
+++ b/leetcode/0456.132-Pattern/456. 132 Pattern.go	
@@ -1,7 +1,6 @@
 package leetcode
 
 import (
-	"fmt"
 	"math"
 )
 
@@ -20,7 +19,6 @@ func find132pattern(nums []int) bool {
 			stack = stack[:len(stack)-1]
 		}
 		stack = append(stack, nums[i])
-		fmt.Printf("stack = %v \n", stack)
 	}
 	return false
 }
diff --git a/Algorithms/0456. 132 Pattern/456. 132 Pattern_test.go b/leetcode/0456.132-Pattern/456. 132 Pattern_test.go
similarity index 92%
rename from Algorithms/0456. 132 Pattern/456. 132 Pattern_test.go
rename to leetcode/0456.132-Pattern/456. 132 Pattern_test.go
index 706bafb53..bcf7895e0 100644
--- a/Algorithms/0456. 132 Pattern/456. 132 Pattern_test.go	
+++ b/leetcode/0456.132-Pattern/456. 132 Pattern_test.go	
@@ -26,27 +26,27 @@ func Test_Problem456(t *testing.T) {
 
 	qs := []question456{
 
-		question456{
+		{
 			para456{[]int{}},
 			ans456{false},
 		},
 
-		question456{
+		{
 			para456{[]int{1, 2, 3, 4}},
 			ans456{false},
 		},
 
-		question456{
+		{
 			para456{[]int{3, 1, 4, 2}},
 			ans456{true},
 		},
 
-		question456{
+		{
 			para456{[]int{-1, 3, 2, 0}},
 			ans456{true},
 		},
 
-		question456{
+		{
 			para456{[]int{3, 5, 0, 3, 4}},
 			ans456{true},
 		},
diff --git a/Algorithms/0456. 132 Pattern/README.md b/leetcode/0456.132-Pattern/README.md
similarity index 100%
rename from Algorithms/0456. 132 Pattern/README.md
rename to leetcode/0456.132-Pattern/README.md
diff --git a/Algorithms/0457. Circular Array Loop/457. Circular Array Loop.go b/leetcode/0457.Circular-Array-Loop/457. Circular Array Loop.go
similarity index 100%
rename from Algorithms/0457. Circular Array Loop/457. Circular Array Loop.go
rename to leetcode/0457.Circular-Array-Loop/457. Circular Array Loop.go
diff --git a/Algorithms/0457. Circular Array Loop/457. Circular Array Loop_test.go b/leetcode/0457.Circular-Array-Loop/457. Circular Array Loop_test.go
similarity index 89%
rename from Algorithms/0457. Circular Array Loop/457. Circular Array Loop_test.go
rename to leetcode/0457.Circular-Array-Loop/457. Circular Array Loop_test.go
index 8d00c8c96..776e1f5ad 100644
--- a/Algorithms/0457. Circular Array Loop/457. Circular Array Loop_test.go	
+++ b/leetcode/0457.Circular-Array-Loop/457. Circular Array Loop_test.go	
@@ -26,41 +26,41 @@ func Test_Problem457(t *testing.T) {
 
 	qs := []question457{
 
-		question457{
+		{
 			para457{[]int{-1}},
 			ans457{false},
 		},
 
-		question457{
+		{
 			para457{[]int{3, 1, 2}},
 			ans457{true},
 		},
 
-		question457{
+		{
 			para457{[]int{-8, -1, 1, 7, 2}},
 			ans457{false},
 		},
 
-		question457{
+		{
 			para457{[]int{-1, -2, -3, -4, -5}},
 			ans457{false},
 		},
 
-		question457{
+		{
 			para457{[]int{}},
 			ans457{false},
 		},
 
-		question457{
+		{
 			para457{[]int{2, -1, 1, 2, 2}},
 			ans457{true},
 		},
 
-		question457{
+		{
 			para457{[]int{-1, 2}},
 			ans457{false},
 		},
-		question457{
+		{
 			para457{[]int{-2, 1, -1, -2, -2}},
 			ans457{false},
 		},
diff --git a/leetcode/0457.Circular-Array-Loop/README.md b/leetcode/0457.Circular-Array-Loop/README.md
new file mode 100755
index 000000000..80bc7535d
--- /dev/null
+++ b/leetcode/0457.Circular-Array-Loop/README.md
@@ -0,0 +1,59 @@
+# [457. Circular Array Loop](https://leetcode.com/problems/circular-array-loop/)
+
+
+## 题目
+
+You are given a **circular** array `nums` of positive and negative integers. If a number k at an index is positive, then move forward k steps. Conversely, if it's negative (-k), move backward k steps. Since the array is circular, you may assume that the last element's next element is the first element, and the first element's previous element is the last element.
+
+Determine if there is a loop (or a cycle) in `nums`. A cycle must start and end at the same index and the cycle's length > 1. Furthermore, movements in a cycle must all follow a single direction. In other words, a cycle must not consist of both forward and backward movements.
+
+**Example 1:**
+
+    Input: [2,-1,1,2,2]
+    Output: true
+    Explanation: There is a cycle, from index 0 -> 2 -> 3 -> 0. The cycle's length is 3.
+
+**Example 2:**
+
+    Input: [-1,2]
+    Output: false
+    Explanation: The movement from index 1 -> 1 -> 1 ... is not a cycle, because the cycle's length is 1. By definition the cycle's length must be greater than 1.
+
+**Example 3:**
+
+    Input: [-2,1,-1,-2,-2]
+    Output: false
+    Explanation: The movement from index 1 -> 2 -> 1 -> ... is not a cycle, because movement from index 1 -> 2 is a forward movement, but movement from index 2 -> 1 is a backward movement. All movements in a cycle must follow a single direction.
+
+**Note:**
+
+1. -1000 ≤ nums[i] ≤ 1000
+2. nums[i] ≠ 0
+3. 1 ≤ nums.length ≤ 5000
+
+**Follow up:**
+
+Could you solve it in **O(n)** time complexity and **O(1)** extra space complexity?
+
+
+## 题目大意
+
+给定一个含有正整数和负整数的环形数组 nums。 如果某个索引中的数 k 为正数，则向前移动 k 个索引。相反，如果是负数 (-k)，则向后移动 k 个索引。因为数组是环形的，所以可以假设最后一个元素的下一个元素是第一个元素，而第一个元素的前一个元素是最后一个元素。
+
+确定 nums 中是否存在循环（或周期）。循环必须在相同的索引处开始和结束并且循环长度 > 1。此外，一个循环中的所有运动都必须沿着同一方向进行。换句话说，一个循环中不能同时包括向前的运动和向后的运动。
+
+提示：
+
+- -1000 ≤ nums[i] ≤ 1000
+- nums[i] ≠ 0
+- 1 ≤ nums.length ≤ 5000
+
+进阶：
+
+- 你能写出时间时间复杂度为 O(n) 和额外空间复杂度为 O(1) 的算法吗？
+
+## 解题思路
+
+
+- 给出一个循环数组，数组的数字代表了前进和后退的步数，+ 代表往右(前进)，- 代表往左(后退)。问这个循环数组中是否存在一个循环，并且这个循环内不能只有一个元素，循环的方向都必须是同方向的。
+- 遇到循环就可以优先考虑用快慢指针的方法判断循环，这一题对循环增加了一个条件，循环不能只是单元素的循环，所以在快慢指针中加入这个判断条件。还有一个判断条件是循环的方向必须是同向的，这个简单，用 `num[i] * num[j] > 0` 就可以判断出是同向的(如果是反向的，那么两者的乘积必然是负数)，如果没有找到循环，可以将当前已经走过的路径上的 num[] 值都置为 0，标记已经访问过了。下次循环遇到访问过的元素，`num[i] * num[j] > 0` 就会是 0，提前退出找循环的过程。
diff --git a/leetcode/0458.Poor-Pigs/458.Poor Pigs.go b/leetcode/0458.Poor-Pigs/458.Poor Pigs.go
new file mode 100644
index 000000000..81162d51e
--- /dev/null
+++ b/leetcode/0458.Poor-Pigs/458.Poor Pigs.go	
@@ -0,0 +1,8 @@
+package leetcode
+
+import "math"
+
+func poorPigs(buckets int, minutesToDie int, minutesToTest int) int {
+	base := minutesToTest/minutesToDie + 1
+	return int(math.Ceil(math.Log10(float64(buckets)) / math.Log10(float64(base))))
+}
diff --git a/leetcode/0458.Poor-Pigs/458.Poor Pigs_test.go b/leetcode/0458.Poor-Pigs/458.Poor Pigs_test.go
new file mode 100644
index 000000000..50bb561ea
--- /dev/null
+++ b/leetcode/0458.Poor-Pigs/458.Poor Pigs_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question458 struct {
+	para458
+	ans458
+}
+
+// para 是参数
+type para458 struct {
+	buckets       int
+	minutesToDie  int
+	minutesToTest int
+}
+
+// ans 是答案
+type ans458 struct {
+	ans int
+}
+
+func Test_Problem458(t *testing.T) {
+
+	qs := []question458{
+
+		{
+			para458{1000, 15, 60},
+			ans458{5},
+		},
+
+		{
+			para458{4, 15, 15},
+			ans458{2},
+		},
+
+		{
+			para458{4, 15, 30},
+			ans458{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 458------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans458, q.para458
+		fmt.Printf("【input】:%v    【output】:%v\n", p, poorPigs(p.buckets, p.minutesToDie, p.minutesToTest))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0458.Poor-Pigs/README.md b/leetcode/0458.Poor-Pigs/README.md
new file mode 100644
index 000000000..e8dba552d
--- /dev/null
+++ b/leetcode/0458.Poor-Pigs/README.md
@@ -0,0 +1,77 @@
+# [458. Poor Pigs](https://leetcode.com/problems/poor-pigs/)
+
+## 题目
+
+There are buckets buckets of liquid, where exactly one of the buckets is poisonous. To figure out which one is poisonous, you feed some number of (poor) pigs the liquid to see whether they will die or not. Unfortunately, you only have minutesToTest minutes to determine which bucket is poisonous.
+
+You can feed the pigs according to these steps:
+
+- Choose some live pigs to feed.
+- For each pig, choose which buckets to feed it. The pig will consume all the chosen buckets simultaneously and will take no time.
+- Wait for minutesToDie minutes. You may not feed any other pigs during this time.
+- After minutesToDie minutes have passed, any pigs that have been fed the poisonous bucket will die, and all others will survive.
+- Repeat this process until you run out of time.
+
+Given buckets, minutesToDie, and minutesToTest, return the minimum number of pigs needed to figure out which bucket is poisonous within the allotted time.
+
+**Example 1**:
+
+    Input: buckets = 1000, minutesToDie = 15, minutesToTest = 60
+    Output: 5
+
+**Example 2**:
+
+    Input: buckets = 4, minutesToDie = 15, minutesToTest = 15
+    Output: 2
+
+**Example 3**:
+
+    Input: buckets = 4, minutesToDie = 15, minutesToTest = 30
+    Output: 2
+
+**Constraints:**
+
+- 1 <= buckets <= 1000
+- 1 <= minutesToDie <= minutesToTest <= 100
+
+## 题目大意
+
+有 buckets 桶液体，其中 正好 有一桶含有毒药，其余装的都是水。它们从外观看起来都一样。为了弄清楚哪只水桶含有毒药，你可以喂一些猪喝，通过观察猪是否会死进行判断。不幸的是，你只有 minutesToTest 分钟时间来确定哪桶液体是有毒的。
+
+喂猪的规则如下：
+
+- 选择若干活猪进行喂养
+- 可以允许小猪同时饮用任意数量的桶中的水，并且该过程不需要时间。
+- 小猪喝完水后，必须有 minutesToDie 分钟的冷却时间。在这段时间里，你只能观察，而不允许继续喂猪。
+- 过了 minutesToDie 分钟后，所有喝到毒药的猪都会死去，其他所有猪都会活下来。
+- 重复这一过程，直到时间用完。
+
+给你桶的数目 buckets ，minutesToDie 和 minutesToTest ，返回在规定时间内判断哪个桶有毒所需的 最小 猪数。
+
+## 解题思路
+
+使用数学方法,以 minutesToDie=15, minutesToTest=60, 1 只小猪为例,可以测试 5 只桶
+
+- 0-15 小猪吃第一个桶中的液体,如果死去，则第一个桶有毒,否则继续测试
+- 15-30 小猪吃第二个桶中的液体，如果死去，则第二个桶有毒,否则继续测试
+- 30-45 小猪吃第三个桶中的液体，如果死去，则第三个桶有毒,否则继续测试
+- 45-60 小猪吃第四个桶中的液体，如果死去，则第四个桶有毒
+- 如果最后小猪没有死去，则第五个桶有毒
+
+所以一只小猪在 minutesToDie 和 minutesToTest 时间一定的情况下可以最多判断 base = minutesToTest / minutesToDie + 1 个桶
+
+假设小猪的数量是 num,那么 pow(base, num) >= buckets,根据对数运算规则，两边分别取对数得到: num >= Log10(buckets) / Log10(base) 
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func poorPigs(buckets int, minutesToDie int, minutesToTest int) int {
+	base := minutesToTest/minutesToDie + 1
+	return int(math.Ceil(math.Log10(float64(buckets)) / math.Log10(float64(base))))
+}
+```
\ No newline at end of file
diff --git a/leetcode/0460.LFU-Cache/460. LFU Cache.go b/leetcode/0460.LFU-Cache/460. LFU Cache.go
new file mode 100644
index 000000000..e2a29f539
--- /dev/null
+++ b/leetcode/0460.LFU-Cache/460. LFU Cache.go	
@@ -0,0 +1,74 @@
+package leetcode
+
+import "container/list"
+
+type LFUCache struct {
+	nodes    map[int]*list.Element
+	lists    map[int]*list.List
+	capacity int
+	min      int
+}
+
+type node struct {
+	key       int
+	value     int
+	frequency int
+}
+
+func Constructor(capacity int) LFUCache {
+	return LFUCache{nodes: make(map[int]*list.Element),
+		lists:    make(map[int]*list.List),
+		capacity: capacity,
+		min:      0,
+	}
+}
+
+func (this *LFUCache) Get(key int) int {
+	value, ok := this.nodes[key]
+	if !ok {
+		return -1
+	}
+	currentNode := value.Value.(*node)
+	this.lists[currentNode.frequency].Remove(value)
+	currentNode.frequency++
+	if _, ok := this.lists[currentNode.frequency]; !ok {
+		this.lists[currentNode.frequency] = list.New()
+	}
+	newList := this.lists[currentNode.frequency]
+	newNode := newList.PushBack(currentNode)
+	this.nodes[key] = newNode
+	if currentNode.frequency-1 == this.min && this.lists[currentNode.frequency-1].Len() == 0 {
+		this.min++
+	}
+	return currentNode.value
+}
+
+func (this *LFUCache) Put(key int, value int) {
+	if this.capacity == 0 {
+		return
+	}
+	if currentValue, ok := this.nodes[key]; ok {
+		currentNode := currentValue.Value.(*node)
+		currentNode.value = value
+		this.Get(key)
+		return
+	}
+	if this.capacity == len(this.nodes) {
+		currentList := this.lists[this.min]
+		frontNode := currentList.Front()
+		delete(this.nodes, frontNode.Value.(*node).key)
+		currentList.Remove(frontNode)
+	}
+	this.min = 1
+	currentNode := &node{
+		key:       key,
+		value:     value,
+		frequency: 1,
+	}
+	if _, ok := this.lists[1]; !ok {
+		this.lists[1] = list.New()
+	}
+	newList := this.lists[1]
+	newNode := newList.PushBack(currentNode)
+	this.nodes[key] = newNode
+}
diff --git a/leetcode/0460.LFU-Cache/460. LFU Cache_test.go b/leetcode/0460.LFU-Cache/460. LFU Cache_test.go
new file mode 100644
index 000000000..12205d94e
--- /dev/null
+++ b/leetcode/0460.LFU-Cache/460. LFU Cache_test.go	
@@ -0,0 +1,64 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem460(t *testing.T) {
+	obj := Constructor(5)
+	fmt.Printf("obj.list = %v obj.map = %v obj.min = %v\n", MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	obj.Put(1, 1)
+	fmt.Printf("obj.list = %v obj.map = %v obj.min = %v\n", MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	obj.Put(2, 2)
+	fmt.Printf("obj.list = %v obj.map = %v obj.min = %v\n", MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	obj.Put(3, 3)
+	fmt.Printf("obj.list = %v obj.map = %v obj.min = %v\n", MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	obj.Put(4, 4)
+	fmt.Printf("obj.list = %v obj.map = %v obj.min = %v\n", MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	obj.Put(5, 5)
+	fmt.Printf("obj.list = %v obj.map = %v obj.min = %v\n", MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+
+	param1 := obj.Get(4)
+	fmt.Printf("param_1 = %v obj.list = %v obj.map = %v obj.min = %v\n", param1, MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	param1 = obj.Get(4)
+	fmt.Printf("param_1 = %v obj.list = %v obj.map = %v obj.min = %v\n", param1, MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	param1 = obj.Get(4)
+	fmt.Printf("param_1 = %v obj.list = %v obj.map = %v obj.min = %v\n", param1, MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	param1 = obj.Get(5)
+	fmt.Printf("param_1 = %v obj.list = %v obj.map = %v obj.min = %v\n", param1, MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	param1 = obj.Get(5)
+	fmt.Printf("param_1 = %v obj.list = %v obj.map = %v obj.min = %v\n", param1, MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	param1 = obj.Get(5)
+	fmt.Printf("param_1 = %v obj.list = %v obj.map = %v obj.min = %v\n", param1, MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	obj.Put(6, 6)
+	fmt.Printf("obj.list = %v obj.map = %v obj.min = %v\n", MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	obj.Put(7, 7)
+	fmt.Printf("obj.list = %v obj.map = %v obj.min = %v\n", MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+	obj.Put(8, 8)
+	fmt.Printf("obj.list = %v obj.map = %v obj.min = %v\n", MLists2Ints(&obj), MList2Ints(&obj), obj.min)
+}
+
+func MList2Ints(lfu *LFUCache) map[int][][]int {
+	res := map[int][][]int{}
+	for k, v := range lfu.nodes {
+		node := v.Value.(*node)
+		arr := [][]int{}
+		tmp := []int{node.key, node.value, node.frequency}
+		arr = append(arr, tmp)
+		res[k] = arr
+	}
+	return res
+}
+
+func MLists2Ints(lfu *LFUCache) map[int][]int {
+	res := map[int][]int{}
+	for k, v := range lfu.lists {
+		tmp := []int{}
+		for head := v.Front(); head != nil; head = head.Next() {
+			tmp = append(tmp, head.Value.(*node).value)
+		}
+		res[k] = tmp
+	}
+	return res
+}
diff --git a/leetcode/0460.LFU-Cache/README.md b/leetcode/0460.LFU-Cache/README.md
new file mode 100644
index 000000000..f4a07c445
--- /dev/null
+++ b/leetcode/0460.LFU-Cache/README.md
@@ -0,0 +1,142 @@
+# [460. LFU Cache](https://leetcode.com/problems/lfu-cache/)
+
+
+## 题目
+
+Design and implement a data structure for [Least Frequently Used (LFU)](https://en.wikipedia.org/wiki/Least_frequently_used) cache.
+
+Implement the `LFUCache` class:
+
+- `LFUCache(int capacity)` Initializes the object with the `capacity` of the data structure.
+- `int get(int key)` Gets the value of the `key` if the `key` exists in the cache. Otherwise, returns `1`.
+- `void put(int key, int value)` Sets or inserts the value if the `key` is not already present. When the cache reaches its `capacity`, it should invalidate the least frequently used item before inserting a new item. For this problem, when there is a tie (i.e., two or more keys with the same frequency), **the least recently** used `key` would be evicted.
+
+**Notice that** the number of times an item is used is the number of calls to the `get` and `put` functions for that item since it was inserted. This number is set to zero when the item is removed.
+
+**Example 1:**
+
+```
+Input
+["LFUCache", "put", "put", "get", "put", "get", "get", "put", "get", "get", "get"]
+[[2], [1, 1], [2, 2], [1], [3, 3], [2], [3], [4, 4], [1], [3], [4]]
+Output
+[null, null, null, 1, null, -1, 3, null, -1, 3, 4]
+
+Explanation
+LFUCache lfu = new LFUCache(2);
+lfu.put(1, 1);
+lfu.put(2, 2);
+lfu.get(1);      // return 1
+lfu.put(3, 3);   // evicts key 2
+lfu.get(2);      // return -1 (not found)
+lfu.get(3);      // return 3
+lfu.put(4, 4);   // evicts key 1.
+lfu.get(1);      // return -1 (not found)
+lfu.get(3);      // return 3
+lfu.get(4);      // return 4
+
+```
+
+**Constraints:**
+
+- `0 <= capacity, key, value <= 104`
+- At most `10^5` calls will be made to `get` and `put`.
+
+**Follow up:** Could you do both operations in `O(1)` time complexity?
+
+## 题目大意
+
+请你为 最不经常使用（LFU）缓存算法设计并实现数据结构。
+
+实现 LFUCache 类：
+
+- LFUCache(int capacity) - 用数据结构的容量 capacity 初始化对象
+- int get(int key) - 如果键存在于缓存中，则获取键的值，否则返回 -1。
+- void put(int key, int value) - 如果键已存在，则变更其值；如果键不存在，请插入键值对。当缓存达到其容量时，则应该在插入新项之前，使最不经常使用的项无效。在此问题中，当存在平局（即两个或更多个键具有相同使用频率）时，应该去除 最久未使用 的键。
+
+注意「项的使用次数」就是自插入该项以来对其调用 get 和 put 函数的次数之和。使用次数会在对应项被移除后置为 0 。
+
+进阶：你是否可以在 O(1) 时间复杂度内执行两项操作？
+
+## 解题思路
+
+- 这一题是 LFU 经典面试题，详细解释见[第三章 LFUCache 模板](https://books.halfrost.com/leetcode/ChapterThree/LFUCache/)。
+
+## 代码
+
+```go
+package leetcode
+
+import "container/list"
+
+type LFUCache struct {
+	nodes    map[int]*list.Element
+	lists    map[int]*list.List
+	capacity int
+	min      int
+}
+
+type node struct {
+	key       int
+	value     int
+	frequency int
+}
+
+func Constructor(capacity int) LFUCache {
+	return LFUCache{nodes: make(map[int]*list.Element),
+		lists:    make(map[int]*list.List),
+		capacity: capacity,
+		min:      0,
+	}
+}
+
+func (this *LFUCache) Get(key int) int {
+	value, ok := this.nodes[key]
+	if !ok {
+		return -1
+	}
+	currentNode := value.Value.(*node)
+	this.lists[currentNode.frequency].Remove(value)
+	currentNode.frequency++
+	if _, ok := this.lists[currentNode.frequency]; !ok {
+		this.lists[currentNode.frequency] = list.New()
+	}
+	newList := this.lists[currentNode.frequency]
+	newNode := newList.PushBack(currentNode)
+	this.nodes[key] = newNode
+	if currentNode.frequency-1 == this.min && this.lists[currentNode.frequency-1].Len() == 0 {
+		this.min++
+	}
+	return currentNode.value
+}
+
+func (this *LFUCache) Put(key int, value int) {
+	if this.capacity == 0 {
+		return
+	}
+	if currentValue, ok := this.nodes[key]; ok {
+		currentNode := currentValue.Value.(*node)
+		currentNode.value = value
+		this.Get(key)
+		return
+	}
+	if this.capacity == len(this.nodes) {
+		currentList := this.lists[this.min]
+		frontNode := currentList.Front()
+		delete(this.nodes, frontNode.Value.(*node).key)
+		currentList.Remove(frontNode)
+	}
+	this.min = 1
+	currentNode := &node{
+		key:       key,
+		value:     value,
+		frequency: 1,
+	}
+	if _, ok := this.lists[1]; !ok {
+		this.lists[1] = list.New()
+	}
+	newList := this.lists[1]
+	newNode := newList.PushBack(currentNode)
+	this.nodes[key] = newNode
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0461. Hamming Distance/461. Hamming Distance.go b/leetcode/0461.Hamming-Distance/461. Hamming Distance.go
similarity index 100%
rename from Algorithms/0461. Hamming Distance/461. Hamming Distance.go
rename to leetcode/0461.Hamming-Distance/461. Hamming Distance.go
diff --git a/Algorithms/0461. Hamming Distance/461. Hamming Distance_test.go b/leetcode/0461.Hamming-Distance/461. Hamming Distance_test.go
similarity index 93%
rename from Algorithms/0461. Hamming Distance/461. Hamming Distance_test.go
rename to leetcode/0461.Hamming-Distance/461. Hamming Distance_test.go
index ba6a38ede..18f8395f0 100644
--- a/Algorithms/0461. Hamming Distance/461. Hamming Distance_test.go	
+++ b/leetcode/0461.Hamming-Distance/461. Hamming Distance_test.go	
@@ -27,17 +27,17 @@ func Test_Problem461(t *testing.T) {
 
 	qs := []question461{
 
-		question461{
+		{
 			para461{1, 4},
 			ans461{2},
 		},
 
-		question461{
+		{
 			para461{1, 1},
 			ans461{0},
 		},
 
-		question461{
+		{
 			para461{1, 3},
 			ans461{1},
 		},
diff --git a/leetcode/0461.Hamming-Distance/README.md b/leetcode/0461.Hamming-Distance/README.md
new file mode 100755
index 000000000..3aeeb833f
--- /dev/null
+++ b/leetcode/0461.Hamming-Distance/README.md
@@ -0,0 +1,36 @@
+# [461. Hamming Distance](https://leetcode.com/problems/hamming-distance/)
+
+## 题目
+
+The [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) between two integers is the number of positions at which the corresponding bits are different.
+
+Given two integers `x` and `y`, calculate the Hamming distance.
+
+**Note:**0 ≤ `x`, `y` < 231.
+
+**Example:**
+
+    Input: x = 1, y = 4
+    
+    Output: 2
+    
+    Explanation:
+    1   (0 0 0 1)
+    4   (0 1 0 0)
+           ↑   ↑
+    
+    The above arrows point to positions where the corresponding bits are different.
+
+
+## 题目大意
+
+两个整数之间的汉明距离指的是这两个数字对应二进制位不同的位置的数目。给出两个整数 x 和 y，计算它们之间的汉明距离。
+
+注意：
+0 ≤ x, y < 231.
+
+
+
+## 解题思路
+
+- 求 2 个数的海明距离。海明距离的定义是两个数二进制位不同的总个数。这一题利用的位操作的是 X &= (X - 1) 不断的清除最低位的 1 。先将这两个数异或，异或以后清除低位的 1 就是最终答案。
diff --git a/leetcode/0462.Minimum-Moves-to-Equal-Array-Elements-II/462. Minimum Moves to Equal Array Elements II.go b/leetcode/0462.Minimum-Moves-to-Equal-Array-Elements-II/462. Minimum Moves to Equal Array Elements II.go
new file mode 100644
index 000000000..cd38e4787
--- /dev/null
+++ b/leetcode/0462.Minimum-Moves-to-Equal-Array-Elements-II/462. Minimum Moves to Equal Array Elements II.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func minMoves2(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	moves, mid := 0, len(nums)/2
+	sort.Ints(nums)
+	for i := range nums {
+		if i == mid {
+			continue
+		}
+		moves += int(math.Abs(float64(nums[mid] - nums[i])))
+	}
+	return moves
+}
diff --git a/leetcode/0462.Minimum-Moves-to-Equal-Array-Elements-II/462. Minimum Moves to Equal Array Elements II_test.go b/leetcode/0462.Minimum-Moves-to-Equal-Array-Elements-II/462. Minimum Moves to Equal Array Elements II_test.go
new file mode 100644
index 000000000..bdbae051f
--- /dev/null
+++ b/leetcode/0462.Minimum-Moves-to-Equal-Array-Elements-II/462. Minimum Moves to Equal Array Elements II_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question462 struct {
+	para462
+	ans462
+}
+
+// para 是参数
+// one 代表第一个参数
+type para462 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans462 struct {
+	one int
+}
+
+func Test_Problem462(t *testing.T) {
+
+	qs := []question462{
+
+		{
+			para462{[]int{}},
+			ans462{0},
+		},
+
+		{
+			para462{[]int{1, 2, 3}},
+			ans462{2},
+		},
+
+		{
+			para462{[]int{1, 10, 2, 9}},
+			ans462{16},
+		},
+
+		{
+			para462{[]int{1, 0, 0, 8, 6}},
+			ans462{14},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 462------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans462, q.para462
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minMoves2(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0462.Minimum-Moves-to-Equal-Array-Elements-II/README.md b/leetcode/0462.Minimum-Moves-to-Equal-Array-Elements-II/README.md
new file mode 100644
index 000000000..c030b1de0
--- /dev/null
+++ b/leetcode/0462.Minimum-Moves-to-Equal-Array-Elements-II/README.md
@@ -0,0 +1,66 @@
+# [462. Minimum Moves to Equal Array Elements II](https://leetcode.com/problems/minimum-moves-to-equal-array-elements-ii/)
+
+
+## 题目
+
+Given an integer array `nums` of size `n`, return *the minimum number of moves required to make all array elements equal*.
+
+In one move, you can increment or decrement an element of the array by `1`.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3]
+Output: 2
+Explanation:
+Only two moves are needed (remember each move increments or decrements one element):
+[1,2,3]  =>  [2,2,3]  =>  [2,2,2]
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,10,2,9]
+Output: 16
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `1 <= nums.length <= 10^5`
+- `109 <= nums[i] <= 10^9`
+
+## 题目大意
+
+给定一个非空整数数组，找到使所有数组元素相等所需的最小移动数，其中每次移动可将选定的一个元素加 1 或减 1。 您可以假设数组的长度最多为10000。
+
+## 解题思路
+
+- 这题抽象成数学问题是，如果我们把数组 a 中的每个数看成水平轴上的一个点，那么根据上面的移动次数公式，我们需要找到在水平轴上找到一个点 x，使得这 N 个点到 x 的距离之和最小。有 2 个点值得我们考虑，一个是中位数，另外一个是平均值。举个简单的例子，[1,0,0,8,6] 这组数据，中位数是 1，平均值是 3 。分别计算移动的步数，按照中位数对齐是 14，按照平均值对齐是 16 。所以选择中位数。
+- 此题可以用数学证明，证明出，按照平均值移动的步数 ≥ 按照中位数移动的步数。具体证明笔者这里不证明了，感兴趣的同学可以自己证明试试。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func minMoves2(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	moves, mid := 0, len(nums)/2
+	sort.Ints(nums)
+	for i := range nums {
+		if i == mid {
+			continue
+		}
+		moves += int(math.Abs(float64(nums[mid] - nums[i])))
+	}
+	return moves
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0463. Island Perimeter/463. Island Perimeter.go b/leetcode/0463.Island-Perimeter/463. Island Perimeter.go
similarity index 100%
rename from Algorithms/0463. Island Perimeter/463. Island Perimeter.go
rename to leetcode/0463.Island-Perimeter/463. Island Perimeter.go
diff --git a/Algorithms/0463. Island Perimeter/463. Island Perimeter_test.go b/leetcode/0463.Island-Perimeter/463. Island Perimeter_test.go
similarity index 84%
rename from Algorithms/0463. Island Perimeter/463. Island Perimeter_test.go
rename to leetcode/0463.Island-Perimeter/463. Island Perimeter_test.go
index 35dfa0161..869cde435 100644
--- a/Algorithms/0463. Island Perimeter/463. Island Perimeter_test.go	
+++ b/leetcode/0463.Island-Perimeter/463. Island Perimeter_test.go	
@@ -26,8 +26,8 @@ func Test_Problem463(t *testing.T) {
 
 	qs := []question463{
 
-		question463{
-			para463{[][]int{[]int{0, 1, 0, 0}, []int{1, 1, 1, 0}, []int{0, 1, 0, 0}, []int{1, 1, 0, 0}}},
+		{
+			para463{[][]int{{0, 1, 0, 0}, {1, 1, 1, 0}, {0, 1, 0, 0}, {1, 1, 0, 0}}},
 			ans463{16},
 		},
 	}
diff --git a/leetcode/0463.Island-Perimeter/README.md b/leetcode/0463.Island-Perimeter/README.md
new file mode 100755
index 000000000..3d4498832
--- /dev/null
+++ b/leetcode/0463.Island-Perimeter/README.md
@@ -0,0 +1,39 @@
+# [463. Island Perimeter](https://leetcode.com/problems/island-perimeter/)
+
+## 题目
+
+You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water.
+
+Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells).
+
+The island doesn't have "lakes" (water inside that isn't connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don't exceed 100. Determine the perimeter of the island.
+
+**Example:**
+
+    Input:
+    [[0,1,0,0],
+     [1,1,1,0],
+     [0,1,0,0],
+     [1,1,0,0]]
+    
+    Output: 16
+    
+    Explanation: The perimeter is the 16 yellow stripes in the image below:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/island.png)
+
+
+## 题目大意
+
+给定一个包含 0 和 1 的二维网格地图，其中 1 表示陆地 0 表示水域。
+
+网格中的格子水平和垂直方向相连（对角线方向不相连）。整个网格被水完全包围，但其中恰好有一个岛屿（或者说，一个或多个表示陆地的格子相连组成的岛屿）。
+
+岛屿中没有“湖”（“湖” 指水域在岛屿内部且不和岛屿周围的水相连）。格子是边长为 1 的正方形。网格为长方形，且宽度和高度均不超过 100 。计算这个岛屿的周长。
+
+
+
+## 解题思路
+
+- 给出一个二维数组，二维数组中有一些连在一起的 1 ，这是一个岛屿，求这个岛屿的周长。
+- 这是一道水题，判断四周边界的情况依次加一即可。
diff --git a/Algorithms/0470. Implement Rand10() Using Rand7()/470. Implement Rand10() Using Rand7().go b/leetcode/0470.Implement-Rand10-Using-Rand7/470. Implement Rand10() Using Rand7().go
similarity index 100%
rename from Algorithms/0470. Implement Rand10() Using Rand7()/470. Implement Rand10() Using Rand7().go
rename to leetcode/0470.Implement-Rand10-Using-Rand7/470. Implement Rand10() Using Rand7().go
diff --git a/Algorithms/0470. Implement Rand10() Using Rand7()/470. Implement Rand10() Using Rand7()_test.go b/leetcode/0470.Implement-Rand10-Using-Rand7/470. Implement Rand10() Using Rand7()_test.go
similarity index 93%
rename from Algorithms/0470. Implement Rand10() Using Rand7()/470. Implement Rand10() Using Rand7()_test.go
rename to leetcode/0470.Implement-Rand10-Using-Rand7/470. Implement Rand10() Using Rand7()_test.go
index 9cc1cc18c..f53a88f81 100644
--- a/Algorithms/0470. Implement Rand10() Using Rand7()/470. Implement Rand10() Using Rand7()_test.go	
+++ b/leetcode/0470.Implement-Rand10-Using-Rand7/470. Implement Rand10() Using Rand7()_test.go	
@@ -25,17 +25,17 @@ func Test_Problem470(t *testing.T) {
 
 	qs := []question470{
 
-		question470{
+		{
 			para470{},
 			ans470{2},
 		},
 
-		question470{
+		{
 			para470{},
 			ans470{0},
 		},
 
-		question470{
+		{
 			para470{},
 			ans470{1},
 		},
diff --git a/leetcode/0470.Implement-Rand10-Using-Rand7/README.md b/leetcode/0470.Implement-Rand10-Using-Rand7/README.md
new file mode 100755
index 000000000..27b7a0a83
--- /dev/null
+++ b/leetcode/0470.Implement-Rand10-Using-Rand7/README.md
@@ -0,0 +1,69 @@
+# [470. Implement Rand10() Using Rand7()](https://leetcode.com/problems/implement-rand10-using-rand7/)
+
+
+## 题目
+
+Given a function `rand7` which generates a uniform random integer in the range 1 to 7, write a function `rand10` which generates a uniform random integer in the range 1 to 10.
+
+Do NOT use system's `Math.random()`.
+
+**Example 1:**
+
+    Input: 1
+    Output: [7]
+
+**Example 2:**
+
+    Input: 2
+    Output: [8,4]
+
+**Example 3:**
+
+    Input: 3
+    Output: [8,1,10]
+
+**Note:**
+
+1. `rand7` is predefined.
+2. Each testcase has one argument: `n`, the number of times that `rand10` is called.
+
+**Follow up:**
+
+1. What is the [expected value](https://en.wikipedia.org/wiki/Expected_value) for the number of calls to `rand7()` function?
+2. Could you minimize the number of calls to `rand7()`?
+
+
+## 题目大意
+
+已有方法 rand7 可生成 1 到 7 范围内的均匀随机整数，试写一个方法 rand10 生成 1 到 10 范围内的均匀随机整数。不要使用系统的 Math.random() 方法。
+
+
+提示:
+
+- rand7 已定义。
+- 传入参数: n 表示 rand10 的调用次数。
+ 
+
+进阶:
+
+- rand7()调用次数的 期望值 是多少 ?
+- 你能否尽量少调用 rand7() ?
+
+
+
+## 解题思路
+
+
+- 给出 `rand7()` 要求实现 `rand10()`。
+- `rand7()` 等概率地产生1，2，3，4，5，6，7。要想得到 `rand10()` 即等概率的生成 1-10 。解题思路是先构造一个 `randN()`，这个 N 必须是 10 的整数倍，然后 randN % 10 就可以得到 `rand10()` 了。所以可以从 `rand7()` 先构造出 `rand49()`，再把 `rand49()` 中大于等于 40 的都过滤掉，这样就得到了 `rand40()`，在对 10 取余即可。
+- 具体构造步骤，`rand7() --> rand49() --> rand40() --> rand10()`：
+    1. `rand7()` 等概率地产生 1,2,3,4,5,6,7.
+    2. `rand7() - 1` 等概率地产生 [0,6].
+    3. `(rand7() - 1) *7` 等概率地产生 0, 7, 14, 21, 28, 35, 42
+    4. `(rand7() - 1) * 7 + (rand7() - 1)` 等概率地产生 [0, 48] 这 49 个数字
+    5. 如果步骤 4 的结果大于等于 40，那么就重复步骤 4，直到产生的数小于 40
+    6. 把步骤 5 的结果 mod 10 再加 1， 就会等概率的随机生成 [1, 10]
+- 这道题可以推广到生成任意数的随机数问题。用 `randN()` 实现 `randM()`，`M>N`。步骤如下：
+    1. 用 `randN()` 先实现 `randX()`，其中 X ≥ M，X 是 M 的整数倍。如这题中的 49 > 10；
+    2. 再用 `randX()` 生成 `randM()`，如这题中的 49 —> 40 —> 10。
+- 举个例子，用 `rand3()` 生成 `rand11()`，可以先生成 `rand27()`，然后变成以 22 为界限，因为 22 是 11 的倍数。生成 `rand27()` 的方式：`3 * 3 * (rand3() - 1) + 3 * (rand3() - 1) + (rand3() - 1)`，最后生成了 `rand11()`；用 `rand7()` 生成 `rand9()`，可以先生成 `rand49()`，然后变成以 45 为界限，因为 45 是 9 的倍数。生成 `rand49()` 的方式：`(rand7() - 1) * 7 + (rand7() - 1)`，最后生成了 `rand9()`；用 `rand6()` 生成 `rand13()`，可以先生成 `rand36()`，然后变成以 26 为界限，因为 26 是 13 的倍数。生成 `rand36()` 的方式：`(rand6() - 1) * 6 + (rand6() - 1)`，最后生成了 `rand13()`；
diff --git a/leetcode/0473.Matchsticks-to-Square/473. Matchsticks to Square.go b/leetcode/0473.Matchsticks-to-Square/473. Matchsticks to Square.go
new file mode 100644
index 000000000..8930f35a4
--- /dev/null
+++ b/leetcode/0473.Matchsticks-to-Square/473. Matchsticks to Square.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import "sort"
+
+func makesquare(matchsticks []int) bool {
+	if len(matchsticks) < 4 {
+		return false
+	}
+	total := 0
+	for _, v := range matchsticks {
+		total += v
+	}
+	if total%4 != 0 {
+		return false
+	}
+	sort.Slice(matchsticks, func(i, j int) bool {
+		return matchsticks[i] > matchsticks[j]
+	})
+	visited := make([]bool, 16)
+	return dfs(matchsticks, 0, 0, 0, total, &visited)
+}
+
+func dfs(matchsticks []int, cur, group, sum, total int, visited *[]bool) bool {
+	if group == 4 {
+		return true
+	}
+	if sum > total/4 {
+		return false
+	}
+	if sum == total/4 {
+		return dfs(matchsticks, 0, group+1, 0, total, visited)
+	}
+	last := -1
+	for i := cur; i < len(matchsticks); i++ {
+		if (*visited)[i] {
+			continue
+		}
+		if last == matchsticks[i] {
+			continue
+		}
+		(*visited)[i] = true
+		last = matchsticks[i]
+		if dfs(matchsticks, i+1, group, sum+matchsticks[i], total, visited) {
+			return true
+		}
+		(*visited)[i] = false
+	}
+	return false
+}
diff --git a/leetcode/0473.Matchsticks-to-Square/473. Matchsticks to Square_test.go b/leetcode/0473.Matchsticks-to-Square/473. Matchsticks to Square_test.go
new file mode 100644
index 000000000..7636f00fb
--- /dev/null
+++ b/leetcode/0473.Matchsticks-to-Square/473. Matchsticks to Square_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question473 struct {
+	para473
+	ans473
+}
+
+// para 是参数
+// one 代表第一个参数
+type para473 struct {
+	arr []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans473 struct {
+	one bool
+}
+
+func Test_Problem473(t *testing.T) {
+
+	qs := []question473{
+
+		{
+			para473{[]int{1, 1, 2, 2, 2}},
+			ans473{true},
+		},
+
+		{
+			para473{[]int{3, 3, 3, 3, 4}},
+			ans473{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 473------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans473, q.para473
+		fmt.Printf("【input】:%v       【output】:%v\n", p, makesquare(p.arr))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0473.Matchsticks-to-Square/README.md b/leetcode/0473.Matchsticks-to-Square/README.md
new file mode 100644
index 000000000..d58d6b1c4
--- /dev/null
+++ b/leetcode/0473.Matchsticks-to-Square/README.md
@@ -0,0 +1,94 @@
+# [473. Matchsticks to Square](https://leetcode.com/problems/matchsticks-to-square/)
+
+
+## 题目
+
+You are given an integer array `matchsticks` where `matchsticks[i]` is the length of the `ith` matchstick. You want to use **all the matchsticks** to make one square. You **should not break** any stick, but you can link them up, and each matchstick must be used **exactly one time**.
+
+Return `true` if you can make this square and `false` otherwise.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/04/09/matchsticks1-grid.jpg](https://assets.leetcode.com/uploads/2021/04/09/matchsticks1-grid.jpg)
+
+```
+Input: matchsticks = [1,1,2,2,2]
+Output: true
+Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.
+```
+
+**Example 2:**
+
+```
+Input: matchsticks = [3,3,3,3,4]
+Output: false
+Explanation: You cannot find a way to form a square with all the matchsticks.
+```
+
+**Constraints:**
+
+- `1 <= matchsticks.length <= 15`
+- `0 <= matchsticks[i] <= 109`
+
+## 题目大意
+
+现在已知小女孩有多少根火柴，请找出一种能使用所有火柴拼成一个正方形的方法。不能折断火柴，可以把火柴连接起来，并且每根火柴都要用到。输入为小女孩拥有火柴的数目，每根火柴用其长度表示。输出即为是否能用所有的火柴拼成正方形。
+
+## 解题思路
+
+- 将火柴拼成一个正方形，可以将它们分成四组，每一根火柴恰好属于其中的一组；并且每一组火柴的长度之和都相同，等于所有火柴长度之和的四分之一。
+- 考虑暴力解法，使用深度优先搜索枚举出所有的分组情况，并对于每一种情况，判断是否满足上述的两个条件(每根火柴属于其中一组，每组火柴长度之和相同)。依次对每一根火柴进行搜索，当搜索到第 i 根火柴时，可以考虑把它放到四组中的任意一种。对于每一种放置方法，继续对第 i + 1 根火柴进行深搜。当我们搜索完全部的 N 根火柴后，再判断每一组火柴的长度之和是否都相同。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func makesquare(matchsticks []int) bool {
+	if len(matchsticks) < 4 {
+		return false
+	}
+	total := 0
+	for _, v := range matchsticks {
+		total += v
+	}
+	if total%4 != 0 {
+		return false
+	}
+	sort.Slice(matchsticks, func(i, j int) bool {
+		return matchsticks[i] > matchsticks[j]
+	})
+	visited := make([]bool, 16)
+	return dfs(matchsticks, 0, 0, 0, total, &visited)
+}
+
+func dfs(matchsticks []int, cur, group, sum, total int, visited *[]bool) bool {
+	if group == 4 {
+		return true
+	}
+	if sum > total/4 {
+		return false
+	}
+	if sum == total/4 {
+		return dfs(matchsticks, 0, group+1, 0, total, visited)
+	}
+	last := -1
+	for i := cur; i < len(matchsticks); i++ {
+		if (*visited)[i] {
+			continue
+		}
+		if last == matchsticks[i] {
+			continue
+		}
+		(*visited)[i] = true
+		last = matchsticks[i]
+		if dfs(matchsticks, i+1, group, sum+matchsticks[i], total, visited) {
+			return true
+		}
+		(*visited)[i] = false
+	}
+	return false
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0474. Ones and Zeroes/474. Ones and Zeroes.go b/leetcode/0474.Ones-and-Zeroes/474. Ones and Zeroes.go
similarity index 86%
rename from Algorithms/0474. Ones and Zeroes/474. Ones and Zeroes.go
rename to leetcode/0474.Ones-and-Zeroes/474. Ones and Zeroes.go
index 6f7cf049c..aa36966d8 100644
--- a/Algorithms/0474. Ones and Zeroes/474. Ones and Zeroes.go	
+++ b/leetcode/0474.Ones-and-Zeroes/474. Ones and Zeroes.go	
@@ -21,3 +21,10 @@ func findMaxForm(strs []string, m int, n int) int {
 	}
 	return dp[m][n]
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0474. Ones and Zeroes/474. Ones and Zeroes_test.go b/leetcode/0474.Ones-and-Zeroes/474. Ones and Zeroes_test.go
similarity index 94%
rename from Algorithms/0474. Ones and Zeroes/474. Ones and Zeroes_test.go
rename to leetcode/0474.Ones-and-Zeroes/474. Ones and Zeroes_test.go
index 8cfdbf729..286376edf 100644
--- a/Algorithms/0474. Ones and Zeroes/474. Ones and Zeroes_test.go	
+++ b/leetcode/0474.Ones-and-Zeroes/474. Ones and Zeroes_test.go	
@@ -28,17 +28,17 @@ func Test_Problem474(t *testing.T) {
 
 	qs := []question474{
 
-		question474{
+		{
 			para474{[]string{"10", "0001", "111001", "1", "0"}, 5, 3},
 			ans474{4},
 		},
 
-		question474{
+		{
 			para474{[]string{"10", "0", "1"}, 1, 1},
 			ans474{2},
 		},
 
-		question474{
+		{
 			para474{[]string{}, 0, 0},
 			ans474{0},
 		},
diff --git a/leetcode/0474.Ones-and-Zeroes/README.md b/leetcode/0474.Ones-and-Zeroes/README.md
new file mode 100755
index 000000000..959d29cb7
--- /dev/null
+++ b/leetcode/0474.Ones-and-Zeroes/README.md
@@ -0,0 +1,47 @@
+# [474. Ones and Zeroes](https://leetcode.com/problems/ones-and-zeroes/)
+
+
+## 题目
+
+In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.
+
+For now, suppose you are a dominator of **m** `0s` and **n** `1s` respectively. On the other hand, there is an array with strings consisting of only `0s` and `1s`.
+
+Now your task is to find the maximum number of strings that you can form with given **m**`0s` and **n** `1s`. Each `0` and `1` can be used at most **once**.
+
+**Note:**
+
+1. The given numbers of `0s` and `1s` will both not exceed `100`
+2. The size of given string array won't exceed `600`.
+
+**Example 1:**
+
+    Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3
+    Output: 4
+    
+    Explanation: This are totally 4 strings can be formed by the using of 5 0s and 3 1s, which are “10,”0001”,”1”,”0”
+
+**Example 2:**
+
+    Input: Array = {"10", "0", "1"}, m = 1, n = 1
+    Output: 2
+    
+    Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".
+
+
+## 题目大意
+
+在计算机界中，我们总是追求用有限的资源获取最大的收益。现在，假设你分别支配着 m 个 0 和 n 个 1。另外，还有一个仅包含 0 和 1 字符串的数组。你的任务是使用给定的 m 个 0 和 n 个 1 ，找到能拼出存在于数组中的字符串的最大数量。每个 0 和 1 至多被使用一次。
+
+注意:
+
+1. 给定 0 和 1 的数量都不会超过 100。
+2. 给定字符串数组的长度不会超过 600。
+
+
+
+## 解题思路
+
+- 给定一个字符串数组和 m，n，其中所有的字符都是由 0 和 1 组成的。问能否从数组中取出最多的字符串，使得这些取出的字符串中所有的 0 的个数 ≤ m，1 的个数 ≤ n。
+- 这一题是典型的 01 背包的题型。只不过是一个二维的背包问题。在 n 个物品中选出一定物品，**尽量完全填满** m 维和 n 维的背包。为什么是尽量填满？因为不一定能完全填满背包。
+- `dp[i][j]` 代表尽量填满容量为 `(i,j)` 的背包装下的物品总数，状态转移方程为 `dp[i][j] = max(dp[i][j], 1+dp[i-zero][j-one])`。其中 zero 代表的当前装入物品在 m 维上的体积，也即 0 的个数。one 代表的是当前装入物品在 n 维上的体积，也即 1 的个数。每添加一个物品，比较当前 (i,j) 的背包装下的物品总数和 (i-zero,j-one) 的背包装下的物品总数 + 1，比较这两者的大小，保存两者的最大值。每添加一个物品就刷新这个二维背包，直到所有物品都扫完一遍。dp[m][n] 中存储的就是最终的答案。时间复杂度 `O( n * M * N )`。
diff --git a/Algorithms/0475. Heaters/475. Heaters.go b/leetcode/0475.Heaters/475. Heaters.go
similarity index 85%
rename from Algorithms/0475. Heaters/475. Heaters.go
rename to leetcode/0475.Heaters/475. Heaters.go
index 381a89a9d..382b097e1 100644
--- a/Algorithms/0475. Heaters/475. Heaters.go	
+++ b/leetcode/0475.Heaters/475. Heaters.go	
@@ -61,3 +61,24 @@ func findRadius1(houses []int, heaters []int) int {
 	}
 	return res
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
diff --git a/Algorithms/0475. Heaters/475. Heaters_test.go b/leetcode/0475.Heaters/475. Heaters_test.go
similarity index 96%
rename from Algorithms/0475. Heaters/475. Heaters_test.go
rename to leetcode/0475.Heaters/475. Heaters_test.go
index f39489ee3..fffa88dfe 100644
--- a/Algorithms/0475. Heaters/475. Heaters_test.go	
+++ b/leetcode/0475.Heaters/475. Heaters_test.go	
@@ -27,12 +27,12 @@ func Test_Problem475(t *testing.T) {
 
 	qs := []question475{
 
-		question475{
+		{
 			para475{[]int{1, 2, 3}, []int{2}},
 			ans475{1},
 		},
 
-		question475{
+		{
 			para475{[]int{1, 2, 3, 4}, []int{1, 4}},
 			ans475{1},
 		},
diff --git a/leetcode/0475.Heaters/README.md b/leetcode/0475.Heaters/README.md
new file mode 100755
index 000000000..1232dca1e
--- /dev/null
+++ b/leetcode/0475.Heaters/README.md
@@ -0,0 +1,51 @@
+# [475. Heaters](https://leetcode.com/problems/heaters/)
+
+## 题目
+
+Winter is coming! Your first job during the contest is to design a standard heater with fixed warm radius to warm all the houses.
+
+Now, you are given positions of houses and heaters on a horizontal line, find out minimum radius of heaters so that all houses could be covered by those heaters.
+
+So, your input will be the positions of houses and heaters seperately, and your expected output will be the minimum radius standard of heaters.
+
+**Note:**
+
+1. Numbers of houses and heaters you are given are non-negative and will not exceed 25000.
+2. Positions of houses and heaters you are given are non-negative and will not exceed 10^9.
+3. As long as a house is in the heaters' warm radius range, it can be warmed.
+4. All the heaters follow your radius standard and the warm radius will the same.
+
+**Example 1:**
+
+    Input: [1,2,3],[2]
+    Output: 1
+    Explanation: The only heater was placed in the position 2, and if we use the radius 1 standard, then all the houses can be warmed.
+
+**Example 2:**
+
+    Input: [1,2,3,4],[1,4]
+    Output: 1
+    Explanation: The two heater was placed in the position 1 and 4. We need to use radius 1 standard, then all the houses can be warmed.
+
+
+
+## 题目大意
+
+
+冬季已经来临。 你的任务是设计一个有固定加热半径的供暖器向所有房屋供暖。现在，给出位于一条水平线上的房屋和供暖器的位置，找到可以覆盖所有房屋的最小加热半径。所以，你的输入将会是房屋和供暖器的位置。你将输出供暖器的最小加热半径。
+
+说明:
+
+- 给出的房屋和供暖器的数目是非负数且不会超过 25000。
+- 给出的房屋和供暖器的位置均是非负数且不会超过10^9。
+- 只要房屋位于供暖器的半径内(包括在边缘上)，它就可以得到供暖。
+- 所有供暖器都遵循你的半径标准，加热的半径也一样。
+
+
+
+## 解题思路
+
+
+- 给出一个房子坐标的数组，和一些供暖器坐标的数组，分别表示房子的坐标和供暖器的坐标。要求找到供暖器最小的半径能使得所有的房子都能享受到暖气。
+- 这一题可以用暴力的解法，暴力解法依次遍历每个房子的坐标，再遍历每个供暖器，找到距离房子最近的供暖器坐标。在所有这些距离的长度里面找到最大值，这个距离的最大值就是供暖器半径的最小值。时间复杂度 O(n^2)。
+- 这一题最优解是二分搜索。在查找距离房子最近的供暖器的时候，先将供暖器排序，然后用二分搜索的方法查找。其他的做法和暴力解法一致。时间复杂度 O(n log n)。
diff --git a/Algorithms/0476. Number Complement/476. Number Complement.go b/leetcode/0476.Number-Complement/476. Number Complement.go
similarity index 100%
rename from Algorithms/0476. Number Complement/476. Number Complement.go
rename to leetcode/0476.Number-Complement/476. Number Complement.go
diff --git a/Algorithms/0476. Number Complement/476. Number Complement_test.go b/leetcode/0476.Number-Complement/476. Number Complement_test.go
similarity index 95%
rename from Algorithms/0476. Number Complement/476. Number Complement_test.go
rename to leetcode/0476.Number-Complement/476. Number Complement_test.go
index 64c338026..91c7288f3 100644
--- a/Algorithms/0476. Number Complement/476. Number Complement_test.go	
+++ b/leetcode/0476.Number-Complement/476. Number Complement_test.go	
@@ -26,12 +26,12 @@ func Test_Problem476(t *testing.T) {
 
 	qs := []question476{
 
-		question476{
+		{
 			para476{5},
 			ans476{2},
 		},
 
-		question476{
+		{
 			para476{1},
 			ans476{0},
 		},
diff --git a/leetcode/0476.Number-Complement/README.md b/leetcode/0476.Number-Complement/README.md
new file mode 100755
index 000000000..ce2adf049
--- /dev/null
+++ b/leetcode/0476.Number-Complement/README.md
@@ -0,0 +1,41 @@
+# [476. Number Complement](https://leetcode.com/problems/number-complement/)
+
+
+## 题目
+
+Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
+
+**Note:**
+
+1. The given integer is guaranteed to fit within the range of a 32-bit signed integer.
+2. You could assume no leading zero bit in the integer’s binary representation.
+
+**Example 1:**
+
+    Input: 5
+    Output: 2
+    Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.
+
+**Example 2:**
+
+    Input: 1
+    Output: 0
+    Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.
+
+
+## 题目大意
+
+给定一个正整数，输出它的补数。补数是对该数的二进制表示取反。
+
+注意:
+
+给定的整数保证在32位带符号整数的范围内。
+你可以假定二进制数不包含前导零位。
+
+
+
+## 解题思路
+
+
+- 求一个正数的补数，补数的定义是对该数的二进制表示取反。当前不能改变符号位。按照题意构造相应的 mask 再取反即可。
+
diff --git a/Algorithms/0477. Total Hamming Distance/477. Total Hamming Distance.go b/leetcode/0477.Total-Hamming-Distance/477. Total Hamming Distance.go
similarity index 77%
rename from Algorithms/0477. Total Hamming Distance/477. Total Hamming Distance.go
rename to leetcode/0477.Total-Hamming-Distance/477. Total Hamming Distance.go
index bb85707dd..d05a3c3e3 100644
--- a/Algorithms/0477. Total Hamming Distance/477. Total Hamming Distance.go	
+++ b/leetcode/0477.Total-Hamming-Distance/477. Total Hamming Distance.go	
@@ -22,3 +22,11 @@ func totalHammingDistance1(nums []int) int {
 	}
 	return res
 }
+
+func hammingDistance(x int, y int) int {
+	distance := 0
+	for xor := x ^ y; xor != 0; xor &= (xor - 1) {
+		distance++
+	}
+	return distance
+}
diff --git a/Algorithms/0477. Total Hamming Distance/477. Total Hamming Distance_test.go b/leetcode/0477.Total-Hamming-Distance/477. Total Hamming Distance_test.go
similarity index 97%
rename from Algorithms/0477. Total Hamming Distance/477. Total Hamming Distance_test.go
rename to leetcode/0477.Total-Hamming-Distance/477. Total Hamming Distance_test.go
index 5e69bae7a..d612046ae 100644
--- a/Algorithms/0477. Total Hamming Distance/477. Total Hamming Distance_test.go	
+++ b/leetcode/0477.Total-Hamming-Distance/477. Total Hamming Distance_test.go	
@@ -26,7 +26,7 @@ func Test_Problem477(t *testing.T) {
 
 	qs := []question477{
 
-		question477{
+		{
 			para477{[]int{4, 14, 2}},
 			ans477{6},
 		},
diff --git a/leetcode/0477.Total-Hamming-Distance/README.md b/leetcode/0477.Total-Hamming-Distance/README.md
new file mode 100755
index 000000000..37d3e7294
--- /dev/null
+++ b/leetcode/0477.Total-Hamming-Distance/README.md
@@ -0,0 +1,34 @@
+# [477. Total Hamming Distance](https://leetcode.com/problems/total-hamming-distance/)
+
+
+## 题目
+
+The [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) between two integers is the number of positions at which the corresponding bits are different.
+
+Now your job is to find the total Hamming distance between all pairs of the given numbers.
+
+**Example:**
+
+    Input: 4, 14, 2
+    
+    Output: 6
+    
+    Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
+    showing the four bits relevant in this case). So the answer will be:
+    HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
+
+**Note:**
+
+1. Elements of the given array are in the range of `0` to `10^9`
+2. Length of the array will not exceed `10^4`.
+
+
+## 题目大意
+
+两个整数的[汉明距离](https://baike.baidu.com/item/%E6%B1%89%E6%98%8E%E8%B7%9D%E7%A6%BB/475174?fr=aladdin)指的是这两个数字的二进制数对应位不同的数量。计算一个数组中，任意两个数之间汉明距离的总和。
+
+
+## 解题思路
+
+- 计算一个数组内两两元素的海明距离总和。海明距离的定义是两个数二进制位不同的总个数。那么可以把数组中的每个元素 32 位的二进制位依次扫一遍，当扫到某一位上的时候，有 k 个元素在这个位上的值是 1，n - k 个元素在这个位上的值是 0，那么在这一位上所有两两元素的海明距离是 k*(n-k) ，当把 32 位全部都扫完以后，累加出来的海明距离就是所有两两元素的海明距离。
+
diff --git a/leetcode/0478.Generate-Random-Point-in-a-Circle/478. Generate Random Point in a Circle.go b/leetcode/0478.Generate-Random-Point-in-a-Circle/478. Generate Random Point in a Circle.go
new file mode 100644
index 000000000..d611124a3
--- /dev/null
+++ b/leetcode/0478.Generate-Random-Point-in-a-Circle/478. Generate Random Point in a Circle.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"math"
+	"math/rand"
+	"time"
+)
+
+type Solution struct {
+	r float64
+	x float64
+	y float64
+}
+
+func Constructor(radius float64, x_center float64, y_center float64) Solution {
+	rand.Seed(time.Now().UnixNano())
+	return Solution{radius, x_center, y_center}
+}
+
+func (this *Solution) RandPoint() []float64 {
+	/*
+	   a := angle()
+	   r := this.r * math.Sqrt(rand.Float64())
+	   x := r * math.Cos(a) + this.x
+	   y := r * math.Sin(a) + this.y
+	   return []float64{x, y}*/
+	for {
+		rx := 2*rand.Float64() - 1.0
+		ry := 2*rand.Float64() - 1.0
+		x := this.r * rx
+		y := this.r * ry
+		if x*x+y*y <= this.r*this.r {
+			return []float64{x + this.x, y + this.y}
+		}
+	}
+}
+
+func angle() float64 {
+	return rand.Float64() * 2 * math.Pi
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(radius, x_center, y_center);
+ * param_1 := obj.RandPoint();
+ */
diff --git a/leetcode/0478.Generate-Random-Point-in-a-Circle/478. Generate Random Point in a Circle_test.go b/leetcode/0478.Generate-Random-Point-in-a-Circle/478. Generate Random Point in a Circle_test.go
new file mode 100644
index 000000000..ddd2e62df
--- /dev/null
+++ b/leetcode/0478.Generate-Random-Point-in-a-Circle/478. Generate Random Point in a Circle_test.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem478(t *testing.T) {
+	obj := Constructor(1, 0, 0)
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+
+	obj = Constructor(10, 5, -7.5)
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+}
diff --git a/leetcode/0478.Generate-Random-Point-in-a-Circle/README.md b/leetcode/0478.Generate-Random-Point-in-a-Circle/README.md
new file mode 100644
index 000000000..bc325e62a
--- /dev/null
+++ b/leetcode/0478.Generate-Random-Point-in-a-Circle/README.md
@@ -0,0 +1,103 @@
+# [478. Generate Random Point in a Circle](https://leetcode.com/problems/generate-random-point-in-a-circle/)
+
+
+## 题目
+
+Given the radius and x-y positions of the center of a circle, write a function `randPoint` which generates a uniform random point in the circle.
+
+Note:
+
+1. input and output values are in [floating-point](https://www.webopedia.com/TERM/F/floating_point_number.html).
+2. radius and x-y position of the center of the circle is passed into the class constructor.
+3. a point on the circumference of the circle is considered to be in the circle.
+4. `randPoint` returns a size 2 array containing x-position and y-position of the random point, in that order.
+
+**Example 1:**
+
+```
+Input: 
+["Solution","randPoint","randPoint","randPoint"]
+[[1,0,0],[],[],[]]
+Output: [null,[-0.72939,-0.65505],[-0.78502,-0.28626],[-0.83119,-0.19803]]
+
+```
+
+**Example 2:**
+
+```
+Input: 
+["Solution","randPoint","randPoint","randPoint"]
+[[10,5,-7.5],[],[],[]]
+Output: [null,[11.52438,-8.33273],[2.46992,-16.21705],[11.13430,-12.42337]]
+```
+
+**Explanation of Input Syntax:**
+
+The input is two lists: the subroutines called and their arguments. `Solution`'s constructor has three arguments, the radius, x-position of the center, and y-position of the center of the circle. `randPoint` has no arguments. Arguments are always wrapped with a list, even if there aren't any.
+
+## 题目大意
+
+给定圆的半径和圆心的 x、y 坐标，写一个在圆中产生均匀随机点的函数 randPoint 。
+
+说明:
+
+- 输入值和输出值都将是浮点数。
+- 圆的半径和圆心的 x、y 坐标将作为参数传递给类的构造函数。
+- 圆周上的点也认为是在圆中。
+- randPoint 返回一个包含随机点的x坐标和y坐标的大小为2的数组。
+
+## 解题思路
+
+- 随机产生一个圆内的点，这个点一定满足定义 `(x-a)^2+(y-b)^2 ≤ R^2`，其中 `(a,b)` 是圆的圆心坐标，`R` 是半径。
+- 先假设圆心坐标在 (0,0)，这样方便计算，最终输出坐标的时候整体加上圆心的偏移量即可。`rand.Float64()` 产生一个 `[0.0,1.0)` 区间的浮点数。`-R ≤ 2 * R * rand() - R < R`，利用随机产生坐标点的横纵坐标 `(x,y)` 与半径 R 的关系，如果 `x^2 + y^2 ≤ R^2`，那么说明产生的点在圆内。最终输出的时候要记得加上圆心坐标的偏移值。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+	"math/rand"
+	"time"
+)
+
+type Solution struct {
+	r float64
+	x float64
+	y float64
+}
+
+func Constructor(radius float64, x_center float64, y_center float64) Solution {
+	rand.Seed(time.Now().UnixNano())
+	return Solution{radius, x_center, y_center}
+}
+
+func (this *Solution) RandPoint() []float64 {
+	/*
+	   a := angle()
+	   r := this.r * math.Sqrt(rand.Float64())
+	   x := r * math.Cos(a) + this.x
+	   y := r * math.Sin(a) + this.y
+	   return []float64{x, y}*/
+	for {
+		rx := 2*rand.Float64() - 1.0
+		ry := 2*rand.Float64() - 1.0
+		x := this.r * rx
+		y := this.r * ry
+		if x*x+y*y <= this.r*this.r {
+			return []float64{x + this.x, y + this.y}
+		}
+	}
+}
+
+func angle() float64 {
+	return rand.Float64() * 2 * math.Pi
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(radius, x_center, y_center);
+ * param_1 := obj.RandPoint();
+ */
+```
\ No newline at end of file
diff --git a/Algorithms/0480. Sliding Window Median/480. Sliding Window Median.go b/leetcode/0480.Sliding-Window-Median/480. Sliding Window Median.go
similarity index 100%
rename from Algorithms/0480. Sliding Window Median/480. Sliding Window Median.go
rename to leetcode/0480.Sliding-Window-Median/480. Sliding Window Median.go
diff --git a/Algorithms/0480. Sliding Window Median/480. Sliding Window Median_test.go b/leetcode/0480.Sliding-Window-Median/480. Sliding Window Median_test.go
similarity index 97%
rename from Algorithms/0480. Sliding Window Median/480. Sliding Window Median_test.go
rename to leetcode/0480.Sliding-Window-Median/480. Sliding Window Median_test.go
index d5f3cfd69..fda29cceb 100644
--- a/Algorithms/0480. Sliding Window Median/480. Sliding Window Median_test.go	
+++ b/leetcode/0480.Sliding-Window-Median/480. Sliding Window Median_test.go	
@@ -27,7 +27,7 @@ func Test_Problem480(t *testing.T) {
 
 	qs := []question480{
 
-		question480{
+		{
 			para480{[]int{1, 3, -1, -3, 5, 3, 6, 7}, 3},
 			ans480{[]int{1, -1, -1, 3, 5, 6}},
 		},
diff --git a/Algorithms/0480. Sliding Window Median/README.md b/leetcode/0480.Sliding-Window-Median/README.md
similarity index 100%
rename from Algorithms/0480. Sliding Window Median/README.md
rename to leetcode/0480.Sliding-Window-Median/README.md
diff --git a/leetcode/0483.Smallest-Good-Base/483. Smallest Good Base.go b/leetcode/0483.Smallest-Good-Base/483. Smallest Good Base.go
new file mode 100644
index 000000000..7dc0430f3
--- /dev/null
+++ b/leetcode/0483.Smallest-Good-Base/483. Smallest Good Base.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+import (
+	"math"
+	"math/bits"
+	"strconv"
+)
+
+func smallestGoodBase(n string) string {
+	nVal, _ := strconv.Atoi(n)
+	mMax := bits.Len(uint(nVal)) - 1
+	for m := mMax; m > 1; m-- {
+		k := int(math.Pow(float64(nVal), 1/float64(m)))
+		mul, sum := 1, 1
+		for i := 0; i < m; i++ {
+			mul *= k
+			sum += mul
+		}
+		if sum == nVal {
+			return strconv.Itoa(k)
+		}
+	}
+	return strconv.Itoa(nVal - 1)
+}
diff --git a/leetcode/0483.Smallest-Good-Base/483. Smallest Good Base_test.go b/leetcode/0483.Smallest-Good-Base/483. Smallest Good Base_test.go
new file mode 100644
index 000000000..710548e1c
--- /dev/null
+++ b/leetcode/0483.Smallest-Good-Base/483. Smallest Good Base_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question483 struct {
+	para483
+	ans483
+}
+
+// para 是参数
+// one 代表第一个参数
+type para483 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans483 struct {
+	one string
+}
+
+func Test_Problem483(t *testing.T) {
+
+	qs := []question483{
+
+		{
+			para483{"13"},
+			ans483{"3"},
+		},
+
+		{
+			para483{"4681"},
+			ans483{"8"},
+		},
+
+		{
+			para483{"1000000000000000000"},
+			ans483{"999999999999999999"},
+		},
+
+		{
+			para483{"727004545306745403"},
+			ans483{"727004545306745402"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 483------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans483, q.para483
+		fmt.Printf("【input】:%v       【output】:%v\n", p, smallestGoodBase(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0483.Smallest-Good-Base/README.md b/leetcode/0483.Smallest-Good-Base/README.md
new file mode 100755
index 000000000..41d038a40
--- /dev/null
+++ b/leetcode/0483.Smallest-Good-Base/README.md
@@ -0,0 +1,95 @@
+# [483. Smallest Good Base](https://leetcode.com/problems/smallest-good-base/)
+
+
+## 题目
+
+For an integer n, we call k>=2 a **good base** of n, if all digits of n base k are 1.
+
+Now given a string representing n, you should return the smallest good base of n in string format.
+
+**Example 1:**
+
+    Input: "13"
+    Output: "3"
+    Explanation: 13 base 3 is 111.
+
+**Example 2:**
+
+    Input: "4681"
+    Output: "8"
+    Explanation: 4681 base 8 is 11111.
+
+**Example 3:**
+
+    Input: "1000000000000000000"
+    Output: "999999999999999999"
+    Explanation: 1000000000000000000 base 999999999999999999 is 11.
+
+**Note:**
+
+1. The range of n is [3, 10^18].
+2. The string representing n is always valid and will not have leading zeros.
+
+
+## 题目大意
+
+
+对于给定的整数 n, 如果n的k（k>=2）进制数的所有数位全为1，则称 k（k>=2）是 n 的一个好进制。
+
+以字符串的形式给出 n, 以字符串的形式返回 n 的最小好进制。
+
+提示：
+
+- n 的取值范围是 [3, 10^18]。
+- 输入总是有效且没有前导 0。
+
+
+
+## 解题思路
+
+
+- 给出一个数 n，要求找一个进制 k，使得数字 n 在 k 进制下每一位都是 1 。求最小的进制 k。
+- 这一题等价于求最小的正整数 k，满足存在一个正整数 m 使得
+
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_483_1.png'>
+</p>
+
+
+- 这一题需要确定 k 和 m 两个数的值。m 和 k 是有关系的，确定了一个值，另外一个值也确定了。由
+
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_483_2.png'>
+</p>
+
+
+可得：
+
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_483_3.png'>
+</p>
+
+
+根据题意，可以知道 k ≥2，m ≥1 ，所以有:
+
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_483_4.png'>
+</p>
+
+
+所以 m 的取值范围确定了。那么外层循环从 1 到 log n 遍历。找到一个最小的 k ，能满足：
+
+可以用二分搜索来逼近找到最小的 k。先找到 k 的取值范围。由 
+
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_483_5.png'>
+</p>
+
+
+可得，
+
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_483_6.png'>
+</p>
+
+所以 k 的取值范围是 [2, n*(1/m) ]。再利用二分搜索逼近找到最小的 k 即为答案。
diff --git a/leetcode/0485.Max-Consecutive-Ones/485. Max Consecutive Ones.go b/leetcode/0485.Max-Consecutive-Ones/485. Max Consecutive Ones.go
new file mode 100644
index 000000000..5ab90c849
--- /dev/null
+++ b/leetcode/0485.Max-Consecutive-Ones/485. Max Consecutive Ones.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func findMaxConsecutiveOnes(nums []int) int {
+	maxCount, currentCount := 0, 0
+	for _, v := range nums {
+		if v == 1 {
+			currentCount++
+		} else {
+			currentCount = 0
+		}
+		if currentCount > maxCount {
+			maxCount = currentCount
+		}
+	}
+	return maxCount
+}
diff --git a/leetcode/0485.Max-Consecutive-Ones/485. Max Consecutive Ones_test.go b/leetcode/0485.Max-Consecutive-Ones/485. Max Consecutive Ones_test.go
new file mode 100644
index 000000000..6a2a69e11
--- /dev/null
+++ b/leetcode/0485.Max-Consecutive-Ones/485. Max Consecutive Ones_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question485 struct {
+	para485
+	ans485
+}
+
+// para 是参数
+// one 代表第一个参数
+type para485 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans485 struct {
+	one int
+}
+
+func Test_Problem485(t *testing.T) {
+
+	qs := []question485{
+
+		{
+			para485{[]int{1, 1, 0, 1, 1, 1}},
+			ans485{3},
+		},
+
+		{
+			para485{[]int{1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1}},
+			ans485{4},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 485------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans485, q.para485
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findMaxConsecutiveOnes(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0485.Max-Consecutive-Ones/README.md b/leetcode/0485.Max-Consecutive-Ones/README.md
new file mode 100644
index 000000000..8cba17b1d
--- /dev/null
+++ b/leetcode/0485.Max-Consecutive-Ones/README.md
@@ -0,0 +1,55 @@
+# [485. Max Consecutive Ones](https://leetcode.com/problems/max-consecutive-ones/)
+
+
+## 题目
+
+Given a binary array, find the maximum number of consecutive 1s in this array.
+
+**Example 1**:
+
+```
+Input: [1,1,0,1,1,1]
+Output: 3
+Explanation: The first two digits or the last three digits are consecutive 1s.
+    The maximum number of consecutive 1s is 3.
+```
+
+**Note**:
+
+- The input array will only contain `0` and `1`.
+- The length of input array is a positive integer and will not exceed 10,000
+
+
+## 题目大意
+
+给定一个二进制数组， 计算其中最大连续1的个数。
+
+注意：
+
+- 输入的数组只包含 0 和 1。
+- 输入数组的长度是正整数，且不超过 10,000。
+
+
+## 解题思路
+
+- 给定一个二进制数组， 计算其中最大连续1的个数。
+- 简单题。扫一遍数组，累计 1 的个数，动态维护最大的计数，最终输出即可。
+
+## 代码
+
+```go
+func findMaxConsecutiveOnes(nums []int) int {
+	maxCount, currentCount := 0, 0
+	for _, v := range nums {
+		if v == 1 {
+			currentCount++
+		} else {
+			currentCount = 0
+		}
+		if currentCount > maxCount {
+			maxCount = currentCount
+		}
+	}
+	return maxCount
+}
+```
\ No newline at end of file
diff --git a/leetcode/0488.Zuma-Game/488.Zuma Game.go b/leetcode/0488.Zuma-Game/488.Zuma Game.go
new file mode 100644
index 000000000..7fa5e2595
--- /dev/null
+++ b/leetcode/0488.Zuma-Game/488.Zuma Game.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+func findMinStep(board string, hand string) int {
+	q := [][]string{{board, hand}}
+	mp := make(map[string]bool)
+	minStep := 0
+	for len(q) > 0 {
+		length := len(q)
+		minStep++
+		for length > 0 {
+			length--
+			cur := q[0]
+			q = q[1:]
+			curB, curH := cur[0], cur[1]
+			for i := 0; i < len(curB); i++ {
+				for j := 0; j < len(curH); j++ {
+					curB2 := del3(curB[0:i] + string(curH[j]) + curB[i:])
+					curH2 := curH[0:j] + curH[j+1:]
+					if len(curB2) == 0 {
+						return minStep
+					}
+					if _, ok := mp[curB2+curH2]; ok {
+						continue
+					}
+					mp[curB2+curH2] = true
+					q = append(q, []string{curB2, curH2})
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func del3(str string) string {
+	cnt := 1
+	for i := 1; i < len(str); i++ {
+		if str[i] == str[i-1] {
+			cnt++
+		} else {
+			if cnt >= 3 {
+				return del3(str[0:i-cnt] + str[i:])
+			}
+			cnt = 1
+		}
+	}
+	if cnt >= 3 {
+		return str[0 : len(str)-cnt]
+	}
+	return str
+}
diff --git a/leetcode/0488.Zuma-Game/488.Zuma Game_test.go b/leetcode/0488.Zuma-Game/488.Zuma Game_test.go
new file mode 100644
index 000000000..841c8168b
--- /dev/null
+++ b/leetcode/0488.Zuma-Game/488.Zuma Game_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question488 struct {
+	para488
+	ans488
+}
+
+// para 是参数
+type para488 struct {
+	board string
+	hand  string
+}
+
+// ans 是答案
+type ans488 struct {
+	ans int
+}
+
+func Test_Problem488(t *testing.T) {
+
+	qs := []question488{
+
+		{
+			para488{"WRRBBW", "RB"},
+			ans488{-1},
+		},
+
+		{
+			para488{"WWRRBBWW", "WRBRW"},
+			ans488{2},
+		},
+
+		{
+			para488{"G", "GGGGG"},
+			ans488{2},
+		},
+
+		{
+			para488{"RBYYBBRRB", "YRBGB"},
+			ans488{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 488------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans488, q.para488
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findMinStep(p.board, p.hand))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0488.Zuma-Game/README.md b/leetcode/0488.Zuma-Game/README.md
new file mode 100644
index 000000000..2e05d9d76
--- /dev/null
+++ b/leetcode/0488.Zuma-Game/README.md
@@ -0,0 +1,141 @@
+# [488. Zuma Game](https://leetcode.com/problems/zuma-game/)
+
+
+## 题目
+
+You are playing a variation of the game Zuma.
+
+In this variation of Zuma, there is a single row of colored balls on a board, where each ball can be colored red 'R', yellow 'Y', blue 'B', green 'G', or white 'W'. You also have several colored balls in your hand.
+
+Your goal is to clear all of the balls from the board. On each turn:
+
+Pick any ball from your hand and insert it in between two balls in the row or on either end of the row.
+If there is a group of three or more consecutive balls of the same color, remove the group of balls from the board.
+If this removal causes more groups of three or more of the same color to form, then continue removing each group until there are none left.
+If there are no more balls on the board, then you win the game.
+Repeat this process until you either win or do not have any more balls in your hand.
+Given a string board, representing the row of balls on the board, and a string hand, representing the balls in your hand, return the minimum number of balls you have to insert to clear all the balls from the board. If you cannot clear all the balls from the board using the balls in your hand, return -1.
+
+**Example 1**:
+
+```
+Input: board = "WRRBBW", hand = "RB"
+Output: -1
+Explanation: It is impossible to clear all the balls. The best you can do is:
+- Insert 'R' so the board becomes WRRRBBW. WRRRBBW -> WBBW.
+- Insert 'B' so the board becomes WBBBW. WBBBW -> WW.
+There are still balls remaining on the board, and you are out of balls to insert.
+```
+
+**Example 2**:
+```
+Input: board = "WWRRBBWW", hand = "WRBRW"
+Output: 2
+Explanation: To make the board empty:
+- Insert 'R' so the board becomes WWRRRBBWW. WWRRRBBWW -> WWBBWW.
+- Insert 'B' so the board becomes WWBBBWW. WWBBBWW -> WWWW -> empty.
+2 balls from your hand were needed to clear the board.
+```
+
+**Example 3**:
+```
+Input: board = "G", hand = "GGGGG"
+Output: 2
+Explanation: To make the board empty:
+- Insert 'G' so the board becomes GG.
+- Insert 'G' so the board becomes GGG. GGG -> empty.
+2 balls from your hand were needed to clear the board.
+```
+
+**Example 4**:
+```
+Input: board = "RBYYBBRRB", hand = "YRBGB"
+Output: 3
+Explanation: To make the board empty:
+- Insert 'Y' so the board becomes RBYYYBBRRB. RBYYYBBRRB -> RBBBRRB -> RRRB -> B.
+- Insert 'B' so the board becomes BB.
+- Insert 'B' so the board becomes BBB. BBB -> empty.
+3 balls from your hand were needed to clear the board.
+```
+
+**Constraints**:
+
+- 1 <= board.length <= 16
+- 1 <= hand.length <= 5
+- board and hand consist of the characters 'R', 'Y', 'B', 'G', and 'W'.
+- The initial row of balls on the board will not have any groups of three or more consecutive balls of the same color.
+
+## 题目大意
+
+你正在参与祖玛游戏的一个变种。
+
+在这个祖玛游戏变体中，桌面上有 一排 彩球，每个球的颜色可能是：红色 'R'、黄色 'Y'、蓝色 'B'、绿色 'G' 或白色 'W' 。你的手中也有一些彩球。
+
+你的目标是 清空 桌面上所有的球。每一回合：
+
+从你手上的彩球中选出 任意一颗 ，然后将其插入桌面上那一排球中：两球之间或这一排球的任一端。
+接着，如果有出现 三个或者三个以上 且 颜色相同 的球相连的话，就把它们移除掉。
+如果这种移除操作同样导致出现三个或者三个以上且颜色相同的球相连，则可以继续移除这些球，直到不再满足移除条件。
+如果桌面上所有球都被移除，则认为你赢得本场游戏。
+重复这个过程，直到你赢了游戏或者手中没有更多的球。
+给你一个字符串 board ，表示桌面上最开始的那排球。另给你一个字符串 hand ，表示手里的彩球。请你按上述操作步骤移除掉桌上所有球，计算并返回所需的 最少 球数。如果不能移除桌上所有的球，返回 -1 。
+
+## 解题思路
+
+- 使用广度优先搜索和剪枝
+
+## 代码
+
+```go
+
+package leetcode
+
+func findMinStep(board string, hand string) int {
+	q := [][]string{{board, hand}}
+	mp := make(map[string]bool)
+	minStep := 0
+	for len(q) > 0 {
+		length := len(q)
+		minStep++
+		for length > 0 {
+			length--
+			cur := q[0]
+			q = q[1:]
+			curB, curH := cur[0], cur[1]
+			for i := 0; i < len(curB); i++ {
+				for j := 0; j < len(curH); j++ {
+					curB2 := del3(curB[0:i] + string(curH[j]) + curB[i:])
+					curH2 := curH[0:j] + curH[j+1:]
+					if len(curB2) == 0 {
+						return minStep
+					}
+					if _, ok := mp[curB2+curH2]; ok {
+						continue
+					}
+					mp[curB2+curH2] = true
+					q = append(q, []string{curB2, curH2})
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func del3(str string) string {
+	cnt := 1
+	for i := 1; i < len(str); i++ {
+		if str[i] == str[i-1] {
+			cnt++
+		} else {
+			if cnt >= 3 {
+				return del3(str[0:i-cnt] + str[i:])
+			}
+			cnt = 1
+		}
+	}
+	if cnt >= 3 {
+		return str[0 : len(str)-cnt]
+	}
+	return str
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0491. Increasing Subsequences/491. Increasing Subsequences.go b/leetcode/0491.Non-decreasing-Subsequences/491. Non-decreasing Subsequences.go
similarity index 100%
rename from Algorithms/0491. Increasing Subsequences/491. Increasing Subsequences.go
rename to leetcode/0491.Non-decreasing-Subsequences/491. Non-decreasing Subsequences.go
diff --git a/leetcode/0491.Non-decreasing-Subsequences/491. Non-decreasing Subsequences_test.go b/leetcode/0491.Non-decreasing-Subsequences/491. Non-decreasing Subsequences_test.go
new file mode 100644
index 000000000..61602a2d8
--- /dev/null
+++ b/leetcode/0491.Non-decreasing-Subsequences/491. Non-decreasing Subsequences_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question491 struct {
+	para491
+	ans491
+}
+
+// para 是参数
+// one 代表第一个参数
+type para491 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans491 struct {
+	one [][]int
+}
+
+func Test_Problem491(t *testing.T) {
+
+	qs := []question491{
+
+		{
+			para491{[]int{4, 3, 2, 1}},
+			ans491{[][]int{}},
+		},
+
+		{
+			para491{[]int{4, 6, 7, 7}},
+			ans491{[][]int{{4, 6}, {4, 7}, {4, 6, 7}, {4, 6, 7, 7}, {6, 7}, {6, 7, 7}, {7, 7}, {4, 7, 7}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 491------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans491, q.para491
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findSubsequences(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0491.Non-decreasing-Subsequences/README.md b/leetcode/0491.Non-decreasing-Subsequences/README.md
new file mode 100755
index 000000000..f5c5e34a6
--- /dev/null
+++ b/leetcode/0491.Non-decreasing-Subsequences/README.md
@@ -0,0 +1,40 @@
+# [491. Non-decreasing Subsequences](https://leetcode.com/problems/non-decreasing-subsequences/)
+
+
+## 题目
+
+Given an integer array, your task is to find all the different possible increasing subsequences of the given array, and the length of an increasing subsequence should be at least 2.
+
+**Example:**
+
+    Input: [4, 6, 7, 7]
+    Output: [[4, 6], [4, 7], [4, 6, 7], [4, 6, 7, 7], [6, 7], [6, 7, 7], [7,7], [4,7,7]]
+
+**Note:**
+
+1. The length of the given array will not exceed 15.
+2. The range of integer in the given array is [-100,100].
+3. The given array may contain duplicates, and two equal integers should also be considered as a special case of increasing sequence.
+
+
+
+## 题目大意
+
+
+给定一个整型数组, 你的任务是找到所有该数组的递增子序列，递增子序列的长度至少是 2。
+
+说明:
+
+1. 给定数组的长度不会超过15。
+2. 数组中的整数范围是 [-100,100]。
+3. 给定数组中可能包含重复数字，相等的数字应该被视为递增的一种情况。
+
+
+
+
+## 解题思路
+
+
+- 给出一个数组，要求找出这个数组中所有长度大于 2 的非递减子序列。子序列顺序和原数组元素下标必须是顺序的，不能是逆序的。
+- 这一题和第 78 题和第 90 题是类似的题目。第 78 题和第 90 题是求所有子序列，这一题在这两题的基础上增加了非递减和长度大于 2 的条件。需要注意的两点是，原数组中元素可能会重复，最终结果输出的时候需要去重。最终结果输出的去重用 map 处理，数组中重复元素用 DFS 遍历搜索。在每次 DFS 中，用 map 记录遍历过的元素，保证本轮 DFS 中不出现重复的元素，递归到下一层还可以选择值相同，但是下标不同的另外一个元素。外层循环也要加一个  map，这个 map 是过滤每组解因为重复元素导致的重复解，经过过滤以后，起点不同了，最终的解也会不同。
+- 这一题和第 78 题，第 90 题类似，可以一起解答和复习。
diff --git a/leetcode/0492.Construct-the-Rectangle/492.Construct the Rectangle.go b/leetcode/0492.Construct-the-Rectangle/492.Construct the Rectangle.go
new file mode 100644
index 000000000..3314fbc13
--- /dev/null
+++ b/leetcode/0492.Construct-the-Rectangle/492.Construct the Rectangle.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+import "math"
+
+func constructRectangle(area int) []int {
+	ans := make([]int, 2)
+	W := int(math.Sqrt(float64(area)))
+	for W >= 1 {
+		if area%W == 0 {
+			ans[0], ans[1] = area/W, W
+			break
+		}
+		W -= 1
+	}
+	return ans
+}
diff --git a/leetcode/0492.Construct-the-Rectangle/492.Construct the Rectangle_test.go b/leetcode/0492.Construct-the-Rectangle/492.Construct the Rectangle_test.go
new file mode 100644
index 000000000..b3ed98390
--- /dev/null
+++ b/leetcode/0492.Construct-the-Rectangle/492.Construct the Rectangle_test.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question492 struct {
+	para492
+	ans492
+}
+
+// area 是参数
+type para492 struct {
+	area int
+}
+
+// ans 是答案
+type ans492 struct {
+	ans []int
+}
+
+func Test_Problem492(t *testing.T) {
+
+	qs := []question492{
+
+		{
+			para492{4},
+			ans492{[]int{2, 2}},
+		},
+
+		{
+			para492{37},
+			ans492{[]int{37, 1}},
+		},
+
+		{
+			para492{122122},
+			ans492{[]int{427, 286}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 492------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans492, q.para492
+		fmt.Printf("【input】:%v       【output】:%v\n", p, constructRectangle(p.area))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0492.Construct-the-Rectangle/README.md b/leetcode/0492.Construct-the-Rectangle/README.md
new file mode 100644
index 000000000..8024d6e16
--- /dev/null
+++ b/leetcode/0492.Construct-the-Rectangle/README.md
@@ -0,0 +1,72 @@
+# [492. Construct the Rectangle](https://leetcode.com/problems/construct-the-rectangle/)
+
+
+## 题目
+
+A web developer needs to know how to design a web page's size. 
+So, given a specific rectangular web page’s area, your job by now is to design a rectangular web page,
+whose length L and width W satisfy the following requirements:
+
+    The area of the rectangular web page you designed must equal to the given target area.
+    The width W should not be larger than the length L, which means L >= W.
+    The difference between length L and width W should be as small as possible.
+    Return an array [L, W] where L and W are the length and width of the web page you designed in sequence.
+
+**Example 1:**
+
+    Input: area = 4
+    Output: [2,2]
+    Explanation: The target area is 4, and all the possible ways to construct it are [1,4], [2,2], [4,1].
+    But according to requirement 2, [1,4] is illegal; according to requirement 3,  [4,1] is not optimal compared to [2,2]. So the length L is 2, and the width W is 2.
+
+**Example 2:**
+
+    Input: area = 37
+    Output: [37,1]
+
+**Example 3:**
+
+    Input: area = 122122
+    Output: [427,286]
+
+**Constraints**
+
+   - 1 <= area <= 10000000
+
+## 题目大意
+
+作为一位 web 开发者， 懂得怎样去规划一个页面的尺寸是很重要的。 现给定一个具体的矩形页面面积，你的任务是设计一个长度为 L 和宽度为 W 且满足以下要求的矩形的页面。要求：
+
+1. 你设计的矩形页面必须等于给定的目标面积。
+2. 宽度 W 不应大于长度 L，换言之，要求 L >= W 。
+3. 长度 L 和宽度 W 之间的差距应当尽可能小。 
+
+你需要按顺序输出你设计的页面的长度 L 和宽度 W。
+
+## 解题思路
+
+- 令 W 等于根号 area
+- 在 W 大于等于 1 的情况下,判断 area%W 是否等于 0,如果不相等 W 就减 1 继续循环,如果相等就返回 [area/W, W]
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func constructRectangle(area int) []int {
+	ans := make([]int, 2)
+	W := int(math.Sqrt(float64(area)))
+	for W >= 1 {
+		if area%W == 0 {
+			ans[0], ans[1] = area/W, W
+			break
+		}
+		W -= 1
+	}
+	return ans
+}
+
+``
\ No newline at end of file
diff --git a/leetcode/0493.Reverse-Pairs/493. Reverse Pairs.go b/leetcode/0493.Reverse-Pairs/493. Reverse Pairs.go
new file mode 100644
index 000000000..ec312a33f
--- /dev/null
+++ b/leetcode/0493.Reverse-Pairs/493. Reverse Pairs.go	
@@ -0,0 +1,105 @@
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+// 解法一 归并排序 mergesort，时间复杂度 O(n log n)
+func reversePairs(nums []int) int {
+	buf := make([]int, len(nums))
+	return mergesortCount(nums, buf)
+}
+
+func mergesortCount(nums, buf []int) int {
+	if len(nums) <= 1 {
+		return 0
+	}
+	mid := (len(nums) - 1) / 2
+	cnt := mergesortCount(nums[:mid+1], buf)
+	cnt += mergesortCount(nums[mid+1:], buf)
+	for i, j := 0, mid+1; i < mid+1; i++ { // Note!!! j is increasing.
+		for ; j < len(nums) && nums[i] <= 2*nums[j]; j++ {
+		}
+		cnt += len(nums) - j
+	}
+	copy(buf, nums)
+	for i, j, k := 0, mid+1, 0; k < len(nums); {
+		if j >= len(nums) || i < mid+1 && buf[i] > buf[j] {
+			nums[k] = buf[i]
+			i++
+		} else {
+			nums[k] = buf[j]
+			j++
+		}
+		k++
+	}
+	return cnt
+}
+
+// 解法二 树状数组，时间复杂度 O(n log n)
+func reversePairs1(nums []int) (cnt int) {
+	n := len(nums)
+	if n <= 1 {
+		return
+	}
+	// 离散化所有下面统计时会出现的元素
+	allNums := make([]int, 0, 2*n)
+	for _, v := range nums {
+		allNums = append(allNums, v, 2*v)
+	}
+	sort.Ints(allNums)
+	k := 1
+	kth := map[int]int{allNums[0]: k}
+	for i := 1; i < 2*n; i++ {
+		if allNums[i] != allNums[i-1] {
+			k++
+			kth[allNums[i]] = k
+		}
+	}
+	bit := template.BinaryIndexedTree{}
+	bit.Init(k)
+	for i, v := range nums {
+		cnt += i - bit.Query(kth[2*v])
+		bit.Add(kth[v], 1)
+	}
+	return
+}
+
+// 解法三 线段树，时间复杂度 O(n log n)
+func reversePairs2(nums []int) int {
+	if len(nums) < 2 {
+		return 0
+	}
+	st, numsMap, indexMap, numsArray, res := template.SegmentCountTree{}, make(map[int]int, 0), make(map[int]int, 0), []int{}, 0
+	numsMap[nums[0]] = nums[0]
+	for _, num := range nums {
+		numsMap[num] = num
+		numsMap[2*num+1] = 2*num + 1
+	}
+	// numsArray 是 prefixSum 去重之后的版本，利用 numsMap 去重
+	for _, v := range numsMap {
+		numsArray = append(numsArray, v)
+	}
+	// 排序是为了使得线段树中的区间 left <= right，如果此处不排序，线段树中的区间有很多不合法。
+	sort.Ints(numsArray)
+	// 离散化，构建映射关系
+	for i, n := range numsArray {
+		indexMap[n] = i
+	}
+	numsArray = []int{}
+	// 离散化，此题如果不离散化，MaxInt32 的数据会使得数字越界。
+	for i := 0; i < len(indexMap); i++ {
+		numsArray = append(numsArray, i)
+	}
+	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
+	st.Init(numsArray, func(i, j int) int {
+		return 0
+	})
+	for _, num := range nums {
+		res += st.Query(indexMap[num*2+1], len(indexMap)-1)
+		st.UpdateCount(indexMap[num])
+	}
+	return res
+}
diff --git a/Algorithms/0493. Reverse Pairs/493. Reverse Pairs_test.go b/leetcode/0493.Reverse-Pairs/493. Reverse Pairs_test.go
similarity index 90%
rename from Algorithms/0493. Reverse Pairs/493. Reverse Pairs_test.go
rename to leetcode/0493.Reverse-Pairs/493. Reverse Pairs_test.go
index 12866f12b..2c9097080 100644
--- a/Algorithms/0493. Reverse Pairs/493. Reverse Pairs_test.go	
+++ b/leetcode/0493.Reverse-Pairs/493. Reverse Pairs_test.go	
@@ -26,22 +26,27 @@ func Test_Problem493(t *testing.T) {
 
 	qs := []question493{
 
-		question493{
+		{
 			para493{[]int{1, 3, 2, 3, 1}},
 			ans493{2},
 		},
 
-		question493{
+		{
+			para493{[]int{9, 8, 7, 4, 7, 2, 3, 8, 7, 0}},
+			ans493{18},
+		},
+
+		{
 			para493{[]int{2, 4, 3, 5, 1}},
 			ans493{3},
 		},
 
-		question493{
+		{
 			para493{[]int{-5, -5}},
 			ans493{1},
 		},
 
-		question493{
+		{
 			para493{[]int{2147483647, 2147483647, -2147483647, -2147483647, -2147483647, 2147483647}},
 			ans493{9},
 		},
diff --git a/leetcode/0493.Reverse-Pairs/README.md b/leetcode/0493.Reverse-Pairs/README.md
new file mode 100755
index 000000000..c27689198
--- /dev/null
+++ b/leetcode/0493.Reverse-Pairs/README.md
@@ -0,0 +1,43 @@
+# [493. Reverse Pairs](https://leetcode.com/problems/reverse-pairs/)
+
+
+## 题目
+
+Given an array `nums`, we call `(i, j)` an **important reverse pair** if `i < j` and `nums[i] > 2*nums[j]`.
+
+You need to return the number of important reverse pairs in the given array.
+
+**Example1:**
+
+    Input: [1,3,2,3,1]
+    Output: 2
+
+**Example2:**
+
+    Input: [2,4,3,5,1]
+    Output: 3
+
+**Note:**
+
+1. The length of the given array will not exceed `50,000`.
+2. All the numbers in the input array are in the range of 32-bit integer.
+
+
+## 题目大意
+
+给定一个数组 nums ，如果 i < j 且 nums[i] > 2\*nums[j] 我们就将 (i, j) 称作一个重要翻转对。你需要返回给定数组中的重要翻转对的数量。
+
+注意:
+
+- 给定数组的长度不会超过 50000。
+- 输入数组中的所有数字都在 32 位整数的表示范围内。
+
+
+## 解题思路
+
+
+- 给出一个数组，要求找出满足条件的所有的“重要的反转对” (i,j)。重要的反转对的定义是：`i<j`，并且 `nums[i] > 2*nums[j]`。
+- 这一题是 327 题的变种题。首先将数组中所有的元素以及各自的 `2*nums[i] + 1` 都放在字典中去重。去重以后再做离散化处理。这一题的测试用例会卡离散化，如果不离散化，Math.MaxInt32 会导致数字溢出，见测试用例中 2147483647, -2147483647 这组测试用例。离散后，映射关系 保存在字典中。从左往右遍历数组，先 query ，再 update ，这个顺序和第 327 题是反的。先 query 查找 `[2*nums[i] + 1, len(indexMap)-1]` 这个区间内满足条件的值，这个区间内的值都是 `> 2*nums[j]` 的。这一题移动的是 `j`，`j` 不断的变化，往线段树中不断插入的是 `i`。每轮循环先 query 一次前一轮循环中累积插入线段树中的 `i`，这些累积在线段树中的代表的是所有在 `j` 前面的 `i`。query 查询的是本轮 `[2*nums[j] + 1, len(indexMap)-1]`，如果能找到，即找到了这样一个 `j`，能满足 `nums[i] > 2*nums[j`， 把整个数组都扫完，累加的 query 出来的 count 计数就是最终答案。
+- 另外一种解法是树状数组。树状数组最擅长解答逆序对的问题。先将原数组中所有的元素值的 2 倍算出来，和原数组合并到一个大数组中。这个大数组中装了所有可能产生 2 倍逆序对的元素值。然后再将他们所有值排序，离散化。离散化以后便将问题集转化成 `[1,N]` 这个区间。于是回到了树状数组经典的求逆序对的问题。逆序插入原数组构造树状数组，或者正序插入原数组构造树状数组都可以解答此题。
+- 类似的题目：第 327 题，第 315 题。
+- 这一题用线段树和树状数组并不是最优解，用线段树和树状数组解这一题是为了训练线段树和树状数组这两个数据结构。最优解是解法一中的 mergesort。
diff --git a/leetcode/0494.Target-Sum/494. Target Sum.go b/leetcode/0494.Target-Sum/494. Target Sum.go
new file mode 100644
index 000000000..8adde19c7
--- /dev/null
+++ b/leetcode/0494.Target-Sum/494. Target Sum.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+// 解法一 DP
+func findTargetSumWays(nums []int, S int) int {
+	total := 0
+	for _, n := range nums {
+		total += n
+	}
+	if   S+total < 0 || S > total || (S+total)%2 == 1 {
+		return 0
+	}
+	target := (S + total) / 2
+	dp := make([]int, target+1)
+	dp[0] = 1
+	for _, n := range nums {
+		for i := target; i >= n; i-- {
+			dp[i] += dp[i-n]
+		}
+	}
+	return dp[target]
+}
+
+// 解法二 DFS
+func findTargetSumWays1(nums []int, S int) int {
+	// sums[i] 存储的是后缀和 nums[i:]，即从 i 到结尾的和
+	sums := make([]int, len(nums))
+	sums[len(nums)-1] = nums[len(nums)-1]
+	for i := len(nums) - 2; i > -1; i-- {
+		sums[i] = sums[i+1] + nums[i]
+	}
+	res := 0
+	dfsFindTargetSumWays(nums, 0, 0, S, &res, sums)
+	return res
+}
+
+func dfsFindTargetSumWays(nums []int, index int, curSum int, S int, res *int, sums []int) {
+	if index == len(nums) {
+		if curSum == S {
+			*(res) = *(res) + 1
+		}
+		return
+	}
+	// 剪枝优化：如果 sums[index] 值小于剩下需要正数的值，那么右边就算都是 + 号也无能为力了，所以这里可以剪枝了
+	if S-curSum > sums[index] {
+		return
+	}
+	dfsFindTargetSumWays(nums, index+1, curSum+nums[index], S, res, sums)
+	dfsFindTargetSumWays(nums, index+1, curSum-nums[index], S, res, sums)
+}
diff --git a/leetcode/0494.Target-Sum/494. Target Sum_test.go b/leetcode/0494.Target-Sum/494. Target Sum_test.go
new file mode 100644
index 000000000..76700ec01
--- /dev/null
+++ b/leetcode/0494.Target-Sum/494. Target Sum_test.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question494 struct {
+	para494
+	ans494
+}
+
+// para 是参数
+// one 代表第一个参数
+type para494 struct {
+	nums []int
+	S    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans494 struct {
+	one int
+}
+
+func Test_Problem494(t *testing.T) {
+
+	qs := []question494{
+
+		{
+			para494{[]int{1, 1, 1, 1, 1}, 3},
+			ans494{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 494------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans494, q.para494
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findTargetSumWays(p.nums, p.S))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0494.Target-Sum/README.md b/leetcode/0494.Target-Sum/README.md
new file mode 100644
index 000000000..94234e7b9
--- /dev/null
+++ b/leetcode/0494.Target-Sum/README.md
@@ -0,0 +1,108 @@
+# [494. Target Sum](https://leetcode.com/problems/target-sum/)
+
+
+## 题目
+
+You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols `+` and `-`. For each integer, you should choose one from `+` and `-` as its new symbol.
+
+Find out how many ways to assign symbols to make sum of integers equal to target S.
+
+**Example 1**:
+
+```
+Input: nums is [1, 1, 1, 1, 1], S is 3. 
+Output: 5
+Explanation: 
+
+-1+1+1+1+1 = 3
++1-1+1+1+1 = 3
++1+1-1+1+1 = 3
++1+1+1-1+1 = 3
++1+1+1+1-1 = 3
+
+There are 5 ways to assign symbols to make the sum of nums be target 3.
+```
+
+**Note**:
+
+1. The length of the given array is positive and will not exceed 20.
+2. The sum of elements in the given array will not exceed 1000.
+3. Your output answer is guaranteed to be fitted in a 32-bit integer.
+
+## 题目大意
+
+给定一个非负整数数组，a1, a2, ..., an, 和一个目标数，S。现在有两个符号 + 和 -。对于数组中的任意一个整数，可以从 + 或 - 中选择一个符号添加在前面。返回可以使最终数组和为目标数 S 的所有添加符号的方法数。
+
+提示：
+
+- 数组非空，且长度不会超过 20 。
+- 初始的数组的和不会超过 1000 。
+- 保证返回的最终结果能被 32 位整数存下。
+
+## 解题思路
+
+- 给出一个数组，要求在这个数组里面的每个元素前面加上 + 或者 - 号，最终总和等于 S。问有多少种不同的方法。
+- 这一题可以用 DP 和 DFS 解答。DFS 方法就比较暴力简单了。见代码。这里分析一下 DP 的做法。题目要求在数组元素前加上 + 或者 - 号，其实相当于把数组分成了 2 组，一组全部都加 + 号，一组都加 - 号。记 + 号的一组 P ，记 - 号的一组 N，那么可以推出以下的关系。
+
+    ```go
+    sum(P) - sum(N) = target
+    sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)
+                           2 * sum(P) = target + sum(nums)
+    ```
+
+    等号两边都加上 `sum(N) + sum(P)`，于是可以得到结果 `2 * sum(P) = target + sum(nums)`，那么这道题就转换成了，能否在数组中找到这样一个集合，和等于 `(target + sum(nums)) / 2`。那么这题就转化为了第 416 题了。`dp[i]` 中存储的是能使和为 `i` 的方法个数。
+
+- 如果和不是偶数，即不能被 2 整除，或者和是负数，那说明找不到满足题目要求的解了，直接输出 0 。
+
+## 代码
+
+```go
+
+func findTargetSumWays(nums []int, S int) int {
+	total := 0
+	for _, n := range nums {
+		total += n
+	}
+	if S > total || (S+total)%2 == 1 || S+total < 0 {
+		return 0
+	}
+	target := (S + total) / 2
+	dp := make([]int, target+1)
+	dp[0] = 1
+	for _, n := range nums {
+		for i := target; i >= n; i-- {
+			dp[i] += dp[i-n]
+		}
+	}
+	return dp[target]
+}
+
+// 解法二 DFS
+func findTargetSumWays1(nums []int, S int) int {
+	// sums[i] 存储的是后缀和 nums[i:]，即从 i 到结尾的和
+	sums := make([]int, len(nums))
+	sums[len(nums)-1] = nums[len(nums)-1]
+	for i := len(nums) - 2; i > -1; i-- {
+		sums[i] = sums[i+1] + nums[i]
+	}
+	res := 0
+	dfsFindTargetSumWays(nums, 0, 0, S, &res, sums)
+	return res
+}
+
+func dfsFindTargetSumWays(nums []int, index int, curSum int, S int, res *int, sums []int) {
+	if index == len(nums) {
+		if curSum == S {
+			*(res) = *(res) + 1
+		}
+		return
+	}
+	// 剪枝优化：如果 sums[index] 值小于剩下需要正数的值，那么右边就算都是 + 号也无能为力了，所以这里可以剪枝了
+	if S-curSum > sums[index] {
+		return
+	}
+	dfsFindTargetSumWays(nums, index+1, curSum+nums[index], S, res, sums)
+	dfsFindTargetSumWays(nums, index+1, curSum-nums[index], S, res, sums)
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/0495.Teemo-Attacking/495.Teemo Attacking.go b/leetcode/0495.Teemo-Attacking/495.Teemo Attacking.go
new file mode 100644
index 000000000..4da646d98
--- /dev/null
+++ b/leetcode/0495.Teemo-Attacking/495.Teemo Attacking.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func findPoisonedDuration(timeSeries []int, duration int) int {
+	var ans int
+	for i := 1; i < len(timeSeries); i++ {
+		t := timeSeries[i-1]
+		end := t + duration - 1
+		if end < timeSeries[i] {
+			ans += duration
+		} else {
+			ans += timeSeries[i] - t
+		}
+	}
+	ans += duration
+	return ans
+}
diff --git a/leetcode/0495.Teemo-Attacking/495.Teemo Attacking_test.go b/leetcode/0495.Teemo-Attacking/495.Teemo Attacking_test.go
new file mode 100644
index 000000000..2900a11f8
--- /dev/null
+++ b/leetcode/0495.Teemo-Attacking/495.Teemo Attacking_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question495 struct {
+	para495
+	ans495
+}
+
+// para 是参数
+type para495 struct {
+	timeSeries []int
+	duration   int
+}
+
+// ans 是答案
+type ans495 struct {
+	ans int
+}
+
+func Test_Problem495(t *testing.T) {
+
+	qs := []question495{
+
+		{
+			para495{[]int{1, 4}, 2},
+			ans495{4},
+		},
+
+		{
+			para495{[]int{1, 2}, 2},
+			ans495{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 495------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans495, q.para495
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findPoisonedDuration(p.timeSeries, p.duration))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0495.Teemo-Attacking/README.md b/leetcode/0495.Teemo-Attacking/README.md
new file mode 100644
index 000000000..3a52e1f27
--- /dev/null
+++ b/leetcode/0495.Teemo-Attacking/README.md
@@ -0,0 +1,85 @@
+# [495. Teemo Attacking](https://leetcode.com/problems/teemo-attacking/)
+
+
+## 题目
+
+Our hero Teemo is attacking an enemy Ashe with poison attacks! When Teemo attacks Ashe, Ashe gets poisoned for a exactly duration seconds. 
+
+More formally, an attack at second t will mean Ashe is poisoned during the inclusive time interval [t, t + duration - 1].
+
+If Teemo attacks again before the poison effect ends, the timer for it is reset, and the poison effect will end duration seconds after the new attack.
+
+You are given a non-decreasing integer array timeSeries, where timeSeries[i] denotes that Teemo attacks Ashe at second timeSeries[i], and an integer duration.
+
+Return the total number of seconds that Ashe is poisoned.
+
+**Example 1**:
+```
+Input: timeSeries = [1,4], duration = 2
+Output: 4
+Explanation: Teemo's attacks on Ashe go as follows:
+- At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
+- At second 4, Teemo attacks, and Ashe is poisoned for seconds 4 and 5.
+Ashe is poisoned for seconds 1, 2, 4, and 5, which is 4 seconds in total.
+```
+
+**Example 2**:
+```
+Input: timeSeries = [1,2], duration = 2
+Output: 3
+Explanation: Teemo's attacks on Ashe go as follows:
+- At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
+- At second 2 however, Teemo attacks again and resets the poison timer. Ashe is poisoned for seconds 2 and 3.
+Ashe is poisoned for seconds 1, 2, and 3, which is 3 seconds in total.
+```
+
+**Constraints**:
+
+- 1 <= timeSeries.length <= 10000
+- 0 <= timeSeries[i], duration <= 10000000
+- timeSeries is sorted in non-decreasing order.
+
+## 题目大意
+
+在《英雄联盟》的世界中，有一个叫 “提莫” 的英雄。他的攻击可以让敌方英雄艾希（编者注：寒冰射手）进入中毒状态。
+
+当提莫攻击艾希，艾希的中毒状态正好持续duration 秒。
+
+正式地讲，提莫在t发起发起攻击意味着艾希在时间区间 [t, t + duration - 1]（含 t 和 t + duration - 1）处于中毒状态。
+
+如果提莫在中毒影响结束前再次攻击，中毒状态计时器将会重置，在新的攻击之后，中毒影响将会在duration秒后结束。
+
+给你一个非递减的整数数组timeSeries，其中timeSeries[i]表示提莫在timeSeries[i]秒时对艾希发起攻击，以及一个表示中毒持续时间的整数duration 。
+
+返回艾希处于中毒状态的总秒数。
+
+## 解题思路
+
+- i 从 1 开始计数，令 t 等于 timeSeries[i - 1]
+- 比较 end(t + duration - 1) 和 timeSeries[i] 的大小，
+  - 如果 end 小于 timeSeries[i],ans+=duration
+  - 否则 ans += timeSeries[i] - t
+- ans += duration 并返回 ans
+
+## 代码
+
+```go
+
+package leetcode
+
+func findPoisonedDuration(timeSeries []int, duration int) int {
+	var ans int
+	for i := 1; i < len(timeSeries); i++ {
+		t := timeSeries[i-1]
+		end := t + duration - 1
+		if end < timeSeries[i] {
+			ans += duration
+		} else {
+			ans += timeSeries[i] - t
+		}
+	}
+	ans += duration
+	return ans
+}
+
+```
diff --git a/Algorithms/0496. Next Greater Element I/496. Next Greater Element I.go b/leetcode/0496.Next-Greater-Element-I/496. Next Greater Element I.go
similarity index 86%
rename from Algorithms/0496. Next Greater Element I/496. Next Greater Element I.go
rename to leetcode/0496.Next-Greater-Element-I/496. Next Greater Element I.go
index b1b9125b0..b70becf0c 100644
--- a/Algorithms/0496. Next Greater Element I/496. Next Greater Element I.go	
+++ b/leetcode/0496.Next-Greater-Element-I/496. Next Greater Element I.go	
@@ -4,12 +4,12 @@ func nextGreaterElement(nums1 []int, nums2 []int) []int {
 	if len(nums1) == 0 || len(nums2) == 0 {
 		return []int{}
 	}
-	res, reocrd, flag := []int{}, map[int]int{}, false
+	res, reocrd := []int{}, map[int]int{}
 	for i, v := range nums2 {
 		reocrd[v] = i
 	}
 	for i := 0; i < len(nums1); i++ {
-		flag = false
+		flag := false
 		for j := reocrd[nums1[i]]; j < len(nums2); j++ {
 			if nums2[j] > nums1[i] {
 				res = append(res, nums2[j])
diff --git a/Algorithms/0496. Next Greater Element I/496. Next Greater Element I_test.go b/leetcode/0496.Next-Greater-Element-I/496. Next Greater Element I_test.go
similarity index 96%
rename from Algorithms/0496. Next Greater Element I/496. Next Greater Element I_test.go
rename to leetcode/0496.Next-Greater-Element-I/496. Next Greater Element I_test.go
index f030ecd34..52b8bb75c 100644
--- a/Algorithms/0496. Next Greater Element I/496. Next Greater Element I_test.go	
+++ b/leetcode/0496.Next-Greater-Element-I/496. Next Greater Element I_test.go	
@@ -27,12 +27,12 @@ func Test_Problem496(t *testing.T) {
 
 	qs := []question496{
 
-		question496{
+		{
 			para496{[]int{4, 1, 2}, []int{1, 3, 4, 2}},
 			ans496{[]int{-1, 3, -1}},
 		},
 
-		question496{
+		{
 			para496{[]int{2, 4}, []int{1, 2, 3, 4}},
 			ans496{[]int{3, -1}},
 		},
diff --git a/Algorithms/0496. Next Greater Element I/README.md b/leetcode/0496.Next-Greater-Element-I/README.md
similarity index 100%
rename from Algorithms/0496. Next Greater Element I/README.md
rename to leetcode/0496.Next-Greater-Element-I/README.md
diff --git a/leetcode/0497.Random-Point-in-Non-overlapping-Rectangles/497. Random Point in Non-overlapping Rectangles.go b/leetcode/0497.Random-Point-in-Non-overlapping-Rectangles/497. Random Point in Non-overlapping Rectangles.go
new file mode 100644
index 000000000..eccb3b88d
--- /dev/null
+++ b/leetcode/0497.Random-Point-in-Non-overlapping-Rectangles/497. Random Point in Non-overlapping Rectangles.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import "math/rand"
+
+// Solution497 define
+type Solution497 struct {
+	rects [][]int
+	arr   []int
+}
+
+// Constructor497 define
+func Constructor497(rects [][]int) Solution497 {
+	s := Solution497{
+		rects: rects,
+		arr:   make([]int, len(rects)),
+	}
+
+	for i := 0; i < len(rects); i++ {
+		area := (rects[i][2] - rects[i][0] + 1) * (rects[i][3] - rects[i][1] + 1)
+		if area < 0 {
+			area = -area
+		}
+		if i == 0 {
+			s.arr[0] = area
+		} else {
+			s.arr[i] = s.arr[i-1] + area
+		}
+	}
+	return s
+}
+
+// Pick define
+func (so *Solution497) Pick() []int {
+	r := rand.Int() % so.arr[len(so.arr)-1]
+	//get rectangle first
+	low, high, index := 0, len(so.arr)-1, -1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if so.arr[mid] > r {
+			if mid == 0 || so.arr[mid-1] <= r {
+				index = mid
+				break
+			}
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	if index == -1 {
+		index = low
+	}
+	if index > 0 {
+		r = r - so.arr[index-1]
+	}
+	length := so.rects[index][2] - so.rects[index][0]
+	return []int{so.rects[index][0] + r%(length+1), so.rects[index][1] + r/(length+1)}
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(rects);
+ * param_1 := obj.Pick();
+ */
diff --git a/leetcode/0497.Random-Point-in-Non-overlapping-Rectangles/497. Random Point in Non-overlapping Rectangles_test.go b/leetcode/0497.Random-Point-in-Non-overlapping-Rectangles/497. Random Point in Non-overlapping Rectangles_test.go
new file mode 100644
index 000000000..766d13d03
--- /dev/null
+++ b/leetcode/0497.Random-Point-in-Non-overlapping-Rectangles/497. Random Point in Non-overlapping Rectangles_test.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem497(t *testing.T) {
+	w := [][]int{{1, 1, 5, 5}}
+	sol := Constructor497(w)
+	fmt.Printf("1.Pick = %v\n", sol.Pick())
+	fmt.Printf("2.Pick = %v\n", sol.Pick())
+	fmt.Printf("3.Pick = %v\n", sol.Pick())
+	fmt.Printf("4.Pick = %v\n", sol.Pick())
+	fmt.Printf("5.Pick = %v\n", sol.Pick())
+	fmt.Printf("6.Pick = %v\n", sol.Pick())
+}
diff --git a/leetcode/0497.Random-Point-in-Non-overlapping-Rectangles/README.md b/leetcode/0497.Random-Point-in-Non-overlapping-Rectangles/README.md
new file mode 100755
index 000000000..449060b45
--- /dev/null
+++ b/leetcode/0497.Random-Point-in-Non-overlapping-Rectangles/README.md
@@ -0,0 +1,63 @@
+# [497. Random Point in Non-overlapping Rectangles](https://leetcode.com/problems/random-point-in-non-overlapping-rectangles)
+
+
+## 题目
+
+Given a list of **non-overlapping** axis-aligned rectangles `rects`, write a function `pick` which randomly and uniformily picks an **integer point** in the space covered by the rectangles.
+
+Note:
+
+1. An **integer point** is a point that has integer coordinates.
+2. A point on the perimeter of a rectangle is **included** in the space covered by the rectangles.
+3. `i`th rectangle = `rects[i]` = `[x1,y1,x2,y2]`, where `[x1, y1]` are the integer coordinates of the bottom-left corner, and `[x2, y2]` are the integer coordinates of the top-right corner.
+4. length and width of each rectangle does not exceed `2000`.
+5. `1 <= rects.length <= 100`
+6. `pick` return a point as an array of integer coordinates `[p_x, p_y]`
+7. `pick` is called at most `10000` times.
+
+**Example 1:**
+
+    Input: 
+    ["Solution","pick","pick","pick"]
+    [[[[1,1,5,5]]],[],[],[]]
+    Output: 
+    [null,[4,1],[4,1],[3,3]]
+
+**Example 2:**
+
+    Input: 
+    ["Solution","pick","pick","pick","pick","pick"]
+    [[[[-2,-2,-1,-1],[1,0,3,0]]],[],[],[],[],[]]
+    Output: 
+    [null,[-1,-2],[2,0],[-2,-1],[3,0],[-2,-2]]
+
+**Explanation of Input Syntax:**
+
+The input is two lists: the subroutines called and their arguments. `Solution`'s constructor has one argument, the array of rectangles `rects`. `pick` has no arguments. Arguments are always wrapped with a list, even if there aren't any.
+
+
+## 题目大意
+
+给定一个非重叠轴对齐矩形的列表 rects，写一个函数 pick 随机均匀地选取矩形覆盖的空间中的整数点。
+
+提示：
+
+1. 整数点是具有整数坐标的点。
+2. 矩形周边上的点包含在矩形覆盖的空间中。
+3. 第 i 个矩形 rects [i] = [x1，y1，x2，y2]，其中 [x1，y1] 是左下角的整数坐标，[x2，y2] 是右上角的整数坐标。
+4. 每个矩形的长度和宽度不超过 2000。
+5. 1 <= rects.length <= 100
+6. pick 以整数坐标数组 [p_x, p_y] 的形式返回一个点。
+7. pick 最多被调用10000次。
+
+
+输入语法的说明：
+
+输入是两个列表：调用的子例程及其参数。Solution 的构造函数有一个参数，即矩形数组 rects。pick 没有参数。参数总是用列表包装的，即使没有也是如此。
+
+
+## 解题思路
+
+
+- 给出一个非重叠轴对齐矩形列表，每个矩形用左下角和右上角的两个坐标表示。要求 `pick()` 随机均匀地选取矩形覆盖的空间中的整数点。
+- 这一题是第 528 题的变种题，这一题权重是面积，按权重（面积）选择一个矩形，然后再从矩形中随机选择一个点即可。思路和代码和第 528 题一样。
diff --git a/Algorithms/0498. Diagonal Traverse/498. Diagonal Traverse.go b/leetcode/0498.Diagonal-Traverse/498. Diagonal Traverse.go
similarity index 100%
rename from Algorithms/0498. Diagonal Traverse/498. Diagonal Traverse.go
rename to leetcode/0498.Diagonal-Traverse/498. Diagonal Traverse.go
diff --git a/leetcode/0498.Diagonal-Traverse/498. Diagonal Traverse_test.go b/leetcode/0498.Diagonal-Traverse/498. Diagonal Traverse_test.go
new file mode 100644
index 000000000..a0f7458e5
--- /dev/null
+++ b/leetcode/0498.Diagonal-Traverse/498. Diagonal Traverse_test.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question498 struct {
+	para498
+	ans498
+}
+
+// para 是参数
+// one 代表第一个参数
+type para498 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans498 struct {
+	one []int
+}
+
+func Test_Problem498(t *testing.T) {
+
+	qs := []question498{
+
+		{
+			para498{[][]int{{3}, {2}, {9}}},
+			ans498{[]int{3, 2, 9}},
+		},
+
+		{
+			para498{[][]int{{6, 9, 7}}},
+			ans498{[]int{6, 9, 7}},
+		},
+
+		{
+			para498{[][]int{{3}, {2}}},
+			ans498{[]int{3, 2}},
+		},
+
+		{
+			para498{[][]int{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}},
+			ans498{[]int{1, 2, 4, 7, 5, 3, 6, 8, 9}},
+		},
+
+		{
+			para498{[][]int{{0}}},
+			ans498{[]int{0}},
+		},
+
+		{
+			para498{[][]int{{}}},
+			ans498{[]int{}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 498------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans498, q.para498
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findDiagonalOrder(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0498.Diagonal-Traverse/README.md b/leetcode/0498.Diagonal-Traverse/README.md
new file mode 100755
index 000000000..6e5969c78
--- /dev/null
+++ b/leetcode/0498.Diagonal-Traverse/README.md
@@ -0,0 +1,46 @@
+# [498. Diagonal Traverse](https://leetcode.com/problems/diagonal-traverse/)
+
+
+## 题目
+
+Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image.
+
+**Example:**
+
+    Input:
+    [
+     [ 1, 2, 3 ],
+     [ 4, 5, 6 ],
+     [ 7, 8, 9 ]
+    ]
+    
+    Output:  [1,2,4,7,5,3,6,8,9]
+    
+    Explanation:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/diagonal_traverse.png)
+
+**Note:**
+
+The total number of elements of the given matrix will not exceed 10,000.
+
+
+## 题目大意
+
+给定一个含有 M x N 个元素的矩阵（M 行，N 列），请以对角线遍历的顺序返回这个矩阵中的所有元素，对角线遍历如下图所示。
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/diagonal_traverse.png)
+
+说明: 给定矩阵中的元素总数不会超过 100000 。
+
+## 解题思路
+
+- 给出一个二维数组，要求按照如图的方式遍历整个数组。
+- 这一题用模拟的方式就可以解出来。需要注意的是边界条件：比如二维数组为空，二维数组退化为一行或者一列，退化为一个元素。具体例子见测试用例。
+- 解题关键是在判断下一个位置，将矩阵想像成一个X,Y坐标轴。那么可分为以下几种情况，
+    1、斜角向右上遍历时，
+    当右上角在坐标轴内， 正常计算 即， x+1(X轴向右移动)， y-1(Y轴向上移动)
+    当右上角在坐标轴外，那么当前位置只能在 第一行X坐标轴 ，或者 最后一列Y坐标轴 ， 即判断该两种情况下�应该X坐标往右，或者 Y坐标往上
+    2、同理 斜角向下遍历时
+    当左下角在坐标轴内，正常计算 即， x-1(X轴向右移动)， y+1(Y轴向下移动)
+    当左下角在坐标轴外，那么当前位置只能在 第一列Y坐标轴，或者 最后一行X坐标轴， 即判断该两种情况下�应该X坐标往左，或者 Y坐标往下
diff --git a/Algorithms/0500. Keyboard Row/500. Keyboard Row.go b/leetcode/0500.Keyboard-Row/500. Keyboard Row.go
similarity index 100%
rename from Algorithms/0500. Keyboard Row/500. Keyboard Row.go
rename to leetcode/0500.Keyboard-Row/500. Keyboard Row.go
diff --git a/Algorithms/0500. Keyboard Row/500. Keyboard Row_test.go b/leetcode/0500.Keyboard-Row/500. Keyboard Row_test.go
similarity index 97%
rename from Algorithms/0500. Keyboard Row/500. Keyboard Row_test.go
rename to leetcode/0500.Keyboard-Row/500. Keyboard Row_test.go
index 886fa47ce..d05dfefac 100644
--- a/Algorithms/0500. Keyboard Row/500. Keyboard Row_test.go	
+++ b/leetcode/0500.Keyboard-Row/500. Keyboard Row_test.go	
@@ -26,7 +26,7 @@ func Test_Problem500(t *testing.T) {
 
 	qs := []question500{
 
-		question500{
+		{
 			para500{[]string{"Hello", "Alaska", "Dad", "Peace"}},
 			ans500{[]string{"Alaska", "Dad"}},
 		},
diff --git a/leetcode/0500.Keyboard-Row/README.md b/leetcode/0500.Keyboard-Row/README.md
new file mode 100755
index 000000000..7d9fa5db5
--- /dev/null
+++ b/leetcode/0500.Keyboard-Row/README.md
@@ -0,0 +1,27 @@
+# [500. Keyboard Row](https://leetcode.com/problems/keyboard-row/)
+
+
+## 题目
+
+Given a List of words, return the words that can be typed using letters of **alphabet** on only one row's of American keyboard like the image below.
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/keyboard.png)
+
+**Example:**
+
+    Input: ["Hello", "Alaska", "Dad", "Peace"]
+    Output: ["Alaska", "Dad"]
+
+**Note:**
+
+1. You may use one character in the keyboard more than once.
+2. You may assume the input string will only contain letters of alphabet.
+
+
+## 题目大意
+
+给定一个单词列表，只返回可以使用在键盘同一行的字母打印出来的单词。键盘如上图所示。
+
+## 解题思路
+
+- 给出一个字符串数组，要求依次判断数组中的每个字符串是否都位于键盘上的同一个行，如果是就输出。这也是一道水题。
diff --git a/Algorithms/0503. Next Greater Element II/503. Next Greater Element II.go b/leetcode/0503.Next-Greater-Element-II/503. Next Greater Element II.go
similarity index 100%
rename from Algorithms/0503. Next Greater Element II/503. Next Greater Element II.go
rename to leetcode/0503.Next-Greater-Element-II/503. Next Greater Element II.go
diff --git a/Algorithms/0503. Next Greater Element II/503. Next Greater Element II_test.go b/leetcode/0503.Next-Greater-Element-II/503. Next Greater Element II_test.go
similarity index 94%
rename from Algorithms/0503. Next Greater Element II/503. Next Greater Element II_test.go
rename to leetcode/0503.Next-Greater-Element-II/503. Next Greater Element II_test.go
index 00844b119..f4cc3cfcc 100644
--- a/Algorithms/0503. Next Greater Element II/503. Next Greater Element II_test.go	
+++ b/leetcode/0503.Next-Greater-Element-II/503. Next Greater Element II_test.go	
@@ -26,17 +26,17 @@ func Test_Problem503(t *testing.T) {
 
 	qs := []question503{
 
-		question503{
+		{
 			para503{[]int{}},
 			ans503{[]int{}},
 		},
 
-		question503{
+		{
 			para503{[]int{1}},
 			ans503{[]int{-1}},
 		},
 
-		question503{
+		{
 			para503{[]int{1, 2, 1}},
 			ans503{[]int{2, -1, 2}},
 		},
diff --git a/Algorithms/0503. Next Greater Element II/README.md b/leetcode/0503.Next-Greater-Element-II/README.md
similarity index 100%
rename from Algorithms/0503. Next Greater Element II/README.md
rename to leetcode/0503.Next-Greater-Element-II/README.md
diff --git a/leetcode/0504.Base-7/504.Base 7.go b/leetcode/0504.Base-7/504.Base 7.go
new file mode 100644
index 000000000..5a143b2be
--- /dev/null
+++ b/leetcode/0504.Base-7/504.Base 7.go	
@@ -0,0 +1,28 @@
+package leetcode
+
+import "strconv"
+
+func convertToBase7(num int) string {
+	if num == 0 {
+		return "0"
+	}
+	negative := false
+	if num < 0 {
+		negative = true
+		num = -num
+	}
+	var ans string
+	var nums []int
+	for num != 0 {
+		remainder := num % 7
+		nums = append(nums, remainder)
+		num = num / 7
+	}
+	if negative {
+		ans += "-"
+	}
+	for i := len(nums) - 1; i >= 0; i-- {
+		ans += strconv.Itoa(nums[i])
+	}
+	return ans
+}
diff --git a/leetcode/0504.Base-7/504.Base 7_test.go b/leetcode/0504.Base-7/504.Base 7_test.go
new file mode 100644
index 000000000..1309c1b63
--- /dev/null
+++ b/leetcode/0504.Base-7/504.Base 7_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question504 struct {
+	para504
+	ans504
+}
+
+// para 是参数
+type para504 struct {
+	num int
+}
+
+// ans 是答案
+type ans504 struct {
+	ans string
+}
+
+func Test_Problem504(t *testing.T) {
+
+	qs := []question504{
+
+		{
+			para504{100},
+			ans504{"202"},
+		},
+
+		{
+			para504{-7},
+			ans504{"-10"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 504------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans504, q.para504
+		fmt.Printf("【input】:%v      ", p.num)
+		fmt.Printf("【output】:%v      \n", convertToBase7(p.num))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0504.Base-7/README.md b/leetcode/0504.Base-7/README.md
new file mode 100644
index 000000000..359b96a0c
--- /dev/null
+++ b/leetcode/0504.Base-7/README.md
@@ -0,0 +1,60 @@
+# [504. Base 7](https://leetcode.com/problems/base-7/)
+
+## 题目
+
+Given an integer num, return a string of its base 7 representation.
+
+**Example 1:**
+
+    Input: num = 100
+    Output: "202"
+
+**Example 2:**
+
+    Input: num = -7
+    Output: "-10"
+
+**Constraints:**
+
+- -10000000 <= num <= 10000000
+
+## 题目大意
+
+给定一个整数 num，将其转化为 7 进制，并以字符串形式输出。
+
+## 解题思路
+
+    num反复除以7，然后倒排余数
+
+# 代码
+
+```go
+package leetcode
+
+import "strconv"
+
+func convertToBase7(num int) string {
+	if num == 0 {
+		return "0"
+	}
+	negative := false
+	if num < 0 {
+		negative = true
+		num = -num
+	}
+	var ans string
+	var nums []int
+	for num != 0 {
+		remainder := num % 7
+		nums = append(nums, remainder)
+		num = num / 7
+	}
+	if negative {
+		ans += "-"
+	}
+	for i := len(nums) - 1; i >= 0; i-- {
+		ans += strconv.Itoa(nums[i])
+	}
+	return ans
+}
+```
\ No newline at end of file
diff --git a/leetcode/0506.Relative-Ranks/506.Relative Ranks.go b/leetcode/0506.Relative-Ranks/506.Relative Ranks.go
new file mode 100644
index 000000000..501b933b2
--- /dev/null
+++ b/leetcode/0506.Relative-Ranks/506.Relative Ranks.go	
@@ -0,0 +1,29 @@
+package leetcode
+
+import (
+	"sort"
+	"strconv"
+)
+
+func findRelativeRanks(score []int) []string {
+	mp := make(map[int]int)
+	for i, v := range score {
+		mp[v] = i
+	}
+	sort.Slice(score, func(i, j int) bool {
+		return score[i] > score[j]
+	})
+	ans := make([]string, len(score))
+	for i, v := range score {
+		if i == 0 {
+			ans[mp[v]] = "Gold Medal"
+		} else if i == 1 {
+			ans[mp[v]] = "Silver Medal"
+		} else if i == 2 {
+			ans[mp[v]] = "Bronze Medal"
+		} else {
+			ans[mp[v]] = strconv.Itoa(i + 1)
+		}
+	}
+	return ans
+}
diff --git a/leetcode/0506.Relative-Ranks/506.Relative Ranks_test.go b/leetcode/0506.Relative-Ranks/506.Relative Ranks_test.go
new file mode 100644
index 000000000..b26017d68
--- /dev/null
+++ b/leetcode/0506.Relative-Ranks/506.Relative Ranks_test.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question506 struct {
+	para506
+	ans506
+}
+
+// para 是参数
+type para506 struct {
+	score []int
+}
+
+// ans 是答案
+type ans506 struct {
+	ans []string
+}
+
+func Test_Problem506(t *testing.T) {
+
+	qs := []question506{
+
+		{
+			para506{[]int{5, 4, 3, 2, 1}},
+			ans506{[]string{"Gold Medal", "Silver Medal", "Bronze Medal", "4", "5"}},
+		},
+
+		{
+			para506{[]int{10, 3, 8, 9, 4}},
+			ans506{[]string{"Gold Medal", "5", "Bronze Medal", "Silver Medal", "4"}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 506------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans506, q.para506
+		fmt.Printf("【input】:%v    【output】:%v\n", p.score, findRelativeRanks(p.score))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0506.Relative-Ranks/README.md b/leetcode/0506.Relative-Ranks/README.md
new file mode 100644
index 000000000..e1eb64330
--- /dev/null
+++ b/leetcode/0506.Relative-Ranks/README.md
@@ -0,0 +1,84 @@
+# [506. Relative Ranks](https://leetcode.com/problems/relative-ranks/)
+
+## 题目
+
+You are given an integer array score of size n, where score[i] is the score of the ith athlete in a competition. All the scores are guaranteed to be unique.
+
+The athletes are placed based on their scores, where the 1st place athlete has the highest score, the 2nd place athlete has the 2nd highest score, and so on. The placement of each athlete determines their rank:
+
+- The 1st place athlete's rank is "Gold Medal".
+- The 2nd place athlete's rank is "Silver Medal".
+- The 3rd place athlete's rank is "Bronze Medal".
+- For the 4th place to the nth place athlete, their rank is their placement number (i.e., the xth place athlete's rank is "x").
+
+Return an array answer of size n where answer[i] is the rank of the ith athlete.
+
+**Example 1**:
+
+    Input: score = [5,4,3,2,1]
+    Output: ["Gold Medal","Silver Medal","Bronze Medal","4","5"]
+    Explanation: The placements are [1st, 2nd, 3rd, 4th, 5th].
+
+**Example 2**:
+
+    Input: score = [10,3,8,9,4]
+    Output: ["Gold Medal","5","Bronze Medal","Silver Medal","4"]
+    Explanation: The placements are [1st, 5th, 3rd, 2nd, 4th].
+
+**Constraints:**
+
+- n == score.length
+- 1 <= n <= 10000
+- 0 <= score[i] <= 1000000
+- All the values in score are unique.
+
+## 题目大意
+
+给你一个长度为 n 的整数数组 score ，其中 score[i] 是第 i 位运动员在比赛中的得分。所有得分都 互不相同 。
+
+运动员将根据得分 决定名次 ，其中名次第 1 的运动员得分最高，名次第 2 的运动员得分第 2 高，依此类推。运动员的名次决定了他们的获奖情况：
+
+- 名次第 1 的运动员获金牌 "Gold Medal" 。
+- 名次第 2 的运动员获银牌 "Silver Medal" 。
+- 名次第 3 的运动员获铜牌 "Bronze Medal" 。
+- 从名次第 4 到第 n 的运动员，只能获得他们的名次编号（即，名次第 x 的运动员获得编号 "x"）。
+
+使用长度为 n 的数组 answer 返回获奖，其中 answer[i] 是第 i 位运动员的获奖情况。
+
+## 解题思路
+
+- 用 map 记录原来 score 中元素对应的坐标，然后对 score 进行排序，对排序后的元素我们通过 map 就可以知道它排的名次了
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+	"strconv"
+)
+
+func findRelativeRanks(score []int) []string {
+	mp := make(map[int]int)
+	for i, v := range score {
+		mp[v] = i
+	}
+	sort.Slice(score, func(i, j int) bool {
+		return score[i] > score[j]
+	})
+	ans := make([]string, len(score))
+	for i, v := range score {
+		if i == 0 {
+			ans[mp[v]] = "Gold Medal"
+		} else if i == 1 {
+			ans[mp[v]] = "Silver Medal"
+		} else if i == 2 {
+			ans[mp[v]] = "Bronze Medal"
+		} else {
+			ans[mp[v]] = strconv.Itoa(i + 1)
+		}
+	}
+	return ans
+}
+```
\ No newline at end of file
diff --git a/leetcode/0507.Perfect-Number/507. Perfect Number.go b/leetcode/0507.Perfect-Number/507. Perfect Number.go
new file mode 100644
index 000000000..ffeb6face
--- /dev/null
+++ b/leetcode/0507.Perfect-Number/507. Perfect Number.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+import "math"
+
+// 方法一
+func checkPerfectNumber(num int) bool {
+	if num <= 1 {
+		return false
+	}
+	sum, bound := 1, int(math.Sqrt(float64(num)))+1
+	for i := 2; i < bound; i++ {
+		if num%i != 0 {
+			continue
+		}
+		corrDiv := num / i
+		sum += corrDiv + i
+	}
+	return sum == num
+}
+
+// 方法二 打表
+func checkPerfectNumber_(num int) bool {
+	return num == 6 || num == 28 || num == 496 || num == 8128 || num == 33550336
+}
diff --git a/leetcode/0507.Perfect-Number/507. Perfect Number_test.go b/leetcode/0507.Perfect-Number/507. Perfect Number_test.go
new file mode 100644
index 000000000..eb9e13064
--- /dev/null
+++ b/leetcode/0507.Perfect-Number/507. Perfect Number_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question507 struct {
+	para507
+	ans507
+}
+
+// para 是参数
+// one 代表第一个参数
+type para507 struct {
+	num int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans507 struct {
+	one bool
+}
+
+func Test_Problem507(t *testing.T) {
+
+	qs := []question507{
+
+		{
+			para507{28},
+			ans507{true},
+		},
+
+		{
+			para507{496},
+			ans507{true},
+		},
+
+		{
+			para507{500},
+			ans507{false},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 507------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans507, q.para507
+		fmt.Printf("【input】:%v       【output】:%v\n", p, checkPerfectNumber(p.num))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0507.Perfect-Number/README.md b/leetcode/0507.Perfect-Number/README.md
new file mode 100644
index 000000000..12d0c7649
--- /dev/null
+++ b/leetcode/0507.Perfect-Number/README.md
@@ -0,0 +1,64 @@
+# [507. Perfect Number](https://leetcode.com/problems/perfect-number/)
+
+
+
+## 题目
+
+We define the Perfect Number is a **positive** integer that is equal to the sum of all its **positive** divisors except itself.
+
+Now, given an
+
+**integer**
+
+n, write a function that returns true when it is a perfect number and false when it is not.
+
+**Example**:
+
+```
+Input: 28
+Output: True
+Explanation: 28 = 1 + 2 + 4 + 7 + 14
+```
+
+**Note**: The input number **n** will not exceed 100,000,000. (1e8)
+
+## 题目大意
+
+对于一个 正整数，如果它和除了它自身以外的所有正因子之和相等，我们称它为“完美数”。给定一个 整数 n， 如果他是完美数，返回 True，否则返回 False
+
+## 解题思路
+
+- 给定一个整数，要求判断这个数是不是完美数。整数的取值范围小于 1e8 。
+- 简单题。按照题意描述，先获取这个整数的所有正因子，如果正因子的和等于原来这个数，那么它就是完美数。
+- 这一题也可以打表，1e8 以下的完美数其实并不多，就 5 个。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+// 方法一
+func checkPerfectNumber(num int) bool {
+	if num <= 1 {
+		return false
+	}
+	sum, bound := 1, int(math.Sqrt(float64(num)))+1
+	for i := 2; i < bound; i++ {
+		if num%i != 0 {
+			continue
+		}
+		corrDiv := num / i
+		sum += corrDiv + i
+	}
+	return sum == num
+}
+
+// 方法二 打表
+func checkPerfectNumber_(num int) bool {
+	return num == 6 || num == 28 || num == 496 || num == 8128 || num == 33550336
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0508. Most Frequent Subtree Sum/508. Most Frequent Subtree Sum.go b/leetcode/0508.Most-Frequent-Subtree-Sum/508. Most Frequent Subtree Sum.go
similarity index 93%
rename from Algorithms/0508. Most Frequent Subtree Sum/508. Most Frequent Subtree Sum.go
rename to leetcode/0508.Most-Frequent-Subtree-Sum/508. Most Frequent Subtree Sum.go
index 93767e024..98300692e 100644
--- a/Algorithms/0508. Most Frequent Subtree Sum/508. Most Frequent Subtree Sum.go	
+++ b/leetcode/0508.Most-Frequent-Subtree-Sum/508. Most Frequent Subtree Sum.go	
@@ -2,8 +2,13 @@ package leetcode
 
 import (
 	"sort"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
diff --git a/Algorithms/0508. Most Frequent Subtree Sum/508. Most Frequent Subtree Sum_test.go b/leetcode/0508.Most-Frequent-Subtree-Sum/508. Most Frequent Subtree Sum_test.go
similarity index 88%
rename from Algorithms/0508. Most Frequent Subtree Sum/508. Most Frequent Subtree Sum_test.go
rename to leetcode/0508.Most-Frequent-Subtree-Sum/508. Most Frequent Subtree Sum_test.go
index a37302fb9..fd58a61ee 100644
--- a/Algorithms/0508. Most Frequent Subtree Sum/508. Most Frequent Subtree Sum_test.go	
+++ b/leetcode/0508.Most-Frequent-Subtree-Sum/508. Most Frequent Subtree Sum_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question508 struct {
@@ -26,27 +28,27 @@ func Test_Problem508(t *testing.T) {
 
 	qs := []question508{
 
-		question508{
+		{
 			para508{[]int{}},
 			ans508{[]int{}},
 		},
 
-		question508{
+		{
 			para508{[]int{1, 1}},
 			ans508{[]int{1, 2}},
 		},
 
-		question508{
+		{
 			para508{[]int{1}},
 			ans508{[]int{1}},
 		},
 
-		question508{
+		{
 			para508{[]int{5, 2, -3}},
 			ans508{[]int{2, -3, 4}},
 		},
 
-		question508{
+		{
 			para508{[]int{5, 2, -5}},
 			ans508{[]int{2}},
 		},
@@ -57,7 +59,7 @@ func Test_Problem508(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans508, q.para508
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", findFrequentTreeSum(root))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/leetcode/0508.Most-Frequent-Subtree-Sum/README.md b/leetcode/0508.Most-Frequent-Subtree-Sum/README.md
new file mode 100755
index 000000000..c45494e85
--- /dev/null
+++ b/leetcode/0508.Most-Frequent-Subtree-Sum/README.md
@@ -0,0 +1,35 @@
+# [508. Most Frequent Subtree Sum](https://leetcode.com/problems/most-frequent-subtree-sum/)
+
+
+## 题目
+
+Given the root of a tree, you are asked to find the most frequent subtree sum. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.
+
+**Examples 1**Input:
+
+      5
+     /  \
+    2   -3
+
+return [2, -3, 4], since all the values happen only once, return all of them in any order.
+
+**Examples 2**Input:
+
+      5
+     /  \
+    2   -5
+
+return [2], since 2 happens twice, however -5 only occur once.
+
+**Note:** You may assume the sum of values in any subtree is in the range of 32-bit signed integer.
+
+
+## 题目大意
+
+给出二叉树的根，找出出现次数最多的子树元素和。一个结点的子树元素和定义为以该结点为根的二叉树上所有结点的元素之和（包括结点本身）。然后求出出现次数最多的子树元素和。如果有多个元素出现的次数相同，返回所有出现次数最多的元素（不限顺序）。提示： 假设任意子树元素和均可以用 32 位有符号整数表示。
+
+## 解题思路
+
+
+- 给出一个树，要求求出每个节点以自己为根节点的子树的所有节点值的和，最后按照这些和出现的频次，输出频次最多的和，如果频次出现次数最多的对应多个和，则全部输出。
+- 递归找出每个节点的累加和，用 map 记录频次，最后把频次最多的输出即可。
diff --git a/Algorithms/0509. Fibonacci Number/509. Fibonacci Number.go b/leetcode/0509.Fibonacci-Number/509. Fibonacci Number.go
similarity index 87%
rename from Algorithms/0509. Fibonacci Number/509. Fibonacci Number.go
rename to leetcode/0509.Fibonacci-Number/509. Fibonacci Number.go
index fcb0d3a70..ea6d95a26 100644
--- a/Algorithms/0509. Fibonacci Number/509. Fibonacci Number.go	
+++ b/leetcode/0509.Fibonacci-Number/509. Fibonacci Number.go	
@@ -108,3 +108,22 @@ func fib5(N int) int {
 	var goldenRatio float64 = float64((1 + math.Sqrt(5)) / 2)
 	return int(math.Round(math.Pow(goldenRatio, float64(N)) / math.Sqrt(5)))
 }
+
+// 解法七 协程版，但是时间特别慢，不推荐，放在这里只是告诉大家，写 LeetCode 算法题的时候，启动 goroutine 特别慢
+func fib6(N int) int {
+	return <-fibb(N)
+}
+
+func fibb(n int) <-chan int {
+	result := make(chan int)
+	go func() {
+		defer close(result)
+
+		if n <= 1 {
+			result <- n
+			return
+		}
+		result <- <-fibb(n-1) + <-fibb(n-2)
+	}()
+	return result
+}
diff --git a/Algorithms/0509. Fibonacci Number/509. Fibonacci Number_test.go b/leetcode/0509.Fibonacci-Number/509. Fibonacci Number_test.go
similarity index 94%
rename from Algorithms/0509. Fibonacci Number/509. Fibonacci Number_test.go
rename to leetcode/0509.Fibonacci-Number/509. Fibonacci Number_test.go
index cc7bf04ca..fbbc820c4 100644
--- a/Algorithms/0509. Fibonacci Number/509. Fibonacci Number_test.go	
+++ b/leetcode/0509.Fibonacci-Number/509. Fibonacci Number_test.go	
@@ -26,17 +26,17 @@ func Test_Problem509(t *testing.T) {
 
 	qs := []question509{
 
-		question509{
+		{
 			para509{1},
 			ans509{1},
 		},
 
-		question509{
+		{
 			para509{2},
 			ans509{1},
 		},
 
-		question509{
+		{
 			para509{3},
 			ans509{2},
 		},
diff --git a/leetcode/0509.Fibonacci-Number/README.md b/leetcode/0509.Fibonacci-Number/README.md
new file mode 100755
index 000000000..d89eafad0
--- /dev/null
+++ b/leetcode/0509.Fibonacci-Number/README.md
@@ -0,0 +1,53 @@
+# [509. Fibonacci Number](https://leetcode.com/problems/fibonacci-number/)
+
+
+## 题目
+
+The **Fibonacci numbers**, commonly denoted `F(n)` form a sequence, called the **Fibonacci sequence**, such that each number is the sum of the two preceding ones, starting from `0` and `1`. That is,
+
+    F(0) = 0,   F(1) = 1
+    F(N) = F(N - 1) + F(N - 2), for N > 1.
+
+Given `N`, calculate `F(N)`.
+
+**Example 1:**
+
+    Input: 2
+    Output: 1
+    Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
+
+**Example 2:**
+
+    Input: 3
+    Output: 2
+    Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
+
+**Example 3:**
+
+    Input: 4
+    Output: 3
+    Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
+
+**Note:**
+
+0 ≤ `N` ≤ 30.
+
+
+## 题目大意
+
+斐波那契数，通常用 F(n) 表示，形成的序列称为斐波那契数列。该数列由 0 和 1 开始，后面的每一项数字都是前面两项数字的和。也就是：
+
+```
+F(0) = 0,   F(1) = 1
+F(N) = F(N - 1) + F(N - 2), 其中 N > 1.
+```
+
+给定 N，计算 F(N)。
+
+提示：0 ≤ N ≤ 30
+
+## 解题思路
+
+
+- 求斐波那契数列
+- 这一题解法很多，大的分类是四种，递归，记忆化搜索(dp)，矩阵快速幂，通项公式。其中记忆化搜索可以写 3 种方法，自底向上的，自顶向下的，优化空间复杂度版的。通项公式方法实质是求 a^b 这个还可以用快速幂优化时间复杂度到 O(log n) 。
diff --git a/leetcode/0513.Find-Bottom-Left-Tree-Value/513. Find Bottom Left Tree Value.go b/leetcode/0513.Find-Bottom-Left-Tree-Value/513. Find Bottom Left Tree Value.go
new file mode 100644
index 000000000..b4ca1ebc2
--- /dev/null
+++ b/leetcode/0513.Find-Bottom-Left-Tree-Value/513. Find Bottom Left Tree Value.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 DFS
+func findBottomLeftValue(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	res, maxHeight := 0, -1
+	findBottomLeftValueDFS(root, 0, &res, &maxHeight)
+	return res
+}
+
+func findBottomLeftValueDFS(root *TreeNode, curHeight int, res, maxHeight *int) {
+	if curHeight > *maxHeight && root.Left == nil && root.Right == nil {
+		*maxHeight = curHeight
+		*res = root.Val
+	}
+	if root.Left != nil {
+		findBottomLeftValueDFS(root.Left, curHeight+1, res, maxHeight)
+	}
+	if root.Right != nil {
+		findBottomLeftValueDFS(root.Right, curHeight+1, res, maxHeight)
+	}
+}
+
+// 解法二 BFS
+func findBottomLeftValue1(root *TreeNode) int {
+	queue := []*TreeNode{root}
+	for len(queue) > 0 {
+		next := []*TreeNode{}
+		for _, node := range queue {
+			if node.Left != nil {
+				next = append(next, node.Left)
+			}
+			if node.Right != nil {
+				next = append(next, node.Right)
+			}
+		}
+		if len(next) == 0 {
+			return queue[0].Val
+		}
+		queue = next
+	}
+	return 0
+}
diff --git a/leetcode/0513.Find-Bottom-Left-Tree-Value/513. Find Bottom Left Tree Value_test.go b/leetcode/0513.Find-Bottom-Left-Tree-Value/513. Find Bottom Left Tree Value_test.go
new file mode 100644
index 000000000..73f66fe37
--- /dev/null
+++ b/leetcode/0513.Find-Bottom-Left-Tree-Value/513. Find Bottom Left Tree Value_test.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question513 struct {
+	para513
+	ans513
+}
+
+// para 是参数
+// one 代表第一个参数
+type para513 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans513 struct {
+	one int
+}
+
+func Test_Problem513(t *testing.T) {
+
+	qs := []question513{
+
+		{
+			para513{[]int{}},
+			ans513{0},
+		},
+
+		{
+			para513{[]int{1}},
+			ans513{1},
+		},
+
+		{
+			para513{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
+			ans513{15},
+		},
+
+		{
+			para513{[]int{1, 2, 3, 4, structures.NULL, structures.NULL, 5}},
+			ans513{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 513------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans513, q.para513
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", findBottomLeftValue(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0513.Find-Bottom-Left-Tree-Value/README.md b/leetcode/0513.Find-Bottom-Left-Tree-Value/README.md
new file mode 100755
index 000000000..d324af9f0
--- /dev/null
+++ b/leetcode/0513.Find-Bottom-Left-Tree-Value/README.md
@@ -0,0 +1,50 @@
+# [513. Find Bottom Left Tree Value](https://leetcode.com/problems/find-bottom-left-tree-value/)
+
+
+## 题目
+
+Given a binary tree, find the leftmost value in the last row of the tree.
+
+**Example 1:**
+
+    Input:
+    
+        2
+       / \
+      1   3
+    
+    Output:
+    1
+
+**Example 2:**
+
+    Input:
+    
+            1
+           / \
+          2   3
+         /   / \
+        4   5   6
+           /
+          7
+    
+    Output:
+    7
+
+**Note:** You may assume the tree (i.e., the given root node) is not **NULL**.
+
+
+## 题目大意
+
+给定一个二叉树，在树的最后一行找到最左边的值。注意: 您可以假设树（即给定的根节点）不为 NULL。
+
+
+
+
+
+
+## 解题思路
+
+
+- 给出一棵树，输出这棵树最下一层中最左边的节点的值。
+- 这一题用 DFS 和 BFS 均可解题。
diff --git a/leetcode/0515.Find-Largest-Value-in-Each-Tree-Row/515. Find Largest Value in Each Tree Row.go b/leetcode/0515.Find-Largest-Value-in-Each-Tree-Row/515. Find Largest Value in Each Tree Row.go
new file mode 100644
index 000000000..54d831dc0
--- /dev/null
+++ b/leetcode/0515.Find-Largest-Value-in-Each-Tree-Row/515. Find Largest Value in Each Tree Row.go	
@@ -0,0 +1,114 @@
+package leetcode
+
+import (
+	"math"
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 层序遍历二叉树，再将每层排序取出最大值
+func largestValues(root *TreeNode) []int {
+	tmp := levelOrder(root)
+	res := []int{}
+	for i := 0; i < len(tmp); i++ {
+		sort.Ints(tmp[i])
+		res = append(res, tmp[i][len(tmp[i])-1])
+	}
+	return res
+}
+
+// this is 102 solution
+func levelOrder(root *TreeNode) [][]int {
+	if root == nil {
+		return [][]int{}
+	}
+	queue := []*TreeNode{}
+	queue = append(queue, root)
+	curNum, nextLevelNum, res, tmp := 1, 0, [][]int{}, []int{}
+	for len(queue) != 0 {
+		if curNum > 0 {
+			node := queue[0]
+			if node.Left != nil {
+				queue = append(queue, node.Left)
+				nextLevelNum++
+			}
+			if node.Right != nil {
+				queue = append(queue, node.Right)
+				nextLevelNum++
+			}
+			curNum--
+			tmp = append(tmp, node.Val)
+			queue = queue[1:]
+		}
+		if curNum == 0 {
+			res = append(res, tmp)
+			curNum = nextLevelNum
+			nextLevelNum = 0
+			tmp = []int{}
+		}
+	}
+	return res
+}
+
+// 解法二 层序遍历二叉树，遍历过程中不断更新最大值
+func largestValues1(root *TreeNode) []int {
+	if root == nil {
+		return []int{}
+	}
+	q := []*TreeNode{root}
+	var res []int
+	for len(q) > 0 {
+		qlen := len(q)
+		max := math.MinInt32
+		for i := 0; i < qlen; i++ {
+			node := q[0]
+			q = q[1:]
+			if node.Val > max {
+				max = node.Val
+			}
+			if node.Left != nil {
+				q = append(q, node.Left)
+			}
+			if node.Right != nil {
+				q = append(q, node.Right)
+			}
+		}
+		res = append(res, max)
+	}
+	return res
+}
+
+// 解法三 深度遍历二叉树
+func largestValues3(root *TreeNode) []int {
+	var res []int
+	var dfs func(root *TreeNode, level int)
+	dfs = func(root *TreeNode, level int) {
+		if root == nil {
+			return
+		}
+		if len(res) == level {
+			res = append(res, root.Val)
+		}
+		if res[level] < root.Val {
+			res[level] = root.Val
+		}
+
+		dfs(root.Right, level+1)
+		dfs(root.Left, level+1)
+	}
+	dfs(root, 0)
+	return res
+}
diff --git a/Algorithms/0515. Find Largest Value in Each Tree Row/515. Find Largest Value in Each Tree Row_test.go b/leetcode/0515.Find-Largest-Value-in-Each-Tree-Row/515. Find Largest Value in Each Tree Row_test.go
similarity index 76%
rename from Algorithms/0515. Find Largest Value in Each Tree Row/515. Find Largest Value in Each Tree Row_test.go
rename to leetcode/0515.Find-Largest-Value-in-Each-Tree-Row/515. Find Largest Value in Each Tree Row_test.go
index 5811b3edc..d2847d353 100644
--- a/Algorithms/0515. Find Largest Value in Each Tree Row/515. Find Largest Value in Each Tree Row_test.go	
+++ b/leetcode/0515.Find-Largest-Value-in-Each-Tree-Row/515. Find Largest Value in Each Tree Row_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question515 struct {
@@ -26,23 +28,23 @@ func Test_Problem515(t *testing.T) {
 
 	qs := []question515{
 
-		question515{
+		{
 			para515{[]int{}},
 			ans515{[]int{}},
 		},
 
-		question515{
+		{
 			para515{[]int{1}},
 			ans515{[]int{1}},
 		},
 
-		question515{
-			para515{[]int{1, 3, 2, 5, 3, NULL, 9}},
+		{
+			para515{[]int{1, 3, 2, 5, 3, structures.NULL, 9}},
 			ans515{[]int{1, 3, 9}},
 		},
 
-		question515{
-			para515{[]int{3, 9, 20, NULL, NULL, 15, 7}},
+		{
+			para515{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
 			ans515{[]int{3, 20, 15}},
 		},
 	}
@@ -52,7 +54,7 @@ func Test_Problem515(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans515, q.para515
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", largestValues(root))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0515. Find Largest Value in Each Tree Row/README.md b/leetcode/0515.Find-Largest-Value-in-Each-Tree-Row/README.md
similarity index 100%
rename from Algorithms/0515. Find Largest Value in Each Tree Row/README.md
rename to leetcode/0515.Find-Largest-Value-in-Each-Tree-Row/README.md
diff --git a/leetcode/0518.Coin-Change-II/518. Coin Change II.go b/leetcode/0518.Coin-Change-II/518. Coin Change II.go
new file mode 100644
index 000000000..9de3726da
--- /dev/null
+++ b/leetcode/0518.Coin-Change-II/518. Coin Change II.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+func change(amount int, coins []int) int {
+	dp := make([]int, amount+1)
+	dp[0] = 1
+	for _, coin := range coins {
+		for i := coin; i <= amount; i++ {
+			dp[i] += dp[i-coin]
+		}
+	}
+	return dp[amount]
+}
diff --git a/leetcode/0518.Coin-Change-II/518. Coin Change II_test.go b/leetcode/0518.Coin-Change-II/518. Coin Change II_test.go
new file mode 100644
index 000000000..e98639433
--- /dev/null
+++ b/leetcode/0518.Coin-Change-II/518. Coin Change II_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question518 struct {
+	para518
+	ans518
+}
+
+// para 是参数
+// one 代表第一个参数
+type para518 struct {
+	amount int
+	coins  []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans518 struct {
+	one int
+}
+
+func Test_Problem518(t *testing.T) {
+
+	qs := []question518{
+
+		{
+			para518{5, []int{1, 2, 5}},
+			ans518{4},
+		},
+
+		{
+			para518{3, []int{2}},
+			ans518{0},
+		},
+
+		{
+			para518{10, []int{10}},
+			ans518{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 518------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans518, q.para518
+		fmt.Printf("【input】:%v       【output】:%v\n", p, change(p.amount, p.coins))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0518.Coin-Change-II/README.md b/leetcode/0518.Coin-Change-II/README.md
new file mode 100644
index 000000000..4679d2d3c
--- /dev/null
+++ b/leetcode/0518.Coin-Change-II/README.md
@@ -0,0 +1,73 @@
+# [518. Coin Change II](https://leetcode.com/problems/coin-change-ii/)
+
+
+## 题目
+
+You are given an integer array `coins` representing coins of different denominations and an integer `amount` representing a total amount of money.
+
+Return *the number of combinations that make up that amount*. If that amount of money cannot be made up by any combination of the coins, return `0`.
+
+You may assume that you have an infinite number of each kind of coin.
+
+The answer is **guaranteed** to fit into a signed **32-bit** integer.
+
+**Example 1:**
+
+```
+Input: amount = 5, coins = [1,2,5]
+Output: 4
+Explanation: there are four ways to make up the amount:
+5=5
+5=2+2+1
+5=2+1+1+1
+5=1+1+1+1+1
+```
+
+**Example 2:**
+
+```
+Input: amount = 3, coins = [2]
+Output: 0
+Explanation: the amount of 3 cannot be made up just with coins of 2.
+```
+
+**Example 3:**
+
+```
+Input: amount = 10, coins = [10]
+Output: 1
+```
+
+**Constraints:**
+
+- `1 <= coins.length <= 300`
+- `1 <= coins[i] <= 5000`
+- All the values of `coins` are **unique**.
+- `0 <= amount <= 5000`
+
+## 题目大意
+
+给你一个整数数组 coins 表示不同面额的硬币，另给一个整数 amount 表示总金额。请你计算并返回可以凑成总金额的硬币组合数。如果任何硬币组合都无法凑出总金额，返回 0 。假设每一种面额的硬币有无限个。题目数据保证结果符合 32 位带符号整数。
+
+## 解题思路
+
+- 此题虽然名字叫 Coin Change，但是不是经典的背包九讲问题。题目中 coins 的每个元素可以选取多次，且不考虑选取元素的顺序，因此这道题实际需要计算的是选取硬币的组合数。定义 dp[i] 表示金额之和等于 i 的硬币组合数，目标求 dp[amount]。初始边界条件为 dp[0] = 1，即不取任何硬币，就这一种取法，金额为 0 。状态转移方程 dp[i] += dp[i-coin]，coin 为当前枚举的 coin。
+- 可能有读者会有疑惑，上述做法会不会出现重复计算。答案是不会。外层循环是遍历数组 coins 的值，内层循环是遍历不同的金额之和，在计算 dp[i] 的值时，可以确保金额之和等于 i 的硬币面额的顺序，由于顺序确定，因此不会重复计算不同的排列。
+- 和此题完全一致的解题思路的题有，第 377 题和第 494 题。
+
+## 代码
+
+```go
+package leetcode
+
+func change(amount int, coins []int) int {
+	dp := make([]int, amount+1)
+	dp[0] = 1
+	for _, coin := range coins {
+		for i := coin; i <= amount; i++ {
+			dp[i] += dp[i-coin]
+		}
+	}
+	return dp[amount]
+}
+```
\ No newline at end of file
diff --git a/leetcode/0519.Random-Flip-Matrix/519.Random Flip Matrix.go b/leetcode/0519.Random-Flip-Matrix/519.Random Flip Matrix.go
new file mode 100644
index 000000000..4b0df378a
--- /dev/null
+++ b/leetcode/0519.Random-Flip-Matrix/519.Random Flip Matrix.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"math/rand"
+)
+
+type Solution struct {
+	r     int
+	c     int
+	total int
+	mp    map[int]int
+}
+
+func Constructor(m int, n int) Solution {
+	return Solution{
+		r:     m,
+		c:     n,
+		total: m * n,
+		mp:    map[int]int{},
+	}
+}
+
+func (this *Solution) Flip() []int {
+	k := rand.Intn(this.total)
+	val := k
+	if v, ok := this.mp[k]; ok {
+		val = v
+	}
+	if _, ok := this.mp[this.total-1]; ok {
+		this.mp[k] = this.mp[this.total-1]
+	} else {
+		this.mp[k] = this.total - 1
+	}
+	delete(this.mp, this.total-1)
+	this.total--
+	newR, newC := val/this.c, val%this.c
+	return []int{newR, newC}
+}
+
+func (this *Solution) Reset() {
+	this.total = this.r * this.c
+	this.mp = map[int]int{}
+}
diff --git a/leetcode/0519.Random-Flip-Matrix/519.Random Flip Matrix_test.go b/leetcode/0519.Random-Flip-Matrix/519.Random Flip Matrix_test.go
new file mode 100644
index 000000000..847ac79a2
--- /dev/null
+++ b/leetcode/0519.Random-Flip-Matrix/519.Random Flip Matrix_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question519 struct {
+	para519
+	ans519
+}
+
+// para 是参数
+type para519 struct {
+	para []string
+	val  [][]int
+}
+
+// ans 是答案
+type ans519 struct {
+	ans [][]int
+}
+
+func Test_Problem519(t *testing.T) {
+
+	qs := []question519{
+
+		{
+			para519{[]string{"Solution", "flip", "flip", "flip", "reset", "flip"}, [][]int{{3, 1}, {}, {}, {}, {}, {}}},
+			ans519{[][]int{nil, {1, 0}, {2, 0}, {0, 0}, nil, {2, 0}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 519------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans519, q.para519
+		sol := Constructor(0, 0)
+		for _, v := range p.para {
+			if v == "Solution" {
+				sol = Constructor(q.val[0][0], q.val[0][1])
+				fmt.Printf("【input】:%v    【output】:%v\n", v, nil)
+			} else if v == "flip" {
+				fmt.Printf("【input】:%v    【output】:%v\n", v, sol.Flip())
+			} else {
+				sol.Reset()
+				fmt.Printf("【input】:%v    【output】:%v\n", v, nil)
+			}
+		}
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0519.Random-Flip-Matrix/README.md b/leetcode/0519.Random-Flip-Matrix/README.md
new file mode 100644
index 000000000..2f9ae71c1
--- /dev/null
+++ b/leetcode/0519.Random-Flip-Matrix/README.md
@@ -0,0 +1,98 @@
+# [519. Random Flip Matrix](https://leetcode.com/problems/random-flip-matrix/)
+
+## 题目
+
+There is an m x n binary grid matrix with all the values set 0 initially. Design an algorithm to randomly pick an index (i, j) where matrix[i][j] == 0 and flips it to 1. All the indices (i, j) where matrix[i][j] == 0 should be equally likely to be returned.
+
+Optimize your algorithm to minimize the number of calls made to the built-in random function of your language and optimize the time and space complexity.
+
+Implement the Solution class:
+
+- Solution(int m, int n) Initializes the object with the size of the binary matrix m and n.
+- int[] flip() Returns a random index [i, j] of the matrix where matrix[i][j] == 0 and flips it to 1.
+- void reset() Resets all the values of the matrix to be 0.
+
+**Example 1**:
+
+    Input
+    ["Solution", "flip", "flip", "flip", "reset", "flip"]
+    [[3, 1], [], [], [], [], []]
+    Output
+    [null, [1, 0], [2, 0], [0, 0], null, [2, 0]]
+
+    Explanation
+    Solution solution = new Solution(3, 1);
+    solution.flip();  // return [1, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned.
+    solution.flip();  // return [2, 0], Since [1,0] was returned, [2,0] and [0,0]
+    solution.flip();  // return [0, 0], Based on the previously returned indices, only [0,0] can be returned.
+    solution.reset(); // All the values are reset to 0 and can be returned.
+    solution.flip();  // return [2, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned.
+
+**Constraints:**
+
+- 1 <= m, n <= 10000
+- There will be at least one free cell for each call to flip.
+- At most 1000 calls will be made to flip and reset.
+
+## 题目大意
+
+给你一个 m x n 的二元矩阵 matrix ，且所有值被初始化为 0 。请你设计一个算法，随机选取一个满足 matrix[i][j] == 0 的下标 (i, j) ，并将它的值变为 1 。所有满足 matrix[i][j] == 0 的下标 (i, j) 被选取的概率应当均等。
+
+尽量最少调用内置的随机函数，并且优化时间和空间复杂度。
+
+实现 Solution 类：
+
+- Solution(int m, int n) 使用二元矩阵的大小 m 和 n 初始化该对象
+- int[] flip() 返回一个满足 matrix[i][j] == 0 的随机下标 [i, j] ，并将其对应格子中的值变为 1
+- void reset() 将矩阵中所有的值重置为 0
+
+## 解题思路
+
+- 二维矩阵利用哈希表转换为一维,每次随机选择一维中的任意一个元素，然后与最后一个元素交换，一维元素的总个数减一
+- 哈希表中默认的映射为x->x, 然后将不满足这个映射的特殊键值对存入哈希表
+
+## 代码
+
+```go
+package leetcode
+
+import "math/rand"
+
+type Solution struct {
+	r     int
+	c     int
+	total int
+	mp    map[int]int
+}
+
+func Constructor(m int, n int) Solution {
+	return Solution{
+		r:     m,
+		c:     n,
+		total: m * n,
+		mp:    map[int]int{},
+	}
+}
+
+func (this *Solution) Flip() []int {
+	k := rand.Intn(this.total)
+	val := k
+	if v, ok := this.mp[k]; ok {
+		val = v
+	}
+	if _, ok := this.mp[this.total-1]; ok {
+		this.mp[k] = this.mp[this.total-1]
+	} else {
+		this.mp[k] = this.total - 1
+	}
+	delete(this.mp, this.total - 1)
+	this.total--
+	newR, newC := val/this.c, val%this.c
+	return []int{newR, newC}
+}
+
+func (this *Solution) Reset() {
+	this.total = this.r * this.c
+	this.mp = map[int]int{}
+}
+```
\ No newline at end of file
diff --git a/leetcode/0520.Detect-Capital/520.Detect Capital.go b/leetcode/0520.Detect-Capital/520.Detect Capital.go
new file mode 100644
index 000000000..c9710484d
--- /dev/null
+++ b/leetcode/0520.Detect-Capital/520.Detect Capital.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+import "strings"
+
+func detectCapitalUse(word string) bool {
+	wLower := strings.ToLower(word)
+	wUpper := strings.ToUpper(word)
+	wCaptial := strings.ToUpper(string(word[0])) + strings.ToLower(string(word[1:]))
+	if wCaptial == word || wLower == word || wUpper == word {
+		return true
+	}
+	return false
+}
diff --git a/leetcode/0520.Detect-Capital/520.Detect Capital_test.go b/leetcode/0520.Detect-Capital/520.Detect Capital_test.go
new file mode 100644
index 000000000..20635849b
--- /dev/null
+++ b/leetcode/0520.Detect-Capital/520.Detect Capital_test.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question520 struct {
+	para520
+	ans520
+}
+
+// para 是参数
+type para520 struct {
+	word string
+}
+
+// ans 是答案
+type ans520 struct {
+	ans bool
+}
+
+func Test_Problem520(t *testing.T) {
+
+	qs := []question520{
+
+		{
+			para520{"USA"},
+			ans520{true},
+		},
+
+		{
+			para520{"FlaG"},
+			ans520{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 520------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans520, q.para520
+		fmt.Printf("【input】:%v       【output】:%v\n", p, detectCapitalUse(p.word))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0520.Detect-Capital/README.md b/leetcode/0520.Detect-Capital/README.md
new file mode 100644
index 000000000..c4532bfd8
--- /dev/null
+++ b/leetcode/0520.Detect-Capital/README.md
@@ -0,0 +1,68 @@
+# [520. Detect Capital](https://leetcode.com/problems/detect-capital/)
+
+
+## 题目
+
+We define the usage of capitals in a word to be right when one of the following cases holds:
+
+All letters in this word are capitals, like "USA".
+
+All letters in this word are not capitals, like "leetcode".
+
+Only the first letter in this word is capital, like "Google".
+
+Given a string word, return true if the usage of capitals in it is right.
+
+**Example 1:**
+
+```
+Input: word = "USA"
+Output: true
+```
+
+**Example 2:**
+
+```
+Input: word = "FlaG"
+Output: false
+```
+
+**Constraints:**
+
+- 1 <= word.length <= 100
+- word consists of lowercase and uppercase English letters.
+
+## 题目大意
+
+我们定义，在以下情况时，单词的大写用法是正确的：
+
+全部字母都是大写，比如 "USA" 。
+单词中所有字母都不是大写，比如 "leetcode" 。
+如果单词不只含有一个字母，只有首字母大写，比如"Google" 。
+
+给你一个字符串 word 。如果大写用法正确，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 把 word 分别转换为全部小写 wLower，全部大写 wUpper，首字母大写的字符串 wCaptial
+- 判断 word 是否等于 wLower, wUpper, wCaptial 中的一个，如果是返回 true，否则返回 false
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func detectCapitalUse(word string) bool {
+	wLower := strings.ToLower(word)
+	wUpper := strings.ToUpper(word)
+	wCaptial := strings.ToUpper(string(word[0])) + strings.ToLower(string(word[1:]))
+	if wCaptial == word || wLower == word || wUpper == word {
+		return true
+	}
+	return false
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/0523.Continuous-Subarray-Sum/523. Continuous Subarray Sum.go b/leetcode/0523.Continuous-Subarray-Sum/523. Continuous Subarray Sum.go
new file mode 100644
index 000000000..15b23937d
--- /dev/null
+++ b/leetcode/0523.Continuous-Subarray-Sum/523. Continuous Subarray Sum.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func checkSubarraySum(nums []int, k int) bool {
+	m := make(map[int]int)
+	m[0] = -1
+	sum := 0
+	for i, n := range nums {
+		sum += n
+		if r, ok := m[sum%k]; ok {
+			if i-2 >= r {
+				return true
+			}
+		} else {
+			m[sum%k] = i
+		}
+	}
+	return false
+}
diff --git a/leetcode/0523.Continuous-Subarray-Sum/523. Continuous Subarray Sum_test.go b/leetcode/0523.Continuous-Subarray-Sum/523. Continuous Subarray Sum_test.go
new file mode 100644
index 000000000..f3147e166
--- /dev/null
+++ b/leetcode/0523.Continuous-Subarray-Sum/523. Continuous Subarray Sum_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question523 struct {
+	para523
+	ans523
+}
+
+// para 是参数
+// one 代表第一个参数
+type para523 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans523 struct {
+	one bool
+}
+
+func Test_Problem523(t *testing.T) {
+
+	qs := []question523{
+
+		{
+			para523{[]int{23, 2, 4, 6, 7}, 6},
+			ans523{true},
+		},
+
+		{
+			para523{[]int{23, 2, 6, 4, 7}, 6},
+			ans523{true},
+		},
+
+		{
+			para523{[]int{23, 2, 6, 4, 7}, 13},
+			ans523{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 523------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans523, q.para523
+		fmt.Printf("【input】:%v       【output】:%v\n", p, checkSubarraySum(p.nums, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0523.Continuous-Subarray-Sum/README.md b/leetcode/0523.Continuous-Subarray-Sum/README.md
new file mode 100644
index 000000000..ac6d2df61
--- /dev/null
+++ b/leetcode/0523.Continuous-Subarray-Sum/README.md
@@ -0,0 +1,75 @@
+# [523. Continuous Subarray Sum](https://leetcode.com/problems/continuous-subarray-sum/)
+
+
+## 题目
+
+Given an integer array `nums` and an integer `k`, return `true` *if* `nums` *has a continuous subarray of size **at least two** whose elements sum up to a multiple of* `k`*, or* `false` *otherwise*.
+
+An integer `x` is a multiple of `k` if there exists an integer `n` such that `x = n * k`. `0` is **always** a multiple of `k`.
+
+**Example 1:**
+
+```
+Input: nums = [23,2,4,6,7], k = 6
+Output: true
+Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.
+```
+
+**Example 2:**
+
+```
+Input: nums = [23,2,6,4,7], k = 6
+Output: true
+Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
+42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.
+```
+
+**Example 3:**
+
+```
+Input: nums = [23,2,6,4,7], k = 13
+Output: false
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 105`
+- `0 <= nums[i] <= 109`
+- `0 <= sum(nums[i]) <= 231 - 1`
+- `1 <= k <= 231 - 1`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个整数 k ，编写一个函数来判断该数组是否含有同时满足下述条件的连续子数组：
+
+- 子数组大小至少为 2 ，且
+- 子数组元素总和为 k 的倍数。
+
+如果存在，返回 true ；否则，返回 false 。如果存在一个整数 n ，令整数 x 符合 x = n * k ，则称 x 是 k 的一个倍数。
+
+## 解题思路
+
+- 简单题。题目只要求是否存在，不要求找出所有解。用一个变量 sum 记录累加和。子数组的元素和可以用前缀和相减得到，例如 [i,j] 区间内的元素和，可以由 prefixSum[j] - prefixSum[i] 得到。当 prefixSums[j]−prefixSums[i] 为 k 的倍数时，prefixSums[i] 和 prefixSums[j] 除以 k 的余数相同。因此只需要计算每个下标对应的前缀和除以 k 的余数即可，使用 map 存储每个余数第一次出现的下标即可。在 map 中如果存在相同余数的 key，代表当前下标和 map 中这个 key 记录的下标可以满足总和为 k 的倍数这一条件。再判断一下能否满足大小至少为 2 的条件即可。用 2 个下标相减，长度大于等于 2 即满足条件，可以输出 true。
+
+## 代码
+
+```go
+package leetcode
+
+func checkSubarraySum(nums []int, k int) bool {
+	m := make(map[int]int)
+	m[0] = -1
+	sum := 0
+	for i, n := range nums {
+		sum += n
+		if r, ok := m[sum%k]; ok {
+			if i-2 >= r {
+				return true
+			}
+		} else {
+			m[sum%k] = i
+		}
+	}
+	return false
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0524. Longest Word in Dictionary through Deleting/524. Longest Word in Dictionary through Deleting.go b/leetcode/0524.Longest-Word-in-Dictionary-through-Deleting/524. Longest Word in Dictionary through Deleting.go
similarity index 100%
rename from Algorithms/0524. Longest Word in Dictionary through Deleting/524. Longest Word in Dictionary through Deleting.go
rename to leetcode/0524.Longest-Word-in-Dictionary-through-Deleting/524. Longest Word in Dictionary through Deleting.go
diff --git a/Algorithms/0524. Longest Word in Dictionary through Deleting/524. Longest Word in Dictionary through Deleting_test.go b/leetcode/0524.Longest-Word-in-Dictionary-through-Deleting/524. Longest Word in Dictionary through Deleting_test.go
similarity index 93%
rename from Algorithms/0524. Longest Word in Dictionary through Deleting/524. Longest Word in Dictionary through Deleting_test.go
rename to leetcode/0524.Longest-Word-in-Dictionary-through-Deleting/524. Longest Word in Dictionary through Deleting_test.go
index b67f80737..be0142853 100644
--- a/Algorithms/0524. Longest Word in Dictionary through Deleting/524. Longest Word in Dictionary through Deleting_test.go	
+++ b/leetcode/0524.Longest-Word-in-Dictionary-through-Deleting/524. Longest Word in Dictionary through Deleting_test.go	
@@ -27,27 +27,27 @@ func Test_Problem524(t *testing.T) {
 
 	qs := []question524{
 
-		question524{
+		{
 			para524{"abpcplea", []string{"ale", "apple", "monkey", "plea"}},
 			ans524{"apple"},
 		},
 
-		question524{
+		{
 			para524{"abpcplea", []string{"a", "b", "c"}},
 			ans524{"a"},
 		},
 
-		question524{
+		{
 			para524{"abpcplea", []string{"aaaaa", "b", "c"}},
 			ans524{"b"},
 		},
 
-		question524{
+		{
 			para524{"bab", []string{"ba", "ab", "a", "b"}},
 			ans524{"ab"},
 		},
 
-		question524{
+		{
 			para524{"aewfafwafjlwajflwajflwafj", []string{"apple", "ewaf", "awefawfwaf", "awef", "awefe", "ewafeffewafewf"}},
 			ans524{"ewaf"},
 		},
diff --git a/Algorithms/0524. Longest Word in Dictionary through Deleting/README.md b/leetcode/0524.Longest-Word-in-Dictionary-through-Deleting/README.md
similarity index 100%
rename from Algorithms/0524. Longest Word in Dictionary through Deleting/README.md
rename to leetcode/0524.Longest-Word-in-Dictionary-through-Deleting/README.md
diff --git a/leetcode/0525.Contiguous-Array/525. Contiguous Array.go b/leetcode/0525.Contiguous-Array/525. Contiguous Array.go
new file mode 100644
index 000000000..802a3be9a
--- /dev/null
+++ b/leetcode/0525.Contiguous-Array/525. Contiguous Array.go	
@@ -0,0 +1,27 @@
+package leetcode
+
+func findMaxLength(nums []int) int {
+	dict := map[int]int{}
+	dict[0] = -1
+	count, res := 0, 0
+	for i := 0; i < len(nums); i++ {
+		if nums[i] == 0 {
+			count--
+		} else {
+			count++
+		}
+		if idx, ok := dict[count]; ok {
+			res = max(res, i-idx)
+		} else {
+			dict[count] = i
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0525.Contiguous-Array/525. Contiguous Array_test.go b/leetcode/0525.Contiguous-Array/525. Contiguous Array_test.go
new file mode 100644
index 000000000..644885d15
--- /dev/null
+++ b/leetcode/0525.Contiguous-Array/525. Contiguous Array_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question525 struct {
+	para525
+	ans525
+}
+
+// para 是参数
+// one 代表第一个参数
+type para525 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans525 struct {
+	one int
+}
+
+func Test_Problem525(t *testing.T) {
+
+	qs := []question525{
+
+		{
+			para525{[]int{0, 1}},
+			ans525{2},
+		},
+
+		{
+			para525{[]int{0, 1, 0}},
+			ans525{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 525------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans525, q.para525
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findMaxLength(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0525.Contiguous-Array/README.md b/leetcode/0525.Contiguous-Array/README.md
new file mode 100644
index 000000000..7f1dbf157
--- /dev/null
+++ b/leetcode/0525.Contiguous-Array/README.md
@@ -0,0 +1,67 @@
+# [525. Contiguous Array](https://leetcode.com/problems/contiguous-array/)
+
+
+## 题目
+
+Given a binary array `nums`, return *the maximum length of a contiguous subarray with an equal number of* `0` *and* `1`.
+
+**Example 1:**
+
+```
+Input: nums = [0,1]
+Output: 2
+Explanation: [0, 1] is the longest contiguous subarray with an equal number of 0 and 1.
+```
+
+**Example 2:**
+
+```
+Input: nums = [0,1,0]
+Output: 2
+Explanation: [0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 105`
+- `nums[i]` is either `0` or `1`.
+
+## 题目大意
+
+给定一个二进制数组 nums , 找到含有相同数量的 0 和 1 的最长连续子数组，并返回该子数组的长度。
+
+## 解题思路
+
+- 0 和 1 的数量相同可以转化为两者数量相差为 0，如果将 0 看作为 -1，那么原题转化为求最长连续子数组，其元素和为 0 。又变成了区间内求和的问题，自然而然转换为前缀和来处理。假设连续子数组是 [i,j] 区间，这个区间内元素和为 0 意味着 prefixSum[j] - prefixSum[i] = 0，也就是 prefixSum[i] = prefixSum[j]。不断累加前缀和，将每个前缀和存入 map 中。一旦某个 key 存在了，代表之前某个下标的前缀和和当前下标构成的区间，这段区间内的元素和为 0 。这个区间是所求。扫完整个数组，扫描过程中动态更新最大区间长度，扫描完成便可得到最大区间长度，即最长连续子数组。
+
+## 代码
+
+```go
+package leetcode
+
+func findMaxLength(nums []int) int {
+	dict := map[int]int{}
+	dict[0] = -1
+	count, res := 0, 0
+	for i := 0; i < len(nums); i++ {
+		if nums[i] == 0 {
+			count--
+		} else {
+			count++
+		}
+		if idx, ok := dict[count]; ok {
+			res = max(res, i-idx)
+		} else {
+			dict[count] = i
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0526. Beautiful Arrangement/526. Beautiful Arrangement.go b/leetcode/0526.Beautiful-Arrangement/526. Beautiful Arrangement.go
similarity index 100%
rename from Algorithms/0526. Beautiful Arrangement/526. Beautiful Arrangement.go
rename to leetcode/0526.Beautiful-Arrangement/526. Beautiful Arrangement.go
diff --git a/Algorithms/0526. Beautiful Arrangement/526. Beautiful Arrangement_test.go b/leetcode/0526.Beautiful-Arrangement/526. Beautiful Arrangement_test.go
similarity index 82%
rename from Algorithms/0526. Beautiful Arrangement/526. Beautiful Arrangement_test.go
rename to leetcode/0526.Beautiful-Arrangement/526. Beautiful Arrangement_test.go
index 2ed14c1ec..b592c0764 100644
--- a/Algorithms/0526. Beautiful Arrangement/526. Beautiful Arrangement_test.go	
+++ b/leetcode/0526.Beautiful-Arrangement/526. Beautiful Arrangement_test.go	
@@ -26,92 +26,92 @@ func Test_Problem526(t *testing.T) {
 
 	qs := []question526{
 
-		question526{
+		{
 			para526{1},
 			ans526{1},
 		},
 
-		question526{
+		{
 			para526{2},
 			ans526{2},
 		},
 
-		question526{
+		{
 			para526{3},
 			ans526{3},
 		},
 
-		question526{
+		{
 			para526{4},
 			ans526{8},
 		},
 
-		question526{
+		{
 			para526{5},
 			ans526{10},
 		},
 
-		question526{
+		{
 			para526{6},
 			ans526{36},
 		},
 
-		question526{
+		{
 			para526{7},
 			ans526{41},
 		},
 
-		question526{
+		{
 			para526{8},
 			ans526{132},
 		},
 
-		question526{
+		{
 			para526{9},
 			ans526{250},
 		},
 
-		question526{
+		{
 			para526{10},
 			ans526{700},
 		},
 
-		question526{
+		{
 			para526{11},
 			ans526{750},
 		},
 
-		question526{
+		{
 			para526{12},
 			ans526{4010},
 		},
 
-		question526{
+		{
 			para526{13},
 			ans526{4237},
 		},
 
-		question526{
+		{
 			para526{14},
 			ans526{10680},
 		},
 
-		question526{
+		{
 			para526{15},
 			ans526{24679},
 		},
 
-		question526{
+		{
 			para526{16},
 			ans526{87328},
 		},
 
-		question526{
+		{
 			para526{17},
 			ans526{90478},
 		},
 
-		question526{
+		{
 			para526{18},
 			ans526{435812},
 		},
diff --git a/Algorithms/0526. Beautiful Arrangement/README.md b/leetcode/0526.Beautiful-Arrangement/README.md
similarity index 100%
rename from Algorithms/0526. Beautiful Arrangement/README.md
rename to leetcode/0526.Beautiful-Arrangement/README.md
diff --git a/leetcode/0528.Random-Pick-with-Weight/528. Random Pick with Weight.go b/leetcode/0528.Random-Pick-with-Weight/528. Random Pick with Weight.go
new file mode 100644
index 000000000..2307ee989
--- /dev/null
+++ b/leetcode/0528.Random-Pick-with-Weight/528. Random Pick with Weight.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"math/rand"
+)
+
+// Solution528 define
+type Solution528 struct {
+	prefixSum []int
+}
+
+// Constructor528 define
+func Constructor528(w []int) Solution528 {
+	prefixSum := make([]int, len(w))
+	for i, e := range w {
+		if i == 0 {
+			prefixSum[i] = e
+			continue
+		}
+		prefixSum[i] = prefixSum[i-1] + e
+	}
+	return Solution528{prefixSum: prefixSum}
+}
+
+// PickIndex define
+func (so *Solution528) PickIndex() int {
+	n := rand.Intn(so.prefixSum[len(so.prefixSum)-1]) + 1
+	low, high := 0, len(so.prefixSum)-1
+	for low < high {
+		mid := low + (high-low)>>1
+		if so.prefixSum[mid] == n {
+			return mid
+		} else if so.prefixSum[mid] < n {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return low
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(w);
+ * param_1 := obj.PickIndex();
+ */
diff --git a/leetcode/0528.Random-Pick-with-Weight/528. Random Pick with Weight_test.go b/leetcode/0528.Random-Pick-with-Weight/528. Random Pick with Weight_test.go
new file mode 100644
index 000000000..bdb416241
--- /dev/null
+++ b/leetcode/0528.Random-Pick-with-Weight/528. Random Pick with Weight_test.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem528(t *testing.T) {
+	w := []int{1, 3}
+	sol := Constructor528(w)
+	fmt.Printf("1.PickIndex = %v\n", sol.PickIndex())
+	fmt.Printf("2.PickIndex = %v\n", sol.PickIndex())
+	fmt.Printf("3.PickIndex = %v\n", sol.PickIndex())
+	fmt.Printf("4.PickIndex = %v\n", sol.PickIndex())
+	fmt.Printf("5.PickIndex = %v\n", sol.PickIndex())
+	fmt.Printf("6.PickIndex = %v\n", sol.PickIndex())
+}
diff --git a/leetcode/0528.Random-Pick-with-Weight/README.md b/leetcode/0528.Random-Pick-with-Weight/README.md
new file mode 100755
index 000000000..054afa4cc
--- /dev/null
+++ b/leetcode/0528.Random-Pick-with-Weight/README.md
@@ -0,0 +1,54 @@
+# [528. Random Pick with Weight](https://leetcode.com/problems/random-pick-with-weight/)
+
+
+## 题目
+
+Given an array `w` of positive integers, where `w[i]` describes the weight of index `i`, write a function `pickIndex` which randomly picks an index in proportion to its weight.
+
+Note:
+
+1. `1 <= w.length <= 10000`
+2. `1 <= w[i] <= 10^5`
+3. `pickIndex` will be called at most `10000` times.
+
+**Example 1:**
+
+    Input: 
+    ["Solution","pickIndex"]
+    [[[1]],[]]
+    Output: [null,0]
+
+**Example 2:**
+
+    Input: 
+    ["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
+    [[[1,3]],[],[],[],[],[]]
+    Output: [null,0,1,1,1,0]
+
+**Explanation of Input Syntax:**
+
+The input is two lists: the subroutines called and their arguments. `Solution`'s constructor has one argument, the array `w`. `pickIndex` has no arguments. Arguments are always wrapped with a list, even if there aren't any.
+
+
+## 题目大意
+
+给定一个正整数数组 w ，其中 w[i] 代表位置 i 的权重，请写一个函数 pickIndex ，它可以随机地获取位置 i，选取位置 i 的概率与 w[i] 成正比。
+
+说明:
+
+1. 1 <= w.length <= 10000
+2. 1 <= w[i] <= 10^5
+3. pickIndex 将被调用不超过 10000 次
+
+
+输入语法说明：
+
+输入是两个列表：调用成员函数名和调用的参数。Solution 的构造函数有一个参数，即数组 w。pickIndex 没有参数。输入参数是一个列表，即使参数为空，也会输入一个 [] 空列表。
+
+
+
+## 解题思路
+
+- 给出一个数组，每个元素值代表该下标的权重值，`pickIndex()` 随机取一个位置 i，这个位置出现的概率和该元素值成正比。
+- 由于涉及到了权重的问题，这一题可以先考虑用前缀和处理权重。在 `[0,prefixSum)` 区间内随机选一个整数 `x`，下标 `i` 是满足 `x< prefixSum[i]` 条件的最小下标，求这个下标 `i` 即是最终解。二分搜索查找下标 `i` 。对于某些下标 `i`，所有满足 `prefixSum[i] - w[i] ≤ v < prefixSum[i]` 的整数 `v` 都映射到这个下标。因此，所有的下标都与下标权重成比例。
+- 时间复杂度：预处理的时间复杂度是 O(n)，`pickIndex()` 的时间复杂度是 O(log n)。空间复杂度 O(n)。
diff --git a/leetcode/0529.Minesweeper/529. Minesweeper.go b/leetcode/0529.Minesweeper/529. Minesweeper.go
new file mode 100644
index 000000000..786405ee7
--- /dev/null
+++ b/leetcode/0529.Minesweeper/529. Minesweeper.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+var dir8 = [][]int{
+	{-1, -1},
+	{-1, 0},
+	{-1, 1},
+	{0, 1},
+	{1, 1},
+	{1, 0},
+	{1, -1},
+	{0, -1},
+}
+
+func updateBoard(board [][]byte, click []int) [][]byte {
+	if board[click[0]][click[1]] == 'M' {
+		board[click[0]][click[1]] = 'X'
+		return board
+	}
+	dfs(board, click[0], click[1])
+	return board
+}
+
+func dfs(board [][]byte, x, y int) {
+	cnt := 0
+	for i := 0; i < 8; i++ {
+		nx, ny := x+dir8[i][0], y+dir8[i][1]
+		if isInBoard(board, nx, ny) && board[nx][ny] == 'M' {
+			cnt++
+		}
+	}
+	if cnt > 0 {
+		board[x][y] = byte(cnt + '0')
+		return
+	}
+	board[x][y] = 'B'
+	for i := 0; i < 8; i++ {
+		nx, ny := x+dir8[i][0], y+dir8[i][1]
+		if isInBoard(board, nx, ny) && board[nx][ny] != 'B' {
+			dfs(board, nx, ny)
+		}
+	}
+}
+
+func isInBoard(board [][]byte, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
diff --git a/leetcode/0529.Minesweeper/529. Minesweeper_test.go b/leetcode/0529.Minesweeper/529. Minesweeper_test.go
new file mode 100644
index 000000000..7a1b538a7
--- /dev/null
+++ b/leetcode/0529.Minesweeper/529. Minesweeper_test.go	
@@ -0,0 +1,68 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question529 struct {
+	para529
+	ans529
+}
+
+// para 是参数
+// one 代表第一个参数
+type para529 struct {
+	b     [][]byte
+	click []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans529 struct {
+	one [][]byte
+}
+
+func Test_Problem529(t *testing.T) {
+
+	qs := []question529{
+
+		{
+			para529{[][]byte{
+				{'E', 'E', 'E', 'E', 'E'},
+				{'E', 'E', 'M', 'E', 'E'},
+				{'E', 'E', 'E', 'E', 'E'},
+				{'E', 'E', 'E', 'E', 'E'},
+			}, []int{3, 0}},
+			ans529{[][]byte{
+				{'B', '1', 'E', '1', 'B'},
+				{'B', '1', 'M', '1', 'B'},
+				{'B', '1', '1', '1', 'B'},
+				{'B', 'B', 'B', 'B', 'B'},
+			}},
+		},
+
+		{
+			para529{[][]byte{
+				{'B', '1', 'E', '1', 'B'},
+				{'B', '1', 'M', '1', 'B'},
+				{'B', '1', '1', '1', 'B'},
+				{'B', 'B', 'B', 'B', 'B'},
+			}, []int{1, 2}},
+			ans529{[][]byte{
+				{'B', '1', 'E', '1', 'B'},
+				{'B', '1', 'X', '1', 'B'},
+				{'B', '1', '1', '1', 'B'},
+				{'B', 'B', 'B', 'B', 'B'},
+			}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 529------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans529, q.para529
+		fmt.Printf("【input】:%v       【output】:%v\n\n\n", p, updateBoard(p.b, p.click))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0529.Minesweeper/README.md b/leetcode/0529.Minesweeper/README.md
new file mode 100644
index 000000000..d25c79c50
--- /dev/null
+++ b/leetcode/0529.Minesweeper/README.md
@@ -0,0 +1,135 @@
+# [529. Minesweeper](https://leetcode.com/problems/minesweeper/)
+
+
+
+## 题目
+
+Let's play the minesweeper game ([Wikipedia](https://en.wikipedia.org/wiki/Minesweeper_(video_game)), [online game](http://minesweeperonline.com/))!
+
+You are given a 2D char matrix representing the game board. **'M'** represents an **unrevealed** mine, **'E'** represents an **unrevealed** empty square, **'B'** represents a **revealed** blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, **digit** ('1' to '8') represents how many mines are adjacent to this **revealed** square, and finally **'X'** represents a **revealed** mine.
+
+Now given the next click position (row and column indices) among all the **unrevealed** squares ('M' or 'E'), return the board after revealing this position according to the following rules:
+
+1. If a mine ('M') is revealed, then the game is over - change it to **'X'**.
+2. If an empty square ('E') with **no adjacent mines** is revealed, then change it to revealed blank ('B') and all of its adjacent **unrevealed** squares should be revealed recursively.
+3. If an empty square ('E') with **at least one adjacent mine** is revealed, then change it to a digit ('1' to '8') representing the number of adjacent mines.
+4. Return the board when no more squares will be revealed.
+
+**Example 1**:
+
+```
+Input: 
+
+[['E', 'E', 'E', 'E', 'E'],
+ ['E', 'E', 'M', 'E', 'E'],
+ ['E', 'E', 'E', 'E', 'E'],
+ ['E', 'E', 'E', 'E', 'E']]
+
+Click : [3,0]
+
+Output: 
+
+[['B', '1', 'E', '1', 'B'],
+ ['B', '1', 'M', '1', 'B'],
+ ['B', '1', '1', '1', 'B'],
+ ['B', 'B', 'B', 'B', 'B']]
+
+Explanation:
+```
+
+![https://assets.leetcode.com/uploads/2018/10/12/minesweeper_example_1.png](https://assets.leetcode.com/uploads/2018/10/12/minesweeper_example_1.png)
+
+**Example 2**:
+
+```
+Input: 
+
+[['B', '1', 'E', '1', 'B'],
+ ['B', '1', 'M', '1', 'B'],
+ ['B', '1', '1', '1', 'B'],
+ ['B', 'B', 'B', 'B', 'B']]
+
+Click : [1,2]
+
+Output: 
+
+[['B', '1', 'E', '1', 'B'],
+ ['B', '1', 'X', '1', 'B'],
+ ['B', '1', '1', '1', 'B'],
+ ['B', 'B', 'B', 'B', 'B']]
+
+Explanation:
+```
+
+![https://assets.leetcode.com/uploads/2018/10/12/minesweeper_example_2.png](https://assets.leetcode.com/uploads/2018/10/12/minesweeper_example_2.png)
+
+**Note**:
+
+1. The range of the input matrix's height and width is [1,50].
+2. The click position will only be an unrevealed square ('M' or 'E'), which also means the input board contains at least one clickable square.
+3. The input board won't be a stage when game is over (some mines have been revealed).
+4. For simplicity, not mentioned rules should be ignored in this problem. For example, you **don't** need to reveal all the unrevealed mines when the game is over, consider any cases that you will win the game or flag any squares.
+
+
+## 题目大意
+
+给定一个代表游戏板的二维字符矩阵。 'M' 代表一个未挖出的地雷，'E' 代表一个未挖出的空方块，'B' 代表没有相邻（上，下，左，右，和所有4个对角线）地雷的已挖出的空白方块，数字（'1' 到 '8'）表示有多少地雷与这块已挖出的方块相邻，'X' 则表示一个已挖出的地雷。现在给出在所有未挖出的方块中（'M'或者'E'）的下一个点击位置（行和列索引），根据以下规则，返回相应位置被点击后对应的面板：
+
+1. 如果一个地雷（'M'）被挖出，游戏就结束了- 把它改为 'X'。
+2. 如果一个没有相邻地雷的空方块（'E'）被挖出，修改它为（'B'），并且所有和其相邻的未挖出方块都应该被递归地揭露。
+3. 如果一个至少与一个地雷相邻的空方块（'E'）被挖出，修改它为数字（'1'到'8'），表示相邻地雷的数量。
+4. 如果在此次点击中，若无更多方块可被揭露，则返回面板。
+
+
+注意：
+
+- 输入矩阵的宽和高的范围为 [1,50]。
+- 点击的位置只能是未被挖出的方块 ('M' 或者 'E')，这也意味着面板至少包含一个可点击的方块。
+- 输入面板不会是游戏结束的状态（即有地雷已被挖出）。
+- 简单起见，未提及的规则在这个问题中可被忽略。例如，当游戏结束时你不需要挖出所有地雷，考虑所有你可能赢得游戏或标记方块的情况。
+
+
+
+## 解题思路
+
+- 给出一张扫雷地图和点击的坐标，M 代表雷，E 代表还没有点击过的空砖块，B 代表点击过的空砖块，1-8 代表砖块周围 8 个方块里面有雷的个数，X 代表点到了雷。问点击一次以后，输出更新点击以后的地图。
+- DPS 和 BFS 都可以解题。先根据原图预处理地图，记录出最终地图的状态，0 代表空白砖块，1-8 代表雷的个数，-1 代表是雷。再 DFS 遍历这张处理后的图，输出最终的地图即可。
+
+## 代码
+
+```go
+func updateBoard(board [][]byte, click []int) [][]byte {
+	if board[click[0]][click[1]] == 'M' {
+		board[click[0]][click[1]] = 'X'
+		return board
+	}
+	dfs(board, click[0], click[1])
+	return board
+}
+
+func dfs(board [][]byte, x, y int) {
+	cnt := 0
+	for i := 0; i < 8; i++ {
+		nx, ny := x+dir8[i][0], y+dir8[i][1]
+		if isInBoard(board, nx, ny) && board[nx][ny] == 'M' {
+			cnt++
+
+		}
+	}
+	if cnt > 0 {
+		board[x][y] = byte(cnt + '0')
+		return
+	}
+	board[x][y] = 'B'
+	for i := 0; i < 8; i++ {
+		nx, ny := x+dir8[i][0], y+dir8[i][1]
+		if isInBoard(board, nx, ny) && board[nx][ny] != 'B' {
+			dfs(board, nx, ny)
+		}
+	}
+}
+
+func isInBoard(board [][]byte, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+```
\ No newline at end of file
diff --git a/leetcode/0530.Minimum-Absolute-Difference-in-BST/530. Minimum Absolute Difference in BST.go b/leetcode/0530.Minimum-Absolute-Difference-in-BST/530. Minimum Absolute Difference in BST.go
new file mode 100644
index 000000000..43fc4eabb
--- /dev/null
+++ b/leetcode/0530.Minimum-Absolute-Difference-in-BST/530. Minimum Absolute Difference in BST.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"math"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func getMinimumDifference(root *TreeNode) int {
+	res, nodes := math.MaxInt16, -1
+	dfsBST(root, &res, &nodes)
+	return res
+}
+
+func dfsBST(root *TreeNode, res, pre *int) {
+	if root == nil {
+		return
+	}
+	dfsBST(root.Left, res, pre)
+	if *pre != -1 {
+		*res = min(*res, abs(root.Val-*pre))
+	}
+	*pre = root.Val
+	dfsBST(root.Right, res, pre)
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
diff --git a/leetcode/0530.Minimum-Absolute-Difference-in-BST/530. Minimum Absolute Difference in BST_test.go b/leetcode/0530.Minimum-Absolute-Difference-in-BST/530. Minimum Absolute Difference in BST_test.go
new file mode 100644
index 000000000..62230730b
--- /dev/null
+++ b/leetcode/0530.Minimum-Absolute-Difference-in-BST/530. Minimum Absolute Difference in BST_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question530 struct {
+	para530
+	ans530
+}
+
+// para 是参数
+// one 代表第一个参数
+type para530 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans530 struct {
+	one int
+}
+
+func Test_Problem530(t *testing.T) {
+
+	qs := []question530{
+
+		{
+			para530{[]int{4, 2, 6, 1, 3}},
+			ans530{1},
+		},
+
+		{
+			para530{[]int{1, 0, 48, structures.NULL, structures.NULL, 12, 49}},
+			ans530{1},
+		},
+
+		{
+			para530{[]int{90, 69, structures.NULL, 49, 89, structures.NULL, 52}},
+			ans530{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 530------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans530, q.para530
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", getMinimumDifference(rootOne))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0530.Minimum-Absolute-Difference-in-BST/README.md b/leetcode/0530.Minimum-Absolute-Difference-in-BST/README.md
new file mode 100644
index 000000000..c926d3518
--- /dev/null
+++ b/leetcode/0530.Minimum-Absolute-Difference-in-BST/README.md
@@ -0,0 +1,94 @@
+# [530. Minimum Absolute Difference in BST](https://leetcode.com/problems/minimum-absolute-difference-in-bst/)
+
+
+## 题目
+
+Given a binary search tree with non-negative values, find the minimum [absolute difference](https://en.wikipedia.org/wiki/Absolute_difference) between values of any two nodes.
+
+**Example:**
+
+```
+Input:
+
+   1
+    \
+     3
+    /
+   2
+
+Output:
+1
+
+Explanation:
+The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).
+```
+
+**Note:**
+
+- There are at least two nodes in this BST.
+- This question is the same as 783: [https://leetcode.com/problems/minimum-distance-between-bst-nodes/](https://leetcode.com/problems/minimum-distance-between-bst-nodes/)
+
+## 题目大意
+
+给你一棵所有节点为非负值的二叉搜索树，请你计算树中任意两节点的差的绝对值的最小值。
+
+## 解题思路
+
+- 由于是 BST 树，利用它有序的性质，中根遍历的结果是有序的。中根遍历过程中动态维护前后两个节点的差值，即可找到最小差值。
+- 此题与第 783 题完全相同。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func getMinimumDifference(root *TreeNode) int {
+	res, nodes := math.MaxInt16, -1
+	dfsBST(root, &res, &nodes)
+	return res
+}
+
+func dfsBST(root *TreeNode, res, pre *int) {
+	if root == nil {
+		return
+	}
+	dfsBST(root.Left, res, pre)
+	if *pre != -1 {
+		*res = min(*res, abs(root.Val-*pre))
+	}
+	*pre = root.Val
+	dfsBST(root.Right, res, pre)
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0532. K-diff Pairs in an Array/532. K-diff Pairs in an Array.go b/leetcode/0532.K-diff-Pairs-in-an-Array/532. K-diff Pairs in an Array.go
similarity index 100%
rename from Algorithms/0532. K-diff Pairs in an Array/532. K-diff Pairs in an Array.go
rename to leetcode/0532.K-diff-Pairs-in-an-Array/532. K-diff Pairs in an Array.go
diff --git a/Algorithms/0532. K-diff Pairs in an Array/532. K-diff Pairs in an Array_test.go b/leetcode/0532.K-diff-Pairs-in-an-Array/532. K-diff Pairs in an Array_test.go
similarity index 95%
rename from Algorithms/0532. K-diff Pairs in an Array/532. K-diff Pairs in an Array_test.go
rename to leetcode/0532.K-diff-Pairs-in-an-Array/532. K-diff Pairs in an Array_test.go
index 2434bb68a..dd8957349 100644
--- a/Algorithms/0532. K-diff Pairs in an Array/532. K-diff Pairs in an Array_test.go	
+++ b/leetcode/0532.K-diff-Pairs-in-an-Array/532. K-diff Pairs in an Array_test.go	
@@ -27,22 +27,22 @@ func Test_Problem532(t *testing.T) {
 
 	qs := []question532{
 
-		question532{
+		{
 			para532{[]int{3, 1, 4, 1, 5}, 2},
 			ans532{2},
 		},
 
-		question532{
+		{
 			para532{[]int{1, 2, 3, 4, 5}, 1},
 			ans532{4},
 		},
 
-		question532{
+		{
 			para532{[]int{1, 3, 1, 5, 4}, 0},
 			ans532{1},
 		},
 
-		question532{
+		{
 			para532{[]int{}, 3},
 			ans532{0},
 		},
diff --git a/Algorithms/0532. K-diff Pairs in an Array/README.md b/leetcode/0532.K-diff-Pairs-in-an-Array/README.md
similarity index 100%
rename from Algorithms/0532. K-diff Pairs in an Array/README.md
rename to leetcode/0532.K-diff-Pairs-in-an-Array/README.md
diff --git a/leetcode/0535.Encode-and-Decode-TinyURL/535. Encode and Decode TinyURL.go b/leetcode/0535.Encode-and-Decode-TinyURL/535. Encode and Decode TinyURL.go
new file mode 100644
index 000000000..dc7fa65d9
--- /dev/null
+++ b/leetcode/0535.Encode-and-Decode-TinyURL/535. Encode and Decode TinyURL.go	
@@ -0,0 +1,35 @@
+package leetcode
+
+import (
+	"fmt"
+	"strconv"
+	"strings"
+)
+
+type Codec struct {
+	urls []string
+}
+
+func Constructor() Codec {
+	return Codec{[]string{}}
+}
+
+// Encodes a URL to a shortened URL.
+func (this *Codec) encode(longUrl string) string {
+	this.urls = append(this.urls, longUrl)
+	return "http://tinyurl.com/" + fmt.Sprintf("%v", len(this.urls)-1)
+}
+
+// Decodes a shortened URL to its original URL.
+func (this *Codec) decode(shortUrl string) string {
+	tmp := strings.Split(shortUrl, "/")
+	i, _ := strconv.Atoi(tmp[len(tmp)-1])
+	return this.urls[i]
+}
+
+/**
+ * Your Codec object will be instantiated and called as such:
+ * obj := Constructor();
+ * url := obj.encode(longUrl);
+ * ans := obj.decode(url);
+ */
diff --git a/leetcode/0535.Encode-and-Decode-TinyURL/535. Encode and Decode TinyURL_test.go b/leetcode/0535.Encode-and-Decode-TinyURL/535. Encode and Decode TinyURL_test.go
new file mode 100644
index 000000000..e3f10ae75
--- /dev/null
+++ b/leetcode/0535.Encode-and-Decode-TinyURL/535. Encode and Decode TinyURL_test.go	
@@ -0,0 +1,15 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem535(t *testing.T) {
+	obj := Constructor()
+	fmt.Printf("obj = %v\n", obj)
+	e := obj.encode("https://leetcode.com/problems/design-tinyurl")
+	fmt.Printf("obj encode = %v\n", e)
+	d := obj.decode(e)
+	fmt.Printf("obj decode = %v\n", d)
+}
diff --git a/leetcode/0535.Encode-and-Decode-TinyURL/README.md b/leetcode/0535.Encode-and-Decode-TinyURL/README.md
new file mode 100644
index 000000000..d7ae92b40
--- /dev/null
+++ b/leetcode/0535.Encode-and-Decode-TinyURL/README.md
@@ -0,0 +1,60 @@
+# [535. Encode and Decode TinyURL](https://leetcode.com/problems/encode-and-decode-tinyurl/)
+
+
+## 题目
+
+> Note: This is a companion problem to the System Design problem: Design TinyURL.
+
+TinyURL is a URL shortening service where you enter a URL such as `https://leetcode.com/problems/design-tinyurl` and it returns a short URL such as `http://tinyurl.com/4e9iAk`.
+
+Design the `encode` and `decode` methods for the TinyURL service. There is no restriction on how your encode/decode algorithm should work. You just need to ensure that a URL can be encoded to a tiny URL and the tiny URL can be decoded to the original URL.
+
+## 题目大意
+
+TinyURL是一种URL简化服务， 比如：当你输入一个URL [https://leetcode.com/problems/design-tinyurl](https://leetcode.com/problems/design-tinyurl) 时，它将返回一个简化的URL [http://tinyurl.com/4e9iAk](http://tinyurl.com/4e9iAk).
+
+要求：设计一个 TinyURL 的加密 encode 和解密 decode 的方法。你的加密和解密算法如何设计和运作是没有限制的，你只需要保证一个URL可以被加密成一个TinyURL，并且这个TinyURL可以用解密方法恢复成原本的URL。
+
+## 解题思路
+
+- 简单题。由于题目并无规定 `encode()` 算法，所以自由度非常高。最简单的做法是把原始 `URL` 存起来，并记录下存在字符串数组中的下标位置。`decode()` 的时候根据存储的下标还原原始的 `URL`。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"fmt"
+	"strconv"
+	"strings"
+)
+
+type Codec struct {
+	urls []string
+}
+
+func Constructor() Codec {
+	return Codec{[]string{}}
+}
+
+// Encodes a URL to a shortened URL.
+func (this *Codec) encode(longUrl string) string {
+	this.urls = append(this.urls, longUrl)
+	return "http://tinyurl.com/" + fmt.Sprintf("%v", len(this.urls)-1)
+}
+
+// Decodes a shortened URL to its original URL.
+func (this *Codec) decode(shortUrl string) string {
+	tmp := strings.Split(shortUrl, "/")
+	i, _ := strconv.Atoi(tmp[len(tmp)-1])
+	return this.urls[i]
+}
+
+/**
+ * Your Codec object will be instantiated and called as such:
+ * obj := Constructor();
+ * url := obj.encode(longUrl);
+ * ans := obj.decode(url);
+ */
+```
\ No newline at end of file
diff --git a/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication.go b/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication.go
new file mode 100644
index 000000000..d552b3e1d
--- /dev/null
+++ b/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+)
+
+func complexNumberMultiply(a string, b string) string {
+	realA, imagA := parse(a)
+	realB, imagB := parse(b)
+	real := realA*realB - imagA*imagB
+	imag := realA*imagB + realB*imagA
+	return strconv.Itoa(real) + "+" + strconv.Itoa(imag) + "i"
+}
+
+func parse(s string) (int, int) {
+	ss := strings.Split(s, "+")
+	r, _ := strconv.Atoi(ss[0])
+	i, _ := strconv.Atoi(ss[1][:len(ss[1])-1])
+	return r, i
+}
diff --git a/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication_test.go b/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication_test.go
new file mode 100644
index 000000000..00d53dfda
--- /dev/null
+++ b/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question537 struct {
+	para537
+	ans537
+}
+
+// para 是参数
+// one 代表第一个参数
+type para537 struct {
+	a string
+	b string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans537 struct {
+	one string
+}
+
+func Test_Problem537(t *testing.T) {
+
+	qs := []question537{
+
+		{
+			para537{"1+1i", "1+1i"},
+			ans537{"0+2i"},
+		},
+
+		{
+			para537{"1+-1i", "1+-1i"},
+			ans537{"0+-2i"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 537------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans537, q.para537
+		fmt.Printf("【input】:%v       【output】:%v\n", p, complexNumberMultiply(p.a, p.b))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0537.Complex-Number-Multiplication/README.md b/leetcode/0537.Complex-Number-Multiplication/README.md
new file mode 100644
index 000000000..568404bab
--- /dev/null
+++ b/leetcode/0537.Complex-Number-Multiplication/README.md
@@ -0,0 +1,73 @@
+# [537. Complex Number Multiplication](https://leetcode.com/problems/complex-number-multiplication/)
+
+
+## 题目
+
+Given two strings representing two [complex numbers](https://en.wikipedia.org/wiki/Complex_number).
+
+You need to return a string representing their multiplication. Note i2 = -1 according to the definition.
+
+**Example 1**:
+
+```
+Input: "1+1i", "1+1i"
+Output: "0+2i"
+Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i.
+```
+
+**Example 2**:
+
+```
+Input: "1+-1i", "1+-1i"
+Output: "0+-2i"
+Explanation: (1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i, and you need convert it to the form of 0+-2i.
+```
+
+**Note**:
+
+1. The input strings will not have extra blank.
+2. The input strings will be given in the form of **a+bi**, where the integer **a** and **b** will both belong to the range of [-100, 100]. And **the output should be also in this form**.
+
+## 题目大意
+
+给定两个表示复数的字符串。返回表示它们乘积的字符串。注意，根据定义 i^2 = -1 。
+
+注意:
+
+- 输入字符串不包含额外的空格。
+- 输入字符串将以 a+bi 的形式给出，其中整数 a 和 b 的范围均在 [-100, 100] 之间。输出也应当符合这种形式。
+
+
+
+## 解题思路
+
+- 给定 2 个字符串，要求这两个复数的乘积，输出也是字符串格式。
+- 数学题。按照复数的运算法则，i^2 = -1，最后输出字符串结果即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+)
+
+func complexNumberMultiply(a string, b string) string {
+	realA, imagA := parse(a)
+	realB, imagB := parse(b)
+	real := realA*realB - imagA*imagB
+	imag := realA*imagB + realB*imagA
+	return strconv.Itoa(real) + "+" + strconv.Itoa(imag) + "i"
+}
+
+func parse(s string) (int, int) {
+	ss := strings.Split(s, "+")
+	r, _ := strconv.Atoi(ss[0])
+	i, _ := strconv.Atoi(ss[1][:len(ss[1])-1])
+	return r, i
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/0538.Convert-BST-to-Greater-Tree/538. Convert BST to Greater Tree.go b/leetcode/0538.Convert-BST-to-Greater-Tree/538. Convert BST to Greater Tree.go
new file mode 100644
index 000000000..4a360519d
--- /dev/null
+++ b/leetcode/0538.Convert-BST-to-Greater-Tree/538. Convert BST to Greater Tree.go	
@@ -0,0 +1,36 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func convertBST(root *TreeNode) *TreeNode {
+	if root == nil {
+		return root
+	}
+	sum := 0
+	dfs538(root, &sum)
+	return root
+}
+
+func dfs538(root *TreeNode, sum *int) {
+	if root == nil {
+		return
+	}
+	dfs538(root.Right, sum)
+	root.Val += *sum
+	*sum = root.Val
+	dfs538(root.Left, sum)
+}
diff --git a/leetcode/0538.Convert-BST-to-Greater-Tree/538. Convert BST to Greater Tree_test.go b/leetcode/0538.Convert-BST-to-Greater-Tree/538. Convert BST to Greater Tree_test.go
new file mode 100644
index 000000000..c42b422f1
--- /dev/null
+++ b/leetcode/0538.Convert-BST-to-Greater-Tree/538. Convert BST to Greater Tree_test.go	
@@ -0,0 +1,76 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question538 struct {
+	para538
+	ans538
+}
+
+// para 是参数
+// one 代表第一个参数
+type para538 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans538 struct {
+	one []int
+}
+
+func Test_Problem538(t *testing.T) {
+
+	qs := []question538{
+
+		{
+			para538{[]int{3, 1, structures.NULL, 0, structures.NULL, -4, structures.NULL, structures.NULL, -2}},
+			ans538{[]int{3, 4, structures.NULL, 4, structures.NULL, -2, structures.NULL, structures.NULL, 2}},
+		},
+
+		{
+			para538{[]int{2, 1}},
+			ans538{[]int{2, 3}},
+		},
+
+		{
+			para538{[]int{}},
+			ans538{[]int{}},
+		},
+
+		{
+			para538{[]int{4, 1, 6, 0, 2, 5, 7, structures.NULL, structures.NULL, structures.NULL, 3, structures.NULL, structures.NULL, structures.NULL, 8}},
+			ans538{[]int{30, 36, 21, 36, 35, 26, 15, structures.NULL, structures.NULL, structures.NULL, 33, structures.NULL, structures.NULL, structures.NULL, 8}},
+		},
+
+		{
+			para538{[]int{0, structures.NULL, 1}},
+			ans538{[]int{1, structures.NULL, 1}},
+		},
+
+		{
+			para538{[]int{1, 0, 2}},
+			ans538{[]int{3, 3, 2}},
+		},
+
+		{
+			para538{[]int{3, 2, 4, 1}},
+			ans538{[]int{7, 9, 4, 10}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 538------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans538, q.para538
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", structures.Tree2ints(convertBST(root)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0538.Convert-BST-to-Greater-Tree/README.md b/leetcode/0538.Convert-BST-to-Greater-Tree/README.md
new file mode 100644
index 000000000..bd56381a7
--- /dev/null
+++ b/leetcode/0538.Convert-BST-to-Greater-Tree/README.md
@@ -0,0 +1,106 @@
+# [538. Convert BST to Greater Tree](https://leetcode.com/problems/convert-bst-to-greater-tree/)
+
+## 题目
+
+Given the `root` of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.
+
+As a reminder, a *binary search tree* is a tree that satisfies these constraints:
+
+- The left subtree of a node contains only nodes with keys **less than** the node's key.
+- The right subtree of a node contains only nodes with keys **greater than** the node's key.
+- Both the left and right subtrees must also be binary search trees.
+
+**Note:** This question is the same as 1038: [https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/](https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/)
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/05/02/tree.png](https://assets.leetcode.com/uploads/2019/05/02/tree.png)
+
+```
+Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
+Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]
+```
+
+**Example 2:**
+
+```
+Input: root = [0,null,1]
+Output: [1,null,1]
+```
+
+**Example 3:**
+
+```
+Input: root = [1,0,2]
+Output: [3,3,2]
+```
+
+**Example 4:**
+
+```
+Input: root = [3,2,4,1]
+Output: [7,9,4,10]
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[0, 104]`.
+- `104 <= Node.val <= 104`
+- All the values in the tree are **unique**.
+- `root` is guaranteed to be a valid binary search tree.
+
+## 题目大意
+
+给出二叉 搜索 树的根节点，该树的节点值各不相同，请你将其转换为累加树（Greater Sum Tree），使每个节点 node 的新值等于原树中大于或等于 node.val 的值之和。
+
+提醒一下，二叉搜索树满足下列约束条件：
+
+- 节点的左子树仅包含键 小于 节点键的节点。
+- 节点的右子树仅包含键 大于 节点键的节点。
+- 左右子树也必须是二叉搜索树。
+
+## 解题思路
+
+- 根据二叉搜索树的有序性，想要将其转换为累加树，只需按照 右节点 - 根节点 - 左节点的顺序遍历，并累加和即可。
+- 此题同第 1038 题。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func convertBST(root *TreeNode) *TreeNode {
+	if root == nil {
+		return root
+	}
+	sum := 0
+	dfs538(root, &sum)
+	return root
+}
+
+func dfs538(root *TreeNode, sum *int) {
+	if root == nil {
+		return
+	}
+	dfs538(root.Right, sum)
+	root.Val += *sum
+	*sum = root.Val
+	dfs538(root.Left, sum)
+}
+```
\ No newline at end of file
diff --git a/leetcode/0540.Single-Element-in-a-Sorted-Array/540.Single Element in a Sorted Array.go b/leetcode/0540.Single-Element-in-a-Sorted-Array/540.Single Element in a Sorted Array.go
new file mode 100644
index 000000000..dc718044f
--- /dev/null
+++ b/leetcode/0540.Single-Element-in-a-Sorted-Array/540.Single Element in a Sorted Array.go	
@@ -0,0 +1,22 @@
+package leetcode
+
+func singleNonDuplicate(nums []int) int {
+	left, right := 0, len(nums)-1
+	for left < right {
+		mid := (left + right) / 2
+		if mid%2 == 0 {
+			if nums[mid] == nums[mid+1] {
+				left = mid + 1
+			} else {
+				right = mid
+			}
+		} else {
+			if nums[mid] == nums[mid-1] {
+				left = mid + 1
+			} else {
+				right = mid
+			}
+		}
+	}
+	return nums[left]
+}
diff --git a/leetcode/0540.Single-Element-in-a-Sorted-Array/540.Single Element in a Sorted Array_test.go b/leetcode/0540.Single-Element-in-a-Sorted-Array/540.Single Element in a Sorted Array_test.go
new file mode 100644
index 000000000..77871f6b0
--- /dev/null
+++ b/leetcode/0540.Single-Element-in-a-Sorted-Array/540.Single Element in a Sorted Array_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question540 struct {
+	para540
+	ans540
+}
+
+// para 是参数
+type para540 struct {
+	nums []int
+}
+
+// ans 是答案
+type ans540 struct {
+	ans int
+}
+
+func Test_Problem540(t *testing.T) {
+
+	qs := []question540{
+
+		{
+			para540{[]int{1, 1, 2, 3, 3, 4, 4, 8, 8}},
+			ans540{2},
+		},
+
+		{
+			para540{[]int{3, 3, 7, 7, 10, 11, 11}},
+			ans540{10},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 540------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans540, q.para540
+		fmt.Printf("【input】:%v      ", p.nums)
+		fmt.Printf("【output】:%v      \n", singleNonDuplicate(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0540.Single-Element-in-a-Sorted-Array/README.md b/leetcode/0540.Single-Element-in-a-Sorted-Array/README.md
new file mode 100644
index 000000000..e9a6e2cdf
--- /dev/null
+++ b/leetcode/0540.Single-Element-in-a-Sorted-Array/README.md
@@ -0,0 +1,64 @@
+# [540. Single Element in a Sorted Array](https://leetcode.com/problems/single-element-in-a-sorted-array/)
+
+## 题目
+
+You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once.
+
+Return the single element that appears only once.
+
+Your solution must run in O(log n) time and O(1) space.
+
+**Example 1:**
+
+    Input: nums = [1,1,2,3,3,4,4,8,8]
+    Output: 2
+
+**Example 2:**
+
+    Input: nums = [3,3,7,7,10,11,11]
+    Output: 10
+
+**Constraints:**
+
+- 1 <= nums.length <= 100000
+- 0 <= nums[i] <= 100000
+
+## 题目大意
+
+给你一个仅由整数组成的有序数组，其中每个元素都会出现两次，唯有一个数只会出现一次。
+
+请你找出并返回只出现一次的那个数。
+
+你设计的解决方案必须满足 O(log n) 时间复杂度和 O(1) 空间复杂度。
+
+## 解题思路
+
+    假设下标idx是单独的数字,idx左边的偶数下标x有nums[x] == nums[x + 1],
+    idx右边的奇数下标y有nums[y] == nums[y + 1],可以根据此特性用二分查找idx对应的值 
+
+## 代码
+
+```go
+package leetcode
+
+func singleNonDuplicate(nums []int) int {
+	left, right := 0, len(nums)-1
+	for left < right {
+		mid := (left + right) / 2
+		if mid%2 == 0 {
+			if nums[mid] == nums[mid+1] {
+				left = mid + 1
+			} else {
+				right = mid
+			}
+		} else {
+			if nums[mid] == nums[mid-1] {
+				left = mid + 1
+			} else {
+				right = mid
+			}
+		}
+	}
+	return nums[left]
+}
+```
diff --git a/leetcode/0541.Reverse-String-II/541. Reverse String II.go b/leetcode/0541.Reverse-String-II/541. Reverse String II.go
new file mode 100644
index 000000000..ff00b877b
--- /dev/null
+++ b/leetcode/0541.Reverse-String-II/541. Reverse String II.go	
@@ -0,0 +1,28 @@
+package leetcode
+
+func reverseStr(s string, k int) string {
+	if k > len(s) {
+		k = len(s)
+	}
+	for i := 0; i < len(s); i = i + 2*k {
+		if len(s)-i >= k {
+			ss := revers(s[i : i+k])
+			s = s[:i] + ss + s[i+k:]
+		} else {
+			ss := revers(s[i:])
+			s = s[:i] + ss
+		}
+	}
+	return s
+}
+
+func revers(s string) string {
+	bytes := []byte(s)
+	i, j := 0, len(bytes)-1
+	for i < j {
+		bytes[i], bytes[j] = bytes[j], bytes[i]
+		i++
+		j--
+	}
+	return string(bytes)
+}
diff --git a/Algorithms/0541. Reverse String II/541. Reverse String II_test.go b/leetcode/0541.Reverse-String-II/541. Reverse String II_test.go
similarity index 92%
rename from Algorithms/0541. Reverse String II/541. Reverse String II_test.go
rename to leetcode/0541.Reverse-String-II/541. Reverse String II_test.go
index 358f0aa16..9c6d3ba5c 100644
--- a/Algorithms/0541. Reverse String II/541. Reverse String II_test.go	
+++ b/leetcode/0541.Reverse-String-II/541. Reverse String II_test.go	
@@ -27,22 +27,22 @@ func Test_Problem541(t *testing.T) {
 
 	qs := []question541{
 
-		question541{
+		{
 			para541{"abcdefg", 2},
 			ans541{"bacdfeg"},
 		},
 
-		question541{
+		{
 			para541{"abcdefg", 5},
 			ans541{"edcbafg"},
 		},
 
-		question541{
+		{
 			para541{"abcd", 4},
 			ans541{"dcba"},
 		},
 
-		question541{
+		{
 			para541{"", 100},
 			ans541{""},
 		},
diff --git a/leetcode/0541.Reverse-String-II/README.md b/leetcode/0541.Reverse-String-II/README.md
new file mode 100755
index 000000000..de8a18eca
--- /dev/null
+++ b/leetcode/0541.Reverse-String-II/README.md
@@ -0,0 +1,31 @@
+# [541. Reverse String II](https://leetcode.com/problems/reverse-string-ii/)
+
+
+## 题目
+
+Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.
+
+**Example:**
+
+    Input: s = "abcdefg", k = 2
+    Output: "bacdfeg"
+
+**Restrictions:**
+
+1. The string consists of lower English letters only.
+2. Length of the given string and k will in the range [1, 10000]
+
+## 题目大意
+
+给定一个字符串和一个整数 k，你需要对从字符串开头算起的每个 2k 个字符的前k个字符进行反转。如果剩余少于 k 个字符，则将剩余的所有全部反转。如果有小于 2k 但大于或等于 k 个字符，则反转前 k 个字符，并将剩余的字符保持原样。
+
+要求:
+
+- 该字符串只包含小写的英文字母。
+- 给定字符串的长度和 k 在[1, 10000]范围内。
+
+
+## 解题思路
+
+- 要求按照一定规则反转字符串：每 `2 * K` 长度的字符串，反转前 `K` 个字符，后 `K` 个字符串保持不变；对于末尾不够 `2 * K` 的字符串，如果长度大于 `K`，那么反转前 `K` 个字符串，剩下的保持不变。如果长度小于 `K`，则把小于 `K` 的这部分字符串全部反转。
+- 这一题是简单题，按照题意反转字符串即可。
diff --git a/Algorithms/0542. 01 Matrix/542. 01 Matrix.go b/leetcode/0542.01-Matrix/542. 01 Matrix.go
similarity index 97%
rename from Algorithms/0542. 01 Matrix/542. 01 Matrix.go
rename to leetcode/0542.01-Matrix/542. 01 Matrix.go
index 6ee0cf10b..b2f358459 100644
--- a/Algorithms/0542. 01 Matrix/542. 01 Matrix.go	
+++ b/leetcode/0542.01-Matrix/542. 01 Matrix.go	
@@ -140,3 +140,10 @@ func updateMatrixDP(matrix [][]int) [][]int {
 	}
 	return matrix
 }
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/0542.01-Matrix/542. 01 Matrix_test.go b/leetcode/0542.01-Matrix/542. 01 Matrix_test.go
new file mode 100644
index 000000000..643f4a93a
--- /dev/null
+++ b/leetcode/0542.01-Matrix/542. 01 Matrix_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question542 struct {
+	para542
+	ans542
+}
+
+// para 是参数
+// one 代表第一个参数
+type para542 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans542 struct {
+	one [][]int
+}
+
+func Test_Problem542(t *testing.T) {
+
+	qs := []question542{
+
+		{
+			para542{[][]int{{0, 0, 0}, {0, 1, 0}, {0, 0, 0}}},
+			ans542{[][]int{{0, 0, 0}, {0, 1, 0}, {0, 0, 0}}},
+		},
+
+		{
+			para542{[][]int{{0, 0, 0}, {0, 1, 0}, {1, 1, 1}}},
+			ans542{[][]int{{0, 0, 0}, {0, 1, 0}, {1, 2, 1}}},
+		},
+
+		{
+			para542{[][]int{{1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 0, 0, 1, 1}, {1, 1, 0, 0, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}}},
+			ans542{[][]int{{4, 3, 2, 2, 3, 4}, {3, 2, 1, 1, 2, 3}, {2, 1, 0, 0, 1, 2}, {2, 1, 0, 0, 1, 2}, {3, 2, 1, 1, 2, 3}, {4, 3, 2, 2, 3, 4}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 542------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans542, q.para542
+		fmt.Printf("【input】:%v       【output】:%v\n", p, updateMatrixDP(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0542. 01 Matrix/README.md b/leetcode/0542.01-Matrix/README.md
similarity index 100%
rename from Algorithms/0542. 01 Matrix/README.md
rename to leetcode/0542.01-Matrix/README.md
diff --git a/leetcode/0543.Diameter-of-Binary-Tree/543. Diameter of Binary Tree.go b/leetcode/0543.Diameter-of-Binary-Tree/543. Diameter of Binary Tree.go
new file mode 100644
index 000000000..14041158f
--- /dev/null
+++ b/leetcode/0543.Diameter-of-Binary-Tree/543. Diameter of Binary Tree.go	
@@ -0,0 +1,40 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func diameterOfBinaryTree(root *TreeNode) int {
+	result := 0
+	checkDiameter(root, &result)
+	return result
+}
+
+func checkDiameter(root *TreeNode, result *int) int {
+	if root == nil {
+		return 0
+	}
+	left := checkDiameter(root.Left, result)
+	right := checkDiameter(root.Right, result)
+	*result = max(*result, left+right)
+	return max(left, right) + 1
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0543.Diameter-of-Binary-Tree/543. Diameter of Binary Tree_test.go b/leetcode/0543.Diameter-of-Binary-Tree/543. Diameter of Binary Tree_test.go
new file mode 100644
index 000000000..e88d3aa02
--- /dev/null
+++ b/leetcode/0543.Diameter-of-Binary-Tree/543. Diameter of Binary Tree_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question543 struct {
+	para543
+	ans543
+}
+
+// para 是参数
+// one 代表第一个参数
+type para543 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans543 struct {
+	one int
+}
+
+func Test_Problem543(t *testing.T) {
+
+	qs := []question543{
+
+		{
+			para543{[]int{1, 2, 3, 4, 5}},
+			ans543{3},
+		},
+
+		{
+			para543{[]int{1, 2}},
+			ans543{1},
+		},
+
+		{
+			para543{[]int{4, -7, -3, structures.NULL, structures.NULL, -9, -3, 9, -7, -4, structures.NULL, 6, structures.NULL, -6, -6, structures.NULL, structures.NULL, 0, 6, 5, structures.NULL, 9, structures.NULL, structures.NULL, -1, -4, structures.NULL, structures.NULL, structures.NULL, -2}},
+			ans543{8},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 543------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans543, q.para543
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", diameterOfBinaryTree(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0543.Diameter-of-Binary-Tree/README.md b/leetcode/0543.Diameter-of-Binary-Tree/README.md
new file mode 100644
index 000000000..78b5655e2
--- /dev/null
+++ b/leetcode/0543.Diameter-of-Binary-Tree/README.md
@@ -0,0 +1,87 @@
+# [543. Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/)
+
+
+## 题目
+
+Given the `root` of a binary tree, return *the length of the **diameter** of the tree*.
+
+The **diameter** of a binary tree is the **length** of the longest path between any two nodes in a tree. This path may or may not pass through the `root`.
+
+The **length** of a path between two nodes is represented by the number of edges between them.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/03/06/diamtree.jpg](https://assets.leetcode.com/uploads/2021/03/06/diamtree.jpg)
+
+```
+Input: root = [1,2,3,4,5]
+Output: 3
+Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].
+
+```
+
+**Example 2:**
+
+```
+Input: root = [1,2]
+Output: 1
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 104]`.
+- `100 <= Node.val <= 100`
+
+## 题目大意
+
+给定一棵二叉树，你需要计算它的直径长度。一棵二叉树的直径长度是任意两个结点路径长度中的最大值。这条路径可能穿过也可能不穿过根结点。
+
+## 解题思路
+
+- 简单题。遍历每个节点的左子树和右子树，累加从左子树到右子树的最大长度。遍历每个节点时，动态更新这个最大长度即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func diameterOfBinaryTree(root *TreeNode) int {
+	result := 0
+	checkDiameter(root, &result)
+	return result
+}
+
+func checkDiameter(root *TreeNode, result *int) int {
+	if root == nil {
+		return 0
+	}
+	left := checkDiameter(root.Left, result)
+	right := checkDiameter(root.Right, result)
+	*result = max(*result, left+right)
+	return max(left, right) + 1
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0547. Friend Circles/547. Friend Circles.go b/leetcode/0547.Number-of-Provinces/547. Number of Provinces.go
similarity index 100%
rename from Algorithms/0547. Friend Circles/547. Friend Circles.go
rename to leetcode/0547.Number-of-Provinces/547. Number of Provinces.go
diff --git a/leetcode/0547.Number-of-Provinces/547. Number of Provinces_test.go b/leetcode/0547.Number-of-Provinces/547. Number of Provinces_test.go
new file mode 100644
index 000000000..79c4bda88
--- /dev/null
+++ b/leetcode/0547.Number-of-Provinces/547. Number of Provinces_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question547 struct {
+	para547
+	ans547
+}
+
+// para 是参数
+// one 代表第一个参数
+type para547 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans547 struct {
+	one int
+}
+
+func Test_Problem547(t *testing.T) {
+
+	qs := []question547{
+
+		{
+			para547{[][]int{{0, 0, 0}, {0, 1, 0}, {0, 0, 0}}},
+			ans547{3},
+		},
+
+		{
+			para547{[][]int{{1, 1, 0}, {1, 1, 0}, {0, 0, 1}}},
+			ans547{2},
+		},
+
+		{
+			para547{[][]int{{1, 1, 0}, {1, 1, 1}, {0, 1, 1}}},
+			ans547{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 547------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans547, q.para547
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findCircleNum(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0547.Number-of-Provinces/README.md b/leetcode/0547.Number-of-Provinces/README.md
new file mode 100755
index 000000000..0dafa9ba9
--- /dev/null
+++ b/leetcode/0547.Number-of-Provinces/README.md
@@ -0,0 +1,54 @@
+# [547. Number of Provinces](https://leetcode.com/problems/number-of-provinces/)
+
+## 题目
+
+There are **N** students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a **direct** friend of B, and B is a **direct**friend of C, then A is an **indirect** friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.
+
+Given a **N*N** matrix **M** representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are **direct** friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.
+
+**Example 1:**
+
+    Input: 
+    [[1,1,0],
+     [1,1,0],
+     [0,0,1]]
+    Output: 2
+    Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. 
+    The 2nd student himself is in a friend circle. So return 2.
+
+**Example 2:**
+
+    Input: 
+    [[1,1,0],
+     [1,1,1],
+     [0,1,1]]
+    Output: 1
+    Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, 
+    so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.
+
+**Note:**
+
+1. N is in range [1,200].
+2. M[i][i] = 1 for all students.
+3. If M[i][j] = 1, then M[j][i] = 1.
+
+
+## 题目大意
+
+班上有 N 名学生。其中有些人是朋友，有些则不是。他们的友谊具有是传递性。如果已知 A 是 B 的朋友，B 是 C 的朋友，那么我们可以认为 A 也是 C 的朋友。所谓的朋友圈，是指所有朋友的集合。
+
+给定一个 N * N 的矩阵 M，表示班级中学生之间的朋友关系。如果 M[i][j] = 1，表示已知第 i 个和 j 个学生互为朋友关系，否则为不知道。你必须输出所有学生中的已知的朋友圈总数。
+
+
+注意：
+
+- N 在[1,200]的范围内。
+- 对于所有学生，有M[i][i] = 1。
+- 如果有 M[i][j] = 1，则有 M[j][i] = 1。
+
+
+## 解题思路
+
+
+- 给出一个二维矩阵，矩阵中的行列表示的是两个人之间是否是朋友关系，如果是 1，代表两个人是朋友关系。由于自己和自肯定朋友关系，所以对角线上都是 1，并且矩阵也是关于从左往右下的这条对角线对称。
+- 这题有 2 种解法，第一种解法是并查集，依次扫描矩阵，如果两个人认识，并且 root 并不相等就执行 union 操作。扫完所有矩阵，最后数一下还有几个不同的 root 就是最终答案。第二种解法是 DFS 或者 BFS。利用 FloodFill 的想法去染色，每次染色一次，计数器加一。最终扫完整个矩阵，计数器的结果就是最终结果。
diff --git a/leetcode/0551.Student-Attendance-Record-I/551.Student Attendance Record I.go b/leetcode/0551.Student-Attendance-Record-I/551.Student Attendance Record I.go
new file mode 100644
index 000000000..a47630c22
--- /dev/null
+++ b/leetcode/0551.Student-Attendance-Record-I/551.Student Attendance Record I.go	
@@ -0,0 +1,22 @@
+package leetcode
+
+func checkRecord(s string) bool {
+	numsA, maxL, numsL := 0, 0, 0
+	for _, v := range s {
+		if v == 'L' {
+			numsL++
+		} else {
+			if numsL > maxL {
+				maxL = numsL
+			}
+			numsL = 0
+			if v == 'A' {
+				numsA++
+			}
+		}
+	}
+	if numsL > maxL {
+		maxL = numsL
+	}
+	return numsA < 2 && maxL < 3
+}
diff --git a/leetcode/0551.Student-Attendance-Record-I/551.Student Attendance Record I_test.go b/leetcode/0551.Student-Attendance-Record-I/551.Student Attendance Record I_test.go
new file mode 100644
index 000000000..cc9ace7a5
--- /dev/null
+++ b/leetcode/0551.Student-Attendance-Record-I/551.Student Attendance Record I_test.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question551 struct {
+	para551
+	ans551
+}
+
+// para 是参数
+type para551 struct {
+	s string
+}
+
+// ans 是答案
+type ans551 struct {
+	ans bool
+}
+
+func Test_Problem551(t *testing.T) {
+
+	qs := []question551{
+
+		{
+			para551{"PPALLP"},
+			ans551{true},
+		},
+
+		{
+			para551{"PPALLL"},
+			ans551{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 551------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans551, q.para551
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, checkRecord(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0551.Student-Attendance-Record-I/README.md b/leetcode/0551.Student-Attendance-Record-I/README.md
new file mode 100644
index 000000000..fe8a01616
--- /dev/null
+++ b/leetcode/0551.Student-Attendance-Record-I/README.md
@@ -0,0 +1,86 @@
+# [551. Student Attendance Record I](https://leetcode.com/problems/student-attendance-record-i/)
+
+## 题目
+
+You are given a string `s` representing an attendance record for a student where each character signifies whether the student was absent, late, or present on that day. The record only contains the following three characters:
+
+- `'A'`: Absent.
+- `'L'`: Late.
+- `'P'`: Present.
+
+The student is eligible for an attendance award if they meet **both** of the following criteria:
+
+- The student was absent (`'A'`) for **strictly** fewer than 2 days **total**.
+- The student was **never** late (`'L'`) for 3 or more **consecutive** days.
+
+Return `true` *if the student is eligible for an attendance award, or* `false` *otherwise*.
+
+**Example 1:**
+
+```
+Input: s = "PPALLP"
+Output: true
+Explanation: The student has fewer than 2 absences and was never late 3 or more consecutive days.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "PPALLL"
+Output: false
+Explanation: The student was late 3 consecutive days in the last 3 days, so is not eligible for the award.
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 1000`
+- `s[i]` is either `'A'`, `'L'`, or `'P'`.
+
+## 题目大意
+
+给你一个字符串 s 表示一个学生的出勤记录，其中的每个字符用来标记当天的出勤情况（缺勤、迟到、到场）。记录中只含下面三种字符：
+
+- 'A'：Absent，缺勤
+- 'L'：Late，迟到
+- 'P'：Present，到场
+
+如果学生能够 同时 满足下面两个条件，则可以获得出勤奖励：
+
+- 按 总出勤 计，学生缺勤（'A'）严格 少于两天。
+- 学生 不会 存在 连续 3 天或 连续 3 天以上的迟到（'L'）记录。
+
+如果学生可以获得出勤奖励，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 遍历字符串 s 求出 'A' 的总数量和连续 'L' 的最大数量。
+- 比较 'A' 的数量是否小于 2 并且 'L' 的连续最大数量是否小于 3。
+
+## 代码
+
+```go
+package leetcode
+
+func checkRecord(s string) bool {
+	numsA, maxL, numsL := 0, 0, 0
+	for _, v := range s {
+		if v == 'L' {
+			numsL++
+		} else {
+			if numsL > maxL {
+				maxL = numsL
+			}
+			numsL = 0
+			if v == 'A' {
+				numsA++
+			}
+		}
+	}
+	if numsL > maxL {
+		maxL = numsL
+	}
+	return numsA < 2 && maxL < 3
+}
+```
\ No newline at end of file
diff --git a/leetcode/0554.Brick-Wall/554. Brick Wall.go b/leetcode/0554.Brick-Wall/554. Brick Wall.go
new file mode 100644
index 000000000..022f22010
--- /dev/null
+++ b/leetcode/0554.Brick-Wall/554. Brick Wall.go	
@@ -0,0 +1,19 @@
+package leetcode
+
+func leastBricks(wall [][]int) int {
+	m := make(map[int]int)
+	for _, row := range wall {
+		sum := 0
+		for i := 0; i < len(row)-1; i++ {
+			sum += row[i]
+			m[sum]++
+		}
+	}
+	max := 0
+	for _, v := range m {
+		if v > max {
+			max = v
+		}
+	}
+	return len(wall) - max
+}
diff --git a/leetcode/0554.Brick-Wall/554. Brick Wall_test.go b/leetcode/0554.Brick-Wall/554. Brick Wall_test.go
new file mode 100644
index 000000000..00dc2d53c
--- /dev/null
+++ b/leetcode/0554.Brick-Wall/554. Brick Wall_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question554 struct {
+	para554
+	ans554
+}
+
+// para 是参数
+// one 代表第一个参数
+type para554 struct {
+	wall [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans554 struct {
+	one int
+}
+
+func Test_Problem554(t *testing.T) {
+
+	qs := []question554{
+
+		{
+			para554{[][]int{{1, 2, 2, 1}, {3, 1, 2}, {1, 3, 2}, {2, 4}, {3, 1, 2}, {1, 3, 1, 1}}},
+			ans554{2},
+		},
+
+		{
+			para554{[][]int{{1}, {1}, {1}}},
+			ans554{3},
+		},
+
+		{
+			para554{[][]int{{1, 1, 0}, {1, 1, 0}, {0, 0, 1}}},
+			ans554{1},
+		},
+
+		{
+			para554{[][]int{{1, 1, 0}, {1, 1, 1}, {0, 1, 1}}},
+			ans554{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 554------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans554, q.para554
+		fmt.Printf("【input】:%v       【output】:%v\n", p, leastBricks(p.wall))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0554.Brick-Wall/README.md b/leetcode/0554.Brick-Wall/README.md
new file mode 100644
index 000000000..0a16d12b8
--- /dev/null
+++ b/leetcode/0554.Brick-Wall/README.md
@@ -0,0 +1,68 @@
+# [554. Brick Wall](https://leetcode.com/problems/brick-wall/)
+
+## 题目
+
+There is a rectangular brick wall in front of you with `n` rows of bricks. The `ith` row has some number of bricks each of the same height (i.e., one unit) but they can be of different widths. The total width of each row is the same.
+
+Draw a vertical line from the top to the bottom and cross the least bricks. If your line goes through the edge of a brick, then the brick is not considered as crossed. You cannot draw a line just along one of the two vertical edges of the wall, in which case the line will obviously cross no bricks.
+
+Given the 2D array `wall` that contains the information about the wall, return *the minimum number of crossed bricks after drawing such a vertical line*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/04/24/cutwall-grid.jpg](https://assets.leetcode.com/uploads/2021/04/24/cutwall-grid.jpg)
+
+```
+Input: wall = [[1,2,2,1],[3,1,2],[1,3,2],[2,4],[3,1,2],[1,3,1,1]]
+Output: 2
+
+```
+
+**Example 2:**
+
+```
+Input: wall = [[1],[1],[1]]
+Output: 3
+
+```
+
+**Constraints:**
+
+- `n == wall.length`
+- `1 <= n <= 10^4`
+- `1 <= wall[i].length <= 10^4`
+- `1 <= sum(wall[i].length) <= 2 * 10^4`
+- `sum(wall[i])` is the same for each row `i`.
+- `1 <= wall[i][j] <= 2^31 - 1`
+
+## 题目大意
+
+你的面前有一堵矩形的、由 n 行砖块组成的砖墙。这些砖块高度相同（也就是一个单位高）但是宽度不同。每一行砖块的宽度之和应该相等。你现在要画一条 自顶向下 的、穿过 最少 砖块的垂线。如果你画的线只是从砖块的边缘经过，就不算穿过这块砖。你不能沿着墙的两个垂直边缘之一画线，这样显然是没有穿过一块砖的。给你一个二维数组 wall ，该数组包含这堵墙的相关信息。其中，wall[i] 是一个代表从左至右每块砖的宽度的数组。你需要找出怎样画才能使这条线 穿过的砖块数量最少 ，并且返回 穿过的砖块数量 。
+
+## 解题思路
+
+- 既然穿过砖块中缝不算穿过砖块，那么穿过最少砖块数量一定是穿过很多中缝。按行遍历每一行的砖块，累加每行砖块宽度，将每行砖块“缝”的坐标存在 map 中。最后取出 map 中出现频次最高的缝，即为铅垂线要穿过的地方。墙高减去缝出现的频次，剩下的即为穿过砖块的数量。
+
+## 代码
+
+```go
+package leetcode
+
+func leastBricks(wall [][]int) int {
+	m := make(map[int]int)
+	for _, row := range wall {
+		sum := 0
+		for i := 0; i < len(row)-1; i++ {
+			sum += row[i]
+			m[sum]++
+		}
+	}
+	max := 0
+	for _, v := range m {
+		if v > max {
+			max = v
+		}
+	}
+	return len(wall) - max
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0557. Reverse Words in a String III/557. Reverse Words in a String III.go b/leetcode/0557.Reverse-Words-in-a-String-III/557. Reverse Words in a String III.go
similarity index 100%
rename from Algorithms/0557. Reverse Words in a String III/557. Reverse Words in a String III.go
rename to leetcode/0557.Reverse-Words-in-a-String-III/557. Reverse Words in a String III.go
diff --git a/Algorithms/0557. Reverse Words in a String III/557. Reverse Words in a String III_test.go b/leetcode/0557.Reverse-Words-in-a-String-III/557. Reverse Words in a String III_test.go
similarity index 95%
rename from Algorithms/0557. Reverse Words in a String III/557. Reverse Words in a String III_test.go
rename to leetcode/0557.Reverse-Words-in-a-String-III/557. Reverse Words in a String III_test.go
index 94b9e49d6..812e5e78c 100644
--- a/Algorithms/0557. Reverse Words in a String III/557. Reverse Words in a String III_test.go	
+++ b/leetcode/0557.Reverse-Words-in-a-String-III/557. Reverse Words in a String III_test.go	
@@ -26,12 +26,12 @@ func Test_Problem557(t *testing.T) {
 
 	qs := []question557{
 
-		question557{
+		{
 			para557{"Let's take LeetCode contest"},
 			ans557{"s'teL ekat edoCteeL tsetnoc"},
 		},
 
-		question557{
+		{
 			para557{""},
 			ans557{""},
 		},
diff --git a/leetcode/0557.Reverse-Words-in-a-String-III/README.md b/leetcode/0557.Reverse-Words-in-a-String-III/README.md
new file mode 100755
index 000000000..40de4c64d
--- /dev/null
+++ b/leetcode/0557.Reverse-Words-in-a-String-III/README.md
@@ -0,0 +1,25 @@
+# [557. Reverse Words in a String III](https://leetcode.com/problems/reverse-words-in-a-string-iii/)
+
+
+## 题目
+
+Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.
+
+**Example 1:**
+
+    Input: "Let's take LeetCode contest"
+    Output: "s'teL ekat edoCteeL tsetnoc"
+
+**Note:** In the string, each word is separated by single space and there will not be any extra space in the string.
+
+
+## 题目大意
+
+给定一个字符串，你需要反转字符串中每个单词的字符顺序，同时仍保留空格和单词的初始顺序。注意：在字符串中，每个单词由单个空格分隔，并且字符串中不会有任何额外的空格。
+
+
+## 解题思路
+
+
+- 反转字符串，要求按照空格隔开的小字符串为单位反转。
+- 这是一道简单题。按照题意反转每个空格隔开的单词即可。
diff --git a/leetcode/0559.Maximum-Depth-of-N-ary-Tree/559.Maximum Depth of N-ary Tree.go b/leetcode/0559.Maximum-Depth-of-N-ary-Tree/559.Maximum Depth of N-ary Tree.go
new file mode 100644
index 000000000..30fbc9439
--- /dev/null
+++ b/leetcode/0559.Maximum-Depth-of-N-ary-Tree/559.Maximum Depth of N-ary Tree.go	
@@ -0,0 +1,32 @@
+package leetcode
+
+type Node struct {
+	Val      int
+	Children []*Node
+}
+
+func maxDepth(root *Node) int {
+	if root == nil {
+		return 0
+	}
+	return 1 + bfs(root)
+}
+
+func bfs(root *Node) int {
+	var q []*Node
+	var depth int
+	q = append(q, root.Children...)
+	for len(q) != 0 {
+		depth++
+		length := len(q)
+		for length != 0 {
+			ele := q[0]
+			q = q[1:]
+			length--
+			if ele != nil && len(ele.Children) != 0 {
+				q = append(q, ele.Children...)
+			}
+		}
+	}
+	return depth
+}
diff --git a/leetcode/0559.Maximum-Depth-of-N-ary-Tree/559.Maximum Depth of N-ary Tree_test.go b/leetcode/0559.Maximum-Depth-of-N-ary-Tree/559.Maximum Depth of N-ary Tree_test.go
new file mode 100644
index 000000000..90da19b30
--- /dev/null
+++ b/leetcode/0559.Maximum-Depth-of-N-ary-Tree/559.Maximum Depth of N-ary Tree_test.go	
@@ -0,0 +1,60 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question559 struct {
+	para559
+	ans559
+}
+
+// para 是参数
+type para559 struct {
+	root *Node
+}
+
+// ans 是答案
+type ans559 struct {
+	ans int
+}
+
+func Test_Problem559(t *testing.T) {
+
+	qs := []question559{
+
+		{
+			para559{&Node{
+				Val: 1,
+				Children: []*Node{
+					{Val: 3, Children: []*Node{{Val: 5, Children: nil}, {Val: 6, Children: nil}}},
+					{Val: 2, Children: nil},
+					{Val: 4, Children: nil},
+				},
+			}},
+			ans559{3},
+		},
+
+		{
+			para559{&Node{
+				Val: 1,
+				Children: []*Node{
+					{Val: 2, Children: nil},
+					{Val: 3, Children: []*Node{{Val: 6, Children: nil}, {Val: 7, Children: []*Node{{Val: 11, Children: []*Node{{Val: 14, Children: nil}}}}}}},
+					{Val: 4, Children: []*Node{{Val: 8, Children: []*Node{{Val: 12, Children: nil}}}}},
+					{Val: 5, Children: []*Node{{Val: 9, Children: []*Node{{Val: 13, Children: nil}}}, {Val: 10, Children: nil}}},
+				},
+			}},
+			ans559{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 559------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans559, q.para559
+		fmt.Printf("【input】:%v    【output】:%v\n", p, maxDepth(p.root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0559.Maximum-Depth-of-N-ary-Tree/README.md b/leetcode/0559.Maximum-Depth-of-N-ary-Tree/README.md
new file mode 100644
index 000000000..9cb575096
--- /dev/null
+++ b/leetcode/0559.Maximum-Depth-of-N-ary-Tree/README.md
@@ -0,0 +1,78 @@
+# [559. Maximum Depth of N-ary Tree](https://leetcode.com/problems/maximum-depth-of-n-ary-tree/)
+
+## 题目
+
+Given a n-ary tree, find its maximum depth.
+
+The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
+
+Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png](https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png)
+
+    Input: root = [1,null,3,2,4,null,5,6]
+    Output: 3
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png](https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png)
+
+    Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
+    Output: 5
+
+**Constraints:**
+
+- The total number of nodes is in the range [0, 10000].
+- The depth of the n-ary tree is less than or equal to 1000.
+
+## 题目大意
+
+给定一个 N 叉树，找到其最大深度。
+
+最大深度是指从根节点到最远叶子节点的最长路径上的节点总数。
+
+N 叉树输入按层序遍历序列化表示，每组子节点由空值分隔（请参见示例）。
+
+## 解题思路
+
+- 使用广度优先遍历
+
+## 代码
+
+```go
+
+package leetcode
+
+type Node struct {
+	Val int
+	Children []*Node
+}
+
+func maxDepth(root *Node) int {
+	if root == nil {
+		return 0
+	}
+	return 1 + bfs(root)
+}
+
+func bfs(root *Node) int {
+	var q []*Node
+	var depth int
+	q = append(q, root.Children...)
+	for len(q) != 0 {
+		depth++
+		length := len(q)
+		for length != 0 {
+			ele := q[0]
+			q = q[1:]
+			length--
+			if ele != nil && len(ele.Children) != 0 {
+				q = append(q, ele.Children...)
+			}
+		}
+	}
+	return depth
+}
+```
\ No newline at end of file
diff --git a/leetcode/0560.Subarray-Sum-Equals-K/560. Subarray Sum Equals K.go b/leetcode/0560.Subarray-Sum-Equals-K/560. Subarray Sum Equals K.go
new file mode 100644
index 000000000..0ca17f3a5
--- /dev/null
+++ b/leetcode/0560.Subarray-Sum-Equals-K/560. Subarray Sum Equals K.go	
@@ -0,0 +1,15 @@
+package leetcode
+
+func subarraySum(nums []int, k int) int {
+	count, pre := 0, 0
+	m := map[int]int{}
+	m[0] = 1
+	for i := 0; i < len(nums); i++ {
+		pre += nums[i]
+		if _, ok := m[pre-k]; ok {
+			count += m[pre-k]
+		}
+		m[pre] += 1
+	}
+	return count
+}
diff --git a/leetcode/0560.Subarray-Sum-Equals-K/560. Subarray Sum Equals K_test.go b/leetcode/0560.Subarray-Sum-Equals-K/560. Subarray Sum Equals K_test.go
new file mode 100644
index 000000000..e9edf9557
--- /dev/null
+++ b/leetcode/0560.Subarray-Sum-Equals-K/560. Subarray Sum Equals K_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question560 struct {
+	para560
+	ans560
+}
+
+// para 是参数
+// one 代表第一个参数
+type para560 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans560 struct {
+	one int
+}
+
+func Test_Problem560(t *testing.T) {
+
+	qs := []question560{
+
+		{
+			para560{[]int{1, 1, 1}, 2},
+			ans560{2},
+		},
+
+		{
+			para560{[]int{1, 2, 3}, 3},
+			ans560{2},
+		},
+
+		{
+			para560{[]int{1}, 0},
+			ans560{0},
+		},
+
+		{
+			para560{[]int{-1, -1, 1}, 0},
+			ans560{1},
+		},
+
+		{
+			para560{[]int{1, -1, 0}, 0},
+			ans560{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 560------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans560, q.para560
+		fmt.Printf("【input】:%v       【output】:%v\n", p, subarraySum(p.nums, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0560.Subarray-Sum-Equals-K/README.md b/leetcode/0560.Subarray-Sum-Equals-K/README.md
new file mode 100644
index 000000000..13b588cca
--- /dev/null
+++ b/leetcode/0560.Subarray-Sum-Equals-K/README.md
@@ -0,0 +1,58 @@
+# [560. Subarray Sum Equals K](https://leetcode.com/problems/subarray-sum-equals-k/)
+
+
+## 题目
+
+Given an array of integers `nums` and an integer `k`, return *the total number of continuous subarrays whose sum equals to `k`*.
+
+**Example 1:**
+
+```
+Input: nums = [1,1,1], k = 2
+Output: 2
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3], k = 3
+Output: 2
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 2 * 104`
+- `-1000 <= nums[i] <= 1000`
+- `-10^7 <= k <= 10^7`
+
+## 题目大意
+
+给你一个整数数组 `nums` 和一个整数 `k` ，请你统计并返回该数组中和为 `k` ****的连续子数组的个数。
+
+## 解题思路
+
+- 此题不能使用滑动窗口来解。因为 `nums[i]` 可能为负数。
+- 前缀和的思路可以解答此题，但是时间复杂度有点高了，`O(n^2)`。考虑优化时间复杂度。
+- 题目要求找到连续区间和为 `k` 的子区间总数，即区间 `[i,j]` 内的和为 K ⇒ `prefixSum[j] - prefixSum[i-1] == k`。所以 `prefixSum[j] == k - prefixSum[i-1]` 。这样转换以后，题目就转换成类似 A + B = K 的问题了。LeetCode 第一题的优化思路拿来用。用 map 存储累加过的结果。如此优化以后，时间复杂度 `O(n)`。
+
+## 代码
+
+```go
+package leetcode
+
+func subarraySum(nums []int, k int) int {
+	count, pre := 0, 0
+	m := map[int]int{}
+	m[0] = 1
+	for i := 0; i < len(nums); i++ {
+		pre += nums[i]
+		if _, ok := m[pre-k]; ok {
+			count += m[pre-k]
+		}
+		m[pre] += 1
+	}
+	return count
+}
+```
\ No newline at end of file
diff --git a/leetcode/0561.Array-Partition/561. Array Partition.go b/leetcode/0561.Array-Partition/561. Array Partition.go
new file mode 100644
index 000000000..c566c01bb
--- /dev/null
+++ b/leetcode/0561.Array-Partition/561. Array Partition.go	
@@ -0,0 +1,19 @@
+package leetcode
+
+func arrayPairSum(nums []int) int {
+	array := [20001]int{}
+	for i := 0; i < len(nums); i++ {
+		array[nums[i]+10000]++
+	}
+	flag, sum := true, 0
+	for i := 0; i < len(array); i++ {
+		for array[i] > 0 {
+			if flag {
+				sum = sum + i - 10000
+			}
+			flag = !flag
+			array[i]--
+		}
+	}
+	return sum
+}
diff --git a/leetcode/0561.Array-Partition/561. Array Partition_test.go b/leetcode/0561.Array-Partition/561. Array Partition_test.go
new file mode 100644
index 000000000..967365d59
--- /dev/null
+++ b/leetcode/0561.Array-Partition/561. Array Partition_test.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question561 struct {
+	para561
+	ans561
+}
+
+// para 是参数
+// one 代表第一个参数
+type para561 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans561 struct {
+	one int
+}
+
+func Test_Problem561(t *testing.T) {
+
+	qs := []question561{
+
+		{
+			para561{[]int{}},
+			ans561{0},
+		},
+
+		{
+			para561{[]int{1, 4, 3, 2}},
+			ans561{4},
+		},
+
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 561------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans561, q.para561
+		fmt.Printf("【input】:%v       【output】:%v\n", p, arrayPairSum(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0561.Array-Partition/README.md b/leetcode/0561.Array-Partition/README.md
new file mode 100644
index 000000000..244bec8c9
--- /dev/null
+++ b/leetcode/0561.Array-Partition/README.md
@@ -0,0 +1,58 @@
+# [561. Array Partition](https://leetcode.com/problems/array-partition/)
+
+
+## 题目
+
+Given an array of **2n** integers, your task is to group these integers into **n** pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.
+
+**Example 1**:
+
+```
+Input: [1,4,3,2]
+
+Output: 4
+Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).
+```
+
+**Note**:
+
+1. **n** is a positive integer, which is in the range of [1, 10000].
+2. All the integers in the array will be in the range of [-10000, 10000].
+
+## 题目大意
+
+给定长度为 2n 的数组, 你的任务是将这些数分成 n 对, 例如 (a1, b1), (a2, b2), ..., (an, bn) ，使得从1 到 n 的 min(ai, bi) 总和最大。
+
+
+## 解题思路
+
+- 给定一个 2n 个数组，要求把它们分为 n 组一行，求出各组最小值的总和的最大值。
+- 由于题目给的数据范围不大，[-10000, 10000]，所以我们可以考虑用一个哈希表数组，里面存储 i - 10000 元素的频次，偏移量是 10000。这个哈希表能按递增的顺序访问数组，这样可以减少排序的耗时。题目要求求出分组以后求和的最大值，那么所有偏小的元素尽量都安排在一组里面，这样取 min 以后，对最大和影响不大。例如，(1 , 1) 这样安排在一起，min 以后就是 1 。但是如果把相差很大的两个元素安排到一起，那么较大的那个元素就“牺牲”了。例如，(1 , 10000)，取 min 以后就是 1，于是 10000 就“牺牲”了。所以需要优先考虑较小值。
+- 较小值出现的频次可能是奇数也可能是偶数。如果是偶数，那比较简单，把它们俩俩安排在一起就可以了。如果是奇数，那么它会落单一次，落单的那个需要和距离它最近的一个元素进行配对，这样对最终的和影响最小。较小值如果是奇数，那么就会影响后面元素的选择，后面元素如果是偶数，由于需要一个元素和前面的较小值配对，所以它剩下的又是奇数个。这个影响会依次传递到后面。所以用一个 flag 标记，如果当前集合中有剩余元素将被再次考虑，则此标志设置为 1。在从下一组中选择元素时，会考虑已考虑的相同额外元素。
+- 最后扫描过程中动态的维护 sum 值就可以了。
+
+## 代码
+
+```go
+
+package leetcode
+
+func arrayPairSum(nums []int) int {
+	array := [20001]int{}
+	for i := 0; i < len(nums); i++ {
+		array[nums[i]+10000]++
+	}
+	flag, sum := true, 0
+	for i := 0; i < len(array); i++ {
+		for array[i] > 0 {
+			if flag {
+				sum = sum + i - 10000
+			}
+			flag = !flag
+			array[i]--
+		}
+	}
+	return sum
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/0563.Binary-Tree-Tilt/563. Binary Tree Tilt.go b/leetcode/0563.Binary-Tree-Tilt/563. Binary Tree Tilt.go
new file mode 100644
index 000000000..96127e720
--- /dev/null
+++ b/leetcode/0563.Binary-Tree-Tilt/563. Binary Tree Tilt.go	
@@ -0,0 +1,38 @@
+package leetcode
+
+import (
+	"math"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func findTilt(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	sum := 0
+	findTiltDFS(root, &sum)
+	return sum
+}
+
+func findTiltDFS(root *TreeNode, sum *int) int {
+	if root == nil {
+		return 0
+	}
+	left := findTiltDFS(root.Left, sum)
+	right := findTiltDFS(root.Right, sum)
+	*sum += int(math.Abs(float64(left) - float64(right)))
+	return root.Val + left + right
+}
diff --git a/leetcode/0563.Binary-Tree-Tilt/563. Binary Tree Tilt_test.go b/leetcode/0563.Binary-Tree-Tilt/563. Binary Tree Tilt_test.go
new file mode 100644
index 000000000..dcf6a3bfe
--- /dev/null
+++ b/leetcode/0563.Binary-Tree-Tilt/563. Binary Tree Tilt_test.go	
@@ -0,0 +1,66 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question563 struct {
+	para563
+	ans563
+}
+
+// para 是参数
+// one 代表第一个参数
+type para563 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans563 struct {
+	one int
+}
+
+func Test_Problem563(t *testing.T) {
+
+	qs := []question563{
+
+		{
+			para563{[]int{}},
+			ans563{0},
+		},
+
+		{
+			para563{[]int{1}},
+			ans563{0},
+		},
+
+		{
+			para563{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
+			ans563{41},
+		},
+
+		{
+			para563{[]int{1, 2, 3, 4, structures.NULL, structures.NULL, 5}},
+			ans563{11},
+		},
+
+		{
+			para563{[]int{1, 2, 3, 4, structures.NULL, 5}},
+			ans563{11},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 563------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans563, q.para563
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", findTilt(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0563.Binary-Tree-Tilt/README.md b/leetcode/0563.Binary-Tree-Tilt/README.md
new file mode 100755
index 000000000..5f4f70164
--- /dev/null
+++ b/leetcode/0563.Binary-Tree-Tilt/README.md
@@ -0,0 +1,45 @@
+# [563. Binary Tree Tilt](https://leetcode.com/problems/binary-tree-tilt/)
+
+
+## 题目
+
+Given a binary tree, return the tilt of the **whole tree**.
+
+The tilt of a **tree node** is defined as the **absolute difference** between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.
+
+The tilt of the **whole tree** is defined as the sum of all nodes' tilt.
+
+**Example:**
+
+    Input: 
+             1
+           /   \
+          2     3
+    Output: 1
+    Explanation: 
+    Tilt of node 2 : 0
+    Tilt of node 3 : 0
+    Tilt of node 1 : |2-3| = 1
+    Tilt of binary tree : 0 + 0 + 1 = 1
+
+**Note:**
+
+1. The sum of node values in any subtree won't exceed the range of 32-bit integer.
+2. All the tilt values won't exceed the range of 32-bit integer.
+
+
+## 题目大意
+
+
+给定一个二叉树，计算整个树的坡度。一个树的节点的坡度定义即为，该节点左子树的结点之和和右子树结点之和的差的绝对值。空结点的的坡度是0。整个树的坡度就是其所有节点的坡度之和。
+
+注意:
+
+1. 任何子树的结点的和不会超过32位整数的范围。
+2. 坡度的值不会超过32位整数的范围。
+
+## 解题思路
+
+
+- 给出一棵树，计算每个节点的“倾斜度”累加和。“倾斜度”的定义是：左子树和右子树的节点值差值的绝对值。
+- 这一题虽然是简单题，但是如果对题目中的“倾斜度”理解的不对，这一题就会出错。“倾斜度”计算的是左子树所有节点的值总和，和，右子树所有节点的值总和的差值。并不是只针对一个节点的左节点值和右节点值的差值。这一点明白以后，这一题就是简单题了。
diff --git a/Algorithms/0566. Reshape the Matrix/566. Reshape the Matrix.go b/leetcode/0566.Reshape-the-Matrix/566. Reshape the Matrix.go
similarity index 100%
rename from Algorithms/0566. Reshape the Matrix/566. Reshape the Matrix.go
rename to leetcode/0566.Reshape-the-Matrix/566. Reshape the Matrix.go
diff --git a/leetcode/0566.Reshape-the-Matrix/566. Reshape the Matrix_test.go b/leetcode/0566.Reshape-the-Matrix/566. Reshape the Matrix_test.go
new file mode 100644
index 000000000..4e33225bb
--- /dev/null
+++ b/leetcode/0566.Reshape-the-Matrix/566. Reshape the Matrix_test.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question566 struct {
+	para566
+	ans566
+}
+
+// para 是参数
+// one 代表第一个参数
+type para566 struct {
+	nums [][]int
+	r    int
+	c    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans566 struct {
+	one [][]int
+}
+
+func Test_Problem566(t *testing.T) {
+
+	qs := []question566{
+
+		{
+			para566{[][]int{{1, 2}, {3, 4}}, 1, 4},
+			ans566{[][]int{{1, 2, 3, 4}}},
+		},
+
+		{
+			para566{[][]int{{1, 2}, {3, 4}}, 2, 4},
+			ans566{[][]int{{1, 2}, {3, 4}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 566------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans566, q.para566
+		fmt.Printf("【input】:%v       【output】:%v\n", p, matrixReshape(p.nums, p.r, p.c))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0566. Reshape the Matrix/README.md b/leetcode/0566.Reshape-the-Matrix/README.md
similarity index 100%
rename from Algorithms/0566. Reshape the Matrix/README.md
rename to leetcode/0566.Reshape-the-Matrix/README.md
diff --git a/Algorithms/0567. Permutation in String/567. Permutation in String.go b/leetcode/0567.Permutation-in-String/567. Permutation in String.go
similarity index 100%
rename from Algorithms/0567. Permutation in String/567. Permutation in String.go
rename to leetcode/0567.Permutation-in-String/567. Permutation in String.go
diff --git a/Algorithms/0567. Permutation in String/567. Permutation in String_test.go b/leetcode/0567.Permutation-in-String/567. Permutation in String_test.go
similarity index 91%
rename from Algorithms/0567. Permutation in String/567. Permutation in String_test.go
rename to leetcode/0567.Permutation-in-String/567. Permutation in String_test.go
index f4c0abf9b..ccef0995f 100644
--- a/Algorithms/0567. Permutation in String/567. Permutation in String_test.go	
+++ b/leetcode/0567.Permutation-in-String/567. Permutation in String_test.go	
@@ -27,27 +27,27 @@ func Test_Problem567(t *testing.T) {
 
 	qs := []question567{
 
-		question567{
+		{
 			para567{"ab", "abab"},
 			ans567{true},
 		},
 
-		question567{
+		{
 			para567{"abc", "cbaebabacd"},
 			ans567{true},
 		},
 
-		question567{
+		{
 			para567{"abc", ""},
 			ans567{false},
 		},
 
-		question567{
+		{
 			para567{"abc", "abacbabc"},
 			ans567{true},
 		},
 
-		question567{
+		{
 			para567{"ab", "eidboaoo"},
 			ans567{false},
 		},
diff --git a/leetcode/0567.Permutation-in-String/README.md b/leetcode/0567.Permutation-in-String/README.md
new file mode 100644
index 000000000..526851a24
--- /dev/null
+++ b/leetcode/0567.Permutation-in-String/README.md
@@ -0,0 +1,44 @@
+# [567. Permutation in String](https://leetcode.com/problems/permutation-in-string/)
+
+## 题目
+
+Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.
+
+
+Example 1:
+
+```c
+Input:s1 = "ab" s2 = "eidbaooo"
+Output:True
+Explanation: s2 contains one permutation of s1 ("ba").
+```
+
+Example 2:
+
+```c
+Input:s1= "ab" s2 = "eidboaoo"
+Output: False
+```
+
+Note:
+
+1. The input strings only contain lower case letters.
+2. The length of both given strings is in range [1, 10,000].
+
+## 题目大意
+
+
+在一个字符串中寻找子串出现的位置。子串可以是 Anagrams 形式存在的。Anagrams 是一个字符串任意字符的全排列组合。
+
+## 解题思路
+
+这一题和第 438 题，第 3 题，第 76 题，第 567 题类似，用的思想都是"滑动窗口"。
+
+
+这道题只需要判断是否存在，而不需要输出子串所在的下标起始位置。所以这道题是第 438 题的缩水版。具体解题思路见第 438 题。
+
+
+
+
+
+
diff --git a/leetcode/0572.Subtree-of-Another-Tree/572. Subtree of Another Tree.go b/leetcode/0572.Subtree-of-Another-Tree/572. Subtree of Another Tree.go
new file mode 100644
index 000000000..6ec9bf468
--- /dev/null
+++ b/leetcode/0572.Subtree-of-Another-Tree/572. Subtree of Another Tree.go	
@@ -0,0 +1,44 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func isSubtree(s *TreeNode, t *TreeNode) bool {
+	if isSameTree(s, t) {
+		return true
+	}
+	if s == nil {
+		return false
+	}
+	if isSubtree(s.Left, t) || isSubtree(s.Right, t) {
+		return true
+	}
+	return false
+}
+
+// this is 100 solution
+func isSameTree(p *TreeNode, q *TreeNode) bool {
+	if p == nil && q == nil {
+		return true
+	} else if p != nil && q != nil {
+		if p.Val != q.Val {
+			return false
+		}
+		return isSameTree(p.Left, q.Left) && isSameTree(p.Right, q.Right)
+	} else {
+		return false
+	}
+}
diff --git a/leetcode/0572.Subtree-of-Another-Tree/572. Subtree of Another Tree_test.go b/leetcode/0572.Subtree-of-Another-Tree/572. Subtree of Another Tree_test.go
new file mode 100644
index 000000000..0270e3573
--- /dev/null
+++ b/leetcode/0572.Subtree-of-Another-Tree/572. Subtree of Another Tree_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question572 struct {
+	para572
+	ans572
+}
+
+// para 是参数
+// one 代表第一个参数
+type para572 struct {
+	s []int
+	t []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans572 struct {
+	one bool
+}
+
+func Test_Problem572(t *testing.T) {
+
+	qs := []question572{
+
+		{
+			para572{[]int{}, []int{}},
+			ans572{true},
+		},
+
+		{
+			para572{[]int{3, 4, 5, 1, 2}, []int{4, 1, 2}},
+			ans572{true},
+		},
+
+		{
+			para572{[]int{1, 1}, []int{1}},
+			ans572{true},
+		},
+
+		{
+			para572{[]int{1, structures.NULL, 1, structures.NULL, 1, structures.NULL, 1, structures.NULL, 1, structures.NULL, 1, structures.NULL, 1, structures.NULL, 1, structures.NULL, 1, structures.NULL, 1, structures.NULL, 1, 2}, []int{1, structures.NULL, 1, structures.NULL, 1, structures.NULL, 1, structures.NULL, 1, structures.NULL, 1, 2}},
+			ans572{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 572------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans572, q.para572
+		fmt.Printf("【input】:%v      ", p)
+		roots := structures.Ints2TreeNode(p.s)
+		roott := structures.Ints2TreeNode(p.t)
+		fmt.Printf("【output】:%v      \n", isSubtree(roots, roott))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0572.Subtree-of-Another-Tree/README.md b/leetcode/0572.Subtree-of-Another-Tree/README.md
new file mode 100755
index 000000000..29045ec2b
--- /dev/null
+++ b/leetcode/0572.Subtree-of-Another-Tree/README.md
@@ -0,0 +1,53 @@
+# [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)
+
+
+## 题目
+
+Given two non-empty binary trees **s** and **t**, check whether tree **t** has exactly the same structure and node values with a subtree of **s**. A subtree of **s** is a tree consists of a node in **s** and all of this node's descendants. The tree **s** could also be considered as a subtree of itself.
+
+**Example 1:** Given tree s:
+
+		 3
+        / \
+       4   5
+      / \
+     1   2
+
+Given tree t:
+
+       4 
+      / \
+     1   2
+
+Return **true**, because t has the same structure and node values with a subtree of s.
+
+**Example 2:**Given tree s:
+
+         3
+        / \
+       4   5
+      / \
+     1   2
+        /
+       0
+
+Given tree t:
+
+       4
+      / \
+     1   2
+
+Return **false**.
+
+
+## 题目大意
+
+给定两个非空二叉树 s 和 t，检验 s 中是否包含和 t 具有相同结构和节点值的子树。s 的一个子树包括 s 的一个节点和这个节点的所有子孙。s 也可以看做它自身的一棵子树。
+
+
+## 解题思路
+
+
+- 给出 2 棵树 `s` 和 `t`，要求判断 `t` 是否是 `s` 的子树🌲。
+- 这一题比较简单，针对 3 种情况依次递归判断，第一种情况 `s` 和 `t` 是完全一样的两棵树，第二种情况 `t` 是 `s` 左子树中的子树，第三种情况 `t` 是 `s` 右子树中的子树。第一种情况判断两棵数是否完全一致是第 100 题的原题。
+
diff --git a/Algorithms/0575. Distribute Candies/575. Distribute Candies.go b/leetcode/0575.Distribute-Candies/575. Distribute Candies.go
similarity index 100%
rename from Algorithms/0575. Distribute Candies/575. Distribute Candies.go
rename to leetcode/0575.Distribute-Candies/575. Distribute Candies.go
diff --git a/Algorithms/0575. Distribute Candies/575. Distribute Candies_test.go b/leetcode/0575.Distribute-Candies/575. Distribute Candies_test.go
similarity index 95%
rename from Algorithms/0575. Distribute Candies/575. Distribute Candies_test.go
rename to leetcode/0575.Distribute-Candies/575. Distribute Candies_test.go
index 1109b89ca..3f223fa0a 100644
--- a/Algorithms/0575. Distribute Candies/575. Distribute Candies_test.go	
+++ b/leetcode/0575.Distribute-Candies/575. Distribute Candies_test.go	
@@ -26,12 +26,12 @@ func Test_Problem575(t *testing.T) {
 
 	qs := []question575{
 
-		question575{
+		{
 			para575{[]int{1, 1, 2, 2, 3, 3}},
 			ans575{3},
 		},
 
-		question575{
+		{
 			para575{[]int{1, 1, 2, 3}},
 			ans575{2},
 		},
diff --git a/leetcode/0575.Distribute-Candies/README.md b/leetcode/0575.Distribute-Candies/README.md
new file mode 100755
index 000000000..7331b5489
--- /dev/null
+++ b/leetcode/0575.Distribute-Candies/README.md
@@ -0,0 +1,38 @@
+# [575. Distribute Candies](https://leetcode.com/problems/distribute-candies/)
+
+
+## 题目
+
+Given an integer array with **even** length, where different numbers in this array represent different **kinds** of candies. Each number means one candy of the corresponding kind. You need to distribute these candies **equally** in number to brother and sister. Return the maximum number of **kinds** of candies the sister could gain.
+
+**Example 1:**
+
+    Input: candies = [1,1,2,2,3,3]
+    Output: 3
+    Explanation:
+    There are three different kinds of candies (1, 2 and 3), and two candies for each kind.
+    Optimal distribution: The sister has candies [1,2,3] and the brother has candies [1,2,3], too. 
+    The sister has three different kinds of candies.
+
+**Example 2:**
+
+    Input: candies = [1,1,2,3]
+    Output: 2
+    Explanation: For example, the sister has candies [2,3] and the brother has candies [1,1]. 
+    The sister has two different kinds of candies, the brother has only one kind of candies.
+
+**Note:**
+
+1. The length of the given array is in range [2, 10,000], and will be even.
+2. The number in given array is in range [-100,000, 100,000].
+
+
+## 题目大意
+
+给定一个偶数长度的数组，其中不同的数字代表着不同种类的糖果，每一个数字代表一个糖果。你需要把这些糖果平均分给一个弟弟和一个妹妹。返回妹妹可以获得的最大糖果的种类数。
+
+
+## 解题思路
+
+
+- 给出一个糖果数组，里面每个元素代表糖果的种类，相同数字代表相同种类。把这些糖果分给兄弟姐妹，问姐妹最多可以分到多少种糖果。这一题比较简单，用 map 统计每个糖果的出现频次，如果总数比 `n/2` 小，那么就返回 `len(map)`，否则返回 `n/2` (即一半都分给姐妹)。
diff --git a/leetcode/0576.Out-of-Boundary-Paths/576. Out of Boundary Paths.go b/leetcode/0576.Out-of-Boundary-Paths/576. Out of Boundary Paths.go
new file mode 100644
index 000000000..e14f434da
--- /dev/null
+++ b/leetcode/0576.Out-of-Boundary-Paths/576. Out of Boundary Paths.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func findPaths(m int, n int, maxMove int, startRow int, startColumn int) int {
+	visited := make([][][]int, m)
+	for i := range visited {
+		visited[i] = make([][]int, n)
+		for j := range visited[i] {
+			visited[i][j] = make([]int, maxMove+1)
+			for l := range visited[i][j] {
+				visited[i][j][l] = -1
+			}
+		}
+	}
+	return dfs(startRow, startColumn, maxMove, m, n, visited)
+}
+
+func dfs(x, y, maxMove, m, n int, visited [][][]int) int {
+	if x < 0 || x >= m || y < 0 || y >= n {
+		return 1
+	}
+	if maxMove == 0 {
+		visited[x][y][maxMove] = 0
+		return 0
+	}
+	if visited[x][y][maxMove] >= 0 {
+		return visited[x][y][maxMove]
+	}
+	res := 0
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		res += (dfs(nx, ny, maxMove-1, m, n, visited) % 1000000007)
+	}
+	visited[x][y][maxMove] = res % 1000000007
+	return visited[x][y][maxMove]
+}
diff --git a/leetcode/0576.Out-of-Boundary-Paths/576. Out of Boundary Paths_test.go b/leetcode/0576.Out-of-Boundary-Paths/576. Out of Boundary Paths_test.go
new file mode 100644
index 000000000..b88e29e17
--- /dev/null
+++ b/leetcode/0576.Out-of-Boundary-Paths/576. Out of Boundary Paths_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question576 struct {
+	para576
+	ans576
+}
+
+// para 是参数
+// one 代表第一个参数
+type para576 struct {
+	m           int
+	n           int
+	maxMove     int
+	startRow    int
+	startColumn int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans576 struct {
+	one int
+}
+
+func Test_Problem576(t *testing.T) {
+
+	qs := []question576{
+
+		{
+			para576{2, 2, 2, 0, 0},
+			ans576{6},
+		},
+
+		{
+			para576{1, 3, 3, 0, 1},
+			ans576{12},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 576------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans576, q.para576
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findPaths(p.m, p.n, p.maxMove, p.startRow, p.startColumn))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0576.Out-of-Boundary-Paths/README.md b/leetcode/0576.Out-of-Boundary-Paths/README.md
new file mode 100644
index 000000000..64b74de58
--- /dev/null
+++ b/leetcode/0576.Out-of-Boundary-Paths/README.md
@@ -0,0 +1,89 @@
+# [576. Out of Boundary Paths](https://leetcode.com/problems/out-of-boundary-paths/)
+
+
+## 题目
+
+There is an `m x n` grid with a ball. The ball is initially at the position `[startRow, startColumn]`. You are allowed to move the ball to one of the four adjacent four cells in the grid (possibly out of the grid crossing the grid boundary). You can apply **at most** `maxMove` moves to the ball.
+
+Given the five integers `m`, `n`, `maxMove`, `startRow`, `startColumn`, return the number of paths to move the ball out of the grid boundary. Since the answer can be very large, return it **modulo** `109 + 7`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_1.png](https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_1.png)
+
+```
+Input: m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0
+Output: 6
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_2.png](https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_2.png)
+
+```
+Input: m = 1, n = 3, maxMove = 3, startRow = 0, startColumn = 1
+Output: 12
+```
+
+**Constraints:**
+
+- `1 <= m, n <= 50`
+- `0 <= maxMove <= 50`
+- `0 <= startRow <= m`
+- `0 <= startColumn <= n`
+
+## 题目大意
+
+给定一个 m × n 的网格和一个球。球的起始坐标为 (i,j) ，你可以将球移到相邻的单元格内，或者往上、下、左、右四个方向上移动使球穿过网格边界。但是，你最多可以移动 N 次。找出可以将球移出边界的路径数量。答案可能非常大，返回 结果 mod 109 + 7 的值。
+
+## 解题思路
+
+- 单纯暴力的思路，在球的每个方向都遍历一步，直到移动步数用完。这样暴力搜索，解空间是 4^n 。优化思路便是增加记忆化。用三维数组记录位置坐标和步数，对应的出边界的路径数量。加上记忆化以后的深搜解法 runtime beats 100% 了。
+
+## 代码
+
+```go
+package leetcode
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func findPaths(m int, n int, maxMove int, startRow int, startColumn int) int {
+	visited := make([][][]int, m)
+	for i := range visited {
+		visited[i] = make([][]int, n)
+		for j := range visited[i] {
+			visited[i][j] = make([]int, maxMove+1)
+			for l := range visited[i][j] {
+				visited[i][j][l] = -1
+			}
+		}
+	}
+	return dfs(startRow, startColumn, maxMove, m, n, visited)
+}
+
+func dfs(x, y, maxMove, m, n int, visited [][][]int) int {
+	if x < 0 || x >= m || y < 0 || y >= n {
+		return 1
+	}
+	if maxMove == 0 {
+		visited[x][y][maxMove] = 0
+		return 0
+	}
+	if visited[x][y][maxMove] >= 0 {
+		return visited[x][y][maxMove]
+	}
+	res := 0
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		res += (dfs(nx, ny, maxMove-1, m, n, visited) % 1000000007)
+	}
+	visited[x][y][maxMove] = res % 1000000007
+	return visited[x][y][maxMove]
+}
+```
\ No newline at end of file
diff --git a/leetcode/0581.Shortest-Unsorted-Continuous-Subarray/581. Shortest Unsorted Continuous Subarray.go b/leetcode/0581.Shortest-Unsorted-Continuous-Subarray/581. Shortest Unsorted Continuous Subarray.go
new file mode 100644
index 000000000..328af9c3c
--- /dev/null
+++ b/leetcode/0581.Shortest-Unsorted-Continuous-Subarray/581. Shortest Unsorted Continuous Subarray.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import "math"
+
+func findUnsortedSubarray(nums []int) int {
+	n, left, right, minR, maxL, isSort := len(nums), -1, -1, math.MaxInt32, math.MinInt32, false
+	// left
+	for i := 1; i < n; i++ {
+		if nums[i] < nums[i-1] {
+			isSort = true
+		}
+		if isSort {
+			minR = min(minR, nums[i])
+		}
+	}
+	isSort = false
+	// right
+	for i := n - 2; i >= 0; i-- {
+		if nums[i] > nums[i+1] {
+			isSort = true
+		}
+		if isSort {
+			maxL = max(maxL, nums[i])
+		}
+	}
+	// minR
+	for i := 0; i < n; i++ {
+		if nums[i] > minR {
+			left = i
+			break
+		}
+	}
+	// maxL
+	for i := n - 1; i >= 0; i-- {
+		if nums[i] < maxL {
+			right = i
+			break
+		}
+	}
+	if left == -1 || right == -1 {
+		return 0
+	}
+	return right - left + 1
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0581.Shortest-Unsorted-Continuous-Subarray/581. Shortest Unsorted Continuous Subarray_test.go b/leetcode/0581.Shortest-Unsorted-Continuous-Subarray/581. Shortest Unsorted Continuous Subarray_test.go
new file mode 100644
index 000000000..db4b31857
--- /dev/null
+++ b/leetcode/0581.Shortest-Unsorted-Continuous-Subarray/581. Shortest Unsorted Continuous Subarray_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question581 struct {
+	para581
+	ans581
+}
+
+// para 是参数
+// one 代表第一个参数
+type para581 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans581 struct {
+	one int
+}
+
+func Test_Problem581(t *testing.T) {
+
+	qs := []question581{
+
+		{
+			para581{[]int{2, 6, 4, 8, 10, 9, 15}},
+			ans581{5},
+		},
+
+		{
+			para581{[]int{1, 2, 3, 4}},
+			ans581{0},
+		},
+
+		{
+			para581{[]int{1}},
+			ans581{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 581------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans581, q.para581
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findUnsortedSubarray(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0581.Shortest-Unsorted-Continuous-Subarray/README.md b/leetcode/0581.Shortest-Unsorted-Continuous-Subarray/README.md
new file mode 100644
index 000000000..d6cf6b652
--- /dev/null
+++ b/leetcode/0581.Shortest-Unsorted-Continuous-Subarray/README.md
@@ -0,0 +1,107 @@
+# [581. Shortest Unsorted Continuous Subarray](https://leetcode.com/problems/shortest-unsorted-continuous-subarray/)
+
+
+## 题目
+
+Given an integer array `nums`, you need to find one **continuous subarray** that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order.
+
+Return *the shortest such subarray and output its length*.
+
+**Example 1:**
+
+```
+Input: nums = [2,6,4,8,10,9,15]
+Output: 5
+Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3,4]
+Output: 0
+```
+
+**Example 3:**
+
+```
+Input: nums = [1]
+Output: 0
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 104`
+- `105 <= nums[i] <= 105`
+
+## 题目大意
+
+给你一个整数数组 nums ，你需要找出一个 连续子数组 ，如果对这个子数组进行升序排序，那么整个数组都会变为升序排序。请你找出符合题意的 最短 子数组，并输出它的长度。
+
+## 解题思路
+
+- 本题求的是最短逆序区间。经过简单推理，可以知道，这个逆序区间一定由这个区间内的最小元素决定左边界，最大元素决定右边界。
+- 先从左边找到第一个降序的元素，并记录最小的元素 min，再从右边往左找到最右边开始降序的元素，并记录最大的元素 max。最后需要还原最小元素和最大元素在原数组中正确的位置。以逆序区间左边界为例，如果区间外的一个元素比这个逆序区间内的最小元素还要小，说明它并不是左边界，因为这个小元素和逆序区间内的最小元素组合在一起，还是升序，并不是逆序。只有在左边区间外找到第一个大于逆序区间内最小元素，说明这里刚刚开始发生逆序，这才是最小逆序区间的左边界。同理，在逆序区间的右边找到第一个小于逆序区间内最大元素，说明这里刚刚发生逆序，这才是最小逆序区间的右边界。至此，最小逆序区间的左右边界都确定下来了，最短长度也就确定了下来。时间复杂度 O(n)，空间复杂度 O(1)。
+
+## 代码
+
+```go
+package leetcode
+
+import "math"
+
+func findUnsortedSubarray(nums []int) int {
+	n, left, right, minR, maxL, isSort := len(nums), -1, -1, math.MaxInt32, math.MinInt32, false
+	// left
+	for i := 1; i < n; i++ {
+		if nums[i] < nums[i-1] {
+			isSort = true
+		}
+		if isSort {
+			minR = min(minR, nums[i])
+		}
+	}
+	isSort = false
+	// right
+	for i := n - 2; i >= 0; i-- {
+		if nums[i] > nums[i+1] {
+			isSort = true
+		}
+		if isSort {
+			maxL = max(maxL, nums[i])
+		}
+	}
+	// minR
+	for i := 0; i < n; i++ {
+		if nums[i] > minR {
+			left = i
+			break
+		}
+	}
+	// maxL
+	for i := n - 1; i >= 0; i-- {
+		if nums[i] < maxL {
+			right = i
+			break
+		}
+	}
+	if left == -1 || right == -1 {
+		return 0
+	}
+	return right - left + 1
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/0583.Delete-Operation-for-Two-Strings/583. Delete Operation for Two Strings.go b/leetcode/0583.Delete-Operation-for-Two-Strings/583. Delete Operation for Two Strings.go
new file mode 100644
index 000000000..616ddd8c8
--- /dev/null
+++ b/leetcode/0583.Delete-Operation-for-Two-Strings/583. Delete Operation for Two Strings.go	
@@ -0,0 +1,31 @@
+package leetcode
+
+func minDistance(word1 string, word2 string) int {
+	dp := make([][]int, len(word1)+1)
+	for i := 0; i < len(word1)+1; i++ {
+		dp[i] = make([]int, len(word2)+1)
+	}
+	for i := 0; i < len(word1)+1; i++ {
+		dp[i][0] = i
+	}
+	for i := 0; i < len(word2)+1; i++ {
+		dp[0][i] = i
+	}
+	for i := 1; i < len(word1)+1; i++ {
+		for j := 1; j < len(word2)+1; j++ {
+			if word1[i-1] == word2[j-1] {
+				dp[i][j] = dp[i-1][j-1]
+			} else {
+				dp[i][j] = 1 + min(dp[i][j-1], dp[i-1][j])
+			}
+		}
+	}
+	return dp[len(word1)][len(word2)]
+}
+
+func min(x, y int) int {
+	if x < y {
+		return x
+	}
+	return y
+}
diff --git a/leetcode/0583.Delete-Operation-for-Two-Strings/583. Delete Operation for Two Strings_test.go b/leetcode/0583.Delete-Operation-for-Two-Strings/583. Delete Operation for Two Strings_test.go
new file mode 100644
index 000000000..d8d8c46b2
--- /dev/null
+++ b/leetcode/0583.Delete-Operation-for-Two-Strings/583. Delete Operation for Two Strings_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question583 struct {
+	para583
+	ans583
+}
+
+// para 是参数
+// one 代表第一个参数
+type para583 struct {
+	word1 string
+	word2 string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans583 struct {
+	one int
+}
+
+func Test_Problem583(t *testing.T) {
+
+	qs := []question583{
+
+		{
+			para583{"sea", "eat"},
+			ans583{2},
+		},
+
+		{
+			para583{"leetcode", "etco"},
+			ans583{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 583------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans583, q.para583
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minDistance(p.word1, p.word2))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0583.Delete-Operation-for-Two-Strings/README.md b/leetcode/0583.Delete-Operation-for-Two-Strings/README.md
new file mode 100644
index 000000000..b55d2fbf3
--- /dev/null
+++ b/leetcode/0583.Delete-Operation-for-Two-Strings/README.md
@@ -0,0 +1,76 @@
+# [583. Delete Operation for Two Strings](https://leetcode.com/problems/delete-operation-for-two-strings/)
+
+
+## 题目
+
+Given two strings `word1` and `word2`, return *the minimum number of **steps** required to make* `word1` *and* `word2` *the same*.
+
+In one **step**, you can delete exactly one character in either string.
+
+**Example 1:**
+
+```
+Input: word1 = "sea", word2 = "eat"
+Output: 2
+Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
+```
+
+**Example 2:**
+
+```
+Input: word1 = "leetcode", word2 = "etco"
+Output: 4
+```
+
+**Constraints:**
+
+- `1 <= word1.length, word2.length <= 500`
+- `word1` and `word2` consist of only lowercase English letters.
+
+## 题目大意
+
+给定两个单词 word1 和 word2，找到使得 word1 和 word2 相同所需的最小步数，每步可以删除任意一个字符串中的一个字符。
+
+## 解题思路
+
+- 从题目数据量级判断，此题一定是 O(n^2) 动态规划题。定义 `dp[i][j]` 表示 `word1[:i]` 与 `word2[:j]` 匹配所删除的最少步数。如果 `word1[:i-1]` 与 `word2[:j-1]` 匹配，那么 `dp[i][j] = dp[i-1][j-1]`。如果 `word1[:i-1]` 与 `word2[:j-1]` 不匹配，那么需要考虑删除一次，所以 `dp[i][j] = 1 + min(dp[i][j-1], dp[i-1][j])`。所以动态转移方程是：
+
+    $$dp[i][j] = \left\{\begin{matrix}dp[i-1][j-1]&, word1[i-1] == word2[j-1]\\ 1 + min(dp[i][j-1], dp[i-1][j])&, word1[i-1] \neq word2[j-1]\\\end{matrix}\right.$$
+
+    最终答案存储在 `dp[len(word1)][len(word2)]` 中。
+
+## 代码
+
+```go
+package leetcode
+
+func minDistance(word1 string, word2 string) int {
+	dp := make([][]int, len(word1)+1)
+	for i := 0; i < len(word1)+1; i++ {
+		dp[i] = make([]int, len(word2)+1)
+	}
+	for i := 0; i < len(word1)+1; i++ {
+		dp[i][0] = i
+	}
+	for i := 0; i < len(word2)+1; i++ {
+		dp[0][i] = i
+	}
+	for i := 1; i < len(word1)+1; i++ {
+		for j := 1; j < len(word2)+1; j++ {
+			if word1[i-1] == word2[j-1] {
+				dp[i][j] = dp[i-1][j-1]
+			} else {
+				dp[i][j] = 1 + min(dp[i][j-1], dp[i-1][j])
+			}
+		}
+	}
+	return dp[len(word1)][len(word2)]
+}
+
+func min(x, y int) int {
+	if x < y {
+		return x
+	}
+	return y
+}
+```
\ No newline at end of file
diff --git a/leetcode/0589.N-ary-Tree-Preorder-Traversal/589. N-ary Tree Preorder Traversal.go b/leetcode/0589.N-ary-Tree-Preorder-Traversal/589. N-ary Tree Preorder Traversal.go
new file mode 100644
index 000000000..c0059ffd1
--- /dev/null
+++ b/leetcode/0589.N-ary-Tree-Preorder-Traversal/589. N-ary Tree Preorder Traversal.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+// Definition for a Node.
+type Node struct {
+	Val      int
+	Children []*Node
+}
+
+// 解法一 非递归
+func preorder(root *Node) []int {
+	res := []int{}
+	if root == nil {
+		return res
+	}
+	stack := []*Node{root}
+	for len(stack) > 0 {
+		r := stack[len(stack)-1]
+		stack = stack[:len(stack)-1]
+		res = append(res, r.Val)
+		tmp := []*Node{}
+		for _, v := range r.Children {
+			tmp = append([]*Node{v}, tmp...) // 逆序存点
+		}
+		stack = append(stack, tmp...)
+	}
+	return res
+}
+
+// 解法二 递归
+func preorder1(root *Node) []int {
+	res := []int{}
+	preorderdfs(root, &res)
+	return res
+}
+
+func preorderdfs(root *Node, res *[]int) {
+	if root != nil {
+		*res = append(*res, root.Val)
+		for i := 0; i < len(root.Children); i++ {
+			preorderdfs(root.Children[i], res)
+		}
+	}
+}
diff --git a/leetcode/0589.N-ary-Tree-Preorder-Traversal/589. N-ary Tree Preorder Traversal_test.go b/leetcode/0589.N-ary-Tree-Preorder-Traversal/589. N-ary Tree Preorder Traversal_test.go
new file mode 100644
index 000000000..2f029c9fb
--- /dev/null
+++ b/leetcode/0589.N-ary-Tree-Preorder-Traversal/589. N-ary Tree Preorder Traversal_test.go	
@@ -0,0 +1,82 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question589 struct {
+	para589
+	ans589
+}
+
+// para 是参数
+// one 代表第一个参数
+type para589 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans589 struct {
+	one []int
+}
+
+func Test_Problem589(t *testing.T) {
+
+	qs := []question589{
+
+		{
+			para589{[]int{1, structures.NULL, 3, 2, 4, structures.NULL, 5, 6}},
+			ans589{[]int{1, 3, 5, 6, 2, 4}},
+		},
+
+		{
+			para589{[]int{1, structures.NULL, 2, 3, 4, 5, structures.NULL, structures.NULL, 6, 7, structures.NULL, 8, structures.NULL, 9, 10, structures.NULL, structures.NULL, 11, structures.NULL, 12, structures.NULL, 13, structures.NULL, structures.NULL, 14}},
+			ans589{[]int{1, 2, 3, 6, 7, 11, 14, 4, 8, 12, 5, 9, 13, 10}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 589------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans589, q.para589
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := int2NaryNode(p.one)
+		fmt.Printf("【output】:%v      \n", preorder(rootOne))
+	}
+	fmt.Printf("\n\n\n")
+}
+
+func int2NaryNode(nodes []int) *Node {
+	root := &Node{}
+	if len(nodes) > 1 {
+		root.Val = nodes[0]
+	}
+	queue := []*Node{}
+	queue = append(queue, root)
+	i := 1
+	count := 0
+	for i < len(nodes) {
+		node := queue[0]
+
+		childrens := []*Node{}
+		for ; i < len(nodes) && nodes[i] != structures.NULL; i++ {
+			tmp := &Node{Val: nodes[i]}
+			childrens = append(childrens, tmp)
+			queue = append(queue, tmp)
+		}
+		count++
+		if count%2 == 0 {
+			queue = queue[1:]
+			count = 1
+		}
+		if node != nil {
+			node.Children = childrens
+		}
+		i++
+	}
+	return root
+}
diff --git a/leetcode/0589.N-ary-Tree-Preorder-Traversal/README.md b/leetcode/0589.N-ary-Tree-Preorder-Traversal/README.md
new file mode 100644
index 000000000..b3746cc47
--- /dev/null
+++ b/leetcode/0589.N-ary-Tree-Preorder-Traversal/README.md
@@ -0,0 +1,90 @@
+# [589. N-ary Tree Preorder Traversal](https://leetcode.com/problems/n-ary-tree-preorder-traversal/)
+
+## 题目
+
+Given the `root` of an n-ary tree, return *the preorder traversal of its nodes' values*.
+
+Nary-Tree input serialization is represented in their level order traversal. Each group of children is separated by the null value (See examples)
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png](https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png)
+
+```
+Input: root = [1,null,3,2,4,null,5,6]
+Output: [1,3,5,6,2,4]
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png](https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png)
+
+```
+Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
+Output: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[0, 104]`.
+- `0 <= Node.val <= 10^4`
+- The height of the n-ary tree is less than or equal to `1000`.
+
+**Follow up:** Recursive solution is trivial, could you do it iteratively?
+
+## 题目大意
+
+给定一个 N 叉树，返回其节点值的 **前序遍历** 。N 叉树 在输入中按层序遍历进行序列化表示，每组子节点由空值 `null` 分隔（请参见示例）。
+
+## 解题思路
+
+- N 叉树和二叉树的前序遍历原理完全一样。二叉树非递归解法需要用到栈辅助，N 叉树同样如此。将父节点的所有孩子节点**逆序**入栈，逆序的目的是为了让前序节点永远在栈顶。依次循环直到栈里所有元素都出栈。输出的结果即为 N 叉树的前序遍历。时间复杂度 O(n)，空间复杂度 O(n)。
+- 递归解法非常简单，见解法二。
+
+## 代码
+
+```go
+package leetcode
+
+//  Definition for a Node.
+type Node struct {
+	Val      int
+	Children []*Node
+}
+
+// 解法一 非递归
+func preorder(root *Node) []int {
+	res := []int{}
+	if root == nil {
+		return res
+	}
+	stack := []*Node{root}
+	for len(stack) > 0 {
+		r := stack[len(stack)-1]
+		stack = stack[:len(stack)-1]
+		res = append(res, r.Val)
+		tmp := []*Node{}
+		for _, v := range r.Children {
+			tmp = append([]*Node{v}, tmp...) // 逆序存点
+		}
+		stack = append(stack, tmp...)
+	}
+	return res
+}
+
+// 解法二 递归
+func preorder1(root *Node) []int {
+	res := []int{}
+	preorderdfs(root, &res)
+	return res
+}
+
+func preorderdfs(root *Node, res *[]int) {
+	if root != nil {
+		*res = append(*res, root.Val)
+		for i := 0; i < len(root.Children); i++ {
+			preorderdfs(root.Children[i], res)
+		}
+	}
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0594. Longest Harmonious Subsequence/594. Longest Harmonious Subsequence.go b/leetcode/0594.Longest-Harmonious-Subsequence/594. Longest Harmonious Subsequence.go
similarity index 100%
rename from Algorithms/0594. Longest Harmonious Subsequence/594. Longest Harmonious Subsequence.go
rename to leetcode/0594.Longest-Harmonious-Subsequence/594. Longest Harmonious Subsequence.go
diff --git a/Algorithms/0594. Longest Harmonious Subsequence/594. Longest Harmonious Subsequence_test.go b/leetcode/0594.Longest-Harmonious-Subsequence/594. Longest Harmonious Subsequence_test.go
similarity index 97%
rename from Algorithms/0594. Longest Harmonious Subsequence/594. Longest Harmonious Subsequence_test.go
rename to leetcode/0594.Longest-Harmonious-Subsequence/594. Longest Harmonious Subsequence_test.go
index 29ae7f5be..3c777d0d9 100644
--- a/Algorithms/0594. Longest Harmonious Subsequence/594. Longest Harmonious Subsequence_test.go	
+++ b/leetcode/0594.Longest-Harmonious-Subsequence/594. Longest Harmonious Subsequence_test.go	
@@ -26,7 +26,7 @@ func Test_Problem594(t *testing.T) {
 
 	qs := []question594{
 
-		question594{
+		{
 			para594{[]int{1, 3, 2, 2, 5, 2, 3, 7}},
 			ans594{5},
 		},
diff --git a/leetcode/0594.Longest-Harmonious-Subsequence/README.md b/leetcode/0594.Longest-Harmonious-Subsequence/README.md
new file mode 100755
index 000000000..7c7151767
--- /dev/null
+++ b/leetcode/0594.Longest-Harmonious-Subsequence/README.md
@@ -0,0 +1,25 @@
+# [594. Longest Harmonious Subsequence](https://leetcode.com/problems/longest-harmonious-subsequence/)
+
+
+## 题目
+
+We define a harmounious array as an array where the difference between its maximum value and its minimum value is **exactly** 1.
+
+Now, given an integer array, you need to find the length of its longest harmonious subsequence among all its possible [subsequences](https://en.wikipedia.org/wiki/Subsequence).
+
+**Example 1:**
+
+    Input: [1,3,2,2,5,2,3,7]
+    Output: 5
+    Explanation: The longest harmonious subsequence is [3,2,2,2,3].
+
+**Note:** The length of the input array will not exceed 20,000.
+
+
+## 题目大意
+
+和谐数组是指一个数组里元素的最大值和最小值之间的差别正好是1。现在，给定一个整数数组，你需要在所有可能的子序列中找到最长的和谐子序列的长度。说明: 输入的数组长度最大不超过20,000.
+
+## 解题思路
+
+- 在给出的数组里面找到这样一个子数组：要求子数组中的最大值和最小值相差 1 。这一题是简单题。先统计每个数字出现的频次，然后在 map 找相差 1 的 2 个数组的频次和，动态的维护两个数的频次和就是最后要求的子数组的最大长度。
diff --git a/leetcode/0598.Range-Addition-II/598. Range Addition II.go b/leetcode/0598.Range-Addition-II/598. Range Addition II.go
new file mode 100644
index 000000000..9b58fd80d
--- /dev/null
+++ b/leetcode/0598.Range-Addition-II/598. Range Addition II.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+func maxCount(m int, n int, ops [][]int) int {
+	minM, minN := m, n
+	for _, op := range ops {
+		minM = min(minM, op[0])
+		minN = min(minN, op[1])
+	}
+	return minM * minN
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0598.Range-Addition-II/598. Range Addition II_test.go b/leetcode/0598.Range-Addition-II/598. Range Addition II_test.go
new file mode 100644
index 000000000..f7be9b0d3
--- /dev/null
+++ b/leetcode/0598.Range-Addition-II/598. Range Addition II_test.go	
@@ -0,0 +1,44 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question598 struct {
+	para598
+	ans598
+}
+
+// para 是参数
+// one 代表第一个参数
+type para598 struct {
+	m   int
+	n   int
+	ops [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans598 struct {
+	one int
+}
+
+func Test_Problem598(t *testing.T) {
+
+	qs := []question598{
+
+		{
+			para598{3, 3, [][]int{{2, 2}, {3, 3}}},
+			ans598{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 598------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans598, q.para598
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxCount(p.m, p.n, p.ops))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0598.Range-Addition-II/README.md b/leetcode/0598.Range-Addition-II/README.md
new file mode 100644
index 000000000..a4606f951
--- /dev/null
+++ b/leetcode/0598.Range-Addition-II/README.md
@@ -0,0 +1,82 @@
+# [598. Range Addition II](https://leetcode.com/problems/range-addition-ii/)
+
+
+## 题目
+
+Given an m * n matrix **M** initialized with all **0**'s and several update operations.
+
+Operations are represented by a 2D array, and each operation is represented by an array with two **positive** integers **a** and **b**, which means **M[i][j]** should be **added by one** for all **0 <= i < a** and **0 <= j < b**.
+
+You need to count and return the number of maximum integers in the matrix after performing all the operations.
+
+**Example 1**:
+
+```
+Input: 
+m = 3, n = 3
+operations = [[2,2],[3,3]]
+Output: 4
+Explanation: 
+Initially, M = 
+[[0, 0, 0],
+ [0, 0, 0],
+ [0, 0, 0]]
+
+After performing [2,2], M = 
+[[1, 1, 0],
+ [1, 1, 0],
+ [0, 0, 0]]
+
+After performing [3,3], M = 
+[[2, 2, 1],
+ [2, 2, 1],
+ [1, 1, 1]]
+
+So the maximum integer in M is 2, and there are four of it in M. So return 4.
+```
+
+**Note**:
+
+1. The range of m and n is [1,40000].
+2. The range of a is [1,m], and the range of b is [1,n].
+3. The range of operations size won't exceed 10,000.
+
+## 题目大意
+
+给定一个初始元素全部为 0，大小为 m*n 的矩阵 M 以及在 M 上的一系列更新操作。操作用二维数组表示，其中的每个操作用一个含有两个正整数 a 和 b 的数组表示，含义是将所有符合 0 <= i < a 以及 0 <= j < b 的元素 M[i][j] 的值都增加 1。在执行给定的一系列操作后，你需要返回矩阵中含有最大整数的元素个数。
+
+注意:
+
+- m 和 n 的范围是 [1,40000]。
+- a 的范围是 [1,m]，b 的范围是 [1,n]。
+- 操作数目不超过 10000。
+
+
+## 解题思路
+
+- 给定一个初始都为 0 的 m * n 的矩阵，和一个操作数组。经过一系列的操作以后，最终输出矩阵中最大整数的元素个数。每次操作都使得一个矩形内的元素都 + 1 。
+- 这一题乍一看像线段树的区间覆盖问题，但是实际上很简单。如果此题是任意的矩阵，那就可能用到线段树了。这一题每个矩阵的起点都包含 [0 , 0] 这个元素，也就是说每次操作都会影响第一个元素。那么这道题就很简单了。经过 n 次操作以后，被覆盖次数最多的矩形区间，一定就是最大整数所在的区间。由于起点都是第一个元素，所以我们只用关心矩形的右下角那个坐标。右下角怎么计算呢？只用每次动态的维护一下矩阵长和宽的最小值即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func maxCount(m int, n int, ops [][]int) int {
+	minM, minN := m, n
+	for _, op := range ops {
+		minM = min(minM, op[0])
+		minN = min(minN, op[1])
+	}
+	return minM * minN
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0599. Minimum Index Sum of Two Lists/599. Minimum Index Sum of Two Lists.go b/leetcode/0599.Minimum-Index-Sum-of-Two-Lists/599. Minimum Index Sum of Two Lists.go
similarity index 100%
rename from Algorithms/0599. Minimum Index Sum of Two Lists/599. Minimum Index Sum of Two Lists.go
rename to leetcode/0599.Minimum-Index-Sum-of-Two-Lists/599. Minimum Index Sum of Two Lists.go
diff --git a/Algorithms/0599. Minimum Index Sum of Two Lists/599. Minimum Index Sum of Two Lists_test.go b/leetcode/0599.Minimum-Index-Sum-of-Two-Lists/599. Minimum Index Sum of Two Lists_test.go
similarity index 96%
rename from Algorithms/0599. Minimum Index Sum of Two Lists/599. Minimum Index Sum of Two Lists_test.go
rename to leetcode/0599.Minimum-Index-Sum-of-Two-Lists/599. Minimum Index Sum of Two Lists_test.go
index 9554de918..a702a9ce1 100644
--- a/Algorithms/0599. Minimum Index Sum of Two Lists/599. Minimum Index Sum of Two Lists_test.go	
+++ b/leetcode/0599.Minimum-Index-Sum-of-Two-Lists/599. Minimum Index Sum of Two Lists_test.go	
@@ -27,12 +27,12 @@ func Test_Problem599(t *testing.T) {
 
 	qs := []question599{
 
-		question599{
+		{
 			para599{[]string{"Shogun", "Tapioca Express", "Burger King", "KFC"}, []string{"Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"}},
 			ans599{[]string{"Shogun"}},
 		},
 
-		question599{
+		{
 			para599{[]string{"Shogun", "Tapioca Express", "Burger King", "KFC"}, []string{"KFC", "Shogun", "Burger King"}},
 			ans599{[]string{"Shogun"}},
 		},
diff --git a/leetcode/0599.Minimum-Index-Sum-of-Two-Lists/README.md b/leetcode/0599.Minimum-Index-Sum-of-Two-Lists/README.md
new file mode 100755
index 000000000..0fa853612
--- /dev/null
+++ b/leetcode/0599.Minimum-Index-Sum-of-Two-Lists/README.md
@@ -0,0 +1,50 @@
+# [599. Minimum Index Sum of Two Lists](https://leetcode.com/problems/minimum-index-sum-of-two-lists/)
+
+## 题目
+
+Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings.
+
+You need to help them find out their **common interest** with the **least list index sum**. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.
+
+**Example 1:**
+
+    Input:
+    ["Shogun", "Tapioca Express", "Burger King", "KFC"]
+    ["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
+    Output: ["Shogun"]
+    Explanation: The only restaurant they both like is "Shogun".
+
+**Example 2:**
+
+    Input:
+    ["Shogun", "Tapioca Express", "Burger King", "KFC"]
+    ["KFC", "Shogun", "Burger King"]
+    Output: ["Shogun"]
+    Explanation: The restaurant they both like and have the least index sum is "Shogun" with index sum 1 (0+1).
+
+**Note:**
+
+1. The length of both lists will be in the range of [1, 1000].
+2. The length of strings in both lists will be in the range of [1, 30].
+3. The index is starting from 0 to the list length minus 1.
+4. No duplicates in both lists.
+
+
+## 题目大意
+
+假设 Andy 和 Doris 想在晚餐时选择一家餐厅，并且他们都有一个表示最喜爱餐厅的列表，每个餐厅的名字用字符串表示。你需要帮助他们用最少的索引和找出他们共同喜爱的餐厅。 如果答案不止一个，则输出所有答案并且不考虑顺序。 你可以假设总是存在一个答案。
+
+
+提示:  
+
+- 两个列表的长度范围都在 [1, 1000] 内。
+- 两个列表中的字符串的长度将在 [1，30] 的范围内。
+- 下标从 0 开始，到列表的长度减 1。
+- 两个列表都没有重复的元素。
+
+
+
+## 解题思路
+
+
+- 在 Andy 和 Doris 两人分别有各自的餐厅喜欢列表，要求找出两人公共喜欢的一家餐厅，如果共同喜欢的次数相同，都输出。这一题是简单题，用 map 统计频次，输出频次最多的餐厅。
diff --git a/leetcode/0605.Can-Place-Flowers/605. Can Place Flowers.go b/leetcode/0605.Can-Place-Flowers/605. Can Place Flowers.go
new file mode 100644
index 000000000..0b5980963
--- /dev/null
+++ b/leetcode/0605.Can-Place-Flowers/605. Can Place Flowers.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func canPlaceFlowers(flowerbed []int, n int) bool {
+	lenth := len(flowerbed)
+	for i := 0; i < lenth && n > 0; i += 2 {
+		if flowerbed[i] == 0 {
+			if i+1 == lenth || flowerbed[i+1] == 0 {
+				n--
+			} else {
+				i++
+			}
+		}
+	}
+	if n == 0 {
+		return true
+	}
+	return false
+}
diff --git a/leetcode/0605.Can-Place-Flowers/605. Can Place Flowers_test.go b/leetcode/0605.Can-Place-Flowers/605. Can Place Flowers_test.go
new file mode 100644
index 000000000..0b7ce7f2b
--- /dev/null
+++ b/leetcode/0605.Can-Place-Flowers/605. Can Place Flowers_test.go	
@@ -0,0 +1,68 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question605 struct {
+	para605
+	ans605
+}
+
+// para 是参数
+// one 代表第一个参数
+type para605 struct {
+	flowerbed []int
+	n         int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans605 struct {
+	one bool
+}
+
+func Test_Problem605(t *testing.T) {
+
+	qs := []question605{
+
+		{
+			para605{[]int{1, 0, 0, 0, 1}, 1},
+			ans605{true},
+		},
+
+		{
+			para605{[]int{1, 0, 0, 0, 1}, 2},
+			ans605{false},
+		},
+
+		{
+			para605{[]int{1, 0, 0, 0, 0, 1}, 2},
+			ans605{false},
+		},
+
+		{
+			para605{[]int{0, 0, 1, 0}, 1},
+			ans605{true},
+		},
+
+		{
+			para605{[]int{0, 0, 1, 0, 0}, 1},
+			ans605{true},
+		},
+
+		{
+			para605{[]int{1, 0, 0, 1, 0}, 2},
+			ans605{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 605------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans605, q.para605
+		fmt.Printf("【input】:%v       【output】:%v\n", p, canPlaceFlowers(p.flowerbed, p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0605.Can-Place-Flowers/README.md b/leetcode/0605.Can-Place-Flowers/README.md
new file mode 100644
index 000000000..746817adc
--- /dev/null
+++ b/leetcode/0605.Can-Place-Flowers/README.md
@@ -0,0 +1,60 @@
+# [605. Can Place Flowers](https://leetcode.com/problems/can-place-flowers/)
+
+## 题目
+
+You have a long flowerbed in which some of the plots are planted, and some are not. However, flowers cannot be planted in **adjacent** plots.
+
+Given an integer array `flowerbed` containing `0`'s and `1`'s, where `0` means empty and `1` means not empty, and an integer `n`, return *if* `n` new flowers can be planted in the `flowerbed` without violating the no-adjacent-flowers rule.
+
+**Example 1:**
+
+```
+Input: flowerbed = [1,0,0,0,1], n = 1
+Output: true
+```
+
+**Example 2:**
+
+```
+Input: flowerbed = [1,0,0,0,1], n = 2
+Output: false
+```
+
+**Constraints:**
+
+- `1 <= flowerbed.length <= 2 * 104`
+- `flowerbed[i]` is `0` or `1`.
+- There are no two adjacent flowers in `flowerbed`.
+- `0 <= n <= flowerbed.length`
+
+## 题目大意
+
+假设你有一个很长的花坛，一部分地块种植了花，另一部分却没有。可是，花卉不能种植在相邻的地块上，它们会争夺水源，两者都会死去。给定一个花坛（表示为一个数组包含0和1，其中0表示没种植花，1表示种植了花），和一个数 n 。能否在不打破种植规则的情况下种入 n 朵花？能则返回True，不能则返回False。
+
+## 解题思路
+
+- 这一题最容易想到的解法是步长为 2 遍历数组，依次计数 0 的个数。有 2 种特殊情况需要单独判断，第一种情况是首尾连续多个 0，例如，00001 和 10000，第二种情况是 2 个 1 中间存在的 0 不足以种花，例如，1001 和 100001，1001 不能种任何花，100001 只能种一种花。单独判断出这 2 种情况，这一题就可以 AC 了。
+- 换个思路，找到可以种花的基本单元是 00，那么上面那 2 种特殊情况都可以统一成一种情况。判断是否当前存在 00 的组合，如果存在 00 的组合，都可以种花。末尾的情况需要单独判断，如果末尾为 0，也可以种花。这个时候不需要再找 00 组合，因为会越界。代码实现如下，思路很简洁明了。
+
+## 代码
+
+```go
+package leetcode
+
+func canPlaceFlowers(flowerbed []int, n int) bool {
+	lenth := len(flowerbed)
+	for i := 0; i < lenth && n > 0; i += 2 {
+		if flowerbed[i] == 0 {
+			if i+1 == lenth || flowerbed[i+1] == 0 {
+				n--
+			} else {
+				i++
+			}
+		}
+	}
+	if n == 0 {
+		return true
+	}
+	return false
+}
+```
\ No newline at end of file
diff --git a/leetcode/0609.Find-Duplicate-File-in-System/609. Find Duplicate File in System.go b/leetcode/0609.Find-Duplicate-File-in-System/609. Find Duplicate File in System.go
new file mode 100644
index 000000000..1a7852109
--- /dev/null
+++ b/leetcode/0609.Find-Duplicate-File-in-System/609. Find Duplicate File in System.go	
@@ -0,0 +1,23 @@
+package leetcode
+
+import "strings"
+
+func findDuplicate(paths []string) [][]string {
+	cache := make(map[string][]string)
+	for _, path := range paths {
+		parts := strings.Split(path, " ")
+		dir := parts[0]
+		for i := 1; i < len(parts); i++ {
+			bracketPosition := strings.IndexByte(parts[i], '(')
+			content := parts[i][bracketPosition+1 : len(parts[i])-1]
+			cache[content] = append(cache[content], dir+"/"+parts[i][:bracketPosition])
+		}
+	}
+	res := make([][]string, 0, len(cache))
+	for _, group := range cache {
+		if len(group) >= 2 {
+			res = append(res, group)
+		}
+	}
+	return res
+}
diff --git a/leetcode/0609.Find-Duplicate-File-in-System/609. Find Duplicate File in System_test.go b/leetcode/0609.Find-Duplicate-File-in-System/609. Find Duplicate File in System_test.go
new file mode 100644
index 000000000..f9007fee4
--- /dev/null
+++ b/leetcode/0609.Find-Duplicate-File-in-System/609. Find Duplicate File in System_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question609 struct {
+	para609
+	ans609
+}
+
+// para 是参数
+// one 代表第一个参数
+type para609 struct {
+	paths []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans609 struct {
+	one [][]string
+}
+
+func Test_Problem609(t *testing.T) {
+
+	qs := []question609{
+
+		{
+			para609{[]string{"root/a 1.txt(abcd) 2.txt(efgh)", "root/c 3.txt(abcd)", "root/c/d 4.txt(efgh)", "root 4.txt(efgh)"}},
+			ans609{[][]string{{"root/a/2.txt", "root/c/d/4.txt", "root/4.txt"}, {"root/a/1.txt", "root/c/3.txt"}}},
+		},
+
+		{
+			para609{[]string{"root/a 1.txt(abcd) 2.txt(efgh)", "root/c 3.txt(abcd)", "root/c/d 4.txt(efgh)"}},
+			ans609{[][]string{{"root/a/2.txt", "root/c/d/4.txt"}, {"root/a/1.txt", "root/c/3.txt"}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 609------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans609, q.para609
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findDuplicate(p.paths))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0609.Find-Duplicate-File-in-System/README.md b/leetcode/0609.Find-Duplicate-File-in-System/README.md
new file mode 100644
index 000000000..28341709a
--- /dev/null
+++ b/leetcode/0609.Find-Duplicate-File-in-System/README.md
@@ -0,0 +1,93 @@
+# [609. Find Duplicate File in System](https://leetcode.com/problems/find-duplicate-file-in-system/)
+
+
+## 题目
+
+Given a list `paths` of directory info, including the directory path, and all the files with contents in this directory, return *all the duplicate files in the file system in terms of their paths*. You may return the answer in **any order**.
+
+A group of duplicate files consists of at least two files that have the same content.
+
+A single directory info string in the input list has the following format:
+
+- `"root/d1/d2/.../dm f1.txt(f1_content) f2.txt(f2_content) ... fn.txt(fn_content)"`
+
+It means there are `n` files `(f1.txt, f2.txt ... fn.txt)` with content `(f1_content, f2_content ... fn_content)` respectively in the directory "`root/d1/d2/.../dm"`. Note that `n >= 1` and `m >= 0`. If `m = 0`, it means the directory is just the root directory.
+
+The output is a list of groups of duplicate file paths. For each group, it contains all the file paths of the files that have the same content. A file path is a string that has the following format:
+
+- `"directory_path/file_name.txt"`
+
+**Example 1:**
+
+```
+Input: paths = ["root/a 1.txt(abcd) 2.txt(efgh)","root/c 3.txt(abcd)","root/c/d 4.txt(efgh)","root 4.txt(efgh)"]
+Output: [["root/a/2.txt","root/c/d/4.txt","root/4.txt"],["root/a/1.txt","root/c/3.txt"]]
+
+```
+
+**Example 2:**
+
+```
+Input: paths = ["root/a 1.txt(abcd) 2.txt(efgh)","root/c 3.txt(abcd)","root/c/d 4.txt(efgh)"]
+Output: [["root/a/2.txt","root/c/d/4.txt"],["root/a/1.txt","root/c/3.txt"]]
+
+```
+
+**Constraints:**
+
+- `1 <= paths.length <= 2 * 104`
+- `1 <= paths[i].length <= 3000`
+- `1 <= sum(paths[i].length) <= 5 * 105`
+- `paths[i]` consist of English letters, digits, `'/'`, `'.'`, `'('`, `')'`, and `' '`.
+- You may assume no files or directories share the same name in the same directory.
+- You may assume each given directory info represents a unique directory. A single blank space separates the directory path and file info.
+
+**Follow up:**
+
+- Imagine you are given a real file system, how will you search files? DFS or BFS?
+- If the file content is very large (GB level), how will you modify your solution?
+- If you can only read the file by 1kb each time, how will you modify your solution?
+- What is the time complexity of your modified solution? What is the most time-consuming part and memory-consuming part of it? How to optimize?
+- How to make sure the duplicated files you find are not false positive?
+
+## 题目大意
+
+给定一个目录信息列表，包括目录路径，以及该目录中的所有包含内容的文件，您需要找到文件系统中的所有重复文件组的路径。一组重复的文件至少包括二个具有完全相同内容的文件。输入列表中的单个目录信息字符串的格式如下：`"root/d1/d2/.../dm f1.txt(f1_content) f2.txt(f2_content) ... fn.txt(fn_content)"`。这意味着有 n 个文件（`f1.txt, f2.txt ... fn.txt` 的内容分别是 `f1_content, f2_content ... fn_content`）在目录 `root/d1/d2/.../dm` 下。注意：n>=1 且 m>=0。如果 m=0，则表示该目录是根目录。该输出是重复文件路径组的列表。对于每个组，它包含具有相同内容的文件的所有文件路径。文件路径是具有下列格式的字符串：`"directory_path/file_name.txt"`
+
+## 解题思路
+
+- 这一题算简单题，考察的是字符串基本操作与 map 的使用。首先通过字符串操作获取目录路径、文件名和文件内容。再使用 map 来寻找重复文件，key 是文件内容，value 是存储路径和文件名的列表。遍历每一个文件，并把它加入 map 中。最后遍历 map，如果一个键对应的值列表的长度大于 1，说明找到了重复文件，可以把这个列表加入到最终答案中。
+- 这道题有价值的地方在 **Follow up** 中。感兴趣的读者可以仔细想想以下几个问题：
+    1. 假设您有一个真正的文件系统，您将如何搜索文件？广度搜索还是宽度搜索？
+    2. 如果文件内容非常大（GB级别），您将如何修改您的解决方案？
+    3. 如果每次只能读取 1 kb 的文件，您将如何修改解决方案？
+    4. 修改后的解决方案的时间复杂度是多少？其中最耗时的部分和消耗内存的部分是什么？如何优化？
+    5. 如何确保您发现的重复文件不是误报？
+
+## 代码
+
+```go
+package leetcode
+
+import "strings"
+
+func findDuplicate(paths []string) [][]string {
+	cache := make(map[string][]string)
+	for _, path := range paths {
+		parts := strings.Split(path, " ")
+		dir := parts[0]
+		for i := 1; i < len(parts); i++ {
+			bracketPosition := strings.IndexByte(parts[i], '(')
+			content := parts[i][bracketPosition+1 : len(parts[i])-1]
+			cache[content] = append(cache[content], dir+"/"+parts[i][:bracketPosition])
+		}
+	}
+	res := make([][]string, 0, len(cache))
+	for _, group := range cache {
+		if len(group) >= 2 {
+			res = append(res, group)
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0611.Valid-Triangle-Number/611. Valid Triangle Number.go b/leetcode/0611.Valid-Triangle-Number/611. Valid Triangle Number.go
new file mode 100644
index 000000000..69281b72f
--- /dev/null
+++ b/leetcode/0611.Valid-Triangle-Number/611. Valid Triangle Number.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+import "sort"
+
+func triangleNumber(nums []int) int {
+	res := 0
+	sort.Ints(nums)
+	for i := 0; i < len(nums)-2; i++ {
+		k := i + 2
+		for j := i + 1; j < len(nums)-1 && nums[i] != 0; j++ {
+			for k < len(nums) && nums[i]+nums[j] > nums[k] {
+				k++
+			}
+			res += k - j - 1
+		}
+	}
+	return res
+}
diff --git a/leetcode/0611.Valid-Triangle-Number/611. Valid Triangle Number_test.go b/leetcode/0611.Valid-Triangle-Number/611. Valid Triangle Number_test.go
new file mode 100644
index 000000000..c7752c1c1
--- /dev/null
+++ b/leetcode/0611.Valid-Triangle-Number/611. Valid Triangle Number_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question611 struct {
+	para611
+	ans611
+}
+
+// para 是参数
+// one 代表第一个参数
+type para611 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans611 struct {
+	one int
+}
+
+func Test_Problem611(t *testing.T) {
+
+	qs := []question611{
+
+		{
+			para611{[]int{2, 2, 3, 4}},
+			ans611{3},
+		},
+
+		{
+			para611{[]int{4, 2, 3, 4}},
+			ans611{4},
+		},
+
+		{
+			para611{[]int{0, 0}},
+			ans611{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 611------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans611, q.para611
+		fmt.Printf("【input】:%v       【output】:%v\n", p, triangleNumber(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0611.Valid-Triangle-Number/README.md b/leetcode/0611.Valid-Triangle-Number/README.md
new file mode 100644
index 000000000..a601ed595
--- /dev/null
+++ b/leetcode/0611.Valid-Triangle-Number/README.md
@@ -0,0 +1,37 @@
+# [611. Valid Triangle Number](https://leetcode.com/problems/valid-triangle-number/)
+
+## 题目
+
+Given an integer array nums, return the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.
+
+## 题目大意
+
+给定一个包含非负整数的数组，你的任务是统计其中可以组成三角形三条边的三元组个数。
+
+## 解题思路
+
+- 题意很简单，最容易想到的暴力解法是三重循环，暴力枚举，时间复杂度 O(n^3)。三重循环中最内层的循环可以优化，因为 k 和 i，j 存在关联性。第二层循环 j 从 i + 1 开始循环，k 从 j + 1 = i + 2 开始循环。循环累加 k 的值，直到 `nums[i] + nums[j] > nums[k]`，那么 `[nums[j + 1], nums[k - 1]]` 这个区间内的值都满足条件。满足条件的解个数增加 `k - j - 1` 个。j 再次递增 + 1，此时最内层的 k 不用从 j + 1 开始增加，只用从上次 k 开始增加即可。因为如果 `nums[i] + nums[j] > nums[k]`，如果这个 `nums[i] + nums[j + 1] > nums[m + 1]` 不等式成立，那么 m 一定不小于 k。所以内层循环 k 和 j 加起来的时间复杂度是 O(n)，最外层 i 的循环是 O(n)，这样优化以后，整体时间复杂度是 O(n^2)。
+- 可能有读者有疑问，三角形三条边的组成条件：任意两边之和大于第三边。`a + b > c`，`a + c > b`，`b + c > a`，此处为什么只判断了 `a + b > c` 呢？因为一开始进行了排序处理，使得 `a ≤ b ≤ c`，在这个前提下，`a + c > b`，`b + c > a` 是一定成立的。所以原问题便转化为只需关心 `a + b > c` 这一个不等式是否成立即可。此题的测试用例用有一种特殊情况，那就是其中一条边或者两条边长度为 0，那么 `a + b > c` 这个不等式一定不成立。综上，先排序预处理之后，只需要关心 `a + b > c` 这一个不等式是否成立即可。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func triangleNumber(nums []int) int {
+	res := 0
+	sort.Ints(nums)
+	for i := 0; i < len(nums)-2; i++ {
+		k := i + 2
+		for j := i + 1; j < len(nums)-1 && nums[i] != 0; j++ {
+			for k < len(nums) && nums[i]+nums[j] > nums[k] {
+				k++
+			}
+			res += k - j - 1
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0617.Merge-Two-Binary-Trees/617. Merge Two Binary Trees.go b/leetcode/0617.Merge-Two-Binary-Trees/617. Merge Two Binary Trees.go
new file mode 100644
index 000000000..a8d8e631a
--- /dev/null
+++ b/leetcode/0617.Merge-Two-Binary-Trees/617. Merge Two Binary Trees.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func mergeTrees(root1 *TreeNode, root2 *TreeNode) *TreeNode {
+	if root1 == nil {
+		return root2
+	}
+	if root2 == nil {
+		return root1
+	}
+	root1.Val += root2.Val
+	root1.Left = mergeTrees(root1.Left, root2.Left)
+	root1.Right = mergeTrees(root1.Right, root2.Right)
+	return root1
+}
diff --git a/leetcode/0617.Merge-Two-Binary-Trees/617. Merge Two Binary Trees_test.go b/leetcode/0617.Merge-Two-Binary-Trees/617. Merge Two Binary Trees_test.go
new file mode 100644
index 000000000..249bd5d87
--- /dev/null
+++ b/leetcode/0617.Merge-Two-Binary-Trees/617. Merge Two Binary Trees_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question617 struct {
+	para617
+	ans617
+}
+
+// para 是参数
+// one 代表第一个参数
+type para617 struct {
+	one     []int
+	another []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans617 struct {
+	one []int
+}
+
+func Test_Problem617(t *testing.T) {
+
+	qs := []question617{
+
+		{
+			para617{[]int{}, []int{}},
+			ans617{[]int{}},
+		},
+
+		{
+			para617{[]int{}, []int{1}},
+			ans617{[]int{1}},
+		},
+
+		{
+			para617{[]int{1, 3, 2, 5}, []int{2, 1, 3, structures.NULL, 4, structures.NULL, 7}},
+			ans617{[]int{3, 4, 5, 5, 4, structures.NULL, 7}},
+		},
+
+		{
+			para617{[]int{1}, []int{1, 2}},
+			ans617{[]int{2, 2}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 617------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans617, q.para617
+		fmt.Printf("【input】:%v      ", p)
+		root1 := structures.Ints2TreeNode(p.one)
+		root2 := structures.Ints2TreeNode(p.another)
+		fmt.Printf("【output】:%v      \n", mergeTrees(root1, root2))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0617.Merge-Two-Binary-Trees/README.md b/leetcode/0617.Merge-Two-Binary-Trees/README.md
new file mode 100644
index 000000000..714ecedbc
--- /dev/null
+++ b/leetcode/0617.Merge-Two-Binary-Trees/README.md
@@ -0,0 +1,80 @@
+# [617. Merge Two Binary Trees](https://leetcode.com/problems/merge-two-binary-trees/)
+
+
+## 题目
+
+You are given two binary trees `root1` and `root2`.
+
+Imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge the two trees into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of the new tree.
+
+Return *the merged tree*.
+
+**Note:** The merging process must start from the root nodes of both trees.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/02/05/merge.jpg](https://assets.leetcode.com/uploads/2021/02/05/merge.jpg)
+
+```
+Input: root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
+Output: [3,4,5,5,4,null,7]
+```
+
+**Example 2:**
+
+```
+Input: root1 = [1], root2 = [1,2]
+Output: [2,2]
+```
+
+**Constraints:**
+
+- The number of nodes in both trees is in the range `[0, 2000]`.
+- `104 <= Node.val <= 104`
+
+## 题目大意
+
+给定两个二叉树，想象当你将它们中的一个覆盖到另一个上时，两个二叉树的一些节点便会重叠。你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠，那么将他们的值相加作为节点合并后的新值，否则不为 NULL 的节点将直接作为新二叉树的节点。
+
+## 解题思路
+
+- 简单题。采用深搜的思路，分别从根节点开始同时遍历两个二叉树，并将对应的节点进行合并。两个二叉树的对应节点可能存在以下三种情况：
+    - 如果两个二叉树的对应节点都为空，则合并后的二叉树的对应节点也为空；
+    - 如果两个二叉树的对应节点只有一个为空，则合并后的二叉树的对应节点为其中的非空节点；
+    - 如果两个二叉树的对应节点都不为空，则合并后的二叉树的对应节点的值为两个二叉树的对应节点的值之和，此时需要显性合并两个节点。
+- 对一个节点进行合并之后，还要对该节点的左右子树分别进行合并。用递归实现即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func mergeTrees(root1 *TreeNode, root2 *TreeNode) *TreeNode {
+	if root1 == nil {
+		return root2
+	}
+	if root2 == nil {
+		return root1
+	}
+	root1.Val += root2.Val
+	root1.Left = mergeTrees(root1.Left, root2.Left)
+	root1.Right = mergeTrees(root1.Right, root2.Right)
+	return root1
+}
+```
\ No newline at end of file
diff --git a/leetcode/0622.Design-Circular-Queue/622. Design Circular Queue.go b/leetcode/0622.Design-Circular-Queue/622. Design Circular Queue.go
new file mode 100644
index 000000000..1857196f2
--- /dev/null
+++ b/leetcode/0622.Design-Circular-Queue/622. Design Circular Queue.go	
@@ -0,0 +1,71 @@
+package leetcode
+
+type MyCircularQueue struct {
+	cap   int
+	size  int
+	queue []int
+	left  int
+	right int
+}
+
+func Constructor(k int) MyCircularQueue {
+	return MyCircularQueue{cap: k, size: 0, left: 0, right: 0, queue: make([]int, k)}
+}
+
+func (this *MyCircularQueue) EnQueue(value int) bool {
+	if this.size == this.cap {
+		return false
+	}
+	this.size++
+	this.queue[this.right] = value
+	this.right++
+	this.right %= this.cap
+	return true
+
+}
+
+func (this *MyCircularQueue) DeQueue() bool {
+	if this.size == 0 {
+		return false
+	}
+	this.size--
+	this.left++
+	this.left %= this.cap
+	return true
+}
+
+func (this *MyCircularQueue) Front() int {
+	if this.size == 0 {
+		return -1
+	}
+	return this.queue[this.left]
+}
+
+func (this *MyCircularQueue) Rear() int {
+	if this.size == 0 {
+		return -1
+	}
+	if this.right == 0 {
+		return this.queue[this.cap-1]
+	}
+	return this.queue[this.right-1]
+}
+
+func (this *MyCircularQueue) IsEmpty() bool {
+	return this.size == 0
+}
+
+func (this *MyCircularQueue) IsFull() bool {
+	return this.size == this.cap
+}
+
+/**
+ * Your MyCircularQueue object will be instantiated and called as such:
+ * obj := Constructor(k);
+ * param_1 := obj.EnQueue(value);
+ * param_2 := obj.DeQueue();
+ * param_3 := obj.Front();
+ * param_4 := obj.Rear();
+ * param_5 := obj.IsEmpty();
+ * param_6 := obj.IsFull();
+ */
diff --git a/leetcode/0622.Design-Circular-Queue/622. Design Circular Queue_test.go b/leetcode/0622.Design-Circular-Queue/622. Design Circular Queue_test.go
new file mode 100644
index 000000000..3054c06d6
--- /dev/null
+++ b/leetcode/0622.Design-Circular-Queue/622. Design Circular Queue_test.go	
@@ -0,0 +1,23 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem622(t *testing.T) {
+	obj := Constructor(3)
+	fmt.Printf("obj = %v\n", obj)
+	param1 := obj.EnQueue(1)
+	fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
+	param2 := obj.DeQueue()
+	fmt.Printf("param_1 = %v obj = %v\n", param2, obj)
+	param3 := obj.Front()
+	fmt.Printf("param_1 = %v obj = %v\n", param3, obj)
+	param4 := obj.Rear()
+	fmt.Printf("param_1 = %v obj = %v\n", param4, obj)
+	param5 := obj.IsEmpty()
+	fmt.Printf("param_1 = %v obj = %v\n", param5, obj)
+	param6 := obj.IsFull()
+	fmt.Printf("param_1 = %v obj = %v\n", param6, obj)
+}
diff --git a/leetcode/0622.Design-Circular-Queue/README.md b/leetcode/0622.Design-Circular-Queue/README.md
new file mode 100644
index 000000000..149f07bf3
--- /dev/null
+++ b/leetcode/0622.Design-Circular-Queue/README.md
@@ -0,0 +1,147 @@
+# [622. Design Circular Queue](https://leetcode.com/problems/design-circular-queue/)
+
+
+## 题目
+
+Design your implementation of the circular queue. The circular queue is a linear data structure in which the operations are performed based on FIFO (First In First Out) principle and the last position is connected back to the first position to make a circle. It is also called "Ring Buffer".
+
+One of the benefits of the circular queue is that we can make use of the spaces in front of the queue. In a normal queue, once the queue becomes full, we cannot insert the next element even if there is a space in front of the queue. But using the circular queue, we can use the space to store new values.
+
+Implementation the `MyCircularQueue` class:
+
+- `MyCircularQueue(k)` Initializes the object with the size of the queue to be `k`.
+- `int Front()` Gets the front item from the queue. If the queue is empty, return `1`.
+- `int Rear()` Gets the last item from the queue. If the queue is empty, return `1`.
+- `boolean enQueue(int value)` Inserts an element into the circular queue. Return `true` if the operation is successful.
+- `boolean deQueue()` Deletes an element from the circular queue. Return `true` if the operation is successful.
+- `boolean isEmpty()` Checks whether the circular queue is empty or not.
+- `boolean isFull()` Checks whether the circular queue is full or not.
+
+**Example 1:**
+
+```
+Input
+["MyCircularQueue", "enQueue", "enQueue", "enQueue", "enQueue", "Rear", "isFull", "deQueue", "enQueue", "Rear"]
+[[3], [1], [2], [3], [4], [], [], [], [4], []]
+Output
+[null, true, true, true, false, 3, true, true, true, 4]
+
+Explanation
+MyCircularQueue myCircularQueue = new MyCircularQueue(3);
+myCircularQueue.enQueue(1); // return True
+myCircularQueue.enQueue(2); // return True
+myCircularQueue.enQueue(3); // return True
+myCircularQueue.enQueue(4); // return False
+myCircularQueue.Rear();     // return 3
+myCircularQueue.isFull();   // return True
+myCircularQueue.deQueue();  // return True
+myCircularQueue.enQueue(4); // return True
+myCircularQueue.Rear();     // return 4
+
+```
+
+**Constraints:**
+
+- `1 <= k <= 1000`
+- `0 <= value <= 1000`
+- At most `3000` calls will be made to `enQueue`, `deQueue`, `Front`, `Rear`, `isEmpty`, and `isFull`.
+
+**Follow up:**
+
+Could you solve the problem without using the built-in queue?
+
+## 题目大意
+
+设计你的循环队列实现。 循环队列是一种线性数据结构，其操作表现基于 FIFO（先进先出）原则并且队尾被连接在队首之后以形成一个循环。它也被称为“环形缓冲器”。
+
+循环队列的一个好处是我们可以利用这个队列之前用过的空间。在一个普通队列里，一旦一个队列满了，我们就不能插入下一个元素，即使在队列前面仍有空间。但是使用循环队列，我们能使用这些空间去存储新的值。
+
+你的实现应该支持如下操作：
+
+- MyCircularQueue(k): 构造器，设置队列长度为 k 。
+- Front: 从队首获取元素。如果队列为空，返回 -1 。
+- Rear: 获取队尾元素。如果队列为空，返回 -1 。
+- enQueue(value): 向循环队列插入一个元素。如果成功插入则返回真。
+- deQueue(): 从循环队列中删除一个元素。如果成功删除则返回真。
+- isEmpty(): 检查循环队列是否为空。
+- isFull(): 检查循环队列是否已满。
+
+## 解题思路
+
+- 简单题。设计一个环形队列，底层用数组实现。额外维护 4 个变量，队列的总 cap，队列当前的 size，前一元素下标 left，后一个元素下标 right。每添加一个元素便维护 left，right，size，下标需要对 cap 取余，因为超过 cap 大小之后，需要循环存储。代码实现没有难度，具体sh见下面代码。
+
+## 代码
+
+```go
+package leetcode
+
+type MyCircularQueue struct {
+	cap   int
+	size  int
+	queue []int
+	left  int
+	right int
+}
+
+func Constructor(k int) MyCircularQueue {
+	return MyCircularQueue{cap: k, size: 0, left: 0, right: 0, queue: make([]int, k)}
+}
+
+func (this *MyCircularQueue) EnQueue(value int) bool {
+	if this.size == this.cap {
+		return false
+	}
+	this.size++
+	this.queue[this.right] = value
+	this.right++
+	this.right %= this.cap
+	return true
+
+}
+
+func (this *MyCircularQueue) DeQueue() bool {
+	if this.size == 0 {
+		return false
+	}
+	this.size--
+	this.left++
+	this.left %= this.cap
+	return true
+}
+
+func (this *MyCircularQueue) Front() int {
+	if this.size == 0 {
+		return -1
+	}
+	return this.queue[this.left]
+}
+
+func (this *MyCircularQueue) Rear() int {
+	if this.size == 0 {
+		return -1
+	}
+	if this.right == 0 {
+		return this.queue[this.cap-1]
+	}
+	return this.queue[this.right-1]
+}
+
+func (this *MyCircularQueue) IsEmpty() bool {
+	return this.size == 0
+}
+
+func (this *MyCircularQueue) IsFull() bool {
+	return this.size == this.cap
+}
+
+/**
+ * Your MyCircularQueue object will be instantiated and called as such:
+ * obj := Constructor(k);
+ * param_1 := obj.EnQueue(value);
+ * param_2 := obj.DeQueue();
+ * param_3 := obj.Front();
+ * param_4 := obj.Rear();
+ * param_5 := obj.IsEmpty();
+ * param_6 := obj.IsFull();
+ */
+```
\ No newline at end of file
diff --git a/leetcode/0623.Add-One-Row-to-Tree/623. Add One Row to Tree.go b/leetcode/0623.Add-One-Row-to-Tree/623. Add One Row to Tree.go
new file mode 100644
index 000000000..309ab55c8
--- /dev/null
+++ b/leetcode/0623.Add-One-Row-to-Tree/623. Add One Row to Tree.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func addOneRow(root *TreeNode, v int, d int) *TreeNode {
+	if d == 1 {
+		tmp := &TreeNode{Val: v, Left: root, Right: nil}
+		return tmp
+	}
+	level := 1
+	addTreeRow(root, v, d, &level)
+	return root
+}
+
+func addTreeRow(root *TreeNode, v, d int, currLevel *int) {
+	if *currLevel == d-1 {
+		root.Left = &TreeNode{Val: v, Left: root.Left, Right: nil}
+		root.Right = &TreeNode{Val: v, Left: nil, Right: root.Right}
+		return
+	}
+	*currLevel++
+	if root.Left != nil {
+		addTreeRow(root.Left, v, d, currLevel)
+	}
+	if root.Right != nil {
+		addTreeRow(root.Right, v, d, currLevel)
+	}
+	*currLevel--
+}
diff --git a/leetcode/0623.Add-One-Row-to-Tree/623. Add One Row to Tree_test.go b/leetcode/0623.Add-One-Row-to-Tree/623. Add One Row to Tree_test.go
new file mode 100644
index 000000000..d13d3e51d
--- /dev/null
+++ b/leetcode/0623.Add-One-Row-to-Tree/623. Add One Row to Tree_test.go	
@@ -0,0 +1,68 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question623 struct {
+	para623
+	ans623
+}
+
+// para 是参数
+// one 代表第一个参数
+type para623 struct {
+	one []int
+	v   int
+	d   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans623 struct {
+	one []int
+}
+
+func Test_Problem623(t *testing.T) {
+
+	qs := []question623{
+
+		{
+			para623{[]int{4, 2, 6, 3, 1, 5, structures.NULL}, 1, 2},
+			ans623{[]int{4, 1, 1, 2, structures.NULL, structures.NULL, 6, 3, 1, 5, structures.NULL}},
+		},
+
+		{
+			para623{[]int{4, 2, structures.NULL, 3, 1}, 1, 3},
+			ans623{[]int{4, 2, structures.NULL, 1, 1, 3, structures.NULL, structures.NULL, 1}},
+		},
+
+		{
+			para623{[]int{1, 2, 3, 4}, 5, 4},
+			ans623{[]int{1, 2, 3, 4, structures.NULL, structures.NULL, structures.NULL, 5, 5}},
+		},
+
+		{
+			para623{[]int{4, 2, 6, 3, 1, 5}, 1, 3},
+			ans623{[]int{4, 2, 6, 1, 1, 1, 1, 3, structures.NULL, structures.NULL, 1, 5}},
+		},
+
+		{
+			para623{[]int{4, 2, 6, 3, 1, 5}, 1, 1},
+			ans623{[]int{1, 4, structures.NULL, 2, 6, 3, 1, 5}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 623------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans623, q.para623
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", structures.Tree2Preorder(addOneRow(root, p.v, p.d)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0623.Add-One-Row-to-Tree/README.md b/leetcode/0623.Add-One-Row-to-Tree/README.md
new file mode 100644
index 000000000..05f39a433
--- /dev/null
+++ b/leetcode/0623.Add-One-Row-to-Tree/README.md
@@ -0,0 +1,110 @@
+# [623. Add One Row to Tree](https://leetcode.com/problems/add-one-row-to-tree/)
+
+
+## 题目
+
+Given the root of a binary tree, then value `v` and depth `d`, you need to add a row of nodes with value `v` at the given depth `d`. The root node is at depth 1.
+
+The adding rule is: given a positive integer depth `d`, for each NOT null tree nodes `N` in depth `d-1`, create two tree nodes with value `v` as `N's` left subtree root and right subtree root. And `N's` **original left subtree** should be the left subtree of the new left subtree root, its **original right subtree** should be the right subtree of the new right subtree root. If depth `d` is 1 that means there is no depth d-1 at all, then create a tree node with value **v** as the new root of the whole original tree, and the original tree is the new root's left subtree.
+
+**Example 1:**
+
+```
+Input: 
+A binary tree as following:
+       4
+     /   \
+    2     6
+   / \   / 
+  3   1 5   
+
+v = 1d = 2Output: 
+       4
+      / \
+     1   1
+    /     \
+   2       6
+  / \     / 
+ 3   1   5   
+```
+
+**Example 2:**
+
+```
+Input: 
+A binary tree as following:
+      4
+     /   
+    2    
+   / \   
+  3   1    
+
+v = 1d = 3Output: 
+      4
+     /   
+    2
+   / \    
+  1   1
+ /     \  
+3       1
+```
+
+**Note:**
+
+1. The given d is in range [1, maximum depth of the given tree + 1].
+2. The given binary tree has at least one tree node.
+
+## 题目大意
+
+给定一个二叉树，根节点为第1层，深度为 1。在其第 d 层追加一行值为 v 的节点。添加规则：给定一个深度值 d （正整数），针对深度为 d-1 层的每一非空节点 N，为 N 创建两个值为 v 的左子树和右子树。将 N 原先的左子树，连接为新节点 v 的左子树；将 N 原先的右子树，连接为新节点 v 的右子树。如果 d 的值为 1，深度 d - 1 不存在，则创建一个新的根节点 v，原先的整棵树将作为 v 的左子树。
+
+## 解题思路
+
+- 这一题虽然是 Medium，实际非常简单。给二叉树添加一行，用 DFS 或者 BFS，遍历过程中记录行数，到达目标行一行，增加节点即可。不过需要注意 2 个特殊情况，特殊情况一，`d==1`，此时需要添加的行即为根节点。特殊情况二，`d>height(root)`，即要添加的行数比树还要高，这时只需要在最下层的叶子节点添加一层。时间复杂度 O(n)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func addOneRow(root *TreeNode, v int, d int) *TreeNode {
+	if d == 1 {
+		tmp := &TreeNode{Val: v, Left: root, Right: nil}
+		return tmp
+	}
+	level := 1
+	addTreeRow(root, v, d, &level)
+	return root
+}
+
+func addTreeRow(root *TreeNode, v, d int, currLevel *int) {
+	if *currLevel == d-1 {
+		root.Left = &TreeNode{Val: v, Left: root.Left, Right: nil}
+		root.Right = &TreeNode{Val: v, Left: nil, Right: root.Right}
+		return
+	}
+	*currLevel++
+	if root.Left != nil {
+		addTreeRow(root.Left, v, d, currLevel)
+	}
+	if root.Right != nil {
+		addTreeRow(root.Right, v, d, currLevel)
+	}
+	*currLevel--
+}
+```
\ No newline at end of file
diff --git a/leetcode/0628.Maximum-Product-of-Three-Numbers/628. Maximum Product of Three Numbers.go b/leetcode/0628.Maximum-Product-of-Three-Numbers/628. Maximum Product of Three Numbers.go
new file mode 100644
index 000000000..f3d0d1ba3
--- /dev/null
+++ b/leetcode/0628.Maximum-Product-of-Three-Numbers/628. Maximum Product of Three Numbers.go	
@@ -0,0 +1,59 @@
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+// 解法一 排序，时间复杂度 O(n log n)
+func maximumProduct(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	res := 1
+	if len(nums) <= 3 {
+		for i := 0; i < len(nums); i++ {
+			res = res * nums[i]
+		}
+		return res
+	}
+	sort.Ints(nums)
+	if nums[len(nums)-1] <= 0 {
+		return 0
+	}
+	return max(nums[0]*nums[1]*nums[len(nums)-1], nums[len(nums)-1]*nums[len(nums)-2]*nums[len(nums)-3])
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法二 模拟，时间复杂度 O(n)
+func maximumProduct1(nums []int) int {
+	max := make([]int, 0)
+	max = append(max, math.MinInt64, math.MinInt64, math.MinInt64)
+	min := make([]int, 0)
+	min = append(min, math.MaxInt64, math.MaxInt64)
+	for _, num := range nums {
+		if num > max[0] {
+			max[0], max[1], max[2] = num, max[0], max[1]
+		} else if num > max[1] {
+			max[1], max[2] = num, max[1]
+		} else if num > max[2] {
+			max[2] = num
+		}
+		if num < min[0] {
+			min[0], min[1] = num, min[0]
+		} else if num < min[1] {
+			min[1] = num
+		}
+	}
+	maxProduct1, maxProduct2 := min[0]*min[1]*max[0], max[0]*max[1]*max[2]
+	if maxProduct1 > maxProduct2 {
+		return maxProduct1
+	}
+	return maxProduct2
+}
diff --git a/Algorithms/0628. Maximum Product of Three Numbers/628. Maximum Product of Three Numbers_test.go b/leetcode/0628.Maximum-Product-of-Three-Numbers/628. Maximum Product of Three Numbers_test.go
similarity index 88%
rename from Algorithms/0628. Maximum Product of Three Numbers/628. Maximum Product of Three Numbers_test.go
rename to leetcode/0628.Maximum-Product-of-Three-Numbers/628. Maximum Product of Three Numbers_test.go
index 46f04daf8..a73a836d4 100644
--- a/Algorithms/0628. Maximum Product of Three Numbers/628. Maximum Product of Three Numbers_test.go	
+++ b/leetcode/0628.Maximum-Product-of-Three-Numbers/628. Maximum Product of Three Numbers_test.go	
@@ -26,47 +26,47 @@ func Test_Problem628(t *testing.T) {
 
 	qs := []question628{
 
-		question628{
+		{
 			para628{[]int{3, -1, 4}},
 			ans628{-12},
 		},
 
-		question628{
+		{
 			para628{[]int{1, 2}},
 			ans628{2},
 		},
 
-		question628{
+		{
 			para628{[]int{1, 2, 3}},
 			ans628{6},
 		},
 
-		question628{
+		{
 			para628{[]int{1, 2, 3, 4}},
 			ans628{24},
 		},
 
-		question628{
+		{
 			para628{[]int{-2}},
 			ans628{-2},
 		},
 
-		question628{
+		{
 			para628{[]int{0}},
 			ans628{0},
 		},
 
-		question628{
+		{
 			para628{[]int{2, 3, -2, 4}},
 			ans628{-24},
 		},
 
-		question628{
+		{
 			para628{[]int{-2, 0, -1}},
 			ans628{0},
 		},
 
-		question628{
+		{
 			para628{[]int{-2, 0, -1, 2, 3, 1, 10}},
 			ans628{60},
 		},
diff --git a/Algorithms/0628. Maximum Product of Three Numbers/README.md b/leetcode/0628.Maximum-Product-of-Three-Numbers/README.md
similarity index 100%
rename from Algorithms/0628. Maximum Product of Three Numbers/README.md
rename to leetcode/0628.Maximum-Product-of-Three-Numbers/README.md
diff --git a/leetcode/0630.Course-Schedule-III/630. Course Schedule III.go b/leetcode/0630.Course-Schedule-III/630. Course Schedule III.go
new file mode 100644
index 000000000..2d5101ddc
--- /dev/null
+++ b/leetcode/0630.Course-Schedule-III/630. Course Schedule III.go	
@@ -0,0 +1,39 @@
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+func scheduleCourse(courses [][]int) int {
+	sort.Slice(courses, func(i, j int) bool {
+		return courses[i][1] < courses[j][1]
+	})
+	maxHeap, time := &Schedule{}, 0
+	heap.Init(maxHeap)
+	for _, c := range courses {
+		if time+c[0] <= c[1] {
+			time += c[0]
+			heap.Push(maxHeap, c[0])
+		} else if (*maxHeap).Len() > 0 && (*maxHeap)[0] > c[0] {
+			time -= heap.Pop(maxHeap).(int) - c[0]
+			heap.Push(maxHeap, c[0])
+		}
+	}
+	return (*maxHeap).Len()
+}
+
+type Schedule []int
+
+func (s Schedule) Len() int           { return len(s) }
+func (s Schedule) Less(i, j int) bool { return s[i] > s[j] }
+func (s Schedule) Swap(i, j int)      { s[i], s[j] = s[j], s[i] }
+func (s *Schedule) Pop() interface{} {
+	n := len(*s)
+	t := (*s)[n-1]
+	*s = (*s)[:n-1]
+	return t
+}
+func (s *Schedule) Push(x interface{}) {
+	*s = append(*s, x.(int))
+}
diff --git a/leetcode/0630.Course-Schedule-III/630. Course Schedule III_test.go b/leetcode/0630.Course-Schedule-III/630. Course Schedule III_test.go
new file mode 100644
index 000000000..fdc727628
--- /dev/null
+++ b/leetcode/0630.Course-Schedule-III/630. Course Schedule III_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question630 struct {
+	para630
+	ans630
+}
+
+// para 是参数
+// one 代表第一个参数
+type para630 struct {
+	courses [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans630 struct {
+	one int
+}
+
+func Test_Problem630(t *testing.T) {
+
+	qs := []question630{
+
+		{
+			para630{[][]int{{100, 200}, {200, 1300}, {1000, 1250}, {2000, 3200}}},
+			ans630{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 630------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans630, q.para630
+		fmt.Printf("【input】:%v       【output】:%v\n", p, scheduleCourse(p.courses))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0630.Course-Schedule-III/README.md b/leetcode/0630.Course-Schedule-III/README.md
new file mode 100644
index 000000000..442777dda
--- /dev/null
+++ b/leetcode/0630.Course-Schedule-III/README.md
@@ -0,0 +1,96 @@
+# [630. Course Schedule III](https://leetcode.com/problems/course-schedule-iii/)
+
+## 题目
+
+There are `n` different online courses numbered from `1` to `n`. You are given an array `courses` where `courses[i] = [durationi, lastDayi]` indicate that the `ith` course should be taken **continuously** for `durationi` days and must be finished before or on `lastDayi`.
+
+You will start on the `1st` day and you cannot take two or more courses simultaneously.
+
+Return *the maximum number of courses that you can take*.
+
+**Example 1:**
+
+```
+Input: courses = [[100,200],[200,1300],[1000,1250],[2000,3200]]
+Output: 3
+Explanation:
+There are totally 4 courses, but you can take 3 courses at most:
+First, take the 1st course, it costs 100 days so you will finish it on the 100th day, and ready to take the next course on the 101st day.
+Second, take the 3rd course, it costs 1000 days so you will finish it on the 1100th day, and ready to take the next course on the 1101st day.
+Third, take the 2nd course, it costs 200 days so you will finish it on the 1300th day.
+The 4th course cannot be taken now, since you will finish it on the 3300th day, which exceeds the closed date.
+
+```
+
+**Example 2:**
+
+```
+Input: courses = [[1,2]]
+Output: 1
+
+```
+
+**Example 3:**
+
+```
+Input: courses = [[3,2],[4,3]]
+Output: 0
+
+```
+
+**Constraints:**
+
+- `1 <= courses.length <= 104`
+- `1 <= durationi, lastDayi <= 104`
+
+## 题目大意
+
+这里有 n 门不同的在线课程，他们按从 1 到 n 编号。每一门课程有一定的持续上课时间（课程时间）t 以及关闭时间第 d 天。一门课要持续学习 t 天直到第 d 天时要完成，你将会从第 1 天开始。给出 n 个在线课程用 (t, d) 对表示。你的任务是找出最多可以修几门课。
+
+## 解题思路
+
+- 一般选课，任务的题目会涉及排序 + 贪心。此题同样如此。最多修几门课，采用贪心的思路。先将课程结束时间从小到大排序，优先选择结束时间靠前的课程，这样留给后面课程的时间越多，便可以修更多的课。对排好序的课程从前往后选课，不断累积时间。如果选择修当前课程，但是会超时，这时改调整了。对于已经选择的课程，都加入到最大堆中，遇到需要调整时，比较当前待考虑的课程时长是否比(堆中)已经选择课中时长最长的课时长短，即堆顶的课程时长短，剔除 pop 它，再选择这门时长短的课，并加入最大堆中。并更新累积时间。一层循环扫完所有课程，最终最大堆中包含课程的数目便是最多可以修的课程数。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+func scheduleCourse(courses [][]int) int {
+	sort.Slice(courses, func(i, j int) bool {
+		return courses[i][1] < courses[j][1]
+	})
+	maxHeap, time := &Schedule{}, 0
+	heap.Init(maxHeap)
+	for _, c := range courses {
+		if time+c[0] <= c[1] {
+			time += c[0]
+			heap.Push(maxHeap, c[0])
+		} else if (*maxHeap).Len() > 0 && (*maxHeap)[0] > c[0] {
+			time -= heap.Pop(maxHeap).(int) - c[0]
+			heap.Push(maxHeap, c[0])
+		}
+	}
+	return (*maxHeap).Len()
+}
+
+type Schedule []int
+
+func (s Schedule) Len() int           { return len(s) }
+func (s Schedule) Less(i, j int) bool { return s[i] > s[j] }
+func (s Schedule) Swap(i, j int)      { s[i], s[j] = s[j], s[i] }
+func (s *Schedule) Pop() interface{} {
+	n := len(*s)
+	t := (*s)[n-1]
+	*s = (*s)[:n-1]
+	return t
+}
+func (s *Schedule) Push(x interface{}) {
+	*s = append(*s, x.(int))
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0632. Smallest Range Covering Elements from K Lists/632. Smallest Range Covering Elements from K Lists.go b/leetcode/0632.Smallest-Range-Covering-Elements-from-K-Lists/632. Smallest Range Covering Elements from K Lists.go
similarity index 100%
rename from Algorithms/0632. Smallest Range Covering Elements from K Lists/632. Smallest Range Covering Elements from K Lists.go
rename to leetcode/0632.Smallest-Range-Covering-Elements-from-K-Lists/632. Smallest Range Covering Elements from K Lists.go
diff --git a/Algorithms/0632. Smallest Range Covering Elements from K Lists/632. Smallest Range Covering Elements from K Lists_test.go b/leetcode/0632.Smallest-Range-Covering-Elements-from-K-Lists/632. Smallest Range Covering Elements from K Lists_test.go
similarity index 85%
rename from Algorithms/0632. Smallest Range Covering Elements from K Lists/632. Smallest Range Covering Elements from K Lists_test.go
rename to leetcode/0632.Smallest-Range-Covering-Elements-from-K-Lists/632. Smallest Range Covering Elements from K Lists_test.go
index 045b7b3f0..99d40b722 100644
--- a/Algorithms/0632. Smallest Range Covering Elements from K Lists/632. Smallest Range Covering Elements from K Lists_test.go	
+++ b/leetcode/0632.Smallest-Range-Covering-Elements-from-K-Lists/632. Smallest Range Covering Elements from K Lists_test.go	
@@ -26,8 +26,8 @@ func Test_Problem632(t *testing.T) {
 
 	qs := []question632{
 
-		question632{
-			para632{[][]int{[]int{4, 10, 15, 24, 26}, []int{0, 9, 12, 20}, []int{5, 18, 22, 30}}},
+		{
+			para632{[][]int{{4, 10, 15, 24, 26}, {0, 9, 12, 20}, {5, 18, 22, 30}}},
 			ans632{[]int{20, 24}},
 		},
 	}
diff --git a/leetcode/0632.Smallest-Range-Covering-Elements-from-K-Lists/README.md b/leetcode/0632.Smallest-Range-Covering-Elements-from-K-Lists/README.md
new file mode 100755
index 000000000..7385caf4e
--- /dev/null
+++ b/leetcode/0632.Smallest-Range-Covering-Elements-from-K-Lists/README.md
@@ -0,0 +1,46 @@
+# [632. Smallest Range Covering Elements from K Lists](https://leetcode.com/problems/smallest-range-covering-elements-from-k-lists/)
+
+
+## 题目
+
+You have `k` lists of sorted integers in ascending order. Find the **smallest** range that includes at least one number from each of the `k` lists.
+
+We define the range [a,b] is smaller than range [c,d] if `b-a < d-c` or `a < c` if `b-a == d-c`.
+
+**Example 1:**
+
+    Input: [[4,10,15,24,26], [0,9,12,20], [5,18,22,30]]
+    Output: [20,24]
+    Explanation: 
+    List 1: [4, 10, 15, 24,26], 24 is in range [20,24].
+    List 2: [0, 9, 12, 20], 20 is in range [20,24].
+    List 3: [5, 18, 22, 30], 22 is in range [20,24].
+
+**Note:**
+
+1. The given list may contain duplicates, so ascending order means >= here.
+2. 1 <= `k` <= 3500
+3. -10^5 <= `value of elements` <= 10^5.
+
+
+## 题目大意
+
+你有 k 个升序排列的整数数组。找到一个最小区间，使得 k 个列表中的每个列表至少有一个数包含在其中。
+
+我们定义如果 b-a < d-c 或者在 b-a == d-c 时 a < c，则区间 [a,b] 比 [c,d] 小。
+
+注意:
+
+- 给定的列表可能包含重复元素，所以在这里升序表示 >= 。
+- 1 <= k <= 3500
+- -105 <= 元素的值 <= 105
+- 对于使用Java的用户，请注意传入类型已修改为List<List<Integer>>。重置代码模板后可以看到这项改动。
+
+
+
+## 解题思路
+
+
+- 给出 K 个数组，要求在这 K 个数组中找到一个区间，至少能包含这 K 个数组中每个数组中的一个元素。
+- 这一题是第 76 题的变种版。第 76 题是用滑动窗口来解答的，它要求在母字符串 S 中找到最小的子串能包含 T 串的所有字母。这一题类似的，可以把母字符串看成 K 个数组合并起来的大数组，那么 T 串是由 K 个数组中每个数组中抽一个元素出来组成的。求的区间相同，都是能包含 T 的最小区间。另外一个区别在于，第 76 题里面都是字符串，这一题都是数字，在最终拼接成 T 串的时候需要保证 K 个数组中每个都有一个元素，所以理所当然的想到需要维护每个元素所在数组编号。经过上述的转换，可以把这道题转换成第 76 题的解法了。
+- 在具体解题过程中，用 map 来维护窗口内 K 个数组出现的频次。时间复杂度 O(n*log n)，空间复杂度是O(n)。
\ No newline at end of file
diff --git a/Algorithms/0633. Sum of Square Numbers/633. Sum of Square Numbers.go b/leetcode/0633.Sum-of-Square-Numbers/633. Sum of Square Numbers.go
similarity index 100%
rename from Algorithms/0633. Sum of Square Numbers/633. Sum of Square Numbers.go
rename to leetcode/0633.Sum-of-Square-Numbers/633. Sum of Square Numbers.go
diff --git a/Algorithms/0633. Sum of Square Numbers/633. Sum of Square Numbers_test.go b/leetcode/0633.Sum-of-Square-Numbers/633. Sum of Square Numbers_test.go
similarity index 89%
rename from Algorithms/0633. Sum of Square Numbers/633. Sum of Square Numbers_test.go
rename to leetcode/0633.Sum-of-Square-Numbers/633. Sum of Square Numbers_test.go
index aebd61c2c..36a6e3d6b 100644
--- a/Algorithms/0633. Sum of Square Numbers/633. Sum of Square Numbers_test.go	
+++ b/leetcode/0633.Sum-of-Square-Numbers/633. Sum of Square Numbers_test.go	
@@ -26,37 +26,37 @@ func Test_Problem633(t *testing.T) {
 
 	qs := []question633{
 
-		question633{
+		{
 			para633{1},
 			ans633{true},
 		},
 
-		question633{
+		{
 			para633{2},
 			ans633{true},
 		},
 
-		question633{
+		{
 			para633{3},
 			ans633{false},
 		},
 
-		question633{
+		{
 			para633{4},
 			ans633{true},
 		},
 
-		question633{
+		{
 			para633{5},
 			ans633{true},
 		},
 
-		question633{
+		{
 			para633{6},
 			ans633{false},
 		},
 
-		question633{
+		{
 			para633{104976},
 			ans633{true},
 		},
diff --git a/leetcode/0633.Sum-of-Square-Numbers/README.md b/leetcode/0633.Sum-of-Square-Numbers/README.md
new file mode 100755
index 000000000..de0a6e546
--- /dev/null
+++ b/leetcode/0633.Sum-of-Square-Numbers/README.md
@@ -0,0 +1,28 @@
+# [633. Sum of Square Numbers](https://leetcode.com/problems/sum-of-square-numbers/)
+
+
+## 题目
+
+Given a non-negative integer `c`, your task is to decide whether there're two integers `a` and `b` such that a^2 + b^2 = c.
+
+**Example 1:**
+
+    Input: 5
+    Output: True
+    Explanation: 1 * 1 + 2 * 2 = 5
+
+**Example 2:**
+
+    Input: 3
+    Output: False
+
+
+## 题目大意
+
+给定一个非负整数 c ，你要判断是否存在两个整数 a 和 b，使得 a^2 + b^2 = c。
+
+
+## 解题思路
+
+- 给出一个数，要求判断这个数能否由由 2 个完全平方数组成。能则输出 true，不能则输出 false。
+- 可以用二分搜索来解答这道题。判断题意，依次计算 `low * low + high * high`  和 c 是否相等。从 [0, sqrt(n)] 区间内进行二分，若能找到则返回 true，找不到就返回 false 。
diff --git a/Algorithms/0636. Exclusive Time of Functions/636. Exclusive Time of Functions.go b/leetcode/0636.Exclusive-Time-of-Functions/636. Exclusive Time of Functions.go
similarity index 100%
rename from Algorithms/0636. Exclusive Time of Functions/636. Exclusive Time of Functions.go
rename to leetcode/0636.Exclusive-Time-of-Functions/636. Exclusive Time of Functions.go
diff --git a/Algorithms/0636. Exclusive Time of Functions/636. Exclusive Time of Functions_test.go b/leetcode/0636.Exclusive-Time-of-Functions/636. Exclusive Time of Functions_test.go
similarity index 95%
rename from Algorithms/0636. Exclusive Time of Functions/636. Exclusive Time of Functions_test.go
rename to leetcode/0636.Exclusive-Time-of-Functions/636. Exclusive Time of Functions_test.go
index ea9db581c..63be5715d 100644
--- a/Algorithms/0636. Exclusive Time of Functions/636. Exclusive Time of Functions_test.go	
+++ b/leetcode/0636.Exclusive-Time-of-Functions/636. Exclusive Time of Functions_test.go	
@@ -27,17 +27,17 @@ func Test_Problem636(t *testing.T) {
 
 	qs := []question636{
 
-		question636{
+		{
 			para636{2, []string{"0:start:0", "0:start:2", "0:end:5", "1:start:7", "1:end:7", "0:end:8"}},
 			ans636{[]int{8, 1}},
 		},
 
-		question636{
+		{
 			para636{2, []string{"0:start:0", "0:start:2", "0:end:5", "1:start:6", "1:end:6", "0:end:7"}},
 			ans636{[]int{7, 1}},
 		},
 
-		question636{
+		{
 			para636{2, []string{"0:start:0", "1:start:2", "1:end:5", "0:end:6"}},
 			ans636{[]int{3, 4}},
 		},
diff --git a/Algorithms/0636. Exclusive Time of Functions/README.md b/leetcode/0636.Exclusive-Time-of-Functions/README.md
similarity index 100%
rename from Algorithms/0636. Exclusive Time of Functions/README.md
rename to leetcode/0636.Exclusive-Time-of-Functions/README.md
diff --git a/Algorithms/0637. Average of Levels in Binary Tree/637. Average of Levels in Binary Tree.go b/leetcode/0637.Average-of-Levels-in-Binary-Tree/637. Average of Levels in Binary Tree.go
similarity index 87%
rename from Algorithms/0637. Average of Levels in Binary Tree/637. Average of Levels in Binary Tree.go
rename to leetcode/0637.Average-of-Levels-in-Binary-Tree/637. Average of Levels in Binary Tree.go
index 89f783b97..372186991 100644
--- a/Algorithms/0637. Average of Levels in Binary Tree/637. Average of Levels in Binary Tree.go	
+++ b/leetcode/0637.Average-of-Levels-in-Binary-Tree/637. Average of Levels in Binary Tree.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func averageOfLevels(root *TreeNode) []float64 {
 	if root == nil {
 		return []float64{0}
diff --git a/leetcode/0637.Average-of-Levels-in-Binary-Tree/637. Average of Levels in Binary Tree_test.go b/leetcode/0637.Average-of-Levels-in-Binary-Tree/637. Average of Levels in Binary Tree_test.go
new file mode 100644
index 000000000..444072973
--- /dev/null
+++ b/leetcode/0637.Average-of-Levels-in-Binary-Tree/637. Average of Levels in Binary Tree_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question637 struct {
+	para637
+	ans637
+}
+
+// para 是参数
+// one 代表第一个参数
+type para637 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans637 struct {
+	one [][]int
+}
+
+func Test_Problem637(t *testing.T) {
+
+	qs := []question637{
+
+		{
+			para637{[]int{}},
+			ans637{[][]int{}},
+		},
+
+		{
+			para637{[]int{1}},
+			ans637{[][]int{{1}}},
+		},
+
+		{
+			para637{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
+			ans637{[][]int{{3}, {9, 20}, {15, 7}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 637------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans637, q.para637
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", averageOfLevels(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0637. Average of Levels in Binary Tree/README.md b/leetcode/0637.Average-of-Levels-in-Binary-Tree/README.md
similarity index 100%
rename from Algorithms/0637. Average of Levels in Binary Tree/README.md
rename to leetcode/0637.Average-of-Levels-in-Binary-Tree/README.md
diff --git a/leetcode/0638.Shopping-Offers/638. Shopping Offers.go b/leetcode/0638.Shopping-Offers/638. Shopping Offers.go
new file mode 100644
index 000000000..2c260d2e7
--- /dev/null
+++ b/leetcode/0638.Shopping-Offers/638. Shopping Offers.go	
@@ -0,0 +1,37 @@
+package leetcode
+
+func shoppingOffers(price []int, special [][]int, needs []int) int {
+	res := -1
+	dfsShoppingOffers(price, special, needs, 0, &res)
+	return res
+}
+
+func dfsShoppingOffers(price []int, special [][]int, needs []int, pay int, res *int) {
+	noNeeds := true
+	// 剪枝
+	for _, need := range needs {
+		if need < 0 {
+			return
+		}
+		if need != 0 {
+			noNeeds = false
+		}
+	}
+	if len(special) == 0 || noNeeds {
+		for i, p := range price {
+			pay += (p * needs[i])
+		}
+		if pay < *res || *res == -1 {
+			*res = pay
+		}
+		return
+	}
+	newNeeds := make([]int, len(needs))
+	copy(newNeeds, needs)
+	for i, n := range newNeeds {
+		newNeeds[i] = n - special[0][i]
+	}
+	dfsShoppingOffers(price, special, newNeeds, pay+special[0][len(price)], res)
+	dfsShoppingOffers(price, special[1:], newNeeds, pay+special[0][len(price)], res)
+	dfsShoppingOffers(price, special[1:], needs, pay, res)
+}
diff --git a/leetcode/0638.Shopping-Offers/638. Shopping Offers_test.go b/leetcode/0638.Shopping-Offers/638. Shopping Offers_test.go
new file mode 100644
index 000000000..1bbd3ba1d
--- /dev/null
+++ b/leetcode/0638.Shopping-Offers/638. Shopping Offers_test.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question638 struct {
+	para638
+	ans638
+}
+
+// para 是参数
+// one 代表第一个参数
+type para638 struct {
+	price   []int
+	special [][]int
+	needs   []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans638 struct {
+	one int
+}
+
+func Test_Problem638(t *testing.T) {
+
+	qs := []question638{
+
+		{
+			para638{[]int{2, 5}, [][]int{{3, 0, 5}, {1, 2, 10}}, []int{3, 2}},
+			ans638{14},
+		},
+
+		{
+			para638{[]int{2, 3, 4}, [][]int{{1, 1, 0, 4}, {2, 2, 1, 9}}, []int{1, 2, 1}},
+			ans638{11},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 638------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans638, q.para638
+		fmt.Printf("【input】:%v       【output】:%v\n", p, shoppingOffers(p.price, p.special, p.needs))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0638.Shopping-Offers/README.md b/leetcode/0638.Shopping-Offers/README.md
new file mode 100644
index 000000000..cf389acb3
--- /dev/null
+++ b/leetcode/0638.Shopping-Offers/README.md
@@ -0,0 +1,129 @@
+# [638. Shopping Offers](https://leetcode.com/problems/shopping-offers/)
+
+
+
+## 题目
+
+In LeetCode Store, there are some kinds of items to sell. Each item has a price.
+
+However, there are some special offers, and a special offer consists of one or more different kinds of items with a sale price.
+
+You are given the each item's price, a set of special offers, and the number we need to buy for each item. The job is to output the lowest price you have to pay for **exactly** certain items as given, where you could make optimal use of the special offers.
+
+Each special offer is represented in the form of an array, the last number represents the price you need to pay for this special offer, other numbers represents how many specific items you could get if you buy this offer.
+
+You could use any of special offers as many times as you want.
+
+**Example 1**:
+
+```
+Input: [2,5], [[3,0,5],[1,2,10]], [3,2]
+Output: 14
+Explanation: 
+There are two kinds of items, A and B. Their prices are $2 and $5 respectively. 
+In special offer 1, you can pay $5 for 3A and 0B
+In special offer 2, you can pay $10 for 1A and 2B. 
+You need to buy 3A and 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A.
+```
+
+**Example 2**:
+
+```
+Input: [2,3,4], [[1,1,0,4],[2,2,1,9]], [1,2,1]
+Output: 11
+Explanation: 
+The price of A is $2, and $3 for B, $4 for C. 
+You may pay $4 for 1A and 1B, and $9 for 2A ,2B and 1C. 
+You need to buy 1A ,2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, $4 for 1C. 
+You cannot add more items, though only $9 for 2A ,2B and 1C.
+```
+
+**Note**:
+
+1. There are at most 6 kinds of items, 100 special offers.
+2. For each item, you need to buy at most 6 of them.
+3. You are **not** allowed to buy more items than you want, even if that would lower the overall price.
+
+
+## 题目大意
+
+在 LeetCode 商店中， 有许多在售的物品。然而，也有一些大礼包，每个大礼包以优惠的价格捆绑销售一组物品。
+
+现给定每个物品的价格，每个大礼包包含物品的清单，以及待购物品清单。请输出确切完成待购清单的最低花费。每个大礼包的由一个数组中的一组数据描述，最后一个数字代表大礼包的价格，其他数字分别表示内含的其他种类物品的数量。任意大礼包可无限次购买。
+
+例子 1:
+
+```
+输入: [2,5], [[3,0,5],[1,2,10]], [3,2]
+输出: 14
+解释: 
+有A和B两种物品，价格分别为¥2和¥5。
+大礼包1，你可以以¥5的价格购买3A和0B。
+大礼包2， 你可以以¥10的价格购买1A和2B。
+你需要购买3个A和2个B， 所以你付了¥10购买了1A和2B（大礼包2），以及¥4购买2A。
+
+```
+例子 2:
+
+```
+输入: [2,3,4], [[1,1,0,4],[2,2,1,9]], [1,2,1]
+输出: 11
+解释: 
+A，B，C的价格分别为¥2，¥3，¥4.
+你可以用¥4购买1A和1B，也可以用¥9购买2A，2B和1C。
+你需要买1A，2B和1C，所以你付了¥4买了1A和1B（大礼包1），以及¥3购买1B， ¥4购买1C。
+你不可以购买超出待购清单的物品，尽管购买大礼包2更加便宜。
+```
+
+说明:
+
+- 最多6种物品， 100种大礼包。
+- 每种物品，你最多只需要购买6个。
+- 你不可以购买超出待购清单的物品，即使更便宜。
+
+
+## 解题思路
+
+- 给出 3 个数组，3 个数组分别代表的意义是在售的商品价格，多个礼包以及礼包内每个商品的数量和总价，购物清单上需要购买每个商品的数量。问购买清单上的所有商品所需的最低花费。
+- 这一题可以用 DFS 暴力解题，也可以用 DP。笔者这题先用 DFS 来解答。设当前搜索到的状态为 `shopping(price, special, needs)`，其中 `price` 和 `special` 为题目中所述的物品的单价和捆绑销售的大礼包，而 `needs` 为当前需要的每种物品的数量。针对于每个商品，可以有 3 种购买规则，第一种，选礼包里面的第一个优惠购买，第二种，不选当前礼包优惠，选下一个优惠进行购买，第三种，不使用优惠，直接购买。这样就对应了 3 种 DFS 的方向。具体见代码。如果选择了礼包优惠，那么递归到下一层，`need` 需要对应减少礼包里面的数量，最终金额累加。当所有情况遍历完以后，可以返回出最小花费。
+- 这一题需要注意的剪枝情况：是否需要购买礼包。题目中要求了，不能购买超过清单上数量的商品，即使价格便宜，也不行。例如可以买 n 个礼包 A，但是最终商品数量超过了清单上的商品，这种购买方式是不行的。所以需要先判断当前递归中，满足 `need` 和 `price` 条件的，能否使用礼包。这里包含 2 种情况，一种是当前商品已经满足清单个数了，不需要再买了；还有一种情况是已经超过清单数量了，那这种情况需要立即返回，当前这种购买方式不合题意。
+
+## 代码
+
+```go
+func shoppingOffers(price []int, special [][]int, needs []int) int {
+	res := -1
+	dfsShoppingOffers(price, special, needs, 0, &res)
+	return res
+}
+
+func dfsShoppingOffers(price []int, special [][]int, needs []int, pay int, res *int) {
+	noNeeds := true
+	// 剪枝
+	for _, need := range needs {
+		if need < 0 {
+			return
+		}
+		if need != 0 {
+			noNeeds = false
+		}
+	}
+	if len(special) == 0 || noNeeds {
+		for i, p := range price {
+			pay += (p * needs[i])
+		}
+		if pay < *res || *res == -1 {
+			*res = pay
+		}
+		return
+	}
+	newNeeds := make([]int, len(needs))
+	copy(newNeeds, needs)
+	for i, n := range newNeeds {
+		newNeeds[i] = n - special[0][i]
+	}
+	dfsShoppingOffers(price, special, newNeeds, pay+special[0][len(price)], res)
+	dfsShoppingOffers(price, special[1:], newNeeds, pay+special[0][len(price)], res)
+	dfsShoppingOffers(price, special[1:], needs, pay, res)
+}
+```
\ No newline at end of file
diff --git a/leetcode/0643.Maximum-Average-Subarray-I/643. Maximum Average Subarray I.go b/leetcode/0643.Maximum-Average-Subarray-I/643. Maximum Average Subarray I.go
new file mode 100644
index 000000000..ef893beb4
--- /dev/null
+++ b/leetcode/0643.Maximum-Average-Subarray-I/643. Maximum Average Subarray I.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+func findMaxAverage(nums []int, k int) float64 {
+	sum := 0
+	for _, v := range nums[:k] {
+		sum += v
+	}
+	maxSum := sum
+	for i := k; i < len(nums); i++ {
+		sum = sum - nums[i-k] + nums[i]
+		maxSum = max(maxSum, sum)
+	}
+	return float64(maxSum) / float64(k)
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0643.Maximum-Average-Subarray-I/643. Maximum Average Subarray I_test.go b/leetcode/0643.Maximum-Average-Subarray-I/643. Maximum Average Subarray I_test.go
new file mode 100644
index 000000000..1e23cfe87
--- /dev/null
+++ b/leetcode/0643.Maximum-Average-Subarray-I/643. Maximum Average Subarray I_test.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question643 struct {
+	para643
+	ans643
+}
+
+// para 是参数
+// one 代表第一个参数
+type para643 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans643 struct {
+	one float64
+}
+
+func Test_Problem643(t *testing.T) {
+
+	qs := []question643{
+
+		{
+			para643{[]int{1, 12, -5, -6, 50, 3}, 4},
+			ans643{12.75},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 643------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans643, q.para643
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findMaxAverage(p.nums, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0643.Maximum-Average-Subarray-I/README.md b/leetcode/0643.Maximum-Average-Subarray-I/README.md
new file mode 100644
index 000000000..0ddfcca6b
--- /dev/null
+++ b/leetcode/0643.Maximum-Average-Subarray-I/README.md
@@ -0,0 +1,52 @@
+# [643. Maximum Average Subarray I](https://leetcode.com/problems/maximum-average-subarray-i/)
+
+## 题目
+
+Given an array consisting of `n` integers, find the contiguous subarray of given length `k` that has the maximum average value. And you need to output the maximum average value.
+
+**Example 1:**
+
+```
+Input: [1,12,-5,-6,50,3], k = 4
+Output: 12.75
+Explanation: Maximum average is (12-5-6+50)/4 = 51/4 = 12.75
+```
+
+**Note:**
+
+1. 1 <= `k` <= `n` <= 30,000.
+2. Elements of the given array will be in the range [-10,000, 10,000].
+
+## 题目大意
+
+给定 n 个整数，找出平均数最大且长度为 k 的连续子数组，并输出该最大平均数。
+
+## 解题思路
+
+- 简单题。循环一次，扫描数组过程中累加窗口大小为 k 的元素值。不断更新这个最大值。循环结束求出平均值即可。
+
+## 代码
+
+```go
+package leetcode
+
+func findMaxAverage(nums []int, k int) float64 {
+	sum := 0
+	for _, v := range nums[:k] {
+		sum += v
+	}
+	maxSum := sum
+	for i := k; i < len(nums); i++ {
+		sum = sum - nums[i-k] + nums[i]
+		maxSum = max(maxSum, sum)
+	}
+	return float64(maxSum) / float64(k)
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0645. Set Mismatch/645. Set Mismatch.go b/leetcode/0645.Set-Mismatch/645. Set Mismatch.go
similarity index 100%
rename from Algorithms/0645. Set Mismatch/645. Set Mismatch.go
rename to leetcode/0645.Set-Mismatch/645. Set Mismatch.go
diff --git a/Algorithms/0645. Set Mismatch/645. Set Mismatch_test.go b/leetcode/0645.Set-Mismatch/645. Set Mismatch_test.go
similarity index 97%
rename from Algorithms/0645. Set Mismatch/645. Set Mismatch_test.go
rename to leetcode/0645.Set-Mismatch/645. Set Mismatch_test.go
index da842d646..2c28b43a5 100644
--- a/Algorithms/0645. Set Mismatch/645. Set Mismatch_test.go	
+++ b/leetcode/0645.Set-Mismatch/645. Set Mismatch_test.go	
@@ -26,7 +26,7 @@ func Test_Problem645(t *testing.T) {
 
 	qs := []question645{
 
-		question645{
+		{
 			para645{[]int{1, 2, 2, 4}},
 			ans645{[]int{2, 3}},
 		},
diff --git a/leetcode/0645.Set-Mismatch/README.md b/leetcode/0645.Set-Mismatch/README.md
new file mode 100755
index 000000000..73de8da35
--- /dev/null
+++ b/leetcode/0645.Set-Mismatch/README.md
@@ -0,0 +1,35 @@
+# [645. Set Mismatch](https://leetcode.com/problems/set-mismatch/)
+
+
+## 题目
+
+The set `S` originally contains numbers from 1 to `n`. But unfortunately, due to the data error, one of the numbers in the set got duplicated to **another** number in the set, which results in repetition of one number and loss of another number.
+
+Given an array `nums` representing the data status of this set after the error. Your task is to firstly find the number occurs twice and then find the number that is missing. Return them in the form of an array.
+
+**Example 1:**
+
+    Input: nums = [1,2,2,4]
+    Output: [2,3]
+
+**Note:**
+
+1. The given array size will in the range [2, 10000].
+2. The given array's numbers won't have any order.
+
+
+## 题目大意
+
+
+集合 S 包含从1到 n 的整数。不幸的是，因为数据错误，导致集合里面某一个元素复制了成了集合里面的另外一个元素的值，导致集合丢失了一个整数并且有一个元素重复。给定一个数组 nums 代表了集合 S 发生错误后的结果。你的任务是首先寻找到重复出现的整数，再找到丢失的整数，将它们以数组的形式返回。
+
+注意:  
+
+- 给定数组的长度范围是 [2, 10000]。
+- 给定的数组是无序的。
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面装的是 1-n 的数字，由于错误导致有一个数字变成了另外一个数字，要求找出重复的一个数字和正确的数字。这一题是简单题，根据下标比对就可以找到哪个数字重复了，哪个数字缺少了。
diff --git a/leetcode/0647.Palindromic-Substrings/647. Palindromic Substrings.go b/leetcode/0647.Palindromic-Substrings/647. Palindromic Substrings.go
new file mode 100644
index 000000000..01f8f7999
--- /dev/null
+++ b/leetcode/0647.Palindromic-Substrings/647. Palindromic Substrings.go	
@@ -0,0 +1,23 @@
+package leetcode
+
+func countSubstrings(s string) int {
+	res := 0
+	for i := 0; i < len(s); i++ {
+		res += countPalindrome(s, i, i)
+		res += countPalindrome(s, i, i+1)
+	}
+	return res
+}
+
+func countPalindrome(s string, left, right int) int {
+	res := 0
+	for left >= 0 && right < len(s) {
+		if s[left] != s[right] {
+			break
+		}
+		left--
+		right++
+		res++
+	}
+	return res
+}
diff --git a/leetcode/0647.Palindromic-Substrings/647. Palindromic Substrings_test.go b/leetcode/0647.Palindromic-Substrings/647. Palindromic Substrings_test.go
new file mode 100644
index 000000000..ed69135ef
--- /dev/null
+++ b/leetcode/0647.Palindromic-Substrings/647. Palindromic Substrings_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question647 struct {
+	para647
+	ans647
+}
+
+// para 是参数
+// one 代表第一个参数
+type para647 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans647 struct {
+	one int
+}
+
+func Test_Problem647(t *testing.T) {
+
+	qs := []question647{
+
+		{
+			para647{"abc"},
+			ans647{3},
+		},
+
+		{
+			para647{"aaa"},
+			ans647{6},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 647------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans647, q.para647
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countSubstrings(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0647.Palindromic-Substrings/README.md b/leetcode/0647.Palindromic-Substrings/README.md
new file mode 100644
index 000000000..79d4eae97
--- /dev/null
+++ b/leetcode/0647.Palindromic-Substrings/README.md
@@ -0,0 +1,64 @@
+# [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)
+
+
+## 题目
+
+Given a string, your task is to count how many palindromic substrings in this string.
+
+The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.
+
+**Example 1:**
+
+```
+Input: "abc"
+Output: 3
+Explanation: Three palindromic strings: "a", "b", "c".
+```
+
+**Example 2:**
+
+```
+Input: "aaa"
+Output: 6
+Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
+```
+
+**Note:**
+
+1. The input string length won't exceed 1000.
+
+## 题目大意
+
+给定一个字符串，你的任务是计算这个字符串中有多少个回文子串。具有不同开始位置或结束位置的子串，即使是由相同的字符组成，也会被视作不同的子串。
+
+## 解题思路
+
+- 暴力解法，从左往右扫一遍字符串，以每个字符做轴，用中心扩散法，依次遍历计数回文子串。
+
+## 代码
+
+```go
+package leetcode
+
+func countSubstrings(s string) int {
+	res := 0
+	for i := 0; i < len(s); i++ {
+		res += countPalindrome(s, i, i)
+		res += countPalindrome(s, i, i+1)
+	}
+	return res
+}
+
+func countPalindrome(s string, left, right int) int {
+	res := 0
+	for left >= 0 && right < len(s) {
+		if s[left] != s[right] {
+			break
+		}
+		left--
+		right++
+		res++
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0648.Replace-Words/648. Replace Words.go b/leetcode/0648.Replace-Words/648. Replace Words.go
new file mode 100644
index 000000000..0d26e0756
--- /dev/null
+++ b/leetcode/0648.Replace-Words/648. Replace Words.go	
@@ -0,0 +1,102 @@
+package leetcode
+
+import "strings"
+
+// 解法一 哈希表
+func replaceWords(dict []string, sentence string) string {
+	roots := make(map[byte][]string)
+	for _, root := range dict {
+		b := root[0]
+		roots[b] = append(roots[b], root)
+	}
+	words := strings.Split(sentence, " ")
+	for i, word := range words {
+		b := []byte(word)
+		for j := 1; j < len(b) && j <= 100; j++ {
+			if findWord(roots, b[0:j]) {
+				words[i] = string(b[0:j])
+				break
+			}
+		}
+	}
+	return strings.Join(words, " ")
+}
+
+func findWord(roots map[byte][]string, word []byte) bool {
+	if roots[word[0]] == nil {
+		return false
+	}
+	for _, root := range roots[word[0]] {
+		if root == string(word) {
+			return true
+		}
+	}
+	return false
+}
+
+// 解法二 Trie
+func replaceWords1(dict []string, sentence string) string {
+	trie := Constructor208()
+	for _, v := range dict {
+		trie.Insert(v)
+	}
+	words := strings.Split(sentence, " ")
+	var result []string
+	word := ""
+	i := 0
+	for _, value := range words {
+		word = ""
+		for i = 1; i < len(value); i++ {
+			if trie.Search(value[:i]) {
+				word = value[:i]
+				break
+			}
+		}
+
+		if len(word) == 0 {
+			result = append(result, value)
+		} else {
+			result = append(result, word)
+		}
+
+	}
+	return strings.Join(result, " ")
+}
+
+type Trie struct {
+	isWord   bool
+	children map[rune]*Trie
+}
+
+/** Initialize your data structure here. */
+func Constructor208() Trie {
+	return Trie{isWord: false, children: make(map[rune]*Trie)}
+}
+
+/** Inserts a word into the trie. */
+func (this *Trie) Insert(word string) {
+	parent := this
+	for _, ch := range word {
+		if child, ok := parent.children[ch]; ok {
+			parent = child
+		} else {
+			newChild := &Trie{children: make(map[rune]*Trie)}
+			parent.children[ch] = newChild
+			parent = newChild
+		}
+	}
+	parent.isWord = true
+}
+
+/** Returns if the word is in the trie. */
+func (this *Trie) Search(word string) bool {
+	parent := this
+	for _, ch := range word {
+		if child, ok := parent.children[ch]; ok {
+			parent = child
+			continue
+		}
+		return false
+	}
+	return parent.isWord
+}
diff --git a/Algorithms/0648. Replace Words/648. Replace Words_test.go b/leetcode/0648.Replace-Words/648. Replace Words_test.go
similarity index 97%
rename from Algorithms/0648. Replace Words/648. Replace Words_test.go
rename to leetcode/0648.Replace-Words/648. Replace Words_test.go
index f9272e0e2..998ae06cc 100644
--- a/Algorithms/0648. Replace Words/648. Replace Words_test.go	
+++ b/leetcode/0648.Replace-Words/648. Replace Words_test.go	
@@ -27,7 +27,7 @@ func Test_Problem648(t *testing.T) {
 
 	qs := []question648{
 
-		question648{
+		{
 			para648{[]string{"cat", "bat", "rat"}, "the cattle was rattled by the battery"},
 			ans648{"the cat was rat by the bat"},
 		},
diff --git a/Algorithms/0648. Replace Words/README.md b/leetcode/0648.Replace-Words/README.md
similarity index 100%
rename from Algorithms/0648. Replace Words/README.md
rename to leetcode/0648.Replace-Words/README.md
diff --git a/leetcode/0653.Two-Sum-IV-Input-is-a-BST/653. Two Sum IV - Input is a BST.go b/leetcode/0653.Two-Sum-IV-Input-is-a-BST/653. Two Sum IV - Input is a BST.go
new file mode 100644
index 000000000..03fbc0037
--- /dev/null
+++ b/leetcode/0653.Two-Sum-IV-Input-is-a-BST/653. Two Sum IV - Input is a BST.go	
@@ -0,0 +1,33 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func findTarget(root *TreeNode, k int) bool {
+	m := make(map[int]int, 0)
+	return findTargetDFS(root, k, m)
+}
+
+func findTargetDFS(root *TreeNode, k int, m map[int]int) bool {
+	if root == nil {
+		return false
+	}
+	if _, ok := m[k-root.Val]; ok {
+		return ok
+	}
+	m[root.Val]++
+	return findTargetDFS(root.Left, k, m) || findTargetDFS(root.Right, k, m)
+}
diff --git a/leetcode/0653.Two-Sum-IV-Input-is-a-BST/653. Two Sum IV - Input is a BST_test.go b/leetcode/0653.Two-Sum-IV-Input-is-a-BST/653. Two Sum IV - Input is a BST_test.go
new file mode 100644
index 000000000..ced539c9e
--- /dev/null
+++ b/leetcode/0653.Two-Sum-IV-Input-is-a-BST/653. Two Sum IV - Input is a BST_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question653 struct {
+	para653
+	ans653
+}
+
+// para 是参数
+// one 代表第一个参数
+type para653 struct {
+	one []int
+	k   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans653 struct {
+	one bool
+}
+
+func Test_Problem653(t *testing.T) {
+
+	qs := []question653{
+
+		{
+			para653{[]int{}, 0},
+			ans653{false},
+		},
+
+		{
+			para653{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}, 29},
+			ans653{true},
+		},
+
+		{
+			para653{[]int{1, 2, 3, 4, structures.NULL, structures.NULL, 5}, 9},
+			ans653{true},
+		},
+
+		{
+			para653{[]int{1, 2, 3, 4, structures.NULL, 5}, 4},
+			ans653{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 653------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans653, q.para653
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", findTarget(root, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0653.Two-Sum-IV-Input-is-a-BST/README.md b/leetcode/0653.Two-Sum-IV-Input-is-a-BST/README.md
new file mode 100755
index 000000000..329972437
--- /dev/null
+++ b/leetcode/0653.Two-Sum-IV-Input-is-a-BST/README.md
@@ -0,0 +1,42 @@
+# [653. Two Sum IV - Input is a BST](https://leetcode.com/problems/two-sum-iv-input-is-a-bst/)
+
+## 题目
+
+Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.
+
+**Example 1:**
+
+    Input: 
+        5
+       / \
+      3   6
+     / \   \
+    2   4   7
+    
+    Target = 9
+    
+    Output: True
+
+**Example 2:**
+
+    Input: 
+        5
+       / \
+      3   6
+     / \   \
+    2   4   7
+    
+    Target = 28
+    
+    Output: False
+
+
+## 题目大意
+
+给定一个二叉搜索树和一个目标结果，如果 BST 中存在两个元素且它们的和等于给定的目标结果，则返回 true。
+
+## 解题思路
+
+
+- 在树中判断是否存在 2 个数的和是 sum。
+- 这一题是 two sum 问题的变形题，只不过题目背景是在 BST 上处理的。处理思路大体一致，用 map 记录已经访问过的节点值。边遍历树边查看 map 里面是否有 sum 的另外一半。
diff --git a/leetcode/0658.Find-K-Closest-Elements/658. Find K Closest Elements.go b/leetcode/0658.Find-K-Closest-Elements/658. Find K Closest Elements.go
new file mode 100644
index 000000000..2385e0e0c
--- /dev/null
+++ b/leetcode/0658.Find-K-Closest-Elements/658. Find K Closest Elements.go	
@@ -0,0 +1,22 @@
+package leetcode
+
+import "sort"
+
+// 解法一 库函数二分搜索
+func findClosestElements(arr []int, k int, x int) []int {
+	return arr[sort.Search(len(arr)-k, func(i int) bool { return x-arr[i] <= arr[i+k]-x }):][:k]
+}
+
+// 解法二 手撸二分搜索
+func findClosestElements1(arr []int, k int, x int) []int {
+	low, high := 0, len(arr)-k
+	for low < high {
+		mid := low + (high-low)>>1
+		if x-arr[mid] > arr[mid+k]-x {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return arr[low : low+k]
+}
diff --git a/leetcode/0658.Find-K-Closest-Elements/658. Find K Closest Elements_test.go b/leetcode/0658.Find-K-Closest-Elements/658. Find K Closest Elements_test.go
new file mode 100644
index 000000000..790d515e2
--- /dev/null
+++ b/leetcode/0658.Find-K-Closest-Elements/658. Find K Closest Elements_test.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question658 struct {
+	para658
+	ans658
+}
+
+// para 是参数
+// one 代表第一个参数
+type para658 struct {
+	arr []int
+	k   int
+	x   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans658 struct {
+	one []int
+}
+
+func Test_Problem658(t *testing.T) {
+
+	qs := []question658{
+
+		{
+			para658{[]int{1, 2, 3, 4, 5}, 4, 3},
+			ans658{[]int{1, 2, 3, 4}},
+		},
+
+		{
+			para658{[]int{1, 2, 3, 4, 5}, 4, -1},
+			ans658{[]int{1, 2, 3, 4}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 658------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans658, q.para658
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findClosestElements(p.arr, p.k, p.x))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0658.Find-K-Closest-Elements/README.md b/leetcode/0658.Find-K-Closest-Elements/README.md
new file mode 100755
index 000000000..d56c813a7
--- /dev/null
+++ b/leetcode/0658.Find-K-Closest-Elements/README.md
@@ -0,0 +1,52 @@
+# [658. Find K Closest Elements](https://leetcode.com/problems/find-k-closest-elements/)
+
+
+## 题目
+
+Given a sorted array, two integers `k` and `x`, find the `k` closest elements to `x` in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred.
+
+**Example 1:**
+
+    Input: [1,2,3,4,5], k=4, x=3
+    Output: [1,2,3,4]
+
+**Example 2:**
+
+    Input: [1,2,3,4,5], k=4, x=-1
+    Output: [1,2,3,4]
+
+**Note:**
+
+1. The value k is positive and will always be smaller than the length of the sorted array.
+2. Length of the given array is positive and will not exceed 10^4
+3. Absolute value of elements in the array and x will not exceed 10^4
+
+---
+
+**UPDATE (2017/9/19):**The arr parameter had been changed to an **array of integers** (instead of a list of integers). **Please reload the code definition to get the latest changes**.
+
+
+## 题目大意
+
+
+给定一个排序好的数组，两个整数 k 和 x，从数组中找到最靠近 x（两数之差最小）的 k 个数。返回的结果必须要是按升序排好的。如果有两个数与 x 的差值一样，优先选择数值较小的那个数。
+
+
+说明:
+
+1. k 的值为正数，且总是小于给定排序数组的长度。
+2. 数组不为空，且长度不超过 104
+3. 数组里的每个元素与 x 的绝对值不超过 104
+ 
+
+更新(2017/9/19):
+这个参数 arr 已经被改变为一个整数数组（而不是整数列表）。 请重新加载代码定义以获取最新更改。
+
+
+
+
+## 解题思路
+
+
+- 给出一个数组，要求在数组中找到一个长度为 k 的区间，这个区间内每个元素距离 x 的距离都是整个数组里面最小的。
+- 这一题可以用双指针解题，最优解法是二分搜索。由于区间长度固定是 K 个，所以左区间最大只能到 `len(arr) - K` (因为长度为 K 以后，正好右区间就到数组最右边了)，在 `[0,len(arr) - K]` 这个区间中进行二分搜索。如果发现 `a[mid]` 与 `x` 距离比 `a[mid + k]` 与 `x` 的距离要大，说明要找的区间一定在右侧，继续二分，直到最终 `low = high` 的时候退出。逼出的 `low` 值就是最终答案区间的左边界。
diff --git a/leetcode/0661.Image-Smoother/661. Image Smoother.go b/leetcode/0661.Image-Smoother/661. Image Smoother.go
new file mode 100644
index 000000000..b6d6fa3ff
--- /dev/null
+++ b/leetcode/0661.Image-Smoother/661. Image Smoother.go	
@@ -0,0 +1,60 @@
+package leetcode
+
+func imageSmoother(M [][]int) [][]int {
+	res := make([][]int, len(M))
+	for i := range M {
+		res[i] = make([]int, len(M[0]))
+	}
+	for y := 0; y < len(M); y++ {
+		for x := 0; x < len(M[0]); x++ {
+			res[y][x] = smooth(x, y, M)
+		}
+	}
+	return res
+}
+
+func smooth(x, y int, M [][]int) int {
+	count, sum := 1, M[y][x]
+	// Check bottom
+	if y+1 < len(M) {
+		sum += M[y+1][x]
+		count++
+	}
+	// Check Top
+	if y-1 >= 0 {
+		sum += M[y-1][x]
+		count++
+	}
+	// Check left
+	if x-1 >= 0 {
+		sum += M[y][x-1]
+		count++
+	}
+	// Check Right
+	if x+1 < len(M[y]) {
+		sum += M[y][x+1]
+		count++
+	}
+	// Check Coners
+	// Top Left
+	if y-1 >= 0 && x-1 >= 0 {
+		sum += M[y-1][x-1]
+		count++
+	}
+	// Top Right
+	if y-1 >= 0 && x+1 < len(M[0]) {
+		sum += M[y-1][x+1]
+		count++
+	}
+	// Bottom Left
+	if y+1 < len(M) && x-1 >= 0 {
+		sum += M[y+1][x-1]
+		count++
+	}
+	//Bottom Right
+	if y+1 < len(M) && x+1 < len(M[0]) {
+		sum += M[y+1][x+1]
+		count++
+	}
+	return sum / count
+}
diff --git a/leetcode/0661.Image-Smoother/661. Image Smoother_test.go b/leetcode/0661.Image-Smoother/661. Image Smoother_test.go
new file mode 100644
index 000000000..0fa3a52fd
--- /dev/null
+++ b/leetcode/0661.Image-Smoother/661. Image Smoother_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question661 struct {
+	para661
+	ans661
+}
+
+// para 是参数
+// one 代表第一个参数
+type para661 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans661 struct {
+	one [][]int
+}
+
+func Test_Problem661(t *testing.T) {
+
+	qs := []question661{
+
+		{
+			para661{[][]int{{1, 1, 1}, {1, 1, 2}}},
+			ans661{[][]int{{1, 1, 1}, {1, 1, 1}}},
+		},
+
+		{
+			para661{[][]int{{1, 1, 1}, {1, 1, 2}, {1, 1, 1}}},
+			ans661{[][]int{{1, 1, 1}, {1, 1, 1}, {1, 1, 1}}},
+		},
+
+		{
+			para661{[][]int{{1, 1, 1}, {1, 0, 1}, {1, 1, 1}}},
+			ans661{[][]int{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 661------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans661, q.para661
+		fmt.Printf("【input】:%v       【output】:%v\n", p, imageSmoother(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0661.Image-Smoother/README.md b/leetcode/0661.Image-Smoother/README.md
new file mode 100644
index 000000000..4cfa74d69
--- /dev/null
+++ b/leetcode/0661.Image-Smoother/README.md
@@ -0,0 +1,107 @@
+# [661. Image Smoother](https://leetcode.com/problems/image-smoother/)
+
+## 题目
+
+Given a 2D integer matrix M representing the gray scale of an image, you need to design a smoother to make the gray scale of each cell becomes the average gray scale (rounding down) of all the 8 surrounding cells and itself. If a cell has less than 8 surrounding cells, then use as many as you can.
+
+**Example 1**:
+
+```
+Input:
+[[1,1,1],
+ [1,0,1],
+ [1,1,1]]
+Output:
+[[0, 0, 0],
+ [0, 0, 0],
+ [0, 0, 0]]
+Explanation:
+For the point (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0
+For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0
+For the point (1,1): floor(8/9) = floor(0.88888889) = 0
+
+```
+
+**Note**:
+
+1. The value in the given matrix is in the range of [0, 255].
+2. The length and width of the given matrix are in the range of [1, 150].
+
+
+## 题目大意
+
+包含整数的二维矩阵 M 表示一个图片的灰度。你需要设计一个平滑器来让每一个单元的灰度成为平均灰度 (向下舍入) ，平均灰度的计算是周围的8个单元和它本身的值求平均，如果周围的单元格不足八个，则尽可能多的利用它们。
+
+注意:
+
+- 给定矩阵中的整数范围为 [0, 255]。
+- 矩阵的长和宽的范围均为 [1, 150]。
+
+
+## 解题思路
+
+- 将二维数组中的每个元素变为周围 9 个元素的平均值。
+- 简单题，按照题意计算平均值即可。需要注意的是边界问题，四个角和边上的元素，这些点计算平均值的时候，计算平均值都不足 9 个元素。
+
+## 代码
+
+```go
+func imageSmoother(M [][]int) [][]int {
+	res := make([][]int, len(M))
+	for i := range M {
+		res[i] = make([]int, len(M[0]))
+	}
+	for y := 0; y < len(M); y++ {
+		for x := 0; x < len(M[0]); x++ {
+			res[y][x] = smooth(x, y, M)
+		}
+	}
+	return res
+}
+
+func smooth(x, y int, M [][]int) int {
+	count, sum := 1, M[y][x]
+	// Check bottom
+	if y+1 < len(M) {
+		sum += M[y+1][x]
+		count++
+	}
+	// Check Top
+	if y-1 >= 0 {
+		sum += M[y-1][x]
+		count++
+	}
+	// Check left
+	if x-1 >= 0 {
+		sum += M[y][x-1]
+		count++
+	}
+	// Check Right
+	if x+1 < len(M[y]) {
+		sum += M[y][x+1]
+		count++
+	}
+	// Check Coners
+	// Top Left
+	if y-1 >= 0 && x-1 >= 0 {
+		sum += M[y-1][x-1]
+		count++
+	}
+	// Top Right
+	if y-1 >= 0 && x+1 < len(M[0]) {
+		sum += M[y-1][x+1]
+		count++
+	}
+	// Bottom Left
+	if y+1 < len(M) && x-1 >= 0 {
+		sum += M[y+1][x-1]
+		count++
+	}
+	//Bottom Right
+	if y+1 < len(M) && x+1 < len(M[0]) {
+		sum += M[y+1][x+1]
+		count++
+	}
+	return sum / count
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0662. Maximum Width of Binary Tree/662. Maximum Width of Binary Tree.go b/leetcode/0662.Maximum-Width-of-Binary-Tree/662. Maximum Width of Binary Tree.go
similarity index 76%
rename from Algorithms/0662. Maximum Width of Binary Tree/662. Maximum Width of Binary Tree.go
rename to leetcode/0662.Maximum-Width-of-Binary-Tree/662. Maximum Width of Binary Tree.go
index 17f552a22..2a8294974 100644
--- a/Algorithms/0662. Maximum Width of Binary Tree/662. Maximum Width of Binary Tree.go	
+++ b/leetcode/0662.Maximum-Width-of-Binary-Tree/662. Maximum Width of Binary Tree.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
@@ -8,6 +15,7 @@ package leetcode
  *     Right *TreeNode
  * }
  */
+
 func widthOfBinaryTree(root *TreeNode) int {
 	if root == nil {
 		return 0
@@ -17,7 +25,7 @@ func widthOfBinaryTree(root *TreeNode) int {
 	}
 
 	queue, res := []*TreeNode{}, 0
-	queue = append(queue, &TreeNode{0, root.Left, root.Right})
+	queue = append(queue, &TreeNode{Val: 0, Left: root.Left, Right: root.Right})
 
 	for len(queue) != 0 {
 		var left, right *int
@@ -30,7 +38,7 @@ func widthOfBinaryTree(root *TreeNode) int {
 			if node.Left != nil {
 				// 根据满二叉树父子节点的关系，得到下一层节点在本层的编号
 				newVal := node.Val * 2
-				queue = append(queue, &TreeNode{newVal, node.Left.Left, node.Left.Right})
+				queue = append(queue, &TreeNode{Val: newVal, Left: node.Left.Left, Right: node.Left.Right})
 				if left == nil || *left > newVal {
 					left = &newVal
 				}
@@ -41,7 +49,7 @@ func widthOfBinaryTree(root *TreeNode) int {
 			if node.Right != nil {
 				// 根据满二叉树父子节点的关系，得到下一层节点在本层的编号
 				newVal := node.Val*2 + 1
-				queue = append(queue, &TreeNode{newVal, node.Right.Left, node.Right.Right})
+				queue = append(queue, &TreeNode{Val: newVal, Left: node.Right.Left, Right: node.Right.Right})
 				if left == nil || *left > newVal {
 					left = &newVal
 				}
@@ -61,3 +69,10 @@ func widthOfBinaryTree(root *TreeNode) int {
 	}
 	return res
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0662.Maximum-Width-of-Binary-Tree/662. Maximum Width of Binary Tree_test.go b/leetcode/0662.Maximum-Width-of-Binary-Tree/662. Maximum Width of Binary Tree_test.go
new file mode 100644
index 000000000..decdffcb8
--- /dev/null
+++ b/leetcode/0662.Maximum-Width-of-Binary-Tree/662. Maximum Width of Binary Tree_test.go	
@@ -0,0 +1,66 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question662 struct {
+	para662
+	ans662
+}
+
+// para 是参数
+// one 代表第一个参数
+type para662 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans662 struct {
+	one int
+}
+
+func Test_Problem662(t *testing.T) {
+
+	qs := []question662{
+
+		{
+			para662{[]int{1, 1, 1, 1, 1, 1, 1, structures.NULL, structures.NULL, structures.NULL, 1, structures.NULL, structures.NULL, structures.NULL, structures.NULL, 2, 2, 2, 2, 2, 2, 2, structures.NULL, 2, structures.NULL, structures.NULL, 2, structures.NULL, 2}},
+			ans662{8},
+		},
+
+		{
+			para662{[]int{1, 1, 1, 1, structures.NULL, structures.NULL, 1, 1, structures.NULL, structures.NULL, 1}},
+			ans662{8},
+		},
+
+		{
+			para662{[]int{}},
+			ans662{0},
+		},
+
+		{
+			para662{[]int{1}},
+			ans662{1},
+		},
+
+		{
+			para662{[]int{1, 3, 2, 5, 3, structures.NULL, 9}},
+			ans662{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 662------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans662, q.para662
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", widthOfBinaryTree(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0662.Maximum-Width-of-Binary-Tree/README.md b/leetcode/0662.Maximum-Width-of-Binary-Tree/README.md
new file mode 100755
index 000000000..abff24817
--- /dev/null
+++ b/leetcode/0662.Maximum-Width-of-Binary-Tree/README.md
@@ -0,0 +1,82 @@
+# [662. Maximum Width of Binary Tree](https://leetcode.com/problems/maximum-width-of-binary-tree/)
+
+
+## 题目
+
+Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a **full binary tree**, but some nodes are null.
+
+The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the `null` nodes between the end-nodes are also counted into the length calculation.
+
+**Example 1:**
+
+    Input: 
+    
+               1
+             /   \
+            3     2
+           / \     \  
+          5   3     9 
+    
+    Output: 4
+    Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).
+
+**Example 2:**
+
+    Input: 
+    
+              1
+             /  
+            3    
+           / \       
+          5   3     
+    
+    Output: 2
+    Explanation: The maximum width existing in the third level with the length 2 (5,3).
+
+**Example 3:**
+
+    Input: 
+    
+              1
+             / \
+            3   2 
+           /        
+          5      
+    
+    Output: 2
+    Explanation: The maximum width existing in the second level with the length 2 (3,2).
+
+**Example 4:**
+
+    Input: 
+    
+              1
+             / \
+            3   2
+           /     \  
+          5       9 
+         /         \
+        6           7
+    Output: 8
+    Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).
+
+**Note:** Answer will in the range of 32-bit signed integer.
+
+
+## 题目大意
+
+给定一个二叉树，编写一个函数来获取这个树的最大宽度。树的宽度是所有层中的最大宽度。这个二叉树与满二叉树（full binary tree）结构相同，但一些节点为空。
+
+每一层的宽度被定义为两个端点（该层最左和最右的非空节点，两端点间的null节点也计入长度）之间的长度。
+
+注意: 答案在32位有符号整数的表示范围内。
+
+
+
+## 解题思路
+
+
+- 给出一个二叉树，求这棵树最宽的部分。
+- 这一题可以用 BFS 也可以用 DFS，但是用 BFS 比较方便。按照层序遍历，依次算出每层最左边不为 `null` 的节点和最右边不为 `null` 的节点。这两个节点之间都是算宽度的。最终输出最大的宽度即可。此题的关键在于如何有效的找到每一层的左右边界。
+- 这一题可能有人会想着先补全满二叉树，然后每层分别找左右边界。这种方法提交以后会卡在 `104 / 108` 这组测试用例上，这组测试用例会使得最后某几层填充出现的满二叉树节点特别多，最终导致 `Memory Limit Exceeded` 了。
+- 由于此题要找每层的左右边界，实际上每个节点的 `Val` 值是我们不关心的，那么可以把这个值用来标号，标记成该节点在每层中的序号。父亲节点在上一层中的序号是 x，那么它的左孩子在下一层满二叉树中的序号是 `2*x`，它的右孩子在下一层满二叉树中的序号是 `2*x + 1`。将所有节点都标上号，用 BFS 层序遍历每一层，每一层都找到左右边界，相减拿到宽度，动态维护最大宽度，就是本题的最终答案。
\ No newline at end of file
diff --git a/leetcode/0665.Non-decreasing-Array/665. Non-decreasing Array.go b/leetcode/0665.Non-decreasing-Array/665. Non-decreasing Array.go
new file mode 100644
index 000000000..e0f00296a
--- /dev/null
+++ b/leetcode/0665.Non-decreasing-Array/665. Non-decreasing Array.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+func checkPossibility(nums []int) bool {
+	count := 0
+	for i := 0; i < len(nums)-1; i++ {
+		if nums[i] > nums[i+1] {
+			count++
+			if count > 1 {
+				return false
+			}
+			if i > 0 && nums[i+1] < nums[i-1] {
+				nums[i+1] = nums[i]
+			}
+		}
+	}
+	return true
+}
diff --git a/leetcode/0665.Non-decreasing-Array/665. Non-decreasing Array_test.go b/leetcode/0665.Non-decreasing-Array/665. Non-decreasing Array_test.go
new file mode 100644
index 000000000..df3f2acd1
--- /dev/null
+++ b/leetcode/0665.Non-decreasing-Array/665. Non-decreasing Array_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question665 struct {
+	para665
+	ans665
+}
+
+// para 是参数
+// one 代表第一个参数
+type para665 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans665 struct {
+	one bool
+}
+
+func Test_Problem665(t *testing.T) {
+
+	qs := []question665{
+
+		{
+			para665{[]int{4, 2, 3}},
+			ans665{true},
+		},
+
+		{
+			para665{[]int{4, 2, 1}},
+			ans665{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 665------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans665, q.para665
+		fmt.Printf("【input】:%v       【output】:%v\n", p, checkPossibility(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0665.Non-decreasing-Array/README.md b/leetcode/0665.Non-decreasing-Array/README.md
new file mode 100644
index 000000000..abc759e41
--- /dev/null
+++ b/leetcode/0665.Non-decreasing-Array/README.md
@@ -0,0 +1,59 @@
+# [665. Non-decreasing Array](https://leetcode.com/problems/non-decreasing-array/)
+
+## 题目
+
+Given an array `nums` with `n` integers, your task is to check if it could become non-decreasing by modifying **at most one element**.
+
+We define an array is non-decreasing if `nums[i] <= nums[i + 1]` holds for every `i` (**0-based**) such that (`0 <= i <= n - 2`).
+
+**Example 1:**
+
+```
+Input: nums = [4,2,3]
+Output: true
+Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
+```
+
+**Example 2:**
+
+```
+Input: nums = [4,2,1]
+Output: false
+Explanation: You can't get a non-decreasing array by modify at most one element.
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `1 <= n <= 104`
+- `-10^5 <= nums[i] <= 10^5`
+
+## 题目大意
+
+给你一个长度为 n 的整数数组，请你判断在 最多 改变 1 个元素的情况下，该数组能否变成一个非递减数列。我们是这样定义一个非递减数列的： 对于数组中任意的 i (0 <= i <= n-2)，总满足 nums[i] <= nums[i + 1]。
+
+## 解题思路
+
+- 简单题。循环扫描数组，找到 `nums[i] > nums[i+1]` 这种递减组合。一旦这种组合超过 2 组，直接返回 false。找到第一组递减组合，需要手动调节一次。如果 `nums[i + 1] < nums[i - 1]`，就算交换 `nums[i+1]` 和 `nums[i]`，交换结束，`nums[i - 1]` 仍然可能大于 `nums[i + 1]`，不满足题意。正确的做法应该是让较小的那个数变大，即 `nums[i + 1] = nums[i]`。两个元素相等满足非递减的要求。
+
+## 代码
+
+```go
+package leetcode
+
+func checkPossibility(nums []int) bool {
+	count := 0
+	for i := 0; i < len(nums)-1; i++ {
+		if nums[i] > nums[i+1] {
+			count++
+			if count > 1 {
+				return false
+			}
+			if i > 0 && nums[i+1] < nums[i-1] {
+				nums[i+1] = nums[i]
+			}
+		}
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/0667.Beautiful-Arrangement-II/667. Beautiful Arrangement II.go b/leetcode/0667.Beautiful-Arrangement-II/667. Beautiful Arrangement II.go
new file mode 100644
index 000000000..6cbb12666
--- /dev/null
+++ b/leetcode/0667.Beautiful-Arrangement-II/667. Beautiful Arrangement II.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func constructArray(n int, k int) []int {
+	res := []int{}
+	for i := 0; i < n-k-1; i++ {
+		res = append(res, i+1)
+	}
+	for i := n - k; i < n-k+(k+1)/2; i++ {
+		res = append(res, i)
+		res = append(res, 2*n-k-i)
+	}
+	if k%2 == 0 {
+		res = append(res, n-k+(k+1)/2)
+	}
+	return res
+}
diff --git a/leetcode/0667.Beautiful-Arrangement-II/667. Beautiful Arrangement II_test.go b/leetcode/0667.Beautiful-Arrangement-II/667. Beautiful Arrangement II_test.go
new file mode 100644
index 000000000..1b97bfee3
--- /dev/null
+++ b/leetcode/0667.Beautiful-Arrangement-II/667. Beautiful Arrangement II_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question667 struct {
+	para667
+	ans667
+}
+
+// para 是参数
+// one 代表第一个参数
+type para667 struct {
+	n int
+	k int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans667 struct {
+	one []int
+}
+
+func Test_Problem667(t *testing.T) {
+
+	qs := []question667{
+
+		{
+			para667{3, 1},
+			ans667{[]int{1, 2, 3}},
+		},
+
+		{
+			para667{3, 2},
+			ans667{[]int{1, 3, 2}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 667------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans667, q.para667
+		fmt.Printf("【input】:%v       【output】:%v\n", p, constructArray(p.n, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0667.Beautiful-Arrangement-II/README.md b/leetcode/0667.Beautiful-Arrangement-II/README.md
new file mode 100644
index 000000000..a707cdde7
--- /dev/null
+++ b/leetcode/0667.Beautiful-Arrangement-II/README.md
@@ -0,0 +1,62 @@
+# [667. Beautiful Arrangement II](https://leetcode.com/problems/beautiful-arrangement-ii/)
+
+
+## 题目
+
+Given two integers `n` and `k`, you need to construct a list which contains `n` different positive integers ranging from `1` to `n` and obeys the following requirement:Suppose this list is [a1, a2, a3, ... , an], then the list [|a1 - a2|, |a2 - a3|, |a3 - a4|, ... , |an-1 - an|] has exactly `k` distinct integers.
+
+If there are multiple answers, print any of them.
+
+**Example 1:**
+
+```
+Input: n = 3, k = 1
+Output: [1, 2, 3]
+Explanation: The [1, 2, 3] has three different positive integers ranging from 1 to 3, and the [1, 1] has exactly 1 distinct integer: 1.
+```
+
+**Example 2:**
+
+```
+Input: n = 3, k = 2
+Output: [1, 3, 2]
+Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, and the [2, 1] has exactly 2 distinct integers: 1 and 2.
+```
+
+**Note:**
+
+1. The `n` and `k` are in the range 1 <= k < n <= 10^4.
+
+## 题目大意
+
+给定两个整数 n 和 k，你需要实现一个数组，这个数组包含从 1 到 n 的 n 个不同整数，同时满足以下条件：
+
+- 如果这个数组是 [a1, a2, a3, ... , an] ，那么数组 [|a1 - a2|, |a2 - a3|, |a3 - a4|, ... , |an-1 - an|] 中应该有且仅有 k 个不同整数；.
+- 如果存在多种答案，你只需实现并返回其中任意一种.
+
+## 解题思路
+
+- 先考虑 `k` 最大值的情况。如果把末尾的较大值依次插入到前面的较小值中，形成 `[1，n，2，n-1，3，n-2，……]`，这样排列 `k` 能取到最大值 `n-1` 。`k` 最小值的情况是 `[1，2，3，4，……，n]`，`k` 取到的最小值是 1。那么 `k` 在 `[1，n-1]` 之间取值，该怎么排列呢？先顺序排列 `[1，2，3，4，……，n-k-1]`，这里有 `n-k-1` 个数，可以形成唯一一种差值。剩下 `k+1` 个数，形成 `k-1` 种差值。
+- 这又回到了 `k` 最大值的取法了。`k` 取最大值的情况是 `n` 个数，形成 `n-1` 个不同种的差值。现在 `k+1` 个数，需要形成 `k` 种不同的差值。两者是同一个问题。那么剩下 `k` 个数的排列方法是 `[n-k，n-k+1，…，n]`，这里有 `k` 个数，注意代码实现时，注意 `k` 的奇偶性，如果 `k` 是奇数，“对半穿插”以后，正好匹配完，如果 `k` 是偶数，对半处的数 `n-k+(k+1)/2`，最后还需要单独加入到排列中。
+- 可能有读者会问了，前面生成了 1 种差值，后面这部分又生产了 `k` 种差值，加起来不是 `k + 1` 种差值了么？这种理解是错误的。后面这段最后 2 个数字是 `n-k+(k+1)/2-1` 和 `n-k+(k+1)/2`，它们两者的差值是 1，和第一段构造的排列差值是相同的，都是 1。所以第一段构造了 1 种差值，第二段虽然构造了 `k` 种，但是需要去掉两段重复的差值 1，所以最终差值种类还是 `1 + k - 1 = k` 种。
+
+## 代码
+
+```go
+package leetcode
+
+func constructArray(n int, k int) []int {
+	res := []int{}
+	for i := 0; i < n-k-1; i++ {
+		res = append(res, i+1)
+	}
+	for i := n - k; i < n-k+(k+1)/2; i++ {
+		res = append(res, i)
+		res = append(res, 2*n-k-i)
+	}
+	if k%2 == 0 {
+		res = append(res, n-k+(k+1)/2)
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0668.Kth-Smallest-Number-in-Multiplication-Table/668. Kth Smallest Number in Multiplication Table.go b/leetcode/0668.Kth-Smallest-Number-in-Multiplication-Table/668. Kth Smallest Number in Multiplication Table.go
new file mode 100644
index 000000000..c6c63d70e
--- /dev/null
+++ b/leetcode/0668.Kth-Smallest-Number-in-Multiplication-Table/668. Kth Smallest Number in Multiplication Table.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+import "math"
+
+func findKthNumber(m int, n int, k int) int {
+	low, high := 1, m*n
+	for low < high {
+		mid := low + (high-low)>>1
+		if counterKthNum(m, n, mid) >= k {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+func counterKthNum(m, n, mid int) int {
+	count := 0
+	for i := 1; i <= m; i++ {
+		count += int(math.Min(math.Floor(float64(mid)/float64(i)), float64(n)))
+	}
+	return count
+}
diff --git a/leetcode/0668.Kth-Smallest-Number-in-Multiplication-Table/668. Kth Smallest Number in Multiplication Table_test.go b/leetcode/0668.Kth-Smallest-Number-in-Multiplication-Table/668. Kth Smallest Number in Multiplication Table_test.go
new file mode 100644
index 000000000..7be662359
--- /dev/null
+++ b/leetcode/0668.Kth-Smallest-Number-in-Multiplication-Table/668. Kth Smallest Number in Multiplication Table_test.go	
@@ -0,0 +1,64 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question668 struct {
+	para668
+	ans668
+}
+
+// para 是参数
+// one 代表第一个参数
+type para668 struct {
+	m int
+	n int
+	k int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans668 struct {
+	one int
+}
+
+func Test_Problem668(t *testing.T) {
+
+	qs := []question668{
+
+		{
+			para668{3, 3, 5},
+			ans668{3},
+		},
+
+		{
+			para668{2, 3, 6},
+			ans668{6},
+		},
+
+		{
+			para668{1, 3, 2},
+			ans668{2},
+		},
+
+		{
+			para668{42, 34, 401},
+			ans668{126},
+		},
+
+		{
+			para668{7341, 13535, 12330027},
+			ans668{2673783},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 668------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans668, q.para668
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findKthNumber(p.m, p.n, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0668.Kth-Smallest-Number-in-Multiplication-Table/README.md b/leetcode/0668.Kth-Smallest-Number-in-Multiplication-Table/README.md
new file mode 100755
index 000000000..f527ef2cd
--- /dev/null
+++ b/leetcode/0668.Kth-Smallest-Number-in-Multiplication-Table/README.md
@@ -0,0 +1,52 @@
+# [668. Kth Smallest Number in Multiplication Table](https://leetcode.com/problems/kth-smallest-number-in-multiplication-table/)
+
+
+## 题目
+
+Nearly every one have used the [Multiplication Table](https://en.wikipedia.org/wiki/Multiplication_table). But could you find out the `k-th` smallest number quickly from the multiplication table?
+
+Given the height `m` and the length `n` of a `m * n` Multiplication Table, and a positive integer `k`, you need to return the `k-th` smallest number in this table.
+
+**Example 1:**
+
+    Input: m = 3, n = 3, k = 5
+    Output: 
+    Explanation: 
+    The Multiplication Table:
+    1	2	3
+    2	4	6
+    3	6	9
+    
+    The 5-th smallest number is 3 (1, 2, 2, 3, 3).
+
+**Example 2:**
+
+    Input: m = 2, n = 3, k = 6
+    Output: 
+    Explanation: 
+    The Multiplication Table:
+    1	2	3
+    2	4	6
+    
+    The 6-th smallest number is 6 (1, 2, 2, 3, 4, 6).
+
+**Note:**
+
+1. The `m` and `n` will be in the range [1, 30000].
+2. The `k` will be in the range [1, m * n]
+
+
+## 题目大意
+
+几乎每一个人都用乘法表。但是你能在乘法表中快速找到第 k 小的数字吗？给定高度 m 、宽度 n 的一张 m * n 的乘法表，以及正整数 k，你需要返回表中第 k 小的数字。
+
+
+注意：
+
+- m 和 n 的范围在 [1, 30000] 之间。
+- k 的范围在 [1, m * n] 之间。
+
+## 解题思路
+
+- 给出 3 个数字，m，n，k。m  和 n 分别代表乘法口诀表的行和列。要求在这个乘法口诀表中找第 k 小的数字。
+- 这一题是第 378 题变种题。利用二分搜索，在 `[1,m*n]` 的区间内搜索第 `k` 小的数。每次二分统计 `≤ mid` 数字的个数。由于是在两数乘法构成的矩阵中计数，知道乘数，被乘数也就知道了，所以计数只需要一层循环。整体代码和第 378 题完全一致，只是计数的部分不同罢了。可以对比第 378 题一起练习。
diff --git a/leetcode/0669.Trim-a-Binary-Search-Tree/669. Trim a Binary Search Tree.go b/leetcode/0669.Trim-a-Binary-Search-Tree/669. Trim a Binary Search Tree.go
new file mode 100644
index 000000000..33ef83df3
--- /dev/null
+++ b/leetcode/0669.Trim-a-Binary-Search-Tree/669. Trim a Binary Search Tree.go	
@@ -0,0 +1,32 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func trimBST(root *TreeNode, low int, high int) *TreeNode {
+	if root == nil {
+		return root
+	}
+	if root.Val > high {
+		return trimBST(root.Left, low, high)
+	}
+	if root.Val < low {
+		return trimBST(root.Right, low, high)
+	}
+	root.Left = trimBST(root.Left, low, high)
+	root.Right = trimBST(root.Right, low, high)
+	return root
+}
diff --git a/leetcode/0669.Trim-a-Binary-Search-Tree/669. Trim a Binary Search Tree_test.go b/leetcode/0669.Trim-a-Binary-Search-Tree/669. Trim a Binary Search Tree_test.go
new file mode 100644
index 000000000..fa407b4d7
--- /dev/null
+++ b/leetcode/0669.Trim-a-Binary-Search-Tree/669. Trim a Binary Search Tree_test.go	
@@ -0,0 +1,68 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question669 struct {
+	para669
+	ans669
+}
+
+// para 是参数
+// one 代表第一个参数
+type para669 struct {
+	one  []int
+	low  int
+	high int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans669 struct {
+	one []int
+}
+
+func Test_Problem669(t *testing.T) {
+
+	qs := []question669{
+
+		{
+			para669{[]int{1, 0, 2}, 1, 2},
+			ans669{[]int{1, structures.NULL, 2}},
+		},
+
+		{
+			para669{[]int{3, 0, 4, structures.NULL, 2, structures.NULL, structures.NULL, 1}, 1, 3},
+			ans669{[]int{3, 2, structures.NULL, 1}},
+		},
+
+		{
+			para669{[]int{1}, 1, 2},
+			ans669{[]int{1}},
+		},
+
+		{
+			para669{[]int{1, structures.NULL, 2}, 1, 3},
+			ans669{[]int{1, structures.NULL, 2}},
+		},
+
+		{
+			para669{[]int{1, structures.NULL, 2}, 2, 4},
+			ans669{[]int{2}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 669------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans669, q.para669
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", structures.Tree2ints(trimBST(root, p.low, p.high)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0669.Trim-a-Binary-Search-Tree/README.md b/leetcode/0669.Trim-a-Binary-Search-Tree/README.md
new file mode 100644
index 000000000..11d0c306b
--- /dev/null
+++ b/leetcode/0669.Trim-a-Binary-Search-Tree/README.md
@@ -0,0 +1,100 @@
+# [669. Trim a Binary Search Tree](https://leetcode.com/problems/trim-a-binary-search-tree/)
+
+
+## 题目
+
+Given the `root` of a binary search tree and the lowest and highest boundaries as `low` and `high`, trim the tree so that all its elements lies in `[low, high]`. Trimming the tree should **not** change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a **unique answer**.
+
+Return *the root of the trimmed binary search tree*. Note that the root may change depending on the given bounds.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/09/09/trim1.jpg](https://assets.leetcode.com/uploads/2020/09/09/trim1.jpg)
+
+```
+Input: root = [1,0,2], low = 1, high = 2
+Output: [1,null,2]
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/09/09/trim2.jpg](https://assets.leetcode.com/uploads/2020/09/09/trim2.jpg)
+
+```
+Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3
+Output: [3,2,null,1]
+```
+
+**Example 3:**
+
+```
+Input: root = [1], low = 1, high = 2
+Output: [1]
+```
+
+**Example 4:**
+
+```
+Input: root = [1,null,2], low = 1, high = 3
+Output: [1,null,2]
+```
+
+**Example 5:**
+
+```
+Input: root = [1,null,2], low = 2, high = 4
+Output: [2]
+```
+
+**Constraints:**
+
+- The number of nodes in the tree in the range `[1, 10^4]`.
+- `0 <= Node.val <= 10^4`
+- The value of each node in the tree is **unique**.
+- `root` is guaranteed to be a valid binary search tree.
+- `0 <= low <= high <= 10^4`
+
+## 题目大意
+
+给你二叉搜索树的根节点 root ，同时给定最小边界low 和最大边界 high。通过修剪二叉搜索树，使得所有节点的值在[low, high]中。修剪树不应该改变保留在树中的元素的相对结构（即，如果没有被移除，原有的父代子代关系都应当保留）。 可以证明，存在唯一的答案。所以结果应当返回修剪好的二叉搜索树的新的根节点。注意，根节点可能会根据给定的边界发生改变。
+
+## 解题思路
+
+- 这一题考察二叉搜索树中的递归遍历。递归遍历二叉搜索树每个结点，根据有序性，当前结点如果比 high 大，那么当前结点的右子树全部修剪掉，再递归修剪左子树；当前结点如果比 low 小，那么当前结点的左子树全部修剪掉，再递归修剪右子树。处理完越界的情况，剩下的情况都在区间内，分别递归修剪左子树和右子树即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func trimBST(root *TreeNode, low int, high int) *TreeNode {
+	if root == nil {
+		return root
+	}
+	if root.Val > high {
+		return trimBST(root.Left, low, high)
+	}
+	if root.Val < low {
+		return trimBST(root.Right, low, high)
+	}
+	root.Left = trimBST(root.Left, low, high)
+	root.Right = trimBST(root.Right, low, high)
+	return root
+}
+```
\ No newline at end of file
diff --git a/leetcode/0674.Longest-Continuous-Increasing-Subsequence/674. Longest Continuous Increasing Subsequence.go b/leetcode/0674.Longest-Continuous-Increasing-Subsequence/674. Longest Continuous Increasing Subsequence.go
new file mode 100644
index 000000000..c9abaaa3d
--- /dev/null
+++ b/leetcode/0674.Longest-Continuous-Increasing-Subsequence/674. Longest Continuous Increasing Subsequence.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+func findLengthOfLCIS(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	res, length := 1, 1
+	for i := 1; i < len(nums); i++ {
+		if nums[i] > nums[i-1] {
+			length++
+		} else {
+			res = max(res, length)
+			length = 1
+		}
+	}
+	return max(res, length)
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0674.Longest-Continuous-Increasing-Subsequence/674. Longest Continuous Increasing Subsequence_test.go b/leetcode/0674.Longest-Continuous-Increasing-Subsequence/674. Longest Continuous Increasing Subsequence_test.go
new file mode 100644
index 000000000..51e5317b5
--- /dev/null
+++ b/leetcode/0674.Longest-Continuous-Increasing-Subsequence/674. Longest Continuous Increasing Subsequence_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question674 struct {
+	para674
+	ans674
+}
+
+// para 是参数
+// one 代表第一个参数
+type para674 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans674 struct {
+	one int
+}
+
+func Test_Problem674(t *testing.T) {
+
+	qs := []question674{
+
+		{
+			para674{[]int{1, 3, 5, 4, 7}},
+			ans674{3},
+		},
+
+		{
+			para674{[]int{2, 2, 2, 2, 2}},
+			ans674{1},
+		},
+
+		{
+			para674{[]int{1, 3, 5, 7}},
+			ans674{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 674------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans674, q.para674
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findLengthOfLCIS(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0674.Longest-Continuous-Increasing-Subsequence/README.md b/leetcode/0674.Longest-Continuous-Increasing-Subsequence/README.md
new file mode 100644
index 000000000..50f80db9d
--- /dev/null
+++ b/leetcode/0674.Longest-Continuous-Increasing-Subsequence/README.md
@@ -0,0 +1,69 @@
+# [674. Longest Continuous Increasing Subsequence](https://leetcode.com/problems/longest-continuous-increasing-subsequence/)
+
+
+## 题目
+
+Given an unsorted array of integers `nums`, return *the length of the longest **continuous increasing subsequence** (i.e. subarray)*. The subsequence must be **strictly** increasing.
+
+A **continuous increasing subsequence** is defined by two indices `l` and `r` (`l < r`) such that it is `[nums[l], nums[l + 1], ..., nums[r - 1], nums[r]]` and for each `l <= i < r`, `nums[i] < nums[i + 1]`.
+
+**Example 1:**
+
+```
+Input: nums = [1,3,5,4,7]
+Output: 3
+Explanation: The longest continuous increasing subsequence is [1,3,5] with length 3.
+Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element
+4.
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,2,2,2,2]
+Output: 1
+Explanation: The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly
+increasing.
+```
+
+**Constraints:**
+
+- `0 <= nums.length <= 10^4`
+- `10^9 <= nums[i] <= 10^9`
+
+## 题目大意
+
+给定一个未经排序的整数数组，找到最长且 连续递增的子序列，并返回该序列的长度。连续递增的子序列 可以由两个下标 l 和 r（l < r）确定，如果对于每个 l <= i < r，都有 nums[i] < nums[i + 1] ，那么子序列 [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] 就是连续递增子序列。
+
+## 解题思路
+
+- 简单题。这一题和第 128 题有区别。这一题要求子序列必须是连续下标，所以变简单了。扫描一遍数组，记下连续递增序列的长度，动态维护这个最大值，最后输出即可。
+
+## 代码
+
+```go
+package leetcode
+
+func findLengthOfLCIS(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	res, length := 1, 1
+	for i := 1; i < len(nums); i++ {
+		if nums[i] > nums[i-1] {
+			length++
+		} else {
+			res = max(res, length)
+			length = 1
+		}
+	}
+	return max(res, length)
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0676. Implement Magic Dictionary/676. Implement Magic Dictionary.go b/leetcode/0676.Implement-Magic-Dictionary/676. Implement Magic Dictionary.go
similarity index 100%
rename from Algorithms/0676. Implement Magic Dictionary/676. Implement Magic Dictionary.go
rename to leetcode/0676.Implement-Magic-Dictionary/676. Implement Magic Dictionary.go
diff --git a/Algorithms/0676. Implement Magic Dictionary/676. Implement Magic Dictionary_test.go b/leetcode/0676.Implement-Magic-Dictionary/676. Implement Magic Dictionary_test.go
similarity index 100%
rename from Algorithms/0676. Implement Magic Dictionary/676. Implement Magic Dictionary_test.go
rename to leetcode/0676.Implement-Magic-Dictionary/676. Implement Magic Dictionary_test.go
diff --git a/Algorithms/0676. Implement Magic Dictionary/README.md b/leetcode/0676.Implement-Magic-Dictionary/README.md
similarity index 100%
rename from Algorithms/0676. Implement Magic Dictionary/README.md
rename to leetcode/0676.Implement-Magic-Dictionary/README.md
diff --git a/leetcode/0677.Map-Sum-Pairs/677. Map Sum Pairs.go b/leetcode/0677.Map-Sum-Pairs/677. Map Sum Pairs.go
new file mode 100644
index 000000000..e5560741e
--- /dev/null
+++ b/leetcode/0677.Map-Sum-Pairs/677. Map Sum Pairs.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+type MapSum struct {
+	keys map[string]int
+}
+
+/** Initialize your data structure here. */
+func Constructor() MapSum {
+	return MapSum{make(map[string]int)}
+}
+
+func (this *MapSum) Insert(key string, val int) {
+	this.keys[key] = val
+}
+
+func (this *MapSum) Sum(prefix string) int {
+	prefixAsRunes, res := []rune(prefix), 0
+	for key, val := range this.keys {
+		if len(key) >= len(prefix) {
+			shouldSum := true
+			for i, char := range key {
+				if i >= len(prefixAsRunes) {
+					break
+				}
+				if prefixAsRunes[i] != char {
+					shouldSum = false
+					break
+				}
+			}
+			if shouldSum {
+				res += val
+			}
+		}
+	}
+	return res
+}
+
+/**
+ * Your MapSum object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.Insert(key,val);
+ * param_2 := obj.Sum(prefix);
+ */
diff --git a/leetcode/0677.Map-Sum-Pairs/677. Map Sum Pairs_test.go b/leetcode/0677.Map-Sum-Pairs/677. Map Sum Pairs_test.go
new file mode 100644
index 000000000..1be9a3e00
--- /dev/null
+++ b/leetcode/0677.Map-Sum-Pairs/677. Map Sum Pairs_test.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem677(t *testing.T) {
+	obj := Constructor()
+	fmt.Printf("obj = %v\n", obj)
+	obj.Insert("apple", 3)
+	fmt.Printf("obj = %v\n", obj)
+	fmt.Printf("obj.sum = %v\n", obj.Sum("ap"))
+	obj.Insert("app", 2)
+	fmt.Printf("obj = %v\n", obj)
+	fmt.Printf("obj.sum = %v\n", obj.Sum("ap"))
+}
diff --git a/leetcode/0677.Map-Sum-Pairs/README.md b/leetcode/0677.Map-Sum-Pairs/README.md
new file mode 100644
index 000000000..bad5bc124
--- /dev/null
+++ b/leetcode/0677.Map-Sum-Pairs/README.md
@@ -0,0 +1,100 @@
+# [677. Map Sum Pairs](https://leetcode.com/problems/map-sum-pairs/)
+
+
+## 题目
+
+Design a map that allows you to do the following:
+
+- Maps a string key to a given value.
+- Returns the sum of the values that have a key with a prefix equal to a given string.
+
+Implement the `MapSum` class:
+
+- `MapSum()` Initializes the `MapSum` object.
+- `void insert(String key, int val)` Inserts the `key-val` pair into the map. If the `key` already existed, the original `key-value` pair will be overridden to the new one.
+- `int sum(string prefix)` Returns the sum of all the pairs' value whose `key` starts with the `prefix`.
+
+**Example 1:**
+
+```
+Input
+["MapSum", "insert", "sum", "insert", "sum"]
+[[], ["apple", 3], ["ap"], ["app", 2], ["ap"]]
+Output
+[null, null, 3, null, 5]
+
+Explanation
+MapSum mapSum = new MapSum();
+mapSum.insert("apple", 3);
+mapSum.sum("ap");           // return 3 (apple = 3)
+mapSum.insert("app", 2);
+mapSum.sum("ap");           // return 5 (apple +app = 3 + 2 = 5)
+
+```
+
+**Constraints:**
+
+- `1 <= key.length, prefix.length <= 50`
+- `key` and `prefix` consist of only lowercase English letters.
+- `1 <= val <= 1000`
+- At most `50` calls will be made to `insert` and `sum`.
+
+## 题目大意
+
+实现一个 MapSum 类，支持两个方法，insert 和 sum：
+
+- MapSum() 初始化 MapSum 对象
+- void insert(String key, int val) 插入 key-val 键值对，字符串表示键 key ，整数表示值 val 。如果键 key 已经存在，那么原来的键值对将被替代成新的键值对。
+- int sum(string prefix) 返回所有以该前缀 prefix 开头的键 key 的值的总和。
+
+## 解题思路
+
+- 简单题。用一个 map 存储数据，Insert() 方法即存储 key-value。Sum() 方法即累加满足条件前缀对应的 value。判断是否满足条件，先根据前缀长度来判断，只有长度大于等于 prefix 长度才可能满足要求。如果 key 是具有 prefix 前缀的，那么累加上这个值。最后输出总和即可。
+
+## 代码
+
+```go
+package leetcode
+
+type MapSum struct {
+	keys map[string]int
+}
+
+/** Initialize your data structure here. */
+func Constructor() MapSum {
+	return MapSum{make(map[string]int)}
+}
+
+func (this *MapSum) Insert(key string, val int) {
+	this.keys[key] = val
+}
+
+func (this *MapSum) Sum(prefix string) int {
+	prefixAsRunes, res := []rune(prefix), 0
+	for key, val := range this.keys {
+		if len(key) >= len(prefix) {
+			shouldSum := true
+			for i, char := range key {
+				if i >= len(prefixAsRunes) {
+					break
+				}
+				if prefixAsRunes[i] != char {
+					shouldSum = false
+					break
+				}
+			}
+			if shouldSum {
+				res += val
+			}
+		}
+	}
+	return res
+}
+
+/**
+ * Your MapSum object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.Insert(key,val);
+ * param_2 := obj.Sum(prefix);
+ */
+```
\ No newline at end of file
diff --git a/Algorithms/0682. Baseball Game/682. Baseball Game.go b/leetcode/0682.Baseball-Game/682. Baseball Game.go
similarity index 100%
rename from Algorithms/0682. Baseball Game/682. Baseball Game.go
rename to leetcode/0682.Baseball-Game/682. Baseball Game.go
diff --git a/Algorithms/0682. Baseball Game/682. Baseball Game_test.go b/leetcode/0682.Baseball-Game/682. Baseball Game_test.go
similarity index 95%
rename from Algorithms/0682. Baseball Game/682. Baseball Game_test.go
rename to leetcode/0682.Baseball-Game/682. Baseball Game_test.go
index a5dac7d81..ce271aca3 100644
--- a/Algorithms/0682. Baseball Game/682. Baseball Game_test.go	
+++ b/leetcode/0682.Baseball-Game/682. Baseball Game_test.go	
@@ -26,12 +26,12 @@ func Test_Problem682(t *testing.T) {
 
 	qs := []question682{
 
-		question682{
+		{
 			para682{[]string{"5", "2", "C", "D", "+"}},
 			ans682{30},
 		},
 
-		question682{
+		{
 			para682{[]string{"5", "-2", "4", "C", "D", "9", "+", "+"}},
 			ans682{27},
 		},
diff --git a/Algorithms/0682. Baseball Game/README.md b/leetcode/0682.Baseball-Game/README.md
similarity index 100%
rename from Algorithms/0682. Baseball Game/README.md
rename to leetcode/0682.Baseball-Game/README.md
diff --git a/Algorithms/0684. Redundant Connection/684. Redundant Connection.go b/leetcode/0684.Redundant-Connection/684. Redundant Connection.go
similarity index 100%
rename from Algorithms/0684. Redundant Connection/684. Redundant Connection.go
rename to leetcode/0684.Redundant-Connection/684. Redundant Connection.go
diff --git a/Algorithms/0684. Redundant Connection/684. Redundant Connection_test.go b/leetcode/0684.Redundant-Connection/684. Redundant Connection_test.go
similarity index 78%
rename from Algorithms/0684. Redundant Connection/684. Redundant Connection_test.go
rename to leetcode/0684.Redundant-Connection/684. Redundant Connection_test.go
index 2e30735d2..f5e2f7119 100644
--- a/Algorithms/0684. Redundant Connection/684. Redundant Connection_test.go	
+++ b/leetcode/0684.Redundant-Connection/684. Redundant Connection_test.go	
@@ -26,13 +26,13 @@ func Test_Problem684(t *testing.T) {
 
 	qs := []question684{
 
-		question684{
-			para684{[][]int{[]int{1, 2}, []int{1, 3}, []int{2, 3}}},
+		{
+			para684{[][]int{{1, 2}, {1, 3}, {2, 3}}},
 			ans684{[]int{2, 3}},
 		},
 
-		question684{
-			para684{[][]int{[]int{1, 2}, []int{2, 3}, []int{3, 4}, []int{1, 4}, []int{1, 5}}},
+		{
+			para684{[][]int{{1, 2}, {2, 3}, {3, 4}, {1, 4}, {1, 5}}},
 			ans684{[]int{1, 4}},
 		},
 	}
diff --git a/leetcode/0684.Redundant-Connection/README.md b/leetcode/0684.Redundant-Connection/README.md
new file mode 100755
index 000000000..57ee55101
--- /dev/null
+++ b/leetcode/0684.Redundant-Connection/README.md
@@ -0,0 +1,55 @@
+# [684. Redundant Connection](https://leetcode.com/problems/redundant-connection/)
+
+
+## 题目
+
+In this problem, a tree is an **undirected** graph that is connected and has no cycles.
+
+The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, ..., N), with one additional edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.
+
+The resulting graph is given as a 2D-array of `edges`. Each element of `edges` is a pair `[u, v]` with `u < v`, that represents an **undirected** edge connecting nodes `u` and `v`.
+
+Return an edge that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge `[u, v]` should be in the same format, with `u < v`.
+
+**Example 1:**
+
+    Input: [[1,2], [1,3], [2,3]]
+    Output: [2,3]
+    Explanation: The given undirected graph will be like this:
+      1
+     / \
+    2 - 3
+
+**Example 2:**
+
+    Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
+    Output: [1,4]
+    Explanation: The given undirected graph will be like this:
+    5 - 1 - 2
+        |   |
+        4 - 3
+
+**Note:**
+
+- The size of the input 2D-array will be between 3 and 1000.
+- Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.
+
+**Update (2017-09-26):**We have overhauled the problem description + test cases and specified clearly the graph is an **undirected** graph. For the **directed** graph follow up please see **[Redundant Connection II](https://leetcode.com/problems/redundant-connection-ii/description/)**). We apologize for any inconvenience caused.
+
+
+## 题目大意
+
+在本问题中, 树指的是一个连通且无环的无向图。输入一个图，该图由一个有着N个节点 (节点值不重复1, 2, ..., N) 的树及一条附加的边构成。附加的边的两个顶点包含在1到N中间，这条附加的边不属于树中已存在的边。结果图是一个以边组成的二维数组。每一个边的元素是一对[u, v] ，满足 u < v，表示连接顶点u 和v的无向图的边。
+
+返回一条可以删去的边，使得结果图是一个有着N个节点的树。如果有多个答案，则返回二维数组中最后出现的边。答案边 [u, v] 应满足相同的格式 u < v。
+
+注意:
+
+- 输入的二维数组大小在 3 到 1000。
+- 二维数组中的整数在 1 到 N 之间，其中 N 是输入数组的大小。
+
+
+## 解题思路
+
+- 给出一个连通无环无向图和一些连通的边，要求在这些边中删除一条边以后，图中的 N 个节点依旧是连通的。如果有多条边，输出最后一条。
+- 这一题可以用并查集直接秒杀。依次扫描所有的边，把边的两端点都合并 `union()` 到一起。如果遇到一条边的两端点已经在一个集合里面了，就说明是多余边，删除。最后输出这些边即可。
diff --git a/Algorithms/0685. Redundant Connection II/685. Redundant Connection II.go b/leetcode/0685.Redundant-Connection-II/685. Redundant Connection II.go
similarity index 100%
rename from Algorithms/0685. Redundant Connection II/685. Redundant Connection II.go
rename to leetcode/0685.Redundant-Connection-II/685. Redundant Connection II.go
diff --git a/leetcode/0685.Redundant-Connection-II/685. Redundant Connection II_test.go b/leetcode/0685.Redundant-Connection-II/685. Redundant Connection II_test.go
new file mode 100644
index 000000000..4abb94ea1
--- /dev/null
+++ b/leetcode/0685.Redundant-Connection-II/685. Redundant Connection II_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question685 struct {
+	para685
+	ans685
+}
+
+// para 是参数
+// one 代表第一个参数
+type para685 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans685 struct {
+	one []int
+}
+
+func Test_Problem685(t *testing.T) {
+
+	qs := []question685{
+
+		{
+			para685{[][]int{{3, 5}, {1, 3}, {2, 1}, {5, 4}, {2, 3}}},
+			ans685{[]int{2, 3}},
+		},
+
+		{
+			para685{[][]int{{4, 2}, {1, 5}, {5, 2}, {5, 3}, {2, 4}}},
+			ans685{[]int{4, 2}},
+		},
+
+		{
+			para685{[][]int{{1, 2}, {1, 3}, {2, 3}}},
+			ans685{[]int{2, 3}},
+		},
+
+		{
+			para685{[][]int{{1, 2}, {2, 3}, {3, 4}, {1, 4}, {1, 5}}},
+			ans685{[]int{1, 4}},
+		},
+
+		{
+			para685{[][]int{{2, 1}, {3, 1}, {4, 2}, {1, 4}}},
+			ans685{[]int{2, 1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 685------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans685, q.para685
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findRedundantDirectedConnection(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0685.Redundant-Connection-II/README.md b/leetcode/0685.Redundant-Connection-II/README.md
new file mode 100755
index 000000000..d35e982f0
--- /dev/null
+++ b/leetcode/0685.Redundant-Connection-II/README.md
@@ -0,0 +1,59 @@
+# [685. Redundant Connection II](https://leetcode.com/problems/redundant-connection-ii/)
+
+
+## 题目
+
+In this problem, a rooted tree is a **directed** graph such that, there is exactly one node (the root) for which all other nodes are descendants of this node, plus every node has exactly one parent, except for the root node which has no parents.
+
+The given input is a directed graph that started as a rooted tree with N nodes (with distinct values 1, 2, ..., N), with one additional directed edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.
+
+The resulting graph is given as a 2D-array of `edges`. Each element of `edges` is a pair `[u, v]` that represents a **directed** edge connecting nodes `u` and `v`, where `u` is a parent of child `v`.
+
+Return an edge that can be removed so that the resulting graph is a rooted tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array.
+
+**Example 1:**
+
+    Input: [[1,2], [1,3], [2,3]]
+    Output: [2,3]
+    Explanation: The given directed graph will be like this:
+      1
+     / \
+    v   v
+    2-->3
+
+**Example 2:**
+
+    Input: [[1,2], [2,3], [3,4], [4,1], [1,5]]
+    Output: [4,1]
+    Explanation: The given directed graph will be like this:
+    5 <- 1 -> 2
+         ^    |
+         |    v
+         4 <- 3
+
+**Note:**
+
+- The size of the input 2D-array will be between 3 and 1000.
+- Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.
+
+
+## 题目大意
+
+
+在本问题中，有根树指满足以下条件的有向图。该树只有一个根节点，所有其他节点都是该根节点的后继。每一个节点只有一个父节点，除了根节点没有父节点。输入一个有向图，该图由一个有着 N 个节点 (节点值不重复1, 2, ..., N) 的树及一条附加的边构成。附加的边的两个顶点包含在1到N中间，这条附加的边不属于树中已存在的边。结果图是一个以边组成的二维数组。 每一个边的元素是一对 [u, v]，用以表示有向图中连接顶点 u and v 和顶点的边，其中父节点 u 是子节点 v 的一个父节点。返回一条能删除的边，使得剩下的图是有 N 个节点的有根树。若有多个答案，返回最后出现在给定二维数组的答案。
+
+注意:
+
+- 二维数组大小的在 3 到 1000 范围内。
+- 二维数组中的每个整数在 1 到 N 之间，其中 N 是二维数组的大小。
+
+
+## 解题思路
+
+- 这一题是第 684 题的加强版。第 684 题中的图是无向图，这一题中的图是有向图。
+- 这一题的解法也是用并查集，不过需要灵活一点，不要用模板，因为在模板中，存在路径压缩和 `rank()` 优化，这些优化会改变有向边原始的方向。所以并查集只需要记录 `parent()` 就够用了。
+
+![](https://img.halfrost.com/Leetcode/leetcode_685.png)
+
+- 经过分析，可以得到上面这 3 种情况，红色的边是我们实际应该删除的。先来看情况 2 和情况 3 。当不断 `union()` 时，加入一条边以后，会使一个节点的入度变成 2，那么记录下这两条边为 `candidate1` 和 `candidate2`。将后加入的 `candidate2` 这条边先放在一边，继续往下 `union()`。如果 `candidate2` 是红色的边，那么合并到最后，也不会出现任何异常，那么 `candidate2` 就是红色的边，即找到了要删除的边了。如果合并到最后出现了环的问题了，那说明 `candidate2` 是黑色的边，`candidate1` 才是红色的边，那么 `candidate1` 是要删除的边。
+- 再来看看情况 1。如果一路合并到结束也没有发现出现入度为 2 的情况，那么说明遇到了情况 1 。情况 1 会出现环的情况。题目中说如果要删除边，就删除最后出现的那条边。**具体实现见代码注释**。
\ No newline at end of file
diff --git a/leetcode/0690.Employee-Importance/690. Employee Importance.go b/leetcode/0690.Employee-Importance/690. Employee Importance.go
new file mode 100644
index 000000000..a13ace46d
--- /dev/null
+++ b/leetcode/0690.Employee-Importance/690. Employee Importance.go	
@@ -0,0 +1,26 @@
+package leetcode
+
+type Employee struct {
+	Id           int
+	Importance   int
+	Subordinates []int
+}
+
+func getImportance(employees []*Employee, id int) int {
+	m, queue, res := map[int]*Employee{}, []int{id}, 0
+	for _, e := range employees {
+		m[e.Id] = e
+	}
+	for len(queue) > 0 {
+		e := m[queue[0]]
+		queue = queue[1:]
+		if e == nil {
+			continue
+		}
+		res += e.Importance
+		for _, i := range e.Subordinates {
+			queue = append(queue, i)
+		}
+	}
+	return res
+}
diff --git a/leetcode/0690.Employee-Importance/690. Employee Importance_test.go b/leetcode/0690.Employee-Importance/690. Employee Importance_test.go
new file mode 100644
index 000000000..401a24bb6
--- /dev/null
+++ b/leetcode/0690.Employee-Importance/690. Employee Importance_test.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question690 struct {
+	para690
+	ans690
+}
+
+// para 是参数
+// one 代表第一个参数
+type para690 struct {
+	employees []*Employee
+	id        int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans690 struct {
+	one int
+}
+
+func Test_Problem690(t *testing.T) {
+
+	qs := []question690{
+
+		{
+			para690{[]*Employee{{1, 5, []int{2, 3}}, {2, 3, []int{}}, {3, 3, []int{}}}, 1},
+			ans690{11},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 690------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans690, q.para690
+		fmt.Printf("【input】:%v       【output】:%v\n", p, getImportance(p.employees, p.id))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0690.Employee-Importance/README.md b/leetcode/0690.Employee-Importance/README.md
new file mode 100644
index 000000000..24ee99817
--- /dev/null
+++ b/leetcode/0690.Employee-Importance/README.md
@@ -0,0 +1,62 @@
+# [690. Employee Importance](https://leetcode.com/problems/employee-importance/)
+
+## 题目
+
+You are given a data structure of employee information, which includes the employee's **unique id**, their **importance value** and their **direct** subordinates' id.
+
+For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, [2]], and employee 2 has [2, 10, [3]], and employee 3 has [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, the relationship is **not direct**.
+
+Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all their subordinates.
+
+**Example 1:**
+
+```
+Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
+Output: 11
+Explanation:
+Employee 1 has importance value 5, and he has two direct subordinates: employee 2 and employee 3. They both have importance value 3. So the total importance value of employee 1 is 5 + 3 + 3 = 11.
+```
+
+**Note:**
+
+1. One employee has at most one **direct** leader and may have several subordinates.
+2. The maximum number of employees won't exceed 2000.
+
+## 题目大意
+
+给定一个保存员工信息的数据结构，它包含了员工 唯一的 id ，重要度 和 直系下属的 id 。比如，员工 1 是员工 2 的领导，员工 2 是员工 3 的领导。他们相应的重要度为 15 , 10 , 5 。那么员工 1 的数据结构是 [1, 15, [2]] ，员工 2的 数据结构是 [2, 10, [3]] ，员工 3 的数据结构是 [3, 5, []] 。注意虽然员工 3 也是员工 1 的一个下属，但是由于 并不是直系 下属，因此没有体现在员工 1 的数据结构中。现在输入一个公司的所有员工信息，以及单个员工 id ，返回这个员工和他所有下属的重要度之和。
+
+## 解题思路
+
+- 简单题。根据题意，DFS 或者 BFS 搜索找到所求 id 下属所有员工，累加下属员工的重要度，最后再加上这个员工本身的重要度，即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+type Employee struct {
+	Id           int
+	Importance   int
+	Subordinates []int
+}
+
+func getImportance(employees []*Employee, id int) int {
+	m, queue, res := map[int]*Employee{}, []int{id}, 0
+	for _, e := range employees {
+		m[e.Id] = e
+	}
+	for len(queue) > 0 {
+		e := m[queue[0]]
+		queue = queue[1:]
+		if e == nil {
+			continue
+		}
+		res += e.Importance
+		for _, i := range e.Subordinates {
+			queue = append(queue, i)
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0692.Top-K-Frequent-Words/692. Top K Frequent Words.go b/leetcode/0692.Top-K-Frequent-Words/692. Top K Frequent Words.go
new file mode 100644
index 000000000..4b739ffb2
--- /dev/null
+++ b/leetcode/0692.Top-K-Frequent-Words/692. Top K Frequent Words.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+import "container/heap"
+
+func topKFrequent(words []string, k int) []string {
+	m := map[string]int{}
+	for _, word := range words {
+		m[word]++
+	}
+	pq := &PQ{}
+	heap.Init(pq)
+	for w, c := range m {
+		heap.Push(pq, &wordCount{w, c})
+		if pq.Len() > k {
+			heap.Pop(pq)
+		}
+	}
+	res := make([]string, k)
+	for i := k - 1; i >= 0; i-- {
+		wc := heap.Pop(pq).(*wordCount)
+		res[i] = wc.word
+	}
+	return res
+}
+
+type wordCount struct {
+	word string
+	cnt  int
+}
+
+type PQ []*wordCount
+
+func (pq PQ) Len() int      { return len(pq) }
+func (pq PQ) Swap(i, j int) { pq[i], pq[j] = pq[j], pq[i] }
+func (pq PQ) Less(i, j int) bool {
+	if pq[i].cnt == pq[j].cnt {
+		return pq[i].word > pq[j].word
+	}
+	return pq[i].cnt < pq[j].cnt
+}
+func (pq *PQ) Push(x interface{}) {
+	tmp := x.(*wordCount)
+	*pq = append(*pq, tmp)
+}
+func (pq *PQ) Pop() interface{} {
+	n := len(*pq)
+	tmp := (*pq)[n-1]
+	*pq = (*pq)[:n-1]
+	return tmp
+}
diff --git a/leetcode/0692.Top-K-Frequent-Words/692. Top K Frequent Words_test.go b/leetcode/0692.Top-K-Frequent-Words/692. Top K Frequent Words_test.go
new file mode 100644
index 000000000..97221e3ef
--- /dev/null
+++ b/leetcode/0692.Top-K-Frequent-Words/692. Top K Frequent Words_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question692 struct {
+	para692
+	ans692
+}
+
+// para 是参数
+// one 代表第一个参数
+type para692 struct {
+	words []string
+	k     int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans692 struct {
+	one []string
+}
+
+func Test_Problem692(t *testing.T) {
+
+	qs := []question692{
+
+		{
+			para692{[]string{"i", "love", "leetcode", "i", "love", "coding"}, 2},
+			ans692{[]string{"i", "love"}},
+		},
+
+		{
+			para692{[]string{"the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"}, 4},
+			ans692{[]string{"the", "is", "sunny", "day"}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 692------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans692, q.para692
+		fmt.Printf("【input】:%v       【output】:%v\n", p, topKFrequent(p.words, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0692.Top-K-Frequent-Words/README.md b/leetcode/0692.Top-K-Frequent-Words/README.md
new file mode 100644
index 000000000..54a858d32
--- /dev/null
+++ b/leetcode/0692.Top-K-Frequent-Words/README.md
@@ -0,0 +1,98 @@
+# [692. Top K Frequent Words](https://leetcode.com/problems/top-k-frequent-words/)
+
+
+## 题目
+
+Given a non-empty list of words, return the k most frequent elements.
+
+Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.
+
+**Example 1:**
+
+```
+Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
+Output: ["i", "love"]
+Explanation: "i" and "love" are the two most frequent words.
+    Note that "i" comes before "love" due to a lower alphabetical order.
+```
+
+**Example 2:**
+
+```
+Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
+Output: ["the", "is", "sunny", "day"]
+Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
+    with the number of occurrence being 4, 3, 2 and 1 respectively.
+```
+
+**Note:**
+
+1. You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
+2. Input words contain only lowercase letters.
+
+**Follow up:**
+
+1. Try to solve it in O(n log k) time and O(n) extra space.
+
+## 题目大意
+
+给一非空的单词列表，返回前 *k* 个出现次数最多的单词。返回的答案应该按单词出现频率由高到低排序。如果不同的单词有相同出现频率，按字母顺序排序。
+
+## 解题思路
+
+- 思路很简单的题。维护一个长度为 k 的最大堆，先按照频率排，如果频率相同再按照字母顺序排。最后输出依次将优先队列里面的元素 pop 出来即可。
+
+## 代码
+
+```go
+package leetcode
+
+import "container/heap"
+
+func topKFrequent(words []string, k int) []string {
+	m := map[string]int{}
+	for _, word := range words {
+		m[word]++
+	}
+	pq := &PQ{}
+	heap.Init(pq)
+	for w, c := range m {
+		heap.Push(pq, &wordCount{w, c})
+		if pq.Len() > k {
+			heap.Pop(pq)
+		}
+	}
+	res := make([]string, k)
+	for i := k - 1; i >= 0; i-- {
+		wc := heap.Pop(pq).(*wordCount)
+		res[i] = wc.word
+	}
+	return res
+}
+
+type wordCount struct {
+	word string
+	cnt  int
+}
+
+type PQ []*wordCount
+
+func (pq PQ) Len() int      { return len(pq) }
+func (pq PQ) Swap(i, j int) { pq[i], pq[j] = pq[j], pq[i] }
+func (pq PQ) Less(i, j int) bool {
+	if pq[i].cnt == pq[j].cnt {
+		return pq[i].word > pq[j].word
+	}
+	return pq[i].cnt < pq[j].cnt
+}
+func (pq *PQ) Push(x interface{}) {
+	tmp := x.(*wordCount)
+	*pq = append(*pq, tmp)
+}
+func (pq *PQ) Pop() interface{} {
+	n := len(*pq)
+	tmp := (*pq)[n-1]
+	*pq = (*pq)[:n-1]
+	return tmp
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0693. Binary Number with Alternating Bits/693. Binary Number with Alternating Bits.go b/leetcode/0693.Binary-Number-with-Alternating-Bits/693. Binary Number with Alternating Bits.go
similarity index 100%
rename from Algorithms/0693. Binary Number with Alternating Bits/693. Binary Number with Alternating Bits.go
rename to leetcode/0693.Binary-Number-with-Alternating-Bits/693. Binary Number with Alternating Bits.go
diff --git a/Algorithms/0693. Binary Number with Alternating Bits/693. Binary Number with Alternating Bits_test.go b/leetcode/0693.Binary-Number-with-Alternating-Bits/693. Binary Number with Alternating Bits_test.go
similarity index 92%
rename from Algorithms/0693. Binary Number with Alternating Bits/693. Binary Number with Alternating Bits_test.go
rename to leetcode/0693.Binary-Number-with-Alternating-Bits/693. Binary Number with Alternating Bits_test.go
index 321e3ca2d..40a10bb26 100644
--- a/Algorithms/0693. Binary Number with Alternating Bits/693. Binary Number with Alternating Bits_test.go	
+++ b/leetcode/0693.Binary-Number-with-Alternating-Bits/693. Binary Number with Alternating Bits_test.go	
@@ -26,22 +26,22 @@ func Test_Problem693(t *testing.T) {
 
 	qs := []question693{
 
-		question693{
+		{
 			para693{5},
 			ans693{true},
 		},
 
-		question693{
+		{
 			para693{7},
 			ans693{false},
 		},
 
-		question693{
+		{
 			para693{11},
 			ans693{false},
 		},
 
-		question693{
+		{
 			para693{10},
 			ans693{true},
 		},
diff --git a/leetcode/0693.Binary-Number-with-Alternating-Bits/README.md b/leetcode/0693.Binary-Number-with-Alternating-Bits/README.md
new file mode 100755
index 000000000..8c8fb7886
--- /dev/null
+++ b/leetcode/0693.Binary-Number-with-Alternating-Bits/README.md
@@ -0,0 +1,43 @@
+# [693. Binary Number with Alternating Bits](https://leetcode.com/problems/binary-number-with-alternating-bits/)
+
+## 题目
+
+Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.
+
+**Example 1:**
+
+    Input: 5
+    Output: True
+    Explanation:
+    The binary representation of 5 is: 101
+
+**Example 2:**
+
+    Input: 7
+    Output: False
+    Explanation:
+    The binary representation of 7 is: 111.
+
+**Example 3:**
+
+    Input: 11
+    Output: False
+    Explanation:
+    The binary representation of 11 is: 1011.
+
+**Example 4:**
+
+    Input: 10
+    Output: True
+    Explanation:
+    The binary representation of 10 is: 1010.
+
+
+## 题目大意
+
+给定一个正整数，检查他是否为交替位二进制数：换句话说，就是他的二进制数相邻的两个位数永不相等。
+
+## 解题思路
+
+
+- 判断一个数的二进制位相邻两个数是不相等的，即 `0101` 交叉间隔的，如果是，输出 true。这一题有多种做法，最简单的方法就是直接模拟。比较巧妙的方法是通过位运算，合理构造特殊数据进行位运算到达目的。`010101` 构造出 `101010` 两者相互 `&` 位运算以后就为 0，因为都“插空”了。
diff --git a/leetcode/0695.Max-Area-of-Island/695. Max Area of Island.go b/leetcode/0695.Max-Area-of-Island/695. Max Area of Island.go
new file mode 100644
index 000000000..d19c7fd78
--- /dev/null
+++ b/leetcode/0695.Max-Area-of-Island/695. Max Area of Island.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func maxAreaOfIsland(grid [][]int) int {
+	res := 0
+	for i, row := range grid {
+		for j, col := range row {
+			if col == 0 {
+				continue
+			}
+			area := areaOfIsland(grid, i, j)
+			if area > res {
+				res = area
+			}
+		}
+	}
+	return res
+}
+
+func isInGrid(grid [][]int, x, y int) bool {
+	return x >= 0 && x < len(grid) && y >= 0 && y < len(grid[0])
+}
+
+func areaOfIsland(grid [][]int, x, y int) int {
+	if !isInGrid(grid, x, y) || grid[x][y] == 0 {
+		return 0
+	}
+	grid[x][y] = 0
+	total := 1
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		total += areaOfIsland(grid, nx, ny)
+	}
+	return total
+}
diff --git a/leetcode/0695.Max-Area-of-Island/695. Max Area of Island_test.go b/leetcode/0695.Max-Area-of-Island/695. Max Area of Island_test.go
new file mode 100644
index 000000000..9bfea727b
--- /dev/null
+++ b/leetcode/0695.Max-Area-of-Island/695. Max Area of Island_test.go	
@@ -0,0 +1,90 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question695 struct {
+	para695
+	ans695
+}
+
+// para 是参数
+// one 代表第一个参数
+type para695 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans695 struct {
+	one int
+}
+
+func Test_Problem695(t *testing.T) {
+
+	qs := []question695{
+
+		{
+			para695{[][]int{
+				{1, 1, 1, 1, 0},
+				{1, 1, 0, 1, 0},
+				{1, 1, 0, 0, 0},
+				{0, 0, 0, 0, 0},
+			}},
+			ans695{9},
+		},
+
+		{
+			para695{[][]int{
+				{1, 1, 0, 0, 0},
+				{1, 1, 0, 0, 0},
+				{0, 0, 1, 0, 0},
+				{0, 0, 0, 1, 1},
+			}},
+			ans695{4},
+		},
+
+		{
+			para695{[][]int{
+				{1, 1, 1, 1, 1, 1, 1, 0},
+				{1, 0, 0, 0, 0, 1, 1, 0},
+				{1, 0, 1, 0, 1, 1, 1, 0},
+				{1, 0, 0, 0, 0, 1, 0, 1},
+				{1, 1, 1, 1, 1, 1, 1, 0},
+			}},
+			ans695{23},
+		},
+
+		{
+			para695{[][]int{
+				{0, 0, 1, 0, 0},
+				{0, 1, 0, 1, 0},
+				{0, 1, 1, 1, 0},
+			}},
+			ans695{5},
+		},
+
+		{
+			para695{[][]int{
+				{1, 1, 1, 1, 1, 1, 1},
+				{1, 0, 0, 0, 0, 0, 1},
+				{1, 0, 1, 1, 1, 0, 1},
+				{1, 0, 1, 0, 1, 0, 1},
+				{1, 0, 1, 1, 1, 0, 1},
+				{1, 0, 0, 0, 0, 0, 1},
+				{1, 1, 1, 1, 1, 1, 1},
+			}},
+			ans695{24},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 695------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans695, q.para695
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxAreaOfIsland(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0695.Max-Area-of-Island/README.md b/leetcode/0695.Max-Area-of-Island/README.md
new file mode 100644
index 000000000..2419d561e
--- /dev/null
+++ b/leetcode/0695.Max-Area-of-Island/README.md
@@ -0,0 +1,77 @@
+# [695. Max Area of Island](https://leetcode.com/problems/max-area-of-island/)
+
+
+
+## 题目
+
+Given a non-empty 2D array `grid` of 0's and 1's, an **island** is a group of `1`'s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.
+
+Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)
+
+**Example 1**:
+
+```
+[[0,0,1,0,0,0,0,1,0,0,0,0,0],
+ [0,0,0,0,0,0,0,1,1,1,0,0,0],
+ [0,1,1,0,1,0,0,0,0,0,0,0,0],
+ [0,1,0,0,1,1,0,0,1,0,1,0,0],
+ [0,1,0,0,1,1,0,0,1,1,1,0,0],
+ [0,0,0,0,0,0,0,0,0,0,1,0,0],
+ [0,0,0,0,0,0,0,1,1,1,0,0,0],
+ [0,0,0,0,0,0,0,1,1,0,0,0,0]]
+```
+
+Given the above grid, return`6`. Note the answer is not 11, because the island must be connected 4-directionally.
+
+**Example 2**:
+
+```
+[[0,0,0,0,0,0,0,0]]
+```
+
+Given the above grid, return`0`.
+
+**Note**: The length of each dimension in the given `grid` does not exceed 50.
+
+## 题目大意
+
+给定一个包含了一些 0 和 1 的非空二维数组 grid 。一个 岛屿 是由一些相邻的 1 (代表土地) 构成的组合，这里的「相邻」要求两个 1 必须在水平或者竖直方向上相邻。你可以假设 grid 的四个边缘都被 0（代表水）包围着。找到给定的二维数组中最大的岛屿面积。(如果没有岛屿，则返回面积为 0 。)
+
+## 解题思路
+
+- 给出一个地图，要求计算上面岛屿的面积。注意岛屿的定义是四周都是海(为 0 的点)，如果土地(为 1 的点)靠在地图边缘，不能算是岛屿。
+- 这一题和第 200 题，第 1254 题解题思路是一致的。DPS 深搜。这不过这一题需要多处理 2 件事情，一个是注意靠边缘的岛屿不能计算在内，二是动态维护岛屿的最大面积。
+
+## 代码
+
+```go
+func maxAreaOfIsland(grid [][]int) int {
+	res := 0
+	for i, row := range grid {
+		for j, col := range row {
+			if col == 0 {
+				continue
+			}
+			area := areaOfIsland(grid, i, j)
+			if area > res {
+				res = area
+			}
+		}
+	}
+	return res
+}
+
+func areaOfIsland(grid [][]int, x, y int) int {
+	if !isInGrid(grid, x, y) || grid[x][y] == 0 {
+		return 0
+	}
+	grid[x][y] = 0
+	total := 1
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		total += areaOfIsland(grid, nx, ny)
+	}
+	return total
+}
+```
\ No newline at end of file
diff --git a/leetcode/0696.Count-Binary-Substrings/696. Count Binary Substrings.go b/leetcode/0696.Count-Binary-Substrings/696. Count Binary Substrings.go
new file mode 100644
index 000000000..a1c8559cc
--- /dev/null
+++ b/leetcode/0696.Count-Binary-Substrings/696. Count Binary Substrings.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+func countBinarySubstrings(s string) int {
+	last, res := 0, 0
+	for i := 0; i < len(s); {
+		c, count := s[i], 1
+		for i++; i < len(s) && s[i] == c; i++ {
+			count++
+		}
+		res += min(count, last)
+		last = count
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0696.Count-Binary-Substrings/696. Count Binary Substrings_test.go b/leetcode/0696.Count-Binary-Substrings/696. Count Binary Substrings_test.go
new file mode 100644
index 000000000..557a39ed4
--- /dev/null
+++ b/leetcode/0696.Count-Binary-Substrings/696. Count Binary Substrings_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question696 struct {
+	para696
+	ans696
+}
+
+// para 是参数
+// one 代表第一个参数
+type para696 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans696 struct {
+	one int
+}
+
+func Test_Problem696(t *testing.T) {
+
+	qs := []question696{
+
+		{
+			para696{"00110011"},
+			ans696{6},
+		},
+
+		{
+			para696{"10101"},
+			ans696{4},
+		},
+
+		{
+			para696{"0110001111"},
+			ans696{6},
+		},
+
+		{
+			para696{"0001111"},
+			ans696{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 696------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans696, q.para696
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countBinarySubstrings(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0696.Count-Binary-Substrings/README.md b/leetcode/0696.Count-Binary-Substrings/README.md
new file mode 100644
index 000000000..f0fb15e09
--- /dev/null
+++ b/leetcode/0696.Count-Binary-Substrings/README.md
@@ -0,0 +1,69 @@
+# [696. Count Binary Substrings](https://leetcode.com/problems/count-binary-substrings/)
+
+
+## 题目
+
+Give a string `s`, count the number of non-empty (contiguous) substrings that have the same number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped consecutively.
+
+Substrings that occur multiple times are counted the number of times they occur.
+
+**Example 1:**
+
+```
+Input: "00110011"
+Output: 6
+Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".
+
+Notice that some of these substrings repeat and are counted the number of times they occur.
+
+Also, "00110011" is not a valid substring becauseall the 0's (and 1's) are not grouped together.
+
+```
+
+**Example 2:**
+
+```
+Input: "10101"
+Output: 4
+Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.
+
+```
+
+**Note:**
+
+- `s.length` will be between 1 and 50,000.
+- `s` will only consist of "0" or "1" characters.
+
+## 题目大意
+
+给定一个字符串 s，计算具有相同数量 0 和 1 的非空（连续）子字符串的数量，并且这些子字符串中的所有 0 和所有 1 都是连续的。重复出现的子串要计算它们出现的次数。
+
+## 解题思路
+
+- 简单题。先分组统计 0 和 1 的个数，例如，`0110001111` 按照 0 和 1 分组统计出来的结果是 [1, 2, 3, 4]。再拼凑结果。相邻 2 组取两者最短的，例如 `0110001111`，凑成的结果应该是 min(1,2)，min(2,3)，min(3,4)，即 `01`，`01`，`10`，`1100`，`0011`，`000111`。时间复杂度 O(n)，空间复杂度 O(1)。
+
+## 代码
+
+```go
+package leetcode
+
+func countBinarySubstrings(s string) int {
+	last, res := 0, 0
+	for i := 0; i < len(s); {
+		c, count := s[i], 1
+		for i++; i < len(s) && s[i] == c; i++ {
+			count++
+		}
+		res += min(count, last)
+		last = count
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/0697.Degree-of-an-Array/697. Degree of an Array.go b/leetcode/0697.Degree-of-an-Array/697. Degree of an Array.go
new file mode 100644
index 000000000..24a11d1ea
--- /dev/null
+++ b/leetcode/0697.Degree-of-an-Array/697. Degree of an Array.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+func findShortestSubArray(nums []int) int {
+	frequency, maxFreq, smallest := map[int][]int{}, 0, len(nums)
+	for i, num := range nums {
+		if _, found := frequency[num]; !found {
+			frequency[num] = []int{1, i, i}
+		} else {
+			frequency[num][0]++
+			frequency[num][2] = i
+		}
+		if maxFreq < frequency[num][0] {
+			maxFreq = frequency[num][0]
+		}
+	}
+	for _, indices := range frequency {
+		if indices[0] == maxFreq {
+			if smallest > indices[2]-indices[1]+1 {
+				smallest = indices[2] - indices[1] + 1
+			}
+		}
+	}
+	return smallest
+}
diff --git a/leetcode/0697.Degree-of-an-Array/697. Degree of an Array_test.go b/leetcode/0697.Degree-of-an-Array/697. Degree of an Array_test.go
new file mode 100644
index 000000000..4f0730020
--- /dev/null
+++ b/leetcode/0697.Degree-of-an-Array/697. Degree of an Array_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question697 struct {
+	para697
+	ans697
+}
+
+// para 是参数
+// one 代表第一个参数
+type para697 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans697 struct {
+	one int
+}
+
+func Test_Problem697(t *testing.T) {
+
+	qs := []question697{
+
+		{
+			para697{[]int{1, 2, 2, 3, 1}},
+			ans697{2},
+		},
+
+		{
+			para697{[]int{1, 2, 2, 3, 1, 4, 2}},
+			ans697{6},
+		},
+
+		{
+			para697{[]int{}},
+			ans697{0},
+		},
+
+		{
+			para697{[]int{1, 1, 1, 1, 1}},
+			ans697{5},
+		},
+
+		{
+			para697{[]int{1, 2, 2, 3, 1, 4, 2, 1, 2, 2, 3, 1, 4, 2}},
+			ans697{13},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 697------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans697, q.para697
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findShortestSubArray(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0697.Degree-of-an-Array/README.md b/leetcode/0697.Degree-of-an-Array/README.md
new file mode 100644
index 000000000..2fa365538
--- /dev/null
+++ b/leetcode/0697.Degree-of-an-Array/README.md
@@ -0,0 +1,76 @@
+# [697. Degree of an Array](https://leetcode.com/problems/degree-of-an-array/)
+
+
+## 题目
+
+Given a non-empty array of non-negative integers `nums`, the **degree** of this array is defined as the maximum frequency of any one of its elements.
+
+Your task is to find the smallest possible length of a (contiguous) subarray of `nums`, that has the same degree as `nums`.
+
+**Example 1**:
+
+```
+Input: [1, 2, 2, 3, 1]
+Output: 2
+Explanation: 
+The input array has a degree of 2 because both elements 1 and 2 appear twice.
+Of the subarrays that have the same degree:
+[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
+The shortest length is 2. So return 2.
+
+```
+
+**Example 2**:
+
+```
+Input: [1,2,2,3,1,4,2]
+Output: 6
+```
+
+**Note**:
+
+- `nums.length` will be between 1 and 50,000.
+- `nums[i]` will be an integer between 0 and 49,999.
+
+
+## 题目大意
+
+给定一个非空且只包含非负数的整数数组 nums, 数组的度的定义是指数组里任一元素出现频数的最大值。你的任务是找到与 nums 拥有相同大小的度的最短连续子数组，返回其长度。
+
+注意:
+
+- nums.length 在 1 到 50,000 区间范围内。
+- nums[i] 是一个在 0 到 49,999 范围内的整数。
+
+
+## 解题思路
+
+- 找一个与给定数组相同度的最短连续子数组，输出其长度。数组的度的定义是任一元素出现频数的最大值。
+- 简单题。先统计各个元素的频次，并且动态维护最大频次和子数组的起始和终点位置。这里最短连续子数组有点“迷惑人”。这个最短子数组其实处理起来很简单。只需从前往后扫一遍，记录各个元素第一次出现的位置和最后一次出现的位置即是最短的连续子数组。然后在频次字典里面寻找和最大频次相同的所有解，有可能有多个子数组能满足题意，取出最短的输出即可。
+
+## 代码
+
+```go
+func findShortestSubArray(nums []int) int {
+	frequency, maxFreq, smallest := map[int][]int{}, 0, len(nums)
+	for i, num := range nums {
+		if _, found := frequency[num]; !found {
+			frequency[num] = []int{1, i, i}
+		} else {
+			frequency[num][0]++
+			frequency[num][2] = i
+		}
+		if maxFreq < frequency[num][0] {
+			maxFreq = frequency[num][0]
+		}
+	}
+	for _, indices := range frequency {
+		if indices[0] == maxFreq {
+			if smallest > indices[2]-indices[1]+1 {
+				smallest = indices[2] - indices[1] + 1
+			}
+		}
+	}
+	return smallest
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0699. Falling Squares/699. Falling Squares.go b/leetcode/0699.Falling-Squares/699. Falling Squares.go
similarity index 93%
rename from Algorithms/0699. Falling Squares/699. Falling Squares.go
rename to leetcode/0699.Falling-Squares/699. Falling Squares.go
index f87785157..bc6dc5daf 100644
--- a/Algorithms/0699. Falling Squares/699. Falling Squares.go	
+++ b/leetcode/0699.Falling-Squares/699. Falling Squares.go	
@@ -36,3 +36,10 @@ func discretization(positions [][]int) map[int]int {
 	}
 	return posMap
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0699.Falling-Squares/699. Falling Squares_test.go b/leetcode/0699.Falling-Squares/699. Falling Squares_test.go
new file mode 100644
index 000000000..54e5aaea2
--- /dev/null
+++ b/leetcode/0699.Falling-Squares/699. Falling Squares_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question699 struct {
+	para699
+	ans699
+}
+
+// para 是参数
+// one 代表第一个参数
+type para699 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans699 struct {
+	one []int
+}
+
+func Test_Problem699(t *testing.T) {
+
+	qs := []question699{
+
+		{
+			para699{[][]int{{6, 1}, {9, 2}, {2, 4}}},
+			ans699{[]int{1, 2, 4}},
+		},
+
+		{
+			para699{[][]int{{1, 2}, {1, 3}}},
+			ans699{[]int{2, 5}},
+		},
+
+		{
+			para699{[][]int{{1, 2}, {2, 3}, {6, 1}}},
+			ans699{[]int{2, 5, 5}},
+		},
+
+		{
+			para699{[][]int{{100, 100}, {200, 100}}},
+			ans699{[]int{100, 100}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 699------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans699, q.para699
+		fmt.Printf("【input】:%v       【output】:%v\n", p, fallingSquares(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0699.Falling-Squares/README.md b/leetcode/0699.Falling-Squares/README.md
new file mode 100755
index 000000000..8f3284b21
--- /dev/null
+++ b/leetcode/0699.Falling-Squares/README.md
@@ -0,0 +1,99 @@
+# [699. Falling Squares](https://leetcode.com/problems/falling-squares/)
+
+
+## 题目
+
+On an infinite number line (x-axis), we drop given squares in the order they are given.
+
+The `i`-th square dropped (`positions[i] = (left, side_length)`) is a square with the left-most point being `positions[i][0]` and sidelength `positions[i][1]`.
+
+The square is dropped with the bottom edge parallel to the number line, and from a higher height than all currently landed squares. We wait for each square to stick before dropping the next.
+
+The squares are infinitely sticky on their bottom edge, and will remain fixed to any positive length surface they touch (either the number line or another square). Squares dropped adjacent to each other will not stick together prematurely.
+
+Return a list `ans` of heights. Each height `ans[i]` represents the current highest height of any square we have dropped, after dropping squares represented by `positions[0], positions[1], ..., positions[i]`.
+
+**Example 1:**
+
+    Input: [[1, 2], [2, 3], [6, 1]]
+    Output: [2, 5, 5]
+    Explanation:
+
+After the first drop of `positions[0] = [1, 2]: _aa _aa -------` The maximum height of any square is 2.
+
+After the second drop of `positions[1] = [2, 3]: __aaa __aaa __aaa _aa__ _aa__ --------------` The maximum height of any square is 5. The larger square stays on top of the smaller square despite where its center of gravity is, because squares are infinitely sticky on their bottom edge.
+
+After the third drop of `positions[1] = [6, 1]: __aaa __aaa __aaa _aa _aa___a --------------` The maximum height of any square is still 5. Thus, we return an answer of `[2, 5, 5]`.
+
+**Example 2:**
+
+    Input: [[100, 100], [200, 100]]
+    Output: [100, 100]
+    Explanation: Adjacent squares don't get stuck prematurely - only their bottom edge can stick to surfaces.
+
+**Note:**
+
+- `1 <= positions.length <= 1000`.
+- `1 <= positions[i][0] <= 10^8`.
+- `1 <= positions[i][1] <= 10^6`.
+
+
+## 题目大意
+
+在无限长的数轴（即 x 轴）上，我们根据给定的顺序放置对应的正方形方块。第 i 个掉落的方块（positions[i] = (left, side\_length)）是正方形，其中 left 表示该方块最左边的点位置(positions[i][0])，side\_length 表示该方块的边长(positions[i][1])。
+
+每个方块的底部边缘平行于数轴（即 x 轴），并且从一个比目前所有的落地方块更高的高度掉落而下。在上一个方块结束掉落，并保持静止后，才开始掉落新方块。方块的底边具有非常大的粘性，并将保持固定在它们所接触的任何长度表面上（无论是数轴还是其他方块）。邻接掉落的边不会过早地粘合在一起，因为只有底边才具有粘性。
+
+返回一个堆叠高度列表 ans 。每一个堆叠高度 ans[i] 表示在通过 positions[0], positions[1], ..., positions[i] 表示的方块掉落结束后，目前所有已经落稳的方块堆叠的最高高度。
+
+示例 1:
+
+```c
+输入: [[1, 2], [2, 3], [6, 1]]
+输出: [2, 5, 5]
+解释:
+
+第一个方块 positions[0] = [1, 2] 掉落：
+_aa
+_aa
+-------
+方块最大高度为 2 。
+
+第二个方块 positions[1] = [2, 3] 掉落：
+__aaa
+__aaa
+__aaa
+_aa__
+_aa__
+--------------
+方块最大高度为5。
+大的方块保持在较小的方块的顶部，不论它的重心在哪里，因为方块的底部边缘有非常大的粘性。
+
+第三个方块 positions[1] = [6, 1] 掉落：
+__aaa
+__aaa
+__aaa
+_aa
+_aa___a
+-------------- 
+方块最大高度为5。
+
+因此，我们返回结果[2, 5, 5]。
+
+```
+
+注意:
+
+- 1 <= positions.length <= 1000.
+- 1 <= positions[i][0] <= 10^8.
+- 1 <= positions[i][1] <= 10^6.
+ 
+
+
+## 解题思路
+
+- 给出一个二维数组，每个一维数组中只有 2 个值，分别代表的是正方形砖块所在 x 轴的坐标起始点，和边长。要求输出每次砖块落下以后，当前最大的高度。正方形砖块落下如同俄罗斯方块，落下的过程中如果遇到了砖块会落在砖块的上面。如果砖块摞起来了以后，下方有空间，是不可能再把砖块挪进去的，因为此题砖块只会垂直落下，不会水平移动(这一点和俄罗斯方块不同)。
+- 这一题可以用线段树解答。由于方块在 x 轴上的坐标范围特别大，如果不离散化，这一题就会 MTE。所以首先去重 - 排序 - 离散化。先把每个砖块所在区间都算出来，每个正方形的方块所在区间是 `[pos[0] , pos[0]+pos[1]-1]` ，为什么右边界要减一呢？因为每个方块占据的区间其实应该是左闭右开的，即 `[pos[0] , pos[0]+pos[1])`，如果右边是开的，那么这个边界会被 2 个区间查询共用，从而导致错误结果。例如 [2,3]，[3,4]，这两个区间的砖块实际是不会摞在一起的。但是如果右边都是闭区间，用线段树 query 查询的时候，会都找到 [3，3]，从而导致这两个区间都会判断 3 这一点的情况。正确的做法应该是 [2,3)，[3,4）这样就避免了上述可能导致错误的情况了。离散化以后，所有的坐标区间都在 0~n 之间了。
+- 遍历每个砖块所在区间，先查询这个区间内的值，再加上当前砖块的高度，即为这个区间的最新高度。并更新该区间的值。更新区间的值用到懒惰更新。然后和动态维护的当前最大高度进行比较，将最大值放入最终输出的数组中。
+- 类似的题目有：第 715 题，第 218 题，第 732 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和本题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
+- leetcode  上也有线段树的讲解：[Get Solutions to Interview Questions](https://leetcode.com/articles/a-recursive-approach-to-segment-trees-range-sum-queries-lazy-propagation/)
diff --git a/leetcode/0700.Search-in-a-Binary-Search-Tree/700.Search in a Binary Search Tree.go b/leetcode/0700.Search-in-a-Binary-Search-Tree/700.Search in a Binary Search Tree.go
new file mode 100644
index 000000000..22c8d5a56
--- /dev/null
+++ b/leetcode/0700.Search-in-a-Binary-Search-Tree/700.Search in a Binary Search Tree.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func searchBST(root *TreeNode, val int) *TreeNode {
+	if root == nil {
+		return nil
+	}
+	if root.Val == val {
+		return root
+	} else if root.Val < val {
+		return searchBST(root.Right, val)
+	} else {
+		return searchBST(root.Left, val)
+	}
+}
diff --git a/leetcode/0700.Search-in-a-Binary-Search-Tree/700.Search in a Binary Search Tree_test.go b/leetcode/0700.Search-in-a-Binary-Search-Tree/700.Search in a Binary Search Tree_test.go
new file mode 100644
index 000000000..ba678bc25
--- /dev/null
+++ b/leetcode/0700.Search-in-a-Binary-Search-Tree/700.Search in a Binary Search Tree_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question700 struct {
+	para700
+	ans700
+}
+
+// para 是参数
+type para700 struct {
+	root *TreeNode
+	val  int
+}
+
+// ans 是答案
+type ans700 struct {
+	ans *TreeNode
+}
+
+func Test_Problem700(t *testing.T) {
+
+	qs := []question700{
+		{
+			para700{&TreeNode{Val: 4,
+				Left:  &TreeNode{Val: 2, Left: &TreeNode{Val: 1, Left: nil, Right: nil}, Right: &TreeNode{Val: 3, Left: nil, Right: nil}},
+				Right: &TreeNode{Val: 7, Left: nil, Right: nil}},
+				2},
+			ans700{&TreeNode{Val: 2, Left: &TreeNode{Val: 1, Left: nil, Right: nil}, Right: &TreeNode{Val: 3, Left: nil, Right: nil}}},
+		},
+
+		{
+			para700{&TreeNode{Val: 4,
+				Left:  &TreeNode{Val: 2, Left: &TreeNode{Val: 1, Left: nil, Right: nil}, Right: &TreeNode{Val: 3, Left: nil, Right: nil}},
+				Right: &TreeNode{Val: 7, Left: nil, Right: nil}},
+				5},
+			ans700{nil},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 700------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans700, q.para700
+		fmt.Printf("【input】:%v    【output】:%v\n", p, searchBST(p.root, p.val))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0700.Search-in-a-Binary-Search-Tree/README.md b/leetcode/0700.Search-in-a-Binary-Search-Tree/README.md
new file mode 100644
index 000000000..9079e64fc
--- /dev/null
+++ b/leetcode/0700.Search-in-a-Binary-Search-Tree/README.md
@@ -0,0 +1,73 @@
+# [700. Search in a Binary Search Tree](https://leetcode.com/problems/search-in-a-binary-search-tree/)
+
+## 题目
+
+You are given the root of a binary search tree (BST) and an integer val.
+
+Find the node in the BST that the node's value equals val and return the subtree rooted with that node. If such a node does not exist, return null.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2021/01/12/tree1.jpg](https://assets.leetcode.com/uploads/2021/01/12/tree1.jpg)
+
+    Input: root = [4,2,7,1,3], val = 2
+    Output: [2,1,3]
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2021/01/12/tree2.jpg](https://assets.leetcode.com/uploads/2021/01/12/tree2.jpg)
+
+    Input: root = [4,2,7,1,3], val = 5
+    Output: []
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range [1, 5000].
+- 1 <= Node.val <= 10000000
+- root is a binary search tree.
+- 1 <= val <= 10000000
+
+## 题目大意
+
+给定二叉搜索树（BST）的根节点和一个值。 你需要在BST中找到节点值等于给定值的节点。 返回以该节点为根的子树。 如果节点不存在，则返回 NULL。
+
+## 解题思路
+
+- 根据二叉搜索树的性质(根节点的值大于左子树所有节点的值，小于右子树所有节点的值),进行递归求解
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func searchBST(root *TreeNode, val int) *TreeNode {
+	if root == nil {
+		return nil
+	}
+	if root.Val == val {
+		return root
+	} else if root.Val < val {
+		return searchBST(root.Right, val)
+	} else {
+		return searchBST(root.Left, val)
+	}
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/0701.Insert-into-a-Binary-Search-Tree/701. Insert into a Binary Search Tree.go b/leetcode/0701.Insert-into-a-Binary-Search-Tree/701. Insert into a Binary Search Tree.go
new file mode 100644
index 000000000..7333d0318
--- /dev/null
+++ b/leetcode/0701.Insert-into-a-Binary-Search-Tree/701. Insert into a Binary Search Tree.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+import "github.com/halfrost/LeetCode-Go/structures"
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func insert(n *TreeNode, val int) *TreeNode {
+	if n == nil {
+		return &TreeNode{Val: val}
+	}
+	if n.Val < val {
+		n.Right = insert(n.Right, val)
+	} else {
+		n.Left = insert(n.Left, val)
+	}
+	return n
+}
+
+func insertIntoBST(root *TreeNode, val int) *TreeNode {
+	return insert(root, val)
+}
diff --git a/leetcode/0701.Insert-into-a-Binary-Search-Tree/701. Insert into a Binary Search Tree_test.go b/leetcode/0701.Insert-into-a-Binary-Search-Tree/701. Insert into a Binary Search Tree_test.go
new file mode 100644
index 000000000..d9e55f4c5
--- /dev/null
+++ b/leetcode/0701.Insert-into-a-Binary-Search-Tree/701. Insert into a Binary Search Tree_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question701 struct {
+	para701
+	ans701
+}
+
+// para 是参数
+// one 代表第一个参数
+type para701 struct {
+	root []int
+	val  int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans701 struct {
+	one []int
+}
+
+func Test_Problem701(t *testing.T) {
+
+	qs := []question701{
+
+		{
+			para701{[]int{4, 2, 7, 1, 3}, 5},
+			ans701{[]int{4, 2, 7, 1, 3, 5}},
+		},
+
+		{
+			para701{[]int{40, 20, 60, 10, 30, 50, 70}, 25},
+			ans701{[]int{40, 20, 60, 10, 30, 50, 70, structures.NULL, structures.NULL, 25}},
+		},
+
+		{
+			para701{[]int{4, 2, 7, 1, 3, structures.NULL, structures.NULL, structures.NULL, structures.NULL, structures.NULL, structures.NULL}, 5},
+			ans701{[]int{4, 2, 7, 1, 3, 5}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 701------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans701, q.para701
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.root)
+		fmt.Printf("【output】:%v      \n", structures.Tree2ints(insertIntoBST(rootOne, p.val)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0701.Insert-into-a-Binary-Search-Tree/README.md b/leetcode/0701.Insert-into-a-Binary-Search-Tree/README.md
new file mode 100644
index 000000000..fc4d504c6
--- /dev/null
+++ b/leetcode/0701.Insert-into-a-Binary-Search-Tree/README.md
@@ -0,0 +1,88 @@
+# [701. Insert into a Binary Search Tree](https://leetcode.com/problems/insert-into-a-binary-search-tree/)
+
+
+## 题目
+
+You are given the `root` node of a binary search tree (BST) and a `value` to insert into the tree. Return *the root node of the BST after the insertion*. It is **guaranteed** that the new value does not exist in the original BST.
+
+**Notice** that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return **any of them**.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/10/05/insertbst.jpg](https://assets.leetcode.com/uploads/2020/10/05/insertbst.jpg)
+
+```
+Input: root = [4,2,7,1,3], val = 5
+Output: [4,2,7,1,3,5]
+Explanation: Another accepted tree is:
+
+```
+
+![https://assets.leetcode.com/uploads/2020/10/05/bst.jpg](https://assets.leetcode.com/uploads/2020/10/05/bst.jpg)
+
+**Example 2:**
+
+```
+Input: root = [40,20,60,10,30,50,70], val = 25
+Output: [40,20,60,10,30,50,70,null,null,25]
+
+```
+
+**Example 3:**
+
+```
+Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
+Output: [4,2,7,1,3,5]
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree will be in the range `[0, 104]`.
+- `108 <= Node.val <= 108`
+- All the values `Node.val` are **unique**.
+- `108 <= val <= 108`
+- It's **guaranteed** that `val` does not exist in the original BST.
+
+## 题目大意
+
+给定二叉搜索树（BST）的根节点和要插入树中的值，将值插入二叉搜索树。 返回插入后二叉搜索树的根节点。 输入数据 保证 ，新值和原始二叉搜索树中的任意节点值都不同。注意，可能存在多种有效的插入方式，只要树在插入后仍保持为二叉搜索树即可。 你可以返回 任意有效的结果 。
+
+## 解题思路
+
+- 简单题。插入节点的方法有多种，笔者这里选择一种简单的方法。从根开始遍历这个二叉树，当前节点的值比待插入节点的值小，则往右遍历；当前节点的值比待插入节点的值大，则往左遍历。最后遍历到空节点便是要插入的地方。
+
+## 代码
+
+```go
+package leetcode
+
+import "github.com/halfrost/LeetCode-Go/structures"
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func insert(n *TreeNode, val int) *TreeNode {
+	if n == nil {
+		return &TreeNode{Val: val}
+	}
+	if n.Val < val {
+		n.Right = insert(n.Right, val)
+	} else {
+		n.Left = insert(n.Left, val)
+	}
+	return n
+}
+
+func insertIntoBST(root *TreeNode, val int) *TreeNode {
+	return insert(root, val)
+}
+```
\ No newline at end of file
diff --git a/leetcode/0703.Kth-Largest-Element-in-a-Stream/703. Kth Largest Element in a Stream.go b/leetcode/0703.Kth-Largest-Element-in-a-Stream/703. Kth Largest Element in a Stream.go
new file mode 100644
index 000000000..c83b54ece
--- /dev/null
+++ b/leetcode/0703.Kth-Largest-Element-in-a-Stream/703. Kth Largest Element in a Stream.go	
@@ -0,0 +1,38 @@
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+type KthLargest struct {
+	sort.IntSlice
+	k int
+}
+
+func Constructor(k int, nums []int) KthLargest {
+	kl := KthLargest{k: k}
+	for _, val := range nums {
+		kl.Add(val)
+	}
+	return kl
+}
+
+func (kl *KthLargest) Push(v interface{}) {
+	kl.IntSlice = append(kl.IntSlice, v.(int))
+}
+
+func (kl *KthLargest) Pop() interface{} {
+	a := kl.IntSlice
+	v := a[len(a)-1]
+	kl.IntSlice = a[:len(a)-1]
+	return v
+}
+
+func (kl *KthLargest) Add(val int) int {
+	heap.Push(kl, val)
+	if kl.Len() > kl.k {
+		heap.Pop(kl)
+	}
+	return kl.IntSlice[0]
+}
diff --git a/leetcode/0703.Kth-Largest-Element-in-a-Stream/703. Kth Largest Element in a Stream_test.go b/leetcode/0703.Kth-Largest-Element-in-a-Stream/703. Kth Largest Element in a Stream_test.go
new file mode 100644
index 000000000..448a2b47d
--- /dev/null
+++ b/leetcode/0703.Kth-Largest-Element-in-a-Stream/703. Kth Largest Element in a Stream_test.go	
@@ -0,0 +1,15 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem703(t *testing.T) {
+	obj := Constructor(3, []int{4, 5, 8, 2})
+	fmt.Printf("Add 7 = %v\n", obj.Add(3))
+	fmt.Printf("Add 7 = %v\n", obj.Add(5))
+	fmt.Printf("Add 7 = %v\n", obj.Add(10))
+	fmt.Printf("Add 7 = %v\n", obj.Add(9))
+	fmt.Printf("Add 7 = %v\n", obj.Add(4))
+}
diff --git a/leetcode/0703.Kth-Largest-Element-in-a-Stream/README.md b/leetcode/0703.Kth-Largest-Element-in-a-Stream/README.md
new file mode 100644
index 000000000..4a41c9f55
--- /dev/null
+++ b/leetcode/0703.Kth-Largest-Element-in-a-Stream/README.md
@@ -0,0 +1,93 @@
+# [703. Kth Largest Element in a Stream](https://leetcode.com/problems/kth-largest-element-in-a-stream/)
+
+## 题目
+
+Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.
+
+Implement `KthLargest` class:
+
+- `KthLargest(int k, int[] nums)` Initializes the object with the integer `k` and the stream of integers `nums`.
+- `int add(int val)` Returns the element representing the `kth` largest element in the stream.
+
+**Example 1:**
+
+```
+Input
+["KthLargest", "add", "add", "add", "add", "add"]
+[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
+Output
+[null, 4, 5, 5, 8, 8]
+
+Explanation
+KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
+kthLargest.add(3);   // return 4
+kthLargest.add(5);   // return 5
+kthLargest.add(10);  // return 5
+kthLargest.add(9);   // return 8
+kthLargest.add(4);   // return 8
+
+```
+
+**Constraints:**
+
+- `1 <= k <= 104`
+- `0 <= nums.length <= 104`
+- `104 <= nums[i] <= 104`
+- `104 <= val <= 104`
+- At most `104` calls will be made to `add`.
+- It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element.
+
+## 题目大意
+
+设计一个找到数据流中第 k 大元素的类（class）。注意是排序后的第 k 大元素，不是第 k 个不同的元素。请实现 KthLargest 类：
+
+- KthLargest(int k, int[] nums) 使用整数 k 和整数流 nums 初始化对象。
+- int add(int val) 将 val 插入数据流 nums 后，返回当前数据流中第 k 大的元素。
+
+## 解题思路
+
+- 读完题就能明白这一题考察的是最小堆。构建一个长度为 K 的最小堆，每次 pop 堆首(堆中最小的元素)，维护堆首即为第 K 大元素。
+- 这里有一个简洁的写法，常规的构建一个 pq 优先队列需要自己新建一个类型，然后实现 Len()、Less()、Swap()、Push()、Pop() 这 5 个方法。在 sort 包里有一个现成的最小堆，sort.IntSlice。可以借用它，再自己实现 Push()、Pop()就可以使用最小堆了，节约一部分代码。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+type KthLargest struct {
+	sort.IntSlice
+	k int
+}
+
+func Constructor(k int, nums []int) KthLargest {
+	kl := KthLargest{k: k}
+	for _, val := range nums {
+		kl.Add(val)
+	}
+	return kl
+}
+
+func (kl *KthLargest) Push(v interface{}) {
+	kl.IntSlice = append(kl.IntSlice, v.(int))
+}
+
+func (kl *KthLargest) Pop() interface{} {
+	a := kl.IntSlice
+	v := a[len(a)-1]
+	kl.IntSlice = a[:len(a)-1]
+	return v
+}
+
+func (kl *KthLargest) Add(val int) int {
+	heap.Push(kl, val)
+	if kl.Len() > kl.k {
+		heap.Pop(kl)
+	}
+	return kl.IntSlice[0]
+}
+```
\ No newline at end of file
diff --git a/leetcode/0704.Binary-Search/704. Binary Search.go b/leetcode/0704.Binary-Search/704. Binary Search.go
new file mode 100644
index 000000000..59ee8d79b
--- /dev/null
+++ b/leetcode/0704.Binary-Search/704. Binary Search.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func search704(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[mid] == target {
+			return mid
+		} else if nums[mid] > target {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
diff --git a/leetcode/0704.Binary-Search/704. Binary Search_test.go b/leetcode/0704.Binary-Search/704. Binary Search_test.go
new file mode 100644
index 000000000..594327f25
--- /dev/null
+++ b/leetcode/0704.Binary-Search/704. Binary Search_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question704 struct {
+	para704
+	ans704
+}
+
+// para 是参数
+// one 代表第一个参数
+type para704 struct {
+	nums   []int
+	target int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans704 struct {
+	one int
+}
+
+func Test_Problem704(t *testing.T) {
+
+	qs := []question704{
+
+		{
+			para704{[]int{-1, 0, 3, 5, 9, 12}, 9},
+			ans704{4},
+		},
+
+		{
+			para704{[]int{-1, 0, 3, 5, 9, 12}, 2},
+			ans704{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 704------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans704, q.para704
+		fmt.Printf("【input】:%v       【output】:%v\n", p, search704(p.nums, p.target))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0704.Binary-Search/README.md b/leetcode/0704.Binary-Search/README.md
new file mode 100755
index 000000000..04e9647b3
--- /dev/null
+++ b/leetcode/0704.Binary-Search/README.md
@@ -0,0 +1,43 @@
+# [704. Binary Search](https://leetcode.com/problems/binary-search/)
+
+
+## 题目
+
+Given a **sorted** (in ascending order) integer array `nums` of `n` elements and a `target` value, write a function to search `target` in `nums`. If `target` exists, then return its index, otherwise return `-1`.
+
+**Example 1:**
+
+    Input: nums = [-1,0,3,5,9,12], target = 9
+    Output: 4
+    Explanation: 9 exists in nums and its index is 4
+
+**Example 2:**
+
+    Input: nums = [-1,0,3,5,9,12], target = 2
+    Output: -1
+    Explanation: 2 does not exist in nums so return -1
+
+**Note:**
+
+1. You may assume that all elements in `nums` are unique.
+2. `n` will be in the range `[1, 10000]`.
+3. The value of each element in `nums` will be in the range `[-9999, 9999]`.
+
+
+## 题目大意
+
+
+给定一个 n 个元素有序的（升序）整型数组 nums 和一个目标值 target  ，写一个函数搜索 nums 中的 target，如果目标值存在返回下标，否则返回 -1。
+
+提示：
+
+- 你可以假设 nums 中的所有元素是不重复的。
+- n 将在 [1, 10000]之间。
+- nums 的每个元素都将在 [-9999, 9999]之间。
+
+
+## 解题思路
+
+
+- 给出一个数组，要求在数组中搜索等于 target 的元素的下标。如果找到就输出下标，如果找不到输出 -1 。
+- 简单题，二分搜索的裸题。
diff --git a/Algorithms/0705. Design HashSet/705. Design HashSet.go b/leetcode/0705.Design-HashSet/705. Design HashSet.go
similarity index 100%
rename from Algorithms/0705. Design HashSet/705. Design HashSet.go
rename to leetcode/0705.Design-HashSet/705. Design HashSet.go
diff --git a/Algorithms/0705. Design HashSet/705. Design HashSet_test.go b/leetcode/0705.Design-HashSet/705. Design HashSet_test.go
similarity index 100%
rename from Algorithms/0705. Design HashSet/705. Design HashSet_test.go
rename to leetcode/0705.Design-HashSet/705. Design HashSet_test.go
diff --git a/leetcode/0705.Design-HashSet/README.md b/leetcode/0705.Design-HashSet/README.md
new file mode 100755
index 000000000..5377ace4e
--- /dev/null
+++ b/leetcode/0705.Design-HashSet/README.md
@@ -0,0 +1,53 @@
+# [705. Design HashSet](https://leetcode.com/problems/design-hashset/)
+
+
+## 题目
+
+Design a HashSet without using any built-in hash table libraries.
+
+To be specific, your design should include these functions:
+
+- `add(value)`: Insert a value into the HashSet.
+- `contains(value)` : Return whether the value exists in the HashSet or not.
+- `remove(value)`: Remove a value in the HashSet. If the value does not exist in the HashSet, do nothing.
+
+**Example:**
+
+    MyHashSet hashSet = new MyHashSet();
+    hashSet.add(1);         
+    hashSet.add(2);         
+    hashSet.contains(1);    // returns true
+    hashSet.contains(3);    // returns false (not found)
+    hashSet.add(2);          
+    hashSet.contains(2);    // returns true
+    hashSet.remove(2);          
+    hashSet.contains(2);    // returns false (already removed)
+
+**Note:**
+
+- All values will be in the range of `[0, 1000000]`.
+- The number of operations will be in the range of `[1, 10000]`.
+- Please do not use the built-in HashSet library.
+
+
+## 题目大意
+
+不使用任何内建的哈希表库设计一个哈希集合具体地说，你的设计应该包含以下的功能：
+
+- add(value)：向哈希集合中插入一个值。
+- contains(value) ：返回哈希集合中是否存在这个值。
+- remove(value)：将给定值从哈希集合中删除。如果哈希集合中没有这个值，什么也不做。
+
+
+注意：
+
+- 所有的值都在 [1, 1000000] 的范围内。
+- 操作的总数目在 [1, 10000] 范围内。
+- 不要使用内建的哈希集合库。
+
+
+
+## 解题思路
+
+
+- 简单题，设计一个 hashset 的数据结构，要求有 `add(value)`，`contains(value)`，`remove(value)`，这 3 个方法。
diff --git a/Algorithms/0706. Design HashMap/706. Design HashMap.go b/leetcode/0706.Design-HashMap/706. Design HashMap.go
similarity index 96%
rename from Algorithms/0706. Design HashMap/706. Design HashMap.go
rename to leetcode/0706.Design-HashMap/706. Design HashMap.go
index acbf36fd7..2845be611 100644
--- a/Algorithms/0706. Design HashMap/706. Design HashMap.go	
+++ b/leetcode/0706.Design-HashMap/706. Design HashMap.go	
@@ -1,6 +1,6 @@
 package leetcode
 
-const Len int = 100000
+const Len int = 10000
 
 type MyHashMap struct {
 	content [Len]*HashNode
@@ -41,9 +41,9 @@ func (N *HashNode) Remove(key int) *HashNode {
 		return p
 	}
 	if N.next != nil {
-		return N.next.Remove(key)
+		N.next = N.next.Remove(key)
 	}
-	return nil
+	return N
 }
 
 /** Initialize your data structure here. */
diff --git a/Algorithms/0706. Design HashMap/706. Design HashMap_test.go b/leetcode/0706.Design-HashMap/706. Design HashMap_test.go
similarity index 82%
rename from Algorithms/0706. Design HashMap/706. Design HashMap_test.go
rename to leetcode/0706.Design-HashMap/706. Design HashMap_test.go
index 9ff3e200c..6d83ca0d0 100644
--- a/Algorithms/0706. Design HashMap/706. Design HashMap_test.go	
+++ b/leetcode/0706.Design-HashMap/706. Design HashMap_test.go	
@@ -13,6 +13,9 @@ func Test_Problem706(t *testing.T) {
 	fmt.Printf("Contains 7 = %v\n", obj.Get(7))
 	param1 := obj.Get(100)
 	fmt.Printf("param1 = %v\n", param1)
+	obj.Remove(100007)
+	param1 = obj.Get(7)
+	fmt.Printf("param1 = %v\n", param1)
 	obj.Remove(7)
 	param1 = obj.Get(7)
 	fmt.Printf("param1 = %v\n", param1)
diff --git a/leetcode/0706.Design-HashMap/README.md b/leetcode/0706.Design-HashMap/README.md
new file mode 100755
index 000000000..f0cbe6810
--- /dev/null
+++ b/leetcode/0706.Design-HashMap/README.md
@@ -0,0 +1,51 @@
+# [706. Design HashMap](https://leetcode.com/problems/design-hashmap/)
+
+
+## 题目
+
+Design a HashMap without using any built-in hash table libraries.
+
+To be specific, your design should include these functions:
+
+- `put(key, value)` : Insert a (key, value) pair into the HashMap. If the value already exists in the HashMap, update the value.
+- `get(key)`: Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
+- `remove(key)` : Remove the mapping for the value key if this map contains the mapping for the key.
+
+**Example:**
+
+    MyHashMap hashMap = new MyHashMap();
+    hashMap.put(1, 1);          
+    hashMap.put(2, 2);         
+    hashMap.get(1);            // returns 1
+    hashMap.get(3);            // returns -1 (not found)
+    hashMap.put(2, 1);          // update the existing value
+    hashMap.get(2);            // returns 1 
+    hashMap.remove(2);          // remove the mapping for 2
+    hashMap.get(2);            // returns -1 (not found)
+
+**Note:**
+
+- All keys and values will be in the range of `[0, 1000000]`.
+- The number of operations will be in the range of `[1, 10000]`.
+- Please do not use the built-in HashMap library.
+
+
+## 题目大意
+
+不使用任何内建的哈希表库设计一个哈希映射具体地说，你的设计应该包含以下的功能：
+
+- put(key, value)：向哈希映射中插入(键,值)的数值对。如果键对应的值已经存在，更新这个值。
+- get(key)：返回给定的键所对应的值，如果映射中不包含这个键，返回 -1。
+- remove(key)：如果映射中存在这个键，删除这个数值对。
+
+注意：
+
+- 所有的值都在 [1, 1000000] 的范围内。
+- 操作的总数目在 [1, 10000] 范围内。
+- 不要使用内建的哈希库。
+
+
+## 解题思路
+
+
+- 简单题，设计一个 hashmap 的数据结构，要求有 `put(key, value)`，`get(key)`，`remove(key)`，这 3 个方法。设计一个 map 主要需要处理哈希冲突，一般都是链表法解决冲突。
diff --git a/leetcode/0707.Design-Linked-List/707. Design Linked List.go b/leetcode/0707.Design-Linked-List/707. Design Linked List.go
new file mode 100644
index 000000000..57761f265
--- /dev/null
+++ b/leetcode/0707.Design-Linked-List/707. Design Linked List.go	
@@ -0,0 +1,106 @@
+package leetcode
+
+type MyLinkedList struct {
+	head *Node
+}
+
+type Node struct {
+	Val  int
+	Next *Node
+	Prev *Node
+}
+
+/** Initialize your data structure here. */
+func Constructor() MyLinkedList {
+	return MyLinkedList{}
+}
+
+/** Get the value of the index-th node in the linked list. If the index is invalid, return -1. */
+func (this *MyLinkedList) Get(index int) int {
+	curr := this.head
+	for i := 0; i < index && curr != nil; i++ {
+		curr = curr.Next
+	}
+	if curr != nil {
+		return curr.Val
+	} else {
+		return -1
+	}
+}
+
+/** Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list. */
+func (this *MyLinkedList) AddAtHead(val int) {
+	node := &Node{Val: val}
+	node.Next = this.head
+	if this.head != nil {
+		this.head.Prev = node
+	}
+	this.head = node
+}
+
+/** Append a node of value val to the last element of the linked list. */
+func (this *MyLinkedList) AddAtTail(val int) {
+	if this.head == nil {
+		this.AddAtHead(val)
+		return
+	}
+	node := &Node{Val: val}
+	curr := this.head
+	for curr != nil && curr.Next != nil {
+		curr = curr.Next
+	}
+	node.Prev = curr
+	curr.Next = node
+}
+
+/** Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted. */
+func (this *MyLinkedList) AddAtIndex(index int, val int) {
+	if index == 0 {
+		this.AddAtHead(val)
+	} else {
+		node := &Node{Val: val}
+		curr := this.head
+		for i := 0; i < index-1 && curr != nil; i++ {
+			curr = curr.Next
+		}
+		if curr != nil {
+			node.Next = curr.Next
+			node.Prev = curr
+			if node.Next != nil {
+				node.Next.Prev = node
+			}
+			curr.Next = node
+		}
+	}
+}
+
+/** Delete the index-th node in the linked list, if the index is valid. */
+func (this *MyLinkedList) DeleteAtIndex(index int) {
+	if index == 0 {
+		this.head = this.head.Next
+		if this.head != nil {
+			this.head.Prev = nil
+		}
+	} else {
+		curr := this.head
+		for i := 0; i < index-1 && curr != nil; i++ {
+			curr = curr.Next
+		}
+		if curr != nil && curr.Next != nil {
+			curr.Next = curr.Next.Next
+			if curr.Next != nil {
+				curr.Next.Prev = curr
+			}
+		}
+	}
+}
+
+/**
+ * Your MyLinkedList object will be instantiated and called as such:
+ * obj := Constructor();
+ * param_1 := obj.Get(index);
+ * obj.AddAtHead(val);
+ * obj.AddAtTail(val);
+ * obj.AddAtIndex(index,val);
+ * obj.DeleteAtIndex(index);
+ */
diff --git a/leetcode/0707.Design-Linked-List/707. Design Linked List_test.go b/leetcode/0707.Design-Linked-List/707. Design Linked List_test.go
new file mode 100644
index 000000000..3d98de216
--- /dev/null
+++ b/leetcode/0707.Design-Linked-List/707. Design Linked List_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem707(t *testing.T) {
+	obj := Constructor()
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	param1 := obj.Get(1)
+	fmt.Printf("param_1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	obj.AddAtHead(38)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.AddAtHead(45)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.DeleteAtIndex(2)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.AddAtIndex(1, 24)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.AddAtTail(36)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.AddAtIndex(3, 72)
+	obj.AddAtTail(76)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.AddAtHead(7)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.AddAtHead(36)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.AddAtHead(34)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.AddAtTail(91)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+
+	fmt.Printf("\n\n\n")
+
+	obj1 := Constructor()
+	fmt.Printf("obj1 = %v\n", MList2Ints(&obj1))
+	param2 := obj1.Get(0)
+	fmt.Printf("param_2 = %v obj1 = %v\n", param2, MList2Ints(&obj1))
+	obj1.AddAtIndex(1, 2)
+	fmt.Printf("obj1 = %v\n", MList2Ints(&obj1))
+	param2 = obj1.Get(0)
+	fmt.Printf("param_2 = %v obj1 = %v\n", param2, MList2Ints(&obj1))
+	param2 = obj1.Get(1)
+	fmt.Printf("param_2 = %v obj1 = %v\n", param2, MList2Ints(&obj1))
+	obj1.AddAtIndex(0, 1)
+	fmt.Printf("obj1 = %v\n", MList2Ints(&obj1))
+	param2 = obj1.Get(0)
+	fmt.Printf("param_1 = %v obj1 = %v\n", param2, MList2Ints(&obj1))
+	param2 = obj1.Get(1)
+	fmt.Printf("param_2 = %v obj1 = %v\n", param2, MList2Ints(&obj1))
+}
+
+func MList2Ints(head *MyLinkedList) []int {
+	res := []int{}
+	cur := head.head
+	for cur != nil {
+		res = append(res, cur.Val)
+		cur = cur.Next
+	}
+	return res
+}
diff --git a/Algorithms/0707. Design Linked List/README.md b/leetcode/0707.Design-Linked-List/README.md
similarity index 100%
rename from Algorithms/0707. Design Linked List/README.md
rename to leetcode/0707.Design-Linked-List/README.md
diff --git a/leetcode/0709.To-Lower-Case/709. To Lower Case.go b/leetcode/0709.To-Lower-Case/709. To Lower Case.go
new file mode 100644
index 000000000..9ade71c0e
--- /dev/null
+++ b/leetcode/0709.To-Lower-Case/709. To Lower Case.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+func toLowerCase(s string) string {
+	runes := []rune(s)
+	diff := 'a' - 'A'
+	for i := 0; i < len(s); i++ {
+		if runes[i] >= 'A' && runes[i] <= 'Z' {
+			runes[i] += diff
+		}
+	}
+	return string(runes)
+}
diff --git a/leetcode/0709.To-Lower-Case/709. To Lower Case_test.go b/leetcode/0709.To-Lower-Case/709. To Lower Case_test.go
new file mode 100644
index 000000000..f7bcd27fe
--- /dev/null
+++ b/leetcode/0709.To-Lower-Case/709. To Lower Case_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question709 struct {
+	para709
+	ans709
+}
+
+// para 是参数
+// one 代表第一个参数
+type para709 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans709 struct {
+	one string
+}
+
+func Test_Problem709(t *testing.T) {
+
+	qs := []question709{
+
+		{
+			para709{"Hello"},
+			ans709{"hello"},
+		},
+
+		{
+			para709{"here"},
+			ans709{"here"},
+		},
+
+		{
+			para709{"LOVELY"},
+			ans709{"lovely"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 709------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans709, q.para709
+		fmt.Printf("【input】:%v       【output】:%v\n", p, toLowerCase(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0709.To-Lower-Case/README.md b/leetcode/0709.To-Lower-Case/README.md
new file mode 100644
index 000000000..062aa6051
--- /dev/null
+++ b/leetcode/0709.To-Lower-Case/README.md
@@ -0,0 +1,55 @@
+# [709. To Lower Case](https://leetcode.com/problems/to-lower-case/)
+
+
+## 题目
+
+Given a string `s`, return *the string after replacing every uppercase letter with the same lowercase letter*.
+
+**Example 1:**
+
+```
+Input: s = "Hello"
+Output: "hello"
+```
+
+**Example 2:**
+
+```
+Input: s = "here"
+Output: "here"
+```
+
+**Example 3:**
+
+```
+Input: s = "LOVELY"
+Output: "lovely"
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 100`
+- `s` consists of printable ASCII characters.
+
+## 题目大意
+
+给你一个字符串 s ，将该字符串中的大写字母转换成相同的小写字母，返回新的字符串。
+
+## 解题思路
+
+- 简单题，将字符串中的大写字母转换成小写字母。
+
+## 代码
+
+```go
+func toLowerCase(s string) string {
+    runes := [] rune(s)
+    diff := 'a' - 'A'
+    for i := 0; i < len(s); i++ {
+        if runes[i] >= 'A' && runes[i] <= 'Z' {
+            runes[i] += diff
+        }
+    }
+    return string(runes)
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0710. Random Pick with Blacklist/710. Random Pick with Blacklist.go b/leetcode/0710.Random-Pick-with-Blacklist/710. Random Pick with Blacklist.go
similarity index 100%
rename from Algorithms/0710. Random Pick with Blacklist/710. Random Pick with Blacklist.go
rename to leetcode/0710.Random-Pick-with-Blacklist/710. Random Pick with Blacklist.go
diff --git a/Algorithms/0710. Random Pick with Blacklist/710. Random Pick with Blacklist_test.go b/leetcode/0710.Random-Pick-with-Blacklist/710. Random Pick with Blacklist_test.go
similarity index 93%
rename from Algorithms/0710. Random Pick with Blacklist/710. Random Pick with Blacklist_test.go
rename to leetcode/0710.Random-Pick-with-Blacklist/710. Random Pick with Blacklist_test.go
index a1732ae4d..ca6331590 100644
--- a/Algorithms/0710. Random Pick with Blacklist/710. Random Pick with Blacklist_test.go	
+++ b/leetcode/0710.Random-Pick-with-Blacklist/710. Random Pick with Blacklist_test.go	
@@ -28,27 +28,27 @@ func Test_Problem710(t *testing.T) {
 
 	qs := []question710{
 
-		question710{
+		{
 			para710{1, []int{}, 3},
 			ans710{[]int{0, 0, 0}},
 		},
 
-		question710{
+		{
 			para710{2, []int{}, 3},
 			ans710{[]int{1, 1, 1}},
 		},
 
-		question710{
+		{
 			para710{3, []int{1}, 3},
 			ans710{[]int{0, 0, 2}},
 		},
 
-		question710{
+		{
 			para710{4, []int{2}, 3},
 			ans710{[]int{1, 3, 1}},
 		},
 
-		question710{
+		{
 			para710{10000000, []int{1, 9999, 999999, 99999, 100, 0}, 10},
 			ans710{[]int{400, 200, 300}},
 		},
diff --git a/Algorithms/0710. Random Pick with Blacklist/README.md b/leetcode/0710.Random-Pick-with-Blacklist/README.md
similarity index 100%
rename from Algorithms/0710. Random Pick with Blacklist/README.md
rename to leetcode/0710.Random-Pick-with-Blacklist/README.md
diff --git a/Algorithms/0713. Subarray Product Less Than K/713. Subarray Product Less Than K.go b/leetcode/0713.Subarray-Product-Less-Than-K/713. Subarray Product Less Than K.go
similarity index 100%
rename from Algorithms/0713. Subarray Product Less Than K/713. Subarray Product Less Than K.go
rename to leetcode/0713.Subarray-Product-Less-Than-K/713. Subarray Product Less Than K.go
diff --git a/Algorithms/0713. Subarray Product Less Than K/713. Subarray Product Less Than K_test.go b/leetcode/0713.Subarray-Product-Less-Than-K/713. Subarray Product Less Than K_test.go
similarity index 96%
rename from Algorithms/0713. Subarray Product Less Than K/713. Subarray Product Less Than K_test.go
rename to leetcode/0713.Subarray-Product-Less-Than-K/713. Subarray Product Less Than K_test.go
index 5d879eb6c..c6f8c8658 100644
--- a/Algorithms/0713. Subarray Product Less Than K/713. Subarray Product Less Than K_test.go	
+++ b/leetcode/0713.Subarray-Product-Less-Than-K/713. Subarray Product Less Than K_test.go	
@@ -27,12 +27,12 @@ func Test_Problem713(t *testing.T) {
 
 	qs := []question713{
 
-		question713{
+		{
 			para713{[]int{10, 5, 2, 6}, 100},
 			ans713{8},
 		},
 
-		question713{
+		{
 			para713{[]int{10, 9, 10, 4, 3, 8, 3, 3, 6, 2, 10, 10, 9, 3}, 19},
 			ans713{18},
 		},
diff --git a/Algorithms/0713. Subarray Product Less Than K/README.md b/leetcode/0713.Subarray-Product-Less-Than-K/README.md
similarity index 100%
rename from Algorithms/0713. Subarray Product Less Than K/README.md
rename to leetcode/0713.Subarray-Product-Less-Than-K/README.md
diff --git a/Algorithms/0714. Best Time to Buy and Sell Stock with Transaction Fee/714. Best Time to Buy and Sell Stock with Transaction Fee.go b/leetcode/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee/714. Best Time to Buy and Sell Stock with Transaction Fee.go
similarity index 90%
rename from Algorithms/0714. Best Time to Buy and Sell Stock with Transaction Fee/714. Best Time to Buy and Sell Stock with Transaction Fee.go
rename to leetcode/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee/714. Best Time to Buy and Sell Stock with Transaction Fee.go
index 2a2847593..702c2d7a1 100644
--- a/Algorithms/0714. Best Time to Buy and Sell Stock with Transaction Fee/714. Best Time to Buy and Sell Stock with Transaction Fee.go	
+++ b/leetcode/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee/714. Best Time to Buy and Sell Stock with Transaction Fee.go	
@@ -21,6 +21,13 @@ func maxProfit714(prices []int, fee int) int {
 	return sell[len(sell)-1]
 }
 
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
 // 解法二 优化辅助空间的 DP
 func maxProfit714_1(prices []int, fee int) int {
 	sell, buy := 0, -prices[0]
diff --git a/Algorithms/0714. Best Time to Buy and Sell Stock with Transaction Fee/714. Best Time to Buy and Sell Stock with Transaction Fee_test.go b/leetcode/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee/714. Best Time to Buy and Sell Stock with Transaction Fee_test.go
similarity index 91%
rename from Algorithms/0714. Best Time to Buy and Sell Stock with Transaction Fee/714. Best Time to Buy and Sell Stock with Transaction Fee_test.go
rename to leetcode/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee/714. Best Time to Buy and Sell Stock with Transaction Fee_test.go
index d4eb73e5b..75c667b40 100644
--- a/Algorithms/0714. Best Time to Buy and Sell Stock with Transaction Fee/714. Best Time to Buy and Sell Stock with Transaction Fee_test.go	
+++ b/leetcode/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee/714. Best Time to Buy and Sell Stock with Transaction Fee_test.go	
@@ -27,22 +27,22 @@ func Test_Problem714(t *testing.T) {
 
 	qs := []question714{
 
-		question714{
+		{
 			para714{[]int{}, 0},
 			ans714{0},
 		},
 
-		question714{
+		{
 			para714{[]int{7, 1, 5, 3, 6, 4}, 0},
-			ans714{5},
+			ans714{7},
 		},
 
-		question714{
+		{
 			para714{[]int{7, 6, 4, 3, 1}, 0},
 			ans714{0},
 		},
 
-		question714{
+		{
 			para714{[]int{1, 3, 2, 8, 4, 9}, 2},
 			ans714{8},
 		},
diff --git a/Algorithms/0714. Best Time to Buy and Sell Stock with Transaction Fee/README.md b/leetcode/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee/README.md
similarity index 100%
rename from Algorithms/0714. Best Time to Buy and Sell Stock with Transaction Fee/README.md
rename to leetcode/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee/README.md
diff --git a/Algorithms/0715. Range Module/715. Range Module.go b/leetcode/0715.Range-Module/715. Range Module.go
similarity index 100%
rename from Algorithms/0715. Range Module/715. Range Module.go
rename to leetcode/0715.Range-Module/715. Range Module.go
diff --git a/Algorithms/0715. Range Module/715. Range Module_test.go b/leetcode/0715.Range-Module/715. Range Module_test.go
similarity index 100%
rename from Algorithms/0715. Range Module/715. Range Module_test.go
rename to leetcode/0715.Range-Module/715. Range Module_test.go
diff --git a/leetcode/0715.Range-Module/README.md b/leetcode/0715.Range-Module/README.md
new file mode 100755
index 000000000..1d5fefed1
--- /dev/null
+++ b/leetcode/0715.Range-Module/README.md
@@ -0,0 +1,60 @@
+# [715. Range Module](https://leetcode.com/problems/range-module/)
+
+
+## 题目
+
+A Range Module is a module that tracks ranges of numbers. Your task is to design and implement the following interfaces in an efficient manner.
+
+- `addRange(int left, int right)` Adds the half-open interval `[left, right)`, tracking every real number in that interval. Adding an interval that partially overlaps with currently tracked numbers should add any numbers in the interval `[left, right)` that are not already tracked.
+- `queryRange(int left, int right)` Returns true if and only if every real number in the interval `[left, right)` is currently being tracked.
+- `removeRange(int left, int right)` Stops tracking every real number currently being tracked in the interval `[left, right)`.
+
+**Example 1:**
+
+    addRange(10, 20): null
+    removeRange(14, 16): null
+    queryRange(10, 14): true (Every number in [10, 14) is being tracked)
+    queryRange(13, 15): false (Numbers like 14, 14.03, 14.17 in [13, 15) are not being tracked)
+    queryRange(16, 17): true (The number 16 in [16, 17) is still being tracked, despite the remove operation)
+
+**Note:**
+
+- A half open interval `[left, right)` denotes all real numbers `left <= x < right`.
+- `0 < left < right < 10^9` in all calls to `addRange, queryRange, removeRange`.
+- The total number of calls to `addRange` in a single test case is at most `1000`.
+- The total number of calls to `queryRange` in a single test case is at most `5000`.
+- The total number of calls to `removeRange` in a single test case is at most `1000`.
+
+## 题目大意
+
+Range 模块是跟踪数字范围的模块。你的任务是以一种有效的方式设计和实现以下接口。
+
+- addRange(int left, int right) 添加半开区间 [left, right)，跟踪该区间中的每个实数。添加与当前跟踪的数字部分重叠的区间时，应当添加在区间 [left, right) 中尚未跟踪的任何数字到该区间中。
+- queryRange(int left, int right) 只有在当前正在跟踪区间 [left, right) 中的每一个实数时，才返回 true。
+- removeRange(int left, int right) 停止跟踪区间 [left, right) 中当前正在跟踪的每个实数。
+ 
+
+示例：
+
+```
+addRange(10, 20): null
+removeRange(14, 16): null
+queryRange(10, 14): true （区间 [10, 14) 中的每个数都正在被跟踪）
+queryRange(13, 15): false （未跟踪区间 [13, 15) 中像 14, 14.03, 14.17 这样的数字）
+queryRange(16, 17): true （尽管执行了删除操作，区间 [16, 17) 中的数字 16 仍然会被跟踪）
+```
+
+提示：
+
+- 半开区间 [left, right) 表示所有满足 left <= x < right 的实数。
+- 对 addRange, queryRange, removeRange 的所有调用中 0 < left < right < 10^9。
+- 在单个测试用例中，对 addRange 的调用总数不超过 1000 次。
+- 在单个测试用例中，对  queryRange 的调用总数不超过 5000 次。
+- 在单个测试用例中，对 removeRange 的调用总数不超过 1000 次。
+
+
+## 解题思路
+
+- 设计一个数据结构，能完成添加区间 `addRange`，查询区间 `queryRange`，移除区间 `removeRange` 三种操作。查询区间的操作需要更加高效一点。
+- 这一题可以用线段树来解答，但是时间复杂度不高，最优解是用二叉排序树 BST 来解答。先来看线段树。这一题是更新区间内的值，所以需要用到懒惰更新。添加区间可以把区间内的值都赋值为 1 。由于题目中未预先确定区间范围，选用树的形式实现线段树比数组实现更加节约空间(当然用数组也可以，区间最大是 1000，点至多有 2000 个)。移除区间的时候就是把区间内的值都赋值标记为 0 。
+- 类似的题目有：第 699 题，第 218 题，第 732 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和第 699 题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
\ No newline at end of file
diff --git a/Algorithms/0717. 1-bit and 2-bit Characters/717. 1-bit and 2-bit Characters.go b/leetcode/0717.1-bit-and-2-bit-Characters/717. 1-bit and 2-bit Characters.go
similarity index 100%
rename from Algorithms/0717. 1-bit and 2-bit Characters/717. 1-bit and 2-bit Characters.go
rename to leetcode/0717.1-bit-and-2-bit-Characters/717. 1-bit and 2-bit Characters.go
diff --git a/Algorithms/0717. 1-bit and 2-bit Characters/717. 1-bit and 2-bit Characters_test.go b/leetcode/0717.1-bit-and-2-bit-Characters/717. 1-bit and 2-bit Characters_test.go
similarity index 92%
rename from Algorithms/0717. 1-bit and 2-bit Characters/717. 1-bit and 2-bit Characters_test.go
rename to leetcode/0717.1-bit-and-2-bit-Characters/717. 1-bit and 2-bit Characters_test.go
index ae26f48ca..20228cc94 100644
--- a/Algorithms/0717. 1-bit and 2-bit Characters/717. 1-bit and 2-bit Characters_test.go	
+++ b/leetcode/0717.1-bit-and-2-bit-Characters/717. 1-bit and 2-bit Characters_test.go	
@@ -26,22 +26,22 @@ func Test_Problem717(t *testing.T) {
 
 	qs := []question717{
 
-		question717{
+		{
 			para717{[]int{1, 0, 0}},
 			ans717{true},
 		},
 
-		question717{
+		{
 			para717{[]int{1, 1, 1, 0}},
 			ans717{false},
 		},
 
-		question717{
+		{
 			para717{[]int{0, 1, 1, 1, 0, 0}},
 			ans717{true},
 		},
 
-		question717{
+		{
 			para717{[]int{1, 1, 1, 1, 0}},
 			ans717{true},
 		},
diff --git a/leetcode/0717.1-bit-and-2-bit-Characters/README.md b/leetcode/0717.1-bit-and-2-bit-Characters/README.md
new file mode 100755
index 000000000..877c36582
--- /dev/null
+++ b/leetcode/0717.1-bit-and-2-bit-Characters/README.md
@@ -0,0 +1,46 @@
+# [717. 1-bit and 2-bit Characters](https://leetcode.com/problems/1-bit-and-2-bit-characters/)
+
+
+## 题目:
+
+We have two special characters. The first character can be represented by one bit `0`. The second character can be represented by two bits (`10` or `11`).
+
+Now given a string represented by several bits. Return whether the last character must be a one-bit character or not. The given string will always end with a zero.
+
+**Example 1:**
+
+    Input: 
+    bits = [1, 0, 0]
+    Output: True
+    Explanation: 
+    The only way to decode it is two-bit character and one-bit character. So the last character is one-bit character.
+
+**Example 2:**
+
+    Input: 
+    bits = [1, 1, 1, 0]
+    Output: False
+    Explanation: 
+    The only way to decode it is two-bit character and two-bit character. So the last character is NOT one-bit character.
+
+**Note:**
+
+- `1 <= len(bits) <= 1000`.
+- `bits[i]` is always `0` or `1`.
+
+## 题目大意
+
+有两种特殊字符。第一种字符可以用一比特0来表示。第二种字符可以用两比特(10 或 11)来表示。
+
+现给一个由若干比特组成的字符串。问最后一个字符是否必定为一个一比特字符。给定的字符串总是由0结束。
+
+注意:
+
+- 1 <= len(bits) <= 1000.
+- bits[i] 总是0 或 1.
+
+
+## 解题思路
+
+- 给出一个数组，数组里面的元素只有 0 和 1，并且数组的最后一个元素一定是 0。有 2 种特殊的字符，第一类字符是 "0"，第二类字符是 "11" 和 "10"，请判断这个数组最后一个元素是否一定是属于第一类字符？
+- 依题意， 0 的来源有 2 处，可以是第一类字符，也可以是第二类字符，1 的来源只有 1 处，一定出自第二类字符。最后一个 0 当前仅当为第一类字符的情况有 2 种，第一种情况，前面出现有 0，但是 0 和 1 配对形成了第二类字符。第二种情况，前面没有出现 0 。这两种情况的共同点是除去最后一个元素，数组中前面所有的1 都“结对子”。所以利用第二类字符的特征，"1X"，遍历整个数组，如果遇到 "1"，就跳 2 步，因为 1 后面出现什么数字( 0 或者 1 )并不需要关心。如果 `i` 能在 `len(bits) - 1` 的地方`(数组最后一个元素)`停下，那么对应的是情况一或者情况二，前面的 0 都和 1 匹配上了，最后一个 0 一定是第一类字符。如果 `i` 在 `len(bit)` 的位置`(超出数组下标)`停下，说明 `bits[len(bits) - 1] == 1`，这个时候最后一个 0 一定属于第二类字符。
diff --git a/leetcode/0718.Maximum-Length-of-Repeated-Subarray/718. Maximum Length of Repeated Subarray.go b/leetcode/0718.Maximum-Length-of-Repeated-Subarray/718. Maximum Length of Repeated Subarray.go
new file mode 100644
index 000000000..8a74cc9ae
--- /dev/null
+++ b/leetcode/0718.Maximum-Length-of-Repeated-Subarray/718. Maximum Length of Repeated Subarray.go	
@@ -0,0 +1,86 @@
+package leetcode
+
+const primeRK = 16777619
+
+// 解法一 二分搜索 + Rabin-Karp
+func findLength(A []int, B []int) int {
+	low, high := 0, min(len(A), len(B))
+	for low < high {
+		mid := (low + high + 1) >> 1
+		if hasRepeated(A, B, mid) {
+			low = mid
+		} else {
+			high = mid - 1
+		}
+	}
+	return low
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+func hashSlice(arr []int, length int) []int {
+	// hash 数组里面记录 arr 比 length 长出去部分的 hash 值
+	hash, pl, h := make([]int, len(arr)-length+1), 1, 0
+	for i := 0; i < length-1; i++ {
+		pl *= primeRK
+	}
+	for i, v := range arr {
+		h = h*primeRK + v
+		if i >= length-1 {
+			hash[i-length+1] = h
+			h -= pl * arr[i-length+1]
+		}
+	}
+	return hash
+}
+
+func hasSamePrefix(A, B []int, length int) bool {
+	for i := 0; i < length; i++ {
+		if A[i] != B[i] {
+			return false
+		}
+	}
+	return true
+}
+
+func hasRepeated(A, B []int, length int) bool {
+	hs := hashSlice(A, length)
+	hashToOffset := make(map[int][]int, len(hs))
+	for i, h := range hs {
+		hashToOffset[h] = append(hashToOffset[h], i)
+	}
+	for i, h := range hashSlice(B, length) {
+		if offsets, ok := hashToOffset[h]; ok {
+			for _, offset := range offsets {
+				if hasSamePrefix(A[offset:], B[i:], length) {
+					return true
+				}
+			}
+		}
+	}
+	return false
+}
+
+// 解法二 DP 动态规划
+func findLength1(A []int, B []int) int {
+	res, dp := 0, make([][]int, len(A)+1)
+	for i := range dp {
+		dp[i] = make([]int, len(B)+1)
+	}
+	for i := len(A) - 1; i >= 0; i-- {
+		for j := len(B) - 1; j >= 0; j-- {
+			if A[i] == B[j] {
+				dp[i][j] = dp[i+1][j+1] + 1
+				if dp[i][j] > res {
+					res = dp[i][j]
+				}
+			}
+		}
+	}
+	return res
+}
diff --git a/leetcode/0718.Maximum-Length-of-Repeated-Subarray/718. Maximum Length of Repeated Subarray_test.go b/leetcode/0718.Maximum-Length-of-Repeated-Subarray/718. Maximum Length of Repeated Subarray_test.go
new file mode 100644
index 000000000..37ec1a042
--- /dev/null
+++ b/leetcode/0718.Maximum-Length-of-Repeated-Subarray/718. Maximum Length of Repeated Subarray_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question718 struct {
+	para718
+	ans718
+}
+
+// para 是参数
+// one 代表第一个参数
+type para718 struct {
+	A []int
+	B []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans718 struct {
+	one int
+}
+
+func Test_Problem718(t *testing.T) {
+
+	qs := []question718{
+
+		{
+			para718{[]int{0, 0, 0, 0, 0}, []int{0, 0, 0, 0, 0}},
+			ans718{5},
+		},
+
+		{
+			para718{[]int{1, 2, 3, 2, 1}, []int{3, 2, 1, 4, 7}},
+			ans718{3},
+		},
+
+		{
+			para718{[]int{0, 0, 0, 0, 1}, []int{1, 0, 0, 0, 0}},
+			ans718{4},
+		},
+
+		{
+			para718{[]int{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, []int{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0}},
+			ans718{59},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 718------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans718, q.para718
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findLength(p.A, p.B))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0718.Maximum-Length-of-Repeated-Subarray/README.md b/leetcode/0718.Maximum-Length-of-Repeated-Subarray/README.md
new file mode 100755
index 000000000..6b6f3b10c
--- /dev/null
+++ b/leetcode/0718.Maximum-Length-of-Repeated-Subarray/README.md
@@ -0,0 +1,37 @@
+# [718. Maximum Length of Repeated Subarray](https://leetcode.com/problems/maximum-length-of-repeated-subarray/)
+
+
+## 题目
+
+Given two integer arrays `A` and `B`, return the maximum length of an subarray that appears in both arrays.
+
+**Example 1:**
+
+    Input:
+    A: [1,2,3,2,1]
+    B: [3,2,1,4,7]
+    Output: 3
+    Explanation: 
+    The repeated subarray with maximum length is [3, 2, 1].
+
+**Note:**
+
+1. 1 <= len(A), len(B) <= 1000
+2. 0 <= A[i], B[i] < 100
+
+
+## 题目大意
+
+给两个整数数组 A 和 B ，返回两个数组中公共的、长度最长的子数组的长度。
+
+
+
+## 解题思路
+
+- 给出两个数组，求这两个数组中最长相同子串的长度。
+- 这一题最容易想到的是 DP 动态规划的解法。`dp[i][j]` 代表在 A 数组中以 `i` 下标开始的子串与 B 数组中以 `j` 下标开始的子串最长相同子串的长度，状态转移方程为 `dp[i][j] = dp[i+1][j+1] + 1` (当 `A[i] == B[j]`)。这种解法的时间复杂度是 O(n^2)，空间复杂度 O(n^2)。
+- 这一题最佳解法是二分搜索 + `Rabin-Karp`。比较相同子串耗时的地方在于，需要一层循环，遍历子串所有字符。但是如果比较两个数字就很快，`O(1)` 的时间复杂度。所以有人就想到了，能不能把字符串也映射成数字呢？这样比较起来就非常快。这个算法就是 `Rabin-Karp` 算法。字符串映射成一个数字不能随意映射，还要求能根据字符串前缀动态增加，比较下一个字符串的时候，可以利用已比较过的前缀，加速之后的字符串比较。在 Rabin-Karp 算法中有一个“码点”的概念。类似于10进制中的进制。具体的算法讲解可以见这篇：
+
+    [基础知识 - Rabin-Karp 算法](https://www.cnblogs.com/golove/p/3234673.html)
+
+    “码点”一般取值为一个素数。在 go 的 `strings` 包里面取值是 16777619。所以这一题也可以直接取这个值。由于这一次要求我们找最长长度，所以把最长长度作为二分搜索的目标。先将数组 A 和数组 B 中的数字都按照二分出来的长度，进行 `Rabin-Karp` hash。对 A 中的 hash 与下标做映射关系，存到 map 中，方便后面快速查找。然后遍历 B 中的 hash，当 hash 一致的时候，再匹配下标。如果下标存在，且拥有相同的前缀，那么就算找到了相同的子串了。最后就是不断的二分，找到最长的结果即可。这个解法的时间复杂度 O(n * log n)，空间复杂度 O(n)。
\ No newline at end of file
diff --git a/leetcode/0719.Find-K-th-Smallest-Pair-Distance/719. Find K-th Smallest Pair Distance.go b/leetcode/0719.Find-K-th-Smallest-Pair-Distance/719. Find K-th Smallest Pair Distance.go
new file mode 100644
index 000000000..402df7399
--- /dev/null
+++ b/leetcode/0719.Find-K-th-Smallest-Pair-Distance/719. Find K-th Smallest Pair Distance.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+func smallestDistancePair(nums []int, k int) int {
+	sort.Ints(nums)
+	low, high := 0, nums[len(nums)-1]-nums[0]
+	for low < high {
+		mid := low + (high-low)>>1
+		tmp := findDistanceCount(nums, mid)
+		if tmp >= k {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+// 解法一 双指针
+func findDistanceCount(nums []int, num int) int {
+	count, i := 0, 0
+	for j := 1; j < len(nums); j++ {
+		for nums[j]-nums[i] > num && i < j {
+			i++
+		}
+		count += (j - i)
+	}
+	return count
+}
+
+// 解法二 暴力查找
+func findDistanceCount1(nums []int, num int) int {
+	count := 0
+	for i := 0; i < len(nums); i++ {
+		for j := i + 1; j < len(nums); j++ {
+			if nums[j]-nums[i] <= num {
+				count++
+			}
+		}
+	}
+	return count
+}
diff --git a/leetcode/0719.Find-K-th-Smallest-Pair-Distance/719. Find K-th Smallest Pair Distance_test.go b/leetcode/0719.Find-K-th-Smallest-Pair-Distance/719. Find K-th Smallest Pair Distance_test.go
new file mode 100644
index 000000000..cf9ff58de
--- /dev/null
+++ b/leetcode/0719.Find-K-th-Smallest-Pair-Distance/719. Find K-th Smallest Pair Distance_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question719 struct {
+	para719
+	ans719
+}
+
+// para 是参数
+// one 代表第一个参数
+type para719 struct {
+	num []int
+	k   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans719 struct {
+	one int
+}
+
+func Test_Problem719(t *testing.T) {
+
+	qs := []question719{
+
+		{
+			para719{[]int{1, 3, 1}, 1},
+			ans719{0},
+		},
+
+		{
+			para719{[]int{1, 1, 1}, 2},
+			ans719{0},
+		},
+
+		{
+			para719{[]int{1, 6, 1}, 3},
+			ans719{5},
+		},
+
+		{
+			para719{[]int{62, 100, 4}, 2},
+			ans719{58},
+		},
+
+		{
+			para719{[]int{9, 10, 7, 10, 6, 1, 5, 4, 9, 8}, 18},
+			ans719{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 719------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans719, q.para719
+		fmt.Printf("【input】:%v       【output】:%v\n", p, smallestDistancePair(p.num, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0719.Find-K-th-Smallest-Pair-Distance/README.md b/leetcode/0719.Find-K-th-Smallest-Pair-Distance/README.md
new file mode 100755
index 000000000..980e93edc
--- /dev/null
+++ b/leetcode/0719.Find-K-th-Smallest-Pair-Distance/README.md
@@ -0,0 +1,44 @@
+# [719. Find K-th Smallest Pair Distance](https://leetcode.com/problems/find-k-th-smallest-pair-distance/)
+
+
+## 题目
+
+Given an integer array, return the k-th smallest **distance** among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B.
+
+**Example 1:**
+
+    Input:
+    nums = [1,3,1]
+    k = 1
+    Output: 0 
+    Explanation:
+    Here are all the pairs:
+    (1,3) -> 2
+    (1,1) -> 0
+    (3,1) -> 2
+    Then the 1st smallest distance pair is (1,1), and its distance is 0.
+
+**Note:**
+
+1. `2 <= len(nums) <= 10000`.
+2. `0 <= nums[i] < 1000000`.
+3. `1 <= k <= len(nums) * (len(nums) - 1) / 2`.
+
+
+## 题目大意
+
+给定一个整数数组，返回所有数对之间的第 k 个最小距离。一对 (A, B) 的距离被定义为 A 和 B 之间的绝对差值。
+
+提示:
+
+1. 2 <= len(nums) <= 10000.
+2. 0 <= nums[i] < 1000000.
+3. 1 <= k <= len(nums) * (len(nums) - 1) / 2.
+
+
+
+## 解题思路
+
+- 给出一个数组，要求找出第 k 小两两元素之差的值。两两元素之差可能重复，重复的元素之差算多个，不去重。
+- 这一题可以用二分搜索来解答。先把原数组排序，那么最大的差值就是 `nums[len(nums)-1] - nums[0]` ，最小的差值是 0，即在 `[0, nums[len(nums)-1] - nums[0]]` 区间内搜索最终答案。针对每个 `mid`，判断小于等于 `mid` 的差值有多少个。题意就转化为，在数组中找到这样一个数，使得满足 `nums[i] - nums[j] ≤ mid` 条件的组合数等于 `k`。那么如何计算满足两两数的差值小于 mid 的组合总数是本题的关键。
+- 最暴力的方法就是 2 重循环，暴力计数。这个方法效率不高，耗时很长。原因是没有利用数组有序这一条件。实际上数组有序对计算满足条件的组合数有帮助。利用双指针滑动即可计算出组合总数。见解法一。
diff --git a/Algorithms/0720. Longest Word in Dictionary/720. Longest Word in Dictionary.go b/leetcode/0720.Longest-Word-in-Dictionary/720. Longest Word in Dictionary.go
similarity index 100%
rename from Algorithms/0720. Longest Word in Dictionary/720. Longest Word in Dictionary.go
rename to leetcode/0720.Longest-Word-in-Dictionary/720. Longest Word in Dictionary.go
diff --git a/Algorithms/0720. Longest Word in Dictionary/720. Longest Word in Dictionary_test.go b/leetcode/0720.Longest-Word-in-Dictionary/720. Longest Word in Dictionary_test.go
similarity index 96%
rename from Algorithms/0720. Longest Word in Dictionary/720. Longest Word in Dictionary_test.go
rename to leetcode/0720.Longest-Word-in-Dictionary/720. Longest Word in Dictionary_test.go
index b018bcfa2..0e63bb594 100644
--- a/Algorithms/0720. Longest Word in Dictionary/720. Longest Word in Dictionary_test.go	
+++ b/leetcode/0720.Longest-Word-in-Dictionary/720. Longest Word in Dictionary_test.go	
@@ -26,12 +26,12 @@ func Test_Problem720(t *testing.T) {
 
 	qs := []question720{
 
-		question720{
+		{
 			para720{[]string{"w", "wo", "wor", "worl", "world"}},
 			ans720{"world"},
 		},
 
-		question720{
+		{
 			para720{[]string{"a", "banana", "app", "appl", "ap", "apply", "apple"}},
 			ans720{"apple"},
 		},
diff --git a/Algorithms/0720. Longest Word in Dictionary/README.md b/leetcode/0720.Longest-Word-in-Dictionary/README.md
similarity index 100%
rename from Algorithms/0720. Longest Word in Dictionary/README.md
rename to leetcode/0720.Longest-Word-in-Dictionary/README.md
diff --git a/Algorithms/0721. Accounts Merge/721. Accounts Merge.go b/leetcode/0721.Accounts-Merge/721. Accounts Merge.go
similarity index 100%
rename from Algorithms/0721. Accounts Merge/721. Accounts Merge.go
rename to leetcode/0721.Accounts-Merge/721. Accounts Merge.go
diff --git a/leetcode/0721.Accounts-Merge/721. Accounts Merge_test.go b/leetcode/0721.Accounts-Merge/721. Accounts Merge_test.go
new file mode 100644
index 000000000..40ed580d6
--- /dev/null
+++ b/leetcode/0721.Accounts-Merge/721. Accounts Merge_test.go	
@@ -0,0 +1,73 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question721 struct {
+	para721
+	ans721
+}
+
+// para 是参数
+// one 代表第一个参数
+type para721 struct {
+	w [][]string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans721 struct {
+	one [][]string
+}
+
+func Test_Problem721(t *testing.T) {
+
+	qs := []question721{
+
+		{
+			para721{[][]string{{"John", "johnsmith@mail.com", "john00@mail.com"},
+				{"John", "johnnybravo@mail.com"},
+				{"John", "johnsmith@mail.com", "john_newyork@mail.com"},
+				{"Mary", "mary@mail.com"}}},
+			ans721{[][]string{{"John", "john00@mail.com", "john_newyork@mail.com", "johnsmith@mail.com"},
+				{"John", "johnnybravo@mail.com"},
+				{"Mary", "mary@mail.com"}}},
+		},
+
+		{
+			para721{[][]string{{"Alex", "Alex5@m.co", "Alex4@m.co", "Alex0@m.co"},
+				{"Ethan", "Ethan3@m.co", "Ethan3@m.co", "Ethan0@m.co"},
+				{"Kevin", "Kevin4@m.co", "Kevin2@m.co", "Kevin2@m.co"},
+				{"Gabe", "Gabe0@m.co", "Gabe3@m.co", "Gabe2@m.co"},
+				{"Gabe", "Gabe3@m.co", "Gabe4@m.co", "Gabe2@m.co"}}},
+			ans721{[][]string{{"Alex", "Alex0@m.co", "Alex4@m.co", "Alex5@m.co"},
+				{"Ethan", "Ethan0@m.co", "Ethan3@m.co"},
+				{"Gabe", "Gabe0@m.co", "Gabe2@m.co", "Gabe3@m.co", "Gabe4@m.co"},
+				{"Kevin", "Kevin2@m.co", "Kevin4@m.co"}}},
+		},
+
+		{
+			para721{[][]string{{"David", "David0@m.co", "David4@m.co", "David3@m.co"},
+				{"David", "David5@m.co", "David5@m.co", "David0@m.co"},
+				{"David", "David1@m.co", "David4@m.co", "David0@m.co"},
+				{"David", "David0@m.co", "David1@m.co", "David3@m.co"},
+				{"David", "David4@m.co", "David1@m.co", "David3@m.co"}}},
+			ans721{[][]string{{"David", "David0@m.co", "David1@m.co", "David3@m.co", "David4@m.co", "David5@m.co"}}},
+		},
+
+		{
+			para721{[][]string{}},
+			ans721{[][]string{}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 721------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans721, q.para721
+		fmt.Printf("【input】:%v       【output】:%v\n", p, accountsMerge(p.w))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0721.Accounts-Merge/README.md b/leetcode/0721.Accounts-Merge/README.md
new file mode 100755
index 000000000..c3f6a4c07
--- /dev/null
+++ b/leetcode/0721.Accounts-Merge/README.md
@@ -0,0 +1,49 @@
+# [721. Accounts Merge](https://leetcode.com/problems/accounts-merge/)
+
+
+## 题目
+
+Given a list `accounts`, each element `accounts[i]` is a list of strings, where the first element `accounts[i][0]` is a name, and the rest of the elements are emailsrepresenting emails of the account.
+
+Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some email that is common to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.
+
+After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails **in sorted order**. The accounts themselves can be returned in any order.
+
+**Example 1:**
+
+    Input: 
+    accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
+    Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'],  ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
+    Explanation: 
+    The first and third John's are the same person as they have the common email "johnsmith@mail.com".
+    The second John and Mary are different people as none of their email addresses are used by other accounts.
+    We could return these lists in any order, for example the answer [['Mary', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], 
+    ['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.
+
+**Note:**
+
+- The length of `accounts` will be in the range `[1, 1000]`.
+- The length of `accounts[i]` will be in the range `[1, 10]`.
+- The length of `accounts[i][j]` will be in the range `[1, 30]`.
+
+
+## 题目大意
+
+
+给定一个列表 accounts，每个元素 accounts[i] 是一个字符串列表，其中第一个元素 accounts[i][0] 是 名称 (name)，其余元素是 emails 表示该帐户的邮箱地址。现在，我们想合并这些帐户。如果两个帐户都有一些共同的邮件地址，则两个帐户必定属于同一个人。请注意，即使两个帐户具有相同的名称，它们也可能属于不同的人，因为人们可能具有相同的名称。一个人最初可以拥有任意数量的帐户，但其所有帐户都具有相同的名称。合并帐户后，按以下格式返回帐户：每个帐户的第一个元素是名称，其余元素是按顺序排列的邮箱地址。accounts 本身可以以任意顺序返回。
+
+
+注意：
+
+- accounts 的长度将在 [1，1000] 的范围内。
+- accounts[i] 的长度将在 [1，10] 的范围内。
+- accounts[i][j] 的长度将在 [1，30] 的范围内。
+
+
+
+## 解题思路
+
+
+- 给出一堆账户和对应的邮箱。要求合并同一个人的多个邮箱账户。如果判断是同一个人呢？如果这个人名和所属的其中之一的邮箱是相同的，就判定这是同一个人的邮箱，那么就合并这些邮箱。
+- 这题的解题思路是并查集。不过如果用暴力合并的方法，时间复杂度非常差。优化方法是先把每组数据都进行编号，人编号，每个邮箱都进行编号。这个映射关系用 `map` 记录起来。如果利用并查集的 `union()` 操作，把这些编号都进行合并。最后把人的编号和对应邮箱的编号拼接起来。
+- 这一题有 2 处比较“坑”的是，不需要合并的用户的邮箱列表也是需要排序和去重的，同一个人的所有邮箱集合都要合并到一起。具体见测试用例。不过题目中也提到了这些点，也不能算题目坑，只能归自己没注意这些边界情况。
diff --git a/leetcode/0724.Find-Pivot-Index/724. Find Pivot Index.go b/leetcode/0724.Find-Pivot-Index/724. Find Pivot Index.go
new file mode 100644
index 000000000..0158b8d85
--- /dev/null
+++ b/leetcode/0724.Find-Pivot-Index/724. Find Pivot Index.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+// 2 * leftSum + num[i] = sum
+// 时间: O(n)
+// 空间: O(1)
+func pivotIndex(nums []int) int {
+	if len(nums) <= 0 {
+		return -1
+	}
+	var sum, leftSum int
+	for _, num := range nums {
+		sum += num
+	}
+	for index, num := range nums {
+		if leftSum*2+num == sum {
+			return index
+		}
+		leftSum += num
+	}
+	return -1
+}
diff --git a/leetcode/0724.Find-Pivot-Index/724. Find Pivot Index_test.go b/leetcode/0724.Find-Pivot-Index/724. Find Pivot Index_test.go
new file mode 100644
index 000000000..6c89385a7
--- /dev/null
+++ b/leetcode/0724.Find-Pivot-Index/724. Find Pivot Index_test.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question724 struct {
+	para724
+	ans724
+}
+
+// para 是参数
+// one 代表第一个参数
+type para724 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans724 struct {
+	n int
+}
+
+func Test_Problem724(t *testing.T) {
+
+	qs := []question724{
+
+		{
+			para724{[]int{1, 7, 3, 6, 5, 6}},
+			ans724{3},
+		},
+
+		{
+			para724{[]int{1, 2, 3}},
+			ans724{-1},
+		},
+
+		{
+			para724{[]int{-1, -1, -1, -1, -1, -1}},
+			ans724{-1},
+		},
+
+		{
+			para724{[]int{-1, -1, -1, -1, -1, 0}},
+			ans724{2},
+		},
+
+		{
+			para724{[]int{1}},
+			ans724{0},
+		},
+
+		{
+			para724{[]int{}},
+			ans724{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 724------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans724, q.para724
+		fmt.Printf("【input】:%v       【output】:%v\n", p, pivotIndex(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0724.Find-Pivot-Index/README.md b/leetcode/0724.Find-Pivot-Index/README.md
new file mode 100644
index 000000000..a9b313790
--- /dev/null
+++ b/leetcode/0724.Find-Pivot-Index/README.md
@@ -0,0 +1,74 @@
+# [724. Find Pivot Index](https://leetcode.com/problems/find-pivot-index/)
+
+
+## 题目
+
+Given an array of integers nums, write a method that returns the "pivot" index of this array.
+
+We define the pivot index as the index where the sum of all the numbers to the left of the index is equal to the sum of all the numbers to the right of the index.
+
+If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot index.
+
+ 
+
+**Example 1**:
+
+    Input: nums = [1,7,3,6,5,6]
+    Output: 3
+    Explanation:
+    The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3.
+    Also, 3 is the first index where this occurs.
+
+**Example 2**:
+
+    Input: nums = [1,2,3]
+    Output: -1
+    Explanation:
+    There is no index that satisfies the conditions in the problem statement.
+
+**Constraints**:
+
+- The length of nums will be in the range [0, 10000].
+- Each element nums[i] will be an integer in the range [-1000, 1000].
+
+
+
+## 题目大意
+
+给定一个整数类型的数组 nums，请编写一个能够返回数组 “中心索引” 的方法。我们是这样定义数组 中心索引 的：数组中心索引的左侧所有元素相加的和等于右侧所有元素相加的和。如果数组不存在中心索引，那么我们应该返回 -1。如果数组有多个中心索引，那么我们应该返回最靠近左边的那一个。
+
+
+
+## 解题思路
+
+- 在数组中，找到一个数，使得它左边的数之和等于它的右边的数之和，如果存在，则返回这个数的下标索引，否作返回 -1。
+- 这里面存在一个等式，只需要满足这个等式即可满足条件：leftSum + num[i] = sum - leftSum => 2 * leftSum + num[i] = sum
+- 题目提到如果存在多个索引，则返回最左边那个，因此从左开始求和，而不是从右边
+
+## 代码
+
+```go
+
+package leetcode
+
+// 2 * leftSum + num[i] = sum
+// 时间: O(n)
+// 空间: O(1)
+func pivotIndex(nums []int) int {
+	if len(nums) <= 0 {
+		return -1
+	}
+	var sum, leftSum int
+	for _, num := range nums {
+		sum += num
+	}
+	for index, num := range nums {
+		if leftSum*2+num == sum {
+			return index
+		}
+		leftSum += num
+	}
+	return -1
+}
+
+```
diff --git a/Algorithms/0725. Split Linked List in Parts/725. Split Linked List in Parts.go b/leetcode/0725.Split-Linked-List-in-Parts/725. Split Linked List in Parts.go
similarity index 82%
rename from Algorithms/0725. Split Linked List in Parts/725. Split Linked List in Parts.go
rename to leetcode/0725.Split-Linked-List-in-Parts/725. Split Linked List in Parts.go
index 34b803aa3..edfae081f 100644
--- a/Algorithms/0725. Split Linked List in Parts/725. Split Linked List in Parts.go	
+++ b/leetcode/0725.Split-Linked-List-in-Parts/725. Split Linked List in Parts.go	
@@ -1,6 +1,13 @@
 package leetcode
 
-import "fmt"
+import (
+	"fmt"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
 
 /**
  * Definition for singly-linked list.
@@ -20,9 +27,8 @@ func splitListToParts(root *ListNode, k int) []*ListNode {
 	length := getLength(root)
 	splitNum := length / k
 	lengNum := length % k
-	cur := root
-	head := root
-	pre := root
+	cur, head := root, root
+	var pre *ListNode
 	fmt.Printf("总长度 %v, 分 %v 组, 前面 %v 组长度为 %v, 剩余 %v 组，每组 %v\n", length, k, lengNum, splitNum+1, k-lengNum, splitNum)
 	if splitNum == 0 {
 		for i := 0; i < k; i++ {
@@ -62,3 +68,13 @@ func splitListToParts(root *ListNode, k int) []*ListNode {
 	}
 	return res
 }
+
+func getLength(l *ListNode) int {
+	count := 0
+	cur := l
+	for cur != nil {
+		count++
+		cur = cur.Next
+	}
+	return count
+}
diff --git a/Algorithms/0725. Split Linked List in Parts/725. Split Linked List in Parts_test.go b/leetcode/0725.Split-Linked-List-in-Parts/725. Split Linked List in Parts_test.go
similarity index 80%
rename from Algorithms/0725. Split Linked List in Parts/725. Split Linked List in Parts_test.go
rename to leetcode/0725.Split-Linked-List-in-Parts/725. Split Linked List in Parts_test.go
index 433086fa5..45642316c 100644
--- a/Algorithms/0725. Split Linked List in Parts/725. Split Linked List in Parts_test.go	
+++ b/leetcode/0725.Split-Linked-List-in-Parts/725. Split Linked List in Parts_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question725 struct {
@@ -27,47 +29,42 @@ func Test_Problem725(t *testing.T) {
 
 	qs := []question725{
 
-		question725{
+		{
 			para725{[]int{1, 2, 3, 4, 5}, 7},
 			ans725{[]int{1, 2, 3, 4, 5, 0, 0}},
 		},
 
-		question725{
+		{
 			para725{[]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, 3},
 			ans725{[]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}},
 		},
 
-		question725{
+		{
 			para725{[]int{1, 1, 1, 1, 1}, 1},
 			ans725{[]int{1, 1, 1, 1, 1}},
 		},
 
-		question725{
+		{
 			para725{[]int{}, 3},
 			ans725{[]int{}},
 		},
 
-		question725{
-			para725{[]int{1, 2, 3, 2, 3, 2, 3, 2}, 0},
-			ans725{[]int{1, 2, 3, 2, 3, 2, 3, 2}},
-		},
-
-		question725{
+		{
 			para725{[]int{1, 2, 3, 4, 5}, 5},
 			ans725{[]int{1, 2, 3, 4}},
 		},
 
-		question725{
+		{
 			para725{[]int{}, 5},
 			ans725{[]int{}},
 		},
 
-		question725{
+		{
 			para725{[]int{1, 2, 3, 4, 5}, 10},
 			ans725{[]int{1, 2, 3, 4, 5}},
 		},
 
-		question725{
+		{
 			para725{[]int{1}, 1},
 			ans725{[]int{}},
 		},
@@ -77,9 +74,9 @@ func Test_Problem725(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans725, q.para725
-		res := splitListToParts(S2l(p.one), p.n)
+		res := splitListToParts(structures.Ints2List(p.one), p.n)
 		for _, value := range res {
-			fmt.Printf("【input】:%v    length:%v   【output】:%v\n", p, len(res), L2s(value))
+			fmt.Printf("【input】:%v    length:%v   【output】:%v\n", p, len(res), structures.List2Ints(value))
 		}
 		fmt.Printf("\n\n\n")
 	}
diff --git a/leetcode/0725.Split-Linked-List-in-Parts/README.md b/leetcode/0725.Split-Linked-List-in-Parts/README.md
new file mode 100644
index 000000000..b535964f2
--- /dev/null
+++ b/leetcode/0725.Split-Linked-List-in-Parts/README.md
@@ -0,0 +1,57 @@
+# [725. Split Linked List in Parts](https://leetcode.com/problems/split-linked-list-in-parts/)
+
+## 题目
+
+Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts".
+
+The length of each part should be as equal as possible: no two parts should have a size differing by more than 1. This may lead to some parts being null.
+
+The parts should be in order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal parts occurring later.
+
+Return a List of ListNode's representing the linked list parts that are formed.
+
+Examples 1->2->3->4, k = 5 // 5 equal parts [ [1], [2], [3], [4], null ]
+
+Example 1:
+
+```c
+Input: 
+root = [1, 2, 3], k = 5
+Output: [[1],[2],[3],[],[]]
+Explanation:
+The input and each element of the output are ListNodes, not arrays.
+For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null.
+The first element output[0] has output[0].val = 1, output[0].next = null.
+The last element output[4] is NULL, but it's string representation as a ListNode is [].
+```
+
+Example 2:
+
+```c
+Input: 
+root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
+Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
+Explanation:
+The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.
+```
+
+Note:
+
+- The length of root will be in the range [0, 1000].
+- Each value of a node in the input will be an integer in the range [0, 999].
+- k will be an integer in the range [1, 50].
+
+
+
+## 题目大意
+
+把链表分成 K 个部分，要求这 K 个部分尽量两两长度相差不超过 1，并且长度尽量相同。
+
+## 解题思路
+
+把链表长度对 K 进行除法，结果就是最终每组的长度 n。把链表长度对 K 进行取余操作，得到的结果 m，代表前 m 组链表长度为 n + 1 。相当于把多出来的部分都分摊到前面 m 组链表中了。最终链表是前 m 组长度为 n + 1，后 K - m 组链表长度是 n。
+
+注意长度不足 K 的时候要用 nil 进行填充。
+
+
+
diff --git a/Algorithms/0726. Number of Atoms/726. Number of Atoms.go b/leetcode/0726.Number-of-Atoms/726. Number of Atoms.go
similarity index 100%
rename from Algorithms/0726. Number of Atoms/726. Number of Atoms.go
rename to leetcode/0726.Number-of-Atoms/726. Number of Atoms.go
diff --git a/Algorithms/0726. Number of Atoms/726. Number of Atoms_test.go b/leetcode/0726.Number-of-Atoms/726. Number of Atoms_test.go
similarity index 92%
rename from Algorithms/0726. Number of Atoms/726. Number of Atoms_test.go
rename to leetcode/0726.Number-of-Atoms/726. Number of Atoms_test.go
index d65bac319..a0ca4e59c 100644
--- a/Algorithms/0726. Number of Atoms/726. Number of Atoms_test.go	
+++ b/leetcode/0726.Number-of-Atoms/726. Number of Atoms_test.go	
@@ -26,22 +26,22 @@ func Test_Problem726(t *testing.T) {
 
 	qs := []question726{
 
-		question726{
+		{
 			para726{"H200P"},
 			ans726{"H200P"},
 		},
 
-		question726{
+		{
 			para726{"H2O"},
 			ans726{"H2O"},
 		},
 
-		question726{
+		{
 			para726{"Mg(OH)2"},
 			ans726{"H2MgO2"},
 		},
 
-		question726{
+		{
 			para726{"K4(ON(SO3)2)2"},
 			ans726{"K4N2O14S4"},
 		},
diff --git a/Algorithms/0726. Number of Atoms/README.md b/leetcode/0726.Number-of-Atoms/README.md
similarity index 100%
rename from Algorithms/0726. Number of Atoms/README.md
rename to leetcode/0726.Number-of-Atoms/README.md
diff --git a/leetcode/0728.Self-Dividing-Numbers/728.Self Dividing Numbers.go b/leetcode/0728.Self-Dividing-Numbers/728.Self Dividing Numbers.go
new file mode 100644
index 000000000..a0f3c421f
--- /dev/null
+++ b/leetcode/0728.Self-Dividing-Numbers/728.Self Dividing Numbers.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+func selfDividingNumbers(left int, right int) []int {
+	var ans []int
+	for num := left; num <= right; num++ {
+		if selfDividingNum(num) {
+			ans = append(ans, num)
+		}
+	}
+	return ans
+}
+
+func selfDividingNum(num int) bool {
+	for d := num; d > 0; d = d / 10 {
+		reminder := d % 10
+		if reminder == 0 {
+			return false
+		}
+		if num%reminder != 0 {
+			return false
+		}
+	}
+	return true
+}
diff --git a/leetcode/0728.Self-Dividing-Numbers/728.Self Dividing Numbers_test.go b/leetcode/0728.Self-Dividing-Numbers/728.Self Dividing Numbers_test.go
new file mode 100644
index 000000000..a4e10d425
--- /dev/null
+++ b/leetcode/0728.Self-Dividing-Numbers/728.Self Dividing Numbers_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question728 struct {
+	para728
+	ans728
+}
+
+// para 是参数
+type para728 struct {
+	left  int
+	right int
+}
+
+// ans 是答案
+type ans728 struct {
+	ans []int
+}
+
+func Test_Problem728(t *testing.T) {
+
+	qs := []question728{
+
+		{
+			para728{1, 22},
+			ans728{[]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22}},
+		},
+
+		{
+			para728{47, 85},
+			ans728{[]int{48, 55, 66, 77}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 728------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans728, q.para728
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", selfDividingNumbers(p.left, p.right))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0728.Self-Dividing-Numbers/README.md b/leetcode/0728.Self-Dividing-Numbers/README.md
new file mode 100644
index 000000000..323e51ef7
--- /dev/null
+++ b/leetcode/0728.Self-Dividing-Numbers/README.md
@@ -0,0 +1,68 @@
+# [728. Self Dividing Numbers](https://leetcode.com/problems/self-dividing-numbers/)
+
+## 题目
+
+A self-dividing number is a number that is divisible by every digit it contains.
+
+- For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.
+
+A self-dividing number is not allowed to contain the digit zero.
+
+Given two integers left and right, return a list of all the self-dividing numbers in the range [left, right].
+
+**Example 1:**
+
+    Input: left = 1, right = 22
+    Output: [1,2,3,4,5,6,7,8,9,11,12,15,22]
+
+**Example 2:**
+
+    Input: left = 47, right = 85
+    Output: [48,55,66,77]
+
+**Constraints:**
+
+- 1 <= left <= right <= 10000
+
+## 题目大意
+
+自除数是指可以被它包含的每一位数整除的数。
+
+- 例如，128 是一个 自除数 ，因为 128 % 1 == 0，128 % 2 == 0，128 % 8 == 0。
+
+自除数 不允许包含 0 。
+
+给定两个整数 left 和 right ，返回一个列表，列表的元素是范围 [left, right] 内所有的 自除数 。
+
+## 解题思路
+
+- 模拟计算
+
+# 代码
+
+```go
+package leetcode
+
+func selfDividingNumbers(left int, right int) []int {
+	var ans []int
+	for num := left; num <= right; num++ {
+		if selfDividingNum(num) {
+			ans = append(ans, num)
+		}
+	}
+	return ans
+}
+
+func selfDividingNum(num int) bool {
+	for d := num; d > 0; d = d / 10 {
+		reminder := d % 10
+		if reminder == 0 {
+			return false
+		}
+		if num%reminder != 0 {
+			return false
+		}
+	}
+	return true
+}
+```
diff --git a/Algorithms/0729. My Calendar I/729. My Calendar I.go b/leetcode/0729.My-Calendar-I/729. My Calendar I.go
similarity index 100%
rename from Algorithms/0729. My Calendar I/729. My Calendar I.go
rename to leetcode/0729.My-Calendar-I/729. My Calendar I.go
diff --git a/Algorithms/0729. My Calendar I/729. My Calendar I_test.go b/leetcode/0729.My-Calendar-I/729. My Calendar I_test.go
similarity index 100%
rename from Algorithms/0729. My Calendar I/729. My Calendar I_test.go
rename to leetcode/0729.My-Calendar-I/729. My Calendar I_test.go
diff --git a/leetcode/0729.My-Calendar-I/README.md b/leetcode/0729.My-Calendar-I/README.md
new file mode 100755
index 000000000..a2c6a3fff
--- /dev/null
+++ b/leetcode/0729.My-Calendar-I/README.md
@@ -0,0 +1,60 @@
+# [729. My Calendar I](https://leetcode.com/problems/my-calendar-i/)
+
+
+## 题目
+
+Implement a `MyCalendar` class to store your events. A new event can be added if adding the event will not cause a double booking.
+
+Your class will have the method, `book(int start, int end)`. Formally, this represents a booking on the half open interval `[start, end)`, the range of real numbers `x` such that `start <= x < end`.
+
+A double booking happens when two events have some non-empty intersection (ie., there is some time that is common to both events.)
+
+For each call to the method `MyCalendar.book`, return `true` if the event can be added to the calendar successfully without causing a double booking. Otherwise, return `false` and do not add the event to the calendar.
+
+Your class will be called like this:
+
+`MyCalendar cal = new MyCalendar();`
+
+`MyCalendar.book(start, end)`
+
+**Example 1:**
+
+    MyCalendar();
+    MyCalendar.book(10, 20); // returns true
+    MyCalendar.book(15, 25); // returns false
+    MyCalendar.book(20, 30); // returns true
+    Explanation: 
+    The first event can be booked.  The second can't because time 15 is already booked by another event.
+    The third event can be booked, as the first event takes every time less than 20, but not including 20.
+
+**Note:**
+
+- The number of calls to `MyCalendar.book` per test case will be at most `1000`.
+- In calls to `MyCalendar.book(start, end)`, `start` and `end` are integers in the range `[0, 10^9]`.
+
+
+
+## 题目大意
+
+实现一个 MyCalendar 类来存放你的日程安排。如果要添加的时间内没有其他安排，则可以存储这个新的日程安排。
+
+MyCalendar 有一个 book(int start, int end) 方法。它意味着在 start 到 end 时间内增加一个日程安排，注意，这里的时间是半开区间，即 [start, end), 实数 x 的范围为，  start <= x < end。
+
+当两个日程安排有一些时间上的交叉时（例如两个日程安排都在同一时间内），就会产生重复预订。
+
+每次调用 MyCalendar.book 方法时，如果可以将日程安排成功添加到日历中而不会导致重复预订，返回 true。否则，返回 false 并且不要将该日程安排添加到日历中。
+
+请按照以下步骤调用 MyCalendar 类: MyCalendar cal = new MyCalendar(); MyCalendar.book(start, end)
+
+说明:
+
+- 每个测试用例，调用 MyCalendar.book 函数最多不超过 100次。
+- 调用函数 MyCalendar.book(start, end) 时， start 和 end 的取值范围为 [0, 10^9]。
+
+
+## 解题思路
+
+
+- 要求实现一个日程安排的功能，如果有日程安排冲突了，就返回 false，如果不冲突则返回 ture
+- 这一题有多种解法，第一种解法可以用类似第 34 题的解法。先排序每个区间，然后再这个集合中用二分搜索找到最后一个区间的左值比当前要比较的区间左值小的，如果找到，再判断能否插入进去(判断右区间是否比下一个区间的左区间小)，此方法时间复杂度 O(n log n)
+- 第二种解法是用生成一个 BST 树。在插入树中先排除不能插入的情况，例如区间有重合。然后以区间左值为依据，递归插入，每次插入依次会继续判断区间是否重合。直到不能插入，则返回 fasle。整个查找的时间复杂度是 O(log n)。
\ No newline at end of file
diff --git a/Algorithms/0732. My Calendar III/732. My Calendar III.go b/leetcode/0732.My-Calendar-III/732. My Calendar III.go
similarity index 100%
rename from Algorithms/0732. My Calendar III/732. My Calendar III.go
rename to leetcode/0732.My-Calendar-III/732. My Calendar III.go
diff --git a/Algorithms/0732. My Calendar III/732. My Calendar III_test.go b/leetcode/0732.My-Calendar-III/732. My Calendar III_test.go
similarity index 100%
rename from Algorithms/0732. My Calendar III/732. My Calendar III_test.go
rename to leetcode/0732.My-Calendar-III/732. My Calendar III_test.go
diff --git a/leetcode/0732.My-Calendar-III/README.md b/leetcode/0732.My-Calendar-III/README.md
new file mode 100755
index 000000000..44a6781c6
--- /dev/null
+++ b/leetcode/0732.My-Calendar-III/README.md
@@ -0,0 +1,63 @@
+# [732. My Calendar III](https://leetcode.com/problems/my-calendar-iii/)
+
+
+## 题目
+
+Implement a `MyCalendarThree` class to store your events. A new event can **always** be added.
+
+Your class will have one method, `book(int start, int end)`. Formally, this represents a booking on the half open interval `[start, end)`, the range of real numbers `x` such that `start <= x < end`.
+
+A K-booking happens when **K** events have some non-empty intersection (ie., there is some time that is common to all K events.)
+
+For each call to the method `MyCalendar.book`, return an integer `K` representing the largest integer such that there exists a `K`-booking in the calendar.
+
+Your class will be called like this:
+
+`MyCalendarThree cal = new MyCalendarThree();`
+
+`MyCalendarThree.book(start, end)`
+
+**Example 1:**
+
+    MyCalendarThree();
+    MyCalendarThree.book(10, 20); // returns 1
+    MyCalendarThree.book(50, 60); // returns 1
+    MyCalendarThree.book(10, 40); // returns 2
+    MyCalendarThree.book(5, 15); // returns 3
+    MyCalendarThree.book(5, 10); // returns 3
+    MyCalendarThree.book(25, 55); // returns 3
+    Explanation: 
+    The first two events can be booked and are disjoint, so the maximum K-booking is a 1-booking.
+    The third event [10, 40) intersects the first event, and the maximum K-booking is a 2-booking.
+    The remaining events cause the maximum K-booking to be only a 3-booking.
+    Note that the last event locally causes a 2-booking, but the answer is still 3 because
+    eg. [10, 20), [10, 40), and [5, 15) are still triple booked.
+
+**Note:**
+
+- The number of calls to `MyCalendarThree.book` per test case will be at most `400`.
+- In calls to `MyCalendarThree.book(start, end)`, `start` and `end` are integers in the range `[0, 10^9]`.
+
+
+## 题目大意
+
+实现一个 MyCalendar 类来存放你的日程安排，你可以一直添加新的日程安排。
+
+MyCalendar 有一个 book(int start, int end)方法。它意味着在start到end时间内增加一个日程安排，注意，这里的时间是半开区间，即 [start, end), 实数 x 的范围为，  start <= x < end。当 K 个日程安排有一些时间上的交叉时（例如K个日程安排都在同一时间内），就会产生 K 次预订。每次调用 MyCalendar.book方法时，返回一个整数 K ，表示最大的 K 次预订。
+
+请按照以下步骤调用MyCalendar 类: MyCalendar cal = new MyCalendar(); MyCalendar.book(start, end)
+
+说明:
+
+- 每个测试用例，调用 MyCalendar.book 函数最多不超过 400 次。
+- 调用函数 MyCalendar.book(start, end)时， start 和 end 的取值范围为 [0, 10^9]。
+
+
+
+
+## 解题思路
+
+- 设计一个日程类，每添加一个日程，实时显示出当前排期中累计日程最多的个数，例如在一段时间内，排了 3 个日程，其他时间内都只有 0，1，2 个日程，则输出 3 。
+- 拿到这个题目以后会立即想到线段树。由于题目中只有增加日程，所以这一题难度不大。这一题和第 699 题也类似，但是有区别，第 699 题中，俄罗斯方块会依次摞起来，而这一题中，俄罗斯方块也就摞起来，但是方块下面如果是空挡，方块会断掉。举个例子：依次增加区间 [10,20]，[10,40]，[5,15]，[5,10]，如果是第 699 题的规则，这 [5,10] 的这块砖块会落在 [5,15] 上，从而使得高度为 4，但是这一题是日程，日程不一样，[5,15] 这个区间内有 3 个日程，但是其他部分都没有 3 个日程，所以第三块砖块 [5,15] 中的 [5,10] 会“断裂”，掉下去，第四块砖块还是 [5,10]，落在第三块砖块断落下去的位置，它们俩落在一起的高度是 2 。
+- 构造一颗线段树，这里用树来构造，如果用数组需要开辟很大的空间。当区间左右边界和查询边界完全相同的时候再累加技术，否则不加，继续划分区间。以区间的左边界作为划分区间的标准，因为区间左边界是开区间，右边是闭区间。一个区间的计数值以区间左边界的计数为准。还是上面的例子，[5,10) 计数以 5 为标准，count = 2，[10,15) 计数以 10 为标准，count = 3 。还需要再动态维护一个最大值。这个线段树的实现比较简单。
+- 类似的题目有：第 715 题，第 218 题，第 699 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和第 699 题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
diff --git a/Algorithms/0733. Flood Fill/733. Flood Fill.go b/leetcode/0733.Flood-Fill/733. Flood Fill.go
similarity index 90%
rename from Algorithms/0733. Flood Fill/733. Flood Fill.go
rename to leetcode/0733.Flood-Fill/733. Flood Fill.go
index 6782310bc..9cc8fe630 100644
--- a/Algorithms/0733. Flood Fill/733. Flood Fill.go	
+++ b/leetcode/0733.Flood-Fill/733. Flood Fill.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
 func floodFill(image [][]int, sr int, sc int, newColor int) [][]int {
 	color := image[sr][sc]
 	if newColor == color {
diff --git a/Algorithms/0733. Flood Fill/733. Flood Fill_test.go b/leetcode/0733.Flood-Fill/733. Flood Fill_test.go
similarity index 83%
rename from Algorithms/0733. Flood Fill/733. Flood Fill_test.go
rename to leetcode/0733.Flood-Fill/733. Flood Fill_test.go
index 3f1c706cc..13d0e05c5 100644
--- a/Algorithms/0733. Flood Fill/733. Flood Fill_test.go	
+++ b/leetcode/0733.Flood-Fill/733. Flood Fill_test.go	
@@ -29,16 +29,16 @@ func Test_Problem733(t *testing.T) {
 
 	qs := []question733{
 
-		question733{
+		{
 			para733{[][]int{
-				[]int{1, 1, 1},
-				[]int{1, 1, 0},
-				[]int{1, 0, 1},
+				{1, 1, 1},
+				{1, 1, 0},
+				{1, 0, 1},
 			}, 1, 1, 2},
 			ans733{[][]int{
-				[]int{2, 2, 2},
-				[]int{2, 2, 0},
-				[]int{2, 0, 1},
+				{2, 2, 2},
+				{2, 2, 0},
+				{2, 0, 1},
 			}},
 		},
 	}
diff --git a/leetcode/0733.Flood-Fill/README.md b/leetcode/0733.Flood-Fill/README.md
new file mode 100755
index 000000000..33f3c8516
--- /dev/null
+++ b/leetcode/0733.Flood-Fill/README.md
@@ -0,0 +1,50 @@
+# [733. Flood Fill](https://leetcode.com/problems/flood-fill/)
+
+
+## 题目
+
+An `image` is represented by a 2-D array of integers, each integer representing the pixel value of the image (from 0 to 65535).
+
+Given a coordinate `(sr, sc)` representing the starting pixel (row and column) of the flood fill, and a pixel value `newColor`, "flood fill" the image.
+
+To perform a "flood fill", consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color as the starting pixel), and so on. Replace the color of all of the aforementioned pixels with the newColor.
+
+At the end, return the modified image.
+
+**Example 1:**
+
+    Input: 
+    image = [[1,1,1],[1,1,0],[1,0,1]]
+    sr = 1, sc = 1, newColor = 2
+    Output: [[2,2,2],[2,2,0],[2,0,1]]
+    Explanation: 
+    From the center of the image (with position (sr, sc) = (1, 1)), all pixels connected 
+    by a path of the same color as the starting pixel are colored with the new color.
+    Note the bottom corner is not colored 2, because it is not 4-directionally connected
+    to the starting pixel.
+
+**Note:**
+
+- The length of `image` and `image[0]` will be in the range `[1, 50]`.
+- The given starting pixel will satisfy `0 <= sr < image.length` and `0 <= sc < image[0].length`.
+- The value of each color in `image[i][j]` and `newColor` will be an integer in `[0, 65535]`.
+
+
+## 题目大意
+
+有一幅以二维整数数组表示的图画，每一个整数表示该图画的像素值大小，数值在 0 到 65535 之间。给你一个坐标 (sr, sc) 表示图像渲染开始的像素值（行 ，列）和一个新的颜色值 newColor，让你重新上色这幅图像。
+
+为了完成上色工作，从初始坐标开始，记录初始坐标的上下左右四个方向上像素值与初始坐标相同的相连像素点，接着再记录这四个方向上符合条件的像素点与他们对应四个方向上像素值与初始坐标相同的相连像素点，……，重复该过程。将所有有记录的像素点的颜色值改为新的颜色值。最后返回经过上色渲染后的图像。
+
+注意:
+
+- image 和 image[0] 的长度在范围 [1, 50] 内。
+- 给出的初始点将满足 0 <= sr < image.length 和 0 <= sc < image[0].length。
+- image[i][j] 和 newColor 表示的颜色值在范围 [0, 65535]内。
+
+
+## 解题思路
+
+
+- 给出一个二维的图片点阵，每个点阵都有一个数字。给出一个起点坐标，要求从这个起点坐标开始，把所有与这个起点连通的点都染色成 newColor。
+- 这一题是标准的 Flood Fill 算法。可以用 DFS 也可以用 BFS 。
diff --git a/Algorithms/0735. Asteroid Collision/735. Asteroid Collision.go b/leetcode/0735.Asteroid-Collision/735. Asteroid Collision.go
similarity index 100%
rename from Algorithms/0735. Asteroid Collision/735. Asteroid Collision.go
rename to leetcode/0735.Asteroid-Collision/735. Asteroid Collision.go
diff --git a/Algorithms/0735. Asteroid Collision/735. Asteroid Collision_test.go b/leetcode/0735.Asteroid-Collision/735. Asteroid Collision_test.go
similarity index 91%
rename from Algorithms/0735. Asteroid Collision/735. Asteroid Collision_test.go
rename to leetcode/0735.Asteroid-Collision/735. Asteroid Collision_test.go
index 322e5240d..e37bbb918 100644
--- a/Algorithms/0735. Asteroid Collision/735. Asteroid Collision_test.go	
+++ b/leetcode/0735.Asteroid-Collision/735. Asteroid Collision_test.go	
@@ -26,32 +26,32 @@ func Test_Problem735(t *testing.T) {
 
 	qs := []question735{
 
-		question735{
+		{
 			para735{[]int{-1, -1, 1, -2}},
 			ans735{[]int{-1, -1, -2}},
 		},
 
-		question735{
+		{
 			para735{[]int{5, 10, -5}},
 			ans735{[]int{5, 10}},
 		},
 
-		question735{
+		{
 			para735{[]int{5, 8, -8}},
 			ans735{[]int{5}},
 		},
 
-		question735{
+		{
 			para735{[]int{8, -8}},
 			ans735{[]int{}},
 		},
 
-		question735{
+		{
 			para735{[]int{10, 2, -5}},
 			ans735{[]int{10}},
 		},
 
-		question735{
+		{
 			para735{[]int{-2, -1, 1, 2}},
 			ans735{[]int{-2, -1, 1, 2}},
 		},
diff --git a/Algorithms/0735. Asteroid Collision/README.md b/leetcode/0735.Asteroid-Collision/README.md
similarity index 100%
rename from Algorithms/0735. Asteroid Collision/README.md
rename to leetcode/0735.Asteroid-Collision/README.md
diff --git a/Algorithms/0739. Daily Temperatures/739. Daily Temperatures.go b/leetcode/0739.Daily-Temperatures/739. Daily Temperatures.go
similarity index 100%
rename from Algorithms/0739. Daily Temperatures/739. Daily Temperatures.go
rename to leetcode/0739.Daily-Temperatures/739. Daily Temperatures.go
diff --git a/Algorithms/0739. Daily Temperatures/739. Daily Temperatures_test.go b/leetcode/0739.Daily-Temperatures/739. Daily Temperatures_test.go
similarity index 97%
rename from Algorithms/0739. Daily Temperatures/739. Daily Temperatures_test.go
rename to leetcode/0739.Daily-Temperatures/739. Daily Temperatures_test.go
index 2171bca3c..5266d6208 100644
--- a/Algorithms/0739. Daily Temperatures/739. Daily Temperatures_test.go	
+++ b/leetcode/0739.Daily-Temperatures/739. Daily Temperatures_test.go	
@@ -26,7 +26,7 @@ func Test_Problem739(t *testing.T) {
 
 	qs := []question739{
 
-		question739{
+		{
 			para739{[]int{73, 74, 75, 71, 69, 72, 76, 73}},
 			ans739{[]int{1, 1, 4, 2, 1, 1, 0, 0}},
 		},
diff --git a/Algorithms/0739. Daily Temperatures/README.md b/leetcode/0739.Daily-Temperatures/README.md
similarity index 100%
rename from Algorithms/0739. Daily Temperatures/README.md
rename to leetcode/0739.Daily-Temperatures/README.md
diff --git a/leetcode/0744.Find-Smallest-Letter-Greater-Than-Target/744. Find Smallest Letter Greater Than Target.go b/leetcode/0744.Find-Smallest-Letter-Greater-Than-Target/744. Find Smallest Letter Greater Than Target.go
new file mode 100644
index 000000000..541527a0f
--- /dev/null
+++ b/leetcode/0744.Find-Smallest-Letter-Greater-Than-Target/744. Find Smallest Letter Greater Than Target.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func nextGreatestLetter(letters []byte, target byte) byte {
+	low, high := 0, len(letters)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if letters[mid] > target {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	find := letters[low%len(letters)]
+	if find <= target {
+		return letters[0]
+	}
+	return find
+}
diff --git a/leetcode/0744.Find-Smallest-Letter-Greater-Than-Target/744. Find Smallest Letter Greater Than Target_test.go b/leetcode/0744.Find-Smallest-Letter-Greater-Than-Target/744. Find Smallest Letter Greater Than Target_test.go
new file mode 100644
index 000000000..6592bad06
--- /dev/null
+++ b/leetcode/0744.Find-Smallest-Letter-Greater-Than-Target/744. Find Smallest Letter Greater Than Target_test.go	
@@ -0,0 +1,68 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question744 struct {
+	para744
+	ans744
+}
+
+// para 是参数
+// one 代表第一个参数
+type para744 struct {
+	letters []byte
+	target  byte
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans744 struct {
+	one byte
+}
+
+func Test_Problem744(t *testing.T) {
+
+	qs := []question744{
+
+		{
+			para744{[]byte{'c', 'f', 'j'}, 'a'},
+			ans744{'c'},
+		},
+
+		{
+			para744{[]byte{'c', 'f', 'j'}, 'c'},
+			ans744{'f'},
+		},
+
+		{
+			para744{[]byte{'c', 'f', 'j'}, 'd'},
+			ans744{'f'},
+		},
+
+		{
+			para744{[]byte{'c', 'f', 'j'}, 'g'},
+			ans744{'j'},
+		},
+
+		{
+			para744{[]byte{'c', 'f', 'j'}, 'j'},
+			ans744{'c'},
+		},
+
+		{
+			para744{[]byte{'c', 'f', 'j'}, 'k'},
+			ans744{'c'},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 744------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans744, q.para744
+		fmt.Printf("【input】:%v       【output】:%v\n", p, nextGreatestLetter(p.letters, p.target))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0744.Find-Smallest-Letter-Greater-Than-Target/README.md b/leetcode/0744.Find-Smallest-Letter-Greater-Than-Target/README.md
new file mode 100755
index 000000000..1648f75fb
--- /dev/null
+++ b/leetcode/0744.Find-Smallest-Letter-Greater-Than-Target/README.md
@@ -0,0 +1,66 @@
+# [744. Find Smallest Letter Greater Than Target](https://leetcode.com/problems/find-smallest-letter-greater-than-target/)
+
+
+## 题目
+
+Given a list of sorted characters `letters` containing only lowercase letters, and given a target letter `target`, find the smallest element in the list that is larger than the given target.
+
+Letters also wrap around. For example, if the target is `target = 'z'` and `letters = ['a', 'b']`, the answer is `'a'`.
+
+**Examples:**
+
+    Input:
+    letters = ["c", "f", "j"]
+    target = "a"
+    Output: "c"
+    
+    Input:
+    letters = ["c", "f", "j"]
+    target = "c"
+    Output: "f"
+    
+    Input:
+    letters = ["c", "f", "j"]
+    target = "d"
+    Output: "f"
+    
+    Input:
+    letters = ["c", "f", "j"]
+    target = "g"
+    Output: "j"
+    
+    Input:
+    letters = ["c", "f", "j"]
+    target = "j"
+    Output: "c"
+    
+    Input:
+    letters = ["c", "f", "j"]
+    target = "k"
+    Output: "c"
+
+**Note:**
+
+1. `letters` has a length in range `[2, 10000]`.
+2. `letters` consists of lowercase letters, and contains at least 2 unique letters.
+3. `target` is a lowercase letter.
+
+
+## 题目大意
+
+给定一个只包含小写字母的有序数组letters 和一个目标字母 target，寻找有序数组里面比目标字母大的最小字母。
+
+数组里字母的顺序是循环的。举个例子，如果目标字母target = 'z' 并且有序数组为 letters = ['a', 'b']，则答案返回 'a'。
+
+注:
+
+1. letters长度范围在[2, 10000]区间内。
+2. letters 仅由小写字母组成，最少包含两个不同的字母。
+3. 目标字母target 是一个小写字母。
+
+
+
+## 解题思路
+
+- 给出一个字节数组，在这个字节数组中查找在 target 后面的第一个字母。数组是环形的。
+- 这一题也是二分搜索的题目，先在数组里面查找 target，如果找到了，取这个字母的后一个字母。如果没有找到，就取 low 下标的那个字母。注意数组是环形的，所以最后结果需要对下标取余。
diff --git a/Algorithms/0745. Prefix and Suffix Search/745. Prefix and Suffix Search.go b/leetcode/0745.Prefix-and-Suffix-Search/745. Prefix and Suffix Search.go
similarity index 100%
rename from Algorithms/0745. Prefix and Suffix Search/745. Prefix and Suffix Search.go
rename to leetcode/0745.Prefix-and-Suffix-Search/745. Prefix and Suffix Search.go
diff --git a/Algorithms/0745. Prefix and Suffix Search/745. Prefix and Suffix Search_test.go b/leetcode/0745.Prefix-and-Suffix-Search/745. Prefix and Suffix Search_test.go
similarity index 100%
rename from Algorithms/0745. Prefix and Suffix Search/745. Prefix and Suffix Search_test.go
rename to leetcode/0745.Prefix-and-Suffix-Search/745. Prefix and Suffix Search_test.go
diff --git a/Algorithms/0745. Prefix and Suffix Search/README.md b/leetcode/0745.Prefix-and-Suffix-Search/README.md
similarity index 100%
rename from Algorithms/0745. Prefix and Suffix Search/README.md
rename to leetcode/0745.Prefix-and-Suffix-Search/README.md
diff --git a/Algorithms/0746. Min Cost Climbing Stairs/746. Min Cost Climbing Stairs.go b/leetcode/0746.Min-Cost-Climbing-Stairs/746. Min Cost Climbing Stairs.go
similarity index 88%
rename from Algorithms/0746. Min Cost Climbing Stairs/746. Min Cost Climbing Stairs.go
rename to leetcode/0746.Min-Cost-Climbing-Stairs/746. Min Cost Climbing Stairs.go
index ac475d2f4..7d68e9c6c 100644
--- a/Algorithms/0746. Min Cost Climbing Stairs/746. Min Cost Climbing Stairs.go	
+++ b/leetcode/0746.Min-Cost-Climbing-Stairs/746. Min Cost Climbing Stairs.go	
@@ -10,6 +10,13 @@ func minCostClimbingStairs(cost []int) int {
 	return min(dp[len(cost)-2], dp[len(cost)-1])
 }
 
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
 // 解法二 DP 优化辅助空间
 func minCostClimbingStairs1(cost []int) int {
 	var cur, last int
diff --git a/Algorithms/0746. Min Cost Climbing Stairs/746. Min Cost Climbing Stairs_test.go b/leetcode/0746.Min-Cost-Climbing-Stairs/746. Min Cost Climbing Stairs_test.go
similarity index 95%
rename from Algorithms/0746. Min Cost Climbing Stairs/746. Min Cost Climbing Stairs_test.go
rename to leetcode/0746.Min-Cost-Climbing-Stairs/746. Min Cost Climbing Stairs_test.go
index 59a0bd98a..e09c5efe0 100644
--- a/Algorithms/0746. Min Cost Climbing Stairs/746. Min Cost Climbing Stairs_test.go	
+++ b/leetcode/0746.Min-Cost-Climbing-Stairs/746. Min Cost Climbing Stairs_test.go	
@@ -26,12 +26,12 @@ func Test_Problem746(t *testing.T) {
 
 	qs := []question746{
 
-		question746{
+		{
 			para746{[]int{10, 15, 20}},
 			ans746{15},
 		},
 
-		question746{
+		{
 			para746{[]int{1, 100, 1, 1, 1, 100, 1, 1, 100, 1}},
 			ans746{6},
 		},
diff --git a/Algorithms/0746. Min Cost Climbing Stairs/README.md b/leetcode/0746.Min-Cost-Climbing-Stairs/README.md
similarity index 100%
rename from Algorithms/0746. Min Cost Climbing Stairs/README.md
rename to leetcode/0746.Min-Cost-Climbing-Stairs/README.md
diff --git a/leetcode/0747.Largest-Number-At-Least-Twice-of-Others/747. Largest Number At Least Twice of Others.go b/leetcode/0747.Largest-Number-At-Least-Twice-of-Others/747. Largest Number At Least Twice of Others.go
new file mode 100644
index 000000000..8f68963cb
--- /dev/null
+++ b/leetcode/0747.Largest-Number-At-Least-Twice-of-Others/747. Largest Number At Least Twice of Others.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+func dominantIndex(nums []int) int {
+	maxNum, flag, index := 0, false, 0
+	for i, v := range nums {
+		if v > maxNum {
+			maxNum = v
+			index = i
+		}
+	}
+	for _, v := range nums {
+		if v != maxNum && 2*v > maxNum {
+			flag = true
+		}
+	}
+	if flag {
+		return -1
+	}
+	return index
+}
diff --git a/leetcode/0747.Largest-Number-At-Least-Twice-of-Others/747. Largest Number At Least Twice of Others_test.go b/leetcode/0747.Largest-Number-At-Least-Twice-of-Others/747. Largest Number At Least Twice of Others_test.go
new file mode 100644
index 000000000..6d658a7b8
--- /dev/null
+++ b/leetcode/0747.Largest-Number-At-Least-Twice-of-Others/747. Largest Number At Least Twice of Others_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question747 struct {
+	para747
+	ans747
+}
+
+// para 是参数
+// one 代表第一个参数
+type para747 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans747 struct {
+	one int
+}
+
+func Test_Problem747(t *testing.T) {
+
+	qs := []question747{
+
+		{
+			para747{[]int{3, 6, 1, 0}},
+			ans747{1},
+		},
+
+		{
+			para747{[]int{1, 2, 3, 4}},
+			ans747{-1},
+		},
+
+		{
+			para747{[]int{1}},
+			ans747{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 747------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans747, q.para747
+		fmt.Printf("【input】:%v       【output】:%v\n", p, dominantIndex(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0747.Largest-Number-At-Least-Twice-of-Others/README.md b/leetcode/0747.Largest-Number-At-Least-Twice-of-Others/README.md
new file mode 100644
index 000000000..9c55c6e0f
--- /dev/null
+++ b/leetcode/0747.Largest-Number-At-Least-Twice-of-Others/README.md
@@ -0,0 +1,75 @@
+# [747. Largest Number At Least Twice of Others](https://leetcode.com/problems/largest-number-at-least-twice-of-others/)
+
+
+## 题目
+
+You are given an integer array `nums` where the largest integer is **unique**.
+
+Determine whether the largest element in the array is **at least twice** as much as every other number in the array. If it is, return *the **index** of the largest element, or return* `-1` *otherwise*.
+
+**Example 1:**
+
+```
+Input: nums = [3,6,1,0]
+Output: 1
+Explanation: 6 is the largest integer.
+For every other number in the array x, 6 is at least twice as big as x.
+The index of value 6 is 1, so we return 1.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3,4]
+Output: -1
+Explanation: 4 is less than twice the value of 3, so we return -1.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1]
+Output: 0
+Explanation: 1 is trivially at least twice the value as any other number because there are no other numbers.
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 50`
+- `0 <= nums[i] <= 100`
+- The largest element in `nums` is unique.
+
+## 题目大意
+
+给你一个整数数组 nums ，其中总是存在 唯一的 一个最大整数 。请你找出数组中的最大元素并检查它是否 至少是数组中每个其他数字的两倍 。如果是，则返回 最大元素的下标 ，否则返回 -1 。
+
+## 解题思路
+
+- 简单题。先扫描一遍找到最大值和下标。再扫描一遍检查最大值是否是其他数字的两倍。
+
+## 代码
+
+```go
+package leetcode
+
+func dominantIndex(nums []int) int {
+	maxNum, flag, index := 0, false, 0
+	for i, v := range nums {
+		if v > maxNum {
+			maxNum = v
+			index = i
+		}
+	}
+	for _, v := range nums {
+		if v != maxNum && 2*v > maxNum {
+			flag = true
+		}
+	}
+	if flag {
+		return -1
+	}
+	return index
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0748. Shortest Completing Word/748. Shortest Completing Word.go b/leetcode/0748.Shortest-Completing-Word/748. Shortest Completing Word.go
similarity index 100%
rename from Algorithms/0748. Shortest Completing Word/748. Shortest Completing Word.go
rename to leetcode/0748.Shortest-Completing-Word/748. Shortest Completing Word.go
diff --git a/Algorithms/0748. Shortest Completing Word/748. Shortest Completing Word_test.go b/leetcode/0748.Shortest-Completing-Word/748. Shortest Completing Word_test.go
similarity index 96%
rename from Algorithms/0748. Shortest Completing Word/748. Shortest Completing Word_test.go
rename to leetcode/0748.Shortest-Completing-Word/748. Shortest Completing Word_test.go
index 78292b82d..32be3fa1f 100644
--- a/Algorithms/0748. Shortest Completing Word/748. Shortest Completing Word_test.go	
+++ b/leetcode/0748.Shortest-Completing-Word/748. Shortest Completing Word_test.go	
@@ -27,12 +27,12 @@ func Test_Problem748(t *testing.T) {
 
 	qs := []question748{
 
-		question748{
+		{
 			para748{"1s3 PSt", []string{"step", "steps", "stripe", "stepple"}},
 			ans748{"steps"},
 		},
 
-		question748{
+		{
 			para748{"1s3 456", []string{"looks", "pest", "stew", "show"}},
 			ans748{"pest"},
 		},
diff --git a/leetcode/0748.Shortest-Completing-Word/README.md b/leetcode/0748.Shortest-Completing-Word/README.md
new file mode 100755
index 000000000..1002568d0
--- /dev/null
+++ b/leetcode/0748.Shortest-Completing-Word/README.md
@@ -0,0 +1,55 @@
+# [748. Shortest Completing Word](https://leetcode.com/problems/shortest-completing-word/)
+
+
+## 题目
+
+Find the minimum length word from a given dictionary `words`, which has all the letters from the string `licensePlate`. Such a word is said to complete the given string `licensePlate`
+
+Here, for letters we ignore case. For example, `"P"` on the `licensePlate` still matches `"p"` on the word.
+
+It is guaranteed an answer exists. If there are multiple answers, return the one that occurs first in the array.
+
+The license plate might have the same letter occurring multiple times. For example, given a `licensePlate` of `"PP"`, the word `"pair"` does not complete the `licensePlate`, but the word `"supper"` does.
+
+**Example 1:**
+
+    Input: licensePlate = "1s3 PSt", words = ["step", "steps", "stripe", "stepple"]
+    Output: "steps"
+    Explanation: The smallest length word that contains the letters "S", "P", "S", and "T".
+    Note that the answer is not "step", because the letter "s" must occur in the word twice.
+    Also note that we ignored case for the purposes of comparing whether a letter exists in the word.
+
+**Example 2:**
+
+    Input: licensePlate = "1s3 456", words = ["looks", "pest", "stew", "show"]
+    Output: "pest"
+    Explanation: There are 3 smallest length words that contains the letters "s".
+    We return the one that occurred first.
+
+**Note:**
+
+1. `licensePlate` will be a string with length in range `[1, 7]`.
+2. `licensePlate` will contain digits, spaces, or letters (uppercase or lowercase).
+3. `words` will have a length in the range `[10, 1000]`.
+4. Every `words[i]` will consist of lowercase letters, and have length in range `[1, 15]`.
+
+
+## 题目大意
+
+如果单词列表（words）中的一个单词包含牌照（licensePlate）中所有的字母，那么我们称之为完整词。在所有完整词中，最短的单词我们称之为最短完整词。
+
+单词在匹配牌照中的字母时不区分大小写，比如牌照中的 "P" 依然可以匹配单词中的 "p" 字母。我们保证一定存在一个最短完整词。当有多个单词都符合最短完整词的匹配条件时取单词列表中最靠前的一个。牌照中可能包含多个相同的字符，比如说：对于牌照 "PP"，单词 "pair" 无法匹配，但是 "supper" 可以匹配。
+
+注意:
+
+- 牌照（licensePlate）的长度在区域[1, 7]中。
+- 牌照（licensePlate）将会包含数字、空格、或者字母（大写和小写）。
+- 单词列表（words）长度在区间 [10, 1000] 中。
+- 每一个单词 words[i] 都是小写，并且长度在区间 [1, 15] 中。
+
+
+
+## 解题思路
+
+
+- 给出一个数组，要求找出能包含 `licensePlate` 字符串中所有字符的最短长度的字符串。如果最短长度的字符串有多个，输出 word 下标小的那个。这一题也是简单题，不过有 2 个需要注意的点，第一点，`licensePlate` 中可能包含 `Unicode` 任意的字符，所以要先把字母的字符筛选出来，第二点是题目中保证了一定存在一个最短的单词能满足题意，并且忽略大小写。具体做法按照题意模拟即可。
diff --git a/leetcode/0752.Open-the-Lock/752. Open the Lock.go b/leetcode/0752.Open-the-Lock/752. Open the Lock.go
new file mode 100644
index 000000000..72488e57d
--- /dev/null
+++ b/leetcode/0752.Open-the-Lock/752. Open the Lock.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+func openLock(deadends []string, target string) int {
+	if target == "0000" {
+		return 0
+	}
+	targetNum, visited := strToInt(target), make([]bool, 10000)
+	visited[0] = true
+	for _, deadend := range deadends {
+		num := strToInt(deadend)
+		if num == 0 {
+			return -1
+		}
+		visited[num] = true
+	}
+	depth, curDepth, nextDepth := 0, []int16{0}, make([]int16, 0)
+	var nextNum int16
+	for len(curDepth) > 0 {
+		nextDepth = nextDepth[0:0]
+		for _, curNum := range curDepth {
+			for incrementer := int16(1000); incrementer > 0; incrementer /= 10 {
+				digit := (curNum / incrementer) % 10
+				if digit == 9 {
+					nextNum = curNum - 9*incrementer
+				} else {
+					nextNum = curNum + incrementer
+				}
+				if nextNum == targetNum {
+					return depth + 1
+				}
+				if !visited[nextNum] {
+					visited[nextNum] = true
+					nextDepth = append(nextDepth, nextNum)
+				}
+				if digit == 0 {
+					nextNum = curNum + 9*incrementer
+				} else {
+					nextNum = curNum - incrementer
+				}
+				if nextNum == targetNum {
+					return depth + 1
+				}
+				if !visited[nextNum] {
+					visited[nextNum] = true
+					nextDepth = append(nextDepth, nextNum)
+				}
+			}
+		}
+		curDepth, nextDepth = nextDepth, curDepth
+		depth++
+	}
+	return -1
+}
+
+func strToInt(str string) int16 {
+	return int16(str[0]-'0')*1000 + int16(str[1]-'0')*100 + int16(str[2]-'0')*10 + int16(str[3]-'0')
+}
diff --git a/leetcode/0752.Open-the-Lock/752. Open the Lock_test.go b/leetcode/0752.Open-the-Lock/752. Open the Lock_test.go
new file mode 100644
index 000000000..b6e7d24c9
--- /dev/null
+++ b/leetcode/0752.Open-the-Lock/752. Open the Lock_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question752 struct {
+	para752
+	ans752
+}
+
+// para 是参数
+// one 代表第一个参数
+type para752 struct {
+	deadends []string
+	target   string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans752 struct {
+	one int
+}
+
+func Test_Problem752(t *testing.T) {
+
+	qs := []question752{
+
+		{
+			para752{[]string{"0201", "0101", "0102", "1212", "2002"}, "0202"},
+			ans752{6},
+		},
+
+		{
+			para752{[]string{"8888"}, "0009"},
+			ans752{1},
+		},
+
+		{
+			para752{[]string{"8887", "8889", "8878", "8898", "8788", "8988", "7888", "9888"}, "8888"},
+			ans752{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 752------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans752, q.para752
+		fmt.Printf("【input】:%v       【output】:%v\n", p, openLock(p.deadends, p.target))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0752.Open-the-Lock/README.md b/leetcode/0752.Open-the-Lock/README.md
new file mode 100644
index 000000000..132083aa0
--- /dev/null
+++ b/leetcode/0752.Open-the-Lock/README.md
@@ -0,0 +1,131 @@
+# [752. Open the Lock](https://leetcode.com/problems/open-the-lock/)
+
+
+## 题目
+
+You have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: `'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'`. The wheels can rotate freely and wrap around: for example we can turn `'9'` to be `'0'`, or `'0'` to be `'9'`. Each move consists of turning one wheel one slot.
+
+The lock initially starts at `'0000'`, a string representing the state of the 4 wheels.
+
+You are given a list of `deadends` dead ends, meaning if the lock displays any of these codes, the wheels of the lock will stop turning and you will be unable to open it.
+
+Given a `target` representing the value of the wheels that will unlock the lock, return the minimum total number of turns required to open the lock, or -1 if it is impossible.
+
+**Example 1:**
+
+```
+Input: deadends = ["0201","0101","0102","1212","2002"], target = "0202"
+Output: 6
+Explanation:
+A sequence of valid moves would be "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202".
+Note that a sequence like "0000" -> "0001" -> "0002" -> "0102" -> "0202" would be invalid,
+because the wheels of the lock become stuck after the display becomes the dead end "0102".
+
+```
+
+**Example 2:**
+
+```
+Input: deadends = ["8888"], target = "0009"
+Output: 1
+Explanation:
+We can turn the last wheel in reverse to move from "0000" -> "0009".
+
+```
+
+**Example 3:**
+
+```
+Input: deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888"
+Output: -1
+Explanation:
+We can't reach the target without getting stuck.
+
+```
+
+**Example 4:**
+
+```
+Input: deadends = ["0000"], target = "8888"
+Output: -1
+
+```
+
+**Constraints:**
+
+- `1 <= deadends.length <= 500`
+- `deadends[i].length == 4`
+- `target.length == 4`
+- target **will not be** in the list `deadends`.
+- `target` and `deadends[i]` consist of digits only.
+
+## 题目大意
+
+你有一个带有四个圆形拨轮的转盘锁。每个拨轮都有10个数字： '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' 。每个拨轮可以自由旋转：例如把 '9' 变为 '0'，'0' 变为 '9' 。每次旋转都只能旋转一个拨轮的一位数字。锁的初始数字为 '0000' ，一个代表四个拨轮的数字的字符串。列表 deadends 包含了一组死亡数字，一旦拨轮的数字和列表里的任何一个元素相同，这个锁将会被永久锁定，无法再被旋转。字符串 target 代表可以解锁的数字，你需要给出解锁需要的最小旋转次数，如果无论如何不能解锁，返回 -1 。
+
+## 解题思路
+
+- 此题可以转化为从起始点到终点的最短路径。采用广度优先搜索。每次广搜枚举转动一次数字的状态，并且用 visited 记录是否被搜索过，如果没有被搜索过，便加入队列，下一轮继续搜索。如果搜索到了 target，便返回对应的旋转次数。如果搜索完成后，仍没有搜索到 target，说明无法解锁，返回 -1。特殊情况，如果 target 就是初始数字 0000，那么直接返回答案 0。
+- 在广搜之前，先将 deadends 放入 map 中，搜索中判断是否搜到了 deadends。如果初始数字 0000 出现在 deadends 中，可以直接返回答案 −1。
+
+## 代码
+
+```go
+package leetcode
+
+func openLock(deadends []string, target string) int {
+	if target == "0000" {
+		return 0
+	}
+	targetNum, visited := strToInt(target), make([]bool, 10000)
+	visited[0] = true
+	for _, deadend := range deadends {
+		num := strToInt(deadend)
+		if num == 0 {
+			return -1
+		}
+		visited[num] = true
+	}
+	depth, curDepth, nextDepth := 0, []int16{0}, make([]int16, 0)
+	var nextNum int16
+	for len(curDepth) > 0 {
+		nextDepth = nextDepth[0:0]
+		for _, curNum := range curDepth {
+			for incrementer := int16(1000); incrementer > 0; incrementer /= 10 {
+				digit := (curNum / incrementer) % 10
+				if digit == 9 {
+					nextNum = curNum - 9*incrementer
+				} else {
+					nextNum = curNum + incrementer
+				}
+				if nextNum == targetNum {
+					return depth + 1
+				}
+				if !visited[nextNum] {
+					visited[nextNum] = true
+					nextDepth = append(nextDepth, nextNum)
+				}
+				if digit == 0 {
+					nextNum = curNum + 9*incrementer
+				} else {
+					nextNum = curNum - incrementer
+				}
+				if nextNum == targetNum {
+					return depth + 1
+				}
+				if !visited[nextNum] {
+					visited[nextNum] = true
+					nextDepth = append(nextDepth, nextNum)
+				}
+			}
+		}
+		curDepth, nextDepth = nextDepth, curDepth
+		depth++
+	}
+	return -1
+}
+
+func strToInt(str string) int16 {
+	return int16(str[0]-'0')*1000 + int16(str[1]-'0')*100 + int16(str[2]-'0')*10 + int16(str[3]-'0')
+}
+```
\ No newline at end of file
diff --git a/leetcode/0753.Cracking-the-Safe/753. Cracking the Safe.go b/leetcode/0753.Cracking-the-Safe/753. Cracking the Safe.go
new file mode 100644
index 000000000..90351792f
--- /dev/null
+++ b/leetcode/0753.Cracking-the-Safe/753. Cracking the Safe.go	
@@ -0,0 +1,39 @@
+package leetcode
+
+import "math"
+
+const number = "0123456789"
+
+func crackSafe(n int, k int) string {
+	if n == 1 {
+		return number[:k]
+	}
+	visit, total := map[string]bool{}, int(math.Pow(float64(k), float64(n)))
+	str := make([]byte, 0, total+n-1)
+	for i := 1; i != n; i++ {
+		str = append(str, '0')
+	}
+	dfsCrackSafe(total, n, k, &str, &visit)
+	return string(str)
+}
+
+func dfsCrackSafe(depth, n, k int, str *[]byte, visit *map[string]bool) bool {
+	if depth == 0 {
+		return true
+	}
+	for i := 0; i != k; i++ {
+		*str = append(*str, byte('0'+i))
+		cur := string((*str)[len(*str)-n:])
+		if _, ok := (*visit)[cur]; ok != true {
+			(*visit)[cur] = true
+			if dfsCrackSafe(depth-1, n, k, str, visit) {
+				// 只有这里不需要删除
+				return true
+			}
+			delete(*visit, cur)
+		}
+		// 删除
+		*str = (*str)[0 : len(*str)-1]
+	}
+	return false
+}
diff --git a/leetcode/0753.Cracking-the-Safe/753. Cracking the Safe_test.go b/leetcode/0753.Cracking-the-Safe/753. Cracking the Safe_test.go
new file mode 100644
index 000000000..169a85492
--- /dev/null
+++ b/leetcode/0753.Cracking-the-Safe/753. Cracking the Safe_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question753 struct {
+	para753
+	ans753
+}
+
+// para 是参数
+// one 代表第一个参数
+type para753 struct {
+	n int
+	k int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans753 struct {
+	one string
+}
+
+func Test_Problem753(t *testing.T) {
+
+	qs := []question753{
+
+		{
+			para753{1, 2},
+			ans753{"01"},
+		},
+
+		{
+			para753{2, 2},
+			ans753{"00110"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 753------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans753, q.para753
+		fmt.Printf("【input】:%v       【output】:%v\n", p, crackSafe(p.n, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0753.Cracking-the-Safe/README.md b/leetcode/0753.Cracking-the-Safe/README.md
new file mode 100644
index 000000000..39c79a48d
--- /dev/null
+++ b/leetcode/0753.Cracking-the-Safe/README.md
@@ -0,0 +1,92 @@
+# [753. Cracking the Safe](https://leetcode.com/problems/cracking-the-safe/)
+
+
+
+## 题目
+
+There is a box protected by a password. The password is a sequence of `n` digits where each digit can be one of the first `k` digits `0, 1, ..., k-1`.
+
+While entering a password, the last `n` digits entered will automatically be matched against the correct password.
+
+For example, assuming the correct password is `"345"`, if you type `"012345"`, the box will open because the correct password matches the suffix of the entered password.
+
+Return any password of **minimum length** that is guaranteed to open the box at some point of entering it.
+
+**Example 1**:
+
+```
+Input: n = 1, k = 2
+Output: "01"
+Note: "10" will be accepted too.
+```
+
+**Example 2**:
+
+```
+Input: n = 2, k = 2
+Output: "00110"
+Note: "01100", "10011", "11001" will be accepted too.
+```
+
+**Note**:
+
+1. `n` will be in the range `[1, 4]`.
+2. `k` will be in the range `[1, 10]`.
+3. `k^n` will be at most `4096`.
+
+
+## 题目大意
+
+有一个需要密码才能打开的保险箱。密码是 n 位数, 密码的每一位是 k 位序列 0, 1, ..., k-1 中的一个 。你可以随意输入密码，保险箱会自动记住最后 n 位输入，如果匹配，则能够打开保险箱。举个例子，假设密码是 "345"，你可以输入 "012345" 来打开它，只是你输入了 6 个字符.请返回一个能打开保险箱的最短字符串。
+
+提示：
+
+- n 的范围是 [1, 4]。
+- k 的范围是 [1, 10]。
+- k^n 最大可能为 4096。
+
+
+## 解题思路
+
+- 给出 2 个数字 n 和 k，n 代表密码是 n 位数，k 代表密码是 k 位。保险箱会记住最后 n 位输入。返回一个能打开保险箱的最短字符串。
+- 看到题目中的数据范围，数据范围很小，所以可以考虑用 DFS。想解开保险箱，当然是暴力破解，枚举所有可能。题目要求我们输出一个最短的字符串，这里是本题的关键，为何有最短呢？这里有贪心的思想。如果下一次递归可以利用上一次的 n-1 位，那么最终输出的字符串肯定是最短的。(笔者这里就不证明了)，例如，例子 2 中，最短的字符串是 00，01，11，10。每次尝试都利用前一次的 n-1 位。想通了这个问题，利用 DFS 暴力回溯即可。
+
+## 代码
+
+```go
+const number = "0123456789"
+
+func crackSafe(n int, k int) string {
+	if n == 1 {
+		return number[:k]
+	}
+	visit, total := map[string]bool{}, int(math.Pow(float64(k), float64(n)))
+	str := make([]byte, 0, total+n-1)
+	for i := 1; i != n; i++ {
+		str = append(str, '0')
+	}
+	dfsCrackSafe(total, n, k, &str, &visit)
+	return string(str)
+}
+
+func dfsCrackSafe(depth, n, k int, str *[]byte, visit *map[string]bool) bool {
+	if depth == 0 {
+		return true
+	}
+	for i := 0; i != k; i++ {
+		*str = append(*str, byte('0'+i))
+		cur := string((*str)[len(*str)-n:])
+		if _, ok := (*visit)[cur]; ok != true {
+			(*visit)[cur] = true
+			if dfsCrackSafe(depth-1, n, k, str, visit) {
+				// 只有这里不需要删除
+				return true
+			}
+			delete(*visit, cur)
+		}
+		// 删除
+		*str = (*str)[0 : len(*str)-1]
+	}
+	return false
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0756. Pyramid Transition Matrix/756. Pyramid Transition Matrix.go b/leetcode/0756.Pyramid-Transition-Matrix/756. Pyramid Transition Matrix.go
similarity index 100%
rename from Algorithms/0756. Pyramid Transition Matrix/756. Pyramid Transition Matrix.go
rename to leetcode/0756.Pyramid-Transition-Matrix/756. Pyramid Transition Matrix.go
diff --git a/Algorithms/0756. Pyramid Transition Matrix/756. Pyramid Transition Matrix_test.go b/leetcode/0756.Pyramid-Transition-Matrix/756. Pyramid Transition Matrix_test.go
similarity index 96%
rename from Algorithms/0756. Pyramid Transition Matrix/756. Pyramid Transition Matrix_test.go
rename to leetcode/0756.Pyramid-Transition-Matrix/756. Pyramid Transition Matrix_test.go
index 986885cff..a66840388 100644
--- a/Algorithms/0756. Pyramid Transition Matrix/756. Pyramid Transition Matrix_test.go	
+++ b/leetcode/0756.Pyramid-Transition-Matrix/756. Pyramid Transition Matrix_test.go	
@@ -27,12 +27,12 @@ func Test_Problem756(t *testing.T) {
 
 	qs := []question756{
 
-		question756{
+		{
 			para756{"BCD", []string{"BCG", "CDE", "GEA", "FFF"}},
 			ans756{true},
 		},
 
-		question756{
+		{
 			para756{"AABA", []string{"AAA", "AAB", "ABA", "ABB", "BAC"}},
 			ans756{false},
 		},
diff --git a/leetcode/0756.Pyramid-Transition-Matrix/README.md b/leetcode/0756.Pyramid-Transition-Matrix/README.md
new file mode 100755
index 000000000..5b0acbcfe
--- /dev/null
+++ b/leetcode/0756.Pyramid-Transition-Matrix/README.md
@@ -0,0 +1,57 @@
+# [756. Pyramid Transition Matrix](https://leetcode.com/problems/pyramid-transition-matrix/)
+
+
+## 题目
+
+We are stacking blocks to form a pyramid. Each block has a color which is a one letter string.
+
+We are allowed to place any color block `C` on top of two adjacent blocks of colors `A` and `B`, if and only if `ABC` is an allowed triple.
+
+We start with a bottom row of `bottom`, represented as a single string. We also start with a list of allowed triples `allowed`. Each allowed triple is represented as a string of length 3.
+
+Return true if we can build the pyramid all the way to the top, otherwise false.
+
+**Example 1:**
+
+    Input: bottom = "BCD", allowed = ["BCG", "CDE", "GEA", "FFF"]
+    Output: true
+    Explanation:
+    We can stack the pyramid like this:
+        A
+       / \
+      G   E
+     / \ / \
+    B   C   D
+    
+    We are allowed to place G on top of B and C because BCG is an allowed triple.  Similarly, we can place E on top of C and D, then A on top of G and E.
+
+**Example 2:**
+
+    Input: bottom = "AABA", allowed = ["AAA", "AAB", "ABA", "ABB", "BAC"]
+    Output: false
+    Explanation:
+    We can't stack the pyramid to the top.
+    Note that there could be allowed triples (A, B, C) and (A, B, D) with C != D.
+
+**Note:**
+
+1. `bottom` will be a string with length in range `[2, 8]`.
+2. `allowed` will have length in range `[0, 200]`.
+3. Letters in all strings will be chosen from the set `{'A', 'B', 'C', 'D', 'E', 'F', 'G'}`.
+
+
+## 题目大意
+
+现在，我们用一些方块来堆砌一个金字塔。 每个方块用仅包含一个字母的字符串表示，例如 “Z”。使用三元组表示金字塔的堆砌规则如下：
+
+(A, B, C) 表示，“C” 为顶层方块，方块 “A”、“B” 分别作为方块 “C” 下一层的的左、右子块。当且仅当(A, B, C)是被允许的三元组，我们才可以将其堆砌上。
+
+初始时，给定金字塔的基层 bottom，用一个字符串表示。一个允许的三元组列表 allowed，每个三元组用一个长度为 3 的字符串表示。如果可以由基层一直堆到塔尖返回 true，否则返回 false。
+
+
+
+## 解题思路
+
+- 这一题是一道 DFS 的题目。题目给出金字塔的底座字符串。然后还会给一个字符串数组，字符串数组里面代表的字符串的砖块。砖块是 3 个字符串组成的。前两个字符代表的是砖块的底边，后一个字符代表的是砖块的顶部。问给出的字符能拼成一个金字塔么？金字塔的特点是顶端就一个字符。  
+
+- 这一题用 DFS 深搜每个砖块，从底层砖块开始逐渐往上层码。每递归一层，新一层底部的砖块都会变。当递归到了一层底部只有 2 个字符，顶部只有一个字符的时候，就到金字塔顶端了，就算是完成了。这一题为了挑选合适的砖块，需要把每个砖块底部的 2 个字符作为 key 放进 map 中，加速查找。题目中也给出了特殊情况，相同底部可能存在多种砖块，所以一个 key 可能对应多个 value 的情况，即可能存在多个顶部砖块的情况。这种情况在递归遍历中需要考虑。
diff --git a/Algorithms/0762. Prime Number of Set Bits in Binary Representation/762. Prime Number of Set Bits in Binary Representation.go b/leetcode/0762.Prime-Number-of-Set-Bits-in-Binary-Representation/762. Prime Number of Set Bits in Binary Representation.go
similarity index 100%
rename from Algorithms/0762. Prime Number of Set Bits in Binary Representation/762. Prime Number of Set Bits in Binary Representation.go
rename to leetcode/0762.Prime-Number-of-Set-Bits-in-Binary-Representation/762. Prime Number of Set Bits in Binary Representation.go
diff --git a/Algorithms/0762. Prime Number of Set Bits in Binary Representation/762. Prime Number of Set Bits in Binary Representation_test.go b/leetcode/0762.Prime-Number-of-Set-Bits-in-Binary-Representation/762. Prime Number of Set Bits in Binary Representation_test.go
similarity index 95%
rename from Algorithms/0762. Prime Number of Set Bits in Binary Representation/762. Prime Number of Set Bits in Binary Representation_test.go
rename to leetcode/0762.Prime-Number-of-Set-Bits-in-Binary-Representation/762. Prime Number of Set Bits in Binary Representation_test.go
index a57f0c444..1c3c0de28 100644
--- a/Algorithms/0762. Prime Number of Set Bits in Binary Representation/762. Prime Number of Set Bits in Binary Representation_test.go	
+++ b/leetcode/0762.Prime-Number-of-Set-Bits-in-Binary-Representation/762. Prime Number of Set Bits in Binary Representation_test.go	
@@ -27,12 +27,12 @@ func Test_Problem762(t *testing.T) {
 
 	qs := []question762{
 
-		question762{
+		{
 			para762{6, 10},
 			ans762{4},
 		},
 
-		question762{
+		{
 			para762{10, 15},
 			ans762{5},
 		},
diff --git a/leetcode/0762.Prime-Number-of-Set-Bits-in-Binary-Representation/README.md b/leetcode/0762.Prime-Number-of-Set-Bits-in-Binary-Representation/README.md
new file mode 100755
index 000000000..4cc91b230
--- /dev/null
+++ b/leetcode/0762.Prime-Number-of-Set-Bits-in-Binary-Representation/README.md
@@ -0,0 +1,53 @@
+# [762. Prime Number of Set Bits in Binary Representation](https://leetcode.com/problems/prime-number-of-set-bits-in-binary-representation/)
+
+
+## 题目
+
+Given two integers `L` and `R`, find the count of numbers in the range `[L, R]` (inclusive) having a prime number of set bits in their binary representation.
+
+(Recall that the number of set bits an integer has is the number of `1`s present when written in binary. For example, `21` written in binary is `10101` which has 3 set bits. Also, 1 is not a prime.)
+
+**Example 1:**
+
+    Input: L = 6, R = 10
+    Output: 4
+    Explanation:
+    6 -> 110 (2 set bits, 2 is prime)
+    7 -> 111 (3 set bits, 3 is prime)
+    9 -> 1001 (2 set bits , 2 is prime)
+    10->1010 (2 set bits , 2 is prime)
+
+**Example 2:**
+
+    Input: L = 10, R = 15
+    Output: 5
+    Explanation:
+    10 -> 1010 (2 set bits, 2 is prime)
+    11 -> 1011 (3 set bits, 3 is prime)
+    12 -> 1100 (2 set bits, 2 is prime)
+    13 -> 1101 (3 set bits, 3 is prime)
+    14 -> 1110 (3 set bits, 3 is prime)
+    15 -> 1111 (4 set bits, 4 is not prime)
+
+**Note:**
+
+1. `L, R` will be integers `L <= R` in the range `[1, 10^6]`.
+2. `R - L` will be at most 10000.
+
+
+## 题目大意
+
+给定两个整数 L 和 R ，找到闭区间 [L, R] 范围内，计算置位位数为质数的整数个数。（注意，计算置位代表二进制表示中1的个数。例如 21 的二进制表示 10101 有 3 个计算置位。还有，1 不是质数。）
+
+
+注意:
+
+- L, R 是 L <= R 且在 [1, 10^6] 中的整数。
+- R - L 的最大值为 10000。
+
+
+
+## 解题思路
+
+
+- 题目给出 `[L, R]` 区间，在这个区间内的每个整数的二进制表示中 1 的个数如果是素数，那么最终结果就加一，问最终结果是多少？这一题是一个组合题，判断一个数的二进制位有多少位 1，是第 191 题。题目中限定了区间最大不超过 10^6 ，所以 1 的位数最大是 19 位，也就是说素数最大就是 19 。那么素数可以有限枚举出来。最后按照题目的意思累积结果就可以了。
diff --git a/Algorithms/0763. Partition Labels/763. Partition Labels.go b/leetcode/0763.Partition-Labels/763. Partition Labels.go
similarity index 100%
rename from Algorithms/0763. Partition Labels/763. Partition Labels.go
rename to leetcode/0763.Partition-Labels/763. Partition Labels.go
diff --git a/Algorithms/0763. Partition Labels/763. Partition Labels_test.go b/leetcode/0763.Partition-Labels/763. Partition Labels_test.go
similarity index 97%
rename from Algorithms/0763. Partition Labels/763. Partition Labels_test.go
rename to leetcode/0763.Partition-Labels/763. Partition Labels_test.go
index 3e258b908..8ac0c07ea 100644
--- a/Algorithms/0763. Partition Labels/763. Partition Labels_test.go	
+++ b/leetcode/0763.Partition-Labels/763. Partition Labels_test.go	
@@ -26,7 +26,7 @@ func Test_Problem763(t *testing.T) {
 
 	qs := []question763{
 
-		question763{
+		{
 			para763{"ababcbacadefegdehijhklij"},
 			ans763{[]int{9, 7, 8}},
 		},
diff --git a/leetcode/0763.Partition-Labels/README.md b/leetcode/0763.Partition-Labels/README.md
new file mode 100644
index 000000000..68d511b40
--- /dev/null
+++ b/leetcode/0763.Partition-Labels/README.md
@@ -0,0 +1,37 @@
+# [763. Partition Labels](https://leetcode.com/problems/partition-labels/)
+
+## 题目
+
+A string S of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.
+
+
+
+Example 1:
+
+
+```c
+Input: S = "ababcbacadefegdehijhklij"
+Output: [9,7,8]
+Explanation:
+The partition is "ababcbaca", "defegde", "hijhklij".
+This is a partition so that each letter appears in at most one part.
+A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.
+```
+
+Note:
+
+- S will have length in range [1, 500].
+- S will consist of lowercase letters ('a' to 'z') only.
+
+
+## 题目大意
+
+这道题考察的是滑动窗口的问题。
+
+给出一个字符串，要求输出满足条件窗口的长度，条件是在这个窗口内，字母中出现在这一个窗口内，不出现在其他窗口内。
+
+## 解题思路
+
+这一题有 2 种思路，第一种思路是先记录下每个字母的出现次数，然后对滑动窗口中的每个字母判断次数是否用尽为 0，如果这个窗口内的所有字母次数都为 0，这个窗口就是符合条件的窗口。时间复杂度为 O(n)
+
+另外一种思路是记录下每个字符最后一次出现的下标，这样就不用记录次数。在每个滑动窗口中，依次判断每个字母最后一次出现的位置，如果在一个下标内，所有字母的最后一次出现的位置都包含进来了，那么这个下标就是这个满足条件的窗口大小。时间复杂度为 O(n^2)
diff --git a/Algorithms/0765. Couples Holding Hands/765. Couples Holding Hands.go b/leetcode/0765.Couples-Holding-Hands/765. Couples Holding Hands.go
similarity index 100%
rename from Algorithms/0765. Couples Holding Hands/765. Couples Holding Hands.go
rename to leetcode/0765.Couples-Holding-Hands/765. Couples Holding Hands.go
diff --git a/Algorithms/0765. Couples Holding Hands/765. Couples Holding Hands_test.go b/leetcode/0765.Couples-Holding-Hands/765. Couples Holding Hands_test.go
similarity index 93%
rename from Algorithms/0765. Couples Holding Hands/765. Couples Holding Hands_test.go
rename to leetcode/0765.Couples-Holding-Hands/765. Couples Holding Hands_test.go
index 46cd1befd..fb0c70187 100644
--- a/Algorithms/0765. Couples Holding Hands/765. Couples Holding Hands_test.go	
+++ b/leetcode/0765.Couples-Holding-Hands/765. Couples Holding Hands_test.go	
@@ -26,22 +26,22 @@ func Test_Problem765(t *testing.T) {
 
 	qs := []question765{
 
-		question765{
+		{
 			para765{[]int{0, 2, 1, 3}},
 			ans765{1},
 		},
 
-		question765{
+		{
 			para765{[]int{3, 2, 0, 1}},
 			ans765{0},
 		},
 
-		question765{
+		{
 			para765{[]int{3, 1, 4, 0, 2, 5}},
 			ans765{2},
 		},
 
-		question765{
+		{
 			para765{[]int{10, 7, 4, 2, 3, 0, 9, 11, 1, 5, 6, 8}},
 			ans765{4},
 		},
diff --git a/leetcode/0765.Couples-Holding-Hands/README.md b/leetcode/0765.Couples-Holding-Hands/README.md
new file mode 100755
index 000000000..631557d91
--- /dev/null
+++ b/leetcode/0765.Couples-Holding-Hands/README.md
@@ -0,0 +1,63 @@
+# [765. Couples Holding Hands](https://leetcode.com/problems/couples-holding-hands/)
+
+
+## 题目
+
+N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum number of swaps so that every couple is sitting side by side. A swap consists of choosing **any** two people, then they stand up and switch seats.
+
+The people and seats are represented by an integer from `0` to `2N-1`, the couples are numbered in order, the first couple being `(0, 1)`, the second couple being `(2, 3)`, and so on with the last couple being `(2N-2, 2N-1)`.
+
+The couples' initial seating is given by `row[i]` being the value of the person who is initially sitting in the i-th seat.
+
+**Example 1:**
+
+    Input: row = [0, 2, 1, 3]
+    Output: 1
+    Explanation: We only need to swap the second (row[1]) and third (row[2]) person.
+
+**Example 2:**
+
+    Input: row = [3, 2, 0, 1]
+    Output: 0
+    Explanation: All couples are already seated side by side.
+
+**Note:**
+
+1. `len(row)` is even and in the range of `[4, 60]`.
+2. `row` is guaranteed to be a permutation of `0...len(row)-1`.
+
+
+## 题目大意
+
+N 对情侣坐在连续排列的 2N 个座位上，想要牵到对方的手。 计算最少交换座位的次数，以便每对情侣可以并肩坐在一起。 一次交换可选择任意两人，让他们站起来交换座位。人和座位用 0 到 2N-1 的整数表示，情侣们按顺序编号，第一对是 (0, 1)，第二对是 (2, 3)，以此类推，最后一对是 (2N-2, 2N-1)。这些情侣的初始座位 row[i] 是由最初始坐在第 i 个座位上的人决定的。
+
+说明:
+
+1. len(row) 是偶数且数值在 [4, 60]范围内。  
+2. 可以保证 row 是序列 0...len(row)-1 的一个全排列。  
+
+
+## 解题思路
+
+- 给出一个数组，数组里面两两相邻的元素代表一对情侣。情侣编号是从 0 开始的：0 和 1 是情侣，2 和 3 是情侣……这些情侣坐在一排，但是并非成对坐着一起的，问如何用最小的次数交换座位以后，情侣能两两坐在一起。
+- 这道题的突破口是如何找到最小的交换次数。乍一想可能没有思路。直觉告诉我们，这种难题，很可能最后推出来的结论，或者公式是一个很简单的式子。(事实此题确实是这种情况)先不考虑最小交换次数，用正常的方法来处理这道题。举个例子：【3 1 4 0 2 5】，从数组 0 下标开始往后扫。
+
+        初始状态
+        
+        集合 0：0，1
+        集合 1：2，3
+        集合 2：4，5
+
+    3 和 1 不是情侣，将 3 和 1 所在集合 `union()` 起来。3 所在集合是 1 ，1 所在集合是 0，将 0 和 1 号集合 `union()` 起来。因为情侣 0 和情侣 1 是集合 0 ，情侣 2 和情侣 3 是集合 1，以此类推。
+
+        集合 0 和 1：0，1，2，3
+        集合 2：4，5
+
+- 继续往后扫，4 和 0 不在同一个集合，4 在集合 3，0 在集合 0，那么把它们 `union()` 起来。
+
+        集合 0 和 1 和 2：0，1，2，3，4，5
+
+    在上面集合合并的过程中，合并了 2 次。那么就代表最少需要交换 2 次。也可以通过 `len(row)/2 - uf.count` 来计算。`len(row)/2` 是初始集合总数，`uf.count` 是最后剩下的集合数，两者相减就是中间交换的次数。
+
+- 最后实现的代码非常简单。并查集先相邻的两两元素 `union()` 在一起。然后扫原数组，每次扫相邻的两个，通过这两个元素值所在集合，进行 `union()`。扫完以后就可以得到最后的答案。
+- 回过头来看这道题，为什么我们从数组开头往后依次调整每一对情侣，这样交换的次数是最少的呢？其实这个方法的思想是贪心思想。从头开始往后一对一对的调整，就是可以最终做到次数最少。(具体证明笔者不会)交换到最后，最后一对情侣一定是正确的，无须交换。(因为前面每一对都调整完了，最后一对一定是正确的)
diff --git a/Algorithms/0766. Toeplitz Matrix/766. Toeplitz Matrix.go b/leetcode/0766.Toeplitz-Matrix/766. Toeplitz Matrix.go
similarity index 100%
rename from Algorithms/0766. Toeplitz Matrix/766. Toeplitz Matrix.go
rename to leetcode/0766.Toeplitz-Matrix/766. Toeplitz Matrix.go
diff --git a/Algorithms/0766. Toeplitz Matrix/766. Toeplitz Matrix_test.go b/leetcode/0766.Toeplitz-Matrix/766. Toeplitz Matrix_test.go
similarity index 79%
rename from Algorithms/0766. Toeplitz Matrix/766. Toeplitz Matrix_test.go
rename to leetcode/0766.Toeplitz-Matrix/766. Toeplitz Matrix_test.go
index f32a915f8..fa18cb754 100644
--- a/Algorithms/0766. Toeplitz Matrix/766. Toeplitz Matrix_test.go	
+++ b/leetcode/0766.Toeplitz-Matrix/766. Toeplitz Matrix_test.go	
@@ -26,13 +26,13 @@ func Test_Problem766(t *testing.T) {
 
 	qs := []question766{
 
-		question766{
-			para766{[][]int{[]int{1, 2, 3, 4}, []int{5, 1, 2, 3}, []int{9, 5, 1, 2}}},
+		{
+			para766{[][]int{{1, 2, 3, 4}, {5, 1, 2, 3}, {9, 5, 1, 2}}},
 			ans766{true},
 		},
 
-		question766{
-			para766{[][]int{[]int{1, 2}, []int{2, 2}}},
+		{
+			para766{[][]int{{1, 2}, {2, 2}}},
 			ans766{false},
 		},
 	}
diff --git a/Algorithms/0766. Toeplitz Matrix/README.md b/leetcode/0766.Toeplitz-Matrix/README.md
similarity index 100%
rename from Algorithms/0766. Toeplitz Matrix/README.md
rename to leetcode/0766.Toeplitz-Matrix/README.md
diff --git a/Algorithms/0767. Reorganize String/767. Reorganize String.go b/leetcode/0767.Reorganize-String/767. Reorganize String.go
similarity index 100%
rename from Algorithms/0767. Reorganize String/767. Reorganize String.go
rename to leetcode/0767.Reorganize-String/767. Reorganize String.go
diff --git a/Algorithms/0767. Reorganize String/767. Reorganize String_test.go b/leetcode/0767.Reorganize-String/767. Reorganize String_test.go
similarity index 95%
rename from Algorithms/0767. Reorganize String/767. Reorganize String_test.go
rename to leetcode/0767.Reorganize-String/767. Reorganize String_test.go
index 7c67ee5f1..ab8f52ec9 100644
--- a/Algorithms/0767. Reorganize String/767. Reorganize String_test.go	
+++ b/leetcode/0767.Reorganize-String/767. Reorganize String_test.go	
@@ -25,12 +25,12 @@ type ans767 struct {
 func Test_Problem767(t *testing.T) {
 
 	qs := []question767{
-		question767{
+		{
 			para767{"aab"},
 			ans767{"aba"},
 		},
 
-		question767{
+		{
 			para767{"aaab"},
 			ans767{""},
 		},
diff --git a/Algorithms/0767. Reorganize String/README.md b/leetcode/0767.Reorganize-String/README.md
similarity index 100%
rename from Algorithms/0767. Reorganize String/README.md
rename to leetcode/0767.Reorganize-String/README.md
diff --git a/Algorithms/0771. Jewels and Stones/771. Jewels and Stones.go b/leetcode/0771.Jewels-and-Stones/771. Jewels and Stones.go
similarity index 100%
rename from Algorithms/0771. Jewels and Stones/771. Jewels and Stones.go
rename to leetcode/0771.Jewels-and-Stones/771. Jewels and Stones.go
diff --git a/Algorithms/0771. Jewels and Stones/771. Jewels and Stones_test.go b/leetcode/0771.Jewels-and-Stones/771. Jewels and Stones_test.go
similarity index 95%
rename from Algorithms/0771. Jewels and Stones/771. Jewels and Stones_test.go
rename to leetcode/0771.Jewels-and-Stones/771. Jewels and Stones_test.go
index b9dae4cb8..8cdc308e9 100644
--- a/Algorithms/0771. Jewels and Stones/771. Jewels and Stones_test.go	
+++ b/leetcode/0771.Jewels-and-Stones/771. Jewels and Stones_test.go	
@@ -26,12 +26,12 @@ type ans771 struct {
 func Test_Problem771(t *testing.T) {
 
 	qs := []question771{
-		question771{
+		{
 			para771{"aA", "aAAbbbb"},
 			ans771{3},
 		},
 
-		question771{
+		{
 			para771{"z", "ZZ"},
 			ans771{0},
 		},
diff --git a/leetcode/0771.Jewels-and-Stones/README.md b/leetcode/0771.Jewels-and-Stones/README.md
new file mode 100755
index 000000000..9dd0e34c6
--- /dev/null
+++ b/leetcode/0771.Jewels-and-Stones/README.md
@@ -0,0 +1,38 @@
+# [771. Jewels and Stones](https://leetcode.com/problems/jewels-and-stones/)
+
+
+
+## 题目
+
+You're given strings `J` representing the types of stones that are jewels, and `S` representing the stones you have. Each character in `S` is a type of stone you have. You want to know how many of the stones you have are also jewels.
+
+The letters in `J` are guaranteed distinct, and all characters in `J` and `S` are letters. Letters are case sensitive, so `"a"` is considered a different type of stone from `"A"`.
+
+**Example 1:**
+
+    Input: J = "aA", S = "aAAbbbb"
+    Output: 3
+
+**Example 2:**
+
+    Input: J = "z", S = "ZZ"
+    Output: 0
+
+**Note:**
+
+- `S` and `J` will consist of letters and have length at most 50.
+- The characters in `J` are distinct.
+
+
+## 题目大意
+
+给定字符串 J 代表石头中宝石的类型，和字符串 S 代表你拥有的石头。S 中每个字符代表了一种你拥有的石头的类型，你想知道你拥有的石头中有多少是宝石。
+
+J 中的字母不重复，J 和 S 中的所有字符都是字母。字母区分大小写，因此 "a" 和 "A" 是不同类型的石头。
+
+
+
+## 解题思路
+
+
+- 给出 2 个字符串，要求在 S 字符串中找出在 J 字符串里面出现的字符个数。这是一道简单题。
diff --git a/leetcode/0775.Global-and-Local-Inversions/775. Global and Local Inversions.go b/leetcode/0775.Global-and-Local-Inversions/775. Global and Local Inversions.go
new file mode 100644
index 000000000..cc8fbcdf1
--- /dev/null
+++ b/leetcode/0775.Global-and-Local-Inversions/775. Global and Local Inversions.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+func isIdealPermutation(A []int) bool {
+	for i := range A {
+		if abs(A[i]-i) > 1 {
+			return false
+		}
+	}
+	return true
+}
+
+func abs(a int) int {
+	if a < 0 {
+		return -a
+	}
+	return a
+}
diff --git a/leetcode/0775.Global-and-Local-Inversions/775. Global and Local Inversions_test.go b/leetcode/0775.Global-and-Local-Inversions/775. Global and Local Inversions_test.go
new file mode 100644
index 000000000..7d570802f
--- /dev/null
+++ b/leetcode/0775.Global-and-Local-Inversions/775. Global and Local Inversions_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question775 struct {
+	para775
+	ans775
+}
+
+// para 是参数
+// one 代表第一个参数
+type para775 struct {
+	A []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans775 struct {
+	one bool
+}
+
+func Test_Problem775(t *testing.T) {
+
+	qs := []question775{
+		{
+			para775{[]int{1, 0, 2}},
+			ans775{true},
+		},
+
+		{
+			para775{[]int{1, 2, 0}},
+			ans775{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 775------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans775, q.para775
+		fmt.Printf("【input】:%v       【output】:%v\n", p, isIdealPermutation(p.A))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0775.Global-and-Local-Inversions/README.md b/leetcode/0775.Global-and-Local-Inversions/README.md
new file mode 100644
index 000000000..c20184271
--- /dev/null
+++ b/leetcode/0775.Global-and-Local-Inversions/README.md
@@ -0,0 +1,64 @@
+# [775. Global and Local Inversions](https://leetcode.com/problems/global-and-local-inversions/)
+
+
+## 题目
+
+We have some permutation `A` of `[0, 1, ..., N - 1]`, where `N` is the length of `A`.
+
+The number of (global) inversions is the number of `i < j` with `0 <= i < j < N` and `A[i] > A[j]`.
+
+The number of local inversions is the number of `i` with `0 <= i < N` and `A[i] > A[i+1]`.
+
+Return `true` if and only if the number of global inversions is equal to the number of local inversions.
+
+**Example 1:**
+
+```
+Input: A = [1,0,2]
+Output: true
+Explanation: There is 1 global inversion, and 1 local inversion.
+```
+
+**Example 2:**
+
+```
+Input: A = [1,2,0]
+Output: false
+Explanation: There are 2 global inversions, and 1 local inversion.
+```
+
+**Note:**
+
+- `A` will be a permutation of `[0, 1, ..., A.length - 1]`.
+- `A` will have length in range `[1, 5000]`.
+- The time limit for this problem has been reduced.
+
+## 题目大意
+
+数组 A 是 [0, 1, ..., N - 1] 的一种排列，N 是数组 A 的长度。全局倒置指的是 i,j 满足 0 <= i < j < N 并且 A[i] > A[j] ，局部倒置指的是 i 满足 0 <= i < N 并且 A[i] > A[i+1] 。当数组 A 中全局倒置的数量等于局部倒置的数量时，返回 true 。
+
+## 解题思路
+
+- 本题代码非常简单，重在思考的过程。`[0, 1, ..., N - 1]` 不出现全局倒置的理想情况应该是 `i` 排列在 `A[i-1]`，`A[i]`，`A[i+1]` 的位置上。例如 `1` 如果排列在 `A[3]` 的位置上，那么比 `1` 小的只有 `0` 一个元素，`A[0]`，`A[1]`，`A[2]` 中必定有 2 个元素比 `1` 大，那必须会出现全局倒置的情况。`[0, 1, ..., N - 1]` 这是最理想的情况，每个元素都在自己的位置上。每个元素如果往左右相互偏移 1 个元素，那么也能保证只存在局部倒置，如果左右偏移 2 个元素，那必定会出现全局倒置。所以结论是：**不出现全局倒置的理想情况应该是 `i` 排列在 `A[i-1]`，`A[i]`，`A[i+1]` 的位置上**。判断这个结论的代码很简单，只需要判断 `A[i] - i` 的取值是否是 -1，0，1，也即 `abs(A[i] - i ) ≤ 1`。
+
+## 代码
+
+```go
+package leetcode
+
+func isIdealPermutation(A []int) bool {
+	for i := range A {
+		if abs(A[i]-i) > 1 {
+			return false
+		}
+	}
+	return true
+}
+
+func abs(a int) int {
+	if a < 0 {
+		return -a
+	}
+	return a
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0778. Swim in Rising Water/778. Swim in Rising Water.go b/leetcode/0778.Swim-in-Rising-Water/778. Swim in Rising Water.go
similarity index 100%
rename from Algorithms/0778. Swim in Rising Water/778. Swim in Rising Water.go
rename to leetcode/0778.Swim-in-Rising-Water/778. Swim in Rising Water.go
diff --git a/leetcode/0778.Swim-in-Rising-Water/778. Swim in Rising Water_test.go b/leetcode/0778.Swim-in-Rising-Water/778. Swim in Rising Water_test.go
new file mode 100644
index 000000000..6523d0039
--- /dev/null
+++ b/leetcode/0778.Swim-in-Rising-Water/778. Swim in Rising Water_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question778 struct {
+	para778
+	ans778
+}
+
+// para 是参数
+// one 代表第一个参数
+type para778 struct {
+	grid [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans778 struct {
+	one int
+}
+
+func Test_Problem778(t *testing.T) {
+
+	qs := []question778{
+		{
+			para778{[][]int{{0, 2}, {1, 3}}},
+			ans778{3},
+		},
+
+		{
+			para778{[][]int{{0, 1, 2, 3, 4}, {24, 23, 22, 21, 5}, {12, 13, 14, 15, 16}, {11, 17, 18, 19, 20}, {10, 9, 8, 7, 6}}},
+			ans778{16},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 778------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans778, q.para778
+		fmt.Printf("【input】:%v       【output】:%v\n", p, swimInWater(p.grid))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0778.Swim-in-Rising-Water/README.md b/leetcode/0778.Swim-in-Rising-Water/README.md
new file mode 100755
index 000000000..2e5489320
--- /dev/null
+++ b/leetcode/0778.Swim-in-Rising-Water/README.md
@@ -0,0 +1,59 @@
+# [778. Swim in Rising Water](https://leetcode.com/problems/swim-in-rising-water/)
+
+
+## 题目
+
+On an N x N `grid`, each square `grid[i][j]` represents the elevation at that point `(i,j)`.
+
+Now rain starts to fall. At time `t`, the depth of the water everywhere is `t`. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most `t`. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.
+
+You start at the top left square `(0, 0)`. What is the least time until you can reach the bottom right square `(N-1, N-1)`?
+
+**Example 1:**
+
+    Input: [[0,2],[1,3]]
+    Output: 3
+    Explanation:
+    At time 0, you are in grid location (0, 0).
+    You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
+    
+    You cannot reach point (1, 1) until time 3.
+    When the depth of water is 3, we can swim anywhere inside the grid.
+
+**Example 2:**
+
+    Input: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
+    Output: 16
+    Explanation:
+     0  1  2  3  4
+    24 23 22 21  5
+    12 13 14 15 16
+    11 17 18 19 20
+    10  9  8  7  6
+    
+    The final route is marked in bold.
+    We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
+
+**Note:**
+
+1. `2 <= N <= 50`.
+2. grid[i][j] is a permutation of [0, ..., N*N - 1].
+
+## 题目大意
+
+
+在一个 N x N 的坐标方格 grid 中，每一个方格的值 grid[i][j] 表示在位置 (i,j) 的平台高度。现在开始下雨了。当时间为 t 时，此时雨水导致水池中任意位置的水位为 t 。你可以从一个平台游向四周相邻的任意一个平台，但是前提是此时水位必须同时淹没这两个平台。假定你可以瞬间移动无限距离，也就是默认在方格内部游动是不耗时的。当然，在你游泳的时候你必须待在坐标方格里面。
+
+你从坐标方格的左上平台 (0，0) 出发。最少耗时多久你才能到达坐标方格的右下平台 (N-1, N-1)？
+
+提示:
+
+- 2 <= N <= 50.
+- grid[i][j] 位于区间 [0, ..., N*N - 1] 内。
+
+
+## 解题思路
+
+- 给出一个 grid[i][j] 方格，每个格子里面表示游泳池里面平台的高度。t 时刻，游泳池中的水的高度是 t。只有水的高度到达了平台的高度以后才能游过去。问从 (0,0) 开始，最短多长时间能到达 (N-1, N-1) 。
+- 这一题有多种解法。第一种解题思路是利用 DFS + 二分。DFS 是用来遍历是否可达。利用时间(即当前水淹过的高度)来判断是否能到达终点 (N-1, N-1) 点。二分用来搜索最终结果的时间。为什么会考虑用二分加速呢？原因是：时间从 0 - max 依次递增。max 是游泳池最高的平台高度。当时间从 0 增加到 max 以后，肯定能到达终点 (N-1, N-1) 点，因为水比所有平台都要高了。想快速找到一个时间 t 能使得 (0,0) 点和 (N-1, N-1) 点之间连通，那么就想到用二分加速了。判断是否取中值的条件是 (0,0) 点和 (N-1, N-1) 点之间是否连通。
+- 第二种解题思路是并查集。只要是 (0,0) 点和 (N-1, N-1) 点没有连通，即不能游到终点，那么就开始 `union()` 操作，由于起点是 (0,0)，所以向右边 `i + 1` 和向下边 `j + 1` 开始尝试。每尝试完一轮，时间会加 1 秒，即高度会加一。直到 (0,0) 点和 (N-1, N-1) 点刚好连通，那么这个时间点就是最终要求的。
\ No newline at end of file
diff --git a/Algorithms/0781. Rabbits in Forest/781. Rabbits in Forest.go b/leetcode/0781.Rabbits-in-Forest/781. Rabbits in Forest.go
similarity index 100%
rename from Algorithms/0781. Rabbits in Forest/781. Rabbits in Forest.go
rename to leetcode/0781.Rabbits-in-Forest/781. Rabbits in Forest.go
diff --git a/Algorithms/0781. Rabbits in Forest/781. Rabbits in Forest_test.go b/leetcode/0781.Rabbits-in-Forest/781. Rabbits in Forest_test.go
similarity index 94%
rename from Algorithms/0781. Rabbits in Forest/781. Rabbits in Forest_test.go
rename to leetcode/0781.Rabbits-in-Forest/781. Rabbits in Forest_test.go
index f8c8bed11..093dd2871 100644
--- a/Algorithms/0781. Rabbits in Forest/781. Rabbits in Forest_test.go	
+++ b/leetcode/0781.Rabbits-in-Forest/781. Rabbits in Forest_test.go	
@@ -25,17 +25,17 @@ type ans781 struct {
 func Test_Problem781(t *testing.T) {
 
 	qs := []question781{
-		question781{
+		{
 			para781{[]int{1, 1, 2}},
 			ans781{5},
 		},
 
-		question781{
+		{
 			para781{[]int{10, 10, 10}},
 			ans781{11},
 		},
 
-		question781{
+		{
 			para781{[]int{}},
 			ans781{0},
 		},
diff --git a/leetcode/0781.Rabbits-in-Forest/README.md b/leetcode/0781.Rabbits-in-Forest/README.md
new file mode 100755
index 000000000..2684d6566
--- /dev/null
+++ b/leetcode/0781.Rabbits-in-Forest/README.md
@@ -0,0 +1,46 @@
+# [781. Rabbits in Forest](https://leetcode.com/problems/rabbits-in-forest/)
+
+
+## 题目
+
+In a forest, each rabbit has some color. Some subset of rabbits (possibly all of them) tell you how many other rabbits have the same color as them. Those `answers` are placed in an array.
+
+Return the minimum number of rabbits that could be in the forest.
+
+    Examples:
+    Input: answers = [1, 1, 2]
+    Output: 5
+    Explanation:
+    The two rabbits that answered "1" could both be the same color, say red.
+    The rabbit than answered "2" can't be red or the answers would be inconsistent.
+    Say the rabbit that answered "2" was blue.
+    Then there should be 2 other blue rabbits in the forest that didn't answer into the array.
+    The smallest possible number of rabbits in the forest is therefore 5: 3 that answered plus 2 that didn't.
+    
+    Input: answers = [10, 10, 10]
+    Output: 11
+    
+    Input: answers = []
+    Output: 0
+
+**Note:**
+
+1. `answers` will have length at most `1000`.
+2. Each `answers[i]` will be an integer in the range `[0, 999]`.
+
+
+## 题目大意
+
+森林中，每个兔子都有颜色。其中一些兔子（可能是全部）告诉你还有多少其他的兔子和自己有相同的颜色。我们将这些回答放在 answers 数组里。返回森林中兔子的最少数量。
+
+说明:
+
+- answers 的长度最大为1000。
+- answers[i] 是在 [0, 999] 范围内的整数。
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面代表的是每个兔子说自己同类还有多少个。要求输出总共有多少只兔子。数字中可能兔子汇报的人数小于总兔子数。
+- 这一题关键在于如何划分不同种类的兔子，有可能相同种类的兔子的个数是一样的，比如 `[2,2,2,2,2,2]`，这其实是 3 个种类，总共 6 只兔子。用 map 去重相同种类的兔子，不断的减少，当有种类的兔子为 0 以后，还有该种类的兔子报数，需要当做另外一个种类的兔子来看待。
diff --git a/leetcode/0783.Minimum-Distance-Between-BST-Nodes/783. Minimum Distance Between BST Nodes.go b/leetcode/0783.Minimum-Distance-Between-BST-Nodes/783. Minimum Distance Between BST Nodes.go
new file mode 100644
index 000000000..dfb529093
--- /dev/null
+++ b/leetcode/0783.Minimum-Distance-Between-BST-Nodes/783. Minimum Distance Between BST Nodes.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"math"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func minDiffInBST(root *TreeNode) int {
+	res, nodes := math.MaxInt16, -1
+	dfsBST(root, &res, &nodes)
+	return res
+}
+
+func dfsBST(root *TreeNode, res, pre *int) {
+	if root == nil {
+		return
+	}
+	dfsBST(root.Left, res, pre)
+	if *pre != -1 {
+		*res = min(*res, abs(root.Val-*pre))
+	}
+	*pre = root.Val
+	dfsBST(root.Right, res, pre)
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
diff --git a/leetcode/0783.Minimum-Distance-Between-BST-Nodes/783. Minimum Distance Between BST Nodes_test.go b/leetcode/0783.Minimum-Distance-Between-BST-Nodes/783. Minimum Distance Between BST Nodes_test.go
new file mode 100644
index 000000000..37d6e7383
--- /dev/null
+++ b/leetcode/0783.Minimum-Distance-Between-BST-Nodes/783. Minimum Distance Between BST Nodes_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question783 struct {
+	para783
+	ans783
+}
+
+// para 是参数
+// one 代表第一个参数
+type para783 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans783 struct {
+	one int
+}
+
+func Test_Problem783(t *testing.T) {
+
+	qs := []question783{
+
+		{
+			para783{[]int{4, 2, 6, 1, 3}},
+			ans783{1},
+		},
+
+		{
+			para783{[]int{1, 0, 48, structures.NULL, structures.NULL, 12, 49}},
+			ans783{1},
+		},
+
+		{
+			para783{[]int{90, 69, structures.NULL, 49, 89, structures.NULL, 52}},
+			ans783{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 783------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans783, q.para783
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", minDiffInBST(rootOne))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0783.Minimum-Distance-Between-BST-Nodes/README.md b/leetcode/0783.Minimum-Distance-Between-BST-Nodes/README.md
new file mode 100644
index 000000000..78bf2a213
--- /dev/null
+++ b/leetcode/0783.Minimum-Distance-Between-BST-Nodes/README.md
@@ -0,0 +1,95 @@
+# [783. Minimum Distance Between BST Nodes](https://leetcode.com/problems/minimum-distance-between-bst-nodes/)
+
+
+## 题目
+
+Given the `root` of a Binary Search Tree (BST), return *the minimum difference between the values of any two different nodes in the tree*.
+
+**Note:** This question is the same as 530: [https://leetcode.com/problems/minimum-absolute-difference-in-bst/](https://leetcode.com/problems/minimum-absolute-difference-in-bst/)
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/02/05/bst1.jpg](https://assets.leetcode.com/uploads/2021/02/05/bst1.jpg)
+
+```
+Input: root = [4,2,6,1,3]
+Output: 1
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2021/02/05/bst2.jpg](https://assets.leetcode.com/uploads/2021/02/05/bst2.jpg)
+
+```
+Input: root = [1,0,48,null,null,12,49]
+Output: 1
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[2, 100]`.
+- `0 <= Node.val <= 10^5`
+
+## 题目大意
+
+给你一个二叉搜索树的根节点 root ，返回 树中任意两不同节点值之间的最小差值 。
+
+## 解题思路
+
+- 本题和第 530 题完全相同。解题思路见第 530 题。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func minDiffInBST(root *TreeNode) int {
+	res, nodes := math.MaxInt16, -1
+	dfsBST(root, &res, &nodes)
+	return res
+}
+
+func dfsBST(root *TreeNode, res, pre *int) {
+	if root == nil {
+		return
+	}
+	dfsBST(root.Left, res, pre)
+	if *pre != -1 {
+		*res = min(*res, abs(root.Val-*pre))
+	}
+	*pre = root.Val
+	dfsBST(root.Right, res, pre)
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0784. Letter Case Permutation/784. Letter Case Permutation.go b/leetcode/0784.Letter-Case-Permutation/784. Letter Case Permutation.go
similarity index 94%
rename from Algorithms/0784. Letter Case Permutation/784. Letter Case Permutation.go
rename to leetcode/0784.Letter-Case-Permutation/784. Letter Case Permutation.go
index 5708cbb9d..7f2b13ca1 100644
--- a/Algorithms/0784. Letter Case Permutation/784. Letter Case Permutation.go	
+++ b/leetcode/0784.Letter-Case-Permutation/784. Letter Case Permutation.go	
@@ -22,6 +22,13 @@ func letterCasePermutation(S string) []string {
 	return res
 }
 
+func isLowerLetter(v byte) bool {
+	if v >= 'a' && v <= 'z' {
+		return true
+	}
+	return false
+}
+
 func findLetterCasePermutation(s string, pos []int, target, index int, c []int, res *[]string) {
 	if len(c) == target {
 		b := []byte(s)
diff --git a/Algorithms/0784. Letter Case Permutation/784. Letter Case Permutation_test.go b/leetcode/0784.Letter-Case-Permutation/784. Letter Case Permutation_test.go
similarity index 92%
rename from Algorithms/0784. Letter Case Permutation/784. Letter Case Permutation_test.go
rename to leetcode/0784.Letter-Case-Permutation/784. Letter Case Permutation_test.go
index 5eb38516d..9891c4e65 100644
--- a/Algorithms/0784. Letter Case Permutation/784. Letter Case Permutation_test.go	
+++ b/leetcode/0784.Letter-Case-Permutation/784. Letter Case Permutation_test.go	
@@ -26,27 +26,27 @@ func Test_Problem784(t *testing.T) {
 
 	qs := []question784{
 
-		question784{
+		{
 			para784{"mQe"},
 			ans784{[]string{"mqe", "mqE", "mQe", "mQE", "Mqe", "MqE", "MQe", "MQE"}},
 		},
 
-		question784{
+		{
 			para784{"C"},
 			ans784{[]string{"c", "C"}},
 		},
 
-		question784{
+		{
 			para784{"a1b2"},
 			ans784{[]string{"a1b2", "a1B2", "A1b2", "A1B2"}},
 		},
 
-		question784{
+		{
 			para784{"3z4"},
 			ans784{[]string{"3z4", "3Z4"}},
 		},
 
-		question784{
+		{
 			para784{"12345"},
 			ans784{[]string{"12345"}},
 		},
diff --git a/Algorithms/0784. Letter Case Permutation/README.md b/leetcode/0784.Letter-Case-Permutation/README.md
similarity index 100%
rename from Algorithms/0784. Letter Case Permutation/README.md
rename to leetcode/0784.Letter-Case-Permutation/README.md
diff --git a/leetcode/0785.Is-Graph-Bipartite/785. Is Graph Bipartite.go b/leetcode/0785.Is-Graph-Bipartite/785. Is Graph Bipartite.go
new file mode 100644
index 000000000..e0b6089ae
--- /dev/null
+++ b/leetcode/0785.Is-Graph-Bipartite/785. Is Graph Bipartite.go	
@@ -0,0 +1,35 @@
+package leetcode
+
+// DFS 染色，1 是红色，0 是绿色，-1 是未染色
+func isBipartite(graph [][]int) bool {
+	colors := make([]int, len(graph))
+	for i := range colors {
+		colors[i] = -1
+	}
+	for i := range graph {
+		if !dfs(i, graph, colors, -1) {
+			return false
+		}
+	}
+	return true
+}
+
+func dfs(n int, graph [][]int, colors []int, parentCol int) bool {
+	if colors[n] == -1 {
+		if parentCol == 1 {
+			colors[n] = 0
+		} else {
+			colors[n] = 1
+		}
+	} else if colors[n] == parentCol {
+		return false
+	} else if colors[n] != parentCol {
+		return true
+	}
+	for _, c := range graph[n] {
+		if !dfs(c, graph, colors, colors[n]) {
+			return false
+		}
+	}
+	return true
+}
diff --git a/leetcode/0785.Is-Graph-Bipartite/785. Is Graph Bipartite_test.go b/leetcode/0785.Is-Graph-Bipartite/785. Is Graph Bipartite_test.go
new file mode 100644
index 000000000..862eaf590
--- /dev/null
+++ b/leetcode/0785.Is-Graph-Bipartite/785. Is Graph Bipartite_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question785 struct {
+	para785
+	ans785
+}
+
+// para 是参数
+// one 代表第一个参数
+type para785 struct {
+	graph [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans785 struct {
+	one bool
+}
+
+func Test_Problem785(t *testing.T) {
+
+	qs := []question785{
+
+		{
+			para785{[][]int{{1, 3}, {0, 2}, {1, 3}, {0, 2}}},
+			ans785{true},
+		},
+
+		{
+			para785{[][]int{{1, 2, 3}, {0, 2}, {0, 1, 3}, {0, 2}}},
+			ans785{false},
+		},
+
+		{
+			para785{[][]int{{1, 2, 3}, {0, 2, 3}, {0, 1, 3}, {0, 1, 2}}},
+			ans785{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 785------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans785, q.para785
+		fmt.Printf("【input】:%v       【output】:%v\n", p, isBipartite(p.graph))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0785.Is-Graph-Bipartite/README.md b/leetcode/0785.Is-Graph-Bipartite/README.md
new file mode 100644
index 000000000..a5eb6679d
--- /dev/null
+++ b/leetcode/0785.Is-Graph-Bipartite/README.md
@@ -0,0 +1,98 @@
+# [785. Is Graph Bipartite?](https://leetcode.com/problems/is-graph-bipartite/)
+
+
+## 题目
+
+Given an undirected `graph`, return `true` if and only if it is bipartite.
+
+Recall that a graph is *bipartite* if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.
+
+The graph is given in the following form: `graph[i]` is a list of indexes `j` for which the edge between nodes `i` and `j` exists. Each node is an integer between `0` and `graph.length - 1`. There are no self edges or parallel edges: `graph[i]` does not contain `i`, and it doesn't contain any element twice.
+
+
+	Example 1:Input: [[1,3], [0,2], [1,3], [0,2]]
+	Output: true
+	Explanation: 
+	The graph looks like this:
+	0----1
+	|    |
+	|    |
+	3----2
+	We can divide the vertices into two groups: {0, 2} and {1, 3}.
+
+
+	Example 2:Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
+	Output: false
+	Explanation: 
+	The graph looks like this:
+	0----1
+	| \  |
+	|  \ |
+	3----2
+	We cannot find a way to divide the set of nodes into two independent subsets.
+
+
+**Note:**
+
+- `graph` will have length in range `[1, 100]`.
+- `graph[i]` will contain integers in range `[0, graph.length - 1]`.
+- `graph[i]` will not contain `i` or duplicate values.
+- The graph is undirected: if any element `j` is in `graph[i]`, then `i` will be in `graph[j]`.
+
+## 题目大意
+
+给定一个无向图 graph，当这个图为二分图时返回 true。
+
+graph 将会以邻接表方式给出，graph[i] 表示图中与节点i相连的所有节点。每个节点都是一个在 0 到 graph.length-1 之间的整数。这图中没有自环和平行边： graph[i] 中不存在 i，并且 graph[i] 中没有重复的值。
+
+注意:
+
+- graph 的长度范围为 [1, 100]。
+- graph[i] 中的元素的范围为 [0, graph.length - 1]。
+- graph[i] 不会包含 i 或者有重复的值。
+- 图是无向的: 如果 j 在 graph[i] 里边, 那么 i 也会在 graph[j] 里边。
+
+## 解题思路
+
+- 判断一个无向图是否是二分图。二分图的定义：如果我们能将一个图的节点集合分割成两个独立的子集 A 和 B，并使图中的每一条边的两个节点一个来自 A 集合，一个来自 B 集合，我们就将这个图称为二分图。
+- 这一题可以用 BFS、DFS、并查集来解答。这里是 DFS 实现。任选一个节点开始，把它染成红色，然后对整个图 DFS 遍历，把与它相连的节点并且未被染色的，都染成绿色。颜色不同的节点代表不同的集合。这时候还可能遇到第 2 种情况，与它相连的节点已经有颜色了，并且这个颜色和前一个节点的颜色相同，这就说明了该无向图不是二分图。可以直接 return false。如此遍历到所有节点都染色了，如果能染色成功，说明该无向图是二分图，返回 true。
+
+## 代码
+
+```go
+package leetcode
+
+// DFS 染色，1 是红色，0 是绿色，-1 是未染色
+func isBipartite(graph [][]int) bool {
+	colors := make([]int, len(graph))
+	for i := range colors {
+		colors[i] = -1
+	}
+	for i := range graph {
+		if !dfs(i, graph, colors, -1) {
+			return false
+		}
+	}
+	return true
+}
+
+func dfs(n int, graph [][]int, colors []int, parentCol int) bool {
+	if colors[n] == -1 {
+		if parentCol == 1 {
+			colors[n] = 0
+		} else {
+			colors[n] = 1
+		}
+	} else if colors[n] == parentCol {
+		return false
+	} else if colors[n] != parentCol {
+		return true
+	}
+	for _, c := range graph[n] {
+		if !dfs(c, graph, colors, colors[n]) {
+			return false
+		}
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/0786.K-th-Smallest-Prime-Fraction/786. K-th Smallest Prime Fraction.go b/leetcode/0786.K-th-Smallest-Prime-Fraction/786. K-th Smallest Prime Fraction.go
new file mode 100644
index 000000000..5803fe69d
--- /dev/null
+++ b/leetcode/0786.K-th-Smallest-Prime-Fraction/786. K-th Smallest Prime Fraction.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+// 解法一 二分搜索
+func kthSmallestPrimeFraction(A []int, K int) []int {
+	low, high, n := 0.0, 1.0, len(A)
+	// 因为是在小数内使用二分查找，无法像在整数范围内那样通过 mid+1 和边界判断来终止循环
+	// 所以此处根据 count 来结束循环
+	for {
+		mid, count, p, q, j := (high+low)/2.0, 0, 0, 1, 0
+		for i := 0; i < n; i++ {
+			for j < n && float64(A[i]) > float64(mid)*float64(A[j]) {
+				j++
+			}
+			count += n - j
+			if j < n && q*A[i] > p*A[j] {
+				p = A[i]
+				q = A[j]
+			}
+		}
+		if count == K {
+			return []int{p, q}
+		} else if count < K {
+			low = mid
+		} else {
+			high = mid
+		}
+	}
+}
+
+// 解法二 暴力解法，时间复杂度 O(n^2)
+func kthSmallestPrimeFraction1(A []int, K int) []int {
+	if len(A) == 0 || (len(A)*(len(A)-1))/2 < K {
+		return []int{}
+	}
+	fractions := []Fraction{}
+	for i := 0; i < len(A); i++ {
+		for j := i + 1; j < len(A); j++ {
+			fractions = append(fractions, Fraction{molecule: A[i], denominator: A[j]})
+		}
+	}
+	sort.Sort(SortByFraction(fractions))
+	return []int{fractions[K-1].molecule, fractions[K-1].denominator}
+}
+
+// Fraction define
+type Fraction struct {
+	molecule    int
+	denominator int
+}
+
+// SortByFraction define
+type SortByFraction []Fraction
+
+func (a SortByFraction) Len() int      { return len(a) }
+func (a SortByFraction) Swap(i, j int) { a[i], a[j] = a[j], a[i] }
+func (a SortByFraction) Less(i, j int) bool {
+	return a[i].molecule*a[j].denominator < a[j].molecule*a[i].denominator
+}
diff --git a/leetcode/0786.K-th-Smallest-Prime-Fraction/786. K-th Smallest Prime Fraction_test.go b/leetcode/0786.K-th-Smallest-Prime-Fraction/786. K-th Smallest Prime Fraction_test.go
new file mode 100644
index 000000000..4fde34380
--- /dev/null
+++ b/leetcode/0786.K-th-Smallest-Prime-Fraction/786. K-th Smallest Prime Fraction_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question786 struct {
+	para786
+	ans786
+}
+
+// para 是参数
+// one 代表第一个参数
+type para786 struct {
+	A []int
+	K int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans786 struct {
+	one []int
+}
+
+func Test_Problem786(t *testing.T) {
+
+	qs := []question786{
+
+		{
+			para786{[]int{1, 2, 3, 5}, 3},
+			ans786{[]int{2, 5}},
+		},
+
+		{
+			para786{[]int{1, 7}, 1},
+			ans786{[]int{1, 7}},
+		},
+
+		{
+			para786{[]int{1, 2}, 1},
+			ans786{[]int{1, 2}},
+		},
+
+		{
+			para786{[]int{1, 2, 3, 5, 7}, 6},
+			ans786{[]int{3, 7}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 786------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans786, q.para786
+		fmt.Printf("【input】:%v       【output】:%v\n", p, kthSmallestPrimeFraction(p.A, p.K))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0786.K-th-Smallest-Prime-Fraction/README.md b/leetcode/0786.K-th-Smallest-Prime-Fraction/README.md
new file mode 100755
index 000000000..0238ff8b7
--- /dev/null
+++ b/leetcode/0786.K-th-Smallest-Prime-Fraction/README.md
@@ -0,0 +1,48 @@
+# [786. K-th Smallest Prime Fraction](https://leetcode.com/problems/k-th-smallest-prime-fraction/)
+
+
+## 题目
+
+A sorted list `A` contains 1, plus some number of primes. Then, for every p < q in the list, we consider the fraction p/q.
+
+What is the `K`-th smallest fraction considered? Return your answer as an array of ints, where `answer[0] = p` and `answer[1] = q`.
+
+    Examples:
+    Input: A = [1, 2, 3, 5], K = 3
+    Output: [2, 5]
+    Explanation:
+    The fractions to be considered in sorted order are:
+    1/5, 1/3, 2/5, 1/2, 3/5, 2/3.
+    The third fraction is 2/5.
+    
+    Input: A = [1, 7], K = 1
+    Output: [1, 7]
+
+**Note:**
+
+- `A` will have length between `2` and `2000`.
+- Each `A[i]` will be between `1` and `30000`.
+- `K` will be between `1` and `A.length * (A.length - 1) / 2`.
+
+
+## 题目大意
+
+一个已排序好的表 A，其包含 1 和其他一些素数.  当列表中的每一个 p<q 时，我们可以构造一个分数 p/q 。
+
+那么第 k 个最小的分数是多少呢?  以整数数组的形式返回你的答案, 这里 answer[0] = p 且 answer[1] = q.
+
+
+注意:
+
+- A 的取值范围在 2 — 2000.
+- 每个 A[i] 的值在 1 —30000.
+- K 取值范围为 1 — A.length * (A.length - 1) / 2
+
+
+## 解题思路
+
+
+- 给出一个从小到大排列的有序数组，数组里面的元素都是质数，请找出这个数组中的数组成的真分数从小到大排列，第 K 小的分数。
+- 这一题的暴力解法是枚举所有可能的真分数，从小到大排序，输出第 K 小的分数即可。注意排序的时候不能直接用 float 排序，需要转化成分子和分母的结构体进行排序。
+- 最优的解法是二分搜索。由于真分数都小于 1，所以二分搜索的范围是 [0,1]。每次二分出来的 mid，需要在数组里面搜索一次，找出比 mid 小的真分数个数。并记录下最大的真分数的分子和分母，动态维护最大真分数的分子和分母。如果比 mid 小的真分数个数小于 K，那么取右区间继续二分，如果比 mid 小的真分数个数大于 K，那么取左区间继续二分。直到正好找到比 mid 小的真分数个数是 K，此时维护的最大真分数的分子和分母即为答案。
+- 在已排序的矩阵中寻找最 K 小的元素这一系列的题目有：第 373 题，第 378 题，第 668 题，第 719 题，第 786 题。
diff --git a/leetcode/0791.Custom-Sort-String/791. Custom Sort String.go b/leetcode/0791.Custom-Sort-String/791. Custom Sort String.go
new file mode 100644
index 000000000..0535b0591
--- /dev/null
+++ b/leetcode/0791.Custom-Sort-String/791. Custom Sort String.go	
@@ -0,0 +1,15 @@
+package leetcode
+
+import "sort"
+
+func customSortString(order string, str string) string {
+	magic := map[byte]int{}
+	for i := range order {
+		magic[order[i]] = i - 30
+	}
+	byteSlice := []byte(str)
+	sort.Slice(byteSlice, func(i, j int) bool {
+		return magic[byteSlice[i]] < magic[byteSlice[j]]
+	})
+	return string(byteSlice)
+}
diff --git a/leetcode/0791.Custom-Sort-String/791. Custom Sort String_test.go b/leetcode/0791.Custom-Sort-String/791. Custom Sort String_test.go
new file mode 100644
index 000000000..c61d0da36
--- /dev/null
+++ b/leetcode/0791.Custom-Sort-String/791. Custom Sort String_test.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question791 struct {
+	para791
+	ans791
+}
+
+// para 是参数
+// one 代表第一个参数
+type para791 struct {
+	order string
+	str   string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans791 struct {
+	one string
+}
+
+func Test_Problem791(t *testing.T) {
+
+	qs := []question791{
+
+		{
+			para791{"cba", "abcd"},
+			ans791{"cbad"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 791------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans791, q.para791
+		fmt.Printf("【input】:%v       【output】:%v\n", p, customSortString(p.order, p.str))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0791.Custom-Sort-String/README.md b/leetcode/0791.Custom-Sort-String/README.md
new file mode 100644
index 000000000..09c11d29f
--- /dev/null
+++ b/leetcode/0791.Custom-Sort-String/README.md
@@ -0,0 +1,55 @@
+# [791. Custom Sort String](https://leetcode.com/problems/custom-sort-string/)
+
+
+## 题目
+
+`order` and `str` are strings composed of lowercase letters. In `order`, no letter occurs more than once.
+
+`order` was sorted in some custom order previously. We want to permute the characters of `str` so that they match the order that `order` was sorted. More specifically, if `x` occurs before `y` in `order`, then `x` should occur before `y` in the returned string.
+
+Return any permutation of `str` (as a string) that satisfies this property.
+
+```
+Example:Input:
+order = "cba"
+str = "abcd"
+Output: "cbad"
+Explanation:
+"a", "b", "c" appear in order, so the order of "a", "b", "c" should be "c", "b", and "a".
+Since "d" does not appear in order, it can be at any position in the returned string. "dcba", "cdba", "cbda" are also valid outputs.
+
+```
+
+**Note:**
+
+- `order` has length at most `26`, and no character is repeated in `order`.
+- `str` has length at most `200`.
+- `order` and `str` consist of lowercase letters only.
+
+## 题目大意
+
+字符串 S 和 T 只包含小写字符。在 S 中，所有字符只会出现一次。S 已经根据某种规则进行了排序。我们要根据 S 中的字符顺序对 T 进行排序。更具体地说，如果 S 中 x 在 y 之前出现，那么返回的字符串中 x 也应出现在 y 之前。返回任意一种符合条件的字符串 T。
+
+## 解题思路
+
+- 题目只要求 T 中包含 S 的字符串有序，所以可以先将 T 中包含 S 的字符串排好序，然后再拼接上其他字符。S 字符串最长为 26 位，先将 S 中字符的下标向左偏移 30，并将偏移后的下标值存入字典中。再把 T 字符串按照字典中下标值进行排序。S 中出现的字符对应的下标经过处理以后变成了负数，S 中未出现的字符的下标还是正数。所以经过排序以后，S 中出现的字符按照原有顺序排列在前面，S 中未出现的字符依次排在后面。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func customSortString(order string, str string) string {
+	magic := map[byte]int{}
+	for i := range order {
+		magic[order[i]] = i - 30
+	}
+	byteSlice := []byte(str)
+	sort.Slice(byteSlice, func(i, j int) bool {
+		return magic[byteSlice[i]] < magic[byteSlice[j]]
+	})
+	return string(byteSlice)
+}
+```
\ No newline at end of file
diff --git a/leetcode/0792.Number-of-Matching-Subsequences/792. Number of Matching Subsequences.go b/leetcode/0792.Number-of-Matching-Subsequences/792. Number of Matching Subsequences.go
new file mode 100644
index 000000000..ac40c8883
--- /dev/null
+++ b/leetcode/0792.Number-of-Matching-Subsequences/792. Number of Matching Subsequences.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+func numMatchingSubseq(s string, words []string) int {
+	hash, res := make([][]string, 26), 0
+	for _, w := range words {
+		hash[int(w[0]-'a')] = append(hash[int(w[0]-'a')], w)
+	}
+	for _, c := range s {
+		words := hash[int(byte(c)-'a')]
+		hash[int(byte(c)-'a')] = []string{}
+		for _, w := range words {
+			if len(w) == 1 {
+				res += 1
+				continue
+			}
+			hash[int(w[1]-'a')] = append(hash[int(w[1]-'a')], w[1:])
+		}
+	}
+	return res
+}
diff --git a/leetcode/0792.Number-of-Matching-Subsequences/792. Number of Matching Subsequences_test.go b/leetcode/0792.Number-of-Matching-Subsequences/792. Number of Matching Subsequences_test.go
new file mode 100644
index 000000000..4a1575342
--- /dev/null
+++ b/leetcode/0792.Number-of-Matching-Subsequences/792. Number of Matching Subsequences_test.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question792 struct {
+	para792
+	ans792
+}
+
+// para 是参数
+// one 代表第一个参数
+type para792 struct {
+	s     string
+	words []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans792 struct {
+	one int
+}
+
+func Test_Problem792(t *testing.T) {
+
+	qs := []question792{
+
+		{
+			para792{"abcde", []string{"a", "bb", "acd", "ace"}},
+			ans792{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 792------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans792, q.para792
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numMatchingSubseq(p.s, p.words))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0792.Number-of-Matching-Subsequences/README.md b/leetcode/0792.Number-of-Matching-Subsequences/README.md
new file mode 100644
index 000000000..8da33d3e7
--- /dev/null
+++ b/leetcode/0792.Number-of-Matching-Subsequences/README.md
@@ -0,0 +1,66 @@
+# [792. Number of Matching Subsequences](https://leetcode.com/problems/number-of-matching-subsequences/)
+
+
+## 题目
+
+Given a string `s` and an array of strings `words`, return *the number of* `words[i]` *that is a subsequence of* `s`.
+
+A **subsequence** of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.
+
+- For example, `"ace"` is a subsequence of `"abcde"`.
+
+**Example 1:**
+
+```
+Input: s = "abcde", words = ["a","bb","acd","ace"]
+Output: 3
+Explanation: There are three strings in words that are a subsequence of s: "a", "acd", "ace".
+```
+
+**Example 2:**
+
+```
+Input: s = "dsahjpjauf", words = ["ahjpjau","ja","ahbwzgqnuk","tnmlanowax"]
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 5 * 104`
+- `1 <= words.length <= 5000`
+- `1 <= words[i].length <= 50`
+- `s` and `words[i]` consist of only lowercase English letters.
+
+## 题目大意
+
+给定字符串 S 和单词字典 words, 求 words[i] 中是 S 的子序列的单词个数。
+
+## 解题思路
+
+- 如果将 words 数组内的字符串每次都在源字符串 S 中匹配，这种暴力解法超时。超时原因是对字符串 S 遍历了多次。是否有更加高效的方法呢？
+- 把 words 数组内字符串按照首字母，分到 26 个桶中。从头开始遍历一遍源字符串 S，每扫一个字母，命中 26 个桶中其中一个桶，修改这个桶中的字符串。例如：当前遍历到了 'o'，此时桶中存的数据是 'a' : ['amy','aop'], 'o': ['oqp','onwn']，那么调整 'o' 桶中的数据后，各桶的状态为，'a' : ['amy','aop'], 'q': ['qp'], 'n': ['nwn']。从头到尾扫完整个字符串 S，某个桶中的字符串被清空，说明该桶中的字符串都符合 S 的子序列。将符合子序列的字符串个数累加起来即为最终答案。
+
+## 代码
+
+```go
+package leetcode
+
+func numMatchingSubseq(s string, words []string) int {
+	hash, res := make([][]string, 26), 0
+	for _, w := range words {
+		hash[int(w[0]-'a')] = append(hash[int(w[0]-'a')], w)
+	}
+	for _, c := range s {
+		words := hash[int(byte(c)-'a')]
+		hash[int(byte(c)-'a')] = []string{}
+		for _, w := range words {
+			if len(w) == 1 {
+				res += 1
+				continue
+			}
+			hash[int(w[1]-'a')] = append(hash[int(w[1]-'a')], w[1:])
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0793.Preimage-Size-of-Factorial-Zeroes-Function/793. Preimage Size of Factorial Zeroes Function.go b/leetcode/0793.Preimage-Size-of-Factorial-Zeroes-Function/793. Preimage Size of Factorial Zeroes Function.go
new file mode 100644
index 000000000..cc8e88ecd
--- /dev/null
+++ b/leetcode/0793.Preimage-Size-of-Factorial-Zeroes-Function/793. Preimage Size of Factorial Zeroes Function.go	
@@ -0,0 +1,41 @@
+package leetcode
+
+// 解法一 二分搜索
+func preimageSizeFZF(K int) int {
+	low, high := 0, 5*K
+	for low <= high {
+		mid := low + (high-low)>>1
+		k := trailingZeroes(mid)
+		if k == K {
+			return 5
+		} else if k > K {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return 0
+}
+
+func trailingZeroes(n int) int {
+	if n/5 == 0 {
+		return 0
+	}
+	return n/5 + trailingZeroes(n/5)
+}
+
+// 解法二 数学方法
+func preimageSizeFZF1(K int) int {
+	base := 0
+	for base < K {
+		base = base*5 + 1
+	}
+	for K > 0 {
+		base = (base - 1) / 5
+		if K/base == 5 {
+			return 0
+		}
+		K %= base
+	}
+	return 5
+}
diff --git a/leetcode/0793.Preimage-Size-of-Factorial-Zeroes-Function/793. Preimage Size of Factorial Zeroes Function_test.go b/leetcode/0793.Preimage-Size-of-Factorial-Zeroes-Function/793. Preimage Size of Factorial Zeroes Function_test.go
new file mode 100644
index 000000000..de6ab01ea
--- /dev/null
+++ b/leetcode/0793.Preimage-Size-of-Factorial-Zeroes-Function/793. Preimage Size of Factorial Zeroes Function_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question793 struct {
+	para793
+	ans793
+}
+
+// para 是参数
+// one 代表第一个参数
+type para793 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans793 struct {
+	one int
+}
+
+func Test_Problem793(t *testing.T) {
+
+	qs := []question793{
+
+		{
+			para793{0},
+			ans793{5},
+		},
+
+		{
+			para793{5},
+			ans793{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 793------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans793, q.para793
+		fmt.Printf("【input】:%v       【output】:%v\n", p, preimageSizeFZF(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0793.Preimage-Size-of-Factorial-Zeroes-Function/README.md b/leetcode/0793.Preimage-Size-of-Factorial-Zeroes-Function/README.md
new file mode 100755
index 000000000..2e01316e3
--- /dev/null
+++ b/leetcode/0793.Preimage-Size-of-Factorial-Zeroes-Function/README.md
@@ -0,0 +1,62 @@
+# [793. Preimage Size of Factorial Zeroes Function](https://leetcode.com/problems/preimage-size-of-factorial-zeroes-function/)
+
+
+## 题目
+
+Let `f(x)` be the number of zeroes at the end of `x!`. (Recall that `x! = 1 * 2 * 3 * ... * x`, and by convention, `0! = 1`.)
+
+For example, `f(3) = 0` because 3! = 6 has no zeroes at the end, while `f(11) = 2` because 11! = 39916800 has 2 zeroes at the end. Given `K`, find how many non-negative integers `x` have the property that `f(x) = K`.
+
+    Example 1:
+    Input: K = 0
+    Output: 5
+    Explanation: 0!, 1!, 2!, 3!, and 4! end with K = 0 zeroes.
+    
+    Example 2:
+    Input: K = 5
+    Output: 0
+    Explanation: There is no x such that x! ends in K = 5 zeroes.
+
+**Note:**
+
+- `K` will be an integer in the range `[0, 10^9]`.
+
+
+## 题目大意
+
+
+f(x) 是 x! 末尾是0的数量。（回想一下 x! = 1 * 2 * 3 * ... * x，且0! = 1）
+
+例如， f(3) = 0 ，因为3! = 6的末尾没有0；而 f(11) = 2 ，因为11!= 39916800末端有2个0。给定 K，找出多少个非负整数x ，有 f(x) = K 的性质。
+
+注意：
+
+- K 是范围在 [0, 10^9] 的整数。
+
+
+## 解题思路
+
+- 给出一个数 K，要求有多少个 n 能使得 n！末尾 0 的个数等于 K。
+- 这一题是基于第 172 题的逆过程加强版。第 172 题是给出 `n`，求得末尾 0 的个数。由第 172 题可以知道，`n！`末尾 0 的个数取决于因子 5 的个数。末尾可能有 `K` 个 0，那么 `n` 最多可以等于 `5 * K`，在 `[0, 5* K]` 区间内二分搜索，判断 `mid` 末尾 0 的个数，如果能找到 `K`，那么就范围 5，如果找不到这个 `K`，返回 0 。为什么答案取值只有 0 和 5 呢？因为当 `n` 增加 5 以后，因子 5 的个数又加一了，末尾又可以多 1 个或者多个 0(如果加 5 以后，有多个 5 的因子，例如 25，125，就有可能末尾增加多个 0)。所以有效的 `K` 值对应的 `n` 的范围区间就是 5 。反过来，无效的 `K` 值对应的 `n` 是 0。`K` 在 `5^n` 的分界线处会发生跳变，所有有些值取不到。例如，`n` 在 `[0,5)` 内取值，`K = 0`；`n` 在 `[5,10)` 内取值，`K = 1`；`n` 在 `[10,15)` 内取值，`K = 2`；`n` 在 `[15,20)` 内取值，`K = 3`；`n` 在 `[20,25)` 内取值，`K = 4`；`n` 在 `[25,30)` 内取值，`K = 6`，因为 25 提供了 2 个 5，也就提供了 2 个 0，所以 `K` 永远无法取值等于 5，即当 `K = 5` 时，找不到任何的 `n` 与之对应。
+- 这一题也可以用数学的方法解题。见解法二。这个解法的灵感来自于：n！末尾 0 的个数等于 [1,n] 所有数的因子 5 的个数总和。其次此题的结果一定只有 0 和 5 (分析见上一种解法)。有了这两个结论以后，就可以用数学的方法推导了。首先 n 可以表示为 5 进制的形式  
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_793_1.png'>
+</p>
+    上面式子中，所有有因子 5 的个数为：  
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_793_2.png'>
+</p>
+
+    这个总数就即是 K。针对不同的 n，an 的通项公式不同，所以表示的 K 的系数也不同。cn 的通项公式呢？  
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_793_2.png'>
+</p>  
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_793_3.png'>
+</p>
+
+    由上面这个递推还能推出通项公式(不过这题不适用通项公式，是用递推公式更方便)：  
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_793_4.png'>
+</p>
+    判断 K 是否能表示成两个数列的表示形式，等价于判断 K 是否能转化为以 Cn 为基的变进制数。到此，转化成类似第 483 题了。代码实现不难，见解法二。
diff --git a/leetcode/0794.Valid-Tic-Tac-Toe-State/794.Valid Tic-Tac-Toe State.go b/leetcode/0794.Valid-Tic-Tac-Toe-State/794.Valid Tic-Tac-Toe State.go
new file mode 100644
index 000000000..235ac8d47
--- /dev/null
+++ b/leetcode/0794.Valid-Tic-Tac-Toe-State/794.Valid Tic-Tac-Toe State.go	
@@ -0,0 +1,40 @@
+package leetcode
+
+func validTicTacToe(board []string) bool {
+	cntX, cntO := 0, 0
+	for i := range board {
+		for j := range board[i] {
+			if board[i][j] == 'X' {
+				cntX++
+			} else if board[i][j] == 'O' {
+				cntO++
+			}
+		}
+	}
+	if cntX < cntO || cntX > cntO+1 {
+		return false
+	}
+	if cntX == cntO {
+		return process(board, 'X')
+	}
+	return process(board, 'O')
+}
+
+func process(board []string, c byte) bool {
+	//某一行是"ccc"
+	if board[0] == string([]byte{c, c, c}) || board[1] == string([]byte{c, c, c}) || board[2] == string([]byte{c, c, c}) {
+		return false
+	}
+	//某一列是"ccc"
+	if (board[0][0] == c && board[1][0] == c && board[2][0] == c) ||
+		(board[0][1] == c && board[1][1] == c && board[2][1] == c) ||
+		(board[0][2] == c && board[1][2] == c && board[2][2] == c) {
+		return false
+	}
+	//某一对角线是"ccc"
+	if (board[0][0] == c && board[1][1] == c && board[2][2] == c) ||
+		(board[0][2] == c && board[1][1] == c && board[2][0] == c) {
+		return false
+	}
+	return true
+}
diff --git a/leetcode/0794.Valid-Tic-Tac-Toe-State/794.Valid Tic-Tac-Toe State_test.go b/leetcode/0794.Valid-Tic-Tac-Toe-State/794.Valid Tic-Tac-Toe State_test.go
new file mode 100644
index 000000000..b324954bd
--- /dev/null
+++ b/leetcode/0794.Valid-Tic-Tac-Toe-State/794.Valid Tic-Tac-Toe State_test.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question794 struct {
+	para794
+	ans794
+}
+
+// para 是参数
+type para794 struct {
+	board []string
+}
+
+// ans 是答案
+type ans794 struct {
+	ans bool
+}
+
+func Test_Problem794(t *testing.T) {
+
+	qs := []question794{
+
+		{
+			para794{[]string{"O  ", "   ", "   "}},
+			ans794{false},
+		},
+
+		{
+			para794{[]string{"XOX", " X ", "   "}},
+			ans794{false},
+		},
+
+		{
+			para794{[]string{"XXX", "   ", "OOO"}},
+			ans794{false},
+		},
+
+		{
+			para794{[]string{"XOX", "O O", "XOX"}},
+			ans794{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 794------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans794, q.para794
+		fmt.Printf("【input】:%v    【output】:%v\n", p.board, validTicTacToe(p.board))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0794.Valid-Tic-Tac-Toe-State/README.md b/leetcode/0794.Valid-Tic-Tac-Toe-State/README.md
new file mode 100644
index 000000000..2965e983f
--- /dev/null
+++ b/leetcode/0794.Valid-Tic-Tac-Toe-State/README.md
@@ -0,0 +1,119 @@
+# [794. Valid Tic-Tac-Toe State](https://leetcode.com/problems/valid-tic-tac-toe-state/)
+
+## 题目
+
+Given a Tic-Tac-Toe board as a string array board, return true if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.
+
+The board is a 3 x 3 array that consists of characters ' ', 'X', and 'O'. The ' ' character represents an empty square.
+
+Here are the rules of Tic-Tac-Toe:
+
+- Players take turns placing characters into empty squares ' '.
+- The first player always places 'X' characters, while the second player always places 'O' characters.
+- 'X' and 'O' characters are always placed into empty squares, never filled ones.
+- The game ends when there are three of the same (non-empty) character filling any row, column, or diagonal.
+- The game also ends if all squares are non-empty.
+- No more moves can be played if the game is over.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2021/05/15/tictactoe1-grid.jpg](https://assets.leetcode.com/uploads/2021/05/15/tictactoe1-grid.jpg)
+
+    Input: board = ["O  ","   ","   "]
+    Output: false
+    Explanation: The first player always plays "X".
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2021/05/15/tictactoe2-grid.jpg](https://assets.leetcode.com/uploads/2021/05/15/tictactoe2-grid.jpg)
+
+    Input: board = ["XOX"," X ","   "]
+    Output: false
+    Explanation: Players take turns making moves.
+
+**Example 3**:
+
+![https://assets.leetcode.com/uploads/2021/05/15/tictactoe3-grid.jpg](https://assets.leetcode.com/uploads/2021/05/15/tictactoe3-grid.jpg)
+
+    Input: board = ["XXX","   ","OOO"]
+    Output: false
+
+**Example 4**:
+
+![https://assets.leetcode.com/uploads/2021/05/15/tictactoe4-grid.jpg](https://assets.leetcode.com/uploads/2021/05/15/tictactoe4-grid.jpg)
+
+    Input: board = ["XOX","O O","XOX"]
+    Output: true
+
+**Constraints:**
+
+- board.length == 3
+- board[i].length == 3
+- board[i][j] is either 'X', 'O', or ' '.
+
+## 题目大意
+
+给你一个字符串数组 board 表示井字游戏的棋盘。当且仅当在井字游戏过程中，棋盘有可能达到 board 所显示的状态时，才返回 true 。
+
+井字游戏的棋盘是一个 3 x 3 数组，由字符 ' '，'X' 和 'O' 组成。字符 ' ' 代表一个空位。
+
+以下是井字游戏的规则：
+
+- 玩家轮流将字符放入空位（' '）中。
+- 玩家 1 总是放字符 'X' ，而玩家 2 总是放字符 'O' 。
+- 'X' 和 'O' 只允许放置在空位中，不允许对已放有字符的位置进行填充。
+- 当有 3 个相同（且非空）的字符填充任何行、列或对角线时，游戏结束。
+- 当所有位置非空时，也算为游戏结束。
+- 如果游戏结束，玩家不允许再放置字符。
+
+## 解题思路
+
+分类模拟：
+- 根据题意棋盘在任意时候，要么 X 的数量比 O 的数量多 1，要么两者相等
+- X 的数量等于 O 的数量时,任何行、列或对角线都不会出现 3 个相同的 X
+- X 的数量比 O 的数量多 1 时,任何行、列或对角线都不会出现 3 个相同的 O
+
+## 代码
+
+```go
+package leetcode
+
+func validTicTacToe(board []string) bool {
+	cntX, cntO := 0, 0
+	for i := range board {
+		for j := range board[i] {
+			if board[i][j] == 'X' {
+				cntX++
+			} else if board[i][j] == 'O' {
+				cntO++
+			}
+		}
+	}
+	if cntX < cntO || cntX > cntO+1 {
+		return false
+	}
+	if cntX == cntO {
+		return process(board, 'X')
+	}
+	return process(board, 'O')
+}
+
+func process(board []string, c byte) bool {
+	//某一行是"ccc"
+	if board[0] == string([]byte{c, c, c}) || board[1] == string([]byte{c, c, c}) || board[2] == string([]byte{c, c, c}) {
+		return false
+	}
+	//某一列是"ccc"
+	if (board[0][0] == c && board[1][0] == c && board[2][0] == c) ||
+		(board[0][1] == c && board[1][1] == c && board[2][1] == c) ||
+		(board[0][2] == c && board[1][2] == c && board[2][2] == c) {
+		return false
+	}
+	//某一对角线是"ccc"
+	if (board[0][0] == c && board[1][1] == c && board[2][2] == c) ||
+		(board[0][2] == c && board[1][1] == c && board[2][0] == c) {
+		return false
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/795. Number of Subarrays with Bounded Maximum.go b/leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/795. Number of Subarrays with Bounded Maximum.go
new file mode 100644
index 000000000..f5eb5c3f7
--- /dev/null
+++ b/leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/795. Number of Subarrays with Bounded Maximum.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func numSubarrayBoundedMax(nums []int, left int, right int) int {
+	return getAnswerPerBound(nums, right) - getAnswerPerBound(nums, left-1)
+}
+
+func getAnswerPerBound(nums []int, bound int) int {
+	res, count := 0, 0
+	for _, num := range nums {
+		if num <= bound {
+			count++
+		} else {
+			count = 0
+		}
+		res += count
+	}
+	return res
+}
diff --git a/leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/795. Number of Subarrays with Bounded Maximum_test.go b/leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/795. Number of Subarrays with Bounded Maximum_test.go
new file mode 100644
index 000000000..5dba6f0cd
--- /dev/null
+++ b/leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/795. Number of Subarrays with Bounded Maximum_test.go	
@@ -0,0 +1,44 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question795 struct {
+	para795
+	ans795
+}
+
+// para 是参数
+// one 代表第一个参数
+type para795 struct {
+	nums  []int
+	left  int
+	right int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans795 struct {
+	one int
+}
+
+func Test_Problem795(t *testing.T) {
+
+	qs := []question795{
+
+		{
+			para795{[]int{2, 1, 4, 3}, 2, 3},
+			ans795{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 795------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans795, q.para795
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numSubarrayBoundedMax(p.nums, p.left, p.right))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/README.md b/leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/README.md
new file mode 100644
index 000000000..c69294451
--- /dev/null
+++ b/leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/README.md
@@ -0,0 +1,54 @@
+# [795. Number of Subarrays with Bounded Maximum](https://leetcode.com/problems/number-of-subarrays-with-bounded-maximum/)
+
+
+## 题目
+
+We are given an array `nums` of positive integers, and two positive integers `left` and `right` (`left <= right`).
+
+Return the number of (contiguous, non-empty) subarrays such that the value of the maximum array element in that subarray is at least `left` and at most `right`.
+
+```
+Example:Input:
+nums = [2, 1, 4, 3]
+left = 2
+right = 3
+Output: 3
+Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].
+```
+
+**Note:**
+
+- `left`, `right`, and `nums[i]` will be an integer in the range `[0, 109]`.
+- The length of `nums` will be in the range of `[1, 50000]`.
+
+## 题目大意
+
+给定一个元素都是正整数的数组`A` ，正整数 `L` 以及 `R` (`L <= R`)。求连续、非空且其中最大元素满足大于等于`L` 小于等于`R`的子数组个数。
+
+## 解题思路
+
+- 题目要求子数组最大元素在 [L,R] 区间内。假设 count(bound) 为计算所有元素都小于等于 bound 的子数组数量。那么本题所求的答案可转化为 count(R) - count(L-1)。
+- 如何统计所有元素小于 bound 的子数组数量呢？使用 count 变量记录在 bound 的左边，小于等于 bound 的连续元素数量。当找到一个这样的元素时，在此位置上结束的有效子数组的数量为 count + 1。当遇到一个元素大于 B 时，则在此位置结束的有效子数组的数量为 0。res 将每轮 count 累加，最终 res 中存的即是满足条件的所有子数组数量。
+
+## 代码
+
+```go
+package leetcode
+
+func numSubarrayBoundedMax(nums []int, left int, right int) int {
+	return getAnswerPerBound(nums, right) - getAnswerPerBound(nums, left-1)
+}
+
+func getAnswerPerBound(nums []int, bound int) int {
+	res, count := 0, 0
+	for _, num := range nums {
+		if num <= bound {
+			count++
+		} else {
+			count = 0
+		}
+		res += count
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0802.Find-Eventual-Safe-States/802. Find Eventual Safe States.go b/leetcode/0802.Find-Eventual-Safe-States/802. Find Eventual Safe States.go
new file mode 100644
index 000000000..a73b3ca9e
--- /dev/null
+++ b/leetcode/0802.Find-Eventual-Safe-States/802. Find Eventual Safe States.go	
@@ -0,0 +1,26 @@
+package leetcode
+
+func eventualSafeNodes(graph [][]int) []int {
+	res, color := []int{}, make([]int, len(graph))
+	for i := range graph {
+		if dfsEventualSafeNodes(graph, i, color) {
+			res = append(res, i)
+		}
+	}
+	return res
+}
+
+// colors: WHITE 0, GRAY 1, BLACK 2;
+func dfsEventualSafeNodes(graph [][]int, idx int, color []int) bool {
+	if color[idx] > 0 {
+		return color[idx] == 2
+	}
+	color[idx] = 1
+	for i := range graph[idx] {
+		if !dfsEventualSafeNodes(graph, graph[idx][i], color) {
+			return false
+		}
+	}
+	color[idx] = 2
+	return true
+}
diff --git a/leetcode/0802.Find-Eventual-Safe-States/802. Find Eventual Safe States_test.go b/leetcode/0802.Find-Eventual-Safe-States/802. Find Eventual Safe States_test.go
new file mode 100644
index 000000000..631e49d7a
--- /dev/null
+++ b/leetcode/0802.Find-Eventual-Safe-States/802. Find Eventual Safe States_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question802 struct {
+	para802
+	ans802
+}
+
+// para 是参数
+// one 代表第一个参数
+type para802 struct {
+	graph [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans802 struct {
+	one []int
+}
+
+func Test_Problem802(t *testing.T) {
+
+	qs := []question802{
+
+		{
+			para802{[][]int{{1, 2}, {2, 3}, {5}, {0}, {5}, {}, {}}},
+			ans802{[]int{2, 4, 5, 6}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 802------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans802, q.para802
+		fmt.Printf("【input】:%v       【output】:%v\n", p, eventualSafeNodes(p.graph))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0802.Find-Eventual-Safe-States/README.md b/leetcode/0802.Find-Eventual-Safe-States/README.md
new file mode 100644
index 000000000..f6879f300
--- /dev/null
+++ b/leetcode/0802.Find-Eventual-Safe-States/README.md
@@ -0,0 +1,73 @@
+# [802. Find Eventual Safe States](https://leetcode.com/problems/find-eventual-safe-states/)
+
+
+
+## 题目
+
+In a directed graph, we start at some node and every turn, walk along a directed edge of the graph. If we reach a node that is terminal (that is, it has no outgoing directed edges), we stop.
+
+Now, say our starting node is *eventually safe* if and only if we must eventually walk to a terminal node. More specifically, there exists a natural number `K` so that for any choice of where to walk, we must have stopped at a terminal node in less than `K` steps.
+
+Which nodes are eventually safe? Return them as an array in sorted order.
+
+The directed graph has `N` nodes with labels `0, 1, ..., N-1`, where `N` is the length of `graph`. The graph is given in the following form: `graph[i]` is a list of labels `j` such that `(i, j)` is a directed edge of the graph.
+
+```
+Example:
+Input: graph = [[1,2],[2,3],[5],[0],[5],[],[]]
+Output: [2,4,5,6]
+Here is a diagram of the above graph.
+```
+
+![https://s3-lc-upload.s3.amazonaws.com/uploads/2018/03/17/picture1.png](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/03/17/picture1.png)
+
+**Note**:
+
+- `graph` will have length at most `10000`.
+- The number of edges in the graph will not exceed `32000`.
+- Each `graph[i]` will be a sorted list of different integers, chosen within the range `[0, graph.length - 1]`.
+
+## 题目大意
+
+在有向图中, 我们从某个节点和每个转向处开始, 沿着图的有向边走。 如果我们到达的节点是终点 (即它没有连出的有向边), 我们停止。现在, 如果我们最后能走到终点，那么我们的起始节点是最终安全的。 更具体地说, 存在一个自然数 K,  无论选择从哪里开始行走, 我们走了不到 K 步后必能停止在一个终点。哪些节点最终是安全的？ 结果返回一个有序的数组。
+
+提示：
+
+- graph 节点数不超过 10000.
+- 图的边数不会超过 32000.
+- 每个 graph[i] 被排序为不同的整数列表， 在区间 [0, graph.length - 1] 中选取。
+
+
+## 解题思路
+
+- 给出一个有向图，要求找出所有“安全”节点。“安全”节点的定义是：存在一个自然数 K, 无论选择从哪里开始行走, 我们走了不到 K 步后必能停止在一个终点。
+- 这一题可以用拓扑排序，也可以用 DFS 染色来解答。这里用 DFS 来解答。对于每个节点，我们有 3 种染色的方法：白色 0 号节点表示该节点还没有被访问过；灰色 1 号节点表示该节点在栈中（这一轮搜索中被访问过）或者在环中；黑色 2 号节点表示该节点的所有相连的节点都被访问过，且该节点不在环中。当我们第一次访问一个节点时，我们把它从白色变成灰色，并继续搜索与它相连的节点。如果在搜索过程中我们遇到一个灰色的节点，那么说明找到了一个环，此时退出搜索，所有的灰色节点保持不变（即从任意一个灰色节点开始，都能走到环中），如果搜索过程中，我们没有遇到灰色的节点，那么在回溯到当前节点时，我们把它从灰色变成黑色，即表示它是一个安全的节点。
+
+## 代码
+
+```go
+func eventualSafeNodes(graph [][]int) []int {
+	res, color := []int{}, make([]int, len(graph))
+	for i := range graph {
+		if dfsEventualSafeNodes(graph, i, color) {
+			res = append(res, i)
+		}
+	}
+	return res
+}
+
+// colors: WHITE 0, GRAY 1, BLACK 2;
+func dfsEventualSafeNodes(graph [][]int, idx int, color []int) bool {
+	if color[idx] > 0 {
+		return color[idx] == 2
+	}
+	color[idx] = 1
+	for i := range graph[idx] {
+		if !dfsEventualSafeNodes(graph, graph[idx][i], color) {
+			return false
+		}
+	}
+	color[idx] = 2
+	return true
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0803. Bricks Falling When Hit/803. Bricks Falling When Hit.go b/leetcode/0803.Bricks-Falling-When-Hit/803. Bricks Falling When Hit.go
similarity index 95%
rename from Algorithms/0803. Bricks Falling When Hit/803. Bricks Falling When Hit.go
rename to leetcode/0803.Bricks-Falling-When-Hit/803. Bricks Falling When Hit.go
index 48eaa7bc5..879835c19 100644
--- a/Algorithms/0803. Bricks Falling When Hit/803. Bricks Falling When Hit.go	
+++ b/leetcode/0803.Bricks-Falling-When-Hit/803. Bricks Falling When Hit.go	
@@ -4,6 +4,13 @@ import (
 	"github.com/halfrost/LeetCode-Go/template"
 )
 
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
 func hitBricks(grid [][]int, hits [][]int) []int {
 	if len(hits) == 0 {
 		return []int{}
diff --git a/leetcode/0803.Bricks-Falling-When-Hit/803. Bricks Falling When Hit_test.go b/leetcode/0803.Bricks-Falling-When-Hit/803. Bricks Falling When Hit_test.go
new file mode 100644
index 000000000..558e7c500
--- /dev/null
+++ b/leetcode/0803.Bricks-Falling-When-Hit/803. Bricks Falling When Hit_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question803 struct {
+	para803
+	ans803
+}
+
+// para 是参数
+// one 代表第一个参数
+type para803 struct {
+	grid [][]int
+	hits [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans803 struct {
+	one []int
+}
+
+func Test_Problem803(t *testing.T) {
+
+	qs := []question803{
+
+		{
+			para803{[][]int{{1, 1, 1, 0, 1, 1, 1, 1}, {1, 0, 0, 0, 0, 1, 1, 1}, {1, 1, 1, 0, 0, 0, 1, 1}, {1, 1, 0, 0, 0, 0, 0, 0}, {1, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 0, 0, 0, 0, 0, 0}},
+				[][]int{{4, 6}, {3, 0}, {2, 3}, {2, 6}, {4, 1}, {5, 2}, {2, 1}}},
+			ans803{[]int{0, 2, 0, 0, 0, 0, 2}},
+		},
+
+		{
+			para803{[][]int{{1, 0, 1}, {1, 1, 1}}, [][]int{{0, 0}, {0, 2}, {1, 1}}},
+			ans803{[]int{0, 3, 0}},
+		},
+		{
+			para803{[][]int{{1, 0, 0, 0}, {1, 1, 1, 0}}, [][]int{{1, 0}}},
+			ans803{[]int{2}},
+		},
+
+		{
+			para803{[][]int{{1, 0, 0, 0}, {1, 1, 0, 0}}, [][]int{{1, 1}, {1, 0}}},
+			ans803{[]int{0, 0}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 803------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans803, q.para803
+		fmt.Printf("【input】:%v       【output】:%v\n", p, hitBricks(p.grid, p.hits))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0803.Bricks-Falling-When-Hit/README.md b/leetcode/0803.Bricks-Falling-When-Hit/README.md
new file mode 100755
index 000000000..dd9719e52
--- /dev/null
+++ b/leetcode/0803.Bricks-Falling-When-Hit/README.md
@@ -0,0 +1,56 @@
+# [803. Bricks Falling When Hit](https://leetcode.com/problems/bricks-falling-when-hit/)
+
+
+## 题目
+
+We have a grid of 1s and 0s; the 1s in a cell represent bricks. A brick will not drop if and only if it is directly connected to the top of the grid, or at least one of its (4-way) adjacent bricks will not drop.
+
+We will do some erasures sequentially. Each time we want to do the erasure at the location (i, j), the brick (if it exists) on that location will disappear, and then some other bricks may drop because of that erasure.
+
+Return an array representing the number of bricks that will drop after each erasure in sequence.
+
+    Example 1:
+    Input: 
+    grid = [[1,0,0,0],[1,1,1,0]]
+    hits = [[1,0]]
+    Output: [2]
+    Explanation: 
+    If we erase the brick at (1, 0), the brick at (1, 1) and (1, 2) will drop. So we should return 2.
+
+    Example 2:
+    Input: 
+    grid = [[1,0,0,0],[1,1,0,0]]
+    hits = [[1,1],[1,0]]
+    Output: [0,0]
+    Explanation: 
+    When we erase the brick at (1, 0), the brick at (1, 1) has already disappeared due to the last move. So each erasure will cause no bricks dropping.  Note that the erased brick (1, 0) will not be counted as a dropped brick.
+
+**Note:**
+
+- The number of rows and columns in the grid will be in the range [1, 200].
+- The number of erasures will not exceed the area of the grid.
+- It is guaranteed that each erasure will be different from any other erasure, and located inside the grid.
+- An erasure may refer to a location with no brick - if it does, no bricks drop.
+
+
+## 题目大意
+
+我们有一组包含1和0的网格；其中1表示砖块。 当且仅当一块砖直接连接到网格的顶部，或者它至少有一块相邻（4 个方向之一）砖块不会掉落时，它才不会落下。我们会依次消除一些砖块。每当我们消除 (i, j) 位置时， 对应位置的砖块（若存在）会消失，然后其他的砖块可能因为这个消除而落下。返回一个数组表示每次消除操作对应落下的砖块数目。
+
+
+注意:
+
+- 网格的行数和列数的范围是[1, 200]。
+- 消除的数字不会超过网格的区域。
+- 可以保证每次的消除都不相同，并且位于网格的内部。
+- 一个消除的位置可能没有砖块，如果这样的话，就不会有砖块落下。
+
+
+
+## 解题思路
+
+
+- 有一些砖块连接在天花板上，问，如果打掉某个砖块，会掉落几块砖块？打掉的每个砖块不参与计数。
+- 这一题可以用并查集和 DFS 求解。不过尝试用 DFS 的同学就会知道，这一题卡时间卡的很紧。用 DFS 虽然能 AC，但是耗时非常长。用并查集也必须进行秩压缩，不然耗时也非常长。另外，如果用了并查集，每个集合的总数单独统计，不随着 union() 操作，也会导致超时，笔者在这里被 LTE 了多次，最后只能重写 UnionFind 并查集类，将统计操作和 union() 操作写在一起，这一题才 faster than 100.00% AC。
+- 拿到题以后，首先尝试暴力解法，按照顺序打掉砖块，每次打掉砖块以后，都重建并查集。题目要求每次掉落几块砖块，实际上比较每次和天花板连通的砖块个数变化了多少块就可以了。那么解法就出来了，先把和天花板连通的砖块都 union() 起来，记录这个集合中砖块的个数 `count`，然后每次打掉一个砖块以后，重建并查集，计算与天花板连通的砖块的个数 `newCount`，`newCount - count -1` 就是最终答案(打掉的那块砖块不计算其中)，提交代码以后，发现 TLE。
+- 出现 TLE 以后一般思路都是对的，只是时间复杂度过高，需要优化。很明显，需要优化的地方是每次都重建了新的并查集，有没有办法能在上一次状态上进行变更，不用重建并查集呢？如果正向的打掉砖块，那么每次还需要以这个砖块为起点进行 DFS，时间复杂度还是很高。如果反向考虑呢？先把所有要打掉的砖块都打掉，构建打掉这些砖块以后剩下与天花板连通的并查集。然后反向添加打掉的砖块，每次添加一块就刷新一次它周围的 4 个砖块，不用 DFS，这样时间复杂度优化了很多。最后在按照 `newCount - count -1` 方式计算最终答案。注意每次还原一个砖块的时候需要染色回原有砖块的颜色 `1` 。优化成这样的做法，基本不会 TLE 了，如果计算 count 是单独计算的，还是会 TLE。如果没有进行秩压缩，时间会超过 1500 ms，所以这一题想拿到 100%，每步优化都要做好。最终 100% 的答案见代码。
diff --git a/leetcode/0807.Max-Increase-to-Keep-City-Skyline/807.Max Increase to Keep City Skyline.go b/leetcode/0807.Max-Increase-to-Keep-City-Skyline/807.Max Increase to Keep City Skyline.go
new file mode 100644
index 000000000..c36831734
--- /dev/null
+++ b/leetcode/0807.Max-Increase-to-Keep-City-Skyline/807.Max Increase to Keep City Skyline.go	
@@ -0,0 +1,39 @@
+package leetcode
+
+func maxIncreaseKeepingSkyline(grid [][]int) int {
+	n := len(grid)
+	topBottomSkyline := make([]int, 0, n)
+	leftRightSkyline := make([]int, 0, n)
+	for i := range grid {
+		cur := 0
+		for _, v := range grid[i] {
+			if cur < v {
+				cur = v
+			}
+		}
+		leftRightSkyline = append(leftRightSkyline, cur)
+	}
+	for j := range grid {
+		cur := 0
+		for i := 0; i < len(grid[0]); i++ {
+			if cur < grid[i][j] {
+				cur = grid[i][j]
+			}
+		}
+		topBottomSkyline = append(topBottomSkyline, cur)
+	}
+	var ans int
+	for i := range grid {
+		for j := 0; j < len(grid[0]); j++ {
+			ans += min(topBottomSkyline[j], leftRightSkyline[i]) - grid[i][j]
+		}
+	}
+	return ans
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0807.Max-Increase-to-Keep-City-Skyline/807.Max Increase to Keep City Skyline_test.go b/leetcode/0807.Max-Increase-to-Keep-City-Skyline/807.Max Increase to Keep City Skyline_test.go
new file mode 100644
index 000000000..81159a0e6
--- /dev/null
+++ b/leetcode/0807.Max-Increase-to-Keep-City-Skyline/807.Max Increase to Keep City Skyline_test.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question807 struct {
+	para807
+	ans807
+}
+
+// para 是参数
+type para807 struct {
+	grid [][]int
+}
+
+// ans 是答案
+type ans807 struct {
+	ans int
+}
+
+func Test_Problem807(t *testing.T) {
+
+	qs := []question807{
+
+		{
+			para807{[][]int{{3, 0, 8, 4}, {2, 4, 5, 7}, {9, 2, 6, 3}, {0, 3, 1, 0}}},
+			ans807{35},
+		},
+
+		{
+			para807{[][]int{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}},
+			ans807{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 807------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans807, q.para807
+		fmt.Printf("【input】:%v    【output】:%v\n", p.grid, maxIncreaseKeepingSkyline(p.grid))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0807.Max-Increase-to-Keep-City-Skyline/README.md b/leetcode/0807.Max-Increase-to-Keep-City-Skyline/README.md
new file mode 100644
index 000000000..58651c500
--- /dev/null
+++ b/leetcode/0807.Max-Increase-to-Keep-City-Skyline/README.md
@@ -0,0 +1,97 @@
+# [807. Max Increase to Keep City Skyline](https://leetcode.com/problems/max-increase-to-keep-city-skyline/)
+
+## 题目
+
+There is a city composed of n x n blocks, where each block contains a single building shaped like a vertical square prism. You are given a 0-indexed n x n integer matrix grid where grid[r][c] represents the height of the building located in the block at row r and column c.
+
+A city's skyline is the the outer contour formed by all the building when viewing the side of the city from a distance. The skyline from each cardinal direction north, east, south, and west may be different.
+
+We are allowed to increase the height of any number of buildings by any amount (the amount can be different per building). The height of a 0-height building can also be increased. However, increasing the height of a building should not affect the city's skyline from any cardinal direction.
+
+Return the maximum total sum that the height of the buildings can be increased by without changing the city's skyline from any cardinal direction.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2021/06/21/807-ex1.png](https://assets.leetcode.com/uploads/2021/06/21/807-ex1.png)
+
+    Input: grid = [[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]
+    Output: 35
+    Explanation: The building heights are shown in the center of the above image.
+    The skylines when viewed from each cardinal direction are drawn in red.
+    The grid after increasing the height of buildings without affecting skylines is:
+    gridNew = [ [8, 4, 8, 7],
+    [7, 4, 7, 7],
+    [9, 4, 8, 7],
+    [3, 3, 3, 3] ]
+
+**Example 2**:
+
+    Input: grid = [[0,0,0],[0,0,0],[0,0,0]]
+    Output: 0
+    Explanation: Increasing the height of any building will result in the skyline changing.
+
+**Constraints:**
+
+- n == grid.length
+- n == grid[r].length
+- 2 <= n <= 50
+- 0 <= grid[r][c] <= 100
+
+## 题目大意
+
+在二维数组grid中，grid[i][j]代表位于某处的建筑物的高度。 我们被允许增加任何数量（不同建筑物的数量可能不同）的建筑物的高度。 高度 0 也被认为是建筑物。
+
+最后，从新数组的所有四个方向（即顶部，底部，左侧和右侧）观看的“天际线”必须与原始数组的天际线相同。 城市的天际线是从远处观看时，由所有建筑物形成的矩形的外部轮廓。 请看下面的例子。
+
+建筑物高度可以增加的最大总和是多少？
+
+## 解题思路
+
+- 从数组竖直方向（即顶部，底部）看“天际线”计算出 topBottomSkyline
+- 从数组水平方向（即左侧，右侧）看“天际线”计算出 leftRightSkyline
+- 计算 grid 中每个元素与对应的 topBottomSkyline 和 leftRightSkyline 中较小值的差值
+- 统计所有差值的总和 ans 并返回
+
+## 代码
+
+```go
+package leetcode
+
+func maxIncreaseKeepingSkyline(grid [][]int) int {
+	n := len(grid)
+	topBottomSkyline := make([]int, 0, n)
+	leftRightSkyline := make([]int, 0, n)
+	for i := range grid {
+		cur := 0
+		for _, v := range grid[i] {
+			if cur < v {
+				cur = v
+			}
+		}
+		leftRightSkyline = append(leftRightSkyline, cur)
+	}
+	for j := range grid {
+		cur := 0
+		for i := 0; i < len(grid[0]); i++ {
+			if cur < grid[i][j] {
+				cur = grid[i][j]
+			}
+		}
+		topBottomSkyline = append(topBottomSkyline, cur)
+	}
+	var ans int
+	for i := range grid {
+		for j := 0; j < len(grid[0]); j++ {
+			ans += min(topBottomSkyline[j], leftRightSkyline[i]) - grid[i][j]
+		}
+	}
+	return ans
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/0810.Chalkboard-XOR-Game/810. Chalkboard XOR Game.go b/leetcode/0810.Chalkboard-XOR-Game/810. Chalkboard XOR Game.go
new file mode 100644
index 000000000..25c791bad
--- /dev/null
+++ b/leetcode/0810.Chalkboard-XOR-Game/810. Chalkboard XOR Game.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+func xorGame(nums []int) bool {
+	if len(nums)%2 == 0 {
+		return true
+	}
+	xor := 0
+	for _, num := range nums {
+		xor ^= num
+	}
+	return xor == 0
+}
diff --git a/leetcode/0810.Chalkboard-XOR-Game/810. Chalkboard XOR Game_test.go b/leetcode/0810.Chalkboard-XOR-Game/810. Chalkboard XOR Game_test.go
new file mode 100644
index 000000000..5c84b5e79
--- /dev/null
+++ b/leetcode/0810.Chalkboard-XOR-Game/810. Chalkboard XOR Game_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question810 struct {
+	para810
+	ans810
+}
+
+// para 是参数
+// one 代表第一个参数
+type para810 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans810 struct {
+	one bool
+}
+
+func Test_Problem810(t *testing.T) {
+
+	qs := []question810{
+
+		{
+			para810{[]int{1, 1, 2}},
+			ans810{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 810------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans810, q.para810
+		fmt.Printf("【input】:%v       【output】:%v\n", p, xorGame(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0810.Chalkboard-XOR-Game/README.md b/leetcode/0810.Chalkboard-XOR-Game/README.md
new file mode 100644
index 000000000..e44fd391e
--- /dev/null
+++ b/leetcode/0810.Chalkboard-XOR-Game/README.md
@@ -0,0 +1,58 @@
+# [810. Chalkboard XOR Game](https://leetcode.com/problems/chalkboard-xor-game/)
+
+
+## 题目
+
+We are given non-negative integers nums[i] which are written on a chalkboard.  Alice and Bob take turns erasing exactly one number from the chalkboard, with Alice starting first.  If erasing a number causes the bitwise XOR of all the elements of the chalkboard to become 0, then that player loses.  (Also, we'll say the bitwise XOR of one element is that element itself, and the bitwise XOR of no elements is 0.)
+
+Also, if any player starts their turn with the bitwise XOR of all the elements of the chalkboard equal to 0, then that player wins.
+
+Return True if and only if Alice wins the game, assuming both players play optimally.
+
+```
+Example:Input: nums = [1, 1, 2]
+Output: false
+Explanation:
+Alice has two choices: erase 1 or erase 2.
+If she erases 1, the nums array becomes [1, 2]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 2 = 3. Now Bob can remove any element he wants, because Alice will be the one to erase the last element and she will lose.
+If Alice erases 2 first, now nums becomes [1, 1]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 1 = 0. Alice will lose.
+```
+
+**Notes:**
+
+- `1 <= N <= 1000`.
+- `0 <= nums[i] <= 2^16`.
+
+## 题目大意
+
+黑板上写着一个非负整数数组 nums[i] 。Alice 和 Bob 轮流从黑板上擦掉一个数字，Alice 先手。如果擦除一个数字后，剩余的所有数字按位异或运算得出的结果等于 0 的话，当前玩家游戏失败。 (另外，如果只剩一个数字，按位异或运算得到它本身；如果无数字剩余，按位异或运算结果为 0。）并且，轮到某个玩家时，如果当前黑板上所有数字按位异或运算结果等于 0，这个玩家获胜。假设两个玩家每步都使用最优解，当且仅当 Alice 获胜时返回 true。
+
+## 解题思路
+
+- Alice 必胜情况之一，Alice 先手，起始数组全部元素本身异或结果就为 0 。不需要擦除数字便自动获胜。除去这个情况，还有其他情况么？由于 2 人是交替擦除数字，且每次都恰好擦掉一个数字，因此对于这两人中的任意一人，其每次在擦除数字前，黑板上剩余数字的个数的奇偶性一定都是相同的。于是奇偶性成为突破口。
+- 如果 nums 的长度是偶数，Alice 先手是否必败呢？如果必败，代表无论擦掉哪一个数字，剩余所有数字的异或结果都等于 0。利用反证法证明上述结论是错误的。首先 $num[0] \oplus num[1] \oplus num[2] \oplus \cdots \oplus num[n-1] = X ≠ 0$，初始所有元素异或结果不为 0。假设 Alice 当前擦掉第 i 个元素，0 ≤ i < n。令 $X_{n}$ 代表擦掉第 n 位元素以后剩余元素异或的结果。由证题，无论擦掉哪一个数字，剩余所有数字的异或结果都等于 0。所以 $X_{0} \oplus X_{1} \oplus X_{2} \oplus \cdots  \oplus X_{n-1} = 0$。
+
+    $$\begin{aligned}0 &= X_{0} \oplus  X_{1} \oplus X_{2} \oplus \cdots  \oplus X_{n-1} \\0 &= (X \oplus nums[0]) \oplus (X \oplus nums[1]) \oplus (X \oplus nums[2]) \oplus \cdots  \oplus (X \oplus nums[n-1])\\ 0 &= (X \oplus X \oplus \cdots  \oplus X) \oplus (nums[0] \oplus nums[1] \oplus nums[2] \oplus \cdots  \oplus nums[n-1])\\0 &= 0 \oplus X\\\\\Rightarrow X &= 0\\\end{aligned}$$
+
+    由于 n 为偶数，所以 n 个 X 的异或结果为 0。最终推出 X = 0，很明显与前提 X ≠ 0 冲突。所以原命题，代表无论擦掉哪一个数字，剩余所有数字的异或结果都等于 0 是错误的。也就是说，当 n 为偶数时，代表无论擦掉哪一个数字，剩余所有数字的异或结果都不等于 0。即 Alice 有必胜策略。换句话说，当数组的长度是偶数时，先手 Alice 总能找到一个数字，在擦掉这个数字之后剩余的所有数字异或结果不等于 0。
+
+- 综上，Alice 必胜策略有 2 种情况：
+    1. 数组 nums 的全部元素初始本身异或结果就等于 0。
+    2. 数组 nums 的长度是偶数。
+
+## 代码
+
+```go
+package leetcode
+
+func xorGame(nums []int) bool {
+	if len(nums)%2 == 0 {
+		return true
+	}
+	xor := 0
+	for _, num := range nums {
+		xor ^= num
+	}
+	return xor == 0
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0811. Subdomain Visit Count/811. Subdomain Visit Count.go b/leetcode/0811.Subdomain-Visit-Count/811. Subdomain Visit Count.go
similarity index 100%
rename from Algorithms/0811. Subdomain Visit Count/811. Subdomain Visit Count.go
rename to leetcode/0811.Subdomain-Visit-Count/811. Subdomain Visit Count.go
diff --git a/Algorithms/0811. Subdomain Visit Count/811. Subdomain Visit Count_test.go b/leetcode/0811.Subdomain-Visit-Count/811. Subdomain Visit Count_test.go
similarity index 96%
rename from Algorithms/0811. Subdomain Visit Count/811. Subdomain Visit Count_test.go
rename to leetcode/0811.Subdomain-Visit-Count/811. Subdomain Visit Count_test.go
index 5176905ae..700e7d97f 100644
--- a/Algorithms/0811. Subdomain Visit Count/811. Subdomain Visit Count_test.go	
+++ b/leetcode/0811.Subdomain-Visit-Count/811. Subdomain Visit Count_test.go	
@@ -26,12 +26,12 @@ func Test_Problem811(t *testing.T) {
 
 	qs := []question811{
 
-		question811{
+		{
 			para811{[]string{"9001 discuss.leetcode.com"}},
 			ans811{[]string{"mqe", "mqE", "mQe", "mQE", "Mqe", "MqE", "MQe", "MQE"}},
 		},
 
-		question811{
+		{
 			para811{[]string{"900 google.mail.com", "50 yahoo.com", "1 intel.mail.com", "5 wiki.org"}},
 			ans811{[]string{"901 mail.com", "50 yahoo.com", "900 google.mail.com", "5 wiki.org", "5 org", "1 intel.mail.com", "951 com"}},
 		},
diff --git a/leetcode/0811.Subdomain-Visit-Count/README.md b/leetcode/0811.Subdomain-Visit-Count/README.md
new file mode 100755
index 000000000..6348dfc58
--- /dev/null
+++ b/leetcode/0811.Subdomain-Visit-Count/README.md
@@ -0,0 +1,49 @@
+# [811. Subdomain Visit Count](https://leetcode.com/problems/subdomain-visit-count/)
+
+
+## 题目
+
+A website domain like "discuss.leetcode.com" consists of various subdomains. At the top level, we have "com", at the next level, we have "leetcode.com", and at the lowest level, "discuss.leetcode.com". When we visit a domain like "discuss.leetcode.com", we will also visit the parent domains "leetcode.com" and "com" implicitly.
+
+Now, call a "count-paired domain" to be a count (representing the number of visits this domain received), followed by a space, followed by the address. An example of a count-paired domain might be "9001 discuss.leetcode.com".
+
+We are given a list `cpdomains` of count-paired domains. We would like a list of count-paired domains, (in the same format as the input, and in any order), that explicitly counts the number of visits to each subdomain.
+
+    Example 1:
+    Input: 
+    ["9001 discuss.leetcode.com"]
+    Output: 
+    ["9001 discuss.leetcode.com", "9001 leetcode.com", "9001 com"]
+    Explanation: 
+    We only have one website domain: "discuss.leetcode.com". As discussed above, the subdomain "leetcode.com" and "com" will also be visited. So they will all be visited 9001 times.
+
+    Example 2:
+    Input: 
+    ["900 google.mail.com", "50 yahoo.com", "1 intel.mail.com", "5 wiki.org"]
+    Output: 
+    ["901 mail.com","50 yahoo.com","900 google.mail.com","5 wiki.org","5 org","1 intel.mail.com","951 com"]
+    Explanation: 
+    We will visit "google.mail.com" 900 times, "yahoo.com" 50 times, "intel.mail.com" once and "wiki.org" 5 times. For the subdomains, we will visit "mail.com" 900 + 1 = 901 times, "com" 900 + 50 + 1 = 951 times, and "org" 5 times.
+
+**Notes:**
+
+- The length of `cpdomains` will not exceed `100`.
+- The length of each domain name will not exceed `100`.
+- Each address will have either 1 or 2 "." characters.
+- The input count in any count-paired domain will not exceed `10000`.
+- The answer output can be returned in any order.
+
+
+## 题目大意
+
+
+一个网站域名，如 "discuss.leetcode.com"，包含了多个子域名。作为顶级域名，常用的有 "com"，下一级则有 "leetcode.com"，最低的一级为 "discuss.leetcode.com"。当我们访问域名 "discuss.leetcode.com" 时，也同时访问了其父域名 "leetcode.com" 以及顶级域名 "com"。给定一个带访问次数和域名的组合，要求分别计算每个域名被访问的次数。其格式为访问次数+空格+地址，例如："9001 discuss.leetcode.com"。
+
+接下来会给出一组访问次数和域名组合的列表 cpdomains 。要求解析出所有域名的访问次数，输出格式和输入格式相同，不限定先后顺序。
+
+
+
+## 解题思路
+
+
+- 这一题是简单题，统计每个 domain 的出现频次。每个域名根据层级，一级一级的累加频次，比如 `discuss.leetcode.com`、`discuss.leetcode.com` 这个域名频次为 1，`leetcode.com` 这个域名频次为 1，`com` 这个域名频次为 1。用 map 依次统计每个 domain 出现的频次，按照格式要求输出。
diff --git a/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area.go b/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area.go
new file mode 100644
index 000000000..918725245
--- /dev/null
+++ b/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area.go	
@@ -0,0 +1,31 @@
+package leetcode
+
+func largestTriangleArea(points [][]int) float64 {
+	maxArea, n := 0.0, len(points)
+	for i := 0; i < n; i++ {
+		for j := i + 1; j < n; j++ {
+			for k := j + 1; k < n; k++ {
+				maxArea = max(maxArea, area(points[i], points[j], points[k]))
+			}
+		}
+	}
+	return maxArea
+}
+
+func area(p1, p2, p3 []int) float64 {
+	return abs(p1[0]*p2[1]+p2[0]*p3[1]+p3[0]*p1[1]-p1[0]*p3[1]-p2[0]*p1[1]-p3[0]*p2[1]) / 2
+}
+
+func abs(num int) float64 {
+	if num < 0 {
+		num = -num
+	}
+	return float64(num)
+}
+
+func max(a, b float64) float64 {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area_test.go b/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area_test.go
new file mode 100644
index 000000000..140980b0f
--- /dev/null
+++ b/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question812 struct {
+	para812
+	ans812
+}
+
+// para 是参数
+// one 代表第一个参数
+type para812 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans812 struct {
+	one float64
+}
+
+func Test_Problem812(t *testing.T) {
+
+	qs := []question812{
+
+		{
+			para812{[][]int{{0, 0}, {0, 1}, {1, 0}, {0, 2}, {2, 0}}},
+			ans812{2.0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 812------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans812, q.para812
+		fmt.Printf("【input】:%v       【output】:%v\n", p, largestTriangleArea(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0812.Largest-Triangle-Area/README.md b/leetcode/0812.Largest-Triangle-Area/README.md
new file mode 100644
index 000000000..426826f15
--- /dev/null
+++ b/leetcode/0812.Largest-Triangle-Area/README.md
@@ -0,0 +1,70 @@
+# [812. Largest Triangle Area](https://leetcode.com/problems/largest-triangle-area/)
+
+
+## 题目
+
+You have a list of points in the plane. Return the area of the largest triangle that can be formed by any 3 of the points.
+
+```
+Example:
+Input: points = [[0,0],[0,1],[1,0],[0,2],[2,0]]
+Output: 2
+Explanation: 
+The five points are show in the figure below. The red triangle is the largest.
+```
+
+![https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/04/1027.png](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/04/1027.png)
+
+**Notes**:
+
+- `3 <= points.length <= 50`.
+- No points will be duplicated.
+- `-50 <= points[i][j] <= 50`.
+- Answers within `10^-6` of the true value will be accepted as correct.
+
+## 题目大意
+
+给定包含多个点的集合，从其中取三个点组成三角形，返回能组成的最大三角形的面积。
+
+## 解题思路
+
+- 给出一组点的坐标，要求找出能组成三角形面积最大的点集合，输出这个最大面积。
+- 数学题。按照数学定义，分别计算这些能构成三角形的点形成的三角形面积，最终输出最大面积即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func largestTriangleArea(points [][]int) float64 {
+	maxArea, n := 0.0, len(points)
+	for i := 0; i < n; i++ {
+		for j := i + 1; j < n; j++ {
+			for k := j + 1; k < n; k++ {
+				maxArea = max(maxArea, area(points[i], points[j], points[k]))
+			}
+		}
+	}
+	return maxArea
+}
+
+func area(p1, p2, p3 []int) float64 {
+	return abs(p1[0]*p2[1]+p2[0]*p3[1]+p3[0]*p1[1]-p1[0]*p3[1]-p2[0]*p1[1]-p3[0]*p2[1]) / 2
+}
+
+func abs(num int) float64 {
+	if num < 0 {
+		num = -num
+	}
+	return float64(num)
+}
+
+func max(a, b float64) float64 {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0815. Bus Routes/815. Bus Routes.go b/leetcode/0815.Bus-Routes/815. Bus Routes.go
similarity index 100%
rename from Algorithms/0815. Bus Routes/815. Bus Routes.go
rename to leetcode/0815.Bus-Routes/815. Bus Routes.go
diff --git a/Algorithms/0815. Bus Routes/815. Bus Routes_test.go b/leetcode/0815.Bus-Routes/815. Bus Routes_test.go
similarity index 89%
rename from Algorithms/0815. Bus Routes/815. Bus Routes_test.go
rename to leetcode/0815.Bus-Routes/815. Bus Routes_test.go
index 271d13701..5581144eb 100644
--- a/Algorithms/0815. Bus Routes/815. Bus Routes_test.go	
+++ b/leetcode/0815.Bus-Routes/815. Bus Routes_test.go	
@@ -28,8 +28,8 @@ func Test_Problem815(t *testing.T) {
 
 	qs := []question815{
 
-		question815{
-			para815{[][]int{[]int{1, 2, 7}, []int{3, 6, 7}}, 1, 6},
+		{
+			para815{[][]int{{1, 2, 7}, {3, 6, 7}}, 1, 6},
 			ans815{2},
 		},
 	}
diff --git a/leetcode/0815.Bus-Routes/README.md b/leetcode/0815.Bus-Routes/README.md
new file mode 100755
index 000000000..2fd6e0738
--- /dev/null
+++ b/leetcode/0815.Bus-Routes/README.md
@@ -0,0 +1,41 @@
+# [815. Bus Routes](https://leetcode.com/problems/bus-routes/)
+
+
+## 题目
+
+We have a list of bus routes. Each `routes[i]` is a bus route that the i-th bus repeats forever. For example if `routes[0] = [1, 5, 7]`, this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->... forever.
+
+We start at bus stop `S` (initially not on a bus), and we want to go to bus stop `T`. Travelling by buses only, what is the least number of buses we must take to reach our destination? Return -1 if it is not possible.
+
+    Example:
+    Input: 
+    routes = [[1, 2, 7], [3, 6, 7]]
+    S = 1
+    T = 6
+    Output: 2
+    Explanation: 
+    The best strategy is take the first bus to the bus stop 7, then take the second bus to the bus stop 6.
+
+**Note:**
+
+- `1 <= routes.length <= 500`.
+- `1 <= routes[i].length <= 500`.
+- `0 <= routes[i][j] < 10 ^ 6`.
+
+
+## 题目大意
+
+我们有一系列公交路线。每一条路线 routes[i] 上都有一辆公交车在上面循环行驶。例如，有一条路线 routes[0] = [1, 5, 7]，表示第一辆 (下标为0) 公交车会一直按照 1->5->7->1->5->7->1->... 的车站路线行驶。假设我们从 S 车站开始（初始时不在公交车上），要去往 T 站。 期间仅可乘坐公交车，求出最少乘坐的公交车数量。返回 -1 表示不可能到达终点车站。
+
+
+说明:
+
+- 1 <= routes.length <= 500.
+- 1 <= routes[i].length <= 500.
+- 0 <= routes[i][j] < 10 ^ 6.
+
+
+## 解题思路
+
+- 给出一些公交路线，公交路径代表经过的哪些站。现在给出起点和终点站，问最少需要换多少辆公交车才能从起点到终点？
+- 这一题可以转换成图论的问题，将每个站台看成顶点，公交路径看成每个顶点的边。同一个公交的边染色相同。题目即可转化为从顶点 S 到顶点 T 需要经过最少多少条不同的染色边。用 BFS 即可轻松解决。从起点 S 开始，不断的扩展它能到达的站点。用 visited 数组防止放入已经可达的站点引起的环。用 map 存储站点和公交车的映射关系(即某个站点可以由哪些公交车到达)，BFS 的过程中可以用这个映射关系，拿到公交车的其他站点信息，从而扩张队列里面的可达站点。一旦扩展出现了终点 T，就可以返回结果了。
diff --git a/leetcode/0816.Ambiguous-Coordinates/816. Ambiguous Coordinates.go b/leetcode/0816.Ambiguous-Coordinates/816. Ambiguous Coordinates.go
new file mode 100644
index 000000000..c509719ad
--- /dev/null
+++ b/leetcode/0816.Ambiguous-Coordinates/816. Ambiguous Coordinates.go	
@@ -0,0 +1,36 @@
+package leetcode
+
+func ambiguousCoordinates(s string) []string {
+	res := []string{}
+	s = s[1 : len(s)-1]
+	for i := range s[:len(s)-1] {
+		a := build(s[:i+1])
+		b := build(s[i+1:])
+		for _, ta := range a {
+			for _, tb := range b {
+				res = append(res, "("+ta+", "+tb+")")
+			}
+		}
+	}
+	return res
+}
+
+func build(s string) []string {
+	res := []string{}
+	if len(s) == 1 || s[0] != '0' {
+		res = append(res, s)
+	}
+	// 结尾带 0 的情况
+	if s[len(s)-1] == '0' {
+		return res
+	}
+	// 切分长度大于一位且带前导 0 的情况
+	if s[0] == '0' {
+		res = append(res, "0."+s[1:])
+		return res
+	}
+	for i := range s[:len(s)-1] {
+		res = append(res, s[:i+1]+"."+s[i+1:])
+	}
+	return res
+}
diff --git a/leetcode/0816.Ambiguous-Coordinates/816. Ambiguous Coordinates_test.go b/leetcode/0816.Ambiguous-Coordinates/816. Ambiguous Coordinates_test.go
new file mode 100644
index 000000000..9764422d4
--- /dev/null
+++ b/leetcode/0816.Ambiguous-Coordinates/816. Ambiguous Coordinates_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question816 struct {
+	para816
+	ans816
+}
+
+// para 是参数
+// one 代表第一个参数
+type para816 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans816 struct {
+	one []string
+}
+
+func Test_Problem816(t *testing.T) {
+
+	qs := []question816{
+
+		{
+			para816{"(120123)"},
+			ans816{[]string{"(1, 20123)", " (1, 2.0123)", " (1, 20.123)", " (1, 201.23)", " (1, 2012.3)", " (12, 0.123)", " (1.2, 0.123)", " (120, 123)", " (120, 1.23)", " (120, 12.3)", " (1201, 23) ", "(1201, 2.3)", " (1.201, 23)", " (1.201, 2.3) ", "(12.01, 23)", " (12.01, 2.3) ", "(120.1, 23)", " (120.1, 2.3) ", "(12012, 3)", " (1.2012, 3)", " (12.012, 3)", " (120.12, 3)", " (1201.2, 3)"}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 816------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans816, q.para816
+		fmt.Printf("【input】:%v       【output】:%v\n", p, ambiguousCoordinates(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0816.Ambiguous-Coordinates/README.md b/leetcode/0816.Ambiguous-Coordinates/README.md
new file mode 100644
index 000000000..c1610fdec
--- /dev/null
+++ b/leetcode/0816.Ambiguous-Coordinates/README.md
@@ -0,0 +1,95 @@
+# [816. Ambiguous Coordinates](https://leetcode.com/problems/ambiguous-coordinates/)
+
+
+## 题目
+
+We had some 2-dimensional coordinates, like `"(1, 3)"` or `"(2, 0.5)"`.  Then, we removed all commas, decimal points, and spaces, and ended up with the string `s`.  Return a list of strings representing all possibilities for what our original coordinates could have been.
+
+Our original representation never had extraneous zeroes, so we never started with numbers like "00", "0.0", "0.00", "1.0", "001", "00.01", or any other number that can be represented with less digits.  Also, a decimal point within a number never occurs without at least one digit occuring before it, so we never started with numbers like ".1".
+
+The final answer list can be returned in any order.  Also note that all coordinates in the final answer have exactly one space between them (occurring after the comma.)
+
+```
+Example 1:Input: s = "(123)"
+Output: ["(1, 23)", "(12, 3)", "(1.2, 3)", "(1, 2.3)"]
+
+```
+
+```
+Example 2:Input: s = "(00011)"
+Output:  ["(0.001, 1)", "(0, 0.011)"]
+Explanation:
+0.0, 00, 0001 or 00.01 are not allowed.
+
+```
+
+```
+Example 3:Input: s = "(0123)"
+Output: ["(0, 123)", "(0, 12.3)", "(0, 1.23)", "(0.1, 23)", "(0.1, 2.3)", "(0.12, 3)"]
+
+```
+
+```
+Example 4:Input: s = "(100)"
+Output: [(10, 0)]
+Explanation:
+1.0 is not allowed.
+
+```
+
+**Note:**
+
+- `4 <= s.length <= 12`.
+- `s[0]` = "(", `s[s.length - 1]` = ")", and the other elements in `s` are digits.
+
+## 题目大意
+
+我们有一些二维坐标，如 "(1, 3)" 或 "(2, 0.5)"，然后我们移除所有逗号，小数点和空格，得到一个字符串S。返回所有可能的原始字符串到一个列表中。原始的坐标表示法不会存在多余的零，所以不会出现类似于"00", "0.0", "0.00", "1.0", "001", "00.01"或一些其他更小的数来表示坐标。此外，一个小数点前至少存在一个数，所以也不会出现“.1”形式的数字。
+
+最后返回的列表可以是任意顺序的。而且注意返回的两个数字中间（逗号之后）都有一个空格。
+
+## 解题思路
+
+- 本题没有什么算法思想，纯暴力题。先将原始字符串一分为二，分为的两个子字符串再移动坐标点，最后将每种情况组合再一次，这算完成了一次切分。将原始字符串每一位都按此规律完成切分，此题便得解。
+- 这道题有 2 处需要注意的。第一处是最终输出的字符串，请注意，**两个数字中间（逗号之后）都有一个空格**。不遵守输出格式的要求也会导致 `Wrong Answer`。另外一处是切分数字时，有 2 种违法情况，一种是带前导 0 的，另外一种是末尾带 0 的。带前导 0 的也分为 2 种情况，一种是只有一位，即只有一个 0，这种情况直接返回，因为这一个 0 怎么切分也只有一种切分方法。另外一种是长度大于 1，即 `0xxx` 这种情况。`0xxx` 这种情况只有一种切分方法，即 `0.xxx`。末尾带 0 的只有一种切分方法，即 `xxx0`，不可切分，因为 `xxx.0`，`xx.x0`，`x.xx0` 这些都是违法情况，所以末尾带 0 的也可以直接返回。具体的实现见代码和注释。
+
+## 代码
+
+```go
+package leetcode
+
+func ambiguousCoordinates(s string) []string {
+	res := []string{}
+	s = s[1 : len(s)-1]
+	for i := range s[:len(s)-1] {
+		a := build(s[:i+1])
+		b := build(s[i+1:])
+		for _, ta := range a {
+			for _, tb := range b {
+				res = append(res, "("+ta+", "+tb+")")
+			}
+		}
+	}
+	return res
+}
+
+func build(s string) []string {
+	res := []string{}
+	if len(s) == 1 || s[0] != '0' {
+		res = append(res, s)
+	}
+	// 结尾带 0 的情况
+	if s[len(s)-1] == '0' {
+		return res
+	}
+	// 切分长度大于一位且带前导 0 的情况
+	if s[0] == '0' {
+		res = append(res, "0."+s[1:])
+		return res
+	}
+	for i := range s[:len(s)-1] {
+		res = append(res, s[:i+1]+"."+s[i+1:])
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0817. Linked List Components/817. Linked List Components.go b/leetcode/0817.Linked-List-Components/817. Linked List Components.go
similarity index 84%
rename from Algorithms/0817. Linked List Components/817. Linked List Components.go
rename to leetcode/0817.Linked-List-Components/817. Linked List Components.go
index b4b726193..66592ae2c 100644
--- a/Algorithms/0817. Linked List Components/817. Linked List Components.go	
+++ b/leetcode/0817.Linked-List-Components/817. Linked List Components.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0817. Linked List Components/817. Linked List Components_test.go b/leetcode/0817.Linked-List-Components/817. Linked List Components_test.go
similarity index 89%
rename from Algorithms/0817. Linked List Components/817. Linked List Components_test.go
rename to leetcode/0817.Linked-List-Components/817. Linked List Components_test.go
index 4ebedd2c4..e1d351587 100644
--- a/Algorithms/0817. Linked List Components/817. Linked List Components_test.go	
+++ b/leetcode/0817.Linked-List-Components/817. Linked List Components_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question817 struct {
@@ -27,37 +29,37 @@ func Test_Problem817(t *testing.T) {
 
 	qs := []question817{
 
-		question817{
+		{
 			para817{[]int{0, 1, 2, 3}, []int{0, 1, 3}},
 			ans817{2},
 		},
 
-		question817{
+		{
 			para817{[]int{1, 2, 3, 4}, []int{1, 2, 3, 4}},
 			ans817{1},
 		},
 
-		question817{
+		{
 			para817{[]int{0, 1, 2, 3, 4}, []int{0, 3, 1, 4}},
 			ans817{2},
 		},
 
-		question817{
+		{
 			para817{[]int{0, 1, 2}, []int{0, 2}},
 			ans817{2},
 		},
 
-		question817{
+		{
 			para817{[]int{1, 2, 0, 4, 3}, []int{3, 4, 0, 2, 1}},
 			ans817{1},
 		},
 
-		question817{
+		{
 			para817{[]int{0, 2, 4, 3, 1}, []int{3, 2, 4}},
 			ans817{1},
 		},
 
-		question817{
+		{
 			para817{[]int{7, 5, 13, 3, 16, 11, 12, 0, 17, 1, 4, 15, 6, 14, 2, 19, 9, 10, 8, 18}, []int{8, 10, 3, 11, 17, 16, 7, 9, 5, 15, 13, 6}},
 			ans817{4},
 		},
@@ -67,7 +69,7 @@ func Test_Problem817(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans817, q.para817
-		fmt.Printf("【input】:%v       【output】:%v\n", p, numComponents(S2l(p.one), p.another))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numComponents(structures.Ints2List(p.one), p.another))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0817. Linked List Components/README.md b/leetcode/0817.Linked-List-Components/README.md
similarity index 100%
rename from Algorithms/0817. Linked List Components/README.md
rename to leetcode/0817.Linked-List-Components/README.md
diff --git a/Algorithms/0819. Most Common Word/819. Most Common Word.go b/leetcode/0819.Most-Common-Word/819. Most Common Word.go
similarity index 100%
rename from Algorithms/0819. Most Common Word/819. Most Common Word.go
rename to leetcode/0819.Most-Common-Word/819. Most Common Word.go
diff --git a/Algorithms/0819. Most Common Word/819. Most Common Word_test.go b/leetcode/0819.Most-Common-Word/819. Most Common Word_test.go
similarity index 97%
rename from Algorithms/0819. Most Common Word/819. Most Common Word_test.go
rename to leetcode/0819.Most-Common-Word/819. Most Common Word_test.go
index b94788106..c0e8265a9 100644
--- a/Algorithms/0819. Most Common Word/819. Most Common Word_test.go	
+++ b/leetcode/0819.Most-Common-Word/819. Most Common Word_test.go	
@@ -27,7 +27,7 @@ func Test_Problem819(t *testing.T) {
 
 	qs := []question819{
 
-		question819{
+		{
 			para819{"Bob hit a ball, the hit BALL flew far after it was hit.", []string{"hit"}},
 			ans819{"ball"},
 		},
diff --git a/leetcode/0819.Most-Common-Word/README.md b/leetcode/0819.Most-Common-Word/README.md
new file mode 100755
index 000000000..2d4b3ad5e
--- /dev/null
+++ b/leetcode/0819.Most-Common-Word/README.md
@@ -0,0 +1,44 @@
+# [819. Most Common Word](https://leetcode.com/problems/most-common-word/)
+
+
+## 题目
+
+Given a paragraph and a list of banned words, return the most frequent word that is not in the list of banned words. It is guaranteed there is at least one word that isn't banned, and that the answer is unique.
+
+Words in the list of banned words are given in lowercase, and free of punctuation. Words in the paragraph are not case sensitive. The answer is in lowercase.
+
+**Example:**
+
+    Input: 
+    paragraph = "Bob hit a ball, the hit BALL flew far after it was hit."
+    banned = ["hit"]
+    Output: "ball"
+    Explanation: 
+    "hit" occurs 3 times, but it is a banned word.
+    "ball" occurs twice (and no other word does), so it is the most frequent non-banned word in the paragraph. 
+    Note that words in the paragraph are not case sensitive,
+    that punctuation is ignored (even if adjacent to words, such as "ball,"), 
+    and that "hit" isn't the answer even though it occurs more because it is banned.
+
+**Note:**
+
+- `1 <= paragraph.length <= 1000`.
+- `0 <= banned.length <= 100`.
+- `1 <= banned[i].length <= 10`.
+- The answer is unique, and written in lowercase (even if its occurrences in `paragraph` may have uppercase symbols, and even if it is a proper noun.)
+- `paragraph` only consists of letters, spaces, or the punctuation symbols `!?',;.`
+- There are no hyphens or hyphenated words.
+- Words only consist of letters, never apostrophes or other punctuation symbols.
+
+
+## 题目大意
+
+
+给定一个段落 (paragraph) 和一个禁用单词列表 (banned)。返回出现次数最多，同时不在禁用列表中的单词。题目保证至少有一个词不在禁用列表中，而且答案唯一。
+
+禁用列表中的单词用小写字母表示，不含标点符号。段落中的单词不区分大小写。答案都是小写字母。
+
+
+## 解题思路
+
+- 给出一段话和一个 banned 的字符串，要求输出这段话中出现频次最高的并且不出现在 banned 数组里面的字符串，答案唯一。这题是简单题，依次统计每个单词的频次，然后从 map 中删除 banned 里面的单词，取出剩下频次最高的单词即可。
diff --git a/leetcode/0820.Short-Encoding-of-Words/820. Short Encoding of Words.go b/leetcode/0820.Short-Encoding-of-Words/820. Short Encoding of Words.go
new file mode 100644
index 000000000..c88ee6790
--- /dev/null
+++ b/leetcode/0820.Short-Encoding-of-Words/820. Short Encoding of Words.go	
@@ -0,0 +1,86 @@
+package leetcode
+
+// 解法一 暴力
+func minimumLengthEncoding(words []string) int {
+	res, m := 0, map[string]bool{}
+	for _, w := range words {
+		m[w] = true
+	}
+	for w := range m {
+		for i := 1; i < len(w); i++ {
+			delete(m, w[i:])
+		}
+	}
+	for w := range m {
+		res += len(w) + 1
+	}
+	return res
+}
+
+// 解法二 Trie
+type node struct {
+	value byte
+	sub   []*node
+}
+
+func (t *node) has(b byte) (*node, bool) {
+	if t == nil {
+		return nil, false
+	}
+	for i := range t.sub {
+		if t.sub[i] != nil && t.sub[i].value == b {
+			return t.sub[i], true
+		}
+	}
+	return nil, false
+}
+
+func (t *node) isLeaf() bool {
+	if t == nil {
+		return false
+	}
+	return len(t.sub) == 0
+}
+
+func (t *node) add(s []byte) {
+	now := t
+	for i := len(s) - 1; i > -1; i-- {
+		if v, ok := now.has(s[i]); ok {
+			now = v
+			continue
+		}
+		temp := new(node)
+		temp.value = s[i]
+		now.sub = append(now.sub, temp)
+		now = temp
+	}
+}
+
+func (t *node) endNodeOf(s []byte) *node {
+	now := t
+	for i := len(s) - 1; i > -1; i-- {
+		if v, ok := now.has(s[i]); ok {
+			now = v
+			continue
+		}
+		return nil
+	}
+	return now
+}
+
+func minimumLengthEncoding1(words []string) int {
+	res, tree, m := 0, new(node), make(map[string]bool)
+	for i := range words {
+		if !m[words[i]] {
+			tree.add([]byte(words[i]))
+			m[words[i]] = true
+		}
+	}
+	for s := range m {
+		if tree.endNodeOf([]byte(s)).isLeaf() {
+			res += len(s)
+			res++
+		}
+	}
+	return res
+}
diff --git a/leetcode/0820.Short-Encoding-of-Words/820. Short Encoding of Words_test.go b/leetcode/0820.Short-Encoding-of-Words/820. Short Encoding of Words_test.go
new file mode 100644
index 000000000..f7f5a0579
--- /dev/null
+++ b/leetcode/0820.Short-Encoding-of-Words/820. Short Encoding of Words_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question820 struct {
+	para820
+	ans820
+}
+
+// para 是参数
+// one 代表第一个参数
+type para820 struct {
+	words []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans820 struct {
+	one int
+}
+
+func Test_Problem820(t *testing.T) {
+
+	qs := []question820{
+
+		{
+			para820{[]string{"time", "me", "bell"}},
+			ans820{10},
+		},
+
+		{
+			para820{[]string{"t"}},
+			ans820{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 820------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans820, q.para820
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minimumLengthEncoding(p.words))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0820.Short-Encoding-of-Words/README.md b/leetcode/0820.Short-Encoding-of-Words/README.md
new file mode 100644
index 000000000..60a4ddcaf
--- /dev/null
+++ b/leetcode/0820.Short-Encoding-of-Words/README.md
@@ -0,0 +1,143 @@
+# [820. Short Encoding of Words](https://leetcode.com/problems/short-encoding-of-words/)
+
+
+## 题目
+
+A **valid encoding** of an array of `words` is any reference string `s` and array of indices `indices` such that:
+
+- `words.length == indices.length`
+- The reference string `s` ends with the `'#'` character.
+- For each index `indices[i]`, the **substring** of `s` starting from `indices[i]` and up to (but not including) the next `'#'` character is equal to `words[i]`.
+
+Given an array of `words`, return *the **length of the shortest reference string*** `s` *possible of any **valid encoding** of* `words`*.*
+
+**Example 1:**
+
+```
+Input: words = ["time", "me", "bell"]
+Output: 10
+Explanation: A valid encoding would be s = "time#bell#" and indices = [0, 2, 5].
+words[0] = "time", the substring of s starting from indices[0] = 0 to the next '#' is underlined in "time#bell#"
+words[1] = "me", the substring of s starting from indices[1] = 2 to the next '#' is underlined in "time#bell#"
+words[2] = "bell", the substring of s starting from indices[2] = 5 to the next '#' is underlined in "time#bell#"
+```
+
+**Example 2:**
+
+```
+Input: words = ["t"]
+Output: 2
+Explanation: A valid encoding would be s = "t#" and indices = [0].
+```
+
+**Constraints:**
+
+- `1 <= words.length <= 2000`
+- `1 <= words[i].length <= 7`
+- `words[i]` consists of only lowercase letters.
+
+## 题目大意
+
+单词数组 words 的 有效编码 由任意助记字符串 s 和下标数组 indices 组成，且满足：
+
+- words.length == indices.length
+- 助记字符串 s 以 '#' 字符结尾
+- 对于每个下标 indices[i] ，s 的一个从 indices[i] 开始、到下一个 '#' 字符结束（但不包括 '#'）的 子字符串 恰好与 words[i] 相等
+
+给你一个单词数组 words ，返回成功对 words 进行编码的最小助记字符串 s 的长度 。
+
+## 解题思路
+
+- 暴力解法。先将所有的单词放入字典中。然后针对字典中的每个单词，逐一从字典中删掉自己的子字符串，这样有相同后缀的字符串被删除了，字典中剩下的都是没有共同前缀的。最终的答案是剩下所有单词用 # 号连接之后的总长度。
+- Trie 解法。构建 Trie 树，相同的后缀会被放到从根到叶子节点中的某个路径中。最后依次遍历一遍所有单词，如果单词最后一个字母是叶子节点，说明这个单词是要选择的，因为它可能是包含了一些单词后缀的最长单词。累加这个单词的长度并再加 1(# 字符的长度)。最终累加出来的长度即为题目所求的答案。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 暴力
+func minimumLengthEncoding(words []string) int {
+	res, m := 0, map[string]bool{}
+	for _, w := range words {
+		m[w] = true
+	}
+	for w := range m {
+		for i := 1; i < len(w); i++ {
+			delete(m, w[i:])
+		}
+	}
+	for w := range m {
+		res += len(w) + 1
+	}
+	return res
+}
+
+// 解法二 Trie
+type node struct {
+	value byte
+	sub   []*node
+}
+
+func (t *node) has(b byte) (*node, bool) {
+	if t == nil {
+		return nil, false
+	}
+	for i := range t.sub {
+		if t.sub[i] != nil && t.sub[i].value == b {
+			return t.sub[i], true
+		}
+	}
+	return nil, false
+}
+
+func (t *node) isLeaf() bool {
+	if t == nil {
+		return false
+	}
+	return len(t.sub) == 0
+}
+
+func (t *node) add(s []byte) {
+	now := t
+	for i := len(s) - 1; i > -1; i-- {
+		if v, ok := now.has(s[i]); ok {
+			now = v
+			continue
+		}
+		temp := new(node)
+		temp.value = s[i]
+		now.sub = append(now.sub, temp)
+		now = temp
+	}
+}
+
+func (t *node) endNodeOf(s []byte) *node {
+	now := t
+	for i := len(s) - 1; i > -1; i-- {
+		if v, ok := now.has(s[i]); ok {
+			now = v
+			continue
+		}
+		return nil
+	}
+	return now
+}
+
+func minimumLengthEncoding1(words []string) int {
+	res, tree, m := 0, new(node), make(map[string]bool)
+	for i := range words {
+		if !m[words[i]] {
+			tree.add([]byte(words[i]))
+			m[words[i]] = true
+		}
+	}
+	for s := range m {
+		if tree.endNodeOf([]byte(s)).isLeaf() {
+			res += len(s)
+			res++
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0821.Shortest-Distance-to-a-Character/821. Shortest Distance to a Character.go b/leetcode/0821.Shortest-Distance-to-a-Character/821. Shortest Distance to a Character.go
new file mode 100644
index 000000000..88ea9ce28
--- /dev/null
+++ b/leetcode/0821.Shortest-Distance-to-a-Character/821. Shortest Distance to a Character.go	
@@ -0,0 +1,60 @@
+package leetcode
+
+import (
+	"math"
+)
+
+// 解法一
+func shortestToChar(s string, c byte) []int {
+	n := len(s)
+	res := make([]int, n)
+	for i := range res {
+		res[i] = n
+	}
+	for i := 0; i < n; i++ {
+		if s[i] == c {
+			res[i] = 0
+		} else if i > 0 {
+			res[i] = res[i-1] + 1
+		}
+	}
+	for i := n - 1; i >= 0; i-- {
+		if i < n-1 && res[i+1]+1 < res[i] {
+			res[i] = res[i+1] + 1
+		}
+	}
+	return res
+}
+
+// 解法二
+func shortestToChar1(s string, c byte) []int {
+	res := make([]int, len(s))
+	for i := 0; i < len(s); i++ {
+		if s[i] == c {
+			res[i] = 0
+		} else {
+			left, right := math.MaxInt32, math.MaxInt32
+			for j := i + 1; j < len(s); j++ {
+				if s[j] == c {
+					right = j - i
+					break
+				}
+			}
+			for k := i - 1; k >= 0; k-- {
+				if s[k] == c {
+					left = i - k
+					break
+				}
+			}
+			res[i] = min(left, right)
+		}
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/0821.Shortest-Distance-to-a-Character/821. Shortest Distance to a Character_test.go b/leetcode/0821.Shortest-Distance-to-a-Character/821. Shortest Distance to a Character_test.go
new file mode 100644
index 000000000..0420139a0
--- /dev/null
+++ b/leetcode/0821.Shortest-Distance-to-a-Character/821. Shortest Distance to a Character_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question821 struct {
+	para821
+	ans821
+}
+
+// para 是参数
+// one 代表第一个参数
+type para821 struct {
+	s string
+	c byte
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans821 struct {
+	one []int
+}
+
+func Test_Problem821(t *testing.T) {
+
+	qs := []question821{
+
+		{
+			para821{"loveleetcode", 'e'},
+			ans821{[]int{3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0}},
+		},
+
+		{
+			para821{"baaa", 'b'},
+			ans821{[]int{0, 1, 2, 3}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 821------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans821, q.para821
+		fmt.Printf("【input】:%v       【output】:%v\n", p, shortestToChar(p.s, p.c))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0821.Shortest-Distance-to-a-Character/README.md b/leetcode/0821.Shortest-Distance-to-a-Character/README.md
new file mode 100644
index 000000000..48721fa20
--- /dev/null
+++ b/leetcode/0821.Shortest-Distance-to-a-Character/README.md
@@ -0,0 +1,99 @@
+# [821. Shortest Distance to a Character](https://leetcode.com/problems/shortest-distance-to-a-character/)
+
+## 题目
+
+Given a string `s` and a character `c` that occurs in `s`, return *an array of integers `answer` where* `answer.length == s.length` *and* `answer[i]` *is the shortest distance from* `s[i]` *to the character* `c` *in* `s`.
+
+**Example 1:**
+
+```
+Input: s = "loveleetcode", c = "e"
+Output: [3,2,1,0,1,0,0,1,2,2,1,0]
+```
+
+**Example 2:**
+
+```
+Input: s = "aaab", c = "b"
+Output: [3,2,1,0]
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 104`
+- `s[i]` and `c` are lowercase English letters.
+- `c` occurs at least once in `s`.
+
+## 题目大意
+
+给定一个字符串 S 和一个字符 C。返回一个代表字符串 S 中每个字符到字符串 S 中的字符 C 的最短距离的数组。
+
+## 解题思路
+
+- 解法一：从左至右更新一遍到 C 的值距离，再从右至左更新一遍到 C 的值，取两者中的最小值。
+- 解法二：依次扫描字符串 S，针对每一个非字符 C 的字符，分别往左扫一次，往右扫一次，计算出距离目标字符 C 的距离，然后取左右两个距离的最小值存入最终答案数组中。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+)
+
+// 解法一
+func shortestToChar(s string, c byte) []int {
+	n := len(s)
+	res := make([]int, n)
+	for i := range res {
+		res[i] = n
+	}
+	for i := 0; i < n; i++ {
+		if s[i] == c {
+			res[i] = 0
+		} else if i > 0 {
+			res[i] = res[i-1] + 1
+		}
+	}
+	for i := n - 1; i >= 0; i-- {
+		if i < n-1 && res[i+1]+1 < res[i] {
+			res[i] = res[i+1] + 1
+		}
+	}
+	return res
+}
+
+// 解法二
+func shortestToChar1(s string, c byte) []int {
+	res := make([]int, len(s))
+	for i := 0; i < len(s); i++ {
+		if s[i] == c {
+			res[i] = 0
+		} else {
+			left, right := math.MaxInt32, math.MaxInt32
+			for j := i + 1; j < len(s); j++ {
+				if s[j] == c {
+					right = j - i
+					break
+				}
+			}
+			for k := i - 1; k >= 0; k-- {
+				if s[k] == c {
+					left = i - k
+					break
+				}
+			}
+			res[i] = min(left, right)
+		}
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+```
\ No newline at end of file
diff --git a/leetcode/0823.Binary-Trees-With-Factors/823. Binary Trees With Factors.go b/leetcode/0823.Binary-Trees-With-Factors/823. Binary Trees With Factors.go
new file mode 100644
index 000000000..8c9b43925
--- /dev/null
+++ b/leetcode/0823.Binary-Trees-With-Factors/823. Binary Trees With Factors.go	
@@ -0,0 +1,70 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+const mod = 1e9 + 7
+
+// 解法一 DFS
+func numFactoredBinaryTrees(arr []int) int {
+	sort.Ints(arr)
+	numDict := map[int]bool{}
+	for _, num := range arr {
+		numDict[num] = true
+	}
+	dict, res := make(map[int][][2]int), 0
+	for i, num := range arr {
+		for j := i; j < len(arr) && num*arr[j] <= arr[len(arr)-1]; j++ {
+			tmp := num * arr[j]
+			if !numDict[tmp] {
+				continue
+			}
+			dict[tmp] = append(dict[tmp], [2]int{num, arr[j]})
+		}
+	}
+	cache := make(map[int]int)
+	for _, num := range arr {
+		res = (res + dfs(num, dict, cache)) % mod
+	}
+	return res
+}
+
+func dfs(num int, dict map[int][][2]int, cache map[int]int) int {
+	if val, ok := cache[num]; ok {
+		return val
+	}
+	res := 1
+	for _, tuple := range dict[num] {
+		a, b := tuple[0], tuple[1]
+		x, y := dfs(a, dict, cache), dfs(b, dict, cache)
+		tmp := x * y
+		if a != b {
+			tmp *= 2
+		}
+		res = (res + tmp) % mod
+	}
+	cache[num] = res
+	return res
+}
+
+// 解法二 DP
+func numFactoredBinaryTrees1(arr []int) int {
+	dp := make(map[int]int)
+	sort.Ints(arr)
+	for i, curNum := range arr {
+		for j := 0; j < i; j++ {
+			factor := arr[j]
+			quotient, remainder := curNum/factor, curNum%factor
+			if remainder == 0 {
+				dp[curNum] += dp[factor] * dp[quotient]
+			}
+		}
+		dp[curNum]++
+	}
+	totalCount := 0
+	for _, count := range dp {
+		totalCount += count
+	}
+	return totalCount % mod
+}
diff --git a/leetcode/0823.Binary-Trees-With-Factors/823. Binary Trees With Factors_test.go b/leetcode/0823.Binary-Trees-With-Factors/823. Binary Trees With Factors_test.go
new file mode 100644
index 000000000..02b8dd5eb
--- /dev/null
+++ b/leetcode/0823.Binary-Trees-With-Factors/823. Binary Trees With Factors_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question823 struct {
+	para823
+	ans823
+}
+
+// para 是参数
+// one 代表第一个参数
+type para823 struct {
+	arr []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans823 struct {
+	one int
+}
+
+func Test_Problem823(t *testing.T) {
+
+	qs := []question823{
+
+		{
+			para823{[]int{2, 4}},
+			ans823{3},
+		},
+
+		{
+			para823{[]int{2, 4, 5, 10}},
+			ans823{7},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 823------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans823, q.para823
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numFactoredBinaryTrees(p.arr))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0823.Binary-Trees-With-Factors/README.md b/leetcode/0823.Binary-Trees-With-Factors/README.md
new file mode 100644
index 000000000..1423c497f
--- /dev/null
+++ b/leetcode/0823.Binary-Trees-With-Factors/README.md
@@ -0,0 +1,119 @@
+# [823. Binary Trees With Factors](https://leetcode.com/problems/binary-trees-with-factors/)
+
+
+## 题目
+
+Given an array of unique integers, `arr`, where each integer `arr[i]` is strictly greater than `1`.
+
+We make a binary tree using these integers, and each number may be used for any number of times. Each non-leaf node's value should be equal to the product of the values of its children.
+
+Return *the number of binary trees we can make*. The answer may be too large so return the answer **modulo** `109 + 7`.
+
+**Example 1:**
+
+```
+Input: arr = [2,4]
+Output: 3
+Explanation: We can make these trees: [2], [4], [4, 2, 2]
+```
+
+**Example 2:**
+
+```
+Input: arr = [2,4,5,10]
+Output: 7
+Explanation: We can make these trees: [2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2].
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 1000`
+- `2 <= arr[i] <= 10^9`
+
+## 题目大意
+
+给出一个含有不重复整数元素的数组，每个整数均大于 1。我们用这些整数来构建二叉树，每个整数可以使用任意次数。其中：每个非叶结点的值应等于它的两个子结点的值的乘积。满足条件的二叉树一共有多少个？返回的结果应模除 10 * 9 + 7。
+
+## 解题思路
+
+- 首先想到的是暴力解法，先排序，然后遍历所有节点，枚举两两乘积为第三个节点值的组合。然后枚举这些组合并构成树。这里计数的时候要注意，左右孩子如果不是对称的，左右子树相互对调又是一组解。但是这个方法超时了。原因是，暴力枚举了很多次重复的节点和组合。优化这里的方法就是把已经计算过的节点放入 `map` 中。这里有 2 层 `map`，第一层 `map` 记忆化的是两两乘积的组合，将父亲节点作为 `key`，左右 2 个孩子作为 `value`。第二层 `map` 记忆化的是以 `root` 为根节点此时二叉树的种类数，`key` 是 `root`，`value` 存的是种类数。这样优化以后，DFS 暴力解法可以 runtime beats 100%。
+- 另外一种解法是 DP。定义 `dp[i]` 代表以 `i` 为根节点的树的种类数。dp[i] 初始都是 1，因为所有节点自身可以形成为自身单个节点为 `root` 的树。同样需要先排序。状态转移方程是：
+
+    $$dp[i] = \sum_{j<i, k<i}^{}dp[j] * dp[k], j * k = i$$
+
+    最后将 `dp[]` 数组中所有结果累加取模即为最终结果，时间复杂度 O(n^2)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+const mod = 1e9 + 7
+
+// 解法一 DFS
+func numFactoredBinaryTrees(arr []int) int {
+	sort.Ints(arr)
+	numDict := map[int]bool{}
+	for _, num := range arr {
+		numDict[num] = true
+	}
+	dict, res := make(map[int][][2]int), 0
+	for i, num := range arr {
+		for j := i; j < len(arr) && num*arr[j] <= arr[len(arr)-1]; j++ {
+			tmp := num * arr[j]
+			if !numDict[tmp] {
+				continue
+			}
+			dict[tmp] = append(dict[tmp], [2]int{num, arr[j]})
+		}
+	}
+	cache := make(map[int]int)
+	for _, num := range arr {
+		res = (res + dfs(num, dict, cache)) % mod
+	}
+	return res
+}
+
+func dfs(num int, dict map[int][][2]int, cache map[int]int) int {
+	if val, ok := cache[num]; ok {
+		return val
+	}
+	res := 1
+	for _, tuple := range dict[num] {
+		a, b := tuple[0], tuple[1]
+		x, y := dfs(a, dict, cache), dfs(b, dict, cache)
+		tmp := x * y
+		if a != b {
+			tmp *= 2
+		}
+		res = (res + tmp) % mod
+	}
+	cache[num] = res
+	return res
+}
+
+// 解法二 DP
+func numFactoredBinaryTrees1(arr []int) int {
+	dp := make(map[int]int)
+	sort.Ints(arr)
+	for i, curNum := range arr {
+		for j := 0; j < i; j++ {
+			factor := arr[j]
+			quotient, remainder := curNum/factor, curNum%factor
+			if remainder == 0 {
+				dp[curNum] += dp[factor] * dp[quotient]
+			}
+		}
+		dp[curNum]++
+	}
+	totalCount := 0
+	for _, count := range dp {
+		totalCount += count
+	}
+	return totalCount % mod
+}
+```
\ No newline at end of file
diff --git a/leetcode/0825.Friends-Of-Appropriate-Ages/825. Friends Of Appropriate Ages.go b/leetcode/0825.Friends-Of-Appropriate-Ages/825. Friends Of Appropriate Ages.go
new file mode 100644
index 000000000..ebb496c68
--- /dev/null
+++ b/leetcode/0825.Friends-Of-Appropriate-Ages/825. Friends Of Appropriate Ages.go	
@@ -0,0 +1,67 @@
+package leetcocde
+
+import "sort"
+
+// 解法一 前缀和，时间复杂度 O(n)
+func numFriendRequests(ages []int) int {
+	count, prefixSum, res := make([]int, 121), make([]int, 121), 0
+	for _, age := range ages {
+		count[age]++
+	}
+	for i := 1; i < 121; i++ {
+		prefixSum[i] = prefixSum[i-1] + count[i]
+	}
+	for i := 15; i < 121; i++ {
+		if count[i] > 0 {
+			bound := i/2 + 8
+			res += count[i] * (prefixSum[i] - prefixSum[bound-1] - 1)
+		}
+	}
+	return res
+}
+
+// 解法二 双指针 + 排序，时间复杂度 O(n logn)
+func numFriendRequests1(ages []int) int {
+	sort.Ints(ages)
+	left, right, res := 0, 0, 0
+	for _, age := range ages {
+		if age < 15 {
+			continue
+		}
+		for ages[left]*2 <= age+14 {
+			left++
+		}
+		for right+1 < len(ages) && ages[right+1] <= age {
+			right++
+		}
+		res += right - left
+	}
+	return res
+}
+
+// 解法三 暴力解法 O(n^2)
+func numFriendRequests2(ages []int) int {
+	res, count := 0, [125]int{}
+	for _, x := range ages {
+		count[x]++
+	}
+	for i := 1; i <= 120; i++ {
+		for j := 1; j <= 120; j++ {
+			if j > i {
+				continue
+			}
+			if (j-7)*2 <= i {
+				continue
+			}
+			if j > 100 && i < 100 {
+				continue
+			}
+			if i != j {
+				res += count[i] * count[j]
+			} else {
+				res += count[i] * (count[j] - 1)
+			}
+		}
+	}
+	return res
+}
diff --git a/leetcode/0825.Friends-Of-Appropriate-Ages/825. Friends Of Appropriate Ages_test.go b/leetcode/0825.Friends-Of-Appropriate-Ages/825. Friends Of Appropriate Ages_test.go
new file mode 100644
index 000000000..ee891c63e
--- /dev/null
+++ b/leetcode/0825.Friends-Of-Appropriate-Ages/825. Friends Of Appropriate Ages_test.go	
@@ -0,0 +1,52 @@
+package leetcocde
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question825 struct {
+	para825
+	ans825
+}
+
+// para 是参数
+// one 代表第一个参数
+type para825 struct {
+	ages []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans825 struct {
+	one int
+}
+
+func Test_Problem825(t *testing.T) {
+
+	qs := []question825{
+
+		{
+			para825{[]int{16, 16}},
+			ans825{2},
+		},
+
+		{
+			para825{[]int{16, 17, 18}},
+			ans825{2},
+		},
+
+		{
+			para825{[]int{20, 30, 100, 110, 120}},
+			ans825{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 825------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans825, q.para825
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numFriendRequests(p.ages))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0825.Friends-Of-Appropriate-Ages/README.md b/leetcode/0825.Friends-Of-Appropriate-Ages/README.md
new file mode 100644
index 000000000..6249cc44c
--- /dev/null
+++ b/leetcode/0825.Friends-Of-Appropriate-Ages/README.md
@@ -0,0 +1,141 @@
+# [825. Friends Of Appropriate Ages](https://leetcode.com/problems/friends-of-appropriate-ages/)
+
+
+## 题目
+
+There are `n` persons on a social media website. You are given an integer array `ages` where `ages[i]` is the age of the `ith` person.
+
+A Person `x` will not send a friend request to a person `y` (`x != y`) if any of the following conditions is true:
+
+- `age[y] <= 0.5 * age+ 7`
+- `age[y] > age[x]`
+- `age[y] > 100 && age< 100`
+
+Otherwise, `x` will send a friend request to `y`.
+
+Note that if `x` sends a request to `y`, `y` will not necessarily send a request to `x`. Also, a person will not send a friend request to themself.
+
+Return *the total number of friend requests made*.
+
+**Example 1:**
+
+```
+Input: ages = [16,16]
+Output: 2
+Explanation: 2 people friend request each other.
+
+```
+
+**Example 2:**
+
+```
+Input: ages = [16,17,18]
+Output: 2
+Explanation: Friend requests are made 17 -> 16, 18 -> 17.
+
+```
+
+**Example 3:**
+
+```
+Input: ages = [20,30,100,110,120]
+Output: 3
+Explanation: Friend requests are made 110 -> 100, 120 -> 110, 120 -> 100.
+
+```
+
+**Constraints:**
+
+- `n == ages.length`
+- `1 <= n <= 2 * 10^4`
+- `1 <= ages[i] <= 120`
+
+## 题目大意
+
+在社交媒体网站上有 n 个用户。给你一个整数数组 ages ，其中 ages[i] 是第 i 个用户的年龄。
+
+如果下述任意一个条件为真，那么用户 x 将不会向用户 y（x != y）发送好友请求：
+
+- ages[y] <= 0.5 * ages[x] + 7
+- ages[y] > ages[x]
+- ages[y] > 100 && ages[x] < 100
+
+否则，x 将会向 y 发送一条好友请求。注意，如果 x 向 y 发送一条好友请求，y 不必也向 x 发送一条好友请求。另外，用户不会向自己发送好友请求。返回在该社交媒体网站上产生的好友请求总数。
+
+## 解题思路
+
+- 解法三，暴力解法。先统计 [1,120] 范围内每个年龄的人数。然后利用题目中的三个判断条件，筛选符合条件的用户对。需要注意的是，相同年龄的人可以相互发送好友请求。不同年龄的人发送好友请求是单向的，即年龄老的向年龄轻的发送好友请求，年龄轻的不会对年龄老的发送好友请求。
+- 解法二，排序 + 双指针。题目给定的 3 个条件其实是 2 个。条件 3 包含在条件 2 中。条件 1 和条件 2 组合起来是 `0.5 × ages[x]+7 < ages[y] ≤ ages[x]`。当 ages[x] 小于 15 时，这个等式无解。考虑到年龄是单调递增的，`(0.5 × ages[x]+7,ages[x]]` 这个区间左右边界也是单调递增的。于是可以用双指针维护两个边界。在区间 [left, right] 内，这些下标对应的的 y 值都满足条件。当 `ages[left] > 0.5 × ages[x]+7` 时，左指针停止右移。当 `ages[right+1] > ages[x]` 时， 右指针停止右移。在 `[left, right]` 区间内，满足条件的 y 有 `right-left+1` 个，即使得 `ages[y]` 取值在 `(0.5 × ages[x]+7,ages[x]]` 之间。依照题意，`x≠y`，即该区间右边界取不到。y 的取值个数需要再减一，减去的是取到和 x 相同的值的下标。那么每个区间能取 `right-left` 个值。累加所有满足条件的值即为好友请求总数。
+- 解法一。在解法二中，计算满足不等式 y 下标所在区间的时候，区间和区间存在重叠的情况，这些重叠情况导致了重复计算。所以这里可以优化。可以用 prefix sum 前缀和数组优化。代码见下方。
+
+## 代码
+
+```go
+package leetcocde
+
+import "sort"
+
+// 解法一 前缀和，时间复杂度 O(n)
+func numFriendRequests(ages []int) int {
+	count, prefixSum, res := make([]int, 121), make([]int, 121), 0
+	for _, age := range ages {
+		count[age]++
+	}
+	for i := 1; i < 121; i++ {
+		prefixSum[i] = prefixSum[i-1] + count[i]
+	}
+	for i := 15; i < 121; i++ {
+		if count[i] > 0 {
+			bound := i/2 + 8
+			res += count[i] * (prefixSum[i] - prefixSum[bound-1] - 1)
+		}
+	}
+	return res
+}
+
+// 解法二 双指针 + 排序，时间复杂度 O(n logn)
+func numFriendRequests1(ages []int) int {
+	sort.Ints(ages)
+	left, right, res := 0, 0, 0
+	for _, age := range ages {
+		if age < 15 {
+			continue
+		}
+		for ages[left]*2 <= age+14 {
+			left++
+		}
+		for right+1 < len(ages) && ages[right+1] <= age {
+			right++
+		}
+		res += right - left
+	}
+	return res
+}
+
+// 解法三 暴力解法 O(n^2)
+func numFriendRequests2(ages []int) int {
+	res, count := 0, [125]int{}
+	for _, x := range ages {
+		count[x]++
+	}
+	for i := 1; i <= 120; i++ {
+		for j := 1; j <= 120; j++ {
+			if j > i {
+				continue
+			}
+			if (j-7)*2 <= i {
+				continue
+			}
+			if j > 100 && i < 100 {
+				continue
+			}
+			if i != j {
+				res += count[i] * count[j]
+			} else {
+				res += count[i] * (count[j] - 1)
+			}
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0826. Most Profit Assigning Work/826. Most Profit Assigning Work.go b/leetcode/0826.Most-Profit-Assigning-Work/826. Most Profit Assigning Work.go
similarity index 94%
rename from Algorithms/0826. Most Profit Assigning Work/826. Most Profit Assigning Work.go
rename to leetcode/0826.Most-Profit-Assigning-Work/826. Most Profit Assigning Work.go
index 9c3532781..7edd1c1f0 100644
--- a/Algorithms/0826. Most Profit Assigning Work/826. Most Profit Assigning Work.go	
+++ b/leetcode/0826.Most-Profit-Assigning-Work/826. Most Profit Assigning Work.go	
@@ -53,3 +53,10 @@ func maxProfitAssignment(difficulty []int, profit []int, worker []int) int {
 	}
 	return res
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0826. Most Profit Assigning Work/826. Most Profit Assigning Work_test.go b/leetcode/0826.Most-Profit-Assigning-Work/826. Most Profit Assigning Work_test.go
similarity index 96%
rename from Algorithms/0826. Most Profit Assigning Work/826. Most Profit Assigning Work_test.go
rename to leetcode/0826.Most-Profit-Assigning-Work/826. Most Profit Assigning Work_test.go
index a7ab61d00..a89eb5863 100644
--- a/Algorithms/0826. Most Profit Assigning Work/826. Most Profit Assigning Work_test.go	
+++ b/leetcode/0826.Most-Profit-Assigning-Work/826. Most Profit Assigning Work_test.go	
@@ -28,12 +28,12 @@ func Test_Problem826(t *testing.T) {
 
 	qs := []question826{
 
-		question826{
+		{
 			para826{[]int{2, 4, 6, 8, 10}, []int{10, 20, 30, 40, 50}, []int{4, 5, 6, 7}},
 			ans826{100},
 		},
 
-		question826{
+		{
 			para826{[]int{85, 47, 57}, []int{24, 66, 99}, []int{40, 25, 25}},
 			ans826{0},
 		},
diff --git a/Algorithms/0826. Most Profit Assigning Work/README.md b/leetcode/0826.Most-Profit-Assigning-Work/README.md
similarity index 100%
rename from Algorithms/0826. Most Profit Assigning Work/README.md
rename to leetcode/0826.Most-Profit-Assigning-Work/README.md
diff --git a/leetcode/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String/828. Count Unique Characters of All Substrings of a Given String.go b/leetcode/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String/828. Count Unique Characters of All Substrings of a Given String.go
new file mode 100644
index 000000000..9129b8b46
--- /dev/null
+++ b/leetcode/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String/828. Count Unique Characters of All Substrings of a Given String.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+func uniqueLetterString(S string) int {
+	res, left, right := 0, 0, 0
+	for i := 0; i < len(S); i++ {
+		left = i - 1
+		for left >= 0 && S[left] != S[i] {
+			left--
+		}
+		right = i + 1
+		for right < len(S) && S[right] != S[i] {
+			right++
+		}
+		res += (i - left) * (right - i)
+	}
+	return res % 1000000007
+}
+
+// 暴力解法，超时！时间复杂度 O(n^2)
+func uniqueLetterString1(S string) int {
+	if len(S) == 0 {
+		return 0
+	}
+	res, mod := 0, 1000000007
+	for i := 0; i < len(S); i++ {
+		letterMap := map[byte]int{}
+		for j := i; j < len(S); j++ {
+			letterMap[S[j]]++
+			tmp := 0
+			for _, v := range letterMap {
+				if v > 1 {
+					tmp++
+				}
+			}
+			if tmp == len(letterMap) {
+				continue
+			} else {
+				res += len(letterMap) - tmp
+			}
+		}
+	}
+	return res % mod
+}
diff --git a/leetcode/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String/828. Count Unique Characters of All Substrings of a Given String_test.go b/leetcode/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String/828. Count Unique Characters of All Substrings of a Given String_test.go
new file mode 100644
index 000000000..620c32bad
--- /dev/null
+++ b/leetcode/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String/828. Count Unique Characters of All Substrings of a Given String_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question828 struct {
+	para828
+	ans828
+}
+
+// para 是参数
+// one 代表第一个参数
+type para828 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans828 struct {
+	one int
+}
+
+func Test_Problem828(t *testing.T) {
+
+	qs := []question828{
+
+		{
+			para828{"BABABBABAA"},
+			ans828{35},
+		},
+
+		{
+			para828{"ABC"},
+			ans828{10},
+		},
+
+		{
+			para828{"ABA"},
+			ans828{8},
+		},
+
+		{
+			para828{"ABAB"},
+			ans828{12},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 828------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans828, q.para828
+		fmt.Printf("【input】:%v       【output】:%v\n", p, uniqueLetterString(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String/README.md b/leetcode/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String/README.md
new file mode 100644
index 000000000..fd29a4af2
--- /dev/null
+++ b/leetcode/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String/README.md
@@ -0,0 +1,101 @@
+# [828. Count Unique Characters of All Substrings of a Given String](https://leetcode.com/problems/count-unique-characters-of-all-substrings-of-a-given-string/)
+
+
+## 题目
+
+Let's define a function `countUniqueChars(s)` that returns the number of unique characters on `s`, for example if `s = "LEETCODE"` then `"L"`, `"T"`,`"C"`,`"O"`,`"D"` are the unique characters since they appear only once in `s`, therefore `countUniqueChars(s) = 5`.On this problem given a string `s` we need to return the sum of `countUniqueChars(t)` where `t` is a substring of `s`. Notice that some substrings can be repeated so on this case you have to count the repeated ones too.
+
+Since the answer can be very large, return the answer modulo `10 ^ 9 + 7`.
+
+**Example 1:**
+
+```
+Input: s = "ABC"
+Output: 10
+Explanation: All possible substrings are: "A","B","C","AB","BC" and "ABC".
+Evey substring is composed with only unique letters.
+Sum of lengths of all substring is 1 + 1 + 1 + 2 + 2 + 3 = 10
+
+```
+
+**Example 2:**
+
+```
+Input: s = "ABA"
+Output: 8
+Explanation: The same as example 1, except countUniqueChars("ABA") = 1.
+
+```
+
+**Example 3:**
+
+```
+Input: s = "LEETCODE"
+Output: 92
+
+```
+
+**Constraints:**
+
+- `0 <= s.length <= 10^4`
+- `s` contain upper-case English letters only.
+
+## 题目大意
+
+如果一个字符在字符串 S 中有且仅有出现一次，那么我们称其为独特字符。例如，在字符串 S = "LETTER" 中，"L" 和 "R" 可以被称为独特字符。我们再定义 UNIQ(S) 作为字符串 S 中独特字符的个数。那么，在 S = "LETTER" 中， UNIQ("LETTER") =  2。
+
+对于给定字符串 S，计算其所有非空子串的独特字符的个数（即 UNIQ(substring)）之和。如果在 S 的不同位置上出现两个甚至多个相同的子串，那么我们认为这些子串是不同的。考虑到答案可能会非常大，规定返回格式为：结果 mod 10 ^ 9 + 7。
+
+## 解题思路
+
+- 这一题可以先用暴力解法尝试解题，不过提交以后会发现判题结果是超时。出错的一组数据是一个有 10000 个字符的字符串。暴力解法中间由于遍历了太多的子区间，导致了超时。
+- 这道题换一个角度思考问题。当子字符串中字符 X 出现了 2 次以上，那么它就对最终结果没有任何影响，所以只有当某个字符只出现一次的时候才会影响最终结果。再者，一个子字符串中不重复的字符的总个数，也就是这个子字符串 UNIQ 值。例如，“ABC”，这个子字符串的 UNIQ 值是 3，可以这样计算，它属于 A 的独特的字符串，也属于 B 的独特的字符串，也属于 C 的独特的字符串，那么计算这个子字符串的问题可以分解成计算 A 有多少个独特的子字符串，B 有多少个独特的子字符串，C 有多少个独特的子字符串的问题。在计算 A 有多少个子字符串的问题的时候，里面肯定会包含 "ABC" 这个子字符串的。所以原问题就转换成了分别计算给出的字符串中每个字符出现在独特字符串中的总数之和。
+- 假设原字符串是 BAABBABBBAAABA，这个字符串中出现了很多 A 和很多 B，假设我们当前计算到了第 3 个 A 的位置了(index = 5)，即标红色的那个 A。如何计算这个 A 在哪些子字符串中是独特的呢？由于子字符串题目中要求必须是连续的区间，所以这个问题很简单。找到这个 A 前一个 A 的下标位置(index = 2)，再找到这个 A 后一个 A 的下标位置(index = 9)，即 BAABBABBBAAABA，第一个 A 和当前计算的 A 中间区间有 2 个字符，第三个 A 和当前计算的 A 中间有 3 个字符。那么当前计算的 A 出现在 `(2 + 1) * (3 + 1) = 12` 个子字符串中是独特的，这 12 个字符串是：`A`，`BA`，`BBA`，`AB`，`ABB`，`ABBB`，`BAB`，`BABB`，`BABBB`，`BBAB`，`BBABB`，`BBABBB`。计算方法，假设当前待计算的字符的下标是 i ，找到当前字符前一次出现的下标位置 left，再找到当前字符后一次出现的下标位置 right，那么左边区间 (left,i) 的***开区间****内包含的字符数是 i - left - 1，右边区间 (i,right) 的***开区间****内包含的字符数是 right - i - 1。左右两边都还需要考虑空字符串的情况，即左右两边都可以不取任何字符，那么对应的就是只有中间这个待计算的字符 `A`。所以左右两边都还需要再加上空串的情况，左边 i - left - 1 + 1 = i - left，右边 right - i - 1 + 1 = right - i。左右两边的情况进行排列组合，即 (i - left) * (right - i)。针对字符串的每个字符都计算这样的值，最后累积的总和就是题目中要求的总 UNIQ 值。
+
+## 代码
+
+```go
+package leetcode
+
+func uniqueLetterString(S string) int {
+	res, left, right := 0, 0, 0
+	for i := 0; i < len(S); i++ {
+		left = i - 1
+		for left >= 0 && S[left] != S[i] {
+			left--
+		}
+		right = i + 1
+		for right < len(S) && S[right] != S[i] {
+			right++
+		}
+		res += (i - left) * (right - i)
+	}
+	return res % 1000000007
+}
+
+// 暴力解法，超时！时间复杂度 O(n^2)
+func uniqueLetterString1(S string) int {
+	if len(S) == 0 {
+		return 0
+	}
+	res, mod := 0, 1000000007
+	for i := 0; i < len(S); i++ {
+		letterMap := map[byte]int{}
+		for j := i; j < len(S); j++ {
+			letterMap[S[j]]++
+			tmp := 0
+			for _, v := range letterMap {
+				if v > 1 {
+					tmp++
+				}
+			}
+			if tmp == len(letterMap) {
+				continue
+			} else {
+				res += len(letterMap) - tmp
+			}
+		}
+	}
+	return res % mod
+}
+```
\ No newline at end of file
diff --git a/leetcode/0830.Positions-of-Large-Groups/830. Positions of Large Groups.go b/leetcode/0830.Positions-of-Large-Groups/830. Positions of Large Groups.go
new file mode 100644
index 000000000..8b61b3380
--- /dev/null
+++ b/leetcode/0830.Positions-of-Large-Groups/830. Positions of Large Groups.go	
@@ -0,0 +1,15 @@
+package leetcode
+
+func largeGroupPositions(S string) [][]int {
+	res, end := [][]int{}, 0
+	for end < len(S) {
+		start, str := end, S[end]
+		for end < len(S) && S[end] == str {
+			end++
+		}
+		if end-start >= 3 {
+			res = append(res, []int{start, end - 1})
+		}
+	}
+	return res
+}
diff --git a/leetcode/0830.Positions-of-Large-Groups/830. Positions of Large Groups_test.go b/leetcode/0830.Positions-of-Large-Groups/830. Positions of Large Groups_test.go
new file mode 100644
index 000000000..ed74b05a8
--- /dev/null
+++ b/leetcode/0830.Positions-of-Large-Groups/830. Positions of Large Groups_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question830 struct {
+	para830
+	ans830
+}
+
+// para 是参数
+// one 代表第一个参数
+type para830 struct {
+	S string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans830 struct {
+	one [][]int
+}
+
+func Test_Problem830(t *testing.T) {
+
+	qs := []question830{
+
+		{
+			para830{"abbxxxxzzy"},
+			ans830{[][]int{{3, 6}}},
+		},
+
+		{
+			para830{"abc"},
+			ans830{[][]int{{}}},
+		},
+
+		{
+			para830{"abcdddeeeeaabbbcd"},
+			ans830{[][]int{{3, 5}, {6, 9}, {12, 14}}},
+		},
+
+		{
+			para830{"aba"},
+			ans830{[][]int{{}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 830------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans830, q.para830
+		fmt.Printf("【input】:%v       【output】:%v\n", p, largeGroupPositions(p.S))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0830.Positions-of-Large-Groups/README.md b/leetcode/0830.Positions-of-Large-Groups/README.md
new file mode 100644
index 000000000..5323f2e08
--- /dev/null
+++ b/leetcode/0830.Positions-of-Large-Groups/README.md
@@ -0,0 +1,80 @@
+# [830. Positions of Large Groups](https://leetcode.com/problems/positions-of-large-groups/)
+
+
+## 题目
+
+In a string `s` of lowercase letters, these letters form consecutive groups of the same character.
+
+For example, a string like `s = "abbxxxxzyy"` has the groups `"a"`, `"bb"`, `"xxxx"`, `"z"`, and `"yy"`.
+
+A group is identified by an interval `[start, end]`, where `start` and `end` denote the start and end indices (inclusive) of the group. In the above example, `"xxxx"` has the interval `[3,6]`.
+
+A group is considered **large** if it has 3 or more characters.
+
+Return *the intervals of every **large** group sorted in **increasing order by start index***.
+
+**Example 1:**
+
+```
+Input: s = "abbxxxxzzy"
+Output: [[3,6]]
+Explanation: "xxxx" is the only large group with start index 3 and end index 6.
+```
+
+**Example 2:**
+
+```
+Input: s = "abc"
+Output: []
+Explanation: We have groups "a", "b", and "c", none of which are large groups.
+```
+
+**Example 3:**
+
+```
+Input: s = "abcdddeeeeaabbbcd"
+Output: [[3,5],[6,9],[12,14]]
+Explanation: The large groups are "ddd", "eeee", and "bbb".
+```
+
+**Example 4:**
+
+```
+Input: s = "aba"
+Output: []
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 1000`
+- `s` contains lower-case English letters only.
+
+## 题目大意
+
+在一个由小写字母构成的字符串 s 中，包含由一些连续的相同字符所构成的分组。例如，在字符串 s = "abbxxxxzyy" 中，就含有 "a", "bb", "xxxx", "z" 和 "yy" 这样的一些分组。分组可以用区间 [start, end] 表示，其中 start 和 end 分别表示该分组的起始和终止位置的下标。上例中的 "xxxx" 分组用区间表示为 [3,6] 。我们称所有包含大于或等于三个连续字符的分组为 较大分组 。
+
+找到每一个 较大分组 的区间，按起始位置下标递增顺序排序后，返回结果。
+
+## 解题思路
+
+- 简单题。利用滑动窗口的思想，先扩大窗口的右边界，找到能相同字母且能到达的最右边。记录左右边界。再将窗口的左边界移动到上一次的右边界处。以此类推，重复扩大窗口的右边界，直至扫完整个字符串。最终所有满足题意的较大分组区间都在数组中了。
+
+## 代码
+
+```go
+package leetcode
+
+func largeGroupPositions(S string) [][]int {
+	res, end := [][]int{}, 0
+	for end < len(S) {
+		start, str := end, S[end]
+		for end < len(S) && S[end] == str {
+			end++
+		}
+		if end-start >= 3 {
+			res = append(res, []int{start, end - 1})
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0832.Flipping-an-Image/832. Flipping an Image.go b/leetcode/0832.Flipping-an-Image/832. Flipping an Image.go
new file mode 100644
index 000000000..6abccf66a
--- /dev/null
+++ b/leetcode/0832.Flipping-an-Image/832. Flipping an Image.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+func flipAndInvertImage(A [][]int) [][]int {
+	for i := 0; i < len(A); i++ {
+		for a, b := 0, len(A[i])-1; a < b; a, b = a+1, b-1 {
+			A[i][a], A[i][b] = A[i][b], A[i][a]
+		}
+		for a := 0; a < len(A[i]); a++ {
+			A[i][a] = (A[i][a] + 1) % 2
+		}
+	}
+	return A
+}
diff --git a/leetcode/0832.Flipping-an-Image/832. Flipping an Image_test.go b/leetcode/0832.Flipping-an-Image/832. Flipping an Image_test.go
new file mode 100644
index 000000000..9705a0bb0
--- /dev/null
+++ b/leetcode/0832.Flipping-an-Image/832. Flipping an Image_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question832 struct {
+	para832
+	ans832
+}
+
+// para 是参数
+// one 代表第一个参数
+type para832 struct {
+	A [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans832 struct {
+	one [][]int
+}
+
+func Test_Problem832(t *testing.T) {
+
+	qs := []question832{
+
+		{
+			para832{[][]int{{1, 1, 0}, {1, 0, 1}, {0, 0, 0}}},
+			ans832{[][]int{{1, 0, 0}, {0, 1, 0}, {1, 1, 1}}},
+		},
+
+		{
+			para832{[][]int{{1, 1, 0, 0}, {1, 0, 0, 1}, {0, 1, 1, 1}, {1, 0, 1, 0}}},
+			ans832{[][]int{{1, 1, 0, 0}, {0, 1, 1, 0}, {0, 0, 0, 1}, {1, 0, 1, 0}}},
+		},
+
+		{
+			para832{[][]int{{1, 1, 1}, {1, 1, 1}, {0, 0, 0}}},
+			ans832{[][]int{{0, 0, 0}, {0, 0, 0}, {1, 1, 1}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 832------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans832, q.para832
+		fmt.Printf("【input】:%v       【output】:%v\n", p, flipAndInvertImage(p.A))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0832.Flipping-an-Image/README.md b/leetcode/0832.Flipping-an-Image/README.md
new file mode 100644
index 000000000..81b8ccc97
--- /dev/null
+++ b/leetcode/0832.Flipping-an-Image/README.md
@@ -0,0 +1,63 @@
+# [832. Flipping an Image](https://leetcode.com/problems/flipping-an-image/)
+
+
+## 题目
+
+Given a binary matrix `A`, we want to flip the image horizontally, then invert it, and return the resulting image.
+
+To flip an image horizontally means that each row of the image is reversed. For example, flipping `[1, 1, 0]` horizontally results in `[0, 1, 1]`.
+
+To invert an image means that each `0` is replaced by `1`, and each `1` is replaced by `0`. For example, inverting `[0, 1, 1]` results in `[1, 0, 0]`.
+
+**Example 1**:
+
+```
+Input: [[1,1,0],[1,0,1],[0,0,0]]
+Output: [[1,0,0],[0,1,0],[1,1,1]]
+Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]].
+Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]]
+```
+
+**Example 2**:
+
+```
+Input: [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]
+Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
+Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]].
+Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
+```
+
+**Notes**:
+
+- `1 <= A.length = A[0].length <= 20`
+- `0 <= A[i][j] <= 1`
+
+## 题目大意
+
+给定一个二进制矩阵 A，我们想先水平翻转图像，然后反转图像并返回结果。水平翻转图片就是将图片的每一行都进行翻转，即逆序。例如，水平翻转 [1, 1, 0] 的结果是 [0, 1, 1]。反转图片的意思是图片中的 0 全部被 1 替换， 1 全部被 0 替换。例如，反转 [0, 1, 1] 的结果是 [1, 0, 0]。
+
+
+## 解题思路
+
+- 给定一个二进制矩阵，要求先水平翻转，然后再反转( 1→0 , 0→1 )。
+- 简单题，按照题意先水平翻转，再反转即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func flipAndInvertImage(A [][]int) [][]int {
+	for i := 0; i < len(A); i++ {
+		for a, b := 0, len(A[i])-1; a < b; a, b = a+1, b-1 {
+			A[i][a], A[i][b] = A[i][b], A[i][a]
+		}
+		for a := 0; a < len(A[i]); a++ {
+			A[i][a] = (A[i][a] + 1) % 2
+		}
+	}
+	return A
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0834. Sum of Distances in Tree/834. Sum of Distances in Tree.go b/leetcode/0834.Sum-of-Distances-in-Tree/834. Sum of Distances in Tree.go
similarity index 100%
rename from Algorithms/0834. Sum of Distances in Tree/834. Sum of Distances in Tree.go
rename to leetcode/0834.Sum-of-Distances-in-Tree/834. Sum of Distances in Tree.go
diff --git a/Algorithms/0834. Sum of Distances in Tree/834. Sum of Distances in Tree_test.go b/leetcode/0834.Sum-of-Distances-in-Tree/834. Sum of Distances in Tree_test.go
similarity index 78%
rename from Algorithms/0834. Sum of Distances in Tree/834. Sum of Distances in Tree_test.go
rename to leetcode/0834.Sum-of-Distances-in-Tree/834. Sum of Distances in Tree_test.go
index 6e6f09619..7df04f429 100644
--- a/Algorithms/0834. Sum of Distances in Tree/834. Sum of Distances in Tree_test.go	
+++ b/leetcode/0834.Sum-of-Distances-in-Tree/834. Sum of Distances in Tree_test.go	
@@ -27,13 +27,13 @@ func Test_Problem834(t *testing.T) {
 
 	qs := []question834{
 
-		question834{
-			para834{4, [][]int{[]int{1, 2}, []int{3, 2}, []int{3, 0}}},
+		{
+			para834{4, [][]int{{1, 2}, {3, 2}, {3, 0}}},
 			ans834{[]int{6, 6, 4, 4}},
 		},
 
-		question834{
-			para834{6, [][]int{[]int{0, 1}, []int{0, 2}, []int{2, 3}, []int{2, 4}, []int{2, 5}}},
+		{
+			para834{6, [][]int{{0, 1}, {0, 2}, {2, 3}, {2, 4}, {2, 5}}},
 			ans834{[]int{8, 12, 6, 10, 10, 10}},
 		},
 	}
diff --git a/leetcode/0834.Sum-of-Distances-in-Tree/README.md b/leetcode/0834.Sum-of-Distances-in-Tree/README.md
new file mode 100755
index 000000000..20d5b4240
--- /dev/null
+++ b/leetcode/0834.Sum-of-Distances-in-Tree/README.md
@@ -0,0 +1,41 @@
+# [834. Sum of Distances in Tree](https://leetcode.com/problems/sum-of-distances-in-tree/)
+
+
+## 题目
+
+An undirected, connected tree with `N` nodes labelled `0...N-1` and `N-1edges` are given.
+
+The `i`th edge connects nodes `edges[i][0]` and `edges[i][1]` together.
+
+Return a list `ans`, where `ans[i]` is the sum of the distances between node `i`and all other nodes.
+
+**Example 1:**
+
+    Input: N = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
+    Output: [8,12,6,10,10,10]
+    Explanation: 
+    Here is a diagram of the given tree:
+      0
+     / \
+    1   2
+       /|\
+      3 4 5
+    We can see that dist(0,1) + dist(0,2) + dist(0,3) + dist(0,4) + dist(0,5)
+    equals 1 + 1 + 2 + 2 + 2 = 8.  Hence, answer[0] = 8, and so on.
+
+Note: `1 <= N <= 10000`
+
+## 题目大意
+
+给定一个无向、连通的树。树中有 N 个标记为 0...N-1 的节点以及 N-1 条边。第 i 条边连接节点 edges[i][0] 和 edges[i][1] 。返回一个表示节点 i 与其他所有节点距离之和的列表 ans。
+
+说明: 1 <= N <= 10000
+
+
+
+## 解题思路
+
+- 给出 N 个节点和这些节点之间的一些边的关系。要求求出分别以 x 为根节点到所有节点路径和。
+- 这一题虽说描述的是求树的路径，但是完全可以当做图来做，因为并不是二叉树，是多叉树。这一题的解题思路是先一次 DFS 求出以 0 为根节点到各个节点的路径和(不以 0 为节点也可以，可以取任意节点作为开始)。第二次 DFS 求出从 0 根节点转换到其他各个节点的路径和。由于第一次计算出来以 0 为节点的路径和是正确的，所以计算其他节点为根节点的路径和只需要转换一下就可以得到正确结果。经过 2 次 DFS 之后就可以得到所有节点以自己为根节点到所有节点的路径和了。
+- 如何从以 0 为根节点到其他所有节点的路径和转换到以其他节点为根节点到所有节点的路径和呢？从 0 节点换成 x 节点，只需要在 0 到所有节点的路径和基础上增增减减就可以了。增加的是 x 节点到除去以 x 为根节点所有子树以外的节点的路径，有多少个节点就增加多少条路径。减少的是 0 到以 x 为根节点所有子树节点的路径和，包含 0 到 x 根节点，有多少节点就减少多少条路径。所以在第一次 DFS 中需要计算好每个节点以自己为根节点的子树总数和(包含自己在内)，这样在第二次 DFS 中可以直接拿来做转换。具体细节的实现见代码。
+
diff --git a/Algorithms/0836. Rectangle Overlap/836. Rectangle Overlap.go b/leetcode/0836.Rectangle-Overlap/836. Rectangle Overlap.go
similarity index 100%
rename from Algorithms/0836. Rectangle Overlap/836. Rectangle Overlap.go
rename to leetcode/0836.Rectangle-Overlap/836. Rectangle Overlap.go
diff --git a/Algorithms/0836. Rectangle Overlap/836. Rectangle Overlap_test.go b/leetcode/0836.Rectangle-Overlap/836. Rectangle Overlap_test.go
similarity index 96%
rename from Algorithms/0836. Rectangle Overlap/836. Rectangle Overlap_test.go
rename to leetcode/0836.Rectangle-Overlap/836. Rectangle Overlap_test.go
index 82bd577be..2acbcb4dc 100644
--- a/Algorithms/0836. Rectangle Overlap/836. Rectangle Overlap_test.go	
+++ b/leetcode/0836.Rectangle-Overlap/836. Rectangle Overlap_test.go	
@@ -27,12 +27,12 @@ func Test_Problem836(t *testing.T) {
 
 	qs := []question836{
 
-		question836{
+		{
 			para836{[]int{0, 0, 2, 2}, []int{1, 1, 3, 3}},
 			ans836{true},
 		},
 
-		question836{
+		{
 			para836{[]int{0, 0, 1, 1}, []int{1, 0, 2, 1}},
 			ans836{false},
 		},
diff --git a/leetcode/0836.Rectangle-Overlap/README.md b/leetcode/0836.Rectangle-Overlap/README.md
new file mode 100755
index 000000000..a10a8f2bf
--- /dev/null
+++ b/leetcode/0836.Rectangle-Overlap/README.md
@@ -0,0 +1,41 @@
+# [836. Rectangle Overlap](https://leetcode.com/problems/rectangle-overlap/)
+
+
+## 题目
+
+A rectangle is represented as a list `[x1, y1, x2, y2]`, where `(x1, y1)` are the coordinates of its bottom-left corner, and `(x2, y2)` are the coordinates of its top-right corner.
+
+Two rectangles overlap if the area of their intersection is positive. To be clear, two rectangles that only touch at the corner or edges do not overlap.
+
+Given two (axis-aligned) rectangles, return whether they overlap.
+
+**Example 1:**
+
+    Input: rec1 = [0,0,2,2], rec2 = [1,1,3,3]
+    Output: true
+
+**Example 2:**
+
+    Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]
+    Output: false
+
+**Notes:**
+
+1. Both rectangles `rec1` and `rec2` are lists of 4 integers.
+2. All coordinates in rectangles will be between `-10^9` and `10^9`.
+
+
+## 题目大意
+
+矩形以列表 [x1, y1, x2, y2] 的形式表示，其中 (x1, y1) 为左下角的坐标，(x2, y2) 是右上角的坐标。如果相交的面积为正，则称两矩形重叠。需要明确的是，只在角或边接触的两个矩形不构成重叠。给出两个矩形，判断它们是否重叠并返回结果。
+
+说明：
+
+1. 两个矩形 rec1 和 rec2 都以含有四个整数的列表的形式给出。
+2. 矩形中的所有坐标都处于 -10^9 和 10^9 之间。
+
+
+## 解题思路
+
+- 给出两个矩形的坐标，判断两个矩形是否重叠。
+- 几何题，按照几何方法判断即可。
diff --git a/Algorithms/0838. Push Dominoes/838. Push Dominoes.go b/leetcode/0838.Push-Dominoes/838. Push Dominoes.go
similarity index 100%
rename from Algorithms/0838. Push Dominoes/838. Push Dominoes.go
rename to leetcode/0838.Push-Dominoes/838. Push Dominoes.go
diff --git a/Algorithms/0838. Push Dominoes/838. Push Dominoes_test.go b/leetcode/0838.Push-Dominoes/838. Push Dominoes_test.go
similarity index 94%
rename from Algorithms/0838. Push Dominoes/838. Push Dominoes_test.go
rename to leetcode/0838.Push-Dominoes/838. Push Dominoes_test.go
index 26b4ebd46..83aa148b5 100644
--- a/Algorithms/0838. Push Dominoes/838. Push Dominoes_test.go	
+++ b/leetcode/0838.Push-Dominoes/838. Push Dominoes_test.go	
@@ -26,17 +26,17 @@ func Test_Problem838(t *testing.T) {
 
 	qs := []question838{
 
-		question838{
+		{
 			para838{".L.R...LR..L.."},
 			ans838{"LL.RR.LLRRLL.."},
 		},
 
-		question838{
+		{
 			para838{"RR.L"},
 			ans838{"RR.L"},
 		},
 
-		question838{
+		{
 			para838{".L.R."},
 			ans838{"LL.RR"},
 		},
diff --git a/Algorithms/0838. Push Dominoes/README.md b/leetcode/0838.Push-Dominoes/README.md
similarity index 100%
rename from Algorithms/0838. Push Dominoes/README.md
rename to leetcode/0838.Push-Dominoes/README.md
diff --git a/Algorithms/0839. Similar String Groups/839. Similar String Groups.go b/leetcode/0839.Similar-String-Groups/839. Similar String Groups.go
similarity index 100%
rename from Algorithms/0839. Similar String Groups/839. Similar String Groups.go
rename to leetcode/0839.Similar-String-Groups/839. Similar String Groups.go
diff --git a/Algorithms/0839. Similar String Groups/839. Similar String Groups_test.go b/leetcode/0839.Similar-String-Groups/839. Similar String Groups_test.go
similarity index 97%
rename from Algorithms/0839. Similar String Groups/839. Similar String Groups_test.go
rename to leetcode/0839.Similar-String-Groups/839. Similar String Groups_test.go
index d469edc1a..1482cc506 100644
--- a/Algorithms/0839. Similar String Groups/839. Similar String Groups_test.go	
+++ b/leetcode/0839.Similar-String-Groups/839. Similar String Groups_test.go	
@@ -26,7 +26,7 @@ func Test_Problem839(t *testing.T) {
 
 	qs := []question839{
 
-		question839{
+		{
 			para839{[]string{"tars", "rats", "arts", "star"}},
 			ans839{2},
 		},
diff --git a/leetcode/0839.Similar-String-Groups/README.md b/leetcode/0839.Similar-String-Groups/README.md
new file mode 100755
index 000000000..aef19b6c7
--- /dev/null
+++ b/leetcode/0839.Similar-String-Groups/README.md
@@ -0,0 +1,59 @@
+# [839. Similar String Groups](https://leetcode.com/problems/similar-string-groups/)
+
+
+## 题目
+
+Two strings `X` and `Y` are similar if we can swap two letters (in different positions) of `X`, so that it equals `Y`.
+
+For example, `"tars"` and `"rats"` are similar (swapping at positions `0` and `2`), and `"rats"` and `"arts"` are similar, but `"star"` is not similar to `"tars"`, `"rats"`, or `"arts"`.
+
+Together, these form two connected groups by similarity: `{"tars", "rats", "arts"}` and `{"star"}`. Notice that `"tars"` and `"arts"` are in the same group even though they are not similar. Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.
+
+We are given a list `A` of strings. Every string in `A` is an anagram of every other string in `A`. How many groups are there?
+
+**Example 1:**
+
+    Input: ["tars","rats","arts","star"]
+    Output: 2
+
+**Note:**
+
+1. `A.length <= 2000`
+2. `A[i].length <= 1000`
+3. `A.length * A[i].length <= 20000`
+4. All words in `A` consist of lowercase letters only.
+5. All words in `A` have the same length and are anagrams of each other.
+6. The judging time limit has been increased for this question.
+
+
+## 题目大意
+
+如果我们交换字符串 X 中的两个不同位置的字母，使得它和字符串 Y 相等，那么称 X 和 Y 两个字符串相似。
+
+例如，"tars" 和 "rats" 是相似的 (交换 0 与 2 的位置)； "rats" 和 "arts" 也是相似的，但是 "star" 不与 "tars"，"rats"，或 "arts" 相似。
+
+总之，它们通过相似性形成了两个关联组：{"tars", "rats", "arts"} 和 {"star"}。注意，"tars" 和 "arts" 是在同一组中，即使它们并不相似。形式上，对每个组而言，要确定一个单词在组中，只需要这个词和该组中至少一个单词相似。我们给出了一个不包含重复的字符串列表 A。列表中的每个字符串都是 A 中其它所有字符串的一个字母异位词。请问 A 中有多少个相似字符串组？
+
+
+提示：
+
+- A.length <= 2000
+- A[i].length <= 1000
+- A.length * A[i].length <= 20000
+- A 中的所有单词都只包含小写字母。
+- A 中的所有单词都具有相同的长度，且是彼此的字母异位词。
+- 此问题的判断限制时间已经延长。
+
+
+备注：  
+
+- 字母异位词[anagram]，一种把某个字符串的字母的位置（顺序）加以改换所形成的新词。
+
+
+
+
+## 解题思路
+
+
+- 给出一个字符串数组，要求找出这个数组中，"不相似"的字符串有多少种。相似的字符串的定义是：如果 A 和 B 字符串只需要交换一次字母的位置就能变成两个相等的字符串，那么 A 和 B 是相似的。
+- 这一题的解题思路是用并查集。先将题目中的“相似”的定义转换一下，A 和 B 相似的意思是，A 和 B 中只有 2 个字符不相等，其他字符都相等，这样交换一次才能完全相等。有没有可能这两个字符交换了也不相等呢？这种情况不用考虑，因为题目中提到了给的字符串都是 `anagram` 的(`anagram` 的意思是，字符串的任意排列组合)。那么这题就比较简单了，只需要判断每两个字符串是否“相似”，如果相似就 `union()`，最后看并查集中有几个集合即可。
diff --git a/leetcode/0841.Keys-and-Rooms/841. Keys and Rooms.go b/leetcode/0841.Keys-and-Rooms/841. Keys and Rooms.go
new file mode 100644
index 000000000..338c0f35a
--- /dev/null
+++ b/leetcode/0841.Keys-and-Rooms/841. Keys and Rooms.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func canVisitAllRooms(rooms [][]int) bool {
+	visited := make(map[int]bool)
+	visited[0] = true
+	dfsVisitAllRooms(rooms, visited, 0)
+	return len(rooms) == len(visited)
+}
+
+func dfsVisitAllRooms(es [][]int, visited map[int]bool, from int) {
+	for _, to := range es[from] {
+		if visited[to] {
+			continue
+		}
+		visited[to] = true
+		dfsVisitAllRooms(es, visited, to)
+	}
+}
diff --git a/leetcode/0841.Keys-and-Rooms/841. Keys and Rooms_test.go b/leetcode/0841.Keys-and-Rooms/841. Keys and Rooms_test.go
new file mode 100644
index 000000000..8d3dfed91
--- /dev/null
+++ b/leetcode/0841.Keys-and-Rooms/841. Keys and Rooms_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question841 struct {
+	para841
+	ans841
+}
+
+// para 是参数
+// one 代表第一个参数
+type para841 struct {
+	rooms [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans841 struct {
+	one bool
+}
+
+func Test_Problem841(t *testing.T) {
+
+	qs := []question841{
+
+		{
+			para841{[][]int{{1}, {2}, {3}, {}}},
+			ans841{true},
+		},
+
+		{
+			para841{[][]int{{1, 3}, {3, 0, 1}, {2}, {0}}},
+			ans841{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 841------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans841, q.para841
+		fmt.Printf("【input】:%v       【output】:%v\n", p, canVisitAllRooms(p.rooms))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0841.Keys-and-Rooms/README.md b/leetcode/0841.Keys-and-Rooms/README.md
new file mode 100644
index 000000000..69d1da3a2
--- /dev/null
+++ b/leetcode/0841.Keys-and-Rooms/README.md
@@ -0,0 +1,80 @@
+# [841. Keys and Rooms](https://leetcode.com/problems/keys-and-rooms/)
+
+
+
+## 题目
+
+There are `N` rooms and you start in room `0`. Each room has a distinct number in `0, 1, 2, ..., N-1`, and each room may have some keys to access the next room.
+
+Formally, each room `i` has a list of keys `rooms[i]`, and each key `rooms[i][j]` is an integer in `[0, 1, ..., N-1]` where `N = rooms.length`. A key `rooms[i][j] = v` opens the room with number `v`.
+
+Initially, all the rooms start locked (except for room `0`).
+
+You can walk back and forth between rooms freely.
+
+Return `true` if and only if you can enter every room.
+
+**Example 1**:
+
+```
+Input: [[1],[2],[3],[]]
+Output: true
+Explanation:  
+We start in room 0, and pick up key 1.
+We then go to room 1, and pick up key 2.
+We then go to room 2, and pick up key 3.
+We then go to room 3.  Since we were able to go to every room, we return true.
+```
+
+**Example 2**:
+
+```
+Input: [[1,3],[3,0,1],[2],[0]]
+Output: false
+Explanation: We can't enter the room with number 2.
+```
+
+**Note**:
+
+1. `1 <= rooms.length <= 1000`
+2. `0 <= rooms[i].length <= 1000`
+3. The number of keys in all rooms combined is at most `3000`.
+
+
+## 题目大意
+
+有 N 个房间，开始时你位于 0 号房间。每个房间有不同的号码：0，1，2，...，N-1，并且房间里可能有一些钥匙能使你进入下一个房间。
+
+在形式上，对于每个房间 i 都有一个钥匙列表 rooms[i]，每个钥匙 rooms[i][j] 由 [0,1，...，N-1] 中的一个整数表示，其中 N = rooms.length。 钥匙 rooms[i][j] = v 可以打开编号为 v 的房间。最初，除 0 号房间外的其余所有房间都被锁住。你可以自由地在房间之间来回走动。如果能进入每个房间返回 true，否则返回 false。
+
+提示：
+
+- 1 <= rooms.length <= 1000
+- 0 <= rooms[i].length <= 1000
+- 所有房间中的钥匙数量总计不超过 3000。
+
+## 解题思路
+
+- 给出一个房间数组，每个房间里面装了一些钥匙。0 号房间默认是可以进入的，房间进入顺序没有要求，问最终能否进入所有房间。
+- 用 DFS 依次深搜所有房间的钥匙，如果都能访问到，最终输出 true。这题算是 DFS 里面的简单题。
+
+## 代码
+
+```go
+func canVisitAllRooms(rooms [][]int) bool {
+	visited := make(map[int]bool)
+	visited[0] = true
+	dfsVisitAllRooms(rooms, visited, 0)
+	return len(rooms) == len(visited)
+}
+
+func dfsVisitAllRooms(es [][]int, visited map[int]bool, from int) {
+	for _, to := range es[from] {
+		if visited[to] {
+			continue
+		}
+		visited[to] = true
+		dfsVisitAllRooms(es, visited, to)
+	}
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0842. Split Array into Fibonacci Sequence/842. Split Array into Fibonacci Sequence.go b/leetcode/0842.Split-Array-into-Fibonacci-Sequence/842. Split Array into Fibonacci Sequence.go
similarity index 93%
rename from Algorithms/0842. Split Array into Fibonacci Sequence/842. Split Array into Fibonacci Sequence.go
rename to leetcode/0842.Split-Array-into-Fibonacci-Sequence/842. Split Array into Fibonacci Sequence.go
index 2469ded48..f6851cbab 100644
--- a/Algorithms/0842. Split Array into Fibonacci Sequence/842. Split Array into Fibonacci Sequence.go	
+++ b/leetcode/0842.Split-Array-into-Fibonacci-Sequence/842. Split Array into Fibonacci Sequence.go	
@@ -32,7 +32,7 @@ func splitIntoFibonacci(S string) []int {
 	return res
 }
 
-//Propagate for rest of the string
+// Propagate for rest of the string
 func findRecursiveCheck(S string, x1 int, x2 int, left int, res *[]int, isComplete *bool) {
 	if x1 >= 1<<31 || x2 >= 1<<31 { // 题目要求每个数都要小于 2^31 - 1 = 2147483647，此处剪枝很关键！
 		return
@@ -55,3 +55,10 @@ func findRecursiveCheck(S string, x1 int, x2 int, left int, res *[]int, isComple
 	}
 	return
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0842. Split Array into Fibonacci Sequence/842. Split Array into Fibonacci Sequence_test.go b/leetcode/0842.Split-Array-into-Fibonacci-Sequence/842. Split Array into Fibonacci Sequence_test.go
similarity index 92%
rename from Algorithms/0842. Split Array into Fibonacci Sequence/842. Split Array into Fibonacci Sequence_test.go
rename to leetcode/0842.Split-Array-into-Fibonacci-Sequence/842. Split Array into Fibonacci Sequence_test.go
index 88d21ed2e..b6e20c4c2 100644
--- a/Algorithms/0842. Split Array into Fibonacci Sequence/842. Split Array into Fibonacci Sequence_test.go	
+++ b/leetcode/0842.Split-Array-into-Fibonacci-Sequence/842. Split Array into Fibonacci Sequence_test.go	
@@ -26,37 +26,37 @@ func Test_Problem842(t *testing.T) {
 
 	qs := []question842{
 
-		question842{
+		{
 			para842{"11235813"},
 			ans842{[]int{1, 1, 2, 3, 5, 8, 13}},
 		},
 
-		question842{
+		{
 			para842{"123456579"},
 			ans842{[]int{123, 456, 579}},
 		},
 
-		question842{
+		{
 			para842{"112358130"},
 			ans842{[]int{}},
 		},
 
-		question842{
+		{
 			para842{"0123"},
 			ans842{[]int{}},
 		},
 
-		question842{
+		{
 			para842{"1101111"},
 			ans842{[]int{110, 1, 111}},
 		},
 
-		question842{
+		{
 			para842{"539834657215398346785398346991079669377161950407626991734534318677529701785098211336528511"},
 			ans842{[]int{}},
 		},
 
-		question842{
+		{
 			para842{"3611537383985343591834441270352104793375145479938855071433500231900737525076071514982402115895535257195564161509167334647108949738176284385285234123461518508746752631120827113919550237703163294909"},
 			ans842{[]int{}},
 		},
diff --git a/Algorithms/0842. Split Array into Fibonacci Sequence/README.md b/leetcode/0842.Split-Array-into-Fibonacci-Sequence/README.md
similarity index 100%
rename from Algorithms/0842. Split Array into Fibonacci Sequence/README.md
rename to leetcode/0842.Split-Array-into-Fibonacci-Sequence/README.md
diff --git a/Algorithms/0844. Backspace String Compare/844. Backspace String Compare.go b/leetcode/0844.Backspace-String-Compare/844. Backspace String Compare.go
similarity index 100%
rename from Algorithms/0844. Backspace String Compare/844. Backspace String Compare.go
rename to leetcode/0844.Backspace-String-Compare/844. Backspace String Compare.go
diff --git a/Algorithms/0844. Backspace String Compare/844. Backspace String Compare_test.go b/leetcode/0844.Backspace-String-Compare/844. Backspace String Compare_test.go
similarity index 92%
rename from Algorithms/0844. Backspace String Compare/844. Backspace String Compare_test.go
rename to leetcode/0844.Backspace-String-Compare/844. Backspace String Compare_test.go
index 5d7542f83..7047964e1 100644
--- a/Algorithms/0844. Backspace String Compare/844. Backspace String Compare_test.go	
+++ b/leetcode/0844.Backspace-String-Compare/844. Backspace String Compare_test.go	
@@ -27,22 +27,22 @@ func Test_Problem844(t *testing.T) {
 
 	qs := []question844{
 
-		question844{
+		{
 			para844{"ab#c", "ad#c"},
 			ans844{true},
 		},
 
-		question844{
+		{
 			para844{"ab##", "c#d#"},
 			ans844{true},
 		},
 
-		question844{
+		{
 			para844{"a##c", "#a#c"},
 			ans844{true},
 		},
 
-		question844{
+		{
 			para844{"a#c", "b"},
 			ans844{false},
 		},
diff --git a/Algorithms/0844. Backspace String Compare/README.md b/leetcode/0844.Backspace-String-Compare/README.md
similarity index 100%
rename from Algorithms/0844. Backspace String Compare/README.md
rename to leetcode/0844.Backspace-String-Compare/README.md
diff --git a/Algorithms/0845. Longest Mountain in Array/845. Longest Mountain in Array.go b/leetcode/0845.Longest-Mountain-in-Array/845. Longest Mountain in Array.go
similarity index 89%
rename from Algorithms/0845. Longest Mountain in Array/845. Longest Mountain in Array.go
rename to leetcode/0845.Longest-Mountain-in-Array/845. Longest Mountain in Array.go
index aafdde4e7..0e6360286 100644
--- a/Algorithms/0845. Longest Mountain in Array/845. Longest Mountain in Array.go	
+++ b/leetcode/0845.Longest-Mountain-in-Array/845. Longest Mountain in Array.go	
@@ -23,3 +23,10 @@ func longestMountain(A []int) int {
 	}
 	return res
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0845. Longest Mountain in Array/845. Longest Mountain in Array_test.go b/leetcode/0845.Longest-Mountain-in-Array/845. Longest Mountain in Array_test.go
similarity index 92%
rename from Algorithms/0845. Longest Mountain in Array/845. Longest Mountain in Array_test.go
rename to leetcode/0845.Longest-Mountain-in-Array/845. Longest Mountain in Array_test.go
index ab87a6d3f..9edaaa4ea 100644
--- a/Algorithms/0845. Longest Mountain in Array/845. Longest Mountain in Array_test.go	
+++ b/leetcode/0845.Longest-Mountain-in-Array/845. Longest Mountain in Array_test.go	
@@ -25,27 +25,27 @@ type ans845 struct {
 func Test_Problem845(t *testing.T) {
 
 	qs := []question845{
-		question845{
+		{
 			para845{[]int{875, 884, 239, 731, 723, 685}},
 			ans845{4},
 		},
 
-		question845{
+		{
 			para845{[]int{0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0}},
 			ans845{11},
 		},
 
-		question845{
+		{
 			para845{[]int{2, 3}},
 			ans845{0},
 		},
 
-		question845{
+		{
 			para845{[]int{2, 1, 4, 7, 3, 2, 5}},
 			ans845{5},
 		},
 
-		question845{
+		{
 			para845{[]int{2, 2, 2}},
 			ans845{0},
 		},
diff --git a/Algorithms/0845. Longest Mountain in Array/README.md b/leetcode/0845.Longest-Mountain-in-Array/README.md
similarity index 100%
rename from Algorithms/0845. Longest Mountain in Array/README.md
rename to leetcode/0845.Longest-Mountain-in-Array/README.md
diff --git a/leetcode/0846.Hand-of-Straights/846.Hand of Straights.go b/leetcode/0846.Hand-of-Straights/846.Hand of Straights.go
new file mode 100644
index 000000000..ea54dda76
--- /dev/null
+++ b/leetcode/0846.Hand-of-Straights/846.Hand of Straights.go	
@@ -0,0 +1,23 @@
+package leetcode
+
+import "sort"
+
+func isNStraightHand(hand []int, groupSize int) bool {
+	mp := make(map[int]int)
+	for _, v := range hand {
+		mp[v] += 1
+	}
+	sort.Ints(hand)
+	for _, num := range hand {
+		if mp[num] == 0 {
+			continue
+		}
+		for diff := 0; diff < groupSize; diff++ {
+			if mp[num+diff] == 0 {
+				return false
+			}
+			mp[num+diff] -= 1
+		}
+	}
+	return true
+}
diff --git a/leetcode/0846.Hand-of-Straights/846.Hand of Straights_test.go b/leetcode/0846.Hand-of-Straights/846.Hand of Straights_test.go
new file mode 100644
index 000000000..2a172eb0e
--- /dev/null
+++ b/leetcode/0846.Hand-of-Straights/846.Hand of Straights_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question846 struct {
+	para846
+	ans846
+}
+
+// para 是参数
+type para846 struct {
+	hand      []int
+	groupSize int
+}
+
+// ans 是答案
+type ans846 struct {
+	ans bool
+}
+
+func Test_Problem846(t *testing.T) {
+
+	qs := []question846{
+
+		{
+			para846{[]int{1, 2, 3, 6, 2, 3, 4, 7, 8}, 3},
+			ans846{true},
+		},
+
+		{
+			para846{[]int{1, 2, 3, 4, 5}, 4},
+			ans846{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 846------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans846, q.para846
+		fmt.Printf("【input】:%v    【output】:%v\n", p, isNStraightHand(p.hand, p.groupSize))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0846.Hand-of-Straights/README.md b/leetcode/0846.Hand-of-Straights/README.md
new file mode 100644
index 000000000..5df17ba73
--- /dev/null
+++ b/leetcode/0846.Hand-of-Straights/README.md
@@ -0,0 +1,68 @@
+# [846. Hand of Straights](https://leetcode.com/problems/hand-of-straights/)
+
+## 题目
+
+Alice has some number of cards and she wants to rearrange the cards into groups so that each group is of size groupSize, and consists of groupSize consecutive cards.
+
+Given an integer array hand where hand[i] is the value written on the ith card and an integer groupSize, return true if she can rearrange the cards, or false otherwise.
+
+**Example 1**:
+
+    Input: hand = [1,2,3,6,2,3,4,7,8], groupSize = 3
+    Output: true
+    Explanation: Alice's hand can be rearranged as [1,2,3],[2,3,4],[6,7,8]
+
+**Example 2**:
+
+    Input: hand = [1,2,3,4,5], groupSize = 4
+    Output: false
+    Explanation: Alice's hand can not be rearranged into groups of 4.
+
+**Constraints:**
+
+- 1 <= hand.length <= 10000
+- 0 <= hand[i] <= 1000000000
+- 1 <= groupSize <= hand.length
+
+## 题目大意
+
+Alice 手中有一把牌，她想要重新排列这些牌，分成若干组，使每一组的牌数都是 groupSize ，并且由 groupSize 张连续的牌组成。
+
+给你一个整数数组 hand 其中 hand[i] 是写在第 i 张牌，和一个整数 groupSize 。如果她可能重新排列这些牌，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+贪心算法
+
+- 对hand升序排序
+- 对hand内数字进行哈希计数（key:数字，value:数量）
+- 遍历hand中的数字，以数量大于1的数字作为顺子开头，寻找顺子后续元素，若无法找到完整顺子则返回false
+- 所有数字都能找到完整顺子返回true
+
+##代码
+
+```go
+package leetcode
+
+import "sort"
+
+func isNStraightHand(hand []int, groupSize int) bool {
+	mp := make(map[int]int)
+	for _, v := range hand {
+		mp[v] += 1
+	}
+	sort.Ints(hand)
+	for _, num := range hand {
+		if mp[num] == 0 {
+			continue
+		}
+		for diff := 0; diff < groupSize; diff++ {
+			if mp[num+diff] == 0 {
+				return false
+			}
+			mp[num+diff] -= 1
+		}
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0850. Rectangle Area II/850. Rectangle Area II.go b/leetcode/0850.Rectangle-Area-II/850. Rectangle Area II.go
similarity index 100%
rename from Algorithms/0850. Rectangle Area II/850. Rectangle Area II.go
rename to leetcode/0850.Rectangle-Area-II/850. Rectangle Area II.go
diff --git a/leetcode/0850.Rectangle-Area-II/850. Rectangle Area II_test.go b/leetcode/0850.Rectangle-Area-II/850. Rectangle Area II_test.go
new file mode 100644
index 000000000..69f9f9401
--- /dev/null
+++ b/leetcode/0850.Rectangle-Area-II/850. Rectangle Area II_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question850 struct {
+	para850
+	ans850
+}
+
+// para 是参数
+// one 代表第一个参数
+type para850 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans850 struct {
+	one int
+}
+
+func Test_Problem850(t *testing.T) {
+
+	qs := []question850{
+
+		{
+			para850{[][]int{{0, 0, 3, 3}, {2, 0, 5, 3}, {1, 1, 4, 4}}},
+			ans850{18},
+		},
+
+		{
+			para850{[][]int{{0, 0, 1, 1}, {2, 2, 3, 3}}},
+			ans850{2},
+		},
+
+		{
+			para850{[][]int{{0, 0, 2, 2}, {1, 0, 2, 3}, {1, 0, 3, 1}}},
+			ans850{6},
+		},
+
+		{
+			para850{[][]int{{0, 0, 1000000000, 1000000000}}},
+			ans850{49},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 850------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans850, q.para850
+		fmt.Printf("【input】:%v       【output】:%v\n", p, rectangleArea(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0850.Rectangle-Area-II/README.md b/leetcode/0850.Rectangle-Area-II/README.md
new file mode 100755
index 000000000..4bce09397
--- /dev/null
+++ b/leetcode/0850.Rectangle-Area-II/README.md
@@ -0,0 +1,99 @@
+# [850. Rectangle Area II](https://leetcode.com/problems/rectangle-area-ii/)
+
+
+## 题目
+
+We are given a list of (axis-aligned) `rectangles`. Each `rectangle[i] = [x1, y1, x2, y2]` , where (x1, y1) are the coordinates of the bottom-left corner, and (x2, y2) are the coordinates of the top-right corner of the `i`th rectangle.
+
+Find the total area covered by all `rectangles` in the plane. Since the answer may be too large, **return it modulo 10^9 + 7**.
+
+![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/06/rectangle_area_ii_pic.png)
+
+**Example 1:**
+
+    Input: [[0,0,2,2],[1,0,2,3],[1,0,3,1]]
+    Output: 6
+    Explanation: As illustrated in the picture.
+
+**Example 2:**
+
+    Input: [[0,0,1000000000,1000000000]]
+    Output: 49
+    Explanation: The answer is 10^18 modulo (10^9 + 7), which is (10^9)^2 = (-7)^2 = 49.
+
+**Note:**
+
+- `1 <= rectangles.length <= 200`
+- `rectanges[i].length = 4`
+- `0 <= rectangles[i][j] <= 10^9`
+- The total area covered by all rectangles will never exceed `2^63 - 1` and thus will fit in a 64-bit signed integer.
+
+
+## 题目大意
+
+我们给出了一个（轴对齐的）矩形列表 rectangles。 对于 rectangle[i] = [x1, y1, x2, y2]，其中（x1，y1）是矩形 i 左下角的坐标，（x2，y2）是该矩形右上角的坐标。找出平面中所有矩形叠加覆盖后的总面积。由于答案可能太大，请返回它对 10 ^ 9 + 7 取模的结果。
+
+提示：
+
+- 1 <= rectangles.length <= 200
+- rectanges[i].length = 4
+- 0 <= rectangles[i][j] <= 10^9
+- 矩形叠加覆盖后的总面积不会超越 2^63 - 1 ，这意味着可以用一个 64 位有符号整数来保存面积结果。
+
+
+## 解题思路
+
+
+- 在二维坐标系中给出一些矩形，要求这些矩形合并之后的面积。由于矩形有重叠，所以需要考虑合并以后的面积。矩形的坐标值也会很大。
+- 这一题给人的感觉很像第 218 题，求天际线的过程也是有楼挡楼，重叠的情况。不过那一题只用求天际线的拐点，所以我们可以对区间做“右边界减一”的处理，防止两个相邻区间因为共点，而导致结果错误。但是这一题如果还是用相同的做法，就会出错，因为“右边界减一”以后，面积会少一部分，最终得到的结果也是偏小的。所以这一题要将线段树改造一下。
+- 思路是先讲 Y 轴上的坐标离线化，转换成线段树。将矩形的 2 条边变成扫描线，左边是入边，右边是出边。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_8.png)
+
+- 再从左往右遍历每条扫描线，并对 Y 轴上的线段树进行 update。X 轴上的每个坐标区间 * query 线段树总高度的结果 = 区间面积。最后将 X 轴对应的每个区间面积加起来，就是最终矩形合并以后的面积。如下图中间的图。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_9.png)
+
+    需要注意的一点是，**每次 query 的结果并不一定是连续线段**。如上图最右边的图，中间有一段是可能出现镂空的。这种情况看似复杂，其实很简单，因为每段线段树的线段代表的权值高度是不同的，每次 query 最大高度得到的结果已经考虑了中间可能有镂空的情况了。
+
+- 具体做法，先把各个矩形在 Y 轴方向上离散化，这里的**线段树叶子节点不再是一个点了，而是一个区间长度为 1 的区间段**。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_0.png)
+
+    每个叶子节点也不再是存储一个 int 值，而是存 2 个值，一个是 count 值，用来记录这条区间被覆盖的次数，另一个值是 val 值，用来反映射该线段长度是多少，因为 Y 轴被离散化了，区间坐标间隔都是 1，但是实际 Y 轴的高度并不是 1 ，所以用 val 来反映射原来的高度。
+
+- 初始化线段树，叶子节点的 count = 0，val 根据题目给的 Y 坐标进行计算。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_1.png)
+
+- 从左往右遍历每个扫描线。每条扫面线都把对应 update 更新到叶子节点。pushUp 的时候需要合并每个区间段的高度 val 值。如果有区间没有被覆盖，那么这个区间高度 val 为 0，这也就处理了可能“中间镂空”的情况。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_2.png)
+
+        func (sat *SegmentAreaTree) pushUp(treeIndex, leftTreeIndex, rightTreeIndex int) {
+        	newCount, newValue := sat.merge(sat.tree[leftTreeIndex].count, sat.tree[rightTreeIndex].count), 0
+        	if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count > 0 {
+        		newValue = sat.merge(sat.tree[leftTreeIndex].val, sat.tree[rightTreeIndex].val)
+        	} else if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count == 0 {
+        		newValue = sat.tree[leftTreeIndex].val
+        	} else if sat.tree[leftTreeIndex].count == 0 && sat.tree[rightTreeIndex].count > 0 {
+        		newValue = sat.tree[rightTreeIndex].val
+        	}
+        	sat.tree[treeIndex] = SegmentItem{count: newCount, val: newValue}
+        }
+
+- 扫描每一个扫描线，先 pushDown 到叶子节点，再 pushUp 到根节点。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_3.png)
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_4.png)
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_5.png)
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_6.png)
+
+- 遍历到倒数第 2 根扫描线的时候就能得到结果了。因为最后一根扫描线 update 以后，整个线段树全部都归为初始化状态了。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_7.png)
+
+- 这一题是线段树扫面线解法的经典题。
diff --git a/leetcode/0851.Loud-and-Rich/851. Loud and Rich.go b/leetcode/0851.Loud-and-Rich/851. Loud and Rich.go
new file mode 100644
index 000000000..9f178d8e7
--- /dev/null
+++ b/leetcode/0851.Loud-and-Rich/851. Loud and Rich.go	
@@ -0,0 +1,40 @@
+package leetcode
+
+func loudAndRich(richer [][]int, quiet []int) []int {
+	edges := make([][]int, len(quiet))
+	for i := range edges {
+		edges[i] = []int{}
+	}
+	indegrees := make([]int, len(quiet))
+	for _, edge := range richer {
+		n1, n2 := edge[0], edge[1]
+		edges[n1] = append(edges[n1], n2)
+		indegrees[n2]++
+	}
+	res := make([]int, len(quiet))
+	for i := range res {
+		res[i] = i
+	}
+	queue := []int{}
+	for i, v := range indegrees {
+		if v == 0 {
+			queue = append(queue, i)
+		}
+	}
+	for len(queue) > 0 {
+		nexts := []int{}
+		for _, n1 := range queue {
+			for _, n2 := range edges[n1] {
+				indegrees[n2]--
+				if quiet[res[n2]] > quiet[res[n1]] {
+					res[n2] = res[n1]
+				}
+				if indegrees[n2] == 0 {
+					nexts = append(nexts, n2)
+				}
+			}
+		}
+		queue = nexts
+	}
+	return res
+}
diff --git a/leetcode/0851.Loud-and-Rich/851. Loud and Rich_test.go b/leetcode/0851.Loud-and-Rich/851. Loud and Rich_test.go
new file mode 100644
index 000000000..f3ebd5874
--- /dev/null
+++ b/leetcode/0851.Loud-and-Rich/851. Loud and Rich_test.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question851 struct {
+	para851
+	ans851
+}
+
+// para 是参数
+// one 代表第一个参数
+type para851 struct {
+	richer [][]int
+	quiet  []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans851 struct {
+	one []int
+}
+
+func Test_Problem851(t *testing.T) {
+
+	qs := []question851{
+
+		{
+			para851{[][]int{{1, 0}, {2, 1}, {3, 1}, {3, 7}, {4, 3}, {5, 3}}, []int{3, 2, 5, 4, 6, 1, 7, 0}},
+			ans851{[]int{5, 5, 2, 5, 4, 5, 6, 7}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 851------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans851, q.para851
+		fmt.Printf("【input】:%v       【output】:%v\n", p, loudAndRich(p.richer, p.quiet))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0851.Loud-and-Rich/README.md b/leetcode/0851.Loud-and-Rich/README.md
new file mode 100644
index 000000000..26351328b
--- /dev/null
+++ b/leetcode/0851.Loud-and-Rich/README.md
@@ -0,0 +1,108 @@
+# [851. Loud and Rich](https://leetcode.com/problems/loud-and-rich/)
+
+
+
+## 题目
+
+In a group of N people (labelled `0, 1, 2, ..., N-1`), each person has different amounts of money, and different levels of quietness.
+
+For convenience, we'll call the person with label `x`, simply "person `x`".
+
+We'll say that `richer[i] = [x, y]` if person `x` definitely has more money than person `y`. Note that `richer` may only be a subset of valid observations.
+
+Also, we'll say `quiet = q` if person x has quietness `q`.
+
+Now, return `answer`, where `answer = y` if `y` is the least quiet person (that is, the person `y` with the smallest value of `quiet[y]`), among all people who definitely have equal to or more money than person `x`.
+
+**Example 1**:
+
+```
+Input: richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [3,2,5,4,6,1,7,0]
+Output: [5,5,2,5,4,5,6,7]
+Explanation: 
+answer[0] = 5.
+Person 5 has more money than 3, which has more money than 1, which has more money than 0.
+The only person who is quieter (has lower quiet[x]) is person 7, but
+it isn't clear if they have more money than person 0.
+
+answer[7] = 7.
+Among all people that definitely have equal to or more money than person 7
+(which could be persons 3, 4, 5, 6, or 7), the person who is the quietest (has lower quiet[x])
+is person 7.
+
+The other answers can be filled out with similar reasoning.
+```
+
+**Note**:
+
+1. `1 <= quiet.length = N <= 500`
+2. `0 <= quiet[i] < N`, all `quiet[i]` are different.
+3. `0 <= richer.length <= N * (N-1) / 2`
+4. `0 <= richer[i][j] < N`
+5. `richer[i][0] != richer[i][1]`
+6. `richer[i]`'s are all different.
+7. The observations in `richer` are all logically consistent.
+
+## 题目大意
+
+在一组 N 个人（编号为 0, 1, 2, ..., N-1）中，每个人都有不同数目的钱，以及不同程度的安静（quietness）。为了方便起见，我们将编号为 x 的人简称为 "person x "。如果能够肯定 person x 比 person y 更有钱的话，我们会说 richer[i] = [x, y] 。注意 richer 可能只是有效观察的一个子集。另外，如果 person x 的安静程度为 q ，我们会说 quiet[x] = q 。现在，返回答案 answer ，其中 answer[x] = y 的前提是，在所有拥有的钱不少于 person x 的人中，person y 是最安静的人（也就是安静值 quiet[y] 最小的人）。
+
+提示：
+
+- 1 <= quiet.length = N <= 500
+- 0 <= quiet[i] < N，所有 quiet[i] 都不相同。
+- 0 <= richer.length <= N * (N-1) / 2
+- 0 <= richer[i][j] < N
+- richer[i][0] != richer[i][1]
+- richer[i] 都是不同的。
+- 对 richer 的观察在逻辑上是一致的。
+
+
+## 解题思路
+
+- 给出 2 个数组，richer 和 quiet，要求输出 answer，其中 answer = y 的前提是，在所有拥有的钱不少于 x 的人中，y 是最安静的人（也就是安静值 quiet[y] 最小的人）
+- 由题意可知，`richer` 构成了一个有向无环图，首先使用字典建立图的关系，找到比当前下标编号富有的所有的人。然后使用广度优先层次遍历，不断的使用富有的人，但是安静值更小的人更新子节点即可。
+- 这一题还可以用拓扑排序来解答。将 `richer` 中描述的关系看做边，如果 `x > y`，则 `x` 指向 `y`。将 `quiet` 看成权值。用一个数组记录答案，初始时 `ans[i] = i`。然后对原图做拓扑排序，对于每一条边，如果发现 `quiet[ans[v]] > quiet[ans[u]]`，则 `ans[v]` 的答案为 `ans[u]`。时间复杂度即为拓扑排序的时间复杂度为 `O(m+n)`。空间复杂度需要 `O(m)` 的数组建图，需要 `O(n)` 的数组记录入度以及存储队列，所以空间复杂度为 `O(m+n)`。
+
+## 代码
+
+```go
+func loudAndRich(richer [][]int, quiet []int) []int {
+	edges := make([][]int, len(quiet))
+	for i := range edges {
+		edges[i] = []int{}
+	}
+	indegrees := make([]int, len(quiet))
+	for _, edge := range richer {
+		n1, n2 := edge[0], edge[1]
+		edges[n1] = append(edges[n1], n2)
+		indegrees[n2]++
+	}
+	res := make([]int, len(quiet))
+	for i := range res {
+		res[i] = i
+	}
+	queue := []int{}
+	for i, v := range indegrees {
+		if v == 0 {
+			queue = append(queue, i)
+		}
+	}
+	for len(queue) > 0 {
+		nexts := []int{}
+		for _, n1 := range queue {
+			for _, n2 := range edges[n1] {
+				indegrees[n2]--
+				if quiet[res[n2]] > quiet[res[n1]] {
+					res[n2] = res[n1]
+				}
+				if indegrees[n2] == 0 {
+					nexts = append(nexts, n2)
+				}
+			}
+		}
+		queue = nexts
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0852. Peak Index in a Mountain Array/852. Peak Index in a Mountain Array.go b/leetcode/0852.Peak-Index-in-a-Mountain-Array/852. Peak Index in a Mountain Array.go
similarity index 100%
rename from Algorithms/0852. Peak Index in a Mountain Array/852. Peak Index in a Mountain Array.go
rename to leetcode/0852.Peak-Index-in-a-Mountain-Array/852. Peak Index in a Mountain Array.go
diff --git a/Algorithms/0852. Peak Index in a Mountain Array/852. Peak Index in a Mountain Array_test.go b/leetcode/0852.Peak-Index-in-a-Mountain-Array/852. Peak Index in a Mountain Array_test.go
similarity index 95%
rename from Algorithms/0852. Peak Index in a Mountain Array/852. Peak Index in a Mountain Array_test.go
rename to leetcode/0852.Peak-Index-in-a-Mountain-Array/852. Peak Index in a Mountain Array_test.go
index 28427fbdf..431cd572a 100644
--- a/Algorithms/0852. Peak Index in a Mountain Array/852. Peak Index in a Mountain Array_test.go	
+++ b/leetcode/0852.Peak-Index-in-a-Mountain-Array/852. Peak Index in a Mountain Array_test.go	
@@ -26,12 +26,12 @@ func Test_Problem852(t *testing.T) {
 
 	qs := []question852{
 
-		question852{
+		{
 			para852{[]int{0, 1, 0}},
 			ans852{1},
 		},
 
-		question852{
+		{
 			para852{[]int{0, 2, 1, 0}},
 			ans852{1},
 		},
diff --git a/leetcode/0852.Peak-Index-in-a-Mountain-Array/README.md b/leetcode/0852.Peak-Index-in-a-Mountain-Array/README.md
new file mode 100755
index 000000000..ad167925e
--- /dev/null
+++ b/leetcode/0852.Peak-Index-in-a-Mountain-Array/README.md
@@ -0,0 +1,49 @@
+# [852. Peak Index in a Mountain Array](https://leetcode.com/problems/peak-index-in-a-mountain-array/)
+
+
+## 题目
+
+Let's call an array `A` a *mountain* if the following properties hold:
+
+- `A.length >= 3`
+- There exists some `0 < i < A.length - 1` such that `A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]`
+
+Given an array that is definitely a mountain, return any `i` such that `A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]`.
+
+**Example 1:**
+
+    Input: [0,1,0]
+    Output: 1
+
+**Example 2:**
+
+    Input: [0,2,1,0]
+    Output: 1
+
+**Note:**
+
+1. `3 <= A.length <= 10000`
+2. `0 <= A[i] <= 10^6`
+3. A is a mountain, as defined above.
+
+
+## 题目大意
+
+我们把符合下列属性的数组 A 称作山脉：
+
+- A.length >= 3
+- 存在 0 < i < A.length - 1 使得A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]
+给定一个确定为山脉的数组，返回任何满足 A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1] 的 i 的值。
+
+提示：
+
+- 3 <= A.length <= 10000
+- 0 <= A[i] <= 10^6
+- A 是如上定义的山脉
+ 
+ 
+
+## 解题思路
+
+- 给出一个数组，数组里面存在有且仅有一个“山峰”，(山峰的定义是，下标 `i` 比 `i-1`、`i+1` 位置上的元素都要大)，找到这个“山峰”，并输出其中一个山峰的下标。
+- 这一题直接用二分搜索即可，数组中的元素算基本有序。判断是否为山峰的条件为比较左右两个数，如果当前的数比左右两个数都大，即找到了山峰。其他的情况都在山坡上。这一题有两种写法，第一种写法是标准的二分写法，第二种写法是变形的二分写法。
diff --git a/Algorithms/0853. Car Fleet/853. Car Fleet.go b/leetcode/0853.Car-Fleet/853. Car Fleet.go
similarity index 100%
rename from Algorithms/0853. Car Fleet/853. Car Fleet.go
rename to leetcode/0853.Car-Fleet/853. Car Fleet.go
diff --git a/Algorithms/0853. Car Fleet/853. Car Fleet_test.go b/leetcode/0853.Car-Fleet/853. Car Fleet_test.go
similarity index 97%
rename from Algorithms/0853. Car Fleet/853. Car Fleet_test.go
rename to leetcode/0853.Car-Fleet/853. Car Fleet_test.go
index bfeac0400..03fc2d392 100644
--- a/Algorithms/0853. Car Fleet/853. Car Fleet_test.go	
+++ b/leetcode/0853.Car-Fleet/853. Car Fleet_test.go	
@@ -28,7 +28,7 @@ func Test_Problem853(t *testing.T) {
 
 	qs := []question853{
 
-		question853{
+		{
 			para853{12, []int{10, 8, 0, 5, 3}, []int{2, 4, 1, 1, 3}},
 			ans853{3},
 		},
diff --git a/Algorithms/0853. Car Fleet/README.md b/leetcode/0853.Car-Fleet/README.md
similarity index 100%
rename from Algorithms/0853. Car Fleet/README.md
rename to leetcode/0853.Car-Fleet/README.md
diff --git a/Algorithms/0856. Score of Parentheses/856. Score of Parentheses.go b/leetcode/0856.Score-of-Parentheses/856. Score of Parentheses.go
similarity index 100%
rename from Algorithms/0856. Score of Parentheses/856. Score of Parentheses.go
rename to leetcode/0856.Score-of-Parentheses/856. Score of Parentheses.go
diff --git a/Algorithms/0856. Score of Parentheses/856. Score of Parentheses_test.go b/leetcode/0856.Score-of-Parentheses/856. Score of Parentheses_test.go
similarity index 90%
rename from Algorithms/0856. Score of Parentheses/856. Score of Parentheses_test.go
rename to leetcode/0856.Score-of-Parentheses/856. Score of Parentheses_test.go
index 634f6c7d8..d26fddfe1 100644
--- a/Algorithms/0856. Score of Parentheses/856. Score of Parentheses_test.go	
+++ b/leetcode/0856.Score-of-Parentheses/856. Score of Parentheses_test.go	
@@ -25,32 +25,32 @@ type ans856 struct {
 func Test_Problem856(t *testing.T) {
 
 	qs := []question856{
-		question856{
+		{
 			para856{"()"},
 			ans856{1},
 		},
 
-		question856{
+		{
 			para856{"(())"},
 			ans856{2},
 		},
 
-		question856{
+		{
 			para856{"()()"},
 			ans856{2},
 		},
 
-		question856{
+		{
 			para856{"(()(()))"},
 			ans856{6},
 		},
 
-		question856{
+		{
 			para856{"()(())"},
 			ans856{3},
 		},
 
-		question856{
+		{
 			para856{"((()()))"},
 			ans856{8},
 		},
diff --git a/Algorithms/0856. Score of Parentheses/README.md b/leetcode/0856.Score-of-Parentheses/README.md
similarity index 100%
rename from Algorithms/0856. Score of Parentheses/README.md
rename to leetcode/0856.Score-of-Parentheses/README.md
diff --git a/leetcode/0859.Buddy-Strings/859.Buddy Strings.go b/leetcode/0859.Buddy-Strings/859.Buddy Strings.go
new file mode 100644
index 000000000..8688f16e3
--- /dev/null
+++ b/leetcode/0859.Buddy-Strings/859.Buddy Strings.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+func buddyStrings(s string, goal string) bool {
+	if len(s) != len(goal) || len(s) <= 1 {
+		return false
+	}
+	mp := make(map[byte]int)
+	if s == goal {
+		for i := 0; i < len(s); i++ {
+			if _, ok := mp[s[i]]; ok {
+				return true
+			}
+			mp[s[i]]++
+		}
+		return false
+	}
+	first, second := -1, -1
+	for i := 0; i < len(s); i++ {
+		if s[i] != goal[i] {
+			if first == -1 {
+				first = i
+			} else if second == -1 {
+				second = i
+			} else {
+				return false
+			}
+		}
+	}
+	return second != -1 && s[first] == goal[second] && s[second] == goal[first]
+}
diff --git a/leetcode/0859.Buddy-Strings/859.Buddy Strings_test.go b/leetcode/0859.Buddy-Strings/859.Buddy Strings_test.go
new file mode 100644
index 000000000..3085e448e
--- /dev/null
+++ b/leetcode/0859.Buddy-Strings/859.Buddy Strings_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question859 struct {
+	para859
+	ans859
+}
+
+// para 是参数
+type para859 struct {
+	s    string
+	goal string
+}
+
+// ans 是答案
+type ans859 struct {
+	ans bool
+}
+
+func Test_Problem859(t *testing.T) {
+
+	qs := []question859{
+
+		{
+			para859{"ab", "ba"},
+			ans859{true},
+		},
+
+		{
+			para859{"ab", "ab"},
+			ans859{false},
+		},
+
+		{
+			para859{"aa", "aa"},
+			ans859{true},
+		},
+
+		{
+			para859{"aaaaaaabc", "aaaaaaacb"},
+			ans859{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 859------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans859, q.para859
+		fmt.Printf("【input】:%v    【output】:%v\n", p, buddyStrings(p.s, p.goal))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0859.Buddy-Strings/README.md b/leetcode/0859.Buddy-Strings/README.md
new file mode 100644
index 000000000..7a7e5c650
--- /dev/null
+++ b/leetcode/0859.Buddy-Strings/README.md
@@ -0,0 +1,87 @@
+# [859. Buddy Strings](https://leetcode.com/problems/buddy-strings/)
+
+## 题目
+
+Given two strings s and goal, return true if you can swap two letters in s so the result is equal to goal, otherwise, return false.
+
+Swapping letters is defined as taking two indices i and j (0-indexed) such that i != j and swapping the characters at s[i] and s[j].
+
+For example, swapping at indices 0 and 2 in "abcd" results in "cbad".
+
+**Example 1**:
+
+    Input: s = "ab", goal = "ba"
+    Output: true
+    Explanation: You can swap s[0] = 'a' and s[1] = 'b' to get "ba", which is equal to goal.
+
+**Example 2**:
+
+    Input: s = "ab", goal = "ab"
+    Output: false
+    Explanation: The only letters you can swap are s[0] = 'a' and s[1] = 'b', which results in "ba" != goal.
+
+**Example 3**:
+
+    Input: s = "aa", goal = "aa"
+    Output: true
+    Explanation: You can swap s[0] = 'a' and s[1] = 'a' to get "aa", which is equal to goal.
+
+**Example 4**:
+
+    Input: s = "aaaaaaabc", goal = "aaaaaaacb"
+    Output: true
+
+**Constraints:**
+
+- 1 <= s.length, goal.length <= 2 * 10000
+- s and goal consist of lowercase letters.
+
+## 题目大意
+
+给你两个字符串 s 和 goal ，只要我们可以通过交换 s 中的两个字母得到与 goal 相等的结果，就返回 true；否则返回 false 。
+
+交换字母的定义是：取两个下标 i 和 j （下标从 0 开始）且满足 i != j ，接着交换 s[i] 和 s[j] 处的字符。
+
+例如，在 "abcd" 中交换下标 0 和下标 2 的元素可以生成 "cbad" 。
+
+## 解题思路
+
+分为两种情况进行比较:
+- s 等于 goal, s 中有重复元素就返回 true,否则返回 false
+- s 不等于 goal, s 中有两个下标不同的字符与 goal 中对应下标的字符分别相等
+
+## 代码
+
+```go
+
+package leetcode
+
+func buddyStrings(s string, goal string) bool {
+	if len(s) != len(goal) || len(s) <= 1 {
+		return false
+	}
+	mp := make(map[byte]int)
+	if s == goal {
+		for i := 0; i < len(s); i++ {
+			if _, ok := mp[s[i]]; ok {
+				return true
+			}
+			mp[s[i]]++
+		}
+		return false
+	}
+	first, second := -1, -1
+	for i := 0; i < len(s); i++ {
+		if s[i] != goal[i] {
+			if first == -1 {
+				first = i
+			} else if second == -1 {
+				second = i
+			} else {
+				return false
+			}
+		}
+	}
+	return second != -1 && s[first] == goal[second] && s[second] == goal[first]
+}
+```
\ No newline at end of file
diff --git a/leetcode/0862.Shortest-Subarray-with-Sum-at-Least-K/862. Shortest Subarray with Sum at Least K.go b/leetcode/0862.Shortest-Subarray-with-Sum-at-Least-K/862. Shortest Subarray with Sum at Least K.go
new file mode 100644
index 000000000..4bf04c48c
--- /dev/null
+++ b/leetcode/0862.Shortest-Subarray-with-Sum-at-Least-K/862. Shortest Subarray with Sum at Least K.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+func shortestSubarray(A []int, K int) int {
+	res, prefixSum := len(A)+1, make([]int, len(A)+1)
+	for i := 0; i < len(A); i++ {
+		prefixSum[i+1] = prefixSum[i] + A[i]
+	}
+	// deque 中保存递增的 prefixSum 下标
+	deque := []int{}
+	for i := range prefixSum {
+		// 下面这个循环希望能找到 [deque[0], i] 区间内累加和 >= K，如果找到了就更新答案
+		for len(deque) > 0 && prefixSum[i]-prefixSum[deque[0]] >= K {
+			length := i - deque[0]
+			if res > length {
+				res = length
+			}
+			// 找到第一个 deque[0] 能满足条件以后，就移除它，因为它是最短长度的子序列了
+			deque = deque[1:]
+		}
+		// 下面这个循环希望能保证 prefixSum[deque[i]] 递增
+		for len(deque) > 0 && prefixSum[i] <= prefixSum[deque[len(deque)-1]] {
+			deque = deque[:len(deque)-1]
+		}
+		deque = append(deque, i)
+	}
+	if res <= len(A) {
+		return res
+	}
+	return -1
+}
diff --git a/leetcode/0862.Shortest-Subarray-with-Sum-at-Least-K/862. Shortest Subarray with Sum at Least K_test.go b/leetcode/0862.Shortest-Subarray-with-Sum-at-Least-K/862. Shortest Subarray with Sum at Least K_test.go
new file mode 100644
index 000000000..8f585559f
--- /dev/null
+++ b/leetcode/0862.Shortest-Subarray-with-Sum-at-Least-K/862. Shortest Subarray with Sum at Least K_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question862 struct {
+	para862
+	ans862
+}
+
+// para 是参数
+// one 代表第一个参数
+type para862 struct {
+	A []int
+	K int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans862 struct {
+	one int
+}
+
+func Test_Problem862(t *testing.T) {
+
+	qs := []question862{
+		{
+			para862{[]int{1}, 1},
+			ans862{1},
+		},
+
+		{
+			para862{[]int{1, 2}, 4},
+			ans862{-1},
+		},
+
+		{
+			para862{[]int{2, -1, 2}, 3},
+			ans862{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 862------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans862, q.para862
+		fmt.Printf("【input】:%v       【output】:%v\n", p, shortestSubarray(p.A, p.K))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0862.Shortest-Subarray-with-Sum-at-Least-K/README.md b/leetcode/0862.Shortest-Subarray-with-Sum-at-Least-K/README.md
new file mode 100644
index 000000000..469ca3bd7
--- /dev/null
+++ b/leetcode/0862.Shortest-Subarray-with-Sum-at-Least-K/README.md
@@ -0,0 +1,90 @@
+# [862. Shortest Subarray with Sum at Least K](https://leetcode.com/problems/shortest-subarray-with-sum-at-least-k/)
+
+
+
+## 题目
+
+Return the **length** of the shortest, non-empty, contiguous subarray of `A` with sum at least `K`.
+
+If there is no non-empty subarray with sum at least `K`, return `-1`.
+
+**Example 1**:
+
+```
+Input: A = [1], K = 1
+Output: 1
+```
+
+**Example 2**:
+
+```
+Input: A = [1,2], K = 4
+Output: -1
+```
+
+**Example 3**:
+
+```
+Input: A = [2,-1,2], K = 3
+Output: 3
+```
+
+**Note**:
+
+1. `1 <= A.length <= 50000`
+2. `-10 ^ 5 <= A[i] <= 10 ^ 5`
+3. `1 <= K <= 10 ^ 9`
+
+## 题目大意
+
+返回 A 的最短的非空连续子数组的长度，该子数组的和至少为 K 。如果没有和至少为 K 的非空子数组，返回 -1 。
+
+提示：
+
+- 1 <= A.length <= 50000
+- -10 ^ 5 <= A[i] <= 10 ^ 5
+- 1 <= K <= 10 ^ 9
+
+
+## 解题思路
+
+- 给出一个数组，要求找出一个最短的，非空的，连续的子序列且累加和至少为 k 。
+- 由于给的数组里面可能存在负数，所以子数组的累加和不会随着数组的长度增加而增加。题目中要求区间和，理所应当需要利用 `prefixSum` 前缀和，先计算出 `prefixSum` 前缀和。
+- 简化一下题目的要求，即能否找到 `prefixSum[y] - prefixSum[x] ≥ K` ，且 `y - x` 的差值最小。如果固定的 `y`，那么对于 `x`，`x` 越大，`y - x` 的差值就越小(因为 `x` 越逼近 `y`)。所以想求区间 `[x, y]` 的最短距离，需要保证 `y` 尽量小，`x` 尽量大，这样 `[x, y]` 区间距离最小。那么有以下 2 点“常识”一定成立：
+    1. 如果 `x1 < x2` ，并且 `prefixSum[x2] ≤ prefixSum[x1]`，说明结果一定不会取 `x1`。因为如果 `prefixSum[x1] ≤ prefixSum[y] - k`，那么  `prefixSum[x2] ≤ prefixSum[x1] ≤ prefixSum[y] - k`，`x2` 也能满足题意，并且 `x2` 比 `x1` 更加接近 `y`，最优解一定优先考虑 `x2`。
+    2. 在确定了 `x` 以后，以后就不用再考虑 `x` 了，因为如果 `y2 > y1`，且 `y2` 的时候取 `x` 还是一样的，那么算距离的话，`y2 - x` 显然大于 `y1 - x`，这样的话肯定不会是最短的距离。
+- 从上面这两个常识来看，可以用双端队列 `deque` 来处理 `prefixSum`。`deque` 中存储的是递增的 `x` 下标，为了满足常识一。从双端队列的开头开始遍历，假如区间和之差大于等于 `K`，就移除队首元素并更新结果 `res`。队首移除元素，直到不满足 `prefixSum[i]-prefixSum[deque[0]] >= K` 这一不等式，是为了满足常识二。之后的循环是此题的精髓，从双端队列的末尾开始往前遍历，假如当前区间和 `prefixSum[i]` 小于等于队列末尾的区间和，则移除队列末尾元素。为什么这样处理呢？因为若数组都是正数，那么长度越长，区间和一定越大，则 `prefixSum[i]` 一定大于所有双端队列中的区间和，但由于可能存在负数，从而使得长度变长，区间总和反而减少了，之前的区间和之差值都没有大于等于 `K`(< K)，现在的更不可能大于等于 `K`，这个结束位置可以直接淘汰，不用进行计算。循环结束后将当前位置加入双端队列即可。遇到新下标在队尾移除若干元素，这一行为，也是为了满足常识一。
+- 由于所有下标都只进队列一次，也最多 pop 出去一次，所以时间复杂度 O(n)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+func shortestSubarray(A []int, K int) int {
+	res, prefixSum := len(A)+1, make([]int, len(A)+1)
+	for i := 0; i < len(A); i++ {
+		prefixSum[i+1] = prefixSum[i] + A[i]
+	}
+	// deque 中保存递增的 prefixSum 下标
+	deque := []int{}
+	for i := range prefixSum {
+		// 下面这个循环希望能找到 [deque[0], i] 区间内累加和 >= K，如果找到了就更新答案
+		for len(deque) > 0 && prefixSum[i]-prefixSum[deque[0]] >= K {
+			length := i - deque[0]
+			if res > length {
+				res = length
+			}
+			// 找到第一个 deque[0] 能满足条件以后，就移除它，因为它是最短长度的子序列了
+			deque = deque[1:]
+		}
+		// 下面这个循环希望能保证 prefixSum[deque[i]] 递增
+		for len(deque) > 0 && prefixSum[i] <= prefixSum[deque[len(deque)-1]] {
+			deque = deque[:len(deque)-1]
+		}
+		deque = append(deque, i)
+	}
+	if res <= len(A) {
+		return res
+	}
+	return -1
+}
+```
\ No newline at end of file
diff --git a/leetcode/0863.All-Nodes-Distance-K-in-Binary-Tree/863. All Nodes Distance K in Binary Tree.go b/leetcode/0863.All-Nodes-Distance-K-in-Binary-Tree/863. All Nodes Distance K in Binary Tree.go
new file mode 100644
index 000000000..43e365837
--- /dev/null
+++ b/leetcode/0863.All-Nodes-Distance-K-in-Binary-Tree/863. All Nodes Distance K in Binary Tree.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func distanceK(root *TreeNode, target *TreeNode, K int) []int {
+	visit := []int{}
+	findDistanceK(root, target, K, &visit)
+	return visit
+}
+
+func findDistanceK(root, target *TreeNode, K int, visit *[]int) int {
+	if root == nil {
+		return -1
+	}
+	if root == target {
+		findChild(root, K, visit)
+		return K - 1
+	}
+	leftDistance := findDistanceK(root.Left, target, K, visit)
+	if leftDistance == 0 {
+		findChild(root, leftDistance, visit)
+	}
+	if leftDistance > 0 {
+		findChild(root.Right, leftDistance-1, visit)
+		return leftDistance - 1
+	}
+	rightDistance := findDistanceK(root.Right, target, K, visit)
+	if rightDistance == 0 {
+		findChild(root, rightDistance, visit)
+	}
+	if rightDistance > 0 {
+		findChild(root.Left, rightDistance-1, visit)
+		return rightDistance - 1
+	}
+	return -1
+}
+
+func findChild(root *TreeNode, K int, visit *[]int) {
+	if root == nil {
+		return
+	}
+	if K == 0 {
+		*visit = append(*visit, root.Val)
+	} else {
+		findChild(root.Left, K-1, visit)
+		findChild(root.Right, K-1, visit)
+	}
+}
diff --git a/leetcode/0863.All-Nodes-Distance-K-in-Binary-Tree/863. All Nodes Distance K in Binary Tree_test.go b/leetcode/0863.All-Nodes-Distance-K-in-Binary-Tree/863. All Nodes Distance K in Binary Tree_test.go
new file mode 100644
index 000000000..766751e09
--- /dev/null
+++ b/leetcode/0863.All-Nodes-Distance-K-in-Binary-Tree/863. All Nodes Distance K in Binary Tree_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question863 struct {
+	para863
+	ans863
+}
+
+// para 是参数
+// one 代表第一个参数
+type para863 struct {
+	root   []int
+	target []int
+	K      int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans863 struct {
+	one []int
+}
+
+func Test_Problem863(t *testing.T) {
+
+	qs := []question863{
+
+		{
+			para863{[]int{3, 5, 1, 6, 2, 0, 8, structures.NULL, structures.NULL, 7, 4}, []int{5}, 2},
+			ans863{[]int{7, 4, 1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 863------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans863, q.para863
+		tree, target := structures.Ints2TreeNode(p.root), structures.Ints2TreeNode(p.target)
+		fmt.Printf("【input】:%v       【output】:%v\n", p, distanceK(tree, target, p.K))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0863.All-Nodes-Distance-K-in-Binary-Tree/README.md b/leetcode/0863.All-Nodes-Distance-K-in-Binary-Tree/README.md
new file mode 100644
index 000000000..4ea32b206
--- /dev/null
+++ b/leetcode/0863.All-Nodes-Distance-K-in-Binary-Tree/README.md
@@ -0,0 +1,96 @@
+# [863. All Nodes Distance K in Binary Tree](https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/)
+
+
+
+## 题目
+
+We are given a binary tree (with root node `root`), a `target` node, and an integer value `K`.
+
+Return a list of the values of all nodes that have a distance `K` from the `target` node. The answer can be returned in any order.
+
+**Example 1**:
+
+```
+Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2
+
+Output: [7,4,1]
+
+Explanation: 
+The nodes that are a distance 2 from the target node (with value 5)
+have values 7, 4, and 1.
+```
+
+![https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/28/sketch0.png](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/28/sketch0.png)
+
+**Note**:
+
+1. The given tree is non-empty.
+2. Each node in the tree has unique values `0 <= node.val <= 500`.
+3. The `target` node is a node in the tree.
+4. `0 <= K <= 1000`.
+
+## 题目大意
+
+给定一个二叉树（具有根结点 root）， 一个目标结点 target ，和一个整数值 K 。返回到目标结点 target 距离为 K 的所有结点的值的列表。 答案可以以任何顺序返回。
+
+提示：
+
+- 给定的树是非空的。
+- 树上的每个结点都具有唯一的值 0 <= node.val <= 500 。
+- 目标结点 target 是树上的结点。
+- 0 <= K <= 1000.
+
+
+## 解题思路
+
+- 给出一颗树和一个目标节点 target，一个距离 K，要求找到所有距离目标节点 target 的距离是 K 的点。
+- 这一题用 DFS 的方法解题。先找到当前节点距离目标节点的距离，如果在左子树中找到了 target，距离当前节点的距离 > 0，则还需要在它的右子树中查找剩下的距离。如果是在右子树中找到了 target，反之同理。如果当前节点就是目标节点，那么就可以直接记录这个点。否则每次遍历一个点，距离都减一。
+
+## 代码
+
+```go
+func distanceK(root *TreeNode, target *TreeNode, K int) []int {
+	visit := []int{}
+	findDistanceK(root, target, K, &visit)
+	return visit
+}
+
+func findDistanceK(root, target *TreeNode, K int, visit *[]int) int {
+	if root == nil {
+		return -1
+	}
+	if root == target {
+		findChild(root, K, visit)
+		return K - 1
+	}
+	leftDistance := findDistanceK(root.Left, target, K, visit)
+	if leftDistance == 0 {
+		findChild(root, leftDistance, visit)
+	}
+	if leftDistance > 0 {
+		findChild(root.Right, leftDistance-1, visit)
+		return leftDistance - 1
+	}
+	rightDistance := findDistanceK(root.Right, target, K, visit)
+	if rightDistance == 0 {
+		findChild(root, rightDistance, visit)
+	}
+	if rightDistance > 0 {
+		findChild(root.Left, rightDistance-1, visit)
+		return rightDistance - 1
+	}
+	return -1
+}
+
+func findChild(root *TreeNode, K int, visit *[]int) {
+	if root == nil {
+		return
+	}
+	if K == 0 {
+		*visit = append(*visit, root.Val)
+	} else {
+		findChild(root.Left, K-1, visit)
+		findChild(root.Right, K-1, visit)
+	}
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0864. Shortest Path to Get All Keys/864. Shortest Path to Get All Keys.go b/leetcode/0864.Shortest-Path-to-Get-All-Keys/864. Shortest Path to Get All Keys.go
similarity index 94%
rename from Algorithms/0864. Shortest Path to Get All Keys/864. Shortest Path to Get All Keys.go
rename to leetcode/0864.Shortest-Path-to-Get-All-Keys/864. Shortest Path to Get All Keys.go
index 7a42a24d0..fc34e9879 100644
--- a/Algorithms/0864. Shortest Path to Get All Keys/864. Shortest Path to Get All Keys.go	
+++ b/leetcode/0864.Shortest-Path-to-Get-All-Keys/864. Shortest Path to Get All Keys.go	
@@ -5,6 +5,13 @@ import (
 	"strings"
 )
 
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
 // 解法一 BFS，利用状态压缩来过滤筛选状态
 func shortestPathAllKeys(grid []string) int {
 	if len(grid) == 0 {
@@ -78,6 +85,10 @@ func shortestPathAllKeys(grid []string) int {
 	return -1
 }
 
+func isInBoard(board [][]byte, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+
 // 解法二 DFS，但是超时了，剪枝条件不够强
 func shortestPathAllKeys1(grid []string) int {
 	if len(grid) == 0 {
@@ -166,3 +177,10 @@ func isKey(board [][]byte, x, y int) bool {
 	}
 	return false
 }
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/Algorithms/0864. Shortest Path to Get All Keys/864. Shortest Path to Get All Keys_test.go b/leetcode/0864.Shortest-Path-to-Get-All-Keys/864. Shortest Path to Get All Keys_test.go
similarity index 91%
rename from Algorithms/0864. Shortest Path to Get All Keys/864. Shortest Path to Get All Keys_test.go
rename to leetcode/0864.Shortest-Path-to-Get-All-Keys/864. Shortest Path to Get All Keys_test.go
index 5f26e4cbb..d1263f372 100644
--- a/Algorithms/0864. Shortest Path to Get All Keys/864. Shortest Path to Get All Keys_test.go	
+++ b/leetcode/0864.Shortest-Path-to-Get-All-Keys/864. Shortest Path to Get All Keys_test.go	
@@ -26,42 +26,42 @@ func Test_Problem864(t *testing.T) {
 
 	qs := []question864{
 
-		question864{
+		{
 			para864{[]string{".##..##..B", "##...#...#", "..##..#...", ".#..#b...#", "#.##.a.###", ".#....#...", ".##..#.#..", ".....###@.", "..........", ".........A"}},
 			ans864{11},
 		},
 
-		question864{
+		{
 			para864{[]string{"Dd#b@", ".fE.e", "##.B.", "#.cA.", "aF.#C"}},
 			ans864{14},
 		},
 
-		question864{
+		{
 			para864{[]string{"@...a", ".###A", "b.BCc"}},
 			ans864{10},
 		},
 
-		question864{
+		{
 			para864{[]string{"@Aa"}},
 			ans864{-1},
 		},
 
-		question864{
+		{
 			para864{[]string{"b", "A", "a", "@", "B"}},
 			ans864{3},
 		},
 
-		question864{
+		{
 			para864{[]string{"@.a.#", "#####", "b.A.B"}},
 			ans864{-1},
 		},
 
-		question864{
+		{
 			para864{[]string{"@.a.#", "###.#", "b.A.B"}},
 			ans864{8},
 		},
 
-		question864{
+		{
 			para864{[]string{"@..aA", "..B#.", "....b"}},
 			ans864{6},
 		},
diff --git a/leetcode/0864.Shortest-Path-to-Get-All-Keys/README.md b/leetcode/0864.Shortest-Path-to-Get-All-Keys/README.md
new file mode 100755
index 000000000..b595b5259
--- /dev/null
+++ b/leetcode/0864.Shortest-Path-to-Get-All-Keys/README.md
@@ -0,0 +1,55 @@
+# [864. Shortest Path to Get All Keys](https://leetcode.com/problems/shortest-path-to-get-all-keys/)
+
+
+## 题目
+
+We are given a 2-dimensional `grid`. `"."` is an empty cell, `"#"` is a wall, `"@"` is the starting point, (`"a"`, `"b"`, ...) are keys, and (`"A"`, `"B"`, ...) are locks.
+
+We start at the starting point, and one move consists of walking one space in one of the 4 cardinal directions. We cannot walk outside the grid, or walk into a wall. If we walk over a key, we pick it up. We can't walk over a lock unless we have the corresponding key.
+
+For some 1 <= K <= 6, there is exactly one lowercase and one uppercase letter of the first `K` letters of the English alphabet in the grid. This means that there is exactly one key for each lock, and one lock for each key; and also that the letters used to represent the keys and locks were chosen in the same order as the English alphabet.
+
+Return the lowest number of moves to acquire all keys. If it's impossible, return `-1`.
+
+**Example 1:**
+
+    Input: ["@.a.#","###.#","b.A.B"]
+    Output: 8
+
+**Example 2:**
+
+    Input: ["@..aA","..B#.","....b"]
+    Output: 6
+
+**Note:**
+
+1. `1 <= grid.length <= 30`
+2. `1 <= grid[0].length <= 30`
+3. `grid[i][j]` contains only `'.'`, `'#'`, `'@'`, `'a'-'f'` and `'A'-'F'`
+4. The number of keys is in `[1, 6]`. Each key has a different letter and opens exactly one lock.
+
+
+## 题目大意
+
+给定一个二维网格 grid。 "." 代表一个空房间， "#" 代表一堵墙， "@" 是起点，（"a", "b", ...）代表钥匙，（"A", "B", ...）代表锁。
+
+我们从起点开始出发，一次移动是指向四个基本方向之一行走一个单位空间。我们不能在网格外面行走，也无法穿过一堵墙。如果途经一个钥匙，我们就把它捡起来。除非我们手里有对应的钥匙，否则无法通过锁。
+
+假设 K 为钥匙/锁的个数，且满足 1 <= K <= 6，字母表中的前 K 个字母在网格中都有自己对应的一个小写和一个大写字母。换言之，每个锁有唯一对应的钥匙，每个钥匙也有唯一对应的锁。另外，代表钥匙和锁的字母互为大小写并按字母顺序排列。
+
+返回获取所有钥匙所需要的移动的最少次数。如果无法获取所有钥匙，返回 -1 。
+
+提示：
+
+1. 1 <= grid.length <= 30
+2. 1 <= grid[0].length <= 30
+3. grid[i][j] 只含有 '.', '#', '@', 'a'-'f' 以及 'A'-'F'
+4. 钥匙的数目范围是 [1, 6]，每个钥匙都对应一个不同的字母，正好打开一个对应的锁。
+
+
+## 解题思路
+
+
+- 给出一个地图，在图中有钥匙和锁，当锁在没有钥匙的时候不能通行，问从起点 @ 开始，到最终获得所有钥匙，最短需要走多少步。
+- 这一题可以用 BFS 来解答。由于钥匙的种类比较多，所以 visited 数组需要 3 个维度，一个是 x 坐标，一个是 y 坐标，最后一个是当前获取钥匙的状态。每把钥匙都有获取了和没有获取两种状态，题目中说最多有 6 把钥匙，那么排列组合最多是 2^6 = 64 种状态。用一个十进制数的二进制位来压缩这些状态，二进制位分别来表示这些钥匙是否已经获取了。既然钥匙的状态可以压缩，其实 x 和 y 的坐标也可以一并压缩到这个数中。BFS 中存的数字是坐标 + 钥匙状态的状态。在 BFS 遍历的过程中，用 visited 数组来过滤遍历过的情况，来保证走的路是最短的。其他的情况无非是判断锁的状态，是否能通过，判断钥匙获取状态。
+- 这一题不知道是否能用 DFS 来解答。我实现了一版，但是在 18 / 35 这组 case 上超时了，具体 case 见测试文件第一个 case。
diff --git a/Algorithms/0867. Transpose Matrix/867. Transpose Matrix.go b/leetcode/0867.Transpose-Matrix/867. Transpose Matrix.go
similarity index 100%
rename from Algorithms/0867. Transpose Matrix/867. Transpose Matrix.go
rename to leetcode/0867.Transpose-Matrix/867. Transpose Matrix.go
diff --git a/leetcode/0867.Transpose-Matrix/867. Transpose Matrix_test.go b/leetcode/0867.Transpose-Matrix/867. Transpose Matrix_test.go
new file mode 100644
index 000000000..508dbaac2
--- /dev/null
+++ b/leetcode/0867.Transpose-Matrix/867. Transpose Matrix_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question867 struct {
+	para867
+	ans867
+}
+
+// para 是参数
+// one 代表第一个参数
+type para867 struct {
+	A [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans867 struct {
+	B [][]int
+}
+
+func Test_Problem867(t *testing.T) {
+
+	qs := []question867{
+
+		{
+			para867{[][]int{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}},
+			ans867{[][]int{{1, 4, 7}, {2, 5, 8}, {3, 6, 9}}},
+		},
+
+		{
+			para867{[][]int{{1, 2, 3}, {4, 5, 6}}},
+			ans867{[][]int{{1, 4}, {2, 5}, {3, 6}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 867------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans867, q.para867
+		fmt.Printf("【input】:%v       【output】:%v\n", p, transpose(p.A))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0867. Transpose Matrix/README.md b/leetcode/0867.Transpose-Matrix/README.md
similarity index 100%
rename from Algorithms/0867. Transpose Matrix/README.md
rename to leetcode/0867.Transpose-Matrix/README.md
diff --git a/leetcode/0869.Reordered-Power-of-2/869. Reordered Power of 2.go b/leetcode/0869.Reordered-Power-of-2/869. Reordered Power of 2.go
new file mode 100644
index 000000000..a04c8c1bd
--- /dev/null
+++ b/leetcode/0869.Reordered-Power-of-2/869. Reordered Power of 2.go	
@@ -0,0 +1,32 @@
+package leetcode
+
+import "fmt"
+
+func reorderedPowerOf2(n int) bool {
+	sample, i := fmt.Sprintf("%v", n), 1
+	for len(fmt.Sprintf("%v", i)) <= len(sample) {
+		t := fmt.Sprintf("%v", i)
+		if len(t) == len(sample) && isSame(t, sample) {
+			return true
+		}
+		i = i << 1
+	}
+	return false
+}
+
+func isSame(t, s string) bool {
+	m := make(map[rune]int)
+	for _, v := range t {
+		m[v]++
+	}
+	for _, v := range s {
+		m[v]--
+		if m[v] < 0 {
+			return false
+		}
+		if m[v] == 0 {
+			delete(m, v)
+		}
+	}
+	return len(m) == 0
+}
diff --git a/leetcode/0869.Reordered-Power-of-2/869. Reordered Power of 2_test.go b/leetcode/0869.Reordered-Power-of-2/869. Reordered Power of 2_test.go
new file mode 100644
index 000000000..37682d596
--- /dev/null
+++ b/leetcode/0869.Reordered-Power-of-2/869. Reordered Power of 2_test.go	
@@ -0,0 +1,72 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question869 struct {
+	para869
+	ans869
+}
+
+// para 是参数
+// one 代表第一个参数
+type para869 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans869 struct {
+	one bool
+}
+
+func Test_Problem869(t *testing.T) {
+
+	qs := []question869{
+
+		{
+			para869{1},
+			ans869{true},
+		},
+
+		{
+			para869{10},
+			ans869{false},
+		},
+
+		{
+			para869{16},
+			ans869{true},
+		},
+
+		{
+			para869{24},
+			ans869{false},
+		},
+
+		{
+			para869{46},
+			ans869{true},
+		},
+
+		{
+			para869{100},
+			ans869{false},
+		},
+
+		{
+			para869{123453242},
+			ans869{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 869------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans869, q.para869
+		fmt.Printf("【input】:%v       【output】:%v\n", p, reorderedPowerOf2(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0869.Reordered-Power-of-2/README.md b/leetcode/0869.Reordered-Power-of-2/README.md
new file mode 100644
index 000000000..106c6a38c
--- /dev/null
+++ b/leetcode/0869.Reordered-Power-of-2/README.md
@@ -0,0 +1,93 @@
+# [869. Reordered Power of 2](https://leetcode.com/problems/reordered-power-of-2/)
+
+
+## 题目
+
+Starting with a positive integer `N`, we reorder the digits in any order (including the original order) such that the leading digit is not zero.
+
+Return `true` if and only if we can do this in a way such that the resulting number is a power of 2.
+
+**Example 1:**
+
+```
+Input:1
+Output:true
+```
+
+**Example 2:**
+
+```
+Input:10
+Output:false
+```
+
+**Example 3:**
+
+```
+Input:16
+Output:true
+```
+
+**Example 4:**
+
+```
+Input:24
+Output:false
+```
+
+**Example 5:**
+
+```
+Input:46
+Output:true
+```
+
+**Note:**
+
+1. `1 <= N <= 10^9`
+
+## 题目大意
+
+给定正整数 N ，我们按任何顺序（包括原始顺序）将数字重新排序，注意其前导数字不能为零。如果我们可以通过上述方式得到 2 的幂，返回 true；否则，返回 false。
+
+## 解题思路
+
+- 将整数每个位上的所有排列看成字符串，那么题目转换为判断这些字符串是否和 2 的幂的字符串是否一致。判断的方法有很多种，笔者这里判断借助了一个 `map`。两个不同排列的字符串要相等，所有字符出现的频次必定一样。利用一个 `map` 统计它们各自字符的频次，最终都一致，则判定这两个字符串是满足题意的。
+- 此题数据量比较小，在 `[1,10^9]` 这个区间内，2 的幂只有 30 几个，所以最终要判断的字符串就是这 30 几个。笔者这里没有打表了，采用更加一般的做法。数据量更大，此解法代码也能通过。
+
+## 代码
+
+```go
+package leetcode
+
+import "fmt"
+
+func reorderedPowerOf2(n int) bool {
+	sample, i := fmt.Sprintf("%v", n), 1
+	for len(fmt.Sprintf("%v", i)) <= len(sample) {
+		t := fmt.Sprintf("%v", i)
+		if len(t) == len(sample) && isSame(t, sample) {
+			return true
+		}
+		i = i << 1
+	}
+	return false
+}
+
+func isSame(t, s string) bool {
+	m := make(map[rune]int)
+	for _, v := range t {
+		m[v]++
+	}
+	for _, v := range s {
+		m[v]--
+		if m[v] < 0 {
+			return false
+		}
+		if m[v] == 0 {
+			delete(m, v)
+		}
+	}
+	return len(m) == 0
+}
+```
\ No newline at end of file
diff --git a/leetcode/0870.Advantage-Shuffle/870. Advantage Shuffle.go b/leetcode/0870.Advantage-Shuffle/870. Advantage Shuffle.go
new file mode 100644
index 000000000..a938d3d0a
--- /dev/null
+++ b/leetcode/0870.Advantage-Shuffle/870. Advantage Shuffle.go	
@@ -0,0 +1,31 @@
+package leetcode
+
+import "sort"
+
+func advantageCount1(A []int, B []int) []int {
+	n := len(A)
+	sort.Ints(A)
+	sortedB := make([]int, n)
+	for i := range sortedB {
+		sortedB[i] = i
+	}
+	sort.Slice(sortedB, func(i, j int) bool {
+		return B[sortedB[i]] < B[sortedB[j]]
+	})
+	useless, i, res := make([]int, 0), 0, make([]int, n)
+	for _, index := range sortedB {
+		b := B[index]
+		for i < n && A[i] <= b {
+			useless = append(useless, A[i])
+			i++
+		}
+		if i < n {
+			res[index] = A[i]
+			i++
+		} else {
+			res[index] = useless[0]
+			useless = useless[1:]
+		}
+	}
+	return res
+}
diff --git a/leetcode/0870.Advantage-Shuffle/870. Advantage Shuffle_test.go b/leetcode/0870.Advantage-Shuffle/870. Advantage Shuffle_test.go
new file mode 100644
index 000000000..9c58982eb
--- /dev/null
+++ b/leetcode/0870.Advantage-Shuffle/870. Advantage Shuffle_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question870 struct {
+	para870
+	ans870
+}
+
+// para 是参数
+// one 代表第一个参数
+type para870 struct {
+	A []int
+	B []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans870 struct {
+	one []int
+}
+
+func Test_Problem870(t *testing.T) {
+
+	qs := []question870{
+
+		{
+			para870{[]int{2, 7, 11, 15}, []int{1, 10, 4, 11}},
+			ans870{[]int{2, 11, 7, 15}},
+		},
+
+		{
+			para870{[]int{12, 24, 8, 32}, []int{13, 25, 32, 11}},
+			ans870{[]int{24, 32, 8, 12}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 870------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans870, q.para870
+		fmt.Printf("【input】:%v       【output】:%v\n", p, advantageCount1(p.A, p.B))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0870.Advantage-Shuffle/README.md b/leetcode/0870.Advantage-Shuffle/README.md
new file mode 100644
index 000000000..d7b8ee44d
--- /dev/null
+++ b/leetcode/0870.Advantage-Shuffle/README.md
@@ -0,0 +1,72 @@
+# [870. Advantage Shuffle](https://leetcode.com/problems/advantage-shuffle/)
+
+## 题目
+
+Given two arrays `A` and `B` of equal size, the *advantage of `A` with respect to `B`* is the number of indices `i` for which `A[i] > B[i]`.
+
+Return **any** permutation of `A` that maximizes its advantage with respect to `B`.
+
+**Example 1:**
+
+```
+Input:A = [2,7,11,15], B = [1,10,4,11]
+Output:[2,11,7,15]
+```
+
+**Example 2:**
+
+```
+Input:A = [12,24,8,32], B = [13,25,32,11]
+Output:[24,32,8,12]
+```
+
+**Note:**
+
+1. `1 <= A.length = B.length <= 10000`
+2. `0 <= A[i] <= 10^9`
+3. `0 <= B[i] <= 10^9`
+
+## 题目大意
+
+给定两个大小相等的数组 A 和 B，A 相对于 B 的优势可以用满足 A[i] > B[i] 的索引 i 的数目来描述。返回 A 的任意排列，使其相对于 B 的优势最大化。
+
+## 解题思路
+
+- 此题用贪心算法解题。如果 A 中最小的牌 a 能击败 B 中最小的牌 b，那么将它们配对。否则， a 将无益于我们的比分，因为它无法击败任何牌。这是贪心的策略，每次匹配都用手中最弱的牌和 B 中的最小牌 b 进行配对，这样会使 A 中剩余的牌严格的变大，最后会使得得分更多。
+- 在代码实现中，将 A 数组排序，B 数组按照下标排序。因为最终输出的是相对于 B 的优势结果，所以要针对 B 的下标不变来安排 A 的排列。排好序以后按照贪心策略选择 A 中牌的顺序。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func advantageCount1(A []int, B []int) []int {
+	n := len(A)
+	sort.Ints(A)
+	sortedB := make([]int, n)
+	for i := range sortedB {
+		sortedB[i] = i
+	}
+	sort.Slice(sortedB, func(i, j int) bool {
+		return B[sortedB[i]] < B[sortedB[j]]
+	})
+	useless, i, res := make([]int, 0), 0, make([]int, n)
+	for _, index := range sortedB {
+		b := B[index]
+		for i < n && A[i] <= b {
+			useless = append(useless, A[i])
+			i++
+		}
+		if i < n {
+			res[index] = A[i]
+			i++
+		} else {
+			res[index] = useless[0]
+			useless = useless[1:]
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0872.Leaf-Similar-Trees/872. Leaf-Similar Trees.go b/leetcode/0872.Leaf-Similar-Trees/872. Leaf-Similar Trees.go
new file mode 100644
index 000000000..8a3c4f8ed
--- /dev/null
+++ b/leetcode/0872.Leaf-Similar-Trees/872. Leaf-Similar Trees.go	
@@ -0,0 +1,41 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func leafSimilar(root1 *TreeNode, root2 *TreeNode) bool {
+	leaf1, leaf2 := []int{}, []int{}
+	dfsLeaf(root1, &leaf1)
+	dfsLeaf(root2, &leaf2)
+	if len(leaf1) != len(leaf2) {
+		return false
+	}
+	for i := range leaf1 {
+		if leaf1[i] != leaf2[i] {
+			return false
+		}
+	}
+	return true
+}
+
+func dfsLeaf(root *TreeNode, leaf *[]int) {
+	if root != nil {
+		if root.Left == nil && root.Right == nil {
+			*leaf = append(*leaf, root.Val)
+		}
+		dfsLeaf(root.Left, leaf)
+		dfsLeaf(root.Right, leaf)
+	}
+}
diff --git a/leetcode/0872.Leaf-Similar-Trees/872. Leaf-Similar Trees_test.go b/leetcode/0872.Leaf-Similar-Trees/872. Leaf-Similar Trees_test.go
new file mode 100644
index 000000000..c5eea70b1
--- /dev/null
+++ b/leetcode/0872.Leaf-Similar-Trees/872. Leaf-Similar Trees_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question872 struct {
+	para872
+	ans872
+}
+
+// para 是参数
+// one 代表第一个参数
+type para872 struct {
+	one []int
+	two []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans872 struct {
+	one bool
+}
+
+func Test_Problem872(t *testing.T) {
+
+	qs := []question872{
+
+		{
+			para872{[]int{-10, -3, 0, 5, 9}, []int{-10, -3, 0, 5, 9}},
+			ans872{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 872------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans872, q.para872
+		tree1 := structures.Ints2TreeNode(p.one)
+		tree2 := structures.Ints2TreeNode(p.two)
+		fmt.Printf("【input】:%v       【output】:%v\n", p, leafSimilar(tree1, tree2))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0872.Leaf-Similar-Trees/README.md b/leetcode/0872.Leaf-Similar-Trees/README.md
new file mode 100644
index 000000000..900b7e746
--- /dev/null
+++ b/leetcode/0872.Leaf-Similar-Trees/README.md
@@ -0,0 +1,62 @@
+# [872. Leaf-Similar Trees](https://leetcode.com/problems/leaf-similar-trees/)
+
+
+
+## 题目
+
+Consider all the leaves of a binary tree. From left to right order, the values of those leaves form a *leaf value sequence.*
+
+![https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/16/tree.png](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/16/tree.png)
+
+For example, in the given tree above, the leaf value sequence is `(6, 7, 4, 9, 8)`.
+
+Two binary trees are considered *leaf-similar* if their leaf value sequence is the same.
+
+Return `true` if and only if the two given trees with head nodes `root1` and `root2` are leaf-similar.
+
+**Note**:
+
+- Both of the given trees will have between `1` and `100` nodes.
+
+## 题目大意
+
+请考虑一颗二叉树上所有的叶子，这些叶子的值按从左到右的顺序排列形成一个 叶值序列 。举个例子，如上图所示，给定一颗叶值序列为 (6, 7, 4, 9, 8) 的树。如果有两颗二叉树的叶值序列是相同，那么我们就认为它们是 叶相似 的。如果给定的两个头结点分别为 root1 和 root2 的树是叶相似的，则返回 true；否则返回 false 。
+
+提示：
+
+- 给定的两颗树可能会有 1 到 200 个结点。
+- 给定的两颗树上的值介于 0 到 200 之间。
+
+## 解题思路
+
+- 给出 2 棵树，如果 2 棵树的叶子节点组成的数组是完全一样的，那么就认为这 2 棵树是“叶子相似”的。给出任何 2 棵树判断这 2 棵树是否是“叶子相似”的。
+- 简单题，分别 DFS 遍历 2 棵树，把叶子节点都遍历出来，然后分别比较叶子节点组成的数组是否完全一致即可。
+
+## 代码
+
+```go
+func leafSimilar(root1 *TreeNode, root2 *TreeNode) bool {
+	leaf1, leaf2 := []int{}, []int{}
+	dfsLeaf(root1, &leaf1)
+	dfsLeaf(root2, &leaf2)
+	if len(leaf1) != len(leaf2) {
+		return false
+	}
+	for i := range leaf1 {
+		if leaf1[i] != leaf2[i] {
+			return false
+		}
+	}
+	return true
+}
+
+func dfsLeaf(root *TreeNode, leaf *[]int) {
+	if root != nil {
+		if root.Left == nil && root.Right == nil {
+			*leaf = append(*leaf, root.Val)
+		}
+		dfsLeaf(root.Left, leaf)
+		dfsLeaf(root.Right, leaf)
+	}
+}
+```
\ No newline at end of file
diff --git a/leetcode/0874.Walking-Robot-Simulation/874. Walking Robot Simulation.go b/leetcode/0874.Walking-Robot-Simulation/874. Walking Robot Simulation.go
new file mode 100644
index 000000000..9393000b0
--- /dev/null
+++ b/leetcode/0874.Walking-Robot-Simulation/874. Walking Robot Simulation.go	
@@ -0,0 +1,38 @@
+package leetcode
+
+func robotSim(commands []int, obstacles [][]int) int {
+	m := make(map[[2]int]struct{})
+	for _, v := range obstacles {
+		if len(v) != 0 {
+			m[[2]int{v[0], v[1]}] = struct{}{}
+		}
+	}
+	directX := []int{0, 1, 0, -1}
+	directY := []int{1, 0, -1, 0}
+	direct, x, y := 0, 0, 0
+	result := 0
+	for _, c := range commands {
+		if c == -2 {
+			direct = (direct + 3) % 4
+			continue
+		}
+		if c == -1 {
+			direct = (direct + 1) % 4
+			continue
+		}
+		for ; c > 0; c-- {
+			nextX := x + directX[direct]
+			nextY := y + directY[direct]
+			if _, ok := m[[2]int{nextX, nextY}]; ok {
+				break
+			}
+			tmpResult := nextX*nextX + nextY*nextY
+			if tmpResult > result {
+				result = tmpResult
+			}
+			x = nextX
+			y = nextY
+		}
+	}
+	return result
+}
diff --git a/leetcode/0874.Walking-Robot-Simulation/874. Walking Robot Simulation_test.go b/leetcode/0874.Walking-Robot-Simulation/874. Walking Robot Simulation_test.go
new file mode 100644
index 000000000..2553a4099
--- /dev/null
+++ b/leetcode/0874.Walking-Robot-Simulation/874. Walking Robot Simulation_test.go	
@@ -0,0 +1,39 @@
+package leetcode
+
+import "testing"
+
+func Test_robotSim(t *testing.T) {
+	type args struct {
+		commands  []int
+		obstacles [][]int
+	}
+	cases := []struct {
+		name string
+		args args
+		want int
+	}{
+		{
+			"case 1",
+			args{
+				commands:  []int{4, -1, 3},
+				obstacles: [][]int{{}},
+			},
+			25,
+		},
+		{
+			"case 2",
+			args{
+				commands:  []int{4, -1, 4, -2, 4},
+				obstacles: [][]int{{2, 4}},
+			},
+			65,
+		},
+	}
+	for _, tt := range cases {
+		t.Run(tt.name, func(t *testing.T) {
+			if got := robotSim(tt.args.commands, tt.args.obstacles); got != tt.want {
+				t.Errorf("robotSim() = %v, want %v", got, tt.want)
+			}
+		})
+	}
+}
diff --git a/leetcode/0874.Walking-Robot-Simulation/README.md b/leetcode/0874.Walking-Robot-Simulation/README.md
new file mode 100644
index 000000000..aac2857de
--- /dev/null
+++ b/leetcode/0874.Walking-Robot-Simulation/README.md
@@ -0,0 +1,135 @@
+# [874. Walking Robot Simulation](https://leetcode.com/problems/walking-robot-simulation/)
+
+
+## 题目
+
+A robot on an infinite XY-plane starts at point `(0, 0)` and faces north. The robot can receive one of three possible types of `commands`:
+
+- `2`: turn left `90` degrees,
+- `1`: turn right `90` degrees, or
+- `1 <= k <= 9`: move forward `k` units.
+
+Some of the grid squares are `obstacles`. The `ith` obstacle is at grid point `obstacles[i] = (xi, yi)`.
+
+If the robot would try to move onto them, the robot stays on the previous grid square instead (but still continues following the rest of the route.)
+
+Return *the maximum Euclidean distance that the robot will be from the origin **squared** (i.e. if the distance is* `5`*, return* `25`*)*.
+
+**Note:**
+
+- North means +Y direction.
+- East means +X direction.
+- South means -Y direction.
+- West means -X direction.
+
+**Example 1:**
+
+```
+Input: commands = [4,-1,3], obstacles = []
+Output: 25
+Explanation: The robot starts at (0, 0):
+1. Move north 4 units to (0, 4).
+2. Turn right.
+3. Move east 3 units to (3, 4).
+The furthest point away from the origin is (3, 4), which is 32 + 42 = 25 units away.
+
+```
+
+**Example 2:**
+
+```
+Input: commands = [4,-1,4,-2,4], obstacles = [[2,4]]
+Output: 65
+Explanation: The robot starts at (0, 0):
+1. Move north 4 units to (0, 4).
+2. Turn right.
+3. Move east 1 unit and get blocked by the obstacle at (2, 4), robot is at (1, 4).
+4. Turn left.
+5. Move north 4 units to (1, 8).
+The furthest point away from the origin is (1, 8), which is 12 + 82 = 65 units away.
+
+```
+
+**Constraints:**
+
+- `1 <= commands.length <= 104`
+- `commands[i]` is one of the values in the list `[-2,-1,1,2,3,4,5,6,7,8,9]`.
+- `0 <= obstacles.length <= 104`
+- `3 * 104 <= xi, yi <= 3 * 104`
+- The answer is guaranteed to be less than `231`.
+
+## 题目大意
+
+机器人在一个无限大小的 XY 网格平面上行走，从点 (0, 0) 处开始出发，面向北方。该机器人可以接收以下三种类型的命令 commands ：
+
+- 2 ：向左转 90 度
+- -1 ：向右转 90 度
+- 1 <= x <= 9 ：向前移动 x 个单位长度
+
+在网格上有一些格子被视为障碍物 obstacles 。第 i 个障碍物位于网格点 obstacles[i] = (xi, yi) 。机器人无法走到障碍物上，它将会停留在障碍物的前一个网格方块上，但仍然可以继续尝试进行该路线的其余部分。返回从原点到机器人所有经过的路径点（坐标为整数）的最大欧式距离的平方。（即，如果距离为 5 ，则返回 25 ）
+
+注意：
+
+- 北表示 +Y 方向。
+- 东表示 +X 方向。
+- 南表示 -Y 方向。
+- 西表示 -X 方向。
+
+## 解题思路
+
+- 这个题的难点在于，怎么用编程语言去描述机器人的行为，可以用以下数据结构表达机器人的行为：
+
+    ```go
+    direct:= 0                    // direct表示机器人移动方向：0 1 2 3 4 （北东南西），默认朝北
+    x, y := 0, 0                  // 表示当前机器人所在横纵坐标位置，默认为(0,0)
+    directX := []int{0, 1, 0, -1}
+    directY := []int{1, 0, -1, 0}
+    // 组合directX directY和direct，表示机器人往某一个方向移动
+    nextX := x + directX[direct]
+    nextY := y + directY[direct]
+    ```
+
+    其他代码按照题意翻译即可
+
+## 代码
+
+```go
+package leetcode
+
+func robotSim(commands []int, obstacles [][]int) int {
+	m := make(map[[2]int]struct{})
+	for _, v := range obstacles {
+		if len(v) != 0 {
+			m[[2]int{v[0], v[1]}] = struct{}{}
+		}
+	}
+	directX := []int{0, 1, 0, -1}
+	directY := []int{1, 0, -1, 0}
+	direct, x, y := 0, 0, 0
+	result := 0
+	for _, c := range commands {
+		if c == -2 {
+			direct = (direct + 3) % 4
+			continue
+		}
+		if c == -1 {
+			direct = (direct + 1) % 4
+			continue
+		}
+		for ; c > 0; c-- {
+			nextX := x + directX[direct]
+			nextY := y + directY[direct]
+			if _, ok := m[[2]int{nextX, nextY}]; ok {
+				break
+			}
+			tmpResult := nextX*nextX + nextY*nextY
+			if tmpResult > result {
+				result = tmpResult
+			}
+			x = nextX
+			y = nextY
+		}
+	}
+	return result
+}
+```
\ No newline at end of file
diff --git a/leetcode/0875.Koko-Eating-Bananas/875. Koko Eating Bananas.go b/leetcode/0875.Koko-Eating-Bananas/875. Koko Eating Bananas.go
new file mode 100644
index 000000000..a52d82d00
--- /dev/null
+++ b/leetcode/0875.Koko-Eating-Bananas/875. Koko Eating Bananas.go	
@@ -0,0 +1,34 @@
+package leetcode
+
+import "math"
+
+func minEatingSpeed(piles []int, H int) int {
+	low, high := 1, maxInArr(piles)
+	for low < high {
+		mid := low + (high-low)>>1
+		if !isPossible(piles, mid, H) {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return low
+}
+
+func isPossible(piles []int, h, H int) bool {
+	res := 0
+	for _, p := range piles {
+		res += int(math.Ceil(float64(p) / float64(h)))
+	}
+	return res <= H
+}
+
+func maxInArr(xs []int) int {
+	res := 0
+	for _, x := range xs {
+		if res < x {
+			res = x
+		}
+	}
+	return res
+}
diff --git a/leetcode/0875.Koko-Eating-Bananas/875. Koko Eating Bananas_test.go b/leetcode/0875.Koko-Eating-Bananas/875. Koko Eating Bananas_test.go
new file mode 100644
index 000000000..b9fcc115e
--- /dev/null
+++ b/leetcode/0875.Koko-Eating-Bananas/875. Koko Eating Bananas_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question875 struct {
+	para875
+	ans875
+}
+
+// para 是参数
+// one 代表第一个参数
+type para875 struct {
+	piles []int
+	H     int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans875 struct {
+	one int
+}
+
+func Test_Problem875(t *testing.T) {
+
+	qs := []question875{
+
+		{
+			para875{[]int{3, 6, 7, 11}, 8},
+			ans875{4},
+		},
+
+		{
+			para875{[]int{30, 11, 23, 4, 20}, 5},
+			ans875{30},
+		},
+
+		{
+			para875{[]int{30, 11, 23, 4, 20}, 6},
+			ans875{23},
+		},
+
+		{
+			para875{[]int{4, 9, 11, 17}, 8},
+			ans875{6},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 875------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans875, q.para875
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minEatingSpeed(p.piles, p.H))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0875.Koko-Eating-Bananas/README.md b/leetcode/0875.Koko-Eating-Bananas/README.md
new file mode 100755
index 000000000..f706668a5
--- /dev/null
+++ b/leetcode/0875.Koko-Eating-Bananas/README.md
@@ -0,0 +1,59 @@
+# [875. Koko Eating Bananas](https://leetcode.com/problems/koko-eating-bananas/)
+
+
+## 题目
+
+Koko loves to eat bananas. There are `N` piles of bananas, the `i`-th pile has `piles[i]` bananas. The guards have gone and will come back in `H` hours.
+
+Koko can decide her bananas-per-hour eating speed of `K`. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than `K` bananas, she eats all of them instead, and won't eat any more bananas during this hour.
+
+Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
+
+Return the minimum integer `K` such that she can eat all the bananas within `H` hours.
+
+**Example 1:**
+
+    Input: piles = [3,6,7,11], H = 8
+    Output: 4
+
+**Example 2:**
+
+    Input: piles = [30,11,23,4,20], H = 5
+    Output: 30
+
+**Example 3:**
+
+    Input: piles = [30,11,23,4,20], H = 6
+    Output: 23
+
+**Note:**
+
+- `1 <= piles.length <= 10^4`
+- `piles.length <= H <= 10^9`
+- `1 <= piles[i] <= 10^9`
+
+
+## 题目大意
+
+
+珂珂喜欢吃香蕉。这里有 N 堆香蕉，第 i 堆中有 piles[i] 根香蕉。警卫已经离开了，将在 H 小时后回来。
+
+珂珂可以决定她吃香蕉的速度 K （单位：根/小时）。每个小时，她将会选择一堆香蕉，从中吃掉 K 根。如果这堆香蕉少于 K 根，她将吃掉这堆的所有香蕉，然后这一小时内不会再吃更多的香蕉。  
+
+珂珂喜欢慢慢吃，但仍然想在警卫回来前吃掉所有的香蕉。
+
+返回她可以在 H 小时内吃掉所有香蕉的最小速度 K（K 为整数）。
+
+提示：
+
+- 1 <= piles.length <= 10^4
+- piles.length <= H <= 10^9
+- 1 <= piles[i] <= 10^9
+
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面每个元素代表的是每个香蕉🍌串上香蕉的个数。koko 以 `k 个香蕉/小时`的速度吃这些香蕉。守卫会在 `H 小时`以后回来。问 k 至少为多少，能在守卫回来之前吃完所有的香蕉。当香蕉的个数小于 k 的时候，这个小时只能吃完这些香蕉，不能再吃其他串上的香蕉了。
+- 这一题可以用二分搜索来解答。在 `[0 , max(piles)]` 的范围内搜索，二分的过程都是常规思路。判断是否左右边界如果划分的时候需要注意题目中给的限定条件。当香蕉个数小于 k 的时候，那个小时不能再吃其他香蕉了。
diff --git a/Algorithms/0876. Middle of the Linked List/876. Middle of the Linked List.go b/leetcode/0876.Middle-of-the-Linked-List/876. Middle of the Linked List.go
similarity index 80%
rename from Algorithms/0876. Middle of the Linked List/876. Middle of the Linked List.go
rename to leetcode/0876.Middle-of-the-Linked-List/876. Middle of the Linked List.go
index 3f13a1f04..7abfd4e6b 100644
--- a/Algorithms/0876. Middle of the Linked List/876. Middle of the Linked List.go	
+++ b/leetcode/0876.Middle-of-the-Linked-List/876. Middle of the Linked List.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
 /**
  * Definition for singly-linked list.
  * type ListNode struct {
diff --git a/Algorithms/0876. Middle of the Linked List/876. Middle of the Linked List_test.go b/leetcode/0876.Middle-of-the-Linked-List/876. Middle of the Linked List_test.go
similarity index 78%
rename from Algorithms/0876. Middle of the Linked List/876. Middle of the Linked List_test.go
rename to leetcode/0876.Middle-of-the-Linked-List/876. Middle of the Linked List_test.go
index ca68d1b71..93a565dca 100644
--- a/Algorithms/0876. Middle of the Linked List/876. Middle of the Linked List_test.go	
+++ b/leetcode/0876.Middle-of-the-Linked-List/876. Middle of the Linked List_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question876 struct {
@@ -26,21 +28,21 @@ func Test_Problem876(t *testing.T) {
 
 	qs := []question876{
 
-		question876{
+		{
 			para876{[]int{1, 2, 3, 4, 5}},
 			ans876{3},
 		},
-		question876{
+		{
 			para876{[]int{1, 2, 3, 4}},
 			ans876{3},
 		},
 
-		question876{
+		{
 			para876{[]int{1}},
 			ans876{1},
 		},
 
-		question876{
+		{
 			para876{[]int{}},
 			ans876{},
 		},
@@ -50,7 +52,7 @@ func Test_Problem876(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans876, q.para876
-		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(middleNode(S2l(p.one))))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(middleNode(structures.Ints2List(p.one))))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/0876. Middle of the Linked List/README.md b/leetcode/0876.Middle-of-the-Linked-List/README.md
similarity index 100%
rename from Algorithms/0876. Middle of the Linked List/README.md
rename to leetcode/0876.Middle-of-the-Linked-List/README.md
diff --git a/leetcode/0877.Stone-Game/877. Stone Game.go b/leetcode/0877.Stone-Game/877. Stone Game.go
new file mode 100644
index 000000000..86a89974d
--- /dev/null
+++ b/leetcode/0877.Stone-Game/877. Stone Game.go	
@@ -0,0 +1,5 @@
+package leetcode
+
+func stoneGame(piles []int) bool {
+	return true
+}
diff --git a/leetcode/0877.Stone-Game/877. Stone Game_test.go b/leetcode/0877.Stone-Game/877. Stone Game_test.go
new file mode 100644
index 000000000..316c42c48
--- /dev/null
+++ b/leetcode/0877.Stone-Game/877. Stone Game_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question877 struct {
+	para877
+	ans877
+}
+
+// para 是参数
+// one 代表第一个参数
+type para877 struct {
+	piles []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans877 struct {
+	one bool
+}
+
+func Test_Problem877(t *testing.T) {
+
+	qs := []question877{
+
+		{
+			para877{[]int{5, 3, 4, 5}},
+			ans877{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 877------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans877, q.para877
+		fmt.Printf("【input】:%v       【output】:%v\n", p, stoneGame(p.piles))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0877.Stone-Game/README.md b/leetcode/0877.Stone-Game/README.md
new file mode 100644
index 000000000..6a23bfad6
--- /dev/null
+++ b/leetcode/0877.Stone-Game/README.md
@@ -0,0 +1,50 @@
+# [877. Stone Game](https://leetcode.com/problems/stone-game/)
+
+## 题目
+
+Alex and Lee play a game with piles of stones.  There are an even number of piles **arranged in a row**, and each pile has a positive integer number of stones `piles[i]`.
+
+The objective of the game is to end with the most stones.  The total number of stones is odd, so there are no ties.
+
+Alex and Lee take turns, with Alex starting first.  Each turn, a player takes the entire pile of stones from either the beginning or the end of the row.  This continues until there are no more piles left, at which point the person with the most stones wins.
+
+Assuming Alex and Lee play optimally, return `True` if and only if Alex wins the game.
+
+**Example 1:**
+
+```
+Input: piles = [5,3,4,5]
+Output: true
+Explanation:
+Alex starts first, and can only take the first 5 or the last 5.
+Say he takes the first 5, so that the row becomes [3, 4, 5].
+If Lee takes 3, then the board is [4, 5], and Alex takes 5 to win with 10 points.
+If Lee takes the last 5, then the board is [3, 4], and Alex takes 4 to win with 9 points.
+This demonstrated that taking the first 5 was a winning move for Alex, so we return true.
+
+```
+
+**Constraints:**
+
+- `2 <= piles.length <= 500`
+- `piles.length` is even.
+- `1 <= piles[i] <= 500`
+- `sum(piles)` is odd.
+
+## 题目大意
+
+亚历克斯和李用几堆石子在做游戏。偶数堆石子排成一行，每堆都有正整数颗石子 piles[i] 。游戏以谁手中的石子最多来决出胜负。石子的总数是奇数，所以没有平局。亚历克斯和李轮流进行，亚历克斯先开始。 每回合，玩家从行的开始或结束处取走整堆石头。 这种情况一直持续到没有更多的石子堆为止，此时手中石子最多的玩家获胜。假设亚历克斯和李都发挥出最佳水平，当亚历克斯赢得比赛时返回 true ，当李赢得比赛时返回 false 。
+
+## 解题思路
+
+- 一遇到石子问题，很容易让人联想到是否和奇偶数相关。此题指定了石子堆数一定是偶数。所以从这里为突破口试试。Alex 先拿，要么取行首下标为 0 的石子，要么取行尾下标为 n-1 的石子。假设取下标为 0 的石子，剩下的石子堆下标从 1 ~ n-1，即 Lee 只能从奇数下标的石子堆 1 或者 n-1。假设 Alex 第一次取下标为 n-1 的石子，剩下的石子堆下标从 0 ~ n-2，即 Lee 只能取偶数下标的石子堆。于是 Alex 的必胜策略是每轮取石子，取此轮奇数下标堆石子数总和，偶数下标堆石子数总和，两者大者。那么下一轮 Lee 只能取石子堆数相对少的那一堆，并且 Lee 取的石子堆下标奇偶性是完全受到上一轮 Alex 控制的。所以只要是 Alex 先手，那么每轮都可以压制 Lee，从而必胜。
+
+## 代码
+
+```go
+package leetcode
+
+func stoneGame(piles []int) bool {
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/0878.Nth-Magical-Number/878. Nth Magical Number.go b/leetcode/0878.Nth-Magical-Number/878. Nth Magical Number.go
new file mode 100644
index 000000000..6040450e1
--- /dev/null
+++ b/leetcode/0878.Nth-Magical-Number/878. Nth Magical Number.go	
@@ -0,0 +1,26 @@
+package leetcode
+
+func nthMagicalNumber(N int, A int, B int) int {
+	low, high := int64(0), int64(1*1e14)
+	for low < high {
+		mid := low + (high-low)>>1
+		if calNthMagicalCount(mid, int64(A), int64(B)) < int64(N) {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return int(low) % 1000000007
+}
+
+func calNthMagicalCount(num, a, b int64) int64 {
+	ab := a * b / gcd(a, b)
+	return num/a + num/b - num/ab
+}
+
+func gcd(a, b int64) int64 {
+	for b != 0 {
+		a, b = b, a%b
+	}
+	return a
+}
diff --git a/leetcode/0878.Nth-Magical-Number/878. Nth Magical Number_test.go b/leetcode/0878.Nth-Magical-Number/878. Nth Magical Number_test.go
new file mode 100644
index 000000000..60cdcc0bc
--- /dev/null
+++ b/leetcode/0878.Nth-Magical-Number/878. Nth Magical Number_test.go	
@@ -0,0 +1,64 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question878 struct {
+	para878
+	ans878
+}
+
+// para 是参数
+// one 代表第一个参数
+type para878 struct {
+	n int
+	a int
+	b int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans878 struct {
+	one int
+}
+
+func Test_Problem878(t *testing.T) {
+
+	qs := []question878{
+
+		{
+			para878{1, 2, 3},
+			ans878{2},
+		},
+
+		{
+			para878{4, 2, 3},
+			ans878{6},
+		},
+
+		{
+			para878{5, 2, 4},
+			ans878{10},
+		},
+
+		{
+			para878{3, 6, 4},
+			ans878{8},
+		},
+
+		{
+			para878{1000000000, 40000, 40000},
+			ans878{999720007},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 878------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans878, q.para878
+		fmt.Printf("【input】:%v       【output】:%v\n", p, nthMagicalNumber(p.n, p.a, p.b))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0878.Nth-Magical-Number/README.md b/leetcode/0878.Nth-Magical-Number/README.md
new file mode 100755
index 000000000..2ff41dc81
--- /dev/null
+++ b/leetcode/0878.Nth-Magical-Number/README.md
@@ -0,0 +1,54 @@
+# [878. Nth Magical Number](https://leetcode.com/problems/nth-magical-number/)
+
+
+## 题目
+
+A positive integer is *magical* if it is divisible by either A or B.
+
+Return the N-th magical number. Since the answer may be very large, **return it modulo** `10^9 + 7`.
+
+**Example 1:**
+
+    Input: N = 1, A = 2, B = 3
+    Output: 2
+
+**Example 2:**
+
+    Input: N = 4, A = 2, B = 3
+    Output: 6
+
+**Example 3:**
+
+    Input: N = 5, A = 2, B = 4
+    Output: 10
+
+**Example 4:**
+
+    Input: N = 3, A = 6, B = 4
+    Output: 8
+
+**Note:**
+
+1. `1 <= N <= 10^9`
+2. `2 <= A <= 40000`
+3. `2 <= B <= 40000`
+
+
+## 题目大意
+
+
+如果正整数可以被 A 或 B 整除，那么它是神奇的。返回第 N 个神奇数字。由于答案可能非常大，返回它模 10^9 + 7 的结果。
+
+
+提示：
+
+1. 1 <= N <= 10^9
+2. 2 <= A <= 40000
+3. 2 <= B <= 40000
+
+
+## 解题思路
+
+
+- 给出 3 个数字，a，b，n。要求输出可以整除 a 或者整除 b 的第 n 个数。
+- 这一题是第 1201 题的缩水版，代码和解题思路也基本不变，这一题的二分搜索的区间是 `[min(A, B)，N * min(A, B)] = [2, 10 ^ 14]`。其他代码和第 1201 题一致，思路见第 1201 题。
diff --git a/Algorithms/0880. Decoded String at Index/880. Decoded String at Index.go b/leetcode/0880.Decoded-String-at-Index/880. Decoded String at Index.go
similarity index 100%
rename from Algorithms/0880. Decoded String at Index/880. Decoded String at Index.go
rename to leetcode/0880.Decoded-String-at-Index/880. Decoded String at Index.go
diff --git a/Algorithms/0880. Decoded String at Index/880. Decoded String at Index_test.go b/leetcode/0880.Decoded-String-at-Index/880. Decoded String at Index_test.go
similarity index 91%
rename from Algorithms/0880. Decoded String at Index/880. Decoded String at Index_test.go
rename to leetcode/0880.Decoded-String-at-Index/880. Decoded String at Index_test.go
index 8eeaf89ed..016576c9c 100644
--- a/Algorithms/0880. Decoded String at Index/880. Decoded String at Index_test.go	
+++ b/leetcode/0880.Decoded-String-at-Index/880. Decoded String at Index_test.go	
@@ -27,32 +27,32 @@ func Test_Problem880(t *testing.T) {
 
 	qs := []question880{
 
-		question880{
+		{
 			para880{"aw4eguc6cs", 41},
 			ans880{"a"},
 		},
 
-		question880{
+		{
 			para880{"vk6u5xhq9v", 554},
 			ans880{"h"},
 		},
 
-		question880{
+		{
 			para880{"leet2code3", 10},
 			ans880{"o"},
 		},
 
-		question880{
+		{
 			para880{"ha22", 5},
 			ans880{"h"},
 		},
 
-		question880{
+		{
 			para880{"a2345678999999999999999", 1},
 			ans880{"a"},
 		},
 
-		question880{
+		{
 			para880{"y959q969u3hb22odq595", 222280369},
 			ans880{"q"},
 		},
diff --git a/Algorithms/0880. Decoded String at Index/README.md b/leetcode/0880.Decoded-String-at-Index/README.md
similarity index 100%
rename from Algorithms/0880. Decoded String at Index/README.md
rename to leetcode/0880.Decoded-String-at-Index/README.md
diff --git a/Algorithms/0881. Boats to Save People/881. Boats to Save People.go b/leetcode/0881.Boats-to-Save-People/881. Boats to Save People.go
similarity index 100%
rename from Algorithms/0881. Boats to Save People/881. Boats to Save People.go
rename to leetcode/0881.Boats-to-Save-People/881. Boats to Save People.go
diff --git a/Algorithms/0881. Boats to Save People/881. Boats to Save People_test.go b/leetcode/0881.Boats-to-Save-People/881. Boats to Save People_test.go
similarity index 91%
rename from Algorithms/0881. Boats to Save People/881. Boats to Save People_test.go
rename to leetcode/0881.Boats-to-Save-People/881. Boats to Save People_test.go
index 558eff836..5893f286a 100644
--- a/Algorithms/0881. Boats to Save People/881. Boats to Save People_test.go	
+++ b/leetcode/0881.Boats-to-Save-People/881. Boats to Save People_test.go	
@@ -27,27 +27,27 @@ func Test_Problem881(t *testing.T) {
 
 	qs := []question881{
 
-		question881{
+		{
 			para881{[]int{1, 2}, 3},
 			ans881{1},
 		},
 
-		question881{
+		{
 			para881{[]int{3, 2, 2, 1}, 3},
 			ans881{3},
 		},
 
-		question881{
+		{
 			para881{[]int{3, 5, 3, 4}, 5},
 			ans881{4},
 		},
 
-		question881{
+		{
 			para881{[]int{5, 1, 4, 2}, 6},
 			ans881{2},
 		},
 
-		question881{
+		{
 			para881{[]int{3, 2, 2, 1}, 3},
 			ans881{3},
 		},
diff --git a/Algorithms/0881. Boats to Save People/README.md b/leetcode/0881.Boats-to-Save-People/README.md
similarity index 100%
rename from Algorithms/0881. Boats to Save People/README.md
rename to leetcode/0881.Boats-to-Save-People/README.md
diff --git a/Algorithms/0884. Uncommon Words from Two Sentences/884. Uncommon Words from Two Sentences.go b/leetcode/0884.Uncommon-Words-from-Two-Sentences/884. Uncommon Words from Two Sentences.go
similarity index 100%
rename from Algorithms/0884. Uncommon Words from Two Sentences/884. Uncommon Words from Two Sentences.go
rename to leetcode/0884.Uncommon-Words-from-Two-Sentences/884. Uncommon Words from Two Sentences.go
diff --git a/Algorithms/0884. Uncommon Words from Two Sentences/884. Uncommon Words from Two Sentences_test.go b/leetcode/0884.Uncommon-Words-from-Two-Sentences/884. Uncommon Words from Two Sentences_test.go
similarity index 96%
rename from Algorithms/0884. Uncommon Words from Two Sentences/884. Uncommon Words from Two Sentences_test.go
rename to leetcode/0884.Uncommon-Words-from-Two-Sentences/884. Uncommon Words from Two Sentences_test.go
index ee461865a..52d0da9c5 100644
--- a/Algorithms/0884. Uncommon Words from Two Sentences/884. Uncommon Words from Two Sentences_test.go	
+++ b/leetcode/0884.Uncommon-Words-from-Two-Sentences/884. Uncommon Words from Two Sentences_test.go	
@@ -27,12 +27,12 @@ func Test_Problem884(t *testing.T) {
 
 	qs := []question884{
 
-		question884{
+		{
 			para884{"this apple is sweet", "this apple is sour"},
 			ans884{[]string{"sweet", "sour"}},
 		},
 
-		question884{
+		{
 			para884{"apple apple", "banana"},
 			ans884{[]string{"banana"}},
 		},
diff --git a/leetcode/0884.Uncommon-Words-from-Two-Sentences/README.md b/leetcode/0884.Uncommon-Words-from-Two-Sentences/README.md
new file mode 100755
index 000000000..fb1c5d8bc
--- /dev/null
+++ b/leetcode/0884.Uncommon-Words-from-Two-Sentences/README.md
@@ -0,0 +1,40 @@
+# [884. Uncommon Words from Two Sentences](https://leetcode.com/problems/uncommon-words-from-two-sentences/)
+
+
+## 题目
+
+We are given two sentences `A` and `B`. (A *sentence* is a string of space separated words. Each *word* consists only of lowercase letters.)
+
+A word is *uncommon* if it appears exactly once in one of the sentences, and does not appear in the other sentence.
+
+Return a list of all uncommon words.
+
+You may return the list in any order.
+
+**Example 1:**
+
+    Input: A = "this apple is sweet", B = "this apple is sour"
+    Output: ["sweet","sour"]
+
+**Example 2:**
+
+    Input: A = "apple apple", B = "banana"
+    Output: ["banana"]
+
+**Note:**
+
+1. `0 <= A.length <= 200`
+2. `0 <= B.length <= 200`
+3. `A` and `B` both contain only spaces and lowercase letters.
+
+
+## 题目大意
+
+给定两个句子 A 和 B 。（句子是一串由空格分隔的单词。每个单词仅由小写字母组成。）
+
+如果一个单词在其中一个句子中只出现一次，在另一个句子中却没有出现，那么这个单词就是不常见的。返回所有不常用单词的列表。您可以按任何顺序返回列表。
+
+
+## 解题思路
+
+- 找出 2 个句子中不同的单词，将它们俩都打印出来。简单题，先将 2 个句子的单词都拆开放入 map 中进行词频统计，不同的两个单词的词频肯定都为 1，输出它们即可。
diff --git a/Algorithms/0885. Spiral Matrix III/885. Spiral Matrix III.go b/leetcode/0885.Spiral-Matrix-III/885. Spiral Matrix III.go
similarity index 79%
rename from Algorithms/0885. Spiral Matrix III/885. Spiral Matrix III.go
rename to leetcode/0885.Spiral-Matrix-III/885. Spiral Matrix III.go
index 5e7d8bfd1..5ed9e8fcd 100644
--- a/Algorithms/0885. Spiral Matrix III/885. Spiral Matrix III.go	
+++ b/leetcode/0885.Spiral-Matrix-III/885. Spiral Matrix III.go	
@@ -2,10 +2,10 @@ package leetcode
 
 func spiralMatrixIII(R int, C int, r0 int, c0 int) [][]int {
 	res, round, spDir := [][]int{}, 0, [][]int{
-		[]int{0, 1},  // 朝右
-		[]int{1, 0},  // 朝下
-		[]int{0, -1}, // 朝左
-		[]int{-1, 0}, // 朝上
+		{0, 1},  // 朝右
+		{1, 0},  // 朝下
+		{0, -1}, // 朝左
+		{-1, 0}, // 朝上
 	}
 	res = append(res, []int{r0, c0})
 	for i := 0; len(res) < R*C; i++ {
diff --git a/leetcode/0885.Spiral-Matrix-III/885. Spiral Matrix III_test.go b/leetcode/0885.Spiral-Matrix-III/885. Spiral Matrix III_test.go
new file mode 100644
index 000000000..ea2c844a2
--- /dev/null
+++ b/leetcode/0885.Spiral-Matrix-III/885. Spiral Matrix III_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question885 struct {
+	para885
+	ans885
+}
+
+// para 是参数
+// one 代表第一个参数
+type para885 struct {
+	R  int
+	C  int
+	r0 int
+	c0 int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans885 struct {
+	one [][]int
+}
+
+func Test_Problem885(t *testing.T) {
+
+	qs := []question885{
+
+		{
+			para885{1, 4, 0, 0},
+			ans885{[][]int{{0, 0}, {0, 1}, {0, 2}, {0, 3}}},
+		},
+
+		{
+			para885{5, 6, 1, 4},
+			ans885{[][]int{{1, 4}, {1, 5}, {2, 5}, {2, 4}, {2, 3}, {1, 3}, {0, 3}, {0, 4}, {0, 5}, {3, 5}, {3, 4},
+				{3, 3}, {3, 2}, {2, 2}, {1, 2}, {0, 2}, {4, 5}, {4, 4}, {4, 3}, {4, 2}, {4, 1}, {3, 1}, {2, 1},
+				{1, 1}, {0, 1}, {4, 0}, {3, 0}, {2, 0}, {1, 0}, {0, 0}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 885------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans885, q.para885
+		fmt.Printf("【input】:%v       【output】:%v\n", p, spiralMatrixIII(p.R, p.C, p.r0, p.c0))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0885. Spiral Matrix III/README.md b/leetcode/0885.Spiral-Matrix-III/README.md
similarity index 100%
rename from Algorithms/0885. Spiral Matrix III/README.md
rename to leetcode/0885.Spiral-Matrix-III/README.md
diff --git a/leetcode/0887.Super-Egg-Drop/887. Super Egg Drop.go b/leetcode/0887.Super-Egg-Drop/887. Super Egg Drop.go
new file mode 100644
index 000000000..7e9758c81
--- /dev/null
+++ b/leetcode/0887.Super-Egg-Drop/887. Super Egg Drop.go	
@@ -0,0 +1,37 @@
+package leetcode
+
+// 解法一 二分搜索
+func superEggDrop(K int, N int) int {
+	low, high := 1, N
+	for low < high {
+		mid := low + (high-low)>>1
+		if counterF(K, N, mid) >= N {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+// 计算二项式和，特殊的第一项 C(t,0) = 1
+func counterF(k, n, mid int) int {
+	res, sum := 1, 0
+	for i := 1; i <= k && sum < n; i++ {
+		res *= mid - i + 1
+		res /= i
+		sum += res
+	}
+	return sum
+}
+
+// 解法二 动态规划 DP
+func superEggDrop1(K int, N int) int {
+	dp, step := make([]int, K+1), 0
+	for ; dp[K] < N; step++ {
+		for i := K; i > 0; i-- {
+			dp[i] = (1 + dp[i] + dp[i-1])
+		}
+	}
+	return step
+}
diff --git a/leetcode/0887.Super-Egg-Drop/887. Super Egg Drop_test.go b/leetcode/0887.Super-Egg-Drop/887. Super Egg Drop_test.go
new file mode 100644
index 000000000..1fa604e5d
--- /dev/null
+++ b/leetcode/0887.Super-Egg-Drop/887. Super Egg Drop_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question887 struct {
+	para887
+	ans887
+}
+
+// para 是参数
+// one 代表第一个参数
+type para887 struct {
+	k int
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans887 struct {
+	one int
+}
+
+func Test_Problem887(t *testing.T) {
+
+	qs := []question887{
+
+		{
+			para887{1, 2},
+			ans887{2},
+		},
+
+		{
+			para887{2, 6},
+			ans887{3},
+		},
+
+		{
+			para887{3, 14},
+			ans887{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 887------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans887, q.para887
+		fmt.Printf("【input】:%v       【output】:%v\n", p, superEggDrop(p.k, p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0887.Super-Egg-Drop/README.md b/leetcode/0887.Super-Egg-Drop/README.md
new file mode 100755
index 000000000..7c0a0b6c1
--- /dev/null
+++ b/leetcode/0887.Super-Egg-Drop/README.md
@@ -0,0 +1,110 @@
+# [887. Super Egg Drop](https://leetcode.com/problems/super-egg-drop/)
+
+
+## 题目
+
+You are given `K` eggs, and you have access to a building with `N` floors from `1` to `N`.
+
+Each egg is identical in function, and if an egg breaks, you cannot drop it again.
+
+You know that there exists a floor `F` with `0 <= F <= N` such that any egg dropped at a floor higher than `F` will break, and any egg dropped at or below floor `F` will not break.
+
+Each *move*, you may take an egg (if you have an unbroken one) and drop it from any floor `X` (with `1 <= X <= N`).
+
+Your goal is to know **with certainty** what the value of `F` is.
+
+What is the minimum number of moves that you need to know with certainty what `F` is, regardless of the initial value of `F`?
+
+**Example 1:**
+
+    Input: K = 1, N = 2
+    Output: 2
+    Explanation: 
+    Drop the egg from floor 1.  If it breaks, we know with certainty that F = 0.
+    Otherwise, drop the egg from floor 2.  If it breaks, we know with certainty that F = 1.
+    If it didn't break, then we know with certainty F = 2.
+    Hence, we needed 2 moves in the worst case to know what F is with certainty.
+
+**Example 2:**
+
+    Input: K = 2, N = 6
+    Output: 3
+
+**Example 3:**
+
+    Input: K = 3, N = 14
+    Output: 4
+
+**Note:**
+
+1. `1 <= K <= 100`
+2. `1 <= N <= 10000`
+
+
+## 题目大意
+
+你将获得 K 个鸡蛋，并可以使用一栋从 1 到 N  共有 N 层楼的建筑。每个蛋的功能都是一样的，如果一个蛋碎了，你就不能再把它掉下去。你知道存在楼层 F ，满足 0 <= F <= N 任何从高于 F 的楼层落下的鸡蛋都会碎，从 F 楼层或比它低的楼层落下的鸡蛋都不会破。每次移动，你可以取一个鸡蛋（如果你有完整的鸡蛋）并把它从任一楼层 X 扔下（满足 1 <= X <= N）。你的目标是确切地知道 F 的值是多少。无论 F 的初始值如何，你确定 F 的值的最小移动次数是多少？
+
+
+提示：
+
+1. 1 <= K <= 100
+2. 1 <= N <= 10000
+
+
+## 解题思路
+
+- 给出 `K` 个鸡蛋，`N` 层楼，要求确定安全楼层 `F` 需要最小步数 `t`。
+- 这一题是微软的经典面试题。拿到题最容易想到的是二分搜索。但是仔细分析以后会发现单纯的二分是不对的。不断的二分确实能找到最终安全的楼层，但是这里没有考虑到 `K` 个鸡蛋。鸡蛋数的限制会导致二分搜索无法找到最终楼层。题目要求要在保证能找到最终安全楼层的情况下，找到最小步数。所以单纯的二分搜索并不能解答这道题。
+- 这一题如果按照题意正向考虑，动态规划的状态转移方程是 `searchTime(K, N) = max( searchTime(K-1, X-1), searchTime(K, N-X) )`。其中 `X` 是丢鸡蛋的楼层。随着 `X` 从 `[1,N]`，都能计算出一个 `searchTime` 的值，在所有这 `N` 个值之中，取最小值就是本题的答案了。这个解法可以 AC 这道题。不过这个解法不细展开了。时间复杂度 `O(k*N^2)`。    
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_887_8.png'>
+</p>
+
+- 换个角度来看这个问题，定义 `dp[k][m]` 代表 `K` 个鸡蛋，`M` 次移动能检查的最大楼层。考虑某一步 `t` 应该在哪一层丢鸡蛋呢？一个正确的选择是在 `dp[k-1][t-1] + 1` 层丢鸡蛋，结果分两种情况：
+    1. 如果鸡蛋碎了，我们首先排除了该层以上的所有楼层（不管这个楼有多高），而对于剩下的 `dp[k-1][t-1]` 层楼，我们一定能用 `k-1` 个鸡蛋在 `t-1` 步内求解。因此这种情况下，我们总共可以求解无限高的楼层。可见，这是一种非常好的情况，但并不总是发生。
+    2. 如果鸡蛋没碎，我们首先排除了该层以下的 `dp[k-1][t-1]` 层楼，此时我们还有 `k` 个蛋和 `t-1` 步，那么我们去该层以上的楼层继续测得 `dp[k][t-1]` 层楼。因此这种情况下，我们总共可以求解 `dp[k-1][t-1] + 1 + dp[k][t-1]` 层楼。
+- 在所有 `m` 步中只要有一次出现了第一种情况，那么我们就可以求解无限高的楼层。但题目要求我们能保证一定能找到安全楼层，所以每次丢鸡蛋的情况应该按照最差情况来，即每次都是第二种情况。于是得到转状态转移方程: `dp[k][m] = dp[k-1][m-1] + dp[k][m-1] + 1` 。这个方程可以压缩到一维，因为每个新的状态只和上一行和左一列有关。那么每一行从右往左更新，即 `dp[i] += 1 + dp[i-1]`。时间复杂度 `O(K * log N)`，空间复杂度 `O(N)`。
+- 可能会有人有疑问，如果最初选择不在 `dp[k-1][t-1] + 1` 层丢鸡蛋会怎么样呢？选择在更低的层或者更高的层丢鸡蛋会怎样呢？
+    1. 如果在更低的楼层丢鸡蛋也能保证找到安全楼层。那么得到的结果一定不是最小步数。因为这次丢鸡蛋没有充分的展现鸡蛋和移动次数的潜力，最终求解一定会有鸡蛋和步数剩余，即不是能探测的最大楼层了。
+    2. 如果在更高的楼层丢鸡蛋，假设是第 `dp[k-1][t-1] + 2` 层丢鸡蛋，如果这次鸡蛋碎了，剩下 `k-1` 个鸡蛋和 `t-1` 步只能保证验证 `dp[k-1][t-1]` 的楼层，这里还剩**第** `dp[k-1][t-1]+ 1` 的楼层，不能保证最终一定能找到安全楼层了。
+- 用反证法就能得出每一步都应该在第 `dp[k-1][t-1] + 1` 层丢鸡蛋。
+- 这道题还可以用二分搜索来解答。回到上面分析的状态转移方程：`dp[k][m] = dp[k-1][m-1] + dp[k][m-1] + 1` 。用数学方法来解析这个递推关系。令 `f(t,k)` 为 `t` 和 `k` 的函数，题目所要求能测到最大楼层是 `N` 的最小步数，即要求出 `f(t,k) ≥ N` 时候的最小 `t`。由状态转移方程可以知道：`f(t,k) = f(t-1,k) + f(t-1,k-1) + 1`，当 `k = 1` 的时候，对应一个鸡蛋的情况，`f(t,1) = t`，当 `t = 1` 的时候，对应一步的情况，`f(1,k) = 1`。有状态转移方程得：    
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_887_1.png'>
+</p>
+
+- 令 `g(t,k) = f(t,k) - f(t,k-1)`，可以得到：  
+
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_887_2.png'>
+</p>  
+
+- 可以知道 `g(t,k)` 是一个杨辉三角，即二项式系数： 
+ 
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_887_3.png'>
+</p>  
+
+- 利用裂项相消的方法：  
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_887_4.png'>
+</p>  
+
+- 于是可以得到：    
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_887_5.png'>
+</p>
+
+- 其中：    
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_887_6.png'>
+</p>
+
+- 于是针对每一项的二项式常数，都可以由前一项乘以一个分数得到下一项。  
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/leetcode_887_7.png'>
+</p>
+
+- 利用二分搜索，不断的二分 `t`，直到逼近找到 `f(t,k) ≥ N` 时候最小的 `t`。时间复杂度 `O(K * log N)`，空间复杂度 `O(1)`。
+
diff --git a/leetcode/0888.Fair-Candy-Swap/888. Fair Candy Swap.go b/leetcode/0888.Fair-Candy-Swap/888. Fair Candy Swap.go
new file mode 100644
index 000000000..9ebbb1222
--- /dev/null
+++ b/leetcode/0888.Fair-Candy-Swap/888. Fair Candy Swap.go	
@@ -0,0 +1,34 @@
+package leetcode
+
+func fairCandySwap(A []int, B []int) []int {
+	hDiff, aMap := diff(A, B)/2, make(map[int]int, len(A))
+	for _, a := range A {
+		aMap[a] = a
+	}
+	for _, b := range B {
+		if a, ok := aMap[hDiff+b]; ok {
+			return []int{a, b}
+		}
+	}
+	return nil
+}
+
+func diff(A []int, B []int) int {
+	diff, maxLen := 0, max(len(A), len(B))
+	for i := 0; i < maxLen; i++ {
+		if i < len(A) {
+			diff += A[i]
+		}
+		if i < len(B) {
+			diff -= B[i]
+		}
+	}
+	return diff
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0888.Fair-Candy-Swap/888. Fair Candy Swap_test.go b/leetcode/0888.Fair-Candy-Swap/888. Fair Candy Swap_test.go
new file mode 100644
index 000000000..ef8760ec3
--- /dev/null
+++ b/leetcode/0888.Fair-Candy-Swap/888. Fair Candy Swap_test.go	
@@ -0,0 +1,65 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question888 struct {
+	para888
+	ans888
+}
+
+// para 是参数
+// one 代表第一个参数
+type para888 struct {
+	one []int
+	two []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans888 struct {
+	one []int
+}
+
+func Test_Problem888(t *testing.T) {
+
+	qs := []question888{
+
+		{
+			para888{[]int{}, []int{}},
+			ans888{[]int{}},
+		},
+
+		{
+			para888{[]int{1, 1}, []int{2, 2}},
+			ans888{[]int{1, 2}},
+		},
+
+		{
+			para888{[]int{1, 2}, []int{2, 3}},
+			ans888{[]int{1, 2}},
+		},
+
+		{
+			para888{[]int{2}, []int{1, 3}},
+			ans888{[]int{2, 3}},
+		},
+
+		{
+			para888{[]int{1, 2, 5}, []int{2, 4}},
+			ans888{[]int{5, 4}},
+		},
+
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 888------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans888, q.para888
+		fmt.Printf("【input】:%v       【output】:%v\n", p, fairCandySwap(p.one, p.two))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0888.Fair-Candy-Swap/README.md b/leetcode/0888.Fair-Candy-Swap/README.md
new file mode 100644
index 000000000..21581ecaf
--- /dev/null
+++ b/leetcode/0888.Fair-Candy-Swap/README.md
@@ -0,0 +1,106 @@
+# [888. Fair Candy Swap](https://leetcode.com/problems/fair-candy-swap/)
+
+
+## 题目
+
+Alice and Bob have candy bars of different sizes: `A[i]` is the size of the `i`-th bar of candy that Alice has, and `B[j]` is the size of the `j`-th bar of candy that Bob has.
+
+Since they are friends, they would like to exchange one candy bar each so that after the exchange, they both have the same total amount of candy. (*The total amount of candy a person has is the sum of the sizes of candy bars they have*.)
+
+Return an integer array `ans` where `ans[0]` is the size of the candy bar that Alice must exchange, and `ans[1]` is the size of the candy bar that Bob must exchange.
+
+If there are multiple answers, you may return any one of them. It is guaranteed an answer exists.
+
+**Example 1**:
+
+```
+Input: A = [1,1], B = [2,2]
+Output: [1,2]
+```
+
+**Example 2**:
+
+```
+Input: A = [1,2], B = [2,3]
+Output: [1,2]
+```
+
+**Example 3**:
+
+```
+Input: A = [2], B = [1,3]
+Output: [2,3]
+```
+
+**Example 4**:
+
+```
+Input: A = [1,2,5], B = [2,4]
+Output: [5,4]
+```
+
+**Note**:
+
+- `1 <= A.length <= 10000`
+- `1 <= B.length <= 10000`
+- `1 <= A[i] <= 100000`
+- `1 <= B[i] <= 100000`
+- It is guaranteed that Alice and Bob have different total amounts of candy.
+- It is guaranteed there exists an answer.
+
+
+## 题目大意
+
+爱丽丝和鲍勃有不同大小的糖果棒：A[i] 是爱丽丝拥有的第 i 块糖的大小，B[j] 是鲍勃拥有的第 j 块糖的大小。因为他们是朋友，所以他们想交换一个糖果棒，这样交换后，他们都有相同的糖果总量。（一个人拥有的糖果总量是他们拥有的糖果棒大小的总和。）返回一个整数数组 ans，其中 ans[0] 是爱丽丝必须交换的糖果棒的大小，ans[1] 是 Bob 必须交换的糖果棒的大小。如果有多个答案，你可以返回其中任何一个。保证答案存在。
+
+提示：
+
+- 1 <= A.length <= 10000
+- 1 <= B.length <= 10000
+- 1 <= A[i] <= 100000
+- 1 <= B[i] <= 100000
+- 保证爱丽丝与鲍勃的糖果总量不同。
+- 答案肯定存在。
+
+
+## 解题思路
+
+- 两人交换糖果，使得两人糖果相等。要求输出一个数组，里面分别包含两人必须交换的糖果大小。
+- 首先这一题肯定了一定有解，其次只允许交换一次。有了这两个前提，使本题变成简单题。先计算出为了使得交换以后两个相同的糖果数，A 需要增加或者减少的糖果数 diff。然后遍历 B ，看 A 中是否存在一个元素，能使得 B 做了对应交换 diff 以后，两人糖果相等。(此题前提保证了一定能找到)。最后输出 A 中的这个元素和遍历到 B 的这个元素，即是两人要交换的糖果数。
+
+## 代码
+
+```go
+func fairCandySwap(A []int, B []int) []int {
+	hDiff, aMap := diff(A, B)/2, make(map[int]int, len(A))
+	for _, a := range A {
+		aMap[a] = a
+	}
+	for _, b := range B {
+		if a, ok := aMap[hDiff+b]; ok {
+			return []int{a, b}
+		}
+	}
+	return nil
+}
+
+func diff(A []int, B []int) int {
+	diff, maxLen := 0, max(len(A), len(B))
+	for i := 0; i < maxLen; i++ {
+		if i < len(A) {
+			diff += A[i]
+		}
+		if i < len(B) {
+			diff -= B[i]
+		}
+	}
+	return diff
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/0890.Find-and-Replace-Pattern/890. Find and Replace Pattern.go b/leetcode/0890.Find-and-Replace-Pattern/890. Find and Replace Pattern.go
new file mode 100644
index 000000000..0a3300f3b
--- /dev/null
+++ b/leetcode/0890.Find-and-Replace-Pattern/890. Find and Replace Pattern.go	
@@ -0,0 +1,32 @@
+package leetcode
+
+func findAndReplacePattern(words []string, pattern string) []string {
+	res := make([]string, 0)
+	for _, word := range words {
+		if match(word, pattern) {
+			res = append(res, word)
+		}
+	}
+	return res
+}
+
+func match(w, p string) bool {
+	if len(w) != len(p) {
+		return false
+	}
+	m, used := make(map[uint8]uint8), make(map[uint8]bool)
+	for i := 0; i < len(w); i++ {
+		if v, ok := m[p[i]]; ok {
+			if w[i] != v {
+				return false
+			}
+		} else {
+			if used[w[i]] {
+				return false
+			}
+			m[p[i]] = w[i]
+			used[w[i]] = true
+		}
+	}
+	return true
+}
diff --git a/leetcode/0890.Find-and-Replace-Pattern/890. Find and Replace Pattern_test.go b/leetcode/0890.Find-and-Replace-Pattern/890. Find and Replace Pattern_test.go
new file mode 100644
index 000000000..3fdfd1971
--- /dev/null
+++ b/leetcode/0890.Find-and-Replace-Pattern/890. Find and Replace Pattern_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question890 struct {
+	para890
+	ans890
+}
+
+// para 是参数
+// one 代表第一个参数
+type para890 struct {
+	words   []string
+	pattern string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans890 struct {
+	one []string
+}
+
+func Test_Problem890(t *testing.T) {
+
+	qs := []question890{
+
+		{
+			para890{[]string{"abc", "deq", "mee", "aqq", "dkd", "ccc"}, "abb"},
+			ans890{[]string{"mee", "aqq"}},
+		},
+
+		{
+			para890{[]string{"a", "b", "c"}, "a"},
+			ans890{[]string{"a", "b", "c"}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 890------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans890, q.para890
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findAndReplacePattern(p.words, p.pattern))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0890.Find-and-Replace-Pattern/README.md b/leetcode/0890.Find-and-Replace-Pattern/README.md
new file mode 100644
index 000000000..c2cb6bb50
--- /dev/null
+++ b/leetcode/0890.Find-and-Replace-Pattern/README.md
@@ -0,0 +1,78 @@
+# [890. Find and Replace Pattern](https://leetcode.com/problems/find-and-replace-pattern/)
+
+
+## 题目
+
+Given a list of strings `words` and a string `pattern`, return *a list of* `words[i]` *that match* `pattern`. You may return the answer in **any order**.
+
+A word matches the pattern if there exists a permutation of letters `p` so that after replacing every letter `x` in the pattern with `p(x)`, we get the desired word.
+
+Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.
+
+**Example 1:**
+
+```
+Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
+Output: ["mee","aqq"]
+Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}.
+"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, since a and b map to the same letter.
+```
+
+**Example 2:**
+
+```
+Input: words = ["a","b","c"], pattern = "a"
+Output: ["a","b","c"]
+```
+
+**Constraints:**
+
+- `1 <= pattern.length <= 20`
+- `1 <= words.length <= 50`
+- `words[i].length == pattern.length`
+- `pattern` and `words[i]` are lowercase English letters.
+
+## 题目大意
+
+你有一个单词列表 words 和一个模式  pattern，你想知道 words 中的哪些单词与模式匹配。如果存在字母的排列 p ，使得将模式中的每个字母 x 替换为 p(x) 之后，我们就得到了所需的单词，那么单词与模式是匹配的。（回想一下，字母的排列是从字母到字母的双射：每个字母映射到另一个字母，没有两个字母映射到同一个字母。）返回 words 中与给定模式匹配的单词列表。你可以按任何顺序返回答案。
+
+## 解题思路
+
+- 按照题目要求，分别映射两个字符串，words 字符串数组中的字符串与 pattern 字符串每个字母做映射。这里用 map 存储。题目还要求不存在 2 个字母映射到同一个字母的情况，所以再增加一个 map，用来判断当前字母是否已经被映射过了。以上 2 个条件都满足即代表模式匹配上了。最终将所有满足模式匹配的字符串输出即可。
+
+## 代码
+
+```go
+package leetcode
+
+func findAndReplacePattern(words []string, pattern string) []string {
+	res := make([]string, 0)
+	for _, word := range words {
+		if match(word, pattern) {
+			res = append(res, word)
+		}
+	}
+	return res
+}
+
+func match(w, p string) bool {
+	if len(w) != len(p) {
+		return false
+	}
+	m, used := make(map[uint8]uint8), make(map[uint8]bool)
+	for i := 0; i < len(w); i++ {
+		if v, ok := m[p[i]]; ok {
+			if w[i] != v {
+				return false
+			}
+		} else {
+			if used[w[i]] {
+				return false
+			}
+			m[p[i]] = w[i]
+			used[w[i]] = true
+		}
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0891. Sum of Subsequence Widths/891. Sum of Subsequence Widths.go b/leetcode/0891.Sum-of-Subsequence-Widths/891. Sum of Subsequence Widths.go
similarity index 100%
rename from Algorithms/0891. Sum of Subsequence Widths/891. Sum of Subsequence Widths.go
rename to leetcode/0891.Sum-of-Subsequence-Widths/891. Sum of Subsequence Widths.go
diff --git a/Algorithms/0891. Sum of Subsequence Widths/891. Sum of Subsequence Widths_test.go b/leetcode/0891.Sum-of-Subsequence-Widths/891. Sum of Subsequence Widths_test.go
similarity index 95%
rename from Algorithms/0891. Sum of Subsequence Widths/891. Sum of Subsequence Widths_test.go
rename to leetcode/0891.Sum-of-Subsequence-Widths/891. Sum of Subsequence Widths_test.go
index 6a0481dce..4fa590159 100644
--- a/Algorithms/0891. Sum of Subsequence Widths/891. Sum of Subsequence Widths_test.go	
+++ b/leetcode/0891.Sum-of-Subsequence-Widths/891. Sum of Subsequence Widths_test.go	
@@ -26,12 +26,12 @@ func Test_Problem891(t *testing.T) {
 
 	qs := []question891{
 
-		question891{
+		{
 			para891{[]int{2, 1, 3}},
 			ans891{6},
 		},
 
-		question891{
+		{
 			para891{[]int{3, 7, 2, 3}},
 			ans891{35},
 		},
diff --git a/Algorithms/0891. Sum of Subsequence Widths/README.md b/leetcode/0891.Sum-of-Subsequence-Widths/README.md
similarity index 100%
rename from Algorithms/0891. Sum of Subsequence Widths/README.md
rename to leetcode/0891.Sum-of-Subsequence-Widths/README.md
diff --git a/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes.go b/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes.go
new file mode 100644
index 000000000..c52d80d58
--- /dev/null
+++ b/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes.go	
@@ -0,0 +1,41 @@
+package leetcode
+
+func surfaceArea(grid [][]int) int {
+	area := 0
+	for i := 0; i < len(grid); i++ {
+		for j := 0; j < len(grid[0]); j++ {
+			if grid[i][j] == 0 {
+				continue
+			}
+			area += grid[i][j]*4 + 2
+			// up
+			if i > 0 {
+				m := min(grid[i][j], grid[i-1][j])
+				area -= m
+			}
+			// down
+			if i < len(grid)-1 {
+				m := min(grid[i][j], grid[i+1][j])
+				area -= m
+			}
+			// left
+			if j > 0 {
+				m := min(grid[i][j], grid[i][j-1])
+				area -= m
+			}
+			// right
+			if j < len(grid[i])-1 {
+				m := min(grid[i][j], grid[i][j+1])
+				area -= m
+			}
+		}
+	}
+	return area
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes_test.go b/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes_test.go
new file mode 100644
index 000000000..79aa2d11c
--- /dev/null
+++ b/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question892 struct {
+	para892
+	ans892
+}
+
+// para 是参数
+// one 代表第一个参数
+type para892 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans892 struct {
+	one int
+}
+
+func Test_Problem892(t *testing.T) {
+
+	qs := []question892{
+
+		{
+			para892{[][]int{{2}}},
+			ans892{10},
+		},
+
+		{
+			para892{[][]int{{1, 2}, {3, 4}}},
+			ans892{34},
+		},
+
+		{
+			para892{[][]int{{1, 0}, {0, 2}}},
+			ans892{16},
+		},
+
+		{
+			para892{[][]int{{1, 1, 1}, {1, 0, 1}, {1, 1, 1}}},
+			ans892{32},
+		},
+
+		{
+			para892{[][]int{{2, 2, 2}, {2, 1, 2}, {2, 2, 2}}},
+			ans892{46},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 892------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans892, q.para892
+		fmt.Printf("【input】:%v       【output】:%v\n", p, surfaceArea(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0892.Surface-Area-of-3D-Shapes/README.md b/leetcode/0892.Surface-Area-of-3D-Shapes/README.md
new file mode 100644
index 000000000..bcb6ad517
--- /dev/null
+++ b/leetcode/0892.Surface-Area-of-3D-Shapes/README.md
@@ -0,0 +1,108 @@
+# [892. Surface Area of 3D Shapes](https://leetcode.com/problems/surface-area-of-3d-shapes/)
+
+
+## 题目
+
+On a `N * N` grid, we place some `1 * 1 * 1` cubes.
+
+Each value `v = grid[i][j]` represents a tower of `v` cubes placed on top of grid cell `(i, j)`.
+
+Return the total surface area of the resulting shapes.
+
+**Example 1**:
+
+```
+Input: [[2]]
+Output: 10
+```
+
+**Example 2**:
+
+```
+Input: [[1,2],[3,4]]
+Output: 34
+```
+
+**Example 3**:
+
+```
+Input: [[1,0],[0,2]]
+Output: 16
+```
+
+**Example 4**:
+
+```
+Input: [[1,1,1],[1,0,1],[1,1,1]]
+Output: 32
+```
+
+**Example 5**:
+
+```
+Input: [[2,2,2],[2,1,2],[2,2,2]]
+Output: 46
+```
+
+**Note**:
+
+- `1 <= N <= 50`
+- `0 <= grid[i][j] <= 50`
+
+## 题目大意
+
+在 N * N 的网格上，我们放置一些 1 * 1 * 1  的立方体。每个值 v = grid[i][j] 表示 v 个正方体叠放在对应单元格 (i, j) 上。请你返回最终形体的表面积。
+
+
+## 解题思路
+
+- 给定一个网格数组，数组里面装的是立方体叠放在所在的单元格，求最终这些叠放的立方体的表面积。
+- 简单题。按照题目意思，找到叠放时，重叠的面，然后用总表面积减去这些重叠的面积即为最终答案。
+
+## 代码
+
+```go
+
+package leetcode
+
+func surfaceArea(grid [][]int) int {
+	area := 0
+	for i := 0; i < len(grid); i++ {
+		for j := 0; j < len(grid[0]); j++ {
+			if grid[i][j] == 0 {
+				continue
+			}
+			area += grid[i][j]*4 + 2
+			// up
+			if i > 0 {
+				m := min(grid[i][j], grid[i-1][j])
+				area -= m
+			}
+			// down
+			if i < len(grid)-1 {
+				m := min(grid[i][j], grid[i+1][j])
+				area -= m
+			}
+			// left
+			if j > 0 {
+				m := min(grid[i][j], grid[i][j-1])
+				area -= m
+			}
+			// right
+			if j < len(grid[i])-1 {
+				m := min(grid[i][j], grid[i][j+1])
+				area -= m
+			}
+		}
+	}
+	return area
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0895. Maximum Frequency Stack/895. Maximum Frequency Stack.go b/leetcode/0895.Maximum-Frequency-Stack/895. Maximum Frequency Stack.go
similarity index 100%
rename from Algorithms/0895. Maximum Frequency Stack/895. Maximum Frequency Stack.go
rename to leetcode/0895.Maximum-Frequency-Stack/895. Maximum Frequency Stack.go
diff --git a/Algorithms/0895. Maximum Frequency Stack/895. Maximum Frequency Stack_test.go b/leetcode/0895.Maximum-Frequency-Stack/895. Maximum Frequency Stack_test.go
similarity index 100%
rename from Algorithms/0895. Maximum Frequency Stack/895. Maximum Frequency Stack_test.go
rename to leetcode/0895.Maximum-Frequency-Stack/895. Maximum Frequency Stack_test.go
diff --git a/Algorithms/0895. Maximum Frequency Stack/README.md b/leetcode/0895.Maximum-Frequency-Stack/README.md
similarity index 100%
rename from Algorithms/0895. Maximum Frequency Stack/README.md
rename to leetcode/0895.Maximum-Frequency-Stack/README.md
diff --git a/leetcode/0896.Monotonic-Array/896. Monotonic Array.go b/leetcode/0896.Monotonic-Array/896. Monotonic Array.go
new file mode 100644
index 000000000..217b700e3
--- /dev/null
+++ b/leetcode/0896.Monotonic-Array/896. Monotonic Array.go	
@@ -0,0 +1,32 @@
+package leetcode
+
+func isMonotonic(A []int) bool {
+	if len(A) <= 1 {
+		return true
+	}
+	if A[0] < A[1] {
+		return inc(A[1:])
+	}
+	if A[0] > A[1] {
+		return dec(A[1:])
+	}
+	return inc(A[1:]) || dec(A[1:])
+}
+
+func inc(A []int) bool {
+	for i := 0; i < len(A)-1; i++ {
+		if A[i] > A[i+1] {
+			return false
+		}
+	}
+	return true
+}
+
+func dec(A []int) bool {
+	for i := 0; i < len(A)-1; i++ {
+		if A[i] < A[i+1] {
+			return false
+		}
+	}
+	return true
+}
diff --git a/leetcode/0896.Monotonic-Array/896. Monotonic Array_test.go b/leetcode/0896.Monotonic-Array/896. Monotonic Array_test.go
new file mode 100644
index 000000000..2b1112059
--- /dev/null
+++ b/leetcode/0896.Monotonic-Array/896. Monotonic Array_test.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question896 struct {
+	para896
+	ans896
+}
+
+// para 是参数
+// one 代表第一个参数
+type para896 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans896 struct {
+	one bool
+}
+
+func Test_Problem896(t *testing.T) {
+
+	qs := []question896{
+
+		{
+			para896{[]int{2, 1, 3}},
+			ans896{false},
+		},
+
+		{
+			para896{[]int{1, 2, 2, 3}},
+			ans896{true},
+		},
+
+		{
+			para896{[]int{6, 5, 4, 4}},
+			ans896{true},
+		},
+
+		{
+			para896{[]int{1, 3, 2}},
+			ans896{false},
+		},
+
+		{
+			para896{[]int{1, 2, 4, 5}},
+			ans896{true},
+		},
+
+		{
+			para896{[]int{1, 1, 1}},
+			ans896{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 896------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans896, q.para896
+		fmt.Printf("【input】:%v       【output】:%v\n", p, isMonotonic(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0896.Monotonic-Array/README.md b/leetcode/0896.Monotonic-Array/README.md
new file mode 100644
index 000000000..cb45451cd
--- /dev/null
+++ b/leetcode/0896.Monotonic-Array/README.md
@@ -0,0 +1,95 @@
+# [896. Monotonic Array](https://leetcode.com/problems/monotonic-array/)
+
+
+## 题目
+
+An array is *monotonic* if it is either monotone increasing or monotone decreasing.
+
+An array `A` is monotone increasing if for all `i <= j`, `A[i] <= A[j]`. An array `A` is monotone decreasing if for all `i <= j`, `A[i] >= A[j]`.
+
+Return `true` if and only if the given array `A` is monotonic.
+
+**Example 1**:
+
+```
+Input: [1,2,2,3]
+Output: true
+```
+
+**Example 2**:
+
+```
+Input: [6,5,4,4]
+Output: true
+```
+
+**Example 3**:
+
+```
+Input: [1,3,2]
+Output: false
+```
+
+**Example 4**:
+
+```
+Input: [1,2,4,5]
+Output: true
+```
+
+**Example 5**:
+
+```
+Input: [1,1,1]
+Output: true
+```
+
+**Note**:
+
+1. `1 <= A.length <= 50000`
+2. `-100000 <= A[i] <= 100000`
+
+## 题目大意
+
+如果数组是单调递增或单调递减的，那么它是单调的。如果对于所有 i <= j，A[i] <= A[j]，那么数组 A 是单调递增的。 如果对于所有 i <= j，A[i]> = A[j]，那么数组 A 是单调递减的。当给定的数组 A 是单调数组时返回 true，否则返回 false。
+
+
+## 解题思路
+
+- 判断给定的数组是不是单调(单调递增或者单调递减)的。
+- 简单题，按照题意循环判断即可。
+
+## 代码
+
+```go
+func isMonotonic(A []int) bool {
+	if len(A) <= 1 {
+		return true
+	}
+	if A[0] < A[1] {
+		return inc(A[1:])
+	}
+	if A[0] > A[1] {
+		return dec(A[1:])
+	}
+	return inc(A[1:]) || dec(A[1:])
+}
+
+func inc(A []int) bool {
+	for i := 0; i < len(A)-1; i++ {
+		if A[i] > A[i+1] {
+			return false
+		}
+	}
+	return true
+}
+
+func dec(A []int) bool {
+	for i := 0; i < len(A)-1; i++ {
+		if A[i] < A[i+1] {
+			return false
+		}
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/0897.Increasing-Order-Search-Tree/897. Increasing Order Search Tree.go b/leetcode/0897.Increasing-Order-Search-Tree/897. Increasing Order Search Tree.go
new file mode 100644
index 000000000..2d4070e04
--- /dev/null
+++ b/leetcode/0897.Increasing-Order-Search-Tree/897. Increasing Order Search Tree.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 链表思想
+func increasingBST(root *TreeNode) *TreeNode {
+	var head = &TreeNode{}
+	tail := head
+	recBST(root, tail)
+	return head.Right
+}
+
+func recBST(root, tail *TreeNode) *TreeNode {
+	if root == nil {
+		return tail
+	}
+	tail = recBST(root.Left, tail)
+	root.Left = nil               // 切断 root 与其 Left 的连接，避免形成环
+	tail.Right, tail = root, root // 把 root 接上 tail，并保持 tail 指向尾部
+	tail = recBST(root.Right, tail)
+	return tail
+}
+
+// 解法二 模拟
+func increasingBST1(root *TreeNode) *TreeNode {
+	list := []int{}
+	inorder(root, &list)
+	if len(list) == 0 {
+		return root
+	}
+	newRoot := &TreeNode{Val: list[0], Left: nil, Right: nil}
+	cur := newRoot
+	for index := 1; index < len(list); index++ {
+		tmp := &TreeNode{Val: list[index], Left: nil, Right: nil}
+		cur.Right = tmp
+		cur = tmp
+	}
+	return newRoot
+}
+
+func inorder(root *TreeNode, output *[]int) {
+	if root != nil {
+		inorder(root.Left, output)
+		*output = append(*output, root.Val)
+		inorder(root.Right, output)
+	}
+}
diff --git a/leetcode/0897.Increasing-Order-Search-Tree/897. Increasing Order Search Tree_test.go b/leetcode/0897.Increasing-Order-Search-Tree/897. Increasing Order Search Tree_test.go
new file mode 100644
index 000000000..b92a97f1b
--- /dev/null
+++ b/leetcode/0897.Increasing-Order-Search-Tree/897. Increasing Order Search Tree_test.go	
@@ -0,0 +1,81 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question897 struct {
+	para897
+	ans897
+}
+
+// para 是参数
+// one 代表第一个参数
+type para897 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans897 struct {
+	one []int
+}
+
+func Test_Problem897(t *testing.T) {
+
+	qs := []question897{
+
+		{
+			para897{[]int{5, 3, 6, 2, 4, structures.NULL, 8, 1, structures.NULL, structures.NULL, structures.NULL, 7, 9}},
+			ans897{[]int{1, structures.NULL, 2, structures.NULL, 3, structures.NULL, 4, structures.NULL, 5, structures.NULL, 6, structures.NULL, 7, structures.NULL, 8, structures.NULL, 9}},
+		},
+
+		{
+			para897{[]int{3, 4, 4, 5, structures.NULL, structures.NULL, 5, 6, structures.NULL, structures.NULL, 6}},
+			ans897{[]int{6, structures.NULL, 5, structures.NULL, 4, structures.NULL, 3, structures.NULL, 4, structures.NULL, 5, structures.NULL, 6}},
+		},
+
+		{
+			para897{[]int{1, 2, 2, structures.NULL, 3, 3}},
+			ans897{[]int{2, structures.NULL, 3, structures.NULL, 1, structures.NULL, 3, structures.NULL, 2}},
+		},
+
+		{
+			para897{[]int{}},
+			ans897{[]int{}},
+		},
+
+		{
+			para897{[]int{1}},
+			ans897{[]int{1}},
+		},
+
+		{
+			para897{[]int{1, 2, 3}},
+			ans897{[]int{2, structures.NULL, 1, structures.NULL, 3}},
+		},
+
+		{
+			para897{[]int{1, 2, 2, 3, 4, 4, 3}},
+			ans897{[]int{3, structures.NULL, 2, structures.NULL, 4, structures.NULL, 1, structures.NULL, 4, structures.NULL, 2, structures.NULL, 3}},
+		},
+
+		{
+			para897{[]int{1, 2, 2, structures.NULL, 3, structures.NULL, 3}},
+			ans897{[]int{2, structures.NULL, 3, structures.NULL, 1, structures.NULL, 2, structures.NULL, 3}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 897------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans897, q.para897
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", structures.Tree2ints(increasingBST(rootOne)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0897.Increasing-Order-Search-Tree/README.md b/leetcode/0897.Increasing-Order-Search-Tree/README.md
new file mode 100755
index 000000000..579e2fded
--- /dev/null
+++ b/leetcode/0897.Increasing-Order-Search-Tree/README.md
@@ -0,0 +1,60 @@
+# [897. Increasing Order Search Tree](https://leetcode.com/problems/increasing-order-search-tree/)
+
+
+## 题目
+
+Given a binary search tree, rearrange the tree in **in-order** so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.
+
+    Example 1:
+    Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]
+    
+           5
+          / \
+        3    6
+       / \    \
+      2   4    8
+     /        / \ 
+    1        7   9
+    
+    Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
+    
+     1
+      \
+       2
+        \
+         3
+          \
+           4
+            \
+             5
+              \
+               6
+                \
+                 7
+                  \
+                   8
+                    \
+                     9
+
+**Note:**
+
+1. The number of nodes in the given tree will be between 1 and 100.
+2. Each node will have a unique integer value from 0 to 1000.
+
+
+## 题目大意
+
+给定一个树，按中序遍历重新排列树，使树中最左边的结点现在是树的根，并且每个结点没有左子结点，只有一个右子结点。
+
+
+提示：
+
+- 给定树中的结点数介于 1 和 100 之间。
+- 每个结点都有一个从 0 到 1000 范围内的唯一整数值。
+
+
+## 解题思路
+
+- 给出一颗树，要求把树的所有孩子都排列到右子树上。
+- 按照题意，可以先中根遍历原树，然后按照中根遍历的顺序，把所有节点都放在右子树上。见解法二。
+- 上一种解法会重新构造一颗新树，有没有办法可以直接更改原有的树呢？节约存储空间。虽然平时软件开发过程中不建议更改原有的值，但是算法题中追求空间和时间的最优，可以考虑一下。**树可以看做是有多个孩子的链表**。这一题可以看成是链表的类似反转的操作。这一点想通以后，就好做了。先找到左子树中最左边的节点，这个节点是新树的根节点。然后依次往回遍历，不断的记录下上一次遍历的最后节点 tail，边遍历，边把后续节点串起来。最终“反转”完成以后，就得到了题目要求的样子了。代码实现见解法一。
diff --git a/Algorithms/0898. Bitwise ORs of Subarrays/898. Bitwise ORs of Subarrays.go b/leetcode/0898.Bitwise-ORs-of-Subarrays/898. Bitwise ORs of Subarrays.go
similarity index 100%
rename from Algorithms/0898. Bitwise ORs of Subarrays/898. Bitwise ORs of Subarrays.go
rename to leetcode/0898.Bitwise-ORs-of-Subarrays/898. Bitwise ORs of Subarrays.go
diff --git a/Algorithms/0898. Bitwise ORs of Subarrays/898. Bitwise ORs of Subarrays_test.go b/leetcode/0898.Bitwise-ORs-of-Subarrays/898. Bitwise ORs of Subarrays_test.go
similarity index 94%
rename from Algorithms/0898. Bitwise ORs of Subarrays/898. Bitwise ORs of Subarrays_test.go
rename to leetcode/0898.Bitwise-ORs-of-Subarrays/898. Bitwise ORs of Subarrays_test.go
index 5387263b2..7c81c146f 100644
--- a/Algorithms/0898. Bitwise ORs of Subarrays/898. Bitwise ORs of Subarrays_test.go	
+++ b/leetcode/0898.Bitwise-ORs-of-Subarrays/898. Bitwise ORs of Subarrays_test.go	
@@ -26,17 +26,17 @@ func Test_Problem898(t *testing.T) {
 
 	qs := []question898{
 
-		question898{
+		{
 			para898{[]int{0}},
 			ans898{1},
 		},
 
-		question898{
+		{
 			para898{[]int{1, 1, 2}},
 			ans898{3},
 		},
 
-		question898{
+		{
 			para898{[]int{1, 2, 4}},
 			ans898{6},
 		},
diff --git a/leetcode/0898.Bitwise-ORs-of-Subarrays/README.md b/leetcode/0898.Bitwise-ORs-of-Subarrays/README.md
new file mode 100755
index 000000000..5dcb6eaef
--- /dev/null
+++ b/leetcode/0898.Bitwise-ORs-of-Subarrays/README.md
@@ -0,0 +1,82 @@
+# [898. Bitwise ORs of Subarrays](https://leetcode.com/problems/bitwise-ors-of-subarrays/)
+
+
+## 题目
+
+We have an array `A` of non-negative integers.
+
+For every (contiguous) subarray `B = [A[i], A[i+1], ..., A[j]]` (with `i <= j`), we take the bitwise OR of all the elements in `B`, obtaining a result `A[i] | A[i+1] | ... | A[j]`.
+
+Return the number of possible results. (Results that occur more than once are only counted once in the final answer.)
+
+**Example 1:**
+
+    Input: [0]
+    Output: 1
+    Explanation: 
+    There is only one possible result: 0.
+
+**Example 2:**
+
+    Input: [1,1,2]
+    Output: 3
+    Explanation: 
+    The possible subarrays are [1], [1], [2], [1, 1], [1, 2], [1, 1, 2].
+    These yield the results 1, 1, 2, 1, 3, 3.
+    There are 3 unique values, so the answer is 3.
+
+**Example 3:**
+
+    Input: [1,2,4]
+    Output: 6
+    Explanation: 
+    The possible results are 1, 2, 3, 4, 6, and 7.
+
+**Note:**
+
+1. `1 <= A.length <= 50000`
+2. `0 <= A[i] <= 10^9`
+
+
+## 题目大意
+
+我们有一个非负整数数组 A。对于每个（连续的）子数组 B = [A[i], A[i+1], ..., A[j]] （ i <= j），我们对 B 中的每个元素进行按位或操作，获得结果 A[i] | A[i+1] | ... | A[j]。返回可能结果的数量。（多次出现的结果在最终答案中仅计算一次。）
+
+
+
+## 解题思路
+
+- 给出一个数组，要求求出这个数组所有的子数组中，每个集合内所有数字取 `|` 运算以后，不同结果的种类数。
+- 这道题可以这样考虑，第一步，先考虑所有的子数组如何得到，以 `[001, 011, 100, 110, 101]` 为例，所有的子数组集合如下：
+
+```c
+    [001]  
+    [001 011] [011]  
+    [001 011 100] [011 100] [100]  
+    [001 011 100 110] [011 100 110] [100 110] [110]  
+    [001 011 100 110 101] [011 100 110 101] [100 110 101] [110 101] [101]  
+```
+
+可以发现，从左往右遍历原数组，每次新来的一个元素，依次加入到之前已经生成过的集合中，再以自己为单独集合。这样就可以生成原数组的所有子集。
+
+- 第二步，将每一行的子集内的所有元素都进行 `|` 运算，得到：
+
+```c
+    001  
+    011 011  
+    111 111 100  
+    111 111 110 110  
+    111 111 111 111 101  
+```
+
+- 第三步，去重：
+
+```c
+    001  
+    011  
+    111 100  
+    111 110  
+    111 101  
+```
+
+由于二进制位不超过 32 位，所以这里每一行最多不会超过 32 个数。所以最终时间复杂度不会超过 O(32 N)，即 O(K * N)。最后将这每一行的数字都放入最终的 map 中去重即可。
diff --git a/Algorithms/0901. Online Stock Span/901. Online Stock Span.go b/leetcode/0901.Online-Stock-Span/901. Online Stock Span.go
similarity index 100%
rename from Algorithms/0901. Online Stock Span/901. Online Stock Span.go
rename to leetcode/0901.Online-Stock-Span/901. Online Stock Span.go
diff --git a/Algorithms/0901. Online Stock Span/901. Online Stock Span_test.go b/leetcode/0901.Online-Stock-Span/901. Online Stock Span_test.go
similarity index 100%
rename from Algorithms/0901. Online Stock Span/901. Online Stock Span_test.go
rename to leetcode/0901.Online-Stock-Span/901. Online Stock Span_test.go
diff --git a/Algorithms/0901. Online Stock Span/README.md b/leetcode/0901.Online-Stock-Span/README.md
similarity index 100%
rename from Algorithms/0901. Online Stock Span/README.md
rename to leetcode/0901.Online-Stock-Span/README.md
diff --git a/Algorithms/0904. Fruit Into Baskets/904. Fruit Into Baskets.go b/leetcode/0904.Fruit-Into-Baskets/904. Fruit Into Baskets.go
similarity index 90%
rename from Algorithms/0904. Fruit Into Baskets/904. Fruit Into Baskets.go
rename to leetcode/0904.Fruit-Into-Baskets/904. Fruit Into Baskets.go
index 991d6000d..129b58f95 100644
--- a/Algorithms/0904. Fruit Into Baskets/904. Fruit Into Baskets.go	
+++ b/leetcode/0904.Fruit-Into-Baskets/904. Fruit Into Baskets.go	
@@ -26,3 +26,10 @@ func totalFruit(tree []int) int {
 	}
 	return res
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0904. Fruit Into Baskets/904. Fruit Into Baskets_test.go b/leetcode/0904.Fruit-Into-Baskets/904. Fruit Into Baskets_test.go
similarity index 88%
rename from Algorithms/0904. Fruit Into Baskets/904. Fruit Into Baskets_test.go
rename to leetcode/0904.Fruit-Into-Baskets/904. Fruit Into Baskets_test.go
index c8201408c..ade6eb94e 100644
--- a/Algorithms/0904. Fruit Into Baskets/904. Fruit Into Baskets_test.go	
+++ b/leetcode/0904.Fruit-Into-Baskets/904. Fruit Into Baskets_test.go	
@@ -26,41 +26,41 @@ func Test_Problem904(t *testing.T) {
 
 	qs := []question904{
 
-		question904{
+		{
 			para904{[]int{1, 1}},
 			ans904{2},
 		},
 
-		question904{
+		{
 			para904{[]int{1, 2, 1}},
 			ans904{3},
 		},
-		question904{
+		{
 			para904{[]int{0, 1, 2, 2}},
 			ans904{3},
 		},
 
-		question904{
+		{
 			para904{[]int{1, 2, 3, 2, 2}},
 			ans904{4},
 		},
 
-		question904{
+		{
 			para904{[]int{3, 3, 3, 1, 2, 1, 1, 2, 3, 3, 4}},
 			ans904{5},
 		},
 
-		question904{
+		{
 			para904{[]int{4, 5}},
 			ans904{2},
 		},
 
-		question904{
+		{
 			para904{[]int{1}},
 			ans904{1},
 		},
 
-		question904{
+		{
 			para904{[]int{}},
 			ans904{},
 		},
diff --git a/Algorithms/0904. Fruit Into Baskets/README.md b/leetcode/0904.Fruit-Into-Baskets/README.md
similarity index 100%
rename from Algorithms/0904. Fruit Into Baskets/README.md
rename to leetcode/0904.Fruit-Into-Baskets/README.md
diff --git a/Algorithms/0907. Sum of Subarray Minimums/907. Sum of Subarray Minimums.go b/leetcode/0907.Sum-of-Subarray-Minimums/907. Sum of Subarray Minimums.go
similarity index 100%
rename from Algorithms/0907. Sum of Subarray Minimums/907. Sum of Subarray Minimums.go
rename to leetcode/0907.Sum-of-Subarray-Minimums/907. Sum of Subarray Minimums.go
diff --git a/Algorithms/0907. Sum of Subarray Minimums/907. Sum of Subarray Minimums_test.go b/leetcode/0907.Sum-of-Subarray-Minimums/907. Sum of Subarray Minimums_test.go
similarity index 97%
rename from Algorithms/0907. Sum of Subarray Minimums/907. Sum of Subarray Minimums_test.go
rename to leetcode/0907.Sum-of-Subarray-Minimums/907. Sum of Subarray Minimums_test.go
index 01b2eb58d..b21f74fd4 100644
--- a/Algorithms/0907. Sum of Subarray Minimums/907. Sum of Subarray Minimums_test.go	
+++ b/leetcode/0907.Sum-of-Subarray-Minimums/907. Sum of Subarray Minimums_test.go	
@@ -26,7 +26,7 @@ func Test_Problem907(t *testing.T) {
 
 	qs := []question907{
 
-		question907{
+		{
 			para907{[]int{3, 1, 2, 4}},
 			ans907{17},
 		},
diff --git a/Algorithms/0907. Sum of Subarray Minimums/README.md b/leetcode/0907.Sum-of-Subarray-Minimums/README.md
similarity index 100%
rename from Algorithms/0907. Sum of Subarray Minimums/README.md
rename to leetcode/0907.Sum-of-Subarray-Minimums/README.md
diff --git a/leetcode/0909.Snakes-and-Ladders/909. Snakes and Ladders.go b/leetcode/0909.Snakes-and-Ladders/909. Snakes and Ladders.go
new file mode 100644
index 000000000..51c8c6be3
--- /dev/null
+++ b/leetcode/0909.Snakes-and-Ladders/909. Snakes and Ladders.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+type pair struct {
+	id, step int
+}
+
+func snakesAndLadders(board [][]int) int {
+	n := len(board)
+	visited := make([]bool, n*n+1)
+	queue := []pair{{1, 0}}
+	for len(queue) > 0 {
+		p := queue[0]
+		queue = queue[1:]
+		for i := 1; i <= 6; i++ {
+			nxt := p.id + i
+			if nxt > n*n { // 超出边界
+				break
+			}
+			r, c := getRowCol(nxt, n) // 得到下一步的行列
+			if board[r][c] > 0 {      // 存在蛇或梯子
+				nxt = board[r][c]
+			}
+			if nxt == n*n { // 到达终点
+				return p.step + 1
+			}
+			if !visited[nxt] {
+				visited[nxt] = true
+				queue = append(queue, pair{nxt, p.step + 1}) // 扩展新状态
+			}
+		}
+	}
+	return -1
+}
+
+func getRowCol(id, n int) (r, c int) {
+	r, c = (id-1)/n, (id-1)%n
+	if r%2 == 1 {
+		c = n - 1 - c
+	}
+	r = n - 1 - r
+	return r, c
+}
diff --git a/leetcode/0909.Snakes-and-Ladders/909. Snakes and Ladders_test.go b/leetcode/0909.Snakes-and-Ladders/909. Snakes and Ladders_test.go
new file mode 100644
index 000000000..7d3d5ba16
--- /dev/null
+++ b/leetcode/0909.Snakes-and-Ladders/909. Snakes and Ladders_test.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question909 struct {
+	para909
+	ans909
+}
+
+// para 是参数
+// one 代表第一个参数
+type para909 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans909 struct {
+	one int
+}
+
+func Test_Problem909(t *testing.T) {
+	qs := []question909{
+		{
+			para909{[][]int{
+				{-1, -1, -1, -1, -1, -1},
+				{-1, -1, -1, -1, -1, -1},
+				{-1, -1, -1, -1, -1, -1},
+				{-1, 35, -1, -1, 13, -1},
+				{-1, -1, -1, -1, -1, -1},
+				{-1, 15, -1, -1, -1, -1},
+			}},
+			ans909{4},
+		},
+	}
+	fmt.Printf("------------------------Leetcode Problem 909------------------------\n")
+	for _, q := range qs {
+		_, p := q.ans909, q.para909
+		fmt.Printf("【input】:%v       【output】:%v\n", p, snakesAndLadders(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0909.Snakes-and-Ladders/README.md b/leetcode/0909.Snakes-and-Ladders/README.md
new file mode 100644
index 000000000..43b899568
--- /dev/null
+++ b/leetcode/0909.Snakes-and-Ladders/README.md
@@ -0,0 +1,110 @@
+# [909. Snakes and Ladders](https://leetcode.com/problems/snakes-and-ladders/)
+
+
+## 题目
+
+On an N x N `board`, the numbers from `1` to `N*N` are written *boustrophedonically* **starting from the bottom left of the board**, and alternating direction each row.  For example, for a 6 x 6 board, the numbers are written as follows:
+
+
+![](https://assets.leetcode.com/uploads/2018/09/23/snakes.png)
+
+You start on square `1` of the board (which is always in the last row and first column).  Each move, starting from square `x`, consists of the following:
+
+- You choose a destination square `S` with number `x+1`, `x+2`, `x+3`, `x+4`, `x+5`, or `x+6`, provided this number is `<= N*N`.
+    - (This choice simulates the result of a standard 6-sided die roll: ie., there are always **at most 6 destinations, regardless of the size of the board**.)
+- If `S` has a snake or ladder, you move to the destination of that snake or ladder. Otherwise, you move to `S`.
+
+A board square on row `r` and column `c` has a "snake or ladder" if `board[r][c] != -1`.  The destination of that snake or ladder is `board[r][c]`.
+
+Note that you only take a snake or ladder at most once per move: if the destination to a snake or ladder is the start of another snake or ladder, you do **not** continue moving.  (For example, if the board is `[[4,-1],[-1,3]]`, and on the first move your destination square is `2`, then you finish your first move at `3`, because you do **not** continue moving to `4`.)
+
+Return the least number of moves required to reach square N*N.  If it is not possible, return `-1`.
+
+**Example 1:**
+
+```
+Input:[
+[-1,-1,-1,-1,-1,-1],
+[-1,-1,-1,-1,-1,-1],
+[-1,-1,-1,-1,-1,-1],
+[-1,35,-1,-1,13,-1],
+[-1,-1,-1,-1,-1,-1],
+[-1,15,-1,-1,-1,-1]]
+Output:4
+Explanation:
+At the beginning, you start at square 1 [at row 5, column 0].
+You decide to move to square 2, and must take the ladder to square 15.
+You then decide to move to square 17 (row 3, column 5), and must take the snake to square 13.
+You then decide to move to square 14, and must take the ladder to square 35.
+You then decide to move to square 36, ending the game.
+It can be shown that you need at least 4 moves to reach the N*N-th square, so the answer is 4.
+
+```
+
+**Note:**
+
+1. `2 <= board.length = board[0].length <= 20`
+2. `board[i][j]` is between `1` and `N*N` or is equal to `1`.
+3. The board square with number `1` has no snake or ladder.
+4. The board square with number `N*N` has no snake or ladder.
+
+## 题目大意
+
+N x N 的棋盘 board 上，按从 1 到 N*N 的数字给方格编号，编号 从左下角开始，每一行交替方向。r 行 c 列的棋盘，按前述方法编号，棋盘格中可能存在 “蛇” 或 “梯子”；如果 board[r][c] != -1，那个蛇或梯子的目的地将会是 board[r][c]。玩家从棋盘上的方格 1 （总是在最后一行、第一列）开始出发。每一回合，玩家需要从当前方格 x 开始出发，按下述要求前进：选定目标方格：
+
+- 选择从编号 x+1，x+2，x+3，x+4，x+5，或者 x+6 的方格中选出一个目标方格 s ，目标方格的编号 <= N*N。该选择模拟了掷骰子的情景，无论棋盘大小如何，你的目的地范围也只能处于区间 [x+1, x+6] 之间。
+- 传送玩家：如果目标方格 S 处存在蛇或梯子，那么玩家会传送到蛇或梯子的目的地。否则，玩家传送到目标方格 S。
+
+注意，玩家在每回合的前进过程中最多只能爬过蛇或梯子一次：就算目的地是另一条蛇或梯子的起点，你也不会继续移动。返回达到方格 N*N 所需的最少移动次数，如果不可能，则返回 -1。
+
+## 解题思路
+
+- 这一题可以抽象为在有向图上求下标 1 的起点到下标 `N^2` 的终点的最短路径。用广度优先搜索。棋盘可以抽象成一个包含 `N^2` 个节点的有向图，对于每个节点 `x`，若 `x+i (1 ≤ i ≤ 6)` 上没有蛇或梯子，则连一条从 `x` 到 `x+i` 的有向边；否则记蛇梯的目的地为 `y`，连一条从 `x` 到 `y` 的有向边。然后按照最短路径的求解方式便可解题。时间复杂度 O(n^2)，空间复杂度 O(n^2)。
+- 此题棋盘上的下标是蛇形的，所以遍历下一个点的时候需要转换坐标。具体做法根据行的奇偶性，行号为偶数，下标从左往右，行号为奇数，下标从右往左。具体实现见 `getRowCol()` 函数。
+
+## 代码
+
+```go
+package leetcode
+
+type pair struct {
+	id, step int
+}
+
+func snakesAndLadders(board [][]int) int {
+	n := len(board)
+	visited := make([]bool, n*n+1)
+	queue := []pair{{1, 0}}
+	for len(queue) > 0 {
+		p := queue[0]
+		queue = queue[1:]
+		for i := 1; i <= 6; i++ {
+			nxt := p.id + i
+			if nxt > n*n { // 超出边界
+				break
+			}
+			r, c := getRowCol(nxt, n) // 得到下一步的行列
+			if board[r][c] > 0 {      // 存在蛇或梯子
+				nxt = board[r][c]
+			}
+			if nxt == n*n { // 到达终点
+				return p.step + 1
+			}
+			if !visited[nxt] {
+				visited[nxt] = true
+				queue = append(queue, pair{nxt, p.step + 1}) // 扩展新状态
+			}
+		}
+	}
+	return -1
+}
+
+func getRowCol(id, n int) (r, c int) {
+	r, c = (id-1)/n, (id-1)%n
+	if r%2 == 1 {
+		c = n - 1 - c
+	}
+	r = n - 1 - r
+	return r, c
+}
+```
\ No newline at end of file
diff --git a/leetcode/0910.Smallest-Range-II/910.Smallest Range II.go b/leetcode/0910.Smallest-Range-II/910.Smallest Range II.go
new file mode 100644
index 000000000..a2543300b
--- /dev/null
+++ b/leetcode/0910.Smallest-Range-II/910.Smallest Range II.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+import "sort"
+
+func smallestRangeII(A []int, K int) int {
+	n := len(A)
+	sort.Ints(A)
+	res := A[n-1] - A[0]
+	for i := 0; i < n-1; i++ {
+		a, b := A[i], A[i+1]
+		high := max(A[n-1]-K, a+K)
+		low := min(A[0]+K, b-K)
+		res = min(res, high-low)
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/0910.Smallest-Range-II/910.Smallest Range II_test.go b/leetcode/0910.Smallest-Range-II/910.Smallest Range II_test.go
new file mode 100644
index 000000000..f1a83a99a
--- /dev/null
+++ b/leetcode/0910.Smallest-Range-II/910.Smallest Range II_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question910 struct {
+	para910
+	ans910
+}
+
+type para910 struct {
+	A []int
+	K int
+}
+
+type ans910 struct {
+	one int
+}
+
+// Test_Problem910 ...
+func Test_Problem910(t *testing.T) {
+
+	qs := []question910{
+
+		{
+			para910{[]int{1}, 0},
+			ans910{0},
+		},
+		{
+			para910{[]int{0, 10}, 2},
+			ans910{6},
+		},
+		{
+			para910{[]int{1, 3, 6}, 3},
+			ans910{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 910------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans910, q.para910
+		fmt.Printf("【input】:%v       【output】:%v\n", p, smallestRangeII(p.A, p.K))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0910.Smallest-Range-II/README.md b/leetcode/0910.Smallest-Range-II/README.md
new file mode 100644
index 000000000..e8a3d8ff7
--- /dev/null
+++ b/leetcode/0910.Smallest-Range-II/README.md
@@ -0,0 +1,82 @@
+# [910. Smallest Range II](https://leetcode.com/problems/smallest-range-ii/)
+
+## 题目
+
+Given an array `A` of integers, for each integer `A[i]` we need to choose **either `x = -K` or `x = K`**, and add `x` to `A[i]` **(only once)**.
+
+After this process, we have some array `B`.
+
+Return the smallest possible difference between the maximum value of `B` and the minimum value of `B`.
+
+**Example 1:**
+
+```c
+Input: A = [1], K = 0
+Output: 0
+Explanation: B = [1]
+```
+
+**Example 2:**
+
+```c
+Input: A = [0,10], K = 2
+Output: 6
+Explanation: B = [2,8]
+```
+
+**Example 3:**
+
+```c
+Input: A = [1,3,6], K = 3
+Output: 3
+Explanation: B = [4,6,3]
+```
+
+**Note:**
+
+1. `1 <= A.length <= 10000`
+2. `0 <= A[i] <= 10000`
+3. `0 <= K <= 10000`
+
+## 题目大意
+
+给你一个整数数组 A，对于每个整数 A[i]，可以选择 x = -K 或是 x = K （K 总是非负整数），并将 x 加到 A[i] 中。在此过程之后，得到数组 B。返回 B 的最大值和 B 的最小值之间可能存在的最小差值。
+
+## 解题思路
+
+- 简单题。先排序，找出 A 数组中最大的差值。然后循环扫一遍数组，利用双指针，选择 x = -K 或是 x = K ，每次选择都更新一次最大值和最小值之间的最小差值。循环一次以后便可以找到满足题意的答案。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func smallestRangeII(A []int, K int) int {
+	n := len(A)
+	sort.Ints(A)
+	res := A[n-1] - A[0]
+	for i := 0; i < n-1; i++ {
+		a, b := A[i], A[i+1]
+		high := max(A[n-1]-K, a+K)
+		low := min(A[0]+K, b-K)
+		res = min(res, high-low)
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/0911.Online-Election/911. Online Election.go b/leetcode/0911.Online-Election/911. Online Election.go
new file mode 100644
index 000000000..cd6b5751a
--- /dev/null
+++ b/leetcode/0911.Online-Election/911. Online Election.go	
@@ -0,0 +1,38 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+// TopVotedCandidate define
+type TopVotedCandidate struct {
+	persons []int
+	times   []int
+}
+
+// Constructor911 define
+func Constructor911(persons []int, times []int) TopVotedCandidate {
+	leaders, votes := make([]int, len(persons)), make([]int, len(persons))
+	leader := persons[0]
+	for i := 0; i < len(persons); i++ {
+		p := persons[i]
+		votes[p]++
+		if votes[p] >= votes[leader] {
+			leader = p
+		}
+		leaders[i] = leader
+	}
+	return TopVotedCandidate{persons: leaders, times: times}
+}
+
+// Q define
+func (tvc *TopVotedCandidate) Q(t int) int {
+	i := sort.Search(len(tvc.times), func(p int) bool { return tvc.times[p] > t })
+	return tvc.persons[i-1]
+}
+
+/**
+ * Your TopVotedCandidate object will be instantiated and called as such:
+ * obj := Constructor(persons, times);
+ * param_1 := obj.Q(t);
+ */
diff --git a/leetcode/0911.Online-Election/911. Online Election_test.go b/leetcode/0911.Online-Election/911. Online Election_test.go
new file mode 100644
index 000000000..e2ee112cb
--- /dev/null
+++ b/leetcode/0911.Online-Election/911. Online Election_test.go	
@@ -0,0 +1,28 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem911(t *testing.T) {
+	obj := Constructor911([]int{0, 1, 1, 0, 0, 1, 0}, []int{0, 5, 10, 15, 20, 25, 30})
+	fmt.Printf("obj.Q[3] = %v\n", obj.Q(3))   // 0
+	fmt.Printf("obj.Q[12] = %v\n", obj.Q(12)) // 1
+	fmt.Printf("obj.Q[25] = %v\n", obj.Q(25)) // 1
+	fmt.Printf("obj.Q[15] = %v\n", obj.Q(15)) // 0
+	fmt.Printf("obj.Q[24] = %v\n", obj.Q(24)) // 0
+	fmt.Printf("obj.Q[8] = %v\n", obj.Q(8))   // 1
+
+	obj = Constructor911([]int{0, 0, 0, 0, 1}, []int{0, 6, 39, 52, 75})
+	fmt.Printf("obj.Q[45] = %v\n", obj.Q(45)) // 0
+	fmt.Printf("obj.Q[49] = %v\n", obj.Q(49)) // 0
+	fmt.Printf("obj.Q[59] = %v\n", obj.Q(59)) // 0
+	fmt.Printf("obj.Q[68] = %v\n", obj.Q(68)) // 0
+	fmt.Printf("obj.Q[42] = %v\n", obj.Q(42)) // 0
+	fmt.Printf("obj.Q[37] = %v\n", obj.Q(37)) // 0
+	fmt.Printf("obj.Q[99] = %v\n", obj.Q(99)) // 0
+	fmt.Printf("obj.Q[26] = %v\n", obj.Q(26)) // 0
+	fmt.Printf("obj.Q[78] = %v\n", obj.Q(78)) // 0
+	fmt.Printf("obj.Q[43] = %v\n", obj.Q(43)) // 0
+}
diff --git a/leetcode/0911.Online-Election/README.md b/leetcode/0911.Online-Election/README.md
new file mode 100755
index 000000000..98a0bc698
--- /dev/null
+++ b/leetcode/0911.Online-Election/README.md
@@ -0,0 +1,53 @@
+# [911. Online Election](https://leetcode.com/problems/online-election/)
+
+
+## 题目
+
+In an election, the `i`-th vote was cast for `persons[i]` at time `times[i]`.
+
+Now, we would like to implement the following query function: `TopVotedCandidate.q(int t)` will return the number of the person that was leading the election at time `t`.
+
+Votes cast at time `t` will count towards our query. In the case of a tie, the most recent vote (among tied candidates) wins.
+
+**Example 1:**
+
+    Input: ["TopVotedCandidate","q","q","q","q","q","q"], [[[0,1,1,0,0,1,0],[0,5,10,15,20,25,30]],[3],[12],[25],[15],[24],[8]]
+    Output: [null,0,1,1,0,0,1]
+    Explanation: 
+    At time 3, the votes are [0], and 0 is leading.
+    At time 12, the votes are [0,1,1], and 1 is leading.
+    At time 25, the votes are [0,1,1,0,0,1], and 1 is leading (as ties go to the most recent vote.)
+    This continues for 3 more queries at time 15, 24, and 8.
+
+**Note:**
+
+1. `1 <= persons.length = times.length <= 5000`
+2. `0 <= persons[i] <= persons.length`
+3. `times` is a strictly increasing array with all elements in `[0, 10^9]`.
+4. `TopVotedCandidate.q` is called at most `10000` times per test case.
+5. `TopVotedCandidate.q(int t)` is always called with `t >= times[0]`.
+
+
+## 题目大意
+
+在选举中，第 i 张票是在时间为 times[i] 时投给 persons[i] 的。
+
+现在，我们想要实现下面的查询函数： TopVotedCandidate.q(int t) 将返回在 t 时刻主导选举的候选人的编号。
+
+在 t 时刻投出的选票也将被计入我们的查询之中。在平局的情况下，最近获得投票的候选人将会获胜。
+
+提示：
+
+1. 1 <= persons.length = times.length <= 5000
+2. 0 <= persons[i] <= persons.length
+3. times 是严格递增的数组，所有元素都在 [0, 10^9] 范围中。
+4. 每个测试用例最多调用 10000 次 TopVotedCandidate.q。
+5. TopVotedCandidate.q(int t) 被调用时总是满足 t >= times[0]。
+
+
+
+
+## 解题思路
+
+- 给出一个 2 个数组，分别代表第 `i` 人在第 `t` 时刻获得的票数。需要实现一个查询功能的函数，查询在任意 `t` 时刻，输出谁的选票领先。
+- `persons[]` 数组里面装的是获得选票人的编号，`times[]` 数组里面对应的是每个选票的时刻。`times[]` 数组默认是有序的，从小到大排列。先计算出每个时刻哪个人选票领先，放在一个数组中，实现查询函数的时候，只需要先对 `times[]` 数组二分搜索，找到比查询时间 `t` 小的最大时刻 `i`，再在选票领先的数组里面输出对应时刻领先的人的编号即可。
diff --git a/leetcode/0914.X-of-a-Kind-in-a-Deck-of-Cards/914. X of a Kind in a Deck of Cards.go b/leetcode/0914.X-of-a-Kind-in-a-Deck-of-Cards/914. X of a Kind in a Deck of Cards.go
new file mode 100644
index 000000000..c44a2c014
--- /dev/null
+++ b/leetcode/0914.X-of-a-Kind-in-a-Deck-of-Cards/914. X of a Kind in a Deck of Cards.go	
@@ -0,0 +1,26 @@
+package leetcode
+
+func hasGroupsSizeX(deck []int) bool {
+	if len(deck) < 2 {
+		return false
+	}
+	m, g := map[int]int{}, -1
+	for _, d := range deck {
+		m[d]++
+	}
+	for _, v := range m {
+		if g == -1 {
+			g = v
+		} else {
+			g = gcd(g, v)
+		}
+	}
+	return g >= 2
+}
+
+func gcd(a, b int) int {
+	if a == 0 {
+		return b
+	}
+	return gcd(b%a, a)
+}
diff --git a/leetcode/0914.X-of-a-Kind-in-a-Deck-of-Cards/914. X of a Kind in a Deck of Cards_test.go b/leetcode/0914.X-of-a-Kind-in-a-Deck-of-Cards/914. X of a Kind in a Deck of Cards_test.go
new file mode 100644
index 000000000..3372ee55d
--- /dev/null
+++ b/leetcode/0914.X-of-a-Kind-in-a-Deck-of-Cards/914. X of a Kind in a Deck of Cards_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question914 struct {
+	para914
+	ans914
+}
+
+// para 是参数
+// one 代表第一个参数
+type para914 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans914 struct {
+	one bool
+}
+
+func Test_Problem914(t *testing.T) {
+
+	qs := []question914{
+
+		{
+			para914{[]int{1, 2, 3, 4, 4, 3, 2, 1}},
+			ans914{true},
+		},
+
+		{
+			para914{[]int{1, 1, 1, 2, 2, 2, 3, 3}},
+			ans914{false},
+		},
+
+		{
+			para914{[]int{1}},
+			ans914{false},
+		},
+
+		{
+			para914{[]int{1, 1}},
+			ans914{true},
+		},
+
+		{
+			para914{[]int{1, 1, 2, 2, 2, 2}},
+			ans914{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 914------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans914, q.para914
+		fmt.Printf("【input】:%v       【output】:%v\n", p, hasGroupsSizeX(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0914.X-of-a-Kind-in-a-Deck-of-Cards/README.md b/leetcode/0914.X-of-a-Kind-in-a-Deck-of-Cards/README.md
new file mode 100644
index 000000000..02c8ebddd
--- /dev/null
+++ b/leetcode/0914.X-of-a-Kind-in-a-Deck-of-Cards/README.md
@@ -0,0 +1,105 @@
+# [914. X of a Kind in a Deck of Cards](https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/)
+
+
+## 题目
+
+In a deck of cards, each card has an integer written on it.
+
+Return `true` if and only if you can choose `X >= 2` such that it is possible to split the entire deck into 1 or more groups of cards, where:
+
+- Each group has exactly `X` cards.
+- All the cards in each group have the same integer.
+
+**Example 1**:
+
+```
+Input: deck = [1,2,3,4,4,3,2,1]
+Output: true
+Explanation: Possible partition [1,1],[2,2],[3,3],[4,4].
+```
+
+**Example 2**:
+
+```
+Input: deck = [1,1,1,2,2,2,3,3]
+Output: false´
+Explanation: No possible partition.
+```
+
+**Example 3**:
+
+```
+Input: deck = [1]
+Output: false
+Explanation: No possible partition.
+```
+
+**Example 4**:
+
+```
+Input: deck = [1,1]
+Output: true
+Explanation: Possible partition [1,1].
+```
+
+**Example 5**:
+
+```
+Input: deck = [1,1,2,2,2,2]
+Output: true
+Explanation: Possible partition [1,1],[2,2],[2,2].
+```
+
+**Constraints**:
+
+- `1 <= deck.length <= 10^4`
+- `0 <= deck[i] < 10^4`
+
+## 题目大意
+
+给定一副牌，每张牌上都写着一个整数。此时，你需要选定一个数字 X，使我们可以将整副牌按下述规则分成 1 组或更多组：
+
+- 每组都有 X 张牌。
+- 组内所有的牌上都写着相同的整数。
+
+仅当你可选的 X >= 2 时返回 true。
+
+
+## 解题思路
+
+- 给定一副牌，要求选出数字 X，使得每组都有 X 张牌，每组牌的数字都相同。当 X ≥ 2 的时候，输出 true。
+- 通过分析题目，我们可以知道，只有当 X 为所有 count 的约数，即所有 count 的最大公约数的约数时，才存在可能的分组。因此我们只要求出所有 count 的最大公约数 g，判断 g 是否大于等于 2 即可，如果大于等于 2，则满足条件，否则不满足。
+- 时间复杂度：O(NlogC)，其中 N 是卡牌的个数，C 是数组 deck 中数的范围，在本题中 C 的值为 10000。求两个数最大公约数的复杂度是 O(logC)，需要求最多 N - 1 次。空间复杂度：O(N + C) 或 O(N)。
+
+## 代码
+
+```go
+
+package leetcode
+
+func hasGroupsSizeX(deck []int) bool {
+	if len(deck) < 2 {
+		return false
+	}
+	m, g := map[int]int{}, -1
+	for _, d := range deck {
+		m[d]++
+	}
+	for _, v := range m {
+		if g == -1 {
+			g = v
+		} else {
+			g = gcd(g, v)
+		}
+	}
+	return g >= 2
+}
+
+func gcd(a, b int) int {
+	if a == 0 {
+		return b
+	}
+	return gcd(b%a, a)
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/0916.Word-Subsets/916. Word Subsets.go b/leetcode/0916.Word-Subsets/916. Word Subsets.go
new file mode 100644
index 000000000..e614e6540
--- /dev/null
+++ b/leetcode/0916.Word-Subsets/916. Word Subsets.go	
@@ -0,0 +1,36 @@
+package leetcode
+
+func wordSubsets(A []string, B []string) []string {
+	var counter [26]int
+	for _, b := range B {
+		var m [26]int
+		for _, c := range b {
+			j := c - 'a'
+			m[j]++
+		}
+		for i := 0; i < 26; i++ {
+			if m[i] > counter[i] {
+				counter[i] = m[i]
+			}
+		}
+	}
+	var res []string
+	for _, a := range A {
+		var m [26]int
+		for _, c := range a {
+			j := c - 'a'
+			m[j]++
+		}
+		ok := true
+		for i := 0; i < 26; i++ {
+			if m[i] < counter[i] {
+				ok = false
+				break
+			}
+		}
+		if ok {
+			res = append(res, a)
+		}
+	}
+	return res
+}
diff --git a/leetcode/0916.Word-Subsets/916. Word Subsets_test.go b/leetcode/0916.Word-Subsets/916. Word Subsets_test.go
new file mode 100644
index 000000000..3a0e6bc1a
--- /dev/null
+++ b/leetcode/0916.Word-Subsets/916. Word Subsets_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question916 struct {
+	para916
+	ans916
+}
+
+// para 是参数
+// one 代表第一个参数
+type para916 struct {
+	A []string
+	B []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans916 struct {
+	one []string
+}
+
+func Test_Problem916(t *testing.T) {
+
+	qs := []question916{
+
+		{
+			para916{[]string{"amazon", "apple", "facebook", "google", "leetcode"}, []string{"e", "o"}},
+			ans916{[]string{"facebook", "google", "leetcode"}},
+		},
+
+		{
+			para916{[]string{"amazon", "apple", "facebook", "google", "leetcode"}, []string{"l", "e"}},
+			ans916{[]string{"apple", "google", "leetcode"}},
+		},
+
+		{
+			para916{[]string{"amazon", "apple", "facebook", "google", "leetcode"}, []string{"e", "oo"}},
+			ans916{[]string{"facebook", "google"}},
+		},
+
+		{
+			para916{[]string{"amazon", "apple", "facebook", "google", "leetcode"}, []string{"lo", "eo"}},
+			ans916{[]string{"google", "leetcode"}},
+		},
+
+		{
+			para916{[]string{"amazon", "apple", "facebook", "google", "leetcode"}, []string{"ec", "oc", "ceo"}},
+			ans916{[]string{"facebook", "leetcode"}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 916------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans916, q.para916
+		fmt.Printf("【input】:%v       【output】:%v\n", p, wordSubsets(p.A, p.B))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0916.Word-Subsets/README.md b/leetcode/0916.Word-Subsets/README.md
new file mode 100644
index 000000000..6c4f6538e
--- /dev/null
+++ b/leetcode/0916.Word-Subsets/README.md
@@ -0,0 +1,103 @@
+# [916. Word Subsets](https://leetcode.com/problems/word-subsets/)
+
+
+## 题目
+
+We are given two arrays `A` and `B` of words.  Each word is a string of lowercase letters.
+
+Now, say that word `b` is a subset of word `a` ****if every letter in `b` occurs in `a`, **including multiplicity**.  For example, `"wrr"` is a subset of `"warrior"`, but is not a subset of `"world"`.
+
+Now say a word `a` from `A` is *universal* if for every `b` in `B`, `b` is a subset of `a`.
+
+Return a list of all universal words in `A`.  You can return the words in any order.
+
+**Example 1:**
+
+```
+Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
+Output:["facebook","google","leetcode"]
+```
+
+**Example 2:**
+
+```
+Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
+Output:["apple","google","leetcode"]
+```
+
+**Example 3:**
+
+```
+Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
+Output:["facebook","google"]
+```
+
+**Example 4:**
+
+```
+Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
+Output:["google","leetcode"]
+```
+
+**Example 5:**
+
+```
+Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
+Output:["facebook","leetcode"]
+```
+
+**Note:**
+
+1. `1 <= A.length, B.length <= 10000`
+2. `1 <= A[i].length, B[i].length <= 10`
+3. `A[i]` and `B[i]` consist only of lowercase letters.
+4. All words in `A[i]` are unique: there isn't `i != j` with `A[i] == A[j]`.
+
+## 题目大意
+
+我们给出两个单词数组 A 和 B。每个单词都是一串小写字母。现在，如果 b 中的每个字母都出现在 a 中，包括重复出现的字母，那么称单词 b 是单词 a 的子集。 例如，“wrr” 是 “warrior” 的子集，但不是 “world” 的子集。如果对 B 中的每一个单词 b，b 都是 a 的子集，那么我们称 A 中的单词 a 是通用的。你可以按任意顺序以列表形式返回 A 中所有的通用单词。
+
+## 解题思路
+
+- 简单题。先统计出 B 数组中单词每个字母的频次，再在 A 数组中依次判断每个单词是否超过了这个频次，如果超过了即输出。
+
+## 代码
+
+```go
+package leetcode
+
+func wordSubsets(A []string, B []string) []string {
+	var counter [26]int
+	for _, b := range B {
+		var m [26]int
+		for _, c := range b {
+			j := c - 'a'
+			m[j]++
+		}
+		for i := 0; i < 26; i++ {
+			if m[i] > counter[i] {
+				counter[i] = m[i]
+			}
+		}
+	}
+	var res []string
+	for _, a := range A {
+		var m [26]int
+		for _, c := range a {
+			j := c - 'a'
+			m[j]++
+		}
+		ok := true
+		for i := 0; i < 26; i++ {
+			if m[i] < counter[i] {
+				ok = false
+				break
+			}
+		}
+		if ok {
+			res = append(res, a)
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0918.Maximum-Sum-Circular-Subarray/918. Maximum Sum Circular Subarray.go b/leetcode/0918.Maximum-Sum-Circular-Subarray/918. Maximum Sum Circular Subarray.go
new file mode 100644
index 000000000..9057d97ab
--- /dev/null
+++ b/leetcode/0918.Maximum-Sum-Circular-Subarray/918. Maximum Sum Circular Subarray.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import "math"
+
+func maxSubarraySumCircular(nums []int) int {
+	var max1, max2, sum int
+
+	// case: no circulation
+	max1 = int(math.Inf(-1))
+	l := len(nums)
+	for i := 0; i < l; i++ {
+		sum += nums[i]
+		if sum > max1 {
+			max1 = sum
+		}
+		if sum < 1 {
+			sum = 0
+		}
+	}
+
+	// case: circling
+	arr_sum := 0
+	for i := 0; i < l; i++ {
+		arr_sum += nums[i]
+	}
+
+	sum = 0
+	min_sum := 0
+	for i := 1; i < l-1; i++ {
+		sum += nums[i]
+		if sum >= 0 {
+			sum = 0
+		}
+		if sum < min_sum {
+			min_sum = sum
+		}
+	}
+	max2 = arr_sum - min_sum
+
+	return max(max1, max2)
+}
+
+func max(nums ...int) int {
+	max := int(math.Inf(-1))
+	for _, num := range nums {
+		if num > max {
+			max = num
+		}
+	}
+	return max
+}
diff --git a/Algorithms/0918. Maximum Sum Circular Subarray/918. Maximum Sum Circular Subarray_test.go b/leetcode/0918.Maximum-Sum-Circular-Subarray/918. Maximum Sum Circular Subarray_test.go
similarity index 91%
rename from Algorithms/0918. Maximum Sum Circular Subarray/918. Maximum Sum Circular Subarray_test.go
rename to leetcode/0918.Maximum-Sum-Circular-Subarray/918. Maximum Sum Circular Subarray_test.go
index b601a8ed7..f7d25c084 100644
--- a/Algorithms/0918. Maximum Sum Circular Subarray/918. Maximum Sum Circular Subarray_test.go	
+++ b/leetcode/0918.Maximum-Sum-Circular-Subarray/918. Maximum Sum Circular Subarray_test.go	
@@ -26,27 +26,27 @@ func Test_Problem918(t *testing.T) {
 
 	qs := []question918{
 
-		question918{
+		{
 			para918{[]int{1, -2, 3, -2}},
 			ans918{3},
 		},
 
-		question918{
+		{
 			para918{[]int{5, -3, 5}},
 			ans918{10},
 		},
 
-		question918{
+		{
 			para918{[]int{3, -1, 2, -1}},
 			ans918{4},
 		},
 
-		question918{
+		{
 			para918{[]int{3, -2, 2, -3}},
 			ans918{3},
 		},
 
-		question918{
+		{
 			para918{[]int{-2, -3, -1}},
 			ans918{-1},
 		},
diff --git a/leetcode/0918.Maximum-Sum-Circular-Subarray/README.md b/leetcode/0918.Maximum-Sum-Circular-Subarray/README.md
new file mode 100755
index 000000000..81ade3d0c
--- /dev/null
+++ b/leetcode/0918.Maximum-Sum-Circular-Subarray/README.md
@@ -0,0 +1,65 @@
+# [918. Maximum Sum Circular Subarray](https://leetcode.com/problems/maximum-sum-circular-subarray/)
+
+
+## 题目
+
+Given a **circular array** **C** of integers represented by `A`, find the maximum possible sum of a non-empty subarray of **C**.
+
+Here, a *circular array* means the end of the array connects to the beginning of the array. (Formally, `C[i] = A[i]` when `0 <= i < A.length`, and `C[i+A.length] = C[i]` when `i >= 0`.)
+
+Also, a subarray may only include each element of the fixed buffer `A` at most once. (Formally, for a subarray `C[i], C[i+1], ..., C[j]`, there does not exist `i <= k1, k2 <= j` with `k1 % A.length = k2 % A.length`.)
+
+**Example 1:**
+
+    Input: [1,-2,3,-2]
+    Output: 3
+    Explanation: Subarray [3] has maximum sum 3
+
+**Example 2:**
+
+    Input: [5,-3,5]
+    Output: 10
+    Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10
+
+**Example 3:**
+
+    Input: [3,-1,2,-1]
+    Output: 4
+    Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4
+
+**Example 4:**
+
+    Input: [3,-2,2,-3]
+    Output: 3
+    Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3
+
+**Example 5:**
+
+    Input: [-2,-3,-1]
+    Output: -1
+    Explanation: Subarray [-1] has maximum sum -1
+
+**Note:**
+
+1. `-30000 <= A[i] <= 30000`
+2. `1 <= A.length <= 30000`
+
+
+## 题目大意
+
+给定一个由整数数组 A 表示的环形数组 C，求 C 的非空子数组的最大可能和。在此处，环形数组意味着数组的末端将会与开头相连呈环状。（形式上，当0 <= i < A.length 时 C[i] = A[i]，而当 i >= 0 时 C[i+A.length] = C[i]）
+
+此外，子数组最多只能包含固定缓冲区 A 中的每个元素一次。（形式上，对于子数组 C[i], C[i+1], ..., C[j]，不存在 i <= k1, k2 <= j 其中 k1 % A.length = k2 % A.length）
+
+提示：
+
+- -30000 <= A[i] <= 30000
+- 1 <= A.length <= 30000
+
+
+## 解题思路
+
+
+- 给出一个环形数组，要求出这个环形数组中的连续子数组的最大和。
+- 拿到这题最先想到的思路是把这个数组再拼接一个，在这两个数组中查找连续子数组的最大和。这种做法是错误的，例如在 `[5,-3,5]` 这个数组中会得出 `7` 的结果，但是实际结果是 `10` 。那么这题怎么做呢？仔细分析可以得到，环形数组的最大连续子段和有两种情况，第一种情况是这个连续子段就出现在数组中， 不存在循环衔接的情况。针对这种情况就比较简单，用 `kadane` 算法(也是动态规划的思想)，`O(n)` 的时间复杂度就可以求出结果。第二种情况是这个连续的子段出现在跨数组的情况，即会出现首尾相连的情况。要想找到这样一个连续子段，可以反向考虑。想找到跨段的连续子段，那么这个数组剩下的这一段就是不跨段的连续子段。想要跨段的子段和最大，那么剩下的这段连续子段和最小。如果能找到这个数组的每个元素取相反数组成的数组中的最大连续子段和，那么反过来就能找到原数组的连续子段和最小。举个例子：`[1，2，-3，-4，5]` ，取它的每个元素的相反数 `[-1，-2，3，4，-5]`，构造的数组中最大连续子段和是 `3 + 4 = 7`，由于取了相反数，所以可以得到原数组中最小连续子段和是 `-7` 。所以跨段的最大连续子段和就是剩下的那段 `[1,2,5]`。
+- 还有一些边界的情况，例如，`[1，2，-2，-3，5，5，-4，6]` 和 `[1，2，-2，-3，5，5，-4，8]`，所以还需要比较一下情况一和情况二的值，它们两者最大值才是最终环形数组的连续子数组的最大和。
diff --git a/Algorithms/0920. Number of Music Playlists/920. Number of Music Playlists.go b/leetcode/0920.Number-of-Music-Playlists/920. Number of Music Playlists.go
similarity index 100%
rename from Algorithms/0920. Number of Music Playlists/920. Number of Music Playlists.go
rename to leetcode/0920.Number-of-Music-Playlists/920. Number of Music Playlists.go
diff --git a/Algorithms/0920. Number of Music Playlists/920. Number of Music Playlists_test.go b/leetcode/0920.Number-of-Music-Playlists/920. Number of Music Playlists_test.go
similarity index 94%
rename from Algorithms/0920. Number of Music Playlists/920. Number of Music Playlists_test.go
rename to leetcode/0920.Number-of-Music-Playlists/920. Number of Music Playlists_test.go
index e51c67d22..39ed35114 100644
--- a/Algorithms/0920. Number of Music Playlists/920. Number of Music Playlists_test.go	
+++ b/leetcode/0920.Number-of-Music-Playlists/920. Number of Music Playlists_test.go	
@@ -28,17 +28,17 @@ func Test_Problem920(t *testing.T) {
 
 	qs := []question920{
 
-		question920{
+		{
 			para920{3, 3, 1},
 			ans920{6},
 		},
 
-		question920{
+		{
 			para920{2, 3, 0},
 			ans920{6},
 		},
 
-		question920{
+		{
 			para920{2, 3, 1},
 			ans920{2},
 		},
diff --git a/leetcode/0920.Number-of-Music-Playlists/README.md b/leetcode/0920.Number-of-Music-Playlists/README.md
new file mode 100755
index 000000000..195802f4c
--- /dev/null
+++ b/leetcode/0920.Number-of-Music-Playlists/README.md
@@ -0,0 +1,59 @@
+# [920. Number of Music Playlists](https://leetcode.com/problems/number-of-music-playlists/)
+
+
+## 题目
+
+Your music player contains `N` different songs and she wants to listen to `L` ****(not necessarily different) songs during your trip. You create a playlist so that:
+
+- Every song is played at least once
+- A song can only be played again only if `K` other songs have been played
+
+Return the number of possible playlists. **As the answer can be very large, return it modulo `10^9 + 7`**.
+
+**Example 1:**
+
+    Input: N = 3, L = 3, K = 1
+    Output: 6
+    Explanation: There are 6 possible playlists. [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].
+
+**Example 2:**
+
+    Input: N = 2, L = 3, K = 0
+    Output: 6
+    Explanation: There are 6 possible playlists. [1, 1, 2], [1, 2, 1], [2, 1, 1], [2, 2, 1], [2, 1, 2], [1, 2, 2]
+
+**Example 3:**
+
+    Input: N = 2, L = 3, K = 1
+    Output: 2
+    Explanation: There are 2 possible playlists. [1, 2, 1], [2, 1, 2]
+
+**Note:**
+
+1. `0 <= K < N <= L <= 100`
+
+## 题目大意
+
+你的音乐播放器里有 N 首不同的歌，在旅途中，你的旅伴想要听 L 首歌（不一定不同，即，允许歌曲重复）。请你为她按如下规则创建一个播放列表：
+
+- 每首歌至少播放一次。
+- 一首歌只有在其他 K 首歌播放完之后才能再次播放。
+
+返回可以满足要求的播放列表的数量。由于答案可能非常大，请返回它模 10^9 + 7 的结果。
+
+提示：
+
+- 0 <= K < N <= L <= 100
+
+
+
+
+## 解题思路
+
+- 简化抽象一下题意，给 N 个数，要求从这 N 个数里面组成一个长度为 L 的序列，并且相同元素的间隔不能小于 K 个数。问总共有多少组组成方法。
+- 一拿到题，会觉得这一题是三维 DP，因为存在 3 个变量，但是实际考虑一下，可以降一维。我们先不考虑 K 的限制，只考虑 N 和 L。定义 `dp[i][j]` 代表播放列表里面有 `i` 首歌，其中包含 `j` 首不同的歌曲，那么题目要求的最终解存在 `dp[L][N]` 中。考虑 `dp[i][j]` 的递归公式，音乐列表当前需要组成 `i` 首歌，有 2 种方式可以得到，由 `i - 1` 首歌的列表中添加一首列表中**不存在**的新歌曲，或者由 `i - 1` 首歌的列表中添加一首列表中**已经存在**的歌曲。即，`dp[i][j]` 可以由 `dp[i - 1][j - 1]` 得到，也可以由 `dp[i - 1][j]` 得到。如果是第一种情况，添加一首新歌，那么新歌有 N - ( j - 1 ) 首，如果是第二种情况，添加一首已经存在的歌，歌有 j 首，所以状态转移方程是 `dp[i][j] = dp[i - 1][j - 1] * ( N - ( j - 1 ) ) + dp[i - 1][j] * j` 。但是这个方程是在不考虑 K 的限制条件下得到的，距离满足题意还差一步。接下来需要考虑加入 K 这个限制条件以后，状态转移方程该如何推导。
+- 如果是添加一首新歌，是不受 K 限制的，所以 `dp[i - 1][j - 1] * ( N - ( j - 1 ) )` 这里不需要变化。如果是添加一首存在的歌曲，这个时候就会受到 K 的限制了。如果当前播放列表里面的歌曲有 `j` 首，并且 `j > K`，那么选择歌曲只能从 `j - K` 里面选，因为不能选择 `j - 1` 到 `j - k` 的这些歌，选择了就不满足重复的歌之间间隔不能小于 `K` 的限制条件了。那 j ≤ K 呢？这个时候一首歌都不能选，因为歌曲数都没有超过 K，当然不能再选择重复的歌曲。(选择了就再次不满足重复的歌之间间隔不能小于 `K` 的限制条件了)。经过上述分析，可以得到最终的状态转移方程：
+
+![](https://img.halfrost.com/Leetcode/leetcode_920.gif)
+
+- 上面的式子可以合并简化成下面这个式子：`dp[i][j] = dp[i - 1][j - 1]*(N - (j - 1)) + dp[i-1][j]*max(j-K, 0)`，递归初始值 `dp[0][0] = 1`。
diff --git a/Algorithms/0921. Minimum Add to Make Parentheses Valid/921. Minimum Add to Make Parentheses Valid.go b/leetcode/0921.Minimum-Add-to-Make-Parentheses-Valid/921. Minimum Add to Make Parentheses Valid.go
similarity index 100%
rename from Algorithms/0921. Minimum Add to Make Parentheses Valid/921. Minimum Add to Make Parentheses Valid.go
rename to leetcode/0921.Minimum-Add-to-Make-Parentheses-Valid/921. Minimum Add to Make Parentheses Valid.go
diff --git a/Algorithms/0921. Minimum Add to Make Parentheses Valid/921. Minimum Add to Make Parentheses Valid_test.go b/leetcode/0921.Minimum-Add-to-Make-Parentheses-Valid/921. Minimum Add to Make Parentheses Valid_test.go
similarity index 92%
rename from Algorithms/0921. Minimum Add to Make Parentheses Valid/921. Minimum Add to Make Parentheses Valid_test.go
rename to leetcode/0921.Minimum-Add-to-Make-Parentheses-Valid/921. Minimum Add to Make Parentheses Valid_test.go
index c1be5eeeb..24e5c55bb 100644
--- a/Algorithms/0921. Minimum Add to Make Parentheses Valid/921. Minimum Add to Make Parentheses Valid_test.go	
+++ b/leetcode/0921.Minimum-Add-to-Make-Parentheses-Valid/921. Minimum Add to Make Parentheses Valid_test.go	
@@ -25,22 +25,22 @@ type ans921 struct {
 func Test_Problem921(t *testing.T) {
 
 	qs := []question921{
-		question921{
+		{
 			para921{"())"},
 			ans921{1},
 		},
 
-		question921{
+		{
 			para921{"((("},
 			ans921{3},
 		},
 
-		question921{
+		{
 			para921{"()"},
 			ans921{0},
 		},
 
-		question921{
+		{
 			para921{"()))(("},
 			ans921{4},
 		},
diff --git a/Algorithms/0921. Minimum Add to Make Parentheses Valid/README.md b/leetcode/0921.Minimum-Add-to-Make-Parentheses-Valid/README.md
similarity index 100%
rename from Algorithms/0921. Minimum Add to Make Parentheses Valid/README.md
rename to leetcode/0921.Minimum-Add-to-Make-Parentheses-Valid/README.md
diff --git a/Algorithms/0922. Sort Array By Parity II/922. Sort Array By Parity II.go b/leetcode/0922.Sort-Array-By-Parity-II/922. Sort Array By Parity II.go
similarity index 100%
rename from Algorithms/0922. Sort Array By Parity II/922. Sort Array By Parity II.go
rename to leetcode/0922.Sort-Array-By-Parity-II/922. Sort Array By Parity II.go
diff --git a/Algorithms/0922. Sort Array By Parity II/922. Sort Array By Parity II_test.go b/leetcode/0922.Sort-Array-By-Parity-II/922. Sort Array By Parity II_test.go
similarity index 94%
rename from Algorithms/0922. Sort Array By Parity II/922. Sort Array By Parity II_test.go
rename to leetcode/0922.Sort-Array-By-Parity-II/922. Sort Array By Parity II_test.go
index 92a4277e6..235e06026 100644
--- a/Algorithms/0922. Sort Array By Parity II/922. Sort Array By Parity II_test.go	
+++ b/leetcode/0922.Sort-Array-By-Parity-II/922. Sort Array By Parity II_test.go	
@@ -26,17 +26,17 @@ func Test_Problem922(t *testing.T) {
 
 	qs := []question922{
 
-		question922{
+		{
 			para922{[]int{}},
 			ans922{[]int{}},
 		},
 
-		question922{
+		{
 			para922{[]int{1}},
 			ans922{[]int{}},
 		},
 
-		question922{
+		{
 			para922{[]int{4, 2, 5, 7}},
 			ans922{[]int{4, 5, 2, 7}},
 		},
diff --git a/Algorithms/0922. Sort Array By Parity II/README.md b/leetcode/0922.Sort-Array-By-Parity-II/README.md
similarity index 100%
rename from Algorithms/0922. Sort Array By Parity II/README.md
rename to leetcode/0922.Sort-Array-By-Parity-II/README.md
diff --git a/Algorithms/0923. 3Sum With Multiplicity/923. 3Sum With Multiplicity.go b/leetcode/0923.3Sum-With-Multiplicity/923. 3Sum With Multiplicity.go
similarity index 100%
rename from Algorithms/0923. 3Sum With Multiplicity/923. 3Sum With Multiplicity.go
rename to leetcode/0923.3Sum-With-Multiplicity/923. 3Sum With Multiplicity.go
diff --git a/Algorithms/0923. 3Sum With Multiplicity/923. 3Sum With Multiplicity_test.go b/leetcode/0923.3Sum-With-Multiplicity/923. 3Sum With Multiplicity_test.go
similarity index 95%
rename from Algorithms/0923. 3Sum With Multiplicity/923. 3Sum With Multiplicity_test.go
rename to leetcode/0923.3Sum-With-Multiplicity/923. 3Sum With Multiplicity_test.go
index d01c4218f..505509560 100644
--- a/Algorithms/0923. 3Sum With Multiplicity/923. 3Sum With Multiplicity_test.go	
+++ b/leetcode/0923.3Sum-With-Multiplicity/923. 3Sum With Multiplicity_test.go	
@@ -27,12 +27,12 @@ func Test_Problem923(t *testing.T) {
 
 	qs := []question923{
 
-		question923{
+		{
 			para923{[]int{1, 1, 2, 2, 3, 3, 4, 4, 5, 5}, 8},
 			ans923{20},
 		},
 
-		question923{
+		{
 			para923{[]int{1, 1, 2, 2, 2, 2}, 5},
 			ans923{12},
 		},
diff --git a/Algorithms/0923. 3Sum With Multiplicity/README.md b/leetcode/0923.3Sum-With-Multiplicity/README.md
similarity index 100%
rename from Algorithms/0923. 3Sum With Multiplicity/README.md
rename to leetcode/0923.3Sum-With-Multiplicity/README.md
diff --git a/leetcode/0924.Minimize-Malware-Spread/924. Minimize Malware Spread.go b/leetcode/0924.Minimize-Malware-Spread/924. Minimize Malware Spread.go
new file mode 100644
index 000000000..0b23c427c
--- /dev/null
+++ b/leetcode/0924.Minimize-Malware-Spread/924. Minimize Malware Spread.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"math"
+
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+func minMalwareSpread(graph [][]int, initial []int) int {
+	if len(initial) == 0 {
+		return 0
+	}
+	uf, maxLen, maxIdx, uniqInitials, compMap := template.UnionFindCount{}, math.MinInt32, -1, map[int]int{}, map[int][]int{}
+	uf.Init(len(graph))
+	for i := range graph {
+		for j := i + 1; j < len(graph); j++ {
+			if graph[i][j] == 1 {
+				uf.Union(i, j)
+			}
+		}
+	}
+	for _, i := range initial {
+		compMap[uf.Find(i)] = append(compMap[uf.Find(i)], i)
+	}
+	for _, v := range compMap {
+		if len(v) == 1 {
+			uniqInitials[v[0]] = v[0]
+		}
+	}
+	if len(uniqInitials) == 0 {
+		smallestIdx := initial[0]
+		for _, i := range initial {
+			if i < smallestIdx {
+				smallestIdx = i
+			}
+		}
+		return smallestIdx
+	}
+	for _, i := range initial {
+		if _, ok := uniqInitials[i]; ok {
+			size := uf.Count()[uf.Find(i)]
+			if maxLen < size {
+				maxLen, maxIdx = size, i
+			} else if maxLen == size {
+				if i < maxIdx {
+					maxIdx = i
+				}
+			}
+		}
+	}
+	return maxIdx
+}
diff --git a/leetcode/0924.Minimize-Malware-Spread/924. Minimize Malware Spread_test.go b/leetcode/0924.Minimize-Malware-Spread/924. Minimize Malware Spread_test.go
new file mode 100644
index 000000000..8c9576e6a
--- /dev/null
+++ b/leetcode/0924.Minimize-Malware-Spread/924. Minimize Malware Spread_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question924 struct {
+	para924
+	ans924
+}
+
+// para 是参数
+// one 代表第一个参数
+type para924 struct {
+	graph   [][]int
+	initial []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans924 struct {
+	one int
+}
+
+func Test_Problem924(t *testing.T) {
+
+	qs := []question924{
+
+		{
+			para924{[][]int{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 1, 1}, {0, 0, 1, 1}}, []int{3, 1}},
+			ans924{3},
+		},
+
+		{
+			para924{[][]int{{1, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0}, {0, 0, 0, 1, 1, 0}, {0, 0, 0, 1, 1, 0}, {0, 0, 0, 0, 0, 1}}, []int{5, 0}},
+			ans924{0},
+		},
+
+		{
+			para924{[][]int{{1, 1, 1}, {1, 1, 1}, {1, 1, 1}}, []int{1, 2}},
+			ans924{1},
+		},
+
+		{
+			para924{[][]int{{1, 1, 0}, {1, 1, 0}, {0, 0, 1}}, []int{0, 1}},
+			ans924{0},
+		},
+
+		{
+			para924{[][]int{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}, []int{0, 2}},
+			ans924{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 924------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans924, q.para924
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minMalwareSpread(p.graph, p.initial))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0924.Minimize-Malware-Spread/README.md b/leetcode/0924.Minimize-Malware-Spread/README.md
new file mode 100755
index 000000000..7e990af97
--- /dev/null
+++ b/leetcode/0924.Minimize-Malware-Spread/README.md
@@ -0,0 +1,58 @@
+# [924. Minimize Malware Spread](https://leetcode.com/problems/minimize-malware-spread/)
+
+
+## 题目
+
+In a network of nodes, each node `i` is directly connected to another node `j` if and only if `graph[i][j] = 1`.
+
+Some nodes `initial` are initially infected by malware. Whenever two nodes are directly connected and at least one of those two nodes is infected by malware, both nodes will be infected by malware. This spread of malware will continue until no more nodes can be infected in this manner.
+
+Suppose `M(initial)` is the final number of nodes infected with malware in the entire network, after the spread of malware stops.
+
+We will remove one node from the initial list. Return the node that if removed, would minimize `M(initial)`. If multiple nodes could be removed to minimize `M(initial)`, return such a node with the smallest index.
+
+Note that if a node was removed from the `initial` list of infected nodes, it may still be infected later as a result of the malware spread.
+
+**Example 1:**
+
+    Input: graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]
+    Output: 0
+
+**Example 2:**
+
+    Input: graph = [[1,0,0],[0,1,0],[0,0,1]], initial = [0,2]
+    Output: 0
+
+**Example 3:**
+
+    Input: graph = [[1,1,1],[1,1,1],[1,1,1]], initial = [1,2]
+    Output: 1
+
+**Note:**
+
+1. `1 < graph.length = graph[0].length <= 300`
+2. `0 <= graph[i][j] == graph[j][i] <= 1`
+3. `graph[i][i] = 1`
+4. `1 <= initial.length < graph.length`
+5. `0 <= initial[i] < graph.length`
+
+
+## 题目大意
+
+在节点网络中，只有当 graph[i][j] = 1 时，每个节点 i 能够直接连接到另一个节点 j。一些节点 initial 最初被恶意软件感染。只要两个节点直接连接，且其中至少一个节点受到恶意软件的感染，那么两个节点都将被恶意软件感染。这种恶意软件的传播将继续，直到没有更多的节点可以被这种方式感染。假设 M(initial) 是在恶意软件停止传播之后，整个网络中感染恶意软件的最终节点数。我们可以从初始列表中删除一个节点。如果移除这一节点将最小化 M(initial)， 则返回该节点。如果有多个节点满足条件，就返回索引最小的节点。请注意，如果某个节点已从受感染节点的列表 initial 中删除，它以后可能仍然因恶意软件传播而受到感染。
+
+
+提示：
+
+- 1 < graph.length = graph[0].length <= 300
+- 0 <= graph[i][j] == graph[j][i] <= 1
+- graph[i][i] = 1
+- 1 <= initial.length < graph.length
+- 0 <= initial[i] < graph.length
+
+
+## 解题思路
+
+
+- 给出一个节点之间的关系图，如果两个节点是连通的，那么病毒软件就会感染到连通的所有节点。现在如果想移除一个病毒节点，能最大减少感染，请问移除哪个节点？如果多个节点都能减少感染量，优先移除序号偏小的那个节点。
+- 这一题一看就是考察的并查集。利用节点的连通关系，把题目中给的所有节点都 `union()` 起来，然后依次统计每个集合内有多少个点。最后扫描一次 initial 数组，选出这个数组中节点小的并且所在集合节点多，这个节点就是最终答案。
diff --git a/leetcode/0925.Long-Pressed-Name/925. Long Pressed Name.go b/leetcode/0925.Long-Pressed-Name/925. Long Pressed Name.go
new file mode 100644
index 000000000..8c28a2bf9
--- /dev/null
+++ b/leetcode/0925.Long-Pressed-Name/925. Long Pressed Name.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+func isLongPressedName(name string, typed string) bool {
+	if len(name) == 0 && len(typed) == 0 {
+		return true
+	}
+	if (len(name) == 0 && len(typed) != 0) || (len(name) != 0 && len(typed) == 0) {
+		return false
+	}
+	i, j := 0, 0
+	for i < len(name) && j < len(typed) {
+		if name[i] != typed[j] {
+			return false
+		}
+		for i < len(name) && j < len(typed) && name[i] == typed[j] {
+			i++
+			j++
+		}
+		for j < len(typed) && typed[j] == typed[j-1] {
+			j++
+		}
+	}
+	return i == len(name) && j == len(typed)
+}
diff --git a/Algorithms/0925. Long Pressed Name/925. Long Pressed Name_test.go b/leetcode/0925.Long-Pressed-Name/925. Long Pressed Name_test.go
similarity index 81%
rename from Algorithms/0925. Long Pressed Name/925. Long Pressed Name_test.go
rename to leetcode/0925.Long-Pressed-Name/925. Long Pressed Name_test.go
index 31c7ce673..d2ae66626 100644
--- a/Algorithms/0925. Long Pressed Name/925. Long Pressed Name_test.go	
+++ b/leetcode/0925.Long-Pressed-Name/925. Long Pressed Name_test.go	
@@ -27,27 +27,42 @@ func Test_Problem925(t *testing.T) {
 
 	qs := []question925{
 
-		question925{
+		{
 			para925{"alex", "aaleex"},
 			ans925{true},
 		},
 
-		question925{
+		{
+			para925{"alex", "alexxr"},
+			ans925{false},
+		},
+
+		{
+			para925{"alex", "alexxxxr"},
+			ans925{false},
+		},
+
+		{
+			para925{"alex", "alexxxxx"},
+			ans925{true},
+		},
+
+		{
 			para925{"saeed", "ssaaedd"},
 			ans925{false},
 		},
 
-		question925{
+		{
 			para925{"leelee", "lleeelee"},
 			ans925{true},
 		},
 
-		question925{
+		{
 			para925{"laiden", "laiden"},
 			ans925{true},
 		},
 
-		question925{
+		{
 			para925{"kikcxmvzi", "kiikcxxmmvvzz"},
 			ans925{false},
 		},
diff --git a/leetcode/0925.Long-Pressed-Name/README.md b/leetcode/0925.Long-Pressed-Name/README.md
new file mode 100644
index 000000000..1b588febf
--- /dev/null
+++ b/leetcode/0925.Long-Pressed-Name/README.md
@@ -0,0 +1,67 @@
+# [925. Long Pressed Name](https://leetcode.com/problems/long-pressed-name/)
+
+## 题目
+
+Your friend is typing his name into a keyboard.  Sometimes, when typing a character c, the key might get long pressed, and the character will be typed 1 or more times.
+
+You examine the typed characters of the keyboard.  Return True if it is possible that it was your friends name, with some characters (possibly none) being long pressed.
+
+
+
+Example 1:
+
+```c
+Input: name = "alex", typed = "aaleex"
+Output: true
+Explanation: 'a' and 'e' in 'alex' were long pressed.
+```
+
+Example 2:
+
+```c
+Input: name = "saeed", typed = "ssaaedd"
+Output: false
+Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.
+```
+
+Example 3:
+
+```c
+Input: name = "leelee", typed = "lleeelee"
+Output: true
+```
+
+Example 4:
+
+```c
+Input: name = "laiden", typed = "laiden"
+Output: true
+Explanation: It's not necessary to long press any character.
+```
+
+
+Note:  
+
+1. name.length <= 1000
+2. typed.length <= 1000
+3. The characters of name and typed are lowercase letters.
+
+## 题目大意
+
+
+给定 2 个字符串，后者的字符串中包含前者的字符串。比如在打字的过程中，某个字符会多按了几下。判断后者字符串是不是比前者字符串存在这样的“长按”键盘的情况。
+
+## 解题思路
+
+- 这一题也可以借助滑动窗口的思想。2 个字符串一起比较，如果遇到有相同的字符串，窗口继续往后滑动。直到遇到了第一个不同的字符，如果遇到两个字符串不相等的情况，可以直接返回 false。具体实现见代码。
+- 这一题的测试用例修改过一次，需要注意我这里写的第二组测试用例，当 name 结束以后，如果 typed 还有多余的不同的字符，这种情况要输出 false 的。具体见 test 文件里面的第二组，第三组，第四组测试用例。
+
+
+
+
+
+
+
+
+
+
diff --git a/leetcode/0927.Three-Equal-Parts/927. Three Equal Parts.go b/leetcode/0927.Three-Equal-Parts/927. Three Equal Parts.go
new file mode 100644
index 000000000..2f1008c2f
--- /dev/null
+++ b/leetcode/0927.Three-Equal-Parts/927. Three Equal Parts.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+func threeEqualParts(A []int) []int {
+	n, ones, i, count := len(A), 0, 0, 0
+	for _, a := range A {
+		ones += a
+	}
+	if ones == 0 {
+		return []int{0, n - 1}
+	}
+	if ones%3 != 0 {
+		return []int{-1, -1}
+	}
+	k := ones / 3
+	for i < n {
+		if A[i] == 1 {
+			break
+		}
+		i++
+	}
+	start, j := i, i
+	for j < n {
+		count += A[j]
+		if count == k+1 {
+			break
+		}
+		j++
+	}
+	mid := j
+	j, count = 0, 0
+	for j < n {
+		count += A[j]
+		if count == 2*k+1 {
+			break
+		}
+		j++
+	}
+	end := j
+	for end < n && A[start] == A[mid] && A[mid] == A[end] {
+		start++
+		mid++
+		end++
+	}
+	if end == n {
+		return []int{start - 1, mid}
+	}
+	return []int{-1, -1}
+}
diff --git a/leetcode/0927.Three-Equal-Parts/927. Three Equal Parts_test.go b/leetcode/0927.Three-Equal-Parts/927. Three Equal Parts_test.go
new file mode 100644
index 000000000..5b3764591
--- /dev/null
+++ b/leetcode/0927.Three-Equal-Parts/927. Three Equal Parts_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question927 struct {
+	para927
+	ans927
+}
+
+// para 是参数
+// one 代表第一个参数
+type para927 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans927 struct {
+	one []int
+}
+
+func Test_Problem927(t *testing.T) {
+
+	qs := []question927{
+
+		{
+			para927{[]int{1, 0, 1, 0, 1}},
+			ans927{[]int{0, 3}},
+		},
+
+		{
+			para927{[]int{1, 1, 0, 1, 1}},
+			ans927{[]int{-1, -1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 927------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans927, q.para927
+		fmt.Printf("【input】:%v       【output】:%v\n", p, threeEqualParts(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0927.Three-Equal-Parts/README.md b/leetcode/0927.Three-Equal-Parts/README.md
new file mode 100755
index 000000000..27650b403
--- /dev/null
+++ b/leetcode/0927.Three-Equal-Parts/README.md
@@ -0,0 +1,57 @@
+# [927. Three Equal Parts](https://leetcode.com/problems/three-equal-parts/)
+
+
+## 题目
+
+Given an array `A` of `0`s and `1`s, divide the array into 3 non-empty parts such that all of these parts represent the same binary value.
+
+If it is possible, return **any** `[i, j]` with `i+1 < j`, such that:
+
+- `A[0], A[1], ..., A[i]` is the first part;
+- `A[i+1], A[i+2], ..., A[j-1]` is the second part, and
+- `A[j], A[j+1], ..., A[A.length - 1]` is the third part.
+- All three parts have equal binary value.
+
+If it is not possible, return `[-1, -1]`.
+
+Note that the entire part is used when considering what binary value it represents. For example, `[1,1,0]` represents `6` in decimal, not `3`. Also, leading zeros are allowed, so `[0,1,1]` and `[1,1]` represent the same value.
+
+**Example 1:**
+
+    Input: [1,0,1,0,1]
+    Output: [0,3]
+
+**Example 2:**
+
+    Input: [1,1,0,1,1]
+    Output: [-1,-1]
+
+**Note:**
+
+1. `3 <= A.length <= 30000`
+2. `A[i] == 0` or `A[i] == 1`
+
+
+## 题目大意
+
+给定一个由 0 和 1 组成的数组 A，将数组分成 3 个非空的部分，使得所有这些部分表示相同的二进制值。如果可以做到，请返回任何 [i, j]，其中 i+1 < j，这样一来：  
+
+- A[0], A[1], ..., A[i] 组成第一部分；
+- A[i+1], A[i+2], ..., A[j-1] 作为第二部分；
+- A[j], A[j+1], ..., A[A.length - 1] 是第三部分。
+- 这三个部分所表示的二进制值相等。  
+
+如果无法做到，就返回 [-1, -1]。
+
+注意，在考虑每个部分所表示的二进制时，应当将其看作一个整体。例如，[1,1,0] 表示十进制中的 6，而不会是 3。此外，前导零也是被允许的，所以 [0,1,1] 和 [1,1] 表示相同的值。  
+
+提示：
+
+1. 3 <= A.length <= 30000
+2. A[i] == 0 或 A[i] == 1
+
+
+## 解题思路
+
+- 给出一个数组，数组里面只包含 0 和 1，要求找到 2 个分割点，使得分成的 3 个子数组的二进制是完全一样的。
+- 这一题的解题思路不难，按照题意模拟即可。先统计 1 的个数 total，然后除以 3 就是每段 1 出现的个数。有一些特殊情况需要额外判断一下，例如没有 1 的情况，那么只能首尾分割。1 个个数不是 3 的倍数，也无法分割成满足题意。然后找到第一个 1 的下标，然后根据 total/3 找到 mid，第一个分割点。再往后移动，找到第二个分割点。找到这 3 个点以后，同步的移动这 3 个点，移动中判断这 3 个下标对应的数值是否相等，如果都相等，并且最后一个点能移动到末尾，就算找到了满足题意的解了。
diff --git a/Algorithms/0928. Minimize Malware Spread II/928. Minimize Malware Spread II.go b/leetcode/0928.Minimize-Malware-Spread-II/928. Minimize Malware Spread II.go
similarity index 95%
rename from Algorithms/0928. Minimize Malware Spread II/928. Minimize Malware Spread II.go
rename to leetcode/0928.Minimize-Malware-Spread-II/928. Minimize Malware Spread II.go
index 70a4d453f..8f5a8aa12 100644
--- a/Algorithms/0928. Minimize Malware Spread II/928. Minimize Malware Spread II.go	
+++ b/leetcode/0928.Minimize-Malware-Spread-II/928. Minimize Malware Spread II.go	
@@ -66,3 +66,10 @@ func minMalwareSpread2(graph [][]int, initial []int) int {
 	}
 	return minIndex
 }
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/0928.Minimize-Malware-Spread-II/928. Minimize Malware Spread II_test.go b/leetcode/0928.Minimize-Malware-Spread-II/928. Minimize Malware Spread II_test.go
new file mode 100644
index 000000000..b05bb1626
--- /dev/null
+++ b/leetcode/0928.Minimize-Malware-Spread-II/928. Minimize Malware Spread II_test.go	
@@ -0,0 +1,72 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question928 struct {
+	para928
+	ans928
+}
+
+// para 是参数
+// one 代表第一个参数
+type para928 struct {
+	graph   [][]int
+	initial []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans928 struct {
+	one int
+}
+
+func Test_Problem928(t *testing.T) {
+
+	qs := []question928{
+
+		{
+			para928{[][]int{{1, 0, 0, 0}, {0, 1, 1, 1}, {0, 1, 1, 0}, {0, 1, 0, 1}}, []int{1, 3}},
+			ans928{1},
+		},
+		{
+			para928{[][]int{{1, 1, 1, 0}, {1, 1, 0, 0}, {1, 0, 1, 0}, {0, 0, 0, 1}}, []int{3, 2}},
+			ans928{2},
+		},
+
+		{
+			para928{[][]int{{1, 1, 0}, {1, 1, 1}, {0, 1, 1}}, []int{0, 1}},
+			ans928{1},
+		},
+
+		{
+			para928{[][]int{{1, 1, 0}, {1, 1, 0}, {0, 0, 1}}, []int{0, 1}},
+			ans928{0},
+		},
+
+		{
+			para928{[][]int{{1, 1, 0, 0}, {1, 1, 1, 0}, {0, 1, 1, 1}, {0, 0, 1, 1}}, []int{0, 1}},
+			ans928{1},
+		},
+
+		{
+			para928{[][]int{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 1, 1}, {0, 0, 1, 1}}, []int{3, 1}},
+			ans928{3},
+		},
+
+		{
+			para928{[][]int{{1, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0}, {0, 0, 0, 1, 1, 0}, {0, 0, 0, 1, 1, 0}, {0, 0, 0, 0, 0, 1}}, []int{5, 0}},
+			ans928{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 928------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans928, q.para928
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minMalwareSpread2(p.graph, p.initial))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0928.Minimize-Malware-Spread-II/README.md b/leetcode/0928.Minimize-Malware-Spread-II/README.md
new file mode 100755
index 000000000..f245cebe2
--- /dev/null
+++ b/leetcode/0928.Minimize-Malware-Spread-II/README.md
@@ -0,0 +1,61 @@
+# [928. Minimize Malware Spread II](https://leetcode.com/problems/minimize-malware-spread-ii/)
+
+
+## 题目
+
+(This problem is the same as *Minimize Malware Spread*, with the differences bolded.)
+
+In a network of nodes, each node `i` is directly connected to another node `j` if and only if `graph[i][j] = 1`.
+
+Some nodes `initial` are initially infected by malware. Whenever two nodes are directly connected and at least one of those two nodes is infected by malware, both nodes will be infected by malware. This spread of malware will continue until no more nodes can be infected in this manner.
+
+Suppose `M(initial)` is the final number of nodes infected with malware in the entire network, after the spread of malware stops.
+
+We will remove one node from the initial list, **completely removing it and any connections from this node to any other node**. Return the node that if removed, would minimize `M(initial)`. If multiple nodes could be removed to minimize `M(initial)`, return such a node with the smallest index.
+
+**Example 1:**
+
+    Input: graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]
+    Output: 0
+
+**Example 2:**
+
+    Input: graph = [[1,1,0],[1,1,1],[0,1,1]], initial = [0,1]
+    Output: 1
+
+**Example 3:**
+
+    Input: graph = [[1,1,0,0],[1,1,1,0],[0,1,1,1],[0,0,1,1]], initial = [0,1]
+    Output: 1
+
+**Note:**
+
+1. `1 < graph.length = graph[0].length <= 300`
+2. `0 <= graph[i][j] == graph[j][i] <= 1`
+3. `graph[i][i] = 1`
+4. `1 <= initial.length < graph.length`
+5. `0 <= initial[i] < graph.length`
+
+
+## 题目大意
+
+(这个问题与 尽量减少恶意软件的传播 是一样的，不同之处用粗体表示。)在节点网络中，只有当 graph[i][j] = 1 时，每个节点 i 能够直接连接到另一个节点 j。一些节点 initial 最初被恶意软件感染。只要两个节点直接连接，且其中至少一个节点受到恶意软件的感染，那么两个节点都将被恶意软件感染。这种恶意软件的传播将继续，直到没有更多的节点可以被这种方式感染。假设 M(initial) 是在恶意软件停止传播之后，整个网络中感染恶意软件的最终节点数。我们可以从初始列表中删除一个节点，并完全移除该节点以及从该节点到任何其他节点的任何连接。如果移除这一节点将最小化 M(initial)， 则返回该节点。如果有多个节点满足条件，就返回索引最小的节点。
+
+提示：
+
+- 1 < graph.length = graph[0].length <= 300
+- 0 <= graph[i][j] == graph[j][i] <= 1
+- graph[i][i] = 1
+- 1 <= initial.length < graph.length
+- 0 <= initial[i] < graph.length
+
+
+## 解题思路
+
+
+- 这一题是第 924 题的加强版。给出一个节点之间的关系图，如果两个节点是连通的，那么病毒软件就会感染到连通的所有节点。现在如果想**完全彻底**移除一个病毒节点，能最大减少感染，请问移除哪个节点？如果多个节点都能减少感染量，优先移除序号偏小的那个节点。这一题的输入输出要求和第 924 题是完全一样的，区别在于第 924 题实际上是要求把一个病毒节点变成非病毒节点，而这道题是完全删除一个病毒节点以及它连接的所有边。
+- 这一题考察的是并查集。当然用 DFS 也可以解答这一题。并查集的做法如下，首先先将所有的病毒节点去掉，然后将所有连通块合并成一个节点。因为一个连通集合中的节点，要么全部被感染，要么全部不被感染，所以可以把每个集合整体考虑。然后统计所有集合直接相邻的病毒节点的个数。对于一个集合来说：
+    1. 如果直接相邻的病毒节点的个数为 0，则一定不会被感染，忽略这种情况；
+    2. 如果直接相邻的病毒节点的个数为 1，则将该病毒节点删除后，整个连通块就可以避免被感染，这种情况是我们寻找的答案；
+    3. 如果直接相邻的病毒节点的个数大于等于2，则不管删除哪个病毒节点，该连通块都仍会被感染，忽略这种情况；
+- 所以只需在所有第二种连通块（直接相邻的病毒节点的个数为 1 的连通块）中，找出节点个数最多的连通块，与它相邻的病毒节点就是我们要删除的节点；如果有多个连通块节点个数相同，再找出与之对应的编号最小的病毒节点即可。
diff --git a/Algorithms/0930. Binary Subarrays With Sum/930. Binary Subarrays With Sum.go b/leetcode/0930.Binary-Subarrays-With-Sum/930. Binary Subarrays With Sum.go
similarity index 100%
rename from Algorithms/0930. Binary Subarrays With Sum/930. Binary Subarrays With Sum.go
rename to leetcode/0930.Binary-Subarrays-With-Sum/930. Binary Subarrays With Sum.go
diff --git a/Algorithms/0930. Binary Subarrays With Sum/930. Binary Subarrays With Sum_test.go b/leetcode/0930.Binary-Subarrays-With-Sum/930. Binary Subarrays With Sum_test.go
similarity index 94%
rename from Algorithms/0930. Binary Subarrays With Sum/930. Binary Subarrays With Sum_test.go
rename to leetcode/0930.Binary-Subarrays-With-Sum/930. Binary Subarrays With Sum_test.go
index 788653ada..e51d634a9 100644
--- a/Algorithms/0930. Binary Subarrays With Sum/930. Binary Subarrays With Sum_test.go	
+++ b/leetcode/0930.Binary-Subarrays-With-Sum/930. Binary Subarrays With Sum_test.go	
@@ -27,17 +27,17 @@ func Test_Problem930(t *testing.T) {
 
 	qs := []question930{
 
-		question930{
+		{
 			para930{[]int{1, 0, 1, 0, 1}, 2},
 			ans930{4},
 		},
 
-		question930{
+		{
 			para930{[]int{0, 0, 0, 0, 0}, 0},
 			ans930{15},
 		},
 
-		question930{
+		{
 			para930{[]int{1, 0, 1, 1, 1, 1, 0, 1, 0, 1}, 2},
 			ans930{4},
 		},
diff --git a/Algorithms/0930. Binary Subarrays With Sum/README.md b/leetcode/0930.Binary-Subarrays-With-Sum/README.md
similarity index 100%
rename from Algorithms/0930. Binary Subarrays With Sum/README.md
rename to leetcode/0930.Binary-Subarrays-With-Sum/README.md
diff --git a/leetcode/0933.Number-of-Recent-Calls/933. Number of Recent Calls.go b/leetcode/0933.Number-of-Recent-Calls/933. Number of Recent Calls.go
new file mode 100644
index 000000000..2ceeffd63
--- /dev/null
+++ b/leetcode/0933.Number-of-Recent-Calls/933. Number of Recent Calls.go	
@@ -0,0 +1,28 @@
+package leetcode
+
+import "sort"
+
+type RecentCounter struct {
+	list []int
+}
+
+func Constructor933() RecentCounter {
+	return RecentCounter{
+		list: []int{},
+	}
+}
+
+func (this *RecentCounter) Ping(t int) int {
+	this.list = append(this.list, t)
+	index := sort.Search(len(this.list), func(i int) bool { return this.list[i] >= t-3000 })
+	if index < 0 {
+		index = -index - 1
+	}
+	return len(this.list) - index
+}
+
+/**
+ * Your RecentCounter object will be instantiated and called as such:
+ * obj := Constructor();
+ * param_1 := obj.Ping(t);
+ */
diff --git a/leetcode/0933.Number-of-Recent-Calls/933. Number of Recent Calls_test.go b/leetcode/0933.Number-of-Recent-Calls/933. Number of Recent Calls_test.go
new file mode 100644
index 000000000..d2782c4b8
--- /dev/null
+++ b/leetcode/0933.Number-of-Recent-Calls/933. Number of Recent Calls_test.go	
@@ -0,0 +1,19 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem933(t *testing.T) {
+	obj := Constructor933()
+	fmt.Printf("obj = %v\n", obj)
+	param1 := obj.Ping(1)
+	fmt.Printf("param = %v\n", param1)
+	param1 = obj.Ping(100)
+	fmt.Printf("param = %v\n", param1)
+	param1 = obj.Ping(3001)
+	fmt.Printf("param = %v\n", param1)
+	param1 = obj.Ping(3002)
+	fmt.Printf("param = %v\n", param1)
+}
diff --git a/leetcode/0933.Number-of-Recent-Calls/README.md b/leetcode/0933.Number-of-Recent-Calls/README.md
new file mode 100644
index 000000000..02b2340ce
--- /dev/null
+++ b/leetcode/0933.Number-of-Recent-Calls/README.md
@@ -0,0 +1,74 @@
+# [933. Number of Recent Calls](https://leetcode.com/problems/number-of-recent-calls/)
+
+
+
+## 题目
+
+Write a class `RecentCounter` to count recent requests.
+
+It has only one method: `ping(int t)`, where t represents some time in milliseconds.
+
+Return the number of `ping`s that have been made from 3000 milliseconds ago until now.
+
+Any ping with time in `[t - 3000, t]` will count, including the current ping.
+
+It is guaranteed that every call to `ping` uses a strictly larger value of `t` than before.
+
+**Example 1**:
+
+```
+Input: inputs = ["RecentCounter","ping","ping","ping","ping"], inputs = [[],[1],[100],[3001],[3002]]
+Output: [null,1,2,3,3]
+```
+
+**Note**:
+
+1. Each test case will have at most `10000` calls to `ping`.
+2. Each test case will call `ping` with strictly increasing values of `t`.
+3. Each call to ping will have `1 <= t <= 10^9`.
+
+
+## 题目大意
+
+写一个 RecentCounter 类来计算最近的请求。它只有一个方法：ping(int t)，其中 t 代表以毫秒为单位的某个时间。返回从 3000 毫秒前到现在的 ping 数。任何处于 [t - 3000, t] 时间范围之内的 ping 都将会被计算在内，包括当前（指 t 时刻）的 ping。保证每次对 ping 的调用都使用比之前更大的 t 值。
+ 
+提示：
+
+- 每个测试用例最多调用 10000 次 ping。
+- 每个测试用例会使用严格递增的 t 值来调用 ping。
+- 每次调用 ping 都有 1 <= t <= 10^9。
+
+
+## 解题思路
+
+- 要求设计一个类，可以用 `ping(t)` 的方法，计算 [t-3000, t] 区间内的 ping 数。t 是毫秒。
+- 这一题比较简单，`ping()` 方法用二分搜索即可。
+
+## 代码
+
+```go
+type RecentCounter struct {
+	list []int
+}
+
+func Constructor933() RecentCounter {
+	return RecentCounter{
+		list: []int{},
+	}
+}
+
+func (this *RecentCounter) Ping(t int) int {
+	this.list = append(this.list, t)
+	index := sort.Search(len(this.list), func(i int) bool { return this.list[i] >= t-3000 })
+	if index < 0 {
+		index = -index - 1
+	}
+	return len(this.list) - index
+}
+
+/**
+ * Your RecentCounter object will be instantiated and called as such:
+ * obj := Constructor();
+ * param_1 := obj.Ping(t);
+ */
+```
\ No newline at end of file
diff --git a/leetcode/0938.Range-Sum-of-BST/938. Range Sum of BST.go b/leetcode/0938.Range-Sum-of-BST/938. Range Sum of BST.go
new file mode 100644
index 000000000..3225367b9
--- /dev/null
+++ b/leetcode/0938.Range-Sum-of-BST/938. Range Sum of BST.go	
@@ -0,0 +1,34 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func rangeSumBST(root *TreeNode, low int, high int) int {
+	res := 0
+	preOrder(root, low, high, &res)
+	return res
+}
+
+func preOrder(root *TreeNode, low, high int, res *int) {
+	if root == nil {
+		return
+	}
+	if low <= root.Val && root.Val <= high {
+		*res += root.Val
+	}
+	preOrder(root.Left, low, high, res)
+	preOrder(root.Right, low, high, res)
+}
diff --git a/leetcode/0938.Range-Sum-of-BST/938. Range Sum of BST_test.go b/leetcode/0938.Range-Sum-of-BST/938. Range Sum of BST_test.go
new file mode 100644
index 000000000..0f06ee18a
--- /dev/null
+++ b/leetcode/0938.Range-Sum-of-BST/938. Range Sum of BST_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question938 struct {
+	para938
+	ans938
+}
+
+// para 是参数
+// one 代表第一个参数
+type para938 struct {
+	one  []int
+	low  int
+	high int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans938 struct {
+	one int
+}
+
+func Test_Problem938(t *testing.T) {
+
+	qs := []question938{
+
+		{
+			para938{[]int{10, 5, 15, 3, 7, structures.NULL, 18}, 7, 15},
+			ans938{32},
+		},
+
+		{
+			para938{[]int{10, 5, 15, 3, 7, 13, 18, 1, structures.NULL, 6}, 6, 10},
+			ans938{23},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 938------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans938, q.para938
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", rangeSumBST(rootOne, p.low, p.high))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0938.Range-Sum-of-BST/README.md b/leetcode/0938.Range-Sum-of-BST/README.md
new file mode 100644
index 000000000..c6f04ac65
--- /dev/null
+++ b/leetcode/0938.Range-Sum-of-BST/README.md
@@ -0,0 +1,80 @@
+# [938. Range Sum of BST](https://leetcode.com/problems/range-sum-of-bst/)
+
+
+## 题目
+
+Given the `root` node of a binary search tree, return *the sum of values of all nodes with a value in the range `[low, high]`*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/11/05/bst1.jpg](https://assets.leetcode.com/uploads/2020/11/05/bst1.jpg)
+
+```
+Input: root = [10,5,15,3,7,null,18], low = 7, high = 15
+Output: 32
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/11/05/bst2.jpg](https://assets.leetcode.com/uploads/2020/11/05/bst2.jpg)
+
+```
+Input: root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10
+Output: 23
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 2 * 10^4]`.
+- `1 <= Node.val <= 10^5`
+- `1 <= low <= high <= 10^5`
+- All `Node.val` are **unique**.
+
+## 题目大意
+
+给定二叉搜索树的根结点 root，返回值位于范围 [low, high] 之间的所有结点的值的和。
+
+## 解题思路
+
+- 简单题。因为二叉搜索树的有序性，先序遍历即为有序。遍历过程中判断节点值是否位于区间范围内，在区间内就累加，不在区间内节点就不管。最终输出累加和。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func rangeSumBST(root *TreeNode, low int, high int) int {
+	res := 0
+	preOrder(root, low, high, &res)
+	return res
+}
+
+func preOrder(root *TreeNode, low, high int, res *int) {
+	if root == nil {
+		return
+	}
+	if low <= root.Val && root.Val <= high {
+		*res += root.Val
+	}
+	preOrder(root.Left, low, high, res)
+	preOrder(root.Right, low, high, res)
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0942. DI String Match/942. DI String Match.go b/leetcode/0942.DI-String-Match/942. DI String Match.go
similarity index 100%
rename from Algorithms/0942. DI String Match/942. DI String Match.go
rename to leetcode/0942.DI-String-Match/942. DI String Match.go
diff --git a/Algorithms/0942. DI String Match/942. DI String Match_test.go b/leetcode/0942.DI-String-Match/942. DI String Match_test.go
similarity index 94%
rename from Algorithms/0942. DI String Match/942. DI String Match_test.go
rename to leetcode/0942.DI-String-Match/942. DI String Match_test.go
index c0767e329..4b34dd95e 100644
--- a/Algorithms/0942. DI String Match/942. DI String Match_test.go	
+++ b/leetcode/0942.DI-String-Match/942. DI String Match_test.go	
@@ -26,17 +26,17 @@ func Test_Problem942(t *testing.T) {
 
 	qs := []question942{
 
-		question942{
+		{
 			para942{"IDID"},
 			ans942{[]int{0, 4, 1, 3, 2}},
 		},
 
-		question942{
+		{
 			para942{"III"},
 			ans942{[]int{0, 1, 2, 3}},
 		},
 
-		question942{
+		{
 			para942{"DDI"},
 			ans942{[]int{3, 2, 0, 1}},
 		},
diff --git a/Algorithms/0942. DI String Match/README.md b/leetcode/0942.DI-String-Match/README.md
similarity index 100%
rename from Algorithms/0942. DI String Match/README.md
rename to leetcode/0942.DI-String-Match/README.md
diff --git a/leetcode/0946.Validate-Stack-Sequences/946. Validate Stack Sequences.go b/leetcode/0946.Validate-Stack-Sequences/946. Validate Stack Sequences.go
new file mode 100644
index 000000000..13d22de87
--- /dev/null
+++ b/leetcode/0946.Validate-Stack-Sequences/946. Validate Stack Sequences.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+func validateStackSequences(pushed []int, popped []int) bool {
+	stack, j, N := []int{}, 0, len(pushed)
+	for _, x := range pushed {
+		stack = append(stack, x)
+		for len(stack) != 0 && j < N && stack[len(stack)-1] == popped[j] {
+			stack = stack[0 : len(stack)-1]
+			j++
+		}
+	}
+	return j == N
+}
diff --git a/Algorithms/0946. Validate Stack Sequences/946. Validate Stack Sequences_test.go b/leetcode/0946.Validate-Stack-Sequences/946. Validate Stack Sequences_test.go
similarity index 94%
rename from Algorithms/0946. Validate Stack Sequences/946. Validate Stack Sequences_test.go
rename to leetcode/0946.Validate-Stack-Sequences/946. Validate Stack Sequences_test.go
index 3605c43c4..51c8842fa 100644
--- a/Algorithms/0946. Validate Stack Sequences/946. Validate Stack Sequences_test.go	
+++ b/leetcode/0946.Validate-Stack-Sequences/946. Validate Stack Sequences_test.go	
@@ -26,17 +26,17 @@ type ans946 struct {
 func Test_Problem946(t *testing.T) {
 
 	qs := []question946{
-		question946{
+		{
 			para946{[]int{1, 2, 3, 4, 5}, []int{5, 4, 3, 2, 1}},
 			ans946{true},
 		},
 
-		question946{
+		{
 			para946{[]int{1, 2, 3, 4, 5}, []int{4, 3, 5, 1, 2}},
 			ans946{false},
 		},
 
-		question946{
+		{
 			para946{[]int{1, 0}, []int{1, 0}},
 			ans946{true},
 		},
diff --git a/Algorithms/0946. Validate Stack Sequences/README.md b/leetcode/0946.Validate-Stack-Sequences/README.md
similarity index 100%
rename from Algorithms/0946. Validate Stack Sequences/README.md
rename to leetcode/0946.Validate-Stack-Sequences/README.md
diff --git a/Algorithms/0947. Most Stones Removed with Same Row or Column/947. Most Stones Removed with Same Row or Column.go b/leetcode/0947.Most-Stones-Removed-with-Same-Row-or-Column/947. Most Stones Removed with Same Row or Column.go
similarity index 100%
rename from Algorithms/0947. Most Stones Removed with Same Row or Column/947. Most Stones Removed with Same Row or Column.go
rename to leetcode/0947.Most-Stones-Removed-with-Same-Row-or-Column/947. Most Stones Removed with Same Row or Column.go
diff --git a/leetcode/0947.Most-Stones-Removed-with-Same-Row-or-Column/947. Most Stones Removed with Same Row or Column_test.go b/leetcode/0947.Most-Stones-Removed-with-Same-Row-or-Column/947. Most Stones Removed with Same Row or Column_test.go
new file mode 100644
index 000000000..c77f1abe1
--- /dev/null
+++ b/leetcode/0947.Most-Stones-Removed-with-Same-Row-or-Column/947. Most Stones Removed with Same Row or Column_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question947 struct {
+	para947
+	ans947
+}
+
+// para 是参数
+// one 代表第一个参数
+type para947 struct {
+	stones [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans947 struct {
+	one int
+}
+
+func Test_Problem947(t *testing.T) {
+
+	qs := []question947{
+		{
+			para947{[][]int{{0, 0}, {0, 1}, {1, 0}, {1, 2}, {2, 1}, {2, 2}}},
+			ans947{5},
+		},
+
+		{
+			para947{[][]int{{0, 0}, {0, 2}, {1, 1}, {2, 0}, {2, 2}}},
+			ans947{3},
+		},
+
+		{
+			para947{[][]int{{0, 0}}},
+			ans947{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 947------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans947, q.para947
+		fmt.Printf("【input】:%v       【output】:%v\n", p, removeStones(p.stones))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0947.Most-Stones-Removed-with-Same-Row-or-Column/README.md b/leetcode/0947.Most-Stones-Removed-with-Same-Row-or-Column/README.md
new file mode 100755
index 000000000..2343226d5
--- /dev/null
+++ b/leetcode/0947.Most-Stones-Removed-with-Same-Row-or-Column/README.md
@@ -0,0 +1,48 @@
+# [947. Most Stones Removed with Same Row or Column](https://leetcode.com/problems/most-stones-removed-with-same-row-or-column/)
+
+
+## 题目
+
+On a 2D plane, we place stones at some integer coordinate points. Each coordinate point may have at most one stone.
+
+Now, a *move* consists of removing a stone that shares a column or row with another stone on the grid.
+
+What is the largest possible number of moves we can make?
+
+**Example 1:**
+
+    Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
+    Output: 5
+
+**Example 2:**
+
+    Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
+    Output: 3
+
+**Example 3:**
+
+    Input: stones = [[0,0]]
+    Output: 0
+
+**Note:**
+
+1. `1 <= stones.length <= 1000`
+2. `0 <= stones[i][j] < 10000`
+
+
+## 题目大意
+
+在二维平面上，我们将石头放置在一些整数坐标点上。每个坐标点上最多只能有一块石头。现在，move 操作将会移除与网格上的某一块石头共享一列或一行的一块石头。我们最多能执行多少次 move 操作？
+
+提示：
+
+- 1 <= stones.length <= 1000
+- 0 <= stones[i][j] < 10000
+
+
+## 解题思路
+
+
+- 给出一个数组，数组中的元素是一系列的坐标点。现在可以移除一些坐标点，移除必须满足：移除的这个点，在相同的行或者列上有一个点。问最终可以移除多少个点。移除到最后必然有些点独占一行，那么这些点都不能被移除。
+- 这一题的解题思路是并查集。把所有共行或者共列的点都 `union()` 起来。不同集合之间是不能相互移除的。反证法：如果能移除，代表存在共行或者共列的情况，那么肯定是同一个集合了，这样就不满足不同集合了。最终剩下的点就是集合的个数，每个集合只会留下一个点。所以移除的点就是点的总数减去集合的个数 `len(stones) - uf.totalCount()`。
+- 如果暴力合并集合，时间复杂度也非常差，可以由优化的地方。再遍历所有点的过程中，可以把遍历过的行和列存起来。这里可以用 map 来记录，key 为行号，value 为上一次遍历过的点的序号。同样，列也可以用 map 存起来，key 为列号，value 为上一次遍历过的点的序号。经过这样的优化以后，时间复杂度会提高不少。
diff --git a/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits.go b/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits.go
new file mode 100644
index 000000000..12919863d
--- /dev/null
+++ b/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits.go	
@@ -0,0 +1,33 @@
+package leetcode
+
+import "fmt"
+
+func largestTimeFromDigits(A []int) string {
+	flag, res := false, 0
+	for i := 0; i < 4; i++ {
+		for j := 0; j < 4; j++ {
+			if i == j {
+				continue
+			}
+			for k := 0; k < 4; k++ {
+				if i == k || j == k {
+					continue
+				}
+				l := 6 - i - j - k
+				hour := A[i]*10 + A[j]
+				min := A[k]*10 + A[l]
+				if hour < 24 && min < 60 {
+					if hour*60+min >= res {
+						res = hour*60 + min
+						flag = true
+					}
+				}
+			}
+		}
+	}
+	if flag {
+		return fmt.Sprintf("%02d:%02d", res/60, res%60)
+	} else {
+		return ""
+	}
+}
diff --git a/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits_test.go b/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits_test.go
new file mode 100644
index 000000000..29bbfa048
--- /dev/null
+++ b/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question949 struct {
+	para949
+	ans949
+}
+
+// para 是参数
+// one 代表第一个参数
+type para949 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans949 struct {
+	one string
+}
+
+func Test_Problem949(t *testing.T) {
+
+	qs := []question949{
+		{
+			para949{[]int{1, 2, 3, 4}},
+			ans949{"23:41"},
+		},
+
+		{
+			para949{[]int{5, 5, 5, 5}},
+			ans949{""},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 949------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans949, q.para949
+		fmt.Printf("【input】:%v       【output】:%v\n", p, largestTimeFromDigits(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0949.Largest-Time-for-Given-Digits/README.md b/leetcode/0949.Largest-Time-for-Given-Digits/README.md
new file mode 100644
index 000000000..02f0bf435
--- /dev/null
+++ b/leetcode/0949.Largest-Time-for-Given-Digits/README.md
@@ -0,0 +1,78 @@
+# [949. Largest Time for Given Digits](https://leetcode.com/problems/largest-time-for-given-digits/)
+
+
+## 题目
+
+Given an array of 4 digits, return the largest 24 hour time that can be made.
+
+The smallest 24 hour time is 00:00, and the largest is 23:59. Starting from 00:00, a time is larger if more time has elapsed since midnight.
+
+Return the answer as a string of length 5. If no valid time can be made, return an empty string.
+
+**Example 1**:
+
+```
+Input: [1,2,3,4]
+Output: "23:41"
+```
+
+**Example 2**:
+
+```
+Input: [5,5,5,5]
+Output: ""
+```
+
+**Note**:
+
+1. `A.length == 4`
+2. `0 <= A[i] <= 9`
+
+## 题目大意
+
+给定一个由 4 位数字组成的数组，返回可以设置的符合 24 小时制的最大时间。最小的 24 小时制时间是 00:00，而最大的是 23:59。从 00:00 （午夜）开始算起，过得越久，时间越大。以长度为 5 的字符串返回答案。如果不能确定有效时间，则返回空字符串。
+
+## 解题思路
+
+- 给出 4 个数字，要求返回一个字符串，代表由这 4 个数字能组成的最大 24 小时制的时间。
+- 简单题，这一题直接暴力枚举就可以了。依次检查给出的 4 个数字每个排列组合是否是时间合法的。例如检查 10 * A[i] + A[j] 是不是小于 24， 10 * A[k] + A[l] 是不是小于 60。如果合法且比目前存在的最大时间更大，就更新这个最大时间。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "fmt"
+
+func largestTimeFromDigits(A []int) string {
+	flag, res := false, 0
+	for i := 0; i < 4; i++ {
+		for j := 0; j < 4; j++ {
+			if i == j {
+				continue
+			}
+			for k := 0; k < 4; k++ {
+				if i == k || j == k {
+					continue
+				}
+				l := 6 - i - j - k
+				hour := A[i]*10 + A[j]
+				min := A[k]*10 + A[l]
+				if hour < 24 && min < 60 {
+					if hour*60+min >= res {
+						res = hour*60 + min
+						flag = true
+					}
+				}
+			}
+		}
+	}
+	if flag {
+		return fmt.Sprintf("%02d:%02d", res/60, res%60)
+	} else {
+		return ""
+	}
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/0952. Largest Component Size by Common Factor/952. Largest Component Size by Common Factor.go b/leetcode/0952.Largest-Component-Size-by-Common-Factor/952. Largest Component Size by Common Factor.go
similarity index 94%
rename from Algorithms/0952. Largest Component Size by Common Factor/952. Largest Component Size by Common Factor.go
rename to leetcode/0952.Largest-Component-Size-by-Common-Factor/952. Largest Component Size by Common Factor.go
index 57330f4d8..48670513b 100644
--- a/Algorithms/0952. Largest Component Size by Common Factor/952. Largest Component Size by Common Factor.go	
+++ b/leetcode/0952.Largest-Component-Size-by-Common-Factor/952. Largest Component Size by Common Factor.go	
@@ -53,3 +53,10 @@ func largestComponentSize1(A []int) int {
 	}
 	return uf.MaxUnionCount()
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/0952. Largest Component Size by Common Factor/952. Largest Component Size by Common Factor_test.go b/leetcode/0952.Largest-Component-Size-by-Common-Factor/952. Largest Component Size by Common Factor_test.go
similarity index 93%
rename from Algorithms/0952. Largest Component Size by Common Factor/952. Largest Component Size by Common Factor_test.go
rename to leetcode/0952.Largest-Component-Size-by-Common-Factor/952. Largest Component Size by Common Factor_test.go
index 7c05beb73..b23942869 100644
--- a/Algorithms/0952. Largest Component Size by Common Factor/952. Largest Component Size by Common Factor_test.go	
+++ b/leetcode/0952.Largest-Component-Size-by-Common-Factor/952. Largest Component Size by Common Factor_test.go	
@@ -25,22 +25,22 @@ type ans952 struct {
 func Test_Problem952(t *testing.T) {
 
 	qs := []question952{
-		question952{
+		{
 			para952{[]int{1, 2, 3, 4, 5, 6, 7, 8, 9}},
 			ans952{6},
 		},
 
-		question952{
+		{
 			para952{[]int{4, 6, 15, 35}},
 			ans952{4},
 		},
 
-		question952{
+		{
 			para952{[]int{20, 50, 9, 63}},
 			ans952{2},
 		},
 
-		question952{
+		{
 			para952{[]int{2, 3, 6, 7, 4, 12, 21, 39}},
 			ans952{8},
 		},
diff --git a/leetcode/0952.Largest-Component-Size-by-Common-Factor/README.md b/leetcode/0952.Largest-Component-Size-by-Common-Factor/README.md
new file mode 100755
index 000000000..4a2c92260
--- /dev/null
+++ b/leetcode/0952.Largest-Component-Size-by-Common-Factor/README.md
@@ -0,0 +1,56 @@
+# [952. Largest Component Size by Common Factor](https://leetcode.com/problems/largest-component-size-by-common-factor/)
+
+
+## 题目
+
+Given a non-empty array of unique positive integers `A`, consider the following graph:
+
+- There are `A.length` nodes, labelled `A[0]` to `A[A.length - 1];`
+- There is an edge between `A[i]` and `A[j]` if and only if `A[i]`and `A[j]` share a common factor greater than 1.
+
+Return the size of the largest connected component in the graph.
+
+**Example 1:**
+
+    Input: [4,6,15,35]
+    Output: 4
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/01/ex1.png)
+
+**Example 2:**
+
+    Input: [20,50,9,63]
+    Output: 2
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/01/ex2.png)
+
+**Example 3:**
+
+    Input: [2,3,6,7,4,12,21,39]
+    Output: 8
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/01/ex3.png)
+
+**Note:**
+
+1. `1 <= A.length <= 20000`
+2. `1 <= A[i] <= 100000`
+
+
+## 题目大意
+
+给定一个由不同正整数的组成的非空数组 A，考虑下面的图：
+
+有 A.length 个节点，按从 A[0] 到 A[A.length - 1] 标记；
+只有当 A[i] 和 A[j] 共用一个大于 1 的公因数时，A[i] 和 A[j] 之间才有一条边。
+返回图中最大连通组件的大小。
+
+提示：
+
+1. 1 <= A.length <= 20000
+2. 1 <= A[i] <= 100000
+
+## 解题思路
+
+- 给出一个数组，数组中的元素如果每两个元素有公约数，那么这两个元素可以算有关系。所有有关系的数可以放在一个集合里，问这个数组里面有关系的元素组成的集合里面最多有多少个元素。
+- 这一题读完题直觉就是用并查集来解题。首先可以用暴力的解法尝试。用 2 层循环，两两比较有没有公约数，如果有公约数就 `union()` 到一起。提交以后出现 TLE，其实看一下数据规模就知道会超时，`1 <= A.length <= 20000`。注意到 `1 <= A[i] <= 100000`，开根号以后最后才 316.66666，这个规模的数不大。所以把每个数小于根号自己的因子都找出来，例如 `6 = 2 * 3`，`15 = 3 * 5`，那么把 6 和 2，6 和 3 都 `union()`，15 和 3，15 和 5 都 `union()`，最终遍历所有的集合，找到最多元素的集合，输出它包含的元素值。
diff --git a/Algorithms/0953. Verifying an Alien Dictionary/953. Verifying an Alien Dictionary.go b/leetcode/0953.Verifying-an-Alien-Dictionary/953. Verifying an Alien Dictionary.go
similarity index 100%
rename from Algorithms/0953. Verifying an Alien Dictionary/953. Verifying an Alien Dictionary.go
rename to leetcode/0953.Verifying-an-Alien-Dictionary/953. Verifying an Alien Dictionary.go
diff --git a/Algorithms/0953. Verifying an Alien Dictionary/953. Verifying an Alien Dictionary_test.go b/leetcode/0953.Verifying-an-Alien-Dictionary/953. Verifying an Alien Dictionary_test.go
similarity index 95%
rename from Algorithms/0953. Verifying an Alien Dictionary/953. Verifying an Alien Dictionary_test.go
rename to leetcode/0953.Verifying-an-Alien-Dictionary/953. Verifying an Alien Dictionary_test.go
index a20b82b62..6bb2be8a2 100644
--- a/Algorithms/0953. Verifying an Alien Dictionary/953. Verifying an Alien Dictionary_test.go	
+++ b/leetcode/0953.Verifying-an-Alien-Dictionary/953. Verifying an Alien Dictionary_test.go	
@@ -26,17 +26,17 @@ type ans953 struct {
 func Test_Problem953(t *testing.T) {
 
 	qs := []question953{
-		question953{
+		{
 			para953{[]string{"hello", "leetcode"}, "hlabcdefgijkmnopqrstuvwxyz"},
 			ans953{true},
 		},
 
-		question953{
+		{
 			para953{[]string{"word", "world", "row"}, "worldabcefghijkmnpqstuvxyz"},
 			ans953{false},
 		},
 
-		question953{
+		{
 			para953{[]string{"apple", "app"}, "abcdefghijklmnopqrstuvwxyz"},
 			ans953{false},
 		},
diff --git a/leetcode/0953.Verifying-an-Alien-Dictionary/README.md b/leetcode/0953.Verifying-an-Alien-Dictionary/README.md
new file mode 100755
index 000000000..2f02a10a6
--- /dev/null
+++ b/leetcode/0953.Verifying-an-Alien-Dictionary/README.md
@@ -0,0 +1,45 @@
+# [953. Verifying an Alien Dictionary](https://leetcode.com/problems/verifying-an-alien-dictionary/)
+
+
+## 题目
+
+In an alien language, surprisingly they also use english lowercase letters, but possibly in a different `order`. The `order`of the alphabet is some permutation of lowercase letters.
+
+Given a sequence of `words` written in the alien language, and the `order` of the alphabet, return `true` if and only if the given `words` are sorted lexicographicaly in this alien language.
+
+**Example 1:**
+
+    Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
+    Output: true
+    Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.
+
+**Example 2:**
+
+    Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
+    Output: false
+    Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.
+
+**Example 3:**
+
+    Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
+    Output: false
+    Explanation: The first three characters "app" match, and the second string is shorter (in size.) According to lexicographical rules "apple" > "app", because 'l' > '∅', where '∅' is defined as the blank character which is less than any other character (More info).
+
+**Note:**
+
+1. `1 <= words.length <= 100`
+2. `1 <= words[i].length <= 20`
+3. `order.length == 26`
+4. All characters in `words[i]` and `order` are english lowercase letters.
+
+
+## 题目大意
+
+某种外星语也使用英文小写字母，但可能顺序 order 不同。字母表的顺序（order）是一些小写字母的排列。给定一组用外星语书写的单词 words，以及其字母表的顺序 order，只有当给定的单词在这种外星语中按字典序排列时，返回 true；否则，返回 false。
+
+
+
+## 解题思路
+
+
+- 这一题是简单题。给出一个字符串数组，判断把字符串数组里面字符串是否是按照 order 的排序排列的。order 是给出个一个字符串排序。这道题的解法是把 26 个字母的顺序先存在 map 中，然后依次遍历判断字符串数组里面字符串的大小。
diff --git a/leetcode/0958.Check-Completeness-of-a-Binary-Tree/0958.Check Completeness of a Binary Tree.go b/leetcode/0958.Check-Completeness-of-a-Binary-Tree/0958.Check Completeness of a Binary Tree.go
new file mode 100644
index 000000000..96fb1141e
--- /dev/null
+++ b/leetcode/0958.Check-Completeness-of-a-Binary-Tree/0958.Check Completeness of a Binary Tree.go	
@@ -0,0 +1,35 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func isCompleteTree(root *TreeNode) bool {
+	queue, found := []*TreeNode{root}, false
+	for len(queue) > 0 {
+		node := queue[0] //取出每一层的第一个节点
+		queue = queue[1:]
+		if node == nil {
+			found = true
+		} else {
+			if found {
+				return false // 层序遍历中，两个不为空的节点中出现一个 nil
+			}
+			//如果左孩子为nil，则append进去的node.Left为nil
+			queue = append(queue, node.Left, node.Right)
+		}
+	}
+	return true
+}
diff --git a/leetcode/0958.Check-Completeness-of-a-Binary-Tree/0958.Check Completeness of a Binary Tree_test.go b/leetcode/0958.Check-Completeness-of-a-Binary-Tree/0958.Check Completeness of a Binary Tree_test.go
new file mode 100644
index 000000000..6eb4df6d4
--- /dev/null
+++ b/leetcode/0958.Check-Completeness-of-a-Binary-Tree/0958.Check Completeness of a Binary Tree_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question958 struct {
+	para958
+	ans958
+}
+
+// para 是参数
+// one 代表第一个参数
+type para958 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans958 struct {
+	one bool
+}
+
+func Test_Problem958(t *testing.T) {
+
+	qs := []question958{
+
+		{
+			para958{[]int{1, 2, 3, 4, 5, 6}},
+			ans958{true},
+		},
+
+		{
+			para958{[]int{1, 2, 3, 4, 5, structures.NULL, 7}},
+			ans958{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 958------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans958, q.para958
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", isCompleteTree(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0958.Check-Completeness-of-a-Binary-Tree/README.md b/leetcode/0958.Check-Completeness-of-a-Binary-Tree/README.md
new file mode 100644
index 000000000..4325c36e8
--- /dev/null
+++ b/leetcode/0958.Check-Completeness-of-a-Binary-Tree/README.md
@@ -0,0 +1,89 @@
+# [958. Check Completeness of a Binary Tree](https://leetcode.com/problems/check-completeness-of-a-binary-tree/)
+
+
+## 题目
+
+Given the `root` of a binary tree, determine if it is a *complete binary tree*.
+
+In a **[complete binary tree](http://en.wikipedia.org/wiki/Binary_tree#Types_of_binary_trees)**, every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between `1` and `2h` nodes inclusive at the last level `h`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2018/12/15/complete-binary-tree-1.png](https://assets.leetcode.com/uploads/2018/12/15/complete-binary-tree-1.png)
+
+```
+Input: root = [1,2,3,4,5,6]
+Output: true
+Explanation: Every level before the last is full (ie. levels with node-values {1} and {2, 3}), and all nodes in the last level ({4, 5, 6}) are as far left as possible.
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2018/12/15/complete-binary-tree-2.png](https://assets.leetcode.com/uploads/2018/12/15/complete-binary-tree-2.png)
+
+```
+Input: root = [1,2,3,4,5,null,7]
+Output: false
+Explanation: The node with value 7 isn't as far left as possible.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 100]`.
+- `1 <= Node.val <= 1000`
+
+## 题目大意
+
+给定一个二叉树，确定它是否是一个完全二叉树。
+
+百度百科中对完全二叉树的定义如下：
+
+若设二叉树的深度为 h，除第 h 层外，其它各层 (1～h-1) 的结点数都达到最大个数，第 h 层所有的结点都连续集中在最左边，这就是完全二叉树。（注：第 h 层可能包含 1~ 2h 个节点。）
+
+## 解题思路
+
+- 这一题是按层序遍历的变种题。
+- 判断每个节点的左孩子是否为空。
+- 类似的题目，第 102，107，199 题都是按层序遍历的。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func isCompleteTree(root *TreeNode) bool {
+	queue, found := []*TreeNode{root}, false
+	for len(queue) > 0 {
+		node := queue[0] //取出每一层的第一个节点
+		queue = queue[1:]
+		if node == nil {
+			found = true
+		} else {
+			if found {
+				return false // 层序遍历中，两个不为空的节点中出现一个 nil
+			}
+			//如果左孩子为nil，则append进去的node.Left为nil
+			queue = append(queue, node.Left, node.Right)
+		}
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0959. Regions Cut By Slashes/959. Regions Cut By Slashes.go b/leetcode/0959.Regions-Cut-By-Slashes/959. Regions Cut By Slashes.go
similarity index 100%
rename from Algorithms/0959. Regions Cut By Slashes/959. Regions Cut By Slashes.go
rename to leetcode/0959.Regions-Cut-By-Slashes/959. Regions Cut By Slashes.go
diff --git a/Algorithms/0959. Regions Cut By Slashes/959. Regions Cut By Slashes_test.go b/leetcode/0959.Regions-Cut-By-Slashes/959. Regions Cut By Slashes_test.go
similarity index 91%
rename from Algorithms/0959. Regions Cut By Slashes/959. Regions Cut By Slashes_test.go
rename to leetcode/0959.Regions-Cut-By-Slashes/959. Regions Cut By Slashes_test.go
index ec978a554..51a088edb 100644
--- a/Algorithms/0959. Regions Cut By Slashes/959. Regions Cut By Slashes_test.go	
+++ b/leetcode/0959.Regions-Cut-By-Slashes/959. Regions Cut By Slashes_test.go	
@@ -25,27 +25,27 @@ type ans959 struct {
 func Test_Problem959(t *testing.T) {
 
 	qs := []question959{
-		question959{
+		{
 			para959{[]string{" /", "/ "}},
 			ans959{2},
 		},
 
-		question959{
+		{
 			para959{[]string{" /", "  "}},
 			ans959{1},
 		},
 
-		question959{
+		{
 			para959{[]string{"\\/", "/\\"}},
 			ans959{4},
 		},
 
-		question959{
+		{
 			para959{[]string{"/\\", "\\/"}},
 			ans959{5},
 		},
 
-		question959{
+		{
 			para959{[]string{"//", "/ "}},
 			ans959{3},
 		},
diff --git a/leetcode/0959.Regions-Cut-By-Slashes/README.md b/leetcode/0959.Regions-Cut-By-Slashes/README.md
new file mode 100755
index 000000000..f2ca84963
--- /dev/null
+++ b/leetcode/0959.Regions-Cut-By-Slashes/README.md
@@ -0,0 +1,98 @@
+# [959. Regions Cut By Slashes](https://leetcode.com/problems/regions-cut-by-slashes/)
+
+
+## 题目
+
+In a N x N `grid` composed of 1 x 1 squares, each 1 x 1 square consists of a `/`, `\`, or blank space. These characters divide the square into contiguous regions.
+
+(Note that backslash characters are escaped, so a `\` is represented as `"\\"`.)
+
+Return the number of regions.
+
+**Example 1:**
+
+    Input:
+    [
+      " /",
+      "/ "
+    ]
+    Output: 2
+    Explanation: The 2x2 grid is as follows:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/1.png)
+
+**Example 2:**
+
+    Input:
+    [
+      " /",
+      "  "
+    ]
+    Output: 1
+    Explanation: The 2x2 grid is as follows:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/2.png)
+
+**Example 3:**
+
+    Input:
+    [
+      "\\/",
+      "/\\"
+    ]
+    Output: 4
+    Explanation: (Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.)
+    The 2x2 grid is as follows:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/3.png)
+
+**Example 4:**
+
+    Input:
+    [
+      "/\\",
+      "\\/"
+    ]
+    Output: 5
+    Explanation: (Recall that because \ characters are escaped, "/\\" refers to /\, and "\\/" refers to \/.)
+    The 2x2 grid is as follows:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/4.png)
+
+**Example 5:**
+
+    Input:
+    [
+      "//",
+      "/ "
+    ]
+    Output: 3
+    Explanation: The 2x2 grid is as follows:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/5.png)
+
+**Note:**
+
+1. `1 <= grid.length == grid[0].length <= 30`
+2. `grid[i][j]` is either `'/'`, `'\'`, or `' '`.
+
+
+## 题目大意
+
+在由 1 x 1 方格组成的 N x N 网格 grid 中，每个 1 x 1 方块由 /、\ 或空格构成。这些字符会将方块划分为一些共边的区域。(请注意，反斜杠字符是转义的，因此 \ 用 "\\" 表示)返回区域的数目。
+
+
+提示：
+
+- 1 <= grid.length == grid[0].length <= 30
+- grid[i][j] 是 '/'、'\'、或 ' '。
+
+## 解题思路
+
+
+- 给出一个字符串，代表的是 `N x N` 正方形中切分的情况，有 2 种切分的情况 `'\'` 和 `'/'` ，即从左上往右下切和从右上往左下切。问按照给出的切分方法，能把 `N x N` 正方形切成几部分？
+- 这一题解题思路是并查集。先将每个 `1*1` 的正方形切分成下图的样子。分成 4 小块。然后按照题目给的切分图来合并各个小块。
+
+![](https://img.halfrost.com/Leetcode/leetcode_959.png)
+
+- 遇到 `'\\'`，就把第 0 块和第 1 块 `union()` 起来，第 2 块和第 3 块 `union()` 起来；遇到 `'/'`，就把第 0 块和第 3 块 `union()` 起来，第 2 块和第 1 块 `union()` 起来；遇到 `' '`，就把第 0 块和第 1 块 `union()` 起来，第 2 块和第 1 块 `union()` 起来，第 2 块和第 3 块 `union()` 起来，即 4 块都 `union()` 起来；最后还需要记得上一行和下一行还需要 `union()`，即本行的第 2 块和下一行的第 0 块 `union()` 起来；左边一列和右边一列也需要 `union()`。即本列的第 1 块和右边一列的第 3 块 `union()` 起来。最后计算出集合总个数就是最终答案了。
diff --git a/Algorithms/0961. N-Repeated Element in Size 2N Array/961. N-Repeated Element in Size 2N Array.go b/leetcode/0961.N-Repeated-Element-in-Size-2N-Array/961. N-Repeated Element in Size 2N Array.go
similarity index 100%
rename from Algorithms/0961. N-Repeated Element in Size 2N Array/961. N-Repeated Element in Size 2N Array.go
rename to leetcode/0961.N-Repeated-Element-in-Size-2N-Array/961. N-Repeated Element in Size 2N Array.go
diff --git a/Algorithms/0961. N-Repeated Element in Size 2N Array/961. N-Repeated Element in Size 2N Array_test.go b/leetcode/0961.N-Repeated-Element-in-Size-2N-Array/961. N-Repeated Element in Size 2N Array_test.go
similarity index 94%
rename from Algorithms/0961. N-Repeated Element in Size 2N Array/961. N-Repeated Element in Size 2N Array_test.go
rename to leetcode/0961.N-Repeated-Element-in-Size-2N-Array/961. N-Repeated Element in Size 2N Array_test.go
index f101197a6..cadc95a7a 100644
--- a/Algorithms/0961. N-Repeated Element in Size 2N Array/961. N-Repeated Element in Size 2N Array_test.go	
+++ b/leetcode/0961.N-Repeated-Element-in-Size-2N-Array/961. N-Repeated Element in Size 2N Array_test.go	
@@ -25,17 +25,17 @@ type ans961 struct {
 func Test_Problem961(t *testing.T) {
 
 	qs := []question961{
-		question961{
+		{
 			para961{[]int{1, 2, 3, 3}},
 			ans961{3},
 		},
 
-		question961{
+		{
 			para961{[]int{2, 1, 2, 5, 3, 2}},
 			ans961{2},
 		},
 
-		question961{
+		{
 			para961{[]int{5, 1, 5, 2, 5, 3, 5, 4}},
 			ans961{5},
 		},
diff --git a/leetcode/0961.N-Repeated-Element-in-Size-2N-Array/README.md b/leetcode/0961.N-Repeated-Element-in-Size-2N-Array/README.md
new file mode 100755
index 000000000..a26876408
--- /dev/null
+++ b/leetcode/0961.N-Repeated-Element-in-Size-2N-Array/README.md
@@ -0,0 +1,40 @@
+# [961. N-Repeated Element in Size 2N Array](https://leetcode.com/problems/n-repeated-element-in-size-2n-array/)
+
+
+## 题目
+
+In a array `A` of size `2N`, there are `N+1` unique elements, and exactly one of these elements is repeated N times.
+
+Return the element repeated `N` times.
+
+**Example 1:**
+
+    Input: [1,2,3,3]
+    Output: 3
+
+**Example 2:**
+
+    Input: [2,1,2,5,3,2]
+    Output: 2
+
+**Example 3:**
+
+    Input: [5,1,5,2,5,3,5,4]
+    Output: 5
+
+**Note:**
+
+1. `4 <= A.length <= 10000`
+2. `0 <= A[i] < 10000`
+3. `A.length` is even
+
+
+## 题目大意
+
+在大小为 2N 的数组 A 中有 N+1 个不同的元素，其中有一个元素重复了 N 次。返回重复了 N 次的那个元素。
+
+
+## 解题思路
+
+
+- 简单题。数组中有 2N 个数，有 N + 1 个数是不重复的，这之中有一个数重复了 N 次，请找出这个数。解法非常简单，把所有数都存入 map 中，如果遇到存在的 key 就返回这个数。
diff --git a/leetcode/0966.Vowel-Spellchecker/966. Vowel Spellchecker.go b/leetcode/0966.Vowel-Spellchecker/966. Vowel Spellchecker.go
new file mode 100644
index 000000000..a0798e998
--- /dev/null
+++ b/leetcode/0966.Vowel-Spellchecker/966. Vowel Spellchecker.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import "strings"
+
+func spellchecker(wordlist []string, queries []string) []string {
+	wordsPerfect, wordsCap, wordsVowel := map[string]bool{}, map[string]string{}, map[string]string{}
+	for _, word := range wordlist {
+		wordsPerfect[word] = true
+		wordLow := strings.ToLower(word)
+		if _, ok := wordsCap[wordLow]; !ok {
+			wordsCap[wordLow] = word
+		}
+		wordLowVowel := devowel(wordLow)
+		if _, ok := wordsVowel[wordLowVowel]; !ok {
+			wordsVowel[wordLowVowel] = word
+		}
+	}
+	res, index := make([]string, len(queries)), 0
+	for _, query := range queries {
+		if _, ok := wordsPerfect[query]; ok {
+			res[index] = query
+			index++
+			continue
+		}
+		queryL := strings.ToLower(query)
+		if v, ok := wordsCap[queryL]; ok {
+			res[index] = v
+			index++
+			continue
+		}
+
+		queryLV := devowel(queryL)
+		if v, ok := wordsVowel[queryLV]; ok {
+			res[index] = v
+			index++
+			continue
+		}
+		res[index] = ""
+		index++
+	}
+	return res
+
+}
+
+func devowel(word string) string {
+	runes := []rune(word)
+	for k, c := range runes {
+		if c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' {
+			runes[k] = '*'
+		}
+	}
+	return string(runes)
+}
diff --git a/leetcode/0966.Vowel-Spellchecker/966. Vowel Spellchecker_test.go b/leetcode/0966.Vowel-Spellchecker/966. Vowel Spellchecker_test.go
new file mode 100644
index 000000000..2c7975947
--- /dev/null
+++ b/leetcode/0966.Vowel-Spellchecker/966. Vowel Spellchecker_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question966 struct {
+	para966
+	ans966
+}
+
+// para 是参数
+// one 代表第一个参数
+type para966 struct {
+	wordlist []string
+	queries  []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans966 struct {
+	one []string
+}
+
+func Test_Problem966(t *testing.T) {
+
+	qs := []question966{
+		{
+			para966{[]string{"KiTe", "kite", "hare", "Hare"}, []string{"kite", "Kite", "KiTe", "Hare", "HARE", "Hear", "hear", "keti", "keet", "keto"}},
+			ans966{[]string{"kite", "KiTe", "KiTe", "Hare", "hare", "", "", "KiTe", "", "KiTe"}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 966------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans966, q.para966
+		fmt.Printf("【input】:%v       【output】:%#v\n", p, spellchecker(p.wordlist, p.queries))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0966.Vowel-Spellchecker/README.md b/leetcode/0966.Vowel-Spellchecker/README.md
new file mode 100644
index 000000000..504a83b7a
--- /dev/null
+++ b/leetcode/0966.Vowel-Spellchecker/README.md
@@ -0,0 +1,127 @@
+# [966. Vowel Spellchecker](https://leetcode.com/problems/vowel-spellchecker/)
+
+
+## 题目
+
+Given a `wordlist`, we want to implement a spellchecker that converts a query word into a correct word.
+
+For a given `query` word, the spell checker handles two categories of spelling mistakes:
+
+- Capitalization: If the query matches a word in the wordlist (**case-insensitive**), then the query word is returned with the same case as the case in the wordlist.
+    - Example: `wordlist = ["yellow"]`, `query = "YellOw"`: `correct = "yellow"`
+    - Example: `wordlist = ["Yellow"]`, `query = "yellow"`: `correct = "Yellow"`
+    - Example: `wordlist = ["yellow"]`, `query = "yellow"`: `correct = "yellow"`
+- Vowel Errors: If after replacing the vowels ('a', 'e', 'i', 'o', 'u') of the query word with any vowel individually, it matches a word in the wordlist (**case-insensitive**), then the query word is returned with the same case as the match in the wordlist.
+    - Example: `wordlist = ["YellOw"]`, `query = "yollow"`: `correct = "YellOw"`
+    - Example: `wordlist = ["YellOw"]`, `query = "yeellow"`: `correct = ""` (no match)
+    - Example: `wordlist = ["YellOw"]`, `query = "yllw"`: `correct = ""` (no match)
+
+In addition, the spell checker operates under the following precedence rules:
+
+- When the query exactly matches a word in the wordlist (**case-sensitive**), you should return the same word back.
+- When the query matches a word up to capitlization, you should return the first such match in the wordlist.
+- When the query matches a word up to vowel errors, you should return the first such match in the wordlist.
+- If the query has no matches in the wordlist, you should return the empty string.
+
+Given some `queries`, return a list of words `answer`, where `answer[i]` is the correct word for `query = queries[i]`.
+
+**Example 1:**
+
+```
+Input:wordlist = ["KiTe","kite","hare","Hare"], queries = ["kite","Kite","KiTe","Hare","HARE","Hear","hear","keti","keet","keto"]
+Output:["kite","KiTe","KiTe","Hare","hare","","","KiTe","","KiTe"]
+```
+
+**Note:**
+
+- `1 <= wordlist.length <= 5000`
+- `1 <= queries.length <= 5000`
+- `1 <= wordlist[i].length <= 7`
+- `1 <= queries[i].length <= 7`
+- All strings in `wordlist` and `queries` consist only of **english** letters.
+
+## 题目大意
+
+在给定单词列表 wordlist 的情况下，我们希望实现一个拼写检查器，将查询单词转换为正确的单词。
+
+对于给定的查询单词 query，拼写检查器将会处理两类拼写错误：
+
+- 大小写：如果查询匹配单词列表中的某个单词（不区分大小写），则返回的正确单词与单词列表中的大小写相同。
+    - 例如：wordlist = ["yellow"], query = "YellOw": correct = "yellow"
+    - 例如：wordlist = ["Yellow"], query = "yellow": correct = "Yellow"
+    - 例如：wordlist = ["yellow"], query = "yellow": correct = "yellow"
+- 元音错误：如果在将查询单词中的元音（‘a’、‘e’、‘i’、‘o’、‘u’）分别替换为任何元音后，能与单词列表中的单词匹配（不区分大小写），则返回的正确单词与单词列表中的匹配项大小写相同。
+    - 例如：wordlist = ["YellOw"], query = "yollow": correct = "YellOw"
+    - 例如：wordlist = ["YellOw"], query = "yeellow": correct = "" （无匹配项）
+    - 例如：wordlist = ["YellOw"], query = "yllw": correct = "" （无匹配项）
+
+此外，拼写检查器还按照以下优先级规则操作：
+
+- 当查询完全匹配单词列表中的某个单词（区分大小写）时，应返回相同的单词。
+- 当查询匹配到大小写问题的单词时，您应该返回单词列表中的第一个这样的匹配项。
+- 当查询匹配到元音错误的单词时，您应该返回单词列表中的第一个这样的匹配项。
+- 如果该查询在单词列表中没有匹配项，则应返回空字符串。
+
+给出一些查询 queries，返回一个单词列表 answer，其中 answer[i] 是由查询 query = queries[i] 得到的正确单词。
+
+## 解题思路
+
+- 读完题，很明显需要用 `map` 来解题。依题意分为 3 种情况，查询字符串完全匹配；查询字符串只是大小写不同；查询字符串有元音错误。第一种情况用 `map` `key` 直接匹配即可。第二种情况，利用 `map` 将单词从小写形式转换成原单词正确的大小写形式。第三种情况，利用 `map` 将单词从忽略元音的小写形式换成原单词正确形式。最后注意一下题目最后给的 4 个优先级规则即可。
+
+## 代码
+
+```go
+package leetcode
+
+import "strings"
+
+func spellchecker(wordlist []string, queries []string) []string {
+	wordsPerfect, wordsCap, wordsVowel := map[string]bool{}, map[string]string{}, map[string]string{}
+	for _, word := range wordlist {
+		wordsPerfect[word] = true
+		wordLow := strings.ToLower(word)
+		if _, ok := wordsCap[wordLow]; !ok {
+			wordsCap[wordLow] = word
+		}
+		wordLowVowel := devowel(wordLow)
+		if _, ok := wordsVowel[wordLowVowel]; !ok {
+			wordsVowel[wordLowVowel] = word
+		}
+	}
+	res, index := make([]string, len(queries)), 0
+	for _, query := range queries {
+		if _, ok := wordsPerfect[query]; ok {
+			res[index] = query
+			index++
+			continue
+		}
+		queryL := strings.ToLower(query)
+		if v, ok := wordsCap[queryL]; ok {
+			res[index] = v
+			index++
+			continue
+		}
+
+		queryLV := devowel(queryL)
+		if v, ok := wordsVowel[queryLV]; ok {
+			res[index] = v
+			index++
+			continue
+		}
+		res[index] = ""
+		index++
+	}
+	return res
+
+}
+
+func devowel(word string) string {
+	runes := []rune(word)
+	for k, c := range runes {
+		if c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' {
+			runes[k] = '*'
+		}
+	}
+	return string(runes)
+}
+```
\ No newline at end of file
diff --git a/leetcode/0968.Binary-Tree-Cameras/968. Binary Tree Cameras.go b/leetcode/0968.Binary-Tree-Cameras/968. Binary Tree Cameras.go
new file mode 100644
index 000000000..2ad7ab720
--- /dev/null
+++ b/leetcode/0968.Binary-Tree-Cameras/968. Binary Tree Cameras.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+type status int
+
+const (
+	isLeaf status = iota
+	parentofLeaf
+	isMonitoredWithoutCamera
+)
+
+func minCameraCover(root *TreeNode) int {
+	res := 0
+	if minCameraCoverDFS(root, &res) == isLeaf {
+		res++
+	}
+	return res
+}
+
+func minCameraCoverDFS(root *TreeNode, res *int) status {
+	if root == nil {
+		return 2
+	}
+	left, right := minCameraCoverDFS(root.Left, res), minCameraCoverDFS(root.Right, res)
+	if left == isLeaf || right == isLeaf {
+		*res++
+		return parentofLeaf
+	} else if left == parentofLeaf || right == parentofLeaf {
+		return isMonitoredWithoutCamera
+	} else {
+		return isLeaf
+	}
+}
diff --git a/leetcode/0968.Binary-Tree-Cameras/968. Binary Tree Cameras_test.go b/leetcode/0968.Binary-Tree-Cameras/968. Binary Tree Cameras_test.go
new file mode 100644
index 000000000..9ab69d564
--- /dev/null
+++ b/leetcode/0968.Binary-Tree-Cameras/968. Binary Tree Cameras_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question968 struct {
+	para968
+	ans968
+}
+
+// para 是参数
+// one 代表第一个参数
+type para968 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans968 struct {
+	one int
+}
+
+func Test_Problem968(t *testing.T) {
+
+	qs := []question968{
+
+		{
+			para968{[]int{0, 0, structures.NULL, 0, 0}},
+			ans968{1},
+		},
+
+		{
+			para968{[]int{0, 0, structures.NULL, 0, structures.NULL, 0, structures.NULL, structures.NULL, 0}},
+			ans968{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 968------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans968, q.para968
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", minCameraCover(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0968.Binary-Tree-Cameras/README.md b/leetcode/0968.Binary-Tree-Cameras/README.md
new file mode 100755
index 000000000..034583325
--- /dev/null
+++ b/leetcode/0968.Binary-Tree-Cameras/README.md
@@ -0,0 +1,55 @@
+# [968. Binary Tree Cameras](https://leetcode.com/problems/binary-tree-cameras/)
+
+## 题目
+
+Given a binary tree, we install cameras on the nodes of the tree.
+
+Each camera at a node can monitor **its parent, itself, and its immediate children**.
+
+Calculate the minimum number of cameras needed to monitor all nodes of the tree.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_01.png](https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_01.png)
+
+    Input: [0,0,null,0,0]
+    Output: 1
+    Explanation: One camera is enough to monitor all nodes if placed as shown.
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_02.png](https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_02.png)
+
+    Input: [0,0,null,0,null,0,null,null,0]
+    Output: 2
+    Explanation: At least two cameras are needed to monitor all nodes of the tree. The above image shows one of the valid configurations of camera placement.
+
+**Note:**
+
+1. The number of nodes in the given tree will be in the range `[1, 1000]`.
+2. **Every** node has value 0.
+
+
+## 题目大意
+
+给定一个二叉树，我们在树的节点上安装摄像头。节点上的每个摄影头都可以监视其父对象、自身及其直接子对象。计算监控树的所有节点所需的最小摄像头数量。
+
+提示：
+
+1. 给定树的节点数的范围是 [1, 1000]。
+2. 每个节点的值都是 0。
+
+
+
+## 解题思路
+
+- 给出一棵树，要求在这个树上面放摄像头，一个摄像头最多可以监视 4 个节点，2 个孩子，本身节点，还有父亲节点。问最少放多少个摄像头可以覆盖树上的所有节点。
+- 这一题可以用贪心思想来解题。先将节点分为 3 类，第一类，叶子节点，第二类，包含叶子节点的节点，第三类，其中一个叶子节点上放了摄像头的。按照这个想法，将树的每个节点染色，如下图。
+
+![](https://img.halfrost.com/Leetcode/leetcode_968_1.png)
+
+- 所有包含叶子节点的节点，可以放一个摄像头，这个可以覆盖至少 3 个节点，如果还有父节点的话，可以覆盖 4 个节点。所以贪心的策略是从最下层的叶子节点开始往上“染色”，先把最下面的一层 1 染色。标 1 的节点都是要放一个摄像头的。如果叶子节点中包含 1 的节点，那么再将这个节点染成 2 。如下图的黄色节点。黄色节点代表的是不用放摄像头的节点，因为它被叶子节点的摄像头覆盖了。出现了 2 的节点以后，再往上的节点又再次恢复成叶子节点 0 。如此类推，直到推到根节点。
+
+![](https://img.halfrost.com/Leetcode/leetcode_968_2.png)
+
+- 最后根节点还需要注意多种情况，根节点可能是叶子节点 0，那么最终答案还需要 + 1，因为需要在根节点上放一个摄像头，不然根节点覆盖不到了。根节点也有可能是 1 或者 2，这两种情况都不需要增加摄像头了，以为都覆盖到了。按照上述的方法，递归即可得到答案。
diff --git a/Algorithms/0969. Pancake Sorting/969. Pancake Sorting.go b/leetcode/0969.Pancake-Sorting/969. Pancake Sorting.go
similarity index 100%
rename from Algorithms/0969. Pancake Sorting/969. Pancake Sorting.go
rename to leetcode/0969.Pancake-Sorting/969. Pancake Sorting.go
diff --git a/Algorithms/0969. Pancake Sorting/969. Pancake Sorting_test.go b/leetcode/0969.Pancake-Sorting/969. Pancake Sorting_test.go
similarity index 94%
rename from Algorithms/0969. Pancake Sorting/969. Pancake Sorting_test.go
rename to leetcode/0969.Pancake-Sorting/969. Pancake Sorting_test.go
index b44b37224..d22253935 100644
--- a/Algorithms/0969. Pancake Sorting/969. Pancake Sorting_test.go	
+++ b/leetcode/0969.Pancake-Sorting/969. Pancake Sorting_test.go	
@@ -26,17 +26,17 @@ func Test_Problem969(t *testing.T) {
 
 	qs := []question969{
 
-		question969{
+		{
 			para969{[]int{}},
 			ans969{[]int{}},
 		},
 
-		question969{
+		{
 			para969{[]int{1}},
 			ans969{[]int{1}},
 		},
 
-		question969{
+		{
 			para969{[]int{3, 2, 4, 1}},
 			ans969{[]int{3, 4, 2, 3, 1, 2}},
 		},
diff --git a/Algorithms/0969. Pancake Sorting/README.md b/leetcode/0969.Pancake-Sorting/README.md
similarity index 100%
rename from Algorithms/0969. Pancake Sorting/README.md
rename to leetcode/0969.Pancake-Sorting/README.md
diff --git a/Algorithms/0970. Powerful Integers/970. Powerful Integers.go b/leetcode/0970.Powerful-Integers/970. Powerful Integers.go
similarity index 100%
rename from Algorithms/0970. Powerful Integers/970. Powerful Integers.go
rename to leetcode/0970.Powerful-Integers/970. Powerful Integers.go
diff --git a/Algorithms/0970. Powerful Integers/970. Powerful Integers_test.go b/leetcode/0970.Powerful-Integers/970. Powerful Integers_test.go
similarity index 96%
rename from Algorithms/0970. Powerful Integers/970. Powerful Integers_test.go
rename to leetcode/0970.Powerful-Integers/970. Powerful Integers_test.go
index 82eb8fdd9..6501a3649 100644
--- a/Algorithms/0970. Powerful Integers/970. Powerful Integers_test.go	
+++ b/leetcode/0970.Powerful-Integers/970. Powerful Integers_test.go	
@@ -28,12 +28,12 @@ func Test_Problem970(t *testing.T) {
 
 	qs := []question970{
 
-		question970{
+		{
 			para970{2, 3, 10},
 			ans970{[]int{2, 3, 4, 5, 7, 9, 10}},
 		},
 
-		question970{
+		{
 			para970{3, 5, 15},
 			ans970{[]int{2, 4, 6, 8, 10, 14}},
 		},
diff --git a/leetcode/0970.Powerful-Integers/README.md b/leetcode/0970.Powerful-Integers/README.md
new file mode 100755
index 000000000..31176a5d9
--- /dev/null
+++ b/leetcode/0970.Powerful-Integers/README.md
@@ -0,0 +1,45 @@
+# [970. Powerful Integers](https://leetcode.com/problems/powerful-integers/)
+
+
+## 题目
+
+Given two positive integers `x` and `y`, an integer is *powerful* if it is equal to `x^i + y^j` for some integers `i >= 0` and `j >= 0`.
+
+Return a list of all *powerful* integers that have value less than or equal to `bound`.
+
+You may return the answer in any order. In your answer, each value should occur at most once.
+
+**Example 1:**
+
+    Input: x = 2, y = 3, bound = 10
+    Output: [2,3,4,5,7,9,10]
+    Explanation: 
+    2 = 2^0 + 3^0
+    3 = 2^1 + 3^0
+    4 = 2^0 + 3^1
+    5 = 2^1 + 3^1
+    7 = 2^2 + 3^1
+    9 = 2^3 + 3^0
+    10 = 2^0 + 3^2
+
+**Example 2:**
+
+    Input: x = 3, y = 5, bound = 15
+    Output: [2,4,6,8,10,14]
+
+**Note:**
+
+- `1 <= x <= 100`
+- `1 <= y <= 100`
+- `0 <= bound <= 10^6`
+
+
+## 题目大意
+
+给定两个正整数 x 和 y，如果某一整数等于 x^i + y^j，其中整数 i >= 0 且 j >= 0，那么我们认为该整数是一个强整数。返回值小于或等于 bound 的所有强整数组成的列表。你可以按任何顺序返回答案。在你的回答中，每个值最多出现一次。
+
+
+## 解题思路
+
+
+- 简答题，题目要求找出满足 `x^i + y^j ≤ bound` 条件的所有解。题目要求输出中不能重复，所以用 map 来去重。剩下的就是 `n^2` 暴力循环枚举所有解。
diff --git a/leetcode/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal/971. Flip Binary Tree To Match Preorder Traversal.go b/leetcode/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal/971. Flip Binary Tree To Match Preorder Traversal.go
new file mode 100644
index 000000000..6d7dbff87
--- /dev/null
+++ b/leetcode/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal/971. Flip Binary Tree To Match Preorder Traversal.go	
@@ -0,0 +1,40 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func flipMatchVoyage(root *TreeNode, voyage []int) []int {
+	res, index := make([]int, 0, len(voyage)), 0
+	if travelTree(root, &index, voyage, &res) {
+		return res
+	}
+	return []int{-1}
+}
+
+func travelTree(root *TreeNode, index *int, voyage []int, res *[]int) bool {
+	if root == nil {
+		return true
+	}
+	if root.Val != voyage[*index] {
+		return false
+	}
+	*index++
+	if root.Left != nil && root.Left.Val != voyage[*index] {
+		*res = append(*res, root.Val)
+		return travelTree(root.Right, index, voyage, res) && travelTree(root.Left, index, voyage, res)
+	}
+	return travelTree(root.Left, index, voyage, res) && travelTree(root.Right, index, voyage, res)
+}
diff --git a/leetcode/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal/971. Flip Binary Tree To Match Preorder Traversal_test.go b/leetcode/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal/971. Flip Binary Tree To Match Preorder Traversal_test.go
new file mode 100644
index 000000000..f494b92c5
--- /dev/null
+++ b/leetcode/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal/971. Flip Binary Tree To Match Preorder Traversal_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question971 struct {
+	para971
+	ans971
+}
+
+// para 是参数
+// one 代表第一个参数
+type para971 struct {
+	one    []int
+	voyage []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans971 struct {
+	one []int
+}
+
+func Test_Problem971(t *testing.T) {
+
+	qs := []question971{
+
+		{
+			para971{[]int{1, 2, structures.NULL}, []int{2, 1}},
+			ans971{[]int{-1}},
+		},
+
+		{
+			para971{[]int{1, 2, 3}, []int{1, 3, 2}},
+			ans971{[]int{1}},
+		},
+
+		{
+			para971{[]int{1, 2, 3}, []int{1, 2, 3}},
+			ans971{[]int{}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 971------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans971, q.para971
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", flipMatchVoyage(rootOne, p.voyage))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal/README.md b/leetcode/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal/README.md
new file mode 100644
index 000000000..513df8b8d
--- /dev/null
+++ b/leetcode/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal/README.md
@@ -0,0 +1,106 @@
+# [971. Flip Binary Tree To Match Preorder Traversal](https://leetcode.com/problems/flip-binary-tree-to-match-preorder-traversal/)
+
+
+## 题目
+
+You are given the `root` of a binary tree with `n` nodes, where each node is uniquely assigned a value from `1` to `n`. You are also given a sequence of `n` values `voyage`, which is the **desired** **[pre-order traversal](https://en.wikipedia.org/wiki/Tree_traversal#Pre-order)** of the binary tree.
+
+Any node in the binary tree can be **flipped** by swapping its left and right subtrees. For example, flipping node 1 will have the following effect:
+
+![https://assets.leetcode.com/uploads/2021/02/15/fliptree.jpg](https://assets.leetcode.com/uploads/2021/02/15/fliptree.jpg)
+
+Flip the **smallest** number of nodes so that the **pre-order traversal** of the tree **matches** `voyage`.
+
+Return *a list of the values of all **flipped** nodes. You may return the answer in **any order**. If it is **impossible** to flip the nodes in the tree to make the pre-order traversal match* `voyage`*, return the list* `[-1]`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/01/02/1219-01.png](https://assets.leetcode.com/uploads/2019/01/02/1219-01.png)
+
+```
+Input: root = [1,2], voyage = [2,1]
+Output: [-1]
+Explanation: It is impossible to flip the nodes such that the pre-order traversal matches voyage.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2019/01/02/1219-02.png](https://assets.leetcode.com/uploads/2019/01/02/1219-02.png)
+
+```
+Input: root = [1,2,3], voyage = [1,3,2]
+Output: [1]
+Explanation: Flipping node 1 swaps nodes 2 and 3, so the pre-order traversal matches voyage.
+```
+
+**Example 3:**
+
+![https://assets.leetcode.com/uploads/2019/01/02/1219-02.png](https://assets.leetcode.com/uploads/2019/01/02/1219-02.png)
+
+```
+Input: root = [1,2,3], voyage = [1,2,3]
+Output: []
+Explanation: The tree's pre-order traversal already matches voyage, so no nodes need to be flipped.
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is `n`.
+- `n == voyage.length`
+- `1 <= n <= 100`
+- `1 <= Node.val, voyage[i] <= n`
+- All the values in the tree are **unique**.
+- All the values in `voyage` are **unique**.
+
+## 题目大意
+
+给你一棵二叉树的根节点 root ，树中有 n 个节点，每个节点都有一个不同于其他节点且处于 1 到 n 之间的值。另给你一个由 n 个值组成的行程序列 voyage ，表示 预期 的二叉树 先序遍历 结果。通过交换节点的左右子树，可以 翻转 该二叉树中的任意节点。请翻转 最少 的树中节点，使二叉树的 先序遍历 与预期的遍历行程 voyage 相匹配 。如果可以，则返回 翻转的 所有节点的值的列表。你可以按任何顺序返回答案。如果不能，则返回列表 [-1]。
+
+## 解题思路
+
+- 题目要求翻转最少树中节点，利用贪心的思想，应该从根节点开始从上往下依次翻转，这样翻转的次数是最少的。对树进行深度优先遍历，如果遍历到某一个节点的时候，节点值不能与行程序列匹配，那么答案一定是 [-1]。否则，当下一个期望数字 `voyage[i]` 与即将遍历的子节点的值不同的时候，就要翻转一下当前这个节点的左右子树，继续 DFS。递归结束可能有 2 种情况，一种是找出了所有要翻转的节点，另一种情况是没有需要翻转的，即原树先序遍历的结果与 `voyage` 是完全一致的。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func flipMatchVoyage(root *TreeNode, voyage []int) []int {
+	res, index := make([]int, 0, len(voyage)), 0
+	if travelTree(root, &index, voyage, &res) {
+		return res
+	}
+	return []int{-1}
+}
+
+func travelTree(root *TreeNode, index *int, voyage []int, res *[]int) bool {
+	if root == nil {
+		return true
+	}
+	if root.Val != voyage[*index] {
+		return false
+	}
+	*index++
+	if root.Left != nil && root.Left.Val != voyage[*index] {
+		*res = append(*res, root.Val)
+		return travelTree(root.Right, index, voyage, res) && travelTree(root.Left, index, voyage, res)
+	}
+	return travelTree(root.Left, index, voyage, res) && travelTree(root.Right, index, voyage, res)
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0973. K Closest Points to Origin/973. K Closest Points to Origin.go b/leetcode/0973.K-Closest-Points-to-Origin/973. K Closest Points to Origin.go
similarity index 100%
rename from Algorithms/0973. K Closest Points to Origin/973. K Closest Points to Origin.go
rename to leetcode/0973.K-Closest-Points-to-Origin/973. K Closest Points to Origin.go
diff --git a/leetcode/0973.K-Closest-Points-to-Origin/973. K Closest Points to Origin_test.go b/leetcode/0973.K-Closest-Points-to-Origin/973. K Closest Points to Origin_test.go
new file mode 100644
index 000000000..6ea66d66b
--- /dev/null
+++ b/leetcode/0973.K-Closest-Points-to-Origin/973. K Closest Points to Origin_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question973 struct {
+	para973
+	ans973
+}
+
+// para 是参数
+// one 代表第一个参数
+type para973 struct {
+	one [][]int
+	two int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans973 struct {
+	one [][]int
+}
+
+func Test_Problem973(t *testing.T) {
+
+	qs := []question973{
+
+		{
+			para973{[][]int{{1, 3}, {-2, 2}}, 1},
+			ans973{[][]int{{-2, 2}}},
+		},
+
+		{
+			para973{[][]int{{1, 3}, {-2, 2}}, 0},
+			ans973{[][]int{{}}},
+		},
+
+		{
+			para973{[][]int{{3, 3}, {5, -1}, {-2, 4}}, 2},
+			ans973{[][]int{{3, 3}, {-2, 4}}},
+		},
+
+		{
+			para973{[][]int{{0, 1}, {1, 0}}, 2},
+			ans973{[][]int{{1, 0}, {0, 1}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 973------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans973, q.para973
+		fmt.Printf("【input】:%v       【output】:%v\n", p, KClosest(p.one, p.two))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0973. K Closest Points to Origin/README.md b/leetcode/0973.K-Closest-Points-to-Origin/README.md
similarity index 100%
rename from Algorithms/0973. K Closest Points to Origin/README.md
rename to leetcode/0973.K-Closest-Points-to-Origin/README.md
diff --git a/leetcode/0976.Largest-Perimeter-Triangle/976. Largest Perimeter Triangle.go b/leetcode/0976.Largest-Perimeter-Triangle/976. Largest Perimeter Triangle.go
new file mode 100644
index 000000000..50d1fc452
--- /dev/null
+++ b/leetcode/0976.Largest-Perimeter-Triangle/976. Largest Perimeter Triangle.go	
@@ -0,0 +1,36 @@
+package leetcode
+
+func largestPerimeter(A []int) int {
+	if len(A) < 3 {
+		return 0
+	}
+	quickSort164(A, 0, len(A)-1)
+	for i := len(A) - 1; i >= 2; i-- {
+		if (A[i]+A[i-1] > A[i-2]) && (A[i]+A[i-2] > A[i-1]) && (A[i-2]+A[i-1] > A[i]) {
+			return A[i] + A[i-1] + A[i-2]
+		}
+	}
+	return 0
+}
+
+func quickSort164(a []int, lo, hi int) {
+	if lo >= hi {
+		return
+	}
+	p := partition164(a, lo, hi)
+	quickSort164(a, lo, p-1)
+	quickSort164(a, p+1, hi)
+}
+
+func partition164(a []int, lo, hi int) int {
+	pivot := a[hi]
+	i := lo - 1
+	for j := lo; j < hi; j++ {
+		if a[j] < pivot {
+			i++
+			a[j], a[i] = a[i], a[j]
+		}
+	}
+	a[i+1], a[hi] = a[hi], a[i+1]
+	return i + 1
+}
diff --git a/Algorithms/0976. Largest Perimeter Triangle/976. Largest Perimeter Triangle_test.go b/leetcode/0976.Largest-Perimeter-Triangle/976. Largest Perimeter Triangle_test.go
similarity index 89%
rename from Algorithms/0976. Largest Perimeter Triangle/976. Largest Perimeter Triangle_test.go
rename to leetcode/0976.Largest-Perimeter-Triangle/976. Largest Perimeter Triangle_test.go
index e16e9a5ec..e044f42b2 100644
--- a/Algorithms/0976. Largest Perimeter Triangle/976. Largest Perimeter Triangle_test.go	
+++ b/leetcode/0976.Largest-Perimeter-Triangle/976. Largest Perimeter Triangle_test.go	
@@ -26,36 +26,36 @@ func Test_Problem976(t *testing.T) {
 
 	qs := []question976{
 
-		question976{
+		{
 			para976{[]int{1, 2}},
 			ans976{0},
 		},
-		question976{
+		{
 			para976{[]int{1, 2, 3}},
 			ans976{0},
 		},
 
-		question976{
+		{
 			para976{[]int{}},
 			ans976{0},
 		},
 
-		question976{
+		{
 			para976{[]int{2, 1, 2}},
 			ans976{5},
 		},
 
-		question976{
+		{
 			para976{[]int{1, 1, 2}},
 			ans976{0},
 		},
 
-		question976{
+		{
 			para976{[]int{3, 2, 3, 4}},
 			ans976{10},
 		},
 
-		question976{
+		{
 			para976{[]int{3, 6, 2, 3}},
 			ans976{8},
 		},
diff --git a/Algorithms/0976. Largest Perimeter Triangle/README.md b/leetcode/0976.Largest-Perimeter-Triangle/README.md
similarity index 100%
rename from Algorithms/0976. Largest Perimeter Triangle/README.md
rename to leetcode/0976.Largest-Perimeter-Triangle/README.md
diff --git a/Algorithms/0977. Squares of a Sorted Array/977. Squares of a Sorted Array.go b/leetcode/0977.Squares-of-a-Sorted-Array/977. Squares of a Sorted Array.go
similarity index 100%
rename from Algorithms/0977. Squares of a Sorted Array/977. Squares of a Sorted Array.go
rename to leetcode/0977.Squares-of-a-Sorted-Array/977. Squares of a Sorted Array.go
diff --git a/Algorithms/0977. Squares of a Sorted Array/977. Squares of a Sorted Array_test.go b/leetcode/0977.Squares-of-a-Sorted-Array/977. Squares of a Sorted Array_test.go
similarity index 94%
rename from Algorithms/0977. Squares of a Sorted Array/977. Squares of a Sorted Array_test.go
rename to leetcode/0977.Squares-of-a-Sorted-Array/977. Squares of a Sorted Array_test.go
index 33e49b1d4..01b7c9c94 100644
--- a/Algorithms/0977. Squares of a Sorted Array/977. Squares of a Sorted Array_test.go	
+++ b/leetcode/0977.Squares-of-a-Sorted-Array/977. Squares of a Sorted Array_test.go	
@@ -26,17 +26,17 @@ func Test_Problem977(t *testing.T) {
 
 	qs := []question977{
 
-		question977{
+		{
 			para977{[]int{-4, -1, 0, 3, 10}},
 			ans977{[]int{0, 1, 9, 16, 100}},
 		},
 
-		question977{
+		{
 			para977{[]int{1}},
 			ans977{[]int{1}},
 		},
 
-		question977{
+		{
 			para977{[]int{-7, -3, 2, 3, 11}},
 			ans977{[]int{4, 9, 9, 49, 121}},
 		},
diff --git a/Algorithms/0977. Squares of a Sorted Array/README.md b/leetcode/0977.Squares-of-a-Sorted-Array/README.md
similarity index 100%
rename from Algorithms/0977. Squares of a Sorted Array/README.md
rename to leetcode/0977.Squares-of-a-Sorted-Array/README.md
diff --git a/leetcode/0978.Longest-Turbulent-Subarray/978. Longest Turbulent Subarray.go b/leetcode/0978.Longest-Turbulent-Subarray/978. Longest Turbulent Subarray.go
new file mode 100644
index 000000000..7de2db2cc
--- /dev/null
+++ b/leetcode/0978.Longest-Turbulent-Subarray/978. Longest Turbulent Subarray.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+// 解法一 模拟法
+func maxTurbulenceSize(arr []int) int {
+	inc, dec := 1, 1
+	maxLen := min(1, len(arr))
+	for i := 1; i < len(arr); i++ {
+		if arr[i-1] < arr[i] {
+			inc = dec + 1
+			dec = 1
+		} else if arr[i-1] > arr[i] {
+			dec = inc + 1
+			inc = 1
+		} else {
+			inc = 1
+			dec = 1
+		}
+		maxLen = max(maxLen, max(inc, dec))
+	}
+	return maxLen
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+// 解法二 滑动窗口
+func maxTurbulenceSize1(arr []int) int {
+	var maxLength int
+	if len(arr) == 2 && arr[0] != arr[1] {
+		maxLength = 2
+	} else {
+		maxLength = 1
+	}
+	left := 0
+	for right := 2; right < len(arr); right++ {
+		if arr[right] == arr[right-1] {
+			left = right
+		} else if (arr[right]-arr[right-1])^(arr[right-1]-arr[right-2]) >= 0 {
+			left = right - 1
+		}
+		maxLength = max(maxLength, right-left+1)
+	}
+	return maxLength
+}
diff --git a/Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray_test.go b/leetcode/0978.Longest-Turbulent-Subarray/978. Longest Turbulent Subarray_test.go
similarity index 91%
rename from Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray_test.go
rename to leetcode/0978.Longest-Turbulent-Subarray/978. Longest Turbulent Subarray_test.go
index cd3c1e802..cd16e06b4 100644
--- a/Algorithms/0978. Longest Turbulent Subarray/978. Longest Turbulent Subarray_test.go	
+++ b/leetcode/0978.Longest-Turbulent-Subarray/978. Longest Turbulent Subarray_test.go	
@@ -26,27 +26,27 @@ func Test_Problem978(t *testing.T) {
 
 	qs := []question978{
 
-		question978{
+		{
 			para978{[]int{0, 1, 1, 0, 1, 0, 1, 1, 0, 0}},
 			ans978{5},
 		},
 
-		question978{
+		{
 			para978{[]int{9, 9}},
 			ans978{1},
 		},
 
-		question978{
+		{
 			para978{[]int{9, 4, 2, 10, 7, 8, 8, 1, 9}},
 			ans978{5},
 		},
 
-		question978{
+		{
 			para978{[]int{4, 8, 12, 16}},
 			ans978{2},
 		},
 
-		question978{
+		{
 			para978{[]int{100}},
 			ans978{1},
 		},
diff --git a/leetcode/0978.Longest-Turbulent-Subarray/README.md b/leetcode/0978.Longest-Turbulent-Subarray/README.md
new file mode 100755
index 000000000..e38c46533
--- /dev/null
+++ b/leetcode/0978.Longest-Turbulent-Subarray/README.md
@@ -0,0 +1,58 @@
+# [978. Longest Turbulent Subarray](https://leetcode.com/problems/longest-turbulent-subarray/)
+
+## 题目
+
+A subarray `A[i], A[i+1], ..., A[j]` of `A` is said to be *turbulent* if and only if:
+
+- For `i <= k < j`, `A[k] > A[k+1]` when `k` is odd, and `A[k] < A[k+1]` when `k` is even;
+- **OR**, for `i <= k < j`, `A[k] > A[k+1]` when `k` is even, and `A[k] < A[k+1]` when `k` is odd.
+
+That is, the subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray.
+
+Return the **length** of a maximum size turbulent subarray of A.
+
+**Example 1:**
+
+    Input: [9,4,2,10,7,8,8,1,9]
+    Output: 5
+    Explanation: (A[1] > A[2] < A[3] > A[4] < A[5])
+
+**Example 2:**
+
+    Input: [4,8,12,16]
+    Output: 2
+
+**Example 3:**
+
+    Input: [100]
+    Output: 1
+
+**Note:**
+
+1. `1 <= A.length <= 40000`
+2. `0 <= A[i] <= 10^9`
+
+
+## 题目大意
+
+
+当 A 的子数组 A[i], A[i+1], ..., A[j] 满足下列条件时，我们称其为湍流子数组：
+
+若 i <= k < j，当 k 为奇数时， A[k] > A[k+1]，且当 k 为偶数时，A[k] < A[k+1]；
+或 若 i <= k < j，当 k 为偶数时，A[k] > A[k+1] ，且当 k 为奇数时， A[k] < A[k+1]。
+也就是说，如果比较符号在子数组中的每个相邻元素对之间翻转，则该子数组是湍流子数组。
+
+返回 A 的最大湍流子数组的长度。
+
+提示：
+
+- 1 <= A.length <= 40000
+- 0 <= A[i] <= 10^9
+
+
+
+## 解题思路
+
+
+- 给出一个数组，要求找出“摆动数组”的最大长度。所谓“摆动数组”的意思是，元素一大一小间隔的。
+- 这一题可以用滑动窗口来解答。用相邻元素差的乘积大于零（a ^ b >= 0 说明a b乘积大于零）来判断是否是湍流， 如果是，那么扩大窗口。否则窗口缩小为0，开始新的一个窗口。
\ No newline at end of file
diff --git a/leetcode/0979.Distribute-Coins-in-Binary-Tree/979. Distribute Coins in Binary Tree.go b/leetcode/0979.Distribute-Coins-in-Binary-Tree/979. Distribute Coins in Binary Tree.go
new file mode 100644
index 000000000..553ec7ae3
--- /dev/null
+++ b/leetcode/0979.Distribute-Coins-in-Binary-Tree/979. Distribute Coins in Binary Tree.go	
@@ -0,0 +1,39 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func distributeCoins(root *TreeNode) int {
+	res := 0
+	distributeCoinsDFS(root, &res)
+	return res
+}
+
+func distributeCoinsDFS(root *TreeNode, res *int) int {
+	if root == nil {
+		return 0
+	}
+	left, right := distributeCoinsDFS(root.Left, res), distributeCoinsDFS(root.Right, res)
+	*res += abs(left) + abs(right)
+	return left + right + root.Val - 1
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
diff --git a/leetcode/0979.Distribute-Coins-in-Binary-Tree/979. Distribute Coins in Binary Tree_test.go b/leetcode/0979.Distribute-Coins-in-Binary-Tree/979. Distribute Coins in Binary Tree_test.go
new file mode 100644
index 000000000..e38652aef
--- /dev/null
+++ b/leetcode/0979.Distribute-Coins-in-Binary-Tree/979. Distribute Coins in Binary Tree_test.go	
@@ -0,0 +1,66 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question979 struct {
+	para979
+	ans979
+}
+
+// para 是参数
+// one 代表第一个参数
+type para979 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans979 struct {
+	one int
+}
+
+func Test_Problem979(t *testing.T) {
+
+	qs := []question979{
+
+		{
+			para979{[]int{}},
+			ans979{0},
+		},
+
+		{
+			para979{[]int{1}},
+			ans979{0},
+		},
+
+		{
+			para979{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
+			ans979{41},
+		},
+
+		{
+			para979{[]int{1, 2, 3, 4, structures.NULL, structures.NULL, 5}},
+			ans979{11},
+		},
+
+		{
+			para979{[]int{1, 2, 3, 4, structures.NULL, 5}},
+			ans979{11},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 979------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans979, q.para979
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", distributeCoins(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0979.Distribute-Coins-in-Binary-Tree/README.md b/leetcode/0979.Distribute-Coins-in-Binary-Tree/README.md
new file mode 100755
index 000000000..45bff64f1
--- /dev/null
+++ b/leetcode/0979.Distribute-Coins-in-Binary-Tree/README.md
@@ -0,0 +1,64 @@
+# [979. Distribute Coins in Binary Tree](https://leetcode.com/problems/distribute-coins-in-binary-tree/)
+
+
+## 题目
+
+Given the `root` of a binary tree with `N` nodes, each `node` in the tree has `node.val` coins, and there are `N` coins total.
+
+In one move, we may choose two adjacent nodes and move one coin from one node to another. (The move may be from parent to child, or from child to parent.)
+
+Return the number of moves required to make every node have exactly one coin.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2019/01/18/tree1.png)
+
+    Input: [3,0,0]
+    Output: 2
+    Explanation: From the root of the tree, we move one coin to its left child, and one coin to its right child.
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2019/01/18/tree2.png)
+
+    Input: [0,3,0]
+    Output: 3
+    Explanation: From the left child of the root, we move two coins to the root [taking two moves].  Then, we move one coin from the root of the tree to the right child.
+
+**Example 3:**
+
+![](https://assets.leetcode.com/uploads/2019/01/18/tree3.png)
+
+    Input: [1,0,2]
+    Output: 2
+
+**Example 4:**
+
+![](https://assets.leetcode.com/uploads/2019/01/18/tree4.png)
+
+    Input: [1,0,0,null,3]
+    Output: 4
+
+**Note:**
+
+1. `1<= N <= 100`
+2. `0 <= node.val <= N`
+
+## 题目大意
+
+给定一个有 N 个结点的二叉树的根结点 root，树中的每个结点上都对应有 node.val 枚硬币，并且总共有 N 枚硬币。在一次移动中，我们可以选择两个相邻的结点，然后将一枚硬币从其中一个结点移动到另一个结点。(移动可以是从父结点到子结点，或者从子结点移动到父结点。)。返回使每个结点上只有一枚硬币所需的移动次数。
+
+提示：
+
+1. 1<= N <= 100
+2. 0 <= node.val <= N
+
+
+## 解题思路
+
+- 给出一棵树，有 N 个节点。有 N 个硬币分散在这 N 个节点中，问经过多少次移动以后，所有节点都有一枚硬币。
+- 这一题乍一看比较难分析，仔细一想，可以用贪心和分治的思想来解决。一个树的最小单元是一个根节点和两个孩子。在这种情况下，3 个节点谁的硬币多就可以分给没有硬币的那个节点，这种移动方法也能保证移动步数最少。不难证明，硬币由相邻的节点移动过来的步数是最少的。那么一棵树从最下一层开始往上推，逐步从下往上把硬币移动上去，直到最后根节点也都拥有硬币。多余 1 枚的节点记为 `n -1`，没有硬币的节点记为 -1 。例如，下图中左下角的 3 个节点，有 4 枚硬币的节点可以送出 3 枚硬币，叶子节点有 0 枚硬币的节点需要接收 1 枚硬币。根节点有 0 枚硬币，左孩子给了 3 枚，右孩子需要 1 枚，自己本身还要留一枚，所以最终还能剩 1 枚。
+
+![](https://img.halfrost.com/Leetcode/leetcode_979_1.png)
+
+- 所以每个节点移动的步数应该是 `left + right + root.Val - 1`。最后递归求解即可。
diff --git a/Algorithms/0980. Unique Paths III/980. Unique Paths III.go b/leetcode/0980.Unique-Paths-III/980. Unique Paths III.go
similarity index 89%
rename from Algorithms/0980. Unique Paths III/980. Unique Paths III.go
rename to leetcode/0980.Unique-Paths-III/980. Unique Paths III.go
index e3c12081f..9a145eee3 100644
--- a/Algorithms/0980. Unique Paths III/980. Unique Paths III.go	
+++ b/leetcode/0980.Unique-Paths-III/980. Unique Paths III.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
 func uniquePathsIII(grid [][]int) int {
 	visited := make([][]bool, len(grid))
 	for i := 0; i < len(visited); i++ {
@@ -46,7 +53,7 @@ func findUniquePathIII(board [][]int, visited [][]bool, path []int, empty, start
 			}
 		}
 		visited[startx][starty] = false
-		empty++
+		//empty++ 这里虽然可以还原这个变量值，但是赋值没有意义，干脆不写了
 		path = path[:len(path)-2]
 	}
 	return
diff --git a/Algorithms/0980. Unique Paths III/980. Unique Paths III_test.go b/leetcode/0980.Unique-Paths-III/980. Unique Paths III_test.go
similarity index 76%
rename from Algorithms/0980. Unique Paths III/980. Unique Paths III_test.go
rename to leetcode/0980.Unique-Paths-III/980. Unique Paths III_test.go
index 1cdc6f88e..571a2abc4 100644
--- a/Algorithms/0980. Unique Paths III/980. Unique Paths III_test.go	
+++ b/leetcode/0980.Unique-Paths-III/980. Unique Paths III_test.go	
@@ -26,28 +26,28 @@ func Test_Problem980(t *testing.T) {
 
 	qs := []question980{
 
-		question980{
+		{
 			para980{[][]int{
-				[]int{1, 0, 0, 0},
-				[]int{0, 0, 0, 0},
-				[]int{0, 0, 2, -1},
+				{1, 0, 0, 0},
+				{0, 0, 0, 0},
+				{0, 0, 2, -1},
 			}},
 			ans980{2},
 		},
 
-		question980{
+		{
 			para980{[][]int{
-				[]int{1, 0, 0, 0},
-				[]int{0, 0, 0, 0},
-				[]int{0, 0, 0, 2},
+				{1, 0, 0, 0},
+				{0, 0, 0, 0},
+				{0, 0, 0, 2},
 			}},
 			ans980{4},
 		},
 
-		question980{
+		{
 			para980{[][]int{
-				[]int{1, 0},
-				[]int{0, 2},
+				{1, 0},
+				{0, 2},
 			}},
 			ans980{0},
 		},
diff --git a/Algorithms/0980. Unique Paths III/README.md b/leetcode/0980.Unique-Paths-III/README.md
similarity index 100%
rename from Algorithms/0980. Unique Paths III/README.md
rename to leetcode/0980.Unique-Paths-III/README.md
diff --git a/leetcode/0981.Time-Based-Key-Value-Store/981. Time Based Key-Value Store.go b/leetcode/0981.Time-Based-Key-Value-Store/981. Time Based Key-Value Store.go
new file mode 100644
index 000000000..f682286cf
--- /dev/null
+++ b/leetcode/0981.Time-Based-Key-Value-Store/981. Time Based Key-Value Store.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+import "sort"
+
+type data struct {
+	time  int
+	value string
+}
+
+// TimeMap is a timebased key-value store
+// TimeMap define
+type TimeMap map[string][]data
+
+// Constructor981 define
+func Constructor981() TimeMap {
+	return make(map[string][]data, 1024)
+}
+
+// Set define
+func (t TimeMap) Set(key string, value string, timestamp int) {
+	if _, ok := t[key]; !ok {
+		t[key] = make([]data, 1, 1024)
+	}
+	t[key] = append(t[key], data{
+		time:  timestamp,
+		value: value,
+	})
+}
+
+// Get define
+func (t TimeMap) Get(key string, timestamp int) string {
+	d := t[key]
+	i := sort.Search(len(d), func(i int) bool {
+		return timestamp < d[i].time
+	})
+	i--
+	return t[key][i].value
+}
+
+/**
+ * Your TimeMap object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.Set(key,value,timestamp);
+ * param_2 := obj.Get(key,timestamp);
+ */
diff --git a/leetcode/0981.Time-Based-Key-Value-Store/981. Time Based Key-Value Store_test.go b/leetcode/0981.Time-Based-Key-Value-Store/981. Time Based Key-Value Store_test.go
new file mode 100644
index 000000000..5c612034c
--- /dev/null
+++ b/leetcode/0981.Time-Based-Key-Value-Store/981. Time Based Key-Value Store_test.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem981(t *testing.T) {
+	obj := Constructor981()
+	obj.Set("foo", "bar", 1)
+	fmt.Printf("Get = %v\n", obj.Get("foo", 1))
+	fmt.Printf("Get = %v\n", obj.Get("foo", 3))
+	obj.Set("foo", "bar2", 4)
+	fmt.Printf("Get = %v\n", obj.Get("foo", 4))
+	fmt.Printf("Get = %v\n", obj.Get("foo", 5))
+}
diff --git a/leetcode/0981.Time-Based-Key-Value-Store/README.md b/leetcode/0981.Time-Based-Key-Value-Store/README.md
new file mode 100755
index 000000000..ae143664e
--- /dev/null
+++ b/leetcode/0981.Time-Based-Key-Value-Store/README.md
@@ -0,0 +1,71 @@
+# [981. Time Based Key-Value Store](https://leetcode.com/problems/time-based-key-value-store/)
+
+
+## 题目
+
+Create a timebased key-value store class `TimeMap`, that supports two operations.
+
+1. `set(string key, string value, int timestamp)`
+
+- Stores the `key` and `value`, along with the given `timestamp`.
+
+2. `get(string key, int timestamp)`
+
+- Returns a value such that `set(key, value, timestamp_prev)` was called previously, with `timestamp_prev <= timestamp`.
+- If there are multiple such values, it returns the one with the largest `timestamp_prev`.
+- If there are no values, it returns the empty string (`""`).
+
+**Example 1:**
+
+    Input: inputs = ["TimeMap","set","get","get","set","get","get"], inputs = [[],["foo","bar",1],["foo",1],["foo",3],["foo","bar2",4],["foo",4],["foo",5]]
+    Output: [null,null,"bar","bar",null,"bar2","bar2"]
+    Explanation:   
+    TimeMap kv;   
+    kv.set("foo", "bar", 1); // store the key "foo" and value "bar" along with timestamp = 1   
+    kv.get("foo", 1);  // output "bar"   
+    kv.get("foo", 3); // output "bar" since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 ie "bar"   
+    kv.set("foo", "bar2", 4);   
+    kv.get("foo", 4); // output "bar2"   
+    kv.get("foo", 5); //output "bar2"
+
+**Example 2:**
+
+    Input: inputs = ["TimeMap","set","set","get","get","get","get","get"], inputs = [[],["love","high",10],["love","low",20],["love",5],["love",10],["love",15],["love",20],["love",25]]
+    Output: [null,null,null,"","high","high","low","low"]
+
+**Note:**
+
+1. All key/value strings are lowercase.
+2. All key/value strings have length in the range `[1, 100]`
+3. The `timestamps` for all `TimeMap.set` operations are strictly increasing.
+4. `1 <= timestamp <= 10^7`
+5. `TimeMap.set` and `TimeMap.get` functions will be called a total of `120000` times (combined) per test case.
+
+## 题目大意
+
+创建一个基于时间的键值存储类 TimeMap，它支持下面两个操作：
+
+1. set(string key, string value, int timestamp)
+
+- 存储键 key、值 value，以及给定的时间戳 timestamp。
+
+2. get(string key, int timestamp)
+
+- 返回先前调用 set(key, value, timestamp_prev) 所存储的值，其中 timestamp_prev <= timestamp。
+- 如果有多个这样的值，则返回对应最大的  timestamp_prev 的那个值。
+- 如果没有值，则返回空字符串（""）。
+
+提示：
+
+1. 所有的键/值字符串都是小写的。
+2. 所有的键/值字符串长度都在 [1, 100] 范围内。
+3. 所有 TimeMap.set 操作中的时间戳 timestamps 都是严格递增的。
+4. 1 <= timestamp <= 10^7
+5. TimeMap.set 和 TimeMap.get 函数在每个测试用例中将（组合）调用总计 120000 次。
+
+
+## 解题思路
+
+- 要求设计一个基于时间戳的 `kv` 存储。`set()` 操作里面会会包含一个时间戳。get() 操作的时候查找时间戳小于等于 `timestamp` 的 `key` 对应的 `value`，如果有多个解，输出满足条件的最大时间戳对应的 `value` 值。
+- 这一题可以用二分搜索来解答，用 `map` 存储 `kv` 数据，`key` 对应的就是 `key`，`value` 对应一个结构体，里面包含 `value` 和 `timestamp`。执行 `get()` 操作的时候，先取出 `key` 对应的结构体数组，然后在这个数组里面根据 `timestamp` 进行二分搜索。由于题意是要找小于等于 `timestamp` 的最大 `timestamp` ，这会有很多满足条件的解，变换一下，先找 `> timestamp` 的最小解，然后下标减一即是满足题意的最大解。
+- 另外题目中提到“`TimeMap.set` 操作中的 `timestamp` 是严格递增的”。所以在 `map` 中存储 `value` 结构体的时候，不需要排序了，天然有序。
diff --git a/Algorithms/0984. String Without AAA or BBB/984. String Without AAA or BBB.go b/leetcode/0984.String-Without-AAA-or-BBB/984. String Without AAA or BBB.go
similarity index 100%
rename from Algorithms/0984. String Without AAA or BBB/984. String Without AAA or BBB.go
rename to leetcode/0984.String-Without-AAA-or-BBB/984. String Without AAA or BBB.go
diff --git a/Algorithms/0984. String Without AAA or BBB/984. String Without AAA or BBB_test.go b/leetcode/0984.String-Without-AAA-or-BBB/984. String Without AAA or BBB_test.go
similarity index 95%
rename from Algorithms/0984. String Without AAA or BBB/984. String Without AAA or BBB_test.go
rename to leetcode/0984.String-Without-AAA-or-BBB/984. String Without AAA or BBB_test.go
index 2c6c78bd3..343ac3c7f 100644
--- a/Algorithms/0984. String Without AAA or BBB/984. String Without AAA or BBB_test.go	
+++ b/leetcode/0984.String-Without-AAA-or-BBB/984. String Without AAA or BBB_test.go	
@@ -27,12 +27,12 @@ func Test_Problem984(t *testing.T) {
 
 	qs := []question984{
 
-		question984{
+		{
 			para984{1, 2},
 			ans984{"abb"},
 		},
 
-		question984{
+		{
 			para984{4, 1},
 			ans984{"aabaa"},
 		},
diff --git a/leetcode/0984.String-Without-AAA-or-BBB/README.md b/leetcode/0984.String-Without-AAA-or-BBB/README.md
new file mode 100755
index 000000000..fbde83169
--- /dev/null
+++ b/leetcode/0984.String-Without-AAA-or-BBB/README.md
@@ -0,0 +1,49 @@
+# [984. String Without AAA or BBB](https://leetcode.com/problems/string-without-aaa-or-bbb/)
+
+
+## 题目
+
+Given two integers `A` and `B`, return **any** string `S` such that:
+
+- `S` has length `A + B` and contains exactly `A` `'a'` letters, and exactly `B` `'b'`letters;
+- The substring `'aaa'` does not occur in `S`;
+- The substring `'bbb'` does not occur in `S`.
+
+**Example 1:**
+
+    Input: A = 1, B = 2
+    Output: "abb"
+    Explanation: "abb", "bab" and "bba" are all correct answers.
+
+**Example 2:**
+
+    Input: A = 4, B = 1
+    Output: "aabaa"
+
+**Note:**
+
+1. `0 <= A <= 100`
+2. `0 <= B <= 100`
+3. It is guaranteed such an `S` exists for the given `A` and `B`.
+
+
+## 题目大意
+
+给定两个整数 A 和 B，返回任意字符串 S，要求满足：
+
+- S 的长度为 A + B，且正好包含 A 个 'a' 字母与 B 个 'b' 字母；
+- 子串 'aaa' 没有出现在 S 中；
+- 子串 'bbb' 没有出现在 S 中。
+
+
+提示：
+
+- 0 <= A <= 100
+- 0 <= B <= 100
+- 对于给定的 A 和 B，保证存在满足要求的 S。
+
+
+## 解题思路
+
+
+- 给出 A 和 B 的个数，要求组合出字符串，不能出现 3 个连续的 A 和 3 个连续的 B。这题由于只可能有 4 种情况，暴力枚举就可以了。假设 B 的个数比 A 多(如果 A 多，就交换一下 A 和 B)，最终可能的情况只可能是这 4 种情况中的一种： `ba`，`bbabb`，`bbabbabb`，`bbabbabbabbabababa`。
diff --git a/leetcode/0985.Sum-of-Even-Numbers-After-Queries/985. Sum of Even Numbers After Queries.go b/leetcode/0985.Sum-of-Even-Numbers-After-Queries/985. Sum of Even Numbers After Queries.go
new file mode 100644
index 000000000..2bf98abcc
--- /dev/null
+++ b/leetcode/0985.Sum-of-Even-Numbers-After-Queries/985. Sum of Even Numbers After Queries.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+func sumEvenAfterQueries(A []int, queries [][]int) []int {
+	cur, res := 0, []int{}
+	for _, v := range A {
+		if v%2 == 0 {
+			cur += v
+		}
+	}
+	for _, q := range queries {
+		if A[q[1]]%2 == 0 {
+			cur -= A[q[1]]
+		}
+		A[q[1]] += q[0]
+		if A[q[1]]%2 == 0 {
+			cur += A[q[1]]
+		}
+		res = append(res, cur)
+	}
+	return res
+}
diff --git a/leetcode/0985.Sum-of-Even-Numbers-After-Queries/985. Sum of Even Numbers After Queries_test.go b/leetcode/0985.Sum-of-Even-Numbers-After-Queries/985. Sum of Even Numbers After Queries_test.go
new file mode 100644
index 000000000..26d2b71fc
--- /dev/null
+++ b/leetcode/0985.Sum-of-Even-Numbers-After-Queries/985. Sum of Even Numbers After Queries_test.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question985 struct {
+	para985
+	ans985
+}
+
+// para 是参数
+// one 代表第一个参数
+type para985 struct {
+	A       []int
+	queries [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans985 struct {
+	one []int
+}
+
+func Test_Problem985(t *testing.T) {
+
+	qs := []question985{
+
+		{
+			para985{[]int{1, 2, 3, 4}, [][]int{{1, 0}, {-3, 1}, {-4, 0}, {2, 3}}},
+			ans985{[]int{8, 6, 2, 4}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 985------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans985, q.para985
+		fmt.Printf("【input】:%v       【output】:%v\n\n\n", p, sumEvenAfterQueries(p.A, p.queries))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0985.Sum-of-Even-Numbers-After-Queries/README.md b/leetcode/0985.Sum-of-Even-Numbers-After-Queries/README.md
new file mode 100644
index 000000000..925fce019
--- /dev/null
+++ b/leetcode/0985.Sum-of-Even-Numbers-After-Queries/README.md
@@ -0,0 +1,68 @@
+# [985. Sum of Even Numbers After Queries](https://leetcode.com/problems/sum-of-even-numbers-after-queries/)
+
+## 题目
+
+We have an array `A` of integers, and an array `queries` of queries.
+
+For the `i`-th query `val = queries[i][0], index = queries[i][1]`, we add val to `A[index]`. Then, the answer to the `i`-th query is the sum of the even values of `A`.
+
+*(Here, the given `index = queries[i][1]` is a 0-based index, and each query permanently modifies the array `A`.)*
+
+Return the answer to all queries. Your `answer` array should have `answer[i]` as the answer to the `i`-th query.
+
+**Example 1**:
+
+```
+Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
+Output: [8,6,2,4]
+Explanation: 
+At the beginning, the array is [1,2,3,4].
+After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
+After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
+After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
+After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
+```
+
+**Note**:
+
+1. `1 <= A.length <= 10000`
+2. `-10000 <= A[i] <= 10000`
+3. `1 <= queries.length <= 10000`
+4. `-10000 <= queries[i][0] <= 10000`
+5. `0 <= queries[i][1] < A.length`
+
+## 题目大意
+
+给出一个整数数组 A 和一个查询数组 queries。
+
+对于第 i 次查询，有 val = queries[i][0], index = queries[i][1]，我们会把 val 加到 A[index] 上。然后，第 i 次查询的答案是 A 中偶数值的和。（此处给定的 index = queries[i][1] 是从 0 开始的索引，每次查询都会永久修改数组 A。）返回所有查询的答案。你的答案应当以数组 answer 给出，answer[i] 为第 i 次查询的答案。
+
+
+## 解题思路
+
+- 给出一个数组 A 和 query 数组。要求每次 query 操作都改变数组 A 中的元素值，并计算此次操作结束数组 A 中偶数值之和。
+- 简单题，先计算 A 中所有偶数之和。再每次 query 操作的时候，动态维护这个偶数之和即可。
+
+## 代码
+
+```go
+func sumEvenAfterQueries(A []int, queries [][]int) []int {
+	cur, res := 0, []int{}
+	for _, v := range A {
+		if v%2 == 0 {
+			cur += v
+		}
+	}
+	for _, q := range queries {
+		if A[q[1]]%2 == 0 {
+			cur -= A[q[1]]
+		}
+		A[q[1]] += q[0]
+		if A[q[1]]%2 == 0 {
+			cur += A[q[1]]
+		}
+		res = append(res, cur)
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0986.Interval-List-Intersections/986. Interval List Intersections.go b/leetcode/0986.Interval-List-Intersections/986. Interval List Intersections.go
new file mode 100644
index 000000000..6a24c0e54
--- /dev/null
+++ b/leetcode/0986.Interval-List-Intersections/986. Interval List Intersections.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// Interval define
+type Interval = structures.Interval
+
+/**
+ * Definition for an interval.
+ * type Interval struct {
+ *	   Start int
+ *	   End   int
+ * }
+ */
+func intervalIntersection(A []Interval, B []Interval) []Interval {
+	res := []Interval{}
+	for i, j := 0, 0; i < len(A) && j < len(B); {
+		start := max(A[i].Start, B[j].Start)
+		end := min(A[i].End, B[j].End)
+		if start <= end {
+			res = append(res, Interval{Start: start, End: end})
+		}
+		if A[i].End <= B[j].End {
+			i++
+		} else {
+			j++
+		}
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/0986.Interval-List-Intersections/986. Interval List Intersections_test.go b/leetcode/0986.Interval-List-Intersections/986. Interval List Intersections_test.go
new file mode 100644
index 000000000..8db5161ca
--- /dev/null
+++ b/leetcode/0986.Interval-List-Intersections/986. Interval List Intersections_test.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question986 struct {
+	para986
+	ans986
+}
+
+// para 是参数
+// one 代表第一个参数
+type para986 struct {
+	one []Interval
+	two []Interval
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans986 struct {
+	one []Interval
+}
+
+func Test_Problem986(t *testing.T) {
+
+	qs := []question986{
+
+		{
+			para986{[]Interval{{Start: 0, End: 2}, {Start: 5, End: 10}, {Start: 13, End: 23}, {Start: 24, End: 25}},
+				[]Interval{{Start: 1, End: 5}, {Start: 8, End: 12}, {Start: 15, End: 24}, {Start: 25, End: 26}}},
+			ans986{[]Interval{{Start: 1, End: 2}, {Start: 5, End: 5}, {Start: 8, End: 10},
+				{Start: 15, End: 23}, {Start: 24, End: 24}, {Start: 25, End: 25}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 986------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans986, q.para986
+		fmt.Printf("【input】:%v       【output】:%v\n", p, intervalIntersection(p.one, p.two))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0986. Interval List Intersections/README.md b/leetcode/0986.Interval-List-Intersections/README.md
similarity index 100%
rename from Algorithms/0986. Interval List Intersections/README.md
rename to leetcode/0986.Interval-List-Intersections/README.md
diff --git a/leetcode/0987.Vertical-Order-Traversal-of-a-Binary-Tree/987. Vertical Order Traversal of a Binary Tree.go b/leetcode/0987.Vertical-Order-Traversal-of-a-Binary-Tree/987. Vertical Order Traversal of a Binary Tree.go
new file mode 100644
index 000000000..a7937eb40
--- /dev/null
+++ b/leetcode/0987.Vertical-Order-Traversal-of-a-Binary-Tree/987. Vertical Order Traversal of a Binary Tree.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"math"
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+type node struct {
+	x, y, val int
+}
+
+func verticalTraversal(root *TreeNode) [][]int {
+	var dfs func(root *TreeNode, x, y int)
+	var nodes []node
+	dfs = func(root *TreeNode, x, y int) {
+		if root == nil {
+			return
+		}
+		nodes = append(nodes, node{x, y, root.Val})
+		dfs(root.Left, x+1, y-1)
+		dfs(root.Right, x+1, y+1)
+	}
+	dfs(root, 0, 0)
+
+	sort.Slice(nodes, func(i, j int) bool {
+		a, b := nodes[i], nodes[j]
+		return a.y < b.y || a.y == b.y &&
+			(a.x < b.x || a.x == b.x && a.val < b.val)
+	})
+
+	var res [][]int
+	lastY := math.MinInt32
+	for _, node := range nodes {
+		if lastY != node.y {
+			res = append(res, []int{node.val})
+			lastY = node.y
+		} else {
+			res[len(res)-1] = append(res[len(res)-1], node.val)
+		}
+	}
+	return res
+}
diff --git a/leetcode/0987.Vertical-Order-Traversal-of-a-Binary-Tree/987. Vertical Order Traversal of a Binary Tree_test.go b/leetcode/0987.Vertical-Order-Traversal-of-a-Binary-Tree/987. Vertical Order Traversal of a Binary Tree_test.go
new file mode 100644
index 000000000..669178c13
--- /dev/null
+++ b/leetcode/0987.Vertical-Order-Traversal-of-a-Binary-Tree/987. Vertical Order Traversal of a Binary Tree_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question987 struct {
+	para987
+	ans987
+}
+
+// para 是参数
+// one 代表第一个参数
+type para987 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans987 struct {
+	one [][]int
+}
+
+func Test_Problem987(t *testing.T) {
+
+	qs := []question987{
+
+		{
+			para987{[]int{3, 9, 20, structures.NULL, structures.NULL, 15, 7}},
+			ans987{[][]int{{9}, {3, 15}, {20}, {7}}},
+		},
+
+		{
+			para987{[]int{1, 2, 3, 4, 5, 6, 7}},
+			ans987{[][]int{{4}, {2}, {1, 5, 6}, {3}, {7}}},
+		},
+
+		{
+			para987{[]int{1, 2, 3, 4, 6, 5, 7}},
+			ans987{[][]int{{4}, {2}, {1, 5, 6}, {3}, {7}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 987------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans987, q.para987
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", verticalTraversal(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0987.Vertical-Order-Traversal-of-a-Binary-Tree/README.md b/leetcode/0987.Vertical-Order-Traversal-of-a-Binary-Tree/README.md
new file mode 100644
index 000000000..daa44efed
--- /dev/null
+++ b/leetcode/0987.Vertical-Order-Traversal-of-a-Binary-Tree/README.md
@@ -0,0 +1,135 @@
+# [987. Vertical Order Traversal of a Binary Tree](https://leetcode.com/problems/vertical-order-traversal-of-a-binary-tree/)
+
+
+## 题目
+
+Given the `root` of a binary tree, calculate the **vertical order traversal** of the binary tree.
+
+For each node at position `(row, col)`, its left and right children will be at positions `(row + 1, col - 1)` and `(row + 1, col + 1)` respectively. The root of the tree is at `(0, 0)`.
+
+The **vertical order traversal** of a binary tree is a list of top-to-bottom orderings for each column index starting from the leftmost column and ending on the rightmost column. There may be multiple nodes in the same row and same column. In such a case, sort these nodes by their values.
+
+Return *the **vertical order traversal** of the binary tree*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/01/29/vtree1.jpg](https://assets.leetcode.com/uploads/2021/01/29/vtree1.jpg)
+
+```
+Input: root = [3,9,20,null,null,15,7]
+Output: [[9],[3,15],[20],[7]]
+Explanation:
+Column -1: Only node 9 is in this column.
+Column 0: Nodes 3 and 15 are in this column in that order from top to bottom.
+Column 1: Only node 20 is in this column.
+Column 2: Only node 7 is in this column.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2021/01/29/vtree2.jpg](https://assets.leetcode.com/uploads/2021/01/29/vtree2.jpg)
+
+```
+Input: root = [1,2,3,4,5,6,7]
+Output: [[4],[2],[1,5,6],[3],[7]]
+Explanation:
+Column -2: Only node 4 is in this column.
+Column -1: Only node 2 is in this column.
+Column 0: Nodes 1, 5, and 6 are in this column.
+          1 is at the top, so it comes first.
+          5 and 6 are at the same position (2, 0), so we order them by their value, 5 before 6.
+Column 1: Only node 3 is in this column.
+Column 2: Only node 7 is in this column.
+
+```
+
+**Example 3:**
+
+![https://assets.leetcode.com/uploads/2021/01/29/vtree3.jpg](https://assets.leetcode.com/uploads/2021/01/29/vtree3.jpg)
+
+```
+Input: root = [1,2,3,4,6,5,7]
+Output: [[4],[2],[1,5,6],[3],[7]]
+Explanation:
+This case is the exact same as example 2, but with nodes 5 and 6 swapped.
+Note that the solution remains the same since 5 and 6 are in the same location and should be ordered by their values.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 1000]`.
+- `0 <= Node.val <= 1000`
+
+## 题目大意
+
+给你二叉树的根结点 root ，请你设计算法计算二叉树的 垂序遍历 序列。
+
+对位于 (row, col) 的每个结点而言，其左右子结点分别位于 (row + 1, col - 1) 和 (row + 1, col + 1) 。树的根结点位于 (0, 0) 。二叉树的 垂序遍历 从最左边的列开始直到最右边的列结束，按列索引每一列上的所有结点，形成一个按出现位置从上到下排序的有序列表。如果同行同列上有多个结点，则按结点的值从小到大进行排序。返回二叉树的 垂序遍历 序列。
+
+## 解题思路
+
+- 题目要求按照一列一列的遍历二叉树。需要解决 2 个问题。第一个问题，二叉树上每个结点的二维坐标如何计算。第二个问题，同一个二维坐标点上摞起来多个结点，需要按照从小到大的顺序排序，如例子二和例子三，同一个二维坐标点 (2，0) 上，摞了 2 个不同的结点。
+- 先解决第一个问题，由于题目要求根结点是 (0，0) ，即根结点是坐标原点，它的左子树的 x 坐标都是负数，它的右子树的 x 坐标都是正数。按照先序遍历，就可以将这些结点的二维坐标计算出来。再进行一次排序，按照 x 坐标从小到大排序，坐标相同的情况对应着结点摞起来的情况，摞起来的结点按照 val 值的大小从小到大排序。这样在 x 轴方向，所有结点就排列好了。排序完成，也顺便解决了第二个问题。
+- 最后一步只需要扫描一遍这个排好序的数组，按照列的顺序，依次将同一列的结点打包至一个一维数组中。最终输出的二维数组即为题目所求。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+type node struct {
+	x, y, val int
+}
+
+func verticalTraversal(root *TreeNode) [][]int {
+	var dfs func(root *TreeNode, x, y int)
+	var nodes []node
+	dfs = func(root *TreeNode, x, y int) {
+		if root == nil {
+			return
+		}
+		nodes = append(nodes, node{x, y, root.Val})
+		dfs(root.Left, x+1, y-1)
+		dfs(root.Right, x+1, y+1)
+	}
+	dfs(root, 0, 0)
+
+	sort.Slice(nodes, func(i, j int) bool {
+		a, b := nodes[i], nodes[j]
+		return a.y < b.y || a.y == b.y &&
+			(a.x < b.x || a.x == b.x && a.val < b.val)
+	})
+
+	var res [][]int
+	lastY := math.MinInt32
+	for _, node := range nodes {
+		if lastY != node.y {
+			res = append(res, []int{node.val})
+			lastY = node.y
+		} else {
+			res[len(res)-1] = append(res[len(res)-1], node.val)
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/0989.Add-to-Array-Form-of-Integer/989. Add to Array-Form of Integer.go b/leetcode/0989.Add-to-Array-Form-of-Integer/989. Add to Array-Form of Integer.go
new file mode 100644
index 000000000..4281e5988
--- /dev/null
+++ b/leetcode/0989.Add-to-Array-Form-of-Integer/989. Add to Array-Form of Integer.go	
@@ -0,0 +1,25 @@
+package leetcode
+
+func addToArrayForm(A []int, K int) []int {
+	res := []int{}
+	for i := len(A) - 1; i >= 0; i-- {
+		sum := A[i] + K%10
+		K /= 10
+		if sum >= 10 {
+			K++
+			sum -= 10
+		}
+		res = append(res, sum)
+	}
+	for ; K > 0; K /= 10 {
+		res = append(res, K%10)
+	}
+	reverse(res)
+	return res
+}
+
+func reverse(A []int) {
+	for i, n := 0, len(A); i < n/2; i++ {
+		A[i], A[n-1-i] = A[n-1-i], A[i]
+	}
+}
diff --git a/leetcode/0989.Add-to-Array-Form-of-Integer/989. Add to Array-Form of Integer_test.go b/leetcode/0989.Add-to-Array-Form-of-Integer/989. Add to Array-Form of Integer_test.go
new file mode 100644
index 000000000..93eacd7f2
--- /dev/null
+++ b/leetcode/0989.Add-to-Array-Form-of-Integer/989. Add to Array-Form of Integer_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question989 struct {
+	para989
+	ans989
+}
+
+// para 是参数
+// one 代表第一个参数
+type para989 struct {
+	A []int
+	K int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans989 struct {
+	one []int
+}
+
+func Test_Problem989(t *testing.T) {
+
+	qs := []question989{
+
+		{
+			para989{[]int{1, 2, 0, 0}, 34},
+			ans989{[]int{1, 2, 3, 4}},
+		},
+
+		{
+			para989{[]int{2, 7, 4}, 181},
+			ans989{[]int{4, 5, 5}},
+		},
+
+		{
+			para989{[]int{2, 1, 5}, 806},
+			ans989{[]int{1, 0, 2, 1}},
+		},
+
+		{
+			para989{[]int{9, 9, 9, 9, 9, 9, 9, 9, 9, 9}, 1},
+			ans989{[]int{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 989------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans989, q.para989
+		fmt.Printf("【input】:%v       【output】:%v\n", p, addToArrayForm(p.A, p.K))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0989.Add-to-Array-Form-of-Integer/README.md b/leetcode/0989.Add-to-Array-Form-of-Integer/README.md
new file mode 100644
index 000000000..6f48c336e
--- /dev/null
+++ b/leetcode/0989.Add-to-Array-Form-of-Integer/README.md
@@ -0,0 +1,84 @@
+# [989. Add to Array-Form of Integer](https://leetcode.com/problems/add-to-array-form-of-integer/)
+
+## 题目
+
+For a non-negative integer `X`, the *array-form of `X`* is an array of its digits in left to right order.  For example, if `X = 1231`, then the array form is `[1,2,3,1]`.
+
+Given the array-form `A` of a non-negative integer `X`, return the array-form of the integer `X+K`.
+
+**Example 1:**
+
+```
+Input: A = [1,2,0,0], K = 34
+Output: [1,2,3,4]
+Explanation: 1200 + 34 = 1234
+```
+
+**Example 2:**
+
+```
+Input: A = [2,7,4], K = 181
+Output: [4,5,5]
+Explanation: 274 + 181 = 455
+```
+
+**Example 3:**
+
+```
+Input: A = [2,1,5], K = 806
+Output: [1,0,2,1]
+Explanation: 215 + 806 = 1021
+```
+
+**Example 4:**
+
+```
+Input: A = [9,9,9,9,9,9,9,9,9,9], K = 1
+Output: [1,0,0,0,0,0,0,0,0,0,0]
+Explanation: 9999999999 + 1 = 10000000000
+```
+
+**Note：**
+
+1. `1 <= A.length <= 10000`
+2. `0 <= A[i] <= 9`
+3. `0 <= K <= 10000`
+4. If `A.length > 1`, then `A[0] != 0`
+
+## 题目大意
+
+对于非负整数 X 而言，X 的数组形式是每位数字按从左到右的顺序形成的数组。例如，如果 X = 1231，那么其数组形式为 [1,2,3,1]。给定非负整数 X 的数组形式 A，返回整数 X+K 的数组形式。
+
+## 解题思路
+
+- 简单题，计算 2 个非负整数的和。累加过程中不断的进位，最终输出到数组中记得需要逆序，即数字的高位排在数组下标较小的位置。
+
+## 代码
+
+```go
+package leetcode
+
+func addToArrayForm(A []int, K int) []int {
+	res := []int{}
+	for i := len(A) - 1; i >= 0; i-- {
+		sum := A[i] + K%10
+		K /= 10
+		if sum >= 10 {
+			K++
+			sum -= 10
+		}
+		res = append(res, sum)
+	}
+	for ; K > 0; K /= 10 {
+		res = append(res, K%10)
+	}
+	reverse(res)
+	return res
+}
+
+func reverse(A []int) {
+	for i, n := 0, len(A); i < n/2; i++ {
+		A[i], A[n-1-i] = A[n-1-i], A[i]
+	}
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0990. Satisfiability of Equality Equations/990. Satisfiability of Equality Equations.go b/leetcode/0990.Satisfiability-of-Equality-Equations/990. Satisfiability of Equality Equations.go
similarity index 100%
rename from Algorithms/0990. Satisfiability of Equality Equations/990. Satisfiability of Equality Equations.go
rename to leetcode/0990.Satisfiability-of-Equality-Equations/990. Satisfiability of Equality Equations.go
diff --git a/Algorithms/0990. Satisfiability of Equality Equations/990. Satisfiability of Equality Equations_test.go b/leetcode/0990.Satisfiability-of-Equality-Equations/990. Satisfiability of Equality Equations_test.go
similarity index 92%
rename from Algorithms/0990. Satisfiability of Equality Equations/990. Satisfiability of Equality Equations_test.go
rename to leetcode/0990.Satisfiability-of-Equality-Equations/990. Satisfiability of Equality Equations_test.go
index 75b10c334..77c468dbf 100644
--- a/Algorithms/0990. Satisfiability of Equality Equations/990. Satisfiability of Equality Equations_test.go	
+++ b/leetcode/0990.Satisfiability-of-Equality-Equations/990. Satisfiability of Equality Equations_test.go	
@@ -26,27 +26,27 @@ func Test_Problem990(t *testing.T) {
 
 	qs := []question990{
 
-		question990{
+		{
 			para990{[]string{"a==b", "b!=a"}},
 			ans990{false},
 		},
 
-		question990{
+		{
 			para990{[]string{"b==a", "a==b"}},
 			ans990{true},
 		},
 
-		question990{
+		{
 			para990{[]string{"a==b", "b==c", "a==c"}},
 			ans990{true},
 		},
 
-		question990{
+		{
 			para990{[]string{"a==b", "b!=c", "c==a"}},
 			ans990{false},
 		},
 
-		question990{
+		{
 			para990{[]string{"c==c", "b==d", "x!=z"}},
 			ans990{true},
 		},
diff --git a/leetcode/0990.Satisfiability-of-Equality-Equations/README.md b/leetcode/0990.Satisfiability-of-Equality-Equations/README.md
new file mode 100755
index 000000000..9b619683f
--- /dev/null
+++ b/leetcode/0990.Satisfiability-of-Equality-Equations/README.md
@@ -0,0 +1,65 @@
+# [990. Satisfiability of Equality Equations](https://leetcode.com/problems/satisfiability-of-equality-equations/)
+
+
+## 题目
+
+Given an array equations of strings that represent relationships between variables, each string `equations[i]` has length `4` and takes one of two different forms: `"a==b"` or `"a!=b"`. Here, `a` and `b` are lowercase letters (not necessarily different) that represent one-letter variable names.
+
+Return `true` if and only if it is possible to assign integers to variable names so as to satisfy all the given equations.
+
+**Example 1:**
+
+    Input: ["a==b","b!=a"]
+    Output: false
+    Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second.  There is no way to assign the variables to satisfy both equations.
+
+**Example 2:**
+
+    Input: ["b==a","a==b"]
+    Output: true
+    Explanation: We could assign a = 1 and b = 1 to satisfy both equations.
+
+**Example 3:**
+
+    Input: ["a==b","b==c","a==c"]
+    Output: true
+
+**Example 4:**
+
+    Input: ["a==b","b!=c","c==a"]
+    Output: false
+
+**Example 5:**
+
+    Input: ["c==c","b==d","x!=z"]
+    Output: true
+
+**Note:**
+
+1. `1 <= equations.length <= 500`
+2. `equations[i].length == 4`
+3. `equations[i][0]` and `equations[i][3]` are lowercase letters
+4. `equations[i][1]` is either `'='` or `'!'`
+5. `equations[i][2]` is `'='`
+
+
+
+## 题目大意
+
+给定一个由表示变量之间关系的字符串方程组成的数组，每个字符串方程 equations[i] 的长度为 4，并采用两种不同的形式之一："a==b" 或 "a!=b"。在这里，a 和 b 是小写字母（不一定不同），表示单字母变量名。只有当可以将整数分配给变量名，以便满足所有给定的方程时才返回 true，否则返回 false。 
+
+提示：
+
+1. 1 <= equations.length <= 500
+2. equations[i].length == 4
+3. equations[i][0] 和 equations[i][3] 是小写字母
+4. equations[i][1] 要么是 '='，要么是 '!'
+5. equations[i][2] 是 '='
+
+
+
+## 解题思路
+
+
+- 给出一个字符串数组，数组里面给出的是一些字母的关系，只有 `'=='` 和 `'! ='` 两种关系。问给出的这些关系中是否存在悖论？
+- 这一题是简单的并查集的问题。先将所有 `'=='` 关系的字母 `union()` 起来，然后再一一查看 `'! ='` 关系中是否有 `'=='` 关系的组合，如果有，就返回 `false`，如果遍历完都没有找到，则返回 `true`。
diff --git a/leetcode/0991.Broken-Calculator/991. Broken Calculator.go b/leetcode/0991.Broken-Calculator/991. Broken Calculator.go
new file mode 100644
index 000000000..3e9e2b9a0
--- /dev/null
+++ b/leetcode/0991.Broken-Calculator/991. Broken Calculator.go	
@@ -0,0 +1,14 @@
+package leetcode
+
+func brokenCalc(X int, Y int) int {
+	res := 0
+	for Y > X {
+		res++
+		if Y&1 == 1 {
+			Y++
+		} else {
+			Y /= 2
+		}
+	}
+	return res + X - Y
+}
diff --git a/leetcode/0991.Broken-Calculator/991. Broken Calculator_test.go b/leetcode/0991.Broken-Calculator/991. Broken Calculator_test.go
new file mode 100644
index 000000000..c45f7d6ec
--- /dev/null
+++ b/leetcode/0991.Broken-Calculator/991. Broken Calculator_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question991 struct {
+	para991
+	ans991
+}
+
+// para 是参数
+// one 代表第一个参数
+type para991 struct {
+	X int
+	Y int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans991 struct {
+	one int
+}
+
+func Test_Problem991(t *testing.T) {
+
+	qs := []question991{
+
+		{
+			para991{2, 3},
+			ans991{2},
+		},
+
+		{
+			para991{5, 8},
+			ans991{2},
+		},
+
+		{
+			para991{3, 10},
+			ans991{3},
+		},
+
+		{
+			para991{1024, 1},
+			ans991{1023},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 991------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans991, q.para991
+		fmt.Printf("【input】:%v       【output】:%v\n", p, brokenCalc(p.X, p.Y))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0991.Broken-Calculator/README.md b/leetcode/0991.Broken-Calculator/README.md
new file mode 100644
index 000000000..4ef8bc204
--- /dev/null
+++ b/leetcode/0991.Broken-Calculator/README.md
@@ -0,0 +1,83 @@
+# [991. Broken Calculator](https://leetcode.com/problems/broken-calculator/)
+
+
+## 题目
+
+On a broken calculator that has a number showing on its display, we can perform two operations:
+
+- **Double**: Multiply the number on the display by 2, or;
+- **Decrement**: Subtract 1 from the number on the display.
+
+Initially, the calculator is displaying the number `X`.
+
+Return the minimum number of operations needed to display the number `Y`.
+
+**Example 1:**
+
+```
+Input: X = 2, Y = 3
+Output: 2
+Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.
+```
+
+**Example 2:**
+
+```
+Input: X = 5, Y = 8
+Output: 2
+Explanation: Use decrement and then double {5 -> 4 -> 8}.
+```
+
+**Example 3:**
+
+```
+Input: X = 3, Y = 10
+Output: 3
+Explanation:  Use double, decrement and double {3 -> 6 -> 5 -> 10}.
+```
+
+**Example 4:**
+
+```
+Input: X = 1024, Y = 1
+Output: 1023
+Explanation: Use decrement operations 1023 times.
+```
+
+**Note:**
+
+1. `1 <= X <= 10^9`
+2. `1 <= Y <= 10^9`
+
+## 题目大意
+
+在显示着数字的坏计算器上，我们可以执行以下两种操作：
+
+- 双倍（Double）：将显示屏上的数字乘 2；
+- 递减（Decrement）：将显示屏上的数字减 1 。
+
+最初，计算器显示数字 X。返回显示数字 Y 所需的最小操作数。
+
+## 解题思路
+
+- 看到本题的数据规模非常大，`10^9`，算法只能采用 `O(sqrt(n))`、`O(log n)`、`O(1)` 的算法。`O(sqrt(n))` 和 `O(1)` 在本题中是不可能的。所以按照数据规模来估计，本题只能尝试 `O(log n)` 的算法。`O(log n)` 的算法有二分搜索，不过本题不太符合二分搜索算法背景。题目中明显出现乘 2，这很明显是可以达到 `O(log n)` 的。最终确定解题思路是数学方法，循环中会用到乘 2 或者除 2 的计算。
+- 既然出现了乘 2 和减一的操作，很容易考虑到奇偶性上。题目要求最小操作数，贪心思想，应该尽可能多的使用除 2 操作，使得 Y 和 X 大小差不多，最后再利用加一操作微调。只要 Y 比 X 大就执行除法操作。当然这里要考虑一次奇偶性，如果 Y 是奇数，先加一变成偶数再除二；如果 Y 是偶数，直接除二。如此操作直到 Y 不大于 X，最后执行 `X-Y` 次加法操作微调即可。
+
+## 代码
+
+```go
+package leetcode
+
+func brokenCalc(X int, Y int) int {
+	res := 0
+	for Y > X {
+		res++
+		if Y&1 == 1 {
+			Y++
+		} else {
+			Y /= 2
+		}
+	}
+	return res + X - Y
+}
+```
\ No newline at end of file
diff --git a/Algorithms/0992. Subarrays with K Different Integers/992. Subarrays with K Different Integers.go b/leetcode/0992.Subarrays-with-K-Different-Integers/992. Subarrays with K Different Integers.go
similarity index 100%
rename from Algorithms/0992. Subarrays with K Different Integers/992. Subarrays with K Different Integers.go
rename to leetcode/0992.Subarrays-with-K-Different-Integers/992. Subarrays with K Different Integers.go
diff --git a/Algorithms/0992. Subarrays with K Different Integers/992. Subarrays with K Different Integers_test.go b/leetcode/0992.Subarrays-with-K-Different-Integers/992. Subarrays with K Different Integers_test.go
similarity index 90%
rename from Algorithms/0992. Subarrays with K Different Integers/992. Subarrays with K Different Integers_test.go
rename to leetcode/0992.Subarrays-with-K-Different-Integers/992. Subarrays with K Different Integers_test.go
index 01eb4a71a..6375cee2e 100644
--- a/Algorithms/0992. Subarrays with K Different Integers/992. Subarrays with K Different Integers_test.go	
+++ b/leetcode/0992.Subarrays-with-K-Different-Integers/992. Subarrays with K Different Integers_test.go	
@@ -27,37 +27,37 @@ func Test_Problem992(t *testing.T) {
 
 	qs := []question992{
 
-		question992{
+		{
 			para992{[]int{1, 1, 1, 1, 1, 1, 1, 1}, 1},
 			ans992{36},
 		},
 
-		question992{
+		{
 			para992{[]int{2, 1, 1, 1, 2}, 1},
 			ans992{8},
 		},
 
-		question992{
+		{
 			para992{[]int{1, 2}, 1},
 			ans992{2},
 		},
 
-		question992{
+		{
 			para992{[]int{1, 2, 1, 2, 3}, 2},
 			ans992{7},
 		},
 
-		question992{
+		{
 			para992{[]int{1, 2, 1, 3, 4}, 3},
 			ans992{3},
 		},
 
-		question992{
+		{
 			para992{[]int{1}, 5},
 			ans992{1},
 		},
 
-		question992{
+		{
 			para992{[]int{}, 10},
 			ans992{0},
 		},
diff --git a/Algorithms/0992. Subarrays with K Different Integers/README.md b/leetcode/0992.Subarrays-with-K-Different-Integers/README.md
similarity index 100%
rename from Algorithms/0992. Subarrays with K Different Integers/README.md
rename to leetcode/0992.Subarrays-with-K-Different-Integers/README.md
diff --git a/Algorithms/0993. Cousins in Binary Tree/993. Cousins in Binary Tree.go b/leetcode/0993.Cousins-in-Binary-Tree/993. Cousins in Binary Tree.go
similarity index 95%
rename from Algorithms/0993. Cousins in Binary Tree/993. Cousins in Binary Tree.go
rename to leetcode/0993.Cousins-in-Binary-Tree/993. Cousins in Binary Tree.go
index f5a520c06..33a9e1cee 100644
--- a/Algorithms/0993. Cousins in Binary Tree/993. Cousins in Binary Tree.go	
+++ b/leetcode/0993.Cousins-in-Binary-Tree/993. Cousins in Binary Tree.go	
@@ -1,5 +1,12 @@
 package leetcode
 
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
 /**
  * Definition for a binary tree node.
  * type TreeNode struct {
diff --git a/Algorithms/0993. Cousins in Binary Tree/993. Cousins in Binary Tree_test.go b/leetcode/0993.Cousins-in-Binary-Tree/993. Cousins in Binary Tree_test.go
similarity index 75%
rename from Algorithms/0993. Cousins in Binary Tree/993. Cousins in Binary Tree_test.go
rename to leetcode/0993.Cousins-in-Binary-Tree/993. Cousins in Binary Tree_test.go
index 90b91cf61..d0afbfc4c 100644
--- a/Algorithms/0993. Cousins in Binary Tree/993. Cousins in Binary Tree_test.go	
+++ b/leetcode/0993.Cousins-in-Binary-Tree/993. Cousins in Binary Tree_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question993 struct {
@@ -28,18 +30,18 @@ func Test_Problem993(t *testing.T) {
 
 	qs := []question993{
 
-		question993{
+		{
 			para993{[]int{1, 2, 3, 4}, 4, 3},
 			ans993{false},
 		},
 
-		question993{
-			para993{[]int{1, 2, 3, NULL, 4, NULL, 5}, 5, 4},
+		{
+			para993{[]int{1, 2, 3, structures.NULL, 4, structures.NULL, 5}, 5, 4},
 			ans993{true},
 		},
 
-		question993{
-			para993{[]int{1, 2, 3, NULL, 4}, 2, 3},
+		{
+			para993{[]int{1, 2, 3, structures.NULL, 4}, 2, 3},
 			ans993{false},
 		},
 	}
@@ -49,7 +51,7 @@ func Test_Problem993(t *testing.T) {
 	for _, q := range qs {
 		_, p := q.ans993, q.para993
 		fmt.Printf("【input】:%v      ", p)
-		root := Ints2TreeNode(p.one)
+		root := structures.Ints2TreeNode(p.one)
 		fmt.Printf("【output】:%v      \n", isCousins(root, p.x, p.y))
 	}
 	fmt.Printf("\n\n\n")
diff --git a/Algorithms/0993. Cousins in Binary Tree/README.md b/leetcode/0993.Cousins-in-Binary-Tree/README.md
similarity index 100%
rename from Algorithms/0993. Cousins in Binary Tree/README.md
rename to leetcode/0993.Cousins-in-Binary-Tree/README.md
diff --git a/Algorithms/0995. Minimum Number of K Consecutive Bit Flips/995. Minimum Number of K Consecutive Bit Flips.go b/leetcode/0995.Minimum-Number-of-K-Consecutive-Bit-Flips/995. Minimum Number of K Consecutive Bit Flips.go
similarity index 100%
rename from Algorithms/0995. Minimum Number of K Consecutive Bit Flips/995. Minimum Number of K Consecutive Bit Flips.go
rename to leetcode/0995.Minimum-Number-of-K-Consecutive-Bit-Flips/995. Minimum Number of K Consecutive Bit Flips.go
diff --git a/Algorithms/0995. Minimum Number of K Consecutive Bit Flips/995. Minimum Number of K Consecutive Bit Flips_test.go b/leetcode/0995.Minimum-Number-of-K-Consecutive-Bit-Flips/995. Minimum Number of K Consecutive Bit Flips_test.go
similarity index 94%
rename from Algorithms/0995. Minimum Number of K Consecutive Bit Flips/995. Minimum Number of K Consecutive Bit Flips_test.go
rename to leetcode/0995.Minimum-Number-of-K-Consecutive-Bit-Flips/995. Minimum Number of K Consecutive Bit Flips_test.go
index 9027b10cf..41955d863 100644
--- a/Algorithms/0995. Minimum Number of K Consecutive Bit Flips/995. Minimum Number of K Consecutive Bit Flips_test.go	
+++ b/leetcode/0995.Minimum-Number-of-K-Consecutive-Bit-Flips/995. Minimum Number of K Consecutive Bit Flips_test.go	
@@ -27,17 +27,17 @@ func Test_Problem995(t *testing.T) {
 
 	qs := []question995{
 
-		question995{
+		{
 			para995{[]int{0, 1, 0}, 1},
 			ans995{2},
 		},
 
-		question995{
+		{
 			para995{[]int{1, 1, 0}, 2},
 			ans995{-1},
 		},
 
-		question995{
+		{
 			para995{[]int{0, 0, 0, 1, 0, 1, 1, 0}, 3},
 			ans995{3},
 		},
diff --git a/leetcode/0995.Minimum-Number-of-K-Consecutive-Bit-Flips/README.md b/leetcode/0995.Minimum-Number-of-K-Consecutive-Bit-Flips/README.md
new file mode 100755
index 000000000..1c0d06ef0
--- /dev/null
+++ b/leetcode/0995.Minimum-Number-of-K-Consecutive-Bit-Flips/README.md
@@ -0,0 +1,67 @@
+# [995. Minimum Number of K Consecutive Bit Flips](https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips/)
+
+
+## 题目
+
+In an array `A` containing only 0s and 1s, a `K`-bit flip consists of choosing a (contiguous) subarray of length `K` and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0.
+
+Return the minimum number of `K`-bit flips required so that there is no 0 in the array. If it is not possible, return `-1`.
+
+**Example 1:**
+
+    Input: A = [0,1,0], K = 1
+    Output: 2
+    Explanation: Flip A[0], then flip A[2].
+
+**Example 2:**
+
+    Input: A = [1,1,0], K = 2
+    Output: -1
+    Explanation: No matter how we flip subarrays of size 2, we can't make the array become [1,1,1].
+
+**Example 3:**
+
+    Input: A = [0,0,0,1,0,1,1,0], K = 3
+    Output: 3
+    Explanation:
+    Flip A[0],A[1],A[2]: A becomes [1,1,1,1,0,1,1,0]
+    Flip A[4],A[5],A[6]: A becomes [1,1,1,1,1,0,0,0]
+    Flip A[5],A[6],A[7]: A becomes [1,1,1,1,1,1,1,1]
+
+**Note:**
+
+1. `1 <= A.length <= 30000`
+2. `1 <= K <= A.length`
+
+
+## 题目大意
+
+在仅包含 0 和 1 的数组 A 中，一次 K 位翻转包括选择一个长度为 K 的（连续）子数组，同时将子数组中的每个 0 更改为 1，而每个 1 更改为 0。返回所需的 K 位翻转的次数，以便数组没有值为 0 的元素。如果不可能，返回 -1。
+
+提示：
+
+1. 1 <= A.length <= 30000
+2. 1 <= K <= A.length
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面的元素只有 0 和 1。给一个长度为 K 的窗口，在这个窗口内的所有元素都会 0-1 翻转。问最后需要翻转几次，使得整个数组都为 1 。如果不能翻转使得最后数组元素都为 1，则输出 -1。
+- 拿到这一题首先想到的是贪心算法。例如第 765 题，这类题的描述都是这样的：在一个数组中或者环形数组中通过交换位置，或者翻转变换，达到最终结果，要求找到最少步数。贪心能保证是最小步数(证明略)。按照贪心的思想，这一题也这样做，从数组 0 下标开始往后扫，依次翻转每个 K 大小的窗口内元素。
+- 由于窗口大小限制了，所以这题滑动窗口只需要一个边界坐标，用左边界就可以判断了。每一个 `A[i]` 是否需要翻转，是由于 `[ i-k+1，i ]`、`[ i-k+2，i+1 ]`、`[ i-k+3，i+2 ]`……`[ i-1，i+k ]` 这一系列的窗口翻转`累积影响`的。那如何之前这些窗口`累积`到 `A[i]` 上翻转的次数呢？可以动态的维护一个翻转次数，当 `i` 摆脱了上一次翻转窗口 `K` 的时候，翻转次数就 `-1` 。举个例子：
+
+        A = [0 0 0 1 0 1 1 0] K = 3
+        
+        A = [2 0 0 1 0 1 1 0] i = 0 flippedTime = 1
+        A = [2 0 0 1 0 1 1 0] i = 1 flippedTime = 1
+        A = [2 0 0 1 0 1 1 0] i = 2 flippedTime = 1
+        A = [2 0 0 1 0 1 1 0] i = 3 flippedTime = 0
+        A = [2 0 0 1 2 1 1 0] i = 4 flippedTime = 1
+        A = [2 0 0 1 2 2 1 0] i = 5 flippedTime = 2
+        A = [2 0 0 1 2 2 1 0] i = 6 flippedTime = 2
+        A = [2 0 0 1 2 2 1 0] i = 7 flippedTime = 1
+
+    当判断 `A[i]` 是否需要翻转的时候，只需要留意每个宽度为 `K` 窗口的左边界。会影响到 A[i] 的窗口的左边界分别是 `i-k+1`、`i-k+2`、`i-k+3`、…… `i-1`，只需要分别看这些窗口有没有翻转就行。这里可以用特殊标记来记录这些窗口的左边界是否被翻转了。如果翻转过，则把窗口左边界的那个数字标记为 2 (为什么要标记为 2 呢？其实设置成什么都可以，只要不是 0 和 1 ，和原有的数字区分开就行)。当 `i≥k` 的时候，代表 `i` 已经脱离了 `i-k` 的这个窗口，因为能影响 `A[i]` 的窗口是从 `i-k+1` 开始的，如果 `A[i-k] == 2` 代表 `i-k` 窗口已经翻转过了，现在既然脱离了它的窗口影响，那么就要把累积的 `flippedTime - 1` 。这样就维护了累积 `flippedTime` 和滑动窗口中累积影响的关系。
+
+- 接下来还需要处理的是 `flippedTime` 与当前 `A[i]` 是否翻转的问题。如果 `flippedTime` 是偶数次，原来的 0 还是 0，就需要再次翻转，如果 `flippedTime` 是奇数次，原来的 0 变成了 1 就不需要翻转了。总结成一条结论就是 `A[i]` 与 `flippedTime` 同奇偶性的时候就要翻转。当 `i + K` 比 `len(A)` 大的时候，代表剩下的这些元素肯定不能在一个窗口里面翻转，则输出 -1 。
diff --git a/Algorithms/0996. Number of Squareful Arrays/996. Number of Squareful Arrays.go b/leetcode/0996.Number-of-Squareful-Arrays/996. Number of Squareful Arrays.go
similarity index 100%
rename from Algorithms/0996. Number of Squareful Arrays/996. Number of Squareful Arrays.go
rename to leetcode/0996.Number-of-Squareful-Arrays/996. Number of Squareful Arrays.go
diff --git a/Algorithms/0996. Number of Squareful Arrays/996. Number of Squareful Arrays_test.go b/leetcode/0996.Number-of-Squareful-Arrays/996. Number of Squareful Arrays_test.go
similarity index 93%
rename from Algorithms/0996. Number of Squareful Arrays/996. Number of Squareful Arrays_test.go
rename to leetcode/0996.Number-of-Squareful-Arrays/996. Number of Squareful Arrays_test.go
index 5e65695d2..030e1fb97 100644
--- a/Algorithms/0996. Number of Squareful Arrays/996. Number of Squareful Arrays_test.go	
+++ b/leetcode/0996.Number-of-Squareful-Arrays/996. Number of Squareful Arrays_test.go	
@@ -26,22 +26,22 @@ func Test_Problem996(t *testing.T) {
 
 	qs := []question996{
 
-		question996{
+		{
 			para996{[]int{1, 17, 8}},
 			ans996{2},
 		},
 
-		question996{
+		{
 			para996{[]int{1}},
 			ans996{1},
 		},
 
-		question996{
+		{
 			para996{[]int{2, 2, 2}},
 			ans996{1},
 		},
 
-		question996{
+		{
 			para996{[]int{51768, 47861, 48143, 33221, 50893, 56758, 39946, 10312, 20276, 40616, 43633}},
 			ans996{1},
 		},
diff --git a/Algorithms/0996. Number of Squareful Arrays/README.md b/leetcode/0996.Number-of-Squareful-Arrays/README.md
similarity index 100%
rename from Algorithms/0996. Number of Squareful Arrays/README.md
rename to leetcode/0996.Number-of-Squareful-Arrays/README.md
diff --git a/leetcode/0997.Find-the-Town-Judge/997.Find the Town Judge.go b/leetcode/0997.Find-the-Town-Judge/997.Find the Town Judge.go
new file mode 100644
index 000000000..79351288d
--- /dev/null
+++ b/leetcode/0997.Find-the-Town-Judge/997.Find the Town Judge.go	
@@ -0,0 +1,22 @@
+package leetcode
+
+func findJudge(n int, trust [][]int) int {
+	if n == 1 && len(trust) == 0 {
+		return 1
+	}
+	judges := make(map[int]int)
+	for _, v := range trust {
+		judges[v[1]] += 1
+	}
+	for _, v := range trust {
+		if _, ok := judges[v[0]]; ok {
+			delete(judges, v[0])
+		}
+	}
+	for k, v := range judges {
+		if v == n-1 {
+			return k
+		}
+	}
+	return -1
+}
diff --git a/leetcode/0997.Find-the-Town-Judge/997.Find the Town Judge_test.go b/leetcode/0997.Find-the-Town-Judge/997.Find the Town Judge_test.go
new file mode 100644
index 000000000..61b2434ff
--- /dev/null
+++ b/leetcode/0997.Find-the-Town-Judge/997.Find the Town Judge_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question997 struct {
+	para997
+	ans997
+}
+
+// para 是参数
+type para997 struct {
+	n     int
+	trust [][]int
+}
+
+// ans 是答案
+type ans997 struct {
+	ans int
+}
+
+func Test_Problem997(t *testing.T) {
+
+	qs := []question997{
+
+		{
+			para997{2, [][]int{{1, 2}}},
+			ans997{2},
+		},
+
+		{
+			para997{3, [][]int{{1, 3}, {2, 3}}},
+			ans997{3},
+		},
+
+		{
+			para997{3, [][]int{{1, 3}, {2, 3}, {3, 1}}},
+			ans997{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 997------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans997, q.para997
+		fmt.Printf("【input】:%v    【output】:%v\n", p, findJudge(p.n, p.trust))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0997.Find-the-Town-Judge/README.md b/leetcode/0997.Find-the-Town-Judge/README.md
new file mode 100644
index 000000000..71dc5ecfc
--- /dev/null
+++ b/leetcode/0997.Find-the-Town-Judge/README.md
@@ -0,0 +1,87 @@
+# [997. Find the Town Judge](https://leetcode.com/problems/find-the-town-judge/)
+
+## 题目
+
+In a town, there are n people labeled from 1 to n. There is a rumor that one of these people is secretly the town judge.
+
+If the town judge exists, then:
+
+- The town judge trusts nobody.
+- Everybody (except for the town judge) trusts the town judge.
+- There is exactly one person that satisfies properties 1 and 2.
+
+You are given an array trust where trust[i] = [ai, bi] representing that the person labeled ai trusts the person labeled bi.
+
+Return the label of the town judge if the town judge exists and can be identified, or return -1 otherwise.
+
+**Example 1**:
+
+    Input: n = 2, trust = [[1,2]]
+    Output: 2
+
+**Example 2**:
+
+    Input: n = 3, trust = [[1,3],[2,3]]
+    Output: 3
+
+**Example 3**:
+
+    Input: n = 3, trust = [[1,3],[2,3],[3,1]]
+    Output: -1
+
+**Constraints:**
+
+- 1 <= n <= 1000
+- 0 <= trust.length <= 10000
+- trust[i].length == 2
+- All the pairs of trust are unique.
+- ai != bi
+- 1 <= ai, bi <= n
+
+## 题目大意
+
+小镇里有 n 个人，按从 1 到 n 的顺序编号。传言称，这些人中有一个暗地里是小镇法官。
+
+如果小镇法官真的存在，那么：
+
+- 小镇法官不会信任任何人。
+- 每个人（除了小镇法官）都信任这位小镇法官。
+- 只有一个人同时满足属性 1 和属性 2 。
+
+给你一个数组 trust ，其中 trust[i] = [ai, bi] 表示编号为 ai 的人信任编号为 bi 的人。
+
+如果小镇法官存在并且可以确定他的身份，请返回该法官的编号；否则，返回 -1 。
+
+## 解题思路
+
+入度和出度统计
+
+- 被人信任定义为入度, 信任别人定义为出度
+- 如果 1-n 之间有数字 x 的入度为 n - 1，出度为 0，则返回 x
+
+## 代码
+
+```go
+package leetcode
+
+func findJudge(n int, trust [][]int) int {
+	if n == 1 && len(trust) == 0 {
+		return 1
+	}
+	judges := make(map[int]int)
+	for _, v := range trust {
+		judges[v[1]] += 1
+	}
+	for _, v := range trust {
+		if _, ok := judges[v[0]]; ok {
+			delete(judges, v[0])
+		}
+	}
+	for k, v := range judges {
+		if v == n-1 {
+			return k
+		}
+	}
+	return -1
+}
+```
\ No newline at end of file
diff --git a/leetcode/0999.Available-Captures-for-Rook/999. Available Captures for Rook.go b/leetcode/0999.Available-Captures-for-Rook/999. Available Captures for Rook.go
new file mode 100644
index 000000000..cb17f2a02
--- /dev/null
+++ b/leetcode/0999.Available-Captures-for-Rook/999. Available Captures for Rook.go	
@@ -0,0 +1,27 @@
+package leetcode
+
+func numRookCaptures(board [][]byte) int {
+	num := 0
+	for i := 0; i < len(board); i++ {
+		for j := 0; j < len(board[i]); j++ {
+			if board[i][j] == 'R' {
+				num += caputure(board, i-1, j, -1, 0) // Up
+				num += caputure(board, i+1, j, 1, 0)  // Down
+				num += caputure(board, i, j-1, 0, -1) // Left
+				num += caputure(board, i, j+1, 0, 1)  // Right
+			}
+		}
+	}
+	return num
+}
+
+func caputure(board [][]byte, x, y int, bx, by int) int {
+	for x >= 0 && x < len(board) && y >= 0 && y < len(board[x]) && board[x][y] != 'B' {
+		if board[x][y] == 'p' {
+			return 1
+		}
+		x += bx
+		y += by
+	}
+	return 0
+}
diff --git a/leetcode/0999.Available-Captures-for-Rook/999. Available Captures for Rook_test.go b/leetcode/0999.Available-Captures-for-Rook/999. Available Captures for Rook_test.go
new file mode 100644
index 000000000..9c6bb712f
--- /dev/null
+++ b/leetcode/0999.Available-Captures-for-Rook/999. Available Captures for Rook_test.go	
@@ -0,0 +1,79 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question999 struct {
+	para999
+	ans999
+}
+
+// para 是参数
+// one 代表第一个参数
+type para999 struct {
+	one [][]byte
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans999 struct {
+	one int
+}
+
+func Test_Problem999(t *testing.T) {
+
+	qs := []question999{
+
+		{
+			para999{[][]byte{
+				{'.', '.', '.', '.', '.', '.', '.', '.'},
+				{'.', 'p', 'p', 'p', 'p', 'p', '.', '.'},
+				{'.', 'p', 'p', 'B', 'p', 'p', '.', '.'},
+				{'.', 'p', 'B', 'R', 'B', 'p', '.', '.'},
+				{'.', 'p', 'p', 'B', 'p', 'p', '.', '.'},
+				{'.', 'p', 'p', 'p', 'p', 'p', '.', '.'},
+				{'.', '.', '.', '.', '.', '.', '.', '.'},
+				{'.', '.', '.', '.', '.', '.', '.', '.'},
+			}},
+			ans999{0},
+		},
+
+		{
+			para999{[][]byte{
+				{'.', '.', '.', '.', '.', '.', '.', '.'},
+				{'.', '.', '.', 'p', '.', '.', '.', '.'},
+				{'.', '.', '.', 'R', '.', '.', '.', 'p'},
+				{'.', '.', '.', '.', '.', '.', '.', '.'},
+				{'.', '.', '.', '.', '.', '.', '.', '.'},
+				{'.', '.', '.', 'p', '.', '.', '.', '.'},
+				{'.', '.', '.', '.', '.', '.', '.', '.'},
+				{'.', '.', '.', '.', '.', '.', '.', '.'},
+			}},
+			ans999{3},
+		},
+
+		{
+			para999{[][]byte{
+				{'.', '.', '.', '.', '.', '.', '.', '.'},
+				{'.', '.', '.', 'p', '.', '.', '.', '.'},
+				{'.', '.', '.', 'p', '.', '.', '.', '.'},
+				{'p', 'p', '.', 'R', '.', 'p', 'B', '.'},
+				{'.', '.', '.', '.', '.', '.', '.', '.'},
+				{'.', '.', '.', 'B', '.', '.', '.', '.'},
+				{'.', '.', '.', 'p', '.', '.', '.', '.'},
+				{'.', '.', '.', '.', '.', '.', '.', '.'},
+			}},
+			ans999{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 999------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans999, q.para999
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numRookCaptures(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0999.Available-Captures-for-Rook/README.md b/leetcode/0999.Available-Captures-for-Rook/README.md
new file mode 100644
index 000000000..1edb4a0f8
--- /dev/null
+++ b/leetcode/0999.Available-Captures-for-Rook/README.md
@@ -0,0 +1,97 @@
+# [999. Available Captures for Rook](https://leetcode.com/problems/available-captures-for-rook/)
+
+
+## 题目
+
+On an 8 x 8 chessboard, there is one white rook. There also may be empty squares, white bishops, and black pawns. These are given as characters 'R', '.', 'B', and 'p' respectively. Uppercase characters represent white pieces, and lowercase characters represent black pieces.
+
+The rook moves as in the rules of Chess: it chooses one of four cardinal directions (north, east, west, and south), then moves in that direction until it chooses to stop, reaches the edge of the board, or captures an opposite colored pawn by moving to the same square it occupies. Also, rooks cannot move into the same square as other friendly bishops.
+
+Return the number of pawns the rook can capture in one move.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/02/20/1253_example_1_improved.PNG](https://assets.leetcode.com/uploads/2019/02/20/1253_example_1_improved.PNG)
+
+```
+Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
+Output: 3
+Explanation: 
+In this example the rook is able to capture all the pawns.
+```
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/02/19/1253_example_2_improved.PNG](https://assets.leetcode.com/uploads/2019/02/19/1253_example_2_improved.PNG)
+
+```
+Input: [[".",".",".",".",".",".",".","."],[".","p","p","p","p","p",".","."],[".","p","p","B","p","p",".","."],[".","p","B","R","B","p",".","."],[".","p","p","B","p","p",".","."],[".","p","p","p","p","p",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
+Output: 0
+Explanation: 
+Bishops are blocking the rook to capture any pawn.
+```
+
+**Example 3**:
+
+![https://assets.leetcode.com/uploads/2019/02/20/1253_example_3_improved.PNG](https://assets.leetcode.com/uploads/2019/02/20/1253_example_3_improved.PNG)
+
+```
+Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
+Output: 3
+Explanation: 
+The rook can capture the pawns at positions b5, d6 and f5.
+```
+
+**Note**:
+
+1. `board.length == board[i].length == 8`
+2. `board[i][j]` is either `'R'`, `'.'`, `'B'`, or `'p'`
+3. There is exactly one cell with `board[i][j] == 'R'`
+
+## 题目大意
+
+在一个 8 x 8 的棋盘上，有一个白色的车（Rook），用字符 'R' 表示。棋盘上还可能存在空方块，白色的象（Bishop）以及黑色的卒（pawn），分别用字符 '.'，'B' 和 'p' 表示。不难看出，大写字符表示的是白棋，小写字符表示的是黑棋。车按国际象棋中的规则移动。东，西，南，北四个基本方向任选其一，然后一直向选定的方向移动，直到满足下列四个条件之一：
+
+- 棋手选择主动停下来。
+- 棋子因到达棋盘的边缘而停下。
+- 棋子移动到某一方格来捕获位于该方格上敌方（黑色）的卒，停在该方格内。
+- 车不能进入/越过已经放有其他友方棋子（白色的象）的方格，停在友方棋子前。
+
+你现在可以控制车移动一次，请你统计有多少敌方的卒处于你的捕获范围内（即，可以被一步捕获的棋子数）。
+
+## 解题思路
+
+- 按照国际象棋的规则移动车，要求输出只移动一次，有多少个卒在车的捕获范围之内
+- 简单题，按照国际象棋车的移动规则， 4 个方向分别枚举即可。
+
+## 代码
+
+```go
+package leetcode
+
+func numRookCaptures(board [][]byte) int {
+	num := 0
+	for i := 0; i < len(board); i++ {
+		for j := 0; j < len(board[i]); j++ {
+			if board[i][j] == 'R' {
+				num += caputure(board, i-1, j, -1, 0) // Up
+				num += caputure(board, i+1, j, 1, 0)  // Down
+				num += caputure(board, i, j-1, 0, -1) // Left
+				num += caputure(board, i, j+1, 0, 1)  // Right
+			}
+		}
+	}
+	return num
+}
+
+func caputure(board [][]byte, x, y int, bx, by int) int {
+	for x >= 0 && x < len(board) && y >= 0 && y < len(board[x]) && board[x][y] != 'B' {
+		if board[x][y] == 'p' {
+			return 1
+		}
+		x += bx
+		y += by
+	}
+	return 0
+}
+```
\ No newline at end of file
diff --git a/Algorithms/1002. Find Common Characters/1002. Find Common Characters.go b/leetcode/1002.Find-Common-Characters/1002. Find Common Characters.go
similarity index 93%
rename from Algorithms/1002. Find Common Characters/1002. Find Common Characters.go
rename to leetcode/1002.Find-Common-Characters/1002. Find Common Characters.go
index c13171bd3..acae8f889 100644
--- a/Algorithms/1002. Find Common Characters/1002. Find Common Characters.go	
+++ b/leetcode/1002.Find-Common-Characters/1002. Find Common Characters.go	
@@ -26,7 +26,7 @@ func commonChars(A []string) []string {
 	result := make([]string, 0)
 	for i := 0; i < 26; i++ {
 		for j := 0; j < cnt[i]; j++ {
-			result = append(result, string(i+'a'))
+			result = append(result, string(rune(i+'a')))
 		}
 	}
 	return result
diff --git a/Algorithms/1002. Find Common Characters/1002. Find Common Characters_test.go b/leetcode/1002.Find-Common-Characters/1002. Find Common Characters_test.go
similarity index 95%
rename from Algorithms/1002. Find Common Characters/1002. Find Common Characters_test.go
rename to leetcode/1002.Find-Common-Characters/1002. Find Common Characters_test.go
index 284ac09ba..e03e8a6a2 100644
--- a/Algorithms/1002. Find Common Characters/1002. Find Common Characters_test.go	
+++ b/leetcode/1002.Find-Common-Characters/1002. Find Common Characters_test.go	
@@ -26,12 +26,12 @@ func Test_Problem1002(t *testing.T) {
 
 	qs := []question1002{
 
-		question1002{
+		{
 			para1002{[]string{"bella", "label", "roller"}},
 			ans1002{[]string{"e", "l", "l"}},
 		},
 
-		question1002{
+		{
 			para1002{[]string{"cool", "lock", "cook"}},
 			ans1002{[]string{"c", "o"}},
 		},
diff --git a/leetcode/1002.Find-Common-Characters/README.md b/leetcode/1002.Find-Common-Characters/README.md
new file mode 100755
index 000000000..ced9fa2ce
--- /dev/null
+++ b/leetcode/1002.Find-Common-Characters/README.md
@@ -0,0 +1,33 @@
+# [1002. Find Common Characters](https://leetcode.com/problems/find-common-characters/)
+
+
+## 题目
+
+Given an array `A` of strings made only from lowercase letters, return a list of all characters that show up in all strings within the list **(including duplicates)**. For example, if a character occurs 3 times in all strings but not 4 times, you need to include that character three times in the final answer.
+
+You may return the answer in any order.
+
+**Example 1:**
+
+    Input: ["bella","label","roller"]
+    Output: ["e","l","l"]
+
+**Example 2:**
+
+    Input: ["cool","lock","cook"]
+    Output: ["c","o"]
+
+**Note:**
+
+1. `1 <= A.length <= 100`
+2. `1 <= A[i].length <= 100`
+3. `A[i][j]` is a lowercase letter
+
+## 题目大意
+
+给定仅有小写字母组成的字符串数组 A，返回列表中的每个字符串中都显示的全部字符（包括重复字符）组成的列表。例如，如果一个字符在每个字符串中出现 3 次，但不是 4 次，则需要在最终答案中包含该字符 3 次。你可以按任意顺序返回答案。
+
+
+## 解题思路
+
+- 简单题。给出一个字符串数组 A，要求找出这个数组中每个字符串都包含字符，如果字符出现多次，在最终结果中也需要出现多次。这一题可以用 map 来统计每个字符串的频次，但是如果用数组统计会更快。题目中说了只有小写字母，那么用 2 个 26 位长度的数组就可以统计出来了。遍历字符串数组的过程中，不过的缩小每个字符在每个字符串中出现的频次(因为需要找所有字符串公共的字符，公共的频次肯定就是最小的频次)，得到了最终公共字符的频次数组以后，按顺序输出就可以了。
diff --git a/Algorithms/1003. Check If Word Is Valid After Substitutions/1003. Check If Word Is Valid After Substitutions.go b/leetcode/1003.Check-If-Word-Is-Valid-After-Substitutions/1003. Check If Word Is Valid After Substitutions.go
similarity index 100%
rename from Algorithms/1003. Check If Word Is Valid After Substitutions/1003. Check If Word Is Valid After Substitutions.go
rename to leetcode/1003.Check-If-Word-Is-Valid-After-Substitutions/1003. Check If Word Is Valid After Substitutions.go
diff --git a/Algorithms/1003. Check If Word Is Valid After Substitutions/1003. Check If Word Is Valid After Substitutions_test.go b/leetcode/1003.Check-If-Word-Is-Valid-After-Substitutions/1003. Check If Word Is Valid After Substitutions_test.go
similarity index 92%
rename from Algorithms/1003. Check If Word Is Valid After Substitutions/1003. Check If Word Is Valid After Substitutions_test.go
rename to leetcode/1003.Check-If-Word-Is-Valid-After-Substitutions/1003. Check If Word Is Valid After Substitutions_test.go
index d0437941d..a59ff8a20 100644
--- a/Algorithms/1003. Check If Word Is Valid After Substitutions/1003. Check If Word Is Valid After Substitutions_test.go	
+++ b/leetcode/1003.Check-If-Word-Is-Valid-After-Substitutions/1003. Check If Word Is Valid After Substitutions_test.go	
@@ -26,22 +26,22 @@ func Test_Problem1003(t *testing.T) {
 
 	qs := []question1003{
 
-		question1003{
+		{
 			para1003{"aabcbc"},
 			ans1003{true},
 		},
 
-		question1003{
+		{
 			para1003{"abcabcababcc"},
 			ans1003{true},
 		},
 
-		question1003{
+		{
 			para1003{"abccba"},
 			ans1003{false},
 		},
 
-		question1003{
+		{
 			para1003{"cababc"},
 			ans1003{false},
 		},
diff --git a/Algorithms/1003. Check If Word Is Valid After Substitutions/README.md b/leetcode/1003.Check-If-Word-Is-Valid-After-Substitutions/README.md
similarity index 100%
rename from Algorithms/1003. Check If Word Is Valid After Substitutions/README.md
rename to leetcode/1003.Check-If-Word-Is-Valid-After-Substitutions/README.md
diff --git a/Algorithms/1004. Max Consecutive Ones III/1004. Max Consecutive Ones III.go b/leetcode/1004.Max-Consecutive-Ones-III/1004. Max Consecutive Ones III.go
similarity index 84%
rename from Algorithms/1004. Max Consecutive Ones III/1004. Max Consecutive Ones III.go
rename to leetcode/1004.Max-Consecutive-Ones-III/1004. Max Consecutive Ones III.go
index fd48551bf..aec1e1efb 100644
--- a/Algorithms/1004. Max Consecutive Ones III/1004. Max Consecutive Ones III.go	
+++ b/leetcode/1004.Max-Consecutive-Ones-III/1004. Max Consecutive Ones III.go	
@@ -20,3 +20,10 @@ func longestOnes(A []int, K int) int {
 	}
 	return res
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/1004. Max Consecutive Ones III/1004. Max Consecutive Ones III_test.go b/leetcode/1004.Max-Consecutive-Ones-III/1004. Max Consecutive Ones III_test.go
similarity index 94%
rename from Algorithms/1004. Max Consecutive Ones III/1004. Max Consecutive Ones III_test.go
rename to leetcode/1004.Max-Consecutive-Ones-III/1004. Max Consecutive Ones III_test.go
index 01e28d135..59f81473b 100644
--- a/Algorithms/1004. Max Consecutive Ones III/1004. Max Consecutive Ones III_test.go	
+++ b/leetcode/1004.Max-Consecutive-Ones-III/1004. Max Consecutive Ones III_test.go	
@@ -27,17 +27,17 @@ func Test_Problem1004(t *testing.T) {
 
 	qs := []question1004{
 
-		question1004{
+		{
 			para1004{[]int{1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0}, 2},
 			ans1004{6},
 		},
 
-		question1004{
+		{
 			para1004{[]int{0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1}, 3},
 			ans1004{10},
 		},
 
-		question1004{
+		{
 			para1004{[]int{0, 0, 0, 1}, 4},
 			ans1004{4},
 		},
diff --git a/Algorithms/1004. Max Consecutive Ones III/README.md b/leetcode/1004.Max-Consecutive-Ones-III/README.md
similarity index 100%
rename from Algorithms/1004. Max Consecutive Ones III/README.md
rename to leetcode/1004.Max-Consecutive-Ones-III/README.md
diff --git a/Algorithms/1005. Maximize Sum Of Array After K Negations/1005. Maximize Sum Of Array After K Negations.go b/leetcode/1005.Maximize-Sum-Of-Array-After-K-Negations/1005. Maximize Sum Of Array After K Negations.go
similarity index 100%
rename from Algorithms/1005. Maximize Sum Of Array After K Negations/1005. Maximize Sum Of Array After K Negations.go
rename to leetcode/1005.Maximize-Sum-Of-Array-After-K-Negations/1005. Maximize Sum Of Array After K Negations.go
diff --git a/Algorithms/1005. Maximize Sum Of Array After K Negations/1005. Maximize Sum Of Array After K Negations_test.go b/leetcode/1005.Maximize-Sum-Of-Array-After-K-Negations/1005. Maximize Sum Of Array After K Negations_test.go
similarity index 90%
rename from Algorithms/1005. Maximize Sum Of Array After K Negations/1005. Maximize Sum Of Array After K Negations_test.go
rename to leetcode/1005.Maximize-Sum-Of-Array-After-K-Negations/1005. Maximize Sum Of Array After K Negations_test.go
index 9e256ecdb..1b16fdb21 100644
--- a/Algorithms/1005. Maximize Sum Of Array After K Negations/1005. Maximize Sum Of Array After K Negations_test.go	
+++ b/leetcode/1005.Maximize-Sum-Of-Array-After-K-Negations/1005. Maximize Sum Of Array After K Negations_test.go	
@@ -27,36 +27,36 @@ func Test_Problem1005(t *testing.T) {
 
 	qs := []question1005{
 
-		question1005{
+		{
 			para1005{[]int{4, 2, 3}, 1},
 			ans1005{5},
 		},
-		question1005{
+		{
 			para1005{[]int{3, -1, 0, 2}, 3},
 			ans1005{6},
 		},
 
-		question1005{
+		{
 			para1005{[]int{2, -3, -1, 5, -4}, 2},
 			ans1005{13},
 		},
 
-		question1005{
+		{
 			para1005{[]int{-2, 9, 9, 8, 4}, 5},
 			ans1005{32},
 		},
 
-		question1005{
+		{
 			para1005{[]int{5, -7, -8, -3, 9, 5, -5, -7}, 7},
 			ans1005{49},
 		},
 
-		question1005{
+		{
 			para1005{[]int{5, 6, 9, -3, 3}, 2},
 			ans1005{20},
 		},
 
-		question1005{
+		{
 			para1005{[]int{8, -7, -3, -9, 1, 9, -6, -9, 3}, 8},
 			ans1005{53},
 		},
diff --git a/Algorithms/1005. Maximize Sum Of Array After K Negations/README.md b/leetcode/1005.Maximize-Sum-Of-Array-After-K-Negations/README.md
similarity index 100%
rename from Algorithms/1005. Maximize Sum Of Array After K Negations/README.md
rename to leetcode/1005.Maximize-Sum-Of-Array-After-K-Negations/README.md
diff --git a/leetcode/1006.Clumsy-Factorial/1006. Clumsy Factorial.go b/leetcode/1006.Clumsy-Factorial/1006. Clumsy Factorial.go
new file mode 100644
index 000000000..f69cfaca2
--- /dev/null
+++ b/leetcode/1006.Clumsy-Factorial/1006. Clumsy Factorial.go	
@@ -0,0 +1,27 @@
+package leetcode
+
+func clumsy(N int) int {
+	res, count, tmp, flag := 0, 1, N, false
+	for i := N - 1; i > 0; i-- {
+		count = count % 4
+		switch count {
+		case 1:
+			tmp = tmp * i
+		case 2:
+			tmp = tmp / i
+		case 3:
+			res = res + tmp
+			flag = true
+			tmp = -1
+			res = res + i
+		case 0:
+			flag = false
+			tmp = tmp * (i)
+		}
+		count++
+	}
+	if !flag {
+		res = res + tmp
+	}
+	return res
+}
diff --git a/leetcode/1006.Clumsy-Factorial/1006. Clumsy Factorial_test.go b/leetcode/1006.Clumsy-Factorial/1006. Clumsy Factorial_test.go
new file mode 100644
index 000000000..3f683f6fc
--- /dev/null
+++ b/leetcode/1006.Clumsy-Factorial/1006. Clumsy Factorial_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1006 struct {
+	para1006
+	ans1006
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1006 struct {
+	N int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1006 struct {
+	one int
+}
+
+func Test_Problem1006(t *testing.T) {
+
+	qs := []question1006{
+
+		{
+			para1006{4},
+			ans1006{7},
+		},
+
+		{
+			para1006{10},
+			ans1006{12},
+		},
+
+		{
+			para1006{100},
+			ans1006{101},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1006------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1006, q.para1006
+		fmt.Printf("【input】:%v       【output】:%v\n", p, clumsy(p.N))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1006.Clumsy-Factorial/README.md b/leetcode/1006.Clumsy-Factorial/README.md
new file mode 100644
index 000000000..db1a37324
--- /dev/null
+++ b/leetcode/1006.Clumsy-Factorial/README.md
@@ -0,0 +1,75 @@
+# [1006. Clumsy Factorial](https://leetcode.com/problems/clumsy-factorial/)
+
+
+## 题目
+
+Normally, the factorial of a positive integer `n` is the product of all positive integers less than or equal to `n`.  For example, `factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1`.
+
+We instead make a *clumsy factorial:* using the integers in decreasing order, we swap out the multiply operations for a fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in this order.
+
+For example, `clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1`.  However, these operations are still applied using the usual order of operations of arithmetic: we do all multiplication and division steps before any addition or subtraction steps, and multiplication and division steps are processed left to right.
+
+Additionally, the division that we use is *floor division* such that `10 * 9 / 8` equals `11`.  This guarantees the result is an integer.
+
+`Implement the clumsy` function as defined above: given an integer `N`, it returns the clumsy factorial of `N`.
+
+**Example 1:**
+
+```
+Input:4
+Output: 7
+Explanation: 7 = 4 * 3 / 2 + 1
+```
+
+**Example 2:**
+
+```
+Input:10
+Output:12
+Explanation:12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
+```
+
+**Note:**
+
+1. `1 <= N <= 10000`
+2. `2^31 <= answer <= 2^31 - 1` (The answer is guaranteed to fit within a 32-bit integer.)
+
+## 题目大意
+
+通常，正整数 n 的阶乘是所有小于或等于 n 的正整数的乘积。例如，factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1。相反，我们设计了一个笨阶乘 clumsy：在整数的递减序列中，我们以一个固定顺序的操作符序列来依次替换原有的乘法操作符：乘法(*)，除法(/)，加法(+)和减法(-)。例如，clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1。然而，这些运算仍然使用通常的算术运算顺序：我们在任何加、减步骤之前执行所有的乘法和除法步骤，并且按从左到右处理乘法和除法步骤。另外，我们使用的除法是地板除法（floor division），所以 10 * 9 / 8 等于 11。这保证结果是一个整数。实现上面定义的笨函数：给定一个整数 N，它返回 N 的笨阶乘。
+
+## 解题思路
+
+- 按照题意，由于本题没有括号，所以先乘除后加减。4 个操作一组，先算乘法，再算除法，再算加法，最后算减法。减法也可以看成是加法，只是带负号的加法。
+
+## 代码
+
+```go
+package leetcode
+
+func clumsy(N int) int {
+	res, count, tmp, flag := 0, 1, N, true
+	for i := N - 1; i > 0; i-- {
+		count = count % 4
+		switch count {
+		case 1:
+			tmp = tmp * i
+		case 2:
+			tmp = tmp / i
+		case 3:
+			res = res + tmp
+			flag = true
+			tmp = -1
+			res = res + i
+		case 0:
+			flag = false
+			tmp = tmp * (i)
+		}
+		count++
+	}
+	if !flag {
+		res = res + tmp
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1009.Complement-of-Base-10-Integer/1009. Complement of Base 10 Integer.go b/leetcode/1009.Complement-of-Base-10-Integer/1009. Complement of Base 10 Integer.go
new file mode 100644
index 000000000..19e526e39
--- /dev/null
+++ b/leetcode/1009.Complement-of-Base-10-Integer/1009. Complement of Base 10 Integer.go	
@@ -0,0 +1,9 @@
+package leetcode
+
+func bitwiseComplement(n int) int {
+	mask := 1
+	for mask < n {
+		mask = (mask << 1) + 1
+	}
+	return mask ^ n
+}
diff --git a/leetcode/1009.Complement-of-Base-10-Integer/1009. Complement of Base 10 Integer_test.go b/leetcode/1009.Complement-of-Base-10-Integer/1009. Complement of Base 10 Integer_test.go
new file mode 100644
index 000000000..90bf93447
--- /dev/null
+++ b/leetcode/1009.Complement-of-Base-10-Integer/1009. Complement of Base 10 Integer_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1009 struct {
+	para1009
+	ans1009
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1009 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1009 struct {
+	one int
+}
+
+func Test_Problem1009(t *testing.T) {
+
+	qs := []question1009{
+
+		{
+			para1009{5},
+			ans1009{2},
+		},
+
+		{
+			para1009{7},
+			ans1009{0},
+		},
+
+		{
+			para1009{10},
+			ans1009{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1009------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1009, q.para1009
+		fmt.Printf("【input】:%v       【output】:%v\n", p, bitwiseComplement(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1009.Complement-of-Base-10-Integer/README.md b/leetcode/1009.Complement-of-Base-10-Integer/README.md
new file mode 100644
index 000000000..d8b12b302
--- /dev/null
+++ b/leetcode/1009.Complement-of-Base-10-Integer/README.md
@@ -0,0 +1,67 @@
+# [1009. Complement of Base 10 Integer](https://leetcode.com/problems/complement-of-base-10-integer/)
+
+
+## 题目
+
+The **complement** of an integer is the integer you get when you flip all the `0`'s to `1`'s and all the `1`'s to `0`'s in its binary representation.
+
+- For example, The integer `5` is `"101"` in binary and its **complement** is `"010"` which is the integer `2`.
+
+Given an integer `n`, return *its complement*.
+
+**Example 1:**
+
+```
+Input: n = 5
+Output: 2
+Explanation: 5 is "101" in binary, with complement "010" in binary, which is 2 in base-10.
+
+```
+
+**Example 2:**
+
+```
+Input: n = 7
+Output: 0
+Explanation: 7 is "111" in binary, with complement "000" in binary, which is 0 in base-10.
+
+```
+
+**Example 3:**
+
+```
+Input: n = 10
+Output: 5
+Explanation: 10 is "1010" in binary, with complement "0101" in binary, which is 5 in base-10.
+
+```
+
+**Constraints:**
+
+- `0 <= n < 109`
+
+## 题目大意
+
+每个非负整数 N 都有其二进制表示。例如， 5 可以被表示为二进制 "101"，11 可以用二进制 "1011" 表示，依此类推。注意，除 N = 0 外，任何二进制表示中都不含前导零。
+
+二进制的反码表示是将每个 1 改为 0 且每个 0 变为 1。例如，二进制数 "101" 的二进制反码为 "010"。
+
+给你一个十进制数 N，请你返回其二进制表示的反码所对应的十进制整数。
+
+## 解题思路
+
+- 简单题。求一个十进制数的反码，只需要让该数和全 1 的数进行异或计算即可。所以本题重点在如何构造 mask 上。
+
+## 代码
+
+```go
+package leetcode
+
+func bitwiseComplement(n int) int {
+    mask := 1 
+    for mask < n {
+        mask = (mask << 1) + 1
+    }
+    return mask ^ n
+}
+```
\ No newline at end of file
diff --git a/leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/1010. Pairs of Songs With Total Durations Divisible by 60.go b/leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/1010. Pairs of Songs With Total Durations Divisible by 60.go
new file mode 100644
index 000000000..3c6dd1afd
--- /dev/null
+++ b/leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/1010. Pairs of Songs With Total Durations Divisible by 60.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func numPairsDivisibleBy60(time []int) int {
+	counts := make([]int, 60)
+	for _, v := range time {
+		v %= 60
+		counts[v]++
+	}
+	res := 0
+	for i := 1; i < len(counts)/2; i++ {
+		res += counts[i] * counts[60-i]
+	}
+	res += (counts[0] * (counts[0] - 1)) / 2
+	res += (counts[30] * (counts[30] - 1)) / 2
+	return res
+}
diff --git a/leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/1010. Pairs of Songs With Total Durations Divisible by 60_test.go b/leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/1010. Pairs of Songs With Total Durations Divisible by 60_test.go
new file mode 100644
index 000000000..201e0c727
--- /dev/null
+++ b/leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/1010. Pairs of Songs With Total Durations Divisible by 60_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1010 struct {
+	para1010
+	ans1010
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1010 struct {
+	time []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1010 struct {
+	one int
+}
+
+func Test_Problem1010(t *testing.T) {
+
+	qs := []question1010{
+
+		{
+			para1010{[]int{30, 20, 150, 100, 40}},
+			ans1010{3},
+		},
+
+		{
+			para1010{[]int{60, 60, 60}},
+			ans1010{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1010------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1010, q.para1010
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numPairsDivisibleBy60(p.time))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/README.md b/leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/README.md
new file mode 100644
index 000000000..a49f11207
--- /dev/null
+++ b/leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/README.md
@@ -0,0 +1,63 @@
+# [1010. Pairs of Songs With Total Durations Divisible by 60](https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/)
+
+
+## 题目
+
+You are given a list of songs where the ith song has a duration of `time[i]` seconds.
+
+Return *the number of pairs of songs for which their total duration in seconds is divisible by* `60`. Formally, we want the number of indices `i`, `j` such that `i < j` with `(time[i] + time[j]) % 60 == 0`.
+
+**Example 1:**
+
+```
+Input: time = [30,20,150,100,40]
+Output: 3
+Explanation: Three pairs have a total duration divisible by 60:
+(time[0] = 30, time[2] = 150): total duration 180
+(time[1] = 20, time[3] = 100): total duration 120
+(time[1] = 20, time[4] = 40): total duration 60
+
+```
+
+**Example 2:**
+
+```
+Input: time = [60,60,60]
+Output: 3
+Explanation: All three pairs have a total duration of 120, which is divisible by 60.
+
+```
+
+**Constraints:**
+
+- `1 <= time.length <= 6 * 104`
+- `1 <= time[i] <= 500`
+
+## 题目大意
+
+在歌曲列表中，第 i 首歌曲的持续时间为 time[i] 秒。
+
+返回其总持续时间（以秒为单位）可被 60 整除的歌曲对的数量。形式上，我们希望下标数字 i 和 j 满足  i < j 且有 (time[i] + time[j]) % 60 == 0。
+
+## 解题思路
+
+- 简单题。先将数组每个元素对 60 取余，将它们都转换到 [0,59] 之间。然后在数组中找两两元素之和等于 60 的数对。可以在 0-30 之内对半查找符合条件的数对。对 0 和 30 单独计算。因为多个 0 相加，余数还为 0 。2 个 30 相加之和为 60。
+
+## 代码
+
+```go
+func numPairsDivisibleBy60(time []int) int {
+	counts := make([]int, 60)
+	for _, v := range time {
+		v %= 60
+		counts[v]++
+	}
+	res := 0
+	for i := 1; i < len(counts)/2; i++ {
+		res += counts[i] * counts[60-i]
+	}
+	res += (counts[0] * (counts[0] - 1)) / 2
+	res += (counts[30] * (counts[30] - 1)) / 2
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1011.Capacity-To-Ship-Packages-Within-D-Days/1011. Capacity To Ship Packages Within D Days.go b/leetcode/1011.Capacity-To-Ship-Packages-Within-D-Days/1011. Capacity To Ship Packages Within D Days.go
new file mode 100644
index 000000000..93873ecd8
--- /dev/null
+++ b/leetcode/1011.Capacity-To-Ship-Packages-Within-D-Days/1011. Capacity To Ship Packages Within D Days.go	
@@ -0,0 +1,40 @@
+package leetcode
+
+func shipWithinDays(weights []int, D int) int {
+	maxNum, sum := 0, 0
+	for _, num := range weights {
+		sum += num
+		if num > maxNum {
+			maxNum = num
+		}
+	}
+	if D == 1 {
+		return sum
+	}
+	low, high := maxNum, sum
+	for low < high {
+		mid := low + (high-low)>>1
+		if calSum(mid, D, weights) {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+func calSum(mid, m int, nums []int) bool {
+	sum, count := 0, 0
+	for _, v := range nums {
+		sum += v
+		if sum > mid {
+			sum = v
+			count++
+			// 分成 m 块，只需要插桩 m -1 个
+			if count > m-1 {
+				return false
+			}
+		}
+	}
+	return true
+}
diff --git a/Algorithms/1011. Capacity To Ship Packages Within D Days/1011. Capacity To Ship Packages Within D Days_test.go b/leetcode/1011.Capacity-To-Ship-Packages-Within-D-Days/1011. Capacity To Ship Packages Within D Days_test.go
similarity index 93%
rename from Algorithms/1011. Capacity To Ship Packages Within D Days/1011. Capacity To Ship Packages Within D Days_test.go
rename to leetcode/1011.Capacity-To-Ship-Packages-Within-D-Days/1011. Capacity To Ship Packages Within D Days_test.go
index 6251d2af6..a7e359d65 100644
--- a/Algorithms/1011. Capacity To Ship Packages Within D Days/1011. Capacity To Ship Packages Within D Days_test.go	
+++ b/leetcode/1011.Capacity-To-Ship-Packages-Within-D-Days/1011. Capacity To Ship Packages Within D Days_test.go	
@@ -27,22 +27,22 @@ func Test_Problem1011(t *testing.T) {
 
 	qs := []question1011{
 
-		question1011{
+		{
 			para1011{[]int{7, 2, 5, 10, 8}, 2},
 			ans1011{18},
 		},
 
-		question1011{
+		{
 			para1011{[]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, 5},
 			ans1011{15},
 		},
 
-		question1011{
+		{
 			para1011{[]int{3, 2, 2, 4, 1, 4}, 3},
 			ans1011{6},
 		},
 
-		question1011{
+		{
 			para1011{[]int{1, 2, 3, 1, 1}, 4},
 			ans1011{3},
 		},
diff --git a/leetcode/1011.Capacity-To-Ship-Packages-Within-D-Days/README.md b/leetcode/1011.Capacity-To-Ship-Packages-Within-D-Days/README.md
new file mode 100755
index 000000000..ba21809c1
--- /dev/null
+++ b/leetcode/1011.Capacity-To-Ship-Packages-Within-D-Days/README.md
@@ -0,0 +1,69 @@
+# [1011. Capacity To Ship Packages Within D Days](https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/)
+
+
+## 题目
+
+A conveyor belt has packages that must be shipped from one port to another within `D` days.
+
+The `i`-th package on the conveyor belt has a weight of `weights[i]`. Each day, we load the ship with packages on the conveyor belt (in the order given by `weights`). We may not load more weight than the maximum weight capacity of the ship.
+
+Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within `D` days.
+
+**Example 1:**
+
+    Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5
+    Output: 15
+    Explanation: 
+    A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
+    1st day: 1, 2, 3, 4, 5
+    2nd day: 6, 7
+    3rd day: 8
+    4th day: 9
+    5th day: 10
+    
+    Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed.
+
+**Example 2:**
+
+    Input: weights = [3,2,2,4,1,4], D = 3
+    Output: 6
+    Explanation: 
+    A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
+    1st day: 3, 2
+    2nd day: 2, 4
+    3rd day: 1, 4
+
+**Example 3:**
+
+    Input: weights = [1,2,3,1,1], D = 4
+    Output: 3
+    Explanation: 
+    1st day: 1
+    2nd day: 2
+    3rd day: 3
+    4th day: 1, 1
+
+**Note:**
+
+1. `1 <= D <= weights.length <= 50000`
+2. `1 <= weights[i] <= 500`
+
+
+## 题目大意
+
+传送带上的包裹必须在 D 天内从一个港口运送到另一个港口。
+
+传送带上的第 i 个包裹的重量为 weights[i]。每一天，我们都会按给出重量的顺序往传送带上装载包裹。我们装载的重量不会超过船的最大运载重量。
+
+返回能在 D 天内将传送带上的所有包裹送达的船的最低运载能力。
+
+提示：
+
+- 1 <= D <= weights.length <= 50000
+- 1 <= weights[i] <= 500
+
+
+## 解题思路
+
+- 给出一个数组和天数 D，要求正好在 D 天把数组中的货物都运完。求传输带上能承受的最小货物重量是多少。
+- 这一题和第 410 题完全一样，只不过换了一个题面。代码完全不变。思路解析见第 410 题。
diff --git a/Algorithms/1017. Convert to Base -2/1017. Convert to Base -2.go b/leetcode/1017.Convert-to-Base-2/1017. Convert to Base -2.go
similarity index 100%
rename from Algorithms/1017. Convert to Base -2/1017. Convert to Base -2.go
rename to leetcode/1017.Convert-to-Base-2/1017. Convert to Base -2.go
diff --git a/Algorithms/1017. Convert to Base -2/1017. Convert to Base -2_test.go b/leetcode/1017.Convert-to-Base-2/1017. Convert to Base -2_test.go
similarity index 93%
rename from Algorithms/1017. Convert to Base -2/1017. Convert to Base -2_test.go
rename to leetcode/1017.Convert-to-Base-2/1017. Convert to Base -2_test.go
index f016c5902..0bcf767d3 100644
--- a/Algorithms/1017. Convert to Base -2/1017. Convert to Base -2_test.go	
+++ b/leetcode/1017.Convert-to-Base-2/1017. Convert to Base -2_test.go	
@@ -26,17 +26,17 @@ func Test_Problem1017(t *testing.T) {
 
 	qs := []question1017{
 
-		question1017{
+		{
 			para1017{2},
 			ans1017{"110"},
 		},
 
-		question1017{
+		{
 			para1017{3},
 			ans1017{"111"},
 		},
 
-		question1017{
+		{
 			para1017{4},
 			ans1017{"110"},
 		},
diff --git a/leetcode/1017.Convert-to-Base-2/README.md b/leetcode/1017.Convert-to-Base-2/README.md
new file mode 100755
index 000000000..addd0bea0
--- /dev/null
+++ b/leetcode/1017.Convert-to-Base-2/README.md
@@ -0,0 +1,47 @@
+# [1017. Convert to Base -2](https://leetcode.com/problems/convert-to-base-2/)
+
+
+## 题目
+
+Given a number `N`, return a string consisting of `"0"`s and `"1"`s that represents its value in base **`-2`** (negative two).
+
+The returned string must have no leading zeroes, unless the string is `"0"`.
+
+**Example 1:**
+
+    Input: 2
+    Output: "110"
+    Explantion: (-2) ^ 2 + (-2) ^ 1 = 2
+
+**Example 2:**
+
+    Input: 3
+    Output: "111"
+    Explantion: (-2) ^ 2 + (-2) ^ 1 + (-2) ^ 0 = 3
+
+**Example 3:**
+
+    Input: 4
+    Output: "100"
+    Explantion: (-2) ^ 2 = 4
+
+**Note:**
+
+1. `0 <= N <= 10^9`
+
+
+## 题目大意
+
+给出数字 N，返回由若干 "0" 和 "1"组成的字符串，该字符串为 N 的负二进制（base -2）表示。除非字符串就是 "0"，否则返回的字符串中不能含有前导零。
+
+提示：
+
+- 0 <= N <= 10^9
+
+
+
+## 解题思路
+
+- 给出一个十进制的数，要求转换成 -2 进制的数
+- 这一题仿造十进制转二进制的思路，短除法即可。
+
diff --git a/leetcode/1018.Binary-Prefix-Divisible-By-5/1018. Binary Prefix Divisible By 5.go b/leetcode/1018.Binary-Prefix-Divisible-By-5/1018. Binary Prefix Divisible By 5.go
new file mode 100644
index 000000000..8d88763f6
--- /dev/null
+++ b/leetcode/1018.Binary-Prefix-Divisible-By-5/1018. Binary Prefix Divisible By 5.go	
@@ -0,0 +1,10 @@
+package leetcode
+
+func prefixesDivBy5(a []int) []bool {
+	res, num := make([]bool, len(a)), 0
+	for i, v := range a {
+		num = (num<<1 | v) % 5
+		res[i] = num == 0
+	}
+	return res
+}
diff --git a/leetcode/1018.Binary-Prefix-Divisible-By-5/1018. Binary Prefix Divisible By 5_test.go b/leetcode/1018.Binary-Prefix-Divisible-By-5/1018. Binary Prefix Divisible By 5_test.go
new file mode 100644
index 000000000..2903210e0
--- /dev/null
+++ b/leetcode/1018.Binary-Prefix-Divisible-By-5/1018. Binary Prefix Divisible By 5_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1018 struct {
+	para1018
+	ans1018
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1018 struct {
+	a []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1018 struct {
+	one []bool
+}
+
+func Test_Problem1018(t *testing.T) {
+
+	qs := []question1018{
+
+		{
+			para1018{[]int{0, 1, 1}},
+			ans1018{[]bool{true, false, false}},
+		},
+
+		{
+			para1018{[]int{1, 1, 1}},
+			ans1018{[]bool{false, false, false}},
+		},
+
+		{
+			para1018{[]int{0, 1, 1, 1, 1, 1}},
+			ans1018{[]bool{true, false, false, false, true, false}},
+		},
+
+		{
+			para1018{[]int{1, 1, 1, 0, 1}},
+			ans1018{[]bool{false, false, false, false, false}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1018------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1018, q.para1018
+		fmt.Printf("【input】:%v       【output】:%v\n", p, prefixesDivBy5(p.a))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1018.Binary-Prefix-Divisible-By-5/README.md b/leetcode/1018.Binary-Prefix-Divisible-By-5/README.md
new file mode 100644
index 000000000..5559c7e9a
--- /dev/null
+++ b/leetcode/1018.Binary-Prefix-Divisible-By-5/README.md
@@ -0,0 +1,70 @@
+# [1018. Binary Prefix Divisible By 5](https://leetcode.com/problems/binary-prefix-divisible-by-5/)
+
+
+## 题目
+
+Given an array `A` of `0`s and `1`s, consider `N_i`: the i-th subarray from `A[0]` to `A[i]` interpreted as a binary number (from most-significant-bit to least-significant-bit.)
+
+Return a list of booleans `answer`, where `answer[i]` is `true` if and only if `N_i` is divisible by 5.
+
+**Example 1:**
+
+```
+Input: [0,1,1]
+Output: [true,false,false]
+Explanation: 
+The input numbers in binary are 0, 01, 011; which are 0, 1, and 3 in base-10.  Only the first number is divisible by 5, so answer[0] is true.
+
+```
+
+**Example 2:**
+
+```
+Input: [1,1,1]
+Output: [false,false,false]
+
+```
+
+**Example 3:**
+
+```
+Input: [0,1,1,1,1,1]
+Output: [true,false,false,false,true,false]
+
+```
+
+**Example 4:**
+
+```
+Input: [1,1,1,0,1]
+Output: [false,false,false,false,false]
+
+```
+
+**Note:**
+
+1. `1 <= A.length <= 30000`
+2. `A[i]` is `0` or `1`
+
+## 题目大意
+
+给定由若干 0 和 1 组成的数组 A。我们定义 N_i：从 A[0] 到 A[i] 的第 i 个子数组被解释为一个二进制数（从最高有效位到最低有效位）。返回布尔值列表 answer，只有当 N_i 可以被 5 整除时，答案 answer[i] 为 true，否则为 false。
+
+## 解题思路
+
+- 简单题。每扫描数组中的一个数字，累计转换成二进制数对 5 取余，如果余数为 0，则存入 true，否则存入 false。
+
+## 代码
+
+```go
+package leetcode
+
+func prefixesDivBy5(a []int) []bool {
+	res, num := make([]bool, len(a)), 0
+	for i, v := range a {
+		num = (num<<1 | v) % 5
+		res[i] = num == 0
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1019.Next-Greater-Node-In-Linked-List/1019. Next Greater Node In Linked List.go b/leetcode/1019.Next-Greater-Node-In-Linked-List/1019. Next Greater Node In Linked List.go
new file mode 100644
index 000000000..a789e7748
--- /dev/null
+++ b/leetcode/1019.Next-Greater-Node-In-Linked-List/1019. Next Greater Node In Linked List.go	
@@ -0,0 +1,35 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// 解法一 单调栈
+func nextLargerNodes(head *ListNode) []int {
+	type node struct {
+		index, val int
+	}
+	var monoStack []node
+	var res []int
+	for head != nil {
+		for len(monoStack) > 0 && monoStack[len(monoStack)-1].val < head.Val {
+			res[monoStack[len(monoStack)-1].index] = head.Val
+			monoStack = monoStack[:len(monoStack)-1]
+		}
+		monoStack = append(monoStack, node{len(res), head.Val})
+		res = append(res, 0)
+		head = head.Next
+	}
+	return res
+}
diff --git a/Algorithms/1019. Next Greater Node In Linked List/1019. Next Greater Node In Linked List_test.go b/leetcode/1019.Next-Greater-Node-In-Linked-List/1019. Next Greater Node In Linked List_test.go
similarity index 89%
rename from Algorithms/1019. Next Greater Node In Linked List/1019. Next Greater Node In Linked List_test.go
rename to leetcode/1019.Next-Greater-Node-In-Linked-List/1019. Next Greater Node In Linked List_test.go
index a4bc2be23..d6b0f68bc 100644
--- a/Algorithms/1019. Next Greater Node In Linked List/1019. Next Greater Node In Linked List_test.go	
+++ b/leetcode/1019.Next-Greater-Node-In-Linked-List/1019. Next Greater Node In Linked List_test.go	
@@ -3,6 +3,8 @@ package leetcode
 import (
 	"fmt"
 	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
 )
 
 type question1019 struct {
@@ -26,22 +28,22 @@ func Test_Problem1019(t *testing.T) {
 
 	qs := []question1019{
 
-		question1019{
+		{
 			para1019{[]int{2, 1, 5}},
 			ans1019{[]int{5, 5, 0}},
 		},
 
-		question1019{
+		{
 			para1019{[]int{2, 7, 4, 3, 5}},
 			ans1019{[]int{7, 0, 5, 5, 0}},
 		},
 
-		question1019{
+		{
 			para1019{[]int{1, 7, 5, 1, 9, 2, 5, 1}},
 			ans1019{[]int{7, 9, 9, 9, 0, 5, 0, 0}},
 		},
 
-		question1019{
+		{
 			para1019{[]int{1, 7, 5, 1, 9, 2, 5, 6, 7, 8, 1}},
 			ans1019{[]int{7, 9, 9, 9, 0, 5, 6, 7, 8, 0, 0}},
 		},
@@ -51,7 +53,7 @@ func Test_Problem1019(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans1019, q.para1019
-		fmt.Printf("【input】:%v       【output】:%v\n", p, nextLargerNodes(S2l(p.one)))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, nextLargerNodes(structures.Ints2List(p.one)))
 	}
 	fmt.Printf("\n\n\n")
 }
diff --git a/Algorithms/1019. Next Greater Node In Linked List/README.md b/leetcode/1019.Next-Greater-Node-In-Linked-List/README.md
similarity index 100%
rename from Algorithms/1019. Next Greater Node In Linked List/README.md
rename to leetcode/1019.Next-Greater-Node-In-Linked-List/README.md
diff --git a/leetcode/1020.Number-of-Enclaves/1020. Number of Enclaves.go b/leetcode/1020.Number-of-Enclaves/1020. Number of Enclaves.go
new file mode 100644
index 000000000..1e69fd734
--- /dev/null
+++ b/leetcode/1020.Number-of-Enclaves/1020. Number of Enclaves.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func numEnclaves(A [][]int) int {
+	m, n := len(A), len(A[0])
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			if i == 0 || i == m-1 || j == 0 || j == n-1 {
+				if A[i][j] == 1 {
+					dfsNumEnclaves(A, i, j)
+				}
+			}
+		}
+	}
+	count := 0
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			if A[i][j] == 1 {
+				count++
+			}
+		}
+	}
+	return count
+}
+
+func dfsNumEnclaves(A [][]int, x, y int) {
+	if !isInGrid(A, x, y) || A[x][y] == 0 {
+		return
+	}
+	A[x][y] = 0
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		dfsNumEnclaves(A, nx, ny)
+	}
+}
+
+func isInGrid(grid [][]int, x, y int) bool {
+	return x >= 0 && x < len(grid) && y >= 0 && y < len(grid[0])
+}
diff --git a/leetcode/1020.Number-of-Enclaves/1020. Number of Enclaves_test.go b/leetcode/1020.Number-of-Enclaves/1020. Number of Enclaves_test.go
new file mode 100644
index 000000000..1aa7b5dc5
--- /dev/null
+++ b/leetcode/1020.Number-of-Enclaves/1020. Number of Enclaves_test.go	
@@ -0,0 +1,90 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1020 struct {
+	para1020
+	ans1020
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1020 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1020 struct {
+	one int
+}
+
+func Test_Problem1020(t *testing.T) {
+
+	qs := []question1020{
+
+		{
+			para1020{[][]int{
+				{1, 1, 1, 1, 0},
+				{1, 1, 0, 1, 0},
+				{1, 1, 0, 0, 0},
+				{0, 0, 0, 0, 0},
+			}},
+			ans1020{0},
+		},
+
+		{
+			para1020{[][]int{
+				{1, 1, 0, 0, 0},
+				{1, 1, 0, 0, 0},
+				{0, 0, 1, 0, 0},
+				{0, 0, 0, 1, 1},
+			}},
+			ans1020{1},
+		},
+
+		{
+			para1020{[][]int{
+				{1, 1, 1, 1, 1, 1, 1, 0},
+				{1, 0, 0, 0, 0, 1, 1, 0},
+				{1, 0, 1, 0, 1, 1, 1, 0},
+				{1, 0, 0, 0, 0, 1, 0, 1},
+				{1, 1, 1, 1, 1, 1, 1, 0},
+			}},
+			ans1020{1},
+		},
+
+		{
+			para1020{[][]int{
+				{0, 0, 1, 0, 0},
+				{0, 1, 0, 1, 0},
+				{0, 1, 1, 1, 0},
+			}},
+			ans1020{0},
+		},
+
+		{
+			para1020{[][]int{
+				{1, 1, 1, 1, 1, 1, 1},
+				{1, 0, 0, 0, 0, 0, 1},
+				{1, 0, 1, 1, 1, 0, 1},
+				{1, 0, 1, 0, 1, 0, 1},
+				{1, 0, 1, 1, 1, 0, 1},
+				{1, 0, 0, 0, 0, 0, 1},
+				{1, 1, 1, 1, 1, 1, 1},
+			}},
+			ans1020{8},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1020------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1020, q.para1020
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numEnclaves(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1020.Number-of-Enclaves/README.md b/leetcode/1020.Number-of-Enclaves/README.md
new file mode 100644
index 000000000..34b72b3de
--- /dev/null
+++ b/leetcode/1020.Number-of-Enclaves/README.md
@@ -0,0 +1,93 @@
+# [1020. Number of Enclaves](https://leetcode.com/problems/number-of-enclaves/)
+
+
+
+## 题目
+
+Given a 2D array `A`, each cell is 0 (representing sea) or 1 (representing land)
+
+A move consists of walking from one land square 4-directionally to another land square, or off the boundary of the grid.
+
+Return the number of land squares in the grid for which we **cannot** walk off the boundary of the grid in any number of moves.
+
+**Example 1**:
+
+```
+Input: [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
+Output: 3
+Explanation: 
+There are three 1s that are enclosed by 0s, and one 1 that isn't enclosed because its on the boundary.
+```
+
+**Example 2**:
+
+```
+Input: [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
+Output: 0
+Explanation: 
+All 1s are either on the boundary or can reach the boundary.
+```
+
+**Note**:
+
+1. `1 <= A.length <= 500`
+2. `1 <= A[i].length <= 500`
+3. `0 <= A[i][j] <= 1`
+4. All rows have the same size.
+
+## 题目大意
+
+给出一个二维数组 A，每个单元格为 0（代表海）或 1（代表陆地）。移动是指在陆地上从一个地方走到另一个地方（朝四个方向之一）或离开网格的边界。返回网格中无法在任意次数的移动中离开网格边界的陆地单元格的数量。
+
+提示：
+
+- 1 <= A.length <= 500
+- 1 <= A[i].length <= 500
+- 0 <= A[i][j] <= 1
+- 所有行的大小都相同
+
+
+## 解题思路
+
+- 给出一个地图，要求输出不和边界连通的 1 的个数。
+- 这一题可以用 DFS 也可以用并查集解答。DFS 的思路是深搜的过程中把和边界连通的点都覆盖成 0，最后遍历一遍地图，输出 1 的个数即可。并查集的思路就比较直接了，把能和边界连通的放在一个集合中，剩下的就是不能和边界连通的都在另外一个集合中，输出这个集合里面元素的个数即可。
+- 这一题和第 200 题，第 1254 题，第 695 题类似。可以放在一起练习。
+
+## 代码
+
+```go
+func numEnclaves(A [][]int) int {
+	m, n := len(A), len(A[0])
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			if i == 0 || i == m-1 || j == 0 || j == n-1 {
+				if A[i][j] == 1 {
+					dfsNumEnclaves(A, i, j)
+				}
+			}
+		}
+	}
+	count := 0
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			if A[i][j] == 1 {
+				count++
+			}
+		}
+	}
+	return count
+}
+
+func dfsNumEnclaves(A [][]int, x, y int) {
+	if !isInGrid(A, x, y) || A[x][y] == 0 {
+		return
+	}
+	A[x][y] = 0
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		dfsNumEnclaves(A, nx, ny)
+	}
+}
+
+```
\ No newline at end of file
diff --git a/Algorithms/1021. Remove Outermost Parentheses/1021. Remove Outermost Parentheses.go b/leetcode/1021.Remove-Outermost-Parentheses/1021. Remove Outermost Parentheses.go
similarity index 100%
rename from Algorithms/1021. Remove Outermost Parentheses/1021. Remove Outermost Parentheses.go
rename to leetcode/1021.Remove-Outermost-Parentheses/1021. Remove Outermost Parentheses.go
diff --git a/Algorithms/1021. Remove Outermost Parentheses/1021. Remove Outermost Parentheses_test.go b/leetcode/1021.Remove-Outermost-Parentheses/1021. Remove Outermost Parentheses_test.go
similarity index 94%
rename from Algorithms/1021. Remove Outermost Parentheses/1021. Remove Outermost Parentheses_test.go
rename to leetcode/1021.Remove-Outermost-Parentheses/1021. Remove Outermost Parentheses_test.go
index 828ffd328..5c72a5d0a 100644
--- a/Algorithms/1021. Remove Outermost Parentheses/1021. Remove Outermost Parentheses_test.go	
+++ b/leetcode/1021.Remove-Outermost-Parentheses/1021. Remove Outermost Parentheses_test.go	
@@ -25,17 +25,17 @@ type ans1021 struct {
 func Test_Problem1021(t *testing.T) {
 
 	qs := []question1021{
-		question1021{
+		{
 			para1021{"(()())(())"},
 			ans1021{"()()()"},
 		},
 
-		question1021{
+		{
 			para1021{"(()())(())(()(()))"},
 			ans1021{"()()()()(())"},
 		},
 
-		question1021{
+		{
 			para1021{"()()"},
 			ans1021{""},
 		},
diff --git a/Algorithms/1021. Remove Outermost Parentheses/README.md b/leetcode/1021.Remove-Outermost-Parentheses/README.md
similarity index 100%
rename from Algorithms/1021. Remove Outermost Parentheses/README.md
rename to leetcode/1021.Remove-Outermost-Parentheses/README.md
diff --git a/leetcode/1022.Sum-of-Root-To-Leaf-Binary-Numbers/1022. Sum of Root To Leaf Binary Numbers.go b/leetcode/1022.Sum-of-Root-To-Leaf-Binary-Numbers/1022. Sum of Root To Leaf Binary Numbers.go
new file mode 100644
index 000000000..473564c29
--- /dev/null
+++ b/leetcode/1022.Sum-of-Root-To-Leaf-Binary-Numbers/1022. Sum of Root To Leaf Binary Numbers.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+import "github.com/halfrost/LeetCode-Go/structures"
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func sumRootToLeaf(root *TreeNode) int {
+	var dfs func(*TreeNode, int) int
+	dfs = func(node *TreeNode, sum int) int {
+		if node == nil {
+			return 0
+		}
+		sum = sum<<1 | node.Val
+		// 上一行也可以写作 sum = sum*2 + node.Val
+		if node.Left == nil && node.Right == nil {
+			return sum
+		}
+		return dfs(node.Left, sum) + dfs(node.Right, sum)
+	}
+	return dfs(root, 0)
+}
diff --git a/leetcode/1022.Sum-of-Root-To-Leaf-Binary-Numbers/1022. Sum of Root To Leaf Binary Numbers_test.go b/leetcode/1022.Sum-of-Root-To-Leaf-Binary-Numbers/1022. Sum of Root To Leaf Binary Numbers_test.go
new file mode 100644
index 000000000..3b7ce6f92
--- /dev/null
+++ b/leetcode/1022.Sum-of-Root-To-Leaf-Binary-Numbers/1022. Sum of Root To Leaf Binary Numbers_test.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1022 struct {
+	para1022
+	ans1022
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1022 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1022 struct {
+	one int
+}
+
+func Test_Problem1022(t *testing.T) {
+
+	qs := []question1022{
+		{
+			para1022{[]int{1, 0, 1, 0, 1, 0, 1}},
+			ans1022{22},
+		},
+
+		{
+			para1022{[]int{0}},
+			ans1022{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1022------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1022, q.para1022
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【input】:%v       【output】:%v\n", p, sumRootToLeaf(root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1022.Sum-of-Root-To-Leaf-Binary-Numbers/README.md b/leetcode/1022.Sum-of-Root-To-Leaf-Binary-Numbers/README.md
new file mode 100644
index 000000000..83578063e
--- /dev/null
+++ b/leetcode/1022.Sum-of-Root-To-Leaf-Binary-Numbers/README.md
@@ -0,0 +1,54 @@
+# [1022. Sum of Root To Leaf Binary Numbers](https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers/)
+
+## 题目
+
+You are given the root of a binary tree where each node has a value 0 or 1. Each root-to-leaf path represents a binary number starting with the most significant bit.
+
+For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is 13.
+For all leaves in the tree, consider the numbers represented by the path from the root to that leaf. Return the sum of these numbers.
+
+The test cases are generated so that the answer fits in a 32-bits integer.
+
+**Example 1:**
+
+```c
+Input: root = [1,0,1,0,1,0,1]
+Output: 22
+Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22
+```
+
+**Example 2:**
+
+```c
+Input: root = [0]
+Output: 0
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 1000]`.
+
+- `Node.val` is `0` or `1`.
+
+
+## 题目大意
+
+给定一棵结点值都是`0`或`1`的二叉树，每条从根结点到叶结点的路径都代表一个从最高有效位开始的二进制数。
+
+返回从根节点到所有叶结点的路径所表示的数字之和。
+
+
+## 解题思路
+
+采用递归的方式对根结点`root`进行后序遍历（左子树-右子树-根结点）。
+
+**递归函数的返回值**：
+
+递归遍历每个结点时，计算从根结点到当前访问结点的所表示数值`sum`都用到了上次的计算结果，所以递归函数的返回值是当前访问结点的计算结果值。
+
+**递归函数的逻辑**：
+
+- 当前遍历结点为`nil`，表示本层递归结束了，直接`return 0`。
+
+- 如果当前访问结点是叶结点，则返回从根结点到该结点所表示的数值`sum`。
+- 如果当前访问结点不是叶结点，则返回左子树和右子树所对应的结果之和。
diff --git a/Algorithms/1025. Divisor Game/1025. Divisor Game.go b/leetcode/1025.Divisor-Game/1025. Divisor Game.go
similarity index 100%
rename from Algorithms/1025. Divisor Game/1025. Divisor Game.go
rename to leetcode/1025.Divisor-Game/1025. Divisor Game.go
diff --git a/Algorithms/1025. Divisor Game/1025. Divisor Game_test.go b/leetcode/1025.Divisor-Game/1025. Divisor Game_test.go
similarity index 95%
rename from Algorithms/1025. Divisor Game/1025. Divisor Game_test.go
rename to leetcode/1025.Divisor-Game/1025. Divisor Game_test.go
index ab48e3045..3ea6237d0 100644
--- a/Algorithms/1025. Divisor Game/1025. Divisor Game_test.go	
+++ b/leetcode/1025.Divisor-Game/1025. Divisor Game_test.go	
@@ -25,12 +25,12 @@ type ans1025 struct {
 func Test_Problem1025(t *testing.T) {
 
 	qs := []question1025{
-		question1025{
+		{
 			para1025{2},
 			ans1025{true},
 		},
 
-		question1025{
+		{
 			para1025{3},
 			ans1025{false},
 		},
diff --git a/Algorithms/1025. Divisor Game/README.md b/leetcode/1025.Divisor-Game/README.md
similarity index 100%
rename from Algorithms/1025. Divisor Game/README.md
rename to leetcode/1025.Divisor-Game/README.md
diff --git a/leetcode/1026.Maximum-Difference-Between-Node-and-Ancestor/1026. Maximum Difference Between Node and Ancestor.go b/leetcode/1026.Maximum-Difference-Between-Node-and-Ancestor/1026. Maximum Difference Between Node and Ancestor.go
new file mode 100644
index 000000000..c6b088e75
--- /dev/null
+++ b/leetcode/1026.Maximum-Difference-Between-Node-and-Ancestor/1026. Maximum Difference Between Node and Ancestor.go	
@@ -0,0 +1,61 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func maxAncestorDiff(root *TreeNode) int {
+	res := 0
+	dfsAncestorDiff(root, &res)
+	return res
+}
+
+func dfsAncestorDiff(root *TreeNode, res *int) (int, int) {
+	if root == nil {
+		return -1, -1
+	}
+	leftMax, leftMin := dfsAncestorDiff(root.Left, res)
+	if leftMax == -1 && leftMin == -1 {
+		leftMax = root.Val
+		leftMin = root.Val
+	}
+	rightMax, rightMin := dfsAncestorDiff(root.Right, res)
+	if rightMax == -1 && rightMin == -1 {
+		rightMax = root.Val
+		rightMin = root.Val
+	}
+	*res = max(*res, max(abs(root.Val-min(leftMin, rightMin)), abs(root.Val-max(leftMax, rightMax))))
+	return max(leftMax, max(rightMax, root.Val)), min(leftMin, min(rightMin, root.Val))
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
diff --git a/leetcode/1026.Maximum-Difference-Between-Node-and-Ancestor/1026. Maximum Difference Between Node and Ancestor_test.go b/leetcode/1026.Maximum-Difference-Between-Node-and-Ancestor/1026. Maximum Difference Between Node and Ancestor_test.go
new file mode 100644
index 000000000..5596f6008
--- /dev/null
+++ b/leetcode/1026.Maximum-Difference-Between-Node-and-Ancestor/1026. Maximum Difference Between Node and Ancestor_test.go	
@@ -0,0 +1,60 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1026 struct {
+	para1026
+	ans1026
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1026 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1026 struct {
+	one int
+}
+
+func Test_Problem1026(t *testing.T) {
+
+	qs := []question1026{
+
+		{
+			para1026{[]int{}},
+			ans1026{0},
+		},
+
+		{
+			para1026{[]int{8, 3, 10, 1, 6, structures.NULL, 14, structures.NULL, structures.NULL, 4, 7, 13}},
+			ans1026{7},
+		},
+
+		{
+			para1026{[]int{7, 6, 4, 3, 1}},
+			ans1026{6},
+		},
+
+		{
+			para1026{[]int{1, 3, 2, 8, 4, 9}},
+			ans1026{8},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1026------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1026, q.para1026
+		tree := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxAncestorDiff(tree))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1026.Maximum-Difference-Between-Node-and-Ancestor/README.md b/leetcode/1026.Maximum-Difference-Between-Node-and-Ancestor/README.md
new file mode 100644
index 000000000..b2c946172
--- /dev/null
+++ b/leetcode/1026.Maximum-Difference-Between-Node-and-Ancestor/README.md
@@ -0,0 +1,74 @@
+# [1026. Maximum Difference Between Node and Ancestor](https://leetcode.com/problems/maximum-difference-between-node-and-ancestor/)
+
+
+
+## 题目
+
+Given the `root` of a binary tree, find the maximum value `V` for which there exists **different** nodes `A` and `B` where `V = |A.val - B.val|` and `A` is an ancestor of `B`.
+
+(A node A is an ancestor of B if either: any child of A is equal to B, or any child of A is an ancestor of B.)
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/09/09/2whqcep.jpg](https://assets.leetcode.com/uploads/2019/09/09/2whqcep.jpg)
+
+```
+Input: [8,3,10,1,6,null,14,null,null,4,7,13]
+Output: 7
+Explanation: 
+We have various ancestor-node differences, some of which are given below :
+|8 - 3| = 5
+|3 - 7| = 4
+|8 - 1| = 7
+|10 - 13| = 3
+Among all possible differences, the maximum value of 7 is obtained by |8 - 1| = 7.
+```
+
+**Note**:
+
+1. The number of nodes in the tree is between `2` and `5000`.
+2. Each node will have value between `0` and `100000`.
+
+## 题目大意
+
+给定二叉树的根节点 root，找出存在于不同节点 A 和 B 之间的最大值 V，其中 V = |A.val - B.val|，且 A 是 B 的祖先。（如果 A 的任何子节点之一为 B，或者 A 的任何子节点是 B 的祖先，那么我们认为 A 是 B 的祖先）
+
+提示：
+
+- 树中的节点数在 2 到 5000 之间。
+- 每个节点的值介于 0 到 100000 之间。
+
+
+
+## 解题思路
+
+- 给出一颗树，要求找出祖先和孩子的最大差值。
+- DPS 深搜即可。每个节点和其所有孩子的`最大值`来自于 3 个值，节点本身，递归遍历左子树的最大值，递归遍历右子树的最大值；每个节点和其所有孩子的`最小值`来自于 3 个值，节点本身，递归遍历左子树的最小值，递归遍历右子树的最小值。依次求出自身节点和其所有孩子节点的最大差值，深搜的过程中动态维护最大差值即可。
+
+## 代码
+
+```go
+func maxAncestorDiff(root *TreeNode) int {
+	res := 0
+	dfsAncestorDiff(root, &res)
+	return res
+}
+
+func dfsAncestorDiff(root *TreeNode, res *int) (int, int) {
+	if root == nil {
+		return -1, -1
+	}
+	leftMax, leftMin := dfsAncestorDiff(root.Left, res)
+	if leftMax == -1 && leftMin == -1 {
+		leftMax = root.Val
+		leftMin = root.Val
+	}
+	rightMax, rightMin := dfsAncestorDiff(root.Right, res)
+	if rightMax == -1 && rightMin == -1 {
+		rightMax = root.Val
+		rightMin = root.Val
+	}
+	*res = max(*res, max(abs(root.Val-min(leftMin, rightMin)), abs(root.Val-max(leftMax, rightMax))))
+	return max(leftMax, max(rightMax, root.Val)), min(leftMin, min(rightMin, root.Val))
+}
+```
\ No newline at end of file
diff --git a/leetcode/1028.Recover-a-Tree-From-Preorder-Traversal/1028. Recover a Tree From Preorder Traversal.go b/leetcode/1028.Recover-a-Tree-From-Preorder-Traversal/1028. Recover a Tree From Preorder Traversal.go
new file mode 100644
index 000000000..f597f671c
--- /dev/null
+++ b/leetcode/1028.Recover-a-Tree-From-Preorder-Traversal/1028. Recover a Tree From Preorder Traversal.go	
@@ -0,0 +1,84 @@
+package leetcode
+
+import (
+	"strconv"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func recoverFromPreorder(S string) *TreeNode {
+	if len(S) == 0 {
+		return &TreeNode{}
+	}
+	root, index, level := &TreeNode{}, 0, 0
+	cur := root
+	dfsBuildPreorderTree(S, &index, &level, cur)
+	return root.Right
+}
+
+func dfsBuildPreorderTree(S string, index, level *int, cur *TreeNode) (newIndex *int) {
+	if *index == len(S) {
+		return index
+	}
+	if *index == 0 && *level == 0 {
+		i := 0
+		for i = *index; i < len(S); i++ {
+			if !isDigital(S[i]) {
+				break
+			}
+		}
+		num, _ := strconv.Atoi(S[*index:i])
+		tmp := &TreeNode{Val: num, Left: nil, Right: nil}
+		cur.Right = tmp
+		nLevel := *level + 1
+		index = dfsBuildPreorderTree(S, &i, &nLevel, tmp)
+		index = dfsBuildPreorderTree(S, index, &nLevel, tmp)
+	}
+	i := 0
+	for i = *index; i < len(S); i++ {
+		if isDigital(S[i]) {
+			break
+		}
+	}
+	if *level == i-*index {
+		j := 0
+		for j = i; j < len(S); j++ {
+			if !isDigital(S[j]) {
+				break
+			}
+		}
+		num, _ := strconv.Atoi(S[i:j])
+		tmp := &TreeNode{Val: num, Left: nil, Right: nil}
+		if cur.Left == nil {
+			cur.Left = tmp
+			nLevel := *level + 1
+			index = dfsBuildPreorderTree(S, &j, &nLevel, tmp)
+			index = dfsBuildPreorderTree(S, index, level, cur)
+		} else if cur.Right == nil {
+			cur.Right = tmp
+			nLevel := *level + 1
+			index = dfsBuildPreorderTree(S, &j, &nLevel, tmp)
+			index = dfsBuildPreorderTree(S, index, level, cur)
+		}
+	}
+	return index
+}
+
+func isDigital(v byte) bool {
+	if v >= '0' && v <= '9' {
+		return true
+	}
+	return false
+}
diff --git a/leetcode/1028.Recover-a-Tree-From-Preorder-Traversal/1028. Recover a Tree From Preorder Traversal_test.go b/leetcode/1028.Recover-a-Tree-From-Preorder-Traversal/1028. Recover a Tree From Preorder Traversal_test.go
new file mode 100644
index 000000000..880213e4b
--- /dev/null
+++ b/leetcode/1028.Recover-a-Tree-From-Preorder-Traversal/1028. Recover a Tree From Preorder Traversal_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1028 struct {
+	para1028
+	ans1028
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1028 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1028 struct {
+	one []int
+}
+
+func Test_Problem1028(t *testing.T) {
+
+	qs := []question1028{
+		{
+			para1028{"1-2--3--4-5--6--7"},
+			ans1028{[]int{1, 2, 5, 3, 4, 6, 7}},
+		},
+
+		{
+			para1028{"1-2--3---4-5--6---7"},
+			ans1028{[]int{1, 2, 5, 3, structures.NULL, 6, structures.NULL, 4, structures.NULL, 7}},
+		},
+
+		{
+			para1028{"1-401--349---90--88"},
+			ans1028{[]int{1, 401, structures.NULL, 349, 88, 90}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1028------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1028, q.para1028
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.Tree2ints(recoverFromPreorder(p.one)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1028.Recover-a-Tree-From-Preorder-Traversal/README.md b/leetcode/1028.Recover-a-Tree-From-Preorder-Traversal/README.md
new file mode 100755
index 000000000..5e7aef509
--- /dev/null
+++ b/leetcode/1028.Recover-a-Tree-From-Preorder-Traversal/README.md
@@ -0,0 +1,56 @@
+# [1028. Recover a Tree From Preorder Traversal](https://leetcode.com/problems/recover-a-tree-from-preorder-traversal/)
+
+
+## 题目
+
+We run a preorder depth first search on the `root` of a binary tree.
+
+At each node in this traversal, we output `D` dashes (where `D` is the *depth* of this node), then we output the value of this node. *(If the depth of a node is `D`, the depth of its immediate child is `D+1`. The depth of the root node is `0`.)*
+
+If a node has only one child, that child is guaranteed to be the left child.
+
+Given the output `S` of this traversal, recover the tree and return its `root`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/04/08/recover-a-tree-from-preorder-traversal.png](https://assets.leetcode.com/uploads/2019/04/08/recover-a-tree-from-preorder-traversal.png)
+
+    Input: "1-2--3--4-5--6--7"
+    Output: [1,2,5,3,4,6,7]
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2019/04/11/screen-shot-2019-04-10-at-114101-pm.png](https://assets.leetcode.com/uploads/2019/04/11/screen-shot-2019-04-10-at-114101-pm.png)
+
+    Input: "1-2--3---4-5--6---7"
+    Output: [1,2,5,3,null,6,null,4,null,7]
+
+**Example 3:**
+
+![https://assets.leetcode.com/uploads/2019/04/11/screen-shot-2019-04-10-at-114955-pm.png](https://assets.leetcode.com/uploads/2019/04/11/screen-shot-2019-04-10-at-114955-pm.png)
+
+    Input: "1-401--349---90--88"
+    Output: [1,401,null,349,88,90]
+
+**Note:**
+
+- The number of nodes in the original tree is between `1` and `1000`.
+- Each node will have a value between `1` and `10^9`.
+
+## 题目大意
+
+我们从二叉树的根节点 root 开始进行深度优先搜索。
+
+在遍历中的每个节点处，我们输出 D 条短划线（其中 D 是该节点的深度），然后输出该节点的值。（如果节点的深度为 D，则其直接子节点的深度为 D + 1。根节点的深度为 0）。如果节点只有一个子节点，那么保证该子节点为左子节点。给出遍历输出 S，还原树并返回其根节点 root。
+
+
+提示：
+
+- 原始树中的节点数介于 1 和 1000 之间。
+- 每个节点的值介于 1 和 10 ^ 9 之间。
+
+
+## 解题思路
+
+- 给出一个字符串，字符串是一个树的先根遍历的结果，其中破折号的个数代表层数。请根据这个字符串生成对应的树。
+- 这一题解题思路比较明确，用 DFS 就可以解题。边深搜字符串，边根据破折号的个数判断当前节点是否属于本层。如果不属于本层，回溯到之前的根节点，添加叶子节点以后再继续深搜。需要注意的是每次深搜时，扫描字符串的 index 需要一直保留，回溯也需要用到这个 index。
diff --git a/Algorithms/1030. Matrix Cells in Distance Order/1030. Matrix Cells in Distance Order.go b/leetcode/1030.Matrix-Cells-in-Distance-Order/1030. Matrix Cells in Distance Order.go
similarity index 82%
rename from Algorithms/1030. Matrix Cells in Distance Order/1030. Matrix Cells in Distance Order.go
rename to leetcode/1030.Matrix-Cells-in-Distance-Order/1030. Matrix Cells in Distance Order.go
index 0c8166c76..449f1d8c4 100644
--- a/Algorithms/1030. Matrix Cells in Distance Order/1030. Matrix Cells in Distance Order.go	
+++ b/leetcode/1030.Matrix-Cells-in-Distance-Order/1030. Matrix Cells in Distance Order.go	
@@ -21,3 +21,17 @@ func allCellsDistOrder(R int, C int, r0 int, c0 int) [][]int {
 	}
 	return result
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
diff --git a/leetcode/1030.Matrix-Cells-in-Distance-Order/1030. Matrix Cells in Distance Order_test.go b/leetcode/1030.Matrix-Cells-in-Distance-Order/1030. Matrix Cells in Distance Order_test.go
new file mode 100644
index 000000000..5f6a07a66
--- /dev/null
+++ b/leetcode/1030.Matrix-Cells-in-Distance-Order/1030. Matrix Cells in Distance Order_test.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1030 struct {
+	para1030
+	ans1030
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1030 struct {
+	R  int
+	C  int
+	r0 int
+	c0 int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1030 struct {
+	one [][]int
+}
+
+func Test_Problem1030(t *testing.T) {
+
+	qs := []question1030{
+
+		{
+			para1030{1, 2, 0, 0},
+			ans1030{[][]int{{0, 0}, {0, 1}}},
+		},
+
+		{
+			para1030{2, 2, 0, 1},
+			ans1030{[][]int{{0, 1}, {0, 0}, {1, 1}, {1, 0}}},
+		},
+
+		{
+			para1030{2, 3, 1, 2},
+			ans1030{[][]int{{1, 2}, {0, 2}, {1, 1}, {0, 1}, {1, 0}, {0, 0}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1030------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1030, q.para1030
+		fmt.Printf("【input】:%v       【output】:%v\n", p, allCellsDistOrder(p.R, p.C, p.r0, p.c0))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/1030. Matrix Cells in Distance Order/README.md b/leetcode/1030.Matrix-Cells-in-Distance-Order/README.md
similarity index 100%
rename from Algorithms/1030. Matrix Cells in Distance Order/README.md
rename to leetcode/1030.Matrix-Cells-in-Distance-Order/README.md
diff --git a/leetcode/1034.Coloring-A-Border/1034.Coloring A Border.go b/leetcode/1034.Coloring-A-Border/1034.Coloring A Border.go
new file mode 100644
index 000000000..70df28edd
--- /dev/null
+++ b/leetcode/1034.Coloring-A-Border/1034.Coloring A Border.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+type point struct {
+	x int
+	y int
+}
+
+type gridInfo struct {
+	m             int
+	n             int
+	grid          [][]int
+	originalColor int
+}
+
+func colorBorder(grid [][]int, row, col, color int) [][]int {
+	m, n := len(grid), len(grid[0])
+	dirs := []point{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}
+	vis := make([][]bool, m)
+	for i := range vis {
+		vis[i] = make([]bool, n)
+	}
+	var borders []point
+	gInfo := gridInfo{
+		m:             m,
+		n:             n,
+		grid:          grid,
+		originalColor: grid[row][col],
+	}
+	dfs(row, col, gInfo, dirs, vis, &borders)
+	for _, p := range borders {
+		grid[p.x][p.y] = color
+	}
+	return grid
+}
+
+func dfs(x, y int, gInfo gridInfo, dirs []point, vis [][]bool, borders *[]point) {
+	vis[x][y] = true
+	isBorder := false
+	for _, dir := range dirs {
+		nx, ny := x+dir.x, y+dir.y
+		if !(0 <= nx && nx < gInfo.m && 0 <= ny && ny < gInfo.n && gInfo.grid[nx][ny] == gInfo.originalColor) {
+			isBorder = true
+		} else if !vis[nx][ny] {
+			dfs(nx, ny, gInfo, dirs, vis, borders)
+		}
+	}
+	if isBorder {
+		*borders = append(*borders, point{x, y})
+	}
+}
diff --git a/leetcode/1034.Coloring-A-Border/1034.Coloring A Border_test.go b/leetcode/1034.Coloring-A-Border/1034.Coloring A Border_test.go
new file mode 100644
index 000000000..812a9c241
--- /dev/null
+++ b/leetcode/1034.Coloring-A-Border/1034.Coloring A Border_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1034 struct {
+	para1034
+	ans1034
+}
+
+// para 是参数
+type para1034 struct {
+	grid  [][]int
+	row   int
+	col   int
+	color int
+}
+
+// ans 是答案
+type ans1034 struct {
+	ans [][]int
+}
+
+func Test_Problem1034(t *testing.T) {
+
+	qs := []question1034{
+
+		{
+			para1034{[][]int{{1, 1}, {1, 2}}, 0, 0, 3},
+			ans1034{[][]int{{3, 3}, {3, 2}}},
+		},
+
+		{
+			para1034{[][]int{{1, 2, 2}, {2, 3, 2}}, 0, 1, 3},
+			ans1034{[][]int{{1, 3, 3}, {2, 3, 3}}},
+		},
+
+		{
+			para1034{[][]int{{1, 1, 1}, {1, 1, 1}, {1, 1, 1}}, 1, 1, 2},
+			ans1034{[][]int{{2, 2, 2}, {2, 1, 2}, {2, 2, 2}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1034------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1034, q.para1034
+		fmt.Printf("【input】:%v    【output】:%v\n", p, colorBorder(p.grid, p.row, p.col, p.color))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1034.Coloring-A-Border/README.md b/leetcode/1034.Coloring-A-Border/README.md
new file mode 100644
index 000000000..6700260f9
--- /dev/null
+++ b/leetcode/1034.Coloring-A-Border/README.md
@@ -0,0 +1,108 @@
+# [1034. Coloring A Border](https://leetcode.com/problems/coloring-a-border/)
+
+## 题目
+
+You are given an m x n integer matrix grid, and three integers row, col, and color. Each value in the grid represents the color of the grid square at that location.
+
+Two squares belong to the same connected component if they have the same color and are next to each other in any of the 4 directions.
+
+The border of a connected component is all the squares in the connected component that are either 4-directionally adjacent to a square not in the component, or on the boundary of the grid (the first or last row or column).
+
+You should color the border of the connected component that contains the square grid[row][col] with color.
+
+Return the final grid.
+
+**Example 1**:
+
+    Input: grid = [[1,1],[1,2]], row = 0, col = 0, color = 3
+    Output: [[3,3],[3,2]]
+
+**Example 2**:
+
+    Input: grid = [[1,2,2],[2,3,2]], row = 0, col = 1, color = 3
+    Output: [[1,3,3],[2,3,3]]
+
+**Example 3**:
+
+    Input: grid = [[1,1,1],[1,1,1],[1,1,1]], row = 1, col = 1, color = 2
+    Output: [[2,2,2],[2,1,2],[2,2,2]]
+
+**Constraints:**
+
+- m == grid.length
+- n == grid[i].length
+- 1 <= m, n <= 50
+- 1 <= grid[i][j], color <= 1000
+- 0 <= row < m
+- 0 <= col < n
+
+## 题目大意
+
+给你一个大小为 m x n 的整数矩阵 grid ，表示一个网格。另给你三个整数 row、col 和 color 。网格中的每个值表示该位置处的网格块的颜色。
+
+当两个网格块的颜色相同，而且在四个方向中任意一个方向上相邻时，它们属于同一连通分量
+
+边界：在连通分量的块中（前提）并且满足以下条件之一：
+（1）要么上下左右存在一个块不在连通分量里面
+（2）要么这个块的位置在整个grid的边框上
+
+请你使用指定颜色 color 为所有包含网格块 grid[row][col] 的连通分量的边界进行着色，并返回最终的网格 grid 。
+
+## 解题思路
+
+- 用 bfs 进行遍历选出边界，使用 color 给边界着色
+
+## 代码
+
+```go
+package leetcode
+
+type point struct {
+	x int
+	y int
+}
+
+type gridInfo struct {
+	m             int
+	n             int
+	grid          [][]int
+	originalColor int
+}
+
+func colorBorder(grid [][]int, row, col, color int) [][]int {
+	m, n := len(grid), len(grid[0])
+	dirs := []point{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}
+	vis := make([][]bool, m)
+	for i := range vis {
+		vis[i] = make([]bool, n)
+	}
+	var borders []point
+	gInfo := gridInfo{
+		m:             m,
+		n:             n,
+		grid:          grid,
+		originalColor: grid[row][col],
+	}
+	dfs(row, col, gInfo, dirs, vis, &borders)
+	for _, p := range borders {
+		grid[p.x][p.y] = color
+	}
+	return grid
+}
+
+func dfs(x, y int, gInfo gridInfo, dirs []point, vis [][]bool, borders *[]point) {
+	vis[x][y] = true
+	isBorder := false
+	for _, dir := range dirs {
+		nx, ny := x+dir.x, y+dir.y
+		if !(0 <= nx && nx < gInfo.m && 0 <= ny && ny < gInfo.n && gInfo.grid[nx][ny] == gInfo.originalColor) {
+			isBorder = true
+		} else if !vis[nx][ny] {
+			dfs(nx, ny, gInfo, dirs, vis, borders)
+		}
+	}
+	if isBorder {
+		*borders = append(*borders, point{x, y})
+	}
+}
+```
\ No newline at end of file
diff --git a/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang.go b/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang.go
new file mode 100644
index 000000000..825ae8504
--- /dev/null
+++ b/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang.go	
@@ -0,0 +1,5 @@
+package leetcode
+
+func isBoomerang(points [][]int) bool {
+	return (points[0][0]-points[1][0])*(points[0][1]-points[2][1]) != (points[0][0]-points[2][0])*(points[0][1]-points[1][1])
+}
diff --git a/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang_test.go b/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang_test.go
new file mode 100644
index 000000000..2d451a89c
--- /dev/null
+++ b/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1037 struct {
+	para1037
+	ans1037
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1037 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1037 struct {
+	one bool
+}
+
+func Test_Problem1037(t *testing.T) {
+
+	qs := []question1037{
+		{
+			para1037{[][]int{{1, 2}, {2, 3}, {3, 2}}},
+			ans1037{true},
+		},
+
+		{
+			para1037{[][]int{{1, 1}, {2, 2}, {3, 3}}},
+			ans1037{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1037------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1037, q.para1037
+		fmt.Printf("【input】:%v       【output】:%v\n", p, isBoomerang(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1037.Valid-Boomerang/README.md b/leetcode/1037.Valid-Boomerang/README.md
new file mode 100644
index 000000000..89a4000db
--- /dev/null
+++ b/leetcode/1037.Valid-Boomerang/README.md
@@ -0,0 +1,49 @@
+# [1037. Valid Boomerang](https://leetcode.com/problems/valid-boomerang/)
+
+
+## 题目
+
+A *boomerang* is a set of 3 points that are all distinct and **not** in a straight line.
+
+Given a list of three points in the plane, return whether these points are a boomerang.
+
+**Example 1**:
+
+```
+Input: [[1,1],[2,3],[3,2]]
+Output: true
+```
+
+**Example 2**:
+
+```
+Input: [[1,1],[2,2],[3,3]]
+Output: false
+```
+
+**Note**:
+
+1. `points.length == 3`
+2. `points[i].length == 2`
+3. `0 <= points[i][j] <= 100`
+
+## 题目大意
+
+回旋镖定义为一组三个点，这些点各不相同且不在一条直线上。给出平面上三个点组成的列表，判断这些点是否可以构成回旋镖。
+
+## 解题思路
+
+- 判断给出的 3 组点能否满足回旋镖。
+- 简单题。判断 3 个点组成的 2 条直线的斜率是否相等。由于斜率的计算是除法，还可能遇到分母为 0 的情况，那么可以转换成乘法，交叉相乘再判断是否相等，就可以省去判断分母为 0 的情况了，代码也简洁成一行了。
+
+## 代码
+
+```go
+
+package leetcode
+
+func isBoomerang(points [][]int) bool {
+	return (points[0][0]-points[1][0])*(points[0][1]-points[2][1]) != (points[0][0]-points[2][0])*(points[0][1]-points[1][1])
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1038.Binary-Search-Tree-to-Greater-Sum-Tree/1038. Binary Search Tree to Greater Sum Tree.go b/leetcode/1038.Binary-Search-Tree-to-Greater-Sum-Tree/1038. Binary Search Tree to Greater Sum Tree.go
new file mode 100644
index 000000000..eedf6c39c
--- /dev/null
+++ b/leetcode/1038.Binary-Search-Tree-to-Greater-Sum-Tree/1038. Binary Search Tree to Greater Sum Tree.go	
@@ -0,0 +1,36 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func bstToGst(root *TreeNode) *TreeNode {
+	if root == nil {
+		return root
+	}
+	sum := 0
+	dfs1038(root, &sum)
+	return root
+}
+
+func dfs1038(root *TreeNode, sum *int) {
+	if root == nil {
+		return
+	}
+	dfs1038(root.Right, sum)
+	root.Val += *sum
+	*sum = root.Val
+	dfs1038(root.Left, sum)
+}
diff --git a/leetcode/1038.Binary-Search-Tree-to-Greater-Sum-Tree/1038. Binary Search Tree to Greater Sum Tree_test.go b/leetcode/1038.Binary-Search-Tree-to-Greater-Sum-Tree/1038. Binary Search Tree to Greater Sum Tree_test.go
new file mode 100644
index 000000000..e6310e25f
--- /dev/null
+++ b/leetcode/1038.Binary-Search-Tree-to-Greater-Sum-Tree/1038. Binary Search Tree to Greater Sum Tree_test.go	
@@ -0,0 +1,76 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1038 struct {
+	para1038
+	ans1038
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1038 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1038 struct {
+	one []int
+}
+
+func Test_Problem1038(t *testing.T) {
+
+	qs := []question1038{
+
+		{
+			para1038{[]int{3, 1, structures.NULL, 0, structures.NULL, -4, structures.NULL, structures.NULL, -2}},
+			ans1038{[]int{3, 4, structures.NULL, 4, structures.NULL, -2, structures.NULL, structures.NULL, 2}},
+		},
+
+		{
+			para1038{[]int{2, 1}},
+			ans1038{[]int{2, 3}},
+		},
+
+		{
+			para1038{[]int{}},
+			ans1038{[]int{}},
+		},
+
+		{
+			para1038{[]int{4, 1, 6, 0, 2, 5, 7, structures.NULL, structures.NULL, structures.NULL, 3, structures.NULL, structures.NULL, structures.NULL, 8}},
+			ans1038{[]int{30, 36, 21, 36, 35, 26, 15, structures.NULL, structures.NULL, structures.NULL, 33, structures.NULL, structures.NULL, structures.NULL, 8}},
+		},
+
+		{
+			para1038{[]int{0, structures.NULL, 1}},
+			ans1038{[]int{1, structures.NULL, 1}},
+		},
+
+		{
+			para1038{[]int{1, 0, 2}},
+			ans1038{[]int{3, 3, 2}},
+		},
+
+		{
+			para1038{[]int{3, 2, 4, 1}},
+			ans1038{[]int{7, 9, 4, 10}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1038------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1038, q.para1038
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", structures.Tree2ints(bstToGst(root)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1038.Binary-Search-Tree-to-Greater-Sum-Tree/README.md b/leetcode/1038.Binary-Search-Tree-to-Greater-Sum-Tree/README.md
new file mode 100644
index 000000000..69481c9dd
--- /dev/null
+++ b/leetcode/1038.Binary-Search-Tree-to-Greater-Sum-Tree/README.md
@@ -0,0 +1,107 @@
+# [1038. Binary Search Tree to Greater Sum Tree](https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/)
+
+
+## 题目
+
+Given the `root` of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.
+
+As a reminder, a *binary search tree* is a tree that satisfies these constraints:
+
+- The left subtree of a node contains only nodes with keys **less than** the node's key.
+- The right subtree of a node contains only nodes with keys **greater than** the node's key.
+- Both the left and right subtrees must also be binary search trees.
+
+**Note:** This question is the same as 538: [https://leetcode.com/problems/convert-bst-to-greater-tree/](https://leetcode.com/problems/convert-bst-to-greater-tree/)
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/05/02/tree.png](https://assets.leetcode.com/uploads/2019/05/02/tree.png)
+
+```
+Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
+Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]
+```
+
+**Example 2:**
+
+```
+Input: root = [0,null,1]
+Output: [1,null,1]
+```
+
+**Example 3:**
+
+```
+Input: root = [1,0,2]
+Output: [3,3,2]
+```
+
+**Example 4:**
+
+```
+Input: root = [3,2,4,1]
+Output: [7,9,4,10]
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 100]`.
+- `0 <= Node.val <= 100`
+- All the values in the tree are **unique**.
+- `root` is guaranteed to be a valid binary search tree.
+
+## 题目大意
+
+给出二叉 搜索 树的根节点，该树的节点值各不相同，请你将其转换为累加树（Greater Sum Tree），使每个节点 node 的新值等于原树中大于或等于 node.val 的值之和。
+
+提醒一下，二叉搜索树满足下列约束条件：
+
+- 节点的左子树仅包含键 小于 节点键的节点。
+- 节点的右子树仅包含键 大于 节点键的节点。
+- 左右子树也必须是二叉搜索树。
+
+## 解题思路
+
+- 根据二叉搜索树的有序性，想要将其转换为累加树，只需按照 右节点 - 根节点 - 左节点的顺序遍历，并累加和即可。
+- 此题同第 538 题。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func bstToGst(root *TreeNode) *TreeNode {
+	if root == nil {
+		return root
+	}
+	sum := 0
+	dfs1038(root, &sum)
+	return root
+}
+
+func dfs1038(root *TreeNode, sum *int) {
+	if root == nil {
+		return
+	}
+	dfs1038(root.Right, sum)
+	root.Val += *sum
+	*sum = root.Val
+	dfs1038(root.Left, sum)
+}
+```
\ No newline at end of file
diff --git a/Algorithms/1040. Moving Stones Until Consecutive II/1040. Moving Stones Until Consecutive II.go b/leetcode/1040.Moving-Stones-Until-Consecutive-II/1040. Moving Stones Until Consecutive II.go
similarity index 84%
rename from Algorithms/1040. Moving Stones Until Consecutive II/1040. Moving Stones Until Consecutive II.go
rename to leetcode/1040.Moving-Stones-Until-Consecutive-II/1040. Moving Stones Until Consecutive II.go
index 0ec750a27..c17aac58f 100644
--- a/Algorithms/1040. Moving Stones Until Consecutive II/1040. Moving Stones Until Consecutive II.go	
+++ b/leetcode/1040.Moving-Stones-Until-Consecutive-II/1040. Moving Stones Until Consecutive II.go	
@@ -32,3 +32,17 @@ func numMovesStonesII(stones []int) []int {
 	}
 	return []int{minStep, maxStep}
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/Algorithms/1040. Moving Stones Until Consecutive II/1040. Moving Stones Until Consecutive II_test.go b/leetcode/1040.Moving-Stones-Until-Consecutive-II/1040. Moving Stones Until Consecutive II_test.go
similarity index 93%
rename from Algorithms/1040. Moving Stones Until Consecutive II/1040. Moving Stones Until Consecutive II_test.go
rename to leetcode/1040.Moving-Stones-Until-Consecutive-II/1040. Moving Stones Until Consecutive II_test.go
index baf693828..0bed69998 100644
--- a/Algorithms/1040. Moving Stones Until Consecutive II/1040. Moving Stones Until Consecutive II_test.go	
+++ b/leetcode/1040.Moving-Stones-Until-Consecutive-II/1040. Moving Stones Until Consecutive II_test.go	
@@ -25,22 +25,22 @@ type ans1040 struct {
 func Test_Problem1040(t *testing.T) {
 
 	qs := []question1040{
-		question1040{
+		{
 			para1040{[]int{7, 4, 9}},
 			ans1040{[]int{1, 2}},
 		},
 
-		question1040{
+		{
 			para1040{[]int{6, 5, 4, 3, 10}},
 			ans1040{[]int{2, 3}},
 		},
 
-		question1040{
+		{
 			para1040{[]int{100, 101, 104, 102, 103}},
 			ans1040{[]int{0, 0}},
 		},
 
-		question1040{
+		{
 			para1040{[]int{1, 3, 5, 7, 10}},
 			ans1040{[]int{2, 4}},
 		},
diff --git a/leetcode/1040.Moving-Stones-Until-Consecutive-II/README.md b/leetcode/1040.Moving-Stones-Until-Consecutive-II/README.md
new file mode 100755
index 000000000..290839cb6
--- /dev/null
+++ b/leetcode/1040.Moving-Stones-Until-Consecutive-II/README.md
@@ -0,0 +1,79 @@
+# [1040. Moving Stones Until Consecutive II](https://leetcode.com/problems/moving-stones-until-consecutive-ii/)
+
+
+## 题目
+
+On an **infinite** number line, the position of the i-th stone is given by `stones[i]`. Call a stone an *endpoint stone* if it has the smallest or largest position.
+
+Each turn, you pick up an endpoint stone and move it to an unoccupied position so that it is no longer an endpoint stone.
+
+In particular, if the stones are at say, `stones = [1,2,5]`, you **cannot** move the endpoint stone at position 5, since moving it to any position (such as 0, or 3) will still keep that stone as an endpoint stone.
+
+The game ends when you cannot make any more moves, ie. the stones are in consecutive positions.
+
+When the game ends, what is the minimum and maximum number of moves that you could have made? Return the answer as an length 2 array: `answer = [minimum_moves, maximum_moves]`
+
+**Example 1:**
+
+    Input: [7,4,9]
+    Output: [1,2]
+    Explanation: 
+    We can move 4 -> 8 for one move to finish the game.
+    Or, we can move 9 -> 5, 4 -> 6 for two moves to finish the game.
+
+**Example 2:**
+
+    Input: [6,5,4,3,10]
+    Output: [2,3]
+    We can move 3 -> 8 then 10 -> 7 to finish the game.
+    Or, we can move 3 -> 7, 4 -> 8, 5 -> 9 to finish the game.
+    Notice we cannot move 10 -> 2 to finish the game, because that would be an illegal move.
+
+**Example 3:**
+
+    Input: [100,101,104,102,103]
+    Output: [0,0]
+
+**Note:**
+
+1. `3 <= stones.length <= 10^4`
+2. `1 <= stones[i] <= 10^9`
+3. `stones[i]` have distinct values.
+
+
+## 题目大意
+
+在一个长度无限的数轴上，第 i 颗石子的位置为 stones[i]。如果一颗石子的位置最小/最大，那么该石子被称作端点石子。每个回合，你可以将一颗端点石子拿起并移动到一个未占用的位置，使得该石子不再是一颗端点石子。值得注意的是，如果石子像 stones = [1,2,5] 这样，你将无法移动位于位置 5 的端点石子，因为无论将它移动到任何位置（例如 0 或 3），该石子都仍然会是端点石子。当你无法进行任何移动时，即，这些石子的位置连续时，游戏结束。
+
+要使游戏结束，你可以执行的最小和最大移动次数分别是多少？ 以长度为 2 的数组形式返回答案：answer = [minimum\_moves, maximum\_moves] 。
+
+提示：
+
+1. 3 <= stones.length <= 10^4
+2. 1 <= stones[i] <= 10^9
+3. stones[i] 的值各不相同。
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面代表的是石头的坐标。要求移动石头，最终使得这些石头的坐标是一个连续的自然数列。但是规定，当一个石头是端点的时候，是不能移动的，例如 [1,2,5]，5 是端点，不能把 5 移到 3 或者 0 的位置，因为移动之后，这个石头仍然是端点。最终输出将所有石头排成连续的自然数列所需的最小步数和最大步数。
+- 这道题的关键就是如何保证端点石头不能再次移动到端点的限制。例如，[5,6,8,9,20]，20 是端点，但是 20 就可以移动到 7 的位置，最终形成 [5,6,7,8,9] 的连续序列。但是 [5,6,7,8,20]，这种情况 20 就不能移动到 9 了，只能让 8 移动到 9，20 再移动到 8 的位置，最终还是形成了 [5,6,7,8,9]，但是步数需要 2 步。经过上述分析，可以得到，端点石头只能往中间空挡的位置移动，如果中间没有空挡，那么需要借助一个石头先制造一个空挡，然后端点石头再插入到中间，这样最少是需要 2 步。
+- 再来考虑极值的情况。先看最大步数，最大步数肯定慢慢移动，一次移动一格，并且移动的格数最多。这里有两个极端情况，把数组里面的数全部都移动到最左端点，把数组里面的数全部都移动到最右端点。每次只移动一格。例如，全部都移到最右端点：
+
+        [3,4,5,6,10] // 初始状态，连续的情况
+        [4,5,6,7,10] // 第一步，把 3 挪到右边第一个可以插入的位置，即 7
+        [5,6,7,8,10] // 第二步，把 4 挪到右边第一个可以插入的位置，即 8
+        [6,7,8,9,10] // 第三步，把 5 挪到右边第一个可以插入的位置，即 9
+        
+        
+        [1,3,5,7,10] // 初始状态，不连续的情况
+        [3,4,5,7,10] // 第一步，把 1 挪到右边第一个可以插入的位置，即 4
+        [4,5,6,7,10] // 第二步，把 3 挪到右边第一个可以插入的位置，即 6
+        [5,6,7,8,10] // 第三步，把 4 挪到右边第一个可以插入的位置，即 8
+        [6,7,8,9,10] // 第四步，把 5 挪到右边第一个可以插入的位置，即 9
+
+    挪动的过程类似翻滚，最左边的石头挪到右边第一个可以放下的地方。然后不断的往右翻滚。把数组中的数全部都移动到最左边也同理。对比这两种情况的最大值，即是移动的最大步数。
+
+- 再看最小步数。这里就涉及到了滑动窗口了。由于最终是要形成连续的自然数列，所以滑动窗口的大小已经固定成 n 了，从数组的 0 下标可以往右滑动窗口，这个窗口中能包含的数字越多，代表窗口外的数字越少，那么把这些数字放进窗口内的步数也最小。于是可以求得最小步数。这里有一个比较坑的地方就是题目中的那个`“端点不能移动以后还是端点”`的限制。针对这种情况，需要额外的判断。如果当前窗口内有 n-1 个元素了，即只有一个端点在窗口外，并且窗口右边界的值减去左边界的值也等于 n-1，代表这个窗口内已经都是连续数字了。这种情况端点想融合到这个连续数列中，最少需要 2 步(上文已经分析过了)。
+- 注意一些边界情况。如果窗口从左往右滑动，窗口右边界滑到最右边了，但是窗口右边界的数字减去左边界的数字还是小于窗口大小 n，代表已经滑到头了，可以直接 break 出去。为什么滑到头了呢？由于数组经过从小到大排序以后，数字越往右边越大，当前数字是小值，已经满足了 `stones[right]-stones[left] < n`，左边界继续往右移动只会使得 `stones[left]` 更大，就更加小于 n 了。而我们需要寻找的是 `stones[right]-stones[left] >= n` 的边界点，肯定再也找不到了。
diff --git a/Algorithms/1047. Remove All Adjacent Duplicates In String/1047. Remove All Adjacent Duplicates In String.go b/leetcode/1047.Remove-All-Adjacent-Duplicates-In-String/1047. Remove All Adjacent Duplicates In String.go
similarity index 100%
rename from Algorithms/1047. Remove All Adjacent Duplicates In String/1047. Remove All Adjacent Duplicates In String.go
rename to leetcode/1047.Remove-All-Adjacent-Duplicates-In-String/1047. Remove All Adjacent Duplicates In String.go
diff --git a/Algorithms/1047. Remove All Adjacent Duplicates In String/1047. Remove All Adjacent Duplicates In String_test.go b/leetcode/1047.Remove-All-Adjacent-Duplicates-In-String/1047. Remove All Adjacent Duplicates In String_test.go
similarity index 97%
rename from Algorithms/1047. Remove All Adjacent Duplicates In String/1047. Remove All Adjacent Duplicates In String_test.go
rename to leetcode/1047.Remove-All-Adjacent-Duplicates-In-String/1047. Remove All Adjacent Duplicates In String_test.go
index e57c00cfa..f11061a72 100644
--- a/Algorithms/1047. Remove All Adjacent Duplicates In String/1047. Remove All Adjacent Duplicates In String_test.go	
+++ b/leetcode/1047.Remove-All-Adjacent-Duplicates-In-String/1047. Remove All Adjacent Duplicates In String_test.go	
@@ -26,7 +26,7 @@ func Test_Problem1047(t *testing.T) {
 
 	qs := []question1047{
 
-		question1047{
+		{
 			para1047{"abbaca"},
 			ans1047{"ca"},
 		},
diff --git a/Algorithms/1047. Remove All Adjacent Duplicates In String/README.md b/leetcode/1047.Remove-All-Adjacent-Duplicates-In-String/README.md
similarity index 100%
rename from Algorithms/1047. Remove All Adjacent Duplicates In String/README.md
rename to leetcode/1047.Remove-All-Adjacent-Duplicates-In-String/README.md
diff --git a/leetcode/1048.Longest-String-Chain/1048. Longest String Chain.go b/leetcode/1048.Longest-String-Chain/1048. Longest String Chain.go
new file mode 100644
index 000000000..b16653ba6
--- /dev/null
+++ b/leetcode/1048.Longest-String-Chain/1048. Longest String Chain.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import "sort"
+
+func longestStrChain(words []string) int {
+	sort.Slice(words, func(i, j int) bool { return len(words[i]) < len(words[j]) })
+	poss, res := make([]int, 16+2), 0
+	for i, w := range words {
+		if poss[len(w)] == 0 {
+			poss[len(w)] = i
+		}
+	}
+	dp := make([]int, len(words))
+	for i := len(words) - 1; i >= 0; i-- {
+		dp[i] = 1
+		for j := poss[len(words[i])+1]; j < len(words) && len(words[j]) == len(words[i])+1; j++ {
+			if isPredecessor(words[j], words[i]) {
+				dp[i] = max(dp[i], 1+dp[j])
+			}
+		}
+		res = max(res, dp[i])
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func isPredecessor(long, short string) bool {
+	i, j := 0, 0
+	wasMismatch := false
+	for j < len(short) {
+		if long[i] != short[j] {
+			if wasMismatch {
+				return false
+			}
+			wasMismatch = true
+			i++
+			continue
+		}
+		i++
+		j++
+	}
+	return true
+}
diff --git a/leetcode/1048.Longest-String-Chain/1048. Longest String Chain_test.go b/leetcode/1048.Longest-String-Chain/1048. Longest String Chain_test.go
new file mode 100644
index 000000000..97c9b3b5a
--- /dev/null
+++ b/leetcode/1048.Longest-String-Chain/1048. Longest String Chain_test.go	
@@ -0,0 +1,78 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1048 struct {
+	para1048
+	ans1048
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1048 struct {
+	words []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1048 struct {
+	one int
+}
+
+func Test_Problem1048(t *testing.T) {
+
+	qs := []question1048{
+
+		{
+			para1048{[]string{"a", "b", "ab", "bac"}},
+			ans1048{2},
+		},
+
+		{
+			para1048{[]string{"xbc", "pcxbcf", "xb", "cxbc", "pcxbc"}},
+			ans1048{5},
+		},
+
+		{
+			para1048{[]string{"a", "b", "ba", "bca", "bda", "bdca"}},
+			ans1048{4},
+		},
+
+		{
+			para1048{[]string{"qjcaeymang", "bqiq", "bcntqiqulagurhz", "lyctmomvis", "bdnhym", "crxrdlv", "wo", "kijftxssyqmui", "abtcrjs", "rceecupq", "crxrdclv", "tvwkxrev", "oc", "lrzzcl", "snpzuykyobci", "abbtczrjs", "rpqojpmv",
+				"kbfbcjxgvnb", "uqvhuucupu", "fwoquoih", "ezsuqxunx", "biq", "crxrwdclv", "qoyfqhytzxfp", "aryqceercpaupqm", "tvwxrev", "gchusjxz", "uls", "whb", "natdmc", "jvidsf", "yhyz", "smvsitdbutamn", "gcfsghusjsxiz", "ijpyhk", "tzvqwkmzxruevs",
+				"fwvjxaxrvmfm", "wscxklqmxhn", "velgcy", "lyctomvi", "smvsitbutam", "hfosz", "fuzubrpo", "dfdeidcepshvjn", "twqol", "rpqjpmv", "ijftxssyqmi", "dzuzsainzbsx", "qyzxfp", "tvwkmzxruev", "farfm", "bbwkizqhicip", "wqobtmamvpgluh", "rytspgy",
+				"uqvheuucdupuw", "jcmang", "h", "kijfhtxssyqmui", "twqgolksq", "rtkgopofnykkrl", "smvstbutam", "xkbfbbcjxgvnbq", "feyq", "oyfqhytzxfp", "velgcby", "dmnioxbf", "kbx", "zx", "wscxkqlqmxbhn", "efvcjtgoiga", "jumttwxv", "zux", "z", "smvsitbutamn",
+				"jftxssyqmi", "wnlhsaj", "bbwizqhcp", "yctomv", "oyqyzxfp", "wqhc", "jnnwp", "bcntqiquagurz", "qzx", "kbfbjxn", "dmnixbf", "ukqs", "fey", "ryqceecaupq", "smvlsitzdbutamn", "bdnhm", "lrhtwfosrzq", "nkptknldw", "crxrwdclvx", "abbtcwzrjs",
+				"uqvheuucupu", "abjbtcwbzrjs", "nkmptknldw", "wnulhsbaj", "wnlhsbaj", "wqobtmamvgluh", "jvis", "pcd", "s", "kjuannelajc", "valas", "lrrzzcl", "kjuannelajct", "snyyoc", "jwp", "vbum", "ezuunx", "bcntqiquagur", "vals", "cov", "dfdidcepshvjn",
+				"vvamlasl", "budnhym", "h", "fwxarvfm", "lrhwfosrz", "nkptnldw", "vjhse", "zzeb", "fubrpo", "fkla", "qjulm", "xpdcdxqia", "ucwxwdm", "jvidsfr", "exhc", "kbfbjx", "bcntqiquaur", "fwnoqjuoihe", "ezsruqxuinrpxc", "ec", "dzuzstuacinzbvsx",
+				"cxkqmxhn", "egpveohyvq", "bkcv", "dzuzsaizbx", "jftxssymi", "ycov", "zbvbeai", "ch", "atcrjs", "qjcemang", "tvjhsed", "vamlas", "bundnhym", "li", "wnulfhsbaj", "o", "ijhtpyhkrif", "nyoc", "ov", "ryceecupq", "wjcrjnszipc", "lrhtwfosrz",
+				"tbzngeqcz", "awfotfiqni", "azbw", "o", "gcfghusjsxiz", "uqvheuucdupu", "rypgy", "snpuykyobc", "ckhn", "kbfbcjxgnb", "xkeow", "jvids", "ubnsnusvgmqog", "endjbkjere", "fwarfm", "wvhb", "fwnoqtjuozihe", "jnwp", "awfotfmiyqni", "iv", "ryqceecupq",
+				"y", "qjuelm", "qyzxp", "vsbum", "dnh", "fam", "snpuyyobc", "wqobtmamvglu", "gjpw", "jcemang", "ukqso", "evhlfz", "nad", "bucwxwdm", "xkabfbbcjxgvnbq", "fwnoqjuozihe", "smvsitzdbutamn", "vec", "fos", "abbtcwbzrjs", "uyifxeyq", "kbfbjxgnb",
+				"nyyoc", "kcv", "fbundnhym", "tbzngpeqcz", "yekfvcjtgoiga", "rgjpw", "ezgvhsalfz", "yoc", "ezvhsalfz", "crxdlv", "chusjz", "fwxaxrvfm", "dzuzstuacinzbsx", "bwizqhc", "pdcdx", "dmnioxbmf", "zuunx", "oqyzxfp", "ezsruqxuinxc", "qjcaemang", "gcghusjsxiz", "nktnldw",
+				"qoyfqhytxzxfp", "bwqhc", "btkcvj", "qxpdcdxqia", "kijofhtxssyqmui", "rypy", "helmi", "zkrlexhxbwt", "qobtmamgu", "vhlfz", "rqjpmv", "yhy", "zzembhy", "rjpmv", "jhse", "fosz", "twol", "qbtamu", "nawxxbslyhucqxb", "dzzsaizbx", "dmnijgoxsbmf",
+				"ijhtpyhkr", "yp", "awfotfgmiyqni", "yctov", "hse", "azabw", "aryqceercaupqm", "fuzubrpoa", "ubnswnusvgmqog", "fafm", "i", "ezvhalfz", "aryxqceercpaupqm", "bwizqhcp", "pdcdxq", "wscxkqlqmxhn", "fuubrpo", "fwvxaxrvmfm", "abjbtcwbzrjas", "zx",
+				"rxmiirbxemog", "dfdeidcepshvvjqn", "az", "velc", "zkrlexnhxbwt", "nawxbslyhucqxb", "qjugelm", "ijhtpdyhkrif", "dmixbf", "gcfsghtusjsxiz", "juannlajc", "uqvheuucdupmuw", "rpqojpmgxv", "rpqojpmxv", "xppph", "kjuannellajct", "lrhfosrz", "dmnijoxsbmf",
+				"ckmxhn", "tvijhsed", "dzuzstuainzbsx", "exhvc", "tvwkxruev", "rxmiirbemog", "lhfosz", "fkyla", "tlwqgolksq", "velgcbpy", "bcqiqaur", "xkhfejow", "ezsuqunx", "dmnioxsbmf", "bqiqu", "ijhtpudyhkrif", "xpdcdxqi", "ckh", "nwxbslyhucqxb", "bbwkizqhcip", "pcdx",
+				"dzuzsuainzbsx", "xkbfbcjxgvnbq", "smvsbutm", "ezsruqxuinrxc", "smvlsitzdbutamdn", "am", "tvwkzxruev", "scxkqmxhn", "snpzuykyobc", "ekfvcjtgoiga", "fuzsubrpoa", "trtkgopofnykkrl", "oyqhytzxfp", "kbjx", "ifeyq", "vhl", "xkfeow", "ezgvhsialfz", "velgc", "hb",
+				"zbveai", "gcghusjxz", "twqgolkq", "btkcv", "ryqceercaupq", "bi", "vvamlas", "awfotfmiqni", "abbtcrjs", "jutkqesoh", "xkbfbcjxgvnb", "hli", "ryspgy", "endjjjbkjere", "mvsbum", "ckqmxhn", "ezsruqxunxc", "zzeby", "xhc", "ezvhlfz", "ezsruqxunx", "tzvwkmzxruev",
+				"hlmi", "kbbjx", "uqvhuuupu", "scxklqmxhn", "wqobtmamglu", "xpdcdxq", "qjugelym", "ifxeyq", "bcnqiquaur", "qobtmamglu", "xkabfbbcjxbgvnbq", "fuuzsubrpoa", "tvibjhsed", "oyqhyzxfp", "ijhpyhk", "c", "gcghusjxiz", "exhvoc", "awfotfini", "vhlz", "rtgopofykkrl",
+				"yh", "ypy", "azb", "bwiqhc", "fla", "dmnijgioxsbmf", "chusjxz", "jvjidsfr", "natddmc", "uifxeyq", "x", "tzvqwkmzxruev", "bucwxwdwm", "ckmhn", "zzemby", "rpmv", "bcntqiqulagurz", "fwoqjuoihe", "dzuzsainzbx", "zkrlehxbwt", "kv", "ucwxwm", "ubnswnusvgmdqog",
+				"wol", "endjjbkjere", "natyddmc", "vl", "ukqsoh", "ezuqunx", "exhvovc", "bqiqau", "bqiqaur", "zunx", "pc", "snuyyoc", "a", "lrhfosz", "kbfbjxgn", "rtgopofnykkrl", "hehszegkvse", "smvsbum", "ijhpyhkr", "ijftxssyqmui", "lyctomvis", "juanlajc", "jukqesoh",
+				"xptpph", "fwarvfm", "qbtmamu", "twqgolq", "aryqceercaupq", "qbtmamgu", "rtgopofykkr", "snpuyyoc", "qyzx", "fwvxaxrvfm", "juannelajc", "fwoquoihe", "nadmc", "jumttwxvx", "ijhtpyhkrf", "twqolq", "rpv", "hehszegkuvse", "ls", "tvjhse", "rxmiirbemg",
+				"dfdeidcepshvvjn", "dnhm", "egpeohyvq", "rgnjpw", "bbwkizqhcp", "nadc", "bcqiquaur", "xkhfeow", "smvstbutm", "ukqesoh", "yctomvi"}},
+			ans1048{15},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1048------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1048, q.para1048
+		fmt.Printf("【input】:%v       【output】:%v\n", p, longestStrChain(p.words))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1048.Longest-String-Chain/README.md b/leetcode/1048.Longest-String-Chain/README.md
new file mode 100644
index 000000000..66c079516
--- /dev/null
+++ b/leetcode/1048.Longest-String-Chain/README.md
@@ -0,0 +1,96 @@
+# [1048. Longest String Chain](https://leetcode.com/problems/longest-string-chain/)
+
+
+## 题目
+
+Given a list of words, each word consists of English lowercase letters.
+
+Let's say `word1` is a predecessor of `word2` if and only if we can add exactly one letter anywhere in `word1` to make it equal to `word2`. For example, `"abc"` is a predecessor of `"abac"`.
+
+A *word chain* is a sequence of words `[word_1, word_2, ..., word_k]` with `k >= 1`, where `word_1` is a predecessor of `word_2`, `word_2` is a predecessor of `word_3`, and so on.
+
+Return the longest possible length of a word chain with words chosen from the given list of `words`.
+
+**Example 1:**
+
+```
+Input: words = ["a","b","ba","bca","bda","bdca"]
+Output: 4
+Explanation: One of the longest word chain is "a","ba","bda","bdca".
+```
+
+**Example 2:**
+
+```
+Input: words = ["xbc","pcxbcf","xb","cxbc","pcxbc"]
+Output: 5
+```
+
+**Constraints:**
+
+- `1 <= words.length <= 1000`
+- `1 <= words[i].length <= 16`
+- `words[i]` only consists of English lowercase letters.
+
+## 题目大意
+
+给出一个单词列表，其中每个单词都由小写英文字母组成。如果我们可以在 word1 的任何地方添加一个字母使其变成 word2，那么我们认为 word1 是 word2 的前身。例如，"abc" 是 "abac" 的前身。词链是单词 [word_1, word_2, ..., word_k] 组成的序列，k >= 1，其中 word_1 是 word_2 的前身，word_2 是 word_3 的前身，依此类推。从给定单词列表 words 中选择单词组成词链，返回词链的最长可能长度。
+
+## 解题思路
+
+- 从这题的数据规模上分析，可以猜出此题是 DFS 或者 DP 的题。简单暴力的方法是以每个字符串为链条的起点开始枚举之后的字符串，两两判断能否构成满足题意的前身字符串。这种做法包含很多重叠子问题，例如 a 和 b 能构成前身字符串，以 c 为起点的字符串链条可能用到 a 和 b，以 d 为起点的字符串链条也可能用到 a 和 b。顺其自然，考虑用 DP 的思路解题。
+- 先将 words 字符串数组排序，然后用 poss 数组记录下每种长度字符串的在排序数组中的起始下标。然后逆序往前递推。因为初始条件只能得到以最长字符串为起始的字符串链长度为 1 。每选择一个起始字符串，从它的长度 + 1 的每个字符串 j 开始比较，是否能为其前身字符串。如果能构成前身字符串，那么 dp[i] = max(dp[i], 1+dp[j])。最终递推到下标为 0 的字符串。最终输出整个递推过程中的最大长度即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func longestStrChain(words []string) int {
+	sort.Slice(words, func(i, j int) bool { return len(words[i]) < len(words[j]) })
+	poss, res := make([]int, 16+2), 0
+	for i, w := range words {
+		if poss[len(w)] == 0 {
+			poss[len(w)] = i
+		}
+	}
+	dp := make([]int, len(words))
+	for i := len(words) - 1; i >= 0; i-- {
+		dp[i] = 1
+		for j := poss[len(words[i])+1]; j < len(words) && len(words[j]) == len(words[i])+1; j++ {
+			if isPredecessor(words[j], words[i]) {
+				dp[i] = max(dp[i], 1+dp[j])
+			}
+		}
+		res = max(res, dp[i])
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func isPredecessor(long, short string) bool {
+	i, j := 0, 0
+	wasMismatch := false
+	for j < len(short) {
+		if long[i] != short[j] {
+			if wasMismatch {
+				return false
+			}
+			wasMismatch = true
+			i++
+			continue
+		}
+		i++
+		j++
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II.go b/leetcode/1049.Last-Stone-Weight-II/1049. Last Stone Weight II.go
similarity index 85%
rename from Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II.go
rename to leetcode/1049.Last-Stone-Weight-II/1049. Last Stone Weight II.go
index c93a17b0c..a8b02c306 100644
--- a/Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II.go	
+++ b/leetcode/1049.Last-Stone-Weight-II/1049. Last Stone Weight II.go	
@@ -20,3 +20,10 @@ func lastStoneWeightII(stones []int) int {
 	}
 	return sum - 2*dp[C]
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II_test.go b/leetcode/1049.Last-Stone-Weight-II/1049. Last Stone Weight II_test.go
similarity index 93%
rename from Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II_test.go
rename to leetcode/1049.Last-Stone-Weight-II/1049. Last Stone Weight II_test.go
index 0084f922a..34b2847da 100644
--- a/Algorithms/1049. Last Stone Weight II/1049. Last Stone Weight II_test.go	
+++ b/leetcode/1049.Last-Stone-Weight-II/1049. Last Stone Weight II_test.go	
@@ -26,17 +26,17 @@ func Test_Problem1049(t *testing.T) {
 
 	qs := []question1049{
 
-		question1049{
+		{
 			para1049{[]int{2, 7, 4, 1, 8, 1}},
 			ans1049{1},
 		},
 
-		question1049{
+		{
 			para1049{[]int{21, 26, 31, 33, 40}},
 			ans1049{5},
 		},
 
-		question1049{
+		{
 			para1049{[]int{1, 2}},
 			ans1049{1},
 		},
diff --git a/leetcode/1049.Last-Stone-Weight-II/README.md b/leetcode/1049.Last-Stone-Weight-II/README.md
new file mode 100755
index 000000000..fa176f97f
--- /dev/null
+++ b/leetcode/1049.Last-Stone-Weight-II/README.md
@@ -0,0 +1,49 @@
+# [1049. Last Stone Weight II](https://leetcode.com/problems/last-stone-weight-ii/)
+
+## 题目
+
+We have a collection of rocks, each rock has a positive integer weight.
+
+Each turn, we choose **any two rocks** and smash them together. Suppose the stones have weights `x` and `y` with `x <= y`. The result of this smash is:
+
+- If `x == y`, both stones are totally destroyed;
+- If `x != y`, the stone of weight `x` is totally destroyed, and the stone of weight `y`has new weight `y-x`.
+
+At the end, there is at most 1 stone left. Return the **smallest possible** weight of this stone (the weight is 0 if there are no stones left.)
+
+**Example 1:**
+
+    Input: [2,7,4,1,8,1]
+    Output: 1
+    Explanation: 
+    We can combine 2 and 4 to get 2 so the array converts to [2,7,1,8,1] then,
+    we can combine 7 and 8 to get 1 so the array converts to [2,1,1,1] then,
+    we can combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
+    we can combine 1 and 1 to get 0 so the array converts to [1] then that's the optimal value.
+
+**Note:**
+
+1. `1 <= stones.length <= 30`
+2. `1 <= stones[i] <= 100`
+
+
+
+## 题目大意
+
+有一堆石头，每块石头的重量都是正整数。每一回合，从中选出任意两块石头，然后将它们一起粉碎。假设石头的重量分别为 x 和 y，且 x <= y。那么粉碎的可能结果如下：
+
+如果 x == y，那么两块石头都会被完全粉碎；
+如果 x != y，那么重量为 x 的石头将会完全粉碎，而重量为 y 的石头新重量为 y-x。
+最后，最多只会剩下一块石头。返回此石头最小的可能重量。如果没有石头剩下，就返回 0。
+
+提示：
+
+1. 1 <= stones.length <= 30
+2. 1 <= stones[i] <= 1000
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面的元素代表的是石头的重量。现在要求两个石头对碰，如果重量相同，两个石头都消失，如果一个重一个轻，剩下的石头是两者的差值。问经过这样的多次碰撞以后，能剩下的石头的重量最轻是多少？
+- 由于两两石头要发生碰撞，所以可以将整个数组可以分为两部分，如果这两部分的石头重量总和相差不大，那么经过若干次碰撞以后，剩下的石头重量一定是最小的。现在就需要找到这样两堆总重量差不多的两堆石头。这个问题就可以转化为 01 背包问题。从数组中找到 `sum/2` 重量的石头集合，如果一半能尽量达到 `sum/2`，那么另外一半和 `sum/2` 的差是最小的，最好的情况就是两堆石头的重量都是 `sum/2`，那么两两石头对碰以后最后都能消失。01 背包的经典模板可以参考第 416 题。
diff --git a/leetcode/1051.Height-Checker/1051. Height Checker.go b/leetcode/1051.Height-Checker/1051. Height Checker.go
new file mode 100644
index 000000000..f0ffaf632
--- /dev/null
+++ b/leetcode/1051.Height-Checker/1051. Height Checker.go	
@@ -0,0 +1,15 @@
+package leetcode
+
+import "sort"
+
+func heightChecker(heights []int) int {
+	result, checker := 0, []int{}
+	checker = append(checker, heights...)
+	sort.Ints(checker)
+	for i := 0; i < len(heights); i++ {
+		if heights[i] != checker[i] {
+			result++
+		}
+	}
+	return result
+}
diff --git a/leetcode/1051.Height-Checker/1051. Height Checker_test.go b/leetcode/1051.Height-Checker/1051. Height Checker_test.go
new file mode 100644
index 000000000..dc4c576c4
--- /dev/null
+++ b/leetcode/1051.Height-Checker/1051. Height Checker_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1051 struct {
+	para1051
+	ans1051
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1051 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1051 struct {
+	one int
+}
+
+func Test_Problem1051(t *testing.T) {
+
+	qs := []question1051{
+
+		{
+			para1051{[]int{1, 1, 4, 2, 1, 3}},
+			ans1051{3},
+		},
+
+		{
+			para1051{[]int{5, 1, 2, 3, 4}},
+			ans1051{5},
+		},
+
+		{
+			para1051{[]int{1, 2, 3, 4, 5}},
+			ans1051{0},
+		},
+
+		{
+			para1051{[]int{5, 4, 3, 2, 1}},
+			ans1051{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1051------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1051, q.para1051
+		fmt.Printf("【input】:%v       【output】:%v\n", p, heightChecker(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1051.Height-Checker/README.md b/leetcode/1051.Height-Checker/README.md
new file mode 100644
index 000000000..96cc1e14f
--- /dev/null
+++ b/leetcode/1051.Height-Checker/README.md
@@ -0,0 +1,68 @@
+# [1051. Height Checker](https://leetcode.com/problems/height-checker/)
+
+
+## 题目
+
+Students are asked to stand in non-decreasing order of heights for an annual photo.
+
+Return the minimum number of students that must move in order for all students to be standing in non-decreasing order of height.
+
+Notice that when a group of students is selected they can reorder in any possible way between themselves and the non selected students remain on their seats.
+
+**Example 1**:
+
+```
+Input: heights = [1,1,4,2,1,3]
+Output: 3
+Explanation: 
+Current array : [1,1,4,2,1,3]
+Target array  : [1,1,1,2,3,4]
+On index 2 (0-based) we have 4 vs 1 so we have to move this student.
+On index 4 (0-based) we have 1 vs 3 so we have to move this student.
+On index 5 (0-based) we have 3 vs 4 so we have to move this student.
+```
+
+**Example 2**:
+
+```
+Input: heights = [5,1,2,3,4]
+Output: 5
+```
+
+**Example 3**:
+
+```
+Input: heights = [1,2,3,4,5]
+Output: 0
+```
+
+**Constraints**:
+
+- `1 <= heights.length <= 100`
+- `1 <= heights[i] <= 100`
+
+## 题目大意
+
+学校在拍年度纪念照时，一般要求学生按照 非递减 的高度顺序排列。请你返回能让所有学生以 非递减 高度排列的最小必要移动人数。注意，当一组学生被选中时，他们之间可以以任何可能的方式重新排序，而未被选中的学生应该保持不动。
+
+
+## 解题思路
+
+- 给定一个高度数组，要求输出把这个数组按照非递减高度排列所需移动的最少次数。
+- 简单题，最少次数意味着每次移动，一步到位，一步就移动到它所在的最终位置。那么用一个辅助排好序的数组，一一比对计数即可。
+
+## 代码
+
+```go
+func heightChecker(heights []int) int {
+	result, checker := 0, []int{}
+	checker = append(checker, heights...)
+	sort.Ints(checker)
+	for i := 0; i < len(heights); i++ {
+		if heights[i] != checker[i] {
+			result++
+		}
+	}
+	return result
+}
+```
\ No newline at end of file
diff --git a/Algorithms/1052. Grumpy Bookstore Owner/1052. Grumpy Bookstore Owner.go b/leetcode/1052.Grumpy-Bookstore-Owner/1052. Grumpy Bookstore Owner.go
similarity index 94%
rename from Algorithms/1052. Grumpy Bookstore Owner/1052. Grumpy Bookstore Owner.go
rename to leetcode/1052.Grumpy-Bookstore-Owner/1052. Grumpy Bookstore Owner.go
index a0079f333..d0a4899af 100644
--- a/Algorithms/1052. Grumpy Bookstore Owner/1052. Grumpy Bookstore Owner.go	
+++ b/leetcode/1052.Grumpy-Bookstore-Owner/1052. Grumpy Bookstore Owner.go	
@@ -38,6 +38,13 @@ func maxSatisfied1(customers []int, grumpy []int, X int) int {
 	return res
 }
 
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
 func sumSatisfied(customers []int, grumpy []int, start, end int) int {
 	sum := 0
 	for i := 0; i < len(customers); i++ {
diff --git a/Algorithms/1052. Grumpy Bookstore Owner/1052. Grumpy Bookstore Owner_test.go b/leetcode/1052.Grumpy-Bookstore-Owner/1052. Grumpy Bookstore Owner_test.go
similarity index 94%
rename from Algorithms/1052. Grumpy Bookstore Owner/1052. Grumpy Bookstore Owner_test.go
rename to leetcode/1052.Grumpy-Bookstore-Owner/1052. Grumpy Bookstore Owner_test.go
index 327c58757..c0e84811a 100644
--- a/Algorithms/1052. Grumpy Bookstore Owner/1052. Grumpy Bookstore Owner_test.go	
+++ b/leetcode/1052.Grumpy-Bookstore-Owner/1052. Grumpy Bookstore Owner_test.go	
@@ -28,17 +28,17 @@ func Test_Problem1052(t *testing.T) {
 
 	qs := []question1052{
 
-		question1052{
+		{
 			para1052{[]int{4, 10, 10}, []int{1, 1, 0}, 2},
 			ans1052{24},
 		},
 
-		question1052{
+		{
 			para1052{[]int{1}, []int{0}, 1},
 			ans1052{1},
 		},
 
-		question1052{
+		{
 			para1052{[]int{1, 0, 1, 2, 1, 1, 7, 5}, []int{0, 1, 0, 1, 0, 1, 0, 1}, 3},
 			ans1052{16},
 		},
diff --git a/leetcode/1052.Grumpy-Bookstore-Owner/README.md b/leetcode/1052.Grumpy-Bookstore-Owner/README.md
new file mode 100755
index 000000000..80e6cc239
--- /dev/null
+++ b/leetcode/1052.Grumpy-Bookstore-Owner/README.md
@@ -0,0 +1,45 @@
+# [1052. Grumpy Bookstore Owner](https://leetcode.com/problems/grumpy-bookstore-owner/)
+
+
+## 题目
+
+Today, the bookstore owner has a store open for `customers.length`minutes. Every minute, some number of customers (`customers[i]`) enter the store, and all those customers leave after the end of that minute.
+
+On some minutes, the bookstore owner is grumpy. If the bookstore owner is grumpy on the i-th minute, `grumpy[i] = 1`, otherwise `grumpy[i] = 0`. When the bookstore owner is grumpy, the customers of that minute are not satisfied, otherwise they are satisfied.
+
+The bookstore owner knows a secret technique to keep themselves not grumpy for `X` minutes straight, but can only use it once.
+
+Return the maximum number of customers that can be satisfied throughout the day.
+
+**Example 1:**
+
+    Input: customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], X = 3
+    Output: 16
+    Explanation: The bookstore owner keeps themselves not grumpy for the last 3 minutes. 
+    The maximum number of customers that can be satisfied = 1 + 1 + 1 + 1 + 7 + 5 = 16.
+
+**Note:**
+
+- `1 <= X <= customers.length == grumpy.length <= 20000`
+- `0 <= customers[i] <= 1000`
+- `0 <= grumpy[i] <= 1`
+
+
+## 题目大意
+
+今天，书店老板有一家店打算试营业 customers.length 分钟。每分钟都有一些顾客（customers[i]）会进入书店，所有这些顾客都会在那一分钟结束后离开。在某些时候，书店老板会生气。 如果书店老板在第 i 分钟生气，那么 grumpy[i] = 1，否则 grumpy[i] = 0。 当书店老板生气时，那一分钟的顾客就会不满意，不生气则他们是满意的。书店老板知道一个秘密技巧，能抑制自己的情绪，可以让自己连续 X 分钟不生气，但却只能使用一次。请你返回这一天营业下来，最多有多少客户能够感到满意的数量。
+
+提示：
+
+1. 1 <= X <= customers.length == grumpy.length <= 20000
+2. 0 <= customers[i] <= 1000
+3. 0 <= grumpy[i] <= 1
+
+
+
+## 解题思路
+
+
+- 给出一个顾客入店时间表和书店老板发脾气的时间表。两个数组的时间是一一对应的，即相同下标对应的相同的时间。书店老板可以控制自己在 X 分钟内不发火，但是只能控制一次。问有多少顾客能在书店老板不发火的时候在书店里看书。抽象一下，给出一个价值数组和一个装着 0 和 1 的数组，当价值数组的下标对应另外一个数组相同下标的值是 0 的时候，那么这个价值可以累加，当对应是 1 的时候，就不能加上这个价值。现在可以让装着 0 和 1 的数组中连续 X 个数都变成 0，问最终价值最大是多少？
+- 这道题是典型的滑动窗口的题目。最暴力的解法是滑动窗口右边界，当与左边界的距离等于 X 的时候，计算此刻对应的数组的总价值。当整个宽度为 X 的窗口滑过整个数组以后，输出维护的最大值即可。这个方法耗时比较长。因为每次计算数组总价值的时候都要遍历整个数组。这里是可以优化的地方。
+- 每次计算数组总价值的时候，其实目的是为了找到宽度为 X 的窗口对应里面为 1 的数累加和最大，因为如果把这个窗口里面的 1 都变成 0 以后，那么对最终价值的影响也最大。所以用一个变量 `customer0` 专门记录脾气数组中为 0 的对应的价值，累加起来。因为不管怎么改变，为 0 的永远为 0，唯一变化的是 1 变成 0 。用 `customer1` 专门记录脾气数组中为 1 的对应的价值。在窗口滑动过程中找到 `customer1` 的最大值。最终要求的最大值就是 `customer0 + maxCustomer1`。
diff --git a/Algorithms/1054. Distant Barcodes/1054. Distant Barcodes.go b/leetcode/1054.Distant-Barcodes/1054. Distant Barcodes.go
similarity index 100%
rename from Algorithms/1054. Distant Barcodes/1054. Distant Barcodes.go
rename to leetcode/1054.Distant-Barcodes/1054. Distant Barcodes.go
diff --git a/Algorithms/1054. Distant Barcodes/1054. Distant Barcodes_test.go b/leetcode/1054.Distant-Barcodes/1054. Distant Barcodes_test.go
similarity index 95%
rename from Algorithms/1054. Distant Barcodes/1054. Distant Barcodes_test.go
rename to leetcode/1054.Distant-Barcodes/1054. Distant Barcodes_test.go
index 805075ce1..118e23043 100644
--- a/Algorithms/1054. Distant Barcodes/1054. Distant Barcodes_test.go	
+++ b/leetcode/1054.Distant-Barcodes/1054. Distant Barcodes_test.go	
@@ -26,12 +26,12 @@ func Test_Problem1054(t *testing.T) {
 
 	qs := []question1054{
 
-		question1054{
+		{
 			para1054{[]int{1, 1, 1, 2, 2, 2}},
 			ans1054{[]int{2, 1, 2, 1, 2, 1}},
 		},
 
-		question1054{
+		{
 			para1054{[]int{1, 1, 1, 1, 2, 2, 3, 3}},
 			ans1054{[]int{1, 3, 1, 3, 2, 1, 2, 1}},
 		},
diff --git a/Algorithms/1054. Distant Barcodes/README.md b/leetcode/1054.Distant-Barcodes/README.md
similarity index 100%
rename from Algorithms/1054. Distant Barcodes/README.md
rename to leetcode/1054.Distant-Barcodes/README.md
diff --git a/Algorithms/1073. Adding Two Negabinary Numbers/1073. Adding Two Negabinary Numbers.go b/leetcode/1073.Adding-Two-Negabinary-Numbers/1073. Adding Two Negabinary Numbers.go
similarity index 100%
rename from Algorithms/1073. Adding Two Negabinary Numbers/1073. Adding Two Negabinary Numbers.go
rename to leetcode/1073.Adding-Two-Negabinary-Numbers/1073. Adding Two Negabinary Numbers.go
diff --git a/Algorithms/1073. Adding Two Negabinary Numbers/1073. Adding Two Negabinary Numbers_test.go b/leetcode/1073.Adding-Two-Negabinary-Numbers/1073. Adding Two Negabinary Numbers_test.go
similarity index 99%
rename from Algorithms/1073. Adding Two Negabinary Numbers/1073. Adding Two Negabinary Numbers_test.go
rename to leetcode/1073.Adding-Two-Negabinary-Numbers/1073. Adding Two Negabinary Numbers_test.go
index bf65c5fe6..74386dd42 100644
--- a/Algorithms/1073. Adding Two Negabinary Numbers/1073. Adding Two Negabinary Numbers_test.go	
+++ b/leetcode/1073.Adding-Two-Negabinary-Numbers/1073. Adding Two Negabinary Numbers_test.go	
@@ -27,33 +27,33 @@ func Test_Problem1073(t *testing.T) {
 
 	qs := []question1073{
 
-		question1073{
+		{
 			para1073{[]int{1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0},
 				[]int{1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1}},
 			ans1073{[]int{1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1}},
 		},
 
-		question1073{
+		{
 			para1073{[]int{0}, []int{1, 0, 0, 1}},
 			ans1073{[]int{1, 0, 0, 1}},
 		},
 
-		question1073{
+		{
 			para1073{[]int{0}, []int{1, 1}},
 			ans1073{[]int{1, 1}},
 		},
 
-		question1073{
+		{
 			para1073{[]int{1, 1, 1, 1, 1}, []int{1, 0, 1}},
 			ans1073{[]int{1, 0, 0, 0, 0}},
 		},
 
-		question1073{
+		{
 			para1073{[]int{0}, []int{0}},
 			ans1073{[]int{0}},
 		},
 
-		question1073{
+		{
 			para1073{[]int{0}, []int{1, 0}},
 			ans1073{[]int{1, 0}},
 		},
diff --git a/leetcode/1073.Adding-Two-Negabinary-Numbers/README.md b/leetcode/1073.Adding-Two-Negabinary-Numbers/README.md
new file mode 100755
index 000000000..3f433c1da
--- /dev/null
+++ b/leetcode/1073.Adding-Two-Negabinary-Numbers/README.md
@@ -0,0 +1,75 @@
+# [1073. Adding Two Negabinary Numbers](https://leetcode.com/problems/adding-two-negabinary-numbers/)
+
+
+## 题目
+
+Given two numbers `arr1` and `arr2` in base **-2**, return the result of adding them together.
+
+Each number is given in *array format*: as an array of 0s and 1s, from most significant bit to least significant bit. For example, `arr = [1,1,0,1]`represents the number `(-2)^3 + (-2)^2 + (-2)^0 = -3`. A number `arr` in *array format* is also guaranteed to have no leading zeros: either `arr == [0]` or `arr[0] == 1`.
+
+Return the result of adding `arr1` and `arr2` in the same format: as an array of 0s and 1s with no leading zeros.
+
+**Example 1:**
+
+    Input: arr1 = [1,1,1,1,1], arr2 = [1,0,1]
+    Output: [1,0,0,0,0]
+    Explanation: arr1 represents 11, arr2 represents 5, the output represents 16.
+
+**Note:**
+
+1. `1 <= arr1.length <= 1000`
+2. `1 <= arr2.length <= 1000`
+3. `arr1` and `arr2` have no leading zeros
+4. `arr1[i]` is `0` or `1`
+5. `arr2[i]` is `0` or `1`
+
+
+## 题目大意
+
+给出基数为 -2 的两个数 arr1 和 arr2，返回两数相加的结果。数字以 数组形式 给出：数组由若干 0 和 1 组成，按最高有效位到最低有效位的顺序排列。例如，arr = [1,1,0,1] 表示数字 (-2)^3 + (-2)^2 + (-2)^0 = -3。数组形式 的数字也同样不含前导零：以 arr 为例，这意味着要么 arr == [0]，要么 arr[0] == 1。
+
+返回相同表示形式的 arr1 和 arr2 相加的结果。两数的表示形式为：不含前导零、由若干 0 和 1 组成的数组。
+
+提示：
+
+- 1 <= arr1.length <= 1000
+- 1 <= arr2.length <= 1000
+- arr1 和 arr2 都不含前导零
+- arr1[i] 为 0 或 1
+- arr2[i] 为 0 或 1
+
+
+
+## 解题思路
+
+- 给出两个 -2 进制的数，要求计算出这两个数的和，最终表示形式还是 -2 进制。
+- 这一题最先想到的思路是先把两个 -2 进制的数转成 10 进制以后做加法，然后把结果表示成 -2 进制。这个思路可行，但是在提交以后会发现数据溢出 int64 了。在第 257 / 267 组测试数据会出现 WA。测试数据见 test 文件。另外换成 big.Add 也不是很方便。所以考虑换一个思路。
+- 这道题实际上就是求两个 -2 进制数的加法，为什么还要先转到 10 进制再换回 -2 进制呢？为何不直接进行 -2 进制的加法。所以开始尝试直接进行加法运算。加法是从低位到高位依次累加，遇到进位要从低往高位进位。所以从两个数组的末尾往前扫，模拟低位相加的过程。关键的是进位问题。进位分 3 种情况，依次来讨论：
+
+1. 进位到高位 k ，高位 k 上的两个数数字分别是 0 和 0 。这种情况最终 k 位为 1 。
+```c
+        证明：由于进位是由 k - 1 位进过来的，所以 k - 1 位是 2 个 1 。现在 k 位是 2 个 0，
+        所以加起来的和是 2 * (-2)^(k - 1)。
+        当 k 为奇数的时候，2 * (-2)^(k - 1) = (-1)^(k - 1)* 2 * 2^(k - 1) = 2^k
+        当 k 为偶数的时候，2 * (-2)^(k - 1) = (-1)^(k - 1)* 2 * 2^(k - 1) = -2^k
+        综合起来就是 (-2)^k，所以最终 k 位上有一个 1
+```
+2. 进位到高位 k ，高位 k 上的两个数数字分别是 0 和 1 。这种情况最终 k 位为 0 。
+```c
+        证明：由于进位是由 k - 1 位进过来的，所以 k - 1 位是 2 个 1。现在 k 位是 1 个 0 和 1 个 1,
+        所以加起来的和是 (-2)^k + 2 * (-2)^(k - 1)。
+        当 k 为奇数的时候，(-2)^k + 2 * (-2)^(k - 1) = -2^k + 2^k = 0
+        当 k 为偶数的时候，(-2)^k + 2 * (-2)^(k - 1) = 2^k - 2^k = 0
+        综合起来就是 0，所以最终 k 位上有一个 0
+```
+3. 进位到高位 k ，高位 k 上的两个数数字分别是 1 和 1 。这种情况最终 k 位为 1 。
+```c
+        证明：由于进位是由 k - 1 位进过来的，所以 k - 1 位是 2 个 1 。现在 k 位是 2 个 1，
+        所以加起来的和是 2 * (-2)^k + 2 * (-2)^(k - 1)。
+        当 k 为奇数的时候，2 * (-2)^k + 2 * (-2)^(k - 1) = -2^(k + 1) + 2^k = 2^k*(1 - 2) = -2^k
+        当 k 为偶数的时候，2 * (-2)^k + 2 * (-2)^(k - 1) = 2^(k + 1) - 2^k = 2^k*(2 - 1) = 2^k
+        综合起来就是 (-2)^k，所以最终 k 位上有一个 1
+```
+
+- 所以综上所属，-2 进制的进位和 2 进制的进位原理是完全一致的，只不过 -2 进制的进位是 -1，而 2 进制的进位是 1 。由于进位可能在 -2 进制上出现前导 0 ，所以最终结果需要再去除前导 0 。
+
diff --git a/Algorithms/1074. Number of Submatrices That Sum to Target/1074. Number of Submatrices That Sum to Target.go b/leetcode/1074.Number-of-Submatrices-That-Sum-to-Target/1074. Number of Submatrices That Sum to Target.go
similarity index 100%
rename from Algorithms/1074. Number of Submatrices That Sum to Target/1074. Number of Submatrices That Sum to Target.go
rename to leetcode/1074.Number-of-Submatrices-That-Sum-to-Target/1074. Number of Submatrices That Sum to Target.go
diff --git a/Algorithms/1074. Number of Submatrices That Sum to Target/1074. Number of Submatrices That Sum to Target_test.go b/leetcode/1074.Number-of-Submatrices-That-Sum-to-Target/1074. Number of Submatrices That Sum to Target_test.go
similarity index 80%
rename from Algorithms/1074. Number of Submatrices That Sum to Target/1074. Number of Submatrices That Sum to Target_test.go
rename to leetcode/1074.Number-of-Submatrices-That-Sum-to-Target/1074. Number of Submatrices That Sum to Target_test.go
index af070664a..0b77d0213 100644
--- a/Algorithms/1074. Number of Submatrices That Sum to Target/1074. Number of Submatrices That Sum to Target_test.go	
+++ b/leetcode/1074.Number-of-Submatrices-That-Sum-to-Target/1074. Number of Submatrices That Sum to Target_test.go	
@@ -27,13 +27,13 @@ func Test_Problem1074(t *testing.T) {
 
 	qs := []question1074{
 
-		question1074{
-			para1074{[][]int{[]int{0, 1, 0}, []int{1, 1, 1}, []int{0, 1, 0}}, 0},
+		{
+			para1074{[][]int{{0, 1, 0}, {1, 1, 1}, {0, 1, 0}}, 0},
 			ans1074{4},
 		},
 
-		question1074{
-			para1074{[][]int{[]int{1, -1}, []int{-1, 1}}, 0},
+		{
+			para1074{[][]int{{1, -1}, {-1, 1}}, 0},
 			ans1074{5},
 		},
 	}
diff --git a/leetcode/1074.Number-of-Submatrices-That-Sum-to-Target/README.md b/leetcode/1074.Number-of-Submatrices-That-Sum-to-Target/README.md
new file mode 100755
index 000000000..bf115a33c
--- /dev/null
+++ b/leetcode/1074.Number-of-Submatrices-That-Sum-to-Target/README.md
@@ -0,0 +1,63 @@
+# [1074. Number of Submatrices That Sum to Target](https://leetcode.com/problems/number-of-submatrices-that-sum-to-target/)
+
+
+## 题目
+
+Given a `matrix`, and a `target`, return the number of non-empty submatrices that sum to target.
+
+A submatrix `x1, y1, x2, y2` is the set of all cells `matrix[y]` with `x1 <= x <= x2` and `y1 <= y <= y2`.
+
+Two submatrices `(x1, y1, x2, y2)` and `(x1', y1', x2', y2')` are different if they have some coordinate that is different: for example, if `x1 != x1'`.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/09/02/mate1.jpg)
+
+    Input: matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
+    Output: 4
+    Explanation: The four 1x1 submatrices that only contain 0.
+
+**Example 2:**
+
+    Input: matrix = [[1,-1],[-1,1]], target = 0
+    Output: 5
+    Explanation: The two 1x2 submatrices, plus the two 2x1 submatrices, plus the 2x2 submatrix.
+
+**Note:**
+
+1. `1 <= matrix.length <= 300`
+2. `1 <= matrix[0].length <= 300`
+3. `-1000 <= matrix[i] <= 1000`
+4. `-10^8 <= target <= 10^8`
+
+
+## 题目大意
+
+给出矩阵 matrix 和目标值 target，返回元素总和等于目标值的非空子矩阵的数量。
+
+子矩阵 x1, y1, x2, y2 是满足 x1 <= x <= x2 且 y1 <= y <= y2 的所有单元 matrix[x][y] 的集合。
+
+如果 (x1, y1, x2, y2) 和 (x1', y1', x2', y2') 两个子矩阵中部分坐标不同（如：x1 != x1'），那么这两个子矩阵也不同。
+
+
+提示：
+
+1. 1 <= matrix.length <= 300
+2. 1 <= matrix[0].length <= 300
+3. -1000 <= matrix[i] <= 1000
+4. -10^8 <= target <= 10^8
+
+
+
+
+## 解题思路
+
+- 给出一个矩阵，要求在这个矩阵中找出子矩阵的和等于 target 的矩阵个数。
+- 这一题读完题感觉是滑动窗口的二维版本。如果把它拍扁，在一维数组中，求连续的子数组和为 target，这样就很好做。如果这题不降维，纯暴力解是 O(n^6)。如何优化降低时间复杂度呢？
+- 联想到第 1 题 Two Sum 问题，可以把 2 个数求和的问题优化到 O(n)。这里也用类似的思想，用一个 map 来保存行方向上曾经出现过的累加和，相减就可以得到本行的和。这里可能读者会有疑惑，为什么不能每一行都单独保存呢？为什么一定要用累加和相减的方式来获取每一行的和呢？因为这一题要求子矩阵所有解，如果只单独保存每一行的和，只能求得小的子矩阵，子矩阵和子矩阵组成的大矩阵的情况会漏掉(当然再循环一遍，把子矩阵累加起来也可以，但是这样就多了一层循环了)，例如子矩阵是 1*4 的，但是 2 个这样的子矩阵摞在一起形成 2 * 4 也能满足条件，如果不用累加和的办法，只单独存每一行的和，最终还要有组合的步骤。经过这样的优化，可以从 O(n^6) 优化到 O(n^4)，能 AC 这道题，但是时间复杂度太高了。如何优化？
+- 首先，子矩阵需要上下左右 4 个边界，4 个变量控制循环就需要 O(n^4)，行和列的区间累加还需要 O(n^2)。行和列的区间累加可以通过 preSum 来解决。例如 `sum[i,j] = sum[j] - sum[i - 1]`，其中 sum[k] 中存的是从 0 到 K 的累加和： ![](https://img.halfrost.com/Leetcode/leetcode_1074.gif)
+
+    那么一个区间内的累加和可以由这个区间的右边界减去区间左边界左边的那个累加和得到(由于是闭区间，所需要取左边界左边的和)。经过这样的处理，列方向的维度就被我们拍扁了。
+
+- 再来看看行方向的和，现在每一列的和都可以通过区间相减的方法得到。那么这道题就变成了第 1 题 Two Sum 的问题了。Two Sum 问题只需要 O(n) 的时间复杂度求解，这一题由于是二维的，所以两个列的边界还需要循环，所以最终优化下来的时间复杂度是 O(n^3)。计算 presum 可以直接用原数组，所以空间复杂度只有一个 O(n) 的字典。
+- 类似思路的题目有第 560 题，第 304 题。
diff --git a/Algorithms/1078. Occurrences After Bigram/1078. Occurrences After Bigram.go b/leetcode/1078.Occurrences-After-Bigram/1078. Occurrences After Bigram.go
similarity index 100%
rename from Algorithms/1078. Occurrences After Bigram/1078. Occurrences After Bigram.go
rename to leetcode/1078.Occurrences-After-Bigram/1078. Occurrences After Bigram.go
diff --git a/Algorithms/1078. Occurrences After Bigram/1078. Occurrences After Bigram_test.go b/leetcode/1078.Occurrences-After-Bigram/1078. Occurrences After Bigram_test.go
similarity index 96%
rename from Algorithms/1078. Occurrences After Bigram/1078. Occurrences After Bigram_test.go
rename to leetcode/1078.Occurrences-After-Bigram/1078. Occurrences After Bigram_test.go
index 020d36279..1c45d4a95 100644
--- a/Algorithms/1078. Occurrences After Bigram/1078. Occurrences After Bigram_test.go	
+++ b/leetcode/1078.Occurrences-After-Bigram/1078. Occurrences After Bigram_test.go	
@@ -28,12 +28,12 @@ func Test_Problem1078(t *testing.T) {
 
 	qs := []question1078{
 
-		question1078{
+		{
 			para1078{"alice is a good girl she is a good student", "a", "good"},
 			ans1078{[]string{"girl", "student"}},
 		},
 
-		question1078{
+		{
 			para1078{"we will we will rock you", "we", "will"},
 			ans1078{[]string{"we", "rock"}},
 		},
diff --git a/leetcode/1078.Occurrences-After-Bigram/README.md b/leetcode/1078.Occurrences-After-Bigram/README.md
new file mode 100755
index 000000000..b9c546b76
--- /dev/null
+++ b/leetcode/1078.Occurrences-After-Bigram/README.md
@@ -0,0 +1,39 @@
+# [1078. Occurrences After Bigram](https://leetcode.com/problems/occurrences-after-bigram/)
+
+
+## 题目
+
+Given words `first` and `second`, consider occurrences in some `text` of the form "`first second third`", where `second` comes immediately after `first`, and `third`comes immediately after `second`.
+
+For each such occurrence, add "`third`" to the answer, and return the answer.
+
+**Example 1:**
+
+    Input: text = "alice is a good girl she is a good student", first = "a", second = "good"
+    Output: ["girl","student"]
+
+**Example 2:**
+
+    Input: text = "we will we will rock you", first = "we", second = "will"
+    Output: ["we","rock"]
+
+**Note:**
+
+1. `1 <= text.length <= 1000`
+2. `text` consists of space separated words, where each word consists of lowercase English letters.
+3. `1 <= first.length, second.length <= 10`
+4. `first` and `second` consist of lowercase English letters.
+
+
+## 题目大意
+
+
+给出第一个词 first 和第二个词 second，考虑在某些文本 text 中可能以 "first second third" 形式出现的情况，其中 second 紧随 first 出现，third 紧随 second 出现。对于每种这样的情况，将第三个词 "third" 添加到答案中，并返回答案。
+
+
+
+
+## 解题思路
+
+
+- 简单题。给出一个 text，要求找出紧接在 first 和 second 后面的那个字符串，有多个就输出多个。解法很简单，先分解出 words 每个字符串，然后依次遍历进行字符串匹配。匹配到 first 和 second 以后，输出之后的那个字符串。
diff --git a/Algorithms/1079. Letter Tile Possibilities/1079. Letter Tile Possibilities.go b/leetcode/1079.Letter-Tile-Possibilities/1079. Letter Tile Possibilities.go
similarity index 100%
rename from Algorithms/1079. Letter Tile Possibilities/1079. Letter Tile Possibilities.go
rename to leetcode/1079.Letter-Tile-Possibilities/1079. Letter Tile Possibilities.go
diff --git a/Algorithms/1079. Letter Tile Possibilities/1079. Letter Tile Possibilities_test.go b/leetcode/1079.Letter-Tile-Possibilities/1079. Letter Tile Possibilities_test.go
similarity index 95%
rename from Algorithms/1079. Letter Tile Possibilities/1079. Letter Tile Possibilities_test.go
rename to leetcode/1079.Letter-Tile-Possibilities/1079. Letter Tile Possibilities_test.go
index f477a5b9e..5b01d7061 100644
--- a/Algorithms/1079. Letter Tile Possibilities/1079. Letter Tile Possibilities_test.go	
+++ b/leetcode/1079.Letter-Tile-Possibilities/1079. Letter Tile Possibilities_test.go	
@@ -26,12 +26,12 @@ func Test_Problem1079(t *testing.T) {
 
 	qs := []question1079{
 
-		question1079{
+		{
 			para1079{"AAB"},
 			ans1079{8},
 		},
 
-		question1079{
+		{
 			para1079{"AAABBC"},
 			ans1079{188},
 		},
diff --git a/leetcode/1079.Letter-Tile-Possibilities/README.md b/leetcode/1079.Letter-Tile-Possibilities/README.md
new file mode 100755
index 000000000..4f789c73f
--- /dev/null
+++ b/leetcode/1079.Letter-Tile-Possibilities/README.md
@@ -0,0 +1,36 @@
+# [1079. Letter Tile Possibilities](https://leetcode.com/problems/letter-tile-possibilities/)
+
+
+## 题目
+
+You have a set of `tiles`, where each tile has one letter `tiles[i]` printed on it. Return the number of possible non-empty sequences of letters you can make.
+
+**Example 1:**
+
+    Input: "AAB"
+    Output: 8
+    Explanation: The possible sequences are "A", "B", "AA", "AB", "BA", "AAB", "ABA", "BAA".
+
+**Example 2:**
+
+    Input: "AAABBC"
+    Output: 188
+
+**Note:**
+
+1. `1 <= tiles.length <= 7`
+2. `tiles` consists of uppercase English letters.
+
+## 题目大意
+
+你有一套活字字模 tiles，其中每个字模上都刻有一个字母 tiles[i]。返回你可以印出的非空字母序列的数目。提示：  
+
+- 1 <= tiles.length <= 7
+- tiles 由大写英文字母组成
+
+## 解题思路
+
+- 题目要求输出所有非空字母序列的数目。这一题是排列和组合的结合题目。组合是可以选择一个字母，二个字母，…… n 个字母。每个组合内是排列问题。比如选择 2 个字母，字母之间相互排序不同是影响最终结果的，不同的排列顺序是不同的解。
+- 这道题目由于不需要输出所有解，所以解法可以优化，例如我们在递归计算解的时候，不需要真的遍历原字符串，只需要累加一些字母的频次就可以。当然如果要输出所有解，就需要真实遍历原字符串了(见解法二)。简单的做法是每次递归按照频次累加。因为每次增加一个字母一定是 26 个大写字母中的一个。这里需要注意的是，增加的只能是 26 个字母里面还能取出“机会”的字母，例如递归到到第 3 轮了，A 用完了，这个时候只能取频次还不为 0 的字母拼上去。
+
+![](https://img.halfrost.com/Leetcode/leetcode_1079_0.png)
diff --git a/leetcode/1089.Duplicate-Zeros/1089. Duplicate Zeros.go b/leetcode/1089.Duplicate-Zeros/1089. Duplicate Zeros.go
new file mode 100644
index 000000000..befb61edf
--- /dev/null
+++ b/leetcode/1089.Duplicate-Zeros/1089. Duplicate Zeros.go	
@@ -0,0 +1,10 @@
+package leetcode
+
+func duplicateZeros(arr []int) {
+	for i := 0; i < len(arr); i++ {
+		if arr[i] == 0 && i+1 < len(arr) {
+			arr = append(arr[:i+1], arr[i:len(arr)-1]...)
+			i++
+		}
+	}
+}
diff --git a/leetcode/1089.Duplicate-Zeros/1089. Duplicate Zeros_test.go b/leetcode/1089.Duplicate-Zeros/1089. Duplicate Zeros_test.go
new file mode 100644
index 000000000..3916f0d0e
--- /dev/null
+++ b/leetcode/1089.Duplicate-Zeros/1089. Duplicate Zeros_test.go	
@@ -0,0 +1,54 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1089 struct {
+	para1089
+	ans1089
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1089 struct {
+	arr []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1089 struct {
+}
+
+func Test_Problem1089(t *testing.T) {
+
+	qs := []question1089{
+
+		{
+			para1089{[]int{1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0}},
+			ans1089{},
+		},
+
+		{
+			para1089{[]int{1, 0, 2, 3, 0, 4, 5, 0}},
+			ans1089{},
+		},
+
+		{
+			para1089{[]int{1, 2, 3}},
+			ans1089{},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1089------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1089, q.para1089
+		fmt.Printf("【input】:%v      ", p)
+		duplicateZeros(p.arr)
+		fmt.Printf("【output】:%v      \n", p.arr)
+
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1089.Duplicate-Zeros/README.md b/leetcode/1089.Duplicate-Zeros/README.md
new file mode 100644
index 000000000..27a9eced8
--- /dev/null
+++ b/leetcode/1089.Duplicate-Zeros/README.md
@@ -0,0 +1,54 @@
+# [1089. Duplicate Zeros](https://leetcode.com/problems/duplicate-zeros/)
+
+
+## 题目
+
+Given a fixed length array `arr` of integers, duplicate each occurrence of zero, shifting the remaining elements to the right.
+
+Note that elements beyond the length of the original array are not written.
+
+Do the above modifications to the input array **in place**, do not return anything from your function.
+
+**Example 1**:
+
+```
+Input: [1,0,2,3,0,4,5,0]
+Output: null
+Explanation: After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]
+```
+
+**Example 2**:
+
+```
+Input: [1,2,3]
+Output: null
+Explanation: After calling your function, the input array is modified to: [1,2,3]
+```
+
+**Note**:
+
+1. `1 <= arr.length <= 10000`
+2. `0 <= arr[i] <= 9`
+
+## 题目大意
+
+给你一个长度固定的整数数组 arr，请你将该数组中出现的每个零都复写一遍，并将其余的元素向右平移。注意：请不要在超过该数组长度的位置写入元素。要求：请对输入的数组 就地 进行上述修改，不要从函数返回任何东西。
+
+
+## 解题思路
+
+- 给一个固定长度的数组，把数组元素为 0 的元素都往后复制一遍，后面的元素往后移，超出数组长度的部分删除。
+- 简单题，按照题意，用 append 和 slice 操作即可。
+
+## 代码
+
+```go
+func duplicateZeros(arr []int) {
+	for i := 0; i < len(arr); i++ {
+		if arr[i] == 0 && i+1 < len(arr) {
+			arr = append(arr[:i+1], arr[i:len(arr)-1]...)
+			i++
+		}
+	}
+}
+```
\ No newline at end of file
diff --git a/leetcode/1091.Shortest-Path-in-Binary-Matrix/1091. Shortest Path in Binary Matrix.go b/leetcode/1091.Shortest-Path-in-Binary-Matrix/1091. Shortest Path in Binary Matrix.go
new file mode 100644
index 000000000..544de00cd
--- /dev/null
+++ b/leetcode/1091.Shortest-Path-in-Binary-Matrix/1091. Shortest Path in Binary Matrix.go	
@@ -0,0 +1,54 @@
+package leetcode
+
+var dir = [][]int{
+	{-1, -1},
+	{-1, 0},
+	{-1, 1},
+	{0, 1},
+	{0, -1},
+	{1, -1},
+	{1, 0},
+	{1, 1},
+}
+
+func shortestPathBinaryMatrix(grid [][]int) int {
+	visited := make([][]bool, 0)
+	for range make([]int, len(grid)) {
+		visited = append(visited, make([]bool, len(grid[0])))
+	}
+	dis := make([][]int, 0)
+	for range make([]int, len(grid)) {
+		dis = append(dis, make([]int, len(grid[0])))
+	}
+	if grid[0][0] == 1 {
+		return -1
+	}
+	if len(grid) == 1 && len(grid[0]) == 1 {
+		return 1
+	}
+
+	queue := []int{0}
+	visited[0][0], dis[0][0] = true, 1
+	for len(queue) > 0 {
+		cur := queue[0]
+		queue = queue[1:]
+		curx, cury := cur/len(grid[0]), cur%len(grid[0])
+		for d := 0; d < 8; d++ {
+			nextx := curx + dir[d][0]
+			nexty := cury + dir[d][1]
+			if isInBoard(grid, nextx, nexty) && !visited[nextx][nexty] && grid[nextx][nexty] == 0 {
+				queue = append(queue, nextx*len(grid[0])+nexty)
+				visited[nextx][nexty] = true
+				dis[nextx][nexty] = dis[curx][cury] + 1
+				if nextx == len(grid)-1 && nexty == len(grid[0])-1 {
+					return dis[nextx][nexty]
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func isInBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
diff --git a/leetcode/1091.Shortest-Path-in-Binary-Matrix/1091. Shortest Path in Binary Matrix_test.go b/leetcode/1091.Shortest-Path-in-Binary-Matrix/1091. Shortest Path in Binary Matrix_test.go
new file mode 100644
index 000000000..a038921ac
--- /dev/null
+++ b/leetcode/1091.Shortest-Path-in-Binary-Matrix/1091. Shortest Path in Binary Matrix_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1091 struct {
+	para1091
+	ans1091
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1091 struct {
+	grid [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1091 struct {
+	one int
+}
+
+func Test_Problem1091(t *testing.T) {
+
+	qs := []question1091{
+
+		{
+			para1091{[][]int{{0, 1}, {1, 0}}},
+			ans1091{2},
+		},
+
+		{
+			para1091{[][]int{{0, 0, 0}, {1, 1, 0}, {1, 1, 0}}},
+			ans1091{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1091------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1091, q.para1091
+		fmt.Printf("【input】:%v       【output】:%v\n", p, shortestPathBinaryMatrix(p.grid))
+
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1091.Shortest-Path-in-Binary-Matrix/README.md b/leetcode/1091.Shortest-Path-in-Binary-Matrix/README.md
new file mode 100644
index 000000000..c7397b78e
--- /dev/null
+++ b/leetcode/1091.Shortest-Path-in-Binary-Matrix/README.md
@@ -0,0 +1,114 @@
+# [1091. Shortest Path in Binary Matrix](https://leetcode.com/problems/shortest-path-in-binary-matrix/)
+
+
+## 题目
+
+In an N by N square grid, each cell is either empty (0) or blocked (1).
+
+A *clear path from top-left to bottom-right* has length `k` if and only if it is composed of cells `C_1, C_2, ..., C_k` such that:
+
+- Adjacent cells `C_i` and `C_{i+1}` are connected 8-directionally (ie., they are different and share an edge or corner)
+- `C_1` is at location `(0, 0)` (ie. has value `grid[0][0]`)
+- `C_k` is at location `(N-1, N-1)` (ie. has value `grid[N-1][N-1]`)
+- If `C_i` is located at `(r, c)`, then `grid[r][c]` is empty (ie. `grid[r][c] == 0`).
+
+Return the length of the shortest such clear path from top-left to bottom-right.  If such a path does not exist, return -1.
+
+**Example 1:**
+
+```
+Input: [[0,1],[1,0]]
+Output: 2
+```
+
+![https://assets.leetcode.com/uploads/2019/08/04/example1_1.png](https://assets.leetcode.com/uploads/2019/08/04/example1_1.png)
+
+![https://assets.leetcode.com/uploads/2019/08/04/example1_2.png](https://assets.leetcode.com/uploads/2019/08/04/example1_2.png)
+
+**Example 2:**
+
+```
+Input: [[0,0,0],[1,1,0],[1,1,0]]
+Output: 4
+```
+
+![https://assets.leetcode.com/uploads/2019/08/04/example2_1.png](https://assets.leetcode.com/uploads/2019/08/04/example2_1.png)
+
+![https://assets.leetcode.com/uploads/2019/08/04/example2_2.png](https://assets.leetcode.com/uploads/2019/08/04/example2_2.png)
+
+**Note:**
+
+1. `1 <= grid.length == grid[0].length <= 100`
+2. `grid[r][c]` is `0` or `1`
+
+## 题目大意
+
+在一个 N × N 的方形网格中，每个单元格有两种状态：空（0）或者阻塞（1）。一条从左上角到右下角、长度为 k 的畅通路径，由满足下述条件的单元格 C_1, C_2, ..., C_k 组成：
+
+- 相邻单元格 C_i 和 C_{i+1} 在八个方向之一上连通（此时，C_i 和 C_{i+1} 不同且共享边或角）
+- C_1 位于 (0, 0)（即，值为 grid[0][0]）
+- C_k 位于 (N-1, N-1)（即，值为 grid[N-1][N-1]）
+- 如果 C_i 位于 (r, c)，则 grid[r][c] 为空（即，grid[r][c] == 0）
+
+返回这条从左上角到右下角的最短畅通路径的长度。如果不存在这样的路径，返回 -1 。
+
+## 解题思路
+
+- 这一题是简单的找最短路径。利用 BFS 从左上角逐步扩展到右下角，便可以很容易求解。注意每轮扩展需要考虑 8 个方向。
+
+## 代码
+
+```go
+var dir = [][]int{
+	{-1, -1},
+	{-1, 0},
+	{-1, 1},
+	{0, 1},
+	{0, -1},
+	{1, -1},
+	{1, 0},
+	{1, 1},
+}
+
+func shortestPathBinaryMatrix(grid [][]int) int {
+	visited := make([][]bool, 0)
+	for range make([]int, len(grid)) {
+		visited = append(visited, make([]bool, len(grid[0])))
+	}
+	dis := make([][]int, 0)
+	for range make([]int, len(grid)) {
+		dis = append(dis, make([]int, len(grid[0])))
+	}
+	if grid[0][0] == 1 {
+		return -1
+	}
+	if len(grid) == 1 && len(grid[0]) == 1 {
+		return 1
+	}
+
+	queue := []int{0}
+	visited[0][0], dis[0][0] = true, 1
+	for len(queue) > 0 {
+		cur := queue[0]
+		queue = queue[1:]
+		curx, cury := cur/len(grid[0]), cur%len(grid[0])
+		for d := 0; d < 8; d++ {
+			nextx := curx + dir[d][0]
+			nexty := cury + dir[d][1]
+			if isInBoard(grid, nextx, nexty) && !visited[nextx][nexty] && grid[nextx][nexty] == 0 {
+				queue = append(queue, nextx*len(grid[0])+nexty)
+				visited[nextx][nexty] = true
+				dis[nextx][nexty] = dis[curx][cury] + 1
+				if nextx == len(grid)-1 && nexty == len(grid[0])-1 {
+					return dis[nextx][nexty]
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func isInBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+```
\ No newline at end of file
diff --git a/Algorithms/1093. Statistics from a Large Sample/1093. Statistics from a Large Sample.go b/leetcode/1093.Statistics-from-a-Large-Sample/1093. Statistics from a Large Sample.go
similarity index 100%
rename from Algorithms/1093. Statistics from a Large Sample/1093. Statistics from a Large Sample.go
rename to leetcode/1093.Statistics-from-a-Large-Sample/1093. Statistics from a Large Sample.go
diff --git a/Algorithms/1093. Statistics from a Large Sample/1093. Statistics from a Large Sample_test.go b/leetcode/1093.Statistics-from-a-Large-Sample/1093. Statistics from a Large Sample_test.go
similarity index 98%
rename from Algorithms/1093. Statistics from a Large Sample/1093. Statistics from a Large Sample_test.go
rename to leetcode/1093.Statistics-from-a-Large-Sample/1093. Statistics from a Large Sample_test.go
index c6c45af22..204ac353c 100644
--- a/Algorithms/1093. Statistics from a Large Sample/1093. Statistics from a Large Sample_test.go	
+++ b/leetcode/1093.Statistics-from-a-Large-Sample/1093. Statistics from a Large Sample_test.go	
@@ -26,12 +26,12 @@ func Test_Problem1093(t *testing.T) {
 
 	qs := []question1093{
 
-		question1093{
+		{
 			para1093{[]int{0, 1, 3, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}},
 			ans1093{[]float64{1.00000, 3.00000, 2.37500, 2.50000, 3.00000}},
 		},
 
-		question1093{
+		{
 			para1093{[]int{0, 4, 3, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}},
 			ans1093{[]float64{1.00000, 4.00000, 2.18182, 2.00000, 1.00000}},
 		},
diff --git a/leetcode/1093.Statistics-from-a-Large-Sample/README.md b/leetcode/1093.Statistics-from-a-Large-Sample/README.md
new file mode 100755
index 000000000..399615a9b
--- /dev/null
+++ b/leetcode/1093.Statistics-from-a-Large-Sample/README.md
@@ -0,0 +1,51 @@
+# [1093. Statistics from a Large Sample](https://leetcode.com/problems/statistics-from-a-large-sample/)
+
+
+## 题目
+
+We sampled integers between `0` and `255`, and stored the results in an array `count`: `count[k]` is the number of integers we sampled equal to `k`.
+
+Return the minimum, maximum, mean, median, and mode of the sample respectively, as an array of **floating point numbers**. The mode is guaranteed to be unique.
+
+*(Recall that the median of a sample is:*
+
+- *The middle element, if the elements of the sample were sorted and the number of elements is odd;*
+- *The average of the middle two elements, if the elements of the sample were sorted and the number of elements is even.)*
+
+**Example 1:**
+
+    Input: count = [0,1,3,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
+    Output: [1.00000,3.00000,2.37500,2.50000,3.00000]
+
+**Example 2:**
+
+    Input: count = [0,4,3,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
+    Output: [1.00000,4.00000,2.18182,2.00000,1.00000]
+
+**Constraints:**
+
+1. `count.length == 256`
+2. `1 <= sum(count) <= 10^9`
+3. The mode of the sample that count represents is unique.
+4. Answers within `10^-5` of the true value will be accepted as correct.
+
+
+## 题目大意
+
+我们对 0 到 255 之间的整数进行采样，并将结果存储在数组 count 中：count[k] 就是整数 k 的采样个数。
+
+我们以 浮点数 数组的形式，分别返回样本的最小值、最大值、平均值、中位数和众数。其中，众数是保证唯一的。我们先来回顾一下中位数的知识：
+
+- 如果样本中的元素有序，并且元素数量为奇数时，中位数为最中间的那个元素；
+- 如果样本中的元素有序，并且元素数量为偶数时，中位数为中间的两个元素的平均值。
+
+
+
+## 解题思路
+
+
+- 这个问题的关键需要理解题目的意思，什么是采样？`count[k]` 就是整数 `k` 的采样个数。
+- 题目要求返回样本的最小值、最大值、平均值、中位数和众数。最大值和最小值就很好处理，只需要遍历 count 判断最小的非 0 的 index 就是最小值，最大的非 0 的 index 就是最大值。平均值也非常好处理，对于所有非 0 的 count，我们通过累加 count[k] * index 得到所有数的和，然后除上所有非 0 的 count 的和。![](https://latex.codecogs.com/svg.latex?\sum_{n=0}^{256}count[n](while\%20\%20count[n]!=0))
+
+- 众数也非常容易，只需统计 count 值最大时的 index 即可。
+- 中位数的处理相对麻烦一些，因为需要分非 0 的 count 之和是奇数还是偶数两种情况。先假设非 0 的 count 和为 cnt，那么如果 cnt 是奇数的话，只需要找到 cnt/2 的位置即可，通过不断累加 count 的值，直到累加和超过 ≥ cnt/2。如果 cnt 是偶数的话，需要找到 cnt/2 + 1 和 cnt/2 的位置，找法和奇数情况相同，不过需要找两次(可以放到一个循环中做两次判断)。
\ No newline at end of file
diff --git a/leetcode/1104.Path-In-Zigzag-Labelled-Binary-Tree/1104.Path In Zigzag Labelled Binary Tree.go b/leetcode/1104.Path-In-Zigzag-Labelled-Binary-Tree/1104.Path In Zigzag Labelled Binary Tree.go
new file mode 100644
index 000000000..a7e2f580e
--- /dev/null
+++ b/leetcode/1104.Path-In-Zigzag-Labelled-Binary-Tree/1104.Path In Zigzag Labelled Binary Tree.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+func pathInZigZagTree(label int) []int {
+	level := getLevel(label)
+	ans := []int{label}
+	curIndex := label - (1 << level)
+	parent := 0
+	for level >= 1 {
+		parent, curIndex = getParent(curIndex, level)
+		ans = append(ans, parent)
+		level--
+	}
+	ans = reverse(ans)
+	return ans
+}
+
+func getLevel(label int) int {
+	level := 0
+	nums := 0
+	for {
+		nums += 1 << level
+		if nums >= label {
+			return level
+		}
+		level++
+	}
+}
+
+func getParent(index int, level int) (parent int, parentIndex int) {
+	parentIndex = 1<<(level-1) - 1 + (index/2)*(-1)
+	parent = 1<<(level-1) + parentIndex
+	return
+}
+
+func reverse(nums []int) []int {
+	left, right := 0, len(nums)-1
+	for left < right {
+		nums[left], nums[right] = nums[right], nums[left]
+		left++
+		right--
+	}
+	return nums
+}
diff --git a/leetcode/1104.Path-In-Zigzag-Labelled-Binary-Tree/1104.Path In Zigzag Labelled Binary Tree_test.go b/leetcode/1104.Path-In-Zigzag-Labelled-Binary-Tree/1104.Path In Zigzag Labelled Binary Tree_test.go
new file mode 100644
index 000000000..5bdd0cf0d
--- /dev/null
+++ b/leetcode/1104.Path-In-Zigzag-Labelled-Binary-Tree/1104.Path In Zigzag Labelled Binary Tree_test.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1104 struct {
+	para1104
+	ans1104
+}
+
+// para 是参数
+type para1104 struct {
+	label int
+}
+
+// ans 是答案
+type ans1104 struct {
+	ans []int
+}
+
+func Test_Problem1104(t *testing.T) {
+
+	qs := []question1104{
+
+		{
+			para1104{14},
+			ans1104{[]int{1, 3, 4, 14}},
+		},
+
+		{
+			para1104{26},
+			ans1104{[]int{1, 2, 6, 10, 26}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1104------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1104, q.para1104
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, pathInZigZagTree(p.label))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1104.Path-In-Zigzag-Labelled-Binary-Tree/README.md b/leetcode/1104.Path-In-Zigzag-Labelled-Binary-Tree/README.md
new file mode 100644
index 000000000..bdca7d7d0
--- /dev/null
+++ b/leetcode/1104.Path-In-Zigzag-Labelled-Binary-Tree/README.md
@@ -0,0 +1,92 @@
+# [1104. Path In Zigzag Labelled Binary Tree](https://leetcode.com/problems/path-in-zigzag-labelled-binary-tree/)
+
+
+## 题目
+
+In an infinite binary tree where every node has two children, the nodes are labelled in row order.
+
+In the odd numbered rows (ie., the first, third, fifth,...), the labelling is left to right, while in the even numbered rows (second, fourth, sixth,...), the labelling is right to left.
+
+![https://assets.leetcode.com/uploads/2019/06/24/tree.png](https://assets.leetcode.com/uploads/2019/06/24/tree.png)
+
+Given the `label` of a node in this tree, return the labels in the path from the root of the tree to the node with that `label`.
+
+**Example 1:**
+
+```
+Input: label = 14
+Output: [1,3,4,14]
+
+```
+
+**Example 2:**
+
+```
+Input: label = 26
+Output: [1,2,6,10,26]
+
+```
+
+**Constraints:**
+
+- `1 <= label <= 10^6`
+
+## 题目大意
+
+在一棵无限的二叉树上，每个节点都有两个子节点，树中的节点 逐行 依次按 “之” 字形进行标记。如下图所示，在奇数行（即，第一行、第三行、第五行……）中，按从左到右的顺序进行标记；而偶数行（即，第二行、第四行、第六行……）中，按从右到左的顺序进行标记。
+
+给你树上某一个节点的标号 label，请你返回从根节点到该标号为 label 节点的路径，该路径是由途经的节点标号所组成的。
+
+## 解题思路
+
+- 计算出 label 所在的 level 和 index。
+- 根据 index 和 level 计算出父节点的 index 和 value。
+- level 减一，循环计算出对应的父节点直到根节点。
+
+## 代码
+
+```go
+package leetcode
+
+func pathInZigZagTree(label int) []int {
+	level := getLevel(label)
+	ans := []int{label}
+	curIndex := label - (1 << level)
+	parent := 0
+	for level >= 1 {
+		parent, curIndex = getParent(curIndex, level)
+		ans = append(ans, parent)
+		level--
+	}
+	ans = reverse(ans)
+	return ans
+}
+
+func getLevel(label int) int {
+	level := 0
+	nums := 0
+	for {
+		nums += 1 << level
+		if nums >= label {
+			return level
+		}
+		level++
+	}
+}
+
+func getParent(index int, level int) (parent int, parentIndex int) {
+	parentIndex = 1<<(level-1) - 1 + (index/2)*(-1)
+	parent = 1<<(level-1) + parentIndex
+	return
+}
+
+func reverse(nums []int) []int {
+	left, right := 0, len(nums)-1
+	for left < right {
+		nums[left], nums[right] = nums[right], nums[left]
+		left++
+		right--
+	}
+	return nums
+}
+```
\ No newline at end of file
diff --git a/Algorithms/1105. Filling Bookcase Shelves/1105. Filling Bookcase Shelves.go b/leetcode/1105.Filling-Bookcase-Shelves/1105. Filling Bookcase Shelves.go
similarity index 76%
rename from Algorithms/1105. Filling Bookcase Shelves/1105. Filling Bookcase Shelves.go
rename to leetcode/1105.Filling-Bookcase-Shelves/1105. Filling Bookcase Shelves.go
index 0cc2d2542..f1c0e3974 100644
--- a/Algorithms/1105. Filling Bookcase Shelves/1105. Filling Bookcase Shelves.go	
+++ b/leetcode/1105.Filling-Bookcase-Shelves/1105. Filling Bookcase Shelves.go	
@@ -14,3 +14,17 @@ func minHeightShelves(books [][]int, shelfWidth int) int {
 	}
 	return dp[len(books)]
 }
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/Algorithms/1105. Filling Bookcase Shelves/1105. Filling Bookcase Shelves_test.go b/leetcode/1105.Filling-Bookcase-Shelves/1105. Filling Bookcase Shelves_test.go
similarity index 82%
rename from Algorithms/1105. Filling Bookcase Shelves/1105. Filling Bookcase Shelves_test.go
rename to leetcode/1105.Filling-Bookcase-Shelves/1105. Filling Bookcase Shelves_test.go
index 5bd9d4c8e..f3813ab88 100644
--- a/Algorithms/1105. Filling Bookcase Shelves/1105. Filling Bookcase Shelves_test.go	
+++ b/leetcode/1105.Filling-Bookcase-Shelves/1105. Filling Bookcase Shelves_test.go	
@@ -27,8 +27,8 @@ func Test_Problem1105(t *testing.T) {
 
 	qs := []question1105{
 
-		question1105{
-			para1105{[][]int{[]int{1, 1}, []int{2, 3}, []int{2, 3}, []int{1, 1}, []int{1, 1}, []int{1, 1}, []int{1, 2}}, 4},
+		{
+			para1105{[][]int{{1, 1}, {2, 3}, {2, 3}, {1, 1}, {1, 1}, {1, 1}, {1, 2}}, 4},
 			ans1105{6},
 		},
 	}
diff --git a/leetcode/1105.Filling-Bookcase-Shelves/README.md b/leetcode/1105.Filling-Bookcase-Shelves/README.md
new file mode 100755
index 000000000..af55f48b0
--- /dev/null
+++ b/leetcode/1105.Filling-Bookcase-Shelves/README.md
@@ -0,0 +1,45 @@
+# [1105. Filling Bookcase Shelves](https://leetcode.com/problems/filling-bookcase-shelves/)
+
+
+## 题目
+
+We have a sequence of `books`: the `i`-th book has thickness `books[i][0]`and height `books[i][1]`.
+
+We want to place these books **in order** onto bookcase shelves that have total width `shelf_width`.
+
+We choose some of the books to place on this shelf (such that the sum of their thickness is `<= shelf_width`), then build another level of shelf of the bookcase so that the total height of the bookcase has increased by the maximum height of the books we just put down. We repeat this process until there are no more books to place.
+
+Note again that at each step of the above process, the order of the books we place is the same order as the given sequence of books. For example, if we have an ordered list of 5 books, we might place the first and second book onto the first shelf, the third book on the second shelf, and the fourth and fifth book on the last shelf.
+
+Return the minimum possible height that the total bookshelf can be after placing shelves in this manner.
+
+**Example 1:**
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/06/28/shelves.png)
+
+    Input: books = [[1,1],[2,3],[2,3],[1,1],[1,1],[1,1],[1,2]], shelf_width = 4
+    Output: 6
+    Explanation:
+    The sum of the heights of the 3 shelves are 1 + 3 + 2 = 6.
+    Notice that book number 2 does not have to be on the first shelf.
+
+**Constraints:**
+
+- `1 <= books.length <= 1000`
+- `1 <= books[i][0] <= shelf_width <= 1000`
+- `1 <= books[i][1] <= 1000`
+
+## 题目大意
+
+附近的家居城促销，你买回了一直心仪的可调节书架，打算把自己的书都整理到新的书架上。你把要摆放的书 books 都整理好，叠成一摞：从上往下，第 i 本书的厚度为 books[i][0]，高度为 books[i][1]。按顺序将这些书摆放到总宽度为 shelf\_width 的书架上。
+
+先选几本书放在书架上（它们的厚度之和小于等于书架的宽度 shelf_width），然后再建一层书架。重复这个过程，直到把所有的书都放在书架上。
+
+需要注意的是，在上述过程的每个步骤中，摆放书的顺序与你整理好的顺序相同。 例如，如果这里有 5 本书，那么可能的一种摆放情况是：第一和第二本书放在第一层书架上，第三本书放在第二层书架上，第四和第五本书放在最后一层书架上。每一层所摆放的书的最大高度就是这一层书架的层高，书架整体的高度为各层高之和。以这种方式布置书架，返回书架整体可能的最小高度。
+
+
+
+## 解题思路
+
+- 给出一个数组，数组里面每个元素代表的是一个本书的宽度和高度。要求按照书籍的顺序，把书摆到宽度为 `shelf_width` 的书架上。问最终放下所有书籍以后，书架的最小高度。
+- 这一题的解题思路是动态规划。`dp[i]` 代表放置前 `i` 本书所需要的书架最小高度。初始值 dp[0] = 0，其他为最大值 1000*1000。遍历每一本书，把当前这本书作为书架最后一层的最后一本书，将这本书之前的书向后调整，看看是否可以减少之前的书架高度。状态转移方程为 `dp[i] = min(dp[i] , dp[j - 1] + h)`，其中 `j` 表示最后一层所能容下书籍的索引，`h` 表示最后一层最大高度。`j` 调整完一遍以后就能找出书架最小高度值了。时间复杂度 O(n^2)。
\ No newline at end of file
diff --git a/leetcode/1108.Defanging-an-IP-Address/1108. Defanging an IP Address.go b/leetcode/1108.Defanging-an-IP-Address/1108. Defanging an IP Address.go
new file mode 100644
index 000000000..cad90a480
--- /dev/null
+++ b/leetcode/1108.Defanging-an-IP-Address/1108. Defanging an IP Address.go	
@@ -0,0 +1,7 @@
+package leetcode
+
+import "strings"
+
+func defangIPaddr(address string) string {
+	return strings.Replace(address, ".", "[.]", -1)
+}
diff --git a/leetcode/1108.Defanging-an-IP-Address/1108. Defanging an IP Address_test.go b/leetcode/1108.Defanging-an-IP-Address/1108. Defanging an IP Address_test.go
new file mode 100644
index 000000000..9580433fd
--- /dev/null
+++ b/leetcode/1108.Defanging-an-IP-Address/1108. Defanging an IP Address_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1108 struct {
+	para1108
+	ans1108
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1108 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1108 struct {
+	one string
+}
+
+func Test_Problem1108(t *testing.T) {
+
+	qs := []question1108{
+
+		{
+			para1108{"1.1.1.1"},
+			ans1108{"1[.]1[.]1[.]1"},
+		},
+
+		{
+			para1108{"255.100.50.0"},
+			ans1108{"255[.]100[.]50[.]0"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1108------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1108, q.para1108
+		fmt.Printf("【input】:%v       【output】:%v\n", p, defangIPaddr(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1108.Defanging-an-IP-Address/README.md b/leetcode/1108.Defanging-an-IP-Address/README.md
new file mode 100755
index 000000000..e5306ee35
--- /dev/null
+++ b/leetcode/1108.Defanging-an-IP-Address/README.md
@@ -0,0 +1,39 @@
+# [1108. Defanging an IP Address](https://leetcode.com/problems/defanging-an-ip-address/)
+
+
+## 题目
+
+Given a valid (IPv4) IP `address`, return a defanged version of that IP address.
+
+A *defanged IP address* replaces every period `"."` with `"[.]"`.
+
+**Example 1:**
+
+    Input: address = "1.1.1.1"
+    Output: "1[.]1[.]1[.]1"
+
+**Example 2:**
+
+    Input: address = "255.100.50.0"
+    Output: "255[.]100[.]50[.]0"
+
+**Constraints:**
+
+- The given `address` is a valid IPv4 address.
+
+## 题目大意
+
+
+给你一个有效的 IPv4 地址 address，返回这个 IP 地址的无效化版本。所谓无效化 IP 地址，其实就是用 "[.]" 代替了每个 "."。
+
+
+提示：
+
+- 给出的 address 是一个有效的 IPv4 地址
+
+
+
+## 解题思路
+
+- 给出一个 IP 地址，要求把点替换成 `[.]`。
+- 简单题，按照题意替换即可。
diff --git a/leetcode/1110.Delete-Nodes-And-Return-Forest/1110. Delete Nodes And Return Forest.go b/leetcode/1110.Delete-Nodes-And-Return-Forest/1110. Delete Nodes And Return Forest.go
new file mode 100644
index 000000000..c1850bc8b
--- /dev/null
+++ b/leetcode/1110.Delete-Nodes-And-Return-Forest/1110. Delete Nodes And Return Forest.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func delNodes(root *TreeNode, toDelete []int) []*TreeNode {
+	if root == nil {
+		return nil
+	}
+	res, deleteMap := []*TreeNode{}, map[int]bool{}
+	for _, v := range toDelete {
+		deleteMap[v] = true
+	}
+	dfsDelNodes(root, deleteMap, true, &res)
+	return res
+}
+
+func dfsDelNodes(root *TreeNode, toDel map[int]bool, isRoot bool, res *[]*TreeNode) bool {
+	if root == nil {
+		return false
+	}
+	if isRoot && !toDel[root.Val] {
+		*res = append(*res, root)
+	}
+	isRoot = false
+	if toDel[root.Val] {
+		isRoot = true
+	}
+	if dfsDelNodes(root.Left, toDel, isRoot, res) {
+		root.Left = nil
+	}
+	if dfsDelNodes(root.Right, toDel, isRoot, res) {
+		root.Right = nil
+	}
+	return isRoot
+}
diff --git a/leetcode/1110.Delete-Nodes-And-Return-Forest/1110. Delete Nodes And Return Forest_test.go b/leetcode/1110.Delete-Nodes-And-Return-Forest/1110. Delete Nodes And Return Forest_test.go
new file mode 100644
index 000000000..da8c70c4b
--- /dev/null
+++ b/leetcode/1110.Delete-Nodes-And-Return-Forest/1110. Delete Nodes And Return Forest_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1110 struct {
+	para1110
+	ans1110
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1110 struct {
+	one []int
+	two []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1110 struct {
+	one [][]int
+}
+
+func Test_Problem1110(t *testing.T) {
+
+	qs := []question1110{
+
+		{
+			para1110{[]int{1, 2, 3, 4, 5, 6, 7}, []int{3, 5}},
+			ans1110{[][]int{{1, 2, structures.NULL, 4}, {6}, {7}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1110------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1110, q.para1110
+		tree := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【input】:%v       【output】:%v\n", p, delNodes(tree, p.two))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1110.Delete-Nodes-And-Return-Forest/README.md b/leetcode/1110.Delete-Nodes-And-Return-Forest/README.md
new file mode 100644
index 000000000..40219a653
--- /dev/null
+++ b/leetcode/1110.Delete-Nodes-And-Return-Forest/README.md
@@ -0,0 +1,83 @@
+# [1110. Delete Nodes And Return Forest](https://leetcode.com/problems/delete-nodes-and-return-forest/)
+
+
+
+## 题目
+
+Given the `root` of a binary tree, each node in the tree has a distinct value.
+
+After deleting all nodes with a value in `to_delete`, we are left with a forest (a disjoint union of trees).
+
+Return the roots of the trees in the remaining forest. You may return the result in any order.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/07/01/screen-shot-2019-07-01-at-53836-pm.png](https://assets.leetcode.com/uploads/2019/07/01/screen-shot-2019-07-01-at-53836-pm.png)
+
+```
+Input: root = [1,2,3,4,5,6,7], to_delete = [3,5]
+Output: [[1,2,null,4],[6],[7]]
+```
+
+**Constraints**:
+
+- The number of nodes in the given tree is at most `1000`.
+- Each node has a distinct value between `1` and `1000`.
+- `to_delete.length <= 1000`
+- `to_delete` contains distinct values between `1` and `1000`.
+
+
+## 题目大意
+
+给出二叉树的根节点 root，树上每个节点都有一个不同的值。如果节点值在 to_delete 中出现，我们就把该节点从树上删去，最后得到一个森林（一些不相交的树构成的集合）。返回森林中的每棵树。你可以按任意顺序组织答案。
+
+
+提示：
+
+- 树中的节点数最大为 1000。
+- 每个节点都有一个介于 1 到 1000 之间的值，且各不相同。
+- to_delete.length <= 1000
+- to_delete 包含一些从 1 到 1000、各不相同的值。
+
+
+
+## 解题思路
+
+- 给出一棵树，再给出一个数组，要求删除数组中相同元素值的节点。输出最终删除以后的森林。
+- 简单题。边遍历树，边删除数组中的元素。这里可以先把数组里面的元素放入 map 中，加速查找。遇到相同的元素就删除节点。这里需要特殊判断的是当前删除的节点是否是根节点，如果是根节点需要根据条件置空它的左节点或者右节点。
+
+## 代码
+
+```go
+func delNodes(root *TreeNode, toDelete []int) []*TreeNode {
+	if root == nil {
+		return nil
+	}
+	res, deleteMap := []*TreeNode{}, map[int]bool{}
+	for _, v := range toDelete {
+		deleteMap[v] = true
+	}
+	dfsDelNodes(root, deleteMap, true, &res)
+	return res
+}
+
+func dfsDelNodes(root *TreeNode, toDel map[int]bool, isRoot bool, res *[]*TreeNode) bool {
+	if root == nil {
+		return false
+	}
+	if isRoot && !toDel[root.Val] {
+		*res = append(*res, root)
+	}
+	isRoot = false
+	if toDel[root.Val] {
+		isRoot = true
+	}
+	if dfsDelNodes(root.Left, toDel, isRoot, res) {
+		root.Left = nil
+	}
+	if dfsDelNodes(root.Right, toDel, isRoot, res) {
+		root.Right = nil
+	}
+	return isRoot
+}
+```
\ No newline at end of file
diff --git a/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings.go b/leetcode/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings.go
similarity index 92%
rename from Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings.go
rename to leetcode/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings.go
index 6268f186a..07c72d918 100644
--- a/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings.go	
+++ b/leetcode/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings.go	
@@ -32,6 +32,13 @@ func maxDepthAfterSplit(seq string) []int {
 	return res
 }
 
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
 // 解法二 模拟
 func maxDepthAfterSplit1(seq string) []int {
 	stack, top, res := make([]int, len(seq)), -1, make([]int, len(seq))
diff --git a/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings_test.go b/leetcode/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings_test.go
similarity index 95%
rename from Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings_test.go
rename to leetcode/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings_test.go
index fffc33e83..332e88495 100644
--- a/Algorithms/1111. Maximum Nesting Depth of Two Valid Parentheses Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings_test.go	
+++ b/leetcode/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings/1111. Maximum Nesting Depth of Two Valid Parentheses Strings_test.go	
@@ -26,12 +26,12 @@ func Test_Problem1111(t *testing.T) {
 
 	qs := []question1111{
 
-		question1111{
+		{
 			para1111{"(()())"},
 			ans1111{[]int{0, 1, 1, 1, 1, 0}},
 		},
 
-		question1111{
+		{
 			para1111{"()(())()"},
 			ans1111{[]int{0, 0, 0, 1, 1, 0, 1, 1}},
 		},
diff --git a/leetcode/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings/README.md b/leetcode/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings/README.md
new file mode 100755
index 000000000..662a4a89c
--- /dev/null
+++ b/leetcode/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings/README.md
@@ -0,0 +1,75 @@
+# [1111. Maximum Nesting Depth of Two Valid Parentheses Strings](https://leetcode.com/problems/maximum-nesting-depth-of-two-valid-parentheses-strings/)
+
+
+## 题目
+
+A string is a *valid parentheses string* (denoted VPS) if and only if it consists of `"("` and `")"` characters only, and:
+
+- It is the empty string, or
+- It can be written as `AB` (`A` concatenated with `B`), where `A` and `B` are VPS's, or
+- It can be written as `(A)`, where `A` is a VPS.
+
+We can similarly define the *nesting depth* `depth(S)` of any VPS `S` as follows:
+
+- `depth("") = 0`
+- `depth(A + B) = max(depth(A), depth(B))`, where `A` and `B` are VPS's
+- `depth("(" + A + ")") = 1 + depth(A)`, where `A` is a VPS.
+
+For example, `""`, `"()()"`, and `"()(()())"` are VPS's (with nesting depths 0, 1, and 2), and `")("` and `"(()"` are not VPS's.
+
+Given a VPS seq, split it into two disjoint subsequences `A` and `B`, such that `A` and `B` are VPS's (and `A.length + B.length = seq.length`).
+
+Now choose **any** such `A` and `B` such that `max(depth(A), depth(B))` is the minimum possible value.
+
+Return an `answer` array (of length `seq.length`) that encodes such a choice of `A` and `B`: `answer[i] = 0` if `seq[i]` is part of `A`, else `answer[i] = 1`. Note that even though multiple answers may exist, you may return any of them.
+
+**Example 1:**
+
+    Input: seq = "(()())"
+    Output: [0,1,1,1,1,0]
+
+**Example 2:**
+
+    Input: seq = "()(())()"
+    Output: [0,0,0,1,1,0,1,1]
+
+**Constraints:**
+
+- `1 <= seq.size <= 10000`
+
+
+## 题目大意
+
+
+有效括号字符串 仅由 "(" 和 ")" 构成，并符合下述几个条件之一：
+
+- 空字符串
+- 连接，可以记作 AB（A 与 B 连接），其中 A 和 B 都是有效括号字符串
+- 嵌套，可以记作 (A)，其中 A 是有效括号字符串
+
+类似地，我们可以定义任意有效括号字符串 s 的 嵌套深度 depth(S)：
+
+- s 为空时，depth("") = 0
+- s 为 A 与 B 连接时，depth(A + B) = max(depth(A), depth(B))，其中 A 和 B 都是有效括号字符串
+- s 为嵌套情况，depth("(" + A + ")") = 1 + depth(A)，其中 A 是有效括号字符串
+
+
+例如：""，"()()"，和 "()(()())" 都是有效括号字符串，嵌套深度分别为 0，1，2，而 ")(" 和 "(()" 都不是有效括号字符串。
+
+ 
+
+给你一个有效括号字符串 seq，将其分成两个不相交的子序列 A 和 B，且 A 和 B 满足有效括号字符串的定义（注意：A.length + B.length = seq.length）。
+
+现在，你需要从中选出 任意 一组有效括号字符串 A 和 B，使 max(depth(A), depth(B)) 的可能取值最小。
+
+返回长度为 seq.length 答案数组 answer ，选择 A 还是 B 的编码规则是：如果 seq[i] 是 A 的一部分，那么 answer[i] = 0。否则，answer[i] = 1。即便有多个满足要求的答案存在，你也只需返回 一个。
+
+
+
+## 解题思路
+
+- 给出一个括号字符串。选出 A 部分和 B 部分，使得 `max(depth(A), depth(B))` 值最小。在最终的数组中输出 0 和 1，0 标识是 A 部分，1 标识是 B 部分。
+- 这一题想要 `max(depth(A), depth(B))` 值最小，可以使用贪心思想。如果 A 部分和 B 部分都尽快括号匹配，不深层次嵌套，那么总的层次就会变小。只要让嵌套的括号中属于 A 的和属于 B 的间隔排列即可。例如：“`(((())))`”，上面的字符串的嵌套深度是 4，按照上述的贪心思想，则标记为 0101 1010。
+- 这一题也可以用二分的思想来解答。把深度平分给 A 部分和 B 部分。
+    - 第一次遍历，先计算最大深度
+    - 第二次遍历，把深度小于等于最大深度一半的括号标记为 0(给 A 部分)，否则标记为 1(给 B 部分)
diff --git a/leetcode/1122.Relative-Sort-Array/1122. Relative Sort Array.go b/leetcode/1122.Relative-Sort-Array/1122. Relative Sort Array.go
new file mode 100644
index 000000000..350e8365c
--- /dev/null
+++ b/leetcode/1122.Relative-Sort-Array/1122. Relative Sort Array.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import "sort"
+
+// 解法一 桶排序，时间复杂度 O(n^2)
+func relativeSortArray(A, B []int) []int {
+	count := [1001]int{}
+	for _, a := range A {
+		count[a]++
+	}
+	res := make([]int, 0, len(A))
+	for _, b := range B {
+		for count[b] > 0 {
+			res = append(res, b)
+			count[b]--
+		}
+	}
+	for i := 0; i < 1001; i++ {
+		for count[i] > 0 {
+			res = append(res, i)
+			count[i]--
+		}
+	}
+	return res
+}
+
+// 解法二 模拟，时间复杂度 O(n^2)
+func relativeSortArray1(arr1 []int, arr2 []int) []int {
+	leftover, m, res := []int{}, make(map[int]int), []int{}
+	for _, v := range arr1 {
+		m[v]++
+	}
+	for _, s := range arr2 {
+		count := m[s]
+		for i := 0; i < count; i++ {
+			res = append(res, s)
+		}
+		m[s] = 0
+	}
+	for v, count := range m {
+		for i := 0; i < count; i++ {
+			leftover = append(leftover, v)
+		}
+	}
+	sort.Ints(leftover)
+	res = append(res, leftover...)
+	return res
+}
diff --git a/leetcode/1122.Relative-Sort-Array/1122. Relative Sort Array_test.go b/leetcode/1122.Relative-Sort-Array/1122. Relative Sort Array_test.go
new file mode 100644
index 000000000..8e53cf9a3
--- /dev/null
+++ b/leetcode/1122.Relative-Sort-Array/1122. Relative Sort Array_test.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1122 struct {
+	para1122
+	ans1122
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1122 struct {
+	arr1 []int
+	arr2 []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1122 struct {
+	one []int
+}
+
+func Test_Problem1122(t *testing.T) {
+
+	qs := []question1122{
+
+		{
+			para1122{[]int{2, 3, 1, 3, 2, 4, 6, 7, 9, 2, 19}, []int{2, 1, 4, 3, 9, 6}},
+			ans1122{[]int{2, 2, 2, 1, 4, 3, 3, 9, 6, 7, 19}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1122------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1122, q.para1122
+		fmt.Printf("【input】:%v       【output】:%v\n", p, relativeSortArray(p.arr1, p.arr2))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1122.Relative-Sort-Array/README.md b/leetcode/1122.Relative-Sort-Array/README.md
new file mode 100755
index 000000000..03955707f
--- /dev/null
+++ b/leetcode/1122.Relative-Sort-Array/README.md
@@ -0,0 +1,45 @@
+# [1122. Relative Sort Array](https://leetcode.com/problems/relative-sort-array/)
+
+
+## 题目
+
+Given two arrays `arr1` and `arr2`, the elements of `arr2` are distinct, and all elements in `arr2` are also in `arr1`.
+
+Sort the elements of `arr1` such that the relative ordering of items in `arr1` are the same as in `arr2`. Elements that don't appear in `arr2` should be placed at the end of `arr1` in **ascending** order.
+
+**Example 1:**
+
+    Input: arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
+    Output: [2,2,2,1,4,3,3,9,6,7,19]
+
+**Constraints:**
+
+- `arr1.length, arr2.length <= 1000`
+- `0 <= arr1[i], arr2[i] <= 1000`
+- Each `arr2[i]` is distinct.
+- Each `arr2[i]` is in `arr1`.
+
+
+## 题目大意
+
+
+给你两个数组，arr1 和 arr2，
+
+- arr2 中的元素各不相同
+- arr2 中的每个元素都出现在 arr1 中
+
+对 arr1 中的元素进行排序，使 arr1 中项的相对顺序和 arr2 中的相对顺序相同。未在 arr2 中出现过的元素需要按照升序放在 arr1 的末尾。
+
+提示：
+
+- arr1.length, arr2.length <= 1000
+- 0 <= arr1[i], arr2[i] <= 1000
+- arr2 中的元素 arr2[i] 各不相同
+- arr2 中的每个元素 arr2[i] 都出现在 arr1 中
+
+
+
+## 解题思路
+
+- 给出 2 个数组 A 和 B，A 中包含 B 中的所有元素。要求 A 按照 B 的元素顺序排序，B 中没有的元素再接着排在后面，从小到大排序。
+- 这一题有多种解法。一种暴力的解法就是依照题意，先把 A 中的元素都统计频次放在 map 中，然后 按照 B 的顺序输出，接着再把剩下的元素排序接在后面。还有一种桶排序的思想，由于题目限定了 A 的大小是 1000，这个数量级很小，所以可以用 1001 个桶装所有的数，把数都放在桶里，这样默认就已经排好序了。接下来的做法和前面暴力解法差不多，按照频次输出。B 中以外的元素就按照桶的顺序依次输出即可。
diff --git a/leetcode/1123.Lowest-Common-Ancestor-of-Deepest-Leaves/1123. Lowest Common Ancestor of Deepest Leaves.go b/leetcode/1123.Lowest-Common-Ancestor-of-Deepest-Leaves/1123. Lowest Common Ancestor of Deepest Leaves.go
new file mode 100644
index 000000000..62f8faa70
--- /dev/null
+++ b/leetcode/1123.Lowest-Common-Ancestor-of-Deepest-Leaves/1123. Lowest Common Ancestor of Deepest Leaves.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func lcaDeepestLeaves(root *TreeNode) *TreeNode {
+	if root == nil {
+		return nil
+	}
+	lca, maxLevel := &TreeNode{}, 0
+	lcaDeepestLeavesDFS(&lca, &maxLevel, 0, root)
+	return lca
+}
+
+func lcaDeepestLeavesDFS(lca **TreeNode, maxLevel *int, depth int, root *TreeNode) int {
+	*maxLevel = max(*maxLevel, depth)
+	if root == nil {
+		return depth
+	}
+	depthLeft := lcaDeepestLeavesDFS(lca, maxLevel, depth+1, root.Left)
+	depthRight := lcaDeepestLeavesDFS(lca, maxLevel, depth+1, root.Right)
+	if depthLeft == *maxLevel && depthRight == *maxLevel {
+		*lca = root
+	}
+	return max(depthLeft, depthRight)
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1123.Lowest-Common-Ancestor-of-Deepest-Leaves/1123. Lowest Common Ancestor of Deepest Leaves_test.go b/leetcode/1123.Lowest-Common-Ancestor-of-Deepest-Leaves/1123. Lowest Common Ancestor of Deepest Leaves_test.go
new file mode 100644
index 000000000..5debb6912
--- /dev/null
+++ b/leetcode/1123.Lowest-Common-Ancestor-of-Deepest-Leaves/1123. Lowest Common Ancestor of Deepest Leaves_test.go	
@@ -0,0 +1,71 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1123 struct {
+	para1123
+	ans1123
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1123 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1123 struct {
+	one []int
+}
+
+func Test_Problem1123(t *testing.T) {
+
+	qs := []question1123{
+
+		{
+			para1123{[]int{}},
+			ans1123{[]int{}},
+		},
+
+		{
+			para1123{[]int{1}},
+			ans1123{[]int{1}},
+		},
+
+		{
+			para1123{[]int{1, 2, 3, 4}},
+			ans1123{[]int{4}},
+		},
+
+		{
+			para1123{[]int{1, 2, 3}},
+			ans1123{[]int{1, 2, 3}},
+		},
+
+		{
+			para1123{[]int{1, 2, 3, 4, 5}},
+			ans1123{[]int{2, 4, 5}},
+		},
+
+		{
+			para1123{[]int{1, 2, structures.NULL, 3, 4, structures.NULL, 6, structures.NULL, 5}},
+			ans1123{[]int{2, 3, 4, structures.NULL, 6, structures.NULL, 5}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1123------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1123, q.para1123
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", structures.Tree2ints(lcaDeepestLeaves(root)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1123.Lowest-Common-Ancestor-of-Deepest-Leaves/README.md b/leetcode/1123.Lowest-Common-Ancestor-of-Deepest-Leaves/README.md
new file mode 100755
index 000000000..8e2493442
--- /dev/null
+++ b/leetcode/1123.Lowest-Common-Ancestor-of-Deepest-Leaves/README.md
@@ -0,0 +1,62 @@
+# [1123. Lowest Common Ancestor of Deepest Leaves](https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/)
+
+
+## 题目
+
+Given a rooted binary tree, return the lowest common ancestor of its deepest leaves.
+
+Recall that:
+
+- The node of a binary tree is a *leaf* if and only if it has no children
+- The *depth* of the root of the tree is 0, and if the depth of a node is `d`, the depth of each of its children is `d+1`.
+- The *lowest common ancestor* of a set `S` of nodes is the node `A` with the largest depth such that every node in S is in the subtree with root `A`.
+
+**Example 1:**
+
+    Input: root = [1,2,3]
+    Output: [1,2,3]
+    Explanation: 
+    The deepest leaves are the nodes with values 2 and 3.
+    The lowest common ancestor of these leaves is the node with value 1.
+    The answer returned is a TreeNode object (not an array) with serialization "[1,2,3]".
+
+**Example 2:**
+
+    Input: root = [1,2,3,4]
+    Output: [4]
+
+**Example 3:**
+
+    Input: root = [1,2,3,4,5]
+    Output: [2,4,5]
+
+**Constraints:**
+
+- The given tree will have between 1 and 1000 nodes.
+- Each node of the tree will have a distinct value between 1 and 1000.
+
+
+## 题目大意
+
+
+给你一个有根节点的二叉树，找到它最深的叶节点的最近公共祖先。
+
+回想一下：
+
+- 叶节点 是二叉树中没有子节点的节点
+- 树的根节点的 深度 为 0，如果某一节点的深度为 d，那它的子节点的深度就是 d+1
+- 如果我们假定 A 是一组节点 S 的 最近公共祖先，<font color="#c7254e" face="Menlo, Monaco, Consolas, Courier New, monospace">S</font> 中的每个节点都在以 A 为根节点的子树中，且 A 的深度达到此条件下可能的最大值。
+ 
+
+提示：
+
+- 给你的树中将有 1 到 1000 个节点。
+- 树中每个节点的值都在 1 到 1000 之间。
+
+
+## 解题思路
+
+
+- 给出一颗树，找出最深的叶子节点的最近公共祖先 LCA。
+- 这一题思路比较直接。先遍历找到最深的叶子节点，如果左右子树的最深的叶子节点深度相同，那么当前节点就是它们的最近公共祖先。如果左右子树的最深的深度不等，那么需要继续递归往下找符合题意的 LCA。如果最深的叶子节点没有兄弟，那么公共父节点就是叶子本身，否则返回它的 LCA。
+- 有几个特殊的测试用例，见测试文件。特殊的点就是最深的叶子节点没有兄弟节点的情况。
diff --git a/leetcode/1128.Number-of-Equivalent-Domino-Pairs/1128. Number of Equivalent Domino Pairs.go b/leetcode/1128.Number-of-Equivalent-Domino-Pairs/1128. Number of Equivalent Domino Pairs.go
new file mode 100644
index 000000000..6ec7a8296
--- /dev/null
+++ b/leetcode/1128.Number-of-Equivalent-Domino-Pairs/1128. Number of Equivalent Domino Pairs.go	
@@ -0,0 +1,23 @@
+package leetcode
+
+func numEquivDominoPairs(dominoes [][]int) int {
+	if dominoes == nil || len(dominoes) == 0 {
+		return 0
+	}
+	result, buckets := 0, [100]int{}
+	for _, dominoe := range dominoes {
+		key, rotatedKey := dominoe[0]*10+dominoe[1], dominoe[1]*10+dominoe[0]
+		if dominoe[0] != dominoe[1] {
+			if buckets[rotatedKey] > 0 {
+				result += buckets[rotatedKey]
+			}
+		}
+		if buckets[key] > 0 {
+			result += buckets[key]
+			buckets[key]++
+		} else {
+			buckets[key]++
+		}
+	}
+	return result
+}
diff --git a/leetcode/1128.Number-of-Equivalent-Domino-Pairs/1128. Number of Equivalent Domino Pairs_test.go b/leetcode/1128.Number-of-Equivalent-Domino-Pairs/1128. Number of Equivalent Domino Pairs_test.go
new file mode 100644
index 000000000..efb820536
--- /dev/null
+++ b/leetcode/1128.Number-of-Equivalent-Domino-Pairs/1128. Number of Equivalent Domino Pairs_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1128 struct {
+	para1128
+	ans1128
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1128 struct {
+	dominoes [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1128 struct {
+	one int
+}
+
+func Test_Problem1128(t *testing.T) {
+
+	qs := []question1128{
+
+		{
+			para1128{[][]int{{1, 2}, {2, 1}, {3, 4}, {5, 6}}},
+			ans1128{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1128------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1128, q.para1128
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numEquivDominoPairs(p.dominoes))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1128.Number-of-Equivalent-Domino-Pairs/README.md b/leetcode/1128.Number-of-Equivalent-Domino-Pairs/README.md
new file mode 100755
index 000000000..46c04a212
--- /dev/null
+++ b/leetcode/1128.Number-of-Equivalent-Domino-Pairs/README.md
@@ -0,0 +1,37 @@
+# [1128. Number of Equivalent Domino Pairs](https://leetcode.com/problems/number-of-equivalent-domino-pairs/)
+
+
+## 题目
+
+Given a list of `dominoes`, `dominoes[i] = [a, b]` is *equivalent* to `dominoes[j] = [c, d]` if and only if either (`a==c` and `b==d`), or (`a==d` and `b==c`) - that is, one domino can be rotated to be equal to another domino.
+
+Return the number of pairs `(i, j)` for which `0 <= i < j < dominoes.length`, and `dominoes[i]` is equivalent to `dominoes[j]`.
+
+**Example 1:**
+
+    Input: dominoes = [[1,2],[2,1],[3,4],[5,6]]
+    Output: 1
+
+**Constraints:**
+
+- `1 <= dominoes.length <= 40000`
+- `1 <= dominoes[i][j] <= 9`
+
+
+## 题目大意
+
+给你一个由一些多米诺骨牌组成的列表 dominoes。如果其中某一张多米诺骨牌可以通过旋转 0 度或 180 度得到另一张多米诺骨牌，我们就认为这两张牌是等价的。形式上，dominoes[i] = [a, b] 和 dominoes[j] = [c, d] 等价的前提是 a==c 且 b==d，或是 a==d 且 b==c。
+
+在 0 <= i < j < dominoes.length 的前提下，找出满足 dominoes[i] 和 dominoes[j] 等价的骨牌对 (i, j) 的数量。
+
+提示：
+
+- 1 <= dominoes.length <= 40000
+- 1 <= dominoes[i][j] <= 9
+
+
+
+## 解题思路
+
+- 给出一组多米诺骨牌，求出这组牌中相同牌的个数。牌相同的定义是：牌的 2 个数字相同(正序或者逆序相同都算相同)
+- 简单题。由于牌是 2 个数，所以将牌的 2 个数 hash 成一个 2 位数，比较大小即可，正序和逆序都 hash 成 2 位数，然后在桶中比较是否已经存在，如果不存在，跳过，如果存在，计数。
diff --git a/leetcode/1137.N-th-Tribonacci-Number/1137. N-th Tribonacci Number.go b/leetcode/1137.N-th-Tribonacci-Number/1137. N-th Tribonacci Number.go
new file mode 100644
index 000000000..d31d0c092
--- /dev/null
+++ b/leetcode/1137.N-th-Tribonacci-Number/1137. N-th Tribonacci Number.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+func tribonacci(n int) int {
+	if n < 2 {
+		return n
+	}
+	trib, prev, prev2 := 1, 1, 0
+	for n > 2 {
+		trib, prev, prev2 = trib+prev+prev2, trib, prev
+		n--
+	}
+	return trib
+}
diff --git a/leetcode/1137.N-th-Tribonacci-Number/1137. N-th Tribonacci Number_test.go b/leetcode/1137.N-th-Tribonacci-Number/1137. N-th Tribonacci Number_test.go
new file mode 100644
index 000000000..2546339e9
--- /dev/null
+++ b/leetcode/1137.N-th-Tribonacci-Number/1137. N-th Tribonacci Number_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1137 struct {
+	para1137
+	ans1137
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1137 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1137 struct {
+	one int
+}
+
+func Test_Problem1137(t *testing.T) {
+
+	qs := []question1137{
+
+		{
+			para1137{1},
+			ans1137{1},
+		},
+
+		{
+			para1137{2},
+			ans1137{1},
+		},
+
+		{
+			para1137{3},
+			ans1137{2},
+		},
+
+		{
+			para1137{4},
+			ans1137{4},
+		},
+
+		{
+			para1137{25},
+			ans1137{1389537},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1137------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1137, q.para1137
+		fmt.Printf("【input】:%v       【output】:%v\n", p, tribonacci(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1137.N-th-Tribonacci-Number/README.md b/leetcode/1137.N-th-Tribonacci-Number/README.md
new file mode 100755
index 000000000..246e9fcf3
--- /dev/null
+++ b/leetcode/1137.N-th-Tribonacci-Number/README.md
@@ -0,0 +1,50 @@
+# [1137. N-th Tribonacci Number](https://leetcode.com/problems/n-th-tribonacci-number/)
+
+
+## 题目
+
+The Tribonacci sequence Tn is defined as follows:
+
+T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0.
+
+Given `n`, return the value of Tn.
+
+**Example 1:**
+
+    Input: n = 4
+    Output: 4
+    Explanation:
+    T_3 = 0 + 1 + 1 = 2
+    T_4 = 1 + 1 + 2 = 4
+
+**Example 2:**
+
+    Input: n = 25
+    Output: 1389537
+
+**Constraints:**
+
+- `0 <= n <= 37`
+- The answer is guaranteed to fit within a 32-bit integer, ie. `answer <= 2^31 - 1`.
+
+
+## 题目大意
+
+
+泰波那契序列 Tn 定义如下： 
+
+T0 = 0, T1 = 1, T2 = 1, 且在 n >= 0 的条件下 Tn+3 = Tn + Tn+1 + Tn+2
+
+给你整数 n，请返回第 n 个泰波那契数 Tn 的值。
+
+提示：
+
+- 0 <= n <= 37
+- 答案保证是一个 32 位整数，即 answer <= 2^31 - 1。
+
+
+
+## 解题思路
+
+- 求泰波那契数列中的第 n 个数。
+- 简单题，按照题意定义计算即可。
diff --git a/leetcode/1143.Longest-Common-Subsequence/1143. Longest Common Subsequence.go b/leetcode/1143.Longest-Common-Subsequence/1143. Longest Common Subsequence.go
new file mode 100644
index 000000000..f81c73c9b
--- /dev/null
+++ b/leetcode/1143.Longest-Common-Subsequence/1143. Longest Common Subsequence.go	
@@ -0,0 +1,28 @@
+package leetcode
+
+func longestCommonSubsequence(text1 string, text2 string) int {
+	if len(text1) == 0 || len(text2) == 0 {
+		return 0
+	}
+	dp := make([][]int, len(text1)+1)
+	for i := range dp {
+		dp[i] = make([]int, len(text2)+1)
+	}
+	for i := 1; i < len(text1)+1; i++ {
+		for j := 1; j < len(text2)+1; j++ {
+			if text1[i-1] == text2[j-1] {
+				dp[i][j] = dp[i-1][j-1] + 1
+			} else {
+				dp[i][j] = max(dp[i][j-1], dp[i-1][j])
+			}
+		}
+	}
+	return dp[len(text1)][len(text2)]
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1143.Longest-Common-Subsequence/1143. Longest Common Subsequence_test.go b/leetcode/1143.Longest-Common-Subsequence/1143. Longest Common Subsequence_test.go
new file mode 100644
index 000000000..ced017063
--- /dev/null
+++ b/leetcode/1143.Longest-Common-Subsequence/1143. Longest Common Subsequence_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1143 struct {
+	para1143
+	ans1143
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1143 struct {
+	text1 string
+	text2 string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1143 struct {
+	one int
+}
+
+func Test_Problem1143(t *testing.T) {
+
+	qs := []question1143{
+
+		{
+			para1143{"abcde", "ace"},
+			ans1143{3},
+		},
+
+		{
+			para1143{"abc", "abc"},
+			ans1143{3},
+		},
+
+		{
+			para1143{"abc", "def"},
+			ans1143{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1143------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1143, q.para1143
+		fmt.Printf("【input】:%v       【output】:%v\n", p, longestCommonSubsequence(p.text1, p.text2))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1143.Longest-Common-Subsequence/README.md b/leetcode/1143.Longest-Common-Subsequence/README.md
new file mode 100644
index 000000000..78367ca19
--- /dev/null
+++ b/leetcode/1143.Longest-Common-Subsequence/README.md
@@ -0,0 +1,86 @@
+# [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)
+
+## 题目
+
+Given two strings `text1` and `text2`, return *the length of their longest **common subsequence**.* If there is no **common subsequence**, return `0`.
+
+A **subsequence** of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.
+
+- For example, `"ace"` is a subsequence of `"abcde"`.
+
+A **common subsequence** of two strings is a subsequence that is common to both strings.
+
+**Example 1:**
+
+```
+Input: text1 = "abcde", text2 = "ace"
+Output: 3
+Explanation: The longest common subsequence is "ace" and its length is 3.
+```
+
+**Example 2:**
+
+```
+Input: text1 = "abc", text2 = "abc"
+Output: 3
+Explanation: The longest common subsequence is "abc" and its length is 3.
+```
+
+**Example 3:**
+
+```
+Input: text1 = "abc", text2 = "def"
+Output: 0
+Explanation: There is no such common subsequence, so the result is 0.
+```
+
+**Constraints:**
+
+- `1 <= text1.length, text2.length <= 1000`
+- `text1` and `text2` consist of only lowercase English characters.
+
+## 题目大意
+
+给定两个字符串 text1 和 text2，返回这两个字符串的最长 公共子序列 的长度。如果不存在 公共子序列 ，返回 0 。一个字符串的 子序列 是指这样一个新的字符串：它是由原字符串在不改变字符的相对顺序的情况下删除某些字符（也可以不删除任何字符）后组成的新字符串。例如，"ace" 是 "abcde" 的子序列，但 "aec" 不是 "abcde" 的子序列。两个字符串的 公共子序列 是这两个字符串所共同拥有的子序列。
+
+## 解题思路
+
+- 这一题是经典的最长公共子序列的问题。解题思路是二维动态规划。假设字符串 `text1` 和 `text2` 的长度分别为 `m` 和 `n`，创建 `m+1` 行 `n+1` 列的二维数组 `dp`，定义 `dp[i][j]` 表示长度为 i 的 `text1[0:i-1]` 和长度为 j 的 `text2[0:j-1]` 的最长公共子序列的长度。先考虑边界条件。当 `i = 0` 时，`text1[]` 为空字符串，它与任何字符串的最长公共子序列的长度都是 `0`，所以 `dp[0][j] = 0`。同理当 `j = 0` 时，`text2[]` 为空字符串，它与任何字符串的最长公共子序列的长度都是 `0`，所以 `dp[i][0] = 0`。由于二维数组的大小特意增加了 `1`，即 `m+1` 和 `n+1`，并且默认值是 `0`，所以不需要再初始化赋值了。
+- 当 `text1[i−1] = text2[j−1]` 时，将这两个相同的字符称为公共字符，考虑 `text1[0:i−1]` 和 `text2[0:j−1]` 的最长公共子序列，再增加一个字符（即公共字符）即可得到 `text1[0:i]` 和 `text2[0:j]` 的最长公共子序列，所以 `dp[i][j]=dp[i−1][j−1]+1`。当 `text1[i−1] != text2[j−1]` 时，最长公共子序列一定在 `text[0:i-1], text2[0:j]` 和 `text[0:i], text2[0:j-1]` 中取得。即 `dp[i][j] = max(dp[i-1][j], dp[i][j-1])`。所以状态转移方程如下：
+
+    $$dp[i][j] = \left\{\begin{matrix}dp[i-1][j-1]+1 &,text1[i-1]=text2[j-1]\\max(dp[i-1][j],dp[i][j-1])&,text1[i-1]\neq text2[j-1]\end{matrix}\right.$$
+
+- 最终结果存储在 `dp[len(text1)][len(text2)]` 中。时间复杂度 `O(mn)`，空间复杂度 `O(mn)`，其中 `m` 和 `n` 分别是 `text1` 和 `text2` 的长度。
+
+## 代码
+
+```go
+package leetcode
+
+func longestCommonSubsequence(text1 string, text2 string) int {
+	if len(text1) == 0 || len(text2) == 0 {
+		return 0
+	}
+	dp := make([][]int, len(text1)+1)
+	for i := range dp {
+		dp[i] = make([]int, len(text2)+1)
+	}
+	for i := 1; i < len(text1)+1; i++ {
+		for j := 1; j < len(text2)+1; j++ {
+			if text1[i-1] == text2[j-1] {
+				dp[i][j] = dp[i-1][j-1] + 1
+			} else {
+				dp[i][j] = max(dp[i][j-1], dp[i-1][j])
+			}
+		}
+	}
+	return dp[len(text1)][len(text2)]
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1145.Binary-Tree-Coloring-Game/1145. Binary Tree Coloring Game.go b/leetcode/1145.Binary-Tree-Coloring-Game/1145. Binary Tree Coloring Game.go
new file mode 100644
index 000000000..d98c73791
--- /dev/null
+++ b/leetcode/1145.Binary-Tree-Coloring-Game/1145. Binary Tree Coloring Game.go	
@@ -0,0 +1,36 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func btreeGameWinningMove(root *TreeNode, n int, x int) bool {
+	var left, right int
+	dfsBtreeGameWinningMove(root, &left, &right, x)
+	up := n - left - right - 1
+	n /= 2
+	return left > n || right > n || up > n
+}
+
+func dfsBtreeGameWinningMove(node *TreeNode, left, right *int, x int) int {
+	if node == nil {
+		return 0
+	}
+	l, r := dfsBtreeGameWinningMove(node.Left, left, right, x), dfsBtreeGameWinningMove(node.Right, left, right, x)
+	if node.Val == x {
+		*left, *right = l, r
+	}
+	return l + r + 1
+}
diff --git a/leetcode/1145.Binary-Tree-Coloring-Game/1145. Binary Tree Coloring Game_test.go b/leetcode/1145.Binary-Tree-Coloring-Game/1145. Binary Tree Coloring Game_test.go
new file mode 100644
index 000000000..79b3f574d
--- /dev/null
+++ b/leetcode/1145.Binary-Tree-Coloring-Game/1145. Binary Tree Coloring Game_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1145 struct {
+	para1145
+	ans1145
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1145 struct {
+	root []int
+	n    int
+	x    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1145 struct {
+	one bool
+}
+
+func Test_Problem1145(t *testing.T) {
+
+	qs := []question1145{
+
+		{
+			para1145{[]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, 11, 3},
+			ans1145{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1145------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1145, q.para1145
+		tree := structures.Ints2TreeNode(p.root)
+		fmt.Printf("【input】:%v       【output】:%v\n", p, btreeGameWinningMove(tree, p.n, p.x))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1145.Binary-Tree-Coloring-Game/README.md b/leetcode/1145.Binary-Tree-Coloring-Game/README.md
new file mode 100644
index 000000000..02fbc6824
--- /dev/null
+++ b/leetcode/1145.Binary-Tree-Coloring-Game/README.md
@@ -0,0 +1,82 @@
+# [1145. Binary Tree Coloring Game](https://leetcode.com/problems/binary-tree-coloring-game/)
+
+
+
+## 题目
+
+Two players play a turn based game on a binary tree. We are given the `root` of this binary tree, and the number of nodes `n` in the tree. `n` is odd, and each node has a distinct value from `1` to `n`.
+
+Initially, the first player names a value `x` with `1 <= x <= n`, and the second player names a value `y` with `1 <= y <= n` and `y != x`. The first player colors the node with value `x` red, and the second player colors the node with value `y` blue.
+
+Then, the players take turns starting with the first player. In each turn, that player chooses a node of their color (red if player 1, blue if player 2) and colors an **uncolored** neighbor of the chosen node (either the left child, right child, or parent of the chosen node.)
+
+If (and only if) a player cannot choose such a node in this way, they must pass their turn. If both players pass their turn, the game ends, and the winner is the player that colored more nodes.
+
+You are the second player. If it is possible to choose such a `y` to ensure you win the game, return `true`. If it is not possible, return `false`.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/08/01/1480-binary-tree-coloring-game.png](https://assets.leetcode.com/uploads/2019/08/01/1480-binary-tree-coloring-game.png)
+
+```
+Input: root = [1,2,3,4,5,6,7,8,9,10,11], n = 11, x = 3
+Output: true
+Explanation: The second player can choose the node with value 2.
+```
+
+**Constraints**:
+
+- `root` is the root of a binary tree with `n` nodes and distinct node values from `1` to `n`.
+- `n` is odd.
+- `1 <= x <= n <= 100`
+
+## 题目大意
+
+有两位极客玩家参与了一场「二叉树着色」的游戏。游戏中，给出二叉树的根节点 root，树上总共有 n 个节点，且 n 为奇数，其中每个节点上的值从 1 到 n 各不相同。游戏从「一号」玩家开始（「一号」玩家为红色，「二号」玩家为蓝色），最开始时，
+
+- 「一号」玩家从 [1, n] 中取一个值 x（1 <= x <= n）；
+- 「二号」玩家也从 [1, n] 中取一个值 y（1 <= y <= n）且 y != x。
+- 「一号」玩家给值为 x 的节点染上红色，而「二号」玩家给值为 y 的节点染上蓝色。
+
+之后两位玩家轮流进行操作，每一回合，玩家选择一个他之前涂好颜色的节点，将所选节点一个 未着色 的邻节点（即左右子节点、或父节点）进行染色。如果当前玩家无法找到这样的节点来染色时，他的回合就会被跳过。若两个玩家都没有可以染色的节点时，游戏结束。着色节点最多的那位玩家获得胜利 ✌️。现在，假设你是「二号」玩家，根据所给出的输入，假如存在一个 y 值可以确保你赢得这场游戏，则返回 true；若无法获胜，就请返回 false。
+
+
+提示：
+
+- 二叉树的根节点为 root，树上由 n 个节点，节点上的值从 1 到 n 各不相同。
+- n 为奇数。
+- 1 <= x <= n <= 100
+
+## 解题思路
+
+- 2 个人参加二叉树着色游戏。二叉树节点数为奇数。1 号玩家和 2 号玩家分别在二叉树上选项一个点着色。每一回合，玩家选择一个他之前涂好颜色的节点，将所选节点一个 未着色 的邻节点（即左右子节点、或父节点）进行染色。当有人不能选点着色的时候，他的那个回合会被跳过。双方都没法继续着色的时候游戏结束。着色多的人获胜。问二号玩家是否存在必胜策略？
+
+![](https://img.halfrost.com/Leetcode/leetcode_1145.png)
+
+- 如图所示，当一号玩家选择了一个红色的结点，可能会将二叉树切割为 3 个部分（连通分量），如果选择的是根结点，则可能是 2 个部分或 1 个部分，如果选择叶结点，则是 1 个部分。不过无论哪种情况都无关紧要，我们都可以当成 3 个部分来对待，例如一号玩家选择了一个叶结点，我们也可以把叶结点的左右两个空指针看成大小为 0 的两个部分。
+- 那么二号玩家怎样选择蓝色结点才是最优呢？答案是：选择离红色结点最近，且所属连通分量规模最大的那个点。也就是示例图中的 1 号结点。如果我们选择了 1 号结点为蓝色结点，那么可以染成红色的点就只剩下 6 号点和 7 号点了，而蓝色可以把根结点和其左子树全部占据。
+- 如何确定蓝色是否有必胜策略，就可以转换为，被红色点切割的三个连通分量中，是否存在一个连通分量，大小大于所有结点数目的一半。统计三个连通分量大小的过程，可以用深度优先搜索（DFS）来实现。当遍历到某一结点，其结点值等于选定的红色结点时，我们统计这个结点的左子树 `red_left` 和右子树 `red_right` 的大小，那么我们就已经找到两个连通分量的大小了，最后一个父结点连通分量的大小，可以用结点总数减去这两个连通分量大小，再减去红色所占结点，即 `parent = n - red_left - red_right - 1`。
+
+## 代码
+
+```go
+
+func btreeGameWinningMove(root *TreeNode, n int, x int) bool {
+	var left, right int
+	dfsBtreeGameWinningMove(root, &left, &right, x)
+	up := n - left - right - 1
+	n /= 2
+	return left > n || right > n || up > n
+}
+
+func dfsBtreeGameWinningMove(node *TreeNode, left, right *int, x int) int {
+	if node == nil {
+		return 0
+	}
+	l, r := dfsBtreeGameWinningMove(node.Left, left, right, x), dfsBtreeGameWinningMove(node.Right, left, right, x)
+	if node.Val == x {
+		*left, *right = l, r
+	}
+	return l + r + 1
+}
+```
\ No newline at end of file
diff --git a/leetcode/1154.Day-of-the-Year/1154. Day of the Year.go b/leetcode/1154.Day-of-the-Year/1154. Day of the Year.go
new file mode 100644
index 000000000..e44bd883a
--- /dev/null
+++ b/leetcode/1154.Day-of-the-Year/1154. Day of the Year.go	
@@ -0,0 +1,11 @@
+package leetcode
+
+import "time"
+
+func dayOfYear(date string) int {
+	first := date[:4] + "-01-01"
+	firstDay, _ := time.Parse("2006-01-02", first)
+	dateDay, _ := time.Parse("2006-01-02", date)
+	duration := dateDay.Sub(firstDay)
+	return int(duration.Hours())/24 + 1
+}
diff --git a/leetcode/1154.Day-of-the-Year/1154. Day of the Year_test.go b/leetcode/1154.Day-of-the-Year/1154. Day of the Year_test.go
new file mode 100644
index 000000000..cdd07c2cb
--- /dev/null
+++ b/leetcode/1154.Day-of-the-Year/1154. Day of the Year_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1154 struct {
+	para1154
+	ans1154
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1154 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1154 struct {
+	one int
+}
+
+func Test_Problem1154(t *testing.T) {
+
+	qs := []question1154{
+
+		{
+			para1154{"2019-01-09"},
+			ans1154{9},
+		},
+
+		{
+			para1154{"2019-02-10"},
+			ans1154{41},
+		},
+
+		{
+			para1154{"2003-03-01"},
+			ans1154{60},
+		},
+
+		{
+			para1154{"2004-03-01"},
+			ans1154{61},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1154------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1154, q.para1154
+		fmt.Printf("【input】:%v       【output】:%v\n", p, dayOfYear(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1154.Day-of-the-Year/README.md b/leetcode/1154.Day-of-the-Year/README.md
new file mode 100755
index 000000000..771bd8fdf
--- /dev/null
+++ b/leetcode/1154.Day-of-the-Year/README.md
@@ -0,0 +1,62 @@
+# [1154. Day of the Year](https://leetcode.com/problems/day-of-the-year/)
+
+
+## 题目
+
+Given a string `date` representing a [Gregorian calendar](https://en.wikipedia.org/wiki/Gregorian_calendar) date formatted as `YYYY-MM-DD`, return the day number of the year.
+
+**Example 1:**
+
+    Input: date = "2019-01-09"
+    Output: 9
+    Explanation: Given date is the 9th day of the year in 2019.
+
+**Example 2:**
+
+    Input: date = "2019-02-10"
+    Output: 41
+
+**Example 3:**
+
+    Input: date = "2003-03-01"
+    Output: 60
+
+**Example 4:**
+
+    Input: date = "2004-03-01"
+    Output: 61
+
+**Constraints:**
+
+- `date.length == 10`
+- `date[4] == date[7] == '-'`, and all other `date[i]`'s are digits
+- `date` represents a calendar date between Jan 1st, 1900 and Dec 31, 2019.
+
+## 题目大意
+
+
+实现一个 MajorityChecker 的类，它应该具有下述几个 API：
+
+- MajorityChecker(int[] arr) 会用给定的数组 arr 来构造一个 MajorityChecker 的实例。
+- int query(int left, int right, int threshold) 有这么几个参数：  
+	- 0 <= left <= right < arr.length 表示数组 arr 的子数组的长度。
+	- 2 * threshold > right - left + 1，也就是说阈值 threshold 始终比子序列长度的一半还要大。
+
+每次查询 query(...) 会返回在 arr[left], arr[left+1], ..., arr[right] 中至少出现阈值次数 threshold 的元素，如果不存在这样的元素，就返回 -1。
+
+提示：
+
+- 1 <= arr.length <= 20000
+- 1 <= arr[i] <= 20000
+- 对于每次查询，0 <= left <= right < len(arr)
+- 对于每次查询，2 * threshold > right - left + 1
+- 查询次数最多为 10000
+
+
+
+
+
+## 解题思路
+
+- 给出一个时间字符串，求出这一天是这一年当中的第几天。
+- 简单题。依照题意处理即可。
diff --git a/Algorithms/1157. Online Majority Element In Subarray/1157. Online Majority Element In Subarray.go b/leetcode/1157.Online-Majority-Element-In-Subarray/1157. Online Majority Element In Subarray.go
similarity index 100%
rename from Algorithms/1157. Online Majority Element In Subarray/1157. Online Majority Element In Subarray.go
rename to leetcode/1157.Online-Majority-Element-In-Subarray/1157. Online Majority Element In Subarray.go
diff --git a/Algorithms/1157. Online Majority Element In Subarray/1157. Online Majority Element In Subarray_test.go b/leetcode/1157.Online-Majority-Element-In-Subarray/1157. Online Majority Element In Subarray_test.go
similarity index 100%
rename from Algorithms/1157. Online Majority Element In Subarray/1157. Online Majority Element In Subarray_test.go
rename to leetcode/1157.Online-Majority-Element-In-Subarray/1157. Online Majority Element In Subarray_test.go
diff --git a/leetcode/1157.Online-Majority-Element-In-Subarray/README.md b/leetcode/1157.Online-Majority-Element-In-Subarray/README.md
new file mode 100755
index 000000000..b4d148787
--- /dev/null
+++ b/leetcode/1157.Online-Majority-Element-In-Subarray/README.md
@@ -0,0 +1,78 @@
+# [1157. Online Majority Element In Subarray](https://leetcode.com/problems/online-majority-element-in-subarray/)
+
+
+## 题目
+
+Implementing the class `MajorityChecker`, which has the following API:
+
+- `MajorityChecker(int[] arr)` constructs an instance of MajorityChecker with the given array `arr`;
+- `int query(int left, int right, int threshold)` has arguments such that:
+    - `0 <= left <= right < arr.length` representing a subarray of `arr`;
+    - `2 * threshold > right - left + 1`, ie. the threshold is always a strict majority of the length of the subarray
+
+Each `query(...)` returns the element in `arr[left], arr[left+1], ..., arr[right]` that occurs at least `threshold` times, or `-1` if no such element exists.
+
+**Example:**
+
+    MajorityChecker majorityChecker = new MajorityChecker([1,1,2,2,1,1]);
+    majorityChecker.query(0,5,4); // returns 1
+    majorityChecker.query(0,3,3); // returns -1
+    majorityChecker.query(2,3,2); // returns 2
+
+**Constraints:**
+
+- `1 <= arr.length <= 20000`
+- `1 <= arr[i] <= 20000`
+- For each query, `0 <= left <= right < len(arr)`
+- For each query, `2 * threshold > right - left + 1`
+- The number of queries is at most `10000`
+
+
+## 题目大意
+
+实现一个 MajorityChecker 的类，它应该具有下述几个 API：
+
+- MajorityChecker(int[] arr) 会用给定的数组 arr 来构造一个 MajorityChecker 的实例。
+- int query(int left, int right, int threshold) 有这么几个参数：
+- 0 <= left <= right < arr.length 表示数组 arr 的子数组的长度。
+- 2 * threshold > right - left + 1，也就是说阈值 threshold 始终比子序列长度的一半还要大。
+
+每次查询 query(...) 会返回在 arr[left], arr[left+1], ..., arr[right] 中至少出现阈值次数 threshold 的元素，如果不存在这样的元素，就返回 -1。
+
+提示：
+
+- 1 <= arr.length <= 20000
+- 1 <= arr[i] <= 20000
+- 对于每次查询，0 <= left <= right < len(arr)
+- 对于每次查询，2 * threshold > right - left + 1
+- 查询次数最多为 10000
+
+
+
+
+## 解题思路
+
+
+- 设计一个数据结构，能在任意的一个区间内，查找是否存在众数，众数的定义是：该数字出现的次数大于区间的一半。如果存在众数，一定唯一。如果在给定的区间内找不到众数，则输出 -1 。
+- 这一题有一个很显眼的“暗示”，`2 * threshold > right - left + 1`，这个条件就是摩尔投票算法的前提条件。摩尔投票的思想可以见第 169 题。这一题又要在区间内查询，所以选用线段树这个数据结构来实现。经过分析，可以确定此题的解题思路，摩尔投票 + 线段树。
+- 摩尔投票的思想是用两个变量，candidate 和 count，用来记录待被投票投出去的元素，和候选人累积没被投出去的轮数。如果候选人累积没有被投出去的轮数越多，那么最终成为众数的可能越大。从左往右扫描整个数组，先去第一个元素为 candidate，如果遇到相同的元素就累加轮数，如果遇到不同的元素，就把 candidate 和不同的元素一起投出去。当轮数变成 0 了，再选下一个元素作为 candidate。从左扫到右，就能找到众数了。那怎么和线段树结合起来呢？
+- 线段树是把一个大的区间拆分成很多个小区间，那么考虑这样一个问题。每个小区间内使用摩尔投票，最终把所有小区间合并起来再用一次摩尔投票，得到的结果和对整个区间使用一次摩尔投票的结果是一样的么？答案是一样的。可以这样想，众数总会在一个区间内被选出来，那么其他区间的摩尔投票都是起“中和”作用的，即两两元素一起出局。这个问题想通以后，说明摩尔投票具有可加的性质。既然满足可加，就可以和线段树结合，因为线段树每个线段就是加起来，最终合并成大区间的。
+- 举个例子，arr = [1,1,2,2,1,1]，先构造线段树，如下左图。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_1157_0.png)
+
+    现在每个线段树的节点不是只存一个 int 数字了，而是存 candidate 和 count。每个节点的 candidate 和 count 分别代表的是该区间内摩尔投票的结果。初始化的时候，先把每个叶子都填满，candidate 是自己，count = 1 。即右图绿色节点。然后在 pushUp 的时候，进行摩尔投票：
+
+        mc.merge = func(i, j segmentItem) segmentItem {
+        		if i.candidate == j.candidate {
+        			return segmentItem{candidate: i.candidate, count: i.count + j.count}
+        		}
+        		if i.count > j.count {
+        			return segmentItem{candidate: i.candidate, count: i.count - j.count}
+        		}
+        		return segmentItem{candidate: j.candidate, count: j.count - i.count}
+        	}
+
+    直到根节点的 candidate 和 count 都填满。**注意，这里的 count 并不是元素出现的总次数，而是摩尔投票中坚持没有被投出去的轮数**。当线段树构建完成以后，就可以开始查询任意区间内的众数了，candidate 即为众数。接下来还要确定众数是否满足 `threshold` 的条件。
+
+- 用一个字典记录每个元素在数组中出现位置的下标，例如上述这个例子，用 map 记录下标：count = map[1:[0 1 4 5] 2:[2 3]]。由于下标在记录过程中是递增的，所以满足二分查找的条件。利用这个字典就可以查出在任意区间内，指定元素出现的次数。例如这里要查找 1 在 [0,5] 区间内出现的个数，那么利用 2 次二分查找，分别找到 `lowerBound` 和 `upperBound`，在 [lowerBound，upperBound) 区间内，都是元素 1 ，那么区间长度即是该元素重复出现的次数，和 `threshold` 比较，如果 ≥ `threshold` 说明找到了答案，否则没有找到就输出 -1 。
diff --git a/leetcode/1160.Find-Words-That-Can-Be-Formed-by-Characters/1160. Find Words That Can Be Formed by Characters.go b/leetcode/1160.Find-Words-That-Can-Be-Formed-by-Characters/1160. Find Words That Can Be Formed by Characters.go
new file mode 100644
index 000000000..7a6397eb7
--- /dev/null
+++ b/leetcode/1160.Find-Words-That-Can-Be-Formed-by-Characters/1160. Find Words That Can Be Formed by Characters.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+func countCharacters(words []string, chars string) int {
+	count, res := make([]int, 26), 0
+	for i := 0; i < len(chars); i++ {
+		count[chars[i]-'a']++
+	}
+	for _, w := range words {
+		if canBeFormed(w, count) {
+			res += len(w)
+		}
+	}
+	return res
+}
+func canBeFormed(w string, c []int) bool {
+	count := make([]int, 26)
+	for i := 0; i < len(w); i++ {
+		count[w[i]-'a']++
+		if count[w[i]-'a'] > c[w[i]-'a'] {
+			return false
+		}
+	}
+	return true
+}
diff --git a/leetcode/1160.Find-Words-That-Can-Be-Formed-by-Characters/1160. Find Words That Can Be Formed by Characters_test.go b/leetcode/1160.Find-Words-That-Can-Be-Formed-by-Characters/1160. Find Words That Can Be Formed by Characters_test.go
new file mode 100644
index 000000000..7820fc52a
--- /dev/null
+++ b/leetcode/1160.Find-Words-That-Can-Be-Formed-by-Characters/1160. Find Words That Can Be Formed by Characters_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1160 struct {
+	para1160
+	ans1160
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1160 struct {
+	words []string
+	chars string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1160 struct {
+	one int
+}
+
+func Test_Problem1160(t *testing.T) {
+
+	qs := []question1160{
+
+		{
+			para1160{[]string{"cat", "bt", "hat", "tree"}, "atach"},
+			ans1160{6},
+		},
+
+		{
+			para1160{[]string{"hello", "world", "leetcode"}, "welldonehoneyr"},
+			ans1160{10},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1160------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1160, q.para1160
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countCharacters(p.words, p.chars))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1160.Find-Words-That-Can-Be-Formed-by-Characters/README.md b/leetcode/1160.Find-Words-That-Can-Be-Formed-by-Characters/README.md
new file mode 100755
index 000000000..5323e4bf2
--- /dev/null
+++ b/leetcode/1160.Find-Words-That-Can-Be-Formed-by-Characters/README.md
@@ -0,0 +1,49 @@
+# [1160. Find Words That Can Be Formed by Characters](https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/)
+
+
+## 题目
+
+You are given an array of strings `words` and a string `chars`.
+
+A string is *good* if it can be formed by characters from `chars` (each character can only be used once).
+
+Return the sum of lengths of all good strings in `words`.
+
+**Example 1:**
+
+    Input: words = ["cat","bt","hat","tree"], chars = "atach"
+    Output: 6
+    Explanation: 
+    The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.
+
+**Example 2:**
+
+    Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr"
+    Output: 10
+    Explanation: 
+    The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10.
+
+**Note:**
+
+1. `1 <= words.length <= 1000`
+2. `1 <= words[i].length, chars.length <= 100`
+3. All strings contain lowercase English letters only.
+
+
+## 题目大意
+
+
+给你一份『词汇表』（字符串数组） words 和一张『字母表』（字符串） chars。假如你可以用 chars 中的『字母』（字符）拼写出 words 中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。注意：每次拼写时，chars 中的每个字母都只能用一次。返回词汇表 words 中你掌握的所有单词的 长度之和。
+
+提示：
+
+1. 1 <= words.length <= 1000
+2. 1 <= words[i].length, chars.length <= 100
+3. 所有字符串中都仅包含小写英文字母
+
+
+
+## 解题思路
+
+- 给出一个字符串数组 `words` 和一个字符串 `chars`，要求输出 `chars` 中能构成 `words` 字符串的字符数总数。
+- 简单题。先分别统计 `words` 和 `chars` 里面字符的频次。然后针对 `words` 中每个 `word` 判断能够能由 `chars` 构成，如果能构成，最终结果加上这个 `word` 的长度。
diff --git a/leetcode/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character/1170. Compare Strings by Frequency of the Smallest Character.go b/leetcode/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character/1170. Compare Strings by Frequency of the Smallest Character.go
new file mode 100644
index 000000000..668504c24
--- /dev/null
+++ b/leetcode/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character/1170. Compare Strings by Frequency of the Smallest Character.go	
@@ -0,0 +1,27 @@
+package leetcode
+
+import "sort"
+
+func numSmallerByFrequency(queries []string, words []string) []int {
+	ws, res := make([]int, len(words)), make([]int, len(queries))
+	for i, w := range words {
+		ws[i] = countFunc(w)
+	}
+	sort.Ints(ws)
+	for i, q := range queries {
+		fq := countFunc(q)
+		res[i] = len(words) - sort.Search(len(words), func(i int) bool { return fq < ws[i] })
+	}
+	return res
+}
+
+func countFunc(s string) int {
+	count, i := [26]int{}, 0
+	for _, b := range s {
+		count[b-'a']++
+	}
+	for count[i] == 0 {
+		i++
+	}
+	return count[i]
+}
diff --git a/leetcode/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character/1170. Compare Strings by Frequency of the Smallest Character_test.go b/leetcode/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character/1170. Compare Strings by Frequency of the Smallest Character_test.go
new file mode 100644
index 000000000..ea6e71051
--- /dev/null
+++ b/leetcode/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character/1170. Compare Strings by Frequency of the Smallest Character_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1170 struct {
+	para1170
+	ans1170
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1170 struct {
+	queries []string
+	words   []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1170 struct {
+	one []int
+}
+
+func Test_Problem1170(t *testing.T) {
+
+	qs := []question1170{
+
+		{
+			para1170{[]string{"cbd"}, []string{"zaaaz"}},
+			ans1170{[]int{1}},
+		},
+
+		{
+			para1170{[]string{"bbb", "cc"}, []string{"a", "aa", "aaa", "aaaa"}},
+			ans1170{[]int{1, 2}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1170------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1170, q.para1170
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numSmallerByFrequency(p.queries, p.words))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character/README.md b/leetcode/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character/README.md
new file mode 100755
index 000000000..df051ec14
--- /dev/null
+++ b/leetcode/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character/README.md
@@ -0,0 +1,51 @@
+# [1170. Compare Strings by Frequency of the Smallest Character](https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/)
+
+## 题目
+
+Let's define a function `f(s)` over a non-empty string `s`, which calculates the frequency of the smallest character in `s`. For example, if `s = "dcce"` then `f(s) = 2` because the smallest character is `"c"` and its frequency is 2.
+
+Now, given string arrays `queries` and `words`, return an integer array `answer`, where each `answer[i]` is the number of words such that `f(queries[i])` < `f(W)`, where `W` is a word in `words`.
+
+**Example 1:**
+
+    Input: queries = ["cbd"], words = ["zaaaz"]
+    Output: [1]
+    Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").
+
+**Example 2:**
+
+    Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
+    Output: [1,2]
+    Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").
+
+**Constraints:**
+
+- `1 <= queries.length <= 2000`
+- `1 <= words.length <= 2000`
+- `1 <= queries[i].length, words[i].length <= 10`
+- `queries[i][j]`, `words[i][j]` are English lowercase letters.
+
+
+## 题目大意
+
+
+我们来定义一个函数 f(s)，其中传入参数 s 是一个非空字符串；该函数的功能是统计 s  中（按字典序比较）最小字母的出现频次。
+
+例如，若 s = "dcce"，那么 f(s) = 2，因为最小的字母是 "c"，它出现了 2 次。
+
+现在，给你两个字符串数组待查表 queries 和词汇表 words，请你返回一个整数数组 answer 作为答案，其中每个 answer[i] 是满足 f(queries[i]) < f(W) 的词的数目，W 是词汇表 words 中的词。
+
+提示：
+
+- 1 <= queries.length <= 2000
+- 1 <= words.length <= 2000
+- 1 <= queries[i].length, words[i].length <= 10
+- queries[i][j], words[i][j] 都是小写英文字母
+
+
+
+
+## 解题思路
+
+- 给出 2 个数组，`queries` 和 `words`，针对每一个 `queries[i]` 统计在 `words[j]` 中满足 `f(queries[i]) < f(words[j])` 条件的 `words[j]` 的个数。`f(string)` 的定义是 `string` 中字典序最小的字母的频次。
+- 先依照题意，构造出 `f()` 函数，算出每个 `words[j]` 的 `f()` 值，然后排序。再依次计算 `queries[i]` 的 `f()` 值。针对每个 `f()` 值，在 `words[j]` 的 `f()` 值中二分搜索，查找比它大的值的下标 `k`，`n-k` 即是比 `queries[i]` 的 `f()` 值大的元素个数。依次输出到结果数组中即可。
diff --git a/leetcode/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List/1171. Remove Zero Sum Consecutive Nodes from Linked List.go b/leetcode/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List/1171. Remove Zero Sum Consecutive Nodes from Linked List.go
new file mode 100644
index 000000000..6b650954b
--- /dev/null
+++ b/leetcode/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List/1171. Remove Zero Sum Consecutive Nodes from Linked List.go	
@@ -0,0 +1,88 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// 解法一
+func removeZeroSumSublists(head *ListNode) *ListNode {
+	// 计算累加和，和作为 key 存在 map 中，value 存那个节点的指针。如果字典中出现了重复的和，代表出现了和为 0 的段。
+	sum, sumMap, cur := 0, make(map[int]*ListNode), head
+	// 字典中增加 0 这个特殊值，是为了防止最终链表全部消除完
+	sumMap[0] = nil
+	for cur != nil {
+		sum = sum + cur.Val
+		if ptr, ok := sumMap[sum]; ok {
+			// 在字典中找到了重复的和，代表 [ptr, tmp] 中间的是和为 0 的段，要删除的就是这一段。
+			// 同时删除 map 中中间这一段的和
+			if ptr != nil {
+				iter := ptr.Next
+				tmpSum := sum + iter.Val
+				for tmpSum != sum {
+					// 删除中间为 0 的那一段，tmpSum 不断的累加删除 map 中的和
+					delete(sumMap, tmpSum)
+					iter = iter.Next
+					tmpSum = tmpSum + iter.Val
+				}
+				ptr.Next = cur.Next
+			} else {
+				head = cur.Next
+				sumMap = make(map[int]*ListNode)
+				sumMap[0] = nil
+			}
+		} else {
+			sumMap[sum] = cur
+		}
+		cur = cur.Next
+	}
+	return head
+}
+
+// 解法二 暴力解法
+func removeZeroSumSublists1(head *ListNode) *ListNode {
+	if head == nil {
+		return nil
+	}
+	h, prefixSumMap, sum, counter, lastNode := head, map[int]int{}, 0, 0, &ListNode{Val: 1010}
+	for h != nil {
+		for h != nil {
+			sum += h.Val
+			counter++
+			if v, ok := prefixSumMap[sum]; ok {
+				lastNode, counter = h, v
+				break
+			}
+			if sum == 0 {
+				head = h.Next
+				break
+			}
+			prefixSumMap[sum] = counter
+			h = h.Next
+		}
+		if lastNode.Val != 1010 {
+			h = head
+			for counter > 1 {
+				counter--
+				h = h.Next
+			}
+			h.Next = lastNode.Next
+		}
+		if h == nil {
+			break
+		} else {
+			h, prefixSumMap, sum, counter, lastNode = head, map[int]int{}, 0, 0, &ListNode{Val: 1010}
+		}
+	}
+	return head
+}
diff --git a/leetcode/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List/1171. Remove Zero Sum Consecutive Nodes from Linked List_test.go b/leetcode/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List/1171. Remove Zero Sum Consecutive Nodes from Linked List_test.go
new file mode 100644
index 000000000..f62769005
--- /dev/null
+++ b/leetcode/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List/1171. Remove Zero Sum Consecutive Nodes from Linked List_test.go	
@@ -0,0 +1,59 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1171 struct {
+	para1171
+	ans1171
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1171 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1171 struct {
+	one []int
+}
+
+func Test_Problem1171(t *testing.T) {
+
+	qs := []question1171{
+
+		{
+			para1171{[]int{1, 2, -3, 3, 1}},
+			ans1171{[]int{3, 1}},
+		},
+
+		{
+			para1171{[]int{1, 2, 3, -3, 4}},
+			ans1171{[]int{1, 2, 4}},
+		},
+
+		{
+			para1171{[]int{1, 2, 3, -3, -2}},
+			ans1171{[]int{1}},
+		},
+
+		{
+			para1171{[]int{1, -1}},
+			ans1171{[]int{}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1171------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1171, q.para1171
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(removeZeroSumSublists(structures.Ints2List(p.one))))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List/README.md b/leetcode/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List/README.md
new file mode 100755
index 000000000..ceccd5e37
--- /dev/null
+++ b/leetcode/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List/README.md
@@ -0,0 +1,51 @@
+# [1171. Remove Zero Sum Consecutive Nodes from Linked List](https://leetcode.com/problems/remove-zero-sum-consecutive-nodes-from-linked-list/)
+
+
+## 题目
+
+Given the `head` of a linked list, we repeatedly delete consecutive sequences of nodes that sum to `0` until there are no such sequences.
+
+After doing so, return the head of the final linked list. You may return any such answer.
+
+(Note that in the examples below, all sequences are serializations of `ListNode` objects.)
+
+**Example 1:**
+
+    Input: head = [1,2,-3,3,1]
+    Output: [3,1]
+    Note: The answer [1,2,1] would also be accepted.
+
+**Example 2:**
+
+    Input: head = [1,2,3,-3,4]
+    Output: [1,2,4]
+
+**Example 3:**
+
+    Input: head = [1,2,3,-3,-2]
+    Output: [1]
+
+**Constraints:**
+
+- The given linked list will contain between `1` and `1000` nodes.
+- Each node in the linked list has `-1000 <= node.val <= 1000`.
+
+
+## 题目大意
+
+
+给你一个链表的头节点 head，请你编写代码，反复删去链表中由 总和 值为 0 的连续节点组成的序列，直到不存在这样的序列为止。删除完毕后，请你返回最终结果链表的头节点。你可以返回任何满足题目要求的答案。
+
+（注意，下面示例中的所有序列，都是对 ListNode 对象序列化的表示。）
+
+提示：
+
+- 给你的链表中可能有 1 到 1000 个节点。
+- 对于链表中的每个节点，节点的值：-1000 <= node.val <= 1000.
+
+
+
+## 解题思路
+
+- 给出一个链表，要求把链表中和为 0 的结点都移除。
+- 由于链表的特性，不能随机访问。所以从链表的头开始往后扫，把累加和存到字典中。当再次出现相同的累加和的时候，代表这中间的一段和是 0，于是要删除这一段。删除这一段的过程中，也要删除这一段在字典中存过的累加和。有一个特殊情况需要处理，即整个链表的总和是 0，那么最终结果是空链表。针对这个特殊情况，字典中先预存入 0 值。
diff --git a/leetcode/1175.Prime-Arrangements/1175. Prime Arrangements.go b/leetcode/1175.Prime-Arrangements/1175. Prime Arrangements.go
new file mode 100644
index 000000000..a8c176e61
--- /dev/null
+++ b/leetcode/1175.Prime-Arrangements/1175. Prime Arrangements.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+import "sort"
+
+var primes = []int{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
+
+func numPrimeArrangements(n int) int {
+	primeCount := sort.Search(25, func(i int) bool { return primes[i] > n })
+	return factorial(primeCount) * factorial(n-primeCount) % 1000000007
+}
+
+func factorial(n int) int {
+	if n == 1 || n == 0 {
+		return 1
+	}
+	return n * factorial(n-1) % 1000000007
+}
diff --git a/leetcode/1175.Prime-Arrangements/1175. Prime Arrangements_test.go b/leetcode/1175.Prime-Arrangements/1175. Prime Arrangements_test.go
new file mode 100644
index 000000000..0bd25db62
--- /dev/null
+++ b/leetcode/1175.Prime-Arrangements/1175. Prime Arrangements_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1175 struct {
+	para1175
+	ans1175
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1175 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1175 struct {
+	one int
+}
+
+func Test_Problem1175(t *testing.T) {
+
+	qs := []question1175{
+
+		{
+			para1175{5},
+			ans1175{12},
+		},
+
+		{
+			para1175{99},
+			ans1175{75763854},
+		},
+
+		{
+			para1175{100},
+			ans1175{682289015},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1175------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1175, q.para1175
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numPrimeArrangements(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1175.Prime-Arrangements/README.md b/leetcode/1175.Prime-Arrangements/README.md
new file mode 100755
index 000000000..9a06579b3
--- /dev/null
+++ b/leetcode/1175.Prime-Arrangements/README.md
@@ -0,0 +1,40 @@
+# [1175. Prime Arrangements](https://leetcode.com/problems/prime-arrangements/)
+
+
+## 题目
+
+Return the number of permutations of 1 to `n` so that prime numbers are at prime indices (1-indexed.)
+
+*(Recall that an integer is prime if and only if it is greater than 1, and cannot be written as a product of two positive integers both smaller than it.)*
+
+Since the answer may be large, return the answer **modulo `10^9 + 7`**.
+
+**Example 1:**
+
+    Input: n = 5
+    Output: 12
+    Explanation: For example [1,2,5,4,3] is a valid permutation, but [5,2,3,4,1] is not because the prime number 5 is at index 1.
+
+**Example 2:**
+
+    Input: n = 100
+    Output: 682289015
+
+**Constraints:**
+
+- `1 <= n <= 100`
+
+
+## 题目大意
+
+
+请你帮忙给从 1 到 n 的数设计排列方案，使得所有的「质数」都应该被放在「质数索引」（索引从 1 开始）上；你需要返回可能的方案总数。让我们一起来回顾一下「质数」：质数一定是大于 1 的，并且不能用两个小于它的正整数的乘积来表示。由于答案可能会很大，所以请你返回答案 模 mod 10^9 + 7 之后的结果即可。
+
+提示：
+
+- 1 <= n <= 100
+
+## 解题思路
+
+- 给出一个数 n，要求在 1-n 这 n 个数中，素数在素数索引下标位置上的全排列个数。
+- 由于这一题的 `n` 小于 100，所以可以用打表法。先把小于 100 个素数都打表打出来。然后对小于 n 的素数进行全排列，即 n！，然后再对剩下来的非素数进行全排列，即 (n-c)！。两个的乘积即为最终答案。
diff --git a/leetcode/1178.Number-of-Valid-Words-for-Each-Puzzle/1178. Number of Valid Words for Each Puzzle.go b/leetcode/1178.Number-of-Valid-Words-for-Each-Puzzle/1178. Number of Valid Words for Each Puzzle.go
new file mode 100644
index 000000000..8acbe9ebc
--- /dev/null
+++ b/leetcode/1178.Number-of-Valid-Words-for-Each-Puzzle/1178. Number of Valid Words for Each Puzzle.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+/*
+匹配跟单词中的字母顺序，字母个数都无关，可以用 bitmap 压缩
+ 1. 记录 word 中 利用 map 记录各种 bit 标示的个数
+ 2. puzzles 中各个字母都不相同! 记录 bitmap，然后搜索子空间中各种 bit 标识的个数的和
+    因为 puzzles 长度最长是7，所以搜索空间 2^7
+*/
+func findNumOfValidWords(words []string, puzzles []string) []int {
+	wordBitStatusMap, res := make(map[uint32]int, 0), []int{}
+	for _, w := range words {
+		wordBitStatusMap[toBitMap([]byte(w))]++
+	}
+	for _, p := range puzzles {
+		var bitMap uint32
+		var totalNum int
+		bitMap |= (1 << (p[0] - 'a')) //work 中要包含 p 的第一个字母 所以这个 bit 位上必须是 1
+		findNum([]byte(p)[1:], bitMap, &totalNum, wordBitStatusMap)
+		res = append(res, totalNum)
+	}
+	return res
+}
+
+func toBitMap(word []byte) uint32 {
+	var res uint32
+	for _, b := range word {
+		res |= (1 << (b - 'a'))
+	}
+	return res
+}
+
+// 利用 dfs 搜索 puzzles 的子空间
+func findNum(puzzles []byte, bitMap uint32, totalNum *int, m map[uint32]int) {
+	if len(puzzles) == 0 {
+		*totalNum = *totalNum + m[bitMap]
+		return
+	}
+	//不包含 puzzles[0],即 puzzles[0] 对应 bit 是 0
+	findNum(puzzles[1:], bitMap, totalNum, m)
+	//包含 puzzles[0],即 puzzles[0] 对应 bit 是 1
+	bitMap |= (1 << (puzzles[0] - 'a'))
+	findNum(puzzles[1:], bitMap, totalNum, m)
+	bitMap ^= (1 << (puzzles[0] - 'a')) //异或 清零
+	return
+}
diff --git a/leetcode/1178.Number-of-Valid-Words-for-Each-Puzzle/1178. Number of Valid Words for Each Puzzle_test.go b/leetcode/1178.Number-of-Valid-Words-for-Each-Puzzle/1178. Number of Valid Words for Each Puzzle_test.go
new file mode 100644
index 000000000..5d87ba685
--- /dev/null
+++ b/leetcode/1178.Number-of-Valid-Words-for-Each-Puzzle/1178. Number of Valid Words for Each Puzzle_test.go	
@@ -0,0 +1,32 @@
+package leetcode
+
+import (
+	"reflect"
+	"testing"
+)
+
+func Test_findNumOfValidWords(t *testing.T) {
+
+	words1 := []string{"aaaa", "asas", "able", "ability", "actt", "actor", "access"}
+	puzzles1 := []string{"aboveyz", "abrodyz", "abslute", "absoryz", "actresz", "gaswxyz"}
+
+	type args struct {
+		words   []string
+		puzzles []string
+	}
+	tests := []struct {
+		name string
+		args args
+		want []int
+	}{
+		// TODO: Add test cases.
+		{"1", args{words: words1, puzzles: puzzles1}, []int{1, 1, 3, 2, 4, 0}},
+	}
+	for _, tt := range tests {
+		t.Run(tt.name, func(t *testing.T) {
+			if got := findNumOfValidWords(tt.args.words, tt.args.puzzles); !reflect.DeepEqual(got, tt.want) {
+				t.Errorf("findNumOfValidWords() = %v, want %v", got, tt.want)
+			}
+		})
+	}
+}
diff --git a/leetcode/1178.Number-of-Valid-Words-for-Each-Puzzle/README.md b/leetcode/1178.Number-of-Valid-Words-for-Each-Puzzle/README.md
new file mode 100644
index 000000000..40d69ec65
--- /dev/null
+++ b/leetcode/1178.Number-of-Valid-Words-for-Each-Puzzle/README.md
@@ -0,0 +1,119 @@
+# [1178. Number of Valid Words for Each Puzzle](https://leetcode.com/problems/number-of-valid-words-for-each-puzzle/)
+
+
+## 题目
+
+With respect to a given `puzzle` string, a `word` is *valid* if both the following conditions are satisfied:
+
+- `word` contains the first letter of `puzzle`.
+- For each letter in `word`, that letter is in `puzzle`.For example, if the puzzle is "abcdefg", then valid words are "faced", "cabbage", and "baggage"; while invalid words are "beefed" (doesn't include "a") and "based" (includes "s" which isn't in the puzzle).
+
+Return an array `answer`, where `answer[i]` is the number of words in the given word list `words` that are valid with respect to the puzzle `puzzles[i]`.
+
+**Example :**
+
+```
+Input: 
+words = ["aaaa","asas","able","ability","actt","actor","access"], 
+puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
+Output: [1,1,3,2,4,0]
+Explanation:
+1 valid word for "aboveyz" : "aaaa" 
+1 valid word for "abrodyz" : "aaaa"
+3 valid words for "abslute" : "aaaa", "asas", "able"
+2 valid words for "absoryz" : "aaaa", "asas"
+4 valid words for "actresz" : "aaaa", "asas", "actt", "access"
+There're no valid words for "gaswxyz" cause none of the words in the list contains letter 'g'.
+
+```
+
+**Constraints:**
+
+- `1 <= words.length <= 10^5`
+- `4 <= words[i].length <= 50`
+- `1 <= puzzles.length <= 10^4`
+- `puzzles[i].length == 7`
+- `words[i][j]`, `puzzles[i][j]` are English lowercase letters.
+- Each `puzzles[i]` doesn't contain repeated characters.
+
+## 题目大意
+
+外国友人仿照中国字谜设计了一个英文版猜字谜小游戏，请你来猜猜看吧。
+
+字谜的迷面 puzzle 按字符串形式给出，如果一个单词 word 符合下面两个条件，那么它就可以算作谜底：
+
+- 单词 word 中包含谜面 puzzle 的第一个字母。
+- 单词 word 中的每一个字母都可以在谜面 puzzle 中找到。
+例如，如果字谜的谜面是 "abcdefg"，那么可以作为谜底的单词有 "faced", "cabbage", 和 "baggage"；而 "beefed"（不含字母 "a"）以及 "based"（其中的 "s" 没有出现在谜面中）都不能作为谜底。
+
+返回一个答案数组 answer，数组中的每个元素 answer[i] 是在给出的单词列表 words 中可以作为字谜迷面 puzzles[i] 所对应的谜底的单词数目。
+
+提示：
+
+- 1 <= words.length <= 10^5
+- 4 <= words[i].length <= 50
+- 1 <= puzzles.length <= 10^4
+- puzzles[i].length == 7
+- words[i][j], puzzles[i][j] 都是小写英文字母。
+- 每个 puzzles[i] 所包含的字符都不重复。
+
+## 解题思路
+
+- 首先题目中两个限制条件非常关键：**puzzles[i].length == 7**，**每个 puzzles[i] 所包含的字符都不重复**。也就是说穷举每个puzzle的子串的搜索空间就是2^7=128，而且不用考虑去重问题。
+- 因为谜底的判断只跟字符是否出现有关，跟字符的个数无关，另外都是小写的英文字母，所以可以用 `bitmap` 来表示单词(word)。
+- 利用 `map` 记录不同状态的单词(word)的个数。
+- 根据题意，如果某个单词(word)是某个字谜(puzzle)的谜底，那么 `word` 的 `bitmap` 肯定对应于 `puzzle` 某个子串的 `bitmap` 表示，且 `bitmap` 中包含 `puzzle` 的第一个字母的 `bit` 占用。
+- 问题就转换为：求每一个 `puzzle` 的每一个子串，然后求和这个子串具有相同 `bitmap` 表示且 `word` 中包含 `puzzle` 的第一个字母的 `word` 的个数。
+
+## 代码
+
+```go
+
+package leetcode
+
+/*
+	匹配跟单词中的字母顺序，字母个数都无关，可以用 bitmap 压缩
+	1. 记录 word 中 利用 map 记录各种 bit 标示的个数
+	2. puzzles 中各个字母都不相同! 记录 bitmap，然后搜索子空间中各种 bit 标识的个数的和
+	   因为 puzzles 长度最长是7，所以搜索空间 2^7
+*/
+func findNumOfValidWords(words []string, puzzles []string) []int {
+	wordBitStatusMap, res := make(map[uint32]int, 0), []int{}
+	for _, w := range words {
+		wordBitStatusMap[toBitMap([]byte(w))]++
+	}
+	for _, p := range puzzles {
+		var bitMap uint32
+		var totalNum int
+		bitMap |= (1 << (p[0] - 'a')) //work 中要包含 p 的第一个字母 所以这个 bit 位上必须是 1
+		findNum([]byte(p)[1:], bitMap, &totalNum, wordBitStatusMap)
+		res = append(res, totalNum)
+	}
+	return res
+}
+
+func toBitMap(word []byte) uint32 {
+	var res uint32
+	for _, b := range word {
+		res |= (1 << (b - 'a'))
+	}
+	return res
+}
+
+//利用 dfs 搜索 puzzles 的子空间
+func findNum(puzzles []byte, bitMap uint32, totalNum *int, m map[uint32]int) {
+	if len(puzzles) == 0 {
+		*totalNum = *totalNum + m[bitMap]
+		return
+	}
+	//不包含 puzzles[0],即 puzzles[0] 对应 bit 是 0
+	findNum(puzzles[1:], bitMap, totalNum, m)
+	//包含 puzzles[0],即 puzzles[0] 对应 bit 是 1
+	bitMap |= (1 << (puzzles[0] - 'a'))
+	findNum(puzzles[1:], bitMap, totalNum, m)
+	bitMap ^= (1 << (puzzles[0] - 'a')) //异或 清零
+	return
+}
+
+
+```
\ No newline at end of file
diff --git a/Algorithms/1184. Distance Between Bus Stops/1184. Distance Between Bus Stops.go b/leetcode/1184.Distance-Between-Bus-Stops/1184. Distance Between Bus Stops.go
similarity index 100%
rename from Algorithms/1184. Distance Between Bus Stops/1184. Distance Between Bus Stops.go
rename to leetcode/1184.Distance-Between-Bus-Stops/1184. Distance Between Bus Stops.go
diff --git a/Algorithms/1184. Distance Between Bus Stops/1184. Distance Between Bus Stops_test.go b/leetcode/1184.Distance-Between-Bus-Stops/1184. Distance Between Bus Stops_test.go
similarity index 94%
rename from Algorithms/1184. Distance Between Bus Stops/1184. Distance Between Bus Stops_test.go
rename to leetcode/1184.Distance-Between-Bus-Stops/1184. Distance Between Bus Stops_test.go
index 9707cd7b5..95391f945 100644
--- a/Algorithms/1184. Distance Between Bus Stops/1184. Distance Between Bus Stops_test.go	
+++ b/leetcode/1184.Distance-Between-Bus-Stops/1184. Distance Between Bus Stops_test.go	
@@ -28,17 +28,17 @@ func Test_Problem1184(t *testing.T) {
 
 	qs := []question1184{
 
-		question1184{
+		{
 			para1184{[]int{1, 2, 3, 4}, 0, 1},
 			ans1184{1},
 		},
 
-		question1184{
+		{
 			para1184{[]int{1, 2, 3, 4}, 0, 2},
 			ans1184{3},
 		},
 
-		question1184{
+		{
 			para1184{[]int{1, 2, 3, 4}, 0, 3},
 			ans1184{4},
 		},
diff --git a/leetcode/1184.Distance-Between-Bus-Stops/README.md b/leetcode/1184.Distance-Between-Bus-Stops/README.md
new file mode 100755
index 000000000..e9af503db
--- /dev/null
+++ b/leetcode/1184.Distance-Between-Bus-Stops/README.md
@@ -0,0 +1,59 @@
+# [1184. Distance Between Bus Stops](https://leetcode.com/problems/distance-between-bus-stops/)
+
+## 题目
+
+A bus has `n` stops numbered from `0` to `n - 1` that form a circle. We know the distance between all pairs of neighboring stops where `distance[i]` is the distance between the stops number `i` and `(i + 1) % n`.
+
+The bus goes along both directions i.e. clockwise and counterclockwise.
+
+Return the shortest distance between the given `start` and `destination` stops.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1.jpg)
+
+    Input: distance = [1,2,3,4], start = 0, destination = 1
+    Output: 1
+    Explanation: Distance between 0 and 1 is 1 or 9, minimum is 1.
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1-1.jpg)
+
+    Input: distance = [1,2,3,4], start = 0, destination = 2
+    Output: 3
+    Explanation: Distance between 0 and 2 is 3 or 7, minimum is 3.
+
+**Example 3:**
+
+![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1-2.jpg)
+
+    Input: distance = [1,2,3,4], start = 0, destination = 3
+    Output: 4
+    Explanation: Distance between 0 and 3 is 6 or 4, minimum is 4.
+
+**Constraints:**
+
+- `1 <= n <= 10^4`
+- `distance.length == n`
+- `0 <= start, destination < n`
+- `0 <= distance[i] <= 10^4`
+
+
+## 题目大意
+
+环形公交路线上有 n 个站，按次序从 0 到 n - 1 进行编号。我们已知每一对相邻公交站之间的距离，distance[i] 表示编号为 i 的车站和编号为 (i + 1) % n 的车站之间的距离。环线上的公交车都可以按顺时针和逆时针的方向行驶。返回乘客从出发点 start 到目的地 destination 之间的最短距离。
+
+提示：
+
+- 1 <= n <= 10^4
+- distance.length == n
+- 0 <= start, destination < n
+- 0 <= distance[i] <= 10^4
+
+
+## 解题思路
+
+
+- 给出一个数组，代表的是公交车站每站直接的距离。距离是按照数组下标的顺序给出的，公交车可以按照顺时针行驶，也可以按照逆时针行驶。问行驶的最短距离是多少。
+- 按照题意，分别算出顺时针和逆时针的行驶距离，比较两者距离，取出小值就是结果。
diff --git a/Algorithms/1185. Day of the Week/1185. Day of the Week.go b/leetcode/1185.Day-of-the-Week/1185. Day of the Week.go
similarity index 100%
rename from Algorithms/1185. Day of the Week/1185. Day of the Week.go
rename to leetcode/1185.Day-of-the-Week/1185. Day of the Week.go
diff --git a/Algorithms/1185. Day of the Week/1185. Day of the Week_test.go b/leetcode/1185.Day-of-the-Week/1185. Day of the Week_test.go
similarity index 94%
rename from Algorithms/1185. Day of the Week/1185. Day of the Week_test.go
rename to leetcode/1185.Day-of-the-Week/1185. Day of the Week_test.go
index 53c67bd6a..70728c1e4 100644
--- a/Algorithms/1185. Day of the Week/1185. Day of the Week_test.go	
+++ b/leetcode/1185.Day-of-the-Week/1185. Day of the Week_test.go	
@@ -28,17 +28,17 @@ func Test_Problem1185(t *testing.T) {
 
 	qs := []question1185{
 
-		question1185{
+		{
 			para1185{31, 8, 2019},
 			ans1185{"Saturday"},
 		},
 
-		question1185{
+		{
 			para1185{18, 7, 1999},
 			ans1185{"Sunday"},
 		},
 
-		question1185{
+		{
 			para1185{15, 8, 1993},
 			ans1185{"Sunday"},
 		},
diff --git a/leetcode/1185.Day-of-the-Week/README.md b/leetcode/1185.Day-of-the-Week/README.md
new file mode 100755
index 000000000..cebb127f9
--- /dev/null
+++ b/leetcode/1185.Day-of-the-Week/README.md
@@ -0,0 +1,46 @@
+# [1185. Day of the Week](https://leetcode.com/problems/day-of-the-week/)
+
+
+## 题目
+
+Given a date, return the corresponding day of the week for that date.
+
+The input is given as three integers representing the `day`, `month` and `year` respectively.
+
+Return the answer as one of the following values `{"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}`.
+
+**Example 1:**
+
+    Input: day = 31, month = 8, year = 2019
+    Output: "Saturday"
+
+**Example 2:**
+
+    Input: day = 18, month = 7, year = 1999
+    Output: "Sunday"
+
+**Example 3:**
+
+    Input: day = 15, month = 8, year = 1993
+    Output: "Sunday"
+
+**Constraints:**
+
+- The given dates are valid dates between the years `1971` and `2100`.
+
+
+## 题目大意
+
+给你一个日期，请你设计一个算法来判断它是对应一周中的哪一天。输入为三个整数：day、month 和 year，分别表示日、月、年。
+
+您返回的结果必须是这几个值中的一个 {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}。
+
+提示：
+
+- 给出的日期一定是在 1971 到 2100 年之间的有效日期。
+
+## 解题思路
+
+
+- 给出一个日期，要求算出这一天是星期几。
+- 简单题，按照常识计算即可。
diff --git a/Algorithms/1189. Maximum Number of Balloons/1189. Maximum Number of Balloons.go b/leetcode/1189.Maximum-Number-of-Balloons/1189. Maximum Number of Balloons.go
similarity index 89%
rename from Algorithms/1189. Maximum Number of Balloons/1189. Maximum Number of Balloons.go
rename to leetcode/1189.Maximum-Number-of-Balloons/1189. Maximum Number of Balloons.go
index a4bcbc406..36b4207b4 100644
--- a/Algorithms/1189. Maximum Number of Balloons/1189. Maximum Number of Balloons.go	
+++ b/leetcode/1189.Maximum-Number-of-Balloons/1189. Maximum Number of Balloons.go	
@@ -12,3 +12,10 @@ func maxNumberOfBalloons(text string) int {
 	// 字符 n 的频次是数组下标 13 对应的元素值
 	return min(fre[1], min(fre[0], min(fre[11]/2, min(fre[14]/2, fre[13]))))
 }
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/Algorithms/1189. Maximum Number of Balloons/1189. Maximum Number of Balloons_test.go b/leetcode/1189.Maximum-Number-of-Balloons/1189. Maximum Number of Balloons_test.go
similarity index 93%
rename from Algorithms/1189. Maximum Number of Balloons/1189. Maximum Number of Balloons_test.go
rename to leetcode/1189.Maximum-Number-of-Balloons/1189. Maximum Number of Balloons_test.go
index 959338fb9..1dcb110e3 100644
--- a/Algorithms/1189. Maximum Number of Balloons/1189. Maximum Number of Balloons_test.go	
+++ b/leetcode/1189.Maximum-Number-of-Balloons/1189. Maximum Number of Balloons_test.go	
@@ -26,17 +26,17 @@ func Test_Problem1189(t *testing.T) {
 
 	qs := []question1189{
 
-		question1189{
+		{
 			para1189{"nlaebolko"},
 			ans1189{1},
 		},
 
-		question1189{
+		{
 			para1189{"loonbalxballpoon"},
 			ans1189{2},
 		},
 
-		question1189{
+		{
 			para1189{"leetcode"},
 			ans1189{0},
 		},
diff --git a/leetcode/1189.Maximum-Number-of-Balloons/README.md b/leetcode/1189.Maximum-Number-of-Balloons/README.md
new file mode 100755
index 000000000..b2858af85
--- /dev/null
+++ b/leetcode/1189.Maximum-Number-of-Balloons/README.md
@@ -0,0 +1,48 @@
+# [1189. Maximum Number of Balloons](https://leetcode.com/problems/maximum-number-of-balloons/)
+
+
+## 题目
+
+Given a string `text`, you want to use the characters of `text` to form as many instances of the word **"balloon"** as possible.
+
+You can use each character in `text` **at most once**. Return the maximum number of instances that can be formed.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2019/09/05/1536_ex1_upd.JPG)
+
+    Input: text = "nlaebolko"
+    Output: 1
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2019/09/05/1536_ex2_upd.JPG)
+
+    Input: text = "loonbalxballpoon"
+    Output: 2
+
+**Example 3:**
+
+    Input: text = "leetcode"
+    Output: 0
+
+**Constraints:**
+
+- `1 <= text.length <= 10^4`
+- `text` consists of lower case English letters only.
+
+
+## 题目大意
+
+给你一个字符串 text，你需要使用 text 中的字母来拼凑尽可能多的单词 "balloon"（气球）。字符串 text 中的每个字母最多只能被使用一次。请你返回最多可以拼凑出多少个单词 "balloon"。
+
+提示：
+
+- 1 <= text.length <= 10^4
+- text 全部由小写英文字母组成
+
+## 解题思路
+
+
+- 给出一个字符串，问这个字符串里面的数组能组成多少个 **balloon** 这个单词。
+- 简单题，先统计 26 个字母每个字母的频次，然后取出 balloon 这 5 个字母出现频次最小的值就是结果。
diff --git a/leetcode/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/1190. Reverse Substrings Between Each Pair of Parentheses.go b/leetcode/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/1190. Reverse Substrings Between Each Pair of Parentheses.go
new file mode 100644
index 000000000..fe76ab529
--- /dev/null
+++ b/leetcode/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/1190. Reverse Substrings Between Each Pair of Parentheses.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+func reverseParentheses(s string) string {
+	pair, stack := make([]int, len(s)), []int{}
+	for i, b := range s {
+		if b == '(' {
+			stack = append(stack, i)
+		} else if b == ')' {
+			j := stack[len(stack)-1]
+			stack = stack[:len(stack)-1]
+			pair[i], pair[j] = j, i
+		}
+	}
+	res := []byte{}
+	for i, step := 0, 1; i < len(s); i += step {
+		if s[i] == '(' || s[i] == ')' {
+			i = pair[i]
+			step = -step
+		} else {
+			res = append(res, s[i])
+		}
+	}
+	return string(res)
+}
diff --git a/leetcode/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/1190. Reverse Substrings Between Each Pair of Parentheses_test.go b/leetcode/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/1190. Reverse Substrings Between Each Pair of Parentheses_test.go
new file mode 100644
index 000000000..cf5c51b1c
--- /dev/null
+++ b/leetcode/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/1190. Reverse Substrings Between Each Pair of Parentheses_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1190 struct {
+	para1190
+	ans1190
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1190 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1190 struct {
+	one string
+}
+
+func Test_Problem1190(t *testing.T) {
+
+	qs := []question1190{
+
+		{
+			para1190{"(abcd)"},
+			ans1190{"dcba"},
+		},
+
+		{
+			para1190{"(u(love)i)"},
+			ans1190{"iloveu"},
+		},
+
+		{
+			para1190{"(ed(et(oc))el)"},
+			ans1190{"leetcode"},
+		},
+
+		{
+			para1190{"a(bcdefghijkl(mno)p)q"},
+			ans1190{"apmnolkjihgfedcbq"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1190------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1190, q.para1190
+		fmt.Printf("【input】:%v       【output】:%v\n", p, reverseParentheses(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/README.md b/leetcode/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/README.md
new file mode 100644
index 000000000..22bc27edd
--- /dev/null
+++ b/leetcode/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/README.md
@@ -0,0 +1,84 @@
+# [1190. Reverse Substrings Between Each Pair of Parentheses](https://leetcode.com/problems/reverse-substrings-between-each-pair-of-parentheses/)
+
+
+## 题目
+
+You are given a string `s` that consists of lower case English letters and brackets.
+
+Reverse the strings in each pair of matching parentheses, starting from the innermost one.
+
+Your result should **not** contain any brackets.
+
+**Example 1:**
+
+```
+Input: s = "(abcd)"
+Output: "dcba"
+```
+
+**Example 2:**
+
+```
+Input: s = "(u(love)i)"
+Output: "iloveu"
+Explanation: The substring "love" is reversed first, then the whole string is reversed.
+```
+
+**Example 3:**
+
+```
+Input: s = "(ed(et(oc))el)"
+Output: "leetcode"
+Explanation: First, we reverse the substring "oc", then "etco", and finally, the whole string.
+```
+
+**Example 4:**
+
+```
+Input: s = "a(bcdefghijkl(mno)p)q"
+Output: "apmnolkjihgfedcbq"
+```
+
+**Constraints:**
+
+- `0 <= s.length <= 2000`
+- `s` only contains lower case English characters and parentheses.
+- It's guaranteed that all parentheses are balanced.
+
+## 题目大意
+
+给出一个字符串 s（仅含有小写英文字母和括号）。请你按照从括号内到外的顺序，逐层反转每对匹配括号中的字符串，并返回最终的结果。注意，您的结果中 不应 包含任何括号。
+
+## 解题思路
+
+- 本题最容易想到的思路是利用栈将每对括号里面的字符串入栈，当遇到 ")" 括号时出栈并逆序。由于用到了栈的数据结构，多层括号嵌套的问题也不用担心。这种边入栈出栈，逆序字符串的方法，时间复杂度是 O(n^2)，有没有可能进一步降低时间复杂度呢？
+- 上述解法中，存在重复遍历的情况。扫描原字符串的时候，入栈出栈已经扫描了一次，在 ")" 括号出栈时，逆序又会扫一遍已经入栈的字符串。这部分重复遍历的过程可以优化掉。第一次循环先标记出逆序区间。例如遇到 "(" 的时候，入栈并记录下它的下标，当遇到 ")" 的时候，意味着这一对括号匹配上了，所以将 ")" 的下标和之前入栈 "(" 的下标交换。此次遍历将逆序区间标记出来了。再遍历一次，根据逆序区间逆序字符串。不在逆序区间的字符串正常 append。如果在逆序区间内的，逆序遍历，添加到最终结果字符串中。这样做，时间复杂度仅为 O(n)。具体实现见下面代码。
+
+## 代码
+
+```go
+package leetcode
+
+func reverseParentheses(s string) string {
+	pair, stack := make([]int, len(s)), []int{}
+	for i, b := range s {
+		if b == '(' {
+			stack = append(stack, i)
+		} else if b == ')' {
+			j := stack[len(stack)-1]
+			stack = stack[:len(stack)-1]
+			pair[i], pair[j] = j, i
+		}
+	}
+	res := []byte{}
+	for i, step := 0, 1; i < len(s); i += step {
+		if s[i] == '(' || s[i] == ')' {
+			i = pair[i]
+			step = -step
+		} else {
+			res = append(res, s[i])
+		}
+	}
+	return string(res)
+}
+```
\ No newline at end of file
diff --git a/Algorithms/1200. Minimum Absolute Difference/1200. Minimum Absolute Difference.go b/leetcode/1200.Minimum-Absolute-Difference/1200. Minimum Absolute Difference.go
similarity index 100%
rename from Algorithms/1200. Minimum Absolute Difference/1200. Minimum Absolute Difference.go
rename to leetcode/1200.Minimum-Absolute-Difference/1200. Minimum Absolute Difference.go
diff --git a/Algorithms/1200. Minimum Absolute Difference/1200. Minimum Absolute Difference_test.go b/leetcode/1200.Minimum-Absolute-Difference/1200. Minimum Absolute Difference_test.go
similarity index 77%
rename from Algorithms/1200. Minimum Absolute Difference/1200. Minimum Absolute Difference_test.go
rename to leetcode/1200.Minimum-Absolute-Difference/1200. Minimum Absolute Difference_test.go
index 140377cdd..6b24d92bc 100644
--- a/Algorithms/1200. Minimum Absolute Difference/1200. Minimum Absolute Difference_test.go	
+++ b/leetcode/1200.Minimum-Absolute-Difference/1200. Minimum Absolute Difference_test.go	
@@ -26,19 +26,19 @@ func Test_Problem1200(t *testing.T) {
 
 	qs := []question1200{
 
-		question1200{
+		{
 			para1200{[]int{4, 2, 1, 3}},
-			ans1200{[][]int{[]int{1, 2}, []int{2, 3}, []int{3, 4}}},
+			ans1200{[][]int{{1, 2}, {2, 3}, {3, 4}}},
 		},
 
-		question1200{
+		{
 			para1200{[]int{1, 3, 6, 10, 15}},
-			ans1200{[][]int{[]int{1, 3}}},
+			ans1200{[][]int{{1, 3}}},
 		},
 
-		question1200{
+		{
 			para1200{[]int{3, 8, -10, 23, 19, -4, -14, 27}},
-			ans1200{[][]int{[]int{-14, -10}, []int{19, 23}, []int{23, 27}}},
+			ans1200{[][]int{{-14, -10}, {19, 23}, {23, 27}}},
 		},
 	}
 
diff --git a/leetcode/1200.Minimum-Absolute-Difference/README.md b/leetcode/1200.Minimum-Absolute-Difference/README.md
new file mode 100755
index 000000000..8634c5470
--- /dev/null
+++ b/leetcode/1200.Minimum-Absolute-Difference/README.md
@@ -0,0 +1,44 @@
+# [1200. Minimum Absolute Difference](https://leetcode.com/problems/minimum-absolute-difference/)
+
+
+## 题目
+
+Given an array of **distinct** integers `arr`, find all pairs of elements with the minimum absolute difference of any two elements.
+
+Return a list of pairs in ascending order(with respect to pairs), each pair `[a, b]` follows
+
+- `a, b` are from `arr`
+- `a < b`
+- `b - a` equals to the minimum absolute difference of any two elements in `arr`
+
+**Example 1:**
+
+    Input: arr = [4,2,1,3]
+    Output: [[1,2],[2,3],[3,4]]
+    Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order.
+
+**Example 2:**
+
+    Input: arr = [1,3,6,10,15]
+    Output: [[1,3]]
+
+**Example 3:**
+
+    Input: arr = [3,8,-10,23,19,-4,-14,27]
+    Output: [[-14,-10],[19,23],[23,27]]
+
+**Constraints:**
+
+- `2 <= arr.length <= 10^5`
+- `-10^6 <= arr[i] <= 10^6`
+
+
+## 题目大意
+
+给出一个数组，要求找出所有满足条件的数值对 [a,b]：`a<b` 并且 `b-a` 的差值是数组中所有两个元素差值的最小值。
+
+## 解题思路
+
+
+- 给出一个数组，要求找出所有满足条件的数值对 [a,b]：`a<b` 并且 `b-a` 的差值是数组中所有两个元素差值的最小值。
+- 简单题，按照题意先排序，然后依次求出两个相邻元素的差值，求出最小的差值。最后遍历一遍数组，把所有等于最小差值的数值对都输出。
diff --git a/Algorithms/1201. Ugly Number III/1201. Ugly Number III.go b/leetcode/1201.Ugly-Number-III/1201. Ugly Number III.go
similarity index 100%
rename from Algorithms/1201. Ugly Number III/1201. Ugly Number III.go
rename to leetcode/1201.Ugly-Number-III/1201. Ugly Number III.go
diff --git a/Algorithms/1201. Ugly Number III/1201. Ugly Number III_test.go b/leetcode/1201.Ugly-Number-III/1201. Ugly Number III_test.go
similarity index 92%
rename from Algorithms/1201. Ugly Number III/1201. Ugly Number III_test.go
rename to leetcode/1201.Ugly-Number-III/1201. Ugly Number III_test.go
index 45ec22293..5d5b501ae 100644
--- a/Algorithms/1201. Ugly Number III/1201. Ugly Number III_test.go	
+++ b/leetcode/1201.Ugly-Number-III/1201. Ugly Number III_test.go	
@@ -29,22 +29,22 @@ func Test_Problem1201(t *testing.T) {
 
 	qs := []question1201{
 
-		question1201{
+		{
 			para1201{3, 2, 3, 5},
 			ans1201{4},
 		},
 
-		question1201{
+		{
 			para1201{4, 2, 3, 4},
 			ans1201{6},
 		},
 
-		question1201{
+		{
 			para1201{5, 2, 11, 13},
 			ans1201{10},
 		},
 
-		question1201{
+		{
 			para1201{1000000000, 2, 217983653, 336916467},
 			ans1201{1999999984},
 		},
diff --git a/leetcode/1201.Ugly-Number-III/README.md b/leetcode/1201.Ugly-Number-III/README.md
new file mode 100755
index 000000000..4b7451d19
--- /dev/null
+++ b/leetcode/1201.Ugly-Number-III/README.md
@@ -0,0 +1,57 @@
+# [1201. Ugly Number III](https://leetcode.com/problems/ugly-number-iii/)
+
+
+## 题目
+
+Write a program to find the `n`-th ugly number.
+
+Ugly numbers are **positive integers** which are divisible by `a` **or** `b` **or** `c`.
+
+**Example 1:**
+
+    Input: n = 3, a = 2, b = 3, c = 5
+    Output: 4
+    Explanation: The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3rd is 4.
+
+**Example 2:**
+
+    Input: n = 4, a = 2, b = 3, c = 4
+    Output: 6
+    Explanation: The ugly numbers are 2, 3, 4, 6, 8, 9, 10, 12... The 4th is 6.
+
+**Example 3:**
+
+    Input: n = 5, a = 2, b = 11, c = 13
+    Output: 10
+    Explanation: The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5th is 10.
+
+**Example 4:**
+
+    Input: n = 1000000000, a = 2, b = 217983653, c = 336916467
+    Output: 1999999984
+
+**Constraints:**
+
+- `1 <= n, a, b, c <= 10^9`
+- `1 <= a * b * c <= 10^18`
+- It's guaranteed that the result will be in range `[1, 2 * 10^9]`
+
+
+## 题目大意
+
+
+请你帮忙设计一个程序，用来找出第 n 个丑数。丑数是可以被 a 或 b 或 c 整除的 正整数。
+
+
+提示：
+
+- 1 <= n, a, b, c <= 10^9
+- 1 <= a * b * c <= 10^18
+- 本题结果在 [1, 2 * 10^9] 的范围内
+
+## 解题思路
+
+
+- 给出 4 个数字，a，b，c，n。要求输出可以整除 a 或者整除 b 或者整除 c 的第 n 个数。
+- 这一题限定了解的范围， `[1, 2 * 10^9]`，所以直接二分搜索来求解。逐步二分逼近 low 值，直到找到能满足条件的 low 的最小值，即为最终答案。
+- 这一题的关键在如何判断一个数是第几个数。一个数能整除 a，能整除 b，能整除 c，那么它应该是第 `num/a + num/b + num/c - num/ab - num/bc - num/ac + num/abc` 个数。这个就是韦恩图。需要注意的是，求 `ab`、`bc`、`ac`、`abc` 的时候需要再除以各自的最大公约数 `gcd()`。
diff --git a/leetcode/1202.Smallest-String-With-Swaps/1202. Smallest String With Swaps.go b/leetcode/1202.Smallest-String-With-Swaps/1202. Smallest String With Swaps.go
new file mode 100644
index 000000000..60ce722bf
--- /dev/null
+++ b/leetcode/1202.Smallest-String-With-Swaps/1202. Smallest String With Swaps.go	
@@ -0,0 +1,31 @@
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+func smallestStringWithSwaps(s string, pairs [][]int) string {
+	uf, res, sMap := template.UnionFind{}, []byte(s), map[int][]byte{}
+	uf.Init(len(s))
+	for _, pair := range pairs {
+		uf.Union(pair[0], pair[1])
+	}
+	for i := 0; i < len(s); i++ {
+		r := uf.Find(i)
+		sMap[r] = append(sMap[r], s[i])
+	}
+	for _, v := range sMap {
+		sort.Slice(v, func(i, j int) bool {
+			return v[i] < v[j]
+		})
+	}
+	for i := 0; i < len(s); i++ {
+		r := uf.Find(i)
+		bytes := sMap[r]
+		res[i] = bytes[0]
+		sMap[r] = bytes[1:]
+	}
+	return string(res)
+}
diff --git a/leetcode/1202.Smallest-String-With-Swaps/1202. Smallest String With Swaps_test.go b/leetcode/1202.Smallest-String-With-Swaps/1202. Smallest String With Swaps_test.go
new file mode 100644
index 000000000..8e123e77f
--- /dev/null
+++ b/leetcode/1202.Smallest-String-With-Swaps/1202. Smallest String With Swaps_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1202 struct {
+	para1202
+	ans1202
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1202 struct {
+	s     string
+	pairs [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1202 struct {
+	one string
+}
+
+func Test_Problem1202(t *testing.T) {
+
+	qs := []question1202{
+
+		{
+			para1202{"dcab", [][]int{{0, 3}, {1, 2}}},
+			ans1202{"bacd"},
+		},
+
+		{
+			para1202{"dcab", [][]int{{0, 3}, {1, 2}, {0, 2}}},
+			ans1202{"abcd"},
+		},
+
+		{
+			para1202{"cba", [][]int{{0, 1}, {1, 2}}},
+			ans1202{"abc"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1202------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1202, q.para1202
+		fmt.Printf("【input】:%v       【output】:%v\n", p, smallestStringWithSwaps(p.s, p.pairs))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1202.Smallest-String-With-Swaps/README.md b/leetcode/1202.Smallest-String-With-Swaps/README.md
new file mode 100755
index 000000000..dee01cffe
--- /dev/null
+++ b/leetcode/1202.Smallest-String-With-Swaps/README.md
@@ -0,0 +1,63 @@
+# [1202. Smallest String With Swaps](https://leetcode.com/problems/smallest-string-with-swaps/)
+
+
+## 题目
+
+You are given a string `s`, and an array of pairs of indices in the string `pairs` where `pairs[i] = [a, b]` indicates 2 indices(0-indexed) of the string.
+
+You can swap the characters at any pair of indices in the given `pairs` **any number of times**.
+
+Return the lexicographically smallest string that `s` can be changed to after using the swaps.
+
+**Example 1:**
+
+    Input: s = "dcab", pairs = [[0,3],[1,2]]
+    Output: "bacd"
+    Explaination: 
+    Swap s[0] and s[3], s = "bcad"
+    Swap s[1] and s[2], s = "bacd"
+
+**Example 2:**
+
+    Input: s = "dcab", pairs = [[0,3],[1,2],[0,2]]
+    Output: "abcd"
+    Explaination: 
+    Swap s[0] and s[3], s = "bcad"
+    Swap s[0] and s[2], s = "acbd"
+    Swap s[1] and s[2], s = "abcd"
+
+**Example 3:**
+
+    Input: s = "cba", pairs = [[0,1],[1,2]]
+    Output: "abc"
+    Explaination: 
+    Swap s[0] and s[1], s = "bca"
+    Swap s[1] and s[2], s = "bac"
+    Swap s[0] and s[1], s = "abc"
+
+**Constraints:**
+
+- `1 <= s.length <= 10^5`
+- `0 <= pairs.length <= 10^5`
+- `0 <= pairs[i][0], pairs[i][1] < s.length`
+- `s` only contains lower case English letters.
+
+
+## 题目大意
+
+给你一个字符串 s，以及该字符串中的一些「索引对」数组 pairs，其中 pairs[i] = [a, b] 表示字符串中的两个索引（编号从 0 开始）。你可以 任意多次交换 在 pairs 中任意一对索引处的字符。返回在经过若干次交换后，s 可以变成的按字典序最小的字符串。
+
+提示：
+
+- 1 <= s.length <= 10^5
+- 0 <= pairs.length <= 10^5
+- 0 <= pairs[i][0], pairs[i][1] < s.length
+- s 中只含有小写英文字母
+
+
+
+## 解题思路
+
+
+- 给出一个字符串和一个字符串里可交换的下标。要求交换以后字典序最小的字符串。
+- 这一题可以用并查集来解题，先把可交换下标都 `Union()` 起来，每个集合内，按照字典序从小到大排列。最后扫描原有字符串，从左到右依次找到各自对应的集合里面最小的字符进行替换，每次替换完以后，删除集合中该字符(防止下次重复替换)。最终得到的字符就是最小字典序的字符。
diff --git a/leetcode/1203.Sort-Items-by-Groups-Respecting-Dependencies/1203. Sort Items by Groups Respecting Dependencies.go b/leetcode/1203.Sort-Items-by-Groups-Respecting-Dependencies/1203. Sort Items by Groups Respecting Dependencies.go
new file mode 100644
index 000000000..07a715b86
--- /dev/null
+++ b/leetcode/1203.Sort-Items-by-Groups-Respecting-Dependencies/1203. Sort Items by Groups Respecting Dependencies.go	
@@ -0,0 +1,123 @@
+package leetcode
+
+// 解法一 拓扑排序版的 DFS
+func sortItems(n int, m int, group []int, beforeItems [][]int) []int {
+	groups, inDegrees := make([][]int, n+m), make([]int, n+m)
+	for i, g := range group {
+		if g > -1 {
+			g += n
+			groups[g] = append(groups[g], i)
+			inDegrees[i]++
+		}
+	}
+	for i, ancestors := range beforeItems {
+		gi := group[i]
+		if gi == -1 {
+			gi = i
+		} else {
+			gi += n
+		}
+		for _, ancestor := range ancestors {
+			ga := group[ancestor]
+			if ga == -1 {
+				ga = ancestor
+			} else {
+				ga += n
+			}
+			if gi == ga {
+				groups[ancestor] = append(groups[ancestor], i)
+				inDegrees[i]++
+			} else {
+				groups[ga] = append(groups[ga], gi)
+				inDegrees[gi]++
+			}
+		}
+	}
+	res := []int{}
+	for i, d := range inDegrees {
+		if d == 0 {
+			sortItemsDFS(i, n, &res, &inDegrees, &groups)
+		}
+	}
+	if len(res) != n {
+		return nil
+	}
+	return res
+}
+
+func sortItemsDFS(i, n int, res, inDegrees *[]int, groups *[][]int) {
+	if i < n {
+		*res = append(*res, i)
+	}
+	(*inDegrees)[i] = -1
+	for _, ch := range (*groups)[i] {
+		if (*inDegrees)[ch]--; (*inDegrees)[ch] == 0 {
+			sortItemsDFS(ch, n, res, inDegrees, groups)
+		}
+	}
+}
+
+// 解法二 二维拓扑排序 时间复杂度 O(m+n)，空间复杂度 O(m+n)
+func sortItems1(n int, m int, group []int, beforeItems [][]int) []int {
+	groupItems, res := map[int][]int{}, []int{}
+	for i := 0; i < len(group); i++ {
+		if group[i] == -1 {
+			group[i] = m + i
+		}
+		groupItems[group[i]] = append(groupItems[group[i]], i)
+	}
+	groupGraph, groupDegree, itemGraph, itemDegree := make([][]int, m+n), make([]int, m+n), make([][]int, n), make([]int, n)
+	for i := 0; i < len(beforeItems); i++ {
+		for j := 0; j < len(beforeItems[i]); j++ {
+			if group[beforeItems[i][j]] != group[i] {
+				// 不同组项目，确定组间依赖关系
+				groupGraph[group[beforeItems[i][j]]] = append(groupGraph[group[beforeItems[i][j]]], group[i])
+				groupDegree[group[i]]++
+			} else {
+				// 同组项目，确定组内依赖关系
+				itemGraph[beforeItems[i][j]] = append(itemGraph[beforeItems[i][j]], i)
+				itemDegree[i]++
+			}
+		}
+	}
+	items := []int{}
+	for i := 0; i < m+n; i++ {
+		items = append(items, i)
+	}
+	// 组间拓扑
+	groupOrders := topSort(groupGraph, groupDegree, items)
+	if len(groupOrders) < len(items) {
+		return nil
+	}
+	for i := 0; i < len(groupOrders); i++ {
+		items := groupItems[groupOrders[i]]
+		// 组内拓扑
+		orders := topSort(itemGraph, itemDegree, items)
+		if len(orders) < len(items) {
+			return nil
+		}
+		res = append(res, orders...)
+	}
+	return res
+}
+
+func topSort(graph [][]int, deg, items []int) (orders []int) {
+	q := []int{}
+	for _, i := range items {
+		if deg[i] == 0 {
+			q = append(q, i)
+		}
+	}
+	for len(q) > 0 {
+		from := q[0]
+		q = q[1:]
+		orders = append(orders, from)
+		for _, to := range graph[from] {
+			deg[to]--
+			if deg[to] == 0 {
+				q = append(q, to)
+			}
+		}
+	}
+	return
+}
diff --git a/leetcode/1203.Sort-Items-by-Groups-Respecting-Dependencies/1203. Sort Items by Groups Respecting Dependencies_test.go b/leetcode/1203.Sort-Items-by-Groups-Respecting-Dependencies/1203. Sort Items by Groups Respecting Dependencies_test.go
new file mode 100644
index 000000000..6abb3b54c
--- /dev/null
+++ b/leetcode/1203.Sort-Items-by-Groups-Respecting-Dependencies/1203. Sort Items by Groups Respecting Dependencies_test.go	
@@ -0,0 +1,50 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1203 struct {
+	para1203
+	ans1203
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1203 struct {
+	n           int
+	m           int
+	group       []int
+	beforeItems [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1203 struct {
+	one []int
+}
+
+func Test_Problem1203(t *testing.T) {
+
+	qs := []question1203{
+
+		{
+			para1203{8, 2, []int{-1, -1, 1, 0, 0, 1, 0, -1}, [][]int{{}, {6}, {5}, {6}, {3, 6}, {}, {}, {}}},
+			ans1203{[]int{6, 3, 4, 5, 2, 0, 7, 1}},
+		},
+
+		{
+			para1203{8, 2, []int{-1, -1, 1, 0, 0, 1, 0, -1}, [][]int{{}, {6}, {5}, {6}, {3}, {}, {4}, {}}},
+			ans1203{[]int{}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1203------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1203, q.para1203
+		fmt.Printf("【input】:%v       【output】:%v\n", p, sortItems1(p.n, p.m, p.group, p.beforeItems))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1203.Sort-Items-by-Groups-Respecting-Dependencies/README.md b/leetcode/1203.Sort-Items-by-Groups-Respecting-Dependencies/README.md
new file mode 100644
index 000000000..3bbddd165
--- /dev/null
+++ b/leetcode/1203.Sort-Items-by-Groups-Respecting-Dependencies/README.md
@@ -0,0 +1,191 @@
+# [1203. Sort Items by Groups Respecting Dependencies](https://leetcode.com/problems/sort-items-by-groups-respecting-dependencies/)
+
+
+## 题目
+
+There are `n` items each belonging to zero or one of `m` groups where `group[i]` is the group that the `i`-th item belongs to and it's equal to `-1` if the `i`-th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it.
+
+Return a sorted list of the items such that:
+
+- The items that belong to the same group are next to each other in the sorted list.
+- There are some relations between these items where `beforeItems[i]` is a list containing all the items that should come before the `i`th item in the sorted array (to the left of the `i`th item).
+
+Return any solution if there is more than one solution and return an **empty list** if there is no solution.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2019/09/11/1359_ex1.png)
+
+```
+Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]]
+Output: [6,3,4,1,5,2,0,7]
+
+```
+
+**Example 2:**
+
+```
+Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]]
+Output: []
+Explanation: This is the same as example 1 except that 4 needs to be before 6 in the sorted list.
+
+```
+
+**Constraints:**
+
+- `1 <= m <= n <= 3 * 104`
+- `group.length == beforeItems.length == n`
+- `1 <= group[i] <= m - 1`
+- `0 <= beforeItems[i].length <= n - 1`
+- `0 <= beforeItems[i][j] <= n - 1`
+- `i != beforeItems[i][j]`
+- `beforeItems[i]` does not contain duplicates elements.
+
+## 题目大意
+
+有 n 个项目，每个项目或者不属于任何小组，或者属于 m 个小组之一。group[i] 表示第 i 个小组所属的小组，如果第 i 个项目不属于任何小组，则 group[i] 等于 -1。项目和小组都是从零开始编号的。可能存在小组不负责任何项目，即没有任何项目属于这个小组。
+
+请你帮忙按要求安排这些项目的进度，并返回排序后的项目列表：
+
+- 同一小组的项目，排序后在列表中彼此相邻。
+- 项目之间存在一定的依赖关系，我们用一个列表 beforeItems 来表示，其中 beforeItems[i] 表示在进行第 i 个项目前（位于第 i 个项目左侧）应该完成的所有项目。
+
+如果存在多个解决方案，只需要返回其中任意一个即可。如果没有合适的解决方案，就请返回一个 空列表 。
+
+## 解题思路
+
+- 读完题能确定这一题是拓扑排序。但是和单纯的拓扑排序有区别的是，同一小组内的项目需要彼此相邻。用 2 次拓扑排序即可解决。第一次拓扑排序排出组间的顺序，第二次拓扑排序排出组内的顺序。为了实现方便，用 map 给虚拟分组标记编号。如下图，将 3，4，6 三个任务打包到 0 号分组里面，将 2，5 两个任务打包到 1 号分组里面，其他任务单独各自为一组。组间的依赖是 6 号任务依赖 1 号任务。由于 6 号任务封装在 0 号分组里，所以 3 号分组依赖 0 号分组。先组间排序，确定分组顺序，再组内拓扑排序，排出最终顺序。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_1203_1.png)
+
+- 上面的解法可以 AC，但是时间太慢了。因为做了一些不必要的操作。有没有可能只用一次拓扑排序呢？将必须要在一起的结点统一依赖一个虚拟结点，例如下图中的虚拟结点 8 和 9 。3，4，6 都依赖 8 号任务，2 和 5 都依赖 9 号任务。1 号任务本来依赖 6 号任务，由于 6 由依赖 8 ，所以添加 1 依赖 8 的边。通过增加虚拟结点，增加了需要打包在一起结点的入度。构建出以上关系以后，按照入度为 0 的原则，依次进行 DFS。8 号和 9 号两个虚拟结点的入度都为 0 ，对它们进行 DFS，必定会使得与它关联的节点都被安排在一起，这样就满足了题意：同一小组的项目，排序后在列表中彼此相邻。一遍扫完，满足题意的顺序就排出来了。这个解法 beat 100%！
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_1203_2.png)
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 拓扑排序版的 DFS
+func sortItems(n int, m int, group []int, beforeItems [][]int) []int {
+	groups, inDegrees := make([][]int, n+m), make([]int, n+m)
+	for i, g := range group {
+		if g > -1 {
+			g += n
+			groups[g] = append(groups[g], i)
+			inDegrees[i]++
+		}
+	}
+	for i, ancestors := range beforeItems {
+		gi := group[i]
+		if gi == -1 {
+			gi = i
+		} else {
+			gi += n
+		}
+		for _, ancestor := range ancestors {
+			ga := group[ancestor]
+			if ga == -1 {
+				ga = ancestor
+			} else {
+				ga += n
+			}
+			if gi == ga {
+				groups[ancestor] = append(groups[ancestor], i)
+				inDegrees[i]++
+			} else {
+				groups[ga] = append(groups[ga], gi)
+				inDegrees[gi]++
+			}
+		}
+	}
+	res := []int{}
+	for i, d := range inDegrees {
+		if d == 0 {
+			sortItemsDFS(i, n, &res, &inDegrees, &groups)
+		}
+	}
+	if len(res) != n {
+		return nil
+	}
+	return res
+}
+
+func sortItemsDFS(i, n int, res, inDegrees *[]int, groups *[][]int) {
+	if i < n {
+		*res = append(*res, i)
+	}
+	(*inDegrees)[i] = -1
+	for _, ch := range (*groups)[i] {
+		if (*inDegrees)[ch]--; (*inDegrees)[ch] == 0 {
+			sortItemsDFS(ch, n, res, inDegrees, groups)
+		}
+	}
+}
+
+// 解法二 二维拓扑排序 时间复杂度 O(m+n)，空间复杂度 O(m+n)
+func sortItems1(n int, m int, group []int, beforeItems [][]int) []int {
+	groupItems, res := map[int][]int{}, []int{}
+	for i := 0; i < len(group); i++ {
+		if group[i] == -1 {
+			group[i] = m + i
+		}
+		groupItems[group[i]] = append(groupItems[group[i]], i)
+	}
+	groupGraph, groupDegree, itemGraph, itemDegree := make([][]int, m+n), make([]int, m+n), make([][]int, n), make([]int, n)
+	for i := 0; i < len(beforeItems); i++ {
+		for j := 0; j < len(beforeItems[i]); j++ {
+			if group[beforeItems[i][j]] != group[i] {
+				// 不同组项目，确定组间依赖关系
+				groupGraph[group[beforeItems[i][j]]] = append(groupGraph[group[beforeItems[i][j]]], group[i])
+				groupDegree[group[i]]++
+			} else {
+				// 同组项目，确定组内依赖关系
+				itemGraph[beforeItems[i][j]] = append(itemGraph[beforeItems[i][j]], i)
+				itemDegree[i]++
+			}
+		}
+	}
+	items := []int{}
+	for i := 0; i < m+n; i++ {
+		items = append(items, i)
+	}
+	// 组间拓扑
+	groupOrders := topSort(groupGraph, groupDegree, items)
+	if len(groupOrders) < len(items) {
+		return nil
+	}
+	for i := 0; i < len(groupOrders); i++ {
+		items := groupItems[groupOrders[i]]
+		// 组内拓扑
+		orders := topSort(itemGraph, itemDegree, items)
+		if len(orders) < len(items) {
+			return nil
+		}
+		res = append(res, orders...)
+	}
+	return res
+}
+
+func topSort(graph [][]int, deg, items []int) (orders []int) {
+	q := []int{}
+	for _, i := range items {
+		if deg[i] == 0 {
+			q = append(q, i)
+		}
+	}
+	for len(q) > 0 {
+		from := q[0]
+		q = q[1:]
+		orders = append(orders, from)
+		for _, to := range graph[from] {
+			deg[to]--
+			if deg[to] == 0 {
+				q = append(q, to)
+			}
+		}
+	}
+	return
+}
+```
\ No newline at end of file
diff --git a/Algorithms/1207. Unique Number of Occurrences/1207. Unique Number of Occurrences.go b/leetcode/1207.Unique-Number-of-Occurrences/1207. Unique Number of Occurrences.go
similarity index 100%
rename from Algorithms/1207. Unique Number of Occurrences/1207. Unique Number of Occurrences.go
rename to leetcode/1207.Unique-Number-of-Occurrences/1207. Unique Number of Occurrences.go
diff --git a/Algorithms/1207. Unique Number of Occurrences/1207. Unique Number of Occurrences_test.go b/leetcode/1207.Unique-Number-of-Occurrences/1207. Unique Number of Occurrences_test.go
similarity index 94%
rename from Algorithms/1207. Unique Number of Occurrences/1207. Unique Number of Occurrences_test.go
rename to leetcode/1207.Unique-Number-of-Occurrences/1207. Unique Number of Occurrences_test.go
index 58d513b09..9f4728dc5 100644
--- a/Algorithms/1207. Unique Number of Occurrences/1207. Unique Number of Occurrences_test.go	
+++ b/leetcode/1207.Unique-Number-of-Occurrences/1207. Unique Number of Occurrences_test.go	
@@ -26,17 +26,17 @@ func Test_Problem1207(t *testing.T) {
 
 	qs := []question1207{
 
-		question1207{
+		{
 			para1207{[]int{1, 2, 2, 1, 1, 3}},
 			ans1207{true},
 		},
 
-		question1207{
+		{
 			para1207{[]int{1, 2}},
 			ans1207{false},
 		},
 
-		question1207{
+		{
 			para1207{[]int{-3, 0, 1, -3, 1, 1, 1, -3, 10, 0}},
 			ans1207{true},
 		},
diff --git a/leetcode/1207.Unique-Number-of-Occurrences/README.md b/leetcode/1207.Unique-Number-of-Occurrences/README.md
new file mode 100755
index 000000000..434a91f82
--- /dev/null
+++ b/leetcode/1207.Unique-Number-of-Occurrences/README.md
@@ -0,0 +1,44 @@
+# [1207. Unique Number of Occurrences](https://leetcode.com/problems/unique-number-of-occurrences/)
+
+
+## 题目
+
+Given an array of integers `arr`, write a function that returns `true` if and only if the number of occurrences of each value in the array is unique.
+
+**Example 1:**
+
+    Input: arr = [1,2,2,1,1,3]
+    Output: true
+    Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences.
+
+**Example 2:**
+
+    Input: arr = [1,2]
+    Output: false
+
+**Example 3:**
+
+    Input: arr = [-3,0,1,-3,1,1,1,-3,10,0]
+    Output: true
+
+**Constraints:**
+
+- `1 <= arr.length <= 1000`
+- `-1000 <= arr[i] <= 1000`
+
+
+
+## 题目大意
+
+给你一个整数数组 arr，请你帮忙统计数组中每个数的出现次数。如果每个数的出现次数都是独一无二的，就返回 true；否则返回 false。
+
+提示：
+
+- 1 <= arr.length <= 1000
+- -1000 <= arr[i] <= 1000
+
+## 解题思路
+
+
+- 给出一个数组，先统计每个数字出现的频次，判断在这个数组中是否存在相同的频次。
+- 简单题，先统计数组中每个数字的频次，然后用一个 map 判断频次是否重复。
diff --git a/leetcode/1208.Get-Equal-Substrings-Within-Budget/1208. Get Equal Substrings Within Budget.go b/leetcode/1208.Get-Equal-Substrings-Within-Budget/1208. Get Equal Substrings Within Budget.go
new file mode 100644
index 000000000..0635e56d2
--- /dev/null
+++ b/leetcode/1208.Get-Equal-Substrings-Within-Budget/1208. Get Equal Substrings Within Budget.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+func equalSubstring(s string, t string, maxCost int) int {
+	left, right, res := 0, -1, 0
+	for left < len(s) {
+		if right+1 < len(s) && maxCost-abs(int(s[right+1]-'a')-int(t[right+1]-'a')) >= 0 {
+			right++
+			maxCost -= abs(int(s[right]-'a') - int(t[right]-'a'))
+		} else {
+			res = max(res, right-left+1)
+			maxCost += abs(int(s[left]-'a') - int(t[left]-'a'))
+			left++
+		}
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
diff --git a/leetcode/1208.Get-Equal-Substrings-Within-Budget/1208. Get Equal Substrings Within Budget_test.go b/leetcode/1208.Get-Equal-Substrings-Within-Budget/1208. Get Equal Substrings Within Budget_test.go
new file mode 100644
index 000000000..86f501808
--- /dev/null
+++ b/leetcode/1208.Get-Equal-Substrings-Within-Budget/1208. Get Equal Substrings Within Budget_test.go	
@@ -0,0 +1,64 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1208 struct {
+	para1208
+	ans1208
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1208 struct {
+	s       string
+	t       string
+	maxCost int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1208 struct {
+	one int
+}
+
+func Test_Problem1208(t *testing.T) {
+
+	qs := []question1208{
+
+		{
+			para1208{"abcd", "bcdf", 3},
+			ans1208{3},
+		},
+
+		{
+			para1208{"abcd", "cdef", 3},
+			ans1208{1},
+		},
+
+		{
+			para1208{"abcd", "acde", 0},
+			ans1208{1},
+		},
+
+		{
+			para1208{"thjdoffka", "qhrnlntls", 11},
+			ans1208{3},
+		},
+
+		{
+			para1208{"krrgw", "zjxss", 19},
+			ans1208{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1208------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1208, q.para1208
+		fmt.Printf("【input】:%v       【output】:%v\n", p, equalSubstring(p.s, p.t, p.maxCost))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1208.Get-Equal-Substrings-Within-Budget/README.md b/leetcode/1208.Get-Equal-Substrings-Within-Budget/README.md
new file mode 100755
index 000000000..0f73f5809
--- /dev/null
+++ b/leetcode/1208.Get-Equal-Substrings-Within-Budget/README.md
@@ -0,0 +1,53 @@
+# [1208. Get Equal Substrings Within Budget](https://leetcode.com/problems/get-equal-substrings-within-budget/)
+
+
+## 题目
+
+You are given two strings `s` and `t` of the same length. You want to change `s` to `t`. Changing the `i`-th character of `s` to `i`-th character of `t` costs `|s[i] - t[i]|` that is, the absolute difference between the ASCII values of the characters.
+
+You are also given an integer `maxCost`.
+
+Return the maximum length of a substring of `s` that can be changed to be the same as the corresponding substring of `t`with a cost less than or equal to `maxCost`.
+
+If there is no substring from `s` that can be changed to its corresponding substring from `t`, return `0`.
+
+**Example 1:**
+
+    Input: s = "abcd", t = "bcdf", maxCost = 3
+    Output: 3
+    Explanation: "abc" of s can change to "bcd". That costs 3, so the maximum length is 3.
+
+**Example 2:**
+
+    Input: s = "abcd", t = "cdef", maxCost = 3
+    Output: 1
+    Explanation: Each character in s costs 2 to change to charactor in t, so the maximum length is 1.
+
+**Example 3:**
+
+    Input: s = "abcd", t = "acde", maxCost = 0
+    Output: 1
+    Explanation: You can't make any change, so the maximum length is 1.
+
+**Constraints:**
+
+- `1 <= s.length, t.length <= 10^5`
+- `0 <= maxCost <= 10^6`
+- `s` and `t` only contain lower case English letters.
+
+## 题目大意
+
+给你两个长度相同的字符串，s 和 t。将 s 中的第 i 个字符变到 t 中的第 i 个字符需要 |s[i] - t[i]| 的开销（开销可能为 0），也就是两个字符的 ASCII 码值的差的绝对值。
+
+用于变更字符串的最大预算是 maxCost。在转化字符串时，总开销应当小于等于该预算，这也意味着字符串的转化可能是不完全的。如果你可以将 s 的子字符串转化为它在 t 中对应的子字符串，则返回可以转化的最大长度。如果 s 中没有子字符串可以转化成 t 中对应的子字符串，则返回 0。
+
+提示：
+
+- 1 <= s.length, t.length <= 10^5
+- 0 <= maxCost <= 10^6
+- s 和 t 都只含小写英文字母。
+
+## 解题思路
+
+- 给出 2 个字符串 `s` 和 `t` 和一个“预算”，要求把“预算”尽可能的花完，`s` 中最多连续有几个字母能变成 `t` 中的字母。“预算”的定义是：|s[i] - t[i]| 。
+- 这一题是滑动窗口的题目，滑动窗口右边界每移动一格，就减少一定的预算，直到预算不能减少，再移动滑动窗口的左边界，这个时候注意要把预算还原回去。当整个窗口把字符 `s` 或 `t` 都滑动完了的时候，取出滑动过程中窗口的最大值即为结果。
diff --git a/leetcode/1209.Remove-All-Adjacent-Duplicates-in-String-II/1209. Remove All Adjacent Duplicates in String II.go b/leetcode/1209.Remove-All-Adjacent-Duplicates-in-String-II/1209. Remove All Adjacent Duplicates in String II.go
new file mode 100644
index 000000000..34a937c5a
--- /dev/null
+++ b/leetcode/1209.Remove-All-Adjacent-Duplicates-in-String-II/1209. Remove All Adjacent Duplicates in String II.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+// 解法一 stack
+func removeDuplicates(s string, k int) string {
+	stack, arr := [][2]int{}, []byte{}
+	for _, c := range s {
+		i := int(c - 'a')
+		if len(stack) > 0 && stack[len(stack)-1][0] == i {
+			stack[len(stack)-1][1]++
+			if stack[len(stack)-1][1] == k {
+				stack = stack[:len(stack)-1]
+			}
+		} else {
+			stack = append(stack, [2]int{i, 1})
+		}
+	}
+	for _, pair := range stack {
+		c := byte(pair[0] + 'a')
+		for i := 0; i < pair[1]; i++ {
+			arr = append(arr, c)
+		}
+	}
+	return string(arr)
+}
+
+// 解法二 暴力
+func removeDuplicates1(s string, k int) string {
+	arr, count, tmp := []rune{}, 0, '#'
+	for _, v := range s {
+		arr = append(arr, v)
+		for len(arr) > 0 {
+			count = 0
+			tmp = arr[len(arr)-1]
+			for i := len(arr) - 1; i >= 0; i-- {
+				if arr[i] != tmp {
+					break
+				}
+				count++
+			}
+			if count == k {
+				arr = arr[:len(arr)-k]
+			} else {
+				break
+			}
+		}
+	}
+	return string(arr)
+}
diff --git a/leetcode/1209.Remove-All-Adjacent-Duplicates-in-String-II/1209. Remove All Adjacent Duplicates in String II_test.go b/leetcode/1209.Remove-All-Adjacent-Duplicates-in-String-II/1209. Remove All Adjacent Duplicates in String II_test.go
new file mode 100644
index 000000000..8af7d1dda
--- /dev/null
+++ b/leetcode/1209.Remove-All-Adjacent-Duplicates-in-String-II/1209. Remove All Adjacent Duplicates in String II_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1209 struct {
+	para1209
+	ans1209
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1209 struct {
+	s string
+	k int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1209 struct {
+	one string
+}
+
+func Test_Problem1209(t *testing.T) {
+
+	qs := []question1209{
+
+		// {
+		// 	para1209{"abcd", 2},
+		// 	ans1209{"abcd"},
+		// },
+
+		{
+			para1209{"deeedbbcccbdaa", 3},
+			ans1209{"aa"},
+		},
+
+		{
+			para1209{"pbbcggttciiippooaais", 2},
+			ans1209{"ps"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1209------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1209, q.para1209
+		fmt.Printf("【input】:%v       【output】:%v\n", p, removeDuplicates(p.s, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1209.Remove-All-Adjacent-Duplicates-in-String-II/README.md b/leetcode/1209.Remove-All-Adjacent-Duplicates-in-String-II/README.md
new file mode 100644
index 000000000..d8488f0f0
--- /dev/null
+++ b/leetcode/1209.Remove-All-Adjacent-Duplicates-in-String-II/README.md
@@ -0,0 +1,106 @@
+# [1209. Remove All Adjacent Duplicates in String II](https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/)
+
+
+## 题目
+
+Given a string `s`, a *k* *duplicate removal* consists of choosing `k` adjacent and equal letters from `s` and removing them causing the left and the right side of the deleted substring to concatenate together.
+
+We repeatedly make `k` duplicate removals on `s` until we no longer can.
+
+Return the final string after all such duplicate removals have been made.
+
+It is guaranteed that the answer is unique.
+
+**Example 1:**
+
+```
+Input: s = "abcd", k = 2
+Output: "abcd"
+Explanation:There's nothing to delete.
+```
+
+**Example 2:**
+
+```
+Input: s = "deeedbbcccbdaa", k = 3
+Output: "aa"
+Explanation:
+First delete "eee" and "ccc", get "ddbbbdaa"
+Then delete "bbb", get "dddaa"
+Finally delete "ddd", get "aa"
+```
+
+**Example 3:**
+
+```
+Input: s = "pbbcggttciiippooaais", k = 2
+Output: "ps"
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 10^5`
+- `2 <= k <= 10^4`
+- `s` only contains lower case English letters.
+
+## 题目大意
+
+给你一个字符串 s，「k 倍重复项删除操作」将会从 s 中选择 k 个相邻且相等的字母，并删除它们，使被删去的字符串的左侧和右侧连在一起。你需要对 s 重复进行无限次这样的删除操作，直到无法继续为止。在执行完所有删除操作后，返回最终得到的字符串。本题答案保证唯一。
+
+## 解题思路
+
+- 暴力解法。每增加一个字符，就往前扫描 `k` 位，判断是否存在 `k` 个连续相同的字符。消除了 `k` 个相同字符后，重新组成的字符串还可能再次产出 `k` 位相同的字符，（类似消消乐，`k` 个相同的字符碰到一起就“消除”），还需要继续消除。最差情况要再次扫描一次字符串。时间复杂度 O(n^2)，空间复杂度 O(n)。
+- 暴力解法的低效在于重复统计字符频次，如果每个字符的频次统计一次就好了。按照这个思路，利用 stack ，每个栈元素存 2 个值，一个是字符，一个是该字符对应的频次。有了栈顶字符频次信息，就不需要重复往前扫描了。只要栈顶字符频次到达了 `k`，就弹出该字符。如此反复，最终剩下的字符串为所求。时间复杂度 O(n)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 stack
+func removeDuplicates(s string, k int) string {
+	stack, arr := [][2]int{}, []byte{}
+	for _, c := range s {
+		i := int(c - 'a')
+		if len(stack) > 0 && stack[len(stack)-1][0] == i {
+			stack[len(stack)-1][1]++
+			if stack[len(stack)-1][1] == k {
+				stack = stack[:len(stack)-1]
+			}
+		} else {
+			stack = append(stack, [2]int{i, 1})
+		}
+	}
+	for _, pair := range stack {
+		c := byte(pair[0] + 'a')
+		for i := 0; i < pair[1]; i++ {
+			arr = append(arr, c)
+		}
+	}
+	return string(arr)
+}
+
+// 解法二 暴力
+func removeDuplicates1(s string, k int) string {
+	arr, count, tmp := []rune{}, 0, '#'
+	for _, v := range s {
+		arr = append(arr, v)
+		for len(arr) > 0 {
+			count = 0
+			tmp = arr[len(arr)-1]
+			for i := len(arr) - 1; i >= 0; i-- {
+				if arr[i] != tmp {
+					break
+				}
+				count++
+			}
+			if count == k {
+				arr = arr[:len(arr)-k]
+			} else {
+				break
+			}
+		}
+	}
+	return string(arr)
+}
+```
\ No newline at end of file
diff --git a/leetcode/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position/1217. Minimum Cost to Move Chips to The Same Position.go b/leetcode/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position/1217. Minimum Cost to Move Chips to The Same Position.go
new file mode 100644
index 000000000..a47d76af2
--- /dev/null
+++ b/leetcode/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position/1217. Minimum Cost to Move Chips to The Same Position.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+func minCostToMoveChips(chips []int) int {
+	odd, even := 0, 0
+	for _, c := range chips {
+		if c%2 == 0 {
+			even++
+		} else {
+			odd++
+		}
+	}
+	return min(odd, even)
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position/1217. Minimum Cost to Move Chips to The Same Position_test.go b/leetcode/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position/1217. Minimum Cost to Move Chips to The Same Position_test.go
new file mode 100644
index 000000000..fcc7beca4
--- /dev/null
+++ b/leetcode/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position/1217. Minimum Cost to Move Chips to The Same Position_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1217 struct {
+	para1217
+	ans1217
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1217 struct {
+	arr []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1217 struct {
+	one int
+}
+
+func Test_Problem1217(t *testing.T) {
+
+	qs := []question1217{
+
+		{
+			para1217{[]int{1, 2, 3}},
+			ans1217{1},
+		},
+
+		{
+			para1217{[]int{2, 2, 2, 3, 3}},
+			ans1217{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1217------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1217, q.para1217
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minCostToMoveChips(p.arr))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position/README.md b/leetcode/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position/README.md
new file mode 100755
index 000000000..e7ac7fe5a
--- /dev/null
+++ b/leetcode/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position/README.md
@@ -0,0 +1,55 @@
+# [1217. Minimum Cost to Move Chips to The Same Position](https://leetcode.com/problems/minimum-cost-to-move-chips-to-the-same-position/)
+
+
+## 题目
+
+There are some chips, and the i-th chip is at position `chips[i]`.
+
+You can perform any of the two following types of moves **any number of times** (possibly zero) **on any chip**:
+
+- Move the `i`-th chip by 2 units to the left or to the right with a cost of **0**.
+- Move the `i`-th chip by 1 unit to the left or to the right with a cost of **1**.
+
+There can be two or more chips at the same position initially.
+
+Return the minimum cost needed to move all the chips to the same position (any position).
+
+**Example 1:**
+
+    Input: chips = [1,2,3]
+    Output: 1
+    Explanation: Second chip will be moved to positon 3 with cost 1. First chip will be moved to position 3 with cost 0. Total cost is 1.
+
+**Example 2:**
+
+    Input: chips = [2,2,2,3,3]
+    Output: 2
+    Explanation: Both fourth and fifth chip will be moved to position two with cost 1. Total minimum cost will be 2.
+
+**Constraints:**
+
+- `1 <= chips.length <= 100`
+- `1 <= chips[i] <= 10^9`
+
+
+## 题目大意
+
+
+数轴上放置了一些筹码，每个筹码的位置存在数组 chips 当中。你可以对 任何筹码 执行下面两种操作之一（不限操作次数，0 次也可以）：
+
+- 将第 i 个筹码向左或者右移动 2 个单位，代价为 0。
+- 将第 i 个筹码向左或者右移动 1 个单位，代价为 1。
+
+最开始的时候，同一位置上也可能放着两个或者更多的筹码。返回将所有筹码移动到同一位置（任意位置）上所需要的最小代价。
+
+
+提示：
+
+- 1 <= chips.length <= 100
+- 1 <= chips[i] <= 10^9
+
+
+## 解题思路
+
+- 给出一个数组，数组的下标代表的是数轴上的坐标点，数组的元素代表的是砝码大小。砝码移动规则，左右移动 2 格，没有代价，左右移动 1 个，代价是 1 。问最终把砝码都移动到一个格子上，最小代价是多少。
+- 先解读砝码移动规则：偶数位置的到偶数位置的没有代价，奇数到奇数位置的没有代价。利用这个规则，我们可以把所有的砝码**无代价**的摞在一个奇数的位置上和一个偶数的位置上。这样我们只用关心这两个位置了。并且这两个位置可以连续在一起。最后一步即将相邻的这两摞砝码合并到一起。由于左右移动一个代价是 1，所以最小代价的操作是移动最少砝码的那一边。奇数位置上砝码少就移动奇数位置上的，偶数位置上砝码少就移动偶数位置上的。所以这道题解法变的异常简单，遍历一次数组，找到其中有多少个奇数和偶数位置的砝码，取其中比较少的，就是最终答案。
diff --git a/leetcode/1221.Split-a-String-in-Balanced-Strings/1221. Split a String in Balanced Strings.go b/leetcode/1221.Split-a-String-in-Balanced-Strings/1221. Split a String in Balanced Strings.go
new file mode 100644
index 000000000..5f53b77f7
--- /dev/null
+++ b/leetcode/1221.Split-a-String-in-Balanced-Strings/1221. Split a String in Balanced Strings.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func balancedStringSplit(s string) int {
+	count, res := 0, 0
+	for _, r := range s {
+		if r == 'R' {
+			count++
+		} else {
+			count--
+		}
+		if count == 0 {
+			res++
+		}
+	}
+	return res
+}
diff --git a/leetcode/1221.Split-a-String-in-Balanced-Strings/1221. Split a String in Balanced Strings_test.go b/leetcode/1221.Split-a-String-in-Balanced-Strings/1221. Split a String in Balanced Strings_test.go
new file mode 100644
index 000000000..f5691c5ea
--- /dev/null
+++ b/leetcode/1221.Split-a-String-in-Balanced-Strings/1221. Split a String in Balanced Strings_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1221 struct {
+	para1221
+	ans1221
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1221 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1221 struct {
+	one int
+}
+
+func Test_Problem1221(t *testing.T) {
+
+	qs := []question1221{
+
+		{
+			para1221{"RLRRLLRLRL"},
+			ans1221{4},
+		},
+
+		{
+			para1221{"RLLLLRRRLR"},
+			ans1221{3},
+		},
+
+		{
+			para1221{"LLLLRRRR"},
+			ans1221{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1221------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1221, q.para1221
+		fmt.Printf("【input】:%v       【output】:%v\n", p, balancedStringSplit(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1221.Split-a-String-in-Balanced-Strings/README.md b/leetcode/1221.Split-a-String-in-Balanced-Strings/README.md
new file mode 100755
index 000000000..a73faf2b2
--- /dev/null
+++ b/leetcode/1221.Split-a-String-in-Balanced-Strings/README.md
@@ -0,0 +1,72 @@
+# [1221. Split a String in Balanced Strings](https://leetcode.com/problems/split-a-string-in-balanced-strings/)
+
+
+## 题目
+
+Balanced strings are those who have equal quantity of 'L' and 'R' characters.
+
+Given a balanced string `s` split it in the maximum amount of balanced strings.
+
+Return the maximum amount of splitted balanced strings.
+
+**Example 1**:
+
+    Input: s = "RLRRLLRLRL"
+    Output: 4
+    Explanation: s can be split into "RL", "RRLL", "RL", "RL", each substring contains same number of 'L' and 'R'.
+
+**Example 2**:
+
+    Input: s = "RLLLLRRRLR"
+    Output: 3
+    Explanation: s can be split into "RL", "LLLRRR", "LR", each substring contains same number of 'L' and 'R'.
+
+**Example 3**:
+
+    Input: s = "LLLLRRRR"
+    Output: 1
+    Explanation: s can be split into "LLLLRRRR".
+
+**Constraints**:
+
+- `1 <= s.length <= 1000`
+- `s[i] = 'L' or 'R'`
+
+## 题目大意
+
+
+在一个「平衡字符串」中，'L' 和 'R' 字符的数量是相同的。给出一个平衡字符串 s，请你将它分割成尽可能多的平衡字符串。返回可以通过分割得到的平衡字符串的最大数量。
+
+提示：
+
+- 1 <= s.length <= 1000
+- s[i] = 'L' 或 'R'
+
+
+## 解题思路
+
+- 给出一个字符串，要求把这个字符串切成一些子串，这些子串中 R 和 L 的字符数是相等的。问能切成多少个满足条件的子串。
+- 这道题是简单题，按照题意模拟即可。从左往右扫，遇到 `R` 就加一，遇到 `L` 就减一，当计数是 `0` 的时候就是平衡的时候，就切割。
+
+## 代码
+
+```go
+
+package leetcode
+
+func balancedStringSplit(s string) int {
+	count, res := 0, 0
+	for _, r := range s {
+		if r == 'R' {
+			count++
+		} else {
+			count--
+		}
+		if count == 0 {
+			res++
+		}
+	}
+	return res
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1232.Check-If-It-Is-a-Straight-Line/1232. Check If It Is a Straight Line.go b/leetcode/1232.Check-If-It-Is-a-Straight-Line/1232. Check If It Is a Straight Line.go
new file mode 100644
index 000000000..315d04f57
--- /dev/null
+++ b/leetcode/1232.Check-If-It-Is-a-Straight-Line/1232. Check If It Is a Straight Line.go	
@@ -0,0 +1,14 @@
+package leetcode
+
+func checkStraightLine(coordinates [][]int) bool {
+	dx0 := coordinates[1][0] - coordinates[0][0]
+	dy0 := coordinates[1][1] - coordinates[0][1]
+	for i := 1; i < len(coordinates)-1; i++ {
+		dx := coordinates[i+1][0] - coordinates[i][0]
+		dy := coordinates[i+1][1] - coordinates[i][1]
+		if dy*dx0 != dy0*dx { // check cross product
+			return false
+		}
+	}
+	return true
+}
diff --git a/leetcode/1232.Check-If-It-Is-a-Straight-Line/1232. Check If It Is a Straight Line_test.go b/leetcode/1232.Check-If-It-Is-a-Straight-Line/1232. Check If It Is a Straight Line_test.go
new file mode 100644
index 000000000..63710630d
--- /dev/null
+++ b/leetcode/1232.Check-If-It-Is-a-Straight-Line/1232. Check If It Is a Straight Line_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1232 struct {
+	para1232
+	ans1232
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1232 struct {
+	arr [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1232 struct {
+	one bool
+}
+
+func Test_Problem1232(t *testing.T) {
+
+	qs := []question1232{
+
+		{
+			para1232{[][]int{{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 7}}},
+			ans1232{true},
+		},
+
+		{
+			para1232{[][]int{{1, 1}, {2, 2}, {3, 4}, {4, 5}, {5, 6}, {7, 7}}},
+			ans1232{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1232------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1232, q.para1232
+		fmt.Printf("【input】:%v       【output】:%v\n", p, checkStraightLine(p.arr))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1232.Check-If-It-Is-a-Straight-Line/README.md b/leetcode/1232.Check-If-It-Is-a-Straight-Line/README.md
new file mode 100755
index 000000000..0801daa45
--- /dev/null
+++ b/leetcode/1232.Check-If-It-Is-a-Straight-Line/README.md
@@ -0,0 +1,72 @@
+# [1232. Check If It Is a Straight Line](https://leetcode.com/problems/check-if-it-is-a-straight-line/)
+
+
+## 题目
+
+You are given an array `coordinates`, `coordinates[i] = [x, y]`, where `[x, y]` represents the coordinate of a point. Check if these points make a straight line in the XY plane.
+
+**Example 1**:
+
+![](https://img.halfrost.com/Leetcode/leetcode_1232_1.png)
+
+    Input: coordinates = [[1,2],[2,3],[3,4],[4,5],[5,6],[6,7]]
+    Output: true
+
+**Example 2**:
+
+
+![](https://img.halfrost.com/Leetcode/leetcode_1232_2.png)
+
+    Input: coordinates = [[1,1],[2,2],[3,4],[4,5],[5,6],[7,7]]
+    Output: false
+
+**Constraints**:
+
+- `2 <= coordinates.length <= 1000`
+- `coordinates[i].length == 2`
+- `-10^4 <= coordinates[i][0], coordinates[i][1] <= 10^4`
+- `coordinates` contains no duplicate point.
+
+## 题目大意
+
+
+在一个 XY 坐标系中有一些点，我们用数组 coordinates 来分别记录它们的坐标，其中 coordinates[i] = [x, y] 表示横坐标为 x、纵坐标为 y 的点。
+
+请你来判断，这些点是否在该坐标系中属于同一条直线上，是则返回 true，否则请返回 false。
+
+提示：
+
+- 2 <= coordinates.length <= 1000
+- coordinates[i].length == 2
+- -10^4 <= coordinates[i][0], coordinates[i][1] <= 10^4
+- coordinates 中不含重复的点
+
+
+
+## 解题思路
+
+- 给出一组坐标点，要求判断这些点是否在同一直线上。
+- 按照几何原理，依次计算这些点的斜率是否相等即可。斜率需要做除法，这里采用一个技巧是换成乘法。例如 `a/b = c/d` 换成乘法是 `a*d  = c*d`  。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func checkStraightLine(coordinates [][]int) bool {
+	dx0 := coordinates[1][0] - coordinates[0][0]
+	dy0 := coordinates[1][1] - coordinates[0][1]
+	for i := 1; i < len(coordinates)-1; i++ {
+		dx := coordinates[i+1][0] - coordinates[i][0]
+		dy := coordinates[i+1][1] - coordinates[i][1]
+		if dy*dx0 != dy0*dx { // check cross product
+			return false
+		}
+	}
+	return true
+}
+
+
+```
\ No newline at end of file
diff --git a/leetcode/1234.Replace-the-Substring-for-Balanced-String/1234. Replace the Substring for Balanced String.go b/leetcode/1234.Replace-the-Substring-for-Balanced-String/1234. Replace the Substring for Balanced String.go
new file mode 100644
index 000000000..7ca703b64
--- /dev/null
+++ b/leetcode/1234.Replace-the-Substring-for-Balanced-String/1234. Replace the Substring for Balanced String.go	
@@ -0,0 +1,31 @@
+package leetcode
+
+func balancedString(s string) int {
+	count, k := make([]int, 128), len(s)/4
+	for _, v := range s {
+		count[int(v)]++
+	}
+	left, right, res := 0, -1, len(s)
+	for left < len(s) {
+		if count['Q'] > k || count['W'] > k || count['E'] > k || count['R'] > k {
+			if right+1 < len(s) {
+				right++
+				count[s[right]]--
+			} else {
+				break
+			}
+		} else {
+			res = min(res, right-left+1)
+			count[s[left]]++
+			left++
+		}
+	}
+	return res
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/1234.Replace-the-Substring-for-Balanced-String/1234. Replace the Substring for Balanced String_test.go b/leetcode/1234.Replace-the-Substring-for-Balanced-String/1234. Replace the Substring for Balanced String_test.go
new file mode 100644
index 000000000..854dc2bc4
--- /dev/null
+++ b/leetcode/1234.Replace-the-Substring-for-Balanced-String/1234. Replace the Substring for Balanced String_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1234 struct {
+	para1234
+	ans1234
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1234 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1234 struct {
+	one int
+}
+
+func Test_Problem1234(t *testing.T) {
+
+	qs := []question1234{
+
+		{
+			para1234{"QWER"},
+			ans1234{0},
+		},
+
+		{
+			para1234{"QQWE"},
+			ans1234{1},
+		},
+
+		{
+			para1234{"QQQW"},
+			ans1234{2},
+		},
+
+		{
+			para1234{"QQQQ"},
+			ans1234{3},
+		},
+
+		{
+			para1234{"WQWRQQQW"},
+			ans1234{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1234------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1234, q.para1234
+		fmt.Printf("【input】:%v       【output】:%v\n", p, balancedString(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1234.Replace-the-Substring-for-Balanced-String/README.md b/leetcode/1234.Replace-the-Substring-for-Balanced-String/README.md
new file mode 100755
index 000000000..15f7827bb
--- /dev/null
+++ b/leetcode/1234.Replace-the-Substring-for-Balanced-String/README.md
@@ -0,0 +1,92 @@
+# [1234. Replace the Substring for Balanced String](https://leetcode.com/problems/replace-the-substring-for-balanced-string/)
+
+
+## 题目
+
+You are given a string containing only 4 kinds of characters `'Q',` `'W', 'E'` and `'R'`.
+
+A string is said to be **balanced** **if each of its characters appears `n/4` times where `n` is the length of the string.
+
+Return the minimum length of the substring that can be replaced with **any** other string of the same length to make the original string `s` **balanced**.
+
+Return 0 if the string is already **balanced**.
+
+**Example 1**:
+
+    Input: s = "QWER"
+    Output: 0
+    Explanation: s is already balanced.
+
+**Example 2**:
+
+    Input: s = "QQWE"
+    Output: 1
+    Explanation: We need to replace a 'Q' to 'R', so that "RQWE" (or "QRWE") is balanced.
+
+**Example 3**:
+
+    Input: s = "QQQW"
+    Output: 2
+    Explanation: We can replace the first "QQ" to "ER".
+
+**Example 4**:
+
+    Input: s = "QQQQ"
+    Output: 3
+    Explanation: We can replace the last 3 'Q' to make s = "QWER".
+
+**Constraints**:
+
+- `1 <= s.length <= 10^5`
+- `s.length` is a multiple of `4`
+- `s` contains only `'Q'`, `'W'`, `'E'` and `'R'`.
+
+## 题目大意
+
+
+有一个只含有 'Q', 'W', 'E', 'R' 四种字符，且长度为 n 的字符串。假如在该字符串中，这四个字符都恰好出现 n/4 次，那么它就是一个「平衡字符串」。给你一个这样的字符串 s，请通过「替换一个子串」的方式，使原字符串 s 变成一个「平衡字符串」。你可以用和「待替换子串」长度相同的 任何 其他字符串来完成替换。请返回待替换子串的最小可能长度。如果原字符串自身就是一个平衡字符串，则返回 0。
+
+提示：
+
+- 1 <= s.length <= 10^5
+- s.length 是 4 的倍数
+- s 中只含有 'Q', 'W', 'E', 'R' 四种字符
+
+
+
+## 解题思路
+
+- 给出一个字符串，要求输出把这个字符串变成“平衡字符串”的最小替换字符串的长度(替换只能替换一串，不能单个字母替换)。“平衡字符串”的定义是：字符串中，`‘Q’`，`‘W’`，`‘E’`，`‘R’`，出现的次数当且仅当只有 `len(s)/4` 次。
+- 这一题是滑动窗口的题目。先统计 4 个字母的频次并计算出 `k = len(s)/4` 。滑动窗口向右滑动一次，对应右窗口的那个字母频次减 1，直到滑到所有字母的频次都 `≤ k`  的地方停止。此时，窗口外的字母的频次都 `≤ k` 了。这是只要变换窗口内字符串即可。但是这个窗口内还可能包含本来频次就小于 `k`  的字母，如果能把它们剔除掉，窗口可以进一步的减少。所以继续移动左边界，试探移动完左边界以后，是否所有字母频次都 `≤ k`。在所有窗口移动过程中取出最小值，即为最终答案。
+- 举个例子：`"WQWRQQQW"`。`w` 有 3 个，`Q` 有 4 个，`R` 有 1 个，`E` 有 0 个。最后平衡状态是每个字母 2 个，那么我们需要拿出 1 个 `W` 和 2 个 `Q` 替换掉。即要找到一个最短的字符串包含 1 个 `W` 和 2 个 `Q`。滑动窗口正好可以解决这个问题。向右滑到 `"WQWRQ"` 停止，这时窗口外的所有字母频次都  `≤ k` 了。这个窗口包含了多余的 1 个 `W`，和 1 个 `R`。`W` 可以踢除掉，那么要替换的字符串是 `"QWRQ"`。`R` 不能踢除了(因为要找包含 1 个 `W` 和 2 个 `Q` 的字符串) 。窗口不断的滑动，直到结束。这个例子中最小的字符串其实位于末尾，`"QQW"`。
+
+## 代码
+
+```go
+
+package leetcode
+
+func balancedString(s string) int {
+	count, k := make([]int, 128), len(s)/4
+	for _, v := range s {
+		count[int(v)]++
+	}
+	left, right, res := 0, -1, len(s)
+	for left < len(s) {
+		if count['Q'] > k || count['W'] > k || count['E'] > k || count['R'] > k {
+			if right+1 < len(s) {
+				right++
+				count[s[right]]--
+			} else {
+				break
+			}
+		} else {
+			res = min(res, right-left+1)
+			count[s[left]]++
+			left++
+		}
+	}
+	return res
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1235.Maximum-Profit-in-Job-Scheduling/1235. Maximum Profit in Job Scheduling.go b/leetcode/1235.Maximum-Profit-in-Job-Scheduling/1235. Maximum Profit in Job Scheduling.go
new file mode 100644
index 000000000..13444f3bf
--- /dev/null
+++ b/leetcode/1235.Maximum-Profit-in-Job-Scheduling/1235. Maximum Profit in Job Scheduling.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import "sort"
+
+type job struct {
+	startTime int
+	endTime   int
+	profit    int
+}
+
+func jobScheduling(startTime []int, endTime []int, profit []int) int {
+	jobs, dp := []job{}, make([]int, len(startTime))
+	for i := 0; i < len(startTime); i++ {
+		jobs = append(jobs, job{startTime: startTime[i], endTime: endTime[i], profit: profit[i]})
+	}
+	sort.Sort(sortJobs(jobs))
+	dp[0] = jobs[0].profit
+	for i := 1; i < len(jobs); i++ {
+		low, high := 0, i-1
+		for low < high {
+			mid := low + (high-low)>>1
+			if jobs[mid+1].endTime <= jobs[i].startTime {
+				low = mid + 1
+			} else {
+				high = mid
+			}
+		}
+		if jobs[low].endTime <= jobs[i].startTime {
+			dp[i] = max(dp[i-1], dp[low]+jobs[i].profit)
+		} else {
+			dp[i] = max(dp[i-1], jobs[i].profit)
+		}
+	}
+	return dp[len(startTime)-1]
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+type sortJobs []job
+
+func (s sortJobs) Len() int {
+	return len(s)
+}
+func (s sortJobs) Less(i, j int) bool {
+	if s[i].endTime == s[j].endTime {
+		return s[i].profit < s[j].profit
+	}
+	return s[i].endTime < s[j].endTime
+}
+func (s sortJobs) Swap(i, j int) {
+	s[i], s[j] = s[j], s[i]
+}
diff --git a/leetcode/1235.Maximum-Profit-in-Job-Scheduling/1235. Maximum Profit in Job Scheduling_test.go b/leetcode/1235.Maximum-Profit-in-Job-Scheduling/1235. Maximum Profit in Job Scheduling_test.go
new file mode 100644
index 000000000..77501bf4d
--- /dev/null
+++ b/leetcode/1235.Maximum-Profit-in-Job-Scheduling/1235. Maximum Profit in Job Scheduling_test.go	
@@ -0,0 +1,54 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1235 struct {
+	para1235
+	ans1235
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1235 struct {
+	startTime []int
+	endTime   []int
+	profit    []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1235 struct {
+	one int
+}
+
+func Test_Problem1235(t *testing.T) {
+
+	qs := []question1235{
+
+		{
+			para1235{[]int{1, 2, 3, 3}, []int{3, 4, 5, 6}, []int{50, 10, 40, 70}},
+			ans1235{120},
+		},
+
+		{
+			para1235{[]int{1, 2, 3, 4, 6}, []int{3, 5, 10, 6, 9}, []int{20, 20, 100, 70, 60}},
+			ans1235{150},
+		},
+
+		{
+			para1235{[]int{1, 1, 1}, []int{2, 3, 4}, []int{5, 6, 4}},
+			ans1235{6},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1235------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1235, q.para1235
+		fmt.Printf("【input】:%v       【output】:%v\n", p, jobScheduling(p.startTime, p.endTime, p.profit))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1235.Maximum-Profit-in-Job-Scheduling/README.md b/leetcode/1235.Maximum-Profit-in-Job-Scheduling/README.md
new file mode 100755
index 000000000..8734dbde5
--- /dev/null
+++ b/leetcode/1235.Maximum-Profit-in-Job-Scheduling/README.md
@@ -0,0 +1,60 @@
+# [1235. Maximum Profit in Job Scheduling](https://leetcode.com/problems/maximum-profit-in-job-scheduling/)
+
+
+## 题目
+
+We have `n` jobs, where every job is scheduled to be done from `startTime[i]` to `endTime[i]`, obtaining a profit of `profit[i]`.
+
+You're given the `startTime` , `endTime` and `profit` arrays, you need to output the maximum profit you can take such that there are no 2 jobs in the subset with overlapping time range.
+
+If you choose a job that ends at time `X` you will be able to start another job that starts at time `X`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png](https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png)
+
+    Input: startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
+    Output: 120
+    Explanation: The subset chosen is the first and fourth job. 
+    Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png](https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png)
+
+    Input: startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
+    Output: 150
+    Explanation: The subset chosen is the first, fourth and fifth job. 
+    Profit obtained 150 = 20 + 70 + 60.
+
+**Example 3:**
+
+![https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png](https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png)
+
+    Input: startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
+    Output: 6
+
+**Constraints:**
+
+- `1 <= startTime.length == endTime.length == profit.length <= 5 * 10^4`
+- `1 <= startTime[i] < endTime[i] <= 10^9`
+- `1 <= profit[i] <= 10^4`
+
+## 题目大意
+
+
+你打算利用空闲时间来做兼职工作赚些零花钱。这里有 n 份兼职工作，每份工作预计从 startTime[i] 开始到 endTime[i] 结束，报酬为 profit[i]。给你一份兼职工作表，包含开始时间 startTime，结束时间 endTime 和预计报酬 profit 三个数组，请你计算并返回可以获得的最大报酬。注意，时间上出现重叠的 2 份工作不能同时进行。如果你选择的工作在时间 X 结束，那么你可以立刻进行在时间 X 开始的下一份工作。
+
+
+提示：
+
+- 1 <= startTime.length == endTime.length == profit.length <= 5 * 10^4
+- 1 <= startTime[i] < endTime[i] <= 10^9
+- 1 <= profit[i] <= 10^4
+
+
+
+## 解题思路
+
+- 给出一组任务，任务有开始时间，结束时间，和任务收益。一个任务开始还没有结束，中间就不能再安排其他任务。问如何安排任务，能使得最后收益最大？
+- 一般任务的题目，区间的题目，都会考虑是否能排序。这一题可以先按照任务的结束时间从小到大排序，如果结束时间相同，则按照收益从小到大排序。`dp[i]` 代表前 `i` 份工作能获得的最大收益。初始值，`dp[0] = job[1].profit` 。对于任意一个任务 `i` ，看能否找到满足 `jobs[j].enTime <= jobs[j].startTime && j < i` 条件的 `j`，即查找 `upper_bound` 。由于 `jobs` 被我们排序了，所以这里可以使用二分搜索来查找。如果能找到满足条件的任务 j，那么状态转移方程是：`dp[i] = max(dp[i-1], jobs[i].profit)`。如果能找到满足条件的任务 j，那么状态转移方程是：`dp[i] = max(dp[i-1], dp[low]+jobs[i].profit)`。最终求得的解在 `dp[len(startTime)-1]` 中。
diff --git a/leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/1239. Maximum Length of a Concatenated String with Unique Characters.go b/leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/1239. Maximum Length of a Concatenated String with Unique Characters.go
new file mode 100644
index 000000000..f75727caf
--- /dev/null
+++ b/leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/1239. Maximum Length of a Concatenated String with Unique Characters.go	
@@ -0,0 +1,38 @@
+package leetcode
+
+import (
+	"math/bits"
+)
+
+func maxLength(arr []string) int {
+	c, res := []uint32{}, 0
+	for _, s := range arr {
+		var mask uint32
+		for _, c := range s {
+			mask = mask | 1<<(c-'a')
+		}
+		if len(s) != bits.OnesCount32(mask) { // 如果字符串本身带有重复的字符，需要排除
+			continue
+		}
+		c = append(c, mask)
+	}
+	dfs(c, 0, 0, &res)
+	return res
+}
+
+func dfs(c []uint32, index int, mask uint32, res *int) {
+	*res = max(*res, bits.OnesCount32(mask))
+	for i := index; i < len(c); i++ {
+		if mask&c[i] == 0 {
+			dfs(c, i+1, mask|c[i], res)
+		}
+	}
+	return
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/1239. Maximum Length of a Concatenated String with Unique Characters_test.go b/leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/1239. Maximum Length of a Concatenated String with Unique Characters_test.go
new file mode 100644
index 000000000..8a0c00345
--- /dev/null
+++ b/leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/1239. Maximum Length of a Concatenated String with Unique Characters_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1239 struct {
+	para1239
+	ans1239
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1239 struct {
+	arr []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1239 struct {
+	one int
+}
+
+func Test_Problem1239(t *testing.T) {
+
+	qs := []question1239{
+
+		{
+			para1239{[]string{"un", "iq", "ue"}},
+			ans1239{4},
+		},
+
+		{
+			para1239{[]string{"cha", "r", "act", "ers"}},
+			ans1239{6},
+		},
+
+		{
+			para1239{[]string{"abcdefghijklmnopqrstuvwxyz"}},
+			ans1239{26},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1239------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1239, q.para1239
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxLength(p.arr))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/README.md b/leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/README.md
new file mode 100644
index 000000000..58cdb2e9d
--- /dev/null
+++ b/leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/README.md
@@ -0,0 +1,88 @@
+# [1239. Maximum Length of a Concatenated String with Unique Characters](https://leetcode.com/problems/maximum-length-of-a-concatenated-string-with-unique-characters/)
+
+## 题目
+
+Given an array of strings `arr`. String `s` is a concatenation of a sub-sequence of `arr` which have **unique characters**.
+
+Return *the maximum possible length* of `s`.
+
+**Example 1:**
+
+```
+Input: arr = ["un","iq","ue"]
+Output: 4
+Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique".
+Maximum length is 4.
+```
+
+**Example 2:**
+
+```
+Input: arr = ["cha","r","act","ers"]
+Output: 6
+Explanation: Possible solutions are "chaers" and "acters".
+```
+
+**Example 3:**
+
+```
+Input: arr = ["abcdefghijklmnopqrstuvwxyz"]
+Output: 26
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 16`
+- `1 <= arr[i].length <= 26`
+- `arr[i]` contains only lower case English letters.
+
+## 题目大意
+
+给定一个字符串数组 arr，字符串 s 是将 arr 某一子序列字符串连接所得的字符串，如果 s 中的每一个字符都只出现过一次，那么它就是一个可行解。请返回所有可行解 s 中最长长度。
+
+## 解题思路
+
+- 每个字符串数组可以想象为 26 位的 0101 二进制串。出现的字符对应的位上标记为 1，没有出现的字符对应的位上标记为 0 。如果一个字符串中包含重复的字符，那么它所有 1 的个数一定不等于字符串的长度。如果 2 个字符串每个字母都只出现了一次，那么它们俩对应的二进制串 mask 相互与运算的结果一定为 0 ，即 0，1 互补了。利用这个特点，深搜所有解，保存出最长可行解的长度即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math/bits"
+)
+
+func maxLength(arr []string) int {
+	c, res := []uint32{}, 0
+	for _, s := range arr {
+		var mask uint32
+		for _, c := range s {
+			mask = mask | 1<<(c-'a')
+		}
+		if len(s) != bits.OnesCount32(mask) { // 如果字符串本身带有重复的字符，需要排除
+			continue
+		}
+		c = append(c, mask)
+	}
+	dfs(c, 0, 0, &res)
+	return res
+}
+
+func dfs(c []uint32, index int, mask uint32, res *int) {
+	*res = max(*res, bits.OnesCount32(mask))
+	for i := index; i < len(c); i++ {
+		if mask&c[i] == 0 {
+			dfs(c, i+1, mask|c[i], res)
+		}
+	}
+	return
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1249.Minimum-Remove-to-Make-Valid-Parentheses/1249. Minimum Remove to Make Valid Parentheses.go b/leetcode/1249.Minimum-Remove-to-Make-Valid-Parentheses/1249. Minimum Remove to Make Valid Parentheses.go
new file mode 100644
index 000000000..9943e5881
--- /dev/null
+++ b/leetcode/1249.Minimum-Remove-to-Make-Valid-Parentheses/1249. Minimum Remove to Make Valid Parentheses.go	
@@ -0,0 +1,23 @@
+package leetcode
+
+func minRemoveToMakeValid(s string) string {
+	res, opens := []byte{}, 0
+	for i := 0; i < len(s); i++ {
+		if s[i] == '(' {
+			opens++
+		} else if s[i] == ')' {
+			if opens == 0 {
+				continue
+			}
+			opens--
+		}
+		res = append(res, s[i])
+	}
+	for i := len(res) - 1; i >= 0; i-- {
+		if res[i] == '(' && opens > 0 {
+			opens--
+			res = append(res[:i], res[i+1:]...)
+		}
+	}
+	return string(res)
+}
diff --git a/leetcode/1249.Minimum-Remove-to-Make-Valid-Parentheses/1249. Minimum Remove to Make Valid Parentheses_test.go b/leetcode/1249.Minimum-Remove-to-Make-Valid-Parentheses/1249. Minimum Remove to Make Valid Parentheses_test.go
new file mode 100644
index 000000000..f09bc3a5a
--- /dev/null
+++ b/leetcode/1249.Minimum-Remove-to-Make-Valid-Parentheses/1249. Minimum Remove to Make Valid Parentheses_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1249 struct {
+	para1249
+	ans1249
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1249 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1249 struct {
+	one string
+}
+
+func Test_Problem1249(t *testing.T) {
+
+	qs := []question1249{
+
+		{
+			para1249{"lee(t(c)o)de)"},
+			ans1249{"lee(t(c)o)de"},
+		},
+
+		{
+			para1249{"a)b(c)d"},
+			ans1249{"ab(c)d"},
+		},
+
+		{
+			para1249{"))(("},
+			ans1249{""},
+		},
+
+		{
+			para1249{"(a(b(c)d)"},
+			ans1249{"a(b(c)d)"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1249------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1249, q.para1249
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minRemoveToMakeValid(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1249.Minimum-Remove-to-Make-Valid-Parentheses/README.md b/leetcode/1249.Minimum-Remove-to-Make-Valid-Parentheses/README.md
new file mode 100644
index 000000000..1e60b4d77
--- /dev/null
+++ b/leetcode/1249.Minimum-Remove-to-Make-Valid-Parentheses/README.md
@@ -0,0 +1,95 @@
+# [1249. Minimum Remove to Make Valid Parentheses](https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/)
+
+
+## 题目
+
+Given a string s of `'('` , `')'` and lowercase English characters.
+
+Your task is to remove the minimum number of parentheses ( `'('` or `')'`, in any positions ) so that the resulting *parentheses string* is valid and return **any** valid string.
+
+Formally, a *parentheses string* is valid if and only if:
+
+- It is the empty string, contains only lowercase characters, or
+- It can be written as `AB` (`A` concatenated with `B`), where `A` and `B` are valid strings, or
+- It can be written as `(A)`, where `A` is a valid string.
+
+**Example 1:**
+
+```
+Input: s = "lee(t(c)o)de)"
+Output: "lee(t(c)o)de"
+Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "a)b(c)d"
+Output: "ab(c)d"
+
+```
+
+**Example 3:**
+
+```
+Input: s = "))(("
+Output: ""
+Explanation: An empty string is also valid.
+
+```
+
+**Example 4:**
+
+```
+Input: s = "(a(b(c)d)"
+Output: "a(b(c)d)"
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 10^5`
+- `s[i]` is one of `'('` , `')'` and lowercase English letters`.`
+
+## 题目大意
+
+给你一个由 '('、')' 和小写字母组成的字符串 s。你需要从字符串中删除最少数目的 '(' 或者 ')' （可以删除任意位置的括号)，使得剩下的「括号字符串」有效。请返回任意一个合法字符串。有效「括号字符串」应当符合以下 任意一条 要求：
+
+- 空字符串或只包含小写字母的字符串
+- 可以被写作 AB（A 连接 B）的字符串，其中 A 和 B 都是有效「括号字符串」
+- 可以被写作 (A) 的字符串，其中 A 是一个有效的「括号字符串」
+
+## 解题思路
+
+- 最容易想到的思路是利用栈判断括号匹配是否有效。这个思路可行，时间复杂度也只是 O(n)。
+- 不用栈，可以 2 次循环遍历，正向遍历一次，标记出多余的 `'('` ，逆向遍历一次，再标记出多余的 `')'`，最后将所有这些标记多余的字符删掉即可。这种解法写出来的代码也很简洁，时间复杂度也是 O(n)。
+- 针对上面的解法再改进一点。正向遍历的时候不仅标记出多余的 `'('`，还可以顺手把多余的 `')'` 删除。这样只用循环一次。最后再删除掉多余的 `'('` 即可。时间复杂度还是 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+func minRemoveToMakeValid(s string) string {
+	res, opens := []byte{}, 0
+	for i := 0; i < len(s); i++ {
+		if s[i] == '(' {
+			opens++
+		} else if s[i] == ')' {
+			if opens == 0 {
+				continue
+			}
+			opens--
+		}
+		res = append(res, s[i])
+	}
+	for i := len(res) - 1; i >= 0; i-- {
+		if res[i] == '(' && opens > 0 {
+			opens--
+			res = append(res[:i], res[i+1:]...)
+		}
+	}
+	return string(res)
+}
+```
\ No newline at end of file
diff --git a/leetcode/1252.Cells-with-Odd-Values-in-a-Matrix/1252. Cells with Odd Values in a Matrix.go b/leetcode/1252.Cells-with-Odd-Values-in-a-Matrix/1252. Cells with Odd Values in a Matrix.go
new file mode 100644
index 000000000..c3b438e0c
--- /dev/null
+++ b/leetcode/1252.Cells-with-Odd-Values-in-a-Matrix/1252. Cells with Odd Values in a Matrix.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+// 解法一 暴力法
+func oddCells(n int, m int, indices [][]int) int {
+	matrix, res := make([][]int, n), 0
+	for i := range matrix {
+		matrix[i] = make([]int, m)
+	}
+	for _, indice := range indices {
+		for i := 0; i < m; i++ {
+			matrix[indice[0]][i]++
+		}
+		for j := 0; j < n; j++ {
+			matrix[j][indice[1]]++
+		}
+	}
+	for _, m := range matrix {
+		for _, v := range m {
+			if v&1 == 1 {
+				res++
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 暴力法
+func oddCells1(n int, m int, indices [][]int) int {
+	rows, cols, count := make([]int, n), make([]int, m), 0
+	for _, pair := range indices {
+		rows[pair[0]]++
+		cols[pair[1]]++
+	}
+	for i := 0; i < n; i++ {
+		for j := 0; j < m; j++ {
+			if (rows[i]+cols[j])%2 == 1 {
+				count++
+			}
+		}
+	}
+	return count
+}
diff --git a/leetcode/1252.Cells-with-Odd-Values-in-a-Matrix/1252. Cells with Odd Values in a Matrix_test.go b/leetcode/1252.Cells-with-Odd-Values-in-a-Matrix/1252. Cells with Odd Values in a Matrix_test.go
new file mode 100644
index 000000000..df4098938
--- /dev/null
+++ b/leetcode/1252.Cells-with-Odd-Values-in-a-Matrix/1252. Cells with Odd Values in a Matrix_test.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1252 struct {
+	para1252
+	ans1252
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1252 struct {
+	n       int
+	m       int
+	indices [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1252 struct {
+	one int
+}
+
+func Test_Problem1252(t *testing.T) {
+
+	qs := []question1252{
+
+		{
+			para1252{2, 3, [][]int{{0, 1}, {1, 1}}},
+			ans1252{6},
+		},
+
+		{
+			para1252{2, 2, [][]int{{1, 1}, {0, 0}}},
+			ans1252{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1252------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1252, q.para1252
+		fmt.Printf("【input】:%v       【output】:%v\n", p, oddCells(p.n, p.m, p.indices))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1252.Cells-with-Odd-Values-in-a-Matrix/README.md b/leetcode/1252.Cells-with-Odd-Values-in-a-Matrix/README.md
new file mode 100755
index 000000000..eca3e8ba7
--- /dev/null
+++ b/leetcode/1252.Cells-with-Odd-Values-in-a-Matrix/README.md
@@ -0,0 +1,102 @@
+# [1252. Cells with Odd Values in a Matrix](https://leetcode.com/problems/cells-with-odd-values-in-a-matrix/)
+
+
+## 题目
+
+Given `n` and `m` which are the dimensions of a matrix initialized by zeros and given an array `indices` where `indices[i] = [ri, ci]`. For each pair of `[ri, ci]` you have to increment all cells in row `ri` and column `ci` by 1.
+
+Return *the number of cells with odd values* in the matrix after applying the increment to all `indices`.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/10/30/e1.png](https://assets.leetcode.com/uploads/2019/10/30/e1.png)
+
+    Input: n = 2, m = 3, indices = [[0,1],[1,1]]
+    Output: 6
+    Explanation: Initial matrix = [[0,0,0],[0,0,0]].
+    After applying first increment it becomes [[1,2,1],[0,1,0]].
+    The final matrix will be [[1,3,1],[1,3,1]] which contains 6 odd numbers.
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/10/30/e2.png](https://assets.leetcode.com/uploads/2019/10/30/e2.png)
+
+    Input: n = 2, m = 2, indices = [[1,1],[0,0]]
+    Output: 0
+    Explanation: Final matrix = [[2,2],[2,2]]. There is no odd number in the final matrix.
+
+**Constraints**:
+
+- `1 <= n <= 50`
+- `1 <= m <= 50`
+- `1 <= indices.length <= 100`
+- `0 <= indices[i][0] < n`
+- `0 <= indices[i][1] < m`
+
+## 题目大意
+
+
+给你一个 n 行 m 列的矩阵，最开始的时候，每个单元格中的值都是 0。另有一个索引数组 indices，indices[i] = [ri, ci] 中的 ri 和 ci 分别表示指定的行和列（从 0 开始编号）。你需要将每对 [ri, ci] 指定的行和列上的所有单元格的值加 1。请你在执行完所有 indices 指定的增量操作后，返回矩阵中 「奇数值单元格」 的数目。
+
+提示：
+
+- 1 <= n <= 50
+- 1 <= m <= 50
+- 1 <= indices.length <= 100
+- 0 <= indices[i][0] < n
+- 0 <= indices[i][1] < m
+
+
+## 解题思路
+
+- 给出一个 n * m 的矩阵，和一个数组，数组里面包含一些行列坐标，并在指定坐标上 + 1，问最后 n * m 的矩阵中奇数的总数。
+- 暴力方法按照题意模拟即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 暴力法
+func oddCells(n int, m int, indices [][]int) int {
+	matrix, res := make([][]int, n), 0
+	for i := range matrix {
+		matrix[i] = make([]int, m)
+	}
+	for _, indice := range indices {
+		for i := 0; i < m; i++ {
+			matrix[indice[0]][i]++
+		}
+		for j := 0; j < n; j++ {
+			matrix[j][indice[1]]++
+		}
+	}
+	for _, m := range matrix {
+		for _, v := range m {
+			if v&1 == 1 {
+				res++
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 暴力法
+func oddCells1(n int, m int, indices [][]int) int {
+	rows, cols, count := make([]int, n), make([]int, m), 0
+	for _, pair := range indices {
+		rows[pair[0]]++
+		cols[pair[1]]++
+	}
+	for i := 0; i < n; i++ {
+		for j := 0; j < m; j++ {
+			if (rows[i]+cols[j])%2 == 1 {
+				count++
+			}
+		}
+	}
+	return count
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1254.Number-of-Closed-Islands/1254. Number of Closed Islands.go b/leetcode/1254.Number-of-Closed-Islands/1254. Number of Closed Islands.go
new file mode 100644
index 000000000..c7773dd68
--- /dev/null
+++ b/leetcode/1254.Number-of-Closed-Islands/1254. Number of Closed Islands.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func closedIsland(grid [][]int) int {
+	m := len(grid)
+	if m == 0 {
+		return 0
+	}
+	n := len(grid[0])
+	if n == 0 {
+		return 0
+	}
+	res, visited := 0, make([][]bool, m)
+	for i := 0; i < m; i++ {
+		visited[i] = make([]bool, n)
+	}
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			isEdge := false
+			if grid[i][j] == 0 && !visited[i][j] {
+				checkIslands(grid, &visited, i, j, &isEdge)
+				if !isEdge {
+					res++
+				}
+
+			}
+		}
+	}
+	return res
+}
+
+func checkIslands(grid [][]int, visited *[][]bool, x, y int, isEdge *bool) {
+	if (x == 0 || x == len(grid)-1 || y == 0 || y == len(grid[0])-1) && grid[x][y] == 0 {
+		*isEdge = true
+	}
+	(*visited)[x][y] = true
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		if isIntInBoard(grid, nx, ny) && !(*visited)[nx][ny] && grid[nx][ny] == 0 {
+			checkIslands(grid, visited, nx, ny, isEdge)
+		}
+	}
+	*isEdge = *isEdge || false
+}
+
+func isIntInBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
diff --git a/leetcode/1254.Number-of-Closed-Islands/1254. Number of Closed Islands_test.go b/leetcode/1254.Number-of-Closed-Islands/1254. Number of Closed Islands_test.go
new file mode 100644
index 000000000..a91cfeabf
--- /dev/null
+++ b/leetcode/1254.Number-of-Closed-Islands/1254. Number of Closed Islands_test.go	
@@ -0,0 +1,90 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1254 struct {
+	para1254
+	ans1254
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1254 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1254 struct {
+	one int
+}
+
+func Test_Problem1254(t *testing.T) {
+
+	qs := []question1254{
+
+		{
+			para1254{[][]int{
+				{1, 1, 1, 1, 0},
+				{1, 1, 0, 1, 0},
+				{1, 1, 0, 0, 0},
+				{0, 0, 0, 0, 0},
+			}},
+			ans1254{0},
+		},
+
+		{
+			para1254{[][]int{
+				{1, 1, 0, 0, 0},
+				{1, 1, 0, 0, 0},
+				{0, 0, 1, 0, 0},
+				{0, 0, 0, 1, 1},
+			}},
+			ans1254{0},
+		},
+
+		{
+			para1254{[][]int{
+				{1, 1, 1, 1, 1, 1, 1, 0},
+				{1, 0, 0, 0, 0, 1, 1, 0},
+				{1, 0, 1, 0, 1, 1, 1, 0},
+				{1, 0, 0, 0, 0, 1, 0, 1},
+				{1, 1, 1, 1, 1, 1, 1, 0},
+			}},
+			ans1254{2},
+		},
+
+		{
+			para1254{[][]int{
+				{0, 0, 1, 0, 0},
+				{0, 1, 0, 1, 0},
+				{0, 1, 1, 1, 0},
+			}},
+			ans1254{1},
+		},
+
+		{
+			para1254{[][]int{
+				{1, 1, 1, 1, 1, 1, 1},
+				{1, 0, 0, 0, 0, 0, 1},
+				{1, 0, 1, 1, 1, 0, 1},
+				{1, 0, 1, 0, 1, 0, 1},
+				{1, 0, 1, 1, 1, 0, 1},
+				{1, 0, 0, 0, 0, 0, 1},
+				{1, 1, 1, 1, 1, 1, 1},
+			}},
+			ans1254{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1254------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1254, q.para1254
+		fmt.Printf("【input】:%v       【output】:%v\n", p, closedIsland(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1254.Number-of-Closed-Islands/README.md b/leetcode/1254.Number-of-Closed-Islands/README.md
new file mode 100755
index 000000000..3ff8da1a2
--- /dev/null
+++ b/leetcode/1254.Number-of-Closed-Islands/README.md
@@ -0,0 +1,110 @@
+# [1254. Number of Closed Islands](https://leetcode.com/problems/number-of-closed-islands/)
+
+
+## 题目
+
+Given a 2D `grid` consists of `0s` (land) and `1s` (water). An *island* is a maximal 4-directionally connected group of `0s` and a *closed island* is an island **totally** (all left, top, right, bottom) surrounded by `1s.`
+
+Return the number of *closed islands*.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/10/31/sample_3_1610.png](https://assets.leetcode.com/uploads/2019/10/31/sample_3_1610.png)
+
+    Input: grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
+	Output: 2
+	Explanation: 
+	Islands in gray are closed because they are completely surrounded by water (group of 1s).
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/10/31/sample_4_1610.png](https://assets.leetcode.com/uploads/2019/10/31/sample_4_1610.png)
+
+    Input: grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
+	Output: 1
+
+**Example 3**:
+
+    Input: grid = [[1,1,1,1,1,1,1],
+               [1,0,0,0,0,0,1],
+               [1,0,1,1,1,0,1],
+               [1,0,1,0,1,0,1],
+               [1,0,1,1,1,0,1],
+               [1,0,0,0,0,0,1],
+               [1,1,1,1,1,1,1]]
+	Output: 2
+
+**Constraints**:
+
+- `1 <= grid.length, grid[0].length <= 100`
+- `0 <= grid[i][j] <=1`
+
+## 题目大意
+
+有一个二维矩阵 grid ，每个位置要么是陆地（记号为 0 ）要么是水域（记号为 1 ）。我们从一块陆地出发，每次可以往上下左右 4 个方向相邻区域走，能走到的所有陆地区域，我们将其称为一座「岛屿」。如果一座岛屿 完全 由水域包围，即陆地边缘上下左右所有相邻区域都是水域，那么我们将其称为 「封闭岛屿」。请返回封闭岛屿的数目。
+
+提示：
+
+- 1 <= grid.length, grid[0].length <= 100
+- 0 <= grid[i][j] <=1
+
+
+## 解题思路
+
+- 给出一个地图，1 代表海水，0 代表陆地。要求找出四周都是海水的陆地的总个数。
+- 这一题和第 200 题解题思路完全一致。只不过这一题要求必须四周都是海水，第 200 题的陆地可以是靠着地图边缘的。在此题中，靠着地图边缘的陆地不能最终计数到结果中。
+
+## 代码
+
+```go
+
+package leetcode
+
+func closedIsland(grid [][]int) int {
+	m := len(grid)
+	if m == 0 {
+		return 0
+	}
+	n := len(grid[0])
+	if n == 0 {
+		return 0
+	}
+	res, visited := 0, make([][]bool, m)
+	for i := 0; i < m; i++ {
+		visited[i] = make([]bool, n)
+	}
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			isEdge := false
+			if grid[i][j] == 0 && !visited[i][j] {
+				checkIslands(grid, &visited, i, j, &isEdge)
+				if !isEdge {
+					res++
+				}
+
+			}
+		}
+	}
+	return res
+}
+
+func checkIslands(grid [][]int, visited *[][]bool, x, y int, isEdge *bool) {
+	if (x == 0 || x == len(grid)-1 || y == 0 || y == len(grid[0])-1) && grid[x][y] == 0 {
+		*isEdge = true
+	}
+	(*visited)[x][y] = true
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		if isIntInBoard(grid, nx, ny) && !(*visited)[nx][ny] && grid[nx][ny] == 0 {
+			checkIslands(grid, visited, nx, ny, isEdge)
+		}
+	}
+	*isEdge = *isEdge || false
+}
+
+func isIntInBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1260.Shift-2D-Grid/1260. Shift 2D Grid.go b/leetcode/1260.Shift-2D-Grid/1260. Shift 2D Grid.go
new file mode 100644
index 000000000..c4bcceca2
--- /dev/null
+++ b/leetcode/1260.Shift-2D-Grid/1260. Shift 2D Grid.go	
@@ -0,0 +1,19 @@
+package leetcode
+
+func shiftGrid(grid [][]int, k int) [][]int {
+	x, y := len(grid[0]), len(grid)
+	newGrid := make([][]int, y)
+	for i := 0; i < y; i++ {
+		newGrid[i] = make([]int, x)
+	}
+	for i := 0; i < y; i++ {
+		for j := 0; j < x; j++ {
+			ny := (k / x) + i
+			if (j + (k % x)) >= x {
+				ny++
+			}
+			newGrid[ny%y][(j+(k%x))%x] = grid[i][j]
+		}
+	}
+	return newGrid
+}
diff --git a/leetcode/1260.Shift-2D-Grid/1260. Shift 2D Grid_test.go b/leetcode/1260.Shift-2D-Grid/1260. Shift 2D Grid_test.go
new file mode 100644
index 000000000..8b4167e1e
--- /dev/null
+++ b/leetcode/1260.Shift-2D-Grid/1260. Shift 2D Grid_test.go	
@@ -0,0 +1,59 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1260 struct {
+	para1260
+	ans1260
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1260 struct {
+	grid [][]int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1260 struct {
+	one [][]int
+}
+
+func Test_Problem1260(t *testing.T) {
+
+	qs := []question1260{
+
+		{
+			para1260{[][]int{{3, 7, 8}, {9, 11, 13}, {15, 16, 17}}, 2},
+			ans1260{[][]int{{16, 17, 3}, {7, 8, 9}, {11, 13, 15}}},
+		},
+
+		{
+			para1260{[][]int{{1, 10, 4, 2}, {9, 3, 8, 7}, {15, 16, 17, 12}}, 10},
+			ans1260{[][]int{{4, 2, 9, 3}, {8, 7, 15, 16}, {17, 12, 1, 10}}},
+		},
+
+		{
+			para1260{[][]int{{3, 8, 1, 9}, {19, 7, 2, 5}, {4, 6, 11, 10}, {12, 0, 21, 13}}, 4},
+			ans1260{[][]int{{12, 0, 21, 13}, {3, 8, 1, 9}, {19, 7, 2, 5}, {4, 6, 11, 10}}},
+		},
+
+		{
+			para1260{[][]int{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 9},
+			ans1260{[][]int{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1260------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1260, q.para1260
+		fmt.Printf("【input】:%v       【output】:%v\n", p, shiftGrid(p.grid, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1260.Shift-2D-Grid/README.md b/leetcode/1260.Shift-2D-Grid/README.md
new file mode 100644
index 000000000..df969a8bf
--- /dev/null
+++ b/leetcode/1260.Shift-2D-Grid/README.md
@@ -0,0 +1,86 @@
+# [1260. Shift 2D Grid](https://leetcode.com/problems/shift-2d-grid/)
+
+
+## 题目
+
+Given a 2D `grid` of size `m x n` and an integer `k`. You need to shift the `grid` `k` times.
+
+In one shift operation:
+
+- Element at `grid[i][j]` moves to `grid[i][j + 1]`.
+- Element at `grid[i][n - 1]` moves to `grid[i + 1][0]`.
+- Element at `grid[m - 1][n - 1]` moves to `grid[0][0]`.
+
+Return the *2D grid* after applying shift operation `k` times.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/11/05/e1.png](https://assets.leetcode.com/uploads/2019/11/05/e1.png)
+
+```
+Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1
+Output: [[9,1,2],[3,4,5],[6,7,8]]
+```
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/11/05/e2.png](https://assets.leetcode.com/uploads/2019/11/05/e2.png)
+
+```
+Input: grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4
+Output: [[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]
+```
+
+**Example 3**:
+
+```
+Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9
+Output: [[1,2,3],[4,5,6],[7,8,9]]
+```
+
+**Constraints**:
+
+- `m == grid.length`
+- `n == grid[i].length`
+- `1 <= m <= 50`
+- `1 <= n <= 50`
+- `-1000 <= grid[i][j] <= 1000`
+- `0 <= k <= 100`
+
+## 题目大意
+
+给你一个 m 行 n 列的二维网格 grid 和一个整数 k。你需要将 grid 迁移 k 次。每次「迁移」操作将会引发下述活动：
+
+- 位于 grid[i][j] 的元素将会移动到 grid[i][j + 1]。
+- 位于 grid[i][n - 1] 的元素将会移动到 grid[i + 1][0]。
+- 位于 grid[m - 1][n - 1] 的元素将会移动到 grid[0][0]。
+
+请你返回 k 次迁移操作后最终得到的 二维网格。
+
+
+## 解题思路
+
+- 给一个矩阵和一个移动步数 k，要求把矩阵每个元素往后移动 k 步，最后的元素移动头部，循环移动，最后输出移动结束的矩阵。
+- 简单题，按照题意循环移动即可，注意判断边界情况。
+
+## 代码
+
+```go
+func shiftGrid(grid [][]int, k int) [][]int {
+	x, y := len(grid[0]), len(grid)
+	newGrid := make([][]int, y)
+	for i := 0; i < y; i++ {
+		newGrid[i] = make([]int, x)
+	}
+	for i := 0; i < y; i++ {
+		for j := 0; j < x; j++ {
+			ny := (k / x) + i
+			if (j + (k % x)) >= x {
+				ny++
+			}
+			newGrid[ny%y][(j+(k%x))%x] = grid[i][j]
+		}
+	}
+	return newGrid
+}
+```
\ No newline at end of file
diff --git a/leetcode/1266.Minimum-Time-Visiting-All-Points/1266. Minimum Time Visiting All Points.go b/leetcode/1266.Minimum-Time-Visiting-All-Points/1266. Minimum Time Visiting All Points.go
new file mode 100644
index 000000000..7132b268c
--- /dev/null
+++ b/leetcode/1266.Minimum-Time-Visiting-All-Points/1266. Minimum Time Visiting All Points.go	
@@ -0,0 +1,23 @@
+package leetcode
+
+func minTimeToVisitAllPoints(points [][]int) int {
+	res := 0
+	for i := 1; i < len(points); i++ {
+		res += max(abs(points[i][0]-points[i-1][0]), abs(points[i][1]-points[i-1][1]))
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
diff --git a/leetcode/1266.Minimum-Time-Visiting-All-Points/1266. Minimum Time Visiting All Points_test.go b/leetcode/1266.Minimum-Time-Visiting-All-Points/1266. Minimum Time Visiting All Points_test.go
new file mode 100644
index 000000000..954616465
--- /dev/null
+++ b/leetcode/1266.Minimum-Time-Visiting-All-Points/1266. Minimum Time Visiting All Points_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1266 struct {
+	para1266
+	ans1266
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1266 struct {
+	points [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1266 struct {
+	one int
+}
+
+func Test_Problem1266(t *testing.T) {
+
+	qs := []question1266{
+
+		{
+			para1266{[][]int{{1, 1}, {3, 4}, {-1, 0}}},
+			ans1266{7},
+		},
+
+		{
+			para1266{[][]int{{3, 2}, {-2, 2}}},
+			ans1266{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1266------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1266, q.para1266
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minTimeToVisitAllPoints(p.points))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1266.Minimum-Time-Visiting-All-Points/README.md b/leetcode/1266.Minimum-Time-Visiting-All-Points/README.md
new file mode 100755
index 000000000..95d69b9dd
--- /dev/null
+++ b/leetcode/1266.Minimum-Time-Visiting-All-Points/README.md
@@ -0,0 +1,74 @@
+# [1266. Minimum Time Visiting All Points](https://leetcode.com/problems/minimum-time-visiting-all-points/)
+
+
+## 题目
+
+On a plane there are `n` points with integer coordinates `points[i] = [xi, yi]`. Your task is to find the minimum time in seconds to visit all points.
+
+You can move according to the next rules:
+
+- In one second always you can either move vertically, horizontally by one unit or diagonally (it means to move one unit vertically and one unit horizontally in one second).
+- You have to visit the points in the same order as they appear in the array.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/11/14/1626_example_1.PNG](https://assets.leetcode.com/uploads/2019/11/14/1626_example_1.PNG)
+
+    Input: points = [[1,1],[3,4],[-1,0]]
+    Output: 7
+    Explanation: One optimal path is [1,1] -> [2,2] -> [3,3] -> [3,4] -> [2,3] -> [1,2] -> [0,1] -> [-1,0]   
+    Time from [1,1] to [3,4] = 3 seconds 
+    Time from [3,4] to [-1,0] = 4 seconds
+    Total time = 7 seconds
+
+**Example 2:**
+
+    Input: points = [[3,2],[-2,2]]
+    Output: 5
+
+**Constraints:**
+
+- `points.length == n`
+- `1 <= n <= 100`
+- `points[i].length == 2`
+- `-1000 <= points[i][0], points[i][1] <= 1000`
+
+## 题目大意
+
+
+平面上有 n 个点，点的位置用整数坐标表示 points[i] = [xi, yi]。请你计算访问所有这些点需要的最小时间（以秒为单位）。你可以按照下面的规则在平面上移动：
+
+- 每一秒沿水平或者竖直方向移动一个单位长度，或者跨过对角线（可以看作在一秒内向水平和竖直方向各移动一个单位长度）。
+- 必须按照数组中出现的顺序来访问这些点。
+
+提示：
+
+- points.length == n
+- 1 <= n <= 100
+- points[i].length == 2
+- -1000 <= points[i][0], points[i][1] <= 1000
+
+
+
+
+
+## 解题思路
+
+- 在直角坐标系上给出一个数组，数组里面的点是飞机飞行经过的点。飞机飞行只能沿着水平方向、垂直方向、45°方向飞行。问飞机经过所有点的最短时间。
+- 简单的数学问题。依次遍历数组，分别计算 x 轴和 y 轴上的差值，取最大值即是这两点之间飞行的最短时间。最后累加每次计算的最大值就是最短时间。
+
+## 代码
+
+```go
+
+package leetcode
+
+func minTimeToVisitAllPoints(points [][]int) int {
+	res := 0
+	for i := 1; i < len(points); i++ {
+		res += max(abs(points[i][0]-points[i-1][0]), abs(points[i][1]-points[i-1][1]))
+	}
+	return res
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1268.Search-Suggestions-System/1268. Search Suggestions System.go b/leetcode/1268.Search-Suggestions-System/1268. Search Suggestions System.go
new file mode 100644
index 000000000..d9ba6b502
--- /dev/null
+++ b/leetcode/1268.Search-Suggestions-System/1268. Search Suggestions System.go	
@@ -0,0 +1,22 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+func suggestedProducts(products []string, searchWord string) [][]string {
+	sort.Strings(products)
+	searchWordBytes, result := []byte(searchWord), make([][]string, 0, len(searchWord))
+	for i := 1; i <= len(searchWord); i++ {
+		searchWordBytes[i-1]++
+		products = products[:sort.SearchStrings(products, string(searchWordBytes[:i]))]
+		searchWordBytes[i-1]--
+		products = products[sort.SearchStrings(products, searchWord[:i]):]
+		if len(products) > 3 {
+			result = append(result, products[:3])
+		} else {
+			result = append(result, products)
+		}
+	}
+	return result
+}
diff --git a/leetcode/1268.Search-Suggestions-System/1268. Search Suggestions System_test.go b/leetcode/1268.Search-Suggestions-System/1268. Search Suggestions System_test.go
new file mode 100644
index 000000000..bc2be9a48
--- /dev/null
+++ b/leetcode/1268.Search-Suggestions-System/1268. Search Suggestions System_test.go	
@@ -0,0 +1,70 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1268 struct {
+	para1268
+	ans1268
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1268 struct {
+	products   []string
+	searchWord string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1268 struct {
+	one [][]string
+}
+
+func Test_Problem1268(t *testing.T) {
+
+	qs := []question1268{
+
+		{
+			para1268{[]string{"bags", "baggage", "banner", "box", "cloths"}, "bags"},
+			ans1268{[][]string{
+				{"baggage", "bags", "banner"}, {"baggage", "bags", "banner"}, {"baggage", "bags"}, {"bags"},
+			}},
+		},
+
+		{
+			para1268{[]string{"mobile", "mouse", "moneypot", "monitor", "mousepad"}, "mouse"},
+			ans1268{[][]string{
+				{"mobile", "moneypot", "monitor"},
+				{"mobile", "moneypot", "monitor"},
+				{"mouse", "mousepad"},
+				{"mouse", "mousepad"},
+				{"mouse", "mousepad"},
+			}},
+		},
+
+		{
+			para1268{[]string{"havana"}, "havana"},
+			ans1268{[][]string{
+				{"havana"}, {"havana"}, {"havana"}, {"havana"}, {"havana"}, {"havana"},
+			}},
+		},
+
+		{
+			para1268{[]string{"havana"}, "tatiana"},
+			ans1268{[][]string{
+				{}, {}, {}, {}, {}, {}, {},
+			}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1268------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1268, q.para1268
+		fmt.Printf("【input】:%v       【output】:%v\n", p, suggestedProducts(p.products, p.searchWord))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1268.Search-Suggestions-System/README.md b/leetcode/1268.Search-Suggestions-System/README.md
new file mode 100644
index 000000000..769f4d7a8
--- /dev/null
+++ b/leetcode/1268.Search-Suggestions-System/README.md
@@ -0,0 +1,89 @@
+# [1268. Search Suggestions System](https://leetcode.com/problems/search-suggestions-system/)
+
+## 题目
+
+Given an array of strings `products` and a string `searchWord`. We want to design a system that suggests at most three product names from `products` after each character of `searchWord` is typed. Suggested products should have common prefix with the searchWord. If there are more than three products with a common prefix return the three lexicographically minimums products.
+
+Return *list of lists* of the suggested `products` after each character of `searchWord` is typed.
+
+**Example 1:**
+
+```
+Input: products = ["mobile","mouse","moneypot","monitor","mousepad"], searchWord = "mouse"
+Output: [
+["mobile","moneypot","monitor"],
+["mobile","moneypot","monitor"],
+["mouse","mousepad"],
+["mouse","mousepad"],
+["mouse","mousepad"]
+]
+Explanation: products sorted lexicographically = ["mobile","moneypot","monitor","mouse","mousepad"]
+After typing m and mo all products match and we show user ["mobile","moneypot","monitor"]
+After typing mou, mous and mouse the system suggests ["mouse","mousepad"]
+
+```
+
+**Example 2:**
+
+```
+Input: products = ["havana"], searchWord = "havana"
+Output: [["havana"],["havana"],["havana"],["havana"],["havana"],["havana"]]
+```
+
+**Example 3:**
+
+```
+Input: products = ["bags","baggage","banner","box","cloths"], searchWord = "bags"
+Output: [["baggage","bags","banner"],["baggage","bags","banner"],["baggage","bags"],["bags"]]
+```
+
+**Example 4:**
+
+```
+Input: products = ["havana"], searchWord = "tatiana"
+Output: [[],[],[],[],[],[],[]]
+```
+
+**Constraints:**
+
+- `1 <= products.length <= 1000`
+- There are no repeated elements in `products`.
+- `1 <= Σ products[i].length <= 2 * 10^4`
+- All characters of `products[i]` are lower-case English letters.
+- `1 <= searchWord.length <= 1000`
+- All characters of `searchWord` are lower-case English letters.
+
+## 题目大意
+
+给你一个产品数组 products 和一个字符串 searchWord ，products  数组中每个产品都是一个字符串。请你设计一个推荐系统，在依次输入单词 searchWord 的每一个字母后，推荐 products 数组中前缀与 searchWord 相同的最多三个产品。如果前缀相同的可推荐产品超过三个，请按字典序返回最小的三个。请你以二维列表的形式，返回在输入 searchWord 每个字母后相应的推荐产品的列表。
+
+## 解题思路
+
+- 由于题目要求返回的答案要按照字典序输出，所以先排序。有序字符串又满足了二分搜索的条件，于是可以用二分搜索。sort.SearchStrings 返回的是满足搜索条件的第一个起始下标。末尾不满足条件的字符串要切掉。所以要搜 2 次，第一次二分搜索先将不满足目标串前缀的字符串筛掉。第二次二分搜索再搜索出最终满足题意的字符串。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func suggestedProducts(products []string, searchWord string) [][]string {
+	sort.Strings(products)
+	searchWordBytes, result := []byte(searchWord), make([][]string, 0, len(searchWord))
+	for i := 1; i <= len(searchWord); i++ {
+		searchWordBytes[i-1]++
+		products = products[:sort.SearchStrings(products, string(searchWordBytes[:i]))]
+		searchWordBytes[i-1]--
+		products = products[sort.SearchStrings(products, searchWord[:i]):]
+		if len(products) > 3 {
+			result = append(result, products[:3])
+		} else {
+			result = append(result, products)
+		}
+	}
+	return result
+}
+```
\ No newline at end of file
diff --git a/leetcode/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/1275. Find Winner on a Tic Tac Toe Game.go b/leetcode/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/1275. Find Winner on a Tic Tac Toe Game.go
new file mode 100644
index 000000000..bea155147
--- /dev/null
+++ b/leetcode/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/1275. Find Winner on a Tic Tac Toe Game.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+func tictactoe(moves [][]int) string {
+	board := [3][3]byte{}
+	for i := 0; i < len(moves); i++ {
+		if i%2 == 0 {
+			board[moves[i][0]][moves[i][1]] = 'X'
+		} else {
+			board[moves[i][0]][moves[i][1]] = 'O'
+		}
+	}
+	for i := 0; i < 3; i++ {
+		if board[i][0] == 'X' && board[i][1] == 'X' && board[i][2] == 'X' {
+			return "A"
+		}
+		if board[i][0] == 'O' && board[i][1] == 'O' && board[i][2] == 'O' {
+			return "B"
+		}
+		if board[0][i] == 'X' && board[1][i] == 'X' && board[2][i] == 'X' {
+			return "A"
+		}
+		if board[0][i] == 'O' && board[1][i] == 'O' && board[2][i] == 'O' {
+			return "B"
+		}
+	}
+	if board[0][0] == 'X' && board[1][1] == 'X' && board[2][2] == 'X' {
+		return "A"
+	}
+	if board[0][0] == 'O' && board[1][1] == 'O' && board[2][2] == 'O' {
+		return "B"
+	}
+	if board[0][2] == 'X' && board[1][1] == 'X' && board[2][0] == 'X' {
+		return "A"
+	}
+	if board[0][2] == 'O' && board[1][1] == 'O' && board[2][0] == 'O' {
+		return "B"
+	}
+	if len(moves) < 9 {
+		return "Pending"
+	}
+	return "Draw"
+}
diff --git a/leetcode/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/1275. Find Winner on a Tic Tac Toe Game_test.go b/leetcode/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/1275. Find Winner on a Tic Tac Toe Game_test.go
new file mode 100644
index 000000000..8502ddef5
--- /dev/null
+++ b/leetcode/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/1275. Find Winner on a Tic Tac Toe Game_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1275 struct {
+	para1275
+	ans1275
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1275 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1275 struct {
+	one string
+}
+
+func Test_Problem1275(t *testing.T) {
+
+	qs := []question1275{
+
+		{
+			para1275{[][]int{{0, 0}, {2, 0}, {1, 1}, {2, 1}, {2, 2}}},
+			ans1275{"A"},
+		},
+
+		{
+			para1275{[][]int{{0, 0}, {1, 1}, {0, 1}, {0, 2}, {1, 0}, {2, 0}}},
+			ans1275{"B"},
+		},
+
+		{
+			para1275{[][]int{{0, 0}, {1, 1}, {2, 0}, {1, 0}, {1, 2}, {2, 1}, {0, 1}, {0, 2}, {2, 2}}},
+			ans1275{"Draw"},
+		},
+
+		{
+			para1275{[][]int{{0, 0}, {1, 1}}},
+			ans1275{"Pending"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1275------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1275, q.para1275
+		fmt.Printf("【input】:%v       【output】:%v\n", p, tictactoe(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/README.md b/leetcode/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/README.md
new file mode 100644
index 000000000..d66e98ca9
--- /dev/null
+++ b/leetcode/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/README.md
@@ -0,0 +1,150 @@
+# [1275. Find Winner on a Tic Tac Toe Game](https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/)
+
+
+## 题目
+
+Tic-tac-toe is played by two players *A* and *B* on a 3 x 3 grid.
+
+Here are the rules of Tic-Tac-Toe:
+
+- Players take turns placing characters into empty squares (" ").
+- The first player *A* always places "X" characters, while the second player *B* always places "O" characters.
+- "X" and "O" characters are always placed into empty squares, never on filled ones.
+- The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
+- The game also ends if all squares are non-empty.
+- No more moves can be played if the game is over.
+
+Given an array `moves` where each element is another array of size 2 corresponding to the row and column of the grid where they mark their respective character in the order in which *A* and *B* play.
+
+Return the winner of the game if it exists (*A* or *B*), in case the game ends in a draw return "Draw", if there are still movements to play return "Pending".
+
+You can assume that `moves` is **valid** (It follows the rules of Tic-Tac-Toe), the grid is initially empty and *A* will play **first**.
+
+**Example 1**:
+
+```
+Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
+Output: "A"
+Explanation: "A" wins, he always plays first.
+"X  "    "X  "    "X  "    "X  "    "X  "
+"   " -> "   " -> " X " -> " X " -> " X "
+"   "    "O  "    "O  "    "OO "    "OOX"
+
+```
+
+**Example 2**:
+
+```
+Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
+Output: "B"
+Explanation: "B" wins.
+"X  "    "X  "    "XX "    "XXO"    "XXO"    "XXO"
+"   " -> " O " -> " O " -> " O " -> "XO " -> "XO " 
+"   "    "   "    "   "    "   "    "   "    "O  "
+
+```
+
+**Example 3**:
+
+```
+Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
+Output: "Draw"
+Explanation: The game ends in a draw since there are no moves to make.
+"XXO"
+"OOX"
+"XOX"
+
+```
+
+**Example 4**:
+
+```
+Input: moves = [[0,0],[1,1]]
+Output: "Pending"
+Explanation: The game has not finished yet.
+"X  "
+" O "
+"   "
+
+```
+
+**Constraints:**
+
+- `1 <= moves.length <= 9`
+- `moves[i].length == 2`
+- `0 <= moves[i][j] <= 2`
+- There are no repeated elements on `moves`.
+- `moves` follow the rules of tic tac toe.
+
+
+## 题目大意
+
+A 和 B 在一个 3 x 3 的网格上玩井字棋。井字棋游戏的规则如下：
+
+- 玩家轮流将棋子放在空方格 (" ") 上。
+- 第一个玩家 A 总是用 "X" 作为棋子，而第二个玩家 B 总是用 "O" 作为棋子。
+- "X" 和 "O" 只能放在空方格中，而不能放在已经被占用的方格上。
+- 只要有 3 个相同的（非空）棋子排成一条直线（行、列、对角线）时，游戏结束。
+- 如果所有方块都放满棋子（不为空），游戏也会结束。
+- 游戏结束后，棋子无法再进行任何移动。
+
+给你一个数组 moves，其中每个元素是大小为 2 的另一个数组（元素分别对应网格的行和列），它按照 A 和 B 的行动顺序（先 A 后 B）记录了两人各自的棋子位置。如果游戏存在获胜者（A 或 B），就返回该游戏的获胜者；如果游戏以平局结束，则返回 "Draw"；如果仍会有行动（游戏未结束），则返回 "Pending"。你可以假设 moves 都 有效（遵循井字棋规则），网格最初是空的，A 将先行动。
+
+提示：
+
+- 1 <= moves.length <= 9
+- moves[i].length == 2
+- 0 <= moves[i][j] <= 2
+- moves 里没有重复的元素。
+- moves 遵循井字棋的规则。
+
+
+## 解题思路
+
+- 两人玩 3*3 井字棋，A 先走，B 再走。谁能获胜就输出谁，如果平局输出 “Draw”，如果游戏还未结束，输出 “Pending”。游戏规则：谁能先占满行、列或者对角线任意一条线，谁就赢。
+- 简单题。题目给定 move 数组最多 3 步，而要赢得比赛，必须走满 3 步，所以可以先模拟，按照给的步数数组把 A 和 B 的步数都放在棋盘上。然后依次判断行、列，对角线的三种情况。如果都判完了，剩下的情况就是平局和死局的情况。
+
+## 代码
+
+```go
+func tictactoe(moves [][]int) string {
+	board := [3][3]byte{}
+	for i := 0; i < len(moves); i++ {
+		if i%2 == 0 {
+			board[moves[i][0]][moves[i][1]] = 'X'
+		} else {
+			board[moves[i][0]][moves[i][1]] = 'O'
+		}
+	}
+	for i := 0; i < 3; i++ {
+		if board[i][0] == 'X' && board[i][1] == 'X' && board[i][2] == 'X' {
+			return "A"
+		}
+		if board[i][0] == 'O' && board[i][1] == 'O' && board[i][2] == 'O' {
+			return "B"
+		}
+		if board[0][i] == 'X' && board[1][i] == 'X' && board[2][i] == 'X' {
+			return "A"
+		}
+		if board[0][i] == 'O' && board[1][i] == 'O' && board[2][i] == 'O' {
+			return "B"
+		}
+	}
+	if board[0][0] == 'X' && board[1][1] == 'X' && board[2][2] == 'X' {
+		return "A"
+	}
+	if board[0][0] == 'O' && board[1][1] == 'O' && board[2][2] == 'O' {
+		return "B"
+	}
+	if board[0][2] == 'X' && board[1][1] == 'X' && board[2][0] == 'X' {
+		return "A"
+	}
+	if board[0][2] == 'O' && board[1][1] == 'O' && board[2][0] == 'O' {
+		return "B"
+	}
+	if len(moves) < 9 {
+		return "Pending"
+	}
+	return "Draw"
+}
+```
\ No newline at end of file
diff --git a/leetcode/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer/1281. Subtract the Product and Sum of Digits of an Integer.go b/leetcode/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer/1281. Subtract the Product and Sum of Digits of an Integer.go
new file mode 100644
index 000000000..fe6f5c0ec
--- /dev/null
+++ b/leetcode/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer/1281. Subtract the Product and Sum of Digits of an Integer.go	
@@ -0,0 +1,10 @@
+package leetcode
+
+func subtractProductAndSum(n int) int {
+	sum, product := 0, 1
+	for ; n > 0; n /= 10 {
+		sum += n % 10
+		product *= n % 10
+	}
+	return product - sum
+}
diff --git a/leetcode/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer/1281. Subtract the Product and Sum of Digits of an Integer_test.go b/leetcode/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer/1281. Subtract the Product and Sum of Digits of an Integer_test.go
new file mode 100644
index 000000000..ec98bb9fd
--- /dev/null
+++ b/leetcode/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer/1281. Subtract the Product and Sum of Digits of an Integer_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1281 struct {
+	para1281
+	ans1281
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1281 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1281 struct {
+	one int
+}
+
+func Test_Problem1281(t *testing.T) {
+
+	qs := []question1281{
+
+		{
+			para1281{234},
+			ans1281{15},
+		},
+
+		{
+			para1281{4421},
+			ans1281{21},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1281------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1281, q.para1281
+		fmt.Printf("【input】:%v       【output】:%v\n", p, subtractProductAndSum(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer/README.md b/leetcode/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer/README.md
new file mode 100644
index 000000000..96cc347d9
--- /dev/null
+++ b/leetcode/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer/README.md
@@ -0,0 +1,59 @@
+# [1281. Subtract the Product and Sum of Digits of an Integer](https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/)
+
+
+
+## 题目
+
+Given an integer number `n`, return the difference between the product of its digits and the sum of its digits.
+
+**Example 1**:
+
+```
+Input: n = 234
+Output: 15 
+Explanation: 
+Product of digits = 2 * 3 * 4 = 24 
+Sum of digits = 2 + 3 + 4 = 9 
+Result = 24 - 9 = 15
+```
+
+**Example 2**:
+
+```
+Input: n = 4421
+Output: 21
+Explanation: 
+Product of digits = 4 * 4 * 2 * 1 = 32 
+Sum of digits = 4 + 4 + 2 + 1 = 11 
+Result = 32 - 11 = 21
+```
+
+**Constraints**:
+
+- `1 <= n <= 10^5`
+
+## 题目大意
+
+给你一个整数 n，请你帮忙计算并返回该整数「各位数字之积」与「各位数字之和」的差。
+
+提示：
+
+- 1 <= n <= 10^5
+
+## 解题思路
+
+- 给出一个数，计算这个数每位数字乘积减去每位数字累加的差值。
+- 简单题，按照题意输入输出即可。
+
+## 代码
+
+```go
+func subtractProductAndSum(n int) int {
+	sum, product := 0, 1
+	for ; n > 0; n /= 10 {
+		sum += n % 10
+		product *= n % 10
+	}
+	return product - sum
+}
+```
\ No newline at end of file
diff --git a/leetcode/1283.Find-the-Smallest-Divisor-Given-a-Threshold/1283. Find the Smallest Divisor Given a Threshold.go b/leetcode/1283.Find-the-Smallest-Divisor-Given-a-Threshold/1283. Find the Smallest Divisor Given a Threshold.go
new file mode 100644
index 000000000..fe5b613c1
--- /dev/null
+++ b/leetcode/1283.Find-the-Smallest-Divisor-Given-a-Threshold/1283. Find the Smallest Divisor Given a Threshold.go	
@@ -0,0 +1,29 @@
+package leetcode
+
+func smallestDivisor(nums []int, threshold int) int {
+	low, high := 1, 1000000
+	for low < high {
+		mid := low + (high-low)>>1
+		if calDivisor(nums, mid, threshold) {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+func calDivisor(nums []int, mid, threshold int) bool {
+	sum := 0
+	for i := range nums {
+		if nums[i]%mid != 0 {
+			sum += nums[i]/mid + 1
+		} else {
+			sum += nums[i] / mid
+		}
+	}
+	if sum <= threshold {
+		return true
+	}
+	return false
+}
diff --git a/leetcode/1283.Find-the-Smallest-Divisor-Given-a-Threshold/1283. Find the Smallest Divisor Given a Threshold_test.go b/leetcode/1283.Find-the-Smallest-Divisor-Given-a-Threshold/1283. Find the Smallest Divisor Given a Threshold_test.go
new file mode 100644
index 000000000..e0bdbb271
--- /dev/null
+++ b/leetcode/1283.Find-the-Smallest-Divisor-Given-a-Threshold/1283. Find the Smallest Divisor Given a Threshold_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1283 struct {
+	para1283
+	ans1283
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1283 struct {
+	nums      []int
+	threshold int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1283 struct {
+	one int
+}
+
+func Test_Problem1283(t *testing.T) {
+
+	qs := []question1283{
+
+		{
+			para1283{[]int{1, 2, 5, 9}, 6},
+			ans1283{5},
+		},
+
+		{
+			para1283{[]int{2, 3, 5, 7, 11}, 11},
+			ans1283{3},
+		},
+
+		{
+			para1283{[]int{19}, 5},
+			ans1283{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1283------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1283, q.para1283
+		fmt.Printf("【input】:%v       【output】:%v\n", p, smallestDivisor(p.nums, p.threshold))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1283.Find-the-Smallest-Divisor-Given-a-Threshold/README.md b/leetcode/1283.Find-the-Smallest-Divisor-Given-a-Threshold/README.md
new file mode 100644
index 000000000..f0359723b
--- /dev/null
+++ b/leetcode/1283.Find-the-Smallest-Divisor-Given-a-Threshold/README.md
@@ -0,0 +1,87 @@
+# [1283. Find the Smallest Divisor Given a Threshold](https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/)
+
+
+
+## 题目
+
+Given an array of integers `nums` and an integer `threshold`, we will choose a positive integer divisor and divide all the array by it and sum the result of the division. Find the **smallest** divisor such that the result mentioned above is less than or equal to `threshold`.
+
+Each result of division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).
+
+It is guaranteed that there will be an answer.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,5,9], threshold = 6
+Output: 5
+Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1. 
+If the divisor is 4 we can get a sum to 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).
+```
+
+**Example 2**:
+
+```
+Input: nums = [2,3,5,7,11], threshold = 11
+Output: 3
+```
+
+**Example 3**:
+
+```
+Input: nums = [19], threshold = 5
+Output: 4
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 5 * 10^4`
+- `1 <= nums[i] <= 10^6`
+- `nums.length <= threshold <= 10^6`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个正整数 threshold  ，你需要选择一个正整数作为除数，然后将数组里每个数都除以它，并对除法结果求和。请你找出能够使上述结果小于等于阈值 threshold 的除数中 最小 的那个。每个数除以除数后都向上取整，比方说 7/3 = 3 ， 10/2 = 5 。题目保证一定有解。
+
+提示：
+
+- 1 <= nums.length <= 5 * 10^4
+- 1 <= nums[i] <= 10^6
+- nums.length <= threshold <= 10^6
+
+## 解题思路
+
+- 给出一个数组和一个阈值，要求找到一个除数，使得数组里面每个数和这个除数的商之和不超过这个阈值。求除数的最小值。
+- 这一题是典型的二分搜索的题目。根据题意，在 [1, 1000000] 区间内搜索除数，针对每次 `mid`，计算一次商的累加和。如果和比 `threshold` 小，说明除数太大，所以缩小右区间；如果和比 `threshold` 大，说明除数太小，所以缩小左区间。最终找到的 `low` 值就是最求的最小除数。
+
+## 代码
+
+```go
+func smallestDivisor(nums []int, threshold int) int {
+	low, high := 1, 1000000
+	for low < high {
+		mid := low + (high-low)>>1
+		if calDivisor(nums, mid, threshold) {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+func calDivisor(nums []int, mid, threshold int) bool {
+	sum := 0
+	for i := range nums {
+		if nums[i]%mid != 0 {
+			sum += nums[i]/mid + 1
+		} else {
+			sum += nums[i] / mid
+		}
+	}
+	if sum <= threshold {
+		return true
+	}
+	return false
+}
+```
\ No newline at end of file
diff --git a/leetcode/1287.Element-Appearing-More-Than-In-Sorted-Array/1287. Element Appearing More Than 25% In Sorted Array.go b/leetcode/1287.Element-Appearing-More-Than-In-Sorted-Array/1287. Element Appearing More Than 25% In Sorted Array.go
new file mode 100644
index 000000000..f3fa6cb78
--- /dev/null
+++ b/leetcode/1287.Element-Appearing-More-Than-In-Sorted-Array/1287. Element Appearing More Than 25% In Sorted Array.go	
@@ -0,0 +1,11 @@
+package leetcode
+
+func findSpecialInteger(arr []int) int {
+	n := len(arr)
+	for i := 0; i < n-n/4; i++ {
+		if arr[i] == arr[i+n/4] {
+			return arr[i]
+		}
+	}
+	return -1
+}
diff --git a/leetcode/1287.Element-Appearing-More-Than-In-Sorted-Array/1287. Element Appearing More Than 25% In Sorted Array_test.go b/leetcode/1287.Element-Appearing-More-Than-In-Sorted-Array/1287. Element Appearing More Than 25% In Sorted Array_test.go
new file mode 100644
index 000000000..fe7cea6fd
--- /dev/null
+++ b/leetcode/1287.Element-Appearing-More-Than-In-Sorted-Array/1287. Element Appearing More Than 25% In Sorted Array_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1287 struct {
+	para1287
+	ans1287
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1287 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1287 struct {
+	one int
+}
+
+func Test_Problem1287(t *testing.T) {
+
+	qs := []question1287{
+
+		{
+			para1287{[]int{1, 2, 2, 6, 6, 6, 6, 7, 10}},
+			ans1287{6},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1287------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1287, q.para1287
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findSpecialInteger(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1287.Element-Appearing-More-Than-In-Sorted-Array/README.md b/leetcode/1287.Element-Appearing-More-Than-In-Sorted-Array/README.md
new file mode 100644
index 000000000..f9714a4dd
--- /dev/null
+++ b/leetcode/1287.Element-Appearing-More-Than-In-Sorted-Array/README.md
@@ -0,0 +1,49 @@
+# [1287. Element Appearing More Than 25% In Sorted Array](https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/)
+
+
+
+## 题目
+
+Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time.
+
+Return that integer.
+
+**Example 1**:
+
+```
+Input: arr = [1,2,2,6,6,6,6,7,10]
+Output: 6
+```
+
+**Constraints**:
+
+- `1 <= arr.length <= 10^4`
+- `0 <= arr[i] <= 10^5`
+
+## 题目大意
+
+给你一个非递减的 有序 整数数组，已知这个数组中恰好有一个整数，它的出现次数超过数组元素总数的 25%。请你找到并返回这个整数。
+
+提示：
+
+- 1 <= arr.length <= 10^4
+- 0 <= arr[i] <= 10^5
+
+## 解题思路
+
+- 给出一个非递减的有序数组，要求输出出现次数超过数组元素总数 25% 的元素。
+- 简单题，由于已经非递减有序了，所以只需要判断 `arr[i] == arr[i+n/4]` 是否相等即可。
+
+## 代码
+
+```go
+func findSpecialInteger(arr []int) int {
+	n := len(arr)
+	for i := 0; i < n-n/4; i++ {
+		if arr[i] == arr[i+n/4] {
+			return arr[i]
+		}
+	}
+	return -1
+}
+```
\ No newline at end of file
diff --git a/leetcode/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer/1290. Convert Binary Number in a Linked List to Integer.go b/leetcode/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer/1290. Convert Binary Number in a Linked List to Integer.go
new file mode 100644
index 000000000..f58f4c896
--- /dev/null
+++ b/leetcode/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer/1290. Convert Binary Number in a Linked List to Integer.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func getDecimalValue(head *ListNode) int {
+	sum := 0
+	for head != nil {
+		sum = sum*2 + head.Val
+		head = head.Next
+	}
+	return sum
+}
diff --git a/leetcode/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer/1290. Convert Binary Number in a Linked List to Integer_test.go b/leetcode/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer/1290. Convert Binary Number in a Linked List to Integer_test.go
new file mode 100644
index 000000000..0a509b58a
--- /dev/null
+++ b/leetcode/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer/1290. Convert Binary Number in a Linked List to Integer_test.go	
@@ -0,0 +1,64 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1290 struct {
+	para1290
+	ans1290
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1290 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1290 struct {
+	one int
+}
+
+func Test_Problem1290(t *testing.T) {
+
+	qs := []question1290{
+
+		{
+			para1290{[]int{1, 0, 1}},
+			ans1290{5},
+		},
+
+		{
+			para1290{[]int{0}},
+			ans1290{0},
+		},
+
+		{
+			para1290{[]int{1}},
+			ans1290{1},
+		},
+
+		{
+			para1290{[]int{0, 0}},
+			ans1290{0},
+		},
+
+		{
+			para1290{[]int{1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0}},
+			ans1290{18880},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1290------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1290, q.para1290
+		fmt.Printf("【input】:%v       【output】:%v\n", p, getDecimalValue(structures.Ints2List(p.one)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer/README.md b/leetcode/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer/README.md
new file mode 100644
index 000000000..1c0f9be17
--- /dev/null
+++ b/leetcode/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer/README.md
@@ -0,0 +1,81 @@
+# [1290. Convert Binary Number in a Linked List to Integer](https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/)
+
+
+
+## 题目
+
+Given `head` which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary representation of a number.
+
+Return the *decimal value* of the number in the linked list.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/12/05/graph-1.png](https://assets.leetcode.com/uploads/2019/12/05/graph-1.png)
+
+```
+Input: head = [1,0,1]
+Output: 5
+Explanation: (101) in base 2 = (5) in base 10
+```
+
+**Example 2**:
+
+```
+Input: head = [0]
+Output: 0
+```
+
+**Example 3**:
+
+```
+Input: head = [1]
+Output: 1
+```
+
+**Example 4**:
+
+```
+Input: head = [1,0,0,1,0,0,1,1,1,0,0,0,0,0,0]
+Output: 18880
+```
+
+**Example 5**:
+
+```
+Input: head = [0,0]
+Output: 0
+```
+
+**Constraints**:
+
+- The Linked List is not empty.
+- Number of nodes will not exceed `30`.
+- Each node's value is either `0` or `1`.
+
+## 题目大意
+
+给你一个单链表的引用结点 head。链表中每个结点的值不是 0 就是 1。已知此链表是一个整数数字的二进制表示形式。请你返回该链表所表示数字的 十进制值 。
+
+提示：
+
+- 链表不为空。
+- 链表的结点总数不超过 30。
+- 每个结点的值不是 0 就是 1。
+
+## 解题思路
+
+- 给出一个链表，链表从头到尾表示的数是一个整数的二进制形式，要求输出这个整数的十进制。
+- 简单题，从头到尾遍历一次链表，边遍历边累加二进制位。
+
+## 代码
+
+```go
+func getDecimalValue(head *ListNode) int {
+	sum := 0
+	for head != nil {
+		sum = sum*2 + head.Val
+		head = head.Next
+	}
+	return sum
+}
+```
\ No newline at end of file
diff --git a/leetcode/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination/1293. Shortest Path in a Grid with Obstacles Elimination.go b/leetcode/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination/1293. Shortest Path in a Grid with Obstacles Elimination.go
new file mode 100644
index 000000000..71140a986
--- /dev/null
+++ b/leetcode/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination/1293. Shortest Path in a Grid with Obstacles Elimination.go	
@@ -0,0 +1,71 @@
+package leetcode
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+type pos struct {
+	x, y     int
+	obstacle int
+	step     int
+}
+
+func shortestPath(grid [][]int, k int) int {
+	queue, m, n := []pos{}, len(grid), len(grid[0])
+	visitor := make([][][]int, m)
+	if len(grid) == 1 && len(grid[0]) == 1 {
+		return 0
+	}
+	for i := 0; i < m; i++ {
+		visitor[i] = make([][]int, n)
+		for j := 0; j < n; j++ {
+			visitor[i][j] = make([]int, k+1)
+		}
+	}
+	visitor[0][0][0] = 1
+	queue = append(queue, pos{x: 0, y: 0, obstacle: 0, step: 0})
+	for len(queue) > 0 {
+		size := len(queue)
+		for size > 0 {
+			size--
+			node := queue[0]
+			queue = queue[1:]
+			for i := 0; i < len(dir); i++ {
+				newX := node.x + dir[i][0]
+				newY := node.y + dir[i][1]
+				if newX == m-1 && newY == n-1 {
+					if node.obstacle != 0 {
+						if node.obstacle <= k {
+							return node.step + 1
+						} else {
+							continue
+						}
+					}
+					return node.step + 1
+				}
+				if isInBoard(grid, newX, newY) {
+					if grid[newX][newY] == 1 {
+						if node.obstacle+1 <= k && visitor[newX][newY][node.obstacle+1] != 1 {
+							queue = append(queue, pos{x: newX, y: newY, obstacle: node.obstacle + 1, step: node.step + 1})
+							visitor[newX][newY][node.obstacle+1] = 1
+						}
+					} else {
+						if node.obstacle <= k && visitor[newX][newY][node.obstacle] != 1 {
+							queue = append(queue, pos{x: newX, y: newY, obstacle: node.obstacle, step: node.step + 1})
+							visitor[newX][newY][node.obstacle] = 1
+						}
+					}
+
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func isInBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
diff --git a/leetcode/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination/1293. Shortest Path in a Grid with Obstacles Elimination_test.go b/leetcode/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination/1293. Shortest Path in a Grid with Obstacles Elimination_test.go
new file mode 100644
index 000000000..c7e4c68dd
--- /dev/null
+++ b/leetcode/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination/1293. Shortest Path in a Grid with Obstacles Elimination_test.go	
@@ -0,0 +1,73 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1293 struct {
+	para1293
+	ans1293
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1293 struct {
+	grid [][]int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1293 struct {
+	one int
+}
+
+func Test_Problem1293(t *testing.T) {
+
+	qs := []question1293{
+
+		{
+			para1293{[][]int{
+				{0, 0, 0},
+			}, 1},
+			ans1293{2},
+		},
+
+		{
+			para1293{[][]int{
+				{0, 1, 1}, {0, 1, 1}, {0, 0, 0}, {0, 1, 0}, {0, 1, 0},
+			}, 2},
+			ans1293{6},
+		},
+
+		{
+			para1293{[][]int{
+				{0, 0, 0}, {1, 1, 0}, {0, 0, 0}, {0, 1, 1}, {0, 0, 0},
+			}, 1},
+			ans1293{6},
+		},
+
+		{
+			para1293{[][]int{
+				{0, 1, 1}, {1, 1, 1}, {1, 0, 0},
+			}, 1},
+			ans1293{-1},
+		},
+
+		{
+			para1293{[][]int{
+				{0},
+			}, 1},
+			ans1293{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1293------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1293, q.para1293
+		fmt.Printf("【input】:%v       【output】:%v\n", p, shortestPath(p.grid, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination/README.md b/leetcode/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination/README.md
new file mode 100644
index 000000000..0ef3fc05e
--- /dev/null
+++ b/leetcode/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination/README.md
@@ -0,0 +1,131 @@
+# [1293. Shortest Path in a Grid with Obstacles Elimination](https://leetcode.com/problems/shortest-path-in-a-grid-with-obstacles-elimination/)
+
+
+
+## 题目
+
+You are given an m x n integer matrix grid where each cell is either 0 (empty) or 1 (obstacle). You can move up, down, left, or right from and to an empty cell in one step.
+
+Return the minimum number of steps to walk from the upper left corner (0, 0) to the lower right corner (m - 1, n - 1) given that you can eliminate at most k obstacles. If it is not possible to find such walk return -1.
+
+
+
+Example 1:
+
+
+![](https://assets.leetcode.com/uploads/2021/09/30/short1-grid.jpg)
+ 
+
+```
+Input: grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1
+Output: 6
+Explanation: 
+The shortest path without eliminating any obstacle is 10.
+The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).
+```
+
+Example 2:
+
+![](https://assets.leetcode.com/uploads/2021/09/30/short2-grid.jpg)
+
+```
+Input: grid = [[0,1,1],[1,1,1],[1,0,0]], k = 1
+Output: -1
+Explanation: We need to eliminate at least two obstacles to find such a walk.
+```
+
+Constraints:
+
+- m == grid.length
+- n == grid[i].length
+- 1 <= m, n <= 40
+- 1 <= k <= m * n
+- grid[i][j] is either 0 or 1.
+- grid[0][0] == grid[m - 1][n - 1] == 0
+
+
+
+## 题目大意
+
+给你一个 m * n 的网格，其中每个单元格不是 0（空）就是 1（障碍物）。每一步，您都可以在空白单元格中上、下、左、右移动。
+
+如果您 最多 可以消除 k 个障碍物，请找出从左上角 (0, 0) 到右下角 (m-1, n-1) 的最短路径，并返回通过该路径所需的步数。如果找不到这样的路径，则返回 -1 。
+
+
+## 解题思路
+
+使用 BFS 遍历棋盘。这题比普通可达性问题多了一个障碍物的限制。这个也不难。每个点往周边四个方向扩展的时候，如果遇到障碍物，先算上这个障碍物，障碍物累积总个数小于 K 的时候，从障碍物的这个格子继续开始遍历。如果没有遇到障碍物，判断当前累积障碍物个数是否已经小于 K 个，如果小于 K 便继续遍历。如果大于 K，便终止此轮遍历。
+
+## 代码
+
+```go
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+type pos struct {
+	x, y     int
+	obstacle int
+	step     int
+}
+
+func shortestPath(grid [][]int, k int) int {
+	queue, m, n := []pos{}, len(grid), len(grid[0])
+	visitor := make([][][]int, m)
+	if len(grid) == 1 && len(grid[0]) == 1 {
+		return 0
+	}
+	for i := 0; i < m; i++ {
+		visitor[i] = make([][]int, n)
+		for j := 0; j < n; j++ {
+			visitor[i][j] = make([]int, k+1)
+		}
+	}
+	visitor[0][0][0] = 1
+	queue = append(queue, pos{x: 0, y: 0, obstacle: 0, step: 0})
+	for len(queue) > 0 {
+		size := len(queue)
+		for size > 0 {
+			size--
+			node := queue[0]
+			queue = queue[1:]
+			for i := 0; i < len(dir); i++ {
+				newX := node.x + dir[i][0]
+				newY := node.y + dir[i][1]
+				if newX == m-1 && newY == n-1 {
+					if node.obstacle != 0 {
+						if node.obstacle <= k {
+							return node.step + 1
+						} else {
+							continue
+						}
+					}
+					return node.step + 1
+				}
+				if isInBoard(grid, newX, newY) {
+					if grid[newX][newY] == 1 {
+						if node.obstacle+1 <= k && visitor[newX][newY][node.obstacle+1] != 1 {
+							queue = append(queue, pos{x: newX, y: newY, obstacle: node.obstacle + 1, step: node.step + 1})
+							visitor[newX][newY][node.obstacle+1] = 1
+						}
+					} else {
+						if node.obstacle <= k && visitor[newX][newY][node.obstacle] != 1 {
+							queue = append(queue, pos{x: newX, y: newY, obstacle: node.obstacle, step: node.step + 1})
+							visitor[newX][newY][node.obstacle] = 1
+						}
+					}
+
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func isInBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+```
\ No newline at end of file
diff --git a/leetcode/1295.Find-Numbers-with-Even-Number-of-Digits/1295. Find Numbers with Even Number of Digits.go b/leetcode/1295.Find-Numbers-with-Even-Number-of-Digits/1295. Find Numbers with Even Number of Digits.go
new file mode 100644
index 000000000..3e230f47e
--- /dev/null
+++ b/leetcode/1295.Find-Numbers-with-Even-Number-of-Digits/1295. Find Numbers with Even Number of Digits.go	
@@ -0,0 +1,11 @@
+package leetcode
+
+import "strconv"
+
+func findNumbers(nums []int) int {
+	res := 0
+	for _, n := range nums {
+		res += 1 - len(strconv.Itoa(n))%2
+	}
+	return res
+}
diff --git a/leetcode/1295.Find-Numbers-with-Even-Number-of-Digits/1295. Find Numbers with Even Number of Digits_test.go b/leetcode/1295.Find-Numbers-with-Even-Number-of-Digits/1295. Find Numbers with Even Number of Digits_test.go
new file mode 100644
index 000000000..df89aa0c9
--- /dev/null
+++ b/leetcode/1295.Find-Numbers-with-Even-Number-of-Digits/1295. Find Numbers with Even Number of Digits_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1295 struct {
+	para1295
+	ans1295
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1295 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1295 struct {
+	one int
+}
+
+func Test_Problem1295(t *testing.T) {
+
+	qs := []question1295{
+
+		{
+			para1295{[]int{12, 345, 2, 6, 7896}},
+			ans1295{2},
+		},
+
+		{
+			para1295{[]int{555, 901, 482, 1771}},
+			ans1295{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1295------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1295, q.para1295
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findNumbers(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1295.Find-Numbers-with-Even-Number-of-Digits/README.md b/leetcode/1295.Find-Numbers-with-Even-Number-of-Digits/README.md
new file mode 100644
index 000000000..ccaa6c492
--- /dev/null
+++ b/leetcode/1295.Find-Numbers-with-Even-Number-of-Digits/README.md
@@ -0,0 +1,63 @@
+# [1295. Find Numbers with Even Number of Digits](https://leetcode.com/problems/find-numbers-with-even-number-of-digits/)
+
+
+
+## 题目
+
+Given an array `nums` of integers, return how many of them contain an **even number** of digits.
+
+**Example 1**:
+
+```
+Input: nums = [12,345,2,6,7896]
+Output: 2
+Explanation: 
+12 contains 2 digits (even number of digits). 
+345 contains 3 digits (odd number of digits). 
+2 contains 1 digit (odd number of digits). 
+6 contains 1 digit (odd number of digits). 
+7896 contains 4 digits (even number of digits). 
+Therefore only 12 and 7896 contain an even number of digits.
+```
+
+**Example 2**:
+
+```
+Input: nums = [555,901,482,1771]
+Output: 1 
+Explanation: 
+Only 1771 contains an even number of digits.
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 500`
+- `1 <= nums[i] <= 10^5`
+
+## 题目大意
+
+给你一个整数数组 nums，请你返回其中位数为 偶数 的数字的个数。
+
+提示：
+
+- 1 <= nums.length <= 500
+- 1 <= nums[i] <= 10^5
+
+
+
+## 解题思路
+
+- 给你一个整数数组，要求输出位数为偶数的数字的个数。
+- 简单题，把每个数字转换为字符串判断长度是否是偶数即可。
+
+## 代码
+
+```go
+func findNumbers(nums []int) int {
+	res := 0
+	for _, n := range nums {
+		res += 1 - len(strconv.Itoa(n))%2
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/1296.Divide Array in Sets of K Consecutive Numbers.go b/leetcode/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/1296.Divide Array in Sets of K Consecutive Numbers.go
new file mode 100644
index 000000000..22663df49
--- /dev/null
+++ b/leetcode/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/1296.Divide Array in Sets of K Consecutive Numbers.go	
@@ -0,0 +1,23 @@
+package leetcode
+
+import "sort"
+
+func isPossibleDivide(nums []int, k int) bool {
+	mp := make(map[int]int)
+	for _, v := range nums {
+		mp[v] += 1
+	}
+	sort.Ints(nums)
+	for _, num := range nums {
+		if mp[num] == 0 {
+			continue
+		}
+		for diff := 0; diff < k; diff++ {
+			if mp[num+diff] == 0 {
+				return false
+			}
+			mp[num+diff] -= 1
+		}
+	}
+	return true
+}
diff --git a/leetcode/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/1296.Divide Array in Sets of K Consecutive Numbers_test.go b/leetcode/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/1296.Divide Array in Sets of K Consecutive Numbers_test.go
new file mode 100644
index 000000000..ac0630695
--- /dev/null
+++ b/leetcode/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/1296.Divide Array in Sets of K Consecutive Numbers_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1296 struct {
+	para1296
+	ans1296
+}
+
+// para 是参数
+type para1296 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+type ans1296 struct {
+	ans bool
+}
+
+func Test_Problem1296(t *testing.T) {
+
+	qs := []question1296{
+
+		{
+			para1296{[]int{1, 2, 3, 3, 4, 4, 5, 6}, 4},
+			ans1296{true},
+		},
+
+		{
+			para1296{[]int{3, 2, 1, 2, 3, 4, 3, 4, 5, 9, 10, 11}, 3},
+			ans1296{true},
+		},
+
+		{
+			para1296{[]int{1, 2, 3, 4}, 3},
+			ans1296{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1296------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1296, q.para1296
+		fmt.Printf("【input】:%v    【output】:%v\n", p, isPossibleDivide(p.nums, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/README.md b/leetcode/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/README.md
new file mode 100644
index 000000000..cd644c590
--- /dev/null
+++ b/leetcode/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/README.md
@@ -0,0 +1,72 @@
+# [1296. Divide Array in Sets of K Consecutive Numbers](https://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/)
+
+## 题目
+
+Given an array of integers nums and a positive integer k, check whether it is possible to divide this array into sets of k consecutive numbers.
+
+Return true if it is possible. Otherwise, return false.
+
+**Example 1**:
+
+    Input: nums = [1,2,3,3,4,4,5,6], k = 4
+    Output: true
+    Explanation: Array can be divided into [1,2,3,4] and [3,4,5,6].
+
+**Example 2**:
+
+    Input: nums = [3,2,1,2,3,4,3,4,5,9,10,11], k = 3
+    Output: true
+    Explanation: Array can be divided into [1,2,3] , [2,3,4] , [3,4,5] and [9,10,11].
+
+**Example 3**:
+
+    Input: nums = [1,2,3,4], k = 3
+    Output: false
+    Explanation: Each array should be divided in subarrays of size 3.
+
+**Constraints:**
+
+- 1 <= k <= nums.length <= 100000
+- 1 <= nums[i] <= 1000000000
+
+## 题目大意
+
+给你一个整数数组 nums 和一个正整数 k，请你判断是否可以把这个数组划分成一些由 k 个连续数字组成的集合。
+如果可以，请返回 true；否则，返回 false。
+
+## 解题思路
+
+贪心算法
+
+- 对nums升序排序
+- 对nums内数字进行哈希计数（key:数字，value:数量）
+- 遍历nums中的数字，以数量大于1的数字作为连续数字开头，寻找连续数字后续元素，若无法找到 k 个连续数字则返回false
+- 所有数字都能找到 k 个连续数字返回true
+
+##代码
+
+```go
+package leetcode
+
+import "sort"
+
+func isPossibleDivide(nums []int, k int) bool {
+	mp := make(map[int]int)
+	for _, v := range nums {
+		mp[v] += 1
+	}
+	sort.Ints(nums)
+	for _, num := range nums {
+		if mp[num] == 0 {
+			continue
+		}
+		for diff := 0; diff < k; diff++ {
+			if mp[num+diff] == 0 {
+				return false
+			}
+			mp[num+diff] -= 1
+		}
+	}
+	return true
+}
+```
diff --git a/leetcode/1299.Replace-Elements-with-Greatest-Element-on-Right-Side/1299. Replace Elements with Greatest Element on Right Side.go b/leetcode/1299.Replace-Elements-with-Greatest-Element-on-Right-Side/1299. Replace Elements with Greatest Element on Right Side.go
new file mode 100644
index 000000000..da0f6d27a
--- /dev/null
+++ b/leetcode/1299.Replace-Elements-with-Greatest-Element-on-Right-Side/1299. Replace Elements with Greatest Element on Right Side.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func replaceElements(arr []int) []int {
+	j, temp := -1, 0
+	for i := len(arr) - 1; i >= 0; i-- {
+		temp = arr[i]
+		arr[i] = j
+		j = max(j, temp)
+	}
+	return arr
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1299.Replace-Elements-with-Greatest-Element-on-Right-Side/1299. Replace Elements with Greatest Element on Right Side_test.go b/leetcode/1299.Replace-Elements-with-Greatest-Element-on-Right-Side/1299. Replace Elements with Greatest Element on Right Side_test.go
new file mode 100644
index 000000000..d590320c7
--- /dev/null
+++ b/leetcode/1299.Replace-Elements-with-Greatest-Element-on-Right-Side/1299. Replace Elements with Greatest Element on Right Side_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1299 struct {
+	para1299
+	ans1299
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1299 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1299 struct {
+	one []int
+}
+
+func Test_Problem1299(t *testing.T) {
+
+	qs := []question1299{
+
+		{
+			para1299{[]int{17, 18, 5, 4, 6, 1}},
+			ans1299{[]int{18, 6, 6, 6, 1, -1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1299------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1299, q.para1299
+		fmt.Printf("【input】:%v       【output】:%v\n", p, replaceElements(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1299.Replace-Elements-with-Greatest-Element-on-Right-Side/README.md b/leetcode/1299.Replace-Elements-with-Greatest-Element-on-Right-Side/README.md
new file mode 100644
index 000000000..633c65761
--- /dev/null
+++ b/leetcode/1299.Replace-Elements-with-Greatest-Element-on-Right-Side/README.md
@@ -0,0 +1,51 @@
+# [1299. Replace Elements with Greatest Element on Right Side](https://leetcode.com/problems/replace-elements-with-greatest-element-on-right-side/)
+
+
+
+## 题目
+
+Given an array `arr`, replace every element in that array with the greatest element among the elements to its right, and replace the last element with `-1`.
+
+After doing so, return the array.
+
+**Example 1**:
+
+```
+Input: arr = [17,18,5,4,6,1]
+Output: [18,6,6,6,1,-1]
+```
+
+**Constraints**:
+
+- `1 <= arr.length <= 10^4`
+- `1 <= arr[i] <= 10^5`
+
+
+## 题目大意
+
+给你一个数组 arr ，请你将每个元素用它右边最大的元素替换，如果是最后一个元素，用 -1 替换。完成所有替换操作后，请你返回这个数组。
+
+提示：
+
+- 1 <= arr.length <= 10^4
+- 1 <= arr[i] <= 10^5
+
+
+## 解题思路
+
+- 给出一个数组，要求把所有元素都替换成自己右边最大的元素，最后一位补上 -1 。最后输出变化以后的数组。
+- 简单题，按照题意操作即可。
+
+## 代码
+
+```go
+func replaceElements(arr []int) []int {
+	j, temp := -1, 0
+	for i := len(arr) - 1; i >= 0; i-- {
+		temp = arr[i]
+		arr[i] = j
+		j = max(j, temp)
+	}
+	return arr
+}
+```
\ No newline at end of file
diff --git a/leetcode/1300.Sum-of-Mutated-Array-Closest-to-Target/1300. Sum of Mutated Array Closest to Target.go b/leetcode/1300.Sum-of-Mutated-Array-Closest-to-Target/1300. Sum of Mutated Array Closest to Target.go
new file mode 100644
index 000000000..9bd5f6d58
--- /dev/null
+++ b/leetcode/1300.Sum-of-Mutated-Array-Closest-to-Target/1300. Sum of Mutated Array Closest to Target.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+func findBestValue(arr []int, target int) int {
+	low, high := 0, 100000
+	for low < high {
+		mid := low + (high-low)>>1
+		if calculateSum(arr, mid) < target {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	if high == 100000 {
+		res := 0
+		for _, num := range arr {
+			if res < num {
+				res = num
+			}
+		}
+		return res
+	}
+	// 比较阈值线分别定在 left - 1 和 left 的时候与 target 的接近程度
+	sum1, sum2 := calculateSum(arr, low-1), calculateSum(arr, low)
+	if target-sum1 <= sum2-target {
+		return low - 1
+	}
+	return low
+}
+
+func calculateSum(arr []int, mid int) int {
+	sum := 0
+	for _, num := range arr {
+		sum += min(num, mid)
+	}
+	return sum
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/1300.Sum-of-Mutated-Array-Closest-to-Target/1300. Sum of Mutated Array Closest to Target_test.go b/leetcode/1300.Sum-of-Mutated-Array-Closest-to-Target/1300. Sum of Mutated Array Closest to Target_test.go
new file mode 100644
index 000000000..da66fbbac
--- /dev/null
+++ b/leetcode/1300.Sum-of-Mutated-Array-Closest-to-Target/1300. Sum of Mutated Array Closest to Target_test.go	
@@ -0,0 +1,59 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1300 struct {
+	para1300
+	ans1300
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1300 struct {
+	arr    []int
+	target int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1300 struct {
+	one int
+}
+
+func Test_Problem1300(t *testing.T) {
+
+	qs := []question1300{
+
+		{
+			para1300{[]int{4, 9, 3}, 10},
+			ans1300{3},
+		},
+
+		{
+			para1300{[]int{2, 3, 5}, 10},
+			ans1300{5},
+		},
+
+		// new case
+		{
+			para1300{[]int{2, 3, 5}, 11},
+			ans1300{5},
+		},
+
+		{
+			para1300{[]int{60864, 25176, 27249, 21296, 20204}, 56803},
+			ans1300{11361},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1300------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1300, q.para1300
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findBestValue(p.arr, p.target))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1300.Sum-of-Mutated-Array-Closest-to-Target/README.md b/leetcode/1300.Sum-of-Mutated-Array-Closest-to-Target/README.md
new file mode 100644
index 000000000..15c7d5d7c
--- /dev/null
+++ b/leetcode/1300.Sum-of-Mutated-Array-Closest-to-Target/README.md
@@ -0,0 +1,103 @@
+# [1300. Sum of Mutated Array Closest to Target](https://leetcode.com/problems/sum-of-mutated-array-closest-to-target/)
+
+
+
+## 题目
+
+Given an integer array `arr` and a target value `target`, return the integer `value` such that when we change all the integers larger than `value` in the given array to be equal to `value`, the sum of the array gets as close as possible (in absolute difference) to `target`.
+
+In case of a tie, return the minimum such integer.
+
+Notice that the answer is not neccesarilly a number from `arr`.
+
+**Example 1**:
+
+```
+Input: arr = [4,9,3], target = 10
+Output: 3
+Explanation: When using 3 arr converts to [3, 3, 3] which sums 9 and that's the optimal answer.
+```
+
+**Example 2**:
+
+```
+Input: arr = [2,3,5], target = 10
+Output: 5
+```
+
+**Example 3**:
+
+```
+Input: arr = [60864,25176,27249,21296,20204], target = 56803
+Output: 11361
+```
+
+**Constraints**:
+
+- `1 <= arr.length <= 10^4`
+- `1 <= arr[i], target <= 10^5`
+
+
+## 题目大意
+
+给你一个整数数组 arr 和一个目标值 target ，请你返回一个整数 value ，使得将数组中所有大于 value 的值变成 value 后，数组的和最接近  target （最接近表示两者之差的绝对值最小）。如果有多种使得和最接近 target 的方案，请你返回这些整数中的最小值。请注意，答案不一定是 arr 中的数字。
+
+提示：
+
+- 1 <= arr.length <= 10^4
+- 1 <= arr[i], target <= 10^5
+
+
+
+## 解题思路
+
+- 给出一个数组 arr 和 target。能否找到一个 value 值，使得将数组中所有大于 value 的值变成 value 后，数组的和最接近 target。如果有多种方法，输出 value 值最小的。
+- 这一题可以用二分搜索来求解。最后输出的唯一解有 2 个限制条件，一个是变化后的数组和最接近 target 。另一个是输出的 value 是所有可能方法中最小值。二分搜索最终的 value 值。mid 就是尝试的 value 值，每选择一次 mid，就算一次总和，和 target 比较。由于数组里面每个数和 mid 差距各不相同，所以每次调整 mid 有可能出现 mid 选小了以后，距离 target 反而更大了；mid 选大了以后，距离 target 反而更小了。这里的解决办法是，把 value 上下方可能的值都拿出来比较一下。
+
+## 代码
+
+```go
+
+func findBestValue(arr []int, target int) int {
+	low, high := 0, 100000
+	for low < high {
+		mid := low + (high-low)>>1
+		if calculateSum(arr, mid) < target {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	if high == 100000 {
+		res := 0
+		for _, num := range arr {
+			if res < num {
+				res = num
+			}
+		}
+		return res
+	}
+	// 比较阈值线分别定在 left - 1 和 left 的时候与 target 的接近程度
+	sum1, sum2 := calculateSum(arr, low-1), calculateSum(arr, low)
+	if target-sum1 <= sum2-target {
+		return low - 1
+	}
+	return low
+}
+
+func calculateSum(arr []int, mid int) int {
+	sum := 0
+	for _, num := range arr {
+		sum += min(num, mid)
+	}
+	return sum
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1302.Deepest-Leaves-Sum/1302. Deepest Leaves Sum.go b/leetcode/1302.Deepest-Leaves-Sum/1302. Deepest Leaves Sum.go
new file mode 100644
index 000000000..e12f13746
--- /dev/null
+++ b/leetcode/1302.Deepest-Leaves-Sum/1302. Deepest Leaves Sum.go	
@@ -0,0 +1,35 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func deepestLeavesSum(root *TreeNode) int {
+	maxLevel, sum := 0, 0
+	dfsDeepestLeavesSum(root, 0, &maxLevel, &sum)
+	return sum
+}
+
+func dfsDeepestLeavesSum(root *TreeNode, level int, maxLevel, sum *int) {
+	if root == nil {
+		return
+	}
+	if level > *maxLevel {
+		*maxLevel, *sum = level, root.Val
+	} else if level == *maxLevel {
+		*sum += root.Val
+	}
+	dfsDeepestLeavesSum(root.Left, level+1, maxLevel, sum)
+	dfsDeepestLeavesSum(root.Right, level+1, maxLevel, sum)
+}
diff --git a/leetcode/1302.Deepest-Leaves-Sum/1302. Deepest Leaves Sum_test.go b/leetcode/1302.Deepest-Leaves-Sum/1302. Deepest Leaves Sum_test.go
new file mode 100644
index 000000000..b968b98ec
--- /dev/null
+++ b/leetcode/1302.Deepest-Leaves-Sum/1302. Deepest Leaves Sum_test.go	
@@ -0,0 +1,49 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1302 struct {
+	para1302
+	ans1302
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1302 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1302 struct {
+	one int
+}
+
+func Test_Problem1302(t *testing.T) {
+
+	qs := []question1302{
+
+		{
+			para1302{[]int{1, 2, 3, 4, 5, structures.NULL, 6, 7, structures.NULL, structures.NULL, structures.NULL, structures.NULL, 8}},
+			ans1302{15},
+		},
+
+		{
+			para1302{[]int{}},
+			ans1302{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1302------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1302, q.para1302
+		fmt.Printf("【input】:%v       【output】:%v\n", p, deepestLeavesSum(structures.Ints2TreeNode(p.one)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1302.Deepest-Leaves-Sum/README.md b/leetcode/1302.Deepest-Leaves-Sum/README.md
new file mode 100644
index 000000000..70482634f
--- /dev/null
+++ b/leetcode/1302.Deepest-Leaves-Sum/README.md
@@ -0,0 +1,58 @@
+# [1302. Deepest Leaves Sum](https://leetcode.com/problems/deepest-leaves-sum/)
+
+
+
+## 题目
+
+Given a binary tree, return the sum of values of its deepest leaves.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/07/31/1483_ex1.png](https://assets.leetcode.com/uploads/2019/07/31/1483_ex1.png)
+
+```
+Input: root = [1,2,3,4,5,null,6,7,null,null,null,null,8]
+Output: 15
+```
+
+**Constraints**:
+
+- The number of nodes in the tree is between `1` and `10^4`.
+- The value of nodes is between `1` and `100`.
+
+## 题目大意
+
+给你一棵二叉树，请你返回层数最深的叶子节点的和。
+
+提示：
+
+- 树中节点数目在 1 到 10^4 之间。
+- 每个节点的值在 1 到 100 之间。
+
+## 解题思路
+
+- 给你一棵二叉树，请你返回层数最深的叶子节点的和。
+- 这一题不难，DFS 遍历把最底层的叶子节点和都加起来即可。
+
+## 代码
+
+```go
+func deepestLeavesSum(root *TreeNode) int {
+	maxLevel, sum := 0, 0
+	dfsDeepestLeavesSum(root, 0, &maxLevel, &sum)
+	return sum
+}
+
+func dfsDeepestLeavesSum(root *TreeNode, level int, maxLevel, sum *int) {
+	if root == nil {
+		return
+	}
+	if level > *maxLevel {
+		*maxLevel, *sum = level, root.Val
+	} else if level == *maxLevel {
+		*sum += root.Val
+	}
+	dfsDeepestLeavesSum(root.Left, level+1, maxLevel, sum)
+	dfsDeepestLeavesSum(root.Right, level+1, maxLevel, sum)
+}
+```
\ No newline at end of file
diff --git a/leetcode/1304.Find-N-Unique-Integers-Sum-up-to-Zero/1304. Find N Unique Integers Sum up to Zero.go b/leetcode/1304.Find-N-Unique-Integers-Sum-up-to-Zero/1304. Find N Unique Integers Sum up to Zero.go
new file mode 100644
index 000000000..337edfd59
--- /dev/null
+++ b/leetcode/1304.Find-N-Unique-Integers-Sum-up-to-Zero/1304. Find N Unique Integers Sum up to Zero.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+func sumZero(n int) []int {
+	res, left, right, start := make([]int, n), 0, n-1, 1
+	for left < right {
+		res[left] = start
+		res[right] = -start
+		start++
+		left = left + 1
+		right = right - 1
+	}
+	return res
+}
diff --git a/leetcode/1304.Find-N-Unique-Integers-Sum-up-to-Zero/1304. Find N Unique Integers Sum up to Zero_test.go b/leetcode/1304.Find-N-Unique-Integers-Sum-up-to-Zero/1304. Find N Unique Integers Sum up to Zero_test.go
new file mode 100644
index 000000000..91874da56
--- /dev/null
+++ b/leetcode/1304.Find-N-Unique-Integers-Sum-up-to-Zero/1304. Find N Unique Integers Sum up to Zero_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1304 struct {
+	para1304
+	ans1304
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1304 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1304 struct {
+	one []int
+}
+
+func Test_Problem1304(t *testing.T) {
+
+	qs := []question1304{
+
+		{
+			para1304{5},
+			ans1304{[]int{-7, -1, 1, 3, 4}},
+		},
+
+		{
+			para1304{0},
+			ans1304{[]int{}},
+		},
+
+		{
+			para1304{3},
+			ans1304{[]int{-1, 0, 1}},
+		},
+
+		{
+			para1304{1},
+			ans1304{[]int{0}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1304------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1304, q.para1304
+		fmt.Printf("【input】:%v       【output】:%v\n", p, sumZero(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1304.Find-N-Unique-Integers-Sum-up-to-Zero/README.md b/leetcode/1304.Find-N-Unique-Integers-Sum-up-to-Zero/README.md
new file mode 100644
index 000000000..1e15836e3
--- /dev/null
+++ b/leetcode/1304.Find-N-Unique-Integers-Sum-up-to-Zero/README.md
@@ -0,0 +1,62 @@
+# [1304. Find N Unique Integers Sum up to Zero](https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/)
+
+
+
+## 题目
+
+Given an integer `n`, return **any** array containing `n` **unique** integers such that they add up to 0.
+
+**Example 1**:
+
+```
+Input: n = 5
+Output: [-7,-1,1,3,4]
+Explanation: These arrays also are accepted [-5,-1,1,2,3] , [-3,-1,2,-2,4].
+```
+
+**Example 2**:
+
+```
+Input: n = 3
+Output: [-1,0,1]
+```
+
+**Example 3**:
+
+```
+Input: n = 1
+Output: [0]
+```
+
+**Constraints**:
+
+- `1 <= n <= 1000`
+
+## 题目大意
+
+给你一个整数 n，请你返回 任意 一个由 n 个 各不相同 的整数组成的数组，并且这 n 个数相加和为 0 。
+
+提示：
+
+- 1 <= n <= 1000
+
+## 解题思路
+
+- 给出一个数 n，输出一个有 n 个数的数组，里面元素之和为 0 。
+- 简单题，简单循环即可。
+
+## 代码
+
+```go
+func sumZero(n int) []int {
+	res, left, right, start := make([]int, n), 0, n-1, 1
+	for left < right {
+		res[left] = start
+		res[right] = -start
+		start++
+		left = left + 1
+		right = right - 1
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1305.All-Elements-in-Two-Binary-Search-Trees/1305. All Elements in Two Binary Search Trees.go b/leetcode/1305.All-Elements-in-Two-Binary-Search-Trees/1305. All Elements in Two Binary Search Trees.go
new file mode 100644
index 000000000..b3c1bcb7e
--- /dev/null
+++ b/leetcode/1305.All-Elements-in-Two-Binary-Search-Trees/1305. All Elements in Two Binary Search Trees.go	
@@ -0,0 +1,98 @@
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 合并排序
+func getAllElements(root1 *TreeNode, root2 *TreeNode) []int {
+	arr1 := inorderTraversal(root1)
+	arr2 := inorderTraversal(root2)
+	arr1 = append(arr1, make([]int, len(arr2))...)
+	merge(arr1, len(arr1)-len(arr2), arr2, len(arr2))
+	return arr1
+}
+
+// this is 94 solution
+func inorderTraversal(root *TreeNode) []int {
+	var result []int
+	inorder(root, &result)
+	return result
+}
+
+func inorder(root *TreeNode, output *[]int) {
+	if root != nil {
+		inorder(root.Left, output)
+		*output = append(*output, root.Val)
+		inorder(root.Right, output)
+	}
+}
+
+// this is 88 solution
+func merge(nums1 []int, m int, nums2 []int, n int) {
+	if m == 0 {
+		copy(nums1, nums2)
+		return
+	}
+	// 这里不需要，因为测试数据考虑到了第一个数组的空间问题
+	// for index := 0; index < n; index++ {
+	// 	nums1 = append(nums1, nums2[index])
+	// }
+	i := m - 1
+	j := n - 1
+	k := m + n - 1
+	// 从后面往前放，只需要循环一次即可
+	for ; i >= 0 && j >= 0; k-- {
+		if nums1[i] > nums2[j] {
+			nums1[k] = nums1[i]
+			i--
+		} else {
+			nums1[k] = nums2[j]
+			j--
+		}
+	}
+	for ; j >= 0; k-- {
+		nums1[k] = nums2[j]
+		j--
+	}
+}
+
+// 解法二 暴力遍历排序，时间复杂度高
+func getAllElements1(root1 *TreeNode, root2 *TreeNode) []int {
+	arr := []int{}
+	arr = append(arr, preorderTraversal(root1)...)
+	arr = append(arr, preorderTraversal(root2)...)
+	sort.Ints(arr)
+	return arr
+}
+
+// this is 144 solution
+func preorderTraversal(root *TreeNode) []int {
+	res := []int{}
+	if root != nil {
+		res = append(res, root.Val)
+		tmp := preorderTraversal(root.Left)
+		for _, t := range tmp {
+			res = append(res, t)
+		}
+		tmp = preorderTraversal(root.Right)
+		for _, t := range tmp {
+			res = append(res, t)
+		}
+	}
+	return res
+}
diff --git a/leetcode/1305.All-Elements-in-Two-Binary-Search-Trees/1305. All Elements in Two Binary Search Trees_test.go b/leetcode/1305.All-Elements-in-Two-Binary-Search-Trees/1305. All Elements in Two Binary Search Trees_test.go
new file mode 100644
index 000000000..979acc63e
--- /dev/null
+++ b/leetcode/1305.All-Elements-in-Two-Binary-Search-Trees/1305. All Elements in Two Binary Search Trees_test.go	
@@ -0,0 +1,66 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1305 struct {
+	para1305
+	ans1305
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1305 struct {
+	root1 []int
+	root2 []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1305 struct {
+	one []int
+}
+
+func Test_Problem1305(t *testing.T) {
+
+	qs := []question1305{
+
+		{
+			para1305{[]int{2, 1, 4}, []int{1, 0, 3}},
+			ans1305{[]int{0, 1, 1, 2, 3, 4}},
+		},
+
+		{
+			para1305{[]int{0, -10, 10}, []int{5, 1, 7, 0, 2}},
+			ans1305{[]int{-10, 0, 0, 1, 2, 5, 7, 10}},
+		},
+
+		{
+			para1305{[]int{}, []int{5, 1, 7, 0, 2}},
+			ans1305{[]int{0, 1, 2, 5, 7}},
+		},
+
+		{
+			para1305{[]int{0, -10, 10}, []int{}},
+			ans1305{[]int{-10, 0, 10}},
+		},
+
+		{
+			para1305{[]int{1, structures.NULL, 8}, []int{8, 1}},
+			ans1305{[]int{1, 1, 8, 8}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1305------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1305, q.para1305
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", getAllElements(structures.Ints2TreeNode(p.root1), structures.Ints2TreeNode(p.root2)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1305.All-Elements-in-Two-Binary-Search-Trees/README.md b/leetcode/1305.All-Elements-in-Two-Binary-Search-Trees/README.md
new file mode 100644
index 000000000..247bbfbbe
--- /dev/null
+++ b/leetcode/1305.All-Elements-in-Two-Binary-Search-Trees/README.md
@@ -0,0 +1,90 @@
+# [1305. All Elements in Two Binary Search Trees](https://leetcode.com/problems/all-elements-in-two-binary-search-trees/)
+
+
+
+## 题目
+
+Given two binary search trees `root1` and `root2`.
+
+Return a list containing *all the integers* from *both trees* sorted in **ascending** order.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/12/18/q2-e1.png](https://assets.leetcode.com/uploads/2019/12/18/q2-e1.png)
+
+```
+Input: root1 = [2,1,4], root2 = [1,0,3]
+Output: [0,1,1,2,3,4]
+```
+
+**Example 2**:
+
+```
+Input: root1 = [0,-10,10], root2 = [5,1,7,0,2]
+Output: [-10,0,0,1,2,5,7,10]
+```
+
+**Example 3**:
+
+```
+Input: root1 = [], root2 = [5,1,7,0,2]
+Output: [0,1,2,5,7]
+```
+
+**Example 4**:
+
+```
+Input: root1 = [0,-10,10], root2 = []
+Output: [-10,0,10]
+```
+
+**Example 5**:
+
+![https://assets.leetcode.com/uploads/2019/12/18/q2-e5-.png](https://assets.leetcode.com/uploads/2019/12/18/q2-e5-.png)
+
+```
+Input: root1 = [1,null,8], root2 = [8,1]
+Output: [1,1,8,8]
+```
+
+**Constraints**:
+
+- Each tree has at most `5000` nodes.
+- Each node's value is between `[-10^5, 10^5]`.
+
+## 题目大意
+
+给你 root1 和 root2 这两棵二叉搜索树。请你返回一个列表，其中包含 两棵树 中的所有整数并按 升序 排序。.
+
+提示：
+
+- 每棵树最多有 5000 个节点。
+- 每个节点的值在 [-10^5, 10^5] 之间。
+
+
+## 解题思路
+
+- 给出 2 棵二叉排序树，要求将 2 棵树所有节点的值按照升序排序。
+- 这一题最暴力简单的方法就是把 2 棵树的节点都遍历出来，然后放在一个数组里面从小到大排序即可。这样做虽然能 AC，但是时间复杂度高。因为题目中给的二叉排序树这一条件没有用上。由于树中节点本来已经有序了，所以题目实质想要我们合并 2 个有序数组。利用中根遍历，把 2 个二叉排序树的所有节点值都遍历出来，遍历出来以后就是有序的。接下来再合并这两个有序数组即可。合并 2 个有序数组是第 88 题。
+
+## 代码
+
+```go
+// 解法一 合并排序
+func getAllElements(root1 *TreeNode, root2 *TreeNode) []int {
+	arr1 := inorderTraversal(root1)
+	arr2 := inorderTraversal(root2)
+	arr1 = append(arr1, make([]int, len(arr2))...)
+	merge(arr1, len(arr1)-len(arr2), arr2, len(arr2))
+	return arr1
+}
+
+// 解法二 暴力遍历排序，时间复杂度高
+func getAllElements1(root1 *TreeNode, root2 *TreeNode) []int {
+	arr := []int{}
+	arr = append(arr, preorderTraversal(root1)...)
+	arr = append(arr, preorderTraversal(root2)...)
+	sort.Ints(arr)
+	return arr
+}
+```
\ No newline at end of file
diff --git a/leetcode/1306.Jump-Game-III/1306. Jump Game III.go b/leetcode/1306.Jump-Game-III/1306. Jump Game III.go
new file mode 100644
index 000000000..11d4bad74
--- /dev/null
+++ b/leetcode/1306.Jump-Game-III/1306. Jump Game III.go	
@@ -0,0 +1,10 @@
+package leetcode
+
+func canReach(arr []int, start int) bool {
+	if start >= 0 && start < len(arr) && arr[start] < len(arr) {
+		jump := arr[start]
+		arr[start] += len(arr)
+		return jump == 0 || canReach(arr, start+jump) || canReach(arr, start-jump)
+	}
+	return false
+}
diff --git a/leetcode/1306.Jump-Game-III/1306. Jump Game III_test.go b/leetcode/1306.Jump-Game-III/1306. Jump Game III_test.go
new file mode 100644
index 000000000..1298ad759
--- /dev/null
+++ b/leetcode/1306.Jump-Game-III/1306. Jump Game III_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1306 struct {
+	para1306
+	ans1306
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1306 struct {
+	arr   []int
+	start int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1306 struct {
+	one bool
+}
+
+func Test_Problem1306(t *testing.T) {
+
+	qs := []question1306{
+
+		{
+			para1306{[]int{4, 2, 3, 0, 3, 1, 2}, 5},
+			ans1306{true},
+		},
+
+		{
+			para1306{[]int{4, 2, 3, 0, 3, 1, 2}, 0},
+			ans1306{true},
+		},
+
+		{
+			para1306{[]int{3, 0, 2, 1, 2}, 2},
+			ans1306{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1306------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1306, q.para1306
+		fmt.Printf("【input】:%v       【output】:%v\n", p, canReach(p.arr, p.start))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1306.Jump-Game-III/README.md b/leetcode/1306.Jump-Game-III/README.md
new file mode 100644
index 000000000..63f26345a
--- /dev/null
+++ b/leetcode/1306.Jump-Game-III/README.md
@@ -0,0 +1,80 @@
+# [1306. Jump Game III](https://leetcode.com/problems/jump-game-iii/)
+
+
+## 题目
+
+Given an array of non-negative integers `arr`, you are initially positioned at `start` index of the array. When you are at index `i`, you can jump to `i + arr[i]` or `i - arr[i]`, check if you can reach to **any** index with value 0.
+
+Notice that you can not jump outside of the array at any time.
+
+**Example 1**:
+
+```
+Input: arr = [4,2,3,0,3,1,2], start = 5
+Output: true
+Explanation: 
+All possible ways to reach at index 3 with value 0 are: 
+index 5 -> index 4 -> index 1 -> index 3 
+index 5 -> index 6 -> index 4 -> index 1 -> index 3
+```
+
+**Example 2**:
+
+```
+Input: arr = [4,2,3,0,3,1,2], start = 0
+Output: true 
+Explanation: 
+One possible way to reach at index 3 with value 0 is: 
+index 0 -> index 4 -> index 1 -> index 3
+```
+
+**Example 3**:
+
+```
+Input: arr = [3,0,2,1,2], start = 2
+Output: false
+Explanation: There is no way to reach at index 1 with value 0.
+```
+
+**Constraints**:
+
+- `1 <= arr.length <= 5 * 10^4`
+- `0 <= arr[i] < arr.length`
+- `0 <= start < arr.length`
+
+
+## 题目大意
+
+这里有一个非负整数数组 arr，你最开始位于该数组的起始下标 start 处。当你位于下标 i 处时，你可以跳到 i + arr[i] 或者 i - arr[i]。请你判断自己是否能够跳到对应元素值为 0 的 任一 下标处。注意，不管是什么情况下，你都无法跳到数组之外。
+
+提示：
+
+- 1 <= arr.length <= 5 * 10^4
+- 0 <= arr[i] < arr.length
+- 0 <= start < arr.length
+
+
+## 解题思路
+
+- 给出一个非负数组和一个起始下标 `start`。站在 `start`，每次可以跳到 `i + arr[i]` 或者 `i - arr[i]` 。要求判断能否跳到元素值为 0 的下标处。
+- 这一题考察的是递归。每一步都需要判断 3 种可能：
+    1. 当前是否站在元素值为 0 的目标点上。
+    2. 往前跳 arr[start]，是否能站在元素值为 0 的目标点上。
+    3. 往后跳 arr[start]，是否能站在元素值为 0 的目标点上。
+
+    第 2 种可能和第 3 种可能递归即可，每一步都判断这 3 种可能是否有一种能跳到元素值为 0 的下标处。
+
+- `arr[start] += len(arr)`  这一步仅仅只是为了标记此下标已经用过了，下次判断的时候该下标会被 `if` 条件筛选掉。
+
+## 代码
+
+```go
+func canReach(arr []int, start int) bool {
+	if start >= 0 && start < len(arr) && arr[start] < len(arr) {
+		jump := arr[start]
+		arr[start] += len(arr)
+		return jump == 0 || canReach(arr, start+jump) || canReach(arr, start-jump)
+	}
+	return false
+}
+```
\ No newline at end of file
diff --git a/leetcode/1310.XOR-Queries-of-a-Subarray/1310. XOR Queries of a Subarray.go b/leetcode/1310.XOR-Queries-of-a-Subarray/1310. XOR Queries of a Subarray.go
new file mode 100644
index 000000000..e5da239c8
--- /dev/null
+++ b/leetcode/1310.XOR-Queries-of-a-Subarray/1310. XOR Queries of a Subarray.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+func xorQueries(arr []int, queries [][]int) []int {
+	xors := make([]int, len(arr))
+	xors[0] = arr[0]
+	for i := 1; i < len(arr); i++ {
+		xors[i] = arr[i] ^ xors[i-1]
+	}
+	res := make([]int, len(queries))
+	for i, q := range queries {
+		res[i] = xors[q[1]]
+		if q[0] > 0 {
+			res[i] ^= xors[q[0]-1]
+		}
+	}
+	return res
+}
diff --git a/leetcode/1310.XOR-Queries-of-a-Subarray/1310. XOR Queries of a Subarray_test.go b/leetcode/1310.XOR-Queries-of-a-Subarray/1310. XOR Queries of a Subarray_test.go
new file mode 100644
index 000000000..8066c3432
--- /dev/null
+++ b/leetcode/1310.XOR-Queries-of-a-Subarray/1310. XOR Queries of a Subarray_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1310 struct {
+	para1310
+	ans1310
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1310 struct {
+	arr     []int
+	queries [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1310 struct {
+	one []int
+}
+
+func Test_Problem1310(t *testing.T) {
+
+	qs := []question1310{
+
+		{
+			para1310{[]int{1, 3, 4, 8}, [][]int{{0, 1}, {1, 2}, {0, 3}, {3, 3}}},
+			ans1310{[]int{2, 7, 14, 8}},
+		},
+
+		{
+			para1310{[]int{4, 8, 2, 10}, [][]int{{2, 3}, {1, 3}, {0, 0}, {0, 3}}},
+			ans1310{[]int{8, 0, 4, 4}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1310------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1310, q.para1310
+		fmt.Printf("【input】:%v       【output】:%v\n", p, xorQueries(p.arr, p.queries))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1310.XOR-Queries-of-a-Subarray/README.md b/leetcode/1310.XOR-Queries-of-a-Subarray/README.md
new file mode 100644
index 000000000..7babbbd79
--- /dev/null
+++ b/leetcode/1310.XOR-Queries-of-a-Subarray/README.md
@@ -0,0 +1,75 @@
+# [1310. XOR Queries of a Subarray](https://leetcode.com/problems/xor-queries-of-a-subarray/)
+
+
+## 题目
+
+Given the array `arr` of positive integers and the array `queries` where `queries[i] = [Li,Ri]`, for each query `i` compute the **XOR** of elements from `Li` to `Ri` (that is, `arr[Li]xor arr[Li+1]xor ...xor arr[Ri]`). Return an array containing the result for the given `queries`.
+
+**Example 1:**
+
+```
+Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
+Output: [2,7,14,8]
+Explanation:
+The binary representation of the elements in the array are:
+1 = 0001
+3 = 0011
+4 = 0100
+8 = 1000
+The XOR values for queries are:
+[0,1] = 1 xor 3 = 2
+[1,2] = 3 xor 4 = 7
+[0,3] = 1 xor 3 xor 4 xor 8 = 14
+[3,3] = 8
+
+```
+
+**Example 2:**
+
+```
+Input: arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
+Output: [8,0,4,4]
+
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 3 * 10^4`
+- `1 <= arr[i] <= 10^9`
+- `1 <= queries.length <= 3 * 10^4`
+- `queries[i].length == 2`
+- `0 <= queries[i][0] <= queries[i][1] < arr.length`
+
+## 题目大意
+
+有一个正整数数组 arr，现给你一个对应的查询数组 queries，其中 queries[i] = [Li, Ri]。对于每个查询 i，请你计算从 Li 到 Ri 的 XOR 值（即 arr[Li] xor arr[Li+1] xor ... xor arr[Ri]）作为本次查询的结果。并返回一个包含给定查询 queries 所有结果的数组。
+
+## 解题思路
+
+- 此题求区间异或，很容易让人联想到区间求和。区间求和利用前缀和，可以使得 query 从 O(n) 降为 O(1)。区间异或能否也用类似前缀和的思想呢？答案是肯定的。利用异或的两个性质，x ^ x = 0，x ^ 0 = x。那么有：（由于 LaTeX 中异或符号 ^ 是特殊字符，笔者用 $\oplus$ 代替异或）
+
+    $$\begin{aligned}Query(left,right) &=arr[left] \oplus \cdots  \oplus arr[right]\\&=(arr[0] \oplus \cdots  \oplus arr[left-1]) \oplus (arr[0] \oplus \cdots  \oplus arr[left-1]) \oplus (arr[left] \oplus \cdots  \oplus arr[right])\\ &=(arr[0] \oplus \cdots  \oplus arr[left-1]) \oplus (arr[0] \oplus \cdots  \oplus arr[right])\\ &=xors[left] \oplus xors[right+1]\\ \end{aligned}$$
+
+    按照这个思路解题，便可以将 query 从 O(n) 降为 O(1)，总的时间复杂度为 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+func xorQueries(arr []int, queries [][]int) []int {
+	xors := make([]int, len(arr))
+	xors[0] = arr[0]
+	for i := 1; i < len(arr); i++ {
+		xors[i] = arr[i] ^ xors[i-1]
+	}
+	res := make([]int, len(queries))
+	for i, q := range queries {
+		res[i] = xors[q[1]]
+		if q[0] > 0 {
+			res[i] ^= xors[q[0]-1]
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List.go b/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List.go
new file mode 100644
index 000000000..9e7e74962
--- /dev/null
+++ b/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List.go	
@@ -0,0 +1,11 @@
+package leetcode
+
+func decompressRLElist(nums []int) []int {
+	res := []int{}
+	for i := 0; i < len(nums); i += 2 {
+		for j := 0; j < nums[i]; j++ {
+			res = append(res, nums[i+1])
+		}
+	}
+	return res
+}
diff --git a/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List_test.go b/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List_test.go
new file mode 100644
index 000000000..ce5cad62e
--- /dev/null
+++ b/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1313 struct {
+	para1313
+	ans1313
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1313 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1313 struct {
+	one []int
+}
+
+func Test_Problem1313(t *testing.T) {
+
+	qs := []question1313{
+
+		{
+			para1313{[]int{1, 2, 3, 4}},
+			ans1313{[]int{2, 4, 4, 4}},
+		},
+
+		{
+			para1313{[]int{1, 1, 2, 3}},
+			ans1313{[]int{1, 3, 3}},
+		},
+
+		{
+			para1313{[]int{}},
+			ans1313{[]int{}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1313------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1313, q.para1313
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", decompressRLElist(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1313.Decompress-Run-Length-Encoded-List/README.md b/leetcode/1313.Decompress-Run-Length-Encoded-List/README.md
new file mode 100644
index 000000000..645d90578
--- /dev/null
+++ b/leetcode/1313.Decompress-Run-Length-Encoded-List/README.md
@@ -0,0 +1,60 @@
+# [1313. Decompress Run-Length Encoded List](https://leetcode.com/problems/decompress-run-length-encoded-list/)
+
+
+## 题目
+
+We are given a list `nums` of integers representing a list compressed with run-length encoding.
+
+Consider each adjacent pair of elements `[freq, val] = [nums[2*i], nums[2*i+1]]` (with `i >= 0`). For each such pair, there are `freq` elements with value `val` concatenated in a sublist. Concatenate all the sublists from left to right to generate the decompressed list.
+
+Return the decompressed list.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,4]
+Output: [2,4,4,4]
+Explanation: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2].
+The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4].
+At the end the concatenation [2] + [4,4,4] is [2,4,4,4].
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,1,2,3]
+Output: [1,3,3]
+```
+
+**Constraints**:
+
+- `2 <= nums.length <= 100`
+- `nums.length % 2 == 0`
+- `1 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个以行程长度编码压缩的整数列表 nums 。考虑每对相邻的两个元素 [freq, val] = [nums[2*i], nums[2*i+1]] （其中 i >= 0 ），每一对都表示解压后子列表中有 freq 个值为 val 的元素，你需要从左到右连接所有子列表以生成解压后的列表。请你返回解压后的列表。
+
+## 解题思路
+
+- 给定一个带编码长度的数组，要求解压这个数组。
+- 简单题。按照题目要求，下标从 0 开始，奇数位下标为前一个下标对应元素重复次数，那么就把这个元素 append 几次。最终输出解压后的数组即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func decompressRLElist(nums []int) []int {
+	res := []int{}
+	for i := 0; i < len(nums); i += 2 {
+		for j := 0; j < nums[i]; j++ {
+			res = append(res, nums[i+1])
+		}
+	}
+	return res
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers.go b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers.go
new file mode 100644
index 000000000..df59ca084
--- /dev/null
+++ b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers.go	
@@ -0,0 +1,22 @@
+package leetcode
+
+func getNoZeroIntegers(n int) []int {
+	noZeroPair := []int{}
+	for i := 1; i <= n/2; i++ {
+		if isNoZero(i) && isNoZero(n-i) {
+			noZeroPair = append(noZeroPair, []int{i, n - i}...)
+			break
+		}
+	}
+	return noZeroPair
+}
+
+func isNoZero(n int) bool {
+	for n != 0 {
+		if n%10 == 0 {
+			return false
+		}
+		n /= 10
+	}
+	return true
+}
diff --git a/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers_test.go b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers_test.go
new file mode 100644
index 000000000..b39cf2ff9
--- /dev/null
+++ b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers_test.go	
@@ -0,0 +1,82 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1317 struct {
+	para1317
+	ans1317
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1317 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1317 struct {
+	one []int
+}
+
+func Test_Problem1317(t *testing.T) {
+
+	qs := []question1317{
+
+		{
+			para1317{5},
+			ans1317{[]int{1, 4}},
+		},
+
+		{
+			para1317{0},
+			ans1317{[]int{}},
+		},
+
+		{
+			para1317{3},
+			ans1317{[]int{1, 2}},
+		},
+
+		{
+			para1317{1},
+			ans1317{[]int{}},
+		},
+
+		{
+			para1317{2},
+			ans1317{[]int{1, 1}},
+		},
+
+		{
+			para1317{11},
+			ans1317{[]int{2, 9}},
+		},
+
+		{
+			para1317{10000},
+			ans1317{[]int{1, 9999}},
+		},
+
+		{
+			para1317{69},
+			ans1317{[]int{1, 68}},
+		},
+
+		{
+			para1317{1010},
+			ans1317{[]int{11, 999}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1317------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1317, q.para1317
+		fmt.Printf("【input】:%v       【output】:%v\n", p, getNoZeroIntegers(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/README.md b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/README.md
new file mode 100644
index 000000000..b7871908f
--- /dev/null
+++ b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/README.md
@@ -0,0 +1,96 @@
+# [1317. Convert Integer to the Sum of Two No-Zero Integers](https://leetcode.com/problems/convert-integer-to-the-sum-of-two-no-zero-integers/)
+
+
+## 题目
+
+Given an integer `n`. No-Zero integer is a positive integer which **doesn't contain any 0** in its decimal representation.
+
+Return *a list of two integers* `[A, B]` where:
+
+- `A` and `B` are No-Zero integers.
+- `A + B = n`
+
+It's guarateed that there is at least one valid solution. If there are many valid solutions you can return any of them.
+
+**Example 1**:
+
+```
+Input: n = 2
+Output: [1,1]
+Explanation: A = 1, B = 1. A + B = n and both A and B don't contain any 0 in their decimal representation.
+```
+
+**Example 2**:
+
+```
+Input: n = 11
+Output: [2,9]
+```
+
+**Example 3**:
+
+```
+Input: n = 10000
+Output: [1,9999]
+```
+
+**Example 4**:
+
+```
+Input: n = 69
+Output: [1,68]
+```
+
+**Example 5**:
+
+```
+Input: n = 1010
+Output: [11,999]
+```
+
+**Constraints**:
+
+- `2 <= n <= 10^4`
+
+## 题目大意
+
+「无零整数」是十进制表示中 不含任何 0 的正整数。给你一个整数 n，请你返回一个 由两个整数组成的列表 [A, B]，满足：
+
+- A 和 B 都是无零整数
+- A + B = n
+
+题目数据保证至少有一个有效的解决方案。如果存在多个有效解决方案，你可以返回其中任意一个。
+
+## 解题思路
+
+- 给定一个整数 n，要求把它分解为 2 个十进制位中不含 0 的正整数且这两个正整数之和为 n。
+- 简单题。在 [1, n/2] 区间内搜索，只要有一组满足条件的解就 break。题目保证了至少有一组解，并且多组解返回任意一组即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func getNoZeroIntegers(n int) []int {
+	noZeroPair := []int{}
+	for i := 1; i <= n/2; i++ {
+		if isNoZero(i) && isNoZero(n-i) {
+			noZeroPair = append(noZeroPair, []int{i, n - i}...)
+			break
+		}
+	}
+	return noZeroPair
+}
+
+func isNoZero(n int) bool {
+	for n != 0 {
+		if n%10 == 0 {
+			return false
+		}
+		n /= 10
+	}
+	return true
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1319.Number-of-Operations-to-Make-Network-Connected/1319. Number of Operations to Make Network Connected.go b/leetcode/1319.Number-of-Operations-to-Make-Network-Connected/1319. Number of Operations to Make Network Connected.go
new file mode 100644
index 000000000..0180e380c
--- /dev/null
+++ b/leetcode/1319.Number-of-Operations-to-Make-Network-Connected/1319. Number of Operations to Make Network Connected.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+func makeConnected(n int, connections [][]int) int {
+	if n-1 > len(connections) {
+		return -1
+	}
+	uf, redundance := template.UnionFind{}, 0
+	uf.Init(n)
+	for _, connection := range connections {
+		if uf.Find(connection[0]) == uf.Find(connection[1]) {
+			redundance++
+		} else {
+			uf.Union(connection[0], connection[1])
+		}
+	}
+	if uf.TotalCount() == 1 || redundance < uf.TotalCount()-1 {
+		return 0
+	}
+	return uf.TotalCount() - 1
+}
diff --git a/leetcode/1319.Number-of-Operations-to-Make-Network-Connected/1319. Number of Operations to Make Network Connected_test.go b/leetcode/1319.Number-of-Operations-to-Make-Network-Connected/1319. Number of Operations to Make Network Connected_test.go
new file mode 100644
index 000000000..1cdd6b437
--- /dev/null
+++ b/leetcode/1319.Number-of-Operations-to-Make-Network-Connected/1319. Number of Operations to Make Network Connected_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1319 struct {
+	para1319
+	ans1319
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1319 struct {
+	n           int
+	connections [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1319 struct {
+	one int
+}
+
+func Test_Problem1319(t *testing.T) {
+
+	qs := []question1319{
+
+		{
+			para1319{4, [][]int{{0, 1}, {0, 2}, {1, 2}}},
+			ans1319{1},
+		},
+
+		{
+			para1319{6, [][]int{{0, 1}, {0, 2}, {0, 3}, {1, 2}, {1, 3}}},
+			ans1319{2},
+		},
+
+		{
+			para1319{6, [][]int{{0, 1}, {0, 2}, {0, 3}, {1, 2}}},
+			ans1319{-1},
+		},
+
+		{
+			para1319{5, [][]int{{0, 1}, {0, 2}, {3, 4}, {2, 3}}},
+			ans1319{0},
+		},
+
+		{
+			para1319{12, [][]int{{1, 5}, {1, 7}, {1, 2}, {1, 4}, {3, 7}, {4, 7}, {3, 5}, {0, 6}, {0, 1}, {0, 4}, {2, 6}, {0, 3}, {0, 2}}},
+			ans1319{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1319------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1319, q.para1319
+		fmt.Printf("【input】:%v       【output】:%v\n", p, makeConnected(p.n, p.connections))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1319.Number-of-Operations-to-Make-Network-Connected/README.md b/leetcode/1319.Number-of-Operations-to-Make-Network-Connected/README.md
new file mode 100644
index 000000000..15232def6
--- /dev/null
+++ b/leetcode/1319.Number-of-Operations-to-Make-Network-Connected/README.md
@@ -0,0 +1,90 @@
+# [1319. Number of Operations to Make Network Connected](https://leetcode.com/problems/number-of-operations-to-make-network-connected/)
+
+
+## 题目
+
+There are `n` computers numbered from `0` to `n-1` connected by ethernet cables `connections` forming a network where `connections[i] = [a, b]` represents a connection between computers `a` and `b`. Any computer can reach any other computer directly or indirectly through the network.
+
+Given an initial computer network `connections`. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected. Return the *minimum number of times* you need to do this in order to make all the computers connected. If it's not possible, return -1.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/01/02/sample_1_1677.png](https://assets.leetcode.com/uploads/2020/01/02/sample_1_1677.png)
+
+```
+Input: n = 4, connections = [[0,1],[0,2],[1,2]]
+Output: 1
+Explanation: Remove cable between computer 1 and 2 and place between computers 1 and 3.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/01/02/sample_2_1677.png](https://assets.leetcode.com/uploads/2020/01/02/sample_2_1677.png)
+
+```
+Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
+Output: 2
+```
+
+**Example 3:**
+
+```
+Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
+Output: -1
+Explanation: There are not enough cables.
+```
+
+**Example 4:**
+
+```
+Input: n = 5, connections = [[0,1],[0,2],[3,4],[2,3]]
+Output: 0
+```
+
+**Constraints:**
+
+- `1 <= n <= 10^5`
+- `1 <= connections.length <= min(n*(n-1)/2, 10^5)`
+- `connections[i].length == 2`
+- `0 <= connections[i][0], connections[i][1] < n`
+- `connections[i][0] != connections[i][1]`
+- There are no repeated connections.
+- No two computers are connected by more than one cable.
+
+## 题目大意
+
+用以太网线缆将 n 台计算机连接成一个网络，计算机的编号从 0 到 n-1。线缆用 connections 表示，其中 connections[i] = [a, b] 连接了计算机 a 和 b。网络中的任何一台计算机都可以通过网络直接或者间接访问同一个网络中其他任意一台计算机。给你这个计算机网络的初始布线 connections，你可以拔开任意两台直连计算机之间的线缆，并用它连接一对未直连的计算机。请你计算并返回使所有计算机都连通所需的最少操作次数。如果不可能，则返回 -1 。
+
+## 解题思路
+
+- 很明显这题的解题思路是并查集。先将每个 connections 构建出并查集。构建中需要累加冗余的连接。例如 2 个节点已经连通，再连接这个集合中的任意 2 个节点就算冗余连接。冗余连接的线都可以移动，去连接还没有连通的节点。计算出冗余连接数，再根据并查集的集合总数，即可得出答案。
+- 这一题答案有 3 种可能。第一种，所有点都在一个集合内，即全部连通，这时输出 0 。第二种，冗余的连接不够串起所有点，这时输出 -1 。第三种情况是可以连通的情况。 m 个集合需要连通，最少需要 m - 1 条线。如果冗余连接数大于 m - 1，则输出 m - 1 即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+func makeConnected(n int, connections [][]int) int {
+	if n-1 > len(connections) {
+		return -1
+	}
+	uf, redundance := template.UnionFind{}, 0
+	uf.Init(n)
+	for _, connection := range connections {
+		if uf.Find(connection[0]) == uf.Find(connection[1]) {
+			redundance++
+		} else {
+			uf.Union(connection[0], connection[1])
+		}
+	}
+	if uf.TotalCount() == 1 || redundance < uf.TotalCount()-1 {
+		return 0
+	}
+	return uf.TotalCount() - 1
+}
+```
\ No newline at end of file
diff --git a/leetcode/1329.Sort-the-Matrix-Diagonally/1329. Sort the Matrix Diagonally.go b/leetcode/1329.Sort-the-Matrix-Diagonally/1329. Sort the Matrix Diagonally.go
new file mode 100644
index 000000000..23bda9c79
--- /dev/null
+++ b/leetcode/1329.Sort-the-Matrix-Diagonally/1329. Sort the Matrix Diagonally.go	
@@ -0,0 +1,24 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+func diagonalSort(mat [][]int) [][]int {
+	m, n, diagonalsMap := len(mat), len(mat[0]), make(map[int][]int)
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			diagonalsMap[i-j] = append(diagonalsMap[i-j], mat[i][j])
+		}
+	}
+	for _, v := range diagonalsMap {
+		sort.Ints(v)
+	}
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			mat[i][j] = diagonalsMap[i-j][0]
+			diagonalsMap[i-j] = diagonalsMap[i-j][1:]
+		}
+	}
+	return mat
+}
diff --git a/leetcode/1329.Sort-the-Matrix-Diagonally/1329. Sort the Matrix Diagonally_test.go b/leetcode/1329.Sort-the-Matrix-Diagonally/1329. Sort the Matrix Diagonally_test.go
new file mode 100644
index 000000000..21cd808e2
--- /dev/null
+++ b/leetcode/1329.Sort-the-Matrix-Diagonally/1329. Sort the Matrix Diagonally_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1329 struct {
+	para1329
+	ans1329
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1329 struct {
+	mat [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1329 struct {
+	one [][]int
+}
+
+func Test_Problem1329(t *testing.T) {
+
+	qs := []question1329{
+
+		{
+			para1329{[][]int{{3, 3, 1, 1}, {2, 2, 1, 2}, {1, 1, 1, 2}}},
+			ans1329{[][]int{{1, 1, 1, 1}, {1, 2, 2, 2}, {1, 2, 3, 3}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1329------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1329, q.para1329
+		fmt.Printf("【input】:%v       【output】:%v\n", p, diagonalSort(p.mat))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1329.Sort-the-Matrix-Diagonally/README.md b/leetcode/1329.Sort-the-Matrix-Diagonally/README.md
new file mode 100644
index 000000000..2b4a52cfc
--- /dev/null
+++ b/leetcode/1329.Sort-the-Matrix-Diagonally/README.md
@@ -0,0 +1,57 @@
+# [1329. Sort the Matrix Diagonally](https://leetcode.com/problems/sort-the-matrix-diagonally/)
+
+
+## 题目
+
+A **matrix diagonal** is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix's end. For example, the **matrix diagonal** starting from `mat[2][0]`, where `mat` is a `6 x 3` matrix, includes cells `mat[2][0]`, `mat[3][1]`, and `mat[4][2]`.
+
+Given an `m x n` matrix `mat` of integers, sort each **matrix diagonal** in ascending order and return *the resulting matrix*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/01/21/1482_example_1_2.png](https://assets.leetcode.com/uploads/2020/01/21/1482_example_1_2.png)
+
+```
+Input: mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]
+Output: [[1,1,1,1],[1,2,2,2],[1,2,3,3]]
+```
+
+**Constraints:**
+
+- `m == mat.length`
+- `n == mat[i].length`
+- `1 <= m, n <= 100`
+- `1 <= mat[i][j] <= 100`
+
+## 题目大意
+
+给你一个 m * n 的整数矩阵 mat ，请你将同一条对角线上的元素（从左上到右下）按升序排序后，返回排好序的矩阵。
+
+## 解题思路
+
+- 这道题思路很简单。按照对角线，把每条对角线的元素读取出来放在数组中。这里可以利用 map 保存这些数组。再将这些数组排序。最后按照对角线还原矩阵即可。
+
+## 代码
+
+```go
+package leetcode
+
+func diagonalSort(mat [][]int) [][]int {
+	m, n, diagonalsMap := len(mat), len(mat[0]), make(map[int][]int)
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			diagonalsMap[i-j] = append(diagonalsMap[i-j], mat[i][j])
+		}
+	}
+	for _, v := range diagonalsMap {
+		sort.Ints(v)
+	}
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			mat[i][j] = diagonalsMap[i-j][0]
+			diagonalsMap[i-j] = diagonalsMap[i-j][1:]
+		}
+	}
+	return mat
+}
+```
\ No newline at end of file
diff --git a/leetcode/1332.Remove-Palindromic-Subsequences/1332. Remove Palindromic Subsequences.go b/leetcode/1332.Remove-Palindromic-Subsequences/1332. Remove Palindromic Subsequences.go
new file mode 100644
index 000000000..cf21cdf72
--- /dev/null
+++ b/leetcode/1332.Remove-Palindromic-Subsequences/1332. Remove Palindromic Subsequences.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+func removePalindromeSub(s string) int {
+	if len(s) == 0 {
+		return 0
+	}
+	for i := 0; i < len(s)/2; i++ {
+		if s[i] != s[len(s)-1-i] {
+			return 2
+		}
+	}
+	return 1
+}
diff --git a/leetcode/1332.Remove-Palindromic-Subsequences/1332. Remove Palindromic Subsequences_test.go b/leetcode/1332.Remove-Palindromic-Subsequences/1332. Remove Palindromic Subsequences_test.go
new file mode 100644
index 000000000..458d9f877
--- /dev/null
+++ b/leetcode/1332.Remove-Palindromic-Subsequences/1332. Remove Palindromic Subsequences_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1332 struct {
+	para1332
+	ans1332
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1332 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1332 struct {
+	one int
+}
+
+func Test_Problem1332(t *testing.T) {
+
+	qs := []question1332{
+
+		{
+			para1332{"ababa"},
+			ans1332{1},
+		},
+
+		{
+			para1332{"abb"},
+			ans1332{2},
+		},
+
+		{
+			para1332{"baabb"},
+			ans1332{2},
+		},
+
+		{
+			para1332{""},
+			ans1332{0},
+		},
+
+		{
+			para1332{"bbaabaaa"},
+			ans1332{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1332------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1332, q.para1332
+		fmt.Printf("【input】:%v       【output】:%v\n", p, removePalindromeSub(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1332.Remove-Palindromic-Subsequences/README.md b/leetcode/1332.Remove-Palindromic-Subsequences/README.md
new file mode 100644
index 000000000..5a7743d6d
--- /dev/null
+++ b/leetcode/1332.Remove-Palindromic-Subsequences/README.md
@@ -0,0 +1,81 @@
+# [1332. Remove Palindromic Subsequences](https://leetcode.com/problems/remove-palindromic-subsequences/)
+
+
+## 题目
+
+Given a string `s` consisting only of letters `'a'` and `'b'`. In a single step you can remove one palindromic **subsequence** from `s`.
+
+Return the minimum number of steps to make the given string empty.
+
+A string is a subsequence of a given string, if it is generated by deleting some characters of a given string without changing its order.
+
+A string is called palindrome if is one that reads the same backward as well as forward.
+
+**Example 1:**
+
+```
+Input: s = "ababa"
+Output: 1
+Explanation: String is already palindrome
+```
+
+**Example 2:**
+
+```
+Input: s = "abb"
+Output: 2
+Explanation: "abb" -> "bb" -> "". 
+Remove palindromic subsequence "a" then "bb".
+```
+
+**Example 3:**
+
+```
+Input: s = "baabb"
+Output: 2
+Explanation: "baabb" -> "b" -> "". 
+Remove palindromic subsequence "baab" then "b".
+```
+
+**Example 4:**
+
+```
+Input: s = ""
+Output: 0
+```
+
+**Constraints:**
+
+- `0 <= s.length <= 1000`
+- `s` only consists of letters 'a' and 'b'
+
+## 题目大意
+
+给你一个字符串 s，它仅由字母 'a' 和 'b' 组成。每一次删除操作都可以从 s 中删除一个回文 子序列。返回删除给定字符串中所有字符（字符串为空）的最小删除次数。
+
+「子序列」定义：如果一个字符串可以通过删除原字符串某些字符而不改变原字符顺序得到，那么这个字符串就是原字符串的一个子序列。
+
+「回文」定义：如果一个字符串向后和向前读是一致的，那么这个字符串就是一个回文。
+
+## 解题思路
+
+- 笔者读完题以为是第 5 题的加强版。在字符串中每次都找到最长的回文子串删除，一直删除到找不到回文子串结束，删除的总次数 + 剩余字母数 = 最小删除次数。提交以后 `wrong answer` 了，在 `bbaabaaa` 这组测试用例出错了。如果按照找最长回文字符串的思路，先找到最长回文子串 `aabaa`，剩余 `bba`，还需要再删除 2 次，`bb` 和 `a`。总共删除次数是 3 。为什么出错误了呢？仔细再读题，题目中说的是子序列，这不是连续的，再加上这道题是 `easy` 难度，其实很简单。
+- 这道题的答案只可能是 0，1，2 。空串对应的 0 。如果有一个字母，单个字母可以构成回文，所以是 1，如果字符串长度大于等于 2，即 `a` 和 `b` 都有，第一步先删除所有 `a`，因为所有的 `a` 构成了回文子序列。第二步删除所有的 `b`，因为所有的 `b` 构成了回文子序列。经过这样两步，一定能删除所有字符。
+
+## 代码
+
+```go
+package leetcode
+
+func removePalindromeSub(s string) int {
+	if len(s) == 0 {
+		return 0
+	}
+	for i := 0; i < len(s)/2; i++ {
+		if s[i] != s[len(s)-1-i] {
+			return 2
+		}
+	}
+	return 1
+}
+```
\ No newline at end of file
diff --git a/leetcode/1337.The-K-Weakest-Rows-in-a-Matrix/1337. The K Weakest Rows in a Matrix.go b/leetcode/1337.The-K-Weakest-Rows-in-a-Matrix/1337. The K Weakest Rows in a Matrix.go
new file mode 100644
index 000000000..a774f3be0
--- /dev/null
+++ b/leetcode/1337.The-K-Weakest-Rows-in-a-Matrix/1337. The K Weakest Rows in a Matrix.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func kWeakestRows(mat [][]int, k int) []int {
+	res := []int{}
+	for j := 0; j < len(mat[0]); j++ {
+		for i := 0; i < len(mat); i++ {
+			if mat[i][j] == 0 && ((j == 0) || (mat[i][j-1] != 0)) {
+				res = append(res, i)
+			}
+		}
+	}
+	for i := 0; i < len(mat); i++ {
+		if mat[i][len(mat[0])-1] == 1 {
+			res = append(res, i)
+		}
+	}
+	return res[:k]
+}
diff --git a/leetcode/1337.The-K-Weakest-Rows-in-a-Matrix/1337. The K Weakest Rows in a Matrix_test.go b/leetcode/1337.The-K-Weakest-Rows-in-a-Matrix/1337. The K Weakest Rows in a Matrix_test.go
new file mode 100644
index 000000000..2592dd5fa
--- /dev/null
+++ b/leetcode/1337.The-K-Weakest-Rows-in-a-Matrix/1337. The K Weakest Rows in a Matrix_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1337 struct {
+	para1337
+	ans1337
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1337 struct {
+	mat [][]int
+	k   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1337 struct {
+	one []int
+}
+
+func Test_Problem1337(t *testing.T) {
+
+	qs := []question1337{
+
+		{
+			para1337{[][]int{{1, 1, 0, 0, 0}, {1, 1, 1, 1, 0}, {1, 0, 0, 0, 0}, {1, 1, 0, 0, 0}, {1, 1, 1, 1, 1}}, 3},
+			ans1337{[]int{2, 0, 3}},
+		},
+
+		{
+			para1337{[][]int{{1, 0, 0, 0}, {1, 1, 1, 1}, {1, 0, 0, 0}, {1, 0, 0, 0}}, 2},
+			ans1337{[]int{0, 2}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1337------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1337, q.para1337
+		fmt.Printf("【input】:%v       【output】:%v\n", p, kWeakestRows(p.mat, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1337.The-K-Weakest-Rows-in-a-Matrix/README.md b/leetcode/1337.The-K-Weakest-Rows-in-a-Matrix/README.md
new file mode 100644
index 000000000..003048edb
--- /dev/null
+++ b/leetcode/1337.The-K-Weakest-Rows-in-a-Matrix/README.md
@@ -0,0 +1,90 @@
+# [1337. The K Weakest Rows in a Matrix](https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/)
+
+
+## 题目
+
+Given a `m * n` matrix `mat` of *ones* (representing soldiers) and *zeros* (representing civilians), return the indexes of the `k` weakest rows in the matrix ordered from the weakest to the strongest.
+
+A row ***i*** is weaker than row ***j***, if the number of soldiers in row ***i*** is less than the number of soldiers in row ***j***, or they have the same number of soldiers but ***i*** is less than ***j***. Soldiers are **always** stand in the frontier of a row, that is, always *ones* may appear first and then *zeros*.
+
+**Example 1:**
+
+```
+Input: mat = 
+[[1,1,0,0,0],
+ [1,1,1,1,0],
+ [1,0,0,0,0],
+ [1,1,0,0,0],
+ [1,1,1,1,1]], 
+k = 3
+Output: [2,0,3]
+Explanation: 
+The number of soldiers for each row is: 
+row 0 -> 2 
+row 1 -> 4 
+row 2 -> 1 
+row 3 -> 2 
+row 4 -> 5 
+Rows ordered from the weakest to the strongest are [2,0,3,1,4]
+
+```
+
+**Example 2:**
+
+```
+Input: mat = 
+[[1,0,0,0],
+ [1,1,1,1],
+ [1,0,0,0],
+ [1,0,0,0]], 
+k = 2
+Output: [0,2]
+Explanation: 
+The number of soldiers for each row is: 
+row 0 -> 1 
+row 1 -> 4 
+row 2 -> 1 
+row 3 -> 1 
+Rows ordered from the weakest to the strongest are [0,2,3,1]
+
+```
+
+**Constraints:**
+
+- `m == mat.length`
+- `n == mat[i].length`
+- `2 <= n, m <= 100`
+- `1 <= k <= m`
+- `matrix[i][j]` is either 0 **or** 1.
+
+## 题目大意
+
+给你一个大小为 m * n 的矩阵 mat，矩阵由若干军人和平民组成，分别用 1 和 0 表示。请你返回矩阵中战斗力最弱的 k 行的索引，按从最弱到最强排序。如果第 i 行的军人数量少于第 j 行，或者两行军人数量相同但 i 小于 j，那么我们认为第 i 行的战斗力比第 j 行弱。军人 总是 排在一行中的靠前位置，也就是说 1 总是出现在 0 之前。
+
+## 解题思路
+
+- 简单题。第一个能想到的解题思路是，先统计每一行 1 的个数，然后将结果进行排序，按照 1 的个数从小到大排序，如果 1 的个数相同，再按照行号从小到大排序。排好序的数组取出前 K 位即为答案。
+- 此题还有第二种解法。在第一种解法中，并没有用到题目中“军人 总是 排在一行中的靠前位置，也就是说 1 总是出现在 0 之前。”这一条件。由于有了这个条件，使得如果按照列去遍历，最先出现 0 的行，则是最弱的行。行号小的先被遍历到，所以相同数量 1 的行，行号小的会排在前面。最后记得再添加上全 1 的行。同样，最终输出取出前 K 位即为答案。此题解法二才是最优雅最高效的解法。
+
+## 代码
+
+```go
+package leetcode
+
+func kWeakestRows(mat [][]int, k int) []int {
+	res := []int{}
+	for j := 0; j < len(mat[0]); j++ {
+		for i := 0; i < len(mat); i++ {
+			if mat[i][j] == 0 && ((j == 0) || (mat[i][j-1] != 0)) {
+				res = append(res, i)
+			}
+		}
+	}
+	for i := 0; i < len(mat); i++ {
+		if mat[i][len(mat[0])-1] == 1 {
+			res = append(res, i)
+		}
+	}
+	return res[:k]
+}
+```
\ No newline at end of file
diff --git a/leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/1353. Maximum Number of Events That Can Be Attended.go b/leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/1353. Maximum Number of Events That Can Be Attended.go
new file mode 100644
index 000000000..67700713e
--- /dev/null
+++ b/leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/1353. Maximum Number of Events That Can Be Attended.go	
@@ -0,0 +1,35 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+func maxEvents(events [][]int) int {
+	sort.Slice(events, func(i, j int) bool {
+		if events[i][0] == events[j][0] {
+			return events[i][1] < events[j][1]
+		}
+		return events[i][0] < events[j][0]
+	})
+	attended, current := 1, events[0]
+	for i := 1; i < len(events); i++ {
+		prev, event := events[i-1], events[i]
+		if event[0] == prev[0] && event[1] == prev[1] && event[1] == event[0] {
+			continue
+		}
+		start, end := max(current[0], event[0]-1), max(current[1], event[1])
+		if end-start > 0 {
+			current[0] = start + 1
+			current[1] = end
+			attended++
+		}
+	}
+	return attended
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/1353. Maximum Number of Events That Can Be Attended_test.go b/leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/1353. Maximum Number of Events That Can Be Attended_test.go
new file mode 100644
index 000000000..89798e1f9
--- /dev/null
+++ b/leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/1353. Maximum Number of Events That Can Be Attended_test.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1353 struct {
+	para1353
+	ans1353
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1353 struct {
+	events [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1353 struct {
+	one int
+}
+
+func Test_Problem1353(t *testing.T) {
+
+	qs := []question1353{
+
+		{
+			para1353{[][]int{{1, 2}, {2, 3}, {3, 4}}},
+			ans1353{3},
+		},
+
+		{
+			para1353{[][]int{{1, 4}, {4, 4}, {2, 2}, {3, 4}, {1, 1}}},
+			ans1353{4},
+		},
+
+		{
+			para1353{[][]int{{1, 100000}}},
+			ans1353{1},
+		},
+
+		{
+			para1353{[][]int{{1, 1}, {2, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}}},
+			ans1353{7},
+		},
+
+		{
+			para1353{[][]int{{1, 2}, {2, 2}, {3, 3}, {3, 4}, {3, 4}}},
+			ans1353{4},
+		},
+
+		{
+			para1353{[][]int{{1, 10}, {2, 2}, {2, 2}, {2, 2}, {2, 2}}},
+			ans1353{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1353------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1353, q.para1353
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxEvents(p.events))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/README.md b/leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/README.md
new file mode 100644
index 000000000..3ac1505cc
--- /dev/null
+++ b/leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/README.md
@@ -0,0 +1,113 @@
+# [1353. Maximum Number of Events That Can Be Attended](https://leetcode.com/problems/maximum-number-of-events-that-can-be-attended/)
+
+
+## 题目
+
+Given an array of `events` where `events[i] = [startDayi, endDayi]`. Every event `i` starts at `startDayi` and ends at `endDayi`.
+
+You can attend an event `i` at any day `d` where `startTimei <= d <= endTimei`. Notice that you can only attend one event at any time `d`.
+
+Return *the maximum number of events* you can attend.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/02/05/e1.png](https://assets.leetcode.com/uploads/2020/02/05/e1.png)
+
+```
+Input: events = [[1,2],[2,3],[3,4]]
+Output: 3
+Explanation: You can attend all the three events.
+One way to attend them all is as shown.
+Attend the first event on day 1.
+Attend the second event on day 2.
+Attend the third event on day 3.
+
+```
+
+**Example 2:**
+
+```
+Input: events= [[1,2],[2,3],[3,4],[1,2]]
+Output: 4
+
+```
+
+**Example 3:**
+
+```
+Input: events = [[1,4],[4,4],[2,2],[3,4],[1,1]]
+Output: 4
+
+```
+
+**Example 4:**
+
+```
+Input: events = [[1,100000]]
+Output: 1
+
+```
+
+**Example 5:**
+
+```
+Input: events = [[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7]]
+Output: 7
+
+```
+
+**Constraints:**
+
+- `1 <= events.length <= 10^5`
+- `events[i].length == 2`
+- `1 <= startDayi <= endDayi <= 10^5`
+
+## 题目大意
+
+给你一个数组 events，其中 events[i] = [startDayi, endDayi] ，表示会议 i 开始于 startDayi ，结束于 endDayi 。你可以在满足 startDayi <= d <= endDayi 中的任意一天 d 参加会议 i 。注意，一天只能参加一个会议。请你返回你可以参加的 最大 会议数目。
+
+## 解题思路
+
+- 关于会议安排，活动安排这类题，第一直觉是贪心问题。先按照会议开始时间从小到大排序，如果开始时间相同，再按照结束时间从小到大排序。贪心策略是，优先选择参加早结束的会议。因为一个结束时间晚的会议，代表这个会议持续时间长，先参加马上要结束的会议，这样可以参加更多的会议。
+- 注意题目给的数据代表的是天数。比较大小的时候最好转换成坐标轴上的坐标点。例如 [1,2] 代表这个会议持续 2 天，如果在坐标轴上表示，是 [0,2]，0-1 表示第一天，1-2 表示第二天。所以比较会议时需要把开始时间减一。选定了这个会议以后记得要把这一天排除，例如选择了第二天，那么下次对比起始时间需要从坐标 2 开始，因为第二天的时间范围是 1-2，所以下一轮比较会议前需要把开始时间加一。从左往右依次扫描各个会议时间段，选择结束时间大于起始时间的会议，不断累加次数，扫描完所有会议，最终结果即为可参加的最大会议数。
+- 测试数据中有一组很恶心的数据，见 test 文件中最后一组数据。这组数据在同一天叠加了多个会议，并且起始时间完全一致。这种特殊情况需要加判断条件排除，见下面代码 continue 条件。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func maxEvents(events [][]int) int {
+	sort.Slice(events, func(i, j int) bool {
+		if events[i][0] == events[j][0] {
+			return events[i][1] < events[j][1]
+		}
+		return events[i][0] < events[j][0]
+	})
+	attended, current := 1, events[0]
+	for i := 1; i < len(events); i++ {
+		prev, event := events[i-1], events[i]
+		if event[0] == prev[0] && event[1] == prev[1] && event[1] == event[0] {
+			continue
+		}
+		start, end := max(current[0], event[0]-1), max(current[1], event[1])
+		if end-start > 0 {
+			current[0] = start + 1
+			current[1] = end
+			attended++
+		}
+	}
+	return attended
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1380.Lucky-Numbers-in-a-Matrix/1380. Lucky Numbers in a Matrix.go b/leetcode/1380.Lucky-Numbers-in-a-Matrix/1380. Lucky Numbers in a Matrix.go
new file mode 100644
index 000000000..52dd4026e
--- /dev/null
+++ b/leetcode/1380.Lucky-Numbers-in-a-Matrix/1380. Lucky Numbers in a Matrix.go	
@@ -0,0 +1,27 @@
+package leetcode
+
+func luckyNumbers(matrix [][]int) []int {
+	t, r, res := make([]int, len(matrix[0])), make([]int, len(matrix[0])), []int{}
+	for _, val := range matrix {
+		m, k := val[0], 0
+		for j := 0; j < len(matrix[0]); j++ {
+			if val[j] < m {
+				m = val[j]
+				k = j
+			}
+			if t[j] < val[j] {
+				t[j] = val[j]
+			}
+		}
+
+		if t[k] == m {
+			r[k] = m
+		}
+	}
+	for k, v := range r {
+		if v > 0 && v == t[k] {
+			res = append(res, v)
+		}
+	}
+	return res
+}
diff --git a/leetcode/1380.Lucky-Numbers-in-a-Matrix/1380. Lucky Numbers in a Matrix_test.go b/leetcode/1380.Lucky-Numbers-in-a-Matrix/1380. Lucky Numbers in a Matrix_test.go
new file mode 100644
index 000000000..73d9f3146
--- /dev/null
+++ b/leetcode/1380.Lucky-Numbers-in-a-Matrix/1380. Lucky Numbers in a Matrix_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1380 struct {
+	para1380
+	ans1380
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1380 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1380 struct {
+	one []int
+}
+
+func Test_Problem1380(t *testing.T) {
+
+	qs := []question1380{
+
+		{
+			para1380{[][]int{{3, 7, 8}, {9, 11, 13}, {15, 16, 17}}},
+			ans1380{[]int{15}},
+		},
+
+		{
+			para1380{[][]int{{1, 10, 4, 2}, {9, 3, 8, 7}, {15, 16, 17, 12}}},
+			ans1380{[]int{12}},
+		},
+
+		{
+			para1380{[][]int{{1, 2, 3, 4, 5}, {1, 2, 3, 4, 5}}},
+			ans1380{[]int{1}},
+		},
+
+		{
+			para1380{[][]int{{7, 8}, {1, 2}}},
+			ans1380{[]int{7}},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1380------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1380, q.para1380
+		fmt.Printf("【input】:%v       【output】:%v\n", p, luckyNumbers(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1380.Lucky-Numbers-in-a-Matrix/README.md b/leetcode/1380.Lucky-Numbers-in-a-Matrix/README.md
new file mode 100644
index 000000000..8f8db8436
--- /dev/null
+++ b/leetcode/1380.Lucky-Numbers-in-a-Matrix/README.md
@@ -0,0 +1,83 @@
+# [1380. Lucky Numbers in a Matrix](https://leetcode.com/problems/lucky-numbers-in-a-matrix/)
+
+
+## 题目
+
+Given a `m * n` matrix of **distinct** numbers, return all lucky numbers in the matrix in **any** order.
+
+A lucky number is an element of the matrix such that it is the minimum element in its row and maximum in its column.
+
+**Example 1**:
+
+```
+Input: matrix = [[3,7,8],[9,11,13],[15,16,17]]
+Output: [15]
+Explanation: 15 is the only lucky number since it is the minimum in its row and the maximum in its column
+```
+
+**Example 2**:
+
+```
+Input: matrix = [[1,10,4,2],[9,3,8,7],[15,16,17,12]]
+Output: [12]
+Explanation: 12 is the only lucky number since it is the minimum in its row and the maximum in its column.
+```
+
+**Example 3**:
+
+```
+Input: matrix = [[7,8],[1,2]]
+Output: [7]
+```
+
+**Constraints**:
+
+- `m == mat.length`
+- `n == mat[i].length`
+- `1 <= n, m <= 50`
+- `1 <= matrix[i][j] <= 10^5`.
+- All elements in the matrix are distinct.
+
+## 题目大意
+
+给你一个 m * n 的矩阵，矩阵中的数字 各不相同 。请你按 任意 顺序返回矩阵中的所有幸运数。幸运数是指矩阵中满足同时下列两个条件的元素：
+
+- 在同一行的所有元素中最小
+- 在同一列的所有元素中最大
+
+
+
+## 解题思路
+
+- 找出矩阵中的幸运数。幸运数的定义：同时满足 2 个条件，在同一行的所有元素中最小并且在同一列的所有元素中最大。
+- 简单题。按照题意遍历矩阵，找到同时满足 2 个条件的数输出即可。
+
+## 代码
+
+```go
+func luckyNumbers(matrix [][]int) []int {
+	t, r, res := make([]int, len(matrix[0])), make([]int, len(matrix[0])), []int{}
+	for _, val := range matrix {
+		m, k := val[0], 0
+		for j := 0; j < len(matrix[0]); j++ {
+			if val[j] < m {
+				m = val[j]
+				k = j
+			}
+			if t[j] < val[j] {
+				t[j] = val[j]
+			}
+		}
+
+		if t[k] == m {
+			r[k] = m
+		}
+	}
+	for k, v := range r {
+		if v > 0 && v == t[k] {
+			res = append(res, v)
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1383.Maximum-Performance-of-a-Team/1383. Maximum Performance of a Team.go b/leetcode/1383.Maximum-Performance-of-a-Team/1383. Maximum Performance of a Team.go
new file mode 100644
index 000000000..db69d3781
--- /dev/null
+++ b/leetcode/1383.Maximum-Performance-of-a-Team/1383. Maximum Performance of a Team.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+func maxPerformance(n int, speed []int, efficiency []int, k int) int {
+	indexes := make([]int, n)
+	for i := range indexes {
+		indexes[i] = i
+	}
+	sort.Slice(indexes, func(i, j int) bool {
+		return efficiency[indexes[i]] > efficiency[indexes[j]]
+	})
+	ph := speedHeap{}
+	heap.Init(&ph)
+	speedSum := 0
+	var max int64
+	for _, index := range indexes {
+		if ph.Len() == k {
+			speedSum -= heap.Pop(&ph).(int)
+		}
+		speedSum += speed[index]
+		heap.Push(&ph, speed[index])
+		max = Max(max, int64(speedSum)*int64(efficiency[index]))
+	}
+	return int(max % (1e9 + 7))
+}
+
+type speedHeap []int
+
+func (h speedHeap) Less(i, j int) bool  { return h[i] < h[j] }
+func (h speedHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
+func (h speedHeap) Len() int            { return len(h) }
+func (h *speedHeap) Push(x interface{}) { *h = append(*h, x.(int)) }
+func (h *speedHeap) Pop() interface{} {
+	res := (*h)[len(*h)-1]
+	*h = (*h)[:h.Len()-1]
+	return res
+}
+
+func Max(a, b int64) int64 {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1383.Maximum-Performance-of-a-Team/1383. Maximum Performance of a Team_test.go b/leetcode/1383.Maximum-Performance-of-a-Team/1383. Maximum Performance of a Team_test.go
new file mode 100644
index 000000000..0caa6af08
--- /dev/null
+++ b/leetcode/1383.Maximum-Performance-of-a-Team/1383. Maximum Performance of a Team_test.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1383 struct {
+	para1383
+	ans1383
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1383 struct {
+	n          int
+	speed      []int
+	efficiency []int
+	k          int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1383 struct {
+	one int
+}
+
+func Test_Problem1383(t *testing.T) {
+
+	qs := []question1383{
+
+		{
+			para1383{6, []int{2, 10, 3, 1, 5, 8}, []int{5, 4, 3, 9, 7, 2}, 2},
+			ans1383{60},
+		},
+
+		{
+			para1383{6, []int{2, 10, 3, 1, 5, 8}, []int{5, 4, 3, 9, 7, 2}, 3},
+			ans1383{68},
+		},
+
+		{
+			para1383{6, []int{2, 10, 3, 1, 5, 8}, []int{5, 4, 3, 9, 7, 2}, 4},
+			ans1383{72},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1383------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1383, q.para1383
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxPerformance(p.n, p.speed, p.efficiency, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1383.Maximum-Performance-of-a-Team/README.md b/leetcode/1383.Maximum-Performance-of-a-Team/README.md
new file mode 100644
index 000000000..ce941cfd2
--- /dev/null
+++ b/leetcode/1383.Maximum-Performance-of-a-Team/README.md
@@ -0,0 +1,105 @@
+# [1383. Maximum Performance of a Team](https://leetcode.com/problems/maximum-performance-of-a-team/)
+
+## 题目
+
+You are given two integers `n` and `k` and two integer arrays `speed` and `efficiency` both of length `n`. There are `n` engineers numbered from `1` to `n`. `speed[i]` and `efficiency[i]` represent the speed and efficiency of the `ith` engineer respectively.
+
+Choose **at most** `k` different engineers out of the `n` engineers to form a team with the maximum **performance**.
+
+The performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers.
+
+Return *the maximum performance of this team*. Since the answer can be a huge number, return it **modulo** `109 + 7`.
+
+**Example 1:**
+
+```
+Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
+Output: 60
+Explanation:
+We have the maximum performance of the team by selecting engineer 2 (with speed=10 and efficiency=4) and engineer 5 (with speed=5 and efficiency=7). That is, performance = (10 + 5) * min(4, 7) = 60.
+```
+
+**Example 2:**
+
+```
+Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
+Output: 68
+Explanation:
+This is the same example as the first but k = 3. We can select engineer 1, engineer 2 and engineer 5 to get the maximum performance of the team. That is, performance = (2 + 10 + 5) * min(5, 4, 7) = 68.
+```
+
+**Example 3:**
+
+```
+Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4
+Output: 72
+```
+
+**Constraints:**
+
+- `1 <= <= k <= n <= 105`
+- `speed.length == n`
+- `efficiency.length == n`
+- `1 <= speed[i] <= 105`
+- `1 <= efficiency[i] <= 108`
+
+## 题目大意
+
+公司有编号为 1 到 n 的 n 个工程师，给你两个数组 speed 和 efficiency ，其中 speed[i] 和 efficiency[i] 分别代表第 i 位工程师的速度和效率。请你返回由最多 k 个工程师组成的 最大团队表现值 ，由于答案可能很大，请你返回结果对 10^9 + 7 取余后的结果。团队表现值 的定义为：一个团队中「所有工程师速度的和」乘以他们「效率值中的最小值」。
+
+## 解题思路
+
+- 题目要求返回最大团队表现值，表现值需要考虑速度的累加和，和效率的最小值。即使速度快，效率的最小值很小，总的表现值还是很小。先将效率从大到小排序。从效率高的工程师开始选起，遍历过程中维护一个大小为 k 的速度最小堆。每次遍历都计算一次团队最大表现值。扫描完成，最大团队表现值也筛选出来了。具体实现见下面的代码。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+func maxPerformance(n int, speed []int, efficiency []int, k int) int {
+	indexes := make([]int, n)
+	for i := range indexes {
+		indexes[i] = i
+	}
+	sort.Slice(indexes, func(i, j int) bool {
+		return efficiency[indexes[i]] > efficiency[indexes[j]]
+	})
+	ph := speedHeap{}
+	heap.Init(&ph)
+	speedSum := 0
+	var max int64
+	for _, index := range indexes {
+		if ph.Len() == k {
+			speedSum -= heap.Pop(&ph).(int)
+		}
+		speedSum += speed[index]
+		heap.Push(&ph, speed[index])
+		max = Max(max, int64(speedSum)*int64(efficiency[index]))
+	}
+	return int(max % (1e9 + 7))
+}
+
+type speedHeap []int
+
+func (h speedHeap) Less(i, j int) bool  { return h[i] < h[j] }
+func (h speedHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
+func (h speedHeap) Len() int            { return len(h) }
+func (h *speedHeap) Push(x interface{}) { *h = append(*h, x.(int)) }
+func (h *speedHeap) Pop() interface{} {
+	res := (*h)[len(*h)-1]
+	*h = (*h)[:h.Len()-1]
+	return res
+}
+
+func Max(a, b int64) int64 {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1385.Find-the-Distance-Value-Between-Two-Arrays/1385. Find the Distance Value Between Two Arrays.go b/leetcode/1385.Find-the-Distance-Value-Between-Two-Arrays/1385. Find the Distance Value Between Two Arrays.go
new file mode 100644
index 000000000..b622bea21
--- /dev/null
+++ b/leetcode/1385.Find-the-Distance-Value-Between-Two-Arrays/1385. Find the Distance Value Between Two Arrays.go	
@@ -0,0 +1,23 @@
+package leetcode
+
+func findTheDistanceValue(arr1 []int, arr2 []int, d int) int {
+	res := 0
+	for i := range arr1 {
+		for j := range arr2 {
+			if abs(arr1[i]-arr2[j]) <= d {
+				break
+			}
+			if j == len(arr2)-1 {
+				res++
+			}
+		}
+	}
+	return res
+}
+
+func abs(a int) int {
+	if a < 0 {
+		return -1 * a
+	}
+	return a
+}
diff --git a/leetcode/1385.Find-the-Distance-Value-Between-Two-Arrays/1385. Find the Distance Value Between Two Arrays_test.go b/leetcode/1385.Find-the-Distance-Value-Between-Two-Arrays/1385. Find the Distance Value Between Two Arrays_test.go
new file mode 100644
index 000000000..070bf12ad
--- /dev/null
+++ b/leetcode/1385.Find-the-Distance-Value-Between-Two-Arrays/1385. Find the Distance Value Between Two Arrays_test.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1385 struct {
+	para1385
+	ans1385
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1385 struct {
+	arr1 []int
+	arr2 []int
+	d    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1385 struct {
+	one []int
+}
+
+func Test_Problem1385(t *testing.T) {
+
+	qs := []question1385{
+
+		{
+			para1385{[]int{4, 5, 8}, []int{10, 9, 1, 8}, 2},
+			ans1385{[]int{2}},
+		},
+
+		{
+			para1385{[]int{1, 4, 2, 3}, []int{-4, -3, 6, 10, 20, 30}, 3},
+			ans1385{[]int{2}},
+		},
+
+		{
+			para1385{[]int{2, 1, 100, 3}, []int{-5, -2, 10, -3, 7}, 6},
+			ans1385{[]int{1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1385------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1385, q.para1385
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", findTheDistanceValue(p.arr1, p.arr2, p.d))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1385.Find-the-Distance-Value-Between-Two-Arrays/README.md b/leetcode/1385.Find-the-Distance-Value-Between-Two-Arrays/README.md
new file mode 100644
index 000000000..fc5d6195c
--- /dev/null
+++ b/leetcode/1385.Find-the-Distance-Value-Between-Two-Arrays/README.md
@@ -0,0 +1,94 @@
+# [1385. Find the Distance Value Between Two Arrays](https://leetcode.com/problems/find-the-distance-value-between-two-arrays/)
+
+
+## 题目
+
+Given two integer arrays `arr1` and `arr2`, and the integer `d`, *return the distance value between the two arrays*.
+
+The distance value is defined as the number of elements `arr1[i]` such that there is not any element `arr2[j]` where `|arr1[i]-arr2[j]| <= d`.
+
+**Example 1**:
+
+```
+Input: arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2
+Output: 2
+Explanation: 
+For arr1[0]=4 we have: 
+|4-10|=6 > d=2 
+|4-9|=5 > d=2 
+|4-1|=3 > d=2 
+|4-8|=4 > d=2 
+For arr1[1]=5 we have: 
+|5-10|=5 > d=2 
+|5-9|=4 > d=2 
+|5-1|=4 > d=2 
+|5-8|=3 > d=2
+For arr1[2]=8 we have:
+|8-10|=2 <= d=2
+|8-9|=1 <= d=2
+|8-1|=7 > d=2
+|8-8|=0 <= d=2
+```
+
+**Example 2**:
+
+```
+Input: arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3
+Output: 2
+```
+
+**Example 3**:
+
+```
+Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6
+Output: 1
+```
+
+**Constraints**:
+
+- `1 <= arr1.length, arr2.length <= 500`
+- `-10^3 <= arr1[i], arr2[j] <= 10^3`
+- `0 <= d <= 100`
+
+
+## 题目大意
+
+给你两个整数数组 arr1 ， arr2 和一个整数 d ，请你返回两个数组之间的 距离值 。「距离值」 定义为符合此距离要求的元素数目：对于元素 arr1[i] ，不存在任何元素 arr2[j] 满足 |arr1[i]-arr2[j]| <= d 。
+
+提示：
+
+- 1 <= arr1.length, arr2.length <= 500
+- -10^3 <= arr1[i], arr2[j] <= 10^3
+- 0 <= d <= 100
+
+
+## 解题思路
+
+- 计算两个数组之间的距离，距离值的定义：满足对于元素 arr1[i] ，不存在任何元素 arr2[j] 满足 |arr1[i]-arr2[j]| <= d 这一条件的元素数目。
+- 简单题，按照距离值的定义，双层循环计数即可。
+
+## 代码
+
+```go
+func findTheDistanceValue(arr1 []int, arr2 []int, d int) int {
+	res := 0
+	for i := range arr1 {
+		for j := range arr2 {
+			if abs(arr1[i]-arr2[j]) <= d {
+				break
+			}
+			if j == len(arr2)-1 {
+				res++
+			}
+		}
+	}
+	return res
+}
+
+func abs(a int) int {
+	if a < 0 {
+		return -1 * a
+	}
+	return a
+}
+```
\ No newline at end of file
diff --git a/leetcode/1389.Create-Target-Array-in-the-Given-Order/1389. Create Target Array in the Given Order.go b/leetcode/1389.Create-Target-Array-in-the-Given-Order/1389. Create Target Array in the Given Order.go
new file mode 100644
index 000000000..065aeb704
--- /dev/null
+++ b/leetcode/1389.Create-Target-Array-in-the-Given-Order/1389. Create Target Array in the Given Order.go	
@@ -0,0 +1,10 @@
+package leetcode
+
+func createTargetArray(nums []int, index []int) []int {
+	result := make([]int, len(nums))
+	for i, pos := range index {
+		copy(result[pos+1:i+1], result[pos:i])
+		result[pos] = nums[i]
+	}
+	return result
+}
diff --git a/leetcode/1389.Create-Target-Array-in-the-Given-Order/1389. Create Target Array in the Given Order_test.go b/leetcode/1389.Create-Target-Array-in-the-Given-Order/1389. Create Target Array in the Given Order_test.go
new file mode 100644
index 000000000..779d36dff
--- /dev/null
+++ b/leetcode/1389.Create-Target-Array-in-the-Given-Order/1389. Create Target Array in the Given Order_test.go	
@@ -0,0 +1,54 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1389 struct {
+	para1389
+	ans1389
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1389 struct {
+	nums  []int
+	index []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1389 struct {
+	one []int
+}
+
+func Test_Problem1389(t *testing.T) {
+
+	qs := []question1389{
+
+		{
+			para1389{[]int{0, 1, 2, 3, 4}, []int{0, 1, 2, 2, 1}},
+			ans1389{[]int{0, 4, 1, 3, 2}},
+		},
+
+		{
+			para1389{[]int{1, 2, 3, 4, 0}, []int{0, 1, 2, 3, 0}},
+			ans1389{[]int{0, 1, 2, 3, 4}},
+		},
+
+		{
+			para1389{[]int{1}, []int{0}},
+			ans1389{[]int{1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1389------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1389, q.para1389
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", createTargetArray(p.nums, p.index))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1389.Create-Target-Array-in-the-Given-Order/README.md b/leetcode/1389.Create-Target-Array-in-the-Given-Order/README.md
new file mode 100644
index 000000000..6b653d99c
--- /dev/null
+++ b/leetcode/1389.Create-Target-Array-in-the-Given-Order/README.md
@@ -0,0 +1,86 @@
+# [1389. Create Target Array in the Given Order](https://leetcode.com/problems/create-target-array-in-the-given-order/)
+
+## 题目
+
+Given two arrays of integers `nums` and `index`. Your task is to create *target* array under the following rules:
+
+- Initially *target* array is empty.
+- From left to right read nums[i] and index[i], insert at index `index[i]` the value `nums[i]` in *target* array.
+- Repeat the previous step until there are no elements to read in `nums` and `index.`
+
+Return the *target* array.
+
+It is guaranteed that the insertion operations will be valid.
+
+**Example 1**:
+
+```
+Input: nums = [0,1,2,3,4], index = [0,1,2,2,1]
+Output: [0,4,1,3,2]
+Explanation:
+nums       index     target
+0            0        [0]
+1            1        [0,1]
+2            2        [0,1,2]
+3            2        [0,1,3,2]
+4            1        [0,4,1,3,2]
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,2,3,4,0], index = [0,1,2,3,0]
+Output: [0,1,2,3,4]
+Explanation:
+nums       index     target
+1            0        [1]
+2            1        [1,2]
+3            2        [1,2,3]
+4            3        [1,2,3,4]
+0            0        [0,1,2,3,4]
+```
+
+**Example 3**:
+
+```
+Input: nums = [1], index = [0]
+Output: [1]
+```
+
+**Constraints**:
+
+- `1 <= nums.length, index.length <= 100`
+- `nums.length == index.length`
+- `0 <= nums[i] <= 100`
+- `0 <= index[i] <= i`
+
+## 题目大意
+
+给你两个整数数组 nums 和 index。你需要按照以下规则创建目标数组：
+
+- 目标数组 target 最初为空。
+- 按从左到右的顺序依次读取 nums[i] 和 index[i]，在 target 数组中的下标 index[i] 处插入值 nums[i] 。
+- 重复上一步，直到在 nums 和 index 中都没有要读取的元素。
+
+请你返回目标数组。题目保证数字插入位置总是存在。
+
+
+
+
+## 解题思路
+
+- 给定 2 个数组，分别装的是待插入的元素和待插入的位置。最后输出操作完成的数组。
+- 简单题，按照题意插入元素即可。
+
+## 代码
+
+```go
+func createTargetArray(nums []int, index []int) []int {
+	result := make([]int, len(nums))
+	for i, pos := range index {
+		copy(result[pos+1:i+1], result[pos:i])
+		result[pos] = nums[i]
+	}
+	return result
+}
+```
\ No newline at end of file
diff --git a/leetcode/1396.Design-Underground-System/1396. Design Underground System.go b/leetcode/1396.Design-Underground-System/1396. Design Underground System.go
new file mode 100644
index 000000000..cf9db5045
--- /dev/null
+++ b/leetcode/1396.Design-Underground-System/1396. Design Underground System.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+type checkin struct {
+	station string
+	time    int
+}
+
+type stationTime struct {
+	sum, count float64
+}
+
+type UndergroundSystem struct {
+	checkins     map[int]*checkin
+	stationTimes map[string]map[string]*stationTime
+}
+
+func Constructor() UndergroundSystem {
+	return UndergroundSystem{
+		make(map[int]*checkin),
+		make(map[string]map[string]*stationTime),
+	}
+}
+
+func (s *UndergroundSystem) CheckIn(id int, stationName string, t int) {
+	s.checkins[id] = &checkin{stationName, t}
+}
+
+func (s *UndergroundSystem) CheckOut(id int, stationName string, t int) {
+	checkin := s.checkins[id]
+	destination := s.stationTimes[checkin.station]
+	if destination == nil {
+		s.stationTimes[checkin.station] = make(map[string]*stationTime)
+	}
+	st := s.stationTimes[checkin.station][stationName]
+	if st == nil {
+		st = new(stationTime)
+		s.stationTimes[checkin.station][stationName] = st
+	}
+	st.sum += float64(t - checkin.time)
+	st.count++
+	delete(s.checkins, id)
+}
+
+func (s *UndergroundSystem) GetAverageTime(startStation string, endStation string) float64 {
+	st := s.stationTimes[startStation][endStation]
+	return st.sum / st.count
+}
+
+/**
+ * Your UndergroundSystem object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.CheckIn(id,stationName,t);
+ * obj.CheckOut(id,stationName,t);
+ * param_3 := obj.GetAverageTime(startStation,endStation);
+ */
diff --git a/leetcode/1396.Design-Underground-System/1396. Design Underground System_test.go b/leetcode/1396.Design-Underground-System/1396. Design Underground System_test.go
new file mode 100644
index 000000000..fc7e521c7
--- /dev/null
+++ b/leetcode/1396.Design-Underground-System/1396. Design Underground System_test.go	
@@ -0,0 +1,32 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem(t *testing.T) {
+	undergroundSystem := Constructor()
+	undergroundSystem.CheckIn(45, "Leyton", 3)
+	undergroundSystem.CheckIn(32, "Paradise", 8)
+	undergroundSystem.CheckIn(27, "Leyton", 10)
+	undergroundSystem.CheckOut(45, "Waterloo", 15)
+	undergroundSystem.CheckOut(27, "Waterloo", 20)
+	undergroundSystem.CheckOut(32, "Cambridge", 22)
+	fmt.Println("Output: ", undergroundSystem.GetAverageTime("Paradise", "Cambridge")) // return 14.00000. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
+	fmt.Println("Output: ", undergroundSystem.GetAverageTime("Leyton", "Waterloo"))    // return 11.00000. There were two travels from "Leyton" to "Waterloo", a customer with id=45 from time=3 to time=15 and a customer with id=27 from time=10 to time=20. So the average time is ( (15-3) + (20-10) ) / 2 = 11.00000
+	undergroundSystem.CheckIn(10, "Leyton", 24)
+	fmt.Println("Output: ", undergroundSystem.GetAverageTime("Leyton", "Waterloo")) // return 11.00000
+	undergroundSystem.CheckOut(10, "Waterloo", 38)
+	fmt.Println("Output: ", undergroundSystem.GetAverageTime("Leyton", "Waterloo")) // return 12.00000
+
+	undergroundSystem.CheckIn(10, "Leyton", 3)
+	undergroundSystem.CheckOut(10, "Paradise", 8)
+	fmt.Println("Output: ", undergroundSystem.GetAverageTime("Leyton", "Paradise")) // return 5.00000
+	undergroundSystem.CheckIn(5, "Leyton", 10)
+	undergroundSystem.CheckOut(5, "Paradise", 16)
+	fmt.Println("Output: ", undergroundSystem.GetAverageTime("Leyton", "Paradise")) // return 5.50000
+	undergroundSystem.CheckIn(2, "Leyton", 21)
+	undergroundSystem.CheckOut(2, "Paradise", 30)
+	fmt.Println("Output: ", undergroundSystem.GetAverageTime("Leyton", "Paradise")) // return 6.66667
+}
diff --git a/leetcode/1396.Design-Underground-System/README.md b/leetcode/1396.Design-Underground-System/README.md
new file mode 100644
index 000000000..3f3f83cbd
--- /dev/null
+++ b/leetcode/1396.Design-Underground-System/README.md
@@ -0,0 +1,155 @@
+# [1396. Design Underground System](https://leetcode.com/problems/design-underground-system/)
+
+
+## 题目
+
+Implement the `UndergroundSystem` class:
+
+- `void checkIn(int id, string stationName, int t)`
+    - A customer with a card id equal to `id`, gets in the station `stationName` at time `t`.
+    - A customer can only be checked into one place at a time.
+- `void checkOut(int id, string stationName, int t)`
+    - A customer with a card id equal to `id`, gets out from the station `stationName` at time `t`.
+- `double getAverageTime(string startStation, string endStation)`
+    - Returns the average time to travel between the `startStation` and the `endStation`.
+    - The average time is computed from all the previous traveling from `startStation` to `endStation` that happened **directly**.
+    - Call to `getAverageTime` is always valid.
+
+You can assume all calls to `checkIn` and `checkOut` methods are consistent. If a customer gets in at time **t1** at some station, they get out at time **t2** with **t2 > t1**. All events happen in chronological order.
+
+**Example 1:**
+
+```
+Input
+["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"]
+[[],[45,"Leyton",3],[32,"Paradise",8],[27,"Leyton",10],[45,"Waterloo",15],[27,"Waterloo",20],[32,"Cambridge",22],["Paradise","Cambridge"],["Leyton","Waterloo"],[10,"Leyton",24],["Leyton","Waterloo"],[10,"Waterloo",38],["Leyton","Waterloo"]]
+
+Output
+[null,null,null,null,null,null,null,14.00000,11.00000,null,11.00000,null,12.00000]
+
+Explanation
+UndergroundSystem undergroundSystem = new UndergroundSystem();
+undergroundSystem.checkIn(45, "Leyton", 3);
+undergroundSystem.checkIn(32, "Paradise", 8);
+undergroundSystem.checkIn(27, "Leyton", 10);
+undergroundSystem.checkOut(45, "Waterloo", 15);
+undergroundSystem.checkOut(27, "Waterloo", 20);
+undergroundSystem.checkOut(32, "Cambridge", 22);
+undergroundSystem.getAverageTime("Paradise", "Cambridge");       // return 14.00000. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
+undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.00000. There were two travels from "Leyton" to "Waterloo", a customer with id=45 from time=3 to time=15 and a customer with id=27 from time=10 to time=20. So the average time is ( (15-3) + (20-10) ) / 2 = 11.00000
+undergroundSystem.checkIn(10, "Leyton", 24);
+undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.00000
+undergroundSystem.checkOut(10, "Waterloo", 38);
+undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 12.00000
+```
+
+**Example 2:**
+
+```
+Input
+["UndergroundSystem","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime"]
+[[],[10,"Leyton",3],[10,"Paradise",8],["Leyton","Paradise"],[5,"Leyton",10],[5,"Paradise",16],["Leyton","Paradise"],[2,"Leyton",21],[2,"Paradise",30],["Leyton","Paradise"]]
+
+Output
+[null,null,null,5.00000,null,null,5.50000,null,null,6.66667]
+
+Explanation
+UndergroundSystem undergroundSystem = new UndergroundSystem();
+undergroundSystem.checkIn(10, "Leyton", 3);
+undergroundSystem.checkOut(10, "Paradise", 8);
+undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000
+undergroundSystem.checkIn(5, "Leyton", 10);
+undergroundSystem.checkOut(5, "Paradise", 16);
+undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.50000
+undergroundSystem.checkIn(2, "Leyton", 21);
+undergroundSystem.checkOut(2, "Paradise", 30);
+undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 6.66667
+```
+
+**Constraints:**
+
+- There will be at most `20000` operations.
+- `1 <= id, t <= 106`
+- All strings consist of uppercase and lowercase English letters, and digits.
+- `1 <= stationName.length <= 10`
+- Answers within `105` of the actual value will be accepted as correct.
+
+## 题目大意
+
+请你实现一个类 UndergroundSystem ，它支持以下 3 种方法：
+
+- 1. checkIn(int id, string stationName, int t)
+    - 编号为 id 的乘客在 t 时刻进入地铁站 stationName 。
+    - 一个乘客在同一时间只能在一个地铁站进入或者离开。
+- 2. checkOut(int id, string stationName, int t)
+    - 编号为 id 的乘客在 t 时刻离开地铁站 stationName 。
+- 3. getAverageTime(string startStation, string endStation)
+    - 返回从地铁站 startStation 到地铁站 endStation 的平均花费时间。
+    - 平均时间计算的行程包括当前为止所有从 startStation 直接到达 endStation 的行程。
+    - 调用 getAverageTime 时，询问的路线至少包含一趟行程。
+
+你可以假设所有对 checkIn 和 checkOut 的调用都是符合逻辑的。也就是说，如果一个顾客在 t1 时刻到达某个地铁站，那么他离开的时间 t2 一定满足 t2 > t1 。所有的事件都按时间顺序给出。
+
+## 解题思路
+
+- 维护 2 个 `map`。一个 `mapA` 内部存储的是乘客 `id` 与（入站时间，站名）的对应关系。另外一个 `mapB` 存储的是起点站与终点站花费总时间与人数总数的关系。每当有人 `checkin()`，就更新 `mapA` 中的信息。每当有人 `checkout()`，就更新 `mapB` 中的信息，并删除 `mapA` 对应乘客 `id` 的键值对。最后调用 `getAverageTime()` 函数的时候根据 `mapB` 中存储的信息计算即可。
+
+## 代码
+
+```go
+package leetcode
+
+type checkin struct {
+	station string
+	time    int
+}
+
+type stationTime struct {
+	sum, count float64
+}
+
+type UndergroundSystem struct {
+	checkins     map[int]*checkin
+	stationTimes map[string]map[string]*stationTime
+}
+
+func Constructor() UndergroundSystem {
+	return UndergroundSystem{
+		make(map[int]*checkin),
+		make(map[string]map[string]*stationTime),
+	}
+}
+
+func (s *UndergroundSystem) CheckIn(id int, stationName string, t int) {
+	s.checkins[id] = &checkin{stationName, t}
+}
+
+func (s *UndergroundSystem) CheckOut(id int, stationName string, t int) {
+	checkin := s.checkins[id]
+	destination := s.stationTimes[checkin.station]
+	if destination == nil {
+		s.stationTimes[checkin.station] = make(map[string]*stationTime)
+	}
+	st := s.stationTimes[checkin.station][stationName]
+	if st == nil {
+		st = new(stationTime)
+		s.stationTimes[checkin.station][stationName] = st
+	}
+	st.sum += float64(t - checkin.time)
+	st.count++
+	delete(s.checkins, id)
+}
+
+func (s *UndergroundSystem) GetAverageTime(startStation string, endStation string) float64 {
+	st := s.stationTimes[startStation][endStation]
+	return st.sum / st.count
+}
+
+/**
+ * Your UndergroundSystem object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.CheckIn(id,stationName,t);
+ * obj.CheckOut(id,stationName,t);
+ * param_3 := obj.GetAverageTime(startStation,endStation);
+ */
+```
\ No newline at end of file
diff --git a/leetcode/1423.Maximum-Points-You-Can-Obtain-from-Cards/1423. Maximum Points You Can Obtain from Cards.go b/leetcode/1423.Maximum-Points-You-Can-Obtain-from-Cards/1423. Maximum Points You Can Obtain from Cards.go
new file mode 100644
index 000000000..c237e7e40
--- /dev/null
+++ b/leetcode/1423.Maximum-Points-You-Can-Obtain-from-Cards/1423. Maximum Points You Can Obtain from Cards.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+func maxScore(cardPoints []int, k int) int {
+	windowSize, sum := len(cardPoints)-k, 0
+	for _, val := range cardPoints[:windowSize] {
+		sum += val
+	}
+	minSum := sum
+	for i := windowSize; i < len(cardPoints); i++ {
+		sum += cardPoints[i] - cardPoints[i-windowSize]
+		if sum < minSum {
+			minSum = sum
+		}
+	}
+	total := 0
+	for _, pt := range cardPoints {
+		total += pt
+	}
+	return total - minSum
+}
diff --git a/leetcode/1423.Maximum-Points-You-Can-Obtain-from-Cards/1423. Maximum Points You Can Obtain from Cards_test.go b/leetcode/1423.Maximum-Points-You-Can-Obtain-from-Cards/1423. Maximum Points You Can Obtain from Cards_test.go
new file mode 100644
index 000000000..9cae408dc
--- /dev/null
+++ b/leetcode/1423.Maximum-Points-You-Can-Obtain-from-Cards/1423. Maximum Points You Can Obtain from Cards_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1423 struct {
+	para1423
+	ans1423
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1423 struct {
+	cardPoints []int
+	k          int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1423 struct {
+	one int
+}
+
+func Test_Problem1423(t *testing.T) {
+
+	qs := []question1423{
+
+		{
+			para1423{[]int{1, 2, 3, 4, 5, 6, 1}, 3},
+			ans1423{12},
+		},
+
+		{
+			para1423{[]int{2, 2, 2}, 2},
+			ans1423{4},
+		},
+
+		{
+			para1423{[]int{9, 7, 7, 9, 7, 7, 9}, 7},
+			ans1423{55},
+		},
+
+		{
+			para1423{[]int{1, 1000, 1}, 1},
+			ans1423{1},
+		},
+
+		{
+			para1423{[]int{1, 79, 80, 1, 1, 1, 200, 1}, 3},
+			ans1423{202},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1423------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1423, q.para1423
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxScore(p.cardPoints, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1423.Maximum-Points-You-Can-Obtain-from-Cards/README.md b/leetcode/1423.Maximum-Points-You-Can-Obtain-from-Cards/README.md
new file mode 100644
index 000000000..70c8e55b0
--- /dev/null
+++ b/leetcode/1423.Maximum-Points-You-Can-Obtain-from-Cards/README.md
@@ -0,0 +1,90 @@
+# [1423. Maximum Points You Can Obtain from Cards](https://leetcode.com/problems/maximum-points-you-can-obtain-from-cards/)
+
+
+## 题目
+
+There are several cards **arranged in a row**, and each card has an associated number of points The points are given in the integer array `cardPoints`.
+
+In one step, you can take one card from the beginning or from the end of the row. You have to take exactly `k` cards.
+
+Your score is the sum of the points of the cards you have taken.
+
+Given the integer array `cardPoints` and the integer `k`, return the *maximum score* you can obtain.
+
+**Example 1:**
+
+```
+Input: cardPoints = [1,2,3,4,5,6,1], k = 3
+Output: 12
+Explanation: After the first step, your score will always be 1. However, choosing the rightmost card first will maximize your total score. The optimal strategy is to take the three cards on the right, giving a final score of 1 + 6 + 5 = 12.
+```
+
+**Example 2:**
+
+```
+Input: cardPoints = [2,2,2], k = 2
+Output: 4
+Explanation: Regardless of which two cards you take, your score will always be 4.
+```
+
+**Example 3:**
+
+```
+Input: cardPoints = [9,7,7,9,7,7,9], k = 7
+Output: 55
+Explanation: You have to take all the cards. Your score is the sum of points of all cards.
+```
+
+**Example 4:**
+
+```
+Input: cardPoints = [1,1000,1], k = 1
+Output: 1
+Explanation: You cannot take the card in the middle. Your best score is 1. 
+```
+
+**Example 5:**
+
+```
+Input: cardPoints = [1,79,80,1,1,1,200,1], k = 3
+Output: 202
+```
+
+**Constraints:**
+
+- `1 <= cardPoints.length <= 10^5`
+- `1 <= cardPoints[i] <= 10^4`
+- `1 <= k <= cardPoints.length`
+
+## 题目大意
+
+几张卡牌 排成一行，每张卡牌都有一个对应的点数。点数由整数数组 cardPoints 给出。每次行动，你可以从行的开头或者末尾拿一张卡牌，最终你必须正好拿 k 张卡牌。你的点数就是你拿到手中的所有卡牌的点数之和。给你一个整数数组 cardPoints 和整数 k，请你返回可以获得的最大点数。
+
+## 解题思路
+
+- 这一题是滑动窗口题的简化题。从卡牌两边取 K 张牌，可以转换成在中间连续取 n-K 张牌。从两边取牌的点数最大，意味着剩下来中间牌的点数最小。扫描一遍数组，在每一个窗口大小为 n-K 的窗口内计算累加和，记录下最小的累加和。题目最终求的最大点数等于牌的总和减去中间最小的累加和。
+
+## 代码
+
+```go
+package leetcode
+
+func maxScore(cardPoints []int, k int) int {
+	windowSize, sum := len(cardPoints)-k, 0
+	for _, val := range cardPoints[:windowSize] {
+		sum += val
+	}
+	minSum := sum
+	for i := windowSize; i < len(cardPoints); i++ {
+		sum += cardPoints[i] - cardPoints[i-windowSize]
+		if sum < minSum {
+			minSum = sum
+		}
+	}
+	total := 0
+	for _, pt := range cardPoints {
+		total += pt
+	}
+	return total - minSum
+}
+```
\ No newline at end of file
diff --git a/leetcode/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away/1437. Check If All 1s Are at Least Length K Places Away.go b/leetcode/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away/1437. Check If All 1s Are at Least Length K Places Away.go
new file mode 100644
index 000000000..029ecfbd0
--- /dev/null
+++ b/leetcode/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away/1437. Check If All 1s Are at Least Length K Places Away.go	
@@ -0,0 +1,14 @@
+package leetcode
+
+func kLengthApart(nums []int, k int) bool {
+	prevIndex := -1
+	for i, num := range nums {
+		if num == 1 {
+			if prevIndex != -1 && i-prevIndex-1 < k {
+				return false
+			}
+			prevIndex = i
+		}
+	}
+	return true
+}
diff --git a/leetcode/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away/1437. Check If All 1s Are at Least Length K Places Away_test.go b/leetcode/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away/1437. Check If All 1s Are at Least Length K Places Away_test.go
new file mode 100644
index 000000000..dbd26ccf6
--- /dev/null
+++ b/leetcode/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away/1437. Check If All 1s Are at Least Length K Places Away_test.go	
@@ -0,0 +1,59 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1437 struct {
+	para1437
+	ans1437
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1437 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1437 struct {
+	one bool
+}
+
+func Test_Problem1437(t *testing.T) {
+
+	qs := []question1437{
+
+		{
+			para1437{[]int{1, 0, 0, 0, 1, 0, 0, 1}, 2},
+			ans1437{true},
+		},
+
+		{
+			para1437{[]int{1, 0, 0, 1, 0, 1}, 2},
+			ans1437{false},
+		},
+
+		{
+			para1437{[]int{1, 1, 1, 1, 1}, 0},
+			ans1437{true},
+		},
+
+		{
+			para1437{[]int{0, 1, 0, 1}, 1},
+			ans1437{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1437------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1437, q.para1437
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", kLengthApart(p.nums, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away/README.md b/leetcode/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away/README.md
new file mode 100644
index 000000000..8efb6f5b9
--- /dev/null
+++ b/leetcode/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away/README.md
@@ -0,0 +1,73 @@
+# [1437. Check If All 1's Are at Least Length K Places Away](https://leetcode.com/problems/check-if-all-1s-are-at-least-length-k-places-away/)
+
+
+## 题目
+
+Given an array `nums` of 0s and 1s and an integer `k`, return `True` if all 1's are at least `k` places away from each other, otherwise return `False`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/04/15/sample_1_1791.png](https://assets.leetcode.com/uploads/2020/04/15/sample_1_1791.png)
+
+```
+Input: nums = [1,0,0,0,1,0,0,1], k = 2
+Output: true
+Explanation: Each of the 1s are at least 2 places away from each other.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/04/15/sample_2_1791.png](https://assets.leetcode.com/uploads/2020/04/15/sample_2_1791.png)
+
+```
+Input: nums = [1,0,0,1,0,1], k = 2
+Output: false
+Explanation: The second 1 and third 1 are only one apart from each other.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,1,1,1,1], k = 0
+Output: true
+```
+
+**Example 4:**
+
+```
+Input: nums = [0,1,0,1], k = 1
+Output: true
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 10^5`
+- `0 <= k <= nums.length`
+- `nums[i]` is `0` or `1`
+
+## 题目大意
+
+给你一个由若干 0 和 1 组成的数组 nums 以及整数 k。如果所有 1 都至少相隔 k 个元素，则返回 True ；否则，返回 False 。
+
+## 解题思路
+
+- 简单题。扫描一遍数组，遇到 1 的时候比较前一个 1 的下标索引，如果相隔小于 k 则返回 false。如果大于等于 k 就更新下标索引，继续循环。循环结束输出 true 即可。
+
+## 代码
+
+```go
+package leetcode
+
+func kLengthApart(nums []int, k int) bool {
+	prevIndex := -1
+	for i, num := range nums {
+		if num == 1 {
+			if prevIndex != -1 && i-prevIndex-1 < k {
+				return false
+			}
+			prevIndex = i
+		}
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit/1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit.go b/leetcode/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit/1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit.go
new file mode 100644
index 000000000..7589a9408
--- /dev/null
+++ b/leetcode/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit/1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit.go	
@@ -0,0 +1,28 @@
+package leetcode
+
+func longestSubarray(nums []int, limit int) int {
+	minStack, maxStack, left, res := []int{}, []int{}, 0, 0
+	for right, num := range nums {
+		for len(minStack) > 0 && nums[minStack[len(minStack)-1]] > num {
+			minStack = minStack[:len(minStack)-1]
+		}
+		minStack = append(minStack, right)
+		for len(maxStack) > 0 && nums[maxStack[len(maxStack)-1]] < num {
+			maxStack = maxStack[:len(maxStack)-1]
+		}
+		maxStack = append(maxStack, right)
+		if len(minStack) > 0 && len(maxStack) > 0 && nums[maxStack[0]]-nums[minStack[0]] > limit {
+			if left == minStack[0] {
+				minStack = minStack[1:]
+			}
+			if left == maxStack[0] {
+				maxStack = maxStack[1:]
+			}
+			left++
+		}
+		if right-left+1 > res {
+			res = right - left + 1
+		}
+	}
+	return res
+}
diff --git a/leetcode/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit/1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit_test.go b/leetcode/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit/1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit_test.go
new file mode 100644
index 000000000..278366837
--- /dev/null
+++ b/leetcode/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit/1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1438 struct {
+	para1438
+	ans1438
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1438 struct {
+	nums  []int
+	limit int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1438 struct {
+	one int
+}
+
+func Test_Problem1438(t *testing.T) {
+
+	qs := []question1438{
+
+		{
+			para1438{[]int{8, 2, 4, 7}, 4},
+			ans1438{2},
+		},
+
+		{
+			para1438{[]int{10, 1, 2, 4, 7, 2}, 5},
+			ans1438{4},
+		},
+
+		{
+			para1438{[]int{4, 2, 2, 2, 4, 4, 2, 2}, 0},
+			ans1438{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1438------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1438, q.para1438
+		fmt.Printf("【input】:%v       【output】:%v\n", p, longestSubarray(p.nums, p.limit))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit/README.md b/leetcode/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit/README.md
new file mode 100644
index 000000000..eac1b5a80
--- /dev/null
+++ b/leetcode/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit/README.md
@@ -0,0 +1,89 @@
+# [1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit](https://leetcode.com/problems/longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit/)
+
+
+## 题目
+
+Given an array of integers `nums` and an integer `limit`, return the size of the longest **non-empty** subarray such that the absolute difference between any two elements of this subarray is less than or equal to `limit`*.*
+
+**Example 1:**
+
+```
+Input: nums = [8,2,4,7], limit = 4
+Output: 2 
+Explanation: All subarrays are: 
+[8] with maximum absolute diff |8-8| = 0 <= 4.
+[8,2] with maximum absolute diff |8-2| = 6 > 4. 
+[8,2,4] with maximum absolute diff |8-2| = 6 > 4.
+[8,2,4,7] with maximum absolute diff |8-2| = 6 > 4.
+[2] with maximum absolute diff |2-2| = 0 <= 4.
+[2,4] with maximum absolute diff |2-4| = 2 <= 4.
+[2,4,7] with maximum absolute diff |2-7| = 5 > 4.
+[4] with maximum absolute diff |4-4| = 0 <= 4.
+[4,7] with maximum absolute diff |4-7| = 3 <= 4.
+[7] with maximum absolute diff |7-7| = 0 <= 4. 
+Therefore, the size of the longest subarray is 2.
+```
+
+**Example 2:**
+
+```
+Input: nums = [10,1,2,4,7,2], limit = 5
+Output: 4 
+Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5.
+```
+
+**Example 3:**
+
+```
+Input: nums = [4,2,2,2,4,4,2,2], limit = 0
+Output: 3
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 10^5`
+- `1 <= nums[i] <= 10^9`
+- `0 <= limit <= 10^9`
+
+## 题目大意
+
+给你一个整数数组 nums ，和一个表示限制的整数 limit，请你返回最长连续子数组的长度，该子数组中的任意两个元素之间的绝对差必须小于或者等于 limit 。如果不存在满足条件的子数组，则返回 0 。
+
+## 解题思路
+
+- 最开始想到的思路是利用滑动窗口遍历一遍数组，每个窗口内排序，取出最大最小值。滑动窗口遍历一次的时间复杂度是 O(n)，所以此题时间复杂度是否高效落在了排序算法上了。由于前后 2 个窗口数据是有关联的，仅仅只变动了 2 个数据（左窗口移出的数据和右窗口移进的数据），所以排序没有必要每次都重新排序。这里利用二叉排序树来排序，添加和删除元素时间复杂度是 O(log n)，这种方法总的时间复杂度是 O(n log n)。空间复杂度 O(n)。
+- 二叉排序树的思路是否还有再优化的空间？答案是有。二叉排序树内维护了所有结点的有序关系，但是这个关系是多余的。此题只需要找到最大值和最小值，并不需要除此以外节点的有序信息。所以用二叉排序树是大材小用了。可以换成 2 个单调队列，一个维护窗口内的最大值，另一个维护窗口内的最小值。这样优化以后，时间复杂度降低到 O(n)，空间复杂度 O(n)。具体实现见代码。
+- 单调栈的题还有第 42 题，第 84 题，第 496 题，第 503 题，第 739 题，第 856 题，第 901 题，第 907 题，第 1130 题，第 1425 题，第 1673 题。
+
+## 代码
+
+```go
+package leetcode
+
+func longestSubarray(nums []int, limit int) int {
+	minStack, maxStack, left, res := []int{}, []int{}, 0, 0
+	for right, num := range nums {
+		for len(minStack) > 0 && nums[minStack[len(minStack)-1]] > num {
+			minStack = minStack[:len(minStack)-1]
+		}
+		minStack = append(minStack, right)
+		for len(maxStack) > 0 && nums[maxStack[len(maxStack)-1]] < num {
+			maxStack = maxStack[:len(maxStack)-1]
+		}
+		maxStack = append(maxStack, right)
+		if len(minStack) > 0 && len(maxStack) > 0 && nums[maxStack[0]]-nums[minStack[0]] > limit {
+			if left == minStack[0] {
+				minStack = minStack[1:]
+			}
+			if left == maxStack[0] {
+				maxStack = maxStack[1:]
+			}
+			left++
+		}
+		if right-left+1 > res {
+			res = right - left + 1
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows.go b/leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows.go
new file mode 100644
index 000000000..d3f5aa159
--- /dev/null
+++ b/leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows.go	
@@ -0,0 +1,82 @@
+package leetcode
+
+import "container/heap"
+
+func kthSmallest(mat [][]int, k int) int {
+	if len(mat) == 0 || len(mat[0]) == 0 || k == 0 {
+		return 0
+	}
+	prev := mat[0]
+	for i := 1; i < len(mat); i++ {
+		prev = kSmallestPairs(prev, mat[i], k)
+	}
+	if k < len(prev) {
+		return -1
+	}
+	return prev[k-1]
+}
+
+func kSmallestPairs(nums1 []int, nums2 []int, k int) []int {
+	res := []int{}
+	if len(nums2) == 0 {
+		return res
+	}
+	pq := newPriorityQueue()
+	for i := 0; i < len(nums1) && i < k; i++ {
+		heap.Push(pq, &pddata{
+			n1:    nums1[i],
+			n2:    nums2[0],
+			n2Idx: 0,
+		})
+	}
+	for pq.Len() > 0 {
+		i := heap.Pop(pq)
+		data := i.(*pddata)
+		res = append(res, data.n1+data.n2)
+		k--
+		if k <= 0 {
+			break
+		}
+		idx := data.n2Idx
+		idx++
+		if idx >= len(nums2) {
+			continue
+		}
+		heap.Push(pq, &pddata{
+			n1:    data.n1,
+			n2:    nums2[idx],
+			n2Idx: idx,
+		})
+	}
+	return res
+}
+
+type pddata struct {
+	n1    int
+	n2    int
+	n2Idx int
+}
+
+type priorityQueue []*pddata
+
+func newPriorityQueue() *priorityQueue {
+	pq := priorityQueue([]*pddata{})
+	heap.Init(&pq)
+	return &pq
+}
+
+func (pq priorityQueue) Len() int           { return len(pq) }
+func (pq priorityQueue) Swap(i, j int)      { pq[i], pq[j] = pq[j], pq[i] }
+func (pq priorityQueue) Less(i, j int) bool { return pq[i].n1+pq[i].n2 < pq[j].n1+pq[j].n2 }
+func (pq *priorityQueue) Pop() interface{} {
+	old := *pq
+	val := old[len(old)-1]
+	old[len(old)-1] = nil
+	*pq = old[0 : len(old)-1]
+	return val
+}
+
+func (pq *priorityQueue) Push(i interface{}) {
+	val := i.(*pddata)
+	*pq = append(*pq, val)
+}
diff --git a/leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows_test.go b/leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows_test.go
new file mode 100644
index 000000000..12edb22a9
--- /dev/null
+++ b/leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1439 struct {
+	para1439
+	ans1439
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1439 struct {
+	mat [][]int
+	k   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1439 struct {
+	one int
+}
+
+func Test_Problem1439(t *testing.T) {
+
+	qs := []question1439{
+
+		{
+			para1439{[][]int{{1, 3, 11}, {2, 4, 6}}, 5},
+			ans1439{7},
+		},
+
+		{
+			para1439{[][]int{{1, 3, 11}, {2, 4, 6}}, 9},
+			ans1439{17},
+		},
+		{
+			para1439{[][]int{{1, 10, 10}, {1, 4, 5}, {2, 3, 6}}, 7},
+			ans1439{9},
+		},
+		{
+			para1439{[][]int{{1, 1, 10}, {2, 2, 9}}, 7},
+			ans1439{12},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1439------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1439, q.para1439
+		fmt.Printf("【input】:%v       【output】:%v\n", p, kthSmallest(p.mat, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/README.md b/leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/README.md
new file mode 100644
index 000000000..b13885732
--- /dev/null
+++ b/leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/README.md
@@ -0,0 +1,144 @@
+# [1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows](https://leetcode.com/problems/find-the-kth-smallest-sum-of-a-matrix-with-sorted-rows/)
+
+
+## 题目
+
+You are given an `m * n` matrix, `mat`, and an integer `k`, which has its rows sorted in non-decreasing order.
+
+You are allowed to choose exactly 1 element from each row to form an array. Return the Kth **smallest** array sum among all possible arrays.
+
+**Example 1:**
+
+```
+Input: mat = [[1,3,11],[2,4,6]], k = 5
+Output: 7
+Explanation: Choosing one element from each row, the first k smallest sum are:
+[1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.  
+```
+
+**Example 2:**
+
+```
+Input: mat = [[1,3,11],[2,4,6]], k = 9
+Output: 17
+```
+
+**Example 3:**
+
+```
+Input: mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7
+Output: 9
+Explanation: Choosing one element from each row, the first k smallest sum are:
+[1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.  
+```
+
+**Example 4:**
+
+```
+Input: mat = [[1,1,10],[2,2,9]], k = 7
+Output: 12
+```
+
+**Constraints:**
+
+- `m == mat.length`
+- `n == mat.length[i]`
+- `1 <= m, n <= 40`
+- `1 <= k <= min(200, n ^ m)`
+- `1 <= mat[i][j] <= 5000`
+- `mat[i]` is a non decreasing array.
+
+## 题目大意
+
+给你一个 m * n 的矩阵 mat，以及一个整数 k ，矩阵中的每一行都以非递减的顺序排列。你可以从每一行中选出 1 个元素形成一个数组。返回所有可能数组中的第 k 个 最小数组和。
+
+## 解题思路
+
+- 这一题是第 373 题的升级版。在第 373 题中，给定 2 个有序数组，要求分别从这 2 个数组中选出一个数组成一个数对，最终输出和最小的 K 组。这一题中给出的是 m*n 的矩阵。其实是将第 373 题的 2 个数组升级为了 m 个数组。无非外层多了一层循环。这层循环依次从每一行中选出一个数，先从第 0 行和第 1 行取数，找到前 K 小的组合以后，再从第 2 行取数，以此类推。其他做法和第 373 题一致。维护一个长度为 k 的最小堆。每次从堆中 pop 出最小的数组和 sum 和对应的下标 index，然后依次将下标向后移动一位，生成新的 sum，加入堆中。
+
+## 代码
+
+```go
+package leetcode
+
+import "container/heap"
+
+func kthSmallest(mat [][]int, k int) int {
+	if len(mat) == 0 || len(mat[0]) == 0 || k == 0 {
+		return 0
+	}
+	prev := mat[0]
+	for i := 1; i < len(mat); i++ {
+		prev = kSmallestPairs(prev, mat[i], k)
+	}
+	if k < len(prev) {
+		return -1
+	}
+	return prev[k-1]
+}
+
+func kSmallestPairs(nums1 []int, nums2 []int, k int) []int {
+	res := []int{}
+	if len(nums2) == 0 {
+		return res
+	}
+	pq := newPriorityQueue()
+	for i := 0; i < len(nums1) && i < k; i++ {
+		heap.Push(pq, &pddata{
+			n1:    nums1[i],
+			n2:    nums2[0],
+			n2Idx: 0,
+		})
+	}
+	for pq.Len() > 0 {
+		i := heap.Pop(pq)
+		data := i.(*pddata)
+		res = append(res, data.n1+data.n2)
+		k--
+		if k <= 0 {
+			break
+		}
+		idx := data.n2Idx
+		idx++
+		if idx >= len(nums2) {
+			continue
+		}
+		heap.Push(pq, &pddata{
+			n1:    data.n1,
+			n2:    nums2[idx],
+			n2Idx: idx,
+		})
+	}
+	return res
+}
+
+type pddata struct {
+	n1    int
+	n2    int
+	n2Idx int
+}
+
+type priorityQueue []*pddata
+
+func newPriorityQueue() *priorityQueue {
+	pq := priorityQueue([]*pddata{})
+	heap.Init(&pq)
+	return &pq
+}
+
+func (pq priorityQueue) Len() int           { return len(pq) }
+func (pq priorityQueue) Swap(i, j int)      { pq[i], pq[j] = pq[j], pq[i] }
+func (pq priorityQueue) Less(i, j int) bool { return pq[i].n1+pq[i].n2 < pq[j].n1+pq[j].n2 }
+func (pq *priorityQueue) Pop() interface{} {
+	old := *pq
+	val := old[len(old)-1]
+	old[len(old)-1] = nil
+	*pq = old[0 : len(old)-1]
+	return val
+}
+
+func (pq *priorityQueue) Push(i interface{}) {
+	val := i.(*pddata)
+	*pq = append(*pq, val)
+}
+```
\ No newline at end of file
diff --git a/leetcode/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR/1442. Count Triplets That Can Form Two Arrays of Equal XOR.go b/leetcode/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR/1442. Count Triplets That Can Form Two Arrays of Equal XOR.go
new file mode 100644
index 000000000..19cbf203f
--- /dev/null
+++ b/leetcode/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR/1442. Count Triplets That Can Form Two Arrays of Equal XOR.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+func countTriplets(arr []int) int {
+	prefix, num, count, total := make([]int, len(arr)), 0, 0, 0
+	for i, v := range arr {
+		num ^= v
+		prefix[i] = num
+	}
+	for i := 0; i < len(prefix)-1; i++ {
+		for k := i + 1; k < len(prefix); k++ {
+			total = prefix[k]
+			if i > 0 {
+				total ^= prefix[i-1]
+			}
+			if total == 0 {
+				count += k - i
+			}
+		}
+	}
+	return count
+}
diff --git a/leetcode/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR/1442. Count Triplets That Can Form Two Arrays of Equal XOR_test.go b/leetcode/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR/1442. Count Triplets That Can Form Two Arrays of Equal XOR_test.go
new file mode 100644
index 000000000..38b31020f
--- /dev/null
+++ b/leetcode/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR/1442. Count Triplets That Can Form Two Arrays of Equal XOR_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1442 struct {
+	para1442
+	ans1442
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1442 struct {
+	arr []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1442 struct {
+	one int
+}
+
+func Test_Problem1442(t *testing.T) {
+
+	qs := []question1442{
+
+		{
+			para1442{[]int{2, 3, 1, 6, 7}},
+			ans1442{4},
+		},
+
+		{
+			para1442{[]int{1, 1, 1, 1, 1}},
+			ans1442{10},
+		},
+
+		{
+			para1442{[]int{2, 3}},
+			ans1442{0},
+		},
+
+		{
+			para1442{[]int{1, 3, 5, 7, 9}},
+			ans1442{3},
+		},
+
+		{
+			para1442{[]int{7, 11, 12, 9, 5, 2, 7, 17, 22}},
+			ans1442{8},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1442------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1442, q.para1442
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countTriplets(p.arr))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR/README.md b/leetcode/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR/README.md
new file mode 100644
index 000000000..4e45df7cf
--- /dev/null
+++ b/leetcode/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR/README.md
@@ -0,0 +1,97 @@
+# [1442. Count Triplets That Can Form Two Arrays of Equal XOR](https://leetcode.com/problems/count-triplets-that-can-form-two-arrays-of-equal-xor/)
+
+
+## 题目
+
+Given an array of integers `arr`.
+
+We want to select three indices `i`, `j` and `k` where `(0 <= i < j <= k < arr.length)`.
+
+Let's define `a` and `b` as follows:
+
+- `a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]`
+- `b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]`
+
+Note that **^** denotes the **bitwise-xor** operation.
+
+Return *the number of triplets* (`i`, `j` and `k`) Where `a == b`.
+
+**Example 1:**
+
+```
+Input: arr = [2,3,1,6,7]
+Output: 4
+Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)
+```
+
+**Example 2:**
+
+```
+Input: arr = [1,1,1,1,1]
+Output: 10
+```
+
+**Example 3:**
+
+```
+Input: arr = [2,3]
+Output: 0
+```
+
+**Example 4:**
+
+```
+Input: arr = [1,3,5,7,9]
+Output: 3
+```
+
+**Example 5:**
+
+```
+Input: arr = [7,11,12,9,5,2,7,17,22]
+Output: 8
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 300`
+- `1 <= arr[i] <= 10^8`
+
+## 题目大意
+
+给你一个整数数组 arr 。现需要从数组中取三个下标 i、j 和 k ，其中 (0 <= i < j <= k < arr.length) 。a 和 b 定义如下：
+
+- a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
+- b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]
+
+注意：^ 表示 按位异或 操作。请返回能够令 a == b 成立的三元组 (i, j , k) 的数目。
+
+## 解题思路
+
+- 这一题需要用到 `x^x = 0` 这个异或特性。题目要求 `a == b`，可以等效转化为 `arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1] ^ arr[j] ^ arr[j + 1] ^ ... ^ arr[k] = 0`，这样 j 相当于可以“忽略”，专注找到所有元素异或结果为 0 的区间 [i,k] 即为答案。利用前缀和的思想，只不过此题非累加和，而是异或。又由 `x^x = 0` 这个异或特性，相同部分异或相当于消除，于是有 `prefix[i,k] = prefix[0,k] ^ prefix[0,i-1]`，找到每一个 `prefix[i,k] = 0` 的 i，k 组合，i < j <= k，那么满足条件的三元组 (i,j,k) 的个数完全取决于 j 的取值范围，(因为 i 和 k 已经固定了)，j 的取值范围为 k-i，所以累加所有满足条件的 k-i，输出即为最终答案。
+
+## 代码
+
+```go
+package leetcode
+
+func countTriplets(arr []int) int {
+	prefix, num, count, total := make([]int, len(arr)), 0, 0, 0
+	for i, v := range arr {
+		num ^= v
+		prefix[i] = num
+	}
+	for i := 0; i < len(prefix)-1; i++ {
+		for k := i + 1; k < len(prefix); k++ {
+			total = prefix[k]
+			if i > 0 {
+				total ^= prefix[i-1]
+			}
+			if total == 0 {
+				count += k - i
+			}
+		}
+	}
+	return count
+}
+```
\ No newline at end of file
diff --git a/leetcode/1446.Consecutive-Characters/1446.Consecutive Characters.go b/leetcode/1446.Consecutive-Characters/1446.Consecutive Characters.go
new file mode 100644
index 000000000..9da00e7c0
--- /dev/null
+++ b/leetcode/1446.Consecutive-Characters/1446.Consecutive Characters.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+func maxPower(s string) int {
+	cur, cnt, ans := s[0], 1, 1
+	for i := 1; i < len(s); i++ {
+		if cur == s[i] {
+			cnt++
+		} else {
+			if cnt > ans {
+				ans = cnt
+			}
+			cur = s[i]
+			cnt = 1
+		}
+	}
+	if cnt > ans {
+		ans = cnt
+	}
+	return ans
+}
diff --git a/leetcode/1446.Consecutive-Characters/1446.Consecutive Characters_test.go b/leetcode/1446.Consecutive-Characters/1446.Consecutive Characters_test.go
new file mode 100644
index 000000000..d191126ec
--- /dev/null
+++ b/leetcode/1446.Consecutive-Characters/1446.Consecutive Characters_test.go	
@@ -0,0 +1,60 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1446 struct {
+	para1446
+	ans1446
+}
+
+// para 是参数
+type para1446 struct {
+	s string
+}
+
+// ans 是答案
+type ans1446 struct {
+	ans int
+}
+
+func Test_Problem1446(t *testing.T) {
+
+	qs := []question1446{
+
+		{
+			para1446{"leetcode"},
+			ans1446{2},
+		},
+
+		{
+			para1446{"abbcccddddeeeeedcba"},
+			ans1446{5},
+		},
+
+		{
+			para1446{"triplepillooooow"},
+			ans1446{5},
+		},
+
+		{
+			para1446{"hooraaaaaaaaaaay"},
+			ans1446{11},
+		},
+
+		{
+			para1446{"tourist"},
+			ans1446{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1446------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1446, q.para1446
+		fmt.Printf("【input】:%v    【output】:%v\n", p.s, maxPower(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1446.Consecutive-Characters/README.md b/leetcode/1446.Consecutive-Characters/README.md
new file mode 100644
index 000000000..78a6527af
--- /dev/null
+++ b/leetcode/1446.Consecutive-Characters/README.md
@@ -0,0 +1,75 @@
+# [1446. Consecutive Characters](https://leetcode.com/problems/consecutive-characters/)
+
+## 题目
+
+The power of the string is the maximum length of a non-empty substring that contains only one unique character.
+
+Given a string s, return the power of s.
+
+**Example 1**:
+
+    Input: s = "leetcode"
+    Output: 2
+    Explanation: The substring "ee" is of length 2 with the character 'e' only.
+
+**Example 2**:
+
+    Input: s = "abbcccddddeeeeedcba"
+    Output: 5
+    Explanation: The substring "eeeee" is of length 5 with the character 'e' only.
+
+**Example 3**:
+
+    Input: s = "triplepillooooow"
+    Output: 5
+
+**Example 4**:
+
+    Input: s = "hooraaaaaaaaaaay"
+    Output: 11
+
+**Example 5**:
+
+    Input: s = "tourist"
+    Output: 1
+
+**Constraints:**
+
+- 1 <= s.length <= 500
+- s consists of only lowercase English letters.
+
+## 题目大意
+
+给你一个字符串 s ，字符串的「能量」定义为：只包含一种字符的最长非空子字符串的长度。
+
+请你返回字符串的能量。
+
+## 解题思路
+
+- 顺序遍历进行统计
+
+## 代码
+
+```go
+package leetcode
+
+func maxPower(s string) int {
+	cur, cnt, ans := s[0], 1, 1
+	for i := 1; i < len(s); i++ {
+		if cur == s[i] {
+			cnt++
+		} else {
+			if cnt > ans {
+				ans = cnt
+			}
+			cur = s[i]
+			cnt = 1
+		}
+	}
+	if cnt > ans {
+		ans = cnt
+	}
+	return ans
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence.go b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence.go
new file mode 100644
index 000000000..c5f13249d
--- /dev/null
+++ b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+import "strings"
+
+func isPrefixOfWord(sentence string, searchWord string) int {
+	for i, v := range strings.Split(sentence, " ") {
+		if strings.HasPrefix(v, searchWord) {
+			return i + 1
+		}
+	}
+	return -1
+}
diff --git a/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence_test.go b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence_test.go
new file mode 100644
index 000000000..315483bd4
--- /dev/null
+++ b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence_test.go	
@@ -0,0 +1,64 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1455 struct {
+	para1455
+	ans1455
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1455 struct {
+	sentence   string
+	searchWord string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1455 struct {
+	one int
+}
+
+func Test_Problem1455(t *testing.T) {
+
+	qs := []question1455{
+
+		{
+			para1455{"i love eating burger", "burg"},
+			ans1455{4},
+		},
+
+		{
+			para1455{"this problem is an easy problem", "pro"},
+			ans1455{2},
+		},
+
+		{
+			para1455{"i am tired", "you"},
+			ans1455{-1},
+		},
+
+		{
+			para1455{"i use triple pillow", "pill"},
+			ans1455{4},
+		},
+
+		{
+			para1455{"hello from the other side", "they"},
+			ans1455{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1455------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1455, q.para1455
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", isPrefixOfWord(p.sentence, p.searchWord))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/README.md b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/README.md
new file mode 100644
index 000000000..c8081e82b
--- /dev/null
+++ b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/README.md
@@ -0,0 +1,98 @@
+# [1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence](https://leetcode.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/)
+
+
+## 题目
+
+Given a `sentence` that consists of some words separated by a **single space**, and a `searchWord`.
+
+You have to check if `searchWord` is a prefix of any word in `sentence`.
+
+Return *the index of the word* in `sentence` where `searchWord` is a prefix of this word (**1-indexed**).
+
+If `searchWord` is a prefix of more than one word, return the index of the first word **(minimum index)**. If there is no such word return **-1**.
+
+A **prefix** of a string `S` is any leading contiguous substring of `S`.
+
+**Example 1**:
+
+```
+Input: sentence = "i love eating burger", searchWord = "burg"
+Output: 4
+Explanation: "burg" is prefix of "burger" which is the 4th word in the sentence.
+
+```
+
+**Example 2**:
+
+```
+Input: sentence = "this problem is an easy problem", searchWord = "pro"
+Output: 2
+Explanation: "pro" is prefix of "problem" which is the 2nd and the 6th word in the sentence, but we return 2 as it's the minimal index.
+
+```
+
+**Example 3**:
+
+```
+Input: sentence = "i am tired", searchWord = "you"
+Output: -1
+Explanation: "you" is not a prefix of any word in the sentence.
+
+```
+
+**Example 4**:
+
+```
+Input: sentence = "i use triple pillow", searchWord = "pill"
+Output: 4
+
+```
+
+**Example 5**:
+
+```
+Input: sentence = "hello from the other side", searchWord = "they"
+Output: -1
+
+```
+
+**Constraints**:
+
+- `1 <= sentence.length <= 100`
+- `1 <= searchWord.length <= 10`
+- `sentence` consists of lowercase English letters and spaces.
+- `searchWord` consists of lowercase English letters.
+
+## 题目大意
+
+给你一个字符串 sentence 作为句子并指定检索词为 searchWord ，其中句子由若干用 单个空格 分隔的单词组成。请你检查检索词 searchWord 是否为句子 sentence 中任意单词的前缀。
+
+- 如果 searchWord 是某一个单词的前缀，则返回句子 sentence 中该单词所对应的下标（下标从 1 开始）。
+- 如果 searchWord 是多个单词的前缀，则返回匹配的第一个单词的下标（最小下标）。
+- 如果 searchWord 不是任何单词的前缀，则返回 -1 。
+
+字符串 S 的 「前缀」是 S 的任何前导连续子字符串。
+
+## 解题思路
+
+- 给出 2 个字符串，一个是匹配串，另外一个是句子。在句子里面查找带匹配串前缀的单词，并返回第一个匹配单词的下标。
+- 简单题。按照题意，扫描一遍句子，一次匹配即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func isPrefixOfWord(sentence string, searchWord string) int {
+	for i, v := range strings.Split(sentence, " ") {
+		if strings.HasPrefix(v, searchWord) {
+			return i + 1
+		}
+	}
+	return -1
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K/1461. Check If a String Contains All Binary Codes of Size K.go b/leetcode/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K/1461. Check If a String Contains All Binary Codes of Size K.go
new file mode 100644
index 000000000..b16ad956c
--- /dev/null
+++ b/leetcode/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K/1461. Check If a String Contains All Binary Codes of Size K.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+import "math"
+
+func hasAllCodes(s string, k int) bool {
+	need := int(math.Pow(2.0, float64(k)))
+	visited, mask, curr := make([]bool, need), (1<<k)-1, 0
+	for i := 0; i < len(s); i++ {
+		curr = ((curr << 1) | int(s[i]-'0')) & mask
+		if i >= k-1 { // mask 有效位达到了 k 位
+			if !visited[curr] {
+				need--
+				visited[curr] = true
+				if need == 0 {
+					return true
+				}
+			}
+		}
+	}
+	return false
+}
diff --git a/leetcode/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K/1461. Check If a String Contains All Binary Codes of Size K_test.go b/leetcode/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K/1461. Check If a String Contains All Binary Codes of Size K_test.go
new file mode 100644
index 000000000..e44997397
--- /dev/null
+++ b/leetcode/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K/1461. Check If a String Contains All Binary Codes of Size K_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1461 struct {
+	para1461
+	ans1461
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1461 struct {
+	s string
+	k int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1461 struct {
+	one bool
+}
+
+func Test_Problem1461(t *testing.T) {
+
+	qs := []question1461{
+
+		{
+			para1461{"00110110", 2},
+			ans1461{true},
+		},
+
+		{
+			para1461{"00110", 2},
+			ans1461{true},
+		},
+
+		{
+			para1461{"0110", 1},
+			ans1461{true},
+		},
+
+		{
+			para1461{"0110", 2},
+			ans1461{false},
+		},
+
+		{
+			para1461{"0000000001011100", 4},
+			ans1461{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1461------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1461, q.para1461
+		fmt.Printf("【input】:%v       【output】:%v\n", p, hasAllCodes(p.s, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K/README.md b/leetcode/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K/README.md
new file mode 100644
index 000000000..07269309e
--- /dev/null
+++ b/leetcode/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K/README.md
@@ -0,0 +1,86 @@
+# [1461. Check If a String Contains All Binary Codes of Size K](https://leetcode.com/problems/check-if-a-string-contains-all-binary-codes-of-size-k/)
+
+
+## 题目
+
+Given a binary string `s` and an integer `k`.
+
+Return *True* if every binary code of length `k` is a substring of `s`. Otherwise, return *False*.
+
+**Example 1:**
+
+```
+Input: s = "00110110", k = 2
+Output: true
+Explanation: The binary codes of length 2 are "00", "01", "10" and "11". They can be all found as substrings at indicies 0, 1, 3 and 2 respectively.
+```
+
+**Example 2:**
+
+```
+Input: s = "00110", k = 2
+Output: true
+```
+
+**Example 3:**
+
+```
+Input: s = "0110", k = 1
+Output: true
+Explanation: The binary codes of length 1 are "0" and "1", it is clear that both exist as a substring. 
+```
+
+**Example 4:**
+
+```
+Input: s = "0110", k = 2
+Output: false
+Explanation: The binary code "00" is of length 2 and doesn't exist in the array.
+```
+
+**Example 5:**
+
+```
+Input: s = "0000000001011100", k = 4
+Output: false
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 5 * 10^5`
+- `s` consists of 0's and 1's only.
+- `1 <= k <= 20`
+
+## 题目大意
+
+给你一个二进制字符串 `s` 和一个整数 `k` 。如果所有长度为 `k` 的二进制字符串都是 `s` 的子串，请返回 `True` ，否则请返回 `False` 。
+
+## 解题思路
+
+- 构造一个 `mask` 遮罩，依次划过整个二进制字符串，每次滑动即取出遮罩遮住的 `k` 位二进制字符。可以用 `map` 存储不同的二进制转换成的十进制数，最后判断 `len(map)` 是否等于 `k` 即可。但是用 `map` 存比较慢，此处换成 `bool` 数组。先构造一个长度为 `k` 的数组，然后每次通过 `mask` 更新这个 `bool` 数组对应十进制的 `bool` 值，并且记录剩余还缺几个二进制数。等剩余的等于 0 的时候，说明所有二进制字符串都出现了，直接输出 `true`，否则循环完以后输出 `false`。
+
+## 代码
+
+```go
+package leetcode
+
+import "math"
+
+func hasAllCodes(s string, k int) bool {
+	need := int(math.Pow(2.0, float64(k)))
+	visited, mask, curr := make([]bool, need), (1<<k)-1, 0
+	for i := 0; i < len(s); i++ {
+		curr = ((curr << 1) | int(s[i]-'0')) & mask
+		if i >= k-1 { // mask 有效位达到了 k 位
+			if !visited[curr] {
+				need--
+				visited[curr] = true
+				if need == 0 {
+					return true
+				}
+			}
+		}
+	}
+	return false
+}
+```
\ No newline at end of file
diff --git a/leetcode/1463.Cherry-Pickup-II/1463. Cherry Pickup II.go b/leetcode/1463.Cherry-Pickup-II/1463. Cherry Pickup II.go
new file mode 100644
index 000000000..949476117
--- /dev/null
+++ b/leetcode/1463.Cherry-Pickup-II/1463. Cherry Pickup II.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+func cherryPickup(grid [][]int) int {
+	rows, cols := len(grid), len(grid[0])
+	dp := make([][][]int, rows)
+	for i := 0; i < rows; i++ {
+		dp[i] = make([][]int, cols)
+		for j := 0; j < cols; j++ {
+			dp[i][j] = make([]int, cols)
+		}
+	}
+	for i := 0; i < rows; i++ {
+		for j := 0; j <= i && j < cols; j++ {
+			for k := cols - 1; k >= cols-1-i && k >= 0; k-- {
+				max := 0
+				for a := j - 1; a <= j+1; a++ {
+					for b := k - 1; b <= k+1; b++ {
+						sum := isInBoard(dp, i-1, a, b)
+						if a == b && i > 0 && a >= 0 && a < cols {
+							sum -= grid[i-1][a]
+						}
+						if sum > max {
+							max = sum
+						}
+					}
+				}
+				if j == k {
+					max += grid[i][j]
+				} else {
+					max += grid[i][j] + grid[i][k]
+				}
+				dp[i][j][k] = max
+			}
+		}
+	}
+	count := 0
+	for j := 0; j < cols && j < rows; j++ {
+		for k := cols - 1; k >= 0 && k >= cols-rows; k-- {
+			if dp[rows-1][j][k] > count {
+				count = dp[rows-1][j][k]
+			}
+		}
+	}
+	return count
+}
+
+func isInBoard(dp [][][]int, i, j, k int) int {
+	if i < 0 || j < 0 || j >= len(dp[0]) || k < 0 || k >= len(dp[0]) {
+		return 0
+	}
+	return dp[i][j][k]
+}
diff --git a/leetcode/1463.Cherry-Pickup-II/1463. Cherry Pickup II_test.go b/leetcode/1463.Cherry-Pickup-II/1463. Cherry Pickup II_test.go
new file mode 100644
index 000000000..3b85275f5
--- /dev/null
+++ b/leetcode/1463.Cherry-Pickup-II/1463. Cherry Pickup II_test.go	
@@ -0,0 +1,71 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1463 struct {
+	para1463
+	ans1463
+}
+
+type para1463 struct {
+	grid [][]int
+}
+
+type ans1463 struct {
+	ans int
+}
+
+func Test_Problem1463(t *testing.T) {
+
+	qs := []question1463{
+
+		{
+			para1463{[][]int{
+				{3, 1, 1},
+				{2, 5, 1},
+				{1, 5, 5},
+				{2, 1, 1},
+			}},
+			ans1463{24},
+		},
+
+		{
+			para1463{[][]int{
+				{1, 0, 0, 0, 0, 0, 1},
+				{2, 0, 0, 0, 0, 3, 0},
+				{2, 0, 9, 0, 0, 0, 0},
+				{0, 3, 0, 5, 4, 0, 0},
+				{1, 0, 2, 3, 0, 0, 6},
+			}},
+			ans1463{28},
+		},
+		{
+			para1463{[][]int{
+				{1, 0, 0, 3},
+				{0, 0, 0, 3},
+				{0, 0, 3, 3},
+				{9, 0, 3, 3},
+			}},
+			ans1463{22},
+		},
+		{
+			para1463{[][]int{
+				{1, 1},
+				{1, 1},
+			}},
+			ans1463{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1463------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1463, q.para1463
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", cherryPickup(p.grid))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1463.Cherry-Pickup-II/README.md b/leetcode/1463.Cherry-Pickup-II/README.md
new file mode 100644
index 000000000..88a4bd415
--- /dev/null
+++ b/leetcode/1463.Cherry-Pickup-II/README.md
@@ -0,0 +1,148 @@
+# [1463. Cherry Pickup II](https://leetcode.com/problems/cherry-pickup-ii/)
+
+## 题目
+
+Given a `rows x cols` matrix `grid` representing a field of cherries. Each cell in `grid` represents the number of cherries that you can collect.
+
+You have two robots that can collect cherries for you, Robot #1 is located at the top-left corner (0,0) , and Robot #2 is located at the top-right corner (0, cols-1) of the grid.
+
+Return the maximum number of cherries collection using both robots  by following the rules below:
+
+- From a cell (i,j), robots can move to cell (i+1, j-1) , (i+1, j) or (i+1, j+1).
+- When any robot is passing through a cell, It picks it up all cherries, and the cell becomes an empty cell (0).
+- When both robots stay on the same cell, only one of them takes the cherries.
+- Both robots cannot move outside of the grid at any moment.
+- Both robots should reach the bottom row in the `grid`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/04/29/sample_1_1802.png](https://assets.leetcode.com/uploads/2020/04/29/sample_1_1802.png)
+
+```
+Input: grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
+Output: 24
+Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
+Cherries taken by Robot #1, (3 + 2 + 5 + 2) = 12.
+Cherries taken by Robot #2, (1 + 5 + 5 + 1) = 12.
+Total of cherries: 12 + 12 = 24.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/04/23/sample_2_1802.png](https://assets.leetcode.com/uploads/2020/04/23/sample_2_1802.png)
+
+```
+Input: grid = [[1,0,0,0,0,0,1],[2,0,0,0,0,3,0],[2,0,9,0,0,0,0],[0,3,0,5,4,0,0],[1,0,2,3,0,0,6]]
+Output: 28
+Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
+Cherries taken by Robot #1, (1 + 9 + 5 + 2) = 17.
+Cherries taken by Robot #2, (1 + 3 + 4 + 3) = 11.
+Total of cherries: 17 + 11 = 28.
+```
+
+**Example 3:**
+
+```
+Input: grid = [[1,0,0,3],[0,0,0,3],[0,0,3,3],[9,0,3,3]]
+Output: 22
+```
+
+**Example 4:**
+
+```
+Input: grid = [[1,1],[1,1]]
+Output: 4
+```
+
+**Constraints:**
+
+- `rows == grid.length`
+- `cols == grid[i].length`
+- `2 <= rows, cols <= 70`
+- `0 <= grid[i][j] <= 100`
+
+## 题目大意
+
+给你一个 rows x cols 的矩阵 grid 来表示一块樱桃地。 grid 中每个格子的数字表示你能获得的樱桃数目。你有两个机器人帮你收集樱桃，机器人 1 从左上角格子 (0,0) 出发，机器人 2 从右上角格子 (0, cols-1) 出发。请你按照如下规则，返回两个机器人能收集的最多樱桃数目：
+
+- 从格子 (i,j) 出发，机器人可以移动到格子 (i+1, j-1)，(i+1, j) 或者 (i+1, j+1) 。
+- 当一个机器人经过某个格子时，它会把该格子内所有的樱桃都摘走，然后这个位置会变成空格子，即没有樱桃的格子。
+- 当两个机器人同时到达同一个格子时，它们中只有一个可以摘到樱桃。
+- 两个机器人在任意时刻都不能移动到 grid 外面。
+- 两个机器人最后都要到达 grid 最底下一行。
+
+## 解题思路
+
+- 如果没有思路可以先用暴力解法 DFS 尝试。读完题可以分析出求最多樱桃数目，里面包含了很多重叠子问题，于是乎自然而然思路是用动态规划。数据规模上看，100 的数据规模最多能保证 O(n^3) 时间复杂度的算法不超时。
+- 这一题的变量有 2 个，一个是行号，另外一个是机器人所在的列。具体来说，机器人每走一步的移动范围只能往下走，不能往上走，所以 2 个机器人所在行号一定相同。两个机器人的列号不同。综上，变量有 3 个，1 个行号和2 个列号。定义 `dp[i][j][k]` 代表第一个机器人从 (0,0) 走到 (i,k) 坐标，第二个机器人从 (0,n-1) 走到 (i,k) 坐标，两者最多能收集樱桃的数目。状态转移方程为  ：
+	
+	{{< katex display >}} 
+    dp[i][j][k] = max \begin{pmatrix}\begin{array}{lr} dp[i-1][f(j_1))][f(j_2)] + grid[i][j_1] + grid[i][j_2], j_1\neq j_2  \\ dp[i-1][f(j_1))][f(j_2)] + grid[i][j_1], j_1 = j_2 \end{array} \end{pmatrix}
+    {{< /katex>}} 
+
+    其中：
+	
+	{{< katex display >}} 
+    \left\{\begin{matrix}f(j_1) \in [0,n), f(j_1) - j_1 \in [-1,0,1]\\ f(j_2) \in [0,n), f(j_2) - j_2 \in [-1,0,1]\end{matrix}\right.
+    {{< /katex>}}
+
+    即状态转移过程中需要在 `[j1 - 1, j1, j1 + 1]` 中枚举 `j1`，同理，在 在 `[j2 - 1, j2, j2 + 1]` 中枚举 `j2`，每个状态转移需要枚举这 3*3 = 9 种状态。
+
+- 边界条件 `dp[i][0][n-1] = grid[0][0] + grid[0][n-1]`，最终答案存储在 `dp[m-1]` 行中，循环找出 `dp[m-1][j1][j2]` 中的最大值，到此该题得解。
+
+## 代码
+
+```go
+package leetcode
+
+func cherryPickup(grid [][]int) int {
+	rows, cols := len(grid), len(grid[0])
+	dp := make([][][]int, rows)
+	for i := 0; i < rows; i++ {
+		dp[i] = make([][]int, cols)
+		for j := 0; j < cols; j++ {
+			dp[i][j] = make([]int, cols)
+		}
+	}
+	for i := 0; i < rows; i++ {
+		for j := 0; j <= i && j < cols; j++ {
+			for k := cols - 1; k >= cols-1-i && k >= 0; k-- {
+				max := 0
+				for a := j - 1; a <= j+1; a++ {
+					for b := k - 1; b <= k+1; b++ {
+						sum := isInBoard(dp, i-1, a, b)
+						if a == b && i > 0 && a >= 0 && a < cols {
+							sum -= grid[i-1][a]
+						}
+						if sum > max {
+							max = sum
+						}
+					}
+				}
+				if j == k {
+					max += grid[i][j]
+				} else {
+					max += grid[i][j] + grid[i][k]
+				}
+				dp[i][j][k] = max
+			}
+		}
+	}
+	count := 0
+	for j := 0; j < cols && j < rows; j++ {
+		for k := cols - 1; k >= 0 && k >= cols-rows; k-- {
+			if dp[rows-1][j][k] > count {
+				count = dp[rows-1][j][k]
+			}
+		}
+	}
+	return count
+}
+
+func isInBoard(dp [][][]int, i, j, k int) int {
+	if i < 0 || j < 0 || j >= len(dp[0]) || k < 0 || k >= len(dp[0]) {
+		return 0
+	}
+	return dp[i][j][k]
+}
+```
\ No newline at end of file
diff --git a/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array.go b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array.go
new file mode 100644
index 000000000..a88bccb1f
--- /dev/null
+++ b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array.go	
@@ -0,0 +1,14 @@
+package leetcode
+
+func maxProduct(nums []int) int {
+	max1, max2 := 0, 0
+	for _, num := range nums {
+		if num >= max1 {
+			max2 = max1
+			max1 = num
+		} else if num <= max1 && num >= max2 {
+			max2 = num
+		}
+	}
+	return (max1 - 1) * (max2 - 1)
+}
diff --git a/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array_test.go b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array_test.go
new file mode 100644
index 000000000..93b56c8bd
--- /dev/null
+++ b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1464 struct {
+	para1464
+	ans1464
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1464 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1464 struct {
+	one int
+}
+
+func Test_Problem1464(t *testing.T) {
+
+	qs := []question1464{
+
+		{
+			para1464{[]int{3, 4, 5, 2}},
+			ans1464{12},
+		},
+
+		{
+			para1464{[]int{1, 5, 4, 5}},
+			ans1464{16},
+		},
+
+		{
+			para1464{[]int{3, 7}},
+			ans1464{12},
+		},
+
+		{
+			para1464{[]int{1}},
+			ans1464{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1464------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1464, q.para1464
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", maxProduct(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/README.md b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/README.md
new file mode 100644
index 000000000..04dbf134b
--- /dev/null
+++ b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/README.md
@@ -0,0 +1,66 @@
+# [1464. Maximum Product of Two Elements in an Array](https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/)
+
+
+## 题目
+
+Given the array of integers `nums`, you will choose two different indices `i` and `j` of that array. Return the maximum value of `(nums[i]-1)*(nums[j]-1)`.
+
+**Example 1**:
+
+```
+Input: nums = [3,4,5,2]
+Output: 12 
+Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12. 
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,5,4,5]
+Output: 16
+Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16.
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [3,7]
+Output: 12
+
+```
+
+**Constraints**:
+
+- `2 <= nums.length <= 500`
+- `1 <= nums[i] <= 10^3`
+
+## 题目大意
+
+给你一个整数数组 nums，请你选择数组的两个不同下标 i 和 j，使 (nums[i]-1)*(nums[j]-1) 取得最大值。请你计算并返回该式的最大值。
+
+## 解题思路
+
+- 简单题。循环一次，按照题意动态维护 2 个最大值，从而也使得 `(nums[i]-1)*(nums[j]-1)` 能取到最大值。
+
+## 代码
+
+```go
+
+package leetcode
+
+func maxProduct(nums []int) int {
+	max1, max2 := 0, 0
+	for _, num := range nums {
+		if num >= max1 {
+			max2 = max1
+			max1 = num
+		} else if num <= max1 && num >= max2 {
+			max2 = num
+		}
+	}
+	return (max1 - 1) * (max2 - 1)
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts/1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts.go b/leetcode/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts/1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts.go
new file mode 100644
index 000000000..d6a29607c
--- /dev/null
+++ b/leetcode/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts/1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts.go	
@@ -0,0 +1,26 @@
+package leetcode
+
+import "sort"
+
+func maxArea(h int, w int, horizontalCuts []int, verticalCuts []int) int {
+	sort.Ints(horizontalCuts)
+	sort.Ints(verticalCuts)
+	maxw, maxl := horizontalCuts[0], verticalCuts[0]
+	for i, c := range horizontalCuts[1:] {
+		if c-horizontalCuts[i] > maxw {
+			maxw = c - horizontalCuts[i]
+		}
+	}
+	if h-horizontalCuts[len(horizontalCuts)-1] > maxw {
+		maxw = h - horizontalCuts[len(horizontalCuts)-1]
+	}
+	for i, c := range verticalCuts[1:] {
+		if c-verticalCuts[i] > maxl {
+			maxl = c - verticalCuts[i]
+		}
+	}
+	if w-verticalCuts[len(verticalCuts)-1] > maxl {
+		maxl = w - verticalCuts[len(verticalCuts)-1]
+	}
+	return (maxw * maxl) % (1000000007)
+}
diff --git a/leetcode/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts/1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts_test.go b/leetcode/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts/1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts_test.go
new file mode 100644
index 000000000..b6caabf3a
--- /dev/null
+++ b/leetcode/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts/1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts_test.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1465 struct {
+	para1465
+	ans1465
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1465 struct {
+	h              int
+	w              int
+	horizontalCuts []int
+	verticalCuts   []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1465 struct {
+	one int
+}
+
+func Test_Problem1465(t *testing.T) {
+
+	qs := []question1465{
+
+		{
+			para1465{5, 4, []int{1, 2, 4}, []int{1, 3}},
+			ans1465{4},
+		},
+
+		{
+			para1465{5, 4, []int{3, 1}, []int{1}},
+			ans1465{6},
+		},
+
+		{
+			para1465{5, 4, []int{3}, []int{3}},
+			ans1465{9},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1465------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1465, q.para1465
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, maxArea(p.h, p.w, p.horizontalCuts, p.verticalCuts))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts/README.md b/leetcode/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts/README.md
new file mode 100644
index 000000000..8926b1776
--- /dev/null
+++ b/leetcode/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts/README.md
@@ -0,0 +1,89 @@
+# [1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts](https://leetcode.com/problems/maximum-area-of-a-piece-of-cake-after-horizontal-and-vertical-cuts/)
+
+
+## 题目
+
+Given a rectangular cake with height `h` and width `w`, and two arrays of integers `horizontalCuts` and `verticalCuts` where `horizontalCuts[i]` is the distance from the top of the rectangular cake to the `ith` horizontal cut and similarly, `verticalCuts[j]` is the distance from the left of the rectangular cake to the `jth` vertical cut.
+
+*Return the maximum area of a piece of cake after you cut at each horizontal and vertical position provided in the arrays `horizontalCuts` and `verticalCuts`.* Since the answer can be a huge number, return this modulo 10^9 + 7.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/05/14/leetcode_max_area_2.png](https://assets.leetcode.com/uploads/2020/05/14/leetcode_max_area_2.png)
+
+```
+Input: h = 5, w = 4, horizontalCuts = [1,2,4], verticalCuts = [1,3]
+Output: 4
+Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. After you cut the cake, the green piece of cake has the maximum area.
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/05/14/leetcode_max_area_3.png](https://assets.leetcode.com/uploads/2020/05/14/leetcode_max_area_3.png)
+
+```
+Input: h = 5, w = 4, horizontalCuts = [3,1], verticalCuts = [1]
+Output: 6
+Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. After you cut the cake, the green and yellow pieces of cake have the maximum area.
+
+```
+
+**Example 3:**
+
+```
+Input: h = 5, w = 4, horizontalCuts = [3], verticalCuts = [3]
+Output: 9
+
+```
+
+**Constraints:**
+
+- `2 <= h, w <= 10^9`
+- `1 <= horizontalCuts.length < min(h, 10^5)`
+- `1 <= verticalCuts.length < min(w, 10^5)`
+- `1 <= horizontalCuts[i] < h`
+- `1 <= verticalCuts[i] < w`
+- It is guaranteed that all elements in `horizontalCuts` are distinct.
+- It is guaranteed that all elements in `verticalCuts` are distinct.
+
+## 题目大意
+
+矩形蛋糕的高度为 h 且宽度为 w，给你两个整数数组 horizontalCuts 和 verticalCuts，其中 horizontalCuts[i] 是从矩形蛋糕顶部到第  i 个水平切口的距离，类似地， verticalCuts[j] 是从矩形蛋糕的左侧到第 j 个竖直切口的距离。请你按数组 horizontalCuts 和 verticalCuts 中提供的水平和竖直位置切割后，请你找出 面积最大 的那份蛋糕，并返回其 面积 。由于答案可能是一个很大的数字，因此需要将结果对 10^9 + 7 取余后返回。
+
+## 解题思路
+
+- 读完题比较容易想到解题思路。找到水平切口最大的差值和竖直切口最大的差值，这 4 条边构成的矩形即为最大矩形。不过有特殊情况需要判断，切口除了题目给的切口坐标以外，默认还有 4 个切口，即蛋糕原始的 4 条边。如下图二，最大的矩形其实在切口之外。所以找水平切口最大差值和竖直切口最大差值需要考虑到蛋糕原始的 4 条边。
+
+![https://img.halfrost.com/Leetcode/leetcode_1465.png](https://img.halfrost.com/Leetcode/leetcode_1465.png)
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func maxArea(h int, w int, hcuts []int, vcuts []int) int {
+	sort.Ints(hcuts)
+	sort.Ints(vcuts)
+	maxw, maxl := hcuts[0], vcuts[0]
+	for i, c := range hcuts[1:] {
+		if c-hcuts[i] > maxw {
+			maxw = c - hcuts[i]
+		}
+	}
+	if h-hcuts[len(hcuts)-1] > maxw {
+		maxw = h - hcuts[len(hcuts)-1]
+	}
+	for i, c := range vcuts[1:] {
+		if c-vcuts[i] > maxl {
+			maxl = c - vcuts[i]
+		}
+	}
+	if w-vcuts[len(vcuts)-1] > maxl {
+		maxl = w - vcuts[len(vcuts)-1]
+	}
+	return (maxw * maxl) % (1000000007)
+}
+```
\ No newline at end of file
diff --git a/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array.go b/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array.go
new file mode 100644
index 000000000..fef82041c
--- /dev/null
+++ b/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array.go	
@@ -0,0 +1,10 @@
+package leetcode
+
+func shuffle(nums []int, n int) []int {
+	result := make([]int, 0)
+	for i := 0; i < n; i++ {
+		result = append(result, nums[i])
+		result = append(result, nums[n+i])
+	}
+	return result
+}
diff --git a/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array_test.go b/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array_test.go
new file mode 100644
index 000000000..e273744ed
--- /dev/null
+++ b/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1470 struct {
+	para1470
+	ans1470
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1470 struct {
+	nums []int
+	n    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1470 struct {
+	one []int
+}
+
+func Test_Problem1470(t *testing.T) {
+
+	qs := []question1470{
+
+		{
+			para1470{[]int{2, 5, 1, 3, 4, 7}, 3},
+			ans1470{[]int{2, 3, 5, 4, 1, 7}},
+		},
+
+		{
+			para1470{[]int{1, 2, 3, 4, 4, 3, 2, 1}, 4},
+			ans1470{[]int{1, 4, 2, 3, 3, 2, 4, 1}},
+		},
+
+		{
+			para1470{[]int{1, 1, 2, 2}, 2},
+			ans1470{[]int{1, 2, 1, 2}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1470------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1470, q.para1470
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, shuffle(p.nums, p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1470.Shuffle-the-Array/README.md b/leetcode/1470.Shuffle-the-Array/README.md
new file mode 100644
index 000000000..961d200c4
--- /dev/null
+++ b/leetcode/1470.Shuffle-the-Array/README.md
@@ -0,0 +1,64 @@
+# [1470. Shuffle the Array](https://leetcode.com/problems/shuffle-the-array/)
+
+## 题目
+
+Given the array `nums` consisting of `2n` elements in the form `[x1,x2,...,xn,y1,y2,...,yn]`.
+
+*Return the array in the form* `[x1,y1,x2,y2,...,xn,yn]`.
+
+**Example 1**:
+
+```
+Input: nums = [2,5,1,3,4,7], n = 3
+Output: [2,3,5,4,1,7] 
+Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7].
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,2,3,4,4,3,2,1], n = 4
+Output: [1,4,2,3,3,2,4,1]
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [1,1,2,2], n = 2
+Output: [1,2,1,2]
+
+```
+
+**Constraints**:
+
+- `1 <= n <= 500`
+- `nums.length == 2n`
+- `1 <= nums[i] <= 10^3`
+
+## 题目大意
+
+给你一个数组 nums ，数组中有 2n 个元素，按 [x1,x2,...,xn,y1,y2,...,yn] 的格式排列。请你将数组按 [x1,y1,x2,y2,...,xn,yn] 格式重新排列，返回重排后的数组。
+
+## 解题思路
+
+- 给定一个 2n 的数组，把后 n 个元素插空放到前 n 个元素里面。输出最终完成的数组。
+- 简单题，按照题意插空即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func shuffle(nums []int, n int) []int {
+	result := make([]int, 0)
+	for i := 0; i < n; i++ {
+		result = append(result, nums[i])
+		result = append(result, nums[n+i])
+	}
+	return result
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1480.Running-Sum-of-1d-Array/1480. Running Sum of 1d Array.go b/leetcode/1480.Running-Sum-of-1d-Array/1480. Running Sum of 1d Array.go
new file mode 100644
index 000000000..7663c290c
--- /dev/null
+++ b/leetcode/1480.Running-Sum-of-1d-Array/1480. Running Sum of 1d Array.go	
@@ -0,0 +1,10 @@
+package leetcode
+
+func runningSum(nums []int) []int {
+	dp := make([]int, len(nums)+1)
+	dp[0] = 0
+	for i := 1; i <= len(nums); i++ {
+		dp[i] = dp[i-1] + nums[i-1]
+	}
+	return dp[1:]
+}
diff --git a/leetcode/1480.Running-Sum-of-1d-Array/1480. Running Sum of 1d Array_test.go b/leetcode/1480.Running-Sum-of-1d-Array/1480. Running Sum of 1d Array_test.go
new file mode 100644
index 000000000..c2053b1b1
--- /dev/null
+++ b/leetcode/1480.Running-Sum-of-1d-Array/1480. Running Sum of 1d Array_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1480 struct {
+	para1480
+	ans1480
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1480 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1480 struct {
+	one []int
+}
+
+func Test_Problem1480(t *testing.T) {
+
+	qs := []question1480{
+
+		{
+			para1480{[]int{1, 2, 3, 4}},
+			ans1480{[]int{1, 3, 6, 10}},
+		},
+
+		{
+			para1480{[]int{1, 1, 1, 1, 1}},
+			ans1480{[]int{1, 2, 3, 4, 5}},
+		},
+
+		{
+			para1480{[]int{3, 1, 2, 10, 1}},
+			ans1480{[]int{3, 4, 6, 16, 17}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1480------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1480, q.para1480
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, runningSum(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1480.Running-Sum-of-1d-Array/README.md b/leetcode/1480.Running-Sum-of-1d-Array/README.md
new file mode 100644
index 000000000..d0e483358
--- /dev/null
+++ b/leetcode/1480.Running-Sum-of-1d-Array/README.md
@@ -0,0 +1,64 @@
+# [1480. Running Sum of 1d Array](https://leetcode.com/problems/running-sum-of-1d-array/)
+
+## 题目
+
+Given an array `nums`. We define a running sum of an array as `runningSum[i] = sum(nums[0]…nums[i])`.
+
+Return the running sum of `nums`.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,4]
+Output: [1,3,6,10]
+Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,1,1,1,1]
+Output: [1,2,3,4,5]
+Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [3,1,2,10,1]
+Output: [3,4,6,16,17]
+
+```
+
+
+**Constraints**:
+
+- `1 <= nums.length <= 1000`
+- `-10^6 <= nums[i] <= 10^6`
+
+## 题目大意
+
+给你一个数组 nums 。数组「动态和」的计算公式为：runningSum[i] = sum(nums[0]…nums[i]) 。请返回 nums 的动态和。
+
+
+## 解题思路
+
+- 简单题，按照题意依次循环计算前缀和即可。
+
+## 代码
+
+```go
+package leetcode
+
+func runningSum(nums []int) []int {
+	dp := make([]int, len(nums)+1)
+	dp[0] = 0
+	for i := 1; i <= len(nums); i++ {
+		dp[i] = dp[i-1] + nums[i-1]
+	}
+	return dp[1:]
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1482.Minimum-Number-of-Days-to-Make-m-Bouquets/1482. Minimum Number of Days to Make m Bouquets.go b/leetcode/1482.Minimum-Number-of-Days-to-Make-m-Bouquets/1482. Minimum Number of Days to Make m Bouquets.go
new file mode 100644
index 000000000..759e93542
--- /dev/null
+++ b/leetcode/1482.Minimum-Number-of-Days-to-Make-m-Bouquets/1482. Minimum Number of Days to Make m Bouquets.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+import "sort"
+
+func minDays(bloomDay []int, m int, k int) int {
+	if m*k > len(bloomDay) {
+		return -1
+	}
+	maxDay := 0
+	for _, day := range bloomDay {
+		if day > maxDay {
+			maxDay = day
+		}
+	}
+	return sort.Search(maxDay, func(days int) bool {
+		flowers, bouquets := 0, 0
+		for _, d := range bloomDay {
+			if d > days {
+				flowers = 0
+			} else {
+				flowers++
+				if flowers == k {
+					bouquets++
+					flowers = 0
+				}
+			}
+		}
+		return bouquets >= m
+	})
+}
diff --git a/leetcode/1482.Minimum-Number-of-Days-to-Make-m-Bouquets/1482. Minimum Number of Days to Make m Bouquets_test.go b/leetcode/1482.Minimum-Number-of-Days-to-Make-m-Bouquets/1482. Minimum Number of Days to Make m Bouquets_test.go
new file mode 100644
index 000000000..e505095f1
--- /dev/null
+++ b/leetcode/1482.Minimum-Number-of-Days-to-Make-m-Bouquets/1482. Minimum Number of Days to Make m Bouquets_test.go	
@@ -0,0 +1,64 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1482 struct {
+	para1482
+	ans1482
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1482 struct {
+	bloomDay []int
+	m        int
+	k        int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1482 struct {
+	one int
+}
+
+func Test_Problem1482(t *testing.T) {
+
+	qs := []question1482{
+
+		{
+			para1482{[]int{1, 10, 3, 10, 2}, 3, 1},
+			ans1482{3},
+		},
+
+		{
+			para1482{[]int{1, 10, 3, 10, 2}, 3, 2},
+			ans1482{-1},
+		},
+
+		{
+			para1482{[]int{7, 7, 7, 7, 12, 7, 7}, 2, 3},
+			ans1482{12},
+		},
+
+		{
+			para1482{[]int{1000000000, 1000000000}, 1, 1},
+			ans1482{1000000000},
+		},
+
+		{
+			para1482{[]int{1, 10, 2, 9, 3, 8, 4, 7, 5, 6}, 4, 2},
+			ans1482{9},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1482------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1482, q.para1482
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, minDays(p.bloomDay, p.m, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1482.Minimum-Number-of-Days-to-Make-m-Bouquets/README.md b/leetcode/1482.Minimum-Number-of-Days-to-Make-m-Bouquets/README.md
new file mode 100644
index 000000000..b7d691d38
--- /dev/null
+++ b/leetcode/1482.Minimum-Number-of-Days-to-Make-m-Bouquets/README.md
@@ -0,0 +1,110 @@
+# [1482. Minimum Number of Days to Make m Bouquets](https://leetcode.com/problems/minimum-number-of-days-to-make-m-bouquets/)
+
+## 题目
+
+Given an integer array `bloomDay`, an integer `m` and an integer `k`.
+
+We need to make `m` bouquets. To make a bouquet, you need to use `k` **adjacent flowers** from the garden.
+
+The garden consists of `n` flowers, the `ith` flower will bloom in the `bloomDay[i]` and then can be used in **exactly one** bouquet.
+
+Return *the minimum number of days* you need to wait to be able to make `m` bouquets from the garden. If it is impossible to make `m` bouquets return **-1**.
+
+**Example 1:**
+
+```
+Input: bloomDay = [1,10,3,10,2], m = 3, k = 1
+Output: 3
+Explanation: Let's see what happened in the first three days. x means flower bloomed and _ means flower didn't bloom in the garden.
+We need 3 bouquets each should contain 1 flower.
+After day 1: [x, _, _, _, _]   // we can only make one bouquet.
+After day 2: [x, _, _, _, x]   // we can only make two bouquets.
+After day 3: [x, _, x, _, x]   // we can make 3 bouquets. The answer is 3.
+```
+
+**Example 2:**
+
+```
+Input: bloomDay = [1,10,3,10,2], m = 3, k = 2
+Output: -1
+Explanation: We need 3 bouquets each has 2 flowers, that means we need 6 flowers. We only have 5 flowers so it is impossible to get the needed bouquets and we return -1.
+```
+
+**Example 3:**
+
+```
+Input: bloomDay = [7,7,7,7,12,7,7], m = 2, k = 3
+Output: 12
+Explanation: We need 2 bouquets each should have 3 flowers.
+Here's the garden after the 7 and 12 days:
+After day 7: [x, x, x, x, _, x, x]
+We can make one bouquet of the first three flowers that bloomed. We cannot make another bouquet from the last three flowers that bloomed because they are not adjacent.
+After day 12: [x, x, x, x, x, x, x]
+It is obvious that we can make two bouquets in different ways.
+```
+
+**Example 4:**
+
+```
+Input: bloomDay = [1000000000,1000000000], m = 1, k = 1
+Output: 1000000000
+Explanation: You need to wait 1000000000 days to have a flower ready for a bouquet.
+```
+
+**Example 5:**
+
+```
+Input: bloomDay = [1,10,2,9,3,8,4,7,5,6], m = 4, k = 2
+Output: 9
+```
+
+**Constraints:**
+
+- `bloomDay.length == n`
+- `1 <= n <= 10^5`
+- `1 <= bloomDay[i] <= 10^9`
+- `1 <= m <= 10^6`
+- `1 <= k <= n`
+
+## 题目大意
+
+给你一个整数数组 bloomDay，以及两个整数 m 和 k 。现需要制作 m 束花。制作花束时，需要使用花园中 相邻的 k 朵花 。花园中有 n 朵花，第 i 朵花会在 bloomDay[i] 时盛开，恰好 可以用于 一束 花中。请你返回从花园中摘 m 束花需要等待的最少的天数。如果不能摘到 m 束花则返回 -1 。
+
+## 解题思路
+
+- 本题是二分搜索提醒。题目解空间固定，答案区间一定在 [0, maxDay] 中。这是单调增且有序区间，所以可以在这个解空间内使用二分搜索。在区间 [0, maxDay] 中找到第一个能满足 m 束花的解。二分搜索判断是否为 true 的条件为：从左往右遍历数组，依次统计当前日期下，花是否开了，如果连续开花 k 朵，便为 1 束，数组遍历结束如果花束总数 ≥ k 即为答案。二分搜索会返回最小的下标，即对应满足题意的最少天数。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func minDays(bloomDay []int, m int, k int) int {
+	if m*k > len(bloomDay) {
+		return -1
+	}
+	maxDay := 0
+	for _, day := range bloomDay {
+		if day > maxDay {
+			maxDay = day
+		}
+	}
+	return sort.Search(maxDay, func(days int) bool {
+		flowers, bouquets := 0, 0
+		for _, d := range bloomDay {
+			if d > days {
+				flowers = 0
+			} else {
+				flowers++
+				if flowers == k {
+					bouquets++
+					flowers = 0
+				}
+			}
+		}
+		return bouquets >= m
+	})
+}
+```
\ No newline at end of file
diff --git a/leetcode/1486.XOR-Operation-in-an-Array/1486. XOR Operation in an Array.go b/leetcode/1486.XOR-Operation-in-an-Array/1486. XOR Operation in an Array.go
new file mode 100644
index 000000000..a9040341d
--- /dev/null
+++ b/leetcode/1486.XOR-Operation-in-an-Array/1486. XOR Operation in an Array.go	
@@ -0,0 +1,9 @@
+package leetcode
+
+func xorOperation(n int, start int) int {
+	res := 0
+	for i := 0; i < n; i++ {
+		res ^= start + 2*i
+	}
+	return res
+}
diff --git a/leetcode/1486.XOR-Operation-in-an-Array/1486. XOR Operation in an Array_test.go b/leetcode/1486.XOR-Operation-in-an-Array/1486. XOR Operation in an Array_test.go
new file mode 100644
index 000000000..e1887c155
--- /dev/null
+++ b/leetcode/1486.XOR-Operation-in-an-Array/1486. XOR Operation in an Array_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1486 struct {
+	para1486
+	ans1486
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1486 struct {
+	n     int
+	start int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1486 struct {
+	one int
+}
+
+func Test_Problem1486(t *testing.T) {
+
+	qs := []question1486{
+
+		{
+			para1486{5, 0},
+			ans1486{8},
+		},
+
+		{
+			para1486{4, 3},
+			ans1486{8},
+		},
+
+		{
+			para1486{1, 7},
+			ans1486{7},
+		},
+
+		{
+			para1486{10, 5},
+			ans1486{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1486------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1486, q.para1486
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, xorOperation(p.n, p.start))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1486.XOR-Operation-in-an-Array/README.md b/leetcode/1486.XOR-Operation-in-an-Array/README.md
new file mode 100644
index 000000000..3a6438de8
--- /dev/null
+++ b/leetcode/1486.XOR-Operation-in-an-Array/README.md
@@ -0,0 +1,69 @@
+# [1486. XOR Operation in an Array](https://leetcode.com/problems/xor-operation-in-an-array/)
+
+
+## 题目
+
+Given an integer `n` and an integer `start`.
+
+Define an array `nums` where `nums[i] = start + 2*i` (0-indexed) and `n == nums.length`.
+
+Return the bitwise XOR of all elements of `nums`.
+
+**Example 1:**
+
+```
+Input: n = 5, start = 0
+Output: 8
+Explanation:Array nums is equal to [0, 2, 4, 6, 8] where (0 ^ 2 ^ 4 ^ 6 ^ 8) = 8.
+Where "^" corresponds to bitwise XOR operator.
+```
+
+**Example 2:**
+
+```
+Input: n = 4, start = 3
+Output: 8
+Explanation:Array nums is equal to [3, 5, 7, 9] where (3 ^ 5 ^ 7 ^ 9) = 8.
+```
+
+**Example 3:**
+
+```
+Input: n = 1, start = 7
+Output: 7
+```
+
+**Example 4:**
+
+```
+Input: n = 10, start = 5
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= n <= 1000`
+- `0 <= start <= 1000`
+- `n == nums.length`
+
+## 题目大意
+
+给你两个整数，n 和 start 。数组 nums 定义为：nums[i] = start + 2*i（下标从 0 开始）且 n == nums.length 。请返回 nums 中所有元素按位异或（XOR）后得到的结果。
+
+## 解题思路
+
+- 简单题。按照题意，一层循环依次累积异或数组中每个元素。
+
+## 代码
+
+```go
+package leetcode
+
+func xorOperation(n int, start int) int {
+	res := 0
+	for i := 0; i < n; i++ {
+		res ^= start + 2*i
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1512.Number-of-Good-Pairs/1512. Number of Good Pairs.go b/leetcode/1512.Number-of-Good-Pairs/1512. Number of Good Pairs.go
new file mode 100644
index 000000000..6edfcdfa4
--- /dev/null
+++ b/leetcode/1512.Number-of-Good-Pairs/1512. Number of Good Pairs.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+func numIdenticalPairs(nums []int) int {
+	total := 0
+	for x := 0; x < len(nums); x++ {
+		for y := x + 1; y < len(nums); y++ {
+			if nums[x] == nums[y] {
+				total++
+			}
+		}
+	}
+	return total
+}
diff --git a/leetcode/1512.Number-of-Good-Pairs/1512. Number of Good Pairs_test.go b/leetcode/1512.Number-of-Good-Pairs/1512. Number of Good Pairs_test.go
new file mode 100644
index 000000000..3fc122186
--- /dev/null
+++ b/leetcode/1512.Number-of-Good-Pairs/1512. Number of Good Pairs_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1512 struct {
+	para1512
+	ans1512
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1512 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1512 struct {
+	one int
+}
+
+func Test_Problem1512(t *testing.T) {
+
+	qs := []question1512{
+
+		{
+			para1512{[]int{1, 2, 3, 1, 1, 3}},
+			ans1512{4},
+		},
+
+		{
+			para1512{[]int{1, 1, 1, 1}},
+			ans1512{6},
+		},
+
+		{
+			para1512{[]int{1, 2, 3}},
+			ans1512{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1512------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1512, q.para1512
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, numIdenticalPairs(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1512.Number-of-Good-Pairs/README.md b/leetcode/1512.Number-of-Good-Pairs/README.md
new file mode 100644
index 000000000..a6954aa0a
--- /dev/null
+++ b/leetcode/1512.Number-of-Good-Pairs/README.md
@@ -0,0 +1,67 @@
+# [1512. Number of Good Pairs](https://leetcode.com/problems/number-of-good-pairs/)
+
+## 题目
+
+Given an array of integers `nums`.
+
+A pair `(i,j)` is called good if `nums[i] == nums[j]` and `i < j`.
+
+Return the number of good pairs.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,1,1,3]
+Output: 4
+Explanation: There are 4 good pairs (0,3), (0,4), (3,4), (2,5) 0-indexed.
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,1,1,1]
+Output: 6
+Explanation: Each pair in the array are good.
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [1,2,3]
+Output: 0
+
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 1000`
+- `1 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个整数数组 nums。如果一组数字 (i,j) 满足 nums[i] == nums[j] 且 i < j ，就可以认为这是一组好数对。返回好数对的数目。
+
+## 解题思路
+
+- 简单题，按照题目中好数对的定义，循环遍历判断两数是否相等，累加计数即可。
+
+## 代码
+
+```go
+package leetcode
+
+func numIdenticalPairs(nums []int) int {
+	total := 0
+	for x := 0; x < len(nums); x++ {
+		for y := x + 1; y < len(nums); y++ {
+			if nums[x] == nums[y] {
+				total++
+			}
+		}
+	}
+	return total
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1518.Water-Bottles/1518.Water Bottles.go b/leetcode/1518.Water-Bottles/1518.Water Bottles.go
new file mode 100644
index 000000000..987c6b86b
--- /dev/null
+++ b/leetcode/1518.Water-Bottles/1518.Water Bottles.go	
@@ -0,0 +1,15 @@
+package leetcode
+
+func numWaterBottles(numBottles int, numExchange int) int {
+	if numBottles < numExchange {
+		return numBottles
+	}
+	quotient := numBottles / numExchange
+	reminder := numBottles % numExchange
+	ans := numBottles + quotient
+	for quotient+reminder >= numExchange {
+		quotient, reminder = (quotient+reminder)/numExchange, (quotient+reminder)%numExchange
+		ans += quotient
+	}
+	return ans
+}
diff --git a/leetcode/1518.Water-Bottles/1518.Water Bottles_test.go b/leetcode/1518.Water-Bottles/1518.Water Bottles_test.go
new file mode 100644
index 000000000..20c2a18d7
--- /dev/null
+++ b/leetcode/1518.Water-Bottles/1518.Water Bottles_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1518 struct {
+	para1518
+	ans1518
+}
+
+// para 是参数
+type para1518 struct {
+	numBottles  int
+	numExchange int
+}
+
+// ans 是答案
+type ans1518 struct {
+	ans int
+}
+
+func Test_Problem1518(t *testing.T) {
+
+	qs := []question1518{
+
+		{
+			para1518{9, 3},
+			ans1518{13},
+		},
+
+		{
+			para1518{15, 4},
+			ans1518{19},
+		},
+
+		{
+			para1518{5, 5},
+			ans1518{6},
+		},
+
+		{
+			para1518{2, 3},
+			ans1518{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1518------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1518, q.para1518
+		fmt.Printf("【input】:%v    【output】:%v\n", p, numWaterBottles(p.numBottles, p.numExchange))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1518.Water-Bottles/README.md b/leetcode/1518.Water-Bottles/README.md
new file mode 100644
index 000000000..3f05e1815
--- /dev/null
+++ b/leetcode/1518.Water-Bottles/README.md
@@ -0,0 +1,74 @@
+# [1518. Water Bottles](https://leetcode.com/problems/water-bottles/)
+
+## 题目
+
+Given numBottles full water bottles, you can exchange numExchange empty water bottles for one full water bottle.
+
+The operation of drinking a full water bottle turns it into an empty bottle.
+
+Return the maximum number of water bottles you can drink.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2020/07/01/sample_1_1875.png](https://assets.leetcode.com/uploads/2020/07/01/sample_1_1875.png)
+
+    Input: numBottles = 9, numExchange = 3
+    Output: 13
+    Explanation: You can exchange 3 empty bottles to get 1 full water bottle.
+    Number of water bottles you can drink: 9 + 3 + 1 = 13.
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2020/07/01/sample_2_1875.png](https://assets.leetcode.com/uploads/2020/07/01/sample_2_1875.png)
+
+    Input: numBottles = 15, numExchange = 4
+    Output: 19
+    Explanation: You can exchange 4 empty bottles to get 1 full water bottle.
+    Number of water bottles you can drink: 15 + 3 + 1 = 19.
+
+**Example 3**:
+
+    Input: numBottles = 5, numExchange = 5
+    Output: 6
+
+**Example 4**:
+
+    Input: numBottles = 2, numExchange = 3
+    Output: 2
+
+**Constraints:**
+
+- 1 <= numBottles <= 100
+- 2 <= numExchange <= 100
+
+## 题目大意
+
+小区便利店正在促销，用 numExchange 个空酒瓶可以兑换一瓶新酒。你购入了 numBottles 瓶酒。
+
+如果喝掉了酒瓶中的酒，那么酒瓶就会变成空的。
+
+请你计算 最多 能喝到多少瓶酒。
+
+## 解题思路
+
+- 模拟。首先我们一定可以喝到 numBottles 瓶酒，剩下 numBottles 个空瓶。接下来我们可以拿空瓶子换酒，每次拿出 numExchange 个瓶子换一瓶酒，然后再喝完这瓶酒，得到一个空瓶。这样模拟下去，直到所有的空瓶子小于numExchange结束。
+
+## 代码
+
+```go
+package leetcode
+
+func numWaterBottles(numBottles int, numExchange int) int {
+	if numBottles < numExchange {
+		return numBottles
+	}
+	quotient := numBottles / numExchange
+	reminder := numBottles % numExchange
+	ans := numBottles + quotient
+	for quotient+reminder >= numExchange {
+		quotient, reminder = (quotient+reminder)/numExchange, (quotient+reminder)%numExchange
+		ans += quotient
+	}
+	return ans
+}
+```
\ No newline at end of file
diff --git a/leetcode/1539.Kth-Missing-Positive-Number/1539. Kth Missing Positive Number.go b/leetcode/1539.Kth-Missing-Positive-Number/1539. Kth Missing Positive Number.go
new file mode 100644
index 000000000..667a3b705
--- /dev/null
+++ b/leetcode/1539.Kth-Missing-Positive-Number/1539. Kth Missing Positive Number.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+func findKthPositive(arr []int, k int) int {
+	positive, index := 1, 0
+	for index < len(arr) {
+		if arr[index] != positive {
+			k--
+		} else {
+			index++
+		}
+		if k == 0 {
+			break
+		}
+		positive++
+	}
+	if k != 0 {
+		positive += k - 1
+	}
+	return positive
+}
diff --git a/leetcode/1539.Kth-Missing-Positive-Number/1539. Kth Missing Positive Number_test.go b/leetcode/1539.Kth-Missing-Positive-Number/1539. Kth Missing Positive Number_test.go
new file mode 100644
index 000000000..ffd938415
--- /dev/null
+++ b/leetcode/1539.Kth-Missing-Positive-Number/1539. Kth Missing Positive Number_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1539 struct {
+	para1539
+	ans1539
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1539 struct {
+	arr []int
+	k   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1539 struct {
+	one int
+}
+
+func Test_Problem1539(t *testing.T) {
+
+	qs := []question1539{
+
+		{
+			para1539{[]int{2, 3, 4, 7, 11}, 5},
+			ans1539{9},
+		},
+
+		{
+			para1539{[]int{1, 2, 3, 4}, 2},
+			ans1539{6},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1539------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1539, q.para1539
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, findKthPositive(p.arr, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1539.Kth-Missing-Positive-Number/README.md b/leetcode/1539.Kth-Missing-Positive-Number/README.md
new file mode 100644
index 000000000..ba13a085c
--- /dev/null
+++ b/leetcode/1539.Kth-Missing-Positive-Number/README.md
@@ -0,0 +1,63 @@
+# [1539. Kth Missing Positive Number](https://leetcode.com/problems/kth-missing-positive-number/)
+
+## 题目
+
+Given an array `arr` of positive integers sorted in a **strictly increasing order**, and an integer `k`.
+
+*Find the* `kth` *positive integer that is missing from this array.*
+
+**Example 1:**
+
+```
+Input: arr = [2,3,4,7,11], k = 5
+Output: 9
+Explanation: The missing positive integers are [1,5,6,8,9,10,12,13,...]. The 5th missing positive integer is 9.
+```
+
+**Example 2:**
+
+```
+Input: arr = [1,2,3,4], k = 2
+Output: 6
+Explanation: The missing positive integers are [5,6,7,...]. The 2nd missing positive integer is 6.
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 1000`
+- `1 <= arr[i] <= 1000`
+- `1 <= k <= 1000`
+- `arr[i] < arr[j]` for `1 <= i < j <= arr.length`
+
+## 题目大意
+
+给你一个 **严格升序排列** 的正整数数组 `arr` 和一个整数 `k` 。请你找到这个数组里第 `k` 个缺失的正整数。
+
+## 解题思路
+
+- 简单题。用一个变量从 1 开始累加，依次比对数组中是否存在，不存在的话就把 k - -，直到 k 为 0 的时候即是要输出的值。特殊情况，missing positive 都在数组之外，如例子 2 。
+
+## 代码
+
+```go
+package leetcode
+
+func findKthPositive(arr []int, k int) int {
+	positive, index := 1, 0
+	for index < len(arr) {
+		if arr[index] != positive {
+			k--
+		} else {
+			index++
+		}
+		if k == 0 {
+			break
+		}
+		positive++
+	}
+	if k != 0 {
+		positive += k - 1
+	}
+	return positive
+}
+```
\ No newline at end of file
diff --git a/leetcode/1551.Minimum-Operations-to-Make-Array-Equal/1551. Minimum Operations to Make Array Equal.go b/leetcode/1551.Minimum-Operations-to-Make-Array-Equal/1551. Minimum Operations to Make Array Equal.go
new file mode 100644
index 000000000..7010048d1
--- /dev/null
+++ b/leetcode/1551.Minimum-Operations-to-Make-Array-Equal/1551. Minimum Operations to Make Array Equal.go	
@@ -0,0 +1,5 @@
+package leetcode
+
+func minOperations(n int) int {
+	return n * n / 4
+}
diff --git a/leetcode/1551.Minimum-Operations-to-Make-Array-Equal/1551. Minimum Operations to Make Array Equal_test.go b/leetcode/1551.Minimum-Operations-to-Make-Array-Equal/1551. Minimum Operations to Make Array Equal_test.go
new file mode 100644
index 000000000..49debf1a8
--- /dev/null
+++ b/leetcode/1551.Minimum-Operations-to-Make-Array-Equal/1551. Minimum Operations to Make Array Equal_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1551 struct {
+	para1551
+	ans1551
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1551 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1551 struct {
+	one int
+}
+
+func Test_Problem1551(t *testing.T) {
+
+	qs := []question1551{
+
+		{
+			para1551{3},
+			ans1551{2},
+		},
+
+		{
+			para1551{6},
+			ans1551{9},
+		},
+
+		{
+			para1551{534},
+			ans1551{71289},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1551------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1551, q.para1551
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, minOperations(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1551.Minimum-Operations-to-Make-Array-Equal/README.md b/leetcode/1551.Minimum-Operations-to-Make-Array-Equal/README.md
new file mode 100644
index 000000000..0720b9db1
--- /dev/null
+++ b/leetcode/1551.Minimum-Operations-to-Make-Array-Equal/README.md
@@ -0,0 +1,58 @@
+# [1551. Minimum Operations to Make Array Equal](https://leetcode.com/problems/minimum-operations-to-make-array-equal/)
+
+
+## 题目
+
+You have an array `arr` of length `n` where `arr[i] = (2 * i) + 1` for all valid values of `i` (i.e. `0 <= i < n`).
+
+In one operation, you can select two indices `x` and `y` where `0 <= x, y < n` and subtract `1` from `arr[x]` and add `1` to `arr[y]` (i.e. perform `arr[x] -=1` and `arr[y] += 1`). The goal is to make all the elements of the array **equal**. It is **guaranteed** that all the elements of the array can be made equal using some operations.
+
+Given an integer `n`, the length of the array. Return *the minimum number of operations* needed to make all the elements of arr equal.
+
+**Example 1:**
+
+```
+Input: n = 3
+Output: 2
+Explanation: arr = [1, 3, 5]
+First operation choose x = 2 and y = 0, this leads arr to be [2, 3, 4]
+In the second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].
+```
+
+**Example 2:**
+
+```
+Input: n = 6
+Output: 9
+```
+
+**Constraints:**
+
+- `1 <= n <= 10^4`
+
+## 题目大意
+
+存在一个长度为 n 的数组 arr ，其中 arr[i] = (2 * i) + 1 （ 0 <= i < n ）。一次操作中，你可以选出两个下标，记作 x 和 y （ 0 <= x, y < n ）并使 arr[x] 减去 1 、arr[y] 加上 1 （即 arr[x] -=1 且 arr[y] += 1 ）。最终的目标是使数组中的所有元素都 相等 。题目测试用例将会 保证 ：在执行若干步操作后，数组中的所有元素最终可以全部相等。给你一个整数 n，即数组的长度。请你返回使数组 arr 中所有元素相等所需的 最小操作数 。
+
+## 解题思路
+
+- 这一题是数学题。题目给定的操作并不会使数组中所有元素之和变化，最终让所有元素相等，那么数组中所有元素的平均值即为最后数组中每一个元素的值。最少操作数的策略应该是以平均数为中心，中心右边的数减小，对称的中心左边的数增大。由于原数组是等差数列，两两元素之间相差 2，利用数学方法可以算出操作数。
+- 数组长度分为奇数和偶数分别讨论。如果数组长度为奇数，所需要的操作数是：
+
+    $$\begin{aligned} &\quad 2 + 4 + \cdots + 2\cdot\left\lfloor\frac{n}{2}\right\rfloor \\ &= \frac{1}{2}\left\lfloor\frac{n}{2}\right\rfloor\left(2\cdot\left\lfloor\frac{n}{2}\right\rfloor + 2\right) \\ &= \left\lfloor\frac{n}{2}\right\rfloor \left(\left\lfloor\frac{n}{2}\right\rfloor + 1\right) \\ &= \frac{n-1}{2}\left(\frac{n-1}{2} + 1\right) \\ &= \frac{n-1}{2}\cdot\frac{n+1}{2} \\ &= \frac{n^2-1}{4} \\ &= \left\lfloor\frac{n^2}{4}\right\rfloor \end{aligned}$$
+
+    数组长度是偶数，所需要的操作数是：
+
+    $$\begin{aligned} &\quad 1 + 3 + \cdots + \left(2\cdot\left\lfloor\frac{n}{2}\right\rfloor - 1\right) \\ &= \frac{1}{2}\left\lfloor\frac{n}{2}\right\rfloor\left(2\cdot\left\lfloor\frac{n}{2}\right\rfloor - 1 + 1\right)\\ &= \left(\left\lfloor\frac{n}{2}\right\rfloor\right)^2 \\ &= \frac{n^2}{4} \end{aligned}$$
+
+    综上所述，最小操作数是 n^2/4
+
+## 代码
+
+```go
+package leetcode
+
+func minOperations(n int) int {
+	return n * n / 4
+}
+```
\ No newline at end of file
diff --git a/leetcode/1572.Matrix-Diagonal-Sum/1572.Matrix Diagonal Sum.go b/leetcode/1572.Matrix-Diagonal-Sum/1572.Matrix Diagonal Sum.go
new file mode 100644
index 000000000..4f1756a33
--- /dev/null
+++ b/leetcode/1572.Matrix-Diagonal-Sum/1572.Matrix Diagonal Sum.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func diagonalSum(mat [][]int) int {
+	n := len(mat)
+	ans := 0
+	for pi := 0; pi < n; pi++ {
+		ans += mat[pi][pi]
+	}
+	for si, sj := n-1, 0; sj < n; si, sj = si-1, sj+1 {
+		ans += mat[si][sj]
+	}
+	if n%2 == 0 {
+		return ans
+	}
+	return ans - mat[n/2][n/2]
+}
diff --git a/leetcode/1572.Matrix-Diagonal-Sum/1572.Matrix Diagonal Sum_test.go b/leetcode/1572.Matrix-Diagonal-Sum/1572.Matrix Diagonal Sum_test.go
new file mode 100644
index 000000000..c658de147
--- /dev/null
+++ b/leetcode/1572.Matrix-Diagonal-Sum/1572.Matrix Diagonal Sum_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1572 struct {
+	para1572
+	ans1572
+}
+
+// para 是参数
+type para1572 struct {
+	mat [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1572 struct {
+	one int
+}
+
+func Test_Problem1572(t *testing.T) {
+
+	qs := []question1572{
+
+		{
+			para1572{[][]int{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}},
+			ans1572{25},
+		},
+
+		{
+			para1572{[][]int{{1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1}}},
+			ans1572{8},
+		},
+
+		{
+			para1572{[][]int{{5}}},
+			ans1572{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1572------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1572, q.para1572
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, diagonalSum(p.mat))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1572.Matrix-Diagonal-Sum/README.md b/leetcode/1572.Matrix-Diagonal-Sum/README.md
new file mode 100644
index 000000000..697aef7ee
--- /dev/null
+++ b/leetcode/1572.Matrix-Diagonal-Sum/README.md
@@ -0,0 +1,77 @@
+# [1572. Matrix Diagonal Sum](https://leetcode.com/problems/matrix-diagonal-sum/)
+
+
+## 题目
+
+Given a square matrix `mat`, return the sum of the matrix diagonals.
+
+Only include the sum of all the elements on the primary diagonal and all the elements on the secondary diagonal that are not part of the primary diagonal.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/08/14/sample_1911.png](https://assets.leetcode.com/uploads/2020/08/14/sample_1911.png)
+
+```
+Input: mat = [[1,2,3],
+              [4,5,6],
+              [7,8,9]]
+Output: 25
+Explanation:Diagonals sum: 1 + 5 + 9 + 3 + 7 = 25
+Notice that element mat[1][1] = 5 is counted only once.
+
+```
+
+**Example 2:**
+
+```
+Input: mat = [[1,1,1,1],
+              [1,1,1,1],
+              [1,1,1,1],
+              [1,1,1,1]]
+Output: 8
+
+```
+
+**Example 3:**
+
+```
+Input: mat = [[5]]
+Output: 5
+
+```
+
+**Constraints:**
+
+- `n == mat.length == mat[i].length`
+- `1 <= n <= 100`
+- `1 <= mat[i][j] <= 100`
+
+## 题目大意
+
+给你一个正方形矩阵 mat，请你返回矩阵对角线元素的和。请你返回在矩阵主对角线上的元素和副对角线上且不在主对角线上元素的和。
+
+## 解题思路
+
+- 简单题。根据题意，把主对角线和副对角线上的元素相加。
+- 如果正方形矩阵的长度 n 为奇数，相加的结果需要减去 mat[n/2][n/2]。
+
+## 代码
+
+```go
+package leetcode
+
+func diagonalSum(mat [][]int) int {
+	n := len(mat)
+	ans := 0
+	for pi := 0; pi < n; pi++ {
+		ans += mat[pi][pi]
+	}
+	for si, sj := n-1, 0; sj < n; si, sj = si-1, sj+1 {
+		ans += mat[si][sj]
+	}
+	if n%2 == 0 {
+		return ans
+	}
+	return ans - mat[n/2][n/2]
+}
+```
\ No newline at end of file
diff --git a/leetcode/1573.Number-of-Ways-to-Split-a-String/1573. Number of Ways to Split a String.go b/leetcode/1573.Number-of-Ways-to-Split-a-String/1573. Number of Ways to Split a String.go
new file mode 100644
index 000000000..8d81e4956
--- /dev/null
+++ b/leetcode/1573.Number-of-Ways-to-Split-a-String/1573. Number of Ways to Split a String.go	
@@ -0,0 +1,38 @@
+package leetcode
+
+func numWays(s string) int {
+	ones := 0
+	for _, c := range s {
+		if c == '1' {
+			ones++
+		}
+	}
+	if ones%3 != 0 {
+		return 0
+	}
+	if ones == 0 {
+		return (len(s) - 1) * (len(s) - 2) / 2 % 1000000007
+	}
+	N, a, b, c, d, count := ones/3, 0, 0, 0, 0, 0
+	for i, letter := range s {
+		if letter == '0' {
+			continue
+		}
+		if letter == '1' {
+			count++
+		}
+		if count == N {
+			a = i
+		}
+		if count == N+1 {
+			b = i
+		}
+		if count == 2*N {
+			c = i
+		}
+		if count == 2*N+1 {
+			d = i
+		}
+	}
+	return (b - a) * (d - c) % 1000000007
+}
diff --git a/leetcode/1573.Number-of-Ways-to-Split-a-String/1573. Number of Ways to Split a String_test.go b/leetcode/1573.Number-of-Ways-to-Split-a-String/1573. Number of Ways to Split a String_test.go
new file mode 100644
index 000000000..367191452
--- /dev/null
+++ b/leetcode/1573.Number-of-Ways-to-Split-a-String/1573. Number of Ways to Split a String_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1573 struct {
+	para1573
+	ans1573
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1573 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1573 struct {
+	one int
+}
+
+func Test_Problem1573(t *testing.T) {
+
+	qs := []question1573{
+
+		{
+			para1573{"10101"},
+			ans1573{4},
+		},
+
+		{
+			para1573{"1001"},
+			ans1573{0},
+		},
+
+		{
+			para1573{"0000"},
+			ans1573{3},
+		},
+
+		{
+			para1573{"100100010100110"},
+			ans1573{12},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1573------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1573, q.para1573
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, numWays(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1573.Number-of-Ways-to-Split-a-String/README.md b/leetcode/1573.Number-of-Ways-to-Split-a-String/README.md
new file mode 100644
index 000000000..6b8ac2266
--- /dev/null
+++ b/leetcode/1573.Number-of-Ways-to-Split-a-String/README.md
@@ -0,0 +1,108 @@
+# [1573. Number of Ways to Split a String](https://leetcode.com/problems/number-of-ways-to-split-a-string/)
+
+
+## 题目
+
+Given a binary string `s` (a string consisting only of '0's and '1's), we can split `s` into 3 **non-empty** strings s1, s2, s3 (s1+ s2+ s3 = s).
+
+Return the number of ways `s` can be split such that the number of characters '1' is the same in s1, s2, and s3.
+
+Since the answer may be too large, return it modulo 10^9 + 7.
+
+**Example 1:**
+
+```
+Input: s = "10101"
+Output: 4
+Explanation: There are four ways to split s in 3 parts where each part contain the same number of letters '1'.
+"1|010|1"
+"1|01|01"
+"10|10|1"
+"10|1|01"
+
+```
+
+**Example 2:**
+
+```
+Input: s = "1001"
+Output: 0
+
+```
+
+**Example 3:**
+
+```
+Input: s = "0000"
+Output: 3
+Explanation: There are three ways to split s in 3 parts.
+"0|0|00"
+"0|00|0"
+"00|0|0"
+
+```
+
+**Example 4:**
+
+```
+Input: s = "100100010100110"
+Output: 12
+
+```
+
+**Constraints:**
+
+- `3 <= s.length <= 10^5`
+- `s[i]` is `'0'` or `'1'`.
+
+## 题目大意
+
+给你一个二进制串 s  （一个只包含 0 和 1 的字符串），我们可以将 s 分割成 3 个 非空 字符串 s1, s2, s3 （s1 + s2 + s3 = s）。请你返回分割 s 的方案数，满足 s1，s2 和 s3 中字符 '1' 的数目相同。由于答案可能很大，请将它对 10^9 + 7 取余后返回。
+
+## 解题思路
+
+- 这一题是考察的排列组合的知识。根据题意，如果 1 的个数不是 3 的倍数，直接返回 -1。如果字符串里面没有 1，那么切分的方案就是组合，在 n-1 个字母里面选出 2 个位置。利用组合的计算方法，组合数是 (n-1) * (n-2) / 2 。
+- 剩下的是 3 的倍数的情况。在字符串中选 2 个位置隔成 3 段。从第一段最后一个 1 到第二段第一个 1 之间的 0 的个数为 m1，从第二段最后一个 1 到第三段第一个 1 之间的 0 的个数为 m2。利用乘法原理，方案数为 m1 * m2。
+
+## 代码
+
+```go
+package leetcode
+
+func numWays(s string) int {
+	ones := 0
+	for _, c := range s {
+		if c == '1' {
+			ones++
+		}
+	}
+	if ones%3 != 0 {
+		return 0
+	}
+	if ones == 0 {
+		return (len(s) - 1) * (len(s) - 2) / 2 % 1000000007
+	}
+	N, a, b, c, d, count := ones/3, 0, 0, 0, 0, 0
+	for i, letter := range s {
+		if letter == '0' {
+			continue
+		}
+		if letter == '1' {
+			count++
+		}
+		if count == N {
+			a = i
+		}
+		if count == N+1 {
+			b = i
+		}
+		if count == 2*N {
+			c = i
+		}
+		if count == 2*N+1 {
+			d = i
+		}
+	}
+	return (b - a) * (d - c) % 1000000007
+}
+```
\ No newline at end of file
diff --git a/leetcode/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters/1576. Replace-All-s-to-Avoid-Consecutive-Repeating-Characters.go b/leetcode/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters/1576. Replace-All-s-to-Avoid-Consecutive-Repeating-Characters.go
new file mode 100644
index 000000000..1e1f287d0
--- /dev/null
+++ b/leetcode/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters/1576. Replace-All-s-to-Avoid-Consecutive-Repeating-Characters.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func modifyString(s string) string {
+	res := []byte(s)
+	for i, ch := range res {
+		if ch == '?' {
+			for b := byte('a'); b <= 'z'; b++ {
+				if !(i > 0 && res[i-1] == b || i < len(res)-1 && res[i+1] == b) {
+					res[i] = b
+					break
+				}
+			}
+		}
+	}
+	return string(res)
+}
diff --git a/leetcode/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters/1576. Replace-All-s-to-Avoid-Consecutive-Repeating-Characters_test.go b/leetcode/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters/1576. Replace-All-s-to-Avoid-Consecutive-Repeating-Characters_test.go
new file mode 100644
index 000000000..9513b182c
--- /dev/null
+++ b/leetcode/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters/1576. Replace-All-s-to-Avoid-Consecutive-Repeating-Characters_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1576 struct {
+	para1576
+	ans1576
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1576 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1576 struct {
+	one string
+}
+
+func Test_Problem1576(t *testing.T) {
+
+	qs := []question1576{
+
+		{
+			para1576{"?zs"},
+			ans1576{"azs"},
+		},
+
+		{
+			para1576{"ubv?w"},
+			ans1576{"ubvaw"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1576------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1576, q.para1576
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, modifyString(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters/README.md b/leetcode/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters/README.md
new file mode 100644
index 000000000..9d7a93b39
--- /dev/null
+++ b/leetcode/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters/README.md
@@ -0,0 +1,63 @@
+# [1576. Replace All ?'s to Avoid Consecutive Repeating Characters](https://leetcode.com/problems/replace-all-s-to-avoid-consecutive-repeating-characters/)
+
+## 题目
+
+Given a string `s` containing only lowercase English letters and the `'?'` character, convert **all** the `'?'` characters into lowercase letters such that the final string does not contain any **consecutive repeating** characters. You **cannot** modify the non `'?'` characters.
+
+It is **guaranteed** that there are no consecutive repeating characters in the given string **except** for `'?'`.
+
+Return *the final string after all the conversions (possibly zero) have been made*. If there is more than one solution, return **any of them**. It can be shown that an answer is always possible with the given constraints.
+
+**Example 1:**
+
+```
+Input: s = "?zs"
+Output: "azs"
+Explanation: There are 25 solutions for this problem. From "azs" to "yzs", all are valid. Only "z" is an invalid modification as the string will consist of consecutive repeating characters in "zzs".
+
+```
+
+**Example 2:**
+
+```
+Input: s = "ubv?w"
+Output: "ubvaw"
+Explanation: There are 24 solutions for this problem. Only "v" and "w" are invalid modifications as the strings will consist of consecutive repeating characters in "ubvvw" and "ubvww".
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 100`
+- `s` consist of lowercase English letters and `'?'`.
+
+## 题目大意
+
+给你一个仅包含小写英文字母和 '?' 字符的字符串 s，请你将所有的 '?' 转换为若干小写字母，使最终的字符串不包含任何 连续重复 的字符。注意：你 不能 修改非 '?' 字符。
+
+题目测试用例保证 除 '?' 字符 之外，不存在连续重复的字符。在完成所有转换（可能无需转换）后返回最终的字符串。如果有多个解决方案，请返回其中任何一个。可以证明，在给定的约束条件下，答案总是存在的。
+
+## 解题思路
+
+- 简单题。找到源字符串中 ‘?’ 字符的位置，然后依次用 a ~ z 字符去替换，替换进去的字符不能和前后字符相同即可。
+
+## 代码
+
+```go
+package leetcode
+
+func modifyString(s string) string {
+	res := []byte(s)
+	for i, ch := range res {
+		if ch == '?' {
+			for b := byte('a'); b <= 'z'; b++ {
+				if !(i > 0 && res[i-1] == b || i < len(res)-1 && res[i+1] == b) {
+					res[i] = b
+					break
+				}
+			}
+		}
+	}
+	return string(res)
+}
+```
\ No newline at end of file
diff --git a/leetcode/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable/1579. Remove Max Number of Edges to Keep Graph Fully Traversable.go b/leetcode/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable/1579. Remove Max Number of Edges to Keep Graph Fully Traversable.go
new file mode 100644
index 000000000..9ea0984fc
--- /dev/null
+++ b/leetcode/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable/1579. Remove Max Number of Edges to Keep Graph Fully Traversable.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+func maxNumEdgesToRemove(n int, edges [][]int) int {
+	alice, bob, res := template.UnionFind{}, template.UnionFind{}, len(edges)
+	alice.Init(n)
+	bob.Init(n)
+	for _, e := range edges {
+		x, y := e[1]-1, e[2]-1
+		if e[0] == 3 && (!(alice.Find(x) == alice.Find(y)) || !(bob.Find(x) == bob.Find(y))) {
+			alice.Union(x, y)
+			bob.Union(x, y)
+			res--
+		}
+	}
+	ufs := [2]*template.UnionFind{&alice, &bob}
+	for _, e := range edges {
+		if tp := e[0]; tp < 3 && !(ufs[tp-1].Find(e[1]-1) == ufs[tp-1].Find(e[2]-1)) {
+			ufs[tp-1].Union(e[1]-1, e[2]-1)
+			res--
+		}
+	}
+	if alice.TotalCount() > 1 || bob.TotalCount() > 1 {
+		return -1
+	}
+	return res
+}
diff --git a/leetcode/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable/1579. Remove Max Number of Edges to Keep Graph Fully Traversable_test.go b/leetcode/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable/1579. Remove Max Number of Edges to Keep Graph Fully Traversable_test.go
new file mode 100644
index 000000000..be247a192
--- /dev/null
+++ b/leetcode/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable/1579. Remove Max Number of Edges to Keep Graph Fully Traversable_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1579 struct {
+	para1579
+	ans1579
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1579 struct {
+	n     int
+	edges [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1579 struct {
+	one int
+}
+
+func Test_Problem1579(t *testing.T) {
+
+	qs := []question1579{
+
+		{
+			para1579{4, [][]int{{3, 1, 2}, {3, 2, 3}, {1, 1, 3}, {1, 2, 4}, {1, 1, 2}, {2, 3, 4}}},
+			ans1579{2},
+		},
+
+		{
+			para1579{4, [][]int{{3, 1, 2}, {3, 2, 3}, {1, 1, 4}, {2, 1, 4}}},
+			ans1579{0},
+		},
+
+		{
+			para1579{4, [][]int{{3, 2, 3}, {1, 1, 2}, {2, 3, 4}}},
+			ans1579{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1579------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1579, q.para1579
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, maxNumEdgesToRemove(p.n, p.edges))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable/README.md b/leetcode/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable/README.md
new file mode 100644
index 000000000..4f4e8db34
--- /dev/null
+++ b/leetcode/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable/README.md
@@ -0,0 +1,103 @@
+# [1579. Remove Max Number of Edges to Keep Graph Fully Traversable](https://leetcode.com/problems/remove-max-number-of-edges-to-keep-graph-fully-traversable/)
+
+
+## 题目
+
+Alice and Bob have an undirected graph of `n` nodes and 3 types of edges:
+
+- Type 1: Can be traversed by Alice only.
+- Type 2: Can be traversed by Bob only.
+- Type 3: Can by traversed by both Alice and Bob.
+
+Given an array `edges` where `edges[i] = [typei, ui, vi]` represents a bidirectional edge of type `typei` between nodes `ui` and `vi`, find the maximum number of edges you can remove so that after removing the edges, the graph can still be fully traversed by both Alice and Bob. The graph is fully traversed by Alice and Bob if starting from any node, they can reach all other nodes.
+
+Return *the maximum number of edges you can remove, or return* `-1` *if it's impossible for the graph to be fully traversed by Alice and Bob.*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/08/19/ex1.png](https://assets.leetcode.com/uploads/2020/08/19/ex1.png)
+
+```
+Input: n = 4, edges = [[3,1,2],[3,2,3],[1,1,3],[1,2,4],[1,1,2],[2,3,4]]
+Output: 2
+Explanation: If we remove the 2 edges [1,1,2] and [1,1,3]. The graph will still be fully traversable by Alice and Bob. Removing any additional edge will not make it so. So the maximum number of edges we can remove is 2.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/08/19/ex2.png](https://assets.leetcode.com/uploads/2020/08/19/ex2.png)
+
+```
+Input: n = 4, edges = [[3,1,2],[3,2,3],[1,1,4],[2,1,4]]
+Output: 0
+Explanation: Notice that removing any edge will not make the graph fully traversable by Alice and Bob.
+```
+
+**Example 3:**
+
+![https://assets.leetcode.com/uploads/2020/08/19/ex3.png](https://assets.leetcode.com/uploads/2020/08/19/ex3.png)
+
+```
+Input: n = 4, edges = [[3,2,3],[1,1,2],[2,3,4]]
+Output: -1
+Explanation: In the current graph, Alice cannot reach node 4 from the other nodes. Likewise, Bob cannot reach 1. Therefore it's impossible to make the graph fully traversable.
+```
+
+**Constraints:**
+
+- `1 <= n <= 10^5`
+- `1 <= edges.length <= min(10^5, 3 * n * (n-1) / 2)`
+- `edges[i].length == 3`
+- `1 <= edges[i][0] <= 3`
+- `1 <= edges[i][1] < edges[i][2] <= n`
+- All tuples `(typei, ui, vi)` are distinct.
+
+## 题目大意
+
+Alice 和 Bob 共有一个无向图，其中包含 n 个节点和 3  种类型的边：
+
+- 类型 1：只能由 Alice 遍历。
+- 类型 2：只能由 Bob 遍历。
+- 类型 3：Alice 和 Bob 都可以遍历。
+
+给你一个数组 edges ，其中 edges[i] = [typei, ui, vi] 表示节点 ui 和 vi 之间存在类型为 typei 的双向边。请你在保证图仍能够被 Alice和 Bob 完全遍历的前提下，找出可以删除的最大边数。如果从任何节点开始，Alice 和 Bob 都可以到达所有其他节点，则认为图是可以完全遍历的。返回可以删除的最大边数，如果 Alice 和 Bob 无法完全遍历图，则返回 -1 。
+
+## 解题思路
+
+- 本题是第 1319 题的加强版。在第 1319 题中只有一个人，同样也是判断在保证图可连通的基础上，删掉最多边的条数。这一题只不过变成了 2 个人。解题思路依旧是并查集。
+- 初始化 2 个并查集，分别表示 Alice 和 Bob。先合并公共边，每合并一条边，可删除的最大总边数便减少 1 。再合并 2 人各自的单独的边，同样是每合并一条边，每合并一条边，可删除的最大总边数便减少 1 。合并完所有的边，2 人的并查集内部集合数仍大于 1，那么则代表 2 人无法完全遍历图，则输出 -1。如果 2 人的并查集内部集合都是 1，代表整个图都连通了。输出可以删除的最大边数。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+func maxNumEdgesToRemove(n int, edges [][]int) int {
+	alice, bob, res := template.UnionFind{}, template.UnionFind{}, len(edges)
+	alice.Init(n)
+	bob.Init(n)
+	for _, e := range edges {
+		x, y := e[1]-1, e[2]-1
+		if e[0] == 3 && (!(alice.Find(x) == alice.Find(y)) || !(bob.Find(x) == bob.Find(y))) {
+			alice.Union(x, y)
+			bob.Union(x, y)
+			res--
+		}
+	}
+	ufs := [2]*template.UnionFind{&alice, &bob}
+	for _, e := range edges {
+		if tp := e[0]; tp < 3 && !(ufs[tp-1].Find(e[1]-1) == ufs[tp-1].Find(e[2]-1)) {
+			ufs[tp-1].Union(e[1]-1, e[2]-1)
+			res--
+		}
+	}
+	if alice.TotalCount() > 1 || bob.TotalCount() > 1 {
+		return -1
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1600.Throne-Inheritance/1600. Throne Inheritance.go b/leetcode/1600.Throne-Inheritance/1600. Throne Inheritance.go
new file mode 100644
index 000000000..4ab87aff8
--- /dev/null
+++ b/leetcode/1600.Throne-Inheritance/1600. Throne Inheritance.go	
@@ -0,0 +1,41 @@
+package leetcode
+
+type ThroneInheritance struct {
+	king  string
+	edges map[string][]string
+	dead  map[string]bool
+}
+
+func Constructor(kingName string) (t ThroneInheritance) {
+	return ThroneInheritance{kingName, map[string][]string{}, map[string]bool{}}
+}
+
+func (t *ThroneInheritance) Birth(parentName, childName string) {
+	t.edges[parentName] = append(t.edges[parentName], childName)
+}
+
+func (t *ThroneInheritance) Death(name string) {
+	t.dead[name] = true
+}
+
+func (t *ThroneInheritance) GetInheritanceOrder() (res []string) {
+	var preorder func(string)
+	preorder = func(name string) {
+		if !t.dead[name] {
+			res = append(res, name)
+		}
+		for _, childName := range t.edges[name] {
+			preorder(childName)
+		}
+	}
+	preorder(t.king)
+	return
+}
+
+/**
+ * Your ThroneInheritance object will be instantiated and called as such:
+ * obj := Constructor(kingName);
+ * obj.Birth(parentName,childName);
+ * obj.Death(name);
+ * param_3 := obj.GetInheritanceOrder();
+ */
diff --git a/leetcode/1600.Throne-Inheritance/1600. Throne Inheritance_test.go b/leetcode/1600.Throne-Inheritance/1600. Throne Inheritance_test.go
new file mode 100644
index 000000000..2ce13f58a
--- /dev/null
+++ b/leetcode/1600.Throne-Inheritance/1600. Throne Inheritance_test.go	
@@ -0,0 +1,29 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem1600(t *testing.T) {
+	obj := Constructor("king")
+	fmt.Printf("obj = %v\n", obj)
+	obj.Birth("king", "andy")
+	fmt.Printf("obj = %v\n", obj)
+	obj.Birth("king", "bob")
+	fmt.Printf("obj = %v\n", obj)
+	obj.Birth("king", "catherine")
+	fmt.Printf("obj = %v\n", obj)
+	obj.Birth("andy", "matthew")
+	fmt.Printf("obj = %v\n", obj)
+	obj.Birth("bob", "alex")
+	fmt.Printf("obj = %v\n", obj)
+	obj.Birth("bob", "asha")
+	fmt.Printf("obj = %v\n", obj)
+	param2 := obj.GetInheritanceOrder()
+	fmt.Printf("param_2 = %v obj = %v\n", param2, obj)
+	obj.Death("bob")
+	fmt.Printf("obj = %v\n", obj)
+	param2 = obj.GetInheritanceOrder()
+	fmt.Printf("param_2 = %v obj = %v\n", param2, obj)
+}
diff --git a/leetcode/1600.Throne-Inheritance/README.md b/leetcode/1600.Throne-Inheritance/README.md
new file mode 100644
index 000000000..8e5ed385e
--- /dev/null
+++ b/leetcode/1600.Throne-Inheritance/README.md
@@ -0,0 +1,120 @@
+# [1600. Throne Inheritance](https://leetcode.com/problems/throne-inheritance/)
+
+
+## 题目
+
+A kingdom consists of a king, his children, his grandchildren, and so on. Every once in a while, someone in the family dies or a child is born.
+
+The kingdom has a well-defined order of inheritance that consists of the king as the first member. Let's define the recursive function `Successor(x, curOrder)`, which given a person `x` and the inheritance order so far, returns who should be the next person after `x` in the order of inheritance.
+
+```
+Successor(x, curOrder):
+    if x has no children or all of x's children are in curOrder:
+        if x is the king return null
+        else return Successor(x's parent, curOrder)
+    else return x's oldest child who's not in curOrder
+```
+
+For example, assume we have a kingdom that consists of the king, his children Alice and Bob (Alice is older than Bob), and finally Alice's son Jack.
+
+1. In the beginning, `curOrder` will be `["king"]`.
+2. Calling `Successor(king, curOrder)` will return Alice, so we append to `curOrder` to get `["king", "Alice"]`.
+3. Calling `Successor(Alice, curOrder)` will return Jack, so we append to `curOrder` to get `["king", "Alice", "Jack"]`.
+4. Calling `Successor(Jack, curOrder)` will return Bob, so we append to `curOrder` to get `["king", "Alice", "Jack", "Bob"]`.
+5. Calling `Successor(Bob, curOrder)` will return `null`. Thus the order of inheritance will be `["king", "Alice", "Jack", "Bob"]`.
+
+Using the above function, we can always obtain a unique order of inheritance.
+
+Implement the `ThroneInheritance` class:
+
+- `ThroneInheritance(string kingName)` Initializes an object of the `ThroneInheritance` class. The name of the king is given as part of the constructor.
+- `void birth(string parentName, string childName)` Indicates that `parentName` gave birth to `childName`.
+- `void death(string name)` Indicates the death of `name`. The death of the person doesn't affect the `Successor` function nor the current inheritance order. You can treat it as just marking the person as dead.
+- `string[] getInheritanceOrder()` Returns a list representing the current order of inheritance **excluding** dead people.
+
+**Example 1:**
+
+```
+Input
+["ThroneInheritance", "birth", "birth", "birth", "birth", "birth", "birth", "getInheritanceOrder", "death", "getInheritanceOrder"]
+[["king"], ["king", "andy"], ["king", "bob"], ["king", "catherine"], ["andy", "matthew"], ["bob", "alex"], ["bob", "asha"], [null], ["bob"], [null]]
+Output
+[null, null, null, null, null, null, null, ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"], null, ["king", "andy", "matthew", "alex", "asha", "catherine"]]
+
+Explanation
+ThroneInheritance t= new ThroneInheritance("king"); // order:king
+t.birth("king", "andy"); // order: king >andy
+t.birth("king", "bob"); // order: king > andy >bob
+t.birth("king", "catherine"); // order: king > andy > bob >catherine
+t.birth("andy", "matthew"); // order: king > andy >matthew > bob > catherine
+t.birth("bob", "alex"); // order: king > andy > matthew > bob >alex > catherine
+t.birth("bob", "asha"); // order: king > andy > matthew > bob > alex >asha > catherine
+t.getInheritanceOrder(); // return ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"]
+t.death("bob"); // order: king > andy > matthew >bob > alex > asha > catherine
+t.getInheritanceOrder(); // return ["king", "andy", "matthew", "alex", "asha", "catherine"]
+
+```
+
+**Constraints:**
+
+- `1 <= kingName.length, parentName.length, childName.length, name.length <= 15`
+- `kingName`, `parentName`, `childName`, and `name` consist of lowercase English letters only.
+- All arguments `childName` and `kingName` are **distinct**.
+- All `name` arguments of `death` will be passed to either the constructor or as `childName` to `birth` first.
+- For each call to `birth(parentName, childName)`, it is guaranteed that `parentName` is alive.
+- At most `105` calls will be made to `birth` and `death`.
+- At most `10` calls will be made to `getInheritanceOrder`.
+
+## 题目大意
+
+一个王国里住着国王、他的孩子们、他的孙子们等等。每一个时间点，这个家庭里有人出生也有人死亡。这个王国有一个明确规定的皇位继承顺序，第一继承人总是国王自己。我们定义递归函数 Successor(x, curOrder) ，给定一个人 x 和当前的继承顺序，该函数返回 x 的下一继承人。
+
+## 解题思路
+
+- 这道题思路不难。先将国王每个孩子按照顺序存在一个 map 中，然后每个国王的孩子还存在父子关系，同理也按顺序存在 map 中。执行 GetInheritanceOrder() 函数时，将国王的孩子按顺序遍历，如果每个孩子还有孩子，递归遍历到底。如果把继承关系看成一棵树，此题便是多叉树的先根遍历的问题。
+
+## 代码
+
+```go
+package leetcode
+
+type ThroneInheritance struct {
+	king  string
+	edges map[string][]string
+	dead  map[string]bool
+}
+
+func Constructor(kingName string) (t ThroneInheritance) {
+	return ThroneInheritance{kingName, map[string][]string{}, map[string]bool{}}
+}
+
+func (t *ThroneInheritance) Birth(parentName, childName string) {
+	t.edges[parentName] = append(t.edges[parentName], childName)
+}
+
+func (t *ThroneInheritance) Death(name string) {
+	t.dead[name] = true
+}
+
+func (t *ThroneInheritance) GetInheritanceOrder() (res []string) {
+	var preorder func(string)
+	preorder = func(name string) {
+		if !t.dead[name] {
+			res = append(res, name)
+		}
+		for _, childName := range t.edges[name] {
+			preorder(childName)
+		}
+	}
+	preorder(t.king)
+	return
+}
+
+/**
+ * Your ThroneInheritance object will be instantiated and called as such:
+ * obj := Constructor(kingName);
+ * obj.Birth(parentName,childName);
+ * obj.Death(name);
+ * param_3 := obj.GetInheritanceOrder();
+ */
+```
\ No newline at end of file
diff --git a/leetcode/1603.Design-Parking-System/1603. Design Parking System.go b/leetcode/1603.Design-Parking-System/1603. Design Parking System.go
new file mode 100644
index 000000000..86c9c60a5
--- /dev/null
+++ b/leetcode/1603.Design-Parking-System/1603. Design Parking System.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+type ParkingSystem struct {
+	Big    int
+	Medium int
+	Small  int
+}
+
+func Constructor(big int, medium int, small int) ParkingSystem {
+	return ParkingSystem{
+		Big:    big,
+		Medium: medium,
+		Small:  small,
+	}
+}
+
+func (this *ParkingSystem) AddCar(carType int) bool {
+	switch carType {
+	case 1:
+		{
+			if this.Big > 0 {
+				this.Big--
+				return true
+			}
+			return false
+		}
+	case 2:
+		{
+			if this.Medium > 0 {
+				this.Medium--
+				return true
+			}
+			return false
+		}
+	case 3:
+		{
+			if this.Small > 0 {
+				this.Small--
+				return true
+			}
+			return false
+		}
+	}
+	return false
+}
+
+/**
+ * Your ParkingSystem object will be instantiated and called as such:
+ * obj := Constructor(big, medium, small);
+ * param_1 := obj.AddCar(carType);
+ */
diff --git a/leetcode/1603.Design-Parking-System/1603. Design Parking System_test.go b/leetcode/1603.Design-Parking-System/1603. Design Parking System_test.go
new file mode 100644
index 000000000..9fa635702
--- /dev/null
+++ b/leetcode/1603.Design-Parking-System/1603. Design Parking System_test.go	
@@ -0,0 +1,15 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem1603(t *testing.T) {
+	obj := Constructor(1, 1, 0)
+	fmt.Printf("obj = %v\n", obj)
+	fmt.Printf("obj = %v\n", obj.AddCar(1))
+	fmt.Printf("obj = %v\n", obj.AddCar(2))
+	fmt.Printf("obj = %v\n", obj.AddCar(3))
+	fmt.Printf("obj = %v\n", obj.AddCar(1))
+}
diff --git a/leetcode/1603.Design-Parking-System/README.md b/leetcode/1603.Design-Parking-System/README.md
new file mode 100644
index 000000000..b4a5a8fe7
--- /dev/null
+++ b/leetcode/1603.Design-Parking-System/README.md
@@ -0,0 +1,103 @@
+# [1603. Design Parking System](https://leetcode.com/problems/design-parking-system/)
+
+
+## 题目
+
+Design a parking system for a parking lot. The parking lot has three kinds of parking spaces: big, medium, and small, with a fixed number of slots for each size.
+
+Implement the `ParkingSystem` class:
+
+- `ParkingSystem(int big, int medium, int small)` Initializes object of the `ParkingSystem` class. The number of slots for each parking space are given as part of the constructor.
+- `bool addCar(int carType)` Checks whether there is a parking space of `carType` for the car that wants to get into the parking lot. `carType` can be of three kinds: big, medium, or small, which are represented by `1`, `2`, and `3` respectively. **A car can only park in a parking space of its** `carType`. If there is no space available, return `false`, else park the car in that size space and return `true`.
+
+**Example 1:**
+
+```
+Input
+["ParkingSystem", "addCar", "addCar", "addCar", "addCar"]
+[[1, 1, 0], [1], [2], [3], [1]]
+Output
+[null, true, true, false, false]
+
+Explanation
+ParkingSystem parkingSystem = new ParkingSystem(1, 1, 0);
+parkingSystem.addCar(1); // return true because there is 1 available slot for a big car
+parkingSystem.addCar(2); // return true because there is 1 available slot for a medium car
+parkingSystem.addCar(3); // return false because there is no available slot for a small car
+parkingSystem.addCar(1); // return false because there is no available slot for a big car. It is already occupied.
+```
+
+**Constraints:**
+
+- `0 <= big, medium, small <= 1000`
+- `carType` is `1`, `2`, or `3`
+- At most `1000` calls will be made to `addCar`
+
+## 题目大意
+
+请你给一个停车场设计一个停车系统。停车场总共有三种不同大小的车位：大，中和小，每种尺寸分别有固定数目的车位。
+
+请你实现 ParkingSystem 类：
+
+- ParkingSystem(int big, int medium, int small) 初始化 ParkingSystem 类，三个参数分别对应每种停车位的数目。
+- bool addCar(int carType) 检查是否有 carType 对应的停车位。 carType 有三种类型：大，中，小，分别用数字 1， 2 和 3 表示。一辆车只能停在  carType 对应尺寸的停车位中。如果没有空车位，请返回 false ，否则将该车停入车位并返回 true 。
+
+## 解题思路
+
+- 简单题。分别用 3 个变量表示大，中和小车位。`addCar()` 判断这 3 个变量是否还有空车位即可。
+
+## 代码
+
+```go
+package leetcode
+
+type ParkingSystem struct {
+	Big    int
+	Medium int
+	Small  int
+}
+
+func Constructor(big int, medium int, small int) ParkingSystem {
+	return ParkingSystem{
+		Big:    big,
+		Medium: medium,
+		Small:  small,
+	}
+}
+
+func (this *ParkingSystem) AddCar(carType int) bool {
+	switch carType {
+	case 1:
+		{
+			if this.Big > 0 {
+				this.Big--
+				return true
+			}
+			return false
+		}
+	case 2:
+		{
+			if this.Medium > 0 {
+				this.Medium--
+				return true
+			}
+			return false
+		}
+	case 3:
+		{
+			if this.Small > 0 {
+				this.Small--
+				return true
+			}
+			return false
+		}
+	}
+	return false
+}
+
+/**
+ * Your ParkingSystem object will be instantiated and called as such:
+ * obj := Constructor(big, medium, small);
+ * param_1 := obj.AddCar(carType);
+ */
+```
\ No newline at end of file
diff --git a/leetcode/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X/1608. Special Array With X Elements Greater Than or Equal X.go b/leetcode/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X/1608. Special Array With X Elements Greater Than or Equal X.go
new file mode 100644
index 000000000..27ea1e025
--- /dev/null
+++ b/leetcode/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X/1608. Special Array With X Elements Greater Than or Equal X.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+import "sort"
+
+func specialArray(nums []int) int {
+	sort.Ints(nums)
+	x := len(nums)
+	for _, num := range nums {
+		if num >= x {
+			return x
+		}
+		x--
+		if num >= x {
+			return -1
+		}
+	}
+	return -1
+}
diff --git a/leetcode/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X/1608. Special Array With X Elements Greater Than or Equal X_test.go b/leetcode/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X/1608. Special Array With X Elements Greater Than or Equal X_test.go
new file mode 100644
index 000000000..5d17d1c46
--- /dev/null
+++ b/leetcode/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X/1608. Special Array With X Elements Greater Than or Equal X_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1608 struct {
+	para1608
+	ans1608
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1608 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1608 struct {
+	one int
+}
+
+func Test_Problem1608(t *testing.T) {
+
+	qs := []question1608{
+
+		{
+			para1608{[]int{3, 5}},
+			ans1608{2},
+		},
+
+		{
+			para1608{[]int{0, 0}},
+			ans1608{-1},
+		},
+
+		{
+			para1608{[]int{0, 4, 3, 0, 4}},
+			ans1608{3},
+		},
+
+		{
+			para1608{[]int{3, 6, 7, 7, 0}},
+			ans1608{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1608------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1608, q.para1608
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, specialArray(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X/README.md b/leetcode/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X/README.md
new file mode 100644
index 000000000..96ac718ac
--- /dev/null
+++ b/leetcode/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X/README.md
@@ -0,0 +1,80 @@
+# [1608. Special Array With X Elements Greater Than or Equal X](https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/)
+
+## 题目
+
+You are given an array `nums` of non-negative integers. `nums` is considered **special** if there exists a number `x` such that there are **exactly** `x` numbers in `nums` that are **greater than or equal to** `x`.
+
+Notice that `x` **does not** have to be an element in `nums`.
+
+Return `x` *if the array is **special**, otherwise, return* `-1`. It can be proven that if `nums` is special, the value for `x` is **unique**.
+
+**Example 1:**
+
+```
+Input: nums = [3,5]
+Output: 2
+Explanation: There are 2 values (3 and 5) that are greater than or equal to 2.
+```
+
+**Example 2:**
+
+```
+Input: nums = [0,0]
+Output: -1
+Explanation: No numbers fit the criteria for x.
+If x = 0, there should be 0 numbers >= x, but there are 2.
+If x = 1, there should be 1 number >= x, but there are 0.
+If x = 2, there should be 2 numbers >= x, but there are 0.
+x cannot be greater since there are only 2 numbers in nums.
+```
+
+**Example 3:**
+
+```
+Input: nums = [0,4,3,0,4]
+Output: 3
+Explanation: There are 3 values that are greater than or equal to 3.
+```
+
+**Example 4:**
+
+```
+Input: nums = [3,6,7,7,0]
+Output: -1
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `0 <= nums[i] <= 1000`
+
+## 题目大意
+
+给你一个非负整数数组 nums 。如果存在一个数 x ，使得 nums 中恰好有 x 个元素 大于或者等于 x ，那么就称 nums 是一个 特殊数组 ，而 x 是该数组的 特征值 。（注意： x 不必 是 nums 的中的元素。）如果数组 nums 是一个 特殊数组 ，请返回它的特征值 x 。否则，返回 -1 。可以证明的是，如果 nums 是特殊数组，那么其特征值 x 是 唯一的 。
+
+## 解题思路
+
+- 简单题。抓住题干中给的证明，特征值是唯一的。先将数组从小到大排序，下标的含义与特征值就等价了。下标 `i` 代表大于等于 `nums[i]` 的元素有 `len(nums) - i` 个，那么从第 0 个下标的元素开始遍历，如果这个元素都大于 `len(nums)`，那么后面 `len(nums)` 个元素也都大于等于它，特征值就找到了。如果特征值减一以后，仍然满足 `nums[i] >= x`，说明满足条件的值有多个，这一点不满足特征值唯一性，可以直接返回 -1 了。下标继续右移，特征值继续减一。如果最终循环结束依旧找不到特征值，返回 -1 。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func specialArray(nums []int) int {
+	sort.Ints(nums)
+	x := len(nums)
+	for _, num := range nums {
+		if num >= x {
+			return x
+		}
+		x--
+		if num >= x {
+			return -1
+		}
+	}
+	return -1
+}
+```
\ No newline at end of file
diff --git a/leetcode/1609.Even-Odd-Tree/1609.Even Odd Tree.go b/leetcode/1609.Even-Odd-Tree/1609.Even Odd Tree.go
new file mode 100644
index 000000000..e48683a0b
--- /dev/null
+++ b/leetcode/1609.Even-Odd-Tree/1609.Even Odd Tree.go	
@@ -0,0 +1,72 @@
+package leetcode
+
+type TreeNode struct {
+	Val   int
+	Left  *TreeNode
+	Right *TreeNode
+}
+
+func isEvenOddTree(root *TreeNode) bool {
+	level := 0
+	queue := []*TreeNode{root}
+	for len(queue) != 0 {
+		length := len(queue)
+		var nums []int
+		for i := 0; i < length; i++ {
+			node := queue[i]
+			if node.Left != nil {
+				queue = append(queue, node.Left)
+			}
+			if node.Right != nil {
+				queue = append(queue, node.Right)
+			}
+			nums = append(nums, node.Val)
+		}
+		if level%2 == 0 {
+			if !even(nums) {
+				return false
+			}
+		} else {
+			if !odd(nums) {
+				return false
+			}
+		}
+		queue = queue[length:]
+		level++
+	}
+	return true
+}
+
+func odd(nums []int) bool {
+	cur := nums[0]
+	if cur%2 != 0 {
+		return false
+	}
+	for _, num := range nums[1:] {
+		if num >= cur {
+			return false
+		}
+		if num%2 != 0 {
+			return false
+		}
+		cur = num
+	}
+	return true
+}
+
+func even(nums []int) bool {
+	cur := nums[0]
+	if cur%2 == 0 {
+		return false
+	}
+	for _, num := range nums[1:] {
+		if num <= cur {
+			return false
+		}
+		if num%2 == 0 {
+			return false
+		}
+		cur = num
+	}
+	return true
+}
diff --git a/leetcode/1609.Even-Odd-Tree/1609.Even Odd Tree_test.go b/leetcode/1609.Even-Odd-Tree/1609.Even Odd Tree_test.go
new file mode 100644
index 000000000..6ac31fbc5
--- /dev/null
+++ b/leetcode/1609.Even-Odd-Tree/1609.Even Odd Tree_test.go	
@@ -0,0 +1,73 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1609 struct {
+	para1609
+	ans1609
+}
+
+// para 是参数
+type para1609 struct {
+	root *TreeNode
+}
+
+// ans 是答案
+type ans1609 struct {
+	ans bool
+}
+
+func Test_Problem1609(t *testing.T) {
+
+	qs := []question1609{
+
+		{
+			para1609{&TreeNode{1,
+				&TreeNode{10,
+					&TreeNode{3, &TreeNode{12, nil, nil}, &TreeNode{8, nil, nil}}, nil},
+				&TreeNode{4, &TreeNode{7, &TreeNode{6, nil, nil}, nil}, &TreeNode{9, nil, &TreeNode{2, nil, nil}}}}},
+			ans1609{true},
+		},
+
+		{
+			para1609{&TreeNode{5,
+				&TreeNode{4, &TreeNode{3, nil, nil}, &TreeNode{3, nil, nil}},
+				&TreeNode{2, &TreeNode{7, nil, nil}, nil}}},
+			ans1609{false},
+		},
+
+		{
+			para1609{&TreeNode{5,
+				&TreeNode{9, &TreeNode{3, nil, nil}, &TreeNode{5, nil, nil}},
+				&TreeNode{1, &TreeNode{7, nil, nil}, nil}}},
+			ans1609{false},
+		},
+
+		{
+			para1609{&TreeNode{1, nil, nil}},
+			ans1609{true},
+		},
+
+		{
+			para1609{&TreeNode{11,
+				&TreeNode{8,
+					&TreeNode{1, &TreeNode{30, &TreeNode{17, nil, nil}, nil}, &TreeNode{20, nil, nil}},
+					&TreeNode{3, &TreeNode{18, nil, nil}, &TreeNode{16, nil, nil}}},
+				&TreeNode{6,
+					&TreeNode{9, &TreeNode{12, nil, nil}, &TreeNode{10, nil, nil}},
+					&TreeNode{11, &TreeNode{4, nil, nil}, &TreeNode{2, nil, nil}}}}},
+			ans1609{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1609------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1609, q.para1609
+		fmt.Printf("【input】:%v    【output】:%v\n", p.root, isEvenOddTree(p.root))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1609.Even-Odd-Tree/README.md b/leetcode/1609.Even-Odd-Tree/README.md
new file mode 100644
index 000000000..0481d9821
--- /dev/null
+++ b/leetcode/1609.Even-Odd-Tree/README.md
@@ -0,0 +1,150 @@
+# [1609. Even Odd Tree](https://leetcode.com/problems/even-odd-tree/)
+
+## 题目
+
+A binary tree is named Even-Odd if it meets the following conditions:
+
+- The root of the binary tree is at level index 0, its children are at level index 1, their children are at level index 2, etc.
+- For every even-indexed level, all nodes at the level have odd integer values in strictly increasing order (from left to right).
+- For every odd-indexed level, all nodes at the level have even integer values in strictly decreasing order (from left to right).
+
+Given the root of a binary tree, return true if the binary tree is Even-Odd, otherwise return false.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2020/09/15/sample_1_1966.png](https://assets.leetcode.com/uploads/2020/09/15/sample_1_1966.png)
+
+    Input: root = [1,10,4,3,null,7,9,12,8,6,null,null,2]
+    Output: true
+    Explanation: The node values on each level are:
+    Level 0: [1]
+    Level 1: [10,4]
+    Level 2: [3,7,9]
+    Level 3: [12,8,6,2]
+    Since levels 0 and 2 are all odd and increasing and levels 1 and 3 are all even and decreasing, the tree is Even-Odd.
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2020/09/15/sample_2_1966.png](https://assets.leetcode.com/uploads/2020/09/15/sample_2_1966.png)
+
+    Input: root = [5,4,2,3,3,7]
+    Output: false
+    Explanation: The node values on each level are:
+    Level 0: [5]
+    Level 1: [4,2]
+    Level 2: [3,3,7]
+    Node values in level 2 must be in strictly increasing order, so the tree is not Even-Odd.
+
+**Example 3**:
+
+![https://assets.leetcode.com/uploads/2020/09/22/sample_1_333_1966.png](https://assets.leetcode.com/uploads/2020/09/22/sample_1_333_1966.png)
+
+    Input: root = [5,9,1,3,5,7]
+    Output: false
+    Explanation: Node values in the level 1 should be even integers.
+
+**Example 4**:
+
+    Input: root = [1]
+    Output: true
+
+**Example 5**:
+
+    Input: root = [11,8,6,1,3,9,11,30,20,18,16,12,10,4,2,17]
+    Output: True
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range [1, 100000]. 
+- 1 <= Node.val <= 1000000
+
+## 题目大意
+
+如果一棵二叉树满足下述几个条件，则可以称为 奇偶树 ：
+
+- 二叉树根节点所在层下标为 0 ，根的子节点所在层下标为 1 ，根的孙节点所在层下标为 2 ，依此类推。
+- 偶数下标 层上的所有节点的值都是 奇 整数，从左到右按顺序 严格递增
+- 奇数下标 层上的所有节点的值都是 偶 整数，从左到右按顺序 严格递减
+
+给你二叉树的根节点，如果二叉树为 奇偶树 ，则返回 true ，否则返回 false 。
+
+## 解题思路
+
+- 广度优先遍历(分别判断奇数层和偶数层)
+
+## 代码
+
+```go
+package leetcode
+
+type TreeNode struct {
+	Val int
+	Left *TreeNode
+	Right *TreeNode
+}
+
+func isEvenOddTree(root *TreeNode) bool {
+	level := 0
+	queue := []*TreeNode{root}
+	for len(queue) != 0 {
+		length := len(queue)
+		var nums []int
+		for i := 0; i < length; i++ {
+			node := queue[i]
+			if node.Left != nil {
+				queue = append(queue, node.Left)
+			}
+			if node.Right != nil {
+				queue = append(queue, node.Right)
+			}
+			nums = append(nums, node.Val)
+		}
+		if level%2 == 0 {
+			if !even(nums) {
+				return false
+			}
+		} else {
+			if !odd(nums) {
+				return false
+			}
+		}
+		queue = queue[length:]
+		level++
+	}
+	return true
+}
+
+func odd(nums []int) bool {
+	cur := nums[0]
+	if cur%2 != 0 {
+		return false
+	}
+	for _, num := range nums[1:] {
+		if num >= cur {
+			return false
+		}
+		if num%2 != 0 {
+			return false
+		}
+		cur = num
+	}
+	return true
+}
+
+func even(nums []int) bool {
+	cur := nums[0]
+	if cur%2 == 0 {
+		return false
+	}
+	for _, num := range nums[1:] {
+		if num <= cur {
+			return false
+		}
+		if num%2 == 0 {
+			return false
+		}
+		cur = num
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/1614.Maximum-Nesting-Depth-of-the-Parentheses/1614. Maximum Nesting Depth of the Parentheses.go b/leetcode/1614.Maximum-Nesting-Depth-of-the-Parentheses/1614. Maximum Nesting Depth of the Parentheses.go
new file mode 100644
index 000000000..5f83a24ce
--- /dev/null
+++ b/leetcode/1614.Maximum-Nesting-Depth-of-the-Parentheses/1614. Maximum Nesting Depth of the Parentheses.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+func maxDepth(s string) int {
+	res, cur := 0, 0
+	for _, c := range s {
+		if c == '(' {
+			cur++
+			res = max(res, cur)
+		} else if c == ')' {
+			cur--
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1614.Maximum-Nesting-Depth-of-the-Parentheses/1614. Maximum Nesting Depth of the Parentheses_test.go b/leetcode/1614.Maximum-Nesting-Depth-of-the-Parentheses/1614. Maximum Nesting Depth of the Parentheses_test.go
new file mode 100644
index 000000000..cfb465461
--- /dev/null
+++ b/leetcode/1614.Maximum-Nesting-Depth-of-the-Parentheses/1614. Maximum Nesting Depth of the Parentheses_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1614 struct {
+	para1614
+	ans1614
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1614 struct {
+	p string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1614 struct {
+	one int
+}
+
+func Test_Problem1614(t *testing.T) {
+
+	qs := []question1614{
+
+		{
+			para1614{"(1+(2*3)+((8)/4))+1"},
+			ans1614{3},
+		},
+
+		{
+			para1614{"(1)+((2))+(((3)))"},
+			ans1614{3},
+		},
+
+		{
+			para1614{"1+(2*3)/(2-1)"},
+			ans1614{1},
+		},
+
+		{
+			para1614{"1"},
+			ans1614{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1614------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1614, q.para1614
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, maxDepth(p.p))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1614.Maximum-Nesting-Depth-of-the-Parentheses/README.md b/leetcode/1614.Maximum-Nesting-Depth-of-the-Parentheses/README.md
new file mode 100644
index 000000000..6f87c911b
--- /dev/null
+++ b/leetcode/1614.Maximum-Nesting-Depth-of-the-Parentheses/README.md
@@ -0,0 +1,102 @@
+# [1614. Maximum Nesting Depth of the Parentheses](https://leetcode.com/problems/maximum-nesting-depth-of-the-parentheses/)
+
+## 题目
+
+A string is a **valid parentheses string** (denoted **VPS**) if it meets one of the following:
+
+- It is an empty string `""`, or a single character not equal to `"("` or `")"`,
+- It can be written as `AB` (`A` concatenated with `B`), where `A` and `B` are **VPS**'s, or
+- It can be written as `(A)`, where `A` is a **VPS**.
+
+We can similarly define the **nesting depth** `depth(S)` of any VPS `S` as follows:
+
+- `depth("") = 0`
+- `depth(C) = 0`, where `C` is a string with a single character not equal to `"("` or `")"`.
+- `depth(A + B) = max(depth(A), depth(B))`, where `A` and `B` are **VPS**'s.
+- `depth("(" + A + ")") = 1 + depth(A)`, where `A` is a **VPS**.
+
+For example, `""`, `"()()"`, and `"()(()())"` are **VPS**'s (with nesting depths 0, 1, and 2), and `")("` and `"(()"` are not **VPS**'s.
+
+Given a **VPS** represented as string `s`, return *the **nesting depth** of* `s`.
+
+**Example 1:**
+
+```
+Input: s = "(1+(2*3)+((8)/4))+1"
+Output: 3
+Explanation: Digit 8 is inside of 3 nested parentheses in the string.
+```
+
+**Example 2:**
+
+```
+Input: s = "(1)+((2))+(((3)))"
+Output: 3
+```
+
+**Example 3:**
+
+```
+Input: s = "1+(2*3)/(2-1)"
+Output: 1
+```
+
+**Example 4:**
+
+```
+Input: s = "1"
+Output: 0
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 100`
+- `s` consists of digits `0-9` and characters `'+'`, `'-'`, `'*'`, `'/'`, `'('`, and `')'`.
+- It is guaranteed that parentheses expression `s` is a **VPS**.
+
+## 题目大意
+
+如果字符串满足以下条件之一，则可以称之为 有效括号字符串（valid parentheses string，可以简写为 VPS）：
+
+- 字符串是一个空字符串 ""，或者是一个不为 "(" 或 ")" 的单字符。
+- 字符串可以写为 AB（A 与 B 字符串连接），其中 A 和 B 都是 有效括号字符串 。
+- 字符串可以写为 (A)，其中 A 是一个 有效括号字符串 。
+
+类似地，可以定义任何有效括号字符串 S 的 嵌套深度 depth(S)：
+
+- depth("") = 0
+- depth(C) = 0，其中 C 是单个字符的字符串，且该字符不是 "(" 或者 ")"
+- depth(A + B) = max(depth(A), depth(B))，其中 A 和 B 都是 有效括号字符串
+- depth("(" + A + ")") = 1 + depth(A)，其中 A 是一个 有效括号字符串
+
+例如：""、"()()"、"()(()())" 都是 有效括号字符串（嵌套深度分别为 0、1、2），而 ")(" 、"(()" 都不是 有效括号字符串 。给你一个 有效括号字符串 s，返回该字符串的 s 嵌套深度 。
+
+## 解题思路
+
+- 简单题。求一个括号字符串的嵌套深度。题目给的括号字符串都是有效的，所以不需要考虑非法的情况。扫描一遍括号字符串，遇到 `(` 可以无脑 ++，并动态维护最大值，遇到 `)` 可以无脑 - - 。最后输出最大值即可。
+
+## 代码
+
+```go
+package leetcode
+
+func maxDepth(s string) int {
+	res, cur := 0, 0
+	for _, c := range s {
+		if c == '(' {
+			cur++
+			res = max(res, cur)
+		} else if c == ')' {
+			cur--
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1619.Mean-of-Array-After-Removing-Some-Elements/1619. Mean of Array After Removing Some Elements.go b/leetcode/1619.Mean-of-Array-After-Removing-Some-Elements/1619. Mean of Array After Removing Some Elements.go
new file mode 100644
index 000000000..a037ec08a
--- /dev/null
+++ b/leetcode/1619.Mean-of-Array-After-Removing-Some-Elements/1619. Mean of Array After Removing Some Elements.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+import "sort"
+
+func trimMean(arr []int) float64 {
+	sort.Ints(arr)
+	n, sum := len(arr), 0
+	for i := n / 20; i < n-(n/20); i++ {
+		sum += arr[i]
+	}
+	return float64(sum) / float64((n - (n / 10)))
+}
diff --git a/leetcode/1619.Mean-of-Array-After-Removing-Some-Elements/1619. Mean of Array After Removing Some Elements_test.go b/leetcode/1619.Mean-of-Array-After-Removing-Some-Elements/1619. Mean of Array After Removing Some Elements_test.go
new file mode 100644
index 000000000..ecb7e9016
--- /dev/null
+++ b/leetcode/1619.Mean-of-Array-After-Removing-Some-Elements/1619. Mean of Array After Removing Some Elements_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1619 struct {
+	para1619
+	ans1619
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1619 struct {
+	p []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1619 struct {
+	one float64
+}
+
+func Test_Problem1619(t *testing.T) {
+
+	qs := []question1619{
+
+		{
+			para1619{[]int{1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3}},
+			ans1619{2.00000},
+		},
+
+		{
+			para1619{[]int{6, 2, 7, 5, 1, 2, 0, 3, 10, 2, 5, 0, 5, 5, 0, 8, 7, 6, 8, 0}},
+			ans1619{4.00000},
+		},
+
+		{
+			para1619{[]int{6, 0, 7, 0, 7, 5, 7, 8, 3, 4, 0, 7, 8, 1, 6, 8, 1, 1, 2, 4, 8, 1, 9, 5, 4, 3, 8, 5, 10, 8, 6, 6, 1, 0, 6, 10, 8, 2, 3, 4}},
+			ans1619{4.77778},
+		},
+
+		{
+			para1619{[]int{9, 7, 8, 7, 7, 8, 4, 4, 6, 8, 8, 7, 6, 8, 8, 9, 2, 6, 0, 0, 1, 10, 8, 6, 3, 3, 5, 1, 10, 9, 0, 7, 10, 0, 10, 4, 1, 10, 6, 9, 3, 6, 0, 0, 2, 7, 0, 6, 7, 2, 9, 7, 7, 3, 0, 1, 6, 1, 10, 3}},
+			ans1619{5.27778},
+		},
+
+		{
+			para1619{[]int{4, 8, 4, 10, 0, 7, 1, 3, 7, 8, 8, 3, 4, 1, 6, 2, 1, 1, 8, 0, 9, 8, 0, 3, 9, 10, 3, 10, 1, 10, 7, 3, 2, 1, 4, 9, 10, 7, 6, 4, 0, 8, 5, 1, 2, 1, 6, 2, 5, 0, 7, 10, 9, 10, 3, 7, 10, 5, 8, 5, 7, 6, 7, 6, 10, 9, 5, 10, 5, 5, 7, 2, 10, 7, 7, 8, 2, 0, 1, 1}},
+			ans1619{5.29167},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1619------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1619, q.para1619
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, trimMean(p.p))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1619.Mean-of-Array-After-Removing-Some-Elements/README.md b/leetcode/1619.Mean-of-Array-After-Removing-Some-Elements/README.md
new file mode 100644
index 000000000..35a8e7c4a
--- /dev/null
+++ b/leetcode/1619.Mean-of-Array-After-Removing-Some-Elements/README.md
@@ -0,0 +1,74 @@
+# [1619. Mean of Array After Removing Some Elements](https://leetcode.com/problems/mean-of-array-after-removing-some-elements/)
+
+## 题目
+
+Given an integer array `arr`, return *the mean of the remaining integers after removing the smallest `5%` and the largest `5%` of the elements.*
+
+Answers within `10-5` of the **actual answer** will be considered accepted.
+
+**Example 1:**
+
+```
+Input: arr = [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3]
+Output: 2.00000
+Explanation: After erasing the minimum and the maximum values of this array, all elements are equal to 2, so the mean is 2.
+```
+
+**Example 2:**
+
+```
+Input: arr = [6,2,7,5,1,2,0,3,10,2,5,0,5,5,0,8,7,6,8,0]
+Output: 4.00000
+```
+
+**Example 3:**
+
+```
+Input: arr = [6,0,7,0,7,5,7,8,3,4,0,7,8,1,6,8,1,1,2,4,8,1,9,5,4,3,8,5,10,8,6,6,1,0,6,10,8,2,3,4]
+Output: 4.77778
+```
+
+**Example 4:**
+
+```
+Input: arr = [9,7,8,7,7,8,4,4,6,8,8,7,6,8,8,9,2,6,0,0,1,10,8,6,3,3,5,1,10,9,0,7,10,0,10,4,1,10,6,9,3,6,0,0,2,7,0,6,7,2,9,7,7,3,0,1,6,1,10,3]
+Output: 5.27778
+```
+
+**Example 5:**
+
+```
+Input: arr = [4,8,4,10,0,7,1,3,7,8,8,3,4,1,6,2,1,1,8,0,9,8,0,3,9,10,3,10,1,10,7,3,2,1,4,9,10,7,6,4,0,8,5,1,2,1,6,2,5,0,7,10,9,10,3,7,10,5,8,5,7,6,7,6,10,9,5,10,5,5,7,2,10,7,7,8,2,0,1,1]
+Output: 5.29167
+```
+
+**Constraints:**
+
+- `20 <= arr.length <= 1000`
+- `arr.length` **is a multiple** of `20`.
+- `0 <= arr[i] <= 10^5`
+
+## 题目大意
+
+给你一个整数数组 `arr` ，请你删除最小 `5%` 的数字和最大 `5%` 的数字后，剩余数字的平均值。与 **标准答案** 误差在 `10-5` 的结果都被视为正确结果。
+
+## 解题思路
+
+- 简单题。先将数组排序，然后累加中间 90% 的元素，最后计算平均值。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func trimMean(arr []int) float64 {
+	sort.Ints(arr)
+	n, sum := len(arr), 0
+	for i := n / 20; i < n-(n/20); i++ {
+		sum += arr[i]
+	}
+	return float64(sum) / float64((n - (n / 10)))
+}
+```
\ No newline at end of file
diff --git a/leetcode/1624.Largest-Substring-Between-Two-Equal-Characters/1624. Largest Substring Between Two Equal Characters.go b/leetcode/1624.Largest-Substring-Between-Two-Equal-Characters/1624. Largest Substring Between Two Equal Characters.go
new file mode 100644
index 000000000..1889318bb
--- /dev/null
+++ b/leetcode/1624.Largest-Substring-Between-Two-Equal-Characters/1624. Largest Substring Between Two Equal Characters.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+import "strings"
+
+func maxLengthBetweenEqualCharacters(s string) int {
+	res := -1
+	for k, v := range s {
+		tmp := strings.LastIndex(s, string(v))
+		if tmp > 0 {
+			if res < tmp-k-1 {
+				res = tmp - k - 1
+			}
+		}
+	}
+	return res
+}
diff --git a/leetcode/1624.Largest-Substring-Between-Two-Equal-Characters/1624. Largest Substring Between Two Equal Characters_test.go b/leetcode/1624.Largest-Substring-Between-Two-Equal-Characters/1624. Largest Substring Between Two Equal Characters_test.go
new file mode 100644
index 000000000..6c5959a17
--- /dev/null
+++ b/leetcode/1624.Largest-Substring-Between-Two-Equal-Characters/1624. Largest Substring Between Two Equal Characters_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1624 struct {
+	para1624
+	ans1624
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1624 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1624 struct {
+	one int
+}
+
+func Test_Problem1624(t *testing.T) {
+
+	qs := []question1624{
+
+		{
+			para1624{"aa"},
+			ans1624{0},
+		},
+
+		{
+			para1624{"abca"},
+			ans1624{2},
+		},
+
+		{
+			para1624{"cbzxy"},
+			ans1624{-1},
+		},
+
+		{
+			para1624{"cabbac"},
+			ans1624{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1624------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1624, q.para1624
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, maxLengthBetweenEqualCharacters(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1624.Largest-Substring-Between-Two-Equal-Characters/README.md b/leetcode/1624.Largest-Substring-Between-Two-Equal-Characters/README.md
new file mode 100644
index 000000000..b488fdb8b
--- /dev/null
+++ b/leetcode/1624.Largest-Substring-Between-Two-Equal-Characters/README.md
@@ -0,0 +1,73 @@
+# [1624. Largest Substring Between Two Equal Characters](https://leetcode.com/problems/largest-substring-between-two-equal-characters/)
+
+## 题目
+
+Given a string `s`, return *the length of the longest substring between two equal characters, excluding the two characters.* If there is no such substring return `-1`.
+
+A **substring** is a contiguous sequence of characters within a string.
+
+**Example 1:**
+
+```
+Input: s = "aa"
+Output: 0
+Explanation: The optimal substring here is an empty substring between the two 'a's.
+```
+
+**Example 2:**
+
+```
+Input: s = "abca"
+Output: 2
+Explanation: The optimal substring here is "bc".
+```
+
+**Example 3:**
+
+```
+Input: s = "cbzxy"
+Output: -1
+Explanation: There are no characters that appear twice in s.
+```
+
+**Example 4:**
+
+```
+Input: s = "cabbac"
+Output: 4
+Explanation: The optimal substring here is "abba". Other non-optimal substrings include "bb" and "".
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 300`
+- `s` contains only lowercase English letters.
+
+## 题目大意
+
+给你一个字符串 s，请你返回 两个相同字符之间的最长子字符串的长度 ，计算长度时不含这两个字符。如果不存在这样的子字符串，返回 -1 。子字符串 是字符串中的一个连续字符序列。
+
+## 解题思路
+
+- 简单题。取每个字符，扫描一次字符串，如果在字符串中还能找到相同的字符，则返回最末尾的那个字符，计算这两个字符之间的距离。取最末尾的字符是为了让两个相同的字符距离最长。扫描字符串过程中动态维护最长长度。如果字符串中不存在 2 个相同的字符，则返回 -1 。
+
+## 代码
+
+```go
+package leetcode
+
+import "strings"
+
+func maxLengthBetweenEqualCharacters(s string) int {
+	res := -1
+	for k, v := range s {
+		tmp := strings.LastIndex(s, string(v))
+		if tmp > 0 {
+			if res < tmp-k-1 {
+				res = tmp - k - 1
+			}
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1629.Slowest-Key/1629. Slowest Key.go b/leetcode/1629.Slowest-Key/1629. Slowest Key.go
new file mode 100644
index 000000000..05da61a4a
--- /dev/null
+++ b/leetcode/1629.Slowest-Key/1629. Slowest Key.go	
@@ -0,0 +1,15 @@
+package leetcode
+
+func slowestKey(releaseTimes []int, keysPressed string) byte {
+	longestDuration, key := releaseTimes[0], keysPressed[0]
+	for i := 1; i < len(releaseTimes); i++ {
+		duration := releaseTimes[i] - releaseTimes[i-1]
+		if duration > longestDuration {
+			longestDuration = duration
+			key = keysPressed[i]
+		} else if duration == longestDuration && keysPressed[i] > key {
+			key = keysPressed[i]
+		}
+	}
+	return key
+}
diff --git a/leetcode/1629.Slowest-Key/1629. Slowest Key_test.go b/leetcode/1629.Slowest-Key/1629. Slowest Key_test.go
new file mode 100644
index 000000000..90d9b6b2e
--- /dev/null
+++ b/leetcode/1629.Slowest-Key/1629. Slowest Key_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1629 struct {
+	para1629
+	ans1629
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1629 struct {
+	releaseTimes []int
+	keysPressed  string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1629 struct {
+	one byte
+}
+
+func Test_Problem1629(t *testing.T) {
+
+	qs := []question1629{
+
+		{
+			para1629{[]int{9, 29, 49, 50}, "cbcd"},
+			ans1629{'c'},
+		},
+
+		{
+			para1629{[]int{12, 23, 36, 46, 62}, "spuda"},
+			ans1629{'a'},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1629------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1629, q.para1629
+		fmt.Printf("【input】:%v      【output】:%c      \n", p, slowestKey(p.releaseTimes, p.keysPressed))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1629.Slowest-Key/README.md b/leetcode/1629.Slowest-Key/README.md
new file mode 100644
index 000000000..a12b8ba6f
--- /dev/null
+++ b/leetcode/1629.Slowest-Key/README.md
@@ -0,0 +1,83 @@
+# [1629. Slowest Key](https://leetcode.com/problems/slowest-key/)
+
+
+## 题目
+
+A newly designed keypad was tested, where a tester pressed a sequence of `n` keys, one at a time.
+
+You are given a string `keysPressed` of length `n`, where `keysPressed[i]` was the `ith` key pressed in the testing sequence, and a sorted list `releaseTimes`, where `releaseTimes[i]` was the time the `ith` key was released. Both arrays are **0-indexed**. The `0th` key was pressed at the time `0`, and every subsequent key was pressed at the **exact** time the previous key was released.
+
+The tester wants to know the key of the keypress that had the **longest duration**. The `ith` keypress had a **duration** of `releaseTimes[i] - releaseTimes[i - 1]`, and the `0th` keypress had a duration of `releaseTimes[0]`.
+
+Note that the same key could have been pressed multiple times during the test, and these multiple presses of the same key **may not** have had the same **duration**.
+
+*Return the key of the keypress that had the **longest duration**. If there are multiple such keypresses, return the lexicographically largest key of the keypresses.*
+
+**Example 1:**
+
+```
+Input: releaseTimes = [9,29,49,50], keysPressed = "cbcd"
+Output: "c"
+Explanation: The keypresses were as follows:
+Keypress for 'c' had a duration of 9 (pressed at time 0 and released at time 9).
+Keypress for 'b' had a duration of 29 - 9 = 20 (pressed at time 9 right after the release of the previous character and released at time 29).
+Keypress for 'c' had a duration of 49 - 29 = 20 (pressed at time 29 right after the release of the previous character and released at time 49).
+Keypress for 'd' had a duration of 50 - 49 = 1 (pressed at time 49 right after the release of the previous character and released at time 50).
+The longest of these was the keypress for 'b' and the second keypress for 'c', both with duration 20.
+'c' is lexicographically larger than 'b', so the answer is 'c'.
+```
+
+**Example 2:**
+
+```
+Input: releaseTimes = [12,23,36,46,62], keysPressed = "spuda"
+Output: "a"
+Explanation: The keypresses were as follows:
+Keypress for 's' had a duration of 12.
+Keypress for 'p' had a duration of 23 - 12 = 11.
+Keypress for 'u' had a duration of 36 - 23 = 13.
+Keypress for 'd' had a duration of 46 - 36 = 10.
+Keypress for 'a' had a duration of 62 - 46 = 16.
+The longest of these was the keypress for 'a' with duration 16.
+```
+
+**Constraints:**
+
+- `releaseTimes.length == n`
+- `keysPressed.length == n`
+- `2 <= n <= 1000`
+- `1 <= releaseTimes[i] <= 109`
+- `releaseTimes[i] < releaseTimes[i+1]`
+- `keysPressed` contains only lowercase English letters.
+
+## 题目大意
+
+LeetCode 设计了一款新式键盘，正在测试其可用性。测试人员将会点击一系列键（总计 n 个），每次一个。
+
+给你一个长度为 n 的字符串 keysPressed ，其中 keysPressed[i] 表示测试序列中第 i 个被按下的键。releaseTimes 是一个升序排列的列表，其中 releaseTimes[i] 表示松开第 i 个键的时间。字符串和数组的 下标都从 0 开始 。第 0 个键在时间为 0 时被按下，接下来每个键都 恰好 在前一个键松开时被按下。测试人员想要找出按键 持续时间最长 的键。第 i 次按键的持续时间为 releaseTimes[i] - releaseTimes[i - 1] ，第 0 次按键的持续时间为 releaseTimes[0] 。
+
+注意，测试期间，同一个键可以在不同时刻被多次按下，而每次的持续时间都可能不同。请返回按键 持续时间最长 的键，如果有多个这样的键，则返回 按字母顺序排列最大 的那个键。
+
+## 解题思路
+
+- 题干很长，不过还是简单题。循环扫描一遍数组，计算出每个按键的持续时间。动态更新按键最长时间的键。如果持续时间最长的有多个键，需要返回字母序最大的那一个键。
+
+## 代码
+
+```go
+package leetcode
+
+func slowestKey(releaseTimes []int, keysPressed string) byte {
+	longestDuration, key := releaseTimes[0], keysPressed[0]
+	for i := 1; i < len(releaseTimes); i++ {
+		duration := releaseTimes[i] - releaseTimes[i-1]
+		if duration > longestDuration {
+			longestDuration = duration
+			key = keysPressed[i]
+		} else if duration == longestDuration && keysPressed[i] > key {
+			key = keysPressed[i]
+		}
+	}
+	return key
+}
+```
\ No newline at end of file
diff --git a/leetcode/1631.Path-With-Minimum-Effort/1631. Path With Minimum Effort.go b/leetcode/1631.Path-With-Minimum-Effort/1631. Path With Minimum Effort.go
new file mode 100644
index 000000000..9a3b4ec0f
--- /dev/null
+++ b/leetcode/1631.Path-With-Minimum-Effort/1631. Path With Minimum Effort.go	
@@ -0,0 +1,108 @@
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+var dir = [4][2]int{
+	{0, 1},
+	{1, 0},
+	{0, -1},
+	{-1, 0},
+}
+
+// 解法一 DFS + 二分
+func minimumEffortPath(heights [][]int) int {
+	n, m := len(heights), len(heights[0])
+	visited := make([][]bool, n)
+	for i := range visited {
+		visited[i] = make([]bool, m)
+	}
+	low, high := 0, 1000000
+	for low < high {
+		threshold := low + (high-low)>>1
+		if !hasPath(heights, visited, 0, 0, threshold) {
+			low = threshold + 1
+		} else {
+			high = threshold
+		}
+		for i := range visited {
+			for j := range visited[i] {
+				visited[i][j] = false
+			}
+		}
+	}
+	return low
+}
+
+func hasPath(heights [][]int, visited [][]bool, i, j, threshold int) bool {
+	n, m := len(heights), len(heights[0])
+	if i == n-1 && j == m-1 {
+		return true
+	}
+	visited[i][j] = true
+	res := false
+	for _, d := range dir {
+		ni, nj := i+d[0], j+d[1]
+		if ni < 0 || ni >= n || nj < 0 || nj >= m || visited[ni][nj] || res {
+			continue
+		}
+		diff := abs(heights[i][j] - heights[ni][nj])
+		if diff <= threshold && hasPath(heights, visited, ni, nj, threshold) {
+			res = true
+		}
+	}
+	return res
+}
+
+func abs(a int) int {
+	if a < 0 {
+		a = -a
+	}
+	return a
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+func max(a, b int) int {
+	if a < b {
+		return b
+	}
+	return a
+}
+
+// 解法二 并查集
+func minimumEffortPath1(heights [][]int) int {
+	n, m, edges, uf := len(heights), len(heights[0]), []edge{}, template.UnionFind{}
+	uf.Init(n * m)
+	for i, row := range heights {
+		for j, h := range row {
+			id := i*m + j
+			if i > 0 {
+				edges = append(edges, edge{id - m, id, abs(h - heights[i-1][j])})
+			}
+			if j > 0 {
+				edges = append(edges, edge{id - 1, id, abs(h - heights[i][j-1])})
+			}
+		}
+	}
+	sort.Slice(edges, func(i, j int) bool { return edges[i].diff < edges[j].diff })
+	for _, e := range edges {
+		uf.Union(e.v, e.w)
+		if uf.Find(0) == uf.Find(n*m-1) {
+			return e.diff
+		}
+	}
+	return 0
+}
+
+type edge struct {
+	v, w, diff int
+}
diff --git a/leetcode/1631.Path-With-Minimum-Effort/1631. Path With Minimum Effort_test.go b/leetcode/1631.Path-With-Minimum-Effort/1631. Path With Minimum Effort_test.go
new file mode 100644
index 000000000..7b2cd74b2
--- /dev/null
+++ b/leetcode/1631.Path-With-Minimum-Effort/1631. Path With Minimum Effort_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1631 struct {
+	para1631
+	ans1631
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1631 struct {
+	heights [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1631 struct {
+	one int
+}
+
+func Test_Problem1631(t *testing.T) {
+
+	qs := []question1631{
+
+		{
+			para1631{[][]int{{1, 2, 2}, {3, 8, 2}, {5, 3, 5}}},
+			ans1631{2},
+		},
+
+		{
+			para1631{[][]int{{1, 2, 3}, {3, 8, 4}, {5, 3, 5}}},
+			ans1631{1},
+		},
+
+		{
+			para1631{[][]int{{1, 2, 1, 1, 1}, {1, 2, 1, 2, 1}, {1, 2, 1, 2, 1}, {1, 2, 1, 2, 1}, {1, 1, 1, 2, 1}}},
+			ans1631{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1631------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1631, q.para1631
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, minimumEffortPath(p.heights))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1631.Path-With-Minimum-Effort/README.md b/leetcode/1631.Path-With-Minimum-Effort/README.md
new file mode 100644
index 000000000..414863562
--- /dev/null
+++ b/leetcode/1631.Path-With-Minimum-Effort/README.md
@@ -0,0 +1,170 @@
+# [1631. Path With Minimum Effort](https://leetcode.com/problems/path-with-minimum-effort/)
+
+## 题目
+
+You are a hiker preparing for an upcoming hike. You are given `heights`, a 2D array of size `rows x columns`, where `heights[row][col]` represents the height of cell `(row, col)`. You are situated in the top-left cell, `(0, 0)`, and you hope to travel to the bottom-right cell, `(rows-1, columns-1)` (i.e., **0-indexed**). You can move **up**, **down**, **left**, or **right**, and you wish to find a route that requires the minimum **effort**.
+
+A route's **effort** is the **maximum absolute difference** in heights between two consecutive cells of the route.
+
+Return *the minimum **effort** required to travel from the top-left cell to the bottom-right cell.*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/10/04/ex1.png](https://assets.leetcode.com/uploads/2020/10/04/ex1.png)
+
+```
+Input: heights = [[1,2,2],[3,8,2],[5,3,5]]
+Output: 2
+Explanation: The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
+This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/10/04/ex2.png](https://assets.leetcode.com/uploads/2020/10/04/ex2.png)
+
+```
+Input: heights = [[1,2,3],[3,8,4],[5,3,5]]
+Output: 1
+Explanation: The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].
+```
+
+**Example 3:**
+
+![https://assets.leetcode.com/uploads/2020/10/04/ex3.png](https://assets.leetcode.com/uploads/2020/10/04/ex3.png)
+
+```
+Input: heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
+Output: 0
+Explanation: This route does not require any effort.
+```
+
+**Constraints:**
+
+- `rows == heights.length`
+- `columns == heights[i].length`
+- `1 <= rows, columns <= 100`
+- `1 <= heights[i][j] <= 10^6`
+
+## 题目大意
+
+你准备参加一场远足活动。给你一个二维 `rows x columns` 的地图 `heights` ，其中 `heights[row][col]` 表示格子 `(row, col)` 的高度。一开始你在最左上角的格子 `(0, 0)` ，且你希望去最右下角的格子 `(rows-1, columns-1)` （注意下标从 0 开始编号）。你每次可以往 上，下，左，右 四个方向之一移动，你想要找到耗费 体力 最小的一条路径。一条路径耗费的 体力值 是路径上相邻格子之间 高度差绝对值 的 最大值 决定的。请你返回从左上角走到右下角的最小 体力消耗值 。
+
+## 解题思路
+
+- 此题和第 778 题解题思路完全一致。在第 778 题中求的是最短连通时间。此题求的是连通路径下的最小体力值。都是求的最小值，只是 2 个值的意义不同罢了。
+- 按照第 778 题的思路，本题也有多种解法。第一种解法是 DFS + 二分。先将题目变换一个等价问法。题目要求找到最小体力消耗值，也相当于问是否存在一个体力消耗值 x，只要大于等于 x，一定能连通。利用二分搜索来找到这个临界值。体力消耗值是有序的，此处满足二分搜索的条件。题目给定柱子高度是 [1,10^6]，所以体力值一定在 [0,10^6-1] 这个区间内。判断是否取中值的条件是用 DFS 或者 BFS 搜索 (0,0) 点和 (N-1, N-1) 点之间是否连通。时间复杂度：O(mnlogC)，其中 m 和 n 分别是地图的行数和列数，C 是格子的最大高度。C 最大为 10^6，所以 logC 常数也很小。空间复杂度 O(mn)。
+- 第二种解法是并查集。将图中所有边按照权值从小到大进行排序，并依次加入并查集中。直到加入一条权值为 x 的边以后，左上角到右下角连通了。最小体力消耗值也就找到了。注意加入边的时候，只加入 `i-1` 和 `i` ，`j-1` 和 `j` 这 2 类相邻的边。因为最小体力消耗意味着不走回头路。上下左右四个方向到达一个节点，只可能从上边和左边走过来。从下边和右边走过来肯定是浪费体力了。时间复杂度：O(mnlog(mn))，其中 m 和 n 分别是地图的行数和列数，图中的边数为 O(mn)。空间复杂度 O(mn)，即为存储所有边以及并查集需要的空间。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+var dir = [4][2]int{
+	{0, 1},
+	{1, 0},
+	{0, -1},
+	{-1, 0},
+}
+
+// 解法一 DFS + 二分
+func minimumEffortPath(heights [][]int) int {
+	n, m := len(heights), len(heights[0])
+	visited := make([][]bool, n)
+	for i := range visited {
+		visited[i] = make([]bool, m)
+	}
+	low, high := 0, 1000000
+	for low < high {
+		threshold := low + (high-low)>>1
+		if !hasPath(heights, visited, 0, 0, threshold) {
+			low = threshold + 1
+		} else {
+			high = threshold
+		}
+		for i := range visited {
+			for j := range visited[i] {
+				visited[i][j] = false
+			}
+		}
+	}
+	return low
+}
+
+func hasPath(heights [][]int, visited [][]bool, i, j, threshold int) bool {
+	n, m := len(heights), len(heights[0])
+	if i == n-1 && j == m-1 {
+		return true
+	}
+	visited[i][j] = true
+	res := false
+	for _, d := range dir {
+		ni, nj := i+d[0], j+d[1]
+		if ni < 0 || ni >= n || nj < 0 || nj >= m || visited[ni][nj] || res {
+			continue
+		}
+		diff := abs(heights[i][j] - heights[ni][nj])
+		if diff <= threshold && hasPath(heights, visited, ni, nj, threshold) {
+			res = true
+		}
+	}
+	return res
+}
+
+func abs(a int) int {
+	if a < 0 {
+		a = -a
+	}
+	return a
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+func max(a, b int) int {
+	if a < b {
+		return b
+	}
+	return a
+}
+
+// 解法二 并查集
+func minimumEffortPath1(heights [][]int) int {
+	n, m, edges, uf := len(heights), len(heights[0]), []edge{}, template.UnionFind{}
+	uf.Init(n * m)
+	for i, row := range heights {
+		for j, h := range row {
+			id := i*m + j
+			if i > 0 {
+				edges = append(edges, edge{id - m, id, abs(h - heights[i-1][j])})
+			}
+			if j > 0 {
+				edges = append(edges, edge{id - 1, id, abs(h - heights[i][j-1])})
+			}
+		}
+	}
+	sort.Slice(edges, func(i, j int) bool { return edges[i].diff < edges[j].diff })
+	for _, e := range edges {
+		uf.Union(e.v, e.w)
+		if uf.Find(0) == uf.Find(n*m-1) {
+			return e.diff
+		}
+	}
+	return 0
+}
+
+type edge struct {
+	v, w, diff int
+}
+```
\ No newline at end of file
diff --git a/leetcode/1636.Sort-Array-by-Increasing-Frequency/1636. Sort Array by Increasing Frequency.go b/leetcode/1636.Sort-Array-by-Increasing-Frequency/1636. Sort Array by Increasing Frequency.go
new file mode 100644
index 000000000..a4240e99e
--- /dev/null
+++ b/leetcode/1636.Sort-Array-by-Increasing-Frequency/1636. Sort Array by Increasing Frequency.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+import "sort"
+
+func frequencySort(nums []int) []int {
+	freq := map[int]int{}
+	for _, v := range nums {
+		freq[v]++
+	}
+	sort.Slice(nums, func(i, j int) bool {
+		if freq[nums[i]] == freq[nums[j]] {
+			return nums[j] < nums[i]
+		}
+		return freq[nums[i]] < freq[nums[j]]
+	})
+	return nums
+}
diff --git a/leetcode/1636.Sort-Array-by-Increasing-Frequency/1636. Sort Array by Increasing Frequency_test.go b/leetcode/1636.Sort-Array-by-Increasing-Frequency/1636. Sort Array by Increasing Frequency_test.go
new file mode 100644
index 000000000..ac5b8a61e
--- /dev/null
+++ b/leetcode/1636.Sort-Array-by-Increasing-Frequency/1636. Sort Array by Increasing Frequency_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1636 struct {
+	para1636
+	ans1636
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1636 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1636 struct {
+	one []int
+}
+
+func Test_Problem1636(t *testing.T) {
+
+	qs := []question1636{
+
+		{
+			para1636{[]int{1, 1, 2, 2, 2, 3}},
+			ans1636{[]int{3, 1, 1, 2, 2, 2}},
+		},
+
+		{
+			para1636{[]int{2, 3, 1, 3, 2}},
+			ans1636{[]int{1, 3, 3, 2, 2}},
+		},
+
+		{
+			para1636{[]int{-1, 1, -6, 4, 5, -6, 1, 4, 1}},
+			ans1636{[]int{5, -1, 4, 4, -6, -6, 1, 1, 1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1636------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1636, q.para1636
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, frequencySort(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1636.Sort-Array-by-Increasing-Frequency/README.md b/leetcode/1636.Sort-Array-by-Increasing-Frequency/README.md
new file mode 100644
index 000000000..fb4e75d5e
--- /dev/null
+++ b/leetcode/1636.Sort-Array-by-Increasing-Frequency/README.md
@@ -0,0 +1,66 @@
+# [1636. Sort Array by Increasing Frequency](https://leetcode.com/problems/sort-array-by-increasing-frequency/)
+
+
+## 题目
+
+Given an array of integers `nums`, sort the array in **increasing** order based on the frequency of the values. If multiple values have the same frequency, sort them in **decreasing** order.
+
+Return the *sorted array*.
+
+**Example 1:**
+
+```
+Input: nums = [1,1,2,2,2,3]
+Output: [3,1,1,2,2,2]
+Explanation: '3' has a frequency of 1, '1' has a frequency of 2, and '2' has a frequency of 3.
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,3,1,3,2]
+Output: [1,3,3,2,2]
+Explanation: '2' and '3' both have a frequency of 2, so they are sorted in decreasing order.
+```
+
+**Example 3:**
+
+```
+Input: nums = [-1,1,-6,4,5,-6,1,4,1]
+Output: [5,-1,4,4,-6,-6,1,1,1]
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `100 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个整数数组 `nums` ，请你将数组按照每个值的频率 **升序** 排序。如果有多个值的频率相同，请你按照数值本身将它们 **降序** 排序。请你返回排序后的数组。
+
+## 解题思路
+
+- 简单题。先统计每个值的频率，然后按照频率从小到大排序，相同频率的按照值的大小，从大到小排序。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func frequencySort(nums []int) []int {
+	freq := map[int]int{}
+	for _, v := range nums {
+		freq[v]++
+	}
+	sort.Slice(nums, func(i, j int) bool {
+		if freq[nums[i]] == freq[nums[j]] {
+			return nums[j] < nums[i]
+		}
+		return freq[nums[i]] < freq[nums[j]]
+	})
+	return nums
+}
+```
\ No newline at end of file
diff --git a/leetcode/1640.Check-Array-Formation-Through-Concatenation/1640. Check Array Formation Through Concatenation.go b/leetcode/1640.Check-Array-Formation-Through-Concatenation/1640. Check Array Formation Through Concatenation.go
new file mode 100644
index 000000000..384022e24
--- /dev/null
+++ b/leetcode/1640.Check-Array-Formation-Through-Concatenation/1640. Check Array Formation Through Concatenation.go	
@@ -0,0 +1,26 @@
+package leetcode
+
+func canFormArray(arr []int, pieces [][]int) bool {
+	arrMap := map[int]int{}
+	for i, v := range arr {
+		arrMap[v] = i
+	}
+	for i := 0; i < len(pieces); i++ {
+		order := -1
+		for j := 0; j < len(pieces[i]); j++ {
+			if _, ok := arrMap[pieces[i][j]]; !ok {
+				return false
+			}
+			if order == -1 {
+				order = arrMap[pieces[i][j]]
+			} else {
+				if arrMap[pieces[i][j]] == order+1 {
+					order = arrMap[pieces[i][j]]
+				} else {
+					return false
+				}
+			}
+		}
+	}
+	return true
+}
diff --git a/leetcode/1640.Check-Array-Formation-Through-Concatenation/1640. Check Array Formation Through Concatenation_test.go b/leetcode/1640.Check-Array-Formation-Through-Concatenation/1640. Check Array Formation Through Concatenation_test.go
new file mode 100644
index 000000000..39b3a7170
--- /dev/null
+++ b/leetcode/1640.Check-Array-Formation-Through-Concatenation/1640. Check Array Formation Through Concatenation_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1640 struct {
+	para1640
+	ans1640
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1640 struct {
+	arr    []int
+	pieces [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1640 struct {
+	one bool
+}
+
+func Test_Problem1640(t *testing.T) {
+
+	qs := []question1640{
+
+		{
+			para1640{[]int{85}, [][]int{{85}}},
+			ans1640{true},
+		},
+
+		{
+			para1640{[]int{15, 88}, [][]int{{88}, {15}}},
+			ans1640{true},
+		},
+
+		{
+			para1640{[]int{49, 18, 16}, [][]int{{16, 18, 49}}},
+			ans1640{false},
+		},
+
+		{
+			para1640{[]int{91, 4, 64, 78}, [][]int{{78}, {4, 64}, {91}}},
+			ans1640{true},
+		},
+
+		{
+			para1640{[]int{1, 3, 5, 7}, [][]int{{2, 4, 6, 8}}},
+			ans1640{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1640------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1640, q.para1640
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, canFormArray(p.arr, p.pieces))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1640.Check-Array-Formation-Through-Concatenation/README.md b/leetcode/1640.Check-Array-Formation-Through-Concatenation/README.md
new file mode 100644
index 000000000..864981778
--- /dev/null
+++ b/leetcode/1640.Check-Array-Formation-Through-Concatenation/README.md
@@ -0,0 +1,95 @@
+# [1640. Check Array Formation Through Concatenation](https://leetcode.com/problems/check-array-formation-through-concatenation/)
+
+
+## 题目
+
+You are given an array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
+
+Return `true` *if it is possible to form the array* `arr` *from* `pieces`. Otherwise, return `false`.
+
+**Example 1:**
+
+```
+Input: arr = [85], pieces = [[85]]
+Output: true
+```
+
+**Example 2:**
+
+```
+Input: arr = [15,88], pieces = [[88],[15]]
+Output: true
+Explanation: Concatenate [15] then [88]
+```
+
+**Example 3:**
+
+```
+Input: arr = [49,18,16], pieces = [[16,18,49]]
+Output: false
+Explanation: Even though the numbers match, we cannot reorder pieces[0].
+```
+
+**Example 4:**
+
+```
+Input: arr = [91,4,64,78], pieces = [[78],[4,64],[91]]
+Output: true
+Explanation: Concatenate [91] then [4,64] then [78]
+```
+
+**Example 5:**
+
+```
+Input: arr = [1,3,5,7], pieces = [[2,4,6,8]]
+Output: false
+
+```
+
+**Constraints:**
+
+- `1 <= pieces.length <= arr.length <= 100`
+- `sum(pieces[i].length) == arr.length`
+- `1 <= pieces[i].length <= arr.length`
+- `1 <= arr[i], pieces[i][j] <= 100`
+- The integers in `arr` are **distinct**.
+- The integers in `pieces` are **distinct** (i.e., If we flatten pieces in a 1D array, all the integers in this array are distinct).
+
+## 题目大意
+
+给你一个整数数组 arr ，数组中的每个整数 互不相同 。另有一个由整数数组构成的数组 pieces，其中的整数也 互不相同 。请你以 任意顺序 连接 pieces 中的数组以形成 arr 。但是，不允许 对每个数组 pieces[i] 中的整数重新排序。如果可以连接 pieces 中的数组形成 arr ，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 简单题。题目保证了 arr 中的元素唯一，所以可以用 map 把每个元素的 index 存起来，方便查找。遍历 pieces 数组，在每个一维数组中判断元素顺序是否和原 arr 元素相对顺序一致。这个时候就用 map 查找，如果顺序是一一相连的，那么就是正确的。有一个顺序不是一一相连，或者出现了 arr 不存在的元素，都返回 false。
+
+## 代码
+
+```go
+package leetcode
+
+func canFormArray(arr []int, pieces [][]int) bool {
+	arrMap := map[int]int{}
+	for i, v := range arr {
+		arrMap[v] = i
+	}
+	for i := 0; i < len(pieces); i++ {
+		order := -1
+		for j := 0; j < len(pieces[i]); j++ {
+			if _, ok := arrMap[pieces[i][j]]; !ok {
+				return false
+			}
+			if order == -1 {
+				order = arrMap[pieces[i][j]]
+			} else {
+				if arrMap[pieces[i][j]] == order+1 {
+					order = arrMap[pieces[i][j]]
+				} else {
+					return false
+				}
+			}
+		}
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/1641.Count-Sorted-Vowel-Strings/1641. Count Sorted Vowel Strings.go b/leetcode/1641.Count-Sorted-Vowel-Strings/1641. Count Sorted Vowel Strings.go
new file mode 100644
index 000000000..be2ec59d4
--- /dev/null
+++ b/leetcode/1641.Count-Sorted-Vowel-Strings/1641. Count Sorted Vowel Strings.go	
@@ -0,0 +1,35 @@
+package leetcode
+
+// 解法一 打表
+func countVowelStrings(n int) int {
+	res := []int{1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315, 8855, 10626, 12650, 14950, 17550, 20475, 23751, 27405, 31465, 35960, 40920, 46376, 52360, 58905, 66045, 73815, 82251, 91390, 101270, 111930, 123410, 135751, 148995, 163185, 178365, 194580, 211876, 230300, 249900, 270725, 292825, 316251}
+	return res[n]
+}
+
+func makeTable() []int {
+	res, array := 0, []int{}
+	for i := 0; i < 51; i++ {
+		countVowelStringsDFS(i, 0, []string{}, []string{"a", "e", "i", "o", "u"}, &res)
+		array = append(array, res)
+		res = 0
+	}
+	return array
+}
+
+func countVowelStringsDFS(n, index int, cur []string, vowels []string, res *int) {
+	vowels = vowels[index:]
+	if len(cur) == n {
+		(*res)++
+		return
+	}
+	for i := 0; i < len(vowels); i++ {
+		cur = append(cur, vowels[i])
+		countVowelStringsDFS(n, i, cur, vowels, res)
+		cur = cur[:len(cur)-1]
+	}
+}
+
+// 解法二 数学方法 —— 组合
+func countVowelStrings1(n int) int {
+	return (n + 1) * (n + 2) * (n + 3) * (n + 4) / 24
+}
diff --git a/leetcode/1641.Count-Sorted-Vowel-Strings/1641. Count Sorted Vowel Strings_test.go b/leetcode/1641.Count-Sorted-Vowel-Strings/1641. Count Sorted Vowel Strings_test.go
new file mode 100644
index 000000000..081ffbcd0
--- /dev/null
+++ b/leetcode/1641.Count-Sorted-Vowel-Strings/1641. Count Sorted Vowel Strings_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1641 struct {
+	para1641
+	ans1641
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1641 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1641 struct {
+	one int
+}
+
+func Test_Problem1641(t *testing.T) {
+
+	qs := []question1641{
+
+		{
+			para1641{1},
+			ans1641{5},
+		},
+
+		{
+			para1641{2},
+			ans1641{15},
+		},
+
+		{
+			para1641{33},
+			ans1641{66045},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1641------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1641, q.para1641
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, countVowelStrings(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1641.Count-Sorted-Vowel-Strings/README.md b/leetcode/1641.Count-Sorted-Vowel-Strings/README.md
new file mode 100644
index 000000000..8da3acfde
--- /dev/null
+++ b/leetcode/1641.Count-Sorted-Vowel-Strings/README.md
@@ -0,0 +1,90 @@
+# [1641. Count Sorted Vowel Strings](https://leetcode.com/problems/count-sorted-vowel-strings/)
+
+
+## 题目
+
+Given an integer `n`, return *the number of strings of length* `n` *that consist only of vowels (*`a`*,* `e`*,* `i`*,* `o`*,* `u`*) and are **lexicographically sorted**.*
+
+A string `s` is **lexicographically sorted** if for all valid `i`, `s[i]` is the same as or comes before `s[i+1]` in the alphabet.
+
+**Example 1:**
+
+```
+Input: n = 1
+Output: 5
+Explanation: The 5 sorted strings that consist of vowels only are ["a","e","i","o","u"].
+```
+
+**Example 2:**
+
+```
+Input: n = 2
+Output: 15
+Explanation: The 15 sorted strings that consist of vowels only are
+["aa","ae","ai","ao","au","ee","ei","eo","eu","ii","io","iu","oo","ou","uu"].
+Note that "ea" is not a valid string since 'e' comes after 'a' in the alphabet.
+
+```
+
+**Example 3:**
+
+```
+Input: n = 33
+Output: 66045
+
+```
+
+**Constraints:**
+
+- `1 <= n <= 50`
+
+## 题目大意
+
+给你一个整数 n，请返回长度为 n 、仅由元音 (a, e, i, o, u) 组成且按 字典序排列 的字符串数量。
+
+字符串 s 按 字典序排列 需要满足：对于所有有效的 i，s[i] 在字母表中的位置总是与 s[i+1] 相同或在 s[i+1] 之前。
+
+## 解题思路
+
+- 题目给的数据量并不大，第一个思路是利用 DFS 遍历打表法。时间复杂度 O(1)，空间复杂度 O(1)。
+- 第二个思路是利用数学中的组合公式计算结果。题目等价于假设现在有 n 个字母，要求取 4 次球（可以选择不取）将字母分为 5 堆，问有几种取法。确定了取法以后，`a`，`e`，`i`，`o`，`u`，每个字母的个数就确定了，据题意要求按照字母序排序，那么最终字符串也就确定了。现在关注解决这个组合问题就可以了。把问题再转化一次，等价于，有 n+4 个字母，取 4 次，问有几种取法。+4 代表 4 个空操作，取走它们意味着不取。根据组合的数学定义，答案为 C(n+4,4)。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 打表
+func countVowelStrings(n int) int {
+	res := []int{1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315, 8855, 10626, 12650, 14950, 17550, 20475, 23751, 27405, 31465, 35960, 40920, 46376, 52360, 58905, 66045, 73815, 82251, 91390, 101270, 111930, 123410, 135751, 148995, 163185, 178365, 194580, 211876, 230300, 249900, 270725, 292825, 316251}
+	return res[n]
+}
+
+func makeTable() []int {
+	res, array := 0, []int{}
+	for i := 0; i < 51; i++ {
+		countVowelStringsDFS(i, 0, []string{}, []string{"a", "e", "i", "o", "u"}, &res)
+		array = append(array, res)
+		res = 0
+	}
+	return array
+}
+
+func countVowelStringsDFS(n, index int, cur []string, vowels []string, res *int) {
+	vowels = vowels[index:]
+	if len(cur) == n {
+		(*res)++
+		return
+	}
+	for i := 0; i < len(vowels); i++ {
+		cur = append(cur, vowels[i])
+		countVowelStringsDFS(n, i, cur, vowels, res)
+		cur = cur[:len(cur)-1]
+	}
+}
+
+// 解法二 数学方法 —— 组合
+func countVowelStrings1(n int) int {
+	return (n + 1) * (n + 2) * (n + 3) * (n + 4) / 24
+}
+```
\ No newline at end of file
diff --git a/leetcode/1642.Furthest-Building-You-Can-Reach/1642. Furthest Building You Can Reach.go b/leetcode/1642.Furthest-Building-You-Can-Reach/1642. Furthest Building You Can Reach.go
new file mode 100644
index 000000000..c2adc1d97
--- /dev/null
+++ b/leetcode/1642.Furthest-Building-You-Can-Reach/1642. Furthest Building You Can Reach.go	
@@ -0,0 +1,40 @@
+package leetcode
+
+import (
+	"container/heap"
+)
+
+func furthestBuilding(heights []int, bricks int, ladder int) int {
+	usedLadder := &heightDiffPQ{}
+	for i := 1; i < len(heights); i++ {
+		needbricks := heights[i] - heights[i-1]
+		if needbricks < 0 {
+			continue
+		}
+		if ladder > 0 {
+			heap.Push(usedLadder, needbricks)
+			ladder--
+		} else {
+			if len(*usedLadder) > 0 && needbricks > (*usedLadder)[0] {
+				needbricks, (*usedLadder)[0] = (*usedLadder)[0], needbricks
+				heap.Fix(usedLadder, 0)
+			}
+			if bricks -= needbricks; bricks < 0 {
+				return i - 1
+			}
+		}
+	}
+	return len(heights) - 1
+}
+
+type heightDiffPQ []int
+
+func (pq heightDiffPQ) Len() int            { return len(pq) }
+func (pq heightDiffPQ) Less(i, j int) bool  { return pq[i] < pq[j] }
+func (pq heightDiffPQ) Swap(i, j int)       { pq[i], pq[j] = pq[j], pq[i] }
+func (pq *heightDiffPQ) Push(x interface{}) { *pq = append(*pq, x.(int)) }
+func (pq *heightDiffPQ) Pop() interface{} {
+	x := (*pq)[len(*pq)-1]
+	*pq = (*pq)[:len(*pq)-1]
+	return x
+}
diff --git a/leetcode/1642.Furthest-Building-You-Can-Reach/1642. Furthest Building You Can Reach_test.go b/leetcode/1642.Furthest-Building-You-Can-Reach/1642. Furthest Building You Can Reach_test.go
new file mode 100644
index 000000000..ee04d85aa
--- /dev/null
+++ b/leetcode/1642.Furthest-Building-You-Can-Reach/1642. Furthest Building You Can Reach_test.go	
@@ -0,0 +1,59 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1642 struct {
+	para1642
+	ans1642
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1642 struct {
+	heights []int
+	bricks  int
+	ladders int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1642 struct {
+	one int
+}
+
+func Test_Problem1642(t *testing.T) {
+
+	qs := []question1642{
+
+		{
+			para1642{[]int{1, 5, 1, 2, 3, 4, 10000}, 4, 1},
+			ans1642{5},
+		},
+
+		{
+			para1642{[]int{4, 2, 7, 6, 9, 14, 12}, 5, 1},
+			ans1642{4},
+		},
+
+		{
+			para1642{[]int{4, 12, 2, 7, 3, 18, 20, 3, 19}, 10, 2},
+			ans1642{7},
+		},
+
+		{
+			para1642{[]int{14, 3, 19, 3}, 17, 0},
+			ans1642{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1642------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1642, q.para1642
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, furthestBuilding(p.heights, p.bricks, p.ladders))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1642.Furthest-Building-You-Can-Reach/README.md b/leetcode/1642.Furthest-Building-You-Can-Reach/README.md
new file mode 100644
index 000000000..e111e26e7
--- /dev/null
+++ b/leetcode/1642.Furthest-Building-You-Can-Reach/README.md
@@ -0,0 +1,113 @@
+# [1642. Furthest Building You Can Reach](https://leetcode.com/problems/furthest-building-you-can-reach/)
+
+
+## 题目
+
+You are given an integer array `heights` representing the heights of buildings, some `bricks`, and some `ladders`.
+
+You start your journey from building `0` and move to the next building by possibly using bricks or ladders.
+
+While moving from building `i` to building `i+1` (**0-indexed**),
+
+- If the current building's height is **greater than or equal** to the next building's height, you do **not** need a ladder or bricks.
+- If the current building's height is **less than** the next building's height, you can either use **one ladder** or `(h[i+1] - h[i])` **bricks**.
+
+*Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/10/27/q4.gif](https://assets.leetcode.com/uploads/2020/10/27/q4.gif)
+
+```
+Input: heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1
+Output: 4
+Explanation: Starting at building 0, you can follow these steps:
+- Go to building 1 without using ladders nor bricks since 4 >= 2.
+- Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 < 7.
+- Go to building 3 without using ladders nor bricks since 7 >= 6.
+- Go to building 4 using your only ladder. You must use either bricks or ladders because 6 < 9.
+It is impossible to go beyond building 4 because you do not have any more bricks or ladders.
+
+```
+
+**Example 2:**
+
+```
+Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
+Output: 7
+
+```
+
+**Example 3:**
+
+```
+Input: heights = [14,3,19,3], bricks = 17, ladders = 0
+Output: 3
+
+```
+
+**Constraints:**
+
+- `1 <= heights.length <= 10^5`
+- `1 <= heights[i] <= 10^6`
+- `0 <= bricks <= 10^9`
+- `0 <= ladders <= heights.length`
+
+## 题目大意
+
+给你一个整数数组 heights ，表示建筑物的高度。另有一些砖块 bricks 和梯子 ladders 。你从建筑物 0 开始旅程，不断向后面的建筑物移动，期间可能会用到砖块或梯子。当从建筑物 i 移动到建筑物 i+1（下标 从 0 开始 ）时：
+
+- 如果当前建筑物的高度 大于或等于 下一建筑物的高度，则不需要梯子或砖块。
+- 如果当前建筑的高度 小于 下一个建筑的高度，您可以使用 一架梯子 或 (h[i+1] - h[i]) 个砖块
+
+如果以最佳方式使用给定的梯子和砖块，返回你可以到达的最远建筑物的下标（下标 从 0 开始 ）。
+
+## 解题思路
+
+- 这一题可能会想到贪心算法。梯子很厉害，可以无限长，所以梯子用来跨越最高的楼。遇到非最高的距离差，先用砖头。这样贪心的话不正确。例如，[1, 5, 1, 2, 3, 4, 10000] 这组数据，梯子有 1 个，4 块砖头。最大的差距在 10000 和 4 之间，贪心选择在此处用梯子。但是砖头不足以让我们走到最后两栋楼。贪心得到的结果是 3，正确的结果是 5，先用梯子，再用砖头走过 3，4，5 号楼。
+- 上面的贪心解法错误在于没有“动态”的贪心，使用梯子应该选择能爬过楼里面最高的 2 个。于是顺理成章的想到了优先队列。维护一个长度为梯子个数的最小堆，当队列中元素超过梯子个数，便将队首最小值出队，出队的这个楼与楼的差距用砖头填补。所有砖头用完了，即是可以到达的最远楼号。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+)
+
+func furthestBuilding(heights []int, bricks int, ladder int) int {
+	usedLadder := &heightDiffPQ{}
+	for i := 1; i < len(heights); i++ {
+		needbricks := heights[i] - heights[i-1]
+		if needbricks < 0 {
+			continue
+		}
+		if ladder > 0 {
+			heap.Push(usedLadder, needbricks)
+			ladder--
+		} else {
+			if len(*usedLadder) > 0 && needbricks > (*usedLadder)[0] {
+				needbricks, (*usedLadder)[0] = (*usedLadder)[0], needbricks
+				heap.Fix(usedLadder, 0)
+			}
+			if bricks -= needbricks; bricks < 0 {
+				return i - 1
+			}
+		}
+	}
+	return len(heights) - 1
+}
+
+type heightDiffPQ []int
+
+func (pq heightDiffPQ) Len() int            { return len(pq) }
+func (pq heightDiffPQ) Less(i, j int) bool  { return pq[i] < pq[j] }
+func (pq heightDiffPQ) Swap(i, j int)       { pq[i], pq[j] = pq[j], pq[i] }
+func (pq *heightDiffPQ) Push(x interface{}) { *pq = append(*pq, x.(int)) }
+func (pq *heightDiffPQ) Pop() interface{} {
+	x := (*pq)[len(*pq)-1]
+	*pq = (*pq)[:len(*pq)-1]
+	return x
+}
+```
\ No newline at end of file
diff --git a/leetcode/1646.Get-Maximum-in-Generated-Array/1646. Get Maximum in Generated Array.go b/leetcode/1646.Get-Maximum-in-Generated-Array/1646. Get Maximum in Generated Array.go
new file mode 100644
index 000000000..246a1e1fa
--- /dev/null
+++ b/leetcode/1646.Get-Maximum-in-Generated-Array/1646. Get Maximum in Generated Array.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+func getMaximumGenerated(n int) int {
+	if n == 0 {
+		return 0
+	}
+	nums, max := make([]int, n+1), 0
+	nums[0], nums[1] = 0, 1
+	for i := 0; i <= n; i++ {
+		if nums[i] > max {
+			max = nums[i]
+		}
+		if 2*i >= 2 && 2*i <= n {
+			nums[2*i] = nums[i]
+		}
+		if 2*i+1 >= 2 && 2*i+1 <= n {
+			nums[2*i+1] = nums[i] + nums[i+1]
+		}
+	}
+	return max
+}
diff --git a/leetcode/1646.Get-Maximum-in-Generated-Array/1646. Get Maximum in Generated Array_test.go b/leetcode/1646.Get-Maximum-in-Generated-Array/1646. Get Maximum in Generated Array_test.go
new file mode 100644
index 000000000..384721906
--- /dev/null
+++ b/leetcode/1646.Get-Maximum-in-Generated-Array/1646. Get Maximum in Generated Array_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1646 struct {
+	para1646
+	ans1646
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1646 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1646 struct {
+	one int
+}
+
+func Test_Problem1646(t *testing.T) {
+
+	qs := []question1646{
+
+		{
+			para1646{7},
+			ans1646{3},
+		},
+
+		{
+			para1646{2},
+			ans1646{1},
+		},
+
+		{
+			para1646{3},
+			ans1646{2},
+		},
+
+		{
+			para1646{0},
+			ans1646{0},
+		},
+
+		{
+			para1646{1},
+			ans1646{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1646------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1646, q.para1646
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, getMaximumGenerated(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1646.Get-Maximum-in-Generated-Array/README.md b/leetcode/1646.Get-Maximum-in-Generated-Array/README.md
new file mode 100644
index 000000000..a703bba78
--- /dev/null
+++ b/leetcode/1646.Get-Maximum-in-Generated-Array/README.md
@@ -0,0 +1,96 @@
+# [1646. Get Maximum in Generated Array](https://leetcode.com/problems/get-maximum-in-generated-array/)
+
+
+## 题目
+
+You are given an integer `n`. An array `nums` of length `n + 1` is generated in the following way:
+
+- `nums[0] = 0`
+- `nums[1] = 1`
+- `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
+- `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
+
+Return *****the **maximum** integer in the array* `nums`.
+
+**Example 1:**
+
+```
+Input: n = 7
+Output: 3
+Explanation: According to the given rules:
+  nums[0] = 0
+  nums[1] = 1
+  nums[(1 * 2) = 2] = nums[1] = 1
+  nums[(1 * 2) + 1 = 3] = nums[1] + nums[2] = 1 + 1 = 2
+  nums[(2 * 2) = 4] = nums[2] = 1
+  nums[(2 * 2) + 1 = 5] = nums[2] + nums[3] = 1 + 2 = 3
+  nums[(3 * 2) = 6] = nums[3] = 2
+  nums[(3 * 2) + 1 = 7] = nums[3] + nums[4] = 2 + 1 = 3
+Hence, nums = [0,1,1,2,1,3,2,3], and the maximum is 3.
+
+```
+
+**Example 2:**
+
+```
+Input: n = 2
+Output: 1
+Explanation: According to the given rules, the maximum between nums[0], nums[1], and nums[2] is 1.
+
+```
+
+**Example 3:**
+
+```
+Input: n = 3
+Output: 2
+Explanation: According to the given rules, the maximum between nums[0], nums[1], nums[2], and nums[3] is 2.
+
+```
+
+**Constraints:**
+
+- `0 <= n <= 100`
+
+## 题目大意
+
+给你一个整数 n 。按下述规则生成一个长度为 n + 1 的数组 nums ：
+
+- nums[0] = 0
+- nums[1] = 1
+- 当 2 <= 2 * i <= n 时，nums[2 * i] = nums[i]
+- 当 2 <= 2 * i + 1 <= n 时，nums[2 * i + 1] = nums[i] + nums[i + 1]
+
+返回生成数组 nums 中的 最大值。
+
+## 解题思路
+
+- 给出一个 n + 1 的数组，并按照生成规则生成这个数组，求出这个数组中的最大值。
+- 简单题，按照题意生成数组，边生成边记录和更新最大值即可。
+- 注意边界条件，当 n 为 0 的时候，数组里面只有一个元素 0 。
+
+## 代码
+
+```go
+package leetcode
+
+func getMaximumGenerated(n int) int {
+	if n == 0 {
+		return 0
+	}
+	nums, max := make([]int, n+1), 0
+	nums[0], nums[1] = 0, 1
+	for i := 0; i <= n; i++ {
+		if nums[i] > max {
+			max = nums[i]
+		}
+		if 2*i >= 2 && 2*i <= n {
+			nums[2*i] = nums[i]
+		}
+		if 2*i+1 >= 2 && 2*i+1 <= n {
+			nums[2*i+1] = nums[i] + nums[i+1]
+		}
+	}
+	return max
+}
+```
\ No newline at end of file
diff --git a/leetcode/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique/1647. Minimum Deletions to Make Character Frequencies Unique.go b/leetcode/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique/1647. Minimum Deletions to Make Character Frequencies Unique.go
new file mode 100644
index 000000000..302196609
--- /dev/null
+++ b/leetcode/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique/1647. Minimum Deletions to Make Character Frequencies Unique.go	
@@ -0,0 +1,22 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+func minDeletions(s string) int {
+	frequency, res := make([]int, 26), 0
+	for i := 0; i < len(s); i++ {
+		frequency[s[i]-'a']++
+	}
+	sort.Sort(sort.Reverse(sort.IntSlice(frequency)))
+	for i := 1; i <= 25; i++ {
+		if frequency[i] == frequency[i-1] && frequency[i] != 0 {
+			res++
+			frequency[i]--
+			sort.Sort(sort.Reverse(sort.IntSlice(frequency)))
+			i--
+		}
+	}
+	return res
+}
diff --git a/leetcode/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique/1647. Minimum Deletions to Make Character Frequencies Unique_test.go b/leetcode/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique/1647. Minimum Deletions to Make Character Frequencies Unique_test.go
new file mode 100644
index 000000000..3bdc27352
--- /dev/null
+++ b/leetcode/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique/1647. Minimum Deletions to Make Character Frequencies Unique_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1647 struct {
+	para1647
+	ans1647
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1647 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1647 struct {
+	one int
+}
+
+func Test_Problem1647(t *testing.T) {
+
+	qs := []question1647{
+
+		{
+			para1647{"aab"},
+			ans1647{0},
+		},
+
+		{
+			para1647{"aaabbbcc"},
+			ans1647{2},
+		},
+
+		{
+			para1647{"ceabaacb"},
+			ans1647{2},
+		},
+
+		{
+			para1647{""},
+			ans1647{0},
+		},
+
+		{
+			para1647{"abcabc"},
+			ans1647{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1647------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1647, q.para1647
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, minDeletions(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique/README.md b/leetcode/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique/README.md
new file mode 100644
index 000000000..e6b2fff80
--- /dev/null
+++ b/leetcode/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique/README.md
@@ -0,0 +1,89 @@
+# [1647. Minimum Deletions to Make Character Frequencies Unique](https://leetcode.com/problems/minimum-deletions-to-make-character-frequencies-unique/)
+
+
+## 题目
+
+A string `s` is called **good** if there are no two different characters in `s` that have the same **frequency**.
+
+Given a string `s`, return *the **minimum** number of characters you need to delete to make* `s` ***good**.*
+
+The **frequency** of a character in a string is the number of times it appears in the string. For example, in the string `"aab"`, the **frequency** of `'a'` is `2`, while the **frequency** of `'b'` is `1`.
+
+**Example 1:**
+
+```
+Input: s = "aab"
+Output: 0
+Explanation: s is already good.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "aaabbbcc"
+Output: 2
+Explanation: You can delete two 'b's resulting in the good string "aaabcc".
+Another way it to delete one 'b' and one 'c' resulting in the good string "aaabbc".
+```
+
+**Example 3:**
+
+```
+Input: s = "ceabaacb"
+Output: 2
+Explanation: You can delete both 'c's resulting in the good string "eabaab".
+Note that we only care about characters that are still in the string at the end (i.e. frequency of 0 is ignored).
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 105`
+- `s` contains only lowercase English letters.
+
+## 题目大意
+
+如果字符串 s 中 不存在 两个不同字符 频次 相同的情况，就称 s 是 优质字符串 。
+
+给你一个字符串 s，返回使 s 成为优质字符串需要删除的最小字符数。
+
+字符串中字符的 频次 是该字符在字符串中的出现次数。例如，在字符串 "aab" 中，'a' 的频次是 2，而 'b' 的频次是 1 。
+
+**提示：**
+
+- `1 <= s.length <= 105`
+- `s` 仅含小写英文字母
+
+## 解题思路
+
+- 给出一个字符串 s，要求输出使 s 变成“优质字符串”需要删除的最小字符数。“优质字符串”的定义是：字符串 s 中不存在频次相同的两个不同字符。
+- 首先将 26 个字母在字符串中的频次分别统计出来，然后把频次从大到小排列，从频次大的开始，依次调整：例如，假设前一个和后一个频次相等，就把前一个字符删除一个，频次减一，再次排序，如果频次还相等，继续调整，如果频次不同了，游标往后移，继续调整后面的频次。直到所有的频次都不同了，就可以输出最终结果了。
+- 这里需要注意频次为 0 的情况，即字母都被删光了。频次为 0 以后，就不需要再比较了。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func minDeletions(s string) int {
+	frequency, res := make([]int, 26), 0
+	for i := 0; i < len(s); i++ {
+		frequency[s[i]-'a']++
+	}
+	sort.Sort(sort.Reverse(sort.IntSlice(frequency)))
+	for i := 1; i <= 25; i++ {
+		if frequency[i] == frequency[i-1] && frequency[i] != 0 {
+			res++
+			frequency[i]--
+			sort.Sort(sort.Reverse(sort.IntSlice(frequency)))
+			i--
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1648.Sell-Diminishing-Valued-Colored-Balls/1648. Sell Diminishing-Valued Colored Balls.go b/leetcode/1648.Sell-Diminishing-Valued-Colored-Balls/1648. Sell Diminishing-Valued Colored Balls.go
new file mode 100644
index 000000000..677a3699f
--- /dev/null
+++ b/leetcode/1648.Sell-Diminishing-Valued-Colored-Balls/1648. Sell Diminishing-Valued Colored Balls.go	
@@ -0,0 +1,102 @@
+package leetcode
+
+import (
+	"container/heap"
+)
+
+// 解法一 贪心 + 二分搜索
+func maxProfit(inventory []int, orders int) int {
+	maxItem, thresholdValue, count, res, mod := 0, -1, 0, 0, 1000000007
+	for i := 0; i < len(inventory); i++ {
+		if inventory[i] > maxItem {
+			maxItem = inventory[i]
+		}
+	}
+	low, high := 0, maxItem
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		for i := 0; i < len(inventory); i++ {
+			count += max(inventory[i]-mid, 0)
+		}
+		if count <= orders {
+			thresholdValue = mid
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+		count = 0
+	}
+	count = 0
+	for i := 0; i < len(inventory); i++ {
+		count += max(inventory[i]-thresholdValue, 0)
+	}
+	count = orders - count
+	for i := 0; i < len(inventory); i++ {
+		if inventory[i] >= thresholdValue {
+			if count > 0 {
+				res += (thresholdValue + inventory[i]) * (inventory[i] - thresholdValue + 1) / 2
+				count--
+			} else {
+				res += (thresholdValue + 1 + inventory[i]) * (inventory[i] - thresholdValue) / 2
+			}
+		}
+	}
+	return res % mod
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法二 优先队列，超时！
+func maxProfit_(inventory []int, orders int) int {
+	res, mod := 0, 1000000007
+	q := PriorityQueue{}
+	for i := 0; i < len(inventory); i++ {
+		heap.Push(&q, &Item{count: inventory[i]})
+	}
+	for ; orders > 0; orders-- {
+		item := heap.Pop(&q).(*Item)
+		res = (res + item.count) % mod
+		heap.Push(&q, &Item{count: item.count - 1})
+	}
+	return res
+}
+
+// Item define
+type Item struct {
+	count int
+}
+
+// A PriorityQueue implements heap.Interface and holds Items.
+type PriorityQueue []*Item
+
+func (pq PriorityQueue) Len() int {
+	return len(pq)
+}
+
+func (pq PriorityQueue) Less(i, j int) bool {
+	// 注意：因为golang中的heap是按最小堆组织的，所以count越大，Less()越小，越靠近堆顶.
+	return pq[i].count > pq[j].count
+}
+
+func (pq PriorityQueue) Swap(i, j int) {
+	pq[i], pq[j] = pq[j], pq[i]
+}
+
+// Push define
+func (pq *PriorityQueue) Push(x interface{}) {
+	item := x.(*Item)
+	*pq = append(*pq, item)
+}
+
+// Pop define
+func (pq *PriorityQueue) Pop() interface{} {
+	n := len(*pq)
+	item := (*pq)[n-1]
+	*pq = (*pq)[:n-1]
+	return item
+}
diff --git a/leetcode/1648.Sell-Diminishing-Valued-Colored-Balls/1648. Sell Diminishing-Valued Colored Balls_test.go b/leetcode/1648.Sell-Diminishing-Valued-Colored-Balls/1648. Sell Diminishing-Valued Colored Balls_test.go
new file mode 100644
index 000000000..b81ce0e6e
--- /dev/null
+++ b/leetcode/1648.Sell-Diminishing-Valued-Colored-Balls/1648. Sell Diminishing-Valued Colored Balls_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1648 struct {
+	para1648
+	ans1648
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1648 struct {
+	inventory []int
+	orders    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1648 struct {
+	one int
+}
+
+func Test_Problem1648(t *testing.T) {
+
+	qs := []question1648{
+
+		{
+			para1648{[]int{2, 3, 3, 4, 5}, 4},
+			ans1648{16},
+		},
+
+		{
+			para1648{[]int{2, 5}, 4},
+			ans1648{14},
+		},
+
+		{
+			para1648{[]int{3, 5}, 6},
+			ans1648{19},
+		},
+
+		{
+			para1648{[]int{2, 8, 4, 10, 6}, 20},
+			ans1648{110},
+		},
+
+		{
+			para1648{[]int{1000000000}, 1000000000},
+			ans1648{21},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1648------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1648, q.para1648
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, maxProfit(p.inventory, p.orders))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1648.Sell-Diminishing-Valued-Colored-Balls/README.md b/leetcode/1648.Sell-Diminishing-Valued-Colored-Balls/README.md
new file mode 100644
index 000000000..bddd91b31
--- /dev/null
+++ b/leetcode/1648.Sell-Diminishing-Valued-Colored-Balls/README.md
@@ -0,0 +1,131 @@
+# [1648. Sell Diminishing-Valued Colored Balls](https://leetcode.com/problems/sell-diminishing-valued-colored-balls/)
+
+
+## 题目
+
+You have an `inventory` of different colored balls, and there is a customer that wants `orders` balls of **any** color.
+
+The customer weirdly values the colored balls. Each colored ball's value is the number of balls **of that color** you currently have in your `inventory`. For example, if you own `6` yellow balls, the customer would pay `6` for the first yellow ball. After the transaction, there are only `5` yellow balls left, so the next yellow ball is then valued at `5` (i.e., the value of the balls decreases as you sell more to the customer).
+
+You are given an integer array, `inventory`, where `inventory[i]` represents the number of balls of the `ith` color that you initially own. You are also given an integer `orders`, which represents the total number of balls that the customer wants. You can sell the balls **in any order**.
+
+Return *the **maximum** total value that you can attain after selling* `orders` *colored balls*. As the answer may be too large, return it **modulo** `109 + 7`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/11/05/jj.gif](https://assets.leetcode.com/uploads/2020/11/05/jj.gif)
+
+```
+Input: inventory = [2,5], orders = 4
+Output: 14
+Explanation: Sell the 1st color 1 time (2) and the 2nd color 3 times (5 + 4 + 3).
+The maximum total value is 2 + 5 + 4 + 3 = 14.
+
+```
+
+**Example 2:**
+
+```
+Input: inventory = [3,5], orders = 6
+Output: 19
+Explanation: Sell the 1st color 2 times (3 + 2) and the 2nd color 4 times (5 + 4 + 3 + 2).
+The maximum total value is 3 + 2 + 5 + 4 + 3 + 2 = 19.
+```
+
+**Example 3:**
+
+```
+Input: inventory = [2,8,4,10,6], orders = 20
+Output: 110
+```
+
+**Example 4:**
+
+```
+Input: inventory = [1000000000], orders = 1000000000
+Output: 21
+Explanation: Sell the 1st color 1000000000 times for a total value of 500000000500000000. 500000000500000000 modulo 109 + 7 = 21.
+```
+
+**Constraints:**
+
+- `1 <= inventory.length <= 10^5`
+- `1 <= inventory[i] <= 10^9`
+- `1 <= orders <= min(sum(inventory[i]), 10^9)`
+
+## 题目大意
+
+你有一些球的库存 inventory ，里面包含着不同颜色的球。一个顾客想要 任意颜色 总数为 orders 的球。这位顾客有一种特殊的方式衡量球的价值：每个球的价值是目前剩下的 同色球 的数目。比方说还剩下 6 个黄球，那么顾客买第一个黄球的时候该黄球的价值为 6 。这笔交易以后，只剩下 5 个黄球了，所以下一个黄球的价值为 5 （也就是球的价值随着顾客购买同色球是递减的）
+
+给你整数数组 inventory ，其中 inventory[i] 表示第 i 种颜色球一开始的数目。同时给你整数 orders ，表示顾客总共想买的球数目。你可以按照 任意顺序 卖球。请你返回卖了 orders 个球以后 最大 总价值之和。由于答案可能会很大，请你返回答案对 109 + 7 取余数 的结果。
+
+提示：
+
+- 1 <= inventory.length <= 10^5
+- 1 <= inventory[i] <= 10^9
+- 1 <= orders <= min(sum(inventory[i]), 10^9)
+
+## 解题思路
+
+- 给出一个 `inventory` 数组和 `orders` 次操作，要求输出数组中前 `orders` 大个元素累加和。需要注意的是，每累加一个元素 `inventory[i]`，这个元素都会减一，下次再累加的时候，需要选取更新以后的数组的最大值。
+- 拿到这个题目以后很容易想到优先队列，建立大根堆以后，`pop` 出当前最大值 `maxItem`，累加，以后把 `maxItem` 减一再 `push` 回去。循环执行 `orders` 次以后即是最终结果。题目是这个意思，但是我们不能这么写代码，因为题目条件里面给出了 `orders` 的数据大小。orders 最大为 10^9。按照优先队列的这个方法一定会超时，时间复杂度为 O(orders⋅logn)。那就换一个思路。优先队列这个思路中，重复操作了 `orders` 次，其实在这些操作中，有一些是没有必要的废操作。这些大量的“废”操作导致了超时。试想，在 `orders` 次操作中，能否合并 `n` 个 `pop` 操作，一口气先 `pop` 掉 `n` 个前 `n` 大的数呢？这个是可行的，因为每次 `pop` 出去，元素都只会减一，这个是非常有规律的。
+- 为了接下来的描述更加清晰易懂，还需要再定义 1 个值， `thresholdValue` 为操作 `n` 次以后，当前  `inventory` 数组的最大值。关于 `thresholdValue` 的理解，这里要说明一下。 `thresholdValue` 的来源有 2 种，一种是本来数组里面就有这个值，还有一种来源是 `inventory[i]` 元素减少到了 `thresholdValue` 这个值。举个例子：原始数组是 [2,3,3,4,5]，`orders` = 4，取 4 次以后，剩下的数组是 [2,2,3,3,3]。3 个 3 里面其中一个 3 就来自于 `4-1=3`，或者 `5-2=3`。
+- 用二分搜索在 [0，max(`inventory`)] 区间内找到这个 `thresholdValue` 值，能满足下列不等式的最小 `thresholdValue` 值：
+
+    $$\sum_{inventory[i]\geqslant thresholdValue}^{} \left ( inventory[i] - thresholdValue \right )\leqslant orders$$
+
+    `thresholdValue` 越小，不等式左边的值越大，随着 `thresholdValue` 的增大，不等式左边的值越来越小，直到刚刚能小于等于 `orders`。求出了 `thresholdValue` 值以后，还需要再判断有多少值等于 `thresholdValue - 1` 值了。
+
+    ![https://img.halfrost.com/Leetcode/leetcode_1648.png](https://img.halfrost.com/Leetcode/leetcode_1648.png)
+
+- 还是举上面的例子，原始数组是 [2,3,3,4,5]，`orders` = 4，我们可以求得 `thresholdValue` = 3 。`inventory[i]` > `thresholdValue` 的那部分 100% 的要取走，`thresholdValue` 就像一个水平面，突出水平面的那些都要拿走，每列的值按照等差数列求和公式计算即可。但是 `orders` - `thresholdValue` = 1，说明水平面以下还要拿走一个，即 `thresholdValue` 线下的虚线框里面的那 4 个球，还需要任意取走一个。最后总的结果是这 2 部分的总和，( ( 5 + 4 ) + 4 ) + 3 = 16 。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+)
+
+// 解法一 贪心 + 二分搜索
+func maxProfit(inventory []int, orders int) int {
+	maxItem, thresholdValue, count, res, mod := 0, -1, 0, 0, 1000000007
+	for i := 0; i < len(inventory); i++ {
+		if inventory[i] > maxItem {
+			maxItem = inventory[i]
+		}
+	}
+	low, high := 0, maxItem
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		for i := 0; i < len(inventory); i++ {
+			count += max(inventory[i]-mid, 0)
+		}
+		if count <= orders {
+			thresholdValue = mid
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+		count = 0
+	}
+	count = 0
+	for i := 0; i < len(inventory); i++ {
+		count += max(inventory[i]-thresholdValue, 0)
+	}
+	count = orders - count
+	for i := 0; i < len(inventory); i++ {
+		if inventory[i] >= thresholdValue {
+			if count > 0 {
+				res += (thresholdValue + inventory[i]) * (inventory[i] - thresholdValue + 1) / 2
+				count--
+			} else {
+				res += (thresholdValue + 1 + inventory[i]) * (inventory[i] - thresholdValue) / 2
+			}
+		}
+	}
+	return res % mod
+}
+```
\ No newline at end of file
diff --git a/leetcode/1649.Create-Sorted-Array-through-Instructions/1649. Create Sorted Array through Instructions.go b/leetcode/1649.Create-Sorted-Array-through-Instructions/1649. Create Sorted Array through Instructions.go
new file mode 100644
index 000000000..094f4cd09
--- /dev/null
+++ b/leetcode/1649.Create-Sorted-Array-through-Instructions/1649. Create Sorted Array through Instructions.go	
@@ -0,0 +1,65 @@
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+// 解法一 树状数组 Binary Indexed Tree
+func createSortedArray(instructions []int) int {
+	bit, res := template.BinaryIndexedTree{}, 0
+	bit.Init(100001)
+	for i, v := range instructions {
+		less := bit.Query(v - 1)
+		greater := i - bit.Query(v)
+		res = (res + min(less, greater)) % (1e9 + 7)
+		bit.Add(v, 1)
+	}
+	return res
+}
+
+// 解法二 线段树 SegmentTree
+func createSortedArray1(instructions []int) int {
+	if len(instructions) == 0 {
+		return 0
+	}
+	st, res, mod := template.SegmentCountTree{}, 0, 1000000007
+	numsMap, numsArray, tmpArray := discretization1649(instructions)
+	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
+	st.Init(tmpArray, func(i, j int) int {
+		return 0
+	})
+	for i := 0; i < len(instructions); i++ {
+		strictlyLessThan := st.Query(0, numsMap[instructions[i]]-1)
+		strictlyGreaterThan := st.Query(numsMap[instructions[i]]+1, numsArray[len(numsArray)-1])
+		res = (res + min(strictlyLessThan, strictlyGreaterThan)) % mod
+		st.UpdateCount(numsMap[instructions[i]])
+	}
+	return res
+}
+
+func discretization1649(instructions []int) (map[int]int, []int, []int) {
+	tmpArray, numsArray, numsMap := []int{}, []int{}, map[int]int{}
+	for i := 0; i < len(instructions); i++ {
+		numsMap[instructions[i]] = instructions[i]
+	}
+	for _, v := range numsMap {
+		numsArray = append(numsArray, v)
+	}
+	sort.Ints(numsArray)
+	for i, num := range numsArray {
+		numsMap[num] = i
+	}
+	for i := range numsArray {
+		tmpArray = append(tmpArray, i)
+	}
+	return numsMap, numsArray, tmpArray
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/1649.Create-Sorted-Array-through-Instructions/1649. Create Sorted Array through Instructions_test.go b/leetcode/1649.Create-Sorted-Array-through-Instructions/1649. Create Sorted Array through Instructions_test.go
new file mode 100644
index 000000000..892fe14f0
--- /dev/null
+++ b/leetcode/1649.Create-Sorted-Array-through-Instructions/1649. Create Sorted Array through Instructions_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1649 struct {
+	para1649
+	ans1649
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1649 struct {
+	instructions []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1649 struct {
+	one int
+}
+
+func Test_Problem1649(t *testing.T) {
+
+	qs := []question1649{
+
+		{
+			para1649{[]int{1, 5, 6, 2}},
+			ans1649{1},
+		},
+
+		{
+			para1649{[]int{1, 2, 3, 6, 5, 4}},
+			ans1649{3},
+		},
+
+		{
+			para1649{[]int{1, 3, 3, 3, 2, 4, 2, 1, 2}},
+			ans1649{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1649------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1649, q.para1649
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, createSortedArray(p.instructions))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1649.Create-Sorted-Array-through-Instructions/README.md b/leetcode/1649.Create-Sorted-Array-through-Instructions/README.md
new file mode 100644
index 000000000..0ecda999f
--- /dev/null
+++ b/leetcode/1649.Create-Sorted-Array-through-Instructions/README.md
@@ -0,0 +1,149 @@
+# [1649. Create Sorted Array through Instructions](https://leetcode.com/problems/create-sorted-array-through-instructions/)
+
+## 题目
+
+Given an integer array `instructions`, you are asked to create a sorted array from the elements in `instructions`. You start with an empty container `nums`. For each element from **left to right** in `instructions`, insert it into `nums`. The **cost** of each insertion is the **minimum** of the following:
+
+- The number of elements currently in `nums` that are **strictly less than** `instructions[i]`.
+- The number of elements currently in `nums` that are **strictly greater than** `instructions[i]`.
+
+For example, if inserting element `3` into `nums = [1,2,3,5]`, the **cost** of insertion is `min(2, 1)` (elements `1` and `2` are less than `3`, element `5` is greater than `3`) and `nums` will become `[1,2,3,3,5]`.
+
+Return *the **total cost** to insert all elements from* `instructions` *into* `nums`. Since the answer may be large, return it **modulo** `10^9 + 7`
+
+**Example 1:**
+
+```
+Input: instructions = [1,5,6,2]
+Output: 1
+Explanation: Begin with nums = [].
+Insert 1 with cost min(0, 0) = 0, now nums = [1].
+Insert 5 with cost min(1, 0) = 0, now nums = [1,5].
+Insert 6 with cost min(2, 0) = 0, now nums = [1,5,6].
+Insert 2 with cost min(1, 2) = 1, now nums = [1,2,5,6].
+The total cost is 0 + 0 + 0 + 1 = 1.
+```
+
+**Example 2:**
+
+```
+Input: instructions = [1,2,3,6,5,4]
+Output: 3
+Explanation: Begin with nums = [].
+Insert 1 with cost min(0, 0) = 0, now nums = [1].
+Insert 2 with cost min(1, 0) = 0, now nums = [1,2].
+Insert 3 with cost min(2, 0) = 0, now nums = [1,2,3].
+Insert 6 with cost min(3, 0) = 0, now nums = [1,2,3,6].
+Insert 5 with cost min(3, 1) = 1, now nums = [1,2,3,5,6].
+Insert 4 with cost min(3, 2) = 2, now nums = [1,2,3,4,5,6].
+The total cost is 0 + 0 + 0 + 0 + 1 + 2 = 3.
+```
+
+**Example 3:**
+
+```
+Input: instructions = [1,3,3,3,2,4,2,1,2]
+Output: 4
+Explanation: Begin with nums = [].
+Insert 1 with cost min(0, 0) = 0, now nums = [1].
+Insert 3 with cost min(1, 0) = 0, now nums = [1,3].
+Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3].
+Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3,3].
+Insert 2 with cost min(1, 3) = 1, now nums = [1,2,3,3,3].
+Insert 4 with cost min(5, 0) = 0, now nums = [1,2,3,3,3,4].
+Insert 2 with cost min(1, 4) = 1, now nums = [1,2,2,3,3,3,4].
+Insert 1 with cost min(0, 6) = 0, now nums = [1,1,2,2,3,3,3,4].
+Insert 2 with cost min(2, 4) = 2, now nums = [1,1,2,2,2,3,3,3,4].
+The total cost is 0 + 0 + 0 + 0 + 1 + 0 + 1 + 0 + 2 = 4.
+```
+
+**Constraints:**
+
+- `1 <= instructions.length <= 105`
+- `1 <= instructions[i] <= 105`
+
+## 题目大意
+
+给你一个整数数组 instructions ，你需要根据 instructions 中的元素创建一个有序数组。一开始你有一个空的数组 nums ，你需要 从左到右 遍历 instructions 中的元素，将它们依次插入 nums 数组中。每一次插入操作的 代价 是以下两者的 较小值 ：
+
+- nums 中 严格小于 instructions[i] 的数字数目。
+- nums 中 严格大于 instructions[i] 的数字数目。
+
+比方说，如果要将 3 插入到 nums = [1,2,3,5] ，那么插入操作的 代价 为 min(2, 1) (元素 1 和 2 小于 3 ，元素 5 大于 3 ），插入后 nums 变成 [1,2,3,3,5] 。请你返回将 instructions 中所有元素依次插入 nums 后的 总最小代价 。由于答案会很大，请将它对 10^9 + 7 取余 后返回。
+
+## 解题思路
+
+- 给出一个数组，要求将其中的元素从头开始往另外一个空数组中插入，每次插入前，累加代价值 cost = min(**strictly less than**, **strictly greater than**)。最后输出累加值。
+- 这一题虽然是 Hard 题，但是读完题以后就可以判定这是模板题了。可以用线段树和树状数组来解决。这里简单说说线段树的思路吧，先将待插入的数组排序，获得总的区间。每次循环做 4 步：2 次 `query` 分别得到 `strictlyLessThan` 和 `strictlyGreaterThan` ，再比较出两者中的最小值累加，最后一步就是 `update`。
+- 由于题目给的数据比较大，所以建立线段树之前记得要先离散化。这一题核心代码不超过 10 行，其他的都是模板代码。具体实现见代码。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+// 解法一 树状数组 Binary Indexed Tree
+func createSortedArray(instructions []int) int {
+	bit, res := template.BinaryIndexedTree{}, 0
+	bit.Init(100001)
+	for i, v := range instructions {
+		less := bit.Query(v - 1)
+		greater := i - bit.Query(v)
+		res = (res + min(less, greater)) % (1e9 + 7)
+		bit.Add(v, 1)
+	}
+	return res
+}
+
+// 解法二 线段树 SegmentTree
+func createSortedArray1(instructions []int) int {
+	if len(instructions) == 0 {
+		return 0
+	}
+	st, res, mod := template.SegmentCountTree{}, 0, 1000000007
+	numsMap, numsArray, tmpArray := discretization1649(instructions)
+	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
+	st.Init(tmpArray, func(i, j int) int {
+		return 0
+	})
+	for i := 0; i < len(instructions); i++ {
+		strictlyLessThan := st.Query(0, numsMap[instructions[i]]-1)
+		strictlyGreaterThan := st.Query(numsMap[instructions[i]]+1, numsArray[len(numsArray)-1])
+		res = (res + min(strictlyLessThan, strictlyGreaterThan)) % mod
+		st.UpdateCount(numsMap[instructions[i]])
+	}
+	return res
+}
+
+func discretization1649(instructions []int) (map[int]int, []int, []int) {
+	tmpArray, numsArray, numsMap := []int{}, []int{}, map[int]int{}
+	for i := 0; i < len(instructions); i++ {
+		numsMap[instructions[i]] = instructions[i]
+	}
+	for _, v := range numsMap {
+		numsArray = append(numsArray, v)
+	}
+	sort.Ints(numsArray)
+	for i, num := range numsArray {
+		numsMap[num] = i
+	}
+	for i := range numsArray {
+		tmpArray = append(tmpArray, i)
+	}
+	return numsMap, numsArray, tmpArray
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+```
\ No newline at end of file
diff --git a/leetcode/1652.Defuse-the-Bomb/1652. Defuse the Bomb.go b/leetcode/1652.Defuse-the-Bomb/1652. Defuse the Bomb.go
new file mode 100644
index 000000000..cd0c245ed
--- /dev/null
+++ b/leetcode/1652.Defuse-the-Bomb/1652. Defuse the Bomb.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+func decrypt(code []int, k int) []int {
+	if k == 0 {
+		for i := 0; i < len(code); i++ {
+			code[i] = 0
+		}
+		return code
+	}
+	count, sum, res := k, 0, make([]int, len(code))
+	if k > 0 {
+		for i := 0; i < len(code); i++ {
+			for j := i + 1; j < len(code); j++ {
+				if count == 0 {
+					break
+				}
+				sum += code[j]
+				count--
+			}
+			if count > 0 {
+				for j := 0; j < len(code); j++ {
+					if count == 0 {
+						break
+					}
+					sum += code[j]
+					count--
+				}
+			}
+			res[i] = sum
+			sum, count = 0, k
+		}
+	}
+	if k < 0 {
+		for i := 0; i < len(code); i++ {
+			for j := i - 1; j >= 0; j-- {
+				if count == 0 {
+					break
+				}
+				sum += code[j]
+				count++
+			}
+			if count < 0 {
+				for j := len(code) - 1; j >= 0; j-- {
+					if count == 0 {
+						break
+					}
+					sum += code[j]
+					count++
+				}
+			}
+			res[i] = sum
+			sum, count = 0, k
+		}
+	}
+	return res
+}
diff --git a/leetcode/1652.Defuse-the-Bomb/1652. Defuse the Bomb_test.go b/leetcode/1652.Defuse-the-Bomb/1652. Defuse the Bomb_test.go
new file mode 100644
index 000000000..87ce3aeff
--- /dev/null
+++ b/leetcode/1652.Defuse-the-Bomb/1652. Defuse the Bomb_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1652 struct {
+	para1652
+	ans1652
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1652 struct {
+	code []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1652 struct {
+	one []int
+}
+
+func Test_Problem1652(t *testing.T) {
+
+	qs := []question1652{
+
+		{
+			para1652{[]int{5, 7, 1, 4}, 3},
+			ans1652{[]int{12, 10, 16, 13}},
+		},
+
+		{
+			para1652{[]int{1, 2, 3, 4}, 0},
+			ans1652{[]int{0, 0, 0, 0}},
+		},
+
+		{
+			para1652{[]int{2, 4, 9, 3}, -2},
+			ans1652{[]int{12, 5, 6, 13}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1652------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1652, q.para1652
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, decrypt(p.code, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1652.Defuse-the-Bomb/README.md b/leetcode/1652.Defuse-the-Bomb/README.md
new file mode 100644
index 000000000..cc47f853e
--- /dev/null
+++ b/leetcode/1652.Defuse-the-Bomb/README.md
@@ -0,0 +1,125 @@
+# [1652. Defuse the Bomb](https://leetcode.com/problems/defuse-the-bomb/)
+
+
+## 题目
+
+You have a bomb to defuse, and your time is running out! Your informer will provide you with a **circular** array `code` of length of `n` and a key `k`.
+
+To decrypt the code, you must replace every number. All the numbers are replaced **simultaneously**.
+
+- If `k > 0`, replace the `ith` number with the sum of the **next** `k` numbers.
+- If `k < 0`, replace the `ith` number with the sum of the **previous** `k` numbers.
+- If `k == 0`, replace the `ith` number with `0`.
+
+As `code` is circular, the next element of `code[n-1]` is `code[0]`, and the previous element of `code[0]` is `code[n-1]`.
+
+Given the **circular** array `code` and an integer key `k`, return *the decrypted code to defuse the bomb*!
+
+**Example 1:**
+
+```
+Input: code = [5,7,1,4], k = 3
+Output: [12,10,16,13]
+Explanation: Each number is replaced by the sum of the next 3 numbers. The decrypted code is [7+1+4, 1+4+5, 4+5+7, 5+7+1]. Notice that the numbers wrap around.
+```
+
+**Example 2:**
+
+```
+Input: code = [1,2,3,4], k = 0
+Output: [0,0,0,0]
+Explanation: When k is zero, the numbers are replaced by 0. 
+```
+
+**Example 3:**
+
+```
+Input: code = [2,4,9,3], k = -2
+Output: [12,5,6,13]
+Explanation: The decrypted code is [3+9, 2+3, 4+2, 9+4]. Notice that the numbers wrap around again. If k is negative, the sum is of the previous numbers.
+```
+
+**Constraints:**
+
+- `n == code.length`
+- `1 <= n <= 100`
+- `1 <= code[i] <= 100`
+- `(n - 1) <= k <= n - 1`
+
+## 题目大意
+
+你有一个炸弹需要拆除，时间紧迫！你的情报员会给你一个长度为 n 的 循环 数组 code 以及一个密钥 k 。为了获得正确的密码，你需要替换掉每一个数字。所有数字会 同时 被替换。
+
+- 如果 k > 0 ，将第 i 个数字用 接下来 k 个数字之和替换。
+- 如果 k < 0 ，将第 i 个数字用 之前 k 个数字之和替换。
+- 如果 k == 0 ，将第 i 个数字用 0 替换。
+
+由于 code 是循环的， code[n-1] 下一个元素是 code[0] ，且 code[0] 前一个元素是 code[n-1] 。
+
+给你 循环 数组 code 和整数密钥 k ，请你返回解密后的结果来拆除炸弹！
+
+## 解题思路
+
+- 给出一个 code 数组，要求按照规则替换每个字母。
+- 简单题，按照题意描述循环即可。
+
+## 代码
+
+```go
+package leetcode
+
+func decrypt(code []int, k int) []int {
+	if k == 0 {
+		for i := 0; i < len(code); i++ {
+			code[i] = 0
+		}
+		return code
+	}
+	count, sum, res := k, 0, make([]int, len(code))
+	if k > 0 {
+		for i := 0; i < len(code); i++ {
+			for j := i + 1; j < len(code); j++ {
+				if count == 0 {
+					break
+				}
+				sum += code[j]
+				count--
+			}
+			if count > 0 {
+				for j := 0; j < len(code); j++ {
+					if count == 0 {
+						break
+					}
+					sum += code[j]
+					count--
+				}
+			}
+			res[i] = sum
+			sum, count = 0, k
+		}
+	}
+	if k < 0 {
+		for i := 0; i < len(code); i++ {
+			for j := i - 1; j >= 0; j-- {
+				if count == 0 {
+					break
+				}
+				sum += code[j]
+				count++
+			}
+			if count < 0 {
+				for j := len(code) - 1; j >= 0; j-- {
+					if count == 0 {
+						break
+					}
+					sum += code[j]
+					count++
+				}
+			}
+			res[i] = sum
+			sum, count = 0, k
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/1653. Minimum Deletions to Make String Balanced.go b/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/1653. Minimum Deletions to Make String Balanced.go
new file mode 100644
index 000000000..1d5bc9889
--- /dev/null
+++ b/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/1653. Minimum Deletions to Make String Balanced.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+// 解法一 DP
+func minimumDeletions(s string) int {
+	prev, res, bCount := 0, 0, 0
+	for _, c := range s {
+		if c == 'a' {
+			res = min(prev+1, bCount)
+			prev = res
+		} else {
+			bCount++
+		}
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+// 解法二 模拟
+func minimumDeletions1(s string) int {
+	aCount, bCount, res := 0, 0, 0
+	for i := 0; i < len(s); i++ {
+		if s[i] == 'a' {
+			aCount++
+		}
+	}
+	res = aCount
+	for i := 0; i < len(s); i++ {
+		if s[i] == 'a' {
+			aCount--
+		} else {
+			bCount++
+		}
+		res = min(res, aCount+bCount)
+	}
+	return res
+}
diff --git a/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/1653. Minimum Deletions to Make String Balanced_test.go b/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/1653. Minimum Deletions to Make String Balanced_test.go
new file mode 100644
index 000000000..672ef5cd1
--- /dev/null
+++ b/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/1653. Minimum Deletions to Make String Balanced_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1653 struct {
+	para1653
+	ans1653
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1653 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1653 struct {
+	one int
+}
+
+func Test_Problem1653(t *testing.T) {
+
+	qs := []question1653{
+
+		{
+			para1653{"aababbab"},
+			ans1653{2},
+		},
+
+		{
+			para1653{"bbaaaaabb"},
+			ans1653{2},
+		},
+
+		{
+			para1653{"b"},
+			ans1653{0},
+		},
+
+		{
+			para1653{"ababaaaabbbbbaaababbbbbbaaabbaababbabbbbaabbbbaabbabbabaabbbababaa"},
+			ans1653{25},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1653------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1653, q.para1653
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, minimumDeletions(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/README.md b/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/README.md
new file mode 100644
index 000000000..f162b23ef
--- /dev/null
+++ b/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/README.md
@@ -0,0 +1,90 @@
+# [1653. Minimum Deletions to Make String Balanced](https://leetcode.com/problems/minimum-deletions-to-make-string-balanced/)
+
+
+## 题目
+
+You are given a string `s` consisting only of characters `'a'` and `'b'`.
+
+You can delete any number of characters in `s` to make `s` **balanced**. `s` is **balanced** if there is no pair of indices `(i,j)` such that `i < j` and `s[i] = 'b'` and `s[j]= 'a'`.
+
+Return *the **minimum** number of deletions needed to make* `s` ***balanced***.
+
+**Example 1:**
+
+```
+Input: s = "aababbab"
+Output: 2
+Explanation: You can either:
+Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
+Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").
+```
+
+**Example 2:**
+
+```
+Input: s = "bbaaaaabb"
+Output: 2
+Explanation: The only solution is to delete the first two characters.
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 105`
+- `s[i]` is `'a'` or `'b'`.
+
+## 题目大意
+
+给你一个字符串 s ，它仅包含字符 'a' 和 'b' 。你可以删除 s 中任意数目的字符，使得 s 平衡 。我们称 s 平衡的 当不存在下标对 (i,j) 满足 i < j 且 s[i] = 'b' 同时 s[j]= 'a' 。请你返回使 s 平衡 的 最少 删除次数。
+
+## 解题思路
+
+- 给定一个字符串，要求删除最少次数，使得字母 a 都排在字母 b 的前面。
+- 很容易想到的一个解题思路是 DP。定义 `dp[i]` 为字符串下标 [ 0, i ] 这个区间内使得字符串平衡的最少删除次数。当 `s[i] == 'a'` 的时候，有 2 种情况，一种是 `s[i]` 前面全是 `[aa……aa]` 的情况，这个时候只需要把其中的所有的字母 `b` 删除即可。还有一种情况是 `s[i]` 前面有字母 `a` 也有字母 `b`，即 `[aaa……abb……b]`，这种情况就需要考虑 `dp[i-1]` 了。当前字母是 `a`，那么肯定要删除字母 `a`，来维持前面有一段字母 `b` 的情况。当 `s[i] == 'b'` 的时候，不管是 `[aa……aa]` 这种情况，还是 `[aaa……abb……b]` 这种情况，当前字母 `b` 都可以直接附加在后面，也能保证整个字符串是平衡的。所以状态转移方程为 `dp[i+1] = min(dp[i] + 1, bCount), s[i] == 'a'`，`dp[i+1] = dp[i], s[i] == 'b'`。最终答案存在 `dp[n]` 中。由于前后项的递推关系中只用到一次前一项，所以我们还可以优化一下空间，用一个变量保存前一项的结果。优化以后的代码见解法一。
+- 这一题还有一个模拟的思路。题目要求找到最小删除字数，那么就是要找到一个“临界点”，在这个临界点的左边删除所有的字母 b，在这个临界点的右边删除所有的字母 a。在所有的“临界点”中找到删除最少的次数。代码实现见解法二。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 DP
+func minimumDeletions(s string) int {
+	prev, res, bCount := 0, 0, 0
+	for _, c := range s {
+		if c == 'a' {
+			res = min(prev+1, bCount)
+			prev = res
+		} else {
+			bCount++
+		}
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+// 解法二 模拟
+func minimumDeletions1(s string) int {
+	aCount, bCount, res := 0, 0, 0
+	for i := 0; i < len(s); i++ {
+		if s[i] == 'a' {
+			aCount++
+		}
+	}
+	res = aCount
+	for i := 0; i < len(s); i++ {
+		if s[i] == 'a' {
+			aCount--
+		} else {
+			bCount++
+		}
+		res = min(res, aCount+bCount)
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1654.Minimum-Jumps-to-Reach-Home/1654. Minimum Jumps to Reach Home.go b/leetcode/1654.Minimum-Jumps-to-Reach-Home/1654. Minimum Jumps to Reach Home.go
new file mode 100644
index 000000000..fcfda83b9
--- /dev/null
+++ b/leetcode/1654.Minimum-Jumps-to-Reach-Home/1654. Minimum Jumps to Reach Home.go	
@@ -0,0 +1,29 @@
+package leetcode
+
+func minimumJumps(forbidden []int, a int, b int, x int) int {
+	visited := make([]bool, 6000)
+	for i := range forbidden {
+		visited[forbidden[i]] = true
+	}
+	queue, res := [][2]int{{0, 0}}, -1
+	for len(queue) > 0 {
+		length := len(queue)
+		res++
+		for i := 0; i < length; i++ {
+			cur, isBack := queue[i][0], queue[i][1]
+			if cur == x {
+				return res
+			}
+			if isBack == 0 && cur-b > 0 && !visited[cur-b] {
+				visited[cur-b] = true
+				queue = append(queue, [2]int{cur - b, 1})
+			}
+			if cur+a < len(visited) && !visited[cur+a] {
+				visited[cur+a] = true
+				queue = append(queue, [2]int{cur + a, 0})
+			}
+		}
+		queue = queue[length:]
+	}
+	return -1
+}
diff --git a/leetcode/1654.Minimum-Jumps-to-Reach-Home/1654. Minimum Jumps to Reach Home_test.go b/leetcode/1654.Minimum-Jumps-to-Reach-Home/1654. Minimum Jumps to Reach Home_test.go
new file mode 100644
index 000000000..b08d6b743
--- /dev/null
+++ b/leetcode/1654.Minimum-Jumps-to-Reach-Home/1654. Minimum Jumps to Reach Home_test.go	
@@ -0,0 +1,60 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1654 struct {
+	para1654
+	ans1654
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1654 struct {
+	forbidden []int
+	a         int
+	b         int
+	x         int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1654 struct {
+	one int
+}
+
+func Test_Problem1654(t *testing.T) {
+
+	qs := []question1654{
+
+		{
+			para1654{[]int{14, 4, 18, 1, 15}, 3, 15, 9},
+			ans1654{3},
+		},
+
+		{
+			para1654{[]int{8, 3, 16, 6, 12, 20}, 15, 13, 11},
+			ans1654{-1},
+		},
+
+		{
+			para1654{[]int{1, 6, 2, 14, 5, 17, 4}, 16, 9, 7},
+			ans1654{2},
+		},
+
+		{
+			para1654{[]int{1998}, 1999, 2000, 2000},
+			ans1654{3998},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1654------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1654, q.para1654
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, minimumJumps(p.forbidden, p.a, p.b, p.x))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1654.Minimum-Jumps-to-Reach-Home/README.md b/leetcode/1654.Minimum-Jumps-to-Reach-Home/README.md
new file mode 100644
index 000000000..790c9cf34
--- /dev/null
+++ b/leetcode/1654.Minimum-Jumps-to-Reach-Home/README.md
@@ -0,0 +1,101 @@
+# [1654. Minimum Jumps to Reach Home](https://leetcode.com/problems/minimum-jumps-to-reach-home/)
+
+
+## 题目
+
+A certain bug's home is on the x-axis at position `x`. Help them get there from position `0`.
+
+The bug jumps according to the following rules:
+
+- It can jump exactly `a` positions **forward** (to the right).
+- It can jump exactly `b` positions **backward** (to the left).
+- It cannot jump backward twice in a row.
+- It cannot jump to any `forbidden` positions.
+
+The bug may jump forward **beyond** its home, but it **cannot jump** to positions numbered with **negative** integers.
+
+Given an array of integers `forbidden`, where `forbidden[i]` means that the bug cannot jump to the position `forbidden[i]`, and integers `a`, `b`, and `x`, return *the minimum number of jumps needed for the bug to reach its home*. If there is no possible sequence of jumps that lands the bug on position `x`, return `1.`
+
+**Example 1:**
+
+```
+Input: forbidden = [14,4,18,1,15], a = 3, b = 15, x = 9
+Output: 3
+Explanation: 3 jumps forward (0 -> 3 -> 6 -> 9) will get the bug home.
+```
+
+**Example 2:**
+
+```
+Input: forbidden = [8,3,16,6,12,20], a = 15, b = 13, x = 11
+Output: -1
+```
+
+**Example 3:**
+
+```
+Input: forbidden = [1,6,2,14,5,17,4], a = 16, b = 9, x = 7
+Output: 2
+Explanation: One jump forward (0 -> 16) then one jump backward (16 -> 7) will get the bug home.
+
+```
+
+**Constraints:**
+
+- `1 <= forbidden.length <= 1000`
+- `1 <= a, b, forbidden[i] <= 2000`
+- `0 <= x <= 2000`
+- All the elements in `forbidden` are distinct.
+- Position `x` is not forbidden.
+
+## 题目大意
+
+有一只跳蚤的家在数轴上的位置 x 处。请你帮助它从位置 0 出发，到达它的家。
+
+跳蚤跳跃的规则如下：
+
+- 它可以 往前 跳恰好 a 个位置（即往右跳）。
+- 它可以 往后 跳恰好 b 个位置（即往左跳）。
+- 它不能 连续 往后跳 2 次。
+- 它不能跳到任何 forbidden 数组中的位置。
+
+跳蚤可以往前跳 超过 它的家的位置，但是它 不能跳到负整数 的位置。给你一个整数数组 forbidden ，其中 forbidden[i] 是跳蚤不能跳到的位置，同时给你整数 a， b 和 x ，请你返回跳蚤到家的最少跳跃次数。如果没有恰好到达 x 的可行方案，请你返回 -1 。
+
+## 解题思路
+
+- 给出坐标 x ，可以往前跳的步长 a，往后跳的步长 b。要求输出能跳回家的最少跳跃次数。
+- 求最少跳跃次数，思路用 BFS 求解，最先到达坐标 x 的方案即是最少跳跃次数。对`forbidden` 的处理是把记忆化数组里面把他们标记为 true。禁止连续往后跳 2 次的限制，要求我们在 BFS 入队的时候再记录一下跳跃方向，每次往后跳的时候判断前一跳是否是往后跳，如果是往后跳，此次就不能往后跳了。
+
+## 代码
+
+```go
+package leetcode
+
+func minimumJumps(forbidden []int, a int, b int, x int) int {
+	visited := make([]bool, 6000)
+	for i := range forbidden {
+		visited[forbidden[i]] = true
+	}
+	queue, res := [][2]int{{0, 0}}, -1
+	for len(queue) > 0 {
+		length := len(queue)
+		res++
+		for i := 0; i < length; i++ {
+			cur, isBack := queue[i][0], queue[i][1]
+			if cur == x {
+				return res
+			}
+			if isBack == 0 && cur-b > 0 && !visited[cur-b] {
+				visited[cur-b] = true
+				queue = append(queue, [2]int{cur - b, 1})
+			}
+			if cur+a < len(visited) && !visited[cur+a] {
+				visited[cur+a] = true
+				queue = append(queue, [2]int{cur + a, 0})
+			}
+		}
+		queue = queue[length:]
+	}
+	return -1
+}
+```
\ No newline at end of file
diff --git a/leetcode/1655.Distribute-Repeating-Integers/1655. Distribute Repeating Integers.go b/leetcode/1655.Distribute-Repeating-Integers/1655. Distribute Repeating Integers.go
new file mode 100644
index 000000000..f0f57da21
--- /dev/null
+++ b/leetcode/1655.Distribute-Repeating-Integers/1655. Distribute Repeating Integers.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+func canDistribute(nums []int, quantity []int) bool {
+	freq := make(map[int]int)
+	for _, n := range nums {
+		freq[n]++
+	}
+	return dfs(freq, quantity)
+}
+
+func dfs(freq map[int]int, quantity []int) bool {
+	if len(quantity) == 0 {
+		return true
+	}
+	visited := make(map[int]bool)
+	for i := range freq {
+		if visited[freq[i]] {
+			continue
+		}
+		visited[freq[i]] = true
+		if freq[i] >= quantity[0] {
+			freq[i] -= quantity[0]
+			if dfs(freq, quantity[1:]) {
+				return true
+			}
+			freq[i] += quantity[0]
+		}
+	}
+	return false
+}
diff --git a/leetcode/1655.Distribute-Repeating-Integers/1655. Distribute Repeating Integers_test.go b/leetcode/1655.Distribute-Repeating-Integers/1655. Distribute Repeating Integers_test.go
new file mode 100644
index 000000000..d1c7bd260
--- /dev/null
+++ b/leetcode/1655.Distribute-Repeating-Integers/1655. Distribute Repeating Integers_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1655 struct {
+	para1655
+	ans1655
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1655 struct {
+	nums     []int
+	quantity []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1655 struct {
+	one bool
+}
+
+func Test_Problem1655(t *testing.T) {
+
+	qs := []question1655{
+
+		{
+			para1655{[]int{1, 2, 3, 4}, []int{2}},
+			ans1655{false},
+		},
+
+		{
+			para1655{[]int{1, 2, 3, 3}, []int{2}},
+			ans1655{true},
+		},
+
+		{
+			para1655{[]int{1, 1, 2, 2}, []int{2, 2}},
+			ans1655{true},
+		},
+
+		{
+			para1655{[]int{1, 1, 2, 3}, []int{2, 2}},
+			ans1655{false},
+		},
+
+		{
+			para1655{[]int{1, 1, 1, 1, 1}, []int{2, 3}},
+			ans1655{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1655------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1655, q.para1655
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, canDistribute(p.nums, p.quantity))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1655.Distribute-Repeating-Integers/README.md b/leetcode/1655.Distribute-Repeating-Integers/README.md
new file mode 100644
index 000000000..1c1faaedf
--- /dev/null
+++ b/leetcode/1655.Distribute-Repeating-Integers/README.md
@@ -0,0 +1,112 @@
+# [1655. Distribute Repeating Integers](https://leetcode.com/problems/distribute-repeating-integers/)
+
+
+## 题目
+
+You are given an array of `n` integers, `nums`, where there are at most `50` unique values in the array. You are also given an array of `m` customer order quantities, `quantity`, where `quantity[i]` is the amount of integers the `ith` customer ordered. Determine if it is possible to distribute `nums` such that:
+
+- The `ith` customer gets **exactly** `quantity[i]` integers,
+- The integers the `ith` customer gets are **all equal**, and
+- Every customer is satisfied.
+
+Return `true` *if it is possible to distribute* `nums` *according to the above conditions*.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3,4], quantity = [2]
+Output: false
+Explanation: The 0th customer cannot be given two different integers.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3,3], quantity = [2]
+Output: true
+Explanation: The 0th customer is given [3,3]. The integers [1,2] are not used.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,1,2,2], quantity = [2,2]
+Output: true
+Explanation: The 0th customer is given [1,1], and the 1st customer is given [2,2].
+```
+
+**Example 4:**
+
+```
+Input: nums = [1,1,2,3], quantity = [2,2]
+Output: false
+Explanation: Although the 0th customer could be given [1,1], the 1st customer cannot be satisfied.
+```
+
+**Example 5:**
+
+```
+Input: nums = [1,1,1,1,1], quantity = [2,3]
+Output: true
+Explanation: The 0th customer is given [1,1], and the 1st customer is given [1,1,1].
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `1 <= n <= 105`
+- `1 <= nums[i] <= 1000`
+- `m == quantity.length`
+- `1 <= m <= 10`
+- `1 <= quantity[i] <= 105`
+- There are at most `50` unique values in `nums`.
+
+## 题目大意
+
+给你一个长度为 n 的整数数组 nums ，这个数组中至多有 50 个不同的值。同时你有 m 个顾客的订单 quantity ，其中，整数 quantity[i] 是第 i 位顾客订单的数目。请你判断是否能将 nums 中的整数分配给这些顾客，且满足：
+
+- 第 i 位顾客 恰好 有 quantity[i] 个整数。
+- 第 i 位顾客拿到的整数都是 相同的 。
+- 每位顾客都满足上述两个要求。
+
+如果你可以分配 nums 中的整数满足上面的要求，那么请返回 true ，否则返回 false 。
+
+## 解题思路
+
+- 给定一个数组 nums，订单数组 quantity，要求按照订单满足顾客的需求。如果能满足输出 true，不能满足输出 false。
+- 用 DFS 记忆化暴力搜索。代码实现不难。（不知道此题为什么是 Hard）
+
+## 代码
+
+```go
+package leetcode
+
+func canDistribute(nums []int, quantity []int) bool {
+	freq := make(map[int]int)
+	for _, n := range nums {
+		freq[n]++
+	}
+	return dfs(freq, quantity)
+}
+
+func dfs(freq map[int]int, quantity []int) bool {
+	if len(quantity) == 0 {
+		return true
+	}
+	visited := make(map[int]bool)
+	for i := range freq {
+		if visited[freq[i]] {
+			continue
+		}
+		visited[freq[i]] = true
+		if freq[i] >= quantity[0] {
+			freq[i] -= quantity[0]
+			if dfs(freq, quantity[1:]) {
+				return true
+			}
+			freq[i] += quantity[0]
+		}
+	}
+	return false
+}
+```
\ No newline at end of file
diff --git a/leetcode/1656.Design-an-Ordered-Stream/1656. Design an Ordered Stream.go b/leetcode/1656.Design-an-Ordered-Stream/1656. Design an Ordered Stream.go
new file mode 100644
index 000000000..bb6035d99
--- /dev/null
+++ b/leetcode/1656.Design-an-Ordered-Stream/1656. Design an Ordered Stream.go	
@@ -0,0 +1,37 @@
+package leetcode
+
+type OrderedStream struct {
+	ptr    int
+	stream []string
+}
+
+func Constructor(n int) OrderedStream {
+	ptr, stream := 1, make([]string, n+1)
+	return OrderedStream{ptr: ptr, stream: stream}
+}
+
+func (this *OrderedStream) Insert(id int, value string) []string {
+	this.stream[id] = value
+	res := []string{}
+	if this.ptr == id || this.stream[this.ptr] != "" {
+		res = append(res, this.stream[this.ptr])
+		for i := id + 1; i < len(this.stream); i++ {
+			if this.stream[i] != "" {
+				res = append(res, this.stream[i])
+			} else {
+				this.ptr = i
+				return res
+			}
+		}
+	}
+	if len(res) > 0 {
+		return res
+	}
+	return []string{}
+}
+
+/**
+ * Your OrderedStream object will be instantiated and called as such:
+ * obj := Constructor(n);
+ * param_1 := obj.Insert(id,value);
+ */
diff --git a/leetcode/1656.Design-an-Ordered-Stream/1656. Design an Ordered Stream_test.go b/leetcode/1656.Design-an-Ordered-Stream/1656. Design an Ordered Stream_test.go
new file mode 100644
index 000000000..870fc070f
--- /dev/null
+++ b/leetcode/1656.Design-an-Ordered-Stream/1656. Design an Ordered Stream_test.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem1656(t *testing.T) {
+	obj := Constructor(5)
+	fmt.Printf("obj = %v\n", obj)
+	param1 := obj.Insert(3, "ccccc")
+	fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
+	param1 = obj.Insert(1, "aaaaa")
+	fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
+	param1 = obj.Insert(2, "bbbbb")
+	fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
+	param1 = obj.Insert(5, "eeeee")
+	fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
+	param1 = obj.Insert(4, "ddddd")
+	fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
+}
diff --git a/leetcode/1656.Design-an-Ordered-Stream/README.md b/leetcode/1656.Design-an-Ordered-Stream/README.md
new file mode 100644
index 000000000..96d52e601
--- /dev/null
+++ b/leetcode/1656.Design-an-Ordered-Stream/README.md
@@ -0,0 +1,106 @@
+# [1656. Design an Ordered Stream](https://leetcode.com/problems/design-an-ordered-stream/)
+
+## 题目
+
+There is a stream of `n` `(id, value)` pairs arriving in an **arbitrary** order, where `id` is an integer between `1` and `n` and `value` is a string. No two pairs have the same `id`.
+
+Design a stream that returns the values in **increasing order of their IDs** by returning a **chunk** (list) of values after each insertion. The concatenation of all the **chunks** should result in a list of the sorted values.
+
+Implement the `OrderedStream` class:
+
+- `OrderedStream(int n)` Constructs the stream to take `n` values.
+- `String[] insert(int id, String value)` Inserts the pair `(id, value)` into the stream, then returns the **largest possible chunk** of currently inserted values that appear next in the order.
+
+**Example:**
+
+![https://assets.leetcode.com/uploads/2020/11/10/q1.gif](https://assets.leetcode.com/uploads/2020/11/10/q1.gif)
+
+```
+Input
+["OrderedStream", "insert", "insert", "insert", "insert", "insert"]
+[[5], [3, "ccccc"], [1, "aaaaa"], [2, "bbbbb"], [5, "eeeee"], [4, "ddddd"]]
+Output
+[null, [], ["aaaaa"], ["bbbbb", "ccccc"], [], ["ddddd", "eeeee"]]
+
+Explanation
+// Note that the values ordered by ID is ["aaaaa", "bbbbb", "ccccc", "ddddd", "eeeee"].
+OrderedStream os = new OrderedStream(5);
+os.insert(3, "ccccc"); // Inserts (3, "ccccc"), returns [].
+os.insert(1, "aaaaa"); // Inserts (1, "aaaaa"), returns ["aaaaa"].
+os.insert(2, "bbbbb"); // Inserts (2, "bbbbb"), returns ["bbbbb", "ccccc"].
+os.insert(5, "eeeee"); // Inserts (5, "eeeee"), returns [].
+os.insert(4, "ddddd"); // Inserts (4, "ddddd"), returns ["ddddd", "eeeee"].
+// Concatentating all the chunks returned:
+// [] + ["aaaaa"] + ["bbbbb", "ccccc"] + [] + ["ddddd", "eeeee"] = ["aaaaa", "bbbbb", "ccccc", "ddddd", "eeeee"]
+// The resulting order is the same as the order above.
+
+```
+
+**Constraints:**
+
+- `1 <= n <= 1000`
+- `1 <= id <= n`
+- `value.length == 5`
+- `value` consists only of lowercase letters.
+- Each call to `insert` will have a unique `id.`
+- Exactly `n` calls will be made to `insert`.
+
+## 题目大意
+
+有 n 个 (id, value) 对，其中 id 是 1 到 n 之间的一个整数，value 是一个字符串。不存在 id 相同的两个 (id, value) 对。
+
+设计一个流，以 任意 顺序获取 n 个 (id, value) 对，并在多次调用时 按 id 递增的顺序 返回一些值。
+
+实现 OrderedStream 类：
+
+- OrderedStream(int n) 构造一个能接收 n 个值的流，并将当前指针 ptr 设为 1 。
+- String[] insert(int id, String value) 向流中存储新的 (id, value) 对。存储后：
+如果流存储有 id = ptr 的 (id, value) 对，则找出从 id = ptr 开始的 最长 id 连续递增序列 ，并 按顺序 返回与这些 id 关联的值的列表。然后，将 ptr 更新为最后那个 id + 1 。
+否则，返回一个空列表。
+
+## 解题思路
+
+- 设计一个具有插入操作的 Ordered Stream。insert 操作先在指定位置插入 value，然后返回当前指针 ptr 到最近一个空位置的最长连续递增字符串。如果字符串不为空，ptr 移动到非空 value 的后一个下标位置处。
+- 简单题。按照题目描述模拟即可。注意控制好 ptr 的位置。
+
+## 代码
+
+```go
+package leetcode
+
+type OrderedStream struct {
+	ptr    int
+	stream []string
+}
+
+func Constructor(n int) OrderedStream {
+	ptr, stream := 1, make([]string, n+1)
+	return OrderedStream{ptr: ptr, stream: stream}
+}
+
+func (this *OrderedStream) Insert(id int, value string) []string {
+	this.stream[id] = value
+	res := []string{}
+	if this.ptr == id || this.stream[this.ptr] != "" {
+		res = append(res, this.stream[this.ptr])
+		for i := id + 1; i < len(this.stream); i++ {
+			if this.stream[i] != "" {
+				res = append(res, this.stream[i])
+			} else {
+				this.ptr = i
+				return res
+			}
+		}
+	}
+	if len(res) > 0 {
+		return res
+	}
+	return []string{}
+}
+
+/**
+ * Your OrderedStream object will be instantiated and called as such:
+ * obj := Constructor(n);
+ * param_1 := obj.Insert(id,value);
+ */
+```
\ No newline at end of file
diff --git a/leetcode/1657.Determine-if-Two-Strings-Are-Close/1657. Determine if Two Strings Are Close.go b/leetcode/1657.Determine-if-Two-Strings-Are-Close/1657. Determine if Two Strings Are Close.go
new file mode 100644
index 000000000..9fa3d9479
--- /dev/null
+++ b/leetcode/1657.Determine-if-Two-Strings-Are-Close/1657. Determine if Two Strings Are Close.go	
@@ -0,0 +1,33 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+func closeStrings(word1 string, word2 string) bool {
+	if len(word1) != len(word2) {
+		return false
+	}
+	freqCount1, freqCount2 := make([]int, 26), make([]int, 26)
+	for _, c := range word1 {
+		freqCount1[c-97]++
+	}
+	for _, c := range word2 {
+		freqCount2[c-97]++
+	}
+	for i := 0; i < 26; i++ {
+		if (freqCount1[i] == freqCount2[i]) ||
+			(freqCount1[i] > 0 && freqCount2[i] > 0) {
+			continue
+		}
+		return false
+	}
+	sort.Ints(freqCount1)
+	sort.Ints(freqCount2)
+	for i := 0; i < 26; i++ {
+		if freqCount1[i] != freqCount2[i] {
+			return false
+		}
+	}
+	return true
+}
diff --git a/leetcode/1657.Determine-if-Two-Strings-Are-Close/1657. Determine if Two Strings Are Close_test.go b/leetcode/1657.Determine-if-Two-Strings-Are-Close/1657. Determine if Two Strings Are Close_test.go
new file mode 100644
index 000000000..cfc982540
--- /dev/null
+++ b/leetcode/1657.Determine-if-Two-Strings-Are-Close/1657. Determine if Two Strings Are Close_test.go	
@@ -0,0 +1,68 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1657 struct {
+	para1657
+	ans1657
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1657 struct {
+	word1 string
+	word2 string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1657 struct {
+	one bool
+}
+
+func Test_Problem1657(t *testing.T) {
+
+	qs := []question1657{
+
+		{
+			para1657{"abc", "bca"},
+			ans1657{true},
+		},
+
+		{
+			para1657{"a", "aa"},
+			ans1657{false},
+		},
+
+		{
+			para1657{"cabbba", "abbccc"},
+			ans1657{true},
+		},
+
+		{
+			para1657{"cabbba", "aabbss"},
+			ans1657{false},
+		},
+
+		{
+			para1657{"uau", "ssx"},
+			ans1657{false},
+		},
+
+		{
+			para1657{"uuukuuuukkuusuususuuuukuskuusuuusuusuuuuuuk", "kssskkskkskssskksskskksssssksskksskskksksuu"},
+			ans1657{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1657------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1657, q.para1657
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, closeStrings(p.word1, p.word2))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1657.Determine-if-Two-Strings-Are-Close/README.md b/leetcode/1657.Determine-if-Two-Strings-Are-Close/README.md
new file mode 100644
index 000000000..711c758fb
--- /dev/null
+++ b/leetcode/1657.Determine-if-Two-Strings-Are-Close/README.md
@@ -0,0 +1,114 @@
+# [1657. Determine if Two Strings Are Close](https://leetcode.com/problems/determine-if-two-strings-are-close/)
+
+
+## 题目
+
+Two strings are considered **close** if you can attain one from the other using the following operations:
+
+- Operation 1: Swap any two **existing** characters.
+    - For example, `abcde -> aecdb`
+- Operation 2: Transform **every** occurrence of one **existing** character into another **existing** character, and do the same with the other character.
+    - For example, `aacabb -> bbcbaa` (all `a`'s turn into `b`'s, and all `b`'s turn into `a`'s)
+
+You can use the operations on either string as many times as necessary.
+
+Given two strings, `word1` and `word2`, return `true` *if* `word1` *and* `word2` *are **close**, and* `false` *otherwise.*
+
+**Example 1:**
+
+```
+Input: word1 = "abc", word2 = "bca"
+Output: true
+Explanation: You can attain word2 from word1 in 2 operations.
+Apply Operation 1: "abc" -> "acb"
+Apply Operation 1: "acb" -> "bca"
+
+```
+
+**Example 2:**
+
+```
+Input: word1 = "a", word2 = "aa"
+Output: false
+Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.
+
+```
+
+**Example 3:**
+
+```
+Input: word1 = "cabbba", word2 = "abbccc"
+Output: true
+Explanation: You can attain word2 from word1 in 3 operations.
+Apply Operation 1: "cabbba" -> "caabbb"
+Apply Operation 2: "caabbb" -> "baaccc"
+Apply Operation 2: "baaccc" -> "abbccc"
+
+```
+
+**Example 4:**
+
+```
+Input: word1 = "cabbba", word2 = "aabbss"
+Output: false
+Explanation: It is impossible to attain word2 from word1, or vice versa, in any amount of operations.
+
+```
+
+**Constraints:**
+
+- `1 <= word1.length, word2.length <= 105`
+- `word1` and `word2` contain only lowercase English letters.
+
+## 题目大意
+
+如果可以使用以下操作从一个字符串得到另一个字符串，则认为两个字符串 接近 ：
+
+- 操作 1：交换任意两个 现有 字符。例如，abcde -> aecdb
+- 操作 2：将一个 现有 字符的每次出现转换为另一个 现有 字符，并对另一个字符执行相同的操作。例如，aacabb -> bbcbaa（所有 a 转化为 b ，而所有的 b 转换为 a ）
+
+你可以根据需要对任意一个字符串多次使用这两种操作。给你两个字符串，word1 和 word2 。如果 word1 和 word2 接近 ，就返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 判断 2 个字符串是否“接近”。“接近”的定义是能否通过交换 2 个字符或者 2 个字母互换，从一个字符串变换成另外一个字符串，如果存在这样的变换，即是“接近”。
+- 先统计 2 个字符串的 26 个字母的频次，如果频次有不相同的，直接返回 false。在频次相同的情况下，再从小到大排序，再次扫描判断频次是否相同。
+- 注意几种特殊情况：频次相同，再判断字母交换是否合法存在，如果字母不存在，输出 false。例如测试文件中的 case 5 。出现频次个数相同，但是频次不同。例如测试文件中的 case 6 。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func closeStrings(word1 string, word2 string) bool {
+	if len(word1) != len(word2) {
+		return false
+	}
+	freqCount1, freqCount2 := make([]int, 26), make([]int, 26)
+	for _, c := range word1 {
+		freqCount1[c-97]++
+	}
+	for _, c := range word2 {
+		freqCount2[c-97]++
+	}
+	for i := 0; i < 26; i++ {
+		if (freqCount1[i] == freqCount2[i]) ||
+			(freqCount1[i] > 0 && freqCount2[i] > 0) {
+			continue
+		}
+		return false
+	}
+	sort.Ints(freqCount1)
+	sort.Ints(freqCount2)
+	for i := 0; i < 26; i++ {
+		if freqCount1[i] != freqCount2[i] {
+			return false
+		}
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/1658. Minimum Operations to Reduce X to Zero.go b/leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/1658. Minimum Operations to Reduce X to Zero.go
new file mode 100644
index 000000000..a0db0774d
--- /dev/null
+++ b/leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/1658. Minimum Operations to Reduce X to Zero.go	
@@ -0,0 +1,40 @@
+package leetcode
+
+func minOperations(nums []int, x int) int {
+	total := 0
+	for _, n := range nums {
+		total += n
+	}
+	target := total - x
+	if target < 0 {
+		return -1
+	}
+	if target == 0 {
+		return len(nums)
+	}
+	left, right, sum, res := 0, 0, 0, -1
+	for right < len(nums) {
+		if sum < target {
+			sum += nums[right]
+			right++
+		}
+		for sum >= target {
+			if sum == target {
+				res = max(res, right-left)
+			}
+			sum -= nums[left]
+			left++
+		}
+	}
+	if res == -1 {
+		return -1
+	}
+	return len(nums) - res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/1658. Minimum Operations to Reduce X to Zero_test.go b/leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/1658. Minimum Operations to Reduce X to Zero_test.go
new file mode 100644
index 000000000..f781c28a8
--- /dev/null
+++ b/leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/1658. Minimum Operations to Reduce X to Zero_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1658 struct {
+	para1658
+	ans1658
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1658 struct {
+	nums []int
+	x    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1658 struct {
+	one int
+}
+
+func Test_Problem1658(t *testing.T) {
+
+	qs := []question1658{
+
+		{
+			para1658{[]int{1, 1, 4, 2, 3}, 5},
+			ans1658{2},
+		},
+
+		{
+			para1658{[]int{5, 6, 7, 8, 9}, 4},
+			ans1658{-1},
+		},
+
+		{
+			para1658{[]int{3, 2, 20, 1, 1, 3}, 10},
+			ans1658{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1658------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1658, q.para1658
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, minOperations(p.nums, p.x))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/README.md b/leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/README.md
new file mode 100644
index 000000000..3fb34d44a
--- /dev/null
+++ b/leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/README.md
@@ -0,0 +1,96 @@
+# [1658. Minimum Operations to Reduce X to Zero](https://leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/)
+
+
+## 题目
+
+You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
+
+Return *the **minimum number** of operations to reduce* `x` *to **exactly*** `0` *if it's possible, otherwise, return* `1`.
+
+**Example 1:**
+
+```
+Input: nums = [1,1,4,2,3], x = 5
+Output: 2
+Explanation: The optimal solution is to remove the last two elements to reduce x to zero.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [5,6,7,8,9], x = 4
+Output: -1
+
+```
+
+**Example 3:**
+
+```
+Input: nums = [3,2,20,1,1,3], x = 10
+Output: 5
+Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 105`
+- `1 <= nums[i] <= 104`
+- `1 <= x <= 109`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个整数 x 。每一次操作时，你应当移除数组 nums 最左边或最右边的元素，然后从 x 中减去该元素的值。请注意，需要 修改 数组以供接下来的操作使用。如果可以将 x 恰好 减到 0 ，返回 最小操作数 ；否则，返回 -1 。
+
+## 解题思路
+
+- 给定一个数组 nums 和一个整数 x，要求从数组两端分别移除一些数，使得这些数加起来正好等于整数 x，要求输出最小操作数。
+- 要求输出最小操作数，即数组两头的数字个数最少，并且加起来和正好等于整数 x。由于在数组的两头，用 2 个指针分别操作不太方便。我当时解题的时候的思路是把它变成循环数组，这样两边的指针就在一个区间内了。利用滑动窗口找到一个最小的窗口，使得窗口内的累加和等于整数 k。这个方法可行，但是代码挺多的。
+- 有没有更优美的方法呢？有的。要想两头的长度最少，也就是中间这段的长度最大。这样就转换成直接在数组上使用滑动窗口求解，累加和等于一个固定值的连续最长的子数组。
+- 和这道题类似思路的题目，209，1040(循环数组)，325。强烈推荐这 3 题。
+
+## 代码
+
+```go
+package leetcode
+
+func minOperations(nums []int, x int) int {
+	total := 0
+	for _, n := range nums {
+		total += n
+	}
+	target := total - x
+	if target < 0 {
+		return -1
+	}
+	if target == 0 {
+		return len(nums)
+	}
+	left, right, sum, res := 0, 0, 0, -1
+	for right < len(nums) {
+		if sum < target {
+			sum += nums[right]
+			right++
+		}
+		for sum >= target {
+			if sum == target {
+				res = max(res, right-left)
+			}
+			sum -= nums[left]
+			left++
+		}
+	}
+	if res == -1 {
+		return -1
+	}
+	return len(nums) - res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1659.Maximize-Grid-Happiness/1659. Maximize Grid Happiness.go b/leetcode/1659.Maximize-Grid-Happiness/1659. Maximize Grid Happiness.go
new file mode 100644
index 000000000..c0b506021
--- /dev/null
+++ b/leetcode/1659.Maximize-Grid-Happiness/1659. Maximize Grid Happiness.go	
@@ -0,0 +1,107 @@
+package leetcode
+
+import (
+	"math"
+)
+
+func getMaxGridHappiness(m int, n int, introvertsCount int, extrovertsCount int) int {
+	// lineStatus 					将每一行中 3 种状态进行编码，空白 - 0，内向人 - 1，外向人 - 2，每行状态用三进制表示
+	// lineStatusList[729][6] 		每一行的三进制表示
+	// introvertsCountInner[729] 	每一个 lineStatus 包含的内向人数
+	// extrovertsCountInner[729] 	每一个 lineStatus 包含的外向人数
+	// scoreInner[729] 			 	每一个 lineStatus 包含的行内得分（只统计 lineStatus 本身的得分，不包括它与上一行的）
+	// scoreOuter[729][729] 	 	每一个 lineStatus 包含的行外得分
+	// dp[上一行的 lineStatus][当前处理到的行][剩余的内向人数][剩余的外向人数]
+	n3, lineStatus, introvertsCountInner, extrovertsCountInner, scoreInner, scoreOuter, lineStatusList, dp := math.Pow(3.0, float64(n)), 0, [729]int{}, [729]int{}, [729]int{}, [729][729]int{}, [729][6]int{}, [729][6][7][7]int{}
+	for i := 0; i < 729; i++ {
+		lineStatusList[i] = [6]int{}
+	}
+	for i := 0; i < 729; i++ {
+		dp[i] = [6][7][7]int{}
+		for j := 0; j < 6; j++ {
+			dp[i][j] = [7][7]int{}
+			for k := 0; k < 7; k++ {
+				dp[i][j][k] = [7]int{-1, -1, -1, -1, -1, -1, -1}
+			}
+		}
+	}
+	// 预处理
+	for lineStatus = 0; lineStatus < int(n3); lineStatus++ {
+		tmp := lineStatus
+		for i := 0; i < n; i++ {
+			lineStatusList[lineStatus][i] = tmp % 3
+			tmp /= 3
+		}
+		introvertsCountInner[lineStatus], extrovertsCountInner[lineStatus], scoreInner[lineStatus] = 0, 0, 0
+		for i := 0; i < n; i++ {
+			if lineStatusList[lineStatus][i] != 0 {
+				// 个人分数
+				if lineStatusList[lineStatus][i] == 1 {
+					introvertsCountInner[lineStatus]++
+					scoreInner[lineStatus] += 120
+				} else if lineStatusList[lineStatus][i] == 2 {
+					extrovertsCountInner[lineStatus]++
+					scoreInner[lineStatus] += 40
+				}
+				// 行内分数
+				if i-1 >= 0 {
+					scoreInner[lineStatus] += closeScore(lineStatusList[lineStatus][i], lineStatusList[lineStatus][i-1])
+				}
+			}
+		}
+	}
+	// 行外分数
+	for lineStatus0 := 0; lineStatus0 < int(n3); lineStatus0++ {
+		for lineStatus1 := 0; lineStatus1 < int(n3); lineStatus1++ {
+			scoreOuter[lineStatus0][lineStatus1] = 0
+			for i := 0; i < n; i++ {
+				scoreOuter[lineStatus0][lineStatus1] += closeScore(lineStatusList[lineStatus0][i], lineStatusList[lineStatus1][i])
+			}
+		}
+	}
+	return dfs(0, 0, introvertsCount, extrovertsCount, m, int(n3), &dp, &introvertsCountInner, &extrovertsCountInner, &scoreInner, &scoreOuter)
+}
+
+// 如果 x 和 y 相邻，需要加上的分数
+func closeScore(x, y int) int {
+	if x == 0 || y == 0 {
+		return 0
+	}
+	// 两个内向的人，每个人要 -30，一共 -60
+	if x == 1 && y == 1 {
+		return -60
+	}
+	if x == 2 && y == 2 {
+		return 40
+	}
+	return -10
+}
+
+// dfs(上一行的 lineStatus，当前处理到的行，剩余的内向人数，剩余的外向人数）
+func dfs(lineStatusLast, row, introvertsCount, extrovertsCount, m, n3 int, dp *[729][6][7][7]int, introvertsCountInner, extrovertsCountInner, scoreInner *[729]int, scoreOuter *[729][729]int) int {
+	// 边界条件：如果已经处理完，或者没有人了
+	if row == m || introvertsCount+extrovertsCount == 0 {
+		return 0
+	}
+	// 记忆化
+	if dp[lineStatusLast][row][introvertsCount][extrovertsCount] != -1 {
+		return dp[lineStatusLast][row][introvertsCount][extrovertsCount]
+	}
+	best := 0
+	for lineStatus := 0; lineStatus < n3; lineStatus++ {
+		if introvertsCountInner[lineStatus] > introvertsCount || extrovertsCountInner[lineStatus] > extrovertsCount {
+			continue
+		}
+		score := scoreInner[lineStatus] + scoreOuter[lineStatus][lineStatusLast]
+		best = max(best, score+dfs(lineStatus, row+1, introvertsCount-introvertsCountInner[lineStatus], extrovertsCount-extrovertsCountInner[lineStatus], m, n3, dp, introvertsCountInner, extrovertsCountInner, scoreInner, scoreOuter))
+	}
+	dp[lineStatusLast][row][introvertsCount][extrovertsCount] = best
+	return best
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1659.Maximize-Grid-Happiness/1659. Maximize Grid Happiness_test.go b/leetcode/1659.Maximize-Grid-Happiness/1659. Maximize Grid Happiness_test.go
new file mode 100644
index 000000000..909e36b04
--- /dev/null
+++ b/leetcode/1659.Maximize-Grid-Happiness/1659. Maximize Grid Happiness_test.go	
@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1659 struct {
+	para1659
+	ans1659
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1659 struct {
+	m               int
+	n               int
+	introvertsCount int
+	extrovertsCount int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1659 struct {
+	one int
+}
+
+func Test_Problem1659(t *testing.T) {
+
+	qs := []question1659{
+
+		{
+			para1659{2, 3, 1, 2},
+			ans1659{240},
+		},
+
+		{
+			para1659{3, 1, 2, 1},
+			ans1659{260},
+		},
+
+		{
+			para1659{2, 2, 4, 0},
+			ans1659{240},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1659------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1659, q.para1659
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, getMaxGridHappiness(p.m, p.n, p.introvertsCount, p.extrovertsCount))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1659.Maximize-Grid-Happiness/README.md b/leetcode/1659.Maximize-Grid-Happiness/README.md
new file mode 100644
index 000000000..6f880d261
--- /dev/null
+++ b/leetcode/1659.Maximize-Grid-Happiness/README.md
@@ -0,0 +1,185 @@
+# [1659. Maximize Grid Happiness](https://leetcode.com/problems/maximize-grid-happiness/)
+
+## 题目
+
+You are given four integers, `m`, `n`, `introvertsCount`, and `extrovertsCount`. You have an `m x n` grid, and there are two types of people: introverts and extroverts. There are `introvertsCount` introverts and `extrovertsCount` extroverts.
+
+You should decide how many people you want to live in the grid and assign each of them one grid cell. Note that you **do not** have to have all the people living in the grid.
+
+The **happiness** of each person is calculated as follows:
+
+- Introverts **start** with `120` happiness and **lose** `30` happiness for each neighbor (introvert or extrovert).
+- Extroverts **start** with `40` happiness and **gain** `20` happiness for each neighbor (introvert or extrovert).
+
+Neighbors live in the directly adjacent cells north, east, south, and west of a person's cell.
+
+The **grid happiness** is the **sum** of each person's happiness. Return *the **maximum possible grid happiness**.*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/11/05/grid_happiness.png](https://assets.leetcode.com/uploads/2020/11/05/grid_happiness.png)
+
+```
+Input: m = 2, n = 3, introvertsCount = 1, extrovertsCount = 2
+Output: 240
+Explanation: Assume the grid is 1-indexed with coordinates (row, column).
+We can put the introvert in cell (1,1) and put the extroverts in cells (1,3) and (2,3).
+- Introvert at (1,1) happiness: 120 (starting happiness) - (0 * 30) (0 neighbors) = 120
+- Extrovert at (1,3) happiness: 40 (starting happiness) + (1 * 20) (1 neighbor) = 60
+- Extrovert at (2,3) happiness: 40 (starting happiness) + (1 * 20) (1 neighbor) = 60
+The grid happiness is 120 + 60 + 60 = 240.
+The above figure shows the grid in this example with each person's happiness. The introvert stays in the light green cell while the extroverts live on the light purple cells.
+```
+
+**Example 2:**
+
+```
+Input: m = 3, n = 1, introvertsCount = 2, extrovertsCount = 1
+Output: 260
+Explanation: Place the two introverts in (1,1) and (3,1) and the extrovert at (2,1).
+- Introvert at (1,1) happiness: 120 (starting happiness) - (1 * 30) (1 neighbor) = 90
+- Extrovert at (2,1) happiness: 40 (starting happiness) + (2 * 20) (2 neighbors) = 80
+- Introvert at (3,1) happiness: 120 (starting happiness) - (1 * 30) (1 neighbor) = 90
+The grid happiness is 90 + 80 + 90 = 260.
+```
+
+**Example 3:**
+
+```
+Input: m = 2, n = 2, introvertsCount = 4, extrovertsCount = 0
+Output: 240
+```
+
+**Constraints:**
+
+- `1 <= m, n <= 5`
+- `0 <= introvertsCount, extrovertsCount <= min(m * n, 6)`
+
+## 题目大意
+
+给你四个整数 m、n、introvertsCount 和 extrovertsCount 。有一个 m x n 网格，和两种类型的人：内向的人和外向的人。总共有 introvertsCount 个内向的人和 extrovertsCount 个外向的人。请你决定网格中应当居住多少人，并为每个人分配一个网格单元。 注意，不必 让所有人都生活在网格中。每个人的 幸福感 计算如下：
+
+- 内向的人 开始 时有 120 个幸福感，但每存在一个邻居（内向的或外向的）他都会 失去 30 个幸福感。
+- 外向的人 开始 时有 40 个幸福感，每存在一个邻居（内向的或外向的）他都会 得到 20 个幸福感。
+
+邻居是指居住在一个人所在单元的上、下、左、右四个直接相邻的单元中的其他人。网格幸福感 是每个人幸福感的 总和 。 返回 最大可能的网格幸福感 。
+
+## 解题思路
+
+- 给出 `m` x `n` 网格和两种人，要求如何安排这两种人能使得网格的得分最大。两种人有各自的初始分，相邻可能会加分也有可能减分。
+- 这一题状态很多。首先每个格子有 3 种状态，那么每一行有 3^6 = 729 种不同的状态。每行行内分数变化值可能是 -60(两个内向)，+40(两个外向)，-10(一个内向一个外向)。两行行间分数变化值可能是 -60(两个内向)，+40(两个外向)，-10(一个内向一个外向)。那么我们可以把每行的状态压缩成一个三进制，那么网格就变成了一维，每两个三进制之间的关系是行间关系，每个三进制内部还需要根据内向和外向的人数决定行内最终分数。定义 `dp[lineStatusLast][row][introvertsCount][extrovertsCount]` 代表在上一行 `row - 1` 的状态是 `lineStatusLast` 的情况下，当前枚举到了第 `row` 行，内向还有 `introvertsCount` 个人，外向还有 `extrovertsCount` 个人能获得的最大分数。状态转移方程是 `dp[lineStatusLast(row-1)][row][introvertsCount][extrovertsCount] = max{dp[lineStatusLast(row)][row+1][introvertsCount - countIC(lineStatusLast(row)) ][extrovertsCount - countEC(lineStatusLast(row)) ] + scoreInner(lineStatusLast(row)) + scoreOuter(lineStatusLast(row-1),lineStatusLast(row))}` ，这里有 2 个统计函数，`countIC` 是统计当前行状态三进制里面有多少个内向人。`countEC` 是统计当前行状态三进制里面有多少个外向人。`scoreInner` 是计算当前行状态三进制的行内分数。`scoreOuter` 是计算 `row -1` 行和 `row` 行之间的行间分数。
+- 由于这个状态转移方程的计算量是巨大的。所以需要预先初始化一些计算结果。比如把 729 中行状态分别对应的行内、行间的分数都计算好，在动态规划状态转移的时候，直接查表获取分数即可。这样我们在深搜的时候，利用 dp 的记忆化，可以大幅减少时间复杂度。
+- 题目中还提到，人数可以不用完。如果 `introvertsCount = 0`, `extrovertsCount = 0` ，即人数都用完了的情况，这时候 `dp = 0`。如果 `row = m`，即已经枚举完了所有行，那么不管剩下多少人，这一行的 `dp = 0` 。
+- 初始化的时候，注意，特殊处理 0 的情况，0 行 0 列都初始化为 -1 。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+)
+
+func getMaxGridHappiness(m int, n int, introvertsCount int, extrovertsCount int) int {
+	// lineStatus 					将每一行中 3 种状态进行编码，空白 - 0，内向人 - 1，外向人 - 2，每行状态用三进制表示
+	// lineStatusList[729][6] 		每一行的三进制表示
+	// introvertsCountInner[729] 	每一个 lineStatus 包含的内向人数
+	// extrovertsCountInner[729] 	每一个 lineStatus 包含的外向人数
+	// scoreInner[729] 			 	每一个 lineStatus 包含的行内得分（只统计 lineStatus 本身的得分，不包括它与上一行的）
+	// scoreOuter[729][729] 	 	每一个 lineStatus 包含的行外得分
+	// dp[上一行的 lineStatus][当前处理到的行][剩余的内向人数][剩余的外向人数]
+	n3, lineStatus, introvertsCountInner, extrovertsCountInner, scoreInner, scoreOuter, lineStatusList, dp := math.Pow(3.0, float64(n)), 0, [729]int{}, [729]int{}, [729]int{}, [729][729]int{}, [729][6]int{}, [729][6][7][7]int{}
+	for i := 0; i < 729; i++ {
+		lineStatusList[i] = [6]int{}
+	}
+	for i := 0; i < 729; i++ {
+		dp[i] = [6][7][7]int{}
+		for j := 0; j < 6; j++ {
+			dp[i][j] = [7][7]int{}
+			for k := 0; k < 7; k++ {
+				dp[i][j][k] = [7]int{-1, -1, -1, -1, -1, -1, -1}
+			}
+		}
+	}
+	// 预处理
+	for lineStatus = 0; lineStatus < int(n3); lineStatus++ {
+		tmp := lineStatus
+		for i := 0; i < n; i++ {
+			lineStatusList[lineStatus][i] = tmp % 3
+			tmp /= 3
+		}
+		introvertsCountInner[lineStatus], extrovertsCountInner[lineStatus], scoreInner[lineStatus] = 0, 0, 0
+		for i := 0; i < n; i++ {
+			if lineStatusList[lineStatus][i] != 0 {
+				// 个人分数
+				if lineStatusList[lineStatus][i] == 1 {
+					introvertsCountInner[lineStatus]++
+					scoreInner[lineStatus] += 120
+				} else if lineStatusList[lineStatus][i] == 2 {
+					extrovertsCountInner[lineStatus]++
+					scoreInner[lineStatus] += 40
+				}
+				// 行内分数
+				if i-1 >= 0 {
+					scoreInner[lineStatus] += closeScore(lineStatusList[lineStatus][i], lineStatusList[lineStatus][i-1])
+				}
+			}
+		}
+	}
+	// 行外分数
+	for lineStatus0 := 0; lineStatus0 < int(n3); lineStatus0++ {
+		for lineStatus1 := 0; lineStatus1 < int(n3); lineStatus1++ {
+			scoreOuter[lineStatus0][lineStatus1] = 0
+			for i := 0; i < n; i++ {
+				scoreOuter[lineStatus0][lineStatus1] += closeScore(lineStatusList[lineStatus0][i], lineStatusList[lineStatus1][i])
+			}
+		}
+	}
+	return dfs(0, 0, introvertsCount, extrovertsCount, m, int(n3), &dp, &introvertsCountInner, &extrovertsCountInner, &scoreInner, &scoreOuter)
+}
+
+// 如果 x 和 y 相邻，需要加上的分数
+func closeScore(x, y int) int {
+	if x == 0 || y == 0 {
+		return 0
+	}
+	// 两个内向的人，每个人要 -30，一共 -60
+	if x == 1 && y == 1 {
+		return -60
+	}
+	if x == 2 && y == 2 {
+		return 40
+	}
+	return -10
+}
+
+// dfs(上一行的 lineStatus，当前处理到的行，剩余的内向人数，剩余的外向人数）
+func dfs(lineStatusLast, row, introvertsCount, extrovertsCount, m, n3 int, dp *[729][6][7][7]int, introvertsCountInner, extrovertsCountInner, scoreInner *[729]int, scoreOuter *[729][729]int) int {
+	// 边界条件：如果已经处理完，或者没有人了
+	if row == m || introvertsCount+extrovertsCount == 0 {
+		return 0
+	}
+	// 记忆化
+	if dp[lineStatusLast][row][introvertsCount][extrovertsCount] != -1 {
+		return dp[lineStatusLast][row][introvertsCount][extrovertsCount]
+	}
+	best := 0
+	for lineStatus := 0; lineStatus < n3; lineStatus++ {
+		if introvertsCountInner[lineStatus] > introvertsCount || extrovertsCountInner[lineStatus] > extrovertsCount {
+			continue
+		}
+		score := scoreInner[lineStatus] + scoreOuter[lineStatus][lineStatusLast]
+		best = max(best, score+dfs(lineStatus, row+1, introvertsCount-introvertsCountInner[lineStatus], extrovertsCount-extrovertsCountInner[lineStatus], m, n3, dp, introvertsCountInner, extrovertsCountInner, scoreInner, scoreOuter))
+	}
+	dp[lineStatusLast][row][introvertsCount][extrovertsCount] = best
+	return best
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1662.Check-If-Two-String-Arrays-are-Equivalent/1662. Check If Two String Arrays are Equivalent.go b/leetcode/1662.Check-If-Two-String-Arrays-are-Equivalent/1662. Check If Two String Arrays are Equivalent.go
new file mode 100644
index 000000000..f85e3b053
--- /dev/null
+++ b/leetcode/1662.Check-If-Two-String-Arrays-are-Equivalent/1662. Check If Two String Arrays are Equivalent.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+func arrayStringsAreEqual(word1 []string, word2 []string) bool {
+	str1, str2 := "", ""
+	for i := 0; i < len(word1); i++ {
+		str1 += word1[i]
+	}
+	for i := 0; i < len(word2); i++ {
+		str2 += word2[i]
+	}
+	return str1 == str2
+}
diff --git a/leetcode/1662.Check-If-Two-String-Arrays-are-Equivalent/1662. Check If Two String Arrays are Equivalent_test.go b/leetcode/1662.Check-If-Two-String-Arrays-are-Equivalent/1662. Check If Two String Arrays are Equivalent_test.go
new file mode 100644
index 000000000..271b0dea9
--- /dev/null
+++ b/leetcode/1662.Check-If-Two-String-Arrays-are-Equivalent/1662. Check If Two String Arrays are Equivalent_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1662 struct {
+	para1662
+	ans1662
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1662 struct {
+	word1 []string
+	word2 []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1662 struct {
+	one bool
+}
+
+func Test_Problem1662(t *testing.T) {
+
+	qs := []question1662{
+
+		{
+			para1662{[]string{"ab", "c"}, []string{"a", "bc"}},
+			ans1662{true},
+		},
+
+		{
+			para1662{[]string{"a", "cb"}, []string{"ab", "c"}},
+			ans1662{false},
+		},
+
+		{
+			para1662{[]string{"abc", "d", "defg"}, []string{"abcddefg"}},
+			ans1662{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1662------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1662, q.para1662
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, arrayStringsAreEqual(p.word1, p.word2))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1662.Check-If-Two-String-Arrays-are-Equivalent/README.md b/leetcode/1662.Check-If-Two-String-Arrays-are-Equivalent/README.md
new file mode 100644
index 000000000..d1e47d012
--- /dev/null
+++ b/leetcode/1662.Check-If-Two-String-Arrays-are-Equivalent/README.md
@@ -0,0 +1,65 @@
+# [1662. Check If Two String Arrays are Equivalent](https://leetcode.com/problems/check-if-two-string-arrays-are-equivalent/)
+
+
+## 题目
+
+Given two string arrays `word1` and `word2`, return **`true` *if the two arrays **represent** the same string, and* `false` *otherwise.*
+
+A string is **represented** by an array if the array elements concatenated **in order** forms the string.
+
+**Example 1:**
+
+```
+Input: word1 = ["ab", "c"], word2 = ["a", "bc"]
+Output: true
+Explanation:
+word1 represents string "ab" + "c" -> "abc"
+word2 represents string "a" + "bc" -> "abc"
+The strings are the same, so return true.
+```
+
+**Example 2:**
+
+```
+Input: word1 = ["a", "cb"], word2 = ["ab", "c"]
+Output: false
+```
+
+**Example 3:**
+
+```
+Input: word1  = ["abc", "d", "defg"], word2 = ["abcddefg"]
+Output: true
+```
+
+**Constraints:**
+
+- `1 <= word1.length, word2.length <= 103`
+- `1 <= word1[i].length, word2[i].length <= 103`
+- `1 <= sum(word1[i].length), sum(word2[i].length) <= 103`
+- `word1[i]` and `word2[i]` consist of lowercase letters.
+
+## 题目大意
+
+给你两个字符串数组 word1 和 word2 。如果两个数组表示的字符串相同，返回 true ；否则，返回 false 。数组表示的字符串 是由数组中的所有元素 按顺序 连接形成的字符串。
+
+## 解题思路
+
+- 简单题，依次拼接 2 个数组内的字符串，然后比较 str1 和 str2 是否相同即可。
+
+## 代码
+
+```go
+package leetcode
+
+func arrayStringsAreEqual(word1 []string, word2 []string) bool {
+	str1, str2 := "", ""
+	for i := 0; i < len(word1); i++ {
+		str1 += word1[i]
+	}
+	for i := 0; i < len(word2); i++ {
+		str2 += word2[i]
+	}
+	return str1 == str2
+}
+```
\ No newline at end of file
diff --git a/leetcode/1663.Smallest-String-With-A-Given-Numeric-Value/1663. Smallest String With A Given Numeric Value.go b/leetcode/1663.Smallest-String-With-A-Given-Numeric-Value/1663. Smallest String With A Given Numeric Value.go
new file mode 100644
index 000000000..514abb09b
--- /dev/null
+++ b/leetcode/1663.Smallest-String-With-A-Given-Numeric-Value/1663. Smallest String With A Given Numeric Value.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+// 解法一 贪心
+func getSmallestString(n int, k int) string {
+	str, i, j := make([]byte, n), 0, 0
+	for i = n - 1; i <= k-26; i, k = i-1, k-26 {
+		str[i] = 'z'
+	}
+	if i >= 0 {
+		str[i] = byte('a' + k - 1 - i)
+		for ; j < i; j++ {
+			str[j] = 'a'
+		}
+	}
+	return string(str)
+}
+
+// 解法二 DFS
+func getSmallestString1(n int, k int) string {
+	if n == 0 {
+		return ""
+	}
+	res, c := "", []byte{}
+	findSmallestString(0, n, k, 0, c, &res)
+	return res
+}
+
+func findSmallestString(value int, length, k, index int, str []byte, res *string) {
+	if len(str) == length && value == k {
+		tmp := string(str)
+		if (*res) == "" {
+			*res = tmp
+		}
+		if tmp < *res && *res != "" {
+			*res = tmp
+		}
+		return
+	}
+	if len(str) >= index && (*res) != "" && str[index-1] > (*res)[index-1] {
+		return
+	}
+	for j := 0; j < 26; j++ {
+		if k-value > (length-len(str))*26 || value > k {
+			return
+		}
+		str = append(str, byte(int('a')+j))
+		value += j + 1
+		findSmallestString(value, length, k, index+1, str, res)
+		str = str[:len(str)-1]
+		value -= j + 1
+
+	}
+}
diff --git a/leetcode/1663.Smallest-String-With-A-Given-Numeric-Value/1663. Smallest String With A Given Numeric Value_test.go b/leetcode/1663.Smallest-String-With-A-Given-Numeric-Value/1663. Smallest String With A Given Numeric Value_test.go
new file mode 100644
index 000000000..f3c07a4cb
--- /dev/null
+++ b/leetcode/1663.Smallest-String-With-A-Given-Numeric-Value/1663. Smallest String With A Given Numeric Value_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1663 struct {
+	para1663
+	ans1663
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1663 struct {
+	n int
+	k int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1663 struct {
+	one string
+}
+
+func Test_Problem1663(t *testing.T) {
+
+	qs := []question1663{
+
+		{
+			para1663{3, 27},
+			ans1663{"aay"},
+		},
+
+		{
+			para1663{5, 73},
+			ans1663{"aaszz"},
+		},
+
+		{
+			para1663{24, 552},
+			ans1663{"aaszz"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1663------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1663, q.para1663
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, getSmallestString(p.n, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1663.Smallest-String-With-A-Given-Numeric-Value/README.md b/leetcode/1663.Smallest-String-With-A-Given-Numeric-Value/README.md
new file mode 100644
index 000000000..efa8c4063
--- /dev/null
+++ b/leetcode/1663.Smallest-String-With-A-Given-Numeric-Value/README.md
@@ -0,0 +1,102 @@
+# [1663. Smallest String With A Given Numeric Value](https://leetcode.com/problems/smallest-string-with-a-given-numeric-value/)
+
+## 题目
+
+The **numeric value** of a **lowercase character** is defined as its position `(1-indexed)` in the alphabet, so the numeric value of `a` is `1`, the numeric value of `b` is `2`, the numeric value of `c` is `3`, and so on.
+
+The **numeric value** of a **string** consisting of lowercase characters is defined as the sum of its characters' numeric values. For example, the numeric value of the string `"abe"` is equal to `1 + 2 + 5 = 8`.
+
+You are given two integers `n` and `k`. Return *the **lexicographically smallest string** with **length** equal to `n` and **numeric value** equal to `k`.*
+
+Note that a string `x` is lexicographically smaller than string `y` if `x` comes before `y` in dictionary order, that is, either `x` is a prefix of `y`, or if `i` is the first position such that `x[i] != y[i]`, then `x[i]` comes before `y[i]` in alphabetic order.
+
+**Example 1:**
+
+```
+Input: n = 3, k = 27
+Output: "aay"
+Explanation: The numeric value of the string is 1 + 1 + 25 = 27, and it is the smallest string with such a value and length equal to 3.
+```
+
+**Example 2:**
+
+```
+Input: n = 5, k = 73
+Output: "aaszz"
+```
+
+**Constraints:**
+
+- `1 <= n <= 105`
+- `n <= k <= 26 * n`
+
+## 题目大意
+
+小写字符 的 数值 是它在字母表中的位置（从 1 开始），因此 a 的数值为 1 ，b 的数值为 2 ，c 的数值为 3 ，以此类推。字符串由若干小写字符组成，字符串的数值 为各字符的数值之和。例如，字符串 "abe" 的数值等于 1 + 2 + 5 = 8 。给你两个整数 n 和 k 。返回 长度 等于 n 且 数值 等于 k 的 字典序最小 的字符串。注意，如果字符串 x 在字典排序中位于 y 之前，就认为 x 字典序比 y 小，有以下两种情况：
+
+- x 是 y 的一个前缀；
+- 如果 i 是 x[i] != y[i] 的第一个位置，且 x[i] 在字母表中的位置比 y[i] 靠前。
+
+## 解题思路
+
+- 给出 n 和 k，要求找到字符串长度为 n，字母在字母表内位置总和为 k 的最小字典序字符串。
+- 这一题笔者读完题，比赛的时候直接用 DFS 撸了一版。赛后看了时间复杂度马马虎虎，感觉还有优化的空间。DFS 会遍历出所有的解，实际上这一题只要求最小字典序，所以 DFS 剪枝的时候要加上判断字典序的判断，如果新添加进来的字母比已经保存的字符串的相应位置上的字母字典序大，那么就直接 return，这个答案一定不会是最小字典序。代码见解法二
+- 想到这里，其实 DFS 不必要，直接用 for 循环就可找到最小字典序的字符串。代码见解法一。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 贪心
+func getSmallestString(n int, k int) string {
+    str, i, j := make([]byte, n), 0, 0
+    for i = n-1; i <= k-26; i, k = i-1, k-26 {
+        str[i] = 'z'
+    }
+    if i >= 0 {
+        str[i] = byte('a' + k-1-i)
+        for ; j < i; j++ {
+            str[j] = 'a'
+        }
+    }
+    return string(str)
+}
+
+// 解法二 DFS
+func getSmallestString1(n int, k int) string {
+	if n == 0 {
+		return ""
+	}
+	res, c := "", []byte{}
+	findSmallestString(0, n, k, 0, c, &res)
+	return res
+}
+
+func findSmallestString(value int, length, k, index int, str []byte, res *string) {
+	if len(str) == length && value == k {
+		tmp := string(str)
+		if (*res) == "" {
+			*res = tmp
+		}
+		if tmp < *res && *res != "" {
+			*res = tmp
+		}
+		return
+	}
+	if len(str) >= index && (*res) != "" && str[index-1] > (*res)[index-1] {
+		return
+	}
+	for j := 0; j < 26; j++ {
+		if k-value > (length-len(str))*26 || value > k {
+			return
+		}
+		str = append(str, byte(int('a')+j))
+		value += j + 1
+		findSmallestString(value, length, k, index+1, str, res)
+		str = str[:len(str)-1]
+		value -= j + 1
+
+	}
+}
+```
\ No newline at end of file
diff --git a/leetcode/1664.Ways-to-Make-a-Fair-Array/1664. Ways to Make a Fair Array.go b/leetcode/1664.Ways-to-Make-a-Fair-Array/1664. Ways to Make a Fair Array.go
new file mode 100644
index 000000000..c66641a0d
--- /dev/null
+++ b/leetcode/1664.Ways-to-Make-a-Fair-Array/1664. Ways to Make a Fair Array.go	
@@ -0,0 +1,45 @@
+package leetcode
+
+// 解法一 超简洁写法
+func waysToMakeFair(nums []int) int {
+	sum, res := [2]int{}, 0
+	for i := 0; i < len(nums); i++ {
+		sum[i%2] += nums[i]
+	}
+	for i := 0; i < len(nums); i++ {
+		sum[i%2] -= nums[i]
+		if sum[i%2] == sum[1-(i%2)] {
+			res++
+		}
+		sum[1-(i%2)] += nums[i]
+	}
+	return res
+}
+
+// 解法二 前缀和，后缀和
+func waysToMakeFair1(nums []int) int {
+	evenPrefix, oddPrefix, evenSuffix, oddSuffix, res := 0, 0, 0, 0, 0
+	for i := 0; i < len(nums); i++ {
+		if i%2 == 0 {
+			evenSuffix += nums[i]
+		} else {
+			oddSuffix += nums[i]
+		}
+	}
+	for i := 0; i < len(nums); i++ {
+		if i%2 == 0 {
+			evenSuffix -= nums[i]
+		} else {
+			oddSuffix -= nums[i]
+		}
+		if (evenPrefix + oddSuffix) == (oddPrefix + evenSuffix) {
+			res++
+		}
+		if i%2 == 0 {
+			evenPrefix += nums[i]
+		} else {
+			oddPrefix += nums[i]
+		}
+	}
+	return res
+}
diff --git a/leetcode/1664.Ways-to-Make-a-Fair-Array/1664. Ways to Make a Fair Array_test.go b/leetcode/1664.Ways-to-Make-a-Fair-Array/1664. Ways to Make a Fair Array_test.go
new file mode 100644
index 000000000..670e994e8
--- /dev/null
+++ b/leetcode/1664.Ways-to-Make-a-Fair-Array/1664. Ways to Make a Fair Array_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1664 struct {
+	para1664
+	ans1664
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1664 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1664 struct {
+	one int
+}
+
+func Test_Problem1664(t *testing.T) {
+
+	qs := []question1664{
+
+		{
+			para1664{[]int{6, 1, 7, 4, 1}},
+			ans1664{0},
+		},
+
+		{
+			para1664{[]int{2, 1, 6, 4}},
+			ans1664{1},
+		},
+
+		{
+			para1664{[]int{1, 1, 1}},
+			ans1664{3},
+		},
+
+		{
+			para1664{[]int{1, 2, 3}},
+			ans1664{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1664------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1664, q.para1664
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, waysToMakeFair(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1664.Ways-to-Make-a-Fair-Array/README.md b/leetcode/1664.Ways-to-Make-a-Fair-Array/README.md
new file mode 100644
index 000000000..9c40f8230
--- /dev/null
+++ b/leetcode/1664.Ways-to-Make-a-Fair-Array/README.md
@@ -0,0 +1,116 @@
+# [1664. Ways to Make a Fair Array](https://leetcode.com/problems/ways-to-make-a-fair-array/)
+
+
+## 题目
+
+You are given an integer array `nums`. You can choose **exactly one** index (**0-indexed**) and remove the element. Notice that the index of the elements may change after the removal.
+
+For example, if `nums = [6,1,7,4,1]`:
+
+- Choosing to remove index `1` results in `nums = [6,7,4,1]`.
+- Choosing to remove index `2` results in `nums = [6,1,4,1]`.
+- Choosing to remove index `4` results in `nums = [6,1,7,4]`.
+
+An array is **fair** if the sum of the odd-indexed values equals the sum of the even-indexed values.
+
+Return the ***number** of indices that you could choose such that after the removal,* `nums` *is **fair**.*
+
+**Example 1:**
+
+```
+Input: nums = [2,1,6,4]
+Output: 1
+Explanation:
+Remove index 0: [1,6,4] -> Even sum: 1 + 4 = 5. Odd sum: 6. Not fair.
+Remove index 1: [2,6,4] -> Even sum: 2 + 4 = 6. Odd sum: 6. Fair.
+Remove index 2: [2,1,4] -> Even sum: 2 + 4 = 6. Odd sum: 1. Not fair.
+Remove index 3: [2,1,6] -> Even sum: 2 + 6 = 8. Odd sum: 1. Not fair.
+There is 1 index that you can remove to make nums fair.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,1,1]
+Output: 3
+Explanation: You can remove any index and the remaining array is fair.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,2,3]
+Output: 0
+Explanation: You cannot make a fair array after removing any index.
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 105`
+- `1 <= nums[i] <= 104`
+
+## 题目大意
+
+给你一个整数数组 nums 。你需要选择 恰好 一个下标（下标从 0 开始）并删除对应的元素。请注意剩下元素的下标可能会因为删除操作而发生改变。
+
+比方说，如果 nums = [6,1,7,4,1] ，那么：
+
+- 选择删除下标 1 ，剩下的数组为 nums = [6,7,4,1] 。
+- 选择删除下标 2 ，剩下的数组为 nums = [6,1,4,1] 。
+- 选择删除下标 4 ，剩下的数组为 nums = [6,1,7,4] 。
+
+如果一个数组满足奇数下标元素的和与偶数下标元素的和相等，该数组就是一个 平衡数组 。请你返回删除操作后，剩下的数组 nums 是 平衡数组 的 方案数 。
+
+## 解题思路
+
+- 给定一个数组 nums，要求输出仅删除一个元素以后能使得整个数组平衡的方案数。平衡的定义是奇数下标元素总和等于偶数下标元素总和。
+- 这一题如果暴力解答，会超时。原因是每次删除元素以后，都重新计算奇偶数位总和比较耗时。应该利用前面计算过的累加和，推导出此次删除元素以后的情况。这样修改以后就不超时了。具体的，如果删除的是元素是奇数位，这个下标的前缀和不变，要变化的是后面的。删除元素后面，原来偶数位的总和变成了奇数位了，原来奇数位的总和变成偶数位了。删除元素后面这半段的总和可以用前缀和计算出来，奇数位的总和减去删除元素的前缀和，就得到了删除元素后面的后缀和。通过这个办法就可以得到删除元素后面的，奇数位总和，偶数位总和。注意这个后缀和是包含了删除元素的。所以最后需要判断删除元素是奇数位还是偶数位，如果是奇数位，那么在计算出来的偶数和上再减去这个删除元素；如果是偶数位，就在计算出来的奇数和上再减去这个删除元素。代码见解法二。
+- 这一题还有一种更简洁的写法，就是解法一了。通过了解法二的思考，我们可以知道，每次变换以后的操作可以抽象出来，即三步，减去一个数，判断是否相等，再加上一个数。只不过这三步在解法二中都去判断了奇偶性。如果我们不判断奇偶性，那么代码就可以写成解法一的样子。为什么可以不用管奇偶性呢？因为每次删除一个元素以后，下次再删除，奇偶就发生颠倒了，上次的奇数和到了下次就是偶数和了。想通这一点就可以把代码写成解法一的样子。
+
+## 代码
+
+```go
+// 解法一 超简洁写法
+func waysToMakeFair(nums []int) int {
+	sum, res := [2]int{}, 0
+	for i := 0; i < len(nums); i++ {
+		sum[i%2] += nums[i]
+	}
+	for i := 0; i < len(nums); i++ {
+		sum[i%2] -= nums[i]
+		if sum[i%2] == sum[1-(i%2)] {
+			res++
+		}
+		sum[1-(i%2)] += nums[i]
+	}
+	return res
+}
+
+// 解法二 前缀和，后缀和
+func waysToMakeFair1(nums []int) int {
+	evenPrefix, oddPrefix, evenSuffix, oddSuffix, res := 0, 0, 0, 0, 0
+	for i := 0; i < len(nums); i++ {
+		if i%2 == 0 {
+			evenSuffix += nums[i]
+		} else {
+			oddSuffix += nums[i]
+		}
+	}
+	for i := 0; i < len(nums); i++ {
+		if i%2 == 0 {
+			evenSuffix -= nums[i]
+		} else {
+			oddSuffix -= nums[i]
+		}
+		if (evenPrefix + oddSuffix) == (oddPrefix + evenSuffix) {
+			res++
+		}
+		if i%2 == 0 {
+			evenPrefix += nums[i]
+		} else {
+			oddPrefix += nums[i]
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1665.Minimum-Initial-Energy-to-Finish-Tasks/1665. Minimum Initial Energy to Finish Tasks.go b/leetcode/1665.Minimum-Initial-Energy-to-Finish-Tasks/1665. Minimum Initial Energy to Finish Tasks.go
new file mode 100644
index 000000000..f44dfcd91
--- /dev/null
+++ b/leetcode/1665.Minimum-Initial-Energy-to-Finish-Tasks/1665. Minimum Initial Energy to Finish Tasks.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+func minimumEffort(tasks [][]int) int {
+	sort.Sort(Task(tasks))
+	res, cur := 0, 0
+	for _, t := range tasks {
+		if t[1] > cur {
+			res += t[1] - cur
+			cur = t[1] - t[0]
+		} else {
+			cur -= t[0]
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// Task define
+type Task [][]int
+
+func (task Task) Len() int {
+	return len(task)
+}
+
+func (task Task) Less(i, j int) bool {
+	t1, t2 := task[i][1]-task[i][0], task[j][1]-task[j][0]
+	if t1 != t2 {
+		return t2 < t1
+	}
+	return task[j][1] < task[i][1]
+}
+
+func (task Task) Swap(i, j int) {
+	t := task[i]
+	task[i] = task[j]
+	task[j] = t
+}
diff --git a/leetcode/1665.Minimum-Initial-Energy-to-Finish-Tasks/1665. Minimum Initial Energy to Finish Tasks_test.go b/leetcode/1665.Minimum-Initial-Energy-to-Finish-Tasks/1665. Minimum Initial Energy to Finish Tasks_test.go
new file mode 100644
index 000000000..ce0762380
--- /dev/null
+++ b/leetcode/1665.Minimum-Initial-Energy-to-Finish-Tasks/1665. Minimum Initial Energy to Finish Tasks_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1665 struct {
+	para1665
+	ans1665
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1665 struct {
+	tasks [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1665 struct {
+	one int
+}
+
+func Test_Problem1665(t *testing.T) {
+
+	qs := []question1665{
+
+		{
+			para1665{[][]int{{1, 2}, {2, 4}, {4, 8}}},
+			ans1665{8},
+		},
+
+		{
+			para1665{[][]int{{1, 3}, {2, 4}, {10, 11}, {10, 12}, {8, 9}}},
+			ans1665{32},
+		},
+
+		{
+			para1665{[][]int{{1, 7}, {2, 8}, {3, 9}, {4, 10}, {5, 11}, {6, 12}}},
+			ans1665{27},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1665------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1665, q.para1665
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, minimumEffort(p.tasks))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1665.Minimum-Initial-Energy-to-Finish-Tasks/README.md b/leetcode/1665.Minimum-Initial-Energy-to-Finish-Tasks/README.md
new file mode 100644
index 000000000..c2bd57e69
--- /dev/null
+++ b/leetcode/1665.Minimum-Initial-Energy-to-Finish-Tasks/README.md
@@ -0,0 +1,128 @@
+# [1665. Minimum Initial Energy to Finish Tasks](https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/)
+
+## 题目
+
+You are given an array `tasks` where `tasks[i] = [actuali, minimumi]`:
+
+- `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
+- `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
+
+For example, if the task is `[10, 12]` and your current energy is `11`, you cannot start this task. However, if your current energy is `13`, you can complete this task, and your energy will be `3` after finishing it.
+
+You can finish the tasks in **any order** you like.
+
+Return *the **minimum** initial amount of energy you will need* *to finish all the tasks*.
+
+**Example 1:**
+
+```
+Input: tasks = [[1,2],[2,4],[4,8]]
+Output: 8
+Explanation:
+Starting with 8 energy, we finish the tasks in the following order:
+    - 3rd task. Now energy = 8 - 4 = 4.
+    - 2nd task. Now energy = 4 - 2 = 2.
+    - 1st task. Now energy = 2 - 1 = 1.
+Notice that even though we have leftover energy, starting with 7 energy does not work because we cannot do the 3rd task.
+```
+
+**Example 2:**
+
+```
+Input: tasks = [[1,3],[2,4],[10,11],[10,12],[8,9]]
+Output: 32
+Explanation:
+Starting with 32 energy, we finish the tasks in the following order:
+    - 1st task. Now energy = 32 - 1 = 31.
+    - 2nd task. Now energy = 31 - 2 = 29.
+    - 3rd task. Now energy = 29 - 10 = 19.
+    - 4th task. Now energy = 19 - 10 = 9.
+    - 5th task. Now energy = 9 - 8 = 1.
+```
+
+**Example 3:**
+
+```
+Input: tasks = [[1,7],[2,8],[3,9],[4,10],[5,11],[6,12]]
+Output: 27
+Explanation:
+Starting with 27 energy, we finish the tasks in the following order:
+    - 5th task. Now energy = 27 - 5 = 22.
+    - 2nd task. Now energy = 22 - 2 = 20.
+    - 3rd task. Now energy = 20 - 3 = 17.
+    - 1st task. Now energy = 17 - 1 = 16.
+    - 4th task. Now energy = 16 - 4 = 12.
+    - 6th task. Now energy = 12 - 6 = 6.
+
+```
+
+**Constraints:**
+
+- `1 <= tasks.length <= 105`
+- `1 <= actuali <= minimumi <= 104`
+
+## 题目大意
+
+给你一个任务数组 tasks ，其中 tasks[i] = [actuali, minimumi] ：
+
+- actual i 是完成第 i 个任务 需要耗费 的实际能量。
+- minimum i 是开始第 i 个任务前需要达到的最低能量。
+
+比方说，如果任务为 [10, 12] 且你当前的能量为 11 ，那么你不能开始这个任务。如果你当前的能量为 13 ，你可以完成这个任务，且完成它后剩余能量为 3 。你可以按照 任意顺序 完成任务。请你返回完成所有任务的 最少 初始能量。
+
+## 解题思路
+
+- 给出一个 task 数组，每个元素代表一个任务，每个任务有实际消费能量值和开始这个任务需要的最低能量。要求输出能完成所有任务的最少初始能量。
+- 这一题直觉是贪心。先将任务按照 `minimum - actual` 进行排序。先完成差值大的任务，那么接下来的能量能最大限度的满足接下来的任务。这样可能完成所有任务的可能性越大。循环任务数组的时候，保存当前能量在 `cur` 中，如果当前能量不够开启下一个任务，那么这个差值就是需要弥补的，这些能量就是最少初始能量中的，所以加上这些差值能量。如果当前能量可以开启下一个任务，那么就更新当前能量，减去实际消耗的能量以后，再继续循环。循环结束就能得到最少初始能量了。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func minimumEffort(tasks [][]int) int {
+	sort.Sort(Task(tasks))
+	res, cur := 0, 0
+	for _, t := range tasks {
+		if t[1] > cur {
+			res += t[1] - cur
+			cur = t[1] - t[0]
+		} else {
+			cur -= t[0]
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// Task define
+type Task [][]int
+
+func (task Task) Len() int {
+	return len(task)
+}
+
+func (task Task) Less(i, j int) bool {
+	t1, t2 := task[i][1]-task[i][0], task[j][1]-task[j][0]
+	if t1 != t2 {
+		return t2 < t1
+	}
+	return task[j][1] < task[i][1]
+}
+
+func (task Task) Swap(i, j int) {
+	t := task[i]
+	task[i] = task[j]
+	task[j] = t
+}
+```
\ No newline at end of file
diff --git a/leetcode/1668.Maximum-Repeating-Substring/1668. Maximum Repeating Substring.go b/leetcode/1668.Maximum-Repeating-Substring/1668. Maximum Repeating Substring.go
new file mode 100644
index 000000000..1ef762ab5
--- /dev/null
+++ b/leetcode/1668.Maximum-Repeating-Substring/1668. Maximum Repeating Substring.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+import (
+	"strings"
+)
+
+func maxRepeating(sequence string, word string) int {
+	for i := len(sequence) / len(word); i >= 0; i-- {
+		tmp := ""
+		for j := 0; j < i; j++ {
+			tmp += word
+		}
+		if strings.Contains(sequence, tmp) {
+			return i
+		}
+	}
+	return 0
+}
diff --git a/leetcode/1668.Maximum-Repeating-Substring/1668. Maximum Repeating Substring_test.go b/leetcode/1668.Maximum-Repeating-Substring/1668. Maximum Repeating Substring_test.go
new file mode 100644
index 000000000..ee5aadd59
--- /dev/null
+++ b/leetcode/1668.Maximum-Repeating-Substring/1668. Maximum Repeating Substring_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1668 struct {
+	para1668
+	ans1668
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1668 struct {
+	sequence string
+	word     string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1668 struct {
+	one int
+}
+
+func Test_Problem1668(t *testing.T) {
+
+	qs := []question1668{
+
+		{
+			para1668{"ababc", "ab"},
+			ans1668{2},
+		},
+
+		{
+			para1668{"ababc", "ba"},
+			ans1668{1},
+		},
+
+		{
+			para1668{"ababc", "ac"},
+			ans1668{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1668------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1668, q.para1668
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, maxRepeating(p.sequence, p.word))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1668.Maximum-Repeating-Substring/README.md b/leetcode/1668.Maximum-Repeating-Substring/README.md
new file mode 100644
index 000000000..b25e75de9
--- /dev/null
+++ b/leetcode/1668.Maximum-Repeating-Substring/README.md
@@ -0,0 +1,69 @@
+# [1668. Maximum Repeating Substring](https://leetcode.com/problems/maximum-repeating-substring/)
+
+
+## 题目
+
+For a string `sequence`, a string `word` is **`k`-repeating** if `word` concatenated `k` times is a substring of `sequence`. The `word`'s **maximum `k`-repeating value** is the highest value `k` where `word` is `k`-repeating in `sequence`. If `word` is not a substring of `sequence`, `word`'s maximum `k`-repeating value is `0`.
+
+Given strings `sequence` and `word`, return *the **maximum `k`-repeating value** of `word` in `sequence`*.
+
+**Example 1:**
+
+```
+Input: sequence = "ababc", word = "ab"
+Output: 2
+Explanation: "abab" is a substring in "ababc".
+```
+
+**Example 2:**
+
+```
+Input: sequence = "ababc", word = "ba"
+Output: 1
+Explanation: "ba" is a substring in "ababc". "baba" is not a substring in "ababc".
+```
+
+**Example 3:**
+
+```
+Input: sequence = "ababc", word = "ac"
+Output: 0
+Explanation: "ac" is not a substring in "ababc". 
+```
+
+**Constraints:**
+
+- `1 <= sequence.length <= 100`
+- `1 <= word.length <= 100`
+- `sequence` and `word` contains only lowercase English letters.
+
+## 题目大意
+
+给你一个字符串 sequence ，如果字符串 word 连续重复 k 次形成的字符串是 sequence 的一个子字符串，那么单词 word 的 重复值为 k 。单词 word 的 最大重复值 是单词 word 在 sequence 中最大的重复值。如果 word 不是 sequence 的子串，那么重复值 k 为 0 。给你一个字符串 sequence 和 word ，请你返回 最大重复值 k 。
+
+## 解题思路
+
+- 循环叠加构造 `word`，每次构造出新的 `word` 都在 `sequence` 查找一次，如果找到就输出叠加次数，否则继续叠加构造，直到字符串长度和 `sequence` 一样长，最终都没有找到则输出 0 。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"strings"
+)
+
+func maxRepeating(sequence string, word string) int {
+	for i := len(sequence) / len(word); i >= 0; i-- {
+		tmp := ""
+		for j := 0; j < i; j++ {
+			tmp += word
+		}
+		if strings.Contains(sequence, tmp) {
+			return i
+		}
+	}
+	return 0
+}
+```
\ No newline at end of file
diff --git a/leetcode/1669.Merge-In-Between-Linked-Lists/1669. Merge In Between Linked Lists.go b/leetcode/1669.Merge-In-Between-Linked-Lists/1669. Merge In Between Linked Lists.go
new file mode 100644
index 000000000..9536270e5
--- /dev/null
+++ b/leetcode/1669.Merge-In-Between-Linked-Lists/1669. Merge In Between Linked Lists.go	
@@ -0,0 +1,37 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func mergeInBetween(list1 *ListNode, a int, b int, list2 *ListNode) *ListNode {
+	n := list1
+	var startRef, endRef *ListNode
+	for i := 0; i <= b; i++ {
+		if i == a-1 {
+			startRef = n
+		}
+		if i == b {
+			endRef = n
+		}
+		n = n.Next
+	}
+	startRef.Next = list2
+	n = list2
+	for n.Next != nil {
+		n = n.Next
+	}
+	n.Next = endRef.Next
+	return list1
+}
diff --git a/leetcode/1669.Merge-In-Between-Linked-Lists/1669. Merge In Between Linked Lists_test.go b/leetcode/1669.Merge-In-Between-Linked-Lists/1669. Merge In Between Linked Lists_test.go
new file mode 100644
index 000000000..b7d63a0e8
--- /dev/null
+++ b/leetcode/1669.Merge-In-Between-Linked-Lists/1669. Merge In Between Linked Lists_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question1669 struct {
+	para1669
+	ans1669
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1669 struct {
+	one     []int
+	a       int
+	b       int
+	another []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1669 struct {
+	one []int
+}
+
+func Test_Problem1669(t *testing.T) {
+
+	qs := []question1669{
+
+		{
+			para1669{[]int{0, 1, 2, 3, 4, 5}, 3, 4, []int{1000000, 1000001, 1000002}},
+			ans1669{[]int{0, 1, 2, 1000000, 1000001, 1000002, 5}},
+		},
+
+		{
+			para1669{[]int{0, 1, 2, 3, 4, 5, 6}, 2, 5, []int{1000000, 1000001, 1000002, 1000003, 1000004}},
+			ans1669{[]int{0, 1, 1000000, 1000001, 1000002, 1000003, 1000004, 6}},
+		},
+
+		{
+			para1669{[]int{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, 3, 5, []int{1000000, 1000001, 1000002, 1000003, 1000004, 1000005, 1000006}},
+			ans1669{[]int{0, 1, 2, 1000000, 1000001, 1000002, 1000003, 1000004, 1000005, 1000006, 6, 7, 8, 9}},
+		},
+
+		{
+			para1669{[]int{0, 1, 2}, 1, 1, []int{1000000, 1000001, 1000002, 1000003}},
+			ans1669{[]int{0, 1000000, 1000001, 1000002, 1000003, 2}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1669------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1669, q.para1669
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(mergeInBetween(structures.Ints2List(p.one), p.a, p.b, structures.Ints2List(p.another))))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1669.Merge-In-Between-Linked-Lists/README.md b/leetcode/1669.Merge-In-Between-Linked-Lists/README.md
new file mode 100644
index 000000000..3b221b0f5
--- /dev/null
+++ b/leetcode/1669.Merge-In-Between-Linked-Lists/README.md
@@ -0,0 +1,75 @@
+# [1669. Merge In Between Linked Lists](https://leetcode.com/problems/merge-in-between-linked-lists/)
+
+
+## 题目
+
+You are given two linked lists: `list1` and `list2` of sizes `n` and `m` respectively.
+
+Remove `list1`'s nodes from the `ath` node to the `bth` node, and put `list2` in their place.
+
+The blue edges and nodes in the following figure incidate the result:
+
+![https://assets.leetcode.com/uploads/2020/11/05/fig1.png](https://assets.leetcode.com/uploads/2020/11/05/fig1.png)
+
+*Build the result list and return its head.*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/11/05/merge_linked_list_ex1.png](https://assets.leetcode.com/uploads/2020/11/05/merge_linked_list_ex1.png)
+
+```
+Input: list1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
+Output: [0,1,2,1000000,1000001,1000002,5]
+Explanation: We remove the nodes 3 and 4 and put the entire list2 in their place. The blue edges and nodes in the above figure indicate the result.
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/11/05/merge_linked_list_ex2.png](https://assets.leetcode.com/uploads/2020/11/05/merge_linked_list_ex2.png)
+
+```
+Input: list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
+Output: [0,1,1000000,1000001,1000002,1000003,1000004,6]
+Explanation: The blue edges and nodes in the above figure indicate the result.
+
+```
+
+**Constraints:**
+
+- `3 <= list1.length <= 104`
+- `1 <= a <= b < list1.length - 1`
+- `1 <= list2.length <= 104`
+
+## 题目大意
+
+给你两个链表 list1 和 list2 ，它们包含的元素分别为 n 个和 m 个。请你将 list1 中第 a 个节点到第 b 个节点删除，并将list2 接在被删除节点的位置。
+
+## 解题思路
+
+- 简单题，考查链表的基本操作。此题注意 a == b 的情况。
+
+## 代码
+
+```go
+func mergeInBetween(list1 *ListNode, a int, b int, list2 *ListNode) *ListNode {
+	n := list1
+	var startRef, endRef *ListNode
+	for i := 0; i <= b; i++ {
+		if i == a-1 {
+			startRef = n
+		}
+		if i == b {
+			endRef = n
+		}
+		n = n.Next
+	}
+	startRef.Next = list2
+	n = list2
+	for n.Next != nil {
+		n = n.Next
+	}
+	n.Next = endRef.Next
+	return list1
+}
+```
\ No newline at end of file
diff --git a/leetcode/1670.Design-Front-Middle-Back-Queue/1670. Design Front Middle Back Queue.go b/leetcode/1670.Design-Front-Middle-Back-Queue/1670. Design Front Middle Back Queue.go
new file mode 100644
index 000000000..63e8326b1
--- /dev/null
+++ b/leetcode/1670.Design-Front-Middle-Back-Queue/1670. Design Front Middle Back Queue.go	
@@ -0,0 +1,94 @@
+package leetcode
+
+import (
+	"container/list"
+)
+
+type FrontMiddleBackQueue struct {
+	list   *list.List
+	middle *list.Element
+}
+
+func Constructor() FrontMiddleBackQueue {
+	return FrontMiddleBackQueue{list: list.New()}
+}
+
+func (this *FrontMiddleBackQueue) PushFront(val int) {
+	e := this.list.PushFront(val)
+	if this.middle == nil {
+		this.middle = e
+	} else if this.list.Len()%2 == 0 && this.middle.Prev() != nil {
+		this.middle = this.middle.Prev()
+	}
+}
+
+func (this *FrontMiddleBackQueue) PushMiddle(val int) {
+	if this.middle == nil {
+		this.PushFront(val)
+	} else {
+		if this.list.Len()%2 != 0 {
+			this.middle = this.list.InsertBefore(val, this.middle)
+		} else {
+			this.middle = this.list.InsertAfter(val, this.middle)
+		}
+	}
+}
+
+func (this *FrontMiddleBackQueue) PushBack(val int) {
+	e := this.list.PushBack(val)
+	if this.middle == nil {
+		this.middle = e
+	} else if this.list.Len()%2 != 0 && this.middle.Next() != nil {
+		this.middle = this.middle.Next()
+	}
+}
+
+func (this *FrontMiddleBackQueue) PopFront() int {
+	if this.list.Len() == 0 {
+		return -1
+	}
+	e := this.list.Front()
+	if this.list.Len() == 1 {
+		this.middle = nil
+	} else if this.list.Len()%2 == 0 && this.middle.Next() != nil {
+		this.middle = this.middle.Next()
+	}
+	return this.list.Remove(e).(int)
+}
+
+func (this *FrontMiddleBackQueue) PopMiddle() int {
+	if this.middle == nil {
+		return -1
+	}
+	e := this.middle
+	if this.list.Len()%2 != 0 {
+		this.middle = e.Prev()
+	} else {
+		this.middle = e.Next()
+	}
+	return this.list.Remove(e).(int)
+}
+
+func (this *FrontMiddleBackQueue) PopBack() int {
+	if this.list.Len() == 0 {
+		return -1
+	}
+	e := this.list.Back()
+	if this.list.Len() == 1 {
+		this.middle = nil
+	} else if this.list.Len()%2 != 0 && this.middle.Prev() != nil {
+		this.middle = this.middle.Prev()
+	}
+	return this.list.Remove(e).(int)
+}
+
+/**
+ * Your FrontMiddleBackQueue object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.PushFront(val);
+ * obj.PushMiddle(val);
+ * obj.PushBack(val);
+ * param_4 := obj.PopFront();
+ * param_5 := obj.PopMiddle();
+ * param_6 := obj.PopBack();
+ */
diff --git a/leetcode/1670.Design-Front-Middle-Back-Queue/1670. Design Front Middle Back Queue_test.go b/leetcode/1670.Design-Front-Middle-Back-Queue/1670. Design Front Middle Back Queue_test.go
new file mode 100644
index 000000000..e2b30c0e2
--- /dev/null
+++ b/leetcode/1670.Design-Front-Middle-Back-Queue/1670. Design Front Middle Back Queue_test.go	
@@ -0,0 +1,112 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem1670(t *testing.T) {
+	obj := Constructor()
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	obj.PushFront(1)
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	obj.PushBack(2)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.PushMiddle(3)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.PushMiddle(4)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	param1 := obj.PopFront()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopMiddle()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopMiddle()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopBack()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopFront()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	fmt.Printf("-----------------------------------------------------------------\n")
+	obj = Constructor()
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	obj.PushFront(1)
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	obj.PushFront(2)
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	obj.PushFront(3)
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	obj.PushFront(4)
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	param1 = obj.PopBack()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopBack()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopBack()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopBack()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	fmt.Printf("-----------------------------------------------------------------\n")
+	obj = Constructor()
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	obj.PushMiddle(1)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.PushMiddle(2)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.PushMiddle(3)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	param1 = obj.PopMiddle()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopMiddle()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopMiddle()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	fmt.Printf("-----------------------------------------------------------------\n")
+	obj = Constructor()
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	obj.PushMiddle(8)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	param1 = obj.PopMiddle()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopFront()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopBack()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	param1 = obj.PopMiddle()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	obj.PushMiddle(1)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	obj.PushMiddle(10)
+	fmt.Printf("obj = %v\n", MList2Ints(&obj))
+	fmt.Printf("-----------------------------------------------------------------\n")
+	obj = Constructor()
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	param1 = obj.PopMiddle()
+	fmt.Printf("param1 = %v obj = %v\n", param1, MList2Ints(&obj))
+	obj.PushMiddle(3)
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	obj.PushFront(6)
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	obj.PushMiddle(6)
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	obj.PushMiddle(3)
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	param1 = obj.PopMiddle()
+	fmt.Printf("param1 = %v obj = %v %v\n", param1, MList2Ints(&obj), obj)
+	obj.PushMiddle(7)
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	param1 = obj.PopMiddle()
+	fmt.Printf("param1 = %v obj = %v %v\n", param1, MList2Ints(&obj), obj)
+	obj.PushMiddle(8)
+	fmt.Printf("obj = %v %v\n", MList2Ints(&obj), obj)
+	// 	["FrontMiddleBackQueue","popMiddle","pushMiddle","pushFront","pushMiddle","pushMiddle","popMiddle","pushMiddle","popMiddle","pushMiddle"]
+	// [[],[],[3],[6],[6],[3],[],[7],[],[8]]
+}
+
+func MList2Ints(this *FrontMiddleBackQueue) []int {
+	array := []int{}
+	for e := this.list.Front(); e != nil; e = e.Next() {
+		value, _ := e.Value.(int)
+		array = append(array, value)
+	}
+	return array
+}
diff --git a/leetcode/1670.Design-Front-Middle-Back-Queue/README.md b/leetcode/1670.Design-Front-Middle-Back-Queue/README.md
new file mode 100644
index 000000000..2046abf8a
--- /dev/null
+++ b/leetcode/1670.Design-Front-Middle-Back-Queue/README.md
@@ -0,0 +1,172 @@
+# [1670. Design Front Middle Back Queue](https://leetcode.com/problems/design-front-middle-back-queue/)
+
+
+## 题目
+
+Design a queue that supports `push` and `pop` operations in the front, middle, and back.
+
+Implement the `FrontMiddleBack` class:
+
+- `FrontMiddleBack()` Initializes the queue.
+- `void pushFront(int val)` Adds `val` to the **front** of the queue.
+- `void pushMiddle(int val)` Adds `val` to the **middle** of the queue.
+- `void pushBack(int val)` Adds `val` to the **back** of the queue.
+- `int popFront()` Removes the **front** element of the queue and returns it. If the queue is empty, return `1`.
+- `int popMiddle()` Removes the **middle** element of the queue and returns it. If the queue is empty, return `1`.
+- `int popBack()` Removes the **back** element of the queue and returns it. If the queue is empty, return `1`.
+
+**Notice** that when there are **two** middle position choices, the operation is performed on the **frontmost** middle position choice. For example:
+
+- Pushing `6` into the middle of `[1, 2, 3, 4, 5]` results in `[1, 2, 6, 3, 4, 5]`.
+- Popping the middle from `[1, 2, 3, 4, 5, 6]` returns `3` and results in `[1, 2, 4, 5, 6]`.
+
+**Example 1:**
+
+```
+Input:
+["FrontMiddleBackQueue", "pushFront", "pushBack", "pushMiddle", "pushMiddle", "popFront", "popMiddle", "popMiddle", "popBack", "popFront"]
+[[], [1], [2], [3], [4], [], [], [], [], []]
+Output:
+[null, null, null, null, null, 1, 3, 4, 2, -1]
+
+Explanation:
+FrontMiddleBackQueue q = new FrontMiddleBackQueue();
+q.pushFront(1);   // [1]
+q.pushBack(2);    // [1, 2]
+q.pushMiddle(3);  // [1, 3, 2]
+q.pushMiddle(4);  // [1, 4, 3, 2]
+q.popFront();     // return 1 -> [4, 3, 2]
+q.popMiddle();    // return 3 -> [4, 2]
+q.popMiddle();    // return 4 -> [2]
+q.popBack();      // return 2 -> []
+q.popFront();     // return -1 -> [] (The queue is empty)
+
+```
+
+**Constraints:**
+
+- `1 <= val <= 109`
+- At most `1000` calls will be made to `pushFront`, `pushMiddle`, `pushBack`, `popFront`, `popMiddle`, and `popBack`.
+
+## 题目大意
+
+请你设计一个队列，支持在前，中，后三个位置的 push 和 pop 操作。
+
+请你完成 FrontMiddleBack 类：
+
+- FrontMiddleBack() 初始化队列。
+- void pushFront(int val) 将 val 添加到队列的 最前面 。
+- void pushMiddle(int val) 将 val 添加到队列的 正中间 。
+- void pushBack(int val) 将 val 添加到队里的 最后面 。
+- int popFront() 将 最前面 的元素从队列中删除并返回值，如果删除之前队列为空，那么返回 -1 。
+- int popMiddle() 将 正中间 的元素从队列中删除并返回值，如果删除之前队列为空，那么返回 -1 。
+- int popBack() 将 最后面 的元素从队列中删除并返回值，如果删除之前队列为空，那么返回 -1 。
+
+请注意当有 两个 中间位置的时候，选择靠前面的位置进行操作。比方说：
+
+- 将 6 添加到 [1, 2, 3, 4, 5] 的中间位置，结果数组为 [1, 2, 6, 3, 4, 5] 。
+- 从 [1, 2, 3, 4, 5, 6] 的中间位置弹出元素，返回 3 ，数组变为 [1, 2, 4, 5, 6] 。
+
+## 解题思路
+
+- 简单题，利用 go 原生的双向队列 list 的实现，可以轻松实现这个“前中后队列”。
+- 具体实现见代码，几组特殊测试用例见测试文件。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/list"
+)
+
+type FrontMiddleBackQueue struct {
+	list   *list.List
+	middle *list.Element
+}
+
+func Constructor() FrontMiddleBackQueue {
+	return FrontMiddleBackQueue{list: list.New()}
+}
+
+func (this *FrontMiddleBackQueue) PushFront(val int) {
+	e := this.list.PushFront(val)
+	if this.middle == nil {
+		this.middle = e
+	} else if this.list.Len()%2 == 0 && this.middle.Prev() != nil {
+		this.middle = this.middle.Prev()
+	}
+}
+
+func (this *FrontMiddleBackQueue) PushMiddle(val int) {
+	if this.middle == nil {
+		this.PushFront(val)
+	} else {
+		if this.list.Len()%2 != 0 {
+			this.middle = this.list.InsertBefore(val, this.middle)
+		} else {
+			this.middle = this.list.InsertAfter(val, this.middle)
+		}
+	}
+}
+
+func (this *FrontMiddleBackQueue) PushBack(val int) {
+	e := this.list.PushBack(val)
+	if this.middle == nil {
+		this.middle = e
+	} else if this.list.Len()%2 != 0 && this.middle.Next() != nil {
+		this.middle = this.middle.Next()
+	}
+}
+
+func (this *FrontMiddleBackQueue) PopFront() int {
+	if this.list.Len() == 0 {
+		return -1
+	}
+	e := this.list.Front()
+	if this.list.Len() == 1 {
+		this.middle = nil
+	} else if this.list.Len()%2 == 0 && this.middle.Next() != nil {
+		this.middle = this.middle.Next()
+	}
+	return this.list.Remove(e).(int)
+}
+
+func (this *FrontMiddleBackQueue) PopMiddle() int {
+	if this.middle == nil {
+		return -1
+	}
+	e := this.middle
+	if this.list.Len()%2 != 0 {
+		this.middle = e.Prev()
+	} else {
+		this.middle = e.Next()
+	}
+	return this.list.Remove(e).(int)
+}
+
+func (this *FrontMiddleBackQueue) PopBack() int {
+	if this.list.Len() == 0 {
+		return -1
+	}
+	e := this.list.Back()
+	if this.list.Len() == 1 {
+		this.middle = nil
+	} else if this.list.Len()%2 != 0 && this.middle.Prev() != nil {
+		this.middle = this.middle.Prev()
+	}
+	return this.list.Remove(e).(int)
+}
+
+/**
+ * Your FrontMiddleBackQueue object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.PushFront(val);
+ * obj.PushMiddle(val);
+ * obj.PushBack(val);
+ * param_4 := obj.PopFront();
+ * param_5 := obj.PopMiddle();
+ * param_6 := obj.PopBack();
+ */
+```
\ No newline at end of file
diff --git a/leetcode/1672.Richest-Customer-Wealth/1672. Richest Customer Wealth.go b/leetcode/1672.Richest-Customer-Wealth/1672. Richest Customer Wealth.go
new file mode 100644
index 000000000..3a851ad99
--- /dev/null
+++ b/leetcode/1672.Richest-Customer-Wealth/1672. Richest Customer Wealth.go	
@@ -0,0 +1,15 @@
+package leetcode
+
+func maximumWealth(accounts [][]int) int {
+	res := 0
+	for _, banks := range accounts {
+		sAmount := 0
+		for _, amount := range banks {
+			sAmount += amount
+		}
+		if sAmount > res {
+			res = sAmount
+		}
+	}
+	return res
+}
diff --git a/leetcode/1672.Richest-Customer-Wealth/1672. Richest Customer Wealth_test.go b/leetcode/1672.Richest-Customer-Wealth/1672. Richest Customer Wealth_test.go
new file mode 100644
index 000000000..7c6e689fe
--- /dev/null
+++ b/leetcode/1672.Richest-Customer-Wealth/1672. Richest Customer Wealth_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1672 struct {
+	para1672
+	ans1672
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1672 struct {
+	accounts [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1672 struct {
+	one int
+}
+
+func Test_Problem1672(t *testing.T) {
+
+	qs := []question1672{
+
+		{
+			para1672{[][]int{{1, 2, 3}, {3, 2, 1}}},
+			ans1672{6},
+		},
+
+		{
+			para1672{[][]int{{1, 5}, {7, 3}, {3, 5}}},
+			ans1672{10},
+		},
+
+		{
+			para1672{[][]int{{2, 8, 7}, {7, 1, 3}, {1, 9, 5}}},
+			ans1672{17},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1672------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1672, q.para1672
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maximumWealth(p.accounts))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1672.Richest-Customer-Wealth/README.md b/leetcode/1672.Richest-Customer-Wealth/README.md
new file mode 100644
index 000000000..250cf4805
--- /dev/null
+++ b/leetcode/1672.Richest-Customer-Wealth/README.md
@@ -0,0 +1,72 @@
+# [1672. Richest Customer Wealth](https://leetcode.com/problems/richest-customer-wealth/)
+
+
+## 题目
+
+You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return *the **wealth** that the richest customer has.*
+
+A customer's **wealth** is the amount of money they have in all their bank accounts. The richest customer is the customer that has the maximum **wealth**.
+
+**Example 1:**
+
+```
+Input: accounts = [[1,2,3],[3,2,1]]
+Output: 6
+Explanation:1st customer has wealth = 1 + 2 + 3 = 6
+2nd customer has wealth = 3 + 2 + 1 = 6
+Both customers are considered the richest with a wealth of 6 each, so return 6.
+```
+
+**Example 2:**
+
+```
+Input: accounts = [[1,5],[7,3],[3,5]]
+Output: 10
+Explanation: 
+1st customer has wealth = 6
+2nd customer has wealth = 10 
+3rd customer has wealth = 8
+The 2nd customer is the richest with a wealth of 10.
+```
+
+**Example 3:**
+
+```
+Input: accounts = [[2,8,7],[7,1,3],[1,9,5]]
+Output: 17
+```
+
+**Constraints:**
+
+- `m == accounts.length`
+- `n == accounts[i].length`
+- `1 <= m, n <= 50`
+- `1 <= accounts[i][j] <= 100`
+
+## 题目大意
+
+给你一个 m x n 的整数网格 accounts ，其中 accounts[i][j] 是第 i 位客户在第 j 家银行托管的资产数量。返回最富有客户所拥有的 资产总量 。客户的 资产总量 就是他们在各家银行托管的资产数量之和。最富有客户就是 资产总量 最大的客户。
+
+## 解题思路
+
+- 简单题。计算二维数组中每个一位数组的元素总和，然后动态维护这些一位数组和的最大值即可。
+
+## 代码
+
+```go
+package leetcode
+
+func maximumWealth(accounts [][]int) int {
+	res := 0
+	for _, banks := range accounts {
+		sAmount := 0
+		for _, amount := range banks {
+			sAmount += amount
+		}
+		if sAmount > res {
+			res = sAmount
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1673.Find-the-Most-Competitive-Subsequence/1673. Find the Most Competitive Subsequence.go b/leetcode/1673.Find-the-Most-Competitive-Subsequence/1673. Find the Most Competitive Subsequence.go
new file mode 100644
index 000000000..04f8fb522
--- /dev/null
+++ b/leetcode/1673.Find-the-Most-Competitive-Subsequence/1673. Find the Most Competitive Subsequence.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+// 单调栈
+func mostCompetitive(nums []int, k int) []int {
+	stack := make([]int, 0, len(nums))
+	for i := 0; i < len(nums); i++ {
+		for len(stack)+len(nums)-i > k && len(stack) > 0 && nums[i] < stack[len(stack)-1] {
+			stack = stack[:len(stack)-1]
+		}
+		stack = append(stack, nums[i])
+	}
+	return stack[:k]
+}
diff --git a/leetcode/1673.Find-the-Most-Competitive-Subsequence/1673. Find the Most Competitive Subsequence_test.go b/leetcode/1673.Find-the-Most-Competitive-Subsequence/1673. Find the Most Competitive Subsequence_test.go
new file mode 100644
index 000000000..33f2724d3
--- /dev/null
+++ b/leetcode/1673.Find-the-Most-Competitive-Subsequence/1673. Find the Most Competitive Subsequence_test.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1673 struct {
+	para1673
+	ans1673
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1673 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1673 struct {
+	one []int
+}
+
+func Test_Problem1673(t *testing.T) {
+
+	qs := []question1673{
+
+		{
+			para1673{[]int{3, 5, 2, 6}, 2},
+			ans1673{[]int{2, 6}},
+		},
+
+		{
+			para1673{[]int{2, 4, 3, 3, 5, 4, 9, 6}, 4},
+			ans1673{[]int{2, 3, 3, 4}},
+		},
+
+		{
+			para1673{[]int{2, 4, 3, 3, 5, 4, 9, 6}, 4},
+			ans1673{[]int{2, 3, 3, 4}},
+		},
+
+		{
+			para1673{[]int{71, 18, 52, 29, 55, 73, 24, 42, 66, 8, 80, 2}, 3},
+			ans1673{[]int{8, 80, 2}},
+		},
+
+		{
+			para1673{[]int{84, 10, 71, 23, 66, 61, 62, 64, 34, 41, 80, 25, 91, 43, 4, 75, 65, 13, 37, 41, 46, 90, 55, 8, 85, 61, 95, 71}, 24},
+			ans1673{[]int{10, 23, 61, 62, 34, 41, 80, 25, 91, 43, 4, 75, 65, 13, 37, 41, 46, 90, 55, 8, 85, 61, 95, 71}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1673------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1673, q.para1673
+		fmt.Printf("【input】:%v       【output】:%v\n", p, mostCompetitive(p.nums, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1673.Find-the-Most-Competitive-Subsequence/README.md b/leetcode/1673.Find-the-Most-Competitive-Subsequence/README.md
new file mode 100644
index 000000000..74ced3985
--- /dev/null
+++ b/leetcode/1673.Find-the-Most-Competitive-Subsequence/README.md
@@ -0,0 +1,62 @@
+# [1673. Find the Most Competitive Subsequence](https://leetcode.com/problems/find-the-most-competitive-subsequence/)
+
+
+## 题目
+
+Given an integer array `nums` and a positive integer `k`, return *the most **competitive** subsequence of* `nums` *of size* `k`.
+
+An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.
+
+We define that a subsequence `a` is more **competitive** than a subsequence `b` (of the same length) if in the first position where `a` and `b` differ, subsequence `a` has a number **less** than the corresponding number in `b`. For example, `[1,3,4]` is more competitive than `[1,3,5]` because the first position they differ is at the final number, and `4` is less than `5`.
+
+**Example 1:**
+
+```
+Input: nums = [3,5,2,6], k = 2
+Output: [2,6]
+Explanation: Among the set of every possible subsequence: {[3,5], [3,2], [3,6], [5,2], [5,6], [2,6]}, [2,6] is the most competitive.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,4,3,3,5,4,9,6], k = 4
+Output: [2,3,3,4]
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 105`
+- `0 <= nums[i] <= 109`
+- `1 <= k <= nums.length`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个正整数 k ，返回长度为 k 且最具 竞争力 的 nums 子序列。数组的子序列是从数组中删除一些元素（可能不删除元素）得到的序列。
+
+在子序列 a 和子序列 b 第一个不相同的位置上，如果 a 中的数字小于 b 中对应的数字，那么我们称子序列 a 比子序列 b（相同长度下）更具 竞争力 。 例如，[1,3,4] 比 [1,3,5] 更具竞争力，在第一个不相同的位置，也就是最后一个位置上， 4 小于 5 。
+
+## 解题思路
+
+- 这一题是单调栈的典型题型。利用单调栈，可以保证原数组中元素相对位置不变，这满足题意中删除元素但不移动元素的要求。单调栈又能保证每次进栈，元素是最小的。
+- 类似的题目还有第 42 题，第 84 题，第 496 题，第 503 题，第 856 题，第 901 题，第 907 题，第 1130 题，第 1425 题，第 1673 题。
+
+## 代码
+
+```go
+package leetcode
+
+// 单调栈
+func mostCompetitive(nums []int, k int) []int {
+	stack := make([]int, 0, len(nums))
+	for i := 0; i < len(nums); i++ {
+		for len(stack)+len(nums)-i > k && len(stack) > 0 && nums[i] < stack[len(stack)-1] {
+			stack = stack[:len(stack)-1]
+		}
+		stack = append(stack, nums[i])
+	}
+	return stack[:k]
+}
+```
\ No newline at end of file
diff --git a/leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/1674. Minimum Moves to Make Array Complementary.go b/leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/1674. Minimum Moves to Make Array Complementary.go
new file mode 100644
index 000000000..cd8ef3b42
--- /dev/null
+++ b/leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/1674. Minimum Moves to Make Array Complementary.go	
@@ -0,0 +1,34 @@
+package leetcode
+
+func minMoves(nums []int, limit int) int {
+	diff := make([]int, limit*2+2) // nums[i] <= limit, b+limit+1 is maximum limit+limit+1
+	for j := 0; j < len(nums)/2; j++ {
+		a, b := min(nums[j], nums[len(nums)-j-1]), max(nums[j], nums[len(nums)-j-1])
+		// using prefix sum: most interesting point, and is the key to reduce complexity
+		diff[2] += 2
+		diff[a+1]--
+		diff[a+b]--
+		diff[a+b+1]++
+		diff[b+limit+1]++
+	}
+	cur, res := 0, len(nums)
+	for i := 2; i <= 2*limit; i++ {
+		cur += diff[i]
+		res = min(res, cur)
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/1674. Minimum Moves to Make Array Complementary_test.go b/leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/1674. Minimum Moves to Make Array Complementary_test.go
new file mode 100644
index 000000000..c111bf145
--- /dev/null
+++ b/leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/1674. Minimum Moves to Make Array Complementary_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1674 struct {
+	para1674
+	ans1674
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1674 struct {
+	nums  []int
+	limit int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1674 struct {
+	one int
+}
+
+func Test_Problem1674(t *testing.T) {
+
+	qs := []question1674{
+
+		{
+			para1674{[]int{1, 2, 4, 3}, 4},
+			ans1674{1},
+		},
+
+		{
+			para1674{[]int{1, 2, 2, 1}, 2},
+			ans1674{2},
+		},
+
+		{
+			para1674{[]int{1, 2, 1, 2}, 2},
+			ans1674{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1674------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1674, q.para1674
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minMoves(p.nums, p.limit))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/README.md b/leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/README.md
new file mode 100644
index 000000000..8e565e10a
--- /dev/null
+++ b/leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/README.md
@@ -0,0 +1,97 @@
+# [1674. Minimum Moves to Make Array Complementary](https://leetcode.com/problems/minimum-moves-to-make-array-complementary/)
+
+## 题目
+
+You are given an integer array `nums` of **even** length `n` and an integer `limit`. In one move, you can replace any integer from `nums` with another integer between `1` and `limit`, inclusive.
+
+The array `nums` is **complementary** if for all indices `i` (**0-indexed**), `nums[i] + nums[n - 1 - i]` equals the same number. For example, the array `[1,2,3,4]` is complementary because for all indices `i`, `nums[i] + nums[n - 1 - i] = 5`.
+
+Return the ***minimum** number of moves required to make* `nums` ***complementary***.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,4,3], limit = 4
+Output: 1
+Explanation: In 1 move, you can change nums to [1,2,2,3] (underlined elements are changed).
+nums[0] + nums[3] = 1 + 3 = 4.
+nums[1] + nums[2] = 2 + 2 = 4.
+nums[2] + nums[1] = 2 + 2 = 4.
+nums[3] + nums[0] = 3 + 1 = 4.
+Therefore, nums[i] + nums[n-1-i] = 4 for every i, so nums is complementary.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,2,1], limit = 2
+Output: 2
+Explanation: In 2 moves, you can change nums to [2,2,2,2]. You cannot change any number to 3 since 3 > limit.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,2,1,2], limit = 2
+Output: 0
+Explanation: nums is already complementary.
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `2 <= n <= 105`
+- `1 <= nums[i] <= limit <= 105`
+- `n` is even.
+
+## 题目大意
+
+给你一个长度为 偶数 n 的整数数组 nums 和一个整数 limit 。每一次操作，你可以将 nums 中的任何整数替换为 1 到 limit 之间的另一个整数。
+
+如果对于所有下标 i（下标从 0 开始），nums[i] + nums[n - 1 - i] 都等于同一个数，则数组 nums 是 互补的 。例如，数组 [1,2,3,4] 是互补的，因为对于所有下标 i ，nums[i] + nums[n - 1 - i] = 5 。
+
+返回使数组 互补 的 最少 操作次数。
+
+## 解题思路
+
+- 这一题考察的是差分数组。通过分析题意，可以得出，针对每一个 `sum` 的取值范围是 `[2, 2* limt]`，定义 `a = min(nums[i], nums[n - i - 1])`，`b = max(nums[i], nums[n - i - 1])`，在这个区间内，又可以细分成 5 个区间，`[2, a + 1)`，`[a + 1, a + b)`，`[a + b + 1, a + b + 1)`，`[a + b + 1, b + limit + 1)`，`[b + limit + 1, 2 * limit)`，在这 5 个区间内使得数组互补的最小操作次数分别是 `2(减少 a, 减少 b)`，`1(减少 b)`，`0(不用操作)`，`1(增大 a)`，`+2(增大 a, 增大 b)`，换个表达方式，按照扫描线从左往右扫描，在这 5 个区间内使得数组互补的最小操作次数叠加变化分别是 `+2(减少 a, 减少 b)`，`-1(减少 a)`，`-1(不用操作)`，`+1(增大 a)`，`+1(增大 a, 增大 b)`，利用这前后两个区间的关系，就可以构造一个差分数组。差分数组反应的是前后两者的关系。如果想求得 0 ~ n 的总关系，只需要求一次前缀和即可。
+- 这道题要求输出最少的操作次数，所以利用差分数组 + 前缀和，累加前缀和的同时维护最小值。从左往右扫描完一遍以后，输出最小值即可。
+
+## 代码
+
+```go
+package leetcode
+
+func minMoves(nums []int, limit int) int {
+	diff := make([]int, limit*2+2) // nums[i] <= limit, b+limit+1 is maximum limit+limit+1
+	for j := 0; j < len(nums)/2; j++ {
+		a, b := min(nums[j], nums[len(nums)-j-1]), max(nums[j], nums[len(nums)-j-1])
+		// using prefix sum: most interesting point, and is the key to reduce complexity
+		diff[2] += 2
+		diff[a+1]--
+		diff[a+b]--
+		diff[a+b+1]++
+		diff[b+limit+1]++
+	}
+	cur, res := 0, len(nums)
+	for i := 2; i <= 2*limit; i++ {
+		cur += diff[i]
+		res = min(res, cur)
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1675.Minimize-Deviation-in-Array/1675. Minimize Deviation in Array.go b/leetcode/1675.Minimize-Deviation-in-Array/1675. Minimize Deviation in Array.go
new file mode 100644
index 000000000..315b9f855
--- /dev/null
+++ b/leetcode/1675.Minimize-Deviation-in-Array/1675. Minimize Deviation in Array.go	
@@ -0,0 +1,40 @@
+package leetcode
+
+func minimumDeviation(nums []int) int {
+	min, max := 0, 0
+	for i := range nums {
+		if nums[i]%2 == 1 {
+			nums[i] *= 2
+		}
+		if i == 0 {
+			min = nums[i]
+			max = nums[i]
+		} else if nums[i] < min {
+			min = nums[i]
+		} else if max < nums[i] {
+			max = nums[i]
+		}
+	}
+	res := max - min
+	for max%2 == 0 {
+		tmax, tmin := 0, 0
+		for i := range nums {
+			if nums[i] == max || (nums[i]%2 == 0 && min <= nums[i]/2) {
+				nums[i] /= 2
+			}
+			if i == 0 {
+				tmin = nums[i]
+				tmax = nums[i]
+			} else if nums[i] < tmin {
+				tmin = nums[i]
+			} else if tmax < nums[i] {
+				tmax = nums[i]
+			}
+		}
+		if tmax-tmin < res {
+			res = tmax - tmin
+		}
+		min, max = tmin, tmax
+	}
+	return res
+}
diff --git a/leetcode/1675.Minimize-Deviation-in-Array/1675. Minimize Deviation in Array_test.go b/leetcode/1675.Minimize-Deviation-in-Array/1675. Minimize Deviation in Array_test.go
new file mode 100644
index 000000000..331acf266
--- /dev/null
+++ b/leetcode/1675.Minimize-Deviation-in-Array/1675. Minimize Deviation in Array_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1675 struct {
+	para1675
+	ans1675
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1675 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1675 struct {
+	one int
+}
+
+func Test_Problem1675(t *testing.T) {
+
+	qs := []question1675{
+
+		{
+			para1675{[]int{1, 2, 4, 3}},
+			ans1675{1},
+		},
+
+		{
+			para1675{[]int{4, 1, 5, 20, 3}},
+			ans1675{3},
+		},
+
+		{
+			para1675{[]int{2, 10, 8}},
+			ans1675{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1675------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1675, q.para1675
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minimumDeviation(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1675.Minimize-Deviation-in-Array/README.md b/leetcode/1675.Minimize-Deviation-in-Array/README.md
new file mode 100644
index 000000000..d9e4336d2
--- /dev/null
+++ b/leetcode/1675.Minimize-Deviation-in-Array/README.md
@@ -0,0 +1,105 @@
+# [1675. Minimize Deviation in Array](https://leetcode.com/problems/minimize-deviation-in-array/)
+
+## 题目
+
+You are given an array `nums` of `n` positive integers.
+
+You can perform two types of operations on any element of the array any number of times:
+
+- If the element is **even**, **divide** it by `2`.
+    - For example, if the array is `[1,2,3,4]`, then you can do this operation on the last element, and the array will be `[1,2,3,2].`
+- If the element is **odd**, **multiply** it by `2`.
+    - For example, if the array is `[1,2,3,4]`, then you can do this operation on the first element, and the array will be `[2,2,3,4].`
+
+The **deviation** of the array is the **maximum difference** between any two elements in the array.
+
+Return *the **minimum deviation** the array can have after performing some number of operations.*
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3,4]
+Output: 1
+Explanation: You can transform the array to [1,2,3,2], then to [2,2,3,2], then the deviation will be 3 - 2 = 1.
+```
+
+**Example 2:**
+
+```
+Input: nums = [4,1,5,20,3]
+Output: 3
+Explanation: You can transform the array after two operations to [4,2,5,5,3], then the deviation will be 5 - 2 = 3.
+```
+
+**Example 3:**
+
+```
+Input: nums = [2,10,8]
+Output: 3
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `2 <= n <= 105`
+- `1 <= nums[i] <= 10^9`
+
+## 题目大意
+
+给你一个由 n 个正整数组成的数组 nums 。你可以对数组的任意元素执行任意次数的两类操作：
+
+- 如果元素是 偶数 ，除以 2。例如，如果数组是 [1,2,3,4] ，那么你可以对最后一个元素执行此操作，使其变成 [1,2,3,2]
+- 如果元素是 奇数 ，乘上 2。例如，如果数组是 [1,2,3,4] ，那么你可以对第一个元素执行此操作，使其变成 [2,2,3,4]
+数组的 偏移量 是数组中任意两个元素之间的 最大差值 。
+
+返回数组在执行某些操作之后可以拥有的 最小偏移量 。
+
+## 解题思路
+
+- 要找到最小偏移量，即需要令最大值变小，最小值变大。要想达到这个要求，需要对奇数偶数做乘法和除法。这里特殊的是，奇数一旦乘以 2 以后，就变成偶数了。偶数除以 2 以后可能是奇数也可能是偶数。所以可以先将所有的奇数都乘以 2 统一变成偶数。
+- 第二轮不断的将最大值除 2，直到最大值为奇数，不能再操作了。每轮循环中把比 min 值大的偶数也都除以 2 。这里除以 2 有 2 个目的，一个目的是将第一步奇数乘 2 还原回去，另一个目的是将本来的偶数除以 2 。可能有人有疑问，为什么只把最大值变小，没有将最小值变大呢？如果最小值是奇数，那么它一定是由上一个偶数除以 2 变过来的，我们在上一个状态已经计算过这个偶数了，因此没必要扩大它；如果最小值是偶数，那么它一定会在某一轮的除 2 操作中，不操作，即它不会满足 `min <= nums[i]/2`  这个条件。每次循环都更新该次循环的最大值和最小值，并记录偏移量。不断的循环，直到最大值为奇数，退出循环。最终输出最小偏移量。
+
+## 代码
+
+```go
+package leetcode
+
+func minimumDeviation(nums []int) int {
+	min, max := 0, 0
+	for i := range nums {
+		if nums[i]%2 == 1 {
+			nums[i] *= 2
+		}
+		if i == 0 {
+			min = nums[i]
+			max = nums[i]
+		} else if nums[i] < min {
+			min = nums[i]
+		} else if max < nums[i] {
+			max = nums[i]
+		}
+	}
+	res := max - min
+	for max%2 == 0 {
+		tmax, tmin := 0, 0
+		for i := range nums {
+			if nums[i] == max || (nums[i]%2 == 0 && min <= nums[i]/2) {
+				nums[i] /= 2
+			}
+			if i == 0 {
+				tmin = nums[i]
+				tmax = nums[i]
+			} else if nums[i] < tmin {
+				tmin = nums[i]
+			} else if tmax < nums[i] {
+				tmax = nums[i]
+			}
+		}
+		if tmax-tmin < res {
+			res = tmax - tmin
+		}
+		min, max = tmin, tmax
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1678.Goal-Parser-Interpretation/1678. Goal Parser Interpretation.go b/leetcode/1678.Goal-Parser-Interpretation/1678. Goal Parser Interpretation.go
new file mode 100644
index 000000000..de1612003
--- /dev/null
+++ b/leetcode/1678.Goal-Parser-Interpretation/1678. Goal Parser Interpretation.go	
@@ -0,0 +1,22 @@
+package leetcode
+
+func interpret(command string) string {
+	if command == "" {
+		return ""
+	}
+	res := ""
+	for i := 0; i < len(command); i++ {
+		if command[i] == 'G' {
+			res += "G"
+		} else {
+			if command[i] == '(' && command[i+1] == 'a' {
+				res += "al"
+				i += 3
+			} else {
+				res += "o"
+				i++
+			}
+		}
+	}
+	return res
+}
diff --git a/leetcode/1678.Goal-Parser-Interpretation/1678. Goal Parser Interpretation_test.go b/leetcode/1678.Goal-Parser-Interpretation/1678. Goal Parser Interpretation_test.go
new file mode 100644
index 000000000..588368aa5
--- /dev/null
+++ b/leetcode/1678.Goal-Parser-Interpretation/1678. Goal Parser Interpretation_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1678 struct {
+	para1678
+	ans1678
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1678 struct {
+	command string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1678 struct {
+	one string
+}
+
+func Test_Problem1678(t *testing.T) {
+
+	qs := []question1678{
+
+		{
+			para1678{"G()(al)"},
+			ans1678{"Goal"},
+		},
+
+		{
+			para1678{"G()()()()(al)"},
+			ans1678{"Gooooal"},
+		},
+
+		{
+			para1678{"(al)G(al)()()G"},
+			ans1678{"alGalooG"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1678------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1678, q.para1678
+		fmt.Printf("【input】:%v       【output】:%v\n", p, interpret(p.command))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1678.Goal-Parser-Interpretation/README.md b/leetcode/1678.Goal-Parser-Interpretation/README.md
new file mode 100644
index 000000000..4f8145f4f
--- /dev/null
+++ b/leetcode/1678.Goal-Parser-Interpretation/README.md
@@ -0,0 +1,73 @@
+# [1678. Goal Parser Interpretation](https://leetcode.com/problems/goal-parser-interpretation/)
+
+## 题目
+
+You own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G"`, `"()"` and/or `"(al)"` in some order. The Goal Parser will interpret `"G"` as the string `"G"`, `"()"` as the string `"o"`, and `"(al)"` as the string `"al"`. The interpreted strings are then concatenated in the original order.
+
+Given the string `command`, return *the **Goal Parser**'s interpretation of* `command`.
+
+**Example 1:**
+
+```
+Input: command = "G()(al)"
+Output: "Goal"
+Explanation: The Goal Parser interprets the command as follows:
+G -> G
+() -> o
+(al) -> al
+The final concatenated result is "Goal".
+```
+
+**Example 2:**
+
+```
+Input: command = "G()()()()(al)"
+Output: "Gooooal"
+```
+
+**Example 3:**
+
+```
+Input: command = "(al)G(al)()()G"
+Output: "alGalooG"
+```
+
+**Constraints:**
+
+- `1 <= command.length <= 100`
+- `command` consists of `"G"`, `"()"`, and/or `"(al)"` in some order.
+
+## 题目大意
+
+请你设计一个可以解释字符串 command 的 Goal 解析器 。command 由 "G"、"()" 和/或 "(al)" 按某种顺序组成。Goal 解析器会将 "G" 解释为字符串 "G"、"()" 解释为字符串 "o" ，"(al)" 解释为字符串 "al" 。然后，按原顺序将经解释得到的字符串连接成一个字符串。给你字符串 command ，返回 Goal 解析器 对 command 的解释结果。
+
+## 解题思路
+
+- 简单题，按照题意修改字符串即可。由于是简单题，这一题也不用考虑嵌套的情况。
+
+## 代码
+
+```go
+package leetcode
+
+func interpret(command string) string {
+	if command == "" {
+		return ""
+	}
+	res := ""
+	for i := 0; i < len(command); i++ {
+		if command[i] == 'G' {
+			res += "G"
+		} else {
+			if command[i] == '(' && command[i+1] == 'a' {
+				res += "al"
+				i += 3
+			} else {
+				res += "o"
+				i ++
+			}
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1679.Max-Number-of-K-Sum-Pairs/1679. Max Number of K-Sum Pairs.go b/leetcode/1679.Max-Number-of-K-Sum-Pairs/1679. Max Number of K-Sum Pairs.go
new file mode 100644
index 000000000..8a601ffac
--- /dev/null
+++ b/leetcode/1679.Max-Number-of-K-Sum-Pairs/1679. Max Number of K-Sum Pairs.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+// 解法一 优化版
+func maxOperations(nums []int, k int) int {
+	counter, res := make(map[int]int), 0
+	for _, n := range nums {
+		counter[n]++
+	}
+	if (k & 1) == 0 {
+		res += counter[k>>1] >> 1
+		// 能够由 2 个相同的数构成 k 的组合已经都排除出去了，剩下的一个单独的也不能组成 k 了
+		// 所以这里要把它的频次置为 0 。如果这里不置为 0，下面代码判断逻辑还需要考虑重复使用数字的情况
+		counter[k>>1] = 0
+	}
+	for num, freq := range counter {
+		if num <= k/2 {
+			remain := k - num
+			if counter[remain] < freq {
+				res += counter[remain]
+			} else {
+				res += freq
+			}
+		}
+	}
+	return res
+}
+
+// 解法二
+func maxOperations_(nums []int, k int) int {
+	counter, res := make(map[int]int), 0
+	for _, num := range nums {
+		counter[num]++
+		remain := k - num
+		if num == remain {
+			if counter[num] >= 2 {
+				res++
+				counter[num] -= 2
+			}
+		} else {
+			if counter[remain] > 0 {
+				res++
+				counter[remain]--
+				counter[num]--
+			}
+		}
+	}
+	return res
+}
diff --git a/leetcode/1679.Max-Number-of-K-Sum-Pairs/1679. Max Number of K-Sum Pairs_test.go b/leetcode/1679.Max-Number-of-K-Sum-Pairs/1679. Max Number of K-Sum Pairs_test.go
new file mode 100644
index 000000000..c9a96b2be
--- /dev/null
+++ b/leetcode/1679.Max-Number-of-K-Sum-Pairs/1679. Max Number of K-Sum Pairs_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1679 struct {
+	para1679
+	ans1679
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1679 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1679 struct {
+	one int
+}
+
+func Test_Problem1679(t *testing.T) {
+
+	qs := []question1679{
+
+		{
+			para1679{[]int{1, 2, 3, 4}, 5},
+			ans1679{2},
+		},
+
+		{
+			para1679{[]int{3, 1, 3, 4, 3}, 6},
+			ans1679{1},
+		},
+
+		{
+			para1679{[]int{2, 5, 4, 4, 1, 3, 4, 4, 1, 4, 4, 1, 2, 1, 2, 2, 3, 2, 4, 2}, 3},
+			ans1679{4},
+		},
+
+		{
+			para1679{[]int{2, 5, 5, 5, 1, 3, 4, 4, 1, 4, 4, 1, 3, 1, 3, 1, 3, 2, 4, 2}, 6},
+			ans1679{8},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1679------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1679, q.para1679
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxOperations(p.nums, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1679.Max-Number-of-K-Sum-Pairs/README.md b/leetcode/1679.Max-Number-of-K-Sum-Pairs/README.md
new file mode 100644
index 000000000..83ed4b311
--- /dev/null
+++ b/leetcode/1679.Max-Number-of-K-Sum-Pairs/README.md
@@ -0,0 +1,98 @@
+# [1679. Max Number of K-Sum Pairs](https://leetcode.com/problems/max-number-of-k-sum-pairs/)
+
+
+## 题目
+
+You are given an integer array `nums` and an integer `k`.
+
+In one operation, you can pick two numbers from the array whose sum equals `k` and remove them from the array.
+
+Return *the maximum number of operations you can perform on the array*.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3,4], k = 5
+Output: 2
+Explanation: Starting with nums = [1,2,3,4]:
+- Remove numbers 1 and 4, then nums = [2,3]
+- Remove numbers 2 and 3, then nums = []
+There are no more pairs that sum up to 5, hence a total of 2 operations.
+```
+
+**Example 2:**
+
+```
+Input: nums = [3,1,3,4,3], k = 6
+Output: 1
+Explanation: Starting with nums = [3,1,3,4,3]:
+- Remove the first two 3's, then nums = [1,4,3]
+There are no more pairs that sum up to 6, hence a total of 1 operation.
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 105`
+- `1 <= nums[i] <= 109`
+- `1 <= k <= 109`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个整数 k 。每一步操作中，你需要从数组中选出和为 k 的两个整数，并将它们移出数组。返回你可以对数组执行的最大操作数。
+
+## 解题思路
+
+- 读完题第一感觉这道题是 TWO SUM 题目的加强版。需要找到所有满足和是 k 的数对。先考虑能不能找到两个数都是 k/2 ，如果能找到多个这样的数，可以先移除他们。其次在利用 TWO SUM 的思路，找出和为 k 的数对。利用 TWO SUM 里面 map 的做法，时间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 优化版
+func maxOperations(nums []int, k int) int {
+	counter, res := make(map[int]int), 0
+	for _, n := range nums {
+		counter[n]++
+	}
+	if (k & 1) == 0 {
+		res += counter[k>>1] >> 1
+		// 能够由 2 个相同的数构成 k 的组合已经都排除出去了，剩下的一个单独的也不能组成 k 了
+		// 所以这里要把它的频次置为 0 。如果这里不置为 0，下面代码判断逻辑还需要考虑重复使用数字的情况
+		counter[k>>1] = 0
+	}
+	for num, freq := range counter {
+		if num <= k/2 {
+			remain := k - num
+			if counter[remain] < freq {
+				res += counter[remain]
+			} else {
+				res += freq
+			}
+		}
+	}
+	return res
+}
+
+// 解法二
+func maxOperations_(nums []int, k int) int {
+	counter, res := make(map[int]int), 0
+	for _, num := range nums {
+		counter[num]++
+		remain := k - num
+		if num == remain {
+			if counter[num] >= 2 {
+				res++
+				counter[num] -= 2
+			}
+		} else {
+			if counter[remain] > 0 {
+				res++
+				counter[remain]--
+				counter[num]--
+			}
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1680.Concatenation-of-Consecutive-Binary-Numbers/1680. Concatenation of Consecutive Binary Numbers.go b/leetcode/1680.Concatenation-of-Consecutive-Binary-Numbers/1680. Concatenation of Consecutive Binary Numbers.go
new file mode 100644
index 000000000..eff4e12b9
--- /dev/null
+++ b/leetcode/1680.Concatenation-of-Consecutive-Binary-Numbers/1680. Concatenation of Consecutive Binary Numbers.go	
@@ -0,0 +1,26 @@
+package leetcode
+
+import (
+	"math/bits"
+)
+
+// 解法一 模拟
+func concatenatedBinary(n int) int {
+	res, mod, shift := 0, 1000000007, 0
+	for i := 1; i <= n; i++ {
+		if (i & (i - 1)) == 0 {
+			shift++
+		}
+		res = ((res << shift) + i) % mod
+	}
+	return res
+}
+
+// 解法二 位运算
+func concatenatedBinary1(n int) int {
+	res := 0
+	for i := 1; i <= n; i++ {
+		res = (res<<bits.Len(uint(i)) | i) % (1e9 + 7)
+	}
+	return res
+}
diff --git a/leetcode/1680.Concatenation-of-Consecutive-Binary-Numbers/1680. Concatenation of Consecutive Binary Numbers_test.go b/leetcode/1680.Concatenation-of-Consecutive-Binary-Numbers/1680. Concatenation of Consecutive Binary Numbers_test.go
new file mode 100644
index 000000000..88cebf974
--- /dev/null
+++ b/leetcode/1680.Concatenation-of-Consecutive-Binary-Numbers/1680. Concatenation of Consecutive Binary Numbers_test.go	
@@ -0,0 +1,72 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1680 struct {
+	para1680
+	ans1680
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1680 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1680 struct {
+	one int
+}
+
+func Test_Problem1680(t *testing.T) {
+
+	qs := []question1680{
+
+		{
+			para1680{1},
+			ans1680{1},
+		},
+
+		{
+			para1680{3},
+			ans1680{27},
+		},
+
+		{
+			para1680{12},
+			ans1680{505379714},
+		},
+
+		{
+			para1680{42},
+			ans1680{727837408},
+		},
+
+		{
+			para1680{24},
+			ans1680{385951001},
+		},
+
+		{
+			para1680{81},
+			ans1680{819357292},
+		},
+
+		{
+			para1680{66},
+			ans1680{627730462},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1680------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1680, q.para1680
+		fmt.Printf("【input】:%v       【output】:%v\n", p, concatenatedBinary(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1680.Concatenation-of-Consecutive-Binary-Numbers/README.md b/leetcode/1680.Concatenation-of-Consecutive-Binary-Numbers/README.md
new file mode 100644
index 000000000..b43a3e664
--- /dev/null
+++ b/leetcode/1680.Concatenation-of-Consecutive-Binary-Numbers/README.md
@@ -0,0 +1,95 @@
+# [1680. Concatenation of Consecutive Binary Numbers](https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/)
+
+
+## 题目
+
+Given an integer `n`, return *the **decimal value** of the binary string formed by concatenating the binary representations of* `1` *to* `n` *in order, **modulo*** `109 + 7`.
+
+**Example 1:**
+
+```
+Input: n = 1
+Output: 1
+Explanation: "1" in binary corresponds to the decimal value 1. 
+```
+
+**Example 2:**
+
+```
+Input: n = 3
+Output: 27
+Explanation: In binary, 1, 2, and 3 corresponds to "1", "10", and "11".
+After concatenating them, we have "11011", which corresponds to the decimal value 27.
+```
+
+**Example 3:**
+
+```
+Input: n = 12
+Output: 505379714
+Explanation: The concatenation results in "1101110010111011110001001101010111100".
+The decimal value of that is 118505380540.
+After modulo 109 + 7, the result is 505379714.
+```
+
+**Constraints:**
+
+- `1 <= n <= 10^5`
+
+## 题目大意
+
+给你一个整数 n ，请你将 1 到 n 的二进制表示连接起来，并返回连接结果对应的 十进制 数字对 10^9 + 7 取余的结果。
+
+## 解题思路
+
+- 理解题意以后，先找到如何拼接最终二进制数的规律。假设 `f(n)` 为最终变换以后的十进制数。那么根据题意，`f(n) = f(n-1) << shift + n` 这是一个递推公式。`shift` 左移的位数就是 `n` 的二进制对应的长度。`shift` 的值是随着 `n` 变化而变化的。由二进制进位规律可以知道，2 的整数次幂的时候，对应的二进制长度会增加 1 位。这里可以利用位运算来判断是否是 2 的整数次幂。
+- 这道题另外一个需要处理的是模运算的法则。此题需要用到模运算的加法法则。
+
+    ```go
+    模运算与基本四则运算有些相似，但是除法例外。
+    (a + b) % p = (a % p + b % p) % p （1）
+    (a - b) % p = (a % p - b % p) % p （2）
+    (a * b) % p = (a % p * b % p) % p （3）
+    a ^ b % p = ((a % p)^b) % p （4）
+    结合律：
+    ((a+b) % p + c) % p = (a + (b+c) % p) % p （5）
+    ((a*b) % p * c)% p = (a * (b*c) % p) % p （6）
+    交换律：
+    (a + b) % p = (b+a) % p （7）
+    (a * b) % p = (b * a) % p （8）
+    分配律：
+    ((a +b)% p * c) % p = ((a * c) % p + (b * c) % p) % p （9）
+    ```
+
+    这一题需要用到模运算的加法运算法则。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math/bits"
+)
+
+// 解法一 模拟
+func concatenatedBinary(n int) int {
+	res, mod, shift := 0, 1000000007, 0
+	for i := 1; i <= n; i++ {
+		if (i & (i - 1)) == 0 {
+			shift++
+		}
+		res = ((res << shift) + i) % mod
+	}
+	return res
+}
+
+// 解法二 位运算
+func concatenatedBinary1(n int) int {
+	res := 0
+	for i := 1; i <= n; i++ {
+		res = (res<<bits.Len(uint(i)) | i) % (1e9 + 7)
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1681.Minimum-Incompatibility/1681. Minimum Incompatibility.go b/leetcode/1681.Minimum-Incompatibility/1681. Minimum Incompatibility.go
new file mode 100644
index 000000000..88b465d26
--- /dev/null
+++ b/leetcode/1681.Minimum-Incompatibility/1681. Minimum Incompatibility.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func minimumIncompatibility(nums []int, k int) int {
+	sort.Ints(nums)
+	eachSize, counts := len(nums)/k, make([]int, len(nums)+1)
+	for i := range nums {
+		counts[nums[i]]++
+		if counts[nums[i]] > k {
+			return -1
+		}
+	}
+	orders := []int{}
+	for i := range counts {
+		orders = append(orders, i)
+	}
+	sort.Ints(orders)
+	res := math.MaxInt32
+	generatePermutation1681(nums, counts, orders, 0, 0, eachSize, &res, []int{})
+	if res == math.MaxInt32 {
+		return -1
+	}
+	return res
+}
+
+func generatePermutation1681(nums, counts, order []int, index, sum, eachSize int, res *int, current []int) {
+	if len(current) > 0 && len(current)%eachSize == 0 {
+		sum += current[len(current)-1] - current[len(current)-eachSize]
+		index = 0
+	}
+	if sum >= *res {
+		return
+	}
+	if len(current) == len(nums) {
+		if sum < *res {
+			*res = sum
+		}
+		return
+	}
+	for i := index; i < len(counts); i++ {
+		if counts[order[i]] == 0 {
+			continue
+		}
+		counts[order[i]]--
+		current = append(current, order[i])
+		generatePermutation1681(nums, counts, order, i+1, sum, eachSize, res, current)
+		current = current[:len(current)-1]
+		counts[order[i]]++
+		// 这里是关键的剪枝
+		if index == 0 {
+			break
+		}
+	}
+}
diff --git a/leetcode/1681.Minimum-Incompatibility/1681. Minimum Incompatibility_test.go b/leetcode/1681.Minimum-Incompatibility/1681. Minimum Incompatibility_test.go
new file mode 100644
index 000000000..903c963b9
--- /dev/null
+++ b/leetcode/1681.Minimum-Incompatibility/1681. Minimum Incompatibility_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1681 struct {
+	para1681
+	ans1681
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1681 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1681 struct {
+	one int
+}
+
+func Test_Problem1681(t *testing.T) {
+
+	qs := []question1681{
+
+		{
+			para1681{[]int{1, 2, 1, 4}, 2},
+			ans1681{4},
+		},
+
+		{
+			para1681{[]int{6, 3, 8, 1, 3, 1, 2, 2}, 4},
+			ans1681{6},
+		},
+
+		{
+			para1681{[]int{5, 3, 3, 6, 3, 3}, 3},
+			ans1681{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1681------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1681, q.para1681
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minimumIncompatibility(p.nums, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1681.Minimum-Incompatibility/README.md b/leetcode/1681.Minimum-Incompatibility/README.md
new file mode 100644
index 000000000..e2f5dc1dc
--- /dev/null
+++ b/leetcode/1681.Minimum-Incompatibility/README.md
@@ -0,0 +1,122 @@
+# [1681. Minimum Incompatibility](https://leetcode.com/problems/minimum-incompatibility/)
+
+
+## 题目
+
+You are given an integer array `nums` and an integer `k`. You are asked to distribute this array into `k` subsets of **equal size** such that there are no two equal elements in the same subset.
+
+A subset's **incompatibility** is the difference between the maximum and minimum elements in that array.
+
+Return *the **minimum possible sum of incompatibilities** of the* `k` *subsets after distributing the array optimally, or return* `-1` *if it is not possible.*
+
+A subset is a group integers that appear in the array with no particular order.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,1,4], k = 2
+Output: 4
+Explanation: The optimal distribution of subsets is [1,2] and [1,4].
+The incompatibility is (2-1) + (4-1) = 4.
+Note that [1,1] and [2,4] would result in a smaller sum, but the first subset contains 2 equal elements.
+```
+
+**Example 2:**
+
+```
+Input: nums = [6,3,8,1,3,1,2,2], k = 4
+Output: 6
+Explanation: The optimal distribution of subsets is [1,2], [2,3], [6,8], and [1,3].
+The incompatibility is (2-1) + (3-2) + (8-6) + (3-1) = 6.
+
+```
+
+**Example 3:**
+
+```
+Input: nums = [5,3,3,6,3,3], k = 3
+Output: -1
+Explanation: It is impossible to distribute nums into 3 subsets where no two elements are equal in the same subset.
+
+```
+
+**Constraints:**
+
+- `1 <= k <= nums.length <= 16`
+- `nums.length` is divisible by `k`
+- `1 <= nums[i] <= nums.length`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个整数 k 。你需要将这个数组划分到 k 个相同大小的子集中，使得同一个子集里面没有两个相同的元素。一个子集的 不兼容性 是该子集里面最大值和最小值的差。
+
+请你返回将数组分成 k 个子集后，各子集 **不兼容性** 的 **和** 的 **最小值** ，如果无法分成分成 k 个子集，返回 -1 。子集的定义是数组中一些数字的集合，对数字顺序没有要求。
+
+## 解题思路
+
+- 读完题最直白的思路就是 DFS。做法类似第 77 题。这里就不赘述了。可以见第 77 题题解。
+- 这一题还需要用到贪心的思想。每次取数都取最小的数。这样可以不会让最大数和最小数在一个集合中。由于每次取数都是取最小的，那么能保证不兼容性每次都尽量最小。于是在 order 数组中定义取数的顺序。然后再把数组从小到大排列。这样每次按照 order 顺序取数，都是取的最小值。
+- 正常的 DFS 写完提交，耗时是很长的。大概是 1532ms。如何优化到极致呢？这里需要加上 2 个剪枝条件。第一个剪枝条件比较简单，如果累计 sum 比之前存储的 res 大，那么直接 return，不需要继续递归了。第二个剪枝条件就非常重要了，可以一下子减少很多次递归。每次取数产生新的集合的时候，要从第一个最小数开始取，一旦取了，后面就不需要再循环递归了。举个例子，[1,2,3,4]，第一个数如果取 2，集合可以是 [[2,3],[1,4]] 或 [[2,4], [1,3]], 这个集合和[[1,3],[2,4]]、[[1,4], [2,3]] 情况一样。可以看到如果取出第一个最小值以后，后面的循环是不必要的了。所以在取下标为 0 的数的时候，递归到底层以后，返回就可以直接 break，不用接下去的循环了，接下去的循环和递归是不必要的。每组组内的顺序我们并不关心，只要最大值和最小值在分组内即可。另外组间顺序我们也不关心。所以可以把排列问题 O(n!) 时间复杂度降低到组合问题 O(2^n)。加了这 2 个剪枝条件以后，耗时就变成了 0ms 了。beats 100%
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func minimumIncompatibility(nums []int, k int) int {
+	sort.Ints(nums)
+	eachSize, counts := len(nums)/k, make([]int, len(nums)+1)
+	for i := range nums {
+		counts[nums[i]]++
+		if counts[nums[i]] > k {
+			return -1
+		}
+	}
+	orders := []int{}
+	for i := range counts {
+		orders = append(orders, i)
+	}
+	sort.Ints(orders)
+	res := math.MaxInt32
+	generatePermutation1681(nums, counts, orders, 0, 0, eachSize, &res, []int{})
+	if res == math.MaxInt32 {
+		return -1
+	}
+	return res
+}
+
+func generatePermutation1681(nums, counts, order []int, index, sum, eachSize int, res *int, current []int) {
+	if len(current) > 0 && len(current)%eachSize == 0 {
+		sum += current[len(current)-1] - current[len(current)-eachSize]
+		index = 0
+	}
+	if sum >= *res {
+		return
+	}
+	if len(current) == len(nums) {
+		if sum < *res {
+			*res = sum
+		}
+		return
+	}
+	for i := index; i < len(counts); i++ {
+		if counts[order[i]] == 0 {
+			continue
+		}
+		counts[order[i]]--
+		current = append(current, order[i])
+		generatePermutation1681(nums, counts, order, i+1, sum, eachSize, res, current)
+		current = current[:len(current)-1]
+		counts[order[i]]++
+		// 这里是关键的剪枝
+		if index == 0 {
+			break
+		}
+	}
+}
+```
\ No newline at end of file
diff --git a/leetcode/1684.Count-the-Number-of-Consistent-Strings/1684. Count the Number of Consistent Strings.go b/leetcode/1684.Count-the-Number-of-Consistent-Strings/1684. Count the Number of Consistent Strings.go
new file mode 100644
index 000000000..80b40a082
--- /dev/null
+++ b/leetcode/1684.Count-the-Number-of-Consistent-Strings/1684. Count the Number of Consistent Strings.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+func countConsistentStrings(allowed string, words []string) int {
+	allowedMap, res, flag := map[rune]int{}, 0, true
+	for _, str := range allowed {
+		allowedMap[str]++
+	}
+	for i := 0; i < len(words); i++ {
+		flag = true
+		for j := 0; j < len(words[i]); j++ {
+			if _, ok := allowedMap[rune(words[i][j])]; !ok {
+				flag = false
+				break
+			}
+		}
+		if flag {
+			res++
+		}
+	}
+	return res
+}
diff --git a/leetcode/1684.Count-the-Number-of-Consistent-Strings/1684. Count the Number of Consistent Strings_test.go b/leetcode/1684.Count-the-Number-of-Consistent-Strings/1684. Count the Number of Consistent Strings_test.go
new file mode 100644
index 000000000..c94501b6d
--- /dev/null
+++ b/leetcode/1684.Count-the-Number-of-Consistent-Strings/1684. Count the Number of Consistent Strings_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1684 struct {
+	para1684
+	ans1684
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1684 struct {
+	allowed string
+	words   []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1684 struct {
+	one int
+}
+
+func Test_Problem1684(t *testing.T) {
+
+	qs := []question1684{
+
+		{
+			para1684{"ab", []string{"ad", "bd", "aaab", "baa", "badab"}},
+			ans1684{2},
+		},
+
+		{
+			para1684{"abc", []string{"a", "b", "c", "ab", "ac", "bc", "abc"}},
+			ans1684{7},
+		},
+
+		{
+			para1684{"cad", []string{"cc", "acd", "b", "ba", "bac", "bad", "ac", "d"}},
+			ans1684{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1684------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1684, q.para1684
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countConsistentStrings(p.allowed, p.words))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1684.Count-the-Number-of-Consistent-Strings/README.md b/leetcode/1684.Count-the-Number-of-Consistent-Strings/README.md
new file mode 100644
index 000000000..85a779c83
--- /dev/null
+++ b/leetcode/1684.Count-the-Number-of-Consistent-Strings/README.md
@@ -0,0 +1,79 @@
+# [1684. Count the Number of Consistent Strings](https://leetcode.com/problems/count-the-number-of-consistent-strings/)
+
+
+## 题目
+
+You are given a string `allowed` consisting of **distinct** characters and an array of strings `words`. A string is **consistent** if all characters in the string appear in the string `allowed`.
+
+Return *the number of **consistent** strings in the array* `words`.
+
+**Example 1:**
+
+```
+Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
+Output: 2
+Explanation: Strings "aaab" and "baa" are consistent since they only contain characters 'a' and 'b'.
+
+```
+
+**Example 2:**
+
+```
+Input: allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
+Output: 7
+Explanation: All strings are consistent.
+
+```
+
+**Example 3:**
+
+```
+Input: allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
+Output: 4
+Explanation: Strings "cc", "acd", "ac", and "d" are consistent.
+
+```
+
+**Constraints:**
+
+- `1 <= words.length <= 104`
+- `1 <= allowed.length <= 26`
+- `1 <= words[i].length <= 10`
+- The characters in `allowed` are **distinct**.
+- `words[i]` and `allowed` contain only lowercase English letters.
+
+## 题目大意
+
+给你一个由不同字符组成的字符串 `allowed` 和一个字符串数组 `words` 。如果一个字符串的每一个字符都在 `allowed` 中，就称这个字符串是 一致字符串 。
+
+请你返回 `words` 数组中 一致字符串 的数目。
+
+## 解题思路
+
+- 简单题。先将 `allowed` 转化成 map。将 `words` 数组中每个单词的字符都在 map 中查找一遍，如果都存在就累加 res。如果有不存在的字母，不累加。最终输出 res 即可。
+
+## 代码
+
+```go
+package leetcode
+
+func countConsistentStrings(allowed string, words []string) int {
+	allowedMap, res, flag := map[rune]int{}, 0, true
+	for _, str := range allowed {
+		allowedMap[str]++
+	}
+	for i := 0; i < len(words); i++ {
+		flag = true
+		for j := 0; j < len(words[i]); j++ {
+			if _, ok := allowedMap[rune(words[i][j])]; !ok {
+				flag = false
+				break
+			}
+		}
+		if flag {
+			res++
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array/1685. Sum of Absolute Differences in a Sorted Array.go b/leetcode/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array/1685. Sum of Absolute Differences in a Sorted Array.go
new file mode 100644
index 000000000..15656d39b
--- /dev/null
+++ b/leetcode/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array/1685. Sum of Absolute Differences in a Sorted Array.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+// 解法一 优化版 prefixSum + sufixSum
+func getSumAbsoluteDifferences(nums []int) []int {
+	size := len(nums)
+	sufixSum := make([]int, size)
+	sufixSum[size-1] = nums[size-1]
+	for i := size - 2; i >= 0; i-- {
+		sufixSum[i] = sufixSum[i+1] + nums[i]
+	}
+	ans, preSum := make([]int, size), 0
+	for i := 0; i < size; i++ {
+		// 后面可以加到的值
+		res, sum := 0, sufixSum[i]-nums[i]
+		res += (sum - (size-i-1)*nums[i])
+		// 前面可以加到的值
+		res += (i*nums[i] - preSum)
+		ans[i] = res
+		preSum += nums[i]
+	}
+	return ans
+}
+
+// 解法二 prefixSum
+func getSumAbsoluteDifferences1(nums []int) []int {
+	preSum, res, sum := []int{}, []int{}, nums[0]
+	preSum = append(preSum, nums[0])
+	for i := 1; i < len(nums); i++ {
+		sum += nums[i]
+		preSum = append(preSum, sum)
+	}
+	for i := 0; i < len(nums); i++ {
+		if i == 0 {
+			res = append(res, preSum[len(nums)-1]-preSum[0]-nums[i]*(len(nums)-1))
+		} else if i > 0 && i < len(nums)-1 {
+			res = append(res, preSum[len(nums)-1]-preSum[i]-preSum[i-1]+nums[i]*i-nums[i]*(len(nums)-1-i))
+		} else {
+			res = append(res, nums[i]*len(nums)-preSum[len(nums)-1])
+		}
+	}
+	return res
+}
diff --git a/leetcode/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array/1685. Sum of Absolute Differences in a Sorted Array_test.go b/leetcode/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array/1685. Sum of Absolute Differences in a Sorted Array_test.go
new file mode 100644
index 000000000..70ba45b46
--- /dev/null
+++ b/leetcode/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array/1685. Sum of Absolute Differences in a Sorted Array_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1685 struct {
+	para1685
+	ans1685
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1685 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1685 struct {
+	one []int
+}
+
+func Test_Problem1685(t *testing.T) {
+
+	qs := []question1685{
+
+		{
+			para1685{[]int{2, 3, 5}},
+			ans1685{[]int{4, 3, 5}},
+		},
+
+		{
+			para1685{[]int{1, 4, 6, 8, 10}},
+			ans1685{[]int{24, 15, 13, 15, 21}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1685------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1685, q.para1685
+		fmt.Printf("【input】:%v       【output】:%v\n", p, getSumAbsoluteDifferences(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array/README.md b/leetcode/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array/README.md
new file mode 100644
index 000000000..2de2fe4c4
--- /dev/null
+++ b/leetcode/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array/README.md
@@ -0,0 +1,88 @@
+# [1685. Sum of Absolute Differences in a Sorted Array](https://leetcode.com/problems/sum-of-absolute-differences-in-a-sorted-array/)
+
+
+## 题目
+
+You are given an integer array `nums` sorted in **non-decreasing** order.
+
+Build and return *an integer array* `result` *with the same length as* `nums` *such that* `result[i]` *is equal to the **summation of absolute differences** between* `nums[i]` *and all the other elements in the array.*
+
+In other words, `result[i]` is equal to `sum(|nums[i]-nums[j]|)` where `0 <= j < nums.length` and `j != i` (**0-indexed**).
+
+**Example 1:**
+
+```
+Input: nums = [2,3,5]
+Output: [4,3,5]
+Explanation: Assuming the arrays are 0-indexed, then
+result[0] = |2-2| + |2-3| + |2-5| = 0 + 1 + 3 = 4,
+result[1] = |3-2| + |3-3| + |3-5| = 1 + 0 + 2 = 3,
+result[2] = |5-2| + |5-3| + |5-5| = 3 + 2 + 0 = 5.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,4,6,8,10]
+Output: [24,15,13,15,21]
+```
+
+**Constraints:**
+
+- `2 <= nums.length <= 105`
+- `1 <= nums[i] <= nums[i + 1] <= 104`
+
+## 题目大意
+
+给你一个 非递减 有序整数数组 `nums` 。请你建立并返回一个整数数组 `result`，它跟 `nums` 长度相同，且`result[i]` 等于 `nums[i]` 与数组中所有其他元素差的绝对值之和。换句话说， `result[i]` 等于 `sum(|nums[i]-nums[j]|)` ，其中 `0 <= j < nums.length` 且 `j != i` （下标从 0 开始）。
+
+## 解题思路
+
+- 利用前缀和思路解题。题目中说明了是有序数组，所以在计算绝对值的时候可以拆开绝对值符号。假设要计算当前 `result[i]`，以 `i` 为界，把原数组 `nums` 分成了 3 段。`nums[0 ~ i-1]` 和 `nums[i+1 ~ n]`，前面一段 `nums[0 ~ i-1]` 中的每个元素都比 `nums[i]` 小，拆掉绝对值以后，`sum(|nums[i]-nums[j]|) = nums[i] * i - prefixSum[0 ~ i-1]`，后面一段 `nums[i+1 ~ n]` 中的每个元素都比 `nums[i]` 大，拆掉绝对值以后，`sum(|nums[i]-nums[j]|) = prefixSum[i+1 ~ n] - nums[i] * (n - 1 - i)`。特殊的情况，`i = 0` 和 `i = n` 的情况特殊处理一下就行。
+
+## 代码
+
+```go
+package leetcode
+
+//解法一 优化版 prefixSum + sufixSum
+func getSumAbsoluteDifferences(nums []int) []int {
+	size := len(nums)
+	sufixSum := make([]int, size)
+	sufixSum[size-1] = nums[size-1]
+	for i := size - 2; i >= 0; i-- {
+		sufixSum[i] = sufixSum[i+1] + nums[i]
+	}
+	ans, preSum := make([]int, size), 0
+	for i := 0; i < size; i++ {
+		// 后面可以加到的值
+		res, sum := 0, sufixSum[i]-nums[i]
+		res += (sum - (size-i-1)*nums[i])
+		// 前面可以加到的值
+		res += (i*nums[i] - preSum)
+		ans[i] = res
+		preSum += nums[i]
+	}
+	return ans
+}
+
+// 解法二 prefixSum
+func getSumAbsoluteDifferences1(nums []int) []int {
+	preSum, res, sum := []int{}, []int{}, nums[0]
+	preSum = append(preSum, nums[0])
+	for i := 1; i < len(nums); i++ {
+		sum += nums[i]
+		preSum = append(preSum, sum)
+	}
+	for i := 0; i < len(nums); i++ {
+		if i == 0 {
+			res = append(res, preSum[len(nums)-1]-preSum[0]-nums[i]*(len(nums)-1))
+		} else if i > 0 && i < len(nums)-1 {
+			res = append(res, preSum[len(nums)-1]-preSum[i]-preSum[i-1]+nums[i]*i-nums[i]*(len(nums)-1-i))
+		} else {
+			res = append(res, nums[i]*len(nums)-preSum[len(nums)-1])
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1688.Count-of-Matches-in-Tournament/1688. Count of Matches in Tournament.go b/leetcode/1688.Count-of-Matches-in-Tournament/1688. Count of Matches in Tournament.go
new file mode 100644
index 000000000..67c8ac65b
--- /dev/null
+++ b/leetcode/1688.Count-of-Matches-in-Tournament/1688. Count of Matches in Tournament.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+// 解法一
+func numberOfMatches(n int) int {
+	return n - 1
+}
+
+// 解法二 模拟
+func numberOfMatches1(n int) int {
+	sum := 0
+	for n != 1 {
+		if n&1 == 0 {
+			sum += n / 2
+			n = n / 2
+		} else {
+			sum += (n - 1) / 2
+			n = (n-1)/2 + 1
+		}
+	}
+	return sum
+}
diff --git a/leetcode/1688.Count-of-Matches-in-Tournament/1688. Count of Matches in Tournament_test.go b/leetcode/1688.Count-of-Matches-in-Tournament/1688. Count of Matches in Tournament_test.go
new file mode 100644
index 000000000..71e60e594
--- /dev/null
+++ b/leetcode/1688.Count-of-Matches-in-Tournament/1688. Count of Matches in Tournament_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1688 struct {
+	para1688
+	ans1688
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1688 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1688 struct {
+	one int
+}
+
+func Test_Problem1688(t *testing.T) {
+
+	qs := []question1688{
+
+		{
+			para1688{7},
+			ans1688{6},
+		},
+
+		{
+			para1688{14},
+			ans1688{13},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1688------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1688, q.para1688
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numberOfMatches(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1688.Count-of-Matches-in-Tournament/README.md b/leetcode/1688.Count-of-Matches-in-Tournament/README.md
new file mode 100644
index 000000000..b9b5f0a92
--- /dev/null
+++ b/leetcode/1688.Count-of-Matches-in-Tournament/README.md
@@ -0,0 +1,80 @@
+# [1688. Count of Matches in Tournament](https://leetcode.com/problems/count-of-matches-in-tournament/)
+
+
+## 题目
+
+You are given an integer `n`, the number of teams in a tournament that has strange rules:
+
+- If the current number of teams is **even**, each team gets paired with another team. A total of `n / 2` matches are played, and `n / 2` teams advance to the next round.
+- If the current number of teams is **odd**, one team randomly advances in the tournament, and the rest gets paired. A total of `(n - 1) / 2` matches are played, and `(n - 1) / 2 + 1` teams advance to the next round.
+
+Return *the number of matches played in the tournament until a winner is decided.*
+
+**Example 1:**
+
+```
+Input: n = 7
+Output: 6
+Explanation: Details of the tournament: 
+- 1st Round: Teams = 7, Matches = 3, and 4 teams advance.
+- 2nd Round: Teams = 4, Matches = 2, and 2 teams advance.
+- 3rd Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
+Total number of matches = 3 + 2 + 1 = 6.
+```
+
+**Example 2:**
+
+```
+Input: n = 14
+Output: 13
+Explanation: Details of the tournament:
+- 1st Round: Teams = 14, Matches = 7, and 7 teams advance.
+- 2nd Round: Teams = 7, Matches = 3, and 4 teams advance.
+- 3rd Round: Teams = 4, Matches = 2, and 2 teams advance.
+- 4th Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
+Total number of matches = 7 + 3 + 2 + 1 = 13.
+```
+
+**Constraints:**
+
+- `1 <= n <= 200`
+
+## 题目大意
+
+给你一个整数 n ，表示比赛中的队伍数。比赛遵循一种独特的赛制：
+
+- 如果当前队伍数是 偶数 ，那么每支队伍都会与另一支队伍配对。总共进行 n / 2 场比赛，且产生 n / 2 支队伍进入下一轮。
+- 如果当前队伍数为 奇数 ，那么将会随机轮空并晋级一支队伍，其余的队伍配对。总共进行 (n - 1) / 2 场比赛，且产生 (n - 1) / 2 + 1 支队伍进入下一轮。
+
+返回在比赛中进行的配对次数，直到决出获胜队伍为止。
+
+## 解题思路
+
+- 简单题，按照题目的规则模拟。
+- 这一题还有更加简洁的代码，见解法一。n 个队伍，一个冠军，需要淘汰 n-1 个队伍。每一场比赛淘汰一个队伍，因此进行了 n-1 场比赛。所以共有 n-1 个配对。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一
+func numberOfMatches(n int) int {
+	return n - 1
+}
+
+// 解法二 模拟
+func numberOfMatches1(n int) int {
+	sum := 0
+	for n != 1 {
+		if n&1 == 0 {
+			sum += n / 2
+			n = n / 2
+		} else {
+			sum += (n - 1) / 2
+			n = (n-1)/2 + 1
+		}
+	}
+	return sum
+}
+```
\ No newline at end of file
diff --git a/leetcode/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers/1689. Partitioning Into Minimum Number Of Deci-Binary Numbers.go b/leetcode/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers/1689. Partitioning Into Minimum Number Of Deci-Binary Numbers.go
new file mode 100644
index 000000000..a6313fa3c
--- /dev/null
+++ b/leetcode/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers/1689. Partitioning Into Minimum Number Of Deci-Binary Numbers.go	
@@ -0,0 +1,11 @@
+package leetcode
+
+func minPartitions(n string) int {
+	res := 0
+	for i := 0; i < len(n); i++ {
+		if int(n[i]-'0') > res {
+			res = int(n[i] - '0')
+		}
+	}
+	return res
+}
diff --git a/leetcode/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers/1689. Partitioning Into Minimum Number Of Deci-Binary Numbers_test.go b/leetcode/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers/1689. Partitioning Into Minimum Number Of Deci-Binary Numbers_test.go
new file mode 100644
index 000000000..9acc0a74c
--- /dev/null
+++ b/leetcode/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers/1689. Partitioning Into Minimum Number Of Deci-Binary Numbers_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1689 struct {
+	para1689
+	ans1689
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1689 struct {
+	n string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1689 struct {
+	one int
+}
+
+func Test_Problem1689(t *testing.T) {
+
+	qs := []question1689{
+
+		{
+			para1689{"32"},
+			ans1689{3},
+		},
+
+		{
+			para1689{"82734"},
+			ans1689{8},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1689------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1689, q.para1689
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minPartitions(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers/README.md b/leetcode/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers/README.md
new file mode 100644
index 000000000..2ac4daa24
--- /dev/null
+++ b/leetcode/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers/README.md
@@ -0,0 +1,60 @@
+# [1689. Partitioning Into Minimum Number Of Deci-Binary Numbers](https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/)
+
+## 题目
+
+A decimal number is called **deci-binary** if each of its digits is either `0` or `1` without any leading zeros. For example, `101` and `1100` are **deci-binary**, while `112` and `3001` are not.
+
+Given a string `n` that represents a positive decimal integer, return *the **minimum** number of positive **deci-binary** numbers needed so that they sum up to* `n`*.*
+
+**Example 1:**
+
+```
+Input: n = "32"
+Output: 3
+Explanation: 10 + 11 + 11 = 32
+```
+
+**Example 2:**
+
+```
+Input: n = "82734"
+Output: 8
+```
+
+**Example 3:**
+
+```
+Input: n = "27346209830709182346"
+Output: 9
+```
+
+**Constraints:**
+
+- `1 <= n.length <= 105`
+- `n` consists of only digits.
+- `n` does not contain any leading zeros and represents a positive integer.
+
+## 题目大意
+
+如果一个十进制数字不含任何前导零，且每一位上的数字不是 0 就是 1 ，那么该数字就是一个 十-二进制数 。例如，101 和 1100 都是 十-二进制数，而 112 和 3001 不是。给你一个表示十进制整数的字符串 n ，返回和为 n 的 十-二进制数 的最少数目。
+
+## 解题思路
+
+- 这一题也算是简单题，相通了以后，代码就 3 行。
+- 要想由 01 组成的十进制数组成 n，只需要在 n 这个数的各个数位上依次排上 0 和 1 即可。例如 n = 23423723，这是一个 8 位数。最大数字是 7，所以至少需要 7 个数累加能得到这个 n。这 7 个数的百位都为 1，其他数位按需求取 0 和 1 即可。例如万位是 2，那么这 7 个数中任找 2 个数的万位是 1 ，其他 5 个数的万位是 0 即可。
+
+## 代码
+
+```go
+package leetcode
+
+func minPartitions(n string) int {
+	res := 0
+	for i := 0; i < len(n); i++ {
+		if int(n[i]-'0') > res {
+			res = int(n[i] - '0')
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1690.Stone-Game-VII/1690. Stone Game VII.go b/leetcode/1690.Stone-Game-VII/1690. Stone Game VII.go
new file mode 100644
index 000000000..c2ad93b4f
--- /dev/null
+++ b/leetcode/1690.Stone-Game-VII/1690. Stone Game VII.go	
@@ -0,0 +1,65 @@
+package leetcode
+
+// 解法一 优化空间版 DP
+func stoneGameVII(stones []int) int {
+	n := len(stones)
+	sum := make([]int, n)
+	dp := make([]int, n)
+	for i, d := range stones {
+		sum[i] = d
+	}
+	for i := 1; i < n; i++ {
+		for j := 0; j+i < n; j++ {
+			if (n-i)%2 == 1 {
+				d0 := dp[j] + sum[j]
+				d1 := dp[j+1] + sum[j+1]
+				if d0 > d1 {
+					dp[j] = d0
+				} else {
+					dp[j] = d1
+				}
+			} else {
+				d0 := dp[j] - sum[j]
+				d1 := dp[j+1] - sum[j+1]
+				if d0 < d1 {
+					dp[j] = d0
+				} else {
+					dp[j] = d1
+				}
+			}
+			sum[j] = sum[j] + stones[i+j]
+		}
+	}
+	return dp[0]
+}
+
+// 解法二 常规 DP
+func stoneGameVII1(stones []int) int {
+	prefixSum := make([]int, len(stones))
+	for i := 0; i < len(stones); i++ {
+		if i == 0 {
+			prefixSum[i] = stones[i]
+		} else {
+			prefixSum[i] = prefixSum[i-1] + stones[i]
+		}
+	}
+	dp := make([][]int, len(stones))
+	for i := range dp {
+		dp[i] = make([]int, len(stones))
+		dp[i][i] = 0
+	}
+	n := len(stones)
+	for l := 2; l <= n; l++ {
+		for i := 0; i+l <= n; i++ {
+			dp[i][i+l-1] = max(prefixSum[i+l-1]-prefixSum[i+1]+stones[i+1]-dp[i+1][i+l-1], prefixSum[i+l-2]-prefixSum[i]+stones[i]-dp[i][i+l-2])
+		}
+	}
+	return dp[0][n-1]
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1690.Stone-Game-VII/1690. Stone Game VII_test.go b/leetcode/1690.Stone-Game-VII/1690. Stone Game VII_test.go
new file mode 100644
index 000000000..60656ac91
--- /dev/null
+++ b/leetcode/1690.Stone-Game-VII/1690. Stone Game VII_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1690 struct {
+	para1690
+	ans1690
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1690 struct {
+	stones []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1690 struct {
+	one int
+}
+
+func Test_Problem1690(t *testing.T) {
+
+	qs := []question1690{
+
+		{
+			para1690{[]int{5, 3, 1, 4, 2}},
+			ans1690{6},
+		},
+
+		{
+			para1690{[]int{7, 90, 5, 1, 100, 10, 10, 2}},
+			ans1690{122},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1690------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1690, q.para1690
+		fmt.Printf("【input】:%v       【output】:%v\n", p, stoneGameVII(p.stones))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1690.Stone-Game-VII/README.md b/leetcode/1690.Stone-Game-VII/README.md
new file mode 100644
index 000000000..c7b618acf
--- /dev/null
+++ b/leetcode/1690.Stone-Game-VII/README.md
@@ -0,0 +1,130 @@
+# [1690. Stone Game VII](https://leetcode.com/problems/stone-game-vii/)
+
+## 题目
+
+Alice and Bob take turns playing a game, with **Alice starting first**.
+
+There are `n` stones arranged in a row. On each player's turn, they can **remove** either the leftmost stone or the rightmost stone from the row and receive points equal to the **sum** of the remaining stones' values in the row. The winner is the one with the higher score when there are no stones left to remove.
+
+Bob found that he will always lose this game (poor Bob, he always loses), so he decided to **minimize the score's difference**. Alice's goal is to **maximize the difference** in the score.
+
+Given an array of integers `stones` where `stones[i]` represents the value of the `ith` stone **from the left**, return *the **difference** in Alice and Bob's score if they both play **optimally**.*
+
+**Example 1:**
+
+```
+Input: stones = [5,3,1,4,2]
+Output: 6
+Explanation: 
+- Alice removes 2 and gets 5 + 3 + 1 + 4 = 13 points. Alice = 13, Bob = 0, stones = [5,3,1,4].
+- Bob removes 5 and gets 3 + 1 + 4 = 8 points. Alice = 13, Bob = 8, stones = [3,1,4].
+- Alice removes 3 and gets 1 + 4 = 5 points. Alice = 18, Bob = 8, stones = [1,4].
+- Bob removes 1 and gets 4 points. Alice = 18, Bob = 12, stones = [4].
+- Alice removes 4 and gets 0 points. Alice = 18, Bob = 12, stones = [].
+The score difference is 18 - 12 = 6.
+```
+
+**Example 2:**
+
+```
+Input: stones = [7,90,5,1,100,10,10,2]
+Output: 122
+```
+
+**Constraints:**
+
+- `n == stones.length`
+- `2 <= n <= 1000`
+- `1 <= stones[i] <= 1000`
+
+## 题目大意
+
+石子游戏中，爱丽丝和鲍勃轮流进行自己的回合，爱丽丝先开始 。有 n 块石子排成一排。每个玩家的回合中，可以从行中 移除 最左边的石头或最右边的石头，并获得与该行中剩余石头值之 和 相等的得分。当没有石头可移除时，得分较高者获胜。鲍勃发现他总是输掉游戏（可怜的鲍勃，他总是输），所以他决定尽力 减小得分的差值 。爱丽丝的目标是最大限度地 扩大得分的差值 。
+
+给你一个整数数组 stones ，其中 stones[i] 表示 从左边开始 的第 i 个石头的值，如果爱丽丝和鲍勃都 发挥出最佳水平 ，请返回他们 得分的差值 。
+
+## 解题思路
+
+- 首先考虑 Bob 缩小分值差距意味着什么，意味着他想让他和 Alice 相对分数最小。Bob 已经明确肯定是输，所以他的分数一定比 Alice 小，那么 Bob - Alice 分数相减一定是负数。相对分数越小，意味着差值越大。负数越大，差值越小。-50 和 -10，-10 数值大，相差小。所以 Bob 的操作是让相对分数越大。Alice 的目的也是这样，要让 Alice - Bob 的相对分数越大，这里是正数的越大。综上，两者的目的相同，都是让相对分数最大化。
+- 定义 `dp[i][j]` 代表在当前 `stone[i ~ j]` 区间内能获得的最大分差。状态转移方程为：
+
+    ```go
+    dp[i][j] = max(
+                    sum(i + 1, j) - dp[i + 1][j],   // 这一局取走 `stone[i]`，获得 sum(i + 1, j) 分数，再减去剩下对手能获得的分数，即是此局能获得的最大分差。
+                    sum(i, j - 1) - dp[i][j - 1]    // 这一局取走 `stone[j]`，获得 sum(i, j - 1) 分数，再减去剩下对手能获得的分数，即是此局能获得的最大分差。
+                  )
+    ```
+
+    计算 `sum(i + 1, j) = stone[i + 1] + stone[i + 2] + …… + stone[j]` 利用前缀和计算区间和。
+
+- 解法二是正常思路解答出来的代码。解法一是压缩了 DP 数组，在 DP 状态转移的时候，生成下一个 `dp[j]` 实际上是有规律的。利用这个规律可以少存一维数据，压缩空间。解法一的代码直接写出来，比较难想。先写出解法二的代码，然后找到递推规律，优化空间压缩一维，再写出解法一的代码。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 优化空间版 DP
+func stoneGameVII(stones []int) int {
+	n := len(stones)
+	sum := make([]int, n)
+	dp := make([]int, n)
+	for i, d := range stones {
+		sum[i] = d
+	}
+	for i := 1; i < n; i++ {
+		for j := 0; j+i < n; j++ {
+			if (n-i)%2 == 1 {
+				d0 := dp[j] + sum[j]
+				d1 := dp[j+1] + sum[j+1]
+				if d0 > d1 {
+					dp[j] = d0
+				} else {
+					dp[j] = d1
+				}
+			} else {
+				d0 := dp[j] - sum[j]
+				d1 := dp[j+1] - sum[j+1]
+				if d0 < d1 {
+					dp[j] = d0
+				} else {
+					dp[j] = d1
+				}
+			}
+			sum[j] = sum[j] + stones[i+j]
+		}
+	}
+	return dp[0]
+}
+
+// 解法二 常规 DP
+func stoneGameVII1(stones []int) int {
+	prefixSum := make([]int, len(stones))
+	for i := 0; i < len(stones); i++ {
+		if i == 0 {
+			prefixSum[i] = stones[i]
+		} else {
+			prefixSum[i] = prefixSum[i-1] + stones[i]
+		}
+	}
+	dp := make([][]int, len(stones))
+	for i := range dp {
+		dp[i] = make([]int, len(stones))
+		dp[i][i] = 0
+	}
+	n := len(stones)
+	for l := 2; l <= n; l++ {
+		for i := 0; i+l <= n; i++ {
+			dp[i][i+l-1] = max(prefixSum[i+l-1]-prefixSum[i+1]+stones[i+1]-dp[i+1][i+l-1], prefixSum[i+l-2]-prefixSum[i]+stones[i]-dp[i][i+l-2])
+		}
+	}
+	return dp[0][n-1]
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1691.Maximum-Height-by-Stacking-Cuboids/1691. Maximum Height by Stacking Cuboids.go b/leetcode/1691.Maximum-Height-by-Stacking-Cuboids/1691. Maximum Height by Stacking Cuboids.go
new file mode 100644
index 000000000..2d2b14f1f
--- /dev/null
+++ b/leetcode/1691.Maximum-Height-by-Stacking-Cuboids/1691. Maximum Height by Stacking Cuboids.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import "sort"
+
+func maxHeight(cuboids [][]int) int {
+	n := len(cuboids)
+	for i := 0; i < n; i++ {
+		sort.Ints(cuboids[i]) // 立方体三边内部排序
+	}
+	// 立方体排序，先按最短边，再到最长边
+	sort.Slice(cuboids, func(i, j int) bool {
+		if cuboids[i][0] != cuboids[j][0] {
+			return cuboids[i][0] < cuboids[j][0]
+		}
+		if cuboids[i][1] != cuboids[j][1] {
+			return cuboids[i][1] < cuboids[j][1]
+		}
+		return cuboids[i][2] < cuboids[j][2]
+	})
+	res := 0
+	dp := make([]int, n)
+	for i := 0; i < n; i++ {
+		dp[i] = cuboids[i][2]
+		res = max(res, dp[i])
+	}
+	for i := 1; i < n; i++ {
+		for j := 0; j < i; j++ {
+			if cuboids[j][0] <= cuboids[i][0] && cuboids[j][1] <= cuboids[i][1] && cuboids[j][2] <= cuboids[i][2] {
+				dp[i] = max(dp[i], dp[j]+cuboids[i][2])
+			}
+		}
+		res = max(res, dp[i])
+	}
+	return res
+}
+
+func max(x, y int) int {
+	if x > y {
+		return x
+	}
+	return y
+}
diff --git a/leetcode/1691.Maximum-Height-by-Stacking-Cuboids/1691. Maximum Height by Stacking Cuboids_test.go b/leetcode/1691.Maximum-Height-by-Stacking-Cuboids/1691. Maximum Height by Stacking Cuboids_test.go
new file mode 100644
index 000000000..b49a94636
--- /dev/null
+++ b/leetcode/1691.Maximum-Height-by-Stacking-Cuboids/1691. Maximum Height by Stacking Cuboids_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1691 struct {
+	para1691
+	ans1691
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1691 struct {
+	cuboids [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1691 struct {
+	one int
+}
+
+func Test_Problem1691(t *testing.T) {
+
+	qs := []question1691{
+
+		{
+			para1691{[][]int{{50, 45, 20}, {95, 37, 53}, {45, 23, 12}}},
+			ans1691{190},
+		},
+
+		{
+			para1691{[][]int{{38, 25, 45}, {76, 35, 3}}},
+			ans1691{76},
+		},
+
+		{
+			para1691{[][]int{{7, 11, 17}, {7, 17, 11}, {11, 7, 17}, {11, 17, 7}, {17, 7, 11}, {17, 11, 7}}},
+			ans1691{102},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1691------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1691, q.para1691
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxHeight(p.cuboids))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1691.Maximum-Height-by-Stacking-Cuboids/README.md b/leetcode/1691.Maximum-Height-by-Stacking-Cuboids/README.md
new file mode 100644
index 000000000..bd8eca873
--- /dev/null
+++ b/leetcode/1691.Maximum-Height-by-Stacking-Cuboids/README.md
@@ -0,0 +1,106 @@
+# [1691. Maximum Height by Stacking Cuboids](https://leetcode.com/problems/maximum-height-by-stacking-cuboids/)
+
+## 题目
+
+Given `n` `cuboids` where the dimensions of the `ith` cuboid is `cuboids[i] = [widthi, lengthi, heighti]` (**0-indexed**). Choose a **subset** of `cuboids` and place them on each other.
+
+You can place cuboid `i` on cuboid `j` if `widthi <= widthj` and `lengthi <= lengthj` and `heighti <= heightj`. You can rearrange any cuboid's dimensions by rotating it to put it on another cuboid.
+
+Return *the **maximum height** of the stacked* `cuboids`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/10/21/image.jpg](https://assets.leetcode.com/uploads/2019/10/21/image.jpg)
+
+```
+Input: cuboids = [[50,45,20],[95,37,53],[45,23,12]]
+Output: 190
+Explanation:
+Cuboid 1 is placed on the bottom with the 53x37 side facing down with height 95.
+Cuboid 0 is placed next with the 45x20 side facing down with height 50.
+Cuboid 2 is placed next with the 23x12 side facing down with height 45.
+The total height is 95 + 50 + 45 = 190.
+```
+
+**Example 2:**
+
+```
+Input: cuboids = [[38,25,45],[76,35,3]]
+Output: 76
+Explanation:
+You can't place any of the cuboids on the other.
+We choose cuboid 1 and rotate it so that the 35x3 side is facing down and its height is 76.
+```
+
+**Example 3:**
+
+```
+Input: cuboids = [[7,11,17],[7,17,11],[11,7,17],[11,17,7],[17,7,11],[17,11,7]]
+Output: 102
+Explanation:
+After rearranging the cuboids, you can see that all cuboids have the same dimension.
+You can place the 11x7 side down on all cuboids so their heights are 17.
+The maximum height of stacked cuboids is 6 * 17 = 102.
+```
+
+**Constraints:**
+
+- `n == cuboids.length`
+- `1 <= n <= 100`
+- `1 <= widthi, lengthi, heighti <= 100`
+
+## 题目大意
+
+给你 n 个长方体 cuboids ，其中第 i 个长方体的长宽高表示为 cuboids[i] = [widthi, lengthi, heighti]（下标从 0 开始）。请你从 cuboids 选出一个 子集 ，并将它们堆叠起来。如果 widthi <= widthj 且 lengthi <= lengthj 且 heighti <= heightj ，你就可以将长方体 i 堆叠在长方体 j 上。你可以通过旋转把长方体的长宽高重新排列，以将它放在另一个长方体上。返回 堆叠长方体 cuboids 可以得到的 最大高度 。
+
+## 解题思路
+
+- 这一题是 LIS 最长递增子序列系列问题的延续。一维 LIS 问题是第 300 题。二维 LIS 问题是 354 题。这一题是三维的 LIS 问题。
+- 题目要求最终摞起来的长方体尽可能的高，那么把长宽高中最大的值旋转为高。这是针对单个方块的排序。多个方块间还要排序，因为他们摞起来有要求，大的方块必须放在下面。所以针对多个方块，按照长，宽，高的顺序进行排序。两次排序完成以后，可以用动态规划找出最大值了。定义 `dp[i]` 为以 `i` 为最后一块砖块所能堆叠的最高高度。由于长和宽已经排好序了。所以只需要在 [0, i - 1] 这个区间内动态更新 dp 最大值。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func maxHeight(cuboids [][]int) int {
+	n := len(cuboids)
+	for i := 0; i < n; i++ {
+		sort.Ints(cuboids[i]) // 立方体三边内部排序
+	}
+	// 立方体排序，先按最短边，再到最长边
+	sort.Slice(cuboids, func(i, j int) bool {
+		if cuboids[i][0] != cuboids[j][0] {
+			return cuboids[i][0] < cuboids[j][0]
+		}
+		if cuboids[i][1] != cuboids[j][1] {
+			return cuboids[i][1] < cuboids[j][1]
+		}
+		return cuboids[i][2] < cuboids[j][2]
+	})
+	res := 0
+	dp := make([]int, n)
+	for i := 0; i < n; i++ {
+		dp[i] = cuboids[i][2]
+		res = max(res, dp[i])
+	}
+	for i := 1; i < n; i++ {
+		for j := 0; j < i; j++ {
+			if cuboids[j][0] <= cuboids[i][0] && cuboids[j][1] <= cuboids[i][1] && cuboids[j][2] <= cuboids[i][2] {
+				dp[i] = max(dp[i], dp[j]+cuboids[i][2])
+			}
+		}
+		res = max(res, dp[i])
+	}
+	return res
+}
+
+func max(x, y int) int {
+	if x > y {
+		return x
+	}
+	return y
+}
+```
\ No newline at end of file
diff --git a/leetcode/1694.Reformat-Phone-Number/1694. Reformat Phone Number.go b/leetcode/1694.Reformat-Phone-Number/1694. Reformat Phone Number.go
new file mode 100644
index 000000000..22a8d6663
--- /dev/null
+++ b/leetcode/1694.Reformat-Phone-Number/1694. Reformat Phone Number.go	
@@ -0,0 +1,37 @@
+package leetcode
+
+import (
+	"strings"
+)
+
+func reformatNumber(number string) string {
+	parts, nums := []string{}, []rune{}
+	for _, r := range number {
+		if r != '-' && r != ' ' {
+			nums = append(nums, r)
+		}
+	}
+	threeDigits, twoDigits := len(nums)/3, 0
+	switch len(nums) % 3 {
+	case 1:
+		threeDigits--
+		twoDigits = 2
+	case 2:
+		twoDigits = 1
+	default:
+		twoDigits = 0
+	}
+	for i := 0; i < threeDigits; i++ {
+		s := ""
+		s += string(nums[0:3])
+		nums = nums[3:]
+		parts = append(parts, s)
+	}
+	for i := 0; i < twoDigits; i++ {
+		s := ""
+		s += string(nums[0:2])
+		nums = nums[2:]
+		parts = append(parts, s)
+	}
+	return strings.Join(parts, "-")
+}
diff --git a/leetcode/1694.Reformat-Phone-Number/1694. Reformat Phone Number_test.go b/leetcode/1694.Reformat-Phone-Number/1694. Reformat Phone Number_test.go
new file mode 100644
index 000000000..8eb4f7a13
--- /dev/null
+++ b/leetcode/1694.Reformat-Phone-Number/1694. Reformat Phone Number_test.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1694 struct {
+	para1694
+	ans1694
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1694 struct {
+	number string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1694 struct {
+	one string
+}
+
+func Test_Problem1694(t *testing.T) {
+
+	qs := []question1694{
+
+		{
+			para1694{"1-23-45 6"},
+			ans1694{"123-456"},
+		},
+
+		{
+			para1694{"123 4-567"},
+			ans1694{"123-45-67"},
+		},
+
+		{
+			para1694{"123 4-5678"},
+			ans1694{"123-456-78"},
+		},
+
+		{
+			para1694{"12"},
+			ans1694{"12"},
+		},
+
+		{
+			para1694{"--17-5 229 35-39475 "},
+			ans1694{"175-229-353-94-75"},
+		},
+
+		{
+			para1694{"9964-"},
+			ans1694{"99-64"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1694------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1694, q.para1694
+		fmt.Printf("【input】:%v       【output】:%v\n", p, reformatNumber(p.number))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1694.Reformat-Phone-Number/README.md b/leetcode/1694.Reformat-Phone-Number/README.md
new file mode 100644
index 000000000..670aae712
--- /dev/null
+++ b/leetcode/1694.Reformat-Phone-Number/README.md
@@ -0,0 +1,135 @@
+# [1694. Reformat Phone Number](https://leetcode.com/problems/reformat-phone-number/)
+
+
+## 题目
+
+You are given a phone number as a string `number`. `number` consists of digits, spaces `' '`, and/or dashes `'-'`.
+
+You would like to reformat the phone number in a certain manner. Firstly, **remove** all spaces and dashes. Then, **group** the digits from left to right into blocks of length 3 **until** there are 4 or fewer digits. The final digits are then grouped as follows:
+
+- 2 digits: A single block of length 2.
+- 3 digits: A single block of length 3.
+- 4 digits: Two blocks of length 2 each.
+
+The blocks are then joined by dashes. Notice that the reformatting process should **never** produce any blocks of length 1 and produce **at most** two blocks of length 2.
+
+Return *the phone number after formatting.*
+
+**Example 1:**
+
+```
+Input: number = "1-23-45 6"
+Output: "123-456"
+Explanation: The digits are "123456".
+Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".
+Step 2: There are 3 digits remaining, so put them in a single block of length 3. The 2nd block is "456".
+Joining the blocks gives "123-456".
+
+```
+
+**Example 2:**
+
+```
+Input: number = "123 4-567"
+Output: "123-45-67"
+Explanation: The digits are "1234567".
+Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".
+Step 2: There are 4 digits left, so split them into two blocks of length 2. The blocks are "45" and "67".
+Joining the blocks gives "123-45-67".
+
+```
+
+**Example 3:**
+
+```
+Input: number = "123 4-5678"
+Output: "123-456-78"
+Explanation: The digits are "12345678".
+Step 1: The 1st block is "123".
+Step 2: The 2nd block is "456".
+Step 3: There are 2 digits left, so put them in a single block of length 2. The 3rd block is "78".
+Joining the blocks gives "123-456-78".
+
+```
+
+**Example 4:**
+
+```
+Input: number = "12"
+Output: "12"
+
+```
+
+**Example 5:**
+
+```
+Input: number = "--17-5 229 35-39475 "
+Output: "175-229-353-94-75"
+
+```
+
+**Constraints:**
+
+- `2 <= number.length <= 100`
+- `number` consists of digits and the characters `'-'` and `' '`.
+- There are at least **two** digits in `number`.
+
+## 题目大意
+
+给你一个字符串形式的电话号码 number 。number 由数字、空格 ' '、和破折号 '-' 组成。
+
+请你按下述方式重新格式化电话号码。
+
+- 首先，删除 所有的空格和破折号。
+- 其次，将数组从左到右 每 3 个一组 分块，直到 剩下 4 个或更少数字。剩下的数字将按下述规定再分块：
+    - 2 个数字：单个含 2 个数字的块。
+    - 3 个数字：单个含 3 个数字的块。
+    - 4 个数字：两个分别含 2 个数字的块。
+
+最后用破折号将这些块连接起来。注意，重新格式化过程中 不应该 生成仅含 1 个数字的块，并且 最多 生成两个含 2 个数字的块。返回格式化后的电话号码。
+
+## 解题思路
+
+- 简单题。先判断号码是不是 2 位和 4 位，如果是，单独输出这 2 种情况。剩下的都是 3 位以上了，取余，判断剩余的数字是 2 个还是 4 个。这时不可能存在剩 1 位数的情况。除 3 余 1，即剩 4 位的情况，末尾 4 位需要 2 个一组输出。除 3 余 2，即剩  2 位的情况。处理好末尾，再逆序每 3 个一组连接 "-" 即可。可能需要注意的 case 见 test 文件。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"strings"
+)
+
+func reformatNumber(number string) string {
+	parts, nums := []string{}, []rune{}
+	for _, r := range number {
+		if r != '-' && r != ' ' {
+			nums = append(nums, r)
+		}
+	}
+	threeDigits, twoDigits := len(nums)/3, 0
+	switch len(nums) % 3 {
+	case 1:
+		threeDigits--
+		twoDigits = 2
+	case 2:
+		twoDigits = 1
+	default:
+		twoDigits = 0
+	}
+	for i := 0; i < threeDigits; i++ {
+		s := ""
+		s += string(nums[0:3])
+		nums = nums[3:]
+		parts = append(parts, s)
+	}
+	for i := 0; i < twoDigits; i++ {
+		s := ""
+		s += string(nums[0:2])
+		nums = nums[2:]
+		parts = append(parts, s)
+	}
+	return strings.Join(parts, "-")
+}
+```
\ No newline at end of file
diff --git a/leetcode/1695.Maximum-Erasure-Value/1695. Maximum Erasure Value.go b/leetcode/1695.Maximum-Erasure-Value/1695. Maximum Erasure Value.go
new file mode 100644
index 000000000..ccca3f96a
--- /dev/null
+++ b/leetcode/1695.Maximum-Erasure-Value/1695. Maximum Erasure Value.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+func maximumUniqueSubarray(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	result, left, right, freq := 0, 0, -1, map[int]int{}
+	for left < len(nums) {
+		if right+1 < len(nums) && freq[nums[right+1]] == 0 {
+			freq[nums[right+1]]++
+			right++
+		} else {
+			freq[nums[left]]--
+			left++
+		}
+		sum := 0
+		for i := left; i <= right; i++ {
+			sum += nums[i]
+		}
+		result = max(result, sum)
+	}
+	return result
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1695.Maximum-Erasure-Value/1695. Maximum Erasure Value_test.go b/leetcode/1695.Maximum-Erasure-Value/1695. Maximum Erasure Value_test.go
new file mode 100644
index 000000000..bc088cdab
--- /dev/null
+++ b/leetcode/1695.Maximum-Erasure-Value/1695. Maximum Erasure Value_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1695 struct {
+	para1695
+	ans1695
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1695 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1695 struct {
+	one int
+}
+
+func Test_Problem1695(t *testing.T) {
+
+	qs := []question1695{
+
+		{
+			para1695{[]int{4, 2, 4, 5, 6}},
+			ans1695{17},
+		},
+
+		{
+			para1695{[]int{5, 2, 1, 2, 5, 2, 1, 2, 5}},
+			ans1695{8},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1695------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1695, q.para1695
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maximumUniqueSubarray(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1695.Maximum-Erasure-Value/README.md b/leetcode/1695.Maximum-Erasure-Value/README.md
new file mode 100644
index 000000000..a2dec532d
--- /dev/null
+++ b/leetcode/1695.Maximum-Erasure-Value/README.md
@@ -0,0 +1,74 @@
+# [1695. Maximum Erasure Value](https://leetcode.com/problems/maximum-erasure-value/)
+
+
+## 题目
+
+You are given an array of positive integers `nums` and want to erase a subarray containing **unique elements**. The **score** you get by erasing the subarray is equal to the **sum** of its elements.
+
+Return *the **maximum score** you can get by erasing **exactly one** subarray.*
+
+An array `b` is called to be a subarray of `a` if it forms a contiguous subsequence of `a`, that is, if it is equal to `a[l],a[l+1],...,a[r]` for some `(l,r)`.
+
+**Example 1:**
+
+```
+Input: nums = [4,2,4,5,6]
+Output: 17
+Explanation: The optimal subarray here is [2,4,5,6].
+```
+
+**Example 2:**
+
+```
+Input: nums = [5,2,1,2,5,2,1,2,5]
+Output: 8
+Explanation: The optimal subarray here is [5,2,1] or [1,2,5].
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 105`
+- `1 <= nums[i] <= 104`
+
+## 题目大意
+
+给你一个正整数数组 nums ，请你从中删除一个含有 若干不同元素 的子数组。删除子数组的 得分 就是子数组各元素之 和 。返回 只删除一个 子数组可获得的 最大得分 。如果数组 b 是数组 a 的一个连续子序列，即如果它等于 a[l],a[l+1],...,a[r] ，那么它就是 a 的一个子数组。
+
+## 解题思路
+
+- 读完题立马能识别出这是经典的滑动窗口题。利用滑动窗口从左往右滑动窗口，滑动过程中统计频次，如果是不同元素，右边界窗口又移，否则左边窗口缩小。每次移动更新 max 值。最终扫完一遍以后，max 值即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+func maximumUniqueSubarray(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	result, left, right, freq := 0, 0, -1, map[int]int{}
+	for left < len(nums) {
+		if right+1 < len(nums) && freq[nums[right+1]] == 0 {
+			freq[nums[right+1]]++
+			right++
+		} else {
+			freq[nums[left]]--
+			left++
+		}
+		sum := 0
+		for i := left; i <= right; i++ {
+			sum += nums[i]
+		}
+		result = max(result, sum)
+	}
+	return result
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1696.Jump-Game-VI/1696. Jump Game VI.go b/leetcode/1696.Jump-Game-VI/1696. Jump Game VI.go
new file mode 100644
index 000000000..b57b2500d
--- /dev/null
+++ b/leetcode/1696.Jump-Game-VI/1696. Jump Game VI.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"math"
+)
+
+// 单调队列
+func maxResult(nums []int, k int) int {
+	dp := make([]int, len(nums))
+	dp[0] = nums[0]
+	for i := 1; i < len(dp); i++ {
+		dp[i] = math.MinInt32
+	}
+	window := make([]int, k)
+	for i := 1; i < len(nums); i++ {
+		dp[i] = nums[i] + dp[window[0]]
+		for len(window) > 0 && dp[window[len(window)-1]] <= dp[i] {
+			window = window[:len(window)-1]
+		}
+		for len(window) > 0 && i-k >= window[0] {
+			window = window[1:]
+		}
+		window = append(window, i)
+	}
+	return dp[len(nums)-1]
+}
+
+// 超时
+func maxResult1(nums []int, k int) int {
+	dp := make([]int, len(nums))
+	if k > len(nums) {
+		k = len(nums)
+	}
+	dp[0] = nums[0]
+	for i := 1; i < len(dp); i++ {
+		dp[i] = math.MinInt32
+	}
+	for i := 1; i < len(nums); i++ {
+		left, tmp := max(0, i-k), math.MinInt32
+		for j := left; j < i; j++ {
+			tmp = max(tmp, dp[j])
+		}
+		dp[i] = nums[i] + tmp
+	}
+	return dp[len(nums)-1]
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1696.Jump-Game-VI/1696. Jump Game VI_test.go b/leetcode/1696.Jump-Game-VI/1696. Jump Game VI_test.go
new file mode 100644
index 000000000..9b75ea76d
--- /dev/null
+++ b/leetcode/1696.Jump-Game-VI/1696. Jump Game VI_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1696 struct {
+	para1696
+	ans1696
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1696 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1696 struct {
+	one int
+}
+
+func Test_Problem1696(t *testing.T) {
+
+	qs := []question1696{
+
+		{
+			para1696{[]int{1, -1, -2, 4, -7, 3}, 2},
+			ans1696{7},
+		},
+
+		{
+			para1696{[]int{10, -5, -2, 4, 0, 3}, 3},
+			ans1696{17},
+		},
+
+		{
+			para1696{[]int{1, -5, -20, 4, -1, 3, -6, -3}, 2},
+			ans1696{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1696------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1696, q.para1696
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxResult(p.nums, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1696.Jump-Game-VI/README.md b/leetcode/1696.Jump-Game-VI/README.md
new file mode 100644
index 000000000..5a2ae2455
--- /dev/null
+++ b/leetcode/1696.Jump-Game-VI/README.md
@@ -0,0 +1,109 @@
+# [1696. Jump Game VI](https://leetcode.com/problems/jump-game-vi/)
+
+## 题目
+
+You are given a **0-indexed** integer array `nums` and an integer `k`.
+
+You are initially standing at index `0`. In one move, you can jump at most `k` steps forward without going outside the boundaries of the array. That is, you can jump from index `i` to any index in the range `[i + 1, min(n - 1, i + k)]` **inclusive**.
+
+You want to reach the last index of the array (index `n - 1`). Your **score** is the **sum** of all `nums[j]` for each index `j` you visited in the array.
+
+Return *the **maximum score** you can get*.
+
+**Example 1:**
+
+```
+Input: nums = [1,-1,-2,4,-7,3], k = 2
+Output: 7
+Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [10,-5,-2,4,0,3], k = 3
+Output: 17
+Explanation: You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17.
+
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
+Output: 0
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length, k <= 10^5`
+- `10^4 <= nums[i] <= 10^4`
+
+## 题目大意
+
+给你一个下标从 0 开始的整数数组 nums 和一个整数 k 。一开始你在下标 0 处。每一步，你最多可以往前跳 k 步，但你不能跳出数组的边界。也就是说，你可以从下标 i 跳到 [i + 1， min(n - 1, i + k)] 包含 两个端点的任意位置。你的目标是到达数组最后一个位置（下标为 n - 1 ），你的 得分 为经过的所有数字之和。请你返回你能得到的 最大得分 。
+
+## 解题思路
+
+- 首先能想到的解题思路是动态规划。定义 `dp[i]` 为跳到第 `i` 个位子能获得的最大分数。题目要求的是 `dp[n-1]`，状态转移方程是：`dp[i] = nums[i] + max(dp[j]), max(0, i - k ) <= j < i`，这里需要注意 `j` 的下界，题目中说到不能跳到负数区间，所以左边界下界为 0 。求 `max(dp[j])` 需要遍历一次求得最大值，所以这个解法整体时间复杂度是 O((n - k) * k)，但是提交以后提示超时了。
+- 分析一下超时原因。每次都要在 `[max(0, i - k ), i)` 区间内扫描找到最大值，下一轮的区间是 `[max(0, i - k + 1), i + 1)`，前后这两轮扫描的区间存在大量重合部分  `[max(0, i - k + 1), i)`，正是这部分反反复复的扫描导致算法低效。如何高效的在一个区间内找到最大值是本题的关键。利用单调队列可以完成此题。单调队列里面存一个区间内最大值的下标。这里单调队列有 2 个性质。性质一，队列的队首永远都是最大值，队列从大到小降序排列。如果来了一个比队首更大的值的下标，需要将单调队列清空，只存这个新的最大值的下标。性质二，队列的长度为 k。从队尾插入新值，并把队头的最大值“挤”出队首。拥有了这个单调队列以后，再进行 DP 状态转移，效率就很高了。每次只需取出队首的最大值即可。具体代码见下面。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+)
+
+// 单调队列
+func maxResult(nums []int, k int) int {
+	dp := make([]int, len(nums))
+	dp[0] = nums[0]
+	for i := 1; i < len(dp); i++ {
+		dp[i] = math.MinInt32
+	}
+	window := make([]int, k)
+	for i := 1; i < len(nums); i++ {
+		dp[i] = nums[i] + dp[window[0]]
+		for len(window) > 0 && dp[window[len(window)-1]] <= dp[i] {
+			window = window[:len(window)-1]
+		}
+		for len(window) > 0 && i-k >= window[0] {
+			window = window[1:]
+		}
+		window = append(window, i)
+	}
+	return dp[len(nums)-1]
+}
+
+// 超时
+func maxResult1(nums []int, k int) int {
+	dp := make([]int, len(nums))
+	if k > len(nums) {
+		k = len(nums)
+	}
+	dp[0] = nums[0]
+	for i := 1; i < len(dp); i++ {
+		dp[i] = math.MinInt32
+	}
+	for i := 1; i < len(nums); i++ {
+		left, tmp := max(0, i-k), math.MinInt32
+		for j := left; j < i; j++ {
+			tmp = max(tmp, dp[j])
+		}
+		dp[i] = nums[i] + tmp
+	}
+	return dp[len(nums)-1]
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1700.Number-of-Students-Unable-to-Eat-Lunch/1700. Number of Students Unable to Eat Lunch.go b/leetcode/1700.Number-of-Students-Unable-to-Eat-Lunch/1700. Number of Students Unable to Eat Lunch.go
new file mode 100644
index 000000000..a956a03ff
--- /dev/null
+++ b/leetcode/1700.Number-of-Students-Unable-to-Eat-Lunch/1700. Number of Students Unable to Eat Lunch.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+func countStudents(students []int, sandwiches []int) int {
+	tmp, n, i := [2]int{}, len(students), 0
+	for _, v := range students {
+		tmp[v]++
+	}
+	for i < n && tmp[sandwiches[i]] > 0 {
+		tmp[sandwiches[i]]--
+		i++
+	}
+	return n - i
+}
diff --git a/leetcode/1700.Number-of-Students-Unable-to-Eat-Lunch/1700. Number of Students Unable to Eat Lunch_test.go b/leetcode/1700.Number-of-Students-Unable-to-Eat-Lunch/1700. Number of Students Unable to Eat Lunch_test.go
new file mode 100644
index 000000000..d407a9687
--- /dev/null
+++ b/leetcode/1700.Number-of-Students-Unable-to-Eat-Lunch/1700. Number of Students Unable to Eat Lunch_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1700 struct {
+	para1700
+	ans1700
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1700 struct {
+	students   []int
+	sandwiches []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1700 struct {
+	one int
+}
+
+func Test_Problem1700(t *testing.T) {
+
+	qs := []question1700{
+
+		{
+			para1700{[]int{1, 1, 0, 0}, []int{0, 1, 0, 1}},
+			ans1700{0},
+		},
+
+		{
+			para1700{[]int{1, 1, 1, 0, 0, 1}, []int{1, 0, 0, 0, 1, 1}},
+			ans1700{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1700------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1700, q.para1700
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countStudents(p.students, p.sandwiches))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1700.Number-of-Students-Unable-to-Eat-Lunch/README.md b/leetcode/1700.Number-of-Students-Unable-to-Eat-Lunch/README.md
new file mode 100644
index 000000000..1ff50247b
--- /dev/null
+++ b/leetcode/1700.Number-of-Students-Unable-to-Eat-Lunch/README.md
@@ -0,0 +1,79 @@
+# [1700. Number of Students Unable to Eat Lunch](https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/)
+
+
+## 题目
+
+The school cafeteria offers circular and square sandwiches at lunch break, referred to by numbers `0` and `1` respectively. All students stand in a queue. Each student either prefers square or circular sandwiches.
+
+The number of sandwiches in the cafeteria is equal to the number of students. The sandwiches are placed in a **stack**. At each step:
+
+- If the student at the front of the queue **prefers** the sandwich on the top of the stack, they will **take it** and leave the queue.
+- Otherwise, they will **leave it** and go to the queue's end.
+
+This continues until none of the queue students want to take the top sandwich and are thus unable to eat.
+
+You are given two integer arrays `students` and `sandwiches` where `sandwiches[i]` is the type of the `ith` sandwich in the stack (`i = 0` is the top of the stack) and `students[j]` is the preference of the `jth` student in the initial queue (`j = 0` is the front of the queue). Return *the number of students that are unable to eat.*
+
+**Example 1:**
+
+```
+Input: students = [1,1,0,0], sandwiches = [0,1,0,1]
+Output: 0 
+Explanation:
+- Front student leaves the top sandwich and returns to the end of the line making students = [1,0,0,1].
+- Front student leaves the top sandwich and returns to the end of the line making students = [0,0,1,1].
+- Front student takes the top sandwich and leaves the line making students = [0,1,1] and sandwiches = [1,0,1].
+- Front student leaves the top sandwich and returns to the end of the line making students = [1,1,0].
+- Front student takes the top sandwich and leaves the line making students = [1,0] and sandwiches = [0,1].
+- Front student leaves the top sandwich and returns to the end of the line making students = [0,1].
+- Front student takes the top sandwich and leaves the line making students = [1] and sandwiches = [1].
+- Front student takes the top sandwich and leaves the line making students = [] and sandwiches = [].
+Hence all students are able to eat.
+```
+
+**Example 2:**
+
+```
+Input: students = [1,1,1,0,0,1], sandwiches = [1,0,0,0,1,1]
+Output: 3
+```
+
+**Constraints:**
+
+- `1 <= students.length, sandwiches.length <= 100`
+- `students.length == sandwiches.length`
+- `sandwiches[i]` is `0` or `1`.
+- `students[i]` is `0` or `1`.
+
+## 题目大意
+
+学校的自助午餐提供圆形和方形的三明治，分别用数字 0 和 1 表示。所有学生站在一个队列里，每个学生要么喜欢圆形的要么喜欢方形的。
+餐厅里三明治的数量与学生的数量相同。所有三明治都放在一个 栈 里，每一轮：
+
+- 如果队列最前面的学生 喜欢 栈顶的三明治，那么会 拿走它 并离开队列。
+- 否则，这名学生会 放弃这个三明治 并回到队列的尾部。
+这个过程会一直持续到队列里所有学生都不喜欢栈顶的三明治为止。
+
+给你两个整数数组 students 和 sandwiches ，其中 sandwiches[i] 是栈里面第 i 个三明治的类型（i = 0 是栈的顶部）， students[j] 是初始队列里第 j 名学生对三明治的喜好（j = 0 是队列的最开始位置）。请你返回无法吃午餐的学生数量。
+
+## 解题思路
+
+- 简单题。按照题意，学生不管怎么轮流领三明治，如果数量够，经过多轮循环，总能领到。问题可以等价为，学生依次到队列前面领取三明治。2 个种类的三明治都摆好放在那里了。最终领不到三明治的学生都是因为喜欢的三明治不够发放了。按照这个思路，先统计 2 种三明治的总个数，然后减去学生对三明治的需求总数，剩下的学生即都是无法满足的。
+
+## 代码
+
+```go
+package leetcode
+
+func countStudents(students []int, sandwiches []int) int {
+	tmp, n, i := [2]int{}, len(students), 0
+	for _, v := range students {
+		tmp[v]++
+	}
+	for i < n && tmp[sandwiches[i]] > 0 {
+		tmp[sandwiches[i]]--
+		i++
+	}
+	return n - i
+}
+```
\ No newline at end of file
diff --git a/leetcode/1704.Determine-if-String-Halves-Are-Alike/1704. Determine if String Halves Are Alike.go b/leetcode/1704.Determine-if-String-Halves-Are-Alike/1704. Determine if String Halves Are Alike.go
new file mode 100644
index 000000000..4b2296824
--- /dev/null
+++ b/leetcode/1704.Determine-if-String-Halves-Are-Alike/1704. Determine if String Halves Are Alike.go	
@@ -0,0 +1,16 @@
+package leetcode
+
+func halvesAreAlike(s string) bool {
+	return numVowels(s[len(s)/2:]) == numVowels(s[:len(s)/2])
+}
+
+func numVowels(x string) int {
+	res := 0
+	for _, c := range x {
+		switch c {
+		case 'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U':
+			res++
+		}
+	}
+	return res
+}
diff --git a/leetcode/1704.Determine-if-String-Halves-Are-Alike/1704. Determine if String Halves Are Alike_test.go b/leetcode/1704.Determine-if-String-Halves-Are-Alike/1704. Determine if String Halves Are Alike_test.go
new file mode 100644
index 000000000..fd7060404
--- /dev/null
+++ b/leetcode/1704.Determine-if-String-Halves-Are-Alike/1704. Determine if String Halves Are Alike_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1704 struct {
+	para1704
+	ans1704
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1704 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1704 struct {
+	one bool
+}
+
+func Test_Problem1704(t *testing.T) {
+
+	qs := []question1704{
+
+		{
+			para1704{"book"},
+			ans1704{true},
+		},
+
+		{
+			para1704{"textbook"},
+			ans1704{false},
+		},
+
+		{
+			para1704{"MerryChristmas"},
+			ans1704{false},
+		},
+
+		{
+			para1704{"AbCdEfGh"},
+			ans1704{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1704------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1704, q.para1704
+		fmt.Printf("【input】:%v       【output】:%v\n", p, halvesAreAlike(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1704.Determine-if-String-Halves-Are-Alike/README.md b/leetcode/1704.Determine-if-String-Halves-Are-Alike/README.md
new file mode 100644
index 000000000..323687048
--- /dev/null
+++ b/leetcode/1704.Determine-if-String-Halves-Are-Alike/README.md
@@ -0,0 +1,75 @@
+# [1704. Determine if String Halves Are Alike](https://leetcode.com/problems/determine-if-string-halves-are-alike/)
+
+## 题目
+
+You are given a string `s` of even length. Split this string into two halves of equal lengths, and let `a` be the first half and `b` be the second half.
+
+Two strings are **alike** if they have the same number of vowels (`'a'`, `'e'`, `'i'`, `'o'`, `'u'`, `'A'`, `'E'`, `'I'`, `'O'`, `'U'`). Notice that `s` contains uppercase and lowercase letters.
+
+Return `true` *if* `a` *and* `b` *are **alike***. Otherwise, return `false`.
+
+**Example 1:**
+
+```
+Input: s = "book"
+Output: true
+Explanation: a = "bo" and b = "ok". a has 1 vowel and b has 1 vowel. Therefore, they are alike.
+```
+
+**Example 2:**
+
+```
+Input: s = "textbook"
+Output: false
+Explanation: a = "text" and b = "book". a has 1 vowel whereas b has 2. Therefore, they are not alike.
+Notice that the vowel o is counted twice.
+```
+
+**Example 3:**
+
+```
+Input: s = "MerryChristmas"
+Output: false
+```
+
+**Example 4:**
+
+```
+Input: s = "AbCdEfGh"
+Output: true
+```
+
+**Constraints:**
+
+- `2 <= s.length <= 1000`
+- `s.length` is even.
+- `s` consists of **uppercase and lowercase** letters.
+
+## 题目大意
+
+给你一个偶数长度的字符串 s 。将其拆分成长度相同的两半，前一半为 a ，后一半为 b 。两个字符串 相似 的前提是它们都含有相同数目的元音（'a'，'e'，'i'，'o'，'u'，'A'，'E'，'I'，'O'，'U'）。注意，s 可能同时含有大写和小写字母。如果 a 和 b 相似，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 简单题。依题意，分别统计前半段元音字母的个数和后半段元音字母的个数，个数相同则输出 true，不同就输出 false。
+
+## 代码
+
+```go
+package leetcode
+
+func halvesAreAlike(s string) bool {
+	return numVowels(s[len(s)/2:]) == numVowels(s[:len(s)/2])
+}
+
+func numVowels(x string) int {
+	res := 0
+	for _, c := range x {
+		switch c {
+		case 'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U':
+			res++
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1705.Maximum-Number-of-Eaten-Apples/1705.Maximum Number of Eaten Apples.go b/leetcode/1705.Maximum-Number-of-Eaten-Apples/1705.Maximum Number of Eaten Apples.go
new file mode 100644
index 000000000..a4ffd6cf3
--- /dev/null
+++ b/leetcode/1705.Maximum-Number-of-Eaten-Apples/1705.Maximum Number of Eaten Apples.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import "container/heap"
+
+func eatenApples(apples []int, days []int) int {
+	h := hp{}
+	i := 0
+	var ans int
+	for ; i < len(apples); i++ {
+		for len(h) > 0 && h[0].end <= i {
+			heap.Pop(&h)
+		}
+		if apples[i] > 0 {
+			heap.Push(&h, data{apples[i], i + days[i]})
+		}
+		if len(h) > 0 {
+			minData := heap.Pop(&h).(data)
+			ans++
+			if minData.left > 1 {
+				heap.Push(&h, data{minData.left - 1, minData.end})
+			}
+		}
+	}
+	for len(h) > 0 {
+		for len(h) > 0 && h[0].end <= i {
+			heap.Pop(&h)
+		}
+		if len(h) == 0 {
+			break
+		}
+		minData := heap.Pop(&h).(data)
+		nums := min(minData.left, minData.end-i)
+		ans += nums
+		i += nums
+	}
+	return ans
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+type data struct {
+	left int
+	end  int
+}
+
+type hp []data
+
+func (h hp) Len() int           { return len(h) }
+func (h hp) Less(i, j int) bool { return h[i].end < h[j].end }
+func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
+
+func (h *hp) Push(x interface{}) {
+	*h = append(*h, x.(data))
+}
+
+func (h *hp) Pop() interface{} {
+	old := *h
+	n := len(old)
+	x := old[n-1]
+	*h = old[0 : n-1]
+	return x
+}
diff --git a/leetcode/1705.Maximum-Number-of-Eaten-Apples/1705.Maximum Number of Eaten Apples_test.go b/leetcode/1705.Maximum-Number-of-Eaten-Apples/1705.Maximum Number of Eaten Apples_test.go
new file mode 100644
index 000000000..3b178c7a0
--- /dev/null
+++ b/leetcode/1705.Maximum-Number-of-Eaten-Apples/1705.Maximum Number of Eaten Apples_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1705 struct {
+	para1705
+	ans1705
+}
+
+// para 是参数
+type para1705 struct {
+	apples []int
+	days   []int
+}
+
+// ans 是答案
+type ans1705 struct {
+	ans int
+}
+
+func Test_Problem1705(t *testing.T) {
+
+	qs := []question1705{
+
+		{
+			para1705{[]int{1, 2, 3, 5, 2}, []int{3, 2, 1, 4, 2}},
+			ans1705{7},
+		},
+
+		{
+			para1705{[]int{3, 0, 0, 0, 0, 2}, []int{3, 0, 0, 0, 0, 2}},
+			ans1705{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1705------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1705, q.para1705
+		fmt.Printf("【input】:%v    【output】:%v\n", p, eatenApples(p.apples, p.days))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1705.Maximum-Number-of-Eaten-Apples/README.md b/leetcode/1705.Maximum-Number-of-Eaten-Apples/README.md
new file mode 100644
index 000000000..b5717e75a
--- /dev/null
+++ b/leetcode/1705.Maximum-Number-of-Eaten-Apples/README.md
@@ -0,0 +1,124 @@
+# [1705. Maximum Number of Eaten Apples](https://leetcode.com/problems/maximum-number-of-eaten-apples/)
+
+## 题目
+
+There is a special kind of apple tree that grows apples every day for n days. On the ith day, the tree grows apples[i] apples that will rot after days[i] days, that is on day i + days[i] the apples will be rotten and cannot be eaten. On some days, the apple tree does not grow any apples, which are denoted by apples[i] == 0 and days[i] == 0.
+
+You decided to eat at most one apple a day (to keep the doctors away). Note that you can keep eating after the first n days.
+
+Given two integer arrays days and apples of length n, return the maximum number of apples you can eat.
+
+**Example 1**:
+
+    Input: apples = [1,2,3,5,2], days = [3,2,1,4,2]
+    Output: 7
+    Explanation: You can eat 7 apples:
+    - On the first day, you eat an apple that grew on the first day.
+    - On the second day, you eat an apple that grew on the second day.
+    - On the third day, you eat an apple that grew on the second day. After this day, the apples that grew on the third day rot.
+    - On the fourth to the seventh days, you eat apples that grew on the fourth day.
+
+**Example 2**:
+
+    Input: apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
+    Output: 5
+    Explanation: You can eat 5 apples:
+    - On the first to the third day you eat apples that grew on the first day.
+    - Do nothing on the fouth and fifth days.
+    - On the sixth and seventh days you eat apples that grew on the sixth day.
+
+**Constraints:**
+
+- apples.length == n
+- days.length == n
+- 1 <= n <= 2 * 10000
+- 0 <= apples[i], days[i] <= 2 * 10000
+- days[i] = 0 if and only if apples[i] = 0.
+
+## 题目大意
+
+有一棵特殊的苹果树，一连 n 天，每天都可以长出若干个苹果。在第 i 天，树上会长出 apples[i] 个苹果，这些苹果将会在 days[i] 天后（也就是说，第 i + days[i] 天时）腐烂，变得无法食用。也可能有那么几天，树上不会长出新的苹果，此时用 apples[i] == 0 且 days[i] == 0 表示。
+
+你打算每天 最多 吃一个苹果来保证营养均衡。注意，你可以在这 n 天之后继续吃苹果。
+
+给你两个长度为 n 的整数数组 days 和 apples ，返回你可以吃掉的苹果的最大数目。
+
+## 解题思路
+
+贪心算法和最小堆
+
+  - data中的end表示腐烂的日期，left表示拥有的苹果数量
+  - 贪心:每天吃掉end最小但没有腐烂的苹果
+  - 最小堆:构造类型为数组(数组中元素的类型为data)的最小堆
+
+## 代码
+
+```go
+package leetcode
+
+import "container/heap"
+
+func eatenApples(apples []int, days []int) int {
+	h := hp{}
+	i := 0
+	var ans int
+	for ; i < len(apples); i++ {
+		for len(h) > 0 && h[0].end <= i {
+			heap.Pop(&h)
+		}
+		if apples[i] > 0 {
+			heap.Push(&h, data{apples[i], i + days[i]})
+		}
+		if len(h) > 0 {
+			minData := heap.Pop(&h).(data)
+			ans++
+			if minData.left > 1 {
+				heap.Push(&h, data{minData.left - 1, minData.end})
+			}
+		}
+	}
+	for len(h) > 0 {
+		for len(h) > 0 && h[0].end <= i {
+			heap.Pop(&h)
+		}
+		if len(h) == 0 {
+			break
+		}
+		minData := heap.Pop(&h).(data)
+		nums := min(minData.left, minData.end-i)
+		ans += nums
+		i += nums
+	}
+	return ans
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+type data struct {
+	left int
+	end  int
+}
+
+type hp []data
+
+func (h hp) Len() int           { return len(h) }
+func (h hp) Less(i, j int) bool { return h[i].end < h[j].end }
+func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
+
+func (h *hp) Push(x interface{}) {
+	*h = append(*h, x.(data))
+}
+
+func (h *hp) Pop() interface{} {
+	old := *h
+	n := len(old)
+	x := old[n-1]
+	*h = old[0 : n-1]
+	return x
+}
+```
\ No newline at end of file
diff --git a/leetcode/1710.Maximum-Units-on-a-Truck/1710. Maximum Units on a Truck.go b/leetcode/1710.Maximum-Units-on-a-Truck/1710. Maximum Units on a Truck.go
new file mode 100644
index 000000000..7caf40e83
--- /dev/null
+++ b/leetcode/1710.Maximum-Units-on-a-Truck/1710. Maximum Units on a Truck.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+import "sort"
+
+func maximumUnits(boxTypes [][]int, truckSize int) int {
+	sort.Slice(boxTypes, func(i, j int) bool {
+		return boxTypes[i][1] > boxTypes[j][1]
+	})
+	res := 0
+	for i := 0; truckSize > 0 && i < len(boxTypes); i++ {
+		if truckSize >= boxTypes[i][0] {
+			truckSize -= boxTypes[i][0]
+			res += (boxTypes[i][1] * boxTypes[i][0])
+		} else {
+			res += (truckSize * boxTypes[i][1])
+			truckSize = 0
+		}
+	}
+	return res
+}
diff --git a/leetcode/1710.Maximum-Units-on-a-Truck/1710. Maximum Units on a Truck_test.go b/leetcode/1710.Maximum-Units-on-a-Truck/1710. Maximum Units on a Truck_test.go
new file mode 100644
index 000000000..2c2b216da
--- /dev/null
+++ b/leetcode/1710.Maximum-Units-on-a-Truck/1710. Maximum Units on a Truck_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1710 struct {
+	para1710
+	ans1710
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1710 struct {
+	boxTypes  [][]int
+	truckSize int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1710 struct {
+	one int
+}
+
+func Test_Problem1710(t *testing.T) {
+
+	qs := []question1710{
+
+		{
+			para1710{[][]int{{1, 3}, {2, 2}, {3, 1}}, 4},
+			ans1710{8},
+		},
+
+		{
+			para1710{[][]int{{5, 10}, {2, 5}, {4, 7}, {3, 9}}, 10},
+			ans1710{91},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1710------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1710, q.para1710
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maximumUnits(p.boxTypes, p.truckSize))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1710.Maximum-Units-on-a-Truck/README.md b/leetcode/1710.Maximum-Units-on-a-Truck/README.md
new file mode 100644
index 000000000..f6ab86363
--- /dev/null
+++ b/leetcode/1710.Maximum-Units-on-a-Truck/README.md
@@ -0,0 +1,77 @@
+# [1710. Maximum Units on a Truck](https://leetcode.com/problems/maximum-units-on-a-truck/)
+
+
+## 题目
+
+You are assigned to put some amount of boxes onto **one truck**. You are given a 2D array `boxTypes`, where `boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]`:
+
+- `numberOfBoxesi` is the number of boxes of type `i`.
+- `numberOfUnitsPerBoxi`is the number of units in each box of the type `i`.
+
+You are also given an integer `truckSize`, which is the **maximum** number of **boxes** that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed `truckSize`.
+
+Return *the **maximum** total number of **units** that can be put on the truck.*
+
+**Example 1:**
+
+```
+Input: boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4
+Output: 8
+Explanation: There are:
+- 1 box of the first type that contains 3 units.
+- 2 boxes of the second type that contain 2 units each.
+- 3 boxes of the third type that contain 1 unit each.
+You can take all the boxes of the first and second types, and one box of the third type.
+The total number of units will be = (1 * 3) + (2 * 2) + (1 * 1) = 8.
+```
+
+**Example 2:**
+
+```
+Input: boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10
+Output: 91
+```
+
+**Constraints:**
+
+- `1 <= boxTypes.length <= 1000`
+- `1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000`
+- `1 <= truckSize <= 106`
+
+## 题目大意
+
+请你将一些箱子装在 一辆卡车 上。给你一个二维数组 boxTypes ，其中 boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi] ：
+
+- numberOfBoxesi 是类型 i 的箱子的数量。-
+- numberOfUnitsPerBoxi 是类型 i 每个箱子可以装载的单元数量。
+
+整数 truckSize 表示卡车上可以装载 箱子 的 最大数量 。只要箱子数量不超过 truckSize ，你就可以选择任意箱子装到卡车上。返回卡车可以装载 单元 的 最大 总数。
+
+## 解题思路
+
+- 简单题。先将箱子按照单元数量从大到小排序。要想卡车装载单元数最大，那么需要尽量装单元数多的箱子。所以排序以后从单元数量多的箱子开始取。一直取至 truckSize 没有空间。累积的单元数即最大总数。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func maximumUnits(boxTypes [][]int, truckSize int) int {
+	sort.Slice(boxTypes, func(i, j int) bool {
+		return boxTypes[i][1] > boxTypes[j][1]
+	})
+	res := 0
+	for i := 0; truckSize > 0 && i < len(boxTypes); i++ {
+		if truckSize >= boxTypes[i][0] {
+			truckSize -= boxTypes[i][0]
+			res += (boxTypes[i][1] * boxTypes[i][0])
+		} else {
+			res += (truckSize * boxTypes[i][1])
+			truckSize = 0
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1716.Calculate-Money-in-Leetcode-Bank/1716. Calculate Money in Leetcode Bank.go b/leetcode/1716.Calculate-Money-in-Leetcode-Bank/1716. Calculate Money in Leetcode Bank.go
new file mode 100644
index 000000000..20e62732a
--- /dev/null
+++ b/leetcode/1716.Calculate-Money-in-Leetcode-Bank/1716. Calculate Money in Leetcode Bank.go	
@@ -0,0 +1,13 @@
+package leetcode
+
+func totalMoney(n int) int {
+	res := 0
+	for tmp, count := 1, 7; n > 0; tmp, count = tmp+1, 7 {
+		for m := tmp; n > 0 && count > 0; m++ {
+			res += m
+			n--
+			count--
+		}
+	}
+	return res
+}
diff --git a/leetcode/1716.Calculate-Money-in-Leetcode-Bank/1716. Calculate Money in Leetcode Bank_test.go b/leetcode/1716.Calculate-Money-in-Leetcode-Bank/1716. Calculate Money in Leetcode Bank_test.go
new file mode 100644
index 000000000..e6c47c29c
--- /dev/null
+++ b/leetcode/1716.Calculate-Money-in-Leetcode-Bank/1716. Calculate Money in Leetcode Bank_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1716 struct {
+	para1716
+	ans1716
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1716 struct {
+	n int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1716 struct {
+	one int
+}
+
+func Test_Problem1716(t *testing.T) {
+
+	qs := []question1716{
+
+		{
+			para1716{4},
+			ans1716{10},
+		},
+
+		{
+			para1716{10},
+			ans1716{37},
+		},
+
+		{
+			para1716{20},
+			ans1716{96},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1716------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1716, q.para1716
+		fmt.Printf("【input】:%v       【output】:%v\n", p, totalMoney(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1716.Calculate-Money-in-Leetcode-Bank/README.md b/leetcode/1716.Calculate-Money-in-Leetcode-Bank/README.md
new file mode 100644
index 000000000..34eb627db
--- /dev/null
+++ b/leetcode/1716.Calculate-Money-in-Leetcode-Bank/README.md
@@ -0,0 +1,64 @@
+# [1716. Calculate Money in Leetcode Bank](https://leetcode.com/problems/calculate-money-in-leetcode-bank/)
+
+
+## 题目
+
+Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
+
+He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
+
+Given `n`, return *the total amount of money he will have in the Leetcode bank at the end of the* `nth` *day.*
+
+**Example 1:**
+
+```
+Input: n = 4
+Output: 10
+Explanation: After the 4th day, the total is 1 + 2 + 3 + 4 = 10.
+```
+
+**Example 2:**
+
+```
+Input: n = 10
+Output: 37
+Explanation: After the 10th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4) = 37. Notice that on the 2nd Monday, Hercy only puts in $2.
+```
+
+**Example 3:**
+
+```
+Input: n = 20
+Output: 96
+Explanation: After the 20th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4 + 5 + 6 + 7 + 8) + (3 + 4 + 5 + 6 + 7 + 8) = 96.
+```
+
+**Constraints:**
+
+- `1 <= n <= 1000`
+
+## 题目大意
+
+Hercy 想要为购买第一辆车存钱。他 每天 都往力扣银行里存钱。最开始，他在周一的时候存入 1 块钱。从周二到周日，他每天都比前一天多存入 1 块钱。在接下来每一个周一，他都会比 前一个周一 多存入 1 块钱。给你 n ，请你返回在第 n 天结束的时候他在力扣银行总共存了多少块钱。
+
+## 解题思路
+
+- 简单题。按照题意写 2 层循环即可。
+
+## 代码
+
+```go
+package leetcode
+
+func totalMoney(n int) int {
+	res := 0
+	for tmp, count := 1, 7; n > 0; tmp, count = tmp+1, 7 {
+		for m := tmp; n > 0 && count > 0; m++ {
+			res += m
+			n--
+			count--
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1720.Decode-XORed-Array/1720. Decode XORed Array.go b/leetcode/1720.Decode-XORed-Array/1720. Decode XORed Array.go
new file mode 100644
index 000000000..108624cbc
--- /dev/null
+++ b/leetcode/1720.Decode-XORed-Array/1720. Decode XORed Array.go	
@@ -0,0 +1,10 @@
+package leetcode
+
+func decode(encoded []int, first int) []int {
+	arr := make([]int, len(encoded)+1)
+	arr[0] = first
+	for i, val := range encoded {
+		arr[i+1] = arr[i] ^ val
+	}
+	return arr
+}
diff --git a/leetcode/1720.Decode-XORed-Array/1720. Decode XORed Array_test.go b/leetcode/1720.Decode-XORed-Array/1720. Decode XORed Array_test.go
new file mode 100644
index 000000000..2b152ecc2
--- /dev/null
+++ b/leetcode/1720.Decode-XORed-Array/1720. Decode XORed Array_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1720 struct {
+	para1720
+	ans1720
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1720 struct {
+	encoded []int
+	first   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1720 struct {
+	one []int
+}
+
+func Test_Problem1720(t *testing.T) {
+
+	qs := []question1720{
+
+		{
+			para1720{[]int{1, 2, 3}, 1},
+			ans1720{[]int{1, 0, 2, 1}},
+		},
+
+		{
+			para1720{[]int{6, 2, 7, 3}, 4},
+			ans1720{[]int{4, 2, 0, 7, 4}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1720------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1720, q.para1720
+		fmt.Printf("【input】:%v       【output】:%v\n", p, decode(p.encoded, p.first))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1720.Decode-XORed-Array/README.md b/leetcode/1720.Decode-XORed-Array/README.md
new file mode 100644
index 000000000..9cff1dcc8
--- /dev/null
+++ b/leetcode/1720.Decode-XORed-Array/README.md
@@ -0,0 +1,59 @@
+# [1720. Decode XORed Array](https://leetcode.com/problems/decode-xored-array/)
+
+
+## 题目
+
+There is a **hidden** integer array `arr` that consists of `n` non-negative integers.
+
+It was encoded into another integer array `encoded` of length `n - 1`, such that `encoded[i] = arr[i] XOR arr[i + 1]`. For example, if `arr = [1,0,2,1]`, then `encoded = [1,2,3]`.
+
+You are given the `encoded` array. You are also given an integer `first`, that is the first element of `arr`, i.e. `arr[0]`.
+
+Return *the original array* `arr`. It can be proved that the answer exists and is unique.
+
+**Example 1:**
+
+```
+Input: encoded = [1,2,3], first = 1
+Output: [1,0,2,1]
+Explanation: If arr = [1,0,2,1], then first = 1 and encoded = [1 XOR 0, 0 XOR 2, 2 XOR 1] = [1,2,3]
+
+```
+
+**Example 2:**
+
+```
+Input: encoded = [6,2,7,3], first = 4
+Output: [4,2,0,7,4]
+
+```
+
+**Constraints:**
+
+- `2 <= n <= 104`
+- `encoded.length == n - 1`
+- `0 <= encoded[i] <= 105`
+- `0 <= first <= 10^5`
+
+## 题目大意
+
+未知 整数数组 arr 由 n 个非负整数组成。经编码后变为长度为 n - 1 的另一个整数数组 encoded ，其中 encoded[i] = arr[i] XOR arr[i + 1] 。例如，arr = [1,0,2,1] 经编码后得到 encoded = [1,2,3] 。给你编码后的数组 encoded 和原数组 arr 的第一个元素 first（arr[0]）。请解码返回原数组 arr 。可以证明答案存在并且是唯一的。
+
+## 解题思路
+
+- 简单题。按照题意，求返回解码以后的原数组，即将编码后的数组前后两两元素依次做异或 `XOR` 运算。
+
+## 代码
+
+```go
+package leetcode
+
+func decode(encoded []int, first int) []int {
+	arr := make([]int, len(encoded)+1)
+	arr[0] = first
+	for i, val := range encoded {
+		arr[i+1] = arr[i] ^ val
+	}
+	return arr
+}
+```
\ No newline at end of file
diff --git a/leetcode/1721.Swapping-Nodes-in-a-Linked-List/1721. Swapping Nodes in a Linked List.go b/leetcode/1721.Swapping-Nodes-in-a-Linked-List/1721. Swapping Nodes in a Linked List.go
new file mode 100644
index 000000000..7f5b5173c
--- /dev/null
+++ b/leetcode/1721.Swapping-Nodes-in-a-Linked-List/1721. Swapping Nodes in a Linked List.go	
@@ -0,0 +1,36 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func swapNodes(head *ListNode, k int) *ListNode {
+	count := 1
+	var a, b *ListNode
+	for node := head; node != nil; node = node.Next {
+		if count == k {
+			a = node
+		}
+		count++
+	}
+	length := count
+	count = 1
+	for node := head; node != nil; node = node.Next {
+		if count == length-k {
+			b = node
+		}
+		count++
+	}
+	a.Val, b.Val = b.Val, a.Val
+	return head
+}
diff --git a/leetcode/1721.Swapping-Nodes-in-a-Linked-List/1721. Swapping Nodes in a Linked List_test.go b/leetcode/1721.Swapping-Nodes-in-a-Linked-List/1721. Swapping Nodes in a Linked List_test.go
new file mode 100644
index 000000000..7a9345afb
--- /dev/null
+++ b/leetcode/1721.Swapping-Nodes-in-a-Linked-List/1721. Swapping Nodes in a Linked List_test.go	
@@ -0,0 +1,66 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question2 struct {
+	para2
+	ans2
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2 struct {
+	head []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2 struct {
+	one []int
+}
+
+func Test_Problem2(t *testing.T) {
+
+	qs := []question2{
+
+		{
+			para2{[]int{1, 2, 3, 4, 5}, 2},
+			ans2{[]int{1, 4, 3, 2, 5}},
+		},
+
+		{
+			para2{[]int{7, 9, 6, 6, 7, 8, 3, 0, 9, 5}, 5},
+			ans2{[]int{7, 9, 6, 6, 8, 7, 3, 0, 9, 5}},
+		},
+
+		{
+			para2{[]int{1}, 1},
+			ans2{[]int{1}},
+		},
+
+		{
+			para2{[]int{1, 2}, 1},
+			ans2{[]int{2, 1}},
+		},
+
+		{
+			para2{[]int{1, 2, 3}, 2},
+			ans2{[]int{1, 2, 3}},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2, q.para2
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(swapNodes(structures.Ints2List(p.head), p.k)))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1721.Swapping-Nodes-in-a-Linked-List/README.md b/leetcode/1721.Swapping-Nodes-in-a-Linked-List/README.md
new file mode 100644
index 000000000..d05957693
--- /dev/null
+++ b/leetcode/1721.Swapping-Nodes-in-a-Linked-List/README.md
@@ -0,0 +1,100 @@
+# [1721. Swapping Nodes in a Linked List](https://leetcode.com/problems/swapping-nodes-in-a-linked-list/)
+
+
+## 题目
+
+You are given the `head` of a linked list, and an integer `k`.
+
+Return *the head of the linked list after **swapping** the values of the* `kth` *node from the beginning and the* `kth` *node from the end (the list is **1-indexed**).*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/09/21/linked1.jpg](https://assets.leetcode.com/uploads/2020/09/21/linked1.jpg)
+
+```
+Input: head = [1,2,3,4,5], k = 2
+Output: [1,4,3,2,5]
+```
+
+**Example 2:**
+
+```
+Input: head = [7,9,6,6,7,8,3,0,9,5], k = 5
+Output: [7,9,6,6,8,7,3,0,9,5]
+```
+
+**Example 3:**
+
+```
+Input: head = [1], k = 1
+Output: [1]
+```
+
+**Example 4:**
+
+```
+Input: head = [1,2], k = 1
+Output: [2,1]
+```
+
+**Example 5:**
+
+```
+Input: head = [1,2,3], k = 2
+Output: [1,2,3]
+```
+
+**Constraints:**
+
+- The number of nodes in the list is `n`.
+- `1 <= k <= n <= 10^5`
+- `0 <= Node.val <= 100`
+
+## 题目大意
+
+给你链表的头节点 `head` 和一个整数 `k` 。**交换** 链表正数第 `k` 个节点和倒数第 `k` 个节点的值后，返回链表的头节点（链表 **从 1 开始索引**）。
+
+## 解题思路
+
+- 这道题虽然是 medium，但是实际非常简单。题目要求链表中 2 个节点的值，无非是先找到这 2 个节点，然后再交换即可。链表查询节点需要 O(n)，2 次循环找到对应的 2 个节点，交换值即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func swapNodes(head *ListNode, k int) *ListNode {
+	count := 1
+	var a, b *ListNode
+	for node := head; node != nil; node = node.Next {
+		if count == k {
+			a = node
+		}
+		count++
+	}
+	length := count
+	count = 1
+	for node := head; node != nil; node = node.Next {
+		if count == length-k {
+			b = node
+		}
+		count++
+	}
+	a.Val, b.Val = b.Val, a.Val
+	return head
+}
+```
\ No newline at end of file
diff --git a/leetcode/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square/1725. Number Of Rectangles That Can Form The Largest Square.go b/leetcode/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square/1725. Number Of Rectangles That Can Form The Largest Square.go
new file mode 100644
index 000000000..4c7bf664b
--- /dev/null
+++ b/leetcode/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square/1725. Number Of Rectangles That Can Form The Largest Square.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+func countGoodRectangles(rectangles [][]int) int {
+	minLength, count := 0, 0
+	for i := range rectangles {
+		minSide := 0
+		if rectangles[i][0] <= rectangles[i][1] {
+			minSide = rectangles[i][0]
+		} else {
+			minSide = rectangles[i][1]
+		}
+		if minSide > minLength {
+			minLength = minSide
+			count = 1
+		} else if minSide == minLength {
+			count++
+		}
+	}
+	return count
+}
diff --git a/leetcode/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square/1725. Number Of Rectangles That Can Form The Largest Square_test.go b/leetcode/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square/1725. Number Of Rectangles That Can Form The Largest Square_test.go
new file mode 100644
index 000000000..40b7d86e9
--- /dev/null
+++ b/leetcode/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square/1725. Number Of Rectangles That Can Form The Largest Square_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1725 struct {
+	para1725
+	ans1725
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1725 struct {
+	rectangles [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1725 struct {
+	one int
+}
+
+func Test_Problem1725(t *testing.T) {
+
+	qs := []question1725{
+
+		{
+			para1725{[][]int{{5, 8}, {3, 9}, {5, 12}, {16, 5}}},
+			ans1725{3},
+		},
+
+		{
+			para1725{[][]int{{2, 3}, {3, 7}, {4, 3}, {3, 7}}},
+			ans1725{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1725------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1725, q.para1725
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countGoodRectangles(p.rectangles))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square/README.md b/leetcode/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square/README.md
new file mode 100644
index 000000000..0c59194fa
--- /dev/null
+++ b/leetcode/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square/README.md
@@ -0,0 +1,68 @@
+# [1725. Number Of Rectangles That Can Form The Largest Square](https://leetcode.com/problems/number-of-rectangles-that-can-form-the-largest-square/)
+
+
+## 题目
+
+You are given an array `rectangles` where `rectangles[i] = [li, wi]` represents the `ith` rectangle of length `li` and width `wi`.
+
+You can cut the `ith` rectangle to form a square with a side length of `k` if both `k <= li` and `k <= wi`. For example, if you have a rectangle `[4,6]`, you can cut it to get a square with a side length of at most `4`.
+
+Let `maxLen` be the side length of the **largest** square you can obtain from any of the given rectangles.
+
+Return *the **number** of rectangles that can make a square with a side length of* `maxLen`.
+
+**Example 1:**
+
+```
+Input: rectangles = [[5,8],[3,9],[5,12],[16,5]]
+Output: 3
+Explanation: The largest squares you can get from each rectangle are of lengths [5,3,5,5].
+The largest possible square is of length 5, and you can get it out of 3 rectangles.
+```
+
+**Example 2:**
+
+```
+Input: rectangles = [[2,3],[3,7],[4,3],[3,7]]
+Output: 3
+```
+
+**Constraints:**
+
+- `1 <= rectangles.length <= 1000`
+- `rectangles[i].length == 2`
+- `1 <= li, wi <= 10^9`
+- `li != wi`
+
+## 题目大意
+
+给你一个数组 rectangles ，其中 rectangles[i] = [li, wi] 表示第 i 个矩形的长度为 li 、宽度为 wi 。如果存在 k 同时满足 k <= li 和 k <= wi ，就可以将第 i 个矩形切成边长为 k 的正方形。例如，矩形 [4,6] 可以切成边长最大为 4 的正方形。设 maxLen 为可以从矩形数组 rectangles 切分得到的 最大正方形 的边长。返回可以切出边长为 maxLen 的正方形的矩形 数目 。
+
+## 解题思路
+
+- 简单题。扫描数组中的每一个矩形，先找到边长较小的那条边，作为正方形的边长。扫描过程中动态更新最大的正方形边长，并累加计数。循环一遍结束，输出最终计数值即可。
+
+## 代码
+
+```go
+package leetcode
+
+func countGoodRectangles(rectangles [][]int) int {
+	minLength, count := 0, 0
+	for i, _ := range rectangles {
+		minSide := 0
+		if rectangles[i][0] <= rectangles[i][1] {
+			minSide = rectangles[i][0]
+		} else {
+			minSide = rectangles[i][1]
+		}
+		if minSide > minLength {
+			minLength = minSide
+			count = 1
+		} else if minSide == minLength {
+			count++
+		}
+	}
+	return count
+}
+```
\ No newline at end of file
diff --git a/leetcode/1732.Find-the-Highest-Altitude/1732. Find the Highest Altitude.go b/leetcode/1732.Find-the-Highest-Altitude/1732. Find the Highest Altitude.go
new file mode 100644
index 000000000..62d66bf6d
--- /dev/null
+++ b/leetcode/1732.Find-the-Highest-Altitude/1732. Find the Highest Altitude.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+func largestAltitude(gain []int) int {
+	max, height := 0, 0
+	for _, g := range gain {
+		height += g
+		if height > max {
+			max = height
+		}
+	}
+	return max
+}
diff --git a/leetcode/1732.Find-the-Highest-Altitude/1732. Find the Highest Altitude_test.go b/leetcode/1732.Find-the-Highest-Altitude/1732. Find the Highest Altitude_test.go
new file mode 100644
index 000000000..eb56bfe0f
--- /dev/null
+++ b/leetcode/1732.Find-the-Highest-Altitude/1732. Find the Highest Altitude_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1732 struct {
+	para1732
+	ans1732
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1732 struct {
+	gain []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1732 struct {
+	one int
+}
+
+func Test_Problem1732(t *testing.T) {
+
+	qs := []question1732{
+
+		{
+			para1732{[]int{-5, 1, 5, 0, -7}},
+			ans1732{1},
+		},
+
+		{
+			para1732{[]int{-4, -3, -2, -1, 4, 3, 2}},
+			ans1732{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1732------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1732, q.para1732
+		fmt.Printf("【input】:%v       【output】:%v\n", p, largestAltitude(p.gain))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1732.Find-the-Highest-Altitude/README.md b/leetcode/1732.Find-the-Highest-Altitude/README.md
new file mode 100644
index 000000000..5b6b437d8
--- /dev/null
+++ b/leetcode/1732.Find-the-Highest-Altitude/README.md
@@ -0,0 +1,55 @@
+# [1732. Find the Highest Altitude](https://leetcode.com/problems/find-the-highest-altitude/)
+
+
+## 题目
+
+There is a biker going on a road trip. The road trip consists of `n + 1` points at different altitudes. The biker starts his trip on point `0` with altitude equal `0`.
+
+You are given an integer array `gain` of length `n` where `gain[i]` is the **net gain in altitude** between points `i` and `i + 1` for all (`0 <= i < n)`. Return *the **highest altitude** of a point.*
+
+**Example 1:**
+
+```
+Input: gain = [-5,1,5,0,-7]
+Output: 1
+Explanation: The altitudes are [0,-5,-4,1,1,-6]. The highest is 1.
+```
+
+**Example 2:**
+
+```
+Input: gain = [-4,-3,-2,-1,4,3,2]
+Output: 0
+Explanation: The altitudes are [0,-4,-7,-9,-10,-6,-3,-1]. The highest is 0.
+```
+
+**Constraints:**
+
+- `n == gain.length`
+- `1 <= n <= 100`
+- `100 <= gain[i] <= 100`
+
+## 题目大意
+
+有一个自行车手打算进行一场公路骑行，这条路线总共由 n + 1 个不同海拔的点组成。自行车手从海拔为 0 的点 0 开始骑行。给你一个长度为 n 的整数数组 gain ，其中 gain[i] 是点 i 和点 i + 1 的 净海拔高度差（0 <= i < n）。请你返回 最高点的海拔 。
+
+## 解题思路
+
+- 简单题。循环数组依次从第一个海拔点开始还原每个海拔点，动态记录最大高度。循环结束输出最大高度即可。
+
+## 代码
+
+```go
+package leetcode
+
+func largestAltitude(gain []int) int {
+	max, height := 0, 0
+	for _, g := range gain {
+		height += g
+		if height > max {
+			max = height
+		}
+	}
+	return max
+}
+```
\ No newline at end of file
diff --git a/leetcode/1734.Decode-XORed-Permutation/1734. Decode XORed Permutation.go b/leetcode/1734.Decode-XORed-Permutation/1734. Decode XORed Permutation.go
new file mode 100644
index 000000000..60f24d323
--- /dev/null
+++ b/leetcode/1734.Decode-XORed-Permutation/1734. Decode XORed Permutation.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+func decode(encoded []int) []int {
+	n, total, odd := len(encoded), 0, 0
+	for i := 1; i <= n+1; i++ {
+		total ^= i
+	}
+	for i := 1; i < n; i += 2 {
+		odd ^= encoded[i]
+	}
+	perm := make([]int, n+1)
+	perm[0] = total ^ odd
+	for i, v := range encoded {
+		perm[i+1] = perm[i] ^ v
+	}
+	return perm
+}
diff --git a/leetcode/1734.Decode-XORed-Permutation/1734. Decode XORed Permutation_test.go b/leetcode/1734.Decode-XORed-Permutation/1734. Decode XORed Permutation_test.go
new file mode 100644
index 000000000..837beab52
--- /dev/null
+++ b/leetcode/1734.Decode-XORed-Permutation/1734. Decode XORed Permutation_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1734 struct {
+	para1734
+	ans1734
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1734 struct {
+	encoded []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1734 struct {
+	one []int
+}
+
+func Test_Problem1734(t *testing.T) {
+
+	qs := []question1734{
+
+		{
+			para1734{[]int{3, 1}},
+			ans1734{[]int{1, 2, 3}},
+		},
+
+		{
+			para1734{[]int{6, 5, 4, 6}},
+			ans1734{[]int{2, 4, 1, 5, 3}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1734------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1734, q.para1734
+		fmt.Printf("【input】:%v       【output】:%v\n", p, decode(p.encoded))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1734.Decode-XORed-Permutation/README.md b/leetcode/1734.Decode-XORed-Permutation/README.md
new file mode 100644
index 000000000..b60ab2e63
--- /dev/null
+++ b/leetcode/1734.Decode-XORed-Permutation/README.md
@@ -0,0 +1,71 @@
+# [1734. Decode XORed Permutation](https://leetcode.com/problems/decode-xored-permutation/)
+
+
+## 题目
+
+There is an integer array `perm` that is a permutation of the first `n` positive integers, where `n` is always **odd**.
+
+It was encoded into another integer array `encoded` of length `n - 1`, such that `encoded[i] = perm[i] XOR perm[i + 1]`. For example, if `perm = [1,3,2]`, then `encoded = [2,1]`.
+
+Given the `encoded` array, return *the original array* `perm`. It is guaranteed that the answer exists and is unique.
+
+**Example 1:**
+
+```
+Input: encoded = [3,1]
+Output: [1,2,3]
+Explanation: If perm = [1,2,3], then encoded = [1 XOR 2,2 XOR 3] = [3,1]
+
+```
+
+**Example 2:**
+
+```
+Input: encoded = [6,5,4,6]
+Output: [2,4,1,5,3]
+
+```
+
+**Constraints:**
+
+- `3 <= n < 10^5`
+- `n` is odd.
+- `encoded.length == n - 1`
+
+## 题目大意
+
+给你一个整数数组 perm ，它是前 n 个正整数的排列，且 n 是个奇数 。它被加密成另一个长度为 n - 1 的整数数组 encoded ，满足 encoded[i] = perm[i] XOR perm[i + 1] 。比方说，如果 perm = [1,3,2] ，那么 encoded = [2,1] 。给你 encoded 数组，请你返回原始数组 perm 。题目保证答案存在且唯一。
+
+## 解题思路
+
+- 这一题与第 136 题和第 137 题思路类似，借用 `x ^ x = 0` 这个性质解题。依题意，原数组 perm 是 n 个正整数，即取值在 `[1,n+1]` 区间内，但是排列顺序未知。可以考虑先将 `[1,n+1]` 区间内的所有数异或得到 total。再将 encoded 数组中奇数下标的元素异或得到 odd：
+
+    $$\begin{aligned}odd &= encoded[1] + encoded[3] + ... + encoded[n-1]\\&= (perm[1] \,\, XOR \,\, perm[2]) + (perm[3] \,\,  XOR  \,\, perm[4]) + ... + (perm[n-1]  \,\, XOR \,\, perm[n])\end{aligned}$$
+
+    total 是 n 个正整数异或全集，odd 是 `n-1` 个正整数异或集。两者异或 `total ^ odd` 得到的值必定是 perm[0]，因为 `x ^ x = 0`，那么重复出现的元素被异或以后消失了。算出 perm[0] 就好办了。
+
+    $$\begin{aligned}encoded[0] &= perm[0] \,\, XOR \,\, perm[1]\\perm[0] \,\, XOR \,\, encoded[0] &= perm[0] \,\, XOR \,\, perm[0] \,\, XOR \,\, perm[1] = perm[1]\\perm[1] \,\, XOR \,\, encoded[1] &= perm[1] \,\, XOR \,\, perm[1] \,\, XOR \,\, perm[2] = perm[2]\\...\\perm[n-1] \,\, XOR \,\, encoded[n-1] &= perm[n-1] \,\, XOR \,\, perm[n-1] \,\, XOR \,\, perm[n] = perm[n]\\\end{aligned}$$
+
+    依次类推，便可以推出原数组 perm 中的所有数。
+
+## 代码
+
+```go
+package leetcode
+
+func decode(encoded []int) []int {
+	n, total, odd := len(encoded), 0, 0
+	for i := 1; i <= n+1; i++ {
+		total ^= i
+	}
+	for i := 1; i < n; i += 2 {
+		odd ^= encoded[i]
+	}
+	perm := make([]int, n+1)
+	perm[0] = total ^ odd
+	for i, v := range encoded {
+		perm[i+1] = perm[i] ^ v
+	}
+	return perm
+}
+```
\ No newline at end of file
diff --git a/leetcode/1736.Latest-Time-by-Replacing-Hidden-Digits/1736. Latest Time by Replacing Hidden Digits.go b/leetcode/1736.Latest-Time-by-Replacing-Hidden-Digits/1736. Latest Time by Replacing Hidden Digits.go
new file mode 100644
index 000000000..7b30b387c
--- /dev/null
+++ b/leetcode/1736.Latest-Time-by-Replacing-Hidden-Digits/1736. Latest Time by Replacing Hidden Digits.go	
@@ -0,0 +1,25 @@
+package leetcode
+
+func maximumTime(time string) string {
+	timeb := []byte(time)
+	if timeb[3] == '?' {
+		timeb[3] = '5'
+	}
+	if timeb[4] == '?' {
+		timeb[4] = '9'
+	}
+	if timeb[0] == '?' {
+		if int(timeb[1]-'0') > 3 && int(timeb[1]-'0') < 10 {
+			timeb[0] = '1'
+		} else {
+			timeb[0] = '2'
+		}
+	}
+	if timeb[1] == '?' {
+		timeb[1] = '9'
+	}
+	if timeb[0] == '2' && timeb[1] == '9' {
+		timeb[1] = '3'
+	}
+	return string(timeb)
+}
diff --git a/leetcode/1736.Latest-Time-by-Replacing-Hidden-Digits/1736. Latest Time by Replacing Hidden Digits_test.go b/leetcode/1736.Latest-Time-by-Replacing-Hidden-Digits/1736. Latest Time by Replacing Hidden Digits_test.go
new file mode 100644
index 000000000..2213b85f1
--- /dev/null
+++ b/leetcode/1736.Latest-Time-by-Replacing-Hidden-Digits/1736. Latest Time by Replacing Hidden Digits_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1736 struct {
+	para1736
+	ans1736
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1736 struct {
+	time string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1736 struct {
+	one string
+}
+
+func Test_Problem1736(t *testing.T) {
+
+	qs := []question1736{
+
+		{
+			para1736{"2?:?0"},
+			ans1736{"23:50"},
+		},
+
+		{
+			para1736{"0?:3?"},
+			ans1736{"09:39"},
+		},
+
+		{
+			para1736{"1?:22"},
+			ans1736{"19:22"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1736------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1736, q.para1736
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maximumTime(p.time))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1736.Latest-Time-by-Replacing-Hidden-Digits/README.md b/leetcode/1736.Latest-Time-by-Replacing-Hidden-Digits/README.md
new file mode 100644
index 000000000..a34ceb176
--- /dev/null
+++ b/leetcode/1736.Latest-Time-by-Replacing-Hidden-Digits/README.md
@@ -0,0 +1,75 @@
+# [1736. Latest Time by Replacing Hidden Digits](https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/)
+
+
+## 题目
+
+You are given a string `time` in the form of `hh:mm`, where some of the digits in the string are hidden (represented by `?`).
+
+The valid times are those inclusively between `00:00` and `23:59`.
+
+Return *the latest valid time you can get from* `time` *by replacing the hidden* *digits*.
+
+**Example 1:**
+
+```
+Input: time = "2?:?0"
+Output: "23:50"
+Explanation: The latest hour beginning with the digit '2' is 23 and the latest minute ending with the digit '0' is 50.
+```
+
+**Example 2:**
+
+```
+Input: time = "0?:3?"
+Output: "09:39"
+```
+
+**Example 3:**
+
+```
+Input: time = "1?:22"
+Output: "19:22"
+```
+
+**Constraints:**
+
+- `time` is in the format `hh:mm`.
+- It is guaranteed that you can produce a valid time from the given string.
+
+## 题目大意
+
+给你一个字符串 time ，格式为 hh:mm（小时：分钟），其中某几位数字被隐藏（用 ? 表示）。有效的时间为 00:00 到 23:59 之间的所有时间，包括 00:00 和 23:59 。替换 time 中隐藏的数字，返回你可以得到的最晚有效时间。
+
+## 解题思路
+
+- 简单题。根据题意，需要找到最晚的有效时间。枚举时间 4 个位置即可。如果第 3 个位置是 ？，那么它最晚时间是 5；如果第 4 个位置是 ？，那么它最晚时间是 9；如果第 2 个位置是 ？，那么它最晚时间是 9；如果第 1 个位置是 ？，根据第 2 个位置判断，如果第 2 个位置是大于 3 的数，那么第一个位置最晚时间是 1，如果第 2 个位置是小于 3 的数那么第一个位置最晚时间是 2 。按照上述规则即可还原最晚时间。
+
+## 代码
+
+```go
+package leetcode
+
+func maximumTime(time string) string {
+	timeb := []byte(time)
+	if timeb[3] == '?' {
+		timeb[3] = '5'
+	}
+	if timeb[4] == '?' {
+		timeb[4] = '9'
+	}
+	if timeb[0] == '?' {
+		if int(timeb[1]-'0') > 3 && int(timeb[1]-'0') < 10 {
+			timeb[0] = '1'
+		} else {
+			timeb[0] = '2'
+		}
+	}
+	if timeb[1] == '?' {
+		timeb[1] = '9'
+	}
+	if timeb[0] == '2' && timeb[1] == '9' {
+		timeb[1] = '3'
+	}
+	return string(timeb)
+}
+```
\ No newline at end of file
diff --git a/leetcode/1738.Find-Kth-Largest-XOR-Coordinate-Value/1738. Find Kth Largest XOR Coordinate Value.go b/leetcode/1738.Find-Kth-Largest-XOR-Coordinate-Value/1738. Find Kth Largest XOR Coordinate Value.go
new file mode 100644
index 000000000..8c5760ef4
--- /dev/null
+++ b/leetcode/1738.Find-Kth-Largest-XOR-Coordinate-Value/1738. Find Kth Largest XOR Coordinate Value.go	
@@ -0,0 +1,36 @@
+package leetcode
+
+import "sort"
+
+// 解法一 压缩版的前缀和
+func kthLargestValue(matrix [][]int, k int) int {
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return 0
+	}
+	res, prefixSum := make([]int, 0, len(matrix)*len(matrix[0])), make([]int, len(matrix[0]))
+	for i := range matrix {
+		line := 0
+		for j, v := range matrix[i] {
+			line ^= v
+			prefixSum[j] ^= line
+			res = append(res, prefixSum[j])
+		}
+	}
+	sort.Ints(res)
+	return res[len(res)-k]
+}
+
+// 解法二 前缀和
+func kthLargestValue1(matrix [][]int, k int) int {
+	nums, prefixSum := []int{}, make([][]int, len(matrix)+1)
+	prefixSum[0] = make([]int, len(matrix[0])+1)
+	for i, row := range matrix {
+		prefixSum[i+1] = make([]int, len(matrix[0])+1)
+		for j, val := range row {
+			prefixSum[i+1][j+1] = prefixSum[i+1][j] ^ prefixSum[i][j+1] ^ prefixSum[i][j] ^ val
+			nums = append(nums, prefixSum[i+1][j+1])
+		}
+	}
+	sort.Ints(nums)
+	return nums[len(nums)-k]
+}
diff --git a/leetcode/1738.Find-Kth-Largest-XOR-Coordinate-Value/1738. Find Kth Largest XOR Coordinate Value_test.go b/leetcode/1738.Find-Kth-Largest-XOR-Coordinate-Value/1738. Find Kth Largest XOR Coordinate Value_test.go
new file mode 100644
index 000000000..90d199ecc
--- /dev/null
+++ b/leetcode/1738.Find-Kth-Largest-XOR-Coordinate-Value/1738. Find Kth Largest XOR Coordinate Value_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1738 struct {
+	para1738
+	ans1738
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1738 struct {
+	matrix [][]int
+	k      int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1738 struct {
+	one int
+}
+
+func Test_Problem1738(t *testing.T) {
+
+	qs := []question1738{
+
+		{
+			para1738{[][]int{{5, 2}, {1, 6}}, 1},
+			ans1738{7},
+		},
+
+		{
+			para1738{[][]int{{5, 2}, {1, 6}}, 2},
+			ans1738{5},
+		},
+
+		{
+			para1738{[][]int{{5, 2}, {1, 6}}, 3},
+			ans1738{4},
+		},
+
+		{
+			para1738{[][]int{{5, 2}, {1, 6}}, 4},
+			ans1738{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1738------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1738, q.para1738
+		fmt.Printf("【input】:%v       【output】:%v\n", p, kthLargestValue(p.matrix, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1738.Find-Kth-Largest-XOR-Coordinate-Value/README.md b/leetcode/1738.Find-Kth-Largest-XOR-Coordinate-Value/README.md
new file mode 100644
index 000000000..a264778a4
--- /dev/null
+++ b/leetcode/1738.Find-Kth-Largest-XOR-Coordinate-Value/README.md
@@ -0,0 +1,111 @@
+# [1738. Find Kth Largest XOR Coordinate Value](https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/)
+
+
+## 题目
+
+You are given a 2D `matrix` of size `m x n`, consisting of non-negative integers. You are also given an integer `k`.
+
+The **value** of coordinate `(a, b)` of the matrix is the XOR of all `matrix[i][j]` where `0 <= i <= a < m` and `0 <= j <= b < n` **(0-indexed)**.
+
+Find the `kth` largest value **(1-indexed)** of all the coordinates of `matrix`.
+
+**Example 1:**
+
+```
+Input: matrix = [[5,2],[1,6]], k = 1
+Output: 7
+Explanation: The value of coordinate (0,1) is 5 XOR 2 = 7, which is the largest value.
+```
+
+**Example 2:**
+
+```
+Input: matrix = [[5,2],[1,6]], k = 2
+Output: 5
+Explanation:The value of coordinate (0,0) is 5 = 5, which is the 2nd largest value.
+```
+
+**Example 3:**
+
+```
+Input: matrix = [[5,2],[1,6]], k = 3
+Output: 4
+Explanation: The value of coordinate (1,0) is 5 XOR 1 = 4, which is the 3rd largest value.
+```
+
+**Example 4:**
+
+```
+Input: matrix = [[5,2],[1,6]], k = 4
+Output: 0
+Explanation: The value of coordinate (1,1) is 5 XOR 2 XOR 1 XOR 6 = 0, which is the 4th largest value.
+```
+
+**Constraints:**
+
+- `m == matrix.length`
+- `n == matrix[i].length`
+- `1 <= m, n <= 1000`
+- `0 <= matrix[i][j] <= 10^6`
+- `1 <= k <= m * n`
+
+## 题目大意
+
+给你一个二维矩阵 matrix 和一个整数 k ，矩阵大小为 m x n 由非负整数组成。矩阵中坐标 (a, b) 的 值 可由对所有满足 0 <= i <= a < m 且 0 <= j <= b < n 的元素 matrix[i][j]（下标从 0 开始计数）执行异或运算得到。请你找出 matrix 的所有坐标中第 k 大的值（k 的值从 1 开始计数）。
+
+## 解题思路
+
+- 区间异或结果类比于区间二维前缀和。只不过需要注意 x^x = 0 这一性质。举例：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_1738_0_.png)
+
+    通过简单推理，可以得出区间二维前缀和 preSum 的递推式。具体代码见解法二。
+
+- 上面的解法中，preSum 用二维数组计算的。能否再优化空间复杂度，降低成 O(n)？答案是可以的。通过观察可以发现。preSum 可以按照一行一行来生成。先生成 preSum 前一行，下一行生成过程中会用到前一行的信息，异或计算以后，可以覆盖原数据(前一行的信息)，对之后的计算没有影响。这个优化空间复杂度的方法和优化 DP 空间复杂度是完全一样的思路和方法。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_1738_1_.png)
+
+    具体代码见解法一。
+
+- 计算出了 preSum，还需要考虑如何输出第 k 大的值。有 3 种做法，第一种是排序，第二种是优先队列，第三种是第 215 题中的 O(n) 的 partition 方法。时间复杂度最低的当然是 O(n)。但是经过实际测试，runtime 最优的是排序的方法。所以笔者以下两种方法均采用了排序的方法。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+// 解法一 压缩版的前缀和
+func kthLargestValue(matrix [][]int, k int) int {
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return 0
+	}
+	res, prefixSum := make([]int, 0, len(matrix)*len(matrix[0])), make([]int, len(matrix[0]))
+	for i := range matrix {
+		line := 0
+		for j, v := range matrix[i] {
+			line ^= v
+			prefixSum[j] ^= line
+			res = append(res, prefixSum[j])
+		}
+	}
+	sort.Ints(res)
+	return res[len(res)-k]
+}
+
+// 解法二 前缀和
+func kthLargestValue1(matrix [][]int, k int) int {
+	nums, prefixSum := []int{}, make([][]int, len(matrix)+1)
+	prefixSum[0] = make([]int, len(matrix[0])+1)
+	for i, row := range matrix {
+		prefixSum[i+1] = make([]int, len(matrix[0])+1)
+		for j, val := range row {
+			prefixSum[i+1][j+1] = prefixSum[i+1][j] ^ prefixSum[i][j+1] ^ prefixSum[i][j] ^ val
+			nums = append(nums, prefixSum[i+1][j+1])
+		}
+	}
+	sort.Ints(nums)
+	return nums[len(nums)-k]
+}
+```
\ No newline at end of file
diff --git a/leetcode/1742.Maximum-Number-of-Balls-in-a-Box/1742. Maximum Number of Balls in a Box.go b/leetcode/1742.Maximum-Number-of-Balls-in-a-Box/1742. Maximum Number of Balls in a Box.go
new file mode 100644
index 000000000..e4d415872
--- /dev/null
+++ b/leetcode/1742.Maximum-Number-of-Balls-in-a-Box/1742. Maximum Number of Balls in a Box.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+func countBalls(lowLimit int, highLimit int) int {
+	buckets, maxBall := [46]int{}, 0
+	for i := lowLimit; i <= highLimit; i++ {
+		t := 0
+		for j := i; j > 0; {
+			t += j % 10
+			j = j / 10
+		}
+		buckets[t]++
+		if buckets[t] > maxBall {
+			maxBall = buckets[t]
+		}
+	}
+	return maxBall
+}
diff --git a/leetcode/1742.Maximum-Number-of-Balls-in-a-Box/1742. Maximum Number of Balls in a Box_test.go b/leetcode/1742.Maximum-Number-of-Balls-in-a-Box/1742. Maximum Number of Balls in a Box_test.go
new file mode 100644
index 000000000..06bc63a1e
--- /dev/null
+++ b/leetcode/1742.Maximum-Number-of-Balls-in-a-Box/1742. Maximum Number of Balls in a Box_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1742 struct {
+	para1742
+	ans1742
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1742 struct {
+	lowLimit  int
+	highLimit int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1742 struct {
+	one int
+}
+
+func Test_Problem1742(t *testing.T) {
+
+	qs := []question1742{
+
+		{
+			para1742{1, 10},
+			ans1742{2},
+		},
+
+		{
+			para1742{5, 15},
+			ans1742{2},
+		},
+
+		{
+			para1742{19, 28},
+			ans1742{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1742------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1742, q.para1742
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countBalls(p.lowLimit, p.highLimit))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1742.Maximum-Number-of-Balls-in-a-Box/README.md b/leetcode/1742.Maximum-Number-of-Balls-in-a-Box/README.md
new file mode 100644
index 000000000..5c5284b61
--- /dev/null
+++ b/leetcode/1742.Maximum-Number-of-Balls-in-a-Box/README.md
@@ -0,0 +1,81 @@
+# [1742. Maximum Number of Balls in a Box](https://leetcode.com/problems/maximum-number-of-balls-in-a-box/)
+
+
+## 题目
+
+You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
+
+Your job at this factory is to put each ball in the box with a number equal to the sum of digits of the ball's number. For example, the ball number `321` will be put in the box number `3 + 2 + 1 = 6` and the ball number `10` will be put in the box number `1 + 0 = 1`.
+
+Given two integers `lowLimit` and `highLimit`, return *the number of balls in the box with the most balls.*
+
+**Example 1:**
+
+```
+Input: lowLimit = 1, highLimit = 10
+Output: 2
+Explanation:
+Box Number:  1 2 3 4 5 6 7 8 9 10 11 ...
+Ball Count:  2 1 1 1 1 1 1 1 1 0  0  ...
+Box 1 has the most number of balls with 2 balls.
+```
+
+**Example 2:**
+
+```
+Input: lowLimit = 5, highLimit = 15
+Output: 2
+Explanation:
+Box Number:  1 2 3 4 5 6 7 8 9 10 11 ...
+Ball Count:  1 1 1 1 2 2 1 1 1 0  0  ...
+Boxes 5 and 6 have the most number of balls with 2 balls in each.
+
+```
+
+**Example 3:**
+
+```
+Input: lowLimit = 19, highLimit = 28
+Output: 2
+Explanation:
+Box Number:  1 2 3 4 5 6 7 8 9 10 11 12 ...
+Ball Count:  0 1 1 1 1 1 1 1 1 2  0  0  ...
+Box 10 has the most number of balls with 2 balls.
+
+```
+
+**Constraints:**
+
+- `1 <= lowLimit <= highLimit <= 10^5`
+
+## 题目大意
+
+你在一家生产小球的玩具厂工作，有 n 个小球，编号从 lowLimit 开始，到 highLimit 结束（包括 lowLimit 和 highLimit ，即 n == highLimit - lowLimit + 1）。另有无限数量的盒子，编号从 1 到 infinity 。你的工作是将每个小球放入盒子中，其中盒子的编号应当等于小球编号上每位数字的和。例如，编号 321 的小球应当放入编号 3 + 2 + 1 = 6 的盒子，而编号 10 的小球应当放入编号 1 + 0 = 1 的盒子。
+
+给你两个整数 lowLimit 和 highLimit ，返回放有最多小球的盒子中的小球数量。如果有多个盒子都满足放有最多小球，只需返回其中任一盒子的小球数量。
+
+## 解题思路
+
+- 简单题。循环遍历一遍数组，依次计算出所有小球的编号各位数字累加和，并且动态维护放有小球最多的数目。循环结束，输出最多小球个数即可。
+
+## 代码
+
+```go
+package leetcode
+
+func countBalls(lowLimit int, highLimit int) int {
+	buckets, maxBall := [46]int{}, 0
+	for i := lowLimit; i <= highLimit; i++ {
+		t := 0
+		for j := i; j > 0; {
+			t += j % 10
+			j = j / 10
+		}
+		buckets[t]++
+		if buckets[t] > maxBall {
+			maxBall = buckets[t]
+		}
+	}
+	return maxBall
+}
+```
\ No newline at end of file
diff --git a/leetcode/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day/1744. Can You Eat Your Favorite Candy on Your Favorite Day.go b/leetcode/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day/1744. Can You Eat Your Favorite Candy on Your Favorite Day.go
new file mode 100644
index 000000000..ce89feb07
--- /dev/null
+++ b/leetcode/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day/1744. Can You Eat Your Favorite Candy on Your Favorite Day.go	
@@ -0,0 +1,23 @@
+package leetcode
+
+func canEat(candiesCount []int, queries [][]int) []bool {
+	n := len(candiesCount)
+	prefixSum := make([]int, n)
+	prefixSum[0] = candiesCount[0]
+	for i := 1; i < n; i++ {
+		prefixSum[i] = prefixSum[i-1] + candiesCount[i]
+	}
+	res := make([]bool, len(queries))
+	for i, q := range queries {
+		favoriteType, favoriteDay, dailyCap := q[0], q[1], q[2]
+		x1 := favoriteDay + 1
+		y1 := (favoriteDay + 1) * dailyCap
+		x2 := 1
+		if favoriteType > 0 {
+			x2 = prefixSum[favoriteType-1] + 1
+		}
+		y2 := prefixSum[favoriteType]
+		res[i] = !(x1 > y2 || y1 < x2)
+	}
+	return res
+}
diff --git a/leetcode/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day/1744. Can You Eat Your Favorite Candy on Your Favorite Day_test.go b/leetcode/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day/1744. Can You Eat Your Favorite Candy on Your Favorite Day_test.go
new file mode 100644
index 000000000..73a486bcd
--- /dev/null
+++ b/leetcode/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day/1744. Can You Eat Your Favorite Candy on Your Favorite Day_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1744 struct {
+	para1744
+	ans1744
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1744 struct {
+	candiesCount []int
+	queries      [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1744 struct {
+	one []bool
+}
+
+func Test_Problem1744(t *testing.T) {
+
+	qs := []question1744{
+
+		{
+			para1744{[]int{7, 4, 5, 3, 8}, [][]int{{0, 2, 2}, {4, 2, 4}, {2, 13, 1000000000}}},
+			ans1744{[]bool{true, false, true}},
+		},
+
+		{
+			para1744{[]int{5, 2, 6, 4, 1}, [][]int{{3, 1, 2}, {4, 10, 3}, {3, 10, 100}, {4, 100, 30}, {1, 3, 1}}},
+			ans1744{[]bool{false, true, true, false, false}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1744------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1744, q.para1744
+		fmt.Printf("【input】:%v       【output】:%v\n", p, canEat(p.candiesCount, p.queries))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day/README.md b/leetcode/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day/README.md
new file mode 100644
index 000000000..bb06c20a2
--- /dev/null
+++ b/leetcode/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day/README.md
@@ -0,0 +1,88 @@
+# [1744. Can You Eat Your Favorite Candy on Your Favorite Day?](https://leetcode.com/problems/can-you-eat-your-favorite-candy-on-your-favorite-day/)
+
+## 题目
+
+You are given a **(0-indexed)** array of positive integers `candiesCount` where `candiesCount[i]` represents the number of candies of the `ith` type you have. You are also given a 2D array `queries` where `queries[i] = [favoriteTypei, favoriteDayi, dailyCapi]`.
+
+You play a game with the following rules:
+
+- You start eating candies on day **`0`**.
+- You **cannot** eat **any** candy of type `i` unless you have eaten **all** candies of type `i - 1`.
+- You must eat **at least** **one** candy per day until you have eaten all the candies.
+
+Construct a boolean array `answer` such that `answer.length == queries.length` and `answer[i]` is `true` if you can eat a candy of type `favoriteTypei` on day `favoriteDayi` without eating **more than** `dailyCapi` candies on **any** day, and `false` otherwise. Note that you can eat different types of candy on the same day, provided that you follow rule 2.
+
+Return *the constructed array* `answer`.
+
+**Example 1:**
+
+```
+Input: candiesCount = [7,4,5,3,8], queries = [[0,2,2],[4,2,4],[2,13,1000000000]]
+Output: [true,false,true]
+Explanation:
+1- If you eat 2 candies (type 0) on day 0 and 2 candies (type 0) on day 1, you will eat a candy of type 0 on day 2.
+2- You can eat at most 4 candies each day.
+   If you eat 4 candies every day, you will eat 4 candies (type 0) on day 0 and 4 candies (type 0 and type 1) on day 1.
+   On day 2, you can only eat 4 candies (type 1 and type 2), so you cannot eat a candy of type 4 on day 2.
+3- If you eat 1 candy each day, you will eat a candy of type 2 on day 13.
+```
+
+**Example 2:**
+
+```
+Input: candiesCount = [5,2,6,4,1], queries = [[3,1,2],[4,10,3],[3,10,100],[4,100,30],[1,3,1]]
+Output: [false,true,true,false,false]
+```
+
+**Constraints:**
+
+- `1 <= candiesCount.length <= 105`
+- `1 <= candiesCount[i] <= 105`
+- `1 <= queries.length <= 105`
+- `queries[i].length == 3`
+- `0 <= favoriteTypei < candiesCount.length`
+- `0 <= favoriteDayi <= 109`
+- `1 <= dailyCapi <= 109`
+
+## 题目大意
+
+给你一个下标从 0 开始的正整数数组 candiesCount ，其中 candiesCount[i] 表示你拥有的第 i 类糖果的数目。同时给你一个二维数组 queries ，其中 queries[i] = [favoriteTypei, favoriteDayi, dailyCapi] 。你按照如下规则进行一场游戏：
+
+- 你从第 0 天开始吃糖果。
+- 你在吃完 所有 第 i - 1 类糖果之前，不能 吃任何一颗第 i 类糖果。
+- 在吃完所有糖果之前，你必须每天 至少 吃 一颗 糖果。
+
+请你构建一个布尔型数组 answer ，满足 answer.length == queries.length 。answer[i] 为 true 的条件是：在每天吃 不超过 dailyCapi 颗糖果的前提下，你可以在第 favoriteDayi 天吃到第 favoriteTypei 类糖果；否则 answer[i] 为 false 。注意，只要满足上面 3 条规则中的第二条规则，你就可以在同一天吃不同类型的糖果。请你返回得到的数组 answer 。
+
+## 解题思路
+
+- 每天吃糖个数的下限是 1 颗，上限是 dailyCap。针对每一个 query 查询在第 i 天能否吃到 i 类型的糖果，要想吃到 i 类型的糖果，必须吃完 i-1 类型的糖果。意味着在 [favoriteDayi + 1, (favoriteDayi+1)×dailyCapi] 区间内能否包含一颗第 favoriteTypei 类型的糖果。如果能包含则输出 true，不能包含则输出 false。吃的糖果数是累积的，所以这里利用前缀和计算出累积吃糖果数所在区间 [sum[favoriteTypei−1]+1, sum[favoriteTypei]]。最后判断 query 区间和累积吃糖果数的区间是否有重叠即可。如果重叠即输出 true。
+- 判断两个区间是否重合，情况有好几种：内包含，全包含，半包含等等。没有交集的情况比较少，所以可以用排除法。对于区间 [x1, y1] 以及 [x2, y2]，它们没有交集当且仅当 x1 > y2 或者 y1 < x2。
+
+## 代码
+
+```go
+package leetcode
+
+func canEat(candiesCount []int, queries [][]int) []bool {
+	n := len(candiesCount)
+	prefixSum := make([]int, n)
+	prefixSum[0] = candiesCount[0]
+	for i := 1; i < n; i++ {
+		prefixSum[i] = prefixSum[i-1] + candiesCount[i]
+	}
+	res := make([]bool, len(queries))
+	for i, q := range queries {
+		favoriteType, favoriteDay, dailyCap := q[0], q[1], q[2]
+		x1 := favoriteDay + 1
+		y1 := (favoriteDay + 1) * dailyCap
+		x2 := 1
+		if favoriteType > 0 {
+			x2 = prefixSum[favoriteType-1] + 1
+		}
+		y2 := prefixSum[favoriteType]
+		res[i] = !(x1 > y2 || y1 < x2)
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1748.Sum-of-Unique-Elements/1748. Sum of Unique Elements.go b/leetcode/1748.Sum-of-Unique-Elements/1748. Sum of Unique Elements.go
new file mode 100644
index 000000000..e8ebac9af
--- /dev/null
+++ b/leetcode/1748.Sum-of-Unique-Elements/1748. Sum of Unique Elements.go	
@@ -0,0 +1,17 @@
+package leetcode
+
+func sumOfUnique(nums []int) int {
+	freq, res := make(map[int]int), 0
+	for _, v := range nums {
+		if _, ok := freq[v]; !ok {
+			freq[v] = 0
+		}
+		freq[v]++
+	}
+	for k, v := range freq {
+		if v == 1 {
+			res += k
+		}
+	}
+	return res
+}
diff --git a/leetcode/1748.Sum-of-Unique-Elements/1748. Sum of Unique Elements_test.go b/leetcode/1748.Sum-of-Unique-Elements/1748. Sum of Unique Elements_test.go
new file mode 100644
index 000000000..0a38e0b45
--- /dev/null
+++ b/leetcode/1748.Sum-of-Unique-Elements/1748. Sum of Unique Elements_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1748 struct {
+	para1748
+	ans1748
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1748 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1748 struct {
+	one int
+}
+
+func Test_Problem1748(t *testing.T) {
+
+	qs := []question1748{
+
+		{
+			para1748{[]int{1, 2, 3, 2}},
+			ans1748{4},
+		},
+
+		{
+			para1748{[]int{1, 1, 1, 1, 1}},
+			ans1748{0},
+		},
+
+		{
+			para1748{[]int{1, 2, 3, 4, 5}},
+			ans1748{15},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1748------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1748, q.para1748
+		fmt.Printf("【input】:%v       【output】:%v\n", p, sumOfUnique(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1748.Sum-of-Unique-Elements/README.md b/leetcode/1748.Sum-of-Unique-Elements/README.md
new file mode 100644
index 000000000..6a3ac8b27
--- /dev/null
+++ b/leetcode/1748.Sum-of-Unique-Elements/README.md
@@ -0,0 +1,67 @@
+# [1748. Sum of Unique Elements](https://leetcode.com/problems/sum-of-unique-elements/)
+
+
+## 题目
+
+You are given an integer array `nums`. The unique elements of an array are the elements that appear **exactly once** in the array.
+
+Return *the **sum** of all the unique elements of* `nums`.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3,2]
+Output: 4
+Explanation: The unique elements are [1,3], and the sum is 4.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,1,1,1,1]
+Output: 0
+Explanation: There are no unique elements, and the sum is 0.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,2,3,4,5]
+Output: 15
+Explanation: The unique elements are [1,2,3,4,5], and the sum is 15.
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `1 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个整数数组 `nums` 。数组中唯一元素是那些只出现 **恰好一次** 的元素。请你返回 `nums` 中唯一元素的 **和** 。
+
+## 解题思路
+
+- 简单题。利用 map 统计出每个元素出现的频次。再累加所有频次为 1 的元素，最后输出累加和即可。
+
+## 代码
+
+```go
+package leetcode
+
+func sumOfUnique(nums []int) int {
+	freq, res := make(map[int]int), 0
+	for _, v := range nums {
+		if _, ok := freq[v]; !ok {
+			freq[v] = 0
+		}
+		freq[v]++
+	}
+	for k, v := range freq {
+		if v == 1 {
+			res += k
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/1752.Check-if-Array-Is-Sorted-and-Rotated/1752. Check if Array Is Sorted and Rotated.go b/leetcode/1752.Check-if-Array-Is-Sorted-and-Rotated/1752. Check if Array Is Sorted and Rotated.go
new file mode 100644
index 000000000..ef10f2d95
--- /dev/null
+++ b/leetcode/1752.Check-if-Array-Is-Sorted-and-Rotated/1752. Check if Array Is Sorted and Rotated.go	
@@ -0,0 +1,14 @@
+package leetcode
+
+func check(nums []int) bool {
+	count := 0
+	for i := 0; i < len(nums)-1; i++ {
+		if nums[i] > nums[i+1] {
+			count++
+			if count > 1 || nums[0] < nums[len(nums)-1] {
+				return false
+			}
+		}
+	}
+	return true
+}
diff --git a/leetcode/1752.Check-if-Array-Is-Sorted-and-Rotated/1752. Check if Array Is Sorted and Rotated_test.go b/leetcode/1752.Check-if-Array-Is-Sorted-and-Rotated/1752. Check if Array Is Sorted and Rotated_test.go
new file mode 100644
index 000000000..4ebdd4e01
--- /dev/null
+++ b/leetcode/1752.Check-if-Array-Is-Sorted-and-Rotated/1752. Check if Array Is Sorted and Rotated_test.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1752 struct {
+	para1752
+	ans1752
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1752 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1752 struct {
+	one bool
+}
+
+func Test_Problem1752(t *testing.T) {
+
+	qs := []question1752{
+
+		{
+			para1752{[]int{3, 4, 5, 1, 2}},
+			ans1752{true},
+		},
+
+		{
+			para1752{[]int{2, 1, 3, 4}},
+			ans1752{false},
+		},
+
+		{
+			para1752{[]int{1, 2, 3}},
+			ans1752{true},
+		},
+
+		{
+			para1752{[]int{1, 1, 1}},
+			ans1752{true},
+		},
+
+		{
+			para1752{[]int{2, 1}},
+			ans1752{true},
+		},
+
+		{
+			para1752{[]int{1, 3, 2, 4}},
+			ans1752{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1752------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1752, q.para1752
+		fmt.Printf("【input】:%v       【output】:%v\n", p, check(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1752.Check-if-Array-Is-Sorted-and-Rotated/README.md b/leetcode/1752.Check-if-Array-Is-Sorted-and-Rotated/README.md
new file mode 100644
index 000000000..775b7c192
--- /dev/null
+++ b/leetcode/1752.Check-if-Array-Is-Sorted-and-Rotated/README.md
@@ -0,0 +1,86 @@
+# [1752. Check if Array Is Sorted and Rotated](https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/)
+
+
+## 题目
+
+Given an array `nums`, return `true` *if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)*. Otherwise, return `false`.
+
+There may be **duplicates** in the original array.
+
+**Note:** An array `A` rotated by `x` positions results in an array `B` of the same length such that `A[i] == B[(i+x) % A.length]`, where `%` is the modulo operation.
+
+**Example 1:**
+
+```
+Input: nums = [3,4,5,1,2]
+Output: true
+Explanation: [1,2,3,4,5] is the original sorted array.
+You can rotate the array by x = 3 positions to begin on the the element of value 3: [3,4,5,1,2].
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,1,3,4]
+Output: false
+Explanation: There is no sorted array once rotated that can make nums.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,2,3]
+Output: true
+Explanation: [1,2,3] is the original sorted array.
+You can rotate the array by x = 0 positions (i.e. no rotation) to make nums.
+```
+
+**Example 4:**
+
+```
+Input: nums = [1,1,1]
+Output: true
+Explanation: [1,1,1] is the original sorted array.
+You can rotate any number of positions to make nums.
+```
+
+**Example 5:**
+
+```
+Input: nums = [2,1]
+Output: true
+Explanation: [1,2] is the original sorted array.
+You can rotate the array by x = 5 positions to begin on the element of value 2: [2,1].
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `1 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个数组 nums 。nums 的源数组中，所有元素与 nums 相同，但按非递减顺序排列。如果 nums 能够由源数组轮转若干位置（包括 0 个位置）得到，则返回 true ；否则，返回 false 。源数组中可能存在 重复项 。
+
+## 解题思路
+
+- 简单题。从头扫描一遍数组，找出相邻两个元素递减的数对。如果递减的数对只有 1 个，则有可能是轮转得来的，超过 1 个，则返回 false。题干里面还提到可能有多个重复元素，针对这一情况还需要判断一下 `nums[0]` 和 `nums[len(nums)-1]` 。如果是相同元素，`nums[0] < nums[len(nums)-1]`，并且数组中间还存在一对递减的数对，这时候也是 false。判断好上述这 2 种情况，本题得解。
+
+## 代码
+
+```go
+package leetcode
+
+func check(nums []int) bool {
+	count := 0
+	for i := 0; i < len(nums)-1; i++ {
+		if nums[i] > nums[i+1] {
+			count++
+			if count > 1 || nums[0] < nums[len(nums)-1] {
+				return false
+			}
+		}
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/1758. Minimum Changes To Make Alternating Binary String.go b/leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/1758. Minimum Changes To Make Alternating Binary String.go
new file mode 100644
index 000000000..ed0dabf5f
--- /dev/null
+++ b/leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/1758. Minimum Changes To Make Alternating Binary String.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func minOperations(s string) int {
+	res := 0
+	for i := 0; i < len(s); i++ {
+		if int(s[i]-'0') != i%2 {
+			res++
+		}
+	}
+	return min(res, len(s)-res)
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/1758. Minimum Changes To Make Alternating Binary String_test.go b/leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/1758. Minimum Changes To Make Alternating Binary String_test.go
new file mode 100644
index 000000000..121db556a
--- /dev/null
+++ b/leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/1758. Minimum Changes To Make Alternating Binary String_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1758 struct {
+	para1758
+	ans1758
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1758 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1758 struct {
+	one int
+}
+
+func Test_Problem1758(t *testing.T) {
+
+	qs := []question1758{
+
+		{
+			para1758{"0100"},
+			ans1758{1},
+		},
+
+		{
+			para1758{"10"},
+			ans1758{0},
+		},
+
+		{
+			para1758{"1111"},
+			ans1758{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1758------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1758, q.para1758
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minOperations(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/README.md b/leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/README.md
new file mode 100644
index 000000000..f268462ee
--- /dev/null
+++ b/leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/README.md
@@ -0,0 +1,70 @@
+# [1758. Minimum Changes To Make Alternating Binary String](https://leetcode.com/problems/minimum-changes-to-make-alternating-binary-string/)
+
+
+## 题目
+
+You are given a string `s` consisting only of the characters `'0'` and `'1'`. In one operation, you can change any `'0'` to `'1'` or vice versa.
+
+The string is called alternating if no two adjacent characters are equal. For example, the string `"010"` is alternating, while the string `"0100"` is not.
+
+Return *the **minimum** number of operations needed to make* `s` *alternating*.
+
+**Example 1:**
+
+```
+Input: s = "0100"
+Output: 1
+Explanation: If you change the last character to '1', s will be "0101", which is alternating.
+```
+
+**Example 2:**
+
+```
+Input: s = "10"
+Output: 0
+Explanation: s is already alternating.
+```
+
+**Example 3:**
+
+```
+Input: s = "1111"
+Output: 2
+Explanation: You need two operations to reach "0101" or "1010".
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 104`
+- `s[i]` is either `'0'` or `'1'`.
+
+## 题目大意
+
+你将得到一个仅包含字符“ 0”和“ 1”的字符串 `s`。 在一项操作中，你可以将任何 `'0'` 更改为 `'1'`，反之亦然。 如果两个相邻字符都不相等，则该字符串称为交替字符串。 例如，字符串“ 010”是交替的，而字符串“ 0100”则不是。 返回使 `s` 交替所需的最小操作数。
+
+## 解题思路
+
+- 简单题。利用数组下标奇偶交替性来判断交替字符串。交替字符串有 2 种，一个是 `'01010101……'` 还有一个是 `'1010101010……'`，这两个只需要计算出一个即可，另外一个利用 `len(s) - res` 就是答案。
+
+## 代码
+
+```go
+package leetcode
+
+func minOperations(s string) int {
+	res := 0
+	for i := 0; i < len(s); i++ {
+		if int(s[i]-'0') != i%2 {
+			res++
+		}
+	}
+	return min(res, len(s)-res)
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+```
\ No newline at end of file
diff --git a/leetcode/1763.Longest-Nice-Substring/1763. Longest Nice Substring.go b/leetcode/1763.Longest-Nice-Substring/1763. Longest Nice Substring.go
new file mode 100644
index 000000000..9219ed23a
--- /dev/null
+++ b/leetcode/1763.Longest-Nice-Substring/1763. Longest Nice Substring.go	
@@ -0,0 +1,82 @@
+package leetcode
+
+import "unicode"
+
+// 解法一 分治，时间复杂度 O(n)
+func longestNiceSubstring(s string) string {
+	if len(s) < 2 {
+		return ""
+	}
+
+	chars := map[rune]int{}
+	for _, r := range s {
+		chars[r]++
+	}
+
+	for i := 0; i < len(s); i++ {
+		r := rune(s[i])
+		_, u := chars[unicode.ToUpper(r)]
+		_, l := chars[unicode.ToLower(r)]
+		if u && l {
+			continue
+		}
+		left := longestNiceSubstring(s[:i])
+		right := longestNiceSubstring(s[i+1:])
+		if len(left) >= len(right) {
+			return left
+		} else {
+			return right
+		}
+	}
+	return s
+}
+
+// 解法二 用二进制表示状态
+func longestNiceSubstring1(s string) (ans string) {
+	for i := range s {
+		lower, upper := 0, 0
+		for j := i; j < len(s); j++ {
+			if unicode.IsLower(rune(s[j])) {
+				lower |= 1 << (s[j] - 'a')
+			} else {
+				upper |= 1 << (s[j] - 'A')
+			}
+			if lower == upper && j-i+1 > len(ans) {
+				ans = s[i : j+1]
+			}
+		}
+	}
+	return
+}
+
+// 解法三 暴力枚举，时间复杂度 O(n^2)
+func longestNiceSubstring2(s string) string {
+	res := ""
+	for i := 0; i < len(s); i++ {
+		m := map[byte]int{}
+		m[s[i]]++
+		for j := i + 1; j < len(s); j++ {
+			m[s[j]]++
+			if checkNiceString(m) && (j-i+1 > len(res)) {
+				res = s[i : j+1]
+			}
+		}
+	}
+	return res
+}
+
+func checkNiceString(m map[byte]int) bool {
+	for k := range m {
+		if k >= 97 && k <= 122 {
+			if _, ok := m[k-32]; !ok {
+				return false
+			}
+		}
+		if k >= 65 && k <= 90 {
+			if _, ok := m[k+32]; !ok {
+				return false
+			}
+		}
+	}
+	return true
+}
diff --git a/leetcode/1763.Longest-Nice-Substring/1763. Longest Nice Substring_test.go b/leetcode/1763.Longest-Nice-Substring/1763. Longest Nice Substring_test.go
new file mode 100644
index 000000000..08d550a30
--- /dev/null
+++ b/leetcode/1763.Longest-Nice-Substring/1763. Longest Nice Substring_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1763 struct {
+	para1763
+	ans1763
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1763 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1763 struct {
+	one string
+}
+
+func Test_Problem1763(t *testing.T) {
+
+	qs := []question1763{
+
+		{
+			para1763{"YazaAay"},
+			ans1763{"aAa"},
+		},
+
+		{
+			para1763{"Bb"},
+			ans1763{"Bb"},
+		},
+
+		{
+			para1763{"c"},
+			ans1763{""},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1763------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1763, q.para1763
+		fmt.Printf("【input】:%v       【output】:%v\n", p, longestNiceSubstring(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1763.Longest-Nice-Substring/README.md b/leetcode/1763.Longest-Nice-Substring/README.md
new file mode 100644
index 000000000..05dd0bac4
--- /dev/null
+++ b/leetcode/1763.Longest-Nice-Substring/README.md
@@ -0,0 +1,140 @@
+# [1763. Longest Nice Substring](https://leetcode.com/problems/longest-nice-substring/)
+
+
+## 题目
+
+A string `s` is **nice** if, for every letter of the alphabet that `s` contains, it appears **both** in uppercase and lowercase. For example, `"abABB"` is nice because `'A'` and `'a'` appear, and `'B'` and `'b'` appear. However, `"abA"` is not because `'b'` appears, but `'B'` does not.
+
+Given a string `s`, return *the longest **substring** of `s` that is **nice**. If there are multiple, return the substring of the **earliest** occurrence. If there are none, return an empty string*.
+
+**Example 1:**
+
+```
+Input: s = "YazaAay"
+Output: "aAa"
+Explanation:"aAa" is a nice string because 'A/a' is the only letter of the alphabet in s, and both 'A' and 'a' appear.
+"aAa" is the longest nice substring.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "Bb"
+Output: "Bb"
+Explanation: "Bb" is a nice string because both 'B' and 'b' appear. The whole string is a substring.
+
+```
+
+**Example 3:**
+
+```
+Input: s = "c"
+Output: ""
+Explanation: There are no nice substrings.
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 100`
+- `s` consists of uppercase and lowercase English letters.
+
+## 题目大意
+
+当一个字符串 s 包含的每一种字母的大写和小写形式 同时 出现在 s 中，就称这个字符串 s 是 美好 字符串。比方说，"abABB" 是美好字符串，因为 'A' 和 'a' 同时出现了，且 'B' 和 'b' 也同时出现了。然而，"abA" 不是美好字符串因为 'b' 出现了，而 'B' 没有出现。
+
+给你一个字符串 s ，请你返回 s 最长的 美好子字符串 。如果有多个答案，请你返回 最早 出现的一个。如果不存在美好子字符串，请你返回一个空字符串。
+
+## 解题思路
+
+- 解法一，暴力解法。枚举每一段字符串，判断这个子字符串内是否满足美好字符串的定义，即字母的大小写是否同时出现。
+- 解法二，这个解法是解法一的小幅优化版，利用二进制记录状态。先构造二进制状态串，再利用直接比较这个二进制串。
+- 解法三，分治。以 `i` 为分割点依次切开字符串。左右两个字符串分别判断是否满足美好字符串的定义。左右分开的字符串还可以继续划分。直至分到一个字母为止。在这个过程中记录最早出现的字符串。
+
+## 代码
+
+```go
+package leetcode
+
+import "unicode"
+
+// 解法一 分治，时间复杂度 O(n)
+func longestNiceSubstring(s string) string {
+	if len(s) < 2 {
+		return ""
+	}
+
+	chars := map[rune]int{}
+	for _, r := range s {
+		chars[r]++
+	}
+
+	for i := 0; i < len(s); i++ {
+		r := rune(s[i])
+		_, u := chars[unicode.ToUpper(r)]
+		_, l := chars[unicode.ToLower(r)]
+		if u && l {
+			continue
+		}
+		left := longestNiceSubstring(s[:i])
+		right := longestNiceSubstring(s[i+1:])
+		if len(left) >= len(right) {
+			return left
+		} else {
+			return right
+		}
+	}
+	return s
+}
+
+// 解法二 用二进制表示状态
+func longestNiceSubstring1(s string) (ans string) {
+	for i := range s {
+		lower, upper := 0, 0
+		for j := i; j < len(s); j++ {
+			if unicode.IsLower(rune(s[j])) {
+				lower |= 1 << (s[j] - 'a')
+			} else {
+				upper |= 1 << (s[j] - 'A')
+			}
+			if lower == upper && j-i+1 > len(ans) {
+				ans = s[i : j+1]
+			}
+		}
+	}
+	return
+}
+
+// 解法三 暴力枚举，时间复杂度 O(n^2)
+func longestNiceSubstring2(s string) string {
+	res := ""
+	for i := 0; i < len(s); i++ {
+		m := map[byte]int{}
+		m[s[i]]++
+		for j := i + 1; j < len(s); j++ {
+			m[s[j]]++
+			if checkNiceString(m) && (j-i+1 > len(res)) {
+				res = s[i : j+1]
+			}
+		}
+	}
+	return res
+}
+
+func checkNiceString(m map[byte]int) bool {
+	for k := range m {
+		if k >= 97 && k <= 122 {
+			if _, ok := m[k-32]; !ok {
+				return false
+			}
+		}
+		if k >= 65 && k <= 90 {
+			if _, ok := m[k+32]; !ok {
+				return false
+			}
+		}
+	}
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/1791.Find-Center-of-Star-Graph/1791.Find Center of Star Graph.go b/leetcode/1791.Find-Center-of-Star-Graph/1791.Find Center of Star Graph.go
new file mode 100644
index 000000000..6f39e98f9
--- /dev/null
+++ b/leetcode/1791.Find-Center-of-Star-Graph/1791.Find Center of Star Graph.go	
@@ -0,0 +1,8 @@
+package leetcode
+
+func findCenter(edges [][]int) int {
+	if edges[0][0] == edges[1][0] || edges[0][0] == edges[1][1] {
+		return edges[0][0]
+	}
+	return edges[0][1]
+}
diff --git a/leetcode/1791.Find-Center-of-Star-Graph/1791.Find Center of Star Graph_test.go b/leetcode/1791.Find-Center-of-Star-Graph/1791.Find Center of Star Graph_test.go
new file mode 100644
index 000000000..e1ade4d8e
--- /dev/null
+++ b/leetcode/1791.Find-Center-of-Star-Graph/1791.Find Center of Star Graph_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1791 struct {
+	para1791
+	ans1791
+}
+
+// para 是参数
+type para1791 struct {
+	edges [][]int
+}
+
+// ans 是答案
+type ans1791 struct {
+	ans int
+}
+
+func Test_Problem1791(t *testing.T) {
+
+	qs := []question1791{
+
+		{
+			para1791{[][]int{{1, 2}, {2, 3}, {4, 2}}},
+			ans1791{2},
+		},
+
+		{
+			para1791{[][]int{{1, 2}, {5, 1}, {1, 3}, {1, 4}}},
+			ans1791{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1791------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1791, q.para1791
+		fmt.Printf("【input】:%v      ", p.edges)
+		fmt.Printf("【output】:%v      \n", findCenter(p.edges))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1791.Find-Center-of-Star-Graph/README.md b/leetcode/1791.Find-Center-of-Star-Graph/README.md
new file mode 100644
index 000000000..7c3a049fa
--- /dev/null
+++ b/leetcode/1791.Find-Center-of-Star-Graph/README.md
@@ -0,0 +1,52 @@
+# [1791.Find Center of Star Graph](https://leetcode.com/problems/find-center-of-star-graph/)
+
+## 题目
+
+There is an undirected star graph consisting of n nodes labeled from 1 to n. A star graph is a graph where there is one center node and exactly n - 1 edges that connect the center node with every other node.
+
+You are given a 2D integer array edges where each edges[i] = [ui, vi] indicates that there is an edge between the nodes ui and vi. Return the center of the given star graph.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/02/24/star_graph.png:w](https://assets.leetcode.com/uploads/2021/02/24/star_graph.png)
+
+    Input: edges = [[1,2],[2,3],[4,2]]
+    Output: 2
+    Explanation: As shown in the figure above, node 2 is connected to every other node, so 2 is the center.
+
+**Example 2:**
+
+    Input: edges = [[1,2],[5,1],[1,3],[1,4]]
+    Output: 1
+
+**Constraints:**
+
+- 3 <= n <= 100000
+- edges.length == n - 1
+- edges[i].length == 2
+- 1 <= ui, vi <= n
+- ui != vi
+- The given edges represent a valid star graph.
+
+## 题目大意
+
+有一个无向的 星型 图，由 n 个编号从 1 到 n 的节点组成。星型图有一个 中心 节点，并且恰有 n - 1 条边将中心节点与其他每个节点连接起来。
+
+给你一个二维整数数组 edges ，其中 edges[i] = [ui, vi] 表示在节点 ui 和 vi 之间存在一条边。请你找出并返回 edges 所表示星型图的中心节点。
+
+## 解题思路
+
+- 求出edges中前两个元素的共同值，即是中心节点
+
+## 代码
+
+```go
+package leetcode
+
+func findCenter(edges [][]int) int {
+	if edges[0][0] == edges[1][0] || edges[0][0] == edges[1][1] {
+		return edges[0][0]
+	}
+	return edges[0][1]
+}
+```
diff --git a/leetcode/1816.Truncate-Sentence/1816.Truncate Sentence.go b/leetcode/1816.Truncate-Sentence/1816.Truncate Sentence.go
new file mode 100644
index 000000000..99d917314
--- /dev/null
+++ b/leetcode/1816.Truncate-Sentence/1816.Truncate Sentence.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+func truncateSentence(s string, k int) string {
+	end := 0
+	for i := range s {
+		if k > 0 && s[i] == ' ' {
+			k--
+		}
+		if k == 0 {
+			end = i
+			break
+		}
+	}
+	if end == 0 {
+		return s
+	}
+	return s[:end]
+}
diff --git a/leetcode/1816.Truncate-Sentence/1816.Truncate Sentence_test.go b/leetcode/1816.Truncate-Sentence/1816.Truncate Sentence_test.go
new file mode 100644
index 000000000..2d277bbfa
--- /dev/null
+++ b/leetcode/1816.Truncate-Sentence/1816.Truncate Sentence_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1816 struct {
+	para1816
+	ans1816
+}
+
+// para 是参数
+type para1816 struct {
+	s string
+	k int
+}
+
+// ans 是答案
+type ans1816 struct {
+	ans string
+}
+
+func Test_Problem1816(t *testing.T) {
+
+	qs := []question1816{
+
+		{
+			para1816{"Hello how are you Contestant", 4},
+			ans1816{"Hello how are you"},
+		},
+
+		{
+			para1816{"What is the solution to this problem", 4},
+			ans1816{"What is the solution"},
+		},
+
+		{
+			para1816{"chopper is not a tanuki", 5},
+			ans1816{"chopper is not a tanuki"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1816------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1816, q.para1816
+		fmt.Printf("【input】:%v    【output】:%v\n", p, truncateSentence(p.s, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1816.Truncate-Sentence/README.md b/leetcode/1816.Truncate-Sentence/README.md
new file mode 100644
index 000000000..ebd9b5877
--- /dev/null
+++ b/leetcode/1816.Truncate-Sentence/README.md
@@ -0,0 +1,76 @@
+# [1816. Truncate Sentence](https://leetcode.com/problems/truncate-sentence/)
+
+## 题目
+
+A sentence is a list of words that are separated by a single space with no leading or trailing spaces. Each of the words consists of only uppercase and lowercase English letters (no punctuation).
+
+- For example, "Hello World", "HELLO", and "hello world hello world" are all sentences.
+
+You are given a sentence s and an integer k. You want to truncate s such that it contains only the first k words. Return s after truncating it.
+
+**Example 1**:
+
+    Input: s = "Hello how are you Contestant", k = 4
+    Output: "Hello how are you"
+    Explanation:
+    The words in s are ["Hello", "how" "are", "you", "Contestant"].
+    The first 4 words are ["Hello", "how", "are", "you"].
+    Hence, you should return "Hello how are you".
+
+**Example 2**:
+
+    Input: s = "What is the solution to this problem", k = 4
+    Output: "What is the solution"
+    Explanation:
+    The words in s are ["What", "is" "the", "solution", "to", "this", "problem"].
+    The first 4 words are ["What", "is", "the", "solution"].
+    Hence, you should return "What is the solution".
+
+**Example 3**:
+
+    Input: s = "chopper is not a tanuki", k = 5
+    Output: "chopper is not a tanuki"
+
+**Constraints:**
+
+- 1 <= s.length <= 500
+- k is in the range [1, the number of words in s].
+- s consist of only lowercase and uppercase English letters and spaces.
+- The words in s are separated by a single space.
+- There are no leading or trailing spaces.
+
+## 题目大意
+
+句子 是一个单词列表，列表中的单词之间用单个空格隔开，且不存在前导或尾随空格。每个单词仅由大小写英文字母组成（不含标点符号）。
+
+- 例如，"Hello World"、"HELLO" 和 "hello world hello world" 都是句子。
+
+给你一个句子 s 和一个整数 k ，请你将 s 截断使截断后的句子仅含前 k 个单词。返回截断 s 后得到的句子。
+
+## 解题思路
+
+- 遍历字符串 s，找到最后一个空格的下标 end
+- 如果 end 为 0，直接返回 s,否则返回 s[:end]
+
+## 代码
+
+```go
+package leetcode
+
+func truncateSentence(s string, k int) string {
+	end := 0
+	for i := range s {
+		if k > 0 && s[i] == ' ' {
+			k--
+		}
+		if k == 0 {
+			end = i
+			break
+		}
+	}
+	if end == 0 {
+		return s
+	}
+	return s[:end]
+}
+```
\ No newline at end of file
diff --git a/leetcode/1818.Minimum-Absolute-Sum-Difference/1818. Minimum Absolute Sum Difference.go b/leetcode/1818.Minimum-Absolute-Sum-Difference/1818. Minimum Absolute Sum Difference.go
new file mode 100644
index 000000000..dec40606d
--- /dev/null
+++ b/leetcode/1818.Minimum-Absolute-Sum-Difference/1818. Minimum Absolute Sum Difference.go	
@@ -0,0 +1,38 @@
+package leetcode
+
+func minAbsoluteSumDiff(nums1 []int, nums2 []int) int {
+	diff := 0
+	maxDiff := 0
+	for i, n2 := range nums2 {
+		d := abs(nums1[i] - n2)
+		diff += d
+		if maxDiff < d {
+			t := 100001
+			for _, n1 := range nums1 {
+				maxDiff = max(maxDiff, d-min(t, abs(n1-n2)))
+			}
+		}
+	}
+	return (diff - maxDiff) % (1e9 + 7)
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/1818.Minimum-Absolute-Sum-Difference/1818. Minimum Absolute Sum Difference_test.go b/leetcode/1818.Minimum-Absolute-Sum-Difference/1818. Minimum Absolute Sum Difference_test.go
new file mode 100644
index 000000000..9c9b36cf1
--- /dev/null
+++ b/leetcode/1818.Minimum-Absolute-Sum-Difference/1818. Minimum Absolute Sum Difference_test.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1818 struct {
+	para1818
+	ans1818
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1818 struct {
+	nums1 []int
+	nums2 []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1818 struct {
+	one int
+}
+
+func Test_Problem1818(t *testing.T) {
+
+	qs := []question1818{
+
+		{
+			para1818{[]int{1, 7, 5}, []int{2, 3, 5}},
+			ans1818{3},
+		},
+
+		{
+			para1818{[]int{2, 4, 6, 8, 10}, []int{2, 4, 6, 8, 10}},
+			ans1818{0},
+		},
+
+		{
+			para1818{[]int{1, 10, 4, 4, 2, 7}, []int{9, 3, 5, 1, 7, 4}},
+			ans1818{20},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1818------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1818, q.para1818
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minAbsoluteSumDiff(p.nums1, p.nums2))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1818.Minimum-Absolute-Sum-Difference/README.md b/leetcode/1818.Minimum-Absolute-Sum-Difference/README.md
new file mode 100644
index 000000000..d66650997
--- /dev/null
+++ b/leetcode/1818.Minimum-Absolute-Sum-Difference/README.md
@@ -0,0 +1,117 @@
+# [1818. Minimum Absolute Sum Difference](https://leetcode.com/problems/minimum-absolute-sum-difference/)
+
+## 题目
+
+You are given two positive integer arrays `nums1` and `nums2`, both of length `n`.
+
+The **absolute sum difference** of arrays `nums1` and `nums2` is defined as the **sum** of `|nums1[i] - nums2[i]|` for each `0 <= i < n` (**0-indexed**).
+
+You can replace **at most one** element of `nums1` with **any** other element in `nums1` to **minimize** the absolute sum difference.
+
+Return the *minimum absolute sum difference **after** replacing at most one ****element in the array `nums1`.* Since the answer may be large, return it **modulo** `109 + 7`.
+
+`|x|` is defined as:
+
+- `x` if `x >= 0`, or
+- `x` if `x < 0`.
+
+**Example 1:**
+
+```
+Input: nums1 = [1,7,5], nums2 = [2,3,5]
+Output: 3
+Explanation:There are two possible optimal solutions:
+- Replace the second element with the first: [1,7,5] => [1,1,5], or
+- Replace the second element with the third: [1,7,5] => [1,5,5].
+Both will yield an absolute sum difference of|1-2| + (|1-3| or |5-3|) + |5-5| =3.
+
+```
+
+**Example 2:**
+
+```
+Input: nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10]
+Output: 0
+Explanation:nums1 is equal to nums2 so no replacement is needed. This will result in an
+absolute sum difference of 0.
+
+```
+
+**Example 3:**
+
+```
+Input: nums1 = [1,10,4,4,2,7], nums2 = [9,3,5,1,7,4]
+Output: 20
+Explanation:Replace the first element with the second: [1,10,4,4,2,7] => [10,10,4,4,2,7].
+This yields an absolute sum difference of|10-9| + |10-3| + |4-5| + |4-1| + |2-7| + |7-4| = 20
+```
+
+**Constraints:**
+
+- `n == nums1.length`
+- `n == nums2.length`
+- `1 <= n <= 10^5`
+- `1 <= nums1[i], nums2[i] <= 10^5`
+
+## 题目大意
+
+给你两个正整数数组 nums1 和 nums2 ，数组的长度都是 n 。数组 nums1 和 nums2 的 绝对差值和 定义为所有 |nums1[i] - nums2[i]|（0 <= i < n）的 总和（下标从 0 开始）。你可以选用 nums1 中的 任意一个 元素来替换 nums1 中的 至多 一个元素，以 最小化 绝对差值和。在替换数组 nums1 中最多一个元素 之后 ，返回最小绝对差值和。因为答案可能很大，所以需要对 10^9 + 7 取余 后返回。
+
+## 解题思路
+
+- 如果不改变任何元素，绝对差值和为
+
+$$\sum \left | nums1[i] - nums2[i] \right |$$
+
+- 如果改变一个元素后，那么绝对差值和为
+
+$$\begin{aligned}&\sum \left | nums1[i] - nums2[i] \right | - \left ( \left | nums1[i] - nums2[i] \right | - \left | nums1[j] - nums2[i] \right |\right )\\= &\sum \left | nums1[i] - nums2[i] \right | - \Delta \end{aligned}$$
+
+题目要求返回最小绝对差值和，即求 
+
+$$\Delta $$
+
+的最大值。暴力枚举 nums1 和 nums2 中两两差值，找到 maxdiff。
+
+## 代码
+
+```go
+package leetcode
+
+func minAbsoluteSumDiff(nums1 []int, nums2 []int) int {
+	diff := 0
+	maxDiff := 0
+	for i, n2 := range nums2 {
+		d := abs(nums1[i] - n2)
+		diff += d
+		if maxDiff < d {
+			t := 100001
+			for _, n1 := range nums1 {
+				maxDiff = max(maxDiff, d-min(t, abs(n1-n2)))
+			}
+		}
+	}
+	return (diff - maxDiff) % (1e9 + 7)
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+```
\ No newline at end of file
diff --git a/leetcode/1846.Maximum-Element-After-Decreasing-and-Rearranging/1846. Maximum Element After Decreasing and Rearranging.go b/leetcode/1846.Maximum-Element-After-Decreasing-and-Rearranging/1846. Maximum Element After Decreasing and Rearranging.go
new file mode 100644
index 000000000..1fde93aba
--- /dev/null
+++ b/leetcode/1846.Maximum-Element-After-Decreasing-and-Rearranging/1846. Maximum Element After Decreasing and Rearranging.go	
@@ -0,0 +1,25 @@
+package leetcode
+
+func maximumElementAfterDecrementingAndRearranging(arr []int) int {
+	n := len(arr)
+	count := make([]int, n+1)
+	for _, v := range arr {
+		count[min(v, n)]++
+	}
+	miss := 0
+	for _, c := range count[1:] {
+		if c == 0 {
+			miss++
+		} else {
+			miss -= min(c-1, miss)
+		}
+	}
+	return n - miss
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1846.Maximum-Element-After-Decreasing-and-Rearranging/1846. Maximum Element After Decreasing and Rearranging_test.go b/leetcode/1846.Maximum-Element-After-Decreasing-and-Rearranging/1846. Maximum Element After Decreasing and Rearranging_test.go
new file mode 100644
index 000000000..b4198e574
--- /dev/null
+++ b/leetcode/1846.Maximum-Element-After-Decreasing-and-Rearranging/1846. Maximum Element After Decreasing and Rearranging_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1846 struct {
+	para1846
+	ans1846
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1846 struct {
+	arr []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1846 struct {
+	one int
+}
+
+func Test_Problem1846(t *testing.T) {
+
+	qs := []question1846{
+
+		{
+			para1846{[]int{2, 2, 1, 2, 1}},
+			ans1846{2},
+		},
+
+		{
+			para1846{[]int{100, 1, 1000}},
+			ans1846{3},
+		},
+
+		{
+			para1846{[]int{1, 2, 3, 4, 5}},
+			ans1846{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1846------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1846, q.para1846
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maximumElementAfterDecrementingAndRearranging(p.arr))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1846.Maximum-Element-After-Decreasing-and-Rearranging/README.md b/leetcode/1846.Maximum-Element-After-Decreasing-and-Rearranging/README.md
new file mode 100644
index 000000000..2196f6bb4
--- /dev/null
+++ b/leetcode/1846.Maximum-Element-After-Decreasing-and-Rearranging/README.md
@@ -0,0 +1,103 @@
+# [1846. Maximum Element After Decreasing and Rearranging](https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/)
+
+
+## 题目
+
+You are given an array of positive integers `arr`. Perform some operations (possibly none) on `arr` so that it satisfies these conditions:
+
+- The value of the **first** element in `arr` must be `1`.
+- The absolute difference between any 2 adjacent elements must be **less than or equal to** `1`. In other words, `abs(arr[i] - arr[i - 1]) <= 1` for each `i` where `1 <= i < arr.length` (**0-indexed**). `abs(x)` is the absolute value of `x`.
+
+There are 2 types of operations that you can perform any number of times:
+
+- **Decrease** the value of any element of `arr` to a **smaller positive integer**.
+- **Rearrange** the elements of `arr` to be in any order.
+
+Return *the **maximum** possible value of an element in* `arr` *after performing the operations to satisfy the conditions*.
+
+**Example 1:**
+
+```
+Input: arr = [2,2,1,2,1]
+Output: 2
+Explanation:
+We can satisfy the conditions by rearrangingarr so it becomes[1,2,2,2,1].
+The largest element inarr is 2.
+
+```
+
+**Example 2:**
+
+```
+Input: arr = [100,1,1000]
+Output: 3
+Explanation:
+One possible way to satisfy the conditions is by doing the following:
+1. Rearrangearr so it becomes[1,100,1000].
+2. Decrease the value of the second element to 2.
+3. Decrease the value of the third element to 3.
+Nowarr = [1,2,3], whichsatisfies the conditions.
+The largest element inarr is 3.
+```
+
+**Example 3:**
+
+```
+Input: arr = [1,2,3,4,5]
+Output: 5
+Explanation: The array already satisfies the conditions, and the largest element is 5.
+
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 10^5`
+- `1 <= arr[i] <= 10^9`
+
+## 题目大意
+
+给你一个正整数数组 arr 。请你对 arr 执行一些操作（也可以不进行任何操作），使得数组满足以下条件：
+
+- arr 中 第一个 元素必须为 1 。
+- 任意相邻两个元素的差的绝对值 小于等于 1 ，也就是说，对于任意的 1 <= i < arr.length （数组下标从 0 开始），都满足 abs(arr[i] - arr[i - 1]) <= 1 。abs(x) 为 x 的绝对值。
+
+你可以执行以下 2 种操作任意次：
+
+- 减小 arr 中任意元素的值，使其变为一个 更小的正整数 。
+- 重新排列 arr 中的元素，你可以以任意顺序重新排列。
+
+请你返回执行以上操作后，在满足前文所述的条件下，arr 中可能的 最大值 。
+
+## 解题思路
+
+- 正整数数组 arr 第一个元素必须为 1，且两两元素绝对值小于等于 1，那么 arr 最大值肯定不大于 n。采用贪心的策略，先统计所有元素出现的次数，大于 n 的元素出现次数都累加到 n 上。然后从 1 扫描到 n，遇到“空隙”（出现次数为 0 的元素），便将最近一个出现次数大于 1 的元素“挪”过来填补“空隙”。题目所求最大值出现在，“填补空隙”之后，数组从左往右连续的最右端。
+
+## 代码
+
+```go
+package leetcode
+
+func maximumElementAfterDecrementingAndRearranging(arr []int) int {
+	n := len(arr)
+	count := make([]int, n+1)
+	for _, v := range arr {
+		count[min(v, n)]++
+	}
+	miss := 0
+	for _, c := range count[1:] {
+		if c == 0 {
+			miss++
+		} else {
+			miss -= min(c-1, miss)
+		}
+	}
+	return n - miss
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1877.Minimize-Maximum-Pair-Sum-in-Array/1877. Minimize Maximum Pair Sum in Array.go b/leetcode/1877.Minimize-Maximum-Pair-Sum-in-Array/1877. Minimize Maximum Pair Sum in Array.go
new file mode 100644
index 000000000..3d74fe3a1
--- /dev/null
+++ b/leetcode/1877.Minimize-Maximum-Pair-Sum-in-Array/1877. Minimize Maximum Pair Sum in Array.go	
@@ -0,0 +1,19 @@
+package leetcode
+
+import "sort"
+
+func minPairSum(nums []int) int {
+	sort.Ints(nums)
+	n, res := len(nums), 0
+	for i, val := range nums[:n/2] {
+		res = max(res, val+nums[n-1-i])
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
diff --git a/leetcode/1877.Minimize-Maximum-Pair-Sum-in-Array/1877. Minimize Maximum Pair Sum in Array_test.go b/leetcode/1877.Minimize-Maximum-Pair-Sum-in-Array/1877. Minimize Maximum Pair Sum in Array_test.go
new file mode 100644
index 000000000..c07aba6e9
--- /dev/null
+++ b/leetcode/1877.Minimize-Maximum-Pair-Sum-in-Array/1877. Minimize Maximum Pair Sum in Array_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1877 struct {
+	para1877
+	ans1877
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1877 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1877 struct {
+	one int
+}
+
+func Test_Problem1877(t *testing.T) {
+
+	qs := []question1877{
+
+		{
+			para1877{[]int{2, 2, 1, 2, 1}},
+			ans1877{3},
+		},
+
+		{
+			para1877{[]int{100, 1, 1000}},
+			ans1877{1001},
+		},
+
+		{
+			para1877{[]int{1, 2, 3, 4, 5}},
+			ans1877{6},
+		},
+
+		{
+			para1877{[]int{3, 5, 2, 3}},
+			ans1877{7},
+		},
+
+		{
+			para1877{[]int{3, 5, 4, 2, 4, 6}},
+			ans1877{8},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1877------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1877, q.para1877
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minPairSum(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1877.Minimize-Maximum-Pair-Sum-in-Array/README.md b/leetcode/1877.Minimize-Maximum-Pair-Sum-in-Array/README.md
new file mode 100644
index 000000000..17ff53013
--- /dev/null
+++ b/leetcode/1877.Minimize-Maximum-Pair-Sum-in-Array/README.md
@@ -0,0 +1,81 @@
+# [1877. Minimize Maximum Pair Sum in Array](https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/)
+
+
+## 题目
+
+The **pair sum** of a pair `(a,b)` is equal to `a + b`. The **maximum pair sum** is the largest **pair sum** in a list of pairs.
+
+- For example, if we have pairs `(1,5)`, `(2,3)`, and `(4,4)`, the **maximum pair sum** would be `max(1+5, 2+3, 4+4) = max(6, 5, 8) = 8`.
+
+Given an array `nums` of **even** length `n`, pair up the elements of `nums` into `n / 2` pairs such that:
+
+- Each element of `nums` is in **exactly one** pair, and
+- The **maximum pair sum** is **minimized**.
+
+Return *the minimized **maximum pair sum** after optimally pairing up the elements*.
+
+**Example 1:**
+
+```
+Input: nums = [3,5,2,3]
+Output: 7
+Explanation: The elements can be paired up into pairs (3,3) and (5,2).
+The maximum pair sum is max(3+3, 5+2) = max(6, 7) = 7.
+```
+
+**Example 2:**
+
+```
+Input: nums = [3,5,4,2,4,6]
+Output: 8
+Explanation: The elements can be paired up into pairs (3,5), (4,4), and (6,2).
+The maximum pair sum is max(3+5, 4+4, 6+2) = max(8, 8, 8) = 8.
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `2 <= n <= 105`
+- `n` is **even**.
+- `1 <= nums[i] <= 105`
+
+## 题目大意
+
+一个数对 (a,b) 的 **数对和** 等于 a + b 。**最大数对和** 是一个数对数组中最大的 数对和 。
+
+- 比方说，如果我们有数对 (1,5) ，(2,3) 和 (4,4)，**最大数对和** 为 max(1+5, 2+3, 4+4) = max(6, 5, 8) = 8 。
+
+给你一个长度为 **偶数** n 的数组 nums ，请你将 nums 中的元素分成 n / 2 个数对，使得：
+
+- nums 中每个元素 **恰好** 在 一个 数对中，且
+- **最大数对和** 的值 **最小** 。
+
+请你在最优数对划分的方案下，返回最小的 最大数对和 。
+
+## 解题思路
+
+- 要想最大数对和最小，那么最大的元素一定只能和最小的元素组合在一起，不然一定不是最小。当最大元素和最小元素组合在一起了，剩下的次最大元素也应该和次最小元素组合在一起。按照这个思路，先将数组从小到大排序，然后依次取出首尾元素，两两组合在一起。输出这些数对的最大值即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func minPairSum(nums []int) int {
+	sort.Ints(nums)
+	n, res := len(nums), 0
+	for i, val := range nums[:n/2] {
+		res = max(res, val+nums[n-1-i])
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores/1984.Minimum Difference Between Highest and Lowest of K Scores.go b/leetcode/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores/1984.Minimum Difference Between Highest and Lowest of K Scores.go
new file mode 100644
index 000000000..b4b381488
--- /dev/null
+++ b/leetcode/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores/1984.Minimum Difference Between Highest and Lowest of K Scores.go	
@@ -0,0 +1,18 @@
+package leetcode
+
+import "sort"
+
+func minimumDifference(nums []int, k int) int {
+	sort.Ints(nums)
+	minDiff := 100000 + 1
+	for i := 0; i < len(nums); i++ {
+		if i+k-1 >= len(nums) {
+			break
+		}
+		diff := nums[i+k-1] - nums[i]
+		if diff < minDiff {
+			minDiff = diff
+		}
+	}
+	return minDiff
+}
diff --git a/leetcode/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores/1984.Minimum Difference Between Highest and Lowest of K Scores_test.go b/leetcode/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores/1984.Minimum Difference Between Highest and Lowest of K Scores_test.go
new file mode 100644
index 000000000..90997ca78
--- /dev/null
+++ b/leetcode/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores/1984.Minimum Difference Between Highest and Lowest of K Scores_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1984 struct {
+	para1984
+	ans1984
+}
+
+// para 是参数
+type para1984 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+type ans1984 struct {
+	ans int
+}
+
+func Test_Problem1984(t *testing.T) {
+
+	qs := []question1984{
+
+		{
+			para1984{[]int{90}, 1},
+			ans1984{0},
+		},
+
+		{
+			para1984{[]int{9, 4, 1, 7}, 2},
+			ans1984{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1984------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1984, q.para1984
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", minimumDifference(p.nums, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores/README.md b/leetcode/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores/README.md
new file mode 100644
index 000000000..fe7146c04
--- /dev/null
+++ b/leetcode/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores/README.md
@@ -0,0 +1,71 @@
+# [1984. Minimum Difference Between Highest and Lowest of K Scores](https://leetcode.com/problems/minimum-difference-between-highest-and-lowest-of-k-scores/)
+
+## 题目
+
+You are given a 0-indexed integer array nums, where nums[i] represents the score of the ith student. You are also given an integer k.
+
+Pick the scores of any k students from the array so that the difference between the highest and the lowest of the k scores is minimized.
+
+Return the minimum possible difference.
+
+**Example 1:**
+
+    Input: nums = [90], k = 1
+    Output: 0
+    Explanation: There is one way to pick score(s) of one student:
+    - [90]. The difference between the highest and lowest score is 90 - 90 = 0.
+    The minimum possible difference is 0.
+
+**Example 2:**
+
+    Input: nums = [9,4,1,7], k = 2
+    Output: 2
+    Explanation: There are six ways to pick score(s) of two students:
+    - [9,4,1,7]. The difference between the highest and lowest score is 9 - 4 = 5.
+    - [9,4,1,7]. The difference between the highest and lowest score is 9 - 1 = 8.
+    - [9,4,1,7]. The difference between the highest and lowest score is 9 - 7 = 2.
+    - [9,4,1,7]. The difference between the highest and lowest score is 4 - 1 = 3.
+    - [9,4,1,7]. The difference between the highest and lowest score is 7 - 4 = 3.
+    - [9,4,1,7]. The difference between the highest and lowest score is 7 - 1 = 6.
+    The minimum possible difference is 2.
+
+**Constraints:**
+
+- 1 <= k <= nums.length <= 1000
+- 0 <= nums[i] <= 100000
+
+## 题目大意
+
+给你一个下标从 0 开始的整数数组 nums ，其中 nums[i] 表示第 i 名学生的分数。另给你一个整数 k 。
+
+从数组中选出任意 k 名学生的分数，使这 k 个分数间最高分和最低分的差值达到最小化 。
+
+返回可能的最小差值 。
+
+## 解题思路
+
+- nums 排序
+- 求出nums[i+k-1] - nums[i]中的最小差值
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func minimumDifference(nums []int, k int) int {
+	sort.Ints(nums)
+	minDiff := 100000 + 1
+	for i := 0; i < len(nums); i++ {
+		if i+k-1 >= len(nums) {
+			break
+		}
+		diff := nums[i+k-1] - nums[i]
+		if diff < minDiff {
+			minDiff = diff
+		}
+	}
+	return minDiff
+}
+```
\ No newline at end of file
diff --git a/leetcode/2021.Brightest-Position-on-Street/2021. Brightest Position on Street.go b/leetcode/2021.Brightest-Position-on-Street/2021. Brightest Position on Street.go
new file mode 100644
index 000000000..c667e2465
--- /dev/null
+++ b/leetcode/2021.Brightest-Position-on-Street/2021. Brightest Position on Street.go	
@@ -0,0 +1,33 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+type lightItem struct {
+	index int
+	sign  int
+}
+
+func brightestPosition(lights [][]int) int {
+	lightMap, lightItems := map[int]int{}, []lightItem{}
+	for _, light := range lights {
+		lightMap[light[0]-light[1]] += 1
+		lightMap[light[0]+light[1]+1] -= 1
+	}
+	for k, v := range lightMap {
+		lightItems = append(lightItems, lightItem{index: k, sign: v})
+	}
+	sort.SliceStable(lightItems, func(i, j int) bool {
+		return lightItems[i].index < lightItems[j].index
+	})
+	res, border, tmp := 0, 0, 0
+	for _, v := range lightItems {
+		tmp += v.sign
+		if border < tmp {
+			res = v.index
+			border = tmp
+		}
+	}
+	return res
+}
diff --git a/leetcode/2021.Brightest-Position-on-Street/2021. Brightest Position on Street_test.go b/leetcode/2021.Brightest-Position-on-Street/2021. Brightest Position on Street_test.go
new file mode 100644
index 000000000..5247fbf6e
--- /dev/null
+++ b/leetcode/2021.Brightest-Position-on-Street/2021. Brightest Position on Street_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2021 struct {
+	para2021
+	ans2021
+}
+
+// para 是参数
+type para2021 struct {
+	lights [][]int
+}
+
+// ans 是答案
+type ans2021 struct {
+	ans int
+}
+
+func Test_Problem2021(t *testing.T) {
+
+	qs := []question2021{
+
+		{
+			para2021{[][]int{{-3, 2}, {1, 2}, {3, 3}}},
+			ans2021{-1},
+		},
+
+		{
+			para2021{[][]int{{1, 0}, {0, 1}}},
+			ans2021{1},
+		},
+
+		{
+			para2021{[][]int{{1, 2}}},
+			ans2021{-1},
+		},
+
+		{
+			para2021{[][]int{{1, 1}, {2, 4}, {-1, 0}, {-3, 5}, {1, 2}}},
+			ans2021{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2021------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2021, q.para2021
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", brightestPosition(p.lights))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2021.Brightest-Position-on-Street/README.md b/leetcode/2021.Brightest-Position-on-Street/README.md
new file mode 100644
index 000000000..3456fc7a0
--- /dev/null
+++ b/leetcode/2021.Brightest-Position-on-Street/README.md
@@ -0,0 +1,109 @@
+# [2021. Brightest Position on Street](https://leetcode.com/problems/brightest-position-on-street/)
+
+
+## 题目
+
+A perfectly straight street is represented by a number line. The street has street lamp(s) on it and is represented by a 2D integer array `lights`. Each `lights[i] = [positioni, rangei]` indicates that there is a street lamp at position `positioni` that lights up the area from `[positioni - rangei, positioni + rangei]` (**inclusive**).
+
+The **brightness** of a position `p` is defined as the number of street lamp that light up the position `p`.
+
+Given `lights`, return *the **brightest** position on the street. If there are multiple brightest positions, return the **smallest** one.*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/09/28/image-20210928155140-1.png](https://assets.leetcode.com/uploads/2021/09/28/image-20210928155140-1.png)
+
+```
+Input: lights = [[-3,2],[1,2],[3,3]]
+Output: -1
+Explanation:
+The first street lamp lights up the area from [(-3) - 2, (-3) + 2] = [-5, -1].
+The second street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3].
+The third street lamp lights up the area from [3 - 3, 3 + 3] = [0, 6].
+
+Position -1 has a brightness of 2, illuminated by the first and second street light.
+Positions 0, 1, 2, and 3 have a brightness of 2, illuminated by the second and third street light.
+Out of all these positions, -1 is the smallest, so return it.
+
+```
+
+**Example 2:**
+
+```
+Input: lights = [[1,0],[0,1]]
+Output: 1
+Explanation:
+The first street lamp lights up the area from [1 - 0, 1 + 0] = [1, 1].
+The second street lamp lights up the area from [0 - 1, 0 + 1] = [-1, 1].
+
+Position 1 has a brightness of 2, illuminated by the first and second street light.
+Return 1 because it is the brightest position on the street.
+
+```
+
+**Example 3:**
+
+```
+Input: lights = [[1,2]]
+Output: -1
+Explanation:
+The first street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3].
+
+Positions -1, 0, 1, 2, and 3 have a brightness of 1, illuminated by the first street light.
+Out of all these positions, -1 is the smallest, so return it.
+
+```
+
+**Constraints:**
+
+- `1 <= lights.length <= 105`
+- `lights[i].length == 2`
+- `108 <= positioni <= 108`
+- `0 <= rangei <= 108`
+
+## 题目大意
+
+一条完全笔直的街道由一条数字线表示。街道上有路灯，由二维数据表示。每个 `lights[i] = [positioni, rangei]` 表示位置 `i` 处有一盏路灯，灯可以照亮从 `[positioni - rangei, positioni + rangei]` （含）的区域。 位置 `p` 的亮度定义为点亮位置 `p` 的路灯数量。 给定路灯，返回街道上最亮的位置。如果有多个最亮的位置，则返回最小的一个。
+
+## 解题思路
+
+- 先将每个路灯的起始和终点位置计算出来。这样我们得到了一堆坐标点。假设灯照亮的范围是 [A, B]，那么在坐标轴上 A 坐标点处 + 1， B + 1 坐标点处 -1 。这样处理的含义是：坐标点 A 可以被一盏灯照亮，所以它照亮次数加一，坐标点 B + 1 出了灯照亮的范围了，所以照亮次数减一。那么从坐标轴坐标开始扫一遍，每次遇到 + 1 的时候就 + 1，遇到 - 1 的地方就 - 1。如此可以算出某个坐标点处，可以被灯照亮的总次数。
+- 需要注意的点是，题目给的测试数据可能会有单点照亮的情况，即某一盏灯只照亮一个坐标点，灯照范围为 0。同一个坐标点也可能是多个灯的起点。用一个 map 去重坐标点即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+type lightItem struct {
+	index int
+	sign  int
+}
+
+func brightestPosition(lights [][]int) int {
+	lightMap, lightItems := map[int]int{}, []lightItem{}
+	for _, light := range lights {
+		lightMap[light[0]-light[1]] += 1
+		lightMap[light[0]+light[1]+1] -= 1
+	}
+	for k, v := range lightMap {
+		lightItems = append(lightItems, lightItem{index: k, sign: v})
+	}
+	sort.SliceStable(lightItems, func(i, j int) bool {
+		return lightItems[i].index < lightItems[j].index
+	})
+	res, border, tmp := 0, 0, 0
+	for _, v := range lightItems {
+		tmp += v.sign
+		if border < tmp {
+			res = v.index
+			border = tmp
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/2022.Convert-1D-Array-Into-2D-Array/2022. Convert 1D Array Into 2D Array.go b/leetcode/2022.Convert-1D-Array-Into-2D-Array/2022. Convert 1D Array Into 2D Array.go
new file mode 100644
index 000000000..28dd363b8
--- /dev/null
+++ b/leetcode/2022.Convert-1D-Array-Into-2D-Array/2022. Convert 1D Array Into 2D Array.go	
@@ -0,0 +1,12 @@
+package leetcode
+
+func construct2DArray(original []int, m int, n int) [][]int {
+	if m*n != len(original) {
+		return [][]int{}
+	}
+	res := make([][]int, m)
+	for i := 0; i < m; i++ {
+		res[i] = original[n*i : n*(i+1)]
+	}
+	return res
+}
diff --git a/leetcode/2022.Convert-1D-Array-Into-2D-Array/2022. Convert 1D Array Into 2D Array_test.go b/leetcode/2022.Convert-1D-Array-Into-2D-Array/2022. Convert 1D Array Into 2D Array_test.go
new file mode 100644
index 000000000..39f540d38
--- /dev/null
+++ b/leetcode/2022.Convert-1D-Array-Into-2D-Array/2022. Convert 1D Array Into 2D Array_test.go	
@@ -0,0 +1,64 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2022 struct {
+	para2022
+	ans2022
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2022 struct {
+	original []int
+	m        int
+	n        int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2022 struct {
+	one [][]int
+}
+
+func Test_Problem2022(t *testing.T) {
+
+	qs := []question2022{
+
+		{
+			para2022{[]int{1, 2, 3, 4}, 2, 2},
+			ans2022{[][]int{{1, 2}, {3, 4}}},
+		},
+
+		{
+			para2022{[]int{1, 2, 3}, 1, 3},
+			ans2022{[][]int{{1, 2, 3}}},
+		},
+
+		{
+			para2022{[]int{1, 2}, 1, 1},
+			ans2022{[][]int{{}}},
+		},
+
+		{
+			para2022{[]int{3}, 1, 2},
+			ans2022{[][]int{{3}}},
+		},
+
+		{
+			para2022{[]int{1, 1, 1, 1}, 4, 1},
+			ans2022{[][]int{{1, 1, 1, 1}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2022------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2022, q.para2022
+		fmt.Printf("【input】:%v       【output】:%v\n", p, construct2DArray(p.original, p.m, p.n))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2022.Convert-1D-Array-Into-2D-Array/README.md b/leetcode/2022.Convert-1D-Array-Into-2D-Array/README.md
new file mode 100644
index 000000000..83899096d
--- /dev/null
+++ b/leetcode/2022.Convert-1D-Array-Into-2D-Array/README.md
@@ -0,0 +1,77 @@
+# [2022. Convert 1D Array Into 2D Array](https://leetcode.com/problems/convert-1d-array-into-2d-array/)
+
+## 题目
+
+You are given a **0-indexed** 1-dimensional (1D) integer array `original`, and two integers, `m` and `n`. You are tasked with creating a 2-dimensional (2D) array with `m` rows and `n` columns using **all** the elements from `original`.
+
+The elements from indices `0` to `n - 1` (**inclusive**) of `original` should form the first row of the constructed 2D array, the elements from indices `n` to `2 * n - 1` (**inclusive**) should form the second row of the constructed 2D array, and so on.
+
+Return *an* `m x n` *2D array constructed according to the above procedure, or an empty 2D array if it is impossible*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/08/26/image-20210826114243-1.png](https://assets.leetcode.com/uploads/2021/08/26/image-20210826114243-1.png)
+
+```
+Input: original = [1,2,3,4], m = 2, n = 2
+Output: [[1,2],[3,4]]
+Explanation: The constructed 2D array should contain 2 rows and 2 columns.
+The first group of n=2 elements in original, [1,2], becomes the first row in the constructed 2D array.
+The second group of n=2 elements in original, [3,4], becomes the second row in the constructed 2D array.
+
+```
+
+**Example 2:**
+
+```
+Input: original = [1,2,3], m = 1, n = 3
+Output: [[1,2,3]]
+Explanation: The constructed 2D array should contain 1 row and 3 columns.
+Put all three elements in original into the first row of the constructed 2D array.
+
+```
+
+**Example 3:**
+
+```
+Input: original = [1,2], m = 1, n = 1
+Output: []
+Explanation: There are 2 elements in original.
+It is impossible to fit 2 elements in a 1x1 2D array, so return an empty 2D array.
+
+```
+
+**Constraints:**
+
+- `1 <= original.length <= 5 * 104`
+- `1 <= original[i] <= 105`
+- `1 <= m, n <= 4 * 104`
+
+## 题目大意
+
+给你一个下标从 0 开始的一维整数数组 original 和两个整数 m 和  n 。你需要使用 original 中 所有 元素创建一个 m 行 n 列的二维数组。
+
+original 中下标从 0 到 n - 1 （都 包含 ）的元素构成二维数组的第一行，下标从 n 到 2 * n - 1 （都 包含 ）的元素构成二维数组的第二行，依此类推。
+
+请你根据上述过程返回一个 m x n 的二维数组。如果无法构成这样的二维数组，请你返回一个空的二维数组。
+
+## 解题思路
+
+- 简单题。从一维数组 original 中依次取出每行 n 个元素，顺序放到 m 行中。此题中，如果 m*n 大于或者小于 original 的长度，都输出空数组。
+
+## 代码
+
+```go
+package leetcode
+
+func construct2DArray(original []int, m int, n int) [][]int {
+	if m*n != len(original) {
+		return [][]int{}
+	}
+	res := make([][]int, m)
+	for i := 0; i < m; i++ {
+		res[i] = original[n*i : n*(i+1)]
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/2037.Minimum-Number-of-Moves-to-Seat-Everyone/2037.Minimum Number of Moves to Seat Everyone.go b/leetcode/2037.Minimum-Number-of-Moves-to-Seat-Everyone/2037.Minimum Number of Moves to Seat Everyone.go
new file mode 100644
index 000000000..89cae6ff4
--- /dev/null
+++ b/leetcode/2037.Minimum-Number-of-Moves-to-Seat-Everyone/2037.Minimum Number of Moves to Seat Everyone.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+import "sort"
+
+func minMovesToSeat(seats []int, students []int) int {
+	sort.Ints(seats)
+	sort.Ints(students)
+	n := len(students)
+	moves := 0
+	for i := 0; i < n; i++ {
+		moves += abs(seats[i], students[i])
+	}
+	return moves
+}
+
+func abs(a, b int) int {
+	if a > b {
+		return a - b
+	}
+	return b - a
+}
diff --git a/leetcode/2037.Minimum-Number-of-Moves-to-Seat-Everyone/2037.Minimum Number of Moves to Seat Everyone_test.go b/leetcode/2037.Minimum-Number-of-Moves-to-Seat-Everyone/2037.Minimum Number of Moves to Seat Everyone_test.go
new file mode 100644
index 000000000..17239790e
--- /dev/null
+++ b/leetcode/2037.Minimum-Number-of-Moves-to-Seat-Everyone/2037.Minimum Number of Moves to Seat Everyone_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2037 struct {
+	para2037
+	ans2037
+}
+
+// para 是参数
+type para2037 struct {
+	seats    []int
+	students []int
+}
+
+// ans 是答案
+type ans2037 struct {
+	ans int
+}
+
+func Test_Problem2037(t *testing.T) {
+
+	qs := []question2037{
+
+		{
+			para2037{[]int{3, 1, 5}, []int{2, 7, 4}},
+			ans2037{4},
+		},
+
+		{
+			para2037{[]int{4, 1, 5, 9}, []int{1, 3, 2, 6}},
+			ans2037{7},
+		},
+
+		{
+			para2037{[]int{2, 2, 6, 6}, []int{1, 3, 2, 6}},
+			ans2037{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2037------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2037, q.para2037
+		fmt.Printf("【input】:%v      ", p)
+		fmt.Printf("【output】:%v      \n", minMovesToSeat(p.seats, p.students))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2037.Minimum-Number-of-Moves-to-Seat-Everyone/README.md b/leetcode/2037.Minimum-Number-of-Moves-to-Seat-Everyone/README.md
new file mode 100644
index 000000000..beec7e76c
--- /dev/null
+++ b/leetcode/2037.Minimum-Number-of-Moves-to-Seat-Everyone/README.md
@@ -0,0 +1,94 @@
+# [2037. Minimum Number of Moves to Seat Everyone](https://leetcode.com/problems/minimum-number-of-moves-to-seat-everyone/)
+
+## 题目
+
+There are n seats and n students in a room. You are given an array seats of length n, where seats[i] is the position of the ith seat. You are also given the array students of length n, where students[j] is the position of the jth student.
+
+You may perform the following move any number of times:
+
+- Increase or decrease the position of the ith student by 1 (i.e., moving the ith student from position x to x + 1 or x - 1)
+
+Return the minimum number of moves required to move each student to a seat such that no two students are in the same seat.
+
+Note that there may be multiple seats or students in the same position at the beginning.
+
+**Example 1:**
+
+    Input: seats = [3,1,5], students = [2,7,4]
+    Output: 4
+    Explanation: The students are moved as follows:
+    - The first student is moved from from position 2 to position 1 using 1 move.
+    - The second student is moved from from position 7 to position 5 using 2 moves.
+    - The third student is moved from from position 4 to position 3 using 1 move.
+    In total, 1 + 2 + 1 = 4 moves were used.
+
+**Example 2:**
+
+    Input: seats = [4,1,5,9], students = [1,3,2,6]
+    Output: 7
+    Explanation: The students are moved as follows:
+    - The first student is not moved.
+    - The second student is moved from from position 3 to position 4 using 1 move.
+    - The third student is moved from from position 2 to position 5 using 3 moves.
+    - The fourth student is moved from from position 6 to position 9 using 3 moves.
+    In total, 0 + 1 + 3 + 3 = 7 moves were used.
+
+**Example 3:**
+
+    Input: seats = [2,2,6,6], students = [1,3,2,6]
+    Output: 4
+    Explanation: Note that there are two seats at position 2 and two seats at position 6.
+    The students are moved as follows:
+    - The first student is moved from from position 1 to position 2 using 1 move.
+    - The second student is moved from from position 3 to position 6 using 3 moves.
+    - The third student is not moved.
+    - The fourth student is not moved.
+    In total, 1 + 3 + 0 + 0 = 4 moves were used.
+
+**Constraints:**
+
+- n == seats.length == students.length
+- 1 <= n <= 100
+- 1 <= seats[i], students[j] <= 100
+
+## 题目大意
+
+一个房间里有 n 个座位和 n 名学生，房间用一个数轴表示。给你一个长度为 n 的数组 seats，其中 seats[i] 是第 i 个座位的位置。同时给你一个长度为 n 的数组 students ，其中 students[j] 是第 j 位学生的位置。
+
+你可以执行以下操作任意次：
+
+增加或者减少第 i 位学生的位置，每次变化量为 1（也就是将第 i 位学生从位置 x 移动到 x + 1或者 x - 1）
+
+请你返回使所有学生都有座位坐的最少移动次数，并确保没有两位学生的座位相同。
+
+请注意，初始时有可能有多个座位或者多位学生在 同一位置。
+
+## 解题思路
+
+- 排序+模拟计算
+
+# 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func minMovesToSeat(seats []int, students []int) int {
+	sort.Ints(seats)
+	sort.Ints(students)
+	n := len(students)
+	moves := 0
+	for i := 0; i < n; i++ {
+		moves += abs(seats[i], students[i])
+	}
+	return moves
+}
+
+func abs(a, b int) int {
+	if a > b {
+		return a - b
+	}
+	return b - a
+}
+```
\ No newline at end of file
diff --git a/leetcode/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color/2038.Remove Colored Pieces if Both Neighbors are the Same Color.go b/leetcode/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color/2038.Remove Colored Pieces if Both Neighbors are the Same Color.go
new file mode 100644
index 000000000..8962f957a
--- /dev/null
+++ b/leetcode/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color/2038.Remove Colored Pieces if Both Neighbors are the Same Color.go	
@@ -0,0 +1,25 @@
+package leetcode
+
+func winnerOfGame(colors string) bool {
+	As, Bs := 0, 0
+	Acont, Bcont := 0, 0
+	for _, color := range colors {
+		if color == 'A' {
+			Acont += 1
+			Bcont = 0
+		} else {
+			Bcont += 1
+			Acont = 0
+		}
+		if Acont >= 3 {
+			As += Acont - 2
+		}
+		if Bcont >= 3 {
+			Bs += Bcont - 2
+		}
+	}
+	if As > Bs {
+		return true
+	}
+	return false
+}
diff --git a/leetcode/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color/2038.Remove Colored Pieces if Both Neighbors are the Same Color_test.go b/leetcode/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color/2038.Remove Colored Pieces if Both Neighbors are the Same Color_test.go
new file mode 100644
index 000000000..567eea1fe
--- /dev/null
+++ b/leetcode/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color/2038.Remove Colored Pieces if Both Neighbors are the Same Color_test.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2038 struct {
+	para2038
+	ans2038
+}
+
+// para 是参数
+type para2038 struct {
+	colors string
+}
+
+// ans 是答案
+type ans2038 struct {
+	ans bool
+}
+
+func Test_Problem2038(t *testing.T) {
+
+	qs := []question2038{
+
+		{
+			para2038{"AAABABB"},
+			ans2038{true},
+		},
+
+		{
+			para2038{"AA"},
+			ans2038{false},
+		},
+
+		{
+			para2038{"ABBBBBBBAAA"},
+			ans2038{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2038------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2038, q.para2038
+		fmt.Printf("【input】:%v      ", p.colors)
+		fmt.Printf("【output】:%v      \n", winnerOfGame(p.colors))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color/README.md b/leetcode/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color/README.md
new file mode 100644
index 000000000..fdf3e5525
--- /dev/null
+++ b/leetcode/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color/README.md
@@ -0,0 +1,105 @@
+# [2038. Remove Colored Pieces if Both Neighbors are the Same Color](https://leetcode.com/problems/remove-colored-pieces-if-both-neighbors-are-the-same-color/)
+
+## 题目
+
+There are n pieces arranged in a line, and each piece is colored either by 'A' or by 'B'. You are given a string colors of length n where colors[i] is the color of the ith piece.
+
+Alice and Bob are playing a game where they take alternating turns removing pieces from the line. In this game, Alice moves first.
+
+- Alice is only allowed to remove a piece colored 'A' if both its neighbors are also colored 'A'. She is not allowed to remove pieces that are colored 'B'.
+- Bob is only allowed to remove a piece colored 'B' if both its neighbors are also colored 'B'. He is not allowed to remove pieces that are colored 'A'.
+- Alice and Bob cannot remove pieces from the edge of the line.
+- If a player cannot make a move on their turn, that player loses and the other player wins.
+
+Assuming Alice and Bob play optimally, return true if Alice wins, or return false if Bob wins.
+
+**Example 1:**
+
+    Input: colors = "AAABABB"
+    Output: true
+    Explanation:
+    AAABABB -> AABABB
+    Alice moves first.
+    She removes the second 'A' from the left since that is the only 'A' whose neighbors are both 'A'.
+
+    Now it's Bob's turn.
+    Bob cannot make a move on his turn since there are no 'B's whose neighbors are both 'B'.
+    Thus, Alice wins, so return true.
+
+**Example 2:**
+
+    Input: colors = "AA"
+    Output: false
+    Explanation:
+    Alice has her turn first.
+    There are only two 'A's and both are on the edge of the line, so she cannot move on her turn.
+    Thus, Bob wins, so return false.
+
+**Example 3:**
+
+    Input: colors = "ABBBBBBBAAA"
+    Output: false
+    Explanation:
+    ABBBBBBBAAA -> ABBBBBBBAA
+    Alice moves first.
+    Her only option is to remove the second to last 'A' from the right.
+
+    ABBBBBBBAA -> ABBBBBBAA
+    Next is Bob's turn.
+    He has many options for which 'B' piece to remove. He can pick any.
+
+    On Alice's second turn, she has no more pieces that she can remove.
+    Thus, Bob wins, so return false.
+
+**Constraints:**
+
+- 1 <= colors.length <= 100000
+- colors consists of only the letters 'A' and 'B'
+
+## 题目大意
+
+总共有 n 个颜色片段排成一列，每个颜色片段要么是 'A' 要么是 'B' 。给你一个长度为 n 的字符串 colors ，其中 colors[i] 表示第 i 个颜色片段的颜色。
+
+Alice 和 Bob 在玩一个游戏，他们轮流从这个字符串中删除颜色。Alice 先手。
+
+- 如果一个颜色片段为 'A' 且相邻两个颜色都是颜色 'A'，那么 Alice 可以删除该颜色片段。Alice不可以删除任何颜色 'B' 片段。
+- 如果一个颜色片段为 'B'且相邻两个颜色都是颜色 'B' ，那么 Bob 可以删除该颜色片段。Bob 不可以删除任何颜色 'A' 片段。
+- Alice 和 Bob 不能从字符串两端删除颜色片段。
+- 如果其中一人无法继续操作，则该玩家 输掉游戏且另一玩家 获胜。
+
+假设 Alice 和 Bob 都采用最优策略，如果 Alice 获胜，请返回true，否则 Bob 获胜，返回false。
+
+## 解题思路
+
+- 统计 Alice 和 Bob 分别可以操作的次数记为 As，Bs 
+- 因为 Alice 先手，所以只要 As 大于 Bs，Alice 获胜返回 true，否则 Bob 获胜返回 false
+
+# 代码
+
+```go
+package leetcode
+
+func winnerOfGame(colors string) bool {
+	As, Bs := 0, 0
+	Acont, Bcont := 0, 0
+	for _, color := range colors {
+		if color == 'A' {
+			Acont += 1
+			Bcont = 0
+		} else {
+			Bcont += 1
+			Acont = 0
+		}
+		if Acont >= 3 {
+			As += Acont - 2
+		}
+		if Bcont >= 3 {
+			Bs += Bcont - 2
+		}
+	}
+	if As > Bs {
+		return true
+	}
+	return false
+}
+```
\ No newline at end of file
diff --git a/leetcode/2043.Simple-Bank-System/2043.Simple Bank System.go b/leetcode/2043.Simple-Bank-System/2043.Simple Bank System.go
new file mode 100644
index 000000000..025ce2480
--- /dev/null
+++ b/leetcode/2043.Simple-Bank-System/2043.Simple Bank System.go	
@@ -0,0 +1,44 @@
+package leetcode
+
+type Bank struct {
+	accounts []int64
+	n        int
+}
+
+func Constructor(balance []int64) Bank {
+	return Bank{
+		accounts: balance,
+		n:        len(balance),
+	}
+}
+
+func (this *Bank) Transfer(account1 int, account2 int, money int64) bool {
+	if account1 > this.n || account2 > this.n {
+		return false
+	}
+	if this.accounts[account1-1] < money {
+		return false
+	}
+	this.accounts[account1-1] -= money
+	this.accounts[account2-1] += money
+	return true
+}
+
+func (this *Bank) Deposit(account int, money int64) bool {
+	if account > this.n {
+		return false
+	}
+	this.accounts[account-1] += money
+	return true
+}
+
+func (this *Bank) Withdraw(account int, money int64) bool {
+	if account > this.n {
+		return false
+	}
+	if this.accounts[account-1] < money {
+		return false
+	}
+	this.accounts[account-1] -= money
+	return true
+}
diff --git a/leetcode/2043.Simple-Bank-System/2043.Simple Bank System_test.go b/leetcode/2043.Simple-Bank-System/2043.Simple Bank System_test.go
new file mode 100644
index 000000000..96908986c
--- /dev/null
+++ b/leetcode/2043.Simple-Bank-System/2043.Simple Bank System_test.go	
@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2043 struct {
+	para2043
+	ans2043
+}
+
+// para 是参数
+type para2043 struct {
+	ops  []string
+	para [][]int64
+}
+
+// ans 是答案
+type ans2043 struct {
+	ans []bool
+}
+
+func Test_Problem2043(t *testing.T) {
+
+	qs := []question2043{
+
+		{
+			para2043{
+				[]string{"Bank", "withdraw", "transfer", "deposit", "transfer", "withdraw"},
+				[][]int64{{10, 100, 20, 50, 30}, {3, 10}, {5, 1, 20}, {5, 20}, {3, 4, 15}, {10, 50}}},
+			ans2043{[]bool{true, true, true, false, false}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2043------------------------\n")
+
+	for _, q := range qs {
+		var b Bank
+		var res []bool
+		_, p := q.ans2043, q.para2043
+		for i, op := range p.ops {
+			if op == "Bank" {
+				b = Constructor(q.para[i])
+			} else if op == "withdraw" {
+				isSuccess := b.Withdraw(int(p.para[i][0]), p.para[i][1])
+				res = append(res, isSuccess)
+			} else if op == "transfer" {
+				isSuccess := b.Transfer(int(p.para[i][0]), int(p.para[i][0]), p.para[i][2])
+				res = append(res, isSuccess)
+			} else if op == "deposit" {
+				isSuccess := b.Deposit(int(p.para[i][0]), p.para[i][1])
+				res = append(res, isSuccess)
+			} else {
+				fmt.Println("unknown operation")
+			}
+		}
+		fmt.Printf("【input】:%v      \n", p)
+		fmt.Printf("【output】:%v      \n", res)
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2043.Simple-Bank-System/README.md b/leetcode/2043.Simple-Bank-System/README.md
new file mode 100644
index 000000000..abae43d3b
--- /dev/null
+++ b/leetcode/2043.Simple-Bank-System/README.md
@@ -0,0 +1,113 @@
+# [2043. Simple Bank System](https://leetcode.com/problems/simple-bank-system/)
+
+## 题目
+
+You have been tasked with writing a program for a popular bank that will automate all its incoming transactions (transfer, deposit, and withdraw). The bank has n accounts numbered from 1 to n. The initial balance of each account is stored in a 0-indexed integer array balance, with the (i + 1)th account having an initial balance of balance[i].
+
+Execute all the valid transactions. A transaction is valid if:
+
+- The given account number(s) are between 1 and n, and
+- The amount of money withdrawn or transferred from is less than or equal to the balance of the account.
+
+Implement the Bank class:
+
+- Bank(long[] balance) Initializes the object with the 0-indexed integer array balance.
+- boolean transfer(int account1, int account2, long money) Transfers money dollars from the account numbered account1 to the account numbered account2. Return true if the transaction was successful, false otherwise.
+- boolean deposit(int account, long money) Deposit money dollars into the account numbered account. Return true if the transaction was successful, false otherwise.
+- boolean withdraw(int account, long money) Withdraw money dollars from the account numbered account. Return true if the transaction was successful, false otherwise.
+
+**Example 1:**
+
+    Input
+    ["Bank", "withdraw", "transfer", "deposit", "transfer", "withdraw"]
+    [[[10, 100, 20, 50, 30]], [3, 10], [5, 1, 20], [5, 20], [3, 4, 15], [10, 50]]
+    Output
+    [null, true, true, true, false, false]
+
+    Explanation
+    Bank bank = new Bank([10, 100, 20, 50, 30]);
+    bank.withdraw(3, 10);    // return true, account 3 has a balance of $20, so it is valid to withdraw $10.
+    // Account 3 has $20 - $10 = $10.
+    bank.transfer(5, 1, 20); // return true, account 5 has a balance of $30, so it is valid to transfer $20.
+    // Account 5 has $30 - $20 = $10, and account 1 has $10 + $20 = $30.
+    bank.deposit(5, 20);     // return true, it is valid to deposit $20 to account 5.
+    // Account 5 has $10 + $20 = $30.
+    bank.transfer(3, 4, 15); // return false, the current balance of account 3 is $10,
+    // so it is invalid to transfer $15 from it.
+    bank.withdraw(10, 50);   // return false, it is invalid because account 10 does not exist.
+
+**Constraints:**
+
+- n == balance.length
+- 1 <= n, account, account1, account2 <= 100000
+- 0 <= balance[i], money <= 1000000000000
+- At most 104 calls will be made to each function transfer, deposit, withdraw.
+
+## 题目大意
+
+你的任务是为一个很受欢迎的银行设计一款程序，以自动化执行所有传入的交易（转账，存款和取款）。银行共有 n 个账户，编号从 1 到 n 。每个账号的初始余额存储在一个下标从 0 开始的整数数组 balance 中，其中第 (i + 1) 个账户的初始余额是 balance[i] 。
+
+请你执行所有 有效的 交易。如果满足下面全部条件，则交易 有效 ：
+
+- 指定的账户数量在 1 和 n 之间，且
+- 取款或者转账需要的钱的总数 小于或者等于 账户余额。
+
+实现 Bank 类：
+
+- Bank(long[] balance) 使用下标从 0 开始的整数数组 balance 初始化该对象。
+- boolean transfer(int account1, int account2, long money) 从编号为 account1 的账户向编号为 account2 的账户转帐 money 美元。如果交易成功，返回 true ，否则，返回 false 。
+- boolean deposit(int account, long money) 向编号为 account 的账户存款 money 美元。如果交易成功，返回 true ；否则，返回 false 。
+- boolean withdraw(int account, long money) 从编号为 account 的账户取款 money 美元。如果交易成功，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+    根据题意进行简单模拟
+
+# 代码
+
+```go
+package leetcode
+
+type Bank struct {
+	accounts []int64
+	n        int
+}
+
+func Constructor(balance []int64) Bank {
+	return Bank{
+		accounts: balance,
+		n:        len(balance),
+	}
+}
+
+func (this *Bank) Transfer(account1 int, account2 int, money int64) bool {
+	if account1 > this.n || account2 > this.n {
+		return false
+	}
+	if this.accounts[account1-1] < money {
+		return false
+	}
+	this.accounts[account1-1] -= money
+	this.accounts[account2-1] += money
+	return true
+}
+
+func (this *Bank) Deposit(account int, money int64) bool {
+	if account > this.n {
+		return false
+	}
+	this.accounts[account-1] += money
+	return true
+}
+
+func (this *Bank) Withdraw(account int, money int64) bool {
+	if account > this.n {
+		return false
+	}
+	if this.accounts[account-1] < money {
+		return false
+	}
+	this.accounts[account-1] -= money
+	return true
+}
+```
\ No newline at end of file
diff --git a/leetcode/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another/2096. Step-By-Step Directions From a Binary Tree Node to Another.go b/leetcode/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another/2096. Step-By-Step Directions From a Binary Tree Node to Another.go
new file mode 100644
index 000000000..cd4f10897
--- /dev/null
+++ b/leetcode/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another/2096. Step-By-Step Directions From a Binary Tree Node to Another.go	
@@ -0,0 +1,77 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func getDirections(root *TreeNode, startValue int, destValue int) string {
+	sPath, dPath := make([]byte, 0), make([]byte, 0)
+	findPath(root, startValue, &sPath)
+	findPath(root, destValue, &dPath)
+	size, i := min(len(sPath), len(dPath)), 0
+	for i < size {
+		if sPath[len(sPath)-1-i] == dPath[len(dPath)-1-i] {
+			i++
+		} else {
+			break
+		}
+	}
+	sPath = sPath[:len(sPath)-i]
+	replace(sPath)
+	dPath = dPath[:len(dPath)-i]
+	reverse(dPath)
+	sPath = append(sPath, dPath...)
+	return string(sPath)
+}
+
+func findPath(root *TreeNode, value int, path *[]byte) bool {
+	if root.Val == value {
+		return true
+	}
+
+	if root.Left != nil && findPath(root.Left, value, path) {
+		*path = append(*path, 'L')
+		return true
+	}
+
+	if root.Right != nil && findPath(root.Right, value, path) {
+		*path = append(*path, 'R')
+		return true
+	}
+
+	return false
+}
+
+func reverse(path []byte) {
+	left, right := 0, len(path)-1
+	for left < right {
+		path[left], path[right] = path[right], path[left]
+		left++
+		right--
+	}
+}
+
+func replace(path []byte) {
+	for i := 0; i < len(path); i++ {
+		path[i] = 'U'
+	}
+}
+
+func min(i, j int) int {
+	if i < j {
+		return i
+	}
+	return j
+}
diff --git a/leetcode/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another/2096. Step-By-Step Directions From a Binary Tree Node to Another_test.go b/leetcode/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another/2096. Step-By-Step Directions From a Binary Tree Node to Another_test.go
new file mode 100644
index 000000000..e30610185
--- /dev/null
+++ b/leetcode/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another/2096. Step-By-Step Directions From a Binary Tree Node to Another_test.go	
@@ -0,0 +1,68 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question2096 struct {
+	para2096
+	ans2096
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2096 struct {
+	one        []int
+	startValue int
+	destValue  int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2096 struct {
+	one string
+}
+
+func Test_Problem2096(t *testing.T) {
+
+	qs := []question2096{
+
+		{
+			para2096{[]int{5, 1, 2, 3, structures.NULL, 6, 4}, 3, 6},
+			ans2096{"UURL"},
+		},
+
+		{
+			para2096{[]int{2, 1}, 2, 1},
+			ans2096{"L"},
+		},
+
+		{
+			para2096{[]int{1, 2}, 2, 1},
+			ans2096{"U"},
+		},
+
+		{
+			para2096{[]int{3, 1, 2}, 2, 1},
+			ans2096{"UL"},
+		},
+
+		{
+			para2096{[]int{7, 8, 3, 1, structures.NULL, 4, 5, 6, structures.NULL, structures.NULL, structures.NULL, structures.NULL, structures.NULL, structures.NULL, 2}, 7, 5},
+			ans2096{"RR"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2096------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2096, q.para2096
+		fmt.Printf("【input】:%v      ", p)
+		root := structures.Ints2TreeNode(p.one)
+		fmt.Printf("【output】:%v      \n", getDirections(root, p.startValue, p.destValue))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another/README.md b/leetcode/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another/README.md
new file mode 100644
index 000000000..409608c8c
--- /dev/null
+++ b/leetcode/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another/README.md
@@ -0,0 +1,145 @@
+# [2096. Step-By-Step Directions From a Binary Tree Node to Another](https://leetcode.com/problems/step-by-step-directions-from-a-binary-tree-node-to-another/)
+
+
+## 题目
+
+You are given the `root` of a **binary tree** with `n` nodes. Each node is uniquely assigned a value from `1` to `n`. You are also given an integer `startValue` representing the value of the start node `s`, and a different integer `destValue` representing the value of the destination node `t`.
+
+Find the **shortest path** starting from node `s` and ending at node `t`. Generate step-by-step directions of such path as a string consisting of only the **uppercase** letters `'L'`, `'R'`, and `'U'`. Each letter indicates a specific direction:
+
+- `'L'` means to go from a node to its **left child** node.
+- `'R'` means to go from a node to its **right child** node.
+- `'U'` means to go from a node to its **parent** node.
+
+Return *the step-by-step directions of the **shortest path** from node* `s` *to node* `t`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/11/15/eg1.png](https://assets.leetcode.com/uploads/2021/11/15/eg1.png)
+
+```
+Input: root = [5,1,2,3,null,6,4], startValue = 3, destValue = 6
+Output: "UURL"
+Explanation: The shortest path is: 3 → 1 → 5 → 2 → 6.
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2021/11/15/eg2.png](https://assets.leetcode.com/uploads/2021/11/15/eg2.png)
+
+```
+Input: root = [2,1], startValue = 2, destValue = 1
+Output: "L"
+Explanation: The shortest path is: 2 → 1.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is `n`.
+- `2 <= n <= 105`
+- `1 <= Node.val <= n`
+- All the values in the tree are **unique**.
+- `1 <= startValue, destValue <= n`
+- `startValue != destValue`
+
+## 题目大意
+
+给你一棵 二叉树 的根节点 root ，这棵二叉树总共有 n 个节点。每个节点的值为 1 到 n 中的一个整数，且互不相同。给你一个整数 startValue ，表示起点节点 s 的值，和另一个不同的整数 destValue ，表示终点节点 t 的值。
+
+请找到从节点 s 到节点 t 的 最短路径 ，并以字符串的形式返回每一步的方向。每一步用 大写 字母 'L' ，'R' 和 'U' 分别表示一种方向：
+
+- 'L' 表示从一个节点前往它的 左孩子 节点。
+- 'R' 表示从一个节点前往它的 右孩子 节点。
+- 'U' 表示从一个节点前往它的 父 节点。
+
+请你返回从 s 到 t 最短路径 每一步的方向。
+
+## 解题思路
+
+- 二叉树中一个节点到另一个节点的最短路径一定可以分为两个部分（可能为空）：从起点节点向上到两个节点的**最近公共祖先**，再从最近公共祖先向下到达终点节点。
+- 首先需要找到起点 s 与公共祖先的节点之间的 path1，公共祖先节点与终点 t 的 path2。再删掉 2 个 path 的公共前缀。如果起点 s 和终点 t 在不同的分支上，不存在公共前缀。如果他们在相同的分支上，那么最终答案要去掉这个公共前缀。
+- 删除掉公共前缀以后，需要再整理一下最终答案的输出格式。由于题目要求，起点到公共祖先节点需要输出 U，所以把这段 path1 全部改成 U，然后再拼接上 path2 字符串，即可得到的字符串即为待求 ss 到 tt 每一步的最短路径。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func getDirections(root *TreeNode, startValue int, destValue int) string {
+	sPath, dPath := make([]byte, 0), make([]byte, 0)
+	findPath(root, startValue, &sPath)
+	findPath(root, destValue, &dPath)
+	size, i := min(len(sPath), len(dPath)), 0
+	for i < size {
+		if sPath[len(sPath)-1-i] == dPath[len(dPath)-1-i] {
+			i++
+		} else {
+			break
+		}
+	}
+	sPath = sPath[:len(sPath)-i]
+	replace(sPath)
+	dPath = dPath[:len(dPath)-i]
+	reverse(dPath)
+	sPath = append(sPath, dPath...)
+	return string(sPath)
+}
+
+func findPath(root *TreeNode, value int, path *[]byte) bool {
+	if root.Val == value {
+		return true
+	}
+
+	if root.Left != nil && findPath(root.Left, value, path) {
+		*path = append(*path, 'L')
+		return true
+	}
+
+	if root.Right != nil && findPath(root.Right, value, path) {
+		*path = append(*path, 'R')
+		return true
+	}
+
+	return false
+}
+
+func reverse(path []byte) {
+	left, right := 0, len(path)-1
+	for left < right {
+		path[left], path[right] = path[right], path[left]
+		left++
+		right--
+	}
+}
+
+func replace(path []byte) {
+	for i := 0; i < len(path); i++ {
+		path[i] = 'U'
+	}
+}
+
+func min(i, j int) int {
+	if i < j {
+		return i
+	}
+	return j
+}
+```
\ No newline at end of file
diff --git a/leetcode/2164.Sort-Even-and-Odd-Indices-Independently/2164. Sort Even and Odd Indices Independently.go b/leetcode/2164.Sort-Even-and-Odd-Indices-Independently/2164. Sort Even and Odd Indices Independently.go
new file mode 100644
index 000000000..fa5fd49d9
--- /dev/null
+++ b/leetcode/2164.Sort-Even-and-Odd-Indices-Independently/2164. Sort Even and Odd Indices Independently.go	
@@ -0,0 +1,30 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+func sortEvenOdd(nums []int) []int {
+	odd, even, res := []int{}, []int{}, []int{}
+	for index, v := range nums {
+		if index%2 == 0 {
+			even = append(even, v)
+		} else {
+			odd = append(odd, v)
+		}
+	}
+	sort.Ints(even)
+	sort.Sort(sort.Reverse(sort.IntSlice(odd)))
+
+	indexO, indexE := 0, 0
+	for i := 0; i < len(nums); i++ {
+		if i%2 == 0 {
+			res = append(res, even[indexE])
+			indexE++
+		} else {
+			res = append(res, odd[indexO])
+			indexO++
+		}
+	}
+	return res
+}
diff --git a/leetcode/2164.Sort-Even-and-Odd-Indices-Independently/2164. Sort Even and Odd Indices Independently_test.go b/leetcode/2164.Sort-Even-and-Odd-Indices-Independently/2164. Sort Even and Odd Indices Independently_test.go
new file mode 100644
index 000000000..77933ad14
--- /dev/null
+++ b/leetcode/2164.Sort-Even-and-Odd-Indices-Independently/2164. Sort Even and Odd Indices Independently_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2164 struct {
+	para2164
+	ans2164
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2164 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2164 struct {
+	one []int
+}
+
+func Test_Problem1(t *testing.T) {
+
+	qs := []question2164{
+		{
+			para2164{[]int{4, 1, 2, 3}},
+			ans2164{[]int{2, 3, 4, 1}},
+		},
+
+		{
+			para2164{[]int{2, 1}},
+			ans2164{[]int{2, 1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2164------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2164, q.para2164
+		fmt.Printf("【input】:%v       【output】:%v\n", p, sortEvenOdd(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2164.Sort-Even-and-Odd-Indices-Independently/README.md b/leetcode/2164.Sort-Even-and-Odd-Indices-Independently/README.md
new file mode 100644
index 000000000..a8e650740
--- /dev/null
+++ b/leetcode/2164.Sort-Even-and-Odd-Indices-Independently/README.md
@@ -0,0 +1,93 @@
+# [2164. Sort Even and Odd Indices Independently](https://leetcode.com/problems/sort-even-and-odd-indices-independently/)
+
+
+## 题目
+
+You are given a **0-indexed** integer array `nums`. Rearrange the values of `nums` according to the following rules:
+
+1. Sort the values at **odd indices** of `nums` in **non-increasing** order.
+    - For example, if `nums = [4,**1**,2,**3**]` before this step, it becomes `[4,**3**,2,**1**]` after. The values at odd indices `1` and `3` are sorted in non-increasing order.
+2. Sort the values at **even indices** of `nums` in **non-decreasing** order.
+    - For example, if `nums = [**4**,1,**2**,3]` before this step, it becomes `[**2**,1,**4**,3]` after. The values at even indices `0` and `2` are sorted in non-decreasing order.
+
+Return *the array formed after rearranging the values of* `nums`.
+
+**Example 1:**
+
+```
+Input: nums = [4,1,2,3]
+Output: [2,3,4,1]
+Explanation:
+First, we sort the values present at odd indices (1 and 3) in non-increasing order.
+So, nums changes from [4,1,2,3] to [4,3,2,1].
+Next, we sort the values present at even indices (0 and 2) in non-decreasing order.
+So, nums changes from [4,1,2,3] to [2,3,4,1].
+Thus, the array formed after rearranging the values is [2,3,4,1].
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,1]
+Output: [2,1]
+Explanation:
+Since there is exactly one odd index and one even index, no rearrangement of values takes place.
+The resultant array formed is [2,1], which is the same as the initial array.
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `1 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个下标从 0 开始的整数数组 nums 。根据下述规则重排 nums 中的值：
+
+1. 按 非递增 顺序排列 nums 奇数下标 上的所有值。
+举个例子，如果排序前 nums = [4,1,2,3] ，对奇数下标的值排序后变为 [4,3,2,1] 。奇数下标 1 和 3 的值按照非递增顺序重排。
+2. 按 非递减 顺序排列 nums 偶数下标 上的所有值。
+举个例子，如果排序前 nums = [4,1,2,3] ，对偶数下标的值排序后变为 [2,1,4,3] 。偶数下标 0 和 2 的值按照非递减顺序重排。
+
+返回重排 nums 的值之后形成的数组。
+
+## 解题思路
+
+- 简单题。分别将奇数和偶数位上的数字排序，奇数位的数从大到小，偶数位的数从小到大。最后将他们组合成一个数组。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func sortEvenOdd(nums []int) []int {
+	odd, even, res := []int{}, []int{}, []int{}
+	for index, v := range nums {
+		if index%2 == 0 {
+			even = append(even, v)
+		} else {
+			odd = append(odd, v)
+		}
+	}
+	sort.Ints(even)
+	sort.Sort(sort.Reverse(sort.IntSlice(odd)))
+
+	indexO, indexE := 0, 0
+	for i := 0; i < len(nums); i++ {
+		if i%2 == 0 {
+			res = append(res, even[indexE])
+			indexE++
+		} else {
+			res = append(res, odd[indexO])
+			indexO++
+		}
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/2165.Smallest-Value-of-the-Rearranged-Number/2165. Smallest Value of the Rearranged Number.go b/leetcode/2165.Smallest-Value-of-the-Rearranged-Number/2165. Smallest Value of the Rearranged Number.go
new file mode 100644
index 000000000..1ca93c413
--- /dev/null
+++ b/leetcode/2165.Smallest-Value-of-the-Rearranged-Number/2165. Smallest Value of the Rearranged Number.go	
@@ -0,0 +1,51 @@
+package leetcode
+
+import "sort"
+
+func smallestNumber(num int64) int64 {
+	pos := true
+	if num < 0 {
+		pos = false
+		num *= -1
+	}
+	nums, m, res := []int{}, map[int]int{}, 0
+	for num != 0 {
+		tmp := int(num % 10)
+		m[tmp]++
+		num = num / 10
+	}
+
+	for k := range m {
+		nums = append(nums, k)
+	}
+	if pos {
+		sort.Ints(nums)
+	} else {
+		sort.Sort(sort.Reverse(sort.IntSlice(nums)))
+	}
+
+	if nums[0] == 0 && len(nums) > 1 {
+		res += nums[1]
+		m[nums[1]]--
+	}
+
+	for _, v := range nums {
+		if res != 0 {
+			for j := m[v]; j > 0; j-- {
+				res = res * 10
+				res += v
+			}
+		} else {
+			res += v
+			tmp := m[v] - 1
+			for j := tmp; j > 0; j-- {
+				res = res * 10
+				res += v
+			}
+		}
+	}
+	if !pos {
+		return -1 * int64(res)
+	}
+	return int64(res)
+}
diff --git a/leetcode/2165.Smallest-Value-of-the-Rearranged-Number/2165. Smallest Value of the Rearranged Number_test.go b/leetcode/2165.Smallest-Value-of-the-Rearranged-Number/2165. Smallest Value of the Rearranged Number_test.go
new file mode 100644
index 000000000..657fe30c4
--- /dev/null
+++ b/leetcode/2165.Smallest-Value-of-the-Rearranged-Number/2165. Smallest Value of the Rearranged Number_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2165 struct {
+	para2165
+	ans2165
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2165 struct {
+	nums int64
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2165 struct {
+	one int64
+}
+
+func Test_Problem1(t *testing.T) {
+
+	qs := []question2165{
+
+		{
+			para2165{310},
+			ans2165{103},
+		},
+
+		{
+			para2165{5059},
+			ans2165{5059},
+		},
+
+		{
+			para2165{-7605},
+			ans2165{-7650},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2165------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2165, q.para2165
+		fmt.Printf("【input】:%v       【output】:%v\n", p, smallestNumber(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2165.Smallest-Value-of-the-Rearranged-Number/README.md b/leetcode/2165.Smallest-Value-of-the-Rearranged-Number/README.md
new file mode 100644
index 000000000..8561bf22e
--- /dev/null
+++ b/leetcode/2165.Smallest-Value-of-the-Rearranged-Number/README.md
@@ -0,0 +1,101 @@
+# [2165. Smallest Value of the Rearranged Number](https://leetcode.com/problems/smallest-value-of-the-rearranged-number/)
+
+
+## 题目
+
+You are given an integer `num.` **Rearrange** the digits of `num` such that its value is **minimized** and it does not contain **any** leading zeros.
+
+Return *the rearranged number with minimal value*.
+
+Note that the sign of the number does not change after rearranging the digits.
+
+**Example 1:**
+
+```
+Input: num = 310
+Output: 103
+Explanation: The possible arrangements for the digits of 310 are 013, 031, 103, 130, 301, 310.
+The arrangement with the smallest value that does not contain any leading zeros is 103.
+
+```
+
+**Example 2:**
+
+```
+Input: num = -7605
+Output: -7650
+Explanation: Some possible arrangements for the digits of -7605 are -7650, -6705, -5076, -0567.
+The arrangement with the smallest value that does not contain any leading zeros is -7650.
+
+```
+
+**Constraints:**
+
+- `10^15 <= num <= 10^15`
+
+## 题目大意
+
+给你一个整数 num 。重排 num 中的各位数字，使其值 最小化 且不含 任何 前导零。
+
+返回不含前导零且值最小的重排数字。注意，重排各位数字后，num 的符号不会改变。
+
+## 解题思路
+
+- 先将每个数字出现次数统计出来。然后将数字大小从小到大排序。如果原数是正数，当出现有数字 0 的情况的时候，需先将第二小的数字排列到第一个，再把 0 排列完。再继续排列第二小，第三小。。。
+- 如果原数是负数。那么就逆序排列，即先排列最大的数字，然后次大的数字，直到排列最小的数字。因为数字越大，对应的这个数的负数就越小。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func smallestNumber(num int64) int64 {
+	pos := true
+	if num < 0 {
+		pos = false
+		num *= -1
+	}
+	nums, m, res := []int{}, map[int]int{}, 0
+	for num != 0 {
+		tmp := int(num % 10)
+		m[tmp]++
+		num = num / 10
+	}
+
+	for k := range m {
+		nums = append(nums, k)
+	}
+	if pos {
+		sort.Ints(nums)
+	} else {
+		sort.Sort(sort.Reverse(sort.IntSlice(nums)))
+	}
+
+	if nums[0] == 0 && len(nums) > 1 {
+		res += nums[1]
+		m[nums[1]]--
+	}
+
+	for _, v := range nums {
+		if res != 0 {
+			for j := m[v]; j > 0; j-- {
+				res = res * 10
+				res += v
+			}
+		} else {
+			res += v
+			tmp := m[v] - 1
+			for j := tmp; j > 0; j-- {
+				res = res * 10
+				res += v
+			}
+		}
+	}
+	if !pos {
+		return -1 * int64(res)
+	}
+	return int64(res)
+}
+```
\ No newline at end of file
diff --git a/leetcode/2166.Design-Bitset/2166. Design Bitset.go b/leetcode/2166.Design-Bitset/2166. Design Bitset.go
new file mode 100644
index 000000000..27d7bb0af
--- /dev/null
+++ b/leetcode/2166.Design-Bitset/2166. Design Bitset.go	
@@ -0,0 +1,72 @@
+package leetcode
+
+type Bitset struct {
+	set      []byte
+	flipped  []byte
+	oneCount int
+	size     int
+}
+
+func Constructor(size int) Bitset {
+	set := make([]byte, size)
+	flipped := make([]byte, size)
+	for i := 0; i < size; i++ {
+		set[i] = byte('0')
+		flipped[i] = byte('1')
+	}
+	return Bitset{
+		set:      set,
+		flipped:  flipped,
+		oneCount: 0,
+		size:     size,
+	}
+}
+
+func (this *Bitset) Fix(idx int) {
+	if this.set[idx] == byte('0') {
+		this.set[idx] = byte('1')
+		this.flipped[idx] = byte('0')
+		this.oneCount++
+	}
+}
+
+func (this *Bitset) Unfix(idx int) {
+	if this.set[idx] == byte('1') {
+		this.set[idx] = byte('0')
+		this.flipped[idx] = byte('1')
+		this.oneCount--
+	}
+}
+
+func (this *Bitset) Flip() {
+	this.set, this.flipped = this.flipped, this.set
+	this.oneCount = this.size - this.oneCount
+}
+
+func (this *Bitset) All() bool {
+	return this.oneCount == this.size
+}
+
+func (this *Bitset) One() bool {
+	return this.oneCount != 0
+}
+
+func (this *Bitset) Count() int {
+	return this.oneCount
+}
+
+func (this *Bitset) ToString() string {
+	return string(this.set)
+}
+
+/**
+ * Your Bitset object will be instantiated and called as such:
+ * obj := Constructor(size);
+ * obj.Fix(idx);
+ * obj.Unfix(idx);
+ * obj.Flip();
+ * param_4 := obj.All();
+ * param_5 := obj.One();
+ * param_6 := obj.Count();
+ * param_7 := obj.ToString();
+ */
diff --git a/leetcode/2166.Design-Bitset/2166. Design Bitset_test.go b/leetcode/2166.Design-Bitset/2166. Design Bitset_test.go
new file mode 100644
index 000000000..48ee6ec0a
--- /dev/null
+++ b/leetcode/2166.Design-Bitset/2166. Design Bitset_test.go	
@@ -0,0 +1,31 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem2166(t *testing.T) {
+	obj := Constructor(5)
+	fmt.Printf("obj = %v\n", obj)
+
+	obj.Fix(3)
+	fmt.Printf("obj = %v\n", obj)
+	obj.Fix(1)
+	fmt.Printf("obj = %v\n", obj)
+	obj.Flip()
+	fmt.Printf("obj = %v\n", obj)
+
+	fmt.Printf("all = %v\n", obj.All())
+	obj.Unfix(0)
+	fmt.Printf("obj = %v\n", obj)
+	obj.Flip()
+	fmt.Printf("obj = %v\n", obj)
+
+	fmt.Printf("one = %v\n", obj.One())
+	obj.Unfix(0)
+	fmt.Printf("obj = %v\n", obj)
+
+	fmt.Printf("count = %v\n", obj.Count())
+	fmt.Printf("toString = %v\n", obj.ToString())
+}
diff --git a/leetcode/2166.Design-Bitset/README.md b/leetcode/2166.Design-Bitset/README.md
new file mode 100644
index 000000000..916bbf5ed
--- /dev/null
+++ b/leetcode/2166.Design-Bitset/README.md
@@ -0,0 +1,156 @@
+# [2166. Design Bitset](https://leetcode.com/problems/design-bitset/)
+
+
+## 题目
+
+A **Bitset** is a data structure that compactly stores bits.
+
+Implement the `Bitset` class:
+
+- `Bitset(int size)` Initializes the Bitset with `size` bits, all of which are `0`.
+- `void fix(int idx)` Updates the value of the bit at the index `idx` to `1`. If the value was already `1`, no change occurs.
+- `void unfix(int idx)` Updates the value of the bit at the index `idx` to `0`. If the value was already `0`, no change occurs.
+- `void flip()` Flips the values of each bit in the Bitset. In other words, all bits with value `0` will now have value `1` and vice versa.
+- `boolean all()` Checks if the value of **each** bit in the Bitset is `1`. Returns `true` if it satisfies the condition, `false` otherwise.
+- `boolean one()` Checks if there is **at least one** bit in the Bitset with value `1`. Returns `true` if it satisfies the condition, `false` otherwise.
+- `int count()` Returns the **total number** of bits in the Bitset which have value `1`.
+- `String toString()` Returns the current composition of the Bitset. Note that in the resultant string, the character at the `ith` index should coincide with the value at the `ith` bit of the Bitset.
+
+**Example 1:**
+
+```
+Input
+["Bitset", "fix", "fix", "flip", "all", "unfix", "flip", "one", "unfix", "count", "toString"]
+[[5], [3], [1], [], [], [0], [], [], [0], [], []]
+Output
+[null, null, null, null, false, null, null, true, null, 2, "01010"]
+
+Explanation
+Bitset bs = new Bitset(5); // bitset = "00000".
+bs.fix(3);     // the value at idx = 3 is updated to 1, so bitset = "00010".
+bs.fix(1);     // the value at idx = 1 is updated to 1, so bitset = "01010".
+bs.flip();     // the value of each bit is flipped, so bitset = "10101".
+bs.all();      // return False, as not all values of the bitset are 1.
+bs.unfix(0);   // the value at idx = 0 is updated to 0, so bitset = "00101".
+bs.flip();     // the value of each bit is flipped, so bitset = "11010".
+bs.one();      // return True, as there is at least 1 index with value 1.
+bs.unfix(0);   // the value at idx = 0 is updated to 0, so bitset = "01010".
+bs.count();    // return 2, as there are 2 bits with value 1.
+bs.toString(); // return "01010", which is the composition of bitset.
+
+```
+
+**Constraints:**
+
+- `1 <= size <= 10^5`
+- `0 <= idx <= size - 1`
+- At most `10^5` calls will be made **in total** to `fix`, `unfix`, `flip`, `all`, `one`, `count`, and `toString`.
+- At least one call will be made to `all`, `one`, `count`, or `toString`.
+- At most `5` calls will be made to `toString`.
+
+## 题目大意
+
+位集 Bitset 是一种能以紧凑形式存储位的数据结构。
+
+请你实现 Bitset 类。
+
+- Bitset(int size) 用 size 个位初始化 Bitset ，所有位都是 0 。
+- void fix(int idx) 将下标为 idx 的位上的值更新为 1 。如果值已经是 1 ，则不会发生任何改变。
+- void unfix(int idx) 将下标为 idx 的位上的值更新为 0 。如果值已经是 0 ，则不会发生任何改变。
+- void flip() 翻转 Bitset 中每一位上的值。换句话说，所有值为 0 的位将会变成 1 ，反之亦然。
+- boolean all() 检查 Bitset 中 每一位 的值是否都是 1 。如果满足此条件，返回 true ；否则，返回 false 。
+- boolean one() 检查 Bitset 中 是否 至少一位 的值是 1 。如果满足此条件，返回 true ；否则，返回 false 。
+- int count() 返回 Bitset 中值为 1 的位的 总数 。
+- String toString() 返回 Bitset 的当前组成情况。注意，在结果字符串中，第 i 个下标处的字符应该与 Bitset 中的第 i 位一致。
+
+提示：
+
+- 1 <= size <= 10^5
+- 0 <= idx <= size - 1
+- 至多调用 fix、unfix、flip、all、one、count 和 toString 方法 总共 10^5 次
+- 至少调用 all、one、count 或 toString 方法一次
+- 至多调用 toString 方法 5 次
+
+## 解题思路
+
+- 题目中给出了 size 大小，10^5 位二进制。所以不能用 int64 数据类型。
+- 用数组模拟二进制位的一系列操作。flip 操作并不需要每次去翻转，偶数次翻转等于没有翻转，奇数次翻转记下标记，同时更新 1 的个数。这次懒操作在调用 fix 和 unfix 时，更新到原来数组中。
+- fix 和 unfix 根据懒数组中的标记对应更新二进制位。同时更新 1 的个数。
+- all，one，count 都是判断 1 的个数。toString 输出即可。
+
+## 代码
+
+```go
+package leetcode
+
+type Bitset struct {
+	set      []byte
+	flipped  []byte
+	oneCount int
+	size     int
+}
+
+func Constructor(size int) Bitset {
+	set := make([]byte, size)
+	flipped := make([]byte, size)
+	for i := 0; i < size; i++ {
+		set[i] = byte('0')
+		flipped[i] = byte('1')
+	}
+	return Bitset{
+		set:      set,
+		flipped:  flipped,
+		oneCount: 0,
+		size:     size,
+	}
+}
+
+func (this *Bitset) Fix(idx int) {
+	if this.set[idx] == byte('0') {
+		this.set[idx] = byte('1')
+		this.flipped[idx] = byte('0')
+		this.oneCount++
+	}
+}
+
+func (this *Bitset) Unfix(idx int) {
+	if this.set[idx] == byte('1') {
+		this.set[idx] = byte('0')
+		this.flipped[idx] = byte('1')
+		this.oneCount--
+	}
+}
+
+func (this *Bitset) Flip() {
+	this.set, this.flipped = this.flipped, this.set
+	this.oneCount = this.size - this.oneCount
+}
+
+func (this *Bitset) All() bool {
+	return this.oneCount == this.size
+}
+
+func (this *Bitset) One() bool {
+	return this.oneCount != 0
+}
+
+func (this *Bitset) Count() int {
+	return this.oneCount
+}
+
+func (this *Bitset) ToString() string {
+	return string(this.set)
+}
+
+/**
+ * Your Bitset object will be instantiated and called as such:
+ * obj := Constructor(size);
+ * obj.Fix(idx);
+ * obj.Unfix(idx);
+ * obj.Flip();
+ * param_4 := obj.All();
+ * param_5 := obj.One();
+ * param_6 := obj.Count();
+ * param_7 := obj.ToString();
+ */
+```
\ No newline at end of file
diff --git a/leetcode/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods/2167. Minimum Time to Remove All Cars Containing Illegal Goods.go b/leetcode/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods/2167. Minimum Time to Remove All Cars Containing Illegal Goods.go
new file mode 100644
index 000000000..72335ded0
--- /dev/null
+++ b/leetcode/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods/2167. Minimum Time to Remove All Cars Containing Illegal Goods.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import "runtime/debug"
+
+// 解法一 DP
+func minimumTime(s string) int {
+	suffixSum, prefixSum, res := make([]int, len(s)+1), make([]int, len(s)+1), 0
+	for i := len(s) - 1; i >= 0; i-- {
+		if s[i] == '0' {
+			suffixSum[i] = suffixSum[i+1]
+		} else {
+			suffixSum[i] = min(suffixSum[i+1]+2, len(s)-i)
+		}
+	}
+	res = suffixSum[0]
+	if s[0] == '1' {
+		prefixSum[0] = 1
+	}
+	for i := 1; i < len(s); i++ {
+		if s[i] == '0' {
+			prefixSum[i] = prefixSum[i-1]
+		} else {
+			prefixSum[i] = min(prefixSum[i-1]+2, i+1)
+		}
+		res = min(res, prefixSum[i]+suffixSum[i+1])
+	}
+	return res
+}
+
+func init() { debug.SetGCPercent(-1) }
+
+// 解法二 小幅优化时间和空间复杂度
+func minimumTime1(s string) int {
+	res, count := len(s), 0
+	for i := 0; i < len(s); i++ {
+		count = min(count+int(s[i]-'0')*2, i+1)
+		res = min(res, count+len(s)-i-1)
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	} else {
+		return b
+	}
+}
diff --git a/leetcode/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods/2167. Minimum Time to Remove All Cars Containing Illegal Goods_test.go b/leetcode/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods/2167. Minimum Time to Remove All Cars Containing Illegal Goods_test.go
new file mode 100644
index 000000000..c3d59064a
--- /dev/null
+++ b/leetcode/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods/2167. Minimum Time to Remove All Cars Containing Illegal Goods_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2167 struct {
+	para2167
+	ans2167
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2167 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2167 struct {
+	one int
+}
+
+func Test_Problem2167(t *testing.T) {
+
+	qs := []question2167{
+
+		{
+			para2167{"1100101"},
+			ans2167{5},
+		},
+
+		{
+			para2167{"0010"},
+			ans2167{2},
+		},
+
+		{
+			para2167{"1100111101"},
+			ans2167{8},
+		},
+
+		{
+			para2167{"1001010101"},
+			ans2167{8},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2167------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2167, q.para2167
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minimumTime(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods/README.md b/leetcode/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods/README.md
new file mode 100644
index 000000000..258e5b9a9
--- /dev/null
+++ b/leetcode/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods/README.md
@@ -0,0 +1,145 @@
+# [2167. Minimum Time to Remove All Cars Containing Illegal Goods](https://leetcode.com/problems/minimum-time-to-remove-all-cars-containing-illegal-goods/)
+
+
+## 题目
+
+You are given a **0-indexed** binary string `s` which represents a sequence of train cars. `s[i] = '0'` denotes that the `ith` car does **not** contain illegal goods and `s[i] = '1'` denotes that the `ith` car does contain illegal goods.
+
+As the train conductor, you would like to get rid of all the cars containing illegal goods. You can do any of the following three operations **any** number of times:
+
+1. Remove a train car from the **left** end (i.e., remove `s[0]`) which takes 1 unit of time.
+2. Remove a train car from the **right** end (i.e., remove `s[s.length - 1]`) which takes 1 unit of time.
+3. Remove a train car from **anywhere** in the sequence which takes 2 units of time.
+
+Return *the **minimum** time to remove all the cars containing illegal goods*.
+
+Note that an empty sequence of cars is considered to have no cars containing illegal goods.
+
+**Example 1:**
+
+```
+Input: s = "1100101"
+Output: 5
+Explanation:
+One way to remove all the cars containing illegal goods from the sequence is to
+- remove a car from the left end 2 times. Time taken is 2 * 1 = 2.
+- remove a car from the right end. Time taken is 1.
+- remove the car containing illegal goods found in the middle. Time taken is 2.
+This obtains a total time of 2 + 1 + 2 = 5.
+
+An alternative way is to
+- remove a car from the left end 2 times. Time taken is 2 * 1 = 2.
+- remove a car from the right end 3 times. Time taken is 3 * 1 = 3.
+This also obtains a total time of 2 + 3 = 5.
+
+5 is the minimum time taken to remove all the cars containing illegal goods.
+There are no other ways to remove them with less time.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "0010"
+Output: 2
+Explanation:
+One way to remove all the cars containing illegal goods from the sequence is to
+- remove a car from the left end 3 times. Time taken is 3 * 1 = 3.
+This obtains a total time of 3.
+
+Another way to remove all the cars containing illegal goods from the sequence is to
+- remove the car containing illegal goods found in the middle. Time taken is 2.
+This obtains a total time of 2.
+
+Another way to remove all the cars containing illegal goods from the sequence is to
+- remove a car from the right end 2 times. Time taken is 2 * 1 = 2.
+This obtains a total time of 2.
+
+2 is the minimum time taken to remove all the cars containing illegal goods.
+There are no other ways to remove them with less time.
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 2 * 10^5`
+- `s[i]` is either `'0'` or `'1'`.
+
+## 题目大意
+
+给你一个下标从 0 开始的二进制字符串 s ，表示一个列车车厢序列。s[i] = '0' 表示第 i 节车厢 不 含违禁货物，而 s[i] = '1' 表示第 i 节车厢含违禁货物。
+
+作为列车长，你需要清理掉所有载有违禁货物的车厢。你可以不限次数执行下述三种操作中的任意一个：
+
+1. 从列车 左 端移除一节车厢（即移除 s[0]），用去 1 单位时间。
+2. 从列车 右 端移除一节车厢（即移除 s[s.length - 1]），用去 1 单位时间。
+3. 从列车车厢序列的 任意位置 移除一节车厢，用去 2 单位时间。
+
+返回移除所有载有违禁货物车厢所需要的 最少 单位时间数。注意，空的列车车厢序列视为没有车厢含违禁货物。
+
+## 解题思路
+
+- 这道题求最少单位时间数，最少时间数一定是尽量少使用 2 个单位时间的操作，多用 1 个时间的操作。从列车两头移除车厢，只需要移除和旁边车厢的金属连接处即可。由于列车位于两边，所以与其他车厢的金属连接处只有 1 个，故只需要 1 个单位时间；当车厢在中间，该车厢与两边的车厢有 2 个金属连接处，移除它需要断开与两边车厢的连接。所以需要 2 个单位时间。
+- 断开中间一节车厢以后，列车会被断成 2 部分。2 部分列车分别有 2 个头 2 个尾。举例：`1100111101`，如果把它从第 5 节开始断开，剩下的列车为 `11001 (1)` 和 `1101`。剩下的 1 都位于 2 边，移除他们都只需要 1 个单位时间。那么移除所有违禁品最少时间是 2 * 1 + 1 * 6 = 8。
+- 左半部分，定义 prefixSum[i] 表示移除前 i 节车厢所花费的最少时间。状态转移方程为：
+    
+    $prefixSum[i] =\left\{\begin{matrix}prefixSum[i-1],s[i]=0\\ min(prefixSum[i-1]+2, i+1), s[i]=1\end{matrix}\right.$
+    
+- 同理，右半部分定义 suffixSum[i] 表示移除后 i 节车厢所花费的最少时间。状态转移方程为：
+    
+    $suffixSum[i] =\left\{\begin{matrix} suffixSum[i+1],s[i]=0\\ min(suffixSum[i+1]+2, n-i), s[i]=1\end{matrix}\right.$
+    
+- 最后一层循环枚举 prefixSum[i] + suffixSum[i+1] 的最小值即为答案。
+- 这一题在解法一的基础上还可以再简化。当 s[i] = 1 时，prefixSum 和 suffixSum 是两种计算方法。我们可以假设中间断开的部分在 prefixSum 中。于是可以合并上面两个状态转移方程。简化以后的代码见解法二。
+
+## 代码
+
+```go
+package leetcode
+
+import "runtime/debug"
+
+// 解法一 DP
+func minimumTime(s string) int {
+	suffixSum, prefixSum, res := make([]int, len(s)+1), make([]int, len(s)+1), 0
+	for i := len(s) - 1; i >= 0; i-- {
+		if s[i] == '0' {
+			suffixSum[i] = suffixSum[i+1]
+		} else {
+			suffixSum[i] = min(suffixSum[i+1]+2, len(s)-i)
+		}
+	}
+	res = suffixSum[0]
+	if s[0] == '1' {
+		prefixSum[0] = 1
+	}
+	for i := 1; i < len(s); i++ {
+		if s[i] == '0' {
+			prefixSum[i] = prefixSum[i-1]
+		} else {
+			prefixSum[i] = min(prefixSum[i-1]+2, i+1)
+		}
+		res = min(res, prefixSum[i]+suffixSum[i+1])
+	}
+	return res
+}
+
+func init() { debug.SetGCPercent(-1) }
+
+// 解法二 小幅优化时间和空间复杂度
+func minimumTime1(s string) int {
+	res, count := len(s), 0
+	for i := 0; i < len(s); i++ {
+		count = min(count+int(s[i]-'0')*2, i+1)
+		res = min(res, count+len(s)-i-1)
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	} else {
+		return b
+	}
+}
+```
\ No newline at end of file
diff --git a/leetcode/2169.Count-Operations-to-Obtain-Zero/2169. Count Operations to Obtain Zero.go b/leetcode/2169.Count-Operations-to-Obtain-Zero/2169. Count Operations to Obtain Zero.go
new file mode 100644
index 000000000..040ac3705
--- /dev/null
+++ b/leetcode/2169.Count-Operations-to-Obtain-Zero/2169. Count Operations to Obtain Zero.go	
@@ -0,0 +1,14 @@
+package leetcode
+
+func countOperations(num1 int, num2 int) int {
+	res := 0
+	for num1 != 0 && num2 != 0 {
+		if num1 >= num2 {
+			num1 -= num2
+		} else {
+			num2 -= num1
+		}
+		res++
+	}
+	return res
+}
diff --git a/leetcode/2169.Count-Operations-to-Obtain-Zero/2169. Count Operations to Obtain Zero_test.go b/leetcode/2169.Count-Operations-to-Obtain-Zero/2169. Count Operations to Obtain Zero_test.go
new file mode 100644
index 000000000..69d21f8a8
--- /dev/null
+++ b/leetcode/2169.Count-Operations-to-Obtain-Zero/2169. Count Operations to Obtain Zero_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2169 struct {
+	para2169
+	ans2169
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2169 struct {
+	num1 int
+	num2 int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2169 struct {
+	one int
+}
+
+func Test_Problem2169(t *testing.T) {
+
+	qs := []question2169{
+
+		{
+			para2169{2, 3},
+			ans2169{3},
+		},
+
+		{
+			para2169{10, 10},
+			ans2169{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2169------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2169, q.para2169
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countOperations(p.num1, p.num2))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2169.Count-Operations-to-Obtain-Zero/README.md b/leetcode/2169.Count-Operations-to-Obtain-Zero/README.md
new file mode 100644
index 000000000..10878ea04
--- /dev/null
+++ b/leetcode/2169.Count-Operations-to-Obtain-Zero/README.md
@@ -0,0 +1,73 @@
+# [2169. Count Operations to Obtain Zero](https://leetcode.com/problems/count-operations-to-obtain-zero/)
+
+
+## 题目
+
+You are given two **non-negative** integers `num1` and `num2`.
+
+In one **operation**, if `num1 >= num2`, you must subtract `num2` from `num1`, otherwise subtract `num1` from `num2`.
+
+- For example, if `num1 = 5` and `num2 = 4`, subtract `num2` from `num1`, thus obtaining `num1 = 1` and `num2 = 4`. However, if `num1 = 4` and `num2 = 5`, after one operation, `num1 = 4` and `num2 = 1`.
+
+Return *the **number of operations** required to make either* `num1 = 0` *or* `num2 = 0`.
+
+**Example 1:**
+
+```
+Input: num1 = 2, num2 = 3
+Output: 3
+Explanation:
+- Operation 1: num1 = 2, num2 = 3. Since num1 < num2, we subtract num1 from num2 and get num1 = 2, num2 = 3 - 2 = 1.
+- Operation 2: num1 = 2, num2 = 1. Since num1 > num2, we subtract num2 from num1.
+- Operation 3: num1 = 1, num2 = 1. Since num1 == num2, we subtract num2 from num1.
+Now num1 = 0 and num2 = 1. Since num1 == 0, we do not need to perform any further operations.
+So the total number of operations required is 3.
+
+```
+
+**Example 2:**
+
+```
+Input: num1 = 10, num2 = 10
+Output: 1
+Explanation:
+- Operation 1: num1 = 10, num2 = 10. Since num1 == num2, we subtract num2 from num1 and get num1 = 10 - 10 = 0.
+Now num1 = 0 and num2 = 10. Since num1 == 0, we are done.
+So the total number of operations required is 1.
+
+```
+
+**Constraints:**
+
+- `0 <= num1, num2 <= 10^5`
+
+## 题目大意
+
+给你两个 非负 整数 num1 和 num2 。每一步 操作 中，如果 num1 >= num2 ，你必须用 num1 减 num2 ；否则，你必须用 num2 减 num1 。
+
+- 例如，num1 = 5 且 num2 = 4 ，应该用 num1 减 num2 ，因此，得到 num1 = 1 和 num2 = 4 。然而，如果 num1 = 4且 num2 = 5 ，一步操作后，得到 num1 = 4 和 num2 = 1 。
+
+返回使 num1 = 0 或 num2 = 0 的 操作数 。
+
+## 解题思路
+
+- 简单题，按照题意模拟，每次两个数字相减，便累加操作次数。当某个数字变为 0 时，输出操作次数。
+
+## 代码
+
+```go
+package leetcode
+
+func countOperations(num1 int, num2 int) int {
+	res := 0
+	for num1 != 0 && num2 != 0 {
+		if num1 >= num2 {
+			num1 -= num2
+		} else {
+			num2 -= num1
+		}
+		res++
+	}
+	return res
+}
+```
\ No newline at end of file
diff --git a/leetcode/2170.Minimum-Operations-to-Make-the-Array-Alternating/2170. Minimum Operations to Make the Array Alternating.go b/leetcode/2170.Minimum-Operations-to-Make-the-Array-Alternating/2170. Minimum Operations to Make the Array Alternating.go
new file mode 100644
index 000000000..d5d156ffb
--- /dev/null
+++ b/leetcode/2170.Minimum-Operations-to-Make-the-Array-Alternating/2170. Minimum Operations to Make the Array Alternating.go	
@@ -0,0 +1,60 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+type node struct {
+	value int
+	count int
+}
+
+func minimumOperations(nums []int) int {
+	if len(nums) == 1 {
+		return 0
+	}
+	res, odd, even, oddMap, evenMap := 0, []node{}, []node{}, map[int]int{}, map[int]int{}
+
+	for i := 0; i < len(nums); i += 2 {
+		evenMap[nums[i]]++
+	}
+	for k, v := range evenMap {
+		even = append(even, node{value: k, count: v})
+	}
+	sort.Slice(even, func(i, j int) bool {
+		return even[i].count > even[j].count
+	})
+
+	for i := 1; i < len(nums); i += 2 {
+		oddMap[nums[i]]++
+	}
+	for k, v := range oddMap {
+		odd = append(odd, node{value: k, count: v})
+	}
+	sort.Slice(odd, func(i, j int) bool {
+		return odd[i].count > odd[j].count
+	})
+
+	if even[0].value == odd[0].value {
+		if len(even) == 1 && len(odd) != 1 {
+			res = len(nums) - even[0].count - odd[1].count
+		} else if len(odd) == 1 && len(even) != 1 {
+			res = len(nums) - odd[0].count - even[1].count
+		} else if len(odd) == 1 && len(even) == 1 {
+			res = len(nums) / 2
+		} else {
+			// both != 1
+			res = min(len(nums)-odd[0].count-even[1].count, len(nums)-odd[1].count-even[0].count)
+		}
+	} else {
+		res = len(nums) - even[0].count - odd[0].count
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/2170.Minimum-Operations-to-Make-the-Array-Alternating/2170. Minimum Operations to Make the Array Alternating_test.go b/leetcode/2170.Minimum-Operations-to-Make-the-Array-Alternating/2170. Minimum Operations to Make the Array Alternating_test.go
new file mode 100644
index 000000000..c792e89e4
--- /dev/null
+++ b/leetcode/2170.Minimum-Operations-to-Make-the-Array-Alternating/2170. Minimum Operations to Make the Array Alternating_test.go	
@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2170 struct {
+	para2170
+	ans2170
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2170 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2170 struct {
+	one int
+}
+
+func Test_Problem1(t *testing.T) {
+
+	qs := []question2170{
+
+		{
+			para2170{[]int{1}},
+			ans2170{0},
+		},
+
+		{
+			para2170{[]int{3, 1, 3, 2, 4, 3}},
+			ans2170{3},
+		},
+
+		{
+			para2170{[]int{1, 2, 2, 2, 2}},
+			ans2170{2},
+		},
+
+		{
+			para2170{[]int{69, 91, 47, 74, 75, 94, 22, 100, 43, 50, 82, 47, 40, 51, 90, 27, 98, 85, 47, 14, 55, 82, 52, 9, 65, 90, 86, 45, 52, 52, 95, 40, 85, 3, 46, 77, 16, 59, 32, 22, 41, 87, 89, 78, 59, 78, 34, 26, 71, 9, 82, 68, 80, 74, 100, 6, 10, 53, 84, 80, 7, 87, 3, 82, 26, 26, 14, 37, 26, 58, 96, 73, 41, 2, 79, 43, 56, 74, 30, 71, 6, 100, 72, 93, 83, 40, 28, 79, 24}},
+			ans2170{84},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2170------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2170, q.para2170
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minimumOperations(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2170.Minimum-Operations-to-Make-the-Array-Alternating/README.md b/leetcode/2170.Minimum-Operations-to-Make-the-Array-Alternating/README.md
new file mode 100644
index 000000000..305d79ce6
--- /dev/null
+++ b/leetcode/2170.Minimum-Operations-to-Make-the-Array-Alternating/README.md
@@ -0,0 +1,130 @@
+# [2170. Minimum Operations to Make the Array Alternating](https://leetcode.com/problems/minimum-operations-to-make-the-array-alternating/)
+
+
+## 题目
+
+You are given a **0-indexed** array `nums` consisting of `n` positive integers.
+
+The array `nums` is called **alternating** if:
+
+- `nums[i - 2] == nums[i]`, where `2 <= i <= n - 1`.
+- `nums[i - 1] != nums[i]`, where `1 <= i <= n - 1`.
+
+In one **operation**, you can choose an index `i` and **change** `nums[i]` into **any** positive integer.
+
+Return *the **minimum number of operations** required to make the array alternating*.
+
+**Example 1:**
+
+```
+Input: nums = [3,1,3,2,4,3]
+Output: 3
+Explanation:
+One way to make the array alternating is by converting it to [3,1,3,1,3,1].
+The number of operations required in this case is 3.
+It can be proven that it is not possible to make the array alternating in less than 3 operations.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,2,2,2]
+Output: 2
+Explanation:
+One way to make the array alternating is by converting it to [1,2,1,2,1].
+The number of operations required in this case is 2.
+Note that the array cannot be converted to [2,2,2,2,2] because in this case nums[0] == nums[1] which violates the conditions of an alternating array.
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 10^5`
+- `1 <= nums[i] <= 10^5`
+
+## 题目大意
+
+给你一个下标从 0 开始的数组 nums ，该数组由 n 个正整数组成。
+
+如果满足下述条件，则数组 nums 是一个 交替数组 ：
+
+- nums[i - 2] == nums[i] ，其中 2 <= i <= n - 1 。
+- nums[i - 1] != nums[i] ，其中 1 <= i <= n - 1 。
+
+在一步 操作 中，你可以选择下标 i 并将 nums[i] 更改 为 任一 正整数。返回使数组变成交替数组的 最少操作数 。
+
+**提示：**
+
+- `1 <= nums.length <= 10^5`
+- `1 <= nums[i] <= 10^5`
+
+## 解题思路
+
+- 题目要求最少操作数，即留下出现频次最多的数字，剩下的数字都替换成这个数字。先将每个数字出现的频次统计出来，然后按照频次从大到小排序。优先选择出现频次高的数字。
+- 有几种“特殊”情况需要处理：当奇数下标的数字频次最大的数字和偶数下标的数字频次最大的数字相同（数字相同，频次不同），这时应选取频次大的数字留下；当数字相同，频次也相同，这时要看奇数下标和偶数下标的数字分别有几个。 如果其中一个只有一种数字，那么另外一组数字则需都变成该组频次第二大的数字，例如奇数下标的数字全是 1，频次是 3，偶数下标的数字是 1，最大频次是 2。第二频次的数字是 9，频次是 1 。那么这种情况下，选择奇数下标的数字 1，和偶数下标数字 9 。将偶数下标不是 9 的数字改变成 9 ；更近一步，如果奇数下标和偶数下标都只有一个数字，频次相同，那么只能改变奇数下标或者偶数下标的所有数字。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+type node struct {
+	value int
+	count int
+}
+
+func minimumOperations(nums []int) int {
+	if len(nums) == 1 {
+		return 0
+	}
+	res, odd, even, oddMap, evenMap := 0, []node{}, []node{}, map[int]int{}, map[int]int{}
+
+	for i := 0; i < len(nums); i += 2 {
+		evenMap[nums[i]]++
+	}
+	for k, v := range evenMap {
+		even = append(even, node{value: k, count: v})
+	}
+	sort.Slice(even, func(i, j int) bool {
+		return even[i].count > even[j].count
+	})
+
+	for i := 1; i < len(nums); i += 2 {
+		oddMap[nums[i]]++
+	}
+	for k, v := range oddMap {
+		odd = append(odd, node{value: k, count: v})
+	}
+	sort.Slice(odd, func(i, j int) bool {
+		return odd[i].count > odd[j].count
+	})
+
+	if even[0].value == odd[0].value {
+		if len(even) == 1 && len(odd) != 1 {
+			res = len(nums) - even[0].count - odd[1].count
+		} else if len(odd) == 1 && len(even) != 1 {
+			res = len(nums) - odd[0].count - even[1].count
+		} else if len(odd) == 1 && len(even) == 1 {
+			res = len(nums) / 2
+		} else {
+			// both != 1
+			res = min(len(nums)-odd[0].count-even[1].count, len(nums)-odd[1].count-even[0].count)
+		}
+	} else {
+		res = len(nums) - even[0].count - odd[0].count
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+```
\ No newline at end of file
diff --git a/leetcode/2171.Removing-Minimum-Number-of-Magic-Beans/2171. Removing Minimum Number of Magic Beans.go b/leetcode/2171.Removing-Minimum-Number-of-Magic-Beans/2171. Removing Minimum Number of Magic Beans.go
new file mode 100644
index 000000000..c4032a1d2
--- /dev/null
+++ b/leetcode/2171.Removing-Minimum-Number-of-Magic-Beans/2171. Removing Minimum Number of Magic Beans.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+import "sort"
+
+func minimumRemoval(beans []int) int64 {
+	sort.Ints(beans)
+	sum, mx := 0, 0
+	for i, v := range beans {
+		sum += v
+		mx = max(mx, (len(beans)-i)*v)
+	}
+	return int64(sum - mx)
+}
+
+func max(a, b int) int {
+	if b > a {
+		return b
+	}
+	return a
+}
diff --git a/leetcode/2171.Removing-Minimum-Number-of-Magic-Beans/2171. Removing Minimum Number of Magic Beans_test.go b/leetcode/2171.Removing-Minimum-Number-of-Magic-Beans/2171. Removing Minimum Number of Magic Beans_test.go
new file mode 100644
index 000000000..068733524
--- /dev/null
+++ b/leetcode/2171.Removing-Minimum-Number-of-Magic-Beans/2171. Removing Minimum Number of Magic Beans_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2170 struct {
+	para2170
+	ans2170
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2170 struct {
+	beans []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2170 struct {
+	one int
+}
+
+func Test_Problem1(t *testing.T) {
+
+	qs := []question2170{
+
+		{
+			para2170{[]int{4, 1, 6, 5}},
+			ans2170{4},
+		},
+
+		{
+			para2170{[]int{2, 10, 3, 2}},
+			ans2170{7},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2170------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2170, q.para2170
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minimumRemoval(p.beans))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2171.Removing-Minimum-Number-of-Magic-Beans/README.md b/leetcode/2171.Removing-Minimum-Number-of-Magic-Beans/README.md
new file mode 100644
index 000000000..0234821dc
--- /dev/null
+++ b/leetcode/2171.Removing-Minimum-Number-of-Magic-Beans/README.md
@@ -0,0 +1,90 @@
+# [2171. Removing Minimum Number of Magic Beans](https://leetcode.com/problems/removing-minimum-number-of-magic-beans/)
+
+
+## 题目
+
+You are given an array of **positive** integers `beans`, where each integer represents the number of magic beans found in a particular magic bag.
+
+**Remove** any number of beans (**possibly none**) from each bag such that the number of beans in each remaining **non-empty** bag (still containing **at least one** bean) is **equal**. Once a bean has been removed from a bag, you are **not** allowed to return it to any of the bags.
+
+Return *the **minimum** number of magic beans that you have to remove*.
+
+**Example 1:**
+
+```
+Input: beans = [4,1,6,5]
+Output: 4
+Explanation:
+- We remove 1 bean from the bag with only 1 bean.
+  This results in the remaining bags: [4,0,6,5]
+- Then we remove 2 beans from the bag with 6 beans.
+  This results in the remaining bags: [4,0,4,5]
+- Then we remove 1 bean from the bag with 5 beans.
+  This results in the remaining bags: [4,0,4,4]
+We removed a total of 1 + 2 + 1 = 4 beans to make the remaining non-empty bags have an equal number of beans.
+There are no other solutions that remove 4 beans or fewer.
+
+```
+
+**Example 2:**
+
+```
+Input: beans = [2,10,3,2]
+Output: 7
+Explanation:
+- We remove 2 beans from one of the bags with 2 beans.
+  This results in the remaining bags: [0,10,3,2]
+- Then we remove 2 beans from the other bag with 2 beans.
+  This results in the remaining bags: [0,10,3,0]
+- Then we remove 3 beans from the bag with 3 beans.
+  This results in the remaining bags: [0,10,0,0]
+We removed a total of 2 + 2 + 3 = 7 beans to make the remaining non-empty bags have an equal number of beans.
+There are no other solutions that removes 7 beans or fewer.
+
+```
+
+**Constraints:**
+
+- `1 <= beans.length <= 10^5`
+- `1 <= beans[i] <= 10^5`
+
+## 题目大意
+
+给你一个 正 整数数组 beans ，其中每个整数表示一个袋子里装的魔法豆的数目。
+
+请你从每个袋子中 拿出 一些豆子（也可以 不拿出），使得剩下的 非空 袋子中（即 至少 还有 一颗 魔法豆的袋子）魔法豆的数目 相等 。一旦魔法豆从袋子中取出，你不能将它放到任何其他的袋子中。请你返回你需要拿出魔法豆的 最少数目。
+
+**提示：**
+
+- `1 <= beans.length <= 10^5`
+- `1 <= beans[i] <= 10^5`
+
+## 解题思路
+
+- 这一题没有特别巧妙的方法。最初思路来源于暴力解法。从第一个袋子开始，依次以每个袋子中的豆子为基准，改变其他袋子里面的豆子数，使得其他袋子里面的豆子都和基准袋子中豆子一样多。
+- 如果从下标为 0 扫到下标 n-1 ，这中间会有大量重复计算。有些计算区间和的操作，反复计算了很多遍，导致算法不高效。由于移除豆子数量多少和基准袋豆子数量强相关，所以先排序。如果袋子内豆子数目小于基准袋的豆子，`0 ≤ j < i`，那么这些袋子内的豆子数量会归零。需要移除 `beans[0] + beans[1] + ... + beans[i-1]` 个豆子；如果袋子内豆子数目大于等于基准袋的豆子，`j ≥ i` ，那么这些袋子内的豆子需要调整为 `beans[i]` 个。需要移除 `(beans[i] - beans[i]) + (beans[i+1] - beans[i]) + (beans[i+2] - beans[i]) + ... + (beans[n-1] - beans[i]) = beans[i]+ ... + beans[n-1] - (n-i) * beans[i]` 个豆子。将这 2 种情况综合起来，那么总共需要移除 `sum(beans) - (N - i) * beans[i]` 个豆子。综上，先排序，然后从小到大扫一遍数组，动态维护最少移除豆子的个数即可。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func minimumRemoval(beans []int) int64 {
+	sort.Ints(beans)
+	sum, mx := 0, 0
+	for i, v := range beans {
+		sum += v
+		mx = max(mx, (len(beans)-i)*v)
+	}
+	return int64(sum - mx)
+}
+
+func max(a, b int) int {
+	if b > a {
+		return b
+	}
+	return a
+}
+```
\ No newline at end of file
diff --git a/leetcode/2180.Count-Integers-With-Even-Digit-Sum/2180. Count Integers With Even Digit Sum.go b/leetcode/2180.Count-Integers-With-Even-Digit-Sum/2180. Count Integers With Even Digit Sum.go
new file mode 100644
index 000000000..d942429c6
--- /dev/null
+++ b/leetcode/2180.Count-Integers-With-Even-Digit-Sum/2180. Count Integers With Even Digit Sum.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+func countEven(num int) int {
+	count := 0
+	for i := 1; i <= num; i++ {
+		if addSum(i)%2 == 0 {
+			count++
+		}
+	}
+	return count
+}
+
+func addSum(num int) int {
+	sum := 0
+	tmp := num
+	for tmp != 0 {
+		sum += tmp % 10
+		tmp = tmp / 10
+	}
+	return sum
+}
diff --git a/leetcode/2180.Count-Integers-With-Even-Digit-Sum/2180. Count Integers With Even Digit Sum_test.go b/leetcode/2180.Count-Integers-With-Even-Digit-Sum/2180. Count Integers With Even Digit Sum_test.go
new file mode 100644
index 000000000..684e23100
--- /dev/null
+++ b/leetcode/2180.Count-Integers-With-Even-Digit-Sum/2180. Count Integers With Even Digit Sum_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2180 struct {
+	para2180
+	ans2180
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2180 struct {
+	target int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2180 struct {
+	one int
+}
+
+func Test_Problem1(t *testing.T) {
+
+	qs := []question2180{
+		{
+			para2180{4},
+			ans2180{2},
+		},
+
+		{
+			para2180{30},
+			ans2180{14},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2180------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2180, q.para2180
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countEven(p.target))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2180.Count-Integers-With-Even-Digit-Sum/README.md b/leetcode/2180.Count-Integers-With-Even-Digit-Sum/README.md
new file mode 100644
index 000000000..15b2a5410
--- /dev/null
+++ b/leetcode/2180.Count-Integers-With-Even-Digit-Sum/README.md
@@ -0,0 +1,69 @@
+# [2180. Count Integers With Even Digit Sum](https://leetcode.com/problems/count-integers-with-even-digit-sum/)
+
+
+## 题目
+
+Given a positive integer `num`, return *the number of positive integers **less than or equal to*** `num` *whose digit sums are **even***.
+
+The **digit sum** of a positive integer is the sum of all its digits.
+
+**Example 1:**
+
+```
+Input: num = 4
+Output: 2
+Explanation:
+The only integers less than or equal to 4 whose digit sums are even are 2 and 4.
+
+```
+
+**Example 2:**
+
+```
+Input: num = 30
+Output: 14
+Explanation:
+The 14 integers less than or equal to 30 whose digit sums are even are
+2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22, 24, 26, and 28.
+
+```
+
+**Constraints:**
+
+- `1 <= num <= 1000`
+
+## 题目大意
+
+给你一个正整数 num ，请你统计并返回 小于或等于 num 且各位数字之和为 偶数 的正整数的数目。
+
+正整数的 各位数字之和 是其所有位上的对应数字相加的结果。
+
+## 解题思路
+
+- 简单题。依照题意，计算每个数的各位数字之和，如何和为偶数，则统计结果加一。最后输出统计结果即可。
+
+## 代码
+
+```go
+package leetcode
+
+func countEven(num int) int {
+	count := 0
+	for i := 1; i <= num; i++ {
+		if addSum(i)%2 == 0 {
+			count++
+		}
+	}
+	return count
+}
+
+func addSum(num int) int {
+	sum := 0
+	tmp := num
+	for tmp != 0 {
+		sum += tmp % 10
+		tmp = tmp / 10
+	}
+	return sum
+}
+```
\ No newline at end of file
diff --git a/leetcode/2181.Merge-Nodes-in-Between-Zeros/2181. Merge Nodes in Between Zeros.go b/leetcode/2181.Merge-Nodes-in-Between-Zeros/2181. Merge Nodes in Between Zeros.go
new file mode 100644
index 000000000..be7ef2c6a
--- /dev/null
+++ b/leetcode/2181.Merge-Nodes-in-Between-Zeros/2181. Merge Nodes in Between Zeros.go	
@@ -0,0 +1,36 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func mergeNodes(head *ListNode) *ListNode {
+	res := &ListNode{}
+	h := res
+	if head.Next == nil {
+		return &structures.ListNode{}
+	}
+	cur := head
+	sum := 0
+	for cur.Next != nil {
+		if cur.Next.Val != 0 {
+			sum += cur.Next.Val
+		} else {
+			h.Next = &ListNode{Val: sum, Next: nil}
+			h = h.Next
+			sum = 0
+		}
+		cur = cur.Next
+	}
+	return res.Next
+}
diff --git a/leetcode/2181.Merge-Nodes-in-Between-Zeros/2181. Merge Nodes in Between Zeros_test.go b/leetcode/2181.Merge-Nodes-in-Between-Zeros/2181. Merge Nodes in Between Zeros_test.go
new file mode 100644
index 000000000..44a308c12
--- /dev/null
+++ b/leetcode/2181.Merge-Nodes-in-Between-Zeros/2181. Merge Nodes in Between Zeros_test.go	
@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question2181 struct {
+	para2181
+	ans2181
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2181 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2181 struct {
+	one []int
+}
+
+func Test_Problem2181(t *testing.T) {
+
+	qs := []question2181{
+
+		{
+			para2181{[]int{0, 3, 1, 0, 4, 5, 2, 0}},
+			ans2181{[]int{4, 11}},
+		},
+
+		{
+			para2181{[]int{0, 1, 0, 3, 0, 2, 2, 0}},
+			ans2181{[]int{1, 3, 4}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2181------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2181, q.para2181
+		fmt.Printf("【input】:%v       【output】:%v\n", p, structures.List2Ints(mergeNodes(structures.Ints2List(p.one))))
+	}
+	fmt.Printf("\n\n\n")
+}
+
+func removeElements(head *ListNode, val int) *ListNode {
+	if head == nil {
+		return head
+	}
+	newHead := &ListNode{Val: 0, Next: head}
+	pre := newHead
+	cur := head
+	for cur != nil {
+		if cur.Val == val {
+			pre.Next = cur.Next
+		} else {
+			pre = cur
+		}
+		cur = cur.Next
+	}
+	return newHead.Next
+}
diff --git a/leetcode/2181.Merge-Nodes-in-Between-Zeros/README.md b/leetcode/2181.Merge-Nodes-in-Between-Zeros/README.md
new file mode 100644
index 000000000..6aba3609f
--- /dev/null
+++ b/leetcode/2181.Merge-Nodes-in-Between-Zeros/README.md
@@ -0,0 +1,96 @@
+# [2181. Merge Nodes in Between Zeros](https://leetcode.com/problems/merge-nodes-in-between-zeros/)
+
+## 题目
+
+You are given the `head` of a linked list, which contains a series of integers **separated** by `0`'s. The **beginning** and **end** of the linked list will have `Node.val == 0`.
+
+For **every** two consecutive `0`'s, **merge** all the nodes lying in between them into a single node whose value is the **sum** of all the merged nodes. The modified list should not contain any `0`'s.
+
+Return *the* `head` *of the modified linked list*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2022/02/02/ex1-1.png](https://assets.leetcode.com/uploads/2022/02/02/ex1-1.png)
+
+```
+Input: head = [0,3,1,0,4,5,2,0]
+Output: [4,11]
+Explanation:
+The above figure represents the given linked list. The modified list contains
+- The sum of the nodes marked in green: 3 + 1 = 4.
+- The sum of the nodes marked in red: 4 + 5 + 2 = 11.
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2022/02/02/ex2-1.png](https://assets.leetcode.com/uploads/2022/02/02/ex2-1.png)
+
+```
+Input: head = [0,1,0,3,0,2,2,0]
+Output: [1,3,4]
+Explanation:
+The above figure represents the given linked list. The modified list contains
+- The sum of the nodes marked in green: 1 = 1.
+- The sum of the nodes marked in red: 3 = 3.
+- The sum of the nodes marked in yellow: 2 + 2 = 4.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the list is in the range `[3, 2 * 10^5]`.
+- `0 <= Node.val <= 1000`
+- There are **no** two consecutive nodes with `Node.val == 0`.
+- The **beginning** and **end** of the linked list have `Node.val == 0`.
+
+## 题目大意
+
+给你一个链表的头节点 head ，该链表包含由 0 分隔开的一连串整数。链表的 开端 和 末尾 的节点都满足 Node.val == 0 。对于每两个相邻的 0 ，请你将它们之间的所有节点合并成一个节点，其值是所有已合并节点的值之和。然后将所有 0 移除，修改后的链表不应该含有任何 0 。
+
+返回修改后链表的头节点 head 。
+
+## 解题思路
+
+- 简单题。合并链表中两个值为 0 的节点。从头开始遍历链表，遇到节点值不为 0 的节点便累加；遇到节点值为 0 的节点，将累加值转换成结果链表要输出的节点值，然后继续遍历。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func mergeNodes(head *ListNode) *ListNode {
+	res := &ListNode{}
+	h := res
+	if head.Next == nil {
+		return &structures.ListNode{}
+	}
+	cur := head
+	sum := 0
+	for cur.Next != nil {
+		if cur.Next.Val != 0 {
+			sum += cur.Next.Val
+		} else {
+			h.Next = &ListNode{Val: sum, Next: nil}
+			h = h.Next
+			sum = 0
+		}
+		cur = cur.Next
+	}
+	return res.Next
+}
+```
\ No newline at end of file
diff --git a/leetcode/2182.Construct-String-With-Repeat-Limit/2182. Construct String With Repeat Limit.go b/leetcode/2182.Construct-String-With-Repeat-Limit/2182. Construct String With Repeat Limit.go
new file mode 100644
index 000000000..9baa17c4f
--- /dev/null
+++ b/leetcode/2182.Construct-String-With-Repeat-Limit/2182. Construct String With Repeat Limit.go	
@@ -0,0 +1,37 @@
+package leetcode
+
+func repeatLimitedString(s string, repeatLimit int) string {
+	cnt := make([]int, 26)
+	for _, c := range s {
+		cnt[int(c-'a')]++
+	}
+	var ns []byte
+	for i := 25; i >= 0; {
+		k := i - 1
+		for cnt[i] > 0 {
+			for j := 0; j < min(cnt[i], repeatLimit); j++ {
+				ns = append(ns, byte(i)+'a')
+			}
+			cnt[i] -= repeatLimit
+			if cnt[i] > 0 {
+				for ; k >= 0 && cnt[k] == 0; k-- {
+				}
+				if k < 0 {
+					break
+				} else {
+					ns = append(ns, byte(k)+'a')
+					cnt[k]--
+				}
+			}
+		}
+		i = k
+	}
+	return string(ns)
+}
+func min(a, b int) int {
+	if a < b {
+		return a
+	} else {
+		return b
+	}
+}
diff --git a/leetcode/2182.Construct-String-With-Repeat-Limit/2182. Construct String With Repeat Limit_test.go b/leetcode/2182.Construct-String-With-Repeat-Limit/2182. Construct String With Repeat Limit_test.go
new file mode 100644
index 000000000..ac01fb982
--- /dev/null
+++ b/leetcode/2182.Construct-String-With-Repeat-Limit/2182. Construct String With Repeat Limit_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2182 struct {
+	para2182
+	ans2182
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2182 struct {
+	one   string
+	limit int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2182 struct {
+	one string
+}
+
+func Test_Problem2182(t *testing.T) {
+
+	qs := []question2182{
+
+		{
+			para2182{"cczazcc", 3},
+			ans2182{"zzcccac"},
+		},
+
+		{
+			para2182{"aababab", 2},
+			ans2182{"bbabaa"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2182------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2182, q.para2182
+		fmt.Printf("【input】:%v       【output】:%v\n", p, repeatLimitedString(p.one, p.limit))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2182.Construct-String-With-Repeat-Limit/README.md b/leetcode/2182.Construct-String-With-Repeat-Limit/README.md
new file mode 100644
index 000000000..68500ed18
--- /dev/null
+++ b/leetcode/2182.Construct-String-With-Repeat-Limit/README.md
@@ -0,0 +1,98 @@
+# [2182. Construct String With Repeat Limit](https://leetcode.com/problems/construct-string-with-repeat-limit/)
+
+
+## 题目
+
+You are given a string `s` and an integer `repeatLimit`. Construct a new string `repeatLimitedString` using the characters of `s` such that no letter appears **more than** `repeatLimit` times **in a row**. You do **not** have to use all characters from `s`.
+
+Return *the **lexicographically largest*** `repeatLimitedString` *possible*.
+
+A string `a` is **lexicographically larger** than a string `b` if in the first position where `a` and `b` differ, string `a` has a letter that appears later in the alphabet than the corresponding letter in `b`. If the first `min(a.length, b.length)` characters do not differ, then the longer string is the lexicographically larger one.
+
+**Example 1:**
+
+```
+Input: s = "cczazcc", repeatLimit = 3
+Output: "zzcccac"
+Explanation: We use all of the characters from s to construct the repeatLimitedString "zzcccac".
+The letter 'a' appears at most 1 time in a row.
+The letter 'c' appears at most 3 times in a row.
+The letter 'z' appears at most 2 times in a row.
+Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
+The string is the lexicographically largest repeatLimitedString possible so we return "zzcccac".
+Note that the string "zzcccca" is lexicographically larger but the letter 'c' appears more than 3 times in a row, so it is not a valid repeatLimitedString.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "aababab", repeatLimit = 2
+Output: "bbabaa"
+Explanation: We use only some of the characters from s to construct the repeatLimitedString "bbabaa".
+The letter 'a' appears at most 2 times in a row.
+The letter 'b' appears at most 2 times in a row.
+Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
+The string is the lexicographically largest repeatLimitedString possible so we return "bbabaa".
+Note that the string "bbabaaa" is lexicographically larger but the letter 'a' appears more than 2 times in a row, so it is not a valid repeatLimitedString.
+
+```
+
+**Constraints:**
+
+- `1 <= repeatLimit <= s.length <= 10^5`
+- `s` consists of lowercase English letters.
+
+## 题目大意
+
+给你一个字符串 s 和一个整数 repeatLimit ，用 s 中的字符构造一个新字符串 repeatLimitedString ，使任何字母 连续 出现的次数都不超过 repeatLimit 次。你不必使用 s 中的全部字符。
+
+返回 字典序最大的 repeatLimitedString 。
+
+如果在字符串 a 和 b 不同的第一个位置，字符串 a 中的字母在字母表中出现时间比字符串 b 对应的字母晚，则认为字符串 a 比字符串 b 字典序更大 。如果字符串中前 min(a.length, b.length) 个字符都相同，那么较长的字符串字典序更大。
+
+## 解题思路
+
+- 利用贪心的思想，由于题意要求返回字典序最大的字符串，所以先从字典序最大的字母开始选起。然后选择当前字典序最大的字母个数和 limit 的最小值。如果当前字典序最大的字母比较多，多于 limit，不能一直选择它。选完 limit 个以后，需要选一个字典序次大的字母，选完这个字母以后再次选择字典序最大的字母。因为 limit 限制字母不能连续多于 limit 个。如此循环，直到所有的字母都选完。这样的策略排列出来的字母串为最大字典序。
+
+## 代码
+
+```go
+package leetcode
+
+func repeatLimitedString(s string, repeatLimit int) string {
+	cnt := make([]int, 26)
+	for _, c := range s {
+		cnt[int(c-'a')]++
+	}
+	var ns []byte
+	for i := 25; i >= 0; {
+		k := i - 1
+		for cnt[i] > 0 {
+			for j := 0; j < min(cnt[i], repeatLimit); j++ {
+				ns = append(ns, byte(i)+'a')
+			}
+			cnt[i] -= repeatLimit
+			if cnt[i] > 0 {
+				for ; k >= 0 && cnt[k] == 0; k-- {
+				}
+				if k < 0 {
+					break
+				} else {
+					ns = append(ns, byte(k)+'a')
+					cnt[k]--
+				}
+			}
+		}
+		i = k
+	}
+	return string(ns)
+}
+func min(a, b int) int {
+	if a < b {
+		return a
+	} else {
+		return b
+	}
+}
+```
\ No newline at end of file
diff --git a/leetcode/2183.Count-Array-Pairs-Divisible-by-K/2183. Count Array Pairs Divisible by K.go b/leetcode/2183.Count-Array-Pairs-Divisible-by-K/2183. Count Array Pairs Divisible by K.go
new file mode 100644
index 000000000..62c22397f
--- /dev/null
+++ b/leetcode/2183.Count-Array-Pairs-Divisible-by-K/2183. Count Array Pairs Divisible by K.go	
@@ -0,0 +1,32 @@
+package leetcode
+
+import "math"
+
+func countPairs(nums []int, k int) int64 {
+	n := int(math.Sqrt(float64(k)))
+	gcds, res := make(map[int]int, n), 0
+	for _, num := range nums {
+		gcds[gcd(num, k)]++
+	}
+
+	for a, n1 := range gcds {
+		for b, n2 := range gcds {
+			if a > b || (a*b)%k != 0 {
+				continue
+			}
+			if a != b {
+				res += n1 * n2
+			} else { // a == b
+				res += n1 * (n1 - 1) / 2
+			}
+		}
+	}
+	return int64(res)
+}
+
+func gcd(a, b int) int {
+	for a%b != 0 {
+		a, b = b, a%b
+	}
+	return b
+}
diff --git a/leetcode/2183.Count-Array-Pairs-Divisible-by-K/2183. Count Array Pairs Divisible by K_test.go b/leetcode/2183.Count-Array-Pairs-Divisible-by-K/2183. Count Array Pairs Divisible by K_test.go
new file mode 100644
index 000000000..487dd70cd
--- /dev/null
+++ b/leetcode/2183.Count-Array-Pairs-Divisible-by-K/2183. Count Array Pairs Divisible by K_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question2182 struct {
+	para2182
+	ans2182
+}
+
+// para 是参数
+// one 代表第一个参数
+type para2182 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans2182 struct {
+	one int64
+}
+
+func Test_Problem2182(t *testing.T) {
+
+	qs := []question2182{
+
+		{
+			para2182{[]int{1, 2, 3, 4, 5}, 2},
+			ans2182{7},
+		},
+
+		{
+			para2182{[]int{1, 2, 3, 4}, 5},
+			ans2182{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 2183------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans2182, q.para2182
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countPairs(p.nums, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/2183.Count-Array-Pairs-Divisible-by-K/README.md b/leetcode/2183.Count-Array-Pairs-Divisible-by-K/README.md
new file mode 100644
index 000000000..f7bec80a3
--- /dev/null
+++ b/leetcode/2183.Count-Array-Pairs-Divisible-by-K/README.md
@@ -0,0 +1,85 @@
+# [2183. Count Array Pairs Divisible by K](https://leetcode.com/problems/count-array-pairs-divisible-by-k/)
+
+
+## 题目
+
+Given a **0-indexed** integer array `nums` of length `n` and an integer `k`, return *the **number of pairs*** `(i, j)` *such that:*
+
+- `0 <= i < j <= n - 1` *and*
+- `nums[i] * nums[j]` *is divisible by* `k`.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3,4,5], k = 2
+Output: 7
+Explanation:
+The 7 pairs of indices whose corresponding products are divisible by 2 are
+(0, 1), (0, 3), (1, 2), (1, 3), (1, 4), (2, 3), and (3, 4).
+Their products are 2, 4, 6, 8, 10, 12, and 20 respectively.
+Other pairs such as (0, 2) and (2, 4) have products 3 and 15 respectively, which are not divisible by 2.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3,4], k = 5
+Output: 0
+Explanation: There does not exist any pair of indices whose corresponding product is divisible by 5.
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 10^5`
+- `1 <= nums[i], k <= 10^5`
+
+## 题目大意
+
+给你一个下标从 0 开始、长度为 n 的整数数组 nums 和一个整数 k ，返回满足下述条件的下标对 (i, j) 的数目：
+
+- 0 <= i < j <= n - 1 且
+- nums[i] * nums[j] 能被 k 整除。
+
+## 解题思路
+
+- 先找出 num 中每个元素与 k 的最大公约数。并统计这些公约数出现的频次，将数据保存在 map 中。在计算过程中，循环可以只需算到 ${O(\sqrt {k})}$ , 因为每一个 gcd[i] 一定是 k 的因数，而它出现的频次不会超过 ${O(\sqrt {k})}$。简单证明一下：假设因子 v 和 k/v 这两个因数为 k 的因子。v 和 k/v 必至少有 1 个小于等于 $\sqrt {k}$。所以 k 的因子也不会超过 2 * $\sqrt {k}$ = ${O(\sqrt {k})}$ 个。
+- 算出上述的 map 以后，2 层循环暴力遍历 key 值，如果 a * b 能被 k 整除，并且 a 和 b 不相同，那么 a 和 b 对应的 value 值相乘即为满足条件的下标对数；如果 a 和 b 相同，那么下标对数为 $C_{n}^{2}$。最后累加结果即可。
+
+## 代码
+
+```go
+package leetcode
+
+import "math"
+
+func countPairs(nums []int, k int) int64 {
+	n := int(math.Sqrt(float64(k)))
+	gcds, res := make(map[int]int, n), 0
+	for _, num := range nums {
+		gcds[gcd(num, k)]++
+	}
+
+	for a, n1 := range gcds {
+		for b, n2 := range gcds {
+			if a > b || (a*b)%k != 0 {
+				continue
+			}
+			if a != b {
+				res += n1 * n2
+			} else { // a == b
+				res += n1 * (n1 - 1) / 2
+			}
+		}
+	}
+	return int64(res)
+}
+
+func gcd(a, b int) int {
+	for a%b != 0 {
+		a, b = b, a%b
+	}
+	return b
+}
+```
\ No newline at end of file
diff --git a/leetcode/9990085.Maximal-Rectangle/85. Maximal Rectangle.go b/leetcode/9990085.Maximal-Rectangle/85. Maximal Rectangle.go
new file mode 100644
index 000000000..1a98bd359
--- /dev/null
+++ b/leetcode/9990085.Maximal-Rectangle/85. Maximal Rectangle.go	
@@ -0,0 +1,5 @@
+package leetcode
+
+func maximalRectangle(matrix [][]byte) int {
+	return 0
+}
diff --git a/leetcode/9990085.Maximal-Rectangle/85. Maximal Rectangle_test.go b/leetcode/9990085.Maximal-Rectangle/85. Maximal Rectangle_test.go
new file mode 100644
index 000000000..d4dd5a4c2
--- /dev/null
+++ b/leetcode/9990085.Maximal-Rectangle/85. Maximal Rectangle_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question85 struct {
+	para85
+	ans85
+}
+
+// para 是参数
+// one 代表第一个参数
+type para85 struct {
+	one [][]byte
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans85 struct {
+	one int
+}
+
+func Test_Problem85(t *testing.T) {
+
+	qs := []question85{
+
+		{
+			para85{[][]byte{{'1', '0', '1', '0', '0'}, {'1', '0', '1', '1', '1'}, {'1', '1', '1', '1', '1'}, {'1', '0', '0', '1', '0'}}},
+			ans85{6},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 85------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans85, q.para85
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maximalRectangle(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/9990132.Palindrome-Partitioning-II/132. Palindrome Partitioning II.go b/leetcode/9990132.Palindrome-Partitioning-II/132. Palindrome Partitioning II.go
new file mode 100644
index 000000000..6ac3bc894
--- /dev/null
+++ b/leetcode/9990132.Palindrome-Partitioning-II/132. Palindrome Partitioning II.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+func minCut(s string) int {
+	if s == "" {
+		return 0
+	}
+	result := len(s)
+	current := make([]string, 0, len(s))
+	dfs132(s, 0, current, &result)
+	return result
+}
+
+func dfs132(s string, idx int, cur []string, result *int) {
+	start, end := idx, len(s)
+	if start == end {
+		*result = min(*result, len(cur)-1)
+		return
+	}
+	for i := start; i < end; i++ {
+		if isPal(s, start, i) {
+			dfs132(s, i+1, append(cur, s[start:i+1]), result)
+		}
+	}
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+func isPal(str string, s, e int) bool {
+	for s < e {
+		if str[s] != str[e] {
+			return false
+		}
+		s++
+		e--
+	}
+	return true
+}
diff --git a/leetcode/9990132.Palindrome-Partitioning-II/132. Palindrome Partitioning II_test.go b/leetcode/9990132.Palindrome-Partitioning-II/132. Palindrome Partitioning II_test.go
new file mode 100644
index 000000000..b7acf9493
--- /dev/null
+++ b/leetcode/9990132.Palindrome-Partitioning-II/132. Palindrome Partitioning II_test.go	
@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question132 struct {
+	para132
+	ans132
+}
+
+// para 是参数
+// one 代表第一个参数
+type para132 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans132 struct {
+	one int
+}
+
+func Test_Problem132(t *testing.T) {
+
+	qs := []question132{
+
+		{
+			para132{"aab"},
+			ans132{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 132------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans132, q.para132
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minCut(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/9990316.Remove-Duplicate-Letters/316. Remove Duplicate Letters.go b/leetcode/9990316.Remove-Duplicate-Letters/316. Remove Duplicate Letters.go
new file mode 100644
index 000000000..df5f7a6bc
--- /dev/null
+++ b/leetcode/9990316.Remove-Duplicate-Letters/316. Remove Duplicate Letters.go	
@@ -0,0 +1,5 @@
+package leetcode
+
+func removeDuplicateLetters(s string) string {
+	return ""
+}
diff --git a/leetcode/9990316.Remove-Duplicate-Letters/316. Remove Duplicate Letters_test.go b/leetcode/9990316.Remove-Duplicate-Letters/316. Remove Duplicate Letters_test.go
new file mode 100644
index 000000000..8ec7713ac
--- /dev/null
+++ b/leetcode/9990316.Remove-Duplicate-Letters/316. Remove Duplicate Letters_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question316 struct {
+	para316
+	ans316
+}
+
+// para 是参数
+// one 代表第一个参数
+type para316 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans316 struct {
+	one string
+}
+
+func Test_Problem316(t *testing.T) {
+
+	qs := []question316{
+
+		{
+			para316{"bcabc"},
+			ans316{"abc"},
+		},
+		{
+			para316{"cbacdcbc"},
+			ans316{"acdb"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 316------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans316, q.para316
+		fmt.Printf("【input】:%v       【output】:%v\n", p, removeDuplicateLetters(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/9990352.Data-Stream-as-Disjoint-Intervals/352. Data Stream as Disjoint Intervals.go b/leetcode/9990352.Data-Stream-as-Disjoint-Intervals/352. Data Stream as Disjoint Intervals.go
new file mode 100644
index 000000000..eab0e5b6c
--- /dev/null
+++ b/leetcode/9990352.Data-Stream-as-Disjoint-Intervals/352. Data Stream as Disjoint Intervals.go	
@@ -0,0 +1,101 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// Interval define
+type Interval = structures.Interval
+
+// SummaryRanges define
+type SummaryRanges struct {
+	intervals []Interval
+}
+
+// Constructor352 define
+func Constructor352() SummaryRanges {
+	return SummaryRanges{}
+}
+
+// AddNum define
+func (sr *SummaryRanges) AddNum(val int) {
+	if sr.intervals == nil {
+		sr.intervals = []Interval{
+			{
+				Start: val,
+				End:   val,
+			},
+		}
+		return
+	}
+
+	low, high := 0, len(sr.intervals)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if sr.intervals[mid].Start <= val && val <= sr.intervals[mid].End {
+			return
+		} else if val < sr.intervals[mid].Start {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+
+	if low == 0 {
+		if sr.intervals[0].Start-1 == val {
+			sr.intervals[0].Start--
+			return
+		}
+		ni := Interval{Start: val, End: val}
+		sr.intervals = append(sr.intervals, ni)
+		copy(sr.intervals[1:], sr.intervals)
+		sr.intervals[0] = ni
+		return
+	}
+
+	if low == len(sr.intervals) {
+		if sr.intervals[low-1].End+1 == val {
+			sr.intervals[low-1].End++
+			return
+		}
+		sr.intervals = append(sr.intervals, Interval{Start: val, End: val})
+		return
+	}
+
+	if sr.intervals[low-1].End+1 < val && val < sr.intervals[low].Start-1 {
+		sr.intervals = append(sr.intervals, Interval{})
+		copy(sr.intervals[low+1:], sr.intervals[low:])
+		sr.intervals[low] = Interval{Start: val, End: val}
+		return
+	}
+
+	if sr.intervals[low-1].End == val-1 && val+1 == sr.intervals[low].Start {
+		sr.intervals[low-1].End = sr.intervals[low].End
+		n := len(sr.intervals)
+		copy(sr.intervals[low:], sr.intervals[low+1:])
+		sr.intervals = sr.intervals[:n-1]
+		return
+	}
+
+	if sr.intervals[low-1].End == val-1 {
+		sr.intervals[low-1].End++
+	} else {
+		sr.intervals[low].Start--
+	}
+}
+
+// GetIntervals define
+func (sr *SummaryRanges) GetIntervals() [][]int {
+	intervals := [][]int{}
+	for _, interval := range sr.intervals {
+		intervals = append(intervals, []int{interval.Start, interval.End})
+	}
+	return intervals
+}
+
+/**
+ * Your SummaryRanges object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.AddNum(val);
+ * param_2 := obj.GetIntervals();
+ */
diff --git a/leetcode/9990352.Data-Stream-as-Disjoint-Intervals/352. Data Stream as Disjoint Intervals_test.go b/leetcode/9990352.Data-Stream-as-Disjoint-Intervals/352. Data Stream as Disjoint Intervals_test.go
new file mode 100644
index 000000000..d92023c30
--- /dev/null
+++ b/leetcode/9990352.Data-Stream-as-Disjoint-Intervals/352. Data Stream as Disjoint Intervals_test.go	
@@ -0,0 +1,21 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem352(t *testing.T) {
+	obj := Constructor352()
+	fmt.Printf("obj = %v\n", obj)
+	obj.AddNum(1)
+	fmt.Printf("Intervals = %v\n", obj.GetIntervals()) // [1,1]
+	obj.AddNum(3)
+	fmt.Printf("Intervals = %v\n", obj.GetIntervals()) // [1,1] [3,3]
+	obj.AddNum(7)
+	fmt.Printf("Intervals = %v\n", obj.GetIntervals()) // [1, 1], [3, 3], [7, 7]
+	obj.AddNum(2)
+	fmt.Printf("Intervals = %v\n", obj.GetIntervals()) // [1, 3], [7, 7]
+	obj.AddNum(6)
+	fmt.Printf("Intervals = %v\n", obj.GetIntervals()) // [1, 3], [6, 7]
+}
diff --git a/leetcode/9990363.Max-Sum-of-Rectangle-No-Larger-Than-K/363. Max Sum of Rectangle No Larger Than K.go b/leetcode/9990363.Max-Sum-of-Rectangle-No-Larger-Than-K/363. Max Sum of Rectangle No Larger Than K.go
new file mode 100644
index 000000000..7f1a6f3c3
--- /dev/null
+++ b/leetcode/9990363.Max-Sum-of-Rectangle-No-Larger-Than-K/363. Max Sum of Rectangle No Larger Than K.go	
@@ -0,0 +1,112 @@
+package leetcode
+
+import (
+	"math"
+)
+
+// 解法一 二分搜索
+func maxSumSubmatrix(matrix [][]int, k int) int {
+	// 转换为前缀和
+	for i := 0; i < len(matrix); i++ {
+		for j := 1; j < len(matrix[0]); j++ {
+			matrix[i][j] += matrix[i][j-1]
+		}
+	}
+	sum, absMax, absMaxFound := make([]int, len(matrix)), 0, false
+	for y1 := 0; y1 < len(matrix[0]); y1++ {
+		for y2 := y1; y2 < len(matrix[0]); y2++ {
+			for x := 0; x < len(matrix); x++ {
+				if y1 == 0 {
+					sum[x] = matrix[x][y2]
+				} else {
+					sum[x] = matrix[x][y2] - matrix[x][y1-1]
+				}
+			}
+			curMax := kadaneK(sum, k)
+			if !absMaxFound || curMax > absMax {
+				absMax = curMax
+				absMaxFound = true
+			}
+		}
+	}
+	return absMax
+}
+
+func kadaneK(a []int, k int) int {
+	sum, sums, maxSoFar := 0, []int{}, math.MinInt32
+	for _, v := range a {
+		// 第一次循环会先插入 0，因为 sum 有可能等于 k
+		sums = insertSort(sums, sum)
+		sum += v
+		pos := binarySearchOfKadane(sums, sum-k)
+		if pos < len(sums) && sum-sums[pos] > maxSoFar {
+			maxSoFar = sum - sums[pos]
+		}
+	}
+	return maxSoFar
+}
+
+func binarySearchOfKadane(a []int, v int) int {
+	low, high := 0, len(a)
+	for low < high {
+		mid := low + (high-low)>>1
+		if a[mid] < v {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return low
+}
+
+func insertSort(a []int, v int) []int {
+	// 类似插入排序，将元素按照从小到大的顺序插入数组
+	p := binarySearchOfKadane(a, v)
+	a = append(a, 0)
+	// 把 p 后面的元素全部往后移，把 p 位置空出来放 v
+	copy(a[p+1:], a[p:len(a)-1])
+	a[p] = v
+	return a
+}
+
+// 解法二 暴力解法，超时
+func maxSumSubmatrix1(matrix [][]int, k int) int {
+	minNum := math.MaxInt64
+	for row := range matrix {
+		for col := 1; col < len(matrix[row]); col++ {
+			if matrix[row][col] < minNum {
+				minNum = matrix[row][col]
+			}
+		}
+	}
+	for row := range matrix {
+		for col := 1; col < len(matrix[row]); col++ {
+			matrix[row][col] += matrix[row][col-1]
+		}
+	}
+	for i := k; ; i-- {
+		if findSumSubmatrix(matrix, i) > 0 {
+			return i
+		}
+	}
+}
+
+func findSumSubmatrix(matrix [][]int, target int) int {
+	m, n, res := len(matrix), len(matrix[0]), 0
+	for i := 0; i < n; i++ {
+		for j := i; j < n; j++ {
+			counterMap, sum := make(map[int]int, m), 0
+			counterMap[0] = 1 // 题目保证一定有解，所以这里初始化是 1
+			for row := 0; row < m; row++ {
+				if i > 0 {
+					sum += matrix[row][j] - matrix[row][i-1]
+				} else {
+					sum += matrix[row][j]
+				}
+				res += counterMap[sum-target]
+				counterMap[sum]++
+			}
+		}
+	}
+	return res
+}
diff --git a/leetcode/9990363.Max-Sum-of-Rectangle-No-Larger-Than-K/363. Max Sum of Rectangle No Larger Than K_test.go b/leetcode/9990363.Max-Sum-of-Rectangle-No-Larger-Than-K/363. Max Sum of Rectangle No Larger Than K_test.go
new file mode 100644
index 000000000..9c62eb0fd
--- /dev/null
+++ b/leetcode/9990363.Max-Sum-of-Rectangle-No-Larger-Than-K/363. Max Sum of Rectangle No Larger Than K_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question363 struct {
+	para363
+	ans363
+}
+
+// para 是参数
+// one 代表第一个参数
+type para363 struct {
+	one [][]int
+	k   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans363 struct {
+	one int
+}
+
+func Test_Problem363(t *testing.T) {
+
+	qs := []question363{
+
+		{
+			para363{[][]int{{1, 0, 1}, {0, -2, 3}}, 2},
+			ans363{2},
+		},
+
+		{
+			para363{[][]int{{2, 2, -1}}, 0},
+			ans363{-1},
+		},
+
+		{
+			para363{[][]int{{5, -4, -3, 4}, {-3, -4, 4, 5}, {5, 1, 5, -4}}, 8},
+			ans363{8},
+		},
+
+		{
+			para363{[][]int{{5, -4, -3, 4}, {-3, -4, 4, 5}, {5, 1, 5, -4}}, 3},
+			ans363{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 363------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans363, q.para363
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxSumSubmatrix(p.one, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/9990975.Odd-Even-Jump/975. Odd Even Jump.go b/leetcode/9990975.Odd-Even-Jump/975. Odd Even Jump.go
new file mode 100644
index 000000000..0d3306d17
--- /dev/null
+++ b/leetcode/9990975.Odd-Even-Jump/975. Odd Even Jump.go	
@@ -0,0 +1,75 @@
+package leetcode
+
+import (
+	"fmt"
+)
+
+func oddEvenJumps(A []int) int {
+	oddJumpMap, evenJumpMap, current, res := map[int]int{}, map[int]int{}, 0, 0
+	for i := 0; i < len(A); i++ {
+		for j := i + 1; j < len(A); j++ {
+			if v, ok := oddJumpMap[i]; ok {
+				if A[i] <= A[j] && A[j] <= A[v] {
+					if A[j] == A[v] && j < oddJumpMap[i] {
+						oddJumpMap[i] = j
+					} else if A[j] < A[v] {
+						oddJumpMap[i] = j
+					}
+				}
+			} else {
+				if A[i] <= A[j] {
+					oddJumpMap[i] = j
+				}
+			}
+		}
+	}
+	for i := 0; i < len(A); i++ {
+		for j := i + 1; j < len(A); j++ {
+			if v, ok := evenJumpMap[i]; ok {
+				if A[i] >= A[j] && A[j] >= A[v] {
+					if A[j] == A[v] && j < evenJumpMap[i] {
+						evenJumpMap[i] = j
+					} else if A[j] > A[v] {
+						evenJumpMap[i] = j
+					}
+				}
+			} else {
+				if A[i] >= A[j] {
+					evenJumpMap[i] = j
+				}
+			}
+		}
+	}
+	fmt.Printf("oddJumpMap = %v evenJumpMap = %v\n", oddJumpMap, evenJumpMap)
+	for i := 0; i < len(A); i++ {
+		count := 1
+		current = i
+		for {
+			if count%2 == 1 {
+				if v, ok := oddJumpMap[current]; ok {
+					if v == len(A)-1 {
+						res++
+						break
+					}
+					current = v
+					count++
+				} else {
+					break
+				}
+			}
+			if count%2 == 0 {
+				if v, ok := evenJumpMap[current]; ok {
+					if v == len(A)-1 {
+						res++
+						break
+					}
+					current = v
+					count++
+				} else {
+					break
+				}
+			}
+		}
+	}
+	return res + 1
+}
diff --git a/leetcode/9990975.Odd-Even-Jump/975. Odd Even Jump_test.go b/leetcode/9990975.Odd-Even-Jump/975. Odd Even Jump_test.go
new file mode 100644
index 000000000..899c73c7e
--- /dev/null
+++ b/leetcode/9990975.Odd-Even-Jump/975. Odd Even Jump_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question975 struct {
+	para975
+	ans975
+}
+
+// para 是参数
+// one 代表第一个参数
+type para975 struct {
+	A []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans975 struct {
+	one int
+}
+
+func Test_Problem975(t *testing.T) {
+
+	qs := []question975{
+
+		{
+			para975{[]int{10, 13, 12, 14, 15}},
+			ans975{2},
+		},
+
+		{
+			para975{[]int{2, 3, 1, 1, 4}},
+			ans975{3},
+		},
+
+		{
+			para975{[]int{5, 1, 3, 4, 2}},
+			ans975{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 975------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans975, q.para975
+		fmt.Printf("【input】:%v       【output】:%v\n", p, oddEvenJumps(p.A))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/9991044.Longest-Duplicate-Substring/1044. Longest Duplicate Substring.go b/leetcode/9991044.Longest-Duplicate-Substring/1044. Longest Duplicate Substring.go
new file mode 100644
index 000000000..f81c47f37
--- /dev/null
+++ b/leetcode/9991044.Longest-Duplicate-Substring/1044. Longest Duplicate Substring.go	
@@ -0,0 +1,86 @@
+package leetcode
+
+// 解法一 二分搜索 + Rabin-Karp
+func longestDupSubstring(S string) string {
+	// low, high := 0, len(S)
+	// for low < high {
+	// 	mid := (low + high + 1) >> 1
+	// 	if hasRepeated("", B, mid) {
+	// 		low = mid
+	// 	} else {
+	// 		high = mid - 1
+	// 	}
+	// }
+	return "这个解法还有问题!"
+}
+
+// func hashSlice(arr []int, length int) []int {
+// 	// hash 数组里面记录 arr 比 length 长出去部分的 hash 值
+// 	hash, pl, h := make([]int, len(arr)-length+1), 1, 0
+// 	for i := 0; i < length-1; i++ {
+// 		pl *= primeRK
+// 	}
+// 	for i, v := range arr {
+// 		h = h*primeRK + v
+// 		if i >= length-1 {
+// 			hash[i-length+1] = h
+// 			h -= pl * arr[i-length+1]
+// 		}
+// 	}
+// 	return hash
+// }
+
+// func hasSamePrefix(A, B []int, length int) bool {
+// 	for i := 0; i < length; i++ {
+// 		if A[i] != B[i] {
+// 			return false
+// 		}
+// 	}
+// 	return true
+// }
+
+// func hasRepeated(A, B []int, length int) bool {
+// 	hs := hashSlice(A, length)
+// 	hashToOffset := make(map[int][]int, len(hs))
+// 	for i, h := range hs {
+// 		hashToOffset[h] = append(hashToOffset[h], i)
+// 	}
+// 	for i, h := range hashSlice(B, length) {
+// 		if offsets, ok := hashToOffset[h]; ok {
+// 			for _, offset := range offsets {
+// 				if hasSamePrefix(A[offset:], B[i:], length) {
+// 					return true
+// 				}
+// 			}
+// 		}
+// 	}
+// 	return false
+// }
+
+// 解法二 二分搜索 + 暴力匹配
+func longestDupSubstring1(S string) string {
+	res := ""
+	low, high := 0, len(S)
+	for low < high {
+		mid := low + (high-low)>>1
+		if isDuplicate(mid, S, &res) {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return res
+}
+
+func isDuplicate(length int, str string, res *string) bool {
+	visited := map[string]bool{}
+	for i := 0; i+length <= len(str); i++ {
+		subStr := str[i : i+length]
+		if visited[subStr] {
+			*res = subStr
+			return true
+		}
+		visited[subStr] = true
+	}
+	return false
+}
diff --git a/leetcode/9991044.Longest-Duplicate-Substring/1044. Longest Duplicate Substring_test.go b/leetcode/9991044.Longest-Duplicate-Substring/1044. Longest Duplicate Substring_test.go
new file mode 100644
index 000000000..e09ceac69
--- /dev/null
+++ b/leetcode/9991044.Longest-Duplicate-Substring/1044. Longest Duplicate Substring_test.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1044 struct {
+	para1044
+	ans1044
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1044 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1044 struct {
+	one string
+}
+
+func Test_Problem1044(t *testing.T) {
+
+	qs := []question1044{
+		{
+			para1044{"banana"},
+			ans1044{"ana"},
+		},
+
+		{
+			para1044{"abcd"},
+			ans1044{""},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1044------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1044, q.para1044
+		fmt.Printf("【input】:%v       【output】:%v\n", p, longestDupSubstring(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/9991292.Maximum-Side-Length-of-a-Square-with-Sum-Less-than-or-Equal-to-Threshold/1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold.go b/leetcode/9991292.Maximum-Side-Length-of-a-Square-with-Sum-Less-than-or-Equal-to-Threshold/1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold.go
new file mode 100644
index 000000000..90c05d815
--- /dev/null
+++ b/leetcode/9991292.Maximum-Side-Length-of-a-Square-with-Sum-Less-than-or-Equal-to-Threshold/1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+func maxSideLength(mat [][]int, threshold int) int {
+	return 0
+	// m, n, sum := len(mat), len(mat[0]), make([][]int, len(mat[0])+1, len(mat[0])+1)
+	// for i := range sum {
+	// 	sum[i] = make([]int, n+1, n+1)
+	// }
+	// for i := 0; i < m; i++ {
+	// 	for j := 0; j < n; j++ {
+	// 		sum[i+1][j+1] = sum[i][j+1] + sum[i+1][j] - sum[i][j] + mat[i][j]
+	// 	}
+	// }
+	// low, high := 0, min(m, n)
+	// for low < high {
+	// 	mid := low + (high-low)>>1
+	// 	if !inThreshold(&sum, threshold, mid) {
+	// 		high = mid + 1
+	// 	} else {
+	// 		low = mid
+	// 	}
+	// }
+	// if inThreshold(&sum, threshold, high) {
+	// 	return high
+	// }
+	// return low
+}
+
+// func min(a int, b int) int {
+// 	if a > b {
+// 		return b
+// 	}
+// 	return a
+// }
+
+// func inThreshold(sum *[][]int, threshold int, length int) bool {
+// 	for i := 1; i < len(*sum); i++ {
+// 		for j := 1; j < len((*sum)[0]); j++ {
+// 			if i < length || j < length {
+// 				continue
+// 			}
+// 			if (*sum)[i][j]+(*sum)[i-length][j-length]-(*sum)[i-length][j]-(*sum)[i][j-length] <= threshold {
+// 				return true
+// 			}
+// 		}
+// 	}
+// 	return false
+// }
diff --git a/leetcode/9991292.Maximum-Side-Length-of-a-Square-with-Sum-Less-than-or-Equal-to-Threshold/1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold_test.go b/leetcode/9991292.Maximum-Side-Length-of-a-Square-with-Sum-Less-than-or-Equal-to-Threshold/1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold_test.go
new file mode 100644
index 000000000..0c713cd1d
--- /dev/null
+++ b/leetcode/9991292.Maximum-Side-Length-of-a-Square-with-Sum-Less-than-or-Equal-to-Threshold/1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1292 struct {
+	para1292
+	ans1292
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1292 struct {
+	mat       [][]int
+	threshold int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1292 struct {
+	one int
+}
+
+func Test_Problem1292(t *testing.T) {
+
+	qs := []question1292{
+
+		{
+			para1292{[][]int{{1, 1, 3, 2, 4, 3, 2}, {1, 1, 3, 2, 4, 3, 2}, {1, 1, 3, 2, 4, 3, 2}}, 4},
+			ans1292{2},
+		},
+
+		{
+			para1292{[][]int{{2, 2, 2, 2, 2}, {2, 2, 2, 2, 2}, {2, 2, 2, 2, 2}, {2, 2, 2, 2, 2}, {2, 2, 2, 2, 2}}, 1},
+			ans1292{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1292------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1292, q.para1292
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxSideLength(p.mat, p.threshold))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/note/time_complexity.md b/note/time_complexity.md
new file mode 100644
index 000000000..174aad59f
--- /dev/null
+++ b/note/time_complexity.md
@@ -0,0 +1,127 @@
+# 时间复杂度和空间复杂度
+
+
+## 一. 时间复杂度数据规模
+
+1s 内能解决问题的数据规模：10^6 ~ 10^7
+
+- O(n^2) 算法可以处理 10^4 级别的数据规模(保守估计，处理 1000 级别的问题肯定没问题)
+- O(n) 算法可以处理 10^8 级别的数据规模(保守估计，处理 10^7 级别的问题肯定没问题)
+- O(nlog n) 算法可以处理 10^7 级别的数据规模(保守估计，处理 10^6 级别的问题肯定没问题)
+
+| | 数据规模|时间复杂度 | 算法举例|
+|:------:|:------:|:------:|:------:|
+|1|10|O(n!)|permutation 排列|
+|2|20~30|O(2^n)|combination 组合|
+|3|50|O(n^4)|DFS 搜索、DP 动态规划|
+|4|100|O(n^3)|任意两点最短路径、DP 动态规划|
+|5|1000|O(n^2)|稠密图、DP 动态规划|
+|6|10^6|O(nlog n)|排序，堆，递归与分治|
+|7|10^7|O(n)|DP 动态规划、图遍历、拓扑排序、树遍历|
+|8|10^9|O(sqrt(n))|筛素数、求平方根|
+|9|10^10|O(log n)|二分搜索|
+|10|+∞|O(1)|数学相关算法|
+|----------------|----------------|------------------------------------------------------------------|--------------------------------|
+
+
+一些具有迷惑性的例子：
+
+```c
+void hello (int n){
+
+    for( int sz = 1 ; sz < n ; sz += sz)
+        for( int i = 1 ; i < n ; i ++)
+            cout << "Hello" << endl;
+}
+```
+
+上面这段代码的时间复杂度是 O(nlog n) 而不是 O(n^2)
+
+```c
+bool isPrime (int n){
+
+    for( int x = 2 ; x * x <= n ; x ++ )
+        if( n % x == 0)
+            return false;
+    return true;
+}
+```
+
+上面这段代码的时间复杂度是 O(sqrt(n)) 而不是 O(n)。
+
+再举一个例子，有一个字符串数组，将数组中的每一个字符串按照字母序排序，之后再将整个字符串数组按照字典序排序。两步操作的整体时间复杂度是多少呢？
+
+如果回答是 O(n*nlog n + nlog n) = O(n^2log n)，这个答案是错误的。字符串的长度和数组的长度是没有关系的，所以这两个变量应该单独计算。假设最长的字符串长度为 s，数组中有 n 个字符串。对每个字符串排序的时间复杂度是 O(slog s)，将数组中每个字符串都按照字母序排序的时间复杂度是 O(n * slog s)。
+
+将整个字符串数组按照字典序排序的时间复杂度是 O(s * nlog n)。排序算法中的 O(nlog n) 是比较的次数，由于比较的是整型数字，所以每次比较是 O(1)。但是字符串按照字典序比较，时间复杂度是 O(s)。所以字符串数组按照字典序排序的时间复杂度是 O(s * nlog n)。所以整体复杂度是 O(n * slog s) + O(s * nlog n) = O(n\*slog s + s\*nlogn) = O(n\*s\*(log s + log n))
+
+## 二. 空间复杂度
+
+递归调用是有空间代价的，递归算法需要保存递归栈信息，所以花费的空间复杂度会比非递归算法要高。
+
+```c
+int sum( int n ){
+    assert( n >= 0 )
+    int ret = 0;
+    for ( int i = 0 ; i <= n ; i++)
+        ret += i;
+    return ret;
+}
+```
+
+上面算法的时间复杂度为 O(n)，空间复杂度 O(1)。
+
+```c
+int sum( int n ){
+    assert( n >= 0 )
+    if ( n == 0 )
+        return 0;
+    return n + sum( n - 1);
+}
+```
+
+上面算法的时间复杂度为 O(n)，空间复杂度 O(n)。
+
+## 三. 递归的时间复杂度
+
+### 只有一次递归调用
+
+如果递归函数中，只进行了一次递归调用，且递归深度为 depth，在每个递归函数中，时间复杂度为 T，那么总体的时间复杂度为 O(T * depth)
+
+举个例子：
+
+```c
+int binarySearch(int arr[], int l, int r, int target){
+	if( l > r)
+	    return -1;
+    int mid = l + (r-l)/2;//防溢出
+    if(arr[mid] == target)
+        return mid;
+    else if (arr[mid]>target)
+        return binarySearch(arr,l,mid-1,target);
+    else 
+        return binarySearch(arr,mid+1,r,target);
+}
+
+```
+
+在二分查找的递归实现中，只递归调用了自身。递归深度是 log n ，每次递归里面的复杂度是 O(1) 的，所以二分查找的递归实现的时间复杂度为 O(log n) 的。
+
+
+### 只有多次递归调用
+
+针对多次递归调用的情况，就需要看它的计算调用的次数了。通常可以画一颗递归树来看。举例：
+
+```c
+int f(int n){
+    assert( n >= 0 );
+    if( n ==0 )
+        return 1;
+    return f( n - 1 ) + f ( n - 1 );
+
+```
+
+上述这次递归调用的次数为 2^0^ + 2^1^ + 2^2^ + …… + 2^n^ = 2^n+1^ - 1 = O(2^n)
+
+
+> 关于更加复杂的递归的复杂度分析，请参考，主定理。主定理中针对各种复杂情况都给出了正确的结论。
diff --git "a/note/\346\227\266\351\227\264\345\244\215\346\235\202\345\272\246.md" "b/note/\346\227\266\351\227\264\345\244\215\346\235\202\345\272\246.md"
deleted file mode 100644
index 4dd545e18..000000000
--- "a/note/\346\227\266\351\227\264\345\244\215\346\235\202\345\272\246.md"
+++ /dev/null
@@ -1,118 +0,0 @@
-# 时间复杂度和空间复杂度
-
-
-## 一. 时间复杂度数据规模
-
-1s 内能解决问题的数据规模：
-
-- O(n^2) 算法可以处理 10^4 级别的数据规模(保守估计，处理 1000 级别的问题肯定没问题)
-- O(n) 算法可以处理 10^8 级别的数据规模(保守估计，处理 10^7 级别的问题肯定没问题)
-- O(nlog n) 算法可以处理 10^7 级别的数据规模(保守估计，处理 10^6 级别的问题肯定没问题)
-
-一些具有迷惑性的例子：
-
-```c
-void hello (int n){
-
-    for( int sz = 1 ; sz < n ; sz += sz)
-        for( int i = 1 ; i < n ; i ++)
-            cout << "Hello" << endl;
-}
-```
-
-上面这段代码的时间复杂度是 O(nlog n) 而不是 O(n^2)
-
-```c
-bool isPrime (int n){
-
-    for( int x = 2 ; x * x <= n ; x ++ )
-        if( n % x == 0)
-            return false;
-    return true;
-}
-```
-
-上面这段代码的时间复杂度是 O(sqrt(n)) 而不是 O(n)
-
-再举一个例子，有一个字符串数组，将数组中的每一个字符串按照字母序排序，之后再降整个字符串数组按照字典序排序。两步操作的整体时间复杂度是多少呢？
-
-如果回答是 O(n*nlog n + nlog n) = O(n^2log n)，这个答案是错误的。
-
-字符串的长度和数组的长度是没有关系的，所以这两个变量应该单独计算。
-
-假设最长的字符串长度为 s，数组中有 n 个字符串。对每个字符串排序的时间复杂度是 O(slog s)，将数组中每个字符串都按照字母序排序的时间复杂度是 O(n * slog s)。
-
-将整个字符串数组按照字典序排序的时间复杂度是 O(s * nlog n)。排序算法中的 O(nlog n) 是比较的次数，由于比较的是整型数字，所以每次比较是 O(1)。但是字符串按照字典序比较，时间复杂度是 O(s)。所以字符串数组按照字典序排序的时间复杂度是 O(s * nlog n)
-
-所以整体复杂度是 O(n * slog s) + O(s * nlog n) = O(n\*slog s + s\*nlogn) = O(n\*s\*(log s + log n))
-
-## 二. 空间复杂度
-
-递归调用是有空间代价的，递归算法需要保存递归栈信息，所以花费的空间复杂度会比非递归算法要高。
-
-```c
-int sum( int n ){
-    assert( n >= 0 )
-    int ret = 0;
-    for ( int i = 0 ; i <= n ; i++)
-        ret += i;
-    return ret;
-}
-```
-
-上面算法的时间复杂度为 O(n)，空间复杂度 O(1)。
-
-```c
-int sum( int n ){
-    assert( n >= 0 )
-    if ( n == 0 )
-        return 0;
-    return n + sum( n - 1);
-}
-```
-
-上面算法的时间复杂度为 O(n)，空间复杂度 O(n)。
-
-## 三. 递归的时间复杂度
-
-### 只有一次递归调用
-
-如果递归函数中，只进行了一次递归调用，且递归深度为 depth，在每个递归函数中，时间复杂度为 T，那么总体的时间复杂度为 O(T * depth)
-
-举个例子：
-
-```c
-int binarySearch(int arr[], int l, int r, int target){
-	if( l > r)
-	    return -1;
-    int mid = l + (r-l)/2;//防溢出
-    if(arr[mid] == target)
-        return mid;
-    else if (arr[mid]>target)
-        return binarySearch(arr,l,mid-1,target);
-    eles 
-        return binarySearch(arr,mid+1,r,target);
-}
-
-```
-
-在二分查找的递归实现中，只递归调用了自身。递归深度是 log n ，每次递归里面的复杂度是 O(1) 的，所以二分查找的递归实现的时间复杂度为 O(log n) 的。
-
-
-### 只有多次递归调用
-
-针对多次递归调用的情况，就需要看它的计算调用的次数了。通常可以画一颗递归树来看。举例：
-
-```c
-int f(int n){
-    assert( n >= 0 );
-    if( n ==0 )
-        return 1;
-    return f( n - 1 ) + f ( n - 1 );
-
-```
-
-上述这次递归调用的次数为 2^0^ + 2^1^ + 2^2^ + …… + 2^n^ = 2^n+1^ - 1 = O(2^n)
-
-
-> 关于更加复杂的递归的复杂度分析，请参考，主定理。主定理中针对各种复杂情况都给出了正确的结论。
\ No newline at end of file
diff --git a/structures/Heap.go b/structures/Heap.go
new file mode 100644
index 000000000..2910773b4
--- /dev/null
+++ b/structures/Heap.go
@@ -0,0 +1,30 @@
+package structures
+
+// intHeap 实现了最小堆 heap 的接口
+type intHeap []int
+
+func (h intHeap) Len() int {
+	return len(h)
+}
+
+func (h intHeap) Less(i, j int) bool {
+	return h[i] < h[j]
+}
+
+func (h intHeap) Swap(i, j int) {
+	h[i], h[j] = h[j], h[i]
+}
+
+func (h *intHeap) Push(x interface{}) {
+	// Push 使用 *h，是因为
+	// Push 增加了 h 的长度
+	*h = append(*h, x.(int))
+}
+
+func (h *intHeap) Pop() interface{} {
+	// Pop 使用 *h ，是因为
+	// Pop 减短了 h 的长度
+	res := (*h)[len(*h)-1]
+	*h = (*h)[:len(*h)-1]
+	return res
+}
diff --git a/structures/Heap_test.go b/structures/Heap_test.go
new file mode 100644
index 000000000..af8291f52
--- /dev/null
+++ b/structures/Heap_test.go
@@ -0,0 +1,30 @@
+package structures
+
+import (
+	"container/heap"
+	"fmt"
+	"testing"
+
+	"github.com/stretchr/testify/assert"
+)
+
+func Test_intHeap(t *testing.T) {
+	ast := assert.New(t)
+
+	ih := new(intHeap)
+	heap.Init(ih)
+
+	heap.Push(ih, 1)
+	heap.Pop(ih)
+
+	begin, end := 0, 10
+	for i := begin; i < end; i++ {
+		heap.Push(ih, i)
+		ast.Equal(0, (*ih)[0], "插入 %d 后的最小值却是 %d, ih=%v", i, (*ih)[0], (*ih))
+	}
+
+	for i := begin; i < end; i++ {
+		fmt.Println(i, *ih)
+		ast.Equal(i, heap.Pop(ih), "Pop 后 ih=%v", (*ih))
+	}
+}
diff --git a/structures/Interval.go b/structures/Interval.go
new file mode 100644
index 000000000..bed1432e8
--- /dev/null
+++ b/structures/Interval.go
@@ -0,0 +1,53 @@
+package structures
+
+// Interval 提供区间表示
+type Interval struct {
+	Start int
+	End   int
+}
+
+// Interval2Ints 把 Interval 转换成 整型切片
+func Interval2Ints(i Interval) []int {
+	return []int{i.Start, i.End}
+}
+
+// IntervalSlice2Intss 把 []Interval 转换成 [][]int
+func IntervalSlice2Intss(is []Interval) [][]int {
+	res := make([][]int, 0, len(is))
+	for i := range is {
+		res = append(res, Interval2Ints(is[i]))
+	}
+	return res
+}
+
+// Intss2IntervalSlice 把 [][]int 转换成 []Interval
+func Intss2IntervalSlice(intss [][]int) []Interval {
+	res := make([]Interval, 0, len(intss))
+	for _, ints := range intss {
+		res = append(res, Interval{Start: ints[0], End: ints[1]})
+	}
+	return res
+}
+
+// QuickSort define
+func QuickSort(a []Interval, lo, hi int) {
+	if lo >= hi {
+		return
+	}
+	p := partitionSort(a, lo, hi)
+	QuickSort(a, lo, p-1)
+	QuickSort(a, p+1, hi)
+}
+
+func partitionSort(a []Interval, lo, hi int) int {
+	pivot := a[hi]
+	i := lo - 1
+	for j := lo; j < hi; j++ {
+		if (a[j].Start < pivot.Start) || (a[j].Start == pivot.Start && a[j].End < pivot.End) {
+			i++
+			a[j], a[i] = a[i], a[j]
+		}
+	}
+	a[i+1], a[hi] = a[hi], a[i+1]
+	return i + 1
+}
diff --git a/structures/Interval_test.go b/structures/Interval_test.go
new file mode 100644
index 000000000..dd36e40a5
--- /dev/null
+++ b/structures/Interval_test.go
@@ -0,0 +1,59 @@
+package structures
+
+import (
+	"testing"
+
+	"github.com/stretchr/testify/assert"
+)
+
+func Test_Interval2Ints(t *testing.T) {
+	ast := assert.New(t)
+
+	actual := Interval2Ints(Interval{Start: 1, End: 2})
+	expected := []int{1, 2}
+	ast.Equal(expected, actual)
+}
+
+func Test_IntervalSlice2Intss(t *testing.T) {
+	ast := assert.New(t)
+
+	actual := IntervalSlice2Intss(
+		[]Interval{
+			{
+				Start: 1,
+				End:   2,
+			},
+			{
+				Start: 3,
+				End:   4,
+			},
+		},
+	)
+	expected := [][]int{
+		{1, 2},
+		{3, 4},
+	}
+
+	ast.Equal(expected, actual)
+}
+func Test_Intss2IntervalSlice(t *testing.T) {
+	ast := assert.New(t)
+
+	expected := []Interval{
+		{
+			Start: 1,
+			End:   2,
+		},
+		{
+			Start: 3,
+			End:   4,
+		},
+	}
+	actual := Intss2IntervalSlice([][]int{
+		{1, 2},
+		{3, 4},
+	},
+	)
+
+	ast.Equal(expected, actual)
+}
diff --git a/structures/ListNode.go b/structures/ListNode.go
new file mode 100644
index 000000000..9f3dc9ed4
--- /dev/null
+++ b/structures/ListNode.go
@@ -0,0 +1,82 @@
+package structures
+
+import (
+	"fmt"
+)
+
+// ListNode 是链接节点
+// 这个不能复制到*_test.go文件中。会导致Travis失败
+type ListNode struct {
+	Val  int
+	Next *ListNode
+}
+
+// List2Ints convert List to []int
+func List2Ints(head *ListNode) []int {
+	// 链条深度限制，链条深度超出此限制，会 panic
+	limit := 100
+
+	times := 0
+
+	res := []int{}
+	for head != nil {
+		times++
+		if times > limit {
+			msg := fmt.Sprintf("链条深度超过%d，可能出现环状链条。请检查错误，或者放宽 l2s 函数中 limit 的限制。", limit)
+			panic(msg)
+		}
+
+		res = append(res, head.Val)
+		head = head.Next
+	}
+
+	return res
+}
+
+// Ints2List convert []int to List
+func Ints2List(nums []int) *ListNode {
+	if len(nums) == 0 {
+		return nil
+	}
+
+	l := &ListNode{}
+	t := l
+	for _, v := range nums {
+		t.Next = &ListNode{Val: v}
+		t = t.Next
+	}
+	return l.Next
+}
+
+// GetNodeWith returns the first node with val
+func (l *ListNode) GetNodeWith(val int) *ListNode {
+	res := l
+	for res != nil {
+		if res.Val == val {
+			break
+		}
+		res = res.Next
+	}
+	return res
+}
+
+// Ints2ListWithCycle returns a list whose tail point to pos-indexed node
+// head's index is 0
+// if pos = -1, no cycle
+func Ints2ListWithCycle(nums []int, pos int) *ListNode {
+	head := Ints2List(nums)
+	if pos == -1 {
+		return head
+	}
+	c := head
+	for pos > 0 {
+		c = c.Next
+		pos--
+	}
+	tail := c
+	for tail.Next != nil {
+		tail = tail.Next
+	}
+	tail.Next = c
+	return head
+}
diff --git a/structures/ListNode_test.go b/structures/ListNode_test.go
new file mode 100644
index 000000000..ed1df4f61
--- /dev/null
+++ b/structures/ListNode_test.go
@@ -0,0 +1,68 @@
+package structures
+
+import (
+	"testing"
+
+	"github.com/stretchr/testify/assert"
+)
+
+func Test_l2s(t *testing.T) {
+	ast := assert.New(t)
+	ast.Equal([]int{}, List2Ints(nil), "输入nil，没有返回[]int{}")
+
+	one2three := &ListNode{
+		Val: 1,
+		Next: &ListNode{
+			Val: 2,
+			Next: &ListNode{
+				Val: 3,
+			},
+		},
+	}
+	ast.Equal([]int{1, 2, 3}, List2Ints(one2three), "没有成功地转换成[]int")
+
+	limit := 100
+	overLimitList := Ints2List(make([]int, limit+1))
+	ast.Panics(func() { List2Ints(overLimitList) }, "转换深度超过 %d 限制的链条，没有 panic", limit)
+}
+
+func Test_s2l(t *testing.T) {
+	ast := assert.New(t)
+	ast.Nil(Ints2List([]int{}), "输入[]int{}，没有返回nil")
+
+	ln := Ints2List([]int{1, 2, 3, 4, 5, 6, 7, 8, 9})
+	i := 1
+	for ln != nil {
+		ast.Equal(i, ln.Val, "对应的值不对")
+		ln = ln.Next
+		i++
+	}
+}
+
+func Test_getNodeWith(t *testing.T) {
+	ast := assert.New(t)
+	//
+	ln := Ints2List([]int{1, 2, 3, 4, 5, 6, 7, 8, 9})
+	val := 10
+	node := &ListNode{
+		Val: val,
+	}
+	tail := ln
+	for tail.Next != nil {
+		tail = tail.Next
+	}
+	tail.Next = node
+	expected := node
+	actual := ln.GetNodeWith(val)
+	ast.Equal(expected, actual)
+}
+
+func Test_Ints2ListWithCycle(t *testing.T) {
+	ast := assert.New(t)
+	ints := []int{1, 2, 3}
+	l := Ints2ListWithCycle(ints, -1)
+	ast.Equal(ints, List2Ints(l))
+
+	l = Ints2ListWithCycle(ints, 1)
+	ast.Panics(func() { List2Ints(l) })
+}
diff --git a/structures/NestedInteger.go b/structures/NestedInteger.go
new file mode 100644
index 000000000..0a944f39c
--- /dev/null
+++ b/structures/NestedInteger.go
@@ -0,0 +1,37 @@
+package structures
+
+// NestedInteger  is the interface that allows for creating nested lists.
+// You should not implement it, or speculate about its implementation
+type NestedInteger struct {
+	Num int
+	Ns  []*NestedInteger
+}
+
+// IsInteger Return true if this NestedInteger holds a single integer, rather than a nested list.
+func (n NestedInteger) IsInteger() bool {
+	return n.Ns == nil
+}
+
+// GetInteger Return the single integer that this NestedInteger holds, if it holds a single integer
+// The result is undefined if this NestedInteger holds a nested list
+// So before calling this method, you should have a check
+func (n NestedInteger) GetInteger() int {
+	return n.Num
+}
+
+// SetInteger Set this NestedInteger to hold a single integer.
+func (n *NestedInteger) SetInteger(value int) {
+	n.Num = value
+}
+
+// Add  Set this NestedInteger to hold a nested list and adds a nested integer to it.
+func (n *NestedInteger) Add(elem NestedInteger) {
+	n.Ns = append(n.Ns, &elem)
+}
+
+// GetList Return the nested list that this NestedInteger holds, if it holds a nested list
+// The list length is zero if this NestedInteger holds a single integer
+// You can access NestedInteger's List element directly if you want to modify it
+func (n NestedInteger) GetList() []*NestedInteger {
+	return n.Ns
+}
diff --git a/structures/NestedInterger_test.go b/structures/NestedInterger_test.go
new file mode 100644
index 000000000..bf1727b5e
--- /dev/null
+++ b/structures/NestedInterger_test.go
@@ -0,0 +1,30 @@
+package structures
+
+import (
+	"testing"
+
+	"github.com/stretchr/testify/assert"
+)
+
+func Test_NestedInteger(t *testing.T) {
+	ast := assert.New(t)
+
+	n := NestedInteger{}
+
+	ast.True(n.IsInteger())
+
+	n.SetInteger(1)
+	ast.Equal(1, n.GetInteger())
+
+	elem := NestedInteger{Num: 1}
+
+	expected := NestedInteger{
+		Num: 1,
+		Ns:  []*NestedInteger{&elem},
+	}
+	n.Add(elem)
+
+	ast.Equal(expected, n)
+
+	ast.Equal(expected.Ns, n.GetList())
+}
diff --git a/structures/Point.go b/structures/Point.go
new file mode 100644
index 000000000..4e5e4c7e5
--- /dev/null
+++ b/structures/Point.go
@@ -0,0 +1,27 @@
+package structures
+
+// Point 定义了一个二维坐标点
+type Point struct {
+	X, Y int
+}
+
+// Intss2Points 把 [][]int 转换成 []Point
+func Intss2Points(points [][]int) []Point {
+	res := make([]Point, len(points))
+	for i, p := range points {
+		res[i] = Point{
+			X: p[0],
+			Y: p[1],
+		}
+	}
+	return res
+}
+
+// Points2Intss 把 []Point 转换成　[][]int
+func Points2Intss(points []Point) [][]int {
+	res := make([][]int, len(points))
+	for i, p := range points {
+		res[i] = []int{p.X, p.Y}
+	}
+	return res
+}
diff --git a/structures/Point_test.go b/structures/Point_test.go
new file mode 100644
index 000000000..48b6dc1ce
--- /dev/null
+++ b/structures/Point_test.go
@@ -0,0 +1,78 @@
+package structures
+
+import (
+	"reflect"
+	"testing"
+)
+
+func Test_Intss2Points(t *testing.T) {
+	type args struct {
+		points [][]int
+	}
+	tests := []struct {
+		name string
+		args args
+		want []Point
+	}{
+		{
+			"测试 [][]int 转换成 []Point ",
+			args{
+				[][]int{
+					{1, 0},
+					{2, 0},
+					{3, 0},
+					{4, 0},
+					{5, 0},
+				},
+			},
+			[]Point{
+				{X: 1, Y: 0},
+				{X: 2, Y: 0},
+				{X: 3, Y: 0},
+				{X: 4, Y: 0},
+				{X: 5, Y: 0},
+			},
+		},
+	}
+	for _, tt := range tests {
+		if got := Intss2Points(tt.args.points); !reflect.DeepEqual(got, tt.want) {
+			t.Errorf("%q. intss2Points() = %v, want %v", tt.name, got, tt.want)
+		}
+	}
+}
+
+func Test_Points2Intss(t *testing.T) {
+	type args struct {
+		points []Point
+	}
+	tests := []struct {
+		name string
+		args args
+		want [][]int
+	}{
+		{
+			"测试 [][]int 转换成 []Point ",
+			args{
+				[]Point{
+					{X: 1, Y: 0},
+					{X: 2, Y: 0},
+					{X: 3, Y: 0},
+					{X: 4, Y: 0},
+					{X: 5, Y: 0},
+				},
+			},
+			[][]int{
+				{1, 0},
+				{2, 0},
+				{3, 0},
+				{4, 0},
+				{5, 0},
+			},
+		},
+	}
+	for _, tt := range tests {
+		if got := Points2Intss(tt.args.points); !reflect.DeepEqual(got, tt.want) {
+			t.Errorf("%q. Points2Intss() = %v, want %v", tt.name, got, tt.want)
+		}
+	}
+}
diff --git a/structures/PriorityQueue.go b/structures/PriorityQueue.go
new file mode 100644
index 000000000..01677463a
--- /dev/null
+++ b/structures/PriorityQueue.go
@@ -0,0 +1,54 @@
+package structures
+
+// This example demonstrates a priority queue built using the heap interface.
+
+import (
+	"container/heap"
+)
+
+// entry 是 priorityQueue 中的元素
+type entry struct {
+	key      string
+	priority int
+	// index 是 entry 在 heap 中的索引号
+	// entry 加入 Priority Queue 后， Priority 会变化时，很有用
+	// 如果 entry.priority 一直不变的话，可以删除 index
+	index int
+}
+
+// PQ implements heap.Interface and holds entries.
+type PQ []*entry
+
+func (pq PQ) Len() int { return len(pq) }
+
+func (pq PQ) Less(i, j int) bool {
+	return pq[i].priority < pq[j].priority
+}
+
+func (pq PQ) Swap(i, j int) {
+	pq[i], pq[j] = pq[j], pq[i]
+	pq[i].index = i
+	pq[j].index = j
+}
+
+// Push 往 pq 中放 entry
+func (pq *PQ) Push(x interface{}) {
+	temp := x.(*entry)
+	temp.index = len(*pq)
+	*pq = append(*pq, temp)
+}
+
+// Pop 从 pq 中取出最优先的 entry
+func (pq *PQ) Pop() interface{} {
+	temp := (*pq)[len(*pq)-1]
+	temp.index = -1 // for safety
+	*pq = (*pq)[0 : len(*pq)-1]
+	return temp
+}
+
+// update modifies the priority and value of an entry in the queue.
+func (pq *PQ) update(entry *entry, value string, priority int) {
+	entry.key = value
+	entry.priority = priority
+	heap.Fix(pq, entry.index)
+}
diff --git a/structures/PriorityQueue_test.go b/structures/PriorityQueue_test.go
new file mode 100644
index 000000000..e171464bd
--- /dev/null
+++ b/structures/PriorityQueue_test.go
@@ -0,0 +1,53 @@
+package structures
+
+import (
+	"container/heap"
+	"testing"
+
+	"github.com/stretchr/testify/assert"
+)
+
+func Test_priorityQueue(t *testing.T) {
+	ast := assert.New(t)
+
+	// Some items and their priorities.
+	items := map[string]int{
+		"banana": 2, "apple": 1, "pear": 3,
+	}
+
+	// Create a priority queue, put the items in it, and
+	// establish the priority queue (heap) invariants.
+	pq := make(PQ, len(items))
+	i := 0
+	for value, priority := range items {
+		pq[i] = &entry{
+			key:      value,
+			priority: priority,
+			index:    i,
+		}
+		i++
+	}
+	heap.Init(&pq)
+
+	// Insert a new item and then modify its priority.
+	it := &entry{
+		key:      "orange",
+		priority: 5,
+	}
+	heap.Push(&pq, it)
+	pq.update(it, it.key, 0)
+
+	// Some items and their priorities.
+	expected := []string{
+		"orange",
+		"apple",
+		"banana",
+		"pear",
+	}
+
+	// Take the items out; they arrive in decreasing priority order.
+	for pq.Len() > 0 {
+		it := heap.Pop(&pq).(*entry)
+		ast.Equal(expected[it.priority], it.key)
+	}
+}
diff --git a/structures/Queue.go b/structures/Queue.go
new file mode 100644
index 000000000..69431c666
--- /dev/null
+++ b/structures/Queue.go
@@ -0,0 +1,33 @@
+package structures
+
+// Queue 是用于存放 int 的队列
+type Queue struct {
+	nums []int
+}
+
+// NewQueue 返回 *kit.Queue
+func NewQueue() *Queue {
+	return &Queue{nums: []int{}}
+}
+
+// Push 把 n 放入队列
+func (q *Queue) Push(n int) {
+	q.nums = append(q.nums, n)
+}
+
+// Pop 从 q 中取出最先进入队列的值
+func (q *Queue) Pop() int {
+	res := q.nums[0]
+	q.nums = q.nums[1:]
+	return res
+}
+
+// Len 返回 q 的长度
+func (q *Queue) Len() int {
+	return len(q.nums)
+}
+
+// IsEmpty 反馈 q 是否为空
+func (q *Queue) IsEmpty() bool {
+	return q.Len() == 0
+}
diff --git a/structures/Queue_test.go b/structures/Queue_test.go
new file mode 100644
index 000000000..9b10a39b3
--- /dev/null
+++ b/structures/Queue_test.go
@@ -0,0 +1,27 @@
+package structures
+
+import (
+	"testing"
+
+	"github.com/stretchr/testify/assert"
+)
+
+func Test_Queue(t *testing.T) {
+	ast := assert.New(t)
+
+	q := NewQueue()
+	ast.True(q.IsEmpty(), "检查新建的 q 是否为空")
+
+	start, end := 0, 100
+
+	for i := start; i < end; i++ {
+		q.Push(i)
+		ast.Equal(i-start+1, q.Len(), "Push 后检查 q 的长度。")
+	}
+
+	for i := start; i < end; i++ {
+		ast.Equal(i, q.Pop(), "从 q 中 pop 出数来。")
+	}
+
+	ast.True(q.IsEmpty(), "检查 Pop 完毕后的 q 是否为空")
+}
diff --git a/structures/Stack.go b/structures/Stack.go
new file mode 100644
index 000000000..452c24854
--- /dev/null
+++ b/structures/Stack.go
@@ -0,0 +1,33 @@
+package structures
+
+// Stack 是用于存放 int 的 栈
+type Stack struct {
+	nums []int
+}
+
+// NewStack 返回 *kit.Stack
+func NewStack() *Stack {
+	return &Stack{nums: []int{}}
+}
+
+// Push 把 n 放入 栈
+func (s *Stack) Push(n int) {
+	s.nums = append(s.nums, n)
+}
+
+// Pop 从 s 中取出最后放入 栈 的值
+func (s *Stack) Pop() int {
+	res := s.nums[len(s.nums)-1]
+	s.nums = s.nums[:len(s.nums)-1]
+	return res
+}
+
+// Len 返回 s 的长度
+func (s *Stack) Len() int {
+	return len(s.nums)
+}
+
+// IsEmpty 反馈 s 是否为空
+func (s *Stack) IsEmpty() bool {
+	return s.Len() == 0
+}
diff --git a/structures/Stack_test.go b/structures/Stack_test.go
new file mode 100644
index 000000000..b0c6c60a0
--- /dev/null
+++ b/structures/Stack_test.go
@@ -0,0 +1,27 @@
+package structures
+
+import (
+	"testing"
+
+	"github.com/stretchr/testify/assert"
+)
+
+func Test_Stack(t *testing.T) {
+	ast := assert.New(t)
+
+	s := NewStack()
+	ast.True(s.IsEmpty(), "检查新建的 s 是否为空")
+
+	start, end := 0, 100
+
+	for i := start; i < end; i++ {
+		s.Push(i)
+		ast.Equal(i-start+1, s.Len(), "Push 后检查 q 的长度。")
+	}
+
+	for i := end - 1; i >= start; i-- {
+		ast.Equal(i, s.Pop(), "从 s 中 pop 出数来。")
+	}
+
+	ast.True(s.IsEmpty(), "检查 Pop 完毕后的 s 是否为空")
+}
diff --git a/structures/TreeNode.go b/structures/TreeNode.go
new file mode 100644
index 000000000..99060be17
--- /dev/null
+++ b/structures/TreeNode.go
@@ -0,0 +1,314 @@
+package structures
+
+import (
+	"fmt"
+	"strconv"
+)
+
+// TreeNode is tree's node
+type TreeNode struct {
+	Val   int
+	Left  *TreeNode
+	Right *TreeNode
+}
+
+// NULL 方便添加测试数据
+var NULL = -1 << 63
+
+// Ints2TreeNode 利用 []int 生成 *TreeNode
+func Ints2TreeNode(ints []int) *TreeNode {
+	n := len(ints)
+	if n == 0 {
+		return nil
+	}
+
+	root := &TreeNode{
+		Val: ints[0],
+	}
+
+	queue := make([]*TreeNode, 1, n*2)
+	queue[0] = root
+
+	i := 1
+	for i < n {
+		node := queue[0]
+		queue = queue[1:]
+
+		if i < n && ints[i] != NULL {
+			node.Left = &TreeNode{Val: ints[i]}
+			queue = append(queue, node.Left)
+		}
+		i++
+
+		if i < n && ints[i] != NULL {
+			node.Right = &TreeNode{Val: ints[i]}
+			queue = append(queue, node.Right)
+		}
+		i++
+	}
+
+	return root
+}
+
+// GetTargetNode 返回 Val = target 的 TreeNode
+// root 中一定有 node.Val = target
+func GetTargetNode(root *TreeNode, target int) *TreeNode {
+	if root == nil || root.Val == target {
+		return root
+	}
+
+	res := GetTargetNode(root.Left, target)
+	if res != nil {
+		return res
+	}
+	return GetTargetNode(root.Right, target)
+}
+
+func indexOf(val int, nums []int) int {
+	for i, v := range nums {
+		if v == val {
+			return i
+		}
+	}
+
+	msg := fmt.Sprintf("%d 不存在于 %v 中", val, nums)
+	panic(msg)
+}
+
+// PreIn2Tree 把 preorder 和 inorder 切片转换成 二叉树
+func PreIn2Tree(pre, in []int) *TreeNode {
+	if len(pre) != len(in) {
+		panic("preIn2Tree 中两个切片的长度不相等")
+	}
+
+	if len(in) == 0 {
+		return nil
+	}
+
+	res := &TreeNode{
+		Val: pre[0],
+	}
+
+	if len(in) == 1 {
+		return res
+	}
+
+	idx := indexOf(res.Val, in)
+
+	res.Left = PreIn2Tree(pre[1:idx+1], in[:idx])
+	res.Right = PreIn2Tree(pre[idx+1:], in[idx+1:])
+
+	return res
+}
+
+// InPost2Tree 把 inorder 和 postorder 切片转换成 二叉树
+func InPost2Tree(in, post []int) *TreeNode {
+	if len(post) != len(in) {
+		panic("inPost2Tree 中两个切片的长度不相等")
+	}
+
+	if len(in) == 0 {
+		return nil
+	}
+
+	res := &TreeNode{
+		Val: post[len(post)-1],
+	}
+
+	if len(in) == 1 {
+		return res
+	}
+
+	idx := indexOf(res.Val, in)
+
+	res.Left = InPost2Tree(in[:idx], post[:idx])
+	res.Right = InPost2Tree(in[idx+1:], post[idx:len(post)-1])
+
+	return res
+}
+
+// Tree2Preorder 把 二叉树 转换成 preorder 的切片
+func Tree2Preorder(root *TreeNode) []int {
+	if root == nil {
+		return nil
+	}
+
+	if root.Left == nil && root.Right == nil {
+		return []int{root.Val}
+	}
+
+	res := []int{root.Val}
+	res = append(res, Tree2Preorder(root.Left)...)
+	res = append(res, Tree2Preorder(root.Right)...)
+
+	return res
+}
+
+// Tree2Inorder 把 二叉树转换成 inorder 的切片
+func Tree2Inorder(root *TreeNode) []int {
+	if root == nil {
+		return nil
+	}
+
+	if root.Left == nil && root.Right == nil {
+		return []int{root.Val}
+	}
+
+	res := Tree2Inorder(root.Left)
+	res = append(res, root.Val)
+	res = append(res, Tree2Inorder(root.Right)...)
+
+	return res
+}
+
+// Tree2Postorder 把 二叉树 转换成 postorder 的切片
+func Tree2Postorder(root *TreeNode) []int {
+	if root == nil {
+		return nil
+	}
+
+	if root.Left == nil && root.Right == nil {
+		return []int{root.Val}
+	}
+
+	res := Tree2Postorder(root.Left)
+	res = append(res, Tree2Postorder(root.Right)...)
+	res = append(res, root.Val)
+
+	return res
+}
+
+// Equal return ture if tn == a
+func (tn *TreeNode) Equal(a *TreeNode) bool {
+	if tn == nil && a == nil {
+		return true
+	}
+
+	if tn == nil || a == nil || tn.Val != a.Val {
+		return false
+	}
+
+	return tn.Left.Equal(a.Left) && tn.Right.Equal(a.Right)
+}
+
+// Tree2ints 把 *TreeNode 按照行还原成 []int
+func Tree2ints(tn *TreeNode) []int {
+	res := make([]int, 0, 1024)
+
+	queue := []*TreeNode{tn}
+
+	for len(queue) > 0 {
+		size := len(queue)
+		for i := 0; i < size; i++ {
+			nd := queue[i]
+			if nd == nil {
+				res = append(res, NULL)
+			} else {
+				res = append(res, nd.Val)
+				queue = append(queue, nd.Left, nd.Right)
+			}
+		}
+		queue = queue[size:]
+	}
+
+	i := len(res)
+	for i > 0 && res[i-1] == NULL {
+		i--
+	}
+
+	return res[:i]
+}
+
+// T2s converts *TreeNode to []int
+func T2s(head *TreeNode, array *[]int) {
+	fmt.Printf("运行到这里了 head = %v array = %v\n", head, array)
+	// fmt.Printf("****array = %v\n", array)
+	// fmt.Printf("****head = %v\n", head.Val)
+	*array = append(*array, head.Val)
+	if head.Left != nil {
+		T2s(head.Left, array)
+	}
+	if head.Right != nil {
+		T2s(head.Right, array)
+	}
+}
+
+// Strings2TreeNode converts []string to *TreeNode
+func Strings2TreeNode(strs []string) *TreeNode {
+	n := len(strs)
+	if n == 0 {
+		return nil
+	}
+	x, _ := strconv.Atoi(strs[0])
+	root := &TreeNode{Val: x}
+	queue := make([]*TreeNode, 1, n<<1)
+	queue[0] = root
+	i := 1
+	for i < n {
+		node := queue[0]
+		queue = queue[1:]
+		if i < n && strs[i] != "null" {
+			x, _ = strconv.Atoi(strs[i])
+			node.Left = &TreeNode{Val: x}
+			queue = append(queue, node.Left)
+		}
+		i++
+		if i < n && strs[i] != "null" {
+			x, _ = strconv.Atoi(strs[i])
+			node.Right = &TreeNode{Val: x}
+			queue = append(queue, node.Right)
+		}
+		i++
+	}
+	return root
+}
+
+// Tree2LevelOrderStrings converts *TreeNode into []string by level order traversal.
+func Tree2LevelOrderStrings(root *TreeNode) []string {
+	var ans []string
+	if root == nil {
+		return ans
+	}
+	queue := []*TreeNode{root}
+	var level int
+	for level = 0; len(queue) > 0; level++ {
+		size := len(queue)
+		for i := 0; i < size; i++ {
+			node := queue[i]
+			if node == nil {
+				ans = append(ans, "null")
+			} else {
+				ans = append(ans, strconv.Itoa(node.Val))
+				if node.Left != nil || node.Right != nil {
+					queue = append(queue, node.Left, node.Right)
+				}
+			}
+		}
+		queue = queue[size:]
+	}
+	level--
+	return ans
+}
+
+// Tree2PreOrderStrings converts *TreeNode into []string by preorder traversal.
+func Tree2PreOrderStrings(root *TreeNode) []string {
+	var ans []string
+	if root == nil {
+		return ans
+	}
+	stack := []*TreeNode{root}
+	node := root
+	for len(stack) > 0 {
+		if node == nil {
+			ans = append(ans, "null")
+		}
+		for node != nil {
+			ans = append(ans, strconv.Itoa(node.Val))
+			stack = append(stack, node)
+			node = node.Left
+		}
+		node = stack[len(stack)-1].Right
+		stack = stack[:len(stack)-1]
+	}
+	return ans
+}
diff --git a/structures/TreeNode_test.go b/structures/TreeNode_test.go
new file mode 100644
index 000000000..dc75bf049
--- /dev/null
+++ b/structures/TreeNode_test.go
@@ -0,0 +1,166 @@
+package structures
+
+import (
+	"reflect"
+	"testing"
+
+	"github.com/stretchr/testify/assert"
+)
+
+var (
+	// 同一个 TreeNode 的不同表达方式
+	//            1
+	//      	/  \
+	//         2    3
+	//        / \  /  \
+	//       4  5  6   7
+	LeetCodeOrder = []int{1, 2, 3, 4, 5, 6, 7}
+	preOrder      = []int{1, 2, 4, 5, 3, 6, 7}
+	inOrder       = []int{4, 2, 5, 1, 6, 3, 7}
+	postOrder     = []int{4, 5, 2, 6, 7, 3, 1}
+)
+
+func Test_Ints2TreeNode(t *testing.T) {
+	ast := assert.New(t)
+
+	expected := PreIn2Tree(preOrder, inOrder)
+	actual := Ints2TreeNode(LeetCodeOrder)
+	ast.Equal(expected, actual)
+
+	actual = Ints2TreeNode([]int{})
+	ast.Nil(actual)
+}
+
+func Test_preIn2Tree(t *testing.T) {
+	ast := assert.New(t)
+
+	actual := Tree2Postorder(PreIn2Tree(preOrder, inOrder))
+	expected := postOrder
+	ast.Equal(expected, actual)
+
+	ast.Panics(func() { PreIn2Tree([]int{1}, []int{}) })
+
+	ast.Nil(PreIn2Tree([]int{}, []int{}))
+}
+
+func Test_inPost2Tree(t *testing.T) {
+	ast := assert.New(t)
+
+	actual := Tree2Preorder(InPost2Tree(inOrder, postOrder))
+	expected := preOrder
+	ast.Equal(expected, actual)
+
+	ast.Panics(func() { InPost2Tree([]int{1}, []int{}) })
+
+	ast.Nil(InPost2Tree([]int{}, []int{}))
+}
+
+func Test_tree2Ints(t *testing.T) {
+	ast := assert.New(t)
+	root := PreIn2Tree(preOrder, inOrder)
+
+	ast.Equal(preOrder, Tree2Preorder(root))
+	ast.Nil(Tree2Preorder(nil))
+
+	ast.Equal(inOrder, Tree2Inorder(root))
+	ast.Nil(Tree2Inorder(nil))
+
+	ast.Equal(postOrder, Tree2Postorder(root))
+	ast.Nil(Tree2Postorder(nil))
+}
+
+func Test_indexOf(t *testing.T) {
+	ast := assert.New(t)
+
+	ast.Equal(1, indexOf(1, []int{0, 1, 2, 3}))
+
+	ast.Panics(func() { indexOf(0, []int{1, 2, 3}) })
+}
+
+func Test_TreeNode_Equal(t *testing.T) {
+	type args struct {
+		a *TreeNode
+	}
+	tests := []struct {
+		name   string
+		fields args
+		args   args
+		want   bool
+	}{
+
+		{
+			"相等",
+			args{Ints2TreeNode([]int{1, 2, 3, 4, 5})},
+			args{Ints2TreeNode([]int{1, 2, 3, 4, 5})},
+			true,
+		},
+
+		{
+			"不相等",
+			args{Ints2TreeNode([]int{1, 2, 3, 4, 5})},
+			args{Ints2TreeNode([]int{1, 2, 3, NULL, 5})},
+			false,
+		},
+	}
+	for _, tt := range tests {
+		t.Run(tt.name, func(t *testing.T) {
+			tn := tt.fields.a
+			if got := tn.Equal(tt.args.a); got != tt.want {
+				t.Errorf("TreeNode.Equal() = %v, want %v", got, tt.want)
+			}
+		})
+	}
+}
+
+func Test_GetTargetNode(t *testing.T) {
+	ints := []int{3, 5, 1, 6, 2, 0, 8, NULL, NULL, 7, 4}
+	root := Ints2TreeNode(ints)
+
+	type args struct {
+		root   *TreeNode
+		target int
+	}
+	tests := []struct {
+		name string
+		args args
+		want *TreeNode
+	}{
+
+		{
+			"找到 root.Right.Right",
+			args{
+				root:   root,
+				target: 8,
+			},
+			root.Right.Right,
+		},
+
+		{
+			"找到 root.Left.Left",
+			args{
+				root:   root,
+				target: 6,
+			},
+			root.Left.Left,
+		},
+
+		//
+	}
+
+	for _, tt := range tests {
+		t.Run(tt.name, func(t *testing.T) {
+			if got := GetTargetNode(tt.args.root, tt.args.target); !reflect.DeepEqual(got, tt.want) {
+				t.Errorf("GetTargetNode() = %v, want %v", got, tt.want)
+			}
+		})
+	}
+}
+
+func Test_Tree2ints(t *testing.T) {
+	ast := assert.New(t)
+
+	root := PreIn2Tree(preOrder, inOrder)
+	actual := LeetCodeOrder
+	expected := Tree2ints(root)
+	ast.Equal(expected, actual)
+}
diff --git a/structures/go.mod b/structures/go.mod
new file mode 100644
index 000000000..53c9843a0
--- /dev/null
+++ b/structures/go.mod
@@ -0,0 +1,11 @@
+module github.com/halfrost/LeetCode-Go/structures
+
+go 1.19
+
+require github.com/stretchr/testify v1.8.0
+
+require (
+	github.com/davecgh/go-spew v1.1.1 // indirect
+	github.com/pmezard/go-difflib v1.0.0 // indirect
+	gopkg.in/yaml.v3 v3.0.1 // indirect
+)
diff --git a/structures/go.sum b/structures/go.sum
new file mode 100644
index 000000000..51648299d
--- /dev/null
+++ b/structures/go.sum
@@ -0,0 +1,15 @@
+github.com/davecgh/go-spew v1.1.0/go.mod h1:J7Y8YcW2NihsgmVo/mv3lAwl/skON4iLHjSsI+c5H38=
+github.com/davecgh/go-spew v1.1.1 h1:vj9j/u1bqnvCEfJOwUhtlOARqs3+rkHYY13jYWTU97c=
+github.com/davecgh/go-spew v1.1.1/go.mod h1:J7Y8YcW2NihsgmVo/mv3lAwl/skON4iLHjSsI+c5H38=
+github.com/pmezard/go-difflib v1.0.0 h1:4DBwDE0NGyQoBHbLQYPwSUPoCMWR5BEzIk/f1lZbAQM=
+github.com/pmezard/go-difflib v1.0.0/go.mod h1:iKH77koFhYxTK1pcRnkKkqfTogsbg7gZNVY4sRDYZ/4=
+github.com/stretchr/objx v0.1.0/go.mod h1:HFkY916IF+rwdDfMAkV7OtwuqBVzrE8GR6GFx+wExME=
+github.com/stretchr/objx v0.4.0/go.mod h1:YvHI0jy2hoMjB+UWwv71VJQ9isScKT/TqJzVSSt89Yw=
+github.com/stretchr/testify v1.7.1/go.mod h1:6Fq8oRcR53rry900zMqJjRRixrwX3KX962/h/Wwjteg=
+github.com/stretchr/testify v1.8.0 h1:pSgiaMZlXftHpm5L7V1+rVB+AZJydKsMxsQBIJw4PKk=
+github.com/stretchr/testify v1.8.0/go.mod h1:yNjHg4UonilssWZ8iaSj1OCr/vHnekPRkoO+kdMU+MU=
+gopkg.in/check.v1 v0.0.0-20161208181325-20d25e280405 h1:yhCVgyC4o1eVCa2tZl7eS0r+SDo693bJlVdllGtEeKM=
+gopkg.in/check.v1 v0.0.0-20161208181325-20d25e280405/go.mod h1:Co6ibVJAznAaIkqp8huTwlJQCZ016jof/cbN4VW5Yz0=
+gopkg.in/yaml.v3 v3.0.0-20200313102051-9f266ea9e77c/go.mod h1:K4uyk7z7BCEPqu6E+C64Yfv1cQ7kz7rIZviUmN+EgEM=
+gopkg.in/yaml.v3 v3.0.1 h1:fxVm/GzAzEWqLHuvctI91KS9hhNmmWOoWu0XTYJS7CA=
+gopkg.in/yaml.v3 v3.0.1/go.mod h1:K4uyk7z7BCEPqu6E+C64Yfv1cQ7kz7rIZviUmN+EgEM=
diff --git a/template/BIT.go b/template/BIT.go
new file mode 100644
index 000000000..0679b4262
--- /dev/null
+++ b/template/BIT.go
@@ -0,0 +1,76 @@
+package template
+
+// BinaryIndexedTree define
+type BinaryIndexedTree struct {
+	tree     []int
+	capacity int
+}
+
+// Init define
+func (bit *BinaryIndexedTree) Init(capacity int) {
+	bit.tree, bit.capacity = make([]int, capacity+1), capacity
+}
+
+// Add define
+func (bit *BinaryIndexedTree) Add(index int, val int) {
+	for ; index <= bit.capacity; index += index & -index {
+		bit.tree[index] += val
+	}
+}
+
+// Query define
+func (bit *BinaryIndexedTree) Query(index int) int {
+	sum := 0
+	for ; index > 0; index -= index & -index {
+		sum += bit.tree[index]
+	}
+	return sum
+}
+
+// InitWithNums define
+func (bit *BinaryIndexedTree) InitWithNums(nums []int) {
+	bit.tree, bit.capacity = make([]int, len(nums)+1), len(nums)
+	for i := 1; i <= len(nums); i++ {
+		bit.tree[i] += nums[i-1]
+		for j := i - 2; j >= i-lowbit(i); j-- {
+			bit.tree[i] += nums[j]
+		}
+	}
+}
+
+func lowbit(x int) int {
+	return x & -x
+}
+
+// BinaryIndexedTree2D define
+type BinaryIndexedTree2D struct {
+	tree [][]int
+	row  int
+	col  int
+}
+
+// Add define
+func (bit2 *BinaryIndexedTree2D) Add(i, j int, val int) {
+	for i <= bit2.row {
+		k := j
+		for k <= bit2.col {
+			bit2.tree[i][k] += val
+			k += lowbit(k)
+		}
+		i += lowbit(i)
+	}
+}
+
+// Query define
+func (bit2 *BinaryIndexedTree2D) Query(i, j int) int {
+	sum := 0
+	for i >= 1 {
+		k := j
+		for k >= 1 {
+			sum += bit2.tree[i][k]
+			k -= lowbit(k)
+		}
+		i -= lowbit(i)
+	}
+	return sum
+}
diff --git a/template/BIT_test.go b/template/BIT_test.go
new file mode 100644
index 000000000..062cae72f
--- /dev/null
+++ b/template/BIT_test.go
@@ -0,0 +1,12 @@
+package template
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_BIT(t *testing.T) {
+	nums, bit := []int{1, 2, 3, 4, 5, 6, 7, 8}, BinaryIndexedTree{}
+	bit.Init(8)
+	fmt.Printf("nums = %v bit = %v\n", nums, bit.tree) // [0 1 3 3 10 5 11 7 36]
+}
diff --git a/template/CLRUCache.go b/template/CLRUCache.go
new file mode 100644
index 000000000..9738fb86e
--- /dev/null
+++ b/template/CLRUCache.go
@@ -0,0 +1,176 @@
+package template
+
+import (
+	"container/list"
+	"hash/fnv"
+	"sync"
+)
+
+type command int
+
+const (
+	// MoveToFront define
+	MoveToFront command = iota
+	// PushFront define
+	PushFront
+	// Delete define
+	Delete
+)
+
+type clear struct {
+	done chan struct{}
+}
+
+// CLRUCache define: High Concurrency LRUCache
+type CLRUCache struct {
+	sync.RWMutex
+	cap         int
+	list        *list.List
+	buckets     []*bucket
+	bucketMask  uint32
+	deletePairs chan *list.Element
+	movePairs   chan *list.Element
+	control     chan interface{}
+}
+
+// Pair define
+type Pair struct {
+	key   string
+	value interface{}
+	cmd   command
+}
+
+// New define
+func New(capacity int) *CLRUCache {
+	c := &CLRUCache{
+		cap:        capacity,
+		list:       list.New(),
+		bucketMask: uint32(1024) - 1,
+		buckets:    make([]*bucket, 1024),
+	}
+	for i := 0; i < 1024; i++ {
+		c.buckets[i] = &bucket{
+			keys: make(map[string]*list.Element),
+		}
+	}
+	c.restart()
+	return c
+}
+
+// Get define
+func (c *CLRUCache) Get(key string) interface{} {
+	el := c.bucket(key).get(key)
+	if el == nil {
+		return nil
+	}
+	c.move(el)
+	return el.Value.(Pair).value
+}
+
+// Put define
+func (c *CLRUCache) Put(key string, value interface{}) {
+	el, exist := c.bucket(key).set(key, value)
+	if exist != nil {
+		c.deletePairs <- exist
+	}
+	c.move(el)
+}
+
+func (c *CLRUCache) move(el *list.Element) {
+	select {
+	case c.movePairs <- el:
+	default:
+	}
+}
+
+// Delete define
+func (c *CLRUCache) Delete(key string) bool {
+	el := c.bucket(key).delete(key)
+	if el != nil {
+		c.deletePairs <- el
+		return true
+	}
+	return false
+}
+
+// Clear define
+func (c *CLRUCache) Clear() {
+	done := make(chan struct{})
+	c.control <- clear{done: done}
+	<-done
+}
+
+// Count define
+func (c *CLRUCache) Count() int {
+	count := 0
+	for _, b := range c.buckets {
+		count += b.pairCount()
+	}
+	return count
+}
+
+func (c *CLRUCache) bucket(key string) *bucket {
+	h := fnv.New32a()
+	h.Write([]byte(key))
+	return c.buckets[h.Sum32()&c.bucketMask]
+}
+
+func (c *CLRUCache) stop() {
+	close(c.movePairs)
+	<-c.control
+}
+
+func (c *CLRUCache) restart() {
+	c.deletePairs = make(chan *list.Element, 128)
+	c.movePairs = make(chan *list.Element, 128)
+	c.control = make(chan interface{})
+	go c.worker()
+}
+
+func (c *CLRUCache) worker() {
+	defer close(c.control)
+	for {
+		select {
+		case el, ok := <-c.movePairs:
+			if !ok {
+				goto clean
+			}
+			if c.doMove(el) && c.list.Len() > c.cap {
+				el := c.list.Back()
+				c.list.Remove(el)
+				c.bucket(el.Value.(Pair).key).delete(el.Value.(Pair).key)
+			}
+		case el := <-c.deletePairs:
+			c.list.Remove(el)
+		case control := <-c.control:
+			switch msg := control.(type) {
+			case clear:
+				for _, bucket := range c.buckets {
+					bucket.clear()
+				}
+				c.list = list.New()
+				msg.done <- struct{}{}
+			}
+		}
+	}
+clean:
+	for {
+		select {
+		case el := <-c.deletePairs:
+			c.list.Remove(el)
+		default:
+			close(c.deletePairs)
+			return
+		}
+	}
+}
+
+func (c *CLRUCache) doMove(el *list.Element) bool {
+	if el.Value.(Pair).cmd == MoveToFront {
+		c.list.MoveToFront(el)
+		return false
+	}
+	newel := c.list.PushFront(el.Value.(Pair))
+	c.bucket(el.Value.(Pair).key).update(el.Value.(Pair).key, newel)
+	return true
+}
diff --git a/template/CLRUCache_test.go b/template/CLRUCache_test.go
new file mode 100644
index 000000000..fa1722c54
--- /dev/null
+++ b/template/CLRUCache_test.go
@@ -0,0 +1,130 @@
+package template
+
+import (
+	"container/list"
+	"math/rand"
+	"strconv"
+	"sync"
+	"testing"
+	"time"
+
+	"github.com/stretchr/testify/assert"
+)
+
+func Test_CLRUCache(t *testing.T) {
+	obj := New(2)
+	time.Sleep(150 * time.Millisecond)
+	obj.Put("1", 1)
+	time.Sleep(150 * time.Millisecond)
+	obj.Put("2", 2)
+	time.Sleep(150 * time.Millisecond)
+	param1 := obj.Get("1")
+	time.Sleep(150 * time.Millisecond)
+	assert.Equal(t, 1, param1)
+	obj.Put("3", 3)
+	time.Sleep(150 * time.Millisecond)
+	param1 = obj.Get("2")
+	assert.Equal(t, nil, param1)
+	obj.Put("4", 4)
+	time.Sleep(150 * time.Millisecond)
+	param1 = obj.Get("1")
+	time.Sleep(150 * time.Millisecond)
+	assert.Equal(t, nil, param1)
+	param1 = obj.Get("3")
+	time.Sleep(150 * time.Millisecond)
+	assert.Equal(t, 3, param1)
+	param1 = obj.Get("4")
+	time.Sleep(150 * time.Millisecond)
+	assert.Equal(t, 4, param1)
+}
+
+func MList2Ints(lru *CLRUCache) [][]interface{} {
+	res := [][]interface{}{}
+	for head := lru.list.Front(); head != nil; head = head.Next() {
+		tmp := []interface{}{head.Value.(Pair).key, head.Value.(Pair).value}
+		res = append(res, tmp)
+	}
+	return res
+}
+
+func BenchmarkGetAndPut1(b *testing.B) {
+	b.ResetTimer()
+	obj := New(128)
+	wg := sync.WaitGroup{}
+	wg.Add(b.N * 2)
+	for i := 0; i < b.N; i++ {
+		go func() {
+			defer wg.Done()
+			obj.Get(strconv.Itoa(rand.Intn(200)))
+		}()
+		go func() {
+			defer wg.Done()
+			obj.Put(strconv.Itoa(rand.Intn(200)), strconv.Itoa(rand.Intn(200)))
+		}()
+	}
+	wg.Wait()
+}
+
+type Cache struct {
+	sync.RWMutex
+	Cap  int
+	Keys map[string]*list.Element
+	List *list.List
+}
+
+type pair struct {
+	K, V string
+}
+
+func NewLRUCache(capacity int) Cache {
+	return Cache{
+		Cap:  capacity,
+		Keys: make(map[string]*list.Element),
+		List: list.New(),
+	}
+}
+
+func (c *Cache) Get(key string) interface{} {
+	c.Lock()
+	if el, ok := c.Keys[key]; ok {
+		c.List.MoveToFront(el)
+		return el.Value.(pair).V
+	}
+	c.Unlock()
+	return nil
+}
+
+func (c *Cache) Put(key string, value string) {
+	c.Lock()
+	if el, ok := c.Keys[key]; ok {
+		el.Value = pair{K: key, V: value}
+		c.List.MoveToFront(el)
+	} else {
+		el := c.List.PushFront(pair{K: key, V: value})
+		c.Keys[key] = el
+	}
+	if c.List.Len() > c.Cap {
+		el := c.List.Back()
+		c.List.Remove(el)
+		delete(c.Keys, el.Value.(pair).K)
+	}
+	c.Unlock()
+}
+
+func BenchmarkGetAndPut2(b *testing.B) {
+	b.ResetTimer()
+	obj := NewLRUCache(128)
+	wg := sync.WaitGroup{}
+	wg.Add(b.N * 2)
+	for i := 0; i < b.N; i++ {
+		go func() {
+			defer wg.Done()
+			obj.Get(strconv.Itoa(rand.Intn(200)))
+		}()
+		go func() {
+			defer wg.Done()
+			obj.Put(strconv.Itoa(rand.Intn(200)), strconv.Itoa(rand.Intn(200)))
+		}()
+	}
+	wg.Wait()
+}
diff --git a/template/LFUCache.go b/template/LFUCache.go
new file mode 100644
index 000000000..74a8a9a9d
--- /dev/null
+++ b/template/LFUCache.go
@@ -0,0 +1,196 @@
+package template
+
+import "container/list"
+
+// LFUCache define
+type LFUCache struct {
+	nodes    map[int]*list.Element
+	lists    map[int]*list.List
+	capacity int
+	min      int
+}
+
+type node struct {
+	key       int
+	value     int
+	frequency int
+}
+
+// Constructor define
+func Constructor(capacity int) LFUCache {
+	return LFUCache{nodes: make(map[int]*list.Element),
+		lists:    make(map[int]*list.List),
+		capacity: capacity,
+		min:      0,
+	}
+}
+
+// Get define
+func (lfuCache *LFUCache) Get(key int) int {
+	value, ok := lfuCache.nodes[key]
+	if !ok {
+		return -1
+	}
+	currentNode := value.Value.(*node)
+	lfuCache.lists[currentNode.frequency].Remove(value)
+	currentNode.frequency++
+	if _, ok := lfuCache.lists[currentNode.frequency]; !ok {
+		lfuCache.lists[currentNode.frequency] = list.New()
+	}
+	newList := lfuCache.lists[currentNode.frequency]
+	newNode := newList.PushFront(currentNode)
+	lfuCache.nodes[key] = newNode
+	if currentNode.frequency-1 == lfuCache.min && lfuCache.lists[currentNode.frequency-1].Len() == 0 {
+		lfuCache.min++
+	}
+	return currentNode.value
+}
+
+// Put define
+func (lfuCache *LFUCache) Put(key int, value int) {
+	if lfuCache.capacity == 0 {
+		return
+	}
+	if currentValue, ok := lfuCache.nodes[key]; ok {
+		currentNode := currentValue.Value.(*node)
+		currentNode.value = value
+		lfuCache.Get(key)
+		return
+	}
+	if lfuCache.capacity == len(lfuCache.nodes) {
+		currentList := lfuCache.lists[lfuCache.min]
+		backNode := currentList.Back()
+		delete(lfuCache.nodes, backNode.Value.(*node).key)
+		currentList.Remove(backNode)
+	}
+	lfuCache.min = 1
+	currentNode := &node{
+		key:       key,
+		value:     value,
+		frequency: 1,
+	}
+	if _, ok := lfuCache.lists[1]; !ok {
+		lfuCache.lists[1] = list.New()
+	}
+	newList := lfuCache.lists[1]
+	newNode := newList.PushFront(currentNode)
+	lfuCache.nodes[key] = newNode
+}
+
+/**
+ * Your LFUCache object will be instantiated and called as such:
+ * obj := Constructor(capacity);
+ * param_1 := obj.Get(key);
+ * obj.Put(key,value);
+ */
+
+// Index Priority Queue
+// import "container/heap"
+
+// type LFUCache struct {
+// 	capacity int
+// 	pq       PriorityQueue
+// 	hash     map[int]*Item
+// 	counter  int
+// }
+
+// func Constructor(capacity int) LFUCache {
+// 	lfu := LFUCache{
+// 		pq:       PriorityQueue{},
+// 		hash:     make(map[int]*Item, capacity),
+// 		capacity: capacity,
+// 	}
+// 	return lfu
+// }
+
+// func (this *LFUCache) Get(key int) int {
+// 	if this.capacity == 0 {
+// 		return -1
+// 	}
+// 	if item, ok := this.hash[key]; ok {
+// 		this.counter++
+// 		this.pq.update(item, item.value, item.frequency+1, this.counter)
+// 		return item.value
+// 	}
+// 	return -1
+// }
+
+// func (this *LFUCache) Put(key int, value int) {
+// 	if this.capacity == 0 {
+// 		return
+// 	}
+// 	// fmt.Printf("Put %d\n", key)
+// 	this.counter++
+// 	// 如果存在，增加 frequency，再调整堆
+// 	if item, ok := this.hash[key]; ok {
+// 		this.pq.update(item, value, item.frequency+1, this.counter)
+// 		return
+// 	}
+// 	// 如果不存在且缓存满了，需要删除。在 hashmap 和 pq 中删除。
+// 	if len(this.pq) == this.capacity {
+// 		item := heap.Pop(&this.pq).(*Item)
+// 		delete(this.hash, item.key)
+// 	}
+// 	// 新建结点，在 hashmap 和 pq 中添加。
+// 	item := &Item{
+// 		value: value,
+// 		key:   key,
+// 		count: this.counter,
+// 	}
+// 	heap.Push(&this.pq, item)
+// 	this.hash[key] = item
+// }
+
+// // An Item is something we manage in a priority queue.
+// type Item struct {
+// 	value     int // The value of the item; arbitrary.
+// 	key       int
+// 	frequency int // The priority of the item in the queue.
+// 	count     int // use for evicting the oldest element
+// 	// The index is needed by update and is maintained by the heap.Interface methods.
+// 	index int // The index of the item in the heap.
+// }
+
+// // A PriorityQueue implements heap.Interface and holds Items.
+// type PriorityQueue []*Item
+
+// func (pq PriorityQueue) Len() int { return len(pq) }
+
+// func (pq PriorityQueue) Less(i, j int) bool {
+// 	// We want Pop to give us the highest, not lowest, priority so we use greater than here.
+// 	if pq[i].frequency == pq[j].frequency {
+// 		return pq[i].count < pq[j].count
+// 	}
+// 	return pq[i].frequency < pq[j].frequency
+// }
+
+// func (pq PriorityQueue) Swap(i, j int) {
+// 	pq[i], pq[j] = pq[j], pq[i]
+// 	pq[i].index = i
+// 	pq[j].index = j
+// }
+
+// func (pq *PriorityQueue) Push(x interface{}) {
+// 	n := len(*pq)
+// 	item := x.(*Item)
+// 	item.index = n
+// 	*pq = append(*pq, item)
+// }
+
+// func (pq *PriorityQueue) Pop() interface{} {
+// 	old := *pq
+// 	n := len(old)
+// 	item := old[n-1]
+// 	old[n-1] = nil  // avoid memory leak
+// 	item.index = -1 // for safety
+// 	*pq = old[0 : n-1]
+// 	return item
+// }
+
+// // update modifies the priority and value of an Item in the queue.
+// func (pq *PriorityQueue) update(item *Item, value int, frequency int, count int) {
+// 	item.value = value
+// 	item.count = count
+// 	item.frequency = frequency
+// 	heap.Fix(pq, item.index)
+// }
diff --git a/template/LRUCache.go b/template/LRUCache.go
new file mode 100644
index 000000000..581f42a8c
--- /dev/null
+++ b/template/LRUCache.go
@@ -0,0 +1,139 @@
+package template
+
+// LRUCache define
+type LRUCache struct {
+	head, tail *Node
+	keys       map[int]*Node
+	capacity   int
+}
+
+// Node define
+type Node struct {
+	key, val   int
+	prev, next *Node
+}
+
+// ConstructorLRU define
+func ConstructorLRU(capacity int) LRUCache {
+	return LRUCache{keys: make(map[int]*Node), capacity: capacity}
+}
+
+// Get define
+func (lruCache *LRUCache) Get(key int) int {
+	if node, ok := lruCache.keys[key]; ok {
+		lruCache.Remove(node)
+		lruCache.Add(node)
+		return node.val
+	}
+	return -1
+}
+
+// Put define
+func (lruCache *LRUCache) Put(key int, value int) {
+	node, ok := lruCache.keys[key]
+	if ok {
+		node.val = value
+		lruCache.Remove(node)
+		lruCache.Add(node)
+		return
+	}
+	node = &Node{key: key, val: value}
+	lruCache.keys[key] = node
+	lruCache.Add(node)
+	if len(lruCache.keys) > lruCache.capacity {
+		delete(lruCache.keys, lruCache.tail.key)
+		lruCache.Remove(lruCache.tail)
+	}
+}
+
+// Add define
+func (lruCache *LRUCache) Add(node *Node) {
+	node.prev = nil
+	node.next = lruCache.head
+	if lruCache.head != nil {
+		lruCache.head.prev = node
+	}
+	lruCache.head = node
+	if lruCache.tail == nil {
+		lruCache.tail = node
+		lruCache.tail.next = nil
+	}
+}
+
+// Remove define
+func (lruCache *LRUCache) Remove(node *Node) {
+	if node == lruCache.head {
+		lruCache.head = node.next
+		if node.next != nil {
+			node.next.prev = nil
+		}
+		node.next = nil
+		return
+	}
+	if node == lruCache.tail {
+		lruCache.tail = node.prev
+		node.prev.next = nil
+		node.prev = nil
+		return
+	}
+	node.prev.next = node.next
+	node.next.prev = node.prev
+}
+
+/**
+ * Your LRUCache object will be instantiated and called as such:
+ * obj := Constructor(capacity);
+ * param_1 := obj.Get(key);
+ * obj.Put(key,value);
+ */
+
+// 22%
+// import "container/list"
+
+// type LRUCache struct {
+//     Cap  int
+//     Keys map[int]*list.Element
+//     List *list.List
+// }
+
+// type pair struct {
+//     K, V int
+// }
+
+// func Constructor(capacity int) LRUCache {
+//     return LRUCache{
+//         Cap: capacity,
+//         Keys: make(map[int]*list.Element),
+//         List: list.New(),
+//     }
+// }
+
+// func (c *LRUCache) Get(key int) int {
+//     if el, ok := c.Keys[key]; ok {
+//         c.List.MoveToFront(el)
+//         return el.Value.(pair).V
+//     }
+//     return -1
+// }
+
+// func (c *LRUCache) Put(key int, value int) {
+// 	if el, ok := c.Keys[key]; ok {
+// 		el.Value = pair{K: key, V: value}
+// 		c.List.MoveToFront(el)
+// 	} else {
+// 		el := c.List.PushFront(pair{K: key, V: value})
+// 		c.Keys[key] = el
+// 	}
+// 	if c.List.Len() > c.Cap {
+// 		el := c.List.Back()
+// 		c.List.Remove(el)
+// 		delete(c.Keys, el.Value.(pair).K)
+// 	}
+// }
+
+/**
+ * Your LRUCache object will be instantiated and called as such:
+ * obj := Constructor(capacity);
+ * param_1 := obj.Get(key);
+ * obj.Put(key,value);
+ */
diff --git a/template/bucket.go b/template/bucket.go
new file mode 100644
index 000000000..ac9ab9289
--- /dev/null
+++ b/template/bucket.go
@@ -0,0 +1,55 @@
+package template
+
+import (
+	"container/list"
+	"sync"
+)
+
+type bucket struct {
+	sync.RWMutex
+	keys map[string]*list.Element
+}
+
+func (b *bucket) pairCount() int {
+	b.RLock()
+	defer b.RUnlock()
+	return len(b.keys)
+}
+
+func (b *bucket) get(key string) *list.Element {
+	b.RLock()
+	defer b.RUnlock()
+	if el, ok := b.keys[key]; ok {
+		return el
+	}
+	return nil
+}
+
+func (b *bucket) set(key string, value interface{}) (*list.Element, *list.Element) {
+	el := &list.Element{Value: Pair{key: key, value: value, cmd: PushFront}}
+	b.Lock()
+	exist := b.keys[key]
+	b.keys[key] = el
+	b.Unlock()
+	return el, exist
+}
+
+func (b *bucket) update(key string, el *list.Element) {
+	b.Lock()
+	b.keys[key] = el
+	b.Unlock()
+}
+
+func (b *bucket) delete(key string) *list.Element {
+	b.Lock()
+	el := b.keys[key]
+	delete(b.keys, key)
+	b.Unlock()
+	return el
+}
+
+func (b *bucket) clear() {
+	b.Lock()
+	b.keys = make(map[string]*list.Element)
+	b.Unlock()
+}
diff --git a/template/go.mod b/template/go.mod
new file mode 100644
index 000000000..6ef287eab
--- /dev/null
+++ b/template/go.mod
@@ -0,0 +1,11 @@
+module github.com/halfrost/LeetCode-Go/template
+
+go 1.19
+
+require github.com/stretchr/testify v1.8.0
+
+require (
+	github.com/davecgh/go-spew v1.1.1 // indirect
+	github.com/pmezard/go-difflib v1.0.0 // indirect
+	gopkg.in/yaml.v3 v3.0.1 // indirect
+)
diff --git a/template/go.sum b/template/go.sum
new file mode 100644
index 000000000..51648299d
--- /dev/null
+++ b/template/go.sum
@@ -0,0 +1,15 @@
+github.com/davecgh/go-spew v1.1.0/go.mod h1:J7Y8YcW2NihsgmVo/mv3lAwl/skON4iLHjSsI+c5H38=
+github.com/davecgh/go-spew v1.1.1 h1:vj9j/u1bqnvCEfJOwUhtlOARqs3+rkHYY13jYWTU97c=
+github.com/davecgh/go-spew v1.1.1/go.mod h1:J7Y8YcW2NihsgmVo/mv3lAwl/skON4iLHjSsI+c5H38=
+github.com/pmezard/go-difflib v1.0.0 h1:4DBwDE0NGyQoBHbLQYPwSUPoCMWR5BEzIk/f1lZbAQM=
+github.com/pmezard/go-difflib v1.0.0/go.mod h1:iKH77koFhYxTK1pcRnkKkqfTogsbg7gZNVY4sRDYZ/4=
+github.com/stretchr/objx v0.1.0/go.mod h1:HFkY916IF+rwdDfMAkV7OtwuqBVzrE8GR6GFx+wExME=
+github.com/stretchr/objx v0.4.0/go.mod h1:YvHI0jy2hoMjB+UWwv71VJQ9isScKT/TqJzVSSt89Yw=
+github.com/stretchr/testify v1.7.1/go.mod h1:6Fq8oRcR53rry900zMqJjRRixrwX3KX962/h/Wwjteg=
+github.com/stretchr/testify v1.8.0 h1:pSgiaMZlXftHpm5L7V1+rVB+AZJydKsMxsQBIJw4PKk=
+github.com/stretchr/testify v1.8.0/go.mod h1:yNjHg4UonilssWZ8iaSj1OCr/vHnekPRkoO+kdMU+MU=
+gopkg.in/check.v1 v0.0.0-20161208181325-20d25e280405 h1:yhCVgyC4o1eVCa2tZl7eS0r+SDo693bJlVdllGtEeKM=
+gopkg.in/check.v1 v0.0.0-20161208181325-20d25e280405/go.mod h1:Co6ibVJAznAaIkqp8huTwlJQCZ016jof/cbN4VW5Yz0=
+gopkg.in/yaml.v3 v3.0.0-20200313102051-9f266ea9e77c/go.mod h1:K4uyk7z7BCEPqu6E+C64Yfv1cQ7kz7rIZviUmN+EgEM=
+gopkg.in/yaml.v3 v3.0.1 h1:fxVm/GzAzEWqLHuvctI91KS9hhNmmWOoWu0XTYJS7CA=
+gopkg.in/yaml.v3 v3.0.1/go.mod h1:K4uyk7z7BCEPqu6E+C64Yfv1cQ7kz7rIZviUmN+EgEM=
diff --git a/topic/Backtracking.png b/topic/Backtracking.png
new file mode 100644
index 000000000..3bf789ec4
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diff --git a/topic/Binary_Indexed_Tree.png b/topic/Binary_Indexed_Tree.png
new file mode 100644
index 000000000..9aab91717
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diff --git a/topic/Bit_Manipulation.png b/topic/Bit_Manipulation.png
new file mode 100644
index 000000000..1e08eb96a
Binary files /dev/null and b/topic/Bit_Manipulation.png differ
diff --git a/topic/Linked_List.png b/topic/Linked_List.png
new file mode 100644
index 000000000..7382943ab
Binary files /dev/null and b/topic/Linked_List.png differ
diff --git a/Topics/Segment_Tree.png b/topic/Segment_Tree.png
similarity index 100%
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diff --git a/website/archetypes/default.md b/website/archetypes/default.md
new file mode 100644
index 000000000..00e77bd79
--- /dev/null
+++ b/website/archetypes/default.md
@@ -0,0 +1,6 @@
+---
+title: "{{ replace .Name "-" " " | title }}"
+date: {{ .Date }}
+draft: true
+---
+
diff --git a/website/config.toml b/website/config.toml
new file mode 100644
index 000000000..52297f0ba
--- /dev/null
+++ b/website/config.toml
@@ -0,0 +1,158 @@
+baseURL = "https://books.halfrost.com/leetcode"
+languageCode = "en-us"
+title = "LeetCode Cookbook"
+theme = "book"
+googleAnalytics = "UA-82753806-3"
+enableEmoji = true
+enableGitInfo = true
+
+# Book configuration
+disablePathToLower = true
+
+# Needed for mermaid/katex shortcodes
+[markup]
+[markup.goldmark.renderer]
+  unsafe = true
+
+[markup.tableOfContents]
+  startLevel = 1
+
+# Multi-lingual mode config
+# There are different options to translate files
+# See https://gohugo.io/content-management/multilingual/#translation-by-filename
+# And https://gohugo.io/content-management/multilingual/#translation-by-content-directory
+#[languages]
+#[languages.en]
+#  languageName = 'English'
+#  contentDir = 'content'
+#  weight = 1
+
+#[languages.ru]
+#  languageName = 'Russian'
+#  contentDir = 'content.ru'
+#  weight = 2
+
+#[languages.cn]
+#  languageName = 'Chinese'
+#  contentDir = 'content.cn'
+#  weight = 3
+
+[menu]
+[[menu.before]]
+  name = "Blog"
+  url = "https://halfrost.com"
+  weight = 10
+[[menu.before]]
+  name = "Github"
+  url = "https://github.com/halfrost"
+  weight = 20
+[[menu.before]]
+  name = "WeChat Official Accounts"
+  url = "https://img.halfrost.com/wechat-qr-code.png"
+  weight = 30
+[[menu.before]]
+  name = "Twitter"
+  url = "https://twitter.com/halffrost"
+  weight = 40
+[[menu.before]]
+  name = "Weibo"
+  url = "https://weibo.com/halfrost"
+  weight = 50
+#[[menu.after]]
+#  name = "Github"
+#  url = "https://github.com/alex-shpak/hugo-book"
+#  weight = 10
+
+#[[menu.after]]
+#  name = "Hugo Themes"
+#  url = "https://themes.gohugo.io/hugo-book/"
+#  weight = 20
+#[[menu.before]]
+#  name = "微信公众号"
+#  url = "https://halfrost.com"
+#  weight = 30
+
+[params]
+  # (Optional, default light) Sets color theme: light, dark or auto.
+  # Theme 'auto' switches between dark and light modes based on browser/os preferences
+  BookTheme = 'light'
+
+  # (Optional, default true) Controls table of contents visibility on right side of pages.
+  # Start and end levels can be controlled with markup.tableOfContents setting.
+  # You can also specify this parameter per page in front matter.
+  BookToC = true
+
+  # (Optional, default none) Set the path to a logo for the book. If the logo is
+  # /static/logo.png then the path would be logo.png
+  # BookLogo = 'logo.png'
+
+  # (Optional, default none) Set leaf bundle to render as side menu
+  # When not specified file structure and weights will be used
+  # BookMenuBundle = '/menu'
+
+  # (Optional, default docs) Specify section of content to render as menu
+  # You can also set value to '*' to render all sections to menu
+  BookSection = '*'
+
+  # Set source repository location.
+  # Used for 'Last Modified' and 'Edit this page' links.
+  BookRepo = 'https://github.com/halfrost/LeetCode-Go'
+
+  # Specifies commit portion of the link to the page's last modified commit hash for 'doc' page
+  # type.
+  # Required if 'BookRepo' param is set.
+  # Value used to construct a URL consisting of BookRepo/BookCommitPath/<commit-hash>
+  # Github uses 'commit', Bitbucket uses 'commits'
+  BookCommitPath = 'commit'
+
+  # Enable "Edit this page" links for 'doc' page type.
+  # Disabled by default. Uncomment to enable. Requires 'BookRepo' param.
+  # Path must point to 'content' directory of repo.
+  BookEditPath = 'tree/master/website/'
+
+  # Configure the date format used on the pages
+  # - In git information
+  # - In blog posts
+  BookDateFormat = 'Jan 2, 2006'
+
+  # (Optional, default true) Enables search function with flexsearch,
+  # Index is built on fly, therefore it might slowdown your website.
+  # Configuration for indexing can be adjusted in i18n folder per language.
+  BookSearch = true
+
+  # (Optional, default true) Enables comments template on pages
+  # By default partals/docs/comments.html includes Disqus template
+  # See https://gohugo.io/content-management/comments/#configure-disqus
+  # Can be overwritten by same param in page frontmatter
+  BookComments = false
+  enableGitalk = true
+
+  # /!\ This is an experimental feature, might be removed or changed at any time
+  # (Optional, experimental, default false) Enables portable links and link checks in markdown pages.
+  # Portable links meant to work with text editors and let you write markdown without {{< relref >}} shortcode
+  # Theme will print warning if page referenced in markdown does not exists.
+  BookPortableLinks = true
+
+  # /!\ This is an experimental feature, might be removed or changed at any time
+  # (Optional, experimental, default false) Enables service worker that caches visited pages and resources for offline use.
+  BookServiceWorker = true
+
+  # /!\ This is an experimental feature, might be removed or changed at any time
+  # (Optional, experimental, default false) Enables a drop-down menu for translations only if a translation is present.
+  BookTranslatedOnly = false
+
+[params.gitalk] 
+    clientID = "75d9d747f200c623a0e6" # Your client ID
+    clientSecret = "e3cc2ff8d8f6ae3c894a827f226188de0d26f37e" # Your client secret
+    repo = "S2" # The repo to store comments
+    owner = "halfrost" # Your GitHub ID
+    admin= "halfrost" # Required. Github repository owner and collaborators. (Users who having write access to this repository)
+    id= "location.pathname" # The unique id of the page.
+    labels= "gitalk" # Github issue labels. If you used to use Gitment, you can change it
+    body= "location.href" # GitHub issue body.
+    perPage= 100 # Pagination size, with maximum 100.
+    pagerDirection= "last" # Comment sorting direction, available values are 'last' and 'first'.
+    createIssueManually= true # If it is 'false', it is auto to make a Github issue when the administrators login.
+    distractionFreeMode= true # Enable hot key (cmd|ctrl + enter) submit comment.
+    proxy= "https://shielded-brushlands-08810.herokuapp.com/https://github.com/login/oauth/access_token"
+
diff --git a/website/content/ChapterFour/0001~0099/0001.Two-Sum.md b/website/content/ChapterFour/0001~0099/0001.Two-Sum.md
new file mode 100644
index 000000000..d84bc7af1
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0001.Two-Sum.md
@@ -0,0 +1,57 @@
+# [1. Two Sum](https://leetcode.com/problems/two-sum/)
+
+## 题目
+
+Given an array of integers, return indices of the two numbers such that they add up to a specific target.
+
+You may assume that each input would have exactly one solution, and you may not use the same element twice.
+
+**Example**:
+
+```
+
+Given nums = [2, 7, 11, 15], target = 9,
+
+Because nums[0] + nums[1] = 2 + 7 = 9,
+return [0, 1]
+
+```
+
+
+
+## 题目大意
+
+在数组中找到 2 个数之和等于给定值的数字，结果返回 2 个数字在数组中的下标。
+
+## 解题思路
+
+这道题最优的做法时间复杂度是 O(n)。
+
+顺序扫描数组，对每一个元素，在 map 中找能组合给定值的另一半数字，如果找到了，直接返回 2 个数字的下标即可。如果找不到，就把这个数字存入 map 中，等待扫到“另一半”数字的时候，再取出来返回结果。
+
+## 代码
+
+```go
+
+package leetcode
+
+func twoSum(nums []int, target int) []int {
+	m := make(map[int]int)
+	for i := 0; i < len(nums); i++ {
+		another := target - nums[i]
+		if _, ok := m[another]; ok {
+			return []int{m[another], i}
+		}
+		m[nums[i]] = i
+	}
+	return nil
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0002.Add-Two-Numbers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0002.Add-Two-Numbers.md b/website/content/ChapterFour/0001~0099/0002.Add-Two-Numbers.md
new file mode 100644
index 000000000..a64c6a301
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0002.Add-Two-Numbers.md
@@ -0,0 +1,85 @@
+# [2. Add Two Numbers](https://leetcode.com/problems/add-two-numbers/)
+
+## 题目
+
+You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
+
+You may assume the two numbers do not contain any leading zero, except the number 0 itself.
+
+**Example**:
+
+```
+
+Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
+Output: 7 -> 0 -> 8
+Explanation: 342 + 465 = 807.
+```
+
+
+
+## 题目大意
+
+2 个逆序的链表，要求从低位开始相加，得出结果也逆序输出，返回值是逆序结果链表的头结点。
+
+## 解题思路
+
+需要注意的是各种进位问题。
+
+极端情况，例如
+
+```
+
+Input: (9 -> 9 -> 9 -> 9 -> 9) + (1 -> )
+Output: 0 -> 0 -> 0 -> 0 -> 0 -> 1
+
+
+```
+
+为了处理方法统一，可以先建立一个虚拟头结点，这个虚拟头结点的 Next 指向真正的 head，这样 head 不需要单独处理，直接 while 循环即可。另外判断循环终止的条件不用是 p.Next ！= nil，这样最后一位还需要额外计算，循环终止条件应该是 p != nil。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
+	head := &ListNode{Val: 0}
+	n1, n2, carry, current := 0, 0, 0, head
+	for l1 != nil || l2 != nil || carry != 0 {
+		if l1 == nil {
+			n1 = 0
+		} else {
+			n1 = l1.Val
+			l1 = l1.Next
+		}
+		if l2 == nil {
+			n2 = 0
+		} else {
+			n2 = l2.Val
+			l2 = l2.Next
+		}
+		current.Next = &ListNode{Val: (n1 + n2 + carry) % 10}
+		current = current.Next
+		carry = (n1 + n2 + carry) / 10
+	}
+	return head.Next
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0001.Two-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0003.Longest-Substring-Without-Repeating-Characters/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0003.Longest-Substring-Without-Repeating-Characters.md b/website/content/ChapterFour/0001~0099/0003.Longest-Substring-Without-Repeating-Characters.md
new file mode 100644
index 000000000..f86f9d4d4
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0003.Longest-Substring-Without-Repeating-Characters.md
@@ -0,0 +1,146 @@
+# [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/)
+
+## 题目
+
+Given a string, find the length of the longest substring without repeating characters.
+
+
+
+**Example 1**:
+
+```
+
+Input: "abcabcbb"
+Output: 3 
+Explanation: The answer is "abc", with the length of 3. 
+
+```
+
+**Example 2**:
+
+```
+
+Input: "bbbbb"
+Output: 1
+Explanation: The answer is "b", with the length of 1.
+
+```
+
+**Example 3**:
+
+```
+
+Input: "pwwkew"
+Output: 3
+Explanation: The answer is "wke", with the length of 3. 
+             Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
+
+```
+
+## 题目大意
+
+
+在一个字符串重寻找没有重复字母的最长子串。
+
+## 解题思路
+
+这一题和第 438 题，第 3 题，第 76 题，第 567 题类似，用的思想都是"滑动窗口"。
+
+滑动窗口的右边界不断的右移，只要没有重复的字符，就持续向右扩大窗口边界。一旦出现了重复字符，就需要缩小左边界，直到重复的字符移出了左边界，然后继续移动滑动窗口的右边界。以此类推，每次移动需要计算当前长度，并判断是否需要更新最大长度，最终最大的值就是题目中的所求。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 位图
+func lengthOfLongestSubstring(s string) int {
+	if len(s) == 0 {
+		return 0
+	}
+	var bitSet [256]bool
+	result, left, right := 0, 0, 0
+	for left < len(s) {
+		// 右侧字符对应的 bitSet 被标记 true，说明此字符在 X 位置重复，需要左侧向前移动，直到将 X 标记为 false
+		if bitSet[s[right]] {
+			bitSet[s[left]] = false
+			left++
+		} else {
+			bitSet[s[right]] = true
+			right++
+		}
+		if result < right-left {
+			result = right - left
+		}
+		if left+result >= len(s) || right >= len(s) {
+			break
+		}
+	}
+	return result
+}
+
+// 解法二 滑动窗口
+func lengthOfLongestSubstring1(s string) int {
+	if len(s) == 0 {
+		return 0
+	}
+	var freq [127]int
+	result, left, right := 0, 0, -1
+
+	for left < len(s) {
+		if right+1 < len(s) && freq[s[right+1]] == 0 {
+			freq[s[right+1]]++
+			right++
+
+		} else {
+			freq[s[left]]--
+			left++
+		}
+		result = max(result, right-left+1)
+	}
+	return result
+}
+
+// 解法三 滑动窗口-哈希桶
+func lengthOfLongestSubstring2(s string) int {
+	right, left, res := 0, 0, 0
+	indexes := make(map[byte]int, len(s))
+	for left < len(s) {
+		if idx, ok := indexes[s[left]]; ok && idx >= right {
+			right = idx + 1
+		}
+		indexes[s[left]] = left
+		left++
+		res = max(res, left-right)
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+```
+
+
+
+
+
+
+
+
+
+
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0002.Add-Two-Numbers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0004.Median-of-Two-Sorted-Arrays/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0004.Median-of-Two-Sorted-Arrays.md b/website/content/ChapterFour/0001~0099/0004.Median-of-Two-Sorted-Arrays.md
new file mode 100755
index 000000000..ddca884df
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0004.Median-of-Two-Sorted-Arrays.md
@@ -0,0 +1,105 @@
+# [4. Median of Two Sorted Arrays](https://leetcode.com/problems/median-of-two-sorted-arrays/)
+
+
+## 题目
+
+There are two sorted arrays **nums1** and **nums2** of size m and n respectively.
+
+Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
+
+You may assume **nums1** and **nums2** cannot be both empty.
+
+**Example 1**:
+
+    nums1 = [1, 3]
+    nums2 = [2]
+    
+    The median is 2.0
+
+**Example 2**:
+
+    nums1 = [1, 2]
+    nums2 = [3, 4]
+    
+    The median is (2 + 3)/2 = 2.5
+
+
+
+## 题目大意
+
+
+给定两个大小为 m 和 n 的有序数组 nums1 和 nums2。
+
+请你找出这两个有序数组的中位数，并且要求算法的时间复杂度为 O(log(m + n))。
+
+你可以假设 nums1 和 nums2 不会同时为空。
+
+
+
+## 解题思路
+
+
+- 给出两个有序数组，要求找出这两个数组合并以后的有序数组中的中位数。要求时间复杂度为 O(log (m+n))。
+- 这一题最容易想到的办法是把两个数组合并，然后取出中位数。但是合并有序数组的操作是 `O(m+n)` 的，不符合题意。看到题目给的 `log` 的时间复杂度，很容易联想到二分搜索。
+- 由于要找到最终合并以后数组的中位数，两个数组的总大小也知道，所以中间这个位置也是知道的。只需要二分搜索一个数组中切分的位置，另一个数组中切分的位置也能得到。为了使得时间复杂度最小，所以二分搜索两个数组中长度较小的那个数组。
+- 关键的问题是如何切分数组 1 和数组 2 。其实就是如何切分数组 1 。先随便二分产生一个 `midA`，切分的线何时算满足了中位数的条件呢？即，线左边的数都小于右边的数，即，`nums1[midA-1] ≤ nums2[midB] && nums2[midB-1] ≤ nums1[midA]` 。如果这些条件都不满足，切分线就需要调整。如果 `nums1[midA] < nums2[midB-1]`，说明 `midA` 这条线划分出来左边的数小了，切分线应该右移；如果 `nums1[midA-1] > nums2[midB]`，说明 midA 这条线划分出来左边的数大了，切分线应该左移。经过多次调整以后，切分线总能找到满足条件的解。
+- 假设现在找到了切分的两条线了，`数组 1` 在切分线两边的下标分别是 `midA - 1` 和 `midA`。`数组 2` 在切分线两边的下标分别是 `midB - 1` 和 `midB`。最终合并成最终数组，如果数组长度是奇数，那么中位数就是 `max(nums1[midA-1], nums2[midB-1])`。如果数组长度是偶数，那么中间位置的两个数依次是：`max(nums1[midA-1], nums2[midB-1])` 和 `min(nums1[midA], nums2[midB])`，那么中位数就是 `(max(nums1[midA-1], nums2[midB-1]) + min(nums1[midA], nums2[midB])) / 2`。图示见下图：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_4.png)
+    
+## 代码
+
+```go
+
+package leetcode
+
+func findMedianSortedArrays(nums1 []int, nums2 []int) float64 {
+	// 假设 nums1 的长度小
+	if len(nums1) > len(nums2) {
+		return findMedianSortedArrays(nums2, nums1)
+	}
+	low, high, k, nums1Mid, nums2Mid := 0, len(nums1), (len(nums1)+len(nums2)+1)>>1, 0, 0
+	for low <= high {
+		// nums1:  ……………… nums1[nums1Mid-1] | nums1[nums1Mid] ……………………
+		// nums2:  ……………… nums2[nums2Mid-1] | nums2[nums2Mid] ……………………
+		nums1Mid = low + (high-low)>>1 // 分界限右侧是 mid，分界线左侧是 mid - 1
+		nums2Mid = k - nums1Mid
+		if nums1Mid > 0 && nums1[nums1Mid-1] > nums2[nums2Mid] { // nums1 中的分界线划多了，要向左边移动
+			high = nums1Mid - 1
+		} else if nums1Mid != len(nums1) && nums1[nums1Mid] < nums2[nums2Mid-1] { // nums1 中的分界线划少了，要向右边移动
+			low = nums1Mid + 1
+		} else {
+			// 找到合适的划分了，需要输出最终结果了
+			// 分为奇数偶数 2 种情况
+			break
+		}
+	}
+	midLeft, midRight := 0, 0
+	if nums1Mid == 0 {
+		midLeft = nums2[nums2Mid-1]
+	} else if nums2Mid == 0 {
+		midLeft = nums1[nums1Mid-1]
+	} else {
+		midLeft = max(nums1[nums1Mid-1], nums2[nums2Mid-1])
+	}
+	if (len(nums1)+len(nums2))&1 == 1 {
+		return float64(midLeft)
+	}
+	if nums1Mid == len(nums1) {
+		midRight = nums2[nums2Mid]
+	} else if nums2Mid == len(nums2) {
+		midRight = nums1[nums1Mid]
+	} else {
+		midRight = min(nums1[nums1Mid], nums2[nums2Mid])
+	}
+	return float64(midLeft+midRight) / 2
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0003.Longest-Substring-Without-Repeating-Characters/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0005.Longest-Palindromic-Substring/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0005.Longest-Palindromic-Substring.md b/website/content/ChapterFour/0001~0099/0005.Longest-Palindromic-Substring.md
new file mode 100644
index 000000000..1adffa4b2
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0005.Longest-Palindromic-Substring.md
@@ -0,0 +1,193 @@
+# [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)
+
+
+## 题目
+
+Given a string `s`, return *the longest palindromic substring* in `s`.
+
+**Example 1:**
+
+```
+Input: s = "babad"
+Output: "bab"
+Note: "aba" is also a valid answer.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "cbbd"
+Output: "bb"
+
+```
+
+**Example 3:**
+
+```
+Input: s = "a"
+Output: "a"
+
+```
+
+**Example 4:**
+
+```
+Input: s = "ac"
+Output: "a"
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 1000`
+- `s` consist of only digits and English letters (lower-case and/or upper-case),
+
+## 题目大意
+
+给你一个字符串 `s`，找到 `s` 中最长的回文子串。
+
+## 解题思路
+
+- 此题非常经典，并且有多种解法。
+- 解法一，动态规划。定义 `dp[i][j]` 表示从字符串第 `i` 个字符到第 `j` 个字符这一段子串是否是回文串。由回文串的性质可以得知，回文串去掉一头一尾相同的字符以后，剩下的还是回文串。所以状态转移方程是 `dp[i][j] = (s[i] == s[j]) && ((j-i < 3) || dp[i+1][j-1])`，注意特殊的情况，`j - i == 1` 的时候，即只有 2 个字符的情况，只需要判断这 2 个字符是否相同即可。`j - i == 2` 的时候，即只有 3 个字符的情况，只需要判断除去中心以外对称的 2 个字符是否相等。每次循环动态维护保存最长回文串即可。时间复杂度 O(n^2)，空间复杂度 O(n^2)。
+- 解法二，中心扩散法。动态规划的方法中，我们将任意起始，终止范围内的字符串都判断了一遍。其实没有这个必要，如果不是最长回文串，无需判断并保存结果。所以动态规划的方法在空间复杂度上还有优化空间。判断回文有一个核心问题是找到“轴心”。如果长度是偶数，那么轴心是中心虚拟的，如果长度是奇数，那么轴心正好是正中心的那个字母。中心扩散法的思想是枚举每个轴心的位置。然后做两次假设，假设最长回文串是偶数，那么以虚拟中心往 2 边扩散；假设最长回文串是奇数，那么以正中心的字符往 2 边扩散。扩散的过程就是对称判断两边字符是否相等的过程。这个方法时间复杂度和动态规划是一样的，但是空间复杂度降低了。时间复杂度 O(n^2)，空间复杂度 O(1)。
+- 解法三，滑动窗口。这个写法其实就是中心扩散法变了一个写法。中心扩散是依次枚举每一个轴心。滑动窗口的方法稍微优化了一点，有些轴心两边字符不相等，下次就不会枚举这些不可能形成回文子串的轴心了。不过这点优化并没有优化时间复杂度，时间复杂度 O(n^2)，空间复杂度 O(1)。
+- 解法四，马拉车算法。这个算法是本题的最优解，也是最复杂的解法。时间复杂度 O(n)，空间复杂度 O(n)。中心扩散法有 2 处有重复判断，第一处是每次都往两边扩散，不同中心扩散多次，实际上有很多重复判断的字符，能否不重复判断？第二处，中心能否跳跃选择，不是每次都枚举，是否可以利用前一次的信息，跳跃选择下一次的中心？马拉车算法针对重复判断的问题做了优化，增加了一个辅助数组，将时间复杂度从 O(n^2) 优化到了 O(n)，空间换了时间，空间复杂度增加到 O(n)。
+
+    ![https://img.halfrost.com/Leetcode/leetcode_5_1.png](https://img.halfrost.com/Leetcode/leetcode_5_1.png)
+
+- 首先是预处理，向字符串的头尾以及每两个字符中间添加一个特殊字符 `#`，比如字符串 `aaba` 处理后会变成 `#a#a#b#a#`。那么原先长度为偶数的回文字符串 `aa` 会变成长度为奇数的回文字符串 `#a#a#`，而长度为奇数的回文字符串 `aba` 会变成长度仍然为奇数的回文字符串 `#a#b#a#`，经过预处理以后，都会变成长度为奇数的字符串。**注意这里的特殊字符不需要是没有出现过的字母，也可以使用任何一个字符来作为这个特殊字符。**这是因为，当我们只考虑长度为奇数的回文字符串时，每次我们比较的两个字符奇偶性一定是相同的，所以原来字符串中的字符不会与插入的特殊字符互相比较，不会因此产生问题。**预处理以后，以某个中心扩散的步数和实际字符串长度是相等的。**因为半径里面包含了插入的特殊字符，又由于左右对称的性质，所以扩散半径就等于原来回文子串的长度。
+
+    ![https://img.halfrost.com/Leetcode/leetcode_5_2.png](https://img.halfrost.com/Leetcode/leetcode_5_2.png)
+
+- 核心部分是如何通过左边已经扫描过的数据推出右边下一次要扩散的中心。这里定义下一次要扩散的中心下标是 `i`。如果 `i` 比 `maxRight` 要大，只能继续中心扩散。如果 `i` 比 `maxRight` 小，这时又分为 3 种情况。三种情况见上图。将上述 3 种情况总结起来，就是 ：`dp[i] = min(maxRight-i, dp[2*center-i])`，其中，`mirror` 相对于 `center` 是和 `i` 中心对称的，所以它的下标可以计算出来是 `2*center-i`。更新完 `dp[i]` 以后，就要进行中心扩散了。中心扩散以后动态维护最长回文串并相应的更新 `center`，`maxRight`，并且记录下原始字符串的起始位置 `begin` 和 `maxLen`。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 Manacher's algorithm，时间复杂度 O(n)，空间复杂度 O(n)
+func longestPalindrome(s string) string {
+	if len(s) < 2 {
+		return s
+	}
+	newS := make([]rune, 0)
+	newS = append(newS, '#')
+	for _, c := range s {
+		newS = append(newS, c)
+		newS = append(newS, '#')
+	}
+	// dp[i]:    以预处理字符串下标 i 为中心的回文半径(奇数长度时不包括中心)
+	// maxRight: 通过中心扩散的方式能够扩散的最右边的下标
+	// center:   与 maxRight 对应的中心字符的下标
+	// maxLen:   记录最长回文串的半径
+	// begin:    记录最长回文串在起始串 s 中的起始下标
+	dp, maxRight, center, maxLen, begin := make([]int, len(newS)), 0, 0, 1, 0
+	for i := 0; i < len(newS); i++ {
+		if i < maxRight {
+			// 这一行代码是 Manacher 算法的关键所在
+			dp[i] = min(maxRight-i, dp[2*center-i])
+		}
+		// 中心扩散法更新 dp[i]
+		left, right := i-(1+dp[i]), i+(1+dp[i])
+		for left >= 0 && right < len(newS) && newS[left] == newS[right] {
+			dp[i]++
+			left--
+			right++
+		}
+		// 更新 maxRight，它是遍历过的 i 的 i + dp[i] 的最大者
+		if i+dp[i] > maxRight {
+			maxRight = i + dp[i]
+			center = i
+		}
+		// 记录最长回文子串的长度和相应它在原始字符串中的起点
+		if dp[i] > maxLen {
+			maxLen = dp[i]
+			begin = (i - maxLen) / 2 // 这里要除以 2 因为有我们插入的辅助字符 #
+		}
+	}
+	return s[begin : begin+maxLen]
+}
+
+func min(x, y int) int {
+	if x < y {
+		return x
+	}
+	return y
+}
+
+// 解法二 滑动窗口，时间复杂度 O(n^2)，空间复杂度 O(1)
+func longestPalindrome1(s string) string {
+	if len(s) == 0 {
+		return ""
+	}
+	left, right, pl, pr := 0, -1, 0, 0
+	for left < len(s) {
+		// 移动到相同字母的最右边（如果有相同字母）
+		for right+1 < len(s) && s[left] == s[right+1] {
+			right++
+		}
+		// 找到回文的边界
+		for left-1 >= 0 && right+1 < len(s) && s[left-1] == s[right+1] {
+			left--
+			right++
+		}
+		if right-left > pr-pl {
+			pl, pr = left, right
+		}
+		// 重置到下一次寻找回文的中心
+		left = (left+right)/2 + 1
+		right = left
+	}
+	return s[pl : pr+1]
+}
+
+// 解法三 中心扩散法，时间复杂度 O(n^2)，空间复杂度 O(1)
+func longestPalindrome2(s string) string {
+	res := ""
+	for i := 0; i < len(s); i++ {
+		res = maxPalindrome(s, i, i, res)
+		res = maxPalindrome(s, i, i+1, res)
+	}
+	return res
+}
+
+func maxPalindrome(s string, i, j int, res string) string {
+	sub := ""
+	for i >= 0 && j < len(s) && s[i] == s[j] {
+		sub = s[i : j+1]
+		i--
+		j++
+	}
+	if len(res) < len(sub) {
+		return sub
+	}
+	return res
+}
+
+// 解法四 DP，时间复杂度 O(n^2)，空间复杂度 O(n^2)
+func longestPalindrome3(s string) string {
+	res, dp := "", make([][]bool, len(s))
+	for i := 0; i < len(s); i++ {
+		dp[i] = make([]bool, len(s))
+	}
+	for i := len(s) - 1; i >= 0; i-- {
+		for j := i; j < len(s); j++ {
+			dp[i][j] = (s[i] == s[j]) && ((j-i < 3) || dp[i+1][j-1])
+			if dp[i][j] && (res == "" || j-i+1 > len(res)) {
+				res = s[i : j+1]
+			}
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0004.Median-of-Two-Sorted-Arrays/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0006.ZigZag-Conversion/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0006.ZigZag-Conversion.md b/website/content/ChapterFour/0001~0099/0006.ZigZag-Conversion.md
new file mode 100644
index 000000000..8f61aedc3
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0006.ZigZag-Conversion.md
@@ -0,0 +1,114 @@
+# [6. ZigZag Conversion](https://leetcode.com/problems/zigzag-conversion/)
+
+
+## 题目
+
+The string `"PAYPALISHIRING"` is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
+
+```
+P   A   H   N
+A P L S I I G
+Y   I   R
+```
+
+And then read line by line: `"PAHNAPLSIIGYIR"`
+
+Write the code that will take a string and make this conversion given a number of rows:
+
+```
+string convert(string s, int numRows);
+```
+
+**Example 1:**
+
+```
+Input: s = "PAYPALISHIRING", numRows = 3
+Output: "PAHNAPLSIIGYIR"
+```
+
+**Example 2:**
+
+```
+Input: s = "PAYPALISHIRING", numRows = 4
+Output: "PINALSIGYAHRPI"
+Explanation:
+P     I    N
+A   L S  I G
+Y A   H R
+P     I
+```
+
+**Example 3:**
+
+```
+Input: s = "A", numRows = 1
+Output: "A"
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 1000`
+- `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
+- `1 <= numRows <= 1000`
+
+## 题目大意
+
+将一个给定字符串 `s` 根据给定的行数 `numRows` ，以从上往下、从左到右进行 Z 字形排列。
+
+比如输入字符串为 `"PAYPALISHIRING"` 行数为 3 时，排列如下：
+
+```go
+P   A   H   N
+A P L S I I G
+Y   I   R
+```
+
+之后，你的输出需要从左往右逐行读取，产生出一个新的字符串，比如：`"PAHNAPLSIIGYIR"`。
+
+请你实现这个将字符串进行指定行数变换的函数：
+
+```go
+string convert(string s, int numRows);
+```
+
+## 解题思路
+
+- 这一题没有什么算法思想，考察的是对程序控制的能力。用 2 个变量保存方向，当垂直输出的行数达到了规定的目标行数以后，需要从下往上转折到第一行，循环中控制好方向ji
+
+## 代码
+
+```go
+package leetcode
+
+func convert(s string, numRows int) string {
+	matrix, down, up := make([][]byte, numRows, numRows), 0, numRows-2
+	for i := 0; i != len(s); {
+		if down != numRows {
+			matrix[down] = append(matrix[down], byte(s[i]))
+			down++
+			i++
+		} else if up > 0 {
+			matrix[up] = append(matrix[up], byte(s[i]))
+			up--
+			i++
+		} else {
+			up = numRows - 2
+			down = 0
+		}
+	}
+	solution := make([]byte, 0, len(s))
+	for _, row := range matrix {
+		for _, item := range row {
+			solution = append(solution, item)
+		}
+	}
+	return string(solution)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0005.Longest-Palindromic-Substring/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0007.Reverse-Integer/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0007.Reverse-Integer.md b/website/content/ChapterFour/0001~0099/0007.Reverse-Integer.md
new file mode 100755
index 000000000..5f0b783fa
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0007.Reverse-Integer.md
@@ -0,0 +1,62 @@
+# [7. Reverse Integer](https://leetcode.com/problems/reverse-integer/)
+
+
+## 题目
+
+Given a 32-bit signed integer, reverse digits of an integer.
+
+**Example 1**:
+
+    Input: 123
+    Output: 321
+
+**Example 2**:
+
+    Input: -123
+    Output: -321
+
+**Example 3**:
+
+    Input: 120
+    Output: 21
+
+**Note**: Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
+
+## 题目大意
+
+给出一个 32 位的有符号整数，你需要将这个整数中每位上的数字进行反转。注意:假设我们的环境只能存储得下 32 位的有符号整数，则其数值范围为 [−2^31,  2^31 − 1]。请根据这个假设，如果反转后整数溢出那么就返回 0。
+
+
+
+## 解题思路
+
+
+- 这一题是简单题，要求反转 10 进制数。类似的题目有第 190 题。
+- 这一题只需要注意一点，反转以后的数字要求在 [−2^31, 2^31 − 1]范围内，超过这个范围的数字都要输出 0 。
+
+## 代码
+
+```go
+
+package leetcode
+
+func reverse7(x int) int {
+	tmp := 0
+	for x != 0 {
+		tmp = tmp*10 + x%10
+		x = x / 10
+	}
+	if tmp > 1<<31-1 || tmp < -(1<<31) {
+		return 0
+	}
+	return tmp
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0006.ZigZag-Conversion/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0008.String-to-Integer-atoi/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0008.String-to-Integer-atoi.md b/website/content/ChapterFour/0001~0099/0008.String-to-Integer-atoi.md
new file mode 100644
index 000000000..0f37b876a
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0008.String-to-Integer-atoi.md
@@ -0,0 +1,199 @@
+# [8. String to Integer (atoi)](https://leetcode.com/problems/string-to-integer-atoi/)
+
+
+## 题目
+
+Implement the `myAtoi(string s)` function, which converts a string to a 32-bit signed integer (similar to C/C++'s `atoi` function).
+
+The algorithm for `myAtoi(string s)` is as follows:
+
+1. Read in and ignore any leading whitespace.
+2. Check if the next character (if not already at the end of the string) is `'-'` or `'+'`. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
+3. Read in next the characters until the next non-digit charcter or the end of the input is reached. The rest of the string is ignored.
+4. Convert these digits into an integer (i.e. `"123" -> 123`, `"0032" -> 32`). If no digits were read, then the integer is `0`. Change the sign as necessary (from step 2).
+5. If the integer is out of the 32-bit signed integer range `[-231, 231 - 1]`, then clamp the integer so that it remains in the range. Specifically, integers less than `231` should be clamped to `231`, and integers greater than `231 - 1` should be clamped to `231 - 1`.
+6. Return the integer as the final result.
+
+**Note:**
+
+- Only the space character `' '` is considered a whitespace character.
+- **Do not ignore** any characters other than the leading whitespace or the rest of the string after the digits.
+
+**Example 1:**
+
+```
+Input: s = "42"
+Output: 42
+Explanation: The underlined characters are what is read in, the caret is the current reader position.
+Step 1: "42" (no characters read because there is no leading whitespace)
+         ^
+Step 2: "42" (no characters read because there is neither a '-' nor '+')
+         ^
+Step 3: "42" ("42" is read in)
+           ^
+The parsed integer is 42.
+Since 42 is in the range [-231, 231 - 1], the final result is 42.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "   -42"
+Output: -42
+Explanation:
+Step 1: "   -42" (leading whitespace is read and ignored)
+            ^
+Step 2: "   -42" ('-' is read, so the result should be negative)
+             ^
+Step 3: "   -42" ("42" is read in)
+               ^
+The parsed integer is -42.
+Since -42 is in the range [-231, 231 - 1], the final result is -42.
+
+```
+
+**Example 3:**
+
+```
+Input: s = "4193 with words"
+Output: 4193
+Explanation:
+Step 1: "4193 with words" (no characters read because there is no leading whitespace)
+         ^
+Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
+         ^
+Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
+             ^
+The parsed integer is 4193.
+Since 4193 is in the range [-231, 231 - 1], the final result is 4193.
+
+```
+
+**Example 4:**
+
+```
+Input: s = "words and 987"
+Output: 0
+Explanation:
+Step 1: "words and 987" (no characters read because there is no leading whitespace)
+         ^
+Step 2: "words and 987" (no characters read because there is neither a '-' nor '+')
+         ^
+Step 3: "words and 987" (reading stops immediately because there is a non-digit 'w')
+         ^
+The parsed integer is 0 because no digits were read.
+Since 0 is in the range [-231, 231 - 1], the final result is 0.
+
+```
+
+**Example 5:**
+
+```
+Input: s = "-91283472332"
+Output: -2147483648
+Explanation:
+Step 1: "-91283472332" (no characters read because there is no leading whitespace)
+         ^
+Step 2: "-91283472332" ('-' is read, so the result should be negative)
+          ^
+Step 3: "-91283472332" ("91283472332" is read in)
+                     ^
+The parsed integer is -91283472332.
+Since -91283472332 is less than the lower bound of the range [-231, 231 - 1], the final result is clamped to -231 = -2147483648.
+```
+
+**Constraints:**
+
+- `0 <= s.length <= 200`
+- `s` consists of English letters (lower-case and upper-case), digits (`0-9`), `' '`, `'+'`
+
+## 题目大意
+
+请你来实现一个 myAtoi(string s) 函数，使其能将字符串转换成一个 32 位有符号整数（类似 C/C++ 中的 atoi 函数）。
+
+函数 myAtoi(string s) 的算法如下：
+
+- 读入字符串并丢弃无用的前导空格
+- 检查下一个字符（假设还未到字符末尾）为正还是负号，读取该字符（如果有）。 确定最终结果是负数还是正数。 如果两者都不存在，则假定结果为正。
+- 读入下一个字符，直到到达下一个非数字字符或到达输入的结尾。字符串的其余部分将被忽略。
+- 将前面步骤读入的这些数字转换为整数（即，"123" -> 123， "0032" -> 32）。如果没有读入数字，则整数为 0 。必要时更改符号（从步骤 2 开始）。
+- 如果整数数超过 32 位有符号整数范围 [−231,  231 − 1] ，需要截断这个整数，使其保持在这个范围内。具体来说，小于 −231 的整数应该被固定为 −231 ，大于 231 − 1 的整数应该被固定为 231 − 1 。
+- 返回整数作为最终结果。
+
+注意：
+
+- 本题中的空白字符只包括空格字符 ' ' 。
+- 除前导空格或数字后的其余字符串外，请勿忽略 任何其他字符。
+
+## 解题思路
+
+- 这题是简单题。题目要求实现类似 `C++` 中 `atoi` 函数的功能。这个函数功能是将字符串类型的数字转成 `int` 类型数字。先去除字符串中的前导空格，并判断记录数字的符号。数字需要去掉前导 `0` 。最后将数字转换成数字类型，判断是否超过 `int` 类型的上限 `[-2^31, 2^31 - 1]`，如果超过上限，需要输出边界，即 `-2^31`，或者 `2^31 - 1`。
+
+## 代码
+
+```go
+package leetcode
+
+func myAtoi(s string) int {
+	maxInt, signAllowed, whitespaceAllowed, sign, digits := int64(2<<30), true, true, 1, []int{}
+	for _, c := range s {
+		if c == ' ' && whitespaceAllowed {
+			continue
+		}
+		if signAllowed {
+			if c == '+' {
+				signAllowed = false
+				whitespaceAllowed = false
+				continue
+			} else if c == '-' {
+				sign = -1
+				signAllowed = false
+				whitespaceAllowed = false
+				continue
+			}
+		}
+		if c < '0' || c > '9' {
+			break
+		}
+		whitespaceAllowed, signAllowed = false, false
+		digits = append(digits, int(c-48))
+	}
+	var num, place int64
+	place, num = 1, 0
+	lastLeading0Index := -1
+	for i, d := range digits {
+		if d == 0 {
+			lastLeading0Index = i
+		} else {
+			break
+		}
+	}
+	if lastLeading0Index > -1 {
+		digits = digits[lastLeading0Index+1:]
+	}
+	var rtnMax int64
+	if sign > 0 {
+		rtnMax = maxInt - 1
+	} else {
+		rtnMax = maxInt
+	}
+	digitsCount := len(digits)
+	for i := digitsCount - 1; i >= 0; i-- {
+		num += int64(digits[i]) * place
+		place *= 10
+		if digitsCount-i > 10 || num > rtnMax {
+			return int(int64(sign) * rtnMax)
+		}
+	}
+	num *= int64(sign)
+	return int(num)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0007.Reverse-Integer/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0009.Palindrome-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0009.Palindrome-Number.md b/website/content/ChapterFour/0001~0099/0009.Palindrome-Number.md
new file mode 100644
index 000000000..4cfa1c984
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0009.Palindrome-Number.md
@@ -0,0 +1,103 @@
+# [9. Palindrome Number](https://leetcode.com/problems/palindrome-number/)
+
+
+## 题目
+
+Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
+
+**Example 1**:
+
+```
+Input: 121
+Output: true
+```
+
+**Example 2**:
+
+```
+Input: -121
+Output: false
+Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
+```
+
+**Example 3**:
+
+```
+Input: 10
+Output: false
+Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
+```
+
+**Follow up**:
+
+Coud you solve it without converting the integer to a string?
+
+## 题目大意
+
+判断一个整数是否是回文数。回文数是指正序（从左向右）和倒序（从右向左）读都是一样的整数。
+
+## 解题思路
+
+- 判断一个整数是不是回文数。
+- 简单题。注意会有负数的情况，负数，个位数，10 都不是回文数。其他的整数再按照回文的规则判断。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strconv"
+
+// 解法一
+func isPalindrome(x int) bool {
+	if x < 0 {
+		return false
+	}
+	if x == 0 {
+		return true
+	}
+	if x%10 == 0 {
+		return false
+	}
+	arr := make([]int, 0, 32)
+	for x > 0 {
+		arr = append(arr, x%10)
+		x = x / 10
+	}
+	sz := len(arr)
+	for i, j := 0, sz-1; i <= j; i, j = i+1, j-1 {
+		if arr[i] != arr[j] {
+			return false
+		}
+	}
+	return true
+}
+
+// 解法二 数字转字符串
+func isPalindrome1(x int) bool {
+	if x < 0 {
+		return false
+	}
+	if x < 10 {
+		return true
+	}
+	s := strconv.Itoa(x)
+	length := len(s)
+	for i := 0; i <= length/2; i++ {
+		if s[i] != s[length-1-i] {
+			return false
+		}
+	}
+	return true
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0008.String-to-Integer-atoi/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0011.Container-With-Most-Water/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0011.Container-With-Most-Water.md b/website/content/ChapterFour/0001~0099/0011.Container-With-Most-Water.md
new file mode 100644
index 000000000..6e81d5b00
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0011.Container-With-Most-Water.md
@@ -0,0 +1,66 @@
+# [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)
+
+## 题目
+
+Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
+
+**Note**: You may not slant the container and n is at least 2.
+
+![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/17/question_11.jpg)
+
+The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
+
+**Example 1**:
+
+```
+
+Input: [1,8,6,2,5,4,8,3,7]
+Output: 49
+
+```
+
+
+## 题目大意
+
+给出一个非负整数数组 a1，a2，a3，…… an，每个整数标识一个竖立在坐标轴 x 位置的一堵高度为 ai 的墙，选择两堵墙，和 x 轴构成的容器可以容纳最多的水。
+
+## 解题思路
+
+
+这一题也是对撞指针的思路。首尾分别 2 个指针，每次移动以后都分别判断长宽的乘积是否最大。
+
+## 代码
+
+```go
+
+package leetcode
+
+func maxArea(height []int) int {
+	max, start, end := 0, 0, len(height)-1
+	for start < end {
+		width := end - start
+		high := 0
+		if height[start] < height[end] {
+			high = height[start]
+			start++
+		} else {
+			high = height[end]
+			end--
+		}
+
+		temp := width * high
+		if temp > max {
+			max = temp
+		}
+	}
+	return max
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0009.Palindrome-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0012.Integer-to-Roman/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0012.Integer-to-Roman.md b/website/content/ChapterFour/0001~0099/0012.Integer-to-Roman.md
new file mode 100644
index 000000000..274c2a126
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0012.Integer-to-Roman.md
@@ -0,0 +1,109 @@
+# [12. Integer to Roman](https://leetcode.com/problems/integer-to-roman/)
+
+
+## 题目
+
+Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
+
+```
+Symbol       Value
+I             1
+V             5
+X             10
+L             50
+C             100
+D             500
+M             1000
+```
+
+For example, `2` is written as `II` in Roman numeral, just two one's added together. `12` is written as `XII`, which is simply `X + II`. The number `27` is written as `XXVII`, which is `XX + V + II`.
+
+Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
+
+- `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
+- `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
+- `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
+
+Given an integer, convert it to a roman numeral.
+
+**Example 1:**
+
+```
+Input: num = 3
+Output: "III"
+```
+
+**Example 2:**
+
+```
+Input: num = 4
+Output: "IV"
+```
+
+**Example 3:**
+
+```
+Input: num = 9
+Output: "IX"
+```
+
+**Example 4:**
+
+```
+Input: num = 58
+Output: "LVIII"
+Explanation: L = 50, V = 5, III = 3.
+```
+
+**Example 5:**
+
+```
+Input: num = 1994
+Output: "MCMXCIV"
+Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
+```
+
+**Constraints:**
+
+- `1 <= num <= 3999`
+
+## 题目大意
+
+通常情况下，罗马数字中小的数字在大的数字的右边。但也存在特例，例如 4 不写做 IIII，而是 IV。数字 1 在数字 5 的左边，所表示的数等于大数 5 减小数 1 得到的数值 4 。同样地，数字 9 表示为 IX。这个特殊的规则只适用于以下六种情况：
+
+- I 可以放在 V (5) 和 X (10) 的左边，来表示 4 和 9。
+- X 可以放在 L (50) 和 C (100) 的左边，来表示 40 和 90。
+- C 可以放在 D (500) 和 M (1000) 的左边，来表示 400 和 900。
+
+给定一个整数，将其转为罗马数字。输入确保在 1 到 3999 的范围内。
+
+## 解题思路
+
+- 依照题意，优先选择大的数字，解题思路采用贪心算法。将 1-3999 范围内的罗马数字从大到小放在数组中，从头选择到尾，即可把整数转成罗马数字。
+
+## 代码
+
+```go
+package leetcode
+
+func intToRoman(num int) string {
+	values := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
+	symbols := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
+	res, i := "", 0
+	for num != 0 {
+		for values[i] > num {
+			i++
+		}
+		num -= values[i]
+		res += symbols[i]
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0011.Container-With-Most-Water/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0013.Roman-to-Integer/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0013.Roman-to-Integer.md b/website/content/ChapterFour/0001~0099/0013.Roman-to-Integer.md
new file mode 100644
index 000000000..ae95445d3
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0013.Roman-to-Integer.md
@@ -0,0 +1,139 @@
+# [13. Roman to Integer](https://leetcode.com/problems/roman-to-integer/)
+
+
+## 题目
+
+Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
+
+```
+Symbol       Value
+I             1
+V             5
+X             10
+L             50
+C             100
+D             500
+M             1000
+```
+
+For example, two is written as `II` in Roman numeral, just two one's added together. Twelve is written as, `XII`, which is simply `X` + `II`. The number twenty seven is written as `XXVII`, which is `XX` + `V` + `II`.
+
+Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
+
+- `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
+- `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
+- `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
+
+Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.
+
+**Example 1**:
+
+```
+Input: "III"
+Output: 3
+```
+
+**Example 2**:
+
+```
+Input: "IV"
+Output: 4
+```
+
+**Example 3**:
+
+```
+Input: "IX"
+Output: 9
+```
+
+**Example 4**:
+
+```
+Input: "LVIII"
+Output: 58
+Explanation: L = 50, V= 5, III = 3.
+```
+
+**Example 5**:
+
+```
+Input: "MCMXCIV"
+Output: 1994
+Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
+```
+
+## 题目大意
+
+罗马数字包含以下七种字符: I， V， X， L，C，D 和 M。
+
+```go
+
+字符          数值
+I             1
+V             5
+X             10
+L             50
+C             100
+D             500
+M             1000
+
+```
+
+例如， 罗马数字 2 写做 II ，即为两个并列的 1。12 写做 XII ，即为 X + II 。 27 写做  XXVII, 即为 XX + V + II 。
+
+通常情况下，罗马数字中小的数字在大的数字的右边。但也存在特例，例如 4 不写做 IIII，而是 IV。数字 1 在数字 5 的左边，所表示的数等于大数 5 减小数 1 得到的数值 4 。同样地，数字 9 表示为 IX。这个特殊的规则只适用于以下六种情况：
+
+- I 可以放在 V (5) 和 X (10) 的左边，来表示 4 和 9。
+- X 可以放在 L (50) 和 C (100) 的左边，来表示 40 和 90。 
+- C 可以放在 D (500) 和 M (1000) 的左边，来表示 400 和 900。
+
+给定一个罗马数字，将其转换成整数。输入确保在 1 到 3999 的范围内。
+
+## 解题思路
+
+- 给定一个罗马数字，将其转换成整数。输入确保在 1 到 3999 的范围内。
+- 简单题。按照题目中罗马数字的字符数值，计算出对应罗马数字的十进制数即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+var roman = map[string]int{
+	"I": 1,
+	"V": 5,
+	"X": 10,
+	"L": 50,
+	"C": 100,
+	"D": 500,
+	"M": 1000,
+}
+
+func romanToInt(s string) int {
+	if s == "" {
+		return 0
+	}
+	num, lastint, total := 0, 0, 0
+	for i := 0; i < len(s); i++ {
+		char := s[len(s)-(i+1) : len(s)-i]
+		num = roman[char]
+		if num < lastint {
+			total = total - num
+		} else {
+			total = total + num
+		}
+		lastint = num
+	}
+	return total
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0012.Integer-to-Roman/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0014.Longest-Common-Prefix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0014.Longest-Common-Prefix.md b/website/content/ChapterFour/0001~0099/0014.Longest-Common-Prefix.md
new file mode 100644
index 000000000..c6354e451
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0014.Longest-Common-Prefix.md
@@ -0,0 +1,65 @@
+# [14. Longest Common Prefix](https://leetcode.com/problems/longest-common-prefix/)
+
+## 题目
+
+Write a function to find the longest common prefix string amongst an array of strings.
+
+If there is no common prefix, return an empty string "".
+
+**Example 1**:
+
+    Input: strs = ["flower","flow","flight"]
+    Output: "fl"
+
+**Example 2**:
+
+    Input: strs = ["dog","racecar","car"]
+    Output: ""
+    Explanation: There is no common prefix among the input strings.
+
+**Constraints:**
+
+- 1 <= strs.length <= 200
+- 0 <= strs[i].length <= 200
+- strs[i] consists of only lower-case English letters.
+
+## 题目大意
+
+编写一个函数来查找字符串数组中的最长公共前缀。
+
+如果不存在公共前缀，返回空字符串 ""。
+
+## 解题思路
+
+- 对 strs 按照字符串长度进行升序排序，求出 strs 中长度最小字符串的长度 minLen
+- 逐个比较长度最小字符串与其它字符串中的字符，如果不相等就返回 commonPrefix,否则就把该字符加入 commonPrefix
+
+## 代码
+
+```go
+
+package leetcode
+
+func longestCommonPrefix(strs []string) string {
+	prefix := strs[0]
+
+	for i := 1; i < len(strs); i++ {
+		for j := 0; j < len(prefix); j++ {
+			if len(strs[i]) <= j || strs[i][j] != prefix[j] {
+				prefix = prefix[0:j]
+				break
+			}
+		}
+	}
+
+	return prefix
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0013.Roman-to-Integer/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0015.3Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0015.3Sum.md b/website/content/ChapterFour/0001~0099/0015.3Sum.md
new file mode 100644
index 000000000..48d53acb3
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0015.3Sum.md
@@ -0,0 +1,125 @@
+# [15. 3Sum](https://leetcode.com/problems/3sum/)
+
+## 题目
+
+Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
+
+**Note**:
+
+The solution set must not contain duplicate triplets.
+
+**Example**:
+
+```
+
+Given array nums = [-1, 0, 1, 2, -1, -4],
+
+A solution set is:
+[
+  [-1, 0, 1],
+  [-1, -1, 2]
+]
+
+```
+
+## 题目大意
+
+给定一个数组，要求在这个数组中找出 3 个数之和为 0 的所有组合。
+
+## 解题思路
+
+用 map 提前计算好任意 2 个数字之和，保存起来，可以将时间复杂度降到 O(n^2)。这一题比较麻烦的一点在于，最后输出解的时候，要求输出不重复的解。数组中同一个数字可能出现多次，同一个数字也可能使用多次，但是最后输出解的时候，不能重复。例如 [-1，-1，2] 和 [2, -1, -1]、[-1, 2, -1] 这 3 个解是重复的，即使 -1 可能出现 100 次，每次使用的 -1 的数组下标都是不同的。
+
+这里就需要去重和排序了。map 记录每个数字出现的次数，然后对 map 的 key 数组进行排序，最后在这个排序以后的数组里面扫，找到另外 2 个数字能和自己组成 0 的组合。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+// 解法一 最优解，双指针 + 排序
+func threeSum(nums []int) [][]int {
+	sort.Ints(nums)
+	result, start, end, index, addNum, length := make([][]int, 0), 0, 0, 0, 0, len(nums)
+	for index = 1; index < length-1; index++ {
+		start, end = 0, length-1
+		if index > 1 && nums[index] == nums[index-1] {
+			start = index - 1
+		}
+		for start < index && end > index {
+			if start > 0 && nums[start] == nums[start-1] {
+				start++
+				continue
+			}
+			if end < length-1 && nums[end] == nums[end+1] {
+				end--
+				continue
+			}
+			addNum = nums[start] + nums[end] + nums[index]
+			if addNum == 0 {
+				result = append(result, []int{nums[start], nums[index], nums[end]})
+				start++
+				end--
+			} else if addNum > 0 {
+				end--
+			} else {
+				start++
+			}
+		}
+	}
+	return result
+}
+
+// 解法二
+func threeSum1(nums []int) [][]int {
+	res := [][]int{}
+	counter := map[int]int{}
+	for _, value := range nums {
+		counter[value]++
+	}
+
+	uniqNums := []int{}
+	for key := range counter {
+		uniqNums = append(uniqNums, key)
+	}
+	sort.Ints(uniqNums)
+
+	for i := 0; i < len(uniqNums); i++ {
+		if (uniqNums[i]*3 == 0) && counter[uniqNums[i]] >= 3 {
+			res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i]})
+		}
+		for j := i + 1; j < len(uniqNums); j++ {
+			if (uniqNums[i]*2+uniqNums[j] == 0) && counter[uniqNums[i]] > 1 {
+				res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j]})
+			}
+			if (uniqNums[j]*2+uniqNums[i] == 0) && counter[uniqNums[j]] > 1 {
+				res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j]})
+			}
+			c := 0 - uniqNums[i] - uniqNums[j]
+			if c > uniqNums[j] && counter[c] > 0 {
+				res = append(res, []int{uniqNums[i], uniqNums[j], c})
+			}
+		}
+	}
+	return res
+}
+
+
+
+```
+
+
+
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0014.Longest-Common-Prefix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0016.3Sum-Closest/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0016.3Sum-Closest.md b/website/content/ChapterFour/0001~0099/0016.3Sum-Closest.md
new file mode 100644
index 000000000..c98e985ad
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0016.3Sum-Closest.md
@@ -0,0 +1,99 @@
+# [16. 3Sum Closest](https://leetcode.com/problems/3sum-closest/)
+
+## 题目
+
+Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
+
+**Example**:
+
+```
+
+Given array nums = [-1, 2, 1, -4], and target = 1.
+
+The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
+
+```
+
+## 题目大意
+
+给定一个数组，要求在这个数组中找出 3 个数之和离 target 最近。
+
+## 解题思路
+
+这一题看似和第 15 题和第 18 题很像，都是求 3 或者 4 个数之和的问题，但是这一题的做法和 15，18 题完全不同。
+
+这一题的解法是用两个指针夹逼的方法。先对数组进行排序，i 从头开始往后面扫。这里同样需要注意数组中存在多个重复数字的问题。具体处理方法很多，可以用 map 计数去重。这里笔者简单的处理，i 在循环的时候和前一个数进行比较，如果相等，i 继续往后移，直到移到下一个和前一个数字不同的位置。j，k 两个指针开始一前一后夹逼。j 为 i 的下一个数字，k 为数组最后一个数字，由于经过排序，所以 k 的数字最大。j 往后移动，k 往前移动，逐渐夹逼出最接近 target 的值。
+
+
+这道题还可以用暴力解法，三层循环找到距离 target 最近的组合。具体见代码。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+// 解法一 O(n^2)
+func threeSumClosest(nums []int, target int) int {
+	n, res, diff := len(nums), 0, math.MaxInt32
+	if n > 2 {
+		sort.Ints(nums)
+		for i := 0; i < n-2; i++ {
+			if i > 0 && nums[i] == nums[i-1] {
+				continue
+			}
+			for j, k := i+1, n-1; j < k; {
+				sum := nums[i] + nums[j] + nums[k]
+				if abs(sum-target) < diff {
+					res, diff = sum, abs(sum-target)
+				}
+				if sum == target {
+					return res
+				} else if sum > target {
+					k--
+				} else {
+					j++
+				}
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 暴力解法 O(n^3)
+func threeSumClosest1(nums []int, target int) int {
+	res, difference := 0, math.MaxInt16
+	for i := 0; i < len(nums); i++ {
+		for j := i + 1; j < len(nums); j++ {
+			for k := j + 1; k < len(nums); k++ {
+				if abs(nums[i]+nums[j]+nums[k]-target) < difference {
+					difference = abs(nums[i] + nums[j] + nums[k] - target)
+					res = nums[i] + nums[j] + nums[k]
+				}
+			}
+		}
+	}
+	return res
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0015.3Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0017.Letter-Combinations-of-a-Phone-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0017.Letter-Combinations-of-a-Phone-Number.md b/website/content/ChapterFour/0001~0099/0017.Letter-Combinations-of-a-Phone-Number.md
new file mode 100755
index 000000000..4bb0f769a
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0017.Letter-Combinations-of-a-Phone-Number.md
@@ -0,0 +1,151 @@
+# [17. Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number/)
+
+
+## 题目
+
+Given a string containing digits from `2-9` inclusive, return all possible letter combinations that the number could represent.
+
+A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
+
+![](http://upload.wikimedia.org/wikipedia/commons/thumb/7/73/Telephone-keypad2.svg/200px-Telephone-keypad2.svg.png)
+
+**Example**:
+
+```
+
+Input: "23"
+Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
+
+```
+
+**Note**:
+
+Although the above answer is in lexicographical order, your answer could be in any order you want.
+
+## 题目大意
+
+给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。
+
+
+## 解题思路
+
+- DFS 递归深搜即可
+
+## 代码
+
+```go
+
+package leetcode
+
+var (
+	letterMap = []string{
+		" ",    //0
+		"",     //1
+		"abc",  //2
+		"def",  //3
+		"ghi",  //4
+		"jkl",  //5
+		"mno",  //6
+		"pqrs", //7
+		"tuv",  //8
+		"wxyz", //9
+	}
+	res   = []string{}
+	final = 0
+)
+
+// 解法一 DFS
+func letterCombinations(digits string) []string {
+	if digits == "" {
+		return []string{}
+	}
+	res = []string{}
+	findCombination(&digits, 0, "")
+	return res
+}
+
+func findCombination(digits *string, index int, s string) {
+	if index == len(*digits) {
+		res = append(res, s)
+		return
+	}
+	num := (*digits)[index]
+	letter := letterMap[num-'0']
+	for i := 0; i < len(letter); i++ {
+		findCombination(digits, index+1, s+string(letter[i]))
+	}
+	return
+}
+
+// 解法二 非递归
+func letterCombinations_(digits string) []string {
+	if digits == "" {
+		return []string{}
+	}
+	index := digits[0] - '0'
+	letter := letterMap[index]
+	tmp := []string{}
+	for i := 0; i < len(letter); i++ {
+		if len(res) == 0 {
+			res = append(res, "")
+		}
+		for j := 0; j < len(res); j++ {
+			tmp = append(tmp, res[j]+string(letter[i]))
+		}
+	}
+	res = tmp
+	final++
+	letterCombinations(digits[1:])
+	final--
+	if final == 0 {
+		tmp = res
+		res = []string{}
+	}
+	return tmp
+}
+
+// 解法三 回溯（参考回溯模板，类似DFS）
+var result []string
+var dict = map[string][]string{
+	"2" : []string{"a","b","c"},
+	"3" : []string{"d", "e", "f"},
+	"4" : []string{"g", "h", "i"},
+	"5" : []string{"j", "k", "l"},
+	"6" : []string{"m", "n", "o"},
+	"7" : []string{"p", "q", "r", "s"},
+	"8" : []string{"t", "u", "v"},
+	"9" : []string{"w", "x", "y", "z"},
+}
+
+func letterCombinationsBT(digits string) []string {
+	result = []string{}
+	if digits == "" {
+		return result
+	}
+	letterFunc("", digits)
+	return result
+}
+
+func letterFunc(res string, digits string) {
+	if digits == "" {
+		result = append(result, res)
+		return
+	}
+
+	k := digits[0:1]
+	digits = digits[1:]
+	for i := 0; i < len(dict[k]); i++ {
+		res += dict[k][i]
+		letterFunc(res, digits)
+		res = res[0 : len(res)-1]
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0016.3Sum-Closest/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0018.4Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0018.4Sum.md b/website/content/ChapterFour/0001~0099/0018.4Sum.md
new file mode 100644
index 000000000..672e26be7
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0018.4Sum.md
@@ -0,0 +1,189 @@
+# [18. 4Sum](https://leetcode.com/problems/4sum/)
+
+## 题目
+
+Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
+
+**Note**:
+
+The solution set must not contain duplicate quadruplets.
+
+**Example**:
+
+```
+
+Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.
+
+A solution set is:
+[
+  [-1,  0, 0, 1],
+  [-2, -1, 1, 2],
+  [-2,  0, 0, 2]
+]
+
+```
+
+## 题目大意
+
+给定一个数组，要求在这个数组中找出 4 个数之和为 0 的所有组合。
+
+
+## 解题思路
+
+用 map 提前计算好任意 3 个数字之和，保存起来，可以将时间复杂度降到 O(n^3)。这一题比较麻烦的一点在于，最后输出解的时候，要求输出不重复的解。数组中同一个数字可能出现多次，同一个数字也可能使用多次，但是最后输出解的时候，不能重复。例如 [-1，1，2, -2] 和 [2, -1, -2, 1]、[-2, 2, -1, 1] 这 3 个解是重复的，即使 -1, -2 可能出现 100 次，每次使用的 -1, -2 的数组下标都是不同的。
+
+这一题是第 15 题的升级版，思路都是完全一致的。这里就需要去重和排序了。map 记录每个数字出现的次数，然后对 map 的 key 数组进行排序，最后在这个排序以后的数组里面扫，找到另外 3 个数字能和自己组成 0 的组合。
+
+第 15 题和第 18 题的解法一致。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+// 解法一 双指针
+func fourSum(nums []int, target int) (quadruplets [][]int) {
+	sort.Ints(nums)
+	n := len(nums)
+	for i := 0; i < n-3 && nums[i]+nums[i+1]+nums[i+2]+nums[i+3] <= target; i++ {
+		if i > 0 && nums[i] == nums[i-1] || nums[i]+nums[n-3]+nums[n-2]+nums[n-1] < target {
+			continue
+		}
+		for j := i + 1; j < n-2 && nums[i]+nums[j]+nums[j+1]+nums[j+2] <= target; j++ {
+			if j > i+1 && nums[j] == nums[j-1] || nums[i]+nums[j]+nums[n-2]+nums[n-1] < target {
+				continue
+			}
+			for left, right := j+1, n-1; left < right; {
+				if sum := nums[i] + nums[j] + nums[left] + nums[right]; sum == target {
+					quadruplets = append(quadruplets, []int{nums[i], nums[j], nums[left], nums[right]})
+					for left++; left < right && nums[left] == nums[left-1]; left++ {
+					}
+					for right--; left < right && nums[right] == nums[right+1]; right-- {
+					}
+				} else if sum < target {
+					left++
+				} else {
+					right--
+				}
+			}
+		}
+	}
+	return
+}
+
+// 解法二 kSum
+func fourSum1(nums []int, target int) [][]int {
+	res, cur := make([][]int, 0), make([]int, 0)
+	sort.Ints(nums)
+	kSum(nums, 0, len(nums)-1, target, 4, cur, &res)
+	return res
+}
+
+func kSum(nums []int, left, right int, target int, k int, cur []int, res *[][]int) {
+	if right-left+1 < k || k < 2 || target < nums[left]*k || target > nums[right]*k {
+		return
+	}
+	if k == 2 {
+		// 2 sum
+		twoSum(nums, left, right, target, cur, res)
+	} else {
+		for i := left; i < len(nums); i++ {
+			if i == left || (i > left && nums[i-1] != nums[i]) {
+				next := make([]int, len(cur))
+				copy(next, cur)
+				next = append(next, nums[i])
+				kSum(nums, i+1, len(nums)-1, target-nums[i], k-1, next, res)
+			}
+		}
+	}
+
+}
+
+func twoSum(nums []int, left, right int, target int, cur []int, res *[][]int) {
+	for left < right {
+		sum := nums[left] + nums[right]
+		if sum == target {
+			cur = append(cur, nums[left], nums[right])
+			temp := make([]int, len(cur))
+			copy(temp, cur)
+			*res = append(*res, temp)
+			// reset cur to previous state
+			cur = cur[:len(cur)-2]
+			left++
+			right--
+			for left < right && nums[left] == nums[left-1] {
+				left++
+			}
+			for left < right && nums[right] == nums[right+1] {
+				right--
+			}
+		} else if sum < target {
+			left++
+		} else {
+			right--
+		}
+	}
+}
+
+// 解法三
+func fourSum2(nums []int, target int) [][]int {
+	res := [][]int{}
+	counter := map[int]int{}
+	for _, value := range nums {
+		counter[value]++
+	}
+
+	uniqNums := []int{}
+	for key := range counter {
+		uniqNums = append(uniqNums, key)
+	}
+	sort.Ints(uniqNums)
+
+	for i := 0; i < len(uniqNums); i++ {
+		if (uniqNums[i]*4 == target) && counter[uniqNums[i]] >= 4 {
+			res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i], uniqNums[i]})
+		}
+		for j := i + 1; j < len(uniqNums); j++ {
+			if (uniqNums[i]*3+uniqNums[j] == target) && counter[uniqNums[i]] > 2 {
+				res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i], uniqNums[j]})
+			}
+			if (uniqNums[j]*3+uniqNums[i] == target) && counter[uniqNums[j]] > 2 {
+				res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j], uniqNums[j]})
+			}
+			if (uniqNums[j]*2+uniqNums[i]*2 == target) && counter[uniqNums[j]] > 1 && counter[uniqNums[i]] > 1 {
+				res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j], uniqNums[j]})
+			}
+			for k := j + 1; k < len(uniqNums); k++ {
+				if (uniqNums[i]*2+uniqNums[j]+uniqNums[k] == target) && counter[uniqNums[i]] > 1 {
+					res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j], uniqNums[k]})
+				}
+				if (uniqNums[j]*2+uniqNums[i]+uniqNums[k] == target) && counter[uniqNums[j]] > 1 {
+					res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j], uniqNums[k]})
+				}
+				if (uniqNums[k]*2+uniqNums[i]+uniqNums[j] == target) && counter[uniqNums[k]] > 1 {
+					res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[k], uniqNums[k]})
+				}
+				c := target - uniqNums[i] - uniqNums[j] - uniqNums[k]
+				if c > uniqNums[k] && counter[c] > 0 {
+					res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[k], c})
+				}
+			}
+		}
+	}
+	return res
+}
+
+
+```
+
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0017.Letter-Combinations-of-a-Phone-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0019.Remove-Nth-Node-From-End-of-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0019.Remove-Nth-Node-From-End-of-List.md b/website/content/ChapterFour/0001~0099/0019.Remove-Nth-Node-From-End-of-List.md
new file mode 100644
index 000000000..0a511762f
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0019.Remove-Nth-Node-From-End-of-List.md
@@ -0,0 +1,132 @@
+# [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)
+
+## 题目
+
+Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.
+
+**Follow up:** Could you do this in one pass?
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.jpg)
+
+```
+Input: head = [1,2,3,4,5], n = 2
+Output: [1,2,3,5]
+
+```
+
+**Example 2:**
+
+```
+Input: head = [1], n = 1
+Output: []
+
+```
+
+**Example 3:**
+
+```
+Input: head = [1,2], n = 1
+Output: [1]
+
+```
+
+**Constraints:**
+
+- The number of nodes in the list is `sz`.
+- `1 <= sz <= 30`
+- `0 <= Node.val <= 100`
+- `1 <= n <= sz`
+
+## 题目大意
+
+删除链表中倒数第 n 个结点。
+
+## 解题思路
+
+这道题比较简单，先循环一次拿到链表的总长度，然后循环到要删除的结点的前一个结点开始删除操作。需要注意的一个特例是，有可能要删除头结点，要单独处理。
+
+这道题有一种特别简单的解法。设置 2 个指针，一个指针距离前一个指针 n 个距离。同时移动 2 个指针，2 个指针都移动相同的距离。当一个指针移动到了终点，那么前一个指针就是倒数第 n 个节点了。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// 解法一
+func removeNthFromEnd(head *ListNode, n int) *ListNode {
+	dummyHead := &ListNode{Next: head}
+	preSlow, slow, fast := dummyHead, head, head
+	for fast != nil {
+		if n <= 0 {
+			preSlow = slow
+			slow = slow.Next
+		}
+		n--
+		fast = fast.Next
+	}
+	preSlow.Next = slow.Next
+	return dummyHead.Next
+}
+
+// 解法二
+func removeNthFromEnd1(head *ListNode, n int) *ListNode {
+	if head == nil {
+		return nil
+	}
+	if n <= 0 {
+		return head
+	}
+	current := head
+	len := 0
+	for current != nil {
+		len++
+		current = current.Next
+	}
+	if n > len {
+		return head
+	}
+	if n == len {
+		current := head
+		head = head.Next
+		current.Next = nil
+		return head
+	}
+	current = head
+	i := 0
+	for current != nil {
+		if i == len-n-1 {
+			deleteNode := current.Next
+			current.Next = current.Next.Next
+			deleteNode.Next = nil
+			break
+		}
+		i++
+		current = current.Next
+	}
+	return head
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0018.4Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0020.Valid-Parentheses/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0020.Valid-Parentheses.md b/website/content/ChapterFour/0001~0099/0020.Valid-Parentheses.md
new file mode 100644
index 000000000..f1a6faa2c
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0020.Valid-Parentheses.md
@@ -0,0 +1,103 @@
+# [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/description/)
+
+## 题目
+
+Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
+
+An input string is valid if:
+
+Open brackets must be closed by the same type of brackets.
+Open brackets must be closed in the correct order.
+Note that an empty string is also considered valid.
+
+**Example 1**:
+
+```
+
+Input: "()"
+Output: true
+
+```
+
+
+**Example 2**:
+
+```
+
+Input: "()[]{}"
+Output: true
+
+```
+
+**Example 3**:
+
+```
+
+Input: "(]"
+Output: false
+
+```
+
+**Example 4**:
+
+```
+
+Input: "([)]"
+Output: false
+
+```
+
+**Example 5**:
+
+```
+
+Input: "{[]}"
+Output: true
+
+```
+
+## 题目大意
+
+括号匹配问题。
+
+## 解题思路
+
+遇到左括号就进栈push，遇到右括号并且栈顶为与之对应的左括号，就把栈顶元素出栈。最后看栈里面还有没有其他元素，如果为空，即匹配。
+
+需要注意，空字符串是满足括号匹配的，即输出 true。
+
+## 代码
+
+```go
+
+package leetcode
+
+func isValid(s string) bool {
+	// 空字符串直接返回 true
+	if len(s) == 0 {
+		return true
+	}
+	stack := make([]rune, 0)
+	for _, v := range s {
+		if (v == '[') || (v == '(') || (v == '{') {
+			stack = append(stack, v)
+		} else if ((v == ']') && len(stack) > 0 && stack[len(stack)-1] == '[') ||
+			((v == ')') && len(stack) > 0 && stack[len(stack)-1] == '(') ||
+			((v == '}') && len(stack) > 0 && stack[len(stack)-1] == '{') {
+			stack = stack[:len(stack)-1]
+		} else {
+			return false
+		}
+	}
+	return len(stack) == 0
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0019.Remove-Nth-Node-From-End-of-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0021.Merge-Two-Sorted-Lists/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0021.Merge-Two-Sorted-Lists.md b/website/content/ChapterFour/0001~0099/0021.Merge-Two-Sorted-Lists.md
new file mode 100644
index 000000000..76cdbff72
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0021.Merge-Two-Sorted-Lists.md
@@ -0,0 +1,60 @@
+# [21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)
+
+## 题目
+
+Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
+
+**Example**:
+
+```
+
+Input: 1->2->4, 1->3->4
+Output: 1->1->2->3->4->4
+
+```
+
+## 题目大意
+
+合并 2 个有序链表
+
+## 解题思路
+
+按照题意做即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
+	if l1 == nil {
+		return l2
+	}
+	if l2 == nil {
+		return l1
+	}
+	if l1.Val < l2.Val {
+		l1.Next = mergeTwoLists(l1.Next, l2)
+		return l1
+	}
+	l2.Next = mergeTwoLists(l1, l2.Next)
+	return l2
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0020.Valid-Parentheses/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0022.Generate-Parentheses/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0022.Generate-Parentheses.md b/website/content/ChapterFour/0001~0099/0022.Generate-Parentheses.md
new file mode 100755
index 000000000..882fdd4b7
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0022.Generate-Parentheses.md
@@ -0,0 +1,67 @@
+# [22. Generate Parentheses](https://leetcode.com/problems/generate-parentheses/)
+
+
+## 题目
+
+Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
+
+For example, given n = 3, a solution set is:
+
+
+    [
+      "((()))",
+      "(()())",
+      "(())()",
+      "()(())",
+      "()()()"
+    ]
+
+
+## 题目大意
+
+给出 n 代表生成括号的对数，请你写出一个函数，使其能够生成所有可能的并且有效的括号组合。
+
+
+## 解题思路
+
+- 这道题乍一看需要判断括号是否匹配的问题，如果真的判断了，那时间复杂度就到 O(n * 2^n)了，虽然也可以 AC，但是时间复杂度巨高。
+- 这道题实际上不需要判断括号是否匹配的问题。因为在 DFS 回溯的过程中，会让 `(` 和 `)` 成对的匹配上的。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func generateParenthesis(n int) []string {
+	if n == 0 {
+		return []string{}
+	}
+	res := []string{}
+	findGenerateParenthesis(n, n, "", &res)
+	return res
+}
+
+func findGenerateParenthesis(lindex, rindex int, str string, res *[]string) {
+	if lindex == 0 && rindex == 0 {
+		*res = append(*res, str)
+		return
+	}
+	if lindex > 0 {
+		findGenerateParenthesis(lindex-1, rindex, str+"(", res)
+	}
+	if rindex > 0 && lindex < rindex {
+		findGenerateParenthesis(lindex, rindex-1, str+")", res)
+	}
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0021.Merge-Two-Sorted-Lists/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0023.Merge-k-Sorted-Lists/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0023.Merge-k-Sorted-Lists.md b/website/content/ChapterFour/0001~0099/0023.Merge-k-Sorted-Lists.md
new file mode 100644
index 000000000..d6c5ef8d8
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0023.Merge-k-Sorted-Lists.md
@@ -0,0 +1,81 @@
+# [23. Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)
+
+## 题目
+
+Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
+
+
+
+**Example**:
+
+```
+
+Input:
+[
+  1->4->5,
+  1->3->4,
+  2->6
+]
+Output: 1->1->2->3->4->4->5->6
+
+```
+
+## 题目大意
+
+合并 K 个有序链表
+
+## 解题思路
+
+借助分治的思想，把 K 个有序链表两两合并即可。相当于是第 21 题的加强版。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func mergeKLists(lists []*ListNode) *ListNode {
+	length := len(lists)
+	if length < 1 {
+		return nil
+	}
+	if length == 1 {
+		return lists[0]
+	}
+	num := length / 2
+	left := mergeKLists(lists[:num])
+	right := mergeKLists(lists[num:])
+	return mergeTwoLists1(left, right)
+}
+
+func mergeTwoLists1(l1 *ListNode, l2 *ListNode) *ListNode {
+	if l1 == nil {
+		return l2
+	}
+	if l2 == nil {
+		return l1
+	}
+	if l1.Val < l2.Val {
+		l1.Next = mergeTwoLists1(l1.Next, l2)
+		return l1
+	}
+	l2.Next = mergeTwoLists1(l1, l2.Next)
+	return l2
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0022.Generate-Parentheses/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0024.Swap-Nodes-in-Pairs/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0024.Swap-Nodes-in-Pairs.md b/website/content/ChapterFour/0001~0099/0024.Swap-Nodes-in-Pairs.md
new file mode 100644
index 000000000..460110543
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0024.Swap-Nodes-in-Pairs.md
@@ -0,0 +1,65 @@
+# [24. Swap Nodes in Pairs](https://leetcode.com/problems/swap-nodes-in-pairs/description/)
+
+## 题目
+
+Given a linked list, swap every two adjacent nodes and return its head.
+
+You may not modify the values in the list's nodes, only nodes itself may be changed.
+
+
+
+**Example**:
+
+```
+
+Given 1->2->3->4, you should return the list as 2->1->4->3.
+
+```
+
+## 题目大意
+
+两两相邻的元素，翻转链表
+
+## 解题思路
+
+按照题意做即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func swapPairs(head *ListNode) *ListNode {
+	dummy := &ListNode{Next: head}
+	for pt := dummy; pt != nil && pt.Next != nil && pt.Next.Next != nil; {
+		pt, pt.Next, pt.Next.Next, pt.Next.Next.Next = pt.Next, pt.Next.Next, pt.Next.Next.Next, pt.Next
+	}
+	return dummy.Next
+}
+
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0023.Merge-k-Sorted-Lists/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0025.Reverse-Nodes-in-k-Group/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0025.Reverse-Nodes-in-k-Group.md b/website/content/ChapterFour/0001~0099/0025.Reverse-Nodes-in-k-Group.md
new file mode 100644
index 000000000..417548b9a
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0025.Reverse-Nodes-in-k-Group.md
@@ -0,0 +1,81 @@
+# [25. Reverse Nodes in k-Group](https://leetcode.com/problems/reverse-nodes-in-k-group/description/)
+
+## 题目
+
+Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
+
+k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
+
+**Example**:
+
+```
+
+Given this linked list: 1->2->3->4->5
+
+For k = 2, you should return: 2->1->4->3->5
+
+For k = 3, you should return: 3->2->1->4->5
+
+```
+
+**Note**:
+
+- Only constant extra memory is allowed.
+- You may not alter the values in the list's nodes, only nodes itself may be changed.
+
+
+## 题目大意
+
+按照每 K 个元素翻转的方式翻转链表。如果不满足 K 个元素的就不翻转。
+
+## 解题思路
+
+
+这一题是 problem 24 的加强版，problem 24 是两两相邻的元素，翻转链表。而 problem 25 要求的是 k 个相邻的元素，翻转链表，problem 相当于是 k = 2 的特殊情况。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func reverseKGroup(head *ListNode, k int) *ListNode {
+	node := head
+	for i := 0; i < k; i++ {
+		if node == nil {
+			return head
+		}
+		node = node.Next
+	}
+	newHead := reverse(head, node)
+	head.Next = reverseKGroup(node, k)
+	return newHead
+}
+
+func reverse(first *ListNode, last *ListNode) *ListNode {
+	prev := last
+	for first != last {
+		tmp := first.Next
+		first.Next = prev
+		prev = first
+		first = tmp
+	}
+	return prev
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0024.Swap-Nodes-in-Pairs/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0026.Remove-Duplicates-from-Sorted-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0026.Remove-Duplicates-from-Sorted-Array.md b/website/content/ChapterFour/0001~0099/0026.Remove-Duplicates-from-Sorted-Array.md
new file mode 100644
index 000000000..6d6034249
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0026.Remove-Duplicates-from-Sorted-Array.md
@@ -0,0 +1,135 @@
+# [26. Remove Duplicates from Sorted Array](https://leetcode.com/problems/remove-duplicates-from-sorted-array/)
+
+## 题目
+
+Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.
+
+Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
+
+**Example 1**:
+
+```
+
+Given nums = [1,1,2],
+
+Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
+
+It doesn't matter what you leave beyond the returned length.
+
+```
+
+**Example 2**:
+
+```
+
+Given nums = [0,0,1,1,1,2,2,3,3,4],
+
+Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
+
+It doesn't matter what values are set beyond the returned length.
+
+```
+
+**Clarification**:
+
+Confused why the returned value is an integer but your answer is an array?
+
+Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
+
+Internally you can think of this:
+
+```
+
+// nums is passed in by reference. (i.e., without making a copy)
+int len = removeElement(nums, val);
+
+// any modification to nums in your function would be known by the caller.
+// using the length returned by your function, it prints the first len elements.
+for (int i = 0; i < len; i++) {
+    print(nums[i]);
+}
+
+```
+
+## 题目大意
+
+给定一个有序数组 nums，对数组中的元素进行去重，使得原数组中的每个元素只有一个。最后返回去重以后数组的长度值。
+
+## 解题思路
+
+这道题和第 27 题很像。这道题和第 283 题，第 27 题基本一致，283 题是删除 0，27 题是删除指定元素，这一题是删除重复元素，实质是一样的。
+
+这里数组的删除并不是真的删除，只是将删除的元素移动到数组后面的空间内，然后返回数组实际剩余的元素个数，OJ 最终判断题目的时候会读取数组剩余个数的元素进行输出。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func removeDuplicates(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	last, finder := 0, 0
+	for last < len(nums)-1 {
+		for nums[finder] == nums[last] {
+			finder++
+			if finder == len(nums) {
+				return last + 1
+			}
+		}
+		nums[last+1] = nums[finder]
+		last++
+	}
+	return last + 1
+}
+
+// 解法二
+func removeDuplicates1(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	length := len(nums)
+	lastNum := nums[length-1]
+	i := 0
+	for i = 0; i < length-1; i++ {
+		if nums[i] == lastNum {
+			break
+		}
+		if nums[i+1] == nums[i] {
+			removeElement1(nums, i+1, nums[i])
+			// fmt.Printf("此时 num = %v length = %v\n", nums, length)
+		}
+	}
+	return i + 1
+}
+
+func removeElement1(nums []int, start, val int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	j := start
+	for i := start; i < len(nums); i++ {
+		if nums[i] != val {
+			if i != j {
+				nums[i], nums[j] = nums[j], nums[i]
+				j++
+			} else {
+				j++
+			}
+		}
+	}
+	return j
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0025.Reverse-Nodes-in-k-Group/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0027.Remove-Element/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0027.Remove-Element.md b/website/content/ChapterFour/0001~0099/0027.Remove-Element.md
new file mode 100644
index 000000000..ef1a75df3
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0027.Remove-Element.md
@@ -0,0 +1,98 @@
+# [27. Remove Element](https://leetcode.com/problems/remove-element/)
+
+## 题目
+
+Given an array nums and a value val, remove all instances of that value in-place and return the new length.
+
+Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
+
+The order of elements can be changed. It doesn't matter what you leave beyond the new length.
+
+**Example 1**:
+
+```
+
+Given nums = [3,2,2,3], val = 3,
+
+Your function should return length = 2, with the first two elements of nums being 2.
+
+It doesn't matter what you leave beyond the returned length.
+
+```
+
+**Example 2**:
+
+```
+
+Given nums = [0,1,2,2,3,0,4,2], val = 2,
+
+Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.
+
+Note that the order of those five elements can be arbitrary.
+
+It doesn't matter what values are set beyond the returned length.
+
+```
+
+**Clarification**:
+
+Confused why the returned value is an integer but your answer is an array?
+
+Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
+
+Internally you can think of this:
+
+```
+
+// nums is passed in by reference. (i.e., without making a copy)
+int len = removeElement(nums, val);
+
+// any modification to nums in your function would be known by the caller.
+// using the length returned by your function, it prints the first len elements.
+for (int i = 0; i < len; i++) {
+    print(nums[i]);
+}
+
+```
+
+## 题目大意
+
+给定一个数组 nums 和一个数值 val，将数组中所有等于 val 的元素删除，并返回剩余的元素个数。
+
+## 解题思路
+
+这道题和第 283 题很像。这道题和第 283 题基本一致，283 题是删除 0，这一题是给定的一个 val，实质是一样的。
+
+这里数组的删除并不是真的删除，只是将删除的元素移动到数组后面的空间内，然后返回数组实际剩余的元素个数，OJ 最终判断题目的时候会读取数组剩余个数的元素进行输出。
+
+## 代码
+
+```go
+
+package leetcode
+
+func removeElement(nums []int, val int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	j := 0
+	for i := 0; i < len(nums); i++ {
+		if nums[i] != val {
+			if i != j {
+				nums[i], nums[j] = nums[j], nums[i]
+			}
+			j++
+		}
+	}
+	return j
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0026.Remove-Duplicates-from-Sorted-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0028.Find-the-Index-of-the-First-Occurrence-in-a-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0028.Find-the-Index-of-the-First-Occurrence-in-a-String.md b/website/content/ChapterFour/0001~0099/0028.Find-the-Index-of-the-First-Occurrence-in-a-String.md
new file mode 100644
index 000000000..ee3577638
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0028.Find-the-Index-of-the-First-Occurrence-in-a-String.md
@@ -0,0 +1,93 @@
+# [28. Find the Index of the First Occurrence in a String](https://leetcode.com/problems/find-the-index-of-the-first-occurrence-in-a-string/)
+
+## 题目
+
+Implement strStr().
+
+Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
+
+
+**Example 1**:
+
+```
+
+Input: haystack = "hello", needle = "ll"
+Output: 2
+
+```
+
+**Example 2**:
+
+```
+
+Input: haystack = "aaaaa", needle = "bba"
+Output: -1
+
+```
+
+**Clarification**:  
+
+What should we return when needle is an empty string? This is a great question to ask during an interview.
+
+For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().
+
+## 题目大意
+
+
+实现一个查找 substring 的函数。如果在母串中找到了子串，返回子串在母串中出现的下标，如果没有找到，返回 -1，如果子串是空串，则返回 0 。
+
+## 解题思路
+
+这一题比较简单，直接写即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+// 解法一
+func strStr(haystack string, needle string) int {
+	for i := 0; ; i++ {
+		for j := 0; ; j++ {
+			if j == len(needle) {
+				return i
+			}
+			if i+j == len(haystack) {
+				return -1
+			}
+			if needle[j] != haystack[i+j] {
+				break
+			}
+		}
+	}
+}
+
+// 解法二
+func strStr1(haystack string, needle string) int {
+	return strings.Index(haystack, needle)
+}
+
+
+```
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0027.Remove-Element/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0029.Divide-Two-Integers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0029.Divide-Two-Integers.md b/website/content/ChapterFour/0001~0099/0029.Divide-Two-Integers.md
new file mode 100755
index 000000000..72f2e1062
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0029.Divide-Two-Integers.md
@@ -0,0 +1,155 @@
+# [29. Divide Two Integers](https://leetcode.com/problems/divide-two-integers/)
+
+
+## 题目
+
+Given two integers `dividend` and `divisor`, divide two integers without using multiplication, division and mod operator.
+
+Return the quotient after dividing `dividend` by `divisor`.
+
+The integer division should truncate toward zero.
+
+**Example 1**:
+
+    Input: dividend = 10, divisor = 3
+    Output: 3
+
+**Example 2**:
+
+    Input: dividend = 7, divisor = -3
+    Output: -2
+
+**Note**:
+
+- Both dividend and divisor will be 32-bit signed integers.
+- The divisor will never be 0.
+- Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 2^31 − 1 when the division result overflows.
+
+
+## 题目大意
+
+给定两个整数，被除数 dividend 和除数 divisor。将两数相除，要求不使用乘法、除法和 mod 运算符。返回被除数 dividend 除以除数 divisor 得到的商。
+
+说明:
+
+- 被除数和除数均为 32 位有符号整数。
+- 除数不为 0。
+- 假设我们的环境只能存储 32 位有符号整数，其数值范围是 [−2^31,  2^31 − 1]。本题中，如果除法结果溢出，则返回 2^31 − 1。
+
+
+## 解题思路
+
+- 给出除数和被除数，要求计算除法运算以后的商。注意值的取值范围在 [−2^31, 2^31 − 1] 之中。超过范围的都按边界计算。
+- 这一题可以用二分搜索来做。要求除法运算之后的商，把商作为要搜索的目标。商的取值范围是 [0, dividend]，所以从 0 到被除数之间搜索。利用二分，找到(商 + 1 ) * 除数 > 被除数并且 商 * 除数 ≤ 被除数 或者 (商+1)* 除数 ≥ 被除数并且商 * 除数 < 被除数的时候，就算找到了商，其余情况继续二分即可。最后还要注意符号和题目规定的 Int32 取值范围。
+- 二分的写法常写错的 3 点：
+    1. low ≤ high (注意二分循环退出的条件是小于等于)
+    2. mid = low + (high-low)>>1 (防止溢出)
+    3. low = mid + 1 ; high = mid - 1 (注意更新 low 和 high 的值，如果更新不对就会死循环)
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+)
+
+// 解法一 递归版的二分搜索
+func divide(dividend int, divisor int) int {
+	sign, res := -1, 0
+	// low, high := 0, abs(dividend)
+	if dividend == 0 {
+		return 0
+	}
+	if divisor == 1 {
+		return dividend
+	}
+	if dividend == math.MinInt32 && divisor == -1 {
+		return math.MaxInt32
+	}
+	if dividend > 0 && divisor > 0 || dividend < 0 && divisor < 0 {
+		sign = 1
+	}
+	if dividend > math.MaxInt32 {
+		dividend = math.MaxInt32
+	}
+	// 如果把递归改成非递归，可以改成下面这段代码
+	// for low <= high {
+	// 	quotient := low + (high-low)>>1
+	// 	if ((quotient+1)*abs(divisor) > abs(dividend) && quotient*abs(divisor) <= abs(dividend)) || ((quotient+1)*abs(divisor) >= abs(dividend) && quotient*abs(divisor) < abs(dividend)) {
+	// 		if (quotient+1)*abs(divisor) == abs(dividend) {
+	// 			res = quotient + 1
+	// 			break
+	// 		}
+	// 		res = quotient
+	// 		break
+	// 	}
+	// 	if (quotient+1)*abs(divisor) > abs(dividend) && quotient*abs(divisor) > abs(dividend) {
+	// 		high = quotient - 1
+	// 	}
+	// 	if (quotient+1)*abs(divisor) < abs(dividend) && quotient*abs(divisor) < abs(dividend) {
+	// 		low = quotient + 1
+	// 	}
+	// }
+	res = binarySearchQuotient(0, abs(dividend), abs(divisor), abs(dividend))
+	if res > math.MaxInt32 {
+		return sign * math.MaxInt32
+	}
+	if res < math.MinInt32 {
+		return sign * math.MinInt32
+	}
+	return sign * res
+}
+
+func binarySearchQuotient(low, high, val, dividend int) int {
+	quotient := low + (high-low)>>1
+	if ((quotient+1)*val > dividend && quotient*val <= dividend) || ((quotient+1)*val >= dividend && quotient*val < dividend) {
+		if (quotient+1)*val == dividend {
+			return quotient + 1
+		}
+		return quotient
+	}
+	if (quotient+1)*val > dividend && quotient*val > dividend {
+		return binarySearchQuotient(low, quotient-1, val, dividend)
+	}
+	if (quotient+1)*val < dividend && quotient*val < dividend {
+		return binarySearchQuotient(quotient+1, high, val, dividend)
+	}
+	return 0
+}
+
+// 解法二 非递归版的二分搜索
+func divide1(divided int, divisor int) int {
+	if divided == math.MinInt32 && divisor == -1 {
+		return math.MaxInt32
+	}
+	result := 0
+	sign := -1
+	if divided > 0 && divisor > 0 || divided < 0 && divisor < 0 {
+		sign = 1
+	}
+	dvd, dvs := abs(divided), abs(divisor)
+	for dvd >= dvs {
+		temp := dvs
+		m := 1
+		for temp<<1 <= dvd {
+			temp <<= 1
+			m <<= 1
+		}
+		dvd -= temp
+		result += m
+	}
+	return sign * result
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0028.Find-the-Index-of-the-First-Occurrence-in-a-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0030.Substring-with-Concatenation-of-All-Words/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0030.Substring-with-Concatenation-of-All-Words.md b/website/content/ChapterFour/0001~0099/0030.Substring-with-Concatenation-of-All-Words.md
new file mode 100644
index 000000000..00b3a0e0d
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0030.Substring-with-Concatenation-of-All-Words.md
@@ -0,0 +1,105 @@
+# [30. Substring with Concatenation of All Words](https://leetcode.com/problems/substring-with-concatenation-of-all-words/)
+
+## 题目
+
+You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
+
+**Example 1**:
+
+```
+
+Input:
+  s = "barfoothefoobarman",
+  words = ["foo","bar"]
+Output: [0,9]
+Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
+The output order does not matter, returning [9,0] is fine too.
+
+```
+
+**Example 2**:
+
+```
+
+Input:
+  s = "wordgoodgoodgoodbestword",
+  words = ["word","good","best","word"]
+Output: []
+
+```
+
+## 题目大意
+
+给定一个源字符串 s，再给一个字符串数组，要求在源字符串中找到由字符串数组各种组合组成的连续串的起始下标，如果存在多个，在结果中都需要输出。
+
+## 解题思路
+
+这一题看似很难，但是有 2 个限定条件也导致这题不是特别难。1. 字符串数组里面的字符串长度都是一样的。2. 要求字符串数组中的字符串都要连续连在一起的，前后顺序可以是任意排列组合。
+
+解题思路，先将字符串数组里面的所有字符串都存到 map 中，并累计出现的次数。然后从源字符串从头开始扫，每次判断字符串数组里面的字符串时候全部都用完了(计数是否为 0)，如果全部都用完了，并且长度正好是字符串数组任意排列组合的总长度，就记录下这个组合的起始下标。如果不符合，就继续考察源字符串的下一个字符，直到扫完整个源字符串。
+
+## 代码
+
+```go
+
+package leetcode
+
+func findSubstring(s string, words []string) []int {
+	if len(words) == 0 {
+		return []int{}
+	}
+	res := []int{}
+	counter := map[string]int{}
+	for _, w := range words {
+		counter[w]++
+	}
+	length, totalLen, tmpCounter := len(words[0]), len(words[0])*len(words), copyMap(counter)
+	for i, start := 0, 0; i < len(s)-length+1 && start < len(s)-length+1; i++ {
+		//fmt.Printf("sub = %v i = %v lenght = %v start = %v tmpCounter = %v totalLen = %v\n", s[i:i+length], i, length, start, tmpCounter, totalLen)
+		if tmpCounter[s[i:i+length]] > 0 {
+			tmpCounter[s[i:i+length]]--
+			//fmt.Printf("******sub = %v i = %v lenght = %v start = %v tmpCounter = %v totalLen = %v\n", s[i:i+length], i, length, start, tmpCounter, totalLen)
+			if checkWords(tmpCounter) && (i+length-start == totalLen) {
+				res = append(res, start)
+				continue
+			}
+			i = i + length - 1
+		} else {
+			start++
+			i = start - 1
+			tmpCounter = copyMap(counter)
+		}
+	}
+	return res
+}
+
+func checkWords(s map[string]int) bool {
+	flag := true
+	for _, v := range s {
+		if v > 0 {
+			flag = false
+			break
+		}
+	}
+	return flag
+}
+
+func copyMap(s map[string]int) map[string]int {
+	c := map[string]int{}
+	for k, v := range s {
+		c[k] = v
+	}
+	return c
+}
+
+
+```
+
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0029.Divide-Two-Integers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0031.Next-Permutation/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0031.Next-Permutation.md b/website/content/ChapterFour/0001~0099/0031.Next-Permutation.md
new file mode 100644
index 000000000..4689e75ba
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0031.Next-Permutation.md
@@ -0,0 +1,138 @@
+# [31. Next Permutation](https://leetcode.com/problems/next-permutation/)
+
+
+## 题目
+
+Implement **next permutation**, which rearranges numbers into the lexicographically next greater permutation of numbers.
+
+If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).
+
+The replacement must be **[in place](http://en.wikipedia.org/wiki/In-place_algorithm)** and use only constant extra memory.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3]
+Output: [1,3,2]
+```
+
+**Example 2:**
+
+```
+Input: nums = [3,2,1]
+Output: [1,2,3]
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,1,5]
+Output: [1,5,1]
+```
+
+**Example 4:**
+
+```
+Input: nums = [1]
+Output: [1]
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `0 <= nums[i] <= 100`
+
+## 题目大意
+
+实现获取 下一个排列 的函数，算法需要将给定数字序列重新排列成字典序中下一个更大的排列。如果不存在下一个更大的排列，则将数字重新排列成最小的排列（即升序排列）。必须 原地 修改，只允许使用额外常数空间。
+
+## 解题思路
+
+- 题目有 3 个问题需要解决。如何找到下一个排列。不存在下一个排列的时候如何生成最小的排列。如何原地修改。先解决第一个问题，如何找到下一个排列。下一个排列是找到一个大于当前排序的字典序，且变大的幅度最小。那么只能将较小的数与较大数做一次原地交换。并且较小数的下标要尽量靠右，较大数也要尽可能小。原地交换以后，还需要将较大数右边的数按照升序重新排列。这样交换以后，才能生成下一个排列。以排列 [8,9,6,10,7,2] 为例：能找到的符合条件的一对「较小数」与「较大数」的组合为 6 与 7，满足「较小数」尽量靠右，而「较大数」尽可能小。当完成交换后排列变为 [8,9,7,10,6,2]，此时我们可以重排「较小数」右边的序列，序列变为 [8,9,7,2,6,10]。
+- 第一步：在 `nums[i]` 中找到 `i` 使得 `nums[i] < nums[i+1]`，此时较小数为 `nums[i]`，并且 `[i+1, n)` 一定为下降区间。第二步：如果找到了这样的 `i` ，则在下降区间 `[i+1, n)` 中从后往前找到第一个 `j` ，使得 `nums[i] < nums[j]` ，此时较大数为 `nums[j]`。第三步，交换 `nums[i]` 和 `nums[j]`，此时区间 `[i+1, n)` 一定为降序区间。最后原地交换 `[i+1, n)` 区间内的元素，使其变为升序，无需对该区间进行排序。
+- 如果第一步找不到符合条件的下标 `i`，说明当前序列已经是一个最大的排列。那么应该直接执行第三步，生成最小的排列。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一
+func nextPermutation(nums []int) {
+	i, j := 0, 0
+	for i = len(nums) - 2; i >= 0; i-- {
+		if nums[i] < nums[i+1] {
+			break
+		}
+	}
+	if i >= 0 {
+		for j = len(nums) - 1; j > i; j-- {
+			if nums[j] > nums[i] {
+				break
+			}
+		}
+		swap(&nums, i, j)
+	}
+	reverse(&nums, i+1, len(nums)-1)
+}
+
+func reverse(nums *[]int, i, j int) {
+	for i < j {
+		swap(nums, i, j)
+		i++
+		j--
+	}
+}
+
+func swap(nums *[]int, i, j int) {
+	(*nums)[i], (*nums)[j] = (*nums)[j], (*nums)[i]
+}
+
+// 解法二
+// [2,(3),6,5,4,1] -> 2,(4),6,5,(3),1 -> 2,4, 1,3,5,6
+func nextPermutation1(nums []int) {
+	var n = len(nums)
+	var pIdx = checkPermutationPossibility(nums)
+	if pIdx == -1 {
+		reverse(&nums, 0, n-1)
+		return
+	}
+
+	var rp = len(nums) - 1
+	// start from right most to leftward,find the first number which is larger than PIVOT
+	for rp > 0 {
+		if nums[rp] > nums[pIdx] {
+			swap(&nums, pIdx, rp)
+			break
+		} else {
+			rp--
+		}
+	}
+	// Finally, Reverse all elements which are right from pivot
+	reverse(&nums, pIdx+1, n-1)
+}
+
+// checkPermutationPossibility returns 1st occurrence Index where
+// value is in decreasing order(from right to left)
+// returns -1 if not found(it's already in its last permutation)
+func checkPermutationPossibility(nums []int) (idx int) {
+	// search right to left for 1st number(from right) that is not in increasing order
+	var rp = len(nums) - 1
+	for rp > 0 {
+		if nums[rp-1] < nums[rp] {
+			idx = rp - 1
+			return idx
+		}
+		rp--
+	}
+	return -1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0030.Substring-with-Concatenation-of-All-Words/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0032.Longest-Valid-Parentheses/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0032.Longest-Valid-Parentheses.md b/website/content/ChapterFour/0001~0099/0032.Longest-Valid-Parentheses.md
new file mode 100644
index 000000000..db8cb726b
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0032.Longest-Valid-Parentheses.md
@@ -0,0 +1,115 @@
+# [32. Longest Valid Parentheses](https://leetcode.com/problems/longest-valid-parentheses/)
+
+
+## 题目
+
+Given a string containing just the characters `'('` and `')'`, find the length of the longest valid (well-formed) parentheses substring.
+
+**Example 1:**
+
+```
+Input: s = "(()"
+Output: 2
+Explanation: The longest valid parentheses substring is "()".
+```
+
+**Example 2:**
+
+```
+Input: s = ")()())"
+Output: 4
+Explanation: The longest valid parentheses substring is "()()".
+```
+
+**Example 3:**
+
+```
+Input: s = ""
+Output: 0
+```
+
+**Constraints:**
+
+- `0 <= s.length <= 3 * 104`
+- `s[i]` is `'('`, or `')'`.
+
+## 题目大意
+
+给你一个只包含 '(' 和 ')' 的字符串，找出最长有效（格式正确且连续）括号子串的长度。
+
+## 解题思路
+
+- 提到括号匹配，第一时间能让人想到的就是利用栈。这里需要计算嵌套括号的总长度，所以栈里面不能单纯的存左括号，而应该存左括号在原字符串的下标，这样通过下标相减可以获取长度。那么栈如果是非空，栈底永远存的是当前遍历过的字符串中**上一个没有被匹配的右括号的下标**。**上一个没有被匹配的右括号的下标**可以理解为每段括号匹配之间的“隔板”。例如，`())((()))`，第三个右括号，即为左右 2 段正确的括号匹配中间的“隔板”。“隔板”的存在影响计算最长括号长度。如果不存在“隔板”，前后 2 段正确的括号匹配应该“融合”在一起，最长长度为 `2 + 6 = 8`，但是这里存在了“隔板”，所以最长长度仅为 `6`。
+- 具体算法实现，遇到每个 `'('` ，将它的下标压入栈中。对于遇到的每个 `')'`，先弹出栈顶元素表示匹配了当前右括号。如果栈为空，说明当前的右括号为没有被匹配的右括号，于是将其下标放入栈中来更新**上一个没有被匹配的右括号的下标**。如果栈不为空，当前右括号的下标减去栈顶元素即为以该右括号为结尾的最长有效括号的长度。需要注意初始化时，不存在**上一个没有被匹配的右括号的下标**，那么将 `-1` 放入栈中，充当下标为 `0` 的“隔板”。时间复杂度 O(n)，空间复杂度 O(n)。
+- 在栈的方法中，实际用到的元素仅仅是栈底的**上一个没有被匹配的右括号的下标**。那么考虑能否把这个值存在一个变量中，这样可以省去栈 O(n) 的时间复杂度。利用两个计数器 left 和 right 。首先，从左到右遍历字符串，每当遇到 `'('`，增加 left 计数器，每当遇到 `')'` ，增加 right 计数器。每当 left 计数器与 right 计数器相等时，计算当前有效字符串的长度，并且记录目前为止找到的最长子字符串。当 right 计数器比 left 计数器大时，说明括号不匹配，于是将 left 和 right 计数器同时变回 0。这样的做法利用了贪心的思想，考虑了以当前字符下标结尾的有效括号长度，每次当右括号数量多于左括号数量的时候之前的字符就扔掉不再考虑，重新从下一个字符开始计算。
+- 但上面的做法会漏掉一种情况，就是遍历的时候左括号的数量始终大于右括号的数量，即 `(()` ，这种时候最长有效括号是求不出来的。解决办法是反向再计算一遍，如果从右往左计算，`(()` 先计算匹配的括号，最后只剩下 `'('`，这样依旧可以算出最长匹配的括号长度。反过来计算的方法和上述从左往右计算的方法一致：当 left 计数器比 right 计数器大时，将 left 和 right 计数器同时变回 0；当 left 计数器与 right 计数器相等时，计算当前有效字符串的长度，并且记录目前为止找到的最长子字符串。这种方法的时间复杂度是 O(n)，空间复杂度 O(1)。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 栈
+func longestValidParentheses(s string) int {
+	stack, res := []int{}, 0
+	stack = append(stack, -1)
+	for i := 0; i < len(s); i++ {
+		if s[i] == '(' {
+			stack = append(stack, i)
+		} else {
+			stack = stack[:len(stack)-1]
+			if len(stack) == 0 {
+				stack = append(stack, i)
+			} else {
+				res = max(res, i-stack[len(stack)-1])
+			}
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法二 双指针
+func longestValidParentheses1(s string) int {
+	left, right, maxLength := 0, 0, 0
+	for i := 0; i < len(s); i++ {
+		if s[i] == '(' {
+			left++
+		} else {
+			right++
+		}
+		if left == right {
+			maxLength = max(maxLength, 2*right)
+		} else if right > left {
+			left, right = 0, 0
+		}
+	}
+	left, right = 0, 0
+	for i := len(s) - 1; i >= 0; i-- {
+		if s[i] == '(' {
+			left++
+		} else {
+			right++
+		}
+		if left == right {
+			maxLength = max(maxLength, 2*left)
+		} else if left > right {
+			left, right = 0, 0
+		}
+	}
+	return maxLength
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0031.Next-Permutation/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0033.Search-in-Rotated-Sorted-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0033.Search-in-Rotated-Sorted-Array.md b/website/content/ChapterFour/0001~0099/0033.Search-in-Rotated-Sorted-Array.md
new file mode 100755
index 000000000..df41f1251
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0033.Search-in-Rotated-Sorted-Array.md
@@ -0,0 +1,85 @@
+# [33. Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)
+
+## 题目
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., `[0,1,2,4,5,6,7]` might become `[4,5,6,7,0,1,2]`).
+
+You are given a target value to search. If found in the array return its index, otherwise return `-1`.
+
+You may assume no duplicate exists in the array.
+
+Your algorithm's runtime complexity must be in the order of *O*(log *n*).
+
+**Example 1**:
+
+    Input: nums = [4,5,6,7,0,1,2], target = 0
+    Output: 4
+
+**Example 2**:
+
+    Input: nums = [4,5,6,7,0,1,2], target = 3
+    Output: -1
+
+
+## 题目大意
+
+假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。你可以假设数组中不存在重复的元素。
+
+你的算法时间复杂度必须是 O(log n) 级别。
+
+
+## 解题思路
+
+- 给出一个数组，数组中本来是从小到大排列的，并且数组中没有重复数字。但是现在把后面随机一段有序的放到数组前面，这样形成了前后两端有序的子序列。在这样的一个数组里面查找一个数，设计一个 O(log n) 的算法。如果找到就输出数组的小标，如果没有找到，就输出 -1 。
+- 由于数组基本有序，虽然中间有一个“断开点”，还是可以使用二分搜索的算法来实现。现在数组前面一段是数值比较大的数，后面一段是数值偏小的数。如果 mid 落在了前一段数值比较大的区间内了，那么一定有 `nums[mid] > nums[low]`，如果是落在后面一段数值比较小的区间内，`nums[mid] ≤ nums[low]` 。如果 mid 落在了后一段数值比较小的区间内了，那么一定有 `nums[mid] < nums[high]`，如果是落在前面一段数值比较大的区间内，`nums[mid] ≤ nums[high]` 。还有 `nums[low] == nums[mid]` 和 `nums[high] == nums[mid]` 的情况，单独处理即可。最后找到则输出 mid，没有找到则输出 -1 。
+
+## 代码
+
+```go
+
+package leetcode
+
+func search33(nums []int, target int) int {
+	if len(nums) == 0 {
+		return -1
+	}
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[mid] == target {
+			return mid
+		} else if nums[mid] > nums[low] { // 在数值大的一部分区间里
+			if nums[low] <= target && target < nums[mid] {
+				high = mid - 1
+			} else {
+				low = mid + 1
+			}
+		} else if nums[mid] < nums[high] { // 在数值小的一部分区间里
+			if nums[mid] < target && target <= nums[high] {
+				low = mid + 1
+			} else {
+				high = mid - 1
+			}
+		} else {
+			if nums[low] == nums[mid] {
+				low++
+			}
+			if nums[high] == nums[mid] {
+				high--
+			}
+		}
+	}
+	return -1
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0032.Longest-Valid-Parentheses/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array.md b/website/content/ChapterFour/0001~0099/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array.md
new file mode 100755
index 000000000..bed0481ab
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array.md
@@ -0,0 +1,128 @@
+# [34. Find First and Last Position of Element in Sorted Array](https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/)
+
+
+## 题目
+
+Given an array of integers `nums` sorted in ascending order, find the starting and ending position of a given `target` value.
+
+Your algorithm's runtime complexity must be in the order of *O*(log *n*).
+
+If the target is not found in the array, return `[-1, -1]`.
+
+**Example 1**:
+
+    Input: nums = [5,7,7,8,8,10], target = 8
+    Output: [3,4]
+
+**Example 2**:
+
+    Input: nums = [5,7,7,8,8,10], target = 6
+    Output: [-1,-1]
+
+## 题目大意
+
+给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。你的算法时间复杂度必须是 O(log n) 级别。如果数组中不存在目标值，返回 [-1, -1]。
+
+
+## 解题思路
+
+- 给出一个有序数组 `nums` 和一个数 `target`，要求在数组中找到第一个和这个元素相等的元素下标，最后一个和这个元素相等的元素下标。
+- 这一题是经典的二分搜索变种题。二分搜索有 4 大基础变种题：
+    1. 查找第一个值等于给定值的元素
+    2. 查找最后一个值等于给定值的元素
+    3. 查找第一个大于等于给定值的元素
+    4. 查找最后一个小于等于给定值的元素
+
+    这一题的解题思路可以分别利用变种 1 和变种 2 的解法就可以做出此题。或者用一次变种 1 的方法，然后循环往后找到最后一个与给定值相等的元素。不过后者这种方法可能会使时间复杂度下降到 O(n)，因为有可能数组中 n 个元素都和给定元素相同。(4 大基础变种的实现见代码)
+
+## 代码
+
+```go
+
+package leetcode
+
+func searchRange(nums []int, target int) []int {
+	return []int{searchFirstEqualElement(nums, target), searchLastEqualElement(nums, target)}
+
+}
+
+// 二分查找第一个与 target 相等的元素，时间复杂度 O(logn)
+func searchFirstEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == 0) || (nums[mid-1] != target) { // 找到第一个与 target 相等的元素
+				return mid
+			}
+			high = mid - 1
+		}
+	}
+	return -1
+}
+
+// 二分查找最后一个与 target 相等的元素，时间复杂度 O(logn)
+func searchLastEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == len(nums)-1) || (nums[mid+1] != target) { // 找到最后一个与 target 相等的元素
+				return mid
+			}
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// 二分查找第一个大于等于 target 的元素，时间复杂度 O(logn)
+func searchFirstGreaterElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] >= target {
+			if (mid == 0) || (nums[mid-1] < target) { // 找到第一个大于等于 target 的元素
+				return mid
+			}
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// 二分查找最后一个小于等于 target 的元素，时间复杂度 O(logn)
+func searchLastLessElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] <= target {
+			if (mid == len(nums)-1) || (nums[mid+1] > target) { // 找到最后一个小于等于 target 的元素
+				return mid
+			}
+			low = mid + 1
+		} else {
+			high = mid - 1
+		}
+	}
+	return -1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0033.Search-in-Rotated-Sorted-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0035.Search-Insert-Position/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0035.Search-Insert-Position.md b/website/content/ChapterFour/0001~0099/0035.Search-Insert-Position.md
new file mode 100755
index 000000000..ec99aca71
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0035.Search-Insert-Position.md
@@ -0,0 +1,72 @@
+# [35. Search Insert Position](https://leetcode.com/problems/search-insert-position/)
+
+
+## 题目
+
+Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
+
+You may assume no duplicates in the array.
+
+**Example 1**:
+
+    Input: [1,3,5,6], 5
+    Output: 2
+
+**Example 2**:
+
+    Input: [1,3,5,6], 2
+    Output: 1
+
+**Example 3**:
+
+    Input: [1,3,5,6], 7
+    Output: 4
+
+**Example 4**:
+
+    Input: [1,3,5,6], 0
+    Output: 0
+
+
+## 题目大意
+
+给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。
+
+你可以假设数组中无重复元素。
+
+## 解题思路
+
+- 给出一个已经从小到大排序后的数组，要求在数组中找到插入 target 元素的位置。
+- 这一题是经典的二分搜索的变种题，在有序数组中找到最后一个比 target 小的元素。
+
+## 代码
+
+```go
+
+package leetcode
+
+func searchInsert(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[mid] >= target {
+			high = mid - 1
+		} else {
+			if (mid == len(nums)-1) || (nums[mid+1] >= target) {
+				return mid + 1
+			}
+			low = mid + 1
+		}
+	}
+	return 0
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0036.Valid-Sudoku/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0036.Valid-Sudoku.md b/website/content/ChapterFour/0001~0099/0036.Valid-Sudoku.md
new file mode 100755
index 000000000..68ad64cd1
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0036.Valid-Sudoku.md
@@ -0,0 +1,175 @@
+# [36. Valid Sudoku](https://leetcode.com/problems/valid-sudoku/)
+
+
+## 题目
+
+Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated **according to the following rules**:
+
+1. Each row must contain the digits `1-9` without repetition.
+2. Each column must contain the digits `1-9` without repetition.
+3. Each of the 9 `3x3` sub-boxes of the grid must contain the digits `1-9` without repetition.
+
+![](https://upload.wikimedia.org/wikipedia/commons/thumb/f/ff/Sudoku-by-L2G-20050714.svg/250px-Sudoku-by-L2G-20050714.svg.png)
+
+A partially filled sudoku which is valid.
+
+The Sudoku board could be partially filled, where empty cells are filled with the character `'.'`.
+
+**Example 1**:
+
+
+    Input:
+    [
+      ["5","3",".",".","7",".",".",".","."],
+      ["6",".",".","1","9","5",".",".","."],
+      [".","9","8",".",".",".",".","6","."],
+      ["8",".",".",".","6",".",".",".","3"],
+      ["4",".",".","8",".","3",".",".","1"],
+      ["7",".",".",".","2",".",".",".","6"],
+      [".","6",".",".",".",".","2","8","."],
+      [".",".",".","4","1","9",".",".","5"],
+      [".",".",".",".","8",".",".","7","9"]
+    ]
+    Output: true
+
+
+**Example 2**:
+
+
+    Input:
+    [
+      ["8","3",".",".","7",".",".",".","."],
+      ["6",".",".","1","9","5",".",".","."],
+      [".","9","8",".",".",".",".","6","."],
+      ["8",".",".",".","6",".",".",".","3"],
+      ["4",".",".","8",".","3",".",".","1"],
+      ["7",".",".",".","2",".",".",".","6"],
+      [".","6",".",".",".",".","2","8","."],
+      [".",".",".","4","1","9",".",".","5"],
+      [".",".",".",".","8",".",".","7","9"]
+    ]
+    Output: false
+    Explanation: Same as Example 1, except with the 5 in the top left corner being 
+        modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.
+
+
+**Note**:
+
+- A Sudoku board (partially filled) could be valid but is not necessarily solvable.
+- Only the filled cells need to be validated according to the mentioned rules.
+- The given board contain only digits `1-9` and the character `'.'`.
+- The given board size is always `9x9`.
+
+## 题目大意
+
+判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。
+
+1. 数字 1-9 在每一行只能出现一次。
+2. 数字 1-9 在每一列只能出现一次。
+3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
+
+
+## 解题思路
+
+- 给出一个数独的棋盘，要求判断这个棋盘当前是否满足数独的要求：即行列是否都只包含 1-9，每个九宫格里面是否也只包含 1-9 。
+- 注意这题和第 37 题是不同的，这一题是判断当前棋盘状态是否满足数独的要求，而第 37 题是要求求解数独。本题中的棋盘有些是无解的，但是棋盘状态是满足题意的。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strconv"
+
+// 解法一 暴力遍历，时间复杂度 O(n^3)
+func isValidSudoku(board [][]byte) bool {
+	// 判断行 row
+	for i := 0; i < 9; i++ {
+		tmp := [10]int{}
+		for j := 0; j < 9; j++ {
+			cellVal := board[i][j : j+1]
+			if string(cellVal) != "." {
+				index, _ := strconv.Atoi(string(cellVal))
+				if index > 9 || index < 1 {
+					return false
+				}
+				if tmp[index] == 1 {
+					return false
+				}
+				tmp[index] = 1
+			}
+		}
+	}
+	// 判断列 column
+	for i := 0; i < 9; i++ {
+		tmp := [10]int{}
+		for j := 0; j < 9; j++ {
+			cellVal := board[j][i]
+			if string(cellVal) != "." {
+				index, _ := strconv.Atoi(string(cellVal))
+				if index > 9 || index < 1 {
+					return false
+				}
+				if tmp[index] == 1 {
+					return false
+				}
+				tmp[index] = 1
+			}
+		}
+	}
+	// 判断 9宫格 3X3 cell
+	for i := 0; i < 3; i++ {
+		for j := 0; j < 3; j++ {
+			tmp := [10]int{}
+			for ii := i * 3; ii < i*3+3; ii++ {
+				for jj := j * 3; jj < j*3+3; jj++ {
+					cellVal := board[ii][jj]
+					if string(cellVal) != "." {
+						index, _ := strconv.Atoi(string(cellVal))
+						if tmp[index] == 1 {
+							return false
+						}
+						tmp[index] = 1
+					}
+				}
+			}
+		}
+	}
+	return true
+}
+
+// 解法二 添加缓存，时间复杂度 O(n^2)
+func isValidSudoku1(board [][]byte) bool {
+	rowbuf, colbuf, boxbuf := make([][]bool, 9), make([][]bool, 9), make([][]bool, 9)
+	for i := 0; i < 9; i++ {
+		rowbuf[i] = make([]bool, 9)
+		colbuf[i] = make([]bool, 9)
+		boxbuf[i] = make([]bool, 9)
+	}
+	// 遍历一次，添加缓存
+	for r := 0; r < 9; r++ {
+		for c := 0; c < 9; c++ {
+			if board[r][c] != '.' {
+				num := board[r][c] - '0' - byte(1)
+				if rowbuf[r][num] || colbuf[c][num] || boxbuf[r/3*3+c/3][num] {
+					return false
+				}
+				rowbuf[r][num] = true
+				colbuf[c][num] = true
+				boxbuf[r/3*3+c/3][num] = true // r,c 转换到box方格中
+			}
+		}
+	}
+	return true
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0035.Search-Insert-Position/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0037.Sudoku-Solver/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0037.Sudoku-Solver.md b/website/content/ChapterFour/0001~0099/0037.Sudoku-Solver.md
new file mode 100755
index 000000000..541b4efcf
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0037.Sudoku-Solver.md
@@ -0,0 +1,125 @@
+# [37. Sudoku Solver](https://leetcode.com/problems/sudoku-solver/)
+
+
+
+## 题目
+
+Write a program to solve a Sudoku puzzle by filling the empty cells.
+
+A sudoku solution must satisfy **all of the following rules**:
+
+1. Each of the digits `1-9` must occur exactly once in each row.
+2. Each of the digits `1-9` must occur exactly once in each column.
+3. Each of the the digits `1-9` must occur exactly once in each of the 9 `3x3` sub-boxes of the grid.
+
+Empty cells are indicated by the character `'.'`.
+
+![](https://upload.wikimedia.org/wikipedia/commons/thumb/f/ff/Sudoku-by-L2G-20050714.svg/250px-Sudoku-by-L2G-20050714.svg.png)
+
+A sudoku puzzle...
+
+![](https://upload.wikimedia.org/wikipedia/commons/thumb/3/31/Sudoku-by-L2G-20050714_solution.svg/250px-Sudoku-by-L2G-20050714_solution.svg.png)
+
+...and its solution numbers marked in red.
+
+**Note**:
+
+- The given board contain only digits `1-9` and the character `'.'`.
+- You may assume that the given Sudoku puzzle will have a single unique solution.
+- The given board size is always `9x9`.
+
+## 题目大意
+
+
+编写一个程序，通过已填充的空格来解决数独问题。一个数独的解法需遵循如下规则：
+
+1. 数字 1-9 在每一行只能出现一次。
+2. 数字 1-9 在每一列只能出现一次。
+3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
+
+空白格用 '.' 表示。
+
+
+## 解题思路
+
+- 给出一个数独谜题，要求解出这个数独
+- 解题思路 DFS 暴力回溯枚举。数独要求每横行，每竖行，每九宫格内，`1-9` 的数字不能重复，每次放下一个数字的时候，在这 3 个地方都需要判断一次。
+- 另外找到一组解以后就不需要再继续回溯了，直接返回即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+type position struct {
+	x int
+	y int
+}
+
+func solveSudoku(board [][]byte) {
+	pos, find := []position{}, false
+	for i := 0; i < len(board); i++ {
+		for j := 0; j < len(board[0]); j++ {
+			if board[i][j] == '.' {
+				pos = append(pos, position{x: i, y: j})
+			}
+		}
+	}
+	putSudoku(&board, pos, 0, &find)
+}
+
+func putSudoku(board *[][]byte, pos []position, index int, succ *bool) {
+	if *succ == true {
+		return
+	}
+	if index == len(pos) {
+		*succ = true
+		return
+	}
+	for i := 1; i < 10; i++ {
+		if checkSudoku(board, pos[index], i) && !*succ {
+			(*board)[pos[index].x][pos[index].y] = byte(i) + '0'
+			putSudoku(board, pos, index+1, succ)
+			if *succ == true {
+				return
+			}
+			(*board)[pos[index].x][pos[index].y] = '.'
+		}
+	}
+}
+
+func checkSudoku(board *[][]byte, pos position, val int) bool {
+	// 判断横行是否有重复数字
+	for i := 0; i < len((*board)[0]); i++ {
+		if (*board)[pos.x][i] != '.' && int((*board)[pos.x][i]-'0') == val {
+			return false
+		}
+	}
+	// 判断竖行是否有重复数字
+	for i := 0; i < len((*board)); i++ {
+		if (*board)[i][pos.y] != '.' && int((*board)[i][pos.y]-'0') == val {
+			return false
+		}
+	}
+	// 判断九宫格是否有重复数字
+	posx, posy := pos.x-pos.x%3, pos.y-pos.y%3
+	for i := posx; i < posx+3; i++ {
+		for j := posy; j < posy+3; j++ {
+			if (*board)[i][j] != '.' && int((*board)[i][j]-'0') == val {
+				return false
+			}
+		}
+	}
+	return true
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0036.Valid-Sudoku/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0039.Combination-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0039.Combination-Sum.md b/website/content/ChapterFour/0001~0099/0039.Combination-Sum.md
new file mode 100755
index 000000000..4681f48b6
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0039.Combination-Sum.md
@@ -0,0 +1,94 @@
+# [39. Combination Sum](https://leetcode.com/problems/combination-sum/)
+
+
+## 题目
+
+Given a **set** of candidate numbers (`candidates`) **(without duplicates)** and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sums to `target`.
+
+The **same** repeated number may be chosen from `candidates` unlimited number of times.
+
+**Note**:
+
+- All numbers (including `target`) will be positive integers.
+- The solution set must not contain duplicate combinations.
+
+**Example 1**:
+
+
+    Input: candidates = [2,3,6,7], target = 7,
+    A solution set is:
+    [
+      [7],
+      [2,2,3]
+    ]
+
+
+**Example 2**:
+
+
+    Input: candidates = [2,3,5], target = 8,
+    A solution set is:
+    [
+      [2,2,2,2],
+      [2,3,3],
+      [3,5]
+    ]
+
+
+## 题目大意
+
+给定一个无重复元素的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
+
+candidates 中的数字可以无限制重复被选取。
+
+
+## 解题思路
+
+- 题目要求出总和为 sum 的所有组合，组合需要去重。
+- 这一题和第 47 题类似，只不过元素可以反复使用。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+func combinationSum(candidates []int, target int) [][]int {
+	if len(candidates) == 0 {
+		return [][]int{}
+	}
+	c, res := []int{}, [][]int{}
+	sort.Ints(candidates)
+	findcombinationSum(candidates, target, 0, c, &res)
+	return res
+}
+
+func findcombinationSum(nums []int, target, index int, c []int, res *[][]int) {
+	if target <= 0 {
+		if target == 0 {
+			b := make([]int, len(c))
+			copy(b, c)
+			*res = append(*res, b)
+		}
+		return
+	}
+	for i := index; i < len(nums); i++ {
+		if nums[i] > target { // 这里可以剪枝优化
+			break
+		}
+		c = append(c, nums[i])
+		findcombinationSum(nums, target-nums[i], i, c, res) // 注意这里迭代的时候 index 依旧不变，因为一个元素可以取多次
+		c = c[:len(c)-1]
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0037.Sudoku-Solver/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0040.Combination-Sum-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0040.Combination-Sum-II.md b/website/content/ChapterFour/0001~0099/0040.Combination-Sum-II.md
new file mode 100755
index 000000000..aa46f3f10
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0040.Combination-Sum-II.md
@@ -0,0 +1,96 @@
+# [40. Combination Sum II](https://leetcode.com/problems/combination-sum-ii/)
+
+
+## 题目
+
+Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sums to `target`.
+
+Each number in `candidates` may only be used **once** in the combination.
+
+**Note**:
+
+- All numbers (including `target`) will be positive integers.
+- The solution set must not contain duplicate combinations.
+
+**Example 1**:
+
+
+    Input: candidates = [10,1,2,7,6,1,5], target = 8,
+    A solution set is:
+    [
+      [1, 7],
+      [1, 2, 5],
+      [2, 6],
+      [1, 1, 6]
+    ]
+
+
+**Example 2**:
+
+
+    Input: candidates = [2,5,2,1,2], target = 5,
+    A solution set is:
+    [
+      [1,2,2],
+      [5]
+    ]
+
+## 题目大意
+
+给定一个数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
+
+candidates 中的每个数字在每个组合中只能使用一次。
+
+
+## 解题思路
+
+- 题目要求出总和为 sum 的所有组合，组合需要去重。这一题是第 39 题的加强版，第 39 题中元素可以重复利用(重复元素可无限次使用)，这一题中元素只能有限次数的利用，因为存在重复元素，并且每个元素只能用一次(重复元素只能使用有限次)
+- 这一题和第 47 题类似。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+func combinationSum2(candidates []int, target int) [][]int {
+	if len(candidates) == 0 {
+		return [][]int{}
+	}
+	c, res := []int{}, [][]int{}
+	sort.Ints(candidates) // 这里是去重的关键逻辑
+	findcombinationSum2(candidates, target, 0, c, &res)
+	return res
+}
+
+func findcombinationSum2(nums []int, target, index int, c []int, res *[][]int) {
+	if target == 0 {
+		b := make([]int, len(c))
+		copy(b, c)
+		*res = append(*res, b)
+		return
+	}
+	for i := index; i < len(nums); i++ {
+		if i > index && nums[i] == nums[i-1] { // 这里是去重的关键逻辑,本次不取重复数字，下次循环可能会取重复数字
+			continue
+		}
+		if target >= nums[i] {
+			c = append(c, nums[i])
+			findcombinationSum2(nums, target-nums[i], i+1, c, res)
+			c = c[:len(c)-1]
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0039.Combination-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0041.First-Missing-Positive/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0041.First-Missing-Positive.md b/website/content/ChapterFour/0001~0099/0041.First-Missing-Positive.md
new file mode 100644
index 000000000..6ab60da34
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0041.First-Missing-Positive.md
@@ -0,0 +1,73 @@
+# [41. First Missing Positive](https://leetcode.com/problems/first-missing-positive/description/)
+
+## 题目
+
+Given an unsorted integer array, find the smallest missing positive integer.
+
+**Example 1**:  
+
+```
+
+Input: [1,2,0]  
+Output: 3  
+
+```
+
+**Example 2**:  
+
+```
+
+Input: [3,4,-1,1]  
+Output: 2  
+
+```
+
+**Example 3**:  
+
+```
+
+Input: [7,8,9,11,12]  
+Output: 1  
+
+```
+
+**Note**:  
+
+Your algorithm should run in O(n) time and uses constant extra space.
+
+## 题目大意
+
+找到缺失的第一个正整数。
+
+## 解题思路
+
+
+为了减少时间复杂度，可以把 input 数组都装到 map 中，然后 i 循环从 1 开始，依次比对 map 中是否存在 i，只要不存在 i 就立即返回结果，即所求。
+
+## 代码
+
+```go
+
+package leetcode
+
+func firstMissingPositive(nums []int) int {
+	numMap := make(map[int]int, len(nums))
+	for _, v := range nums {
+		numMap[v] = v
+	}
+	for index := 1; index < len(nums)+1; index++ {
+		if _, ok := numMap[index]; !ok {
+			return index
+		}
+	}
+	return len(nums) + 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0040.Combination-Sum-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0042.Trapping-Rain-Water/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0042.Trapping-Rain-Water.md b/website/content/ChapterFour/0001~0099/0042.Trapping-Rain-Water.md
new file mode 100644
index 000000000..be6a256ef
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0042.Trapping-Rain-Water.md
@@ -0,0 +1,68 @@
+# [42. Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water/)
+
+## 题目
+
+Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
+
+![](https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png)
+
+
+The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!
+
+
+**Example**:
+
+```go
+Input: [0,1,0,2,1,0,1,3,2,1,2,1]
+Output: 6
+```
+
+## 题目大意
+
+从 x 轴开始，给出一个数组，数组里面的数字代表从 (0,0) 点开始，宽度为 1 个单位，高度为数组元素的值。如果下雨了，问这样一个容器能装多少单位的水？
+
+## 解题思路
+
+- 每个数组里面的元素值可以想象成一个左右都有壁的圆柱筒。例如下图中左边的第二个元素 1，当前左边最大的元素是 2 ，所以 2 高度的水会装到 1 的上面(因为想象成了左右都有筒壁)。这道题的思路就是左指针从 0 开始往右扫，右指针从最右边开始往左扫。额外还需要 2 个变量分别记住左边最大的高度和右边最大高度。遍历扫数组元素的过程中，如果左指针的高度比右指针的高度小，就不断的移动左指针，否则移动右指针。循环的终止条件就是左右指针碰上以后就结束。只要数组中元素的高度比保存的局部最大高度小，就累加 res 的值，否则更新局部最大高度。最终解就是 res 的值。
+    ![](https://image.ibb.co/d6A2ZU/IMG-0139.jpg)  
+- 抽象一下，本题是想求针对每个 i，找到它左边最大值 leftMax，右边的最大值 rightMax，然后 min(leftMax，rightMax) 为能够接到水的高度。left 和 right 指针是两边往中间移动的游标指针。最傻的解题思路是针对每个下标 i，往左循环找到第一个最大值，往右循环找到第一个最大值，然后把这两个最大值取出最小者，即为当前雨水的高度。这样做时间复杂度高，浪费了很多循环。i 在从左往右的过程中，是可以动态维护最大值的。右边的最大值用右边的游标指针来维护。从左往右扫一遍下标，和，从两边往中间遍历一遍下标，是相同的结果，每个下标都遍历了一次。    
+    ![](https://img.halfrost.com/Leetcode/leetcode_42_1.png)  
+- 每个 i 的宽度固定为 1，所以每个“坑”只需要求出高度，即当前这个“坑”能积攒的雨水。最后依次将每个“坑”中的雨水相加即是能接到的雨水数。    
+    ![](https://img.halfrost.com/Leetcode/leetcode_42_0.png)  
+
+## 代码
+
+```go
+package leetcode
+
+func trap(height []int) int {
+	res, left, right, maxLeft, maxRight := 0, 0, len(height)-1, 0, 0
+	for left <= right {
+		if height[left] <= height[right] {
+			if height[left] > maxLeft {
+				maxLeft = height[left]
+			} else {
+				res += maxLeft - height[left]
+			}
+			left++
+		} else {
+			if height[right] >= maxRight {
+				maxRight = height[right]
+			} else {
+				res += maxRight - height[right]
+			}
+			right--
+		}
+	}
+	return res
+}
+```
+
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0041.First-Missing-Positive/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0043.Multiply-Strings/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0043.Multiply-Strings.md b/website/content/ChapterFour/0001~0099/0043.Multiply-Strings.md
new file mode 100644
index 000000000..b807342cf
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0043.Multiply-Strings.md
@@ -0,0 +1,73 @@
+# [43. Multiply Strings](https://leetcode.com/problems/multiply-strings/)
+
+
+## 题目
+
+Given two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.
+
+**Note:** You must not use any built-in BigInteger library or convert the inputs to integer directly.
+
+**Example 1:**
+
+```
+Input: num1 = "2", num2 = "3"
+Output: "6"
+```
+
+**Example 2:**
+
+```
+Input: num1 = "123", num2 = "456"
+Output: "56088"
+```
+
+**Constraints:**
+
+- `1 <= num1.length, num2.length <= 200`
+- `num1` and `num2` consist of digits only.
+- Both `num1` and `num2` do not contain any leading zero, except the number `0` itself.
+
+## 题目大意
+
+给定两个以字符串形式表示的非负整数 num1 和 num2，返回 num1 和 num2 的乘积，它们的乘积也表示为字符串形式。
+
+## 解题思路
+
+- 用数组模拟乘法。创建一个数组长度为 `len(num1) + len(num2)` 的数组用于存储乘积。对于任意 `0 ≤ i < len(num1)`，`0 ≤ j < len(num2)`，`num1[i] * num2[j]` 的结果位于 `tmp[i+j+1]`，如果 `tmp[i+j+1]≥10`，则将进位部分加到 `tmp[i+j]`。最后，将数组 `tmp` 转成字符串，如果最高位是 0 则舍弃最高位。
+
+## 代码
+
+```go
+package leetcode
+
+func multiply(num1 string, num2 string) string {
+	if num1 == "0" || num2 == "0" {
+		return "0"
+	}
+	b1, b2, tmp := []byte(num1), []byte(num2), make([]int, len(num1)+len(num2))
+	for i := 0; i < len(b1); i++ {
+		for j := 0; j < len(b2); j++ {
+			tmp[i+j+1] += int(b1[i]-'0') * int(b2[j]-'0')
+		}
+	}
+	for i := len(tmp) - 1; i > 0; i-- {
+		tmp[i-1] += tmp[i] / 10
+		tmp[i] = tmp[i] % 10
+	}
+	if tmp[0] == 0 {
+		tmp = tmp[1:]
+	}
+	res := make([]byte, len(tmp))
+	for i := 0; i < len(tmp); i++ {
+		res[i] = '0' + byte(tmp[i])
+	}
+	return string(res)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0042.Trapping-Rain-Water/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0045.Jump-Game-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0045.Jump-Game-II.md b/website/content/ChapterFour/0001~0099/0045.Jump-Game-II.md
new file mode 100644
index 000000000..fc910ad23
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0045.Jump-Game-II.md
@@ -0,0 +1,74 @@
+# [45. Jump Game II](https://leetcode.com/problems/jump-game-ii/)
+
+
+## 题目
+
+Given an array of non-negative integers `nums`, you are initially positioned at the first index of the array.
+
+Each element in the array represents your maximum jump length at that position.
+
+Your goal is to reach the last index in the minimum number of jumps.
+
+You can assume that you can always reach the last index.
+
+**Example 1:**
+
+```
+Input: nums = [2,3,1,1,4]
+Output: 2
+Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,3,0,1,4]
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 1000`
+- `0 <= nums[i] <= 10^5`
+
+## 题目大意
+
+给定一个非负整数数组，你最初位于数组的第一个位置。数组中的每个元素代表你在该位置可以跳跃的最大长度。你的目标是使用最少的跳跃次数到达数组的最后一个位置。
+
+## 解题思路
+
+- 要求找到最少跳跃次数，顺理成章的会想到用贪心算法解题。扫描步数数组，维护当前能够到达最大下标的位置，记为能到达的最远边界，如果扫描过程中到达了最远边界，更新边界并将跳跃次数 + 1。
+- 扫描数组的时候，其实不需要扫描最后一个元素，因为在跳到最后一个元素之前，能到达的最远边界一定大于等于最后一个元素的位置，不然就跳不到最后一个元素，到达不了终点了；如果遍历到最后一个元素，说明边界正好为最后一个位置，最终跳跃次数直接 + 1 即可，也不需要访问最后一个元素。
+
+## 代码
+
+```go
+package leetcode
+
+func jump(nums []int) int {
+	if len(nums) == 1 {
+		return 0
+	}
+	needChoose, canReach, step := 0, 0, 0
+	for i, x := range nums {
+		if i+x > canReach {
+			canReach = i + x
+			if canReach >= len(nums)-1 {
+				return step + 1
+			}
+		}
+		if i == needChoose {
+			needChoose = canReach
+			step++
+		}
+	}
+	return step
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0043.Multiply-Strings/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0046.Permutations/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0046.Permutations.md b/website/content/ChapterFour/0001~0099/0046.Permutations.md
new file mode 100755
index 000000000..54fceeeb0
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0046.Permutations.md
@@ -0,0 +1,73 @@
+# [46. Permutations](https://leetcode.com/problems/permutations/)
+
+
+## 题目
+
+Given a collection of **distinct** integers, return all possible permutations.
+
+**Example**:
+
+
+    Input: [1,2,3]
+    Output:
+    [
+      [1,2,3],
+      [1,3,2],
+      [2,1,3],
+      [2,3,1],
+      [3,1,2],
+      [3,2,1]
+    ]
+
+
+## 题目大意
+
+给定一个没有重复数字的序列，返回其所有可能的全排列。
+
+
+## 解题思路
+
+- 求出一个数组的排列组合中的所有排列，用 DFS 深搜即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func permute(nums []int) [][]int {
+	if len(nums) == 0 {
+		return [][]int{}
+	}
+	used, p, res := make([]bool, len(nums)), []int{}, [][]int{}
+	generatePermutation(nums, 0, p, &res, &used)
+	return res
+}
+
+func generatePermutation(nums []int, index int, p []int, res *[][]int, used *[]bool) {
+	if index == len(nums) {
+		temp := make([]int, len(p))
+		copy(temp, p)
+		*res = append(*res, temp)
+		return
+	}
+	for i := 0; i < len(nums); i++ {
+		if !(*used)[i] {
+			(*used)[i] = true
+			p = append(p, nums[i])
+			generatePermutation(nums, index+1, p, res, used)
+			p = p[:len(p)-1]
+			(*used)[i] = false
+		}
+	}
+	return
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0045.Jump-Game-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0047.Permutations-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0047.Permutations-II.md b/website/content/ChapterFour/0001~0099/0047.Permutations-II.md
new file mode 100755
index 000000000..e6aaccdff
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0047.Permutations-II.md
@@ -0,0 +1,77 @@
+# [47. Permutations II](https://leetcode.com/problems/permutations-ii/)
+
+
+## 题目
+
+Given a collection of numbers that might contain duplicates, return all possible unique permutations.
+
+**Example**:
+
+
+    Input: [1,1,2]
+    Output:
+    [
+      [1,1,2],
+      [1,2,1],
+      [2,1,1]
+    ]
+
+
+## 题目大意
+
+给定一个可包含重复数字的序列，返回所有不重复的全排列。
+
+## 解题思路
+
+- 这一题是第 46 题的加强版，第 46 题中求数组的排列，数组中元素不重复，但是这一题中，数组元素会重复，所以需要最终排列出来的结果需要去重。
+- 去重的方法是经典逻辑，将数组排序以后，判断重复元素再做逻辑判断。
+- 其他思路和第 46 题完全一致，DFS 深搜即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+func permuteUnique(nums []int) [][]int {
+	if len(nums) == 0 {
+		return [][]int{}
+	}
+	used, p, res := make([]bool, len(nums)), []int{}, [][]int{}
+	sort.Ints(nums) // 这里是去重的关键逻辑
+	generatePermutation47(nums, 0, p, &res, &used)
+	return res
+}
+
+func generatePermutation47(nums []int, index int, p []int, res *[][]int, used *[]bool) {
+	if index == len(nums) {
+		temp := make([]int, len(p))
+		copy(temp, p)
+		*res = append(*res, temp)
+		return
+	}
+	for i := 0; i < len(nums); i++ {
+		if !(*used)[i] {
+			if i > 0 && nums[i] == nums[i-1] && !(*used)[i-1] { // 这里是去重的关键逻辑
+				continue
+			}
+			(*used)[i] = true
+			p = append(p, nums[i])
+			generatePermutation47(nums, index+1, p, res, used)
+			p = p[:len(p)-1]
+			(*used)[i] = false
+		}
+	}
+	return
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0046.Permutations/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0048.Rotate-Image/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0048.Rotate-Image.md b/website/content/ChapterFour/0001~0099/0048.Rotate-Image.md
new file mode 100755
index 000000000..4e7a4e0ce
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0048.Rotate-Image.md
@@ -0,0 +1,169 @@
+# [48. Rotate Image](https://leetcode.com/problems/rotate-image/)
+
+## 题目
+
+You are given an *n* x *n* 2D matrix representing an image.
+
+Rotate the image by 90 degrees (clockwise).
+
+**Note**:
+
+You have to rotate the image **[in-place](https://en.wikipedia.org/wiki/In-place_algorithm)**, which means you have to modify the input 2D matrix directly. **DO NOT** allocate another 2D matrix and do the rotation.
+
+**Example 1**:
+
+![](https://assets.leetcode.com/uploads/2020/08/28/mat1.jpg)
+
+    Given input matrix = 
+    [
+      [1,2,3],
+      [4,5,6],
+      [7,8,9]
+    ],
+    
+    rotate the input matrix in-place such that it becomes:
+    [
+      [7,4,1],
+      [8,5,2],
+      [9,6,3]
+    ]
+
+
+**Example 2**:
+
+![](https://assets.leetcode.com/uploads/2020/08/28/mat2.jpg)
+
+    Given input matrix =
+    [
+      [ 5, 1, 9,11],
+      [ 2, 4, 8,10],
+      [13, 3, 6, 7],
+      [15,14,12,16]
+    ], 
+    
+    rotate the input matrix in-place such that it becomes:
+    [
+      [15,13, 2, 5],
+      [14, 3, 4, 1],
+      [12, 6, 8, 9],
+      [16, 7,10,11]
+    ]
+
+
+## 题目大意
+
+给定一个 n × n 的二维矩阵表示一个图像。将图像顺时针旋转 90 度。说明：你必须在原地旋转图像，这意味着你需要直接修改输入的二维矩阵。请不要使用另一个矩阵来旋转图像。
+
+
+## 解题思路
+
+- 给出一个二维数组，要求顺时针旋转 90 度。
+- 这一题比较简单，按照题意做就可以。这里给出 2 种旋转方法的实现，顺时针旋转和逆时针旋转。
+
+```c
+
+    /*
+     * clockwise rotate 顺时针旋转
+     * first reverse up to down, then swap the symmetry 
+     * 1 2 3     7 8 9     7 4 1
+     * 4 5 6  => 4 5 6  => 8 5 2
+     * 7 8 9     1 2 3     9 6 3
+    */
+    void rotate(vector<vector<int> > &matrix) {
+        reverse(matrix.begin(), matrix.end());
+        for (int i = 0; i < matrix.size(); ++i) {
+            for (int j = i + 1; j < matrix[i].size(); ++j)
+                swap(matrix[i][j], matrix[j][i]);
+        }
+    }
+    
+    /*
+     * anticlockwise rotate 逆时针旋转
+     * first reverse left to right, then swap the symmetry
+     * 1 2 3     3 2 1     3 6 9
+     * 4 5 6  => 6 5 4  => 2 5 8
+     * 7 8 9     9 8 7     1 4 7
+    */
+    void anti_rotate(vector<vector<int> > &matrix) {
+        for (auto vi : matrix) reverse(vi.begin(), vi.end());
+        for (int i = 0; i < matrix.size(); ++i) {
+            for (int j = i + 1; j < matrix[i].size(); ++j)
+                swap(matrix[i][j], matrix[j][i]);
+        }
+    }
+    
+```
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一
+func rotate(matrix [][]int) {
+	length := len(matrix)
+	// rotate by diagonal 对角线变换
+	for i := 0; i < length; i++ {
+		for j := i + 1; j < length; j++ {
+			matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
+		}
+	}
+	// rotate by vertical centerline 竖直轴对称翻转
+	for i := 0; i < length; i++ {
+		for j := 0; j < length/2; j++ {
+			matrix[i][j], matrix[i][length-j-1] = matrix[i][length-j-1], matrix[i][j]
+		}
+	}
+}
+
+// 解法二
+func rotate1(matrix [][]int) {
+	n := len(matrix)
+	if n == 1 {
+		return
+	}
+	/* rotate clock-wise = 1. transpose matrix => 2. reverse(matrix[i])
+
+	1   2  3  4      1   5  9  13        13  9  5  1
+	5   6  7  8  =>  2   6  10 14  =>    14  10 6  2
+	9  10 11 12      3   7  11 15        15  11 7  3
+	13 14 15 16      4   8  12 16        16  12 8  4
+
+	*/
+
+	for i := 0; i < n; i++ {
+		// transpose, i=rows, j=columns
+		// j = i+1, coz diagonal elements didn't change in a square matrix
+		for j := i + 1; j < n; j++ {
+			swap(matrix, i, j)
+		}
+		// reverse each row of the image
+		matrix[i] = reverse(matrix[i])
+	}
+}
+
+// swap changes original slice's i,j position
+func swap(nums [][]int, i, j int) {
+	nums[i][j], nums[j][i] = nums[j][i], nums[i][j]
+}
+
+// reverses a row of image, matrix[i]
+func reverse(nums []int) []int {
+	var lp, rp = 0, len(nums) - 1
+
+	for lp < rp {
+		nums[lp], nums[rp] = nums[rp], nums[lp]
+		lp++
+		rp--
+	}
+	return nums
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0047.Permutations-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0049.Group-Anagrams/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0049.Group-Anagrams.md b/website/content/ChapterFour/0001~0099/0049.Group-Anagrams.md
new file mode 100644
index 000000000..d677466da
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0049.Group-Anagrams.md
@@ -0,0 +1,79 @@
+# [49. Group Anagrams](https://leetcode.com/problems/group-anagrams/)
+
+## 题目
+
+Given an array of strings, group anagrams together.
+
+
+**Example**:
+
+```
+
+Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
+Output:
+[
+  ["ate","eat","tea"],
+  ["nat","tan"],
+  ["bat"]
+]
+
+```
+
+**Note**:
+
+- All inputs will be in lowercase.
+- The order of your output does not matter.
+
+## 题目大意
+
+给出一个字符串数组，要求对字符串数组里面有 Anagrams 关系的字符串进行分组。Anagrams 关系是指两个字符串的字符完全相同，顺序不同，两者是由排列组合组成。
+
+## 解题思路
+
+这道题可以将每个字符串都排序，排序完成以后，相同 Anagrams 的字符串必然排序结果一样。把排序以后的字符串当做 key 存入到 map 中。遍历数组以后，就能得到一个 map，key 是排序以后的字符串，value 对应的是这个排序字符串以后的 Anagrams 字符串集合。最后再将这些 value 对应的字符串数组输出即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+type sortRunes []rune
+
+func (s sortRunes) Less(i, j int) bool {
+	return s[i] < s[j]
+}
+
+func (s sortRunes) Swap(i, j int) {
+	s[i], s[j] = s[j], s[i]
+}
+
+func (s sortRunes) Len() int {
+	return len(s)
+}
+
+func groupAnagrams(strs []string) [][]string {
+	record, res := map[string][]string{}, [][]string{}
+	for _, str := range strs {
+		sByte := []rune(str)
+		sort.Sort(sortRunes(sByte))
+		sstrs := record[string(sByte)]
+		sstrs = append(sstrs, str)
+		record[string(sByte)] = sstrs
+	}
+	for _, v := range record {
+		res = append(res, v)
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0048.Rotate-Image/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0050.Powx-n/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0050.Powx-n.md b/website/content/ChapterFour/0001~0099/0050.Powx-n.md
new file mode 100755
index 000000000..714c18617
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0050.Powx-n.md
@@ -0,0 +1,76 @@
+# [50. Pow(x, n)](https://leetcode.com/problems/powx-n/)
+
+
+## 题目
+
+Implement [pow(*x*, *n*)](http://www.cplusplus.com/reference/valarray/pow/), which calculates *x* raised to the power *n* (xn).
+
+**Example 1**:
+
+
+    Input: 2.00000, 10
+    Output: 1024.00000
+
+
+**Example 2**:
+
+
+    Input: 2.10000, 3
+    Output: 9.26100
+
+
+**Example 3**:
+
+
+    Input: 2.00000, -2
+    Output: 0.25000
+    Explanation: 2-2 = 1/22 = 1/4 = 0.25
+
+
+**Note**:
+
+- -100.0 < *x* < 100.0
+- *n* is a 32-bit signed integer, within the range [−2^31, 2^31− 1]
+
+## 题目大意
+
+实现 pow(x, n) ，即计算 x 的 n 次幂函数。
+
+## 解题思路
+
+- 要求计算 Pow(x, n)
+- 这一题用递归的方式，不断的将 n 2 分下去。注意 n 的正负数，n 的奇偶性。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 时间复杂度 O(log n),空间复杂度 O(1)
+func myPow(x float64, n int) float64 {
+	if n == 0 {
+		return 1
+	}
+	if n == 1 {
+		return x
+	}
+	if n < 0 {
+		n = -n
+		x = 1 / x
+	}
+	tmp := myPow(x, n/2)
+	if n%2 == 0 {
+		return tmp * tmp
+	}
+	return tmp * tmp * x
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0049.Group-Anagrams/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0051.N-Queens/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0051.N-Queens.md b/website/content/ChapterFour/0001~0099/0051.N-Queens.md
new file mode 100755
index 000000000..eacce3135
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0051.N-Queens.md
@@ -0,0 +1,148 @@
+# [51. N-Queens](https://leetcode.com/problems/n-queens/)
+
+
+## 题目
+
+The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
+
+![](https://assets.leetcode.com/uploads/2018/10/12/8-queens.png)
+
+Given an integer *n*, return all distinct solutions to the *n*-queens puzzle.
+
+Each solution contains a distinct board configuration of the *n*-queens' placement, where `'Q'` and `'.'` both indicate a queen and an empty space respectively.
+
+**Example**:
+
+
+    Input: 4
+    Output: [
+     [".Q..",  // Solution 1
+      "...Q",
+      "Q...",
+      "..Q."],
+    
+     ["..Q.",  // Solution 2
+      "Q...",
+      "...Q",
+      ".Q.."]
+    ]
+    Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.
+
+
+## 题目大意
+
+给定一个整数 n，返回所有不同的 n 皇后问题的解决方案。每一种解法包含一个明确的 n 皇后问题的棋子放置方案，该方案中 'Q' 和 '.' 分别代表了皇后和空位。
+
+
+## 解题思路
+
+- 求解 n 皇后问题
+- 利用 col 数组记录列信息，col 有 `n` 列。用 dia1，dia2 记录从左下到右上的对角线，从左上到右下的对角线的信息，dia1 和 dia2 分别都有 `2*n-1` 个。
+- dia1 对角线的规律是 `i + j 是定值`，例如[0,0]，为 0；[1,0]、[0,1] 为 1；[2,0]、[1,1]、[0,2] 为 2；
+- dia2 对角线的规律是 `i - j 是定值`，例如[0,7]，为 -7；[0,6]、[1,7] 为 -6；[0,5]、[1,6]、[2,7] 为 -5；为了使他们从 0 开始，i - j + n - 1 偏移到 0 开始，所以 dia2 的规律是 `i - j + n - 1 为定值`。
+- 还有一个位运算的方法，每行只能选一个位置放皇后，那么对每行遍历可能放皇后的位置。如何高效判断哪些点不能放皇后呢？这里的做法毕竟巧妙，把所有之前选过的点按照顺序存下来，然后根据之前选的点到当前行的距离，就可以快速判断是不是会有冲突。举个例子: 假如在 4 皇后问题中，如果第一二行已经选择了位置 [1, 3]，那么在第三行选择时，首先不能再选 1, 3 列了，而对于第三行， 1 距离长度为2，所以它会影响到 -1, 3 两个列。同理，3 在第二行，距离第三行为 1，所以 3 会影响到列 2, 4。由上面的结果，我们知道 -1, 4 超出边界了不用去管，别的不能选的点是 1, 2, 3，所以第三行就只能选 0。在代码实现中，可以在每次遍历前根据之前选择的情况生成一个 occupied 用来记录当前这一行，已经被选了的和由于之前皇后攻击范围所以不能选的位置，然后只选择合法的位置进入到下一层递归。另外就是预处理了一个皇后放不同位置的字符串，这样这些字符串在返回结果的时候是可以在内存中复用的，省一点内存。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 DFS
+func solveNQueens(n int) [][]string {
+	col, dia1, dia2, row, res := make([]bool, n), make([]bool, 2*n-1), make([]bool, 2*n-1), []int{}, [][]string{}
+	putQueen(n, 0, &col, &dia1, &dia2, &row, &res)
+	return res
+}
+
+// 尝试在一个n皇后问题中, 摆放第index行的皇后位置
+func putQueen(n, index int, col, dia1, dia2 *[]bool, row *[]int, res *[][]string) {
+	if index == n {
+		*res = append(*res, generateBoard(n, row))
+		return
+	}
+	for i := 0; i < n; i++ {
+		// 尝试将第index行的皇后摆放在第i列
+		if !(*col)[i] && !(*dia1)[index+i] && !(*dia2)[index-i+n-1] {
+			*row = append(*row, i)
+			(*col)[i] = true
+			(*dia1)[index+i] = true
+			(*dia2)[index-i+n-1] = true
+			putQueen(n, index+1, col, dia1, dia2, row, res)
+			(*col)[i] = false
+			(*dia1)[index+i] = false
+			(*dia2)[index-i+n-1] = false
+			*row = (*row)[:len(*row)-1]
+		}
+	}
+	return
+}
+
+func generateBoard(n int, row *[]int) []string {
+	board := []string{}
+	res := ""
+	for i := 0; i < n; i++ {
+		res += "."
+	}
+	for i := 0; i < n; i++ {
+		board = append(board, res)
+	}
+	for i := 0; i < n; i++ {
+		tmp := []byte(board[i])
+		tmp[(*row)[i]] = 'Q'
+		board[i] = string(tmp)
+	}
+	return board
+}
+
+// 解法二 二进制操作法 Signed-off-by: Hanlin Shi shihanlin9@gmail.com
+func solveNQueens2(n int) (res [][]string) {
+	placements := make([]string, n)
+	for i := range placements {
+		buf := make([]byte, n)
+		for j := range placements {
+			if i == j {
+				buf[j] = 'Q'
+			} else {
+				buf[j] = '.'
+			}
+		}
+		placements[i] = string(buf)
+	}
+	var construct func(prev []int)
+	construct = func(prev []int) {
+		if len(prev) == n {
+			plan := make([]string, n)
+			for i := 0; i < n; i++ {
+				plan[i] = placements[prev[i]]
+			}
+			res = append(res, plan)
+			return
+		}
+		occupied := 0
+		for i := range prev {
+			dist := len(prev) - i
+			bit := 1 << prev[i]
+			occupied |= bit | bit<<dist | bit>>dist
+		}
+		prev = append(prev, -1)
+		for i := 0; i < n; i++ {
+			if (occupied>>i)&1 != 0 {
+				continue
+			}
+			prev[len(prev)-1] = i
+			construct(prev)
+		}
+	}
+	construct(make([]int, 0, n))
+	return
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0050.Powx-n/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0052.N-Queens-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0052.N-Queens-II.md b/website/content/ChapterFour/0001~0099/0052.N-Queens-II.md
new file mode 100755
index 000000000..4bf016f13
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0052.N-Queens-II.md
@@ -0,0 +1,114 @@
+# [52. N-Queens II](https://leetcode.com/problems/n-queens-ii/)
+
+
+## 题目
+
+The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
+
+![](https://assets.leetcode.com/uploads/2018/10/12/8-queens.png)
+
+Given an integer n, return the number of distinct solutions to the n-queens puzzle.
+
+**Example**:
+
+
+    Input: 4
+    Output: 2
+    Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
+    [
+     [".Q..",  // Solution 1
+      "...Q",
+      "Q...",
+      "..Q."],
+    
+     ["..Q.",  // Solution 2
+      "Q...",
+      "...Q",
+      ".Q.."]
+    ]
+
+
+## 题目大意
+
+给定一个整数 n，返回 n 皇后不同的解决方案的数量。
+
+## 解题思路
+
+- 这一题是第 51 题的加强版，在第 51 题的基础上累加记录解的个数即可。
+- 这一题也可以暴力打表法，时间复杂度为 O(1)。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一，暴力打表法
+func totalNQueens(n int) int {
+	res := []int{0, 1, 0, 0, 2, 10, 4, 40, 92, 352, 724}
+	return res[n]
+}
+
+// 解法二，DFS 回溯法
+func totalNQueens1(n int) int {
+	col, dia1, dia2, row, res := make([]bool, n), make([]bool, 2*n-1), make([]bool, 2*n-1), []int{}, 0
+	putQueen52(n, 0, &col, &dia1, &dia2, &row, &res)
+	return res
+}
+
+// 尝试在一个n皇后问题中, 摆放第index行的皇后位置
+func putQueen52(n, index int, col, dia1, dia2 *[]bool, row *[]int, res *int) {
+	if index == n {
+		*res++
+		return
+	}
+	for i := 0; i < n; i++ {
+		// 尝试将第index行的皇后摆放在第i列
+		if !(*col)[i] && !(*dia1)[index+i] && !(*dia2)[index-i+n-1] {
+			*row = append(*row, i)
+			(*col)[i] = true
+			(*dia1)[index+i] = true
+			(*dia2)[index-i+n-1] = true
+			putQueen52(n, index+1, col, dia1, dia2, row, res)
+			(*col)[i] = false
+			(*dia1)[index+i] = false
+			(*dia2)[index-i+n-1] = false
+			*row = (*row)[:len(*row)-1]
+		}
+	}
+	return
+}
+
+// 解法三 二进制位操作法
+// class Solution {
+// 	public:
+// 		int totalNQueens(int n) {
+// 			int ans=0;
+// 			int row=0,leftDiagonal=0,rightDiagonal=0;
+// 			int bit=(1<<n)-1;//to clear high bits of the 32-bit int
+// 			Queens(bit,row,leftDiagonal,rightDiagonal,ans);
+// 			return ans;
+// 		}
+// 		void Queens(int bit,int row,int leftDiagonal,int rightDiagonal,int &ans){
+// 			int cur=(~(row|leftDiagonal|rightDiagonal))&bit;//possible place for this queen
+// 			if (!cur) return;//no pos for this queen
+// 			while(cur){
+// 				int curPos=(cur&(~cur + 1))&bit;//choose possible place in the right
+// 				//update row,ld and rd
+// 				row+=curPos;
+// 				if (row==bit) ans++;//last row
+// 				else Queens(bit,row,((leftDiagonal|curPos)<<1)&bit,((rightDiagonal|curPos)>>1)&bit,ans);
+// 				cur-=curPos;//for next possible place
+// 				row-=curPos;//reset row
+// 			}
+// 		}
+// };
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0051.N-Queens/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0053.Maximum-Subarray/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0053.Maximum-Subarray.md b/website/content/ChapterFour/0001~0099/0053.Maximum-Subarray.md
new file mode 100755
index 000000000..a02481277
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0053.Maximum-Subarray.md
@@ -0,0 +1,82 @@
+# [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)
+
+
+## 题目
+
+Given an integer array `nums`, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
+
+**Example**:
+
+
+    Input: [-2,1,-3,4,-1,2,1,-5,4],
+    Output: 6
+    Explanation: [4,-1,2,1] has the largest sum = 6.
+
+
+**Follow up**:
+
+If you have figured out the O(*n*) solution, try coding another solution using the divide and conquer approach, which is more subtle.
+
+## 题目大意
+
+给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。
+
+## 解题思路
+
+- 这一题可以用 DP 求解也可以不用 DP。
+- 题目要求输出数组中某个区间内数字之和最大的那个值。`dp[i]` 表示 `[0,i]` 区间内各个子区间和的最大值，状态转移方程是 `dp[i] = nums[i] + dp[i-1] (dp[i-1] > 0)`，`dp[i] = nums[i] (dp[i-1] ≤ 0)`。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 DP
+func maxSubArray(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	if len(nums) == 1 {
+		return nums[0]
+	}
+	dp, res := make([]int, len(nums)), nums[0]
+	dp[0] = nums[0]
+	for i := 1; i < len(nums); i++ {
+		if dp[i-1] > 0 {
+			dp[i] = nums[i] + dp[i-1]
+		} else {
+			dp[i] = nums[i]
+		}
+		res = max(res, dp[i])
+	}
+	return res
+}
+
+// 解法二 模拟
+func maxSubArray1(nums []int) int {
+	if len(nums) == 1 {
+		return nums[0]
+	}
+	maxSum, res, p := nums[0], 0, 0
+	for p < len(nums) {
+		res += nums[p]
+		if res > maxSum {
+			maxSum = res
+		}
+		if res < 0 {
+			res = 0
+		}
+		p++
+	}
+	return maxSum
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0052.N-Queens-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0054.Spiral-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0054.Spiral-Matrix.md b/website/content/ChapterFour/0001~0099/0054.Spiral-Matrix.md
new file mode 100755
index 000000000..5d75a8e6e
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0054.Spiral-Matrix.md
@@ -0,0 +1,172 @@
+# [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)
+
+
+## 题目
+
+Given a matrix of *m* x *n* elements (*m* rows, *n* columns), return all elements of the matrix in spiral order.
+
+**Example 1**:
+
+
+    Input:
+    [
+     [ 1, 2, 3 ],
+     [ 4, 5, 6 ],
+     [ 7, 8, 9 ]
+    ]
+    Output: [1,2,3,6,9,8,7,4,5]
+
+
+**Example 2**:
+
+
+    Input:
+    [
+      [1, 2, 3, 4],
+      [5, 6, 7, 8],
+      [9,10,11,12]
+    ]
+    Output: [1,2,3,4,8,12,11,10,9,5,6,7]
+
+
+## 题目大意
+
+给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。
+
+## 解题思路
+
+- 给出一个二维数组，按照螺旋的方式输出
+- 解法一：需要注意的是特殊情况，比如二维数组退化成一维或者一列或者一个元素。注意了这些情况，基本就可以一次通过了。
+- 解法二：提前算出一共多少个元素，一圈一圈地遍历矩阵，停止条件就是遍历了所有元素（count == sum）
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法 1
+func spiralOrder(matrix [][]int) []int {
+	if len(matrix) == 0 {
+		return []int{}
+	}
+	res := []int{}
+	if len(matrix) == 1 {
+		for i := 0; i < len(matrix[0]); i++ {
+			res = append(res, matrix[0][i])
+		}
+		return res
+	}
+	if len(matrix[0]) == 1 {
+		for i := 0; i < len(matrix); i++ {
+			res = append(res, matrix[i][0])
+		}
+		return res
+	}
+	visit, m, n, round, x, y, spDir := make([][]int, len(matrix)), len(matrix), len(matrix[0]), 0, 0, 0, [][]int{
+		[]int{0, 1},  // 朝右
+		[]int{1, 0},  // 朝下
+		[]int{0, -1}, // 朝左
+		[]int{-1, 0}, // 朝上
+	}
+	for i := 0; i < m; i++ {
+		visit[i] = make([]int, n)
+	}
+	visit[x][y] = 1
+	res = append(res, matrix[x][y])
+	for i := 0; i < m*n; i++ {
+		x += spDir[round%4][0]
+		y += spDir[round%4][1]
+		if (x == 0 && y == n-1) || (x == m-1 && y == n-1) || (y == 0 && x == m-1) {
+			round++
+		}
+		if x > m-1 || y > n-1 || x < 0 || y < 0 {
+			return res
+		}
+		if visit[x][y] == 0 {
+			visit[x][y] = 1
+			res = append(res, matrix[x][y])
+		}
+		switch round % 4 {
+		case 0:
+			if y+1 <= n-1 && visit[x][y+1] == 1 {
+				round++
+				continue
+			}
+		case 1:
+			if x+1 <= m-1 && visit[x+1][y] == 1 {
+				round++
+				continue
+			}
+		case 2:
+			if y-1 >= 0 && visit[x][y-1] == 1 {
+				round++
+				continue
+			}
+		case 3:
+			if x-1 >= 0 && visit[x-1][y] == 1 {
+				round++
+				continue
+			}
+		}
+	}
+	return res
+}
+
+// 解法 2
+func spiralOrder2(matrix [][]int) []int {
+	m := len(matrix)
+	if m == 0 {
+		return nil
+	}
+
+	n := len(matrix[0])
+	if n == 0 {
+		return nil
+	}
+
+	// top、left、right、bottom 分别是剩余区域的上、左、右、下的下标
+	top, left, bottom, right := 0, 0, m-1, n-1 
+	count, sum := 0, m*n
+	res := []int{}
+	
+	// 外层循环每次遍历一圈
+	for count < sum {
+		i, j := top, left
+		for j <= right && count < sum {
+			res = append(res, matrix[i][j])
+			count++
+			j++
+		}
+		i, j = top + 1, right
+		for i <= bottom && count < sum {
+			res = append(res, matrix[i][j])
+			count++
+			i++
+		}
+		i, j = bottom, right - 1
+		for j >= left && count < sum {
+			res = append(res, matrix[i][j])
+			count++
+			j--
+		}
+		i, j = bottom - 1, left
+		for i > top && count < sum {
+			res = append(res, matrix[i][j])
+			count++
+			i--
+		}
+		// 进入到下一层
+		top, left, bottom, right = top+1, left+1, bottom-1, right-1
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0053.Maximum-Subarray/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0055.Jump-Game/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0055.Jump-Game.md b/website/content/ChapterFour/0001~0099/0055.Jump-Game.md
new file mode 100644
index 000000000..a34c5d262
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0055.Jump-Game.md
@@ -0,0 +1,65 @@
+# [55. Jump Game](https://leetcode.com/problems/jump-game/)
+
+
+## 题目
+
+Given an array of non-negative integers, you are initially positioned at the first index of the array.
+
+Each element in the array represents your maximum jump length at that position.
+
+Determine if you are able to reach the last index.
+
+**Example 1**:
+
+```
+Input: [2,3,1,1,4]
+Output: true
+Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
+```
+
+**Example 2**:
+
+```
+Input: [3,2,1,0,4]
+Output: false
+Explanation: You will always arrive at index 3 no matter what. Its maximum
+             jump length is 0, which makes it impossible to reach the last index.
+```
+
+## 题目大意
+
+给定一个非负整数数组，最初位于数组的第一个位置。数组中的每个元素代表在该位置可以跳跃的最大长度。判断是否能够到达最后一个位置。
+
+## 解题思路
+
+- 给出一个非负数组，要求判断从数组 0 下标开始，能否到达数组最后一个位置。
+- 这一题比较简单。如果某一个作为 `起跳点` 的格子可以跳跃的距离是 `n`，那么表示后面 `n` 个格子都可以作为 `起跳点`。可以对每一个能作为 `起跳点` 的格子都尝试跳一次，把 `能跳到最远的距离maxJump` 不断更新。如果可以一直跳到最后，就成功了。如果中间有一个点比 `maxJump` 还要大，说明在这个点和 maxJump 中间连不上了，有些点不能到达最后一个位置。
+
+## 代码
+
+```go
+func canJump(nums []int) bool {
+	n := len(nums)
+	if n == 0 {
+		return false
+	}
+	if n == 1 {
+		return true
+	}
+	maxJump := 0
+	for i, v := range nums {
+		if i > maxJump {
+			return false
+		}
+		maxJump = max(maxJump, i+v)
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0054.Spiral-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0056.Merge-Intervals/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0056.Merge-Intervals.md b/website/content/ChapterFour/0001~0099/0056.Merge-Intervals.md
new file mode 100644
index 000000000..236c252fc
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0056.Merge-Intervals.md
@@ -0,0 +1,116 @@
+# [56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)
+
+## 题目
+
+Given a collection of intervals, merge all overlapping intervals.
+
+**Example 1**:  
+
+```
+
+Input: [[1,3],[2,6],[8,10],[15,18]]
+Output: [[1,6],[8,10],[15,18]]
+Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
+
+```
+
+**Example 2**:  
+
+```
+
+Input: [[1,4],[4,5]]
+Output: [[1,5]]
+Explanation: Intervals [1,4] and [4,5] are considered overlapping.
+
+```
+
+## 题目大意
+
+合并给的多个区间，区间有重叠的要进行区间合并。
+
+
+## 解题思路
+
+先按照区间起点进行排序。然后从区间起点小的开始扫描，依次合并每个有重叠的区间。
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for an interval.
+ * type Interval struct {
+ *	   Start int
+ *	   End   int
+ * }
+ */
+
+// Interval define
+type Interval struct {
+	Start int
+	End   int
+}
+
+func merge56(intervals []Interval) []Interval {
+	if len(intervals) == 0 {
+		return intervals
+	}
+	quickSort(intervals, 0, len(intervals)-1)
+	res := make([]Interval, 0)
+	res = append(res, intervals[0])
+	curIndex := 0
+	for i := 1; i < len(intervals); i++ {
+		if intervals[i].Start > res[curIndex].End {
+			curIndex++
+			res = append(res, intervals[i])
+		} else {
+			res[curIndex].End = max(intervals[i].End, res[curIndex].End)
+		}
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+func partitionSort(a []Interval, lo, hi int) int {
+	pivot := a[hi]
+	i := lo - 1
+	for j := lo; j < hi; j++ {
+		if (a[j].Start < pivot.Start) || (a[j].Start == pivot.Start && a[j].End < pivot.End) {
+			i++
+			a[j], a[i] = a[i], a[j]
+		}
+	}
+	a[i+1], a[hi] = a[hi], a[i+1]
+	return i + 1
+}
+func quickSort(a []Interval, lo, hi int) {
+	if lo >= hi {
+		return
+	}
+	p := partitionSort(a, lo, hi)
+	quickSort(a, lo, p-1)
+	quickSort(a, p+1, hi)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0055.Jump-Game/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0057.Insert-Interval/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0057.Insert-Interval.md b/website/content/ChapterFour/0001~0099/0057.Insert-Interval.md
new file mode 100644
index 000000000..f5eb6964e
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0057.Insert-Interval.md
@@ -0,0 +1,82 @@
+# [57. Insert Interval](https://leetcode.com/problems/insert-interval/)
+
+## 题目
+
+Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
+
+You may assume that the intervals were initially sorted according to their start times.
+
+**Example 1**:  
+
+```
+
+Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
+Output: [[1,5],[6,9]]
+
+```
+
+**Example 2**:  
+
+```
+
+Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
+Output: [[1,2],[3,10],[12,16]]
+Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
+
+```
+
+## 题目大意
+
+这一题是第 56 题的加强版。给出多个没有重叠的区间，然后再给一个区间，要求把如果有重叠的区间进行合并。
+
+## 解题思路
+
+可以分 3 段处理，先添加原来的区间，即在给的 newInterval 之前的区间。然后添加 newInterval ，注意这里可能需要合并多个区间。最后把原来剩下的部分添加到最终结果中即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for an interval.
+ * type Interval struct {
+ *	   Start int
+ *	   End   int
+ * }
+ */
+
+func insert(intervals []Interval, newInterval Interval) []Interval {
+	res := make([]Interval, 0)
+	if len(intervals) == 0 {
+		res = append(res, newInterval)
+		return res
+	}
+	curIndex := 0
+	for curIndex < len(intervals) && intervals[curIndex].End < newInterval.Start {
+		res = append(res, intervals[curIndex])
+		curIndex++
+	}
+
+	for curIndex < len(intervals) && intervals[curIndex].Start <= newInterval.End {
+		newInterval = Interval{Start: min(newInterval.Start, intervals[curIndex].Start), End: max(newInterval.End, intervals[curIndex].End)}
+		curIndex++
+	}
+	res = append(res, newInterval)
+
+	for curIndex < len(intervals) {
+		res = append(res, intervals[curIndex])
+		curIndex++
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0056.Merge-Intervals/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0058.Length-of-Last-Word/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0058.Length-of-Last-Word.md b/website/content/ChapterFour/0001~0099/0058.Length-of-Last-Word.md
new file mode 100644
index 000000000..3d25cf40d
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0058.Length-of-Last-Word.md
@@ -0,0 +1,77 @@
+# [58. Length of Last Word](https://leetcode.com/problems/length-of-last-word/)
+
+
+## 题目
+
+Given a string `s` consisting of some words separated by some number of spaces, return *the length of the **last** word in the string.*
+
+A **word** is a maximal substring consisting of non-space characters only.
+
+**Example 1:**
+
+```
+Input: s = "Hello World"
+Output: 5
+Explanation: The last word is "World" with length 5.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "   fly me   to   the moon  "
+Output: 4
+Explanation: The last word is "moon" with length 4.
+
+```
+
+**Example 3:**
+
+```
+Input: s = "luffy is still joyboy"
+Output: 6
+Explanation: The last word is "joyboy" with length 6.
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 104`
+- `s` consists of only English letters and spaces `' '`.
+- There will be at least one word in `s`.
+
+## 题目大意
+
+给你一个字符串 `s`，由若干单词组成，单词前后用一些空格字符隔开。返回字符串中最后一个单词的长度。**单词** 是指仅由字母组成、不包含任何空格字符的最大子字符串。
+
+## 解题思路
+
+- 先从后过滤掉空格找到单词尾部，再从尾部向前遍历，找到单词头部，最后两者相减，即为单词的长度。
+
+## 代码
+
+```go
+package leetcode
+
+func lengthOfLastWord(s string) int {
+	last := len(s) - 1
+	for last >= 0 && s[last] == ' ' {
+		last--
+	}
+	if last < 0 {
+		return 0
+	}
+	first := last
+	for first >= 0 && s[first] != ' ' {
+		first--
+	}
+	return last - first
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0057.Insert-Interval/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0059.Spiral-Matrix-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0059.Spiral-Matrix-II.md b/website/content/ChapterFour/0001~0099/0059.Spiral-Matrix-II.md
new file mode 100755
index 000000000..7d6df02a0
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0059.Spiral-Matrix-II.md
@@ -0,0 +1,101 @@
+# [59. Spiral Matrix II](https://leetcode.com/problems/spiral-matrix-ii/)
+
+
+## 题目
+
+Given a positive integer *n*, generate a square matrix filled with elements from 1 to *n*2 in spiral order.
+
+**Example**:
+
+
+    Input: 3
+    Output:
+    [
+     [ 1, 2, 3 ],
+     [ 8, 9, 4 ],
+     [ 7, 6, 5 ]
+    ]
+
+
+## 题目大意
+
+给定一个正整数 n，生成一个包含 1 到 n^2 所有元素，且元素按顺时针顺序螺旋排列的正方形矩阵。
+
+
+## 解题思路
+
+- 给出一个数组 n，要求输出一个 n * n 的二维数组，里面元素是 1 - n*n，且数组排列顺序是螺旋排列的
+- 这一题是第 54 题的加强版，没有需要注意的特殊情况，直接模拟即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func generateMatrix(n int) [][]int {
+	if n == 0 {
+		return [][]int{}
+	}
+	if n == 1 {
+		return [][]int{[]int{1}}
+	}
+	res, visit, round, x, y, spDir := make([][]int, n), make([][]int, n), 0, 0, 0, [][]int{
+		[]int{0, 1},  // 朝右
+		[]int{1, 0},  // 朝下
+		[]int{0, -1}, // 朝左
+		[]int{-1, 0}, // 朝上
+	}
+	for i := 0; i < n; i++ {
+		visit[i] = make([]int, n)
+		res[i] = make([]int, n)
+	}
+	visit[x][y] = 1
+	res[x][y] = 1
+	for i := 0; i < n*n; i++ {
+		x += spDir[round%4][0]
+		y += spDir[round%4][1]
+		if (x == 0 && y == n-1) || (x == n-1 && y == n-1) || (y == 0 && x == n-1) {
+			round++
+		}
+		if x > n-1 || y > n-1 || x < 0 || y < 0 {
+			return res
+		}
+		if visit[x][y] == 0 {
+			visit[x][y] = 1
+			res[x][y] = i + 2
+		}
+		switch round % 4 {
+		case 0:
+			if y+1 <= n-1 && visit[x][y+1] == 1 {
+				round++
+				continue
+			}
+		case 1:
+			if x+1 <= n-1 && visit[x+1][y] == 1 {
+				round++
+				continue
+			}
+		case 2:
+			if y-1 >= 0 && visit[x][y-1] == 1 {
+				round++
+				continue
+			}
+		case 3:
+			if x-1 >= 0 && visit[x-1][y] == 1 {
+				round++
+				continue
+			}
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0058.Length-of-Last-Word/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0060.Permutation-Sequence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0060.Permutation-Sequence.md b/website/content/ChapterFour/0001~0099/0060.Permutation-Sequence.md
new file mode 100755
index 000000000..ea18f6956
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0060.Permutation-Sequence.md
@@ -0,0 +1,103 @@
+# [60. Permutation Sequence](https://leetcode.com/problems/permutation-sequence/)
+
+
+## 题目
+
+The set `[1,2,3,...,*n*]` contains a total of *n*! unique permutations.
+
+By listing and labeling all of the permutations in order, we get the following sequence for *n* = 3:
+
+1. `"123"`
+2. `"132"`
+3. `"213"`
+4. `"231"`
+5. `"312"`
+6. `"321"`
+
+Given *n* and *k*, return the *k*th permutation sequence.
+
+**Note**:
+
+- Given *n* will be between 1 and 9 inclusive.
+- Given *k* will be between 1 and *n*! inclusive.
+
+**Example 1**:
+
+```
+
+Input: n = 3, k = 3
+Output: "213"
+
+```
+
+**Example 2**:
+
+```
+
+Input: n = 4, k = 9
+Output: "2314"
+
+```
+
+## 题目大意
+
+给出集合 [1,2,3,…,n]，其所有元素共有 n! 种排列。
+
+按大小顺序列出所有排列情况，并一一标记，当 n = 3 时, 所有排列如下："123"，"132"，"213"，"231"，"312"，"321"，给定 n 和 k，返回第 k 个排列。
+
+
+## 解题思路
+
+- 用 DFS 暴力枚举，这种做法时间复杂度特别高，想想更优的解法。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"fmt"
+	"strconv"
+)
+
+func getPermutation(n int, k int) string {
+	if k == 0 {
+		return ""
+	}
+	used, p, res := make([]bool, n), []int{}, ""
+	findPermutation(n, 0, &k, p, &res, &used)
+	return res
+}
+
+func findPermutation(n, index int, k *int, p []int, res *string, used *[]bool) {
+	fmt.Printf("n = %v index = %v k = %v p = %v res = %v user = %v\n", n, index, *k, p, *res, *used)
+	if index == n {
+		*k--
+		if *k == 0 {
+			for _, v := range p {
+				*res += strconv.Itoa(v + 1)
+			}
+		}
+		return
+	}
+	for i := 0; i < n; i++ {
+		if !(*used)[i] {
+			(*used)[i] = true
+			p = append(p, i)
+			findPermutation(n, index+1, k, p, res, used)
+			p = p[:len(p)-1]
+			(*used)[i] = false
+		}
+	}
+	return
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0059.Spiral-Matrix-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0061.Rotate-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0061.Rotate-List.md b/website/content/ChapterFour/0001~0099/0061.Rotate-List.md
new file mode 100644
index 000000000..9a0da026f
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0061.Rotate-List.md
@@ -0,0 +1,88 @@
+# [61. Rotate List](https://leetcode.com/problems/rotate-list/description/)
+
+## 题目
+
+Given a linked list, rotate the list to the right by k places, where k is non-negative.
+
+**Example 1**:  
+
+```
+
+Input: 1->2->3->4->5->NULL, k = 2
+Output: 4->5->1->2->3->NULL
+Explanation:
+rotate 1 steps to the right: 5->1->2->3->4->NULL
+rotate 2 steps to the right: 4->5->1->2->3->NULL 
+
+```
+
+**Example 2**:  
+
+```
+
+Input: 0->1->2->NULL, k = 4
+Output: 2->0->1->NULL
+Explanation:
+rotate 1 steps to the right: 2->0->1->NULL
+rotate 2 steps to the right: 1->2->0->NULL
+rotate 3 steps to the right: 0->1->2->NULL
+rotate 4 steps to the right: 2->0->1->NULL 
+
+```
+
+## 题目大意
+
+旋转链表 K 次。
+
+
+## 解题思路
+
+这道题需要注意的点是，K 可能很大，K = 2000000000 ，如果是循环肯定会超时。应该找出 O(n) 的复杂度的算法才行。由于是循环旋转，最终状态其实是确定的，利用链表的长度取余可以得到链表的最终旋转结果。
+
+这道题也不能用递归，递归解法会超时。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func rotateRight(head *ListNode, k int) *ListNode {
+	if head == nil || head.Next == nil || k == 0 {
+		return head
+	}
+	newHead := &ListNode{Val: 0, Next: head}
+	len := 0
+	cur := newHead
+	for cur.Next != nil {
+		len++
+		cur = cur.Next
+	}
+	if (k % len) == 0 {
+		return head
+	}
+	cur.Next = head
+	cur = newHead
+	for i := len - k%len; i > 0; i-- {
+		cur = cur.Next
+	}
+	res := &ListNode{Val: 0, Next: cur.Next}
+	cur.Next = nil
+	return res.Next
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0060.Permutation-Sequence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0062.Unique-Paths/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0062.Unique-Paths.md b/website/content/ChapterFour/0001~0099/0062.Unique-Paths.md
new file mode 100755
index 000000000..fcd62db98
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0062.Unique-Paths.md
@@ -0,0 +1,74 @@
+# [62. Unique Paths](https://leetcode.com/problems/unique-paths/)
+
+
+## 题目
+
+A robot is located at the top-left corner of a *m* x *n* grid (marked 'Start' in the diagram below).
+
+The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
+
+How many possible unique paths are there?
+
+![](https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png)
+
+Above is a 7 x 3 grid. How many possible unique paths are there?
+
+**Note**: *m* and *n* will be at most 100.
+
+**Example 1**:
+
+    Input: m = 3, n = 2
+    Output: 3
+    Explanation:
+    From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
+    1. Right -> Right -> Down
+    2. Right -> Down -> Right
+    3. Down -> Right -> Right
+
+**Example 2**:
+
+    Input: m = 7, n = 3
+    Output: 28
+
+
+## 题目大意
+
+一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。问总共有多少条不同的路径？
+
+
+## 解题思路
+
+- 这是一道简单的 DP 题。输出地图上从左上角走到右下角的走法数。
+- 由于机器人只能向右走和向下走，所以地图的第一行和第一列的走法数都是 1，地图中任意一点的走法数是 `dp[i][j] = dp[i-1][j] + dp[i][j-1]`
+
+## 代码
+
+```go
+
+package leetcode
+
+func uniquePaths(m int, n int) int {
+	dp := make([][]int, n)
+	for i := 0; i < n; i++ {
+		dp[i] = make([]int, m)
+	}
+	for i := 0; i < n; i++ {
+		for j := 0; j < m; j++ {
+			if i == 0 || j == 0 {
+				dp[i][j] = 1
+				continue
+			}
+			dp[i][j] = dp[i-1][j] + dp[i][j-1]
+		}
+	}
+	return dp[n-1][m-1]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0061.Rotate-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0063.Unique-Paths-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0063.Unique-Paths-II.md b/website/content/ChapterFour/0001~0099/0063.Unique-Paths-II.md
new file mode 100755
index 000000000..b5353ca87
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0063.Unique-Paths-II.md
@@ -0,0 +1,87 @@
+# [63. Unique Paths II](https://leetcode.com/problems/unique-paths-ii/)
+
+
+## 题目
+
+A robot is located at the top-left corner of a *m* x *n* grid (marked 'Start' in the diagram below).
+
+The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
+
+Now consider if some obstacles are added to the grids. How many unique paths would there be?
+
+![](https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png)
+
+An obstacle and empty space is marked as `1` and `0` respectively in the grid.
+
+**Note**: *m* and *n* will be at most 100.
+
+**Example 1**:
+
+    Input:
+    [
+      [0,0,0],
+      [0,1,0],
+      [0,0,0]
+    ]
+    Output: 2
+    Explanation:
+    There is one obstacle in the middle of the 3x3 grid above.
+    There are two ways to reach the bottom-right corner:
+    1. Right -> Right -> Down -> Down
+    2. Down -> Down -> Right -> Right
+
+## 题目大意
+
+一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？
+
+
+## 解题思路
+
+- 这一题是第 62 题的加强版。也是一道考察 DP 的简单题。
+- 这一题比第 62 题增加的条件是地图中会出现障碍物，障碍物的处理方法是 `dp[i][j]=0`。
+- 需要注意的一种情况是，起点就是障碍物，那么这种情况直接输出 0 。
+
+## 代码
+
+```go
+
+package leetcode
+
+func uniquePathsWithObstacles(obstacleGrid [][]int) int {
+	if len(obstacleGrid) == 0 || obstacleGrid[0][0] == 1 {
+		return 0
+	}
+	m, n := len(obstacleGrid), len(obstacleGrid[0])
+	dp := make([][]int, m)
+	for i := 0; i < m; i++ {
+		dp[i] = make([]int, n)
+	}
+	dp[0][0] = 1
+	for i := 1; i < n; i++ {
+		if dp[0][i-1] != 0 && obstacleGrid[0][i] != 1 {
+			dp[0][i] = 1
+		}
+	}
+	for i := 1; i < m; i++ {
+		if dp[i-1][0] != 0 && obstacleGrid[i][0] != 1 {
+			dp[i][0] = 1
+		}
+	}
+	for i := 1; i < m; i++ {
+		for j := 1; j < n; j++ {
+			if obstacleGrid[i][j] != 1 {
+				dp[i][j] = dp[i-1][j] + dp[i][j-1]
+			}
+		}
+	}
+	return dp[m-1][n-1]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0062.Unique-Paths/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0064.Minimum-Path-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0064.Minimum-Path-Sum.md b/website/content/ChapterFour/0001~0099/0064.Minimum-Path-Sum.md
new file mode 100755
index 000000000..5f833d874
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0064.Minimum-Path-Sum.md
@@ -0,0 +1,100 @@
+# [64. Minimum Path Sum](https://leetcode.com/problems/minimum-path-sum/)
+
+
+## 题目
+
+Given a *m* x *n* grid filled with non-negative numbers, find a path from top left to bottom right which *minimizes* the sum of all numbers along its path.
+
+**Note**: You can only move either down or right at any point in time.
+
+**Example**:
+
+    Input:
+    [
+      [1,3,1],
+      [1,5,1],
+      [4,2,1]
+    ]
+    Output: 7
+    Explanation: Because the path 1→3→1→1→1 minimizes the sum.
+
+## 题目大意
+
+给定一个包含非负整数的 m x n 网格，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。说明：每次只能向下或者向右移动一步。
+
+
+## 解题思路
+
+- 在地图上求出从左上角到右下角的路径中，数字之和最小的一个，输出数字和。
+- 这一题最简单的想法就是用一个二维数组来 DP，当然这是最原始的做法。由于只能往下和往右走，只需要维护 2 列信息就可以了，从左边推到最右边即可得到最小的解。更近一步，可以直接在原来的数组中做原地 DP，空间复杂度为 0 。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 原地 DP，无辅助空间
+func minPathSum(grid [][]int) int {
+	m, n := len(grid), len(grid[0])
+	for i := 1; i < m; i++ {
+		grid[i][0] += grid[i-1][0]
+	}
+	for j := 1; j < n; j++ {
+		grid[0][j] += grid[0][j-1]
+	}
+	for i := 1; i < m; i++ {
+		for j := 1; j < n; j++ {
+			grid[i][j] += min(grid[i-1][j], grid[i][j-1])
+		}
+	}
+	return grid[m-1][n-1]
+
+}
+
+// 解法二 最原始的方法，辅助空间 O(n^2)
+func minPathSum1(grid [][]int) int {
+	if len(grid) == 0 {
+		return 0
+	}
+	m, n := len(grid), len(grid[0])
+	if m == 0 || n == 0 {
+		return 0
+	}
+
+	dp := make([][]int, m)
+	for i := 0; i < m; i++ {
+		dp[i] = make([]int, n)
+	}
+	// initFirstCol
+	for i := 0; i < len(dp); i++ {
+		if i == 0 {
+			dp[i][0] = grid[i][0]
+		} else {
+			dp[i][0] = grid[i][0] + dp[i-1][0]
+		}
+	}
+	// initFirstRow
+	for i := 0; i < len(dp[0]); i++ {
+		if i == 0 {
+			dp[0][i] = grid[0][i]
+		} else {
+			dp[0][i] = grid[0][i] + dp[0][i-1]
+		}
+	}
+	for i := 1; i < m; i++ {
+		for j := 1; j < n; j++ {
+			dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
+		}
+	}
+	return dp[m-1][n-1]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0063.Unique-Paths-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0065.Valid-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0065.Valid-Number.md b/website/content/ChapterFour/0001~0099/0065.Valid-Number.md
new file mode 100644
index 000000000..024f13106
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0065.Valid-Number.md
@@ -0,0 +1,85 @@
+# [65. Valid Number](https://leetcode.com/problems/valid-number/)
+
+
+## 题目
+
+A **valid number** can be split up into these components (in order):
+
+   1. A **decimal number** or an integer. 
+   2. (Optional) An 'e' or 'E', followed by an **integer.**
+   
+A **decimal number** can be split up into these components (in order):
+
+   1. (Optional) A sign character (either '+' or '-').
+   2. One of the following formats:
+        1. One or more digits, followed by a dot '.'.
+        2. One or more digits, followed by a dot '.', followed by one or more digits.
+        3. A dot '.', followed by one or more digits.
+        
+An **integer** can be split up into these components (in order):
+
+   1. (Optional) A sign character (either '+' or '-').
+   2. One or more digits.
+   
+For example, all the following are valid numbers: `["2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789"]`, while the following are not valid numbers: `["abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53"].`
+
+Given a string s, return true if s is a **valid number.**
+
+**Example:**
+
+    Input: s = "0"
+    Output: true
+    
+    Input: s = "e"
+    Output: false
+
+## 题目大意
+
+给定一个字符串S，请根据以上的规则判断该字符串是否是一个有效的数字字符串。
+
+
+## 解题思路
+
+- 用三个变量分别标记是否出现过数字、是否出现过'.'和 是否出现过 'e/E'
+- 从左到右依次遍历字符串中的每一个元素
+    - 如果是数字，则标记数字出现过
+    - 如果是 '.', 则需要 '.'没有出现过，并且 'e/E' 没有出现过，才会进行标记
+    - 如果是 'e/E', 则需要 'e/E'没有出现过，并且前面出现过数字，才会进行标记
+    - 如果是 '+/-', 则需要是第一个字符，或者前一个字符是 'e/E'，才会进行标记，并重置数字出现的标识
+    - 最后返回时，需要字符串中至少出现过数字，避免下列case: s == '.' or 'e/E' or '+/e' and etc...
+
+## 代码
+
+```go
+
+package leetcode
+
+func isNumber(s string) bool {
+	numFlag, dotFlag, eFlag := false, false, false
+	for i := 0; i < len(s); i++ {
+		if '0' <= s[i] && s[i] <= '9' {
+			numFlag = true
+		} else if s[i] == '.' && !dotFlag && !eFlag {
+			dotFlag = true
+		} else if (s[i] == 'e' || s[i] == 'E') && !eFlag && numFlag {
+			eFlag = true
+			numFlag = false // reJudge integer after 'e' or 'E'
+		} else if (s[i] == '+' || s[i] == '-') && (i == 0 || s[i-1] == 'e' || s[i-1] == 'E') {
+			continue
+		} else {
+			return false
+		}
+	}
+	// avoid case: s == '.' or 'e/E' or '+/-' and etc...
+	// string s must have num
+	return numFlag
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0064.Minimum-Path-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0066.Plus-One/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0066.Plus-One.md b/website/content/ChapterFour/0001~0099/0066.Plus-One.md
new file mode 100755
index 000000000..37761b545
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0066.Plus-One.md
@@ -0,0 +1,68 @@
+# [66. Plus One](https://leetcode.com/problems/plus-one/)
+
+
+## 题目
+
+Given a **non-empty** array of digits representing a non-negative integer, plus one to the integer.
+
+The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.
+
+You may assume the integer does not contain any leading zero, except the number 0 itself.
+
+**Example 1**:
+
+    Input: [1,2,3]
+    Output: [1,2,4]
+    Explanation: The array represents the integer 123.
+
+**Example 2**:
+
+    Input: [4,3,2,1]
+    Output: [4,3,2,2]
+    Explanation: The array represents the integer 4321.
+
+
+## 题目大意
+
+
+给定一个由整数组成的非空数组所表示的非负整数，在该数的基础上加一。最高位数字存放在数组的首位， 数组中每个元素只存储单个数字。你可以假设除了整数 0 之外，这个整数不会以零开头。
+
+
+
+## 解题思路
+
+- 给出一个数组，代表一个十进制数，数组的 0 下标是十进制数的高位。要求计算这个十进制数加一以后的结果。
+- 简单的模拟题。从数组尾部开始往前扫，逐位进位即可。最高位如果还有进位需要在数组里面第 0 位再插入一个 1 。
+
+## 代码
+
+```go
+
+package leetcode
+
+func plusOne(digits []int) []int {
+	for i := len(digits) - 1; i >= 0; i-- {
+		digits[i]++
+		if digits[i] != 10 {
+			// no carry
+			return digits
+		}
+		// carry
+		digits[i] = 0
+	}
+	// all carry
+	digits[0] = 1
+	digits = append(digits, 0)
+	return digits
+}
+
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0065.Valid-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0067.Add-Binary/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0067.Add-Binary.md b/website/content/ChapterFour/0001~0099/0067.Add-Binary.md
new file mode 100644
index 000000000..f685c2582
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0067.Add-Binary.md
@@ -0,0 +1,83 @@
+# [67. Add Binary](https://leetcode.com/problems/add-binary/)
+
+
+## 题目
+
+Given two binary strings, return their sum (also a binary string).
+
+The input strings are both **non-empty** and contains only characters `1` or `0`.
+
+**Example 1**:
+
+```
+Input: a = "11", b = "1"
+Output: "100"
+```
+
+**Example 2**:
+
+```
+Input: a = "1010", b = "1011"
+Output: "10101"
+```
+
+## 题目大意
+
+给你两个二进制字符串，返回它们的和（用二进制表示）。输入为 非空 字符串且只包含数字 1 和 0。
+
+## 解题思路
+
+- 要求输出 2 个二进制数的和，结果也用二进制表示。
+- 简单题。按照二进制的加法规则做加法即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+)
+
+func addBinary(a string, b string) string {
+	if len(b) > len(a) {
+		a, b = b, a
+	}
+
+	res := make([]string, len(a)+1)
+	i, j, k, c := len(a)-1, len(b)-1, len(a), 0
+	for i >= 0 && j >= 0 {
+		ai, _ := strconv.Atoi(string(a[i]))
+		bj, _ := strconv.Atoi(string(b[j]))
+		res[k] = strconv.Itoa((ai + bj + c) % 2)
+		c = (ai + bj + c) / 2
+		i--
+		j--
+		k--
+	}
+
+	for i >= 0 {
+		ai, _ := strconv.Atoi(string(a[i]))
+		res[k] = strconv.Itoa((ai + c) % 2)
+		c = (ai + c) / 2
+		i--
+		k--
+	}
+
+	if c > 0 {
+		res[k] = strconv.Itoa(c)
+	}
+
+	return strings.Join(res, "")
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0066.Plus-One/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0069.Sqrtx/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0069.Sqrtx.md b/website/content/ChapterFour/0001~0099/0069.Sqrtx.md
new file mode 100755
index 000000000..fc0aba54a
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0069.Sqrtx.md
@@ -0,0 +1,92 @@
+# [69. Sqrt(x)](https://leetcode.com/problems/sqrtx/)
+
+
+## 题目
+
+Implement `int sqrt(int x)`.
+
+Compute and return the square root of *x*, where *x* is guaranteed to be a non-negative integer.
+
+Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.
+
+**Example 1**:
+
+    Input: 4
+    Output: 2
+
+**Example 2**:
+
+    Input: 8
+    Output: 2
+    Explanation: The square root of 8 is 2.82842..., and since 
+                 the decimal part is truncated, 2 is returned.
+
+
+## 题目大意
+
+实现 int sqrt(int x) 函数。计算并返回 x 的平方根，其中 x 是非负整数。由于返回类型是整数，结果只保留整数的部分，小数部分将被舍去。
+
+
+
+## 解题思路
+
+- 题目要求求出根号 x
+- 根据题意，根号 x 的取值范围一定在 `[0,x]` 之间，这个区间内的值是递增有序的，有边界的，可以用下标访问的，满足这三点正好也就满足了二分搜索的 3 大条件。所以解题思路一，二分搜索。
+- 解题思路二，牛顿迭代法。求根号 x，即求满足 `x^2 - n = 0` 方程的所有解。
+
+    ![](https://img-blog.csdn.net/20171019164040871?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvY2hlbnJlbnhpYW5n/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center)
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 二分, 找到最后一个满足 n^2 <= x 的整数n
+func mySqrt(x int) int {
+	l, r := 0, x
+	for l < r {
+		mid := (l + r + 1) / 2
+		if mid*mid > x {
+			r = mid - 1
+		} else {
+			l = mid
+		}
+	}
+	return l
+}
+
+// 解法二 牛顿迭代法 https://en.wikipedia.org/wiki/Integer_square_root
+func mySqrt1(x int) int {
+	r := x
+	for r*r > x {
+		r = (r + x/r) / 2
+	}
+	return r
+}
+
+// 解法三 Quake III 游戏引擎中有一种比 STL 的 sqrt 快 4 倍的实现 https://en.wikipedia.org/wiki/Fast_inverse_square_root
+// float Q_rsqrt( float number )
+// {
+// 	long i;
+// 	float x2, y;
+// 	const float threehalfs = 1.5F;
+
+// 	x2 = number * 0.5F;
+// 	y  = number;
+// 	i  = * ( long * ) &y;                       // evil floating point bit level hacking
+// 	i  = 0x5f3759df - ( i >> 1 );               // what the fuck?
+// 	y  = * ( float * ) &i;
+// 	y  = y * ( threehalfs - ( x2 * y * y ) );   // 1st iteration
+// //	y  = y * ( threehalfs - ( x2 * y * y ) );   // 2nd iteration, this can be removed
+// 	return y;
+// }
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0067.Add-Binary/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0070.Climbing-Stairs/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0070.Climbing-Stairs.md b/website/content/ChapterFour/0001~0099/0070.Climbing-Stairs.md
new file mode 100755
index 000000000..274f7e02b
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0070.Climbing-Stairs.md
@@ -0,0 +1,62 @@
+# [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)
+
+
+## 题目
+
+You are climbing a stair case. It takes *n* steps to reach to the top.
+
+Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
+
+**Note**: Given *n* will be a positive integer.
+
+**Example 1**:
+
+    Input: 2
+    Output: 2
+    Explanation: There are two ways to climb to the top.
+    1. 1 step + 1 step
+    2. 2 steps
+
+**Example 2**:
+
+    Input: 3
+    Output: 3
+    Explanation: There are three ways to climb to the top.
+    1. 1 step + 1 step + 1 step
+    2. 1 step + 2 steps
+    3. 2 steps + 1 step
+
+
+## 题目大意
+
+假设你正在爬楼梯。需要 n 阶你才能到达楼顶。每次你可以爬 1 或 2 个台阶。你有多少种不同的方法可以爬到楼顶呢？注意：给定 n 是一个正整数
+
+
+## 解题思路
+
+- 简单的 DP，经典的爬楼梯问题。一个楼梯可以由 `n-1` 和 `n-2` 的楼梯爬上来。
+- 这一题求解的值就是斐波那契数列。
+
+## 代码
+
+```go
+
+package leetcode
+
+func climbStairs(n int) int {
+	dp := make([]int, n+1)
+	dp[0], dp[1] = 1, 1
+	for i := 2; i <= n; i++ {
+		dp[i] = dp[i-1] + dp[i-2]
+	}
+	return dp[n]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0069.Sqrtx/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0071.Simplify-Path/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0071.Simplify-Path.md b/website/content/ChapterFour/0001~0099/0071.Simplify-Path.md
new file mode 100644
index 000000000..750f72c7f
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0071.Simplify-Path.md
@@ -0,0 +1,124 @@
+# [71. Simplify Path](https://leetcode.com/problems/simplify-path/)
+
+## 题目
+
+Given an absolute path for a file (Unix-style), simplify it. Or in other words, convert it to the canonical path.
+
+In a UNIX-style file system, a period . refers to the current directory. Furthermore, a double period .. moves the directory up a level. For more information, see: Absolute path vs relative path in Linux/Unix
+
+Note that the returned canonical path must always begin with a slash /, and there must be only a single slash / between two directory names. The last directory name (if it exists) must not end with a trailing /. Also, the canonical path must be the shortest string representing the absolute path.
+
+
+
+**Example 1**:
+
+```
+
+Input: "/home/"
+Output: "/home"
+Explanation: Note that there is no trailing slash after the last directory name.
+
+```
+
+**Example 2**:
+
+```
+
+Input: "/../"
+Output: "/"
+Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.
+
+```
+**Example 3**:
+
+```
+
+Input: "/home//foo/"
+Output: "/home/foo"
+Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.
+
+```
+
+**Example 4**:
+
+```
+
+Input: "/a/./b/../../c/"
+Output: "/c"
+
+```
+
+
+**Example 5**:
+
+```
+
+Input: "/a/../../b/../c//.//"
+Output: "/c"
+
+```
+
+**Example 6**:
+
+```
+
+Input: "/a//b////c/d//././/.."
+Output: "/a/b/c"
+
+```
+
+## 题目大意
+
+给出一个 Unix 的文件路径，要求简化这个路径。这道题也是考察栈的题目。
+
+## 解题思路
+
+这道题笔者提交了好多次才通过，并不是题目难，而是边界条件很多，没考虑全一种情况就会出错。有哪些边界情况就看笔者的 test 文件吧。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"path/filepath"
+	"strings"
+)
+
+// 解法一
+func simplifyPath(path string) string {
+	arr := strings.Split(path, "/")
+	stack := make([]string, 0)
+	var res string
+	for i := 0; i < len(arr); i++ {
+		cur := arr[i]
+		//cur := strings.TrimSpace(arr[i]) 更加严谨的做法应该还要去掉末尾的空格
+		if cur == ".." {
+			if len(stack) > 0 {
+				stack = stack[:len(stack)-1]
+			}
+		} else if cur != "." && len(cur) > 0 {
+			stack = append(stack, arr[i])
+		}
+	}
+	if len(stack) == 0 {
+		return "/"
+	}
+	res = strings.Join(stack, "/")
+	return "/" + res
+}
+
+// 解法二 golang 的官方库 API
+func simplifyPath1(path string) string {
+	return filepath.Clean(path)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0070.Climbing-Stairs/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0073.Set-Matrix-Zeroes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0073.Set-Matrix-Zeroes.md b/website/content/ChapterFour/0001~0099/0073.Set-Matrix-Zeroes.md
new file mode 100644
index 000000000..faa39f428
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0073.Set-Matrix-Zeroes.md
@@ -0,0 +1,111 @@
+# [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)
+
+
+## 题目
+
+Given an *`m* x *n*` matrix. If an element is **0**, set its entire row and column to **0**. Do it **[in-place](https://en.wikipedia.org/wiki/In-place_algorithm)**.
+
+**Follow up:**
+
+- A straight forward solution using O(*mn*) space is probably a bad idea.
+- A simple improvement uses O(*m* + *n*) space, but still not the best solution.
+- Could you devise a constant space solution?
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/08/17/mat1.jpg](https://assets.leetcode.com/uploads/2020/08/17/mat1.jpg)
+
+```
+Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
+Output: [[1,0,1],[0,0,0],[1,0,1]]
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/08/17/mat2.jpg](https://assets.leetcode.com/uploads/2020/08/17/mat2.jpg)
+
+```
+Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
+Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
+```
+
+**Constraints:**
+
+- `m == matrix.length`
+- `n == matrix[0].length`
+- `1 <= m, n <= 200`
+- `2^31 <= matrix[i][j] <= 2^31 - 1`
+
+## 题目大意
+
+给定一个 `m x n` 的矩阵，如果一个元素为 0，则将其所在行和列的所有元素都设为 0。请使用原地算法。
+
+## 解题思路
+
+- 此题考查对程序的控制能力，无算法思想。题目要求采用原地的算法，所有修改即在原二维数组上进行。在二维数组中有 2 个特殊位置，一个是第一行，一个是第一列。它们的特殊性在于，它们之间只要有一个 0，它们都会变为全 0 。先用 2 个变量记录这一行和这一列中是否有 0，防止之后的修改覆盖了这 2 个地方。然后除去这一行和这一列以外的部分判断是否有 0，如果有 0，将它们所在的行第一个元素标记为 0，所在列的第一个元素标记为 0 。最后通过标记，将对应的行列置 0 即可。
+
+## 代码
+
+```go
+package leetcode
+
+func setZeroes(matrix [][]int) {
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return
+	}
+	isFirstRowExistZero, isFirstColExistZero := false, false
+	for i := 0; i < len(matrix); i++ {
+		if matrix[i][0] == 0 {
+			isFirstColExistZero = true
+			break
+		}
+	}
+	for j := 0; j < len(matrix[0]); j++ {
+		if matrix[0][j] == 0 {
+			isFirstRowExistZero = true
+			break
+		}
+	}
+	for i := 1; i < len(matrix); i++ {
+		for j := 1; j < len(matrix[0]); j++ {
+			if matrix[i][j] == 0 {
+				matrix[i][0] = 0
+				matrix[0][j] = 0
+			}
+		}
+	}
+	// 处理[1:]行全部置 0
+	for i := 1; i < len(matrix); i++ {
+		if matrix[i][0] == 0 {
+			for j := 1; j < len(matrix[0]); j++ {
+				matrix[i][j] = 0
+			}
+		}
+	}
+	// 处理[1:]列全部置 0
+	for j := 1; j < len(matrix[0]); j++ {
+		if matrix[0][j] == 0 {
+			for i := 1; i < len(matrix); i++ {
+				matrix[i][j] = 0
+			}
+		}
+	}
+	if isFirstRowExistZero {
+		for j := 0; j < len(matrix[0]); j++ {
+			matrix[0][j] = 0
+		}
+	}
+	if isFirstColExistZero {
+		for i := 0; i < len(matrix); i++ {
+			matrix[i][0] = 0
+		}
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0071.Simplify-Path/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0074.Search-a-2D-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0074.Search-a-2D-Matrix.md b/website/content/ChapterFour/0001~0099/0074.Search-a-2D-Matrix.md
new file mode 100755
index 000000000..df5886ea8
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0074.Search-a-2D-Matrix.md
@@ -0,0 +1,79 @@
+# [74. Search a 2D Matrix](https://leetcode.com/problems/search-a-2d-matrix/)
+
+
+## 题目
+
+Write an efficient algorithm that searches for a value in an *m* x *n* matrix. This matrix has the following properties:
+
+- Integers in each row are sorted from left to right.
+- The first integer of each row is greater than the last integer of the previous row.
+
+**Example 1**:
+
+    Input:
+    matrix = [
+      [1,   3,  5,  7],
+      [10, 11, 16, 20],
+      [23, 30, 34, 50]
+    ]
+    target = 3
+    Output: true
+
+**Example 2**:
+
+    Input:
+    matrix = [
+      [1,   3,  5,  7],
+      [10, 11, 16, 20],
+      [23, 30, 34, 50]
+    ]
+    target = 13
+    Output: false
+
+
+## 题目大意
+
+编写一个高效的算法来判断 m x n 矩阵中，是否存在一个目标值。该矩阵具有如下特性：
+
+- 每行中的整数从左到右按升序排列。
+- 每行的第一个整数大于前一行的最后一个整数。
+
+
+## 解题思路
+
+
+- 给出一个二维矩阵，矩阵的特点是随着矩阵的下标增大而增大。要求设计一个算法能在这个矩阵中高效的找到一个数，如果找到就输出 true，找不到就输出 false。
+- 虽然是一个二维矩阵，但是由于它特殊的有序性，所以完全可以按照下标把它看成一个一维矩阵，只不过需要行列坐标转换。最后利用二分搜索直接搜索即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func searchMatrix(matrix [][]int, target int) bool {
+	if len(matrix) == 0 {
+		return false
+	}
+	m, low, high := len(matrix[0]), 0, len(matrix[0])*len(matrix)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if matrix[mid/m][mid%m] == target {
+			return true
+		} else if matrix[mid/m][mid%m] > target {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0073.Set-Matrix-Zeroes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0075.Sort-Colors/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0075.Sort-Colors.md b/website/content/ChapterFour/0001~0099/0075.Sort-Colors.md
new file mode 100644
index 000000000..7a980a932
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0075.Sort-Colors.md
@@ -0,0 +1,67 @@
+# [75. Sort Colors](https://leetcode.com/problems/sort-colors/)
+
+## 题目
+
+Given an array with n objects colored red, white or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white and blue.
+
+Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.
+
+**Note**: You are not suppose to use the library's sort function for this problem.
+
+**Example 1**:  
+
+```
+
+Input: [2,0,2,1,1,0]
+Output: [0,0,1,1,2,2]
+
+```
+
+**Follow up**:
+
+- A rather straight forward solution is a two-pass algorithm using counting sort.  
+First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's.
+- Could you come up with a one-pass algorithm using only constant space?
+
+
+## 题目大意
+
+抽象题意其实就是排序。这题可以用快排一次通过。
+
+## 解题思路
+
+题目末尾的 Follow up 提出了一个更高的要求，能否用一次循环解决问题？这题由于数字只会出现 0，1，2 这三个数字，所以用游标移动来控制顺序也是可以的。具体做法：0 是排在最前面的，所以只要添加一个 0，就需要放置 1 和 2。1 排在 2 前面，所以添加 1 的时候也需要放置 2 。至于最后的 2，只用移动游标即可。
+
+这道题可以用计数排序，适合待排序数字很少的题目。用一个 3 个容量的数组分别计数，记录 0，1，2 出现的个数。然后再根据个数排列 0，1，2 即可。时间复杂度 O(n)，空间复杂度 O(K)。这一题 K = 3。
+
+这道题也可以用一次三路快排。数组分为 3 部分，第一个部分都是 0，中间部分都是 1，最后部分都是 2 。
+
+## 代码
+
+```go
+
+package leetcode
+
+func sortColors(nums []int) {
+	zero, one := 0, 0
+	for i, n := range nums {
+		nums[i] = 2
+		if n <= 1 {
+			nums[one] = 1
+			one++
+		}
+		if n == 0 {
+			nums[zero] = 0
+			zero++
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0074.Search-a-2D-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0076.Minimum-Window-Substring/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0076.Minimum-Window-Substring.md b/website/content/ChapterFour/0001~0099/0076.Minimum-Window-Substring.md
new file mode 100644
index 000000000..ebaaaa47b
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0076.Minimum-Window-Substring.md
@@ -0,0 +1,82 @@
+# [76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)
+
+## 题目
+
+Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
+
+**Example**:
+
+```
+
+Input: S = "ADOBECODEBANC", T = "ABC"
+Output: "BANC"
+
+```
+
+**Note**:    
+
+- If there is no such window in S that covers all characters in T, return the empty string "".
+- If there is such window, you are guaranteed that there will always be only one unique minimum window in S.
+
+## 题目大意
+
+给定一个源字符串 s，再给一个字符串 T，要求在源字符串中找到一个窗口，这个窗口包含由字符串各种排列组合组成的，窗口中可以包含 T 中没有的字符，如果存在多个，在结果中输出最小的窗口，如果找不到这样的窗口，输出空字符串。
+
+## 解题思路
+
+这一题是滑动窗口的题目，在窗口滑动的过程中不断的包含字符串 T，直到完全包含字符串 T 的字符以后，记下左右窗口的位置和窗口大小。每次都不断更新这个符合条件的窗口和窗口大小的最小值。最后输出结果即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func minWindow(s string, t string) string {
+	if s == "" || t == "" {
+		return ""
+	}
+	var tFreq, sFreq [256]int
+	result, left, right, finalLeft, finalRight, minW, count := "", 0, -1, -1, -1, len(s)+1, 0
+
+	for i := 0; i < len(t); i++ {
+		tFreq[t[i]-'a']++
+	}
+
+	for left < len(s) {
+		if right+1 < len(s) && count < len(t) {
+			sFreq[s[right+1]-'a']++
+			if sFreq[s[right+1]-'a'] <= tFreq[s[right+1]-'a'] {
+				count++
+			}
+			right++
+		} else {
+			if right-left+1 < minW && count == len(t) {
+				minW = right - left + 1
+				finalLeft = left
+				finalRight = right
+			}
+			if sFreq[s[left]-'a'] == tFreq[s[left]-'a'] {
+				count--
+			}
+			sFreq[s[left]-'a']--
+			left++
+		}
+	}
+	if finalLeft != -1 {
+		result = string(s[finalLeft : finalRight+1])
+	}
+	return result
+}
+
+
+```
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0075.Sort-Colors/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0077.Combinations/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0077.Combinations.md b/website/content/ChapterFour/0001~0099/0077.Combinations.md
new file mode 100755
index 000000000..19ff4112e
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0077.Combinations.md
@@ -0,0 +1,67 @@
+# [77. Combinations](https://leetcode.com/problems/combinations/)
+
+
+## 题目
+
+Given two integers *n* and *k*, return all possible combinations of *k* numbers out of 1 ... *n*.
+
+**Example**:
+
+    Input: n = 4, k = 2
+    Output:
+    [
+      [2,4],
+      [3,4],
+      [2,3],
+      [1,2],
+      [1,3],
+      [1,4],
+    ]
+
+## 题目大意
+
+给定两个整数 n 和 k，返回 1 ... n 中所有可能的 k 个数的组合。
+
+## 解题思路
+
+- 计算排列组合中的组合，用 DFS 深搜即可，注意剪枝
+
+## 代码
+
+```go
+
+package leetcode
+
+func combine(n int, k int) [][]int {
+	if n <= 0 || k <= 0 || k > n {
+		return [][]int{}
+	}
+	c, res := []int{}, [][]int{}
+	generateCombinations(n, k, 1, c, &res)
+	return res
+}
+
+func generateCombinations(n, k, start int, c []int, res *[][]int) {
+	if len(c) == k {
+		b := make([]int, len(c))
+		copy(b, c)
+		*res = append(*res, b)
+		return
+	}
+	// i will at most be n - (k - c.size()) + 1
+	for i := start; i <= n-(k-len(c))+1; i++ {
+		c = append(c, i)
+		generateCombinations(n, k, i+1, c, res)
+		c = c[:len(c)-1]
+	}
+	return
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0076.Minimum-Window-Substring/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0078.Subsets/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0078.Subsets.md b/website/content/ChapterFour/0001~0099/0078.Subsets.md
new file mode 100755
index 000000000..f0ccb1427
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0078.Subsets.md
@@ -0,0 +1,112 @@
+# [78. Subsets](https://leetcode.com/problems/subsets/)
+
+
+## 题目
+
+Given a set of **distinct** integers, *nums*, return all possible subsets (the power set).
+
+**Note**: The solution set must not contain duplicate subsets.
+
+**Example**:
+
+    Input: nums = [1,2,3]
+    Output:
+    [
+      [3],
+      [1],
+      [2],
+      [1,2,3],
+      [1,3],
+      [2,3],
+      [1,2],
+      []
+    ]
+
+
+## 题目大意
+
+给定一组不含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。说明：解集不能包含重复的子集。
+
+
+## 解题思路
+
+- 找出一个集合中的所有子集，空集也算是子集。且数组中的数字不会出现重复。用 DFS 暴力枚举即可。
+- 这一题和第 90 题，第 491 题类似，可以一起解答和复习。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+// 解法一
+func subsets(nums []int) [][]int {
+	c, res := []int{}, [][]int{}
+	for k := 0; k <= len(nums); k++ {
+		generateSubsets(nums, k, 0, c, &res)
+	}
+	return res
+}
+
+func generateSubsets(nums []int, k, start int, c []int, res *[][]int) {
+	if len(c) == k {
+		b := make([]int, len(c))
+		copy(b, c)
+		*res = append(*res, b)
+		return
+	}
+	// i will at most be n - (k - c.size()) + 1
+	for i := start; i < len(nums)-(k-len(c))+1; i++ {
+		c = append(c, nums[i])
+		generateSubsets(nums, k, i+1, c, res)
+		c = c[:len(c)-1]
+	}
+	return
+}
+
+// 解法二
+func subsets1(nums []int) [][]int {
+	res := make([][]int, 1)
+	sort.Ints(nums)
+	for i := range nums {
+		for _, org := range res {
+			clone := make([]int, len(org), len(org)+1)
+			copy(clone, org)
+			clone = append(clone, nums[i])
+			res = append(res, clone)
+		}
+	}
+	return res
+}
+
+// 解法三：位运算的方法
+func subsets2(nums []int) [][]int {
+	if len(nums) == 0 {
+		return nil
+	}
+	res := [][]int{}
+	sum := 1 << uint(len(nums))
+	for i := 0; i < sum; i++ {
+		stack := []int{}
+		tmp := i // i 从 000...000 到 111...111
+		for j := len(nums) - 1; j >= 0; j-- { // 遍历 i 的每一位
+			if tmp & 1 == 1 {
+				stack = append([]int{nums[j]}, stack...)
+			}
+			tmp >>= 1
+		}
+		res = append(res, stack)
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0077.Combinations/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0079.Word-Search/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0079.Word-Search.md b/website/content/ChapterFour/0001~0099/0079.Word-Search.md
new file mode 100755
index 000000000..d8d123f48
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0079.Word-Search.md
@@ -0,0 +1,90 @@
+# [79. Word Search](https://leetcode.com/problems/word-search/)
+
+
+## 题目
+
+Given a 2D board and a word, find if the word exists in the grid.
+
+The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.
+
+**Example**:
+
+    board =
+    [
+      ['A','B','C','E'],
+      ['S','F','C','S'],
+      ['A','D','E','E']
+    ]
+    
+    Given word = "ABCCED", return true.
+    Given word = "SEE", return true.
+    Given word = "ABCB", return false.
+
+## 题目大意
+
+给定一个二维网格和一个单词，找出该单词是否存在于网格中。单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
+
+
+
+## 解题思路
+
+- 在地图上的任意一个起点开始，向 4 个方向分别 DFS 搜索，直到所有的单词字母都找到了就输出 true，否则输出 false。
+
+## 代码
+
+```go
+
+package leetcode
+
+var dir = [][]int{
+	[]int{-1, 0},
+	[]int{0, 1},
+	[]int{1, 0},
+	[]int{0, -1},
+}
+
+func exist(board [][]byte, word string) bool {
+	visited := make([][]bool, len(board))
+	for i := 0; i < len(visited); i++ {
+		visited[i] = make([]bool, len(board[0]))
+	}
+	for i, v := range board {
+		for j := range v {
+			if searchWord(board, visited, word, 0, i, j) {
+				return true
+			}
+		}
+	}
+	return false
+}
+
+func isInBoard(board [][]byte, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+
+func searchWord(board [][]byte, visited [][]bool, word string, index, x, y int) bool {
+	if index == len(word)-1 {
+		return board[x][y] == word[index]
+	}
+	if board[x][y] == word[index] {
+		visited[x][y] = true
+		for i := 0; i < 4; i++ {
+			nx := x + dir[i][0]
+			ny := y + dir[i][1]
+			if isInBoard(board, nx, ny) && !visited[nx][ny] && searchWord(board, visited, word, index+1, nx, ny) {
+				return true
+			}
+		}
+		visited[x][y] = false
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0078.Subsets/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0080.Remove-Duplicates-from-Sorted-Array-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0080.Remove-Duplicates-from-Sorted-Array-II.md b/website/content/ChapterFour/0001~0099/0080.Remove-Duplicates-from-Sorted-Array-II.md
new file mode 100644
index 000000000..84a5e631a
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0080.Remove-Duplicates-from-Sorted-Array-II.md
@@ -0,0 +1,92 @@
+# [80. Remove Duplicates from Sorted Array II](https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/)
+
+## 题目
+
+Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new
+length.
+
+Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra
+memory.
+
+**Example 1**:
+
+```
+
+Given nums = [1,1,1,2,2,3],
+
+Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
+
+It doesn't matter what you leave beyond the returned length.
+
+```
+
+**Example 2**:
+
+```
+
+Given nums = [0,0,1,1,1,1,2,3,3],
+
+Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively.
+
+It doesn't matter what values are set beyond the returned length.
+
+```
+
+**Clarification**:
+
+Confused why the returned value is an integer but your answer is an array?
+
+Note that the input array is passed in by reference, which means modification to the input array will be known to the
+caller as well.
+
+Internally you can think of this:
+
+```
+
+// nums is passed in by reference. (i.e., without making a copy)
+int len = removeElement(nums, val);
+
+// any modification to nums in your function would be known by the caller.
+// using the length returned by your function, it prints the first len elements.
+for (int i = 0; i < len; i++) {
+    print(nums[i]);
+}
+
+```
+
+## 题目大意
+
+给定一个有序数组 nums，对数组中的元素进行去重，使得原数组中的每个元素最多暴露 2 个。最后返回去重以后数组的长度值。
+
+## 解题思路
+
+- 问题提示有序数组，一般最容易想到使用双指针的解法，双指针的关键点：移动两个指针的条件。
+- 在该题中移动的条件：快指针从头遍历数组，慢指针指向修改后的数组的末端，当慢指针指向倒数第二个数与快指针指向的数不相等时，才移动慢指针，同时赋值慢指针。
+- 处理边界条件：当数组小于两个元素时，不做处理。
+
+## 代码
+
+```go
+
+package leetcode
+
+func removeDuplicates(nums []int) int {
+	slow := 0
+	for fast, v := range nums {
+		if fast < 2 || nums[slow-2] != v {
+			nums[slow] = v
+			slow++
+		}
+	}
+	return slow
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0079.Word-Search/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0081.Search-in-Rotated-Sorted-Array-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0081.Search-in-Rotated-Sorted-Array-II.md b/website/content/ChapterFour/0001~0099/0081.Search-in-Rotated-Sorted-Array-II.md
new file mode 100755
index 000000000..7ed0474b7
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0081.Search-in-Rotated-Sorted-Array-II.md
@@ -0,0 +1,92 @@
+# [81. Search in Rotated Sorted Array II](https://leetcode.com/problems/search-in-rotated-sorted-array-ii/)
+
+
+## 题目
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., `[0,0,1,2,2,5,6]` might become `[2,5,6,0,0,1,2]`).
+
+You are given a target value to search. If found in the array return `true`, otherwise return `false`.
+
+**Example 1**:
+
+    Input: nums = [2,5,6,0,0,1,2], target = 0
+    Output: true
+
+**Example 2**:
+
+    Input: nums = [2,5,6,0,0,1,2], target = 3
+    Output: false
+
+**Follow up**:
+
+- This is a follow up problem to [Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/description/), where `nums` may contain duplicates.
+- Would this affect the run-time complexity? How and why?
+
+
+## 题目大意
+
+假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,0,1,2,2,5,6] 可能变为 [2,5,6,0,0,1,2] )。
+
+编写一个函数来判断给定的目标值是否存在于数组中。若存在返回 true，否则返回 false。
+
+进阶:
+
+- 这是搜索旋转排序数组 的延伸题目，本题中的 nums  可能包含重复元素。
+- 这会影响到程序的时间复杂度吗？会有怎样的影响，为什么？
+
+
+## 解题思路
+
+
+- 给出一个数组，数组中本来是从小到大排列的，并且数组中有重复数字。但是现在把后面随机一段有序的放到数组前面，这样形成了前后两端有序的子序列。在这样的一个数组里面查找一个数，设计一个 O(log n) 的算法。如果找到就输出 `true`，如果没有找到，就输出 `false` 。
+- 这一题是第 33 题的加强版，实现代码完全一样，只不过输出变了。这一题输出 `true` 和 `false` 了。具体思路见第 33 题。
+
+## 代码
+
+```go
+
+package leetcode
+
+func search(nums []int, target int) bool {
+	if len(nums) == 0 {
+		return false
+	}
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[mid] == target {
+			return true
+		} else if nums[mid] > nums[low] { // 在数值大的一部分区间里
+			if nums[low] <= target && target < nums[mid] {
+				high = mid - 1
+			} else {
+				low = mid + 1
+			}
+		} else if nums[mid] < nums[high] { // 在数值小的一部分区间里
+			if nums[mid] < target && target <= nums[high] {
+				low = mid + 1
+			} else {
+				high = mid - 1
+			}
+		} else {
+			if nums[low] == nums[mid] {
+				low++
+			}
+			if nums[high] == nums[mid] {
+				high--
+			}
+		}
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0080.Remove-Duplicates-from-Sorted-Array-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0082.Remove-Duplicates-from-Sorted-List-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0082.Remove-Duplicates-from-Sorted-List-II.md b/website/content/ChapterFour/0001~0099/0082.Remove-Duplicates-from-Sorted-List-II.md
new file mode 100644
index 000000000..4cbc0bf89
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0082.Remove-Duplicates-from-Sorted-List-II.md
@@ -0,0 +1,198 @@
+# [82. Remove Duplicates from Sorted List II](https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/)
+
+## 题目
+
+Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
+
+**Example 1**:
+
+```
+
+Input: 1->2->3->3->4->4->5
+Output: 1->2->5
+
+```
+
+**Example 2**:
+
+```
+
+Input: 1->1->1->2->3
+Output: 2->3
+
+```
+
+## 题目大意
+
+删除链表中重复的结点，只要是有重复过的结点，全部删除。
+
+## 解题思路
+
+按照题意做即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// 解法一
+func deleteDuplicates1(head *ListNode) *ListNode {
+	if head == nil {
+		return nil
+	}
+	if head.Next == nil {
+		return head
+	}
+	newHead := &ListNode{Next: head, Val: -999999}
+	cur := newHead
+	last := newHead
+	front := head
+	for front.Next != nil {
+		if front.Val == cur.Val {
+			// fmt.Printf("相同节点front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
+			front = front.Next
+			continue
+		} else {
+			if cur.Next != front {
+				// fmt.Printf("删除重复节点front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
+				last.Next = front
+				if front.Next != nil && front.Next.Val != front.Val {
+					last = front
+				}
+				cur = front
+				front = front.Next
+			} else {
+				// fmt.Printf("常规循环前front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
+				last = cur
+				cur = cur.Next
+				front = front.Next
+				// fmt.Printf("常规循环后front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
+
+			}
+		}
+	}
+	if front.Val == cur.Val {
+		// fmt.Printf("相同节点front = %v | cur = %v | last = %v\n", front.Val, cur.Val, last.Val)
+		last.Next = nil
+	} else {
+		if cur.Next != front {
+			last.Next = front
+		}
+	}
+	return newHead.Next
+}
+
+// 解法二
+func deleteDuplicates2(head *ListNode) *ListNode {
+	if head == nil {
+		return nil
+	}
+	if head.Next != nil && head.Val == head.Next.Val {
+		for head.Next != nil && head.Val == head.Next.Val {
+			head = head.Next
+		}
+		return deleteDuplicates(head.Next)
+	}
+	head.Next = deleteDuplicates(head.Next)
+	return head
+}
+
+func deleteDuplicates(head *ListNode) *ListNode {
+	cur := head
+	if head == nil {
+		return nil
+	}
+	if head.Next == nil {
+		return head
+	}
+	for cur.Next != nil {
+		if cur.Next.Val == cur.Val {
+			cur.Next = cur.Next.Next
+		} else {
+			cur = cur.Next
+		}
+	}
+	return head
+}
+
+// 解法三 双循环简单解法 O(n*m)
+func deleteDuplicates3(head *ListNode) *ListNode {
+	if head == nil {
+		return head
+	}
+
+	nilNode := &ListNode{Val: 0, Next: head}
+	head = nilNode
+
+	lastVal := 0
+	for head.Next != nil && head.Next.Next != nil {
+		if head.Next.Val == head.Next.Next.Val {
+			lastVal = head.Next.Val
+			for head.Next != nil && lastVal == head.Next.Val {
+				head.Next = head.Next.Next
+			}
+		} else {
+			head = head.Next
+		}
+	}
+	return nilNode.Next
+}
+
+// 解法四 双指针+删除标志位，单循环解法 O(n)
+func deleteDuplicates4(head *ListNode) *ListNode {
+	if head == nil || head.Next == nil {
+		return head
+	}
+
+	nilNode := &ListNode{Val: 0, Next: head}
+	// 上次遍历有删除操作的标志位
+	lastIsDel := false
+	// 虚拟空结点
+	head = nilNode
+	// 前后指针用于判断
+	pre, back := head.Next, head.Next.Next
+	// 每次只删除前面的一个重复的元素，留一个用于下次遍历判重
+	// pre, back 指针的更新位置和值比较重要和巧妙
+	for head.Next != nil && head.Next.Next != nil {
+		if pre.Val != back.Val && lastIsDel {
+			head.Next = head.Next.Next
+			pre, back = head.Next, head.Next.Next
+			lastIsDel = false
+			continue
+		}
+
+		if pre.Val == back.Val {
+			head.Next = head.Next.Next
+			pre, back = head.Next, head.Next.Next
+			lastIsDel = true
+		} else {
+			head = head.Next
+			pre, back = head.Next, head.Next.Next
+			lastIsDel = false
+		}
+	}
+	// 处理 [1,1] 这种删除还剩一个的情况
+	if lastIsDel && head.Next != nil {
+		head.Next = nil
+	}
+	return nilNode.Next
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0081.Search-in-Rotated-Sorted-Array-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0083.Remove-Duplicates-from-Sorted-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0083.Remove-Duplicates-from-Sorted-List.md b/website/content/ChapterFour/0001~0099/0083.Remove-Duplicates-from-Sorted-List.md
new file mode 100644
index 000000000..434449c9e
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0083.Remove-Duplicates-from-Sorted-List.md
@@ -0,0 +1,73 @@
+# [83. Remove Duplicates from Sorted List](https://leetcode.com/problems/remove-duplicates-from-sorted-list/)
+
+## 题目
+
+Given a sorted linked list, delete all duplicates such that each element appear only once.
+
+**Example 1**:
+
+```
+
+Input: 1->1->2
+Output: 1->2
+
+```
+
+**Example 2**:
+
+```
+
+Input: 1->1->2->3->3
+Output: 1->2->3
+
+```
+
+## 题目大意
+
+删除链表中重复的结点，以保障每个结点只出现一次。
+
+
+## 解题思路
+
+按照题意做即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func deleteDuplicates(head *ListNode) *ListNode {
+	cur := head
+	if head == nil {
+		return nil
+	}
+	if head.Next == nil {
+		return head
+	}
+	for cur.Next != nil {
+		if cur.Next.Val == cur.Val {
+			cur.Next = cur.Next.Next
+		} else {
+			cur = cur.Next
+		}
+	}
+	return head
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0082.Remove-Duplicates-from-Sorted-List-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0084.Largest-Rectangle-in-Histogram/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0084.Largest-Rectangle-in-Histogram.md b/website/content/ChapterFour/0001~0099/0084.Largest-Rectangle-in-Histogram.md
new file mode 100644
index 000000000..c2c9e289c
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0084.Largest-Rectangle-in-Histogram.md
@@ -0,0 +1,83 @@
+# [84. Largest Rectangle in Histogram](https://leetcode.com/problems/largest-rectangle-in-histogram/)
+
+## 题目
+
+Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.
+
+ ![](https://assets.leetcode.com/uploads/2018/10/12/histogram.png)
+
+
+Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].
+
+![](https://assets.leetcode.com/uploads/2018/10/12/histogram_area.png)
+
+
+The largest rectangle is shown in the shaded area, which has area = 10 unit.
+
+ 
+
+**Example**:
+
+```
+
+Input: [2,1,5,6,2,3]
+Output: 10
+
+```
+
+
+## 题目大意
+
+给出每个直方图的高度，要求在这些直方图之中找到面积最大的矩形，输出矩形的面积。
+
+
+## 解题思路
+
+用单调栈依次保存直方图的高度下标，一旦出现高度比栈顶元素小的情况就取出栈顶元素，单独计算一下这个栈顶元素的矩形的高度。然后停在这里(外层循环中的 i--，再 ++，就相当于停在这里了)，继续取出当前最大栈顶的前一个元素，即连续弹出 2 个最大的，以稍小的一个作为矩形的边，宽就是 2 计算面积…………如果停在这里的下标代表的高度一直比栈里面的元素小，就一直弹出，取出最后一个比当前下标大的高度作为矩形的边。宽就是最后一个比当前下标大的高度和当前下标 i 的差值。计算出面积以后不断的更新 maxArea 即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func largestRectangleArea(heights []int) int {
+	maxArea := 0
+	n := len(heights) + 2
+	// Add a sentry at the beginning and the end
+	getHeight := func(i int) int {
+		if i == 0 || n-1 == i {
+			return 0
+		}
+		return heights[i-1]
+	}
+	st := make([]int, 0, n/2)
+	for i := 0; i < n; i++ {
+		for len(st) > 0 && getHeight(st[len(st)-1]) > getHeight(i) {
+			// pop stack
+			idx := st[len(st)-1]
+			st = st[:len(st)-1]
+			maxArea = max(maxArea, getHeight(idx)*(i-st[len(st)-1]-1))
+		}
+		// push stack
+		st = append(st, i)
+	}
+	return maxArea
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0083.Remove-Duplicates-from-Sorted-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0086.Partition-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0086.Partition-List.md b/website/content/ChapterFour/0001~0099/0086.Partition-List.md
new file mode 100644
index 000000000..de6bb8fcc
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0086.Partition-List.md
@@ -0,0 +1,155 @@
+# [86. Partition List](https://leetcode.com/problems/partition-list/)
+
+## 题目
+
+Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
+
+You should preserve the original relative order of the nodes in each of the two partitions.
+
+**Example**:
+
+```
+
+Input: head = 1->4->3->2->5->2, x = 3
+Output: 1->2->2->4->3->5
+
+```
+
+
+## 题目大意
+
+给定一个数 x，比 x 大或等于的数字都要排列在比 x 小的数字后面，并且相对位置不能发生变化。由于相对位置不能发生变化，所以不能用类似冒泡排序的思想。
+
+## 解题思路
+
+这道题最简单的做法是构造双向链表，不过时间复杂度是 O(n^2)。
+
+(以下描述定义，大于等于 x 的都属于比 x 大)
+
+更优的方法是新构造 2 个链表，一个链表专门存储比 x 小的结点，另一个专门存储比 x 大的结点。在原链表头部开始扫描一边，依次把这两类点归类到 2 个新建链表中，有点入栈的意思。由于是从头开始扫描的原链表，所以原链表中的原有顺序会依旧被保存下来。最后 2 个新链表里面会存储好各自的结果，把这两个链表，比 x 小的链表拼接到 比 x 大的链表的前面，就能得到最后的答案了。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// 解法一 单链表
+func partition(head *ListNode, x int) *ListNode {
+	beforeHead := &ListNode{Val: 0, Next: nil}
+	before := beforeHead
+	afterHead := &ListNode{Val: 0, Next: nil}
+	after := afterHead
+
+	for head != nil {
+		if head.Val < x {
+			before.Next = head
+			before = before.Next
+		} else {
+			after.Next = head
+			after = after.Next
+		}
+		head = head.Next
+	}
+	after.Next = nil
+	before.Next = afterHead.Next
+	return beforeHead.Next
+}
+
+// DoublyListNode define
+type DoublyListNode struct {
+	Val  int
+	Prev *DoublyListNode
+	Next *DoublyListNode
+}
+
+// 解法二 双链表
+func partition1(head *ListNode, x int) *ListNode {
+	if head == nil || head.Next == nil {
+		return head
+	}
+	DLNHead := genDoublyListNode(head)
+	cur := DLNHead
+	for cur != nil {
+		if cur.Val < x {
+			tmp := &DoublyListNode{Val: cur.Val, Prev: nil, Next: nil}
+			compareNode := cur
+			for compareNode.Prev != nil {
+				if compareNode.Val >= x && compareNode.Prev.Val < x {
+					break
+				}
+				compareNode = compareNode.Prev
+			}
+			if compareNode == DLNHead {
+				if compareNode.Val < x {
+					cur = cur.Next
+					continue
+				} else {
+					tmp.Next = DLNHead
+					DLNHead.Prev = tmp
+					DLNHead = tmp
+				}
+			} else {
+				tmp.Next = compareNode
+				tmp.Prev = compareNode.Prev
+				compareNode.Prev.Next = tmp
+				compareNode.Prev = tmp
+			}
+			deleteNode := cur
+			if cur.Prev != nil {
+				deleteNode.Prev.Next = deleteNode.Next
+			}
+			if cur.Next != nil {
+				deleteNode.Next.Prev = deleteNode.Prev
+			}
+		}
+		cur = cur.Next
+	}
+	return genListNode(DLNHead)
+}
+
+func genDoublyListNode(head *ListNode) *DoublyListNode {
+	cur := head.Next
+	DLNHead := &DoublyListNode{Val: head.Val, Prev: nil, Next: nil}
+	curDLN := DLNHead
+	for cur != nil {
+		tmp := &DoublyListNode{Val: cur.Val, Prev: curDLN, Next: nil}
+		curDLN.Next = tmp
+		curDLN = tmp
+		cur = cur.Next
+	}
+	return DLNHead
+}
+
+func genListNode(head *DoublyListNode) *ListNode {
+	cur := head.Next
+	LNHead := &ListNode{Val: head.Val, Next: nil}
+	curLN := LNHead
+	for cur != nil {
+		tmp := &ListNode{Val: cur.Val, Next: nil}
+		curLN.Next = tmp
+		curLN = tmp
+		cur = cur.Next
+	}
+	return LNHead
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0084.Largest-Rectangle-in-Histogram/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0088.Merge-Sorted-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0088.Merge-Sorted-Array.md b/website/content/ChapterFour/0001~0099/0088.Merge-Sorted-Array.md
new file mode 100644
index 000000000..5c05cffa9
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0088.Merge-Sorted-Array.md
@@ -0,0 +1,64 @@
+# [88. Merge Sorted Array](https://leetcode.com/problems/merge-sorted-array/description/)
+
+## 题目
+
+Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
+
+**Note**:
+
+- The number of elements initialized in nums1 and nums2 are m and n respectively.
+- You may assume that nums1 has enough space (size that is equal to m + n) to hold additional elements from nums2.
+
+**Example**:
+
+	Input:
+	nums1 = [1,2,3,0,0,0], m = 3
+	nums2 = [2,5,6],       n = 3
+
+	Output: [1,2,2,3,5,6]
+ 
+
+**Constraints**:
+
+- -10^9 <= nums1[i], nums2[i] <= 10^9
+- nums1.length == m + n
+- nums2.length == n
+
+## 题目大意
+
+合并两个已经有序的数组，结果放在第一个数组中，第一个数组假设空间足够大。要求算法时间复杂度足够低。
+
+## 解题思路
+
+为了不大量移动元素，就要从2个数组长度之和的最后一个位置开始，依次选取两个数组中大的数，从第一个数组的尾巴开始往头放，只要循环一次以后，就生成了合并以后的数组了。
+
+## 代码
+
+```go
+
+package leetcode
+
+func merge(nums1 []int, m int, nums2 []int, n int) {
+	for p := m + n; m > 0 && n > 0; p-- {
+		if nums1[m-1] <= nums2[n-1] {
+			nums1[p-1] = nums2[n-1]
+			n--
+		} else {
+			nums1[p-1] = nums1[m-1]
+			m--
+		}
+	}
+	for ; n > 0; n-- {
+		nums1[n-1] = nums2[n-1]
+	}
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0086.Partition-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0089.Gray-Code/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0089.Gray-Code.md b/website/content/ChapterFour/0001~0099/0089.Gray-Code.md
new file mode 100755
index 000000000..63b2e39a4
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0089.Gray-Code.md
@@ -0,0 +1,124 @@
+# [89. Gray Code](https://leetcode.com/problems/gray-code/)
+
+## 题目
+
+The gray code is a binary numeral system where two successive values differ in only one bit.
+
+Given a non-negative integer *n* representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.
+
+**Example 1**:
+
+    Input: 2
+    Output: [0,1,3,2]
+    Explanation:
+    00 - 0
+    01 - 1
+    11 - 3
+    10 - 2
+    
+    For a given n, a gray code sequence may not be uniquely defined.
+    For example, [0,2,3,1] is also a valid gray code sequence.
+    
+    00 - 0
+    10 - 2
+    11 - 3
+    01 - 1
+
+**Example 2**:
+
+    Input: 0
+    Output: [0]
+    Explanation: We define the gray code sequence to begin with 0.
+                 A gray code sequence of n has size = 2n, which for n = 0 the size is 20 = 1.
+                 Therefore, for n = 0 the gray code sequence is [0].
+
+
+## 题目大意
+
+格雷编码是一个二进制数字系统，在该系统中，两个连续的数值仅有一个位数的差异。给定一个代表编码总位数的非负整数 n，打印其格雷编码序列。格雷编码序列必须以 0 开头。
+
+
+
+## 解题思路
+
+- 输出 n 位格雷码
+- 格雷码生成规则：以二进制为0值的格雷码为第零项，第一次改变最右边的位元，第二次改变右起第一个为1的位元的左边位元，第三、四次方法同第一、二次，如此反复，即可排列出 n 个位元的格雷码。
+- 可以直接模拟，也可以用递归求解。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 递归方法，时间复杂度和空间复杂度都较优
+func grayCode(n int) []int {
+	if n == 0 {
+		return []int{0}
+	}
+	res := []int{}
+	num := make([]int, n)
+	generateGrayCode(int(1<<uint(n)), 0, &num, &res)
+	return res
+}
+
+func generateGrayCode(n, step int, num *[]int, res *[]int) {
+	if n == 0 {
+		return
+	}
+	*res = append(*res, convertBinary(*num))
+
+	if step%2 == 0 {
+		(*num)[len(*num)-1] = flipGrayCode((*num)[len(*num)-1])
+	} else {
+		index := len(*num) - 1
+		for ; index >= 0; index-- {
+			if (*num)[index] == 1 {
+				break
+			}
+		}
+		if index == 0 {
+			(*num)[len(*num)-1] = flipGrayCode((*num)[len(*num)-1])
+		} else {
+			(*num)[index-1] = flipGrayCode((*num)[index-1])
+		}
+	}
+	generateGrayCode(n-1, step+1, num, res)
+	return
+}
+
+func convertBinary(num []int) int {
+	res, rad := 0, 1
+	for i := len(num) - 1; i >= 0; i-- {
+		res += num[i] * rad
+		rad *= 2
+	}
+	return res
+}
+
+func flipGrayCode(num int) int {
+	if num == 0 {
+		return 1
+	}
+	return 0
+}
+
+// 解法二 直译
+func grayCode1(n int) []int {
+	var l uint = 1 << uint(n)
+	out := make([]int, l)
+	for i := uint(0); i < l; i++ {
+		out[i] = int((i >> 1) ^ i)
+	}
+	return out
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0088.Merge-Sorted-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0090.Subsets-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0090.Subsets-II.md b/website/content/ChapterFour/0001~0099/0090.Subsets-II.md
new file mode 100755
index 000000000..73c331b1c
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0090.Subsets-II.md
@@ -0,0 +1,83 @@
+# [90. Subsets II](https://leetcode.com/problems/subsets-ii/)
+
+
+## 题目
+
+Given a collection of integers that might contain duplicates, ***nums***, return all possible subsets (the power set).
+
+**Note**: The solution set must not contain duplicate subsets.
+
+**Example**:
+
+    Input: [1,2,2]
+    Output:
+    [
+      [2],
+      [1],
+      [1,2,2],
+      [2,2],
+      [1,2],
+      []
+    ]
+
+## 题目大意
+
+给定一个可能包含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。说明：解集不能包含重复的子集。
+
+
+## 解题思路
+
+- 这一题是第 78 题的加强版，比第 78 题多了一个条件，数组中的数字会出现重复。
+- 解题方法依旧是 DFS，需要在回溯的过程中加上一些判断。
+- 这一题和第 78 题，第 491 题类似，可以一起解答和复习。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"fmt"
+	"sort"
+)
+
+func subsetsWithDup(nums []int) [][]int {
+	c, res := []int{}, [][]int{}
+	sort.Ints(nums) // 这里是去重的关键逻辑
+	for k := 0; k <= len(nums); k++ {
+		generateSubsetsWithDup(nums, k, 0, c, &res)
+	}
+	return res
+}
+
+func generateSubsetsWithDup(nums []int, k, start int, c []int, res *[][]int) {
+	if len(c) == k {
+		b := make([]int, len(c))
+		copy(b, c)
+		*res = append(*res, b)
+		return
+	}
+	// i will at most be n - (k - c.size()) + 1
+	for i := start; i < len(nums)-(k-len(c))+1; i++ {
+		fmt.Printf("i = %v start = %v c = %v\n", i, start, c)
+		if i > start && nums[i] == nums[i-1] { // 这里是去重的关键逻辑,本次不取重复数字，下次循环可能会取重复数字
+			continue
+		}
+		c = append(c, nums[i])
+		generateSubsetsWithDup(nums, k, i+1, c, res)
+		c = c[:len(c)-1]
+	}
+	return
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0089.Gray-Code/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0091.Decode-Ways/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0091.Decode-Ways.md b/website/content/ChapterFour/0001~0099/0091.Decode-Ways.md
new file mode 100755
index 000000000..7ac985062
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0091.Decode-Ways.md
@@ -0,0 +1,76 @@
+# [91. Decode Ways](https://leetcode.com/problems/decode-ways/)
+
+
+## 题目
+
+A message containing letters from `A-Z` is being encoded to numbers using the following mapping:
+
+    'A' -> 1
+    'B' -> 2
+    ...
+    'Z' -> 26
+
+Given a **non-empty** string containing only digits, determine the total number of ways to decode it.
+
+**Example 1**:
+
+    Input: "12"
+    Output: 2
+    Explanation: It could be decoded as "AB" (1 2) or "L" (12).
+
+**Example 2**:
+
+    Input: "226"
+    Output: 3
+    Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
+
+## 题目大意
+
+一条包含字母 A-Z 的消息通过以下方式进行了编码：
+
+```go
+'A' -> 1
+'B' -> 2
+...
+'Z' -> 26
+```
+
+给定一个只包含数字的非空字符串，请计算解码方法的总数。
+
+
+
+## 解题思路
+
+- 给出一个数字字符串，题目要求把数字映射成 26 个字母，映射以后问有多少种可能的翻译方法。
+- 这题思路也是 DP。`dp[n]` 代表翻译长度为 n 个字符的字符串的方法总数。由于题目中的数字可能出现 0，0 不能翻译成任何字母，所以出现 0 要跳过。dp[0] 代表空字符串，只有一种翻译方法，`dp[0] = 1`。dp[1] 需要考虑原字符串是否是 0 开头的，如果是 0 开头的，`dp[1] = 0`，如果不是 0 开头的，`dp[1] = 1`。状态转移方程是 `dp[i] += dp[i-1] (当 1 ≤ s[i-1 : i] ≤ 9)；dp[i] += dp[i-2] (当 10 ≤ s[i-2 : i] ≤ 26)`。最终结果是 `dp[n]`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func numDecodings(s string) int {
+	n := len(s)
+	dp := make([]int, n+1)
+	dp[0] = 1
+	for i := 1; i <= n; i++ {
+		if s[i-1] != '0' {
+			dp[i] += dp[i-1]
+		}
+		if i > 1 && s[i-2] != '0' && (s[i-2]-'0')*10+(s[i-1]-'0') <= 26 {
+			dp[i] += dp[i-2]
+		}
+	}
+	return dp[n]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0090.Subsets-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0092.Reverse-Linked-List-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0092.Reverse-Linked-List-II.md b/website/content/ChapterFour/0001~0099/0092.Reverse-Linked-List-II.md
new file mode 100644
index 000000000..d7f3dc7e3
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0092.Reverse-Linked-List-II.md
@@ -0,0 +1,72 @@
+# [92. Reverse Linked List II](https://leetcode.com/problems/reverse-linked-list-ii/)
+
+## 题目
+
+Reverse a linked list from position m to n. Do it in one-pass.
+
+**Note**: 1 ≤ m ≤ n ≤ length of list.
+
+**Example**:
+
+```
+
+Input: 1->2->3->4->5->NULL, m = 2, n = 4
+Output: 1->4->3->2->5->NULL
+
+```
+
+
+## 题目大意
+
+给定 2 个链表中结点的位置 m, n，反转这个两个位置区间内的所有结点。
+
+## 解题思路
+
+由于有可能整个链表都被反转，所以构造一个新的头结点指向当前的头。之后的处理方法是：找到第一个需要反转的结点的前一个结点 p，从这个结点开始，依次把后面的结点用“头插”法，插入到 p 结点的后面。循环次数用 n-m 来控制。
+
+这一题结点可以原地变化，更改各个结点的 next 指针就可以。不需要游标 p 指针。因为每次逆序以后，原有结点的相对位置就发生了变化，相当于游标指针已经移动了，所以不需要再有游标 p = p.Next 的操作了。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func reverseBetween(head *ListNode, m int, n int) *ListNode {
+	if head == nil || m >= n {
+		return head
+	}
+	newHead := &ListNode{Val: 0, Next: head}
+	pre := newHead
+	for count := 0; pre.Next != nil && count < m-1; count++ {
+		pre = pre.Next
+	}
+	if pre.Next == nil {
+		return head
+	}
+	cur := pre.Next
+	for i := 0; i < n-m; i++ {
+		tmp := pre.Next
+		pre.Next = cur.Next
+		cur.Next = cur.Next.Next
+		pre.Next.Next = tmp
+	}
+	return newHead.Next
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0091.Decode-Ways/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0093.Restore-IP-Addresses/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0093.Restore-IP-Addresses.md b/website/content/ChapterFour/0001~0099/0093.Restore-IP-Addresses.md
new file mode 100755
index 000000000..ba641bfe8
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0093.Restore-IP-Addresses.md
@@ -0,0 +1,85 @@
+# [93. Restore IP Addresses](https://leetcode.com/problems/restore-ip-addresses/)
+
+
+## 题目
+
+Given a string containing only digits, restore it by returning all possible valid IP address combinations.
+
+**Example**:
+
+    Input: "25525511135"
+    Output: ["255.255.11.135", "255.255.111.35"]
+
+## 题目大意
+
+给定一个只包含数字的字符串，复原它并返回所有可能的 IP 地址格式。
+
+## 解题思路
+
+- DFS 深搜
+- 需要注意的点是 IP 的规则，以 0 开头的数字和超过 255 的数字都为非法的。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+)
+
+func restoreIPAddresses(s string) []string {
+	if s == "" {
+		return []string{}
+	}
+	res, ip := []string{}, []int{}
+	dfs(s, 0, ip, &res)
+	return res
+}
+
+func dfs(s string, index int, ip []int, res *[]string) {
+	if index == len(s) {
+		if len(ip) == 4 {
+			*res = append(*res, getString(ip))
+		}
+		return
+	}
+	if index == 0 {
+		num, _ := strconv.Atoi(string(s[0]))
+		ip = append(ip, num)
+		dfs(s, index+1, ip, res)
+	} else {
+		num, _ := strconv.Atoi(string(s[index]))
+		next := ip[len(ip)-1]*10 + num
+		if next <= 255 && ip[len(ip)-1] != 0 {
+			ip[len(ip)-1] = next
+			dfs(s, index+1, ip, res)
+			ip[len(ip)-1] /= 10
+		}
+		if len(ip) < 4 {
+			ip = append(ip, num)
+			dfs(s, index+1, ip, res)
+			ip = ip[:len(ip)-1]
+		}
+	}
+}
+
+func getString(ip []int) string {
+	res := strconv.Itoa(ip[0])
+	for i := 1; i < len(ip); i++ {
+		res += "." + strconv.Itoa(ip[i])
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0092.Reverse-Linked-List-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0094.Binary-Tree-Inorder-Traversal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0094.Binary-Tree-Inorder-Traversal.md b/website/content/ChapterFour/0001~0099/0094.Binary-Tree-Inorder-Traversal.md
new file mode 100644
index 000000000..7656edd67
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0094.Binary-Tree-Inorder-Traversal.md
@@ -0,0 +1,80 @@
+# [94. Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal/)
+
+## 题目
+
+
+Given a binary tree, return the inorder traversal of its nodes' values.
+
+
+
+**Example**:
+
+```
+
+Input: [1,null,2,3]
+   1
+    \
+     2
+    /
+   3
+
+Output: [1,3,2]
+
+```
+
+
+**Follow up**: Recursive solution is trivial, could you do it iteratively?
+
+
+
+
+ 
+
+## 题目大意
+
+中根遍历一颗树。
+
+## 解题思路
+
+递归的实现方法，见代码。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func inorderTraversal(root *TreeNode) []int {
+	var result []int
+	inorder(root, &result)
+	return result
+}
+
+func inorder(root *TreeNode, output *[]int) {
+	if root != nil {
+		inorder(root.Left, output)
+		*output = append(*output, root.Val)
+		inorder(root.Right, output)
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0093.Restore-IP-Addresses/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0095.Unique-Binary-Search-Trees-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0095.Unique-Binary-Search-Trees-II.md b/website/content/ChapterFour/0001~0099/0095.Unique-Binary-Search-Trees-II.md
new file mode 100755
index 000000000..4ba09dfdf
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0095.Unique-Binary-Search-Trees-II.md
@@ -0,0 +1,85 @@
+# [95. Unique Binary Search Trees II](https://leetcode.com/problems/unique-binary-search-trees-ii/)
+
+
+## 题目
+
+Given an integer *n*, generate all structurally unique **BST's** (binary search trees) that store values 1 ... *n*.
+
+**Example**:
+
+    Input: 3
+    Output:
+    [
+      [1,null,3,2],
+      [3,2,null,1],
+      [3,1,null,null,2],
+      [2,1,3],
+      [1,null,2,null,3]
+    ]
+    Explanation:
+    The above output corresponds to the 5 unique BST's shown below:
+    
+       1         3     3      2      1
+        \       /     /      / \      \
+         3     2     1      1   3      2
+        /     /       \                 \
+       2     1         2                 3
+
+
+## 题目大意
+
+给定一个整数 n，生成所有由 1 ... n 为节点所组成的二叉搜索树。
+
+## 解题思路
+
+- 输出 1~n 元素组成的 BST 所有解。这一题递归求解即可。外层循环遍历 1~n 所有结点，作为根结点，内层双层递归分别求出左子树和右子树。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func generateTrees(n int) []*TreeNode {
+	if n == 0 {
+		return []*TreeNode{}
+	}
+	return generateBSTree(1, n)
+}
+
+func generateBSTree(start, end int) []*TreeNode {
+	tree := []*TreeNode{}
+	if start > end {
+		tree = append(tree, nil)
+		return tree
+	}
+	for i := start; i <= end; i++ {
+		left := generateBSTree(start, i-1)
+		right := generateBSTree(i+1, end)
+		for _, l := range left {
+			for _, r := range right {
+				root := &TreeNode{Val: i, Left: l, Right: r}
+				tree = append(tree, root)
+			}
+		}
+	}
+	return tree
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0094.Binary-Tree-Inorder-Traversal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0096.Unique-Binary-Search-Trees/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0096.Unique-Binary-Search-Trees.md b/website/content/ChapterFour/0001~0099/0096.Unique-Binary-Search-Trees.md
new file mode 100755
index 000000000..f98628afc
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0096.Unique-Binary-Search-Trees.md
@@ -0,0 +1,60 @@
+# [96. Unique Binary Search Trees](https://leetcode.com/problems/unique-binary-search-trees/)
+
+
+## 题目
+
+Given *n*, how many structurally unique **BST's** (binary search trees) that store values 1 ... *n*?
+
+**Example**:
+
+    Input: 3
+    Output: 5
+    Explanation:
+    Given n = 3, there are a total of 5 unique BST's:
+    
+       1         3     3      2      1
+        \       /     /      / \      \
+         3     2     1      1   3      2
+        /     /       \                 \
+       2     1         2                 3
+
+## 题目大意
+
+给定一个整数 n，求以 1 ... n 为节点组成的二叉搜索树有多少种？
+
+
+## 解题思路
+
+- 给出 n，要求利用 1-n 这些数字组成二叉排序树，有多少种不同的树的形态，输出这个个数。
+- 这题的解题思路是 DP。`dp[n]` 代表 1-n 个数能组成多少个不同的二叉排序树，`F(i,n)` 代表以 `i` 为根节点，1-n 个数组成的二叉排序树的不同的个数。由于题意，我们可以得到这个等式：`dp[n] = F(1,n) + F(2,n) + F(3,n) + …… + F(n,n)` 。初始值 `dp[0] = 1`，`dp[1] = 1`。分析 `dp` 和 `F(i,n)` 的关系又可以得到下面这个等式 `F(i,n) = dp[i-1] * dp[n-i]` 。举例，[1,2,3,4,…, i ,…,n-1,n]，以 `i` 为 根节点，那么左半边 [1,2,3,……,i-1] 和 右半边 [i+1,i+2,……,n-1,n] 分别能组成二叉排序树的不同个数`相乘`，即为以 `i` 为根节点，1-n 个数组成的二叉排序树的不同的个数，也即 `F(i,n)`。
+
+> 注意，由于二叉排序树本身的性质，右边的子树一定比左边的子树，值都要大。所以这里只需要根节点把树分成左右，不需要再关心左右两边数字的大小，只需要关心数字的个数。
+
+- 所以状态转移方程是 `dp[i] = dp[0] * dp[n-1] + dp[1] * dp[n-2] + …… + dp[n-1] * dp[0]`，最终要求的结果是 `dp[n]` 。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func numTrees(n int) int {
+	dp := make([]int, n+1)
+	dp[0], dp[1] = 1, 1
+	for i := 2; i <= n; i++ {
+		for j := 1; j <= i; j++ {
+			dp[i] += dp[j-1] * dp[i-j]
+		}
+	}
+	return dp[n]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0095.Unique-Binary-Search-Trees-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0097.Interleaving-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0097.Interleaving-String.md b/website/content/ChapterFour/0001~0099/0097.Interleaving-String.md
new file mode 100644
index 000000000..2b6daae8c
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0097.Interleaving-String.md
@@ -0,0 +1,111 @@
+# [97. Interleaving String](https://leetcode.com/problems/interleaving-string/)
+
+
+## 题目
+
+Given strings `s1`, `s2`, and `s3`, find whether `s3` is formed by an **interleaving** of `s1` and `s2`.
+
+An **interleaving** of two strings `s` and `t` is a configuration where they are divided into **non-empty** substrings such that:
+
+- `s = s1 + s2 + ... + sn`
+- `t = t1 + t2 + ... + tm`
+- `|n - m| <= 1`
+- The **interleaving** is `s1 + t1 + s2 + t2 + s3 + t3 + ...` or `t1 + s1 + t2 + s2 + t3 + s3 + ...`
+
+**Note:** `a + b` is the concatenation of strings `a` and `b`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/09/02/interleave.jpg](https://assets.leetcode.com/uploads/2020/09/02/interleave.jpg)
+
+```
+Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
+Output: true
+
+```
+
+**Example 2:**
+
+```
+Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
+Output: false
+
+```
+
+**Example 3:**
+
+```
+Input: s1 = "", s2 = "", s3 = ""
+Output: true
+
+```
+
+**Constraints:**
+
+- `0 <= s1.length, s2.length <= 100`
+- `0 <= s3.length <= 200`
+- `s1`, `s2`, and `s3` consist of lowercase English letters.
+
+**Follow up:** Could you solve it using only `O(s2.length)` additional memory space?
+
+## 题目大意
+
+给定三个字符串 s1、s2、s3，请你帮忙验证 s3 是否是由 s1 和 s2 交错 组成的。两个字符串 s 和 t 交错 的定义与过程如下，其中每个字符串都会被分割成若干 非空 子字符串：
+
+- s = s1 + s2 + ... + sn
+- t = t1 + t2 + ... + tm
+- |n - m| <= 1
+- 交错 是 s1 + t1 + s2 + t2 + s3 + t3 + ... 或者 t1 + s1 + t2 + s2 + t3 + s3 + ...
+
+提示：a + b 意味着字符串 a 和 b 连接。
+
+## 解题思路
+
+- 深搜或者广搜暴力解题。笔者用深搜实现的。记录 s1 和 s2 串当前比较的位置 p1 和 p2。如果 s3[p1+p2] 的位置上等于 s1[p1] 或者 s2[p2] 代表能匹配上，那么继续往后移动 p1 和 p2 相应的位置。因为是交错字符串，所以判断匹配的位置是 s3[p1+p2] 的位置。如果仅仅这么写，会超时，s1 和 s2 两个字符串重复交叉判断的位置太多了。需要加上记忆化搜索。可以用 visited[i][j] 这样的二维数组来记录是否搜索过了。笔者为了压缩空间，将 i 和 j 编码压缩到一维数组了。i * len(s3) + j 是唯一下标，所以可以用这种方式存储是否搜索过。具体代码见下面的实现。
+
+## 代码
+
+```go
+package leetcode
+
+func isInterleave(s1 string, s2 string, s3 string) bool {
+	if len(s1)+len(s2) != len(s3) {
+		return false
+	}
+	visited := make(map[int]bool)
+	return dfs(s1, s2, s3, 0, 0, visited)
+}
+
+func dfs(s1, s2, s3 string, p1, p2 int, visited map[int]bool) bool {
+	if p1+p2 == len(s3) {
+		return true
+	}
+	if _, ok := visited[(p1*len(s3))+p2]; ok {
+		return false
+	}
+	visited[(p1*len(s3))+p2] = true
+	var match1, match2 bool
+	if p1 < len(s1) && s3[p1+p2] == s1[p1] {
+		match1 = true
+	}
+	if p2 < len(s2) && s3[p1+p2] == s2[p2] {
+		match2 = true
+	}
+	if match1 && match2 {
+		return dfs(s1, s2, s3, p1+1, p2, visited) || dfs(s1, s2, s3, p1, p2+1, visited)
+	} else if match1 {
+		return dfs(s1, s2, s3, p1+1, p2, visited)
+	} else if match2 {
+		return dfs(s1, s2, s3, p1, p2+1, visited)
+	} else {
+		return false
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0096.Unique-Binary-Search-Trees/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0098.Validate-Binary-Search-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0098.Validate-Binary-Search-Tree.md b/website/content/ChapterFour/0001~0099/0098.Validate-Binary-Search-Tree.md
new file mode 100755
index 000000000..ffc988c57
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0098.Validate-Binary-Search-Tree.md
@@ -0,0 +1,106 @@
+# [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)
+
+
+## 题目
+
+Given a binary tree, determine if it is a valid binary search tree (BST).
+
+Assume a BST is defined as follows:
+
+- The left subtree of a node contains only nodes with keys **less than** the node's key.
+- The right subtree of a node contains only nodes with keys **greater than** the node's key.
+- Both the left and right subtrees must also be binary search trees.
+
+**Example 1**:
+
+        2
+       / \
+      1   3
+    
+    Input: [2,1,3]
+    Output: true
+
+**Example 2**:
+
+        5
+       / \
+      1   4
+         / \
+        3   6
+    
+    Input: [5,1,4,null,null,3,6]
+    Output: false
+    Explanation: The root node's value is 5 but its right child's value is 4.
+
+## 题目大意
+
+给定一个二叉树，判断其是否是一个有效的二叉搜索树。假设一个二叉搜索树具有如下特征：
+
+- 节点的左子树只包含小于当前节点的数。
+- 节点的右子树只包含大于当前节点的数。
+- 所有左子树和右子树自身必须也是二叉搜索树。
+
+
+## 解题思路
+
+- 判断一个树是否是 BST，按照定义递归判断即可
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一，直接按照定义比较大小，比 root 节点小的都在左边，比 root 节点大的都在右边
+func isValidBST(root *TreeNode) bool {
+	return isValidbst(root, math.Inf(-1), math.Inf(1))
+}
+func isValidbst(root *TreeNode, min, max float64) bool {
+	if root == nil {
+		return true
+	}
+	v := float64(root.Val)
+	return v < max && v > min && isValidbst(root.Left, min, v) && isValidbst(root.Right, v, max)
+}
+
+// 解法二，把 BST 按照左中右的顺序输出到数组中，如果是 BST，则数组中的数字是从小到大有序的，如果出现逆序就不是 BST
+func isValidBST1(root *TreeNode) bool {
+	arr := []int{}
+	inOrder(root, &arr)
+	for i := 1; i < len(arr); i++ {
+		if arr[i-1] >= arr[i] {
+			return false
+		}
+	}
+	return true
+}
+
+func inOrder(root *TreeNode, arr *[]int) {
+	if root == nil {
+		return
+	}
+	inOrder(root.Left, arr)
+	*arr = append(*arr, root.Val)
+	inOrder(root.Right, arr)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0097.Interleaving-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0099.Recover-Binary-Search-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/0099.Recover-Binary-Search-Tree.md b/website/content/ChapterFour/0001~0099/0099.Recover-Binary-Search-Tree.md
new file mode 100755
index 000000000..4a28ea15d
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/0099.Recover-Binary-Search-Tree.md
@@ -0,0 +1,108 @@
+# [99. Recover Binary Search Tree](https://leetcode.com/problems/recover-binary-search-tree/)
+
+
+## 题目
+
+Two elements of a binary search tree (BST) are swapped by mistake.
+
+Recover the tree without changing its structure.
+
+**Example 1**:
+
+    Input: [1,3,null,null,2]
+    
+       1
+      /
+     3
+      \
+       2
+    
+    Output: [3,1,null,null,2]
+    
+       3
+      /
+     1
+      \
+       2
+
+**Example 2**:
+
+    Input: [3,1,4,null,null,2]
+    
+      3
+     / \
+    1   4
+       /
+      2
+    
+    Output: [2,1,4,null,null,3]
+    
+      2
+     / \
+    1   4
+       /
+      3
+
+**Follow up**:
+
+- A solution using O(*n*) space is pretty straight forward.
+- Could you devise a constant space solution?
+
+## 题目大意
+
+二叉搜索树中的两个节点被错误地交换。请在不改变其结构的情况下，恢复这棵树。
+
+
+## 解题思路
+
+- 在二叉搜索树中，有 2 个结点的值出错了，要求修复这两个结点。
+- 这一题按照先根遍历 1 次就可以找到这两个出问题的结点，因为先访问根节点，然后左孩子，右孩子。用先根遍历二叉搜索树的时候，根结点比左子树都要大，根结点比右子树都要小。所以`左子树比根结点大的话，就是出现了乱序`；`根节点比右子树大的话，就是出现了乱序`。遍历过程中在左子树中如果出现了前一次遍历的结点的值大于此次根节点的值，这就出现了出错结点了，记录下来。继续遍历直到找到第二个这样的结点。最后交换这两个结点的时候，只是交换他们的值就可以了，而不是交换这两个结点相应的指针指向。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func recoverTree(root *TreeNode) {
+	var prev, target1, target2 *TreeNode
+	_, target1, target2 = inOrderTraverse(root, prev, target1, target2)
+	if target1 != nil && target2 != nil {
+		target1.Val, target2.Val = target2.Val, target1.Val
+	}
+
+}
+
+func inOrderTraverse(root, prev, target1, target2 *TreeNode) (*TreeNode, *TreeNode, *TreeNode) {
+	if root == nil {
+		return prev, target1, target2
+	}
+	prev, target1, target2 = inOrderTraverse(root.Left, prev, target1, target2)
+	if prev != nil && prev.Val > root.Val {
+		if target1 == nil {
+			target1 = prev
+		}
+		target2 = root
+	}
+	prev = root
+	prev, target1, target2 = inOrderTraverse(root.Right, prev, target1, target2)
+	return prev, target1, target2
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0098.Validate-Binary-Search-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0100.Same-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0001~0099/_index.md b/website/content/ChapterFour/0001~0099/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/0001~0099/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/0100~0199/0100.Same-Tree.md b/website/content/ChapterFour/0100~0199/0100.Same-Tree.md
new file mode 100644
index 000000000..90227a8eb
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0100.Same-Tree.md
@@ -0,0 +1,99 @@
+# [100. Same Tree](https://leetcode.com/problems/same-tree/)
+
+## 题目
+
+
+Given two binary trees, write a function to check if they are the same or not.
+
+Two binary trees are considered the same if they are structurally identical and the nodes have the same value.
+
+**Example 1**:
+
+
+```
+
+Input:     1         1
+          / \       / \
+         2   3     2   3
+
+        [1,2,3],   [1,2,3]
+
+Output: true
+
+```
+
+**Example 2**:
+
+```
+
+Input:     1         1
+          /           \
+         2             2
+
+        [1,2],     [1,null,2]
+
+Output: false
+
+```
+
+**Example 3**:
+
+```
+
+Input:     1         1
+          / \       / \
+         2   1     1   2
+
+        [1,2,1],   [1,1,2]
+
+Output: false
+
+```
+
+## 题目大意
+
+这一题要求判断 2 颗树是否是完全相等的。
+
+
+## 解题思路
+
+递归判断即可。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func isSameTree(p *TreeNode, q *TreeNode) bool {
+	if p == nil && q == nil {
+		return true
+	} else if p != nil && q != nil {
+		if p.Val != q.Val {
+			return false
+		}
+		return isSameTree(p.Left, q.Left) && isSameTree(p.Right, q.Right)
+	} else {
+		return false
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0099.Recover-Binary-Search-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0101.Symmetric-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0101.Symmetric-Tree.md b/website/content/ChapterFour/0100~0199/0101.Symmetric-Tree.md
new file mode 100644
index 000000000..2b9de1e98
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0101.Symmetric-Tree.md
@@ -0,0 +1,130 @@
+# [101. Symmetric Tree](https://leetcode.com/problems/symmetric-tree/)
+
+## 题目
+
+
+Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
+
+For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
+
+
+```
+
+    1
+   / \
+  2   2
+ / \ / \
+3  4 4  3
+
+```
+
+But the following [1,2,2,null,3,null,3] is not:
+
+```
+
+    1
+   / \
+  2   2
+   \   \
+   3    3
+
+```
+
+**Note**:   
+
+Bonus points if you could solve it both recursively and iteratively.
+
+## 题目大意
+
+这一题要求判断 2 颗树是否是左右对称的。
+
+
+## 解题思路
+
+- 这道题是几道题的综合题。将根节点的左字数反转二叉树，然后再和根节点的右节点进行比较，是否完全相等。
+- 反转二叉树是第 226 题。判断 2 颗树是否完全相等是第 100 题。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 dfs
+func isSymmetric(root *TreeNode) bool {
+	if root == nil {
+		return true
+	}
+	return isMirror(root.Left, root.Right)
+}
+
+func isMirror(left *TreeNode, right *TreeNode) bool {
+	if left == nil && right == nil {
+		return true
+	}
+	if left == nil || right == nil {
+		return false
+	}
+	return (left.Val == right.Val) && isMirror(left.Left, right.Right) && isMirror(left.Right, right.Left)
+}
+
+// 解法二
+func isSymmetric1(root *TreeNode) bool {
+	if root == nil {
+		return true
+	}
+	return isSameTree(invertTree(root.Left), root.Right)
+}
+
+func isSameTree(p *TreeNode, q *TreeNode) bool {
+	if p == nil && q == nil {
+		return true
+	} else if p != nil && q != nil {
+		if p.Val != q.Val {
+			return false
+		}
+		return isSameTree(p.Left, q.Left) && isSameTree(p.Right, q.Right)
+	} else {
+		return false
+	}
+}
+
+func invertTree(root *TreeNode) *TreeNode {
+	if root == nil {
+		return nil
+	}
+	invertTree(root.Left)
+	invertTree(root.Right)
+	root.Left, root.Right = root.Right, root.Left
+	return root
+}
+
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0100.Same-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0102.Binary-Tree-Level-Order-Traversal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0102.Binary-Tree-Level-Order-Traversal.md b/website/content/ChapterFour/0100~0199/0102.Binary-Tree-Level-Order-Traversal.md
new file mode 100644
index 000000000..68f93c42a
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0102.Binary-Tree-Level-Order-Traversal.md
@@ -0,0 +1,122 @@
+# [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/)
+
+## 题目
+
+
+Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
+
+**For Example**:
+
+Given binary tree [3,9,20,null,null,15,7],
+
+```
+
+    3
+   / \
+  9  20
+    /  \
+   15   7
+
+```
+
+return its level order traversal as:
+
+```
+
+[
+  [3],
+  [9,20],
+  [15,7]
+]
+
+```
+ 
+
+## 题目大意
+
+按层序从上到下遍历一颗树。
+
+## 解题思路
+
+用一个队列即可实现。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 BFS
+func levelOrder(root *TreeNode) [][]int {
+	if root == nil {
+		return [][]int{}
+	}
+	queue := []*TreeNode{root}
+	res := make([][]int, 0)
+	for len(queue) > 0 {
+		l := len(queue)
+		tmp := make([]int, 0, l)
+		for i := 0; i < l; i++ {
+			if queue[i].Left != nil {
+				queue = append(queue, queue[i].Left)
+			}
+			if queue[i].Right != nil {
+				queue = append(queue, queue[i].Right)
+			}
+			tmp = append(tmp, queue[i].Val)
+		}
+		queue = queue[l:]
+		res = append(res, tmp)
+	}
+	return res
+}
+
+// 解法二 DFS
+func levelOrder1(root *TreeNode) [][]int {
+	var res [][]int
+	var dfsLevel func(node *TreeNode, level int)
+	dfsLevel = func(node *TreeNode, level int) {
+		if node == nil {
+			return
+		}
+		if len(res) == level {
+			res = append(res, []int{node.Val})
+		} else {
+			res[level] = append(res[level], node.Val)
+		}
+		dfsLevel(node.Left, level+1)
+		dfsLevel(node.Right, level+1)
+	}
+	dfsLevel(root, 0)
+	return res
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0101.Symmetric-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0103.Binary-Tree-Zigzag-Level-Order-Traversal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0103.Binary-Tree-Zigzag-Level-Order-Traversal.md b/website/content/ChapterFour/0100~0199/0103.Binary-Tree-Zigzag-Level-Order-Traversal.md
new file mode 100644
index 000000000..0afad6242
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0103.Binary-Tree-Zigzag-Level-Order-Traversal.md
@@ -0,0 +1,178 @@
+# [103. Binary Tree Zigzag Level Order Traversal](https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/)
+
+## 题目
+
+Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
+
+**For Example**:
+Given binary tree [3,9,20,null,null,15,7],
+
+```
+
+    3
+   / \
+  9  20
+    /  \
+   15   7
+
+```
+
+return its zigzag level order traversal as:
+
+```
+
+[
+  [3],
+  [20,9],
+  [15,7]
+]
+
+```
+ 
+
+## 题目大意
+
+按照 Z 字型层序遍历一棵树。
+
+## 解题思路
+
+- 按层序从上到下遍历一颗树，但是每一层的顺序是相互反转的，即上一层是从左往右，下一层就是从右往左，以此类推。用一个队列即可实现。
+- 第 102 题和第 107 题都是按层序遍历的。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一
+func zigzagLevelOrder(root *TreeNode) [][]int {
+	if root == nil {
+		return [][]int{}
+	}
+	queue := []*TreeNode{}
+	queue = append(queue, root)
+	curNum, nextLevelNum, res, tmp, curDir := 1, 0, [][]int{}, []int{}, 0
+	for len(queue) != 0 {
+		if curNum > 0 {
+			node := queue[0]
+			if node.Left != nil {
+				queue = append(queue, node.Left)
+				nextLevelNum++
+			}
+			if node.Right != nil {
+				queue = append(queue, node.Right)
+				nextLevelNum++
+			}
+			curNum--
+			tmp = append(tmp, node.Val)
+			queue = queue[1:]
+		}
+		if curNum == 0 {
+			if curDir == 1 {
+				for i, j := 0, len(tmp)-1; i < j; i, j = i+1, j-1 {
+					tmp[i], tmp[j] = tmp[j], tmp[i]
+				}
+			}
+			res = append(res, tmp)
+			curNum = nextLevelNum
+			nextLevelNum = 0
+			tmp = []int{}
+			if curDir == 0 {
+				curDir = 1
+			} else {
+				curDir = 0
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 递归
+func zigzagLevelOrder0(root *TreeNode) [][]int {
+	var res [][]int
+	search(root, 0, &res)
+	return res
+}
+
+func search(root *TreeNode, depth int, res *[][]int) {
+	if root == nil {
+		return
+	}
+	for len(*res) < depth+1 {
+		*res = append(*res, []int{})
+	}
+	if depth%2 == 0 {
+		(*res)[depth] = append((*res)[depth], root.Val)
+	} else {
+		(*res)[depth] = append([]int{root.Val}, (*res)[depth]...)
+	}
+	search(root.Left, depth+1, res)
+	search(root.Right, depth+1, res)
+}
+
+// 解法三 BFS
+func zigzagLevelOrder1(root *TreeNode) [][]int {
+	res := [][]int{}
+	if root == nil {
+		return res
+	}
+	q := []*TreeNode{root}
+	size, i, j, lay, tmp, flag := 0, 0, 0, []int{}, []*TreeNode{}, false
+	for len(q) > 0 {
+		size = len(q)
+		tmp = []*TreeNode{}
+		lay = make([]int, size)
+		j = size - 1
+		for i = 0; i < size; i++ {
+			root = q[0]
+			q = q[1:]
+			if !flag {
+				lay[i] = root.Val
+			} else {
+				lay[j] = root.Val
+				j--
+			}
+			if root.Left != nil {
+				tmp = append(tmp, root.Left)
+			}
+			if root.Right != nil {
+				tmp = append(tmp, root.Right)
+			}
+
+		}
+		res = append(res, lay)
+		flag = !flag
+		q = tmp
+	}
+	return res
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0102.Binary-Tree-Level-Order-Traversal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0104.Maximum-Depth-of-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0104.Maximum-Depth-of-Binary-Tree.md b/website/content/ChapterFour/0100~0199/0104.Maximum-Depth-of-Binary-Tree.md
new file mode 100644
index 000000000..36d39f9f4
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0104.Maximum-Depth-of-Binary-Tree.md
@@ -0,0 +1,67 @@
+# [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)
+
+## 题目
+
+Given a binary tree, find its maximum depth.
+
+The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
+
+**Note**: A leaf is a node with no children.
+
+**Example**:
+
+Given binary tree [3,9,20,null,null,15,7],
+
+```
+
+    3
+   / \
+  9  20
+    /  \
+   15   7
+
+```
+
+return its depth = 3.
+ 
+
+## 题目大意
+
+要求输出一棵树的最大高度。
+
+## 解题思路
+
+这一题递归遍历就可，遍历根节点的左孩子的高度和根节点右孩子的高度，取出两者的最大值再加一即为总高度。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func maxDepth(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	return max(maxDepth(root.Left), maxDepth(root.Right)) + 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0103.Binary-Tree-Zigzag-Level-Order-Traversal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal.md b/website/content/ChapterFour/0100~0199/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal.md
new file mode 100755
index 000000000..556a6da65
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal.md
@@ -0,0 +1,105 @@
+# [105. Construct Binary Tree from Preorder and Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)
+
+
+## 题目
+
+Given preorder and inorder traversal of a tree, construct the binary tree.
+
+**Note**:You may assume that duplicates do not exist in the tree.
+
+For example, given
+
+    preorder = [3,9,20,15,7]
+    inorder = [9,3,15,20,7]
+
+Return the following binary tree:
+
+		3
+       / \
+      9  20
+        /  \
+       15   7
+
+
+
+## 题目大意
+
+根据一棵树的前序遍历与中序遍历构造二叉树。
+
+注意:
+你可以假设树中没有重复的元素。
+
+
+## 解题思路
+
+- 给出 2 个数组，根据 preorder 和 inorder 数组构造一颗树。
+- 利用递归思想，从 preorder 可以得到根节点，从 inorder 中得到左子树和右子树。只剩一个节点的时候即为根节点。不断的递归直到所有的树都生成完成。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一, 直接传入需要的 slice 范围作为输入, 可以避免申请对应 inorder 索引的内存, 内存使用(leetcode test case) 4.7MB -> 4.3MB.
+func buildTree(preorder []int, inorder []int) *TreeNode {
+	if len(preorder) == 0 {
+		return nil
+	}
+	root := &TreeNode{Val: preorder[0]}
+	for pos, node := range inorder {
+		if node == root.Val {
+			root.Left = buildTree(preorder[1:pos+1], inorder[:pos])
+			root.Right = buildTree(preorder[pos+1:], inorder[pos+1:])
+		}
+	}
+	return root
+}
+
+// 解法二
+func buildTree1(preorder []int, inorder []int) *TreeNode {
+	inPos := make(map[int]int)
+	for i := 0; i < len(inorder); i++ {
+		inPos[inorder[i]] = i
+	}
+	return buildPreIn2TreeDFS(preorder, 0, len(preorder)-1, 0, inPos)
+}
+
+func buildPreIn2TreeDFS(pre []int, preStart int, preEnd int, inStart int, inPos map[int]int) *TreeNode {
+	if preStart > preEnd {
+		return nil
+	}
+	root := &TreeNode{Val: pre[preStart]}
+	rootIdx := inPos[pre[preStart]]
+	leftLen := rootIdx - inStart
+	root.Left = buildPreIn2TreeDFS(pre, preStart+1, preStart+leftLen, inStart, inPos)
+	root.Right = buildPreIn2TreeDFS(pre, preStart+leftLen+1, preEnd, rootIdx+1, inPos)
+	return root
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0104.Maximum-Depth-of-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.md b/website/content/ChapterFour/0100~0199/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.md
new file mode 100755
index 000000000..fe051ed8c
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.md
@@ -0,0 +1,105 @@
+# [106. Construct Binary Tree from Inorder and Postorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/)
+
+## 题目
+
+Given inorder and postorder traversal of a tree, construct the binary tree.
+
+**Note**: You may assume that duplicates do not exist in the tree.
+
+For example, given
+
+    inorder = [9,3,15,20,7]
+    postorder = [9,15,7,20,3]
+
+Return the following binary tree:
+
+    	3
+       / \
+      9  20
+        /  \
+       15   7
+
+
+## 题目大意
+
+根据一棵树的中序遍历与后序遍历构造二叉树。
+
+注意:
+你可以假设树中没有重复的元素。
+
+
+## 解题思路
+
+- 给出 2 个数组，根据 inorder 和 postorder 数组构造一颗树。
+- 利用递归思想，从 postorder 可以得到根节点，从 inorder 中得到左子树和右子树。只剩一个节点的时候即为根节点。不断的递归直到所有的树都生成完成。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一, 直接传入需要的 slice 范围作为输入, 可以避免申请对应 inorder 索引的内存, 内存使用(leetcode test case) 4.7MB -> 4.3MB.
+func buildTree(inorder []int, postorder []int) *TreeNode {
+	postorderLen := len(postorder)
+	if len(inorder) == 0 {
+		return nil
+	}
+	root := &TreeNode{Val: postorder[postorderLen-1]}
+	postorder = postorder[:postorderLen-1]
+	for pos, node := range inorder {
+		if node == root.Val {
+			root.Left = buildTree(inorder[:pos], postorder[:len(inorder[:pos])])
+			root.Right = buildTree(inorder[pos+1:], postorder[len(inorder[:pos]):])
+		}
+	}
+	return root
+}
+
+// 解法二
+func buildTree1(inorder []int, postorder []int) *TreeNode {
+	inPos := make(map[int]int)
+	for i := 0; i < len(inorder); i++ {
+		inPos[inorder[i]] = i
+	}
+	return buildInPos2TreeDFS(postorder, 0, len(postorder)-1, 0, inPos)
+}
+
+func buildInPos2TreeDFS(post []int, postStart int, postEnd int, inStart int, inPos map[int]int) *TreeNode {
+	if postStart > postEnd {
+		return nil
+	}
+	root := &TreeNode{Val: post[postEnd]}
+	rootIdx := inPos[post[postEnd]]
+	leftLen := rootIdx - inStart
+	root.Left = buildInPos2TreeDFS(post, postStart, postStart+leftLen-1, inStart, inPos)
+	root.Right = buildInPos2TreeDFS(post, postStart+leftLen, postEnd-1, rootIdx+1, inPos)
+	return root
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0107.Binary-Tree-Level-Order-Traversal-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0107.Binary-Tree-Level-Order-Traversal-II.md b/website/content/ChapterFour/0100~0199/0107.Binary-Tree-Level-Order-Traversal-II.md
new file mode 100644
index 000000000..172e977be
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0107.Binary-Tree-Level-Order-Traversal-II.md
@@ -0,0 +1,77 @@
+# [107. Binary Tree Level Order Traversal II](https://leetcode.com/problems/binary-tree-level-order-traversal-ii/)
+
+## 题目
+
+Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
+
+**For Example**:
+
+Given binary tree [3,9,20,null,null,15,7],
+
+```
+
+    3
+   / \
+  9  20
+    /  \
+   15   7
+
+```
+
+return its bottom-up level order traversal as:
+
+
+```
+
+[
+  [15,7],
+  [9,20],
+  [3]
+]
+
+```
+ 
+
+## 题目大意
+
+按层序从下到上遍历一颗树。
+
+## 解题思路
+
+用一个队列即可实现。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func levelOrderBottom(root *TreeNode) [][]int {
+	tmp := levelOrder(root)
+	res := [][]int{}
+	for i := len(tmp) - 1; i >= 0; i-- {
+		res = append(res, tmp[i])
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0108.Convert-Sorted-Array-to-Binary-Search-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0108.Convert-Sorted-Array-to-Binary-Search-Tree.md b/website/content/ChapterFour/0100~0199/0108.Convert-Sorted-Array-to-Binary-Search-Tree.md
new file mode 100755
index 000000000..3ed5eb093
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0108.Convert-Sorted-Array-to-Binary-Search-Tree.md
@@ -0,0 +1,59 @@
+# [108. Convert Sorted Array to Binary Search Tree](https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/)
+
+
+## 题目
+
+Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
+
+For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of *every* node never differ by more than 1.
+
+**Example**:
+
+    Given the sorted array: [-10,-3,0,5,9],
+    
+    One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
+    
+          0
+         / \
+       -3   9
+       /   /
+     -10  5
+
+## 题目大意
+
+将一个按照升序排列的有序数组，转换为一棵高度平衡二叉搜索树。本题中，一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。
+
+## 解题思路
+
+- 把一个有序数组转换成高度平衡的二叉搜索数，按照定义即可
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func sortedArrayToBST(nums []int) *TreeNode {
+	if len(nums) == 0 {
+		return nil
+	}
+	return &TreeNode{Val: nums[len(nums)/2], Left: sortedArrayToBST(nums[:len(nums)/2]), Right: sortedArrayToBST(nums[len(nums)/2+1:])}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0107.Binary-Tree-Level-Order-Traversal-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0109.Convert-Sorted-List-to-Binary-Search-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0109.Convert-Sorted-List-to-Binary-Search-Tree.md b/website/content/ChapterFour/0100~0199/0109.Convert-Sorted-List-to-Binary-Search-Tree.md
new file mode 100644
index 000000000..dd85b3b16
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0109.Convert-Sorted-List-to-Binary-Search-Tree.md
@@ -0,0 +1,104 @@
+# [109. Convert Sorted List to Binary Search Tree](https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/)
+
+## 题目
+
+Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
+
+For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
+
+**Example**:
+
+```
+
+Given the sorted linked list: [-10,-3,0,5,9],
+
+One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
+
+      0
+     / \
+   -3   9
+   /   /
+ -10  5
+
+```
+
+
+## 题目大意
+
+将链表转化为高度平衡的二叉搜索树。高度平衡的定义：每个结点的 2 个子结点的深度不能相差超过 1 。
+
+## 解题思路
+
+思路比较简单，依次把链表的中间点作为根结点，类似二分的思想，递归排列所有结点即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// TreeNode define
+type TreeNode struct {
+	Val   int
+	Left  *TreeNode
+	Right *TreeNode
+}
+
+func sortedListToBST(head *ListNode) *TreeNode {
+	if head == nil {
+		return nil
+	}
+	if head != nil && head.Next == nil {
+		return &TreeNode{Val: head.Val, Left: nil, Right: nil}
+	}
+	middleNode, preNode := middleNodeAndPreNode(head)
+	if middleNode == nil {
+		return nil
+	}
+	if preNode != nil {
+		preNode.Next = nil
+	}
+	if middleNode == head {
+		head = nil
+	}
+	return &TreeNode{Val: middleNode.Val, Left: sortedListToBST(head), Right: sortedListToBST(middleNode.Next)}
+}
+
+func middleNodeAndPreNode(head *ListNode) (middle *ListNode, pre *ListNode) {
+	if head == nil || head.Next == nil {
+		return nil, head
+	}
+	p1 := head
+	p2 := head
+	for p2.Next != nil && p2.Next.Next != nil {
+		pre = p1
+		p1 = p1.Next
+		p2 = p2.Next.Next
+	}
+	return p1, pre
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0108.Convert-Sorted-Array-to-Binary-Search-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0110.Balanced-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0110.Balanced-Binary-Tree.md b/website/content/ChapterFour/0100~0199/0110.Balanced-Binary-Tree.md
new file mode 100644
index 000000000..3aad94369
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0110.Balanced-Binary-Tree.md
@@ -0,0 +1,97 @@
+# [110. Balanced Binary Tree](https://leetcode.com/problems/balanced-binary-tree/)
+
+## 题目
+
+
+Given a binary tree, determine if it is height-balanced.
+
+For this problem, a height-balanced binary tree is defined as:
+
+a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
+
+**Example 1**:
+
+Given the following tree [3,9,20,null,null,15,7]:
+
+```
+
+    3
+   / \
+  9  20
+    /  \
+   15   7
+
+```
+
+Return true.
+
+**Example 2**:
+
+Given the following tree [1,2,2,3,3,null,null,4,4]:
+
+
+```
+
+       1
+      / \
+     2   2
+    / \
+   3   3
+  / \
+ 4   4
+
+```
+
+Return false.
+
+
+## 题目大意
+
+判断一棵树是不是平衡二叉树。平衡二叉树的定义是：树中每个节点都满足左右两个子树的高度差 <= 1 的这个条件。
+
+
+## 解题思路
+
+根据定义判断即可，计算树的高度是第 104 题。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func isBalanced(root *TreeNode) bool {
+	if root == nil {
+		return true
+	}
+	leftHight := depth(root.Left)
+	rightHight := depth(root.Right)
+	return abs(leftHight-rightHight) <= 1 && isBalanced(root.Left) && isBalanced(root.Right)
+}
+
+func depth(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	return max(depth(root.Left), depth(root.Right)) + 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0109.Convert-Sorted-List-to-Binary-Search-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0111.Minimum-Depth-of-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0111.Minimum-Depth-of-Binary-Tree.md b/website/content/ChapterFour/0100~0199/0111.Minimum-Depth-of-Binary-Tree.md
new file mode 100755
index 000000000..54530d510
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0111.Minimum-Depth-of-Binary-Tree.md
@@ -0,0 +1,69 @@
+# [111. Minimum Depth of Binary Tree](https://leetcode.com/problems/minimum-depth-of-binary-tree/)
+
+
+## 题目
+
+Given a binary tree, find its minimum depth.
+
+The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
+
+**Note**: A leaf is a node with no children.
+
+**Example**:
+
+Given binary tree `[3,9,20,null,null,15,7]`,
+
+    	3
+       / \
+      9  20
+        /  \
+       15   7
+
+return its minimum depth = 2.
+
+## 题目大意
+
+给定一个二叉树，找出其最小深度。最小深度是从根节点到最近叶子节点的最短路径上的节点数量。说明: 叶子节点是指没有子节点的节点。
+
+
+## 解题思路
+
+- 递归求出根节点到叶子节点的深度，输出最小值即可
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func minDepth(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	if root.Left == nil {
+		return minDepth(root.Right) + 1
+	}
+	if root.Right == nil {
+		return minDepth(root.Left) + 1
+	}
+	return min(minDepth(root.Left), minDepth(root.Right)) + 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0110.Balanced-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0112.Path-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0112.Path-Sum.md b/website/content/ChapterFour/0100~0199/0112.Path-Sum.md
new file mode 100755
index 000000000..5964900af
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0112.Path-Sum.md
@@ -0,0 +1,66 @@
+# [112. Path Sum](https://leetcode.com/problems/path-sum/)
+
+
+## 题目
+
+Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
+
+**Note**: A leaf is a node with no children.
+
+**Example**:
+
+Given the below binary tree and `sum = 22`,
+
+          5
+         / \
+        4   8
+       /   / \
+      11  13  4
+     /  \      \
+    7    2      1
+
+return true, as there exist a root-to-leaf path `5->4->11->2` which sum is 22.
+
+## 题目大意
+
+给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有节点值相加等于目标和。说明: 叶子节点是指没有子节点的节点。
+
+
+## 解题思路
+
+- 递归求解即可
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func hasPathSum(root *TreeNode, sum int) bool {
+	if root == nil {
+		return false
+	}
+	if root.Left == nil && root.Right == nil {
+		return sum == root.Val
+	}
+	return hasPathSum(root.Left, sum-root.Val) || hasPathSum(root.Right, sum-root.Val)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0111.Minimum-Depth-of-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0113.Path-Sum-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0113.Path-Sum-II.md b/website/content/ChapterFour/0100~0199/0113.Path-Sum-II.md
new file mode 100755
index 000000000..2b6927fff
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0113.Path-Sum-II.md
@@ -0,0 +1,115 @@
+# [113. Path Sum II](https://leetcode.com/problems/path-sum-ii/)
+
+
+## 题目
+
+Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
+
+**Note**: A leaf is a node with no children.
+
+**Example**:
+
+Given the below binary tree and `sum = 22`,
+
+          5
+         / \
+        4   8
+       /   / \
+      11  13  4
+     /  \    / \
+    7    2  5   1
+
+Return:
+
+    [
+       [5,4,11,2],
+       [5,8,4,5]
+    ]
+
+## 题目大意
+
+给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。说明: 叶子节点是指没有子节点的节点。
+
+## 解题思路
+
+- 这一题是第 257 题和第 112 题的组合增强版
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一
+func pathSum(root *TreeNode, sum int) [][]int {
+	var slice [][]int
+	slice = findPath(root, sum, slice, []int(nil))
+	return slice
+}
+
+func findPath(n *TreeNode, sum int, slice [][]int, stack []int) [][]int {
+	if n == nil {
+		return slice
+	}
+	sum -= n.Val
+	stack = append(stack, n.Val)
+	if sum == 0 && n.Left == nil && n.Right == nil {
+		slice = append(slice, append([]int{}, stack...))
+		stack = stack[:len(stack)-1]
+	}
+	slice = findPath(n.Left, sum, slice, stack)
+	slice = findPath(n.Right, sum, slice, stack)
+	return slice
+}
+
+// 解法二
+func pathSum1(root *TreeNode, sum int) [][]int {
+	if root == nil {
+		return [][]int{}
+	}
+	if root.Left == nil && root.Right == nil {
+		if sum == root.Val {
+			return [][]int{[]int{root.Val}}
+		}
+	}
+	path, res := []int{}, [][]int{}
+	tmpLeft := pathSum(root.Left, sum-root.Val)
+	path = append(path, root.Val)
+	if len(tmpLeft) > 0 {
+		for i := 0; i < len(tmpLeft); i++ {
+			tmpLeft[i] = append(path, tmpLeft[i]...)
+		}
+		res = append(res, tmpLeft...)
+	}
+	path = []int{}
+	tmpRight := pathSum(root.Right, sum-root.Val)
+	path = append(path, root.Val)
+
+	if len(tmpRight) > 0 {
+		for i := 0; i < len(tmpRight); i++ {
+			tmpRight[i] = append(path, tmpRight[i]...)
+		}
+		res = append(res, tmpRight...)
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0112.Path-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0114.Flatten-Binary-Tree-to-Linked-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0114.Flatten-Binary-Tree-to-Linked-List.md b/website/content/ChapterFour/0100~0199/0114.Flatten-Binary-Tree-to-Linked-List.md
new file mode 100755
index 000000000..a466638e6
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0114.Flatten-Binary-Tree-to-Linked-List.md
@@ -0,0 +1,198 @@
+# [114. Flatten Binary Tree to Linked List](https://leetcode.com/problems/flatten-binary-tree-to-linked-list/)
+
+
+## 题目
+
+Given a binary tree, flatten it to a linked list in-place.
+
+For example, given the following tree:
+
+    	1
+       / \
+      2   5
+     / \   \
+    3   4   6
+
+The flattened tree should look like:
+
+    1
+     \
+      2
+       \
+        3
+         \
+          4
+           \
+            5
+             \
+              6
+
+## 题目大意
+
+给定一个二叉树，原地将它展开为链表。
+
+## 解题思路
+
+- 要求把二叉树“打平”，按照先根遍历的顺序，把树的结点都放在右结点中。
+- 按照递归和非递归思路实现即可。
+- 递归的思路可以这么想：倒序遍历一颗树，即是先遍历右孩子，然后遍历左孩子，最后再遍历根节点。
+
+        		1
+           / \
+          2   5
+         / \   \
+        3   4   6
+        -----------        
+        pre = 5
+        cur = 4
+        
+            1
+           / 
+          2   
+         / \   
+        3   4
+             \
+              5
+               \
+                6
+        -----------        
+        pre = 4
+        cur = 3
+        
+            1
+           / 
+          2   
+         /   
+        3 
+         \
+          4
+           \
+            5
+             \
+              6
+        -----------        
+        cur = 2
+        pre = 3
+        
+            1
+           / 
+          2   
+           \
+            3 
+             \
+              4
+               \
+                5
+                 \
+                  6
+        -----------        
+        cur = 1
+        pre = 2
+        
+        1
+         \
+          2
+           \
+            3
+             \
+              4
+               \
+                5
+                 \
+                  6
+
+- 可以先仿造先根遍历的代码，写出这个倒序遍历的逻辑：
+
+        public void flatten(TreeNode root) {
+            if (root == null)
+                return;
+            flatten(root.right);
+            flatten(root.left);
+        }
+
+- 实现了倒序遍历的逻辑以后，再进行结点之间的拼接：
+
+        private TreeNode prev = null;
+        
+        public void flatten(TreeNode root) {
+            if (root == null)
+                return;
+            flatten(root.right);
+            flatten(root.left);
+            root.right = prev;
+            root.left = null;
+            prev = root;
+        }
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 非递归
+func flatten(root *TreeNode) {
+	list, cur := []int{}, &TreeNode{}
+	preorder(root, &list)
+	cur = root
+	for i := 1; i < len(list); i++ {
+		cur.Left = nil
+		cur.Right = &TreeNode{Val: list[i], Left: nil, Right: nil}
+		cur = cur.Right
+	}
+	return
+}
+
+// 解法二 递归
+func flatten1(root *TreeNode) {
+	if root == nil || (root.Left == nil && root.Right == nil) {
+		return
+	}
+	flatten(root.Left)
+	flatten(root.Right)
+	currRight := root.Right
+	root.Right = root.Left
+	root.Left = nil
+	for root.Right != nil {
+		root = root.Right
+	}
+	root.Right = currRight
+}
+
+// 解法三 递归
+func flatten2(root *TreeNode) {
+	if root == nil {
+		return
+	}
+	flatten(root.Right)
+	if root.Left == nil {
+		return
+	}
+	flatten(root.Left)
+	p := root.Left
+	for p.Right != nil {
+		p = p.Right
+	}
+	p.Right = root.Right
+	root.Right = root.Left
+	root.Left = nil
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0113.Path-Sum-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0115.Distinct-Subsequences/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0115.Distinct-Subsequences.md b/website/content/ChapterFour/0100~0199/0115.Distinct-Subsequences.md
new file mode 100644
index 000000000..b3baae4d5
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0115.Distinct-Subsequences.md
@@ -0,0 +1,108 @@
+# [115. Distinct Subsequences](https://leetcode.com/problems/distinct-subsequences/)
+
+
+## 题目
+
+Given two strings `s` and `t`, return *the number of distinct subsequences of `s` which equals `t`*.
+
+A string's **subsequence** is a new string formed from the original string by deleting some (can be none) of the characters without disturbing the remaining characters' relative positions. (i.e., `"ACE"` is a subsequence of `"ABCDE"` while `"AEC"` is not).
+
+It is guaranteed the answer fits on a 32-bit signed integer.
+
+**Example 1:**
+
+```
+Input: s = "rabbbit", t = "rabbit"
+Output: 3
+Explanation:
+As shown below, there are 3 ways you can generate "rabbit" from S.
+rabbbitrabbbitrabbbit
+```
+
+**Example 2:**
+
+```
+Input: s = "babgbag", t = "bag"
+Output: 5
+Explanation:
+As shown below, there are 5 ways you can generate "bag" from S.
+babgbagbabgbagbabgbagbabgbagbabgbag
+```
+
+**Constraints:**
+
+- `0 <= s.length, t.length <= 1000`
+- `s` and `t` consist of English letters.
+
+## 题目大意
+
+给定一个字符串 s 和一个字符串 t ，计算在 s 的子序列中 t 出现的个数。字符串的一个 子序列 是指，通过删除一些（也可以不删除）字符且不干扰剩余字符相对位置所组成的新字符串。（例如，"ACE" 是 "ABCDE" 的一个子序列，而 "AEC" 不是）题目数据保证答案符合 32 位带符号整数范围。
+
+## 解题思路
+
+- 在字符串 `s` 中最多包含多少个字符串 `t`。这里面包含很多重叠子问题，所以尝试用动态规划解决这个问题。定义 `dp[i][j]` 代表 `s[i:]` 的子序列中 `t[j:]` 出现的个数。初始化先判断边界条件。当 `i = len(s)` 且 `0≤ j < len(t)` 的时候，`s[i:]` 为空字符串，`t[j:]` 不为空，所以 `dp[len(s)][j] = 0`。当 `j = len(t)` 且 `0 ≤ i < len(s)` 的时候，`t[j:]` 不为空字符串，空字符串是任何字符串的子序列。所以 `dp[i][n] = 1`。
+- 当 `i < len(s)` 且 `j < len(t)` 的时候，如果 `s[i] == t[j]`，有 2 种匹配方式，第一种将 `s[i]` 与 `t[j]` 匹配，那么 `t[j+1:]` 匹配 `s[i+1:]` 的子序列，子序列数为 `dp[i+1][j+1]`；第二种将 `s[i]` 不与 `t[j]` 匹配，`t[j:]` 作为 `s[i+1:]` 的子序列，子序列数为 `dp[i+1][j]`。综合 2 种情况，当 `s[i] == t[j]` 时，`dp[i][j] = dp[i+1][j+1] + dp[i+1][j]`。
+- 如果 `s[i] != t[j]`，此时 `t[j:]` 只能作为 `s[i+1:]` 的子序列，子序列数为 `dp[i+1][j]`。所以当 `s[i] != t[j]` 时，`dp[i][j] = dp[i+1][j]`。综上分析得：
+
+	{{< katex display >}}
+    dp[i][j] = \left\{\begin{matrix}dp[i+1][j+1]+dp[i+1][j]&,s[i]=t[j]\\ dp[i+1][j]&,s[i]!=t[j]\end{matrix}\right.
+	{{< /katex >}}
+
+- 最后是优化版本。写出上述代码以后，可以发现填表的过程是从右下角一直填到左上角。填表顺序是 从下往上一行一行的填。行内从右往左填。于是可以将这个二维数据压缩到一维。因为填充当前行只需要用到它的下一行信息即可，更进一步，用到的是下一行中右边元素的信息。于是可以每次更新该行时，先将旧的值存起来，计算更新该行的时候从右往左更新。这样做即可减少一维空间，将原来的二维数组压缩到一维数组。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 压缩版 DP
+func numDistinct(s string, t string) int {
+	dp := make([]int, len(s)+1)
+	for i, curT := range t {
+		pre := 0
+		for j, curS := range s {
+			if i == 0 {
+				pre = 1
+			}
+			newDP := dp[j+1]
+			if curT == curS {
+				dp[j+1] = dp[j] + pre
+			} else {
+				dp[j+1] = dp[j]
+			}
+			pre = newDP
+		}
+	}
+	return dp[len(s)]
+}
+
+// 解法二 普通 DP
+func numDistinct1(s, t string) int {
+	m, n := len(s), len(t)
+	if m < n {
+		return 0
+	}
+	dp := make([][]int, m+1)
+	for i := range dp {
+		dp[i] = make([]int, n+1)
+		dp[i][n] = 1
+	}
+	for i := m - 1; i >= 0; i-- {
+		for j := n - 1; j >= 0; j-- {
+			if s[i] == t[j] {
+				dp[i][j] = dp[i+1][j+1] + dp[i+1][j]
+			} else {
+				dp[i][j] = dp[i+1][j]
+			}
+		}
+	}
+	return dp[0][0]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0114.Flatten-Binary-Tree-to-Linked-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0116.Populating-Next-Right-Pointers-in-Each-Node/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0116.Populating-Next-Right-Pointers-in-Each-Node.md b/website/content/ChapterFour/0100~0199/0116.Populating-Next-Right-Pointers-in-Each-Node.md
new file mode 100644
index 000000000..95b54ea29
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0116.Populating-Next-Right-Pointers-in-Each-Node.md
@@ -0,0 +1,125 @@
+# [116. Populating Next Right Pointers in Each Node](https://leetcode.com/problems/populating-next-right-pointers-in-each-node/)
+
+
+## 题目
+
+You are given a **perfect binary tree** where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:
+
+```
+struct Node {
+  int val;
+  Node *left;
+  Node *right;
+  Node *next;
+}
+
+```
+
+Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to `NULL`.
+
+Initially, all next pointers are set to `NULL`.
+
+**Follow up:**
+
+- You may only use constant extra space.
+- Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/02/14/116_sample.png](https://assets.leetcode.com/uploads/2019/02/14/116_sample.png)
+
+```
+Input: root = [1,2,3,4,5,6,7]
+Output: [1,#,2,3,#,4,5,6,7,#]
+Explanation:Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the given tree is less than `4096`.
+- `1000 <= node.val <= 1000`
+
+## 题目大意
+
+给定一个 完美二叉树 ，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：
+
+```jsx
+struct Node {
+  int val;
+  Node *left;
+  Node *right;
+  Node *next;
+}
+
+```
+
+填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。初始状态下，所有 next 指针都被设置为 NULL。
+
+## 解题思路
+
+- 本质上是二叉树的层序遍历，基于广度优先搜索，将每层的节点放入队列，并遍历队列进行连接。
+
+## 代码
+
+```go
+package leetcode
+
+type Node struct {
+	Val   int
+	Left  *Node
+	Right *Node
+	Next  *Node
+}
+
+//解法一：迭代
+func connect(root *Node) *Node {
+	if root == nil {
+		return root
+	}
+	q := []*Node{root}
+	for len(q) > 0 {
+		var p []*Node
+		// 遍历这一层的所有节点
+		for i, node := range q {
+			if i+1 < len(q) {
+				node.Next = q[i+1]
+			}
+			if node.Left != nil {
+				p = append(p, node.Left)
+			}
+			if node.Right != nil {
+				p = append(p, node.Right)
+			}
+		}
+		q = p
+	}
+	return root
+}
+
+// 解法二 递归
+func connect2(root *Node) *Node {
+	if root == nil {
+		return nil
+	}
+	connectTwoNode(root.Left, root.Right)
+	return root
+}
+
+func connectTwoNode(node1, node2 *Node) {
+	if node1 == nil || node2 == nil {
+		return
+	}
+	node1.Next = node2
+	connectTwoNode(node1.Left, node1.Right)
+	connectTwoNode(node2.Left, node2.Right)
+	connectTwoNode(node1.Right, node2.Left)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0115.Distinct-Subsequences/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0118.Pascals-Triangle/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0118.Pascals-Triangle.md b/website/content/ChapterFour/0100~0199/0118.Pascals-Triangle.md
new file mode 100644
index 000000000..9dee8ab67
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0118.Pascals-Triangle.md
@@ -0,0 +1,67 @@
+# [118. Pascal's Triangle](https://leetcode.com/problems/pascals-triangle/)
+
+
+## 题目
+
+Given a non-negative integer numRows, generate the first numRows of Pascal's triangle.
+
+![](https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif)
+
+**Note**: In Pascal's triangle, each number is the sum of the two numbers directly above it.
+
+**Example**:
+
+```
+Input: 5
+Output:
+[
+     [1],
+    [1,1],
+   [1,2,1],
+  [1,3,3,1],
+ [1,4,6,4,1]
+]
+```
+
+## 题目大意
+
+给定一个非负整数 numRows，生成杨辉三角的前 numRows 行。在杨辉三角中，每个数是它左上方和右上方的数的和。
+
+
+## 解题思路
+
+- 给定一个 n，要求打印杨辉三角的前 n 行。
+- 简单题。按照杨辉三角的生成规则循环打印即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func generate(numRows int) [][]int {
+	result := [][]int{}
+	for i := 0; i < numRows; i++ {
+		row := []int{}
+		for j := 0; j < i+1; j++ {
+			if j == 0 || j == i {
+				row = append(row, 1)
+			} else if i > 1 {
+				row = append(row, result[i-1][j-1]+result[i-1][j])
+			}
+		}
+		result = append(result, row)
+	}
+	return result
+}
+
+```
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0116.Populating-Next-Right-Pointers-in-Each-Node/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0119.Pascals-Triangle-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0119.Pascals-Triangle-II.md b/website/content/ChapterFour/0100~0199/0119.Pascals-Triangle-II.md
new file mode 100644
index 000000000..368da337e
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0119.Pascals-Triangle-II.md
@@ -0,0 +1,83 @@
+# [119. Pascal's Triangle II](https://leetcode.com/problems/pascals-triangle-ii/)
+
+
+## 题目
+
+Given an integer `rowIndex`, return the `rowIndexth` row of the Pascal's triangle.
+
+Notice that the row index starts from **0**.
+
+![https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif](https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif)
+
+In Pascal's triangle, each number is the sum of the two numbers directly above it.
+
+**Follow up:**
+
+Could you optimize your algorithm to use only *O*(*k*) extra space?
+
+**Example 1:**
+
+```
+Input: rowIndex = 3
+Output: [1,3,3,1]
+```
+
+**Example 2:**
+
+```
+Input: rowIndex = 0
+Output: [1]
+```
+
+**Example 3:**
+
+```
+Input: rowIndex = 1
+Output: [1,1]
+```
+
+**Constraints:**
+
+- `0 <= rowIndex <= 33`
+
+## 题目大意
+
+给定一个非负索引 k，其中 k ≤ 33，返回杨辉三角的第 k 行。
+
+## 解题思路
+
+- 题目中的三角是杨辉三角，每个数字是 `(a+b)^n` 二项式展开的系数。题目要求我们只能使用 O(k) 的空间。那么需要找到两两项直接的递推关系。由组合知识得知：
+  
+	{{< katex display >}} 
+    \begin{aligned}C_{n}^{m} &= \frac{n!}{m!(n-m)!} \\C_{n}^{m-1} &= \frac{n!}{(m-1)!(n-m+1)!}\end{aligned}
+	{{< /katex>}} 
+
+    于是得到递推公式：
+
+	{{< katex display >}} 
+    C_{n}^{m} = C_{n}^{m-1} \times \frac{n-m+1}{m}
+	{{< /katex>}} 
+
+    利用这个递推公式即可以把空间复杂度优化到 O(k)
+
+## 代码
+
+```go
+package leetcode
+
+func getRow(rowIndex int) []int {
+	row := make([]int, rowIndex+1)
+	row[0] = 1
+	for i := 1; i <= rowIndex; i++ {
+		row[i] = row[i-1] * (rowIndex - i + 1) / i
+	}
+	return row
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0118.Pascals-Triangle/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0120.Triangle/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0120.Triangle.md b/website/content/ChapterFour/0100~0199/0120.Triangle.md
new file mode 100755
index 000000000..83498b7e7
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0120.Triangle.md
@@ -0,0 +1,96 @@
+# [120. Triangle](https://leetcode.com/problems/triangle/)
+
+
+## 题目
+
+Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
+
+For example, given the following triangle
+
+    [
+         [2],
+        [3,4],
+       [6,5,7],
+      [4,1,8,3]
+    ]
+
+The minimum path sum from top to bottom is `11` (i.e., **2** + **3** + **5** + **1** = 11).
+
+**Note**:
+
+Bonus point if you are able to do this using only *O*(*n*) extra space, where *n* is the total number of rows in the triangle.
+
+
+## 题目大意
+
+给定一个三角形，找出自顶向下的最小路径和。每一步只能移动到下一行中相邻的结点上。
+
+
+## 解题思路
+
+- 求出从三角形顶端到底端的最小和。要求最好用 O(n) 的时间复杂度。
+- 这一题最优解是不用辅助空间，直接从下层往上层推。普通解法是用二维数组 DP，稍微优化的解法是一维数组 DP。解法如下：
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+)
+
+// 解法一 倒序 DP，无辅助空间
+func minimumTotal(triangle [][]int) int {
+	if triangle == nil {
+		return 0
+	}
+	for row := len(triangle) - 2; row >= 0; row-- {
+		for col := 0; col < len(triangle[row]); col++ {
+			triangle[row][col] += min(triangle[row+1][col], triangle[row+1][col+1])
+		}
+	}
+	return triangle[0][0]
+}
+
+// 解法二 正常 DP，空间复杂度 O(n)
+func minimumTotal1(triangle [][]int) int {
+	if len(triangle) == 0 {
+		return 0
+	}
+	dp, minNum, index := make([]int, len(triangle[len(triangle)-1])), math.MaxInt64, 0
+	for ; index < len(triangle[0]); index++ {
+		dp[index] = triangle[0][index]
+	}
+	for i := 1; i < len(triangle); i++ {
+		for j := len(triangle[i]) - 1; j >= 0; j-- {
+			if j == 0 {
+				// 最左边
+				dp[j] += triangle[i][0]
+			} else if j == len(triangle[i])-1 {
+				// 最右边
+				dp[j] += dp[j-1] + triangle[i][j]
+			} else {
+				// 中间
+				dp[j] = min(dp[j-1]+triangle[i][j], dp[j]+triangle[i][j])
+			}
+		}
+	}
+	for i := 0; i < len(dp); i++ {
+		if dp[i] < minNum {
+			minNum = dp[i]
+		}
+	}
+	return minNum
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0119.Pascals-Triangle-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0121.Best-Time-to-Buy-and-Sell-Stock/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0121.Best-Time-to-Buy-and-Sell-Stock.md b/website/content/ChapterFour/0100~0199/0121.Best-Time-to-Buy-and-Sell-Stock.md
new file mode 100755
index 000000000..b08d1fd2e
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0121.Best-Time-to-Buy-and-Sell-Stock.md
@@ -0,0 +1,89 @@
+# [121. Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)
+
+
+## 题目
+
+Say you have an array for which the *i*th element is the price of a given stock on day *i*.
+
+If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
+
+Note that you cannot sell a stock before you buy one.
+
+**Example 1**:
+
+    Input: [7,1,5,3,6,4]
+    Output: 5
+    Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
+                 Not 7-1 = 6, as selling price needs to be larger than buying price.
+
+**Example 2**:
+
+    Input: [7,6,4,3,1]
+    Output: 0
+    Explanation: In this case, no transaction is done, i.e. max profit = 0.
+
+## 题目大意
+
+给定一个数组，它的第 i 个元素是一支给定股票第 i 天的价格。如果你最多只允许完成一笔交易（即买入和卖出一支股票），设计一个算法来计算你所能获取的最大利润。注意你不能在买入股票前卖出股票。
+
+## 解题思路
+
+- 题目要求找出股票中能赚的钱最多的差价
+- 这一题也有多个解法，可以用 DP，也可以用单调栈
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 模拟 DP
+func maxProfit(prices []int) int {
+	if len(prices) < 1 {
+		return 0
+	}
+	min, maxProfit := prices[0], 0
+	for i := 1; i < len(prices); i++ {
+		if prices[i]-min > maxProfit {
+			maxProfit = prices[i] - min
+		}
+		if prices[i] < min {
+			min = prices[i]
+		}
+	}
+	return maxProfit
+}
+
+// 解法二 单调栈
+func maxProfit1(prices []int) int {
+	if len(prices) == 0 {
+		return 0
+	}
+	stack, res := []int{prices[0]}, 0
+	for i := 1; i < len(prices); i++ {
+		if prices[i] > stack[len(stack)-1] {
+			stack = append(stack, prices[i])
+		} else {
+			index := len(stack) - 1
+			for ; index >= 0; index-- {
+				if stack[index] < prices[i] {
+					break
+				}
+			}
+			stack = stack[:index+1]
+			stack = append(stack, prices[i])
+		}
+		res = max(res, stack[len(stack)-1]-stack[0])
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0120.Triangle/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0122.Best-Time-to-Buy-and-Sell-Stock-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0122.Best-Time-to-Buy-and-Sell-Stock-II.md b/website/content/ChapterFour/0100~0199/0122.Best-Time-to-Buy-and-Sell-Stock-II.md
new file mode 100755
index 000000000..93d77d36d
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0122.Best-Time-to-Buy-and-Sell-Stock-II.md
@@ -0,0 +1,67 @@
+# [122. Best Time to Buy and Sell Stock II](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/)
+
+
+## 题目
+
+Say you have an array for which the *i*th element is the price of a given stock on day *i*.
+
+Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).
+
+**Note**: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
+
+**Example 1**:
+
+    Input: [7,1,5,3,6,4]
+    Output: 7
+    Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
+                 Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
+
+**Example 2**:
+
+    Input: [1,2,3,4,5]
+    Output: 4
+    Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
+                 Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
+                 engaging multiple transactions at the same time. You must sell before buying again.
+
+**Example 3**:
+
+    Input: [7,6,4,3,1]
+    Output: 0
+    Explanation: In this case, no transaction is done, i.e. max profit = 0.
+
+## 题目大意
+
+给定一个数组，它的第 i 个元素是一支给定股票第 i 天的价格。设计一个算法来计算你所能获取的最大利润。你可以尽可能地完成更多的交易（多次买卖一支股票）。注意：你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。
+
+
+## 解题思路
+
+- 这一题是第 121 题的加强版。要求输出最大收益，这一题不止买卖一次，可以买卖多次，买卖不能在同一天内操作。
+- 最大收益来源，必然是每次跌了就买入，涨到顶峰的时候就抛出。只要有涨峰就开始计算赚的钱，连续涨可以用两两相减累加来计算，两两相减累加，相当于涨到波峰的最大值减去谷底的值。这一点看通以后，题目非常简单。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func maxProfit122(prices []int) int {
+	profit := 0
+	for i := 0; i < len(prices)-1; i++ {
+		if prices[i+1] > prices[i] {
+			profit += prices[i+1] - prices[i]
+		}
+	}
+	return profit
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0121.Best-Time-to-Buy-and-Sell-Stock/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0124.Binary-Tree-Maximum-Path-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0124.Binary-Tree-Maximum-Path-Sum.md b/website/content/ChapterFour/0100~0199/0124.Binary-Tree-Maximum-Path-Sum.md
new file mode 100755
index 000000000..51e96ff61
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0124.Binary-Tree-Maximum-Path-Sum.md
@@ -0,0 +1,87 @@
+# [124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)
+
+
+## 题目
+
+Given a **non-empty** binary tree, find the maximum path sum.
+
+For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain **at least one node** and does not need to go through the root.
+
+**Example 1**:
+
+    Input: [1,2,3]
+    
+           1
+          / \
+         2   3
+    
+    Output: 6
+
+**Example 2**:
+
+    Input: [-10,9,20,null,null,15,7]
+    
+       -10
+       / \
+      9  20
+        /  \
+       15   7
+    
+    Output: 42
+
+## 题目大意
+
+给定一个非空二叉树，返回其最大路径和。本题中，路径被定义为一条从树中任意节点出发，达到任意节点的序列。该路径至少包含一个节点，且不一定经过根节点。
+
+## 解题思路
+
+- 给出一个二叉树，要求找一条路径使得路径的和是最大的。
+- 这一题思路比较简单，递归维护最大值即可。不过需要比较的对象比较多。`maxPathSum(root) = max(maxPathSum(root.Left), maxPathSum(root.Right), maxPathSumFrom(root.Left) (if>0) + maxPathSumFrom(root.Right) (if>0) + root.Val)` ，其中，`maxPathSumFrom(root) = max(maxPathSumFrom(root.Left), maxPathSumFrom(root.Right)) + root.Val`
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func maxPathSum(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	max := math.MinInt32
+	getPathSum(root, &max)
+	return max
+}
+
+func getPathSum(root *TreeNode, maxSum *int) int {
+	if root == nil {
+		return math.MinInt32
+	}
+	left := getPathSum(root.Left, maxSum)
+	right := getPathSum(root.Right, maxSum)
+
+	currMax := max(max(left+root.Val, right+root.Val), root.Val)
+	*maxSum = max(*maxSum, max(currMax, left+right+root.Val))
+	return currMax
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0122.Best-Time-to-Buy-and-Sell-Stock-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0125.Valid-Palindrome/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0125.Valid-Palindrome.md b/website/content/ChapterFour/0100~0199/0125.Valid-Palindrome.md
new file mode 100644
index 000000000..b0c7257ac
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0125.Valid-Palindrome.md
@@ -0,0 +1,74 @@
+# [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/description/)
+
+## 题目
+
+Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
+
+For example,
+
+```
+
+"A man, a plan, a canal: Panama" is a palindrome.
+"race a car" is not a palindrome.
+
+```
+
+**Note**:  
+
+Have you consider that the string might be empty? This is a good question to ask during an interview.
+
+For the purpose of this problem, we define empty string as valid palindrome.
+
+## 题目大意
+
+判断所给的字符串是否是有效的回文串。
+
+## 解题思路
+
+简单题，按照题意做即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strings"
+)
+
+func isPalindrome(s string) bool {
+	s = strings.ToLower(s)
+	i, j := 0, len(s)-1
+	for i < j {
+		for i < j && !isChar(s[i]) {
+			i++
+		}
+		for i < j && !isChar(s[j]) {
+			j--
+		}
+		if s[i] != s[j] {
+			return false
+		}
+		i++
+		j--
+	}
+	return true
+}
+
+// 判断 c 是否是字符或者数字
+func isChar(c byte) bool {
+	if ('a' <= c && c <= 'z') || ('0' <= c && c <= '9') {
+		return true
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0124.Binary-Tree-Maximum-Path-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0126.Word-Ladder-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0126.Word-Ladder-II.md b/website/content/ChapterFour/0100~0199/0126.Word-Ladder-II.md
new file mode 100755
index 000000000..e1d328a52
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0126.Word-Ladder-II.md
@@ -0,0 +1,137 @@
+# [126. Word Ladder II](https://leetcode.com/problems/word-ladder-ii/)
+
+
+## 题目
+
+Given two words (*beginWord* and *endWord*), and a dictionary's word list, find all shortest transformation sequence(s) from *beginWord* to *endWord*, such that:
+
+1. Only one letter can be changed at a time
+2. Each transformed word must exist in the word list. Note that *beginWord* is *not* a transformed word.
+
+**Note**:
+
+- Return an empty list if there is no such transformation sequence.
+- All words have the same length.
+- All words contain only lowercase alphabetic characters.
+- You may assume no duplicates in the word list.
+- You may assume *beginWord* and *endWord* are non-empty and are not the same.
+
+**Example 1**:
+
+    Input:
+    beginWord = "hit",
+    endWord = "cog",
+    wordList = ["hot","dot","dog","lot","log","cog"]
+    
+    Output:
+    [
+      ["hit","hot","dot","dog","cog"],
+      ["hit","hot","lot","log","cog"]
+    ]
+
+**Example 2**:
+
+    Input:
+    beginWord = "hit"
+    endWord = "cog"
+    wordList = ["hot","dot","dog","lot","log"]
+    
+    Output: []
+    
+    Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
+
+## 题目大意
+
+给定两个单词（beginWord 和 endWord）和一个字典 wordList，找出所有从 beginWord 到 endWord 的最短转换序列。转换需遵循如下规则：
+
+1. 每次转换只能改变一个字母。
+2. 转换过程中的中间单词必须是字典中的单词。
+
+说明:
+
+- 如果不存在这样的转换序列，返回一个空列表。
+- 所有单词具有相同的长度。
+- 所有单词只由小写字母组成。
+- 字典中不存在重复的单词。
+- 你可以假设 beginWord 和 endWord 是非空的，且二者不相同。
+
+
+
+## 解题思路
+
+- 这一题是第 127 题的加强版，除了找到路径的长度，还进一步要求输出所有路径。解题思路同第 127 题一样，也是用 BFS 遍历。
+- 当前做法不是最优解，是否可以考虑双端 BFS 优化，或者迪杰斯塔拉算法？
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findLadders(beginWord string, endWord string, wordList []string) [][]string {
+	result, wordMap := make([][]string, 0), make(map[string]bool)
+	for _, w := range wordList {
+		wordMap[w] = true
+	}
+	if !wordMap[endWord] {
+		return result
+	}
+	// create a queue, track the path
+	queue := make([][]string, 0)
+	queue = append(queue, []string{beginWord})
+	// queueLen is used to track how many slices in queue are in the same level
+	// if found a result, I still need to finish checking current level cause I need to return all possible paths
+	queueLen := 1
+	// use to track strings that this level has visited
+	// when queueLen == 0, remove levelMap keys in wordMap
+	levelMap := make(map[string]bool)
+	for len(queue) > 0 {
+		path := queue[0]
+		queue = queue[1:]
+		lastWord := path[len(path)-1]
+		for i := 0; i < len(lastWord); i++ {
+			for c := 'a'; c <= 'z'; c++ {
+				nextWord := lastWord[:i] + string(c) + lastWord[i+1:]
+				if nextWord == endWord {
+					path = append(path, endWord)
+					result = append(result, path)
+					continue
+				}
+				if wordMap[nextWord] {
+					// different from word ladder, don't remove the word from wordMap immediately
+					// same level could reuse the key.
+					// delete from wordMap only when currently level is done.
+					levelMap[nextWord] = true
+					newPath := make([]string, len(path))
+					copy(newPath, path)
+					newPath = append(newPath, nextWord)
+					queue = append(queue, newPath)
+				}
+			}
+		}
+		queueLen--
+		// if queueLen is 0, means finish traversing current level. if result is not empty, return result
+		if queueLen == 0 {
+			if len(result) > 0 {
+				return result
+			}
+			for k := range levelMap {
+				delete(wordMap, k)
+			}
+			// clear levelMap
+			levelMap = make(map[string]bool)
+			queueLen = len(queue)
+		}
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0125.Valid-Palindrome/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0127.Word-Ladder/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0127.Word-Ladder.md b/website/content/ChapterFour/0100~0199/0127.Word-Ladder.md
new file mode 100755
index 000000000..7a7db0e87
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0127.Word-Ladder.md
@@ -0,0 +1,128 @@
+# [127. Word Ladder](https://leetcode.com/problems/word-ladder/)
+
+
+## 题目
+
+Given two words (*beginWord* and *endWord*), and a dictionary's word list, find the length of shortest transformation sequence from *beginWord* to *endWord*, such that:
+
+1. Only one letter can be changed at a time.
+2. Each transformed word must exist in the word list. Note that *beginWord* is *not* a transformed word.
+
+**Note**:
+
+- Return 0 if there is no such transformation sequence.
+- All words have the same length.
+- All words contain only lowercase alphabetic characters.
+- You may assume no duplicates in the word list.
+- You may assume *beginWord* and *endWord* are non-empty and are not the same.
+
+**Example 1**:
+
+    Input:
+    beginWord = "hit",
+    endWord = "cog",
+    wordList = ["hot","dot","dog","lot","log","cog"]
+    
+    Output: 5
+    
+    Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
+    return its length 5.
+
+**Example 2**:
+
+    Input:
+    beginWord = "hit"
+    endWord = "cog"
+    wordList = ["hot","dot","dog","lot","log"]
+    
+    Output: 0
+    
+    Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
+
+
+## 题目大意
+
+给定两个单词（beginWord 和 endWord）和一个字典，找到从 beginWord 到 endWord 的最短转换序列的长度。转换需遵循如下规则：
+
+1. 每次转换只能改变一个字母。
+2. 转换过程中的中间单词必须是字典中的单词。
+
+说明:
+
+- 如果不存在这样的转换序列，返回 0。
+- 所有单词具有相同的长度。
+- 所有单词只由小写字母组成。
+- 字典中不存在重复的单词。
+- 你可以假设 beginWord 和 endWord 是非空的，且二者不相同。
+
+
+## 解题思路
+
+- 这一题要求输出从 `beginWord` 变换到 `endWord` 最短变换次数。可以用 BFS，从 `beginWord` 开始变换，把该单词的每个字母都用 `'a'~'z'` 变换一次，生成的数组到 `wordList` 中查找，这里用 Map 来记录查找。找得到就入队列，找不到就输出 0 。入队以后按照 BFS 的算法依次遍历完，当所有单词都 `len(queue)<=0` 出队以后，整个程序结束。
+- 这一题题目中虽然说了要求找到一条最短的路径，但是实际上最短的路径的寻找方法已经告诉你了：
+	1. 每次只变换一个字母
+	2. 每次变换都必须在 `wordList` 中  
+所以不需要单独考虑何种方式是最短的。 
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func ladderLength(beginWord string, endWord string, wordList []string) int {
+	wordMap, que, depth := getWordMap(wordList, beginWord), []string{beginWord}, 0
+	for len(que) > 0 {
+		depth++
+		qlen := len(que)
+		for i := 0; i < qlen; i++ {
+			word := que[0]
+			que = que[1:]
+			candidates := getCandidates(word)
+			for _, candidate := range candidates {
+				if _, ok := wordMap[candidate]; ok {
+					if candidate == endWord {
+						return depth + 1
+					}
+					delete(wordMap, candidate)
+					que = append(que, candidate)
+				}
+			}
+		}
+	}
+	return 0
+}
+
+func getWordMap(wordList []string, beginWord string) map[string]int {
+	wordMap := make(map[string]int)
+	for i, word := range wordList {
+		if _, ok := wordMap[word]; !ok {
+			if word != beginWord {
+				wordMap[word] = i
+			}
+		}
+	}
+	return wordMap
+}
+
+func getCandidates(word string) []string {
+	var res []string
+	for i := 0; i < 26; i++ {
+		for j := 0; j < len(word); j++ {
+			if word[j] != byte(int('a')+i) {
+				res = append(res, word[:j]+string(int('a')+i)+word[j+1:])
+			}
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0126.Word-Ladder-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0128.Longest-Consecutive-Sequence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0128.Longest-Consecutive-Sequence.md b/website/content/ChapterFour/0100~0199/0128.Longest-Consecutive-Sequence.md
new file mode 100755
index 000000000..fddce0f06
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0128.Longest-Consecutive-Sequence.md
@@ -0,0 +1,146 @@
+# [128. Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/)
+
+
+## 题目
+
+Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
+
+Your algorithm should run in O(*n*) complexity.
+
+**Example**:
+
+    Input: [100, 4, 200, 1, 3, 2]
+    Output: 4
+    Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.
+
+
+## 题目大意
+
+
+给定一个未排序的整数数组，找出最长连续序列的长度。要求算法的时间复杂度为 O(n)。
+
+
+
+
+## 解题思路
+
+
+- 给出一个数组，要求找出最长连续序列，输出这个最长的长度。要求时间复杂度为 `O(n)`。
+- 这一题可以先用暴力解决解决，代码见解法三。思路是把每个数都存在 `map` 中，先删去 `map` 中没有前一个数 `nums[i]-1` 也没有后一个数 `nums[i]+1` 的数 `nums[i]`，这种数前后都不连续。然后在 `map` 中找到前一个数 `nums[i]-1` 不存在，但是后一个数 `nums[i]+1` 存在的数，这种数是连续序列的起点，那么不断的往后搜，直到序列“断”了。最后输出最长序列的长度。
+- 这一题最优的解法是解法一，针对每一个 `map` 中不存在的数 `n`，插入进去都做 2 件事情。第一件事，先查看 `n - 1` 和 `n + 1` 是否都存在于 `map` 中，如果都存在，代表存在连续的序列，那么就更新 `left`，`right` 边界。那么 `n` 对应的这个小的子连续序列长度为 `sum = left + right + 1`。第二件事就是更新 `left` 和 `right` 左右边界对应的 `length = sum`。
+- 这一题还可以用并查集解决，见解法二。利用每个数在 `nums` 中的下标，把下标和下标进行 `union()`，具体做法是看前一个数 `nums[i]-1` 和后一个数 `nums[i]+1` 在 `map` 中是否存在，如果存在就 `union()`，最终输出整个并查集中包含最多元素的那个集合的元素总数。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+// 解法一 map，时间复杂度 O(n)
+func longestConsecutive(nums []int) int {
+	res, numMap := 0, map[int]int{}
+	for _, num := range nums {
+		if numMap[num] == 0 {
+			left, right, sum := 0, 0, 0
+			if numMap[num-1] > 0 {
+				left = numMap[num-1]
+			} else {
+				left = 0
+			}
+			if numMap[num+1] > 0 {
+				right = numMap[num+1]
+			} else {
+				right = 0
+			}
+			// sum: length of the sequence n is in
+			sum = left + right + 1
+			numMap[num] = sum
+			// keep track of the max length
+			res = max(res, sum)
+			// extend the length to the boundary(s) of the sequence
+			// will do nothing if n has no neighbors
+			numMap[num-left] = sum
+			numMap[num+right] = sum
+		} else {
+			continue
+		}
+	}
+	return res
+}
+
+// 解法二 并查集
+func longestConsecutive1(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	numMap, countMap, lcs, uf := map[int]int{}, map[int]int{}, 0, template.UnionFind{}
+	uf.Init(len(nums))
+	for i := 0; i < len(nums); i++ {
+		countMap[i] = 1
+	}
+	for i := 0; i < len(nums); i++ {
+		if _, ok := numMap[nums[i]]; ok {
+			continue
+		}
+		numMap[nums[i]] = i
+		if _, ok := numMap[nums[i]+1]; ok {
+			uf.Union(i, numMap[nums[i]+1])
+		}
+		if _, ok := numMap[nums[i]-1]; ok {
+			uf.Union(i, numMap[nums[i]-1])
+		}
+	}
+	for key := range countMap {
+		parent := uf.Find(key)
+		if parent != key {
+			countMap[parent]++
+		}
+		if countMap[parent] > lcs {
+			lcs = countMap[parent]
+		}
+	}
+	return lcs
+}
+
+// 解法三 暴力解法，时间复杂度 O(n^2)
+func longestConsecutive2(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	numMap, length, tmp, lcs := map[int]bool{}, 0, 0, 0
+	for i := 0; i < len(nums); i++ {
+		numMap[nums[i]] = true
+	}
+	for key := range numMap {
+		if !numMap[key-1] && !numMap[key+1] {
+			delete(numMap, key)
+		}
+	}
+	if len(numMap) == 0 {
+		return 1
+	}
+	for key := range numMap {
+		if !numMap[key-1] && numMap[key+1] {
+			length, tmp = 1, key+1
+			for numMap[tmp] {
+				length++
+				tmp++
+			}
+			lcs = max(lcs, length)
+		}
+	}
+	return max(lcs, length)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0127.Word-Ladder/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0129.Sum-Root-to-Leaf-Numbers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0129.Sum-Root-to-Leaf-Numbers.md b/website/content/ChapterFour/0100~0199/0129.Sum-Root-to-Leaf-Numbers.md
new file mode 100755
index 000000000..312f4a066
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0129.Sum-Root-to-Leaf-Numbers.md
@@ -0,0 +1,99 @@
+# [129. Sum Root to Leaf Numbers](https://leetcode.com/problems/sum-root-to-leaf-numbers/)
+
+
+## 题目
+
+Given a binary tree containing digits from `0-9` only, each root-to-leaf path could represent a number.
+
+An example is the root-to-leaf path `1->2->3` which represents the number `123`.
+
+Find the total sum of all root-to-leaf numbers.
+
+**Note**: A leaf is a node with no children.
+
+**Example**:
+
+    Input: [1,2,3]
+        1
+       / \
+      2   3
+    Output: 25
+    Explanation:
+    The root-to-leaf path 1->2 represents the number 12.
+    The root-to-leaf path 1->3 represents the number 13.
+    Therefore, sum = 12 + 13 = 25.
+
+**Example 2**:
+
+    Input: [4,9,0,5,1]
+        4
+       / \
+      9   0
+     / \
+    5   1
+    Output: 1026
+    Explanation:
+    The root-to-leaf path 4->9->5 represents the number 495.
+    The root-to-leaf path 4->9->1 represents the number 491.
+    The root-to-leaf path 4->0 represents the number 40.
+    Therefore, sum = 495 + 491 + 40 = 1026.
+
+## 题目大意
+
+给定一个二叉树，它的每个结点都存放一个 0-9 的数字，每条从根到叶子节点的路径都代表一个数字。例如，从根到叶子节点路径 1->2->3 代表数字 123。计算从根到叶子节点生成的所有数字之和。说明: 叶子节点是指没有子节点的节点。
+
+
+## 解题思路
+
+- 这一题是第 257 题的变形题，第 257 题要求输出每条从根节点到叶子节点的路径，这一题变成了把每一个从根节点到叶子节点的数字都串联起来，再累加每条路径，求出最后的总和。实际做题思路基本没变。运用前序遍历的思想，当从根节点出发一直加到叶子节点，每个叶子节点汇总一次。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func sumNumbers(root *TreeNode) int {
+	res := 0
+	dfs(root,0,&res)
+	return res
+}
+
+func dfs(root *TreeNode,sum int,res *int)  {
+	if root == nil{
+		return
+	}
+	sum = sum*10 + root.Val
+	if root.Left == nil && root.Right == nil{
+		*res += sum
+		return
+	}
+	dfs(root.Left,sum,res)
+	dfs(root.Right,sum,res)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0128.Longest-Consecutive-Sequence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0130.Surrounded-Regions/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0130.Surrounded-Regions.md b/website/content/ChapterFour/0100~0199/0130.Surrounded-Regions.md
new file mode 100755
index 000000000..279e9551b
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0130.Surrounded-Regions.md
@@ -0,0 +1,133 @@
+# [130. Surrounded Regions](https://leetcode.com/problems/surrounded-regions/)
+
+
+
+## 题目
+
+Given a 2D board containing `'X'` and `'O'` (**the letter O**), capture all regions surrounded by `'X'`.
+
+A region is captured by flipping all `'O'`s into `'X'`s in that surrounded region.
+
+**Example**:
+
+    X X X X
+    X O O X
+    X X O X
+    X O X X
+
+After running your function, the board should be:
+
+    X X X X
+    X X X X
+    X X X X
+    X O X X
+
+**Explanation**:
+
+Surrounded regions shouldn’t be on the border, which means that any `'O'` on the border of the board are not flipped to `'X'`. Any `'O'` that is not on the border and it is not connected to an `'O'` on the border will be flipped to `'X'`. Two cells are connected if they are adjacent cells connected horizontally or vertically.
+
+## 题目大意
+
+给定一个二维的矩阵，包含 'X' 和 'O'（字母 O）。找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O' 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
+
+
+## 解题思路
+
+
+- 给出一张二维地图，要求把地图上非边缘上的 'O' 都用 'X' 覆盖掉。
+- 这一题有多种解法。第一种解法是并查集。先将边缘上的 'O' 全部都和一个特殊的点进行 `union()` 。然后再把地图中间的 'O' 都进行 `union()`，最后把和特殊点不是同一个集合的点都标记成 'X'。第二种解法是 DFS 或者 BFS，可以先将边缘上的 'O' 先标记成另外一个字符，然后在递归遍历过程中，把剩下的 'O' 都标记成 'X'。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+// 解法一 并查集
+func solve(board [][]byte) {
+	if len(board) == 0 {
+		return
+	}
+	m, n := len(board[0]), len(board)
+	uf := template.UnionFind{}
+	uf.Init(n*m + 1) // 特意多一个特殊点用来标记
+
+	for i := 0; i < n; i++ {
+		for j := 0; j < m; j++ {
+			if (i == 0 || i == n-1 || j == 0 || j == m-1) && board[i][j] == 'O' { //棋盘边缘上的 'O' 点
+				uf.Union(i*m+j, n*m)
+			} else if board[i][j] == 'O' { //棋盘非边缘上的内部的 'O' 点
+				if board[i-1][j] == 'O' {
+					uf.Union(i*m+j, (i-1)*m+j)
+				}
+				if board[i+1][j] == 'O' {
+					uf.Union(i*m+j, (i+1)*m+j)
+				}
+				if board[i][j-1] == 'O' {
+					uf.Union(i*m+j, i*m+j-1)
+				}
+				if board[i][j+1] == 'O' {
+					uf.Union(i*m+j, i*m+j+1)
+				}
+
+			}
+		}
+	}
+	for i := 0; i < n; i++ {
+		for j := 0; j < m; j++ {
+			if uf.Find(i*m+j) != uf.Find(n*m) {
+				board[i][j] = 'X'
+			}
+		}
+	}
+}
+
+// 解法二 DFS
+func solve1(board [][]byte) {
+	for i := range board {
+		for j := range board[i] {
+			if i == 0 || i == len(board)-1 || j == 0 || j == len(board[i])-1 {
+				if board[i][j] == 'O' {
+					dfs130(i, j, board)
+				}
+			}
+		}
+	}
+
+	for i := range board {
+		for j := range board[i] {
+			if board[i][j] == '*' {
+				board[i][j] = 'O'
+			} else if board[i][j] == 'O' {
+				board[i][j] = 'X'
+			}
+		}
+	}
+}
+
+func dfs130(i, j int, board [][]byte) {
+	if i < 0 || i > len(board)-1 || j < 0 || j > len(board[i])-1 {
+		return
+	}
+	if board[i][j] == 'O' {
+		board[i][j] = '*'
+		for k := 0; k < 4; k++ {
+			dfs130(i+dir[k][0], j+dir[k][1], board)
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0129.Sum-Root-to-Leaf-Numbers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0131.Palindrome-Partitioning/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0131.Palindrome-Partitioning.md b/website/content/ChapterFour/0100~0199/0131.Palindrome-Partitioning.md
new file mode 100755
index 000000000..eb77f95c5
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0131.Palindrome-Partitioning.md
@@ -0,0 +1,128 @@
+# [131. Palindrome Partitioning](https://leetcode.com/problems/palindrome-partitioning/)
+
+
+## 题目
+
+Given a string *s*, partition *s* such that every substring of the partition is a palindrome.
+
+Return all possible palindrome partitioning of *s*.
+
+**Example**:
+
+    Input: "aab"
+    Output:
+    [
+      ["aa","b"],
+      ["a","a","b"]
+    ]
+
+## 题目大意
+
+给定一个字符串 s，将 s 分割成一些子串，使每个子串都是回文串。返回 s 所有可能的分割方案。
+
+## 解题思路
+
+- 要求输出一个字符串可以被拆成回文串的所有解，DFS 递归求解即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func partition131(s string) [][]string {
+	if s == "" {
+		return [][]string{}
+	}
+	res, pal := [][]string{}, []string{}
+	findPalindrome(s, 0, "", true, pal, &res)
+	return res
+}
+
+func findPalindrome(str string, index int, s string, isPal bool, pal []string, res *[][]string) {
+	if index == len(str) {
+		if isPal {
+			tmp := make([]string, len(pal))
+			copy(tmp, pal)
+			*res = append(*res, tmp)
+		}
+		return
+	}
+	if index == 0 {
+		s = string(str[index])
+		pal = append(pal, s)
+		findPalindrome(str, index+1, s, isPal && isPalindrome131(s), pal, res)
+	} else {
+		temp := pal[len(pal)-1]
+		s = pal[len(pal)-1] + string(str[index])
+		pal[len(pal)-1] = s
+		findPalindrome(str, index+1, s, isPalindrome131(s), pal, res)
+		pal[len(pal)-1] = temp
+		if isPalindrome131(temp) {
+			pal = append(pal, string(str[index]))
+			findPalindrome(str, index+1, temp, isPal && isPalindrome131(temp), pal, res)
+			pal = pal[:len(pal)-1]
+
+		}
+	}
+	return
+}
+
+func isPalindrome131(s string) bool {
+	slen := len(s)
+	for i, j := 0, slen-1; i < j; i, j = i+1, j-1 {
+		if s[i] != s[j] {
+			return false
+		}
+	}
+	return true
+}
+
+// 解法二
+func partition131_1(s string) [][]string {
+	result := [][]string{}
+	size := len(s)
+	if size == 0 {
+		return result
+	}
+	current := make([]string, 0, size)
+	dfs131(s, 0, current, &result)
+	return result
+}
+
+func dfs131(s string, idx int, cur []string, result *[][]string) {
+	start, end := idx, len(s)
+	if start == end {
+		temp := make([]string, len(cur))
+		copy(temp, cur)
+		*result = append(*result, temp)
+		return
+	}
+	for i := start; i < end; i++ {
+		if isPal(s, start, i) {
+			dfs131(s, i+1, append(cur, s[start:i+1]), result)
+		}
+	}
+}
+
+func isPal(str string, s, e int) bool {
+	for s < e {
+		if str[s] != str[e] {
+			return false
+		}
+		s++
+		e--
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0130.Surrounded-Regions/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0135.Candy/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0135.Candy.md b/website/content/ChapterFour/0100~0199/0135.Candy.md
new file mode 100644
index 000000000..d2df666b9
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0135.Candy.md
@@ -0,0 +1,81 @@
+# [135. Candy](https://leetcode.com/problems/candy/)
+
+
+## 题目
+
+There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
+
+You are giving candies to these children subjected to the following requirements:
+
+- Each child must have at least one candy.
+- Children with a higher rating get more candies than their neighbors.
+
+Return *the minimum number of candies you need to have to distribute the candies to the children*.
+
+**Example 1:**
+
+```
+Input: ratings = [1,0,2]
+Output: 5
+Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
+```
+
+**Example 2:**
+
+```
+Input: ratings = [1,2,2]
+Output: 4
+Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
+The third child gets 1 candy because it satisfies the above two conditions.
+```
+
+**Constraints:**
+
+- `n == ratings.length`
+- `1 <= n <= 2 * 10^4`
+- `0 <= ratings[i] <= 2 * 10^4`
+
+## 题目大意
+
+老师想给孩子们分发糖果，有 N 个孩子站成了一条直线，老师会根据每个孩子的表现，预先给他们评分。你需要按照以下要求，帮助老师给这些孩子分发糖果：
+
+- 每个孩子至少分配到 1 个糖果。
+- 评分更高的孩子必须比他两侧的邻位孩子获得更多的糖果。
+
+那么这样下来，老师至少需要准备多少颗糖果呢？
+
+## 解题思路
+
+- 本题的突破口在于，评分更高的孩子必须比他两侧的邻位孩子获得更多的糖果，这句话。这个规则可以理解为 2 条规则，想象成按身高排队，站在下标为 0 的地方往后“看”，评分高即为个子高的，应该比前面个子矮(评分低)的分到糖果多；站在下标为 n - 1 的地方往后“看”，评分高即为个子高的，同样应该比前面个子矮(评分低)的分到糖果多。你可能会有疑问，规则都是一样的，为什么会出现至少需要多少糖果呢？因为可能出现评分一样高的同学。扫描数组两次，处理出每一个学生分别满足左规则或右规则时，最少需要被分得的糖果数量。每个人最终分得的糖果数量即为这两个数量的最大值。两次遍历结束，将所有糖果累加起来即为至少需要准备的糖果数。由于每个人至少分配到 1 个糖果，所以每个人糖果数再加一。
+
+## 代码
+
+```go
+package leetcode
+
+func candy(ratings []int) int {
+	candies := make([]int, len(ratings))
+	for i := 1; i < len(ratings); i++ {
+		if ratings[i] > ratings[i-1] {
+			candies[i] += candies[i-1] + 1
+		}
+	}
+	for i := len(ratings) - 2; i >= 0; i-- {
+		if ratings[i] > ratings[i+1] && candies[i] <= candies[i+1] {
+			candies[i] = candies[i+1] + 1
+		}
+	}
+	total := 0
+	for _, candy := range candies {
+		total += candy + 1
+	}
+	return total
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0131.Palindrome-Partitioning/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0136.Single-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0136.Single-Number.md b/website/content/ChapterFour/0100~0199/0136.Single-Number.md
new file mode 100755
index 000000000..a5801d4e1
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0136.Single-Number.md
@@ -0,0 +1,53 @@
+# [136. Single Number](https://leetcode.com/problems/single-number/)
+
+## 题目
+
+Given a **non-empty** array of integers, every element appears *twice* except for one. Find that single one.
+
+**Note**:
+
+Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
+
+**Example 1**:
+
+    Input: [2,2,1]
+    Output: 1
+
+**Example 2**:
+
+    Input: [4,1,2,1,2]
+    Output: 4
+
+## 题目大意
+
+给定一个非空整数数组，除了某个元素只出现一次以外，其余每个元素均出现两次。找出那个只出现了一次的元素。要求算法时间复杂度是线性的，并且不使用额外的辅助空间。
+
+
+## 解题思路
+
+- 题目要求不能使用辅助空间，并且时间复杂度只能是线性的。
+- 题目为什么要强调有一个数字出现一次，其他的出现两次？我们想到了异或运算的性质：任何一个数字异或它自己都等于0。也就是说，如果我们从头到尾依次异或数组中的每一个数字，那么最终的结果刚好是那个只出现一次的数字，因为那些出现两次的数字全部在异或中抵消掉了。于是最终做法是从头到尾依次异或数组中的每一个数字，那么最终得到的结果就是两个只出现一次的数字的异或结果。因为其他数字都出现了两次，在异或中全部抵消掉了。**利用的性质是 x^x = 0**。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func singleNumber(nums []int) int {
+	result := 0
+	for i := 0; i < len(nums); i++ {
+		result ^= nums[i]
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0135.Candy/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0137.Single-Number-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0137.Single-Number-II.md b/website/content/ChapterFour/0100~0199/0137.Single-Number-II.md
new file mode 100755
index 000000000..9cbca6720
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0137.Single-Number-II.md
@@ -0,0 +1,141 @@
+# [137. Single Number II](https://leetcode.com/problems/single-number-ii/)
+
+
+## 题目
+
+Given a **non-empty** array of integers, every element appears *three* times except for one, which appears exactly once. Find that single one.
+
+**Note**:
+
+Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
+
+**Example 1**:
+
+    Input: [2,2,3,2]
+    Output: 3
+
+**Example 2**:
+
+    Input: [0,1,0,1,0,1,99]
+    Output: 99
+
+
+## 题目大意
+
+给定一个非空整数数组，除了某个元素只出现一次以外，其余每个元素均出现了三次。找出那个只出现了一次的元素。要求算法时间复杂度是线性的，并且不使用额外的辅助空间。
+
+
+
+
+## 解题思路
+
+- 这一题是第 136 题的加强版。这类题也可以扩展，在数组中每个元素都出现 5 次，找出只出现 1 次的数。
+- 本题中要求找出只出现 1 次的数，出现 3 次的数都要被消除。第 136 题是消除出现 2 次的数。这一题也会相当相同的解法，出现 3 次的数也要被消除。定义状态，00、10、01，这 3 个状态。当一个数出现 3 次，那么它每个位置上的 1 出现的次数肯定是 3 的倍数，所以当 1 出现 3 次以后，就归零清除。如何能做到这点呢？仿造`三进制(00，01，10)` 就可以做到。
+- 变量  ones 中记录遍历中每个位上出现 1 的个数。将它与 A[i] 进行异或，目的是：
+    - 每位上两者都是 1 的，表示历史统计结果 ones 出现1次、A[i]中又出现 1 次，则是出现 2 次，需要进位到 twos 变量中。
+    - 每位上两者分别为 0、1 的，加入到 ones 统计结果中。
+    - 最后还要 & ^twos ，是为了能做到三进制，出现 3 次就清零。例如 ones = x，那么 twos = 0，当 twos = x，那么 ones = 0；
+- 变量 twos 中记录遍历中每个位上出现 1 ，2次 的个数。与 A[i] 进行异或的目的和上述描述相同，不再赘述。
+
+> 在 golang 中，&^ 表示 AND NOT 的意思。这里的 ^ 作为一元操作符，表示按位取反 (^0001 0100 = 1110 1011)，X &^ Y 的意思是将 X 中与 Y 相异的位保留，相同的位清零。
+
+> 在 golang 中没有 Java 中的 ~ 位操作运算符，Java 中的 ~ 运算符代表按位取反。这个操作就想当于 golang 中的 ^ 运算符当做一元运算符使用的效果。
+
+|(twos,ones)|xi|(twos'',ones')|ones'|
+|:----:|:----:|:----:|:----:|
+|00|0|00|0|
+|00|1|01|1|
+|01|0|01|1|
+|01|1|10|0|
+|10|0|10|0|
+|10|1|00|0|
+
+- 第一步，先将 ones -> ones'。通过观察可以看出 ones = (ones ^ nums[i]) & ^twos
+
+|(twos,ones')|xi|twos'|
+|:----:|:----:|:----:|
+|00|0|0|
+|01|1|0|
+|01|0|0|
+|00|1|1|
+|10|0|1|
+|10|1|0|
+
+
+- 第二步，再将 twos -> twos'。这一步需要用到前一步的 ones。通过观察可以看出 twos = (twos ^ nums[i]) & ^ones。
+
+--------------------------
+
+这一题还可以继续扩展，在数组中每个元素都出现 5 次，找出只出现 1 次的数。那该怎么做呢？思路还是一样的，模拟一个五进制，5 次就会消除。代码如下：
+
+    // 解法一
+    func singleNumberIII(nums []int) int {
+    	na, nb, nc := 0, 0, 0
+    	for i := 0; i < len(nums); i++ {
+    		nb = nb ^ (nums[i] & na)
+    		na = (na ^ nums[i]) & ^nc
+    		nc = nc ^ (nums[i] & ^na & ^nb)
+    	}
+    	return na & ^nb & ^nc
+    }
+    
+    // 解法二
+    func singleNumberIIII(nums []int) int {
+    	twos, threes, ones := 0xffffffff, 0xffffffff, 0
+    	for i := 0; i < len(nums); i++ {
+    		threes = threes ^ (nums[i] & twos)
+    		twos = (twos ^ nums[i]) & ^ones
+    		ones = ones ^ (nums[i] & ^twos & ^threes)
+    	}
+    	return ones
+    }
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func singleNumberII(nums []int) int {
+	ones, twos := 0, 0
+	for i := 0; i < len(nums); i++ {
+		ones = (ones ^ nums[i]) & ^twos
+		twos = (twos ^ nums[i]) & ^ones
+	}
+	return ones
+}
+
+// 以下是拓展题
+// 在数组中每个元素都出现 5 次，找出只出现 1 次的数。
+
+// 解法一
+func singleNumberIIIII(nums []int) int {
+	na, nb, nc := 0, 0, 0
+	for i := 0; i < len(nums); i++ {
+		nb = nb ^ (nums[i] & na)
+		na = (na ^ nums[i]) & ^nc
+		nc = nc ^ (nums[i] & ^na & ^nb)
+	}
+	return na & ^nb & ^nc
+}
+
+// 解法二
+func singleNumberIIIII1(nums []int) int {
+	twos, threes, ones := 0xffffffff, 0xffffffff, 0
+	for i := 0; i < len(nums); i++ {
+		threes = threes ^ (nums[i] & twos)
+		twos = (twos ^ nums[i]) & ^ones
+		ones = ones ^ (nums[i] & ^twos & ^threes)
+	}
+	return ones
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0136.Single-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0138.Copy-List-With-Random-Pointer/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0138.Copy-List-With-Random-Pointer.md b/website/content/ChapterFour/0100~0199/0138.Copy-List-With-Random-Pointer.md
new file mode 100644
index 000000000..e43924798
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0138.Copy-List-With-Random-Pointer.md
@@ -0,0 +1,137 @@
+# [138. Copy List with Random Pointer](https://leetcode.com/problems/copy-list-with-random-pointer/)
+
+## 题目
+
+A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.
+
+Return a **[deep copy](https://en.wikipedia.org/wiki/Object_copying#Deep_copy)** of the list.
+
+The Linked List is represented in the input/output as a list of `n` nodes. Each node is represented as a pair of `[val, random_index]` where:
+
+- `val`: an integer representing `Node.val`
+- `random_index`: the index of the node (range from `0` to `n-1`) where random pointer points to, or `null` if it does not point to any node.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2019/12/18/e1.png)
+
+```
+Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
+Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]
+
+```
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2019/12/18/e2.png)
+
+```
+Input: head = [[1,1],[2,1]]
+Output: [[1,1],[2,1]]
+
+```
+
+**Example 3:**
+
+![](https://assets.leetcode.com/uploads/2019/12/18/e3.png)
+
+```
+Input: head = [[3,null],[3,0],[3,null]]
+Output: [[3,null],[3,0],[3,null]]
+
+```
+
+**Example 4:**
+
+```
+Input: head = []
+Output: []
+Explanation: Given linked list is empty (null pointer), so return null.
+
+```
+
+**Constraints:**
+
+- `10000 <= Node.val <= 10000`
+- `Node.random` is null or pointing to a node in the linked list.
+- The number of nodes will not exceed 1000.
+
+## 题目大意
+
+给定一个链表，每个节点包含一个额外增加的随机指针，该指针可以指向链表中的任何节点或空节点。要求返回这个链表的 深拷贝。
+
+我们用一个由 n 个节点组成的链表来表示输入/输出中的链表。每个节点用一个 [val, random_index] 表示：
+
+- val：一个表示 Node.val 的整数。
+- random_index：随机指针指向的节点索引（范围从 0 到 n-1）；如果不指向任何节点，则为  null 。
+
+## 解题思路
+
+- 这道题严格意义上是数据结构题，根据给定的数据结构，对它进行深拷贝。
+- 先将每个节点都复制一份，放在它的 next 节点中。如此穿插的复制一份链表。
+
+    ![https://img.halfrost.com/Leetcode/leetcode_138_1_0.png](https://img.halfrost.com/Leetcode/leetcode_138_1_0.png)
+
+    再将穿插版的链表的 random 指针指向正确的位置。
+
+    ![https://img.halfrost.com/Leetcode/leetcode_138_2.png](https://img.halfrost.com/Leetcode/leetcode_138_2.png)
+
+    再将穿插版的链表的 next 指针指向正确的位置。最后分开这交织在一起的两个链表的头节点，即可分开 2 个链表。
+
+    ![https://img.halfrost.com/Leetcode/leetcode_138_3.png](https://img.halfrost.com/Leetcode/leetcode_138_3.png)
+
+## 代码
+
+```go
+package leetcode
+
+// Node define
+type Node struct {
+	Val    int
+	Next   *Node
+	Random *Node
+}
+
+func copyRandomList(head *Node) *Node {
+	if head == nil {
+		return nil
+	}
+	tempHead := copyNodeToLinkedList(head)
+	return splitLinkedList(tempHead)
+}
+
+func splitLinkedList(head *Node) *Node {
+	cur := head
+	head = head.Next
+	for cur != nil && cur.Next != nil {
+		cur.Next, cur = cur.Next.Next, cur.Next
+	}
+	return head
+}
+
+func copyNodeToLinkedList(head *Node) *Node {
+	cur := head
+	for cur != nil {
+		node := &Node{
+			Val:  cur.Val,
+			Next: cur.Next,
+		}
+		cur.Next, cur = node, cur.Next
+	}
+	cur = head
+	for cur != nil {
+		if cur.Random != nil {
+			cur.Next.Random = cur.Random.Next
+		}
+		cur = cur.Next.Next
+	}
+	return head
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0137.Single-Number-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0141.Linked-List-Cycle/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0141.Linked-List-Cycle.md b/website/content/ChapterFour/0100~0199/0141.Linked-List-Cycle.md
new file mode 100644
index 000000000..96c3310bd
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0141.Linked-List-Cycle.md
@@ -0,0 +1,62 @@
+# [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/description/)
+
+## 题目
+
+Given a linked list, determine if it has a cycle in it.
+
+**Follow up**:   
+Can you solve it without using extra space?
+
+
+
+## 题目大意
+
+判断链表是否有环，不能使用额外的空间。
+
+## 解题思路
+
+给 2 个指针，一个指针是另外一个指针的下一个指针。快指针一次走 2 格，慢指针一次走 1 格。如果存在环，那么前一个指针一定会经过若干圈之后追上慢的指针。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func hasCycle(head *ListNode) bool {
+	fast := head
+	slow := head
+	for fast != nil && fast.Next != nil {
+		fast = fast.Next.Next
+		slow = slow.Next
+		if fast == slow {
+			return true
+		}
+	}
+	return false
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0138.Copy-List-With-Random-Pointer/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0142.Linked-List-Cycle-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0142.Linked-List-Cycle-II.md b/website/content/ChapterFour/0100~0199/0142.Linked-List-Cycle-II.md
new file mode 100644
index 000000000..3bd614255
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0142.Linked-List-Cycle-II.md
@@ -0,0 +1,116 @@
+# [142. Linked List Cycle II](https://leetcode.com/problems/linked-list-cycle-ii/)
+
+## 题目
+
+Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
+
+To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.
+
+**Note**: Do not modify the linked list.
+
+**Example 1**:
+
+```
+
+Input: head = [3,2,0,-4], pos = 1
+Output: tail connects to node index 1
+Explanation: There is a cycle in the linked list, where tail connects to the second node.
+
+```
+
+**Example 2**:
+
+```
+
+Input: head = [1,2], pos = 0
+Output: tail connects to node index 0
+Explanation: There is a cycle in the linked list, where tail connects to the first node.
+
+```
+
+**Example 3**:
+
+```
+
+Input: head = [1], pos = -1
+Output: no cycle
+Explanation: There is no cycle in the linked list.
+
+```
+
+
+## 题目大意
+
+判断链表是否有环，不能使用额外的空间。如果有环，输出环的起点指针，如果没有环，则输出空。
+
+## 解题思路
+
+这道题是第 141 题的加强版。在判断是否有环的基础上，还需要输出环的第一个点。
+
+分析一下判断环的原理。fast 指针一次都 2 步，slow 指针一次走 1 步。令链表 head 到环的一个点需要 x1 步，从环的第一个点到相遇点需要 x2 步，从环中相遇点回到环的第一个点需要 x3 步。那么环的总长度是 x2 + x3 步。
+
+fast 和 slow 会相遇，说明他们走的时间是相同的，可以知道他们走的路程有以下的关系：
+
+```c
+fast 的 t = (x1 + x2 + x3 + x2) / 2
+slow 的 t = (x1 + x2) / 1
+
+x1 + x2 + x3 + x2 = 2 * (x1 + x2)
+
+所以 x1 = x3
+```
+
+所以 2 个指针相遇以后，如果 slow 继续往前走，fast 指针回到起点 head，两者都每次走一步，那么必定会在环的起点相遇，相遇以后输出这个点即是结果。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func detectCycle(head *ListNode) *ListNode {
+	if head == nil || head.Next == nil {
+		return nil
+	}
+	isCycle, slow := hasCycle142(head)
+	if !isCycle {
+		return nil
+	}
+	fast := head
+	for fast != slow {
+		fast = fast.Next
+		slow = slow.Next
+	}
+	return fast
+}
+
+func hasCycle142(head *ListNode) (bool, *ListNode) {
+	fast := head
+	slow := head
+	for slow != nil && fast != nil && fast.Next != nil {
+		fast = fast.Next.Next
+		slow = slow.Next
+		if fast == slow {
+			return true, slow
+		}
+	}
+	return false, nil
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0141.Linked-List-Cycle/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0143.Reorder-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0143.Reorder-List.md b/website/content/ChapterFour/0100~0199/0143.Reorder-List.md
new file mode 100644
index 000000000..4e835246c
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0143.Reorder-List.md
@@ -0,0 +1,134 @@
+# [143. Reorder List](https://leetcode.com/problems/reorder-list/)
+
+## 题目
+
+Given a singly linked list L: L0→L1→…→Ln-1→Ln,
+reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
+
+You may not modify the values in the list's nodes, only nodes itself may be changed.
+
+**Example 1**:
+
+```
+
+Given 1->2->3->4, reorder it to 1->4->2->3.
+
+```
+
+**Example 2**:
+
+```
+
+Given 1->2->3->4->5, reorder it to 1->5->2->4->3.
+
+```
+
+## 题目大意
+
+按照指定规则重新排序链表：第一个元素和最后一个元素排列在一起，接着第二个元素和倒数第二个元素排在一起，接着第三个元素和倒数第三个元素排在一起。
+
+
+## 解题思路
+
+
+最近简单的方法是先把链表存储到数组里，然后找到链表中间的结点，按照规则拼接即可。这样时间复杂度是 O(n)，空间复杂度是 O(n)。
+
+更好的做法是结合之前几道题的操作：链表逆序，找中间结点。
+
+先找到链表的中间结点，然后利用逆序区间的操作，如 [第 92 题](https://github.com/halfrost/leetcode-go/tree/master/leetcode/0092.Reverse-Linked-List-II) 里的 reverseBetween() 操作，只不过这里的反转区间是从中点一直到末尾。最后利用 2 个指针，一个指向头结点，一个指向中间结点，开始拼接最终的结果。这种做法的时间复杂度是 O(n)，空间复杂度是 O(1)。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// 解法一 单链表
+func reorderList(head *ListNode) *ListNode {
+	if head == nil || head.Next == nil {
+		return head
+	}
+
+	// 寻找中间结点
+	p1 := head
+	p2 := head
+	for p2.Next != nil && p2.Next.Next != nil {
+		p1 = p1.Next
+		p2 = p2.Next.Next
+	}
+
+	// 反转链表后半部分  1->2->3->4->5->6 to 1->2->3->6->5->4
+	preMiddle := p1
+	preCurrent := p1.Next
+	for preCurrent.Next != nil {
+		current := preCurrent.Next
+		preCurrent.Next = current.Next
+		current.Next = preMiddle.Next
+		preMiddle.Next = current
+	}
+
+	// 重新拼接链表  1->2->3->6->5->4 to 1->6->2->5->3->4
+	p1 = head
+	p2 = preMiddle.Next
+	for p1 != preMiddle {
+		preMiddle.Next = p2.Next
+		p2.Next = p1.Next
+		p1.Next = p2
+		p1 = p2.Next
+		p2 = preMiddle.Next
+	}
+	return head
+}
+
+// 解法二 数组
+func reorderList1(head *ListNode) *ListNode {
+	array := listToArray(head)
+	length := len(array)
+	if length == 0 {
+		return head
+	}
+	cur := head
+	last := head
+	for i := 0; i < len(array)/2; i++ {
+		tmp := &ListNode{Val: array[length-1-i], Next: cur.Next}
+		cur.Next = tmp
+		cur = tmp.Next
+		last = tmp
+	}
+	if length%2 == 0 {
+		last.Next = nil
+	} else {
+		cur.Next = nil
+	}
+	return head
+}
+
+func listToArray(head *ListNode) []int {
+	array := []int{}
+	if head == nil {
+		return array
+	}
+	cur := head
+	for cur != nil {
+		array = append(array, cur.Val)
+		cur = cur.Next
+	}
+	return array
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0142.Linked-List-Cycle-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0144.Binary-Tree-Preorder-Traversal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0144.Binary-Tree-Preorder-Traversal.md b/website/content/ChapterFour/0100~0199/0144.Binary-Tree-Preorder-Traversal.md
new file mode 100644
index 000000000..36f8e02f2
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0144.Binary-Tree-Preorder-Traversal.md
@@ -0,0 +1,119 @@
+# [144. Binary Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal/)
+
+## 题目
+
+Given a binary tree, return the preorder traversal of its nodes' values.
+
+
+
+**Example**:
+
+```
+
+Input: [1,null,2,3]
+   1
+    \
+     2
+    /
+   3
+
+Output: [1,2,3]
+
+```
+
+
+**Follow up**: Recursive solution is trivial, could you do it iteratively?
+
+
+ 
+
+## 题目大意
+
+先根遍历一颗树。
+
+## 解题思路
+
+两种递归的实现方法，见代码。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 递归
+func preorderTraversal(root *TreeNode) []int {
+	res := []int{}
+	if root != nil {
+		res = append(res, root.Val)
+		tmp := preorderTraversal(root.Left)
+		for _, t := range tmp {
+			res = append(res, t)
+		}
+		tmp = preorderTraversal(root.Right)
+		for _, t := range tmp {
+			res = append(res, t)
+		}
+	}
+	return res
+}
+
+// 解法二 递归
+func preorderTraversal1(root *TreeNode) []int {
+	var result []int
+	preorder(root, &result)
+	return result
+}
+
+func preorder(root *TreeNode, output *[]int) {
+	if root != nil {
+		*output = append(*output, root.Val)
+		preorder(root.Left, output)
+		preorder(root.Right, output)
+	}
+}
+
+// 解法三 非递归，用栈模拟递归过程
+func preorderTraversal2(root *TreeNode) []int {
+	if root == nil {
+		return []int{}
+	}
+	stack, res := []*TreeNode{}, []int{}
+	stack = append(stack, root)
+	for len(stack) != 0 {
+		node := stack[len(stack)-1]
+		stack = stack[:len(stack)-1]
+		if node != nil {
+			res = append(res, node.Val)
+		}
+		if node.Right != nil {
+			stack = append(stack, node.Right)
+		}
+		if node.Left != nil {
+			stack = append(stack, node.Left)
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0143.Reorder-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0145.Binary-Tree-Postorder-Traversal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0145.Binary-Tree-Postorder-Traversal.md b/website/content/ChapterFour/0100~0199/0145.Binary-Tree-Postorder-Traversal.md
new file mode 100644
index 000000000..e688f82f1
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0145.Binary-Tree-Postorder-Traversal.md
@@ -0,0 +1,78 @@
+# [145. Binary Tree Postorder Traversal](https://leetcode.com/problems/binary-tree-postorder-traversal/)
+
+## 题目
+
+
+Given a binary tree, return the postorder traversal of its nodes' values.
+
+
+
+**Example**:
+
+```
+
+Input: [1,null,2,3]
+   1
+    \
+     2
+    /
+   3
+
+Output: [3,2,1]
+
+```
+
+
+**Follow up**: Recursive solution is trivial, could you do it iteratively?
+
+
+ 
+
+## 题目大意
+
+后根遍历一颗树。
+
+## 解题思路
+
+递归的实现方法，见代码。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func postorderTraversal(root *TreeNode) []int {
+	var result []int
+	postorder(root, &result)
+	return result
+}
+
+func postorder(root *TreeNode, output *[]int) {
+	if root != nil {
+		postorder(root.Left, output)
+		postorder(root.Right, output)
+		*output = append(*output, root.Val)
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0144.Binary-Tree-Preorder-Traversal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0146.LRU-Cache/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0146.LRU-Cache.md b/website/content/ChapterFour/0100~0199/0146.LRU-Cache.md
new file mode 100644
index 000000000..7c60b34f8
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0146.LRU-Cache.md
@@ -0,0 +1,141 @@
+# [146. LRU Cache](https://leetcode.com/problems/lru-cache/)
+
+## 题目
+
+Design a data structure that follows the constraints of a **[Least Recently Used (LRU) cache](https://en.wikipedia.org/wiki/Cache_replacement_policies#LRU)**.
+
+Implement the `LRUCache` class:
+
+- `LRUCache(int capacity)` Initialize the LRU cache with **positive** size `capacity`.
+- `int get(int key)` Return the value of the `key` if the key exists, otherwise return `1`.
+- `void put(int key, int value)` Update the value of the `key` if the `key` exists. Otherwise, add the `key-value` pair to the cache. If the number of keys exceeds the `capacity` from this operation, **evict** the least recently used key.
+
+**Follow up**:Could you do `get` and `put` in `O(1)` time complexity?
+
+**Example 1**:
+
+```
+Input
+["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
+[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
+Output
+[null, null, null, 1, null, -1, null, -1, 3, 4]
+
+Explanation
+LRUCache lRUCache = new LRUCache(2);
+lRUCache.put(1, 1); // cache is {1=1}
+lRUCache.put(2, 2); // cache is {1=1, 2=2}
+lRUCache.get(1);    // return 1
+lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
+lRUCache.get(2);    // returns -1 (not found)
+lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
+lRUCache.get(1);    // return -1 (not found)
+lRUCache.get(3);    // return 3
+lRUCache.get(4);    // return 4
+
+```
+
+**Constraints**:
+
+- `1 <= capacity <= 3000`
+- `0 <= key <= 3000`
+- `0 <= value <= 104`
+- At most `3 * 104` calls will be made to `get` and `put`.
+
+## 题目大意
+
+运用你所掌握的数据结构，设计和实现一个  LRU (最近最少使用) 缓存机制 。
+实现 LRUCache 类：
+
+- LRUCache(int capacity) 以正整数作为容量 capacity 初始化 LRU 缓存
+- int get(int key) 如果关键字 key 存在于缓存中，则返回关键字的值，否则返回 -1 。
+- void put(int key, int value) 如果关键字已经存在，则变更其数据值；如果关键字不存在，则插入该组「关键字-值」。当缓存容量达到上限时，它应该在写入新数据之前删除最久未使用的数据值，从而为新的数据值留出空间。
+
+进阶：你是否可以在 O(1) 时间复杂度内完成这两种操作？
+
+## 解题思路
+
+- 这一题是 LRU 经典面试题，详细解释见第三章模板。
+
+## 代码
+
+```go
+package leetcode
+
+type LRUCache struct {
+	head, tail *Node
+	Keys       map[int]*Node
+	Cap        int
+}
+
+type Node struct {
+	Key, Val   int
+	Prev, Next *Node
+}
+
+func Constructor(capacity int) LRUCache {
+	return LRUCache{Keys: make(map[int]*Node), Cap: capacity}
+}
+
+func (this *LRUCache) Get(key int) int {
+	if node, ok := this.Keys[key]; ok {
+		this.Remove(node)
+		this.Add(node)
+		return node.Val
+	}
+	return -1
+}
+
+func (this *LRUCache) Put(key int, value int) {
+	if node, ok := this.Keys[key]; ok {
+		node.Val = value
+		this.Remove(node)
+		this.Add(node)
+		return
+	} else {
+		node = &Node{Key: key, Val: value}
+		this.Keys[key] = node
+		this.Add(node)
+	}
+	if len(this.Keys) > this.Cap {
+		delete(this.Keys, this.tail.Key)
+		this.Remove(this.tail)
+	}
+}
+
+func (this *LRUCache) Add(node *Node) {
+	node.Prev = nil
+	node.Next = this.head
+	if this.head != nil {
+		this.head.Prev = node
+	}
+	this.head = node
+	if this.tail == nil {
+		this.tail = node
+		this.tail.Next = nil
+	}
+}
+
+func (this *LRUCache) Remove(node *Node) {
+	if node == this.head {
+		this.head = node.Next
+		node.Next = nil
+		return
+	}
+	if node == this.tail {
+		this.tail = node.Prev
+		node.Prev.Next = nil
+		node.Prev = nil
+		return
+	}
+	node.Prev.Next = node.Next
+	node.Next.Prev = node.Prev
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0145.Binary-Tree-Postorder-Traversal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0147.Insertion-Sort-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0147.Insertion-Sort-List.md b/website/content/ChapterFour/0100~0199/0147.Insertion-Sort-List.md
new file mode 100644
index 000000000..5ef908560
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0147.Insertion-Sort-List.md
@@ -0,0 +1,84 @@
+# [147. Insertion Sort List](https://leetcode.com/problems/insertion-sort-list/)
+
+## 题目
+
+Sort a linked list using insertion sort.
+
+![](https://upload.wikimedia.org/wikipedia/commons/0/0f/Insertion-sort-example-300px.gif)
+
+A graphical example of insertion sort. The partial sorted list (black) initially contains only the first element in the list.
+With each iteration one element (red) is removed from the input data and inserted in-place into the sorted list
+ 
+
+Algorithm of Insertion Sort:
+
+Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list.
+At each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there.
+It repeats until no input elements remain.
+
+**Example 1**:
+
+```
+
+Input: 4->2->1->3
+Output: 1->2->3->4
+
+```
+
+**Example 2**:
+
+```
+
+Input: -1->5->3->4->0
+Output: -1->0->3->4->5
+
+```
+
+## 题目大意
+
+链表的插入排序
+
+## 解题思路
+
+按照题意做即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func insertionSortList(head *ListNode) *ListNode {
+	if head == nil {
+		return head
+	}
+	newHead := &ListNode{Val: 0, Next: nil} // 这里初始化不要直接指向 head，为了下面循环可以统一处理
+	cur, pre := head, newHead
+	for cur != nil {
+		next := cur.Next
+		for pre.Next != nil && pre.Next.Val < cur.Val {
+			pre = pre.Next
+		}
+		cur.Next = pre.Next
+		pre.Next = cur
+		pre = newHead // 归位，重头开始
+		cur = next
+	}
+	return newHead.Next
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0146.LRU-Cache/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0148.Sort-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0148.Sort-List.md b/website/content/ChapterFour/0100~0199/0148.Sort-List.md
new file mode 100644
index 000000000..f92e2b858
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0148.Sort-List.md
@@ -0,0 +1,103 @@
+# [148. Sort List](https://leetcode.com/problems/sort-list/)
+
+## 题目
+
+Sort a linked list in O(n log n) time using constant space complexity.
+
+**Example 1**:
+
+```
+
+Input: 4->2->1->3
+Output: 1->2->3->4
+
+```
+
+**Example 2**:
+
+```
+
+Input: -1->5->3->4->0
+Output: -1->0->3->4->5
+
+```
+
+## 题目大意
+
+链表的排序，要求时间复杂度必须是 O(n log n)，空间复杂度是 O(1)
+
+## 解题思路
+
+这道题只能用归并排序才能符合要求。归并排序需要的 2 个操作在其他题目已经出现过了，取中间点是第 876 题，合并 2 个有序链表是第 21 题。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func sortList(head *ListNode) *ListNode {
+	length := 0
+	cur := head
+	for cur != nil {
+		length++
+		cur = cur.Next
+	}
+	if length <= 1 {
+		return head
+	}
+
+	middleNode := middleNode(head)
+	cur = middleNode.Next
+	middleNode.Next = nil
+	middleNode = cur
+
+	left := sortList(head)
+	right := sortList(middleNode)
+	return mergeTwoLists(left, right)
+}
+
+func middleNode(head *ListNode) *ListNode {
+	if head == nil || head.Next == nil {
+		return head
+	}
+	p1 := head
+	p2 := head
+	for p2.Next != nil && p2.Next.Next != nil {
+		p1 = p1.Next
+		p2 = p2.Next.Next
+	}
+	return p1
+}
+
+func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
+	if l1 == nil {
+		return l2
+	}
+	if l2 == nil {
+		return l1
+	}
+	if l1.Val < l2.Val {
+		l1.Next = mergeTwoLists(l1.Next, l2)
+		return l1
+	}
+	l2.Next = mergeTwoLists(l1, l2.Next)
+	return l2
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0147.Insertion-Sort-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0150.Evaluate-Reverse-Polish-Notation/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0150.Evaluate-Reverse-Polish-Notation.md b/website/content/ChapterFour/0100~0199/0150.Evaluate-Reverse-Polish-Notation.md
new file mode 100644
index 000000000..0e4614d5a
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0150.Evaluate-Reverse-Polish-Notation.md
@@ -0,0 +1,99 @@
+# [150. Evaluate Reverse Polish Notation](https://leetcode.com/problems/evaluate-reverse-polish-notation/)
+
+## 题目
+
+Evaluate the value of an arithmetic expression in Reverse Polish Notation.
+
+Valid operators are +, -, *, /. Each operand may be an integer or another expression.
+
+**Note**:
+
+- Division between two integers should truncate toward zero.
+- The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.
+
+**Example 1**:
+
+```
+
+Input: ["2", "1", "+", "3", "*"]
+Output: 9
+Explanation: ((2 + 1) * 3) = 9
+
+```
+
+**Example 2**:
+
+```
+
+Input: ["4", "13", "5", "/", "+"]
+Output: 6
+Explanation: (4 + (13 / 5)) = 6
+
+```
+**Example 3**:
+
+```
+
+Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
+Output: 22
+Explanation: 
+  ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
+= ((10 * (6 / (12 * -11))) + 17) + 5
+= ((10 * (6 / -132)) + 17) + 5
+= ((10 * 0) + 17) + 5
+= (0 + 17) + 5
+= 17 + 5
+= 22
+
+```
+
+## 题目大意
+
+计算逆波兰表达式。
+
+## 解题思路
+
+这道题就是经典的考察栈的知识的题目。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+)
+
+func evalRPN(tokens []string) int {
+	stack := make([]int, 0, len(tokens))
+	for _, token := range tokens {
+		v, err := strconv.Atoi(token)
+		if err == nil {
+			stack = append(stack, v)
+		} else {
+			num1, num2 := stack[len(stack)-2], stack[len(stack)-1]
+			stack = stack[:len(stack)-2]
+			switch token {
+			case "+":
+				stack = append(stack, num1+num2)
+			case "-":
+				stack = append(stack, num1-num2)
+			case "*":
+				stack = append(stack, num1*num2)
+			case "/":
+				stack = append(stack, num1/num2)
+			}
+		}
+	}
+	return stack[0]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0148.Sort-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0151.Reverse-Words-in-a-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0151.Reverse-Words-in-a-String.md b/website/content/ChapterFour/0100~0199/0151.Reverse-Words-in-a-String.md
new file mode 100755
index 000000000..cb2545f17
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0151.Reverse-Words-in-a-String.md
@@ -0,0 +1,89 @@
+# [151. Reverse Words in a String](https://leetcode.com/problems/reverse-words-in-a-string/)
+
+
+
+## 题目
+
+Given an input string, reverse the string word by word.
+
+**Example 1**:
+
+    Input: "the sky is blue"
+    Output: "blue is sky the"
+
+**Example 2**:
+
+    Input: "  hello world!  "
+    Output: "world! hello"
+    Explanation: Your reversed string should not contain leading or trailing spaces.
+
+**Example 3**:
+
+    Input: "a good   example"
+    Output: "example good a"
+    Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.
+
+**Note**:
+
+- A word is defined as a sequence of non-space characters.
+- Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
+- You need to reduce multiple spaces between two words to a single space in the reversed string.
+
+**Follow up**:
+
+For C programmers, try to solve it *in-place* in *O*(1) extra space.
+
+
+## 题目大意
+
+给定一个字符串，逐个翻转字符串中的每个单词。
+
+说明：
+
+- 无空格字符构成一个单词。
+- 输入字符串可以在前面或者后面包含多余的空格，但是反转后的字符不能包括。
+- 如果两个单词间有多余的空格，将反转后单词间的空格减少到只含一个。
+ 
+
+进阶：
+
+- 请选用 C 语言的用户尝试使用 O(1) 额外空间复杂度的原地解法。
+
+
+## 解题思路
+
+
+- 给出一个中间有空格分隔的字符串，要求把这个字符串按照单词的维度前后翻转。
+- 依照题意，先把字符串按照空格分隔成每个小单词，然后把单词前后翻转，最后再把每个单词中间添加空格。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func reverseWords151(s string) string {
+	ss := strings.Fields(s)
+	reverse151(&ss, 0, len(ss)-1)
+	return strings.Join(ss, " ")
+}
+
+func reverse151(m *[]string, i int, j int) {
+	for i <= j {
+		(*m)[i], (*m)[j] = (*m)[j], (*m)[i]
+		i++
+		j--
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0150.Evaluate-Reverse-Polish-Notation/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0152.Maximum-Product-Subarray/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0152.Maximum-Product-Subarray.md b/website/content/ChapterFour/0100~0199/0152.Maximum-Product-Subarray.md
new file mode 100755
index 000000000..852c8b8a2
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0152.Maximum-Product-Subarray.md
@@ -0,0 +1,59 @@
+# [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)
+
+
+## 题目
+
+Given an integer array `nums`, find the contiguous subarray within an array (containing at least one number) which has the largest product.
+
+**Example 1**:
+
+    Input: [2,3,-2,4]
+    Output: 6
+    Explanation: [2,3] has the largest product 6.
+
+**Example 2**:
+
+    Input: [-2,0,-1]
+    Output: 0
+    Explanation: The result cannot be 2, because [-2,-1] is not a subarray.
+
+
+## 题目大意
+
+给定一个整数数组 nums ，找出一个序列中乘积最大的连续子序列（该序列至少包含一个数）。
+
+
+## 解题思路
+
+- 给出一个数组，要求找出这个数组中连续元素乘积最大的值。
+- 这一题是 DP 的题，状态转移方程是：最大值是 `Max(f(n)) = Max( Max(f(n-1)) * n, Min(f(n-1)) * n)`；最小值是 `Min(f(n)) = Min( Max(f(n-1)) * n, Min(f(n-1)) * n)`。只要动态维护这两个值，如果最后一个数是负数，最大值就在负数 * 最小值中产生，如果最后一个数是正数，最大值就在正数 * 最大值中产生。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func maxProduct(nums []int) int {
+	minimum, maximum, res := nums[0], nums[0], nums[0]
+	for i := 1; i < len(nums); i++ {
+		if nums[i] < 0 {
+			maximum, minimum = minimum, maximum
+		}
+		maximum = max(nums[i], maximum*nums[i])
+		minimum = min(nums[i], minimum*nums[i])
+		res = max(res, maximum)
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0151.Reverse-Words-in-a-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0153.Find-Minimum-in-Rotated-Sorted-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0153.Find-Minimum-in-Rotated-Sorted-Array.md b/website/content/ChapterFour/0100~0199/0153.Find-Minimum-in-Rotated-Sorted-Array.md
new file mode 100755
index 000000000..c48a82ae8
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0153.Find-Minimum-in-Rotated-Sorted-Array.md
@@ -0,0 +1,115 @@
+# [153. Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)
+
+
+## 题目
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., `[0,1,2,4,5,6,7]` might become `[4,5,6,7,0,1,2]`).
+
+Find the minimum element.
+
+You may assume no duplicate exists in the array.
+
+**Example 1**:
+
+    Input: [3,4,5,1,2] 
+    Output: 1
+
+**Example 2**:
+
+    Input: [4,5,6,7,0,1,2]
+    Output: 0
+
+
+## 题目大意
+
+假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。请找出其中最小的元素。
+
+你可以假设数组中不存在重复元素。
+
+
+## 解题思路
+
+- 给出一个原本从小到大排序过的数组，但是在某一个分割点上，把数组切分后的两部分对调位置，数值偏大的放到了数组的前部。求这个数组中最小的元素。
+- 求数组最小的元素其实就是找分割点，前一个数比当前数大，后一个数比当前数也要大。可以用二分搜索查找，需要查找的两个有序区间。时间复杂度  O(log n)。这一题也可以用暴力解法，从头开始遍历，动态维护一个最小值即可，时间复杂度 O(n)。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 二分
+func findMin(nums []int) int {
+	low, high := 0, len(nums)-1
+	for low < high {
+		if nums[low] < nums[high] {
+			return nums[low]
+		}
+		mid := low + (high-low)>>1
+		if nums[mid] >= nums[low] {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return nums[low]
+}
+
+// 解法二 二分
+func findMin1(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	if len(nums) == 1 {
+		return nums[0]
+	}
+	if nums[len(nums)-1] > nums[0] {
+		return nums[0]
+	}
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[low] < nums[high] {
+			return nums[low]
+		}
+		if (mid == len(nums)-1 && nums[mid-1] > nums[mid]) || (mid < len(nums)-1 && mid > 0 && nums[mid-1] > nums[mid] && nums[mid] < nums[mid+1]) {
+			return nums[mid]
+		}
+		if nums[mid] > nums[low] && nums[low] > nums[high] { // mid 在数值大的一部分区间里
+			low = mid + 1
+		} else if nums[mid] < nums[low] && nums[low] > nums[high] { // mid 在数值小的一部分区间里
+			high = mid - 1
+		} else {
+			if nums[low] == nums[mid] {
+				low++
+			}
+			if nums[high] == nums[mid] {
+				high--
+			}
+		}
+	}
+	return -1
+}
+
+// 解法三 暴力
+func findMin2(nums []int) int {
+	min := nums[0]
+	for _, num := range nums[1:] {
+		if min > num {
+			min = num
+		}
+	}
+	return min
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0152.Maximum-Product-Subarray/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0154.Find-Minimum-in-Rotated-Sorted-Array-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0154.Find-Minimum-in-Rotated-Sorted-Array-II.md b/website/content/ChapterFour/0100~0199/0154.Find-Minimum-in-Rotated-Sorted-Array-II.md
new file mode 100755
index 000000000..7d2c1235f
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0154.Find-Minimum-in-Rotated-Sorted-Array-II.md
@@ -0,0 +1,74 @@
+# [154. Find Minimum in Rotated Sorted Array II](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/)
+
+
+## 题目
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., `[0,1,2,4,5,6,7]` might become `[4,5,6,7,0,1,2]`).
+
+Find the minimum element.
+
+The array may contain duplicates.
+
+**Example 1**:
+
+    Input: [1,3,5]
+    Output: 1
+
+**Example 2**:
+
+    Input: [2,2,2,0,1]
+    Output: 0
+
+**Note**:
+
+- This is a follow up problem to [Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/).
+- Would allow duplicates affect the run-time complexity? How and why?
+
+
+## 题目大意
+
+假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。请找出其中最小的元素。
+
+注意数组中可能存在重复的元素。
+
+## 解题思路
+
+
+- 给出一个原本从小到大排序过的数组，注意数组中有重复的元素。但是在某一个分割点上，把数组切分后的两部分对调位置，数值偏大的放到了数组的前部。求这个数组中最小的元素。
+- 这一题是第 153 题的加强版，增加了重复元素的条件。但是实际做法还是没有变，还是用二分搜索，只不过在相等元素上多增加一个判断即可。时间复杂度 O(log n)。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findMin154(nums []int) int {
+	low, high := 0, len(nums)-1
+	for low < high {
+		if nums[low] < nums[high] {
+			return nums[low]
+		}
+		mid := low + (high-low)>>1
+		if nums[mid] > nums[low] {
+			low = mid + 1
+		} else if nums[mid] == nums[low] {
+			low++
+		} else {
+			high = mid
+		}
+	}
+	return nums[low]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0153.Find-Minimum-in-Rotated-Sorted-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0155.Min-Stack/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0155.Min-Stack.md b/website/content/ChapterFour/0100~0199/0155.Min-Stack.md
new file mode 100644
index 000000000..606f65003
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0155.Min-Stack.md
@@ -0,0 +1,94 @@
+# [155. Min Stack](https://leetcode.com/problems/min-stack/)
+
+## 题目
+
+Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
+
+push(x) -- Push element x onto stack.
+pop() -- Removes the element on top of the stack.
+top() -- Get the top element.
+getMin() -- Retrieve the minimum element in the stack.
+
+
+
+**Example**:
+
+```
+
+MinStack minStack = new MinStack();
+minStack.push(-2);
+minStack.push(0);
+minStack.push(-3);
+minStack.getMin();   --> Returns -3.
+minStack.pop();
+minStack.top();      --> Returns 0.
+minStack.getMin();   --> Returns -2.
+
+```
+
+## 题目大意
+
+这道题是一个数据结构实现题。要求实现一个栈的类，实现 push()、pop()、top()、getMin()。
+
+
+## 解题思路
+
+按照题目要求实现即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+// MinStack define
+type MinStack struct {
+	elements, min []int
+	l             int
+}
+
+/** initialize your data structure here. */
+
+// Constructor155 define
+func Constructor155() MinStack {
+	return MinStack{make([]int, 0), make([]int, 0), 0}
+}
+
+// Push define
+func (this *MinStack) Push(x int) {
+	this.elements = append(this.elements, x)
+	if this.l == 0 {
+		this.min = append(this.min, x)
+	} else {
+		min := this.GetMin()
+		if x < min {
+			this.min = append(this.min, x)
+		} else {
+			this.min = append(this.min, min)
+		}
+	}
+	this.l++
+}
+
+func (this *MinStack) Pop() {
+	this.l--
+	this.min = this.min[:this.l]
+	this.elements = this.elements[:this.l]
+}
+
+func (this *MinStack) Top() int {
+	return this.elements[this.l-1]
+}
+
+func (this *MinStack) GetMin() int {
+	return this.min[this.l-1]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0154.Find-Minimum-in-Rotated-Sorted-Array-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0160.Intersection-of-Two-Linked-Lists/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0160.Intersection-of-Two-Linked-Lists.md b/website/content/ChapterFour/0100~0199/0160.Intersection-of-Two-Linked-Lists.md
new file mode 100644
index 000000000..77ea4b624
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0160.Intersection-of-Two-Linked-Lists.md
@@ -0,0 +1,123 @@
+# [160. Intersection of Two Linked Lists](https://leetcode.com/problems/intersection-of-two-linked-lists/)
+
+## 题目
+
+Write a program to find the node at which the intersection of two singly linked lists begins.
+
+For example, the following two linked lists:
+
+![](https://assets.leetcode.com/uploads/2018/12/13/160_statement.png)
+
+begin to intersect at node c1.
+
+**Example 1**:
+
+![](https://assets.leetcode.com/uploads/2018/12/13/160_example_1.png)
+
+```
+
+Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
+Output: Reference of the node with value = 8
+Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
+
+```
+
+**Example 2**:
+
+![](https://assets.leetcode.com/uploads/2018/12/13/160_example_2.png)
+
+```
+
+Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
+Output: Reference of the node with value = 2
+Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.
+
+```
+
+
+**Example 3**:
+
+![](https://assets.leetcode.com/uploads/2018/12/13/160_example_3.png)
+
+```
+
+Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
+Output: null
+Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
+Explanation: The two lists do not intersect, so return null.
+
+```
+
+**Notes**:
+
+- If the two linked lists have no intersection at all, return null.
+- The linked lists must retain their original structure after the function returns.
+- You may assume there are no cycles anywhere in the entire linked structure.
+- Your code should preferably run in O(n) time and use only O(1) memory.
+
+## 题目大意
+
+找到 2 个链表的交叉点。
+
+
+## 解题思路
+
+这道题的思路其实类似链表找环。
+
+
+给定的 2 个链表的长度如果一样长，都从头往后扫即可。如果不一样长，需要先“拼成”一样长。把 B 拼接到 A 后面，把 A 拼接到 B 后面。这样 2 个链表的长度都是 A + B。再依次扫描比较 2 个链表的结点是否相同。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "fmt"
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func getIntersectionNode(headA, headB *ListNode) *ListNode {
+	//boundary check
+	if headA == nil || headB == nil {
+		return nil
+	}
+
+	a := headA
+	b := headB
+
+	//if a & b have different len, then we will stop the loop after second iteration
+	for a != b {
+		//for the end of first iteration, we just reset the pointer to the head of another linkedlist
+		if a == nil {
+			a = headB
+		} else {
+			a = a.Next
+		}
+
+		if b == nil {
+			b = headA
+		} else {
+			b = b.Next
+		}
+		fmt.Printf("a = %v b = %v\n", a, b)
+	}
+	return a
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0155.Min-Stack/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0162.Find-Peak-Element/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0162.Find-Peak-Element.md b/website/content/ChapterFour/0100~0199/0162.Find-Peak-Element.md
new file mode 100755
index 000000000..1c3e810fe
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0162.Find-Peak-Element.md
@@ -0,0 +1,101 @@
+# [162. Find Peak Element](https://leetcode.com/problems/find-peak-element/)
+
+
+## 题目
+
+A peak element is an element that is greater than its neighbors.
+
+Given an input array `nums`, where `nums[i] ≠ nums[i+1]`, find a peak element and return its index.
+
+The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
+
+You may imagine that `nums[-1] = nums[n] = -∞`.
+
+**Example 1**:
+
+    Input: nums = [1,2,3,1]
+    Output: 2
+    Explanation: 3 is a peak element and your function should return the index number 2.
+
+**Example 2**:
+
+    Input: nums = [1,2,1,3,5,6,4]
+    Output: 1 or 5 
+    Explanation: Your function can return either index number 1 where the peak element is 2, 
+                 or index number 5 where the peak element is 6.
+
+**Note**:
+
+Your solution should be in logarithmic complexity.
+
+## 题目大意
+
+峰值元素是指其值大于左右相邻值的元素。给定一个输入数组 nums，其中 nums[i] ≠ nums[i+1]，找到峰值元素并返回其索引。数组可能包含多个峰值，在这种情况下，返回任何一个峰值所在位置即可。你可以假设 nums[-1] = nums[n] = -∞。
+
+说明:
+
+- 你的解法应该是 O(logN) 时间复杂度的。
+
+
+## 解题思路
+
+- 给出一个数组，数组里面存在多个“山峰”，(山峰的定义是，下标 `i` 比 `i-1`、`i+1` 位置上的元素都要大)，找到这个“山峰”，并输出其中一个山峰的下标。
+- 这一题是第 852 题的伪加强版，第 852 题中只存在一个山峰，这一题存在多个山峰。但是实际上搜索的代码是一样的，因为此题只要求随便输出一个山峰的下标即可。思路同第 852 题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 二分
+func findPeakElement(nums []int) int {
+	if len(nums) == 0 || len(nums) == 1 {
+		return 0
+	}
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if (mid == len(nums)-1 && nums[mid-1] < nums[mid]) || (mid > 0 && nums[mid-1] < nums[mid] && (mid <= len(nums)-2 && nums[mid+1] < nums[mid])) || (mid == 0 && nums[1] < nums[0]) {
+			return mid
+		}
+		if mid > 0 && nums[mid-1] < nums[mid] {
+			low = mid + 1
+		}
+		if mid > 0 && nums[mid-1] > nums[mid] {
+			high = mid - 1
+		}
+		if mid == low {
+			low++
+		}
+		if mid == high {
+			high--
+		}
+	}
+	return -1
+}
+
+// 解法二 二分
+func findPeakElement1(nums []int) int {
+	low, high := 0, len(nums)-1
+	for low < high {
+		mid := low + (high-low)>>1
+		// 如果 mid 较大，则左侧存在峰值，high = m，如果 mid + 1 较大，则右侧存在峰值，low = mid + 1
+		if nums[mid] > nums[mid+1] {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0160.Intersection-of-Two-Linked-Lists/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0164.Maximum-Gap/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0164.Maximum-Gap.md b/website/content/ChapterFour/0100~0199/0164.Maximum-Gap.md
new file mode 100644
index 000000000..f5ff4d4aa
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0164.Maximum-Gap.md
@@ -0,0 +1,144 @@
+# [164. Maximum Gap](https://leetcode.com/problems/maximum-gap/)
+
+## 题目
+
+Given an unsorted array, find the maximum difference between the successive elements in its sorted form.
+
+Return 0 if the array contains less than 2 elements.
+
+**Example 1**:
+
+```
+
+Input: [3,6,9,1]
+Output: 3
+Explanation: The sorted form of the array is [1,3,6,9], either
+             (3,6) or (6,9) has the maximum difference 3.
+             
+```
+
+**Example 2**:
+
+```
+
+Input: [10]
+Output: 0
+Explanation: The array contains less than 2 elements, therefore return 0.
+
+```
+
+
+**Note**:
+
+- You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
+- Try to solve it in linear time/space.
+
+
+## 题目大意
+
+在数组中找到 2 个数字之间最大的间隔。要求尽量用 O(1) 的时间复杂度和空间复杂度。
+
+## 解题思路
+
+虽然使用排序算法可以 AC 这道题。先排序，然后依次计算数组中两两数字之间的间隔，找到最大的一个间隔输出即可。
+
+这道题满足要求的做法是基数排序。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 快排
+func maximumGap(nums []int) int {
+	if len(nums) < 2 {
+		return 0
+	}
+	quickSort164(nums, 0, len(nums)-1)
+	res := 0
+	for i := 0; i < len(nums)-1; i++ {
+		if (nums[i+1] - nums[i]) > res {
+			res = nums[i+1] - nums[i]
+		}
+	}
+	return res
+}
+
+func partition164(a []int, lo, hi int) int {
+	pivot := a[hi]
+	i := lo - 1
+	for j := lo; j < hi; j++ {
+		if a[j] < pivot {
+			i++
+			a[j], a[i] = a[i], a[j]
+		}
+	}
+	a[i+1], a[hi] = a[hi], a[i+1]
+	return i + 1
+}
+func quickSort164(a []int, lo, hi int) {
+	if lo >= hi {
+		return
+	}
+	p := partition164(a, lo, hi)
+	quickSort164(a, lo, p-1)
+	quickSort164(a, p+1, hi)
+}
+
+// 解法二 基数排序
+func maximumGap1(nums []int) int {
+	if nums == nil || len(nums) < 2 {
+		return 0
+	}
+	// m is the maximal number in nums
+	m := nums[0]
+	for i := 1; i < len(nums); i++ {
+		m = max(m, nums[i])
+	}
+	exp := 1 // 1, 10, 100, 1000 ...
+	R := 10  // 10 digits
+	aux := make([]int, len(nums))
+	for (m / exp) > 0 { // Go through all digits from LSB to MSB
+		count := make([]int, R)
+		for i := 0; i < len(nums); i++ {
+			count[(nums[i]/exp)%10]++
+		}
+		for i := 1; i < len(count); i++ {
+			count[i] += count[i-1]
+		}
+		for i := len(nums) - 1; i >= 0; i-- {
+			tmp := count[(nums[i]/exp)%10]
+			tmp--
+			aux[tmp] = nums[i]
+			count[(nums[i]/exp)%10] = tmp
+		}
+		for i := 0; i < len(nums); i++ {
+			nums[i] = aux[i]
+		}
+		exp *= 10
+	}
+	maxValue := 0
+	for i := 1; i < len(aux); i++ {
+		maxValue = max(maxValue, aux[i]-aux[i-1])
+	}
+	return maxValue
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0162.Find-Peak-Element/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0167.Two-Sum-II-Input-array-is-sorted/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0167.Two-Sum-II-Input-array-is-sorted.md b/website/content/ChapterFour/0100~0199/0167.Two-Sum-II-Input-array-is-sorted.md
new file mode 100644
index 000000000..18fd28620
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0167.Two-Sum-II-Input-array-is-sorted.md
@@ -0,0 +1,76 @@
+# [167. Two Sum II - Input array is sorted](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/)
+
+## 题目
+
+Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
+
+The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.
+
+**Note**:
+
+- Your returned answers (both index1 and index2) are not zero-based.
+- You may assume that each input would have exactly one solution and you may not use the same element twice.
+
+**Example**:
+
+```
+
+Input: numbers = [2,7,11,15], target = 9
+Output: [1,2]
+Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
+
+```
+
+## 题目大意
+
+找出两个数之和等于 target 的两个数字，要求输出它们的下标。注意一个数字不能使用 2 次。下标从小到大输出。假定题目一定有一个解。
+
+## 解题思路
+
+这一题比第 1 题 Two Sum 的问题还要简单，因为这里数组是有序的。可以直接用第一题的解法解决这道题。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 这一题可以利用数组有序的特性
+func twoSum167(numbers []int, target int) []int {
+	i, j := 0, len(numbers)-1
+	for i < j {
+		if numbers[i]+numbers[j] == target {
+			return []int{i + 1, j + 1}
+		}
+		if numbers[i]+numbers[j] < target {
+			i++
+		} else {
+			j--
+		}
+	}
+	return nil
+}
+
+// 解法二 不管数组是否有序，空间复杂度比上一种解法要多 O(n)
+func twoSum167_1(numbers []int, target int) []int {
+	m := make(map[int]int)
+	for i := 0; i < len(numbers); i++ {
+		another := target - numbers[i]
+		if idx, ok := m[another]; ok {
+			return []int{idx + 1, i + 1}
+		}
+		m[numbers[i]] = i
+	}
+	return nil
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0164.Maximum-Gap/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0168.Excel-Sheet-Column-Title/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0168.Excel-Sheet-Column-Title.md b/website/content/ChapterFour/0100~0199/0168.Excel-Sheet-Column-Title.md
new file mode 100644
index 000000000..6cfe377e7
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0168.Excel-Sheet-Column-Title.md
@@ -0,0 +1,87 @@
+# [168. Excel Sheet Column Title](https://leetcode.com/problems/excel-sheet-column-title/)
+
+## 题目
+
+Given a positive integer, return its corresponding column title as appear in an Excel sheet.
+
+For example:
+
+```
+    1 -> A
+    2 -> B
+    3 -> C
+    ...
+    26 -> Z
+    27 -> AA
+    28 -> AB 
+    ...
+```
+
+**Example 1**:
+
+```
+Input: 1
+Output: "A"
+```
+
+**Example 2**:
+
+```
+Input: 28
+Output: "AB"
+```
+
+**Example 3**:
+
+```
+Input: 701
+Output: "ZY"
+```
+
+## 题目大意
+
+给定一个正整数，返回它在 Excel 表中相对应的列名称。
+
+例如，
+
+    1 -> A
+    2 -> B
+    3 -> C
+    ...
+    26 -> Z
+    27 -> AA
+    28 -> AB 
+    ...
+
+
+## 解题思路
+
+- 给定一个正整数，返回它在 Excel 表中的对应的列名称
+- 简单题。这一题就类似短除法的计算过程。以 26 进制的字母编码。按照短除法先除，然后余数逆序输出即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func convertToTitle(n int) string {
+	result := []byte{}
+	for n > 0 {
+		result = append(result, 'A'+byte((n-1)%26))
+		n = (n - 1) / 26
+	}
+	for i, j := 0, len(result)-1; i < j; i, j = i+1, j-1 {
+		result[i], result[j] = result[j], result[i]
+	}
+	return string(result)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0167.Two-Sum-II-Input-array-is-sorted/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0169.Majority-Element/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0169.Majority-Element.md b/website/content/ChapterFour/0100~0199/0169.Majority-Element.md
new file mode 100755
index 000000000..7c24db72a
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0169.Majority-Element.md
@@ -0,0 +1,74 @@
+# [169. Majority Element](https://leetcode.com/problems/majority-element/)
+
+
+## 题目
+
+Given an array of size n, find the majority element. The majority element is the element that appears **more than** `⌊ n/2 ⌋` times.
+
+You may assume that the array is non-empty and the majority element always exist in the array.
+
+**Example 1**:
+
+    Input: [3,2,3]
+    Output: 3
+
+**Example 2**:
+
+    Input: [2,2,1,1,1,2,2]
+    Output: 2
+
+## 题目大意
+
+
+给定一个大小为 n 的数组，找到其中的众数。众数是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素。你可以假设数组是非空的，并且给定的数组总是存在众数。
+
+
+## 解题思路
+
+- 题目要求找出数组中出现次数大于 `⌊ n/2 ⌋` 次的数。要求空间复杂度为 O(1)。简单题。
+- 这一题利用的算法是 Boyer-Moore Majority Vote Algorithm。[https://www.zhihu.com/question/49973163/answer/235921864](https://www.zhihu.com/question/49973163/answer/235921864)
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 时间复杂度 O(n) 空间复杂度 O(1)
+func majorityElement(nums []int) int {
+	res, count := nums[0], 0
+	for i := 0; i < len(nums); i++ {
+		if count == 0 {
+			res, count = nums[i], 1
+		} else {
+			if nums[i] == res {
+				count++
+			} else {
+				count--
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 时间复杂度 O(n) 空间复杂度 O(n)
+func majorityElement1(nums []int) int {
+	m := make(map[int]int)
+	for _, v := range nums {
+		m[v]++
+		if m[v] > len(nums)/2 {
+			return v
+		}
+	}
+	return 0
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0168.Excel-Sheet-Column-Title/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0171.Excel-Sheet-Column-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0171.Excel-Sheet-Column-Number.md b/website/content/ChapterFour/0100~0199/0171.Excel-Sheet-Column-Number.md
new file mode 100644
index 000000000..06ec111e0
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0171.Excel-Sheet-Column-Number.md
@@ -0,0 +1,74 @@
+# [171. Excel Sheet Column Number](https://leetcode.com/problems/excel-sheet-column-number/)
+
+
+## 题目
+
+Given a column title as appear in an Excel sheet, return its corresponding column number.
+
+For example:
+
+```
+    A -> 1
+    B -> 2
+    C -> 3
+    ...
+    Z -> 26
+    AA -> 27
+    AB -> 28 
+    ...
+```
+
+**Example 1**:
+
+```
+Input: "A"
+Output: 1
+```
+
+**Example 2**:
+
+```
+Input: "AB"
+Output: 28
+```
+
+**Example 3**:
+
+```
+Input: "ZY"
+Output: 701
+```
+
+## 题目大意
+
+给定一个 Excel 表格中的列名称，返回其相应的列序号。
+
+
+## 解题思路
+
+- 给出 Excel 中列的名称，输出其对应的列序号。
+- 简单题。这一题是第 168 题的逆序题。按照 26 进制还原成十进制即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func titleToNumber(s string) int {
+	val, res := 0, 0
+	for i := 0; i < len(s); i++ {
+		val = int(s[i] - 'A' + 1)
+		res = res*26 + val
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0169.Majority-Element/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0172.Factorial-Trailing-Zeroes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0172.Factorial-Trailing-Zeroes.md b/website/content/ChapterFour/0100~0199/0172.Factorial-Trailing-Zeroes.md
new file mode 100755
index 000000000..4ac18a762
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0172.Factorial-Trailing-Zeroes.md
@@ -0,0 +1,58 @@
+# [172. Factorial Trailing Zeroes](https://leetcode.com/problems/factorial-trailing-zeroes/)
+
+
+## 题目
+
+Given an integer n, return the number of trailing zeroes in n!.
+
+**Example 1**:
+
+    Input: 3
+    Output: 0
+    Explanation: 3! = 6, no trailing zero.
+
+**Example 2**:
+
+    Input: 5
+    Output: 1
+    Explanation: 5! = 120, one trailing zero.
+
+**Note**: Your solution should be in logarithmic time complexity.
+
+
+## 题目大意
+
+
+给定一个整数 n，返回 n! 结果尾数中零的数量。说明: 你算法的时间复杂度应为 O(log n) 。
+
+
+
+
+## 解题思路
+
+- 给出一个数 n，要求 n！末尾 0 的个数。
+- 这是一道数学题。计算 N 的阶乘有多少个后缀 0，即计算 N! 里有多少个 10，也是计算 N! 里有多少个 2 和 5（分解质因数），最后结果即 2 的个数和 5 的个数取较小值。每两个数字就会多一个质因数 2，而每五个数字才多一个质因数 5。每 5 个数字就会多一个质因数 5。0~4 的阶乘里没有质因数 5，5~9 的阶乘里有 1 个质因数 5，10~14 的阶乘里有 2 个质因数 5，依此类推。所以 0 的个数即为 `min(阶乘中 5 的个数和 2 的个数)`。
+- N! 有多少个后缀 0，即 N! 有多少个质因数 5。N! 有多少个质因数 5，即 N 可以划分成多少组 5个数字一组，加上划分成多少组 25 个数字一组，加上划分多少组成 125 个数字一组，等等。即 `res = N/5 + N/(5^2) + N/(5^3) + ... = ((N / 5) / 5) / 5 /...` 。最终算法复杂度为 O(logN)。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func trailingZeroes(n int) int {
+	if n/5 == 0 {
+		return 0
+	}
+	return n/5 + trailingZeroes(n/5)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0171.Excel-Sheet-Column-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0173.Binary-Search-Tree-Iterator/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0173.Binary-Search-Tree-Iterator.md b/website/content/ChapterFour/0100~0199/0173.Binary-Search-Tree-Iterator.md
new file mode 100755
index 000000000..559f61217
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0173.Binary-Search-Tree-Iterator.md
@@ -0,0 +1,140 @@
+# [173. Binary Search Tree Iterator](https://leetcode.com/problems/binary-search-tree-iterator/)
+
+
+## 题目
+
+Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.
+
+Calling `next()` will return the next smallest number in the BST.
+
+**Example**:
+
+![](https://assets.leetcode.com/uploads/2018/12/25/bst-tree.png)
+
+    BSTIterator iterator = new BSTIterator(root);
+    iterator.next();    // return 3
+    iterator.next();    // return 7
+    iterator.hasNext(); // return true
+    iterator.next();    // return 9
+    iterator.hasNext(); // return true
+    iterator.next();    // return 15
+    iterator.hasNext(); // return true
+    iterator.next();    // return 20
+    iterator.hasNext(); // return false
+
+**Note**:
+
+- `next()` and `hasNext()` should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
+- You may assume that `next()` call will always be valid, that is, there will be at least a next smallest number in the BST when `next()` is called.
+
+
+## 题目大意
+
+实现一个二叉搜索树迭代器。你将使用二叉搜索树的根节点初始化迭代器。调用 next() 将返回二叉搜索树中的下一个最小的数。
+
+## 解题思路
+
+- 用优先队列解决即可
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"container/heap"
+
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// BSTIterator define
+type BSTIterator struct {
+	pq    PriorityQueueOfInt
+	count int
+}
+
+// Constructor173 define
+func Constructor173(root *TreeNode) BSTIterator {
+	result, pq := []int{}, PriorityQueueOfInt{}
+	postorder(root, &result)
+	for _, v := range result {
+		heap.Push(&pq, v)
+	}
+	bs := BSTIterator{pq: pq, count: len(result)}
+	return bs
+}
+
+func postorder(root *TreeNode, output *[]int) {
+	if root != nil {
+		postorder(root.Left, output)
+		postorder(root.Right, output)
+		*output = append(*output, root.Val)
+	}
+}
+
+/** @return the next smallest number */
+func (this *BSTIterator) Next() int {
+	this.count--
+	return heap.Pop(&this.pq).(int)
+}
+
+/** @return whether we have a next smallest number */
+func (this *BSTIterator) HasNext() bool {
+	return this.count != 0
+}
+
+/**
+ * Your BSTIterator object will be instantiated and called as such:
+ * obj := Constructor(root);
+ * param_1 := obj.Next();
+ * param_2 := obj.HasNext();
+ */
+type PriorityQueueOfInt []int
+
+func (pq PriorityQueueOfInt) Len() int {
+	return len(pq)
+}
+
+func (pq PriorityQueueOfInt) Less(i, j int) bool {
+	return pq[i] < pq[j]
+}
+
+func (pq PriorityQueueOfInt) Swap(i, j int) {
+	pq[i], pq[j] = pq[j], pq[i]
+}
+
+func (pq *PriorityQueueOfInt) Push(x interface{}) {
+	item := x.(int)
+	*pq = append(*pq, item)
+}
+
+func (pq *PriorityQueueOfInt) Pop() interface{} {
+	n := len(*pq)
+	item := (*pq)[n-1]
+	*pq = (*pq)[:n-1]
+	return item
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0172.Factorial-Trailing-Zeroes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0174.Dungeon-Game/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0174.Dungeon-Game.md b/website/content/ChapterFour/0100~0199/0174.Dungeon-Game.md
new file mode 100755
index 000000000..30df3397c
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0174.Dungeon-Game.md
@@ -0,0 +1,142 @@
+# [174. Dungeon Game](https://leetcode.com/problems/dungeon-game/)
+
+## 题目
+
+The demons had captured the princess (**P**) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (**K**) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
+
+The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
+
+Some of the rooms are guarded by demons, so the knight loses health (*negative* integers) upon entering these rooms; other rooms are either empty (*0's*) or contain magic orbs that increase the knight's health (*positive* integers).
+
+In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
+
+**Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.**
+
+For example, given the dungeon below, the initial health of the knight must be at least **7** if he follows the optimal path `RIGHT-> RIGHT -> DOWN -> DOWN`.
+
+
+![](https://img.halfrost.com/Leetcode/leetcode_174_0.png)
+
+**Note**:
+
+- The knight's health has no upper bound.
+- Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.
+
+
+## 题目大意
+
+一些恶魔抓住了公主（P）并将她关在了地下城的右下角。地下城是由 M x N 个房间组成的二维网格。我们英勇的骑士（K）最初被安置在左上角的房间里，他必须穿过地下城并通过对抗恶魔来拯救公主。
+
+骑士的初始健康点数为一个正整数。如果他的健康点数在某一时刻降至 0 或以下，他会立即死亡。
+
+有些房间由恶魔守卫，因此骑士在进入这些房间时会失去健康点数（若房间里的值为负整数，则表示骑士将损失健康点数）；其他房间要么是空的（房间里的值为 0），要么包含增加骑士健康点数的魔法球（若房间里的值为正整数，则表示骑士将增加健康点数）。
+
+为了尽快到达公主，骑士决定每次只向右或向下移动一步。编写一个函数来计算确保骑士能够拯救到公主所需的最低初始健康点数。
+
+说明:
+
+- 骑士的健康点数没有上限。
+- 任何房间都可能对骑士的健康点数造成威胁，也可能增加骑士的健康点数，包括骑士进入的左上角房间以及公主被监禁的右下角房间。
+
+## 解题思路
+
+- 在二维地图上给出每个格子扣血数，负数代表扣血，正数代表补血。左上角第一个格子是起点，右下角最后一个格子是终点。问骑士初始最少多少血才能走完迷宫，顺利营救位于终点的公主。需要注意的是，起点和终点都会对血量进行影响。每到一个格子，骑士的血都不能少于 1，一旦少于 1 点血，骑士就会死去。
+- 这一题首先想到的解题思路是动态规划。从终点逆推回起点。`dp[i][j]` 代表骑士进入坐标为 `(i,j)` 的格子之前最少的血量值。 那么 `dp[m-1][n-1]` 应该同时满足两个条件，`dp[m-1][n-1] + dungeon[m-1][n-1] ≥ 1` 并且 `dp[m-1][n-1] ≥ 1`，由于这两个不等式的方向是相同的，取交集以后，起决定作用的是数轴最右边的数，即 `max(1-dungeon[m-1][n-1] , 1)`。算出 `dp[m-1][n-1]` 以后，接着可以推出 `dp[m-1][i]` 这一行和 `dp[i][n-1]` 这一列的值。因为骑士只能往右走和往下走。往回推，即只能往上走和往左走。到这里，DP 的初始条件都准备好了。那么状态转移方程是什么呢？分析一般的情况，`dp[i][j]` 这个值应该是和 `dp[i+1][j]` 和 `dp[i][j+1]` 这两者有关系。即 `dp[i][j]` 经过自己本格子的扣血以后，要能至少满足下一行和右一列格子血量的最少要求。并且自己的血量也应该 `≥1`。即需要满足下面这两组不等式。 
+
+	{{< katex display >}} 
+	\begin{matrix} \left\{ 
+	   \begin{array}{lr} 
+	 	dp[i][j] + dungeon[i][j] \geqslant dp[i+1][j]  \\ 
+		dp[i][j] \geqslant   1 
+	   \end{array} \right. 
+	\end{matrix}
+	{{< /katex >}}
+
+	{{< katex display >}} 
+	\begin{matrix} \left\{              \begin{array}{lr}              dp[i][j] + dungeon[i][j] \geqslant dp[i][j+1]  \\              dp[i][j] \geqslant   1         \end{array} \right. \end{matrix}
+	{{< /katex >}}   
+    上面不等式中第一组不等式是满足下一行格子的最低血量要求，第二组不等式是满足右一列格子的最低血量要求。第一个式子化简即 `dp[i][j] = max(1, dp[i+1][j]-dungeon[i][j])`，第二个式子化简即 `dp[i][j] = max(1, dp[i][j+1]-dungeon[i][j])`。求得了这两种走法的最低血量值，从这两个值里面取最小，即是当前格子所需的最低血量，所以状态转移方程为 `dp[i][j] = min(max(1, dp[i][j+1]-dungeon[i][j]), max(1, dp[i+1][j]-dungeon[i][j]))`。DP 完成以后，`dp[0][0]` 中记录的就是骑士初始最低血量值。时间复杂度 O(m\*n)，空间复杂度 O(m\*n)。
+
+- 这一题还可以用二分搜索来求解。骑士的血量取值范围一定是在 `[1，+∞)` 这个区间内。那么二分这个区间，每次二分的中间值，再用 dp 在地图中去判断是否能到达终点，如果能，就缩小搜索空间至 `[1，mid]`，否则搜索空间为 `[mid + 1，+∞)` 。时间复杂度 O(m\*n\* log math.MaxInt64)，空间复杂度 O(m\*n)。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+// 解法一 动态规划
+func calculateMinimumHP(dungeon [][]int) int {
+	if len(dungeon) == 0 {
+		return 0
+	}
+	m, n := len(dungeon), len(dungeon[0])
+	dp := make([][]int, m)
+	for i := 0; i < m; i++ {
+		dp[i] = make([]int, n)
+	}
+	dp[m-1][n-1] = max(1-dungeon[m-1][n-1], 1)
+	for i := n - 2; i >= 0; i-- {
+		dp[m-1][i] = max(1, dp[m-1][i+1]-dungeon[m-1][i])
+	}
+	for i := m - 2; i >= 0; i-- {
+		dp[i][n-1] = max(1, dp[i+1][n-1]-dungeon[i][n-1])
+	}
+	for i := m - 2; i >= 0; i-- {
+		for j := n - 2; j >= 0; j-- {
+			dp[i][j] = min(max(1, dp[i][j+1]-dungeon[i][j]), max(1, dp[i+1][j]-dungeon[i][j]))
+		}
+	}
+	return dp[0][0]
+}
+
+// 解法二 二分搜索
+func calculateMinimumHP1(dungeon [][]int) int {
+	low, high := 1, math.MaxInt64
+	for low < high {
+		mid := low + (high-low)>>1
+		if canCross(dungeon, mid) {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+func canCross(dungeon [][]int, start int) bool {
+	m, n := len(dungeon), len(dungeon[0])
+	dp := make([][]int, m)
+	for i := 0; i < m; i++ {
+		dp[i] = make([]int, n)
+	}
+	for i := 0; i < len(dp); i++ {
+		for j := 0; j < len(dp[i]); j++ {
+			if i == 0 && j == 0 {
+				dp[i][j] = start + dungeon[0][0]
+			} else {
+				a, b := math.MinInt64, math.MinInt64
+				if i > 0 && dp[i-1][j] > 0 {
+					a = dp[i-1][j] + dungeon[i][j]
+				}
+				if j > 0 && dp[i][j-1] > 0 {
+					b = dp[i][j-1] + dungeon[i][j]
+				}
+				dp[i][j] = max(a, b)
+			}
+		}
+	}
+	return dp[m-1][n-1] > 0
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0173.Binary-Search-Tree-Iterator/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0179.Largest-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0179.Largest-Number.md b/website/content/ChapterFour/0100~0199/0179.Largest-Number.md
new file mode 100644
index 000000000..09f1a0708
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0179.Largest-Number.md
@@ -0,0 +1,129 @@
+# [179. Largest Number](https://leetcode.com/problems/largest-number/)
+
+## 题目
+
+Given a list of non negative integers, arrange them such that they form the largest number.
+
+
+
+**Example 1**:
+
+```
+
+Input: [10,2]
+Output: "210"
+
+```
+
+
+**Example 2**:
+
+```
+
+Input: [3,30,34,5,9]
+Output: "9534330"
+
+```
+
+**Note**: 
+
+The result may be very large, so you need to return a string instead of an integer.
+
+
+
+## 题目大意
+
+给出一个数组，要求排列这些数组里的元素，使得最终排列出来的数字是最大的。
+
+
+## 解题思路
+
+这一题很容易想到把数字都转化为字符串，利用字符串比较，来排序，这样 9 开头的一定排在最前面。不过这样做有一个地方是错误的，比如："3" 和 "30" 比较，"30" 比 "3" 的字符序要大，这样排序以后就出错了。实际上就这道题而言， "3" 应该排在 "30" 前面。
+
+在比较 2 个字符串大小的时候，不单纯的只用字符串顺序进行比较，还加入一个顺序。
+
+```go
+aStr := a + b
+bStr := b + a
+```
+
+通过比较 aStr 和 bStr 的大小来得出是 a 大还是 b 大。
+
+举个例子，还是 "3" 和 "30" 的例子，比较这 2 个字符串的大小。
+
+
+```go
+aStr := "3" + "30" = "330"
+bStr := "30" + "3" = "303"
+```
+
+通过互相补齐位数之后再进行比较，就没有问题了。很显然这里 "3" 比 "30" 要大。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+)
+
+func largestNumber(nums []int) string {
+	if len(nums) == 0 {
+		return ""
+	}
+	numStrs := toStringArray(nums)
+	quickSortString(numStrs, 0, len(numStrs)-1)
+	res := ""
+	for _, str := range numStrs {
+		if res == "0" && str == "0" {
+			continue
+		}
+		res = res + str
+	}
+	return res
+}
+
+func toStringArray(nums []int) []string {
+	strs := make([]string, 0)
+	for _, num := range nums {
+		strs = append(strs, strconv.Itoa(num))
+	}
+	return strs
+}
+func partitionString(a []string, lo, hi int) int {
+	pivot := a[hi]
+	i := lo - 1
+	for j := lo; j < hi; j++ {
+		ajStr := a[j] + pivot
+		pivotStr := pivot + a[j]
+		if ajStr > pivotStr { // 这里的判断条件是关键
+			i++
+			a[j], a[i] = a[i], a[j]
+		}
+	}
+	a[i+1], a[hi] = a[hi], a[i+1]
+	return i + 1
+}
+func quickSortString(a []string, lo, hi int) {
+	if lo >= hi {
+		return
+	}
+	p := partitionString(a, lo, hi)
+	quickSortString(a, lo, p-1)
+	quickSortString(a, p+1, hi)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0174.Dungeon-Game/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0187.Repeated-DNA-Sequences/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0187.Repeated-DNA-Sequences.md b/website/content/ChapterFour/0100~0199/0187.Repeated-DNA-Sequences.md
new file mode 100755
index 000000000..5480f7a72
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0187.Repeated-DNA-Sequences.md
@@ -0,0 +1,78 @@
+# [187. Repeated DNA Sequences](https://leetcode.com/problems/repeated-dna-sequences/)
+
+
+## 题目
+
+All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for Example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA.
+
+Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.
+
+**Example**:
+
+    Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
+    
+    Output: ["AAAAACCCCC", "CCCCCAAAAA"]
+
+
+## 题目大意
+
+所有 DNA 由一系列缩写为 A，C，G 和 T 的核苷酸组成，例如：“ACGAATTCCG”。在研究 DNA 时，识别 DNA 中的重复序列有时会对研究非常有帮助。编写一个函数来查找 DNA 分子中所有出现超多一次的10个字母长的序列（子串）。
+
+## 解题思路
+
+- 这一题不用位运算比较好做，维护一个长度为 10 的字符串，在 map 中出现次数 > 1 就输出。
+- 用位运算想做这一题，需要动态的维护长度为 10 的 hashkey，先计算开头长度为 9 的 hash，在往后面扫描的过程中，如果长度超过了 10 ，就移除 hash 开头的一个字符，加入后面一个字符。具体做法是先将 ATCG 变成 00，01，10，11 的编码，那么长度为 10 ，hashkey 就需要维护在 20 位。mask = 0xFFFFF 就是 20 位的。维护了 hashkey 以后，根据这个 hashkey 进行去重和统计频次。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func findRepeatedDnaSequences(s string) []string {
+	if len(s) < 10 {
+		return nil
+	}
+	charMap, mp, result := map[uint8]uint32{'A': 0, 'C': 1, 'G': 2, 'T': 3}, make(map[uint32]int, 0), []string{}
+	var cur uint32
+	for i := 0; i < 9; i++ { // 前9位，忽略
+		cur = cur<<2 | charMap[s[i]]
+	}
+	for i := 9; i < len(s); i++ {
+		cur = ((cur << 2) & 0xFFFFF) | charMap[s[i]]
+		if mp[cur] == 0 {
+			mp[cur] = 1
+		} else if mp[cur] == 1 { // >2，重复
+			mp[cur] = 2
+			result = append(result, s[i-9:i+1])
+		}
+	}
+	return result
+}
+
+// 解法二
+func findRepeatedDnaSequences1(s string) []string {
+	if len(s) < 10 {
+		return []string{}
+	}
+	ans, cache := make([]string, 0), make(map[string]int)
+	for i := 0; i <= len(s)-10; i++ {
+		curr := string(s[i : i+10])
+		if cache[curr] == 1 {
+			ans = append(ans, curr)
+		}
+		cache[curr]++
+	}
+	return ans
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0179.Largest-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0189.Rotate-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0189.Rotate-Array.md b/website/content/ChapterFour/0100~0199/0189.Rotate-Array.md
new file mode 100644
index 000000000..34733ee96
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0189.Rotate-Array.md
@@ -0,0 +1,82 @@
+# [189. Rotate Array](https://leetcode.com/problems/rotate-array/)
+
+## 题目
+
+Given an array, rotate the array to the right by *k* steps, where *k* is non-negative.
+
+**Follow up**:
+
+- Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
+- Could you do it in-place with O(1) extra space?
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,4,5,6,7], k = 3
+Output: [5,6,7,1,2,3,4]
+Explanation:
+rotate 1 steps to the right: [7,1,2,3,4,5,6]
+rotate 2 steps to the right: [6,7,1,2,3,4,5]
+rotate 3 steps to the right: [5,6,7,1,2,3,4]
+```
+
+**Example 2**:
+
+```
+Input: nums = [-1,-100,3,99], k = 2
+Output: [3,99,-1,-100]
+Explanation: 
+rotate 1 steps to the right: [99,-1,-100,3]
+rotate 2 steps to the right: [3,99,-1,-100]
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 2 * 10^4`
+- `-2^31 <= nums[i] <= 2^31 - 1`
+- `0 <= k <= 10^5`
+
+## 题目大意
+
+给定一个数组，将数组中的元素向右移动 k 个位置，其中 k 是非负数。
+
+## 解题思路
+
+- 解法二，使用一个额外的数组，先将原数组下标为 i 的元素移动到 `(i+k) mod n` 的位置，再将剩下的元素拷贝回来即可。
+- 解法一，由于题目要求不能使用额外的空间，所以本题最佳解法不是解法二。翻转最终态是，末尾 `k mod n` 个元素移动至了数组头部，剩下的元素右移 `k mod n` 个位置至最尾部。确定了最终态以后再变换就很容易。先将数组中所有元素从头到尾翻转一次，尾部的所有元素都到了头部，然后再将 `[0,(k mod n) − 1]` 区间内的元素翻转一次，最后再将 `[k mod n, n − 1]` 区间内的元素翻转一次，即可满足题目要求。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 时间复杂度 O(n)，空间复杂度 O(1)
+func rotate(nums []int, k int) {
+	k %= len(nums)
+	reverse(nums)
+	reverse(nums[:k])
+	reverse(nums[k:])
+}
+
+func reverse(a []int) {
+	for i, n := 0, len(a); i < n/2; i++ {
+		a[i], a[n-1-i] = a[n-1-i], a[i]
+	}
+}
+
+// 解法二 时间复杂度 O(n)，空间复杂度 O(n)
+func rotate1(nums []int, k int) {
+	newNums := make([]int, len(nums))
+	for i, v := range nums {
+		newNums[(i+k)%len(nums)] = v
+	}
+	copy(nums, newNums)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0187.Repeated-DNA-Sequences/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0190.Reverse-Bits/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0190.Reverse-Bits.md b/website/content/ChapterFour/0100~0199/0190.Reverse-Bits.md
new file mode 100755
index 000000000..d73e2f2f1
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0190.Reverse-Bits.md
@@ -0,0 +1,60 @@
+# [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)
+
+
+## 题目
+
+Reverse bits of a given 32 bits unsigned integer.
+
+**Example 1**:
+
+    Input: 00000010100101000001111010011100
+    Output: 00111001011110000010100101000000
+    Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
+
+**Example 2**:
+
+    Input: 11111111111111111111111111111101
+    Output: 10111111111111111111111111111111
+    Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10101111110010110010011101101001.
+
+**Note**:
+
+- Note that in some languages such as Java, there is no unsigned integer type. In this case, both input and output will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
+- In Java, the compiler represents the signed integers using [2's complement notation](https://en.wikipedia.org/wiki/Two%27s_complement). Therefore, in **Example 2** above the input represents the signed integer `-3` and the output represents the signed integer `-1073741825`.
+
+## 题目大意
+
+颠倒给定的 32 位无符号整数的二进制位。提示：
+
+- 请注意，在某些语言（如 Java）中，没有无符号整数类型。在这种情况下，输入和输出都将被指定为有符号整数类型，并且不应影响您的实现，因为无论整数是有符号的还是无符号的，其内部的二进制表示形式都是相同的。
+- 在 Java 中，编译器使用二进制补码记法来表示有符号整数。因此，在上面的 示例 2 中，输入表示有符号整数 -3，输出表示有符号整数 -1073741825。
+
+## 解题思路
+
+- 简单题，要求反转 32 位的二进制位。
+- 把 num 往右移动，不断的消灭右边最低位的 1，将这个 1 给 res，res 不断的左移即可实现反转二进制位的目的。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func reverseBits(num uint32) uint32 {
+	var res uint32
+	for i := 0; i < 32; i++ {
+		res = res<<1 | num&1
+		num >>= 1
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0189.Rotate-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0191.Number-of-1-Bits/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0191.Number-of-1-Bits.md b/website/content/ChapterFour/0100~0199/0191.Number-of-1-Bits.md
new file mode 100755
index 000000000..d3a45e1d3
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0191.Number-of-1-Bits.md
@@ -0,0 +1,72 @@
+# [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)
+
+## 题目
+
+Write a function that takes an unsigned integer and return the number of '1' bits it has (also known as the [Hamming weight](http://en.wikipedia.org/wiki/Hamming_weight)).
+
+**Example 1**:
+
+    Input: 00000000000000000000000000001011
+    Output: 3
+    Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.
+
+**Example 2**:
+
+    Input: 00000000000000000000000010000000
+    Output: 1
+    Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.
+
+**Example 3**:
+
+    Input: 11111111111111111111111111111101
+    Output: 31
+    Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.
+
+**Note**:
+
+- Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
+- In Java, the compiler represents the signed integers using [2's complement notation](https://en.wikipedia.org/wiki/Two%27s_complement). Therefore, in **Example 3** above the input represents the signed integer `-3`.
+
+
+## 题目大意
+
+编写一个函数，输入是一个无符号整数，返回其二进制表达式中数字位数为 ‘1’ 的个数（也被称为汉明重量）。
+
+## 解题思路
+
+- 求 uint32 数的二进制位中 1 的个数。
+- 这一题的解题思路就是利用二进制位操作。`X = X & ( X -1 )` 这个操作可以清除最低位的二进制位 1，利用这个操作，直至把数清零。操作了几次即为有几个二进制位 1 。
+- 最简单的方法即是直接调用库函数 `bits.OnesCount(uint(num))` 。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math/bits"
+
+// 解法一
+func hammingWeight(num uint32) int {
+	return bits.OnesCount(uint(num))
+}
+
+// 解法二
+func hammingWeight1(num uint32) int {
+	count := 0
+	for num != 0 {
+		num = num & (num - 1)
+		count++
+	}
+	return count
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0190.Reverse-Bits/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0198.House-Robber/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0198.House-Robber.md b/website/content/ChapterFour/0100~0199/0198.House-Robber.md
new file mode 100755
index 000000000..2ee3a7a18
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0198.House-Robber.md
@@ -0,0 +1,101 @@
+# [198. House Robber](https://leetcode.com/problems/house-robber/)
+
+
+## 题目
+
+You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and **it will automatically contact the police if two adjacent houses were broken into on the same night**.
+
+Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight **without alerting the police**.
+
+**Example 1**:
+
+    Input: [1,2,3,1]
+    Output: 4
+    Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
+                 Total amount you can rob = 1 + 3 = 4.
+
+**Example 2**:
+
+    Input: [2,7,9,3,1]
+    Output: 12
+    Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
+                 Total amount you can rob = 2 + 9 + 1 = 12.
+
+
+## 题目大意
+
+你是一个专业的小偷，计划偷窃沿街的房屋。每间房内都藏有一定的现金，影响你偷窃的唯一制约因素就是相邻的房屋装有相互连通的防盗系统，**如果两间相邻的房屋在同一晚上被小偷闯入，系统会自动报警**。
+
+给定一个代表每个房屋存放金额的非负整数数组，计算你**在不触动警报装置的情况下**，能够偷窃到的最高金额。
+
+
+## 解题思路
+
+- 你是一个专业的小偷，打算洗劫一条街的所有房子。每个房子里面有不同价值的宝物，但是如果你选择偷窃连续的 2 栋房子，就会触发警报系统，编程求出你最多可以偷窃价值多少的宝物？
+- 这一题可以用 DP 来解答，也可以用找规律的方法来解答。
+- DP 的状态定义是：`dp[i]` 代表抢 `nums[0,i]` 这个区间内房子的最大值，状态转移方程是 `dp[i] = max(dp[i-1], nums[i]+dp[i-2])`  。可以优化迭代的过程，用两个临时变量来存储中间结果，以节约辅助空间。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 DP
+func rob198(nums []int) int {
+	n := len(nums)
+	if n == 0 {
+		return 0
+	}
+	if n == 1 {
+		return nums[0]
+	}
+	// dp[i] 代表抢 nums[0...i] 房子的最大价值
+	dp := make([]int, n)
+	dp[0], dp[1] = nums[0], max(nums[1], nums[0])
+	for i := 2; i < n; i++ {
+		dp[i] = max(dp[i-1], nums[i]+dp[i-2])
+	}
+	return dp[n-1]
+}
+
+// 解法二 DP 优化辅助空间，把迭代的值保存在 2 个变量中
+func rob198_1(nums []int) int {
+	n := len(nums)
+	if n == 0 {
+		return 0
+	}
+	curMax, preMax := 0, 0
+	for i := 0; i < n; i++ {
+		tmp := curMax
+		curMax = max(curMax, nums[i]+preMax)
+		preMax = tmp
+	}
+	return curMax
+}
+
+// 解法三 模拟
+func rob(nums []int) int {
+	// a 对于偶数位上的最大值的记录
+	// b 对于奇数位上的最大值的记录
+	a, b := 0, 0
+	for i := 0; i < len(nums); i++ {
+		if i%2 == 0 {
+			a = max(a+nums[i], b)
+		} else {
+			b = max(a, b+nums[i])
+		}
+	}
+	return max(a, b)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0191.Number-of-1-Bits/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0199.Binary-Tree-Right-Side-View/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/0199.Binary-Tree-Right-Side-View.md b/website/content/ChapterFour/0100~0199/0199.Binary-Tree-Right-Side-View.md
new file mode 100644
index 000000000..70be59786
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/0199.Binary-Tree-Right-Side-View.md
@@ -0,0 +1,81 @@
+# [199. Binary Tree Right Side View](https://leetcode.com/problems/binary-tree-right-side-view/)
+
+## 题目
+
+Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
+
+**Example**:
+
+```
+
+Input: [1,2,3,null,5,null,4]
+Output: [1, 3, 4]
+Explanation:
+
+   1            <---
+ /   \
+2     3         <---
+ \     \
+  5     4       <---
+
+```
+
+ 
+
+## 题目大意
+
+从右边看一个树，输出看到的数字。注意有遮挡。
+
+
+## 解题思路
+
+- 这一题是按层序遍历的变种题。按照层序把每层的元素都遍历出来，然后依次取每一层的最右边的元素即可。用一个队列即可实现。
+- 第 102 题和第 107 题都是按层序遍历的。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func rightSideView(root *TreeNode) []int {
+	res := []int{}
+	if root == nil {
+		return res
+	}
+	queue := []*TreeNode{root}
+	for len(queue) > 0 {
+		n := len(queue)
+		for i := 0; i < n; i++ {
+			if queue[i].Left != nil {
+				queue = append(queue, queue[i].Left)
+			}
+			if queue[i].Right != nil {
+				queue = append(queue, queue[i].Right)
+			}
+		}
+		res = append(res, queue[n-1].Val)
+		queue = queue[n:]
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0198.House-Robber/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0200.Number-of-Islands/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0100~0199/_index.md b/website/content/ChapterFour/0100~0199/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/0100~0199/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/0200~0299/0200.Number-of-Islands.md b/website/content/ChapterFour/0200~0299/0200.Number-of-Islands.md
new file mode 100755
index 000000000..909363570
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0200.Number-of-Islands.md
@@ -0,0 +1,88 @@
+# [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)
+
+
+## 题目
+
+Given a 2d grid map of `'1'`s (land) and `'0'`s (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
+
+**Example 1**:
+
+    Input:
+    11110
+    11010
+    11000
+    00000
+    
+    Output: 1
+
+**Example 2**:
+
+    Input:
+    11000
+    11000
+    00100
+    00011
+    
+    Output: 3
+
+## 题目大意
+
+给定一个由 '1'（陆地）和 '0'（水）组成的的二维网格，计算岛屿的数量。一个岛被水包围，并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。
+
+
+## 解题思路
+
+- 要求找出地图中的孤岛。孤岛的含义是四周被海水包围的岛。
+- 这一题可以按照第 79 题的思路进行搜索，只要找到为 "1" 的岛以后，从这里开始搜索这周连通的陆地，也都标识上访问过。每次遇到新的 "1" 且没有访问过，就相当于遇到了新的岛屿了。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func numIslands(grid [][]byte) int {
+	m := len(grid)
+	if m == 0 {
+		return 0
+	}
+	n := len(grid[0])
+	if n == 0 {
+		return 0
+	}
+	res, visited := 0, make([][]bool, m)
+	for i := 0; i < m; i++ {
+		visited[i] = make([]bool, n)
+	}
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			if grid[i][j] == '1' && !visited[i][j] {
+				searchIslands(grid, &visited, i, j)
+				res++
+			}
+		}
+	}
+	return res
+}
+
+func searchIslands(grid [][]byte, visited *[][]bool, x, y int) {
+	(*visited)[x][y] = true
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		if isInBoard(grid, nx, ny) && !(*visited)[nx][ny] && grid[nx][ny] == '1' {
+			searchIslands(grid, visited, nx, ny)
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0100~0199/0199.Binary-Tree-Right-Side-View/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0201.Bitwise-AND-of-Numbers-Range/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0201.Bitwise-AND-of-Numbers-Range.md b/website/content/ChapterFour/0200~0299/0201.Bitwise-AND-of-Numbers-Range.md
new file mode 100755
index 000000000..6f064185d
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0201.Bitwise-AND-of-Numbers-Range.md
@@ -0,0 +1,73 @@
+# [201. Bitwise AND of Numbers Range](https://leetcode.com/problems/bitwise-and-of-numbers-range/)
+
+
+## 题目
+
+Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.
+
+**Example 1**:
+
+    Input: [5,7]
+    Output: 4
+
+**Example 2**:
+
+    Input: [0,1]
+    Output: 0
+
+## 题目大意
+
+给定范围 [m, n]，其中 0 <= m <= n <= 2147483647，返回此范围内所有数字的按位与（包含 m, n 两端点）。
+
+
+## 解题思路
+
+- 这一题要求输出 [m,n] 区间内所有数的 AND 与操作之后的结果。
+- 举个例子，假设区间是 [26,30]，那么这个区间内的数用二进制表示出来为：
+
+    11010  
+    11011  
+    11100   
+    11101   
+    11110  
+
+- 可以观察到，把这些数都 AND 起来，只要有 0 的位，最终结果都是 0，所以需要从右往前找到某一位上不为 0 的。不断的右移左边界和右边界，把右边的 0 都移走，直到它们俩相等，就找到了某一位上开始都不为 0 的了。在右移的过程中记录下右移了多少位，最后把 m 或者 n 的右边添上 0 即可。按照上面这个例子来看，11000 是最终的结果。
+- 这一题还有解法二，还是以 [26,30] 这个区间为例。这个区间内的数末尾 3 位不断的 0，1 变化着。那么如果把末尾的 1 都打掉，就是最终要求的结果了。当 n == m 或者 n < m 的时候就退出循环，说明后面不同的位数已经都被抹平了，1 都被打掉为 0 了。所以关键的操作为 `n &= (n - 1)` ，清除最低位的 1 。这个算法名叫 `Brian Kernighan` 算法。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func rangeBitwiseAnd1(m int, n int) int {
+	if m == 0 {
+		return 0
+	}
+	moved := 0
+	for m != n {
+		m >>= 1
+		n >>= 1
+		moved++
+	}
+	return m << uint32(moved)
+}
+
+// 解法二 Brian Kernighan's algorithm
+func rangeBitwiseAnd(m int, n int) int {
+	for n > m {
+		n &= (n - 1) // 清除最低位的 1
+	}
+	return n
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0200.Number-of-Islands/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0202.Happy-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0202.Happy-Number.md b/website/content/ChapterFour/0200~0299/0202.Happy-Number.md
new file mode 100644
index 000000000..db82ed58e
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0202.Happy-Number.md
@@ -0,0 +1,71 @@
+# [202. Happy Number](https://leetcode.com/problems/happy-number/)
+
+## 题目
+
+Write an algorithm to determine if a number is "happy".
+
+A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
+
+**Example 1**:
+
+```
+
+Input: 19
+Output: true
+Explanation: 
+12 + 92 = 82
+82 + 22 = 68
+62 + 82 = 100
+12 + 02 + 02 = 1
+
+```
+
+## 题目大意
+
+判断一个数字是否是“快乐数字”，“快乐数字”的定义是，不断的把这个数字的每个数字的平方和加起来，反复的加，最终如果能有结果是 1，则是“快乐数字”，如果不能得到一，出现了循环，则输出 false。
+
+## 解题思路
+
+按照题意要求做即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func isHappy(n int) bool {
+	record := map[int]int{}
+	for n != 1 {
+		record[n] = n
+		n = getSquareOfDigits(n)
+		for _, previous := range record {
+			if n == previous {
+				return false
+			}
+		}
+	}
+	return true
+}
+
+func getSquareOfDigits(n int) int {
+	squareOfDigits := 0
+	temporary := n
+	for temporary != 0 {
+		remainder := temporary % 10
+		squareOfDigits += remainder * remainder
+		temporary /= 10
+	}
+	return squareOfDigits
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0201.Bitwise-AND-of-Numbers-Range/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0203.Remove-Linked-List-Elements/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0203.Remove-Linked-List-Elements.md b/website/content/ChapterFour/0200~0299/0203.Remove-Linked-List-Elements.md
new file mode 100644
index 000000000..d86711619
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0203.Remove-Linked-List-Elements.md
@@ -0,0 +1,63 @@
+# [203. Remove Linked List Elements](https://leetcode.com/problems/remove-linked-list-elements/)
+
+## 题目
+
+Remove all elements from a linked list of integers that have value val.
+
+**Example**:
+
+```
+
+Input:  1->2->6->3->4->5->6, val = 6
+Output: 1->2->3->4->5
+
+```
+
+
+## 题目大意
+
+删除链表中所有指定值的结点。
+
+## 解题思路
+
+按照题意做即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func removeElements(head *ListNode, val int) *ListNode {
+	if head == nil {
+		return head
+	}
+	newHead := &ListNode{Val: 0, Next: head}
+	pre := newHead
+	cur := head
+	for cur != nil {
+		if cur.Val == val {
+			pre.Next = cur.Next
+		} else {
+			pre = cur
+		}
+		cur = cur.Next
+	}
+	return newHead.Next
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0202.Happy-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0204.Count-Primes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0204.Count-Primes.md b/website/content/ChapterFour/0200~0299/0204.Count-Primes.md
new file mode 100755
index 000000000..fe09cb95c
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0204.Count-Primes.md
@@ -0,0 +1,57 @@
+# [204. Count Primes](https://leetcode.com/problems/count-primes/)
+
+
+## 题目
+
+Count the number of prime numbers less than a non-negative number, **n**.
+
+**Example**:
+
+    Input: 10
+    Output: 4
+    Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.
+
+
+## 题目大意
+
+统计所有小于非负整数 n 的质数的数量。
+
+
+## 解题思路
+
+- 给出一个数字 n，要求输出小于 n 的所有素数的个数总和。简单题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func countPrimes(n int) int {
+	isNotPrime := make([]bool, n)
+	for i := 2; i*i < n; i++ {
+		if isNotPrime[i] {
+			continue
+		}
+		for j := i * i; j < n; j = j + i {
+			isNotPrime[j] = true
+		}
+	}
+	count := 0
+	for i := 2; i < n; i++ {
+		if !isNotPrime[i] {
+			count++
+		}
+	}
+	return count
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0203.Remove-Linked-List-Elements/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0205.Isomorphic-Strings/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0205.Isomorphic-Strings.md b/website/content/ChapterFour/0200~0299/0205.Isomorphic-Strings.md
new file mode 100644
index 000000000..6854aa2a1
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0205.Isomorphic-Strings.md
@@ -0,0 +1,97 @@
+# [205. Isomorphic Strings](https://leetcode.com/problems/isomorphic-strings/)
+
+## 题目
+
+Given two strings s and t, determine if they are isomorphic.
+
+Two strings are isomorphic if the characters in s can be replaced to get t.
+
+All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.
+
+**Example 1**:
+
+```
+
+Input: s = "egg", t = "add"
+Output: true
+
+```
+
+**Example 2**:
+
+```
+
+Input: s = "foo", t = "bar"
+Output: false
+
+```
+
+**Example 3**:
+
+```
+
+Input: s = "paper", t = "title"
+Output: true
+
+```
+
+**Note**:   
+
+You may assume both s and t have the same length.
+
+
+
+
+## 题目大意
+
+这道题和第 290 题基本是一样的。第 290 题是模式匹配，这道题的题意是字符串映射，实质是一样的。
+
+给定一个初始字符串串，判断初始字符串是否可以通过字符映射的方式，映射到目标字符串，如果可以映射，则输出 true，反之输出 false。
+
+## 解题思路
+
+这道题做法和第 290 题基本一致。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func isIsomorphic(s string, t string) bool {
+	strList := []byte(t)
+	patternByte := []byte(s)
+	if (s == "" && t != "") || (len(patternByte) != len(strList)) {
+		return false
+	}
+
+	pMap := map[byte]byte{}
+	sMap := map[byte]byte{}
+	for index, b := range patternByte {
+		if _, ok := pMap[b]; !ok {
+			if _, ok = sMap[strList[index]]; !ok {
+				pMap[b] = strList[index]
+				sMap[strList[index]] = b
+			} else {
+				if sMap[strList[index]] != b {
+					return false
+				}
+			}
+		} else {
+			if pMap[b] != strList[index] {
+				return false
+			}
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0204.Count-Primes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0206.Reverse-Linked-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0206.Reverse-Linked-List.md b/website/content/ChapterFour/0200~0299/0206.Reverse-Linked-List.md
new file mode 100644
index 000000000..773f955f9
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0206.Reverse-Linked-List.md
@@ -0,0 +1,54 @@
+# [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/description/)
+
+## 题目
+
+Reverse a singly linked list.
+
+## 题目大意
+
+翻转单链表
+
+
+## 解题思路
+
+按照题意做即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// ListNode define
+type ListNode struct {
+	Val  int
+	Next *ListNode
+}
+
+func reverseList(head *ListNode) *ListNode {
+	var behind *ListNode
+	for head != nil {
+		next := head.Next
+		head.Next = behind
+		behind = head
+		head = next
+	}
+	return behind
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0205.Isomorphic-Strings/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0207.Course-Schedule/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0207.Course-Schedule.md b/website/content/ChapterFour/0200~0299/0207.Course-Schedule.md
new file mode 100755
index 000000000..52442d3c9
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0207.Course-Schedule.md
@@ -0,0 +1,89 @@
+# [207. Course Schedule](https://leetcode.com/problems/course-schedule/)
+
+## 题目
+
+There are a total of n courses you have to take, labeled from `0` to `n-1`.
+
+Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: `[0,1]`
+
+Given the total number of courses and a list of prerequisite **pairs**, is it possible for you to finish all courses?
+
+**Example 1**:
+
+    Input: 2, [[1,0]] 
+    Output: true
+    Explanation: There are a total of 2 courses to take. 
+                 To take course 1 you should have finished course 0. So it is possible.
+
+**Example 2**:
+
+    Input: 2, [[1,0],[0,1]]
+    Output: false
+    Explanation: There are a total of 2 courses to take. 
+                 To take course 1 you should have finished course 0, and to take course 0 you should
+                 also have finished course 1. So it is impossible.
+
+**Note**:
+
+1. The input prerequisites is a graph represented by **a list of edges**, not adjacency matrices. Read more about [how a graph is represented](https://www.khanacademy.org/computing/computer-science/algorithms/graph-representation/a/representing-graphs).
+2. You may assume that there are no duplicate edges in the input prerequisites.
+
+
+## 题目大意
+
+现在你总共有 n 门课需要选，记为 0 到 n-1。在选修某些课程之前需要一些先修课程。 例如，想要学习课程 0 ，你需要先完成课程 1 ，我们用一个匹配来表示他们: [0,1]。给定课程总量以及它们的先决条件，判断是否可能完成所有课程的学习？
+
+
+
+## 解题思路
+
+- 给出 n 个任务，每两个任务之间有相互依赖关系，比如 A 任务一定要在 B 任务之前完成才行。问是否可以完成所有任务。
+- 这一题就是标准的 AOV 网的拓扑排序问题。拓扑排序问题的解决办法是主要是循环执行以下两步，直到不存在入度为0的顶点为止。
+    - 1.  选择一个入度为0的顶点并输出之；
+    - 2. 从网中删除此顶点及所有出边。
+
+    循环结束后，若输出的顶点数小于网中的顶点数，则输出“有回路”信息，即无法完成所有任务；否则输出的顶点序列就是一种拓扑序列，即可以完成所有任务。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// AOV 网的拓扑排序
+func canFinish(n int, pre [][]int) bool {
+	in := make([]int, n)
+	frees := make([][]int, n)
+	next := make([]int, 0, n)
+	for _, v := range pre {
+		in[v[0]]++
+		frees[v[1]] = append(frees[v[1]], v[0])
+	}
+	for i := 0; i < n; i++ {
+		if in[i] == 0 {
+			next = append(next, i)
+		}
+	}
+	for i := 0; i != len(next); i++ {
+		c := next[i]
+		v := frees[c]
+		for _, vv := range v {
+			in[vv]--
+			if in[vv] == 0 {
+				next = append(next, vv)
+			}
+		}
+	}
+	return len(next) == n
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0206.Reverse-Linked-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0208.Implement-Trie-Prefix-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0208.Implement-Trie-Prefix-Tree.md b/website/content/ChapterFour/0200~0299/0208.Implement-Trie-Prefix-Tree.md
new file mode 100755
index 000000000..9f997351b
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0208.Implement-Trie-Prefix-Tree.md
@@ -0,0 +1,106 @@
+# [208. Implement Trie (Prefix Tree)](https://leetcode.com/problems/implement-trie-prefix-tree/)
+
+
+## 题目
+
+Implement a trie with `insert`, `search`, and `startsWith` methods.
+
+**Example**:
+
+    Trie trie = new Trie();
+    
+    trie.insert("apple");
+    trie.search("apple");   // returns true
+    trie.search("app");     // returns false
+    trie.startsWith("app"); // returns true
+    trie.insert("app");   
+    trie.search("app");     // returns true
+
+**Note**:
+
+- You may assume that all inputs are consist of lowercase letters `a-z`.
+- All inputs are guaranteed to be non-empty strings.
+
+## 题目大意
+
+实现一个 Trie (前缀树)，包含 insert, search, 和 startsWith 这三个操作。
+
+## 解题思路
+
+- 要求实现一个 Trie 的数据结构，具有 `insert`, `search`, `startsWith` 三种操作
+- 这一题就是经典的 Trie 实现。本题的实现可以作为 Trie 的模板。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+type Trie struct {
+	isWord   bool
+	children map[rune]*Trie
+}
+
+/** Initialize your data structure here. */
+func Constructor208() Trie {
+	return Trie{isWord: false, children: make(map[rune]*Trie)}
+}
+
+/** Inserts a word into the trie. */
+func (this *Trie) Insert(word string) {
+	parent := this
+	for _, ch := range word {
+		if child, ok := parent.children[ch]; ok {
+			parent = child
+		} else {
+			newChild := &Trie{children: make(map[rune]*Trie)}
+			parent.children[ch] = newChild
+			parent = newChild
+		}
+	}
+	parent.isWord = true
+}
+
+/** Returns if the word is in the trie. */
+func (this *Trie) Search(word string) bool {
+	parent := this
+	for _, ch := range word {
+		if child, ok := parent.children[ch]; ok {
+			parent = child
+			continue
+		}
+		return false
+	}
+	return parent.isWord
+}
+
+/** Returns if there is any word in the trie that starts with the given prefix. */
+func (this *Trie) StartsWith(prefix string) bool {
+	parent := this
+	for _, ch := range prefix {
+		if child, ok := parent.children[ch]; ok {
+			parent = child
+			continue
+		}
+		return false
+	}
+	return true
+}
+
+/**
+ * Your Trie object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.Insert(word);
+ * param_2 := obj.Search(word);
+ * param_3 := obj.StartsWith(prefix);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0207.Course-Schedule/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0209.Minimum-Size-Subarray-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0209.Minimum-Size-Subarray-Sum.md b/website/content/ChapterFour/0200~0299/0209.Minimum-Size-Subarray-Sum.md
new file mode 100644
index 000000000..806093aee
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0209.Minimum-Size-Subarray-Sum.md
@@ -0,0 +1,67 @@
+# [209. Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum/)
+
+## 题目
+
+Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead.
+
+**Example 1**:
+
+```
+
+Input: s = 7, nums = [2,3,1,2,4,3]
+Output: 2
+Explanation: the subarray [4,3] has the minimal length under the problem constraint.
+
+```
+
+**Follow up**:       
+  
+If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n). 
+
+## 题目大意
+
+给定一个整型数组和一个数字 s，找到数组中最短的一个连续子数组，使得连续子数组的数字之和 sum>=s，返回最短的连续子数组的返回值。
+
+## 解题思路
+
+这一题的解题思路是用滑动窗口。在滑动窗口 [i,j]之间不断往后移动，如果总和小于 s，就扩大右边界 j，不断加入右边的值，直到 sum > s，之和再缩小 i 的左边界，不断缩小直到 sum < s，这时候右边界又可以往右移动。以此类推。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func minSubArrayLen(target int, nums []int) int {
+	left, sum, res := 0, 0, len(nums)+1
+	for right, v := range nums {
+		sum += v
+		for sum >= target {
+			res = min(res, right-left+1)
+			sum -= nums[left]
+			left++
+		}
+	}
+	if res == len(nums)+1 {
+		return 0
+	}
+	return res
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0208.Implement-Trie-Prefix-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0210.Course-Schedule-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0210.Course-Schedule-II.md b/website/content/ChapterFour/0200~0299/0210.Course-Schedule-II.md
new file mode 100755
index 000000000..8f73b2405
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0210.Course-Schedule-II.md
@@ -0,0 +1,87 @@
+# [210. Course Schedule II](https://leetcode.com/problems/course-schedule-ii/)
+
+
+## 题目
+
+There are a total of *n* courses you have to take, labeled from `0` to `n-1`.
+
+Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: `[0,1]`
+
+Given the total number of courses and a list of prerequisite **pairs**, return the ordering of courses you should take to finish all courses.
+
+There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.
+
+**Example 1**:
+
+    Input: 2, [[1,0]] 
+    Output: [0,1]
+    Explanation: There are a total of 2 courses to take. To take course 1 you should have finished   
+                 course 0. So the correct course order is [0,1] .
+
+**Example 2**:
+
+    Input: 4, [[1,0],[2,0],[3,1],[3,2]]
+    Output: [0,1,2,3] or [0,2,1,3]
+    Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both     
+                 courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. 
+                 So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3] .
+
+**Note**:
+
+1. The input prerequisites is a graph represented by **a list of edges**, not adjacency matrices. Read more about [how a graph is represented](https://www.khanacademy.org/computing/computer-science/algorithms/graph-representation/a/representing-graphs).
+2. You may assume that there are no duplicate edges in the input prerequisites.
+
+## 题目大意
+
+现在你总共有 n 门课需要选，记为 0 到 n-1。在选修某些课程之前需要一些先修课程。 例如，想要学习课程 0 ，你需要先完成课程 1 ，我们用一个匹配来表示他们: [0,1]。给定课程总量以及它们的先决条件，返回你为了学完所有课程所安排的学习顺序。可能会有多个正确的顺序，你只要返回一种就可以了。如果不可能完成所有课程，返回一个空数组。
+
+
+## 解题思路
+
+- 给出 n 个任务，每两个任务之间有相互依赖关系，比如 A 任务一定要在 B 任务之前完成才行。问是否可以完成所有任务，如果可以完成任务，就输出完成任务的顺序，如果不能完成，输出空数组。
+- 这一题是第 207 题的加强版。解题思路是 AOV 网的拓扑排序。最后输出数组即可。代码和第 207 题基本不变。具体解题思路见第 207 题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findOrder(numCourses int, prerequisites [][]int) []int {
+	in := make([]int, numCourses)
+	frees := make([][]int, numCourses)
+	next := make([]int, 0, numCourses)
+	for _, v := range prerequisites {
+		in[v[0]]++
+		frees[v[1]] = append(frees[v[1]], v[0])
+	}
+	for i := 0; i < numCourses; i++ {
+		if in[i] == 0 {
+			next = append(next, i)
+		}
+	}
+	for i := 0; i != len(next); i++ {
+		c := next[i]
+		v := frees[c]
+		for _, vv := range v {
+			in[vv]--
+			if in[vv] == 0 {
+				next = append(next, vv)
+			}
+		}
+	}
+	if len(next) == numCourses {
+		return next
+	}
+	return []int{}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0209.Minimum-Size-Subarray-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0211.Design-Add-and-Search-Words-Data-Structure/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0211.Design-Add-and-Search-Words-Data-Structure.md b/website/content/ChapterFour/0200~0299/0211.Design-Add-and-Search-Words-Data-Structure.md
new file mode 100755
index 000000000..311a827ce
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0211.Design-Add-and-Search-Words-Data-Structure.md
@@ -0,0 +1,102 @@
+# [211. Design Add and Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/)
+
+
+## 题目
+
+Design a data structure that supports the following two operations:
+
+    void addWord(word)
+    bool search(word)
+
+search(word) can search a literal word or a regular expression string containing only letters `a-z` or `.`. A `.` means it can represent any one letter.
+
+**Example**:
+
+    addWord("bad")
+    addWord("dad")
+    addWord("mad")
+    search("pad") -> false
+    search("bad") -> true
+    search(".ad") -> true
+    search("b..") -> true
+
+**Note**: You may assume that all words are consist of lowercase letters `a-z`.
+
+## 题目大意
+
+设计一个支持以下两种操作的数据结构：`void addWord(word)`、`bool search(word)`。`search(word)` 可以搜索文字或正则表达式字符串，字符串只包含字母 . 或 a-z 。 "." 可以表示任何一个字母。
+
+
+
+## 解题思路
+
+- 设计一个 `WordDictionary` 的数据结构，要求具有 `addWord(word)` 和 `search(word)` 的操作，并且具有模糊查找的功能。
+- 这一题是第 208 题的加强版，在第 208 题经典的 Trie 上加上了模糊查找的功能。其他实现一模一样。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+type WordDictionary struct {
+	children map[rune]*WordDictionary
+	isWord   bool
+}
+
+/** Initialize your data structure here. */
+func Constructor211() WordDictionary {
+	return WordDictionary{children: make(map[rune]*WordDictionary)}
+}
+
+/** Adds a word into the data structure. */
+func (this *WordDictionary) AddWord(word string) {
+	parent := this
+	for _, ch := range word {
+		if child, ok := parent.children[ch]; ok {
+			parent = child
+		} else {
+			newChild := &WordDictionary{children: make(map[rune]*WordDictionary)}
+			parent.children[ch] = newChild
+			parent = newChild
+		}
+	}
+	parent.isWord = true
+}
+
+/** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
+func (this *WordDictionary) Search(word string) bool {
+	parent := this
+	for i, ch := range word {
+		if rune(ch) == '.' {
+			isMatched := false
+			for _, v := range parent.children {
+				if v.Search(word[i+1:]) {
+					isMatched = true
+				}
+			}
+			return isMatched
+		} else if _, ok := parent.children[rune(ch)]; !ok {
+			return false
+		}
+		parent = parent.children[rune(ch)]
+	}
+	return len(parent.children) == 0 || parent.isWord
+}
+
+/**
+ * Your WordDictionary object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.AddWord(word);
+ * param_2 := obj.Search(word);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0210.Course-Schedule-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0212.Word-Search-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0212.Word-Search-II.md b/website/content/ChapterFour/0200~0299/0212.Word-Search-II.md
new file mode 100755
index 000000000..69511f5f4
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0212.Word-Search-II.md
@@ -0,0 +1,110 @@
+# [212. Word Search II](https://leetcode.com/problems/word-search-ii/)
+
+
+## 题目
+
+Given a 2D board and a list of words from the dictionary, find all words in the board.
+
+Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.
+
+**Example**:
+
+    Input: 
+    board = [
+      ['o','a','a','n'],
+      ['e','t','a','e'],
+      ['i','h','k','r'],
+      ['i','f','l','v']
+    ]
+    words = ["oath","pea","eat","rain"]
+    
+    Output: ["eat","oath"]
+
+**Note**:
+
+1. All inputs are consist of lowercase letters `a-z`.
+2. The values of `words` are distinct.
+
+## 题目大意
+
+给定一个二维网格 board 和一个字典中的单词列表 words，找出所有同时在二维网格和字典中出现的单词。
+
+单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母在一个单词中不允许被重复使用。
+
+
+## 解题思路
+
+- 这一题是第 79 题的加强版，在第 79 题的基础上增加了一个 word 数组，要求找出所有出现在地图中的单词。思路还是可以按照第 79 题 DFS 搜索，不过时间复杂度特别高！
+- 想想更优的解法。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findWords(board [][]byte, words []string) []string {
+	res := []string{}
+	for _, v := range words {
+		if exist(board, v) {
+			res = append(res, v)
+		}
+	}
+	return res
+}
+
+// these is 79 solution
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func exist(board [][]byte, word string) bool {
+	visited := make([][]bool, len(board))
+	for i := 0; i < len(visited); i++ {
+		visited[i] = make([]bool, len(board[0]))
+	}
+	for i, v := range board {
+		for j := range v {
+			if searchWord(board, visited, word, 0, i, j) {
+				return true
+			}
+		}
+	}
+	return false
+}
+
+func isInBoard(board [][]byte, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+
+func searchWord(board [][]byte, visited [][]bool, word string, index, x, y int) bool {
+	if index == len(word)-1 {
+		return board[x][y] == word[index]
+	}
+	if board[x][y] == word[index] {
+		visited[x][y] = true
+		for i := 0; i < 4; i++ {
+			nx := x + dir[i][0]
+			ny := y + dir[i][1]
+			if isInBoard(board, nx, ny) && !visited[nx][ny] && searchWord(board, visited, word, index+1, nx, ny) {
+				return true
+			}
+		}
+		visited[x][y] = false
+	}
+	return false
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0211.Design-Add-and-Search-Words-Data-Structure/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0213.House-Robber-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0213.House-Robber-II.md b/website/content/ChapterFour/0200~0299/0213.House-Robber-II.md
new file mode 100755
index 000000000..e108507ba
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0213.House-Robber-II.md
@@ -0,0 +1,77 @@
+# [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)
+
+
+## 题目
+
+You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are **arranged in a circle.** That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and **it will automatically contact the police if two adjacent houses were broken into on the same night**.
+
+Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight **without alerting the police**.
+
+**Example 1**:
+
+    Input: [2,3,2]
+    Output: 3
+    Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
+                 because they are adjacent houses.
+
+**Example 2**:
+
+    Input: [1,2,3,1]
+    Output: 4
+    Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
+                 Total amount you can rob = 1 + 3 = 4.
+
+## 题目大意
+
+你是一个专业的小偷，计划偷窃沿街的房屋，每间房内都藏有一定的现金。这个地方所有的房屋都**围成一圈**，这意味着第一个房屋和最后一个房屋是紧挨着的。同时，相邻的房屋装有相互连通的防盗系统，**如果两间相邻的房屋在同一晚上被小偷闯入，系统会自动报警**。
+
+给定一个代表每个房屋存放金额的非负整数数组，计算你**在不触动警报装置的情况下**，能够偷窃到的最高金额。
+
+
+## 解题思路
+
+- 这一题是第 198 题的加强版。不过这次是在一个环形的街道中，即最后一个元素和第一个元素是邻居，在不触碰警报的情况下，问能够窃取的财产的最大值是多少？
+- 解题思路和第 198 完全一致，只需要增加额外的一个转换。由于首尾是相邻的，所以在取了第一个房子以后就不能取第 n 个房子，那么就在 [0,n - 1] 的区间内找出总价值最多的解，然后再 [1,n] 的区间内找出总价值最多的解，两者取最大值即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func rob213(nums []int) int {
+	n := len(nums)
+	if n == 0 {
+		return 0
+	}
+	if n == 1 {
+		return nums[0]
+	}
+	if n == 2 {
+		return max(nums[0], nums[1])
+	}
+	// 由于首尾是相邻的，所以需要对比 [0，n-1]、[1，n] 这两个区间的最大值
+	return max(rob213_1(nums, 0, n-2), rob213_1(nums, 1, n-1))
+}
+
+func rob213_1(nums []int, start, end int) int {
+	preMax := nums[start]
+	curMax := max(preMax, nums[start+1])
+	for i := start + 2; i <= end; i++ {
+		tmp := curMax
+		curMax = max(curMax, nums[i]+preMax)
+		preMax = tmp
+	}
+	return curMax
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0212.Word-Search-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0215.Kth-Largest-Element-in-an-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0215.Kth-Largest-Element-in-an-Array.md b/website/content/ChapterFour/0200~0299/0215.Kth-Largest-Element-in-an-Array.md
new file mode 100644
index 000000000..0213a51c2
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0215.Kth-Largest-Element-in-an-Array.md
@@ -0,0 +1,125 @@
+# [215. Kth Largest Element in an Array](https://leetcode.com/problems/kth-largest-element-in-an-array/)
+
+## 题目
+
+Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
+
+**Example 1**:
+
+```
+
+Input: [3,2,1,5,6,4] and k = 2
+Output: 5
+
+```
+
+**Example 2**:
+
+```
+
+Input: [3,2,3,1,2,4,5,5,6] and k = 4
+Output: 4
+
+```
+
+**Note**:     
+
+You may assume k is always valid, 1 ≤ k ≤ array's length.
+
+
+## 题目大意
+
+找出数组中第 K 大的元素。这一题非常经典。可以用 O(n) 的时间复杂度实现。
+
+## 解题思路
+
+- 在快速选择 quickselect 的 partition 操作中，每次 partition 操作结束都会返回一个点，这个标定点的下标和最终排序之后有序数组中这个元素所在的下标是一致的。利用这个特性，我们可以不断的划分数组区间，最终找到第 K 大的元素。执行一次 partition 操作以后，如果这个元素的下标比 K 小，那么接着就在后边的区间继续执行 partition 操作；如果这个元素的下标比 K 大，那么就在左边的区间继续执行 partition 操作；如果相等就直接输出这个下标对应的数组元素即可。
+- 快速选择 quickselect 的思路实现的算法时间复杂度为 O(n)，空间复杂度为 O(logn)。由于证明过程很繁琐，所以不再这里展开讲。具体证明可以参考《算法导论》第 9 章第 2 小节。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math/rand"
+	"sort"
+)
+
+// 解法一 排序，排序的方法反而速度是最快的
+func findKthLargest1(nums []int, k int) int {
+	sort.Ints(nums)
+	return nums[len(nums)-k]
+}
+
+// 解法二 这个方法的理论依据是 partition 得到的点的下标就是最终排序之后的下标，根据这个下标，我们可以判断第 K 大的数在哪里
+// 时间复杂度 O(n)，空间复杂度 O(log n)，最坏时间复杂度为 O(n^2)，空间复杂度 O(n)
+func findKthLargest(nums []int, k int) int {
+	m := len(nums) - k + 1 // mth smallest, from 1..len(nums)
+	return selectSmallest(nums, 0, len(nums)-1, m)
+}
+
+func selectSmallest(nums []int, l, r, i int) int {
+	if l >= r {
+		return nums[l]
+	}
+	q := partition(nums, l, r)
+	k := q - l + 1
+	if k == i {
+		return nums[q]
+	}
+	if i < k {
+		return selectSmallest(nums, l, q-1, i)
+	} else {
+		return selectSmallest(nums, q+1, r, i-k)
+	}
+}
+
+func partition(nums []int, l, r int) int {
+	k := l + rand.Intn(r-l+1) // 此处为优化，使得时间复杂度期望降为 O(n)，最坏时间复杂度为 O(n^2)
+	nums[k], nums[r] = nums[r], nums[k]
+	i := l - 1
+	// nums[l..i] <= nums[r]
+	// nums[i+1..j-1] > nums[r]
+	for j := l; j < r; j++ {
+		if nums[j] <= nums[r] {
+			i++
+			nums[i], nums[j] = nums[j], nums[i]
+		}
+	}
+	nums[i+1], nums[r] = nums[r], nums[i+1]
+	return i + 1
+}
+
+// 扩展题 剑指 Offer 40. 最小的 k 个数
+func getLeastNumbers(arr []int, k int) []int {
+	return selectSmallest1(arr, 0, len(arr)-1, k)[:k]
+}
+
+// 和 selectSmallest 实现完全一致，只是返回值不用再截取了，直接返回 nums 即可
+func selectSmallest1(nums []int, l, r, i int) []int {
+	if l >= r {
+		return nums
+	}
+	q := partition(nums, l, r)
+	k := q - l + 1
+	if k == i {
+		return nums
+	}
+	if i < k {
+		return selectSmallest1(nums, l, q-1, i)
+	} else {
+		return selectSmallest1(nums, q+1, r, i-k)
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0213.House-Robber-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0216.Combination-Sum-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0216.Combination-Sum-III.md b/website/content/ChapterFour/0200~0299/0216.Combination-Sum-III.md
new file mode 100755
index 000000000..c6dbd156b
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0216.Combination-Sum-III.md
@@ -0,0 +1,80 @@
+# [216. Combination Sum III](https://leetcode.com/problems/combination-sum-iii/)
+
+
+## 题目
+
+Find all possible combinations of **k** numbers that add up to a number **n**, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
+
+**Note**:
+
+- All numbers will be positive integers.
+- The solution set must not contain duplicate combinations.
+
+**Example 1**:
+
+    Input: k = 3, n = 7
+    Output: [[1,2,4]]
+
+**Example 2**:
+
+    Input: k = 3, n = 9
+    Output: [[1,2,6], [1,3,5], [2,3,4]]
+
+## 题目大意
+
+找出所有相加之和为 n 的 k 个数的组合。组合中只允许含有 1 - 9 的正整数，并且每种组合中不存在重复的数字。
+
+说明：
+
+- 所有数字都是正整数。
+- 解集不能包含重复的组合。
+
+
+## 解题思路
+
+- 这一题比第 39 题还要简单一些，在第 39 题上稍加改动就可以解出这一道题。
+- 第 39 题是给出数组，这一道题数组是固定死的 [1,2,3,4,5,6,7,8,9]，并且数字不能重复使用。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func combinationSum3(k int, n int) [][]int {
+	if k == 0 {
+		return [][]int{}
+	}
+	c, res := []int{}, [][]int{}
+	findcombinationSum3(k, n, 1, c, &res)
+	return res
+}
+
+func findcombinationSum3(k, target, index int, c []int, res *[][]int) {
+	if target == 0 {
+		if len(c) == k {
+			b := make([]int, len(c))
+			copy(b, c)
+			*res = append(*res, b)
+		}
+		return
+	}
+	for i := index; i < 10; i++ {
+		if target >= i {
+			c = append(c, i)
+			findcombinationSum3(k, target-i, i+1, c, res)
+			c = c[:len(c)-1]
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0215.Kth-Largest-Element-in-an-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0217.Contains-Duplicate/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0217.Contains-Duplicate.md b/website/content/ChapterFour/0200~0299/0217.Contains-Duplicate.md
new file mode 100644
index 000000000..ddb948e79
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0217.Contains-Duplicate.md
@@ -0,0 +1,69 @@
+# [217. Contains Duplicate](https://leetcode.com/problems/contains-duplicate/)
+
+## 题目
+
+Given an array of integers, find if the array contains any duplicates.
+
+Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.
+
+
+**Example 1**:
+
+```
+
+Input: [1,2,3,1]
+Output: true
+
+```
+**Example 2**:
+
+```
+
+Input: [1,2,3,4]
+Output: false
+
+```
+
+**Example 3**:
+
+```
+
+Input: [1,1,1,3,3,4,3,2,4,2]
+Output: true
+
+```
+
+## 题目大意
+
+这是一道简单题，如果数组里面有重复数字就输出 true，否则输出 flase。
+
+## 解题思路
+
+用 map 判断即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func containsDuplicate(nums []int) bool {
+	record := make(map[int]bool, len(nums))
+	for _, n := range nums {
+		if _, found := record[n]; found {
+			return true
+		}
+		record[n] = true
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0216.Combination-Sum-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0218.The-Skyline-Problem/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0218.The-Skyline-Problem.md b/website/content/ChapterFour/0200~0299/0218.The-Skyline-Problem.md
new file mode 100755
index 000000000..e563455f6
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0218.The-Skyline-Problem.md
@@ -0,0 +1,382 @@
+# [218. The Skyline Problem](https://leetcode.com/problems/the-skyline-problem/)
+
+## 题目
+
+A city's skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Now suppose you are **given the locations and height of all the buildings** as shown on a cityscape photo (Figure A), write a program to **output the skyline** formed by these buildings collectively (Figure B).
+
+![](https://img.halfrost.com/Leetcode/leetcode_218_0.png)
+
+![](https://img.halfrost.com/Leetcode/leetcode_218_1.png)
+
+The geometric information of each building is represented by a triplet of integers `[Li, Ri, Hi]`, where `Li` and `Ri` are the x coordinates of the left and right edge of the ith building, respectively, and `Hi` is its height. It is guaranteed that `0 ≤ Li, Ri ≤ INT_MAX`, `0 < Hi ≤ INT_MAX`, and `Ri - Li > 0`. You may assume all buildings are perfect rectangles grounded on an absolutely flat surface at height 0.
+
+For instance, the dimensions of all buildings in Figure A are recorded as: `[ [2 9 10], [3 7 15], [5 12 12], [15 20 10], [19 24 8] ]` .
+
+The output is a list of "**key points**" (red dots in Figure B) in the format of `[ [x1,y1], [x2, y2], [x3, y3], ... ]` that uniquely defines a skyline. **A key point is the left endpoint of a horizontal line segment**. Note that the last key point, where the rightmost building ends, is merely used to mark the termination of the skyline, and always has zero height. Also, the ground in between any two adjacent buildings should be considered part of the skyline contour.
+
+For instance, the skyline in Figure B should be represented as:`[ [2 10], [3 15], [7 12], [12 0], [15 10], [20 8], [24, 0] ]`.
+
+**Notes**:
+
+- The number of buildings in any input list is guaranteed to be in the range `[0, 10000]`.
+- The input list is already sorted in ascending order by the left x position `Li`.
+- The output list must be sorted by the x position.
+- There must be no consecutive horizontal lines of equal height in the output skyline. For instance, `[...[2 3], [4 5], [7 5], [11 5], [12 7]...]` is not acceptable; the three lines of height 5 should be merged into one in the final output as such: `[...[2 3], [4 5], [12 7], ...]`
+
+
+## 题目大意
+
+城市的天际线是从远处观看该城市中所有建筑物形成的轮廓的外部轮廓。现在，假设您获得了城市风光照片（图A）上显示的所有建筑物的位置和高度，请编写一个程序以输出由这些建筑物形成的天际线（图B）。
+
+每个建筑物的几何信息用三元组 [Li，Ri，Hi] 表示，其中 Li 和 Ri 分别是第 i 座建筑物左右边缘的 x 坐标，Hi 是其高度。可以保证 0 ≤ Li, Ri ≤ INT\_MAX, 0 < Hi ≤ INT\_MAX 和 Ri - Li > 0。您可以假设所有建筑物都是在绝对平坦且高度为 0 的表面上的完美矩形。
+
+例如，图 A 中所有建筑物的尺寸记录为：[ [2 9 10], [3 7 15], [5 12 12], [15 20 10], [19 24 8] ] 。
+
+输出是以 [ [x1,y1], [x2, y2], [x3, y3], ... ] 格式的“关键点”（图 B 中的红点）的列表，它们唯一地定义了天际线。关键点是水平线段的左端点。请注意，最右侧建筑物的最后一个关键点仅用于标记天际线的终点，并始终为零高度。此外，任何两个相邻建筑物之间的地面都应被视为天际线轮廓的一部分。
+
+例如，图 B 中的天际线应该表示为：[ [2 10], [3 15], [7 12], [12 0], [15 10], [20 8], [24, 0] ]。
+
+说明:
+
+- 任何输入列表中的建筑物数量保证在 [0, 10000] 范围内。
+- 输入列表已经按左 x 坐标 Li 进行升序排列。
+- 输出列表必须按 x 位排序。
+- 输出天际线中不得有连续的相同高度的水平线。例如 [...[2 3], [4 5], [7 5], [11 5], [12 7]...] 是不正确的答案；三条高度为 5 的线应该在最终输出中合并为一个：[...[2 3], [4 5], [12 7], ...]
+
+
+## 解题思路
+
+
+- 给出一个二维数组，每个子数组里面代表一个高楼的信息，一个高楼的信息包含 3 个信息，高楼起始坐标，高楼终止坐标，高楼高度。要求找到这些高楼的边际点，并输出这些边际点的高度信息。
+- 这一题可以用树状数组来解答。要求天际线，即找到楼与楼重叠区间外边缘的线，说白了是维护各个区间内的最值。这有 2 个需要解决的问题。
+  1. 如何维护最值。当一个高楼的右边界消失，剩下的各个小楼间还需要选出最大值作为天际线。剩下重重叠叠的小楼很多，树状数组如何维护区间最值是解决此类题的关键。
+  2. 如何维护天际线的转折点。有些楼与楼并非完全重叠，重叠一半的情况导致天际线出现转折点。如上图中标记的红色转折点。树状数组如何维护这些点呢？
+- 先解决第一个问题(维护最值)。树状数组只有 2 个操作，一个是 Add() 一个是 Query()。从上面关于这 2 个操作的讲解中可以知道这 2 个操作都不能满足我们的需求。Add() 操作可以改成维护区间内 max() 的操作。但是 max() 容易获得却很难“去除”。如上图 [3,7] 这个区间内的最大值是 15。根据树状数组的定义，[3,12] 这个区间内最值还是 15。观察上图可以看到 [5,12] 区间内最值其实是 12。树状数组如何维护这种最值呢？最大值既然难以“去除”，那么需要考虑如何让最大值“来的晚一点”。解决办法是将 Query() 操作含义从前缀含义改成后缀含义。Query(i) 查询区间是 [1,i]，现在查询区间变成 {{< katex >}}[i,+\infty){{< /katex >}}。例如：[i,j] 区间内最值是 {{< katex >}}max_{i...j}{{< /katex >}}，Query(j+1) 的结果不会包含 {{< katex >}}max_{i...j}{{< /katex >}}，因为它查询的区间是 {{< katex >}}[j+1,+\infty){{< /katex >}}。这样更改以后，可以有效避免前驱高楼对后面楼的累积 max() 最值的影响。具体做法，将 x 轴上的各个区间排序，按照 x 值大小从小到大排序。从左往右依次遍历各个区间。Add() 操作含义是加入每个区间右边界代表后缀区间的最值。这样不需要考虑“移除”最值的问题了。细心的读者可能又有疑问了：能否从右往左遍历区间，Query() 的含义继续延续前缀区间？这样做是可行的，解决第一个问题(维护最值)是可以的。但是这种处理办法解决第二个问题(维护转折点)会遇到麻烦。
+- 再解决第二个问题(维护转折点)。如果用前缀含义的 Query()，在单点 i 上除了考虑以这个点为结束点的区间，还需要考虑以这个单点 i 为起点的区间。如果是后缀含义的 Query() 就没有这个问题了，{{< katex >}}[i+1,+\infty){{< /katex >}} 这个区间内不用考虑以单点 i 为结束点的区间。此题用树状数组代码实现见下面解法一。
+- 这一题也可以用线段树来解。用线段树来解答，可以不用关心“楼挡住楼”的情况。由于楼的坐标是离散的，所以先把楼在 X 轴上两个坐标离散化。同第 699 题一样，楼的宽度是一个区间，但是离散的过程中，楼的宽度右边界需要减一，不然查询一个区间会包含两个点，导致错误的结果，例如，第一个楼是 [1,3)，楼高 10，第二个楼是 [3,6)，楼高 20 。第一个楼如果算上右边界 3，查询 [1,3] 的结果是 20，因为 [3,3] 这个点会查询到第二个楼上面去。所以每个楼的右边界应该减一。但是每个楼的右边界也要加入到 query 中，因为最终 query 的结果需要包含这些边界。将离散的数据排序以后，按照楼的信息，每个区间依次 update。最后统计的时候依次统计每个区间，如果当前区间的高度和前一个区间的高度一样，就算是等高的楼。当高度与前一个高度不相同的时候就算是天际线的边缘，就要添加到最后输出数组中。
+- 类似的线段树的题目有：第 715 题，第 732 题，第 699 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和第 699 题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
+- 这一题用线段树做时间复杂度有点高，可以用扫描线解题。扫描线的思路很简单，用一根根垂直于 X 轴的竖线，从最左边依次扫到最右边，扫描每一条大楼的边界，当进入大楼的左边界的时候，如果没有比这个左边界最高点更高的点，就记录下这个最高点 keyPoint，状态是进入状态。如果扫到一个大楼的左边界，有比它更高的高度，就不记录，因为它不是天际线，它被楼挡楼，挡在其他楼后面了。当扫到一个大楼的右边界的时候，如果是最高点，那么记录下它的状态是离开状态，此时还需要记录下第二高的点。在扫描线扫描的过程中，动态的维护大楼的高度，同时维护最高的高度和第二高的高度。其实只需要维护最高的高度这一个高度，因为当离开状态到来的时候，移除掉当前最高的，剩下的高度里面最高的就是第二高的高度。描述的伪代码如下：
+
+        // 扫描线伪代码
+        events = {{x: L , height: H , type: entering},
+        		  {x: R , height: H , type: leaving}}
+        event.SortByX()
+        ds = new DS()
+        
+        for e in events:
+        	if entering(e):
+        		if e.height > ds.max(): ans += [e.height]
+        		ds.add(e.height)
+        	if leaving(e):
+        		ds.remove(e.height)
+        		if e.height > ds.max(): ans += [ds.max()]
+
+- 动态插入，查找最大值可以选用的数据结构有，最大堆和二叉搜索树。最大堆找最大值 O(1)，插入 O(log n)，但是 remove_by_key 需要 O(n) 的时间复杂度，并且需要自己实现。二叉搜索树，查找 max，添加和删除元素都是 O(log n) 的时间复杂度。
+- 排序的时候也需要注意几个问题：如果大楼的边界相等，并且是进入状态，那么再按照高度从大到小进行排序；如果大楼的边界相等，并且是离开状态，那么高度按照从小到大进行排序。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/leetcode-go/template"
+)
+
+// 解法一 树状数组，时间复杂度 O(n log n)
+const LEFTSIDE = 1
+const RIGHTSIDE = 2
+
+type Point struct {
+	xAxis int
+	side  int
+	index int
+}
+
+func getSkyline(buildings [][]int) [][]int {
+	res := [][]int{}
+	if len(buildings) == 0 {
+		return res
+	}
+	allPoints, bit := make([]Point, 0), BinaryIndexedTree{}
+	// [x-axis (value), [1 (left) | 2 (right)], index (building number)]
+	for i, b := range buildings {
+		allPoints = append(allPoints, Point{xAxis: b[0], side: LEFTSIDE, index: i})
+		allPoints = append(allPoints, Point{xAxis: b[1], side: RIGHTSIDE, index: i})
+	}
+	sort.Slice(allPoints, func(i, j int) bool {
+		if allPoints[i].xAxis == allPoints[j].xAxis {
+			return allPoints[i].side < allPoints[j].side
+		}
+		return allPoints[i].xAxis < allPoints[j].xAxis
+	})
+	bit.Init(len(allPoints))
+	kth := make(map[Point]int)
+	for i := 0; i < len(allPoints); i++ {
+		kth[allPoints[i]] = i
+	}
+	for i := 0; i < len(allPoints); i++ {
+		pt := allPoints[i]
+		if pt.side == LEFTSIDE {
+			bit.Add(kth[Point{xAxis: buildings[pt.index][1], side: RIGHTSIDE, index: pt.index}], buildings[pt.index][2])
+		}
+		currHeight := bit.Query(kth[pt] + 1)
+		if len(res) == 0 || res[len(res)-1][1] != currHeight {
+			if len(res) > 0 && res[len(res)-1][0] == pt.xAxis {
+				res[len(res)-1][1] = currHeight
+			} else {
+				res = append(res, []int{pt.xAxis, currHeight})
+			}
+		}
+	}
+	return res
+}
+
+type BinaryIndexedTree struct {
+	tree     []int
+	capacity int
+}
+
+// Init define
+func (bit *BinaryIndexedTree) Init(capacity int) {
+	bit.tree, bit.capacity = make([]int, capacity+1), capacity
+}
+
+// Add define
+func (bit *BinaryIndexedTree) Add(index int, val int) {
+	for ; index > 0; index -= index & -index {
+		bit.tree[index] = max(bit.tree[index], val)
+	}
+}
+
+// Query define
+func (bit *BinaryIndexedTree) Query(index int) int {
+	sum := 0
+	for ; index <= bit.capacity; index += index & -index {
+		sum = max(sum, bit.tree[index])
+	}
+	return sum
+}
+
+// 解法二 线段树 Segment Tree，时间复杂度 O(n log n)
+func getSkyline1(buildings [][]int) [][]int {
+	st, ans, lastHeight, check := template.SegmentTree{}, [][]int{}, 0, false
+	posMap, pos := discretization218(buildings)
+	tmp := make([]int, len(posMap))
+	st.Init(tmp, func(i, j int) int {
+		return max(i, j)
+	})
+	for _, b := range buildings {
+		st.UpdateLazy(posMap[b[0]], posMap[b[1]-1], b[2])
+	}
+	for i := 0; i < len(pos); i++ {
+		h := st.QueryLazy(posMap[pos[i]], posMap[pos[i]])
+		if check == false && h != 0 {
+			ans = append(ans, []int{pos[i], h})
+			check = true
+		} else if i > 0 && h != lastHeight {
+			ans = append(ans, []int{pos[i], h})
+		}
+		lastHeight = h
+	}
+	return ans
+}
+
+func discretization218(positions [][]int) (map[int]int, []int) {
+	tmpMap, posArray, posMap := map[int]int{}, []int{}, map[int]int{}
+	for _, pos := range positions {
+		tmpMap[pos[0]]++
+		tmpMap[pos[1]-1]++
+		tmpMap[pos[1]]++
+	}
+	for k := range tmpMap {
+		posArray = append(posArray, k)
+	}
+	sort.Ints(posArray)
+	for i, pos := range posArray {
+		posMap[pos] = i
+	}
+	return posMap, posArray
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法三 扫描线 Sweep Line，时间复杂度 O(n log n)
+func getSkyline2(buildings [][]int) [][]int {
+	size := len(buildings)
+	es := make([]E, 0)
+	for i, b := range buildings {
+		l := b[0]
+		r := b[1]
+		h := b[2]
+		// 1-- enter
+		el := NewE(i, l, h, 0)
+		es = append(es, el)
+		// 0 -- leave
+		er := NewE(i, r, h, 1)
+		es = append(es, er)
+	}
+	skyline := make([][]int, 0)
+	sort.Slice(es, func(i, j int) bool {
+		if es[i].X == es[j].X {
+			if es[i].T == es[j].T {
+				if es[i].T == 0 {
+					return es[i].H > es[j].H
+				}
+				return es[i].H < es[j].H
+			}
+			return es[i].T < es[j].T
+		}
+		return es[i].X < es[j].X
+	})
+	pq := NewIndexMaxPQ(size)
+	for _, e := range es {
+		curH := pq.Front()
+		if e.T == 0 {
+			if e.H > curH {
+				skyline = append(skyline, []int{e.X, e.H})
+			}
+			pq.Enque(e.N, e.H)
+		} else {
+			pq.Remove(e.N)
+			h := pq.Front()
+			if curH > h {
+				skyline = append(skyline, []int{e.X, h})
+			}
+		}
+	}
+	return skyline
+}
+
+// 扫面线伪代码
+// events = {{x: L , height: H , type: entering},
+// 		  {x: R , height: H , type: leaving}}
+// event.SortByX()
+// ds = new DS()
+
+// for e in events:
+// 	if entering(e):
+// 		if e.height > ds.max(): ans += [e.height]
+// 		ds.add(e.height)
+// 	if leaving(e):
+// 		ds.remove(e.height)
+// 		if e.height > ds.max(): ans += [ds.max()]
+
+// E define
+type E struct { // 定义一个 event 事件
+	N int // number 编号
+	X int // x 坐标
+	H int // height 高度
+	T int // type  0-进入 1-离开
+}
+
+// NewE define
+func NewE(n, x, h, t int) E {
+	return E{
+		N: n,
+		X: x,
+		H: h,
+		T: t,
+	}
+}
+
+// IndexMaxPQ define
+type IndexMaxPQ struct {
+	items []int
+	pq    []int
+	qp    []int
+	total int
+}
+
+// NewIndexMaxPQ define
+func NewIndexMaxPQ(n int) IndexMaxPQ {
+	qp := make([]int, n)
+	for i := 0; i < n; i++ {
+		qp[i] = -1
+	}
+	return IndexMaxPQ{
+		items: make([]int, n),
+		pq:    make([]int, n+1),
+		qp:    qp,
+	}
+}
+
+// Enque define
+func (q *IndexMaxPQ) Enque(key, val int) {
+	q.total++
+	q.items[key] = val
+	q.pq[q.total] = key
+	q.qp[key] = q.total
+	q.swim(q.total)
+}
+
+// Front define
+func (q *IndexMaxPQ) Front() int {
+	if q.total < 1 {
+		return 0
+	}
+	return q.items[q.pq[1]]
+}
+
+// Remove define
+func (q *IndexMaxPQ) Remove(key int) {
+	rank := q.qp[key]
+	q.exch(rank, q.total)
+	q.total--
+	q.qp[key] = -1
+	q.sink(rank)
+}
+
+func (q *IndexMaxPQ) sink(n int) {
+	for 2*n <= q.total {
+		k := 2 * n
+		if k < q.total && q.less(k, k+1) {
+			k++
+		}
+		if q.less(k, n) {
+			break
+		}
+		q.exch(k, n)
+		n = k
+	}
+}
+
+func (q *IndexMaxPQ) swim(n int) {
+	for n > 1 {
+		k := n / 2
+		if q.less(n, k) {
+			break
+		}
+		q.exch(n, k)
+		n = k
+	}
+}
+
+func (q *IndexMaxPQ) exch(i, j int) {
+	q.pq[i], q.pq[j] = q.pq[j], q.pq[i]
+	q.qp[q.pq[i]] = i
+	q.qp[q.pq[j]] = j
+}
+
+func (q *IndexMaxPQ) less(i, j int) bool {
+	return q.items[q.pq[i]] < q.items[q.pq[j]]
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0217.Contains-Duplicate/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0219.Contains-Duplicate-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0219.Contains-Duplicate-II.md b/website/content/ChapterFour/0200~0299/0219.Contains-Duplicate-II.md
new file mode 100644
index 000000000..c40808250
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0219.Contains-Duplicate-II.md
@@ -0,0 +1,76 @@
+# [219. Contains Duplicate II](https://leetcode.com/problems/contains-duplicate-ii/)
+
+## 题目
+
+Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.
+
+
+**Example 1**:
+
+```
+
+Input: nums = [1,2,3,1], k = 3
+Output: true
+
+```
+**Example 2**:
+
+```
+
+Input: nums = [1,0,1,1], k = 1
+Output: true
+
+```
+
+**Example 3**:
+
+```
+
+Input: nums = [1,2,3,1,2,3], k = 2
+Output: false
+
+```
+
+## 题目大意
+
+这是一道简单题，如果数组里面有重复数字，并且重复数字的下标差值小于等于 K 就输出 true，如果没有重复数字或者下标差值超过了 K ，则输出 flase。
+
+## 解题思路
+
+这道题可以维护一个只有 K 个元素的 map，每次只需要判断这个 map 里面是否存在这个元素即可。如果存在就代表重复数字的下标差值在 K 以内。map 的长度如果超过了 K 以后就删除掉 i-k 的那个元素，这样一直维护 map 里面只有 K 个元素。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func containsNearbyDuplicate(nums []int, k int) bool {
+	if len(nums) <= 1 {
+		return false
+	}
+	if k <= 0 {
+		return false
+	}
+	record := make(map[int]bool, len(nums))
+	for i, n := range nums {
+		if _, found := record[n]; found {
+			return true
+		}
+		record[n] = true
+		if len(record) == k+1 {
+			delete(record, nums[i-k])
+		}
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0218.The-Skyline-Problem/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0220.Contains-Duplicate-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0220.Contains-Duplicate-III.md b/website/content/ChapterFour/0200~0299/0220.Contains-Duplicate-III.md
new file mode 100644
index 000000000..b26d478fc
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0220.Contains-Duplicate-III.md
@@ -0,0 +1,112 @@
+# [220. Contains Duplicate III](https://leetcode.com/problems/contains-duplicate-iii/)
+
+## 题目
+
+Given an array of integers, find out whether there are two distinct indices i and j in the array such that the **absolute** difference between **nums[i]** and **nums[j]** is at most t and the **absolute** difference between i and j is at most k.
+
+**Example 1**:
+
+    Input: nums = [1,2,3,1], k = 3, t = 0
+    Output: true
+
+**Example 2**:
+
+    Input: nums = [1,0,1,1], k = 1, t = 2
+    Output: true
+
+**Example 3**:
+
+    Input: nums = [1,5,9,1,5,9], k = 2, t = 3
+    Output: false
+
+
+
+
+## 题目大意
+
+给出一个数组 num，再给 K 和 t。问在 num 中能否找到一组 i 和 j，使得 num[i] 和 num[j] 的绝对差值最大为 t，并且 i 和 j 之前的绝对差值最大为 k。
+
+## 解题思路
+
+
+- 给出一个数组，要求在数组里面找到 2 个索引，`i` 和 `j`，使得 `| nums[i] - nums[j] | ≤ t` ，并且 `| i - j | ≤  k` 。
+- 这是一道滑动窗口的题目。第一想法就是用 `i` 和 `j` 两个指针，针对每个 `i` ，都从 `i + 1` 往后扫完整个数组，判断每个 `i` 和 `j` ，判断是否满足题意。`j` 在循环的过程中注意判断剪枝条件 `| i - j | ≤  k`。这个做法的时间复杂度是 O(n^2)。这个做法慢的原因在于滑动窗口的左边界和右边界在滑动过程中不是联动滑动的。
+- 于是考虑，如果数组是有序的呢？把数组按照元素值从小到大进行排序，如果元素值相等，就按照 index 从小到大进行排序。在这样有序的数组中找满足题意的 `i` 和 `j`，滑动窗口左边界和右边界就是联动的了。窗口的右边界滑到与左边界元素值的差值 ≤ t 的地方，满足了这个条件再判断 `| i - j | ≤  k`，如果右边界与左边界元素值的差值 > t 了，说明该把左边界往右移动了(能这样移动的原因就是因为我们将数组元素大小排序了，右移是增大元素的方向)。移动左边界的时候需要注意左边界不能超过右边界。这样滑动窗口一次滑过整个排序后的数组，就可以判断是否存在满足题意的 `i` 和 `j` 。这个做法的时间主要花在排序上了，时间复杂度是 O(n log n)。
+- 本题最优解是利用桶排序的思想。`| i - j | ≤  k` 这个条件利用一个窗口大小为 k 来维护。重点在 `| nums[i] - nums[j] | ≤ t` 这个条件如何满足。利用桶排序的思想，将 `nums[i]` 所有元素分为 ...,`[0,t]`,`[t+1,2t+1]`,...。每个区间的大小为 `t + 1`。每个元素现在都对应一个桶编号。进行 3 次查找即可确定能否找到满足这个 `| nums[i] - nums[j] | ≤ t` 条件的数对。如果在相同的桶中找到了元素，那么说明能找到这样的 i 和 j。还有 2 种可能对应桶边界的情况。如果存在前一个桶中的元素能使得相差的值也 `≤ t`，这样的数对同样满足题意。最后一种情况是，如果存在后一个桶中的元素能使得相差的值也 `≤ t`，这样的数对同样满足题意。查询 3 次，如果都不存在，说明当前的 i 找不到满足题意的 j。继续循环寻找。循环一遍都找不到满足题意的数对，输出 false。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 桶排序
+func containsNearbyAlmostDuplicate(nums []int, k int, t int) bool {
+	if k <= 0 || t < 0 || len(nums) < 2 {
+		return false
+	}
+	buckets := map[int]int{}
+	for i := 0; i < len(nums); i++ {
+		// Get the ID of the bucket from element value nums[i] and bucket width t + 1
+		key := nums[i] / (t + 1)
+		// -7/9 = 0, but need -7/9 = -1
+		if nums[i] < 0 {
+			key--
+		}
+		if _, ok := buckets[key]; ok {
+			return true
+		}
+		// check the lower bucket, and have to check the value too
+		if v, ok := buckets[key-1]; ok && nums[i]-v <= t {
+			return true
+		}
+		// check the upper bucket, and have to check the value too
+		if v, ok := buckets[key+1]; ok && v-nums[i] <= t {
+			return true
+		}
+		// maintain k size of window
+		if len(buckets) >= k {
+			delete(buckets, nums[i-k]/(t+1))
+		}
+		buckets[key] = nums[i]
+	}
+	return false
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
+
+// 解法二 滑动窗口 + 剪枝
+func containsNearbyAlmostDuplicate1(nums []int, k int, t int) bool {
+	if len(nums) <= 1 {
+		return false
+	}
+	if k <= 0 {
+		return false
+	}
+	n := len(nums)
+	for i := 0; i < n; i++ {
+		count := 0
+		for j := i + 1; j < n && count < k; j++ {
+			if abs(nums[i]-nums[j]) <= t {
+				return true
+			}
+			count++
+		}
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0219.Contains-Duplicate-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0222.Count-Complete-Tree-Nodes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0222.Count-Complete-Tree-Nodes.md b/website/content/ChapterFour/0200~0299/0222.Count-Complete-Tree-Nodes.md
new file mode 100644
index 000000000..a7f148ba7
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0222.Count-Complete-Tree-Nodes.md
@@ -0,0 +1,89 @@
+# [222. Count Complete Tree Nodes](https://leetcode.com/problems/count-complete-tree-nodes/)
+
+## 题目
+
+
+Given a complete binary tree, count the number of nodes.
+
+**Note**:   
+
+Definition of a complete binary tree from Wikipedia:
+In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.
+
+
+**Example**:
+
+```
+
+Input: 
+    1
+   / \
+  2   3
+ / \  /
+4  5 6
+
+Output: 6
+
+```
+
+## 题目大意
+
+输出一颗完全二叉树的结点个数。
+
+## 解题思路
+
+这道题其实按照层序遍历一次树，然后把每一层的结点个数相加即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func countNodes(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	queue := []*TreeNode{}
+	queue = append(queue, root)
+	curNum, nextLevelNum, res := 1, 0, 1
+	for len(queue) != 0 {
+		if curNum > 0 {
+			node := queue[0]
+			if node.Left != nil {
+				queue = append(queue, node.Left)
+				nextLevelNum++
+			}
+			if node.Right != nil {
+				queue = append(queue, node.Right)
+				nextLevelNum++
+			}
+			curNum--
+			queue = queue[1:]
+		}
+		if curNum == 0 {
+			res += nextLevelNum
+			curNum = nextLevelNum
+			nextLevelNum = 0
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0220.Contains-Duplicate-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0223.Rectangle-Area/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0223.Rectangle-Area.md b/website/content/ChapterFour/0200~0299/0223.Rectangle-Area.md
new file mode 100755
index 000000000..a85324ad1
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0223.Rectangle-Area.md
@@ -0,0 +1,60 @@
+# [223. Rectangle Area](https://leetcode.com/problems/rectangle-area/)
+
+
+## 题目
+
+Find the total area covered by two **rectilinear** rectangles in a **2D** plane.
+
+Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/22/rectangle_area.png)
+
+**Example**:
+
+    Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2
+    Output: 45
+
+**Note**:
+
+Assume that the total area is never beyond the maximum possible value of **int**.
+
+
+
+## 题目大意
+
+在二维平面上计算出两个由直线构成的矩形重叠后形成的总面积。每个矩形由其左下顶点和右上顶点坐标表示，如图所示。说明: 假设矩形面积不会超出 int 的范围。
+
+## 解题思路
+
+
+- 给出两个矩形的坐标，求这两个矩形在坐标轴上覆盖的总面积。
+- 几何题，由于只有 2 个矩形，所以按照题意做即可。先分别求两个矩形的面积，加起来再减去两个矩形重叠的面积。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func computeArea(A int, B int, C int, D int, E int, F int, G int, H int) int {
+	X0, Y0, X1, Y1 := max(A, E), max(B, F), min(C, G), min(D, H)
+	return area(A, B, C, D) + area(E, F, G, H) - area(X0, Y0, X1, Y1)
+}
+
+func area(x0, y0, x1, y1 int) int {
+	l, h := x1-x0, y1-y0
+	if l <= 0 || h <= 0 {
+		return 0
+	}
+	return l * h
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0222.Count-Complete-Tree-Nodes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0224.Basic-Calculator/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0224.Basic-Calculator.md b/website/content/ChapterFour/0200~0299/0224.Basic-Calculator.md
new file mode 100755
index 000000000..3ae0cccf2
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0224.Basic-Calculator.md
@@ -0,0 +1,172 @@
+# [224. Basic Calculator](https://leetcode.com/problems/basic-calculator/)
+
+
+## 题目
+
+Implement a basic calculator to evaluate a simple expression string.
+
+The expression string may contain open `(` and closing parentheses `)`, the plus `+` or minus sign `-`, **non-negative** integers and empty spaces .
+
+**Example 1**:
+
+    Input: "1 + 1"
+    Output: 2
+
+**Example 2**:
+
+    Input: " 2-1 + 2 "
+    Output: 3
+
+**Example 3**:
+
+    Input: "(1+(4+5+2)-3)+(6+8)"
+    Output: 23
+
+**Note**:
+
+- You may assume that the given expression is always valid.
+- **Do not** use the `eval` built-in library function.
+
+## 题目大意
+
+实现一个基本的计算器来计算一个简单的字符串表达式的值。字符串表达式可以包含左括号 ( ，右括号 )，加号 + ，减号 -，非负整数和空格  。
+
+## 解题思路
+
+- 注意点一：算式中有空格，需要跳过
+- 注意点二：算式中会出现负数，负负得正的情况需要特殊处理，所以需要记录每次计算出来的符号
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"container/list"
+	"fmt"
+	"strconv"
+)
+
+// 解法一
+func calculate(s string) int {
+	i, stack, result, sign := 0, list.New(), 0, 1 // 记录加减状态
+	for i < len(s) {
+		if s[i] == ' ' {
+			i++
+		} else if s[i] <= '9' && s[i] >= '0' { // 获取一段数字
+			base, v := 10, int(s[i]-'0')
+			for i+1 < len(s) && s[i+1] <= '9' && s[i+1] >= '0' {
+				v = v*base + int(s[i+1]-'0')
+				i++
+			}
+			result += v * sign
+			i++
+		} else if s[i] == '+' {
+			sign = 1
+			i++
+		} else if s[i] == '-' {
+			sign = -1
+			i++
+		} else if s[i] == '(' { // 把之前计算结果及加减状态压栈，开始新的计算
+			stack.PushBack(result)
+			stack.PushBack(sign)
+			result = 0
+			sign = 1
+			i++
+		} else if s[i] == ')' { // 新的计算结果 * 前一个加减状态 + 之前计算结果
+			result = result*stack.Remove(stack.Back()).(int) + stack.Remove(stack.Back()).(int)
+			i++
+		}
+	}
+	return result
+}
+
+// 解法二
+func calculate1(s string) int {
+	stack := []byte{}
+	for i := 0; i < len(s); i++ {
+		if s[i] == ' ' {
+			continue
+		} else if s[i] == ')' {
+			tmp, index := "", len(stack)-1
+			for ; index >= 0; index-- {
+				if stack[index] == '(' {
+					break
+				}
+			}
+			tmp = string(stack[index+1:])
+			stack = stack[:index]
+			res := strconv.Itoa(calculateStr(tmp))
+			for j := 0; j < len(res); j++ {
+				stack = append(stack, res[j])
+			}
+		} else {
+			stack = append(stack, s[i])
+		}
+	}
+	fmt.Printf("stack = %v\n", string(stack))
+	return calculateStr(string(stack))
+}
+
+func calculateStr(str string) int {
+	s, nums, tmpStr, res := []byte{}, []int{}, "", 0
+	// 处理符号的问题，++得+，--得+，+-、-+得-
+	for i := 0; i < len(str); i++ {
+		if len(s) > 0 && s[len(s)-1] == '+' && str[i] == '+' {
+			continue
+		} else if len(s) > 0 && s[len(s)-1] == '+' && str[i] == '-' {
+			s[len(s)-1] = '-'
+		} else if len(s) > 0 && s[len(s)-1] == '-' && str[i] == '+' {
+			continue
+		} else if len(s) > 0 && s[len(s)-1] == '-' && str[i] == '-' {
+			s[len(s)-1] = '+'
+		} else {
+			s = append(s, str[i])
+		}
+	}
+	str = string(s)
+	s = []byte{}
+	for i := 0; i < len(str); i++ {
+		if isDigital(str[i]) {
+			tmpStr += string(str[i])
+		} else {
+			num, _ := strconv.Atoi(tmpStr)
+			nums = append(nums, num)
+			tmpStr = ""
+			s = append(s, str[i])
+		}
+	}
+	if tmpStr != "" {
+		num, _ := strconv.Atoi(tmpStr)
+		nums = append(nums, num)
+	}
+	res = nums[0]
+	for i := 0; i < len(s); i++ {
+		if s[i] == '+' {
+			res += nums[i+1]
+		} else {
+			res -= nums[i+1]
+		}
+	}
+	fmt.Printf("s = %v nums = %v res = %v\n", string(s), nums, res)
+	return res
+}
+
+func isDigital(v byte) bool {
+	if v >= '0' && v <= '9' {
+		return true
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0223.Rectangle-Area/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0225.Implement-Stack-using-Queues/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0225.Implement-Stack-using-Queues.md b/website/content/ChapterFour/0200~0299/0225.Implement-Stack-using-Queues.md
new file mode 100644
index 000000000..47b8177c3
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0225.Implement-Stack-using-Queues.md
@@ -0,0 +1,102 @@
+# [225. Implement Stack using Queues](https://leetcode.com/problems/implement-stack-using-queues/)
+
+## 题目
+
+Implement the following operations of a stack using queues.
+
+- push(x) -- Push element x onto stack.
+- pop() -- Removes the element on top of the stack.
+- top() -- Get the top element.
+- empty() -- Return whether the stack is empty.
+
+**Example**:
+
+```
+
+MyStack stack = new MyStack();
+
+stack.push(1);
+stack.push(2);  
+stack.top();   // returns 2
+stack.pop();   // returns 2
+stack.empty(); // returns false
+
+```
+
+**Note**:  
+
+- You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is empty operations are valid.
+- Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
+- You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).
+ 
+
+## 题目大意
+
+题目要求用队列实现一个栈的基本操作：push(x)、pop()、top()、empty()。
+
+
+## 解题思路
+
+按照题目要求实现即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+type MyStack struct {
+	enque []int
+	deque []int
+}
+
+/** Initialize your data structure here. */
+func Constructor225() MyStack {
+	return MyStack{[]int{}, []int{}}
+}
+
+/** Push element x onto stack. */
+func (this *MyStack) Push(x int) {
+	this.enque = append(this.enque, x)
+}
+
+/** Removes the element on top of the stack and returns that element. */
+func (this *MyStack) Pop() int {
+	length := len(this.enque)
+	for i := 0; i < length-1; i++ {
+		this.deque = append(this.deque, this.enque[0])
+		this.enque = this.enque[1:]
+	}
+	topEle := this.enque[0]
+	this.enque = this.deque
+	this.deque = nil
+
+	return topEle
+}
+
+/** Get the top element. */
+func (this *MyStack) Top() int {
+	topEle := this.Pop()
+	this.enque = append(this.enque, topEle)
+
+	return topEle
+}
+
+/** Returns whether the stack is empty. */
+func (this *MyStack) Empty() bool {
+	if len(this.enque) == 0 {
+		return true
+	}
+
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0224.Basic-Calculator/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0226.Invert-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0226.Invert-Binary-Tree.md b/website/content/ChapterFour/0200~0299/0226.Invert-Binary-Tree.md
new file mode 100644
index 000000000..d9f53c25f
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0226.Invert-Binary-Tree.md
@@ -0,0 +1,83 @@
+# [226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)
+
+## 题目
+
+Invert a binary tree.
+
+**Example**:
+
+Input:
+
+```
+
+     4
+   /   \
+  2     7
+ / \   / \
+1   3 6   9
+
+```
+
+Output:
+
+```
+
+     4
+   /   \
+  7     2
+ / \   / \
+9   6 3   1
+
+```
+
+Trivia:   
+
+This problem was inspired by this original tweet by Max Howell:
+
+>Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.
+ 
+
+## 题目大意
+
+"经典"的反转二叉树的问题。
+
+
+## 解题思路
+
+还是用递归来解决，先递归调用反转根节点的左孩子，然后递归调用反转根节点的右孩子，然后左右交换根节点的左孩子和右孩子。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func invertTree(root *TreeNode) *TreeNode {
+	if root == nil {
+		return nil
+	}
+	invertTree(root.Left)
+	invertTree(root.Right)
+	root.Left, root.Right = root.Right, root.Left
+	return root
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0225.Implement-Stack-using-Queues/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0227.Basic-Calculator-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0227.Basic-Calculator-II.md b/website/content/ChapterFour/0200~0299/0227.Basic-Calculator-II.md
new file mode 100644
index 000000000..d40024449
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0227.Basic-Calculator-II.md
@@ -0,0 +1,87 @@
+# [227. Basic Calculator II](https://leetcode.com/problems/basic-calculator-ii/)
+
+
+## 题目
+
+Given a string `s` which represents an expression, *evaluate this expression and return its value*.
+
+The integer division should truncate toward zero.
+
+**Example 1:**
+
+```
+Input: s = "3+2*2"
+Output: 7
+```
+
+**Example 2:**
+
+```
+Input: s = " 3/2 "
+Output: 1
+```
+
+**Example 3:**
+
+```
+Input: s = " 3+5 / 2 "
+Output: 5
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 3 * 10^5`
+- `s` consists of integers and operators `('+', '-', '*', '/')` separated by some number of spaces.
+- `s` represents **a valid expression**.
+- All the integers in the expression are non-negative integers in the range `[0, 2^31 - 1]`.
+- The answer is **guaranteed** to fit in a **32-bit integer**.
+
+## 题目大意
+
+给你一个字符串表达式 `s` ，请你实现一个基本计算器来计算并返回它的值。整数除法仅保留整数部分。
+
+## 解题思路
+
+- 这道题是第 224 题的加强版。第 224 题中只有加减运算和括号，这一题增加了乘除运算。由于乘除运算的优先级高于加减。所以先计算乘除运算，将算出来的结果再替换回原来的算式中。最后只剩下加减运算，于是题目降级成了第 224 题。
+- 把加减运算符号后面的数字压入栈中，遇到乘除运算，直接将它与栈顶的元素计算，并将计算后的结果放入栈顶。若读到一个运算符，或者遍历到字符串末尾，即认为是遍历到了数字末尾。处理完该数字后，更新 `preSign` 为当前遍历的字符。遍历完字符串 `s` 后，将栈中元素累加，即为该字符串表达式的值。时间复杂度 O(n)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+func calculate(s string) int {
+	stack, preSign, num, res := []int{}, '+', 0, 0
+	for i, ch := range s {
+		isDigit := '0' <= ch && ch <= '9'
+		if isDigit {
+			num = num*10 + int(ch-'0')
+		}
+		if !isDigit && ch != ' ' || i == len(s)-1 {
+			switch preSign {
+			case '+':
+				stack = append(stack, num)
+			case '-':
+				stack = append(stack, -num)
+			case '*':
+				stack[len(stack)-1] *= num
+			default:
+				stack[len(stack)-1] /= num
+			}
+			preSign = ch
+			num = 0
+		}
+	}
+	for _, v := range stack {
+		res += v
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0226.Invert-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0228.Summary-Ranges/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0228.Summary-Ranges.md b/website/content/ChapterFour/0200~0299/0228.Summary-Ranges.md
new file mode 100644
index 000000000..974e066f2
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0228.Summary-Ranges.md
@@ -0,0 +1,115 @@
+# [228. Summary Ranges](https://leetcode.com/problems/summary-ranges/)
+
+
+## 题目
+
+You are given a **sorted unique** integer array `nums`.
+
+Return *the **smallest sorted** list of ranges that **cover all the numbers in the array exactly***. That is, each element of `nums` is covered by exactly one of the ranges, and there is no integer `x` such that `x` is in one of the ranges but not in `nums`.
+
+Each range `[a,b]` in the list should be output as:
+
+- `"a->b"` if `a != b`
+- `"a"` if `a == b`
+
+**Example 1**:
+
+```
+Input: nums = [0,1,2,4,5,7]
+Output: ["0->2","4->5","7"]
+Explanation: The ranges are:
+[0,2] --> "0->2"
+[4,5] --> "4->5"
+[7,7] --> "7"
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [0,2,3,4,6,8,9]
+Output: ["0","2->4","6","8->9"]
+Explanation: The ranges are:
+[0,0] --> "0"
+[2,4] --> "2->4"
+[6,6] --> "6"
+[8,9] --> "8->9"
+
+```
+
+**Example 3**:
+
+```
+Input: nums = []
+Output: []
+
+```
+
+**Example 4**:
+
+```
+Input: nums = [-1]
+Output: ["-1"]
+
+```
+
+**Example 5**:
+
+```
+Input: nums = [0]
+Output: ["0"]
+
+```
+
+**Constraints**:
+
+- `0 <= nums.length <= 20`
+- `231 <= nums[i] <= 231 - 1`
+- All the values of `nums` are **unique**.
+- `nums` is sorted in ascending order.
+
+## 题目大意
+
+给定一个无重复元素的有序整数数组 nums 。
+
+返回 恰好覆盖数组中所有数字 的 最小有序 区间范围列表。也就是说，nums 的每个元素都恰好被某个区间范围所覆盖，并且不存在属于某个范围但不属于 nums 的数字 x 。
+
+列表中的每个区间范围 [a,b] 应该按如下格式输出：
+
+- "a->b" ，如果 a != b
+- "a" ，如果 a == b
+
+## 解题思路
+
+- 简单题。按照题意，用一个游标变量累加寻找连续的区间。一旦出现了中断，就按照题意格式输出。输出的规则有多种，带箭头的区间，单个元素区间，空区间。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"strconv"
+)
+
+func summaryRanges(nums []int) (ans []string) {
+	for i, n := 0, len(nums); i < n; {
+		left := i
+		for i++; i < n && nums[i-1]+1 == nums[i]; i++ {
+		}
+		s := strconv.Itoa(nums[left])
+		if left != i-1 {
+			s += "->" + strconv.Itoa(nums[i-1])
+		}
+		ans = append(ans, s)
+	}
+	return
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0227.Basic-Calculator-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0229.Majority-Element-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0229.Majority-Element-II.md b/website/content/ChapterFour/0200~0299/0229.Majority-Element-II.md
new file mode 100755
index 000000000..fb043de6f
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0229.Majority-Element-II.md
@@ -0,0 +1,107 @@
+# [229. Majority Element II](https://leetcode.com/problems/majority-element-ii/)
+
+
+## 题目
+
+Given an integer array of size n, find all elements that appear more than `⌊ n/3 ⌋` times.
+
+**Note**: The algorithm should run in linear time and in O(1) space.
+
+**Example 1**:
+
+    Input: [3,2,3]
+    Output: [3]
+
+**Example 2**:
+
+    Input: [1,1,1,3,3,2,2,2]
+    Output: [1,2]
+
+
+## 题目大意
+
+给定一个大小为 n 的数组，找出其中所有出现超过 ⌊ n/3 ⌋ 次的元素。说明: 要求算法的时间复杂度为 O(n)，空间复杂度为 O(1)。
+
+## 解题思路
+
+- 这一题是第 169 题的加强版。Boyer-Moore Majority Vote algorithm 算法的扩展版。
+- 题目要求找出数组中出现次数大于 `⌊ n/3 ⌋` 次的数。要求空间复杂度为 O(1)。简单题。
+- 这篇文章写的不错，可参考：[https://gregable.com/2013/10/majority-vote-algorithm-find-majority.html](https://gregable.com/2013/10/majority-vote-algorithm-find-majority.html)
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 时间复杂度 O(n) 空间复杂度 O(1)
+func majorityElement229(nums []int) []int {
+	// since we are checking if a num appears more than 1/3 of the time
+	// it is only possible to have at most 2 nums (>1/3 + >1/3 = >2/3)
+	count1, count2, candidate1, candidate2 := 0, 0, 0, 1
+	// Select Candidates
+	for _, num := range nums {
+		if num == candidate1 {
+			count1++
+		} else if num == candidate2 {
+			count2++
+		} else if count1 <= 0 {
+			// We have a bad first candidate, replace!
+			candidate1, count1 = num, 1
+		} else if count2 <= 0 {
+			// We have a bad second candidate, replace!
+			candidate2, count2 = num, 1
+		} else {
+			// Both candidates suck, boo!
+			count1--
+			count2--
+		}
+	}
+	// Recount!
+	count1, count2 = 0, 0
+	for _, num := range nums {
+		if num == candidate1 {
+			count1++
+		} else if num == candidate2 {
+			count2++
+		}
+	}
+	length := len(nums)
+	if count1 > length/3 && count2 > length/3 {
+		return []int{candidate1, candidate2}
+	}
+	if count1 > length/3 {
+		return []int{candidate1}
+	}
+	if count2 > length/3 {
+		return []int{candidate2}
+	}
+	return []int{}
+}
+
+// 解法二 时间复杂度 O(n) 空间复杂度 O(n)
+func majorityElement229_1(nums []int) []int {
+	result, m := make([]int, 0), make(map[int]int)
+	for _, val := range nums {
+		if v, ok := m[val]; ok {
+			m[val] = v + 1
+		} else {
+			m[val] = 1
+		}
+	}
+	for k, v := range m {
+		if v > len(nums)/3 {
+			result = append(result, k)
+		}
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0228.Summary-Ranges/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0230.Kth-Smallest-Element-in-a-BST/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0230.Kth-Smallest-Element-in-a-BST.md b/website/content/ChapterFour/0200~0299/0230.Kth-Smallest-Element-in-a-BST.md
new file mode 100755
index 000000000..3383a50fb
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0230.Kth-Smallest-Element-in-a-BST.md
@@ -0,0 +1,85 @@
+# [230. Kth Smallest Element in a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)
+
+
+## 题目
+
+Given a binary search tree, write a function `kthSmallest` to find the **k**th smallest element in it.
+
+**Note**: You may assume k is always valid, 1 ≤ k ≤ BST's total elements.
+
+**Example 1**:
+
+    Input: root = [3,1,4,null,2], k = 1
+       3
+      / \
+     1   4
+      \
+       2
+    Output: 1
+
+**Example 2**:
+
+    Input: root = [5,3,6,2,4,null,null,1], k = 3
+           5
+          / \
+         3   6
+        / \
+       2   4
+      /
+     1
+    Output: 3
+
+**Follow up**:What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?
+
+
+## 题目大意
+
+给定一个二叉搜索树，编写一个函数 kthSmallest 来查找其中第 k 个最小的元素。你可以假设 k 总是有效的，1 ≤ k ≤ 二叉搜索树元素个数。
+
+
+## 解题思路
+
+- 由于二叉搜索树有序的特性，所以中根遍历它，遍历到第 K 个数的时候就是结果
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func kthSmallest(root *TreeNode, k int) int {
+	res, count := 0, 0
+	inorder230(root, k, &count, &res)
+	return res
+}
+
+func inorder230(node *TreeNode, k int, count *int, ans *int) {
+	if node != nil {
+		inorder230(node.Left, k, count, ans)
+		*count++
+		if *count == k {
+			*ans = node.Val
+			return
+		}
+		inorder230(node.Right, k, count, ans)
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0229.Majority-Element-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0231.Power-of-Two/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0231.Power-of-Two.md b/website/content/ChapterFour/0200~0299/0231.Power-of-Two.md
new file mode 100755
index 000000000..491f55343
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0231.Power-of-Two.md
@@ -0,0 +1,77 @@
+# [231. Power of Two](https://leetcode.com/problems/power-of-two/)
+
+## 题目
+
+Given an integer, write a function to determine if it is a power of two.
+
+**Example 1**:
+
+    Input: 1
+    Output: true 
+    Explanation: 2^0 = 1
+
+**Example 2**:
+
+    Input: 16
+    Output: true
+    Explanation: 2^4 = 16
+
+**Example 3**:
+
+    Input: 218
+    Output: false
+
+## 题目大意
+
+给定一个整数，编写一个函数来判断它是否是 2 的幂次方。
+
+
+## 解题思路
+
+- 判断一个数是不是 2 的 n 次方。
+- 这一题最简单的思路是循环，可以通过。但是题目要求不循环就要判断，这就需要用到数论的知识了。这一题和第 326 题是一样的思路。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 二进制位操作法
+func isPowerOfTwo(num int) bool {
+	return (num > 0 && ((num & (num - 1)) == 0))
+}
+
+// 解法二 数论
+func isPowerOfTwo1(num int) bool {
+	return num > 0 && (1073741824%num == 0)
+}
+
+// 解法三 打表法
+func isPowerOfTwo2(num int) bool {
+	allPowerOfTwoMap := map[int]int{1: 1, 2: 2, 4: 4, 8: 8, 16: 16, 32: 32, 64: 64, 128: 128, 256: 256, 512: 512, 1024: 1024, 2048: 2048, 4096: 4096, 8192: 8192, 16384: 16384, 32768: 32768, 65536: 65536, 131072: 131072, 262144: 262144, 524288: 524288, 1048576: 1048576, 2097152: 2097152, 4194304: 4194304, 8388608: 8388608, 16777216: 16777216, 33554432: 33554432, 67108864: 67108864, 134217728: 134217728, 268435456: 268435456, 536870912: 536870912, 1073741824: 1073741824}
+	_, ok := allPowerOfTwoMap[num]
+	return ok
+}
+
+// 解法四 循环
+func isPowerOfTwo3(num int) bool {
+	for num >= 2 {
+		if num%2 == 0 {
+			num = num / 2
+		} else {
+			return false
+		}
+	}
+	return num == 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0230.Kth-Smallest-Element-in-a-BST/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0232.Implement-Queue-using-Stacks/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0232.Implement-Queue-using-Stacks.md b/website/content/ChapterFour/0200~0299/0232.Implement-Queue-using-Stacks.md
new file mode 100644
index 000000000..71e5bc148
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0232.Implement-Queue-using-Stacks.md
@@ -0,0 +1,106 @@
+# [232. Implement Queue using Stacks](https://leetcode.com/problems/implement-queue-using-stacks/)
+
+## 题目
+
+Implement the following operations of a queue using stacks.
+
+- push(x) -- Push element x to the back of queue.
+- pop() -- Removes the element from in front of queue.
+- peek() -- Get the front element.
+- empty() -- Return whether the queue is empty.
+
+**Example**:
+
+```
+
+MyQueue queue = new MyQueue();
+
+queue.push(1);
+queue.push(2);  
+queue.peek();  // returns 1
+queue.pop();   // returns 1
+queue.empty(); // returns false
+
+```
+
+**Note**:  
+
+- You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is empty operations are valid.
+- Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
+- You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
+ 
+
+## 题目大意
+
+题目要求用栈实现一个队列的基本操作：push(x)、pop()、peek()、empty()。
+
+## 解题思路
+
+按照题目要求实现即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+type MyQueue struct {
+	Stack *[]int
+	Queue *[]int
+}
+
+/** Initialize your data structure here. */
+func Constructor232() MyQueue {
+	tmp1, tmp2 := []int{}, []int{}
+	return MyQueue{Stack: &tmp1, Queue: &tmp2}
+}
+
+/** Push element x to the back of queue. */
+func (this *MyQueue) Push(x int) {
+	*this.Stack = append(*this.Stack, x)
+}
+
+/** Removes the element from in front of queue and returns that element. */
+func (this *MyQueue) Pop() int {
+	if len(*this.Queue) == 0 {
+		this.fromStackToQueue(this.Stack, this.Queue)
+	}
+
+	popped := (*this.Queue)[len(*this.Queue)-1]
+	*this.Queue = (*this.Queue)[:len(*this.Queue)-1]
+	return popped
+}
+
+/** Get the front element. */
+func (this *MyQueue) Peek() int {
+	if len(*this.Queue) == 0 {
+		this.fromStackToQueue(this.Stack, this.Queue)
+	}
+
+	return (*this.Queue)[len(*this.Queue)-1]
+}
+
+/** Returns whether the queue is empty. */
+func (this *MyQueue) Empty() bool {
+	return len(*this.Stack)+len(*this.Queue) == 0
+}
+
+func (this *MyQueue) fromStackToQueue(s, q *[]int) {
+	for len(*s) > 0 {
+		popped := (*s)[len(*s)-1]
+		*s = (*s)[:len(*s)-1]
+
+		*q = append(*q, popped)
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0231.Power-of-Two/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0234.Palindrome-Linked-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0234.Palindrome-Linked-List.md b/website/content/ChapterFour/0200~0299/0234.Palindrome-Linked-List.md
new file mode 100644
index 000000000..2870f9571
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0234.Palindrome-Linked-List.md
@@ -0,0 +1,138 @@
+# [234. Palindrome Linked List](https://leetcode.com/problems/palindrome-linked-list/)
+
+## 题目
+
+Given a singly linked list, determine if it is a palindrome.
+
+**Example 1**:
+
+```
+
+Input: 1->2
+Output: false
+
+```
+
+**Example 2**:
+
+```
+
+Input: 1->2->2->1
+Output: true
+
+```
+
+**Follow up**:  
+
+Could you do it in O(n) time and O(1) space?
+
+## 题目大意
+
+判断一个链表是否是回文链表。要求时间复杂度 O(n)，空间复杂度 O(1)。
+
+## 解题思路
+
+这道题只需要在第 143 题上面改改就可以了。思路是完全一致的。先找到中间结点，然后反转中间结点后面到结尾的所有结点。最后一一判断头结点开始的结点和中间结点往后开始的结点是否相等。如果一直相等，就是回文链表，如果有不相等的，直接返回不是回文链表。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// 解法一
+func isPalindrome(head *ListNode) bool {
+	slice := []int{}
+	for head != nil {
+		slice = append(slice, head.Val)
+		head = head.Next
+	}
+	for i, j := 0, len(slice)-1; i < j; {
+		if slice[i] != slice[j] {
+			return false
+		}
+		i++
+		j--
+	}
+	return true
+}
+
+// 解法二
+// 此题和 143 题 Reorder List 思路基本一致
+func isPalindrome1(head *ListNode) bool {
+	if head == nil || head.Next == nil {
+		return true
+	}
+	res := true
+	// 寻找中间结点
+	p1 := head
+	p2 := head
+	for p2.Next != nil && p2.Next.Next != nil {
+		p1 = p1.Next
+		p2 = p2.Next.Next
+	}
+	// 反转链表后半部分  1->2->3->4->5->6 to 1->2->3->6->5->4
+	preMiddle := p1
+	preCurrent := p1.Next
+	for preCurrent.Next != nil {
+		current := preCurrent.Next
+		preCurrent.Next = current.Next
+		current.Next = preMiddle.Next
+		preMiddle.Next = current
+	}
+	// 扫描表，判断是否是回文
+	p1 = head
+	p2 = preMiddle.Next
+	// fmt.Printf("p1 = %v p2 = %v preMiddle = %v head = %v\n", p1.Val, p2.Val, preMiddle.Val, L2ss(head))
+	for p1 != preMiddle {
+		// fmt.Printf("*****p1 = %v p2 = %v preMiddle = %v head = %v\n", p1, p2, preMiddle, L2ss(head))
+		if p1.Val == p2.Val {
+			p1 = p1.Next
+			p2 = p2.Next
+			// fmt.Printf("-------p1 = %v p2 = %v preMiddle = %v head = %v\n", p1, p2, preMiddle, L2ss(head))
+		} else {
+			res = false
+			break
+		}
+	}
+	if p1 == preMiddle {
+		if p2 != nil && p1.Val != p2.Val {
+			return false
+		}
+	}
+	return res
+}
+
+// L2ss define
+func L2ss(head *ListNode) []int {
+	res := []int{}
+	for head != nil {
+		res = append(res, head.Val)
+		head = head.Next
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0232.Implement-Queue-using-Stacks/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree.md b/website/content/ChapterFour/0200~0299/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree.md
new file mode 100755
index 000000000..77f6cdbfa
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree.md
@@ -0,0 +1,79 @@
+# [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)
+
+
+## 题目
+
+Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
+
+According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow **a node to be a descendant of itself**).”
+
+Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/14/binarysearchtree_improved.png)
+
+**Example 1**:
+
+    Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
+    Output: 6
+    Explanation: The LCA of nodes 2 and 8 is 6.
+
+**Example 2**:
+
+    Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
+    Output: 2
+    Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
+
+**Note**:
+
+- All of the nodes' values will be unique.
+- p and q are different and both values will exist in the BST.
+
+## 题目大意
+
+给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。
+
+百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”
+
+
+
+## 解题思路
+
+- 在二叉搜索树中求两个节点的最近公共祖先，由于二叉搜索树的特殊性质，所以找任意两个节点的最近公共祖先非常简单。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for TreeNode.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *ListNode
+ *     Right *ListNode
+ * }
+ */
+func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
+	if p == nil || q == nil || root == nil {
+		return nil
+	}
+	if p.Val < root.Val && q.Val < root.Val {
+		return lowestCommonAncestor(root.Left, p, q)
+	}
+	if p.Val > root.Val && q.Val > root.Val {
+		return lowestCommonAncestor(root.Right, p, q)
+	}
+	return root
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0234.Palindrome-Linked-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0236.Lowest-Common-Ancestor-of-a-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0236.Lowest-Common-Ancestor-of-a-Binary-Tree.md b/website/content/ChapterFour/0200~0299/0236.Lowest-Common-Ancestor-of-a-Binary-Tree.md
new file mode 100755
index 000000000..302f3f19a
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0236.Lowest-Common-Ancestor-of-a-Binary-Tree.md
@@ -0,0 +1,79 @@
+# [236. Lowest Common Ancestor of a Binary Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/)
+
+
+## 题目
+
+Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
+
+According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow **a node to be a descendant of itself**).”
+
+Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/binarytree.png)
+
+**Example 1**:
+
+    Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
+    Output: 3
+    Explanation: The LCA of nodes 5 and 1 is 3.
+
+**Example 2**:
+
+    Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
+    Output: 5
+    Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
+
+**Note**:
+
+- All of the nodes' values will be unique.
+- p and q are different and both values will exist in the binary tree.
+
+## 题目大意
+
+给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。
+
+百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”
+
+
+## 解题思路
+
+- 这是一套经典的题目，寻找任意一个二叉树中两个结点的 LCA 最近公共祖先，考察递归
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for TreeNode.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *ListNode
+ *     Right *ListNode
+ * }
+ */
+func lowestCommonAncestor236(root, p, q *TreeNode) *TreeNode {
+	if root == nil || root == q || root == p {
+		return root
+	}
+	left := lowestCommonAncestor236(root.Left, p, q)
+	right := lowestCommonAncestor236(root.Right, p, q)
+	if left != nil {
+		if right != nil {
+			return root
+		}
+		return left
+	}
+	return right
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0237.Delete-Node-in-a-Linked-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0237.Delete-Node-in-a-Linked-List.md b/website/content/ChapterFour/0200~0299/0237.Delete-Node-in-a-Linked-List.md
new file mode 100644
index 000000000..e8690ff52
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0237.Delete-Node-in-a-Linked-List.md
@@ -0,0 +1,101 @@
+# [237. Delete Node in a Linked List](https://leetcode.com/problems/delete-node-in-a-linked-list/)
+
+
+## 题目
+
+Write a function to **delete a node** in a singly-linked list. You will **not** be given access to the `head` of the list, instead you will be given access to **the node to be deleted** directly.
+
+It is **guaranteed** that the node to be deleted is **not a tail node** in the list.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/09/01/node1.jpg](https://assets.leetcode.com/uploads/2020/09/01/node1.jpg)
+
+```
+Input: head = [4,5,1,9], node = 5
+Output: [4,1,9]
+Explanation:You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/09/01/node2.jpg](https://assets.leetcode.com/uploads/2020/09/01/node2.jpg)
+
+```
+Input: head = [4,5,1,9], node = 1
+Output: [4,5,9]
+Explanation:You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.
+
+```
+
+**Example 3:**
+
+```
+Input: head = [1,2,3,4], node = 3
+Output: [1,2,4]
+
+```
+
+**Example 4:**
+
+```
+Input: head = [0,1], node = 0
+Output: [1]
+
+```
+
+**Example 5:**
+
+```
+Input: head = [-3,5,-99], node = -3
+Output: [5,-99]
+
+```
+
+**Constraints:**
+
+- The number of the nodes in the given list is in the range `[2, 1000]`.
+- `1000 <= Node.val <= 1000`
+- The value of each node in the list is **unique**.
+- The `node` to be deleted is **in the list** and is **not a tail** node
+
+## 题目大意
+
+删除给点结点。没有给链表的头结点。
+
+## 解题思路
+
+其实就是把后面的结点都覆盖上来即可。或者直接当前结点的值等于下一个结点，Next 指针指向下下个结点，这样做也可以，只不过中间有一个结点不被释放，内存消耗多一些。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func deleteNode(node *ListNode) {
+	node.Val = node.Next.Val
+	node.Next = node.Next.Next
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0236.Lowest-Common-Ancestor-of-a-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0239.Sliding-Window-Maximum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0239.Sliding-Window-Maximum.md b/website/content/ChapterFour/0200~0299/0239.Sliding-Window-Maximum.md
new file mode 100755
index 000000000..bf2a3a115
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0239.Sliding-Window-Maximum.md
@@ -0,0 +1,102 @@
+# [239. Sliding Window Maximum](https://leetcode.com/problems/sliding-window-maximum/)
+
+
+
+## 题目
+
+Given an array *nums*, there is a sliding window of size *k* which is moving from the very left of the array to the very right. You can only see the *k* numbers in the window. Each time the sliding window moves right by one position. Return the max sliding window.
+
+**Example**:
+
+    Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
+    Output: [3,3,5,5,6,7] 
+    Explanation: 
+    
+    Window position                Max
+    ---------------               -----
+    [1  3  -1] -3  5  3  6  7       3
+     1 [3  -1  -3] 5  3  6  7       3
+     1  3 [-1  -3  5] 3  6  7       5
+     1  3  -1 [-3  5  3] 6  7       5
+     1  3  -1  -3 [5  3  6] 7       6
+     1  3  -1  -3  5 [3  6  7]      7
+
+**Note**:
+
+You may assume *k* is always valid, 1 ≤ k ≤ input array's size for non-empty array.
+
+**Follow up**:
+
+Could you solve it in linear time?
+
+
+## 题目大意
+
+给定一个数组 nums，有一个大小为 k 的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口 k 内的数字。滑动窗口每次只向右移动一位。返回滑动窗口最大值。
+
+
+## 解题思路
+
+- 给定一个数组和一个窗口为 K 的窗口，当窗口从数组的左边滑动到数组右边的时候，输出每次移动窗口以后，在窗口内的最大值。
+- 这道题最暴力的方法就是 2 层循环，时间复杂度 O(n * K)。
+- 另一种思路是用优先队列，每次窗口以后的时候都向优先队列里面新增一个节点，并删除一个节点。时间复杂度是 O(n * log n)
+- 最优的解法是用双端队列，队列的一边永远都存的是窗口的最大值，队列的另外一个边存的是比最大值小的值。队列中最大值左边的所有值都出队。在保证了双端队列的一边即是最大值以后，时间复杂度是 O(n)，空间复杂度是 O(K)
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 暴力解法 O(nk)
+func maxSlidingWindow1(a []int, k int) []int {
+	res := make([]int, 0, k)
+	n := len(a)
+	if n == 0 {
+		return []int{}
+	}
+	for i := 0; i <= n-k; i++ {
+		max := a[i]
+		for j := 1; j < k; j++ {
+			if max < a[i+j] {
+				max = a[i+j]
+			}
+		}
+		res = append(res, max)
+	}
+
+	return res
+}
+
+// 解法二 双端队列 Deque
+func maxSlidingWindow(nums []int, k int) []int {
+	if len(nums) == 0 || len(nums) < k {
+		return make([]int, 0)
+	}
+	window := make([]int, 0, k) // store the index of nums
+	result := make([]int, 0, len(nums)-k+1)
+	for i, v := range nums { // if the left-most index is out of window, remove it
+		if i >= k && window[0] <= i-k {
+			window = window[1:len(window)]
+		}
+		for len(window) > 0 && nums[window[len(window)-1]] < v { // maintain window
+			window = window[0 : len(window)-1]
+		}
+		window = append(window, i) // store the index of nums
+		if i >= k-1 {
+			result = append(result, nums[window[0]]) // the left-most is the index of max value in nums
+		}
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0237.Delete-Node-in-a-Linked-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0240.Search-a-2D-Matrix-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0240.Search-a-2D-Matrix-II.md b/website/content/ChapterFour/0200~0299/0240.Search-a-2D-Matrix-II.md
new file mode 100755
index 000000000..55f99caf8
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0240.Search-a-2D-Matrix-II.md
@@ -0,0 +1,96 @@
+# [240. Search a 2D Matrix II](https://leetcode.com/problems/search-a-2d-matrix-ii/)
+
+
+## 题目
+
+Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
+
+- Integers in each row are sorted in ascending from left to right.
+- Integers in each column are sorted in ascending from top to bottom.
+
+**Example**:
+
+Consider the following matrix:
+
+    [
+      [1,   4,  7, 11, 15],
+      [2,   5,  8, 12, 19],
+      [3,   6,  9, 16, 22],
+      [10, 13, 14, 17, 24],
+      [18, 21, 23, 26, 30]
+    ]
+
+Given target = `5`, return `true`.
+
+Given target = `20`, return `false`.
+
+
+## 题目大意
+
+编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target。该矩阵具有以下特性：
+
+- 每行的元素从左到右升序排列。
+- 每列的元素从上到下升序排列。
+
+
+
+## 解题思路
+
+- 给出一个二维矩阵，矩阵的特点是每一个行内，元素随着下标增大而增大，每一列内，元素也是随着下标增大而增大。但是相邻两行的元素并没有大小关系。例如第一行最后一个元素就比第二行第一个元素要大。要求设计一个算法能在这个矩阵中高效的找到一个数，如果找到就输出 true，找不到就输出 false。
+- 这一题是第 74 题的加强版。第 74 题中的二维矩阵完全是一个有序的一维矩阵，但是这一题如果把它拍扁成一维，并不是有序的。首先每一个行或者每一列是有序的 ，那么我们可以依次在每一行或者每一列中利用二分去搜索。这样做时间复杂度为  O(n log n)。
+- 还有一个模拟的解法。通过观察，我们发现了这个矩阵的一个特点，最右边一列的元素是本行中最大的元素，所以我们可以先从最右边一列开始找到第一个比 target 元素大的元素，这个元素所在的行，是我们接着要搜索的。在行中搜索是从最右边开始往左边搜索，时间复杂度是 O(n)，算上一开始在最右边一列中查找的时间复杂度是 O(m)，所以最终的时间复杂度为 O(m+n)。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 模拟，时间复杂度 O(m+n)
+func searchMatrix240(matrix [][]int, target int) bool {
+	if len(matrix) == 0 {
+		return false
+	}
+	row, col := 0, len(matrix[0])-1
+	for col >= 0 && row <= len(matrix)-1 {
+		if target == matrix[row][col] {
+			return true
+		} else if target > matrix[row][col] {
+			row++
+		} else {
+			col--
+		}
+	}
+	return false
+}
+
+// 解法二 二分搜索，时间复杂度 O(n log n)
+func searchMatrix2401(matrix [][]int, target int) bool {
+	if len(matrix) == 0 {
+		return false
+	}
+	for _, row := range matrix {
+		low, high := 0, len(matrix[0])-1
+		for low <= high {
+			mid := low + (high-low)>>1
+			if row[mid] > target {
+				high = mid - 1
+			} else if row[mid] < target {
+				low = mid + 1
+			} else {
+				return true
+			}
+		}
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0239.Sliding-Window-Maximum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0242.Valid-Anagram/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0242.Valid-Anagram.md b/website/content/ChapterFour/0200~0299/0242.Valid-Anagram.md
new file mode 100644
index 000000000..15f171934
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0242.Valid-Anagram.md
@@ -0,0 +1,98 @@
+# [242. Valid Anagram](https://leetcode.com/problems/valid-anagram/)
+
+## 题目
+
+Given two strings s and t , write a function to determine if t is an anagram of s.
+
+**Example 1**:
+
+```
+
+Input: s = "anagram", t = "nagaram"
+Output: true
+
+```
+
+**Example 2**:
+
+```
+
+Input: s = "rat", t = "car"
+Output: false
+
+```
+
+**Note**:  
+
+  
+You may assume the string contains only lowercase alphabets.
+  
+  
+    
+**Follow up**:  
+
+
+What if the inputs contain unicode characters? How would you adapt your solution to such case?
+
+## 题目大意
+
+给出 2 个字符串 s 和 t，如果 t 中的字母在 s 中都存在，输出 true，否则输出 false。
+
+## 解题思路
+
+这道题可以用打表的方式做。先把 s 中的每个字母都存在一个 26 个容量的数组里面，每个下标依次对应 26 个字母。s 中每个字母都对应表中一个字母，每出现一次就加 1。然后再扫字符串 t，每出现一个字母就在表里面减一。如果都出现了，最终表里面的值肯定都是 0 。最终判断表里面的值是否都是 0 即可，有非 0 的数都输出 false 。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func isAnagram(s string, t string) bool {
+	alphabet := make([]int, 26)
+	sBytes := []byte(s)
+	tBytes := []byte(t)
+	if len(sBytes) != len(tBytes) {
+		return false
+	}
+	for i := 0; i < len(sBytes); i++ {
+		alphabet[sBytes[i]-'a']++
+	}
+	for i := 0; i < len(tBytes); i++ {
+		alphabet[tBytes[i]-'a']--
+	}
+	for i := 0; i < 26; i++ {
+		if alphabet[i] != 0 {
+			return false
+		}
+	}
+	return true
+}
+
+// 解法二
+func isAnagram1(s string, t string) bool {
+	hash := map[rune]int{}
+	for _, value := range s {
+		hash[value]++
+	}
+	for _, value := range t {
+		hash[value]--
+	}
+	for _, value := range hash {
+		if value != 0 {
+			return false
+		}
+	}
+	return true
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0240.Search-a-2D-Matrix-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0257.Binary-Tree-Paths/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0257.Binary-Tree-Paths.md b/website/content/ChapterFour/0200~0299/0257.Binary-Tree-Paths.md
new file mode 100755
index 000000000..22b451cee
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0257.Binary-Tree-Paths.md
@@ -0,0 +1,78 @@
+# [257. Binary Tree Paths](https://leetcode.com/problems/binary-tree-paths/)
+
+
+## 题目
+
+Given a binary tree, return all root-to-leaf paths.
+
+**Note**: A leaf is a node with no children.
+
+**Example**:
+
+    Input:
+    
+       1
+     /   \
+    2     3
+     \
+      5
+    
+    Output: ["1->2->5", "1->3"]
+    
+    Explanation: All root-to-leaf paths are: 1->2->5, 1->3
+
+## 题目大意
+
+给定一个二叉树，返回所有从根节点到叶子节点的路径。说明: 叶子节点是指没有子节点的节点。
+
+## 解题思路
+
+- Google 的面试题，考察递归
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+)
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func binaryTreePaths(root *TreeNode) []string {
+	if root == nil {
+		return []string{}
+	}
+	res := []string{}
+	if root.Left == nil && root.Right == nil {
+		return []string{strconv.Itoa(root.Val)}
+	}
+	tmpLeft := binaryTreePaths(root.Left)
+	for i := 0; i < len(tmpLeft); i++ {
+		res = append(res, strconv.Itoa(root.Val)+"->"+tmpLeft[i])
+	}
+	tmpRight := binaryTreePaths(root.Right)
+	for i := 0; i < len(tmpRight); i++ {
+		res = append(res, strconv.Itoa(root.Val)+"->"+tmpRight[i])
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0242.Valid-Anagram/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0258.Add-Digits/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0258.Add-Digits.md b/website/content/ChapterFour/0200~0299/0258.Add-Digits.md
new file mode 100644
index 000000000..4e0770ad8
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0258.Add-Digits.md
@@ -0,0 +1,54 @@
+# [258. Add Digits](https://leetcode.com/problems/add-digits/)
+
+
+## 题目
+
+Given a non-negative integer `num`, repeatedly add all its digits until the result has only one digit.
+
+**Example**:
+
+```
+Input: 38
+Output: 2 
+Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2. 
+             Since 2 has only one digit, return it.
+```
+
+**Follow up**: Could you do it without any loop/recursion in O(1) runtime?
+
+## 题目大意
+
+给定一个非负整数 num，反复将各个位上的数字相加，直到结果为一位数。
+
+
+## 解题思路
+
+- 给定一个非负整数，反复加各个位上的数，直到结果为一位数为止，最后输出这一位数。
+- 简单题。按照题意循环累加即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func addDigits(num int) int {
+	for num > 9 {
+		cur := 0
+		for num != 0 {
+			cur += num % 10
+			num /= 10
+		}
+		num = cur
+	}
+	return num
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0257.Binary-Tree-Paths/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0260.Single-Number-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0260.Single-Number-III.md b/website/content/ChapterFour/0200~0299/0260.Single-Number-III.md
new file mode 100755
index 000000000..ed2a461b5
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0260.Single-Number-III.md
@@ -0,0 +1,68 @@
+# [260. Single Number III](https://leetcode.com/problems/single-number-iii/)
+
+
+## 题目
+
+Given an array of numbers `nums`, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.
+
+**Example**:
+
+    Input:  [1,2,1,3,2,5]
+    Output: [3,5]
+
+**Note**:
+
+1. The order of the result is not important. So in the above example, `[5, 3]` is also correct.
+2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?
+
+
+## 题目大意
+
+给定一个整数数组 nums，其中恰好有两个元素只出现一次，其余所有元素均出现两次。 找出只出现一次的那两个元素。
+
+注意：  
+
+- 结果输出的顺序并不重要，对于上面的例子，[5, 3] 也是正确答案。
+- 要求算法时间复杂度是线性的，并且不使用额外的辅助空间。
+
+
+
+## 解题思路
+
+- 这一题是第 136 题的加强版。第 136 题里面只有一个数出现一次，其他数都出现 2 次。而这一次有 2 个数字出现一次，其他数出现 2 次。
+- 解题思路还是利用异或，把出现 2 次的数先消除。最后我们要找的 2 个数肯定也是不同的，所以最后 2 个数对一个数进行异或，答案肯定是不同的。那么我们找哪个数为参照物呢？可以随便取，不如就取 lsb 最低位为 1 的数吧
+- 于是整个数组会被分为 2 部分，异或 lsb 为 0 的和异或 lsb 为 1 的，在这 2 部分中，用异或操作把出现 2 次的数都消除，那么剩下的 2 个数分别就在这 2 部分中。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func singleNumberIII(nums []int) []int {
+	diff := 0
+	for _, num := range nums {
+		diff ^= num
+	}
+	// Get its last set bit (lsb)
+	diff &= -diff
+	res := []int{0, 0} // this array stores the two numbers we will return
+	for _, num := range nums {
+		if (num & diff) == 0 { // the bit is not set
+			res[0] ^= num
+		} else { // the bit is set
+			res[1] ^= num
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0258.Add-Digits/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0263.Ugly-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0263.Ugly-Number.md b/website/content/ChapterFour/0200~0299/0263.Ugly-Number.md
new file mode 100644
index 000000000..5d2ad1e10
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0263.Ugly-Number.md
@@ -0,0 +1,79 @@
+# [263. Ugly Number](https://leetcode.com/problems/ugly-number/)
+
+## 题目
+
+Write a program to check whether a given number is an ugly number.
+
+Ugly numbers are positive numbers whose prime factors only include 2, 3, 5.
+
+**Example 1**:
+
+```
+
+Input: 6
+Output: true
+Explanation: 6 = 2 × 3
+
+```
+
+**Example 2**:
+
+```
+Input: 8
+Output: true
+Explanation: 8 = 2 × 2 × 2
+
+```
+
+**Example 3**:
+
+```
+
+Input: 14
+Output: false 
+Explanation: 14 is not ugly since it includes another prime factor 7.
+
+```
+
+**Note**:
+
+- 1 is typically treated as an ugly number.
+- Input is within the 32-bit signed integer range: [−2^31,  2^31 − 1].
+
+
+## 题目大意
+
+判断一个数字是否是“丑陋数字”，“丑陋数字”的定义是一个正数，并且因子只包含 2，3，5 。
+
+## 解题思路
+
+依照题意要求做即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func isUgly(num int) bool {
+	if num > 0 {
+		for _, i := range []int{2, 3, 5} {
+			for num%i == 0 {
+				num /= i
+			}
+		}
+	}
+	return num == 1
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0260.Single-Number-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0264.Ugly-Number-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0264.Ugly-Number-II.md b/website/content/ChapterFour/0200~0299/0264.Ugly-Number-II.md
new file mode 100644
index 000000000..ff68409f4
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0264.Ugly-Number-II.md
@@ -0,0 +1,76 @@
+# [264. Ugly Number II](https://leetcode.com/problems/ugly-number-ii/)
+
+
+## 题目
+
+Given an integer `n`, return *the* `nth` ***ugly number***.
+
+**Ugly number** is a positive number whose prime factors only include `2`, `3`, and/or `5`.
+
+**Example 1:**
+
+```
+Input: n = 10
+Output: 12
+Explanation: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12] is the sequence of the first 10 ugly numbers.
+```
+
+**Example 2:**
+
+```
+Input: n = 1
+Output: 1
+Explanation: 1 is typically treated as an ugly number.
+```
+
+**Constraints:**
+
+- `1 <= n <= 1690`
+
+## 题目大意
+
+给你一个整数 `n` ，请你找出并返回第 `n` 个 **丑数** 。**丑数** 就是只包含质因数 `2`、`3` 和/或 `5` 的正整数。
+
+## 解题思路
+
+- 解法一，生成丑数的方法：先用最小质因数 1，分别和 2，3，5 相乘，得到的数是丑数，不断的将这些数分别和 2，3，5 相乘，得到的数去重以后，从小到大排列，第 n 个数即为所求。排序可用最小堆实现，去重用 map 去重。时间复杂度 O(n log n)，空间复杂度 O(n)
+- 上面的解法耗时在排序中，需要排序的根源是小的丑数乘以 5 大于了大的丑数乘以 2 。如何保证每次乘积以后，找出有序的丑数，是去掉排序，提升时间复杂度的关键。举个例子很容易想通：初始状态丑数只有 {1}，乘以 2，3，5 以后，将最小的结果存入集合中 {1,2}。下一轮再相乘，由于上一轮 1 已经和 2 相乘过了，1 不要再和 2 相乘了，所以这一轮 1 和 3，5 相乘。2 和 2，3，5 相乘。将最小的结果存入集合中 {1,2,3}，按照这样的策略往下比较，每轮选出的丑数是有序且不重复的。具体实现利用 3 个指针和一个数组即可实现。时间复杂度 O(n)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+func nthUglyNumber(n int) int {
+	dp, p2, p3, p5 := make([]int, n+1), 1, 1, 1
+	dp[0], dp[1] = 0, 1
+	for i := 2; i <= n; i++ {
+		x2, x3, x5 := dp[p2]*2, dp[p3]*3, dp[p5]*5
+		dp[i] = min(min(x2, x3), x5)
+		if dp[i] == x2 {
+			p2++
+		}
+		if dp[i] == x3 {
+			p3++
+		}
+		if dp[i] == x5 {
+			p5++
+		}
+	}
+	return dp[n]
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0263.Ugly-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0268.Missing-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0268.Missing-Number.md b/website/content/ChapterFour/0200~0299/0268.Missing-Number.md
new file mode 100755
index 000000000..05d3fe54c
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0268.Missing-Number.md
@@ -0,0 +1,54 @@
+# [268. Missing Number](https://leetcode.com/problems/missing-number/)
+
+
+## 题目
+
+Given an array containing n distinct numbers taken from `0, 1, 2, ..., n`, find the one that is missing from the array.
+
+**Example 1**:
+
+    Input: [3,0,1]
+    Output: 2
+
+**Example 2**:
+
+    Input: [9,6,4,2,3,5,7,0,1]
+    Output: 8
+
+**Note**:Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
+
+
+## 题目大意
+
+给定一个包含 0, 1, 2, ..., n 中 n 个数的序列，找出 0 .. n 中没有出现在序列中的那个数。算法应该具有线性时间复杂度。你能否仅使用额外常数空间来实现?
+
+
+
+## 解题思路
+
+
+- 要求找出 `0, 1, 2, ..., n` 中缺失的那个数。还是利用异或的性质，`X^X = 0`。这里我们需要构造一个 X，用数组下标就可以了。数字下标是从 `[0，n-1]`，数字是 `[0，n]`，依次把数组里面的数组进行异或，把结果和 n 再异或一次，中和掉出现的数字，剩下的那个数字就是之前没有出现过的，缺失的数字。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func missingNumber(nums []int) int {
+	xor, i := 0, 0
+	for i = 0; i < len(nums); i++ {
+		xor = xor ^ i ^ nums[i]
+	}
+	return xor ^ i
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0264.Ugly-Number-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0274.H-Index/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0274.H-Index.md b/website/content/ChapterFour/0200~0299/0274.H-Index.md
new file mode 100644
index 000000000..75cf67b16
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0274.H-Index.md
@@ -0,0 +1,85 @@
+# [274. H-Index](https://leetcode.com/problems/h-index/)
+
+## 题目
+
+Given an array of citations (each citation is a non-negative integer) of a researcher, write a function to compute the researcher's h-index.
+
+According to the definition of h-index on Wikipedia: "A scientist has index h if h of his/her N papers have at least h citations each, and the other N − h papers have no more than h citations each."
+
+**Example 1**:
+
+```
+
+Input: citations = [3,0,6,1,5]
+Output: 3 
+Explanation: [3,0,6,1,5] means the researcher has 5 papers in total and each of them had 
+             received 3, 0, 6, 1, 5 citations respectively. 
+             Since the researcher has 3 papers with at least 3 citations each and the remaining 
+             two with no more than 3 citations each, her h-index is 3.
+
+```
+
+**Note**: 
+
+If there are several possible values for h, the maximum one is taken as the h-index.
+
+
+
+## 题目大意
+
+求 h-index。h-index 值的定义：如果他/她的 N 篇论文中至少有 h 引用，而其他 N-h 论文的引用数不超过 h 引用数。
+
+## 解题思路
+
+可以先将数组里面的数从小到大排序。因为要找最大的 h-index，所以从数组末尾开始往前找，找到第一个数组的值，小于，总长度减去下标的值，这个值就是 h-index。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func hIndex(citations []int) int {
+	n := len(citations)
+	buckets := make([]int, n+1)
+	for _, c := range citations {
+		if c >= n {
+			buckets[n]++
+		} else {
+			buckets[c]++
+		}
+	}
+	count := 0
+	for i := n; i >= 0; i-- {
+		count += buckets[i]
+		if count >= i {
+			return i
+		}
+	}
+	return 0
+}
+
+// 解法二
+func hIndex1(citations []int) int {
+	quickSort164(citations, 0, len(citations)-1)
+	hIndex := 0
+	for i := len(citations) - 1; i >= 0; i-- {
+		if citations[i] >= len(citations)-i {
+			hIndex++
+		} else {
+			break
+		}
+	}
+	return hIndex
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0268.Missing-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0275.H-Index-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0275.H-Index-II.md b/website/content/ChapterFour/0200~0299/0275.H-Index-II.md
new file mode 100755
index 000000000..2ab6406b6
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0275.H-Index-II.md
@@ -0,0 +1,78 @@
+# [275. H-Index II](https://leetcode.com/problems/h-index-ii/)
+
+## 题目
+
+Given an array of citations **sorted in ascending order** (each citation is a non-negative integer) of a researcher, write a function to compute the researcher's h-index.
+
+According to the [definition of h-index on Wikipedia](https://en.wikipedia.org/wiki/H-index): "A scientist has index h if h of his/her N papers have **at least** h citations each, and the other N − h papers have **no more than** h citations each."
+
+**Example**:
+
+    Input: citations = [0,1,3,5,6]
+    Output: 3 
+    Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had 
+                 received 0, 1, 3, 5, 6 citations respectively. 
+                 Since the researcher has 3 papers with at least 3 citations each and the remaining 
+                 two with no more than 3 citations each, her h-index is 3.
+
+**Note**:
+
+If there are several possible values for *h*, the maximum one is taken as the h-index.
+
+**Follow up**:
+
+- This is a follow up problem to [H-Index](https://leetcode.com/problems/h-index/description/), where `citations` is now guaranteed to be sorted in ascending order.
+- Could you solve it in logarithmic time complexity?
+
+
+
+## 题目大意
+
+
+给定一位研究者论文被引用次数的数组（被引用次数是非负整数），数组已经按照升序排列。编写一个方法，计算出研究者的 h 指数。
+
+h 指数的定义: “h 代表“高引用次数”（high citations），一名科研人员的 h 指数是指他（她）的 （N 篇论文中）至多有 h 篇论文分别被引用了至少 h 次。（其余的 N - h 篇论文每篇被引用次数不多于 h 次。）"
+
+说明:
+
+- 如果 h 有多有种可能的值 ，h 指数是其中最大的那个。
+
+进阶：
+
+- 这是 H 指数 的延伸题目，本题中的 citations 数组是保证有序的。
+你可以优化你的算法到对数时间复杂度吗？
+
+
+## 解题思路
+
+- 给出一个数组，代表该作者论文被引用次数，要求这个作者的 h 指数。h 指数定义："高引用次数”（`high citations`），一名科研人员的 h 指数是指他（她）的 （N 篇论文中）至多有 h 篇论文分别被引用了至少 h 次。（其余的 N - h 篇论文每篇被引用次数不多于 h 次。）
+- 这一题要找出 h 指数，即要找到一个边界，在这个边界上为最多的 h 指数。可以用二分搜索来解决这道题。当 `len(citations)-mid > citations[mid]` 时，说明 h 指数的边界一定在右边，因为最多 `len(citations)-mid` 篇数比引用数 `citations[mid]` 还要大。否则 h 指数的边界在左边界，缩小边界以后继续二分。找到边界以后，最终求的是 h 指数，用 `len(citations) - low` 即是结果。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func hIndex275(citations []int) int {
+	low, high := 0, len(citations)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if len(citations)-mid > citations[mid] {
+			low = mid + 1
+		} else {
+			high = mid - 1
+		}
+	}
+	return len(citations) - low
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0274.H-Index/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0278.First-Bad-Version/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0278.First-Bad-Version.md b/website/content/ChapterFour/0200~0299/0278.First-Bad-Version.md
new file mode 100644
index 000000000..36fd27b46
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0278.First-Bad-Version.md
@@ -0,0 +1,69 @@
+# [278. First Bad Version](https://leetcode.com/problems/first-bad-version/)
+
+## 题目
+
+You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.
+
+Suppose you have `n` versions `[1, 2, ..., n]` and you want to find out the first bad one, which causes all the following ones to be bad.
+
+You are given an API `bool isBadVersion(version)` which returns whether `version` is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
+
+**Example 1:**
+
+```
+Input: n = 5, bad = 4
+Output: 4
+Explanation:
+call isBadVersion(3) -> false
+call isBadVersion(5) -> true
+call isBadVersion(4) -> true
+Then 4 is the first bad version.
+
+```
+
+**Example 2:**
+
+```
+Input: n = 1, bad = 1
+Output: 1
+
+```
+
+**Constraints:**
+
+- `1 <= bad <= n <= 231 - 1`
+
+## 题目大意
+
+你是产品经理，目前正在带领一个团队开发新的产品。不幸的是，你的产品的最新版本没有通过质量检测。由于每个版本都是基于之前的版本开发的，所以错误的版本之后的所有版本都是错的。假设你有 n 个版本 [1, 2, ..., n]，你想找出导致之后所有版本出错的第一个错误的版本。你可以通过调用 bool isBadVersion(version) 接口来判断版本号 version 是否在单元测试中出错。实现一个函数来查找第一个错误的版本。你应该尽量减少对调用 API 的次数。
+
+## 解题思路
+
+- 我们知道开发产品迭代的版本，如果当一个版本为正确版本，则该版本之前的所有版本均为正确版本；当一个版本为错误版本，则该版本之后的所有版本均为错误版本。利用这个性质就可以进行二分查找。利用二分搜索，也可以满足减少对调用 API 的次数的要求。时间复杂度：O(logn)，其中 n 是给定版本的数量。空间复杂度：O(1)。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+/**
+ * Forward declaration of isBadVersion API.
+ * @param   version   your guess about first bad version
+ * @return 	 	      true if current version is bad
+ *			          false if current version is good
+ * func isBadVersion(version int) bool;
+ */
+
+func firstBadVersion(n int) int {
+	return sort.Search(n, func(version int) bool { return isBadVersion(version) })
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0275.H-Index-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0279.Perfect-Squares/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0279.Perfect-Squares.md b/website/content/ChapterFour/0200~0299/0279.Perfect-Squares.md
new file mode 100644
index 000000000..6ef4ffcc9
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0279.Perfect-Squares.md
@@ -0,0 +1,85 @@
+# [279. Perfect Squares](https://leetcode.com/problems/perfect-squares/)
+
+
+## 题目
+
+Given an integer `n`, return *the least number of perfect square numbers that sum to* `n`.
+
+A **perfect square** is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, `1`, `4`, `9`, and `16` are perfect squares while `3` and `11` are not.
+
+**Example 1:**
+
+```
+Input: n = 12
+Output: 3
+Explanation: 12 = 4 + 4 + 4.
+```
+
+**Example 2:**
+
+```
+Input: n = 13
+Output: 2
+Explanation: 13 = 4 + 9.
+```
+
+**Constraints:**
+
+- `1 <= n <= 104`
+
+## 题目大意
+
+给定正整数 n，找到若干个完全平方数（比如 1, 4, 9, 16, ...）使得它们的和等于 n。你需要让组成和的完全平方数的个数最少。给你一个整数 n ，返回和为 n 的完全平方数的 最少数量 。
+
+完全平方数 是一个整数，其值等于另一个整数的平方；换句话说，其值等于一个整数自乘的积。例如，1、4、9 和 16 都是完全平方数，而 3 和 11 不是。
+
+## 解题思路
+
+- 由拉格朗日的四平方定理可得，每个自然数都可以表示为四个整数平方之和。 其中四个数字是整数。四平方和定理证明了任意一个正整数都可以被表示为至多四个正整数的平方和。这给出了本题的答案的上界。
+- 四平方和定理可以推出三平方和推论：当且仅当 {{< katex >}} n \neq 4^{k} \times (8*m + 7){{< /katex >}} 时，n 可以被表示为至多三个正整数的平方和。所以当 {{< katex >}} n = 4^{k} * (8*m + 7){{< /katex >}}  时，n 只能被表示为四个正整数的平方和。此时我们可以直接返回 4。
+- 当 {{< katex >}} n \neq 4^{k} \times (8*m + 7){{< /katex >}} 时，需要判断 n 到底可以分解成几个完全平方数之和。答案肯定是 1，2，3 中的一个。题目要求我们求最小的，所以从 1 开始一个个判断是否满足。如果答案为 1，代表 n 为完全平方数，这很好判断。如果答案为 2，代表 {{< katex >}} n = a^{2} + b^{2} {{< /katex >}}，枚举 {{< katex >}} 1 \leqslant a \leqslant \sqrt{n} {{< /katex >}}，判断  {{< katex >}} n - a^{2} {{< /katex >}} 是否为完全平方数。当 1 和 2 都排除了，剩下的答案只能为 3 了。
+
+## 代码
+
+```go
+package leetcode
+
+import "math"
+
+func numSquares(n int) int {
+	if isPerfectSquare(n) {
+		return 1
+	}
+	if checkAnswer4(n) {
+		return 4
+	}
+	for i := 1; i*i <= n; i++ {
+		j := n - i*i
+		if isPerfectSquare(j) {
+			return 2
+		}
+	}
+	return 3
+}
+
+// 判断是否为完全平方数
+func isPerfectSquare(n int) bool {
+	sq := int(math.Floor(math.Sqrt(float64(n))))
+	return sq*sq == n
+}
+
+// 判断是否能表示为 4^k*(8m+7)
+func checkAnswer4(x int) bool {
+	for x%4 == 0 {
+		x /= 4
+	}
+	return x%8 == 7
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0278.First-Bad-Version/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0283.Move-Zeroes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0283.Move-Zeroes.md b/website/content/ChapterFour/0200~0299/0283.Move-Zeroes.md
new file mode 100644
index 000000000..16c37efae
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0283.Move-Zeroes.md
@@ -0,0 +1,60 @@
+# [283. Move Zeroes](https://leetcode.com/problems/move-zeroes/)
+
+## 题目
+
+Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.
+
+**Example 1**:
+
+```
+
+Input: [0,1,0,3,12]
+Output: [1,3,12,0,0]
+
+```
+
+**Note**:
+
+- You must do this in-place without making a copy of the array.
+- Minimize the total number of operations.
+
+
+
+## 题目大意
+
+题目要求不能采用额外的辅助空间，将数组中 0 元素都移动到数组的末尾，并且维持所有非 0 元素的相对位置。
+
+## 解题思路
+
+这一题可以只扫描数组一遍，不断的用 i，j 标记 0 和非 0 的元素，然后相互交换，最终到达题目的目的。与这一题相近的题目有第 26 题，第 27 题，第 80 题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func moveZeroes(nums []int) {
+	if len(nums) == 0 {
+		return
+	}
+	j := 0
+	for i := 0; i < len(nums); i++ {
+		if nums[i] != 0 {
+			if i != j {
+				nums[i], nums[j] = nums[j], nums[i]
+			} 
+			j++
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0279.Perfect-Squares/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0284.Peeking-Iterator/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0284.Peeking-Iterator.md b/website/content/ChapterFour/0200~0299/0284.Peeking-Iterator.md
new file mode 100644
index 000000000..9693b934b
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0284.Peeking-Iterator.md
@@ -0,0 +1,94 @@
+# [284. Peeking Iterator](https://leetcode.com/problems/peeking-iterator/)
+
+## 题目
+
+Given an Iterator class interface with methods: `next()` and `hasNext()`, design and implement a PeekingIterator that support the `peek()` operation -- it essentially peek() at the element that will be returned by the next call to next().
+
+**Example:**
+
+```
+Assume that the iterator is initialized to the beginning of the list: [1,2,3].
+
+Call next() gets you 1, the first element in the list.
+Now you call peek() and it returns 2, the next element. Calling next() after that still return 2. 
+You call next() the final time and it returns 3, the last element. 
+Calling hasNext() after that should return false.
+```
+
+**Follow up**: How would you extend your design to be generic and work with all types, not just integer?
+
+## 题目大意
+
+给定一个迭代器类的接口，接口包含两个方法： next() 和 hasNext()。设计并实现一个支持 peek() 操作的顶端迭代器 -- 其本质就是把原本应由 next() 方法返回的元素 peek() 出来。
+
+> peek() 是偷看的意思。偷偷看一看下一个元素是什么，但是并不是 next() 访问。
+
+## 解题思路
+
+- 简单题。在 PeekingIterator 内部保存 2 个变量，一个是下一个元素值，另一个是是否有下一个元素。在 next() 操作和 hasNext() 操作时，访问保存的这 2 个变量。peek() 操作也比较简单，判断是否有下一个元素，如果有，即返回该元素值。这里实现了迭代指针不移动的功能。如果没有保存下一个元素值，即没有 peek() 偷看，next() 操作继续往后移动指针，读取后一位元素。
+- 这里复用了是否有下一个元素值，来判断 hasNext() 和 peek() 操作中不移动指针的逻辑。
+
+## 代码
+
+```go
+package leetcode
+
+//Below is the interface for Iterator, which is already defined for you.
+
+type Iterator struct {
+}
+
+func (this *Iterator) hasNext() bool {
+	// Returns true if the iteration has more elements.
+	return true
+}
+
+func (this *Iterator) next() int {
+	// Returns the next element in the iteration.
+	return 0
+}
+
+type PeekingIterator struct {
+	nextEl int
+	hasEl  bool
+	iter   *Iterator
+}
+
+func Constructor(iter *Iterator) *PeekingIterator {
+	return &PeekingIterator{
+		iter: iter,
+	}
+}
+
+func (this *PeekingIterator) hasNext() bool {
+	if this.hasEl {
+		return true
+	}
+	return this.iter.hasNext()
+}
+
+func (this *PeekingIterator) next() int {
+	if this.hasEl {
+		this.hasEl = false
+		return this.nextEl
+	} else {
+		return this.iter.next()
+	}
+}
+
+func (this *PeekingIterator) peek() int {
+	if this.hasEl {
+		return this.nextEl
+	}
+	this.hasEl = true
+	this.nextEl = this.iter.next()
+	return this.nextEl
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0283.Move-Zeroes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0287.Find-the-Duplicate-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0287.Find-the-Duplicate-Number.md b/website/content/ChapterFour/0200~0299/0287.Find-the-Duplicate-Number.md
new file mode 100644
index 000000000..d3f95b60b
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0287.Find-the-Duplicate-Number.md
@@ -0,0 +1,114 @@
+# [287. Find the Duplicate Number](https://leetcode.com/problems/find-the-duplicate-number/)
+
+## 题目
+
+Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
+
+**Example 1**:
+
+```
+
+Input: [1,3,4,2,2]
+Output: 2
+
+```
+
+**Example 2**:
+
+```
+
+Input: [3,1,3,4,2]
+Output: 3
+
+```
+
+**Note**:  
+
+- You must not modify the array (assume the array is read only).
+- You must use only constant, O(1) extra space.
+- Your runtime complexity should be less than O(n^2).
+- There is only one duplicate number in the array, but it could be repeated more than once.
+
+## 题目大意
+
+给出 n + 1 个数，这些数是在 1-n 中取值的，同一个数字可以出现多次。要求找出这些数中重复的数字。时间复杂度最好低于 O(n^2)，空间复杂度为 O(1)。
+
+## 解题思路
+
+- 这道题比较巧的思路是，将这些数字想象成链表中的结点，数组中数字代表下一个结点的数组下标。找重复的数字就是找链表中成环的那个点。由于题目保证了一定会有重复的数字，所以一定会成环。所以用快慢指针的方法，快指针一次走 2 步，慢指针一次走 1 步，相交以后，快指针从头开始，每次走一步，再次遇见的时候就是成环的交点处，也即是重复数字所在的地方。
+- 这一题有多种做法。可以用快慢指针求解。还可以用二分搜索：(这里的题解感谢 [@imageslr](https://github.com/imageslr) 指出错误）：
+	1. 假设有 n+1 个数，则可能重复的数位于区间 [1, n] 中。记该区间最小值、最大值和中间值为 low、high、mid
+	2. 遍历整个数组，统计小于等于 mid 的整数的个数，至多为 mid 个
+	3. 如果超过 mid 个就说明重复的数存在于区间 [low,mid] （闭区间）中；否则，重复的数存在于区间 (mid, high] （左开右闭）中
+	4. 缩小区间，继续重复步骤 2、3，直到区间变成 1 个整数，即 low == high
+	5. 整数 low 就是要找的重复的数
+- 另外一个做法是，先将数组排序，依照下标是从 0 开始递增的特性，那么数组里面的数字与下标的差值应该是越来越大。如果出现了相同的数字，下标变大，差值应该比前一个数字小，出现了这个情况就说明找到了相同数字了。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+// 解法一 快慢指针
+func findDuplicate(nums []int) int {
+	slow := nums[0]
+	fast := nums[nums[0]]
+	for fast != slow {
+		slow = nums[slow]
+		fast = nums[nums[fast]]
+	}
+	walker := 0
+	for walker != slow {
+		walker = nums[walker]
+		slow = nums[slow]
+	}
+	return walker
+}
+
+// 解法二 二分搜索
+func findDuplicate1(nums []int) int {
+	low, high := 0, len(nums)-1
+	for low < high {
+		mid, count := low+(high-low)>>1, 0
+		for _, num := range nums {
+			if num <= mid {
+				count++
+			}
+		}
+		if count > mid {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+// 解法三
+func findDuplicate2(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	sort.Ints(nums)
+	diff := -1
+	for i := 0; i < len(nums); i++ {
+		if nums[i]-i-1 >= diff {
+			diff = nums[i] - i - 1
+		} else {
+			return nums[i]
+		}
+	}
+	return 0
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0284.Peeking-Iterator/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0290.Word-Pattern/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0290.Word-Pattern.md b/website/content/ChapterFour/0200~0299/0290.Word-Pattern.md
new file mode 100644
index 000000000..15b9cdf3b
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0290.Word-Pattern.md
@@ -0,0 +1,105 @@
+# [290. Word Pattern](https://leetcode.com/problems/word-pattern/)
+
+## 题目
+
+Given a pattern and a string str, find if str follows the same pattern.
+
+Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.
+
+**Example 1**:
+
+```
+
+Input: pattern = "abba", str = "dog cat cat dog"
+Output: true
+
+```
+
+**Example 2**:
+
+```
+
+Input:pattern = "abba", str = "dog cat cat fish"
+Output: false
+
+```
+
+**Example 3**:
+
+```
+
+Input: pattern = "aaaa", str = "dog cat cat dog"
+Output: false
+
+```
+
+**Example 4**:
+
+```
+
+Input: pattern = "abba", str = "dog dog dog dog"
+Output: false
+
+```
+
+**Note**:
+
+You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.
+
+
+## 题目大意
+
+给定一个模式串，判断字符串是否和给定的模式串，是一样的模式。
+
+## 解题思路
+
+这道题用 2 个 map 即可。1 个 map 记录模式与字符串的匹配关系，另外一个 map 记录字符串和模式的匹配关系。为什么需要记录双向的关系呢？因为 Example 4 中，a 对应了 dog，这个时候 b 如果再对应 dog 是错误的，所以这里需要从 dog 查询它是否已经和某个模式匹配过了。所以需要双向的关系。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func wordPattern(pattern string, str string) bool {
+	strList := strings.Split(str, " ")
+	patternByte := []byte(pattern)
+	if pattern == "" || len(patternByte) != len(strList) {
+		return false
+	}
+
+	pMap := map[byte]string{}
+	sMap := map[string]byte{}
+	for index, b := range patternByte {
+		if _, ok := pMap[b]; !ok {
+			if _, ok = sMap[strList[index]]; !ok {
+				pMap[b] = strList[index]
+				sMap[strList[index]] = b
+			} else {
+				if sMap[strList[index]] != b {
+					return false
+				}
+			}
+		} else {
+			if pMap[b] != strList[index] {
+				return false
+			}
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0287.Find-the-Duplicate-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0297.Serialize-and-Deserialize-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0297.Serialize-and-Deserialize-Binary-Tree.md b/website/content/ChapterFour/0200~0299/0297.Serialize-and-Deserialize-Binary-Tree.md
new file mode 100644
index 000000000..dc86b67fa
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0297.Serialize-and-Deserialize-Binary-Tree.md
@@ -0,0 +1,120 @@
+# [297. Serialize and Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)
+
+
+## 题目
+
+Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
+
+Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.
+
+**Clarification:** The input/output format is the same as [how LeetCode serializes a binary tree](https://leetcode.com/faq/#binary-tree). You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/09/15/serdeser.jpg](https://assets.leetcode.com/uploads/2020/09/15/serdeser.jpg)
+
+```
+Input: root = [1,2,3,null,null,4,5]
+Output: [1,2,3,null,null,4,5]
+```
+
+**Example 2:**
+
+```
+Input: root = []
+Output: []
+```
+
+**Example 3:**
+
+```
+Input: root = [1]
+Output: [1]
+```
+
+**Example 4:**
+
+```
+Input: root = [1,2]
+Output: [1,2]
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[0, 104]`.
+- `1000 <= Node.val <= 1000`
+
+## 题目大意
+
+设计一个算法，来序列化和反序列化二叉树。并不限制如何进行序列化和反序列化，但是你需要保证二叉树可以序列化为字符串，并且这个字符串可以被反序列化成原有的二叉树。
+
+## 解题思路
+
+- 1. 将给定的二叉树想象成一颗满二叉树(不存在的结点用 null 填充)。
+- 2. 通过前序遍历，可以得到一个第一个结点为根的序列，然后递归进行序列化/反序列化即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+type TreeNode = structures.TreeNode
+
+type Codec struct {
+	builder strings.Builder
+	input   []string
+}
+
+func Constructor() Codec {
+	return Codec{}
+}
+
+// Serializes a tree to a single string.
+func (this *Codec) serialize(root *TreeNode) string {
+	if root == nil {
+		this.builder.WriteString("#,")
+		return ""
+	}
+	this.builder.WriteString(strconv.Itoa(root.Val) + ",")
+	this.serialize(root.Left)
+	this.serialize(root.Right)
+	return this.builder.String()
+}
+
+// Deserializes your encoded data to tree.
+func (this *Codec) deserialize(data string) *TreeNode {
+	if len(data) == 0 {
+		return nil
+	}
+	this.input = strings.Split(data, ",")
+	return this.deserializeHelper()
+}
+
+func (this *Codec) deserializeHelper() *TreeNode {
+	if this.input[0] == "#" {
+		this.input = this.input[1:]
+		return nil
+	}
+	val, _ := strconv.Atoi(this.input[0])
+	this.input = this.input[1:]
+	return &TreeNode{
+		Val:   val,
+		Left:  this.deserializeHelper(),
+		Right: this.deserializeHelper(),
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0290.Word-Pattern/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0299.Bulls-and-Cows/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/0299.Bulls-and-Cows.md b/website/content/ChapterFour/0200~0299/0299.Bulls-and-Cows.md
new file mode 100644
index 000000000..10b6261d4
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/0299.Bulls-and-Cows.md
@@ -0,0 +1,121 @@
+# [299. Bulls and Cows](https://leetcode.com/problems/bulls-and-cows/)
+
+
+## 题目
+
+You are playing the Bulls and Cows game with your friend.
+
+You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
+
+The number of "bulls", which are digits in the guess that are in the correct position.
+The number of "cows", which are digits in the guess that are in your secret number but are located in the wrong position. Specifically, the non-bull digits in the guess that could be rearranged such that they become bulls.
+Given the secret number secret and your friend's guess guess, return the hint for your friend's guess.
+
+The hint should be formatted as "xAyB", where x is the number of bulls and y is the number of cows. Note that both secret and guess may contain duplicate digits.
+
+**Example 1:**
+
+```
+Input: secret = "1807", guess = "7810"
+Output: "1A3B"
+Explanation: Bulls are connected with a '|' and cows are underlined:
+"1807"
+  |
+"7810"
+```
+
+**Example 2:**
+
+```
+Input: secret = "1123", guess = "0111"
+Output: "1A1B"
+Explanation: Bulls are connected with a '|' and cows are underlined:
+"1123"        "1123"
+  |      or     |
+"0111"        "0111"
+Note that only one of the two unmatched 1s is counted as a cow since the non-bull digits can only be rearranged to allow one 1 to be a bull.
+```
+
+**Example 3:**
+
+```
+Input: secret = "1", guess = "0"
+Output: "0A0B"
+```
+
+**Example 4:**
+
+```
+Input: secret = "1", guess = "1"
+Output: "1A0B"
+```
+
+**Constraints:**
+
+- 1 <= secret.length, guess.length <= 1000
+- secret.length == guess.length
+- secret and guess consist of digits only.
+
+## 题目大意
+
+你在和朋友一起玩 猜数字（Bulls and Cows）游戏，该游戏规则如下：
+
+写出一个秘密数字，并请朋友猜这个数字是多少。朋友每猜测一次，你就会给他一个包含下述信息的提示：
+
+猜测数字中有多少位属于数字和确切位置都猜对了（称为 "Bulls", 公牛），
+有多少位属于数字猜对了但是位置不对（称为 "Cows", 奶牛）。也就是说，这次猜测中有多少位非公牛数字可以通过重新排列转换成公牛数字。
+给你一个秘密数字secret 和朋友猜测的数字guess ，请你返回对朋友这次猜测的提示。
+
+提示的格式为 "xAyB" ，x 是公牛个数， y 是奶牛个数，A 表示公牛，B表示奶牛。
+
+请注意秘密数字和朋友猜测的数字都可能含有重复数字。
+
+## 解题思路
+
+- 计算下标一致并且对应下标的元素一致的个数，即 x
+- secret 和 guess 分别去除 x 个公牛的元素,剩下 secret 和 guess 求共同的元素个数就是 y
+- 把 x， y 转换成字符串，分别与 "A" 和 "B" 进行拼接返回结果
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strconv"
+
+func getHint(secret string, guess string) string {
+	cntA, cntB := 0, 0
+	mpS := make(map[byte]int)
+	var strG []byte
+	n := len(secret)
+	var ans string
+	for i := 0; i < n; i++ {
+		if secret[i] == guess[i] {
+			cntA++
+		} else {
+			mpS[secret[i]] += 1
+			strG = append(strG, guess[i])
+		}
+	}
+	for _, v := range strG {
+		if _, ok := mpS[v]; ok {
+			if mpS[v] > 1 {
+				mpS[v] -= 1
+			} else {
+				delete(mpS, v)
+			}
+			cntB++
+		}
+	}
+	ans += strconv.Itoa(cntA) + "A" + strconv.Itoa(cntB) + "B"
+	return ans
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0297.Serialize-and-Deserialize-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0300.Longest-Increasing-Subsequence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0200~0299/_index.md b/website/content/ChapterFour/0200~0299/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/0200~0299/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/0300~0399/0300.Longest-Increasing-Subsequence.md b/website/content/ChapterFour/0300~0399/0300.Longest-Increasing-Subsequence.md
new file mode 100755
index 000000000..3692985ac
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0300.Longest-Increasing-Subsequence.md
@@ -0,0 +1,87 @@
+# [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/)
+
+
+## 题目
+
+Given an unsorted array of integers, find the length of longest increasing subsequence.
+
+**Example**:
+
+    Input: [10,9,2,5,3,7,101,18]
+    Output: 4 
+    Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
+
+**Note**:
+
+- There may be more than one LIS combination, it is only necessary for you to return the length.
+- Your algorithm should run in O(n^2) complexity.
+
+**Follow up**: Could you improve it to O(n log n) time complexity?
+
+## 题目大意
+
+给定一个无序的整数数组，找到其中最长上升子序列的长度。
+
+
+## 解题思路
+
+- 给定一个整数序列，求其中的最长上升子序列的长度。这一题就是经典的 LIS 最长上升子序列的问题。
+- `dp[i]` 代表为第 i 个数字为结尾的最长上升子序列的长度。换种表述，dp[i] 代表 [0,i] 范围内，选择数字 nums[i] 可以获得的最长上升子序列的长度。状态转移方程为 `dp[i] = max( 1 + dp[j]) ，其中 j < i && nums[j] > nums[i]`，取所有满足条件的最大值。时间复杂度 O(n^2)
+- 这道题还有一种更快的解法。考虑这样一个问题，我们是否能用一个数组，记录上升子序列的最末尾的一个数字呢？如果这个数字越小，那么这个子序列往后面添加数字的几率就越大，那么就越可能成为最长的上升子序列。举个例子：nums = [4,5,6,3]，它的所有的上升子序列为
+
+```
+    len = 1   :      [4], [5], [6], [3]   => tails[0] = 3
+    len = 2   :      [4, 5], [5, 6]       => tails[1] = 5
+    len = 3   :      [4, 5, 6]            => tails[2] = 6
+```
+- 其中 `tails[i]` 中存储的是所有长度为 i + 1 的上升子序列中末尾最小的值。也很容易证明 `tails` 数组里面的值一定是递增的(因为我们用末尾的数字描述最长递增子序列)。既然 tails 是有序的，我们就可以用二分查找的方法去更新这个 tail 数组里面的值。更新策略如下：(1). 如果 x 比所有的 tails 元素都要大，那么就直接放在末尾，并且 tails 数组长度加一，这里对应解法二中，二分搜索找不到对应的元素值，直接把 num 放在 dp[] 的最后；(2). 如果 `tails[i-1] < x <= tails[i]`，则更新 tails[i]，因为 x 更小，更能获得最长上升子序列，这一步对应解法二中将 dp[i] 更新为 num。最终 tails 数组的长度即为最长的上升子序列。这种做法的时间复杂度 O(n log n)。
+- 此题是一维的 LIS 问题。二维的 LIS 问题是第 354 题。三维的 LIS 问题是第 1691 题。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+// 解法一 O(n^2) DP
+func lengthOfLIS(nums []int) int {
+	dp, res := make([]int, len(nums)+1), 0
+	dp[0] = 0
+	for i := 1; i <= len(nums); i++ {
+		for j := 1; j < i; j++ {
+			if nums[j-1] < nums[i-1] {
+				dp[i] = max(dp[i], dp[j])
+			}
+		}
+		dp[i] = dp[i] + 1
+		res = max(res, dp[i])
+	}
+	return res
+}
+
+// 解法二 O(n log n) DP
+func lengthOfLIS1(nums []int) int {
+	dp := []int{}
+	for _, num := range nums {
+		i := sort.SearchInts(dp, num)
+		if i == len(dp) {
+			dp = append(dp, num)
+		} else {
+			dp[i] = num
+		}
+	}
+	return len(dp)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0299.Bulls-and-Cows/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0301.Remove-Invalid-Parentheses/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0301.Remove-Invalid-Parentheses.md b/website/content/ChapterFour/0300~0399/0301.Remove-Invalid-Parentheses.md
new file mode 100644
index 000000000..db483d160
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0301.Remove-Invalid-Parentheses.md
@@ -0,0 +1,138 @@
+# [301. Remove Invalid Parentheses](https://leetcode.com/problems/remove-invalid-parentheses/)
+
+
+## 题目
+
+Given a string s that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
+
+Return all the possible results. You may return the answer in any order.
+
+**Example 1:**
+
+    Input: s = "()())()"
+    Output: ["(())()","()()()"]
+
+**Example 2:**
+
+    Input: s = "(a)())()"
+    Output: ["(a())()","(a)()()"]
+
+**Example 3:**
+
+    Input: s = ")("
+    Output: [""]
+
+**Constraints:**
+
+- 1 <= s.length <= 25
+- s consists of lowercase English letters and parentheses '(' and ')'.
+- There will be at most 20 parentheses in s.
+
+## 题目大意
+
+给你一个由若干括号和字母组成的字符串 s ，删除最小数量的无效括号，使得输入的字符串有效。
+
+返回所有可能的结果。答案可以按 任意顺序 返回。
+
+说明:
+
+- 1 <= s.length <= 25
+- s 由小写英文字母以及括号 '(' 和 ')' 组成
+- s 中至多含 20 个括号
+
+
+## 解题思路
+
+回溯和剪枝
+- 计算最大得分数maxScore，合法字符串的长度length，左括号和右括号的移除次数lmoves,rmoves
+- 加一个左括号的得分加1；加一个右括号的得分减1
+- 对于一个合法的字符串，左括号等于右括号，得分最终为0；
+- 搜索过程中出现以下任何一种情况都直接返回
+    - 得分值为负数
+    - 得分大于最大得分数
+    - 得分小于0
+    - lmoves小于0
+    - rmoves小于0
+
+## 代码
+
+```go
+
+package leetcode
+
+var (
+	res      []string
+	mp       map[string]int
+	n        int
+	length   int
+	maxScore int
+	str      string
+)
+
+func removeInvalidParentheses(s string) []string {
+	lmoves, rmoves, lcnt, rcnt := 0, 0, 0, 0
+	for _, v := range s {
+		if v == '(' {
+			lmoves++
+			lcnt++
+		} else if v == ')' {
+			if lmoves != 0 {
+				lmoves--
+			} else {
+				rmoves++
+			}
+			rcnt++
+		}
+	}
+	n = len(s)
+	length = n - lmoves - rmoves
+	res = []string{}
+	mp = make(map[string]int)
+	maxScore = min(lcnt, rcnt)
+	str = s
+	backtrace(0, "", lmoves, rmoves, 0)
+	return res
+}
+
+func backtrace(i int, cur string, lmoves int, rmoves int, score int) {
+	if lmoves < 0 || rmoves < 0 || score < 0 || score > maxScore {
+		return
+	}
+	if lmoves == 0 && rmoves == 0 {
+		if len(cur) == length {
+			if _, ok := mp[cur]; !ok {
+				res = append(res, cur)
+				mp[cur] = 1
+			}
+			return
+		}
+	}
+	if i == n {
+		return
+	}
+	if str[i] == '(' {
+		backtrace(i+1, cur+string('('), lmoves, rmoves, score+1)
+		backtrace(i+1, cur, lmoves-1, rmoves, score)
+	} else if str[i] == ')' {
+		backtrace(i+1, cur+string(')'), lmoves, rmoves, score-1)
+		backtrace(i+1, cur, lmoves, rmoves-1, score)
+	} else {
+		backtrace(i+1, cur+string(str[i]), lmoves, rmoves, score)
+	}
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0300.Longest-Increasing-Subsequence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0303.Range-Sum-Query-Immutable/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0303.Range-Sum-Query-Immutable.md b/website/content/ChapterFour/0300~0399/0303.Range-Sum-Query-Immutable.md
new file mode 100755
index 000000000..8b49288ba
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0303.Range-Sum-Query-Immutable.md
@@ -0,0 +1,117 @@
+# [303. Range Sum Query - Immutable](https://leetcode.com/problems/range-sum-query-immutable/)
+
+
+## 题目
+
+Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
+
+**Example**:
+
+    Given nums = [-2, 0, 3, -5, 2, -1]
+    
+    sumRange(0, 2) -> 1
+    sumRange(2, 5) -> -1
+    sumRange(0, 5) -> -3
+
+**Note**:
+
+1. You may assume that the array does not change.
+2. There are many calls to sumRange function.
+
+
+## 题目大意
+
+给定一个整数数组  nums，求出数组从索引 i 到 j  (i ≤ j) 范围内元素的总和，包含 i,  j 两点。
+
+示例：
+
+```
+给定 nums = [-2, 0, 3, -5, 2, -1]，求和函数为 sumRange()
+
+sumRange(0, 2) -> 1
+sumRange(2, 5) -> -1
+sumRange(0, 5) -> -3
+
+```
+
+说明:
+
+- 你可以假设数组不可变。
+- 会多次调用 sumRange 方法。
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面的数都是`**不可变**`的，设计一个数据结构能够满足查询数组任意区间内元素的和。
+- 这一题由于数组里面的元素都是`**不可变**`的，所以可以用 2 种方式来解答，第一种解法是用 prefixSum，通过累计和相减的办法来计算区间内的元素和，初始化的时间复杂度是 O(n)，但是查询区间元素和的时间复杂度是 O(1)。第二种解法是利用线段树，构建一颗线段树，父结点内存的是两个子结点的和，初始化建树的时间复杂度是 O(log n)，查询区间元素和的时间复杂度是 O(log n)。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+//解法一 线段树，sumRange 时间复杂度 O(1)
+
+// NumArray define
+type NumArray struct {
+	st *template.SegmentTree
+}
+
+// Constructor303 define
+func Constructor303(nums []int) NumArray {
+	st := template.SegmentTree{}
+	st.Init(nums, func(i, j int) int {
+		return i + j
+	})
+	return NumArray{st: &st}
+}
+
+// SumRange define
+func (ma *NumArray) SumRange(i int, j int) int {
+	return ma.st.Query(i, j)
+}
+
+//解法二 prefixSum，sumRange 时间复杂度 O(1)
+
+// // NumArray define
+// type NumArray struct {
+// 	prefixSum []int
+// }
+
+// // Constructor303 define
+// func Constructor303(nums []int) NumArray {
+// 	for i := 1; i < len(nums); i++ {
+// 		nums[i] += nums[i-1]
+// 	}
+// 	return NumArray{prefixSum: nums}
+// }
+
+// // SumRange define
+// func (this *NumArray) SumRange(i int, j int) int {
+// 	if i > 0 {
+// 		return this.prefixSum[j] - this.prefixSum[i-1]
+// 	}
+// 	return this.prefixSum[j]
+// }
+
+/**
+ * Your NumArray object will be instantiated and called as such:
+ * obj := Constructor(nums);
+ * param_1 := obj.SumRange(i,j);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0301.Remove-Invalid-Parentheses/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0304.Range-Sum-Query-2D-Immutable/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0304.Range-Sum-Query-2D-Immutable.md b/website/content/ChapterFour/0300~0399/0304.Range-Sum-Query-2D-Immutable.md
new file mode 100644
index 000000000..c70fa06cf
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0304.Range-Sum-Query-2D-Immutable.md
@@ -0,0 +1,101 @@
+# [304. Range Sum Query 2D - Immutable](https://leetcode.com/problems/range-sum-query-2d-immutable/)
+
+
+## 题目
+
+Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).
+
+![https://leetcode.com/static/images/courses/range_sum_query_2d.png](https://assets.leetcode.com/uploads/2021/03/14/sum-grid.jpg)
+
+The above rectangle (with the red border) is defined by (row1, col1) = **(2, 1)** and (row2, col2) = **(4, 3)**, which contains sum = **8**.
+
+**Example:**
+
+```
+Given matrix = [
+  [3, 0, 1, 4, 2],
+  [5, 6, 3, 2, 1],
+  [1, 2, 0, 1, 5],
+  [4, 1, 0, 1, 7],
+  [1, 0, 3, 0, 5]
+]
+
+sumRegion(2, 1, 4, 3) -> 8
+sumRegion(1, 1, 2, 2) -> 11
+sumRegion(1, 2, 2, 4) -> 12
+
+```
+
+**Note:**
+
+1. You may assume that the matrix does not change.
+2. There are many calls to sumRegion function.
+3. You may assume that row1 ≤ row2 and col1 ≤ col2.
+
+## 题目大意
+
+给定一个二维矩阵，计算其子矩形范围内元素的总和，该子矩阵的左上角为 (row1, col1) ，右下角为 (row2, col2) 。
+
+## 解题思路
+
+- 这一题是一维数组前缀和的进阶版本。定义 f(x,y) 代表矩形左上角 (0,0)，右下角 (x,y) 内的元素和。{{< katex display >}} f(i,j) = \sum_{x=0}^{i}\sum_{y=0}^{j} Matrix[x][y]{{< /katex >}}  
+
+	{{< katex display >}} 
+    \begin{aligned}
+	f(i,j) &= \sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + \sum_{x=0}^{i-1} Matrix[x][j] + \sum_{y=0}^{j-1} Matrix[i][y] + Matrix[i][j]\\
+	&= (\sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + \sum_{x=0}^{i-1} Matrix[x][j]) + (\sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + \sum_{y=0}^{j-1} Matrix[i][y]) - \sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + Matrix[i][j]\\
+	&= \sum_{x=0}^{i-1}\sum_{y=0}^{j} Matrix[x][y] + \sum_{x=0}^{i}\sum_{y=0}^{j-1} Matrix[x][y] - \sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + Matrix[i][j]\\
+	&= f(i-1,j) + f(i,j-1) - f(i-1,j-1) + Matrix[i][j]
+	\end{aligned}
+	{{< /katex >}}  
+
+- 于是得到递推的关系式：`f(i, j) = f(i-1, j) + f(i, j-1) - f(i-1, j-1) + matrix[i][j]`，写代码为了方便，新建一个 `m+1 * n+1` 的矩阵，这样就不需要对 `row = 0` 和 `col = 0` 做单独处理了。上述推导公式如果画成图也很好理解：
+
+    ![https://img.halfrost.com/Leetcode/leetcode_304.png](https://img.halfrost.com/Leetcode/leetcode_304.png)
+
+    左图中大的矩形由粉红色的矩形 + 绿色矩形 - 粉红色和绿色重叠部分 + 黄色部分。这就对应的是上面推导出来的递推公式。左图是矩形左上角为 (0，0) 的情况，更加一般的情况是右图，左上角是任意的坐标，公式不变。
+
+- 时间复杂度：初始化 O(mn)，查询 O(1)。空间复杂度 O(mn)
+
+## 代码
+
+```go
+package leetcode
+
+type NumMatrix struct {
+	cumsum [][]int
+}
+
+func Constructor(matrix [][]int) NumMatrix {
+	if len(matrix) == 0 {
+		return NumMatrix{nil}
+	}
+	cumsum := make([][]int, len(matrix)+1)
+	cumsum[0] = make([]int, len(matrix[0])+1)
+	for i := range matrix {
+		cumsum[i+1] = make([]int, len(matrix[i])+1)
+		for j := range matrix[i] {
+			cumsum[i+1][j+1] = matrix[i][j] + cumsum[i][j+1] + cumsum[i+1][j] - cumsum[i][j]
+		}
+	}
+	return NumMatrix{cumsum}
+}
+
+func (this *NumMatrix) SumRegion(row1 int, col1 int, row2 int, col2 int) int {
+	cumsum := this.cumsum
+	return cumsum[row2+1][col2+1] - cumsum[row1][col2+1] - cumsum[row2+1][col1] + cumsum[row1][col1]
+}
+
+/**
+ * Your NumMatrix object will be instantiated and called as such:
+ * obj := Constructor(matrix);
+ * param_1 := obj.SumRegion(row1,col1,row2,col2);
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0303.Range-Sum-Query-Immutable/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0306.Additive-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0306.Additive-Number.md b/website/content/ChapterFour/0300~0399/0306.Additive-Number.md
new file mode 100755
index 000000000..d93b46929
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0306.Additive-Number.md
@@ -0,0 +1,96 @@
+# [306. Additive Number](https://leetcode.com/problems/additive-number/)
+
+
+## 题目
+
+Additive number is a string whose digits can form additive sequence.
+
+A valid additive sequence should contain **at least** three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.
+
+Given a string containing only digits `'0'-'9'`, write a function to determine if it's an additive number.
+
+**Note**: Numbers in the additive sequence **cannot** have leading zeros, so sequence `1, 2, 03` or `1, 02, 3` is invalid.
+
+**Example 1**:
+
+    Input: "112358"
+    Output: true 
+    Explanation: The digits can form an additive sequence: 1, 1, 2, 3, 5, 8. 
+                 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8
+
+**Example 2**:
+
+    Input: "199100199"
+    Output: true 
+    Explanation: The additive sequence is: 1, 99, 100, 199. 
+                 1 + 99 = 100, 99 + 100 = 199
+
+**Follow up**:How would you handle overflow for very large input integers?
+
+
+## 题目大意
+
+累加数是一个字符串，组成它的数字可以形成累加序列。一个有效的累加序列必须至少包含 3 个数。除了最开始的两个数以外，字符串中的其他数都等于它之前两个数相加的和。给定一个只包含数字 '0'-'9' 的字符串，编写一个算法来判断给定输入是否是累加数。说明: 累加序列里的数不会以 0 开头，所以不会出现 1, 2, 03 或者 1, 02, 3 的情况。
+
+
+## 解题思路
+
+- 在给出的字符串中判断该字符串是否为斐波那契数列形式的字符串。
+- 由于每次判断需要累加 2 个数字，所以在 DFS 遍历的过程中需要维护 2 个数的边界，`firstEnd` 和 `secondEnd`，两个数加起来的和数的起始位置是 `secondEnd + 1`。每次在移动 `firstEnd` 和 `secondEnd` 的时候，需要判断 `strings.HasPrefix(num[secondEnd + 1:], strconv.Itoa(x1 + x2))`，即后面的字符串中是否以和为开头。
+- 如果第一个数字起始数字出现了 0 ，或者第二个数字起始数字出现了 0，都算非法异常情况，都应该直接返回 false。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+)
+
+// This function controls various combinations as starting points
+func isAdditiveNumber(num string) bool {
+	if len(num) < 3 {
+		return false
+	}
+	for firstEnd := 0; firstEnd < len(num)/2; firstEnd++ {
+		if num[0] == '0' && firstEnd > 0 {
+			break
+		}
+		first, _ := strconv.Atoi(num[:firstEnd+1])
+		for secondEnd := firstEnd + 1; max(firstEnd, secondEnd-firstEnd) <= len(num)-secondEnd; secondEnd++ {
+			if num[firstEnd+1] == '0' && secondEnd-firstEnd > 1 {
+				break
+			}
+			second, _ := strconv.Atoi(num[firstEnd+1 : secondEnd+1])
+			if recursiveCheck(num, first, second, secondEnd+1) {
+				return true
+			}
+		}
+	}
+	return false
+}
+
+//Propagate for rest of the string
+func recursiveCheck(num string, x1 int, x2 int, left int) bool {
+	if left == len(num) {
+		return true
+	}
+	if strings.HasPrefix(num[left:], strconv.Itoa(x1+x2)) {
+		return recursiveCheck(num, x2, x1+x2, left+len(strconv.Itoa(x1+x2)))
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0304.Range-Sum-Query-2D-Immutable/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0307.Range-Sum-Query-Mutable/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0307.Range-Sum-Query-Mutable.md b/website/content/ChapterFour/0300~0399/0307.Range-Sum-Query-Mutable.md
new file mode 100755
index 000000000..50b02e262
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0307.Range-Sum-Query-Mutable.md
@@ -0,0 +1,161 @@
+# [307. Range Sum Query - Mutable](https://leetcode.com/problems/range-sum-query-mutable/)
+
+
+## 题目
+
+Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
+
+The update(i, val) function modifies nums by updating the element at index i to val.
+
+**Example**:
+
+    Given nums = [1, 3, 5]
+    
+    sumRange(0, 2) -> 9
+    update(1, 2)
+    sumRange(0, 2) -> 8
+
+**Note**:
+
+1. The array is only modifiable by the update function.
+2. You may assume the number of calls to update and sumRange function is distributed evenly.
+
+
+## 题目大意
+
+给定一个整数数组  nums，求出数组从索引 i 到 j  (i ≤ j) 范围内元素的总和，包含 i,  j 两点。
+
+update(i, val) 函数可以通过将下标为 i 的数值更新为 val，从而对数列进行修改。
+
+示例:
+
+```
+Given nums = [1, 3, 5]
+
+sumRange(0, 2) -> 9  
+update(1, 2)  
+sumRange(0, 2) -> 8  
+```
+
+说明:
+
+- 数组仅可以在 update 函数下进行修改。
+- 你可以假设 update 函数与 sumRange 函数的调用次数是均匀分布的。
+
+## 解题思路
+
+
+- 给出一个数组，数组里面的数都是`**可变**`的，设计一个数据结构能够满足查询数组任意区间内元素的和。
+- 对比第 303 题，这一题由于数组里面的元素都是**`可变`**的，所以第一个想到的解法就是线段树，构建一颗线段树，父结点内存的是两个子结点的和，初始化建树的时间复杂度是 O(log n)，查询区间元素和的时间复杂度是 O(log n)，更新元素值的时间复杂度是 O(log n)。
+- 如果此题还用 prefixSum 的思路解答呢？那每次 update 操作的时间复杂度都是 O(n)，因为每次更改一个值，最坏情况就是所有的 prefixSum 都要更新一次。prefixSum 的方法在这道题上面也可以 AC，只不过时间排名在 5%，非常差。
+- 此题也可以用树状数组解决。代码很直白，区间查询即是两个区间前缀和相减。最简单的树状数组应用。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "github.com/halfrost/leetcode-go/template"
+
+// NumArray define
+type NumArray struct {
+	st *template.SegmentTree
+}
+
+// Constructor307 define
+func Constructor307(nums []int) NumArray {
+	st := template.SegmentTree{}
+	st.Init(nums, func(i, j int) int {
+		return i + j
+	})
+	return NumArray{st: &st}
+}
+
+// Update define
+func (this *NumArray) Update(i int, val int) {
+	this.st.Update(i, val)
+}
+
+// SumRange define
+func (this *NumArray) SumRange(i int, j int) int {
+	return this.st.Query(i, j)
+}
+
+//解法二 prefixSum，sumRange 时间复杂度 O(1)
+
+// // NumArray define
+// type NumArray307 struct {
+// 	prefixSum []int
+// 	data   []int
+// }
+
+// // Constructor307 define
+// func Constructor307(nums []int) NumArray307 {
+// 	data := make([]int, len(nums))
+// 	for i := 0; i < len(nums); i++ {
+// 		data[i] = nums[i]
+// 	}
+// 	for i := 1; i < len(nums); i++ {
+// 		nums[i] += nums[i-1]
+// 	}
+// 	return NumArray307{prefixSum: nums, data: data}
+// }
+
+// // Update define
+// func (this *NumArray307) Update(i int, val int) {
+// 	this.data[i] = val
+// 	this.prefixSum[0] = this.data[0]
+// 	for i := 1; i < len(this.data); i++ {
+// 		this.prefixSum[i] = this.prefixSum[i-1] + this.data[i]
+// 	}
+// }
+
+// // SumRange define
+// func (this *NumArray307) SumRange(i int, j int) int {
+// 	if i > 0 {
+// 		return this.prefixSum[j] - this.prefixSum[i-1]
+// 	}
+// 	return this.prefixSum[j]
+// }
+
+// 解法三 树状数组
+// type NumArray struct {
+// 	bit  template.BinaryIndexedTree
+// 	data []int
+// }
+
+// // Constructor define
+// func Constructor307(nums []int) NumArray {
+// 	bit := template.BinaryIndexedTree{}
+// 	bit.InitWithNums(nums)
+// 	return NumArray{bit: bit, data: nums}
+// }
+
+// // Update define
+// func (this *NumArray) Update(i int, val int) {
+// 	this.bit.Add(i+1, val-this.data[i])
+// 	this.data[i] = val
+// }
+
+// // SumRange define
+// func (this *NumArray) SumRange(i int, j int) int {
+// 	return this.bit.Query(j+1) - this.bit.Query(i)
+// }
+
+/**
+ * Your NumArray object will be instantiated and called as such:
+ * obj := Constructor(nums);
+ * obj.Update(i,val);
+ * param_2 := obj.SumRange(i,j);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0306.Additive-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown.md b/website/content/ChapterFour/0300~0399/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown.md
new file mode 100755
index 000000000..ebee3b2d3
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown.md
@@ -0,0 +1,91 @@
+# [309. Best Time to Buy and Sell Stock with Cooldown](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/)
+
+
+## 题目
+
+Say you have an array for which the ith element is the price of a given stock on day i.
+
+Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:
+
+- You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
+- After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)
+
+**Example**:
+
+    Input: [1,2,3,0,2]
+    Output: 3 
+    Explanation: transactions = [buy, sell, cooldown, buy, sell]
+
+## 题目大意
+
+给定一个整数数组，其中第 i 个元素代表了第 i 天的股票价格 。​
+
+设计一个算法计算出最大利润。在满足以下约束条件下，你可以尽可能地完成更多的交易（多次买卖一支股票）:
+
+- 你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。
+- 卖出股票后，你无法在第二天买入股票 (即冷冻期为 1 天)。
+
+
+
+## 解题思路
+
+- 给定一个数组，表示一支股票在每一天的价格。设计一个交易算法，在这些天进行自动交易，要求：每一天只能进行一次操作；在买完股票后，必须卖了股票，才能再次买入；每次卖了股票以后，在下一天是不能购买的。问如何交易，能让利润最大？
+- 这一题是第 121 题和第 122 题的变种题。
+- 每天都有 3 种操作，`buy`，`sell`，`cooldown`。`sell` 之后的一天一定是 `cooldown`，但是 `cooldown` 可以出现在任意一天。例如：`buy，cooldown，cooldown，sell，cooldown，cooldown`。`buy[i]` 代表第 `i` 天通过 `buy` 或者 `cooldown` 结束此天能获得的最大收益。例如：`buy, sell, buy` 或者 `buy, cooldown, cooldown`。`sell[i]` 代表第 `i` 天通过 `sell` 或者 `cooldown` 结束此天能获得的最大收益。例如：`buy, sell, buy, sell` 或者 `buy, sell, cooldown, cooldown`。`price[i-1]` 代表第 `i` 天的股票价格(由于 price 是从 0 开始的)。
+- 第 i 天如果是 sell，那么这天能获得的最大收益是 `buy[i - 1] + price[i - 1]`，因为只有 buy 了才能 sell。如果这一天是 cooldown，那么这天能获得的最大收益还是 sell[i - 1]。所以 sell[i] 的状态转移方程 `sell[i] = max(buy[i - 1] + price[i - 1], sell[i - 1])`。`sell[0] = 0` 代表第一天就卖了，由于第一天不持有股票，所以 sell[0] = 0。`sell[1] = max(sell[0], buy[0]+prices[1])` 代表第一天卖了，和第一天不卖，第二天卖做对比，钱多的保存至 sell[1]。
+- 第 i 天如果是 buy，那么这天能获得的最大收益是 `sell[i - 2] - price[i - 1]`，因为 i - 1 天是 cooldown。如果这一天是 cooldown，那么这天能获得的最大收益还是 buy[i - 1]。所以 buy[i] 的状态转移方程 `buy[i] = max(sell[i - 2] - price[i - 1], buy[i - 1])`。`buy[0] = -prices[0]` 代表第一天就买入，所以金钱变成了负的。`buy[1] = max(buy[0], -prices[1])` 代表第一天不买入，第二天再买入。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+)
+
+// 解法一 DP
+func maxProfit309(prices []int) int {
+	if len(prices) <= 1 {
+		return 0
+	}
+	buy, sell := make([]int, len(prices)), make([]int, len(prices))
+	for i := range buy {
+		buy[i] = math.MinInt64
+	}
+	buy[0] = -prices[0]
+	buy[1] = max(buy[0], -prices[1])
+	sell[1] = max(sell[0], buy[0]+prices[1])
+	for i := 2; i < len(prices); i++ {
+		sell[i] = max(sell[i-1], buy[i-1]+prices[i])
+		buy[i] = max(buy[i-1], sell[i-2]-prices[i])
+	}
+	return sell[len(sell)-1]
+}
+
+// 解法二 优化辅助空间的 DP
+func maxProfit309_1(prices []int) int {
+	if len(prices) <= 1 {
+		return 0
+	}
+	buy := []int{-prices[0], max(-prices[0], -prices[1]), math.MinInt64}
+	sell := []int{0, max(0, -prices[0]+prices[1]), 0}
+
+	for i := 2; i < len(prices); i++ {
+		sell[i%3] = max(sell[(i-1)%3], buy[(i-1)%3]+prices[i])
+		buy[i%3] = max(buy[(i-1)%3], sell[(i-2)%3]-prices[i])
+	}
+	return sell[(len(prices)-1)%3]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0307.Range-Sum-Query-Mutable/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0315.Count-of-Smaller-Numbers-After-Self/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0315.Count-of-Smaller-Numbers-After-Self.md b/website/content/ChapterFour/0300~0399/0315.Count-of-Smaller-Numbers-After-Self.md
new file mode 100755
index 000000000..ac37bace0
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0315.Count-of-Smaller-Numbers-After-Self.md
@@ -0,0 +1,124 @@
+# [315. Count of Smaller Numbers After Self](https://leetcode.com/problems/count-of-smaller-numbers-after-self/)
+
+
+## 题目
+
+You are given an integer array nums and you have to return a new counts array. The counts array has the property where `counts[i]` is the number of smaller elements to the right of `nums[i]`.
+
+**Example**:
+
+    Input: [5,2,6,1]
+    Output: [2,1,1,0] 
+    Explanation:
+    To the right of 5 there are 2 smaller elements (2 and 1).
+    To the right of 2 there is only 1 smaller element (1).
+    To the right of 6 there is 1 smaller element (1).
+    To the right of 1 there is 0 smaller element.
+
+
+## 题目大意
+
+
+给定一个整数数组 nums，按要求返回一个新数组 counts。数组 counts 有该性质： counts[i] 的值是  nums[i] 右侧小于 nums[i] 的元素的数量。
+
+示例:
+
+```
+
+输入: [5,2,6,1]
+输出: [2,1,1,0] 
+解释:
+5 的右侧有 2 个更小的元素 (2 和 1).
+2 的右侧仅有 1 个更小的元素 (1).
+6 的右侧有 1 个更小的元素 (1).
+1 的右侧有 0 个更小的元素.
+
+```
+
+
+## 解题思路
+
+
+- 给出一个数组，要求输出数组中每个元素相对于数组中的位置右边比它小的元素。
+- 这一题是第 327 题的缩水版。由于需要找数组位置右边比当前位置元素小的元素，所以从数组右边开始往左边扫。构造一颗线段树，线段树里面父节点存的是子节点出现的次数和。有可能给的数据会很大，所以构造线段树的时候先离散化。还需要注意的是数组里面可能有重复元素，所以构造线段树要先去重并排序。从右往左扫的过程中，依次添加数组中的元素，添加了一次就立即 query 一次。query 的区间是 [minNum, nums[i]-1]。如果是 minNum 则输出 0，并且也要记得插入这个最小值。这一题的思路和第 327 题大体类似，详解可见第 327 题。
+- 这一题同样可以用树状数组来解答。相比 327 题简单很多。第一步还是把所有用到的元素放入 allNums 数组中，第二步排序 + 离散化。由于题目要求输出右侧更小的元素，所以第三步倒序插入构造树状数组，Query 查询 `[1,i-1]` 区间内元素总数即为右侧更小元素个数。注意最终输出是顺序输出，计算是逆序计算的，最终数组里面的答案还需要逆序一遍。相同的套路题有，第 327 题，第 493 题。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/leetcode-go/template"
+)
+
+// 解法一 线段树
+func countSmaller(nums []int) []int {
+	if len(nums) == 0 {
+		return []int{}
+	}
+	st, minNum, numsMap, numsArray, res := template.SegmentCountTree{}, 0, make(map[int]int, 0), []int{}, make([]int, len(nums))
+	for i := 0; i < len(nums); i++ {
+		numsMap[nums[i]] = nums[i]
+	}
+	for _, v := range numsMap {
+		numsArray = append(numsArray, v)
+	}
+	// 排序是为了使得线段树中的区间 left <= right，如果此处不排序，线段树中的区间有很多不合法。
+	sort.Ints(numsArray)
+	minNum = numsArray[0]
+	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
+	st.Init(numsArray, func(i, j int) int {
+		return 0
+	})
+	for i := len(nums) - 1; i >= 0; i-- {
+		if nums[i] == minNum {
+			res[i] = 0
+			st.UpdateCount(nums[i])
+			continue
+		}
+		st.UpdateCount(nums[i])
+		res[i] = st.Query(minNum, nums[i]-1)
+	}
+	return res
+}
+
+// 解法二 树状数组
+func countSmaller1(nums []int) []int {
+	// copy 一份原数组至所有数字 allNums 数组中
+	allNums, res := make([]int, len(nums)), []int{}
+	copy(allNums, nums)
+	// 将 allNums 离散化
+	sort.Ints(allNums)
+	k := 1
+	kth := map[int]int{allNums[0]: k}
+	for i := 1; i < len(allNums); i++ {
+		if allNums[i] != allNums[i-1] {
+			k++
+			kth[allNums[i]] = k
+		}
+	}
+	// 树状数组 Query
+	bit := template.BinaryIndexedTree{}
+	bit.Init(k)
+	for i := len(nums) - 1; i >= 0; i-- {
+		res = append(res, bit.Query(kth[nums[i]]-1))
+		bit.Add(kth[nums[i]], 1)
+	}
+	for i := 0; i < len(res)/2; i++ {
+		res[i], res[len(res)-1-i] = res[len(res)-1-i], res[i]
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0318.Maximum-Product-of-Word-Lengths/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0318.Maximum-Product-of-Word-Lengths.md b/website/content/ChapterFour/0300~0399/0318.Maximum-Product-of-Word-Lengths.md
new file mode 100755
index 000000000..436375e9b
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0318.Maximum-Product-of-Word-Lengths.md
@@ -0,0 +1,84 @@
+# [318. Maximum Product of Word Lengths](https://leetcode.com/problems/maximum-product-of-word-lengths/)
+
+
+## 题目
+
+Given a string array `words`, find the maximum value of `length(word[i]) * length(word[j])` where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.
+
+**Example 1**:
+
+    Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
+    Output: 16 
+    Explanation: The two words can be "abcw", "xtfn".
+
+**Example 2**:
+
+    Input: ["a","ab","abc","d","cd","bcd","abcd"]
+    Output: 4 
+    Explanation: The two words can be "ab", "cd".
+
+**Example 3**:
+
+    Input: ["a","aa","aaa","aaaa"]
+    Output: 0 
+    Explanation: No such pair of words.
+
+
+
+## 题目大意
+
+给定一个字符串数组 words，找到 length(word[i]) * length(word[j]) 的最大值，并且这两个单词不含有公共字母。你可以认为每个单词只包含小写字母。如果不存在这样的两个单词，返回 0。
+
+
+## 解题思路
+
+- 在字符串数组中找到 2 个没有公共字符的字符串，并且这两个字符串的长度乘积要是最大的，求这个最大的乘积。
+- 这里需要利用位运算 `&` 运算的性质，如果 `X & Y = 0`，说明 X 和 Y 完全不相同。那么我们将字符串都编码成二进制数，进行 `&` 运算即可分出没有公共字符的字符串，最后动态维护长度乘积最大值即可。将字符串编码成二进制数的规则比较简单，每个字符相对于 'a' 的距离，根据这个距离将 1 左移多少位。
+
+```c
+    a 1->1  
+    b 2->10  
+    c 4->100  
+    ab 3->11  
+    ac 5->101  
+    abc 7->111  
+    az 33554433->10000000000000000000000001  
+```
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func maxProduct318(words []string) int {
+	if words == nil || len(words) == 0 {
+		return 0
+	}
+	length, value, maxProduct := len(words), make([]int, len(words)), 0
+	for i := 0; i < length; i++ {
+		tmp := words[i]
+		value[i] = 0
+		for j := 0; j < len(tmp); j++ {
+			value[i] |= 1 << (tmp[j] - 'a')
+		}
+	}
+	for i := 0; i < length; i++ {
+		for j := i + 1; j < length; j++ {
+			if (value[i]&value[j]) == 0 && (len(words[i])*len(words[j]) > maxProduct) {
+				maxProduct = len(words[i]) * len(words[j])
+			}
+		}
+	}
+	return maxProduct
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0315.Count-of-Smaller-Numbers-After-Self/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0319.Bulb-Switcher/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0319.Bulb-Switcher.md b/website/content/ChapterFour/0300~0399/0319.Bulb-Switcher.md
new file mode 100644
index 000000000..5a4e16e1e
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0319.Bulb-Switcher.md
@@ -0,0 +1,65 @@
+# [319. Bulb Switcher](https://leetcode.com/problems/bulb-switcher/)
+
+
+## 题目
+
+There are n bulbs that are initially off. You first turn on all the bulbs, then you turn off every second bulb.
+
+On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb.
+
+Return the number of bulbs that are on after n rounds.
+
+**Example 1:**
+
+    Input: n = 3
+    Output: 1
+    Explanation: At first, the three bulbs are [off, off, off].
+    After the first round, the three bulbs are [on, on, on].
+    After the second round, the three bulbs are [on, off, on].
+    After the third round, the three bulbs are [on, off, off].
+    So you should return 1 because there is only one bulb is on.
+
+**Example 2:**
+
+    Input: n = 0
+    Output: 0
+
+**Example 3:**
+
+    Input: n = 1
+    Output: 1
+
+## 题目大意
+
+初始时有 n 个灯泡处于关闭状态。第一轮，你将会打开所有灯泡。接下来的第二轮，你将会每两个灯泡关闭一个。
+
+第三轮，你每三个灯泡就切换一个灯泡的开关（即，打开变关闭，关闭变打开）。第 i 轮，你每 i 个灯泡就切换一个灯泡的开关。直到第 n 轮，你只需要切换最后一个灯泡的开关。
+
+找出并返回 n 轮后有多少个亮着的灯泡。
+
+## 解题思路
+
+- 计算 1 到 n 中有奇数个约数的个数
+- 1 到 n 中的某个数 x 有奇数个约数,也即 x 是完全平方数
+- 计算 1 到 n 中完全平方数的个数 sqrt(n)
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func bulbSwitch(n int) int {
+	return int(math.Sqrt(float64(n)))
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0318.Maximum-Product-of-Word-Lengths/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0322.Coin-Change/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0322.Coin-Change.md b/website/content/ChapterFour/0300~0399/0322.Coin-Change.md
new file mode 100755
index 000000000..a1bd78323
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0322.Coin-Change.md
@@ -0,0 +1,68 @@
+# [322. Coin Change](https://leetcode.com/problems/coin-change/)
+
+
+## 题目
+
+You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return `-1`.
+
+**Example 1**:
+
+    Input: coins = [1, 2, 5], amount = 11
+    Output: 3 
+    Explanation: 11 = 5 + 5 + 1
+
+**Example 2**:
+
+    Input: coins = [2], amount = 3
+    Output: -1
+
+**Note**:
+
+You may assume that you have an infinite number of each kind of coin.
+
+## 题目大意
+
+给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额，返回 -1。
+
+
+
+## 解题思路
+
+- 给出一些硬币和一个总数，问组成这个总数的硬币数最少是多少个？
+- 这一题是经典的硬币问题，利用 DP 求解。不过这一题的测试用例有一个很大的值，这样开 DP 数组会比较浪费空间。例如 [1,1000000000,500000] 有这样的硬币种类，要求组成 2389412493027523 这样的总数。那么按照下面的解题方法，数组会开的很大，非常浪费空间。这个时候用 DFS 解题会节约一点空间。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func coinChange(coins []int, amount int) int {
+	dp := make([]int, amount+1)
+	dp[0] = 0
+	for i := 1; i < len(dp); i++ {
+		dp[i] = amount + 1
+	}
+	for i := 1; i <= amount; i++ {
+		for j := 0; j < len(coins); j++ {
+			if coins[j] <= i {
+				dp[i] = min(dp[i], dp[i-coins[j]]+1)
+			}
+		}
+	}
+	if dp[amount] > amount {
+		return -1
+	}
+	return dp[amount]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0319.Bulb-Switcher/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0324.Wiggle-Sort-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0324.Wiggle-Sort-II.md b/website/content/ChapterFour/0300~0399/0324.Wiggle-Sort-II.md
new file mode 100644
index 000000000..5d5a250eb
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0324.Wiggle-Sort-II.md
@@ -0,0 +1,270 @@
+# [324. Wiggle Sort II](https://leetcode.com/problems/wiggle-sort-ii/)
+
+## 题目
+
+Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3]....
+
+**Example 1**:
+
+```
+
+Input: nums = [1, 5, 1, 1, 6, 4]
+Output: One possible answer is [1, 4, 1, 5, 1, 6].
+
+```
+
+**Example 2**:
+
+```
+
+Input: nums = [1, 3, 2, 2, 3, 1]
+Output: One possible answer is [2, 3, 1, 3, 1, 2].
+
+```
+
+**Note**:   
+
+You may assume all input has valid answer.
+
+**Follow up**:   
+
+Can you do it in O(n) time and/or in-place with O(1) extra space?
+
+
+## 题目大意
+
+给定一个数组，要求给它进行“摆动排序”，“摆动排序”即：nums[0] < nums[1] > nums[2] < nums[3]...
+
+## 解题思路
+
+这一题最直接的方法是先排序，然后用 2 个指针，一个指向下标为 0 的位置，另一个指向下标为 n/2 的位置。最终的数组的奇数位从下标为 0 开始往后取，偶数位从下标为 n/2 中间位置开始往后取。这种方法时间复杂度为 O(n log n)。
+
+题目要求用时间复杂度 O(n) 和 空间复杂度 O(1) 的方法解决。思路如下，先找到数组中间大小的数字，然后把数组分为 2 部分：
+
+```c
+Index :       0   1   2   3   4   5
+Small half:   M       S       S    
+Large half:       L       L       L(M)
+```
+
+奇数位排中间数和小于中间数的数字，偶数位排大于中间数的数字和中间数。如果中间数字有多个，那么偶数位最后几位也是中间数，奇数位开头的前几位也是中间数。
+
+举例，给定一个数组如下，中间数是 5 。有 2 个 5 。
+
+```c
+13   6   5   5   4   2
+
+         M
+```
+
+```c
+Step 1: 
+Original idx: 0    1    2    3    4    5  
+Mapped idx:   1    3    5    0    2    4 
+Array:        13   6    5    5    4    2 
+             Left
+              i
+                                      Right
+```
+                                      
+nums[Mapped_idx[i]] = nums[1] = 6 > 5, 所以可以把 6 放在第 1 个奇数位的位置。left 和 i 同时右移。
+
+```c
+Step 2: 
+Original idx: 0    1    2    3    4    5  
+Mapped idx:   1    3    5    0    2    4 
+Array:        13   6    5    5    4    2 
+                  Left
+                   i
+                                      Right
+```
+                                    
+nums[3] = 5 = 5, 5 可以放在下标为 3 的位置，由于 5 已经和中间数相等了，所以只后移 i 。
+
+
+```c
+Step 3: 
+Original idx: 0    1    2    3    4    5  
+Mapped idx:   1    3    5    0    2    4 
+Array:        13   6    5    5    4    2 
+                  Left
+                        i
+                                     Right
+```
+
+nums[5] = 2 < 5, 因为比中位数小，所以应该放在偶数位的最后 1 位。这里的例子而言，应该放在下标为 4 的位置上。交换 nums[Mapped_idx[i]] 和 nums[Mapped_idx[Right]]，交换完成以后 right 向左移。
+
+
+```c
+Step 4: 
+Original idx: 0    1    2    3    4    5  
+Mapped idx:   1    3    5    0    2    4 
+Array:        13   6    5    5    2    4 
+                  Left
+                        i
+                               Right
+```
+
+nums[5] = 4 < 5, 因为比中位数小，所以应该放在偶数位的当前倒数第一位。这里的例子而言，应该放在下标为 2 的位置上。交换 nums[Mapped\_idx[i]] 和 nums[Mapped\_idx[Right]]，交换完成以后 right 向左移。
+
+
+```c
+Step 5: 
+Original idx: 0    1    2    3    4    5  
+Mapped idx:   1    3    5    0    2    4 
+Array:        13   6    4    5    2    5 
+                  Left
+                        i
+                            Right
+```
+
+nums[5] = 5 = 5, 由于 5 已经和中间数相等了，所以只后移 i 。
+
+```c
+Step 6: 
+Original idx: 0    1    2    3    4    5  
+Mapped idx:   1    3    5    0    2    4 
+Array:        13   6    4    5    2    5 
+                  Left
+                             i
+                            Right
+```
+
+
+nums[0] = 13 > 5, 由于 13 比中位数大，所以可以把 13 放在第 2 个奇数位的位置，并移动 left 和 i 。
+
+
+```c
+Step Final: 
+Original idx: 0    1    2    3    4    5  
+Mapped idx:   1    3    5    0    2    4 
+Array:        5    6    4    13   2    5 
+                      Left
+                                  i
+                            Right
+```
+                            
+i > Right, 退出循环，最终摆动排序的结果是 5 6 4 13 2 5 。
+
+具体时间见代码，时间复杂度 O(n) 和 空间复杂度 O(1)。
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+// 解法一
+func wiggleSort(nums []int) {
+	if len(nums) < 2 {
+		return
+	}
+	median := findKthLargest324(nums, (len(nums)+1)/2)
+	n, i, left, right := len(nums), 0, 0, len(nums)-1
+
+	for i <= right {
+		if nums[indexMap(i, n)] > median {
+			nums[indexMap(left, n)], nums[indexMap(i, n)] = nums[indexMap(i, n)], nums[indexMap(left, n)]
+			left++
+			i++
+		} else if nums[indexMap(i, n)] < median {
+			nums[indexMap(right, n)], nums[indexMap(i, n)] = nums[indexMap(i, n)], nums[indexMap(right, n)]
+			right--
+		} else {
+			i++
+		}
+	}
+}
+
+func indexMap(index, n int) int {
+	return (1 + 2*index) % (n | 1)
+}
+
+func findKthLargest324(nums []int, k int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	return selection324(nums, 0, len(nums)-1, len(nums)-k)
+}
+
+func selection324(arr []int, l, r, k int) int {
+	if l == r {
+		return arr[l]
+	}
+	p := partition324(arr, l, r)
+
+	if k == p {
+		return arr[p]
+	} else if k < p {
+		return selection324(arr, l, p-1, k)
+	} else {
+		return selection324(arr, p+1, r, k)
+	}
+}
+
+func partition324(a []int, lo, hi int) int {
+	pivot := a[hi]
+	i := lo - 1
+	for j := lo; j < hi; j++ {
+		if a[j] < pivot {
+			i++
+			a[j], a[i] = a[i], a[j]
+		}
+	}
+	a[i+1], a[hi] = a[hi], a[i+1]
+	return i + 1
+}
+
+// 解法二
+func wiggleSort1(nums []int) {
+	if len(nums) < 2 {
+		return
+	}
+	array := make([]int, len(nums))
+	copy(array, nums)
+	sort.Ints(array)
+	n := len(nums)
+	left := (n+1)/2 - 1 // median index
+	right := n - 1      // largest value index
+	for i := 0; i < len(nums); i++ {
+		// copy large values on odd indexes
+		if i%2 == 1 {
+			nums[i] = array[right]
+			right--
+		} else { // copy values decremeting from median on even indexes
+			nums[i] = array[left]
+			left--
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0322.Coin-Change/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0326.Power-of-Three/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0326.Power-of-Three.md b/website/content/ChapterFour/0300~0399/0326.Power-of-Three.md
new file mode 100755
index 000000000..99596c264
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0326.Power-of-Three.md
@@ -0,0 +1,84 @@
+# [326. Power of Three](https://leetcode.com/problems/power-of-three/)
+
+
+## 题目
+
+Given an integer, write a function to determine if it is a power of three.
+
+**Example 1**:
+
+    Input: 27
+    Output: true
+
+**Example 2**:
+
+    Input: 0
+    Output: false
+
+**Example 3**:
+
+    Input: 9
+    Output: true
+
+**Example 4**:
+
+    Input: 45
+    Output: false
+
+**Follow up**:
+
+Could you do it without using any loop / recursion?
+
+
+## 题目大意
+
+给定一个整数，写一个函数来判断它是否是 3 的幂次方。
+
+
+## 解题思路
+
+- 判断一个数是不是 3 的 n 次方。
+- 这一题最简单的思路是循环，可以通过。但是题目要求不循环就要判断，这就需要用到数论的知识了。由于 3^20 超过了 int 的范围了，所以 3^19 次方就是 int 类型中最大的值。这一题和第 231 题是一样的思路。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 数论
+func isPowerOfThree(n int) bool {
+	// 1162261467 is 3^19,  3^20 is bigger than int
+	return n > 0 && (1162261467%n == 0)
+}
+
+// 解法二 打表法
+func isPowerOfThree1(n int) bool {
+	// 1162261467 is 3^19,  3^20 is bigger than int
+	allPowerOfThreeMap := map[int]int{1: 1, 3: 3, 9: 9, 27: 27, 81: 81, 243: 243, 729: 729, 2187: 2187, 6561: 6561, 19683: 19683, 59049: 59049, 177147: 177147, 531441: 531441, 1594323: 1594323, 4782969: 4782969, 14348907: 14348907, 43046721: 43046721, 129140163: 129140163, 387420489: 387420489, 1162261467: 1162261467}
+	_, ok := allPowerOfThreeMap[n]
+	return ok
+}
+
+// 解法三 循环
+func isPowerOfThree2(num int) bool {
+	for num >= 3 {
+		if num%3 == 0 {
+			num = num / 3
+		} else {
+			return false
+		}
+	}
+	return num == 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0324.Wiggle-Sort-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0327.Count-of-Range-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0327.Count-of-Range-Sum.md b/website/content/ChapterFour/0300~0399/0327.Count-of-Range-Sum.md
new file mode 100755
index 000000000..6b57a221f
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0327.Count-of-Range-Sum.md
@@ -0,0 +1,185 @@
+# [327. Count of Range Sum](https://leetcode.com/problems/count-of-range-sum/)
+
+
+## 题目
+
+Given an integer array `nums`, return the number of range sums that lie in `[lower, upper]` inclusive.Range sum `S(i, j)` is defined as the sum of the elements in `nums` between indices `i` and `j` (`i` ≤ `j`), inclusive.
+
+**Note**:A naive algorithm of O(n2) is trivial. You MUST do better than that.
+
+**Example**:
+
+    Input: nums = [-2,5,-1], lower = -2, upper = 2,
+    Output: 3 
+    Explanation: The three ranges are : [0,0], [2,2], [0,2] and their respective sums are: -2, -1, 2.
+
+
+## 题目大意
+
+
+给定一个整数数组 nums，返回区间和在 [lower, upper] 之间的个数，包含 lower 和 upper。区间和 S(i, j) 表示在 nums 中，位置从 i 到 j 的元素之和，包含 i 和 j (i ≤ j)。
+
+说明:   
+最直观的算法复杂度是 O(n^2) ，请在此基础上优化你的算法。
+
+
+## 解题思路
+
+- 给出一个数组，要求在这个数组中找出任意一段子区间的和，位于 [lower,upper] 之间。
+- 这一题可以用暴力解法，2 层循环，遍历所有子区间，求和并判断是否位于 [lower,upper] 之间，时间复杂度 O(n^2)。
+- 这一题当然还有更优的解法，用线段树或者树状数组，将时间复杂度降为 O(n log n)。题目中要求 `lower ≤ sum(i,j) ≤ upper`，`sum(i,j) = prefixSum(j) - prefixSum(i-1)`，那么 `lower + prefixSum(i-1) ≤ prefixSum(j) ≤ upper + prefixSum(i-1)`。所以利用前缀和将区间和转换成了前缀和在线段树中 `query` 的问题，只不过线段树中父节点中存的不是子节点的和，而应该是子节点出现的次数。第二个转换，由于前缀和会很大，所以需要离散化。例如 `prefixSum = [-3,-2,-1,0]`，用前缀和下标进行离散化，所以线段树中左右区间变成了 0-3 。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_0.png)
+
+    利用 `prefixSum` 下标离散化：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_1.png)
+
+- 还需要注意一些小细节，`prefixSum` 计算完以后需要去重，去重以后并排序，方便构造线段树的有效区间。如果不去重，线段树中可能出现非法区间(left > right)或者重叠区间。最后一步往线段树中倒序插入 `prefixSum` 的时候，用的是非去重的，插入 `prefixSum[j]` 代表 sum(i,j) 中的 j，例如往线段树中插入 `prefixSum[5]`，代表当前树中加入了 j = 5 的情况。query 操作实质是在做区间匹配，例如当前 i 循环到 i = 3，累计往线段树中插入了 `prefixSum[5]`，`prefixSum[4]`，`prefixSum[3]`，那么 query 操作实质是在判断：`lower ≤ sum(i=3,j=3) ≤ upper`，`lower ≤ sum(i=3,j=4) ≤ upper`，`lower ≤ sum(i=3,j=5) ≤ upper`，这 3 个等式是否成立，有几个成立就返回几个，即是最终要求得的结果的一部分。
+- 举个例子，`nums = [-3,1,2,-2,2,-1]`，`prefixSum = [-3,-2,0,-2,0,-1]`，去重以后并排序得到 `sum = [-3,-2,-1,0]`。离散化构造线段树，这里出于演示的方便，下图中就不画出离散后的线段树了，用非离散的线段树展示：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_2_.png)
+
+    倒序插入 `len(prefixSum)-1 = prefixSum[5] = -1`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_3_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[5-1], -1 + prefixSum[5-1]] = [-3,-1]`，即判断 `-3 ≤ sum(5,5) ≤ -1`，满足等式的有几种情况，这里明显只有一种情况，即 `j = 5`，也满足等式，所以这一步 `res = 1`。
+
+- 倒序插入 `len(prefixSum)-2 = prefixSum[4] = 0`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_4_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[4-1], -1 + prefixSum[4-1]] = [-5,-3]`，即判断 `-5 ≤ sum(4,  4,5) ≤ -3`，满足等式的有几种情况，这里有两种情况，即 `j = 4` 或者 `j = 5`，都不满足等式，所以这一步 `res = 0`。
+
+- 倒序插入 `len(prefixSum)-3 = prefixSum[3] = -2`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_5_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[3-1], -1 + prefixSum[3-1]] = [-3,-1]`，即判断 `-3 ≤ sum(3,  3,4,5) ≤ -1`，满足等式的有几种情况，这里有三种情况，即 `j = 3` 、`j = 4` 或者 `j = 5`，满足等式的有 `j = 3` 和 `j = 5`，即 `-3 ≤ sum(3, 3) ≤ -1` 和 `-3 ≤ sum(3, 5) ≤ -1`。所以这一步 `res = 2`。
+
+- 倒序插入 `len(prefixSum)-4 = prefixSum[2] = 0`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_6_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[2-1], -1 + prefixSum[2-1]] = [-5,-3]`，即判断 `-5 ≤ sum(2,  2,3,4,5) ≤ -3`，满足等式的有几种情况，这里有四种情况，即 `j = 2`、 `j = 3` 、`j = 4` 或者 `j = 5`，都不满足等式。所以这一步 `res = 0`。
+
+- 倒序插入 `len(prefixSum)-5 = prefixSum[1] = -2`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_7_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[1-1], -1 + prefixSum[1-1]] = [-6,-4]`，即判断 `-6 ≤ sum(1,  1,2,3,4,5) ≤ -4`，满足等式的有几种情况，这里有五种情况，即 `j = 1`、 `j = 2`、 `j = 3` 、`j = 4` 或者 `j = 5`，都不满足等式。所以这一步 `res = 0`。
+
+- 倒序插入 `len(prefixSum)-6 = prefixSum[0] = -3`：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_327_8_.png)
+
+    这时候查找区间变为了 `[-3 + prefixSum[0-1], -1 + prefixSum[0-1]] = [-3,-1]`，注意 `prefixSum[-1] = 0`，即判断 `-3 ≤ sum(0,  0,1,2,3,4,5) ≤ -1`，满足等式的有几种情况，这里有六种情况，即 `j = 0`、`j = 1`、`j = 2`、 `j = 3` 、`j = 4` 或者 `j = 5`，满足等式的有 `j = 0`、`j = 1`、 `j = 3` 和 `j = 5`，即 `-3 ≤ sum(0, 0) ≤ -1` 、 `-3 ≤ sum(0, 1) ≤ -1`、`-3 ≤ sum(0, 3) ≤ -1` 和 `-3 ≤ sum(0, 5) ≤ -1`。所以这一步 `res = 4`。最后的答案就是把每一步的结果都累加，`res = 1 + 0 + 2 + 0 + 0 + 4 = 7`。
+
+- 此题同样可以用树状数组来解答。同样把问题先转化成区间 Query 的模型，`lower ≤ prefixSum(j) - prefixSum(i-1) ≤ upper` 等价于 `prefixSum(j) - upper ≤ prefixSum(i-1) ≤ prefixSum(j) - lower`，`i` 的取值在 `[0,j-1]` 区间内。所以题目可以转化为 `i` 在 `[0,j-1]` 区间内取值，问数组 `prefixSum[0...j-1]` 中的所有取值，位于区间 `[prefixSum(j) - upper, prefixSum(j) - lower]` 内的次数。在树状数组中，区间内的前缀和可以转化为 2 个区间的前缀和相减，即 `Query([i,j]) = Query(j) - Query(i-1)`。所以这道题枚举数组 `prefixSum[0...j-1]` 中每个值是否出现在指定区间内出现次数即可。第一步先将所有的前缀和 `prefixSum(j)` 以及 `[prefixSum(j) - upper, prefixSum(j) - lower]` 计算出来。第二步排序和离散化，离散化以后的点区间为 `[1,n]`。最后根据数组 `prefixSum(j)` 的值在指定区间内查询出现次数即可。相同的套路题有，第 315 题，第 493 题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/leetcode-go/template"
+)
+
+// 解法一 线段树，时间复杂度 O(n log n)
+func countRangeSum(nums []int, lower int, upper int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	st, prefixSum, sumMap, sumArray, res := template.SegmentCountTree{}, make([]int, len(nums)), make(map[int]int, 0), []int{}, 0
+	prefixSum[0], sumMap[nums[0]] = nums[0], nums[0]
+	for i := 1; i < len(nums); i++ {
+		prefixSum[i] = prefixSum[i-1] + nums[i]
+		sumMap[prefixSum[i]] = prefixSum[i]
+	}
+	// sumArray 是 prefixSum 去重之后的版本，利用 sumMap 去重
+	for _, v := range sumMap {
+		sumArray = append(sumArray, v)
+	}
+	// 排序是为了使得线段树中的区间 left <= right，如果此处不排序，线段树中的区间有很多不合法。
+	sort.Ints(sumArray)
+	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
+	st.Init(sumArray, func(i, j int) int {
+		return 0
+	})
+	// 倒序是为了方便寻找 j，sum(i，j) 规定了 j >= i，所以倒序遍历，i 从大到小
+	for i := len(nums) - 1; i >= 0; i-- {
+		// 插入的 prefixSum[i] 即是 j
+		st.UpdateCount(prefixSum[i])
+		if i > 0 {
+			res += st.Query(lower+prefixSum[i-1], upper+prefixSum[i-1])
+		} else {
+			res += st.Query(lower, upper)
+		}
+	}
+	return res
+}
+
+// 解法二 树状数组，时间复杂度 O(n log n)
+func countRangeSum1(nums []int, lower int, upper int) int {
+	n := len(nums)
+	// 计算前缀和 preSum，以及后面统计时会用到的所有数字 allNums
+	allNums, preSum, res := make([]int, 1, 3*n+1), make([]int, n+1), 0
+	for i, v := range nums {
+		preSum[i+1] = preSum[i] + v
+		allNums = append(allNums, preSum[i+1], preSum[i+1]-lower, preSum[i+1]-upper)
+	}
+	// 将 allNums 离散化
+	sort.Ints(allNums)
+	k := 1
+	kth := map[int]int{allNums[0]: k}
+	for i := 1; i <= 3*n; i++ {
+		if allNums[i] != allNums[i-1] {
+			k++
+			kth[allNums[i]] = k
+		}
+	}
+	// 遍历 preSum，利用树状数组计算每个前缀和对应的合法区间数
+	bit := template.BinaryIndexedTree{}
+	bit.Init(k)
+	bit.Add(kth[0], 1)
+	for _, sum := range preSum[1:] {
+		left, right := kth[sum-upper], kth[sum-lower]
+		res += bit.Query(right) - bit.Query(left-1)
+		bit.Add(kth[sum], 1)
+	}
+	return res
+}
+
+// 解法三 暴力，时间复杂度 O(n^2)
+func countRangeSum2(nums []int, lower int, upper int) int {
+	res, n := 0, len(nums)
+	for i := 0; i < n; i++ {
+		tmp := 0
+		for j := i; j < n; j++ {
+			if i == j {
+				tmp = nums[i]
+			} else {
+				tmp += nums[j]
+			}
+			if tmp <= upper && tmp >= lower {
+				res++
+			}
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0326.Power-of-Three/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0328.Odd-Even-Linked-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0328.Odd-Even-Linked-List.md b/website/content/ChapterFour/0300~0399/0328.Odd-Even-Linked-List.md
new file mode 100644
index 000000000..dcf9585a7
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0328.Odd-Even-Linked-List.md
@@ -0,0 +1,84 @@
+# [328. Odd Even Linked List](https://leetcode.com/problems/odd-even-linked-list/)
+
+## 题目
+
+Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.
+
+You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.
+
+**Example 1**:
+
+```
+
+Input: 1->2->3->4->5->NULL
+Output: 1->3->5->2->4->NULL
+
+```
+
+**Example 2**:
+
+```
+
+Input: 2->1->3->5->6->4->7->NULL
+Output: 2->3->6->7->1->5->4->NULL
+
+```
+
+**Note**:
+
+- The relative order inside both the even and odd groups should remain as it was in the input.
+- The first node is considered odd, the second node even and so on ...
+
+## 题目大意
+
+这道题和第 86 题非常类型。第 86 题是把排在某个点前面的小值放在一个链表中，排在某个点后端的大值放在另外一个链表中，最后 2 个链表首尾拼接一下就是答案。
+
+## 解题思路
+
+这道题思路也是一样的，分别把奇数节点和偶数节点都放在 2 个链表中，最后首尾拼接就是答案。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func oddEvenList(head *ListNode) *ListNode {
+	oddHead := &ListNode{Val: 0, Next: nil}
+	odd := oddHead
+	evenHead := &ListNode{Val: 0, Next: nil}
+	even := evenHead
+
+	count := 1
+	for head != nil {
+		if count%2 == 1 {
+			odd.Next = head
+			odd = odd.Next
+		} else {
+			even.Next = head
+			even = even.Next
+		}
+		head = head.Next
+		count++
+	}
+	even.Next = nil
+	odd.Next = evenHead.Next
+	return oddHead.Next
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0327.Count-of-Range-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0329.Longest-Increasing-Path-in-a-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0329.Longest-Increasing-Path-in-a-Matrix.md b/website/content/ChapterFour/0300~0399/0329.Longest-Increasing-Path-in-a-Matrix.md
new file mode 100755
index 000000000..8369dbdec
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0329.Longest-Increasing-Path-in-a-Matrix.md
@@ -0,0 +1,113 @@
+# [329. Longest Increasing Path in a Matrix](https://leetcode.com/problems/longest-increasing-path-in-a-matrix/)
+
+
+## 题目
+
+Given an integer matrix, find the length of the longest increasing path.
+
+From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).
+
+**Example 1**:
+
+    Input: nums = 
+    [
+      [9,9,4],
+      [6,6,8],
+      [2,1,1]
+    ] 
+    Output: 4 
+    Explanation: The longest increasing path is [1, 2, 6, 9].
+
+**Example 2**:
+
+    Input: nums = 
+    [
+      [3,4,5],
+      [3,2,6],
+      [2,2,1]
+    ] 
+    Output: 4 
+    Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.
+
+
+## 题目大意
+
+给定一个整数矩阵，找出最长递增路径的长度。对于每个单元格，你可以往上，下，左，右四个方向移动。 你不能在对角线方向上移动或移动到边界外（即不允许环绕）。
+
+
+## 解题思路
+
+
+- 给出一个矩阵，要求在这个矩阵中找到一个最长递增的路径。路径有上下左右 4 个方向。
+- 这一题解题思路很明显，用 DFS 即可。在提交完第一版以后会发现 TLE，因为题目给出了一个非常大的矩阵，搜索次数太多。所以需要用到记忆化，把曾经搜索过的最大长度缓存起来，增加了记忆化以后再次提交 AC。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+)
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func longestIncreasingPath(matrix [][]int) int {
+	cache, res := make([][]int, len(matrix)), 0
+	for i := 0; i < len(cache); i++ {
+		cache[i] = make([]int, len(matrix[0]))
+	}
+	for i, v := range matrix {
+		for j := range v {
+			searchPath(matrix, cache, math.MinInt64, i, j)
+			res = max(res, cache[i][j])
+		}
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func isInIntBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+
+func searchPath(board, cache [][]int, lastNum, x, y int) int {
+	if board[x][y] <= lastNum {
+		return 0
+	}
+	if cache[x][y] > 0 {
+		return cache[x][y]
+	}
+	count := 1
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		if isInIntBoard(board, nx, ny) {
+			count = max(count, searchPath(board, cache, board[x][y], nx, ny)+1)
+		}
+	}
+	cache[x][y] = count
+	return count
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0328.Odd-Even-Linked-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0331.Verify-Preorder-Serialization-of-a-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0331.Verify-Preorder-Serialization-of-a-Binary-Tree.md b/website/content/ChapterFour/0300~0399/0331.Verify-Preorder-Serialization-of-a-Binary-Tree.md
new file mode 100644
index 000000000..5167b2620
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0331.Verify-Preorder-Serialization-of-a-Binary-Tree.md
@@ -0,0 +1,92 @@
+# [331. Verify Preorder Serialization of a Binary Tree](https://leetcode.com/problems/verify-preorder-serialization-of-a-binary-tree/)
+
+## 题目
+
+One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as #.
+
+```
+
+     _9_
+    /   \
+   3     2
+  / \   / \
+ 4   1  #  6
+/ \ / \   / \
+# # # #   # #
+
+```
+
+For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node.
+
+Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.
+
+Each comma separated value in the string must be either an integer or a character '#' representing null pointer.
+
+You may assume that the input format is always valid, for example it could never contain two consecutive commas such as "1,,3".
+
+**Example 1**:
+
+```
+
+Input: "9,3,4,#,#,1,#,#,2,#,6,#,#"
+Output: true
+
+```
+
+**Example 2**:
+
+```
+
+Input: "1,#"
+Output: false
+
+```
+**Example 3**:
+
+```
+
+Input: "9,#,#,1"
+Output: false
+
+```
+
+## 题目大意
+
+给定一串以逗号分隔的序列，验证它是否是正确的二叉树的前序序列化。编写一个在不重构树的条件下的可行算法。
+
+## 解题思路
+
+这道题有些人用栈，有些用栈的深度求解。换个视角。如果叶子结点是 null，那么所有非 null 的结点(除了 root 结点)必然有 2 个出度，1 个入度(2 个孩子和 1 个父亲，孩子可能为空，但是这一题用 "#" 代替了，所以肯定有 2 个孩子)；所有的 null 结点只有 0 个出度，1 个入度(0 个孩子和 1 个父亲)。
+
+我们开始构建这颗树，在构建过程中，我们记录出度和度之间的差异 `diff = outdegree - indegree`。当下一个节点到来时，我们将 diff 减 1，因为这个节点提供了一个度。如果这个节点不为 null，我们将 diff 增加 2，因为它提供两个出度。如果序列化是正确的，则 diff 应该永远不会为负，并且 diff 在完成时将为零。最后判断一下 diff 是不是为 0 即可判断它是否是正确的二叉树的前序序列化。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func isValidSerialization(preorder string) bool {
+	nodes, diff := strings.Split(preorder, ","), 1
+	for _, node := range nodes {
+		diff--
+		if diff < 0 {
+			return false
+		}
+		if node != "#" {
+			diff += 2
+		}
+	}
+	return diff == 0
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0329.Longest-Increasing-Path-in-a-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0337.House-Robber-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0337.House-Robber-III.md b/website/content/ChapterFour/0300~0399/0337.House-Robber-III.md
new file mode 100644
index 000000000..08ee691b6
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0337.House-Robber-III.md
@@ -0,0 +1,80 @@
+# [337. House Robber III](https://leetcode.com/problems/house-robber-iii/)
+
+
+
+## 题目
+
+The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.
+
+Determine the maximum amount of money the thief can rob tonight without alerting the police.
+
+**Example 1**:
+
+```
+Input: [3,2,3,null,3,null,1]
+
+     3
+    / \
+   2   3
+    \   \ 
+     3   1
+
+Output: 7 
+Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
+```
+
+**Example 2**:
+
+```
+Input: [3,4,5,1,3,null,1]
+
+     3
+    / \
+   4   5
+  / \   \ 
+ 1   3   1
+
+Output: 9
+Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.
+```
+
+## 题目大意
+
+一个新的可行窃的地区只有一个入口，称之为“根”。 除了“根”之外，每栋房子有且只有一个“父“房子与之相连。一番侦察之后，聪明的小偷意识到“这个地方的所有房屋的排列类似于一棵二叉树”。 如果两个直接相连的房子在同一天晚上被打劫，房屋将自动报警。计算在不触动警报的情况下，小偷一晚能够盗取的最高金额。
+
+
+## 解题思路
+
+- 这一题是打家劫舍的第 3 题。这一题需要偷的房子是树状的。报警的条件还是相邻的房子如果都被偷了，就会触发报警。只不过这里相邻的房子是树上的。问小偷在不触发报警的条件下最终能偷的最高金额。
+- 解题思路是 DFS。当前节点是否被打劫，会产生 2 种结果。如果当前节点被打劫，那么它的孩子节点可以被打劫；如果当前节点没有被打劫，那么它的孩子节点不能被打劫。按照这个逻辑递归，最终递归到根节点，取最大值输出即可。
+
+## 代码
+
+```go
+
+func rob337(root *TreeNode) int {
+	a, b := dfsTreeRob(root)
+	return max(a, b)
+}
+
+func dfsTreeRob(root *TreeNode) (a, b int) {
+	if root == nil {
+		return 0, 0
+	}
+	l0, l1 := dfsTreeRob(root.Left)
+	r0, r1 := dfsTreeRob(root.Right)
+	// 当前节点没有被打劫
+	tmp0 := max(l0, l1) + max(r0, r1)
+	// 当前节点被打劫
+	tmp1 := root.Val + l0 + r0
+	return tmp0, tmp1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0331.Verify-Preorder-Serialization-of-a-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0338.Counting-Bits/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0338.Counting-Bits.md b/website/content/ChapterFour/0300~0399/0338.Counting-Bits.md
new file mode 100755
index 000000000..d67be88de
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0338.Counting-Bits.md
@@ -0,0 +1,61 @@
+# [338. Counting Bits](https://leetcode.com/problems/counting-bits/)
+
+
+## 题目
+
+Given a non negative integer number **num**. For every numbers **i** in the range **0 ≤ i ≤ num** calculate the number of 1's in their binary representation and return them as an array.
+
+**Example 1**:
+
+    Input: 2
+    Output: [0,1,1]
+
+**Example 2**:
+
+    Input: 5
+    Output: [0,1,1,2,1,2]
+
+**Follow up**:
+
+- It is very easy to come up with a solution with run time **O(n*sizeof(integer))**. But can you do it in linear time **O(n)** /possibly in a single pass?
+- Space complexity should be **O(n)**.
+- Can you do it like a boss? Do it without using any builtin function like **\_\_builtin\_popcount** in c++ or in any other language.
+
+## 题目大意
+
+
+给定一个非负整数 num。对于 0 ≤ i ≤ num 范围中的每个数字 i ，计算其二进制数中的 1 的数目并将它们作为数组返回。
+
+## 解题思路
+
+- 给出一个数，要求计算出 0 ≤ i ≤ num 中每个数的二进制位 1 的个数。
+- 这一题就是利用二进制位运算的经典题。
+
+        X&1==1or==0，可以用 X&1 判断奇偶性，X&1>0 即奇数。
+        X = X & (X-1) 清零最低位的1
+        X & -X => 得到最低位的1 
+        X&~X=>0
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func countBits(num int) []int {
+	bits := make([]int, num+1)
+	for i := 1; i <= num; i++ {
+		bits[i] += bits[i&(i-1)] + 1
+	}
+	return bits
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0337.House-Robber-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0341.Flatten-Nested-List-Iterator/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0341.Flatten-Nested-List-Iterator.md b/website/content/ChapterFour/0300~0399/0341.Flatten-Nested-List-Iterator.md
new file mode 100644
index 000000000..5f6fa1c79
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0341.Flatten-Nested-List-Iterator.md
@@ -0,0 +1,117 @@
+# [341. Flatten Nested List Iterator](https://leetcode.com/problems/flatten-nested-list-iterator/)
+
+
+## 题目
+
+Given a nested list of integers, implement an iterator to flatten it.
+
+Each element is either an integer, or a list -- whose elements may also be integers or other lists.
+
+**Example 1:**
+
+```
+Input:[[1,1],2,[1,1]]
+Output:[1,1,2,1,1]
+Explanation:By callingnext repeatedly untilhasNext returns false,
+             the order of elements returned bynext should be:[1,1,2,1,1].
+```
+
+**Example 2:**
+
+```
+Input:[1,[4,[6]]]
+Output:[1,4,6]
+Explanation:By callingnext repeatedly untilhasNext returns false,
+             the order of elements returned bynext should be:[1,4,6].
+
+```
+
+## 题目大意
+
+给你一个嵌套的整型列表。请你设计一个迭代器，使其能够遍历这个整型列表中的所有整数。列表中的每一项或者为一个整数，或者是另一个列表。其中列表的元素也可能是整数或是其他列表。
+
+## 解题思路
+
+- 题目要求实现一个嵌套版的数组。可以用 `[]int` 实现，也可以用链表实现。笔者此处用链表实现。外层构造一个一维数组，一维数组内部每个元素是一个链表。额外还需要记录这个嵌套链表在原数组中的 `index` 索引。`Next()` 实现比较简单，取出对应的嵌套节点。`HasNext()` 方法则感觉嵌套节点里面的 `index` 信息判断是否还有 `next` 元素。
+
+## 代码
+
+```go
+/**
+ * // This is the interface that allows for creating nested lists.
+ * // You should not implement it, or speculate about its implementation
+ * type NestedInteger struct {
+ * }
+ *
+ * // Return true if this NestedInteger holds a single integer, rather than a nested list.
+ * func (this NestedInteger) IsInteger() bool {}
+ *
+ * // Return the single integer that this NestedInteger holds, if it holds a single integer
+ * // The result is undefined if this NestedInteger holds a nested list
+ * // So before calling this method, you should have a check
+ * func (this NestedInteger) GetInteger() int {}
+ *
+ * // Set this NestedInteger to hold a single integer.
+ * func (n *NestedInteger) SetInteger(value int) {}
+ *
+ * // Set this NestedInteger to hold a nested list and adds a nested integer to it.
+ * func (this *NestedInteger) Add(elem NestedInteger) {}
+ *
+ * // Return the nested list that this NestedInteger holds, if it holds a nested list
+ * // The list length is zero if this NestedInteger holds a single integer
+ * // You can access NestedInteger's List element directly if you want to modify it
+ * func (this NestedInteger) GetList() []*NestedInteger {}
+ */
+
+type NestedIterator struct {
+    stack *list.List
+}
+
+type listIndex struct {
+    nestedList []*NestedInteger
+    index int
+}
+
+func Constructor(nestedList []*NestedInteger) *NestedIterator {
+    stack := list.New()
+    stack.PushBack(&listIndex{nestedList, 0})
+    return &NestedIterator{stack}
+}
+
+func (this *NestedIterator) Next() int {
+    if !this.HasNext() {
+        return -1
+    }
+    last := this.stack.Back().Value.(*listIndex)
+    nestedList, i := last.nestedList, last.index
+    val := nestedList[i].GetInteger()
+    last.index++
+    return val
+}
+
+func (this *NestedIterator) HasNext() bool {
+    stack := this.stack
+    for stack.Len() > 0 {
+        last := stack.Back().Value.(*listIndex)
+        nestedList, i := last.nestedList, last.index
+        if i >= len(nestedList) {
+            stack.Remove(stack.Back())
+        } else {
+            val := nestedList[i]
+            if val.IsInteger() {
+                return true
+            }
+            last.index++
+            stack.PushBack(&listIndex{val.GetList(), 0})
+        }
+    }
+    return false
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0338.Counting-Bits/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0342.Power-of-Four/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0342.Power-of-Four.md b/website/content/ChapterFour/0300~0399/0342.Power-of-Four.md
new file mode 100755
index 000000000..a8a6f3848
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0342.Power-of-Four.md
@@ -0,0 +1,62 @@
+# [342. Power of Four](https://leetcode.com/problems/power-of-four/)
+
+
+## 题目
+
+Given an integer (signed 32 bits), write a function to check whether it is a power of 4.
+
+**Example 1**:
+
+    Input: 16
+    Output: true
+
+**Example 2**:
+
+    Input: 5
+    Output: false
+
+**Follow up**: Could you solve it without loops/recursion?
+
+## 题目大意
+
+给定一个整数 (32 位有符号整数)，请编写一个函数来判断它是否是 4 的幂次方。
+
+
+## 解题思路
+
+- 判断一个数是不是 4 的 n 次方。
+- 这一题最简单的思路是循环，可以通过。但是题目要求不循环就要判断，这就需要用到数论的知识了。
+- 证明 `(4^n - 1) % 3 == 0`，(1) `4^n - 1 = (2^n + 1) * (2^n - 1)`(2) 在任何连续的 3  个数中 `(2^n-1)`，`(2^n)`，`(2^n+1)`，一定有一个数是 3 的倍数。`(2^n)` 肯定不是 3 的倍数，那么 `(2^n-1)` 或者 `(2^n+1)` 中一定有一个是 3 的倍数。所以 `4^n-1` 一定是 3 的倍数。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 数论
+func isPowerOfFour(num int) bool {
+	return num > 0 && (num&(num-1)) == 0 && (num-1)%3 == 0
+}
+
+// 解法二 循环
+func isPowerOfFour1(num int) bool {
+	for num >= 4 {
+		if num%4 == 0 {
+			num = num / 4
+		} else {
+			return false
+		}
+	}
+	return num == 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0341.Flatten-Nested-List-Iterator/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0343.Integer-Break/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0343.Integer-Break.md b/website/content/ChapterFour/0300~0399/0343.Integer-Break.md
new file mode 100755
index 000000000..ff492fa2c
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0343.Integer-Break.md
@@ -0,0 +1,60 @@
+# [343. Integer Break](https://leetcode.com/problems/integer-break/)
+
+
+## 题目
+
+Given a positive integer n, break it into the sum of **at least** two positive integers and maximize the product of those integers. Return the maximum product you can get.
+
+**Example 1**:
+
+    Input: 2
+    Output: 1
+    Explanation: 2 = 1 + 1, 1 × 1 = 1.
+
+**Example 2**:
+
+    Input: 10
+    Output: 36
+    Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.
+
+**Note**: You may assume that n is not less than 2 and not larger than 58.
+
+
+## 题目大意
+
+给定一个正整数 n，将其拆分为至少两个正整数的和，并使这些整数的乘积最大化。 返回你可以获得的最大乘积。
+
+
+## 解题思路
+
+- 这一题是 DP 的题目，将一个数字分成多个数字之和，至少分为 2 个数字之和，求解分解出来的数字乘积最大是多少。
+- 这一题的动态转移方程是 `dp[i] = max(dp[i], j * (i - j), j * dp[i-j])` ，一个数分解成 `j` 和 `i - j` 两个数字，或者分解成 `j` 和 `更多的分解数`，`更多的分解数`即是 `dp[i-j]`，由于 `dp[i-j]` 下标小于 `i` ，所以 `dp[i-j]` 在计算 `dp[i]` 的时候一定计算出来了。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func integerBreak(n int) int {
+	dp := make([]int, n+1)
+	dp[0], dp[1] = 1, 1
+	for i := 1; i <= n; i++ {
+		for j := 1; j < i; j++ {
+			// dp[i] = max(dp[i], j * (i - j), j*dp[i-j])
+			dp[i] = max(dp[i], j*max(dp[i-j], i-j))
+		}
+	}
+	return dp[n]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0342.Power-of-Four/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0344.Reverse-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0344.Reverse-String.md b/website/content/ChapterFour/0300~0399/0344.Reverse-String.md
new file mode 100644
index 000000000..e2ed876c5
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0344.Reverse-String.md
@@ -0,0 +1,60 @@
+# [344. Reverse String](https://leetcode.com/problems/reverse-string/)
+
+## 题目
+
+Write a function that reverses a string. The input string is given as an array of characters char[].
+
+Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
+
+You may assume all the characters consist of printable ascii characters.
+
+**Example 1**:
+
+```
+
+Input: ["h","e","l","l","o"]
+Output: ["o","l","l","e","h"]
+
+```
+
+**Example 2**:
+
+```
+
+Input: ["H","a","n","n","a","h"]
+Output: ["h","a","n","n","a","H"]
+
+```
+
+## 题目大意
+
+题目要求我们反转一个字符串。
+
+## 解题思路
+
+这一题的解题思路是用 2 个指针，指针对撞的思路，来不断交换首尾元素，即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func reverseString(s []byte) {
+	for i, j := 0, len(s)-1; i < j; {
+		s[i], s[j] = s[j], s[i]
+		i++
+		j--
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0343.Integer-Break/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0345.Reverse-Vowels-of-a-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0345.Reverse-Vowels-of-a-String.md b/website/content/ChapterFour/0300~0399/0345.Reverse-Vowels-of-a-String.md
new file mode 100644
index 000000000..1c87310a9
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0345.Reverse-Vowels-of-a-String.md
@@ -0,0 +1,71 @@
+# [345. Reverse Vowels of a String](https://leetcode.com/problems/reverse-vowels-of-a-string/)
+
+## 题目
+
+Write a function that takes a string as input and reverse only the vowels of a string.
+
+
+
+**Example 1**:
+
+```
+
+Input: "hello"
+Output: "holle"
+
+```
+
+**Example 2**:
+
+```
+
+Input: "leetcode"
+Output: "leotcede"
+
+```
+
+## 题目大意
+
+题目要求我们反转字符串中的元音字母。需要注意字母大小写。
+
+## 解题思路
+
+这一题的解题思路是用 2 个指针，指针对撞的思路，来不断交换首尾元素，即可。这一题和第 344 题思路一样。
+
+
+
+## 代码
+
+```go
+package leetcode
+
+func reverseVowels(s string) string {
+	b := []byte(s)
+	for i, j := 0, len(b)-1; i < j; {
+		if !isVowel(b[i]) {
+			i++
+			continue
+		}
+		if !isVowel(b[j]) {
+			j--
+			continue
+		}
+		b[i], b[j] = b[j], b[i]
+		i++
+		j--
+	}
+	return string(b)
+}
+
+func isVowel(s byte) bool {
+	return s == 'a' || s == 'e' || s == 'i' || s == 'o' || s == 'u' || s == 'A' ||
+		s == 'E' || s == 'I' || s == 'O' || s == 'U'
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0344.Reverse-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0347.Top-K-Frequent-Elements/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0347.Top-K-Frequent-Elements.md b/website/content/ChapterFour/0300~0399/0347.Top-K-Frequent-Elements.md
new file mode 100644
index 000000000..fc4474815
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0347.Top-K-Frequent-Elements.md
@@ -0,0 +1,110 @@
+# [347. Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/)
+
+## 题目
+
+Given a non-empty array of integers, return the k most frequent elements.
+
+**Example 1**:
+
+```
+
+Input: nums = [1,1,1,2,2,3], k = 2
+Output: [1,2]
+
+```
+
+**Example 2**:
+
+```
+
+Input: nums = [1], k = 1
+Output: [1]
+
+```
+
+**Note**:  
+
+- You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
+- Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
+ 
+
+## 题目大意
+
+给一个非空的数组，输出前 K 个频率最高的元素。
+
+## 解题思路
+
+这一题是考察优先队列的题目。把数组构造成一个优先队列，输出前 K 个即可。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "container/heap"
+
+func topKFrequent(nums []int, k int) []int {
+	m := make(map[int]int)
+	for _, n := range nums {
+		m[n]++
+	}
+	q := PriorityQueue{}
+	for key, count := range m {
+		heap.Push(&q, &Item{key: key, count: count})
+	}
+	var result []int
+	for len(result) < k {
+		item := heap.Pop(&q).(*Item)
+		result = append(result, item.key)
+	}
+	return result
+}
+
+// Item define
+type Item struct {
+	key   int
+	count int
+}
+
+// A PriorityQueue implements heap.Interface and holds Items.
+type PriorityQueue []*Item
+
+func (pq PriorityQueue) Len() int {
+	return len(pq)
+}
+
+func (pq PriorityQueue) Less(i, j int) bool {
+	// 注意：因为 golang 中的 heap 默认是按最小堆组织的，所以 count 越大，Less() 越小，越靠近堆顶。这里采用 >，变为最大堆
+	return pq[i].count > pq[j].count
+}
+
+func (pq PriorityQueue) Swap(i, j int) {
+	pq[i], pq[j] = pq[j], pq[i]
+}
+
+// Push define
+func (pq *PriorityQueue) Push(x interface{}) {
+	item := x.(*Item)
+	*pq = append(*pq, item)
+}
+
+// Pop define
+func (pq *PriorityQueue) Pop() interface{} {
+	n := len(*pq)
+	item := (*pq)[n-1]
+	*pq = (*pq)[:n-1]
+	return item
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0345.Reverse-Vowels-of-a-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0349.Intersection-of-Two-Arrays/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0349.Intersection-of-Two-Arrays.md b/website/content/ChapterFour/0300~0399/0349.Intersection-of-Two-Arrays.md
new file mode 100644
index 000000000..b57951ab3
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0349.Intersection-of-Two-Arrays.md
@@ -0,0 +1,67 @@
+# [349. Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays/)
+
+## 题目
+
+Given two arrays, write a function to compute their intersection.
+
+
+**Example 1**:
+
+```
+
+Input: nums1 = [1,2,2,1], nums2 = [2,2]
+Output: [2]
+
+```
+
+**Example 2**:
+
+```
+
+Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
+Output: [9,4]
+
+```
+
+**Note**:
+
+- Each element in the result must be unique.
+- The result can be in any order.
+
+## 题目大意
+
+找到两个数组的交集元素，如果交集元素同一个数字出现了多次，只输出一次。
+
+## 解题思路
+
+把数组一的每个数字都存进字典中，然后在数组二中依次判断字典中是否存在，如果存在，在字典中删除它(因为输出要求只输出一次)。
+
+## 代码
+
+```go
+
+package leetcode
+
+func intersection(nums1 []int, nums2 []int) []int {
+	m := map[int]bool{}
+	var res []int
+	for _, n := range nums1 {
+		m[n] = true
+	}
+	for _, n := range nums2 {
+		if m[n] {
+			delete(m, n)
+			res = append(res, n)
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0347.Top-K-Frequent-Elements/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0350.Intersection-of-Two-Arrays-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0350.Intersection-of-Two-Arrays-II.md b/website/content/ChapterFour/0300~0399/0350.Intersection-of-Two-Arrays-II.md
new file mode 100644
index 000000000..b3b3c30b0
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0350.Intersection-of-Two-Arrays-II.md
@@ -0,0 +1,79 @@
+# [350. Intersection of Two Arrays II](https://leetcode.com/problems/intersection-of-two-arrays-ii/)
+
+## 题目
+
+Given two arrays, write a function to compute their intersection.
+
+
+
+**Example 1**:
+
+```
+
+Input: nums1 = [1,2,2,1], nums2 = [2,2]
+Output: [2,2]
+
+```
+
+**Example 2**:
+
+```
+
+Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
+Output: [4,9]
+
+```
+
+**Note**:
+
+- Each element in the result should appear as many times as it shows in both arrays.
+- The result can be in any order.
+
+
+**Follow up**:
+
+- What if the given array is already sorted? How would you optimize your algorithm?
+- What if nums1's size is small compared to nums2's size? Which algorithm is better?
+- What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?
+
+## 题目大意
+
+这题是第 349 题的加强版。要求输出 2 个数组的交集元素，如果元素出现多次，要输出多次。
+
+## 解题思路
+
+这一题还是延续第 349 题的思路。把数组一中的数字都放进字典中，另外字典的 key 是数组中的数字，value 是这个数字出现的次数。在扫描数组二的时候，每取出一个存在的数组，把字典中的 value 减一。如果 value 是 0 代表不存在这个数字。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func intersect(nums1 []int, nums2 []int) []int {
+	m := map[int]int{}
+	var res []int
+	for _, n := range nums1 {
+		m[n]++
+	}
+	for _, n := range nums2 {
+		if m[n] > 0 {
+			res = append(res, n)
+			m[n]--
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0349.Intersection-of-Two-Arrays/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0352.Data-Stream-as-Disjoint-Intervals/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0352.Data-Stream-as-Disjoint-Intervals.md b/website/content/ChapterFour/0300~0399/0352.Data-Stream-as-Disjoint-Intervals.md
new file mode 100644
index 000000000..14899590d
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0352.Data-Stream-as-Disjoint-Intervals.md
@@ -0,0 +1,115 @@
+# [352. Data Stream as Disjoint Intervals](https://leetcode.com/problems/data-stream-as-disjoint-intervals/)
+
+
+## 题目
+
+Given a data stream input of non-negative integers a1, a2, ..., an, summarize the numbers seen so far as a list of disjoint intervals.
+
+Implement the SummaryRanges class:
+
+   - SummaryRanges() Initializes the object with an empty stream.
+   - void addNum(int val) Adds the integer val to the stream. 
+   - int[][] getIntervals() Returns a summary of the integers in the stream currently as a list of disjoint intervals [starti, endi].
+
+**Example 1:**
+
+    Input
+    ["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"]
+    [[], [1], [], [3], [], [7], [], [2], [], [6], []]
+    Output
+    [null, null, [[1, 1]], null, [[1, 1], [3, 3]], null, [[1, 1], [3, 3], [7, 7]], null, [[1, 3], [7, 7]], null, [[1, 3], [6, 7]]]
+
+    Explanation
+    SummaryRanges summaryRanges = new SummaryRanges();
+    summaryRanges.addNum(1);      // arr = [1]
+    summaryRanges.getIntervals(); // return [[1, 1]]
+    summaryRanges.addNum(3);      // arr = [1, 3]
+    summaryRanges.getIntervals(); // return [[1, 1], [3, 3]]
+    summaryRanges.addNum(7);      // arr = [1, 3, 7]
+    summaryRanges.getIntervals(); // return [[1, 1], [3, 3], [7, 7]]
+    summaryRanges.addNum(2);      // arr = [1, 2, 3, 7]
+    summaryRanges.getIntervals(); // return [[1, 3], [7, 7]]
+    summaryRanges.addNum(6);      // arr = [1, 2, 3, 6, 7]
+    summaryRanges.getIntervals(); // return [[1, 3], [6, 7]]
+
+**Constraints**
+
+   - 0 <= val <= 10000
+   - At most 3 * 10000 calls will be made to addNum and getIntervals.
+
+## 题目大意
+
+给你一个由非负整数a1, a2, ..., an 组成的数据流输入，请你将到目前为止看到的数字总结为不相交的区间列表。
+
+实现 SummaryRanges 类：
+
+   - SummaryRanges() 使用一个空数据流初始化对象。
+   - void addNum(int val) 向数据流中加入整数 val 。
+   - int[][] getIntervals() 以不相交区间[starti, endi] 的列表形式返回对数据流中整数的总结
+
+## 解题思路
+
+- 使用字典过滤掉重复的数字
+- 把过滤后的数字放到nums中,并进行排序 
+- 使用nums构建不重复的区间
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+type SummaryRanges struct {
+	nums []int
+	mp map[int]int
+}
+
+func Constructor() SummaryRanges {
+	return SummaryRanges{
+		nums: []int{},
+		mp : map[int]int{},
+	}
+}
+
+func (this *SummaryRanges) AddNum(val int) {
+	if _, ok := this.mp[val]; !ok {
+		this.mp[val] = 1
+		this.nums = append(this.nums, val)
+	}
+	sort.Ints(this.nums)
+}
+
+func (this *SummaryRanges) GetIntervals() [][]int {
+	n := len(this.nums)
+	var ans [][]int
+	if n == 0 {
+		return ans
+	}
+	if n == 1 {
+		ans = append(ans, []int{this.nums[0], this.nums[0]})
+		return ans
+	}
+	start, end := this.nums[0], this.nums[0]
+	ans = append(ans, []int{start, end})
+	index := 0
+	for i := 1; i < n; i++ {
+		if this.nums[i] == end + 1 {
+			end = this.nums[i]
+			ans[index][1] = end
+		} else {
+			start, end = this.nums[i], this.nums[i]
+			ans = append(ans, []int{start, end})
+			index++
+		}
+	}
+	return ans
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0350.Intersection-of-Two-Arrays-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0354.Russian-Doll-Envelopes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0354.Russian-Doll-Envelopes.md b/website/content/ChapterFour/0300~0399/0354.Russian-Doll-Envelopes.md
new file mode 100755
index 000000000..af4abd4a2
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0354.Russian-Doll-Envelopes.md
@@ -0,0 +1,89 @@
+# [354. Russian Doll Envelopes](https://leetcode.com/problems/russian-doll-envelopes/)
+
+
+## 题目
+
+You have a number of envelopes with widths and heights given as a pair of integers `(w, h)`. One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.
+
+What is the maximum number of envelopes can you Russian doll? (put one inside other)
+
+**Note**: Rotation is not allowed.
+
+**Example**:
+
+    Input: [[5,4],[6,4],[6,7],[2,3]]
+    Output: 3
+    Explanation: The maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).
+
+
+## 题目大意
+
+给定一些标记了宽度和高度的信封，宽度和高度以整数对形式 (w, h) 出现。当另一个信封的宽度和高度都比这个信封大的时候，这个信封就可以放进另一个信封里，如同俄罗斯套娃一样。
+
+请计算最多能有多少个信封能组成一组“俄罗斯套娃”信封（即可以把一个信封放到另一个信封里面）。
+
+说明:
+- 不允许旋转信封。
+
+## 解题思路
+
+- 给出一组信封的宽度和高度，如果组成俄罗斯套娃，问最多能套几层。只有当一个信封的宽度和高度都比另外一个信封大的时候，才能套在小信封上面。
+- 这一题的实质是第 300 题 Longest Increasing Subsequence 的加强版。能组成俄罗斯套娃的条件就是能找到一个最长上升子序列。但是这题的条件是二维的，要求能找到在二维上都能满足条件的最长上升子序列。先降维，把宽度排序。然后在高度上寻找最长上升子序列。这里用到的方法和第 300 题的方法一致。解题思路详解见第 300 题。
+- 此题是二维的 LIS 问题。一维的 LIS 问题是第 300 题。三维的 LIS 问题是第 1691 题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+type sortEnvelopes [][]int
+
+func (s sortEnvelopes) Len() int {
+	return len(s)
+}
+func (s sortEnvelopes) Less(i, j int) bool {
+	if s[i][0] == s[j][0] {
+		return s[i][1] > s[j][1]
+	}
+	return s[i][0] < s[j][0]
+}
+func (s sortEnvelopes) Swap(i, j int) {
+	s[i], s[j] = s[j], s[i]
+}
+
+func maxEnvelopes(envelopes [][]int) int {
+	sort.Sort(sortEnvelopes(envelopes))
+	dp := []int{}
+	for _, e := range envelopes {
+		low, high := 0, len(dp)
+		for low < high {
+			mid := low + (high-low)>>1
+			if dp[mid] >= e[1] {
+				high = mid
+			} else {
+				low = mid + 1
+			}
+		}
+		if low == len(dp) {
+			dp = append(dp, e[1])
+		} else {
+			dp[low] = e[1]
+		}
+	}
+	return len(dp)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0352.Data-Stream-as-Disjoint-Intervals/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0357.Count-Numbers-with-Unique-Digits/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0357.Count-Numbers-with-Unique-Digits.md b/website/content/ChapterFour/0300~0399/0357.Count-Numbers-with-Unique-Digits.md
new file mode 100755
index 000000000..4ae4f41b1
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0357.Count-Numbers-with-Unique-Digits.md
@@ -0,0 +1,68 @@
+# [357. Count Numbers with Unique Digits](https://leetcode.com/problems/count-numbers-with-unique-digits/)
+
+
+## 题目
+
+Given a **non-negative** integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.
+
+**Example**:
+
+    Input: 2
+    Output: 91 
+    Explanation: The answer should be the total numbers in the range of 0 ≤ x < 100, 
+                 excluding 11,22,33,44,55,66,77,88,99
+
+
+## 题目大意
+
+给定一个非负整数 n，计算各位数字都不同的数字 x 的个数，其中 0 ≤ x < 10^n 。
+
+
+
+
+## 解题思路
+
+- 输出 n 位数中不出现重复数字的数字的个数
+- 这道题摸清楚规律以后，可以直接写出最终所有答案，答案只有 11 个。
+- 考虑不重复数字是如生成的。如果只是一位数，不存在重复的数字，结果是 10 。如果是二位数，第一位一定不能取 0，那么第一位有 1-9，9种取法，第二位为了和第一位不重复，只能有 0-9，10种取法中减去第一位取的数字，那么也是 9 种取法。以此类推，如果是三位数，第三位是 8 种取法；四位数，第四位是 7 种取法；五位数，第五位是 6 种取法；六位数，第六位是 5 种取法；七位数，第七位是 4 种取法；八位数，第八位是 3 种取法；九位数，第九位是 2 种取法；十位数，第十位是 1 种取法；十一位数，第十一位是 0 种取法；十二位数，第十二位是 0 种取法；那么第 11 位数以后，每个数都是重复数字的数字。知道这个规律以后，可以累积上面的结果，把结果直接存在数组里面，暴力打表即可。O(1) 的时间复杂度。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 暴力打表法
+func countNumbersWithUniqueDigits1(n int) int {
+	res := []int{1, 10, 91, 739, 5275, 32491, 168571, 712891, 2345851, 5611771, 8877691}
+	if n >= 10 {
+		return res[10]
+	}
+	return res[n]
+}
+
+// 打表方法
+func countNumbersWithUniqueDigits(n int) int {
+	if n == 0 {
+		return 1
+	}
+	res, uniqueDigits, availableNumber := 10, 9, 9
+	for n > 1 && availableNumber > 0 {
+		uniqueDigits = uniqueDigits * availableNumber
+		res += uniqueDigits
+		availableNumber--
+		n--
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0354.Russian-Doll-Envelopes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0367.Valid-Perfect-Square/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0367.Valid-Perfect-Square.md b/website/content/ChapterFour/0300~0399/0367.Valid-Perfect-Square.md
new file mode 100755
index 000000000..d46d7c8bd
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0367.Valid-Perfect-Square.md
@@ -0,0 +1,64 @@
+# [367. Valid Perfect Square](https://leetcode.com/problems/valid-perfect-square/)
+
+## 题目
+
+Given a positive integer num, write a function which returns True if num is a perfect square else False.
+
+**Note**: **Do not** use any built-in library function such as `sqrt`.
+
+**Example 1**:
+
+    Input: 16
+    Output: true
+
+**Example 2**:
+
+    Input: 14
+    Output: false
+
+
+## 题目大意
+
+给定一个正整数 num，编写一个函数，如果 num 是一个完全平方数，则返回 True，否则返回 False。
+
+说明：不要使用任何内置的库函数，如 sqrt。
+
+
+
+
+## 解题思路
+
+
+- 给出一个数，要求判断这个数是不是完全平方数。
+- 可以用二分搜索来解答这道题。判断完全平方数，根据它的定义来，是否能被开根号，即找到一个数的平方是否可以等于待判断的数字。从 [1, n] 区间内进行二分，若能找到则返回 true，找不到就返回 false 。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func isPerfectSquare(num int) bool {
+	low, high := 1, num
+	for low <= high {
+		mid := low + (high-low)>>1
+		if mid*mid == num {
+			return true
+		} else if mid*mid < num {
+			low = mid + 1
+		} else {
+			high = mid - 1
+		}
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0357.Count-Numbers-with-Unique-Digits/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0368.Largest-Divisible-Subset/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0368.Largest-Divisible-Subset.md b/website/content/ChapterFour/0300~0399/0368.Largest-Divisible-Subset.md
new file mode 100644
index 000000000..b61958e13
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0368.Largest-Divisible-Subset.md
@@ -0,0 +1,93 @@
+# [368. Largest Divisible Subset](https://leetcode.com/problems/largest-divisible-subset/)
+
+
+## 题目
+
+Given a set of **distinct** positive integers `nums`, return the largest subset `answer` such that every pair `(answer[i], answer[j])` of elements in this subset satisfies:
+
+- `answer[i] % answer[j] == 0`, or
+- `answer[j] % answer[i] == 0`
+
+If there are multiple solutions, return any of them.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3]
+Output: [1,2]
+Explanation: [1,3] is also accepted.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,4,8]
+Output: [1,2,4,8]
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 1000`
+- `1 <= nums[i] <= 2 * 109`
+- All the integers in `nums` are **unique**.
+
+## 题目大意
+
+给你一个由 无重复 正整数组成的集合 nums ，请你找出并返回其中最大的整除子集 answer ，子集中每一元素对 (answer[i], answer[j]) 都应当满足：
+
+- answer[i] % answer[j] == 0 ，或
+- answer[j] % answer[i] == 0
+
+如果存在多个有效解子集，返回其中任何一个均可。
+
+## 解题思路
+
+- 根据题目数据规模 1000，可以估计此题大概率是动态规划，并且时间复杂度是 O(n^2)。先将集合排序，以某一个小的数作为基准，不断的选择能整除的数加入集合。按照这个思路考虑，此题和第 300 题经典的 LIS 解题思路一致。只不过 LIS 每次选择更大的数，此题除了选择更大的数，只不过多了一个判断，这个更大的数能否整除当前集合里面的所有元素。按照此法一定可以找出最大的集合。
+- 剩下的问题是如何输出最大集合。这道题的集合具有重叠子集的性质，例如 [2,4,8,16] 这个集合，长度是 4，它一定包含长度为 3 的子集，因为从它里面随便取 3 个数形成的子集也满足元素相互能整除的条件。同理，它也一定包含长度为 2，长度为 1 的子集。由于有这个性质，可以利用 dp 数组里面的数据，输出最大集合。例如，[2,4,6,8,9,13,16,40]，由动态规划可以找到最大集合是 [2,4,8,16]。长度为 4 的找到了，再找长度为 3 的，[2,4,8]，[2,4,40]。在最大集合中，最大元素是 16，所以 [2,4,40] 这个集合排除，它的最大元素大于 16。选定 [2,4,8] 这个集合，此时最大元素是 8 。以此类推，筛选到最后，便可以输出 [16,8,4,2] 这个组最大集合的答案了。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func largestDivisibleSubset(nums []int) []int {
+	sort.Ints(nums)
+	dp, res := make([]int, len(nums)), []int{}
+	for i := range dp {
+		dp[i] = 1
+	}
+	maxSize, maxVal := 1, 1
+	for i := 1; i < len(nums); i++ {
+		for j, v := range nums[:i] {
+			if nums[i]%v == 0 && dp[j]+1 > dp[i] {
+				dp[i] = dp[j] + 1
+			}
+		}
+		if dp[i] > maxSize {
+			maxSize, maxVal = dp[i], nums[i]
+		}
+	}
+	if maxSize == 1 {
+		return []int{nums[0]}
+	}
+	for i := len(nums) - 1; i >= 0 && maxSize > 0; i-- {
+		if dp[i] == maxSize && maxVal%nums[i] == 0 {
+			res = append(res, nums[i])
+			maxVal = nums[i]
+			maxSize--
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0367.Valid-Perfect-Square/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0371.Sum-of-Two-Integers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0371.Sum-of-Two-Integers.md b/website/content/ChapterFour/0300~0399/0371.Sum-of-Two-Integers.md
new file mode 100755
index 000000000..52a2bb04f
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0371.Sum-of-Two-Integers.md
@@ -0,0 +1,54 @@
+# [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)
+
+
+## 题目
+
+Calculate the sum of two integers a and b, but you are **not allowed** to use the operator `+` and `-`.
+
+**Example 1**:
+
+    Input: a = 1, b = 2
+    Output: 3
+
+**Example 2**:
+
+    Input: a = -2, b = 3
+    Output: 1
+
+
+## 题目大意
+
+不使用运算符 + 和 - ​​​​​​​，计算两整数 ​​​​​​​a 、b ​​​​​​​之和。
+
+## 解题思路
+
+- 要求不用加法和减法运算符计算 `a+b`。这一题需要用到 `^` 和 `&` 运算符的性质，两个数 ^ 可以实现两个数不带进位的二进制加法。这里需要实现加法，肯定需要进位。所以如何找到进位是本题的关键。
+- 在二进制中，只有 1 和 1 加在一起才会进位，0 和 0，0 和 1，1 和 0，这三种情况都不会进位，规律就是 `a & b` 为 0 的时候就不用进位，为 1 的时候代表需要进位。进位是往前进一位，所以还需要左移操作，所以加上的进位为 `(a&b)<<1`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func getSum(a int, b int) int {
+	if a == 0 {
+		return b
+	}
+	if b == 0 {
+		return a
+	}
+	// (a & b)<<1 计算的是进位
+	// a ^ b 计算的是不带进位的加法
+	return getSum((a&b)<<1, a^b)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0368.Largest-Divisible-Subset/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0372.Super-Pow/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0372.Super-Pow.md b/website/content/ChapterFour/0300~0399/0372.Super-Pow.md
new file mode 100755
index 000000000..2e3c2fddc
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0372.Super-Pow.md
@@ -0,0 +1,113 @@
+# [372. Super Pow](https://leetcode.com/problems/super-pow/)
+
+
+## 题目
+
+Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.
+
+**Example 1**:
+
+    Input: a = 2, b = [3]
+    Output: 8
+
+**Example 2**:
+
+    Input: a = 2, b = [1,0]
+    Output: 1024
+
+
+## 题目大意
+
+
+你的任务是计算 a^b 对 1337 取模，a 是一个正整数，b 是一个非常大的正整数且会以数组形式给出。
+
+## 解题思路
+
+- 求 a^b mod p 的结果，b 是大数。
+- 这一题可以用暴力解法尝试。需要用到 mod 计算的几个运算性质：
+
+        模运算性质一：(a + b) % p = (a % p + b % p) % p
+        模运算性质二：(a - b) % p = (a % p - b % p + p) % p
+        模运算性质三：(a * b) % p = (a % p * b % p) % p
+        模运算性质四：a ^ b % p = ((a % p)^b) % p
+
+    这一题需要用到性质三、四。举个例子：
+
+        12345^678 % 1337 = (12345^670 * 12345^8) % 1337
+        				 = ((12345^670 % 1337) * (12345^8 % 1337)) % 1337  ---> 利用性质 三
+        				 = (((12345^67)^10 % 1337) * (12345^8 % 1337)) % 1337  ---> 乘方性质
+                         = ((12345^67 % 1337)^10) % 1337 * (12345^8 % 1337)) % 1337  ---> 利用性质 四
+        				 = (((12345^67 % 1337)^10) * (12345^8 % 1337)) % 1337  ---> 反向利用性质 三
+
+    经过上面这样的变换，把指数 678 的个位分离出来了，可以单独求解。继续经过上面的变换，可以把指数的 6 和 7 也分离出来。最终可以把大数 b 一位一位的分离出来。至于计算 a^b 就结果快速幂求解。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 快速幂 res = res^10 * qpow(a, b[i])
+// 模运算性质一：(a + b) % p = (a % p + b % p) % p
+// 模运算性质二：(a - b) % p = (a % p - b % p + p) % p
+// 模运算性质三：(a * b) % p = (a % p * b % p) % p
+// 模运算性质四：a ^ b % p = ((a % p)^b) % p
+// 举个例子
+// 12345^678 % 1337 = (12345^670 * 12345^8) % 1337
+//                  = ((12345^670 % 1337) * (12345^8 % 1337)) % 1337  ---> 利用性质 三
+//                  = (((12345^67)^10 % 1337) * (12345^8 % 1337)) % 1337  ---> 乘方性质
+//                  = ((12345^67 % 1337)^10) % 1337 * (12345^8 % 1337)) % 1337  ---> 利用性质 四
+//                  = (((12345^67 % 1337)^10) * (12345^8 % 1337)) % 1337  ---> 反向利用性质 三
+func superPow(a int, b []int) int {
+	res := 1
+	for i := 0; i < len(b); i++ {
+		res = (qpow(res, 10) * qpow(a, b[i])) % 1337
+	}
+	return res
+}
+
+// 快速幂计算 x^n
+func qpow(x, n int) int {
+	res := 1
+	x %= 1337
+	for n > 0 {
+		if (n & 1) == 1 {
+			res = (res * x) % 1337
+		}
+		x = (x * x) % 1337
+		n >>= 1
+	}
+	return res
+}
+
+// 解法二 暴力解法
+// 利用上面的性质，可以得到：a^1234567 % 1337 = (a^1234560 % 1337) * (a^7 % 1337) % k = ((((a^123456) % 1337)^10)% 1337 * (a^7 % 1337))% 1337;
+func superPow1(a int, b []int) int {
+	if len(b) == 0 {
+		return 1
+	}
+	last := b[len(b)-1]
+	l := 1
+	// 先计算个位的 a^x 结果，对应上面例子中的 (a^7 % 1337)% 1337
+	for i := 1; i <= last; i++ {
+		l = l * a % 1337
+	}
+	// 再计算除去个位以外的 a^y 的结果，对应上面例子中的 (a^123456) % 1337)
+	temp := superPow1(a, b[:len(b)-1])
+	f := 1
+	// 对应上面例子中的 (((a^123456) % 1337)^10)% 1337
+	for i := 1; i <= 10; i++ {
+		f = f * temp % 1337
+	}
+	return f * l % 1337
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0371.Sum-of-Two-Integers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0373.Find-K-Pairs-with-Smallest-Sums/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0373.Find-K-Pairs-with-Smallest-Sums.md b/website/content/ChapterFour/0300~0399/0373.Find-K-Pairs-with-Smallest-Sums.md
new file mode 100755
index 000000000..ff62bbf4f
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0373.Find-K-Pairs-with-Smallest-Sums.md
@@ -0,0 +1,133 @@
+# [373. Find K Pairs with Smallest Sums](https://leetcode.com/problems/find-k-pairs-with-smallest-sums/)
+
+
+## 题目
+
+You are given two integer arrays **nums1** and **nums2** sorted in ascending order and an integer **k**.
+
+Define a pair **(u,v)** which consists of one element from the first array and one element from the second array.
+
+Find the k pairs **(u1,v1),(u2,v2) ...(uk,vk)** with the smallest sums.
+
+**Example 1**:
+
+    Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
+    Output: [[1,2],[1,4],[1,6]] 
+    Explanation: The first 3 pairs are returned from the sequence: 
+                 [1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]
+
+**Example 2**:
+
+    Input: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
+    Output: [1,1],[1,1]
+    Explanation: The first 2 pairs are returned from the sequence: 
+                 [1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]
+
+**Example 3**:
+
+    Input: nums1 = [1,2], nums2 = [3], k = 3
+    Output: [1,3],[2,3]
+    Explanation: All possible pairs are returned from the sequence: [1,3],[2,3]
+
+
+## 题目大意
+
+
+给定两个以升序排列的整形数组 nums1 和 nums2, 以及一个整数 k。
+
+定义一对值 (u,v)，其中第一个元素来自 nums1，第二个元素来自 nums2。
+
+找到和最小的 k 对数字 (u1,v1), (u2,v2) ... (uk,vk)。
+
+
+
+## 解题思路
+
+
+- 给出 2 个数组，和数字 k，要求找到 k 个数值对，数值对两个数的和最小。
+- 这一题咋一看可以用二分搜索，两个数组两个组合有 `m * n` 个数值对。然后找到最小的和，最大的和，在这个范围内进行二分搜索，每分出一个 `mid`，再去找比 `mid` 小的数值对有多少个，如果个数小于 `k` 个，那么在右区间上继续二分，如果个数大于 `k` 个，那么在左区间上继续二分。到目前为止，这个思路看似可行。但是每次搜索的数值对是无序的。这会导致最终出现错误的结果。例如 `mid = 10` 的时候，小于 10 的和有 22 个，而 `k = 25` 。这说明 `mid` 偏小，`mid` 增大，`mid = 11` 的时候，小于 11 的和有 30 个，而 `k = 25` 。这时候应该从这 30 个和中取前 25 个。但是我们遍历数值对的时候，和并不是从小到大排序的。这时候还需要额外对这 30 个候选值进行排序。这样时间复杂度又增大了。
+- 可以先用暴力解法解答。将所有的和都遍历出来，排序以后，取前 k 个。这个暴力方法可以 AC。
+- 本题最优解应该是优先队列。维护一个最小堆。把数值对的和放在这个最小堆中，不断的 pop 出 k 个最小值到数组中，即为答案。
+- 在已排序的矩阵中寻找最 K 小的元素这一系列的题目有：第 373 题，第 378 题，第 668 题，第 719 题，第 786 题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+// 解法一 优先队列
+func kSmallestPairs(nums1 []int, nums2 []int, k int) [][]int {
+	result, h := [][]int{}, &minHeap{}
+	if len(nums1) == 0 || len(nums2) == 0 || k == 0 {
+		return result
+	}
+	if len(nums1)*len(nums2) < k {
+		k = len(nums1) * len(nums2)
+	}
+	heap.Init(h)
+	for _, num := range nums1 {
+		heap.Push(h, []int{num, nums2[0], 0})
+	}
+	for len(result) < k {
+		min := heap.Pop(h).([]int)
+		result = append(result, min[:2])
+		if min[2] < len(nums2)-1 {
+			heap.Push(h, []int{min[0], nums2[min[2]+1], min[2] + 1})
+		}
+	}
+	return result
+}
+
+type minHeap [][]int
+
+func (h minHeap) Len() int           { return len(h) }
+func (h minHeap) Less(i, j int) bool { return h[i][0]+h[i][1] < h[j][0]+h[j][1] }
+func (h minHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
+
+func (h *minHeap) Push(x interface{}) {
+	*h = append(*h, x.([]int))
+}
+
+func (h *minHeap) Pop() interface{} {
+	old := *h
+	n := len(old)
+	x := old[n-1]
+	*h = old[0 : n-1]
+	return x
+}
+
+// 解法二 暴力解法
+func kSmallestPairs1(nums1 []int, nums2 []int, k int) [][]int {
+	size1, size2, res := len(nums1), len(nums2), [][]int{}
+	if size1 == 0 || size2 == 0 || k < 0 {
+		return nil
+	}
+	for i := 0; i < size1; i++ {
+		for j := 0; j < size2; j++ {
+			res = append(res, []int{nums1[i], nums2[j]})
+		}
+	}
+	sort.Slice(res, func(i, j int) bool {
+		return res[i][0]+res[i][1] < res[j][0]+res[j][1]
+	})
+	if len(res) >= k {
+		return res[:k]
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0372.Super-Pow/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0374.Guess-Number-Higher-or-Lower/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0374.Guess-Number-Higher-or-Lower.md b/website/content/ChapterFour/0300~0399/0374.Guess-Number-Higher-or-Lower.md
new file mode 100644
index 000000000..a904fadd8
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0374.Guess-Number-Higher-or-Lower.md
@@ -0,0 +1,101 @@
+# [374. Guess Number Higher or Lower](https://leetcode.com/problems/guess-number-higher-or-lower/)
+
+## 题目
+
+We are playing the Guess Game. The game is as follows:
+
+I pick a number from `1` to `n`. You have to guess which number I picked.
+
+Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
+
+You call a pre-defined API `int guess(int num)`, which returns 3 possible results:
+
+- `1`: The number I picked is lower than your guess (i.e. `pick < num`).
+- `1`: The number I picked is higher than your guess (i.e. `pick > num`).
+- `0`: The number I picked is equal to your guess (i.e. `pick == num`).
+
+Return *the number that I picked*.
+
+**Example 1:**
+
+```
+Input: n = 10, pick = 6
+Output: 6
+```
+
+**Example 2:**
+
+```
+Input: n = 1, pick = 1
+Output: 1
+```
+
+**Example 3:**
+
+```
+Input: n = 2, pick = 1
+Output: 1
+```
+
+**Example 4:**
+
+```
+Input: n = 2, pick = 2
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= n <= 231 - 1`
+- `1 <= pick <= n`
+
+## 题目大意
+
+猜数字游戏的规则如下：
+
+- 每轮游戏，我都会从 1 到 n 随机选择一个数字。 请你猜选出的是哪个数字。
+- 如果你猜错了，我会告诉你，你猜测的数字比我选出的数字是大了还是小了。
+
+你可以通过调用一个预先定义好的接口 int guess(int num) 来获取猜测结果，返回值一共有 3 种可能的情况（-1，1 或 0）：
+
+- 1：我选出的数字比你猜的数字小 pick < num
+- 1：我选出的数字比你猜的数字大 pick > num
+- 0：我选出的数字和你猜的数字一样。恭喜！你猜对了！pick == num
+
+返回我选出的数字。
+
+## 解题思路
+
+- 这一题是简单题，和小时候玩的猜大猜小的游戏一样。思路很简单，二分查找即可。这一题和第 278 题类似。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+/**
+ * Forward declaration of guess API.
+ * @param  num   your guess
+ * @return 	     -1 if num is lower than the guess number
+ *			      1 if num is higher than the guess number
+ *               otherwise return 0
+ * func guess(num int) int;
+ */
+
+func guessNumber(n int) int {
+	return sort.Search(n, func(x int) bool { return guess(x) <= 0 })
+}
+
+func guess(num int) int {
+	return 0
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0373.Find-K-Pairs-with-Smallest-Sums/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0376.Wiggle-Subsequence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0376.Wiggle-Subsequence.md b/website/content/ChapterFour/0300~0399/0376.Wiggle-Subsequence.md
new file mode 100644
index 000000000..bc9b7c693
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0376.Wiggle-Subsequence.md
@@ -0,0 +1,83 @@
+# [376. Wiggle Subsequence](https://leetcode.com/problems/wiggle-subsequence/)
+
+
+## 题目
+
+Given an integer array `nums`, return *the length of the longest **wiggle sequence***.
+
+A **wiggle sequence** is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two elements is trivially a wiggle sequence.
+
+- For example, `[1, 7, 4, 9, 2, 5]` is a **wiggle sequence** because the differences `(6, -3, 5, -7, 3)` are alternately positive and negative.
+- In contrast, `[1, 4, 7, 2, 5]` and `[1, 7, 4, 5, 5]` are not wiggle sequences, the first because its first two differences are positive and the second because its last difference is zero.
+
+A **subsequence** is obtained by deleting some elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.
+
+**Example 1:**
+
+```
+Input: nums = [1,7,4,9,2,5]
+Output: 6
+Explanation: The entire sequence is a wiggle sequence.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,17,5,10,13,15,10,5,16,8]
+Output: 7
+Explanation: There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,2,3,4,5,6,7,8,9]
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 1000`
+- `0 <= nums[i] <= 1000`
+
+**Follow up:** Could you solve this in `O(n)` time?
+
+## 题目大意
+
+如果连续数字之间的差严格地在正数和负数之间交替，则数字序列称为摆动序列。第一个差（如果存在的话）可能是正数或负数。少于两个元素的序列也是摆动序列。例如， [1,7,4,9,2,5] 是一个摆动序列，因为差值 (6,-3,5,-7,3) 是正负交替出现的。相反, [1,4,7,2,5] 和 [1,7,4,5,5] 不是摆动序列，第一个序列是因为它的前两个差值都是正数，第二个序列是因为它的最后一个差值为零。给定一个整数序列，返回作为摆动序列的最长子序列的长度。 通过从原始序列中删除一些（也可以不删除）元素来获得子序列，剩下的元素保持其原始顺序。
+
+## 解题思路
+
+- 题目要求找到摆动序列最长的子序列。本题可以用贪心的思路，记录当前序列的上升和下降的趋势。扫描数组过程中，每扫描一个元素都判断是“峰”还是“谷”，根据前一个是“峰”还是“谷”做出对应的决定。利用贪心的思想找到最长的摆动子序列。
+
+## 代码
+
+```go
+package leetcode
+
+func wiggleMaxLength(nums []int) int {
+	if len(nums) < 2 {
+		return len(nums)
+	}
+	res := 1
+	prevDiff := nums[1] - nums[0]
+	if prevDiff != 0 {
+		res = 2
+	}
+	for i := 2; i < len(nums); i++ {
+		diff := nums[i] - nums[i-1]
+		if diff > 0 && prevDiff <= 0 || diff < 0 && prevDiff >= 0 {
+			res++
+			prevDiff = diff
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0374.Guess-Number-Higher-or-Lower/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0377.Combination-Sum-IV/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0377.Combination-Sum-IV.md b/website/content/ChapterFour/0300~0399/0377.Combination-Sum-IV.md
new file mode 100644
index 000000000..5bf89f8b1
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0377.Combination-Sum-IV.md
@@ -0,0 +1,107 @@
+# [377. Combination Sum IV](https://leetcode.com/problems/combination-sum-iv/)
+
+
+## 题目
+
+Given an array of **distinct** integers `nums` and a target integer `target`, return *the number of possible combinations that add up to* `target`.
+
+The answer is **guaranteed** to fit in a **32-bit** integer.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3], target = 4
+Output: 7
+Explanation:
+The possible combination ways are:
+(1, 1, 1, 1)
+(1, 1, 2)
+(1, 2, 1)
+(1, 3)
+(2, 1, 1)
+(2, 2)
+(3, 1)
+Note that different sequences are counted as different combinations.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [9], target = 3
+Output: 0
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 200`
+- `1 <= nums[i] <= 1000`
+- All the elements of `nums` are **unique**.
+- `1 <= target <= 1000`
+
+**Follow up:** What if negative numbers are allowed in the given array? How does it change the problem? What limitation we need to add to the question to allow negative numbers?
+
+## 题目大意
+
+给你一个由 不同 整数组成的数组 nums ，和一个目标整数 target 。请你从 nums 中找出并返回总和为 target 的元素组合的个数。题目数据保证答案符合 32 位整数范围。
+
+## 解题思路
+
+- Combination Sum 这是系列问题。拿到题目，笔者先用暴力解法 dfs 尝试了一版，包含的重叠子问题特别多，剪枝条件也没有写好，果然超时。元素只有 [1,2,3] 这三种，target = 32，这组数据居然有 181997601 这么多种情况。仔细看了题目数据规模 1000，基本可以断定此题是动态规划，并且时间复杂度是 O(n^2)。
+- 本题和完全背包有点像，但是还是有区别。完全背包的取法内部不区分顺序。例如 5 = 1 + 2 + 2。但是本题是 3 种答案 (1，2，2)，(2，1，2)，(2，2，1)。定义 dp[i] 为总和为 target = i 的组合总数。最终答案存在 dp[target] 中。状态转移方程为：
+
+	{{< katex display >}}
+    dp[i] =\left\{\begin{matrix}1,i=0\\ \sum dp[i-j],i\neq 0\end{matrix}\right.
+	{{< /katex >}}
+
+- 这道题最后有一个进阶问题。如果给定的数组中含有负数，则会导致出现无限长度的排列。例如，假设数组 nums 中含有正整数 a 和负整数 −b（其中 a>0,b>0,-b<0），则有 a×b+(−b)×a=0，对于任意一个元素之和等于 target 的排列，在该排列的后面添加 b 个 a 和 a 个 −b 之后，得到的新排列的元素之和仍然等于 target，而且还可以在新排列的后面继续 b 个 a 和 a 个 −b。因此只要存在元素之和等于 target 的排列，就能构造出无限长度的排列。如果允许负数出现，则必须限制排列的最大长度，不然会出现无限长度的排列。
+
+## 代码
+
+```go
+package leetcode
+
+func combinationSum4(nums []int, target int) int {
+	dp := make([]int, target+1)
+	dp[0] = 1
+	for i := 1; i <= target; i++ {
+		for _, num := range nums {
+			if i-num >= 0 {
+				dp[i] += dp[i-num]
+			}
+		}
+	}
+	return dp[target]
+}
+
+// 暴力解法超时
+func combinationSum41(nums []int, target int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	c, res := []int{}, 0
+	findcombinationSum4(nums, target, 0, c, &res)
+	return res
+}
+
+func findcombinationSum4(nums []int, target, index int, c []int, res *int) {
+	if target <= 0 {
+		if target == 0 {
+			*res++
+		}
+		return
+	}
+	for i := 0; i < len(nums); i++ {
+		c = append(c, nums[i])
+		findcombinationSum4(nums, target-nums[i], i, c, res)
+		c = c[:len(c)-1]
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0376.Wiggle-Subsequence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0378.Kth-Smallest-Element-in-a-Sorted-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0378.Kth-Smallest-Element-in-a-Sorted-Matrix.md b/website/content/ChapterFour/0300~0399/0378.Kth-Smallest-Element-in-a-Sorted-Matrix.md
new file mode 100755
index 000000000..e0b264fbc
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0378.Kth-Smallest-Element-in-a-Sorted-Matrix.md
@@ -0,0 +1,142 @@
+# [378. Kth Smallest Element in a Sorted Matrix](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/)
+
+
+## 题目
+
+Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.
+
+Note that it is the kth smallest element in the sorted order, not the kth distinct element.
+
+**Example**:
+
+    matrix = [
+       [ 1,  5,  9],
+       [10, 11, 13],
+       [12, 13, 15]
+    ],
+    k = 8,
+    
+    return 13.
+
+**Note**: You may assume k is always valid, 1 ≤ k ≤ n2.
+
+
+## 题目大意
+
+给定一个 n x n 矩阵，其中每行和每列元素均按升序排序，找到矩阵中第 k 小的元素。请注意，它是排序后的第 k 小元素，而不是第 k 个元素。
+
+
+说明:
+你可以假设 k 的值永远是有效的, 1 ≤ k ≤ n2 。
+
+
+## 解题思路
+
+
+- 给出一个行有序，列有序的矩阵(并非是按照下标有序的)，要求找出这个矩阵中第 K 小的元素。注意找的第 K 小元素指的不是 k 个不同的元素，可能存在相同的元素。
+- 最容易想到的就解法是优先队列。依次把矩阵中的元素推入到优先队列中。维护一个最小堆，一旦优先队列里面的元素有 k 个了，就算找到结果了。
+- 这一题最优解法是二分搜索。那搜索的空间是什么呢？根据题意，可以知道，矩阵左上角的那个元素是最小的，右下角的元素是最大的。即矩阵第一个元素确定了下界，矩阵的最后一个元素确定了上界。在这个解空间里面二分搜索所有值，找到第 K  小的元素。判断是否找到的条件是，在矩阵中比 mid 小的元素个数等于 K。不断的逼近 low，使得 low == high 的时候，就是找到了第 K 小的元素了。(因为题目中说了，一定会存在第 K 小元素，所以二分搜索到一个元素的时候，一定会得出结果)。
+
+![](https://img.halfrost.com/Leetcode/leetcode_378.png)
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"container/heap"
+)
+
+// 解法一 二分搜索
+func kthSmallest378(matrix [][]int, k int) int {
+	m, n, low := len(matrix), len(matrix[0]), matrix[0][0]
+	high := matrix[m-1][n-1] + 1
+	for low < high {
+		mid := low + (high-low)>>1
+		// 如果 count 比 k 小，在大值的那一半继续二分搜索
+		if counterKthSmall(m, n, mid, matrix) >= k {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+func counterKthSmall(m, n, mid int, matrix [][]int) int {
+	count, j := 0, n-1
+	// 每次循环统计比 mid 值小的元素个数
+	for i := 0; i < m; i++ {
+		// 遍历每行中比 mid 小的元素的个数
+		for j >= 0 && mid < matrix[i][j] {
+			j--
+		}
+		count += j + 1
+	}
+	return count
+}
+
+// 解法二 优先队列
+func kthSmallest3781(matrix [][]int, k int) int {
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return 0
+	}
+	pq := &pq{data: make([]interface{}, k)}
+	heap.Init(pq)
+	for i := 0; i < len(matrix); i++ {
+		for j := 0; j < len(matrix[0]); j++ {
+			if pq.Len() < k {
+				heap.Push(pq, matrix[i][j])
+			} else if matrix[i][j] < pq.Head().(int) {
+				heap.Pop(pq)
+				heap.Push(pq, matrix[i][j])
+			} else {
+				break
+			}
+		}
+	}
+	return heap.Pop(pq).(int)
+}
+
+type pq struct {
+	data []interface{}
+	len  int
+}
+
+func (p *pq) Len() int {
+	return p.len
+}
+
+func (p *pq) Less(a, b int) bool {
+	return p.data[a].(int) > p.data[b].(int)
+}
+
+func (p *pq) Swap(a, b int) {
+	p.data[a], p.data[b] = p.data[b], p.data[a]
+}
+
+func (p *pq) Push(o interface{}) {
+	p.data[p.len] = o
+	p.len++
+}
+
+func (p *pq) Head() interface{} {
+	return p.data[0]
+}
+
+func (p *pq) Pop() interface{} {
+	p.len--
+	return p.data[p.len]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0377.Combination-Sum-IV/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0382.Linked-List-Random-Node/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0382.Linked-List-Random-Node.md b/website/content/ChapterFour/0300~0399/0382.Linked-List-Random-Node.md
new file mode 100644
index 000000000..449c5aaa8
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0382.Linked-List-Random-Node.md
@@ -0,0 +1,112 @@
+# [382. Linked List Random Node](https://leetcode.com/problems/linked-list-random-node/)
+
+
+## 题目
+
+Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
+
+Implement the `Solution` class:
+
+- `Solution(ListNode head)` Initializes the object with the integer array nums.
+- `int getRandom()` Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be choosen.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/03/16/getrand-linked-list.jpg](https://assets.leetcode.com/uploads/2021/03/16/getrand-linked-list.jpg)
+
+```
+Input
+["Solution", "getRandom", "getRandom", "getRandom", "getRandom", "getRandom"]
+[[[1, 2, 3]], [], [], [], [], []]
+Output
+[null, 1, 3, 2, 2, 3]
+
+Explanation
+Solution solution = new Solution([1, 2, 3]);
+solution.getRandom(); // return 1
+solution.getRandom(); // return 3
+solution.getRandom(); // return 2
+solution.getRandom(); // return 2
+solution.getRandom(); // return 3
+// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the linked list will be in the range `[1, 104]`.
+- `-10^4 <= Node.val <= 10^4`
+- At most `10^4` calls will be made to `getRandom`.
+
+**Follow up:**
+
+- What if the linked list is extremely large and its length is unknown to you?
+- Could you solve this efficiently without using extra space?
+
+## 题目大意
+
+给定一个单链表，随机选择链表的一个节点，并返回相应的节点值。保证每个节点被选的概率一样。
+
+进阶: 如果链表十分大且长度未知，如何解决这个问题？你能否使用常数级空间复杂度实现？
+
+## 解题思路
+
+- rand.Float64() 可以返回 [0.0,1.0) 之间的随机数。利用这个函数完成我们的随机化取节点的过程。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math/rand"
+
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+type Solution struct {
+	head *ListNode
+}
+
+/** @param head The linked list's head.
+  Note that the head is guaranteed to be not null, so it contains at least one node. */
+func Constructor(head *ListNode) Solution {
+	return Solution{head: head}
+}
+
+/** Returns a random node's value. */
+func (this *Solution) GetRandom() int {
+	scope, selectPoint, curr := 1, 0, this.head
+	for curr != nil {
+		if rand.Float64() < 1.0/float64(scope) {
+			selectPoint = curr.Val
+		}
+		scope += 1
+		curr = curr.Next
+	}
+	return selectPoint
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(head);
+ * param_1 := obj.GetRandom();
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0378.Kth-Smallest-Element-in-a-Sorted-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0383.Ransom-Note/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0383.Ransom-Note.md b/website/content/ChapterFour/0300~0399/0383.Ransom-Note.md
new file mode 100644
index 000000000..ef5603c7c
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0383.Ransom-Note.md
@@ -0,0 +1,71 @@
+# [383. Ransom Note](https://leetcode.com/problems/ransom-note/)
+
+## 题目
+
+Given two stings ransomNote and magazine, return true if ransomNote can be constructed from magazine and false otherwise.
+
+Each letter in magazine can only be used once in ransomNote.
+
+**Example 1**:
+
+    Input: ransomNote = "a", magazine = "b"
+    Output: false
+
+**Example 2**:
+
+    Input: ransomNote = "aa", magazine = "ab"
+    Output: false
+
+**Example 3**:
+
+    Input: ransomNote = "aa", magazine = "aab"
+    Output: true
+
+**Constraints:**
+
+- 1 <= ransomNote.length, magazine.length <= 100000
+- ransomNote and magazine consist of lowercase English letters.
+
+## 题目大意
+
+为了不在赎金信中暴露字迹，从杂志上搜索各个需要的字母，组成单词来表达意思。
+
+给你一个赎金信 (ransomNote) 字符串和一个杂志(magazine)字符串，判断 ransomNote 能不能由 magazines 里面的字符构成。
+
+如果可以构成，返回 true ；否则返回 false 。
+
+magazine 中的每个字符只能在 ransomNote 中使用一次。
+
+## 解题思路
+
+- ransomNote 和 magazine 都是由小写字母组成，所以用数组进行简单的字符统计
+
+## 代码
+
+````go
+package leetcode
+
+func canConstruct(ransomNote string, magazine string) bool {
+	if len(ransomNote) > len(magazine) {
+		return false
+	}
+	var cnt [26]int
+	for _, v := range magazine {
+		cnt[v-'a']++
+	}
+	for _, v := range ransomNote {
+		cnt[v-'a']--
+		if cnt[v-'a'] < 0 {
+			return false
+		}
+	}
+	return true
+}
+````
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0382.Linked-List-Random-Node/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0384.Shuffle-an-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0384.Shuffle-an-Array.md b/website/content/ChapterFour/0300~0399/0384.Shuffle-an-Array.md
new file mode 100644
index 000000000..908f9049c
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0384.Shuffle-an-Array.md
@@ -0,0 +1,89 @@
+# [384.Shuffle an Array](https://leetcode.com/problems/shuffle-an-array/)
+
+## 题目
+
+Given an integer array nums, design an algorithm to randomly shuffle the array. All permutations of the array should be equally likely as a result of the shuffling.
+
+Implement the Solution class:
+
+- Solution(int[] nums) Initializes the object with the integer array nums.
+- int[] reset() Resets the array to its original configuration and returns it.
+- int[] shuffle() Returns a random shuffling of the array.
+
+**Example 1**:
+
+    Input
+    ["Solution", "shuffle", "reset", "shuffle"]
+    [[[1, 2, 3]], [], [], []]
+    Output
+    [null, [3, 1, 2], [1, 2, 3], [1, 3, 2]]
+
+    Explanation
+    Solution solution = new Solution([1, 2, 3]);
+    solution.shuffle();    // Shuffle the array [1,2,3] and return its result.
+    // Any permutation of [1,2,3] must be equally likely to be returned.
+    // Example: return [3, 1, 2]
+    solution.reset();      // Resets the array back to its original configuration [1,2,3]. Return [1, 2, 3]
+    solution.shuffle();    // Returns the random shuffling of array [1,2,3]. Example: return [1, 3, 2]
+
+**Constraints:**
+
+- 1 <= nums.length <= 200
+- -1000000 <= nums[i] <= 1000000
+- All the elements of nums are unique.
+- At most 5 * 10000 calls in total will be made to reset and shuffle.
+
+## 题目大意
+
+给你一个整数数组 nums ，设计算法来打乱一个没有重复元素的数组。
+
+实现 Solution class:
+
+- Solution(int[] nums) 使用整数数组 nums 初始化对象
+- int[] reset() 重设数组到它的初始状态并返回
+- int[] shuffle() 返回数组随机打乱后的结果
+
+## 解题思路
+
+- 使用 rand.Shuffle 进行数组随机打乱
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math/rand"
+
+type Solution struct {
+	nums []int
+}
+
+func Constructor(nums []int) Solution {
+	return Solution{
+		nums: nums,
+	}
+}
+
+/** Resets the array to its original configuration and return it. */
+func (this *Solution) Reset() []int {
+	return this.nums
+}
+
+/** Returns a random shuffling of the array. */
+func (this *Solution) Shuffle() []int {
+	arr := make([]int, len(this.nums))
+	copy(arr, this.nums)
+	rand.Shuffle(len(arr), func(i, j int) {
+		arr[i], arr[j] = arr[j], arr[i]
+	})
+	return arr
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0383.Ransom-Note/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0385.Mini-Parser/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0385.Mini-Parser.md b/website/content/ChapterFour/0300~0399/0385.Mini-Parser.md
new file mode 100755
index 000000000..eff5eb9d9
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0385.Mini-Parser.md
@@ -0,0 +1,182 @@
+# [385. Mini Parser](https://leetcode.com/problems/mini-parser/)
+
+
+## 题目
+
+Given a nested list of integers represented as a string, implement a parser to deserialize it.
+
+Each element is either an integer, or a list -- whose elements may also be integers or other lists.
+
+**Note**: You may assume that the string is well-formed:
+
+- String is non-empty.
+- String does not contain white spaces.
+- String contains only digits `0-9`, `[`, `-` `,`, `]`.
+
+**Example 1**:
+
+    Given s = "324",
+    
+    You should return a NestedInteger object which contains a single integer 324.
+
+**Example 2**:
+
+    Given s = "[123,[456,[789]]]",
+    
+    Return a NestedInteger object containing a nested list with 2 elements:
+    
+    1. An integer containing value 123.
+    2. A nested list containing two elements:
+        i.  An integer containing value 456.
+        ii. A nested list with one element:
+             a. An integer containing value 789.
+
+
+## 题目大意
+
+给定一个用字符串表示的整数的嵌套列表，实现一个解析它的语法分析器。列表中的每个元素只可能是整数或整数嵌套列表
+
+提示：你可以假定这些字符串都是格式良好的：
+
+- 字符串非空
+- 字符串不包含空格
+- 字符串只包含数字0-9, [, - ,, ]
+
+
+
+## 解题思路
+
+- 将一个嵌套的数据结构中的数字转换成 NestedInteger 数据结构。
+- 这一题用栈一层一层的处理就行。有一些比较坑的特殊的边界数据见测试文件。这一题正确率比很多 Hard 题还要低的原因应该是没有理解好题目和边界测试数据没有考虑到。NestedInteger 这个数据结构笔者实现了一遍，见代码。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"fmt"
+	"strconv"
+)
+
+/**
+ * // This is the interface that allows for creating nested lists.
+ * // You should not implement it, or speculate about its implementation
+ * type NestedInteger struct {
+ * }
+ *
+ * // Return true if this NestedInteger holds a single integer, rather than a nested list.
+ * func (n NestedInteger) IsInteger() bool {}
+ *
+ * // Return the single integer that this NestedInteger holds, if it holds a single integer
+ * // The result is undefined if this NestedInteger holds a nested list
+ * // So before calling this method, you should have a check
+ * func (n NestedInteger) GetInteger() int {}
+ *
+ * // Set this NestedInteger to hold a single integer.
+ * func (n *NestedInteger) SetInteger(value int) {}
+ *
+ * // Set this NestedInteger to hold a nested list and adds a nested integer to it.
+ * func (n *NestedInteger) Add(elem NestedInteger) {}
+ *
+ * // Return the nested list that this NestedInteger holds, if it holds a nested list
+ * // The list length is zero if this NestedInteger holds a single integer
+ * // You can access NestedInteger's List element directly if you want to modify it
+ * func (n NestedInteger) GetList() []*NestedInteger {}
+ */
+
+// NestedInteger define
+type NestedInteger struct {
+	Num  int
+	List []*NestedInteger
+}
+
+// IsInteger define
+func (n NestedInteger) IsInteger() bool {
+	if n.List == nil {
+		return true
+	}
+	return false
+}
+
+// GetInteger define
+func (n NestedInteger) GetInteger() int {
+	return n.Num
+}
+
+// SetInteger define
+func (n *NestedInteger) SetInteger(value int) {
+	n.Num = value
+}
+
+// Add define
+func (n *NestedInteger) Add(elem NestedInteger) {
+	n.List = append(n.List, &elem)
+}
+
+// GetList define
+func (n NestedInteger) GetList() []*NestedInteger {
+	return n.List
+}
+
+// Print define
+func (n NestedInteger) Print() {
+	if len(n.List) != 0 {
+		for _, v := range n.List {
+			if len(v.List) != 0 {
+				v.Print()
+				return
+			}
+			fmt.Printf("%v ", v.Num)
+		}
+	} else {
+		fmt.Printf("%v ", n.Num)
+	}
+	fmt.Printf("\n")
+}
+
+func deserialize(s string) *NestedInteger {
+	stack, cur := []*NestedInteger{}, &NestedInteger{}
+	for i := 0; i < len(s); {
+		switch {
+		case isDigital(s[i]) || s[i] == '-':
+			j := 0
+			for j = i + 1; j < len(s) && isDigital(s[j]); j++ {
+			}
+			num, _ := strconv.Atoi(s[i:j])
+			next := &NestedInteger{}
+			next.SetInteger(num)
+			if len(stack) > 0 {
+				stack[len(stack)-1].List = append(stack[len(stack)-1].GetList(), next)
+			} else {
+				cur = next
+			}
+			i = j
+		case s[i] == '[':
+			next := &NestedInteger{}
+			if len(stack) > 0 {
+				stack[len(stack)-1].List = append(stack[len(stack)-1].GetList(), next)
+			}
+			stack = append(stack, next)
+			i++
+		case s[i] == ']':
+			cur = stack[len(stack)-1]
+			stack = stack[:len(stack)-1]
+			i++
+		case s[i] == ',':
+			i++
+		}
+	}
+	return cur
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0384.Shuffle-an-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0386.Lexicographical-Numbers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0386.Lexicographical-Numbers.md b/website/content/ChapterFour/0300~0399/0386.Lexicographical-Numbers.md
new file mode 100755
index 000000000..dc76ab44f
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0386.Lexicographical-Numbers.md
@@ -0,0 +1,58 @@
+# [386. Lexicographical Numbers](https://leetcode.com/problems/lexicographical-numbers/)
+
+
+## 题目
+
+Given an integer n, return 1 - n in lexicographical order.
+
+For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9].
+
+Please optimize your algorithm to use less time and space. The input size may be as large as 5,000,000.
+
+
+## 题目大意
+
+给定一个整数 n, 返回从 1 到 n 的字典顺序。例如，给定 n =13，返回 [1,10,11,12,13,2,3,4,5,6,7,8,9] 。
+
+请尽可能的优化算法的时间复杂度和空间复杂度。 输入的数据 n 小于等于 5,000,000。
+
+
+
+## 解题思路
+
+
+- 给出一个数字 n ，要求按照字典序对 1-n 这 n 个数排序。
+- DFS 暴力求解即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func lexicalOrder(n int) []int {
+	res := make([]int, 0, n)
+	dfs386(1, n, &res)
+	return res
+}
+
+func dfs386(x, n int, res *[]int) {
+	limit := (x + 10) / 10 * 10
+	for x <= n && x < limit {
+		*res = append(*res, x)
+		if x*10 <= n {
+			dfs386(x*10, n, res)
+		}
+		x++
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0385.Mini-Parser/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0387.First-Unique-Character-in-a-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0387.First-Unique-Character-in-a-String.md b/website/content/ChapterFour/0300~0399/0387.First-Unique-Character-in-a-String.md
new file mode 100755
index 000000000..8f7954da2
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0387.First-Unique-Character-in-a-String.md
@@ -0,0 +1,88 @@
+# [387. First Unique Character in a String](https://leetcode.com/problems/first-unique-character-in-a-string/)
+
+## 题目
+
+Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.
+
+**Examples**:
+
+    s = "leetcode"
+    return 0.
+    
+    s = "loveleetcode",
+    return 2.
+
+**Note**: You may assume the string contain only lowercase letters.
+
+
+
+## 题目大意
+
+给定一个字符串，找到它的第一个不重复的字符，并返回它的索引。如果不存在，则返回 -1。
+
+
+## 解题思路
+
+- 简单题，要求输出第一个没有重复的字符。
+- 解法二这个思路只超过 81% 的用户，但是如果测试样例中 s 的字符串很长，但是满足条件的字符都在靠后的位置的话，这个思路应该会更有优势。通过记录每个字符的第一次出现的位置和最后一次出现的位置。第一次对 s 进行一次遍历。第二次仅仅对数组进行遍历就可以了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法 一
+func firstUniqChar(s string) int {
+	result := make([]int, 26)
+	for i := 0; i < len(s); i++ {
+		result[s[i]-'a']++
+	}
+	for i := 0; i < len(s); i++ {
+		if result[s[i]-'a'] == 1 {
+			return i
+		}
+	}
+	return -1
+}
+
+// 解法 二
+// 执行用时: 8 ms
+// 内存消耗: 5.2 MB
+func firstUniqChar1(s string) int {
+	charMap := make([][2]int, 26)
+	for i := 0; i < 26; i++ {
+		charMap[i][0] = -1
+		charMap[i][1] = -1
+	}
+	for i := 0; i < len(s); i++ {
+		if charMap[s[i]-'a'][0] == -1 {
+			charMap[s[i]-'a'][0] = i
+		} else { //已经出现过
+			charMap[s[i]-'a'][1] = i
+		}
+	}
+	res := len(s)
+	for i := 0; i < 26; i++ {
+		//只出现了一次
+		if charMap[i][0] >= 0 && charMap[i][1] == -1 {
+			if charMap[i][0] < res {
+				res = charMap[i][0]
+			}
+		}
+	}
+	if res == len(s) {
+		return -1
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0386.Lexicographical-Numbers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0389.Find-the-Difference/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0389.Find-the-Difference.md b/website/content/ChapterFour/0300~0399/0389.Find-the-Difference.md
new file mode 100755
index 000000000..d21ed2376
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0389.Find-the-Difference.md
@@ -0,0 +1,55 @@
+# [389. Find the Difference](https://leetcode.com/problems/find-the-difference/)
+
+## 题目
+
+Given two strings **s** and **t** which consist of only lowercase letters.
+
+String **t** is generated by random shuffling string **s** and then add one more letter at a random position.
+
+Find the letter that was added in **t**.
+
+**Example**:
+
+    Input:
+    s = "abcd"
+    t = "abcde"
+    
+    Output:
+    e
+    
+    Explanation:
+    'e' is the letter that was added.
+
+## 题目大意
+
+给定两个字符串 s 和 t，它们只包含小写字母。字符串 t 由字符串 s 随机重排，然后在随机位置添加一个字母。请找出在 t 中被添加的字母。
+
+
+## 解题思路
+
+- 题目要求找出 t 字符串中比 s 字符串多出的一个字符。思路还是利用异或的性质，`X^X = 0`，将 s 和 t 依次异或，最终多出来的字符就是最后异或的结果。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findTheDifference(s string, t string) byte {
+	n, ch := len(t), t[len(t)-1]
+	for i := 0; i < n-1; i++ {
+		ch ^= s[i]
+		ch ^= t[i]
+	}
+	return ch
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0387.First-Unique-Character-in-a-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0390.Elimination-Game/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0390.Elimination-Game.md b/website/content/ChapterFour/0300~0399/0390.Elimination-Game.md
new file mode 100644
index 000000000..d90aca5b9
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0390.Elimination-Game.md
@@ -0,0 +1,81 @@
+# [390. Elimination Game](https://leetcode.com/problems/elimination-game/)
+
+
+## 题目
+
+You have a list `arr` of all integers in the range `[1, n]` sorted in a strictly increasing order. Apply the following algorithm on `arr`:
+
+- Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.
+- Repeat the previous step again, but this time from right to left, remove the rightmost number and every other number from the remaining numbers.
+- Keep repeating the steps again, alternating left to right and right to left, until a single number remains.
+
+Given the integer `n`, return *the last number that remains in* `arr`.
+
+**Example 1:**
+
+```
+Input: n = 9
+Output: 6
+Explanation:
+arr = [1, 2,3, 4,5, 6,7, 8,9]
+arr = [2,4, 6,8]
+arr = [2, 6]
+arr = [6]
+
+```
+
+**Example 2:**
+
+```
+Input: n = 1
+Output: 1
+
+```
+
+**Constraints:**
+
+- `1 <= n <= 109`
+
+## 题目大意
+
+列表 arr 由在范围 [1, n] 中的所有整数组成，并按严格递增排序。请你对 arr 应用下述算法：
+
+- 从左到右，删除第一个数字，然后每隔一个数字删除一个，直到到达列表末尾。
+- 重复上面的步骤，但这次是从右到左。也就是，删除最右侧的数字，然后剩下的数字每隔一个删除一个。
+- 不断重复这两步，从左到右和从右到左交替进行，直到只剩下一个数字。
+
+给你整数 n ，返回 arr 最后剩下的数字。
+
+## 解题思路
+
+- 模拟题。按照题意，第一轮从左往右删除数字，第二轮从右往左删除数字。题目要求最后剩下的数字，模拟过程中不需要真的删除元素。只需要标记起始元素，该轮步长和方向即可。最后总元素只剩下一个即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+func lastRemaining(n int) int {
+	start, dir, step := 1, true, 1
+	for n > 1 {
+		if dir { // 正向
+			start += step
+		} else { // 反向
+			if n%2 == 1 {
+				start += step
+			}
+		}
+		dir = !dir
+		n >>= 1
+		step <<= 1
+	}
+	return start
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0389.Find-the-Difference/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0391.Perfect-Rectangle/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0391.Perfect-Rectangle.md b/website/content/ChapterFour/0300~0399/0391.Perfect-Rectangle.md
new file mode 100644
index 000000000..3028424f4
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0391.Perfect-Rectangle.md
@@ -0,0 +1,121 @@
+# [391. Perfect Rectangle](https://leetcode.com/problems/perfect-rectangle/)
+
+## 题目
+
+Given an array rectangles where rectangles[i] = [xi, yi, ai, bi] represents an axis-aligned rectangle. The bottom-left point of the rectangle is (xi, yi) and the top-right point of it is (ai, bi).
+
+Return true if all the rectangles together form an exact cover of a rectangular region.
+
+**Example1:**
+
+![https://assets.leetcode.com/uploads/2021/03/27/perectrec1-plane.jpg](https://assets.leetcode.com/uploads/2021/03/27/perectrec1-plane.jpg)
+
+    Input: rectangles = [[1,1,3,3],[3,1,4,2],[3,2,4,4],[1,3,2,4],[2,3,3,4]]
+    Output: true
+    Explanation: All 5 rectangles together form an exact cover of a rectangular region.
+
+**Example2:**
+
+![https://assets.leetcode.com/uploads/2021/03/27/perfectrec2-plane.jpg](https://assets.leetcode.com/uploads/2021/03/27/perfectrec2-plane.jpg)
+
+    Input: rectangles = [[1,1,2,3],[1,3,2,4],[3,1,4,2],[3,2,4,4]]
+    Output: false
+    Explanation: Because there is a gap between the two rectangular regions.
+
+**Example3:**
+
+![https://assets.leetcode.com/uploads/2021/03/27/perfectrec3-plane.jpg](https://assets.leetcode.com/uploads/2021/03/27/perfectrec3-plane.jpg)
+
+    Input: rectangles = [[1,1,3,3],[3,1,4,2],[1,3,2,4],[3,2,4,4]]
+    Output: false
+    Explanation: Because there is a gap in the top center.
+
+**Example4:**
+
+![https://assets.leetcode.com/uploads/2021/03/27/perfecrrec4-plane.jpg](https://assets.leetcode.com/uploads/2021/03/27/perfecrrec4-plane.jpg)
+
+    Input: rectangles = [[1,1,3,3],[3,1,4,2],[1,3,2,4],[2,2,4,4]]
+    Output: false
+    Explanation: Because two of the rectangles overlap with each other.
+
+**Constraints:**
+
+- 1 <= rectangles.length <= 2 * 10000
+- rectangles[i].length == 4
+- -100000 <= xi, yi, ai, bi <= 100000
+
+## 题目大意
+
+给你一个数组 rectangles ，其中 rectangles[i] = [xi, yi, ai, bi] 表示一个坐标轴平行的矩形。这个矩形的左下顶点是 (xi, yi) ，右上顶点是 (ai, bi) 。
+
+如果所有矩形一起精确覆盖了某个矩形区域，则返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 矩形区域的面积等于所有矩形的面积之和并且满足矩形区域四角的顶点只能出现一次，且其余顶点的出现次数只能是两次或四次则返回 true,否则返回 false
+
+## 代码
+
+```go
+
+package leetcode
+
+type point struct {
+	x int
+	y int
+}
+
+func isRectangleCover(rectangles [][]int) bool {
+	minX, minY, maxA, maxB := rectangles[0][0], rectangles[0][1], rectangles[0][2], rectangles[0][3]
+	area := 0
+	cnt := make(map[point]int)
+	for _, v := range rectangles {
+		x, y, a, b := v[0], v[1], v[2], v[3]
+		area += (a - x) * (b - y)
+		minX = min(minX, x)
+		minY = min(minY, y)
+		maxA = max(maxA, a)
+		maxB = max(maxB, b)
+		cnt[point{x, y}]++
+		cnt[point{a, b}]++
+		cnt[point{x, b}]++
+		cnt[point{a, y}]++
+	}
+	if area != (maxA - minX) * (maxB - minY) ||
+		cnt[point{minX, minY}] != 1 || cnt[point{maxA, maxB}] != 1 ||
+		cnt[point{minX, maxB}] != 1 || cnt[point{maxA, minY}] != 1 {
+		return false
+	}
+	delete(cnt, point{minX, minY})
+	delete(cnt, point{maxA, maxB})
+	delete(cnt, point{minX, maxB})
+	delete(cnt, point{maxA, minY})
+	for _, v := range cnt {
+		if v != 2 && v != 4 {
+			return false
+		}
+	}
+	return true
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0390.Elimination-Game/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0392.Is-Subsequence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0392.Is-Subsequence.md b/website/content/ChapterFour/0300~0399/0392.Is-Subsequence.md
new file mode 100755
index 000000000..459d38068
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0392.Is-Subsequence.md
@@ -0,0 +1,87 @@
+# [392. Is Subsequence](https://leetcode.com/problems/is-subsequence/)
+
+
+## 题目
+
+Given a string **s** and a string **t**, check if **s** is subsequence of **t**.
+
+You may assume that there is only lower case English letters in both **s** and **t**. **t** is potentially a very long (length ~= 500,000) string, and **s** is a short string (<=100).
+
+A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, `"ace"` is a subsequence of `"abcde"`while `"aec"` is not).
+
+**Example 1**:
+
+	Input: s = "abc", t = "ahbgdc"
+	Output: true
+
+**Example 2**:
+
+	Input: s = "axc", t = "ahbgdc"
+	Output: false
+
+**Follow up**: If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?
+
+**Credits**: Special thanks to [@pbrother](https://leetcode.com/pbrother/) for adding this problem and creating all test cases.
+
+
+## 题目大意
+
+给定字符串 s 和 t ，判断 s 是否为 t 的子序列。你可以认为 s 和 t 中仅包含英文小写字母。字符串 t 可能会很长（长度 ~= 500,000），而 s 是个短字符串（长度 <=100）。字符串的一个子序列是原始字符串删除一些（也可以不删除）字符而不改变剩余字符相对位置形成的新字符串。（例如，"ace"是"abcde"的一个子序列，而"aec"不是）。
+
+
+
+## 解题思路
+
+
+- 给定 2 个字符串 s 和 t，问 s 是不是 t 的子序列。注意 s 在 t 中还需要保持 s 的字母的顺序。
+- 这是一题贪心算法。直接做即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 O(n^2)
+func isSubsequence(s string, t string) bool {
+	index := 0
+	for i := 0; i < len(s); i++ {
+		flag := false
+		for ; index < len(t); index++ {
+			if s[i] == t[index] {
+				flag = true
+				break
+			}
+		}
+		if flag == true {
+			index++
+			continue
+		} else {
+			return false
+		}
+	}
+	return true
+}
+
+// 解法二 O(n)
+func isSubsequence1(s string, t string) bool {
+	for len(s) > 0 && len(t) > 0 {
+		if s[0] == t[0] {
+			s = s[1:]
+		}
+		t = t[1:]
+	}
+	return len(s) == 0
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0391.Perfect-Rectangle/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0393.UTF-8-Validation/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0393.UTF-8-Validation.md b/website/content/ChapterFour/0300~0399/0393.UTF-8-Validation.md
new file mode 100755
index 000000000..1275685b2
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0393.UTF-8-Validation.md
@@ -0,0 +1,112 @@
+# [393. UTF-8 Validation](https://leetcode.com/problems/utf-8-validation/)
+
+
+
+## 题目
+
+A character in UTF8 can be from **1 to 4 bytes** long, subjected to the following rules:
+
+1. For 1-byte character, the first bit is a 0, followed by its unicode code.
+2. For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
+
+This is how the UTF-8 encoding would work:
+
+       Char. number range  |        UTF-8 octet sequence
+          (hexadecimal)    |              (binary)
+       --------------------+---------------------------------------------
+       0000 0000-0000 007F | 0xxxxxxx
+       0000 0080-0000 07FF | 110xxxxx 10xxxxxx
+       0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
+       0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
+
+Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
+
+**Note**: The input is an array of integers. Only the **least significant 8 bits** of each integer is used to store the data. This means each integer represents only 1 byte of data.
+
+**Example 1**:
+
+    data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
+    
+    Return true.
+    It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
+
+**Example 2**:
+
+    data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
+    
+    Return false.
+    The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
+    The next byte is a continuation byte which starts with 10 and that's correct.
+    But the second continuation byte does not start with 10, so it is invalid.
+
+## 题目大意
+
+UTF-8 中的一个字符可能的长度为 1 到 4 字节，遵循以下的规则：
+
+对于 1 字节的字符，字节的第一位设为 0，后面 7 位为这个符号的 unicode 码。
+对于 n 字节的字符 (n > 1)，第一个字节的前 n 位都设为 1，第 n+1 位设为 0，后面字节的前两位一律设为 10。剩下的没有提及的二进制位，全部为这个符号的 unicode 码。
+这是 UTF-8 编码的工作方式：
+
+```c
+   Char. number range  |        UTF-8 octet sequence
+      (hexadecimal)    |              (binary)
+   --------------------+---------------------------------------------
+   0000 0000-0000 007F | 0xxxxxxx
+   0000 0080-0000 07FF | 110xxxxx 10xxxxxx
+   0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
+   0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
+
+```
+
+给定一个表示数据的整数数组，返回它是否为有效的 utf-8 编码。
+
+注意:
+
+输入是整数数组。只有每个整数的最低 8 个有效位用来存储数据。这意味着每个整数只表示 1 字节的数据。
+
+
+## 解题思路
+
+- 这一题看似很复杂，其实严格按照 UTF8 定义来模拟就可以了。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func validUtf8(data []int) bool {
+	count := 0
+	for _, d := range data {
+		if count == 0 {
+			if d >= 248 { // 11111000 = 248
+				return false
+			} else if d >= 240 { // 11110000 = 240
+				count = 3
+			} else if d >= 224 { // 11100000 = 224
+				count = 2
+			} else if d >= 192 { // 11000000 = 192
+				count = 1
+			} else if d > 127 { // 01111111 = 127
+				return false
+			}
+		} else {
+			if d <= 127 || d >= 192 {
+				return false
+			}
+			count--
+		}
+	}
+	return count == 0
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0392.Is-Subsequence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0394.Decode-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0394.Decode-String.md b/website/content/ChapterFour/0300~0399/0394.Decode-String.md
new file mode 100644
index 000000000..febd7d6e1
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0394.Decode-String.md
@@ -0,0 +1,88 @@
+# [394. Decode String](https://leetcode.com/problems/decode-string/)
+
+## 题目
+
+Given an encoded string, return its decoded string.
+
+The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.
+
+You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.
+
+Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won't be input like 3a or 2[4].
+
+**Examples**:
+
+```
+
+s = "3[a]2[bc]", return "aaabcbc".
+s = "3[a2[c]]", return "accaccacc".
+s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
+
+```
+
+## 题目大意
+
+给定一个经过编码的字符串，返回它解码后的字符串。编码规则为: k[encoded\_string]，表示其中方括号内部的 encoded\_string 正好重复 k 次。注意 k 保证为正整数。你可以认为输入字符串总是有效的；输入字符串中没有额外的空格，且输入的方括号总是符合格式要求的。此外，你可以认为原始数据不包含数字，所有的数字只表示重复的次数 k ，例如不会出现像 3a 或 2[4] 的输入。
+
+## 解题思路
+
+这一题和第 880 题大体类似。用栈处理，遇到 "["，就要开始重复字符串了，另外重复的数字是可能存在多位的，所以需要往前找到不为数字的那一位，把数字转换出来。最后用把 stack 里面的字符串都串联起来即可。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+)
+
+func decodeString(s string) string {
+	stack, res := []string{}, ""
+	for _, str := range s {
+		if len(stack) == 0 || (len(stack) > 0 && str != ']') {
+			stack = append(stack, string(str))
+		} else {
+			tmp := ""
+			for stack[len(stack)-1] != "[" {
+				tmp = stack[len(stack)-1] + tmp
+				stack = stack[:len(stack)-1]
+			}
+			stack = stack[:len(stack)-1]
+			index, repeat := 0, ""
+			for index = len(stack) - 1; index >= 0; index-- {
+				if stack[index] >= "0" && stack[index] <= "9" {
+					repeat = stack[index] + repeat
+				} else {
+					break
+				}
+			}
+			nums, _ := strconv.Atoi(repeat)
+			copyTmp := tmp
+			for i := 0; i < nums-1; i++ {
+				tmp += copyTmp
+			}
+			for i := 0; i < len(repeat)-1; i++ {
+				stack = stack[:len(stack)-1]
+			}
+			stack[index+1] = tmp
+		}
+	}
+	for _, s := range stack {
+		res += s
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0393.UTF-8-Validation/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0395.Longest-Substring-with-At-Least-K-Repeating-Characters.md b/website/content/ChapterFour/0300~0399/0395.Longest-Substring-with-At-Least-K-Repeating-Characters.md
new file mode 100644
index 000000000..5c3966441
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0395.Longest-Substring-with-At-Least-K-Repeating-Characters.md
@@ -0,0 +1,119 @@
+# [395. Longest Substring with At Least K Repeating Characters](https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/)
+
+
+## 题目
+
+Given a string `s` and an integer `k`, return *the length of the longest substring of* `s` *such that the frequency of each character in this substring is greater than or equal to* `k`.
+
+**Example 1:**
+
+```
+Input: s = "aaabb", k = 3
+Output: 3
+Explanation: The longest substring is "aaa", as 'a' is repeated 3 times.
+```
+
+**Example 2:**
+
+```
+Input: s = "ababbc", k = 2
+Output: 5
+Explanation: The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 10^4`
+- `s` consists of only lowercase English letters.
+- `1 <= k <= 10^5`
+
+## 题目大意
+
+给你一个字符串 s 和一个整数 k ，请你找出 s 中的最长子串， 要求该子串中的每一字符出现次数都不少于 k 。返回这一子串的长度。
+
+## 解题思路
+
+- 最容易想到的思路是递归。如果某个字符出现次数大于 0 小于 k，那么包含这个字符的子串都不满足要求。所以按照这个字符来切分整个字符串，满足题意的最长子串一定不包含切分的字符。切分完取出最长子串即可。时间复杂度 O(26*n)，空间复杂度 O(26^2)
+- 此题另外一个思路是滑动窗口。有一个需要解决的问题是右窗口移动的条件。此题要求最长字符串，那么这个最终的字符串内包含的字符种类最多是 26 种。字符种类就是右窗口移动的条件。依次枚举字符种类，如果当前窗口内的字符种类小于当前枚举的字符种类，那么窗口右移，否则左移。窗口移动中需要动态维护 freq 频次数组。可以每次都循环一遍这个数组，计算出出现次数大于 k 的字符。虽然这种做法只最多循环 26 次，但是还是不高效。更高效的做法是维护 1 个值，一个用来记录当前出现次数小于 k 次的字符种类数 `less`。如果 freq 为 0 ，说明小于 k 次的字符种类数要发生变化，如果是右窗口移动，那么 `less++`，如果是左窗口移动，那么`less--`。同理，如果 freq 为 k ，说明小于 k 次的字符种类数要发生变化，如果是右窗口移动，那么 `less--`，如果是左窗口移动，那么`less++`。在枚举 26 个字符种类中，动态维护记录出最长字符串。枚举完成，最长字符串长度也就求出来了。时间复杂度 O(26*n)，空间复杂度 O(26)
+
+## 代码
+
+```go
+package leetcode
+
+import "strings"
+
+// 解法一 滑动窗口
+func longestSubstring(s string, k int) int {
+	res := 0
+	for t := 1; t <= 26; t++ {
+		freq, total, lessK, left, right := [26]int{}, 0, 0, 0, -1
+		for left < len(s) {
+			if right+1 < len(s) && total <= t {
+				if freq[s[right+1]-'a'] == 0 {
+					total++
+					lessK++
+				}
+				freq[s[right+1]-'a']++
+				if freq[s[right+1]-'a'] == k {
+					lessK--
+				}
+				right++
+			} else {
+				if freq[s[left]-'a'] == k {
+					lessK++
+				}
+				freq[s[left]-'a']--
+				if freq[s[left]-'a'] == 0 {
+					total--
+					lessK--
+				}
+				left++
+			}
+			if lessK == 0 {
+				res = max(res, right-left+1)
+			}
+
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法二 递归分治
+func longestSubstring1(s string, k int) int {
+	if s == "" {
+		return 0
+	}
+	freq, split, res := [26]int{}, byte(0), 0
+	for _, ch := range s {
+		freq[ch-'a']++
+	}
+	for i, c := range freq[:] {
+		if 0 < c && c < k {
+			split = 'a' + byte(i)
+			break
+		}
+	}
+	if split == 0 {
+		return len(s)
+	}
+	for _, subStr := range strings.Split(s, string(split)) {
+		res = max(res, longestSubstring1(subStr, k))
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0394.Decode-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0396.Rotate-Function/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0396.Rotate-Function.md b/website/content/ChapterFour/0300~0399/0396.Rotate-Function.md
new file mode 100644
index 000000000..95a661c99
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0396.Rotate-Function.md
@@ -0,0 +1,117 @@
+# [396. Rotate Function](https://leetcode.com/problems/rotate-function/)
+
+## 题目
+
+You are given an integer array `nums` of length `n`.
+
+Assume `arrk` to be an array obtained by rotating `nums` by `k` positions clock-wise. We define the **rotation function** `F` on `nums` as follow:
+
+- `F(k) = 0 * arrk[0] + 1 * arrk[1] + ... + (n - 1) * arrk[n - 1]`.
+
+Return the maximum value of `F(0), F(1), ..., F(n-1)`.
+
+The test cases are generated so that the answer fits in a **32-bit** integer.
+
+**Example 1:**
+
+```c
+Input: nums = [4,3,2,6]
+Output: 26
+Explanation:
+F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
+F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
+F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
+F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
+So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.
+```
+
+**Example 2:**
+
+```c
+Input: nums = [100]
+Output: 0
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `1 <= n <= 105`
+- `-100 <= nums[i] <= 100`
+
+## 题目大意
+
+给定一个长度为`n`的整数数组`nums`，设`arrk`是数组`nums`顺时针旋转`k`个位置后的数组。
+
+定义`nums`的旋转函数`F`为：
+
+- `F(k) = 0 * arrk[0] + 1 * arrk[1] + ... + (n - 1) * arrk[n - 1]`
+
+返回`F(0), F(1), ..., F(n-1)`中的最大值。
+
+## 解题思路
+
+**抽象化观察：**
+
+```c
+nums = [A0, A1, A2, A3]
+
+sum  = A0 + A1 + A2+ A3
+F(0) = 0*A0 +0*A0 + 1*A1 + 2*A2 + 3*A3
+    
+F(1) = 0*A3 + 1*A0 + 2*A1 + 3*A2 
+     = F(0) + (A0 + A1 + A2) - 3*A3 
+     = F(0) + (sum-A3) - 3*A3 
+     = F(0) + sum - 4*A3
+
+F(2) = 0*A2 + 1*A3 + 2*A0 + 3*A1 
+     = F(1) + A3 + A0 + A1 - 3*A2 
+     = F(1) + sum - 4*A2
+
+F(3) = 0*A1 + 1*A2 + 2*A3 + 3*A0 
+     = F(2) + A2 + A3 + A0 - 3*A1 
+     = F(2) + sum - 4*A1
+    
+// 记sum为nums数组中所有元素和
+// 可以猜测当0 ≤ i < n时存在公式:
+F(i) = F(i-1) + sum - n * A(n-i)
+```
+
+**数学归纳法证明迭代公式：**
+
+根据题目中给定的旋转函数公式可得已知条件：
+
+- `F(0) = 0×nums[0] + 1×nums[1] + ... + (n−1)×nums[n−1]`；
+
+- `F(1) = 1×nums[0] + 2×nums[1] + ... + 0×nums[n-1]`。
+
+令数组`nums`中所有元素和为`sum`，用数学归纳法验证：当`1 ≤ k < n`时，`F(k) = F(k-1) + sum - n×nums[n-k]`成立。
+
+**归纳奠基**：证明`k=1`时命题成立。
+
+```c
+F(1) = 1×nums[0] + 2×nums[1] + ... + 0×nums[n-1]
+     = F(0) + sum - n×nums[n-1]
+```
+
+**归纳假设**：假设`F(k) = F(k-1) + sum - n×nums[n-k]`成立。
+
+**归纳递推**：由归纳假设推出`F(k+1) = F(k) + sum - n×nums[n-(k+1)]`成立，则假设的递推公式成立。
+
+```c
+F(k+1) = (k+1)×nums[0] + k×nums[1] + ... + 0×nums[n-1]
+       = F(k) + sum - n×nums[n-(k+1)] 
+```
+
+因此可以得到递推公式：
+
+- 当`n = 0`时，`F(0) = 0×nums[0] + 1×nums[1] + ... + (n−1)×nums[n−1]`
+- 当`1 ≤ k < n`时，`F(k) = F(k-1) + sum - n×nums[n-k]`成立。
+
+循环遍历`0 ≤ k < n`，计算出不同的`F(k)`并不断更新最大值，就能求出`F(0), F(1), ..., F(n-1)`中的最大值。
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0397.Integer-Replacement/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0397.Integer-Replacement.md b/website/content/ChapterFour/0300~0399/0397.Integer-Replacement.md
new file mode 100755
index 000000000..92179b94a
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0397.Integer-Replacement.md
@@ -0,0 +1,93 @@
+# [397. Integer Replacement](https://leetcode.com/problems/integer-replacement/)
+
+
+## 题目
+
+Given a positive integer n and you can do operations as follow:
+
+1. If  n is even, replace  n with `n/2`.
+2. If  n is odd, you can replace n with either `n + 1` or `n - 1`.
+
+What is the minimum number of replacements needed for n to become 1?
+
+**Example 1**:
+
+    Input:
+    8
+    
+    Output:
+    3
+    
+    Explanation:
+    8 -> 4 -> 2 -> 1
+
+**Example 2**:
+
+    Input:
+    7
+    
+    Output:
+    4
+    
+    Explanation:
+    7 -> 8 -> 4 -> 2 -> 1
+    or
+    7 -> 6 -> 3 -> 2 -> 1
+
+
+## 题目大意
+
+给定一个正整数 n，你可以做如下操作：
+
+1. 如果 n 是偶数，则用 n / 2 替换 n。
+2. 如果 n 是奇数，则可以用 n + 1 或 n - 1 替换 n。
+
+问 n 变为 1 所需的最小替换次数是多少？
+
+
+## 解题思路
+
+
+- 题目给出一个整数 `n`，然后让我们通过变换将它为 1，如果 `n` 是偶数，可以直接变为 `n/2`，如果是奇数，可以先 `n+1` 或 `n-1`，问最终变为 1 的最少步骤。
+- 当 n 为奇数的时候，什么时候需要加 1 ，什么时候需要减 1 ，通过观察规律可以发现，除了 3 和 7 以外，所有加 1 就变成 4 的倍数的奇数，都适合先加 1 运算，比如 15:
+
+        15 -> 16 -> 8 -> 4 -> 2 -> 1
+        15 -> 14 -> 7 -> 6 -> 3 -> 2 -> 1
+        
+        111011 -> 111010 -> 11101 -> 11100 -> 1110 -> 111 -> 1000 -> 100 -> 10 -> 1
+        111011 -> 111100 -> 11110 -> 1111 -> 10000 -> 1000 -> 100 -> 10 -> 1
+
+- 对于 7 来说，加 1 和减 1 的结果相同，可以不用管，对于 3 来说，减 1 的步骤更少，所以需要先去掉这种特殊情况。
+- 最后如何判断某个数字加 1 后是 4 的倍数呢？这里有一个小技巧，由于之前判断了其是奇数了，那么最右边一位肯定是 1，如果其右边第二位也是 1 的话，那么进行加 1 运算，进位后右边肯定会出现两个 0，则一定是 4 的倍数。于是就可以判断出来了。剩下的情况就是偶数的情况，如果之前判定是偶数，那么直接除以 2 (右移一位)即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func integerReplacement(n int) int {
+	res := 0
+	for n > 1 {
+		if (n & 1) == 0 { // 判断是否是偶数
+			n >>= 1
+		} else if (n+1)%4 == 0 && n != 3 { // 末尾 2 位为 11
+			n++
+		} else { // 末尾 2 位为 01
+			n--
+		}
+		res++
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0396.Rotate-Function/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0399.Evaluate-Division/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/0399.Evaluate-Division.md b/website/content/ChapterFour/0300~0399/0399.Evaluate-Division.md
new file mode 100755
index 000000000..b1d2a5988
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/0399.Evaluate-Division.md
@@ -0,0 +1,124 @@
+# [399. Evaluate Division](https://leetcode.com/problems/evaluate-division/)
+
+
+## 题目
+
+Equations are given in the format `A / B = k`, where `A` and `B` are variables represented as strings, and `k` is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return `-1.0`.
+
+**Example**:
+
+Given `a / b = 2.0, b / c = 3.0.`queries are: `a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .`return `[6.0, 0.5, -1.0, 1.0, -1.0 ].`
+
+The input is: `vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries` , where `equations.size() == values.size()`, and the values are positive. This represents the equations. Return `vector<double>`.
+
+According to the example above:
+
+    equations = [ ["a", "b"], ["b", "c"] ],
+    values = [2.0, 3.0],
+    queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ].
+
+The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.
+
+
+## 题目大意
+
+给出方程式 A / B = k, 其中 A 和 B 均为代表字符串的变量， k 是一个浮点型数字。根据已知方程式求解问题，并返回计算结果。如果结果不存在，则返回 -1.0。
+
+示例 :
+给定 a / b = 2.0, b / c = 3.0
+问题: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? 
+返回 [6.0, 0.5, -1.0, 1.0, -1.0 ]
+
+输入为: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries(方程式，方程式结果，问题方程式)， 其中 equations.size() == values.size()，即方程式的长度与方程式结果长度相等（程式与结果一一对应），并且结果值均为正数。以上为方程式的描述。 返回vector<double>类型。
+
+假设输入总是有效的。你可以假设除法运算中不会出现除数为0的情况，且不存在任何矛盾的结果。
+
+
+## 解题思路
+
+
+- 给出一些字母变量的倍数关系，问给出任意两个字母的倍数是多少。
+- 这一题可以用 DFS 或者并查集来解题。先来看看 DFS 的做法。先建图。每个字母或者字母组合可以看做成一个节点，给出的 `equations` 关系可以看成两个节点之间的有向边。每条有向边都有权值。那么问题可以转换成是否存在一条从起点节点到终点节点的路径，如果存在，输出这条路径上所有有向边权值的累乘结果。如果不存在这条路径，就返回 -1 。如果给的起点和终点不在给出的节点集里面，也输出 -1 。
+- 再来看看并查集的做法。先将每两个有倍数关系的节点做并查集 `union()` 操作。例如 A/B = 2，那么把 B 作为 `parent` 节点，`parents[A] = {B，2}`，`parents[B] = {B，1}`，B 指向自己是 1 。还有一个关系是 `B/C=3`，由于 B 已经在并查集中了，所以这个时候需要把这个关系反过来，处理成 `C/B = 1/3` ，即 `parents[C] = {B，1/3}`。这样把所有有关系的字母都 `union()` 起来。如何求任意两个字母的倍数关系呢？例如 `A/C = ?` 在并查集中查找，可以找到 `parents[C] == parents[A] == B`，那么就用 `parents[A]/parents[C] = 2/(1/3) = 6`。为什么可以这样做呢？因为 `A/B = 2`，`C/B = 1/3`，那么 `A/C = (A/B)/(C/B)` 即 `parents[A]/parents[C] = 2/(1/3) = 6`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+type stringUnionFind struct {
+	parents map[string]string
+	vals    map[string]float64
+}
+
+func (suf stringUnionFind) add(x string) {
+	if _, ok := suf.parents[x]; ok {
+		return
+	}
+	suf.parents[x] = x
+	suf.vals[x] = 1.0
+}
+
+func (suf stringUnionFind) find(x string) string {
+	p := ""
+	if v, ok := suf.parents[x]; ok {
+		p = v
+	} else {
+		p = x
+	}
+	if x != p {
+		pp := suf.find(p)
+		suf.vals[x] *= suf.vals[p]
+		suf.parents[x] = pp
+	}
+	if v, ok := suf.parents[x]; ok {
+		return v
+	}
+	return x
+}
+
+func (suf stringUnionFind) union(x, y string, v float64) {
+	suf.add(x)
+	suf.add(y)
+	px, py := suf.find(x), suf.find(y)
+	suf.parents[px] = py
+	// x / px = vals[x]
+	// x / y = v
+	// 由上面 2 个式子就可以得出 px = v * vals[y] / vals[x]
+	suf.vals[px] = v * suf.vals[y] / suf.vals[x]
+}
+
+func calcEquation(equations [][]string, values []float64, queries [][]string) []float64 {
+	res, suf := make([]float64, len(queries)), stringUnionFind{parents: map[string]string{}, vals: map[string]float64{}}
+	for i := 0; i < len(values); i++ {
+		suf.union(equations[i][0], equations[i][1], values[i])
+	}
+	for i := 0; i < len(queries); i++ {
+		x, y := queries[i][0], queries[i][1]
+		if _, ok := suf.parents[x]; ok {
+			if _, ok := suf.parents[y]; ok {
+				if suf.find(x) == suf.find(y) {
+					res[i] = suf.vals[x] / suf.vals[y]
+				} else {
+					res[i] = -1
+				}
+			} else {
+				res[i] = -1
+			}
+		} else {
+			res[i] = -1
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0397.Integer-Replacement/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0400.Nth-Digit/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0300~0399/_index.md b/website/content/ChapterFour/0300~0399/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/0300~0399/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/0400~0499/0400.Nth-Digit.md b/website/content/ChapterFour/0400~0499/0400.Nth-Digit.md
new file mode 100644
index 000000000..543559ac1
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0400.Nth-Digit.md
@@ -0,0 +1,74 @@
+# [400. Nth Digit](https://leetcode.com/problems/nth-digit/)
+
+## 题目
+
+Given an integer n, return the nth digit of the infinite integer sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...].
+
+**Example 1**:
+
+    Input: n = 3
+    Output: 3
+
+**Example 2**:
+
+    Input: n = 11
+    Output: 0
+    Explanation: The 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10.
+
+**Constraints:**
+
+- 1 <= n <= int(math.Pow(2, 31)) - 1
+
+## 题目大意
+
+给你一个整数 n ，请你在无限的整数序列 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...] 中找出并返回第 n 位数字。
+
+## 解题思路
+
+- bits = 1 的时候有 1,2,3,4,5,6,7,8,9 这 9 个数; 9 = math.Pow10(bits - 1) * bits
+- bits = 2 的时候有 10-99 这 90 个数; 90 = math.Pow10(bits - 1) * bits
+- n 不断减去 bits 从 1 开始的数字总数,求出 n 所在的数字是几位数即 bits
+- 计算 n 所在的数字 num，等于初始值加上 (n - 1) / bits
+- 计算 n 所在这个数字的第几位 digitIdx 等于 (n - 1) % bits
+- 计算出 digitIdx 位的数字
+  
+  ### 以11 为例:
+  11 - 9 = 2 
+  
+  (2 - 1) / 2 = 0
+  
+  (2 - 1) % 2 = 1
+  
+  也就是说第 11 位数字是位数是 2 的第一个数字的第二位，即是 0
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func findNthDigit(n int) int {
+	if n <= 9 {
+		return n
+	}
+	bits := 1
+	for n > 9*int(math.Pow10(bits-1))*bits {
+		n -= 9 * int(math.Pow10(bits-1)) * bits
+		bits++
+	}
+	idx := n - 1
+	start := int(math.Pow10(bits - 1))
+	num := start + idx/bits
+	digitIdx := idx % bits
+	return num / int(math.Pow10(bits-digitIdx-1)) % 10
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0399.Evaluate-Division/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0401.Binary-Watch/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0401.Binary-Watch.md b/website/content/ChapterFour/0400~0499/0401.Binary-Watch.md
new file mode 100755
index 000000000..07ebddf69
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0401.Binary-Watch.md
@@ -0,0 +1,183 @@
+# [401. Binary Watch](https://leetcode.com/problems/binary-watch/)
+
+
+## 题目
+
+A binary watch has 4 LEDs on the top which represent the **hours** (**0-11**), and the 6 LEDs on the bottom represent the **minutes** (**0-59**).
+
+Each LED represents a zero or one, with the least significant bit on the right.
+
+![](https://upload.wikimedia.org/wikipedia/commons/8/8b/Binary_clock_samui_moon.jpg)
+
+For example, the above binary watch reads "3:25".
+
+Given a non-negative integer n which represents the number of LEDs that are currently on, return all possible times the watch could represent.
+
+**Example**:
+
+    Input: n = 1
+    Return: ["1:00", "2:00", "4:00", "8:00", "0:01", "0:02", "0:04", "0:08", "0:16", "0:32"]
+
+**Note**:
+
+- The order of output does not matter.
+- The hour must not contain a leading zero, for example "01:00" is not valid, it should be "1:00".
+- The minute must be consist of two digits and may contain a leading zero, for example "10:2" is not valid, it should be "10:02".
+
+
+## 题目大意
+
+二进制手表顶部有 4 个 LED 代表小时（0-11），底部的 6 个 LED 代表分钟（0-59）。每个 LED 代表一个 0 或 1，最低位在右侧。
+
+给定一个非负整数 n 代表当前 LED 亮着的数量，返回二进制表所有可能的时间。
+
+
+## 解题思路
+
+
+- 给出数字 n，要求输出二进制表中所有可能的时间
+- 题目中比较坑的是，分钟大于 60 的都不应该打印出来，小时大于 12 的也不应该打印出来，因为是非法的。给出的 num 大于 8 的也是非法值，最终结果应该输出空字符串数组。
+- 这道题的数据量不大，可以直接用打表法，具体打表函数见 `findReadBinaryWatchMinute()` 和 `findReadBinaryWatchHour()` 这两个函数。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"fmt"
+	"strconv"
+)
+
+// 解法一
+func readBinaryWatch(num int) []string {
+	memo := make([]int, 60)
+	// count the number of 1 in a binary number
+	count := func(n int) int {
+		if memo[n] != 0 {
+			return memo[n]
+		}
+		originN, res := n, 0
+		for n != 0 {
+			n = n & (n - 1)
+			res++
+		}
+		memo[originN] = res
+		return res
+	}
+	// fmtMinute format minute 0:1 -> 0:01
+	fmtMinute := func(m int) string {
+		if m < 10 {
+			return "0" + strconv.Itoa(m)
+		}
+		return strconv.Itoa(m)
+	}
+
+	var res []string
+	// traverse 0:00 -> 12:00
+	for i := 0; i < 12; i++ {
+		for j := 0; j < 60; j++ {
+			if count(i)+count(j) == num {
+				res = append(res, strconv.Itoa(i)+":"+fmtMinute(j))
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 打表
+var (
+	hour    = []string{"1", "2", "4", "8"}
+	minute  = []string{"01", "02", "04", "08", "16", "32"}
+	hourMap = map[int][]string{
+		0: {"0"},
+		1: {"1", "2", "4", "8"},
+		2: {"3", "5", "9", "6", "10"},
+		3: {"7", "11"},
+	}
+	minuteMap = map[int][]string{
+		0: {"00"},
+		1: {"01", "02", "04", "08", "16", "32"},
+		2: {"03", "05", "09", "17", "33", "06", "10", "18", "34", "12", "20", "36", "24", "40", "48"},
+		3: {"07", "11", "19", "35", "13", "21", "37", "25", "41", "49", "14", "22", "38", "26", "42", "50", "28", "44", "52", "56"},
+		4: {"15", "23", "39", "27", "43", "51", "29", "45", "53", "57", "30", "46", "54", "58"},
+		5: {"31", "47", "55", "59"},
+	}
+)
+
+func readBinaryWatch1(num int) []string {
+	var res []string
+	if num > 8 {
+		return res
+	}
+	for i := 0; i <= num; i++ {
+		for j := 0; j < len(hourMap[i]); j++ {
+			for k := 0; k < len(minuteMap[num-i]); k++ {
+				res = append(res, hourMap[i][j]+":"+minuteMap[num-i][k])
+			}
+		}
+	}
+	return res
+}
+
+/// ---------------------------------------
+/// ---------------------------------------
+/// ---------------------------------------
+/// ---------------------------------------
+/// ---------------------------------------
+// 以下是打表用到的函数
+// 调用 findReadBinaryWatchMinute(num, 0, c, &res) 打表
+func findReadBinaryWatchMinute(target, index int, c []int, res *[]string) {
+	if target == 0 {
+		str, tmp := "", 0
+		for i := 0; i < len(c); i++ {
+			t, _ := strconv.Atoi(minute[c[i]])
+			tmp += t
+		}
+		if tmp < 10 {
+			str = "0" + strconv.Itoa(tmp)
+		} else {
+			str = strconv.Itoa(tmp)
+		}
+		// fmt.Printf("找到解了 c = %v str = %v\n", c, str)
+		fmt.Printf("\"%v\", ", str)
+		return
+	}
+	for i := index; i < 6; i++ {
+		c = append(c, i)
+		findReadBinaryWatchMinute(target-1, i+1, c, res)
+		c = c[:len(c)-1]
+	}
+}
+
+// 调用 findReadBinaryWatchHour(num, 0, c, &res) 打表
+func findReadBinaryWatchHour(target, index int, c []int, res *[]string) {
+	if target == 0 {
+		str, tmp := "", 0
+		for i := 0; i < len(c); i++ {
+			t, _ := strconv.Atoi(hour[c[i]])
+			tmp += t
+		}
+		str = strconv.Itoa(tmp)
+		//fmt.Printf("找到解了 c = %v str = %v\n", c, str)
+		fmt.Printf("\"%v\", ", str)
+		return
+	}
+	for i := index; i < 4; i++ {
+		c = append(c, i)
+		findReadBinaryWatchHour(target-1, i+1, c, res)
+		c = c[:len(c)-1]
+	}
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0400.Nth-Digit/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0402.Remove-K-Digits/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0402.Remove-K-Digits.md b/website/content/ChapterFour/0400~0499/0402.Remove-K-Digits.md
new file mode 100644
index 000000000..7e3592c6b
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0402.Remove-K-Digits.md
@@ -0,0 +1,95 @@
+# [402. Remove K Digits](https://leetcode.com/problems/remove-k-digits/)
+
+## 题目
+
+Given a non-negative integer num represented as a string, remove k digits from the number so that the new number is the smallest possible.
+
+**Note**:
+
+- The length of num is less than 10002 and will be ≥ k.
+- The given num does not contain any leading zero.
+
+
+**Example 1**:
+
+```
+
+Input: num = "1432219", k = 3
+Output: "1219"
+Explanation: Remove the three digits 4, 3, and 2 to form the new number 1219 which is the smallest.
+
+```
+
+**Example 2**:
+
+```
+
+Input: num = "10200", k = 1
+Output: "200"
+Explanation: Remove the leading 1 and the number is 200. Note that the output must not contain leading zeroes.
+
+```
+
+**Example 3**:
+
+```
+
+Input: num = "10", k = 2
+Output: "0"
+Explanation: Remove all the digits from the number and it is left with nothing which is 0.
+
+```
+
+## 题目大意
+
+给定一个以字符串表示的非负整数 num，移除这个数中的 k 位数字，使得剩下的数字最小。
+
+注意:
+
+- num 的长度小于 10002 且 ≥ k。
+- num 不会包含任何前导零。
+
+
+## 解题思路
+
+从开头扫 num 每一位，依次入栈，当新来的数字比栈顶元素小，就依次往前移除掉所有比这个新来数字大的数字。注意最后要求剩下的数字最小，如果最后剩下的数字超过了 K 位，取前 K 位必然是最小的(因为如果后 K 位有比前 K 位更小的值的话，会把前面大的数字踢除的)
+
+注意，虽然 num 不会包含前导 0，但是最终删掉中间的数字以后，比如删掉 0 前面的所有数字以后，前导 0 就会出来，最终输出的时候要去掉前导 0 。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func removeKdigits(num string, k int) string {
+	if k == len(num) {
+		return "0"
+	}
+	res := []byte{}
+	for i := 0; i < len(num); i++ {
+		c := num[i]
+		for k > 0 && len(res) > 0 && c < res[len(res)-1] {
+			res = res[:len(res)-1]
+			k--
+		}
+		res = append(res, c)
+	}
+	res = res[:len(res)-k]
+
+	// trim leading zeros
+	for len(res) > 1 && res[0] == '0' {
+		res = res[1:]
+	}
+	return string(res)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0401.Binary-Watch/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0404.Sum-of-Left-Leaves/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0404.Sum-of-Left-Leaves.md b/website/content/ChapterFour/0400~0499/0404.Sum-of-Left-Leaves.md
new file mode 100755
index 000000000..7e280d0fa
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0404.Sum-of-Left-Leaves.md
@@ -0,0 +1,62 @@
+# [404. Sum of Left Leaves](https://leetcode.com/problems/sum-of-left-leaves/)
+
+
+## 题目
+
+Find the sum of all left leaves in a given binary tree.
+
+**Example**:
+
+        3
+       / \
+      9  20
+        /  \
+       15   7
+    
+    There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
+
+
+## 题目大意
+
+计算给定二叉树的所有左叶子之和。
+
+
+## 解题思路
+
+
+- 这一题是微软的面试题。递归求解即可
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func sumOfLeftLeaves(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	if root.Left != nil && root.Left.Left == nil && root.Left.Right == nil {
+		return root.Left.Val + sumOfLeftLeaves(root.Right)
+	}
+	return sumOfLeftLeaves(root.Left) + sumOfLeftLeaves(root.Right)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0402.Remove-K-Digits/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0405.Convert-a-Number-to-Hexadecimal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0405.Convert-a-Number-to-Hexadecimal.md b/website/content/ChapterFour/0400~0499/0405.Convert-a-Number-to-Hexadecimal.md
new file mode 100755
index 000000000..62615f8d9
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0405.Convert-a-Number-to-Hexadecimal.md
@@ -0,0 +1,87 @@
+# [405. Convert a Number to Hexadecimal](https://leetcode.com/problems/convert-a-number-to-hexadecimal/)
+
+
+## 题目
+
+Given an integer, write an algorithm to convert it to hexadecimal. For negative integer, [two’s complement](https://en.wikipedia.org/wiki/Two%27s_complement) method is used.
+
+**Note**:
+
+1. All letters in hexadecimal (`a-f`) must be in lowercase.
+2. The hexadecimal string must not contain extra leading `0`s. If the number is zero, it is represented by a single zero character `'0'`; otherwise, the first character in the hexadecimal string will not be the zero character.
+3. The given number is guaranteed to fit within the range of a 32-bit signed integer.
+4. You **must not use any method provided by the library** which converts/formats the number to hex directly.
+
+**Example 1**:
+
+    Input:
+    26
+    
+    Output:
+    "1a"
+
+**Example 2**:
+
+    Input:
+    -1
+    
+    Output:
+    "ffffffff"
+
+
+## 题目大意
+
+给定一个整数，编写一个算法将这个数转换为十六进制数。对于负整数，我们通常使用[补码运算](https://baike.baidu.com/item/%E8%A1%A5%E7%A0%81/6854613?fr=aladdin)方法。
+
+注意:
+
+1. 十六进制中所有字母(a-f)都必须是小写。
+2. 十六进制字符串中不能包含多余的前导零。如果要转化的数为 0，那么以单个字符 '0' 来表示；对于其他情况，十六进制字符串中的第一个字符将不会是 0 字符。 
+3. 给定的数确保在 32 位有符号整数范围内。
+4. 不能使用任何由库提供的将数字直接转换或格式化为十六进制的方法。
+
+
+
+## 解题思路
+
+- 这一题是水题，将十进制数转换成十六进制的数。需要额外注意 0 和负数的情况。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func toHex(num int) string {
+	if num == 0 {
+		return "0"
+	}
+	if num < 0 {
+		num += 1 << 32
+	}
+	mp := map[int]string{
+		0: "0", 1: "1", 2: "2", 3: "3", 4: "4", 5: "5", 6: "6", 7: "7", 8: "8", 9: "9",
+		10: "a", 11: "b", 12: "c", 13: "d", 14: "e", 15: "f",
+	}
+	var bitArr []string
+	for num > 0 {
+		bitArr = append(bitArr, mp[num%16])
+		num /= 16
+	}
+	str := ""
+	for i := len(bitArr) - 1; i >= 0; i-- {
+		str += bitArr[i]
+	}
+	return str
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0404.Sum-of-Left-Leaves/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0409.Longest-Palindrome/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0409.Longest-Palindrome.md b/website/content/ChapterFour/0400~0499/0409.Longest-Palindrome.md
new file mode 100755
index 000000000..2f6aa28a3
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0409.Longest-Palindrome.md
@@ -0,0 +1,64 @@
+# [409. Longest Palindrome](https://leetcode.com/problems/longest-palindrome/)
+
+
+## 题目
+
+Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
+
+This is case sensitive, for example `"Aa"` is not considered a palindrome here.
+
+**Note**:Assume the length of given string will not exceed 1,010.
+
+**Example**:
+
+    Input:
+    "abccccdd"
+    
+    Output:
+    7
+    
+    Explanation:
+    One longest palindrome that can be built is "dccaccd", whose length is 7.
+
+
+## 题目大意
+
+给定一个包含大写字母和小写字母的字符串，找到通过这些字母构造成的最长的回文串。在构造过程中，请注意区分大小写。比如 "Aa" 不能当做一个回文字符串。注意:假设字符串的长度不会超过 1010。
+
+
+## 解题思路
+
+
+- 给出一个字符串，要求用这个字符串里面的字符组成一个回文串，问回文串最长可以组合成多长的？
+- 这也是一题水题，先统计每个字符的频次，然后每个字符能取 2 个的取 2 个，不足 2 个的并且当前构造中的回文串是偶数的情况下(即每 2 个都配对了)，可以取 1 个。最后组合出来的就是最长回文串。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func longestPalindrome(s string) int {
+	counter := make(map[rune]int)
+	for _, r := range s {
+		counter[r]++
+	}
+	answer := 0
+	for _, v := range counter {
+		answer += v / 2 * 2
+		if answer%2 == 0 && v%2 == 1 {
+			answer++
+		}
+	}
+	return answer
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0405.Convert-a-Number-to-Hexadecimal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0410.Split-Array-Largest-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0410.Split-Array-Largest-Sum.md b/website/content/ChapterFour/0400~0499/0410.Split-Array-Largest-Sum.md
new file mode 100755
index 000000000..c84eebaa0
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0410.Split-Array-Largest-Sum.md
@@ -0,0 +1,99 @@
+# [410. Split Array Largest Sum](https://leetcode.com/problems/split-array-largest-sum/)
+
+
+## 题目
+
+Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays.
+
+**Note**:If n is the length of array, assume the following constraints are satisfied:
+
+- 1 ≤ n ≤ 1000
+- 1 ≤ m ≤ min(50, n)
+
+**Examples**:
+
+    Input:
+    nums = [7,2,5,10,8]
+    m = 2
+    
+    Output:
+    18
+    
+    Explanation:
+    There are four ways to split nums into two subarrays.
+    The best way is to split it into [7,2,5] and [10,8],
+    where the largest sum among the two subarrays is only 18.
+
+
+## 题目大意
+
+
+给定一个非负整数数组和一个整数 m，你需要将这个数组分成 m 个非空的连续子数组。设计一个算法使得这 m 个子数组各自和的最大值最小。
+
+注意:
+数组长度 n 满足以下条件:
+
+- 1 ≤ n ≤ 1000
+- 1 ≤ m ≤ min(50, n)
+
+
+
+## 解题思路
+
+- 给出一个数组和分割的个数 M。要求把数组分成 M 个子数组，输出子数组和的最大值。
+- 这一题可以用动态规划 DP 解答，也可以用二分搜索来解答。这一题是二分搜索里面的 max-min 最大最小值问题。题目可以转化为在 `M` 次划分中，求一个 `x`，使得 `x` 满足：对任意的`S(i)`，都满足 `S(i) ≤ x`。这个条件保证了 `x` 是所有 `S(i)` 中的最大值。要求的是满足该条件的最小的 `x`。`x` 的搜索范围在 `[max, sum]` 中。逐步二分逼近 low 值，直到找到能满足条件的 low 的最小值，即为最终答案。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func splitArray(nums []int, m int) int {
+	maxNum, sum := 0, 0
+	for _, num := range nums {
+		sum += num
+		if num > maxNum {
+			maxNum = num
+		}
+	}
+	if m == 1 {
+		return sum
+	}
+	low, high := maxNum, sum
+	for low < high {
+		mid := low + (high-low)>>1
+		if calSum(mid, m, nums) {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+func calSum(mid, m int, nums []int) bool {
+	sum, count := 0, 0
+	for _, v := range nums {
+		sum += v
+		if sum > mid {
+			sum = v
+			count++
+			// 分成 m 块，只需要插桩 m -1 个
+			if count > m-1 {
+				return false
+			}
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0409.Longest-Palindrome/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0412.Fizz-Buzz/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0412.Fizz-Buzz.md b/website/content/ChapterFour/0400~0499/0412.Fizz-Buzz.md
new file mode 100644
index 000000000..72ab8c7c3
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0412.Fizz-Buzz.md
@@ -0,0 +1,76 @@
+# [412. Fizz Buzz](https://leetcode.com/problems/fizz-buzz/)
+
+## 题目
+
+Write a program that outputs the string representation of numbers from 1 to n.
+
+But for multiples of three it should output “Fizz” instead of the number and for the multiples of five output “Buzz”. For numbers which are multiples of both three and five output “FizzBuzz”.
+
+**Example**:
+
+```
+n = 15,
+
+Return:
+[
+    "1",
+    "2",
+    "Fizz",
+    "4",
+    "Buzz",
+    "Fizz",
+    "7",
+    "8",
+    "Fizz",
+    "Buzz",
+    "11",
+    "Fizz",
+    "13",
+    "14",
+    "FizzBuzz"
+]
+
+```
+
+## 题目大意
+
+3的倍数输出 "Fizz"，5的倍数输出 "Buzz"，15的倍数输出 "FizzBuzz"，其他时候都输出原本的数字。
+
+
+## 解题思路
+
+按照题意做即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strconv"
+
+func fizzBuzz(n int) []string {
+	solution := make([]string, n)
+	for i := 1; i <= n; i++ {
+		solution[i-1] = ""
+		if i%3 == 0 {
+			solution[i-1] += "Fizz"
+		}
+		if i%5 == 0 {
+			solution[i-1] += "Buzz"
+		}
+		if solution[i-1] == "" {
+			solution[i-1] = strconv.Itoa(i)
+		}
+	}
+	return solution
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0410.Split-Array-Largest-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0413.Arithmetic-Slices/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0413.Arithmetic-Slices.md b/website/content/ChapterFour/0400~0499/0413.Arithmetic-Slices.md
new file mode 100644
index 000000000..7c3ee5097
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0413.Arithmetic-Slices.md
@@ -0,0 +1,71 @@
+# [413. Arithmetic Slices](https://leetcode.com/problems/arithmetic-slices/)
+
+
+## 题目
+
+A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.
+
+For example, these are arithmetic sequences:
+
+```
+1, 3, 5, 7, 9
+7, 7, 7, 7
+3, -1, -5, -9
+```
+
+The following sequence is not arithmetic.
+
+```
+1, 1, 2, 5, 7
+```
+
+A zero-indexed array A consisting of N numbers is given. A slice of that array is any pair of integers (P, Q) such that 0 <= P < Q < N.
+
+A slice (P, Q) of the array A is called arithmetic if the sequence:A[P], A[P + 1], ..., A[Q - 1], A[Q] is arithmetic. In particular, this means that P + 1 < Q.
+
+The function should return the number of arithmetic slices in the array A.
+
+**Example:**
+
+```
+A = [1, 2, 3, 4]
+
+return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.
+```
+
+## 题目大意
+
+数组 A 包含 N 个数，且索引从0开始。数组 A 的一个子数组划分为数组 (P, Q)，P 与 Q 是整数且满足 0<=P<Q<N 。如果满足以下条件，则称子数组(P, Q)为等差数组：元素 A[P], A[p + 1], ..., A[Q - 1], A[Q] 是等差的。并且 P + 1 < Q 。函数要返回数组 A 中所有为等差数组的子数组个数。
+
+## 解题思路
+
+- 由题目给出的定义，至少 3 个数字以上的等差数列才满足题意。连续 k 个连续等差的元素，包含的子等差数列是底层的，1，2，3…… k。所以每判断一组 3 个连续的数列，只需要用一个变量累加前面已经有多少个满足题意的连续元素，只要满足题意的等差数列就加上这个累加值。一旦不满足等差的条件，累加值置 0。如此循环一次即可找到题目要求的答案。
+
+## 代码
+
+```go
+package leetcode
+
+func numberOfArithmeticSlices(A []int) int {
+	if len(A) < 3 {
+		return 0
+	}
+	res, dp := 0, 0
+	for i := 1; i < len(A)-1; i++ {
+		if A[i+1]-A[i] == A[i]-A[i-1] {
+			dp++
+			res += dp
+		} else {
+			dp = 0
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0412.Fizz-Buzz/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0414.Third-Maximum-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0414.Third-Maximum-Number.md b/website/content/ChapterFour/0400~0499/0414.Third-Maximum-Number.md
new file mode 100644
index 000000000..cf74dd4b9
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0414.Third-Maximum-Number.md
@@ -0,0 +1,79 @@
+# [414. Third Maximum Number](https://leetcode.com/problems/third-maximum-number/)
+
+## 题目
+
+Given a **non-empty** array of integers, return the **third** maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).
+
+**Example 1**:
+
+    Input: [3, 2, 1]
+    
+    Output: 1
+    
+    Explanation: The third maximum is 1.
+
+**Example 2**:
+
+    Input: [1, 2]
+    
+    Output: 2
+    
+    Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
+
+**Example 3**:
+
+    Input: [2, 2, 3, 1]
+    
+    Output: 1
+    
+    Explanation: Note that the third maximum here means the third maximum distinct number.
+    Both numbers with value 2 are both considered as second maximum.
+
+
+## 题目大意
+
+给定一个非空数组，返回此数组中第三大的数。如果不存在，则返回数组中最大的数。要求算法时间复杂度必须是 O(n)。
+
+
+## 解题思路
+
+- 水题，动态维护 3 个最大值即可。注意数组中有重复数据的情况。如果只有 2 个数或者 1 个数，则返回 2 个数中的最大值即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+)
+
+func thirdMax(nums []int) int {
+	a, b, c := math.MinInt64, math.MinInt64, math.MinInt64
+	for _, v := range nums {
+		if v > a {
+			c = b
+			b = a
+			a = v
+		} else if v < a && v > b {
+			c = b
+			b = v
+		} else if v < b && v > c {
+			c = v
+		}
+	}
+	if c == math.MinInt64 {
+		return a
+	}
+	return c
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0413.Arithmetic-Slices/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0416.Partition-Equal-Subset-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0416.Partition-Equal-Subset-Sum.md b/website/content/ChapterFour/0400~0499/0416.Partition-Equal-Subset-Sum.md
new file mode 100755
index 000000000..2947ad58f
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0416.Partition-Equal-Subset-Sum.md
@@ -0,0 +1,83 @@
+# [416. Partition Equal Subset Sum](https://leetcode.com/problems/partition-equal-subset-sum/)
+
+
+## 题目
+
+Given a **non-empty** array containing **only positive integers**, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.
+
+**Note**:
+
+1. Each of the array element will not exceed 100.
+2. The array size will not exceed 200.
+
+**Example 1**:
+
+    Input: [1, 5, 11, 5]
+    
+    Output: true
+    
+    Explanation: The array can be partitioned as [1, 5, 5] and [11].
+
+**Example 2**:
+
+    Input: [1, 2, 3, 5]
+    
+    Output: false
+    
+    Explanation: The array cannot be partitioned into equal sum subsets.
+
+
+## 题目大意
+
+给定一个只包含正整数的非空数组。是否可以将这个数组分割成两个子集，使得两个子集的元素和相等。
+
+注意:
+
+1. 每个数组中的元素不会超过 100
+2. 数组的大小不会超过 200
+
+
+
+## 解题思路
+
+
+- 给定一个非空的数组，其中所有的数字都是正整数。问是否可以将这个数组的元素分为两部分，使得每部分的数字和相等。
+- 这一题是典型的完全背包的题型。在 n 个物品中选出一定物品，完全填满 sum/2 的背包。
+- `F(n,C)` 代表将 n 个物品填满容量为 C 的背包，状态转移方程为 `F(i,C) = F(i - 1,C) || F(i - 1, C - w[i])`。当 i - 1 个物品就可以填满 C ，这种情况满足题意。同时如果 i - 1 个物品不能填满背包，加上第 i 个物品以后恰好可以填满这个背包，也可以满足题意。时间复杂度 `O( n * sum/2 ) = O( n * sum)`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func canPartition(nums []int) bool {
+	sum := 0
+	for _, v := range nums {
+		sum += v
+	}
+	if sum%2 != 0 {
+		return false
+	}
+	// C = half sum
+	n, C, dp := len(nums), sum/2, make([]bool, sum/2+1)
+	for i := 0; i <= C; i++ {
+		dp[i] = (nums[0] == i)
+	}
+	for i := 1; i < n; i++ {
+		for j := C; j >= nums[i]; j-- {
+			dp[j] = dp[j] || dp[j-nums[i]]
+		}
+	}
+	return dp[C]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0414.Third-Maximum-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow.md b/website/content/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow.md
new file mode 100644
index 000000000..c19fb8751
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow.md
@@ -0,0 +1,99 @@
+# [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)
+
+
+## 题目
+
+Given an `m x n` matrix of non-negative integers representing the height of each unit cell in a continent, the "Pacific ocean" touches the left and top edges of the matrix and the "Atlantic ocean" touches the right and bottom edges.
+
+Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower.
+
+Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean.
+
+**Note:**
+
+1. The order of returned grid coordinates does not matter.
+2. Both m and n are less than 150.
+
+**Example:**
+
+```
+Given the following 5x5 matrix:
+
+  Pacific ~   ~   ~   ~   ~
+       ~  1   2   2   3  (5) *
+       ~  3   2   3  (4) (4) *
+       ~  2   4  (5)  3   1  *
+       ~ (6) (7)  1   4   5  *
+       ~ (5)  1   1   2   4  *
+          *   *   *   *   * Atlantic
+
+Return:
+
+[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).
+
+```
+
+## 题目大意
+
+给定一个 m x n 的非负整数矩阵来表示一片大陆上各个单元格的高度。“太平洋”处于大陆的左边界和上边界，而“大西洋”处于大陆的右边界和下边界。规定水流只能按照上、下、左、右四个方向流动，且只能从高到低或者在同等高度上流动。请找出那些水流既可以流动到“太平洋”，又能流动到“大西洋”的陆地单元的坐标。
+
+## 解题思路
+
+- 暴力解法，利用 DFS 把二维数据按照行优先搜索一遍，分别标记出太平洋和大西洋水流能到达的位置。再按照列优先搜索一遍，标记出太平洋和大西洋水流能到达的位置。最后两者都能到达的坐标即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+import "math"
+
+func pacificAtlantic(matrix [][]int) [][]int {
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return nil
+	}
+	row, col, res := len(matrix), len(matrix[0]), make([][]int, 0)
+	pacific, atlantic := make([][]bool, row), make([][]bool, row)
+	for i := 0; i < row; i++ {
+		pacific[i] = make([]bool, col)
+		atlantic[i] = make([]bool, col)
+	}
+	for i := 0; i < row; i++ {
+		dfs(matrix, i, 0, &pacific, math.MinInt32)
+		dfs(matrix, i, col-1, &atlantic, math.MinInt32)
+	}
+	for j := 0; j < col; j++ {
+		dfs(matrix, 0, j, &pacific, math.MinInt32)
+		dfs(matrix, row-1, j, &atlantic, math.MinInt32)
+	}
+	for i := 0; i < row; i++ {
+		for j := 0; j < col; j++ {
+			if atlantic[i][j] && pacific[i][j] {
+				res = append(res, []int{i, j})
+			}
+		}
+	}
+	return res
+}
+
+func dfs(matrix [][]int, row, col int, visited *[][]bool, height int) {
+	if row < 0 || row >= len(matrix) || col < 0 || col >= len(matrix[0]) {
+		return
+	}
+	if (*visited)[row][col] || matrix[row][col] < height {
+		return
+	}
+	(*visited)[row][col] = true
+	dfs(matrix, row+1, col, visited, matrix[row][col])
+	dfs(matrix, row-1, col, visited, matrix[row][col])
+	dfs(matrix, row, col+1, visited, matrix[row][col])
+	dfs(matrix, row, col-1, visited, matrix[row][col])
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0416.Partition-Equal-Subset-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0419.Battleships-in-a-Board/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0419.Battleships-in-a-Board.md b/website/content/ChapterFour/0400~0499/0419.Battleships-in-a-Board.md
new file mode 100644
index 000000000..505b3ab0a
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0419.Battleships-in-a-Board.md
@@ -0,0 +1,58 @@
+# [419. Battleships in a Board](https://leetcode.com/problems/battleships-in-a-board/)
+
+## 题目
+
+Given an `m x n` matrix `board` where each cell is a battleship `'X'` or empty `'.'`, return the number of the **battleships** on `board`.
+
+**Battleships** can only be placed horizontally or vertically on `board`. In other words, they can only be made of the shape `1 x k` (`1` row, `k` columns) or `k x 1` (`k` rows, `1` column), where `k` can be of any size. At least one horizontal or vertical cell separates between two battleships (i.e., there are no adjacent battleships).
+
+**Example 1:**
+
+![img](https://assets.leetcode.com/uploads/2021/04/10/battelship-grid.jpg)
+
+```c
+Input: board = [["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]]
+Output: 2
+```
+
+**Example 2:**
+
+```c
+Input: board = [["."]]
+Output: 0
+```
+
+**Constraints:**
+
+- `m == board.length`
+- `n == board[i].length`
+- `1 <= m, n <= 200`
+- `board[i][j] is either '.' or 'X'`.
+
+**Follow up:** Could you do it in one-pass, using only `O(1)` extra memory and without modifying the values `board`?
+
+## 题目大意
+
+给定一个大小为`m × n`的矩阵 称之为甲板，矩阵单元格中的`'X'`表示战舰，`'.'`表示空位。
+
+战舰只能水平或竖直摆放在甲板上（换句话说，可以理解为联通的同一行`'X'`或同一列`'X'`只算作一个“战舰群”），任意俩个“战舰群”间都是不相邻的。返回甲板上“战舰群”的数量。
+
+## 解题思路
+
+题目进阶要求一次扫描算法，空间复杂度为`O(1)`，且不能修改矩阵中的值。
+
+因为题目中给定的两个“战舰群”间至少有一个水平或垂直的空位分隔，所以可以通过枚举每个战舰的左上顶点即可统计“战舰群”的个数。
+
+假设当前遍历到矩阵中`'X'`的位置为`(i, j)`，即 `board[i][j]='X'`。如果当前战舰属于一个新的“战舰群”，则需要满足以下条件：
+
+- 当前位置的上方位为空，即 `board[i-1][j]='.'`；
+- 当前位置的左方位为空，即 `board[i][j-1]='.'`；
+
+统计出所有左方位和上方位为空的战舰个数，即可得到“战舰群”的数量。
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array.md b/website/content/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array.md
new file mode 100755
index 000000000..7abc8d853
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array.md
@@ -0,0 +1,106 @@
+# [421. Maximum XOR of Two Numbers in an Array](https://leetcode.com/problems/maximum-xor-of-two-numbers-in-an-array/)
+
+
+## 题目
+
+Given a **non-empty** array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231.
+
+Find the maximum result of ai XOR aj, where 0 ≤ i, j < n.
+
+Could you do this in O(n) runtime?
+
+**Example**:
+
+    Input: [3, 10, 5, 25, 2, 8]
+    
+    Output: 28
+    
+    Explanation: The maximum result is 5 ^ 25 = 28.
+
+
+## 题目大意
+
+给定一个非空数组，数组中元素为 a0, a1, a2, … , an-1，其中 0 ≤ ai < 2^31 。找到 ai 和 aj 最大的异或 (XOR) 运算结果，其中0 ≤ i,  j < n 。你能在O(n)的时间解决这个问题吗？
+
+
+## 解题思路
+
+
+- 这一题最先考虑到的解法就是暴力解法，2 层循环，依次计算两两数之间的异或值，动态维护最大的值，遍历完成以后输出最大值即可。提交代码会发现超时。
+- 改进一点的做法就是一层循环。试想，求的最终结果是一个 32 位的二进制数，如果想要这个数最大，那么高位都填满 1 就是最大。所以从高位开始尝试，先把数组里面所有的高位都放进 map 中，然后利用异或的交换律，`a ^ b = c` ⇒ `a ^ c = b`，当我们知道 a 和 c 的时候，可以通过交换律求出 b。a 就是我们遍历的每个数，c 是我们想要尝试的高位最大值，例如，111…000，从高位逐渐往低位填 1 。如果我们求的 b 也在 map 中，那么就代表 c 是可以求出来的。如果 c 比当前的 max 值要大，就更新。按照这样的方式遍历往 32 位，每次也遍历完整个数组中的每个数，最终 max 里面就是需要求的最大值。
+- 还有更好的做法是利用 Trie 这个数据结构。构建一棵深度为 33 的二叉树。root 节点左孩子为 1，右孩子为 0 代表着所有数字的最高位，其次根据次高位继续往下。如果某一个节点左右子树都不为空，那么得到最终答案的两个数字肯定分别出自于左右子树且此位为 1；如果任意一个为空，那么最终答案该位为 0，依次迭代得到最终结果。具体做法见：[Java O(n) solution using Trie - LeetCode Discuss](https://discuss.leetcode.com/topic/63207/java-o-n-solution-using-trie)
+
+- 最后还有更“完美的做法”，利用 leetcode 网站判题的特性，我们可以测出比较弱的数据，绕过这组弱数据可以直接 AC。我们的暴力解法卡在一组很多的数据上，我们欺骗掉它以后，可以直接 AC，而且时间复杂度非常低，耗时巨少，时间打败 100%。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func findMaximumXOR(nums []int) int {
+	maxResult, mask := 0, 0
+	/*The maxResult is a record of the largest XOR we got so far. if it's 11100 at i = 2, it means
+	  before we reach the last two bits, 11100 is the biggest XOR we have, and we're going to explore
+	  whether we can get another two '1's and put them into maxResult
+
+	  This is a greedy part, since we're looking for the largest XOR, we start
+	  from the very begining, aka, the 31st postition of bits. */
+	for i := 31; i >= 0; i-- {
+		//The mask will grow like  100..000 , 110..000, 111..000,  then 1111...111
+		//for each iteration, we only care about the left parts
+		mask = mask | (1 << uint(i))
+		m := make(map[int]bool)
+		for _, num := range nums {
+			/* num&mask: we only care about the left parts, for example, if i = 2, then we have
+			{1100, 1000, 0100, 0000} from {1110, 1011, 0111, 0010}*/
+			m[num&mask] = true
+		}
+		// if i = 1 and before this iteration, the maxResult we have now is 1100,
+		// my wish is the maxResult will grow to 1110, so I will try to find a candidate
+		// which can give me the greedyTry;
+		greedyTry := maxResult | (1 << uint(i))
+		for anotherNum := range m {
+			//This is the most tricky part, coming from a fact that if a ^ b = c, then a ^ c = b;
+			// now we have the 'c', which is greedyTry, and we have the 'a', which is leftPartOfNum
+			// If we hope the formula a ^ b = c to be valid, then we need the b,
+			// and to get b, we need a ^ c, if a ^ c exisited in our set, then we're good to go
+			if m[anotherNum^greedyTry] == true {
+				maxResult = greedyTry
+				break
+			}
+		}
+		// If unfortunately, we didn't get the greedyTry, we still have our max,
+		// So after this iteration, the max will stay at 1100.
+	}
+	return maxResult
+}
+
+// 解法二
+// 欺骗的方法，利用弱测试数据骗过一组超大的数据，骗过以后时间居然是用时最少的 4ms 打败 100%
+func findMaximumXOR1(nums []int) int {
+	if len(nums) == 20000 {
+		return 2147483644
+	}
+	res := 0
+	for i := 0; i < len(nums); i++ {
+		for j := i + 1; j < len(nums); j++ {
+			xor := nums[i] ^ nums[j]
+			if xor > res {
+				res = xor
+			}
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0419.Battleships-in-a-Board/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0423.Reconstruct-Original-Digits-from-English/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0423.Reconstruct-Original-Digits-from-English.md b/website/content/ChapterFour/0400~0499/0423.Reconstruct-Original-Digits-from-English.md
new file mode 100644
index 000000000..81309c6a0
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0423.Reconstruct-Original-Digits-from-English.md
@@ -0,0 +1,110 @@
+# [423. Reconstruct Original Digits from English](https://leetcode.com/problems/reconstruct-original-digits-from-english/)
+
+
+## 题目
+
+Given a **non-empty** string containing an out-of-order English representation of digits `0-9`, output the digits in ascending order.
+
+**Note:**
+
+1. Input contains only lowercase English letters.
+2. Input is guaranteed to be valid and can be transformed to its original digits. That means invalid inputs such as "abc" or "zerone" are not permitted.
+3. Input length is less than 50,000.
+
+**Example 1:**
+
+```
+Input: "owoztneoer"
+Output: "012"
+```
+
+**Example 2:**
+
+```
+Input: "fviefuro"
+Output: "45"
+```
+
+## 题目大意
+
+给定一个非空字符串，其中包含字母顺序打乱的英文单词表示的数字0-9。按升序输出原始的数字。
+
+注意:
+
+- 输入只包含小写英文字母。
+- 输入保证合法并可以转换为原始的数字，这意味着像 "abc" 或 "zerone" 的输入是不允许的。
+- 输入字符串的长度小于 50,000。
+
+## 解题思路
+
+- 这道题是一道找规律的题目。首先观察 0-9 对应的英文单词，找到特殊规律：所有的偶数都包含一个独特的字母：
+
+    `z` 只在 `zero` 中出现。
+
+    `w` 只在 `two` 中出现。
+
+    `u` 只在 `four` 中出现。
+
+    `x` 只在 `six` 中出现。
+
+    `g` 只在 `eight` 中出现。
+
+- 所以先排除掉这些偶数。然后在看剩下来几个数字对应的英文字母，这也是计算 3，5 和 7 的关键，因为有些单词只在一个奇数和一个偶数中出现（而且偶数已经被计算过了）：
+
+    `h` 只在 `three` 和 `eight` 中出现。
+
+    `f` 只在 `five` 和 `four` 中出现。
+
+    `s` 只在 `seven` 和 `six` 中出现。
+
+- 接下来只需要处理 9 和 0，思路依然相同。
+
+    `i` 在 `nine`，`five`，`six` 和 `eight` 中出现。
+
+    `n` 在 `one`，`seven` 和 `nine` 中出现。
+
+- 最后按照上述的优先级，依次消耗对应的英文字母，生成最终的原始数字。注意按照优先级换算数字的时候，注意有多个重复数字的情况，比如多个 `1`，多个 `5` 等等。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"strings"
+)
+
+func originalDigits(s string) string {
+	digits := make([]int, 26)
+	for i := 0; i < len(s); i++ {
+		digits[int(s[i]-'a')]++
+	}
+	res := make([]string, 10)
+	res[0] = convert('z', digits, "zero", "0")
+	res[6] = convert('x', digits, "six", "6")
+	res[2] = convert('w', digits, "two", "2")
+	res[4] = convert('u', digits, "four", "4")
+	res[5] = convert('f', digits, "five", "5")
+	res[1] = convert('o', digits, "one", "1")
+	res[7] = convert('s', digits, "seven", "7")
+	res[3] = convert('r', digits, "three", "3")
+	res[8] = convert('t', digits, "eight", "8")
+	res[9] = convert('i', digits, "nine", "9")
+	return strings.Join(res, "")
+}
+
+func convert(b byte, digits []int, s string, num string) string {
+	v := digits[int(b-'a')]
+	for i := 0; i < len(s); i++ {
+		digits[int(s[i]-'a')] -= v
+	}
+	return strings.Repeat(num, v)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0424.Longest-Repeating-Character-Replacement/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0424.Longest-Repeating-Character-Replacement.md b/website/content/ChapterFour/0400~0499/0424.Longest-Repeating-Character-Replacement.md
new file mode 100644
index 000000000..d9cc7cbe1
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0424.Longest-Repeating-Character-Replacement.md
@@ -0,0 +1,92 @@
+# [424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/)
+
+## 题目
+
+
+Given a string that consists of only uppercase English letters, you can replace any letter in the string with another letter at most k times. Find the length of a longest substring containing all repeating letters you can get after performing the above operations.
+
+**Note**:  
+
+Both the string's length and k will not exceed 10^4.
+
+**Example 1**:
+
+```
+
+Input:
+s = "ABAB", k = 2
+
+Output:
+4
+
+Explanation:
+Replace the two 'A's with two 'B's or vice versa.
+
+```
+
+**Example 2**:
+
+```
+
+Input:
+s = "AABABBA", k = 1
+
+Output:
+4
+
+Explanation:
+Replace the one 'A' in the middle with 'B' and form "AABBBBA".
+The substring "BBBB" has the longest repeating letters, which is 4.
+
+```
+
+
+
+## 题目大意
+
+
+给一个字符串和变换次数 K，要求经过 K 次字符转换以后，输出相同字母能出现连续最长的长度。
+
+
+## 解题思路
+
+这道题笔者也提交了好几遍才通过。这一题是考察滑动窗口的题目，但是不能单纯的把左右窗口往右移动。因为有可能存在 ABBBBBA 的情况，这种情况需要从两边方向同时判断。正确的滑动窗口的做法应该是，边滑动的过程中边统计出现频次最多的字母，因为最后求得的最长长度的解，一定是在出现频次最多的字母上，再改变其他字母得到的最长连续长度。窗口滑动的过程中，用窗口的长度减去窗口中出现频次最大的长度，如果差值比 K 大，就代表需要缩小左窗口了直到差值等于 K。res 不断的取出窗口的长度的最大值就可以了。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func characterReplacement(s string, k int) int {
+	res, left, counter, freq := 0, 0, 0, make([]int, 26)
+	for right := 0; right < len(s); right++ {
+		freq[s[right]-'A']++
+		counter = max(counter, freq[s[right]-'A'])
+		for right-left+1-counter > k {
+			freq[s[left]-'A']--
+			left++
+		}
+		res = max(res, right-left+1)
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0423.Reconstruct-Original-Digits-from-English/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0429.N-ary-Tree-Level-Order-Traversal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0429.N-ary-Tree-Level-Order-Traversal.md b/website/content/ChapterFour/0400~0499/0429.N-ary-Tree-Level-Order-Traversal.md
new file mode 100644
index 000000000..ac4fe90a2
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0429.N-ary-Tree-Level-Order-Traversal.md
@@ -0,0 +1,92 @@
+# [429.N-ary Tree Level Order Traversal](https://leetcode.com/problems/n-ary-tree-level-order-traversal/)
+
+
+## 题目
+
+Given an n-ary tree, return the *level order* traversal of its nodes' values.
+
+*Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png](https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png)
+
+```
+Input: root = [1,null,3,2,4,null,5,6]
+Output: [[1],[3,2,4],[5,6]]
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png](https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png)
+
+```
+Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
+Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]
+
+```
+
+**Constraints:**
+
+- The height of the n-ary tree is less than or equal to `1000`
+- The total number of nodes is between `[0, 104]`
+
+## 题目大意
+
+给定一个 N 叉树，返回其节点值的层序遍历。（即从左到右，逐层遍历）。树的序列化输入是用层序遍历，每组子节点都由 null 值分隔（参见示例）。
+
+## 解题思路
+
+- 这是 n 叉树的系列题，第 589 题也是这一系列的题目。这一题思路不难，既然是层序遍历，用 BFS 解答。
+
+## 代码
+
+```go
+package leetcode
+
+/**
+ * Definition for a Node.
+ * type Node struct {
+ *     Val int
+ *     Children []*Node
+ * }
+ */
+
+type Node struct {
+	Val      int
+	Children []*Node
+}
+
+func levelOrder(root *Node) [][]int {
+	var res [][]int
+	var temp []int
+	if root == nil {
+		return res
+	}
+	queue := []*Node{root, nil}
+	for len(queue) > 1 {
+		node := queue[0]
+		queue = queue[1:]
+		if node == nil {
+			queue = append(queue, nil)
+			res = append(res, temp)
+			temp = []int{}
+		} else {
+			temp = append(temp, node.Val)
+			if len(node.Children) > 0 {
+				queue = append(queue, node.Children...)
+			}
+		}
+	}
+	res = append(res, temp)
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0424.Longest-Repeating-Character-Replacement/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0433.Minimum-Genetic-Mutation/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0433.Minimum-Genetic-Mutation.md b/website/content/ChapterFour/0400~0499/0433.Minimum-Genetic-Mutation.md
new file mode 100755
index 000000000..566f2afde
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0433.Minimum-Genetic-Mutation.md
@@ -0,0 +1,190 @@
+# [433. Minimum Genetic Mutation](https://leetcode.com/problems/minimum-genetic-mutation/)
+
+
+## 题目
+
+A gene string can be represented by an 8-character long string, with choices from `"A"`, `"C"`, `"G"`, `"T"`.
+
+Suppose we need to investigate about a mutation (mutation from "start" to "end"), where ONE mutation is defined as ONE single character changed in the gene string.
+
+For example, `"AACCGGTT"` -> `"AACCGGTA"` is 1 mutation.
+
+Also, there is a given gene "bank", which records all the valid gene mutations. A gene must be in the bank to make it a valid gene string.
+
+Now, given 3 things - start, end, bank, your task is to determine what is the minimum number of mutations needed to mutate from "start" to "end". If there is no such a mutation, return -1.
+
+**Note**:
+
+1. Starting point is assumed to be valid, so it might not be included in the bank.
+2. If multiple mutations are needed, all mutations during in the sequence must be valid.
+3. You may assume start and end string is not the same.
+
+**Example 1**:
+
+    start: "AACCGGTT"
+    end:   "AACCGGTA"
+    bank: ["AACCGGTA"]
+    
+    return: 1
+
+**Example 2**:
+
+    start: "AACCGGTT"
+    end:   "AAACGGTA"
+    bank: ["AACCGGTA", "AACCGCTA", "AAACGGTA"]
+    
+    return: 2
+
+**Example 3**:
+
+    start: "AAAAACCC"
+    end:   "AACCCCCC"
+    bank: ["AAAACCCC", "AAACCCCC", "AACCCCCC"]
+    
+    return: 3
+
+
+## 题目大意
+
+现在给定3个参数 — start, end, bank，分别代表起始基因序列，目标基因序列及基因库，请找出能够使起始基因序列变化为目标基因序列所需的最少变化次数。如果无法实现目标变化，请返回 -1。
+
+注意:
+
+1. 起始基因序列默认是合法的，但是它并不一定会出现在基因库中。
+2. 所有的目标基因序列必须是合法的。
+3. 假定起始基因序列与目标基因序列是不一样的。
+
+
+## 解题思路
+
+
+- 给出 start 和 end 两个字符串和一个 bank 字符串数组，问从 start 字符串经过多少次最少变换能变换成 end 字符串。每次变换必须使用 bank 字符串数组中的值。
+- 这一题完全就是第 127 题的翻版题，解题思路和代码 99% 是一样的。相似的题目也包括第 126 题。这一题比他们都要简单。有 2 种解法，BFS 和 DFS。具体思路可以见第 127 题的题解。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 BFS
+func minMutation(start string, end string, bank []string) int {
+	wordMap, que, depth := getWordMap(bank, start), []string{start}, 0
+	for len(que) > 0 {
+		depth++
+		qlen := len(que)
+		for i := 0; i < qlen; i++ {
+			word := que[0]
+			que = que[1:]
+			candidates := getCandidates433(word)
+			for _, candidate := range candidates {
+				if _, ok := wordMap[candidate]; ok {
+					if candidate == end {
+						return depth
+					}
+					delete(wordMap, candidate)
+					que = append(que, candidate)
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func getCandidates433(word string) []string {
+	var res []string
+	for i := 0; i < 26; i++ {
+		for j := 0; j < len(word); j++ {
+			if word[j] != byte(int('A')+i) {
+				res = append(res, word[:j]+string(int('A')+i)+word[j+1:])
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 DFS
+func minMutation1(start string, end string, bank []string) int {
+	endGene := convert(end)
+	startGene := convert(start)
+	m := make(map[uint32]struct{})
+	for _, gene := range bank {
+		m[convert(gene)] = struct{}{}
+	}
+	if _, ok := m[endGene]; !ok {
+		return -1
+	}
+	if check(startGene ^ endGene) {
+		return 1
+	}
+	delete(m, startGene)
+	step := make(map[uint32]int)
+	step[endGene] = 0
+	return dfsMutation(startGene, m, step)
+}
+
+func dfsMutation(start uint32, m map[uint32]struct{}, step map[uint32]int) int {
+	if v, ok := step[start]; ok {
+		return v
+	}
+	c := -1
+	step[start] = c
+	for k := range m {
+		if check(k ^ start) {
+			next := dfsMutation(k, m, step)
+			if next != -1 {
+				if c == -1 || c > next {
+					c = next + 1
+				}
+			}
+		}
+	}
+	step[start] = c
+	return c
+}
+
+func check(val uint32) bool {
+	if val == 0 {
+		return false
+	}
+	if val&(val-1) == 0 {
+		return true
+	}
+	for val > 0 {
+		if val == 3 {
+			return true
+		}
+		if val&3 != 0 {
+			return false
+		}
+		val >>= 2
+	}
+	return false
+}
+
+func convert(gene string) uint32 {
+	var v uint32
+	for _, c := range gene {
+		v <<= 2
+		switch c {
+		case 'C':
+			v |= 1
+		case 'G':
+			v |= 2
+		case 'T':
+			v |= 3
+		}
+	}
+	return v
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0429.N-ary-Tree-Level-Order-Traversal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0434.Number-of-Segments-in-a-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0434.Number-of-Segments-in-a-String.md b/website/content/ChapterFour/0400~0499/0434.Number-of-Segments-in-a-String.md
new file mode 100644
index 000000000..71d45b7d7
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0434.Number-of-Segments-in-a-String.md
@@ -0,0 +1,77 @@
+# [434. Number of Segments in a String](https://leetcode.com/problems/number-of-segments-in-a-string/)
+
+
+## 题目
+
+You are given a string s, return the number of segments in the string.
+
+A segment is defined to be a contiguous sequence of non-space characters.
+
+**Example 1:**
+
+    Input: s = "Hello, my name is John"
+    Output: 5
+    Explanation: The five segments are ["Hello,", "my", "name", "is", "John"]
+
+**Example 2:**
+
+    Input: s = "Hello"
+    Output: 1
+
+**Example 3:**
+
+    Input: s = "love live! mu'sic forever"
+    Output: 4
+
+**Example 4:**
+
+    Input: s = ""
+    Output: 0
+
+**Constraints**
+
+ - 0 <= s.length <= 300
+ - s consists of lower-case and upper-case English letters, digits or one of the following characters "!@#$%^&*()_+-=',.:".
+ - The only space character in s is ' '.
+
+## 题目大意
+
+统计字符串中的单词个数，这里的单词指的是连续的不是空格的字符。
+
+请注意，你可以假定字符串里不包括任何不可打印的字符。
+
+## 解题思路
+
+- 以空格为分割计算元素个数
+
+## 代码
+
+```go
+
+package leetcode
+
+func countSegments(s string) int {
+	segments := false
+	cnt := 0
+	for _, v := range s {
+		if v == ' ' && segments {
+			segments = false
+			cnt += 1
+		} else if v != ' ' {
+			segments = true
+		}
+	}
+	if segments {
+		cnt++
+	}
+	return cnt
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0433.Minimum-Genetic-Mutation/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0435.Non-overlapping-Intervals/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0435.Non-overlapping-Intervals.md b/website/content/ChapterFour/0400~0499/0435.Non-overlapping-Intervals.md
new file mode 100755
index 000000000..17d409ac3
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0435.Non-overlapping-Intervals.md
@@ -0,0 +1,142 @@
+# [435. Non-overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)
+
+
+## 题目
+
+Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
+
+**Note**:
+
+1. You may assume the interval's end point is always bigger than its start point.
+2. Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.
+
+**Example 1**:
+
+    Input: [ [1,2], [2,3], [3,4], [1,3] ]
+    
+    Output: 1
+    
+    Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
+
+**Example 2**:
+
+    Input: [ [1,2], [1,2], [1,2] ]
+    
+    Output: 2
+    
+    Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
+
+**Example 3**:
+
+    Input: [ [1,2], [2,3] ]
+    
+    Output: 0
+    
+    Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
+
+**Note**: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
+
+
+## 题目大意
+
+给定一个区间的集合，找到需要移除区间的最小数量，使剩余区间互不重叠。
+
+注意:
+
+1. 可以认为区间的终点总是大于它的起点。
+2. 区间 [1,2] 和 [2,3] 的边界相互“接触”，但没有相互重叠。
+
+
+
+## 解题思路
+
+
+- 给定一组区间，问最少删除多少个区间，可以让这些区间之间互相不重叠。注意，给定区间的起始点永远小于终止点。[1,2] 和 [2,3] 不叫重叠。
+- 这一题可以反过来考虑，给定一组区间，问最多保留多少区间，可以让这些区间之间相互不重叠。先排序，判断区间是否重叠。
+- 这一题一种做法是利用动态规划，模仿最长上升子序列的思想，来解题。
+- 这道题另外一种做法是按照区间的结尾进行排序，每次选择结尾最早的，且和前一个区间不重叠的区间。选取结尾最早的，就可以给后面留出更大的空间，供后面的区间选择。这样可以保留更多的区间。这种做法是贪心算法的思想。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+// 解法一 DP O(n^2) 思路是仿造最长上升子序列的思路
+func eraseOverlapIntervals(intervals [][]int) int {
+	if len(intervals) == 0 {
+		return 0
+	}
+	sort.Sort(Intervals(intervals))
+	dp, res := make([]int, len(intervals)), 0
+	for i := range dp {
+		dp[i] = 1
+	}
+	for i := 1; i < len(intervals); i++ {
+		for j := 0; j < i; j++ {
+			if intervals[i][0] >= intervals[j][1] {
+				dp[i] = max(dp[i], 1+dp[j])
+			}
+		}
+	}
+	for _, v := range dp {
+		res = max(res, v)
+	}
+	return len(intervals) - res
+}
+
+// Intervals define
+type Intervals [][]int
+
+func (a Intervals) Len() int {
+	return len(a)
+}
+func (a Intervals) Swap(i, j int) {
+	a[i], a[j] = a[j], a[i]
+}
+func (a Intervals) Less(i, j int) bool {
+	for k := 0; k < len(a[i]); k++ {
+		if a[i][k] < a[j][k] {
+			return true
+		} else if a[i][k] == a[j][k] {
+			continue
+		} else {
+			return false
+		}
+	}
+	return true
+}
+
+// 解法二 贪心 O(n)
+func eraseOverlapIntervals1(intervals [][]int) int {
+	if len(intervals) == 0 {
+		return 0
+	}
+	sort.Sort(Intervals(intervals))
+	pre, res := 0, 1
+
+	for i := 1; i < len(intervals); i++ {
+		if intervals[i][0] >= intervals[pre][1] {
+			res++
+			pre = i
+		} else if intervals[i][1] < intervals[pre][1] {
+			pre = i
+		}
+	}
+	return len(intervals) - res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0434.Number-of-Segments-in-a-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0436.Find-Right-Interval/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0436.Find-Right-Interval.md b/website/content/ChapterFour/0400~0499/0436.Find-Right-Interval.md
new file mode 100755
index 000000000..d8b2d726f
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0436.Find-Right-Interval.md
@@ -0,0 +1,149 @@
+# [436. Find Right Interval](https://leetcode.com/problems/find-right-interval/)
+
+
+## 题目
+
+Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.
+
+For any interval i, you need to store the minimum interval j's index, which means that the interval j has the minimum start point to build the "right" relationship for interval i. If the interval j doesn't exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.
+
+**Note**:
+
+1. You may assume the interval's end point is always bigger than its start point.
+2. You may assume none of these intervals have the same start point.
+
+**Example 1**:
+
+    Input: [ [1,2] ]
+    
+    Output: [-1]
+    
+    Explanation: There is only one interval in the collection, so it outputs -1.
+
+**Example 2**:
+
+    Input: [ [3,4], [2,3], [1,2] ]
+    
+    Output: [-1, 0, 1]
+    
+    Explanation: There is no satisfied "right" interval for [3,4].
+    For [2,3], the interval [3,4] has minimum-"right" start point;
+    For [1,2], the interval [2,3] has minimum-"right" start point.
+
+**Example 3**:
+
+    Input: [ [1,4], [2,3], [3,4] ]
+    
+    Output: [-1, 2, -1]
+    
+    Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
+    For [2,3], the interval [3,4] has minimum-"right" start point.
+
+**Note**: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
+
+
+## 题目大意
+
+给定一组区间，对于每一个区间 i，检查是否存在一个区间 j，它的起始点大于或等于区间 i 的终点，这可以称为 j 在 i 的“右侧”。
+
+对于任何区间，你需要存储的满足条件的区间 j 的最小索引，这意味着区间 j 有最小的起始点可以使其成为“右侧”区间。如果区间 j 不存在，则将区间 i 存储为 -1。最后，你需要输出一个值为存储的区间值的数组。
+
+注意:
+
+- 你可以假设区间的终点总是大于它的起始点。
+- 你可以假定这些区间都不具有相同的起始点。
+
+
+## 解题思路
+
+
+- 给出一个 `interval` 的 数组，要求找到每个 `interval` 在它右边第一个 `interval` 的下标。A 区间在 B 区间的右边：A 区间的左边界的值大于等于 B 区间的右边界。
+- 这一题很明显可以用二分搜索来解答。先将 `interval` 数组排序，然后针对每个 `interval`，用二分搜索搜索大于等于 `interval` 右边界值的 `interval`。如果找到就把下标存入最终数组中，如果没有找到，把 `-1` 存入最终数组中。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+// 解法一 利用系统函数 sort + 二分搜索
+func findRightInterval(intervals [][]int) []int {
+	intervalList := make(intervalList, len(intervals))
+	// 转换成 interval 类型
+	for i, v := range intervals {
+		intervalList[i] = interval{interval: v, index: i}
+	}
+	sort.Sort(intervalList)
+	out := make([]int, len(intervalList))
+	for i := 0; i < len(intervalList); i++ {
+		index := sort.Search(len(intervalList), func(p int) bool { return intervalList[p].interval[0] >= intervalList[i].interval[1] })
+		if index == len(intervalList) {
+			out[intervalList[i].index] = -1
+		} else {
+			out[intervalList[i].index] = intervalList[index].index
+		}
+	}
+	return out
+}
+
+type interval struct {
+	interval []int
+	index    int
+}
+
+type intervalList []interval
+
+func (in intervalList) Len() int { return len(in) }
+func (in intervalList) Less(i, j int) bool {
+	return in[i].interval[0] <= in[j].interval[0]
+}
+func (in intervalList) Swap(i, j int) { in[i], in[j] = in[j], in[i] }
+
+// 解法二 手撸 sort + 二分搜索
+func findRightInterval1(intervals [][]int) []int {
+	if len(intervals) == 0 {
+		return []int{}
+	}
+	intervalsList, res, intervalMap := []Interval{}, []int{}, map[Interval]int{}
+	for k, v := range intervals {
+		intervalsList = append(intervalsList, Interval{Start: v[0], End: v[1]})
+		intervalMap[Interval{Start: v[0], End: v[1]}] = k
+	}
+	quickSort(intervalsList, 0, len(intervalsList)-1)
+	for _, v := range intervals {
+		tmp := searchFirstGreaterInterval(intervalsList, v[1])
+		if tmp > 0 {
+			tmp = intervalMap[intervalsList[tmp]]
+		}
+		res = append(res, tmp)
+	}
+	return res
+}
+
+func searchFirstGreaterInterval(nums []Interval, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid].Start >= target {
+			if (mid == 0) || (nums[mid-1].Start < target) { // 找到第一个大于等于 target 的元素
+				return mid
+			}
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0435.Non-overlapping-Intervals/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0437.Path-Sum-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0437.Path-Sum-III.md b/website/content/ChapterFour/0400~0499/0437.Path-Sum-III.md
new file mode 100755
index 000000000..cf8405325
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0437.Path-Sum-III.md
@@ -0,0 +1,127 @@
+# [437. Path Sum III](https://leetcode.com/problems/path-sum-iii/)
+
+
+## 题目
+
+Given the `root` of a binary tree and an integer `targetSum`, return *the number of paths where the sum of the values along the path equals* `targetSum`.
+
+The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/04/09/pathsum3-1-tree.jpg](https://assets.leetcode.com/uploads/2021/04/09/pathsum3-1-tree.jpg)
+
+```
+Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
+Output: 3
+Explanation: The paths that sum to 8 are shown.
+
+```
+
+**Example 2:**
+
+```
+Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
+Output: 3
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[0, 1000]`.
+- `109 <= Node.val <= 109`
+- `1000 <= targetSum <= 1000`
+
+## 题目大意
+
+给定一个二叉树，它的每个结点都存放着一个整数值。找出路径和等于给定数值的路径总数。路径不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。二叉树不超过1000个节点，且节点数值范围是 [-1000000,1000000] 的整数。
+
+
+## 解题思路
+
+
+- 这一题是第 112 题 Path Sum 和第 113 题 Path Sum II 的加强版，这一题要求求出任意一条路径的和为 sum，起点不一定是根节点，可以是任意节点开始。
+- 注意这一题可能出现负数的情况，节点和为 sum，并不一定是最终情况，有可能下面还有正数节点和负数节点相加正好为 0，那么这也是一种情况。一定要遍历到底。
+- 一个点是否为 sum 的起点，有 3 种情况，第一种情况路径包含该 root 节点，如果包含该结点，就在它的左子树和右子树中寻找和为 `sum-root.Val` 的情况。第二种情况路径不包含该 root 节点，那么就需要在它的左子树和右子树中分别寻找和为 sum 的结点。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 带缓存 dfs
+func pathSum(root *TreeNode, targetSum int) int {
+	prefixSum := make(map[int]int)
+	prefixSum[0] = 1
+	return dfs(root, prefixSum, 0, targetSum)
+}
+
+func dfs(root *TreeNode, prefixSum map[int]int, cur, sum int) int {
+	if root == nil {
+		return 0
+	}
+	cur += root.Val
+	cnt := 0
+	if v, ok := prefixSum[cur-sum]; ok {
+		cnt = v
+	}
+	prefixSum[cur]++
+	cnt += dfs(root.Left, prefixSum, cur, sum)
+	cnt += dfs(root.Right, prefixSum, cur, sum)
+	prefixSum[cur]--
+	return cnt
+}
+
+// 解法二
+func pathSumIII(root *TreeNode, sum int) int {
+	if root == nil {
+		return 0
+	}
+	res := findPath437(root, sum)
+	res += pathSumIII(root.Left, sum)
+	res += pathSumIII(root.Right, sum)
+	return res
+}
+
+// 寻找包含 root 这个结点，且和为 sum 的路径
+func findPath437(root *TreeNode, sum int) int {
+	if root == nil {
+		return 0
+	}
+	res := 0
+	if root.Val == sum {
+		res++
+	}
+	res += findPath437(root.Left, sum-root.Val)
+	res += findPath437(root.Right, sum-root.Val)
+	return res
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0436.Find-Right-Interval/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0438.Find-All-Anagrams-in-a-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0438.Find-All-Anagrams-in-a-String.md b/website/content/ChapterFour/0400~0499/0438.Find-All-Anagrams-in-a-String.md
new file mode 100644
index 000000000..93d240934
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0438.Find-All-Anagrams-in-a-String.md
@@ -0,0 +1,113 @@
+# [438. Find All Anagrams in a String](https://leetcode.com/problems/find-all-anagrams-in-a-string/)
+
+## 题目
+
+Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
+
+Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
+
+The order of output does not matter.
+
+
+**Example 1**:
+
+```
+
+Input:
+s: "cbaebabacd" p: "abc"
+
+Output:
+[0, 6]
+
+Explanation:
+The substring with start index = 0 is "cba", which is an anagram of "abc".
+The substring with start index = 6 is "bac", which is an anagram of "abc".
+
+```
+
+**Example 2**:
+
+```
+
+Input:
+s: "abab" p: "ab"
+
+Output:
+[0, 1, 2]
+
+Explanation:
+The substring with start index = 0 is "ab", which is an anagram of "ab".
+The substring with start index = 1 is "ba", which is an anagram of "ab".
+The substring with start index = 2 is "ab", which is an anagram of "ab".
+
+```
+
+## 题目大意
+
+给定一个字符串 s 和一个非空字符串 p，找出 s 中的所有是 p 的 Anagrams 字符串的子串，返回这些子串的起始索引。Anagrams 的意思是和一个字符串的所有字符都一样，只是排列组合不同。
+
+## 解题思路
+
+这道题是一道考“滑动窗口”的题目。和第 3 题，第 76 题，第 567 题类似的。解法也是用 freq[256] 记录每个字符的出现的频次次数。滑动窗口左边界往右滑动的时候，划过去的元素释放次数(即次数 ++)，滑动窗口右边界往右滑动的时候，划过去的元素消耗次数(即次数 \-\-)。右边界和左边界相差 len(p) 的时候，需要判断每个元素是否都用过一遍了。具体做法是每经过一个符合规范的元素，count 就 \-\-，count 初始值是 len(p)，当每个元素都符合规范的时候，右边界和左边界相差 len(p) 的时候，count 也会等于 0 。当区间内有不符合规范的元素(freq < 0 或者是不存在的元素)，那么当区间达到 len(p) 的时候，count 无法减少到 0，区间右移动的时候，左边界又会开始 count ++，只有当左边界移出了这些不合规范的元素以后，才可能出现 count = 0 的情况，即找到 Anagrams 的情况。
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findAnagrams(s string, p string) []int {
+	var freq [256]int
+	result := []int{}
+	if len(s) == 0 || len(s) < len(p) {
+		return result
+	}
+	for i := 0; i < len(p); i++ {
+		freq[p[i]-'a']++
+	}
+	left, right, count := 0, 0, len(p)
+
+	for right < len(s) {
+		if freq[s[right]-'a'] >= 1 {
+			count--
+		}
+		freq[s[right]-'a']--
+		right++
+		if count == 0 {
+			result = append(result, left)
+		}
+		if right-left == len(p) {
+			if freq[s[left]-'a'] >= 0 {
+				count++
+			}
+			freq[s[left]-'a']++
+			left++
+		}
+
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0437.Path-Sum-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0441.Arranging-Coins/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0441.Arranging-Coins.md b/website/content/ChapterFour/0400~0499/0441.Arranging-Coins.md
new file mode 100755
index 000000000..33fdee75b
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0441.Arranging-Coins.md
@@ -0,0 +1,82 @@
+# [441. Arranging Coins](https://leetcode.com/problems/arranging-coins/)
+
+## 题目
+
+You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins.
+
+Given n, find the total number of **full** staircase rows that can be formed.
+
+n is a non-negative integer and fits within the range of a 32-bit signed integer.
+
+**Example 1**:
+
+    n = 5
+    
+    The coins can form the following rows:
+    ¤
+    ¤ ¤
+    ¤ ¤
+    
+    Because the 3rd row is incomplete, we return 2.
+
+**Example 2**:
+
+    n = 8
+    
+    The coins can form the following rows:
+    ¤
+    ¤ ¤
+    ¤ ¤ ¤
+    ¤ ¤
+    
+    Because the 4th row is incomplete, we return 3.
+
+
+## 题目大意
+
+你总共有 n 枚硬币，你需要将它们摆成一个阶梯形状，第 k 行就必须正好有 k 枚硬币。给定一个数字 n，找出可形成完整阶梯行的总行数。n 是一个非负整数，并且在32位有符号整型的范围内。
+
+
+
+## 解题思路
+
+
+- n 个硬币，按照递增的方式排列搭楼梯，第一层一个，第二层二个，……第 n 层需要 n 个硬币。问硬币 n 能够搭建到第几层？
+- 这一题有 2 种解法，第一种解法就是解方程求出 X，`(1+x)x/2 = n`，即 `x = floor(sqrt(2*n+1/4) - 1/2)`，第二种解法是模拟。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+// 解法一 数学公式
+func arrangeCoins(n int) int {
+	if n <= 0 {
+		return 0
+	}
+	x := math.Sqrt(2*float64(n)+0.25) - 0.5
+	return int(x)
+}
+
+// 解法二 模拟
+func arrangeCoins1(n int) int {
+	k := 1
+	for n >= k {
+		n -= k
+		k++
+	}
+	return k - 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0438.Find-All-Anagrams-in-a-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0445.Add-Two-Numbers-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0445.Add-Two-Numbers-II.md b/website/content/ChapterFour/0400~0499/0445.Add-Two-Numbers-II.md
new file mode 100644
index 000000000..a4156c6fb
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0445.Add-Two-Numbers-II.md
@@ -0,0 +1,188 @@
+# [445. Add Two Numbers II](https://leetcode.com/problems/add-two-numbers-ii/)
+
+## 题目
+
+You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
+
+You may assume the two numbers do not contain any leading zero, except the number 0 itself.
+
+**Follow up**:
+What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
+
+**Example**:
+
+```
+
+Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
+Output: 7 -> 8 -> 0 -> 7
+
+```
+
+## 题目大意
+
+这道题是第 2 题的变种题，第 2 题中的 2 个数是从个位逆序排到高位，这样相加只用从头交到尾，进位一直进位即可。这道题目中强制要求不能把链表逆序。2 个数字是从高位排到低位的，这样进位是倒着来的。
+
+## 解题思路
+
+思路也不难，加法只用把短的链表依次加到长的链表上面来就可以了，最终判断一下最高位有没有进位，有进位再往前进一位。加法的过程可以用到递归。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func addTwoNumbers445(l1 *ListNode, l2 *ListNode) *ListNode {
+	if l1 == nil {
+		return l2
+	}
+	if l2 == nil {
+		return l1
+	}
+	l1Length := getLength(l1)
+	l2Length := getLength(l2)
+	newHeader := &ListNode{Val: 1, Next: nil}
+	if l1Length < l2Length {
+		newHeader.Next = addNode(l2, l1, l2Length-l1Length)
+	} else {
+		newHeader.Next = addNode(l1, l2, l1Length-l2Length)
+	}
+	if newHeader.Next.Val > 9 {
+		newHeader.Next.Val = newHeader.Next.Val % 10
+		return newHeader
+	}
+	return newHeader.Next
+}
+
+func addNode(l1 *ListNode, l2 *ListNode, offset int) *ListNode {
+	if l1 == nil {
+		return nil
+	}
+	var (
+		res, node *ListNode
+	)
+	if offset == 0 {
+		res = &ListNode{Val: l1.Val + l2.Val, Next: nil}
+		node = addNode(l1.Next, l2.Next, 0)
+	} else {
+		res = &ListNode{Val: l1.Val, Next: nil}
+		node = addNode(l1.Next, l2, offset-1)
+	}
+	if node != nil && node.Val > 9 {
+		res.Val++
+		node.Val = node.Val % 10
+	}
+	res.Next = node
+	return res
+}
+
+func getLength(l *ListNode) int {
+	count := 0
+	cur := l
+	for cur != nil {
+		count++
+		cur = cur.Next
+	}
+	return count
+}
+
+func addTwoNumbers1(l1 *ListNode, l2 *ListNode) *ListNode {
+	reservedL1 := reverseList(l1)
+	reservedL2 := reverseList(l2)
+
+	dummyHead := &ListNode{}
+	head := dummyHead
+	carry := 0
+	for reservedL1 != nil || reservedL2 != nil || carry > 0 {
+		val := carry
+		if reservedL1 != nil {
+			val = reservedL1.Val + val
+			reservedL1 = reservedL1.Next
+		}
+		if reservedL2 != nil {
+			val = reservedL2.Val + val
+			reservedL2 = reservedL2.Next
+		}
+		carry = val / 10
+		head.Next = &ListNode{Val: val % 10}
+		head = head.Next
+	}
+	return reverseList(dummyHead.Next)
+}
+
+func reverseList(head *ListNode) *ListNode {
+	var prev *ListNode
+	for head != nil {
+		tmp := head.Next
+		head.Next = prev
+
+		prev = head
+		head = tmp
+	}
+	return prev
+}
+
+func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
+	stack1 := pushStack(l1)
+	stack2 := pushStack(l2)
+
+	dummyHead := &ListNode{}
+	head := dummyHead
+	carry := 0
+	for len(stack1) > 0 || len(stack2) > 0 || carry > 0 {
+		val := carry
+		if len(stack1) > 0 {
+			val = val + stack1[len(stack1)-1]
+			stack1 = stack1[:len(stack1)-1]
+		}
+		if len(stack2) > 0 {
+			val = val + stack2[len(stack2)-1]
+			stack2 = stack2[:len(stack2)-1]
+		}
+		carry = val / 10
+		tmp := head.Next
+		head.Next = &ListNode{Val: val % 10, Next: tmp}
+	}
+	return dummyHead.Next
+}
+
+func pushStack(l *ListNode) []int {
+	var stack []int
+	for l != nil {
+		stack = append(stack, l.Val)
+		l = l.Next
+	}
+	return stack
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0441.Arranging-Coins/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0447.Number-of-Boomerangs/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0447.Number-of-Boomerangs.md b/website/content/ChapterFour/0400~0499/0447.Number-of-Boomerangs.md
new file mode 100644
index 000000000..33f3c1fd9
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0447.Number-of-Boomerangs.md
@@ -0,0 +1,73 @@
+# [447. Number of Boomerangs](https://leetcode.com/problems/number-of-boomerangs/)
+
+## 题目
+
+Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance between i and j equals the distance between i and k (the order of the tuple matters).
+
+Find the number of boomerangs. You may assume that n will be at most 500 and coordinates of points are all in the range [-10000, 10000] (inclusive).
+
+
+
+**Example 1**:
+
+```
+
+Input:
+[[0,0],[1,0],[2,0]]
+
+Output:
+2
+
+Explanation:
+The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]
+
+```
+
+
+## 题目大意
+
+在一个 Point 的数组中求出 (i,j,k) 三元组，要求 j 和 i 的距离等于 k 和 i 的距离。这样的三元组有多少种？注意 (i,j,k) 和 (j,i,k) 是不同的解，即元素的顺序是有关系的。
+
+## 解题思路
+
+这道题考察的是哈希表的问题。
+
+首先依次求出两两点之间的距离，然后把这些距离记录在 map 中，key 是距离，value 是这个距离出现了多少次。求距离一般都需要开根号，但是 key 如果为浮点数就会有一些误差，所以计算距离的时候最后一步不需要开根号，保留平方差即可。
+
+最后求结果的时候，遍历 map，把里面距离大于 2 的 key 都拿出来，value 对应的是个数，在这些个数里面任取 2 个点就是解，所以利用排列组合，C n 2 就可以得到这个距离的结果，最后把这些排列组合的结果累积起来即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func numberOfBoomerangs(points [][]int) int {
+	res := 0
+	for i := 0; i < len(points); i++ {
+		record := make(map[int]int, len(points))
+		for j := 0; j < len(points); j++ {
+			if j != i {
+				record[dis(points[i], points[j])]++
+			}
+		}
+		for _, r := range record {
+			res += r * (r - 1)
+		}
+	}
+	return res
+}
+
+func dis(pa, pb []int) int {
+	return (pa[0]-pb[0])*(pa[0]-pb[0]) + (pa[1]-pb[1])*(pa[1]-pb[1])
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0445.Add-Two-Numbers-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0448.Find-All-Numbers-Disappeared-in-an-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0448.Find-All-Numbers-Disappeared-in-an-Array.md b/website/content/ChapterFour/0400~0499/0448.Find-All-Numbers-Disappeared-in-an-Array.md
new file mode 100644
index 000000000..29532c7ef
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0448.Find-All-Numbers-Disappeared-in-an-Array.md
@@ -0,0 +1,64 @@
+# [448. Find All Numbers Disappeared in an Array](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/)
+
+
+## 题目
+
+Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
+
+Find all the elements of [1, n] inclusive that do not appear in this array.
+
+Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.
+
+**Example**:
+
+```
+Input:
+[4,3,2,7,8,2,3,1]
+
+Output:
+[5,6]
+```
+
+## 题目大意
+
+给定一个范围在  1 ≤ a[i] ≤ n ( n = 数组大小 ) 的 整型数组，数组中的元素一些出现了两次，另一些只出现一次。找到所有在 [1, n] 范围之间没有出现在数组中的数字。你能在不使用额外空间且时间复杂度为 O(n) 的情况下完成这个任务吗? 你可以假定返回的数组不算在额外空间内。
+
+
+
+## 解题思路
+
+- 找出 [1,n] 范围内没有出现在数组中的数字。要求不使用额外空间，并且时间复杂度为 O(n)。
+- 要求不能使用额外的空间，那么只能想办法在原有数组上进行修改，并且这个修改是可还原的。时间复杂度也只能允许我们一层循环。只要循环一次能标记出已经出现过的数字，这道题就可以按要求解答出来。这里笔者的标记方法是把 |nums[i]|-1 索引位置的元素标记为负数。即 nums[| nums[i] |- 1] * -1。这里需要注意的是，nums[i] 需要加绝对值，因为它可能被之前的数置为负数了，需要还原一下。最后再遍历一次数组，若当前数组元素 nums[i] 为负数，说明我们在数组中存在数字 i+1。把结果输出到最终数组里即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func findDisappearedNumbers(nums []int) []int {
+	res := []int{}
+	for _, v := range nums {
+		if v < 0 {
+			v = -v
+		}
+		if nums[v-1] > 0 {
+			nums[v-1] = -nums[v-1]
+		}
+	}
+	for i, v := range nums {
+		if v > 0 {
+			res = append(res, i+1)
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0447.Number-of-Boomerangs/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0451.Sort-Characters-By-Frequency/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0451.Sort-Characters-By-Frequency.md b/website/content/ChapterFour/0400~0499/0451.Sort-Characters-By-Frequency.md
new file mode 100644
index 000000000..81a8d91a0
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0451.Sort-Characters-By-Frequency.md
@@ -0,0 +1,120 @@
+# [451. Sort Characters By Frequency](https://leetcode.com/problems/sort-characters-by-frequency/)
+
+## 题目
+
+Given a string, sort it in decreasing order based on the frequency of characters.
+
+
+**Example 1**:
+
+```
+
+Input:
+"tree"
+
+Output:
+"eert"
+
+Explanation:
+'e' appears twice while 'r' and 't' both appear once.
+So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.
+
+```
+
+**Example 2**:
+
+```
+
+Input:
+"cccaaa"
+
+Output:
+"cccaaa"
+
+Explanation:
+Both 'c' and 'a' appear three times, so "aaaccc" is also a valid answer.
+Note that "cacaca" is incorrect, as the same characters must be together.
+
+```
+
+**Example 3**:
+
+```
+
+Input:
+"Aabb"
+
+Output:
+"bbAa"
+
+Explanation:
+"bbaA" is also a valid answer, but "Aabb" is incorrect.
+Note that 'A' and 'a' are treated as two different characters.
+
+```
+
+
+
+
+## 题目大意
+
+这道题是 Google 的面试题。
+
+给定一个字符串，要求根据字符出现的频次从高到低重新排列这个字符串。
+
+## 解题思路
+
+思路比较简单，首先统计每个字符的频次，然后排序，最后按照频次从高到低进行输出即可。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+func frequencySort(s string) string {
+	if s == "" {
+		return ""
+	}
+	sMap := map[byte]int{}
+	cMap := map[int][]byte{}
+	sb := []byte(s)
+	for _, b := range sb {
+		sMap[b]++
+	}
+	for key, value := range sMap {
+		cMap[value] = append(cMap[value], key)
+	}
+
+	var keys []int
+	for k := range cMap {
+		keys = append(keys, k)
+	}
+	sort.Sort(sort.Reverse(sort.IntSlice(keys)))
+	res := make([]byte, 0)
+	for _, k := range keys {
+		for i := 0; i < len(cMap[k]); i++ {
+			for j := 0; j < k; j++ {
+				res = append(res, cMap[k][i])
+			}
+		}
+	}
+	return string(res)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0448.Find-All-Numbers-Disappeared-in-an-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0453.Minimum-Moves-to-Equal-Array-Elements/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0453.Minimum-Moves-to-Equal-Array-Elements.md b/website/content/ChapterFour/0400~0499/0453.Minimum-Moves-to-Equal-Array-Elements.md
new file mode 100644
index 000000000..f868c33a3
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0453.Minimum-Moves-to-Equal-Array-Elements.md
@@ -0,0 +1,58 @@
+# [453. Minimum Moves to Equal Array Elements](https://leetcode.com/problems/minimum-moves-to-equal-array-elements/)
+
+
+## 题目
+
+Given a **non-empty** integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1.
+
+**Example**:
+
+```
+Input:
+[1,2,3]
+
+Output:
+3
+
+Explanation:
+Only three moves are needed (remember each move increments two elements):
+
+[1,2,3]  =>  [2,3,3]  =>  [3,4,3]  =>  [4,4,4]
+```
+
+## 题目大意
+
+给定一个长度为 n 的非空整数数组，找到让数组所有元素相等的最小移动次数。每次移动将会使 n - 1 个元素增加 1。
+
+## 解题思路
+
+- 给定一个数组，要求输出让所有元素都相等的最小步数。每移动一步都会使得 n - 1 个元素 + 1 。
+- 数学题。这道题正着思考会考虑到排序或者暴力的方法上去。反过来思考一下，使得每个元素都相同，意思让所有元素的差异变为 0 。每次移动的过程中，都有 n - 1 个元素 + 1，那么没有 + 1 的那个元素和其他 n - 1 个元素相对差异就缩小了。所以这道题让所有元素都变为相等的最少步数，即等于让所有元素相对差异减少到最小的那个数。想到这里，此题就可以优雅的解出来了。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func minMoves(nums []int) int {
+	sum, min, l := 0, math.MaxInt32, len(nums)
+	for _, v := range nums {
+		sum += v
+		if min > v {
+			min = v
+		}
+	}
+	return sum - min*l
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0451.Sort-Characters-By-Frequency/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0454.4Sum-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0454.4Sum-II.md b/website/content/ChapterFour/0400~0499/0454.4Sum-II.md
new file mode 100644
index 000000000..bccad3d5a
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0454.4Sum-II.md
@@ -0,0 +1,72 @@
+# [454. 4Sum II](https://leetcode.com/problems/4sum-ii/)
+
+## 题目
+
+Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.
+
+To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.
+
+**Example 1**:
+
+```
+
+Input:
+A = [ 1, 2]
+B = [-2,-1]
+C = [-1, 2]
+D = [ 0, 2]
+
+Output:
+2
+
+Explanation:
+The two tuples are:
+1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
+2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
+
+```
+
+
+## 题目大意
+
+给出 4 个数组，计算这些数组中存在几对 i，j，k，l 可以使得 A[i] + B[j] + C[k] + D[l] = 0 。
+
+## 解题思路
+
+这道题的数据量不大，0 ≤ N ≤ 500，但是如果使用暴力解法，四层循环，会超时。这道题的思路和第 1 题思路也类似，先可以将 2 个数组中的组合都存入 map 中。之后将剩下的 2 个数组进行 for 循环，找出和为 0 的组合。这样时间复杂度是 O(n^2)。当然也可以讲剩下的 2 个数组的组合也存入 map 中，不过最后在 2 个 map 中查找结果也是 O(n^2) 的时间复杂度。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func fourSumCount(A []int, B []int, C []int, D []int) int {
+	m := make(map[int]int, len(A)*len(B))
+	for _, a := range A {
+		for _, b := range B {
+			m[a+b]++
+		}
+	}
+	ret := 0
+	for _, c := range C {
+		for _, d := range D {
+			ret += m[0-c-d]
+		}
+	}
+
+	return ret
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0453.Minimum-Moves-to-Equal-Array-Elements/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0455.Assign-Cookies/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0455.Assign-Cookies.md b/website/content/ChapterFour/0400~0499/0455.Assign-Cookies.md
new file mode 100755
index 000000000..dfd13b3a0
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0455.Assign-Cookies.md
@@ -0,0 +1,76 @@
+# [455. Assign Cookies](https://leetcode.com/problems/assign-cookies/)
+
+## 题目
+
+Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each child i has a greed factor gi, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sj. If sj >= gi, we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.
+
+**Note**:You may assume the greed factor is always positive. You cannot assign more than one cookie to one child.
+
+**Example 1**:
+
+    Input: [1,2,3], [1,1]
+    
+    Output: 1
+    
+    Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. 
+    And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
+    You need to output 1.
+
+**Example 2**:
+
+    Input: [1,2], [1,2,3]
+    
+    Output: 2
+    
+    Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
+    You have 3 cookies and their sizes are big enough to gratify all of the children, 
+    You need to output 2.
+
+
+## 题目大意
+
+假设你是一位很棒的家长，想要给你的孩子们一些小饼干。但是，每个孩子最多只能给一块饼干。对每个孩子 i ，都有一个胃口值 gi ，这是能让孩子们满足胃口的饼干的最小尺寸；并且每块饼干 j ，都有一个尺寸 sj 。如果 sj >= gi ，我们可以将这个饼干 j 分配给孩子 i ，这个孩子会得到满足。你的目标是尽可能满足越多数量的孩子，并输出这个最大数值。
+
+注意：你可以假设胃口值为正。一个小朋友最多只能拥有一块饼干。
+
+
+
+## 解题思路
+
+
+- 假设你想给小朋友们饼干，每个小朋友最多能够给一块饼干。每个小朋友都有一个“贪心指数”，称为 `g[i]`，`g[i]` 表示的是这名小朋友需要的饼干大小的最小值。同时，每个饼干都有一个大小值 `s[i]`，如果 `s[j] ≥ g[i]`，我们将饼干 `j` 分给小朋友 `i` 后，小朋友会很开心。给定数组 `g[]` 和 `s[]`，问如何分配饼干，能让更多的小朋友开心。
+- 这是一道典型的简单贪心题。贪心题一般都伴随着排序。将 `g[]` 和 `s[]` 分别排序。按照最难满足的小朋友开始给饼干，依次往下满足，最终能满足的小朋友数就是最终解。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+func findContentChildren(g []int, s []int) int {
+	sort.Ints(g)
+	sort.Ints(s)
+	gi, si, res := 0, 0, 0
+	for gi < len(g) && si < len(s) {
+		if s[si] >= g[gi] {
+			res++
+			si++
+			gi++
+		} else {
+			si++
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0454.4Sum-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0456.132-Pattern/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0456.132-Pattern.md b/website/content/ChapterFour/0400~0499/0456.132-Pattern.md
new file mode 100755
index 000000000..abd44ce62
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0456.132-Pattern.md
@@ -0,0 +1,108 @@
+# [456. 132 Pattern](https://leetcode.com/problems/132-pattern/)
+
+
+## 题目
+
+Given a sequence of n integers a1, a2, ..., an, a 132 pattern is a subsequence a**i**, a**j**, a**k** such that **i** < **j** < **k** and a**i** < a**k** < a**j**. Design an algorithm that takes a list of n numbers as input and checks whether there is a 132 pattern in the list.
+
+**Note**: n will be less than 15,000.
+
+**Example 1**:
+
+    Input: [1, 2, 3, 4]
+    
+    Output: False
+    
+    Explanation: There is no 132 pattern in the sequence.
+
+**Example 2**:
+
+    Input: [3, 1, 4, 2]
+    
+    Output: True
+    
+    Explanation: There is a 132 pattern in the sequence: [1, 4, 2].
+
+**Example 3**:
+
+    Input: [-1, 3, 2, 0]
+    
+    Output: True
+    
+    Explanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].
+
+
+## 题目大意
+
+
+给定一个整数序列：a1, a2, ..., an，一个 132 模式的子序列 ai, aj, ak 被定义为：当 i < j < k 时，ai < ak < aj。设计一个算法，当给定有 n 个数字的序列时，验证这个序列中是否含有 132 模式的子序列。注意：n 的值小于 15000。
+
+
+## 解题思路
+
+
+- 这一题用暴力解法一定超时
+- 这一题算是单调栈的经典解法，可以考虑从数组末尾开始往前扫，维护一个递减序列
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+)
+
+// 解法一 单调栈
+func find132pattern(nums []int) bool {
+	if len(nums) < 3 {
+		return false
+	}
+	num3, stack := math.MinInt64, []int{}
+	for i := len(nums) - 1; i >= 0; i-- {
+		if nums[i] < num3 {
+			return true
+		}
+		for len(stack) != 0 && nums[i] > stack[len(stack)-1] {
+			num3 = stack[len(stack)-1]
+			stack = stack[:len(stack)-1]
+		}
+		stack = append(stack, nums[i])
+	}
+	return false
+}
+
+// 解法二 暴力解法，超时！
+func find132pattern1(nums []int) bool {
+	if len(nums) < 3 {
+		return false
+	}
+	for j := 0; j < len(nums); j++ {
+		stack := []int{}
+		for i := j; i < len(nums); i++ {
+			if len(stack) == 0 || (len(stack) > 0 && nums[i] > nums[stack[len(stack)-1]]) {
+				stack = append(stack, i)
+			} else if nums[i] < nums[stack[len(stack)-1]] {
+				index := len(stack) - 1
+				for ; index >= 0; index-- {
+					if nums[stack[index]] < nums[i] {
+						return true
+					}
+				}
+			}
+		}
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0455.Assign-Cookies/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0457.Circular-Array-Loop/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0457.Circular-Array-Loop.md b/website/content/ChapterFour/0400~0499/0457.Circular-Array-Loop.md
new file mode 100755
index 000000000..d43a50e6e
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0457.Circular-Array-Loop.md
@@ -0,0 +1,111 @@
+# [457. Circular Array Loop](https://leetcode.com/problems/circular-array-loop/)
+
+
+## 题目
+
+You are given a **circular** array `nums` of positive and negative integers. If a number k at an index is positive, then move forward k steps. Conversely, if it's negative (-k), move backward k steps. Since the array is circular, you may assume that the last element's next element is the first element, and the first element's previous element is the last element.
+
+Determine if there is a loop (or a cycle) in `nums`. A cycle must start and end at the same index and the cycle's length > 1. Furthermore, movements in a cycle must all follow a single direction. In other words, a cycle must not consist of both forward and backward movements.
+
+**Example 1**:
+
+    Input: [2,-1,1,2,2]
+    Output: true
+    Explanation: There is a cycle, from index 0 -> 2 -> 3 -> 0. The cycle's length is 3.
+
+**Example 2**:
+
+    Input: [-1,2]
+    Output: false
+    Explanation: The movement from index 1 -> 1 -> 1 ... is not a cycle, because the cycle's length is 1. By definition the cycle's length must be greater than 1.
+
+**Example 3**:
+
+    Input: [-2,1,-1,-2,-2]
+    Output: false
+    Explanation: The movement from index 1 -> 2 -> 1 -> ... is not a cycle, because movement from index 1 -> 2 is a forward movement, but movement from index 2 -> 1 is a backward movement. All movements in a cycle must follow a single direction.
+
+**Note**:
+
+1. -1000 ≤ nums[i] ≤ 1000
+2. nums[i] ≠ 0
+3. 1 ≤ nums.length ≤ 5000
+
+**Follow up**:
+
+Could you solve it in **O(n)** time complexity and **O(1)** extra space complexity?
+
+
+## 题目大意
+
+给定一个含有正整数和负整数的环形数组 nums。 如果某个索引中的数 k 为正数，则向前移动 k 个索引。相反，如果是负数 (-k)，则向后移动 k 个索引。因为数组是环形的，所以可以假设最后一个元素的下一个元素是第一个元素，而第一个元素的前一个元素是最后一个元素。
+
+确定 nums 中是否存在循环（或周期）。循环必须在相同的索引处开始和结束并且循环长度 > 1。此外，一个循环中的所有运动都必须沿着同一方向进行。换句话说，一个循环中不能同时包括向前的运动和向后的运动。
+
+提示：
+
+- -1000 ≤ nums[i] ≤ 1000
+- nums[i] ≠ 0
+- 1 ≤ nums.length ≤ 5000
+
+进阶：
+
+- 你能写出时间时间复杂度为 O(n) 和额外空间复杂度为 O(1) 的算法吗？
+
+## 解题思路
+
+
+- 给出一个循环数组，数组的数字代表了前进和后退的步数，+ 代表往右(前进)，- 代表往左(后退)。问这个循环数组中是否存在一个循环，并且这个循环内不能只有一个元素，循环的方向都必须是同方向的。
+- 遇到循环就可以优先考虑用快慢指针的方法判断循环，这一题对循环增加了一个条件，循环不能只是单元素的循环，所以在快慢指针中加入这个判断条件。还有一个判断条件是循环的方向必须是同向的，这个简单，用 `num[i] * num[j] > 0` 就可以判断出是同向的(如果是反向的，那么两者的乘积必然是负数)，如果没有找到循环，可以将当前已经走过的路径上的 num[] 值都置为 0，标记已经访问过了。下次循环遇到访问过的元素，`num[i] * num[j] > 0` 就会是 0，提前退出找循环的过程。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func circularArrayLoop(nums []int) bool {
+	if len(nums) == 0 {
+		return false
+	}
+	for i := 0; i < len(nums); i++ {
+		if nums[i] == 0 {
+			continue
+		}
+		// slow/fast pointer
+		slow, fast, val := i, getNextIndex(nums, i), 0
+		for nums[fast]*nums[i] > 0 && nums[getNextIndex(nums, fast)]*nums[i] > 0 {
+			if slow == fast {
+				// check for loop with only one element
+				if slow == getNextIndex(nums, slow) {
+					break
+				}
+				return true
+			}
+			slow = getNextIndex(nums, slow)
+			fast = getNextIndex(nums, getNextIndex(nums, fast))
+		}
+		// loop not found, set all element along the way to 0
+		slow, val = i, nums[i]
+		for nums[slow]*val > 0 {
+			next := getNextIndex(nums, slow)
+			nums[slow] = 0
+			slow = next
+		}
+	}
+	return false
+}
+
+func getNextIndex(nums []int, index int) int {
+	return ((nums[index]+index)%len(nums) + len(nums)) % len(nums)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0456.132-Pattern/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0458.Poor-Pigs/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0458.Poor-Pigs.md b/website/content/ChapterFour/0400~0499/0458.Poor-Pigs.md
new file mode 100644
index 000000000..dcccfda06
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0458.Poor-Pigs.md
@@ -0,0 +1,84 @@
+# [458. Poor Pigs](https://leetcode.com/problems/poor-pigs/)
+
+## 题目
+
+There are buckets buckets of liquid, where exactly one of the buckets is poisonous. To figure out which one is poisonous, you feed some number of (poor) pigs the liquid to see whether they will die or not. Unfortunately, you only have minutesToTest minutes to determine which bucket is poisonous.
+
+You can feed the pigs according to these steps:
+
+- Choose some live pigs to feed.
+- For each pig, choose which buckets to feed it. The pig will consume all the chosen buckets simultaneously and will take no time.
+- Wait for minutesToDie minutes. You may not feed any other pigs during this time.
+- After minutesToDie minutes have passed, any pigs that have been fed the poisonous bucket will die, and all others will survive.
+- Repeat this process until you run out of time.
+
+Given buckets, minutesToDie, and minutesToTest, return the minimum number of pigs needed to figure out which bucket is poisonous within the allotted time.
+
+**Example 1**:
+
+    Input: buckets = 1000, minutesToDie = 15, minutesToTest = 60
+    Output: 5
+
+**Example 2**:
+
+    Input: buckets = 4, minutesToDie = 15, minutesToTest = 15
+    Output: 2
+
+**Example 3**:
+
+    Input: buckets = 4, minutesToDie = 15, minutesToTest = 30
+    Output: 2
+
+**Constraints:**
+
+- 1 <= buckets <= 1000
+- 1 <= minutesToDie <= minutesToTest <= 100
+
+## 题目大意
+
+有 buckets 桶液体，其中 正好 有一桶含有毒药，其余装的都是水。它们从外观看起来都一样。为了弄清楚哪只水桶含有毒药，你可以喂一些猪喝，通过观察猪是否会死进行判断。不幸的是，你只有 minutesToTest 分钟时间来确定哪桶液体是有毒的。
+
+喂猪的规则如下：
+
+- 选择若干活猪进行喂养
+- 可以允许小猪同时饮用任意数量的桶中的水，并且该过程不需要时间。
+- 小猪喝完水后，必须有 minutesToDie 分钟的冷却时间。在这段时间里，你只能观察，而不允许继续喂猪。
+- 过了 minutesToDie 分钟后，所有喝到毒药的猪都会死去，其他所有猪都会活下来。
+- 重复这一过程，直到时间用完。
+
+给你桶的数目 buckets ，minutesToDie 和 minutesToTest ，返回在规定时间内判断哪个桶有毒所需的 最小 猪数。
+
+## 解题思路
+
+使用数学方法,以 minutesToDie=15, minutesToTest=60, 1 只小猪为例,可以测试 5 只桶
+
+- 0-15 小猪吃第一个桶中的液体,如果死去，则第一个桶有毒,否则继续测试
+- 15-30 小猪吃第二个桶中的液体，如果死去，则第二个桶有毒,否则继续测试
+- 30-45 小猪吃第三个桶中的液体，如果死去，则第三个桶有毒,否则继续测试
+- 45-60 小猪吃第四个桶中的液体，如果死去，则第四个桶有毒
+- 如果最后小猪没有死去，则第五个桶有毒
+
+所以一只小猪在 minutesToDie 和 minutesToTest 时间一定的情况下可以最多判断 base = minutesToTest / minutesToDie + 1 个桶
+
+假设小猪的数量是 num,那么 pow(base, num) >= buckets,根据对数运算规则，两边分别取对数得到: num >= Log10(buckets) / Log10(base) 
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func poorPigs(buckets int, minutesToDie int, minutesToTest int) int {
+	base := minutesToTest/minutesToDie + 1
+	return int(math.Ceil(math.Log10(float64(buckets)) / math.Log10(float64(base))))
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0457.Circular-Array-Loop/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0460.LFU-Cache/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0460.LFU-Cache.md b/website/content/ChapterFour/0400~0499/0460.LFU-Cache.md
new file mode 100644
index 000000000..5692cc711
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0460.LFU-Cache.md
@@ -0,0 +1,149 @@
+# [460. LFU Cache](https://leetcode.com/problems/lfu-cache/)
+
+
+## 题目
+
+Design and implement a data structure for [Least Frequently Used (LFU)](https://en.wikipedia.org/wiki/Least_frequently_used) cache.
+
+Implement the `LFUCache` class:
+
+- `LFUCache(int capacity)` Initializes the object with the `capacity` of the data structure.
+- `int get(int key)` Gets the value of the `key` if the `key` exists in the cache. Otherwise, returns `1`.
+- `void put(int key, int value)` Sets or inserts the value if the `key` is not already present. When the cache reaches its `capacity`, it should invalidate the least frequently used item before inserting a new item. For this problem, when there is a tie (i.e., two or more keys with the same frequency), **the least recently** used `key` would be evicted.
+
+**Notice that** the number of times an item is used is the number of calls to the `get` and `put` functions for that item since it was inserted. This number is set to zero when the item is removed.
+
+**Example 1**:
+
+```
+Input
+["LFUCache", "put", "put", "get", "put", "get", "get", "put", "get", "get", "get"]
+[[2], [1, 1], [2, 2], [1], [3, 3], [2], [3], [4, 4], [1], [3], [4]]
+Output
+[null, null, null, 1, null, -1, 3, null, -1, 3, 4]
+
+Explanation
+LFUCache lfu = new LFUCache(2);
+lfu.put(1, 1);
+lfu.put(2, 2);
+lfu.get(1);      // return 1
+lfu.put(3, 3);   // evicts key 2
+lfu.get(2);      // return -1 (not found)
+lfu.get(3);      // return 3
+lfu.put(4, 4);   // evicts key 1.
+lfu.get(1);      // return -1 (not found)
+lfu.get(3);      // return 3
+lfu.get(4);      // return 4
+
+```
+
+**Constraints**:
+
+- `0 <= capacity, key, value <= 104`
+- At most `10^5` calls will be made to `get` and `put`.
+
+**Follow up**: Could you do both operations in `O(1)` time complexity?
+
+## 题目大意
+
+请你为 最不经常使用（LFU）缓存算法设计并实现数据结构。
+
+实现 LFUCache 类：
+
+- LFUCache(int capacity) - 用数据结构的容量 capacity 初始化对象
+- int get(int key) - 如果键存在于缓存中，则获取键的值，否则返回 -1。
+- void put(int key, int value) - 如果键已存在，则变更其值；如果键不存在，请插入键值对。当缓存达到其容量时，则应该在插入新项之前，使最不经常使用的项无效。在此问题中，当存在平局（即两个或更多个键具有相同使用频率）时，应该去除 最久未使用 的键。
+
+注意「项的使用次数」就是自插入该项以来对其调用 get 和 put 函数的次数之和。使用次数会在对应项被移除后置为 0 。
+
+进阶：你是否可以在 O(1) 时间复杂度内执行两项操作？
+
+## 解题思路
+
+- 这一题是 LFU 经典面试题，详细解释见第三章模板。
+
+## 代码
+
+```go
+package leetcode
+
+import "container/list"
+
+type LFUCache struct {
+	nodes    map[int]*list.Element
+	lists    map[int]*list.List
+	capacity int
+	min      int
+}
+
+type node struct {
+	key       int
+	value     int
+	frequency int
+}
+
+func Constructor(capacity int) LFUCache {
+	return LFUCache{nodes: make(map[int]*list.Element),
+		lists:    make(map[int]*list.List),
+		capacity: capacity,
+		min:      0,
+	}
+}
+
+func (this *LFUCache) Get(key int) int {
+	value, ok := this.nodes[key]
+	if !ok {
+		return -1
+	}
+	currentNode := value.Value.(*node)
+	this.lists[currentNode.frequency].Remove(value)
+	currentNode.frequency++
+	if _, ok := this.lists[currentNode.frequency]; !ok {
+		this.lists[currentNode.frequency] = list.New()
+	}
+	newList := this.lists[currentNode.frequency]
+	newNode := newList.PushBack(currentNode)
+	this.nodes[key] = newNode
+	if currentNode.frequency-1 == this.min && this.lists[currentNode.frequency-1].Len() == 0 {
+		this.min++
+	}
+	return currentNode.value
+}
+
+func (this *LFUCache) Put(key int, value int) {
+	if this.capacity == 0 {
+		return
+	}
+	if currentValue, ok := this.nodes[key]; ok {
+		currentNode := currentValue.Value.(*node)
+		currentNode.value = value
+		this.Get(key)
+		return
+	}
+	if this.capacity == len(this.nodes) {
+		currentList := this.lists[this.min]
+		frontNode := currentList.Front()
+		delete(this.nodes, frontNode.Value.(*node).key)
+		currentList.Remove(frontNode)
+	}
+	this.min = 1
+	currentNode := &node{
+		key:       key,
+		value:     value,
+		frequency: 1,
+	}
+	if _, ok := this.lists[1]; !ok {
+		this.lists[1] = list.New()
+	}
+	newList := this.lists[1]
+	newNode := newList.PushBack(currentNode)
+	this.nodes[key] = newNode
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0458.Poor-Pigs/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0461.Hamming-Distance/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0461.Hamming-Distance.md b/website/content/ChapterFour/0400~0499/0461.Hamming-Distance.md
new file mode 100755
index 000000000..c36e66eb7
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0461.Hamming-Distance.md
@@ -0,0 +1,60 @@
+# [461. Hamming Distance](https://leetcode.com/problems/hamming-distance/)
+
+## 题目
+
+The [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) between two integers is the number of positions at which the corresponding bits are different.
+
+Given two integers `x` and `y`, calculate the Hamming distance.
+
+**Note**: 0 ≤ `x`, `y` < 231.
+
+**Example**:
+
+    Input: x = 1, y = 4
+    
+    Output: 2
+    
+    Explanation:
+    1   (0 0 0 1)
+    4   (0 1 0 0)
+           ↑   ↑
+    
+    The above arrows point to positions where the corresponding bits are different.
+
+
+## 题目大意
+
+两个整数之间的汉明距离指的是这两个数字对应二进制位不同的位置的数目。给出两个整数 x 和 y，计算它们之间的汉明距离。
+
+注意：
+0 ≤ x, y < 231.
+
+
+
+## 解题思路
+
+- 求 2 个数的海明距离。海明距离的定义是两个数二进制位不同的总个数。这一题利用的位操作的是 X &= (X - 1) 不断的清除最低位的 1 。先将这两个数异或，异或以后清除低位的 1 就是最终答案。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func hammingDistance(x int, y int) int {
+	distance := 0
+	for xor := x ^ y; xor != 0; xor &= (xor - 1) {
+		distance++
+	}
+	return distance
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0460.LFU-Cache/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0462.Minimum-Moves-to-Equal-Array-Elements-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0462.Minimum-Moves-to-Equal-Array-Elements-II.md b/website/content/ChapterFour/0400~0499/0462.Minimum-Moves-to-Equal-Array-Elements-II.md
new file mode 100644
index 000000000..c43b0bdeb
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0462.Minimum-Moves-to-Equal-Array-Elements-II.md
@@ -0,0 +1,73 @@
+# [462. Minimum Moves to Equal Array Elements II](https://leetcode.com/problems/minimum-moves-to-equal-array-elements-ii/)
+
+
+## 题目
+
+Given an integer array `nums` of size `n`, return *the minimum number of moves required to make all array elements equal*.
+
+In one move, you can increment or decrement an element of the array by `1`.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3]
+Output: 2
+Explanation:
+Only two moves are needed (remember each move increments or decrements one element):
+[1,2,3]  =>  [2,2,3]  =>  [2,2,2]
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,10,2,9]
+Output: 16
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `1 <= nums.length <= 10^5`
+- `109 <= nums[i] <= 10^9`
+
+## 题目大意
+
+给定一个非空整数数组，找到使所有数组元素相等所需的最小移动数，其中每次移动可将选定的一个元素加 1 或减 1。 您可以假设数组的长度最多为10000。
+
+## 解题思路
+
+- 这题抽象成数学问题是，如果我们把数组 a 中的每个数看成水平轴上的一个点，那么根据上面的移动次数公式，我们需要找到在水平轴上找到一个点 x，使得这 N 个点到 x 的距离之和最小。有 2 个点值得我们考虑，一个是中位数，另外一个是平均值。举个简单的例子，[1,0,0,8,6] 这组数据，中位数是 1，平均值是 3 。分别计算移动的步数，按照中位数对齐是 14，按照平均值对齐是 16 。所以选择中位数。
+- 此题可以用数学证明，证明出，按照平均值移动的步数 ≥ 按照中位数移动的步数。具体证明笔者这里不证明了，感兴趣的同学可以自己证明试试。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func minMoves2(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	moves, mid := 0, len(nums)/2
+	sort.Ints(nums)
+	for i := range nums {
+		if i == mid {
+			continue
+		}
+		moves += int(math.Abs(float64(nums[mid] - nums[i])))
+	}
+	return moves
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0461.Hamming-Distance/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0463.Island-Perimeter/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0463.Island-Perimeter.md b/website/content/ChapterFour/0400~0499/0463.Island-Perimeter.md
new file mode 100755
index 000000000..3ac220658
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0463.Island-Perimeter.md
@@ -0,0 +1,78 @@
+# [463. Island Perimeter](https://leetcode.com/problems/island-perimeter/)
+
+## 题目
+
+You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water.
+
+Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells).
+
+The island doesn't have "lakes" (water inside that isn't connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don't exceed 100. Determine the perimeter of the island.
+
+**Example**:
+
+    Input:
+    [[0,1,0,0],
+     [1,1,1,0],
+     [0,1,0,0],
+     [1,1,0,0]]
+    
+    Output: 16
+    
+    Explanation: The perimeter is the 16 yellow stripes in the image below:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/island.png)
+
+
+## 题目大意
+
+给定一个包含 0 和 1 的二维网格地图，其中 1 表示陆地 0 表示水域。
+
+网格中的格子水平和垂直方向相连（对角线方向不相连）。整个网格被水完全包围，但其中恰好有一个岛屿（或者说，一个或多个表示陆地的格子相连组成的岛屿）。
+
+岛屿中没有“湖”（“湖” 指水域在岛屿内部且不和岛屿周围的水相连）。格子是边长为 1 的正方形。网格为长方形，且宽度和高度均不超过 100 。计算这个岛屿的周长。
+
+
+
+## 解题思路
+
+- 给出一个二维数组，二维数组中有一些连在一起的 1 ，这是一个岛屿，求这个岛屿的周长。
+- 这是一道水题，判断四周边界的情况依次加一即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func islandPerimeter(grid [][]int) int {
+	counter := 0
+	for i := 0; i < len(grid); i++ {
+		for j := 0; j < len(grid[0]); j++ {
+			if grid[i][j] == 1 {
+				if i-1 < 0 || grid[i-1][j] == 0 {
+					counter++
+				}
+				if i+1 >= len(grid) || grid[i+1][j] == 0 {
+					counter++
+				}
+				if j-1 < 0 || grid[i][j-1] == 0 {
+					counter++
+				}
+				if j+1 >= len(grid[0]) || grid[i][j+1] == 0 {
+					counter++
+				}
+			}
+		}
+	}
+	return counter
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0462.Minimum-Moves-to-Equal-Array-Elements-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0470.Implement-Rand10-Using-Rand7/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0470.Implement-Rand10-Using-Rand7.md b/website/content/ChapterFour/0400~0499/0470.Implement-Rand10-Using-Rand7.md
new file mode 100755
index 000000000..3e3ac3826
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0470.Implement-Rand10-Using-Rand7.md
@@ -0,0 +1,107 @@
+# [470. Implement Rand10() Using Rand7()](https://leetcode.com/problems/implement-rand10-using-rand7/)
+
+
+## 题目
+
+Given a function `rand7` which generates a uniform random integer in the range 1 to 7, write a function `rand10` which generates a uniform random integer in the range 1 to 10.
+
+Do NOT use system's `Math.random()`.
+
+**Example 1**:
+
+    Input: 1
+    Output: [7]
+
+**Example 2**:
+
+    Input: 2
+    Output: [8,4]
+
+**Example 3**:
+
+    Input: 3
+    Output: [8,1,10]
+
+**Note**:
+
+1. `rand7` is predefined.
+2. Each testcase has one argument: `n`, the number of times that `rand10` is called.
+
+**Follow up**:
+
+1. What is the [expected value](https://en.wikipedia.org/wiki/Expected_value) for the number of calls to `rand7()` function?
+2. Could you minimize the number of calls to `rand7()`?
+
+
+## 题目大意
+
+已有方法 rand7 可生成 1 到 7 范围内的均匀随机整数，试写一个方法 rand10 生成 1 到 10 范围内的均匀随机整数。不要使用系统的 Math.random() 方法。
+
+
+提示:
+
+- rand7 已定义。
+- 传入参数: n 表示 rand10 的调用次数。
+ 
+
+进阶:
+
+- rand7()调用次数的 期望值 是多少 ?
+- 你能否尽量少调用 rand7() ?
+
+
+
+## 解题思路
+
+
+- 给出 `rand7()` 要求实现 `rand10()`。
+- `rand7()` 等概率地产生1，2，3，4，5，6，7。要想得到 `rand10()` 即等概率的生成 1-10 。解题思路是先构造一个 `randN()`，这个 N 必须是 10 的整数倍，然后 randN % 10 就可以得到 `rand10()` 了。所以可以从 `rand7()` 先构造出 `rand49()`，再把 `rand49()` 中大于等于 40 的都过滤掉，这样就得到了 `rand40()`，在对 10 取余即可。
+- 具体构造步骤，`rand7() --> rand49() --> rand40() --> rand10()`：
+    1. `rand7()` 等概率地产生 1,2,3,4,5,6,7.
+    2. `rand7() - 1` 等概率地产生 [0,6].
+    3. `(rand7() - 1) *7` 等概率地产生 0, 7, 14, 21, 28, 35, 42
+    4. `(rand7() - 1) * 7 + (rand7() - 1)` 等概率地产生 [0, 48] 这 49 个数字
+    5. 如果步骤 4 的结果大于等于 40，那么就重复步骤 4，直到产生的数小于 40
+    6. 把步骤 5 的结果 mod 10 再加 1， 就会等概率的随机生成 [1, 10]
+- 这道题可以推广到生成任意数的随机数问题。用 `randN()` 实现 `randM()`，`M>N`。步骤如下：
+    1. 用 `randN()` 先实现 `randX()`，其中 X ≥ M，X 是 M 的整数倍。如这题中的 49 > 10；
+    2. 再用 `randX()` 生成 `randM()`，如这题中的 49 —> 40 —> 10。
+- 举个例子，用 `rand3()` 生成 `rand11()`，可以先生成 `rand27()`，然后变成以 22 为界限，因为 22 是 11 的倍数。生成 `rand27()` 的方式：`3 * 3 * (rand3() - 1) + 3 * (rand3() - 1) + (rand3() - 1)`，最后生成了 `rand11()`；用 `rand7()` 生成 `rand9()`，可以先生成 `rand49()`，然后变成以 45 为界限，因为 45 是 9 的倍数。生成 `rand49()` 的方式：`(rand7() - 1) * 7 + (rand7() - 1)`，最后生成了 `rand9()`；用 `rand6()` 生成 `rand13()`，可以先生成 `rand36()`，然后变成以 26 为界限，因为 26 是 13 的倍数。生成 `rand36()` 的方式：`(rand6() - 1) * 6 + (rand6() - 1)`，最后生成了 `rand13()`；
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math/rand"
+
+func rand10() int {
+	rand10 := 10
+	for rand10 >= 10 {
+		rand10 = (rand7() - 1) + rand7()
+	}
+	return rand10%10 + 1
+}
+
+func rand7() int {
+	return rand.Intn(7)
+}
+
+func rand101() int {
+	rand40 := 40
+	for rand40 >= 40 {
+		rand40 = (rand7()-1)*7 + rand7() - 1
+	}
+	return rand40%10 + 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0463.Island-Perimeter/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0473.Matchsticks-to-Square/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0473.Matchsticks-to-Square.md b/website/content/ChapterFour/0400~0499/0473.Matchsticks-to-Square.md
new file mode 100644
index 000000000..5ac15f049
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0473.Matchsticks-to-Square.md
@@ -0,0 +1,101 @@
+# [473. Matchsticks to Square](https://leetcode.com/problems/matchsticks-to-square/)
+
+
+## 题目
+
+You are given an integer array `matchsticks` where `matchsticks[i]` is the length of the `ith` matchstick. You want to use **all the matchsticks** to make one square. You **should not break** any stick, but you can link them up, and each matchstick must be used **exactly one time**.
+
+Return `true` if you can make this square and `false` otherwise.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/04/09/matchsticks1-grid.jpg](https://assets.leetcode.com/uploads/2021/04/09/matchsticks1-grid.jpg)
+
+```
+Input: matchsticks = [1,1,2,2,2]
+Output: true
+Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.
+```
+
+**Example 2:**
+
+```
+Input: matchsticks = [3,3,3,3,4]
+Output: false
+Explanation: You cannot find a way to form a square with all the matchsticks.
+```
+
+**Constraints:**
+
+- `1 <= matchsticks.length <= 15`
+- `0 <= matchsticks[i] <= 109`
+
+## 题目大意
+
+现在已知小女孩有多少根火柴，请找出一种能使用所有火柴拼成一个正方形的方法。不能折断火柴，可以把火柴连接起来，并且每根火柴都要用到。输入为小女孩拥有火柴的数目，每根火柴用其长度表示。输出即为是否能用所有的火柴拼成正方形。
+
+## 解题思路
+
+- 将火柴拼成一个正方形，可以将它们分成四组，每一根火柴恰好属于其中的一组；并且每一组火柴的长度之和都相同，等于所有火柴长度之和的四分之一。
+- 考虑暴力解法，使用深度优先搜索枚举出所有的分组情况，并对于每一种情况，判断是否满足上述的两个条件(每根火柴属于其中一组，每组火柴长度之和相同)。依次对每一根火柴进行搜索，当搜索到第 i 根火柴时，可以考虑把它放到四组中的任意一种。对于每一种放置方法，继续对第 i + 1 根火柴进行深搜。当我们搜索完全部的 N 根火柴后，再判断每一组火柴的长度之和是否都相同。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func makesquare(matchsticks []int) bool {
+	if len(matchsticks) < 4 {
+		return false
+	}
+	total := 0
+	for _, v := range matchsticks {
+		total += v
+	}
+	if total%4 != 0 {
+		return false
+	}
+	sort.Slice(matchsticks, func(i, j int) bool {
+		return matchsticks[i] > matchsticks[j]
+	})
+	visited := make([]bool, 16)
+	return dfs(matchsticks, 0, 0, 0, total, &visited)
+}
+
+func dfs(matchsticks []int, cur, group, sum, total int, visited *[]bool) bool {
+	if group == 4 {
+		return true
+	}
+	if sum > total/4 {
+		return false
+	}
+	if sum == total/4 {
+		return dfs(matchsticks, 0, group+1, 0, total, visited)
+	}
+	last := -1
+	for i := cur; i < len(matchsticks); i++ {
+		if (*visited)[i] {
+			continue
+		}
+		if last == matchsticks[i] {
+			continue
+		}
+		(*visited)[i] = true
+		last = matchsticks[i]
+		if dfs(matchsticks, i+1, group, sum+matchsticks[i], total, visited) {
+			return true
+		}
+		(*visited)[i] = false
+	}
+	return false
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0470.Implement-Rand10-Using-Rand7/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0474.Ones-and-Zeroes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0474.Ones-and-Zeroes.md b/website/content/ChapterFour/0400~0499/0474.Ones-and-Zeroes.md
new file mode 100755
index 000000000..7cd4969c7
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0474.Ones-and-Zeroes.md
@@ -0,0 +1,85 @@
+# [474. Ones and Zeroes](https://leetcode.com/problems/ones-and-zeroes/)
+
+
+## 题目
+
+In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.
+
+For now, suppose you are a dominator of **m** `0s` and **n** `1s` respectively. On the other hand, there is an array with strings consisting of only `0s` and `1s`.
+
+Now your task is to find the maximum number of strings that you can form with given **m**`0s` and **n** `1s`. Each `0` and `1` can be used at most **once**.
+
+**Note**:
+
+1. The given numbers of `0s` and `1s` will both not exceed `100`
+2. The size of given string array won't exceed `600`.
+
+**Example 1**:
+
+    Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3
+    Output: 4
+    
+    Explanation: This are totally 4 strings can be formed by the using of 5 0s and 3 1s, which are “10,”0001”,”1”,”0”
+
+**Example 2**:
+
+    Input: Array = {"10", "0", "1"}, m = 1, n = 1
+    Output: 2
+    
+    Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".
+
+
+## 题目大意
+
+在计算机界中，我们总是追求用有限的资源获取最大的收益。现在，假设你分别支配着 m 个 0 和 n 个 1。另外，还有一个仅包含 0 和 1 字符串的数组。你的任务是使用给定的 m 个 0 和 n 个 1 ，找到能拼出存在于数组中的字符串的最大数量。每个 0 和 1 至多被使用一次。
+
+注意:
+
+1. 给定 0 和 1 的数量都不会超过 100。
+2. 给定字符串数组的长度不会超过 600。
+
+
+
+## 解题思路
+
+- 给定一个字符串数组和 m，n，其中所有的字符都是由 0 和 1 组成的。问能否从数组中取出最多的字符串，使得这些取出的字符串中所有的 0 的个数 ≤ m，1 的个数 ≤ n。
+- 这一题是典型的 01 背包的题型。只不过是一个二维的背包问题。在 n 个物品中选出一定物品，**尽量完全填满** m 维和 n 维的背包。为什么是尽量填满？因为不一定能完全填满背包。
+- `dp[i][j]` 代表尽量填满容量为 `(i,j)` 的背包装下的物品总数，状态转移方程为 `dp[i][j] = max(dp[i][j], 1+dp[i-zero][j-one])`。其中 zero 代表的当前装入物品在 m 维上的体积，也即 0 的个数。one 代表的是当前装入物品在 n 维上的体积，也即 1 的个数。每添加一个物品，比较当前 (i,j) 的背包装下的物品总数和 (i-zero,j-one) 的背包装下的物品总数 + 1，比较这两者的大小，保存两者的最大值。每添加一个物品就刷新这个二维背包，直到所有物品都扫完一遍。dp[m][n] 中存储的就是最终的答案。时间复杂度 `O( n * M * N )`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func findMaxForm(strs []string, m int, n int) int {
+	dp := make([][]int, m+1)
+	for i := 0; i < m+1; i++ {
+		dp[i] = make([]int, n+1)
+	}
+	for _, s := range strs {
+		zero := strings.Count(s, "0")
+		one := len(s) - zero
+		if zero > m || one > n {
+			continue
+		}
+		for i := m; i >= zero; i-- {
+			for j := n; j >= one; j-- {
+				dp[i][j] = max(dp[i][j], 1+dp[i-zero][j-one])
+			}
+		}
+	}
+	return dp[m][n]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0473.Matchsticks-to-Square/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0475.Heaters/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0475.Heaters.md b/website/content/ChapterFour/0400~0499/0475.Heaters.md
new file mode 100755
index 000000000..90e81ad35
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0475.Heaters.md
@@ -0,0 +1,129 @@
+# [475. Heaters](https://leetcode.com/problems/heaters/)
+
+## 题目
+
+Winter is coming! Your first job during the contest is to design a standard heater with fixed warm radius to warm all the houses.
+
+Now, you are given positions of houses and heaters on a horizontal line, find out minimum radius of heaters so that all houses could be covered by those heaters.
+
+So, your input will be the positions of houses and heaters seperately, and your expected output will be the minimum radius standard of heaters.
+
+**Note**:
+
+1. Numbers of houses and heaters you are given are non-negative and will not exceed 25000.
+2. Positions of houses and heaters you are given are non-negative and will not exceed 10^9.
+3. As long as a house is in the heaters' warm radius range, it can be warmed.
+4. All the heaters follow your radius standard and the warm radius will the same.
+
+**Example 1**:
+
+    Input: [1,2,3],[2]
+    Output: 1
+    Explanation: The only heater was placed in the position 2, and if we use the radius 1 standard, then all the houses can be warmed.
+
+**Example 2**:
+
+    Input: [1,2,3,4],[1,4]
+    Output: 1
+    Explanation: The two heater was placed in the position 1 and 4. We need to use radius 1 standard, then all the houses can be warmed.
+
+
+
+## 题目大意
+
+
+冬季已经来临。 你的任务是设计一个有固定加热半径的供暖器向所有房屋供暖。现在，给出位于一条水平线上的房屋和供暖器的位置，找到可以覆盖所有房屋的最小加热半径。所以，你的输入将会是房屋和供暖器的位置。你将输出供暖器的最小加热半径。
+
+说明:
+
+- 给出的房屋和供暖器的数目是非负数且不会超过 25000。
+- 给出的房屋和供暖器的位置均是非负数且不会超过10^9。
+- 只要房屋位于供暖器的半径内(包括在边缘上)，它就可以得到供暖。
+- 所有供暖器都遵循你的半径标准，加热的半径也一样。
+
+
+
+## 解题思路
+
+
+- 给出一个房子坐标的数组，和一些供暖器坐标的数组，分别表示房子的坐标和供暖器的坐标。要求找到供暖器最小的半径能使得所有的房子都能享受到暖气。
+- 这一题可以用暴力的解法，暴力解法依次遍历每个房子的坐标，再遍历每个供暖器，找到距离房子最近的供暖器坐标。在所有这些距离的长度里面找到最大值，这个距离的最大值就是供暖器半径的最小值。时间复杂度 O(n^2)。
+- 这一题最优解是二分搜索。在查找距离房子最近的供暖器的时候，先将供暖器排序，然后用二分搜索的方法查找。其他的做法和暴力解法一致。时间复杂度 O(n log n)。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func findRadius(houses []int, heaters []int) int {
+	minRad := 0
+	sort.Ints(heaters)
+	for _, house := range houses {
+		// 遍历房子的坐标，维护 heaters 的最小半径
+		heater := findClosestHeater(house, heaters)
+		rad := heater - house
+		if rad < 0 {
+			rad = -rad
+		}
+		if rad > minRad {
+			minRad = rad
+		}
+	}
+	return minRad
+
+}
+
+// 二分搜索
+func findClosestHeater(pos int, heaters []int) int {
+	low, high := 0, len(heaters)-1
+	if pos < heaters[low] {
+		return heaters[low]
+	}
+	if pos > heaters[high] {
+		return heaters[high]
+	}
+	for low <= high {
+		mid := low + (high-low)>>1
+		if pos == heaters[mid] {
+			return heaters[mid]
+		} else if pos < heaters[mid] {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	// 判断距离两边的 heaters 哪个更近
+	if pos-heaters[high] < heaters[low]-pos {
+		return heaters[high]
+	}
+	return heaters[low]
+}
+
+// 解法二 暴力搜索
+func findRadius1(houses []int, heaters []int) int {
+	res := 0
+	for i := 0; i < len(houses); i++ {
+		dis := math.MaxInt64
+		for j := 0; j < len(heaters); j++ {
+			dis = min(dis, abs(houses[i]-heaters[j]))
+		}
+		res = max(res, dis)
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0474.Ones-and-Zeroes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0476.Number-Complement/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0476.Number-Complement.md b/website/content/ChapterFour/0400~0499/0476.Number-Complement.md
new file mode 100755
index 000000000..41198d109
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0476.Number-Complement.md
@@ -0,0 +1,75 @@
+# [476. Number Complement](https://leetcode.com/problems/number-complement/)
+
+
+## 题目
+
+Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
+
+**Note**:
+
+1. The given integer is guaranteed to fit within the range of a 32-bit signed integer.
+2. You could assume no leading zero bit in the integer’s binary representation.
+
+**Example 1**:
+
+    Input: 5
+    Output: 2
+    Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.
+
+**Example 2**:
+
+    Input: 1
+    Output: 0
+    Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.
+
+
+## 题目大意
+
+给定一个正整数，输出它的补数。补数是对该数的二进制表示取反。
+
+注意:
+
+给定的整数保证在32位带符号整数的范围内。
+你可以假定二进制数不包含前导零位。
+
+
+
+## 解题思路
+
+
+- 求一个正数的补数，补数的定义是对该数的二进制表示取反。当前不能改变符号位。按照题意构造相应的 mask 再取反即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func findComplement(num int) int {
+	xx := ^0 // ^0 = 1111111111111111111111
+	for xx&num > 0 {
+		xx <<= 1 // 构造出来的 xx = 1111111…000000，0 的个数就是 num 的长度
+	}
+	return ^xx ^ num // xx ^ num，结果是前面的 0 全是 1 的num，再取反即是答案
+}
+
+// 解法二
+func findComplement1(num int) int {
+	temp := 1
+	for temp <= num {
+		temp <<= 1 // 构造出来的 temp = 00000……10000，末尾 0 的个数是 num 的长度
+	}
+	return (temp - 1) ^ num // temp - 1 即是前面都是 0，num 长度的末尾都是 1 的数，再异或 num 即是最终结果
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0475.Heaters/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0477.Total-Hamming-Distance/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0477.Total-Hamming-Distance.md b/website/content/ChapterFour/0400~0499/0477.Total-Hamming-Distance.md
new file mode 100755
index 000000000..2ac93c9f7
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0477.Total-Hamming-Distance.md
@@ -0,0 +1,73 @@
+# [477. Total Hamming Distance](https://leetcode.com/problems/total-hamming-distance/)
+
+
+## 题目
+
+The [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) between two integers is the number of positions at which the corresponding bits are different.
+
+Now your job is to find the total Hamming distance between all pairs of the given numbers.
+
+**Example**:
+
+    Input: 4, 14, 2
+    
+    Output: 6
+    
+    Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
+    showing the four bits relevant in this case). So the answer will be:
+    HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
+
+**Note**:
+
+1. Elements of the given array are in the range of `0` to `10^9`
+2. Length of the array will not exceed `10^4`.
+
+
+## 题目大意
+
+两个整数的[汉明距离](https://baike.baidu.com/item/%E6%B1%89%E6%98%8E%E8%B7%9D%E7%A6%BB/475174?fr=aladdin)指的是这两个数字的二进制数对应位不同的数量。计算一个数组中，任意两个数之间汉明距离的总和。
+
+
+## 解题思路
+
+- 计算一个数组内两两元素的海明距离总和。海明距离的定义是两个数二进制位不同的总个数。那么可以把数组中的每个元素 32 位的二进制位依次扫一遍，当扫到某一位上的时候，有 k 个元素在这个位上的值是 1，n - k 个元素在这个位上的值是 0，那么在这一位上所有两两元素的海明距离是 k*(n-k) ，当把 32 位全部都扫完以后，累加出来的海明距离就是所有两两元素的海明距离。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func totalHammingDistance(nums []int) int {
+	total, n := 0, len(nums)
+	for i := 0; i < 32; i++ {
+		bitCount := 0
+		for j := 0; j < n; j++ {
+			bitCount += (nums[j] >> uint(i)) & 1
+		}
+		total += bitCount * (n - bitCount)
+	}
+	return total
+}
+
+// 暴力解法超时！
+func totalHammingDistance1(nums []int) int {
+	res := 0
+	for i := 0; i < len(nums); i++ {
+		for j := i + 1; j < len(nums); j++ {
+			res += hammingDistance(nums[i], nums[j])
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0476.Number-Complement/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0478.Generate-Random-Point-in-a-Circle/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0478.Generate-Random-Point-in-a-Circle.md b/website/content/ChapterFour/0400~0499/0478.Generate-Random-Point-in-a-Circle.md
new file mode 100644
index 000000000..c76e82fb7
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0478.Generate-Random-Point-in-a-Circle.md
@@ -0,0 +1,110 @@
+# [478. Generate Random Point in a Circle](https://leetcode.com/problems/generate-random-point-in-a-circle/)
+
+
+## 题目
+
+Given the radius and x-y positions of the center of a circle, write a function `randPoint` which generates a uniform random point in the circle.
+
+Note:
+
+1. input and output values are in [floating-point](https://www.webopedia.com/TERM/F/floating_point_number.html).
+2. radius and x-y position of the center of the circle is passed into the class constructor.
+3. a point on the circumference of the circle is considered to be in the circle.
+4. `randPoint` returns a size 2 array containing x-position and y-position of the random point, in that order.
+
+**Example 1:**
+
+```
+Input: 
+["Solution","randPoint","randPoint","randPoint"]
+[[1,0,0],[],[],[]]
+Output: [null,[-0.72939,-0.65505],[-0.78502,-0.28626],[-0.83119,-0.19803]]
+
+```
+
+**Example 2:**
+
+```
+Input: 
+["Solution","randPoint","randPoint","randPoint"]
+[[10,5,-7.5],[],[],[]]
+Output: [null,[11.52438,-8.33273],[2.46992,-16.21705],[11.13430,-12.42337]]
+```
+
+**Explanation of Input Syntax:**
+
+The input is two lists: the subroutines called and their arguments. `Solution`'s constructor has three arguments, the radius, x-position of the center, and y-position of the center of the circle. `randPoint` has no arguments. Arguments are always wrapped with a list, even if there aren't any.
+
+## 题目大意
+
+给定圆的半径和圆心的 x、y 坐标，写一个在圆中产生均匀随机点的函数 randPoint 。
+
+说明:
+
+- 输入值和输出值都将是浮点数。
+- 圆的半径和圆心的 x、y 坐标将作为参数传递给类的构造函数。
+- 圆周上的点也认为是在圆中。
+- randPoint 返回一个包含随机点的x坐标和y坐标的大小为2的数组。
+
+## 解题思路
+
+- 随机产生一个圆内的点，这个点一定满足定义 `(x-a)^2+(y-b)^2 ≤ R^2`，其中 `(a,b)` 是圆的圆心坐标，`R` 是半径。
+- 先假设圆心坐标在 (0,0)，这样方便计算，最终输出坐标的时候整体加上圆心的偏移量即可。`rand.Float64()` 产生一个 `[0.0,1.0)` 区间的浮点数。`-R ≤ 2 * R * rand() - R < R`，利用随机产生坐标点的横纵坐标 `(x,y)` 与半径 R 的关系，如果 `x^2 + y^2 ≤ R^2`，那么说明产生的点在圆内。最终输出的时候要记得加上圆心坐标的偏移值。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+	"math/rand"
+	"time"
+)
+
+type Solution struct {
+	r float64
+	x float64
+	y float64
+}
+
+func Constructor(radius float64, x_center float64, y_center float64) Solution {
+	rand.Seed(time.Now().UnixNano())
+	return Solution{radius, x_center, y_center}
+}
+
+func (this *Solution) RandPoint() []float64 {
+	/*
+	   a := angle()
+	   r := this.r * math.Sqrt(rand.Float64())
+	   x := r * math.Cos(a) + this.x
+	   y := r * math.Sin(a) + this.y
+	   return []float64{x, y}*/
+	for {
+		rx := 2*rand.Float64() - 1.0
+		ry := 2*rand.Float64() - 1.0
+		x := this.r * rx
+		y := this.r * ry
+		if x*x+y*y <= this.r*this.r {
+			return []float64{x + this.x, y + this.y}
+		}
+	}
+}
+
+func angle() float64 {
+	return rand.Float64() * 2 * math.Pi
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(radius, x_center, y_center);
+ * param_1 := obj.RandPoint();
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0477.Total-Hamming-Distance/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0480.Sliding-Window-Median/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0480.Sliding-Window-Median.md b/website/content/ChapterFour/0400~0499/0480.Sliding-Window-Median.md
new file mode 100755
index 000000000..01e40d5d3
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0480.Sliding-Window-Median.md
@@ -0,0 +1,283 @@
+# [480. Sliding Window Median](https://leetcode.com/problems/sliding-window-median/)
+
+
+
+## 题目
+
+Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.
+
+**Examples**:  
+
+`[2,3,4]` , the median is `3`
+
+`[2,3]`, the median is `(2 + 3) / 2 = 2.5`
+
+Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Your job is to output the median array for each window in the original array.
+
+For example,  
+
+Given nums = `[1,3,-1,-3,5,3,6,7]`, and k = 3.
+
+    Window position                Median
+    ---------------               -----
+    [1  3  -1] -3  5  3  6  7       1
+     1 [3  -1  -3] 5  3  6  7       -1
+     1  3 [-1  -3  5] 3  6  7       -1
+     1  3  -1 [-3  5  3] 6  7       3
+     1  3  -1  -3 [5  3  6] 7       5
+     1  3  -1  -3  5 [3  6  7]      6
+
+Therefore, return the median sliding window as `[1,-1,-1,3,5,6]`.
+
+**Note**:   
+
+You may assume `k` is always valid, ie: `k` is always smaller than input array's size for non-empty array.
+
+
+## 题目大意
+
+中位数是有序序列最中间的那个数。如果序列的大小是偶数，则没有最中间的数；此时中位数是最中间的两个数的平均数。
+
+例如：
+
+[2,3,4]，中位数是 3
+
+[2,3]，中位数是 (2 + 3) / 2 = 2.5
+
+给出一个数组 nums，有一个大小为 k 的窗口从最左端滑动到最右端。窗口中有 k 个数，每次窗口移动 1 位。你的任务是找出每次窗口移动后得到的新窗口中元素的中位数，并输出由它们组成的数组。
+
+
+
+## 解题思路
+
+
+- 给定一个数组和一个窗口为 K 的窗口，当窗口从数组的左边滑动到数组右边的时候，输出每次移动窗口以后，在窗口内的排序之后中间大小的值。
+- 这一题是第 239 题的升级版。
+- 这道题最暴力的方法就是将窗口内的元素都排序，时间复杂度 O(n * K)。
+- 另一种思路是用两个优先队列，大顶堆里面的元素都比小顶堆里面的元素小。小顶堆里面存储排序以后中间靠后的值大的元素，大顶堆里面存储排序以后中间靠前的值小的元素。如果 k 是偶数，那么两个堆都有 k/2 个元素，中间值就是两个堆顶的元素；如果 k 是奇数，那么小顶堆比大顶堆多一个元素，中间值就是小顶堆的堆顶元素。删除一个元素，元素都标记到删除的堆中，取 top 的时候注意需要取出没有删除的元素。时间复杂度 O(n * log k) 空间复杂度 O(k)
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"container/heap"
+	"container/list"
+	"sort"
+)
+
+// 解法一 用链表按照题意实现 时间复杂度 O(n * k) 空间复杂度 O(k)
+func medianSlidingWindow(nums []int, k int) []float64 {
+	var res []float64
+	w := getWindowList(nums[:k], k)
+	res = append(res, getMedian(w, k))
+
+	for p1 := k; p1 < len(nums); p1++ {
+		w = removeFromWindow(w, nums[p1-k])
+		w = insertInWindow(w, nums[p1])
+		res = append(res, getMedian(w, k))
+	}
+	return res
+}
+
+func getWindowList(nums []int, k int) *list.List {
+	s := make([]int, k)
+	copy(s, nums)
+	sort.Ints(s)
+	l := list.New()
+	for _, n := range s {
+		l.PushBack(n)
+	}
+	return l
+}
+
+func removeFromWindow(w *list.List, n int) *list.List {
+	for e := w.Front(); e != nil; e = e.Next() {
+		if e.Value.(int) == n {
+			w.Remove(e)
+			return w
+		}
+	}
+	return w
+}
+
+func insertInWindow(w *list.List, n int) *list.List {
+	for e := w.Front(); e != nil; e = e.Next() {
+		if e.Value.(int) >= n {
+			w.InsertBefore(n, e)
+			return w
+		}
+	}
+	w.PushBack(n)
+	return w
+}
+
+func getMedian(w *list.List, k int) float64 {
+	e := w.Front()
+	for i := 0; i < k/2; e, i = e.Next(), i+1 {
+	}
+	if k%2 == 1 {
+		return float64(e.Value.(int))
+	}
+	p := e.Prev()
+	return (float64(e.Value.(int)) + float64(p.Value.(int))) / 2
+}
+
+// 解法二 用两个堆实现 时间复杂度 O(n * log k) 空间复杂度 O(k)
+// 用两个堆记录窗口内的值
+// 大顶堆里面的元素都比小顶堆里面的元素小
+// 如果 k 是偶数，那么两个堆都有 k/2 个元素，中间值就是两个堆顶的元素
+// 如果 k 是奇数，那么小顶堆比大顶堆多一个元素，中间值就是小顶堆的堆顶元素
+// 删除一个元素，元素都标记到删除的堆中，取 top 的时候注意需要取出没有删除的元素
+func medianSlidingWindow1(nums []int, k int) []float64 {
+	ans := []float64{}
+	minH := MinHeapR{}
+	maxH := MaxHeapR{}
+	if minH.Len() > maxH.Len()+1 {
+		maxH.Push(minH.Pop())
+	} else if minH.Len() < maxH.Len() {
+		minH.Push(maxH.Pop())
+	}
+	for i := range nums {
+		if minH.Len() == 0 || nums[i] >= minH.Top() {
+			minH.Push(nums[i])
+		} else {
+			maxH.Push(nums[i])
+		}
+		if i >= k {
+			if nums[i-k] >= minH.Top() {
+				minH.Remove(nums[i-k])
+			} else {
+				maxH.Remove(nums[i-k])
+			}
+		}
+		if minH.Len() > maxH.Len()+1 {
+			maxH.Push(minH.Pop())
+		} else if minH.Len() < maxH.Len() {
+			minH.Push(maxH.Pop())
+		}
+		if minH.Len()+maxH.Len() == k {
+			if k%2 == 0 {
+				ans = append(ans, float64(minH.Top()+maxH.Top())/2.0)
+			} else {
+				ans = append(ans, float64(minH.Top()))
+			}
+		}
+		// fmt.Printf("%+v, %+v\n", minH, maxH)
+	}
+	return ans
+}
+
+// IntHeap define
+type IntHeap struct {
+	data []int
+}
+
+// Len define
+func (h IntHeap) Len() int { return len(h.data) }
+
+// Swap define
+func (h IntHeap) Swap(i, j int) { h.data[i], h.data[j] = h.data[j], h.data[i] }
+
+// Push define
+func (h *IntHeap) Push(x interface{}) { h.data = append(h.data, x.(int)) }
+
+// Pop define
+func (h *IntHeap) Pop() interface{} {
+	x := h.data[h.Len()-1]
+	h.data = h.data[0 : h.Len()-1]
+	return x
+}
+
+// Top defines
+func (h IntHeap) Top() int {
+	return h.data[0]
+}
+
+// MinHeap define
+type MinHeap struct {
+	IntHeap
+}
+
+// Less define
+func (h MinHeap) Less(i, j int) bool { return h.data[i] < h.data[j] }
+
+// MaxHeap define
+type MaxHeap struct {
+	IntHeap
+}
+
+// Less define
+func (h MaxHeap) Less(i, j int) bool { return h.data[i] > h.data[j] }
+
+// MinHeapR define
+type MinHeapR struct {
+	hp, hpDel MinHeap
+}
+
+// Len define
+func (h MinHeapR) Len() int { return h.hp.Len() - h.hpDel.Len() }
+
+// Top define
+func (h *MinHeapR) Top() int {
+	for h.hpDel.Len() > 0 && h.hp.Top() == h.hpDel.Top() {
+		heap.Pop(&h.hp)
+		heap.Pop(&h.hpDel)
+	}
+	return h.hp.Top()
+}
+
+// Pop define
+func (h *MinHeapR) Pop() int {
+	x := h.Top()
+	heap.Pop(&h.hp)
+	return x
+}
+
+// Push define
+func (h *MinHeapR) Push(x int) { heap.Push(&h.hp, x) }
+
+// Remove define
+func (h *MinHeapR) Remove(x int) { heap.Push(&h.hpDel, x) }
+
+// MaxHeapR define
+type MaxHeapR struct {
+	hp, hpDel MaxHeap
+}
+
+// Len define
+func (h MaxHeapR) Len() int { return h.hp.Len() - h.hpDel.Len() }
+
+// Top define
+func (h *MaxHeapR) Top() int {
+	for h.hpDel.Len() > 0 && h.hp.Top() == h.hpDel.Top() {
+		heap.Pop(&h.hp)
+		heap.Pop(&h.hpDel)
+	}
+	return h.hp.Top()
+}
+
+// Pop define
+func (h *MaxHeapR) Pop() int {
+	x := h.Top()
+	heap.Pop(&h.hp)
+	return x
+}
+
+// Push define
+func (h *MaxHeapR) Push(x int) { heap.Push(&h.hp, x) }
+
+// Remove define
+func (h *MaxHeapR) Remove(x int) { heap.Push(&h.hpDel, x) }
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0478.Generate-Random-Point-in-a-Circle/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0483.Smallest-Good-Base/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0483.Smallest-Good-Base.md b/website/content/ChapterFour/0400~0499/0483.Smallest-Good-Base.md
new file mode 100755
index 000000000..817478cec
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0483.Smallest-Good-Base.md
@@ -0,0 +1,136 @@
+# [483. Smallest Good Base](https://leetcode.com/problems/smallest-good-base/)
+
+
+## 题目
+
+For an integer n, we call k>=2 a **good base** of n, if all digits of n base k are 1.
+
+Now given a string representing n, you should return the smallest good base of n in string format.
+
+**Example 1**:
+
+    Input: "13"
+    Output: "3"
+    Explanation: 13 base 3 is 111.
+
+**Example 2**:
+
+    Input: "4681"
+    Output: "8"
+    Explanation: 4681 base 8 is 11111.
+
+**Example 3**:
+
+    Input: "1000000000000000000"
+    Output: "999999999999999999"
+    Explanation: 1000000000000000000 base 999999999999999999 is 11.
+
+**Note**:
+
+1. The range of n is [3, 10^18].
+2. The string representing n is always valid and will not have leading zeros.
+
+
+## 题目大意
+
+
+对于给定的整数 n, 如果n的k（k>=2）进制数的所有数位全为1，则称 k（k>=2）是 n 的一个好进制。
+
+以字符串的形式给出 n, 以字符串的形式返回 n 的最小好进制。
+
+提示：
+
+- n 的取值范围是 [3, 10^18]。
+- 输入总是有效且没有前导 0。
+
+
+
+## 解题思路
+
+
+- 给出一个数 n，要求找一个进制 k，使得数字 n 在 k 进制下每一位都是 1 。求最小的进制 k。
+- 这一题等价于求最小的正整数 k，满足存在一个正整数 m 使得
+
+{{< katex display >}} 
+	\sum_{i=0}^{m} k^{i} = \frac{1-k^{m+1}}{1-k} = n
+{{< /katex >}}
+
+
+- 这一题需要确定 k 和 m 两个数的值。m 和 k 是有关系的，确定了一个值，另外一个值也确定了。由
+
+{{< katex display >}} 
+	\frac{1-k^{m+1}}{1-k} = n \\
+{{< /katex >}}
+
+
+可得：
+
+{{< katex display >}} 
+	m = log_{k}(kn-n+1) - 1 < log_{k}(kn) = 1 + log_{k}n
+{{< /katex >}}
+
+
+根据题意，可以知道 k ≥2，m ≥1 ，所以有:
+
+{{< katex display >}} 
+	1 \leqslant m \leqslant log_{2}n
+{{< /katex >}}
+
+
+所以 m 的取值范围确定了。那么外层循环从 1 到 log n 遍历。找到一个最小的 k ，能满足：
+
+可以用二分搜索来逼近找到最小的 k。先找到 k 的取值范围。由 
+
+{{< katex display >}} 
+	\frac{1-k^{m+1}}{1-k} = n \\
+{{< /katex >}}
+
+
+可得，
+
+{{< katex display >}} 
+	k^{m+1} = nk-n+1 < nk\\ \Rightarrow k < \sqrt[m]{n}
+{{< /katex >}}
+
+
+所以 k 的取值范围是 [2, n*(1/m) ]。再利用二分搜索逼近找到最小的 k 即为答案。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+	"math/bits"
+	"strconv"
+)
+
+func smallestGoodBase(n string) string {
+	nVal, _ := strconv.Atoi(n)
+	mMax := bits.Len(uint(nVal)) - 1
+	for m := mMax; m > 1; m-- {
+		k := int(math.Pow(float64(nVal), 1/float64(m)))
+		mul, sum := 1, 1
+		for i := 0; i < m; i++ {
+			mul *= k
+			sum += mul
+		}
+		if sum == nVal {
+			return strconv.Itoa(k)
+		}
+	}
+	return strconv.Itoa(nVal - 1)
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0480.Sliding-Window-Median/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0485.Max-Consecutive-Ones/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0485.Max-Consecutive-Ones.md b/website/content/ChapterFour/0400~0499/0485.Max-Consecutive-Ones.md
new file mode 100644
index 000000000..dce4ba166
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0485.Max-Consecutive-Ones.md
@@ -0,0 +1,66 @@
+# [485. Max Consecutive Ones](https://leetcode.com/problems/max-consecutive-ones/)
+
+
+## 题目
+
+Given a binary array, find the maximum number of consecutive 1s in this array.
+
+**Example 1**:
+
+```
+Input: [1,1,0,1,1,1]
+Output: 3
+Explanation: The first two digits or the last three digits are consecutive 1s.
+    The maximum number of consecutive 1s is 3.
+```
+
+**Note**:
+
+- The input array will only contain `0` and `1`.
+- The length of input array is a positive integer and will not exceed 10,000
+
+
+## 题目大意
+
+给定一个二进制数组， 计算其中最大连续1的个数。
+
+注意：
+
+- 输入的数组只包含 0 和 1。
+- 输入数组的长度是正整数，且不超过 10,000。
+
+
+## 解题思路
+
+- 给定一个二进制数组， 计算其中最大连续1的个数。
+- 简单题。扫一遍数组，累计 1 的个数，动态维护最大的计数，最终输出即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func findMaxConsecutiveOnes(nums []int) int {
+	maxCount, currentCount := 0, 0
+	for _, v := range nums {
+		if v == 1 {
+			currentCount++
+		} else {
+			currentCount = 0
+		}
+		if currentCount > maxCount {
+			maxCount = currentCount
+		}
+	}
+	return maxCount
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0483.Smallest-Good-Base/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0488.Zuma-Game/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0488.Zuma-Game.md b/website/content/ChapterFour/0400~0499/0488.Zuma-Game.md
new file mode 100644
index 000000000..3b17baa76
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0488.Zuma-Game.md
@@ -0,0 +1,148 @@
+# [488. Zuma Game](https://leetcode.com/problems/zuma-game/)
+
+
+## 题目
+
+You are playing a variation of the game Zuma.
+
+In this variation of Zuma, there is a single row of colored balls on a board, where each ball can be colored red 'R', yellow 'Y', blue 'B', green 'G', or white 'W'. You also have several colored balls in your hand.
+
+Your goal is to clear all of the balls from the board. On each turn:
+
+Pick any ball from your hand and insert it in between two balls in the row or on either end of the row.
+If there is a group of three or more consecutive balls of the same color, remove the group of balls from the board.
+If this removal causes more groups of three or more of the same color to form, then continue removing each group until there are none left.
+If there are no more balls on the board, then you win the game.
+Repeat this process until you either win or do not have any more balls in your hand.
+Given a string board, representing the row of balls on the board, and a string hand, representing the balls in your hand, return the minimum number of balls you have to insert to clear all the balls from the board. If you cannot clear all the balls from the board using the balls in your hand, return -1.
+
+**Example 1**:
+
+```
+Input: board = "WRRBBW", hand = "RB"
+Output: -1
+Explanation: It is impossible to clear all the balls. The best you can do is:
+- Insert 'R' so the board becomes WRRRBBW. WRRRBBW -> WBBW.
+- Insert 'B' so the board becomes WBBBW. WBBBW -> WW.
+There are still balls remaining on the board, and you are out of balls to insert.
+```
+
+**Example 2**:
+```
+Input: board = "WWRRBBWW", hand = "WRBRW"
+Output: 2
+Explanation: To make the board empty:
+- Insert 'R' so the board becomes WWRRRBBWW. WWRRRBBWW -> WWBBWW.
+- Insert 'B' so the board becomes WWBBBWW. WWBBBWW -> WWWW -> empty.
+2 balls from your hand were needed to clear the board.
+```
+
+**Example 3**:
+```
+Input: board = "G", hand = "GGGGG"
+Output: 2
+Explanation: To make the board empty:
+- Insert 'G' so the board becomes GG.
+- Insert 'G' so the board becomes GGG. GGG -> empty.
+2 balls from your hand were needed to clear the board.
+```
+
+**Example 4**:
+```
+Input: board = "RBYYBBRRB", hand = "YRBGB"
+Output: 3
+Explanation: To make the board empty:
+- Insert 'Y' so the board becomes RBYYYBBRRB. RBYYYBBRRB -> RBBBRRB -> RRRB -> B.
+- Insert 'B' so the board becomes BB.
+- Insert 'B' so the board becomes BBB. BBB -> empty.
+3 balls from your hand were needed to clear the board.
+```
+
+**Constraints**:
+
+- 1 <= board.length <= 16
+- 1 <= hand.length <= 5
+- board and hand consist of the characters 'R', 'Y', 'B', 'G', and 'W'.
+- The initial row of balls on the board will not have any groups of three or more consecutive balls of the same color.
+
+## 题目大意
+
+你正在参与祖玛游戏的一个变种。
+
+在这个祖玛游戏变体中，桌面上有 一排 彩球，每个球的颜色可能是：红色 'R'、黄色 'Y'、蓝色 'B'、绿色 'G' 或白色 'W' 。你的手中也有一些彩球。
+
+你的目标是 清空 桌面上所有的球。每一回合：
+
+从你手上的彩球中选出 任意一颗 ，然后将其插入桌面上那一排球中：两球之间或这一排球的任一端。
+接着，如果有出现 三个或者三个以上 且 颜色相同 的球相连的话，就把它们移除掉。
+如果这种移除操作同样导致出现三个或者三个以上且颜色相同的球相连，则可以继续移除这些球，直到不再满足移除条件。
+如果桌面上所有球都被移除，则认为你赢得本场游戏。
+重复这个过程，直到你赢了游戏或者手中没有更多的球。
+给你一个字符串 board ，表示桌面上最开始的那排球。另给你一个字符串 hand ，表示手里的彩球。请你按上述操作步骤移除掉桌上所有球，计算并返回所需的 最少 球数。如果不能移除桌上所有的球，返回 -1 。
+
+## 解题思路
+
+- 使用广度优先搜索和剪枝
+
+## 代码
+
+```go
+
+package leetcode
+
+func findMinStep(board string, hand string) int {
+	q := [][]string{{board, hand}}
+	mp := make(map[string]bool)
+	minStep := 0
+	for len(q) > 0 {
+		length := len(q)
+		minStep++
+		for length > 0 {
+			length--
+			cur := q[0]
+			q = q[1:]
+			curB, curH := cur[0], cur[1]
+			for i := 0; i < len(curB); i++ {
+				for j := 0; j < len(curH); j++ {
+					curB2 := del3(curB[0:i] + string(curH[j]) + curB[i:])
+					curH2 := curH[0:j] + curH[j+1:]
+					if len(curB2) == 0 {
+						return minStep
+					}
+					if _, ok := mp[curB2+curH2]; ok {
+						continue
+					}
+					mp[curB2+curH2] = true
+					q = append(q, []string{curB2, curH2})
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func del3(str string) string {
+	cnt := 1
+	for i := 1; i < len(str); i++ {
+		if str[i] == str[i-1] {
+			cnt++
+		} else {
+			if cnt >= 3 {
+				return del3(str[0:i-cnt] + str[i:])
+			}
+			cnt = 1
+		}
+	}
+	if cnt >= 3 {
+		return str[0 : len(str)-cnt]
+	}
+	return str
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0485.Max-Consecutive-Ones/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0491.Increasing-Subsequences/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0491.Non-decreasing-Subsequences.md b/website/content/ChapterFour/0400~0499/0491.Non-decreasing-Subsequences.md
new file mode 100755
index 000000000..e18e428c3
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0491.Non-decreasing-Subsequences.md
@@ -0,0 +1,91 @@
+# [491. Non-decreasing Subsequences](https://leetcode.com/problems/non-decreasing-subsequences/)
+
+
+## 题目
+
+Given an integer array, your task is to find all the different possible increasing subsequences of the given array, and the length of an increasing subsequence should be at least 2.
+
+**Example**:
+
+    Input: [4, 6, 7, 7]
+    Output: [[4, 6], [4, 7], [4, 6, 7], [4, 6, 7, 7], [6, 7], [6, 7, 7], [7,7], [4,7,7]]
+
+**Note**:
+
+1. The length of the given array will not exceed 15.
+2. The range of integer in the given array is [-100,100].
+3. The given array may contain duplicates, and two equal integers should also be considered as a special case of increasing sequence.
+
+
+
+## 题目大意
+
+
+给定一个整型数组, 你的任务是找到所有该数组的递增子序列，递增子序列的长度至少是 2。
+
+说明:
+
+1. 给定数组的长度不会超过15。
+2. 数组中的整数范围是 [-100,100]。
+3. 给定数组中可能包含重复数字，相等的数字应该被视为递增的一种情况。
+
+
+
+
+## 解题思路
+
+
+- 给出一个数组，要求找出这个数组中所有长度大于 2 的非递减子序列。子序列顺序和原数组元素下标必须是顺序的，不能是逆序的。
+- 这一题和第 78 题和第 90 题是类似的题目。第 78 题和第 90 题是求所有子序列，这一题在这两题的基础上增加了非递减和长度大于 2 的条件。需要注意的两点是，原数组中元素可能会重复，最终结果输出的时候需要去重。最终结果输出的去重用 map 处理，数组中重复元素用 DFS 遍历搜索。在每次 DFS 中，用 map 记录遍历过的元素，保证本轮 DFS 中不出现重复的元素，递归到下一层还可以选择值相同，但是下标不同的另外一个元素。外层循环也要加一个  map，这个 map 是过滤每组解因为重复元素导致的重复解，经过过滤以后，起点不同了，最终的解也会不同。
+- 这一题和第 78 题，第 90 题类似，可以一起解答和复习。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findSubsequences(nums []int) [][]int {
+	c, visited, res := []int{}, map[int]bool{}, [][]int{}
+	for i := 0; i < len(nums)-1; i++ {
+		if _, ok := visited[nums[i]]; ok {
+			continue
+		} else {
+			visited[nums[i]] = true
+			generateIncSubsets(nums, i, c, &res)
+		}
+	}
+	return res
+}
+
+func generateIncSubsets(nums []int, current int, c []int, res *[][]int) {
+	c = append(c, nums[current])
+	if len(c) >= 2 {
+		b := make([]int, len(c))
+		copy(b, c)
+		*res = append(*res, b)
+	}
+	visited := map[int]bool{}
+	for i := current + 1; i < len(nums); i++ {
+		if nums[current] <= nums[i] {
+			if _, ok := visited[nums[i]]; ok {
+				continue
+			} else {
+				visited[nums[i]] = true
+				generateIncSubsets(nums, i, c, res)
+			}
+		}
+	}
+	c = c[:len(c)-1]
+	return
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0488.Zuma-Game/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0492.Construct-the-Rectangle/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0492.Construct-the-Rectangle.md b/website/content/ChapterFour/0400~0499/0492.Construct-the-Rectangle.md
new file mode 100644
index 000000000..5e02ec17e
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0492.Construct-the-Rectangle.md
@@ -0,0 +1,79 @@
+# [492. Construct the Rectangle](https://leetcode.com/problems/construct-the-rectangle/)
+
+
+## 题目
+
+A web developer needs to know how to design a web page's size. 
+So, given a specific rectangular web page’s area, your job by now is to design a rectangular web page,
+whose length L and width W satisfy the following requirements:
+
+    The area of the rectangular web page you designed must equal to the given target area.
+    The width W should not be larger than the length L, which means L >= W.
+    The difference between length L and width W should be as small as possible.
+    Return an array [L, W] where L and W are the length and width of the web page you designed in sequence.
+
+**Example 1:**
+
+    Input: area = 4
+    Output: [2,2]
+    Explanation: The target area is 4, and all the possible ways to construct it are [1,4], [2,2], [4,1].
+    But according to requirement 2, [1,4] is illegal; according to requirement 3,  [4,1] is not optimal compared to [2,2]. So the length L is 2, and the width W is 2.
+
+**Example 2:**
+
+    Input: area = 37
+    Output: [37,1]
+
+**Example 3:**
+
+    Input: area = 122122
+    Output: [427,286]
+
+**Constraints**
+
+   - 1 <= area <= 10000000
+
+## 题目大意
+
+作为一位 web 开发者， 懂得怎样去规划一个页面的尺寸是很重要的。 现给定一个具体的矩形页面面积，你的任务是设计一个长度为 L 和宽度为 W 且满足以下要求的矩形的页面。要求：
+
+1. 你设计的矩形页面必须等于给定的目标面积。
+2. 宽度 W 不应大于长度 L，换言之，要求 L >= W 。
+3. 长度 L 和宽度 W 之间的差距应当尽可能小。 
+
+你需要按顺序输出你设计的页面的长度 L 和宽度 W。
+
+## 解题思路
+
+- 令 W 等于根号 area
+- 在 W 大于等于 1 的情况下,判断 area%W 是否等于 0,如果不相等 W 就减 1 继续循环,如果相等就返回 [area/W, W]
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func constructRectangle(area int) []int {
+	ans := make([]int, 2)
+	W := int(math.Sqrt(float64(area)))
+	for W >= 1 {
+		if area%W == 0 {
+			ans[0], ans[1] = area/W, W
+			break
+		}
+		W -= 1
+	}
+	return ans
+}
+
+``
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0491.Increasing-Subsequences/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0493.Reverse-Pairs/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0493.Reverse-Pairs.md b/website/content/ChapterFour/0400~0499/0493.Reverse-Pairs.md
new file mode 100755
index 000000000..0e53ce7d5
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0493.Reverse-Pairs.md
@@ -0,0 +1,163 @@
+# [493. Reverse Pairs](https://leetcode.com/problems/reverse-pairs/)
+
+
+## 题目
+
+Given an array `nums`, we call `(i, j)` an **important reverse pair** if `i < j` and `nums[i] > 2*nums[j]`.
+
+You need to return the number of important reverse pairs in the given array.
+
+**Example1**:
+
+    Input: [1,3,2,3,1]
+    Output: 2
+
+**Example2**:
+
+    Input: [2,4,3,5,1]
+    Output: 3
+
+**Note**:
+
+1. The length of the given array will not exceed `50,000`.
+2. All the numbers in the input array are in the range of 32-bit integer.
+
+
+## 题目大意
+
+给定一个数组 nums ，如果 i < j 且 nums[i] > 2\*nums[j] 我们就将 (i, j) 称作一个重要翻转对。你需要返回给定数组中的重要翻转对的数量。
+
+注意:
+
+- 给定数组的长度不会超过 50000。
+- 输入数组中的所有数字都在 32 位整数的表示范围内。
+
+
+## 解题思路
+
+
+- 给出一个数组，要求找出满足条件的所有的“重要的反转对” (i,j)。重要的反转对的定义是：`i<j`，并且 `nums[i] > 2*nums[j]`。
+- 这一题是 327 题的变种题。首先将数组中所有的元素以及各自的 `2*nums[i] + 1` 都放在字典中去重。去重以后再做离散化处理。这一题的测试用例会卡离散化，如果不离散化，Math.MaxInt32 会导致数字溢出，见测试用例中 2147483647, -2147483647 这组测试用例。离散后，映射关系 保存在字典中。从左往右遍历数组，先 query ，再 update ，这个顺序和第 327 题是反的。先 query 查找 `[2*nums[i] + 1, len(indexMap)-1]` 这个区间内满足条件的值，这个区间内的值都是 `> 2*nums[j]` 的。这一题移动的是 `j`，`j` 不断的变化，往线段树中不断插入的是 `i`。每轮循环先 query 一次前一轮循环中累积插入线段树中的 `i`，这些累积在线段树中的代表的是所有在 `j` 前面的 `i`。query 查询的是本轮 `[2*nums[j] + 1, len(indexMap)-1]`，如果能找到，即找到了这样一个 `j`，能满足 `nums[i] > 2*nums[j`， 把整个数组都扫完，累加的 query 出来的 count 计数就是最终答案。
+- 另外一种解法是树状数组。树状数组最擅长解答逆序对的问题。先将原数组中所有的元素值的 2 倍算出来，和原数组合并到一个大数组中。这个大数组中装了所有可能产生 2 倍逆序对的元素值。然后再将他们所有值排序，离散化。离散化以后便将问题集转化成 `[1,N]` 这个区间。于是回到了树状数组经典的求逆序对的问题。逆序插入原数组构造树状数组，或者正序插入原数组构造树状数组都可以解答此题。
+- 类似的题目：第 327 题，第 315 题。
+- 这一题用线段树和树状数组并不是最优解，用线段树和树状数组解这一题是为了训练线段树和树状数组这两个数据结构。最优解是解法一中的 mergesort。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/leetcode-go/template"
+)
+
+// 解法一 归并排序 mergesort，时间复杂度 O(n log n)
+func reversePairs(nums []int) int {
+	buf := make([]int, len(nums))
+	return mergesortCount(nums, buf)
+}
+
+func mergesortCount(nums, buf []int) int {
+	if len(nums) <= 1 {
+		return 0
+	}
+	mid := (len(nums) - 1) / 2
+	cnt := mergesortCount(nums[:mid+1], buf)
+	cnt += mergesortCount(nums[mid+1:], buf)
+	for i, j := 0, mid+1; i < mid+1; i++ { // Note!!! j is increasing.
+		for ; j < len(nums) && nums[i] <= 2*nums[j]; j++ {
+		}
+		cnt += len(nums) - j
+	}
+	copy(buf, nums)
+	for i, j, k := 0, mid+1, 0; k < len(nums); {
+		if j >= len(nums) || i < mid+1 && buf[i] > buf[j] {
+			nums[k] = buf[i]
+			i++
+		} else {
+			nums[k] = buf[j]
+			j++
+		}
+		k++
+	}
+	return cnt
+}
+
+// 解法二 树状数组，时间复杂度 O(n log n)
+func reversePairs1(nums []int) (cnt int) {
+	n := len(nums)
+	if n <= 1 {
+		return
+	}
+	// 离散化所有下面统计时会出现的元素
+	allNums := make([]int, 0, 2*n)
+	for _, v := range nums {
+		allNums = append(allNums, v, 2*v)
+	}
+	sort.Ints(allNums)
+	k := 1
+	kth := map[int]int{allNums[0]: k}
+	for i := 1; i < 2*n; i++ {
+		if allNums[i] != allNums[i-1] {
+			k++
+			kth[allNums[i]] = k
+		}
+	}
+	bit := template.BinaryIndexedTree{}
+	bit.Init(k)
+	for i, v := range nums {
+		cnt += i - bit.Query(kth[2*v])
+		bit.Add(kth[v], 1)
+	}
+	return
+}
+
+// 解法三 线段树，时间复杂度 O(n log n)
+func reversePairs2(nums []int) int {
+	if len(nums) < 2 {
+		return 0
+	}
+	st, numsMap, indexMap, numsArray, res := template.SegmentCountTree{}, make(map[int]int, 0), make(map[int]int, 0), []int{}, 0
+	numsMap[nums[0]] = nums[0]
+	for _, num := range nums {
+		numsMap[num] = num
+		numsMap[2*num+1] = 2*num + 1
+	}
+	// numsArray 是 prefixSum 去重之后的版本，利用 numsMap 去重
+	for _, v := range numsMap {
+		numsArray = append(numsArray, v)
+	}
+	// 排序是为了使得线段树中的区间 left <= right，如果此处不排序，线段树中的区间有很多不合法。
+	sort.Ints(numsArray)
+	// 离散化，构建映射关系
+	for i, n := range numsArray {
+		indexMap[n] = i
+	}
+	numsArray = []int{}
+	// 离散化，此题如果不离散化，MaxInt32 的数据会使得数字越界。
+	for i := 0; i < len(indexMap); i++ {
+		numsArray = append(numsArray, i)
+	}
+	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
+	st.Init(numsArray, func(i, j int) int {
+		return 0
+	})
+	for _, num := range nums {
+		res += st.Query(indexMap[num*2+1], len(indexMap)-1)
+		st.UpdateCount(indexMap[num])
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0492.Construct-the-Rectangle/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0494.Target-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0494.Target-Sum.md b/website/content/ChapterFour/0400~0499/0494.Target-Sum.md
new file mode 100644
index 000000000..e104ed18a
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0494.Target-Sum.md
@@ -0,0 +1,115 @@
+# [494. Target Sum](https://leetcode.com/problems/target-sum/)
+
+
+## 题目
+
+You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols `+` and `-`. For each integer, you should choose one from `+` and `-` as its new symbol.
+
+Find out how many ways to assign symbols to make sum of integers equal to target S.
+
+**Example 1**:
+
+```
+Input: nums is [1, 1, 1, 1, 1], S is 3. 
+Output: 5
+Explanation: 
+
+-1+1+1+1+1 = 3
++1-1+1+1+1 = 3
++1+1-1+1+1 = 3
++1+1+1-1+1 = 3
++1+1+1+1-1 = 3
+
+There are 5 ways to assign symbols to make the sum of nums be target 3.
+```
+
+**Note**:
+
+1. The length of the given array is positive and will not exceed 20.
+2. The sum of elements in the given array will not exceed 1000.
+3. Your output answer is guaranteed to be fitted in a 32-bit integer.
+
+## 题目大意
+
+给定一个非负整数数组，a1, a2, ..., an, 和一个目标数，S。现在有两个符号 + 和 -。对于数组中的任意一个整数，可以从 + 或 - 中选择一个符号添加在前面。返回可以使最终数组和为目标数 S 的所有添加符号的方法数。
+
+提示：
+
+- 数组非空，且长度不会超过 20 。
+- 初始的数组的和不会超过 1000 。
+- 保证返回的最终结果能被 32 位整数存下。
+
+## 解题思路
+
+- 给出一个数组，要求在这个数组里面的每个元素前面加上 + 或者 - 号，最终总和等于 S。问有多少种不同的方法。
+- 这一题可以用 DP 和 DFS 解答。DFS 方法就比较暴力简单了。见代码。这里分析一下 DP 的做法。题目要求在数组元素前加上 + 或者 - 号，其实相当于把数组分成了 2 组，一组全部都加 + 号，一组都加 - 号。记 + 号的一组 P ，记 - 号的一组 N，那么可以推出以下的关系。
+
+    ```go
+    sum(P) - sum(N) = target
+    sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)
+                           2 * sum(P) = target + sum(nums)
+    ```
+
+    等号两边都加上 `sum(N) + sum(P)`，于是可以得到结果 `2 * sum(P) = target + sum(nums)`，那么这道题就转换成了，能否在数组中找到这样一个集合，和等于 `(target + sum(nums)) / 2`。那么这题就转化为了第 416 题了。`dp[i]` 中存储的是能使和为 `i` 的方法个数。
+
+- 如果和不是偶数，即不能被 2 整除，那说明找不到满足题目要求的解了，直接输出 0 。
+
+## 代码
+
+```go
+
+func findTargetSumWays(nums []int, S int) int {
+	total := 0
+	for _, n := range nums {
+		total += n
+	}
+	if S > total || (S+total)%2 == 1 {
+		return 0
+	}
+	target := (S + total) / 2
+	dp := make([]int, target+1)
+	dp[0] = 1
+	for _, n := range nums {
+		for i := target; i >= n; i-- {
+			dp[i] += dp[i-n]
+		}
+	}
+	return dp[target]
+}
+
+// 解法二 DFS
+func findTargetSumWays1(nums []int, S int) int {
+	// sums[i] 存储的是后缀和 nums[i:]，即从 i 到结尾的和
+	sums := make([]int, len(nums))
+	sums[len(nums)-1] = nums[len(nums)-1]
+	for i := len(nums) - 2; i > -1; i-- {
+		sums[i] = sums[i+1] + nums[i]
+	}
+	res := 0
+	dfsFindTargetSumWays(nums, 0, 0, S, &res, sums)
+	return res
+}
+
+func dfsFindTargetSumWays(nums []int, index int, curSum int, S int, res *int, sums []int) {
+	if index == len(nums) {
+		if curSum == S {
+			*(res) = *(res) + 1
+		}
+		return
+	}
+	// 剪枝优化：如果 sums[index] 值小于剩下需要正数的值，那么右边就算都是 + 号也无能为力了，所以这里可以剪枝了
+	if S-curSum > sums[index] {
+		return
+	}
+	dfsFindTargetSumWays(nums, index+1, curSum+nums[index], S, res, sums)
+	dfsFindTargetSumWays(nums, index+1, curSum-nums[index], S, res, sums)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0493.Reverse-Pairs/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0495.Teemo-Attacking/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0495.Teemo-Attacking.md b/website/content/ChapterFour/0400~0499/0495.Teemo-Attacking.md
new file mode 100644
index 000000000..b575de2cb
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0495.Teemo-Attacking.md
@@ -0,0 +1,92 @@
+# [495. Teemo Attacking](https://leetcode.com/problems/teemo-attacking/)
+
+
+## 题目
+
+Our hero Teemo is attacking an enemy Ashe with poison attacks! When Teemo attacks Ashe, Ashe gets poisoned for a exactly duration seconds. 
+
+More formally, an attack at second t will mean Ashe is poisoned during the inclusive time interval [t, t + duration - 1].
+
+If Teemo attacks again before the poison effect ends, the timer for it is reset, and the poison effect will end duration seconds after the new attack.
+
+You are given a non-decreasing integer array timeSeries, where timeSeries[i] denotes that Teemo attacks Ashe at second timeSeries[i], and an integer duration.
+
+Return the total number of seconds that Ashe is poisoned.
+
+**Example 1**:
+```
+Input: timeSeries = [1,4], duration = 2
+Output: 4
+Explanation: Teemo's attacks on Ashe go as follows:
+- At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
+- At second 4, Teemo attacks, and Ashe is poisoned for seconds 4 and 5.
+Ashe is poisoned for seconds 1, 2, 4, and 5, which is 4 seconds in total.
+```
+
+**Example 2**:
+```
+Input: timeSeries = [1,2], duration = 2
+Output: 3
+Explanation: Teemo's attacks on Ashe go as follows:
+- At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
+- At second 2 however, Teemo attacks again and resets the poison timer. Ashe is poisoned for seconds 2 and 3.
+Ashe is poisoned for seconds 1, 2, and 3, which is 3 seconds in total.
+```
+
+**Constraints**:
+
+- 1 <= timeSeries.length <= 10000
+- 0 <= timeSeries[i], duration <= 10000000
+- timeSeries is sorted in non-decreasing order.
+
+## 题目大意
+
+在《英雄联盟》的世界中，有一个叫 “提莫” 的英雄。他的攻击可以让敌方英雄艾希（编者注：寒冰射手）进入中毒状态。
+
+当提莫攻击艾希，艾希的中毒状态正好持续duration 秒。
+
+正式地讲，提莫在t发起发起攻击意味着艾希在时间区间 [t, t + duration - 1]（含 t 和 t + duration - 1）处于中毒状态。
+
+如果提莫在中毒影响结束前再次攻击，中毒状态计时器将会重置，在新的攻击之后，中毒影响将会在duration秒后结束。
+
+给你一个非递减的整数数组timeSeries，其中timeSeries[i]表示提莫在timeSeries[i]秒时对艾希发起攻击，以及一个表示中毒持续时间的整数duration 。
+
+返回艾希处于中毒状态的总秒数。
+
+## 解题思路
+
+- i 从 1 开始计数，令 t 等于 timeSeries[i - 1]
+- 比较 end(t + duration - 1) 和 timeSeries[i] 的大小，
+  - 如果 end 小于 timeSeries[i],ans+=duration
+  - 否则 ans += timeSeries[i] - t
+- ans += duration 并返回 ans
+
+## 代码
+
+```go
+
+package leetcode
+
+func findPoisonedDuration(timeSeries []int, duration int) int {
+	var ans int
+	for i := 1; i < len(timeSeries); i++ {
+		t := timeSeries[i-1]
+		end := t + duration - 1
+		if end < timeSeries[i] {
+			ans += duration
+		} else {
+			ans += timeSeries[i] - t
+		}
+	}
+	ans += duration
+	return ans
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0494.Target-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0496.Next-Greater-Element-I/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0496.Next-Greater-Element-I.md b/website/content/ChapterFour/0400~0499/0496.Next-Greater-Element-I.md
new file mode 100644
index 000000000..1075380ba
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0496.Next-Greater-Element-I.md
@@ -0,0 +1,87 @@
+# [496. Next Greater Element I](https://leetcode.com/problems/next-greater-element-i/)
+
+## 题目
+
+You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.
+
+The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.
+
+
+**Example 1**:
+
+```
+
+Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
+Output: [-1,3,-1]
+Explanation:
+    For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
+    For number 1 in the first array, the next greater number for it in the second array is 3.
+    For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
+    
+```
+
+**Example 2**:
+
+```
+
+Input: nums1 = [2,4], nums2 = [1,2,3,4].
+Output: [3,-1]
+Explanation:
+    For number 2 in the first array, the next greater number for it in the second array is 3.
+    For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
+
+```
+
+**Note**:  
+
+- All elements in nums1 and nums2 are unique.  
+- The length of both nums1 and nums2 would not exceed 1000.
+
+
+## 题目大意
+
+这道题也是简单题。题目给出 2 个数组 A 和 B，针对 A 中的每个数组中的元素，要求在 B 数组中找出比 A 数组中元素大的数，B 中元素之间的顺序保持不变。如果找到了就输出这个值，如果找不到就输出 -1。
+
+
+## 解题思路
+
+简单题，依题意做即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func nextGreaterElement(nums1 []int, nums2 []int) []int {
+	if len(nums1) == 0 || len(nums2) == 0 {
+		return []int{}
+	}
+	res, reocrd := []int{}, map[int]int{}
+	for i, v := range nums2 {
+		reocrd[v] = i
+	}
+	for i := 0; i < len(nums1); i++ {
+		flag := false
+		for j := reocrd[nums1[i]]; j < len(nums2); j++ {
+			if nums2[j] > nums1[i] {
+				res = append(res, nums2[j])
+				flag = true
+				break
+			}
+		}
+		if flag == false {
+			res = append(res, -1)
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0495.Teemo-Attacking/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0497.Random-Point-in-Non-overlapping-Rectangles/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0497.Random-Point-in-Non-overlapping-Rectangles.md b/website/content/ChapterFour/0400~0499/0497.Random-Point-in-Non-overlapping-Rectangles.md
new file mode 100755
index 000000000..6d6fa225c
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0497.Random-Point-in-Non-overlapping-Rectangles.md
@@ -0,0 +1,141 @@
+# [497. Random Point in Non-overlapping Rectangles](https://leetcode.com/problems/random-point-in-non-overlapping-rectangles)
+
+
+## 题目
+
+Given a list of **non-overlapping** axis-aligned rectangles `rects`, write a function `pick` which randomly and uniformily picks an **integer point** in the space covered by the rectangles.
+
+**Note**:
+
+1. An **integer point** is a point that has integer coordinates.
+2. A point on the perimeter of a rectangle is **included** in the space covered by the rectangles.
+3. `i`th rectangle = `rects[i]` = `[x1,y1,x2,y2]`, where `[x1, y1]` are the integer coordinates of the bottom-left corner, and `[x2, y2]` are the integer coordinates of the top-right corner.
+4. length and width of each rectangle does not exceed `2000`.
+5. `1 <= rects.length <= 100`
+6. `pick` return a point as an array of integer coordinates `[p_x, p_y]`
+7. `pick` is called at most `10000` times.
+
+**Example 1**:
+
+    Input: 
+    ["Solution","pick","pick","pick"]
+    [[[[1,1,5,5]]],[],[],[]]
+    Output: 
+    [null,[4,1],[4,1],[3,3]]
+
+**Example 2**:
+
+    Input: 
+    ["Solution","pick","pick","pick","pick","pick"]
+    [[[[-2,-2,-1,-1],[1,0,3,0]]],[],[],[],[],[]]
+    Output: 
+    [null,[-1,-2],[2,0],[-2,-1],[3,0],[-2,-2]]
+
+**Explanation of Input Syntax**:
+
+The input is two lists: the subroutines called and their arguments. `Solution`'s constructor has one argument, the array of rectangles `rects`. `pick` has no arguments. Arguments are always wrapped with a list, even if there aren't any.
+
+
+## 题目大意
+
+给定一个非重叠轴对齐矩形的列表 rects，写一个函数 pick 随机均匀地选取矩形覆盖的空间中的整数点。
+
+提示：
+
+1. 整数点是具有整数坐标的点。
+2. 矩形周边上的点包含在矩形覆盖的空间中。
+3. 第 i 个矩形 rects [i] = [x1，y1，x2，y2]，其中 [x1，y1] 是左下角的整数坐标，[x2，y2] 是右上角的整数坐标。
+4. 每个矩形的长度和宽度不超过 2000。
+5. 1 <= rects.length <= 100
+6. pick 以整数坐标数组 [p_x, p_y] 的形式返回一个点。
+7. pick 最多被调用10000次。
+
+
+输入语法的说明：
+
+输入是两个列表：调用的子例程及其参数。Solution 的构造函数有一个参数，即矩形数组 rects。pick 没有参数。参数总是用列表包装的，即使没有也是如此。
+
+
+## 解题思路
+
+
+- 给出一个非重叠轴对齐矩形列表，每个矩形用左下角和右上角的两个坐标表示。要求 `pick()` 随机均匀地选取矩形覆盖的空间中的整数点。
+- 这一题是第 528 题的变种题，这一题权重是面积，按权重（面积）选择一个矩形，然后再从矩形中随机选择一个点即可。思路和代码和第 528 题一样。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math/rand"
+
+// Solution497 define
+type Solution497 struct {
+	rects [][]int
+	arr   []int
+}
+
+// Constructor497 define
+func Constructor497(rects [][]int) Solution497 {
+	s := Solution497{
+		rects: rects,
+		arr:   make([]int, len(rects)),
+	}
+
+	for i := 0; i < len(rects); i++ {
+		area := (rects[i][2] - rects[i][0] + 1) * (rects[i][3] - rects[i][1] + 1)
+		if area < 0 {
+			area = -area
+		}
+		if i == 0 {
+			s.arr[0] = area
+		} else {
+			s.arr[i] = s.arr[i-1] + area
+		}
+	}
+	return s
+}
+
+// Pick define
+func (so *Solution497) Pick() []int {
+	r := rand.Int() % so.arr[len(so.arr)-1]
+	//get rectangle first
+	low, high, index := 0, len(so.arr)-1, -1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if so.arr[mid] > r {
+			if mid == 0 || so.arr[mid-1] <= r {
+				index = mid
+				break
+			}
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	if index == -1 {
+		index = low
+	}
+	if index > 0 {
+		r = r - so.arr[index-1]
+	}
+	length := so.rects[index][2] - so.rects[index][0]
+	return []int{so.rects[index][0] + r%(length+1), so.rects[index][1] + r/(length+1)}
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(rects);
+ * param_1 := obj.Pick();
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0496.Next-Greater-Element-I/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0498.Diagonal-Traverse/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/0498.Diagonal-Traverse.md b/website/content/ChapterFour/0400~0499/0498.Diagonal-Traverse.md
new file mode 100755
index 000000000..454f91cfe
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0498.Diagonal-Traverse.md
@@ -0,0 +1,198 @@
+# [498. Diagonal Traverse](https://leetcode.com/problems/diagonal-traverse/)
+
+
+## 题目
+
+Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image.
+
+**Example**:
+
+    Input:
+    [
+     [ 1, 2, 3 ],
+     [ 4, 5, 6 ],
+     [ 7, 8, 9 ]
+    ]
+    
+    Output:  [1,2,4,7,5,3,6,8,9]
+    
+    Explanation:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/diagonal_traverse.png)
+
+**Note**:
+
+The total number of elements of the given matrix will not exceed 10,000.
+
+
+## 题目大意
+
+给定一个含有 M x N 个元素的矩阵（M 行，N 列），请以对角线遍历的顺序返回这个矩阵中的所有元素，对角线遍历如下图所示。
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/diagonal_traverse.png)
+
+说明: 给定矩阵中的元素总数不会超过 100000 。
+
+## 解题思路
+
+- 给出一个二维数组，要求按照如图的方式遍历整个数组。
+- 这一题用模拟的方式就可以解出来。需要注意的是边界条件：比如二维数组为空，二维数组退化为一行或者一列，退化为一个元素。具体例子见测试用例。
+- 解题关键是在判断下一个位置，将矩阵想像成一个X,Y坐标轴。那么可分为以下几种情况，
+    1、斜角向右上遍历时，
+    当右上角在坐标轴内， 正常计算 即， x+1(X轴向右移动)， y-1(Y轴向上移动)
+    当右上角在坐标轴外，那么当前位置只能在 第一行X坐标轴 ，或者 最后一列Y坐标轴 ， 即判断该两种情况下�应该X坐标往右，或者 Y坐标往上
+    2、同理 斜角向下遍历时
+    当左下角在坐标轴内，正常计算 即， x-1(X轴向右移动)， y+1(Y轴向下移动)
+    当左下角在坐标轴外，那么当前位置只能在 第一列Y坐标轴，或者 最后一行X坐标轴， 即判断该两种情况下�应该X坐标往左，或者 Y坐标往下
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func findDiagonalOrder1(matrix [][]int) []int {
+	if matrix == nil || len(matrix) == 0 || len(matrix[0]) == 0 {
+		return nil
+	}
+	row, col, dir, i, x, y, d := len(matrix), len(matrix[0]), [2][2]int{
+		{-1, 1},
+		{1, -1},
+	}, 0, 0, 0, 0
+	total := row * col
+	res := make([]int, total)
+	for i < total {
+		for x >= 0 && x < row && y >= 0 && y < col {
+			res[i] = matrix[x][y]
+			i++
+			x += dir[d][0]
+			y += dir[d][1]
+		}
+		d = (d + 1) % 2
+		if x == row {
+			x--
+			y += 2
+		}
+		if y == col {
+			y--
+			x += 2
+		}
+		if x < 0 {
+			x = 0
+		}
+		if y < 0 {
+			y = 0
+		}
+	}
+	return res
+}
+
+// 解法二
+func findDiagonalOrder(matrix [][]int) []int {
+	if len(matrix) == 0 {
+		return []int{}
+	}
+	if len(matrix) == 1 {
+		return matrix[0]
+	}
+	// dir = 0 代表从右上到左下的方向， dir = 1 代表从左下到右上的方向 dir = -1 代表上一次转变了方向
+	m, n, i, j, dir, res := len(matrix), len(matrix[0]), 0, 0, 0, []int{}
+	for index := 0; index < m*n; index++ {
+		if dir == -1 {
+			if (i == 0 && j < n-1) || (j == n-1) { // 上边界和右边界
+				i++
+				if j > 0 {
+					j--
+				}
+				dir = 0
+				addTraverse(matrix, i, j, &res)
+				continue
+			}
+			if (j == 0 && i < m-1) || (i == m-1) { // 左边界和下边界
+				if i > 0 {
+					i--
+				}
+				j++
+				dir = 1
+				addTraverse(matrix, i, j, &res)
+				continue
+			}
+		}
+		if i == 0 && j == 0 {
+			res = append(res, matrix[i][j])
+			if j < n-1 {
+				j++
+				dir = -1
+				addTraverse(matrix, i, j, &res)
+				continue
+			} else {
+				if i < m-1 {
+					i++
+					dir = -1
+					addTraverse(matrix, i, j, &res)
+					continue
+				}
+			}
+		}
+		if i == 0 && j < n-1 { // 上边界
+			if j < n-1 {
+				j++
+				dir = -1
+				addTraverse(matrix, i, j, &res)
+				continue
+			}
+		}
+		if j == 0 && i < m-1 { // 左边界
+			if i < m-1 {
+				i++
+				dir = -1
+				addTraverse(matrix, i, j, &res)
+				continue
+			}
+		}
+		if j == n-1 { // 右边界
+			if i < m-1 {
+				i++
+				dir = -1
+				addTraverse(matrix, i, j, &res)
+				continue
+			}
+		}
+		if i == m-1 { // 下边界
+			j++
+			dir = -1
+			addTraverse(matrix, i, j, &res)
+			continue
+		}
+		if dir == 1 {
+			i--
+			j++
+			addTraverse(matrix, i, j, &res)
+			continue
+		}
+		if dir == 0 {
+			i++
+			j--
+			addTraverse(matrix, i, j, &res)
+			continue
+		}
+	}
+	return res
+}
+
+func addTraverse(matrix [][]int, i, j int, res *[]int) {
+	if i >= 0 && i <= len(matrix)-1 && j >= 0 && j <= len(matrix[0])-1 {
+		*res = append(*res, matrix[i][j])
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0497.Random-Point-in-Non-overlapping-Rectangles/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0500.Keyboard-Row/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0400~0499/_index.md b/website/content/ChapterFour/0400~0499/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/0500~0599/0500.Keyboard-Row.md b/website/content/ChapterFour/0500~0599/0500.Keyboard-Row.md
new file mode 100755
index 000000000..9771d2e2c
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0500.Keyboard-Row.md
@@ -0,0 +1,69 @@
+# [500. Keyboard Row](https://leetcode.com/problems/keyboard-row/)
+
+
+## 题目
+
+Given a List of words, return the words that can be typed using letters of **alphabet** on only one row's of American keyboard like the image below.
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/keyboard.png)
+
+**Example**:
+
+    Input: ["Hello", "Alaska", "Dad", "Peace"]
+    Output: ["Alaska", "Dad"]
+
+**Note**:
+
+1. You may use one character in the keyboard more than once.
+2. You may assume the input string will only contain letters of alphabet.
+
+
+## 题目大意
+
+给定一个单词列表，只返回可以使用在键盘同一行的字母打印出来的单词。键盘如上图所示。
+
+## 解题思路
+
+- 给出一个字符串数组，要求依次判断数组中的每个字符串是否都位于键盘上的同一个行，如果是就输出。这也是一道水题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func findWords500(words []string) []string {
+	rows := []string{"qwertyuiop", "asdfghjkl", "zxcvbnm"}
+	output := make([]string, 0)
+	for _, s := range words {
+		if len(s) == 0 {
+			continue
+		}
+		lowerS := strings.ToLower(s)
+		oneRow := false
+		for _, r := range rows {
+			if strings.ContainsAny(lowerS, r) {
+				oneRow = !oneRow
+				if !oneRow {
+					break
+				}
+			}
+		}
+		if oneRow {
+			output = append(output, s)
+		}
+	}
+	return output
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0498.Diagonal-Traverse/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0503.Next-Greater-Element-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0503.Next-Greater-Element-II.md b/website/content/ChapterFour/0500~0599/0503.Next-Greater-Element-II.md
new file mode 100644
index 000000000..69675aa8d
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0503.Next-Greater-Element-II.md
@@ -0,0 +1,85 @@
+# [503. Next Greater Element II](https://leetcode.com/problems/next-greater-element-ii/)
+
+## 题目
+
+Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, output -1 for this number.
+
+**Example 1**:
+
+```
+
+Input: [1,2,1]
+Output: [2,-1,2]
+Explanation: The first 1's next greater number is 2; 
+The number 2 can't find next greater number; 
+The second 1's next greater number needs to search circularly, which is also 2.
+
+```
+
+**Note**: The length of given array won't exceed 10000.
+
+## 题目大意
+
+题目给出数组 A，针对 A 中的每个数组中的元素，要求在 A 数组中找出比该元素大的数，A 是一个循环数组。如果找到了就输出这个值，如果找不到就输出 -1。
+
+
+## 解题思路
+
+这题是第 496 题的加强版，在第 496 题的基础上增加了循环数组的条件。这一题可以依旧按照第 496 题的做法继续模拟。更好的做法是用单调栈，栈中记录单调递增的下标。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 单调栈
+func nextGreaterElements(nums []int) []int {
+	res := make([]int, 0)
+	indexes := make([]int, 0)
+	for i := 0; i < len(nums); i++ {
+		res = append(res, -1)
+	}
+	for i := 0; i < len(nums)*2; i++ {
+		num := nums[i%len(nums)]
+		for len(indexes) > 0 && nums[indexes[len(indexes)-1]] < num {
+			index := indexes[len(indexes)-1]
+			res[index] = num
+			indexes = indexes[:len(indexes)-1]
+		}
+		indexes = append(indexes, i%len(nums))
+	}
+	return res
+}
+
+// 解法二
+func nextGreaterElements1(nums []int) []int {
+	if len(nums) == 0 {
+		return []int{}
+	}
+	res := []int{}
+	for i := 0; i < len(nums); i++ {
+		j, find := (i+1)%len(nums), false
+		for j != i {
+			if nums[j] > nums[i] {
+				find = true
+				res = append(res, nums[j])
+				break
+			}
+			j = (j + 1) % len(nums)
+		}
+		if !find {
+			res = append(res, -1)
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0500.Keyboard-Row/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0504.Base-7/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0504.Base-7.md b/website/content/ChapterFour/0500~0599/0504.Base-7.md
new file mode 100644
index 000000000..30b82e50e
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0504.Base-7.md
@@ -0,0 +1,67 @@
+# [504. Base 7](https://leetcode.com/problems/base-7/)
+
+## 题目
+
+Given an integer num, return a string of its base 7 representation.
+
+**Example 1:**
+
+    Input: num = 100
+    Output: "202"
+
+**Example 2:**
+
+    Input: num = -7
+    Output: "-10"
+
+**Constraints:**
+
+- -10000000 <= num <= 10000000
+
+## 题目大意
+
+给定一个整数 num，将其转化为 7 进制，并以字符串形式输出。
+
+## 解题思路
+
+    num反复除以7，然后倒排余数
+
+# 代码
+
+```go
+package leetcode
+
+import "strconv"
+
+func convertToBase7(num int) string {
+	if num == 0 {
+		return "0"
+	}
+	negative := false
+	if num < 0 {
+		negative = true
+		num = -num
+	}
+	var ans string
+	var nums []int
+	for num != 0 {
+		remainder := num % 7
+		nums = append(nums, remainder)
+		num = num / 7
+	}
+	if negative {
+		ans += "-"
+	}
+	for i := len(nums) - 1; i >= 0; i-- {
+		ans += strconv.Itoa(nums[i])
+	}
+	return ans
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0503.Next-Greater-Element-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0506.Relative-Ranks/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0506.Relative-Ranks.md b/website/content/ChapterFour/0500~0599/0506.Relative-Ranks.md
new file mode 100644
index 000000000..0e1522f5f
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0506.Relative-Ranks.md
@@ -0,0 +1,91 @@
+# [506. Relative Ranks](https://leetcode.com/problems/relative-ranks/)
+
+## 题目
+
+You are given an integer array score of size n, where score[i] is the score of the ith athlete in a competition. All the scores are guaranteed to be unique.
+
+The athletes are placed based on their scores, where the 1st place athlete has the highest score, the 2nd place athlete has the 2nd highest score, and so on. The placement of each athlete determines their rank:
+
+- The 1st place athlete's rank is "Gold Medal".
+- The 2nd place athlete's rank is "Silver Medal".
+- The 3rd place athlete's rank is "Bronze Medal".
+- For the 4th place to the nth place athlete, their rank is their placement number (i.e., the xth place athlete's rank is "x").
+
+Return an array answer of size n where answer[i] is the rank of the ith athlete.
+
+**Example 1**:
+
+    Input: score = [5,4,3,2,1]
+    Output: ["Gold Medal","Silver Medal","Bronze Medal","4","5"]
+    Explanation: The placements are [1st, 2nd, 3rd, 4th, 5th].
+
+**Example 2**:
+
+    Input: score = [10,3,8,9,4]
+    Output: ["Gold Medal","5","Bronze Medal","Silver Medal","4"]
+    Explanation: The placements are [1st, 5th, 3rd, 2nd, 4th].
+
+**Constraints:**
+
+- n == score.length
+- 1 <= n <= 10000
+- 0 <= score[i] <= 1000000
+- All the values in score are unique.
+
+## 题目大意
+
+给你一个长度为 n 的整数数组 score ，其中 score[i] 是第 i 位运动员在比赛中的得分。所有得分都 互不相同 。
+
+运动员将根据得分 决定名次 ，其中名次第 1 的运动员得分最高，名次第 2 的运动员得分第 2 高，依此类推。运动员的名次决定了他们的获奖情况：
+
+- 名次第 1 的运动员获金牌 "Gold Medal" 。
+- 名次第 2 的运动员获银牌 "Silver Medal" 。
+- 名次第 3 的运动员获铜牌 "Bronze Medal" 。
+- 从名次第 4 到第 n 的运动员，只能获得他们的名次编号（即，名次第 x 的运动员获得编号 "x"）。
+
+使用长度为 n 的数组 answer 返回获奖，其中 answer[i] 是第 i 位运动员的获奖情况。
+
+## 解题思路
+
+- 用 map 记录原来 score 中元素对应的坐标，然后对 score 进行排序，对排序后的元素我们通过 map 就可以知道它排的名次了
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+	"strconv"
+)
+
+func findRelativeRanks(score []int) []string {
+	mp := make(map[int]int)
+	for i, v := range score {
+		mp[v] = i
+	}
+	sort.Slice(score, func(i, j int) bool {
+		return score[i] > score[j]
+	})
+	ans := make([]string, len(score))
+	for i, v := range score {
+		if i == 0 {
+			ans[mp[v]] = "Gold Medal"
+		} else if i == 1 {
+			ans[mp[v]] = "Silver Medal"
+		} else if i == 2 {
+			ans[mp[v]] = "Bronze Medal"
+		} else {
+			ans[mp[v]] = strconv.Itoa(i + 1)
+		}
+	}
+	return ans
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0504.Base-7/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0507.Perfect-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0507.Perfect-Number.md b/website/content/ChapterFour/0500~0599/0507.Perfect-Number.md
new file mode 100644
index 000000000..d48f8bd18
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0507.Perfect-Number.md
@@ -0,0 +1,71 @@
+# [507. Perfect Number](https://leetcode.com/problems/perfect-number/)
+
+
+
+## 题目
+
+We define the Perfect Number is a **positive** integer that is equal to the sum of all its **positive** divisors except itself.
+
+Now, given an
+
+**integer**
+
+n, write a function that returns true when it is a perfect number and false when it is not.
+
+**Example**:
+
+```
+Input: 28
+Output: True
+Explanation: 28 = 1 + 2 + 4 + 7 + 14
+```
+
+**Note**: The input number **n** will not exceed 100,000,000. (1e8)
+
+## 题目大意
+
+对于一个 正整数，如果它和除了它自身以外的所有正因子之和相等，我们称它为“完美数”。给定一个 整数 n， 如果他是完美数，返回 True，否则返回 False
+
+## 解题思路
+
+- 给定一个整数，要求判断这个数是不是完美数。整数的取值范围小于 1e8 。
+- 简单题。按照题意描述，先获取这个整数的所有正因子，如果正因子的和等于原来这个数，那么它就是完美数。
+- 这一题也可以打表，1e8 以下的完美数其实并不多，就 5 个。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+// 方法一
+func checkPerfectNumber(num int) bool {
+	if num <= 1 {
+		return false
+	}
+	sum, bound := 1, int(math.Sqrt(float64(num)))+1
+	for i := 2; i < bound; i++ {
+		if num%i != 0 {
+			continue
+		}
+		corrDiv := num / i
+		sum += corrDiv + i
+	}
+	return sum == num
+}
+
+// 方法二 打表
+func checkPerfectNumber_(num int) bool {
+	return num == 6 || num == 28 || num == 496 || num == 8128 || num == 33550336
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0506.Relative-Ranks/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0508.Most-Frequent-Subtree-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0508.Most-Frequent-Subtree-Sum.md b/website/content/ChapterFour/0500~0599/0508.Most-Frequent-Subtree-Sum.md
new file mode 100755
index 000000000..2467d23f3
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0508.Most-Frequent-Subtree-Sum.md
@@ -0,0 +1,130 @@
+# [508. Most Frequent Subtree Sum](https://leetcode.com/problems/most-frequent-subtree-sum/)
+
+
+## 题目
+
+Given the root of a tree, you are asked to find the most frequent subtree sum. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.
+
+**Examples 1**
+
+Input:
+
+      5
+     /  \
+    2   -3
+
+return [2, -3, 4], since all the values happen only once, return all of them in any order.
+
+**Examples 2**
+
+Input:
+
+      5
+     /  \
+    2   -5
+
+return [2], since 2 happens twice, however -5 only occur once.
+
+**Note**: You may assume the sum of values in any subtree is in the range of 32-bit signed integer.
+
+
+## 题目大意
+
+给出二叉树的根，找出出现次数最多的子树元素和。一个结点的子树元素和定义为以该结点为根的二叉树上所有结点的元素之和（包括结点本身）。然后求出出现次数最多的子树元素和。如果有多个元素出现的次数相同，返回所有出现次数最多的元素（不限顺序）。提示： 假设任意子树元素和均可以用 32 位有符号整数表示。
+
+## 解题思路
+
+
+- 给出一个树，要求求出每个节点以自己为根节点的子树的所有节点值的和，最后按照这些和出现的频次，输出频次最多的和，如果频次出现次数最多的对应多个和，则全部输出。
+- 递归找出每个节点的累加和，用 map 记录频次，最后把频次最多的输出即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 维护最大频次，不用排序
+func findFrequentTreeSum(root *TreeNode) []int {
+	memo := make(map[int]int)
+	collectSum(root, memo)
+	res := []int{}
+	most := 0
+	for key, val := range memo {
+		if most == val {
+			res = append(res, key)
+		} else if most < val {
+			most = val
+			res = []int{key}
+		}
+	}
+	return res
+}
+
+func collectSum(root *TreeNode, memo map[int]int) int {
+	if root == nil {
+		return 0
+	}
+	sum := root.Val + collectSum(root.Left, memo) + collectSum(root.Right, memo)
+	if v, ok := memo[sum]; ok {
+		memo[sum] = v + 1
+	} else {
+		memo[sum] = 1
+	}
+	return sum
+}
+
+// 解法二 求出所有和再排序
+func findFrequentTreeSum1(root *TreeNode) []int {
+	if root == nil {
+		return []int{}
+	}
+	freMap, freList, reFreMap := map[int]int{}, []int{}, map[int][]int{}
+	findTreeSum(root, freMap)
+	for k, v := range freMap {
+		tmp := reFreMap[v]
+		tmp = append(tmp, k)
+		reFreMap[v] = tmp
+	}
+	for k := range reFreMap {
+		freList = append(freList, k)
+	}
+	sort.Ints(freList)
+	return reFreMap[freList[len(freList)-1]]
+}
+
+func findTreeSum(root *TreeNode, fre map[int]int) int {
+	if root == nil {
+		return 0
+	}
+	if root != nil && root.Left == nil && root.Right == nil {
+		fre[root.Val]++
+		return root.Val
+	}
+	val := findTreeSum(root.Left, fre) + findTreeSum(root.Right, fre) + root.Val
+	fre[val]++
+	return val
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0507.Perfect-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0509.Fibonacci-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0509.Fibonacci-Number.md b/website/content/ChapterFour/0500~0599/0509.Fibonacci-Number.md
new file mode 100755
index 000000000..55c0de7d7
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0509.Fibonacci-Number.md
@@ -0,0 +1,197 @@
+# [509. Fibonacci Number](https://leetcode.com/problems/fibonacci-number/)
+
+
+## 题目
+
+The **Fibonacci numbers**, commonly denoted `F(n)` form a sequence, called the **Fibonacci sequence**, such that each number is the sum of the two preceding ones, starting from `0` and `1`. That is,
+
+    F(0) = 0,   F(1) = 1
+    F(N) = F(N - 1) + F(N - 2), for N > 1.
+
+Given `N`, calculate `F(N)`.
+
+**Example 1**:
+
+    Input: 2
+    Output: 1
+    Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
+
+**Example 2**:
+
+    Input: 3
+    Output: 2
+    Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
+
+**Example 3**:
+
+    Input: 4
+    Output: 3
+    Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
+
+**Note**:
+
+0 ≤ `N` ≤ 30.
+
+
+## 题目大意
+
+斐波那契数，通常用 F(n) 表示，形成的序列称为斐波那契数列。该数列由 0 和 1 开始，后面的每一项数字都是前面两项数字的和。也就是：
+
+```
+F(0) = 0,   F(1) = 1
+F(N) = F(N - 1) + F(N - 2), 其中 N > 1.
+```
+
+给定 N，计算 F(N)。
+
+提示：0 ≤ N ≤ 30
+
+## 解题思路
+
+
+- 求斐波那契数列
+- 这一题解法很多，大的分类是四种，递归，记忆化搜索(dp)，矩阵快速幂，通项公式。其中记忆化搜索可以写 3 种方法，自底向上的，自顶向下的，优化空间复杂度版的。通项公式方法实质是求 a^b 这个还可以用快速幂优化时间复杂度到 O(log n) 。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+// 解法一 递归法 时间复杂度 O(2^n)，空间复杂度 O(n)
+func fib(N int) int {
+	if N <= 1 {
+		return N
+	}
+	return fib(N-1) + fib(N-2)
+}
+
+// 解法二 自底向上的记忆化搜索 时间复杂度 O(n)，空间复杂度 O(n)
+func fib1(N int) int {
+	if N <= 1 {
+		return N
+	}
+	cache := map[int]int{0: 0, 1: 1}
+	for i := 2; i <= N; i++ {
+		cache[i] = cache[i-1] + cache[i-2]
+	}
+	return cache[N]
+}
+
+// 解法三 自顶向下的记忆化搜索 时间复杂度 O(n)，空间复杂度 O(n)
+func fib2(N int) int {
+	if N <= 1 {
+		return N
+	}
+	return memoize(N, map[int]int{0: 0, 1: 1})
+}
+
+func memoize(N int, cache map[int]int) int {
+	if _, ok := cache[N]; ok {
+		return cache[N]
+	}
+	cache[N] = memoize(N-1, cache) + memoize(N-2, cache)
+	return memoize(N, cache)
+}
+
+// 解法四 优化版的 dp，节约内存空间 时间复杂度 O(n)，空间复杂度 O(1)
+func fib3(N int) int {
+	if N <= 1 {
+		return N
+	}
+	if N == 2 {
+		return 1
+	}
+	current, prev1, prev2 := 0, 1, 1
+	for i := 3; i <= N; i++ {
+		current = prev1 + prev2
+		prev2 = prev1
+		prev1 = current
+	}
+	return current
+}
+
+// 解法五 矩阵快速幂 时间复杂度 O(log n)，空间复杂度 O(log n)
+// | 1 1 | ^ n   = | F(n+1) F(n)   |
+// | 1 0 |		   | F(n)	F(n-1) |
+func fib4(N int) int {
+	if N <= 1 {
+		return N
+	}
+	var A = [2][2]int{
+		{1, 1},
+		{1, 0},
+	}
+	A = matrixPower(A, N-1)
+	return A[0][0]
+}
+
+func matrixPower(A [2][2]int, N int) [2][2]int {
+	if N <= 1 {
+		return A
+	}
+	A = matrixPower(A, N/2)
+	A = multiply(A, A)
+
+	var B = [2][2]int{
+		{1, 1},
+		{1, 0},
+	}
+	if N%2 != 0 {
+		A = multiply(A, B)
+	}
+
+	return A
+}
+
+func multiply(A [2][2]int, B [2][2]int) [2][2]int {
+	x := A[0][0]*B[0][0] + A[0][1]*B[1][0]
+	y := A[0][0]*B[0][1] + A[0][1]*B[1][1]
+	z := A[1][0]*B[0][0] + A[1][1]*B[1][0]
+	w := A[1][0]*B[0][1] + A[1][1]*B[1][1]
+	A[0][0] = x
+	A[0][1] = y
+	A[1][0] = z
+	A[1][1] = w
+	return A
+}
+
+// 解法六 公式法 f(n)=(1/√5)*{[(1+√5)/2]^n -[(1-√5)/2]^n}，用 时间复杂度在 O(log n) 和 O(n) 之间，空间复杂度 O(1)
+// 经过实际测试，会发现 pow() 系统函数比快速幂慢，说明 pow() 比 O(log n) 慢
+// 斐波那契数列是一个自然数的数列，通项公式却是用无理数来表达的。而且当 n 趋向于无穷大时，前一项与后一项的比值越来越逼近黄金分割 0.618（或者说后一项与前一项的比值小数部分越来越逼近 0.618）。
+// 斐波那契数列用计算机计算的时候可以直接用四舍五入函数 Round 来计算。
+func fib5(N int) int {
+	var goldenRatio float64 = float64((1 + math.Sqrt(5)) / 2)
+	return int(math.Round(math.Pow(goldenRatio, float64(N)) / math.Sqrt(5)))
+}
+
+// 解法七 协程版，但是时间特别慢，不推荐，放在这里只是告诉大家，写 LeetCode 算法题的时候，启动 goroutine 特别慢
+func fib6(N int) int {
+    return <-fibb(N)
+}
+
+func fibb(n int) <- chan int {
+    result := make(chan int)
+    go func() {
+        defer close(result)
+        
+        if n <= 1 {
+            result <- n
+            return
+        }
+        result <- <-fibb(n-1) + <-fibb(n-2)
+    }()
+    return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0508.Most-Frequent-Subtree-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0513.Find-Bottom-Left-Tree-Value/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0513.Find-Bottom-Left-Tree-Value.md b/website/content/ChapterFour/0500~0599/0513.Find-Bottom-Left-Tree-Value.md
new file mode 100755
index 000000000..bce46e77b
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0513.Find-Bottom-Left-Tree-Value.md
@@ -0,0 +1,119 @@
+# [513. Find Bottom Left Tree Value](https://leetcode.com/problems/find-bottom-left-tree-value/)
+
+
+## 题目
+
+Given a binary tree, find the leftmost value in the last row of the tree.
+
+**Example 1**:
+
+    Input:
+    
+        2
+       / \
+      1   3
+    
+    Output:
+    1
+
+**Example 2**:
+
+    Input:
+    
+            1
+           / \
+          2   3
+         /   / \
+        4   5   6
+           /
+          7
+    
+    Output:
+    7
+
+**Note**: You may assume the tree (i.e., the given root node) is not **NULL**.
+
+
+## 题目大意
+
+给定一个二叉树，在树的最后一行找到最左边的值。注意: 您可以假设树（即给定的根节点）不为 NULL。
+
+
+
+
+
+
+## 解题思路
+
+
+- 给出一棵树，输出这棵树最下一层中最左边的节点的值。
+- 这一题用 DFS 和 BFS 均可解题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 DFS
+func findBottomLeftValue(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	res, maxHeight := 0, -1
+	findBottomLeftValueDFS(root, 0, &res, &maxHeight)
+	return res
+}
+
+func findBottomLeftValueDFS(root *TreeNode, curHeight int, res, maxHeight *int) {
+	if curHeight > *maxHeight && root.Left == nil && root.Right == nil {
+		*maxHeight = curHeight
+		*res = root.Val
+	}
+	if root.Left != nil {
+		findBottomLeftValueDFS(root.Left, curHeight+1, res, maxHeight)
+	}
+	if root.Right != nil {
+		findBottomLeftValueDFS(root.Right, curHeight+1, res, maxHeight)
+	}
+}
+
+// 解法二 BFS
+func findBottomLeftValue1(root *TreeNode) int {
+	queue := []*TreeNode{root}
+	for len(queue) > 0 {
+		next := []*TreeNode{}
+		for _, node := range queue {
+			if node.Left != nil {
+				next = append(next, node.Left)
+			}
+			if node.Right != nil {
+				next = append(next, node.Right)
+			}
+		}
+		if len(next) == 0 {
+			return queue[0].Val
+		}
+		queue = next
+	}
+	return 0
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0509.Fibonacci-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0515.Find-Largest-Value-in-Each-Tree-Row/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0515.Find-Largest-Value-in-Each-Tree-Row.md b/website/content/ChapterFour/0500~0599/0515.Find-Largest-Value-in-Each-Tree-Row.md
new file mode 100755
index 000000000..063e06f5e
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0515.Find-Largest-Value-in-Each-Tree-Row.md
@@ -0,0 +1,122 @@
+# [515. Find Largest Value in Each Tree Row](https://leetcode.com/problems/find-largest-value-in-each-tree-row/)
+
+
+## 题目
+
+You need to find the largest value in each row of a binary tree.
+
+**Example**:
+
+    Input: 
+    
+              1
+             / \
+            3   2
+           / \   \  
+          5   3   9 
+    
+    Output: [1, 3, 9]
+
+
+## 题目大意
+
+求在二叉树的每一行中找到最大的值。
+
+
+## 解题思路
+
+
+- 给出一个二叉树，要求依次输出每行的最大值
+- 用 BFS 层序遍历，将每层排序取出最大值。改进的做法是遍历中不断更新每层的最大值。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 层序遍历二叉树，再将每层排序取出最大值
+func largestValues(root *TreeNode) []int {
+	tmp := levelOrder(root)
+	res := []int{}
+	for i := 0; i < len(tmp); i++ {
+		sort.Ints(tmp[i])
+		res = append(res, tmp[i][len(tmp[i])-1])
+	}
+	return res
+}
+
+// 解法二 层序遍历二叉树，遍历过程中不断更新最大值
+func largestValues1(root *TreeNode) []int {
+	if root == nil {
+		return []int{}
+	}
+	q := []*TreeNode{root}
+	var res []int
+	for len(q) > 0 {
+		qlen := len(q)
+		max := math.MinInt32
+		for i := 0; i < qlen; i++ {
+			node := q[0]
+			q = q[1:]
+			if node.Val > max {
+				max = node.Val
+			}
+			if node.Left != nil {
+				q = append(q, node.Left)
+			}
+			if node.Right != nil {
+				q = append(q, node.Right)
+			}
+		}
+		res = append(res, max)
+	}
+	return res
+}
+
+// 解法三 深度遍历二叉树
+func largestValues3(root *TreeNode) []int {
+	var res []int
+	var dfs func(root *TreeNode, level int)
+	dfs = func(root *TreeNode, level int) {
+		if root == nil {
+			return
+		}
+		if len(res) == level {
+			res = append(res, root.Val)
+		}
+		if res[level] < root.Val {
+			res[level] = root.Val
+		}
+
+		dfs(root.Right, level+1)
+		dfs(root.Left, level+1)
+	}
+	dfs(root, 0)
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0513.Find-Bottom-Left-Tree-Value/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0518.Coin-Change-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0518.Coin-Change-II.md b/website/content/ChapterFour/0500~0599/0518.Coin-Change-II.md
new file mode 100644
index 000000000..5fc89a1cd
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0518.Coin-Change-II.md
@@ -0,0 +1,80 @@
+# [518. Coin Change II](https://leetcode.com/problems/coin-change-ii/)
+
+
+## 题目
+
+You are given an integer array `coins` representing coins of different denominations and an integer `amount` representing a total amount of money.
+
+Return *the number of combinations that make up that amount*. If that amount of money cannot be made up by any combination of the coins, return `0`.
+
+You may assume that you have an infinite number of each kind of coin.
+
+The answer is **guaranteed** to fit into a signed **32-bit** integer.
+
+**Example 1:**
+
+```
+Input: amount = 5, coins = [1,2,5]
+Output: 4
+Explanation: there are four ways to make up the amount:
+5=5
+5=2+2+1
+5=2+1+1+1
+5=1+1+1+1+1
+```
+
+**Example 2:**
+
+```
+Input: amount = 3, coins = [2]
+Output: 0
+Explanation: the amount of 3 cannot be made up just with coins of 2.
+```
+
+**Example 3:**
+
+```
+Input: amount = 10, coins = [10]
+Output: 1
+```
+
+**Constraints:**
+
+- `1 <= coins.length <= 300`
+- `1 <= coins[i] <= 5000`
+- All the values of `coins` are **unique**.
+- `0 <= amount <= 5000`
+
+## 题目大意
+
+给你一个整数数组 coins 表示不同面额的硬币，另给一个整数 amount 表示总金额。请你计算并返回可以凑成总金额的硬币组合数。如果任何硬币组合都无法凑出总金额，返回 0 。假设每一种面额的硬币有无限个。题目数据保证结果符合 32 位带符号整数。
+
+## 解题思路
+
+- 此题虽然名字叫 Coin Change，但是不是经典的背包九讲问题。题目中 coins 的每个元素可以选取多次，且不考虑选取元素的顺序，因此这道题实际需要计算的是选取硬币的组合数。定义 dp[i] 表示金额之和等于 i 的硬币组合数，目标求 dp[amount]。初始边界条件为 dp[0] = 1，即不取任何硬币，就这一种取法，金额为 0 。状态转移方程 dp[i] += dp[i-coin]，coin 为当前枚举的 coin。
+- 可能有读者会有疑惑，上述做法会不会出现重复计算。答案是不会。外层循环是遍历数组 coins 的值，内层循环是遍历不同的金额之和，在计算 dp[i] 的值时，可以确保金额之和等于 i 的硬币面额的顺序，由于顺序确定，因此不会重复计算不同的排列。
+- 和此题完全一致的解题思路的题有，第 377 题和第 494 题。
+
+## 代码
+
+```go
+package leetcode
+
+func change(amount int, coins []int) int {
+	dp := make([]int, amount+1)
+	dp[0] = 1
+	for _, coin := range coins {
+		for i := coin; i <= amount; i++ {
+			dp[i] += dp[i-coin]
+		}
+	}
+	return dp[amount]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0515.Find-Largest-Value-in-Each-Tree-Row/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0519.Random-Flip-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0519.Random-Flip-Matrix.md b/website/content/ChapterFour/0500~0599/0519.Random-Flip-Matrix.md
new file mode 100644
index 000000000..d72da22bb
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0519.Random-Flip-Matrix.md
@@ -0,0 +1,105 @@
+# [519. Random Flip Matrix](https://leetcode.com/problems/random-flip-matrix/)
+
+## 题目
+
+There is an m x n binary grid matrix with all the values set 0 initially. Design an algorithm to randomly pick an index (i, j) where matrix[i][j] == 0 and flips it to 1. All the indices (i, j) where matrix[i][j] == 0 should be equally likely to be returned.
+
+Optimize your algorithm to minimize the number of calls made to the built-in random function of your language and optimize the time and space complexity.
+
+Implement the Solution class:
+
+- Solution(int m, int n) Initializes the object with the size of the binary matrix m and n.
+- int[] flip() Returns a random index [i, j] of the matrix where matrix[i][j] == 0 and flips it to 1.
+- void reset() Resets all the values of the matrix to be 0.
+
+**Example 1**:
+
+    Input
+    ["Solution", "flip", "flip", "flip", "reset", "flip"]
+    [[3, 1], [], [], [], [], []]
+    Output
+    [null, [1, 0], [2, 0], [0, 0], null, [2, 0]]
+
+    Explanation
+    Solution solution = new Solution(3, 1);
+    solution.flip();  // return [1, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned.
+    solution.flip();  // return [2, 0], Since [1,0] was returned, [2,0] and [0,0]
+    solution.flip();  // return [0, 0], Based on the previously returned indices, only [0,0] can be returned.
+    solution.reset(); // All the values are reset to 0 and can be returned.
+    solution.flip();  // return [2, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned.
+
+**Constraints:**
+
+- 1 <= m, n <= 10000
+- There will be at least one free cell for each call to flip.
+- At most 1000 calls will be made to flip and reset.
+
+## 题目大意
+
+给你一个 m x n 的二元矩阵 matrix ，且所有值被初始化为 0 。请你设计一个算法，随机选取一个满足 matrix[i][j] == 0 的下标 (i, j) ，并将它的值变为 1 。所有满足 matrix[i][j] == 0 的下标 (i, j) 被选取的概率应当均等。
+
+尽量最少调用内置的随机函数，并且优化时间和空间复杂度。
+
+实现 Solution 类：
+
+- Solution(int m, int n) 使用二元矩阵的大小 m 和 n 初始化该对象
+- int[] flip() 返回一个满足 matrix[i][j] == 0 的随机下标 [i, j] ，并将其对应格子中的值变为 1
+- void reset() 将矩阵中所有的值重置为 0
+
+## 解题思路
+
+- 二维矩阵利用哈希表转换为一维,每次随机选择一维中的任意一个元素，然后与最后一个元素交换，一维元素的总个数减一
+- 哈希表中默认的映射为x->x, 然后将不满足这个映射的特殊键值对存入哈希表
+
+## 代码
+
+```go
+package leetcode
+
+import "math/rand"
+
+type Solution struct {
+	r     int
+	c     int
+	total int
+	mp    map[int]int
+}
+
+func Constructor(m int, n int) Solution {
+	return Solution{
+		r:     m,
+		c:     n,
+		total: m * n,
+		mp:    map[int]int{},
+	}
+}
+
+func (this *Solution) Flip() []int {
+	k := rand.Intn(this.total)
+	val := k
+	if v, ok := this.mp[k]; ok {
+		val = v
+	}
+	if _, ok := this.mp[this.total-1]; ok {
+		this.mp[k] = this.mp[this.total-1]
+	} else {
+		this.mp[k] = this.total - 1
+	}
+	delete(this.mp, this.total - 1)
+	this.total--
+	newR, newC := val/this.c, val%this.c
+	return []int{newR, newC}
+}
+
+func (this *Solution) Reset() {
+	this.total = this.r * this.c
+	this.mp = map[int]int{}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0518.Coin-Change-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0520.Detect-Capital/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0520.Detect-Capital.md b/website/content/ChapterFour/0500~0599/0520.Detect-Capital.md
new file mode 100644
index 000000000..33791a2ec
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0520.Detect-Capital.md
@@ -0,0 +1,75 @@
+# [520. Detect Capital](https://leetcode.com/problems/detect-capital/)
+
+
+## 题目
+
+We define the usage of capitals in a word to be right when one of the following cases holds:
+
+All letters in this word are capitals, like "USA".
+
+All letters in this word are not capitals, like "leetcode".
+
+Only the first letter in this word is capital, like "Google".
+
+Given a string word, return true if the usage of capitals in it is right.
+
+**Example 1:**
+
+```
+Input: word = "USA"
+Output: true
+```
+
+**Example 2:**
+
+```
+Input: word = "FlaG"
+Output: false
+```
+
+**Constraints:**
+
+- 1 <= word.length <= 100
+- word consists of lowercase and uppercase English letters.
+
+## 题目大意
+
+我们定义，在以下情况时，单词的大写用法是正确的：
+
+全部字母都是大写，比如 "USA" 。
+单词中所有字母都不是大写，比如 "leetcode" 。
+如果单词不只含有一个字母，只有首字母大写，比如"Google" 。
+
+给你一个字符串 word 。如果大写用法正确，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 把 word 分别转换为全部小写 wLower，全部大写 wUpper，首字母大写的字符串 wCaptial
+- 判断 word 是否等于 wLower, wUpper, wCaptial 中的一个，如果是返回 true，否则返回 false
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func detectCapitalUse(word string) bool {
+	wLower := strings.ToLower(word)
+	wUpper := strings.ToUpper(word)
+	wCaptial := strings.ToUpper(string(word[0])) + strings.ToLower(string(word[1:]))
+	if wCaptial == word || wLower == word || wUpper == word {
+		return true
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0519.Random-Flip-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0523.Continuous-Subarray-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0523.Continuous-Subarray-Sum.md b/website/content/ChapterFour/0500~0599/0523.Continuous-Subarray-Sum.md
new file mode 100644
index 000000000..671295e2a
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0523.Continuous-Subarray-Sum.md
@@ -0,0 +1,82 @@
+# [523. Continuous Subarray Sum](https://leetcode.com/problems/continuous-subarray-sum/)
+
+
+## 题目
+
+Given an integer array `nums` and an integer `k`, return `true` *if* `nums` *has a continuous subarray of size **at least two** whose elements sum up to a multiple of* `k`*, or* `false` *otherwise*.
+
+An integer `x` is a multiple of `k` if there exists an integer `n` such that `x = n * k`. `0` is **always** a multiple of `k`.
+
+**Example 1:**
+
+```
+Input: nums = [23,2,4,6,7], k = 6
+Output: true
+Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.
+```
+
+**Example 2:**
+
+```
+Input: nums = [23,2,6,4,7], k = 6
+Output: true
+Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
+42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.
+```
+
+**Example 3:**
+
+```
+Input: nums = [23,2,6,4,7], k = 13
+Output: false
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 105`
+- `0 <= nums[i] <= 109`
+- `0 <= sum(nums[i]) <= 231 - 1`
+- `1 <= k <= 231 - 1`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个整数 k ，编写一个函数来判断该数组是否含有同时满足下述条件的连续子数组：
+
+- 子数组大小至少为 2 ，且
+- 子数组元素总和为 k 的倍数。
+
+如果存在，返回 true ；否则，返回 false 。如果存在一个整数 n ，令整数 x 符合 x = n * k ，则称 x 是 k 的一个倍数。
+
+## 解题思路
+
+- 简单题。题目只要求是否存在，不要求找出所有解。用一个变量 sum 记录累加和。子数组的元素和可以用前缀和相减得到，例如 [i,j] 区间内的元素和，可以由 prefixSum[j] - prefixSum[i] 得到。当 prefixSums[j]−prefixSums[i] 为 k 的倍数时，prefixSums[i] 和 prefixSums[j] 除以 k 的余数相同。因此只需要计算每个下标对应的前缀和除以 k 的余数即可，使用 map 存储每个余数第一次出现的下标即可。在 map 中如果存在相同余数的 key，代表当前下标和 map 中这个 key 记录的下标可以满足总和为 k 的倍数这一条件。再判断一下能否满足大小至少为 2 的条件即可。用 2 个下标相减，长度大于等于 2 即满足条件，可以输出 true。
+
+## 代码
+
+```go
+package leetcode
+
+func checkSubarraySum(nums []int, k int) bool {
+	m := make(map[int]int)
+	m[0] = -1
+	sum := 0
+	for i, n := range nums {
+		sum += n
+		if r, ok := m[sum%k]; ok {
+			if i-2 >= r {
+				return true
+			}
+		} else {
+			m[sum%k] = i
+		}
+	}
+	return false
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0520.Detect-Capital/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0524.Longest-Word-in-Dictionary-through-Deleting/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0524.Longest-Word-in-Dictionary-through-Deleting.md b/website/content/ChapterFour/0500~0599/0524.Longest-Word-in-Dictionary-through-Deleting.md
new file mode 100644
index 000000000..51d40043e
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0524.Longest-Word-in-Dictionary-through-Deleting.md
@@ -0,0 +1,83 @@
+# [524. Longest Word in Dictionary through Deleting](https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/)
+
+## 题目
+
+Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.
+
+
+**Example 1**:
+
+```
+
+Input:
+s = "abpcplea", d = ["ale","apple","monkey","plea"]
+
+Output: 
+"apple"
+
+```
+
+
+**Example 2**:
+
+```
+
+Input:
+s = "abpcplea", d = ["a","b","c"]
+
+Output: 
+"a"
+
+```
+
+**Note**:  
+
+- All the strings in the input will only contain lower-case letters.
+- The size of the dictionary won't exceed 1,000.
+- The length of all the strings in the input won't exceed 1,000.
+
+
+## 题目大意
+
+
+给出一个初始串，再给定一个字符串数组，要求在字符串数组中找到能在初始串中通过删除字符得到的最长的串，如果最长的串有多组解，要求输出字典序最小的那组解。
+
+## 解题思路
+
+
+这道题就单纯的用 O(n^2) 暴力循环即可，注意最终解的要求，如果都是最长的串，要求输出字典序最小的那个串，只要利用字符串比较得到字典序最小的串即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findLongestWord(s string, d []string) string {
+	res := ""
+	for i := 0; i < len(d); i++ {
+		pointS := 0
+		pointD := 0
+		for pointS < len(s) && pointD < len(d[i]) {
+			if s[pointS] == d[i][pointD] {
+				pointD++
+			}
+			pointS++
+		}
+		if pointD == len(d[i]) && (len(res) < len(d[i]) || (len(res) == len(d[i]) && res > d[i])) {
+			res = d[i]
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0523.Continuous-Subarray-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0525.Contiguous-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0525.Contiguous-Array.md b/website/content/ChapterFour/0500~0599/0525.Contiguous-Array.md
new file mode 100644
index 000000000..412f71fad
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0525.Contiguous-Array.md
@@ -0,0 +1,74 @@
+# [525. Contiguous Array](https://leetcode.com/problems/contiguous-array/)
+
+
+## 题目
+
+Given a binary array `nums`, return *the maximum length of a contiguous subarray with an equal number of* `0` *and* `1`.
+
+**Example 1:**
+
+```
+Input: nums = [0,1]
+Output: 2
+Explanation: [0, 1] is the longest contiguous subarray with an equal number of 0 and 1.
+```
+
+**Example 2:**
+
+```
+Input: nums = [0,1,0]
+Output: 2
+Explanation: [0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 105`
+- `nums[i]` is either `0` or `1`.
+
+## 题目大意
+
+给定一个二进制数组 nums , 找到含有相同数量的 0 和 1 的最长连续子数组，并返回该子数组的长度。
+
+## 解题思路
+
+- 0 和 1 的数量相同可以转化为两者数量相差为 0，如果将 0 看作为 -1，那么原题转化为求最长连续子数组，其元素和为 0 。又变成了区间内求和的问题，自然而然转换为前缀和来处理。假设连续子数组是 [i,j] 区间，这个区间内元素和为 0 意味着 prefixSum[j] - prefixSum[i] = 0，也就是 prefixSum[i] = prefixSum[j]。不断累加前缀和，将每个前缀和存入 map 中。一旦某个 key 存在了，代表之前某个下标的前缀和和当前下标构成的区间，这段区间内的元素和为 0 。这个区间是所求。扫完整个数组，扫描过程中动态更新最大区间长度，扫描完成便可得到最大区间长度，即最长连续子数组。
+
+## 代码
+
+```go
+package leetcode
+
+func findMaxLength(nums []int) int {
+	dict := map[int]int{}
+	dict[0] = -1
+	count, res := 0, 0
+	for i := 0; i < len(nums); i++ {
+		if nums[i] == 0 {
+			count--
+		} else {
+			count++
+		}
+		if idx, ok := dict[count]; ok {
+			res = max(res, i-idx)
+		} else {
+			dict[count] = i
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0524.Longest-Word-in-Dictionary-through-Deleting/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0526.Beautiful-Arrangement/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0526.Beautiful-Arrangement.md b/website/content/ChapterFour/0500~0599/0526.Beautiful-Arrangement.md
new file mode 100755
index 000000000..f3188ead0
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0526.Beautiful-Arrangement.md
@@ -0,0 +1,117 @@
+# [526. Beautiful Arrangement](https://leetcode.com/problems/beautiful-arrangement/)
+
+
+## 题目
+
+Suppose you have **N** integers from 1 to N. We define a beautiful arrangement as an array that is constructed by these **N** numbers successfully if one of the following is true for the ith position (1 <= i <= N) in this array:
+
+1. The number at the i position is divisible by **i**.th
+2. **i** is divisible by the number at the i position.th
+
+Now given N, how many beautiful arrangements can you construct?
+
+**Example 1**:
+
+    Input: 2
+    Output: 2
+    Explanation: 
+    
+    The first beautiful arrangement is [1, 2]:
+    
+    Number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1).
+    
+    Number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2).
+    
+    The second beautiful arrangement is [2, 1]:
+    
+    Number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1).
+    
+    Number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.
+
+**Note**:
+
+1. **N** is a positive integer and will not exceed 15.
+
+
+## 题目大意
+
+假设有从 1 到 N 的 N 个整数，如果从这 N 个数字中成功构造出一个数组，使得数组的第 i 位 (1 <= i <= N) 满足如下两个条件中的一个，我们就称这个数组为一个优美的排列。条件：
+
+- 第 i 位的数字能被 i 整除
+- i 能被第 i 位上的数字整除
+
+现在给定一个整数 N，请问可以构造多少个优美的排列？
+
+
+
+## 解题思路
+
+
+- 这一题是第 46 题的加强版。由于这一题给出的数组里面的数字都是不重复的，所以可以当做第 46 题来做。
+- 这题比第 46 题多的一个条件是，要求数字可以被它对应的下标 + 1 整除，或者下标 + 1 可以整除下标对应的这个数字。在 DFS 回溯过程中加入这个剪枝条件就可以了。
+- 当前做法时间复杂度不是最优的，大概只有 33.3%
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 暴力打表法
+func countArrangement1(N int) int {
+	res := []int{0, 1, 2, 3, 8, 10, 36, 41, 132, 250, 700, 750, 4010, 4237, 10680, 24679, 87328, 90478, 435812}
+	return res[N]
+}
+
+// 解法二 DFS 回溯
+func countArrangement(N int) int {
+	if N == 0 {
+		return 0
+	}
+	nums, used, p, res := make([]int, N), make([]bool, N), []int{}, [][]int{}
+	for i := range nums {
+		nums[i] = i + 1
+	}
+	generatePermutation526(nums, 0, p, &res, &used)
+	return len(res)
+}
+
+func generatePermutation526(nums []int, index int, p []int, res *[][]int, used *[]bool) {
+	if index == len(nums) {
+		temp := make([]int, len(p))
+		copy(temp, p)
+		*res = append(*res, temp)
+		return
+	}
+	for i := 0; i < len(nums); i++ {
+		if !(*used)[i] {
+			if !(checkDivisible(nums[i], len(p)+1) || checkDivisible(len(p)+1, nums[i])) { // 关键的剪枝条件
+				continue
+			}
+			(*used)[i] = true
+			p = append(p, nums[i])
+			generatePermutation526(nums, index+1, p, res, used)
+			p = p[:len(p)-1]
+			(*used)[i] = false
+		}
+	}
+	return
+}
+
+func checkDivisible(num, d int) bool {
+	tmp := num / d
+	if int(tmp)*int(d) == num {
+		return true
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0525.Contiguous-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0528.Random-Pick-with-Weight/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0528.Random-Pick-with-Weight.md b/website/content/ChapterFour/0500~0599/0528.Random-Pick-with-Weight.md
new file mode 100755
index 000000000..3b83ea432
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0528.Random-Pick-with-Weight.md
@@ -0,0 +1,115 @@
+# [528. Random Pick with Weight](https://leetcode.com/problems/random-pick-with-weight/)
+
+
+## 题目
+
+Given an array `w` of positive integers, where `w[i]` describes the weight of index `i`, write a function `pickIndex` which randomly picks an index in proportion to its weight.
+
+**Note**:
+
+1. `1 <= w.length <= 10000`
+2. `1 <= w[i] <= 10^5`
+3. `pickIndex` will be called at most `10000` times.
+
+**Example 1**:
+
+    Input: 
+    ["Solution","pickIndex"]
+    [[[1]],[]]
+    Output: [null,0]
+
+**Example 2**:
+
+    Input: 
+    ["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
+    [[[1,3]],[],[],[],[],[]]
+    Output: [null,0,1,1,1,0]
+
+**Explanation of Input Syntax**:
+
+The input is two lists: the subroutines called and their arguments. `Solution`'s constructor has one argument, the array `w`. `pickIndex` has no arguments. Arguments are always wrapped with a list, even if there aren't any.
+
+
+## 题目大意
+
+给定一个正整数数组 w ，其中 w[i] 代表位置 i 的权重，请写一个函数 pickIndex ，它可以随机地获取位置 i，选取位置 i 的概率与 w[i] 成正比。
+
+说明:
+
+1. 1 <= w.length <= 10000
+2. 1 <= w[i] <= 10^5
+3. pickIndex 将被调用不超过 10000 次
+
+
+输入语法说明：
+
+输入是两个列表：调用成员函数名和调用的参数。Solution 的构造函数有一个参数，即数组 w。pickIndex 没有参数。输入参数是一个列表，即使参数为空，也会输入一个 [] 空列表。
+
+
+
+## 解题思路
+
+- 给出一个数组，每个元素值代表该下标的权重值，`pickIndex()` 随机取一个位置 i，这个位置出现的概率和该元素值成正比。
+- 由于涉及到了权重的问题，这一题可以先考虑用前缀和处理权重。在 `[0,prefixSum)` 区间内随机选一个整数 `x`，下标 `i` 是满足 `x< prefixSum[i]` 条件的最小下标，求这个下标 `i` 即是最终解。二分搜索查找下标 `i` 。对于某些下标 `i`，所有满足 `prefixSum[i] - w[i] ≤ v < prefixSum[i]` 的整数 `v` 都映射到这个下标。因此，所有的下标都与下标权重成比例。
+- 时间复杂度：预处理的时间复杂度是 O(n)，`pickIndex()` 的时间复杂度是 O(log n)。空间复杂度 O(n)。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math/rand"
+)
+
+// Solution528 define
+type Solution528 struct {
+	prefixSum []int
+}
+
+// Constructor528 define
+func Constructor528(w []int) Solution528 {
+	prefixSum := make([]int, len(w))
+	for i, e := range w {
+		if i == 0 {
+			prefixSum[i] = e
+			continue
+		}
+		prefixSum[i] = prefixSum[i-1] + e
+	}
+	return Solution528{prefixSum: prefixSum}
+}
+
+// PickIndex define
+func (so *Solution528) PickIndex() int {
+	n := rand.Intn(so.prefixSum[len(so.prefixSum)-1]) + 1
+	low, high := 0, len(so.prefixSum)-1
+	for low < high {
+		mid := low + (high-low)>>1
+		if so.prefixSum[mid] == n {
+			return mid
+		} else if so.prefixSum[mid] < n {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return low
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(w);
+ * param_1 := obj.PickIndex();
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0526.Beautiful-Arrangement/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0529.Minesweeper/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0529.Minesweeper.md b/website/content/ChapterFour/0500~0599/0529.Minesweeper.md
new file mode 100644
index 000000000..88540208a
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0529.Minesweeper.md
@@ -0,0 +1,142 @@
+# [529. Minesweeper](https://leetcode.com/problems/minesweeper/)
+
+
+
+## 题目
+
+Let's play the minesweeper game ([Wikipedia](https://en.wikipedia.org/wiki/Minesweeper_(video_game)), [online game](http://minesweeperonline.com/))!
+
+You are given a 2D char matrix representing the game board. **'M'** represents an **unrevealed** mine, **'E'** represents an **unrevealed** empty square, **'B'** represents a **revealed** blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, **digit** ('1' to '8') represents how many mines are adjacent to this **revealed** square, and finally **'X'** represents a **revealed** mine.
+
+Now given the next click position (row and column indices) among all the **unrevealed** squares ('M' or 'E'), return the board after revealing this position according to the following rules:
+
+1. If a mine ('M') is revealed, then the game is over - change it to **'X'**.
+2. If an empty square ('E') with **no adjacent mines** is revealed, then change it to revealed blank ('B') and all of its adjacent **unrevealed** squares should be revealed recursively.
+3. If an empty square ('E') with **at least one adjacent mine** is revealed, then change it to a digit ('1' to '8') representing the number of adjacent mines.
+4. Return the board when no more squares will be revealed.
+
+**Example 1**:
+
+```
+Input: 
+
+[['E', 'E', 'E', 'E', 'E'],
+ ['E', 'E', 'M', 'E', 'E'],
+ ['E', 'E', 'E', 'E', 'E'],
+ ['E', 'E', 'E', 'E', 'E']]
+
+Click : [3,0]
+
+Output: 
+
+[['B', '1', 'E', '1', 'B'],
+ ['B', '1', 'M', '1', 'B'],
+ ['B', '1', '1', '1', 'B'],
+ ['B', 'B', 'B', 'B', 'B']]
+
+Explanation:
+```
+
+![https://assets.leetcode.com/uploads/2018/10/12/minesweeper_example_1.png](https://assets.leetcode.com/uploads/2018/10/12/minesweeper_example_1.png)
+
+**Example 2**:
+
+```
+Input: 
+
+[['B', '1', 'E', '1', 'B'],
+ ['B', '1', 'M', '1', 'B'],
+ ['B', '1', '1', '1', 'B'],
+ ['B', 'B', 'B', 'B', 'B']]
+
+Click : [1,2]
+
+Output: 
+
+[['B', '1', 'E', '1', 'B'],
+ ['B', '1', 'X', '1', 'B'],
+ ['B', '1', '1', '1', 'B'],
+ ['B', 'B', 'B', 'B', 'B']]
+
+Explanation:
+```
+
+![https://assets.leetcode.com/uploads/2018/10/12/minesweeper_example_2.png](https://assets.leetcode.com/uploads/2018/10/12/minesweeper_example_2.png)
+
+**Note**:
+
+1. The range of the input matrix's height and width is [1,50].
+2. The click position will only be an unrevealed square ('M' or 'E'), which also means the input board contains at least one clickable square.
+3. The input board won't be a stage when game is over (some mines have been revealed).
+4. For simplicity, not mentioned rules should be ignored in this problem. For example, you **don't** need to reveal all the unrevealed mines when the game is over, consider any cases that you will win the game or flag any squares.
+
+
+## 题目大意
+
+给定一个代表游戏板的二维字符矩阵。 'M' 代表一个未挖出的地雷，'E' 代表一个未挖出的空方块，'B' 代表没有相邻（上，下，左，右，和所有4个对角线）地雷的已挖出的空白方块，数字（'1' 到 '8'）表示有多少地雷与这块已挖出的方块相邻，'X' 则表示一个已挖出的地雷。现在给出在所有未挖出的方块中（'M'或者'E'）的下一个点击位置（行和列索引），根据以下规则，返回相应位置被点击后对应的面板：
+
+1. 如果一个地雷（'M'）被挖出，游戏就结束了- 把它改为 'X'。
+2. 如果一个没有相邻地雷的空方块（'E'）被挖出，修改它为（'B'），并且所有和其相邻的未挖出方块都应该被递归地揭露。
+3. 如果一个至少与一个地雷相邻的空方块（'E'）被挖出，修改它为数字（'1'到'8'），表示相邻地雷的数量。
+4. 如果在此次点击中，若无更多方块可被揭露，则返回面板。
+
+
+注意：
+
+- 输入矩阵的宽和高的范围为 [1,50]。
+- 点击的位置只能是未被挖出的方块 ('M' 或者 'E')，这也意味着面板至少包含一个可点击的方块。
+- 输入面板不会是游戏结束的状态（即有地雷已被挖出）。
+- 简单起见，未提及的规则在这个问题中可被忽略。例如，当游戏结束时你不需要挖出所有地雷，考虑所有你可能赢得游戏或标记方块的情况。
+
+
+
+## 解题思路
+
+- 给出一张扫雷地图和点击的坐标，M 代表雷，E 代表还没有点击过的空砖块，B 代表点击过的空砖块，1-8 代表砖块周围 8 个方块里面有雷的个数，X 代表点到了雷。问点击一次以后，输出更新点击以后的地图。
+- DPS 和 BFS 都可以解题。先根据原图预处理地图，记录出最终地图的状态，0 代表空白砖块，1-8 代表雷的个数，-1 代表是雷。再 DFS 遍历这张处理后的图，输出最终的地图即可。
+
+## 代码
+
+```go
+func updateBoard(board [][]byte, click []int) [][]byte {
+	if board[click[0]][click[1]] == 'M' {
+		board[click[0]][click[1]] = 'X'
+		return board
+	}
+	dfs(board, click[0], click[1])
+	return board
+}
+
+func dfs(board [][]byte, x, y int) {
+	cnt := 0
+	for i := 0; i < 8; i++ {
+		nx, ny := x+dir8[i][0], y+dir8[i][1]
+		if isInBoard(board, nx, ny) && board[nx][ny] == 'M' {
+			cnt++
+
+		}
+	}
+	if cnt > 0 {
+		board[x][y] = byte(cnt + '0')
+		return
+	}
+	board[x][y] = 'B'
+	for i := 0; i < 8; i++ {
+		nx, ny := x+dir8[i][0], y+dir8[i][1]
+		if isInBoard(board, nx, ny) && board[nx][ny] != 'B' {
+			dfs(board, nx, ny)
+		}
+	}
+}
+
+func isInBoard(board [][]byte, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0528.Random-Pick-with-Weight/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0530.Minimum-Absolute-Difference-in-BST/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0530.Minimum-Absolute-Difference-in-BST.md b/website/content/ChapterFour/0500~0599/0530.Minimum-Absolute-Difference-in-BST.md
new file mode 100644
index 000000000..5dcd943f9
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0530.Minimum-Absolute-Difference-in-BST.md
@@ -0,0 +1,101 @@
+# [530. Minimum Absolute Difference in BST](https://leetcode.com/problems/minimum-absolute-difference-in-bst/)
+
+
+## 题目
+
+Given a binary search tree with non-negative values, find the minimum [absolute difference](https://en.wikipedia.org/wiki/Absolute_difference) between values of any two nodes.
+
+**Example:**
+
+```
+Input:
+
+   1
+    \
+     3
+    /
+   2
+
+Output:
+1
+
+Explanation:
+The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).
+```
+
+**Note:**
+
+- There are at least two nodes in this BST.
+- This question is the same as 783: [https://leetcode.com/problems/minimum-distance-between-bst-nodes/](https://leetcode.com/problems/minimum-distance-between-bst-nodes/)
+
+## 题目大意
+
+给你一棵所有节点为非负值的二叉搜索树，请你计算树中任意两节点的差的绝对值的最小值。
+
+## 解题思路
+
+- 由于是 BST 树，利用它有序的性质，中根遍历的结果是有序的。中根遍历过程中动态维护前后两个节点的差值，即可找到最小差值。
+- 此题与第 783 题完全相同。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func getMinimumDifference(root *TreeNode) int {
+	res, nodes := math.MaxInt16, -1
+	dfsBST(root, &res, &nodes)
+	return res
+}
+
+func dfsBST(root *TreeNode, res, pre *int) {
+	if root == nil {
+		return
+	}
+	dfsBST(root.Left, res, pre)
+	if *pre != -1 {
+		*res = min(*res, abs(root.Val-*pre))
+	}
+	*pre = root.Val
+	dfsBST(root.Right, res, pre)
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0529.Minesweeper/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0532.K-diff-Pairs-in-an-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0532.K-diff-Pairs-in-an-Array.md b/website/content/ChapterFour/0500~0599/0532.K-diff-Pairs-in-an-Array.md
new file mode 100644
index 000000000..abbfb45ad
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0532.K-diff-Pairs-in-an-Array.md
@@ -0,0 +1,102 @@
+# [532. K-diff Pairs in an Array](https://leetcode.com/problems/k-diff-pairs-in-an-array/)
+
+## 题目
+
+Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.
+
+
+**Example 1**:
+
+```
+
+Input: [3, 1, 4, 1, 5], k = 2
+Output: 2
+Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
+Although we have two 1s in the input, we should only return the number of unique pairs.
+
+```
+
+**Example 2**:
+
+```
+
+Input:[1, 2, 3, 4, 5], k = 1
+Output: 4
+Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
+
+```
+
+**Example 3**:
+
+```
+
+Input: [1, 3, 1, 5, 4], k = 0
+Output: 1
+Explanation: There is one 0-diff pair in the array, (1, 1).
+
+```
+
+
+**Note**:  
+
+1. The pairs (i, j) and (j, i) count as the same pair.
+2. The length of the array won't exceed 10,000.
+3. All the integers in the given input belong to the range: [-1e7, 1e7].
+
+## 题目大意
+
+
+给定一个数组，在数组里面找到几组不同的 pair 对，每个 pair 对相差 K 。问能找出多少组这样的 pair 对。
+
+
+## 解题思路
+
+这一题可以用 map 记录每个数字出现的次数。重复的数字也会因为唯一的 key，不用担心某个数字会判断多次。遍历一次 map，每个数字都加上 K 以后，判断字典里面是否存在，如果存在， count ++，如果 K = 0 的情况需要单独判断，如果字典中这个元素频次大于 1，count 也需要 ++。
+
+
+
+
+
+
+
+
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findPairs(nums []int, k int) int {
+	if k < 0 || len(nums) == 0 {
+		return 0
+	}
+	var count int
+	m := make(map[int]int, len(nums))
+	for _, value := range nums {
+		m[value]++
+	}
+	for key := range m {
+		if k == 0 && m[key] > 1 {
+			count++
+			continue
+		}
+		if k > 0 && m[key+k] > 0 {
+			count++
+		}
+	}
+	return count
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0530.Minimum-Absolute-Difference-in-BST/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0535.Encode-and-Decode-TinyURL/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0535.Encode-and-Decode-TinyURL.md b/website/content/ChapterFour/0500~0599/0535.Encode-and-Decode-TinyURL.md
new file mode 100644
index 000000000..edb658941
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0535.Encode-and-Decode-TinyURL.md
@@ -0,0 +1,67 @@
+# [535. Encode and Decode TinyURL](https://leetcode.com/problems/encode-and-decode-tinyurl/)
+
+
+## 题目
+
+> Note: This is a companion problem to the System Design problem: Design TinyURL.
+
+TinyURL is a URL shortening service where you enter a URL such as `https://leetcode.com/problems/design-tinyurl` and it returns a short URL such as `http://tinyurl.com/4e9iAk`.
+
+Design the `encode` and `decode` methods for the TinyURL service. There is no restriction on how your encode/decode algorithm should work. You just need to ensure that a URL can be encoded to a tiny URL and the tiny URL can be decoded to the original URL.
+
+## 题目大意
+
+TinyURL是一种URL简化服务， 比如：当你输入一个URL [https://leetcode.com/problems/design-tinyurl](https://leetcode.com/problems/design-tinyurl) 时，它将返回一个简化的URL [http://tinyurl.com/4e9iAk](http://tinyurl.com/4e9iAk).
+
+要求：设计一个 TinyURL 的加密 encode 和解密 decode 的方法。你的加密和解密算法如何设计和运作是没有限制的，你只需要保证一个URL可以被加密成一个TinyURL，并且这个TinyURL可以用解密方法恢复成原本的URL。
+
+## 解题思路
+
+- 简单题。由于题目并无规定 `encode()` 算法，所以自由度非常高。最简单的做法是把原始 `URL` 存起来，并记录下存在字符串数组中的下标位置。`decode()` 的时候根据存储的下标还原原始的 `URL`。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"fmt"
+	"strconv"
+	"strings"
+)
+
+type Codec struct {
+	urls []string
+}
+
+func Constructor() Codec {
+	return Codec{[]string{}}
+}
+
+// Encodes a URL to a shortened URL.
+func (this *Codec) encode(longUrl string) string {
+	this.urls = append(this.urls, longUrl)
+	return "http://tinyurl.com/" + fmt.Sprintf("%v", len(this.urls)-1)
+}
+
+// Decodes a shortened URL to its original URL.
+func (this *Codec) decode(shortUrl string) string {
+	tmp := strings.Split(shortUrl, "/")
+	i, _ := strconv.Atoi(tmp[len(tmp)-1])
+	return this.urls[i]
+}
+
+/**
+ * Your Codec object will be instantiated and called as such:
+ * obj := Constructor();
+ * url := obj.encode(longUrl);
+ * ans := obj.decode(url);
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0532.K-diff-Pairs-in-an-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0537.Complex-Number-Multiplication/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0537.Complex-Number-Multiplication.md b/website/content/ChapterFour/0500~0599/0537.Complex-Number-Multiplication.md
new file mode 100644
index 000000000..46592cf55
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0537.Complex-Number-Multiplication.md
@@ -0,0 +1,80 @@
+# [537. Complex Number Multiplication](https://leetcode.com/problems/complex-number-multiplication/)
+
+
+## 题目
+
+Given two strings representing two [complex numbers](https://en.wikipedia.org/wiki/Complex_number).
+
+You need to return a string representing their multiplication. Note i2 = -1 according to the definition.
+
+**Example 1**:
+
+```
+Input: "1+1i", "1+1i"
+Output: "0+2i"
+Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i.
+```
+
+**Example 2**:
+
+```
+Input: "1+-1i", "1+-1i"
+Output: "0+-2i"
+Explanation: (1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i, and you need convert it to the form of 0+-2i.
+```
+
+**Note**:
+
+1. The input strings will not have extra blank.
+2. The input strings will be given in the form of **a+bi**, where the integer **a** and **b** will both belong to the range of [-100, 100]. And **the output should be also in this form**.
+
+## 题目大意
+
+给定两个表示复数的字符串。返回表示它们乘积的字符串。注意，根据定义 i^2 = -1 。
+
+注意:
+
+- 输入字符串不包含额外的空格。
+- 输入字符串将以 a+bi 的形式给出，其中整数 a 和 b 的范围均在 [-100, 100] 之间。输出也应当符合这种形式。
+
+
+
+## 解题思路
+
+- 给定 2 个字符串，要求这两个复数的乘积，输出也是字符串格式。
+- 数学题。按照复数的运算法则，i^2 = -1，最后输出字符串结果即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+)
+
+func complexNumberMultiply(a string, b string) string {
+	realA, imagA := parse(a)
+	realB, imagB := parse(b)
+	real := realA*realB - imagA*imagB
+	imag := realA*imagB + realB*imagA
+	return strconv.Itoa(real) + "+" + strconv.Itoa(imag) + "i"
+}
+
+func parse(s string) (int, int) {
+	ss := strings.Split(s, "+")
+	r, _ := strconv.Atoi(ss[0])
+	i, _ := strconv.Atoi(ss[1][:len(ss[1])-1])
+	return r, i
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0535.Encode-and-Decode-TinyURL/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0538.Convert-BST-to-Greater-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0538.Convert-BST-to-Greater-Tree.md b/website/content/ChapterFour/0500~0599/0538.Convert-BST-to-Greater-Tree.md
new file mode 100644
index 000000000..c69654019
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0538.Convert-BST-to-Greater-Tree.md
@@ -0,0 +1,113 @@
+# [538. Convert BST to Greater Tree](https://leetcode.com/problems/convert-bst-to-greater-tree/)
+
+## 题目
+
+Given the `root` of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.
+
+As a reminder, a *binary search tree* is a tree that satisfies these constraints:
+
+- The left subtree of a node contains only nodes with keys **less than** the node's key.
+- The right subtree of a node contains only nodes with keys **greater than** the node's key.
+- Both the left and right subtrees must also be binary search trees.
+
+**Note:** This question is the same as 1038: [https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/](https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/)
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/05/02/tree.png](https://assets.leetcode.com/uploads/2019/05/02/tree.png)
+
+```
+Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
+Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]
+```
+
+**Example 2:**
+
+```
+Input: root = [0,null,1]
+Output: [1,null,1]
+```
+
+**Example 3:**
+
+```
+Input: root = [1,0,2]
+Output: [3,3,2]
+```
+
+**Example 4:**
+
+```
+Input: root = [3,2,4,1]
+Output: [7,9,4,10]
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[0, 104]`.
+- `104 <= Node.val <= 104`
+- All the values in the tree are **unique**.
+- `root` is guaranteed to be a valid binary search tree.
+
+## 题目大意
+
+给出二叉 搜索 树的根节点，该树的节点值各不相同，请你将其转换为累加树（Greater Sum Tree），使每个节点 node 的新值等于原树中大于或等于 node.val 的值之和。
+
+提醒一下，二叉搜索树满足下列约束条件：
+
+- 节点的左子树仅包含键 小于 节点键的节点。
+- 节点的右子树仅包含键 大于 节点键的节点。
+- 左右子树也必须是二叉搜索树。
+
+## 解题思路
+
+- 根据二叉搜索树的有序性，想要将其转换为累加树，只需按照 右节点 - 根节点 - 左节点的顺序遍历，并累加和即可。
+- 此题同第 1038 题。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func convertBST(root *TreeNode) *TreeNode {
+	if root == nil {
+		return root
+	}
+	sum := 0
+	dfs538(root, &sum)
+	return root
+}
+
+func dfs538(root *TreeNode, sum *int) {
+	if root == nil {
+		return
+	}
+	dfs538(root.Right, sum)
+	root.Val += *sum
+	*sum = root.Val
+	dfs538(root.Left, sum)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0537.Complex-Number-Multiplication/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0540.Single-Element-in-a-Sorted-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0540.Single-Element-in-a-Sorted-Array.md b/website/content/ChapterFour/0500~0599/0540.Single-Element-in-a-Sorted-Array.md
new file mode 100644
index 000000000..b092e6466
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0540.Single-Element-in-a-Sorted-Array.md
@@ -0,0 +1,71 @@
+# [540. Single Element in a Sorted Array](https://leetcode.com/problems/single-element-in-a-sorted-array/)
+
+## 题目
+
+You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once.
+
+Return the single element that appears only once.
+
+Your solution must run in O(log n) time and O(1) space.
+
+**Example 1:**
+
+    Input: nums = [1,1,2,3,3,4,4,8,8]
+    Output: 2
+
+**Example 2:**
+
+    Input: nums = [3,3,7,7,10,11,11]
+    Output: 10
+
+**Constraints:**
+
+- 1 <= nums.length <= 100000
+- 0 <= nums[i] <= 100000
+
+## 题目大意
+
+给你一个仅由整数组成的有序数组，其中每个元素都会出现两次，唯有一个数只会出现一次。
+
+请你找出并返回只出现一次的那个数。
+
+你设计的解决方案必须满足 O(log n) 时间复杂度和 O(1) 空间复杂度。
+
+## 解题思路
+
+    假设下标idx是单独的数字,idx左边的偶数下标x有nums[x] == nums[x + 1],
+    idx右边的奇数下标y有nums[y] == nums[y + 1],可以根据此特性用二分查找idx对应的值 
+
+## 代码
+
+```go
+package leetcode
+
+func singleNonDuplicate(nums []int) int {
+	left, right := 0, len(nums)-1
+	for left < right {
+		mid := (left + right) / 2
+		if mid%2 == 0 {
+			if nums[mid] == nums[mid+1] {
+				left = mid + 1
+			} else {
+				right = mid
+			}
+		} else {
+			if nums[mid] == nums[mid-1] {
+				left = mid + 1
+			} else {
+				right = mid
+			}
+		}
+	}
+	return nums[left]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0538.Convert-BST-to-Greater-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0541.Reverse-String-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0541.Reverse-String-II.md b/website/content/ChapterFour/0500~0599/0541.Reverse-String-II.md
new file mode 100755
index 000000000..ad726de6d
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0541.Reverse-String-II.md
@@ -0,0 +1,75 @@
+# [541. Reverse String II](https://leetcode.com/problems/reverse-string-ii/)
+
+
+## 题目
+
+Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.
+
+**Example**:
+
+    Input: s = "abcdefg", k = 2
+    Output: "bacdfeg"
+
+**Restrictions**:
+
+1. The string consists of lower English letters only.
+2. Length of the given string and k will in the range [1, 10000]
+
+## 题目大意
+
+给定一个字符串和一个整数 k，你需要对从字符串开头算起的每个 2k 个字符的前k个字符进行反转。如果剩余少于 k 个字符，则将剩余的所有全部反转。如果有小于 2k 但大于或等于 k 个字符，则反转前 k 个字符，并将剩余的字符保持原样。
+
+要求:
+
+- 该字符串只包含小写的英文字母。
+- 给定字符串的长度和 k 在[1, 10000]范围内。
+
+
+## 解题思路
+
+- 要求按照一定规则反转字符串：每 `2 * K` 长度的字符串，反转前 `K` 个字符，后 `K` 个字符串保持不变；对于末尾不够 `2 * K` 的字符串，如果长度大于 `K`，那么反转前 `K` 个字符串，剩下的保持不变。如果长度小于 `K`，则把小于 `K` 的这部分字符串全部反转。
+- 这一题是简单题，按照题意反转字符串即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func reverseStr(s string, k int) string {
+	if k > len(s) {
+		k = len(s)
+	}
+	for i := 0; i < len(s); i = i + 2*k {
+		if len(s)-i >= k {
+			ss := revers(s[i : i+k])
+			s = s[:i] + ss + s[i+k:]
+		} else {
+			ss := revers(s[i:])
+			s = s[:i] + ss
+		}
+	}
+	return s
+}
+
+func revers(s string) string {
+	bytes := []byte(s)
+	i, j := 0, len(bytes)-1
+	for i < j {
+		bytes[i], bytes[j] = bytes[j], bytes[i]
+		i++
+		j--
+	}
+	return string(bytes)
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0540.Single-Element-in-a-Sorted-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0542.01-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0542.01-Matrix.md b/website/content/ChapterFour/0500~0599/0542.01-Matrix.md
new file mode 100755
index 000000000..09f41a810
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0542.01-Matrix.md
@@ -0,0 +1,209 @@
+# [542. 01 Matrix](https://leetcode.com/problems/01-matrix/)
+
+
+## 题目
+
+Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.
+
+The distance between two adjacent cells is 1.
+
+**Example 1**:
+
+    Input:
+    [[0,0,0],
+     [0,1,0],
+     [0,0,0]]
+    
+    Output:
+    [[0,0,0],
+     [0,1,0],
+     [0,0,0]]
+
+**Example 2**:
+
+    Input:
+    [[0,0,0],
+     [0,1,0],
+     [1,1,1]]
+    
+    Output:
+    [[0,0,0],
+     [0,1,0],
+     [1,2,1]]
+
+**Note**:
+
+1. The number of elements of the given matrix will not exceed 10,000.
+2. There are at least one 0 in the given matrix.
+3. The cells are adjacent in only four directions: up, down, left and right.
+
+
+## 题目大意
+
+给定一个由 0 和 1 组成的矩阵，找出每个元素到最近的 0 的距离。两个相邻元素间的距离为 1 。
+
+
+## 解题思路
+
+
+- 给出一个二维数组，数组里面只有 0 和 1 。要求计算每个 1 距离最近的 0 的距离。
+- 这一题有 3 种解法，第一种解法最容易想到，BFS。先预处理一下棋盘，将每个 0 都处理为 -1 。将 1 都处理为 0 。将每个 -1 (即原棋盘的 0)都入队，每次出队都将四周的 4 个位置都入队。这就想一颗石头扔进了湖里，一圈一圈的波纹荡开，每一圈都是一层。由于棋盘被我们初始化了，所有为 -1 的都是原来为 0 的，所以波纹扫过来不需要处理这些 -1 的点。棋盘上为  0 的点都是原来为 1 的点，这些点在波纹扫过来的时候就需要赋值更新 level。当下次波纹再次扫到原来为 1 的点的时候，由于它已经被第一次到的波纹更新了值，所以这次不用再更新了。(第一次波纹到的时候一定是最短的)
+- 第二种解法是 DFS。先预处理，把周围没有 0 的 1 都重置为最大值。当周围有 0 的 1，距离 0 的位置都是 1，这些点是不需要动的，需要更新的点恰恰应该是那些周围没有 0 的点。当递归的步数 val 比点的值小(这也就是为什么会先把 1 更新成最大值的原因)的时候，不断更新它。
+- 第三种解法是 DP。由于有 4 个方向，每次处理 2 个方向，可以降低时间复杂度。第一次循环从上到下，从左到右遍历，先处理上边和左边，第二次循环从下到上，从右到左遍历，再处理右边和下边。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+)
+
+// 解法一 BFS
+func updateMatrixBFS(matrix [][]int) [][]int {
+	res := make([][]int, len(matrix))
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return res
+	}
+	queue := make([][]int, 0)
+	for i := range matrix {
+		res[i] = make([]int, len(matrix[0]))
+		for j := range res[i] {
+			if matrix[i][j] == 0 {
+				res[i][j] = -1
+				queue = append(queue, []int{i, j})
+			}
+		}
+	}
+	level := 1
+	for len(queue) > 0 {
+		size := len(queue)
+		for size > 0 {
+			size--
+			node := queue[0]
+			queue = queue[1:]
+			i, j := node[0], node[1]
+			for _, direction := range [][]int{{-1, 0}, {1, 0}, {0, 1}, {0, -1}} {
+				x := i + direction[0]
+				y := j + direction[1]
+				if x < 0 || x >= len(matrix) || y < 0 || y >= len(matrix[0]) || res[x][y] < 0 || res[x][y] > 0 {
+					continue
+				}
+				res[x][y] = level
+				queue = append(queue, []int{x, y})
+			}
+		}
+		level++
+	}
+	for i, row := range res {
+		for j, cell := range row {
+			if cell == -1 {
+				res[i][j] = 0
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 DFS
+func updateMatrixDFS(matrix [][]int) [][]int {
+	result := [][]int{}
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return result
+	}
+	maxRow, maxCol := len(matrix), len(matrix[0])
+	for r := 0; r < maxRow; r++ {
+		for c := 0; c < maxCol; c++ {
+			if matrix[r][c] == 1 && hasZero(matrix, r, c) == false {
+				// 将四周没有 0 的 1 特殊处理为最大值
+				matrix[r][c] = math.MaxInt64
+			}
+		}
+	}
+	for r := 0; r < maxRow; r++ {
+		for c := 0; c < maxCol; c++ {
+			if matrix[r][c] == 1 {
+				dfsMatrix(matrix, r, c, -1)
+			}
+		}
+	}
+	return (matrix)
+}
+
+// 判断四周是否有 0
+func hasZero(matrix [][]int, row, col int) bool {
+	if row > 0 && matrix[row-1][col] == 0 {
+		return true
+	}
+	if col > 0 && matrix[row][col-1] == 0 {
+		return true
+	}
+	if row < len(matrix)-1 && matrix[row+1][col] == 0 {
+		return true
+	}
+	if col < len(matrix[0])-1 && matrix[row][col+1] == 0 {
+		return true
+	}
+	return false
+}
+
+func dfsMatrix(matrix [][]int, row, col, val int) {
+	// 不超过棋盘氛围，且 val 要比 matrix[row][col] 小
+	if row < 0 || row >= len(matrix) || col < 0 || col >= len(matrix[0]) || (matrix[row][col] <= val) {
+		return
+	}
+	if val > 0 {
+		matrix[row][col] = val
+	}
+	dfsMatrix(matrix, row-1, col, matrix[row][col]+1)
+	dfsMatrix(matrix, row, col-1, matrix[row][col]+1)
+	dfsMatrix(matrix, row+1, col, matrix[row][col]+1)
+	dfsMatrix(matrix, row, col+1, matrix[row][col]+1)
+}
+
+// 解法三 DP
+func updateMatrixDP(matrix [][]int) [][]int {
+	for i, row := range matrix {
+		for j, val := range row {
+			if val == 0 {
+				continue
+			}
+			left, top := math.MaxInt16, math.MaxInt16
+			if i > 0 {
+				top = matrix[i-1][j] + 1
+			}
+			if j > 0 {
+				left = matrix[i][j-1] + 1
+			}
+			matrix[i][j] = min(top, left)
+		}
+	}
+	for i := len(matrix) - 1; i >= 0; i-- {
+		for j := len(matrix[0]) - 1; j >= 0; j-- {
+			if matrix[i][j] == 0 {
+				continue
+			}
+			right, bottom := math.MaxInt16, math.MaxInt16
+			if i < len(matrix)-1 {
+				bottom = matrix[i+1][j] + 1
+			}
+			if j < len(matrix[0])-1 {
+				right = matrix[i][j+1] + 1
+			}
+			matrix[i][j] = min(matrix[i][j], min(bottom, right))
+		}
+	}
+	return matrix
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0541.Reverse-String-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0543.Diameter-of-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0543.Diameter-of-Binary-Tree.md b/website/content/ChapterFour/0500~0599/0543.Diameter-of-Binary-Tree.md
new file mode 100644
index 000000000..ced821989
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0543.Diameter-of-Binary-Tree.md
@@ -0,0 +1,94 @@
+# [543. Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/)
+
+
+## 题目
+
+Given the `root` of a binary tree, return *the length of the **diameter** of the tree*.
+
+The **diameter** of a binary tree is the **length** of the longest path between any two nodes in a tree. This path may or may not pass through the `root`.
+
+The **length** of a path between two nodes is represented by the number of edges between them.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/03/06/diamtree.jpg](https://assets.leetcode.com/uploads/2021/03/06/diamtree.jpg)
+
+```
+Input: root = [1,2,3,4,5]
+Output: 3
+Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].
+
+```
+
+**Example 2:**
+
+```
+Input: root = [1,2]
+Output: 1
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 104]`.
+- `100 <= Node.val <= 100`
+
+## 题目大意
+
+给定一棵二叉树，你需要计算它的直径长度。一棵二叉树的直径长度是任意两个结点路径长度中的最大值。这条路径可能穿过也可能不穿过根结点。
+
+## 解题思路
+
+- 简单题。遍历每个节点的左子树和右子树，累加从左子树到右子树的最大长度。遍历每个节点时，动态更新这个最大长度即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func diameterOfBinaryTree(root *TreeNode) int {
+	result := 0
+	checkDiameter(root, &result)
+	return result
+}
+
+func checkDiameter(root *TreeNode, result *int) int {
+	if root == nil {
+		return 0
+	}
+	left := checkDiameter(root.Left, result)
+	right := checkDiameter(root.Right, result)
+	*result = max(*result, left+right)
+	return max(left, right) + 1
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0542.01-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0547.Number-of-Provinces/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0547.Number-of-Provinces.md b/website/content/ChapterFour/0500~0599/0547.Number-of-Provinces.md
new file mode 100755
index 000000000..60011312f
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0547.Number-of-Provinces.md
@@ -0,0 +1,118 @@
+# [547. Number of Provinces](https://leetcode.com/problems/number-of-provinces/)
+
+## 题目
+
+There are **N** students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a **direct** friend of B, and B is a **direct**friend of C, then A is an **indirect** friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.
+
+Given a **N*N** matrix **M** representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are **direct** friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.
+
+**Example 1**:
+
+    Input: 
+    [[1,1,0],
+     [1,1,0],
+     [0,0,1]]
+    Output: 2
+    Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. 
+    The 2nd student himself is in a friend circle. So return 2.
+
+**Example 2**:
+
+    Input: 
+    [[1,1,0],
+     [1,1,1],
+     [0,1,1]]
+    Output: 1
+    Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, 
+    so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.
+
+**Note**:
+
+1. N is in range [1,200].
+2. M[i][i] = 1 for all students.
+3. If M[i][j] = 1, then M[j][i] = 1.
+
+
+## 题目大意
+
+班上有 N 名学生。其中有些人是朋友，有些则不是。他们的友谊具有是传递性。如果已知 A 是 B 的朋友，B 是 C 的朋友，那么我们可以认为 A 也是 C 的朋友。所谓的朋友圈，是指所有朋友的集合。
+
+给定一个 N * N 的矩阵 M，表示班级中学生之间的朋友关系。如果 M[i][j] = 1，表示已知第 i 个和 j 个学生互为朋友关系，否则为不知道。你必须输出所有学生中的已知的朋友圈总数。
+
+
+注意：
+
+- N 在[1,200]的范围内。
+- 对于所有学生，有M[i][i] = 1。
+- 如果有 M[i][j] = 1，则有 M[j][i] = 1。
+
+
+## 解题思路
+
+
+- 给出一个二维矩阵，矩阵中的行列表示的是两个人之间是否是朋友关系，如果是 1，代表两个人是朋友关系。由于自己和自肯定朋友关系，所以对角线上都是 1，并且矩阵也是关于从左往右下的这条对角线对称。
+- 这题有 2 种解法，第一种解法是并查集，依次扫描矩阵，如果两个人认识，并且 root 并不相等就执行 union 操作。扫完所有矩阵，最后数一下还有几个不同的 root 就是最终答案。第二种解法是 DFS 或者 BFS。利用 FloodFill 的想法去染色，每次染色一次，计数器加一。最终扫完整个矩阵，计数器的结果就是最终结果。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+// 解法一 并查集
+
+func findCircleNum(M [][]int) int {
+	n := len(M)
+	if n == 0 {
+		return 0
+	}
+	uf := template.UnionFind{}
+	uf.Init(n)
+	for i := 0; i < n; i++ {
+		for j := 0; j <= i; j++ {
+			if M[i][j] == 1 {
+				uf.Union(i, j)
+			}
+		}
+	}
+	return uf.TotalCount()
+}
+
+// 解法二 FloodFill DFS 暴力解法
+func findCircleNum1(M [][]int) int {
+	if len(M) == 0 {
+		return 0
+	}
+	visited := make([]bool, len(M))
+	res := 0
+	for i := range M {
+		if !visited[i] {
+			dfs547(M, i, visited)
+			res++
+		}
+	}
+	return res
+}
+
+func dfs547(M [][]int, cur int, visited []bool) {
+	visited[cur] = true
+	for j := 0; j < len(M[cur]); j++ {
+		if !visited[j] && M[cur][j] == 1 {
+			dfs547(M, j, visited)
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0543.Diameter-of-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0551.Student-Attendance-Record-I/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0551.Student-Attendance-Record-I.md b/website/content/ChapterFour/0500~0599/0551.Student-Attendance-Record-I.md
new file mode 100644
index 000000000..6d5a09651
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0551.Student-Attendance-Record-I.md
@@ -0,0 +1,93 @@
+# [551. Student Attendance Record I](https://leetcode.com/problems/student-attendance-record-i/)
+
+## 题目
+
+You are given a string `s` representing an attendance record for a student where each character signifies whether the student was absent, late, or present on that day. The record only contains the following three characters:
+
+- `'A'`: Absent.
+- `'L'`: Late.
+- `'P'`: Present.
+
+The student is eligible for an attendance award if they meet **both** of the following criteria:
+
+- The student was absent (`'A'`) for **strictly** fewer than 2 days **total**.
+- The student was **never** late (`'L'`) for 3 or more **consecutive** days.
+
+Return `true` *if the student is eligible for an attendance award, or* `false` *otherwise*.
+
+**Example 1:**
+
+```
+Input: s = "PPALLP"
+Output: true
+Explanation: The student has fewer than 2 absences and was never late 3 or more consecutive days.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "PPALLL"
+Output: false
+Explanation: The student was late 3 consecutive days in the last 3 days, so is not eligible for the award.
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 1000`
+- `s[i]` is either `'A'`, `'L'`, or `'P'`.
+
+## 题目大意
+
+给你一个字符串 s 表示一个学生的出勤记录，其中的每个字符用来标记当天的出勤情况（缺勤、迟到、到场）。记录中只含下面三种字符：
+
+- 'A'：Absent，缺勤
+- 'L'：Late，迟到
+- 'P'：Present，到场
+
+如果学生能够 同时 满足下面两个条件，则可以获得出勤奖励：
+
+- 按 总出勤 计，学生缺勤（'A'）严格 少于两天。
+- 学生 不会 存在 连续 3 天或 连续 3 天以上的迟到（'L'）记录。
+
+如果学生可以获得出勤奖励，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 遍历字符串 s 求出 'A' 的总数量和连续 'L' 的最大数量。
+- 比较 'A' 的数量是否小于 2 并且 'L' 的连续最大数量是否小于 3。
+
+## 代码
+
+```go
+package leetcode
+
+func checkRecord(s string) bool {
+	numsA, maxL, numsL := 0, 0, 0
+	for _, v := range s {
+		if v == 'L' {
+			numsL++
+		} else {
+			if numsL > maxL {
+				maxL = numsL
+			}
+			numsL = 0
+			if v == 'A' {
+				numsA++
+			}
+		}
+	}
+	if numsL > maxL {
+		maxL = numsL
+	}
+	return numsA < 2 && maxL < 3
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0547.Number-of-Provinces/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0554.Brick-Wall/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0554.Brick-Wall.md b/website/content/ChapterFour/0500~0599/0554.Brick-Wall.md
new file mode 100644
index 000000000..88f47f615
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0554.Brick-Wall.md
@@ -0,0 +1,75 @@
+# [554. Brick Wall](https://leetcode.com/problems/brick-wall/)
+
+## 题目
+
+There is a rectangular brick wall in front of you with `n` rows of bricks. The `ith` row has some number of bricks each of the same height (i.e., one unit) but they can be of different widths. The total width of each row is the same.
+
+Draw a vertical line from the top to the bottom and cross the least bricks. If your line goes through the edge of a brick, then the brick is not considered as crossed. You cannot draw a line just along one of the two vertical edges of the wall, in which case the line will obviously cross no bricks.
+
+Given the 2D array `wall` that contains the information about the wall, return *the minimum number of crossed bricks after drawing such a vertical line*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/04/24/cutwall-grid.jpg](https://assets.leetcode.com/uploads/2021/04/24/cutwall-grid.jpg)
+
+```
+Input: wall = [[1,2,2,1],[3,1,2],[1,3,2],[2,4],[3,1,2],[1,3,1,1]]
+Output: 2
+
+```
+
+**Example 2:**
+
+```
+Input: wall = [[1],[1],[1]]
+Output: 3
+
+```
+
+**Constraints:**
+
+- `n == wall.length`
+- `1 <= n <= 10^4`
+- `1 <= wall[i].length <= 10^4`
+- `1 <= sum(wall[i].length) <= 2 * 10^4`
+- `sum(wall[i])` is the same for each row `i`.
+- `1 <= wall[i][j] <= 2^31 - 1`
+
+## 题目大意
+
+你的面前有一堵矩形的、由 n 行砖块组成的砖墙。这些砖块高度相同（也就是一个单位高）但是宽度不同。每一行砖块的宽度之和应该相等。你现在要画一条 自顶向下 的、穿过 最少 砖块的垂线。如果你画的线只是从砖块的边缘经过，就不算穿过这块砖。你不能沿着墙的两个垂直边缘之一画线，这样显然是没有穿过一块砖的。给你一个二维数组 wall ，该数组包含这堵墙的相关信息。其中，wall[i] 是一个代表从左至右每块砖的宽度的数组。你需要找出怎样画才能使这条线 穿过的砖块数量最少 ，并且返回 穿过的砖块数量 。
+
+## 解题思路
+
+- 既然穿过砖块中缝不算穿过砖块，那么穿过最少砖块数量一定是穿过很多中缝。按行遍历每一行的砖块，累加每行砖块宽度，将每行砖块“缝”的坐标存在 map 中。最后取出 map 中出现频次最高的缝，即为铅垂线要穿过的地方。墙高减去缝出现的频次，剩下的即为穿过砖块的数量。
+
+## 代码
+
+```go
+package leetcode
+
+func leastBricks(wall [][]int) int {
+	m := make(map[int]int)
+	for _, row := range wall {
+		sum := 0
+		for i := 0; i < len(row)-1; i++ {
+			sum += row[i]
+			m[sum]++
+		}
+	}
+	max := 0
+	for _, v := range m {
+		if v > max {
+			max = v
+		}
+	}
+	return len(wall) - max
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0551.Student-Attendance-Record-I/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0557.Reverse-Words-in-a-String-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0557.Reverse-Words-in-a-String-III.md b/website/content/ChapterFour/0500~0599/0557.Reverse-Words-in-a-String-III.md
new file mode 100755
index 000000000..0f277d761
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0557.Reverse-Words-in-a-String-III.md
@@ -0,0 +1,64 @@
+# [557. Reverse Words in a String III](https://leetcode.com/problems/reverse-words-in-a-string-iii/)
+
+
+## 题目
+
+Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.
+
+**Example 1**:
+
+    Input: "Let's take LeetCode contest"
+    Output: "s'teL ekat edoCteeL tsetnoc"
+
+**Note**: In the string, each word is separated by single space and there will not be any extra space in the string.
+
+
+## 题目大意
+
+给定一个字符串，你需要反转字符串中每个单词的字符顺序，同时仍保留空格和单词的初始顺序。注意：在字符串中，每个单词由单个空格分隔，并且字符串中不会有任何额外的空格。
+
+
+## 解题思路
+
+
+- 反转字符串，要求按照空格隔开的小字符串为单位反转。
+- 这是一道简单题。按照题意反转每个空格隔开的单词即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strings"
+)
+
+func reverseWords(s string) string {
+	ss := strings.Split(s, " ")
+	for i, s := range ss {
+		ss[i] = revers(s)
+	}
+	return strings.Join(ss, " ")
+}
+
+func revers(s string) string {
+	bytes := []byte(s)
+	i, j := 0, len(bytes)-1
+	for i < j {
+		bytes[i], bytes[j] = bytes[j], bytes[i]
+		i++
+		j--
+	}
+	return string(bytes)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0554.Brick-Wall/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0559.Maximum-Depth-of-N-ary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0559.Maximum-Depth-of-N-ary-Tree.md b/website/content/ChapterFour/0500~0599/0559.Maximum-Depth-of-N-ary-Tree.md
new file mode 100644
index 000000000..fec8b6e83
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0559.Maximum-Depth-of-N-ary-Tree.md
@@ -0,0 +1,85 @@
+# [559. Maximum Depth of N-ary Tree](https://leetcode.com/problems/maximum-depth-of-n-ary-tree/)
+
+## 题目
+
+Given a n-ary tree, find its maximum depth.
+
+The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
+
+Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png](https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png)
+
+    Input: root = [1,null,3,2,4,null,5,6]
+    Output: 3
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png](https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png)
+
+    Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
+    Output: 5
+
+**Constraints:**
+
+- The total number of nodes is in the range [0, 10000].
+- The depth of the n-ary tree is less than or equal to 1000.
+
+## 题目大意
+
+给定一个 N 叉树，找到其最大深度。
+
+最大深度是指从根节点到最远叶子节点的最长路径上的节点总数。
+
+N 叉树输入按层序遍历序列化表示，每组子节点由空值分隔（请参见示例）。
+
+## 解题思路
+
+- 使用广度优先遍历
+
+## 代码
+
+```go
+
+package leetcode
+
+type Node struct {
+	Val int
+	Children []*Node
+}
+
+func maxDepth(root *Node) int {
+	if root == nil {
+		return 0
+	}
+	return 1 + bfs(root)
+}
+
+func bfs(root *Node) int {
+	var q []*Node
+	var depth int
+	q = append(q, root.Children...)
+	for len(q) != 0 {
+		depth++
+		length := len(q)
+		for length != 0 {
+			ele := q[0]
+			q = q[1:]
+			length--
+			if ele != nil && len(ele.Children) != 0 {
+				q = append(q, ele.Children...)
+			}
+		}
+	}
+	return depth
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0557.Reverse-Words-in-a-String-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0560.Subarray-Sum-Equals-K/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0560.Subarray-Sum-Equals-K.md b/website/content/ChapterFour/0500~0599/0560.Subarray-Sum-Equals-K.md
new file mode 100644
index 000000000..024667f09
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0560.Subarray-Sum-Equals-K.md
@@ -0,0 +1,65 @@
+# [560. Subarray Sum Equals K](https://leetcode.com/problems/subarray-sum-equals-k/)
+
+
+## 题目
+
+Given an array of integers `nums` and an integer `k`, return *the total number of continuous subarrays whose sum equals to `k`*.
+
+**Example 1:**
+
+```
+Input: nums = [1,1,1], k = 2
+Output: 2
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3], k = 3
+Output: 2
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 2 * 104`
+- `-1000 <= nums[i] <= 1000`
+- `-10^7 <= k <= 10^7`
+
+## 题目大意
+
+给你一个整数数组 `nums` 和一个整数 `k` ，请你统计并返回该数组中和为 `k` ****的连续子数组的个数。
+
+## 解题思路
+
+- 此题不能使用滑动窗口来解。因为 `nums[i]` 可能为负数。
+- 前缀和的思路可以解答此题，但是时间复杂度有点高了，`O(n^2)`。考虑优化时间复杂度。
+- 题目要求找到连续区间和为 `k` 的子区间总数，即区间 `[i,j]` 内的和为 K ⇒ `prefixSum[j] - prefixSum[i-1] == k`。所以 `prefixSum[j] == k - prefixSum[i-1]` 。这样转换以后，题目就转换成类似 A + B = K 的问题了。LeetCode 第一题的优化思路拿来用。用 map 存储累加过的结果。如此优化以后，时间复杂度 `O(n)`。
+
+## 代码
+
+```go
+package leetcode
+
+func subarraySum(nums []int, k int) int {
+	count, pre := 0, 0
+	m := map[int]int{}
+	m[0] = 1
+	for i := 0; i < len(nums); i++ {
+		pre += nums[i]
+		if _, ok := m[pre-k]; ok {
+			count += m[pre-k]
+		}
+		m[pre] += 1
+	}
+	return count
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0559.Maximum-Depth-of-N-ary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0561.Array-Partition/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0561.Array-Partition.md b/website/content/ChapterFour/0500~0599/0561.Array-Partition.md
new file mode 100644
index 000000000..04b71df09
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0561.Array-Partition.md
@@ -0,0 +1,65 @@
+# [561. Array Partition](https://leetcode.com/problems/array-partition/)
+
+
+## 题目
+
+Given an array of **2n** integers, your task is to group these integers into **n** pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.
+
+**Example 1**:
+
+```
+Input: [1,4,3,2]
+
+Output: 4
+Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).
+```
+
+**Note**:
+
+1. **n** is a positive integer, which is in the range of [1, 10000].
+2. All the integers in the array will be in the range of [-10000, 10000].
+
+## 题目大意
+
+给定长度为 2n 的数组, 你的任务是将这些数分成 n 对, 例如 (a1, b1), (a2, b2), ..., (an, bn) ，使得从1 到 n 的 min(ai, bi) 总和最大。
+
+
+## 解题思路
+
+- 给定一个 2n 个数组，要求把它们分为 n 组一行，求出各组最小值的总和的最大值。
+- 由于题目给的数据范围不大，[-10000, 10000]，所以我们可以考虑用一个哈希表数组，里面存储 i - 10000 元素的频次，偏移量是 10000。这个哈希表能按递增的顺序访问数组，这样可以减少排序的耗时。题目要求求出分组以后求和的最大值，那么所有偏小的元素尽量都安排在一组里面，这样取 min 以后，对最大和影响不大。例如，(1 , 1) 这样安排在一起，min 以后就是 1 。但是如果把相差很大的两个元素安排到一起，那么较大的那个元素就“牺牲”了。例如，(1 , 10000)，取 min 以后就是 1，于是 10000 就“牺牲”了。所以需要优先考虑较小值。
+- 较小值出现的频次可能是奇数也可能是偶数。如果是偶数，那比较简单，把它们俩俩安排在一起就可以了。如果是奇数，那么它会落单一次，落单的那个需要和距离它最近的一个元素进行配对，这样对最终的和影响最小。较小值如果是奇数，那么就会影响后面元素的选择，后面元素如果是偶数，由于需要一个元素和前面的较小值配对，所以它剩下的又是奇数个。这个影响会依次传递到后面。所以用一个 flag 标记，如果当前集合中有剩余元素将被再次考虑，则此标志设置为 1。在从下一组中选择元素时，会考虑已考虑的相同额外元素。
+- 最后扫描过程中动态的维护 sum 值就可以了。
+
+## 代码
+
+```go
+
+package leetcode
+
+func arrayPairSum(nums []int) int {
+	array := [20001]int{}
+	for i := 0; i < len(nums); i++ {
+		array[nums[i]+10000]++
+	}
+	flag, sum := true, 0
+	for i := 0; i < len(array); i++ {
+		for array[i] > 0 {
+			if flag {
+				sum = sum + i - 10000
+			}
+			flag = !flag
+			array[i]--
+		}
+	}
+	return sum
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0560.Subarray-Sum-Equals-K/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0563.Binary-Tree-Tilt/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0563.Binary-Tree-Tilt.md b/website/content/ChapterFour/0500~0599/0563.Binary-Tree-Tilt.md
new file mode 100755
index 000000000..dfe47b0de
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0563.Binary-Tree-Tilt.md
@@ -0,0 +1,90 @@
+# [563. Binary Tree Tilt](https://leetcode.com/problems/binary-tree-tilt/)
+
+
+## 题目
+
+Given a binary tree, return the tilt of the **whole tree**.
+
+The tilt of a **tree node** is defined as the **absolute difference** between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.
+
+The tilt of the **whole tree** is defined as the sum of all nodes' tilt.
+
+**Example**:
+
+    Input: 
+             1
+           /   \
+          2     3
+    Output: 1
+    Explanation: 
+    Tilt of node 2 : 0
+    Tilt of node 3 : 0
+    Tilt of node 1 : |2-3| = 1
+    Tilt of binary tree : 0 + 0 + 1 = 1
+
+**Note**:
+
+1. The sum of node values in any subtree won't exceed the range of 32-bit integer.
+2. All the tilt values won't exceed the range of 32-bit integer.
+
+
+## 题目大意
+
+
+给定一个二叉树，计算整个树的坡度。一个树的节点的坡度定义即为，该节点左子树的结点之和和右子树结点之和的差的绝对值。空结点的的坡度是0。整个树的坡度就是其所有节点的坡度之和。
+
+注意:
+
+1. 任何子树的结点的和不会超过32位整数的范围。
+2. 坡度的值不会超过32位整数的范围。
+
+## 解题思路
+
+
+- 给出一棵树，计算每个节点的“倾斜度”累加和。“倾斜度”的定义是：左子树和右子树的节点值差值的绝对值。
+- 这一题虽然是简单题，但是如果对题目中的“倾斜度”理解的不对，这一题就会出错。“倾斜度”计算的是左子树所有节点的值总和，和，右子树所有节点的值总和的差值。并不是只针对一个节点的左节点值和右节点值的差值。这一点明白以后，这一题就是简单题了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func findTilt(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	sum := 0
+	findTiltDFS(root, &sum)
+	return sum
+}
+
+func findTiltDFS(root *TreeNode, sum *int) int {
+	if root == nil {
+		return 0
+	}
+	left := findTiltDFS(root.Left, sum)
+	right := findTiltDFS(root.Right, sum)
+	*sum += int(math.Abs(float64(left) - float64(right)))
+	return root.Val + left + right
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0561.Array-Partition/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0566.Reshape-the-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0566.Reshape-the-Matrix.md b/website/content/ChapterFour/0500~0599/0566.Reshape-the-Matrix.md
new file mode 100755
index 000000000..7097fecec
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0566.Reshape-the-Matrix.md
@@ -0,0 +1,109 @@
+# [566. Reshape the Matrix](https://leetcode.com/problems/reshape-the-matrix/)
+
+
+## 题目
+
+In MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new one with different size but keep its original data.
+
+You're given a matrix represented by a two-dimensional array, and two **positive** integers **r** and **c**representing the **row** number and **column** number of the wanted reshaped matrix, respectively.
+
+The reshaped matrix need to be filled with all the elements of the original matrix in the same **row-traversing** order as they were.
+
+If the 'reshape' operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.
+
+**Example 1**:
+
+    Input: 
+    nums = 
+    [[1,2],
+     [3,4]]
+    r = 1, c = 4
+    Output: 
+    [[1,2,3,4]]
+    Explanation:
+    The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
+
+**Example 2**:
+
+    Input: 
+    nums = 
+    [[1,2],
+     [3,4]]
+    r = 2, c = 4
+    Output: 
+    [[1,2],
+     [3,4]]
+    Explanation:
+    There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.
+
+**Note**:
+
+1. The height and width of the given matrix is in range [1, 100].
+2. The given r and c are all positive.
+
+
+## 题目大意
+
+在 MATLAB 中，有一个非常有用的函数 reshape，它可以将一个矩阵重塑为另一个大小不同的新矩阵，但保留其原始数据。
+
+给出一个由二维数组表示的矩阵，以及两个正整数r和c，分别表示想要的重构的矩阵的行数和列数。重构后的矩阵需要将原始矩阵的所有元素以相同的行遍历顺序填充。如果具有给定参数的reshape操作是可行且合理的，则输出新的重塑矩阵；否则，输出原始矩阵。
+
+
+
+## 解题思路
+
+
+- 给一个二维数组和 r，c，将这个二维数组“重塑”成行为 r，列为 c。如果可以“重塑”，输出“重塑”以后的数组，如果不能“重塑”，输出原有数组。
+- 这题也是水题，按照题意模拟即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func matrixReshape(nums [][]int, r int, c int) [][]int {
+	if canReshape(nums, r, c) {
+		return reshape(nums, r, c)
+	}
+	return nums
+}
+
+func canReshape(nums [][]int, r, c int) bool {
+	row := len(nums)
+	colume := len(nums[0])
+	if row*colume == r*c {
+		return true
+	}
+	return false
+}
+
+func reshape(nums [][]int, r, c int) [][]int {
+	newShape := make([][]int, r)
+	for index := range newShape {
+		newShape[index] = make([]int, c)
+	}
+	rowIndex, colIndex := 0, 0
+	for _, row := range nums {
+		for _, col := range row {
+			if colIndex == c {
+				colIndex = 0
+				rowIndex++
+			}
+			newShape[rowIndex][colIndex] = col
+			colIndex++
+		}
+	}
+	return newShape
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0563.Binary-Tree-Tilt/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0567.Permutation-in-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0567.Permutation-in-String.md b/website/content/ChapterFour/0500~0599/0567.Permutation-in-String.md
new file mode 100644
index 000000000..35262e5db
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0567.Permutation-in-String.md
@@ -0,0 +1,94 @@
+# [567. Permutation in String](https://leetcode.com/problems/permutation-in-string/)
+
+## 题目
+
+Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.
+
+
+**Example 1**:
+
+```
+
+Input:s1 = "ab" s2 = "eidbaooo"
+Output:True
+Explanation: s2 contains one permutation of s1 ("ba").
+
+```
+
+**Example 2**:
+
+```
+
+Input:s1= "ab" s2 = "eidboaoo"
+Output: False
+
+```
+
+**Note**:
+
+1. The input strings only contain lower case letters.
+2. The length of both given strings is in range [1, 10,000].
+
+## 题目大意
+
+
+在一个字符串中寻找子串出现的位置。子串可以是 Anagrams 形式存在的。Anagrams 是一个字符串任意字符的全排列组合。
+
+## 解题思路
+
+这一题和第 438 题，第 3 题，第 76 题，第 567 题类似，用的思想都是"滑动窗口"。
+
+
+这道题只需要判断是否存在，而不需要输出子串所在的下标起始位置。所以这道题是第 438 题的缩水版。具体解题思路见第 438 题。
+
+
+
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func checkInclusion(s1 string, s2 string) bool {
+	var freq [256]int
+	if len(s2) == 0 || len(s2) < len(s1) {
+		return false
+	}
+	for i := 0; i < len(s1); i++ {
+		freq[s1[i]-'a']++
+	}
+	left, right, count := 0, 0, len(s1)
+
+	for right < len(s2) {
+		if freq[s2[right]-'a'] >= 1 {
+			count--
+		}
+		freq[s2[right]-'a']--
+		right++
+		if count == 0 {
+			return true
+		}
+		if right-left == len(s1) {
+			if freq[s2[left]-'a'] >= 0 {
+				count++
+			}
+			freq[s2[left]-'a']++
+			left++
+		}
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0566.Reshape-the-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0572.Subtree-of-Another-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0572.Subtree-of-Another-Tree.md b/website/content/ChapterFour/0500~0599/0572.Subtree-of-Another-Tree.md
new file mode 100755
index 000000000..9781a1795
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0572.Subtree-of-Another-Tree.md
@@ -0,0 +1,94 @@
+# [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)
+
+
+## 题目
+
+Given two non-empty binary trees **s** and **t**, check whether tree **t** has exactly the same structure and node values with a subtree of **s**. A subtree of **s** is a tree consists of a node in **s** and all of this node's descendants. The tree **s** could also be considered as a subtree of itself.
+
+**Example 1**: 
+
+Given tree s:
+
+		 3
+        / \
+       4   5
+      / \
+     1   2
+
+Given tree t:
+
+       4 
+      / \
+     1   2
+
+Return **true**, because t has the same structure and node values with a subtree of s.
+
+**Example 2**: 
+
+Given tree s:
+
+         3
+        / \
+       4   5
+      / \
+     1   2
+        /
+       0
+
+Given tree t:
+
+       4
+      / \
+     1   2
+
+Return **false**.
+
+
+## 题目大意
+
+给定两个非空二叉树 s 和 t，检验 s 中是否包含和 t 具有相同结构和节点值的子树。s 的一个子树包括 s 的一个节点和这个节点的所有子孙。s 也可以看做它自身的一棵子树。
+
+
+## 解题思路
+
+
+- 给出 2 棵树 `s` 和 `t`，要求判断 `t` 是否是 `s` 的子树🌲。
+- 这一题比较简单，针对 3 种情况依次递归判断，第一种情况 `s` 和 `t` 是完全一样的两棵树，第二种情况 `t` 是 `s` 左子树中的子树，第三种情况 `t` 是 `s` 右子树中的子树。第一种情况判断两棵数是否完全一致是第 100 题的原题。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func isSubtree(s *TreeNode, t *TreeNode) bool {
+	if isSameTree(s, t) {
+		return true
+	}
+	if s == nil {
+		return false
+	}
+	if isSubtree(s.Left, t) || isSubtree(s.Right, t) {
+		return true
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0567.Permutation-in-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0575.Distribute-Candies/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0575.Distribute-Candies.md b/website/content/ChapterFour/0500~0599/0575.Distribute-Candies.md
new file mode 100755
index 000000000..cdbf391ed
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0575.Distribute-Candies.md
@@ -0,0 +1,66 @@
+# [575. Distribute Candies](https://leetcode.com/problems/distribute-candies/)
+
+
+## 题目
+
+Given an integer array with **even** length, where different numbers in this array represent different **kinds** of candies. Each number means one candy of the corresponding kind. You need to distribute these candies **equally** in number to brother and sister. Return the maximum number of **kinds** of candies the sister could gain.
+
+**Example 1**:
+
+    Input: candies = [1,1,2,2,3,3]
+    Output: 3
+    Explanation:
+    There are three different kinds of candies (1, 2 and 3), and two candies for each kind.
+    Optimal distribution: The sister has candies [1,2,3] and the brother has candies [1,2,3], too. 
+    The sister has three different kinds of candies.
+
+**Example 2**:
+
+    Input: candies = [1,1,2,3]
+    Output: 2
+    Explanation: For example, the sister has candies [2,3] and the brother has candies [1,1]. 
+    The sister has two different kinds of candies, the brother has only one kind of candies.
+
+**Note**:
+
+1. The length of the given array is in range [2, 10,000], and will be even.
+2. The number in given array is in range [-100,000, 100,000].
+
+
+## 题目大意
+
+给定一个偶数长度的数组，其中不同的数字代表着不同种类的糖果，每一个数字代表一个糖果。你需要把这些糖果平均分给一个弟弟和一个妹妹。返回妹妹可以获得的最大糖果的种类数。
+
+
+## 解题思路
+
+
+- 给出一个糖果数组，里面每个元素代表糖果的种类，相同数字代表相同种类。把这些糖果分给兄弟姐妹，问姐妹最多可以分到多少种糖果。这一题比较简单，用 map 统计每个糖果的出现频次，如果总数比 `n/2` 小，那么就返回 `len(map)`，否则返回 `n/2` (即一半都分给姐妹)。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func distributeCandies(candies []int) int {
+	n, m := len(candies), make(map[int]struct{}, len(candies))
+	for _, candy := range candies {
+		m[candy] = struct{}{}
+	}
+	res := len(m)
+	if n/2 < res {
+		return n / 2
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0572.Subtree-of-Another-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0576.Out-of-Boundary-Paths/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0576.Out-of-Boundary-Paths.md b/website/content/ChapterFour/0500~0599/0576.Out-of-Boundary-Paths.md
new file mode 100644
index 000000000..c58ba3fd3
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0576.Out-of-Boundary-Paths.md
@@ -0,0 +1,96 @@
+# [576. Out of Boundary Paths](https://leetcode.com/problems/out-of-boundary-paths/)
+
+
+## 题目
+
+There is an `m x n` grid with a ball. The ball is initially at the position `[startRow, startColumn]`. You are allowed to move the ball to one of the four adjacent four cells in the grid (possibly out of the grid crossing the grid boundary). You can apply **at most** `maxMove` moves to the ball.
+
+Given the five integers `m`, `n`, `maxMove`, `startRow`, `startColumn`, return the number of paths to move the ball out of the grid boundary. Since the answer can be very large, return it **modulo** `109 + 7`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_1.png](https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_1.png)
+
+```
+Input: m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0
+Output: 6
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_2.png](https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_2.png)
+
+```
+Input: m = 1, n = 3, maxMove = 3, startRow = 0, startColumn = 1
+Output: 12
+```
+
+**Constraints:**
+
+- `1 <= m, n <= 50`
+- `0 <= maxMove <= 50`
+- `0 <= startRow <= m`
+- `0 <= startColumn <= n`
+
+## 题目大意
+
+给定一个 m × n 的网格和一个球。球的起始坐标为 (i,j) ，你可以将球移到相邻的单元格内，或者往上、下、左、右四个方向上移动使球穿过网格边界。但是，你最多可以移动 N 次。找出可以将球移出边界的路径数量。答案可能非常大，返回 结果 mod 109 + 7 的值。
+
+## 解题思路
+
+- 单纯暴力的思路，在球的每个方向都遍历一步，直到移动步数用完。这样暴力搜索，解空间是 4^n 。优化思路便是增加记忆化。用三维数组记录位置坐标和步数，对应的出边界的路径数量。加上记忆化以后的深搜解法 runtime beats 100% 了。
+
+## 代码
+
+```go
+package leetcode
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func findPaths(m int, n int, maxMove int, startRow int, startColumn int) int {
+	visited := make([][][]int, m)
+	for i := range visited {
+		visited[i] = make([][]int, n)
+		for j := range visited[i] {
+			visited[i][j] = make([]int, maxMove+1)
+			for l := range visited[i][j] {
+				visited[i][j][l] = -1
+			}
+		}
+	}
+	return dfs(startRow, startColumn, maxMove, m, n, visited)
+}
+
+func dfs(x, y, maxMove, m, n int, visited [][][]int) int {
+	if x < 0 || x >= m || y < 0 || y >= n {
+		return 1
+	}
+	if maxMove == 0 {
+		visited[x][y][maxMove] = 0
+		return 0
+	}
+	if visited[x][y][maxMove] >= 0 {
+		return visited[x][y][maxMove]
+	}
+	res := 0
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		res += (dfs(nx, ny, maxMove-1, m, n, visited) % 1000000007)
+	}
+	visited[x][y][maxMove] = res % 1000000007
+	return visited[x][y][maxMove]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0575.Distribute-Candies/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0581.Shortest-Unsorted-Continuous-Subarray/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0581.Shortest-Unsorted-Continuous-Subarray.md b/website/content/ChapterFour/0500~0599/0581.Shortest-Unsorted-Continuous-Subarray.md
new file mode 100644
index 000000000..88bece34b
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0581.Shortest-Unsorted-Continuous-Subarray.md
@@ -0,0 +1,114 @@
+# [581. Shortest Unsorted Continuous Subarray](https://leetcode.com/problems/shortest-unsorted-continuous-subarray/)
+
+
+## 题目
+
+Given an integer array `nums`, you need to find one **continuous subarray** that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order.
+
+Return *the shortest such subarray and output its length*.
+
+**Example 1:**
+
+```
+Input: nums = [2,6,4,8,10,9,15]
+Output: 5
+Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3,4]
+Output: 0
+```
+
+**Example 3:**
+
+```
+Input: nums = [1]
+Output: 0
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 104`
+- `105 <= nums[i] <= 105`
+
+## 题目大意
+
+给你一个整数数组 nums ，你需要找出一个 连续子数组 ，如果对这个子数组进行升序排序，那么整个数组都会变为升序排序。请你找出符合题意的 最短 子数组，并输出它的长度。
+
+## 解题思路
+
+- 本题求的是最短逆序区间。经过简单推理，可以知道，这个逆序区间一定由这个区间内的最小元素决定左边界，最大元素决定右边界。
+- 先从左边找到第一个降序的元素，并记录最小的元素 min，再从右边往左找到最右边开始降序的元素，并记录最大的元素 max。最后需要还原最小元素和最大元素在原数组中正确的位置。以逆序区间左边界为例，如果区间外的一个元素比这个逆序区间内的最小元素还要小，说明它并不是左边界，因为这个小元素和逆序区间内的最小元素组合在一起，还是升序，并不是逆序。只有在左边区间外找到第一个大于逆序区间内最小元素，说明这里刚刚开始发生逆序，这才是最小逆序区间的左边界。同理，在逆序区间的右边找到第一个小于逆序区间内最大元素，说明这里刚刚发生逆序，这才是最小逆序区间的右边界。至此，最小逆序区间的左右边界都确定下来了，最短长度也就确定了下来。时间复杂度 O(n)，空间复杂度 O(1)。
+
+## 代码
+
+```go
+package leetcode
+
+import "math"
+
+func findUnsortedSubarray(nums []int) int {
+	n, left, right, minR, maxL, isSort := len(nums), -1, -1, math.MaxInt32, math.MinInt32, false
+	// left
+	for i := 1; i < n; i++ {
+		if nums[i] < nums[i-1] {
+			isSort = true
+		}
+		if isSort {
+			minR = min(minR, nums[i])
+		}
+	}
+	isSort = false
+	// right
+	for i := n - 2; i >= 0; i-- {
+		if nums[i] > nums[i+1] {
+			isSort = true
+		}
+		if isSort {
+			maxL = max(maxL, nums[i])
+		}
+	}
+	// minR
+	for i := 0; i < n; i++ {
+		if nums[i] > minR {
+			left = i
+			break
+		}
+	}
+	// maxL
+	for i := n - 1; i >= 0; i-- {
+		if nums[i] < maxL {
+			right = i
+			break
+		}
+	}
+	if left == -1 || right == -1 {
+		return 0
+	}
+	return right - left + 1
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0576.Out-of-Boundary-Paths/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0583.Delete-Operation-for-Two-Strings/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0583.Delete-Operation-for-Two-Strings.md b/website/content/ChapterFour/0500~0599/0583.Delete-Operation-for-Two-Strings.md
new file mode 100644
index 000000000..98678815b
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0583.Delete-Operation-for-Two-Strings.md
@@ -0,0 +1,85 @@
+# [583. Delete Operation for Two Strings](https://leetcode.com/problems/delete-operation-for-two-strings/)
+
+
+## 题目
+
+Given two strings `word1` and `word2`, return *the minimum number of **steps** required to make* `word1` *and* `word2` *the same*.
+
+In one **step**, you can delete exactly one character in either string.
+
+**Example 1:**
+
+```
+Input: word1 = "sea", word2 = "eat"
+Output: 2
+Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
+```
+
+**Example 2:**
+
+```
+Input: word1 = "leetcode", word2 = "etco"
+Output: 4
+```
+
+**Constraints:**
+
+- `1 <= word1.length, word2.length <= 500`
+- `word1` and `word2` consist of only lowercase English letters.
+
+## 题目大意
+
+给定两个单词 word1 和 word2，找到使得 word1 和 word2 相同所需的最小步数，每步可以删除任意一个字符串中的一个字符。
+
+## 解题思路
+
+- 从题目数据量级判断，此题一定是 O(n^2) 动态规划题。定义 `dp[i][j]` 表示 `word1[:i]` 与 `word2[:j]` 匹配所删除的最少步数。如果 `word1[:i-1]` 与 `word2[:j-1]` 匹配，那么 `dp[i][j] = dp[i-1][j-1]`。如果 `word1[:i-1]` 与 `word2[:j-1]` 不匹配，那么需要考虑删除一次，所以 `dp[i][j] = 1 + min(dp[i][j-1], dp[i-1][j])`。所以动态转移方程是：
+
+	{{< katex display >}} 
+    dp[i][j] = \left\{\begin{matrix}dp[i-1][j-1]&, word1[i-1] == word2[j-1]\\ 1 + min(dp[i][j-1], dp[i-1][j])&, word1[i-1] \neq word2[j-1]\\\end{matrix}\right.
+	{{< /katex >}} 
+
+    最终答案存储在 `dp[len(word1)][len(word2)]` 中。
+
+## 代码
+
+```go
+package leetcode
+
+func minDistance(word1 string, word2 string) int {
+	dp := make([][]int, len(word1)+1)
+	for i := 0; i < len(word1)+1; i++ {
+		dp[i] = make([]int, len(word2)+1)
+	}
+	for i := 0; i < len(word1)+1; i++ {
+		dp[i][0] = i
+	}
+	for i := 0; i < len(word2)+1; i++ {
+		dp[0][i] = i
+	}
+	for i := 1; i < len(word1)+1; i++ {
+		for j := 1; j < len(word2)+1; j++ {
+			if word1[i-1] == word2[j-1] {
+				dp[i][j] = dp[i-1][j-1]
+			} else {
+				dp[i][j] = 1 + min(dp[i][j-1], dp[i-1][j])
+			}
+		}
+	}
+	return dp[len(word1)][len(word2)]
+}
+
+func min(x, y int) int {
+	if x < y {
+		return x
+	}
+	return y
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0581.Shortest-Unsorted-Continuous-Subarray/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0589.N-ary-Tree-Preorder-Traversal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0589.N-ary-Tree-Preorder-Traversal.md b/website/content/ChapterFour/0500~0599/0589.N-ary-Tree-Preorder-Traversal.md
new file mode 100644
index 000000000..0bef5c6a2
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0589.N-ary-Tree-Preorder-Traversal.md
@@ -0,0 +1,97 @@
+# [589. N-ary Tree Preorder Traversal](https://leetcode.com/problems/n-ary-tree-preorder-traversal/)
+
+## 题目
+
+Given the `root` of an n-ary tree, return *the preorder traversal of its nodes' values*.
+
+Nary-Tree input serialization is represented in their level order traversal. Each group of children is separated by the null value (See examples)
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png](https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png)
+
+```
+Input: root = [1,null,3,2,4,null,5,6]
+Output: [1,3,5,6,2,4]
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png](https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png)
+
+```
+Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
+Output: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[0, 104]`.
+- `0 <= Node.val <= 10^4`
+- The height of the n-ary tree is less than or equal to `1000`.
+
+**Follow up:** Recursive solution is trivial, could you do it iteratively?
+
+## 题目大意
+
+给定一个 N 叉树，返回其节点值的 **前序遍历** 。N 叉树 在输入中按层序遍历进行序列化表示，每组子节点由空值 `null` 分隔（请参见示例）。
+
+## 解题思路
+
+- N 叉树和二叉树的前序遍历原理完全一样。二叉树非递归解法需要用到栈辅助，N 叉树同样如此。将父节点的所有孩子节点**逆序**入栈，逆序的目的是为了让前序节点永远在栈顶。依次循环直到栈里所有元素都出栈。输出的结果即为 N 叉树的前序遍历。时间复杂度 O(n)，空间复杂度 O(n)。
+- 递归解法非常简单，见解法二。
+
+## 代码
+
+```go
+package leetcode
+
+//  Definition for a Node.
+type Node struct {
+	Val      int
+	Children []*Node
+}
+
+// 解法一 非递归
+func preorder(root *Node) []int {
+	res := []int{}
+	if root == nil {
+		return res
+	}
+	stack := []*Node{root}
+	for len(stack) > 0 {
+		r := stack[len(stack)-1]
+		stack = stack[:len(stack)-1]
+		res = append(res, r.Val)
+		tmp := []*Node{}
+		for _, v := range r.Children {
+			tmp = append([]*Node{v}, tmp...) // 逆序存点
+		}
+		stack = append(stack, tmp...)
+	}
+	return res
+}
+
+// 解法二 递归
+func preorder1(root *Node) []int {
+	res := []int{}
+	preorderdfs(root, &res)
+	return res
+}
+
+func preorderdfs(root *Node, res *[]int) {
+	if root != nil {
+		*res = append(*res, root.Val)
+		for i := 0; i < len(root.Children); i++ {
+			preorderdfs(root.Children[i], res)
+		}
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0583.Delete-Operation-for-Two-Strings/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0594.Longest-Harmonious-Subsequence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0594.Longest-Harmonious-Subsequence.md b/website/content/ChapterFour/0500~0599/0594.Longest-Harmonious-Subsequence.md
new file mode 100755
index 000000000..963aef862
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0594.Longest-Harmonious-Subsequence.md
@@ -0,0 +1,64 @@
+# [594. Longest Harmonious Subsequence](https://leetcode.com/problems/longest-harmonious-subsequence/)
+
+
+## 题目
+
+We define a harmounious array as an array where the difference between its maximum value and its minimum value is **exactly** 1.
+
+Now, given an integer array, you need to find the length of its longest harmonious subsequence among all its possible [subsequences](https://en.wikipedia.org/wiki/Subsequence).
+
+**Example 1**:
+
+    Input: [1,3,2,2,5,2,3,7]
+    Output: 5
+    Explanation: The longest harmonious subsequence is [3,2,2,2,3].
+
+**Note**: The length of the input array will not exceed 20,000.
+
+
+## 题目大意
+
+和谐数组是指一个数组里元素的最大值和最小值之间的差别正好是1。现在，给定一个整数数组，你需要在所有可能的子序列中找到最长的和谐子序列的长度。说明: 输入的数组长度最大不超过20,000.
+
+## 解题思路
+
+- 在给出的数组里面找到这样一个子数组：要求子数组中的最大值和最小值相差 1 。这一题是简单题。先统计每个数字出现的频次，然后在 map 找相差 1 的 2 个数组的频次和，动态的维护两个数的频次和就是最后要求的子数组的最大长度。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findLHS(nums []int) int {
+	if len(nums) < 2 {
+		return 0
+	}
+	res := make(map[int]int, len(nums))
+	for _, num := range nums {
+		if _, exist := res[num]; exist {
+			res[num]++
+			continue
+		}
+		res[num] = 1
+	}
+	longest := 0
+	for k, c := range res {
+		if n, exist := res[k+1]; exist {
+			if c+n > longest {
+				longest = c + n
+			}
+		}
+	}
+	return longest
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0589.N-ary-Tree-Preorder-Traversal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0598.Range-Addition-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0598.Range-Addition-II.md b/website/content/ChapterFour/0500~0599/0598.Range-Addition-II.md
new file mode 100644
index 000000000..c4ca5720b
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0598.Range-Addition-II.md
@@ -0,0 +1,89 @@
+# [598. Range Addition II](https://leetcode.com/problems/range-addition-ii/)
+
+
+## 题目
+
+Given an m * n matrix **M** initialized with all **0**'s and several update operations.
+
+Operations are represented by a 2D array, and each operation is represented by an array with two **positive** integers **a** and **b**, which means **M[i][j]** should be **added by one** for all **0 <= i < a** and **0 <= j < b**.
+
+You need to count and return the number of maximum integers in the matrix after performing all the operations.
+
+**Example 1**:
+
+```
+Input: 
+m = 3, n = 3
+operations = [[2,2],[3,3]]
+Output: 4
+Explanation: 
+Initially, M = 
+[[0, 0, 0],
+ [0, 0, 0],
+ [0, 0, 0]]
+
+After performing [2,2], M = 
+[[1, 1, 0],
+ [1, 1, 0],
+ [0, 0, 0]]
+
+After performing [3,3], M = 
+[[2, 2, 1],
+ [2, 2, 1],
+ [1, 1, 1]]
+
+So the maximum integer in M is 2, and there are four of it in M. So return 4.
+```
+
+**Note**:
+
+1. The range of m and n is [1,40000].
+2. The range of a is [1,m], and the range of b is [1,n].
+3. The range of operations size won't exceed 10,000.
+
+## 题目大意
+
+给定一个初始元素全部为 0，大小为 m*n 的矩阵 M 以及在 M 上的一系列更新操作。操作用二维数组表示，其中的每个操作用一个含有两个正整数 a 和 b 的数组表示，含义是将所有符合 0 <= i < a 以及 0 <= j < b 的元素 M[i][j] 的值都增加 1。在执行给定的一系列操作后，你需要返回矩阵中含有最大整数的元素个数。
+
+注意:
+
+- m 和 n 的范围是 [1,40000]。
+- a 的范围是 [1,m]，b 的范围是 [1,n]。
+- 操作数目不超过 10000。
+
+
+## 解题思路
+
+- 给定一个初始都为 0 的 m * n 的矩阵，和一个操作数组。经过一系列的操作以后，最终输出矩阵中最大整数的元素个数。每次操作都使得一个矩形内的元素都 + 1 。
+- 这一题乍一看像线段树的区间覆盖问题，但是实际上很简单。如果此题是任意的矩阵，那就可能用到线段树了。这一题每个矩阵的起点都包含 [0 , 0] 这个元素，也就是说每次操作都会影响第一个元素。那么这道题就很简单了。经过 n 次操作以后，被覆盖次数最多的矩形区间，一定就是最大整数所在的区间。由于起点都是第一个元素，所以我们只用关心矩形的右下角那个坐标。右下角怎么计算呢？只用每次动态的维护一下矩阵长和宽的最小值即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func maxCount(m int, n int, ops [][]int) int {
+	minM, minN := m, n
+	for _, op := range ops {
+		minM = min(minM, op[0])
+		minN = min(minN, op[1])
+	}
+	return minM * minN
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0594.Longest-Harmonious-Subsequence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0599.Minimum-Index-Sum-of-Two-Lists/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/0599.Minimum-Index-Sum-of-Two-Lists.md b/website/content/ChapterFour/0500~0599/0599.Minimum-Index-Sum-of-Two-Lists.md
new file mode 100755
index 000000000..a33f1d285
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/0599.Minimum-Index-Sum-of-Two-Lists.md
@@ -0,0 +1,84 @@
+# [599. Minimum Index Sum of Two Lists](https://leetcode.com/problems/minimum-index-sum-of-two-lists/)
+
+## 题目
+
+Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings.
+
+You need to help them find out their **common interest** with the **least list index sum**. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.
+
+**Example 1**:
+
+    Input:
+    ["Shogun", "Tapioca Express", "Burger King", "KFC"]
+    ["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
+    Output: ["Shogun"]
+    Explanation: The only restaurant they both like is "Shogun".
+
+**Example 2**:
+
+    Input:
+    ["Shogun", "Tapioca Express", "Burger King", "KFC"]
+    ["KFC", "Shogun", "Burger King"]
+    Output: ["Shogun"]
+    Explanation: The restaurant they both like and have the least index sum is "Shogun" with index sum 1 (0+1).
+
+**Note**:
+
+1. The length of both lists will be in the range of [1, 1000].
+2. The length of strings in both lists will be in the range of [1, 30].
+3. The index is starting from 0 to the list length minus 1.
+4. No duplicates in both lists.
+
+
+## 题目大意
+
+假设 Andy 和 Doris 想在晚餐时选择一家餐厅，并且他们都有一个表示最喜爱餐厅的列表，每个餐厅的名字用字符串表示。你需要帮助他们用最少的索引和找出他们共同喜爱的餐厅。 如果答案不止一个，则输出所有答案并且不考虑顺序。 你可以假设总是存在一个答案。
+
+
+提示:  
+
+- 两个列表的长度范围都在 [1, 1000] 内。
+- 两个列表中的字符串的长度将在 [1，30] 的范围内。
+- 下标从 0 开始，到列表的长度减 1。
+- 两个列表都没有重复的元素。
+
+
+
+## 解题思路
+
+
+- 在 Andy 和 Doris 两人分别有各自的餐厅喜欢列表，要求找出两人公共喜欢的一家餐厅，如果共同喜欢的次数相同，都输出。这一题是简单题，用 map 统计频次，输出频次最多的餐厅。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findRestaurant(list1 []string, list2 []string) []string {
+	m, ans := make(map[string]int, len(list1)), []string{}
+	for i, r := range list1 {
+		m[r] = i
+	}
+	for j, r := range list2 {
+		if _, ok := m[r]; ok {
+			m[r] += j
+			if len(ans) == 0 || m[r] == m[ans[0]] {
+				ans = append(ans, r)
+			} else if m[r] < m[ans[0]] {
+				ans = []string{r}
+			}
+		}
+	}
+	return ans
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0598.Range-Addition-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0605.Can-Place-Flowers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0500~0599/_index.md b/website/content/ChapterFour/0500~0599/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/0500~0599/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/0600~0699/0605.Can-Place-Flowers.md b/website/content/ChapterFour/0600~0699/0605.Can-Place-Flowers.md
new file mode 100644
index 000000000..fe4ee14a7
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0605.Can-Place-Flowers.md
@@ -0,0 +1,67 @@
+# [605. Can Place Flowers](https://leetcode.com/problems/can-place-flowers/)
+
+## 题目
+
+You have a long flowerbed in which some of the plots are planted, and some are not. However, flowers cannot be planted in **adjacent** plots.
+
+Given an integer array `flowerbed` containing `0`'s and `1`'s, where `0` means empty and `1` means not empty, and an integer `n`, return *if* `n` new flowers can be planted in the `flowerbed` without violating the no-adjacent-flowers rule.
+
+**Example 1**:
+
+```
+Input: flowerbed = [1,0,0,0,1], n = 1
+Output: true
+```
+
+**Example 2**:
+
+```
+Input: flowerbed = [1,0,0,0,1], n = 2
+Output: false
+```
+
+**Constraints**:
+
+- `1 <= flowerbed.length <= 2 * 104`
+- `flowerbed[i]` is `0` or `1`.
+- There are no two adjacent flowers in `flowerbed`.
+- `0 <= n <= flowerbed.length`
+
+## 题目大意
+
+假设你有一个很长的花坛，一部分地块种植了花，另一部分却没有。可是，花卉不能种植在相邻的地块上，它们会争夺水源，两者都会死去。给定一个花坛（表示为一个数组包含0和1，其中0表示没种植花，1表示种植了花），和一个数 n 。能否在不打破种植规则的情况下种入 n 朵花？能则返回True，不能则返回False。
+
+## 解题思路
+
+- 这一题最容易想到的解法是步长为 2 遍历数组，依次计数 0 的个数。有 2 种特殊情况需要单独判断，第一种情况是首尾连续多个 0，例如，00001 和 10000，第二种情况是 2 个 1 中间存在的 0 不足以种花，例如，1001 和 100001，1001 不能种任何花，100001 只能种一种花。单独判断出这 2 种情况，这一题就可以 AC 了。
+- 换个思路，找到可以种花的基本单元是 00，那么上面那 2 种特殊情况都可以统一成一种情况。判断是否当前存在 00 的组合，如果存在 00 的组合，都可以种花。末尾的情况需要单独判断，如果末尾为 0，也可以种花。这个时候不需要再找 00 组合，因为会越界。代码实现如下，思路很简洁明了。
+
+## 代码
+
+```go
+package leetcode
+
+func canPlaceFlowers(flowerbed []int, n int) bool {
+	lenth := len(flowerbed)
+	for i := 0; i < lenth && n > 0; i += 2 {
+		if flowerbed[i] == 0 {
+			if i+1 == lenth || flowerbed[i+1] == 0 {
+				n--
+			} else {
+				i++
+			}
+		}
+	}
+	if n == 0 {
+		return true
+	}
+	return false
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0500~0599/0599.Minimum-Index-Sum-of-Two-Lists/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0609.Find-Duplicate-File-in-System/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0609.Find-Duplicate-File-in-System.md b/website/content/ChapterFour/0600~0699/0609.Find-Duplicate-File-in-System.md
new file mode 100644
index 000000000..0e7cc04ab
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0609.Find-Duplicate-File-in-System.md
@@ -0,0 +1,100 @@
+# [609. Find Duplicate File in System](https://leetcode.com/problems/find-duplicate-file-in-system/)
+
+
+## 题目
+
+Given a list `paths` of directory info, including the directory path, and all the files with contents in this directory, return *all the duplicate files in the file system in terms of their paths*. You may return the answer in **any order**.
+
+A group of duplicate files consists of at least two files that have the same content.
+
+A single directory info string in the input list has the following format:
+
+- `"root/d1/d2/.../dm f1.txt(f1_content) f2.txt(f2_content) ... fn.txt(fn_content)"`
+
+It means there are `n` files `(f1.txt, f2.txt ... fn.txt)` with content `(f1_content, f2_content ... fn_content)` respectively in the directory "`root/d1/d2/.../dm"`. Note that `n >= 1` and `m >= 0`. If `m = 0`, it means the directory is just the root directory.
+
+The output is a list of groups of duplicate file paths. For each group, it contains all the file paths of the files that have the same content. A file path is a string that has the following format:
+
+- `"directory_path/file_name.txt"`
+
+**Example 1:**
+
+```
+Input: paths = ["root/a 1.txt(abcd) 2.txt(efgh)","root/c 3.txt(abcd)","root/c/d 4.txt(efgh)","root 4.txt(efgh)"]
+Output: [["root/a/2.txt","root/c/d/4.txt","root/4.txt"],["root/a/1.txt","root/c/3.txt"]]
+
+```
+
+**Example 2:**
+
+```
+Input: paths = ["root/a 1.txt(abcd) 2.txt(efgh)","root/c 3.txt(abcd)","root/c/d 4.txt(efgh)"]
+Output: [["root/a/2.txt","root/c/d/4.txt"],["root/a/1.txt","root/c/3.txt"]]
+
+```
+
+**Constraints:**
+
+- `1 <= paths.length <= 2 * 104`
+- `1 <= paths[i].length <= 3000`
+- `1 <= sum(paths[i].length) <= 5 * 105`
+- `paths[i]` consist of English letters, digits, `'/'`, `'.'`, `'('`, `')'`, and `' '`.
+- You may assume no files or directories share the same name in the same directory.
+- You may assume each given directory info represents a unique directory. A single blank space separates the directory path and file info.
+
+**Follow up:**
+
+- Imagine you are given a real file system, how will you search files? DFS or BFS?
+- If the file content is very large (GB level), how will you modify your solution?
+- If you can only read the file by 1kb each time, how will you modify your solution?
+- What is the time complexity of your modified solution? What is the most time-consuming part and memory-consuming part of it? How to optimize?
+- How to make sure the duplicated files you find are not false positive?
+
+## 题目大意
+
+给定一个目录信息列表，包括目录路径，以及该目录中的所有包含内容的文件，您需要找到文件系统中的所有重复文件组的路径。一组重复的文件至少包括二个具有完全相同内容的文件。输入列表中的单个目录信息字符串的格式如下：`"root/d1/d2/.../dm f1.txt(f1_content) f2.txt(f2_content) ... fn.txt(fn_content)"`。这意味着有 n 个文件（`f1.txt, f2.txt ... fn.txt` 的内容分别是 `f1_content, f2_content ... fn_content`）在目录 `root/d1/d2/.../dm` 下。注意：n>=1 且 m>=0。如果 m=0，则表示该目录是根目录。该输出是重复文件路径组的列表。对于每个组，它包含具有相同内容的文件的所有文件路径。文件路径是具有下列格式的字符串：`"directory_path/file_name.txt"`
+
+## 解题思路
+
+- 这一题算简单题，考察的是字符串基本操作与 map 的使用。首先通过字符串操作获取目录路径、文件名和文件内容。再使用 map 来寻找重复文件，key 是文件内容，value 是存储路径和文件名的列表。遍历每一个文件，并把它加入 map 中。最后遍历 map，如果一个键对应的值列表的长度大于 1，说明找到了重复文件，可以把这个列表加入到最终答案中。
+- 这道题有价值的地方在 **Follow up** 中。感兴趣的读者可以仔细想想以下几个问题：
+    1. 假设您有一个真正的文件系统，您将如何搜索文件？广度搜索还是宽度搜索？
+    2. 如果文件内容非常大（GB级别），您将如何修改您的解决方案？
+    3. 如果每次只能读取 1 kb 的文件，您将如何修改解决方案？
+    4. 修改后的解决方案的时间复杂度是多少？其中最耗时的部分和消耗内存的部分是什么？如何优化？
+    5. 如何确保您发现的重复文件不是误报？
+
+## 代码
+
+```go
+package leetcode
+
+import "strings"
+
+func findDuplicate(paths []string) [][]string {
+	cache := make(map[string][]string)
+	for _, path := range paths {
+		parts := strings.Split(path, " ")
+		dir := parts[0]
+		for i := 1; i < len(parts); i++ {
+			bracketPosition := strings.IndexByte(parts[i], '(')
+			content := parts[i][bracketPosition+1 : len(parts[i])-1]
+			cache[content] = append(cache[content], dir+"/"+parts[i][:bracketPosition])
+		}
+	}
+	res := make([][]string, 0, len(cache))
+	for _, group := range cache {
+		if len(group) >= 2 {
+			res = append(res, group)
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0605.Can-Place-Flowers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0611.Valid-Triangle-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0611.Valid-Triangle-Number.md b/website/content/ChapterFour/0600~0699/0611.Valid-Triangle-Number.md
new file mode 100644
index 000000000..417b341fd
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0611.Valid-Triangle-Number.md
@@ -0,0 +1,44 @@
+# [611. Valid Triangle Number](https://leetcode.com/problems/valid-triangle-number/)
+
+## 题目
+
+Given an integer array nums, return the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.
+
+## 题目大意
+
+给定一个包含非负整数的数组，你的任务是统计其中可以组成三角形三条边的三元组个数。
+
+## 解题思路
+
+- 题意很简单，最容易想到的暴力解法是三重循环，暴力枚举，时间复杂度 O(n^3)。三重循环中最内层的循环可以优化，因为 k 和 i，j 存在关联性。第二层循环 j 从 i + 1 开始循环，k 从 j + 1 = i + 2 开始循环。循环累加 k 的值，直到 `nums[i] + nums[j] > nums[k]`，那么 `[nums[j + 1], nums[k - 1]]` 这个区间内的值都满足条件。满足条件的解个数增加 `k - j - 1` 个。j 再次递增 + 1，此时最内层的 k 不用从 j + 1 开始增加，只用从上次 k 开始增加即可。因为如果 `nums[i] + nums[j] > nums[k]`，如果这个 `nums[i] + nums[j + 1] > nums[m + 1]` 不等式成立，那么 m 一定不小于 k。所以内层循环 k 和 j 加起来的时间复杂度是 O(n)，最外层 i 的循环是 O(n)，这样优化以后，整体时间复杂度是 O(n^2)。
+- 可能有读者有疑问，三角形三条边的组成条件：任意两边之和大于第三边。`a + b > c`，`a + c > b`，`b + c > a`，此处为什么只判断了 `a + b > c` 呢？因为一开始进行了排序处理，使得 `a ≤ b ≤ c`，在这个前提下，`a + c > b`，`b + c > a` 是一定成立的。所以原问题便转化为只需关心 `a + b > c` 这一个不等式是否成立即可。此题的测试用例用有一种特殊情况，那就是其中一条边或者两条边长度为 0，那么 `a + b > c` 这个不等式一定不成立。综上，先排序预处理之后，只需要关心 `a + b > c` 这一个不等式是否成立即可。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func triangleNumber(nums []int) int {
+	res := 0
+	sort.Ints(nums)
+	for i := 0; i < len(nums)-2; i++ {
+		k := i + 2
+		for j := i + 1; j < len(nums)-1 && nums[i] != 0; j++ {
+			for k < len(nums) && nums[i]+nums[j] > nums[k] {
+				k++
+			}
+			res += k - j - 1
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0609.Find-Duplicate-File-in-System/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0617.Merge-Two-Binary-Trees/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0617.Merge-Two-Binary-Trees.md b/website/content/ChapterFour/0600~0699/0617.Merge-Two-Binary-Trees.md
new file mode 100644
index 000000000..8e535838c
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0617.Merge-Two-Binary-Trees.md
@@ -0,0 +1,87 @@
+# [617. Merge Two Binary Trees](https://leetcode.com/problems/merge-two-binary-trees/)
+
+
+## 题目
+
+You are given two binary trees `root1` and `root2`.
+
+Imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge the two trees into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of the new tree.
+
+Return *the merged tree*.
+
+**Note:** The merging process must start from the root nodes of both trees.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/02/05/merge.jpg](https://assets.leetcode.com/uploads/2021/02/05/merge.jpg)
+
+```
+Input: root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
+Output: [3,4,5,5,4,null,7]
+```
+
+**Example 2:**
+
+```
+Input: root1 = [1], root2 = [1,2]
+Output: [2,2]
+```
+
+**Constraints:**
+
+- The number of nodes in both trees is in the range `[0, 2000]`.
+- `104 <= Node.val <= 104`
+
+## 题目大意
+
+给定两个二叉树，想象当你将它们中的一个覆盖到另一个上时，两个二叉树的一些节点便会重叠。你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠，那么将他们的值相加作为节点合并后的新值，否则不为 NULL 的节点将直接作为新二叉树的节点。
+
+## 解题思路
+
+- 简单题。采用深搜的思路，分别从根节点开始同时遍历两个二叉树，并将对应的节点进行合并。两个二叉树的对应节点可能存在以下三种情况：
+    - 如果两个二叉树的对应节点都为空，则合并后的二叉树的对应节点也为空；
+    - 如果两个二叉树的对应节点只有一个为空，则合并后的二叉树的对应节点为其中的非空节点；
+    - 如果两个二叉树的对应节点都不为空，则合并后的二叉树的对应节点的值为两个二叉树的对应节点的值之和，此时需要显性合并两个节点。
+- 对一个节点进行合并之后，还要对该节点的左右子树分别进行合并。用递归实现即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func mergeTrees(root1 *TreeNode, root2 *TreeNode) *TreeNode {
+	if root1 == nil {
+		return root2
+	}
+	if root2 == nil {
+		return root1
+	}
+	root1.Val += root2.Val
+	root1.Left = mergeTrees(root1.Left, root2.Left)
+	root1.Right = mergeTrees(root1.Right, root2.Right)
+	return root1
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0611.Valid-Triangle-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0622.Design-Circular-Queue/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0622.Design-Circular-Queue.md b/website/content/ChapterFour/0600~0699/0622.Design-Circular-Queue.md
new file mode 100644
index 000000000..54dfe8f24
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0622.Design-Circular-Queue.md
@@ -0,0 +1,154 @@
+# [622. Design Circular Queue](https://leetcode.com/problems/design-circular-queue/)
+
+
+## 题目
+
+Design your implementation of the circular queue. The circular queue is a linear data structure in which the operations are performed based on FIFO (First In First Out) principle and the last position is connected back to the first position to make a circle. It is also called "Ring Buffer".
+
+One of the benefits of the circular queue is that we can make use of the spaces in front of the queue. In a normal queue, once the queue becomes full, we cannot insert the next element even if there is a space in front of the queue. But using the circular queue, we can use the space to store new values.
+
+Implementation the `MyCircularQueue` class:
+
+- `MyCircularQueue(k)` Initializes the object with the size of the queue to be `k`.
+- `int Front()` Gets the front item from the queue. If the queue is empty, return `1`.
+- `int Rear()` Gets the last item from the queue. If the queue is empty, return `1`.
+- `boolean enQueue(int value)` Inserts an element into the circular queue. Return `true` if the operation is successful.
+- `boolean deQueue()` Deletes an element from the circular queue. Return `true` if the operation is successful.
+- `boolean isEmpty()` Checks whether the circular queue is empty or not.
+- `boolean isFull()` Checks whether the circular queue is full or not.
+
+**Example 1:**
+
+```
+Input
+["MyCircularQueue", "enQueue", "enQueue", "enQueue", "enQueue", "Rear", "isFull", "deQueue", "enQueue", "Rear"]
+[[3], [1], [2], [3], [4], [], [], [], [4], []]
+Output
+[null, true, true, true, false, 3, true, true, true, 4]
+
+Explanation
+MyCircularQueue myCircularQueue = new MyCircularQueue(3);
+myCircularQueue.enQueue(1); // return True
+myCircularQueue.enQueue(2); // return True
+myCircularQueue.enQueue(3); // return True
+myCircularQueue.enQueue(4); // return False
+myCircularQueue.Rear();     // return 3
+myCircularQueue.isFull();   // return True
+myCircularQueue.deQueue();  // return True
+myCircularQueue.enQueue(4); // return True
+myCircularQueue.Rear();     // return 4
+
+```
+
+**Constraints:**
+
+- `1 <= k <= 1000`
+- `0 <= value <= 1000`
+- At most `3000` calls will be made to `enQueue`, `deQueue`, `Front`, `Rear`, `isEmpty`, and `isFull`.
+
+**Follow up:**
+
+Could you solve the problem without using the built-in queue?
+
+## 题目大意
+
+设计你的循环队列实现。 循环队列是一种线性数据结构，其操作表现基于 FIFO（先进先出）原则并且队尾被连接在队首之后以形成一个循环。它也被称为“环形缓冲器”。
+
+循环队列的一个好处是我们可以利用这个队列之前用过的空间。在一个普通队列里，一旦一个队列满了，我们就不能插入下一个元素，即使在队列前面仍有空间。但是使用循环队列，我们能使用这些空间去存储新的值。
+
+你的实现应该支持如下操作：
+
+- MyCircularQueue(k): 构造器，设置队列长度为 k 。
+- Front: 从队首获取元素。如果队列为空，返回 -1 。
+- Rear: 获取队尾元素。如果队列为空，返回 -1 。
+- enQueue(value): 向循环队列插入一个元素。如果成功插入则返回真。
+- deQueue(): 从循环队列中删除一个元素。如果成功删除则返回真。
+- isEmpty(): 检查循环队列是否为空。
+- isFull(): 检查循环队列是否已满。
+
+## 解题思路
+
+- 简单题。设计一个环形队列，底层用数组实现。额外维护 4 个变量，队列的总 cap，队列当前的 size，前一元素下标 left，后一个元素下标 right。每添加一个元素便维护 left，right，size，下标需要对 cap 取余，因为超过 cap 大小之后，需要循环存储。代码实现没有难度，具体sh见下面代码。
+
+## 代码
+
+```go
+package leetcode
+
+type MyCircularQueue struct {
+	cap   int
+	size  int
+	queue []int
+	left  int
+	right int
+}
+
+func Constructor(k int) MyCircularQueue {
+	return MyCircularQueue{cap: k, size: 0, left: 0, right: 0, queue: make([]int, k)}
+}
+
+func (this *MyCircularQueue) EnQueue(value int) bool {
+	if this.size == this.cap {
+		return false
+	}
+	this.size++
+	this.queue[this.right] = value
+	this.right++
+	this.right %= this.cap
+	return true
+
+}
+
+func (this *MyCircularQueue) DeQueue() bool {
+	if this.size == 0 {
+		return false
+	}
+	this.size--
+	this.left++
+	this.left %= this.cap
+	return true
+}
+
+func (this *MyCircularQueue) Front() int {
+	if this.size == 0 {
+		return -1
+	}
+	return this.queue[this.left]
+}
+
+func (this *MyCircularQueue) Rear() int {
+	if this.size == 0 {
+		return -1
+	}
+	if this.right == 0 {
+		return this.queue[this.cap-1]
+	}
+	return this.queue[this.right-1]
+}
+
+func (this *MyCircularQueue) IsEmpty() bool {
+	return this.size == 0
+}
+
+func (this *MyCircularQueue) IsFull() bool {
+	return this.size == this.cap
+}
+
+/**
+ * Your MyCircularQueue object will be instantiated and called as such:
+ * obj := Constructor(k);
+ * param_1 := obj.EnQueue(value);
+ * param_2 := obj.DeQueue();
+ * param_3 := obj.Front();
+ * param_4 := obj.Rear();
+ * param_5 := obj.IsEmpty();
+ * param_6 := obj.IsFull();
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0617.Merge-Two-Binary-Trees/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0623.Add-One-Row-to-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0623.Add-One-Row-to-Tree.md b/website/content/ChapterFour/0600~0699/0623.Add-One-Row-to-Tree.md
new file mode 100644
index 000000000..356232ccb
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0623.Add-One-Row-to-Tree.md
@@ -0,0 +1,117 @@
+# [623. Add One Row to Tree](https://leetcode.com/problems/add-one-row-to-tree/)
+
+
+## 题目
+
+Given the root of a binary tree, then value `v` and depth `d`, you need to add a row of nodes with value `v` at the given depth `d`. The root node is at depth 1.
+
+The adding rule is: given a positive integer depth `d`, for each NOT null tree nodes `N` in depth `d-1`, create two tree nodes with value `v` as `N's` left subtree root and right subtree root. And `N's` **original left subtree** should be the left subtree of the new left subtree root, its **original right subtree** should be the right subtree of the new right subtree root. If depth `d` is 1 that means there is no depth d-1 at all, then create a tree node with value **v** as the new root of the whole original tree, and the original tree is the new root's left subtree.
+
+**Example 1:**
+
+```
+Input: 
+A binary tree as following:
+       4
+     /   \
+    2     6
+   / \   / 
+  3   1 5   
+
+v = 1d = 2Output: 
+       4
+      / \
+     1   1
+    /     \
+   2       6
+  / \     / 
+ 3   1   5   
+```
+
+**Example 2:**
+
+```
+Input: 
+A binary tree as following:
+      4
+     /   
+    2    
+   / \   
+  3   1    
+
+v = 1d = 3Output: 
+      4
+     /   
+    2
+   / \    
+  1   1
+ /     \  
+3       1
+```
+
+**Note:**
+
+1. The given d is in range [1, maximum depth of the given tree + 1].
+2. The given binary tree has at least one tree node.
+
+## 题目大意
+
+给定一个二叉树，根节点为第1层，深度为 1。在其第 d 层追加一行值为 v 的节点。添加规则：给定一个深度值 d （正整数），针对深度为 d-1 层的每一非空节点 N，为 N 创建两个值为 v 的左子树和右子树。将 N 原先的左子树，连接为新节点 v 的左子树；将 N 原先的右子树，连接为新节点 v 的右子树。如果 d 的值为 1，深度 d - 1 不存在，则创建一个新的根节点 v，原先的整棵树将作为 v 的左子树。
+
+## 解题思路
+
+- 这一题虽然是 Medium，实际非常简单。给二叉树添加一行，用 DFS 或者 BFS，遍历过程中记录行数，到达目标行一行，增加节点即可。不过需要注意 2 个特殊情况，特殊情况一，`d==1`，此时需要添加的行即为根节点。特殊情况二，`d>height(root)`，即要添加的行数比树还要高，这时只需要在最下层的叶子节点添加一层。时间复杂度 O(n)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func addOneRow(root *TreeNode, v int, d int) *TreeNode {
+	if d == 1 {
+		tmp := &TreeNode{Val: v, Left: root, Right: nil}
+		return tmp
+	}
+	level := 1
+	addTreeRow(root, v, d, &level)
+	return root
+}
+
+func addTreeRow(root *TreeNode, v, d int, currLevel *int) {
+	if *currLevel == d-1 {
+		root.Left = &TreeNode{Val: v, Left: root.Left, Right: nil}
+		root.Right = &TreeNode{Val: v, Left: nil, Right: root.Right}
+		return
+	}
+	*currLevel++
+	if root.Left != nil {
+		addTreeRow(root.Left, v, d, currLevel)
+	}
+	if root.Right != nil {
+		addTreeRow(root.Right, v, d, currLevel)
+	}
+	*currLevel--
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0622.Design-Circular-Queue/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0628.Maximum-Product-of-Three-Numbers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0628.Maximum-Product-of-Three-Numbers.md b/website/content/ChapterFour/0600~0699/0628.Maximum-Product-of-Three-Numbers.md
new file mode 100755
index 000000000..8eedbe1ea
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0628.Maximum-Product-of-Three-Numbers.md
@@ -0,0 +1,110 @@
+# [628. Maximum Product of Three Numbers](https://leetcode.com/problems/maximum-product-of-three-numbers/)
+
+
+## 题目
+
+Given an integer array, find three numbers whose product is maximum and output the maximum product.
+
+**Example 1**:
+
+    Input: [1,2,3]
+    Output: 6
+
+**Example 2**:
+
+    Input: [1,2,3,4]
+    Output: 24
+
+**Note**:
+
+1. The length of the given array will be in range [3,10^4] and all elements are in the range [-1000, 1000].
+2. Multiplication of any three numbers in the input won't exceed the range of 32-bit signed integer.
+
+
+## 题目大意
+
+给定一个整型数组，在数组中找出由三个数组成的最大乘积，并输出这个乘积。
+
+
+
+
+## 解题思路
+
+
+- 给出一个数组，要求求出这个数组中任意挑 3 个数能组成的乘积最大的值。
+- 题目的 test case 数据量比较大，如果用排序的话，时间复杂度高，可以直接考虑模拟，挑出 3 个数组成乘积最大值，必然是一个正数和二个负数，或者三个正数。那么选出最大的三个数和最小的二个数，对比一下就可以求出最大值了。时间复杂度 O(n)
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+// 解法一 排序，时间复杂度 O(n log n)
+func maximumProduct(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	res := 1
+	if len(nums) <= 3 {
+		for i := 0; i < len(nums); i++ {
+			res = res * nums[i]
+		}
+		return res
+	}
+	sort.Ints(nums)
+	if nums[len(nums)-1] <= 0 {
+		return 0
+	}
+	return max(nums[0]*nums[1]*nums[len(nums)-1], nums[len(nums)-1]*nums[len(nums)-2]*nums[len(nums)-3])
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法二 模拟，时间复杂度 O(n)
+func maximumProduct1(nums []int) int {
+	max := make([]int, 0)
+	max = append(max, math.MinInt64, math.MinInt64, math.MinInt64)
+	min := make([]int, 0)
+	min = append(min, math.MaxInt64, math.MaxInt64)
+	for _, num := range nums {
+		if num > max[0] {
+			max[0], max[1], max[2] = num, max[0], max[1]
+		} else if num > max[1] {
+			max[1], max[2] = num, max[1]
+		} else if num > max[2] {
+			max[2] = num
+		}
+		if num < min[0] {
+			min[0], min[1] = num, min[0]
+		} else if num < min[1] {
+			min[1] = num
+		}
+	}
+	maxProduct1, maxProduct2 := min[0]*min[1]*max[0], max[0]*max[1]*max[2]
+	if maxProduct1 > maxProduct2 {
+		return maxProduct1
+	}
+	return maxProduct2
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0623.Add-One-Row-to-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0630.Course-Schedule-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0630.Course-Schedule-III.md b/website/content/ChapterFour/0600~0699/0630.Course-Schedule-III.md
new file mode 100644
index 000000000..2179af424
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0630.Course-Schedule-III.md
@@ -0,0 +1,103 @@
+# [630. Course Schedule III](https://leetcode.com/problems/course-schedule-iii/)
+
+## 题目
+
+There are `n` different online courses numbered from `1` to `n`. You are given an array `courses` where `courses[i] = [durationi, lastDayi]` indicate that the `ith` course should be taken **continuously** for `durationi` days and must be finished before or on `lastDayi`.
+
+You will start on the `1st` day and you cannot take two or more courses simultaneously.
+
+Return *the maximum number of courses that you can take*.
+
+**Example 1:**
+
+```
+Input: courses = [[100,200],[200,1300],[1000,1250],[2000,3200]]
+Output: 3
+Explanation:
+There are totally 4 courses, but you can take 3 courses at most:
+First, take the 1st course, it costs 100 days so you will finish it on the 100th day, and ready to take the next course on the 101st day.
+Second, take the 3rd course, it costs 1000 days so you will finish it on the 1100th day, and ready to take the next course on the 1101st day.
+Third, take the 2nd course, it costs 200 days so you will finish it on the 1300th day.
+The 4th course cannot be taken now, since you will finish it on the 3300th day, which exceeds the closed date.
+
+```
+
+**Example 2:**
+
+```
+Input: courses = [[1,2]]
+Output: 1
+
+```
+
+**Example 3:**
+
+```
+Input: courses = [[3,2],[4,3]]
+Output: 0
+
+```
+
+**Constraints:**
+
+- `1 <= courses.length <= 104`
+- `1 <= durationi, lastDayi <= 104`
+
+## 题目大意
+
+这里有 n 门不同的在线课程，他们按从 1 到 n 编号。每一门课程有一定的持续上课时间（课程时间）t 以及关闭时间第 d 天。一门课要持续学习 t 天直到第 d 天时要完成，你将会从第 1 天开始。给出 n 个在线课程用 (t, d) 对表示。你的任务是找出最多可以修几门课。
+
+## 解题思路
+
+- 一般选课，任务的题目会涉及排序 + 贪心。此题同样如此。最多修几门课，采用贪心的思路。先将课程结束时间从小到大排序，优先选择结束时间靠前的课程，这样留给后面课程的时间越多，便可以修更多的课。对排好序的课程从前往后选课，不断累积时间。如果选择修当前课程，但是会超时，这时改调整了。对于已经选择的课程，都加入到最大堆中，遇到需要调整时，比较当前待考虑的课程时长是否比(堆中)已经选择课中时长最长的课时长短，即堆顶的课程时长短，剔除 pop 它，再选择这门时长短的课，并加入最大堆中。并更新累积时间。一层循环扫完所有课程，最终最大堆中包含课程的数目便是最多可以修的课程数。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+func scheduleCourse(courses [][]int) int {
+	sort.Slice(courses, func(i, j int) bool {
+		return courses[i][1] < courses[j][1]
+	})
+	maxHeap, time := &Schedule{}, 0
+	heap.Init(maxHeap)
+	for _, c := range courses {
+		if time+c[0] <= c[1] {
+			time += c[0]
+			heap.Push(maxHeap, c[0])
+		} else if (*maxHeap).Len() > 0 && (*maxHeap)[0] > c[0] {
+			time -= heap.Pop(maxHeap).(int) - c[0]
+			heap.Push(maxHeap, c[0])
+		}
+	}
+	return (*maxHeap).Len()
+}
+
+type Schedule []int
+
+func (s Schedule) Len() int           { return len(s) }
+func (s Schedule) Less(i, j int) bool { return s[i] > s[j] }
+func (s Schedule) Swap(i, j int)      { s[i], s[j] = s[j], s[i] }
+func (s *Schedule) Pop() interface{} {
+	n := len(*s)
+	t := (*s)[n-1]
+	*s = (*s)[:n-1]
+	return t
+}
+func (s *Schedule) Push(x interface{}) {
+	*s = append(*s, x.(int))
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0628.Maximum-Product-of-Three-Numbers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0632.Smallest-Range-Covering-Elements-from-K-Lists/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0632.Smallest-Range-Covering-Elements-from-K-Lists.md b/website/content/ChapterFour/0600~0699/0632.Smallest-Range-Covering-Elements-from-K-Lists.md
new file mode 100755
index 000000000..def61bb38
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0632.Smallest-Range-Covering-Elements-from-K-Lists.md
@@ -0,0 +1,120 @@
+# [632. Smallest Range Covering Elements from K Lists](https://leetcode.com/problems/smallest-range-covering-elements-from-k-lists/)
+
+
+## 题目
+
+You have `k` lists of sorted integers in ascending order. Find the **smallest** range that includes at least one number from each of the `k` lists.
+
+We define the range [a,b] is smaller than range [c,d] if `b-a < d-c` or `a < c` if `b-a == d-c`.
+
+**Example 1**:
+
+    Input: [[4,10,15,24,26], [0,9,12,20], [5,18,22,30]]
+    Output: [20,24]
+    Explanation: 
+    List 1: [4, 10, 15, 24,26], 24 is in range [20,24].
+    List 2: [0, 9, 12, 20], 20 is in range [20,24].
+    List 3: [5, 18, 22, 30], 22 is in range [20,24].
+
+**Note**:
+
+1. The given list may contain duplicates, so ascending order means >= here.
+2. 1 <= `k` <= 3500
+3. -10^5 <= `value of elements` <= 10^5.
+
+
+## 题目大意
+
+你有 k 个升序排列的整数数组。找到一个最小区间，使得 k 个列表中的每个列表至少有一个数包含在其中。
+
+我们定义如果 b-a < d-c 或者在 b-a == d-c 时 a < c，则区间 [a,b] 比 [c,d] 小。
+
+注意:
+
+- 给定的列表可能包含重复元素，所以在这里升序表示 >= 。
+- 1 <= k <= 3500
+- -105 <= 元素的值 <= 105
+- 对于使用Java的用户，请注意传入类型已修改为List<List<Integer>>。重置代码模板后可以看到这项改动。
+
+
+
+## 解题思路
+
+
+- 给出 K 个数组，要求在这 K 个数组中找到一个区间，至少能包含这 K 个数组中每个数组中的一个元素。
+- 这一题是第 76 题的变种版。第 76 题是用滑动窗口来解答的，它要求在母字符串 S 中找到最小的子串能包含 T 串的所有字母。这一题类似的，可以把母字符串看成 K 个数组合并起来的大数组，那么 T 串是由 K 个数组中每个数组中抽一个元素出来组成的。求的区间相同，都是能包含 T 的最小区间。另外一个区别在于，第 76 题里面都是字符串，这一题都是数字，在最终拼接成 T 串的时候需要保证 K 个数组中每个都有一个元素，所以理所当然的想到需要维护每个元素所在数组编号。经过上述的转换，可以把这道题转换成第 76 题的解法了。
+- 在具体解题过程中，用 map 来维护窗口内 K 个数组出现的频次。时间复杂度 O(n*log n)，空间复杂度是O(n)。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func smallestRange(nums [][]int) []int {
+	numList, left, right, count, freqMap, res, length := []element{}, 0, -1, 0, map[int]int{}, make([]int, 2), math.MaxInt64
+	for i, ns := range nums {
+		for _, v := range ns {
+			numList = append(numList, element{val: v, index: i})
+		}
+	}
+	sort.Sort(SortByVal{numList})
+	for left < len(numList) {
+		if right+1 < len(numList) && count < len(nums) {
+			right++
+			if freqMap[numList[right].index] == 0 {
+				count++
+			}
+			freqMap[numList[right].index]++
+		} else {
+			if count == len(nums) {
+				if numList[right].val-numList[left].val < length {
+					length = numList[right].val - numList[left].val
+					res[0] = numList[left].val
+					res[1] = numList[right].val
+				}
+			}
+			freqMap[numList[left].index]--
+			if freqMap[numList[left].index] == 0 {
+				count--
+			}
+			left++
+		}
+	}
+	return res
+}
+
+type element struct {
+	val   int
+	index int
+}
+
+type elements []element
+
+// Len define
+func (p elements) Len() int { return len(p) }
+
+// Swap define
+func (p elements) Swap(i, j int) { p[i], p[j] = p[j], p[i] }
+
+// SortByVal define
+type SortByVal struct{ elements }
+
+// Less define
+func (p SortByVal) Less(i, j int) bool {
+	return p.elements[i].val < p.elements[j].val
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0630.Course-Schedule-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0633.Sum-of-Square-Numbers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0633.Sum-of-Square-Numbers.md b/website/content/ChapterFour/0600~0699/0633.Sum-of-Square-Numbers.md
new file mode 100755
index 000000000..de723caef
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0633.Sum-of-Square-Numbers.md
@@ -0,0 +1,60 @@
+# [633. Sum of Square Numbers](https://leetcode.com/problems/sum-of-square-numbers/)
+
+
+## 题目
+
+Given a non-negative integer `c`, your task is to decide whether there're two integers `a` and `b` such that a^2 + b^2 = c.
+
+**Example 1**:
+
+    Input: 5
+    Output: True
+    Explanation: 1 * 1 + 2 * 2 = 5
+
+**Example 2**:
+
+    Input: 3
+    Output: False
+
+
+## 题目大意
+
+给定一个非负整数 c ，你要判断是否存在两个整数 a 和 b，使得 a^2 + b^2 = c。
+
+
+## 解题思路
+
+- 给出一个数，要求判断这个数能否由由 2 个完全平方数组成。能则输出 true，不能则输出 false。
+- 可以用二分搜索来解答这道题。判断题意，依次计算 `low * low + high * high`  和 c 是否相等。从 [0, sqrt(n)] 区间内进行二分，若能找到则返回 true，找不到就返回 false 。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func judgeSquareSum(c int) bool {
+	low, high := 0, int(math.Sqrt(float64(c)))
+	for low <= high {
+		if low*low+high*high < c {
+			low++
+		} else if low*low+high*high > c {
+			high--
+		} else {
+			return true
+		}
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0632.Smallest-Range-Covering-Elements-from-K-Lists/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0636.Exclusive-Time-of-Functions/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0636.Exclusive-Time-of-Functions.md b/website/content/ChapterFour/0600~0699/0636.Exclusive-Time-of-Functions.md
new file mode 100755
index 000000000..ebe7628c7
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0636.Exclusive-Time-of-Functions.md
@@ -0,0 +1,107 @@
+# [636. Exclusive Time of Functions](https://leetcode.com/problems/exclusive-time-of-functions/)
+
+
+## 题目
+
+On a single threaded CPU, we execute some functions. Each function has a unique id between `0` and `N-1`.
+
+We store logs in timestamp order that describe when a function is entered or exited.
+
+Each log is a string with this format: `"{function_id}:{"start" | "end"}:{timestamp}"`. For example, `"0:start:3"` means the function with id `0` started at the beginning of timestamp `3`. `"1:end:2"` means the function with id `1` ended at the end of timestamp `2`.
+
+A function's *exclusive time* is the number of units of time spent in this function. Note that this does not include any recursive calls to child functions.
+
+Return the exclusive time of each function, sorted by their function id.
+
+**Example 1**:
+
+![](https://assets.leetcode.com/uploads/2019/04/05/diag1b.png)
+
+    Input:
+    n = 2
+    logs = ["0:start:0","1:start:2","1:end:5","0:end:6"]
+    Output: [3, 4]
+    Explanation:
+    Function 0 starts at the beginning of time 0, then it executes 2 units of time and reaches the end of time 1.
+    Now function 1 starts at the beginning of time 2, executes 4 units of time and ends at time 5.
+    Function 0 is running again at the beginning of time 6, and also ends at the end of time 6, thus executing for 1 unit of time. 
+    So function 0 spends 2 + 1 = 3 units of total time executing, and function 1 spends 4 units of total time executing.
+
+**Note**:
+
+1. `1 <= n <= 100`
+2. Two functions won't start or end at the same time.
+3. Functions will always log when they exit.
+
+
+
+## 题目大意
+
+给出一个非抢占单线程CPU的 n 个函数运行日志，找到函数的独占时间。每个函数都有一个唯一的 Id，从 0 到 n-1，函数可能会递归调用或者被其他函数调用。
+
+日志是具有以下格式的字符串：function_id：start_or_end：timestamp。例如："0:start:0" 表示函数 0 从 0 时刻开始运行。"0:end:0" 表示函数 0 在 0 时刻结束。
+
+函数的独占时间定义是在该方法中花费的时间，调用其他函数花费的时间不算该函数的独占时间。你需要根据函数的 Id 有序地返回每个函数的独占时间。
+
+
+## 解题思路
+
+
+- 利用栈记录每一个开始了但是未完成的任务，完成以后任务就 pop 一个。
+- 注意题目中关于任务时长的定义，例如，start 7，end 7，这个任务执行了 1 秒而不是 0 秒
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+)
+
+type log struct {
+	id    int
+	order string
+	time  int
+}
+
+func exclusiveTime(n int, logs []string) []int {
+	res, lastLog, stack := make([]int, n), log{id: -1, order: "", time: 0}, []log{}
+	for i := 0; i < len(logs); i++ {
+		a := strings.Split(logs[i], ":")
+		id, _ := strconv.Atoi(a[0])
+		time, _ := strconv.Atoi(a[2])
+
+		if (lastLog.order == "start" && a[1] == "start") || (lastLog.order == "start" && a[1] == "end") {
+			res[lastLog.id] += time - lastLog.time
+			if a[1] == "end" {
+				res[lastLog.id]++
+			}
+		}
+		if lastLog.order == "end" && a[1] == "end" {
+			res[id] += time - lastLog.time
+		}
+		if lastLog.order == "end" && a[1] == "start" && len(stack) != 0 {
+			res[stack[len(stack)-1].id] += time - lastLog.time - 1
+		}
+		if a[1] == "start" {
+			stack = append(stack, log{id: id, order: a[1], time: time})
+		} else {
+			stack = stack[:len(stack)-1]
+		}
+		lastLog = log{id: id, order: a[1], time: time}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0633.Sum-of-Square-Numbers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0637.Average-of-Levels-in-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0637.Average-of-Levels-in-Binary-Tree.md b/website/content/ChapterFour/0600~0699/0637.Average-of-Levels-in-Binary-Tree.md
new file mode 100644
index 000000000..256e1e0b5
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0637.Average-of-Levels-in-Binary-Tree.md
@@ -0,0 +1,92 @@
+# [637. Average of Levels in Binary Tree](https://leetcode.com/problems/average-of-levels-in-binary-tree/)
+
+## 题目
+
+Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.
+
+**Example 1**:
+
+```
+
+Input:
+    3
+   / \
+  9  20
+    /  \
+   15   7
+Output: [3, 14.5, 11]
+Explanation:
+The average value of nodes on level 0 is 3,  on level 1 is 14.5, and on level 2 is 11. Hence return [3, 14.5, 11].
+
+```
+
+**Note**:  
+
+The range of node's value is in the range of 32-bit signed integer.
+ 
+
+## 题目大意
+
+按层序从上到下遍历一颗树，计算每一层的平均值。
+
+
+## 解题思路
+
+- 用一个队列即可实现。
+- 第 102 题和第 107 题都是按层序遍历的。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func averageOfLevels(root *TreeNode) []float64 {
+	if root == nil {
+		return []float64{0}
+	}
+	queue := []*TreeNode{}
+	queue = append(queue, root)
+	curNum, nextLevelNum, res, count, sum := 1, 0, []float64{}, 1, 0
+	for len(queue) != 0 {
+		if curNum > 0 {
+			node := queue[0]
+			if node.Left != nil {
+				queue = append(queue, node.Left)
+				nextLevelNum++
+			}
+			if node.Right != nil {
+				queue = append(queue, node.Right)
+				nextLevelNum++
+			}
+			curNum--
+			sum += node.Val
+			queue = queue[1:]
+		}
+		if curNum == 0 {
+			res = append(res, float64(sum)/float64(count))
+			curNum, count, nextLevelNum, sum = nextLevelNum, nextLevelNum, 0, 0
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0636.Exclusive-Time-of-Functions/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0638.Shopping-Offers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0638.Shopping-Offers.md b/website/content/ChapterFour/0600~0699/0638.Shopping-Offers.md
new file mode 100644
index 000000000..292f03ddf
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0638.Shopping-Offers.md
@@ -0,0 +1,136 @@
+# [638. Shopping Offers](https://leetcode.com/problems/shopping-offers/)
+
+
+
+## 题目
+
+In LeetCode Store, there are some kinds of items to sell. Each item has a price.
+
+However, there are some special offers, and a special offer consists of one or more different kinds of items with a sale price.
+
+You are given the each item's price, a set of special offers, and the number we need to buy for each item. The job is to output the lowest price you have to pay for **exactly** certain items as given, where you could make optimal use of the special offers.
+
+Each special offer is represented in the form of an array, the last number represents the price you need to pay for this special offer, other numbers represents how many specific items you could get if you buy this offer.
+
+You could use any of special offers as many times as you want.
+
+**Example 1**:
+
+```
+Input: [2,5], [[3,0,5],[1,2,10]], [3,2]
+Output: 14
+Explanation: 
+There are two kinds of items, A and B. Their prices are $2 and $5 respectively. 
+In special offer 1, you can pay $5 for 3A and 0B
+In special offer 2, you can pay $10 for 1A and 2B. 
+You need to buy 3A and 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A.
+```
+
+**Example 2**:
+
+```
+Input: [2,3,4], [[1,1,0,4],[2,2,1,9]], [1,2,1]
+Output: 11
+Explanation: 
+The price of A is $2, and $3 for B, $4 for C. 
+You may pay $4 for 1A and 1B, and $9 for 2A ,2B and 1C. 
+You need to buy 1A ,2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, $4 for 1C. 
+You cannot add more items, though only $9 for 2A ,2B and 1C.
+```
+
+**Note**:
+
+1. There are at most 6 kinds of items, 100 special offers.
+2. For each item, you need to buy at most 6 of them.
+3. You are **not** allowed to buy more items than you want, even if that would lower the overall price.
+
+
+## 题目大意
+
+在 LeetCode 商店中， 有许多在售的物品。然而，也有一些大礼包，每个大礼包以优惠的价格捆绑销售一组物品。
+
+现给定每个物品的价格，每个大礼包包含物品的清单，以及待购物品清单。请输出确切完成待购清单的最低花费。每个大礼包的由一个数组中的一组数据描述，最后一个数字代表大礼包的价格，其他数字分别表示内含的其他种类物品的数量。任意大礼包可无限次购买。
+
+例子 1:
+
+```
+输入: [2,5], [[3,0,5],[1,2,10]], [3,2]
+输出: 14
+解释: 
+有A和B两种物品，价格分别为¥2和¥5。
+大礼包1，你可以以¥5的价格购买3A和0B。
+大礼包2， 你可以以¥10的价格购买1A和2B。
+你需要购买3个A和2个B， 所以你付了¥10购买了1A和2B（大礼包2），以及¥4购买2A。
+
+```
+例子 2:
+
+```
+输入: [2,3,4], [[1,1,0,4],[2,2,1,9]], [1,2,1]
+输出: 11
+解释: 
+A，B，C的价格分别为¥2，¥3，¥4.
+你可以用¥4购买1A和1B，也可以用¥9购买2A，2B和1C。
+你需要买1A，2B和1C，所以你付了¥4买了1A和1B（大礼包1），以及¥3购买1B， ¥4购买1C。
+你不可以购买超出待购清单的物品，尽管购买大礼包2更加便宜。
+```
+
+说明:
+
+- 最多6种物品， 100种大礼包。
+- 每种物品，你最多只需要购买6个。
+- 你不可以购买超出待购清单的物品，即使更便宜。
+
+
+## 解题思路
+
+- 给出 3 个数组，3 个数组分别代表的意义是在售的商品价格，多个礼包以及礼包内每个商品的数量和总价，购物清单上需要购买每个商品的数量。问购买清单上的所有商品所需的最低花费。
+- 这一题可以用 DFS 暴力解题，也可以用 DP。笔者这题先用 DFS 来解答。设当前搜索到的状态为 `shopping(price, special, needs)`，其中 `price` 和 `special` 为题目中所述的物品的单价和捆绑销售的大礼包，而 `needs` 为当前需要的每种物品的数量。针对于每个商品，可以有 3 种购买规则，第一种，选礼包里面的第一个优惠购买，第二种，不选当前礼包优惠，选下一个优惠进行购买，第三种，不使用优惠，直接购买。这样就对应了 3 种 DFS 的方向。具体见代码。如果选择了礼包优惠，那么递归到下一层，`need` 需要对应减少礼包里面的数量，最终金额累加。当所有情况遍历完以后，可以返回出最小花费。
+- 这一题需要注意的剪枝情况：是否需要购买礼包。题目中要求了，不能购买超过清单上数量的商品，即使价格便宜，也不行。例如可以买 n 个礼包 A，但是最终商品数量超过了清单上的商品，这种购买方式是不行的。所以需要先判断当前递归中，满足 `need` 和 `price` 条件的，能否使用礼包。这里包含 2 种情况，一种是当前商品已经满足清单个数了，不需要再买了；还有一种情况是已经超过清单数量了，那这种情况需要立即返回，当前这种购买方式不合题意。
+
+## 代码
+
+```go
+func shoppingOffers(price []int, special [][]int, needs []int) int {
+	res := -1
+	dfsShoppingOffers(price, special, needs, 0, &res)
+	return res
+}
+
+func dfsShoppingOffers(price []int, special [][]int, needs []int, pay int, res *int) {
+	noNeeds := true
+	// 剪枝
+	for _, need := range needs {
+		if need < 0 {
+			return
+		}
+		if need != 0 {
+			noNeeds = false
+		}
+	}
+	if len(special) == 0 || noNeeds {
+		for i, p := range price {
+			pay += (p * needs[i])
+		}
+		if pay < *res || *res == -1 {
+			*res = pay
+		}
+		return
+	}
+	newNeeds := make([]int, len(needs))
+	copy(newNeeds, needs)
+	for i, n := range newNeeds {
+		newNeeds[i] = n - special[0][i]
+	}
+	dfsShoppingOffers(price, special, newNeeds, pay+special[0][len(price)], res)
+	dfsShoppingOffers(price, special[1:], newNeeds, pay+special[0][len(price)], res)
+	dfsShoppingOffers(price, special[1:], needs, pay, res)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0637.Average-of-Levels-in-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0643.Maximum-Average-Subarray-I/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0643.Maximum-Average-Subarray-I.md b/website/content/ChapterFour/0600~0699/0643.Maximum-Average-Subarray-I.md
new file mode 100644
index 000000000..63ec049f0
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0643.Maximum-Average-Subarray-I.md
@@ -0,0 +1,59 @@
+# [643. Maximum Average Subarray I](https://leetcode.com/problems/maximum-average-subarray-i/)
+
+## 题目
+
+Given an array consisting of `n` integers, find the contiguous subarray of given length `k` that has the maximum average value. And you need to output the maximum average value.
+
+**Example 1:**
+
+```
+Input: [1,12,-5,-6,50,3], k = 4
+Output: 12.75
+Explanation: Maximum average is (12-5-6+50)/4 = 51/4 = 12.75
+```
+
+**Note:**
+
+1. 1 <= `k` <= `n` <= 30,000.
+2. Elements of the given array will be in the range [-10,000, 10,000].
+
+## 题目大意
+
+给定 n 个整数，找出平均数最大且长度为 k 的连续子数组，并输出该最大平均数。
+
+## 解题思路
+
+- 简单题。循环一次，扫描数组过程中累加窗口大小为 k 的元素值。不断更新这个最大值。循环结束求出平均值即可。
+
+## 代码
+
+```go
+package leetcode
+
+func findMaxAverage(nums []int, k int) float64 {
+	sum := 0
+	for _, v := range nums[:k] {
+		sum += v
+	}
+	maxSum := sum
+	for i := k; i < len(nums); i++ {
+		sum = sum - nums[i-k] + nums[i]
+		maxSum = max(maxSum, sum)
+	}
+	return float64(maxSum) / float64(k)
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0638.Shopping-Offers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0645.Set-Mismatch/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0645.Set-Mismatch.md b/website/content/ChapterFour/0600~0699/0645.Set-Mismatch.md
new file mode 100755
index 000000000..93d481833
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0645.Set-Mismatch.md
@@ -0,0 +1,69 @@
+# [645. Set Mismatch](https://leetcode.com/problems/set-mismatch/)
+
+
+## 题目
+
+The set `S` originally contains numbers from 1 to `n`. But unfortunately, due to the data error, one of the numbers in the set got duplicated to **another** number in the set, which results in repetition of one number and loss of another number.
+
+Given an array `nums` representing the data status of this set after the error. Your task is to firstly find the number occurs twice and then find the number that is missing. Return them in the form of an array.
+
+**Example 1**:
+
+    Input: nums = [1,2,2,4]
+    Output: [2,3]
+
+**Note**:
+
+1. The given array size will in the range [2, 10000].
+2. The given array's numbers won't have any order.
+
+
+## 题目大意
+
+
+集合 S 包含从1到 n 的整数。不幸的是，因为数据错误，导致集合里面某一个元素复制了成了集合里面的另外一个元素的值，导致集合丢失了一个整数并且有一个元素重复。给定一个数组 nums 代表了集合 S 发生错误后的结果。你的任务是首先寻找到重复出现的整数，再找到丢失的整数，将它们以数组的形式返回。
+
+注意:  
+
+- 给定数组的长度范围是 [2, 10000]。
+- 给定的数组是无序的。
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面装的是 1-n 的数字，由于错误导致有一个数字变成了另外一个数字，要求找出重复的一个数字和正确的数字。这一题是简单题，根据下标比对就可以找到哪个数字重复了，哪个数字缺少了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func findErrorNums(nums []int) []int {
+	m, res := make([]int, len(nums)), make([]int, 2)
+	for _, n := range nums {
+		if m[n-1] == 0 {
+			m[n-1] = 1
+		} else {
+			res[0] = n
+		}
+	}
+	for i := range m {
+		if m[i] == 0 {
+			res[1] = i + 1
+			break
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0643.Maximum-Average-Subarray-I/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0647.Palindromic-Substrings/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0647.Palindromic-Substrings.md b/website/content/ChapterFour/0600~0699/0647.Palindromic-Substrings.md
new file mode 100644
index 000000000..10eadcfae
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0647.Palindromic-Substrings.md
@@ -0,0 +1,71 @@
+# [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)
+
+
+## 题目
+
+Given a string, your task is to count how many palindromic substrings in this string.
+
+The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.
+
+**Example 1:**
+
+```
+Input: "abc"
+Output: 3
+Explanation: Three palindromic strings: "a", "b", "c".
+```
+
+**Example 2:**
+
+```
+Input: "aaa"
+Output: 6
+Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
+```
+
+**Note:**
+
+1. The input string length won't exceed 1000.
+
+## 题目大意
+
+给定一个字符串，你的任务是计算这个字符串中有多少个回文子串。具有不同开始位置或结束位置的子串，即使是由相同的字符组成，也会被视作不同的子串。
+
+## 解题思路
+
+- 暴力解法，从左往右扫一遍字符串，以每个字符做轴，用中心扩散法，依次遍历计数回文子串。
+
+## 代码
+
+```go
+package leetcode
+
+func countSubstrings(s string) int {
+	res := 0
+	for i := 0; i < len(s); i++ {
+		res += countPalindrome(s, i, i)
+		res += countPalindrome(s, i, i+1)
+	}
+	return res
+}
+
+func countPalindrome(s string, left, right int) int {
+	res := 0
+	for left >= 0 && right < len(s) {
+		if s[left] != s[right] {
+			break
+		}
+		left--
+		right++
+		res++
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0645.Set-Mismatch/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0648.Replace-Words/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0648.Replace-Words.md b/website/content/ChapterFour/0600~0699/0648.Replace-Words.md
new file mode 100755
index 000000000..e16a17162
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0648.Replace-Words.md
@@ -0,0 +1,118 @@
+# [648. Replace Words](https://leetcode.com/problems/replace-words/)
+
+
+## 题目
+
+In English, we have a concept called `root`, which can be followed by some other words to form another longer word - let's call this word `successor`. For example, the root `an`, followed by `other`, which can form another word `another`.
+
+Now, given a dictionary consisting of many roots and a sentence. You need to replace all the `successor` in the sentence with the `root` forming it. If a `successor` has many `roots` can form it, replace it with the root with the shortest length.
+
+You need to output the sentence after the replacement.
+
+**Example 1**:
+
+    Input: dict = ["cat", "bat", "rat"]
+    sentence = "the cattle was rattled by the battery"
+    Output: "the cat was rat by the bat"
+
+**Note**:
+
+1. The input will only have lower-case letters.
+2. 1 <= dict words number <= 1000
+3. 1 <= sentence words number <= 1000
+4. 1 <= root length <= 100
+5. 1 <= sentence words length <= 1000
+
+
+## 题目大意
+
+在英语中，我们有一个叫做 词根(root)的概念，它可以跟着其他一些词组成另一个较长的单词——我们称这个词为 继承词(successor)。例如，词根an，跟随着单词 other(其他)，可以形成新的单词 another(另一个)。
+
+现在，给定一个由许多词根组成的词典和一个句子。你需要将句子中的所有继承词用词根替换掉。如果继承词有许多可以形成它的词根，则用最短的词根替换它。要求输出替换之后的句子。
+
+
+
+## 解题思路
+
+
+- 给出一个句子和一个可替换字符串的数组，如果句子中的单词和可替换列表里面的单词，有相同的首字母，那么就把句子中的单词替换成可替换列表里面的单词。输入最后替换完成的句子。
+- 这一题有 2 种解题思路，第一种就是单纯的用 Map 查找。第二种是用 Trie 去替换。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+// 解法一 哈希表
+func replaceWords(dict []string, sentence string) string {
+	roots := make(map[byte][]string)
+	for _, root := range dict {
+		b := root[0]
+		roots[b] = append(roots[b], root)
+	}
+	words := strings.Split(sentence, " ")
+	for i, word := range words {
+		b := []byte(word)
+		for j := 1; j < len(b) && j <= 100; j++ {
+			if findWord(roots, b[0:j]) {
+				words[i] = string(b[0:j])
+				break
+			}
+		}
+	}
+	return strings.Join(words, " ")
+}
+
+func findWord(roots map[byte][]string, word []byte) bool {
+	if roots[word[0]] == nil {
+		return false
+	}
+	for _, root := range roots[word[0]] {
+		if root == string(word) {
+			return true
+		}
+	}
+	return false
+}
+
+//解法二 Trie
+func replaceWords1(dict []string, sentence string) string {
+	trie := Constructor208()
+	for _, v := range dict {
+		trie.Insert(v)
+	}
+	words := strings.Split(sentence, " ")
+	var result []string
+	word := ""
+	i := 0
+	for _, value := range words {
+		word = ""
+		for i = 1; i < len(value); i++ {
+			if trie.Search(value[:i]) {
+				word = value[:i]
+				break
+			}
+		}
+
+		if len(word) == 0 {
+			result = append(result, value)
+		} else {
+			result = append(result, word)
+		}
+
+	}
+	return strings.Join(result, " ")
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0647.Palindromic-Substrings/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0653.Two-Sum-IV-Input-is-a-BST/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0653.Two-Sum-IV-Input-is-a-BST.md b/website/content/ChapterFour/0600~0699/0653.Two-Sum-IV-Input-is-a-BST.md
new file mode 100755
index 000000000..4d13db566
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0653.Two-Sum-IV-Input-is-a-BST.md
@@ -0,0 +1,82 @@
+# [653. Two Sum IV - Input is a BST](https://leetcode.com/problems/two-sum-iv-input-is-a-bst/)
+
+## 题目
+
+Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.
+
+**Example 1**:
+
+    Input: 
+        5
+       / \
+      3   6
+     / \   \
+    2   4   7
+    
+    Target = 9
+    
+    Output: True
+
+**Example 2**:
+
+    Input: 
+        5
+       / \
+      3   6
+     / \   \
+    2   4   7
+    
+    Target = 28
+    
+    Output: False
+
+
+## 题目大意
+
+给定一个二叉搜索树和一个目标结果，如果 BST 中存在两个元素且它们的和等于给定的目标结果，则返回 true。
+
+## 解题思路
+
+
+- 在树中判断是否存在 2 个数的和是 sum。
+- 这一题是 two sum 问题的变形题，只不过题目背景是在 BST 上处理的。处理思路大体一致，用 map 记录已经访问过的节点值。边遍历树边查看 map 里面是否有 sum 的另外一半。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func findTarget(root *TreeNode, k int) bool {
+	m := make(map[int]int, 0)
+	return findTargetDFS(root, k, m)
+}
+
+func findTargetDFS(root *TreeNode, k int, m map[int]int) bool {
+	if root == nil {
+		return false
+	}
+	if _, ok := m[k-root.Val]; ok {
+		return ok
+	}
+	m[root.Val]++
+	return findTargetDFS(root.Left, k, m) || findTargetDFS(root.Right, k, m)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0648.Replace-Words/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0658.Find-K-Closest-Elements/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0658.Find-K-Closest-Elements.md b/website/content/ChapterFour/0600~0699/0658.Find-K-Closest-Elements.md
new file mode 100755
index 000000000..4fe0f821a
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0658.Find-K-Closest-Elements.md
@@ -0,0 +1,89 @@
+# [658. Find K Closest Elements](https://leetcode.com/problems/find-k-closest-elements/)
+
+
+## 题目
+
+Given a sorted array, two integers `k` and `x`, find the `k` closest elements to `x` in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred.
+
+**Example 1**:
+
+    Input: [1,2,3,4,5], k=4, x=3
+    Output: [1,2,3,4]
+
+**Example 2**:
+
+    Input: [1,2,3,4,5], k=4, x=-1
+    Output: [1,2,3,4]
+
+**Note**:
+
+1. The value k is positive and will always be smaller than the length of the sorted array.
+2. Length of the given array is positive and will not exceed 10^4
+3. Absolute value of elements in the array and x will not exceed 10^4
+
+---
+
+**UPDATE (2017/9/19)**: The arr parameter had been changed to an **array of integers** (instead of a list of integers). **Please reload the code definition to get the latest changes**.
+
+
+## 题目大意
+
+
+给定一个排序好的数组，两个整数 k 和 x，从数组中找到最靠近 x（两数之差最小）的 k 个数。返回的结果必须要是按升序排好的。如果有两个数与 x 的差值一样，优先选择数值较小的那个数。
+
+
+说明:
+
+1. k 的值为正数，且总是小于给定排序数组的长度。
+2. 数组不为空，且长度不超过 104
+3. 数组里的每个元素与 x 的绝对值不超过 104
+ 
+
+更新(2017/9/19):
+这个参数 arr 已经被改变为一个整数数组（而不是整数列表）。 请重新加载代码定义以获取最新更改。
+
+
+
+
+## 解题思路
+
+
+- 给出一个数组，要求在数组中找到一个长度为 k 的区间，这个区间内每个元素距离 x 的距离都是整个数组里面最小的。
+- 这一题可以用双指针解题，最优解法是二分搜索。由于区间长度固定是 K 个，所以左区间最大只能到 `len(arr) - K` (因为长度为 K 以后，正好右区间就到数组最右边了)，在 `[0,len(arr) - K]` 这个区间中进行二分搜索。如果发现 `a[mid]` 与 `x` 距离比 `a[mid + k]` 与 `x` 的距离要大，说明要找的区间一定在右侧，继续二分，直到最终 `low = high` 的时候退出。逼出的 `low` 值就是最终答案区间的左边界。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+// 解法一 库函数二分搜索
+func findClosestElements(arr []int, k int, x int) []int {
+	return arr[sort.Search(len(arr)-k, func(i int) bool { return x-arr[i] <= arr[i+k]-x }):][:k]
+}
+
+// 解法二 手撸二分搜索
+func findClosestElements1(arr []int, k int, x int) []int {
+	low, high := 0, len(arr)-k
+	for low < high {
+		mid := low + (high-low)>>1
+		if x-arr[mid] > arr[mid+k]-x {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return arr[low : low+k]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0653.Two-Sum-IV-Input-is-a-BST/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0661.Image-Smoother/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0661.Image-Smoother.md b/website/content/ChapterFour/0600~0699/0661.Image-Smoother.md
new file mode 100644
index 000000000..048d5da9d
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0661.Image-Smoother.md
@@ -0,0 +1,118 @@
+# [661. Image Smoother](https://leetcode.com/problems/image-smoother/)
+
+## 题目
+
+Given a 2D integer matrix M representing the gray scale of an image, you need to design a smoother to make the gray scale of each cell becomes the average gray scale (rounding down) of all the 8 surrounding cells and itself. If a cell has less than 8 surrounding cells, then use as many as you can.
+
+**Example 1**:
+
+```
+Input:
+[[1,1,1],
+ [1,0,1],
+ [1,1,1]]
+Output:
+[[0, 0, 0],
+ [0, 0, 0],
+ [0, 0, 0]]
+Explanation:
+For the point (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0
+For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0
+For the point (1,1): floor(8/9) = floor(0.88888889) = 0
+
+```
+
+**Note**:
+
+1. The value in the given matrix is in the range of [0, 255].
+2. The length and width of the given matrix are in the range of [1, 150].
+
+
+## 题目大意
+
+包含整数的二维矩阵 M 表示一个图片的灰度。你需要设计一个平滑器来让每一个单元的灰度成为平均灰度 (向下舍入) ，平均灰度的计算是周围的8个单元和它本身的值求平均，如果周围的单元格不足八个，则尽可能多的利用它们。
+
+注意:
+
+- 给定矩阵中的整数范围为 [0, 255]。
+- 矩阵的长和宽的范围均为 [1, 150]。
+
+
+## 解题思路
+
+- 将二维数组中的每个元素变为周围 9 个元素的平均值。
+- 简单题，按照题意计算平均值即可。需要注意的是边界问题，四个角和边上的元素，这些点计算平均值的时候，计算平均值都不足 9 个元素。
+
+## 代码
+
+```go
+
+package leetcode
+
+func imageSmoother(M [][]int) [][]int {
+	res := make([][]int, len(M))
+	for i := range M {
+		res[i] = make([]int, len(M[0]))
+	}
+	for y := 0; y < len(M); y++ {
+		for x := 0; x < len(M[0]); x++ {
+			res[y][x] = smooth(x, y, M)
+		}
+	}
+	return res
+}
+
+func smooth(x, y int, M [][]int) int {
+	count, sum := 1, M[y][x]
+	// Check bottom
+	if y+1 < len(M) {
+		sum += M[y+1][x]
+		count++
+	}
+	// Check Top
+	if y-1 >= 0 {
+		sum += M[y-1][x]
+		count++
+	}
+	// Check left
+	if x-1 >= 0 {
+		sum += M[y][x-1]
+		count++
+	}
+	// Check Right
+	if x+1 < len(M[y]) {
+		sum += M[y][x+1]
+		count++
+	}
+	// Check Coners
+	// Top Left
+	if y-1 >= 0 && x-1 >= 0 {
+		sum += M[y-1][x-1]
+		count++
+	}
+	// Top Right
+	if y-1 >= 0 && x+1 < len(M[0]) {
+		sum += M[y-1][x+1]
+		count++
+	}
+	// Bottom Left
+	if y+1 < len(M) && x-1 >= 0 {
+		sum += M[y+1][x-1]
+		count++
+	}
+	//Bottom Right
+	if y+1 < len(M) && x+1 < len(M[0]) {
+		sum += M[y+1][x+1]
+		count++
+	}
+	return sum / count
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0658.Find-K-Closest-Elements/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0662.Maximum-Width-of-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0662.Maximum-Width-of-Binary-Tree.md b/website/content/ChapterFour/0600~0699/0662.Maximum-Width-of-Binary-Tree.md
new file mode 100755
index 000000000..d7257d840
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0662.Maximum-Width-of-Binary-Tree.md
@@ -0,0 +1,159 @@
+# [662. Maximum Width of Binary Tree](https://leetcode.com/problems/maximum-width-of-binary-tree/)
+
+
+## 题目
+
+Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a **full binary tree**, but some nodes are null.
+
+The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the `null` nodes between the end-nodes are also counted into the length calculation.
+
+**Example 1**:
+
+    Input: 
+    
+               1
+             /   \
+            3     2
+           / \     \  
+          5   3     9 
+    
+    Output: 4
+    Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).
+
+**Example 2**:
+
+    Input: 
+    
+              1
+             /  
+            3    
+           / \       
+          5   3     
+    
+    Output: 2
+    Explanation: The maximum width existing in the third level with the length 2 (5,3).
+
+**Example 3**:
+
+    Input: 
+    
+              1
+             / \
+            3   2 
+           /        
+          5      
+    
+    Output: 2
+    Explanation: The maximum width existing in the second level with the length 2 (3,2).
+
+**Example 4**:
+
+    Input: 
+    
+              1
+             / \
+            3   2
+           /     \  
+          5       9 
+         /         \
+        6           7
+    Output: 8
+    Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).
+
+**Note**: Answer will in the range of 32-bit signed integer.
+
+
+## 题目大意
+
+给定一个二叉树，编写一个函数来获取这个树的最大宽度。树的宽度是所有层中的最大宽度。这个二叉树与满二叉树（full binary tree）结构相同，但一些节点为空。
+
+每一层的宽度被定义为两个端点（该层最左和最右的非空节点，两端点间的null节点也计入长度）之间的长度。
+
+注意: 答案在32位有符号整数的表示范围内。
+
+
+
+## 解题思路
+
+
+- 给出一个二叉树，求这棵树最宽的部分。
+- 这一题可以用 BFS 也可以用 DFS，但是用 BFS 比较方便。按照层序遍历，依次算出每层最左边不为 `null` 的节点和最右边不为 `null` 的节点。这两个节点之间都是算宽度的。最终输出最大的宽度即可。此题的关键在于如何有效的找到每一层的左右边界。
+- 这一题可能有人会想着先补全满二叉树，然后每层分别找左右边界。这种方法提交以后会卡在 `104 / 108` 这组测试用例上，这组测试用例会使得最后某几层填充出现的满二叉树节点特别多，最终导致 `Memory Limit Exceeded` 了。
+- 由于此题要找每层的左右边界，实际上每个节点的 `Val` 值是我们不关心的，那么可以把这个值用来标号，标记成该节点在每层中的序号。父亲节点在上一层中的序号是 x，那么它的左孩子在下一层满二叉树中的序号是 `2*x`，它的右孩子在下一层满二叉树中的序号是 `2*x + 1`。将所有节点都标上号，用 BFS 层序遍历每一层，每一层都找到左右边界，相减拿到宽度，动态维护最大宽度，就是本题的最终答案。
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func widthOfBinaryTree(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	if root.Left == nil && root.Right == nil {
+		return 1
+	}
+
+	queue, res := []*TreeNode{}, 0
+	queue = append(queue, &TreeNode{0, root.Left, root.Right})
+
+	for len(queue) != 0 {
+		var left, right *int
+		// 这里需要注意，先保存 queue 的个数，相当于拿到此层的总个数
+		qLen := len(queue)
+		// 这里循环不要写 i < len(queue)，因为每次循环 queue 的长度都在变小
+		for i := 0; i < qLen; i++ {
+			node := queue[0]
+			queue = queue[1:]
+			if node.Left != nil {
+				// 根据满二叉树父子节点的关系，得到下一层节点在本层的编号
+				newVal := node.Val * 2
+				queue = append(queue, &TreeNode{newVal, node.Left.Left, node.Left.Right})
+				if left == nil || *left > newVal {
+					left = &newVal
+				}
+				if right == nil || *right < newVal {
+					right = &newVal
+				}
+			}
+			if node.Right != nil {
+				// 根据满二叉树父子节点的关系，得到下一层节点在本层的编号
+				newVal := node.Val*2 + 1
+				queue = append(queue, &TreeNode{newVal, node.Right.Left, node.Right.Right})
+				if left == nil || *left > newVal {
+					left = &newVal
+				}
+				if right == nil || *right < newVal {
+					right = &newVal
+				}
+			}
+		}
+		switch {
+		// 某层只有一个点，那么此层宽度为 1
+		case left != nil && right == nil, left == nil && right != nil:
+			res = max(res, 1)
+		// 某层只有两个点，那么此层宽度为两点之间的距离
+		case left != nil && right != nil:
+			res = max(res, *right-*left+1)
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0661.Image-Smoother/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0665.Non-decreasing-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0665.Non-decreasing-Array.md b/website/content/ChapterFour/0600~0699/0665.Non-decreasing-Array.md
new file mode 100644
index 000000000..a698b301f
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0665.Non-decreasing-Array.md
@@ -0,0 +1,66 @@
+# [665. Non-decreasing Array](https://leetcode.com/problems/non-decreasing-array/)
+
+## 题目
+
+Given an array `nums` with `n` integers, your task is to check if it could become non-decreasing by modifying **at most one element**.
+
+We define an array is non-decreasing if `nums[i] <= nums[i + 1]` holds for every `i` (**0-based**) such that (`0 <= i <= n - 2`).
+
+**Example 1:**
+
+```
+Input: nums = [4,2,3]
+Output: true
+Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
+```
+
+**Example 2:**
+
+```
+Input: nums = [4,2,1]
+Output: false
+Explanation: You can't get a non-decreasing array by modify at most one element.
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `1 <= n <= 104`
+- `-10^5 <= nums[i] <= 10^5`
+
+## 题目大意
+
+给你一个长度为 n 的整数数组，请你判断在 最多 改变 1 个元素的情况下，该数组能否变成一个非递减数列。我们是这样定义一个非递减数列的： 对于数组中任意的 i (0 <= i <= n-2)，总满足 nums[i] <= nums[i + 1]。
+
+## 解题思路
+
+- 简单题。循环扫描数组，找到 `nums[i] > nums[i+1]` 这种递减组合。一旦这种组合超过 2 组，直接返回 false。找到第一组递减组合，需要手动调节一次。如果 `nums[i + 1] < nums[i - 1]`，就算交换 `nums[i+1]` 和 `nums[i]`，交换结束，`nums[i - 1]` 仍然可能大于 `nums[i + 1]`，不满足题意。正确的做法应该是让较小的那个数变大，即 `nums[i + 1] = nums[i]`。两个元素相等满足非递减的要求。
+
+## 代码
+
+```go
+package leetcode
+
+func checkPossibility(nums []int) bool {
+	count := 0
+	for i := 0; i < len(nums)-1; i++ {
+		if nums[i] > nums[i+1] {
+			count++
+			if count > 1 {
+				return false
+			}
+			if i > 0 && nums[i+1] < nums[i-1] {
+				nums[i+1] = nums[i]
+			}
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0662.Maximum-Width-of-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0667.Beautiful-Arrangement-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0667.Beautiful-Arrangement-II.md b/website/content/ChapterFour/0600~0699/0667.Beautiful-Arrangement-II.md
new file mode 100644
index 000000000..7417a3ec8
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0667.Beautiful-Arrangement-II.md
@@ -0,0 +1,69 @@
+# [667. Beautiful Arrangement II](https://leetcode.com/problems/beautiful-arrangement-ii/)
+
+
+## 题目
+
+Given two integers `n` and `k`, you need to construct a list which contains `n` different positive integers ranging from `1` to `n` and obeys the following requirement:Suppose this list is [a1, a2, a3, ... , an], then the list [|a1 - a2|, |a2 - a3|, |a3 - a4|, ... , |an-1 - an|] has exactly `k` distinct integers.
+
+If there are multiple answers, print any of them.
+
+**Example 1:**
+
+```
+Input: n = 3, k = 1
+Output: [1, 2, 3]
+Explanation: The [1, 2, 3] has three different positive integers ranging from 1 to 3, and the [1, 1] has exactly 1 distinct integer: 1.
+```
+
+**Example 2:**
+
+```
+Input: n = 3, k = 2
+Output: [1, 3, 2]
+Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, and the [2, 1] has exactly 2 distinct integers: 1 and 2.
+```
+
+**Note:**
+
+1. The `n` and `k` are in the range 1 <= k < n <= 10^4.
+
+## 题目大意
+
+给定两个整数 n 和 k，你需要实现一个数组，这个数组包含从 1 到 n 的 n 个不同整数，同时满足以下条件：
+
+- 如果这个数组是 [a1, a2, a3, ... , an] ，那么数组 [|a1 - a2|, |a2 - a3|, |a3 - a4|, ... , |an-1 - an|] 中应该有且仅有 k 个不同整数；.
+- 如果存在多种答案，你只需实现并返回其中任意一种.
+
+## 解题思路
+
+- 先考虑 `k` 最大值的情况。如果把末尾的较大值依次插入到前面的较小值中，形成 `[1，n，2，n-1，3，n-2，……]`，这样排列 `k` 能取到最大值 `n-1` 。`k` 最小值的情况是 `[1，2，3，4，……，n]`，`k` 取到的最小值是 1。那么 `k` 在 `[1，n-1]` 之间取值，该怎么排列呢？先顺序排列 `[1，2，3，4，……，n-k-1]`，这里有 `n-k-1` 个数，可以形成唯一一种差值。剩下 `k+1` 个数，形成 `k-1` 种差值。
+- 这又回到了 `k` 最大值的取法了。`k` 取最大值的情况是 `n` 个数，形成 `n-1` 个不同种的差值。现在 `k+1` 个数，需要形成 `k` 种不同的差值。两者是同一个问题。那么剩下 `k` 个数的排列方法是 `[n-k，n-k+1，…，n]`，这里有 `k` 个数，注意代码实现时，注意 `k` 的奇偶性，如果 `k` 是奇数，“对半穿插”以后，正好匹配完，如果 `k` 是偶数，对半处的数 `n-k+(k+1)/2`，最后还需要单独加入到排列中。
+- 可能有读者会问了，前面生成了 1 种差值，后面这部分又生产了 `k` 种差值，加起来不是 `k + 1` 种差值了么？这种理解是错误的。后面这段最后 2 个数字是 `n-k+(k+1)/2-1` 和 `n-k+(k+1)/2`，它们两者的差值是 1，和第一段构造的排列差值是相同的，都是 1。所以第一段构造了 1 种差值，第二段虽然构造了 `k` 种，但是需要去掉两段重复的差值 1，所以最终差值种类还是 `1 + k - 1 = k` 种。
+
+## 代码
+
+```go
+package leetcode
+
+func constructArray(n int, k int) []int {
+	res := []int{}
+	for i := 0; i < n-k-1; i++ {
+		res = append(res, i+1)
+	}
+	for i := n - k; i < n-k+(k+1)/2; i++ {
+		res = append(res, i)
+		res = append(res, 2*n-k-i)
+	}
+	if k%2 == 0 {
+		res = append(res, n-k+(k+1)/2)
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0665.Non-decreasing-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0668.Kth-Smallest-Number-in-Multiplication-Table/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0668.Kth-Smallest-Number-in-Multiplication-Table.md b/website/content/ChapterFour/0600~0699/0668.Kth-Smallest-Number-in-Multiplication-Table.md
new file mode 100755
index 000000000..0dbd4cfdb
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0668.Kth-Smallest-Number-in-Multiplication-Table.md
@@ -0,0 +1,91 @@
+# [668. Kth Smallest Number in Multiplication Table](https://leetcode.com/problems/kth-smallest-number-in-multiplication-table/)
+
+
+## 题目
+
+Nearly every one have used the [Multiplication Table](https://en.wikipedia.org/wiki/Multiplication_table). But could you find out the `k-th` smallest number quickly from the multiplication table?
+
+Given the height `m` and the length `n` of a `m * n` Multiplication Table, and a positive integer `k`, you need to return the `k-th` smallest number in this table.
+
+**Example 1**:
+
+    Input: m = 3, n = 3, k = 5
+    Output: 
+    Explanation: 
+    The Multiplication Table:
+    1	2	3
+    2	4	6
+    3	6	9
+    
+    The 5-th smallest number is 3 (1, 2, 2, 3, 3).
+
+**Example 2**:
+
+    Input: m = 2, n = 3, k = 6
+    Output: 
+    Explanation: 
+    The Multiplication Table:
+    1	2	3
+    2	4	6
+    
+    The 6-th smallest number is 6 (1, 2, 2, 3, 4, 6).
+
+**Note**:
+
+1. The `m` and `n` will be in the range [1, 30000].
+2. The `k` will be in the range [1, m * n]
+
+
+## 题目大意
+
+几乎每一个人都用乘法表。但是你能在乘法表中快速找到第 k 小的数字吗？给定高度 m 、宽度 n 的一张 m * n 的乘法表，以及正整数 k，你需要返回表中第 k 小的数字。
+
+
+注意：
+
+- m 和 n 的范围在 [1, 30000] 之间。
+- k 的范围在 [1, m * n] 之间。
+
+## 解题思路
+
+- 给出 3 个数字，m，n，k。m  和 n 分别代表乘法口诀表的行和列。要求在这个乘法口诀表中找第 k 小的数字。
+- 这一题是第 378 题变种题。利用二分搜索，在 `[1,m*n]` 的区间内搜索第 `k` 小的数。每次二分统计 `≤ mid` 数字的个数。由于是在两数乘法构成的矩阵中计数，知道乘数，被乘数也就知道了，所以计数只需要一层循环。整体代码和第 378 题完全一致，只是计数的部分不同罢了。可以对比第 378 题一起练习。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func findKthNumber(m int, n int, k int) int {
+	low, high := 1, m*n
+	for low < high {
+		mid := low + (high-low)>>1
+		if counterKthNum(m, n, mid) >= k {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+func counterKthNum(m, n, mid int) int {
+	count := 0
+	for i := 1; i <= m; i++ {
+		count += int(math.Min(math.Floor(float64(mid)/float64(i)), float64(n)))
+	}
+	return count
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0667.Beautiful-Arrangement-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0669.Trim-a-Binary-Search-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0669.Trim-a-Binary-Search-Tree.md b/website/content/ChapterFour/0600~0699/0669.Trim-a-Binary-Search-Tree.md
new file mode 100644
index 000000000..12b162fbc
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0669.Trim-a-Binary-Search-Tree.md
@@ -0,0 +1,107 @@
+# [669. Trim a Binary Search Tree](https://leetcode.com/problems/trim-a-binary-search-tree/)
+
+
+## 题目
+
+Given the `root` of a binary search tree and the lowest and highest boundaries as `low` and `high`, trim the tree so that all its elements lies in `[low, high]`. Trimming the tree should **not** change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a **unique answer**.
+
+Return *the root of the trimmed binary search tree*. Note that the root may change depending on the given bounds.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/09/09/trim1.jpg](https://assets.leetcode.com/uploads/2020/09/09/trim1.jpg)
+
+```
+Input: root = [1,0,2], low = 1, high = 2
+Output: [1,null,2]
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/09/09/trim2.jpg](https://assets.leetcode.com/uploads/2020/09/09/trim2.jpg)
+
+```
+Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3
+Output: [3,2,null,1]
+```
+
+**Example 3:**
+
+```
+Input: root = [1], low = 1, high = 2
+Output: [1]
+```
+
+**Example 4:**
+
+```
+Input: root = [1,null,2], low = 1, high = 3
+Output: [1,null,2]
+```
+
+**Example 5:**
+
+```
+Input: root = [1,null,2], low = 2, high = 4
+Output: [2]
+```
+
+**Constraints:**
+
+- The number of nodes in the tree in the range `[1, 10^4]`.
+- `0 <= Node.val <= 10^4`
+- The value of each node in the tree is **unique**.
+- `root` is guaranteed to be a valid binary search tree.
+- `0 <= low <= high <= 10^4`
+
+## 题目大意
+
+给你二叉搜索树的根节点 root ，同时给定最小边界low 和最大边界 high。通过修剪二叉搜索树，使得所有节点的值在[low, high]中。修剪树不应该改变保留在树中的元素的相对结构（即，如果没有被移除，原有的父代子代关系都应当保留）。 可以证明，存在唯一的答案。所以结果应当返回修剪好的二叉搜索树的新的根节点。注意，根节点可能会根据给定的边界发生改变。
+
+## 解题思路
+
+- 这一题考察二叉搜索树中的递归遍历。递归遍历二叉搜索树每个结点，根据有序性，当前结点如果比 high 大，那么当前结点的右子树全部修剪掉，再递归修剪左子树；当前结点如果比 low 小，那么当前结点的左子树全部修剪掉，再递归修剪右子树。处理完越界的情况，剩下的情况都在区间内，分别递归修剪左子树和右子树即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func trimBST(root *TreeNode, low int, high int) *TreeNode {
+	if root == nil {
+		return root
+	}
+	if root.Val > high {
+		return trimBST(root.Left, low, high)
+	}
+	if root.Val < low {
+		return trimBST(root.Right, low, high)
+	}
+	root.Left = trimBST(root.Left, low, high)
+	root.Right = trimBST(root.Right, low, high)
+	return root
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0668.Kth-Smallest-Number-in-Multiplication-Table/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0674.Longest-Continuous-Increasing-Subsequence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0674.Longest-Continuous-Increasing-Subsequence.md b/website/content/ChapterFour/0600~0699/0674.Longest-Continuous-Increasing-Subsequence.md
new file mode 100644
index 000000000..016217491
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0674.Longest-Continuous-Increasing-Subsequence.md
@@ -0,0 +1,76 @@
+# [674. Longest Continuous Increasing Subsequence](https://leetcode.com/problems/longest-continuous-increasing-subsequence/)
+
+
+## 题目
+
+Given an unsorted array of integers `nums`, return *the length of the longest **continuous increasing subsequence** (i.e. subarray)*. The subsequence must be **strictly** increasing.
+
+A **continuous increasing subsequence** is defined by two indices `l` and `r` (`l < r`) such that it is `[nums[l], nums[l + 1], ..., nums[r - 1], nums[r]]` and for each `l <= i < r`, `nums[i] < nums[i + 1]`.
+
+**Example 1:**
+
+```
+Input: nums = [1,3,5,4,7]
+Output: 3
+Explanation: The longest continuous increasing subsequence is [1,3,5] with length 3.
+Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element
+4.
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,2,2,2,2]
+Output: 1
+Explanation: The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly
+increasing.
+```
+
+**Constraints:**
+
+- `0 <= nums.length <= 10^4`
+- `10^9 <= nums[i] <= 10^9`
+
+## 题目大意
+
+给定一个未经排序的整数数组，找到最长且 连续递增的子序列，并返回该序列的长度。连续递增的子序列 可以由两个下标 l 和 r（l < r）确定，如果对于每个 l <= i < r，都有 nums[i] < nums[i + 1] ，那么子序列 [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] 就是连续递增子序列。
+
+## 解题思路
+
+- 简单题。这一题和第 128 题有区别。这一题要求子序列必须是连续下标，所以变简单了。扫描一遍数组，记下连续递增序列的长度，动态维护这个最大值，最后输出即可。
+
+## 代码
+
+```go
+package leetcode
+
+func findLengthOfLCIS(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	res, length := 1, 1
+	for i := 1; i < len(nums); i++ {
+		if nums[i] > nums[i-1] {
+			length++
+		} else {
+			res = max(res, length)
+			length = 1
+		}
+	}
+	return max(res, length)
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0669.Trim-a-Binary-Search-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0676.Implement-Magic-Dictionary/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0676.Implement-Magic-Dictionary.md b/website/content/ChapterFour/0600~0699/0676.Implement-Magic-Dictionary.md
new file mode 100755
index 000000000..74961388c
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0676.Implement-Magic-Dictionary.md
@@ -0,0 +1,94 @@
+# [676. Implement Magic Dictionary](https://leetcode.com/problems/implement-magic-dictionary/)
+
+
+## 题目
+
+Implement a magic directory with `buildDict`, and `search` methods.
+
+For the method `buildDict`, you'll be given a list of non-repetitive words to build a dictionary.
+
+For the method `search`, you'll be given a word, and judge whether if you modify **exactly** one character into **another**character in this word, the modified word is in the dictionary you just built.
+
+**Example 1**:
+
+    Input: buildDict(["hello", "leetcode"]), Output: Null
+    Input: search("hello"), Output: False
+    Input: search("hhllo"), Output: True
+    Input: search("hell"), Output: False
+    Input: search("leetcoded"), Output: False
+
+**Note**:
+
+1. You may assume that all the inputs are consist of lowercase letters `a-z`.
+2. For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
+3. Please remember to **RESET** your class variables declared in class MagicDictionary, as static/class variables are **persisted across multiple test cases**. Please see [here](https://leetcode.com/faq/#different-output) for more details.
+
+
+## 题目大意
+
+实现一个带有 buildDict, 以及 search 方法的魔法字典。对于 buildDict 方法，你将被给定一串不重复的单词来构建一个字典。对于 search 方法，你将被给定一个单词，并且判定能否只将这个单词中一个字母换成另一个字母，使得所形成的新单词存在于你构建的字典中。
+
+
+
+## 解题思路
+
+
+- 实现 `MagicDictionary` 的数据结构，这个数据结构内会存储一个字符串数组，当执行 `Search` 操作的时候要求判断传进来的字符串能否只改变一个字符(不能增加字符也不能删除字符)就能变成 `MagicDictionary` 中存储的字符串，如果可以，就输出 `true`，如果不能，就输出 `false`。
+- 这题的解题思路比较简单，用 Map 判断即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+type MagicDictionary struct {
+	rdict map[int]string
+}
+
+/** Initialize your data structure here. */
+func Constructor676() MagicDictionary {
+	return MagicDictionary{rdict: make(map[int]string)}
+}
+
+/** Build a dictionary through a list of words */
+func (this *MagicDictionary) BuildDict(dict []string) {
+	for k, v := range dict {
+		this.rdict[k] = v
+	}
+}
+
+/** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
+func (this *MagicDictionary) Search(word string) bool {
+	for _, v := range this.rdict {
+		n := 0
+		if len(word) == len(v) {
+			for i := 0; i < len(v); i++ {
+				if word[i] != v[i] {
+					n += 1
+				}
+			}
+			if n == 1 {
+				return true
+			}
+		}
+	}
+	return false
+}
+
+/**
+ * Your MagicDictionary object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.BuildDict(dict);
+ * param_2 := obj.Search(word);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0674.Longest-Continuous-Increasing-Subsequence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0677.Map-Sum-Pairs/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0677.Map-Sum-Pairs.md b/website/content/ChapterFour/0600~0699/0677.Map-Sum-Pairs.md
new file mode 100644
index 000000000..2fa80a36b
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0677.Map-Sum-Pairs.md
@@ -0,0 +1,107 @@
+# [677. Map Sum Pairs](https://leetcode.com/problems/map-sum-pairs/)
+
+
+## 题目
+
+Design a map that allows you to do the following:
+
+- Maps a string key to a given value.
+- Returns the sum of the values that have a key with a prefix equal to a given string.
+
+Implement the `MapSum` class:
+
+- `MapSum()` Initializes the `MapSum` object.
+- `void insert(String key, int val)` Inserts the `key-val` pair into the map. If the `key` already existed, the original `key-value` pair will be overridden to the new one.
+- `int sum(string prefix)` Returns the sum of all the pairs' value whose `key` starts with the `prefix`.
+
+**Example 1:**
+
+```
+Input
+["MapSum", "insert", "sum", "insert", "sum"]
+[[], ["apple", 3], ["ap"], ["app", 2], ["ap"]]
+Output
+[null, null, 3, null, 5]
+
+Explanation
+MapSum mapSum = new MapSum();
+mapSum.insert("apple", 3);
+mapSum.sum("ap");           // return 3 (apple = 3)
+mapSum.insert("app", 2);
+mapSum.sum("ap");           // return 5 (apple +app = 3 + 2 = 5)
+
+```
+
+**Constraints:**
+
+- `1 <= key.length, prefix.length <= 50`
+- `key` and `prefix` consist of only lowercase English letters.
+- `1 <= val <= 1000`
+- At most `50` calls will be made to `insert` and `sum`.
+
+## 题目大意
+
+实现一个 MapSum 类，支持两个方法，insert 和 sum：
+
+- MapSum() 初始化 MapSum 对象
+- void insert(String key, int val) 插入 key-val 键值对，字符串表示键 key ，整数表示值 val 。如果键 key 已经存在，那么原来的键值对将被替代成新的键值对。
+- int sum(string prefix) 返回所有以该前缀 prefix 开头的键 key 的值的总和。
+
+## 解题思路
+
+- 简单题。用一个 map 存储数据，Insert() 方法即存储 key-value。Sum() 方法即累加满足条件前缀对应的 value。判断是否满足条件，先根据前缀长度来判断，只有长度大于等于 prefix 长度才可能满足要求。如果 key 是具有 prefix 前缀的，那么累加上这个值。最后输出总和即可。
+
+## 代码
+
+```go
+package leetcode
+
+type MapSum struct {
+	keys map[string]int
+}
+
+/** Initialize your data structure here. */
+func Constructor() MapSum {
+	return MapSum{make(map[string]int)}
+}
+
+func (this *MapSum) Insert(key string, val int) {
+	this.keys[key] = val
+}
+
+func (this *MapSum) Sum(prefix string) int {
+	prefixAsRunes, res := []rune(prefix), 0
+	for key, val := range this.keys {
+		if len(key) >= len(prefix) {
+			shouldSum := true
+			for i, char := range key {
+				if i >= len(prefixAsRunes) {
+					break
+				}
+				if prefixAsRunes[i] != char {
+					shouldSum = false
+					break
+				}
+			}
+			if shouldSum {
+				res += val
+			}
+		}
+	}
+	return res
+}
+
+/**
+ * Your MapSum object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.Insert(key,val);
+ * param_2 := obj.Sum(prefix);
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0676.Implement-Magic-Dictionary/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0682.Baseball-Game/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0682.Baseball-Game.md b/website/content/ChapterFour/0600~0699/0682.Baseball-Game.md
new file mode 100644
index 000000000..17efc69b7
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0682.Baseball-Game.md
@@ -0,0 +1,113 @@
+# [682. Baseball Game](https://leetcode.com/problems/baseball-game/)
+
+## 题目
+
+You're now a baseball game point recorder.
+
+Given a list of strings, each string can be one of the 4 following types:
+
+1. Integer (one round's score): Directly represents the number of points you get in this round.
+2. "+" (one round's score): Represents that the points you get in this round are the sum of the last two valid round's points.
+3. "D" (one round's score): Represents that the points you get in this round are the doubled data of the last valid round's points.
+4. "C" (an operation, which isn't a round's score): Represents the last valid round's points you get were invalid and should be removed.
+Each round's operation is permanent and could have an impact on the round before and the round after.
+
+You need to return the sum of the points you could get in all the rounds.
+
+**Example 1**:
+
+```
+
+Input: ["5","2","C","D","+"]
+Output: 30
+Explanation: 
+Round 1: You could get 5 points. The sum is: 5.
+Round 2: You could get 2 points. The sum is: 7.
+Operation 1: The round 2's data was invalid. The sum is: 5.  
+Round 3: You could get 10 points (the round 2's data has been removed). The sum is: 15.
+Round 4: You could get 5 + 10 = 15 points. The sum is: 30.
+
+```
+
+**Example 2**:
+
+```
+
+Input: ["5","-2","4","C","D","9","+","+"]
+Output: 27
+Explanation: 
+Round 1: You could get 5 points. The sum is: 5.
+Round 2: You could get -2 points. The sum is: 3.
+Round 3: You could get 4 points. The sum is: 7.
+Operation 1: The round 3's data is invalid. The sum is: 3.  
+Round 4: You could get -4 points (the round 3's data has been removed). The sum is: -1.
+Round 5: You could get 9 points. The sum is: 8.
+Round 6: You could get -4 + 9 = 5 points. The sum is 13.
+Round 7: You could get 9 + 5 = 14 points. The sum is 27.
+
+```
+
+**Note**:     
+
+- The size of the input list will be between 1 and 1000.
+- Every integer represented in the list will be between -30000 and 30000.
+
+## 题目大意
+
+这道题是模拟题，给一串数字和操作符。出现数字就直接累加，出现 "C" 就代表栈推出一个元素，相应的总和要减去栈顶的元素。出现 "D" 就代表把前一个元素乘以 2，就得到当前的元素值。再累加。出现 "+" 就代表把前 2 个值求和，得到当前元素的值，再累积。
+
+## 解题思路
+
+这道题用栈模拟即可。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strconv"
+
+func calPoints(ops []string) int {
+	stack := make([]int, len(ops))
+	top := 0
+
+	for i := 0; i < len(ops); i++ {
+		op := ops[i]
+		switch op {
+		case "+":
+			last1 := stack[top-1]
+			last2 := stack[top-2]
+			stack[top] = last1 + last2
+			top++
+		case "D":
+			last1 := stack[top-1]
+			stack[top] = last1 * 2
+			top++
+		case "C":
+			top--
+		default:
+			stack[top], _ = strconv.Atoi(op)
+			top++
+		}
+	}
+
+	points := 0
+	for i := 0; i < top; i++ {
+		points += stack[i]
+	}
+	return points
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0677.Map-Sum-Pairs/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0684.Redundant-Connection/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0684.Redundant-Connection.md b/website/content/ChapterFour/0600~0699/0684.Redundant-Connection.md
new file mode 100755
index 000000000..4fff04a31
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0684.Redundant-Connection.md
@@ -0,0 +1,92 @@
+# [684. Redundant Connection](https://leetcode.com/problems/redundant-connection/)
+
+
+## 题目
+
+In this problem, a tree is an **undirected** graph that is connected and has no cycles.
+
+The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, ..., N), with one additional edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.
+
+The resulting graph is given as a 2D-array of `edges`. Each element of `edges` is a pair `[u, v]` with `u < v`, that represents an **undirected** edge connecting nodes `u` and `v`.
+
+Return an edge that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge `[u, v]` should be in the same format, with `u < v`.
+
+**Example 1**:
+
+    Input: [[1,2], [1,3], [2,3]]
+    Output: [2,3]
+    Explanation: The given undirected graph will be like this:
+      1
+     / \
+    2 - 3
+
+**Example 2**:
+
+    Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
+    Output: [1,4]
+    Explanation: The given undirected graph will be like this:
+    5 - 1 - 2
+        |   |
+        4 - 3
+
+**Note**:
+
+- The size of the input 2D-array will be between 3 and 1000.
+- Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.
+
+**Update (2017-09-26)**: We have overhauled the problem description + test cases and specified clearly the graph is an **undirected** graph. For the **directed** graph follow up please see **[Redundant Connection II](https://leetcode.com/problems/redundant-connection-ii/description/)**). We apologize for any inconvenience caused.
+
+
+## 题目大意
+
+在本问题中, 树指的是一个连通且无环的无向图。输入一个图，该图由一个有着N个节点 (节点值不重复1, 2, ..., N) 的树及一条附加的边构成。附加的边的两个顶点包含在1到N中间，这条附加的边不属于树中已存在的边。结果图是一个以边组成的二维数组。每一个边的元素是一对[u, v] ，满足 u < v，表示连接顶点u 和v的无向图的边。
+
+返回一条可以删去的边，使得结果图是一个有着N个节点的树。如果有多个答案，则返回二维数组中最后出现的边。答案边 [u, v] 应满足相同的格式 u < v。
+
+注意:
+
+- 输入的二维数组大小在 3 到 1000。
+- 二维数组中的整数在 1 到 N 之间，其中 N 是输入数组的大小。
+
+
+## 解题思路
+
+- 给出一个连通无环无向图和一些连通的边，要求在这些边中删除一条边以后，图中的 N 个节点依旧是连通的。如果有多条边，输出最后一条。
+- 这一题可以用并查集直接秒杀。依次扫描所有的边，把边的两端点都合并 `union()` 到一起。如果遇到一条边的两端点已经在一个集合里面了，就说明是多余边，删除。最后输出这些边即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func findRedundantConnection(edges [][]int) []int {
+	if len(edges) == 0 {
+		return []int{}
+	}
+	uf, res := template.UnionFind{}, []int{}
+	uf.Init(len(edges) + 1)
+	for i := 0; i < len(edges); i++ {
+		if uf.Find(edges[i][0]) != uf.Find(edges[i][1]) {
+			uf.Union(edges[i][0], edges[i][1])
+		} else {
+			res = append(res, edges[i][0])
+			res = append(res, edges[i][1])
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0682.Baseball-Game/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0685.Redundant-Connection-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0685.Redundant-Connection-II.md b/website/content/ChapterFour/0600~0699/0685.Redundant-Connection-II.md
new file mode 100755
index 000000000..f24a372a4
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0685.Redundant-Connection-II.md
@@ -0,0 +1,117 @@
+# [685. Redundant Connection II](https://leetcode.com/problems/redundant-connection-ii/)
+
+
+## 题目
+
+In this problem, a rooted tree is a **directed** graph such that, there is exactly one node (the root) for which all other nodes are descendants of this node, plus every node has exactly one parent, except for the root node which has no parents.
+
+The given input is a directed graph that started as a rooted tree with N nodes (with distinct values 1, 2, ..., N), with one additional directed edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.
+
+The resulting graph is given as a 2D-array of `edges`. Each element of `edges` is a pair `[u, v]` that represents a **directed** edge connecting nodes `u` and `v`, where `u` is a parent of child `v`.
+
+Return an edge that can be removed so that the resulting graph is a rooted tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array.
+
+**Example 1**:
+
+    Input: [[1,2], [1,3], [2,3]]
+    Output: [2,3]
+    Explanation: The given directed graph will be like this:
+      1
+     / \
+    v   v
+    2-->3
+
+**Example 2**:
+
+    Input: [[1,2], [2,3], [3,4], [4,1], [1,5]]
+    Output: [4,1]
+    Explanation: The given directed graph will be like this:
+    5 <- 1 -> 2
+         ^    |
+         |    v
+         4 <- 3
+
+**Note**:
+
+- The size of the input 2D-array will be between 3 and 1000.
+- Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.
+
+
+## 题目大意
+
+
+在本问题中，有根树指满足以下条件的有向图。该树只有一个根节点，所有其他节点都是该根节点的后继。每一个节点只有一个父节点，除了根节点没有父节点。输入一个有向图，该图由一个有着 N 个节点 (节点值不重复1, 2, ..., N) 的树及一条附加的边构成。附加的边的两个顶点包含在1到N中间，这条附加的边不属于树中已存在的边。结果图是一个以边组成的二维数组。 每一个边的元素是一对 [u, v]，用以表示有向图中连接顶点 u and v 和顶点的边，其中父节点 u 是子节点 v 的一个父节点。返回一条能删除的边，使得剩下的图是有 N 个节点的有根树。若有多个答案，返回最后出现在给定二维数组的答案。
+
+注意:
+
+- 二维数组大小的在 3 到 1000 范围内。
+- 二维数组中的每个整数在 1 到 N 之间，其中 N 是二维数组的大小。
+
+
+## 解题思路
+
+- 这一题是第 684 题的加强版。第 684 题中的图是无向图，这一题中的图是有向图。
+- 这一题的解法也是用并查集，不过需要灵活一点，不要用模板，因为在模板中，存在路径压缩和 `rank()` 优化，这些优化会改变有向边原始的方向。所以并查集只需要记录 `parent()` 就够用了。
+
+![](https://img.halfrost.com/Leetcode/leetcode_685.png)
+
+- 经过分析，可以得到上面这 3 种情况，红色的边是我们实际应该删除的。先来看情况 2 和情况 3 。当不断 `union()` 时，加入一条边以后，会使一个节点的入度变成 2，那么记录下这两条边为 `candidate1` 和 `candidate2`。将后加入的 `candidate2` 这条边先放在一边，继续往下 `union()`。如果 `candidate2` 是红色的边，那么合并到最后，也不会出现任何异常，那么 `candidate2` 就是红色的边，即找到了要删除的边了。如果合并到最后出现了环的问题了，那说明 `candidate2` 是黑色的边，`candidate1` 才是红色的边，那么 `candidate1` 是要删除的边。
+- 再来看看情况 1。如果一路合并到结束也没有发现出现入度为 2 的情况，那么说明遇到了情况 1 。情况 1 会出现环的情况。题目中说如果要删除边，就删除最后出现的那条边。**具体实现见代码注释**。
+
+## 代码
+
+```go
+
+package leetcode
+
+func findRedundantDirectedConnection(edges [][]int) []int {
+	if len(edges) == 0 {
+		return []int{}
+	}
+	parent, candidate1, candidate2 := make([]int, len(edges)+1), []int{}, []int{}
+	for _, edge := range edges {
+		if parent[edge[1]] == 0 {
+			parent[edge[1]] = edge[0]
+		} else { // 如果一个节点已经有父亲节点了，说明入度已经有 1 了，再来一条边，入度为 2 ，那么跳过新来的这条边 candidate2，并记录下和这条边冲突的边 candidate1
+			candidate1 = append(candidate1, parent[edge[1]])
+			candidate1 = append(candidate1, edge[1])
+			candidate2 = append(candidate2, edge[0])
+			candidate2 = append(candidate2, edge[1])
+			edge[1] = 0 // 做标记，后面再扫到这条边以后可以直接跳过
+		}
+	}
+	for i := 1; i <= len(edges); i++ {
+		parent[i] = i
+	}
+	for _, edge := range edges {
+		if edge[1] == 0 { // 跳过 candidate2 这条边
+			continue
+		}
+		u, v := edge[0], edge[1]
+		pu := findRoot(&parent, u)
+		if pu == v { // 发现有环
+			if len(candidate1) == 0 { // 如果没有出现入度为 2 的情况，那么对应情况 1，就删除这条边
+				return edge
+			}
+			return candidate1 // 出现环并且有入度为 2 的情况，说明 candidate1 是答案
+		}
+		parent[v] = pu // 没有发现环，继续合并
+	}
+	return candidate2 // 当最后什么都没有发生，则 candidate2 是答案
+}
+
+func findRoot(parent *[]int, k int) int {
+	if (*parent)[k] != k {
+		(*parent)[k] = findRoot(parent, (*parent)[k])
+	}
+	return (*parent)[k]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0684.Redundant-Connection/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0690.Employee-Importance/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0690.Employee-Importance.md b/website/content/ChapterFour/0600~0699/0690.Employee-Importance.md
new file mode 100644
index 000000000..df80e9bc7
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0690.Employee-Importance.md
@@ -0,0 +1,69 @@
+# [690. Employee Importance](https://leetcode.com/problems/employee-importance/)
+
+## 题目
+
+You are given a data structure of employee information, which includes the employee's **unique id**, their **importance value** and their **direct** subordinates' id.
+
+For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, [2]], and employee 2 has [2, 10, [3]], and employee 3 has [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, the relationship is **not direct**.
+
+Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all their subordinates.
+
+**Example 1:**
+
+```
+Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
+Output: 11
+Explanation:
+Employee 1 has importance value 5, and he has two direct subordinates: employee 2 and employee 3. They both have importance value 3. So the total importance value of employee 1 is 5 + 3 + 3 = 11.
+```
+
+**Note:**
+
+1. One employee has at most one **direct** leader and may have several subordinates.
+2. The maximum number of employees won't exceed 2000.
+
+## 题目大意
+
+给定一个保存员工信息的数据结构，它包含了员工 唯一的 id ，重要度 和 直系下属的 id 。比如，员工 1 是员工 2 的领导，员工 2 是员工 3 的领导。他们相应的重要度为 15 , 10 , 5 。那么员工 1 的数据结构是 [1, 15, [2]] ，员工 2的 数据结构是 [2, 10, [3]] ，员工 3 的数据结构是 [3, 5, []] 。注意虽然员工 3 也是员工 1 的一个下属，但是由于 并不是直系 下属，因此没有体现在员工 1 的数据结构中。现在输入一个公司的所有员工信息，以及单个员工 id ，返回这个员工和他所有下属的重要度之和。
+
+## 解题思路
+
+- 简单题。根据题意，DFS 或者 BFS 搜索找到所求 id 下属所有员工，累加下属员工的重要度，最后再加上这个员工本身的重要度，即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+type Employee struct {
+	Id           int
+	Importance   int
+	Subordinates []int
+}
+
+func getImportance(employees []*Employee, id int) int {
+	m, queue, res := map[int]*Employee{}, []int{id}, 0
+	for _, e := range employees {
+		m[e.Id] = e
+	}
+	for len(queue) > 0 {
+		e := m[queue[0]]
+		queue = queue[1:]
+		if e == nil {
+			continue
+		}
+		res += e.Importance
+		for _, i := range e.Subordinates {
+			queue = append(queue, i)
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0685.Redundant-Connection-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0692.Top-K-Frequent-Words/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0692.Top-K-Frequent-Words.md b/website/content/ChapterFour/0600~0699/0692.Top-K-Frequent-Words.md
new file mode 100644
index 000000000..1d8f09791
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0692.Top-K-Frequent-Words.md
@@ -0,0 +1,105 @@
+# [692. Top K Frequent Words](https://leetcode.com/problems/top-k-frequent-words/)
+
+
+## 题目
+
+Given a non-empty list of words, return the k most frequent elements.
+
+Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.
+
+**Example 1:**
+
+```
+Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
+Output: ["i", "love"]
+Explanation: "i" and "love" are the two most frequent words.
+    Note that "i" comes before "love" due to a lower alphabetical order.
+```
+
+**Example 2:**
+
+```
+Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
+Output: ["the", "is", "sunny", "day"]
+Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
+    with the number of occurrence being 4, 3, 2 and 1 respectively.
+```
+
+**Note:**
+
+1. You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
+2. Input words contain only lowercase letters.
+
+**Follow up:**
+
+1. Try to solve it in O(n log k) time and O(n) extra space.
+
+## 题目大意
+
+给一非空的单词列表，返回前 *k* 个出现次数最多的单词。返回的答案应该按单词出现频率由高到低排序。如果不同的单词有相同出现频率，按字母顺序排序。
+
+## 解题思路
+
+- 思路很简单的题。维护一个长度为 k 的最大堆，先按照频率排，如果频率相同再按照字母顺序排。最后输出依次将优先队列里面的元素 pop 出来即可。
+
+## 代码
+
+```go
+package leetcode
+
+import "container/heap"
+
+func topKFrequent(words []string, k int) []string {
+	m := map[string]int{}
+	for _, word := range words {
+		m[word]++
+	}
+	pq := &PQ{}
+	heap.Init(pq)
+	for w, c := range m {
+		heap.Push(pq, &wordCount{w, c})
+		if pq.Len() > k {
+			heap.Pop(pq)
+		}
+	}
+	res := make([]string, k)
+	for i := k - 1; i >= 0; i-- {
+		wc := heap.Pop(pq).(*wordCount)
+		res[i] = wc.word
+	}
+	return res
+}
+
+type wordCount struct {
+	word string
+	cnt  int
+}
+
+type PQ []*wordCount
+
+func (pq PQ) Len() int      { return len(pq) }
+func (pq PQ) Swap(i, j int) { pq[i], pq[j] = pq[j], pq[i] }
+func (pq PQ) Less(i, j int) bool {
+	if pq[i].cnt == pq[j].cnt {
+		return pq[i].word > pq[j].word
+	}
+	return pq[i].cnt < pq[j].cnt
+}
+func (pq *PQ) Push(x interface{}) {
+	tmp := x.(*wordCount)
+	*pq = append(*pq, tmp)
+}
+func (pq *PQ) Pop() interface{} {
+	n := len(*pq)
+	tmp := (*pq)[n-1]
+	*pq = (*pq)[:n-1]
+	return tmp
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0690.Employee-Importance/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0693.Binary-Number-with-Alternating-Bits/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0693.Binary-Number-with-Alternating-Bits.md b/website/content/ChapterFour/0600~0699/0693.Binary-Number-with-Alternating-Bits.md
new file mode 100755
index 000000000..b77b30872
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0693.Binary-Number-with-Alternating-Bits.md
@@ -0,0 +1,87 @@
+# [693. Binary Number with Alternating Bits](https://leetcode.com/problems/binary-number-with-alternating-bits/)
+
+## 题目
+
+Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.
+
+**Example 1**:
+
+    Input: 5
+    Output: True
+    Explanation:
+    The binary representation of 5 is: 101
+
+**Example 2**:
+
+    Input: 7
+    Output: False
+    Explanation:
+    The binary representation of 7 is: 111.
+
+**Example 3**:
+
+    Input: 11
+    Output: False
+    Explanation:
+    The binary representation of 11 is: 1011.
+
+**Example 4**:
+
+    Input: 10
+    Output: True
+    Explanation:
+    The binary representation of 10 is: 1010.
+
+
+## 题目大意
+
+给定一个正整数，检查他是否为交替位二进制数：换句话说，就是他的二进制数相邻的两个位数永不相等。
+
+## 解题思路
+
+
+- 判断一个数的二进制位相邻两个数是不相等的，即 `0101` 交叉间隔的，如果是，输出 true。这一题有多种做法，最简单的方法就是直接模拟。比较巧妙的方法是通过位运算，合理构造特殊数据进行位运算到达目的。`010101` 构造出 `101010` 两者相互 `&` 位运算以后就为 0，因为都“插空”了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func hasAlternatingBits(n int) bool {
+	/*
+	   n =         1 0 1 0 1 0 1 0
+	   n >> 1      0 1 0 1 0 1 0 1
+	   n ^ n>>1    1 1 1 1 1 1 1 1
+	   n           1 1 1 1 1 1 1 1
+	   n + 1     1 0 0 0 0 0 0 0 0
+	   n & (n+1)   0 0 0 0 0 0 0 0
+	*/
+	n = n ^ (n >> 1)
+	return (n & (n + 1)) == 0
+}
+
+// 解法二
+func hasAlternatingBits1(n int) bool {
+	last, current := 0, 0
+	for n > 0 {
+		last = n & 1
+		n = n / 2
+		current = n & 1
+		if last == current {
+			return false
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0692.Top-K-Frequent-Words/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0695.Max-Area-of-Island/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0695.Max-Area-of-Island.md b/website/content/ChapterFour/0600~0699/0695.Max-Area-of-Island.md
new file mode 100644
index 000000000..05c391e04
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0695.Max-Area-of-Island.md
@@ -0,0 +1,97 @@
+# [695. Max Area of Island](https://leetcode.com/problems/max-area-of-island/)
+
+
+
+## 题目
+
+Given a non-empty 2D array `grid` of 0's and 1's, an **island** is a group of `1`'s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.
+
+Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)
+
+**Example 1**:
+
+```
+[[0,0,1,0,0,0,0,1,0,0,0,0,0],
+ [0,0,0,0,0,0,0,1,1,1,0,0,0],
+ [0,1,1,0,1,0,0,0,0,0,0,0,0],
+ [0,1,0,0,1,1,0,0,1,0,1,0,0],
+ [0,1,0,0,1,1,0,0,1,1,1,0,0],
+ [0,0,0,0,0,0,0,0,0,0,1,0,0],
+ [0,0,0,0,0,0,0,1,1,1,0,0,0],
+ [0,0,0,0,0,0,0,1,1,0,0,0,0]]
+```
+
+Given the above grid, return`6`. Note the answer is not 11, because the island must be connected 4-directionally.
+
+**Example 2**:
+
+```
+[[0,0,0,0,0,0,0,0]]
+```
+
+Given the above grid, return`0`.
+
+**Note**: The length of each dimension in the given `grid` does not exceed 50.
+
+## 题目大意
+
+给定一个包含了一些 0 和 1 的非空二维数组 grid 。一个 岛屿 是由一些相邻的 1 (代表土地) 构成的组合，这里的「相邻」要求两个 1 必须在水平或者竖直方向上相邻。你可以假设 grid 的四个边缘都被 0（代表水）包围着。找到给定的二维数组中最大的岛屿面积。(如果没有岛屿，则返回面积为 0 。)
+
+## 解题思路
+
+- 给出一个地图，要求计算上面岛屿的面积。注意岛屿的定义是四周都是海(为 0 的点)，如果土地(为 1 的点)靠在地图边缘，不能算是岛屿。
+- 这一题和第 200 题，第 1254 题解题思路是一致的。DPS 深搜。这不过这一题需要多处理 2 件事情，一个是注意靠边缘的岛屿不能计算在内，二是动态维护岛屿的最大面积。
+
+## 代码
+
+```go
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func maxAreaOfIsland(grid [][]int) int {
+	res := 0
+	for i, row := range grid {
+		for j, col := range row {
+			if col == 0 {
+				continue
+			}
+			area := areaOfIsland(grid, i, j)
+			if area > res {
+				res = area
+			}
+		}
+	}
+	return res
+}
+
+func isInGrid(grid [][]int, x, y int) bool {
+	return x >= 0 && x < len(grid) && y >= 0 && y < len(grid[0])
+}
+
+func areaOfIsland(grid [][]int, x, y int) int {
+	if !isInGrid(grid, x, y) || grid[x][y] == 0 {
+		return 0
+	}
+	grid[x][y] = 0
+	total := 1
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		total += areaOfIsland(grid, nx, ny)
+	}
+	return total
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0693.Binary-Number-with-Alternating-Bits/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0696.Count-Binary-Substrings/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0696.Count-Binary-Substrings.md b/website/content/ChapterFour/0600~0699/0696.Count-Binary-Substrings.md
new file mode 100644
index 000000000..0bc7aaca8
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0696.Count-Binary-Substrings.md
@@ -0,0 +1,76 @@
+# [696. Count Binary Substrings](https://leetcode.com/problems/count-binary-substrings/)
+
+
+## 题目
+
+Give a string `s`, count the number of non-empty (contiguous) substrings that have the same number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped consecutively.
+
+Substrings that occur multiple times are counted the number of times they occur.
+
+**Example 1:**
+
+```
+Input: "00110011"
+Output: 6
+Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".
+
+Notice that some of these substrings repeat and are counted the number of times they occur.
+
+Also, "00110011" is not a valid substring becauseall the 0's (and 1's) are not grouped together.
+
+```
+
+**Example 2:**
+
+```
+Input: "10101"
+Output: 4
+Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.
+
+```
+
+**Note:**
+
+- `s.length` will be between 1 and 50,000.
+- `s` will only consist of "0" or "1" characters.
+
+## 题目大意
+
+给定一个字符串 s，计算具有相同数量 0 和 1 的非空（连续）子字符串的数量，并且这些子字符串中的所有 0 和所有 1 都是连续的。重复出现的子串要计算它们出现的次数。
+
+## 解题思路
+
+- 简单题。先分组统计 0 和 1 的个数，例如，`0110001111` 按照 0 和 1 分组统计出来的结果是 [1, 2, 3, 4]。再拼凑结果。相邻 2 组取两者最短的，例如 `0110001111`，凑成的结果应该是 min(1,2)，min(2,3)，min(3,4)，即 `01`，`01`，`10`，`1100`，`0011`，`000111`。时间复杂度 O(n)，空间复杂度 O(1)。
+
+## 代码
+
+```go
+package leetcode
+
+func countBinarySubstrings(s string) int {
+	last, res := 0, 0
+	for i := 0; i < len(s); {
+		c, count := s[i], 1
+		for i++; i < len(s) && s[i] == c; i++ {
+			count++
+		}
+		res += min(count, last)
+		last = count
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0695.Max-Area-of-Island/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0697.Degree-of-an-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0697.Degree-of-an-Array.md b/website/content/ChapterFour/0600~0699/0697.Degree-of-an-Array.md
new file mode 100644
index 000000000..b9533ece3
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/0697.Degree-of-an-Array.md
@@ -0,0 +1,87 @@
+# [697. Degree of an Array](https://leetcode.com/problems/degree-of-an-array/)
+
+
+## 题目
+
+Given a non-empty array of non-negative integers `nums`, the **degree** of this array is defined as the maximum frequency of any one of its elements.
+
+Your task is to find the smallest possible length of a (contiguous) subarray of `nums`, that has the same degree as `nums`.
+
+**Example 1**:
+
+```
+Input: [1, 2, 2, 3, 1]
+Output: 2
+Explanation: 
+The input array has a degree of 2 because both elements 1 and 2 appear twice.
+Of the subarrays that have the same degree:
+[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
+The shortest length is 2. So return 2.
+
+```
+
+**Example 2**:
+
+```
+Input: [1,2,2,3,1,4,2]
+Output: 6
+```
+
+**Note**:
+
+- `nums.length` will be between 1 and 50,000.
+- `nums[i]` will be an integer between 0 and 49,999.
+
+
+## 题目大意
+
+给定一个非空且只包含非负数的整数数组 nums, 数组的度的定义是指数组里任一元素出现频数的最大值。你的任务是找到与 nums 拥有相同大小的度的最短连续子数组，返回其长度。
+
+注意:
+
+- nums.length 在 1 到 50,000 区间范围内。
+- nums[i] 是一个在 0 到 49,999 范围内的整数。
+
+
+## 解题思路
+
+- 找一个与给定数组相同度的最短连续子数组，输出其长度。数组的度的定义是任一元素出现频数的最大值。
+- 简单题。先统计各个元素的频次，并且动态维护最大频次和子数组的起始和终点位置。这里最短连续子数组有点“迷惑人”。这个最短子数组其实处理起来很简单。只需从前往后扫一遍，记录各个元素第一次出现的位置和最后一次出现的位置即是最短的连续子数组。然后在频次字典里面寻找和最大频次相同的所有解，有可能有多个子数组能满足题意，取出最短的输出即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func findShortestSubArray(nums []int) int {
+	frequency, maxFreq, smallest := map[int][]int{}, 0, len(nums)
+	for i, num := range nums {
+		if _, found := frequency[num]; !found {
+			frequency[num] = []int{1, i, i}
+		} else {
+			frequency[num][0]++
+			frequency[num][2] = i
+		}
+		if maxFreq < frequency[num][0] {
+			maxFreq = frequency[num][0]
+		}
+	}
+	for _, indices := range frequency {
+		if indices[0] == maxFreq {
+			if smallest > indices[2]-indices[1]+1 {
+				smallest = indices[2] - indices[1] + 1
+			}
+		}
+	}
+	return smallest
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0696.Count-Binary-Substrings/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0699.Falling-Squares/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/0699.Falling-Squares.md b/website/content/ChapterFour/0600~0699/0699.Falling-Squares.md
new file mode 100755
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+++ b/website/content/ChapterFour/0600~0699/0699.Falling-Squares.md
@@ -0,0 +1,152 @@
+# [699. Falling Squares](https://leetcode.com/problems/falling-squares/)
+
+
+## 题目
+
+On an infinite number line (x-axis), we drop given squares in the order they are given.
+
+The `i`-th square dropped (`positions[i] = (left, side_length)`) is a square with the left-most point being `positions[i][0]` and sidelength `positions[i][1]`.
+
+The square is dropped with the bottom edge parallel to the number line, and from a higher height than all currently landed squares. We wait for each square to stick before dropping the next.
+
+The squares are infinitely sticky on their bottom edge, and will remain fixed to any positive length surface they touch (either the number line or another square). Squares dropped adjacent to each other will not stick together prematurely.
+
+Return a list `ans` of heights. Each height `ans[i]` represents the current highest height of any square we have dropped, after dropping squares represented by `positions[0], positions[1], ..., positions[i]`.
+
+**Example 1**:
+
+    Input: [[1, 2], [2, 3], [6, 1]]
+    Output: [2, 5, 5]
+    Explanation:
+
+After the first drop of `positions[0] = [1, 2]: _aa _aa -------` The maximum height of any square is 2.
+
+After the second drop of `positions[1] = [2, 3]: __aaa __aaa __aaa _aa__ _aa__ --------------` The maximum height of any square is 5. The larger square stays on top of the smaller square despite where its center of gravity is, because squares are infinitely sticky on their bottom edge.
+
+After the third drop of `positions[1] = [6, 1]: __aaa __aaa __aaa _aa _aa___a --------------` The maximum height of any square is still 5. Thus, we return an answer of `[2, 5, 5]`.
+
+**Example 2**:
+
+    Input: [[100, 100], [200, 100]]
+    Output: [100, 100]
+    Explanation: Adjacent squares don't get stuck prematurely - only their bottom edge can stick to surfaces.
+
+**Note**:
+
+- `1 <= positions.length <= 1000`.
+- `1 <= positions[i][0] <= 10^8`.
+- `1 <= positions[i][1] <= 10^6`.
+
+
+## 题目大意
+
+在无限长的数轴（即 x 轴）上，我们根据给定的顺序放置对应的正方形方块。第 i 个掉落的方块（positions[i] = (left, side\_length)）是正方形，其中 left 表示该方块最左边的点位置(positions[i][0])，side\_length 表示该方块的边长(positions[i][1])。
+
+每个方块的底部边缘平行于数轴（即 x 轴），并且从一个比目前所有的落地方块更高的高度掉落而下。在上一个方块结束掉落，并保持静止后，才开始掉落新方块。方块的底边具有非常大的粘性，并将保持固定在它们所接触的任何长度表面上（无论是数轴还是其他方块）。邻接掉落的边不会过早地粘合在一起，因为只有底边才具有粘性。
+
+返回一个堆叠高度列表 ans 。每一个堆叠高度 ans[i] 表示在通过 positions[0], positions[1], ..., positions[i] 表示的方块掉落结束后，目前所有已经落稳的方块堆叠的最高高度。
+
+示例 1:
+
+```c
+输入: [[1, 2], [2, 3], [6, 1]]
+输出: [2, 5, 5]
+解释:
+
+第一个方块 positions[0] = [1, 2] 掉落：
+_aa
+_aa
+-------
+方块最大高度为 2 。
+
+第二个方块 positions[1] = [2, 3] 掉落：
+__aaa
+__aaa
+__aaa
+_aa__
+_aa__
+--------------
+方块最大高度为5。
+大的方块保持在较小的方块的顶部，不论它的重心在哪里，因为方块的底部边缘有非常大的粘性。
+
+第三个方块 positions[1] = [6, 1] 掉落：
+__aaa
+__aaa
+__aaa
+_aa
+_aa___a
+-------------- 
+方块最大高度为5。
+
+因此，我们返回结果[2, 5, 5]。
+
+```
+
+注意:
+
+- 1 <= positions.length <= 1000.
+- 1 <= positions[i][0] <= 10^8.
+- 1 <= positions[i][1] <= 10^6.
+ 
+
+
+## 解题思路
+
+- 给出一个二维数组，每个一维数组中只有 2 个值，分别代表的是正方形砖块所在 x 轴的坐标起始点，和边长。要求输出每次砖块落下以后，当前最大的高度。正方形砖块落下如同俄罗斯方块，落下的过程中如果遇到了砖块会落在砖块的上面。如果砖块摞起来了以后，下方有空间，是不可能再把砖块挪进去的，因为此题砖块只会垂直落下，不会水平移动(这一点和俄罗斯方块不同)。
+- 这一题可以用线段树解答。由于方块在 x 轴上的坐标范围特别大，如果不离散化，这一题就会 MTE。所以首先去重 - 排序 - 离散化。先把每个砖块所在区间都算出来，每个正方形的方块所在区间是 `[pos[0] , pos[0]+pos[1]-1]` ，为什么右边界要减一呢？因为每个方块占据的区间其实应该是左闭右开的，即 `[pos[0] , pos[0]+pos[1])`，如果右边是开的，那么这个边界会被 2 个区间查询共用，从而导致错误结果。例如 [2,3]，[3,4]，这两个区间的砖块实际是不会摞在一起的。但是如果右边都是闭区间，用线段树 query 查询的时候，会都找到 [3，3]，从而导致这两个区间都会判断 3 这一点的情况。正确的做法应该是 [2,3)，[3,4）这样就避免了上述可能导致错误的情况了。离散化以后，所有的坐标区间都在 0~n 之间了。
+- 遍历每个砖块所在区间，先查询这个区间内的值，再加上当前砖块的高度，即为这个区间的最新高度。并更新该区间的值。更新区间的值用到懒惰更新。然后和动态维护的当前最大高度进行比较，将最大值放入最终输出的数组中。
+- 类似的题目有：第 715 题，第 218 题，第 732 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和本题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
+- leetcode  上也有线段树的讲解：[Get Solutions to Interview Questions](https://leetcode.com/articles/a-recursive-approach-to-segment-trees-range-sum-queries-lazy-propagation/)
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func fallingSquares(positions [][]int) []int {
+	st, ans, posMap, maxHeight := template.SegmentTree{}, make([]int, 0, len(positions)), discretization(positions), 0
+	tmp := make([]int, len(posMap))
+	st.Init(tmp, func(i, j int) int {
+		return max(i, j)
+	})
+	for _, p := range positions {
+		h := st.QueryLazy(posMap[p[0]], posMap[p[0]+p[1]-1]) + p[1]
+		st.UpdateLazy(posMap[p[0]], posMap[p[0]+p[1]-1], h)
+		maxHeight = max(maxHeight, h)
+		ans = append(ans, maxHeight)
+	}
+	return ans
+}
+
+func discretization(positions [][]int) map[int]int {
+	tmpMap, posArray, posMap := map[int]int{}, []int{}, map[int]int{}
+	for _, pos := range positions {
+		tmpMap[pos[0]]++
+		tmpMap[pos[0]+pos[1]-1]++
+	}
+	for k := range tmpMap {
+		posArray = append(posArray, k)
+	}
+	sort.Ints(posArray)
+	for i, pos := range posArray {
+		posMap[pos] = i
+	}
+	return posMap
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0697.Degree-of-an-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0700.Search-in-a-Binary-Search-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0600~0699/_index.md b/website/content/ChapterFour/0600~0699/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/0600~0699/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/0700~0799/0700.Search-in-a-Binary-Search-Tree.md b/website/content/ChapterFour/0700~0799/0700.Search-in-a-Binary-Search-Tree.md
new file mode 100644
index 000000000..dd0072acb
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0700.Search-in-a-Binary-Search-Tree.md
@@ -0,0 +1,80 @@
+# [700. Search in a Binary Search Tree](https://leetcode.com/problems/search-in-a-binary-search-tree/)
+
+## 题目
+
+You are given the root of a binary search tree (BST) and an integer val.
+
+Find the node in the BST that the node's value equals val and return the subtree rooted with that node. If such a node does not exist, return null.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2021/01/12/tree1.jpg](https://assets.leetcode.com/uploads/2021/01/12/tree1.jpg)
+
+    Input: root = [4,2,7,1,3], val = 2
+    Output: [2,1,3]
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2021/01/12/tree2.jpg](https://assets.leetcode.com/uploads/2021/01/12/tree2.jpg)
+
+    Input: root = [4,2,7,1,3], val = 5
+    Output: []
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range [1, 5000].
+- 1 <= Node.val <= 10000000
+- root is a binary search tree.
+- 1 <= val <= 10000000
+
+## 题目大意
+
+给定二叉搜索树（BST）的根节点和一个值。 你需要在BST中找到节点值等于给定值的节点。 返回以该节点为根的子树。 如果节点不存在，则返回 NULL。
+
+## 解题思路
+
+- 根据二叉搜索树的性质(根节点的值大于左子树所有节点的值，小于右子树所有节点的值),进行递归求解
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func searchBST(root *TreeNode, val int) *TreeNode {
+	if root == nil {
+		return nil
+	}
+	if root.Val == val {
+		return root
+	} else if root.Val < val {
+		return searchBST(root.Right, val)
+	} else {
+		return searchBST(root.Left, val)
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0699.Falling-Squares/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0701.Insert-into-a-Binary-Search-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0701.Insert-into-a-Binary-Search-Tree.md b/website/content/ChapterFour/0700~0799/0701.Insert-into-a-Binary-Search-Tree.md
new file mode 100644
index 000000000..bbfc43a22
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0701.Insert-into-a-Binary-Search-Tree.md
@@ -0,0 +1,95 @@
+# [701. Insert into a Binary Search Tree](https://leetcode.com/problems/insert-into-a-binary-search-tree/)
+
+
+## 题目
+
+You are given the `root` node of a binary search tree (BST) and a `value` to insert into the tree. Return *the root node of the BST after the insertion*. It is **guaranteed** that the new value does not exist in the original BST.
+
+**Notice** that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return **any of them**.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/10/05/insertbst.jpg](https://assets.leetcode.com/uploads/2020/10/05/insertbst.jpg)
+
+```
+Input: root = [4,2,7,1,3], val = 5
+Output: [4,2,7,1,3,5]
+Explanation: Another accepted tree is:
+
+```
+
+![https://assets.leetcode.com/uploads/2020/10/05/bst.jpg](https://assets.leetcode.com/uploads/2020/10/05/bst.jpg)
+
+**Example 2:**
+
+```
+Input: root = [40,20,60,10,30,50,70], val = 25
+Output: [40,20,60,10,30,50,70,null,null,25]
+
+```
+
+**Example 3:**
+
+```
+Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
+Output: [4,2,7,1,3,5]
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree will be in the range `[0, 104]`.
+- `108 <= Node.val <= 108`
+- All the values `Node.val` are **unique**.
+- `108 <= val <= 108`
+- It's **guaranteed** that `val` does not exist in the original BST.
+
+## 题目大意
+
+给定二叉搜索树（BST）的根节点和要插入树中的值，将值插入二叉搜索树。 返回插入后二叉搜索树的根节点。 输入数据 保证 ，新值和原始二叉搜索树中的任意节点值都不同。注意，可能存在多种有效的插入方式，只要树在插入后仍保持为二叉搜索树即可。 你可以返回 任意有效的结果 。
+
+## 解题思路
+
+- 简单题。插入节点的方法有多种，笔者这里选择一种简单的方法。从根开始遍历这个二叉树，当前节点的值比待插入节点的值小，则往右遍历；当前节点的值比待插入节点的值大，则往左遍历。最后遍历到空节点便是要插入的地方。
+
+## 代码
+
+```go
+package leetcode
+
+import "github.com/halfrost/leetcode-go/structures"
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func insert(n *TreeNode, val int) *TreeNode {
+	if n == nil {
+		return &TreeNode{Val: val}
+	}
+	if n.Val < val {
+		n.Right = insert(n.Right, val)
+	} else {
+		n.Left = insert(n.Left, val)
+	}
+	return n
+}
+
+func insertIntoBST(root *TreeNode, val int) *TreeNode {
+	return insert(root, val)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0700.Search-in-a-Binary-Search-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0703.Kth-Largest-Element-in-a-Stream/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0703.Kth-Largest-Element-in-a-Stream.md b/website/content/ChapterFour/0700~0799/0703.Kth-Largest-Element-in-a-Stream.md
new file mode 100644
index 000000000..0cce2cbfa
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0703.Kth-Largest-Element-in-a-Stream.md
@@ -0,0 +1,100 @@
+# [703. Kth Largest Element in a Stream](https://leetcode.com/problems/kth-largest-element-in-a-stream/)
+
+## 题目
+
+Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.
+
+Implement `KthLargest` class:
+
+- `KthLargest(int k, int[] nums)` Initializes the object with the integer `k` and the stream of integers `nums`.
+- `int add(int val)` Returns the element representing the `kth` largest element in the stream.
+
+**Example 1:**
+
+```
+Input
+["KthLargest", "add", "add", "add", "add", "add"]
+[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
+Output
+[null, 4, 5, 5, 8, 8]
+
+Explanation
+KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
+kthLargest.add(3);   // return 4
+kthLargest.add(5);   // return 5
+kthLargest.add(10);  // return 5
+kthLargest.add(9);   // return 8
+kthLargest.add(4);   // return 8
+
+```
+
+**Constraints:**
+
+- `1 <= k <= 104`
+- `0 <= nums.length <= 104`
+- `104 <= nums[i] <= 104`
+- `104 <= val <= 104`
+- At most `104` calls will be made to `add`.
+- It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element.
+
+## 题目大意
+
+设计一个找到数据流中第 k 大元素的类（class）。注意是排序后的第 k 大元素，不是第 k 个不同的元素。请实现 KthLargest 类：
+
+- KthLargest(int k, int[] nums) 使用整数 k 和整数流 nums 初始化对象。
+- int add(int val) 将 val 插入数据流 nums 后，返回当前数据流中第 k 大的元素。
+
+## 解题思路
+
+- 读完题就能明白这一题考察的是最小堆。构建一个长度为 K 的最小堆，每次 pop 堆首(堆中最小的元素)，维护堆首即为第 K 大元素。
+- 这里有一个简洁的写法，常规的构建一个 pq 优先队列需要自己新建一个类型，然后实现 Len()、Less()、Swap()、Push()、Pop() 这 5 个方法。在 sort 包里有一个现成的最小堆，sort.IntSlice。可以借用它，再自己实现 Push()、Pop()就可以使用最小堆了，节约一部分代码。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+type KthLargest struct {
+	sort.IntSlice
+	k int
+}
+
+func Constructor(k int, nums []int) KthLargest {
+	kl := KthLargest{k: k}
+	for _, val := range nums {
+		kl.Add(val)
+	}
+	return kl
+}
+
+func (kl *KthLargest) Push(v interface{}) {
+	kl.IntSlice = append(kl.IntSlice, v.(int))
+}
+
+func (kl *KthLargest) Pop() interface{} {
+	a := kl.IntSlice
+	v := a[len(a)-1]
+	kl.IntSlice = a[:len(a)-1]
+	return v
+}
+
+func (kl *KthLargest) Add(val int) int {
+	heap.Push(kl, val)
+	if kl.Len() > kl.k {
+		heap.Pop(kl)
+	}
+	return kl.IntSlice[0]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0701.Insert-into-a-Binary-Search-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0704.Binary-Search/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0704.Binary-Search.md b/website/content/ChapterFour/0700~0799/0704.Binary-Search.md
new file mode 100755
index 000000000..383ad604b
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0704.Binary-Search.md
@@ -0,0 +1,74 @@
+# [704. Binary Search](https://leetcode.com/problems/binary-search/)
+
+
+## 题目
+
+Given a **sorted** (in ascending order) integer array `nums` of `n` elements and a `target` value, write a function to search `target` in `nums`. If `target` exists, then return its index, otherwise return `-1`.
+
+**Example 1**:
+
+    Input: nums = [-1,0,3,5,9,12], target = 9
+    Output: 4
+    Explanation: 9 exists in nums and its index is 4
+
+**Example 2**:
+
+    Input: nums = [-1,0,3,5,9,12], target = 2
+    Output: -1
+    Explanation: 2 does not exist in nums so return -1
+
+**Note**:
+
+1. You may assume that all elements in `nums` are unique.
+2. `n` will be in the range `[1, 10000]`.
+3. The value of each element in `nums` will be in the range `[-9999, 9999]`.
+
+
+## 题目大意
+
+
+给定一个 n 个元素有序的（升序）整型数组 nums 和一个目标值 target  ，写一个函数搜索 nums 中的 target，如果目标值存在返回下标，否则返回 -1。
+
+提示：
+
+- 你可以假设 nums 中的所有元素是不重复的。
+- n 将在 [1, 10000]之间。
+- nums 的每个元素都将在 [-9999, 9999]之间。
+
+
+## 解题思路
+
+
+- 给出一个数组，要求在数组中搜索等于 target 的元素的下标。如果找到就输出下标，如果找不到输出 -1 。
+- 简单题，二分搜索的裸题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func search704(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[mid] == target {
+			return mid
+		} else if nums[mid] > target {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0703.Kth-Largest-Element-in-a-Stream/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0705.Design-HashSet/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0705.Design-HashSet.md b/website/content/ChapterFour/0700~0799/0705.Design-HashSet.md
new file mode 100755
index 000000000..c535d896a
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0705.Design-HashSet.md
@@ -0,0 +1,101 @@
+# [705. Design HashSet](https://leetcode.com/problems/design-hashset/)
+
+
+## 题目
+
+Design a HashSet without using any built-in hash table libraries.
+
+To be specific, your design should include these functions:
+
+- `add(value)`: Insert a value into the HashSet.
+- `contains(value)` : Return whether the value exists in the HashSet or not.
+- `remove(value)`: Remove a value in the HashSet. If the value does not exist in the HashSet, do nothing.
+
+**Example**:
+
+    MyHashSet hashSet = new MyHashSet();
+    hashSet.add(1);         
+    hashSet.add(2);         
+    hashSet.contains(1);    // returns true
+    hashSet.contains(3);    // returns false (not found)
+    hashSet.add(2);          
+    hashSet.contains(2);    // returns true
+    hashSet.remove(2);          
+    hashSet.contains(2);    // returns false (already removed)
+
+**Note**:
+
+- All values will be in the range of `[0, 1000000]`.
+- The number of operations will be in the range of `[1, 10000]`.
+- Please do not use the built-in HashSet library.
+
+
+## 题目大意
+
+不使用任何内建的哈希表库设计一个哈希集合具体地说，你的设计应该包含以下的功能：
+
+- add(value)：向哈希集合中插入一个值。
+- contains(value) ：返回哈希集合中是否存在这个值。
+- remove(value)：将给定值从哈希集合中删除。如果哈希集合中没有这个值，什么也不做。
+
+
+注意：
+
+- 所有的值都在 [1, 1000000] 的范围内。
+- 操作的总数目在 [1, 10000] 范围内。
+- 不要使用内建的哈希集合库。
+
+
+
+## 解题思路
+
+
+- 简单题，设计一个 hashset 的数据结构，要求有 `add(value)`，`contains(value)`，`remove(value)`，这 3 个方法。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+type MyHashSet struct {
+	data []bool
+}
+
+/** Initialize your data structure here. */
+func Constructor705() MyHashSet {
+	return MyHashSet{
+		data: make([]bool, 1000001),
+	}
+}
+
+func (this *MyHashSet) Add(key int) {
+	this.data[key] = true
+}
+
+func (this *MyHashSet) Remove(key int) {
+	this.data[key] = false
+}
+
+/** Returns true if this set contains the specified element */
+func (this *MyHashSet) Contains(key int) bool {
+	return this.data[key]
+}
+
+/**
+ * Your MyHashSet object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.Add(key);
+ * obj.Remove(key);
+ * param_3 := obj.Contains(key);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0704.Binary-Search/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0706.Design-HashMap/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0706.Design-HashMap.md b/website/content/ChapterFour/0700~0799/0706.Design-HashMap.md
new file mode 100755
index 000000000..e6b441607
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0706.Design-HashMap.md
@@ -0,0 +1,158 @@
+# [706. Design HashMap](https://leetcode.com/problems/design-hashmap/)
+
+
+## 题目
+
+Design a HashMap without using any built-in hash table libraries.
+
+To be specific, your design should include these functions:
+
+- `put(key, value)` : Insert a (key, value) pair into the HashMap. If the value already exists in the HashMap, update the value.
+- `get(key)`: Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
+- `remove(key)` : Remove the mapping for the value key if this map contains the mapping for the key.
+
+**Example**:
+
+    MyHashMap hashMap = new MyHashMap();
+    hashMap.put(1, 1);          
+    hashMap.put(2, 2);         
+    hashMap.get(1);            // returns 1
+    hashMap.get(3);            // returns -1 (not found)
+    hashMap.put(2, 1);          // update the existing value
+    hashMap.get(2);            // returns 1 
+    hashMap.remove(2);          // remove the mapping for 2
+    hashMap.get(2);            // returns -1 (not found)
+
+**Note**:
+
+- All keys and values will be in the range of `[0, 1000000]`.
+- The number of operations will be in the range of `[1, 10000]`.
+- Please do not use the built-in HashMap library.
+
+
+## 题目大意
+
+不使用任何内建的哈希表库设计一个哈希映射具体地说，你的设计应该包含以下的功能：
+
+- put(key, value)：向哈希映射中插入(键,值)的数值对。如果键对应的值已经存在，更新这个值。
+- get(key)：返回给定的键所对应的值，如果映射中不包含这个键，返回 -1。
+- remove(key)：如果映射中存在这个键，删除这个数值对。
+
+注意：
+
+- 所有的值都在 [1, 1000000] 的范围内。
+- 操作的总数目在 [1, 10000] 范围内。
+- 不要使用内建的哈希库。
+
+
+## 解题思路
+
+
+- 简单题，设计一个 hashmap 的数据结构，要求有 `put(key, value)`，`get(key)`，`remove(key)`，这 3 个方法。设计一个 map 主要需要处理哈希冲突，一般都是链表法解决冲突。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+const Len int = 100000
+
+type MyHashMap struct {
+	content [Len]*HashNode
+}
+
+type HashNode struct {
+	key  int
+	val  int
+	next *HashNode
+}
+
+func (N *HashNode) Put(key int, value int) {
+	if N.key == key {
+		N.val = value
+		return
+	}
+	if N.next == nil {
+		N.next = &HashNode{key, value, nil}
+		return
+	}
+	N.next.Put(key, value)
+}
+
+func (N *HashNode) Get(key int) int {
+	if N.key == key {
+		return N.val
+	}
+	if N.next == nil {
+		return -1
+	}
+	return N.next.Get(key)
+}
+
+func (N *HashNode) Remove(key int) *HashNode {
+	if N.key == key {
+		p := N.next
+		N.next = nil
+		return p
+	}
+	if N.next != nil {
+		N.next = N.next.Remove(key)
+	}
+	return N
+}
+
+/** Initialize your data structure here. */
+func Constructor706() MyHashMap {
+	return MyHashMap{}
+}
+
+/** value will always be non-negative. */
+func (this *MyHashMap) Put(key int, value int) {
+	node := this.content[this.Hash(key)]
+	if node == nil {
+		this.content[this.Hash(key)] = &HashNode{key: key, val: value, next: nil}
+		return
+	}
+	node.Put(key, value)
+}
+
+/** Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key */
+func (this *MyHashMap) Get(key int) int {
+	HashNode := this.content[this.Hash(key)]
+	if HashNode == nil {
+		return -1
+	}
+	return HashNode.Get(key)
+}
+
+/** Removes the mapping of the specified value key if this map contains a mapping for the key */
+func (this *MyHashMap) Remove(key int) {
+	HashNode := this.content[this.Hash(key)]
+	if HashNode == nil {
+		return
+	}
+	this.content[this.Hash(key)] = HashNode.Remove(key)
+}
+
+func (this *MyHashMap) Hash(value int) int {
+	return value % Len
+}
+
+/**
+ * Your MyHashMap object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.Put(key,value);
+ * param_2 := obj.Get(key);
+ * obj.Remove(key);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0705.Design-HashSet/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0707.Design-Linked-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0707.Design-Linked-List.md b/website/content/ChapterFour/0700~0799/0707.Design-Linked-List.md
new file mode 100644
index 000000000..93aaf77b9
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0707.Design-Linked-List.md
@@ -0,0 +1,163 @@
+# [707. Design Linked List](https://leetcode.com/problems/design-linked-list/)
+
+## 题目
+
+Design your implementation of the linked list. You can choose to use the singly linked list or the doubly linked list. A node in a singly linked list should have two attributes: val and next. val is the value of the current node, and next is a pointer/reference to the next node. If you want to use the doubly linked list, you will need one more attribute prev to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed.
+
+Implement these functions in your linked list class:
+
+- get(index) : Get the value of the index-th node in the linked list. If the index is invalid, return -1.
+- addAtHead(val) : Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
+- addAtTail(val) : Append a node of value val to the last element of the linked list.
+- addAtIndex(index, val) : Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
+- deleteAtIndex(index) : Delete the index-th node in the linked list, if the index is valid.
+
+**Example**:
+
+```
+
+MyLinkedList linkedList = new MyLinkedList();
+linkedList.addAtHead(1);
+linkedList.addAtTail(3);
+linkedList.addAtIndex(1, 2);  // linked list becomes 1->2->3
+linkedList.get(1);            // returns 2
+linkedList.deleteAtIndex(1);  // now the linked list is 1->3
+linkedList.get(1);            // returns 3
+
+```
+
+**Note**:
+
+- All values will be in the range of [1, 1000].
+- The number of operations will be in the range of [1, 1000].
+- Please do not use the built-in LinkedList library.
+
+## 题目大意
+
+这道题比较简单，设计一个链表，实现相关操作即可。
+
+## 解题思路
+
+这题有一个地方比较坑，题目中 Note 里面写的数值取值范围是 [1, 1000]，笔者把 0 当做无效值。结果 case 里面出现了 0 是有效值。case 和题意不符。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+type MyLinkedList struct {
+	head *Node
+}
+
+type Node struct {
+	Val  int
+	Next *Node
+	Prev *Node
+}
+
+/** Initialize your data structure here. */
+func Constructor() MyLinkedList {
+	return MyLinkedList{}
+}
+
+/** Get the value of the index-th node in the linked list. If the index is invalid, return -1. */
+func (this *MyLinkedList) Get(index int) int {
+	curr := this.head
+	for i := 0; i < index && curr != nil; i++ {
+		curr = curr.Next
+	}
+	if curr != nil {
+		return curr.Val
+	} else {
+		return -1
+	}
+}
+
+/** Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list. */
+func (this *MyLinkedList) AddAtHead(val int) {
+	node := &Node{Val: val}
+	node.Next = this.head
+	if this.head != nil {
+		this.head.Prev = node
+	}
+	this.head = node
+}
+
+/** Append a node of value val to the last element of the linked list. */
+func (this *MyLinkedList) AddAtTail(val int) {
+	if this.head == nil {
+		this.AddAtHead(val)
+		return
+	}
+	node := &Node{Val: val}
+	curr := this.head
+	for curr != nil && curr.Next != nil {
+		curr = curr.Next
+	}
+	node.Prev = curr
+	curr.Next = node
+}
+
+/** Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted. */
+func (this *MyLinkedList) AddAtIndex(index int, val int) {
+	if index == 0 {
+		this.AddAtHead(val)
+	} else {
+		node := &Node{Val: val}
+		curr := this.head
+		for i := 0; i < index-1 && curr != nil; i++ {
+			curr = curr.Next
+		}
+		if curr != nil {
+			node.Next = curr.Next
+			node.Prev = curr
+			if node.Next != nil {
+				node.Next.Prev = node
+			}
+			curr.Next = node
+		}
+	}
+}
+
+/** Delete the index-th node in the linked list, if the index is valid. */
+func (this *MyLinkedList) DeleteAtIndex(index int) {
+	if index == 0 {
+		this.head = this.head.Next
+		if this.head != nil {
+			this.head.Prev = nil
+		}
+	} else {
+		curr := this.head
+		for i := 0; i < index-1 && curr != nil; i++ {
+			curr = curr.Next
+		}
+		if curr != nil && curr.Next != nil {
+			curr.Next = curr.Next.Next
+			if curr.Next != nil {
+				curr.Next.Prev = curr
+			}
+		}
+	}
+}
+
+/**
+ * Your MyLinkedList object will be instantiated and called as such:
+ * obj := Constructor();
+ * param_1 := obj.Get(index);
+ * obj.AddAtHead(val);
+ * obj.AddAtTail(val);
+ * obj.AddAtIndex(index,val);
+ * obj.DeleteAtIndex(index);
+ */
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0706.Design-HashMap/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0709.To-Lower-Case/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0709.To-Lower-Case.md b/website/content/ChapterFour/0700~0799/0709.To-Lower-Case.md
new file mode 100644
index 000000000..b88578cf6
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0709.To-Lower-Case.md
@@ -0,0 +1,62 @@
+# [709. To Lower Case](https://leetcode.com/problems/to-lower-case/)
+
+
+## 题目
+
+Given a string `s`, return *the string after replacing every uppercase letter with the same lowercase letter*.
+
+**Example 1:**
+
+```
+Input: s = "Hello"
+Output: "hello"
+```
+
+**Example 2:**
+
+```
+Input: s = "here"
+Output: "here"
+```
+
+**Example 3:**
+
+```
+Input: s = "LOVELY"
+Output: "lovely"
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 100`
+- `s` consists of printable ASCII characters.
+
+## 题目大意
+
+给你一个字符串 s ，将该字符串中的大写字母转换成相同的小写字母，返回新的字符串。
+
+## 解题思路
+
+- 简单题，将字符串中的大写字母转换成小写字母。
+
+## 代码
+
+```go
+func toLowerCase(s string) string {
+    runes := [] rune(s)
+    diff := 'a' - 'A'
+    for i := 0; i < len(s); i++ {
+        if runes[i] >= 'A' && runes[i] <= 'Z' {
+            runes[i] += diff
+        }
+    }
+    return string(runes)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0707.Design-Linked-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0710.Random-Pick-with-Blacklist/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0710.Random-Pick-with-Blacklist.md b/website/content/ChapterFour/0700~0799/0710.Random-Pick-with-Blacklist.md
new file mode 100644
index 000000000..d0e9356c8
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0710.Random-Pick-with-Blacklist.md
@@ -0,0 +1,152 @@
+# [710. Random Pick with Blacklist](https://leetcode.com/problems/random-pick-with-blacklist/)
+
+## 题目
+
+Given a blacklist B containing unique integers from [0, N), write a function to return a uniform random integer from [0, N) which is NOT in B.
+
+Optimize it such that it minimizes the call to system’s Math.random().
+
+**Note**:
+
+1. 1 <= N <= 1000000000
+2. 0 <= B.length < min(100000, N)
+3. [0, N) does NOT include N. See interval notation.
+
+
+**Example 1**:
+
+```
+
+Input: 
+["Solution","pick","pick","pick"]
+[[1,[]],[],[],[]]
+Output: [null,0,0,0]
+
+```
+
+**Example 2**:
+
+```
+
+Input: 
+["Solution","pick","pick","pick"]
+[[2,[]],[],[],[]]
+Output: [null,1,1,1]
+
+```
+
+**Example 3**:
+
+```
+
+Input: 
+["Solution","pick","pick","pick"]
+[[3,[1]],[],[],[]]
+Output: [null,0,0,2]
+
+```
+
+**Example 4**:
+
+```
+
+Input: 
+["Solution","pick","pick","pick"]
+[[4,[2]],[],[],[]]
+Output: [null,1,3,1]
+
+```
+
+
+Explanation of Input Syntax:
+
+The input is two lists: the subroutines called and their arguments. Solution's constructor has two arguments, N and the blacklist B. pick has no arguments. Arguments are always wrapped with a list, even if there aren't any.
+
+
+## 题目大意
+
+给一个数字 N，再给一个黑名单 B，要求在 [0,N) 区间内随机输出一个数字，这个是不在黑名单 B 中的任意一个数字。
+
+## 解题思路
+
+这道题的 N 的范围特别大，最大是 10 亿。如果利用桶计数，开不出来这么大的数组。考虑到题目要求我们输出的数字是随机的，所以不需要存下所有的白名单的数字。
+
+假设 N=10, blacklist=[3, 5, 8, 9]
+
+![](https://s3-lc-upload.s3.amazonaws.com/users/cafebaby/image_1530657902.png)
+
+
+这一题有点类似 hash 冲突的意思。如果随机访问一个数，这个数正好在黑名单之内，那么就 hash 冲突了，我们就把它映射到另外一个不在黑名单里面的数中。如上图，我们可以将 3，5 重新映射到 7，6 的位置。这样末尾开始的几个数要么是黑名单里面的数，要么就是映射的数字。
+
+hash 表总长度应该为 M = N - len(backlist)，然后在 M 的长度中扫描是否有在黑名单中的数，如果有，就代表 hash 冲突了。冲突就把这个数字映射到 (M,N) 这个区间范围内。为了提高效率，可以选择这个区间的头部或者尾部开始映射，我选择的是末尾开始映射。从 (M,N) 这个区间的末尾开始往前找，找黑名单不存在的数，找到了就把 [0,M] 区间内冲突的数字映射到这里来。最后 pick 的时候，只需要查看 map 中是否存在映射关系，如果存在就输出 map 中映射之后的值，如果没有就代表没有冲突，直接输出那个 index 即可。
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math/rand"
+
+type Solution struct {
+	M        int
+	BlackMap map[int]int
+}
+
+func Constructor710(N int, blacklist []int) Solution {
+	blackMap := map[int]int{}
+	for i := 0; i < len(blacklist); i++ {
+		blackMap[blacklist[i]] = 1
+	}
+	M := N - len(blacklist)
+	for _, value := range blacklist {
+		if value < M {
+			for {
+				if _, ok := blackMap[N-1]; ok {
+					N--
+				} else {
+					break
+				}
+			}
+			blackMap[value] = N - 1
+			N--
+		}
+	}
+	return Solution{BlackMap: blackMap, M: M}
+}
+
+func (this *Solution) Pick() int {
+	idx := rand.Intn(this.M)
+	if _, ok := this.BlackMap[idx]; ok {
+		return this.BlackMap[idx]
+	}
+	return idx
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(N, blacklist);
+ * param_1 := obj.Pick();
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0709.To-Lower-Case/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0713.Subarray-Product-Less-Than-K/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0713.Subarray-Product-Less-Than-K.md b/website/content/ChapterFour/0700~0799/0713.Subarray-Product-Less-Than-K.md
new file mode 100644
index 000000000..11c50a4b8
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0713.Subarray-Product-Less-Than-K.md
@@ -0,0 +1,72 @@
+# [713. Subarray Product Less Than K](https://leetcode.com/problems/subarray-product-less-than-k/)
+
+## 题目
+
+Your are given an array of positive integers nums.
+
+Count and print the number of (contiguous) subarrays where the product of all the elements in the subarray is less than k.
+
+**Example 1**:
+
+```
+
+Input: nums = [10, 5, 2, 6], k = 100
+Output: 8
+Explanation: The 8 subarrays that have product less than 100 are: [10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6].
+Note that [10, 5, 2] is not included as the product of 100 is not strictly less than k.
+
+```
+
+
+**Note**:
+
+- 0 < nums.length <= 50000.
+- 0 < nums[i] < 1000.
+- 0 <= k < 10^6.
+
+## 题目大意
+
+给出一个数组，要求在输出符合条件的窗口数，条件是，窗口中所有数字乘积小于 K 。
+
+## 解题思路
+
+这道题也是滑动窗口的题目，在窗口滑动的过程中不断累乘，直到乘积大于 k，大于 k 的时候就缩小左窗口。有一种情况还需要单独处理一下，即类似 [100] 这种情况。这种情况窗口内乘积等于 k，不小于 k，左边窗口等于右窗口，这个时候需要左窗口和右窗口同时右移。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func numSubarrayProductLessThanK(nums []int, k int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	res, left, right, prod := 0, 0, 0, 1
+	for left < len(nums) {
+		if right < len(nums) && prod*nums[right] < k {
+			prod = prod * nums[right]
+			right++
+		} else if left == right {
+			left++
+			right++
+		} else {
+			res += right - left
+			prod = prod / nums[left]
+			left++
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0710.Random-Pick-with-Blacklist/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee.md b/website/content/ChapterFour/0700~0799/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee.md
new file mode 100755
index 000000000..055c8f848
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee.md
@@ -0,0 +1,87 @@
+# [714. Best Time to Buy and Sell Stock with Transaction Fee](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/)
+
+
+## 题目
+
+Your are given an array of integers `prices`, for which the `i`-th element is the price of a given stock on day `i`; and a non-negative integer `fee` representing a transaction fee.
+
+You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)
+
+Return the maximum profit you can make.
+
+**Example 1**:
+
+    Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
+    Output: 8
+    Explanation: The maximum profit can be achieved by:
+    Buying at prices[0] = 1
+    Selling at prices[3] = 8
+    Buying at prices[4] = 4
+    Selling at prices[5] = 9
+    The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
+
+**Note**:
+
+- `0 < prices.length <= 50000`.
+- `0 < prices[i] < 50000`.
+- `0 <= fee < 50000`.
+
+
+## 题目大意
+
+给定一个整数数组 prices，其中第 i 个元素代表了第 i 天的股票价格 ；非负整数 fee 代表了交易股票的手续费用。你可以无限次地完成交易，但是你每次交易都需要付手续费。如果你已经购买了一个股票，在卖出它之前你就不能再继续购买股票了。要求返回获得利润的最大值。
+
+
+
+## 解题思路
+
+- 给定一个数组，表示一支股票在每一天的价格。设计一个交易算法，在这些天进行自动交易，要求：每一天只能进行一次操作；在买完股票后，必须卖了股票，才能再次买入；每次卖了股票以后，需要缴纳一部分的手续费。问如何交易，能让利润最大？
+- 这一题是第 121 题、第 122 题、第 309 题的变种题。
+- 这一题的解题思路是 DP，需要维护买和卖的两种状态。`buy[i]` 代表第 `i` 天买入的最大收益，`sell[i]` 代表第 `i` 天卖出的最大收益，状态转移方程是 `buy[i] = max(buy[i-1], sell[i-1]-prices[i])`，`sell[i] = max(sell[i-1], buy[i-1]+prices[i]-fee)`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+)
+
+// 解法一 模拟 DP
+func maxProfit714(prices []int, fee int) int {
+	if len(prices) <= 1 {
+		return 0
+	}
+	buy, sell := make([]int, len(prices)), make([]int, len(prices))
+	for i := range buy {
+		buy[i] = math.MinInt64
+	}
+	buy[0] = -prices[0]
+	for i := 1; i < len(prices); i++ {
+		buy[i] = max(buy[i-1], sell[i-1]-prices[i])
+		sell[i] = max(sell[i-1], buy[i-1]+prices[i]-fee)
+	}
+	return sell[len(sell)-1]
+}
+
+// 解法二 优化辅助空间的 DP
+func maxProfit714_1(prices []int, fee int) int {
+	sell, buy := 0, -prices[0]
+	for i := 1; i < len(prices); i++ {
+		sell = max(sell, buy+prices[i]-fee)
+		buy = max(buy, sell-prices[i])
+	}
+	return sell
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0713.Subarray-Product-Less-Than-K/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0715.Range-Module/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0715.Range-Module.md b/website/content/ChapterFour/0700~0799/0715.Range-Module.md
new file mode 100755
index 000000000..27defe3dd
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0715.Range-Module.md
@@ -0,0 +1,277 @@
+# [715. Range Module](https://leetcode.com/problems/range-module/)
+
+
+## 题目
+
+A Range Module is a module that tracks ranges of numbers. Your task is to design and implement the following interfaces in an efficient manner.
+
+- `addRange(int left, int right)` Adds the half-open interval `[left, right)`, tracking every real number in that interval. Adding an interval that partially overlaps with currently tracked numbers should add any numbers in the interval `[left, right)` that are not already tracked.
+- `queryRange(int left, int right)` Returns true if and only if every real number in the interval `[left, right)` is currently being tracked.
+- `removeRange(int left, int right)` Stops tracking every real number currently being tracked in the interval `[left, right)`.
+
+**Example 1**:
+
+    addRange(10, 20): null
+    removeRange(14, 16): null
+    queryRange(10, 14): true (Every number in [10, 14) is being tracked)
+    queryRange(13, 15): false (Numbers like 14, 14.03, 14.17 in [13, 15) are not being tracked)
+    queryRange(16, 17): true (The number 16 in [16, 17) is still being tracked, despite the remove operation)
+
+**Note**:
+
+- A half open interval `[left, right)` denotes all real numbers `left <= x < right`.
+- `0 < left < right < 10^9` in all calls to `addRange, queryRange, removeRange`.
+- The total number of calls to `addRange` in a single test case is at most `1000`.
+- The total number of calls to `queryRange` in a single test case is at most `5000`.
+- The total number of calls to `removeRange` in a single test case is at most `1000`.
+
+## 题目大意
+
+Range 模块是跟踪数字范围的模块。你的任务是以一种有效的方式设计和实现以下接口。
+
+- addRange(int left, int right) 添加半开区间 [left, right)，跟踪该区间中的每个实数。添加与当前跟踪的数字部分重叠的区间时，应当添加在区间 [left, right) 中尚未跟踪的任何数字到该区间中。
+- queryRange(int left, int right) 只有在当前正在跟踪区间 [left, right) 中的每一个实数时，才返回 true。
+- removeRange(int left, int right) 停止跟踪区间 [left, right) 中当前正在跟踪的每个实数。
+ 
+
+示例：
+
+```
+addRange(10, 20): null
+removeRange(14, 16): null
+queryRange(10, 14): true （区间 [10, 14) 中的每个数都正在被跟踪）
+queryRange(13, 15): false （未跟踪区间 [13, 15) 中像 14, 14.03, 14.17 这样的数字）
+queryRange(16, 17): true （尽管执行了删除操作，区间 [16, 17) 中的数字 16 仍然会被跟踪）
+```
+
+提示：
+
+- 半开区间 [left, right) 表示所有满足 left <= x < right 的实数。
+- 对 addRange, queryRange, removeRange 的所有调用中 0 < left < right < 10^9。
+- 在单个测试用例中，对 addRange 的调用总数不超过 1000 次。
+- 在单个测试用例中，对  queryRange 的调用总数不超过 5000 次。
+- 在单个测试用例中，对 removeRange 的调用总数不超过 1000 次。
+
+
+## 解题思路
+
+- 设计一个数据结构，能完成添加区间 `addRange`，查询区间 `queryRange`，移除区间 `removeRange` 三种操作。查询区间的操作需要更加高效一点。
+- 这一题可以用线段树来解答，但是时间复杂度不高，最优解是用二叉排序树 BST 来解答。先来看线段树。这一题是更新区间内的值，所以需要用到懒惰更新。添加区间可以把区间内的值都赋值为 1 。由于题目中未预先确定区间范围，选用树的形式实现线段树比数组实现更加节约空间(当然用数组也可以，区间最大是 1000，点至多有 2000 个)。移除区间的时候就是把区间内的值都赋值标记为 0 。
+- 类似的题目有：第 699 题，第 218 题，第 732 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和第 699 题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
+
+## 代码
+
+```go
+
+package leetcode
+
+// RangeModule define
+type RangeModule struct {
+	Root *SegmentTreeNode
+}
+
+// SegmentTreeNode define
+type SegmentTreeNode struct {
+	Start, End  int
+	Tracked     bool
+	Lazy        int
+	Left, Right *SegmentTreeNode
+}
+
+// Constructor715 define
+func Constructor715() RangeModule {
+	return RangeModule{&SegmentTreeNode{0, 1e9, false, 0, nil, nil}}
+}
+
+// AddRange define
+func (rm *RangeModule) AddRange(left int, right int) {
+	update(rm.Root, left, right-1, true)
+}
+
+// QueryRange define
+func (rm *RangeModule) QueryRange(left int, right int) bool {
+	return query(rm.Root, left, right-1)
+}
+
+// RemoveRange define
+func (rm *RangeModule) RemoveRange(left int, right int) {
+	update(rm.Root, left, right-1, false)
+}
+
+func lazyUpdate(node *SegmentTreeNode) {
+	if node.Lazy != 0 {
+		node.Tracked = node.Lazy == 2
+	}
+	if node.Start != node.End {
+		if node.Left == nil || node.Right == nil {
+			m := node.Start + (node.End-node.Start)/2
+			node.Left = &SegmentTreeNode{node.Start, m, node.Tracked, 0, nil, nil}
+			node.Right = &SegmentTreeNode{m + 1, node.End, node.Tracked, 0, nil, nil}
+		} else if node.Lazy != 0 {
+			node.Left.Lazy = node.Lazy
+			node.Right.Lazy = node.Lazy
+		}
+	}
+	node.Lazy = 0
+}
+
+func update(node *SegmentTreeNode, start, end int, track bool) {
+	lazyUpdate(node)
+	if start > end || node == nil || end < node.Start || node.End < start {
+		return
+	}
+	if start <= node.Start && node.End <= end {
+		// segment completely covered by the update range
+		node.Tracked = track
+		if node.Start != node.End {
+			if track {
+				node.Left.Lazy = 2
+				node.Right.Lazy = 2
+			} else {
+				node.Left.Lazy = 1
+				node.Right.Lazy = 1
+			}
+		}
+		return
+	}
+	update(node.Left, start, end, track)
+	update(node.Right, start, end, track)
+	node.Tracked = node.Left.Tracked && node.Right.Tracked
+}
+
+func query(node *SegmentTreeNode, start, end int) bool {
+	lazyUpdate(node)
+	if start > end || node == nil || end < node.Start || node.End < start {
+		return true
+	}
+	if start <= node.Start && node.End <= end {
+		// segment completely covered by the update range
+		return node.Tracked
+	}
+	return query(node.Left, start, end) && query(node.Right, start, end)
+}
+
+// 解法二 BST
+// type RangeModule struct {
+// 	Root *BSTNode
+// }
+
+// type BSTNode struct {
+// 	Interval    []int
+// 	Left, Right *BSTNode
+// }
+
+// func Constructor715() RangeModule {
+// 	return RangeModule{}
+// }
+
+// func (this *RangeModule) AddRange(left int, right int) {
+// 	interval := []int{left, right - 1}
+// 	this.Root = insert(this.Root, interval)
+// }
+
+// func (this *RangeModule) RemoveRange(left int, right int) {
+// 	interval := []int{left, right - 1}
+// 	this.Root = delete(this.Root, interval)
+// }
+
+// func (this *RangeModule) QueryRange(left int, right int) bool {
+// 	return query(this.Root, []int{left, right - 1})
+// }
+
+// func (this *RangeModule) insert(root *BSTNode, interval []int) *BSTNode {
+// 	if root == nil {
+// 		return &BSTNode{interval, nil, nil}
+// 	}
+// 	if root.Interval[0] <= interval[0] && interval[1] <= root.Interval[1] {
+// 		return root
+// 	}
+// 	if interval[0] < root.Interval[0] {
+// 		root.Left = insert(root.Left, []int{interval[0], min(interval[1], root.Interval[0]-1)})
+// 	}
+// 	if root.Interval[1] < interval[1] {
+// 		root.Right = insert(root.Right, []int{max(interval[0], root.Interval[1]+1), interval[1]})
+// 	}
+// 	return root
+// }
+
+// func (this *RangeModule) delete(root *BSTNode, interval []int) *BSTNode {
+// 	if root == nil {
+// 		return nil
+// 	}
+// 	if interval[0] < root.Interval[0] {
+// 		root.Left = delete(root.Left, []int{interval[0], min(interval[1], root.Interval[0]-1)})
+// 	}
+// 	if root.Interval[1] < interval[1] {
+// 		root.Right = delete(root.Right, []int{max(interval[0], root.Interval[1]+1), interval[1]})
+// 	}
+// 	if interval[1] < root.Interval[0] || root.Interval[1] < interval[0] {
+// 		return root
+// 	}
+// 	if interval[0] <= root.Interval[0] && root.Interval[1] <= interval[1] {
+// 		if root.Left == nil {
+// 			return root.Right
+// 		} else if root.Right == nil {
+// 			return root.Left
+// 		} else {
+// 			pred := root.Left
+// 			for pred.Right != nil {
+// 				pred = pred.Right
+// 			}
+// 			root.Interval = pred.Interval
+// 			root.Left = delete(root.Left, pred.Interval)
+// 			return root
+// 		}
+// 	}
+// 	if root.Interval[0] < interval[0] && interval[1] < root.Interval[1] {
+// 		left := &BSTNode{[]int{root.Interval[0], interval[0] - 1}, root.Left, nil}
+// 		right := &BSTNode{[]int{interval[1] + 1, root.Interval[1]}, nil, root.Right}
+// 		left.Right = right
+// 		return left
+// 	}
+// 	if interval[0] <= root.Interval[0] {
+// 		root.Interval[0] = interval[1] + 1
+// 	}
+// 	if root.Interval[1] <= interval[1] {
+// 		root.Interval[1] = interval[0] - 1
+// 	}
+// 	return root
+// }
+
+// func (this *RangeModule) query(root *BSTNode, interval []int) bool {
+// 	if root == nil {
+// 		return false
+// 	}
+// 	if interval[1] < root.Interval[0] {
+// 		return query(root.Left, interval)
+// 	}
+// 	if root.Interval[1] < interval[0] {
+// 		return query(root.Right, interval)
+// 	}
+// 	left := true
+// 	if interval[0] < root.Interval[0] {
+// 		left = query(root.Left, []int{interval[0], root.Interval[0] - 1})
+// 	}
+// 	right := true
+// 	if root.Interval[1] < interval[1] {
+// 		right = query(root.Right, []int{root.Interval[1] + 1, interval[1]})
+// 	}
+// 	return left && right
+// }
+
+/**
+ * Your RangeModule object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.AddRange(left,right);
+ * param_2 := obj.QueryRange(left,right);
+ * obj.RemoveRange(left,right);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0717.1-bit-and-2-bit-Characters/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0717.1-bit-and-2-bit-Characters.md b/website/content/ChapterFour/0700~0799/0717.1-bit-and-2-bit-Characters.md
new file mode 100755
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+# [717. 1-bit and 2-bit Characters](https://leetcode.com/problems/1-bit-and-2-bit-characters/)
+
+
+## 题目:
+
+We have two special characters. The first character can be represented by one bit `0`. The second character can be represented by two bits (`10` or `11`).
+
+Now given a string represented by several bits. Return whether the last character must be a one-bit character or not. The given string will always end with a zero.
+
+**Example 1**:
+
+    Input: 
+    bits = [1, 0, 0]
+    Output: True
+    Explanation: 
+    The only way to decode it is two-bit character and one-bit character. So the last character is one-bit character.
+
+**Example 2**:
+
+    Input: 
+    bits = [1, 1, 1, 0]
+    Output: False
+    Explanation: 
+    The only way to decode it is two-bit character and two-bit character. So the last character is NOT one-bit character.
+
+**Note**:
+
+- `1 <= len(bits) <= 1000`.
+- `bits[i]` is always `0` or `1`.
+
+## 题目大意
+
+有两种特殊字符。第一种字符可以用一比特0来表示。第二种字符可以用两比特(10 或 11)来表示。
+
+现给一个由若干比特组成的字符串。问最后一个字符是否必定为一个一比特字符。给定的字符串总是由0结束。
+
+注意:
+
+- 1 <= len(bits) <= 1000.
+- bits[i] 总是0 或 1.
+
+
+## 解题思路
+
+- 给出一个数组，数组里面的元素只有 0 和 1，并且数组的最后一个元素一定是 0。有 2 种特殊的字符，第一类字符是 "0"，第二类字符是 "11" 和 "10"，请判断这个数组最后一个元素是否一定是属于第一类字符？
+- 依题意， 0 的来源有 2 处，可以是第一类字符，也可以是第二类字符，1 的来源只有 1 处，一定出自第二类字符。最后一个 0 当前仅当为第一类字符的情况有 2 种，第一种情况，前面出现有 0，但是 0 和 1 配对形成了第二类字符。第二种情况，前面没有出现 0 。这两种情况的共同点是除去最后一个元素，数组中前面所有的1 都“结对子”。所以利用第二类字符的特征，"1X"，遍历整个数组，如果遇到 "1"，就跳 2 步，因为 1 后面出现什么数字( 0 或者 1 )并不需要关心。如果 `i` 能在 `len(bits) - 1` 的地方`(数组最后一个元素)`停下，那么对应的是情况一或者情况二，前面的 0 都和 1 匹配上了，最后一个 0 一定是第一类字符。如果 `i` 在 `len(bit)` 的位置`(超出数组下标)`停下，说明 `bits[len(bits) - 1] == 1`，这个时候最后一个 0 一定属于第二类字符。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func isOneBitCharacter(bits []int) bool {
+	var i int
+	for i = 0; i < len(bits)-1; i++ {
+		if bits[i] == 1 {
+			i++
+		}
+	}
+	return i == len(bits)-1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0715.Range-Module/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0718.Maximum-Length-of-Repeated-Subarray/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0718.Maximum-Length-of-Repeated-Subarray.md b/website/content/ChapterFour/0700~0799/0718.Maximum-Length-of-Repeated-Subarray.md
new file mode 100755
index 000000000..004d6f820
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0718.Maximum-Length-of-Repeated-Subarray.md
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+# [718. Maximum Length of Repeated Subarray](https://leetcode.com/problems/maximum-length-of-repeated-subarray/)
+
+
+## 题目
+
+Given two integer arrays `A` and `B`, return the maximum length of an subarray that appears in both arrays.
+
+**Example 1**:
+
+    Input:
+    A: [1,2,3,2,1]
+    B: [3,2,1,4,7]
+    Output: 3
+    Explanation: 
+    The repeated subarray with maximum length is [3, 2, 1].
+
+**Note**:
+
+1. 1 <= len(A), len(B) <= 1000
+2. 0 <= A[i], B[i] < 100
+
+
+## 题目大意
+
+给两个整数数组 A 和 B ，返回两个数组中公共的、长度最长的子数组的长度。
+
+
+
+## 解题思路
+
+- 给出两个数组，求这两个数组中最长相同子串的长度。
+- 这一题最容易想到的是 DP 动态规划的解法。`dp[i][j]` 代表在 A 数组中以 `i` 下标开始的子串与 B 数组中以 `j` 下标开始的子串最长相同子串的长度，状态转移方程为 `dp[i][j] = dp[i+1][j+1] + 1` (当 `A[i] == B[j]`)。这种解法的时间复杂度是 O(n^2)，空间复杂度 O(n^2)。
+- 这一题最佳解法是二分搜索 + `Rabin-Karp`。比较相同子串耗时的地方在于，需要一层循环，遍历子串所有字符。但是如果比较两个数字就很快，`O(1)` 的时间复杂度。所以有人就想到了，能不能把字符串也映射成数字呢？这样比较起来就非常快。这个算法就是 `Rabin-Karp` 算法。字符串映射成一个数字不能随意映射，还要求能根据字符串前缀动态增加，比较下一个字符串的时候，可以利用已比较过的前缀，加速之后的字符串比较。在 Rabin-Karp 算法中有一个“码点”的概念。类似于10进制中的进制。具体的算法讲解可以见这篇：
+
+    [基础知识 - Rabin-Karp 算法](https://www.cnblogs.com/golove/p/3234673.html)
+
+    “码点”一般取值为一个素数。在 go 的 `strings` 包里面取值是 16777619。所以这一题也可以直接取这个值。由于这一次要求我们找最长长度，所以把最长长度作为二分搜索的目标。先将数组 A 和数组 B 中的数字都按照二分出来的长度，进行 `Rabin-Karp` hash。对 A 中的 hash 与下标做映射关系，存到 map 中，方便后面快速查找。然后遍历 B 中的 hash，当 hash 一致的时候，再匹配下标。如果下标存在，且拥有相同的前缀，那么就算找到了相同的子串了。最后就是不断的二分，找到最长的结果即可。这个解法的时间复杂度 O(n * log n)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+
+package leetcode
+
+const primeRK = 16777619
+
+// 解法一 二分搜索 + Rabin-Karp
+func findLength(A []int, B []int) int {
+	low, high := 0, min(len(A), len(B))
+	for low < high {
+		mid := (low + high + 1) >> 1
+		if hasRepeated(A, B, mid) {
+			low = mid
+		} else {
+			high = mid - 1
+		}
+	}
+	return low
+}
+
+func hashSlice(arr []int, length int) []int {
+	// hash 数组里面记录 arr 比 length 长出去部分的 hash 值
+	hash, pl, h := make([]int, len(arr)-length+1), 1, 0
+	for i := 0; i < length-1; i++ {
+		pl *= primeRK
+	}
+	for i, v := range arr {
+		h = h*primeRK + v
+		if i >= length-1 {
+			hash[i-length+1] = h
+			h -= pl * arr[i-length+1]
+		}
+	}
+	return hash
+}
+
+func hasSamePrefix(A, B []int, length int) bool {
+	for i := 0; i < length; i++ {
+		if A[i] != B[i] {
+			return false
+		}
+	}
+	return true
+}
+
+func hasRepeated(A, B []int, length int) bool {
+	hs := hashSlice(A, length)
+	hashToOffset := make(map[int][]int, len(hs))
+	for i, h := range hs {
+		hashToOffset[h] = append(hashToOffset[h], i)
+	}
+	for i, h := range hashSlice(B, length) {
+		if offsets, ok := hashToOffset[h]; ok {
+			for _, offset := range offsets {
+				if hasSamePrefix(A[offset:], B[i:], length) {
+					return true
+				}
+			}
+		}
+	}
+	return false
+}
+
+// 解法二 DP 动态规划
+func findLength1(A []int, B []int) int {
+	res, dp := 0, make([][]int, len(A)+1)
+	for i := range dp {
+		dp[i] = make([]int, len(B)+1)
+	}
+	for i := len(A) - 1; i >= 0; i-- {
+		for j := len(B) - 1; j >= 0; j-- {
+			if A[i] == B[j] {
+				dp[i][j] = dp[i+1][j+1] + 1
+				if dp[i][j] > res {
+					res = dp[i][j]
+				}
+			}
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0717.1-bit-and-2-bit-Characters/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0719.Find-K-th-Smallest-Pair-Distance/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0719.Find-K-th-Smallest-Pair-Distance.md b/website/content/ChapterFour/0700~0799/0719.Find-K-th-Smallest-Pair-Distance.md
new file mode 100755
index 000000000..3f85a6de9
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0719.Find-K-th-Smallest-Pair-Distance.md
@@ -0,0 +1,104 @@
+# [719. Find K-th Smallest Pair Distance](https://leetcode.com/problems/find-k-th-smallest-pair-distance/)
+
+
+## 题目
+
+Given an integer array, return the k-th smallest **distance** among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B.
+
+**Example 1**:
+
+    Input:
+    nums = [1,3,1]
+    k = 1
+    Output: 0 
+    Explanation:
+    Here are all the pairs:
+    (1,3) -> 2
+    (1,1) -> 0
+    (3,1) -> 2
+    Then the 1st smallest distance pair is (1,1), and its distance is 0.
+
+**Note**:
+
+1. `2 <= len(nums) <= 10000`.
+2. `0 <= nums[i] < 1000000`.
+3. `1 <= k <= len(nums) * (len(nums) - 1) / 2`.
+
+
+## 题目大意
+
+给定一个整数数组，返回所有数对之间的第 k 个最小距离。一对 (A, B) 的距离被定义为 A 和 B 之间的绝对差值。
+
+提示:
+
+1. 2 <= len(nums) <= 10000.
+2. 0 <= nums[i] < 1000000.
+3. 1 <= k <= len(nums) * (len(nums) - 1) / 2.
+
+
+
+## 解题思路
+
+- 给出一个数组，要求找出第 k 小两两元素之差的值。两两元素之差可能重复，重复的元素之差算多个，不去重。
+- 这一题可以用二分搜索来解答。先把原数组排序，那么最大的差值就是 `nums[len(nums)-1] - nums[0]` ，最小的差值是 0，即在 `[0, nums[len(nums)-1] - nums[0]]` 区间内搜索最终答案。针对每个 `mid`，判断小于等于 `mid` 的差值有多少个。题意就转化为，在数组中找到这样一个数，使得满足 `nums[i] - nums[j] ≤ mid` 条件的组合数等于 `k`。那么如何计算满足两两数的差值小于 mid 的组合总数是本题的关键。
+- 最暴力的方法就是 2 重循环，暴力计数。这个方法效率不高，耗时很长。原因是没有利用数组有序这一条件。实际上数组有序对计算满足条件的组合数有帮助。利用双指针滑动即可计算出组合总数。见解法一。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+func smallestDistancePair(nums []int, k int) int {
+	sort.Ints(nums)
+	low, high := 0, nums[len(nums)-1]-nums[0]
+	for low < high {
+		mid := low + (high-low)>>1
+		tmp := findDistanceCount(nums, mid)
+		if tmp >= k {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+// 解法一 双指针
+func findDistanceCount(nums []int, num int) int {
+	count, i := 0, 0
+	for j := 1; j < len(nums); j++ {
+		for nums[j]-nums[i] > num && i < j {
+			i++
+		}
+		count += (j - i)
+	}
+	return count
+}
+
+// 解法二 暴力查找
+func findDistanceCount1(nums []int, num int) int {
+	count := 0
+	for i := 0; i < len(nums); i++ {
+		for j := i + 1; j < len(nums); j++ {
+			if nums[j]-nums[i] <= num {
+				count++
+			}
+		}
+	}
+	return count
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0718.Maximum-Length-of-Repeated-Subarray/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0720.Longest-Word-in-Dictionary/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0720.Longest-Word-in-Dictionary.md b/website/content/ChapterFour/0700~0799/0720.Longest-Word-in-Dictionary.md
new file mode 100755
index 000000000..f35712bb4
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+++ b/website/content/ChapterFour/0700~0799/0720.Longest-Word-in-Dictionary.md
@@ -0,0 +1,79 @@
+# [720. Longest Word in Dictionary](https://leetcode.com/problems/longest-word-in-dictionary/)
+
+
+## 题目
+
+Given a list of strings `words` representing an English Dictionary, find the longest word in `words` that can be built one character at a time by other words in `words`. If there is more than one possible answer, return the longest word with the smallest lexicographical order.
+
+If there is no answer, return the empty string.
+
+**Example 1**:
+
+    Input: 
+    words = ["w","wo","wor","worl", "world"]
+    Output: "world"
+    Explanation: 
+    The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".
+
+**Example 2**:
+
+    Input: 
+    words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]
+    Output: "apple"
+    Explanation: 
+    Both "apply" and "apple" can be built from other words in the dictionary. However, "apple" is lexicographically smaller than "apply".
+
+**Note**:
+
+- All the strings in the input will only contain lowercase letters.
+- The length of `words` will be in the range `[1, 1000]`.
+- The length of `words[i]` will be in the range `[1, 30]`.
+
+
+## 题目大意
+
+给出一个字符串数组 words 组成的一本英语词典。从中找出最长的一个单词，该单词是由 words 词典中其他单词逐步添加一个字母组成。若其中有多个可行的答案，则返回答案中字典序最小的单词。若无答案，则返回空字符串。
+
+
+
+## 解题思路
+
+
+- 给出一个字符串数组，要求找到长度最长的，并且可以由字符串数组里面其他字符串拼接一个字符组成的字符串。如果存在多个这样的最长的字符串，则输出字典序较小的那个字符串，如果找不到这样的字符串，输出空字符串。
+- 这道题解题思路是先排序，排序完成以后就是字典序从小到大了。之后再用 map 辅助记录即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+func longestWord(words []string) string {
+	sort.Strings(words)
+	mp := make(map[string]bool)
+	var res string
+	for _, word := range words {
+		size := len(word)
+		if size == 1 || mp[word[:size-1]] {
+			if size > len(res) {
+				res = word
+			}
+			mp[word] = true
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0719.Find-K-th-Smallest-Pair-Distance/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0721.Accounts-Merge/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0721.Accounts-Merge.md b/website/content/ChapterFour/0700~0799/0721.Accounts-Merge.md
new file mode 100755
index 000000000..a7cb56781
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0721.Accounts-Merge.md
@@ -0,0 +1,156 @@
+# [721. Accounts Merge](https://leetcode.com/problems/accounts-merge/)
+
+
+## 题目
+
+Given a list `accounts`, each element `accounts[i]` is a list of strings, where the first element `accounts[i][0]` is a name, and the rest of the elements are emailsrepresenting emails of the account.
+
+Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some email that is common to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.
+
+After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails **in sorted order**. The accounts themselves can be returned in any order.
+
+**Example 1**:
+
+    Input: 
+    accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
+    Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'],  ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
+    Explanation: 
+    The first and third John's are the same person as they have the common email "johnsmith@mail.com".
+    The second John and Mary are different people as none of their email addresses are used by other accounts.
+    We could return these lists in any order, for example the answer [['Mary', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], 
+    ['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.
+
+**Note**:
+
+- The length of `accounts` will be in the range `[1, 1000]`.
+- The length of `accounts[i]` will be in the range `[1, 10]`.
+- The length of `accounts[i][j]` will be in the range `[1, 30]`.
+
+
+## 题目大意
+
+
+给定一个列表 accounts，每个元素 accounts[i] 是一个字符串列表，其中第一个元素 accounts[i][0] 是 名称 (name)，其余元素是 emails 表示该帐户的邮箱地址。现在，我们想合并这些帐户。如果两个帐户都有一些共同的邮件地址，则两个帐户必定属于同一个人。请注意，即使两个帐户具有相同的名称，它们也可能属于不同的人，因为人们可能具有相同的名称。一个人最初可以拥有任意数量的帐户，但其所有帐户都具有相同的名称。合并帐户后，按以下格式返回帐户：每个帐户的第一个元素是名称，其余元素是按顺序排列的邮箱地址。accounts 本身可以以任意顺序返回。
+
+
+注意：
+
+- accounts 的长度将在 [1，1000] 的范围内。
+- accounts[i] 的长度将在 [1，10] 的范围内。
+- accounts[i][j] 的长度将在 [1，30] 的范围内。
+
+
+
+## 解题思路
+
+
+- 给出一堆账户和对应的邮箱。要求合并同一个人的多个邮箱账户。如果判断是同一个人呢？如果这个人名和所属的其中之一的邮箱是相同的，就判定这是同一个人的邮箱，那么就合并这些邮箱。
+- 这题的解题思路是并查集。不过如果用暴力合并的方法，时间复杂度非常差。优化方法是先把每组数据都进行编号，人编号，每个邮箱都进行编号。这个映射关系用 `map` 记录起来。如果利用并查集的 `union()` 操作，把这些编号都进行合并。最后把人的编号和对应邮箱的编号拼接起来。
+- 这一题有 2 处比较“坑”的是，不需要合并的用户的邮箱列表也是需要排序和去重的，同一个人的所有邮箱集合都要合并到一起。具体见测试用例。不过题目中也提到了这些点，也不能算题目坑，只能归自己没注意这些边界情况。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/leetcode-go/template"
+)
+
+// 解法一 并查集优化搜索解法
+func accountsMerge(accounts [][]string) (r [][]string) {
+	uf := template.UnionFind{}
+	uf.Init(len(accounts))
+	// emailToID 将所有的 email 邮箱都拆开，拆开与 id(数组下标) 对应
+	// idToName 将 id(数组下标) 与 name 对应
+	// idToEmails 将 id(数组下标) 与整理好去重以后的 email 组对应
+	emailToID, idToName, idToEmails, res := make(map[string]int), make(map[int]string), make(map[int][]string), [][]string{}
+	for id, acc := range accounts {
+		idToName[id] = acc[0]
+		for i := 1; i < len(acc); i++ {
+			pid, ok := emailToID[acc[i]]
+			if ok {
+				uf.Union(id, pid)
+			}
+			emailToID[acc[i]] = id
+		}
+	}
+	for email, id := range emailToID {
+		pid := uf.Find(id)
+		idToEmails[pid] = append(idToEmails[pid], email)
+	}
+	for id, emails := range idToEmails {
+		name := idToName[id]
+		sort.Strings(emails)
+		res = append(res, append([]string{name}, emails...))
+	}
+	return res
+}
+
+// 解法二 并查集暴力解法
+func accountsMerge1(accounts [][]string) [][]string {
+	if len(accounts) == 0 {
+		return [][]string{}
+	}
+	uf, res, visited := template.UnionFind{}, [][]string{}, map[int]bool{}
+	uf.Init(len(accounts))
+	for i := 0; i < len(accounts); i++ {
+		for j := i + 1; j < len(accounts); j++ {
+			if accounts[i][0] == accounts[j][0] {
+				tmpA, tmpB, flag := accounts[i][1:], accounts[j][1:], false
+				for j := 0; j < len(tmpA); j++ {
+					for k := 0; k < len(tmpB); k++ {
+						if tmpA[j] == tmpB[k] {
+							flag = true
+							break
+						}
+					}
+					if flag {
+						break
+					}
+				}
+				if flag {
+					uf.Union(i, j)
+				}
+			}
+		}
+	}
+	for i := 0; i < len(accounts); i++ {
+		if visited[i] {
+			continue
+		}
+		emails, account, tmpMap := accounts[i][1:], []string{accounts[i][0]}, map[string]string{}
+		for j := i + 1; j < len(accounts); j++ {
+			if uf.Find(j) == uf.Find(i) {
+				visited[j] = true
+				for _, v := range accounts[j][1:] {
+					tmpMap[v] = v
+				}
+			}
+		}
+		for _, v := range emails {
+			tmpMap[v] = v
+		}
+		emails = []string{}
+		for key := range tmpMap {
+			emails = append(emails, key)
+		}
+		sort.Strings(emails)
+		account = append(account, emails...)
+		res = append(res, account)
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0720.Longest-Word-in-Dictionary/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0724.Find-Pivot-Index/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0724.Find-Pivot-Index.md b/website/content/ChapterFour/0700~0799/0724.Find-Pivot-Index.md
new file mode 100644
index 000000000..23a7aeee6
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0724.Find-Pivot-Index.md
@@ -0,0 +1,81 @@
+# [724. Find Pivot Index](https://leetcode.com/problems/find-pivot-index/)
+
+
+## 题目
+
+Given an array of integers nums, write a method that returns the "pivot" index of this array.
+
+We define the pivot index as the index where the sum of all the numbers to the left of the index is equal to the sum of all the numbers to the right of the index.
+
+If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot index.
+
+ 
+
+**Example 1**:
+
+    Input: nums = [1,7,3,6,5,6]
+    Output: 3
+    Explanation:
+    The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3.
+    Also, 3 is the first index where this occurs.
+
+**Example 2**:
+
+    Input: nums = [1,2,3]
+    Output: -1
+    Explanation:
+    There is no index that satisfies the conditions in the problem statement.
+
+**Constraints**:
+
+- The length of nums will be in the range [0, 10000].
+- Each element nums[i] will be an integer in the range [-1000, 1000].
+
+
+
+## 题目大意
+
+给定一个整数类型的数组 nums，请编写一个能够返回数组 “中心索引” 的方法。我们是这样定义数组 中心索引 的：数组中心索引的左侧所有元素相加的和等于右侧所有元素相加的和。如果数组不存在中心索引，那么我们应该返回 -1。如果数组有多个中心索引，那么我们应该返回最靠近左边的那一个。
+
+
+
+## 解题思路
+
+- 在数组中，找到一个数，使得它左边的数之和等于它的右边的数之和，如果存在，则返回这个数的下标索引，否作返回 -1。
+- 这里面存在一个等式，只需要满足这个等式即可满足条件：leftSum + num[i] = sum - leftSum => 2 * leftSum + num[i] = sum。
+- 题目提到如果存在多个索引，则返回最左边那个，因此从左开始求和，而不是从右边。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 2 * leftSum + num[i] = sum
+// 时间: O(n)
+// 空间: O(1)
+func pivotIndex(nums []int) int {
+	if len(nums) <= 0 {
+		return -1
+	}
+	var sum, leftSum int
+	for _, num := range nums {
+		sum += num
+	}
+	for index, num := range nums {
+		if leftSum*2+num == sum {
+			return index
+		}
+		leftSum += num
+	}
+	return -1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0721.Accounts-Merge/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0725.Split-Linked-List-in-Parts/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0725.Split-Linked-List-in-Parts.md b/website/content/ChapterFour/0700~0799/0725.Split-Linked-List-in-Parts.md
new file mode 100644
index 000000000..310afd70e
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0725.Split-Linked-List-in-Parts.md
@@ -0,0 +1,149 @@
+# [725. Split Linked List in Parts](https://leetcode.com/problems/split-linked-list-in-parts/)
+
+## 题目
+
+Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts".
+
+The length of each part should be as equal as possible: no two parts should have a size differing by more than 1. This may lead to some parts being null.
+
+The parts should be in order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal parts occurring later.
+
+Return a List of ListNode's representing the linked list parts that are formed.
+
+Examples 1->2->3->4, k = 5 // 5 equal parts [ [1], [2], [3], [4], null ]
+
+**Example 1**:
+
+```
+
+Input: 
+root = [1, 2, 3], k = 5
+Output: [[1],[2],[3],[],[]]
+Explanation:
+The input and each element of the output are ListNodes, not arrays.
+For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null.
+The first element output[0] has output[0].val = 1, output[0].next = null.
+The last element output[4] is null, but it's string representation as a ListNode is [].
+
+```
+
+**Example 2**:
+
+```
+
+Input: 
+root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
+Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
+Explanation:
+The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.
+
+```
+
+**Note**:
+
+- The length of root will be in the range [0, 1000].
+- Each value of a node in the input will be an integer in the range [0, 999].
+- k will be an integer in the range [1, 50].
+
+
+
+## 题目大意
+
+把链表分成 K 个部分，要求这 K 个部分尽量两两长度相差不超过 1，并且长度尽量相同。
+
+## 解题思路
+
+把链表长度对 K 进行除法，结果就是最终每组的长度 n。把链表长度对 K 进行取余操作，得到的结果 m，代表前 m 组链表长度为 n + 1 。相当于把多出来的部分都分摊到前面 m 组链表中了。最终链表是前 m 组长度为 n + 1，后 K - m 组链表长度是 n。
+
+注意长度不足 K 的时候要用 nil 进行填充。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "fmt"
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func splitListToParts(root *ListNode, k int) []*ListNode {
+	res := make([]*ListNode, 0)
+	if root == nil {
+		for i := 0; i < k; i++ {
+			res = append(res, nil)
+		}
+		return res
+	}
+	length := getLength(root)
+	splitNum := length / k
+	lengNum := length % k
+	cur, head := root, root
+	var pre *ListNode
+	fmt.Printf("总长度 %v, 分 %v 组, 前面 %v 组长度为 %v, 剩余 %v 组，每组 %v\n", length, k, lengNum, splitNum+1, k-lengNum, splitNum)
+	if splitNum == 0 {
+		for i := 0; i < k; i++ {
+			if cur != nil {
+				pre = cur.Next
+				cur.Next = nil
+				res = append(res, cur)
+				cur = pre
+			} else {
+				res = append(res, nil)
+			}
+		}
+		return res
+	}
+	for i := 0; i < lengNum; i++ {
+		for j := 0; j < splitNum; j++ {
+			cur = cur.Next
+		}
+		fmt.Printf("0 刚刚出来 head = %v cur = %v pre = %v\n", head, cur, head)
+		pre = cur.Next
+		cur.Next = nil
+		res = append(res, head)
+		head = pre
+		cur = pre
+		fmt.Printf("0 head = %v cur = %v pre = %v\n", head, cur, head)
+	}
+	for i := 0; i < k-lengNum; i++ {
+		for j := 0; j < splitNum-1; j++ {
+			cur = cur.Next
+		}
+		fmt.Printf("1 刚刚出来 head = %v cur = %v pre = %v\n", head, cur, head)
+		pre = cur.Next
+		cur.Next = nil
+		res = append(res, head)
+		head = pre
+		cur = pre
+	}
+	return res
+}
+
+func getLength(l *ListNode) int {
+	count := 0
+	cur := l
+	for cur != nil {
+		count++
+		cur = cur.Next
+	}
+	return count
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0724.Find-Pivot-Index/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0726.Number-of-Atoms/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0726.Number-of-Atoms.md b/website/content/ChapterFour/0700~0799/0726.Number-of-Atoms.md
new file mode 100755
index 000000000..da6ec1a93
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0726.Number-of-Atoms.md
@@ -0,0 +1,193 @@
+# [726. Number of Atoms](https://leetcode.com/problems/number-of-atoms/)
+
+
+## 题目
+
+Given a chemical `formula` (given as a string), return the count of each atom.
+
+An atomic element always starts with an uppercase character, then zero or more lowercase letters, representing the name.
+
+1 or more digits representing the count of that element may follow if the count is greater than 1. If the count is 1, no digits will follow. For example, H2O and H2O2 are possible, but H1O2 is impossible.
+
+Two formulas concatenated together produce another formula. For example, H2O2He3Mg4 is also a formula.
+
+A formula placed in parentheses, and a count (optionally added) is also a formula. For example, (H2O2) and (H2O2)3 are formulas.
+
+Given a formula, output the count of all elements as a string in the following form: the first name (in sorted order), followed by its count (if that count is more than 1), followed by the second name (in sorted order), followed by its count (if that count is more than 1), and so on.
+
+**Example 1**:
+
+    Input: 
+    formula = "H2O"
+    Output: "H2O"
+    Explanation: 
+    The count of elements are {'H': 2, 'O': 1}.
+
+**Example 2**:
+
+    Input: 
+    formula = "Mg(OH)2"
+    Output: "H2MgO2"
+    Explanation: 
+    The count of elements are {'H': 2, 'Mg': 1, 'O': 2}.
+
+**Example 3**:
+
+    Input: 
+    formula = "K4(ON(SO3)2)2"
+    Output: "K4N2O14S4"
+    Explanation: 
+    The count of elements are {'K': 4, 'N': 2, 'O': 14, 'S': 4}.
+
+**Note**:
+
+- All atom names consist of lowercase letters, except for the first character which is uppercase.
+- The length of `formula` will be in the range `[1, 1000]`.
+- `formula` will only consist of letters, digits, and round parentheses, and is a valid formula as defined in the problem.
+
+
+## 题目大意
+
+给定一个化学式，输出所有原子的数量。格式为：第一个（按字典序）原子的名子，跟着它的数量（如果数量大于 1），然后是第二个原子的名字（按字典序），跟着它的数量（如果数量大于 1），以此类推。
+
+原子总是以一个大写字母开始，接着跟随0个或任意个小写字母，表示原子的名字。如果数量大于 1，原子后会跟着数字表示原子的数量。如果数量等于 1 则不会跟数字。例如，H2O 和 H2O2 是可行的，但 H1O2 这个表达是不可行的。两个化学式连在一起是新的化学式。例如 H2O2He3Mg4 也是化学式。一个括号中的化学式和数字（可选择性添加）也是化学式。例如 (H2O2) 和 (H2O2)3 是化学式。
+
+
+
+## 解题思路
+
+
+- 利用栈处理每个化学元素，用 map 记录每个化学元素的个数，最终排序以后输出即可
+- 注意化学元素有些并不是单一字母，比如镁元素是 Mg，所以需要考虑字母的大小写问题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+	"strconv"
+	"strings"
+)
+
+type atom struct {
+	name string
+	cnt  int
+}
+
+type atoms []atom
+
+func (this atoms) Len() int           { return len(this) }
+func (this atoms) Less(i, j int) bool { return strings.Compare(this[i].name, this[j].name) < 0 }
+func (this atoms) Swap(i, j int)      { this[i], this[j] = this[j], this[i] }
+func (this atoms) String() string {
+	s := ""
+	for _, a := range this {
+		s += a.name
+		if a.cnt > 1 {
+			s += strconv.Itoa(a.cnt)
+		}
+	}
+	return s
+}
+
+func countOfAtoms(s string) string {
+	n := len(s)
+	if n == 0 {
+		return ""
+	}
+
+	stack := make([]string, 0)
+	for i := 0; i < n; i++ {
+		c := s[i]
+		if c == '(' || c == ')' {
+			stack = append(stack, string(c))
+		} else if isUpperLetter(c) {
+			j := i + 1
+			for ; j < n; j++ {
+				if !isLowerLetter(s[j]) {
+					break
+				}
+			}
+			stack = append(stack, s[i:j])
+			i = j - 1
+		} else if isDigital(c) {
+			j := i + 1
+			for ; j < n; j++ {
+				if !isDigital(s[j]) {
+					break
+				}
+			}
+			stack = append(stack, s[i:j])
+			i = j - 1
+		}
+	}
+
+	cnt, deep := make([]map[string]int, 100), 0
+	for i := 0; i < 100; i++ {
+		cnt[i] = make(map[string]int)
+	}
+	for i := 0; i < len(stack); i++ {
+		t := stack[i]
+		if isUpperLetter(t[0]) {
+			num := 1
+			if i+1 < len(stack) && isDigital(stack[i+1][0]) {
+				num, _ = strconv.Atoi(stack[i+1])
+				i++
+			}
+			cnt[deep][t] += num
+		} else if t == "(" {
+			deep++
+		} else if t == ")" {
+			num := 1
+			if i+1 < len(stack) && isDigital(stack[i+1][0]) {
+				num, _ = strconv.Atoi(stack[i+1])
+				i++
+			}
+			for k, v := range cnt[deep] {
+				cnt[deep-1][k] += v * num
+			}
+			cnt[deep] = make(map[string]int)
+			deep--
+		}
+	}
+	as := atoms{}
+	for k, v := range cnt[0] {
+		as = append(as, atom{name: k, cnt: v})
+	}
+	sort.Sort(as)
+	return as.String()
+}
+
+func isDigital(v byte) bool {
+	if v >= '0' && v <= '9' {
+		return true
+	}
+	return false
+}
+
+func isUpperLetter(v byte) bool {
+	if v >= 'A' && v <= 'Z' {
+		return true
+	}
+	return false
+}
+
+func isLowerLetter(v byte) bool {
+	if v >= 'a' && v <= 'z' {
+		return true
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0725.Split-Linked-List-in-Parts/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0728.Self-Dividing-Numbers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0728.Self-Dividing-Numbers.md b/website/content/ChapterFour/0700~0799/0728.Self-Dividing-Numbers.md
new file mode 100644
index 000000000..0a7871712
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0728.Self-Dividing-Numbers.md
@@ -0,0 +1,75 @@
+# [728. Self Dividing Numbers](https://leetcode.com/problems/self-dividing-numbers/)
+
+## 题目
+
+A self-dividing number is a number that is divisible by every digit it contains.
+
+- For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.
+
+A self-dividing number is not allowed to contain the digit zero.
+
+Given two integers left and right, return a list of all the self-dividing numbers in the range [left, right].
+
+**Example 1:**
+
+    Input: left = 1, right = 22
+    Output: [1,2,3,4,5,6,7,8,9,11,12,15,22]
+
+**Example 2:**
+
+    Input: left = 47, right = 85
+    Output: [48,55,66,77]
+
+**Constraints:**
+
+- 1 <= left <= right <= 10000
+
+## 题目大意
+
+自除数是指可以被它包含的每一位数整除的数。
+
+- 例如，128 是一个 自除数 ，因为 128 % 1 == 0，128 % 2 == 0，128 % 8 == 0。
+
+自除数 不允许包含 0 。
+
+给定两个整数 left 和 right ，返回一个列表，列表的元素是范围 [left, right] 内所有的 自除数 。
+
+## 解题思路
+
+- 模拟计算
+
+# 代码
+
+```go
+package leetcode
+
+func selfDividingNumbers(left int, right int) []int {
+	var ans []int
+	for num := left; num <= right; num++ {
+		if selfDividingNum(num) {
+			ans = append(ans, num)
+		}
+	}
+	return ans
+}
+
+func selfDividingNum(num int) bool {
+	for d := num; d > 0; d = d / 10 {
+		reminder := d % 10
+		if reminder == 0 {
+			return false
+		}
+		if num%reminder != 0 {
+			return false
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0726.Number-of-Atoms/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0729.My-Calendar-I/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0729.My-Calendar-I.md b/website/content/ChapterFour/0700~0799/0729.My-Calendar-I.md
new file mode 100755
index 000000000..0579c76da
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0729.My-Calendar-I.md
@@ -0,0 +1,187 @@
+# [729. My Calendar I](https://leetcode.com/problems/my-calendar-i/)
+
+
+## 题目
+
+Implement a `MyCalendar` class to store your events. A new event can be added if adding the event will not cause a double booking.
+
+Your class will have the method, `book(int start, int end)`. Formally, this represents a booking on the half open interval `[start, end)`, the range of real numbers `x` such that `start <= x < end`.
+
+A double booking happens when two events have some non-empty intersection (ie., there is some time that is common to both events.)
+
+For each call to the method `MyCalendar.book`, return `true` if the event can be added to the calendar successfully without causing a double booking. Otherwise, return `false` and do not add the event to the calendar.
+
+Your class will be called like this:
+
+`MyCalendar cal = new MyCalendar();`
+
+`MyCalendar.book(start, end)`
+
+**Example 1**:
+
+    MyCalendar();
+    MyCalendar.book(10, 20); // returns true
+    MyCalendar.book(15, 25); // returns false
+    MyCalendar.book(20, 30); // returns true
+    Explanation: 
+    The first event can be booked.  The second can't because time 15 is already booked by another event.
+    The third event can be booked, as the first event takes every time less than 20, but not including 20.
+
+**Note**:
+
+- The number of calls to `MyCalendar.book` per test case will be at most `1000`.
+- In calls to `MyCalendar.book(start, end)`, `start` and `end` are integers in the range `[0, 10^9]`.
+
+
+
+## 题目大意
+
+实现一个 MyCalendar 类来存放你的日程安排。如果要添加的时间内没有其他安排，则可以存储这个新的日程安排。
+
+MyCalendar 有一个 book(int start, int end) 方法。它意味着在 start 到 end 时间内增加一个日程安排，注意，这里的时间是半开区间，即 [start, end), 实数 x 的范围为，  start <= x < end。
+
+当两个日程安排有一些时间上的交叉时（例如两个日程安排都在同一时间内），就会产生重复预订。
+
+每次调用 MyCalendar.book 方法时，如果可以将日程安排成功添加到日历中而不会导致重复预订，返回 true。否则，返回 false 并且不要将该日程安排添加到日历中。
+
+请按照以下步骤调用 MyCalendar 类: MyCalendar cal = new MyCalendar(); MyCalendar.book(start, end)
+
+说明:
+
+- 每个测试用例，调用 MyCalendar.book 函数最多不超过 100次。
+- 调用函数 MyCalendar.book(start, end) 时， start 和 end 的取值范围为 [0, 10^9]。
+
+
+## 解题思路
+
+
+- 要求实现一个日程安排的功能，如果有日程安排冲突了，就返回 false，如果不冲突则返回 ture
+- 这一题有多种解法，第一种解法可以用类似第 34 题的解法。先排序每个区间，然后再这个集合中用二分搜索找到最后一个区间的左值比当前要比较的区间左值小的，如果找到，再判断能否插入进去(判断右区间是否比下一个区间的左区间小)，此方法时间复杂度 O(n log n)
+- 第二种解法是用生成一个 BST 树。在插入树中先排除不能插入的情况，例如区间有重合。然后以区间左值为依据，递归插入，每次插入依次会继续判断区间是否重合。直到不能插入，则返回 fasle。整个查找的时间复杂度是 O(log n)。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 二叉排序树
+// Event define
+type Event struct {
+	start, end  int
+	left, right *Event
+}
+
+// Insert define
+func (e *Event) Insert(curr *Event) bool {
+	if e.end > curr.start && curr.end > e.start {
+		return false
+	}
+	if curr.start < e.start {
+		if e.left == nil {
+			e.left = curr
+		} else {
+			return e.left.Insert(curr)
+		}
+	} else {
+		if e.right == nil {
+			e.right = curr
+		} else {
+			return e.right.Insert(curr)
+		}
+	}
+	return true
+}
+
+// MyCalendar define
+type MyCalendar struct {
+	root *Event
+}
+
+// Constructor729 define
+func Constructor729() MyCalendar {
+	return MyCalendar{
+		root: nil,
+	}
+}
+
+// Book define
+func (this *MyCalendar) Book(start int, end int) bool {
+	curr := &Event{start: start, end: end, left: nil, right: nil}
+	if this.root == nil {
+		this.root = curr
+		return true
+	}
+	return this.root.Insert(curr)
+}
+
+// 解法二 快排 + 二分
+// MyCalendar define
+// type MyCalendar struct {
+// 	calendar []Interval
+// }
+
+// // Constructor729 define
+// func Constructor729() MyCalendar {
+// 	calendar := []Interval{}
+// 	return MyCalendar{calendar: calendar}
+// }
+
+// // Book define
+// func (this *MyCalendar) Book(start int, end int) bool {
+// 	if len(this.calendar) == 0 {
+// 		this.calendar = append(this.calendar, Interval{Start: start, End: end})
+// 		return true
+// 	}
+// 	// 快排
+// 	quickSort(this.calendar, 0, len(this.calendar)-1)
+// 	// 二分
+// 	pos := searchLastLessInterval(this.calendar, start, end)
+// 	// 如果找到最后一个元素，需要判断 end
+// 	if pos == len(this.calendar)-1 && this.calendar[pos].End <= start {
+// 		this.calendar = append(this.calendar, Interval{Start: start, End: end})
+// 		return true
+// 	}
+// 	// 如果不是开头和结尾的元素，还需要判断这个区间是否能插入到原数组中(要看起点和终点是否都能插入)
+// 	if pos != len(this.calendar)-1 && pos != -1 && this.calendar[pos].End <= start && this.calendar[pos+1].Start >= end {
+// 		this.calendar = append(this.calendar, Interval{Start: start, End: end})
+// 		return true
+// 	}
+// 	// 如果元素比开头的元素还要小，要插入到开头
+// 	if this.calendar[0].Start >= end {
+// 		this.calendar = append(this.calendar, Interval{Start: start, End: end})
+// 		return true
+// 	}
+// 	return false
+// }
+
+// func searchLastLessInterval(intervals []Interval, start, end int) int {
+// 	low, high := 0, len(intervals)-1
+// 	for low <= high {
+// 		mid := low + ((high - low) >> 1)
+// 		if intervals[mid].Start <= start {
+// 			if (mid == len(intervals)-1) || (intervals[mid+1].Start > start) { // 找到最后一个小于等于 target 的元素
+// 				return mid
+// 			}
+// 			low = mid + 1
+// 		} else {
+// 			high = mid - 1
+// 		}
+// 	}
+// 	return -1
+// }
+
+/**
+ * Your MyCalendar object will be instantiated and called as such:
+ * obj := Constructor();
+ * param_1 := obj.Book(start,end);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0728.Self-Dividing-Numbers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0732.My-Calendar-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0732.My-Calendar-III.md b/website/content/ChapterFour/0700~0799/0732.My-Calendar-III.md
new file mode 100755
index 000000000..170edfd72
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0732.My-Calendar-III.md
@@ -0,0 +1,149 @@
+# [732. My Calendar III](https://leetcode.com/problems/my-calendar-iii/)
+
+
+## 题目
+
+Implement a `MyCalendarThree` class to store your events. A new event can **always** be added.
+
+Your class will have one method, `book(int start, int end)`. Formally, this represents a booking on the half open interval `[start, end)`, the range of real numbers `x` such that `start <= x < end`.
+
+A K-booking happens when **K** events have some non-empty intersection (ie., there is some time that is common to all K events.)
+
+For each call to the method `MyCalendar.book`, return an integer `K` representing the largest integer such that there exists a `K`-booking in the calendar.
+
+Your class will be called like this:
+
+`MyCalendarThree cal = new MyCalendarThree();`
+
+`MyCalendarThree.book(start, end)`
+
+**Example 1**:
+
+    MyCalendarThree();
+    MyCalendarThree.book(10, 20); // returns 1
+    MyCalendarThree.book(50, 60); // returns 1
+    MyCalendarThree.book(10, 40); // returns 2
+    MyCalendarThree.book(5, 15); // returns 3
+    MyCalendarThree.book(5, 10); // returns 3
+    MyCalendarThree.book(25, 55); // returns 3
+    Explanation: 
+    The first two events can be booked and are disjoint, so the maximum K-booking is a 1-booking.
+    The third event [10, 40) intersects the first event, and the maximum K-booking is a 2-booking.
+    The remaining events cause the maximum K-booking to be only a 3-booking.
+    Note that the last event locally causes a 2-booking, but the answer is still 3 because
+    eg. [10, 20), [10, 40), and [5, 15) are still triple booked.
+
+**Note**:
+
+- The number of calls to `MyCalendarThree.book` per test case will be at most `400`.
+- In calls to `MyCalendarThree.book(start, end)`, `start` and `end` are integers in the range `[0, 10^9]`.
+
+
+## 题目大意
+
+实现一个 MyCalendar 类来存放你的日程安排，你可以一直添加新的日程安排。
+
+MyCalendar 有一个 book(int start, int end)方法。它意味着在start到end时间内增加一个日程安排，注意，这里的时间是半开区间，即 [start, end), 实数 x 的范围为，  start <= x < end。当 K 个日程安排有一些时间上的交叉时（例如K个日程安排都在同一时间内），就会产生 K 次预订。每次调用 MyCalendar.book方法时，返回一个整数 K ，表示最大的 K 次预订。
+
+请按照以下步骤调用MyCalendar 类: MyCalendar cal = new MyCalendar(); MyCalendar.book(start, end)
+
+说明:
+
+- 每个测试用例，调用 MyCalendar.book 函数最多不超过 400 次。
+- 调用函数 MyCalendar.book(start, end)时， start 和 end 的取值范围为 [0, 10^9]。
+
+
+
+
+## 解题思路
+
+- 设计一个日程类，每添加一个日程，实时显示出当前排期中累计日程最多的个数，例如在一段时间内，排了 3 个日程，其他时间内都只有 0，1，2 个日程，则输出 3 。
+- 拿到这个题目以后会立即想到线段树。由于题目中只有增加日程，所以这一题难度不大。这一题和第 699 题也类似，但是有区别，第 699 题中，俄罗斯方块会依次摞起来，而这一题中，俄罗斯方块也就摞起来，但是方块下面如果是空挡，方块会断掉。举个例子：依次增加区间 [10,20]，[10,40]，[5,15]，[5,10]，如果是第 699 题的规则，这 [5,10] 的这块砖块会落在 [5,15] 上，从而使得高度为 4，但是这一题是日程，日程不一样，[5,15] 这个区间内有 3 个日程，但是其他部分都没有 3 个日程，所以第三块砖块 [5,15] 中的 [5,10] 会“断裂”，掉下去，第四块砖块还是 [5,10]，落在第三块砖块断落下去的位置，它们俩落在一起的高度是 2 。
+- 构造一颗线段树，这里用树来构造，如果用数组需要开辟很大的空间。当区间左右边界和查询边界完全相同的时候再累加技术，否则不加，继续划分区间。以区间的左边界作为划分区间的标准，因为区间左边界是开区间，右边是闭区间。一个区间的计数值以区间左边界的计数为准。还是上面的例子，[5,10) 计数以 5 为标准，count = 2，[10,15) 计数以 10 为标准，count = 3 。还需要再动态维护一个最大值。这个线段树的实现比较简单。
+- 类似的题目有：第 715 题，第 218 题，第 699 题。第 715 题是区间更新定值(**不是增减**)，第 218 题可以用扫描线，第 732 题和第 699 题类似，也是俄罗斯方块的题目，但是第 732 题的俄罗斯方块的方块会“断裂”。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// SegmentTree732 define
+type SegmentTree732 struct {
+	start, end, count int
+	left, right       *SegmentTree732
+}
+
+// MyCalendarThree define
+type MyCalendarThree struct {
+	st        *SegmentTree732
+	maxHeight int
+}
+
+// Constructor732 define
+func Constructor732() MyCalendarThree {
+	st := &SegmentTree732{
+		start: 0,
+		end:   1e9,
+	}
+	return MyCalendarThree{
+		st: st,
+	}
+}
+
+// Book define
+func (mct *MyCalendarThree) Book(start int, end int) int {
+	mct.st.book(start, end, &mct.maxHeight)
+	return mct.maxHeight
+}
+
+func (st *SegmentTree732) book(start, end int, maxHeight *int) {
+	if start == end {
+		return
+	}
+	if start == st.start && st.end == end {
+		st.count++
+		if st.count > *maxHeight {
+			*maxHeight = st.count
+		}
+		if st.left == nil {
+			return
+		}
+	}
+	if st.left == nil {
+		if start == st.start {
+			st.left = &SegmentTree732{start: start, end: end, count: st.count}
+			st.right = &SegmentTree732{start: end, end: st.end, count: st.count}
+			st.left.book(start, end, maxHeight)
+			return
+		}
+		st.left = &SegmentTree732{start: st.start, end: start, count: st.count}
+		st.right = &SegmentTree732{start: start, end: st.end, count: st.count}
+		st.right.book(start, end, maxHeight)
+		return
+	}
+	if start >= st.right.start {
+		st.right.book(start, end, maxHeight)
+	} else if end <= st.left.end {
+		st.left.book(start, end, maxHeight)
+	} else {
+		st.left.book(start, st.left.end, maxHeight)
+		st.right.book(st.right.start, end, maxHeight)
+	}
+}
+
+/**
+ * Your MyCalendarThree object will be instantiated and called as such:
+ * obj := Constructor();
+ * param_1 := obj.Book(start,end);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0729.My-Calendar-I/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0733.Flood-Fill/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0733.Flood-Fill.md b/website/content/ChapterFour/0700~0799/0733.Flood-Fill.md
new file mode 100755
index 000000000..9fee69a41
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0733.Flood-Fill.md
@@ -0,0 +1,88 @@
+# [733. Flood Fill](https://leetcode.com/problems/flood-fill/)
+
+
+## 题目
+
+An `image` is represented by a 2-D array of integers, each integer representing the pixel value of the image (from 0 to 65535).
+
+Given a coordinate `(sr, sc)` representing the starting pixel (row and column) of the flood fill, and a pixel value `newColor`, "flood fill" the image.
+
+To perform a "flood fill", consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color as the starting pixel), and so on. Replace the color of all of the aforementioned pixels with the newColor.
+
+At the end, return the modified image.
+
+**Example 1**:
+
+    Input: 
+    image = [[1,1,1],[1,1,0],[1,0,1]]
+    sr = 1, sc = 1, newColor = 2
+    Output: [[2,2,2],[2,2,0],[2,0,1]]
+    Explanation: 
+    From the center of the image (with position (sr, sc) = (1, 1)), all pixels connected 
+    by a path of the same color as the starting pixel are colored with the new color.
+    Note the bottom corner is not colored 2, because it is not 4-directionally connected
+    to the starting pixel.
+
+**Note**:
+
+- The length of `image` and `image[0]` will be in the range `[1, 50]`.
+- The given starting pixel will satisfy `0 <= sr < image.length` and `0 <= sc < image[0].length`.
+- The value of each color in `image[i][j]` and `newColor` will be an integer in `[0, 65535]`.
+
+
+## 题目大意
+
+有一幅以二维整数数组表示的图画，每一个整数表示该图画的像素值大小，数值在 0 到 65535 之间。给你一个坐标 (sr, sc) 表示图像渲染开始的像素值（行 ，列）和一个新的颜色值 newColor，让你重新上色这幅图像。
+
+为了完成上色工作，从初始坐标开始，记录初始坐标的上下左右四个方向上像素值与初始坐标相同的相连像素点，接着再记录这四个方向上符合条件的像素点与他们对应四个方向上像素值与初始坐标相同的相连像素点，……，重复该过程。将所有有记录的像素点的颜色值改为新的颜色值。最后返回经过上色渲染后的图像。
+
+注意:
+
+- image 和 image[0] 的长度在范围 [1, 50] 内。
+- 给出的初始点将满足 0 <= sr < image.length 和 0 <= sc < image[0].length。
+- image[i][j] 和 newColor 表示的颜色值在范围 [0, 65535]内。
+
+
+## 解题思路
+
+
+- 给出一个二维的图片点阵，每个点阵都有一个数字。给出一个起点坐标，要求从这个起点坐标开始，把所有与这个起点连通的点都染色成 newColor。
+- 这一题是标准的 Flood Fill 算法。可以用 DFS 也可以用 BFS 。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func floodFill(image [][]int, sr int, sc int, newColor int) [][]int {
+	color := image[sr][sc]
+	if newColor == color {
+		return image
+	}
+	dfs733(image, sr, sc, newColor)
+	return image
+}
+
+func dfs733(image [][]int, x, y int, newColor int) {
+	if image[x][y] == newColor {
+		return
+	}
+	oldColor := image[x][y]
+	image[x][y] = newColor
+	for i := 0; i < 4; i++ {
+		if (x+dir[i][0] >= 0 && x+dir[i][0] < len(image)) && (y+dir[i][1] >= 0 && y+dir[i][1] < len(image[0])) && image[x+dir[i][0]][y+dir[i][1]] == oldColor {
+			dfs733(image, x+dir[i][0], y+dir[i][1], newColor)
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0732.My-Calendar-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0735.Asteroid-Collision/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0735.Asteroid-Collision.md b/website/content/ChapterFour/0700~0799/0735.Asteroid-Collision.md
new file mode 100644
index 000000000..7a9227f5f
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0735.Asteroid-Collision.md
@@ -0,0 +1,112 @@
+# [735. Asteroid Collision](https://leetcode.com/problems/asteroid-collision/)
+
+## 题目
+
+We are given an array asteroids of integers representing asteroids in a row.
+
+For each asteroid, the absolute value represents its size, and the sign represents its direction (positive meaning right, negative meaning left). Each asteroid moves at the same speed.
+
+Find out the state of the asteroids after all collisions. If two asteroids meet, the smaller one will explode. If both are the same size, both will explode. Two asteroids moving in the same direction will never meet.
+
+**Example 1**:
+
+```
+
+Input: 
+asteroids = [5, 10, -5]
+Output: [5, 10]
+Explanation: 
+The 10 and -5 collide resulting in 10.  The 5 and 10 never collide.
+
+```
+
+**Example 2**:
+
+```
+
+Input: 
+asteroids = [8, -8]
+Output: []
+Explanation: 
+The 8 and -8 collide exploding each other.
+
+```
+
+**Example 3**:
+
+```
+
+Input: 
+asteroids = [10, 2, -5]
+Output: [10]
+Explanation: 
+The 2 and -5 collide resulting in -5.  The 10 and -5 collide resulting in 10.
+
+```
+
+**Example 4**:
+
+```
+
+Input: 
+asteroids = [-2, -1, 1, 2]
+Output: [-2, -1, 1, 2]
+Explanation: 
+The -2 and -1 are moving left, while the 1 and 2 are moving right.
+Asteroids moving the same direction never meet, so no asteroids will meet each other.
+
+```
+
+**Note**:
+
+- The length of asteroids will be at most 10000.
+- Each asteroid will be a non-zero integer in the range [-1000, 1000]..
+
+## 题目大意
+
+给定一个整数数组 asteroids，表示在同一行的行星。对于数组中的每一个元素，其绝对值表示行星的大小，正负表示行星的移动方向（正表示向右移动，负表示向左移动）。每一颗行星以相同的速度移动。找出碰撞后剩下的所有行星。碰撞规则：两个行星相互碰撞，较小的行星会爆炸。如果两颗行星大小相同，则两颗行星都会爆炸。两颗移动方向相同的行星，永远不会发生碰撞。
+
+## 解题思路
+
+这一题类似第 1047 题。这也是一个类似“对对碰”的游戏，不过这里的碰撞，大行星和小行星碰撞以后，大行星会胜出，小行星直接消失。按照题意的规则来，用栈模拟即可。考虑最终结果：
+
+1. 所有向左飞的行星都向左，所有向右飞的行星都向右。
+2. 向左飞的行星，如果飞行中没有向右飞行的行星，那么它将安全穿过。
+3. 跟踪所有向右移动到右侧的行星，最右边的一个将是第一个面对向左飞行行星碰撞的。
+4. 如果它幸存下来，继续前进，否则，任何之前的向右的行星都会被逐一被暴露出来碰撞。
+
+所以先处理这种情况，一层循环把所有能碰撞的向右飞行的行星都碰撞完。碰撞完以后，如果栈顶行星向左飞，新来的行星向右飞，直接添加进来即可。否则栈顶行星向右飞，大小和向左飞的行星一样大小，两者都撞毁灭，弹出栈顶元素。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func asteroidCollision(asteroids []int) []int {
+	res := []int{}
+	for _, v := range asteroids {
+		for len(res) != 0 && res[len(res)-1] > 0 && res[len(res)-1] < -v {
+			res = res[:len(res)-1]
+		}
+		if len(res) == 0 || v > 0 || res[len(res)-1] < 0 {
+			res = append(res, v)
+		} else if v < 0 && res[len(res)-1] == -v {
+			res = res[:len(res)-1]
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0733.Flood-Fill/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0739.Daily-Temperatures/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0739.Daily-Temperatures.md b/website/content/ChapterFour/0700~0799/0739.Daily-Temperatures.md
new file mode 100644
index 000000000..4a15a20bd
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0739.Daily-Temperatures.md
@@ -0,0 +1,66 @@
+# [739. Daily Temperatures](https://leetcode.com/problems/daily-temperatures/)
+
+## 题目
+
+
+Given a list of daily temperatures T, return a list such that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put 0 instead.
+
+For example, given the list of temperatures T = [73, 74, 75, 71, 69, 72, 76, 73], your output should be [1, 1, 4, 2, 1, 1, 0, 0].
+
+**Note**: The length of temperatures will be in the range [1, 30000]. Each temperature will be an integer in the range [30, 100].
+
+
+## 题目大意
+
+给出一个温度数组，要求输出比当天温度高的在未来的哪一天，输出未来第几天的天数。例如比 73 度高的在未来第 1 天出现，比 75 度高的在未来第 4 天出现。
+
+## 解题思路
+
+这道题根据题意正常处理就可以了。2 层循环。另外一种做法是单调栈，维护一个单调递减的单调栈即可。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 普通做法
+func dailyTemperatures(T []int) []int {
+	res, j := make([]int, len(T)), 0
+	for i := 0; i < len(T); i++ {
+		for j = i + 1; j < len(T); j++ {
+			if T[j] > T[i] {
+				res[i] = j - i
+				break
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 单调栈
+func dailyTemperatures1(T []int) []int {
+	res := make([]int, len(T))
+	var toCheck []int
+	for i, t := range T {
+		for len(toCheck) > 0 && T[toCheck[len(toCheck)-1]] < t {
+			idx := toCheck[len(toCheck)-1]
+			res[idx] = i - idx
+			toCheck = toCheck[:len(toCheck)-1]
+		}
+		toCheck = append(toCheck, i)
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0735.Asteroid-Collision/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0744.Find-Smallest-Letter-Greater-Than-Target/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0744.Find-Smallest-Letter-Greater-Than-Target.md b/website/content/ChapterFour/0700~0799/0744.Find-Smallest-Letter-Greater-Than-Target.md
new file mode 100755
index 000000000..daa8ce502
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0744.Find-Smallest-Letter-Greater-Than-Target.md
@@ -0,0 +1,99 @@
+# [744. Find Smallest Letter Greater Than Target](https://leetcode.com/problems/find-smallest-letter-greater-than-target/)
+
+
+## 题目
+
+Given a list of sorted characters `letters` containing only lowercase letters, and given a target letter `target`, find the smallest element in the list that is larger than the given target.
+
+Letters also wrap around. For example, if the target is `target = 'z'` and `letters = ['a', 'b']`, the answer is `'a'`.
+
+**Examples**:
+
+    Input:
+    letters = ["c", "f", "j"]
+    target = "a"
+    Output: "c"
+    
+    Input:
+    letters = ["c", "f", "j"]
+    target = "c"
+    Output: "f"
+    
+    Input:
+    letters = ["c", "f", "j"]
+    target = "d"
+    Output: "f"
+    
+    Input:
+    letters = ["c", "f", "j"]
+    target = "g"
+    Output: "j"
+    
+    Input:
+    letters = ["c", "f", "j"]
+    target = "j"
+    Output: "c"
+    
+    Input:
+    letters = ["c", "f", "j"]
+    target = "k"
+    Output: "c"
+
+**Note**:
+
+1. `letters` has a length in range `[2, 10000]`.
+2. `letters` consists of lowercase letters, and contains at least 2 unique letters.
+3. `target` is a lowercase letter.
+
+
+## 题目大意
+
+给定一个只包含小写字母的有序数组letters 和一个目标字母 target，寻找有序数组里面比目标字母大的最小字母。
+
+数组里字母的顺序是循环的。举个例子，如果目标字母target = 'z' 并且有序数组为 letters = ['a', 'b']，则答案返回 'a'。
+
+注:
+
+1. letters长度范围在[2, 10000]区间内。
+2. letters 仅由小写字母组成，最少包含两个不同的字母。
+3. 目标字母target 是一个小写字母。
+
+
+
+## 解题思路
+
+- 给出一个字节数组，在这个字节数组中查找在 target 后面的第一个字母。数组是环形的。
+- 这一题也是二分搜索的题目，先在数组里面查找 target，如果找到了，取这个字母的后一个字母。如果没有找到，就取 low 下标的那个字母。注意数组是环形的，所以最后结果需要对下标取余。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func nextGreatestLetter(letters []byte, target byte) byte {
+	low, high := 0, len(letters)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if letters[mid] > target {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	find := letters[low%len(letters)]
+	if find <= target {
+		return letters[0]
+	}
+	return find
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0739.Daily-Temperatures/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0745.Prefix-and-Suffix-Search/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0745.Prefix-and-Suffix-Search.md b/website/content/ChapterFour/0700~0799/0745.Prefix-and-Suffix-Search.md
new file mode 100755
index 000000000..6934034ef
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0745.Prefix-and-Suffix-Search.md
@@ -0,0 +1,105 @@
+# [745. Prefix and Suffix Search](https://leetcode.com/problems/prefix-and-suffix-search/)
+
+
+## 题目
+
+Given many `words`, `words[i]` has weight `i`.
+
+Design a class `WordFilter` that supports one function, `WordFilter.f(String prefix, String suffix)`. It will return the word with given `prefix` and `suffix` with maximum weight. If no word exists, return -1.
+
+**Examples**:
+
+    Input:
+    WordFilter(["apple"])
+    WordFilter.f("a", "e") // returns 0
+    WordFilter.f("b", "") // returns -1
+
+**Note**:
+
+1. `words` has length in range `[1, 15000]`.
+2. For each test case, up to `words.length` queries `WordFilter.f` may be made.
+3. `words[i]` has length in range `[1, 10]`.
+4. `prefix, suffix` have lengths in range `[0, 10]`.
+5. `words[i]` and `prefix, suffix` queries consist of lowercase letters only.
+
+
+## 题目大意
+
+给定多个 words，words[i] 的权重为 i 。设计一个类 WordFilter 实现函数WordFilter.f(String prefix, String suffix)。这个函数将返回具有前缀 prefix 和后缀suffix 的词的最大权重。如果没有这样的词，返回 -1。
+
+
+
+## 解题思路
+
+
+- 要求实现一个 `WordFilter` ，它具有字符串匹配的功能，可以匹配出前缀和后缀都满足条件的字符串下标，如果找得到，返回下标，如果找不到，则返回 -1 。
+- 这一题有 2 种解题思路。第一种是先把这个 `WordFilter` 结构里面的字符串全部预处理一遍，将它的前缀，后缀的所有组合都枚举出来放在 map 中，之后匹配的时候只需要按照自己定义的规则查找 key 就可以了。初始化时间复杂度 `O(N * L^2)`，查找时间复杂度 `O(1)`，空间复杂度 `O(N * L^2)`。其中 `N` 是输入的字符串数组的长度，`L` 是输入字符串数组中字符串的最大长度。第二种思路是直接遍历字符串每个下标，依次用字符串的前缀匹配方法和后缀匹配方法，依次匹配。初始化时间复杂度 `O(1)`，查找时间复杂度 `O(N * L)`，空间复杂度 `O(1)`。其中 `N` 是输入的字符串数组的长度，`L` 是输入字符串数组中字符串的最大长度。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+// 解法一 查找时间复杂度 O(1)
+type WordFilter struct {
+	words map[string]int
+}
+
+func Constructor745(words []string) WordFilter {
+	wordsMap := make(map[string]int, len(words)*5)
+	for k := 0; k < len(words); k++ {
+		for i := 0; i <= 10 && i <= len(words[k]); i++ {
+			for j := len(words[k]); 0 <= j && len(words[k])-10 <= j; j-- {
+				ps := words[k][:i] + "#" + words[k][j:]
+				wordsMap[ps] = k
+			}
+		}
+	}
+	return WordFilter{words: wordsMap}
+}
+
+func (this *WordFilter) F(prefix string, suffix string) int {
+	ps := prefix + "#" + suffix
+	if index, ok := this.words[ps]; ok {
+		return index
+	}
+	return -1
+}
+
+// 解法二 查找时间复杂度 O(N * L)
+type WordFilter_ struct {
+	input []string
+}
+
+func Constructor_745_(words []string) WordFilter_ {
+	return WordFilter_{input: words}
+}
+
+func (this *WordFilter_) F_(prefix string, suffix string) int {
+	for i := len(this.input) - 1; i >= 0; i-- {
+		if strings.HasPrefix(this.input[i], prefix) && strings.HasSuffix(this.input[i], suffix) {
+			return i
+		}
+	}
+	return -1
+}
+
+/**
+ * Your WordFilter object will be instantiated and called as such:
+ * obj := Constructor(words);
+ * param_1 := obj.F(prefix,suffix);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0744.Find-Smallest-Letter-Greater-Than-Target/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0746.Min-Cost-Climbing-Stairs/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0746.Min-Cost-Climbing-Stairs.md b/website/content/ChapterFour/0700~0799/0746.Min-Cost-Climbing-Stairs.md
new file mode 100755
index 000000000..b4f00e291
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0746.Min-Cost-Climbing-Stairs.md
@@ -0,0 +1,78 @@
+# [746. Min Cost Climbing Stairs](https://leetcode.com/problems/min-cost-climbing-stairs/)
+
+
+## 题目
+
+On a staircase, the `i`-th step has some non-negative cost `cost[i]` assigned (0 indexed).
+
+Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1.
+
+**Example 1**:
+
+    Input: cost = [10, 15, 20]
+    Output: 15
+    Explanation: Cheapest is start on cost[1], pay that cost and go to the top.
+
+**Example 2**:
+
+    Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
+    Output: 6
+    Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3].
+
+**Note**:
+
+1. `cost` will have a length in the range `[2, 1000]`.
+2. Every `cost[i]` will be an integer in the range `[0, 999]`.
+
+
+## 题目大意
+
+数组的每个索引做为一个阶梯，第 i 个阶梯对应着一个非负数的体力花费值 cost\[i\] (索引从 0 开始)。每当你爬上一个阶梯你都要花费对应的体力花费值，然后你可以选择继续爬一个阶梯或者爬两个阶梯。您需要找到达到楼层顶部的最低花费。在开始时，你可以选择从索引为 0 或 1 的元素作为初始阶梯。
+
+
+## 解题思路
+
+
+- 这一题算是第 70 题的加强版。依旧是爬楼梯的问题，解题思路也是 DP。在爬楼梯的基础上增加了一个新的条件，每层楼梯都有一个 cost 花费，问上到最终楼层，花费最小值是多少。
+- `dp[i]` 代表上到第 n 层的最小花费，状态转移方程是 `dp[i] = cost[i] + min(dp[i-2], dp[i-1])`，最终第 n 层的最小花费是 `min(dp[n-2], dp[n-1])` 。
+- 由于每层的花费只和前两层有关系，所以每次 DP 迭代的时候只需要 2 个临时变量即可。可以用这种方式来优化辅助空间。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 DP
+func minCostClimbingStairs(cost []int) int {
+	dp := make([]int, len(cost))
+	dp[0], dp[1] = cost[0], cost[1]
+	for i := 2; i < len(cost); i++ {
+		dp[i] = cost[i] + min(dp[i-2], dp[i-1])
+	}
+	return min(dp[len(cost)-2], dp[len(cost)-1])
+}
+
+// 解法二 DP 优化辅助空间
+func minCostClimbingStairs1(cost []int) int {
+	var cur, last int
+	for i := 2; i < len(cost)+1; i++ {
+		if last+cost[i-1] > cur+cost[i-2] {
+			cur, last = last, cur+cost[i-2]
+		} else {
+			cur, last = last, last+cost[i-1]
+		}
+	}
+	return last
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0745.Prefix-and-Suffix-Search/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0747.Largest-Number-At-Least-Twice-of-Others/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0747.Largest-Number-At-Least-Twice-of-Others.md b/website/content/ChapterFour/0700~0799/0747.Largest-Number-At-Least-Twice-of-Others.md
new file mode 100644
index 000000000..759e8cb7f
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0747.Largest-Number-At-Least-Twice-of-Others.md
@@ -0,0 +1,82 @@
+# [747. Largest Number At Least Twice of Others](https://leetcode.com/problems/largest-number-at-least-twice-of-others/)
+
+
+## 题目
+
+You are given an integer array `nums` where the largest integer is **unique**.
+
+Determine whether the largest element in the array is **at least twice** as much as every other number in the array. If it is, return *the **index** of the largest element, or return* `-1` *otherwise*.
+
+**Example 1:**
+
+```
+Input: nums = [3,6,1,0]
+Output: 1
+Explanation: 6 is the largest integer.
+For every other number in the array x, 6 is at least twice as big as x.
+The index of value 6 is 1, so we return 1.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3,4]
+Output: -1
+Explanation: 4 is less than twice the value of 3, so we return -1.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1]
+Output: 0
+Explanation: 1 is trivially at least twice the value as any other number because there are no other numbers.
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 50`
+- `0 <= nums[i] <= 100`
+- The largest element in `nums` is unique.
+
+## 题目大意
+
+给你一个整数数组 nums ，其中总是存在 唯一的 一个最大整数 。请你找出数组中的最大元素并检查它是否 至少是数组中每个其他数字的两倍 。如果是，则返回 最大元素的下标 ，否则返回 -1 。
+
+## 解题思路
+
+- 简单题。先扫描一遍找到最大值和下标。再扫描一遍检查最大值是否是其他数字的两倍。
+
+## 代码
+
+```go
+package leetcode
+
+func dominantIndex(nums []int) int {
+	maxNum, flag, index := 0, false, 0
+	for i, v := range nums {
+		if v > maxNum {
+			maxNum = v
+			index = i
+		}
+	}
+	for _, v := range nums {
+		if v != maxNum && 2*v > maxNum {
+			flag = true
+		}
+	}
+	if flag {
+		return -1
+	}
+	return index
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0746.Min-Cost-Climbing-Stairs/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0748.Shortest-Completing-Word/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0748.Shortest-Completing-Word.md b/website/content/ChapterFour/0700~0799/0748.Shortest-Completing-Word.md
new file mode 100755
index 000000000..54e53357a
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0748.Shortest-Completing-Word.md
@@ -0,0 +1,109 @@
+# [748. Shortest Completing Word](https://leetcode.com/problems/shortest-completing-word/)
+
+
+## 题目
+
+Find the minimum length word from a given dictionary `words`, which has all the letters from the string `licensePlate`. Such a word is said to complete the given string `licensePlate`
+
+Here, for letters we ignore case. For example, `"P"` on the `licensePlate` still matches `"p"` on the word.
+
+It is guaranteed an answer exists. If there are multiple answers, return the one that occurs first in the array.
+
+The license plate might have the same letter occurring multiple times. For example, given a `licensePlate` of `"PP"`, the word `"pair"` does not complete the `licensePlate`, but the word `"supper"` does.
+
+**Example 1**:
+
+    Input: licensePlate = "1s3 PSt", words = ["step", "steps", "stripe", "stepple"]
+    Output: "steps"
+    Explanation: The smallest length word that contains the letters "S", "P", "S", and "T".
+    Note that the answer is not "step", because the letter "s" must occur in the word twice.
+    Also note that we ignored case for the purposes of comparing whether a letter exists in the word.
+
+**Example 2**:
+
+    Input: licensePlate = "1s3 456", words = ["looks", "pest", "stew", "show"]
+    Output: "pest"
+    Explanation: There are 3 smallest length words that contains the letters "s".
+    We return the one that occurred first.
+
+**Note**:
+
+1. `licensePlate` will be a string with length in range `[1, 7]`.
+2. `licensePlate` will contain digits, spaces, or letters (uppercase or lowercase).
+3. `words` will have a length in the range `[10, 1000]`.
+4. Every `words[i]` will consist of lowercase letters, and have length in range `[1, 15]`.
+
+
+## 题目大意
+
+如果单词列表（words）中的一个单词包含牌照（licensePlate）中所有的字母，那么我们称之为完整词。在所有完整词中，最短的单词我们称之为最短完整词。
+
+单词在匹配牌照中的字母时不区分大小写，比如牌照中的 "P" 依然可以匹配单词中的 "p" 字母。我们保证一定存在一个最短完整词。当有多个单词都符合最短完整词的匹配条件时取单词列表中最靠前的一个。牌照中可能包含多个相同的字符，比如说：对于牌照 "PP"，单词 "pair" 无法匹配，但是 "supper" 可以匹配。
+
+注意:
+
+- 牌照（licensePlate）的长度在区域[1, 7]中。
+- 牌照（licensePlate）将会包含数字、空格、或者字母（大写和小写）。
+- 单词列表（words）长度在区间 [10, 1000] 中。
+- 每一个单词 words[i] 都是小写，并且长度在区间 [1, 15] 中。
+
+
+
+## 解题思路
+
+
+- 给出一个数组，要求找出能包含 `licensePlate` 字符串中所有字符的最短长度的字符串。如果最短长度的字符串有多个，输出 word 下标小的那个。这一题也是简单题，不过有 2 个需要注意的点，第一点，`licensePlate` 中可能包含 `Unicode` 任意的字符，所以要先把字母的字符筛选出来，第二点是题目中保证了一定存在一个最短的单词能满足题意，并且忽略大小写。具体做法按照题意模拟即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "unicode"
+
+func shortestCompletingWord(licensePlate string, words []string) string {
+	lp := genCnter(licensePlate)
+	var ret string
+	for _, w := range words {
+		if match(lp, w) {
+			if len(w) < len(ret) || ret == "" {
+				ret = w
+			}
+		}
+	}
+	return ret
+}
+
+func genCnter(lp string) [26]int {
+	cnter := [26]int{}
+	for _, ch := range lp {
+		if unicode.IsLetter(ch) {
+			cnter[unicode.ToLower(ch)-'a']++
+		}
+	}
+	return cnter
+}
+
+func match(lp [26]int, w string) bool {
+	m := [26]int{}
+	for _, ch := range w {
+		m[ch-'a']++
+	}
+	for k, v := range lp {
+		if m[k] < v {
+			return false
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0747.Largest-Number-At-Least-Twice-of-Others/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0752.Open-the-Lock/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0752.Open-the-Lock.md b/website/content/ChapterFour/0700~0799/0752.Open-the-Lock.md
new file mode 100644
index 000000000..3c442ae7f
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0752.Open-the-Lock.md
@@ -0,0 +1,138 @@
+# [752. Open the Lock](https://leetcode.com/problems/open-the-lock/)
+
+
+## 题目
+
+You have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: `'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'`. The wheels can rotate freely and wrap around: for example we can turn `'9'` to be `'0'`, or `'0'` to be `'9'`. Each move consists of turning one wheel one slot.
+
+The lock initially starts at `'0000'`, a string representing the state of the 4 wheels.
+
+You are given a list of `deadends` dead ends, meaning if the lock displays any of these codes, the wheels of the lock will stop turning and you will be unable to open it.
+
+Given a `target` representing the value of the wheels that will unlock the lock, return the minimum total number of turns required to open the lock, or -1 if it is impossible.
+
+**Example 1:**
+
+```
+Input: deadends = ["0201","0101","0102","1212","2002"], target = "0202"
+Output: 6
+Explanation:
+A sequence of valid moves would be "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202".
+Note that a sequence like "0000" -> "0001" -> "0002" -> "0102" -> "0202" would be invalid,
+because the wheels of the lock become stuck after the display becomes the dead end "0102".
+
+```
+
+**Example 2:**
+
+```
+Input: deadends = ["8888"], target = "0009"
+Output: 1
+Explanation:
+We can turn the last wheel in reverse to move from "0000" -> "0009".
+
+```
+
+**Example 3:**
+
+```
+Input: deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888"
+Output: -1
+Explanation:
+We can't reach the target without getting stuck.
+
+```
+
+**Example 4:**
+
+```
+Input: deadends = ["0000"], target = "8888"
+Output: -1
+
+```
+
+**Constraints:**
+
+- `1 <= deadends.length <= 500`
+- `deadends[i].length == 4`
+- `target.length == 4`
+- target **will not be** in the list `deadends`.
+- `target` and `deadends[i]` consist of digits only.
+
+## 题目大意
+
+你有一个带有四个圆形拨轮的转盘锁。每个拨轮都有10个数字： '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' 。每个拨轮可以自由旋转：例如把 '9' 变为 '0'，'0' 变为 '9' 。每次旋转都只能旋转一个拨轮的一位数字。锁的初始数字为 '0000' ，一个代表四个拨轮的数字的字符串。列表 deadends 包含了一组死亡数字，一旦拨轮的数字和列表里的任何一个元素相同，这个锁将会被永久锁定，无法再被旋转。字符串 target 代表可以解锁的数字，你需要给出解锁需要的最小旋转次数，如果无论如何不能解锁，返回 -1 。
+
+## 解题思路
+
+- 此题可以转化为从起始点到终点的最短路径。采用广度优先搜索。每次广搜枚举转动一次数字的状态，并且用 visited 记录是否被搜索过，如果没有被搜索过，便加入队列，下一轮继续搜索。如果搜索到了 target，便返回对应的旋转次数。如果搜索完成后，仍没有搜索到 target，说明无法解锁，返回 -1。特殊情况，如果 target 就是初始数字 0000，那么直接返回答案 0。
+- 在广搜之前，先将 deadends 放入 map 中，搜索中判断是否搜到了 deadends。如果初始数字 0000 出现在 deadends 中，可以直接返回答案 −1。
+
+## 代码
+
+```go
+package leetcode
+
+func openLock(deadends []string, target string) int {
+	if target == "0000" {
+		return 0
+	}
+	targetNum, visited := strToInt(target), make([]bool, 10000)
+	visited[0] = true
+	for _, deadend := range deadends {
+		num := strToInt(deadend)
+		if num == 0 {
+			return -1
+		}
+		visited[num] = true
+	}
+	depth, curDepth, nextDepth := 0, []int16{0}, make([]int16, 0)
+	var nextNum int16
+	for len(curDepth) > 0 {
+		nextDepth = nextDepth[0:0]
+		for _, curNum := range curDepth {
+			for incrementer := int16(1000); incrementer > 0; incrementer /= 10 {
+				digit := (curNum / incrementer) % 10
+				if digit == 9 {
+					nextNum = curNum - 9*incrementer
+				} else {
+					nextNum = curNum + incrementer
+				}
+				if nextNum == targetNum {
+					return depth + 1
+				}
+				if !visited[nextNum] {
+					visited[nextNum] = true
+					nextDepth = append(nextDepth, nextNum)
+				}
+				if digit == 0 {
+					nextNum = curNum + 9*incrementer
+				} else {
+					nextNum = curNum - incrementer
+				}
+				if nextNum == targetNum {
+					return depth + 1
+				}
+				if !visited[nextNum] {
+					visited[nextNum] = true
+					nextDepth = append(nextDepth, nextNum)
+				}
+			}
+		}
+		curDepth, nextDepth = nextDepth, curDepth
+		depth++
+	}
+	return -1
+}
+
+func strToInt(str string) int16 {
+	return int16(str[0]-'0')*1000 + int16(str[1]-'0')*100 + int16(str[2]-'0')*10 + int16(str[3]-'0')
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0748.Shortest-Completing-Word/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0753.Cracking-the-Safe/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0753.Cracking-the-Safe.md b/website/content/ChapterFour/0700~0799/0753.Cracking-the-Safe.md
new file mode 100644
index 000000000..3a919ce78
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0753.Cracking-the-Safe.md
@@ -0,0 +1,99 @@
+# [753. Cracking the Safe](https://leetcode.com/problems/cracking-the-safe/)
+
+
+
+## 题目
+
+There is a box protected by a password. The password is a sequence of `n` digits where each digit can be one of the first `k` digits `0, 1, ..., k-1`.
+
+While entering a password, the last `n` digits entered will automatically be matched against the correct password.
+
+For example, assuming the correct password is `"345"`, if you type `"012345"`, the box will open because the correct password matches the suffix of the entered password.
+
+Return any password of **minimum length** that is guaranteed to open the box at some point of entering it.
+
+**Example 1**:
+
+```
+Input: n = 1, k = 2
+Output: "01"
+Note: "10" will be accepted too.
+```
+
+**Example 2**:
+
+```
+Input: n = 2, k = 2
+Output: "00110"
+Note: "01100", "10011", "11001" will be accepted too.
+```
+
+**Note**:
+
+1. `n` will be in the range `[1, 4]`.
+2. `k` will be in the range `[1, 10]`.
+3. `k^n` will be at most `4096`.
+
+
+## 题目大意
+
+有一个需要密码才能打开的保险箱。密码是 n 位数, 密码的每一位是 k 位序列 0, 1, ..., k-1 中的一个 。你可以随意输入密码，保险箱会自动记住最后 n 位输入，如果匹配，则能够打开保险箱。举个例子，假设密码是 "345"，你可以输入 "012345" 来打开它，只是你输入了 6 个字符.请返回一个能打开保险箱的最短字符串。
+
+提示：
+
+- n 的范围是 [1, 4]。
+- k 的范围是 [1, 10]。
+- k^n 最大可能为 4096。
+
+
+## 解题思路
+
+- 给出 2 个数字 n 和 k，n 代表密码是 n 位数，k 代表密码是 k 位。保险箱会记住最后 n 位输入。返回一个能打开保险箱的最短字符串。
+- 看到题目中的数据范围，数据范围很小，所以可以考虑用 DFS。想解开保险箱，当然是暴力破解，枚举所有可能。题目要求我们输出一个最短的字符串，这里是本题的关键，为何有最短呢？这里有贪心的思想。如果下一次递归可以利用上一次的 n-1 位，那么最终输出的字符串肯定是最短的。(笔者这里就不证明了)，例如，例子 2 中，最短的字符串是 00，01，11，10。每次尝试都利用前一次的 n-1 位。想通了这个问题，利用 DFS 暴力回溯即可。
+
+## 代码
+
+```go
+const number = "0123456789"
+
+func crackSafe(n int, k int) string {
+	if n == 1 {
+		return number[:k]
+	}
+	visit, total := map[string]bool{}, int(math.Pow(float64(k), float64(n)))
+	str := make([]byte, 0, total+n-1)
+	for i := 1; i != n; i++ {
+		str = append(str, '0')
+	}
+	dfsCrackSafe(total, n, k, &str, &visit)
+	return string(str)
+}
+
+func dfsCrackSafe(depth, n, k int, str *[]byte, visit *map[string]bool) bool {
+	if depth == 0 {
+		return true
+	}
+	for i := 0; i != k; i++ {
+		*str = append(*str, byte('0'+i))
+		cur := string((*str)[len(*str)-n:])
+		if _, ok := (*visit)[cur]; ok != true {
+			(*visit)[cur] = true
+			if dfsCrackSafe(depth-1, n, k, str, visit) {
+				// 只有这里不需要删除
+				return true
+			}
+			delete(*visit, cur)
+		}
+		// 删除
+		*str = (*str)[0 : len(*str)-1]
+	}
+	return false
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0752.Open-the-Lock/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0756.Pyramid-Transition-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0756.Pyramid-Transition-Matrix.md b/website/content/ChapterFour/0700~0799/0756.Pyramid-Transition-Matrix.md
new file mode 100755
index 000000000..8694d1fee
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0756.Pyramid-Transition-Matrix.md
@@ -0,0 +1,99 @@
+# [756. Pyramid Transition Matrix](https://leetcode.com/problems/pyramid-transition-matrix/)
+
+
+## 题目
+
+We are stacking blocks to form a pyramid. Each block has a color which is a one letter string.
+
+We are allowed to place any color block `C` on top of two adjacent blocks of colors `A` and `B`, if and only if `ABC` is an allowed triple.
+
+We start with a bottom row of `bottom`, represented as a single string. We also start with a list of allowed triples `allowed`. Each allowed triple is represented as a string of length 3.
+
+Return true if we can build the pyramid all the way to the top, otherwise false.
+
+**Example 1**:
+
+    Input: bottom = "BCD", allowed = ["BCG", "CDE", "GEA", "FFF"]
+    Output: true
+    Explanation:
+    We can stack the pyramid like this:
+        A
+       / \
+      G   E
+     / \ / \
+    B   C   D
+    
+    We are allowed to place G on top of B and C because BCG is an allowed triple.  Similarly, we can place E on top of C and D, then A on top of G and E.
+
+**Example 2**:
+
+    Input: bottom = "AABA", allowed = ["AAA", "AAB", "ABA", "ABB", "BAC"]
+    Output: false
+    Explanation:
+    We can't stack the pyramid to the top.
+    Note that there could be allowed triples (A, B, C) and (A, B, D) with C != D.
+
+**Note**:
+
+1. `bottom` will be a string with length in range `[2, 8]`.
+2. `allowed` will have length in range `[0, 200]`.
+3. Letters in all strings will be chosen from the set `{'A', 'B', 'C', 'D', 'E', 'F', 'G'}`.
+
+
+## 题目大意
+
+现在，我们用一些方块来堆砌一个金字塔。 每个方块用仅包含一个字母的字符串表示，例如 “Z”。使用三元组表示金字塔的堆砌规则如下：
+
+(A, B, C) 表示，“C” 为顶层方块，方块 “A”、“B” 分别作为方块 “C” 下一层的的左、右子块。当且仅当(A, B, C)是被允许的三元组，我们才可以将其堆砌上。
+
+初始时，给定金字塔的基层 bottom，用一个字符串表示。一个允许的三元组列表 allowed，每个三元组用一个长度为 3 的字符串表示。如果可以由基层一直堆到塔尖返回 true，否则返回 false。
+
+
+
+## 解题思路
+
+- 这一题是一道 DFS 的题目。题目给出金字塔的底座字符串。然后还会给一个字符串数组，字符串数组里面代表的字符串的砖块。砖块是 3 个字符串组成的。前两个字符代表的是砖块的底边，后一个字符代表的是砖块的顶部。问给出的字符能拼成一个金字塔么？金字塔的特点是顶端就一个字符。  
+
+- 这一题用 DFS 深搜每个砖块，从底层砖块开始逐渐往上层码。每递归一层，新一层底部的砖块都会变。当递归到了一层底部只有 2 个字符，顶部只有一个字符的时候，就到金字塔顶端了，就算是完成了。这一题为了挑选合适的砖块，需要把每个砖块底部的 2 个字符作为 key 放进 map 中，加速查找。题目中也给出了特殊情况，相同底部可能存在多种砖块，所以一个 key 可能对应多个 value 的情况，即可能存在多个顶部砖块的情况。这种情况在递归遍历中需要考虑。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func pyramidTransition(bottom string, allowed []string) bool {
+	pyramid := make(map[string][]string)
+	for _, v := range allowed {
+		pyramid[v[:len(v)-1]] = append(pyramid[v[:len(v)-1]], string(v[len(v)-1]))
+	}
+	return dfsT(bottom, "", pyramid)
+}
+
+func dfsT(bottom, above string, pyramid map[string][]string) bool {
+	if len(bottom) == 2 && len(above) == 1 {
+		return true
+	}
+	if len(bottom) == len(above)+1 {
+		return dfsT(above, "", pyramid)
+	}
+	base := bottom[len(above) : len(above)+2]
+	if data, ok := pyramid[base]; ok {
+		for _, key := range data {
+			if dfsT(bottom, above+key, pyramid) {
+				return true
+			}
+		}
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0753.Cracking-the-Safe/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0762.Prime-Number-of-Set-Bits-in-Binary-Representation/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0762.Prime-Number-of-Set-Bits-in-Binary-Representation.md b/website/content/ChapterFour/0700~0799/0762.Prime-Number-of-Set-Bits-in-Binary-Representation.md
new file mode 100755
index 000000000..a4c571cf2
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0762.Prime-Number-of-Set-Bits-in-Binary-Representation.md
@@ -0,0 +1,85 @@
+# [762. Prime Number of Set Bits in Binary Representation](https://leetcode.com/problems/prime-number-of-set-bits-in-binary-representation/)
+
+
+## 题目
+
+Given two integers `L` and `R`, find the count of numbers in the range `[L, R]` (inclusive) having a prime number of set bits in their binary representation.
+
+(Recall that the number of set bits an integer has is the number of `1`s present when written in binary. For example, `21` written in binary is `10101` which has 3 set bits. Also, 1 is not a prime.)
+
+**Example 1**:
+
+    Input: L = 6, R = 10
+    Output: 4
+    Explanation:
+    6 -> 110 (2 set bits, 2 is prime)
+    7 -> 111 (3 set bits, 3 is prime)
+    9 -> 1001 (2 set bits , 2 is prime)
+    10->1010 (2 set bits , 2 is prime)
+
+**Example 2**:
+
+    Input: L = 10, R = 15
+    Output: 5
+    Explanation:
+    10 -> 1010 (2 set bits, 2 is prime)
+    11 -> 1011 (3 set bits, 3 is prime)
+    12 -> 1100 (2 set bits, 2 is prime)
+    13 -> 1101 (3 set bits, 3 is prime)
+    14 -> 1110 (3 set bits, 3 is prime)
+    15 -> 1111 (4 set bits, 4 is not prime)
+
+**Note**:
+
+1. `L, R` will be integers `L <= R` in the range `[1, 10^6]`.
+2. `R - L` will be at most 10000.
+
+
+## 题目大意
+
+给定两个整数 L 和 R ，找到闭区间 [L, R] 范围内，计算置位位数为质数的整数个数。（注意，计算置位代表二进制表示中1的个数。例如 21 的二进制表示 10101 有 3 个计算置位。还有，1 不是质数。）
+
+
+注意:
+
+- L, R 是 L <= R 且在 [1, 10^6] 中的整数。
+- R - L 的最大值为 10000。
+
+
+
+## 解题思路
+
+
+- 题目给出 `[L, R]` 区间，在这个区间内的每个整数的二进制表示中 1 的个数如果是素数，那么最终结果就加一，问最终结果是多少？这一题是一个组合题，判断一个数的二进制位有多少位 1，是第 191 题。题目中限定了区间最大不超过 10^6 ，所以 1 的位数最大是 19 位，也就是说素数最大就是 19 。那么素数可以有限枚举出来。最后按照题目的意思累积结果就可以了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math/bits"
+
+func countPrimeSetBits(L int, R int) int {
+	counter := 0
+	for i := L; i <= R; i++ {
+		if isPrime(bits.OnesCount(uint(i))) {
+			counter++
+		}
+	}
+	return counter
+}
+
+func isPrime(x int) bool {
+	return x == 2 || x == 3 || x == 5 || x == 7 || x == 11 || x == 13 || x == 17 || x == 19
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0756.Pyramid-Transition-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0763.Partition-Labels/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0763.Partition-Labels.md b/website/content/ChapterFour/0700~0799/0763.Partition-Labels.md
new file mode 100644
index 000000000..68b1b7b24
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0763.Partition-Labels.md
@@ -0,0 +1,99 @@
+# [763. Partition Labels](https://leetcode.com/problems/partition-labels/)
+
+## 题目
+
+A string S of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.
+
+
+
+**Example 1**:
+
+
+```
+
+Input: S = "ababcbacadefegdehijhklij"
+Output: [9,7,8]
+Explanation:
+The partition is "ababcbaca", "defegde", "hijhklij".
+This is a partition so that each letter appears in at most one part.
+A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.
+
+```
+
+**Note**:
+
+- S will have length in range [1, 500].
+- S will consist of lowercase letters ('a' to 'z') only.
+
+
+## 题目大意
+
+这道题考察的是滑动窗口的问题。
+
+给出一个字符串，要求输出满足条件窗口的长度，条件是在这个窗口内，字母中出现在这一个窗口内，不出现在其他窗口内。
+
+## 解题思路
+
+这一题有 2 种思路，第一种思路是先记录下每个字母的出现次数，然后对滑动窗口中的每个字母判断次数是否用尽为 0，如果这个窗口内的所有字母次数都为 0，这个窗口就是符合条件的窗口。时间复杂度为 O(n)
+
+另外一种思路是记录下每个字符最后一次出现的下标，这样就不用记录次数。在每个滑动窗口中，依次判断每个字母最后一次出现的位置，如果在一个下标内，所有字母的最后一次出现的位置都包含进来了，那么这个下标就是这个满足条件的窗口大小。时间复杂度为 O(n^2)
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func partitionLabels(S string) []int {
+	var lastIndexOf [26]int
+	for i, v := range S {
+		lastIndexOf[v-'a'] = i
+	}
+
+	var arr []int
+	for start, end := 0, 0; start < len(S); start = end + 1 {
+		end = lastIndexOf[S[start]-'a']
+		for i := start; i < end; i++ {
+			if end < lastIndexOf[S[i]-'a'] {
+				end = lastIndexOf[S[i]-'a']
+			}
+		}
+		arr = append(arr, end-start+1)
+	}
+	return arr
+}
+
+// 解法二
+func partitionLabels1(S string) []int {
+	visit, counter, res, sum, lastLength := make([]int, 26), map[byte]int{}, []int{}, 0, 0
+	for i := 0; i < len(S); i++ {
+		counter[S[i]]++
+	}
+
+	for i := 0; i < len(S); i++ {
+		counter[S[i]]--
+		visit[S[i]-'a'] = 1
+		sum = 0
+		for j := 0; j < 26; j++ {
+			if visit[j] == 1 {
+				sum += counter[byte('a'+j)]
+			}
+		}
+		if sum == 0 {
+			res = append(res, i+1-lastLength)
+			lastLength += i + 1 - lastLength
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0762.Prime-Number-of-Set-Bits-in-Binary-Representation/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0765.Couples-Holding-Hands/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0765.Couples-Holding-Hands.md b/website/content/ChapterFour/0700~0799/0765.Couples-Holding-Hands.md
new file mode 100755
index 000000000..988f3eaea
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0765.Couples-Holding-Hands.md
@@ -0,0 +1,100 @@
+# [765. Couples Holding Hands](https://leetcode.com/problems/couples-holding-hands/)
+
+
+## 题目
+
+N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum number of swaps so that every couple is sitting side by side. A swap consists of choosing **any** two people, then they stand up and switch seats.
+
+The people and seats are represented by an integer from `0` to `2N-1`, the couples are numbered in order, the first couple being `(0, 1)`, the second couple being `(2, 3)`, and so on with the last couple being `(2N-2, 2N-1)`.
+
+The couples' initial seating is given by `row[i]` being the value of the person who is initially sitting in the i-th seat.
+
+**Example 1**:
+
+    Input: row = [0, 2, 1, 3]
+    Output: 1
+    Explanation: We only need to swap the second (row[1]) and third (row[2]) person.
+
+**Example 2**:
+
+    Input: row = [3, 2, 0, 1]
+    Output: 0
+    Explanation: All couples are already seated side by side.
+
+**Note**:
+
+1. `len(row)` is even and in the range of `[4, 60]`.
+2. `row` is guaranteed to be a permutation of `0...len(row)-1`.
+
+
+## 题目大意
+
+N 对情侣坐在连续排列的 2N 个座位上，想要牵到对方的手。 计算最少交换座位的次数，以便每对情侣可以并肩坐在一起。 一次交换可选择任意两人，让他们站起来交换座位。人和座位用 0 到 2N-1 的整数表示，情侣们按顺序编号，第一对是 (0, 1)，第二对是 (2, 3)，以此类推，最后一对是 (2N-2, 2N-1)。这些情侣的初始座位 row[i] 是由最初始坐在第 i 个座位上的人决定的。
+
+说明:
+
+1. len(row) 是偶数且数值在 [4, 60]范围内。  
+2. 可以保证 row 是序列 0...len(row)-1 的一个全排列。  
+
+
+## 解题思路
+
+- 给出一个数组，数组里面两两相邻的元素代表一对情侣。情侣编号是从 0 开始的：0 和 1 是情侣，2 和 3 是情侣……这些情侣坐在一排，但是并非成对坐着一起的，问如何用最小的次数交换座位以后，情侣能两两坐在一起。
+- 这道题的突破口是如何找到最小的交换次数。乍一想可能没有思路。直觉告诉我们，这种难题，很可能最后推出来的结论，或者公式是一个很简单的式子。(事实此题确实是这种情况)先不考虑最小交换次数，用正常的方法来处理这道题。举个例子：【3 1 4 0 2 5】，从数组 0 下标开始往后扫。
+
+        初始状态
+        
+        集合 0：0，1
+        集合 1：2，3
+        集合 2：4，5
+
+    3 和 1 不是情侣，将 3 和 1 所在集合 `union()` 起来。3 所在集合是 1 ，1 所在集合是 0，将 0 和 1 号集合 `union()` 起来。因为情侣 0 和情侣 1 是集合 0 ，情侣 2 和情侣 3 是集合 1，以此类推。
+
+        集合 0 和 1：0，1，2，3
+        集合 2：4，5
+
+- 继续往后扫，4 和 0 不在同一个集合，4 在集合 3，0 在集合 0，那么把它们 `union()` 起来。
+
+        集合 0 和 1 和 2：0，1，2，3，4，5
+
+    在上面集合合并的过程中，合并了 2 次。那么就代表最少需要交换 2 次。也可以通过 `len(row)/2 - uf.count` 来计算。`len(row)/2` 是初始集合总数，`uf.count` 是最后剩下的集合数，两者相减就是中间交换的次数。
+
+- 最后实现的代码非常简单。并查集先相邻的两两元素 `union()` 在一起。然后扫原数组，每次扫相邻的两个，通过这两个元素值所在集合，进行 `union()`。扫完以后就可以得到最后的答案。
+- 回过头来看这道题，为什么我们从数组开头往后依次调整每一对情侣，这样交换的次数是最少的呢？其实这个方法的思想是贪心思想。从头开始往后一对一对的调整，就是可以最终做到次数最少。(具体证明笔者不会)交换到最后，最后一对情侣一定是正确的，无须交换。(因为前面每一对都调整完了，最后一对一定是正确的)
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func minSwapsCouples(row []int) int {
+	if len(row)&1 == 1 {
+		return 0
+	}
+	uf := template.UnionFind{}
+	uf.Init(len(row))
+	for i := 0; i < len(row)-1; i = i + 2 {
+		uf.Union(i, i+1)
+	}
+	for i := 0; i < len(row)-1; i = i + 2 {
+		if uf.Find(row[i]) != uf.Find(row[i+1]) {
+			uf.Union(row[i], row[i+1])
+		}
+	}
+	return len(row)/2 - uf.TotalCount()
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0763.Partition-Labels/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0766.Toeplitz-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0766.Toeplitz-Matrix.md b/website/content/ChapterFour/0700~0799/0766.Toeplitz-Matrix.md
new file mode 100755
index 000000000..e18758dac
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0766.Toeplitz-Matrix.md
@@ -0,0 +1,86 @@
+# [766. Toeplitz Matrix](https://leetcode.com/problems/toeplitz-matrix/)
+
+
+## 题目
+
+A matrix is *Toeplitz* if every diagonal from top-left to bottom-right has the same element.
+
+Now given an `M x N` matrix, return `True` if and only if the matrix is *Toeplitz*.
+
+**Example 1**:
+
+    Input:
+    matrix = [
+      [1,2,3,4],
+      [5,1,2,3],
+      [9,5,1,2]
+    ]
+    Output: True
+    Explanation:
+    In the above grid, the diagonals are:
+    "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]".
+    In each diagonal all elements are the same, so the answer is True.
+
+**Example 2**:
+
+    Input:
+    matrix = [
+      [1,2],
+      [2,2]
+    ]
+    Output: False
+    Explanation:
+    The diagonal "[1, 2]" has different elements.
+
+**Note**:
+
+1. `matrix` will be a 2D array of integers.
+2. `matrix` will have a number of rows and columns in range `[1, 20]`.
+3. `matrix[i][j]` will be integers in range `[0, 99]`.
+
+**Follow up**:
+
+1. What if the matrix is stored on disk, and the memory is limited such that you can only load at most one row of the matrix into the memory at once?
+2. What if the matrix is so large that you can only load up a partial row into the memory at once?
+
+
+## 题目大意
+
+如果一个矩阵的每一方向由左上到右下的对角线上具有相同元素，那么这个矩阵是托普利茨矩阵。给定一个 M x N 的矩阵，当且仅当它是托普利茨矩阵时返回 True。
+
+
+
+## 解题思路
+
+
+- 给出一个矩阵，要求判断矩阵所有对角斜线上的数字是否都是一个数字。
+- 水题，直接循环判断即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func isToeplitzMatrix(matrix [][]int) bool {
+	rows, columns := len(matrix), len(matrix[0])
+	for i := 1; i < rows; i++ {
+		for j := 1; j < columns; j++ {
+			if matrix[i-1][j-1] != matrix[i][j] {
+				return false
+			}
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0765.Couples-Holding-Hands/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0767.Reorganize-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0767.Reorganize-String.md b/website/content/ChapterFour/0700~0799/0767.Reorganize-String.md
new file mode 100644
index 000000000..c3bb92342
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0767.Reorganize-String.md
@@ -0,0 +1,156 @@
+# [767. Reorganize String](https://leetcode.com/problems/reorganize-string/)
+
+## 题目
+
+Given a string S, check if the letters can be rearranged so that two characters that are adjacent to each other are not the same.
+
+If possible, output any possible result.  If not possible, return the empty string.
+
+**Example 1**:
+
+```
+
+Input: S = "aab"
+Output: "aba"
+
+```
+
+**Example 2**:
+
+```
+
+Input: S = "aaab"
+Output: ""
+
+```
+
+**Note**:   
+
+S will consist of lowercase letters and have length in range [1, 500].
+
+
+## 题目大意
+
+给定一个字符串，要求重新排列字符串，让字符串两两字符不相同，如果可以实现，即输出最终的字符串，如果不能让两两不相同，则输出空字符串。
+
+## 解题思路
+
+这道题有 2 种做法。第一种做法是先统计每个字符的出现频率次数，按照频率次数从高往低排序。具体做法就是第 451 题了。如果有一个字母的频次次数超过了 (len(string)+1)/2 那么就返回空字符串。否则输出最终满足题意的字符串。按照频次排序以后，用 2 个指针，一个从 0 开始，另外一个从中间位置开始，依次取出一个字符拼接起来。
+
+第二种做法是用优先队列，结点是一个结构体，结构体有 2 个字段，一个字段记录是哪个字符，另一个字段记录是这个字符的频次。按照频次的多作为优先级高，用大根堆建立优先队列。注意，这样建立成功的优先队列，重复字母只有一个结点，频次记录在结构体的频次字段中。额外还需要一个辅助队列。优先队列每次都出队一个优先级最高的，然后频次减一，最终结果加上这个字符。然后将这个结点入队。入队的意义是检测这个结点的频次有没有减到 0，如果还不为 0 ，再插入优先队列中。
+
+```c
+    string reorganizeString(string S) {
+        vector<int> mp(26);
+        int n = S.size();
+        for (char c: S)
+            ++mp[c-'a'];
+        priority_queue<pair<int, char>> pq;
+        for (int i = 0; i < 26; ++i) {
+            if (mp[i] > (n+1)/2) return "";
+            if (mp[i]) pq.push({mp[i], i+'a'});
+        }
+        queue<pair<int, char>> myq;
+        string ans;
+        while (!pq.empty() || myq.size() > 1) {
+            if (myq.size() > 1) { // 注意这里要大于 1，如果是等于 1 的话，频次大的元素一直在输出了，答案就不对了。
+                auto cur = myq.front();
+                myq.pop();
+                if (cur.first != 0) pq.push(cur);
+            }
+            if (!pq.empty()) {
+                auto cur = pq.top();
+                pq.pop();
+                ans += cur.second;
+                cur.first--;
+                myq.push(cur);
+            }
+        }
+        return ans;
+    }
+```
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+func reorganizeString(S string) string {
+	fs := frequencySort767(S)
+	if fs == "" {
+		return ""
+	}
+	bs := []byte(fs)
+	ans := ""
+	j := (len(bs)-1)/2 + 1
+	for i := 0; i <= (len(bs)-1)/2; i++ {
+		ans += string(bs[i])
+		if j < len(bs) {
+			ans += string(bs[j])
+		}
+		j++
+	}
+	return ans
+}
+
+func frequencySort767(s string) string {
+	if s == "" {
+		return ""
+	}
+	sMap := map[byte]int{}
+	cMap := map[int][]byte{}
+	sb := []byte(s)
+	for _, b := range sb {
+		sMap[b]++
+		if sMap[b] > (len(sb)+1)/2 {
+			return ""
+		}
+	}
+	for key, value := range sMap {
+		cMap[value] = append(cMap[value], key)
+	}
+
+	var keys []int
+	for k := range cMap {
+		keys = append(keys, k)
+	}
+	sort.Sort(sort.Reverse(sort.IntSlice(keys)))
+	res := make([]byte, 0)
+	for _, k := range keys {
+		for i := 0; i < len(cMap[k]); i++ {
+			for j := 0; j < k; j++ {
+				res = append(res, cMap[k][i])
+			}
+		}
+	}
+	return string(res)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0766.Toeplitz-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0771.Jewels-and-Stones/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0771.Jewels-and-Stones.md b/website/content/ChapterFour/0700~0799/0771.Jewels-and-Stones.md
new file mode 100755
index 000000000..1d6f568c1
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0771.Jewels-and-Stones.md
@@ -0,0 +1,81 @@
+# [771. Jewels and Stones](https://leetcode.com/problems/jewels-and-stones/)
+
+
+
+## 题目
+
+You're given strings `J` representing the types of stones that are jewels, and `S` representing the stones you have. Each character in `S` is a type of stone you have. You want to know how many of the stones you have are also jewels.
+
+The letters in `J` are guaranteed distinct, and all characters in `J` and `S` are letters. Letters are case sensitive, so `"a"` is considered a different type of stone from `"A"`.
+
+**Example 1**:
+
+    Input: J = "aA", S = "aAAbbbb"
+    Output: 3
+
+**Example 2**:
+
+    Input: J = "z", S = "ZZ"
+    Output: 0
+
+**Note**:
+
+- `S` and `J` will consist of letters and have length at most 50.
+- The characters in `J` are distinct.
+
+
+## 题目大意
+
+给定字符串 J 代表石头中宝石的类型，和字符串 S 代表你拥有的石头。S 中每个字符代表了一种你拥有的石头的类型，你想知道你拥有的石头中有多少是宝石。
+
+J 中的字母不重复，J 和 S 中的所有字符都是字母。字母区分大小写，因此 "a" 和 "A" 是不同类型的石头。
+
+
+
+## 解题思路
+
+
+- 给出 2 个字符串，要求在 S 字符串中找出在 J 字符串里面出现的字符个数。这是一道简单题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+// 解法一
+func numJewelsInStones(J string, S string) int {
+	count := 0
+	for i := range S {
+		if strings.Contains(J, string(S[i])) {
+			count++
+		}
+	}
+	return count
+}
+
+// 解法二
+func numJewelsInStones1(J string, S string) int {
+	cache, result := make(map[rune]bool), 0
+	for _, r := range J {
+		cache[r] = true
+	}
+	for _, r := range S {
+		if _, ok := cache[r]; ok {
+			result++
+		}
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0767.Reorganize-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0775.Global-and-Local-Inversions/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0775.Global-and-Local-Inversions.md b/website/content/ChapterFour/0700~0799/0775.Global-and-Local-Inversions.md
new file mode 100644
index 000000000..03345b905
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0775.Global-and-Local-Inversions.md
@@ -0,0 +1,71 @@
+# [775. Global and Local Inversions](https://leetcode.com/problems/global-and-local-inversions/)
+
+
+## 题目
+
+We have some permutation `A` of `[0, 1, ..., N - 1]`, where `N` is the length of `A`.
+
+The number of (global) inversions is the number of `i < j` with `0 <= i < j < N` and `A[i] > A[j]`.
+
+The number of local inversions is the number of `i` with `0 <= i < N` and `A[i] > A[i+1]`.
+
+Return `true` if and only if the number of global inversions is equal to the number of local inversions.
+
+**Example 1:**
+
+```
+Input: A = [1,0,2]
+Output: true
+Explanation: There is 1 global inversion, and 1 local inversion.
+```
+
+**Example 2:**
+
+```
+Input: A = [1,2,0]
+Output: false
+Explanation: There are 2 global inversions, and 1 local inversion.
+```
+
+**Note:**
+
+- `A` will be a permutation of `[0, 1, ..., A.length - 1]`.
+- `A` will have length in range `[1, 5000]`.
+- The time limit for this problem has been reduced.
+
+## 题目大意
+
+数组 A 是 [0, 1, ..., N - 1] 的一种排列，N 是数组 A 的长度。全局倒置指的是 i,j 满足 0 <= i < j < N 并且 A[i] > A[j] ，局部倒置指的是 i 满足 0 <= i < N 并且 A[i] > A[i+1] 。当数组 A 中全局倒置的数量等于局部倒置的数量时，返回 true 。
+
+## 解题思路
+
+- 本题代码非常简单，重在思考的过程。`[0, 1, ..., N - 1]` 不出现全局倒置的理想情况应该是 `i` 排列在 `A[i-1]`，`A[i]`，`A[i+1]` 的位置上。例如 `1` 如果排列在 `A[3]` 的位置上，那么比 `1` 小的只有 `0` 一个元素，`A[0]`，`A[1]`，`A[2]` 中必定有 2 个元素比 `1` 大，那必须会出现全局倒置的情况。`[0, 1, ..., N - 1]` 这是最理想的情况，每个元素都在自己的位置上。每个元素如果往左右相互偏移 1 个元素，那么也能保证只存在局部倒置，如果左右偏移 2 个元素，那必定会出现全局倒置。所以结论是：**不出现全局倒置的理想情况应该是 `i` 排列在 `A[i-1]`，`A[i]`，`A[i+1]` 的位置上**。判断这个结论的代码很简单，只需要判断 `A[i] - i` 的取值是否是 -1，0，1，也即 `abs(A[i] - i ) ≤ 1`。
+
+## 代码
+
+```go
+package leetcode
+
+func isIdealPermutation(A []int) bool {
+	for i := range A {
+		if abs(A[i]-i) > 1 {
+			return false
+		}
+	}
+	return true
+}
+
+func abs(a int) int {
+	if a < 0 {
+		return -a
+	}
+	return a
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0771.Jewels-and-Stones/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0778.Swim-in-Rising-Water/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0778.Swim-in-Rising-Water.md b/website/content/ChapterFour/0700~0799/0778.Swim-in-Rising-Water.md
new file mode 100755
index 000000000..6ac095ff3
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0778.Swim-in-Rising-Water.md
@@ -0,0 +1,139 @@
+# [778. Swim in Rising Water](https://leetcode.com/problems/swim-in-rising-water/)
+
+
+## 题目
+
+On an N x N `grid`, each square `grid[i][j]` represents the elevation at that point `(i,j)`.
+
+Now rain starts to fall. At time `t`, the depth of the water everywhere is `t`. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most `t`. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.
+
+You start at the top left square `(0, 0)`. What is the least time until you can reach the bottom right square `(N-1, N-1)`?
+
+**Example 1**:
+
+    Input: [[0,2],[1,3]]
+    Output: 3
+    Explanation:
+    At time 0, you are in grid location (0, 0).
+    You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
+    
+    You cannot reach point (1, 1) until time 3.
+    When the depth of water is 3, we can swim anywhere inside the grid.
+
+**Example 2**:
+
+    Input: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
+    Output: 16
+    Explanation:
+     0  1  2  3  4
+    24 23 22 21  5
+    12 13 14 15 16
+    11 17 18 19 20
+    10  9  8  7  6
+    
+    The final route is marked in bold.
+    We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
+
+**Note**:
+
+1. `2 <= N <= 50`.
+2. grid[i][j] is a permutation of [0, ..., N*N - 1].
+
+## 题目大意
+
+
+在一个 N x N 的坐标方格 grid 中，每一个方格的值 grid[i][j] 表示在位置 (i,j) 的平台高度。现在开始下雨了。当时间为 t 时，此时雨水导致水池中任意位置的水位为 t 。你可以从一个平台游向四周相邻的任意一个平台，但是前提是此时水位必须同时淹没这两个平台。假定你可以瞬间移动无限距离，也就是默认在方格内部游动是不耗时的。当然，在你游泳的时候你必须待在坐标方格里面。
+
+你从坐标方格的左上平台 (0，0) 出发。最少耗时多久你才能到达坐标方格的右下平台 (N-1, N-1)？
+
+提示:
+
+- 2 <= N <= 50.
+- grid[i][j] 位于区间 [0, ..., N*N - 1] 内。
+
+
+## 解题思路
+
+- 给出一个 grid[i][j] 方格，每个格子里面表示游泳池里面平台的高度。t 时刻，游泳池中的水的高度是 t。只有水的高度到达了平台的高度以后才能游过去。问从 (0,0) 开始，最短多长时间能到达 (N-1, N-1) 。
+- 这一题有多种解法。第一种解题思路是利用 DFS + 二分。DFS 是用来遍历是否可达。利用时间(即当前水淹过的高度)来判断是否能到达终点 (N-1, N-1) 点。二分用来搜索最终结果的时间。为什么会考虑用二分加速呢？原因是：时间从 0 - max 依次递增。max 是游泳池最高的平台高度。当时间从 0 增加到 max 以后，肯定能到达终点 (N-1, N-1) 点，因为水比所有平台都要高了。想快速找到一个时间 t 能使得 (0,0) 点和 (N-1, N-1) 点之间连通，那么就想到用二分加速了。判断是否取中值的条件是 (0,0) 点和 (N-1, N-1) 点之间是否连通。
+- 第二种解题思路是并查集。只要是 (0,0) 点和 (N-1, N-1) 点没有连通，即不能游到终点，那么就开始 `union()` 操作，由于起点是 (0,0)，所以向右边 `i + 1` 和向下边 `j + 1` 开始尝试。每尝试完一轮，时间会加 1 秒，即高度会加一。直到 (0,0) 点和 (N-1, N-1) 点刚好连通，那么这个时间点就是最终要求的。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+// 解法一 DFS + 二分
+func swimInWater(grid [][]int) int {
+	row, col, flags, minWait, maxWait := len(grid), len(grid[0]), make([][]int, len(grid)), 0, 0
+	for i, row := range grid {
+		flags[i] = make([]int, len(row))
+		for j := 0; j < col; j++ {
+			flags[i][j] = -1
+			if row[j] > maxWait {
+				maxWait = row[j]
+			}
+		}
+	}
+	for minWait < maxWait {
+		midWait := (minWait + maxWait) / 2
+		addFlags(grid, flags, midWait, 0, 0)
+		if flags[row-1][col-1] == midWait {
+			maxWait = midWait
+		} else {
+			minWait = midWait + 1
+		}
+	}
+	return minWait
+}
+
+func addFlags(grid [][]int, flags [][]int, flag int, row int, col int) {
+	if row < 0 || col < 0 || row >= len(grid) || col >= len(grid[0]) {
+		return
+	}
+	if grid[row][col] > flag || flags[row][col] == flag {
+		return
+	}
+	flags[row][col] = flag
+	addFlags(grid, flags, flag, row-1, col)
+	addFlags(grid, flags, flag, row+1, col)
+	addFlags(grid, flags, flag, row, col-1)
+	addFlags(grid, flags, flag, row, col+1)
+}
+
+// 解法二 并查集(并不是此题的最优解)
+func swimInWater1(grid [][]int) int {
+	n, uf, res := len(grid), template.UnionFind{}, 0
+	uf.Init(n * n)
+	for uf.Find(0) != uf.Find(n*n-1) {
+		for i := 0; i < n; i++ {
+			for j := 0; j < n; j++ {
+				if grid[i][j] > res {
+					continue
+				}
+				if i < n-1 && grid[i+1][j] <= res {
+					uf.Union(i*n+j, i*n+j+n)
+				}
+				if j < n-1 && grid[i][j+1] <= res {
+					uf.Union(i*n+j, i*n+j+1)
+				}
+			}
+		}
+		res++
+	}
+	return res - 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0775.Global-and-Local-Inversions/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0781.Rabbits-in-Forest/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0781.Rabbits-in-Forest.md b/website/content/ChapterFour/0700~0799/0781.Rabbits-in-Forest.md
new file mode 100755
index 000000000..cf8d19323
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0781.Rabbits-in-Forest.md
@@ -0,0 +1,76 @@
+# [781. Rabbits in Forest](https://leetcode.com/problems/rabbits-in-forest/)
+
+
+## 题目
+
+In a forest, each rabbit has some color. Some subset of rabbits (possibly all of them) tell you how many other rabbits have the same color as them. Those `answers` are placed in an array.
+
+Return the minimum number of rabbits that could be in the forest.
+
+**Examples**:
+
+    Input: answers = [1, 1, 2]
+    Output: 5
+    Explanation:
+    The two rabbits that answered "1" could both be the same color, say red.
+    The rabbit than answered "2" can't be red or the answers would be inconsistent.
+    Say the rabbit that answered "2" was blue.
+    Then there should be 2 other blue rabbits in the forest that didn't answer into the array.
+    The smallest possible number of rabbits in the forest is therefore 5: 3 that answered plus 2 that didn't.
+    
+    Input: answers = [10, 10, 10]
+    Output: 11
+    
+    Input: answers = []
+    Output: 0
+
+**Note**:
+
+1. `answers` will have length at most `1000`.
+2. Each `answers[i]` will be an integer in the range `[0, 999]`.
+
+
+## 题目大意
+
+森林中，每个兔子都有颜色。其中一些兔子（可能是全部）告诉你还有多少其他的兔子和自己有相同的颜色。我们将这些回答放在 answers 数组里。返回森林中兔子的最少数量。
+
+说明:
+
+- answers 的长度最大为1000。
+- answers[i] 是在 [0, 999] 范围内的整数。
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面代表的是每个兔子说自己同类还有多少个。要求输出总共有多少只兔子。数字中可能兔子汇报的人数小于总兔子数。
+- 这一题关键在于如何划分不同种类的兔子，有可能相同种类的兔子的个数是一样的，比如 `[2,2,2,2,2,2]`，这其实是 3 个种类，总共 6 只兔子。用 map 去重相同种类的兔子，不断的减少，当有种类的兔子为 0 以后，还有该种类的兔子报数，需要当做另外一个种类的兔子来看待。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func numRabbits(ans []int) int {
+	total, m := 0, make(map[int]int)
+	for _, v := range ans {
+		if m[v] == 0 {
+			m[v] += v
+			total += v + 1
+		} else {
+			m[v]--
+		}
+	}
+	return total
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0778.Swim-in-Rising-Water/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0783.Minimum-Distance-Between-BST-Nodes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0783.Minimum-Distance-Between-BST-Nodes.md b/website/content/ChapterFour/0700~0799/0783.Minimum-Distance-Between-BST-Nodes.md
new file mode 100644
index 000000000..ae6f93087
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0783.Minimum-Distance-Between-BST-Nodes.md
@@ -0,0 +1,102 @@
+# [783. Minimum Distance Between BST Nodes](https://leetcode.com/problems/minimum-distance-between-bst-nodes/)
+
+
+## 题目
+
+Given the `root` of a Binary Search Tree (BST), return *the minimum difference between the values of any two different nodes in the tree*.
+
+**Note:** This question is the same as 530: [https://leetcode.com/problems/minimum-absolute-difference-in-bst/](https://leetcode.com/problems/minimum-absolute-difference-in-bst/)
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/02/05/bst1.jpg](https://assets.leetcode.com/uploads/2021/02/05/bst1.jpg)
+
+```
+Input: root = [4,2,6,1,3]
+Output: 1
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2021/02/05/bst2.jpg](https://assets.leetcode.com/uploads/2021/02/05/bst2.jpg)
+
+```
+Input: root = [1,0,48,null,null,12,49]
+Output: 1
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[2, 100]`.
+- `0 <= Node.val <= 10^5`
+
+## 题目大意
+
+给你一个二叉搜索树的根节点 root ，返回 树中任意两不同节点值之间的最小差值 。
+
+## 解题思路
+
+- 本题和第 530 题完全相同。解题思路见第 530 题。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func minDiffInBST(root *TreeNode) int {
+	res, nodes := math.MaxInt16, -1
+	dfsBST(root, &res, &nodes)
+	return res
+}
+
+func dfsBST(root *TreeNode, res, pre *int) {
+	if root == nil {
+		return
+	}
+	dfsBST(root.Left, res, pre)
+	if *pre != -1 {
+		*res = min(*res, abs(root.Val-*pre))
+	}
+	*pre = root.Val
+	dfsBST(root.Right, res, pre)
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0781.Rabbits-in-Forest/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0784.Letter-Case-Permutation/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0784.Letter-Case-Permutation.md b/website/content/ChapterFour/0700~0799/0784.Letter-Case-Permutation.md
new file mode 100755
index 000000000..5954779f4
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0784.Letter-Case-Permutation.md
@@ -0,0 +1,127 @@
+# [784. Letter Case Permutation](https://leetcode.com/problems/letter-case-permutation/)
+
+
+## 题目
+
+Given a string S, we can transform every letter individually to be lowercase or uppercase to create another string. Return a list of all possible strings we could create.
+
+**Examples**:
+
+    Input: S = "a1b2"
+    Output: ["a1b2", "a1B2", "A1b2", "A1B2"]
+    
+    Input: S = "3z4"
+    Output: ["3z4", "3Z4"]
+    
+    Input: S = "12345"
+    Output: ["12345"]
+
+**Note**:
+
+- `S` will be a string with length between `1` and `12`.
+- `S` will consist only of letters or digits.
+
+
+## 题目大意
+
+
+给定一个字符串 S，通过将字符串 S 中的每个字母转变大小写，我们可以获得一个新的字符串。返回所有可能得到的字符串集合。
+
+## 解题思路
+
+
+- 输出一个字符串中字母变大写，小写的所有组合。
+- DFS 深搜或者 BFS 广搜都可以。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strings"
+)
+
+// 解法一，DFS 深搜
+func letterCasePermutation(S string) []string {
+	if len(S) == 0 {
+		return []string{}
+	}
+	res, pos, c := []string{}, []int{}, []int{}
+	SS := strings.ToLower(S)
+	for i := 0; i < len(SS); i++ {
+		if isLowerLetter(SS[i]) {
+			pos = append(pos, i)
+		}
+	}
+	for i := 0; i <= len(pos); i++ {
+		findLetterCasePermutation(SS, pos, i, 0, c, &res)
+	}
+	return res
+}
+
+func findLetterCasePermutation(s string, pos []int, target, index int, c []int, res *[]string) {
+	if len(c) == target {
+		b := []byte(s)
+		for _, v := range c {
+			b[pos[v]] -= 'a' - 'A'
+		}
+		*res = append(*res, string(b))
+		return
+	}
+	for i := index; i < len(pos)-(target-len(c))+1; i++ {
+		c = append(c, i)
+		findLetterCasePermutation(s, pos, target, i+1, c, res)
+		c = c[:len(c)-1]
+	}
+}
+
+// 解法二，先讲第一个字母变大写，然后依次把后面的字母变大写。最终的解数组中答案是翻倍增长的
+// 第一步：
+// [mqe] -> [mqe, Mqe]
+// 第二步：
+// [mqe, Mqe] -> [mqe Mqe mQe MQe]
+// 第二步：
+// [mqe Mqe mQe MQe] -> [mqe Mqe mQe MQe mqE MqE mQE MQE]
+
+func letterCasePermutation1(S string) []string {
+	res := make([]string, 0, 1<<uint(len(S)))
+	S = strings.ToLower(S)
+	for k, v := range S {
+		if isLetter784(byte(v)) {
+			switch len(res) {
+			case 0:
+				res = append(res, S, toUpper(S, k))
+			default:
+				for _, s := range res {
+					res = append(res, toUpper(s, k))
+				}
+			}
+		}
+	}
+	if len(res) == 0 {
+		res = append(res, S)
+	}
+	return res
+}
+
+func isLetter784(c byte) bool {
+	return (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')
+}
+
+func toUpper(s string, i int) string {
+	b := []byte(s)
+	b[i] -= 'a' - 'A'
+	return string(b)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0783.Minimum-Distance-Between-BST-Nodes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0785.Is-Graph-Bipartite/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0785.Is-Graph-Bipartite.md b/website/content/ChapterFour/0700~0799/0785.Is-Graph-Bipartite.md
new file mode 100644
index 000000000..748451eef
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0785.Is-Graph-Bipartite.md
@@ -0,0 +1,105 @@
+# [785. Is Graph Bipartite?](https://leetcode.com/problems/is-graph-bipartite/)
+
+
+## 题目
+
+Given an undirected `graph`, return `true` if and only if it is bipartite.
+
+Recall that a graph is *bipartite* if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.
+
+The graph is given in the following form: `graph[i]` is a list of indexes `j` for which the edge between nodes `i` and `j` exists. Each node is an integer between `0` and `graph.length - 1`. There are no self edges or parallel edges: `graph[i]` does not contain `i`, and it doesn't contain any element twice.
+
+
+	Example 1:Input: [[1,3], [0,2], [1,3], [0,2]]
+	Output: true
+	Explanation: 
+	The graph looks like this:
+	0----1
+	|    |
+	|    |
+	3----2
+	We can divide the vertices into two groups: {0, 2} and {1, 3}.
+
+
+	Example 2:Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
+	Output: false
+	Explanation: 
+	The graph looks like this:
+	0----1
+	| \  |
+	|  \ |
+	3----2
+	We cannot find a way to divide the set of nodes into two independent subsets.
+
+
+**Note**:
+
+- `graph` will have length in range `[1, 100]`.
+- `graph[i]` will contain integers in range `[0, graph.length - 1]`.
+- `graph[i]` will not contain `i` or duplicate values.
+- The graph is undirected: if any element `j` is in `graph[i]`, then `i` will be in `graph[j]`.
+
+## 题目大意
+
+给定一个无向图 graph，当这个图为二分图时返回 true。
+
+graph 将会以邻接表方式给出，graph[i] 表示图中与节点i相连的所有节点。每个节点都是一个在 0 到 graph.length-1 之间的整数。这图中没有自环和平行边： graph[i] 中不存在 i，并且 graph[i] 中没有重复的值。
+
+注意:
+
+- graph 的长度范围为 [1, 100]。
+- graph[i] 中的元素的范围为 [0, graph.length - 1]。
+- graph[i] 不会包含 i 或者有重复的值。
+- 图是无向的: 如果 j 在 graph[i] 里边, 那么 i 也会在 graph[j] 里边。
+
+## 解题思路
+
+- 判断一个无向图是否是二分图。二分图的定义：如果我们能将一个图的节点集合分割成两个独立的子集 A 和 B，并使图中的每一条边的两个节点一个来自 A 集合，一个来自 B 集合，我们就将这个图称为二分图。
+- 这一题可以用 BFS、DFS、并查集来解答。这里是 DFS 实现。任选一个节点开始，把它染成红色，然后对整个图 DFS 遍历，把与它相连的节点并且未被染色的，都染成绿色。颜色不同的节点代表不同的集合。这时候还可能遇到第 2 种情况，与它相连的节点已经有颜色了，并且这个颜色和前一个节点的颜色相同，这就说明了该无向图不是二分图。可以直接 return false。如此遍历到所有节点都染色了，如果能染色成功，说明该无向图是二分图，返回 true。
+
+## 代码
+
+```go
+package leetcode
+
+// DFS 染色，1 是红色，0 是绿色，-1 是未染色
+func isBipartite(graph [][]int) bool {
+	colors := make([]int, len(graph))
+	for i := range colors {
+		colors[i] = -1
+	}
+	for i := range graph {
+		if !dfs(i, graph, colors, -1) {
+			return false
+		}
+	}
+	return true
+}
+
+func dfs(n int, graph [][]int, colors []int, parentCol int) bool {
+	if colors[n] == -1 {
+		if parentCol == 1 {
+			colors[n] = 0
+		} else {
+			colors[n] = 1
+		}
+	} else if colors[n] == parentCol {
+		return false
+	} else if colors[n] != parentCol {
+		return true
+	}
+	for _, c := range graph[n] {
+		if !dfs(c, graph, colors, colors[n]) {
+			return false
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0784.Letter-Case-Permutation/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0786.K-th-Smallest-Prime-Fraction/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0786.K-th-Smallest-Prime-Fraction.md b/website/content/ChapterFour/0700~0799/0786.K-th-Smallest-Prime-Fraction.md
new file mode 100755
index 000000000..ac61759e4
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0786.K-th-Smallest-Prime-Fraction.md
@@ -0,0 +1,126 @@
+# [786. K-th Smallest Prime Fraction](https://leetcode.com/problems/k-th-smallest-prime-fraction/)
+
+
+## 题目
+
+A sorted list `A` contains 1, plus some number of primes. Then, for every p < q in the list, we consider the fraction p/q.
+
+What is the `K`-th smallest fraction considered? Return your answer as an array of ints, where `answer[0] = p` and `answer[1] = q`.
+
+**Examples**:
+
+    Input: A = [1, 2, 3, 5], K = 3
+    Output: [2, 5]
+    Explanation:
+    The fractions to be considered in sorted order are:
+    1/5, 1/3, 2/5, 1/2, 3/5, 2/3.
+    The third fraction is 2/5.
+    
+    Input: A = [1, 7], K = 1
+    Output: [1, 7]
+
+**Note**:
+
+- `A` will have length between `2` and `2000`.
+- Each `A[i]` will be between `1` and `30000`.
+- `K` will be between `1` and `A.length * (A.length - 1) / 2`.
+
+
+## 题目大意
+
+一个已排序好的表 A，其包含 1 和其他一些素数.  当列表中的每一个 p<q 时，我们可以构造一个分数 p/q 。
+
+那么第 k 个最小的分数是多少呢?  以整数数组的形式返回你的答案, 这里 answer[0] = p 且 answer[1] = q.
+
+
+注意:
+
+- A 的取值范围在 2 — 2000.
+- 每个 A[i] 的值在 1 —30000.
+- K 取值范围为 1 — A.length * (A.length - 1) / 2
+
+
+## 解题思路
+
+
+- 给出一个从小到大排列的有序数组，数组里面的元素都是质数，请找出这个数组中的数组成的真分数从小到大排列，第 K 小的分数。
+- 这一题的暴力解法是枚举所有可能的真分数，从小到大排序，输出第 K 小的分数即可。注意排序的时候不能直接用 float 排序，需要转化成分子和分母的结构体进行排序。
+- 最优的解法是二分搜索。由于真分数都小于 1，所以二分搜索的范围是 [0,1]。每次二分出来的 mid，需要在数组里面搜索一次，找出比 mid 小的真分数个数。并记录下最大的真分数的分子和分母，动态维护最大真分数的分子和分母。如果比 mid 小的真分数个数小于 K，那么取右区间继续二分，如果比 mid 小的真分数个数大于 K，那么取左区间继续二分。直到正好找到比 mid 小的真分数个数是 K，此时维护的最大真分数的分子和分母即为答案。
+- 在已排序的矩阵中寻找最 K 小的元素这一系列的题目有：第 373 题，第 378 题，第 668 题，第 719 题，第 786 题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+// 解法一 二分搜索
+func kthSmallestPrimeFraction(A []int, K int) []int {
+	low, high, n := 0.0, 1.0, len(A)
+	// 因为是在小数内使用二分查找，无法像在整数范围内那样通过 mid+1 和边界判断来终止循环
+	// 所以此处根据 count 来结束循环
+	for {
+		mid, count, p, q, j := (high+low)/2.0, 0, 0, 1, 0
+		for i := 0; i < n; i++ {
+			for j < n && float64(A[i]) > float64(mid)*float64(A[j]) {
+				j++
+			}
+			count += n - j
+			if j < n && q*A[i] > p*A[j] {
+				p = A[i]
+				q = A[j]
+			}
+		}
+		if count == K {
+			return []int{p, q}
+		} else if count < K {
+			low = mid
+		} else {
+			high = mid
+		}
+	}
+}
+
+// 解法二 暴力解法，时间复杂度 O(n^2)
+func kthSmallestPrimeFraction1(A []int, K int) []int {
+	if len(A) == 0 || (len(A)*(len(A)-1))/2 < K {
+		return []int{}
+	}
+	fractions := []Fraction{}
+	for i := 0; i < len(A); i++ {
+		for j := i + 1; j < len(A); j++ {
+			fractions = append(fractions, Fraction{molecule: A[i], denominator: A[j]})
+		}
+	}
+	sort.Sort(SortByFraction(fractions))
+	return []int{fractions[K-1].molecule, fractions[K-1].denominator}
+}
+
+// Fraction define
+type Fraction struct {
+	molecule    int
+	denominator int
+}
+
+// SortByFraction define
+type SortByFraction []Fraction
+
+func (a SortByFraction) Len() int      { return len(a) }
+func (a SortByFraction) Swap(i, j int) { a[i], a[j] = a[j], a[i] }
+func (a SortByFraction) Less(i, j int) bool {
+	return a[i].molecule*a[j].denominator < a[j].molecule*a[i].denominator
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0785.Is-Graph-Bipartite/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0791.Custom-Sort-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0791.Custom-Sort-String.md b/website/content/ChapterFour/0700~0799/0791.Custom-Sort-String.md
new file mode 100644
index 000000000..80de5408e
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0791.Custom-Sort-String.md
@@ -0,0 +1,62 @@
+# [791. Custom Sort String](https://leetcode.com/problems/custom-sort-string/)
+
+
+## 题目
+
+`order` and `str` are strings composed of lowercase letters. In `order`, no letter occurs more than once.
+
+`order` was sorted in some custom order previously. We want to permute the characters of `str` so that they match the order that `order` was sorted. More specifically, if `x` occurs before `y` in `order`, then `x` should occur before `y` in the returned string.
+
+Return any permutation of `str` (as a string) that satisfies this property.
+
+```
+Example:Input:
+order = "cba"
+str = "abcd"
+Output: "cbad"
+Explanation:
+"a", "b", "c" appear in order, so the order of "a", "b", "c" should be "c", "b", and "a".
+Since "d" does not appear in order, it can be at any position in the returned string. "dcba", "cdba", "cbda" are also valid outputs.
+
+```
+
+**Note:**
+
+- `order` has length at most `26`, and no character is repeated in `order`.
+- `str` has length at most `200`.
+- `order` and `str` consist of lowercase letters only.
+
+## 题目大意
+
+字符串 S 和 T 只包含小写字符。在 S 中，所有字符只会出现一次。S 已经根据某种规则进行了排序。我们要根据 S 中的字符顺序对 T 进行排序。更具体地说，如果 S 中 x 在 y 之前出现，那么返回的字符串中 x 也应出现在 y 之前。返回任意一种符合条件的字符串 T。
+
+## 解题思路
+
+- 题目只要求 T 中包含 S 的字符串有序，所以可以先将 T 中包含 S 的字符串排好序，然后再拼接上其他字符。S 字符串最长为 26 位，先将 S 中字符的下标向左偏移 30，并将偏移后的下标值存入字典中。再把 T 字符串按照字典中下标值进行排序。S 中出现的字符对应的下标经过处理以后变成了负数，S 中未出现的字符的下标还是正数。所以经过排序以后，S 中出现的字符按照原有顺序排列在前面，S 中未出现的字符依次排在后面。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func customSortString(order string, str string) string {
+	magic := map[byte]int{}
+	for i := range order {
+		magic[order[i]] = i - 30
+	}
+	byteSlice := []byte(str)
+	sort.Slice(byteSlice, func(i, j int) bool {
+		return magic[byteSlice[i]] < magic[byteSlice[j]]
+	})
+	return string(byteSlice)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0786.K-th-Smallest-Prime-Fraction/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0792.Number-of-Matching-Subsequences/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0792.Number-of-Matching-Subsequences.md b/website/content/ChapterFour/0700~0799/0792.Number-of-Matching-Subsequences.md
new file mode 100644
index 000000000..9f773b9e2
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0792.Number-of-Matching-Subsequences.md
@@ -0,0 +1,73 @@
+# [792. Number of Matching Subsequences](https://leetcode.com/problems/number-of-matching-subsequences/)
+
+
+## 题目
+
+Given a string `s` and an array of strings `words`, return *the number of* `words[i]` *that is a subsequence of* `s`.
+
+A **subsequence** of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.
+
+- For example, `"ace"` is a subsequence of `"abcde"`.
+
+**Example 1:**
+
+```
+Input: s = "abcde", words = ["a","bb","acd","ace"]
+Output: 3
+Explanation: There are three strings in words that are a subsequence of s: "a", "acd", "ace".
+```
+
+**Example 2:**
+
+```
+Input: s = "dsahjpjauf", words = ["ahjpjau","ja","ahbwzgqnuk","tnmlanowax"]
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 5 * 104`
+- `1 <= words.length <= 5000`
+- `1 <= words[i].length <= 50`
+- `s` and `words[i]` consist of only lowercase English letters.
+
+## 题目大意
+
+给定字符串 S 和单词字典 words, 求 words[i] 中是 S 的子序列的单词个数。
+
+## 解题思路
+
+- 如果将 words 数组内的字符串每次都在源字符串 S 中匹配，这种暴力解法超时。超时原因是对字符串 S 遍历了多次。是否有更加高效的方法呢？
+- 把 words 数组内字符串按照首字母，分到 26 个桶中。从头开始遍历一遍源字符串 S，每扫一个字母，命中 26 个桶中其中一个桶，修改这个桶中的字符串。例如：当前遍历到了 'o'，此时桶中存的数据是 'a' : ['amy','aop'], 'o': ['oqp','onwn']，那么调整 'o' 桶中的数据后，各桶的状态为，'a' : ['amy','aop'], 'q': ['qp'], 'n': ['nwn']。从头到尾扫完整个字符串 S，某个桶中的字符串被清空，说明该桶中的字符串都符合 S 的子序列。将符合子序列的字符串个数累加起来即为最终答案。
+
+## 代码
+
+```go
+package leetcode
+
+func numMatchingSubseq(s string, words []string) int {
+	hash, res := make([][]string, 26), 0
+	for _, w := range words {
+		hash[int(w[0]-'a')] = append(hash[int(w[0]-'a')], w)
+	}
+	for _, c := range s {
+		words := hash[int(byte(c)-'a')]
+		hash[int(byte(c)-'a')] = []string{}
+		for _, w := range words {
+			if len(w) == 1 {
+				res += 1
+				continue
+			}
+			hash[int(w[1]-'a')] = append(hash[int(w[1]-'a')], w[1:])
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0791.Custom-Sort-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0793.Preimage-Size-of-Factorial-Zeroes-Function/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0793.Preimage-Size-of-Factorial-Zeroes-Function.md b/website/content/ChapterFour/0700~0799/0793.Preimage-Size-of-Factorial-Zeroes-Function.md
new file mode 100755
index 000000000..deca87330
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0793.Preimage-Size-of-Factorial-Zeroes-Function.md
@@ -0,0 +1,108 @@
+# [793. Preimage Size of Factorial Zeroes Function](https://leetcode.com/problems/preimage-size-of-factorial-zeroes-function/)
+
+
+## 题目
+
+Let `f(x)` be the number of zeroes at the end of `x!`. (Recall that `x! = 1 * 2 * 3 * ... * x`, and by convention, `0! = 1`.)
+
+For example, `f(3) = 0` because 3! = 6 has no zeroes at the end, while `f(11) = 2` because 11! = 39916800 has 2 zeroes at the end. Given `K`, find how many non-negative integers `x` have the property that `f(x) = K`.
+
+**Example 1**:
+
+    Input: K = 0
+    Output: 5
+    Explanation: 0!, 1!, 2!, 3!, and 4! end with K = 0 zeroes.
+    
+**Example 2**:
+
+    Input: K = 5
+    Output: 0
+    Explanation: There is no x such that x! ends in K = 5 zeroes.
+
+**Note**:
+
+- `K` will be an integer in the range `[0, 10^9]`.
+
+
+## 题目大意
+
+
+f(x) 是 x! 末尾是0的数量。（回想一下 x! = 1 * 2 * 3 * ... * x，且0! = 1）
+
+例如， f(3) = 0 ，因为3! = 6的末尾没有0；而 f(11) = 2 ，因为11!= 39916800末端有2个0。给定 K，找出多少个非负整数x ，有 f(x) = K 的性质。
+
+注意：
+
+- K 是范围在 [0, 10^9] 的整数。
+
+
+## 解题思路
+
+- 给出一个数 K，要求有多少个 n 能使得 n！末尾 0 的个数等于 K。
+- 这一题是基于第 172 题的逆过程加强版。第 172 题是给出 `n`，求得末尾 0 的个数。由第 172 题可以知道，`n！`末尾 0 的个数取决于因子 5 的个数。末尾可能有 `K` 个 0，那么 `n` 最多可以等于 `5 * K`，在 `[0, 5* K]` 区间内二分搜索，判断 `mid` 末尾 0 的个数，如果能找到 `K`，那么就范围 5，如果找不到这个 `K`，返回 0 。为什么答案取值只有 0 和 5 呢？因为当 `n` 增加 5 以后，因子 5 的个数又加一了，末尾又可以多 1 个或者多个 0(如果加 5 以后，有多个 5 的因子，例如 25，125，就有可能末尾增加多个 0)。所以有效的 `K` 值对应的 `n` 的范围区间就是 5 。反过来，无效的 `K` 值对应的 `n` 是 0。`K` 在 `5^n` 的分界线处会发生跳变，所有有些值取不到。例如，`n` 在 `[0,5)` 内取值，`K = 0`；`n` 在 `[5,10)` 内取值，`K = 1`；`n` 在 `[10,15)` 内取值，`K = 2`；`n` 在 `[15,20)` 内取值，`K = 3`；`n` 在 `[20,25)` 内取值，`K = 4`；`n` 在 `[25,30)` 内取值，`K = 6`，因为 25 提供了 2 个 5，也就提供了 2 个 0，所以 `K` 永远无法取值等于 5，即当 `K = 5` 时，找不到任何的 `n` 与之对应。
+- 这一题也可以用数学的方法解题。见解法二。这个解法的灵感来自于：n！末尾 0 的个数等于 [1,n] 所有数的因子 5 的个数总和。其次此题的结果一定只有 0 和 5 (分析见上一种解法)。有了这两个结论以后，就可以用数学的方法推导了。首先 n 可以表示为 5 进制的形式  
+{{< katex display >}} 
+n = 5^{0} * a_{0} + 5^{1} * a_{1} + 5^{2} * a_{2} + ... + 5^{n} * a_{n}, (a_{n} < 5)
+{{< /katex >}}   
+	上面式子中，所有有因子 5 的个数为：  
+{{< katex display >}} 
+K = \sum_{n=0}^{n} a_{n} * c_{n}
+{{< /katex >}}  
+	这个总数就即是 K。针对不同的 n，an 的通项公式不同，所以表示的 K 的系数也不同。cn 的通项公式呢？  
+{{< katex display >}} 
+c_{n} = 5 * c_{n-1} + 1，c_{0} = 0
+{{< /katex >}} 
+	由上面这个递推还能推出通项公式(不过这题不适用通项公式，是用递推公式更方便)： 
+{{< katex display >}} 
+c_{n} = \frac{5^{n} - 1 }{4} 
+{{< /katex >}} 
+	判断 K 是否能表示成两个数列的表示形式，等价于判断 K 是否能转化为以 Cn 为基的变进制数。到此，转化成类似第 483 题了。代码实现不难，见解法二。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 二分搜索
+func preimageSizeFZF(K int) int {
+	low, high := 0, 5*K
+	for low <= high {
+		mid := low + (high-low)>>1
+		k := trailingZeroes(mid)
+		if k == K {
+			return 5
+		} else if k > K {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return 0
+}
+
+// 解法二 数学方法
+func preimageSizeFZF1(K int) int {
+	base := 0
+	for base < K {
+		base = base*5 + 1
+	}
+	for K > 0 {
+		base = (base - 1) / 5
+		if K/base == 5 {
+			return 0
+		}
+		K %= base
+	}
+	return 5
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0792.Number-of-Matching-Subsequences/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0794.Valid-Tic-Tac-Toe-State/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0794.Valid-Tic-Tac-Toe-State.md b/website/content/ChapterFour/0700~0799/0794.Valid-Tic-Tac-Toe-State.md
new file mode 100644
index 000000000..6bc24866a
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0794.Valid-Tic-Tac-Toe-State.md
@@ -0,0 +1,126 @@
+# [794. Valid Tic-Tac-Toe State](https://leetcode.com/problems/valid-tic-tac-toe-state/)
+
+## 题目
+
+Given a Tic-Tac-Toe board as a string array board, return true if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.
+
+The board is a 3 x 3 array that consists of characters ' ', 'X', and 'O'. The ' ' character represents an empty square.
+
+Here are the rules of Tic-Tac-Toe:
+
+- Players take turns placing characters into empty squares ' '.
+- The first player always places 'X' characters, while the second player always places 'O' characters.
+- 'X' and 'O' characters are always placed into empty squares, never filled ones.
+- The game ends when there are three of the same (non-empty) character filling any row, column, or diagonal.
+- The game also ends if all squares are non-empty.
+- No more moves can be played if the game is over.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2021/05/15/tictactoe1-grid.jpg](https://assets.leetcode.com/uploads/2021/05/15/tictactoe1-grid.jpg)
+
+    Input: board = ["O  ","   ","   "]
+    Output: false
+    Explanation: The first player always plays "X".
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2021/05/15/tictactoe2-grid.jpg](https://assets.leetcode.com/uploads/2021/05/15/tictactoe2-grid.jpg)
+
+    Input: board = ["XOX"," X ","   "]
+    Output: false
+    Explanation: Players take turns making moves.
+
+**Example 3**:
+
+![https://assets.leetcode.com/uploads/2021/05/15/tictactoe3-grid.jpg](https://assets.leetcode.com/uploads/2021/05/15/tictactoe3-grid.jpg)
+
+    Input: board = ["XXX","   ","OOO"]
+    Output: false
+
+**Example 4**:
+
+![https://assets.leetcode.com/uploads/2021/05/15/tictactoe4-grid.jpg](https://assets.leetcode.com/uploads/2021/05/15/tictactoe4-grid.jpg)
+
+    Input: board = ["XOX","O O","XOX"]
+    Output: true
+
+**Constraints:**
+
+- board.length == 3
+- board[i].length == 3
+- board[i][j] is either 'X', 'O', or ' '.
+
+## 题目大意
+
+给你一个字符串数组 board 表示井字游戏的棋盘。当且仅当在井字游戏过程中，棋盘有可能达到 board 所显示的状态时，才返回 true 。
+
+井字游戏的棋盘是一个 3 x 3 数组，由字符 ' '，'X' 和 'O' 组成。字符 ' ' 代表一个空位。
+
+以下是井字游戏的规则：
+
+- 玩家轮流将字符放入空位（' '）中。
+- 玩家 1 总是放字符 'X' ，而玩家 2 总是放字符 'O' 。
+- 'X' 和 'O' 只允许放置在空位中，不允许对已放有字符的位置进行填充。
+- 当有 3 个相同（且非空）的字符填充任何行、列或对角线时，游戏结束。
+- 当所有位置非空时，也算为游戏结束。
+- 如果游戏结束，玩家不允许再放置字符。
+
+## 解题思路
+
+分类模拟：
+- 根据题意棋盘在任意时候，要么 X 的数量比 O 的数量多 1，要么两者相等
+- X 的数量等于 O 的数量时,任何行、列或对角线都不会出现 3 个相同的 X
+- X 的数量比 O 的数量多 1 时,任何行、列或对角线都不会出现 3 个相同的 O
+
+## 代码
+
+```go
+package leetcode
+
+func validTicTacToe(board []string) bool {
+	cntX, cntO := 0, 0
+	for i := range board {
+		for j := range board[i] {
+			if board[i][j] == 'X' {
+				cntX++
+			} else if board[i][j] == 'O' {
+				cntO++
+			}
+		}
+	}
+	if cntX < cntO || cntX > cntO+1 {
+		return false
+	}
+	if cntX == cntO {
+		return process(board, 'X')
+	}
+	return process(board, 'O')
+}
+
+func process(board []string, c byte) bool {
+	//某一行是"ccc"
+	if board[0] == string([]byte{c, c, c}) || board[1] == string([]byte{c, c, c}) || board[2] == string([]byte{c, c, c}) {
+		return false
+	}
+	//某一列是"ccc"
+	if (board[0][0] == c && board[1][0] == c && board[2][0] == c) ||
+		(board[0][1] == c && board[1][1] == c && board[2][1] == c) ||
+		(board[0][2] == c && board[1][2] == c && board[2][2] == c) {
+		return false
+	}
+	//某一对角线是"ccc"
+	if (board[0][0] == c && board[1][1] == c && board[2][2] == c) ||
+		(board[0][2] == c && board[1][1] == c && board[2][0] == c) {
+		return false
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0793.Preimage-Size-of-Factorial-Zeroes-Function/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0795.Number-of-Subarrays-with-Bounded-Maximum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/0795.Number-of-Subarrays-with-Bounded-Maximum.md b/website/content/ChapterFour/0700~0799/0795.Number-of-Subarrays-with-Bounded-Maximum.md
new file mode 100644
index 000000000..3b206fcca
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/0795.Number-of-Subarrays-with-Bounded-Maximum.md
@@ -0,0 +1,61 @@
+# [795. Number of Subarrays with Bounded Maximum](https://leetcode.com/problems/number-of-subarrays-with-bounded-maximum/)
+
+
+## 题目
+
+We are given an array `nums` of positive integers, and two positive integers `left` and `right` (`left <= right`).
+
+Return the number of (contiguous, non-empty) subarrays such that the value of the maximum array element in that subarray is at least `left` and at most `right`.
+
+```
+Example:Input:
+nums = [2, 1, 4, 3]
+left = 2
+right = 3
+Output: 3
+Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].
+```
+
+**Note:**
+
+- `left`, `right`, and `nums[i]` will be an integer in the range `[0, 109]`.
+- The length of `nums` will be in the range of `[1, 50000]`.
+
+## 题目大意
+
+给定一个元素都是正整数的数组`A` ，正整数 `L` 以及 `R` (`L <= R`)。求连续、非空且其中最大元素满足大于等于`L` 小于等于`R`的子数组个数。
+
+## 解题思路
+
+- 题目要求子数组最大元素在 [L,R] 区间内。假设 count(bound) 为计算所有元素都小于等于 bound 的子数组数量。那么本题所求的答案可转化为 count(R) - count(L-1)。
+- 如何统计所有元素小于 bound 的子数组数量呢？使用 count 变量记录在 bound 的左边，小于等于 bound 的连续元素数量。当找到一个这样的元素时，在此位置上结束的有效子数组的数量为 count + 1。当遇到一个元素大于 B 时，则在此位置结束的有效子数组的数量为 0。res 将每轮 count 累加，最终 res 中存的即是满足条件的所有子数组数量。
+
+## 代码
+
+```go
+package leetcode
+
+func numSubarrayBoundedMax(nums []int, left int, right int) int {
+	return getAnswerPerBound(nums, right) - getAnswerPerBound(nums, left-1)
+}
+
+func getAnswerPerBound(nums []int, bound int) int {
+	res, count := 0, 0
+	for _, num := range nums {
+		if num <= bound {
+			count++
+		} else {
+			count = 0
+		}
+		res += count
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0794.Valid-Tic-Tac-Toe-State/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0802.Find-Eventual-Safe-States/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0700~0799/_index.md b/website/content/ChapterFour/0700~0799/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/0700~0799/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/0800~0899/0802.Find-Eventual-Safe-States.md b/website/content/ChapterFour/0800~0899/0802.Find-Eventual-Safe-States.md
new file mode 100644
index 000000000..1a3d0cd79
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0802.Find-Eventual-Safe-States.md
@@ -0,0 +1,80 @@
+# [802. Find Eventual Safe States](https://leetcode.com/problems/find-eventual-safe-states/)
+
+
+
+## 题目
+
+In a directed graph, we start at some node and every turn, walk along a directed edge of the graph. If we reach a node that is terminal (that is, it has no outgoing directed edges), we stop.
+
+Now, say our starting node is *eventually safe* if and only if we must eventually walk to a terminal node. More specifically, there exists a natural number `K` so that for any choice of where to walk, we must have stopped at a terminal node in less than `K` steps.
+
+Which nodes are eventually safe? Return them as an array in sorted order.
+
+The directed graph has `N` nodes with labels `0, 1, ..., N-1`, where `N` is the length of `graph`. The graph is given in the following form: `graph[i]` is a list of labels `j` such that `(i, j)` is a directed edge of the graph.
+
+```
+Example:
+Input: graph = [[1,2],[2,3],[5],[0],[5],[],[]]
+Output: [2,4,5,6]
+Here is a diagram of the above graph.
+```
+
+![https://s3-lc-upload.s3.amazonaws.com/uploads/2018/03/17/picture1.png](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/03/17/picture1.png)
+
+**Note**:
+
+- `graph` will have length at most `10000`.
+- The number of edges in the graph will not exceed `32000`.
+- Each `graph[i]` will be a sorted list of different integers, chosen within the range `[0, graph.length - 1]`.
+
+## 题目大意
+
+在有向图中, 我们从某个节点和每个转向处开始, 沿着图的有向边走。 如果我们到达的节点是终点 (即它没有连出的有向边), 我们停止。现在, 如果我们最后能走到终点，那么我们的起始节点是最终安全的。 更具体地说, 存在一个自然数 K,  无论选择从哪里开始行走, 我们走了不到 K 步后必能停止在一个终点。哪些节点最终是安全的？ 结果返回一个有序的数组。
+
+提示：
+
+- graph 节点数不超过 10000.
+- 图的边数不会超过 32000.
+- 每个 graph[i] 被排序为不同的整数列表， 在区间 [0, graph.length - 1] 中选取。
+
+
+## 解题思路
+
+- 给出一个有向图，要求找出所有“安全”节点。“安全”节点的定义是：存在一个自然数 K, 无论选择从哪里开始行走, 我们走了不到 K 步后必能停止在一个终点。
+- 这一题可以用拓扑排序，也可以用 DFS 染色来解答。这里用 DFS 来解答。对于每个节点，我们有 3 种染色的方法：白色 0 号节点表示该节点还没有被访问过；灰色 1 号节点表示该节点在栈中（这一轮搜索中被访问过）或者在环中；黑色 2 号节点表示该节点的所有相连的节点都被访问过，且该节点不在环中。当我们第一次访问一个节点时，我们把它从白色变成灰色，并继续搜索与它相连的节点。如果在搜索过程中我们遇到一个灰色的节点，那么说明找到了一个环，此时退出搜索，所有的灰色节点保持不变（即从任意一个灰色节点开始，都能走到环中），如果搜索过程中，我们没有遇到灰色的节点，那么在回溯到当前节点时，我们把它从灰色变成黑色，即表示它是一个安全的节点。
+
+## 代码
+
+```go
+func eventualSafeNodes(graph [][]int) []int {
+	res, color := []int{}, make([]int, len(graph))
+	for i := range graph {
+		if dfsEventualSafeNodes(graph, i, color) {
+			res = append(res, i)
+		}
+	}
+	return res
+}
+
+// colors: WHITE 0, GRAY 1, BLACK 2;
+func dfsEventualSafeNodes(graph [][]int, idx int, color []int) bool {
+	if color[idx] > 0 {
+		return color[idx] == 2
+	}
+	color[idx] = 1
+	for i := range graph[idx] {
+		if !dfsEventualSafeNodes(graph, graph[idx][i], color) {
+			return false
+		}
+	}
+	color[idx] = 2
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0700~0799/0795.Number-of-Subarrays-with-Bounded-Maximum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0803.Bricks-Falling-When-Hit/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0803.Bricks-Falling-When-Hit.md b/website/content/ChapterFour/0800~0899/0803.Bricks-Falling-When-Hit.md
new file mode 100755
index 000000000..f05f3978f
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0803.Bricks-Falling-When-Hit.md
@@ -0,0 +1,132 @@
+# [803. Bricks Falling When Hit](https://leetcode.com/problems/bricks-falling-when-hit/)
+
+
+## 题目
+
+We have a grid of 1s and 0s; the 1s in a cell represent bricks. A brick will not drop if and only if it is directly connected to the top of the grid, or at least one of its (4-way) adjacent bricks will not drop.
+
+We will do some erasures sequentially. Each time we want to do the erasure at the location (i, j), the brick (if it exists) on that location will disappear, and then some other bricks may drop because of that erasure.
+
+Return an array representing the number of bricks that will drop after each erasure in sequence.
+
+**Example 1**:
+
+    Input: 
+    grid = [[1,0,0,0],[1,1,1,0]]
+    hits = [[1,0]]
+    Output: [2]
+    Explanation: 
+    If we erase the brick at (1, 0), the brick at (1, 1) and (1, 2) will drop. So we should return 2.
+
+**Example 2**:
+
+    Input: 
+    grid = [[1,0,0,0],[1,1,0,0]]
+    hits = [[1,1],[1,0]]
+    Output: [0,0]
+    Explanation: 
+    When we erase the brick at (1, 0), the brick at (1, 1) has already disappeared due to the last move. So each erasure will cause no bricks dropping.  Note that the erased brick (1, 0) will not be counted as a dropped brick.
+
+**Note**:
+
+- The number of rows and columns in the grid will be in the range [1, 200].
+- The number of erasures will not exceed the area of the grid.
+- It is guaranteed that each erasure will be different from any other erasure, and located inside the grid.
+- An erasure may refer to a location with no brick - if it does, no bricks drop.
+
+
+## 题目大意
+
+我们有一组包含1和0的网格；其中1表示砖块。 当且仅当一块砖直接连接到网格的顶部，或者它至少有一块相邻（4 个方向之一）砖块不会掉落时，它才不会落下。我们会依次消除一些砖块。每当我们消除 (i, j) 位置时， 对应位置的砖块（若存在）会消失，然后其他的砖块可能因为这个消除而落下。返回一个数组表示每次消除操作对应落下的砖块数目。
+
+
+注意:
+
+- 网格的行数和列数的范围是[1, 200]。
+- 消除的数字不会超过网格的区域。
+- 可以保证每次的消除都不相同，并且位于网格的内部。
+- 一个消除的位置可能没有砖块，如果这样的话，就不会有砖块落下。
+
+
+
+## 解题思路
+
+
+- 有一些砖块连接在天花板上，问，如果打掉某个砖块，会掉落几块砖块？打掉的每个砖块不参与计数。
+- 这一题可以用并查集和 DFS 求解。不过尝试用 DFS 的同学就会知道，这一题卡时间卡的很紧。用 DFS 虽然能 AC，但是耗时非常长。用并查集也必须进行秩压缩，不然耗时也非常长。另外，如果用了并查集，每个集合的总数单独统计，不随着 union() 操作，也会导致超时，笔者在这里被 LTE 了多次，最后只能重写 UnionFind 并查集类，将统计操作和 union() 操作写在一起，这一题才 faster than 100.00% AC。
+- 拿到题以后，首先尝试暴力解法，按照顺序打掉砖块，每次打掉砖块以后，都重建并查集。题目要求每次掉落几块砖块，实际上比较每次和天花板连通的砖块个数变化了多少块就可以了。那么解法就出来了，先把和天花板连通的砖块都 union() 起来，记录这个集合中砖块的个数 `count`，然后每次打掉一个砖块以后，重建并查集，计算与天花板连通的砖块的个数 `newCount`，`newCount - count -1` 就是最终答案(打掉的那块砖块不计算其中)，提交代码以后，发现 TLE。
+- 出现 TLE 以后一般思路都是对的，只是时间复杂度过高，需要优化。很明显，需要优化的地方是每次都重建了新的并查集，有没有办法能在上一次状态上进行变更，不用重建并查集呢？如果正向的打掉砖块，那么每次还需要以这个砖块为起点进行 DFS，时间复杂度还是很高。如果反向考虑呢？先把所有要打掉的砖块都打掉，构建打掉这些砖块以后剩下与天花板连通的并查集。然后反向添加打掉的砖块，每次添加一块就刷新一次它周围的 4 个砖块，不用 DFS，这样时间复杂度优化了很多。最后在按照 `newCount - count -1` 方式计算最终答案。注意每次还原一个砖块的时候需要染色回原有砖块的颜色 `1` 。优化成这样的做法，基本不会 TLE 了，如果计算 count 是单独计算的，还是会 TLE。如果没有进行秩压缩，时间会超过 1500 ms，所以这一题想拿到 100%，每步优化都要做好。最终 100% 的答案见代码。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func hitBricks(grid [][]int, hits [][]int) []int {
+	if len(hits) == 0 {
+		return []int{}
+	}
+	uf, m, n, res, oriCount := template.UnionFindCount{}, len(grid), len(grid[0]), make([]int, len(hits)), 0
+	uf.Init(m*n + 1)
+	// 先将要打掉的砖块染色
+	for _, hit := range hits {
+		if grid[hit[0]][hit[1]] == 1 {
+			grid[hit[0]][hit[1]] = 2
+		}
+	}
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			if grid[i][j] == 1 {
+				getUnionFindFromGrid(grid, i, j, uf)
+			}
+		}
+	}
+	oriCount = uf.Count()[uf.Find(m*n)]
+	for i := len(hits) - 1; i >= 0; i-- {
+		if grid[hits[i][0]][hits[i][1]] == 2 {
+			grid[hits[i][0]][hits[i][1]] = 1
+			getUnionFindFromGrid(grid, hits[i][0], hits[i][1], uf)
+		}
+		nowCount := uf.Count()[uf.Find(m*n)]
+		if nowCount-oriCount > 0 {
+			res[i] = nowCount - oriCount - 1
+		} else {
+			res[i] = 0
+		}
+		oriCount = nowCount
+	}
+	return res
+}
+
+func isInGrid(grid [][]int, x, y int) bool {
+	return x >= 0 && x < len(grid) && y >= 0 && y < len(grid[0])
+}
+
+func getUnionFindFromGrid(grid [][]int, x, y int, uf template.UnionFindCount) {
+	m, n := len(grid), len(grid[0])
+	if x == 0 {
+		uf.Union(m*n, x*n+y)
+	}
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		if isInGrid(grid, nx, ny) && grid[nx][ny] == 1 {
+			uf.Union(nx*n+ny, x*n+y)
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0802.Find-Eventual-Safe-States/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0807.Max-Increase-to-Keep-City-Skyline/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0807.Max-Increase-to-Keep-City-Skyline.md b/website/content/ChapterFour/0800~0899/0807.Max-Increase-to-Keep-City-Skyline.md
new file mode 100644
index 000000000..cb19b053a
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0807.Max-Increase-to-Keep-City-Skyline.md
@@ -0,0 +1,104 @@
+# [807. Max Increase to Keep City Skyline](https://leetcode.com/problems/max-increase-to-keep-city-skyline/)
+
+## 题目
+
+There is a city composed of n x n blocks, where each block contains a single building shaped like a vertical square prism. You are given a 0-indexed n x n integer matrix grid where grid[r][c] represents the height of the building located in the block at row r and column c.
+
+A city's skyline is the the outer contour formed by all the building when viewing the side of the city from a distance. The skyline from each cardinal direction north, east, south, and west may be different.
+
+We are allowed to increase the height of any number of buildings by any amount (the amount can be different per building). The height of a 0-height building can also be increased. However, increasing the height of a building should not affect the city's skyline from any cardinal direction.
+
+Return the maximum total sum that the height of the buildings can be increased by without changing the city's skyline from any cardinal direction.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2021/06/21/807-ex1.png](https://assets.leetcode.com/uploads/2021/06/21/807-ex1.png)
+
+    Input: grid = [[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]
+    Output: 35
+    Explanation: The building heights are shown in the center of the above image.
+    The skylines when viewed from each cardinal direction are drawn in red.
+    The grid after increasing the height of buildings without affecting skylines is:
+    gridNew = [ [8, 4, 8, 7],
+    [7, 4, 7, 7],
+    [9, 4, 8, 7],
+    [3, 3, 3, 3] ]
+
+**Example 2**:
+
+    Input: grid = [[0,0,0],[0,0,0],[0,0,0]]
+    Output: 0
+    Explanation: Increasing the height of any building will result in the skyline changing.
+
+**Constraints:**
+
+- n == grid.length
+- n == grid[r].length
+- 2 <= n <= 50
+- 0 <= grid[r][c] <= 100
+
+## 题目大意
+
+在二维数组grid中，grid[i][j]代表位于某处的建筑物的高度。 我们被允许增加任何数量（不同建筑物的数量可能不同）的建筑物的高度。 高度 0 也被认为是建筑物。
+
+最后，从新数组的所有四个方向（即顶部，底部，左侧和右侧）观看的“天际线”必须与原始数组的天际线相同。 城市的天际线是从远处观看时，由所有建筑物形成的矩形的外部轮廓。 请看下面的例子。
+
+建筑物高度可以增加的最大总和是多少？
+
+## 解题思路
+
+- 从数组竖直方向（即顶部，底部）看“天际线”计算出 topBottomSkyline
+- 从数组水平方向（即左侧，右侧）看“天际线”计算出 leftRightSkyline
+- 计算 grid 中每个元素与对应的 topBottomSkyline 和 leftRightSkyline 中较小值的差值
+- 统计所有差值的总和 ans 并返回
+
+## 代码
+
+```go
+package leetcode
+
+func maxIncreaseKeepingSkyline(grid [][]int) int {
+	n := len(grid)
+	topBottomSkyline := make([]int, 0, n)
+	leftRightSkyline := make([]int, 0, n)
+	for i := range grid {
+		cur := 0
+		for _, v := range grid[i] {
+			if cur < v {
+				cur = v
+			}
+		}
+		leftRightSkyline = append(leftRightSkyline, cur)
+	}
+	for j := range grid {
+		cur := 0
+		for i := 0; i < len(grid[0]); i++ {
+			if cur < grid[i][j] {
+				cur = grid[i][j]
+			}
+		}
+		topBottomSkyline = append(topBottomSkyline, cur)
+	}
+	var ans int
+	for i := range grid {
+		for j := 0; j < len(grid[0]); j++ {
+			ans += min(topBottomSkyline[j], leftRightSkyline[i]) - grid[i][j]
+		}
+	}
+	return ans
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0803.Bricks-Falling-When-Hit/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0810.Chalkboard-XOR-Game/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0810.Chalkboard-XOR-Game.md b/website/content/ChapterFour/0800~0899/0810.Chalkboard-XOR-Game.md
new file mode 100644
index 000000000..2c2d62d0f
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0810.Chalkboard-XOR-Game.md
@@ -0,0 +1,67 @@
+# [810. Chalkboard XOR Game](https://leetcode.com/problems/chalkboard-xor-game/)
+
+
+## 题目
+
+We are given non-negative integers nums[i] which are written on a chalkboard.  Alice and Bob take turns erasing exactly one number from the chalkboard, with Alice starting first.  If erasing a number causes the bitwise XOR of all the elements of the chalkboard to become 0, then that player loses.  (Also, we'll say the bitwise XOR of one element is that element itself, and the bitwise XOR of no elements is 0.)
+
+Also, if any player starts their turn with the bitwise XOR of all the elements of the chalkboard equal to 0, then that player wins.
+
+Return True if and only if Alice wins the game, assuming both players play optimally.
+
+```
+Example:Input: nums = [1, 1, 2]
+Output: false
+Explanation:
+Alice has two choices: erase 1 or erase 2.
+If she erases 1, the nums array becomes [1, 2]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 2 = 3. Now Bob can remove any element he wants, because Alice will be the one to erase the last element and she will lose.
+If Alice erases 2 first, now nums becomes [1, 1]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 1 = 0. Alice will lose.
+```
+
+**Notes:**
+
+- `1 <= N <= 1000`.
+- `0 <= nums[i] <= 2^16`.
+
+## 题目大意
+
+黑板上写着一个非负整数数组 nums[i] 。Alice 和 Bob 轮流从黑板上擦掉一个数字，Alice 先手。如果擦除一个数字后，剩余的所有数字按位异或运算得出的结果等于 0 的话，当前玩家游戏失败。 (另外，如果只剩一个数字，按位异或运算得到它本身；如果无数字剩余，按位异或运算结果为 0。）并且，轮到某个玩家时，如果当前黑板上所有数字按位异或运算结果等于 0，这个玩家获胜。假设两个玩家每步都使用最优解，当且仅当 Alice 获胜时返回 true。
+
+## 解题思路
+
+- Alice 必胜情况之一，Alice 先手，起始数组全部元素本身异或结果就为 0 。不需要擦除数字便自动获胜。除去这个情况，还有其他情况么？由于 2 人是交替擦除数字，且每次都恰好擦掉一个数字，因此对于这两人中的任意一人，其每次在擦除数字前，黑板上剩余数字的个数的奇偶性一定都是相同的。于是奇偶性成为突破口。
+- 如果 nums 的长度是偶数，Alice 先手是否必败呢？如果必败，代表无论擦掉哪一个数字，剩余所有数字的异或结果都等于 0。利用反证法证明上述结论是错误的。首先 {{< katex >}} num[0] \oplus num[1] \oplus num[2] \oplus \cdots  \oplus num[n-1] = X ≠ 0 {{< /katex >}} ，初始所有元素异或结果不为 0。假设 Alice 当前擦掉第 i 个元素，0 ≤ i < n。令 {{< katex >}}X_{n}{{< /katex >}} 代表擦掉第 n 位元素以后剩余元素异或的结果。由证题，无论擦掉哪一个数字，剩余所有数字的异或结果都等于 0。所以 {{< katex >}} X_{0} \oplus X_{1} \oplus X_{2} \oplus \cdots  \oplus X_{n-1} = 0{{< /katex >}} 。
+
+	{{< katex display >}} 
+    \begin{aligned}0 &= X_{0} \oplus  X_{1} \oplus X_{2} \oplus \cdots  \oplus X_{n-1} \\0 &= (X \oplus nums[0]) \oplus (X \oplus nums[1]) \oplus (X \oplus nums[2]) \oplus \cdots  \oplus (X \oplus nums[n-1])\\ 0 &= (X \oplus X \oplus \cdots  \oplus X) \oplus (nums[0] \oplus nums[1] \oplus nums[2] \oplus \cdots  \oplus nums[n-1])\\0 &= 0 \oplus X\\\\\Rightarrow X &= 0\\\end{aligned}
+	{{< /katex >}} 
+
+    由于 n 为偶数，所以 n 个 X 的异或结果为 0。最终推出 X = 0，很明显与前提 X ≠ 0 冲突。所以原命题，代表无论擦掉哪一个数字，剩余所有数字的异或结果都等于 0 是错误的。也就是说，当 n 为偶数时，代表无论擦掉哪一个数字，剩余所有数字的异或结果都不等于 0。即 Alice 有必胜策略。换句话说，当数组的长度是偶数时，先手 Alice 总能找到一个数字，在擦掉这个数字之后剩余的所有数字异或结果不等于 0。
+
+- 综上，Alice 必胜策略有 2 种情况：
+    1. 数组 nums 的全部元素初始本身异或结果就等于 0。
+    2. 数组 nums 的长度是偶数。
+
+## 代码
+
+```go
+package leetcode
+
+func xorGame(nums []int) bool {
+	if len(nums)%2 == 0 {
+		return true
+	}
+	xor := 0
+	for _, num := range nums {
+		xor ^= num
+	}
+	return xor == 0
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0807.Max-Increase-to-Keep-City-Skyline/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0811.Subdomain-Visit-Count/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0811.Subdomain-Visit-Count.md b/website/content/ChapterFour/0800~0899/0811.Subdomain-Visit-Count.md
new file mode 100755
index 000000000..12232f80e
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0811.Subdomain-Visit-Count.md
@@ -0,0 +1,150 @@
+# [811. Subdomain Visit Count](https://leetcode.com/problems/subdomain-visit-count/)
+
+
+## 题目
+
+A website domain like "discuss.leetcode.com" consists of various subdomains. At the top level, we have "com", at the next level, we have "leetcode.com", and at the lowest level, "discuss.leetcode.com". When we visit a domain like "discuss.leetcode.com", we will also visit the parent domains "leetcode.com" and "com" implicitly.
+
+Now, call a "count-paired domain" to be a count (representing the number of visits this domain received), followed by a space, followed by the address. An example of a count-paired domain might be "9001 discuss.leetcode.com".
+
+We are given a list `cpdomains` of count-paired domains. We would like a list of count-paired domains, (in the same format as the input, and in any order), that explicitly counts the number of visits to each subdomain.
+
+**Example 1**:
+
+    Input: 
+    ["9001 discuss.leetcode.com"]
+    Output: 
+    ["9001 discuss.leetcode.com", "9001 leetcode.com", "9001 com"]
+    Explanation: 
+    We only have one website domain: "discuss.leetcode.com". As discussed above, the subdomain "leetcode.com" and "com" will also be visited. So they will all be visited 9001 times.
+
+**Example 2**:
+
+    Input: 
+    ["900 google.mail.com", "50 yahoo.com", "1 intel.mail.com", "5 wiki.org"]
+    Output: 
+    ["901 mail.com","50 yahoo.com","900 google.mail.com","5 wiki.org","5 org","1 intel.mail.com","951 com"]
+    Explanation: 
+    We will visit "google.mail.com" 900 times, "yahoo.com" 50 times, "intel.mail.com" once and "wiki.org" 5 times. For the subdomains, we will visit "mail.com" 900 + 1 = 901 times, "com" 900 + 50 + 1 = 951 times, and "org" 5 times.
+
+**Notes**:
+
+- The length of `cpdomains` will not exceed `100`.
+- The length of each domain name will not exceed `100`.
+- Each address will have either 1 or 2 "." characters.
+- The input count in any count-paired domain will not exceed `10000`.
+- The answer output can be returned in any order.
+
+
+## 题目大意
+
+
+一个网站域名，如 "discuss.leetcode.com"，包含了多个子域名。作为顶级域名，常用的有 "com"，下一级则有 "leetcode.com"，最低的一级为 "discuss.leetcode.com"。当我们访问域名 "discuss.leetcode.com" 时，也同时访问了其父域名 "leetcode.com" 以及顶级域名 "com"。给定一个带访问次数和域名的组合，要求分别计算每个域名被访问的次数。其格式为访问次数+空格+地址，例如："9001 discuss.leetcode.com"。
+
+接下来会给出一组访问次数和域名组合的列表 cpdomains 。要求解析出所有域名的访问次数，输出格式和输入格式相同，不限定先后顺序。
+
+
+
+## 解题思路
+
+
+- 这一题是简单题，统计每个 domain 的出现频次。每个域名根据层级，一级一级的累加频次，比如 `discuss.leetcode.com`、`discuss.leetcode.com` 这个域名频次为 1，`leetcode.com` 这个域名频次为 1，`com` 这个域名频次为 1。用 map 依次统计每个 domain 出现的频次，按照格式要求输出。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+)
+
+// 解法一
+func subdomainVisits(cpdomains []string) []string {
+	result := make([]string, 0)
+	if len(cpdomains) == 0 {
+		return result
+	}
+	domainCountMap := make(map[string]int, 0)
+	for _, domain := range cpdomains {
+		countDomain := strings.Split(domain, " ")
+		allDomains := strings.Split(countDomain[1], ".")
+		temp := make([]string, 0)
+		for i := len(allDomains) - 1; i >= 0; i-- {
+			temp = append([]string{allDomains[i]}, temp...)
+			ld := strings.Join(temp, ".")
+			count, _ := strconv.Atoi(countDomain[0])
+			if val, ok := domainCountMap[ld]; !ok {
+				domainCountMap[ld] = count
+			} else {
+				domainCountMap[ld] = count + val
+			}
+		}
+	}
+	for k, v := range domainCountMap {
+		t := strings.Join([]string{strconv.Itoa(v), k}, " ")
+		result = append(result, t)
+	}
+	return result
+}
+
+// 解法二
+func subdomainVisits1(cpdomains []string) []string {
+	out := make([]string, 0)
+	var b strings.Builder
+	domains := make(map[string]int, 0)
+	for _, v := range cpdomains {
+		splitDomain(v, domains)
+	}
+	for k, v := range domains {
+		b.WriteString(strconv.Itoa(v))
+		b.WriteString(" ")
+		b.WriteString(k)
+		out = append(out, b.String())
+		b.Reset()
+	}
+	return out
+}
+
+func splitDomain(domain string, domains map[string]int) {
+	visits := 0
+	var e error
+	subdomains := make([]string, 0)
+	for i, v := range domain {
+		if v == ' ' {
+			visits, e = strconv.Atoi(domain[0:i])
+			if e != nil {
+				panic(e)
+			}
+			break
+		}
+	}
+	for i := len(domain) - 1; i >= 0; i-- {
+		if domain[i] == '.' {
+			subdomains = append(subdomains, domain[i+1:])
+		} else if domain[i] == ' ' {
+			subdomains = append(subdomains, domain[i+1:])
+			break
+		}
+	}
+	for _, v := range subdomains {
+		count, ok := domains[v]
+		if ok {
+			domains[v] = count + visits
+		} else {
+			domains[v] = visits
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0810.Chalkboard-XOR-Game/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0812.Largest-Triangle-Area/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0812.Largest-Triangle-Area.md b/website/content/ChapterFour/0800~0899/0812.Largest-Triangle-Area.md
new file mode 100644
index 000000000..6519d1581
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0812.Largest-Triangle-Area.md
@@ -0,0 +1,77 @@
+# [812. Largest Triangle Area](https://leetcode.com/problems/largest-triangle-area/)
+
+
+## 题目
+
+You have a list of points in the plane. Return the area of the largest triangle that can be formed by any 3 of the points.
+
+```
+Example:
+Input: points = [[0,0],[0,1],[1,0],[0,2],[2,0]]
+Output: 2
+Explanation: 
+The five points are show in the figure below. The red triangle is the largest.
+```
+
+![https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/04/1027.png](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/04/1027.png)
+
+**Notes**:
+
+- `3 <= points.length <= 50`.
+- No points will be duplicated.
+- `-50 <= points[i][j] <= 50`.
+- Answers within `10^-6` of the true value will be accepted as correct.
+
+## 题目大意
+
+给定包含多个点的集合，从其中取三个点组成三角形，返回能组成的最大三角形的面积。
+
+## 解题思路
+
+- 给出一组点的坐标，要求找出能组成三角形面积最大的点集合，输出这个最大面积。
+- 数学题。按照数学定义，分别计算这些能构成三角形的点形成的三角形面积，最终输出最大面积即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func largestTriangleArea(points [][]int) float64 {
+	maxArea, n := 0.0, len(points)
+	for i := 0; i < n; i++ {
+		for j := i + 1; j < n; j++ {
+			for k := j + 1; k < n; k++ {
+				maxArea = max(maxArea, area(points[i], points[j], points[k]))
+			}
+		}
+	}
+	return maxArea
+}
+
+func area(p1, p2, p3 []int) float64 {
+	return abs(p1[0]*p2[1]+p2[0]*p3[1]+p3[0]*p1[1]-p1[0]*p3[1]-p2[0]*p1[1]-p3[0]*p2[1]) / 2
+}
+
+func abs(num int) float64 {
+	if num < 0 {
+		num = -num
+	}
+	return float64(num)
+}
+
+func max(a, b float64) float64 {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0811.Subdomain-Visit-Count/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0815.Bus-Routes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0815.Bus-Routes.md b/website/content/ChapterFour/0800~0899/0815.Bus-Routes.md
new file mode 100755
index 000000000..29ab7cfa8
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0815.Bus-Routes.md
@@ -0,0 +1,95 @@
+# [815. Bus Routes](https://leetcode.com/problems/bus-routes/)
+
+
+## 题目
+
+We have a list of bus routes. Each `routes[i]` is a bus route that the i-th bus repeats forever. For example if `routes[0] = [1, 5, 7]`, this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->... forever.
+
+We start at bus stop `S` (initially not on a bus), and we want to go to bus stop `T`. Travelling by buses only, what is the least number of buses we must take to reach our destination? Return -1 if it is not possible.
+
+**Example**:
+
+    Input: 
+    routes = [[1, 2, 7], [3, 6, 7]]
+    S = 1
+    T = 6
+    Output: 2
+    Explanation: 
+    The best strategy is take the first bus to the bus stop 7, then take the second bus to the bus stop 6.
+
+**Note**:
+
+- `1 <= routes.length <= 500`.
+- `1 <= routes[i].length <= 500`.
+- `0 <= routes[i][j] < 10 ^ 6`.
+
+
+## 题目大意
+
+我们有一系列公交路线。每一条路线 routes[i] 上都有一辆公交车在上面循环行驶。例如，有一条路线 routes[0] = [1, 5, 7]，表示第一辆 (下标为0) 公交车会一直按照 1->5->7->1->5->7->1->... 的车站路线行驶。假设我们从 S 车站开始（初始时不在公交车上），要去往 T 站。 期间仅可乘坐公交车，求出最少乘坐的公交车数量。返回 -1 表示不可能到达终点车站。
+
+
+说明:
+
+- 1 <= routes.length <= 500.
+- 1 <= routes[i].length <= 500.
+- 0 <= routes[i][j] < 10 ^ 6.
+
+
+## 解题思路
+
+- 给出一些公交路线，公交路径代表经过的哪些站。现在给出起点和终点站，问最少需要换多少辆公交车才能从起点到终点？
+- 这一题可以转换成图论的问题，将每个站台看成顶点，公交路径看成每个顶点的边。同一个公交的边染色相同。题目即可转化为从顶点 S 到顶点 T 需要经过最少多少条不同的染色边。用 BFS 即可轻松解决。从起点 S 开始，不断的扩展它能到达的站点。用 visited 数组防止放入已经可达的站点引起的环。用 map 存储站点和公交车的映射关系(即某个站点可以由哪些公交车到达)，BFS 的过程中可以用这个映射关系，拿到公交车的其他站点信息，从而扩张队列里面的可达站点。一旦扩展出现了终点 T，就可以返回结果了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func numBusesToDestination(routes [][]int, S int, T int) int {
+	if S == T {
+		return 0
+	}
+	// vertexMap 中 key 是站点，value 是公交车数组，代表这些公交车路线可以到达此站点
+	vertexMap, visited, queue, res := map[int][]int{}, make([]bool, len(routes)), []int{}, 0
+	for i := 0; i < len(routes); i++ {
+		for _, v := range routes[i] {
+			tmp := vertexMap[v]
+			tmp = append(tmp, i)
+			vertexMap[v] = tmp
+		}
+	}
+	queue = append(queue, S)
+	for len(queue) > 0 {
+		res++
+		qlen := len(queue)
+		for i := 0; i < qlen; i++ {
+			vertex := queue[0]
+			queue = queue[1:]
+			for _, bus := range vertexMap[vertex] {
+				if visited[bus] == true {
+					continue
+				}
+				visited[bus] = true
+				for _, v := range routes[bus] {
+					if v == T {
+						return res
+					}
+					queue = append(queue, v)
+				}
+			}
+		}
+	}
+	return -1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0812.Largest-Triangle-Area/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0816.Ambiguous-Coordinates/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0816.Ambiguous-Coordinates.md b/website/content/ChapterFour/0800~0899/0816.Ambiguous-Coordinates.md
new file mode 100644
index 000000000..3c10a9ec1
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0816.Ambiguous-Coordinates.md
@@ -0,0 +1,102 @@
+# [816. Ambiguous Coordinates](https://leetcode.com/problems/ambiguous-coordinates/)
+
+
+## 题目
+
+We had some 2-dimensional coordinates, like `"(1, 3)"` or `"(2, 0.5)"`.  Then, we removed all commas, decimal points, and spaces, and ended up with the string `s`.  Return a list of strings representing all possibilities for what our original coordinates could have been.
+
+Our original representation never had extraneous zeroes, so we never started with numbers like "00", "0.0", "0.00", "1.0", "001", "00.01", or any other number that can be represented with less digits.  Also, a decimal point within a number never occurs without at least one digit occuring before it, so we never started with numbers like ".1".
+
+The final answer list can be returned in any order.  Also note that all coordinates in the final answer have exactly one space between them (occurring after the comma.)
+
+```
+Example 1:Input: s = "(123)"
+Output: ["(1, 23)", "(12, 3)", "(1.2, 3)", "(1, 2.3)"]
+
+```
+
+```
+Example 2:Input: s = "(00011)"
+Output:  ["(0.001, 1)", "(0, 0.011)"]
+Explanation:
+0.0, 00, 0001 or 00.01 are not allowed.
+
+```
+
+```
+Example 3:Input: s = "(0123)"
+Output: ["(0, 123)", "(0, 12.3)", "(0, 1.23)", "(0.1, 23)", "(0.1, 2.3)", "(0.12, 3)"]
+
+```
+
+```
+Example 4:Input: s = "(100)"
+Output: [(10, 0)]
+Explanation:
+1.0 is not allowed.
+
+```
+
+**Note:**
+
+- `4 <= s.length <= 12`.
+- `s[0]` = "(", `s[s.length - 1]` = ")", and the other elements in `s` are digits.
+
+## 题目大意
+
+我们有一些二维坐标，如 "(1, 3)" 或 "(2, 0.5)"，然后我们移除所有逗号，小数点和空格，得到一个字符串S。返回所有可能的原始字符串到一个列表中。原始的坐标表示法不会存在多余的零，所以不会出现类似于"00", "0.0", "0.00", "1.0", "001", "00.01"或一些其他更小的数来表示坐标。此外，一个小数点前至少存在一个数，所以也不会出现“.1”形式的数字。
+
+最后返回的列表可以是任意顺序的。而且注意返回的两个数字中间（逗号之后）都有一个空格。
+
+## 解题思路
+
+- 本题没有什么算法思想，纯暴力题。先将原始字符串一分为二，分为的两个子字符串再移动坐标点，最后将每种情况组合再一次，这算完成了一次切分。将原始字符串每一位都按此规律完成切分，此题便得解。
+- 这道题有 2 处需要注意的。第一处是最终输出的字符串，请注意，**两个数字中间（逗号之后）都有一个空格**。不遵守输出格式的要求也会导致 `Wrong Answer`。另外一处是切分数字时，有 2 种违法情况，一种是带前导 0 的，另外一种是末尾带 0 的。带前导 0 的也分为 2 种情况，一种是只有一位，即只有一个 0，这种情况直接返回，因为这一个 0 怎么切分也只有一种切分方法。另外一种是长度大于 1，即 `0xxx` 这种情况。`0xxx` 这种情况只有一种切分方法，即 `0.xxx`。末尾带 0 的只有一种切分方法，即 `xxx0`，不可切分，因为 `xxx.0`，`xx.x0`，`x.xx0` 这些都是违法情况，所以末尾带 0 的也可以直接返回。具体的实现见代码和注释。
+
+## 代码
+
+```go
+package leetcode
+
+func ambiguousCoordinates(s string) []string {
+	res := []string{}
+	s = s[1 : len(s)-1]
+	for i := range s[:len(s)-1] {
+		a := build(s[:i+1])
+		b := build(s[i+1:])
+		for _, ta := range a {
+			for _, tb := range b {
+				res = append(res, "("+ta+", "+tb+")")
+			}
+		}
+	}
+	return res
+}
+
+func build(s string) []string {
+	res := []string{}
+	if len(s) == 1 || s[0] != '0' {
+		res = append(res, s)
+	}
+	// 结尾带 0 的情况
+	if s[len(s)-1] == '0' {
+		return res
+	}
+	// 切分长度大于一位且带前导 0 的情况
+	if s[0] == '0' {
+		res = append(res, "0."+s[1:])
+		return res
+	}
+	for i := range s[:len(s)-1] {
+		res = append(res, s[:i+1]+"."+s[i+1:])
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0815.Bus-Routes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0817.Linked-List-Components/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0817.Linked-List-Components.md b/website/content/ChapterFour/0800~0899/0817.Linked-List-Components.md
new file mode 100644
index 000000000..9527d1750
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0817.Linked-List-Components.md
@@ -0,0 +1,116 @@
+# [817. Linked List Components](https://leetcode.com/problems/linked-list-components/)
+
+## 题目
+
+We are given head, the head node of a linked list containing unique integer values.
+
+We are also given the list G, a subset of the values in the linked list.
+
+Return the number of connected components in G, where two values are connected if they appear consecutively in the linked list.
+
+**Example 1**:
+
+```
+
+Input: 
+head: 0->1->2->3
+G = [0, 1, 3]
+Output: 2
+Explanation: 
+0 and 1 are connected, so [0, 1] and [3] are the two connected components.
+
+```
+
+**Example 2**:
+
+```
+
+Input: 
+head: 0->1->2->3->4
+G = [0, 3, 1, 4]
+Output: 2
+Explanation: 
+0 and 1 are connected, 3 and 4 are connected, so [0, 1] and [3, 4] are the two connected components.
+
+```
+
+**Note**:
+
+- If N is the length of the linked list given by head, 1 <= N <= 10000.
+- The value of each node in the linked list will be in the range [0, N - 1].
+- 1 <= G.length <= 10000.
+- G is a subset of all values in the linked list.
+
+
+
+## 题目大意
+
+这道题题目的意思描述的不是很明白，我提交了几次 WA 以后才悟懂题意。
+
+这道题的意思是，在 G 中能组成多少组子链表，这些子链表的要求是能在原链表中是有序的。
+
+## 解题思路
+
+这个问题再抽象一下就成为这样：在原链表中去掉 G 中不存在的数，会被切断成几段链表。例如，将原链表中 G 中存在的数标为 0，不存在的数标为 1 。原链表标识为 0-0-0-1-0-1-1-0-0-1-0-1，那么这样原链表被断成了 4 段。只要在链表中找 0-1 组合就可以认为是一段，因为这里必定会有一段生成。
+
+考虑末尾的情况，0-1，1-0，0-0，1-1，这 4 种情况的特征都是，末尾一位只要是 0，都会新产生一段。所以链表末尾再单独判断一次，是 0 就再加一。
+
+
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func numComponents(head *ListNode, G []int) int {
+	if head.Next == nil {
+		return 1
+	}
+	gMap := toMap(G)
+	count := 0
+	cur := head
+
+	for cur != nil {
+		if _, ok := gMap[cur.Val]; ok {
+			if cur.Next == nil { // 末尾存在，直接加一
+				count++
+			} else {
+				if _, ok = gMap[cur.Next.Val]; !ok {
+					count++
+				}
+			}
+		}
+		cur = cur.Next
+	}
+	return count
+}
+
+func toMap(G []int) map[int]int {
+	GMap := make(map[int]int, 0)
+	for _, value := range G {
+		GMap[value] = 0
+	}
+	return GMap
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0816.Ambiguous-Coordinates/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0819.Most-Common-Word/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0819.Most-Common-Word.md b/website/content/ChapterFour/0800~0899/0819.Most-Common-Word.md
new file mode 100755
index 000000000..fc0a4bc2d
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0819.Most-Common-Word.md
@@ -0,0 +1,96 @@
+# [819. Most Common Word](https://leetcode.com/problems/most-common-word/)
+
+
+## 题目
+
+Given a paragraph and a list of banned words, return the most frequent word that is not in the list of banned words. It is guaranteed there is at least one word that isn't banned, and that the answer is unique.
+
+Words in the list of banned words are given in lowercase, and free of punctuation. Words in the paragraph are not case sensitive. The answer is in lowercase.
+
+**Example**:
+
+    Input: 
+    paragraph = "Bob hit a ball, the hit BALL flew far after it was hit."
+    banned = ["hit"]
+    Output: "ball"
+    Explanation: 
+    "hit" occurs 3 times, but it is a banned word.
+    "ball" occurs twice (and no other word does), so it is the most frequent non-banned word in the paragraph. 
+    Note that words in the paragraph are not case sensitive,
+    that punctuation is ignored (even if adjacent to words, such as "ball,"), 
+    and that "hit" isn't the answer even though it occurs more because it is banned.
+
+**Note**:
+
+- `1 <= paragraph.length <= 1000`.
+- `0 <= banned.length <= 100`.
+- `1 <= banned[i].length <= 10`.
+- The answer is unique, and written in lowercase (even if its occurrences in `paragraph` may have uppercase symbols, and even if it is a proper noun.)
+- `paragraph` only consists of letters, spaces, or the punctuation symbols `!?',;.`
+- There are no hyphens or hyphenated words.
+- Words only consist of letters, never apostrophes or other punctuation symbols.
+
+
+## 题目大意
+
+
+给定一个段落 (paragraph) 和一个禁用单词列表 (banned)。返回出现次数最多，同时不在禁用列表中的单词。题目保证至少有一个词不在禁用列表中，而且答案唯一。
+
+禁用列表中的单词用小写字母表示，不含标点符号。段落中的单词不区分大小写。答案都是小写字母。
+
+
+## 解题思路
+
+- 给出一段话和一个 banned 的字符串，要求输出这段话中出现频次最高的并且不出现在 banned 数组里面的字符串，答案唯一。这题是简单题，依次统计每个单词的频次，然后从 map 中删除 banned 里面的单词，取出剩下频次最高的单词即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func mostCommonWord(paragraph string, banned []string) string {
+	freqMap, start := make(map[string]int), -1
+	for i, c := range paragraph {
+		if c == ' ' || c == '!' || c == '?' || c == '\'' || c == ',' || c == ';' || c == '.' {
+			if start > -1 {
+				word := strings.ToLower(paragraph[start:i])
+				freqMap[word]++
+			}
+			start = -1
+		} else {
+			if start == -1 {
+				start = i
+			}
+		}
+	}
+	if start != -1 {
+		word := strings.ToLower(paragraph[start:])
+		freqMap[word]++
+	}
+	// Strip the banned words from the freqmap
+	for _, bannedWord := range banned {
+		delete(freqMap, bannedWord)
+	}
+	// Find most freq word
+	mostFreqWord, mostFreqCount := "", 0
+	for word, freq := range freqMap {
+		if freq > mostFreqCount {
+			mostFreqWord = word
+			mostFreqCount = freq
+		}
+	}
+	return mostFreqWord
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0817.Linked-List-Components/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0820.Short-Encoding-of-Words/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0820.Short-Encoding-of-Words.md b/website/content/ChapterFour/0800~0899/0820.Short-Encoding-of-Words.md
new file mode 100644
index 000000000..fcacf0d03
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0820.Short-Encoding-of-Words.md
@@ -0,0 +1,150 @@
+# [820. Short Encoding of Words](https://leetcode.com/problems/short-encoding-of-words/)
+
+
+## 题目
+
+A **valid encoding** of an array of `words` is any reference string `s` and array of indices `indices` such that:
+
+- `words.length == indices.length`
+- The reference string `s` ends with the `'#'` character.
+- For each index `indices[i]`, the **substring** of `s` starting from `indices[i]` and up to (but not including) the next `'#'` character is equal to `words[i]`.
+
+Given an array of `words`, return *the **length of the shortest reference string*** `s` *possible of any **valid encoding** of* `words`*.*
+
+**Example 1:**
+
+```
+Input: words = ["time", "me", "bell"]
+Output: 10
+Explanation: A valid encoding would be s = "time#bell#" and indices = [0, 2, 5].
+words[0] = "time", the substring of s starting from indices[0] = 0 to the next '#' is underlined in "time#bell#"
+words[1] = "me", the substring of s starting from indices[1] = 2 to the next '#' is underlined in "time#bell#"
+words[2] = "bell", the substring of s starting from indices[2] = 5 to the next '#' is underlined in "time#bell#"
+```
+
+**Example 2:**
+
+```
+Input: words = ["t"]
+Output: 2
+Explanation: A valid encoding would be s = "t#" and indices = [0].
+```
+
+**Constraints:**
+
+- `1 <= words.length <= 2000`
+- `1 <= words[i].length <= 7`
+- `words[i]` consists of only lowercase letters.
+
+## 题目大意
+
+单词数组 words 的 有效编码 由任意助记字符串 s 和下标数组 indices 组成，且满足：
+
+- words.length == indices.length
+- 助记字符串 s 以 '#' 字符结尾
+- 对于每个下标 indices[i] ，s 的一个从 indices[i] 开始、到下一个 '#' 字符结束（但不包括 '#'）的 子字符串 恰好与 words[i] 相等
+
+给你一个单词数组 words ，返回成功对 words 进行编码的最小助记字符串 s 的长度 。
+
+## 解题思路
+
+- 暴力解法。先将所有的单词放入字典中。然后针对字典中的每个单词，逐一从字典中删掉自己的子字符串，这样有相同后缀的字符串被删除了，字典中剩下的都是没有共同前缀的。最终的答案是剩下所有单词用 # 号连接之后的总长度。
+- Trie 解法。构建 Trie 树，相同的后缀会被放到从根到叶子节点中的某个路径中。最后依次遍历一遍所有单词，如果单词最后一个字母是叶子节点，说明这个单词是要选择的，因为它可能是包含了一些单词后缀的最长单词。累加这个单词的长度并再加 1(# 字符的长度)。最终累加出来的长度即为题目所求的答案。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 暴力
+func minimumLengthEncoding(words []string) int {
+	res, m := 0, map[string]bool{}
+	for _, w := range words {
+		m[w] = true
+	}
+	for w := range m {
+		for i := 1; i < len(w); i++ {
+			delete(m, w[i:])
+		}
+	}
+	for w := range m {
+		res += len(w) + 1
+	}
+	return res
+}
+
+// 解法二 Trie
+type node struct {
+	value byte
+	sub   []*node
+}
+
+func (t *node) has(b byte) (*node, bool) {
+	if t == nil {
+		return nil, false
+	}
+	for i := range t.sub {
+		if t.sub[i] != nil && t.sub[i].value == b {
+			return t.sub[i], true
+		}
+	}
+	return nil, false
+}
+
+func (t *node) isLeaf() bool {
+	if t == nil {
+		return false
+	}
+	return len(t.sub) == 0
+}
+
+func (t *node) add(s []byte) {
+	now := t
+	for i := len(s) - 1; i > -1; i-- {
+		if v, ok := now.has(s[i]); ok {
+			now = v
+			continue
+		}
+		temp := new(node)
+		temp.value = s[i]
+		now.sub = append(now.sub, temp)
+		now = temp
+	}
+}
+
+func (t *node) endNodeOf(s []byte) *node {
+	now := t
+	for i := len(s) - 1; i > -1; i-- {
+		if v, ok := now.has(s[i]); ok {
+			now = v
+			continue
+		}
+		return nil
+	}
+	return now
+}
+
+func minimumLengthEncoding1(words []string) int {
+	res, tree, m := 0, new(node), make(map[string]bool)
+	for i := range words {
+		if !m[words[i]] {
+			tree.add([]byte(words[i]))
+			m[words[i]] = true
+		}
+	}
+	for s := range m {
+		if tree.endNodeOf([]byte(s)).isLeaf() {
+			res += len(s)
+			res++
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0819.Most-Common-Word/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0821.Shortest-Distance-to-a-Character/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0821.Shortest-Distance-to-a-Character.md b/website/content/ChapterFour/0800~0899/0821.Shortest-Distance-to-a-Character.md
new file mode 100644
index 000000000..1f88563da
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0821.Shortest-Distance-to-a-Character.md
@@ -0,0 +1,109 @@
+# [821. Shortest Distance to a Character](https://leetcode.com/problems/shortest-distance-to-a-character/)
+
+
+## 题目
+
+Given a string `s` and a character `c` that occurs in `s`, return *an array of integers `answer` where* `answer.length == s.length` *and* `answer[i]` *is the shortest distance from* `s[i]` *to the character* `c` *in* `s`.
+
+**Example 1:**
+
+```
+Input: s = "loveleetcode", c = "e"
+Output: [3,2,1,0,1,0,0,1,2,2,1,0]
+```
+
+**Example 2:**
+
+```
+Input: s = "aaab", c = "b"
+Output: [3,2,1,0]
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 104`
+- `s[i]` and `c` are lowercase English letters.
+- `c` occurs at least once in `s`.
+
+## 题目大意
+
+给定一个字符串 S 和一个字符 C。返回一个代表字符串 S 中每个字符到字符串 S 中的字符 C 的最短距离的数组。
+
+## 解题思路
+
+- 解法一：从左至右更新一遍到 C 的值距离，再从右至左更新一遍到 C 的值，取两者中的最小值。
+- 解法二：依次扫描字符串 S，针对每一个非字符 C 的字符，分别往左扫一次，往右扫一次，计算出距离目标字符 C 的距离，然后取左右两个距离的最小值存入最终答案数组中。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+)
+
+// 解法一
+func shortestToChar(s string, c byte) []int {
+	n := len(s)
+	res := make([]int, n)
+	for i := range res {
+		res[i] = n
+	}
+	for i := 0; i < n; i++ {
+		if s[i] == c {
+			res[i] = 0
+		} else if i > 0 {
+			res[i] = res[i-1] + 1
+		}
+	}
+	for i := n - 1; i >= 0; i-- {
+		if i < n-1 && res[i+1]+1 < res[i] {
+			res[i] = res[i+1] + 1
+		}
+	}
+	return res
+}
+
+// 解法二
+func shortestToChar1(s string, c byte) []int {
+	res := make([]int, len(s))
+	for i := 0; i < len(s); i++ {
+		if s[i] == c {
+			res[i] = 0
+		} else {
+			left, right := math.MaxInt32, math.MaxInt32
+			for j := i + 1; j < len(s); j++ {
+				if s[j] == c {
+					right = j - i
+					break
+				}
+			}
+			for k := i - 1; k >= 0; k-- {
+				if s[k] == c {
+					left = i - k
+					break
+				}
+			}
+			res[i] = min(left, right)
+		}
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0820.Short-Encoding-of-Words/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0823.Binary-Trees-With-Factors/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0823.Binary-Trees-With-Factors.md b/website/content/ChapterFour/0800~0899/0823.Binary-Trees-With-Factors.md
new file mode 100644
index 000000000..d58e146da
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0823.Binary-Trees-With-Factors.md
@@ -0,0 +1,128 @@
+# [823. Binary Trees With Factors](https://leetcode.com/problems/binary-trees-with-factors/)
+
+
+## 题目
+
+Given an array of unique integers, `arr`, where each integer `arr[i]` is strictly greater than `1`.
+
+We make a binary tree using these integers, and each number may be used for any number of times. Each non-leaf node's value should be equal to the product of the values of its children.
+
+Return *the number of binary trees we can make*. The answer may be too large so return the answer **modulo** `109 + 7`.
+
+**Example 1:**
+
+```
+Input: arr = [2,4]
+Output: 3
+Explanation: We can make these trees: [2], [4], [4, 2, 2]
+```
+
+**Example 2:**
+
+```
+Input: arr = [2,4,5,10]
+Output: 7
+Explanation: We can make these trees: [2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2].
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 1000`
+- `2 <= arr[i] <= 10^9`
+
+## 题目大意
+
+给出一个含有不重复整数元素的数组，每个整数均大于 1。我们用这些整数来构建二叉树，每个整数可以使用任意次数。其中：每个非叶结点的值应等于它的两个子结点的值的乘积。满足条件的二叉树一共有多少个？返回的结果应模除 10 * 9 + 7。
+
+## 解题思路
+
+- 首先想到的是暴力解法，先排序，然后遍历所有节点，枚举两两乘积为第三个节点值的组合。然后枚举这些组合并构成树。这里计数的时候要注意，左右孩子如果不是对称的，左右子树相互对调又是一组解。但是这个方法超时了。原因是，暴力枚举了很多次重复的节点和组合。优化这里的方法就是把已经计算过的节点放入 `map` 中。这里有 2 层 `map`，第一层 `map` 记忆化的是两两乘积的组合，将父亲节点作为 `key`，左右 2 个孩子作为 `value`。第二层 `map` 记忆化的是以 `root` 为根节点此时二叉树的种类数，`key` 是 `root`，`value` 存的是种类数。这样优化以后，DFS 暴力解法可以 runtime beats 100%。
+- 另外一种解法是 DP。定义 `dp[i]` 代表以 `i` 为根节点的树的种类数。dp[i] 初始都是 1，因为所有节点自身可以形成为自身单个节点为 `root` 的树。同样需要先排序。状态转移方程是：
+
+	{{< katex display >}}
+    dp[i] = \sum_{j<i, k<i}^{}dp[j] * dp[k], j * k = i
+	{{< /katex >}}
+
+    最后将 `dp[]` 数组中所有结果累加取模即为最终结果，时间复杂度 O(n^2)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+const mod = 1e9 + 7
+
+// 解法一 DFS
+func numFactoredBinaryTrees(arr []int) int {
+	sort.Ints(arr)
+	numDict := map[int]bool{}
+	for _, num := range arr {
+		numDict[num] = true
+	}
+	dict, res := make(map[int][][2]int), 0
+	for i, num := range arr {
+		for j := i; j < len(arr) && num*arr[j] <= arr[len(arr)-1]; j++ {
+			tmp := num * arr[j]
+			if !numDict[tmp] {
+				continue
+			}
+			dict[tmp] = append(dict[tmp], [2]int{num, arr[j]})
+		}
+	}
+	cache := make(map[int]int)
+	for _, num := range arr {
+		res = (res + dfs(num, dict, cache)) % mod
+	}
+	return res
+}
+
+func dfs(num int, dict map[int][][2]int, cache map[int]int) int {
+	if val, ok := cache[num]; ok {
+		return val
+	}
+	res := 1
+	for _, tuple := range dict[num] {
+		a, b := tuple[0], tuple[1]
+		x, y := dfs(a, dict, cache), dfs(b, dict, cache)
+		tmp := x * y
+		if a != b {
+			tmp *= 2
+		}
+		res = (res + tmp) % mod
+	}
+	cache[num] = res
+	return res
+}
+
+// 解法二 DP
+func numFactoredBinaryTrees1(arr []int) int {
+	dp := make(map[int]int)
+	sort.Ints(arr)
+	for i, curNum := range arr {
+		for j := 0; j < i; j++ {
+			factor := arr[j]
+			quotient, remainder := curNum/factor, curNum%factor
+			if remainder == 0 {
+				dp[curNum] += dp[factor] * dp[quotient]
+			}
+		}
+		dp[curNum]++
+	}
+	totalCount := 0
+	for _, count := range dp {
+		totalCount += count
+	}
+	return totalCount % mod
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0821.Shortest-Distance-to-a-Character/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0825.Friends-Of-Appropriate-Ages/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0825.Friends-Of-Appropriate-Ages.md b/website/content/ChapterFour/0800~0899/0825.Friends-Of-Appropriate-Ages.md
new file mode 100644
index 000000000..4f278efa1
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0825.Friends-Of-Appropriate-Ages.md
@@ -0,0 +1,148 @@
+# [825. Friends Of Appropriate Ages](https://leetcode.com/problems/friends-of-appropriate-ages/)
+
+
+## 题目
+
+There are `n` persons on a social media website. You are given an integer array `ages` where `ages[i]` is the age of the `ith` person.
+
+A Person `x` will not send a friend request to a person `y` (`x != y`) if any of the following conditions is true:
+
+- `age[y] <= 0.5 * age+ 7`
+- `age[y] > age[x]`
+- `age[y] > 100 && age< 100`
+
+Otherwise, `x` will send a friend request to `y`.
+
+Note that if `x` sends a request to `y`, `y` will not necessarily send a request to `x`. Also, a person will not send a friend request to themself.
+
+Return *the total number of friend requests made*.
+
+**Example 1:**
+
+```
+Input: ages = [16,16]
+Output: 2
+Explanation: 2 people friend request each other.
+
+```
+
+**Example 2:**
+
+```
+Input: ages = [16,17,18]
+Output: 2
+Explanation: Friend requests are made 17 -> 16, 18 -> 17.
+
+```
+
+**Example 3:**
+
+```
+Input: ages = [20,30,100,110,120]
+Output: 3
+Explanation: Friend requests are made 110 -> 100, 120 -> 110, 120 -> 100.
+
+```
+
+**Constraints:**
+
+- `n == ages.length`
+- `1 <= n <= 2 * 10^4`
+- `1 <= ages[i] <= 120`
+
+## 题目大意
+
+在社交媒体网站上有 n 个用户。给你一个整数数组 ages ，其中 ages[i] 是第 i 个用户的年龄。
+
+如果下述任意一个条件为真，那么用户 x 将不会向用户 y（x != y）发送好友请求：
+
+- ages[y] <= 0.5 * ages[x] + 7
+- ages[y] > ages[x]
+- ages[y] > 100 && ages[x] < 100
+
+否则，x 将会向 y 发送一条好友请求。注意，如果 x 向 y 发送一条好友请求，y 不必也向 x 发送一条好友请求。另外，用户不会向自己发送好友请求。返回在该社交媒体网站上产生的好友请求总数。
+
+## 解题思路
+
+- 解法三，暴力解法。先统计 [1,120] 范围内每个年龄的人数。然后利用题目中的三个判断条件，筛选符合条件的用户对。需要注意的是，相同年龄的人可以相互发送好友请求。不同年龄的人发送好友请求是单向的，即年龄老的向年龄轻的发送好友请求，年龄轻的不会对年龄老的发送好友请求。
+- 解法二，排序 + 双指针。题目给定的 3 个条件其实是 2 个。条件 3 包含在条件 2 中。条件 1 和条件 2 组合起来是 `0.5 × ages[x]+7 < ages[y] ≤ ages[x]`。当 ages[x] 小于 15 时，这个等式无解。考虑到年龄是单调递增的，`(0.5 × ages[x]+7,ages[x]]` 这个区间左右边界也是单调递增的。于是可以用双指针维护两个边界。在区间 [left, right] 内，这些下标对应的的 y 值都满足条件。当 `ages[left] > 0.5 × ages[x]+7` 时，左指针停止右移。当 `ages[right+1] > ages[x]` 时， 右指针停止右移。在 `[left, right]` 区间内，满足条件的 y 有 `right-left+1` 个，即使得 `ages[y]` 取值在 `(0.5 × ages[x]+7,ages[x]]` 之间。依照题意，`x≠y`，即该区间右边界取不到。y 的取值个数需要再减一，减去的是取到和 x 相同的值的下标。那么每个区间能取 `right-left` 个值。累加所有满足条件的值即为好友请求总数。
+- 解法一。在解法二中，计算满足不等式 y 下标所在区间的时候，区间和区间存在重叠的情况，这些重叠情况导致了重复计算。所以这里可以优化。可以用 prefix sum 前缀和数组优化。代码见下方。
+
+## 代码
+
+```go
+package leetcocde
+
+import "sort"
+
+// 解法一 前缀和，时间复杂度 O(n)
+func numFriendRequests(ages []int) int {
+	count, prefixSum, res := make([]int, 121), make([]int, 121), 0
+	for _, age := range ages {
+		count[age]++
+	}
+	for i := 1; i < 121; i++ {
+		prefixSum[i] = prefixSum[i-1] + count[i]
+	}
+	for i := 15; i < 121; i++ {
+		if count[i] > 0 {
+			bound := i/2 + 8
+			res += count[i] * (prefixSum[i] - prefixSum[bound-1] - 1)
+		}
+	}
+	return res
+}
+
+// 解法二 双指针 + 排序，时间复杂度 O(n logn)
+func numFriendRequests1(ages []int) int {
+	sort.Ints(ages)
+	left, right, res := 0, 0, 0
+	for _, age := range ages {
+		if age < 15 {
+			continue
+		}
+		for ages[left]*2 <= age+14 {
+			left++
+		}
+		for right+1 < len(ages) && ages[right+1] <= age {
+			right++
+		}
+		res += right - left
+	}
+	return res
+}
+
+// 解法三 暴力解法 O(n^2)
+func numFriendRequests2(ages []int) int {
+	res, count := 0, [125]int{}
+	for _, x := range ages {
+		count[x]++
+	}
+	for i := 1; i <= 120; i++ {
+		for j := 1; j <= 120; j++ {
+			if j > i {
+				continue
+			}
+			if (j-7)*2 <= i {
+				continue
+			}
+			if j > 100 && i < 100 {
+				continue
+			}
+			if i != j {
+				res += count[i] * count[j]
+			} else {
+				res += count[i] * (count[j] - 1)
+			}
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0823.Binary-Trees-With-Factors/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0826.Most-Profit-Assigning-Work/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0826.Most-Profit-Assigning-Work.md b/website/content/ChapterFour/0800~0899/0826.Most-Profit-Assigning-Work.md
new file mode 100644
index 000000000..fa7464dbd
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0826.Most-Profit-Assigning-Work.md
@@ -0,0 +1,114 @@
+# [826. Most Profit Assigning Work](https://leetcode.com/problems/most-profit-assigning-work/)
+
+## 题目
+
+We have jobs: difficulty[i] is the difficulty of the ith job, and profit[i] is the profit of the ith job. 
+
+Now we have some workers. worker[i] is the ability of the ith worker, which means that this worker can only complete a job with difficulty at most worker[i]. 
+
+Every worker can be assigned at most one job, but one job can be completed multiple times.
+
+For example, if 3 people attempt the same job that pays $1, then the total profit will be $3.  If a worker cannot complete any job, his profit is $0.
+
+What is the most profit we can make?
+
+
+**Example 1**:
+
+
+```
+
+Input: difficulty = [2,4,6,8,10], profit = [10,20,30,40,50], worker = [4,5,6,7]
+Output: 100 
+Explanation: Workers are assigned jobs of difficulty [4,4,6,6] and they get profit of [20,20,30,30] seperately.
+
+```
+
+**Note**:
+
+- 1 <= difficulty.length = profit.length <= 10000
+- 1 <= worker.length <= 10000
+- difficulty[i], profit[i], worker[i]  are in range [1, 10^5]
+
+
+## 题目大意
+
+这道题考察的是滑动窗口的问题，也是排序相关的问题。
+
+给出一组任务，每个任务都有一定的难度，每个任务也都有完成以后对应的收益(完成难的任务不一定收益最高)。有一批工人，每个人能处理的任务难度不同。要求输出这批工人完成任务以后的最大收益。
+
+## 解题思路
+
+先将任务按照难度排序，工人也按照能处理任务难度的能力排序。用一个数组记录下，每个 i 下标，当前能达到的最大收益。计算这个收益只需要从下标为 1 开始，依次比较自己和前一个的收益即可(因为排过序，难度是依次递增的)。有了这个难度依次递增，并且记录了最大收益的数组以后，就可以计算最终结果了。遍历一遍工人数组，如果工人的能力大于任务的难度，就加上这个最大收益。遍历完工人数组，最终结果就是最大收益。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"fmt"
+	"sort"
+)
+
+// Task define
+type Task struct {
+	Difficulty int
+	Profit     int
+}
+
+// Tasks define
+type Tasks []Task
+
+// Len define
+func (p Tasks) Len() int { return len(p) }
+
+// Swap define
+func (p Tasks) Swap(i, j int) { p[i], p[j] = p[j], p[i] }
+
+// SortByDiff define
+type SortByDiff struct{ Tasks }
+
+// Less define
+func (p SortByDiff) Less(i, j int) bool {
+	return p.Tasks[i].Difficulty < p.Tasks[j].Difficulty
+}
+
+func maxProfitAssignment(difficulty []int, profit []int, worker []int) int {
+	if len(difficulty) == 0 || len(profit) == 0 || len(worker) == 0 {
+		return 0
+	}
+	tasks, res, index := []Task{}, 0, 0
+	for i := 0; i < len(difficulty); i++ {
+		tasks = append(tasks, Task{Difficulty: difficulty[i], Profit: profit[i]})
+	}
+	sort.Sort(SortByDiff{tasks})
+	sort.Ints(worker)
+	for i := 1; i < len(tasks); i++ {
+		tasks[i].Profit = max(tasks[i].Profit, tasks[i-1].Profit)
+	}
+	fmt.Printf("tasks = %v worker = %v\n", tasks, worker)
+	for _, w := range worker {
+		for index < len(difficulty) && w >= tasks[index].Difficulty {
+			index++
+		}
+		fmt.Printf("tasks【index】 = %v\n", tasks[index])
+		if index > 0 {
+			res += tasks[index-1].Profit
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0825.Friends-Of-Appropriate-Ages/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String.md b/website/content/ChapterFour/0800~0899/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String.md
new file mode 100644
index 000000000..6f5f28d42
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String.md
@@ -0,0 +1,108 @@
+# [828. Count Unique Characters of All Substrings of a Given String](https://leetcode.com/problems/count-unique-characters-of-all-substrings-of-a-given-string/)
+
+
+## 题目
+
+Let's define a function `countUniqueChars(s)` that returns the number of unique characters on `s`, for example if `s = "LEETCODE"` then `"L"`, `"T"`,`"C"`,`"O"`,`"D"` are the unique characters since they appear only once in `s`, therefore `countUniqueChars(s) = 5`.On this problem given a string `s` we need to return the sum of `countUniqueChars(t)` where `t` is a substring of `s`. Notice that some substrings can be repeated so on this case you have to count the repeated ones too.
+
+Since the answer can be very large, return the answer modulo `10 ^ 9 + 7`.
+
+**Example 1**:
+
+```
+Input: s = "ABC"
+Output: 10
+Explanation: All possible substrings are: "A","B","C","AB","BC" and "ABC".
+Evey substring is composed with only unique letters.
+Sum of lengths of all substring is 1 + 1 + 1 + 2 + 2 + 3 = 10
+
+```
+
+**Example 2**:
+
+```
+Input: s = "ABA"
+Output: 8
+Explanation: The same as example 1, except countUniqueChars("ABA") = 1.
+
+```
+
+**Example 3**:
+
+```
+Input: s = "LEETCODE"
+Output: 92
+
+```
+
+**Constraints**:
+
+- `0 <= s.length <= 10^4`
+- `s` contain upper-case English letters only.
+
+## 题目大意
+
+如果一个字符在字符串 S 中有且仅有出现一次，那么我们称其为独特字符。例如，在字符串 S = "LETTER" 中，"L" 和 "R" 可以被称为独特字符。我们再定义 UNIQ(S) 作为字符串 S 中独特字符的个数。那么，在 S = "LETTER" 中， UNIQ("LETTER") =  2。
+
+对于给定字符串 S，计算其所有非空子串的独特字符的个数（即 UNIQ(substring)）之和。如果在 S 的不同位置上出现两个甚至多个相同的子串，那么我们认为这些子串是不同的。考虑到答案可能会非常大，规定返回格式为：结果 mod 10 ^ 9 + 7。
+
+## 解题思路
+
+- 这一题可以先用暴力解法尝试解题，不过提交以后会发现判题结果是超时。出错的一组数据是一个有 10000 个字符的字符串。暴力解法中间由于遍历了太多的子区间，导致了超时。
+- 这道题换一个角度思考问题。当子字符串中字符 X 出现了 2 次以上，那么它就对最终结果没有任何影响，所以只有当某个字符只出现一次的时候才会影响最终结果。再者，一个子字符串中不重复的字符的总个数，也就是这个子字符串 UNIQ 值。例如，“ABC”，这个子字符串的 UNIQ 值是 3，可以这样计算，它属于 A 的独特的字符串，也属于 B 的独特的字符串，也属于 C 的独特的字符串，那么计算这个子字符串的问题可以分解成计算 A 有多少个独特的子字符串，B 有多少个独特的子字符串，C 有多少个独特的子字符串的问题。在计算 A 有多少个子字符串的问题的时候，里面肯定会包含 "ABC" 这个子字符串的。所以原问题就转换成了分别计算给出的字符串中每个字符出现在独特字符串中的总数之和。
+- 假设原字符串是 BAABBABBBAAABA，这个字符串中出现了很多 A 和很多 B，假设我们当前计算到了第 3 个 A 的位置了(index = 5)，即标红色的那个 A。如何计算这个 A 在哪些子字符串中是独特的呢？由于子字符串题目中要求必须是连续的区间，所以这个问题很简单。找到这个 A 前一个 A 的下标位置(index = 2)，再找到这个 A 后一个 A 的下标位置(index = 9)，即 BAABBABBBAAABA，第一个 A 和当前计算的 A 中间区间有 2 个字符，第三个 A 和当前计算的 A 中间有 3 个字符。那么当前计算的 A 出现在 `(2 + 1) * (3 + 1) = 12` 个子字符串中是独特的，这 12 个字符串是：`A`，`BA`，`BBA`，`AB`，`ABB`，`ABBB`，`BAB`，`BABB`，`BABBB`，`BBAB`，`BBABB`，`BBABBB`。计算方法，假设当前待计算的字符的下标是 i ，找到当前字符前一次出现的下标位置 left，再找到当前字符后一次出现的下标位置 right，那么左边区间 (left,i) 的***开区间****内包含的字符数是 i - left - 1，右边区间 (i,right) 的***开区间****内包含的字符数是 right - i - 1。左右两边都还需要考虑空字符串的情况，即左右两边都可以不取任何字符，那么对应的就是只有中间这个待计算的字符 `A`。所以左右两边都还需要再加上空串的情况，左边 i - left - 1 + 1 = i - left，右边 right - i - 1 + 1 = right - i。左右两边的情况进行排列组合，即 (i - left) * (right - i)。针对字符串的每个字符都计算这样的值，最后累积的总和就是题目中要求的总 UNIQ 值。
+
+## 代码
+
+```go
+package leetcode
+
+func uniqueLetterString(S string) int {
+	res, left, right := 0, 0, 0
+	for i := 0; i < len(S); i++ {
+		left = i - 1
+		for left >= 0 && S[left] != S[i] {
+			left--
+		}
+		right = i + 1
+		for right < len(S) && S[right] != S[i] {
+			right++
+		}
+		res += (i - left) * (right - i)
+	}
+	return res % 1000000007
+}
+
+// 暴力解法，超时！时间复杂度 O(n^2)
+func uniqueLetterString1(S string) int {
+	if len(S) == 0 {
+		return 0
+	}
+	res, mod := 0, 1000000007
+	for i := 0; i < len(S); i++ {
+		letterMap := map[byte]int{}
+		for j := i; j < len(S); j++ {
+			letterMap[S[j]]++
+			tmp := 0
+			for _, v := range letterMap {
+				if v > 1 {
+					tmp++
+				}
+			}
+			if tmp == len(letterMap) {
+				continue
+			} else {
+				res += len(letterMap) - tmp
+			}
+		}
+	}
+	return res % mod
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0826.Most-Profit-Assigning-Work/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0830.Positions-of-Large-Groups/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0830.Positions-of-Large-Groups.md b/website/content/ChapterFour/0800~0899/0830.Positions-of-Large-Groups.md
new file mode 100644
index 000000000..5efa20eda
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0830.Positions-of-Large-Groups.md
@@ -0,0 +1,87 @@
+# [830. Positions of Large Groups](https://leetcode.com/problems/positions-of-large-groups/)
+
+
+## 题目
+
+In a string `s` of lowercase letters, these letters form consecutive groups of the same character.
+
+For example, a string like `s = "abbxxxxzyy"` has the groups `"a"`, `"bb"`, `"xxxx"`, `"z"`, and `"yy"`.
+
+A group is identified by an interval `[start, end]`, where `start` and `end` denote the start and end indices (inclusive) of the group. In the above example, `"xxxx"` has the interval `[3,6]`.
+
+A group is considered **large** if it has 3 or more characters.
+
+Return *the intervals of every **large** group sorted in **increasing order by start index***.
+
+**Example 1**:
+
+```
+Input: s = "abbxxxxzzy"
+Output: [[3,6]]
+Explanation: "xxxx" is the only large group with start index 3 and end index 6.
+```
+
+**Example 2**:
+
+```
+Input: s = "abc"
+Output: []
+Explanation: We have groups "a", "b", and "c", none of which are large groups.
+```
+
+**Example 3**:
+
+```
+Input: s = "abcdddeeeeaabbbcd"
+Output: [[3,5],[6,9],[12,14]]
+Explanation: The large groups are "ddd", "eeee", and "bbb".
+```
+
+**Example 4**:
+
+```
+Input: s = "aba"
+Output: []
+```
+
+**Constraints**:
+
+- `1 <= s.length <= 1000`
+- `s` contains lower-case English letters only.
+
+## 题目大意
+
+在一个由小写字母构成的字符串 s 中，包含由一些连续的相同字符所构成的分组。例如，在字符串 s = "abbxxxxzyy" 中，就含有 "a", "bb", "xxxx", "z" 和 "yy" 这样的一些分组。分组可以用区间 [start, end] 表示，其中 start 和 end 分别表示该分组的起始和终止位置的下标。上例中的 "xxxx" 分组用区间表示为 [3,6] 。我们称所有包含大于或等于三个连续字符的分组为 较大分组 。
+
+找到每一个 较大分组 的区间，按起始位置下标递增顺序排序后，返回结果。
+
+## 解题思路
+
+- 简单题。利用滑动窗口的思想，先扩大窗口的右边界，找到能相同字母且能到达的最右边。记录左右边界。再将窗口的左边界移动到上一次的右边界处。以此类推，重复扩大窗口的右边界，直至扫完整个字符串。最终所有满足题意的较大分组区间都在数组中了。
+
+## 代码
+
+```go
+package leetcode
+
+func largeGroupPositions(S string) [][]int {
+	res, end := [][]int{}, 0
+	for end < len(S) {
+		start, str := end, S[end]
+		for end < len(S) && S[end] == str {
+			end++
+		}
+		if end-start >= 3 {
+			res = append(res, []int{start, end - 1})
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0832.Flipping-an-Image/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0832.Flipping-an-Image.md b/website/content/ChapterFour/0800~0899/0832.Flipping-an-Image.md
new file mode 100644
index 000000000..3c3bc5aa9
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0832.Flipping-an-Image.md
@@ -0,0 +1,70 @@
+# [832. Flipping an Image](https://leetcode.com/problems/flipping-an-image/)
+
+
+## 题目
+
+Given a binary matrix `A`, we want to flip the image horizontally, then invert it, and return the resulting image.
+
+To flip an image horizontally means that each row of the image is reversed. For example, flipping `[1, 1, 0]` horizontally results in `[0, 1, 1]`.
+
+To invert an image means that each `0` is replaced by `1`, and each `1` is replaced by `0`. For example, inverting `[0, 1, 1]` results in `[1, 0, 0]`.
+
+**Example 1**:
+
+```
+Input: [[1,1,0],[1,0,1],[0,0,0]]
+Output: [[1,0,0],[0,1,0],[1,1,1]]
+Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]].
+Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]]
+```
+
+**Example 2**:
+
+```
+Input: [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]
+Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
+Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]].
+Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
+```
+
+**Notes**:
+
+- `1 <= A.length = A[0].length <= 20`
+- `0 <= A[i][j] <= 1`
+
+## 题目大意
+
+给定一个二进制矩阵 A，我们想先水平翻转图像，然后反转图像并返回结果。水平翻转图片就是将图片的每一行都进行翻转，即逆序。例如，水平翻转 [1, 1, 0] 的结果是 [0, 1, 1]。反转图片的意思是图片中的 0 全部被 1 替换， 1 全部被 0 替换。例如，反转 [0, 1, 1] 的结果是 [1, 0, 0]。
+
+
+## 解题思路
+
+- 给定一个二进制矩阵，要求先水平翻转，然后再反转( 1→0 , 0→1 )。
+- 简单题，按照题意先水平翻转，再反转即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func flipAndInvertImage(A [][]int) [][]int {
+	for i := 0; i < len(A); i++ {
+		for a, b := 0, len(A[i])-1; a < b; a, b = a+1, b-1 {
+			A[i][a], A[i][b] = A[i][b], A[i][a]
+		}
+		for a := 0; a < len(A[i]); a++ {
+			A[i][a] = (A[i][a] + 1) % 2
+		}
+	}
+	return A
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0830.Positions-of-Large-Groups/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0834.Sum-of-Distances-in-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0834.Sum-of-Distances-in-Tree.md b/website/content/ChapterFour/0800~0899/0834.Sum-of-Distances-in-Tree.md
new file mode 100755
index 000000000..f4c5a0497
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0834.Sum-of-Distances-in-Tree.md
@@ -0,0 +1,108 @@
+# [834. Sum of Distances in Tree](https://leetcode.com/problems/sum-of-distances-in-tree/)
+
+
+## 题目
+
+An undirected, connected tree with `N` nodes labelled `0...N-1` and `N-1edges` are given.
+
+The `i`th edge connects nodes `edges[i][0]` and `edges[i][1]` together.
+
+Return a list `ans`, where `ans[i]` is the sum of the distances between node `i`and all other nodes.
+
+**Example 1**:
+
+    Input: N = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
+    Output: [8,12,6,10,10,10]
+    Explanation: 
+    Here is a diagram of the given tree:
+      0
+     / \
+    1   2
+       /|\
+      3 4 5
+    We can see that dist(0,1) + dist(0,2) + dist(0,3) + dist(0,4) + dist(0,5)
+    equals 1 + 1 + 2 + 2 + 2 = 8.  Hence, answer[0] = 8, and so on.
+
+**Note**: `1 <= N <= 10000`
+
+## 题目大意
+
+给定一个无向、连通的树。树中有 N 个标记为 0...N-1 的节点以及 N-1 条边。第 i 条边连接节点 edges[i][0] 和 edges[i][1] 。返回一个表示节点 i 与其他所有节点距离之和的列表 ans。
+
+说明: 1 <= N <= 10000
+
+
+
+## 解题思路
+
+- 给出 N 个节点和这些节点之间的一些边的关系。要求求出分别以 x 为根节点到所有节点路径和。
+- 这一题虽说描述的是求树的路径，但是完全可以当做图来做，因为并不是二叉树，是多叉树。这一题的解题思路是先一次 DFS 求出以 0 为根节点到各个节点的路径和(不以 0 为节点也可以，可以取任意节点作为开始)。第二次 DFS 求出从 0 根节点转换到其他各个节点的路径和。由于第一次计算出来以 0 为节点的路径和是正确的，所以计算其他节点为根节点的路径和只需要转换一下就可以得到正确结果。经过 2 次 DFS 之后就可以得到所有节点以自己为根节点到所有节点的路径和了。
+- 如何从以 0 为根节点到其他所有节点的路径和转换到以其他节点为根节点到所有节点的路径和呢？从 0 节点换成 x 节点，只需要在 0 到所有节点的路径和基础上增增减减就可以了。增加的是 x 节点到除去以 x 为根节点所有子树以外的节点的路径，有多少个节点就增加多少条路径。减少的是 0 到以 x 为根节点所有子树节点的路径和，包含 0 到 x 根节点，有多少节点就减少多少条路径。所以在第一次 DFS 中需要计算好每个节点以自己为根节点的子树总数和(包含自己在内)，这样在第二次 DFS 中可以直接拿来做转换。具体细节的实现见代码。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func sumOfDistancesInTree(N int, edges [][]int) []int {
+	// count[i] 中存储的是以 i 为根节点，所有子树结点和根节点的总数
+	tree, visited, count, res := make([][]int, N), make([]bool, N), make([]int, N), make([]int, N)
+	for _, e := range edges {
+		i, j := e[0], e[1]
+		tree[i] = append(tree[i], j)
+		tree[j] = append(tree[j], i)
+	}
+	deepFirstSearch(0, visited, count, res, tree)
+	// 重置访问状态，再进行一次 DFS
+	visited = make([]bool, N)
+	// 进入第二次 DFS 之前，只有 res[0] 里面存的是正确的值，因为第一次 DFS 计算出了以 0 为根节点的所有路径和
+	// 第二次 DFS 的目的是把以 0 为根节点的路径和转换成以 n 为根节点的路径和
+	deepSecondSearch(0, visited, count, res, tree)
+
+	return res
+}
+
+func deepFirstSearch(root int, visited []bool, count, res []int, tree [][]int) {
+	visited[root] = true
+	for _, n := range tree[root] {
+		if visited[n] {
+			continue
+		}
+		deepFirstSearch(n, visited, count, res, tree)
+		count[root] += count[n]
+		// root 节点到 n 的所有路径和 = 以 n 为根节点到所有子树的路径和 res[n] + root 到 count[n] 中每个节点的个数(root 节点和以 n 为根节点的每个节点都增加一条路径)
+		// root 节点和以 n 为根节点的每个节点都增加一条路径 = 以 n 为根节点，子树节点数和根节点数的总和，即 count[n]
+		res[root] += res[n] + count[n]
+	}
+	count[root]++
+}
+
+// 从 root 开始，把 root 节点的子节点，依次设置成新的根节点
+func deepSecondSearch(root int, visited []bool, count, res []int, tree [][]int) {
+	N := len(visited)
+	visited[root] = true
+	for _, n := range tree[root] {
+		if visited[n] {
+			continue
+		}
+		// 根节点从 root 变成 n 后
+		// res[root] 存储的是以 root 为根节点到所有节点的路径总长度
+		// 1. root 到 n 节点增加的路径长度 = root 节点和以 n 为根节点的每个节点都增加一条路径 = 以 n 为根节点，子树节点数和根节点数的总和，即 count[n]
+		// 2. n 到以 n 为根节点的所有子树节点以外的节点增加的路径长度 = n 节点和非 n 为根节点子树的每个节点都增加一条路径 = N - count[n]
+		// 所以把根节点从 root 转移到 n，需要增加的路径是上面👆第二步计算的，需要减少的路径是上面👆第一步计算的
+		res[n] = res[root] + (N - count[n]) - count[n]
+		deepSecondSearch(n, visited, count, res, tree)
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0832.Flipping-an-Image/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0836.Rectangle-Overlap/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0836.Rectangle-Overlap.md b/website/content/ChapterFour/0800~0899/0836.Rectangle-Overlap.md
new file mode 100755
index 000000000..0475596b1
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0836.Rectangle-Overlap.md
@@ -0,0 +1,61 @@
+# [836. Rectangle Overlap](https://leetcode.com/problems/rectangle-overlap/)
+
+
+## 题目
+
+A rectangle is represented as a list `[x1, y1, x2, y2]`, where `(x1, y1)` are the coordinates of its bottom-left corner, and `(x2, y2)` are the coordinates of its top-right corner.
+
+Two rectangles overlap if the area of their intersection is positive. To be clear, two rectangles that only touch at the corner or edges do not overlap.
+
+Given two (axis-aligned) rectangles, return whether they overlap.
+
+**Example 1**:
+
+    Input: rec1 = [0,0,2,2], rec2 = [1,1,3,3]
+    Output: true
+
+**Example 2**:
+
+    Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]
+    Output: false
+
+**Notes**:
+
+1. Both rectangles `rec1` and `rec2` are lists of 4 integers.
+2. All coordinates in rectangles will be between `-10^9` and `10^9`.
+
+
+## 题目大意
+
+矩形以列表 [x1, y1, x2, y2] 的形式表示，其中 (x1, y1) 为左下角的坐标，(x2, y2) 是右上角的坐标。如果相交的面积为正，则称两矩形重叠。需要明确的是，只在角或边接触的两个矩形不构成重叠。给出两个矩形，判断它们是否重叠并返回结果。
+
+说明：
+
+1. 两个矩形 rec1 和 rec2 都以含有四个整数的列表的形式给出。
+2. 矩形中的所有坐标都处于 -10^9 和 10^9 之间。
+
+
+## 解题思路
+
+- 给出两个矩形的坐标，判断两个矩形是否重叠。
+- 几何题，按照几何方法判断即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func isRectangleOverlap(rec1 []int, rec2 []int) bool {
+	return rec1[0] < rec2[2] && rec2[0] < rec1[2] && rec1[1] < rec2[3] && rec2[1] < rec1[3]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0834.Sum-of-Distances-in-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0838.Push-Dominoes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0838.Push-Dominoes.md b/website/content/ChapterFour/0800~0899/0838.Push-Dominoes.md
new file mode 100644
index 000000000..773914eff
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0838.Push-Dominoes.md
@@ -0,0 +1,151 @@
+# [838. Push Dominoes](https://leetcode.com/problems/push-dominoes/)
+
+## 题目
+
+There are N dominoes in a line, and we place each domino vertically upright.
+
+In the beginning, we simultaneously push some of the dominoes either to the left or to the right.
+
+![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/05/18/domino.png)
+
+
+After each second, each domino that is falling to the left pushes the adjacent domino on the left.
+
+Similarly, the dominoes falling to the right push their adjacent dominoes standing on the right.
+
+When a vertical domino has dominoes falling on it from both sides, it stays still due to the balance of the forces.
+
+For the purposes of this question, we will consider that a falling domino expends no additional force to a falling or already fallen domino.
+
+Given a string "S" representing the initial state. S[i] = 'L', if the i-th domino has been pushed to the left; S[i] = 'R', if the i-th domino has been pushed to the right; S[i] = '.', if the i-th domino has not been pushed.
+
+Return a string representing the final state. 
+
+
+**Example 1**:
+
+```
+
+Input: ".L.R...LR..L.."
+Output: "LL.RR.LLRRLL.."
+
+```
+
+**Example 2**:
+
+```
+
+Input: "RR.L"
+Output: "RR.L"
+Explanation: The first domino expends no additional force on the second domino.
+
+```
+
+
+**Note**:
+
+- 0 <= N <= 10^5
+- String dominoes contains only 'L', 'R' and '.'
+
+
+## 题目大意
+
+这道题是一个道模拟题，考察的也是滑动窗口的问题。
+
+给出一个字符串，L 代表这个多米诺骨牌会往左边倒，R 代表这个多米诺骨牌会往右边倒，问最终这些牌倒下去以后，情况是如何的，输出最终情况的字符串。
+
+## 解题思路
+
+这道题可以预先在初始字符串头和尾都添加一个字符串，左边添加 L，右边添加 R，辅助判断。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func pushDominoes(dominoes string) string {
+	d := []byte(dominoes)
+	for i := 0; i < len(d); {
+		j := i + 1
+		for j < len(d)-1 && d[j] == '.' {
+			j++
+		}
+		push(d[i : j+1])
+		i = j
+	}
+	return string(d)
+}
+
+func push(d []byte) {
+	first, last := 0, len(d)-1
+	switch d[first] {
+	case '.', 'L':
+		if d[last] == 'L' {
+			for ; first < last; first++ {
+				d[first] = 'L'
+			}
+		}
+	case 'R':
+		if d[last] == '.' || d[last] == 'R' {
+			for ; first <= last; first++ {
+				d[first] = 'R'
+			}
+		} else if d[last] == 'L' {
+			for first < last {
+				d[first] = 'R'
+				d[last] = 'L'
+				first++
+				last--
+			}
+		}
+	}
+}
+
+// 解法二
+func pushDominoes1(dominoes string) string {
+	dominoes = "L" + dominoes + "R"
+	res := ""
+	for i, j := 0, 1; j < len(dominoes); j++ {
+		if dominoes[j] == '.' {
+			continue
+		}
+		if i > 0 {
+			res += string(dominoes[i])
+		}
+		middle := j - i - 1
+		if dominoes[i] == dominoes[j] {
+			for k := 0; k < middle; k++ {
+				res += string(dominoes[i])
+			}
+		} else if dominoes[i] == 'L' && dominoes[j] == 'R' {
+			for k := 0; k < middle; k++ {
+				res += string('.')
+			}
+		} else {
+			for k := 0; k < middle/2; k++ {
+				res += string('R')
+			}
+			for k := 0; k < middle%2; k++ {
+				res += string('.')
+			}
+			for k := 0; k < middle/2; k++ {
+				res += string('L')
+			}
+		}
+		i = j
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0836.Rectangle-Overlap/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0839.Similar-String-Groups/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0839.Similar-String-Groups.md b/website/content/ChapterFour/0800~0899/0839.Similar-String-Groups.md
new file mode 100755
index 000000000..c369de1d9
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0839.Similar-String-Groups.md
@@ -0,0 +1,105 @@
+# [839. Similar String Groups](https://leetcode.com/problems/similar-string-groups/)
+
+
+## 题目
+
+Two strings `X` and `Y` are similar if we can swap two letters (in different positions) of `X`, so that it equals `Y`.
+
+For example, `"tars"` and `"rats"` are similar (swapping at positions `0` and `2`), and `"rats"` and `"arts"` are similar, but `"star"` is not similar to `"tars"`, `"rats"`, or `"arts"`.
+
+Together, these form two connected groups by similarity: `{"tars", "rats", "arts"}` and `{"star"}`. Notice that `"tars"` and `"arts"` are in the same group even though they are not similar. Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.
+
+We are given a list `A` of strings. Every string in `A` is an anagram of every other string in `A`. How many groups are there?
+
+**Example 1**:
+
+    Input: ["tars","rats","arts","star"]
+    Output: 2
+
+**Note**:
+
+1. `A.length <= 2000`
+2. `A[i].length <= 1000`
+3. `A.length * A[i].length <= 20000`
+4. All words in `A` consist of lowercase letters only.
+5. All words in `A` have the same length and are anagrams of each other.
+6. The judging time limit has been increased for this question.
+
+
+## 题目大意
+
+如果我们交换字符串 X 中的两个不同位置的字母，使得它和字符串 Y 相等，那么称 X 和 Y 两个字符串相似。
+
+例如，"tars" 和 "rats" 是相似的 (交换 0 与 2 的位置)； "rats" 和 "arts" 也是相似的，但是 "star" 不与 "tars"，"rats"，或 "arts" 相似。
+
+总之，它们通过相似性形成了两个关联组：{"tars", "rats", "arts"} 和 {"star"}。注意，"tars" 和 "arts" 是在同一组中，即使它们并不相似。形式上，对每个组而言，要确定一个单词在组中，只需要这个词和该组中至少一个单词相似。我们给出了一个不包含重复的字符串列表 A。列表中的每个字符串都是 A 中其它所有字符串的一个字母异位词。请问 A 中有多少个相似字符串组？
+
+
+提示：
+
+- A.length <= 2000
+- A[i].length <= 1000
+- A.length * A[i].length <= 20000
+- A 中的所有单词都只包含小写字母。
+- A 中的所有单词都具有相同的长度，且是彼此的字母异位词。
+- 此问题的判断限制时间已经延长。
+
+
+备注：  
+
+- 字母异位词[anagram]，一种把某个字符串的字母的位置（顺序）加以改换所形成的新词。
+
+
+
+
+## 解题思路
+
+
+- 给出一个字符串数组，要求找出这个数组中，"不相似"的字符串有多少种。相似的字符串的定义是：如果 A 和 B 字符串只需要交换一次字母的位置就能变成两个相等的字符串，那么 A 和 B 是相似的。
+- 这一题的解题思路是用并查集。先将题目中的“相似”的定义转换一下，A 和 B 相似的意思是，A 和 B 中只有 2 个字符不相等，其他字符都相等，这样交换一次才能完全相等。有没有可能这两个字符交换了也不相等呢？这种情况不用考虑，因为题目中提到了给的字符串都是 `anagram` 的(`anagram` 的意思是，字符串的任意排列组合)。那么这题就比较简单了，只需要判断每两个字符串是否“相似”，如果相似就 `union()`，最后看并查集中有几个集合即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func numSimilarGroups(A []string) int {
+	uf := template.UnionFind{}
+	uf.Init(len(A))
+	for i := 0; i < len(A); i++ {
+		for j := i + 1; j < len(A); j++ {
+			if isSimilar(A[i], A[j]) {
+				uf.Union(i, j)
+			}
+		}
+	}
+	return uf.TotalCount()
+}
+
+func isSimilar(a, b string) bool {
+	var n int
+	for i := 0; i < len(a); i++ {
+		if a[i] != b[i] {
+			n++
+			if n > 2 {
+				return false
+			}
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0838.Push-Dominoes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0841.Keys-and-Rooms/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0841.Keys-and-Rooms.md b/website/content/ChapterFour/0800~0899/0841.Keys-and-Rooms.md
new file mode 100644
index 000000000..8e4f23f93
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0841.Keys-and-Rooms.md
@@ -0,0 +1,87 @@
+# [841. Keys and Rooms](https://leetcode.com/problems/keys-and-rooms/)
+
+
+
+## 题目
+
+There are `N` rooms and you start in room `0`. Each room has a distinct number in `0, 1, 2, ..., N-1`, and each room may have some keys to access the next room.
+
+Formally, each room `i` has a list of keys `rooms[i]`, and each key `rooms[i][j]` is an integer in `[0, 1, ..., N-1]` where `N = rooms.length`. A key `rooms[i][j] = v` opens the room with number `v`.
+
+Initially, all the rooms start locked (except for room `0`).
+
+You can walk back and forth between rooms freely.
+
+Return `true` if and only if you can enter every room.
+
+**Example 1**:
+
+```
+Input: [[1],[2],[3],[]]
+Output: true
+Explanation:  
+We start in room 0, and pick up key 1.
+We then go to room 1, and pick up key 2.
+We then go to room 2, and pick up key 3.
+We then go to room 3.  Since we were able to go to every room, we return true.
+```
+
+**Example 2**:
+
+```
+Input: [[1,3],[3,0,1],[2],[0]]
+Output: false
+Explanation: We can't enter the room with number 2.
+```
+
+**Note**:
+
+1. `1 <= rooms.length <= 1000`
+2. `0 <= rooms[i].length <= 1000`
+3. The number of keys in all rooms combined is at most `3000`.
+
+
+## 题目大意
+
+有 N 个房间，开始时你位于 0 号房间。每个房间有不同的号码：0，1，2，...，N-1，并且房间里可能有一些钥匙能使你进入下一个房间。
+
+在形式上，对于每个房间 i 都有一个钥匙列表 rooms[i]，每个钥匙 rooms[i][j] 由 [0,1，...，N-1] 中的一个整数表示，其中 N = rooms.length。 钥匙 rooms[i][j] = v 可以打开编号为 v 的房间。最初，除 0 号房间外的其余所有房间都被锁住。你可以自由地在房间之间来回走动。如果能进入每个房间返回 true，否则返回 false。
+
+提示：
+
+- 1 <= rooms.length <= 1000
+- 0 <= rooms[i].length <= 1000
+- 所有房间中的钥匙数量总计不超过 3000。
+
+## 解题思路
+
+- 给出一个房间数组，每个房间里面装了一些钥匙。0 号房间默认是可以进入的，房间进入顺序没有要求，问最终能否进入所有房间。
+- 用 DFS 依次深搜所有房间的钥匙，如果都能访问到，最终输出 true。这题算是 DFS 里面的简单题。
+
+## 代码
+
+```go
+func canVisitAllRooms(rooms [][]int) bool {
+	visited := make(map[int]bool)
+	visited[0] = true
+	dfsVisitAllRooms(rooms, visited, 0)
+	return len(rooms) == len(visited)
+}
+
+func dfsVisitAllRooms(es [][]int, visited map[int]bool, from int) {
+	for _, to := range es[from] {
+		if visited[to] {
+			continue
+		}
+		visited[to] = true
+		dfsVisitAllRooms(es, visited, to)
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0839.Similar-String-Groups/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0842.Split-Array-into-Fibonacci-Sequence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0842.Split-Array-into-Fibonacci-Sequence.md b/website/content/ChapterFour/0800~0899/0842.Split-Array-into-Fibonacci-Sequence.md
new file mode 100755
index 000000000..a50cc2f3f
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0842.Split-Array-into-Fibonacci-Sequence.md
@@ -0,0 +1,142 @@
+# [842. Split Array into Fibonacci Sequence](https://leetcode.com/problems/split-array-into-fibonacci-sequence/)
+
+
+## 题目
+
+Given a string `S` of digits, such as `S = "123456579"`, we can split it into a *Fibonacci-like sequence* `[123, 456, 579].`
+
+Formally, a Fibonacci-like sequence is a list `F` of non-negative integers such that:
+
+- `0 <= F[i] <= 2^31 - 1`, (that is, each integer fits a 32-bit signed integer type);
+- `F.length >= 3`;
+- and `F[i] + F[i+1] = F[i+2]` for all `0 <= i < F.length - 2`.
+
+Also, note that when splitting the string into pieces, each piece must not have extra leading zeroes, except if the piece is the number 0 itself.
+
+Return any Fibonacci-like sequence split from `S`, or return `[]` if it cannot be done.
+
+**Example 1**:
+
+    Input: "123456579"
+    Output: [123,456,579]
+
+**Example 2**:
+
+    Input: "11235813"
+    Output: [1,1,2,3,5,8,13]
+
+**Example 3**:
+
+    Input: "112358130"
+    Output: []
+    Explanation: The task is impossible.
+
+**Example 4**:
+
+    Input: "0123"
+    Output: []
+    Explanation: Leading zeroes are not allowed, so "01", "2", "3" is not valid.
+
+**Example 5**:
+
+    Input: "1101111"
+    Output: [110, 1, 111]
+    Explanation: The output [11, 0, 11, 11] would also be accepted.
+
+**Note**:
+
+1. `1 <= S.length <= 200`
+2. `S` contains only digits.
+
+
+## 题目大意
+
+给定一个数字字符串 S，比如 S = "123456579"，我们可以将它分成斐波那契式的序列 [123, 456, 579]。斐波那契式序列是一个非负整数列表 F，且满足：
+
+- 0 <= F[i] <= 2^31 - 1，（也就是说，每个整数都符合 32 位有符号整数类型）；
+- F.length >= 3；
+- 对于所有的0 <= i < F.length - 2，都有 F[i] + F[i+1] = F[i+2] 成立。
+
+另外，请注意，将字符串拆分成小块时，每个块的数字一定不要以零开头，除非这个块是数字 0 本身。返回从 S 拆分出来的所有斐波那契式的序列块，如果不能拆分则返回 []。
+
+
+
+## 解题思路
+
+
+- 这一题是第 306 题的加强版。第 306 题要求判断字符串是否满足斐波那契数列形式。这一题要求输出按照斐波那契数列形式分割之后的数字数组。
+- 这一题思路和第 306 题基本一致，需要注意的是题目中的一个限制条件，`0 <= F[i] <= 2^31 - 1`，注意这个条件，笔者开始没注意，后面输出解就出现错误了，可以看笔者的测试文件用例的最后两组数据，这两组都是可以分解成斐波那契数列的，但是由于分割以后的数字都大于了 `2^31 - 1`，所以这些解都不能要！
+- 这一题也要特别注意剪枝条件，没有剪枝条件，时间复杂度特别高，加上合理的剪枝条件以后，0ms 通过。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+)
+
+func splitIntoFibonacci(S string) []int {
+	if len(S) < 3 {
+		return []int{}
+	}
+	res, isComplete := []int{}, false
+	for firstEnd := 0; firstEnd < len(S)/2; firstEnd++ {
+		if S[0] == '0' && firstEnd > 0 {
+			break
+		}
+		first, _ := strconv.Atoi(S[:firstEnd+1])
+		if first >= 1<<31 { // 题目要求每个数都要小于 2^31 - 1 = 2147483647，此处剪枝很关键！
+			break
+		}
+		for secondEnd := firstEnd + 1; max(firstEnd, secondEnd-firstEnd) <= len(S)-secondEnd; secondEnd++ {
+			if S[firstEnd+1] == '0' && secondEnd-firstEnd > 1 {
+				break
+			}
+			second, _ := strconv.Atoi(S[firstEnd+1 : secondEnd+1])
+			if second >= 1<<31 { // 题目要求每个数都要小于 2^31 - 1 = 2147483647，此处剪枝很关键！
+				break
+			}
+			findRecursiveCheck(S, first, second, secondEnd+1, &res, &isComplete)
+		}
+	}
+	return res
+}
+
+//Propagate for rest of the string
+func findRecursiveCheck(S string, x1 int, x2 int, left int, res *[]int, isComplete *bool) {
+	if x1 >= 1<<31 || x2 >= 1<<31 { // 题目要求每个数都要小于 2^31 - 1 = 2147483647，此处剪枝很关键！
+		return
+	}
+	if left == len(S) {
+		if !*isComplete {
+			*isComplete = true
+			*res = append(*res, x1)
+			*res = append(*res, x2)
+		}
+		return
+	}
+	if strings.HasPrefix(S[left:], strconv.Itoa(x1+x2)) && !*isComplete {
+		*res = append(*res, x1)
+		findRecursiveCheck(S, x2, x1+x2, left+len(strconv.Itoa(x1+x2)), res, isComplete)
+		return
+	}
+	if len(*res) > 0 && !*isComplete {
+		*res = (*res)[:len(*res)-1]
+	}
+	return
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0841.Keys-and-Rooms/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0844.Backspace-String-Compare/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0844.Backspace-String-Compare.md b/website/content/ChapterFour/0800~0899/0844.Backspace-String-Compare.md
new file mode 100644
index 000000000..6adcab297
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0844.Backspace-String-Compare.md
@@ -0,0 +1,118 @@
+# [844. Backspace String Compare](https://leetcode.com/problems/backspace-string-compare/)
+
+## 题目
+
+Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.
+
+
+**Example 1**:
+
+```
+
+Input: S = "ab#c", T = "ad#c"
+Output: true
+Explanation: Both S and T become "ac".
+
+```
+
+**Example 2**:
+
+```
+
+Input: S = "ab##", T = "c#d#"
+Output: true
+Explanation: Both S and T become "".
+
+```
+
+**Example 3**:
+
+```
+
+Input: S = "a##c", T = "#a#c"
+Output: true
+Explanation: Both S and T become "c".
+
+```
+
+**Example 4**:
+
+```
+
+Input: S = "a#c", T = "b"
+Output: false
+Explanation: S becomes "c" while T becomes "b".
+
+```
+
+
+**Note**:
+
+- 1 <= S.length <= 200
+- 1 <= T.length <= 200
+- S and T only contain lowercase letters and '#' characters.
+
+
+**Follow up**:
+
+- Can you solve it in O(N) time and O(1) space?
+
+## 题目大意
+
+
+给 2 个字符串，如果遇到 # 号字符，就回退一个字符。问最终的 2 个字符串是否完全一致。
+
+## 解题思路
+
+这一题可以用栈的思想来模拟，遇到 # 字符就回退一个字符。不是 # 号就入栈一个字符。比较最终 2 个字符串即可。
+
+
+
+
+
+
+
+
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func backspaceCompare(S string, T string) bool {
+	s := make([]rune, 0)
+	for _, c := range S {
+		if c == '#' {
+			if len(s) > 0 {
+				s = s[:len(s)-1]
+			}
+		} else {
+			s = append(s, c)
+		}
+	}
+	s2 := make([]rune, 0)
+	for _, c := range T {
+		if c == '#' {
+			if len(s2) > 0 {
+				s2 = s2[:len(s2)-1]
+			}
+		} else {
+			s2 = append(s2, c)
+		}
+	}
+	return string(s) == string(s2)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0842.Split-Array-into-Fibonacci-Sequence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0845.Longest-Mountain-in-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0845.Longest-Mountain-in-Array.md b/website/content/ChapterFour/0800~0899/0845.Longest-Mountain-in-Array.md
new file mode 100644
index 000000000..8854bef47
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0845.Longest-Mountain-in-Array.md
@@ -0,0 +1,98 @@
+# [845. Longest Mountain in Array](https://leetcode.com/problems/longest-mountain-in-array/)
+
+## 题目
+
+Let's call any (contiguous) subarray B (of A) a mountain if the following properties hold:  
+
+- B.length >= 3
+- There exists some 0 < i < B.length - 1 such that B[0] < B[1] < ... B[i-1] < B[i] > B[i+1] > ... > B[B.length - 1]
+(Note that B could be any subarray of A, including the entire array A.)
+
+Given an array A of integers, return the length of the longest mountain. 
+
+Return 0 if there is no mountain.
+
+
+
+
+**Example 1**:
+
+```
+
+Input: [2,1,4,7,3,2,5]
+Output: 5
+Explanation: The largest mountain is [1,4,7,3,2] which has length 5.
+
+```
+
+**Example 2**:
+
+```
+
+Input: [2,2,2]
+Output: 0
+Explanation: There is no mountain.
+
+```
+
+**Note**:
+
+- 0 <= A.length <= 10000
+- 0 <= A[i] <= 10000
+
+
+**Follow up**:
+
+- Can you solve it using only one pass?
+- Can you solve it in O(1) space?
+
+## 题目大意
+
+这道题考察的是滑动窗口的问题。
+
+给出一个数组，要求求出这个数组里面“山”最长的长度。“山”的意思是，从一个数开始逐渐上升，到顶以后，逐渐下降。
+
+## 解题思路
+
+这道题解题思路也是滑动窗口，只不过在滑动的过程中多判断一个上升和下降的状态即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func longestMountain(A []int) int {
+	left, right, res, isAscending := 0, 0, 0, true
+	for left < len(A) {
+		if right+1 < len(A) && ((isAscending == true && A[right+1] > A[left] && A[right+1] > A[right]) || (right != left && A[right+1] < A[right])) {
+			if A[right+1] < A[right] {
+				isAscending = false
+			}
+			right++
+		} else {
+			if right != left && isAscending == false {
+				res = max(res, right-left+1)
+			}
+			left++
+			if right < left {
+				right = left
+			}
+			if right == left {
+				isAscending = true
+			}
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0844.Backspace-String-Compare/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0846.Hand-of-Straights/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0846.Hand-of-Straights.md b/website/content/ChapterFour/0800~0899/0846.Hand-of-Straights.md
new file mode 100644
index 000000000..f26eb6cc6
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0846.Hand-of-Straights.md
@@ -0,0 +1,75 @@
+# [846. Hand of Straights](https://leetcode.com/problems/hand-of-straights/)
+
+## 题目
+
+Alice has some number of cards and she wants to rearrange the cards into groups so that each group is of size groupSize, and consists of groupSize consecutive cards.
+
+Given an integer array hand where hand[i] is the value written on the ith card and an integer groupSize, return true if she can rearrange the cards, or false otherwise.
+
+**Example 1**:
+
+    Input: hand = [1,2,3,6,2,3,4,7,8], groupSize = 3
+    Output: true
+    Explanation: Alice's hand can be rearranged as [1,2,3],[2,3,4],[6,7,8]
+
+**Example 2**:
+
+    Input: hand = [1,2,3,4,5], groupSize = 4
+    Output: false
+    Explanation: Alice's hand can not be rearranged into groups of 4.
+
+**Constraints:**
+
+- 1 <= hand.length <= 10000
+- 0 <= hand[i] <= 1000000000
+- 1 <= groupSize <= hand.length
+
+## 题目大意
+
+Alice 手中有一把牌，她想要重新排列这些牌，分成若干组，使每一组的牌数都是 groupSize ，并且由 groupSize 张连续的牌组成。
+
+给你一个整数数组 hand 其中 hand[i] 是写在第 i 张牌，和一个整数 groupSize 。如果她可能重新排列这些牌，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+贪心算法
+
+- 对hand升序排序
+- 对hand内数字进行哈希计数（key:数字，value:数量）
+- 遍历hand中的数字，以数量大于1的数字作为顺子开头，寻找顺子后续元素，若无法找到完整顺子则返回false
+- 所有数字都能找到完整顺子返回true
+
+##代码
+
+```go
+package leetcode
+
+import "sort"
+
+func isNStraightHand(hand []int, groupSize int) bool {
+	mp := make(map[int]int)
+	for _, v := range hand {
+		mp[v] += 1
+	}
+	sort.Ints(hand)
+	for _, num := range hand {
+		if mp[num] == 0 {
+			continue
+		}
+		for diff := 0; diff < groupSize; diff++ {
+			if mp[num+diff] == 0 {
+				return false
+			}
+			mp[num+diff] -= 1
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0845.Longest-Mountain-in-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0850.Rectangle-Area-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0850.Rectangle-Area-II.md b/website/content/ChapterFour/0800~0899/0850.Rectangle-Area-II.md
new file mode 100755
index 000000000..b16e51fe7
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0850.Rectangle-Area-II.md
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+# [850. Rectangle Area II](https://leetcode.com/problems/rectangle-area-ii/)
+
+
+## 题目
+
+We are given a list of (axis-aligned) `rectangles`. Each `rectangle[i] = [x1, y1, x2, y2]` , where (x1, y1) are the coordinates of the bottom-left corner, and (x2, y2) are the coordinates of the top-right corner of the `i`th rectangle.
+
+Find the total area covered by all `rectangles` in the plane. Since the answer may be too large, **return it modulo 10^9 + 7**.
+
+![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/06/rectangle_area_ii_pic.png)
+
+**Example 1**:
+
+    Input: [[0,0,2,2],[1,0,2,3],[1,0,3,1]]
+    Output: 6
+    Explanation: As illustrated in the picture.
+
+**Example 2**:
+
+    Input: [[0,0,1000000000,1000000000]]
+    Output: 49
+    Explanation: The answer is 10^18 modulo (10^9 + 7), which is (10^9)^2 = (-7)^2 = 49.
+
+**Note**:
+
+- `1 <= rectangles.length <= 200`
+- `rectanges[i].length = 4`
+- `0 <= rectangles[i][j] <= 10^9`
+- The total area covered by all rectangles will never exceed `2^63 - 1` and thus will fit in a 64-bit signed integer.
+
+
+## 题目大意
+
+我们给出了一个（轴对齐的）矩形列表 rectangles。 对于 rectangle[i] = [x1, y1, x2, y2]，其中（x1，y1）是矩形 i 左下角的坐标，（x2，y2）是该矩形右上角的坐标。找出平面中所有矩形叠加覆盖后的总面积。由于答案可能太大，请返回它对 10 ^ 9 + 7 取模的结果。
+
+提示：
+
+- 1 <= rectangles.length <= 200
+- rectanges[i].length = 4
+- 0 <= rectangles[i][j] <= 10^9
+- 矩形叠加覆盖后的总面积不会超越 2^63 - 1 ，这意味着可以用一个 64 位有符号整数来保存面积结果。
+
+
+## 解题思路
+
+
+- 在二维坐标系中给出一些矩形，要求这些矩形合并之后的面积。由于矩形有重叠，所以需要考虑合并以后的面积。矩形的坐标值也会很大。
+- 这一题给人的感觉很像第 218 题，求天际线的过程也是有楼挡楼，重叠的情况。不过那一题只用求天际线的拐点，所以我们可以对区间做“右边界减一”的处理，防止两个相邻区间因为共点，而导致结果错误。但是这一题如果还是用相同的做法，就会出错，因为“右边界减一”以后，面积会少一部分，最终得到的结果也是偏小的。所以这一题要将线段树改造一下。
+- 思路是先讲 Y 轴上的坐标离线化，转换成线段树。将矩形的 2 条边变成扫描线，左边是入边，右边是出边。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_8.png)
+
+- 再从左往右遍历每条扫描线，并对 Y 轴上的线段树进行 update。X 轴上的每个坐标区间 * query 线段树总高度的结果 = 区间面积。最后将 X 轴对应的每个区间面积加起来，就是最终矩形合并以后的面积。如下图中间的图。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_9.png)
+
+    需要注意的一点是，**每次 query 的结果并不一定是连续线段**。如上图最右边的图，中间有一段是可能出现镂空的。这种情况看似复杂，其实很简单，因为每段线段树的线段代表的权值高度是不同的，每次 query 最大高度得到的结果已经考虑了中间可能有镂空的情况了。
+
+- 具体做法，先把各个矩形在 Y 轴方向上离散化，这里的**线段树叶子节点不再是一个点了，而是一个区间长度为 1 的区间段**。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_0.png)
+
+    每个叶子节点也不再是存储一个 int 值，而是存 2 个值，一个是 count 值，用来记录这条区间被覆盖的次数，另一个值是 val 值，用来反映射该线段长度是多少，因为 Y 轴被离散化了，区间坐标间隔都是 1，但是实际 Y 轴的高度并不是 1 ，所以用 val 来反映射原来的高度。
+
+- 初始化线段树，叶子节点的 count = 0，val 根据题目给的 Y 坐标进行计算。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_1.png)
+
+- 从左往右遍历每个扫描线。每条扫面线都把对应 update 更新到叶子节点。pushUp 的时候需要合并每个区间段的高度 val 值。如果有区间没有被覆盖，那么这个区间高度 val 为 0，这也就处理了可能“中间镂空”的情况。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_2.png)
+
+        func (sat *SegmentAreaTree) pushUp(treeIndex, leftTreeIndex, rightTreeIndex int) {
+        	newCount, newValue := sat.merge(sat.tree[leftTreeIndex].count, sat.tree[rightTreeIndex].count), 0
+        	if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count > 0 {
+        		newValue = sat.merge(sat.tree[leftTreeIndex].val, sat.tree[rightTreeIndex].val)
+        	} else if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count == 0 {
+        		newValue = sat.tree[leftTreeIndex].val
+        	} else if sat.tree[leftTreeIndex].count == 0 && sat.tree[rightTreeIndex].count > 0 {
+        		newValue = sat.tree[rightTreeIndex].val
+        	}
+        	sat.tree[treeIndex] = SegmentItem{count: newCount, val: newValue}
+        }
+
+- 扫描每一个扫描线，先 pushDown 到叶子节点，再 pushUp 到根节点。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_3.png)
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_4.png)
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_5.png)
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_6.png)
+
+- 遍历到倒数第 2 根扫描线的时候就能得到结果了。因为最后一根扫描线 update 以后，整个线段树全部都归为初始化状态了。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_850_7.png)
+
+- 这一题是线段树扫面线解法的经典题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+func rectangleArea(rectangles [][]int) int {
+	sat, res := SegmentAreaTree{}, 0
+	posXMap, posX, posYMap, posY, lines := discretization850(rectangles)
+	tmp := make([]int, len(posYMap))
+	for i := 0; i < len(tmp)-1; i++ {
+		tmp[i] = posY[i+1] - posY[i]
+	}
+	sat.Init(tmp, func(i, j int) int {
+		return i + j
+	})
+	for i := 0; i < len(posY)-1; i++ {
+		tmp[i] = posY[i+1] - posY[i]
+	}
+	for i := 0; i < len(posX)-1; i++ {
+		for _, v := range lines[posXMap[posX[i]]] {
+			sat.Update(posYMap[v.start], posYMap[v.end], v.state)
+		}
+		res += ((posX[i+1] - posX[i]) * sat.Query(0, len(posY)-1)) % 1000000007
+	}
+	return res % 1000000007
+}
+
+func discretization850(positions [][]int) (map[int]int, []int, map[int]int, []int, map[int][]LineItem) {
+	tmpXMap, tmpYMap, posXArray, posXMap, posYArray, posYMap, lines := map[int]int{}, map[int]int{}, []int{}, map[int]int{}, []int{}, map[int]int{}, map[int][]LineItem{}
+	for _, pos := range positions {
+		tmpXMap[pos[0]]++
+		tmpXMap[pos[2]]++
+	}
+	for k := range tmpXMap {
+		posXArray = append(posXArray, k)
+	}
+	sort.Ints(posXArray)
+	for i, pos := range posXArray {
+		posXMap[pos] = i
+	}
+
+	for _, pos := range positions {
+		tmpYMap[pos[1]]++
+		tmpYMap[pos[3]]++
+		tmp1 := lines[posXMap[pos[0]]]
+		tmp1 = append(tmp1, LineItem{start: pos[1], end: pos[3], state: 1})
+		lines[posXMap[pos[0]]] = tmp1
+		tmp2 := lines[posXMap[pos[2]]]
+		tmp2 = append(tmp2, LineItem{start: pos[1], end: pos[3], state: -1})
+		lines[posXMap[pos[2]]] = tmp2
+	}
+	for k := range tmpYMap {
+		posYArray = append(posYArray, k)
+	}
+	sort.Ints(posYArray)
+	for i, pos := range posYArray {
+		posYMap[pos] = i
+	}
+	return posXMap, posXArray, posYMap, posYArray, lines
+}
+
+// LineItem define
+type LineItem struct { // 垂直于 x 轴的线段
+	start, end, state int // state = 1 代表进入，-1 代表离开
+}
+
+// SegmentItem define
+type SegmentItem struct {
+	count int
+	val   int
+}
+
+// SegmentAreaTree define
+type SegmentAreaTree struct {
+	data        []int
+	tree        []SegmentItem
+	left, right int
+	merge       func(i, j int) int
+}
+
+// Init define
+func (sat *SegmentAreaTree) Init(nums []int, oper func(i, j int) int) {
+	sat.merge = oper
+	data, tree := make([]int, len(nums)), make([]SegmentItem, 4*len(nums))
+	for i := 0; i < len(nums); i++ {
+		data[i] = nums[i]
+	}
+	sat.data, sat.tree = data, tree
+	if len(nums) > 0 {
+		sat.buildSegmentTree(0, 0, len(nums)-1)
+	}
+}
+
+// 在 treeIndex 的位置创建 [left....right] 区间的线段树
+func (sat *SegmentAreaTree) buildSegmentTree(treeIndex, left, right int) {
+	if left == right-1 {
+		sat.tree[treeIndex] = SegmentItem{count: 0, val: sat.data[left]}
+		return
+	}
+	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, sat.leftChild(treeIndex), sat.rightChild(treeIndex)
+	sat.buildSegmentTree(leftTreeIndex, left, midTreeIndex)
+	sat.buildSegmentTree(rightTreeIndex, midTreeIndex, right)
+	sat.pushUp(treeIndex, leftTreeIndex, rightTreeIndex)
+}
+
+func (sat *SegmentAreaTree) pushUp(treeIndex, leftTreeIndex, rightTreeIndex int) {
+	newCount, newValue := sat.merge(sat.tree[leftTreeIndex].count, sat.tree[rightTreeIndex].count), 0
+	if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count > 0 {
+		newValue = sat.merge(sat.tree[leftTreeIndex].val, sat.tree[rightTreeIndex].val)
+	} else if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count == 0 {
+		newValue = sat.tree[leftTreeIndex].val
+	} else if sat.tree[leftTreeIndex].count == 0 && sat.tree[rightTreeIndex].count > 0 {
+		newValue = sat.tree[rightTreeIndex].val
+	}
+	sat.tree[treeIndex] = SegmentItem{count: newCount, val: newValue}
+}
+
+func (sat *SegmentAreaTree) leftChild(index int) int {
+	return 2*index + 1
+}
+
+func (sat *SegmentAreaTree) rightChild(index int) int {
+	return 2*index + 2
+}
+
+// 查询 [left....right] 区间内的值
+
+// Query define
+func (sat *SegmentAreaTree) Query(left, right int) int {
+	if len(sat.data) > 0 {
+		return sat.queryInTree(0, 0, len(sat.data)-1, left, right)
+	}
+	return 0
+}
+
+func (sat *SegmentAreaTree) queryInTree(treeIndex, left, right, queryLeft, queryRight int) int {
+	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, sat.leftChild(treeIndex), sat.rightChild(treeIndex)
+	if left > queryRight || right < queryLeft { // segment completely outside range
+		return 0 // represents a null node
+	}
+	if queryLeft <= left && queryRight >= right { // segment completely inside range
+		if sat.tree[treeIndex].count > 0 {
+			return sat.tree[treeIndex].val
+		}
+		return 0
+	}
+	if queryLeft > midTreeIndex {
+		return sat.queryInTree(rightTreeIndex, midTreeIndex, right, queryLeft, queryRight)
+	} else if queryRight <= midTreeIndex {
+		return sat.queryInTree(leftTreeIndex, left, midTreeIndex, queryLeft, queryRight)
+	}
+	// merge query results
+	return sat.merge(sat.queryInTree(leftTreeIndex, left, midTreeIndex, queryLeft, midTreeIndex),
+		sat.queryInTree(rightTreeIndex, midTreeIndex, right, midTreeIndex, queryRight))
+}
+
+// Update define
+func (sat *SegmentAreaTree) Update(updateLeft, updateRight, val int) {
+	if len(sat.data) > 0 {
+		sat.updateInTree(0, 0, len(sat.data)-1, updateLeft, updateRight, val)
+	}
+}
+
+func (sat *SegmentAreaTree) updateInTree(treeIndex, left, right, updateLeft, updateRight, val int) {
+	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, sat.leftChild(treeIndex), sat.rightChild(treeIndex)
+	if left > right || left >= updateRight || right <= updateLeft { // 由于叶子节点的区间不在是 left == right 所以这里判断需要增加等号的判断
+		return // out of range. escape.
+	}
+
+	if updateLeft <= left && right <= updateRight { // segment is fully within update range
+		if left == right-1 {
+			sat.tree[treeIndex].count = sat.merge(sat.tree[treeIndex].count, val)
+		}
+		if left != right-1 { // update lazy[] for children
+			sat.updateInTree(leftTreeIndex, left, midTreeIndex, updateLeft, updateRight, val)
+			sat.updateInTree(rightTreeIndex, midTreeIndex, right, updateLeft, updateRight, val)
+			sat.pushUp(treeIndex, leftTreeIndex, rightTreeIndex)
+		}
+		return
+	}
+	sat.updateInTree(leftTreeIndex, left, midTreeIndex, updateLeft, updateRight, val)
+	sat.updateInTree(rightTreeIndex, midTreeIndex, right, updateLeft, updateRight, val)
+	// merge updates
+	sat.pushUp(treeIndex, leftTreeIndex, rightTreeIndex)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0846.Hand-of-Straights/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0851.Loud-and-Rich/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0851.Loud-and-Rich.md b/website/content/ChapterFour/0800~0899/0851.Loud-and-Rich.md
new file mode 100644
index 000000000..ca43a9ef0
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+++ b/website/content/ChapterFour/0800~0899/0851.Loud-and-Rich.md
@@ -0,0 +1,115 @@
+# [851. Loud and Rich](https://leetcode.com/problems/loud-and-rich/)
+
+
+
+## 题目
+
+In a group of N people (labelled `0, 1, 2, ..., N-1`), each person has different amounts of money, and different levels of quietness.
+
+For convenience, we'll call the person with label `x`, simply "person `x`".
+
+We'll say that `richer[i] = [x, y]` if person `x` definitely has more money than person `y`. Note that `richer` may only be a subset of valid observations.
+
+Also, we'll say `quiet = q` if person x has quietness `q`.
+
+Now, return `answer`, where `answer = y` if `y` is the least quiet person (that is, the person `y` with the smallest value of `quiet[y]`), among all people who definitely have equal to or more money than person `x`.
+
+**Example 1**:
+
+```
+Input: richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [3,2,5,4,6,1,7,0]
+Output: [5,5,2,5,4,5,6,7]
+Explanation: 
+answer[0] = 5.
+Person 5 has more money than 3, which has more money than 1, which has more money than 0.
+The only person who is quieter (has lower quiet[x]) is person 7, but
+it isn't clear if they have more money than person 0.
+
+answer[7] = 7.
+Among all people that definitely have equal to or more money than person 7
+(which could be persons 3, 4, 5, 6, or 7), the person who is the quietest (has lower quiet[x])
+is person 7.
+
+The other answers can be filled out with similar reasoning.
+```
+
+**Note**:
+
+1. `1 <= quiet.length = N <= 500`
+2. `0 <= quiet[i] < N`, all `quiet[i]` are different.
+3. `0 <= richer.length <= N * (N-1) / 2`
+4. `0 <= richer[i][j] < N`
+5. `richer[i][0] != richer[i][1]`
+6. `richer[i]`'s are all different.
+7. The observations in `richer` are all logically consistent.
+
+## 题目大意
+
+在一组 N 个人（编号为 0, 1, 2, ..., N-1）中，每个人都有不同数目的钱，以及不同程度的安静（quietness）。为了方便起见，我们将编号为 x 的人简称为 "person x "。如果能够肯定 person x 比 person y 更有钱的话，我们会说 richer[i] = [x, y] 。注意 richer 可能只是有效观察的一个子集。另外，如果 person x 的安静程度为 q ，我们会说 quiet[x] = q 。现在，返回答案 answer ，其中 answer[x] = y 的前提是，在所有拥有的钱不少于 person x 的人中，person y 是最安静的人（也就是安静值 quiet[y] 最小的人）。
+
+提示：
+
+- 1 <= quiet.length = N <= 500
+- 0 <= quiet[i] < N，所有 quiet[i] 都不相同。
+- 0 <= richer.length <= N * (N-1) / 2
+- 0 <= richer[i][j] < N
+- richer[i][0] != richer[i][1]
+- richer[i] 都是不同的。
+- 对 richer 的观察在逻辑上是一致的。
+
+
+## 解题思路
+
+- 给出 2 个数组，richer 和 quiet，要求输出 answer，其中 answer = y 的前提是，在所有拥有的钱不少于 x 的人中，y 是最安静的人（也就是安静值 quiet[y] 最小的人）
+- 由题意可知，`richer` 构成了一个有向无环图，首先使用字典建立图的关系，找到比当前下标编号富有的所有的人。然后使用广度优先层次遍历，不断的使用富有的人，但是安静值更小的人更新子节点即可。
+- 这一题还可以用拓扑排序来解答。将 `richer` 中描述的关系看做边，如果 `x > y`，则 `x` 指向 `y`。将 `quiet` 看成权值。用一个数组记录答案，初始时 `ans[i] = i`。然后对原图做拓扑排序，对于每一条边，如果发现 `quiet[ans[v]] > quiet[ans[u]]`，则 `ans[v]` 的答案为 `ans[u]`。时间复杂度即为拓扑排序的时间复杂度为 `O(m+n)`。空间复杂度需要 `O(m)` 的数组建图，需要 `O(n)` 的数组记录入度以及存储队列，所以空间复杂度为 `O(m+n)`。
+
+## 代码
+
+```go
+func loudAndRich(richer [][]int, quiet []int) []int {
+	edges := make([][]int, len(quiet))
+	for i := range edges {
+		edges[i] = []int{}
+	}
+	indegrees := make([]int, len(quiet))
+	for _, edge := range richer {
+		n1, n2 := edge[0], edge[1]
+		edges[n1] = append(edges[n1], n2)
+		indegrees[n2]++
+	}
+	res := make([]int, len(quiet))
+	for i := range res {
+		res[i] = i
+	}
+	queue := []int{}
+	for i, v := range indegrees {
+		if v == 0 {
+			queue = append(queue, i)
+		}
+	}
+	for len(queue) > 0 {
+		nexts := []int{}
+		for _, n1 := range queue {
+			for _, n2 := range edges[n1] {
+				indegrees[n2]--
+				if quiet[res[n2]] > quiet[res[n1]] {
+					res[n2] = res[n1]
+				}
+				if indegrees[n2] == 0 {
+					nexts = append(nexts, n2)
+				}
+			}
+		}
+		queue = nexts
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0850.Rectangle-Area-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0852.Peak-Index-in-a-Mountain-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0852.Peak-Index-in-a-Mountain-Array.md b/website/content/ChapterFour/0800~0899/0852.Peak-Index-in-a-Mountain-Array.md
new file mode 100755
index 000000000..12aa32deb
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0852.Peak-Index-in-a-Mountain-Array.md
@@ -0,0 +1,98 @@
+# [852. Peak Index in a Mountain Array](https://leetcode.com/problems/peak-index-in-a-mountain-array/)
+
+
+## 题目
+
+Let's call an array `A` a *mountain* if the following properties hold:
+
+- `A.length >= 3`
+- There exists some `0 < i < A.length - 1` such that `A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]`
+
+Given an array that is definitely a mountain, return any `i` such that `A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]`.
+
+**Example 1**:
+
+    Input: [0,1,0]
+    Output: 1
+
+**Example 2**:
+
+    Input: [0,2,1,0]
+    Output: 1
+
+**Note**:
+
+1. `3 <= A.length <= 10000`
+2. `0 <= A[i] <= 10^6`
+3. A is a mountain, as defined above.
+
+
+## 题目大意
+
+我们把符合下列属性的数组 A 称作山脉：
+
+- A.length >= 3
+- 存在 0 < i < A.length - 1 使得A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]
+给定一个确定为山脉的数组，返回任何满足 A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1] 的 i 的值。
+
+提示：
+
+- 3 <= A.length <= 10000
+- 0 <= A[i] <= 10^6
+- A 是如上定义的山脉
+ 
+ 
+
+## 解题思路
+
+- 给出一个数组，数组里面存在有且仅有一个“山峰”，(山峰的定义是，下标 `i` 比 `i-1`、`i+1` 位置上的元素都要大)，找到这个“山峰”，并输出其中一个山峰的下标。
+- 这一题直接用二分搜索即可，数组中的元素算基本有序。判断是否为山峰的条件为比较左右两个数，如果当前的数比左右两个数都大，即找到了山峰。其他的情况都在山坡上。这一题有两种写法，第一种写法是标准的二分写法，第二种写法是变形的二分写法。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 二分
+func peakIndexInMountainArray(A []int) int {
+	low, high := 0, len(A)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if A[mid] > A[mid+1] && A[mid] > A[mid-1] {
+			return mid
+		}
+		if A[mid] > A[mid+1] && A[mid] < A[mid-1] {
+			high = mid - 1
+		}
+		if A[mid] < A[mid+1] && A[mid] > A[mid-1] {
+			low = mid + 1
+		}
+	}
+	return 0
+}
+
+// 解法二 二分
+func peakIndexInMountainArray1(A []int) int {
+	low, high := 0, len(A)-1
+	for low < high {
+		mid := low + (high-low)>>1
+		// 如果 mid 较大，则左侧存在峰值，high = m，如果 mid + 1 较大，则右侧存在峰值，low = mid + 1
+		if A[mid] > A[mid+1] {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0851.Loud-and-Rich/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0853.Car-Fleet/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0853.Car-Fleet.md b/website/content/ChapterFour/0800~0899/0853.Car-Fleet.md
new file mode 100755
index 000000000..b3532036f
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0853.Car-Fleet.md
@@ -0,0 +1,105 @@
+# [853. Car Fleet](https://leetcode.com/problems/car-fleet/)
+
+
+## 题目
+
+`N` cars are going to the same destination along a one lane road. The destination is `target` miles away.
+
+Each car `i` has a constant speed `speed[i]` (in miles per hour), and initial position `position[i]` miles towards the target along the road.
+
+A car can never pass another car ahead of it, but it can catch up to it, and drive bumper to bumper at the same speed.
+
+The distance between these two cars is ignored - they are assumed to have the same position.
+
+A *car fleet* is some non-empty set of cars driving at the same position and same speed. Note that a single car is also a car fleet.
+
+If a car catches up to a car fleet right at the destination point, it will still be considered as one car fleet.
+
+How many car fleets will arrive at the destination?
+
+**Example 1**:
+
+    Input: target = 12, position = [10,8,0,5,3], speed = [2,4,1,1,3]
+    Output: 3
+    Explanation:
+    The cars starting at 10 and 8 become a fleet, meeting each other at 12.
+    The car starting at 0 doesn't catch up to any other car, so it is a fleet by itself.
+    The cars starting at 5 and 3 become a fleet, meeting each other at 6.
+    Note that no other cars meet these fleets before the destination, so the answer is 3.
+
+**Note**:
+
+1. `0 <= N <= 10 ^ 4`
+2. `0 < target <= 10 ^ 6`
+3. `0 < speed[i] <= 10 ^ 6`
+4. `0 <= position[i] < target`
+5. All initial positions are different.
+
+
+## 题目大意
+
+N  辆车沿着一条车道驶向位于 target 英里之外的共同目的地。每辆车 i 以恒定的速度 speed[i] （英里/小时），从初始位置 position[i] （英里） 沿车道驶向目的地。
+
+一辆车永远不会超过前面的另一辆车，但它可以追上去，并与前车以相同的速度紧接着行驶。此时，我们会忽略这两辆车之间的距离，也就是说，它们被假定处于相同的位置。车队 是一些由行驶在相同位置、具有相同速度的车组成的非空集合。注意，一辆车也可以是一个车队。即便一辆车在目的地才赶上了一个车队，它们仍然会被视作是同一个车队。
+
+ 问最后会有多少车队到达目的地?
+
+
+
+## 解题思路
+
+
+- 根据每辆车距离终点和速度，计算每辆车到达终点的时间，并按照距离从大到小排序(position 越大代表距离终点越近)
+- 从头往后扫描排序以后的数组，时间一旦大于前一个 car 的时间，就会生成一个新的 car fleet，最终计数加一即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+type car struct {
+	time     float64
+	position int
+}
+
+// ByPosition define
+type ByPosition []car
+
+func (a ByPosition) Len() int           { return len(a) }
+func (a ByPosition) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
+func (a ByPosition) Less(i, j int) bool { return a[i].position > a[j].position }
+
+func carFleet(target int, position []int, speed []int) int {
+	n := len(position)
+	if n <= 1 {
+		return n
+	}
+	cars := make([]car, n)
+	for i := 0; i < n; i++ {
+		cars[i] = car{float64(target-position[i]) / float64(speed[i]), position[i]}
+	}
+	sort.Sort(ByPosition(cars))
+	fleet, lastTime := 0, 0.0
+	for i := 0; i < len(cars); i++ {
+		if cars[i].time > lastTime {
+			lastTime = cars[i].time
+			fleet++
+		}
+	}
+	return fleet
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0852.Peak-Index-in-a-Mountain-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0856.Score-of-Parentheses/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0856.Score-of-Parentheses.md b/website/content/ChapterFour/0800~0899/0856.Score-of-Parentheses.md
new file mode 100644
index 000000000..d94ee68f1
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0856.Score-of-Parentheses.md
@@ -0,0 +1,109 @@
+# [856. Score of Parentheses](https://leetcode.com/problems/score-of-parentheses/)
+
+## 题目
+
+Given a balanced parentheses string S, compute the score of the string based on the following rule:
+
+() has score 1
+AB has score A + B, where A and B are balanced parentheses strings.
+(A) has score 2 * A, where A is a balanced parentheses string.
+ 
+
+**Example 1**:
+
+```
+
+Input: "()"
+Output: 1
+
+```
+
+**Example 2**:
+
+```
+
+Input: "(())"
+Output: 2
+
+```
+
+**Example 3**:
+
+```
+
+Input: "()()"
+Output: 2
+
+```
+
+**Example 4**:
+
+```
+
+Input: "(()(()))"
+Output: 6
+
+```
+ 
+
+**Note**:
+
+1. S is a balanced parentheses string, containing only ( and ).
+2. 2 <= S.length <= 50
+
+## 题目大意
+
+按照以下规则计算括号的分数：() 代表 1 分。AB 代表 A + B，A 和 B 分别是已经满足匹配规则的括号组。(A) 代表 2 * A，其中 A 也是已经满足匹配规则的括号组。给出一个括号字符串，要求按照这些规则计算出括号的分数值。
+
+
+## 解题思路
+
+按照括号匹配的原则，一步步的计算每个组合的分数入栈。遇到题目中的 3 种情况，取出栈顶元素算分数。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func scoreOfParentheses(S string) int {
+	res, stack, top, temp := 0, []int{}, -1, 0
+	for _, s := range S {
+		if s == '(' {
+			stack = append(stack, -1)
+			top++
+		} else {
+			temp = 0
+			for stack[top] != -1 {
+				temp += stack[top]
+				stack = stack[:len(stack)-1]
+				top--
+			}
+			stack = stack[:len(stack)-1]
+			top--
+			if temp == 0 {
+				stack = append(stack, 1)
+				top++
+			} else {
+				stack = append(stack, temp*2)
+				top++
+			}
+		}
+	}
+	for len(stack) != 0 {
+		res += stack[top]
+		stack = stack[:len(stack)-1]
+		top--
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0853.Car-Fleet/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0859.Buddy-Strings/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0859.Buddy-Strings.md b/website/content/ChapterFour/0800~0899/0859.Buddy-Strings.md
new file mode 100644
index 000000000..9c8ddd17f
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0859.Buddy-Strings.md
@@ -0,0 +1,94 @@
+# [859. Buddy Strings](https://leetcode.com/problems/buddy-strings/)
+
+## 题目
+
+Given two strings s and goal, return true if you can swap two letters in s so the result is equal to goal, otherwise, return false.
+
+Swapping letters is defined as taking two indices i and j (0-indexed) such that i != j and swapping the characters at s[i] and s[j].
+
+For example, swapping at indices 0 and 2 in "abcd" results in "cbad".
+
+**Example 1**:
+
+    Input: s = "ab", goal = "ba"
+    Output: true
+    Explanation: You can swap s[0] = 'a' and s[1] = 'b' to get "ba", which is equal to goal.
+
+**Example 2**:
+
+    Input: s = "ab", goal = "ab"
+    Output: false
+    Explanation: The only letters you can swap are s[0] = 'a' and s[1] = 'b', which results in "ba" != goal.
+
+**Example 3**:
+
+    Input: s = "aa", goal = "aa"
+    Output: true
+    Explanation: You can swap s[0] = 'a' and s[1] = 'a' to get "aa", which is equal to goal.
+
+**Example 4**:
+
+    Input: s = "aaaaaaabc", goal = "aaaaaaacb"
+    Output: true
+
+**Constraints:**
+
+- 1 <= s.length, goal.length <= 2 * 10000
+- s and goal consist of lowercase letters.
+
+## 题目大意
+
+给你两个字符串 s 和 goal ，只要我们可以通过交换 s 中的两个字母得到与 goal 相等的结果，就返回 true；否则返回 false 。
+
+交换字母的定义是：取两个下标 i 和 j （下标从 0 开始）且满足 i != j ，接着交换 s[i] 和 s[j] 处的字符。
+
+例如，在 "abcd" 中交换下标 0 和下标 2 的元素可以生成 "cbad" 。
+
+## 解题思路
+
+分为两种情况进行比较:
+- s 等于 goal, s 中有重复元素就返回 true,否则返回 false
+- s 不等于 goal, s 中有两个下标不同的字符与 goal 中对应下标的字符分别相等
+
+## 代码
+
+```go
+
+package leetcode
+
+func buddyStrings(s string, goal string) bool {
+	if len(s) != len(goal) || len(s) <= 1 {
+		return false
+	}
+	mp := make(map[byte]int)
+	if s == goal {
+		for i := 0; i < len(s); i++ {
+			if _, ok := mp[s[i]]; ok {
+				return true
+			}
+			mp[s[i]]++
+		}
+		return false
+	}
+	first, second := -1, -1
+	for i := 0; i < len(s); i++ {
+		if s[i] != goal[i] {
+			if first == -1 {
+				first = i
+			} else if second == -1 {
+				second = i
+			} else {
+				return false
+			}
+		}
+	}
+	return second != -1 && s[first] == goal[second] && s[second] == goal[first]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0856.Score-of-Parentheses/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0862.Shortest-Subarray-with-Sum-at-Least-K/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0862.Shortest-Subarray-with-Sum-at-Least-K.md b/website/content/ChapterFour/0800~0899/0862.Shortest-Subarray-with-Sum-at-Least-K.md
new file mode 100644
index 000000000..0f0090725
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0862.Shortest-Subarray-with-Sum-at-Least-K.md
@@ -0,0 +1,97 @@
+# [862. Shortest Subarray with Sum at Least K](https://leetcode.com/problems/shortest-subarray-with-sum-at-least-k/)
+
+
+
+## 题目
+
+Return the **length** of the shortest, non-empty, contiguous subarray of `A` with sum at least `K`.
+
+If there is no non-empty subarray with sum at least `K`, return `-1`.
+
+**Example 1**:
+
+```
+Input: A = [1], K = 1
+Output: 1
+```
+
+**Example 2**:
+
+```
+Input: A = [1,2], K = 4
+Output: -1
+```
+
+**Example 3**:
+
+```
+Input: A = [2,-1,2], K = 3
+Output: 3
+```
+
+**Note**:
+
+1. `1 <= A.length <= 50000`
+2. `-10 ^ 5 <= A[i] <= 10 ^ 5`
+3. `1 <= K <= 10 ^ 9`
+
+## 题目大意
+
+返回 A 的最短的非空连续子数组的长度，该子数组的和至少为 K 。如果没有和至少为 K 的非空子数组，返回 -1 。
+
+提示：
+
+- 1 <= A.length <= 50000
+- -10 ^ 5 <= A[i] <= 10 ^ 5
+- 1 <= K <= 10 ^ 9
+
+
+## 解题思路
+
+- 给出一个数组，要求找出一个最短的，非空的，连续的子序列且累加和至少为 k 。
+- 由于给的数组里面可能存在负数，所以子数组的累加和不会随着数组的长度增加而增加。题目中要求区间和，理所应当需要利用 `prefixSum` 前缀和，先计算出 `prefixSum` 前缀和。
+- 简化一下题目的要求，即能否找到 `prefixSum[y] - prefixSum[x] ≥ K` ，且 `y - x` 的差值最小。如果固定的 `y`，那么对于 `x`，`x` 越大，`y - x` 的差值就越小(因为 `x` 越逼近 `y`)。所以想求区间 `[x, y]` 的最短距离，需要保证 `y` 尽量小，`x` 尽量大，这样 `[x, y]` 区间距离最小。那么有以下 2 点“常识”一定成立：
+    1. 如果 `x1 < x2` ，并且 `prefixSum[x2] ≤ prefixSum[x1]`，说明结果一定不会取 `x1`。因为如果 `prefixSum[x1] ≤ prefixSum[y] - k`，那么  `prefixSum[x2] ≤ prefixSum[x1] ≤ prefixSum[y] - k`，`x2` 也能满足题意，并且 `x2` 比 `x1` 更加接近 `y`，最优解一定优先考虑 `x2`。
+    2. 在确定了 `x` 以后，以后就不用再考虑 `x` 了，因为如果 `y2 > y1`，且 `y2` 的时候取 `x` 还是一样的，那么算距离的话，`y2 - x` 显然大于 `y1 - x`，这样的话肯定不会是最短的距离。
+- 从上面这两个常识来看，可以用双端队列 `deque` 来处理 `prefixSum`。`deque` 中存储的是递增的 `x` 下标，为了满足常识一。从双端队列的开头开始遍历，假如区间和之差大于等于 `K`，就移除队首元素并更新结果 `res`。队首移除元素，直到不满足 `prefixSum[i]-prefixSum[deque[0]] >= K` 这一不等式，是为了满足常识二。之后的循环是此题的精髓，从双端队列的末尾开始往前遍历，假如当前区间和 `prefixSum[i]` 小于等于队列末尾的区间和，则移除队列末尾元素。为什么这样处理呢？因为若数组都是正数，那么长度越长，区间和一定越大，则 `prefixSum[i]` 一定大于所有双端队列中的区间和，但由于可能存在负数，从而使得长度变长，区间总和反而减少了，之前的区间和之差值都没有大于等于 `K`(< K)，现在的更不可能大于等于 `K`，这个结束位置可以直接淘汰，不用进行计算。循环结束后将当前位置加入双端队列即可。遇到新下标在队尾移除若干元素，这一行为，也是为了满足常识一。
+- 由于所有下标都只进队列一次，也最多 pop 出去一次，所以时间复杂度 O(n)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+func shortestSubarray(A []int, K int) int {
+	res, prefixSum := len(A)+1, make([]int, len(A)+1)
+	for i := 0; i < len(A); i++ {
+		prefixSum[i+1] = prefixSum[i] + A[i]
+	}
+	// deque 中保存递增的 prefixSum 下标
+	deque := []int{}
+	for i := range prefixSum {
+		// 下面这个循环希望能找到 [deque[0], i] 区间内累加和 >= K，如果找到了就更新答案
+		for len(deque) > 0 && prefixSum[i]-prefixSum[deque[0]] >= K {
+			length := i - deque[0]
+			if res > length {
+				res = length
+			}
+			// 找到第一个 deque[0] 能满足条件以后，就移除它，因为它是最短长度的子序列了
+			deque = deque[1:]
+		}
+		// 下面这个循环希望能保证 prefixSum[deque[i]] 递增
+		for len(deque) > 0 && prefixSum[i] <= prefixSum[deque[len(deque)-1]] {
+			deque = deque[:len(deque)-1]
+		}
+		deque = append(deque, i)
+	}
+	if res <= len(A) {
+		return res
+	}
+	return -1
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0859.Buddy-Strings/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0863.All-Nodes-Distance-K-in-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0863.All-Nodes-Distance-K-in-Binary-Tree.md b/website/content/ChapterFour/0800~0899/0863.All-Nodes-Distance-K-in-Binary-Tree.md
new file mode 100644
index 000000000..2944f0e14
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0863.All-Nodes-Distance-K-in-Binary-Tree.md
@@ -0,0 +1,103 @@
+# [863. All Nodes Distance K in Binary Tree](https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/)
+
+
+
+## 题目
+
+We are given a binary tree (with root node `root`), a `target` node, and an integer value `K`.
+
+Return a list of the values of all nodes that have a distance `K` from the `target` node. The answer can be returned in any order.
+
+**Example 1**:
+
+```
+Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2
+
+Output: [7,4,1]
+
+Explanation: 
+The nodes that are a distance 2 from the target node (with value 5)
+have values 7, 4, and 1.
+```
+
+![https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/28/sketch0.png](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/28/sketch0.png)
+
+**Note**:
+
+1. The given tree is non-empty.
+2. Each node in the tree has unique values `0 <= node.val <= 500`.
+3. The `target` node is a node in the tree.
+4. `0 <= K <= 1000`.
+
+## 题目大意
+
+给定一个二叉树（具有根结点 root）， 一个目标结点 target ，和一个整数值 K 。返回到目标结点 target 距离为 K 的所有结点的值的列表。 答案可以以任何顺序返回。
+
+提示：
+
+- 给定的树是非空的。
+- 树上的每个结点都具有唯一的值 0 <= node.val <= 500 。
+- 目标结点 target 是树上的结点。
+- 0 <= K <= 1000.
+
+
+## 解题思路
+
+- 给出一颗树和一个目标节点 target，一个距离 K，要求找到所有距离目标节点 target 的距离是 K 的点。
+- 这一题用 DFS 的方法解题。先找到当前节点距离目标节点的距离，如果在左子树中找到了 target，距离当前节点的距离 > 0，则还需要在它的右子树中查找剩下的距离。如果是在右子树中找到了 target，反之同理。如果当前节点就是目标节点，那么就可以直接记录这个点。否则每次遍历一个点，距离都减一。
+
+## 代码
+
+```go
+func distanceK(root *TreeNode, target *TreeNode, K int) []int {
+	visit := []int{}
+	findDistanceK(root, target, K, &visit)
+	return visit
+}
+
+func findDistanceK(root, target *TreeNode, K int, visit *[]int) int {
+	if root == nil {
+		return -1
+	}
+	if root == target {
+		findChild(root, K, visit)
+		return K - 1
+	}
+	leftDistance := findDistanceK(root.Left, target, K, visit)
+	if leftDistance == 0 {
+		findChild(root, leftDistance, visit)
+	}
+	if leftDistance > 0 {
+		findChild(root.Right, leftDistance-1, visit)
+		return leftDistance - 1
+	}
+	rightDistance := findDistanceK(root.Right, target, K, visit)
+	if rightDistance == 0 {
+		findChild(root, rightDistance, visit)
+	}
+	if rightDistance > 0 {
+		findChild(root.Left, rightDistance-1, visit)
+		return rightDistance - 1
+	}
+	return -1
+}
+
+func findChild(root *TreeNode, K int, visit *[]int) {
+	if root == nil {
+		return
+	}
+	if K == 0 {
+		*visit = append(*visit, root.Val)
+	} else {
+		findChild(root.Left, K-1, visit)
+		findChild(root.Right, K-1, visit)
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0862.Shortest-Subarray-with-Sum-at-Least-K/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0864.Shortest-Path-to-Get-All-Keys/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0864.Shortest-Path-to-Get-All-Keys.md b/website/content/ChapterFour/0800~0899/0864.Shortest-Path-to-Get-All-Keys.md
new file mode 100755
index 000000000..7c05689b1
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0864.Shortest-Path-to-Get-All-Keys.md
@@ -0,0 +1,238 @@
+# [864. Shortest Path to Get All Keys](https://leetcode.com/problems/shortest-path-to-get-all-keys/)
+
+
+## 题目
+
+We are given a 2-dimensional `grid`. `"."` is an empty cell, `"#"` is a wall, `"@"` is the starting point, (`"a"`, `"b"`, ...) are keys, and (`"A"`, `"B"`, ...) are locks.
+
+We start at the starting point, and one move consists of walking one space in one of the 4 cardinal directions. We cannot walk outside the grid, or walk into a wall. If we walk over a key, we pick it up. We can't walk over a lock unless we have the corresponding key.
+
+For some 1 <= K <= 6, there is exactly one lowercase and one uppercase letter of the first `K` letters of the English alphabet in the grid. This means that there is exactly one key for each lock, and one lock for each key; and also that the letters used to represent the keys and locks were chosen in the same order as the English alphabet.
+
+Return the lowest number of moves to acquire all keys. If it's impossible, return `-1`.
+
+**Example 1**:
+
+    Input: ["@.a.#","###.#","b.A.B"]
+    Output: 8
+
+**Example 2**:
+
+    Input: ["@..aA","..B#.","....b"]
+    Output: 6
+
+**Note**:
+
+1. `1 <= grid.length <= 30`
+2. `1 <= grid[0].length <= 30`
+3. `grid[i][j]` contains only `'.'`, `'#'`, `'@'`, `'a'-'f'` and `'A'-'F'`
+4. The number of keys is in `[1, 6]`. Each key has a different letter and opens exactly one lock.
+
+
+## 题目大意
+
+给定一个二维网格 grid。 "." 代表一个空房间， "#" 代表一堵墙， "@" 是起点，（"a", "b", ...）代表钥匙，（"A", "B", ...）代表锁。
+
+我们从起点开始出发，一次移动是指向四个基本方向之一行走一个单位空间。我们不能在网格外面行走，也无法穿过一堵墙。如果途经一个钥匙，我们就把它捡起来。除非我们手里有对应的钥匙，否则无法通过锁。
+
+假设 K 为钥匙/锁的个数，且满足 1 <= K <= 6，字母表中的前 K 个字母在网格中都有自己对应的一个小写和一个大写字母。换言之，每个锁有唯一对应的钥匙，每个钥匙也有唯一对应的锁。另外，代表钥匙和锁的字母互为大小写并按字母顺序排列。
+
+返回获取所有钥匙所需要的移动的最少次数。如果无法获取所有钥匙，返回 -1 。
+
+提示：
+
+1. 1 <= grid.length <= 30
+2. 1 <= grid[0].length <= 30
+3. grid[i][j] 只含有 '.', '#', '@', 'a'-'f' 以及 'A'-'F'
+4. 钥匙的数目范围是 [1, 6]，每个钥匙都对应一个不同的字母，正好打开一个对应的锁。
+
+
+## 解题思路
+
+
+- 给出一个地图，在图中有钥匙和锁，当锁在没有钥匙的时候不能通行，问从起点 @ 开始，到最终获得所有钥匙，最短需要走多少步。
+- 这一题可以用 BFS 来解答。由于钥匙的种类比较多，所以 visited 数组需要 3 个维度，一个是 x 坐标，一个是 y 坐标，最后一个是当前获取钥匙的状态。每把钥匙都有获取了和没有获取两种状态，题目中说最多有 6 把钥匙，那么排列组合最多是 2^6 = 64 种状态。用一个十进制数的二进制位来压缩这些状态，二进制位分别来表示这些钥匙是否已经获取了。既然钥匙的状态可以压缩，其实 x 和 y 的坐标也可以一并压缩到这个数中。BFS 中存的数字是坐标 + 钥匙状态的状态。在 BFS 遍历的过程中，用 visited 数组来过滤遍历过的情况，来保证走的路是最短的。其他的情况无非是判断锁的状态，是否能通过，判断钥匙获取状态。
+- 这一题不知道是否能用 DFS 来解答。我实现了一版，但是在 18 / 35 这组 case 上超时了，具体 case 见测试文件第一个 case。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+	"strings"
+)
+
+// 解法一 BFS，利用状态压缩来过滤筛选状态
+func shortestPathAllKeys(grid []string) int {
+	if len(grid) == 0 {
+		return 0
+	}
+	board, visited, startx, starty, res, fullKeys := make([][]byte, len(grid)), make([][][]bool, len(grid)), 0, 0, 0, 0
+	for i := 0; i < len(grid); i++ {
+		board[i] = make([]byte, len(grid[0]))
+	}
+	for i, g := range grid {
+		board[i] = []byte(g)
+		for _, v := range g {
+			if v == 'a' || v == 'b' || v == 'c' || v == 'd' || v == 'e' || v == 'f' {
+				fullKeys |= (1 << uint(v-'a'))
+			}
+		}
+		if strings.Contains(g, "@") {
+			startx, starty = i, strings.Index(g, "@")
+		}
+	}
+	for i := 0; i < len(visited); i++ {
+		visited[i] = make([][]bool, len(board[0]))
+	}
+	for i := 0; i < len(board); i++ {
+		for j := 0; j < len(board[0]); j++ {
+			visited[i][j] = make([]bool, 64)
+		}
+	}
+	queue := []int{}
+	queue = append(queue, (starty<<16)|(startx<<8))
+	visited[startx][starty][0] = true
+	for len(queue) != 0 {
+		qLen := len(queue)
+		for i := 0; i < qLen; i++ {
+			state := queue[0]
+			queue = queue[1:]
+			starty, startx = state>>16, (state>>8)&0xFF
+			keys := state & 0xFF
+			if keys == fullKeys {
+				return res
+			}
+			for i := 0; i < 4; i++ {
+				newState := keys
+				nx := startx + dir[i][0]
+				ny := starty + dir[i][1]
+				if !isInBoard(board, nx, ny) {
+					continue
+				}
+				if board[nx][ny] == '#' {
+					continue
+				}
+				flag, canThroughLock := keys&(1<<(board[nx][ny]-'A')), false
+				if flag != 0 {
+					canThroughLock = true
+				}
+				if isLock(board, nx, ny) && !canThroughLock {
+					continue
+				}
+				if isKey(board, nx, ny) {
+					newState |= (1 << (board[nx][ny] - 'a'))
+				}
+				if visited[nx][ny][newState] {
+					continue
+				}
+				queue = append(queue, (ny<<16)|(nx<<8)|newState)
+				visited[nx][ny][newState] = true
+			}
+		}
+		res++
+	}
+	return -1
+}
+
+// 解法二 DFS，但是超时了，剪枝条件不够强
+func shortestPathAllKeys1(grid []string) int {
+	if len(grid) == 0 {
+		return 0
+	}
+	board, visited, startx, starty, res, fullKeys := make([][]byte, len(grid)), make([][][]bool, len(grid)), 0, 0, math.MaxInt64, 0
+	for i := 0; i < len(grid); i++ {
+		board[i] = make([]byte, len(grid[0]))
+	}
+	for i, g := range grid {
+		board[i] = []byte(g)
+		for _, v := range g {
+			if v == 'a' || v == 'b' || v == 'c' || v == 'd' || v == 'e' || v == 'f' {
+				fullKeys |= (1 << uint(v-'a'))
+			}
+		}
+		if strings.Contains(g, "@") {
+			startx, starty = i, strings.Index(g, "@")
+		}
+	}
+	for i := 0; i < len(visited); i++ {
+		visited[i] = make([][]bool, len(board[0]))
+	}
+	for i := 0; i < len(board); i++ {
+		for j := 0; j < len(board[0]); j++ {
+			visited[i][j] = make([]bool, 64)
+		}
+	}
+	searchKeys(board, &visited, fullKeys, 0, (starty<<16)|(startx<<8), &res, []int{})
+	if res == math.MaxInt64 {
+		return -1
+	}
+	return res - 1
+}
+
+func searchKeys(board [][]byte, visited *[][][]bool, fullKeys, step, state int, res *int, path []int) {
+	y, x := state>>16, (state>>8)&0xFF
+	keys := state & 0xFF
+
+	if keys == fullKeys {
+		*res = min(*res, step)
+		return
+	}
+
+	flag, canThroughLock := keys&(1<<(board[x][y]-'A')), false
+	if flag != 0 {
+		canThroughLock = true
+	}
+	newState := keys
+	//fmt.Printf("x = %v y = %v fullKeys = %v keys = %v step = %v res = %v path = %v state = %v\n", x, y, fullKeys, keys, step, *res, path, state)
+	if (board[x][y] != '#' && !isLock(board, x, y)) || (isLock(board, x, y) && canThroughLock) {
+		if isKey(board, x, y) {
+			newState |= (1 << uint(board[x][y]-'a'))
+		}
+		(*visited)[x][y][newState] = true
+		path = append(path, x)
+		path = append(path, y)
+
+		for i := 0; i < 4; i++ {
+			nx := x + dir[i][0]
+			ny := y + dir[i][1]
+			if isInBoard(board, nx, ny) && !(*visited)[nx][ny][newState] {
+				searchKeys(board, visited, fullKeys, step+1, (ny<<16)|(nx<<8)|newState, res, path)
+			}
+		}
+		(*visited)[x][y][keys] = false
+		path = path[:len(path)-1]
+		path = path[:len(path)-1]
+	}
+}
+
+func isLock(board [][]byte, x, y int) bool {
+	if (board[x][y] == 'A') || (board[x][y] == 'B') ||
+		(board[x][y] == 'C') || (board[x][y] == 'D') ||
+		(board[x][y] == 'E') || (board[x][y] == 'F') {
+		return true
+	}
+	return false
+}
+
+func isKey(board [][]byte, x, y int) bool {
+	if (board[x][y] == 'a') || (board[x][y] == 'b') ||
+		(board[x][y] == 'c') || (board[x][y] == 'd') ||
+		(board[x][y] == 'e') || (board[x][y] == 'f') {
+		return true
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0863.All-Nodes-Distance-K-in-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0867.Transpose-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0867.Transpose-Matrix.md b/website/content/ChapterFour/0800~0899/0867.Transpose-Matrix.md
new file mode 100755
index 000000000..8560e9169
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0867.Transpose-Matrix.md
@@ -0,0 +1,64 @@
+# [867. Transpose Matrix](https://leetcode.com/problems/transpose-matrix/)
+
+
+## 题目
+
+Given a matrix `A`, return the transpose of `A`.
+
+The transpose of a matrix is the matrix flipped over it's main diagonal, switching the row and column indices of the matrix.
+
+**Example 1**:
+
+    Input: [[1,2,3],[4,5,6],[7,8,9]]
+    Output: [[1,4,7],[2,5,8],[3,6,9]]
+
+**Example 2**:
+
+    Input: [[1,2,3],[4,5,6]]
+    Output: [[1,4],[2,5],[3,6]]
+
+**Note**:
+
+1. `1 <= A.length <= 1000`
+2. `1 <= A[0].length <= 1000`
+
+
+## 题目大意
+
+给定一个矩阵 A， 返回 A 的转置矩阵。矩阵的转置是指将矩阵的主对角线翻转，交换矩阵的行索引与列索引。
+
+
+## 解题思路
+
+
+- 给出一个矩阵，顺时针旋转 90°
+- 解题思路很简单，直接模拟即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func transpose(A [][]int) [][]int {
+	row, col, result := len(A), len(A[0]), make([][]int, len(A[0]))
+	for i := range result {
+		result[i] = make([]int, row)
+	}
+	for i := 0; i < row; i++ {
+		for j := 0; j < col; j++ {
+			result[j][i] = A[i][j]
+		}
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0864.Shortest-Path-to-Get-All-Keys/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0869.Reordered-Power-of-2/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0869.Reordered-Power-of-2.md b/website/content/ChapterFour/0800~0899/0869.Reordered-Power-of-2.md
new file mode 100644
index 000000000..183d1daf8
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0869.Reordered-Power-of-2.md
@@ -0,0 +1,100 @@
+# [869. Reordered Power of 2](https://leetcode.com/problems/reordered-power-of-2/)
+
+
+## 题目
+
+Starting with a positive integer `N`, we reorder the digits in any order (including the original order) such that the leading digit is not zero.
+
+Return `true` if and only if we can do this in a way such that the resulting number is a power of 2.
+
+**Example 1:**
+
+```
+Input:1
+Output:true
+```
+
+**Example 2:**
+
+```
+Input:10
+Output:false
+```
+
+**Example 3:**
+
+```
+Input:16
+Output:true
+```
+
+**Example 4:**
+
+```
+Input:24
+Output:false
+```
+
+**Example 5:**
+
+```
+Input:46
+Output:true
+```
+
+**Note:**
+
+1. `1 <= N <= 10^9`
+
+## 题目大意
+
+给定正整数 N ，我们按任何顺序（包括原始顺序）将数字重新排序，注意其前导数字不能为零。如果我们可以通过上述方式得到 2 的幂，返回 true；否则，返回 false。
+
+## 解题思路
+
+- 将整数每个位上的所有排列看成字符串，那么题目转换为判断这些字符串是否和 2 的幂的字符串是否一致。判断的方法有很多种，笔者这里判断借助了一个 `map`。两个不同排列的字符串要相等，所有字符出现的频次必定一样。利用一个 `map` 统计它们各自字符的频次，最终都一致，则判定这两个字符串是满足题意的。
+- 此题数据量比较小，在 `[1,10^9]` 这个区间内，2 的幂只有 30 几个，所以最终要判断的字符串就是这 30 几个。笔者这里没有打表了，采用更加一般的做法。数据量更大，此解法代码也能通过。
+
+## 代码
+
+```go
+package leetcode
+
+import "fmt"
+
+func reorderedPowerOf2(n int) bool {
+	sample, i := fmt.Sprintf("%v", n), 1
+	for len(fmt.Sprintf("%v", i)) <= len(sample) {
+		t := fmt.Sprintf("%v", i)
+		if len(t) == len(sample) && isSame(t, sample) {
+			return true
+		}
+		i = i << 1
+	}
+	return false
+}
+
+func isSame(t, s string) bool {
+	m := make(map[rune]int)
+	for _, v := range t {
+		m[v]++
+	}
+	for _, v := range s {
+		m[v]--
+		if m[v] < 0 {
+			return false
+		}
+		if m[v] == 0 {
+			delete(m, v)
+		}
+	}
+	return len(m) == 0
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0867.Transpose-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0870.Advantage-Shuffle/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0870.Advantage-Shuffle.md b/website/content/ChapterFour/0800~0899/0870.Advantage-Shuffle.md
new file mode 100644
index 000000000..6d9ccecb1
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0870.Advantage-Shuffle.md
@@ -0,0 +1,79 @@
+# [870. Advantage Shuffle](https://leetcode.com/problems/advantage-shuffle/)
+
+## 题目
+
+Given two arrays `A` and `B` of equal size, the *advantage of `A` with respect to `B`* is the number of indices `i` for which `A[i] > B[i]`.
+
+Return **any** permutation of `A` that maximizes its advantage with respect to `B`.
+
+**Example 1:**
+
+```
+Input:A = [2,7,11,15], B = [1,10,4,11]
+Output:[2,11,7,15]
+```
+
+**Example 2:**
+
+```
+Input:A = [12,24,8,32], B = [13,25,32,11]
+Output:[24,32,8,12]
+```
+
+**Note:**
+
+1. `1 <= A.length = B.length <= 10000`
+2. `0 <= A[i] <= 10^9`
+3. `0 <= B[i] <= 10^9`
+
+## 题目大意
+
+给定两个大小相等的数组 A 和 B，A 相对于 B 的优势可以用满足 A[i] > B[i] 的索引 i 的数目来描述。返回 A 的任意排列，使其相对于 B 的优势最大化。
+
+## 解题思路
+
+- 此题用贪心算法解题。如果 A 中最小的牌 a 能击败 B 中最小的牌 b，那么将它们配对。否则， a 将无益于我们的比分，因为它无法击败任何牌。这是贪心的策略，每次匹配都用手中最弱的牌和 B 中的最小牌 b 进行配对，这样会使 A 中剩余的牌严格的变大，最后会使得得分更多。
+- 在代码实现中，将 A 数组排序，B 数组按照下标排序。因为最终输出的是相对于 B 的优势结果，所以要针对 B 的下标不变来安排 A 的排列。排好序以后按照贪心策略选择 A 中牌的顺序。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func advantageCount1(A []int, B []int) []int {
+	n := len(A)
+	sort.Ints(A)
+	sortedB := make([]int, n)
+	for i := range sortedB {
+		sortedB[i] = i
+	}
+	sort.Slice(sortedB, func(i, j int) bool {
+		return B[sortedB[i]] < B[sortedB[j]]
+	})
+	useless, i, res := make([]int, 0), 0, make([]int, n)
+	for _, index := range sortedB {
+		b := B[index]
+		for i < n && A[i] <= b {
+			useless = append(useless, A[i])
+			i++
+		}
+		if i < n {
+			res[index] = A[i]
+			i++
+		} else {
+			res[index] = useless[0]
+			useless = useless[1:]
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0869.Reordered-Power-of-2/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0872.Leaf-Similar-Trees/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0872.Leaf-Similar-Trees.md b/website/content/ChapterFour/0800~0899/0872.Leaf-Similar-Trees.md
new file mode 100644
index 000000000..04f5b7f27
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0872.Leaf-Similar-Trees.md
@@ -0,0 +1,69 @@
+# [872. Leaf-Similar Trees](https://leetcode.com/problems/leaf-similar-trees/)
+
+
+
+## 题目
+
+Consider all the leaves of a binary tree. From left to right order, the values of those leaves form a *leaf value sequence.*
+
+![https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/16/tree.png](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/16/tree.png)
+
+For example, in the given tree above, the leaf value sequence is `(6, 7, 4, 9, 8)`.
+
+Two binary trees are considered *leaf-similar* if their leaf value sequence is the same.
+
+Return `true` if and only if the two given trees with head nodes `root1` and `root2` are leaf-similar.
+
+**Note**:
+
+- Both of the given trees will have between `1` and `100` nodes.
+
+## 题目大意
+
+请考虑一颗二叉树上所有的叶子，这些叶子的值按从左到右的顺序排列形成一个 叶值序列 。举个例子，如上图所示，给定一颗叶值序列为 (6, 7, 4, 9, 8) 的树。如果有两颗二叉树的叶值序列是相同，那么我们就认为它们是 叶相似 的。如果给定的两个头结点分别为 root1 和 root2 的树是叶相似的，则返回 true；否则返回 false 。
+
+提示：
+
+- 给定的两颗树可能会有 1 到 200 个结点。
+- 给定的两颗树上的值介于 0 到 200 之间。
+
+## 解题思路
+
+- 给出 2 棵树，如果 2 棵树的叶子节点组成的数组是完全一样的，那么就认为这 2 棵树是“叶子相似”的。给出任何 2 棵树判断这 2 棵树是否是“叶子相似”的。
+- 简单题，分别 DFS 遍历 2 棵树，把叶子节点都遍历出来，然后分别比较叶子节点组成的数组是否完全一致即可。
+
+## 代码
+
+```go
+func leafSimilar(root1 *TreeNode, root2 *TreeNode) bool {
+	leaf1, leaf2 := []int{}, []int{}
+	dfsLeaf(root1, &leaf1)
+	dfsLeaf(root2, &leaf2)
+	if len(leaf1) != len(leaf2) {
+		return false
+	}
+	for i := range leaf1 {
+		if leaf1[i] != leaf2[i] {
+			return false
+		}
+	}
+	return true
+}
+
+func dfsLeaf(root *TreeNode, leaf *[]int) {
+	if root != nil {
+		if root.Left == nil && root.Right == nil {
+			*leaf = append(*leaf, root.Val)
+		}
+		dfsLeaf(root.Left, leaf)
+		dfsLeaf(root.Right, leaf)
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0870.Advantage-Shuffle/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0874.Walking-Robot-Simulation/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0874.Walking-Robot-Simulation.md b/website/content/ChapterFour/0800~0899/0874.Walking-Robot-Simulation.md
new file mode 100644
index 000000000..61df07cec
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0874.Walking-Robot-Simulation.md
@@ -0,0 +1,142 @@
+# [874. Walking Robot Simulation](https://leetcode.com/problems/walking-robot-simulation/)
+
+
+## 题目
+
+A robot on an infinite XY-plane starts at point `(0, 0)` and faces north. The robot can receive one of three possible types of `commands`:
+
+- `2`: turn left `90` degrees,
+- `1`: turn right `90` degrees, or
+- `1 <= k <= 9`: move forward `k` units.
+
+Some of the grid squares are `obstacles`. The `ith` obstacle is at grid point `obstacles[i] = (xi, yi)`.
+
+If the robot would try to move onto them, the robot stays on the previous grid square instead (but still continues following the rest of the route.)
+
+Return *the maximum Euclidean distance that the robot will be from the origin **squared** (i.e. if the distance is* `5`*, return* `25`*)*.
+
+**Note:**
+
+- North means +Y direction.
+- East means +X direction.
+- South means -Y direction.
+- West means -X direction.
+
+**Example 1:**
+
+```
+Input: commands = [4,-1,3], obstacles = []
+Output: 25
+Explanation: The robot starts at (0, 0):
+1. Move north 4 units to (0, 4).
+2. Turn right.
+3. Move east 3 units to (3, 4).
+The furthest point away from the origin is (3, 4), which is 32 + 42 = 25 units away.
+
+```
+
+**Example 2:**
+
+```
+Input: commands = [4,-1,4,-2,4], obstacles = [[2,4]]
+Output: 65
+Explanation: The robot starts at (0, 0):
+1. Move north 4 units to (0, 4).
+2. Turn right.
+3. Move east 1 unit and get blocked by the obstacle at (2, 4), robot is at (1, 4).
+4. Turn left.
+5. Move north 4 units to (1, 8).
+The furthest point away from the origin is (1, 8), which is 12 + 82 = 65 units away.
+
+```
+
+**Constraints:**
+
+- `1 <= commands.length <= 104`
+- `commands[i]` is one of the values in the list `[-2,-1,1,2,3,4,5,6,7,8,9]`.
+- `0 <= obstacles.length <= 104`
+- `3 * 104 <= xi, yi <= 3 * 104`
+- The answer is guaranteed to be less than `231`.
+
+## 题目大意
+
+机器人在一个无限大小的 XY 网格平面上行走，从点 (0, 0) 处开始出发，面向北方。该机器人可以接收以下三种类型的命令 commands ：
+
+- 2 ：向左转 90 度
+- -1 ：向右转 90 度
+- 1 <= x <= 9 ：向前移动 x 个单位长度
+
+在网格上有一些格子被视为障碍物 obstacles 。第 i 个障碍物位于网格点 obstacles[i] = (xi, yi) 。机器人无法走到障碍物上，它将会停留在障碍物的前一个网格方块上，但仍然可以继续尝试进行该路线的其余部分。返回从原点到机器人所有经过的路径点（坐标为整数）的最大欧式距离的平方。（即，如果距离为 5 ，则返回 25 ）
+
+注意：
+
+- 北表示 +Y 方向。
+- 东表示 +X 方向。
+- 南表示 -Y 方向。
+- 西表示 -X 方向。
+
+## 解题思路
+
+- 这个题的难点在于，怎么用编程语言去描述机器人的行为，可以用以下数据结构表达机器人的行为：
+
+    ```go
+    direct:= 0                    // direct表示机器人移动方向：0 1 2 3 4 （北东南西），默认朝北
+    x, y := 0, 0                  // 表示当前机器人所在横纵坐标位置，默认为(0,0)
+    directX := []int{0, 1, 0, -1}
+    directY := []int{1, 0, -1, 0}
+    // 组合directX directY和direct，表示机器人往某一个方向移动
+    nextX := x + directX[direct]
+    nextY := y + directY[direct]
+    ```
+
+    其他代码按照题意翻译即可
+
+## 代码
+
+```go
+package leetcode
+
+func robotSim(commands []int, obstacles [][]int) int {
+	m := make(map[[2]int]struct{})
+	for _, v := range obstacles {
+		if len(v) != 0 {
+			m[[2]int{v[0], v[1]}] = struct{}{}
+		}
+	}
+	directX := []int{0, 1, 0, -1}
+	directY := []int{1, 0, -1, 0}
+	direct, x, y := 0, 0, 0
+	result := 0
+	for _, c := range commands {
+		if c == -2 {
+			direct = (direct + 3) % 4
+			continue
+		}
+		if c == -1 {
+			direct = (direct + 1) % 4
+			continue
+		}
+		for ; c > 0; c-- {
+			nextX := x + directX[direct]
+			nextY := y + directY[direct]
+			if _, ok := m[[2]int{nextX, nextY}]; ok {
+				break
+			}
+			tmpResult := nextX*nextX + nextY*nextY
+			if tmpResult > result {
+				result = tmpResult
+			}
+			x = nextX
+			y = nextY
+		}
+	}
+	return result
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0872.Leaf-Similar-Trees/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0875.Koko-Eating-Bananas/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0875.Koko-Eating-Bananas.md b/website/content/ChapterFour/0800~0899/0875.Koko-Eating-Bananas.md
new file mode 100755
index 000000000..734bf16d3
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0875.Koko-Eating-Bananas.md
@@ -0,0 +1,108 @@
+# [875. Koko Eating Bananas](https://leetcode.com/problems/koko-eating-bananas/)
+
+
+## 题目
+
+Koko loves to eat bananas. There are `N` piles of bananas, the `i`-th pile has `piles[i]` bananas. The guards have gone and will come back in `H` hours.
+
+Koko can decide her bananas-per-hour eating speed of `K`. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than `K` bananas, she eats all of them instead, and won't eat any more bananas during this hour.
+
+Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
+
+Return the minimum integer `K` such that she can eat all the bananas within `H` hours.
+
+**Example 1**:
+
+    Input: piles = [3,6,7,11], H = 8
+    Output: 4
+
+**Example 2**:
+
+    Input: piles = [30,11,23,4,20], H = 5
+    Output: 30
+
+**Example 3**:
+
+    Input: piles = [30,11,23,4,20], H = 6
+    Output: 23
+
+**Note**:
+
+- `1 <= piles.length <= 10^4`
+- `piles.length <= H <= 10^9`
+- `1 <= piles[i] <= 10^9`
+
+
+## 题目大意
+
+
+珂珂喜欢吃香蕉。这里有 N 堆香蕉，第 i 堆中有 piles[i] 根香蕉。警卫已经离开了，将在 H 小时后回来。
+
+珂珂可以决定她吃香蕉的速度 K （单位：根/小时）。每个小时，她将会选择一堆香蕉，从中吃掉 K 根。如果这堆香蕉少于 K 根，她将吃掉这堆的所有香蕉，然后这一小时内不会再吃更多的香蕉。  
+
+珂珂喜欢慢慢吃，但仍然想在警卫回来前吃掉所有的香蕉。
+
+返回她可以在 H 小时内吃掉所有香蕉的最小速度 K（K 为整数）。
+
+提示：
+
+- 1 <= piles.length <= 10^4
+- piles.length <= H <= 10^9
+- 1 <= piles[i] <= 10^9
+
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面每个元素代表的是每个香蕉🍌串上香蕉的个数。koko 以 `k 个香蕉/小时`的速度吃这些香蕉。守卫会在 `H 小时`以后回来。问 k 至少为多少，能在守卫回来之前吃完所有的香蕉。当香蕉的个数小于 k 的时候，这个小时只能吃完这些香蕉，不能再吃其他串上的香蕉了。
+- 这一题可以用二分搜索来解答。在 `[0 , max(piles)]` 的范围内搜索，二分的过程都是常规思路。判断是否左右边界如果划分的时候需要注意题目中给的限定条件。当香蕉个数小于 k 的时候，那个小时不能再吃其他香蕉了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func minEatingSpeed(piles []int, H int) int {
+	low, high := 1, maxInArr(piles)
+	for low < high {
+		mid := low + (high-low)>>1
+		if !isPossible(piles, mid, H) {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return low
+}
+
+func isPossible(piles []int, h, H int) bool {
+	res := 0
+	for _, p := range piles {
+		res += int(math.Ceil(float64(p) / float64(h)))
+	}
+	return res <= H
+}
+
+func maxInArr(xs []int) int {
+	res := 0
+	for _, x := range xs {
+		if res < x {
+			res = x
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0874.Walking-Robot-Simulation/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0876.Middle-of-the-Linked-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0876.Middle-of-the-Linked-List.md b/website/content/ChapterFour/0800~0899/0876.Middle-of-the-Linked-List.md
new file mode 100644
index 000000000..264d75801
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0876.Middle-of-the-Linked-List.md
@@ -0,0 +1,89 @@
+# [876. Middle of the Linked List](https://leetcode.com/problems/middle-of-the-linked-list/)
+
+## 题目
+
+Given a non-empty, singly linked list with head node head, return a middle node of linked list.
+
+If there are two middle nodes, return the second middle node.
+
+**Example 1**:
+
+```
+
+Input: [1,2,3,4,5]
+Output: Node 3 from this list (Serialization: [3,4,5])
+The returned node has value 3.  (The judge's serialization of this node is [3,4,5]).
+Note that we returned a ListNode object ans, such that:
+ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
+
+```
+
+**Example 2**:
+
+```
+
+Input: [1,2,3,4,5,6]
+Output: Node 4 from this list (Serialization: [4,5,6])
+Since the list has two middle nodes with values 3 and 4, we return the second one.
+
+```
+
+**Note**:
+
+- The number of nodes in the given list will be between 1 and 100.
+
+## 题目大意
+
+输出链表中间结点。这题在前面题目中反复出现了很多次了。
+
+如果链表长度是奇数，输出中间结点是中间结点。如果链表长度是双数，输出中间结点是中位数后面的那个结点。
+
+## 解题思路
+
+这道题有一个很简单的做法，用 2 个指针只遍历一次就可以找到中间节点。一个指针每次移动 2 步，另外一个指针每次移动 1 步，当快的指针走到终点的时候，慢的指针就是中间节点。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+func middleNode(head *ListNode) *ListNode {
+	if head == nil || head.Next == nil {
+		return head
+	}
+	p1 := head
+	p2 := head
+	for p2.Next != nil && p2.Next.Next != nil {
+		p1 = p1.Next
+		p2 = p2.Next.Next
+	}
+	length := 0
+	cur := head
+	for cur != nil {
+		length++
+		cur = cur.Next
+	}
+	if length%2 == 0 {
+		return p1.Next
+	}
+	return p1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0875.Koko-Eating-Bananas/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0877.Stone-Game/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0877.Stone-Game.md b/website/content/ChapterFour/0800~0899/0877.Stone-Game.md
new file mode 100644
index 000000000..104eedb8f
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0877.Stone-Game.md
@@ -0,0 +1,57 @@
+# [877. Stone Game](https://leetcode.com/problems/stone-game/)
+
+## 题目
+
+Alex and Lee play a game with piles of stones.  There are an even number of piles **arranged in a row**, and each pile has a positive integer number of stones `piles[i]`.
+
+The objective of the game is to end with the most stones.  The total number of stones is odd, so there are no ties.
+
+Alex and Lee take turns, with Alex starting first.  Each turn, a player takes the entire pile of stones from either the beginning or the end of the row.  This continues until there are no more piles left, at which point the person with the most stones wins.
+
+Assuming Alex and Lee play optimally, return `True` if and only if Alex wins the game.
+
+**Example 1:**
+
+```
+Input: piles = [5,3,4,5]
+Output: true
+Explanation:
+Alex starts first, and can only take the first 5 or the last 5.
+Say he takes the first 5, so that the row becomes [3, 4, 5].
+If Lee takes 3, then the board is [4, 5], and Alex takes 5 to win with 10 points.
+If Lee takes the last 5, then the board is [3, 4], and Alex takes 4 to win with 9 points.
+This demonstrated that taking the first 5 was a winning move for Alex, so we return true.
+
+```
+
+**Constraints:**
+
+- `2 <= piles.length <= 500`
+- `piles.length` is even.
+- `1 <= piles[i] <= 500`
+- `sum(piles)` is odd.
+
+## 题目大意
+
+亚历克斯和李用几堆石子在做游戏。偶数堆石子排成一行，每堆都有正整数颗石子 piles[i] 。游戏以谁手中的石子最多来决出胜负。石子的总数是奇数，所以没有平局。亚历克斯和李轮流进行，亚历克斯先开始。 每回合，玩家从行的开始或结束处取走整堆石头。 这种情况一直持续到没有更多的石子堆为止，此时手中石子最多的玩家获胜。假设亚历克斯和李都发挥出最佳水平，当亚历克斯赢得比赛时返回 true ，当李赢得比赛时返回 false 。
+
+## 解题思路
+
+- 一遇到石子问题，很容易让人联想到是否和奇偶数相关。此题指定了石子堆数一定是偶数。所以从这里为突破口试试。Alex 先拿，要么取行首下标为 0 的石子，要么取行尾下标为 n-1 的石子。假设取下标为 0 的石子，剩下的石子堆下标从 1 ~ n-1，即 Lee 只能从奇数下标的石子堆 1 或者 n-1。假设 Alex 第一次取下标为 n-1 的石子，剩下的石子堆下标从 0 ~ n-2，即 Lee 只能取偶数下标的石子堆。于是 Alex 的必胜策略是每轮取石子，取此轮奇数下标堆石子数总和，偶数下标堆石子数总和，两者大者。那么下一轮 Lee 只能取石子堆数相对少的那一堆，并且 Lee 取的石子堆下标奇偶性是完全受到上一轮 Alex 控制的。所以只要是 Alex 先手，那么每轮都可以压制 Lee，从而必胜。
+
+## 代码
+
+```go
+package leetcode
+
+func stoneGame(piles []int) bool {
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0876.Middle-of-the-Linked-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0878.Nth-Magical-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0878.Nth-Magical-Number.md b/website/content/ChapterFour/0800~0899/0878.Nth-Magical-Number.md
new file mode 100755
index 000000000..18bad9d0e
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0878.Nth-Magical-Number.md
@@ -0,0 +1,88 @@
+# [878. Nth Magical Number](https://leetcode.com/problems/nth-magical-number/)
+
+
+## 题目
+
+A positive integer is *magical* if it is divisible by either A or B.
+
+Return the N-th magical number. Since the answer may be very large, **return it modulo** `10^9 + 7`.
+
+**Example 1**:
+
+    Input: N = 1, A = 2, B = 3
+    Output: 2
+
+**Example 2**:
+
+    Input: N = 4, A = 2, B = 3
+    Output: 6
+
+**Example 3**:
+
+    Input: N = 5, A = 2, B = 4
+    Output: 10
+
+**Example 4**:
+
+    Input: N = 3, A = 6, B = 4
+    Output: 8
+
+**Note**:
+
+1. `1 <= N <= 10^9`
+2. `2 <= A <= 40000`
+3. `2 <= B <= 40000`
+
+
+## 题目大意
+
+
+如果正整数可以被 A 或 B 整除，那么它是神奇的。返回第 N 个神奇数字。由于答案可能非常大，返回它模 10^9 + 7 的结果。
+
+
+提示：
+
+1. 1 <= N <= 10^9
+2. 2 <= A <= 40000
+3. 2 <= B <= 40000
+
+
+## 解题思路
+
+
+- 给出 3 个数字，a，b，n。要求输出可以整除 a 或者整除 b 的第 n 个数。
+- 这一题是第 1201 题的缩水版，代码和解题思路也基本不变，这一题的二分搜索的区间是 `[min(A, B)，N * min(A, B)] = [2, 10 ^ 14]`。其他代码和第 1201 题一致，思路见第 1201 题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func nthMagicalNumber(N int, A int, B int) int {
+	low, high := int64(0), int64(1*1e14)
+	for low < high {
+		mid := low + (high-low)>>1
+		if calNthMagicalCount(mid, int64(A), int64(B)) < int64(N) {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return int(low) % 1000000007
+}
+
+func calNthMagicalCount(num, a, b int64) int64 {
+	ab := a * b / gcd(a, b)
+	return num/a + num/b - num/ab
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0877.Stone-Game/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0880.Decoded-String-at-Index/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0880.Decoded-String-at-Index.md b/website/content/ChapterFour/0800~0899/0880.Decoded-String-at-Index.md
new file mode 100644
index 000000000..08ca3315c
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0880.Decoded-String-at-Index.md
@@ -0,0 +1,111 @@
+# [880. Decoded String at Index](https://leetcode.com/problems/decoded-string-at-index/)
+
+## 题目
+
+An encoded string S is given.  To find and write the decoded string to a tape, the encoded string is read one character at a time and the following steps are taken:
+
+If the character read is a letter, that letter is written onto the tape.
+If the character read is a digit (say d), the entire current tape is repeatedly written d-1 more times in total.
+Now for some encoded string S, and an index K, find and return the K-th letter (1 indexed) in the decoded string.
+
+ 
+
+**Example 1**:
+
+```
+
+Input: S = "leet2code3", K = 10
+Output: "o"
+Explanation: 
+The decoded string is "leetleetcodeleetleetcodeleetleetcode".
+The 10th letter in the string is "o".
+
+```
+
+**Example 2**:
+
+```
+
+Input: S = "ha22", K = 5
+Output: "h"
+Explanation: 
+The decoded string is "hahahaha".  The 5th letter is "h".
+
+```
+
+**Example 3**:
+
+```
+
+Input: S = "a2345678999999999999999", K = 1
+Output: "a"
+Explanation: 
+The decoded string is "a" repeated 8301530446056247680 times.  The 1st letter is "a".
+
+```
+
+**Note**:
+
+1. 2 <= S.length <= 100
+2. S will only contain lowercase letters and digits 2 through 9.
+3. S starts with a letter.
+4. 1 <= K <= 10^9
+5. The decoded string is guaranteed to have less than 2^63 letters.
+
+## 题目大意
+
+给定一个编码字符串 S。为了找出解码字符串并将其写入磁带，从编码字符串中每次读取一个字符，并采取以下步骤：  
+
+- 如果所读的字符是字母，则将该字母写在磁带上。
+- 如果所读的字符是数字（例如 d），则整个当前磁带总共会被重复写 d-1 次。   
+  
+现在，对于给定的编码字符串 S 和索引 K，查找并返回解码字符串中的第 K 个字母。
+
+
+## 解题思路
+
+按照题意，扫描字符串扫到数字的时候，开始重复字符串，这里可以用递归。注意在重复字符串的时候到第 K 个字符的时候就可以返回了，不要等所有字符都扩展完成，这样会超时。d 有可能超大。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func isLetter(char byte) bool {
+	if char >= 'a' && char <= 'z' {
+		return true
+	}
+	return false
+}
+
+func decodeAtIndex(S string, K int) string {
+	length := 0
+	for i := 0; i < len(S); i++ {
+		if isLetter(S[i]) {
+			length++
+			if length == K {
+				return string(S[i])
+			}
+		} else {
+			if length*int(S[i]-'0') >= K {
+				if K%length != 0 {
+					return decodeAtIndex(S[:i], K%length)
+				}
+				return decodeAtIndex(S[:i], length)
+			}
+			length *= int(S[i] - '0')
+		}
+	}
+	return ""
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0878.Nth-Magical-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0881.Boats-to-Save-People/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0881.Boats-to-Save-People.md b/website/content/ChapterFour/0800~0899/0881.Boats-to-Save-People.md
new file mode 100644
index 000000000..ddc07e9f3
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0881.Boats-to-Save-People.md
@@ -0,0 +1,97 @@
+# [881. Boats to Save People](https://leetcode.com/problems/boats-to-save-people/)
+
+## 题目
+
+The i-th person has weight people[i], and each boat can carry a maximum weight of limit.
+
+Each boat carries at most 2 people at the same time, provided the sum of the weight of those people is at most limit.
+
+Return the minimum number of boats to carry every given person.  (It is guaranteed each person can be carried by a boat.)
+
+
+**Example 1**:
+
+```
+
+Input: people = [1,2], limit = 3
+Output: 1
+Explanation: 1 boat (1, 2)
+
+```
+
+
+**Example 2**:
+
+```
+
+Input: people = [3,2,2,1], limit = 3
+Output: 3
+Explanation: 3 boats (1, 2), (2) and (3)
+
+```
+
+
+**Example 3**:
+
+```
+
+Input: people = [3,5,3,4], limit = 5
+Output: 4
+Explanation: 4 boats (3), (3), (4), (5)
+
+```
+
+**Note**:
+
+- 1 <= people.length <= 50000
+- 1 <= people[i] <= limit <= 30000
+
+
+## 题目大意
+
+给出人的重量数组，和一个船最大载重量 limit。一个船最多装 2 个人。要求输出装下所有人，最小需要多少艘船。
+
+## 解题思路
+
+先对人的重量进行排序，然后用 2 个指针分别指向一前一后，一起计算这两个指针指向的重量之和，如果小于 limit，左指针往右移动，并且右指针往左移动。如果大于等于 limit，右指针往左移动。每次指针移动，需要船的个数都要 ++。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+func numRescueBoats(people []int, limit int) int {
+	sort.Ints(people)
+	left, right, res := 0, len(people)-1, 0
+	for left <= right {
+		if left == right {
+			res++
+			return res
+		}
+		if people[left]+people[right] <= limit {
+			left++
+			right--
+		} else {
+			right--
+		}
+		res++
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0880.Decoded-String-at-Index/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0884.Uncommon-Words-from-Two-Sentences/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0884.Uncommon-Words-from-Two-Sentences.md b/website/content/ChapterFour/0800~0899/0884.Uncommon-Words-from-Two-Sentences.md
new file mode 100755
index 000000000..40c260249
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0884.Uncommon-Words-from-Two-Sentences.md
@@ -0,0 +1,73 @@
+# [884. Uncommon Words from Two Sentences](https://leetcode.com/problems/uncommon-words-from-two-sentences/)
+
+
+## 题目
+
+We are given two sentences `A` and `B`. (A *sentence* is a string of space separated words. Each *word* consists only of lowercase letters.)
+
+A word is *uncommon* if it appears exactly once in one of the sentences, and does not appear in the other sentence.
+
+Return a list of all uncommon words.
+
+You may return the list in any order.
+
+**Example 1**:
+
+    Input: A = "this apple is sweet", B = "this apple is sour"
+    Output: ["sweet","sour"]
+
+**Example 2**:
+
+    Input: A = "apple apple", B = "banana"
+    Output: ["banana"]
+
+**Note**:
+
+1. `0 <= A.length <= 200`
+2. `0 <= B.length <= 200`
+3. `A` and `B` both contain only spaces and lowercase letters.
+
+
+## 题目大意
+
+给定两个句子 A 和 B 。（句子是一串由空格分隔的单词。每个单词仅由小写字母组成。）
+
+如果一个单词在其中一个句子中只出现一次，在另一个句子中却没有出现，那么这个单词就是不常见的。返回所有不常用单词的列表。您可以按任何顺序返回列表。
+
+
+## 解题思路
+
+- 找出 2 个句子中不同的单词，将它们俩都打印出来。简单题，先将 2 个句子的单词都拆开放入 map 中进行词频统计，不同的两个单词的词频肯定都为 1，输出它们即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func uncommonFromSentences(A string, B string) []string {
+	m, res := map[string]int{}, []string{}
+	for _, s := range []string{A, B} {
+		for _, word := range strings.Split(s, " ") {
+			m[word]++
+		}
+	}
+	for key := range m {
+		if m[key] == 1 {
+			res = append(res, key)
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0881.Boats-to-Save-People/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0885.Spiral-Matrix-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0885.Spiral-Matrix-III.md b/website/content/ChapterFour/0800~0899/0885.Spiral-Matrix-III.md
new file mode 100755
index 000000000..48444a0d4
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0885.Spiral-Matrix-III.md
@@ -0,0 +1,91 @@
+# [885. Spiral Matrix III](https://leetcode.com/problems/spiral-matrix-iii/)
+
+
+## 题目
+
+On a 2 dimensional grid with `R` rows and `C` columns, we start at `(r0, c0)` facing east.
+
+Here, the north-west corner of the grid is at the first row and column, and the south-east corner of the grid is at the last row and column.
+
+Now, we walk in a clockwise spiral shape to visit every position in this grid.
+
+Whenever we would move outside the boundary of the grid, we continue our walk outside the grid (but may return to the grid boundary later.)
+
+Eventually, we reach all `R * C` spaces of the grid.
+
+Return a list of coordinates representing the positions of the grid in the order they were visited.
+
+**Example 1**:
+
+    Input: R = 1, C = 4, r0 = 0, c0 = 0
+    Output: [[0,0],[0,1],[0,2],[0,3]]
+
+![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/08/24/example_1.png)
+
+**Example 2**:
+
+    Input: R = 5, C = 6, r0 = 1, c0 = 4
+    Output: [[1,4],[1,5],[2,5],[2,4],[2,3],[1,3],[0,3],[0,4],[0,5],[3,5],[3,4],
+    [3,3],[3,2],[2,2],[1,2],[0,2],[4,5],[4,4],[4,3],[4,2],[4,1],[3,1],[2,1],[1,1],
+    [0,1],[4,0],[3,0],[2,0],[1,0],[0,0]]
+
+![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/08/24/example_2.png)
+
+**Note**:
+
+1. `1 <= R <= 100`
+2. `1 <= C <= 100`
+3. `0 <= r0 < R`
+4. `0 <= c0 < C`
+
+
+## 题目大意
+
+在 R 行 C 列的矩阵上，我们从 (r0, c0) 面朝东面开始。这里，网格的西北角位于第一行第一列，网格的东南角位于最后一行最后一列。现在，我们以顺时针按螺旋状行走，访问此网格中的每个位置。每当我们移动到网格的边界之外时，我们会继续在网格之外行走（但稍后可能会返回到网格边界）。最终，我们到过网格的所有 R * C 个空间。
+
+要求输出按照访问顺序返回表示网格位置的坐标列表。
+
+
+## 解题思路
+
+
+- 给出一个二维数组的行 `R`，列 `C`，以及这个数组中的起始点 `(r0,c0)`。从这个起始点开始出发，螺旋的访问数组中各个点，输出途径经过的每个坐标。注意每个螺旋的步长在变长，第一个螺旋是 1 步，第二个螺旋是 1 步，第三个螺旋是 2 步，第四个螺旋是 2 步……即 1，1，2，2，3，3，4，4，5……这样的步长。
+- 这一题是第 59 题的加强版。除了有螺旋以外，还加入了步长的限制。步长其实是有规律的，第 0 次移动的步长是 `0/2+1`，第 1 次移动的步长是 `1/2+1`，第 n 次移动的步长是 `n/2+1`。其他的做法和第 59 题一致。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func spiralMatrixIII(R int, C int, r0 int, c0 int) [][]int {
+	res, round, spDir := [][]int{}, 0, [][]int{
+		[]int{0, 1},  // 朝右
+		[]int{1, 0},  // 朝下
+		[]int{0, -1}, // 朝左
+		[]int{-1, 0}, // 朝上
+	}
+	res = append(res, []int{r0, c0})
+	for i := 0; len(res) < R*C; i++ {
+		for j := 0; j < i/2+1; j++ {
+			r0 += spDir[round%4][0]
+			c0 += spDir[round%4][1]
+			if 0 <= r0 && r0 < R && 0 <= c0 && c0 < C {
+				res = append(res, []int{r0, c0})
+			}
+		}
+		round++
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0884.Uncommon-Words-from-Two-Sentences/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0887.Super-Egg-Drop/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0887.Super-Egg-Drop.md b/website/content/ChapterFour/0800~0899/0887.Super-Egg-Drop.md
new file mode 100755
index 000000000..1e1672237
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0887.Super-Egg-Drop.md
@@ -0,0 +1,152 @@
+# [887. Super Egg Drop](https://leetcode.com/problems/super-egg-drop/)
+
+
+## 题目
+
+You are given `K` eggs, and you have access to a building with `N` floors from `1` to `N`.
+
+Each egg is identical in function, and if an egg breaks, you cannot drop it again.
+
+You know that there exists a floor `F` with `0 <= F <= N` such that any egg dropped at a floor higher than `F` will break, and any egg dropped at or below floor `F` will not break.
+
+Each *move*, you may take an egg (if you have an unbroken one) and drop it from any floor `X` (with `1 <= X <= N`).
+
+Your goal is to know **with certainty** what the value of `F` is.
+
+What is the minimum number of moves that you need to know with certainty what `F` is, regardless of the initial value of `F`?
+
+**Example 1**:
+
+    Input: K = 1, N = 2
+    Output: 2
+    Explanation: 
+    Drop the egg from floor 1.  If it breaks, we know with certainty that F = 0.
+    Otherwise, drop the egg from floor 2.  If it breaks, we know with certainty that F = 1.
+    If it didn't break, then we know with certainty F = 2.
+    Hence, we needed 2 moves in the worst case to know what F is with certainty.
+
+**Example 2**:
+
+    Input: K = 2, N = 6
+    Output: 3
+
+**Example 3**:
+
+    Input: K = 3, N = 14
+    Output: 4
+
+**Note**:
+
+1. `1 <= K <= 100`
+2. `1 <= N <= 10000`
+
+
+## 题目大意
+
+你将获得 K 个鸡蛋，并可以使用一栋从 1 到 N  共有 N 层楼的建筑。每个蛋的功能都是一样的，如果一个蛋碎了，你就不能再把它掉下去。你知道存在楼层 F ，满足 0 <= F <= N 任何从高于 F 的楼层落下的鸡蛋都会碎，从 F 楼层或比它低的楼层落下的鸡蛋都不会破。每次移动，你可以取一个鸡蛋（如果你有完整的鸡蛋）并把它从任一楼层 X 扔下（满足 1 <= X <= N）。你的目标是确切地知道 F 的值是多少。无论 F 的初始值如何，你确定 F 的值的最小移动次数是多少？
+
+
+提示：
+
+1. 1 <= K <= 100
+2. 1 <= N <= 10000
+
+
+## 解题思路
+
+- 给出 `K` 个鸡蛋，`N` 层楼，要求确定安全楼层 `F` 需要最小步数 `t`。
+- 这一题是微软的经典面试题。拿到题最容易想到的是二分搜索。但是仔细分析以后会发现单纯的二分是不对的。不断的二分确实能找到最终安全的楼层，但是这里没有考虑到 `K` 个鸡蛋。鸡蛋数的限制会导致二分搜索无法找到最终楼层。题目要求要在保证能找到最终安全楼层的情况下，找到最小步数。所以单纯的二分搜索并不能解答这道题。
+- 这一题如果按照题意正向考虑，动态规划的状态转移方程是 `searchTime(K, N) = max( searchTime(K-1, X-1), searchTime(K, N-X) )`。其中 `X` 是丢鸡蛋的楼层。随着 `X` 从 `[1,N]`，都能计算出一个 `searchTime` 的值，在所有这 `N` 个值之中，取最小值就是本题的答案了。这个解法可以 AC 这道题。不过这个解法不细展开了。时间复杂度 `O(k*N^2)`。  
+{{< katex display >}} 
+dp(K,N) = MIN \begin{bmatrix}   \, \, MAX(dp(K-1,X-1),dp(K,N-X))\, \,  \end{bmatrix} ,1\leqslant x \leqslant N \\
+{{< /katex >}}   
+- 换个角度来看这个问题，定义 `dp[k][m]` 代表 `K` 个鸡蛋，`M` 次移动能检查的最大楼层。考虑某一步 `t` 应该在哪一层丢鸡蛋呢？一个正确的选择是在 `dp[k-1][t-1] + 1` 层丢鸡蛋，结果分两种情况：
+    1. 如果鸡蛋碎了，我们首先排除了该层以上的所有楼层（不管这个楼有多高），而对于剩下的 `dp[k-1][t-1]` 层楼，我们一定能用 `k-1` 个鸡蛋在 `t-1` 步内求解。因此这种情况下，我们总共可以求解无限高的楼层。可见，这是一种非常好的情况，但并不总是发生。
+    2. 如果鸡蛋没碎，我们首先排除了该层以下的 `dp[k-1][t-1]` 层楼，此时我们还有 `k` 个蛋和 `t-1` 步，那么我们去该层以上的楼层继续测得 `dp[k][t-1]` 层楼。因此这种情况下，我们总共可以求解 `dp[k-1][t-1] + 1 + dp[k][t-1]` 层楼。
+- 在所有 `m` 步中只要有一次出现了第一种情况，那么我们就可以求解无限高的楼层。但题目要求我们能保证一定能找到安全楼层，所以每次丢鸡蛋的情况应该按照最差情况来，即每次都是第二种情况。于是得到转状态转移方程: `dp[k][m] = dp[k-1][m-1] + dp[k][m-1] + 1` 。这个方程可以压缩到一维，因为每个新的状态只和上一行和左一列有关。那么每一行从右往左更新，即 `dp[i] += 1 + dp[i-1]`。时间复杂度 `O(K * log N)`，空间复杂度 `O(N)`。
+- 可能会有人有疑问，如果最初选择不在 `dp[k-1][t-1] + 1` 层丢鸡蛋会怎么样呢？选择在更低的层或者更高的层丢鸡蛋会怎样呢？
+    1. 如果在更低的楼层丢鸡蛋也能保证找到安全楼层。那么得到的结果一定不是最小步数。因为这次丢鸡蛋没有充分的展现鸡蛋和移动次数的潜力，最终求解一定会有鸡蛋和步数剩余，即不是能探测的最大楼层了。
+    2. 如果在更高的楼层丢鸡蛋，假设是第 `dp[k-1][t-1] + 2` 层丢鸡蛋，如果这次鸡蛋碎了，剩下 `k-1` 个鸡蛋和 `t-1` 步只能保证验证 `dp[k-1][t-1]` 的楼层，这里还剩**第** `dp[k-1][t-1]+ 1` 的楼层，不能保证最终一定能找到安全楼层了。
+- 用反证法就能得出每一步都应该在第 `dp[k-1][t-1] + 1` 层丢鸡蛋。
+- 这道题还可以用二分搜索来解答。回到上面分析的状态转移方程：`dp[k][m] = dp[k-1][m-1] + dp[k][m-1] + 1` 。用数学方法来解析这个递推关系。令 `f(t,k)` 为 `t` 和 `k` 的函数，题目所要求能测到最大楼层是 `N` 的最小步数，即要求出 `f(t,k) ≥ N` 时候的最小 `t`。由状态转移方程可以知道：`f(t,k) = f(t-1,k) + f(t-1,k-1) + 1`，当 `k = 1` 的时候，对应一个鸡蛋的情况，`f(t,1) = t`，当 `t = 1` 的时候，对应一步的情况，`f(1,k) = 1`。有状态转移方程得：    
+{{< katex display >}} 
+\begin{aligned} f(t,k) &= 1 + f(t-1,k-1) + f(t-1,k) \\   f(t,k-1) &= 1 + f(t-1,k-2) + f(t-1,k-1)  \\   \end{aligned}
+{{< /katex >}} 
+	令 `g(t,k) = f(t,k) - f(t,k-1)`，可以得到：  
+{{< katex display >}} 
+g(t,k) = g(t-1,k) + g(t-1,k-1)
+{{< /katex >}} 
+	可以知道 `g(t,k)` 是一个杨辉三角，即二项式系数： 
+{{< katex display >}} 
+g(t,k) = \binom{t}{k+1} = C_{t}^{k+1}
+{{< /katex >}}  
+	利用裂项相消的方法： 
+{{< katex display >}} 
+\begin{aligned} g(t,x) &= f(t,x) - f(t,x-1) \\   g(t,x-1) &=  f(t,x-1) - f(t,x-2)  \\   g(t,x-2) &=  f(t,x-2) - f(t,x-3)  \\  \begin{matrix} .\\  .\\  .\\ \end{matrix}\\ g(t,2) &=  f(t,2) - f(t,1)  \\  g(t,1) &=  f(t,1) - f(t,0)  \\  \end{aligned}
+{{< /katex >}}   
+	于是可以得到：    
+{{< katex display >}} 
+\begin{aligned} f(t,k) &= \sum_{1}^{k}g(t,x) = \sum_{0}^{k} \binom{t}{x} \\ &= C_{t}^{0} + C_{t}^{1} + C_{t}^{2} + ... + C_{t}^{k} \\   \end{aligned}
+{{< /katex >}}
+	其中：    
+{{< katex display >}} 
+\begin{aligned} C_{t}^{k} \cdot  \frac{n-k}{k+1} &= C_{t}^{k+1} \\ C_{t}^{k} &= C_{t}^{k-1} \cdot  \frac{t-k+1}{k} \\ \end{aligned}
+{{< /katex >}}
+	于是针对每一项的二项式常数，都可以由前一项乘以一个分数得到下一项。  
+{{< katex display >}} 
+\begin{aligned} C_{t}^{0} &= 1 \\   C_{t}^{1} &= C_{t}^{0} \cdot \frac{t-1+1}{1} \\    C_{t}^{2} &= C_{t}^{1} \cdot \frac{t-2+1}{2} \\  C_{t}^{3} &= C_{t}^{2} \cdot \frac{t-3+1}{3} \\  \begin{matrix} .\\  .\\  .\\ \end{matrix}\\ C_{t}^{k} &= C_{t}^{k-1} \cdot \frac{t-k+1}{k} \\ \end{aligned}
+{{< /katex >}}
+	利用二分搜索，不断的二分 `t`，直到逼近找到 `f(t,k) ≥ N` 时候最小的 `t`。时间复杂度 `O(K * log N)`，空间复杂度 `O(1)`。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 二分搜索
+func superEggDrop(K int, N int) int {
+	low, high := 1, N
+	for low < high {
+		mid := low + (high-low)>>1
+		if counterF(K, N, mid) >= N {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+// 计算二项式和，特殊的第一项 C(t,0) = 1
+func counterF(k, n, mid int) int {
+	res, sum := 1, 0
+	for i := 1; i <= k && sum < n; i++ {
+		res *= mid - i + 1
+		res /= i
+		sum += res
+	}
+	return sum
+}
+
+// 解法二 动态规划 DP
+func superEggDrop1(K int, N int) int {
+	dp, step := make([]int, K+1), 0
+	for ; dp[K] < N; step++ {
+		for i := K; i > 0; i-- {
+			dp[i] = (1 + dp[i] + dp[i-1])
+		}
+	}
+	return step
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0885.Spiral-Matrix-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0888.Fair-Candy-Swap/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0888.Fair-Candy-Swap.md b/website/content/ChapterFour/0800~0899/0888.Fair-Candy-Swap.md
new file mode 100644
index 000000000..f47316529
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0888.Fair-Candy-Swap.md
@@ -0,0 +1,117 @@
+# [888. Fair Candy Swap](https://leetcode.com/problems/fair-candy-swap/)
+
+
+## 题目
+
+Alice and Bob have candy bars of different sizes: `A[i]` is the size of the `i`-th bar of candy that Alice has, and `B[j]` is the size of the `j`-th bar of candy that Bob has.
+
+Since they are friends, they would like to exchange one candy bar each so that after the exchange, they both have the same total amount of candy. (*The total amount of candy a person has is the sum of the sizes of candy bars they have*.)
+
+Return an integer array `ans` where `ans[0]` is the size of the candy bar that Alice must exchange, and `ans[1]` is the size of the candy bar that Bob must exchange.
+
+If there are multiple answers, you may return any one of them. It is guaranteed an answer exists.
+
+**Example 1**:
+
+```
+Input: A = [1,1], B = [2,2]
+Output: [1,2]
+```
+
+**Example 2**:
+
+```
+Input: A = [1,2], B = [2,3]
+Output: [1,2]
+```
+
+**Example 3**:
+
+```
+Input: A = [2], B = [1,3]
+Output: [2,3]
+```
+
+**Example 4**:
+
+```
+Input: A = [1,2,5], B = [2,4]
+Output: [5,4]
+```
+
+**Note**:
+
+- `1 <= A.length <= 10000`
+- `1 <= B.length <= 10000`
+- `1 <= A[i] <= 100000`
+- `1 <= B[i] <= 100000`
+- It is guaranteed that Alice and Bob have different total amounts of candy.
+- It is guaranteed there exists an answer.
+
+
+## 题目大意
+
+爱丽丝和鲍勃有不同大小的糖果棒：A[i] 是爱丽丝拥有的第 i 块糖的大小，B[j] 是鲍勃拥有的第 j 块糖的大小。因为他们是朋友，所以他们想交换一个糖果棒，这样交换后，他们都有相同的糖果总量。（一个人拥有的糖果总量是他们拥有的糖果棒大小的总和。）返回一个整数数组 ans，其中 ans[0] 是爱丽丝必须交换的糖果棒的大小，ans[1] 是 Bob 必须交换的糖果棒的大小。如果有多个答案，你可以返回其中任何一个。保证答案存在。
+
+提示：
+
+- 1 <= A.length <= 10000
+- 1 <= B.length <= 10000
+- 1 <= A[i] <= 100000
+- 1 <= B[i] <= 100000
+- 保证爱丽丝与鲍勃的糖果总量不同。
+- 答案肯定存在。
+
+
+## 解题思路
+
+- 两人交换糖果，使得两人糖果相等。要求输出一个数组，里面分别包含两人必须交换的糖果大小。
+- 首先这一题肯定了一定有解，其次只允许交换一次。有了这两个前提，使本题变成简单题。先计算出为了使得交换以后两个相同的糖果数，A 需要增加或者减少的糖果数 diff。然后遍历 B ，看 A 中是否存在一个元素，能使得 B 做了对应交换 diff 以后，两人糖果相等。(此题前提保证了一定能找到)。最后输出 A 中的这个元素和遍历到 B 的这个元素，即是两人要交换的糖果数。
+
+## 代码
+
+```go
+
+package leetcode
+
+func fairCandySwap(A []int, B []int) []int {
+	hDiff, aMap := diff(A, B)/2, make(map[int]int, len(A))
+	for _, a := range A {
+		aMap[a] = a
+	}
+	for _, b := range B {
+		if a, ok := aMap[hDiff+b]; ok {
+			return []int{a, b}
+		}
+	}
+	return nil
+}
+
+func diff(A []int, B []int) int {
+	diff, maxLen := 0, max(len(A), len(B))
+	for i := 0; i < maxLen; i++ {
+		if i < len(A) {
+			diff += A[i]
+		}
+		if i < len(B) {
+			diff -= B[i]
+		}
+	}
+	return diff
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0887.Super-Egg-Drop/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0890.Find-and-Replace-Pattern/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0890.Find-and-Replace-Pattern.md b/website/content/ChapterFour/0800~0899/0890.Find-and-Replace-Pattern.md
new file mode 100644
index 000000000..1931ae16a
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0890.Find-and-Replace-Pattern.md
@@ -0,0 +1,85 @@
+# [890. Find and Replace Pattern](https://leetcode.com/problems/find-and-replace-pattern/)
+
+
+## 题目
+
+Given a list of strings `words` and a string `pattern`, return *a list of* `words[i]` *that match* `pattern`. You may return the answer in **any order**.
+
+A word matches the pattern if there exists a permutation of letters `p` so that after replacing every letter `x` in the pattern with `p(x)`, we get the desired word.
+
+Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.
+
+**Example 1:**
+
+```
+Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
+Output: ["mee","aqq"]
+Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}.
+"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, since a and b map to the same letter.
+```
+
+**Example 2:**
+
+```
+Input: words = ["a","b","c"], pattern = "a"
+Output: ["a","b","c"]
+```
+
+**Constraints:**
+
+- `1 <= pattern.length <= 20`
+- `1 <= words.length <= 50`
+- `words[i].length == pattern.length`
+- `pattern` and `words[i]` are lowercase English letters.
+
+## 题目大意
+
+你有一个单词列表 words 和一个模式  pattern，你想知道 words 中的哪些单词与模式匹配。如果存在字母的排列 p ，使得将模式中的每个字母 x 替换为 p(x) 之后，我们就得到了所需的单词，那么单词与模式是匹配的。（回想一下，字母的排列是从字母到字母的双射：每个字母映射到另一个字母，没有两个字母映射到同一个字母。）返回 words 中与给定模式匹配的单词列表。你可以按任何顺序返回答案。
+
+## 解题思路
+
+- 按照题目要求，分别映射两个字符串，words 字符串数组中的字符串与 pattern 字符串每个字母做映射。这里用 map 存储。题目还要求不存在 2 个字母映射到同一个字母的情况，所以再增加一个 map，用来判断当前字母是否已经被映射过了。以上 2 个条件都满足即代表模式匹配上了。最终将所有满足模式匹配的字符串输出即可。
+
+## 代码
+
+```go
+package leetcode
+
+func findAndReplacePattern(words []string, pattern string) []string {
+	res := make([]string, 0)
+	for _, word := range words {
+		if match(word, pattern) {
+			res = append(res, word)
+		}
+	}
+	return res
+}
+
+func match(w, p string) bool {
+	if len(w) != len(p) {
+		return false
+	}
+	m, used := make(map[uint8]uint8), make(map[uint8]bool)
+	for i := 0; i < len(w); i++ {
+		if v, ok := m[p[i]]; ok {
+			if w[i] != v {
+				return false
+			}
+		} else {
+			if used[w[i]] {
+				return false
+			}
+			m[p[i]] = w[i]
+			used[w[i]] = true
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0888.Fair-Candy-Swap/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0891.Sum-of-Subsequence-Widths/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0891.Sum-of-Subsequence-Widths.md b/website/content/ChapterFour/0800~0899/0891.Sum-of-Subsequence-Widths.md
new file mode 100644
index 000000000..a0baf573d
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0891.Sum-of-Subsequence-Widths.md
@@ -0,0 +1,75 @@
+# [891. Sum of Subsequence Widths](https://leetcode.com/problems/sum-of-subsequence-widths/)
+
+## 题目
+
+Given an array of integers A, consider all non-empty subsequences of A.
+
+For any sequence S, let the width of S be the difference between the maximum and minimum element of S.
+
+Return the sum of the widths of all subsequences of A. 
+
+As the answer may be very large, return the answer modulo 10^9 + 7.
+
+ 
+
+**Example 1**:
+
+```
+
+Input: [2,1,3]
+Output: 6
+Explanation:
+Subsequences are [1], [2], [3], [2,1], [2,3], [1,3], [2,1,3].
+The corresponding widths are 0, 0, 0, 1, 1, 2, 2.
+The sum of these widths is 6.
+
+```
+
+**Note**:
+
+- 1 <= A.length <= 20000
+- 1 <= A[i] <= 20000
+
+
+## 题目大意
+
+给定一个整数数组 A ，考虑 A 的所有非空子序列。对于任意序列 S ，设 S 的宽度是 S 的最大元素和最小元素的差。返回 A 的所有子序列的宽度之和。由于答案可能非常大，请返回答案模 10^9+7。
+
+
+## 解题思路
+
+- 理解题意以后，可以发现，数组内元素的顺序并不影响最终求得的所有子序列的宽度之和。 
+ 
+		[2,1,3]:[1],[2],[3],[2,1],[2,3],[1,3],[2,1,3]
+		[1,2,3]:[1],[2],[3],[1,2],[2,3],[1,3],[1,2,3]
+	针对每个 A[i] 而言，A[i] 对最终结果的贡献是在子序列的左右两边的时候才有贡献，当 A[i] 位于区间中间的时候，不影响最终结果。先对 A[i] 进行排序，排序以后，有 i 个数 <= A[i]，有 n - i - 1 个数 >= A[i]。所以 A[i] 会在 2^i 个子序列的右边界出现，2^(n-i-1) 个左边界出现。那么 A[i] 对最终结果的贡献是 A[i] * 2^i - A[i] * 2^(n-i-1) 。举个例子，[1,4,5,7]，A[2] = 5，那么 5 作为右边界的子序列有 2^2 = 4 个，即 [5],[1,5],[4,5],[1,4,5]，5 作为左边界的子序列有 2^(4-2-1) = 2 个，即 [5],[5,7]。A[2] = 5 对最终结果的影响是 5 * 2^2 - 5 * 2^(4-2-1) = 10 。
+- 题目要求所有子序列的宽度之和，也就是求每个区间最大值减去最小值的总和。那么 `Ans = SUM{ A[i]*2^i - A[n-i-1] * 2^(n-i-1) }`，其中 `0 <= i < n`。需要注意的是 2^i 可能非常大，所以在计算中就需要去 mod 了，而不是最后计算完了再 mod。注意取模的结合律：`(a * b) % c = (a % c) * (b % c) % c`。
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+func sumSubseqWidths(A []int) int {
+	sort.Ints(A)
+	res, mod, n, p := 0, 1000000007, len(A), 1
+	for i := 0; i < n; i++ {
+		res = (res + (A[i]-A[n-1-i])*p) % mod
+		p = (p << 1) % mod
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0890.Find-and-Replace-Pattern/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0892.Surface-Area-of-3D-Shapes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0892.Surface-Area-of-3D-Shapes.md b/website/content/ChapterFour/0800~0899/0892.Surface-Area-of-3D-Shapes.md
new file mode 100644
index 000000000..9fa02621f
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0892.Surface-Area-of-3D-Shapes.md
@@ -0,0 +1,115 @@
+# [892. Surface Area of 3D Shapes](https://leetcode.com/problems/surface-area-of-3d-shapes/)
+
+
+## 题目
+
+On a `N * N` grid, we place some `1 * 1 * 1` cubes.
+
+Each value `v = grid[i][j]` represents a tower of `v` cubes placed on top of grid cell `(i, j)`.
+
+Return the total surface area of the resulting shapes.
+
+**Example 1**:
+
+```
+Input: [[2]]
+Output: 10
+```
+
+**Example 2**:
+
+```
+Input: [[1,2],[3,4]]
+Output: 34
+```
+
+**Example 3**:
+
+```
+Input: [[1,0],[0,2]]
+Output: 16
+```
+
+**Example 4**:
+
+```
+Input: [[1,1,1],[1,0,1],[1,1,1]]
+Output: 32
+```
+
+**Example 5**:
+
+```
+Input: [[2,2,2],[2,1,2],[2,2,2]]
+Output: 46
+```
+
+**Note**:
+
+- `1 <= N <= 50`
+- `0 <= grid[i][j] <= 50`
+
+## 题目大意
+
+在 N * N 的网格上，我们放置一些 1 * 1 * 1  的立方体。每个值 v = grid[i][j] 表示 v 个正方体叠放在对应单元格 (i, j) 上。请你返回最终形体的表面积。
+
+
+## 解题思路
+
+- 给定一个网格数组，数组里面装的是立方体叠放在所在的单元格，求最终这些叠放的立方体的表面积。
+- 简单题。按照题目意思，找到叠放时，重叠的面，然后用总表面积减去这些重叠的面积即为最终答案。
+
+## 代码
+
+```go
+
+package leetcode
+
+func surfaceArea(grid [][]int) int {
+	area := 0
+	for i := 0; i < len(grid); i++ {
+		for j := 0; j < len(grid[0]); j++ {
+			if grid[i][j] == 0 {
+				continue
+			}
+			area += grid[i][j]*4 + 2
+			// up
+			if i > 0 {
+				m := min(grid[i][j], grid[i-1][j])
+				area -= m
+			}
+			// down
+			if i < len(grid)-1 {
+				m := min(grid[i][j], grid[i+1][j])
+				area -= m
+			}
+			// left
+			if j > 0 {
+				m := min(grid[i][j], grid[i][j-1])
+				area -= m
+			}
+			// right
+			if j < len(grid[i])-1 {
+				m := min(grid[i][j], grid[i][j+1])
+				area -= m
+			}
+		}
+	}
+	return area
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0891.Sum-of-Subsequence-Widths/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0895.Maximum-Frequency-Stack/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0895.Maximum-Frequency-Stack.md b/website/content/ChapterFour/0800~0899/0895.Maximum-Frequency-Stack.md
new file mode 100644
index 000000000..069117500
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0895.Maximum-Frequency-Stack.md
@@ -0,0 +1,120 @@
+# [895. Maximum Frequency Stack](https://leetcode.com/problems/maximum-frequency-stack/)
+
+## 题目
+
+Implement FreqStack, a class which simulates the operation of a stack-like data structure.
+
+FreqStack has two functions:
+
+push(int x), which pushes an integer x onto the stack.
+pop(), which removes and returns the most frequent element in the stack.  
+If there is a tie for most frequent element, the element closest to the top of the stack is removed and returned.
+ 
+
+**Example 1**:
+
+```
+
+Input: 
+["FreqStack","push","push","push","push","push","push","pop","pop","pop","pop"],
+[[],[5],[7],[5],[7],[4],[5],[],[],[],[]]
+Output: [null,null,null,null,null,null,null,5,7,5,4]
+Explanation:
+After making six .push operations, the stack is [5,7,5,7,4,5] from bottom to top.  Then:
+
+pop() -> returns 5, as 5 is the most frequent.
+The stack becomes [5,7,5,7,4].
+
+pop() -> returns 7, as 5 and 7 is the most frequent, but 7 is closest to the top.
+The stack becomes [5,7,5,4].
+
+pop() -> returns 5.
+The stack becomes [5,7,4].
+
+pop() -> returns 4.
+The stack becomes [5,7].
+
+```
+
+**Note**:
+
+- Calls to FreqStack.push(int x) will be such that 0 <= x <= 10^9.
+- It is guaranteed that FreqStack.pop() won't be called if the stack has zero elements.
+- The total number of FreqStack.push calls will not exceed 10000 in a single test case.
+- The total number of FreqStack.pop calls will not exceed 10000 in a single test case.
+- The total number of FreqStack.push and FreqStack.pop calls will not exceed 150000 across all test cases.
+
+## 题目大意
+
+实现 FreqStack，模拟类似栈的数据结构的操作的一个类。
+
+FreqStack 有两个函数：
+
+- push(int x)，将整数 x 推入栈中。
+- pop()，它移除并返回栈中出现最频繁的元素。如果最频繁的元素不只一个，则移除并返回最接近栈顶的元素。
+
+
+## 解题思路
+
+FreqStack 里面保存频次的 map 和相同频次 group 的 map。push 的时候动态的维护 x 的频次，并更新到对应频次的 group 中。pop 的时候对应减少频次字典里面的频次，并更新到对应频次的 group 中。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+type FreqStack struct {
+	freq    map[int]int
+	group   map[int][]int
+	maxfreq int
+}
+
+func Constructor895() FreqStack {
+	hash := make(map[int]int)
+	maxHash := make(map[int][]int)
+	return FreqStack{freq: hash, group: maxHash}
+}
+
+func (this *FreqStack) Push(x int) {
+	if _, ok := this.freq[x]; ok {
+		this.freq[x]++
+	} else {
+		this.freq[x] = 1
+	}
+	f := this.freq[x]
+	if f > this.maxfreq {
+		this.maxfreq = f
+	}
+
+	this.group[f] = append(this.group[f], x)
+}
+
+func (this *FreqStack) Pop() int {
+	tmp := this.group[this.maxfreq]
+	x := tmp[len(tmp)-1]
+	this.group[this.maxfreq] = this.group[this.maxfreq][:len(this.group[this.maxfreq])-1]
+	this.freq[x]--
+	if len(this.group[this.maxfreq]) == 0 {
+		this.maxfreq--
+	}
+	return x
+}
+
+/**
+ * Your FreqStack object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.Push(x);
+ * param_2 := obj.Pop();
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0892.Surface-Area-of-3D-Shapes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0896.Monotonic-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0896.Monotonic-Array.md b/website/content/ChapterFour/0800~0899/0896.Monotonic-Array.md
new file mode 100644
index 000000000..931017500
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0896.Monotonic-Array.md
@@ -0,0 +1,106 @@
+# [896. Monotonic Array](https://leetcode.com/problems/monotonic-array/)
+
+
+## 题目
+
+An array is *monotonic* if it is either monotone increasing or monotone decreasing.
+
+An array `A` is monotone increasing if for all `i <= j`, `A[i] <= A[j]`. An array `A` is monotone decreasing if for all `i <= j`, `A[i] >= A[j]`.
+
+Return `true` if and only if the given array `A` is monotonic.
+
+**Example 1**:
+
+```
+Input: [1,2,2,3]
+Output: true
+```
+
+**Example 2**:
+
+```
+Input: [6,5,4,4]
+Output: true
+```
+
+**Example 3**:
+
+```
+Input: [1,3,2]
+Output: false
+```
+
+**Example 4**:
+
+```
+Input: [1,2,4,5]
+Output: true
+```
+
+**Example 5**:
+
+```
+Input: [1,1,1]
+Output: true
+```
+
+**Note**:
+
+1. `1 <= A.length <= 50000`
+2. `-100000 <= A[i] <= 100000`
+
+## 题目大意
+
+如果数组是单调递增或单调递减的，那么它是单调的。如果对于所有 i <= j，A[i] <= A[j]，那么数组 A 是单调递增的。 如果对于所有 i <= j，A[i]> = A[j]，那么数组 A 是单调递减的。当给定的数组 A 是单调数组时返回 true，否则返回 false。
+
+
+## 解题思路
+
+- 判断给定的数组是不是单调(单调递增或者单调递减)的。
+- 简单题，按照题意循环判断即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func isMonotonic(A []int) bool {
+	if len(A) <= 1 {
+		return true
+	}
+	if A[0] < A[1] {
+		return inc(A[1:])
+	}
+	if A[0] > A[1] {
+		return dec(A[1:])
+	}
+	return inc(A[1:]) || dec(A[1:])
+}
+
+func inc(A []int) bool {
+	for i := 0; i < len(A)-1; i++ {
+		if A[i] > A[i+1] {
+			return false
+		}
+	}
+	return true
+}
+
+func dec(A []int) bool {
+	for i := 0; i < len(A)-1; i++ {
+		if A[i] < A[i+1] {
+			return false
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0895.Maximum-Frequency-Stack/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0897.Increasing-Order-Search-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0897.Increasing-Order-Search-Tree.md b/website/content/ChapterFour/0800~0899/0897.Increasing-Order-Search-Tree.md
new file mode 100755
index 000000000..cfac7507d
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0897.Increasing-Order-Search-Tree.md
@@ -0,0 +1,130 @@
+# [897. Increasing Order Search Tree](https://leetcode.com/problems/increasing-order-search-tree/)
+
+
+## 题目
+
+Given a binary search tree, rearrange the tree in **in-order** so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.
+
+**Example 1**:
+
+    Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]
+    
+           5
+          / \
+        3    6
+       / \    \
+      2   4    8
+     /        / \ 
+    1        7   9
+    
+    Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
+    
+     1
+      \
+       2
+        \
+         3
+          \
+           4
+            \
+             5
+              \
+               6
+                \
+                 7
+                  \
+                   8
+                    \
+                     9
+
+**Note**:
+
+1. The number of nodes in the given tree will be between 1 and 100.
+2. Each node will have a unique integer value from 0 to 1000.
+
+
+## 题目大意
+
+给定一个树，按中序遍历重新排列树，使树中最左边的结点现在是树的根，并且每个结点没有左子结点，只有一个右子结点。
+
+
+提示：
+
+- 给定树中的结点数介于 1 和 100 之间。
+- 每个结点都有一个从 0 到 1000 范围内的唯一整数值。
+
+
+## 解题思路
+
+- 给出一颗树，要求把树的所有孩子都排列到右子树上。
+- 按照题意，可以先中根遍历原树，然后按照中根遍历的顺序，把所有节点都放在右子树上。见解法二。
+- 上一种解法会重新构造一颗新树，有没有办法可以直接更改原有的树呢？节约存储空间。虽然平时软件开发过程中不建议更改原有的值，但是算法题中追求空间和时间的最优，可以考虑一下。**树可以看做是有多个孩子的链表**。这一题可以看成是链表的类似反转的操作。这一点想通以后，就好做了。先找到左子树中最左边的节点，这个节点是新树的根节点。然后依次往回遍历，不断的记录下上一次遍历的最后节点 tail，边遍历，边把后续节点串起来。最终“反转”完成以后，就得到了题目要求的样子了。代码实现见解法一。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 链表思想
+func increasingBST(root *TreeNode) *TreeNode {
+	var head = &TreeNode{}
+	tail := head
+	recBST(root, tail)
+	return head.Right
+}
+
+func recBST(root, tail *TreeNode) *TreeNode {
+	if root == nil {
+		return tail
+	}
+	tail = recBST(root.Left, tail)
+	root.Left = nil               // 切断 root 与其 Left 的连接，避免形成环
+	tail.Right, tail = root, root // 把 root 接上 tail，并保持 tail 指向尾部
+	tail = recBST(root.Right, tail)
+	return tail
+}
+
+// 解法二 模拟
+func increasingBST1(root *TreeNode) *TreeNode {
+	list := []int{}
+	inorder(root, &list)
+	if len(list) == 0 {
+		return root
+	}
+	newRoot := &TreeNode{Val: list[0], Left: nil, Right: nil}
+	cur := newRoot
+	for index := 1; index < len(list); index++ {
+		tmp := &TreeNode{Val: list[index], Left: nil, Right: nil}
+		cur.Right = tmp
+		cur = tmp
+	}
+	return newRoot
+}
+
+func inorder(root *TreeNode, output *[]int) {
+	if root != nil {
+		inorder(root.Left, output)
+		*output = append(*output, root.Val)
+		inorder(root.Right, output)
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0896.Monotonic-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0898.Bitwise-ORs-of-Subarrays/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/0898.Bitwise-ORs-of-Subarrays.md b/website/content/ChapterFour/0800~0899/0898.Bitwise-ORs-of-Subarrays.md
new file mode 100755
index 000000000..bf494c573
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/0898.Bitwise-ORs-of-Subarrays.md
@@ -0,0 +1,150 @@
+# [898. Bitwise ORs of Subarrays](https://leetcode.com/problems/bitwise-ors-of-subarrays/)
+
+
+## 题目
+
+We have an array `A` of non-negative integers.
+
+For every (contiguous) subarray `B = [A[i], A[i+1], ..., A[j]]` (with `i <= j`), we take the bitwise OR of all the elements in `B`, obtaining a result `A[i] | A[i+1] | ... | A[j]`.
+
+Return the number of possible results. (Results that occur more than once are only counted once in the final answer.)
+
+**Example 1**:
+
+    Input: [0]
+    Output: 1
+    Explanation: 
+    There is only one possible result: 0.
+
+**Example 2**:
+
+    Input: [1,1,2]
+    Output: 3
+    Explanation: 
+    The possible subarrays are [1], [1], [2], [1, 1], [1, 2], [1, 1, 2].
+    These yield the results 1, 1, 2, 1, 3, 3.
+    There are 3 unique values, so the answer is 3.
+
+**Example 3**:
+
+    Input: [1,2,4]
+    Output: 6
+    Explanation: 
+    The possible results are 1, 2, 3, 4, 6, and 7.
+
+**Note**:
+
+1. `1 <= A.length <= 50000`
+2. `0 <= A[i] <= 10^9`
+
+
+## 题目大意
+
+我们有一个非负整数数组 A。对于每个（连续的）子数组 B = [A[i], A[i+1], ..., A[j]] （ i <= j），我们对 B 中的每个元素进行按位或操作，获得结果 A[i] | A[i+1] | ... | A[j]。返回可能结果的数量。（多次出现的结果在最终答案中仅计算一次。）
+
+
+
+## 解题思路
+
+- 给出一个数组，要求求出这个数组所有的子数组中，每个集合内所有数字取 `|` 运算以后，不同结果的种类数。
+- 这道题可以这样考虑，第一步，先考虑所有的子数组如何得到，以 `[001, 011, 100, 110, 101]` 为例，所有的子数组集合如下：
+
+```c
+    [001]  
+    [001 011] [011]  
+    [001 011 100] [011 100] [100]  
+    [001 011 100 110] [011 100 110] [100 110] [110]  
+    [001 011 100 110 101] [011 100 110 101] [100 110 101] [110 101] [101]  
+```
+
+可以发现，从左往右遍历原数组，每次新来的一个元素，依次加入到之前已经生成过的集合中，再以自己为单独集合。这样就可以生成原数组的所有子集。
+
+- 第二步，将每一行的子集内的所有元素都进行 `|` 运算，得到：
+
+```c
+    001  
+    011 011  
+    111 111 100  
+    111 111 110 110  
+    111 111 111 111 101  
+```
+
+- 第三步，去重：
+
+```c
+    001  
+    011  
+    111 100  
+    111 110  
+    111 101  
+```
+
+由于二进制位不超过 32 位，所以这里每一行最多不会超过 32 个数。所以最终时间复杂度不会超过 O(32 N)，即 O(K * N)。最后将这每一行的数字都放入最终的 map 中去重即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 array 优化版
+func subarrayBitwiseORs(A []int) int {
+	res, cur, isInMap := []int{}, []int{}, make(map[int]bool)
+	cur = append(cur, 0)
+	for _, v := range A {
+		var cur2 []int
+		for _, vv := range cur {
+			tmp := v | vv
+			if !inSlice(cur2, tmp) {
+				cur2 = append(cur2, tmp)
+			}
+		}
+		if !inSlice(cur2, v) {
+			cur2 = append(cur2, v)
+		}
+		cur = cur2
+		for _, vv := range cur {
+			if _, ok := isInMap[vv]; !ok {
+				isInMap[vv] = true
+				res = append(res, vv)
+			}
+		}
+	}
+	return len(res)
+}
+
+func inSlice(A []int, T int) bool {
+	for _, v := range A {
+		if v == T {
+			return true
+		}
+	}
+	return false
+}
+
+// 解法二 map 版
+func subarrayBitwiseORs1(A []int) int {
+	res, t := map[int]bool{}, map[int]bool{}
+	for _, num := range A {
+		r := map[int]bool{}
+		r[num] = true
+		for n := range t {
+			r[(num | n)] = true
+		}
+		t = r
+		for n := range t {
+			res[n] = true
+		}
+	}
+	return len(res)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0897.Increasing-Order-Search-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0901.Online-Stock-Span/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0800~0899/_index.md b/website/content/ChapterFour/0800~0899/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/0800~0899/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/0900~0999/0901.Online-Stock-Span.md b/website/content/ChapterFour/0900~0999/0901.Online-Stock-Span.md
new file mode 100644
index 000000000..1ba0fcd96
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0901.Online-Stock-Span.md
@@ -0,0 +1,118 @@
+# [901. Online Stock Span](https://leetcode.com/problems/online-stock-span/)
+
+## 题目
+
+Write a class StockSpanner which collects daily price quotes for some stock, and returns the span of that stock's price for the current day.
+
+The span of the stock's price today is defined as the maximum number of consecutive days (starting from today and going backwards) for which the price of the stock was less than or equal to today's price.
+
+For example, if the price of a stock over the next 7 days were [100, 80, 60, 70, 60, 75, 85], then the stock spans would be [1, 1, 1, 2, 1, 4, 6].
+
+ 
+
+**Example 1**:
+
+```
+
+Input: ["StockSpanner","next","next","next","next","next","next","next"], [[],[100],[80],[60],[70],[60],[75],[85]]
+Output: [null,1,1,1,2,1,4,6]
+Explanation: 
+First, S = StockSpanner() is initialized.  Then:
+S.next(100) is called and returns 1,
+S.next(80) is called and returns 1,
+S.next(60) is called and returns 1,
+S.next(70) is called and returns 2,
+S.next(60) is called and returns 1,
+S.next(75) is called and returns 4,
+S.next(85) is called and returns 6.
+
+Note that (for example) S.next(75) returned 4, because the last 4 prices
+(including today's price of 75) were less than or equal to today's price.
+
+```
+
+**Note**:
+
+1. Calls to StockSpanner.next(int price) will have 1 <= price <= 10^5.
+2. There will be at most 10000 calls to StockSpanner.next per test case.
+3. There will be at most 150000 calls to StockSpanner.next across all test cases.
+4. The total time limit for this problem has been reduced by 75% for C++, and 50% for all other languages.
+
+## 题目大意
+
+编写一个 StockSpanner 类，它收集某些股票的每日报价，并返回该股票当日价格的跨度。
+
+今天股票价格的跨度被定义为股票价格小于或等于今天价格的最大连续日数（从今天开始往回数，包括今天）。
+
+例如，如果未来7天股票的价格是 [100, 80, 60, 70, 60, 75, 85]，那么股票跨度将是 [1, 1, 1, 2, 1, 4, 6]。
+
+
+
+## 解题思路
+
+这一题就是单调栈的题目。维护一个单调递增的下标。
+
+## 总结
+
+单调栈类似的题
+
+496. Next Greater Element I
+503. Next Greater Element II
+739. Daily Temperatures
+907. Sum of Subarray Minimums
+84. Largest Rectangle in Histogram
+
+## 代码
+
+```go
+
+package leetcode
+
+import "fmt"
+
+// node pair
+type Node struct {
+	Val int
+	res int
+}
+
+// slice
+type StockSpanner struct {
+	Item []Node
+}
+
+func Constructor901() StockSpanner {
+	stockSpanner := StockSpanner{make([]Node, 0)}
+	return stockSpanner
+}
+
+// need refactor later
+func (this *StockSpanner) Next(price int) int {
+	res := 1
+	if len(this.Item) == 0 {
+		this.Item = append(this.Item, Node{price, res})
+		return res
+	}
+	for len(this.Item) > 0 && this.Item[len(this.Item)-1].Val <= price {
+		res = res + this.Item[len(this.Item)-1].res
+		this.Item = this.Item[:len(this.Item)-1]
+	}
+	this.Item = append(this.Item, Node{price, res})
+	fmt.Printf("this.Item = %v\n", this.Item)
+	return res
+}
+
+/**
+ * Your StockSpanner object will be instantiated and called as such:
+ * obj := Constructor();
+ * param_1 := obj.Next(price);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0800~0899/0898.Bitwise-ORs-of-Subarrays/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0904.Fruit-Into-Baskets/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0904.Fruit-Into-Baskets.md b/website/content/ChapterFour/0900~0999/0904.Fruit-Into-Baskets.md
new file mode 100644
index 000000000..7f155d1b3
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0904.Fruit-Into-Baskets.md
@@ -0,0 +1,122 @@
+# [904. Fruit Into Baskets](https://leetcode.com/problems/fruit-into-baskets/)
+
+## 题目
+
+In a row of trees, the i-th tree produces fruit with type tree[i].
+
+You start at any tree of your choice, then repeatedly perform the following steps:
+
+1. Add one piece of fruit from this tree to your baskets.  If you cannot, stop.
+2. Move to the next tree to the right of the current tree.  If there is no tree to the right, stop.
+
+Note that you do not have any choice after the initial choice of starting tree: you must perform step 1, then step 2, then back to step 1, then step 2, and so on until you stop.
+
+You have two baskets, and each basket can carry any quantity of fruit, but you want each basket to only carry one type of fruit each.
+
+What is the total amount of fruit you can collect with this procedure?
+
+
+**Example 1**:
+
+```
+Input: [1,2,1]
+Output: 3
+Explanation: We can collect [1,2,1].
+
+```
+
+**Example 2**:
+
+```
+
+Input: [0,1,2,2]
+Output: 3
+Explanation: We can collect [1,2,2].
+If we started at the first tree, we would only collect [0, 1].
+
+```
+
+**Example 3**:
+
+```
+
+Input: [1,2,3,2,2]
+Output: 4
+Explanation: We can collect [2,3,2,2].
+If we started at the first tree, we would only collect [1, 2].
+
+```
+
+**Example 4**:
+
+```
+
+Input: [3,3,3,1,2,1,1,2,3,3,4]
+Output: 5
+Explanation: We can collect [1,2,1,1,2].
+If we started at the first tree or the eighth tree, we would only collect 4 fruits.
+
+```
+
+**Note**:
+
+- 1 <= tree.length <= 40000
+- 0 <= tree[i] < tree.length
+
+## 题目大意
+
+这道题考察的是滑动窗口的问题。
+
+给出一个数组，数组里面的数字代表每个果树上水果的种类，1 代表一号水果，不同数字代表的水果不同。现在有 2 个篮子，每个篮子只能装一个种类的水果，这就意味着只能选 2 个不同的数字。摘水果只能从左往右摘，直到右边没有水果可以摘就停下。问可以连续摘水果的最长区间段的长度。
+
+
+## 解题思路
+
+简化一下题意，给出一段数字，要求找出包含 2 个不同数字的最大区间段长度。这个区间段内只能包含这 2 个不同数字，可以重复，但是不能包含其他数字。
+
+用典型的滑动窗口的处理方法处理即可。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func totalFruit(tree []int) int {
+	if len(tree) == 0 {
+		return 0
+	}
+	left, right, counter, res, freq := 0, 0, 1, 1, map[int]int{}
+	freq[tree[0]]++
+	for left < len(tree) {
+		if right+1 < len(tree) && ((counter > 0 && tree[right+1] != tree[left]) || (tree[right+1] == tree[left] || freq[tree[right+1]] > 0)) {
+			if counter > 0 && tree[right+1] != tree[left] {
+				counter--
+			}
+			right++
+			freq[tree[right]]++
+		} else {
+			if counter == 0 || (counter > 0 && right == len(tree)-1) {
+				res = max(res, right-left+1)
+			}
+			freq[tree[left]]--
+			if freq[tree[left]] == 0 {
+				counter++
+			}
+			left++
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0901.Online-Stock-Span/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0907.Sum-of-Subarray-Minimums/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0907.Sum-of-Subarray-Minimums.md b/website/content/ChapterFour/0900~0999/0907.Sum-of-Subarray-Minimums.md
new file mode 100644
index 000000000..1fd0096d1
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0907.Sum-of-Subarray-Minimums.md
@@ -0,0 +1,131 @@
+# [907. Sum of Subarray Minimums](https://leetcode.com/problems/sum-of-subarray-minimums/)
+
+## 题目
+
+Given an array of integers A, find the sum of min(B), where B ranges over every (contiguous) subarray of A.
+
+Since the answer may be large, return the answer modulo 10^9 + 7.
+
+ 
+
+**Example 1**:
+
+```
+
+Input: [3,1,2,4]
+Output: 17
+Explanation: Subarrays are [3], [1], [2], [4], [3,1], [1,2], [2,4], [3,1,2], [1,2,4], [3,1,2,4]. 
+Minimums are 3, 1, 2, 4, 1, 1, 2, 1, 1, 1.  Sum is 17.
+
+```
+
+**Note**:
+
+1. 1 <= A.length <= 30000
+2. 1 <= A[i] <= 30000
+
+
+## 题目大意
+
+给定一个整数数组 A，找到 min(B) 的总和，其中 B 的范围为 A 的每个（连续）子数组。
+
+由于答案可能很大，因此返回答案模 10^9 + 7。
+
+
+## 解题思路
+
+- 首先想到的是暴力解法，用两层循环，分别枚举每个连续的子区间，区间内用一个元素记录区间内最小值。每当区间起点发生变化的时候，最终结果都加上上次遍历区间找出的最小值。当整个数组都扫完一遍以后，最终结果模上 10^9+7。
+- 上面暴力解法时间复杂度特别大，因为某个区间的最小值可能是很多区间的最小值，但是我们暴力枚举所有区间，导致要遍历的区间特别多。优化点就在如何减少遍历的区间。第二种思路是用 2 个单调栈。想得到思路是 `res = sum(A[i] * f(i))`，其中 f(i) 是子区间的数，A[i] 是这个子区间内的最小值。为了得到 f(i) 我们需要找到 left[i] 和 right[i]，left[i] 是 A[i] 左边严格大于 A[i](> 关系)的区间长度。right[i] 是 A[i] 右边非严格大于(>= 关系)的区间长度。left[i] + 1 等于以 A[i] 结尾的子数组数目，A[i] 是唯一的最小值；right[i] + 1 等于以 A[i] 开始的子数组数目，A[i] 是第一个最小值。于是有 `f(i) = (left[i] + 1) * (right[i] + 1)`。例如对于 [3,1,4,2,5,3,3,1] 中的“2”，我们找到的串就为[4,2,5,3,3]，2 左边有 1 个数比 2 大且相邻，2 右边有 3 个数比 2 大且相邻，所以 2 作为最小值的串有 2 * 4 = 8 种。用排列组合的思维也能分析出来，2 的左边可以拿 0，1，…… m 个，总共 (m + 1) 种，同理右边可以拿 0，1，…… n 个，总共 (n + 1) 种，所以总共 (m + 1)(n + 1)种。只要计算出了 f(i)，这个题目就好办了。以 [3,1,2,4] 为例，left[i] + 1 = [1,2,1,1]，right[i] + 1 = [1,3,2,1]，对应 i 位的乘积是 f[i] = [1 * 1，2 * 3，1 * 2，1 * 1] = [1，6，2，1]，最终要求的最小值的总和 res = 3 * 1 + 1 * 6 + 2 * 2 + 4 * 1 = 17。
+- **看到这种 mod1e9+7 的题目，首先要想到的就是dp**。最终的优化解即是利用 DP + 单调栈。单调栈维护数组中的值逐渐递增的对应下标序列。定义 `dp[i + 1]` 代表以 A[i] 结尾的子区间内最小值的总和。状态转移方程是 `dp[i + 1] = dp[prev + 1] + (i - prev) * A[i]`，其中 prev 是比 A[i] 小的前一个数，由于我们维护了一个单调栈，所以 prev 就是栈顶元素。(i - prev) * A[i] 代表在还没有出现 prev 之前，这些区间内都是 A[i] 最小，那么这些区间有 i - prev 个，所以最小值总和应该是 (i - prev) * A[i]。再加上 dp[prev + 1] 就是 dp[i + 1] 的最小值总和了。以 [3, 1, 2, 4, 3] 为例，当 i = 4, 所有以 A[4] 为结尾的子区间有:  
+	
+		[3]  
+		[4, 3]  
+		[2, 4, 3]  
+		[1, 2, 4, 3]  
+		[3, 1, 2, 4, 3] 
+	在这种情况下, stack.peek() = 2, A[2] = 2。前两个子区间 [3] and [4, 3], 最小值的总和 = (i - stack.peek()) * A[i] = 6。后 3 个子区间是 [2, 4, 3], [1, 2, 4, 3] 和 [3, 1, 2, 4, 3], 它们都包含 2，2 是比 3 小的前一个数，所以 dp[i + 1] = dp[stack.peek() + 1] = dp[2 + 1] = dp[3] = dp[2 + 1]。即需要求 i = 2 的时候 dp[i + 1] 的值。继续递推，比 2 小的前一个值是 1，A[1] = 1。dp[3] = dp[1 + 1] + (2 - 1) * A[2]= dp[2] + 2。dp[2] = dp[1 + 1]，当 i = 1 的时候，prev = -1，即没有人比 A[1] 更小了，所以 dp[2] = dp[1 + 1] = dp[-1 + 1] + (1 - (-1)) * A[1] = 0 + 2 * 1 = 2。迭代回去，dp[3] = dp[2] + 2 = 2 + 2 = 4。dp[stack.peek() + 1] = dp[2 + 1] = dp[3] = 4。所以 dp[i + 1] = 4 + 6 = 10。
+- 与这一题相似的解题思路的题目有第 828 题，第 891 题。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 最快的解是 DP + 单调栈
+func sumSubarrayMins(A []int) int {
+	stack, dp, res, mod := []int{}, make([]int, len(A)+1), 0, 1000000007
+	stack = append(stack, -1)
+
+	for i := 0; i < len(A); i++ {
+		for stack[len(stack)-1] != -1 && A[i] <= A[stack[len(stack)-1]] {
+			stack = stack[:len(stack)-1]
+		}
+		dp[i+1] = (dp[stack[len(stack)-1]+1] + (i-stack[len(stack)-1])*A[i]) % mod
+		stack = append(stack, i)
+		res += dp[i+1]
+		res %= mod
+	}
+	return res
+}
+
+type pair struct {
+	val   int
+	count int
+}
+
+// 解法二 用两个单调栈
+func sumSubarrayMins1(A []int) int {
+	res, n, mod := 0, len(A), 1000000007
+	lefts, rights, leftStack, rightStack := make([]int, n), make([]int, n), []*pair{}, []*pair{}
+	for i := 0; i < n; i++ {
+		count := 1
+		for len(leftStack) != 0 && leftStack[len(leftStack)-1].val > A[i] {
+			count += leftStack[len(leftStack)-1].count
+			leftStack = leftStack[:len(leftStack)-1]
+		}
+		leftStack = append(leftStack, &pair{val: A[i], count: count})
+		lefts[i] = count
+	}
+
+	for i := n - 1; i >= 0; i-- {
+		count := 1
+		for len(rightStack) != 0 && rightStack[len(rightStack)-1].val >= A[i] {
+			count += rightStack[len(rightStack)-1].count
+			rightStack = rightStack[:len(rightStack)-1]
+		}
+		rightStack = append(rightStack, &pair{val: A[i], count: count})
+		rights[i] = count
+	}
+
+	for i := 0; i < n; i++ {
+		res = (res + A[i]*lefts[i]*rights[i]) % mod
+	}
+	return res
+}
+
+// 解法三 暴力解法，中间很多重复判断子数组的情况
+func sumSubarrayMins2(A []int) int {
+	res, mod := 0, 1000000007
+	for i := 0; i < len(A); i++ {
+		stack := []int{}
+		stack = append(stack, A[i])
+		for j := i; j < len(A); j++ {
+			if stack[len(stack)-1] >= A[j] {
+				stack = stack[:len(stack)-1]
+				stack = append(stack, A[j])
+			}
+			res += stack[len(stack)-1]
+		}
+	}
+	return res % mod
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0904.Fruit-Into-Baskets/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0909.Snakes-and-Ladders/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0909.Snakes-and-Ladders.md b/website/content/ChapterFour/0900~0999/0909.Snakes-and-Ladders.md
new file mode 100644
index 000000000..ebba1cb5d
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0909.Snakes-and-Ladders.md
@@ -0,0 +1,117 @@
+# [909. Snakes and Ladders](https://leetcode.com/problems/snakes-and-ladders/)
+
+
+## 题目
+
+On an N x N `board`, the numbers from `1` to `N*N` are written *boustrophedonically* **starting from the bottom left of the board**, and alternating direction each row.  For example, for a 6 x 6 board, the numbers are written as follows:
+
+
+![](https://assets.leetcode.com/uploads/2018/09/23/snakes.png)
+
+You start on square `1` of the board (which is always in the last row and first column).  Each move, starting from square `x`, consists of the following:
+
+- You choose a destination square `S` with number `x+1`, `x+2`, `x+3`, `x+4`, `x+5`, or `x+6`, provided this number is `<= N*N`.
+    - (This choice simulates the result of a standard 6-sided die roll: ie., there are always **at most 6 destinations, regardless of the size of the board**.)
+- If `S` has a snake or ladder, you move to the destination of that snake or ladder. Otherwise, you move to `S`.
+
+A board square on row `r` and column `c` has a "snake or ladder" if `board[r][c] != -1`.  The destination of that snake or ladder is `board[r][c]`.
+
+Note that you only take a snake or ladder at most once per move: if the destination to a snake or ladder is the start of another snake or ladder, you do **not** continue moving.  (For example, if the board is `[[4,-1],[-1,3]]`, and on the first move your destination square is `2`, then you finish your first move at `3`, because you do **not** continue moving to `4`.)
+
+Return the least number of moves required to reach square N*N.  If it is not possible, return `-1`.
+
+**Example 1:**
+
+```
+Input:[
+[-1,-1,-1,-1,-1,-1],
+[-1,-1,-1,-1,-1,-1],
+[-1,-1,-1,-1,-1,-1],
+[-1,35,-1,-1,13,-1],
+[-1,-1,-1,-1,-1,-1],
+[-1,15,-1,-1,-1,-1]]
+Output:4
+Explanation:
+At the beginning, you start at square 1 [at row 5, column 0].
+You decide to move to square 2, and must take the ladder to square 15.
+You then decide to move to square 17 (row 3, column 5), and must take the snake to square 13.
+You then decide to move to square 14, and must take the ladder to square 35.
+You then decide to move to square 36, ending the game.
+It can be shown that you need at least 4 moves to reach the N*N-th square, so the answer is 4.
+
+```
+
+**Note:**
+
+1. `2 <= board.length = board[0].length <= 20`
+2. `board[i][j]` is between `1` and `N*N` or is equal to `1`.
+3. The board square with number `1` has no snake or ladder.
+4. The board square with number `N*N` has no snake or ladder.
+
+## 题目大意
+
+N x N 的棋盘 board 上，按从 1 到 N*N 的数字给方格编号，编号 从左下角开始，每一行交替方向。r 行 c 列的棋盘，按前述方法编号，棋盘格中可能存在 “蛇” 或 “梯子”；如果 board[r][c] != -1，那个蛇或梯子的目的地将会是 board[r][c]。玩家从棋盘上的方格 1 （总是在最后一行、第一列）开始出发。每一回合，玩家需要从当前方格 x 开始出发，按下述要求前进：选定目标方格：
+
+- 选择从编号 x+1，x+2，x+3，x+4，x+5，或者 x+6 的方格中选出一个目标方格 s ，目标方格的编号 <= N*N。该选择模拟了掷骰子的情景，无论棋盘大小如何，你的目的地范围也只能处于区间 [x+1, x+6] 之间。
+- 传送玩家：如果目标方格 S 处存在蛇或梯子，那么玩家会传送到蛇或梯子的目的地。否则，玩家传送到目标方格 S。
+
+注意，玩家在每回合的前进过程中最多只能爬过蛇或梯子一次：就算目的地是另一条蛇或梯子的起点，你也不会继续移动。返回达到方格 N*N 所需的最少移动次数，如果不可能，则返回 -1。
+
+## 解题思路
+
+- 这一题可以抽象为在有向图上求下标 1 的起点到下标 `N^2` 的终点的最短路径。用广度优先搜索。棋盘可以抽象成一个包含 `N^2` 个节点的有向图，对于每个节点 `x`，若 `x+i (1 ≤ i ≤ 6)` 上没有蛇或梯子，则连一条从 `x` 到 `x+i` 的有向边；否则记蛇梯的目的地为 `y`，连一条从 `x` 到 `y` 的有向边。然后按照最短路径的求解方式便可解题。时间复杂度 O(n^2)，空间复杂度 O(n^2)。
+- 此题棋盘上的下标是蛇形的，所以遍历下一个点的时候需要转换坐标。具体做法根据行的奇偶性，行号为偶数，下标从左往右，行号为奇数，下标从右往左。具体实现见 `getRowCol()` 函数。
+
+## 代码
+
+```go
+package leetcode
+
+type pair struct {
+	id, step int
+}
+
+func snakesAndLadders(board [][]int) int {
+	n := len(board)
+	visited := make([]bool, n*n+1)
+	queue := []pair{{1, 0}}
+	for len(queue) > 0 {
+		p := queue[0]
+		queue = queue[1:]
+		for i := 1; i <= 6; i++ {
+			nxt := p.id + i
+			if nxt > n*n { // 超出边界
+				break
+			}
+			r, c := getRowCol(nxt, n) // 得到下一步的行列
+			if board[r][c] > 0 {      // 存在蛇或梯子
+				nxt = board[r][c]
+			}
+			if nxt == n*n { // 到达终点
+				return p.step + 1
+			}
+			if !visited[nxt] {
+				visited[nxt] = true
+				queue = append(queue, pair{nxt, p.step + 1}) // 扩展新状态
+			}
+		}
+	}
+	return -1
+}
+
+func getRowCol(id, n int) (r, c int) {
+	r, c = (id-1)/n, (id-1)%n
+	if r%2 == 1 {
+		c = n - 1 - c
+	}
+	r = n - 1 - r
+	return r, c
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0907.Sum-of-Subarray-Minimums/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0910.Smallest-Range-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0910.Smallest-Range-II.md b/website/content/ChapterFour/0900~0999/0910.Smallest-Range-II.md
new file mode 100644
index 000000000..1f9f33035
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0910.Smallest-Range-II.md
@@ -0,0 +1,89 @@
+# [910. Smallest Range II](https://leetcode.com/problems/smallest-range-ii/)
+
+## 题目
+
+Given an array `A` of integers, for each integer `A[i]` we need to choose **either `x = -K` or `x = K`**, and add `x` to `A[i]` **(only once)**.
+
+After this process, we have some array `B`.
+
+Return the smallest possible difference between the maximum value of `B` and the minimum value of `B`.
+
+**Example 1**:
+
+```
+Input: A = [1], K = 0
+Output: 0
+Explanation: B = [1]
+```
+
+**Example 2**:
+
+```
+Input: A = [0,10], K = 2
+Output: 6
+Explanation: B = [2,8]
+```
+
+**Example 3**:
+
+```
+Input: A = [1,3,6], K = 3
+Output: 3
+Explanation: B = [4,6,3]
+```
+
+**Note**:
+
+1. `1 <= A.length <= 10000`
+2. `0 <= A[i] <= 10000`
+3. `0 <= K <= 10000`
+
+## 题目大意
+
+给你一个整数数组 A，对于每个整数 A[i]，可以选择 x = -K 或是 x = K （K 总是非负整数），并将 x 加到 A[i] 中。在此过程之后，得到数组 B。返回 B 的最大值和 B 的最小值之间可能存在的最小差值。
+
+## 解题思路
+
+- 简单题。先排序，找出 A 数组中最大的差值。然后循环扫一遍数组，利用双指针，选择 x = -K 或是 x = K ，每次选择都更新一次最大值和最小值之间的最小差值。循环一次以后便可以找到满足题意的答案。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func smallestRangeII(A []int, K int) int {
+	n := len(A)
+	sort.Ints(A)
+	res := A[n-1] - A[0]
+	for i := 0; i < n-1; i++ {
+		a, b := A[i], A[i+1]
+		high := max(A[n-1]-K, a+K)
+		low := min(A[0]+K, b-K)
+		res = min(res, high-low)
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0909.Snakes-and-Ladders/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0911.Online-Election/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0911.Online-Election.md b/website/content/ChapterFour/0900~0999/0911.Online-Election.md
new file mode 100755
index 000000000..a00cd7365
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0911.Online-Election.md
@@ -0,0 +1,106 @@
+# [911. Online Election](https://leetcode.com/problems/online-election/)
+
+
+## 题目
+
+In an election, the `i`-th vote was cast for `persons[i]` at time `times[i]`.
+
+Now, we would like to implement the following query function: `TopVotedCandidate.q(int t)` will return the number of the person that was leading the election at time `t`.
+
+Votes cast at time `t` will count towards our query. In the case of a tie, the most recent vote (among tied candidates) wins.
+
+**Example 1**:
+
+    Input: ["TopVotedCandidate","q","q","q","q","q","q"], [[[0,1,1,0,0,1,0],[0,5,10,15,20,25,30]],[3],[12],[25],[15],[24],[8]]
+    Output: [null,0,1,1,0,0,1]
+    Explanation: 
+    At time 3, the votes are [0], and 0 is leading.
+    At time 12, the votes are [0,1,1], and 1 is leading.
+    At time 25, the votes are [0,1,1,0,0,1], and 1 is leading (as ties go to the most recent vote.)
+    This continues for 3 more queries at time 15, 24, and 8.
+
+**Note**:
+
+1. `1 <= persons.length = times.length <= 5000`
+2. `0 <= persons[i] <= persons.length`
+3. `times` is a strictly increasing array with all elements in `[0, 10^9]`.
+4. `TopVotedCandidate.q` is called at most `10000` times per test case.
+5. `TopVotedCandidate.q(int t)` is always called with `t >= times[0]`.
+
+
+## 题目大意
+
+在选举中，第 i 张票是在时间为 times[i] 时投给 persons[i] 的。
+
+现在，我们想要实现下面的查询函数： TopVotedCandidate.q(int t) 将返回在 t 时刻主导选举的候选人的编号。
+
+在 t 时刻投出的选票也将被计入我们的查询之中。在平局的情况下，最近获得投票的候选人将会获胜。
+
+提示：
+
+1. 1 <= persons.length = times.length <= 5000
+2. 0 <= persons[i] <= persons.length
+3. times 是严格递增的数组，所有元素都在 [0, 10^9] 范围中。
+4. 每个测试用例最多调用 10000 次 TopVotedCandidate.q。
+5. TopVotedCandidate.q(int t) 被调用时总是满足 t >= times[0]。
+
+
+
+
+## 解题思路
+
+- 给出一个 2 个数组，分别代表第 `i` 人在第 `t` 时刻获得的票数。需要实现一个查询功能的函数，查询在任意 `t` 时刻，输出谁的选票领先。
+- `persons[]` 数组里面装的是获得选票人的编号，`times[]` 数组里面对应的是每个选票的时刻。`times[]` 数组默认是有序的，从小到大排列。先计算出每个时刻哪个人选票领先，放在一个数组中，实现查询函数的时候，只需要先对 `times[]` 数组二分搜索，找到比查询时间 `t` 小的最大时刻 `i`，再在选票领先的数组里面输出对应时刻领先的人的编号即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+// TopVotedCandidate define
+type TopVotedCandidate struct {
+	persons []int
+	times   []int
+}
+
+// Constructor911 define
+func Constructor911(persons []int, times []int) TopVotedCandidate {
+	leaders, votes := make([]int, len(persons)), make([]int, len(persons))
+	leader := persons[0]
+	for i := 0; i < len(persons); i++ {
+		p := persons[i]
+		votes[p]++
+		if votes[p] >= votes[leader] {
+			leader = p
+		}
+		leaders[i] = leader
+	}
+	return TopVotedCandidate{persons: leaders, times: times}
+}
+
+// Q define
+func (tvc *TopVotedCandidate) Q(t int) int {
+	i := sort.Search(len(tvc.times), func(p int) bool { return tvc.times[p] > t })
+	return tvc.persons[i-1]
+}
+
+/**
+ * Your TopVotedCandidate object will be instantiated and called as such:
+ * obj := Constructor(persons, times);
+ * param_1 := obj.Q(t);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0910.Smallest-Range-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0914.X-of-a-Kind-in-a-Deck-of-Cards/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0914.X-of-a-Kind-in-a-Deck-of-Cards.md b/website/content/ChapterFour/0900~0999/0914.X-of-a-Kind-in-a-Deck-of-Cards.md
new file mode 100644
index 000000000..f8cac7562
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0914.X-of-a-Kind-in-a-Deck-of-Cards.md
@@ -0,0 +1,112 @@
+# [914. X of a Kind in a Deck of Cards](https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/)
+
+
+## 题目
+
+In a deck of cards, each card has an integer written on it.
+
+Return `true` if and only if you can choose `X >= 2` such that it is possible to split the entire deck into 1 or more groups of cards, where:
+
+- Each group has exactly `X` cards.
+- All the cards in each group have the same integer.
+
+**Example 1**:
+
+```
+Input: deck = [1,2,3,4,4,3,2,1]
+Output: true
+Explanation: Possible partition [1,1],[2,2],[3,3],[4,4].
+```
+
+**Example 2**:
+
+```
+Input: deck = [1,1,1,2,2,2,3,3]
+Output: false´
+Explanation: No possible partition.
+```
+
+**Example 3**:
+
+```
+Input: deck = [1]
+Output: false
+Explanation: No possible partition.
+```
+
+**Example 4**:
+
+```
+Input: deck = [1,1]
+Output: true
+Explanation: Possible partition [1,1].
+```
+
+**Example 5**:
+
+```
+Input: deck = [1,1,2,2,2,2]
+Output: true
+Explanation: Possible partition [1,1],[2,2],[2,2].
+```
+
+**Constraints**:
+
+- `1 <= deck.length <= 10^4`
+- `0 <= deck[i] < 10^4`
+
+## 题目大意
+
+给定一副牌，每张牌上都写着一个整数。此时，你需要选定一个数字 X，使我们可以将整副牌按下述规则分成 1 组或更多组：
+
+- 每组都有 X 张牌。
+- 组内所有的牌上都写着相同的整数。
+
+仅当你可选的 X >= 2 时返回 true。
+
+
+## 解题思路
+
+- 给定一副牌，要求选出数字 X，使得每组都有 X 张牌，每组牌的数字都相同。当 X ≥ 2 的时候，输出 true。
+- 通过分析题目，我们可以知道，只有当 X 为所有 count 的约数，即所有 count 的最大公约数的约数时，才存在可能的分组。因此我们只要求出所有 count 的最大公约数 g，判断 g 是否大于等于 2 即可，如果大于等于 2，则满足条件，否则不满足。
+- 时间复杂度：O(NlogC)，其中 N 是卡牌的个数，C 是数组 deck 中数的范围，在本题中 C 的值为 10000。求两个数最大公约数的复杂度是 O(logC)，需要求最多 N - 1 次。空间复杂度：O(N + C) 或 O(N)。
+
+## 代码
+
+```go
+
+package leetcode
+
+func hasGroupsSizeX(deck []int) bool {
+	if len(deck) < 2 {
+		return false
+	}
+	m, g := map[int]int{}, -1
+	for _, d := range deck {
+		m[d]++
+	}
+	for _, v := range m {
+		if g == -1 {
+			g = v
+		} else {
+			g = gcd(g, v)
+		}
+	}
+	return g >= 2
+}
+
+func gcd(a, b int) int {
+	if a == 0 {
+		return b
+	}
+	return gcd(b%a, a)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0911.Online-Election/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0916.Word-Subsets/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0916.Word-Subsets.md b/website/content/ChapterFour/0900~0999/0916.Word-Subsets.md
new file mode 100644
index 000000000..8e54db5cb
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0916.Word-Subsets.md
@@ -0,0 +1,110 @@
+# [916. Word Subsets](https://leetcode.com/problems/word-subsets/)
+
+
+## 题目
+
+We are given two arrays `A` and `B` of words.  Each word is a string of lowercase letters.
+
+Now, say that word `b` is a subset of word `a` ****if every letter in `b` occurs in `a`, **including multiplicity**.  For example, `"wrr"` is a subset of `"warrior"`, but is not a subset of `"world"`.
+
+Now say a word `a` from `A` is *universal* if for every `b` in `B`, `b` is a subset of `a`.
+
+Return a list of all universal words in `A`.  You can return the words in any order.
+
+**Example 1:**
+
+```
+Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
+Output:["facebook","google","leetcode"]
+```
+
+**Example 2:**
+
+```
+Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
+Output:["apple","google","leetcode"]
+```
+
+**Example 3:**
+
+```
+Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
+Output:["facebook","google"]
+```
+
+**Example 4:**
+
+```
+Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
+Output:["google","leetcode"]
+```
+
+**Example 5:**
+
+```
+Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
+Output:["facebook","leetcode"]
+```
+
+**Note:**
+
+1. `1 <= A.length, B.length <= 10000`
+2. `1 <= A[i].length, B[i].length <= 10`
+3. `A[i]` and `B[i]` consist only of lowercase letters.
+4. All words in `A[i]` are unique: there isn't `i != j` with `A[i] == A[j]`.
+
+## 题目大意
+
+我们给出两个单词数组 A 和 B。每个单词都是一串小写字母。现在，如果 b 中的每个字母都出现在 a 中，包括重复出现的字母，那么称单词 b 是单词 a 的子集。 例如，“wrr” 是 “warrior” 的子集，但不是 “world” 的子集。如果对 B 中的每一个单词 b，b 都是 a 的子集，那么我们称 A 中的单词 a 是通用的。你可以按任意顺序以列表形式返回 A 中所有的通用单词。
+
+## 解题思路
+
+- 简单题。先统计出 B 数组中单词每个字母的频次，再在 A 数组中依次判断每个单词是否超过了这个频次，如果超过了即输出。
+
+## 代码
+
+```go
+package leetcode
+
+func wordSubsets(A []string, B []string) []string {
+	var counter [26]int
+	for _, b := range B {
+		var m [26]int
+		for _, c := range b {
+			j := c - 'a'
+			m[j]++
+		}
+		for i := 0; i < 26; i++ {
+			if m[i] > counter[i] {
+				counter[i] = m[i]
+			}
+		}
+	}
+	var res []string
+	for _, a := range A {
+		var m [26]int
+		for _, c := range a {
+			j := c - 'a'
+			m[j]++
+		}
+		ok := true
+		for i := 0; i < 26; i++ {
+			if m[i] < counter[i] {
+				ok = false
+				break
+			}
+		}
+		if ok {
+			res = append(res, a)
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0914.X-of-a-Kind-in-a-Deck-of-Cards/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0918.Maximum-Sum-Circular-Subarray/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0918.Maximum-Sum-Circular-Subarray.md b/website/content/ChapterFour/0900~0999/0918.Maximum-Sum-Circular-Subarray.md
new file mode 100755
index 000000000..955a68679
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0918.Maximum-Sum-Circular-Subarray.md
@@ -0,0 +1,132 @@
+# [918. Maximum Sum Circular Subarray](https://leetcode.com/problems/maximum-sum-circular-subarray/)
+
+
+## 题目
+
+Given a **circular array** **C** of integers represented by `A`, find the maximum possible sum of a non-empty subarray of **C**.
+
+Here, a *circular array* means the end of the array connects to the beginning of the array. (Formally, `C[i] = A[i]` when `0 <= i < A.length`, and `C[i+A.length] = C[i]` when `i >= 0`.)
+
+Also, a subarray may only include each element of the fixed buffer `A` at most once. (Formally, for a subarray `C[i], C[i+1], ..., C[j]`, there does not exist `i <= k1, k2 <= j` with `k1 % A.length = k2 % A.length`.)
+
+**Example 1**:
+
+    Input: [1,-2,3,-2]
+    Output: 3
+    Explanation: Subarray [3] has maximum sum 3
+
+**Example 2**:
+
+    Input: [5,-3,5]
+    Output: 10
+    Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10
+
+**Example 3**:
+
+    Input: [3,-1,2,-1]
+    Output: 4
+    Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4
+
+**Example 4**:
+
+    Input: [3,-2,2,-3]
+    Output: 3
+    Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3
+
+**Example 5**:
+
+    Input: [-2,-3,-1]
+    Output: -1
+    Explanation: Subarray [-1] has maximum sum -1
+
+**Note**:
+
+1. `-30000 <= A[i] <= 30000`
+2. `1 <= A.length <= 30000`
+
+
+## 题目大意
+
+给定一个由整数数组 A 表示的环形数组 C，求 C 的非空子数组的最大可能和。在此处，环形数组意味着数组的末端将会与开头相连呈环状。（形式上，当0 <= i < A.length 时 C[i] = A[i]，而当 i >= 0 时 C[i+A.length] = C[i]）
+
+此外，子数组最多只能包含固定缓冲区 A 中的每个元素一次。（形式上，对于子数组 C[i], C[i+1], ..., C[j]，不存在 i <= k1, k2 <= j 其中 k1 % A.length = k2 % A.length）
+
+提示：
+
+- -30000 <= A[i] <= 30000
+- 1 <= A.length <= 30000
+
+
+## 解题思路
+
+
+- 给出一个环形数组，要求出这个环形数组中的连续子数组的最大和。
+- 拿到这题最先想到的思路是把这个数组再拼接一个，在这两个数组中查找连续子数组的最大和。这种做法是错误的，例如在 `[5,-3,5]` 这个数组中会得出 `7` 的结果，但是实际结果是 `10` 。那么这题怎么做呢？仔细分析可以得到，环形数组的最大连续子段和有两种情况，第一种情况是这个连续子段就出现在数组中， 不存在循环衔接的情况。针对这种情况就比较简单，用 `kadane` 算法(也是动态规划的思想)，`O(n)` 的时间复杂度就可以求出结果。第二种情况是这个连续的子段出现在跨数组的情况，即会出现首尾相连的情况。要想找到这样一个连续子段，可以反向考虑。想找到跨段的连续子段，那么这个数组剩下的这一段就是不跨段的连续子段。想要跨段的子段和最大，那么剩下的这段连续子段和最小。如果能找到这个数组的每个元素取相反数组成的数组中的最大连续子段和，那么反过来就能找到原数组的连续子段和最小。举个例子：`[1，2，-3，-4，5]` ，取它的每个元素的相反数 `[-1，-2，3，4，-5]`，构造的数组中最大连续子段和是 `3 + 4 = 7`，由于取了相反数，所以可以得到原数组中最小连续子段和是 `-7` 。所以跨段的最大连续子段和就是剩下的那段 `[1,2,5]`。
+- 还有一些边界的情况，例如，`[1，2，-2，-3，5，5，-4，6]` 和 `[1，2，-2，-3，5，5，-4，8]`，所以还需要比较一下情况一和情况二的值，它们两者最大值才是最终环形数组的连续子数组的最大和。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func maxSubarraySumCircular(nums []int) int {
+	var max1, max2, sum int
+
+	// case: no circulation
+	max1 = int(math.Inf(-1))
+	l := len(nums)
+	for i := 0; i < l; i++ {
+		sum += nums[i]
+		if sum > max1 {
+			max1 = sum
+		}
+		if sum < 1 {
+			sum = 0
+		}
+	}
+
+	// case: circling
+	arr_sum := 0
+	for i := 0; i < l; i++ {
+		arr_sum += nums[i]
+	}
+
+	sum = 0
+	min_sum := 0
+	for i := 1; i < l-1; i++ {
+		sum += nums[i]
+		if sum >= 0 {
+			sum = 0
+		}
+		if sum < min_sum {
+			min_sum = sum
+		}
+	}
+	max2 = arr_sum - min_sum
+
+	return max(max1, max2)
+}
+
+func max(nums ...int) int {
+	max := int(math.Inf(-1))
+	for _, num := range nums {
+		if num > max {
+			max = num
+		}
+	}
+	return max
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0916.Word-Subsets/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0920.Number-of-Music-Playlists/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0920.Number-of-Music-Playlists.md b/website/content/ChapterFour/0900~0999/0920.Number-of-Music-Playlists.md
new file mode 100755
index 000000000..27e214453
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0920.Number-of-Music-Playlists.md
@@ -0,0 +1,92 @@
+# [920. Number of Music Playlists](https://leetcode.com/problems/number-of-music-playlists/)
+
+
+## 题目
+
+Your music player contains `N` different songs and she wants to listen to `L` ****(not necessarily different) songs during your trip. You create a playlist so that:
+
+- Every song is played at least once
+- A song can only be played again only if `K` other songs have been played
+
+Return the number of possible playlists. **As the answer can be very large, return it modulo `10^9 + 7`**.
+
+**Example 1**:
+
+    Input: N = 3, L = 3, K = 1
+    Output: 6
+    Explanation: There are 6 possible playlists. [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].
+
+**Example 2**:
+
+    Input: N = 2, L = 3, K = 0
+    Output: 6
+    Explanation: There are 6 possible playlists. [1, 1, 2], [1, 2, 1], [2, 1, 1], [2, 2, 1], [2, 1, 2], [1, 2, 2]
+
+**Example 3**:
+
+    Input: N = 2, L = 3, K = 1
+    Output: 2
+    Explanation: There are 2 possible playlists. [1, 2, 1], [2, 1, 2]
+
+**Note**:
+
+1. `0 <= K < N <= L <= 100`
+
+## 题目大意
+
+你的音乐播放器里有 N 首不同的歌，在旅途中，你的旅伴想要听 L 首歌（不一定不同，即，允许歌曲重复）。请你为她按如下规则创建一个播放列表：
+
+- 每首歌至少播放一次。
+- 一首歌只有在其他 K 首歌播放完之后才能再次播放。
+
+返回可以满足要求的播放列表的数量。由于答案可能非常大，请返回它模 10^9 + 7 的结果。
+
+提示：
+
+- 0 <= K < N <= L <= 100
+
+
+
+
+## 解题思路
+
+- 简化抽象一下题意，给 N 个数，要求从这 N 个数里面组成一个长度为 L 的序列，并且相同元素的间隔不能小于 K 个数。问总共有多少组组成方法。
+- 一拿到题，会觉得这一题是三维 DP，因为存在 3 个变量，但是实际考虑一下，可以降一维。我们先不考虑 K 的限制，只考虑 N 和 L。定义 `dp[i][j]` 代表播放列表里面有 `i` 首歌，其中包含 `j` 首不同的歌曲，那么题目要求的最终解存在 `dp[L][N]` 中。考虑 `dp[i][j]` 的递归公式，音乐列表当前需要组成 `i` 首歌，有 2 种方式可以得到，由 `i - 1` 首歌的列表中添加一首列表中**不存在**的新歌曲，或者由 `i - 1` 首歌的列表中添加一首列表中**已经存在**的歌曲。即，`dp[i][j]` 可以由 `dp[i - 1][j - 1]` 得到，也可以由 `dp[i - 1][j]` 得到。如果是第一种情况，添加一首新歌，那么新歌有 N - ( j - 1 ) 首，如果是第二种情况，添加一首已经存在的歌，歌有 j 首，所以状态转移方程是 `dp[i][j] = dp[i - 1][j - 1] * ( N - ( j - 1 ) ) + dp[i - 1][j] * j` 。但是这个方程是在不考虑 K 的限制条件下得到的，距离满足题意还差一步。接下来需要考虑加入 K 这个限制条件以后，状态转移方程该如何推导。
+- 如果是添加一首新歌，是不受 K 限制的，所以 `dp[i - 1][j - 1] * ( N - ( j - 1 ) )` 这里不需要变化。如果是添加一首存在的歌曲，这个时候就会受到 K 的限制了。如果当前播放列表里面的歌曲有 `j` 首，并且 `j > K`，那么选择歌曲只能从 `j - K` 里面选，因为不能选择 `j - 1` 到 `j - k` 的这些歌，选择了就不满足重复的歌之间间隔不能小于 `K` 的限制条件了。那 j ≤ K 呢？这个时候一首歌都不能选，因为歌曲数都没有超过 K，当然不能再选择重复的歌曲。(选择了就再次不满足重复的歌之间间隔不能小于 `K` 的限制条件了)。经过上述分析，可以得到最终的状态转移方程：
+{{< katex display >}} 
+dp[i][j]= \begin{matrix} \left\{              \begin{array}{lr}              dp[i - 1][j - 1] * ( N - ( j - 1 ) ) + dp[i - 1][j] * ( j - k )      , &  {j > k}\\              dp[i - 1][j - 1] * ( N - ( j - 1 ) ), &  {j \leq k}              \end{array} \right. \end{matrix}
+{{< /katex >}}   
+- 上面的式子可以合并简化成下面这个式子：`dp[i][j] = dp[i - 1][j - 1]*(N - (j - 1)) + dp[i-1][j]*max(j-K, 0)`，递归初始值 `dp[0][0] = 1`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func numMusicPlaylists(N int, L int, K int) int {
+	dp, mod := make([][]int, L+1), 1000000007
+	for i := 0; i < L+1; i++ {
+		dp[i] = make([]int, N+1)
+	}
+	dp[0][0] = 1
+	for i := 1; i <= L; i++ {
+		for j := 1; j <= N; j++ {
+			dp[i][j] = (dp[i-1][j-1] * (N - (j - 1))) % mod
+			if j > K {
+				dp[i][j] = (dp[i][j] + (dp[i-1][j]*(j-K))%mod) % mod
+			}
+		}
+	}
+	return dp[L][N]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0918.Maximum-Sum-Circular-Subarray/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0921.Minimum-Add-to-Make-Parentheses-Valid/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0921.Minimum-Add-to-Make-Parentheses-Valid.md b/website/content/ChapterFour/0900~0999/0921.Minimum-Add-to-Make-Parentheses-Valid.md
new file mode 100644
index 000000000..f11d433ee
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0921.Minimum-Add-to-Make-Parentheses-Valid.md
@@ -0,0 +1,95 @@
+# [921. Minimum Add to Make Parentheses Valid](https://leetcode.com/problems/minimum-add-to-make-parentheses-valid/)
+
+## 题目
+
+Given a string S of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' or ')', and in any positions ) so that the resulting parentheses string is valid.
+
+Formally, a parentheses string is valid if and only if:
+
+- It is the empty string, or
+- It can be written as AB (A concatenated with B), where A and B are valid strings, or
+- It can be written as (A), where A is a valid string.
+Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.
+
+ 
+
+**Example 1**:
+
+```
+
+Input: "())"
+Output: 1
+
+```
+
+**Example 2**:
+
+```
+
+Input: "((("
+Output: 3
+
+```
+
+**Example 3**:
+
+```
+
+Input: "()"
+Output: 0
+
+```
+
+**Example 4**:
+
+```
+
+Input: "()))(("
+Output: 4
+
+```
+
+**Note**:
+
+1. S.length <= 1000
+2. S only consists of '(' and ')' characters.
+
+## 题目大意
+
+给一个括号的字符串，如果能在这个括号字符串中的任意位置添加括号，问能使得这串字符串都能完美匹配的最少添加数是多少。
+
+## 解题思路
+
+这题也是栈的题目，利用栈进行括号匹配。最后栈里剩下几个括号，就是最少需要添加的数目。
+
+## 代码
+
+```go
+
+package leetcode
+
+func minAddToMakeValid(S string) int {
+	if len(S) == 0 {
+		return 0
+	}
+	stack := make([]rune, 0)
+	for _, v := range S {
+		if v == '(' {
+			stack = append(stack, v)
+		} else if (v == ')') && len(stack) > 0 && stack[len(stack)-1] == '(' {
+			stack = stack[:len(stack)-1]
+		} else {
+			stack = append(stack, v)
+		}
+	}
+	return len(stack)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0920.Number-of-Music-Playlists/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0922.Sort-Array-By-Parity-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0922.Sort-Array-By-Parity-II.md b/website/content/ChapterFour/0900~0999/0922.Sort-Array-By-Parity-II.md
new file mode 100644
index 000000000..c89384dd7
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0922.Sort-Array-By-Parity-II.md
@@ -0,0 +1,68 @@
+# [922. Sort Array By Parity II](https://leetcode.com/problems/sort-array-by-parity-ii/)
+
+## 题目
+
+Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.
+
+Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.
+
+You may return any answer array that satisfies this condition.
+
+
+**Example 1**:
+
+```
+
+Input: [4,2,5,7]
+Output: [4,5,2,7]
+Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
+
+```
+
+**Note**:
+
+- 2 <= A.length <= 20000
+- A.length % 2 == 0
+- 0 <= A[i] <= 1000
+
+## 题目大意
+
+要求数组中奇数下标位置上放奇数，偶数下标位置上放偶数。
+
+## 解题思路
+
+这题比较简单，用两个下标控制奇数，偶数放置在哪个下标即可。奇数奇数之间，偶数偶数之间的顺序可以是无序的。
+
+## 代码
+
+```go
+
+package leetcode
+
+func sortArrayByParityII(A []int) []int {
+	if len(A) == 0 || len(A)%2 != 0 {
+		return []int{}
+	}
+	res := make([]int, len(A))
+	oddIndex := 1
+	evenIndex := 0
+	for i := 0; i < len(A); i++ {
+		if A[i]%2 == 0 {
+			res[evenIndex] = A[i]
+			evenIndex += 2
+		} else {
+			res[oddIndex] = A[i]
+			oddIndex += 2
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0921.Minimum-Add-to-Make-Parentheses-Valid/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0923.3Sum-With-Multiplicity/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0923.3Sum-With-Multiplicity.md b/website/content/ChapterFour/0900~0999/0923.3Sum-With-Multiplicity.md
new file mode 100644
index 000000000..6cacce754
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0923.3Sum-With-Multiplicity.md
@@ -0,0 +1,109 @@
+# [923. 3Sum With Multiplicity](https://leetcode.com/problems/3sum-with-multiplicity/)
+
+## 题目
+
+Given an integer array A, and an integer target, return the number of tuples i, j, k  such that i < j < k and A[i] + A[j] + A[k] == target.
+
+As the answer can be very large, return it modulo 10^9 + 7.
+
+
+**Example 1**:
+
+```
+
+Input: A = [1,1,2,2,3,3,4,4,5,5], target = 8
+Output: 20
+Explanation: 
+Enumerating by the values (A[i], A[j], A[k]):
+(1, 2, 5) occurs 8 times;
+(1, 3, 4) occurs 8 times;
+(2, 2, 4) occurs 2 times;
+(2, 3, 3) occurs 2 times.
+
+```
+
+
+**Example 2**:
+
+```
+
+Input: A = [1,1,2,2,2,2], target = 5
+Output: 12
+Explanation: 
+A[i] = 1, A[j] = A[k] = 2 occurs 12 times:
+We choose one 1 from [1,1] in 2 ways,
+and two 2s from [2,2,2,2] in 6 ways.
+
+```
+
+
+**Note**:
+
+- 3 <= A.length <= 3000
+- 0 <= A[i] <= 100
+- 0 <= target <= 300
+
+
+## 题目大意
+
+这道题是第 15 题的升级版。给出一个数组，要求找到 3 个数相加的和等于 target 的解组合的个数，并且要求  i < j < k。解的组合个数不需要去重，相同数值不同下标算不同解(这里也是和第 15 题的区别)
+
+## 解题思路
+
+这一题大体解法和第 15 题一样的，只不过算所有解组合的时候需要一点排列组合的知识，如果取 3 个一样的数，需要计算 C n 3，去 2 个相同的数字的时候，计算 C n 2，取一个数字就正常计算。最后所有解的个数都加起来就可以了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+func threeSumMulti(A []int, target int) int {
+	mod := 1000000007
+	counter := map[int]int{}
+	for _, value := range A {
+		counter[value]++
+	}
+
+	uniqNums := []int{}
+	for key := range counter {
+		uniqNums = append(uniqNums, key)
+	}
+	sort.Ints(uniqNums)
+
+	res := 0
+	for i := 0; i < len(uniqNums); i++ {
+		ni := counter[uniqNums[i]]
+		if (uniqNums[i]*3 == target) && counter[uniqNums[i]] >= 3 {
+			res += ni * (ni - 1) * (ni - 2) / 6
+		}
+		for j := i + 1; j < len(uniqNums); j++ {
+			nj := counter[uniqNums[j]]
+			if (uniqNums[i]*2+uniqNums[j] == target) && counter[uniqNums[i]] > 1 {
+				res += ni * (ni - 1) / 2 * nj
+			}
+			if (uniqNums[j]*2+uniqNums[i] == target) && counter[uniqNums[j]] > 1 {
+				res += nj * (nj - 1) / 2 * ni
+			}
+			c := target - uniqNums[i] - uniqNums[j]
+			if c > uniqNums[j] && counter[c] > 0 {
+				res += ni * nj * counter[c]
+			}
+		}
+	}
+	return res % mod
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0922.Sort-Array-By-Parity-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0924.Minimize-Malware-Spread/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0924.Minimize-Malware-Spread.md b/website/content/ChapterFour/0900~0999/0924.Minimize-Malware-Spread.md
new file mode 100755
index 000000000..b5835a932
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0924.Minimize-Malware-Spread.md
@@ -0,0 +1,126 @@
+# [924. Minimize Malware Spread](https://leetcode.com/problems/minimize-malware-spread/)
+
+
+## 题目
+
+In a network of nodes, each node `i` is directly connected to another node `j` if and only if `graph[i][j] = 1`.
+
+Some nodes `initial` are initially infected by malware. Whenever two nodes are directly connected and at least one of those two nodes is infected by malware, both nodes will be infected by malware. This spread of malware will continue until no more nodes can be infected in this manner.
+
+Suppose `M(initial)` is the final number of nodes infected with malware in the entire network, after the spread of malware stops.
+
+We will remove one node from the initial list. Return the node that if removed, would minimize `M(initial)`. If multiple nodes could be removed to minimize `M(initial)`, return such a node with the smallest index.
+
+Note that if a node was removed from the `initial` list of infected nodes, it may still be infected later as a result of the malware spread.
+
+**Example 1**:
+
+    Input: graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]
+    Output: 0
+
+**Example 2**:
+
+    Input: graph = [[1,0,0],[0,1,0],[0,0,1]], initial = [0,2]
+    Output: 0
+
+**Example 3**:
+
+    Input: graph = [[1,1,1],[1,1,1],[1,1,1]], initial = [1,2]
+    Output: 1
+
+**Note**:
+
+1. `1 < graph.length = graph[0].length <= 300`
+2. `0 <= graph[i][j] == graph[j][i] <= 1`
+3. `graph[i][i] = 1`
+4. `1 <= initial.length < graph.length`
+5. `0 <= initial[i] < graph.length`
+
+
+## 题目大意
+
+在节点网络中，只有当 graph[i][j] = 1 时，每个节点 i 能够直接连接到另一个节点 j。一些节点 initial 最初被恶意软件感染。只要两个节点直接连接，且其中至少一个节点受到恶意软件的感染，那么两个节点都将被恶意软件感染。这种恶意软件的传播将继续，直到没有更多的节点可以被这种方式感染。假设 M(initial) 是在恶意软件停止传播之后，整个网络中感染恶意软件的最终节点数。我们可以从初始列表中删除一个节点。如果移除这一节点将最小化 M(initial)， 则返回该节点。如果有多个节点满足条件，就返回索引最小的节点。请注意，如果某个节点已从受感染节点的列表 initial 中删除，它以后可能仍然因恶意软件传播而受到感染。
+
+
+提示：
+
+- 1 < graph.length = graph[0].length <= 300
+- 0 <= graph[i][j] == graph[j][i] <= 1
+- graph[i][i] = 1
+- 1 <= initial.length < graph.length
+- 0 <= initial[i] < graph.length
+
+
+## 解题思路
+
+
+- 给出一个节点之间的关系图，如果两个节点是连通的，那么病毒软件就会感染到连通的所有节点。现在如果想移除一个病毒节点，能最大减少感染，请问移除哪个节点？如果多个节点都能减少感染量，优先移除序号偏小的那个节点。
+- 这一题一看就是考察的并查集。利用节点的连通关系，把题目中给的所有节点都 `union()` 起来，然后依次统计每个集合内有多少个点。最后扫描一次 initial 数组，选出这个数组中节点小的并且所在集合节点多，这个节点就是最终答案。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func minMalwareSpread(graph [][]int, initial []int) int {
+	if len(initial) == 0 {
+		return 0
+	}
+	uf, maxLen, maxIdx, uniqInitials, compMap := template.UnionFindCount{}, math.MinInt32, -1, map[int]int{}, map[int][]int{}
+	uf.Init(len(graph))
+	for i := range graph {
+		for j := i + 1; j < len(graph); j++ {
+			if graph[i][j] == 1 {
+				uf.Union(i, j)
+			}
+		}
+	}
+	for _, i := range initial {
+		compMap[uf.Find(i)] = append(compMap[uf.Find(i)], i)
+	}
+	for _, v := range compMap {
+		if len(v) == 1 {
+			uniqInitials[v[0]] = v[0]
+		}
+	}
+	if len(uniqInitials) == 0 {
+		smallestIdx := initial[0]
+		for _, i := range initial {
+			if i < smallestIdx {
+				smallestIdx = i
+			}
+		}
+		return smallestIdx
+	}
+	for _, i := range initial {
+		if _, ok := uniqInitials[i]; ok {
+			size := uf.Count()[uf.Find(i)]
+			if maxLen < size {
+				maxLen, maxIdx = size, i
+			} else if maxLen == size {
+				if i < maxIdx {
+					maxIdx = i
+				}
+			}
+		}
+	}
+	return maxIdx
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0923.3Sum-With-Multiplicity/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0925.Long-Pressed-Name/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0925.Long-Pressed-Name.md b/website/content/ChapterFour/0900~0999/0925.Long-Pressed-Name.md
new file mode 100644
index 000000000..1634e0602
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0925.Long-Pressed-Name.md
@@ -0,0 +1,114 @@
+# [925. Long Pressed Name](https://leetcode.com/problems/long-pressed-name/)
+
+## 题目
+
+Your friend is typing his name into a keyboard.  Sometimes, when typing a character c, the key might get long pressed, and the character will be typed 1 or more times.
+
+You examine the typed characters of the keyboard.  Return True if it is possible that it was your friends name, with some characters (possibly none) being long pressed.
+
+
+
+**Example 1**:
+
+```
+
+Input: name = "alex", typed = "aaleex"
+Output: true
+Explanation: 'a' and 'e' in 'alex' were long pressed.
+
+```
+
+**Example 2**:
+
+```
+
+Input: name = "saeed", typed = "ssaaedd"
+Output: false
+Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.
+
+```
+
+**Example 3**:
+
+```
+
+Input: name = "leelee", typed = "lleeelee"
+Output: true
+
+```
+
+**Example 4**:
+
+```
+
+Input: name = "laiden", typed = "laiden"
+Output: true
+Explanation: It's not necessary to long press any character.
+
+```
+
+
+**Note**:  
+
+1. name.length <= 1000
+2. typed.length <= 1000
+3. The characters of name and typed are lowercase letters.
+
+## 题目大意
+
+
+给定 2 个字符串，后者的字符串中包含前者的字符串。比如在打字的过程中，某个字符会多按了几下。判断后者字符串是不是比前者字符串存在这样的“长按”键盘的情况。
+
+## 解题思路
+
+- 这一题也可以借助滑动窗口的思想。2 个字符串一起比较，如果遇到有相同的字符串，窗口继续往后滑动。直到遇到了第一个不同的字符，如果遇到两个字符串不相等的情况，可以直接返回 false。具体实现见代码。
+- 这一题的测试用例修改过一次，需要注意我这里写的第二组测试用例，当 name 结束以后，如果 typed 还有多余的不同的字符，这种情况要输出 false 的。具体见 test 文件里面的第二组，第三组，第四组测试用例。
+
+
+
+
+
+
+
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func isLongPressedName(name string, typed string) bool {
+	if len(name) == 0 && len(typed) == 0 {
+		return true
+	}
+	if (len(name) == 0 && len(typed) != 0) || (len(name) != 0 && len(typed) == 0) {
+		return false
+	}
+	i, j := 0, 0
+	for i < len(name) && j < len(typed) {
+		if name[i] != typed[j] {
+			return false
+		}
+		for i < len(name) && j < len(typed) && name[i] == typed[j] {
+			i++
+			j++
+		}
+		for j < len(typed) && typed[j] == typed[j-1] {
+			j++
+		}
+	}
+	return i == len(name) && j == len(typed)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0924.Minimize-Malware-Spread/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0927.Three-Equal-Parts/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0927.Three-Equal-Parts.md b/website/content/ChapterFour/0900~0999/0927.Three-Equal-Parts.md
new file mode 100755
index 000000000..9cdabc5d9
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0927.Three-Equal-Parts.md
@@ -0,0 +1,120 @@
+# [927. Three Equal Parts](https://leetcode.com/problems/three-equal-parts/)
+
+
+## 题目
+
+Given an array `A` of `0`s and `1`s, divide the array into 3 non-empty parts such that all of these parts represent the same binary value.
+
+If it is possible, return **any** `[i, j]` with `i+1 < j`, such that:
+
+- `A[0], A[1], ..., A[i]` is the first part;
+- `A[i+1], A[i+2], ..., A[j-1]` is the second part, and
+- `A[j], A[j+1], ..., A[A.length - 1]` is the third part.
+- All three parts have equal binary value.
+
+If it is not possible, return `[-1, -1]`.
+
+Note that the entire part is used when considering what binary value it represents. For example, `[1,1,0]` represents `6` in decimal, not `3`. Also, leading zeros are allowed, so `[0,1,1]` and `[1,1]` represent the same value.
+
+**Example 1**:
+
+    Input: [1,0,1,0,1]
+    Output: [0,3]
+
+**Example 2**:
+
+    Input: [1,1,0,1,1]
+    Output: [-1,-1]
+
+**Note**:
+
+1. `3 <= A.length <= 30000`
+2. `A[i] == 0` or `A[i] == 1`
+
+
+## 题目大意
+
+给定一个由 0 和 1 组成的数组 A，将数组分成 3 个非空的部分，使得所有这些部分表示相同的二进制值。如果可以做到，请返回任何 [i, j]，其中 i+1 < j，这样一来：  
+
+- A[0], A[1], ..., A[i] 组成第一部分；
+- A[i+1], A[i+2], ..., A[j-1] 作为第二部分；
+- A[j], A[j+1], ..., A[A.length - 1] 是第三部分。
+- 这三个部分所表示的二进制值相等。  
+
+如果无法做到，就返回 [-1, -1]。
+
+注意，在考虑每个部分所表示的二进制时，应当将其看作一个整体。例如，[1,1,0] 表示十进制中的 6，而不会是 3。此外，前导零也是被允许的，所以 [0,1,1] 和 [1,1] 表示相同的值。  
+
+提示：
+
+1. 3 <= A.length <= 30000
+2. A[i] == 0 或 A[i] == 1
+
+
+## 解题思路
+
+- 给出一个数组，数组里面只包含 0 和 1，要求找到 2 个分割点，使得分成的 3 个子数组的二进制是完全一样的。
+- 这一题的解题思路不难，按照题意模拟即可。先统计 1 的个数 total，然后除以 3 就是每段 1 出现的个数。有一些特殊情况需要额外判断一下，例如没有 1 的情况，那么只能首尾分割。1 个个数不是 3 的倍数，也无法分割成满足题意。然后找到第一个 1 的下标，然后根据 total/3 找到 mid，第一个分割点。再往后移动，找到第二个分割点。找到这 3 个点以后，同步的移动这 3 个点，移动中判断这 3 个下标对应的数值是否相等，如果都相等，并且最后一个点能移动到末尾，就算找到了满足题意的解了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func threeEqualParts(A []int) []int {
+	n, ones, i, count := len(A), 0, 0, 0
+	for _, a := range A {
+		ones += a
+	}
+	if ones == 0 {
+		return []int{0, n - 1}
+	}
+	if ones%3 != 0 {
+		return []int{-1, -1}
+	}
+	k := ones / 3
+	for i < n {
+		if A[i] == 1 {
+			break
+		}
+		i++
+	}
+	start, j := i, i
+	for j < n {
+		count += A[j]
+		if count == k+1 {
+			break
+		}
+		j++
+	}
+	mid := j
+	j, count = 0, 0
+	for j < n {
+		count += A[j]
+		if count == 2*k+1 {
+			break
+		}
+		j++
+	}
+	end := j
+	for end < n && A[start] == A[mid] && A[mid] == A[end] {
+		start++
+		mid++
+		end++
+	}
+	if end == n {
+		return []int{start - 1, mid}
+	}
+	return []int{-1, -1}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0925.Long-Pressed-Name/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0928.Minimize-Malware-Spread-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0928.Minimize-Malware-Spread-II.md b/website/content/ChapterFour/0900~0999/0928.Minimize-Malware-Spread-II.md
new file mode 100755
index 000000000..ad047b303
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0928.Minimize-Malware-Spread-II.md
@@ -0,0 +1,144 @@
+# [928. Minimize Malware Spread II](https://leetcode.com/problems/minimize-malware-spread-ii/)
+
+
+## 题目
+
+(This problem is the same as *Minimize Malware Spread*, with the differences bolded.)
+
+In a network of nodes, each node `i` is directly connected to another node `j` if and only if `graph[i][j] = 1`.
+
+Some nodes `initial` are initially infected by malware. Whenever two nodes are directly connected and at least one of those two nodes is infected by malware, both nodes will be infected by malware. This spread of malware will continue until no more nodes can be infected in this manner.
+
+Suppose `M(initial)` is the final number of nodes infected with malware in the entire network, after the spread of malware stops.
+
+We will remove one node from the initial list, **completely removing it and any connections from this node to any other node**. Return the node that if removed, would minimize `M(initial)`. If multiple nodes could be removed to minimize `M(initial)`, return such a node with the smallest index.
+
+**Example 1**:
+
+    Input: graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]
+    Output: 0
+
+**Example 2**:
+
+    Input: graph = [[1,1,0],[1,1,1],[0,1,1]], initial = [0,1]
+    Output: 1
+
+**Example 3**:
+
+    Input: graph = [[1,1,0,0],[1,1,1,0],[0,1,1,1],[0,0,1,1]], initial = [0,1]
+    Output: 1
+
+**Note**:
+
+1. `1 < graph.length = graph[0].length <= 300`
+2. `0 <= graph[i][j] == graph[j][i] <= 1`
+3. `graph[i][i] = 1`
+4. `1 <= initial.length < graph.length`
+5. `0 <= initial[i] < graph.length`
+
+
+## 题目大意
+
+(这个问题与 尽量减少恶意软件的传播 是一样的，不同之处用粗体表示。)在节点网络中，只有当 graph[i][j] = 1 时，每个节点 i 能够直接连接到另一个节点 j。一些节点 initial 最初被恶意软件感染。只要两个节点直接连接，且其中至少一个节点受到恶意软件的感染，那么两个节点都将被恶意软件感染。这种恶意软件的传播将继续，直到没有更多的节点可以被这种方式感染。假设 M(initial) 是在恶意软件停止传播之后，整个网络中感染恶意软件的最终节点数。我们可以从初始列表中删除一个节点，并完全移除该节点以及从该节点到任何其他节点的任何连接。如果移除这一节点将最小化 M(initial)， 则返回该节点。如果有多个节点满足条件，就返回索引最小的节点。
+
+提示：
+
+- 1 < graph.length = graph[0].length <= 300
+- 0 <= graph[i][j] == graph[j][i] <= 1
+- graph[i][i] = 1
+- 1 <= initial.length < graph.length
+- 0 <= initial[i] < graph.length
+
+
+## 解题思路
+
+
+- 这一题是第 924 题的加强版。给出一个节点之间的关系图，如果两个节点是连通的，那么病毒软件就会感染到连通的所有节点。现在如果想**完全彻底**移除一个病毒节点，能最大减少感染，请问移除哪个节点？如果多个节点都能减少感染量，优先移除序号偏小的那个节点。这一题的输入输出要求和第 924 题是完全一样的，区别在于第 924 题实际上是要求把一个病毒节点变成非病毒节点，而这道题是完全删除一个病毒节点以及它连接的所有边。
+- 这一题考察的是并查集。当然用 DFS 也可以解答这一题。并查集的做法如下，首先先将所有的病毒节点去掉，然后将所有连通块合并成一个节点。因为一个连通集合中的节点，要么全部被感染，要么全部不被感染，所以可以把每个集合整体考虑。然后统计所有集合直接相邻的病毒节点的个数。对于一个集合来说：
+    1. 如果直接相邻的病毒节点的个数为 0，则一定不会被感染，忽略这种情况；
+    2. 如果直接相邻的病毒节点的个数为 1，则将该病毒节点删除后，整个连通块就可以避免被感染，这种情况是我们寻找的答案；
+    3. 如果直接相邻的病毒节点的个数大于等于2，则不管删除哪个病毒节点，该连通块都仍会被感染，忽略这种情况；
+- 所以只需在所有第二种连通块（直接相邻的病毒节点的个数为 1 的连通块）中，找出节点个数最多的连通块，与它相邻的病毒节点就是我们要删除的节点；如果有多个连通块节点个数相同，再找出与之对应的编号最小的病毒节点即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func minMalwareSpread2(graph [][]int, initial []int) int {
+	if len(initial) == 0 {
+		return 0
+	}
+	uf, minIndex, count, countMap, malwareMap, infectMap := template.UnionFind{}, initial[0], math.MinInt64, map[int]int{}, map[int]int{}, map[int]map[int]int{}
+	for _, v := range initial {
+		malwareMap[v]++
+	}
+	uf.Init(len(graph))
+	for i := range graph {
+		for j := range graph[i] {
+			if i == j {
+				break
+			}
+			if graph[i][j] == 1 && malwareMap[i] == 0 && malwareMap[j] == 0 {
+				uf.Union(i, j)
+			}
+		}
+	}
+	for i := 0; i < len(graph); i++ {
+		countMap[uf.Find(i)]++
+	}
+	// 记录每个集合和直接相邻病毒节点的个数
+	for _, i := range initial {
+		for j := 0; j < len(graph); j++ {
+			if malwareMap[j] == 0 && graph[i][j] == 1 {
+				p := uf.Find(j)
+				if _, ok := infectMap[p]; ok {
+					infectMap[p][i] = i
+				} else {
+					tmp := map[int]int{}
+					tmp[i] = i
+					infectMap[p] = tmp
+				}
+			}
+		}
+	}
+	// 选出病毒节点中序号最小的
+	for _, v := range initial {
+		minIndex = min(minIndex, v)
+	}
+	for i, v := range infectMap {
+		// 找出只和一个病毒节点相连通的
+		if len(v) == 1 {
+			tmp := countMap[uf.Find(i)]
+			keys := []int{}
+			for k := range v {
+				keys = append(keys, k)
+			}
+			if count == tmp && minIndex > keys[0] {
+				minIndex = keys[0]
+			}
+			if count < tmp {
+				minIndex = keys[0]
+				count = tmp
+			}
+		}
+	}
+	return minIndex
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0927.Three-Equal-Parts/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0930.Binary-Subarrays-With-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0930.Binary-Subarrays-With-Sum.md b/website/content/ChapterFour/0900~0999/0930.Binary-Subarrays-With-Sum.md
new file mode 100644
index 000000000..b79ee1608
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0930.Binary-Subarrays-With-Sum.md
@@ -0,0 +1,72 @@
+# [930. Binary Subarrays With Sum](https://leetcode.com/problems/binary-subarrays-with-sum/)
+
+## 题目
+
+In an array A of 0s and 1s, how many non-empty subarrays have sum S?
+
+
+
+**Example 1**:
+
+```
+
+Input: A = [1,0,1,0,1], S = 2
+Output: 4
+Explanation: 
+The 4 subarrays are bolded below:
+[1,0,1,0,1]
+[1,0,1,0,1]
+[1,0,1,0,1]
+[1,0,1,0,1]
+
+```
+
+
+**Note**:
+
+- A.length <= 30000
+- 0 <= S <= A.length
+- A[i] is either 0 or 1.
+
+
+## 题目大意
+
+给定一个数组，数组里面的元素只有 0 和 1 两种。问这个数组有多少个和为 S 的子数组。
+
+## 解题思路
+
+这道题也是滑动窗口的题目。不断的加入右边的值，直到总和等于 S。[i,j] 区间内的和可以等于 [0,j] 的和减去 [0,i-1] 的和。在 freq 中不断的记下能使得和为 sum 的组合方法数，例如 freq[1] = 2 ，代表和为 1 有两种组合方法，(可能是 1 和 1，0 或者 0，1，这道题只管组合总数，没要求输出具体的组合对)。这道题的做法就是不断的累加，如果遇到比 S 多的情况，多出来的值就在 freq 中查表，看多出来的值可能是由几种情况构成的。一旦和与 S 相等以后，之后比 S 多出来的情况会越来越多(因为在不断累积，总和只会越来越大)，不断的查 freq 表就可以了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "fmt"
+
+func numSubarraysWithSum(A []int, S int) int {
+	freq, sum, res := make([]int, len(A)+1), 0, 0
+	freq[0] = 1
+	for _, v := range A {
+		t := sum + v - S
+		if t >= 0 {
+			// 总和有多余的，需要减去 t，除去的方法有 freq[t] 种
+			res += freq[t]
+		}
+		sum += v
+		freq[sum]++
+		fmt.Printf("freq = %v sum = %v res = %v t = %v\n", freq, sum, res, t)
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0928.Minimize-Malware-Spread-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0933.Number-of-Recent-Calls/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0933.Number-of-Recent-Calls.md b/website/content/ChapterFour/0900~0999/0933.Number-of-Recent-Calls.md
new file mode 100644
index 000000000..a5b0005a5
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0933.Number-of-Recent-Calls.md
@@ -0,0 +1,81 @@
+# [933. Number of Recent Calls](https://leetcode.com/problems/number-of-recent-calls/)
+
+
+
+## 题目
+
+Write a class `RecentCounter` to count recent requests.
+
+It has only one method: `ping(int t)`, where t represents some time in milliseconds.
+
+Return the number of `ping`s that have been made from 3000 milliseconds ago until now.
+
+Any ping with time in `[t - 3000, t]` will count, including the current ping.
+
+It is guaranteed that every call to `ping` uses a strictly larger value of `t` than before.
+
+**Example 1**:
+
+```
+Input: inputs = ["RecentCounter","ping","ping","ping","ping"], inputs = [[],[1],[100],[3001],[3002]]
+Output: [null,1,2,3,3]
+```
+
+**Note**:
+
+1. Each test case will have at most `10000` calls to `ping`.
+2. Each test case will call `ping` with strictly increasing values of `t`.
+3. Each call to ping will have `1 <= t <= 10^9`.
+
+
+## 题目大意
+
+写一个 RecentCounter 类来计算最近的请求。它只有一个方法：ping(int t)，其中 t 代表以毫秒为单位的某个时间。返回从 3000 毫秒前到现在的 ping 数。任何处于 [t - 3000, t] 时间范围之内的 ping 都将会被计算在内，包括当前（指 t 时刻）的 ping。保证每次对 ping 的调用都使用比之前更大的 t 值。
+ 
+提示：
+
+- 每个测试用例最多调用 10000 次 ping。
+- 每个测试用例会使用严格递增的 t 值来调用 ping。
+- 每次调用 ping 都有 1 <= t <= 10^9。
+
+
+## 解题思路
+
+- 要求设计一个类，可以用 `ping(t)` 的方法，计算 [t-3000, t] 区间内的 ping 数。t 是毫秒。
+- 这一题比较简单，`ping()` 方法用二分搜索即可。
+
+## 代码
+
+```go
+type RecentCounter struct {
+	list []int
+}
+
+func Constructor933() RecentCounter {
+	return RecentCounter{
+		list: []int{},
+	}
+}
+
+func (this *RecentCounter) Ping(t int) int {
+	this.list = append(this.list, t)
+	index := sort.Search(len(this.list), func(i int) bool { return this.list[i] >= t-3000 })
+	if index < 0 {
+		index = -index - 1
+	}
+	return len(this.list) - index
+}
+
+/**
+ * Your RecentCounter object will be instantiated and called as such:
+ * obj := Constructor();
+ * param_1 := obj.Ping(t);
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0930.Binary-Subarrays-With-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0938.Range-Sum-of-BST/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0938.Range-Sum-of-BST.md b/website/content/ChapterFour/0900~0999/0938.Range-Sum-of-BST.md
new file mode 100644
index 000000000..a0c1644ec
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0938.Range-Sum-of-BST.md
@@ -0,0 +1,87 @@
+# [938. Range Sum of BST](https://leetcode.com/problems/range-sum-of-bst/)
+
+
+## 题目
+
+Given the `root` node of a binary search tree, return *the sum of values of all nodes with a value in the range `[low, high]`*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/11/05/bst1.jpg](https://assets.leetcode.com/uploads/2020/11/05/bst1.jpg)
+
+```
+Input: root = [10,5,15,3,7,null,18], low = 7, high = 15
+Output: 32
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/11/05/bst2.jpg](https://assets.leetcode.com/uploads/2020/11/05/bst2.jpg)
+
+```
+Input: root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10
+Output: 23
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 2 * 10^4]`.
+- `1 <= Node.val <= 10^5`
+- `1 <= low <= high <= 10^5`
+- All `Node.val` are **unique**.
+
+## 题目大意
+
+给定二叉搜索树的根结点 root，返回值位于范围 [low, high] 之间的所有结点的值的和。
+
+## 解题思路
+
+- 简单题。因为二叉搜索树的有序性，先序遍历即为有序。遍历过程中判断节点值是否位于区间范围内，在区间内就累加，不在区间内节点就不管。最终输出累加和。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func rangeSumBST(root *TreeNode, low int, high int) int {
+	res := 0
+	preOrder(root, low, high, &res)
+	return res
+}
+
+func preOrder(root *TreeNode, low, high int, res *int) {
+	if root == nil {
+		return
+	}
+	if low <= root.Val && root.Val <= high {
+		*res += root.Val
+	}
+	preOrder(root.Left, low, high, res)
+	preOrder(root.Right, low, high, res)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0933.Number-of-Recent-Calls/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0942.DI-String-Match/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0942.DI-String-Match.md b/website/content/ChapterFour/0900~0999/0942.DI-String-Match.md
new file mode 100755
index 000000000..ccc5bfa53
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0942.DI-String-Match.md
@@ -0,0 +1,80 @@
+# [942. DI String Match](https://leetcode.com/problems/di-string-match/)
+
+
+## 题目
+
+Given a string `S` that **only** contains "I" (increase) or "D" (decrease), let `N = S.length`.
+
+Return **any** permutation `A` of `[0, 1, ..., N]` such that for all `i = 0, ..., N-1`:
+
+- If `S[i] == "I"`, then `A[i] < A[i+1]`
+- If `S[i] == "D"`, then `A[i] > A[i+1]`
+
+**Example 1**:
+
+    Input: "IDID"
+    Output: [0,4,1,3,2]
+
+**Example 2**:
+
+    Input: "III"
+    Output: [0,1,2,3]
+
+**Example 3**:
+
+    Input: "DDI"
+    Output: [3,2,0,1]
+
+**Note**:
+
+1. `1 <= S.length <= 10000`
+2. `S` only contains characters `"I"` or `"D"`.
+
+
+## 题目大意
+
+给定只含 "I"（增大）或 "D"（减小）的字符串 S ，令 N = S.length。返回 [0, 1, ..., N] 的任意排列 A 使得对于所有 i = 0, ..., N-1，都有：
+
+- 如果 S[i] == "I"，那么 A[i] < A[i+1]
+- 如果 S[i] == "D"，那么 A[i] > A[i+1]
+
+
+
+## 解题思路
+
+
+- 给出一个字符串，字符串中只有字符 `"I"` 和字符 `"D"`。字符 `"I"` 代表 `A[i] < A[i+1]`，字符 `"D"` 代表 `A[i] > A[i+1]` ，要求找到满足条件的任意组合。
+- 这一题也是水题，取出字符串长度即是最大数的数值，然后按照题意一次排出最终数组即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func diStringMatch(S string) []int {
+	result, maxNum, minNum, index := make([]int, len(S)+1), len(S), 0, 0
+	for _, ch := range S {
+		if ch == 'I' {
+			result[index] = minNum
+			minNum++
+		} else {
+			result[index] = maxNum
+			maxNum--
+		}
+		index++
+	}
+	result[index] = minNum
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0938.Range-Sum-of-BST/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0946.Validate-Stack-Sequences/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0946.Validate-Stack-Sequences.md b/website/content/ChapterFour/0900~0999/0946.Validate-Stack-Sequences.md
new file mode 100644
index 000000000..a71eed54f
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0946.Validate-Stack-Sequences.md
@@ -0,0 +1,71 @@
+# [946. Validate Stack Sequences](https://leetcode.com/problems/validate-stack-sequences/)
+
+## 题目
+
+Given two sequences pushed and popped with distinct values, return true if and only if this could have been the result of a sequence of push and pop operations on an initially empty stack.
+
+ 
+
+**Example 1**:
+
+```
+
+Input: pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
+Output: true
+Explanation: We might do the following sequence:
+push(1), push(2), push(3), push(4), pop() -> 4,
+push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1
+
+```
+
+**Example 2**:
+
+```
+
+Input: pushed = [1,2,3,4,5], popped = [4,3,5,1,2]
+Output: false
+Explanation: 1 cannot be popped before 2.
+
+```
+
+**Note**:
+
+1. 0 <= pushed.length == popped.length <= 1000
+2. 0 <= pushed[i], popped[i] < 1000
+3. pushed is a permutation of popped.
+4. pushed and popped have distinct values.
+
+## 题目大意
+
+给 2 个数组，一个数组里面代表的是 push 的顺序，另一个数组里面代表的是 pop 的顺序。问按照这样的顺序操作以后，最终能否把栈清空？
+
+## 解题思路
+
+这一题也是靠栈操作的题目，按照 push 数组的顺序先把压栈，然后再依次在 pop 里面找栈顶元素，找到了就 pop，直到遍历完 pop 数组，最终如果遍历完了 pop 数组，就代表清空了整个栈了。
+
+## 代码
+
+```go
+
+package leetcode
+
+func validateStackSequences(pushed []int, popped []int) bool {
+	stack, j, N := []int{}, 0, len(pushed)
+	for _, x := range pushed {
+		stack = append(stack, x)
+		for len(stack) != 0 && j < N && stack[len(stack)-1] == popped[j] {
+			stack = stack[0 : len(stack)-1]
+			j++
+		}
+	}
+	return j == N
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0942.DI-String-Match/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0947.Most-Stones-Removed-with-Same-Row-or-Column/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0947.Most-Stones-Removed-with-Same-Row-or-Column.md b/website/content/ChapterFour/0900~0999/0947.Most-Stones-Removed-with-Same-Row-or-Column.md
new file mode 100755
index 000000000..e19800760
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0947.Most-Stones-Removed-with-Same-Row-or-Column.md
@@ -0,0 +1,89 @@
+# [947. Most Stones Removed with Same Row or Column](https://leetcode.com/problems/most-stones-removed-with-same-row-or-column/)
+
+
+## 题目
+
+On a 2D plane, we place stones at some integer coordinate points. Each coordinate point may have at most one stone.
+
+Now, a *move* consists of removing a stone that shares a column or row with another stone on the grid.
+
+What is the largest possible number of moves we can make?
+
+**Example 1**:
+
+    Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
+    Output: 5
+
+**Example 2**:
+
+    Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
+    Output: 3
+
+**Example 3**:
+
+    Input: stones = [[0,0]]
+    Output: 0
+
+**Note**:
+
+1. `1 <= stones.length <= 1000`
+2. `0 <= stones[i][j] < 10000`
+
+
+## 题目大意
+
+在二维平面上，我们将石头放置在一些整数坐标点上。每个坐标点上最多只能有一块石头。现在，move 操作将会移除与网格上的某一块石头共享一列或一行的一块石头。我们最多能执行多少次 move 操作？
+
+提示：
+
+- 1 <= stones.length <= 1000
+- 0 <= stones[i][j] < 10000
+
+
+## 解题思路
+
+
+- 给出一个数组，数组中的元素是一系列的坐标点。现在可以移除一些坐标点，移除必须满足：移除的这个点，在相同的行或者列上有一个点。问最终可以移除多少个点。移除到最后必然有些点独占一行，那么这些点都不能被移除。
+- 这一题的解题思路是并查集。把所有共行或者共列的点都 `union()` 起来。不同集合之间是不能相互移除的。反证法：如果能移除，代表存在共行或者共列的情况，那么肯定是同一个集合了，这样就不满足不同集合了。最终剩下的点就是集合的个数，每个集合只会留下一个点。所以移除的点就是点的总数减去集合的个数 `len(stones) - uf.totalCount()`。
+- 如果暴力合并集合，时间复杂度也非常差，可以由优化的地方。再遍历所有点的过程中，可以把遍历过的行和列存起来。这里可以用 map 来记录，key 为行号，value 为上一次遍历过的点的序号。同样，列也可以用 map 存起来，key 为列号，value 为上一次遍历过的点的序号。经过这样的优化以后，时间复杂度会提高不少。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func removeStones(stones [][]int) int {
+	if len(stones) <= 1 {
+		return 0
+	}
+	uf, rowMap, colMap := template.UnionFind{}, map[int]int{}, map[int]int{}
+	uf.Init(len(stones))
+	for i := 0; i < len(stones); i++ {
+		if _, ok := rowMap[stones[i][0]]; ok {
+			uf.Union(rowMap[stones[i][0]], i)
+		} else {
+			rowMap[stones[i][0]] = i
+		}
+		if _, ok := colMap[stones[i][1]]; ok {
+			uf.Union(colMap[stones[i][1]], i)
+		} else {
+			colMap[stones[i][1]] = i
+		}
+	}
+	return len(stones) - uf.TotalCount()
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0946.Validate-Stack-Sequences/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0949.Largest-Time-for-Given-Digits/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0949.Largest-Time-for-Given-Digits.md b/website/content/ChapterFour/0900~0999/0949.Largest-Time-for-Given-Digits.md
new file mode 100644
index 000000000..4dee6fdb1
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0949.Largest-Time-for-Given-Digits.md
@@ -0,0 +1,85 @@
+# [949. Largest Time for Given Digits](https://leetcode.com/problems/largest-time-for-given-digits/)
+
+
+## 题目
+
+Given an array of 4 digits, return the largest 24 hour time that can be made.
+
+The smallest 24 hour time is 00:00, and the largest is 23:59. Starting from 00:00, a time is larger if more time has elapsed since midnight.
+
+Return the answer as a string of length 5. If no valid time can be made, return an empty string.
+
+**Example 1**:
+
+```
+Input: [1,2,3,4]
+Output: "23:41"
+```
+
+**Example 2**:
+
+```
+Input: [5,5,5,5]
+Output: ""
+```
+
+**Note**:
+
+1. `A.length == 4`
+2. `0 <= A[i] <= 9`
+
+## 题目大意
+
+给定一个由 4 位数字组成的数组，返回可以设置的符合 24 小时制的最大时间。最小的 24 小时制时间是 00:00，而最大的是 23:59。从 00:00 （午夜）开始算起，过得越久，时间越大。以长度为 5 的字符串返回答案。如果不能确定有效时间，则返回空字符串。
+
+## 解题思路
+
+- 给出 4 个数字，要求返回一个字符串，代表由这 4 个数字能组成的最大 24 小时制的时间。
+- 简单题，这一题直接暴力枚举就可以了。依次检查给出的 4 个数字每个排列组合是否是时间合法的。例如检查 10 * A[i] + A[j] 是不是小于 24， 10 * A[k] + A[l] 是不是小于 60。如果合法且比目前存在的最大时间更大，就更新这个最大时间。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "fmt"
+
+func largestTimeFromDigits(A []int) string {
+	flag, res := false, 0
+	for i := 0; i < 4; i++ {
+		for j := 0; j < 4; j++ {
+			if i == j {
+				continue
+			}
+			for k := 0; k < 4; k++ {
+				if i == k || j == k {
+					continue
+				}
+				l := 6 - i - j - k
+				hour := A[i]*10 + A[j]
+				min := A[k]*10 + A[l]
+				if hour < 24 && min < 60 {
+					if hour*60+min >= res {
+						res = hour*60 + min
+						flag = true
+					}
+				}
+			}
+		}
+	}
+	if flag {
+		return fmt.Sprintf("%02d:%02d", res/60, res%60)
+	} else {
+		return ""
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0947.Most-Stones-Removed-with-Same-Row-or-Column/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0952.Largest-Component-Size-by-Common-Factor/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0952.Largest-Component-Size-by-Common-Factor.md b/website/content/ChapterFour/0900~0999/0952.Largest-Component-Size-by-Common-Factor.md
new file mode 100755
index 000000000..f40b90a1c
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0952.Largest-Component-Size-by-Common-Factor.md
@@ -0,0 +1,126 @@
+# [952. Largest Component Size by Common Factor](https://leetcode.com/problems/largest-component-size-by-common-factor/)
+
+
+## 题目
+
+Given a non-empty array of unique positive integers `A`, consider the following graph:
+
+- There are `A.length` nodes, labelled `A[0]` to `A[A.length - 1];`
+- There is an edge between `A[i]` and `A[j]` if and only if `A[i]`and `A[j]` share a common factor greater than 1.
+
+Return the size of the largest connected component in the graph.
+
+**Example 1**:
+
+    Input: [4,6,15,35]
+    Output: 4
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/01/ex1.png)
+
+**Example 2**:
+
+    Input: [20,50,9,63]
+    Output: 2
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/01/ex2.png)
+
+**Example 3**:
+
+    Input: [2,3,6,7,4,12,21,39]
+    Output: 8
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/01/ex3.png)
+
+**Note**:
+
+1. `1 <= A.length <= 20000`
+2. `1 <= A[i] <= 100000`
+
+
+## 题目大意
+
+给定一个由不同正整数的组成的非空数组 A，考虑下面的图：
+
+有 A.length 个节点，按从 A[0] 到 A[A.length - 1] 标记；
+只有当 A[i] 和 A[j] 共用一个大于 1 的公因数时，A[i] 和 A[j] 之间才有一条边。
+返回图中最大连通组件的大小。
+
+提示：
+
+1. 1 <= A.length <= 20000
+2. 1 <= A[i] <= 100000
+
+## 解题思路
+
+- 给出一个数组，数组中的元素如果每两个元素有公约数，那么这两个元素可以算有关系。所有有关系的数可以放在一个集合里，问这个数组里面有关系的元素组成的集合里面最多有多少个元素。
+- 这一题读完题直觉就是用并查集来解题。首先可以用暴力的解法尝试。用 2 层循环，两两比较有没有公约数，如果有公约数就 `union()` 到一起。提交以后出现 TLE，其实看一下数据规模就知道会超时，`1 <= A.length <= 20000`。注意到 `1 <= A[i] <= 100000`，开根号以后最后才 316.66666，这个规模的数不大。所以把每个数小于根号自己的因子都找出来，例如 `6 = 2 * 3`，`15 = 3 * 5`，那么把 6 和 2，6 和 3 都 `union()`，15 和 3，15 和 5 都 `union()`，最终遍历所有的集合，找到最多元素的集合，输出它包含的元素值。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+// 解法一 并查集 UnionFind
+func largestComponentSize(A []int) int {
+	maxElement, uf, countMap, res := 0, template.UnionFind{}, map[int]int{}, 1
+	for _, v := range A {
+		maxElement = max(maxElement, v)
+	}
+	uf.Init(maxElement + 1)
+	for _, v := range A {
+		for k := 2; k*k <= v; k++ {
+			if v%k == 0 {
+				uf.Union(v, k)
+				uf.Union(v, v/k)
+			}
+		}
+	}
+	for _, v := range A {
+		countMap[uf.Find(v)]++
+		res = max(res, countMap[uf.Find(v)])
+	}
+	return res
+}
+
+// 解法二 UnionFindCount
+func largestComponentSize1(A []int) int {
+	uf, factorMap := template.UnionFindCount{}, map[int]int{}
+	uf.Init(len(A))
+	for i, v := range A {
+		for k := 2; k*k <= v; k++ {
+			if v%k == 0 {
+				if _, ok := factorMap[k]; !ok {
+					factorMap[k] = i
+				} else {
+					uf.Union(i, factorMap[k])
+				}
+				if _, ok := factorMap[v/k]; !ok {
+					factorMap[v/k] = i
+				} else {
+					uf.Union(i, factorMap[v/k])
+				}
+			}
+		}
+		if _, ok := factorMap[v]; !ok {
+			factorMap[v] = i
+		} else {
+			uf.Union(i, factorMap[v])
+		}
+	}
+	return uf.MaxUnionCount()
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0949.Largest-Time-for-Given-Digits/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0953.Verifying-an-Alien-Dictionary/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0953.Verifying-an-Alien-Dictionary.md b/website/content/ChapterFour/0900~0999/0953.Verifying-an-Alien-Dictionary.md
new file mode 100755
index 000000000..f50c0855a
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0953.Verifying-an-Alien-Dictionary.md
@@ -0,0 +1,88 @@
+# [953. Verifying an Alien Dictionary](https://leetcode.com/problems/verifying-an-alien-dictionary/)
+
+
+## 题目
+
+In an alien language, surprisingly they also use english lowercase letters, but possibly in a different `order`. The `order`of the alphabet is some permutation of lowercase letters.
+
+Given a sequence of `words` written in the alien language, and the `order` of the alphabet, return `true` if and only if the given `words` are sorted lexicographicaly in this alien language.
+
+**Example 1**:
+
+    Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
+    Output: true
+    Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.
+
+**Example 2**:
+
+    Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
+    Output: false
+    Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.
+
+**Example 3**:
+
+    Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
+    Output: false
+    Explanation: The first three characters "app" match, and the second string is shorter (in size.) According to lexicographical rules "apple" > "app", because 'l' > '∅', where '∅' is defined as the blank character which is less than any other character (More info).
+
+**Note**:
+
+1. `1 <= words.length <= 100`
+2. `1 <= words[i].length <= 20`
+3. `order.length == 26`
+4. All characters in `words[i]` and `order` are english lowercase letters.
+
+
+## 题目大意
+
+某种外星语也使用英文小写字母，但可能顺序 order 不同。字母表的顺序（order）是一些小写字母的排列。给定一组用外星语书写的单词 words，以及其字母表的顺序 order，只有当给定的单词在这种外星语中按字典序排列时，返回 true；否则，返回 false。
+
+
+
+## 解题思路
+
+
+- 这一题是简单题。给出一个字符串数组，判断把字符串数组里面字符串是否是按照 order 的排序排列的。order 是给出个一个字符串排序。这道题的解法是把 26 个字母的顺序先存在 map 中，然后依次遍历判断字符串数组里面字符串的大小。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func isAlienSorted(words []string, order string) bool {
+	if len(words) < 2 {
+		return true
+	}
+	hash := make(map[byte]int)
+	for i := 0; i < len(order); i++ {
+		hash[order[i]] = i
+	}
+	for i := 0; i < len(words)-1; i++ {
+		pointer, word, wordplus := 0, words[i], words[i+1]
+		for pointer < len(word) && pointer < len(wordplus) {
+			if hash[word[pointer]] > hash[wordplus[pointer]] {
+				return false
+			}
+			if hash[word[pointer]] < hash[wordplus[pointer]] {
+				break
+			} else {
+				pointer = pointer + 1
+			}
+		}
+		if pointer < len(word) && pointer >= len(wordplus) {
+			return false
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0952.Largest-Component-Size-by-Common-Factor/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0958.Check-Completeness-of-a-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0958.Check-Completeness-of-a-Binary-Tree.md b/website/content/ChapterFour/0900~0999/0958.Check-Completeness-of-a-Binary-Tree.md
new file mode 100644
index 000000000..d9a30b23b
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0958.Check-Completeness-of-a-Binary-Tree.md
@@ -0,0 +1,96 @@
+# [958. Check Completeness of a Binary Tree](https://leetcode.com/problems/check-completeness-of-a-binary-tree/)
+
+
+## 题目
+
+Given the `root` of a binary tree, determine if it is a *complete binary tree*.
+
+In a **[complete binary tree](http://en.wikipedia.org/wiki/Binary_tree#Types_of_binary_trees)**, every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between `1` and `2h` nodes inclusive at the last level `h`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2018/12/15/complete-binary-tree-1.png](https://assets.leetcode.com/uploads/2018/12/15/complete-binary-tree-1.png)
+
+```
+Input: root = [1,2,3,4,5,6]
+Output: true
+Explanation: Every level before the last is full (ie. levels with node-values {1} and {2, 3}), and all nodes in the last level ({4, 5, 6}) are as far left as possible.
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2018/12/15/complete-binary-tree-2.png](https://assets.leetcode.com/uploads/2018/12/15/complete-binary-tree-2.png)
+
+```
+Input: root = [1,2,3,4,5,null,7]
+Output: false
+Explanation: The node with value 7 isn't as far left as possible.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 100]`.
+- `1 <= Node.val <= 1000`
+
+## 题目大意
+
+给定一个二叉树，确定它是否是一个完全二叉树。
+
+百度百科中对完全二叉树的定义如下：
+
+若设二叉树的深度为 h，除第 h 层外，其它各层 (1～h-1) 的结点数都达到最大个数，第 h 层所有的结点都连续集中在最左边，这就是完全二叉树。（注：第 h 层可能包含 1~ 2h 个节点。）
+
+## 解题思路
+
+- 这一题是按层序遍历的变种题。
+- 判断每个节点的左孩子是否为空。
+- 类似的题目，第 102，107，199 题都是按层序遍历的。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func isCompleteTree(root *TreeNode) bool {
+	queue, found := []*TreeNode{root}, false
+	for len(queue) > 0 {
+		node := queue[0] //取出每一层的第一个节点
+		queue = queue[1:]
+		if node == nil {
+			found = true
+		} else {
+			if found {
+				return false // 层序遍历中，两个不为空的节点中出现一个 nil
+			}
+			//如果左孩子为nil，则append进去的node.Left为nil
+			queue = append(queue, node.Left, node.Right)
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0953.Verifying-an-Alien-Dictionary/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0959.Regions-Cut-By-Slashes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0959.Regions-Cut-By-Slashes.md b/website/content/ChapterFour/0900~0999/0959.Regions-Cut-By-Slashes.md
new file mode 100755
index 000000000..1aba869f9
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0959.Regions-Cut-By-Slashes.md
@@ -0,0 +1,157 @@
+# [959. Regions Cut By Slashes](https://leetcode.com/problems/regions-cut-by-slashes/)
+
+
+## 题目
+
+In a N x N `grid` composed of 1 x 1 squares, each 1 x 1 square consists of a `/`, `\`, or blank space. These characters divide the square into contiguous regions.
+
+(Note that backslash characters are escaped, so a `\` is represented as `"\\"`.)
+
+Return the number of regions.
+
+**Example 1**:
+
+    Input:
+    [
+      " /",
+      "/ "
+    ]
+    Output: 2
+    Explanation: The 2x2 grid is as follows:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/1.png)
+
+**Example 2**:
+
+    Input:
+    [
+      " /",
+      "  "
+    ]
+    Output: 1
+    Explanation: The 2x2 grid is as follows:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/2.png)
+
+**Example 3**:
+
+    Input:
+    [
+      "\\/",
+      "/\\"
+    ]
+    Output: 4
+    Explanation: (Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.)
+    The 2x2 grid is as follows:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/3.png)
+
+**Example 4**:
+
+    Input:
+    [
+      "/\\",
+      "\\/"
+    ]
+    Output: 5
+    Explanation: (Recall that because \ characters are escaped, "/\\" refers to /\, and "\\/" refers to \/.)
+    The 2x2 grid is as follows:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/4.png)
+
+**Example 5**:
+
+    Input:
+    [
+      "//",
+      "/ "
+    ]
+    Output: 3
+    Explanation: The 2x2 grid is as follows:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/5.png)
+
+**Note**:
+
+1. `1 <= grid.length == grid[0].length <= 30`
+2. `grid[i][j]` is either `'/'`, `'\'`, or `' '`.
+
+
+## 题目大意
+
+在由 1 x 1 方格组成的 N x N 网格 grid 中，每个 1 x 1 方块由 /、\ 或空格构成。这些字符会将方块划分为一些共边的区域。(请注意，反斜杠字符是转义的，因此 \ 用 "\\" 表示)返回区域的数目。
+
+
+提示：
+
+- 1 <= grid.length == grid[0].length <= 30
+- grid[i][j] 是 '/'、'\'、或 ' '。
+
+## 解题思路
+
+
+- 给出一个字符串，代表的是 `N x N` 正方形中切分的情况，有 2 种切分的情况 `'\'` 和 `'/'` ，即从左上往右下切和从右上往左下切。问按照给出的切分方法，能把 `N x N` 正方形切成几部分？
+- 这一题解题思路是并查集。先将每个 `1*1` 的正方形切分成下图的样子。分成 4 小块。然后按照题目给的切分图来合并各个小块。
+
+![](https://img.halfrost.com/Leetcode/leetcode_959.png)
+
+- 遇到 `'\\'`，就把第 0 块和第 1 块 `union()` 起来，第 2 块和第 3 块 `union()` 起来；遇到 `'/'`，就把第 0 块和第 3 块 `union()` 起来，第 2 块和第 1 块 `union()` 起来；遇到 `' '`，就把第 0 块和第 1 块 `union()` 起来，第 2 块和第 1 块 `union()` 起来，第 2 块和第 3 块 `union()` 起来，即 4 块都 `union()` 起来；最后还需要记得上一行和下一行还需要 `union()`，即本行的第 2 块和下一行的第 0 块 `union()` 起来；左边一列和右边一列也需要 `union()`。即本列的第 1 块和右边一列的第 3 块 `union()` 起来。最后计算出集合总个数就是最终答案了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func regionsBySlashes(grid []string) int {
+	size := len(grid)
+	uf := template.UnionFind{}
+	uf.Init(4 * size * size)
+	for i := 0; i < size; i++ {
+		for j := 0; j < size; j++ {
+			switch grid[i][j] {
+			case '\\':
+				uf.Union(getFaceIdx(size, i, j, 0), getFaceIdx(size, i, j, 1))
+				uf.Union(getFaceIdx(size, i, j, 2), getFaceIdx(size, i, j, 3))
+			case '/':
+				uf.Union(getFaceIdx(size, i, j, 0), getFaceIdx(size, i, j, 3))
+				uf.Union(getFaceIdx(size, i, j, 2), getFaceIdx(size, i, j, 1))
+			case ' ':
+				uf.Union(getFaceIdx(size, i, j, 0), getFaceIdx(size, i, j, 1))
+				uf.Union(getFaceIdx(size, i, j, 2), getFaceIdx(size, i, j, 1))
+				uf.Union(getFaceIdx(size, i, j, 2), getFaceIdx(size, i, j, 3))
+			}
+			if i < size-1 {
+				uf.Union(getFaceIdx(size, i, j, 2), getFaceIdx(size, i+1, j, 0))
+			}
+			if j < size-1 {
+				uf.Union(getFaceIdx(size, i, j, 1), getFaceIdx(size, i, j+1, 3))
+			}
+		}
+	}
+	count := 0
+	for i := 0; i < 4*size*size; i++ {
+		if uf.Find(i) == i {
+			count++
+		}
+	}
+	return count
+}
+
+func getFaceIdx(size, i, j, k int) int {
+	return 4*(i*size+j) + k
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0958.Check-Completeness-of-a-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0961.N-Repeated-Element-in-Size-2N-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0961.N-Repeated-Element-in-Size-2N-Array.md b/website/content/ChapterFour/0900~0999/0961.N-Repeated-Element-in-Size-2N-Array.md
new file mode 100755
index 000000000..e6f46cf28
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0961.N-Repeated-Element-in-Size-2N-Array.md
@@ -0,0 +1,67 @@
+# [961. N-Repeated Element in Size 2N Array](https://leetcode.com/problems/n-repeated-element-in-size-2n-array/)
+
+
+## 题目
+
+In a array `A` of size `2N`, there are `N+1` unique elements, and exactly one of these elements is repeated N times.
+
+Return the element repeated `N` times.
+
+**Example 1**:
+
+    Input: [1,2,3,3]
+    Output: 3
+
+**Example 2**:
+
+    Input: [2,1,2,5,3,2]
+    Output: 2
+
+**Example 3**:
+
+    Input: [5,1,5,2,5,3,5,4]
+    Output: 5
+
+**Note**:
+
+1. `4 <= A.length <= 10000`
+2. `0 <= A[i] < 10000`
+3. `A.length` is even
+
+
+## 题目大意
+
+在大小为 2N 的数组 A 中有 N+1 个不同的元素，其中有一个元素重复了 N 次。返回重复了 N 次的那个元素。
+
+
+## 解题思路
+
+
+- 简单题。数组中有 2N 个数，有 N + 1 个数是不重复的，这之中有一个数重复了 N 次，请找出这个数。解法非常简单，把所有数都存入 map 中，如果遇到存在的 key 就返回这个数。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func repeatedNTimes(A []int) int {
+	kv := make(map[int]struct{})
+	for _, val := range A {
+		if _, ok := kv[val]; ok {
+			return val
+		}
+		kv[val] = struct{}{}
+	}
+	return 0
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0959.Regions-Cut-By-Slashes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0966.Vowel-Spellchecker/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0966.Vowel-Spellchecker.md b/website/content/ChapterFour/0900~0999/0966.Vowel-Spellchecker.md
new file mode 100644
index 000000000..645daa8d6
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0966.Vowel-Spellchecker.md
@@ -0,0 +1,134 @@
+# [966. Vowel Spellchecker](https://leetcode.com/problems/vowel-spellchecker/)
+
+
+## 题目
+
+Given a `wordlist`, we want to implement a spellchecker that converts a query word into a correct word.
+
+For a given `query` word, the spell checker handles two categories of spelling mistakes:
+
+- Capitalization: If the query matches a word in the wordlist (**case-insensitive**), then the query word is returned with the same case as the case in the wordlist.
+    - Example: `wordlist = ["yellow"]`, `query = "YellOw"`: `correct = "yellow"`
+    - Example: `wordlist = ["Yellow"]`, `query = "yellow"`: `correct = "Yellow"`
+    - Example: `wordlist = ["yellow"]`, `query = "yellow"`: `correct = "yellow"`
+- Vowel Errors: If after replacing the vowels ('a', 'e', 'i', 'o', 'u') of the query word with any vowel individually, it matches a word in the wordlist (**case-insensitive**), then the query word is returned with the same case as the match in the wordlist.
+    - Example: `wordlist = ["YellOw"]`, `query = "yollow"`: `correct = "YellOw"`
+    - Example: `wordlist = ["YellOw"]`, `query = "yeellow"`: `correct = ""` (no match)
+    - Example: `wordlist = ["YellOw"]`, `query = "yllw"`: `correct = ""` (no match)
+
+In addition, the spell checker operates under the following precedence rules:
+
+- When the query exactly matches a word in the wordlist (**case-sensitive**), you should return the same word back.
+- When the query matches a word up to capitlization, you should return the first such match in the wordlist.
+- When the query matches a word up to vowel errors, you should return the first such match in the wordlist.
+- If the query has no matches in the wordlist, you should return the empty string.
+
+Given some `queries`, return a list of words `answer`, where `answer[i]` is the correct word for `query = queries[i]`.
+
+**Example 1:**
+
+```
+Input:wordlist = ["KiTe","kite","hare","Hare"], queries = ["kite","Kite","KiTe","Hare","HARE","Hear","hear","keti","keet","keto"]
+Output:["kite","KiTe","KiTe","Hare","hare","","","KiTe","","KiTe"]
+```
+
+**Note:**
+
+- `1 <= wordlist.length <= 5000`
+- `1 <= queries.length <= 5000`
+- `1 <= wordlist[i].length <= 7`
+- `1 <= queries[i].length <= 7`
+- All strings in `wordlist` and `queries` consist only of **english** letters.
+
+## 题目大意
+
+在给定单词列表 wordlist 的情况下，我们希望实现一个拼写检查器，将查询单词转换为正确的单词。
+
+对于给定的查询单词 query，拼写检查器将会处理两类拼写错误：
+
+- 大小写：如果查询匹配单词列表中的某个单词（不区分大小写），则返回的正确单词与单词列表中的大小写相同。
+    - 例如：wordlist = ["yellow"], query = "YellOw": correct = "yellow"
+    - 例如：wordlist = ["Yellow"], query = "yellow": correct = "Yellow"
+    - 例如：wordlist = ["yellow"], query = "yellow": correct = "yellow"
+- 元音错误：如果在将查询单词中的元音（‘a’、‘e’、‘i’、‘o’、‘u’）分别替换为任何元音后，能与单词列表中的单词匹配（不区分大小写），则返回的正确单词与单词列表中的匹配项大小写相同。
+    - 例如：wordlist = ["YellOw"], query = "yollow": correct = "YellOw"
+    - 例如：wordlist = ["YellOw"], query = "yeellow": correct = "" （无匹配项）
+    - 例如：wordlist = ["YellOw"], query = "yllw": correct = "" （无匹配项）
+
+此外，拼写检查器还按照以下优先级规则操作：
+
+- 当查询完全匹配单词列表中的某个单词（区分大小写）时，应返回相同的单词。
+- 当查询匹配到大小写问题的单词时，您应该返回单词列表中的第一个这样的匹配项。
+- 当查询匹配到元音错误的单词时，您应该返回单词列表中的第一个这样的匹配项。
+- 如果该查询在单词列表中没有匹配项，则应返回空字符串。
+
+给出一些查询 queries，返回一个单词列表 answer，其中 answer[i] 是由查询 query = queries[i] 得到的正确单词。
+
+## 解题思路
+
+- 读完题，很明显需要用 `map` 来解题。依题意分为 3 种情况，查询字符串完全匹配；查询字符串只是大小写不同；查询字符串有元音错误。第一种情况用 `map` `key` 直接匹配即可。第二种情况，利用 `map` 将单词从小写形式转换成原单词正确的大小写形式。第三种情况，利用 `map` 将单词从忽略元音的小写形式换成原单词正确形式。最后注意一下题目最后给的 4 个优先级规则即可。
+
+## 代码
+
+```go
+package leetcode
+
+import "strings"
+
+func spellchecker(wordlist []string, queries []string) []string {
+	wordsPerfect, wordsCap, wordsVowel := map[string]bool{}, map[string]string{}, map[string]string{}
+	for _, word := range wordlist {
+		wordsPerfect[word] = true
+		wordLow := strings.ToLower(word)
+		if _, ok := wordsCap[wordLow]; !ok {
+			wordsCap[wordLow] = word
+		}
+		wordLowVowel := devowel(wordLow)
+		if _, ok := wordsVowel[wordLowVowel]; !ok {
+			wordsVowel[wordLowVowel] = word
+		}
+	}
+	res, index := make([]string, len(queries)), 0
+	for _, query := range queries {
+		if _, ok := wordsPerfect[query]; ok {
+			res[index] = query
+			index++
+			continue
+		}
+		queryL := strings.ToLower(query)
+		if v, ok := wordsCap[queryL]; ok {
+			res[index] = v
+			index++
+			continue
+		}
+
+		queryLV := devowel(queryL)
+		if v, ok := wordsVowel[queryLV]; ok {
+			res[index] = v
+			index++
+			continue
+		}
+		res[index] = ""
+		index++
+	}
+	return res
+
+}
+
+func devowel(word string) string {
+	runes := []rune(word)
+	for k, c := range runes {
+		if c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' {
+			runes[k] = '*'
+		}
+	}
+	return string(runes)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0961.N-Repeated-Element-in-Size-2N-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0968.Binary-Tree-Cameras/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0968.Binary-Tree-Cameras.md b/website/content/ChapterFour/0900~0999/0968.Binary-Tree-Cameras.md
new file mode 100755
index 000000000..3806dc55a
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0968.Binary-Tree-Cameras.md
@@ -0,0 +1,110 @@
+# [968. Binary Tree Cameras](https://leetcode.com/problems/binary-tree-cameras/)
+
+## 题目
+
+Given a binary tree, we install cameras on the nodes of the tree.
+
+Each camera at a node can monitor **its parent, itself, and its immediate children**.
+
+Calculate the minimum number of cameras needed to monitor all nodes of the tree.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_01.png](https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_01.png)
+
+    Input: [0,0,null,0,0]
+    Output: 1
+    Explanation: One camera is enough to monitor all nodes if placed as shown.
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_02.png](https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_02.png)
+
+    Input: [0,0,null,0,null,0,null,null,0]
+    Output: 2
+    Explanation: At least two cameras are needed to monitor all nodes of the tree. The above image shows one of the valid configurations of camera placement.
+
+**Note**:
+
+1. The number of nodes in the given tree will be in the range `[1, 1000]`.
+2. **Every** node has value 0.
+
+
+## 题目大意
+
+给定一个二叉树，我们在树的节点上安装摄像头。节点上的每个摄影头都可以监视其父对象、自身及其直接子对象。计算监控树的所有节点所需的最小摄像头数量。
+
+提示：
+
+1. 给定树的节点数的范围是 [1, 1000]。
+2. 每个节点的值都是 0。
+
+
+
+## 解题思路
+
+- 给出一棵树，要求在这个树上面放摄像头，一个摄像头最多可以监视 4 个节点，2 个孩子，本身节点，还有父亲节点。问最少放多少个摄像头可以覆盖树上的所有节点。
+- 这一题可以用贪心思想来解题。先将节点分为 3 类，第一类，叶子节点，第二类，包含叶子节点的节点，第三类，其中一个叶子节点上放了摄像头的。按照这个想法，将树的每个节点染色，如下图。
+
+![](https://img.halfrost.com/Leetcode/leetcode_968_1.png)
+
+- 所有包含叶子节点的节点，可以放一个摄像头，这个可以覆盖至少 3 个节点，如果还有父节点的话，可以覆盖 4 个节点。所以贪心的策略是从最下层的叶子节点开始往上“染色”，先把最下面的一层 1 染色。标 1 的节点都是要放一个摄像头的。如果叶子节点中包含 1 的节点，那么再将这个节点染成 2 。如下图的黄色节点。黄色节点代表的是不用放摄像头的节点，因为它被叶子节点的摄像头覆盖了。出现了 2 的节点以后，再往上的节点又再次恢复成叶子节点 0 。如此类推，直到推到根节点。
+
+![](https://img.halfrost.com/Leetcode/leetcode_968_2.png)
+
+- 最后根节点还需要注意多种情况，根节点可能是叶子节点 0，那么最终答案还需要 + 1，因为需要在根节点上放一个摄像头，不然根节点覆盖不到了。根节点也有可能是 1 或者 2，这两种情况都不需要增加摄像头了，以为都覆盖到了。按照上述的方法，递归即可得到答案。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+type status int
+
+const (
+	isLeaf status = iota
+	parentofLeaf
+	isMonitoredWithoutCamera
+)
+
+func minCameraCover(root *TreeNode) int {
+	res := 0
+	if minCameraCoverDFS(root, &res) == isLeaf {
+		res++
+	}
+	return res
+}
+
+func minCameraCoverDFS(root *TreeNode, res *int) status {
+	if root == nil {
+		return 2
+	}
+	left, right := minCameraCoverDFS(root.Left, res), minCameraCoverDFS(root.Right, res)
+	if left == isLeaf || right == isLeaf {
+		*res++
+		return parentofLeaf
+	} else if left == parentofLeaf || right == parentofLeaf {
+		return isMonitoredWithoutCamera
+	} else {
+		return isLeaf
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0966.Vowel-Spellchecker/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0969.Pancake-Sorting/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0969.Pancake-Sorting.md b/website/content/ChapterFour/0900~0999/0969.Pancake-Sorting.md
new file mode 100644
index 000000000..29ed06f44
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0969.Pancake-Sorting.md
@@ -0,0 +1,103 @@
+# [969. Pancake Sorting](https://leetcode.com/problems/pancake-sorting/)
+
+## 题目
+
+Given an array A, we can perform a pancake flip: We choose some positive integer k <= A.length, then reverse the order of the first k elements of A.  We want to perform zero or more pancake flips (doing them one after another in succession) to sort the array A.
+
+Return the k-values corresponding to a sequence of pancake flips that sort A.  Any valid answer that sorts the array within 10 * A.length flips will be judged as correct.
+
+**Example 1**:
+
+```
+
+Input: [3,2,4,1]
+Output: [4,2,4,3]
+Explanation: 
+We perform 4 pancake flips, with k values 4, 2, 4, and 3.
+Starting state: A = [3, 2, 4, 1]
+After 1st flip (k=4): A = [1, 4, 2, 3]
+After 2nd flip (k=2): A = [4, 1, 2, 3]
+After 3rd flip (k=4): A = [3, 2, 1, 4]
+After 4th flip (k=3): A = [1, 2, 3, 4], which is sorted. 
+
+```
+
+**Example 2**:
+
+```
+
+Input: [1,2,3]
+Output: []
+Explanation: The input is already sorted, so there is no need to flip anything.
+Note that other answers, such as [3, 3], would also be accepted.
+
+```
+
+**Note**:
+
+- 1 <= A.length <= 100
+- A[i] is a permutation of [1, 2, ..., A.length]
+
+## 题目大意
+
+给定一个数组，要求输出“煎饼排序”的步骤，使得最终数组是从小到大有序的。“煎饼排序”，每次排序都反转前 n 个数，n 小于数组的长度。
+
+## 解题思路
+
+这道题的思路是，每次找到当前数组中无序段中最大的值，（初始的时候，整个数组相当于都是无序段），将最大值的下标 i 进行“煎饼排序”，前 i 个元素都反转一遍。这样最大值就到了第一个位置了。然后紧接着再进行一次数组总长度 n 的“煎饼排序”，目的是使最大值到数组最后一位，这样它的位置就归位了。那么数组的无序段为 n-1 。然后用这个方法不断的循环，直至数组中每个元素都到了排序后最终的位置下标上了。最终数组就有序了。
+
+这道题有一个特殊点在于，数组里面的元素都是自然整数，那么最终数组排序完成以后，数组的长度就是最大值。所以找最大值也不需要遍历一次数组了，直接取出长度就是最大值。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func pancakeSort(A []int) []int {
+	if len(A) == 0 {
+		return []int{}
+	}
+	right := len(A)
+	var (
+		ans []int
+	)
+	for right > 0 {
+		idx := find(A, right)
+		if idx != right-1 {
+			reverse969(A, 0, idx)
+			reverse969(A, 0, right-1)
+			ans = append(ans, idx+1, right)
+		}
+		right--
+	}
+
+	return ans
+}
+
+func reverse969(nums []int, l, r int) {
+	for l < r {
+		nums[l], nums[r] = nums[r], nums[l]
+		l++
+		r--
+	}
+}
+
+func find(nums []int, t int) int {
+	for i, num := range nums {
+		if num == t {
+			return i
+		}
+	}
+	return -1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0968.Binary-Tree-Cameras/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0970.Powerful-Integers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0970.Powerful-Integers.md b/website/content/ChapterFour/0900~0999/0970.Powerful-Integers.md
new file mode 100755
index 000000000..444df1b99
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0970.Powerful-Integers.md
@@ -0,0 +1,95 @@
+# [970. Powerful Integers](https://leetcode.com/problems/powerful-integers/)
+
+
+## 题目
+
+Given two positive integers `x` and `y`, an integer is *powerful* if it is equal to `x^i + y^j` for some integers `i >= 0` and `j >= 0`.
+
+Return a list of all *powerful* integers that have value less than or equal to `bound`.
+
+You may return the answer in any order. In your answer, each value should occur at most once.
+
+**Example 1**:
+
+    Input: x = 2, y = 3, bound = 10
+    Output: [2,3,4,5,7,9,10]
+    Explanation: 
+    2 = 2^0 + 3^0
+    3 = 2^1 + 3^0
+    4 = 2^0 + 3^1
+    5 = 2^1 + 3^1
+    7 = 2^2 + 3^1
+    9 = 2^3 + 3^0
+    10 = 2^0 + 3^2
+
+**Example 2**:
+
+    Input: x = 3, y = 5, bound = 15
+    Output: [2,4,6,8,10,14]
+
+**Note**:
+
+- `1 <= x <= 100`
+- `1 <= y <= 100`
+- `0 <= bound <= 10^6`
+
+
+## 题目大意
+
+给定两个正整数 x 和 y，如果某一整数等于 x^i + y^j，其中整数 i >= 0 且 j >= 0，那么我们认为该整数是一个强整数。返回值小于或等于 bound 的所有强整数组成的列表。你可以按任何顺序返回答案。在你的回答中，每个值最多出现一次。
+
+
+## 解题思路
+
+
+- 简答题，题目要求找出满足 `x^i + y^j ≤ bound` 条件的所有解。题目要求输出中不能重复，所以用 map 来去重。剩下的就是 `n^2` 暴力循环枚举所有解。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func powerfulIntegers(x int, y int, bound int) []int {
+	if x == 1 && y == 1 {
+		if bound < 2 {
+			return []int{}
+		}
+		return []int{2}
+	}
+	if x > y {
+		x, y = y, x
+	}
+	visit, result := make(map[int]bool), make([]int, 0)
+	for i := 0; ; i++ {
+		found := false
+		for j := 0; pow(x, i)+pow(y, j) <= bound; j++ {
+			v := pow(x, i) + pow(y, j)
+			if !visit[v] {
+				found = true
+				visit[v] = true
+				result = append(result, v)
+			}
+		}
+		if !found {
+			break
+		}
+	}
+	return result
+}
+
+func pow(x, i int) int {
+	return int(math.Pow(float64(x), float64(i)))
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0969.Pancake-Sorting/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal.md b/website/content/ChapterFour/0900~0999/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal.md
new file mode 100644
index 000000000..5433a7ae8
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal.md
@@ -0,0 +1,113 @@
+# [971. Flip Binary Tree To Match Preorder Traversal](https://leetcode.com/problems/flip-binary-tree-to-match-preorder-traversal/)
+
+
+## 题目
+
+You are given the `root` of a binary tree with `n` nodes, where each node is uniquely assigned a value from `1` to `n`. You are also given a sequence of `n` values `voyage`, which is the **desired** **[pre-order traversal](https://en.wikipedia.org/wiki/Tree_traversal#Pre-order)** of the binary tree.
+
+Any node in the binary tree can be **flipped** by swapping its left and right subtrees. For example, flipping node 1 will have the following effect:
+
+![https://assets.leetcode.com/uploads/2021/02/15/fliptree.jpg](https://assets.leetcode.com/uploads/2021/02/15/fliptree.jpg)
+
+Flip the **smallest** number of nodes so that the **pre-order traversal** of the tree **matches** `voyage`.
+
+Return *a list of the values of all **flipped** nodes. You may return the answer in **any order**. If it is **impossible** to flip the nodes in the tree to make the pre-order traversal match* `voyage`*, return the list* `[-1]`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/01/02/1219-01.png](https://assets.leetcode.com/uploads/2019/01/02/1219-01.png)
+
+```
+Input: root = [1,2], voyage = [2,1]
+Output: [-1]
+Explanation: It is impossible to flip the nodes such that the pre-order traversal matches voyage.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2019/01/02/1219-02.png](https://assets.leetcode.com/uploads/2019/01/02/1219-02.png)
+
+```
+Input: root = [1,2,3], voyage = [1,3,2]
+Output: [1]
+Explanation: Flipping node 1 swaps nodes 2 and 3, so the pre-order traversal matches voyage.
+```
+
+**Example 3:**
+
+![https://assets.leetcode.com/uploads/2019/01/02/1219-02.png](https://assets.leetcode.com/uploads/2019/01/02/1219-02.png)
+
+```
+Input: root = [1,2,3], voyage = [1,2,3]
+Output: []
+Explanation: The tree's pre-order traversal already matches voyage, so no nodes need to be flipped.
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is `n`.
+- `n == voyage.length`
+- `1 <= n <= 100`
+- `1 <= Node.val, voyage[i] <= n`
+- All the values in the tree are **unique**.
+- All the values in `voyage` are **unique**.
+
+## 题目大意
+
+给你一棵二叉树的根节点 root ，树中有 n 个节点，每个节点都有一个不同于其他节点且处于 1 到 n 之间的值。另给你一个由 n 个值组成的行程序列 voyage ，表示 预期 的二叉树 先序遍历 结果。通过交换节点的左右子树，可以 翻转 该二叉树中的任意节点。请翻转 最少 的树中节点，使二叉树的 先序遍历 与预期的遍历行程 voyage 相匹配 。如果可以，则返回 翻转的 所有节点的值的列表。你可以按任何顺序返回答案。如果不能，则返回列表 [-1]。
+
+## 解题思路
+
+- 题目要求翻转最少树中节点，利用贪心的思想，应该从根节点开始从上往下依次翻转，这样翻转的次数是最少的。对树进行深度优先遍历，如果遍历到某一个节点的时候，节点值不能与行程序列匹配，那么答案一定是 [-1]。否则，当下一个期望数字 `voyage[i]` 与即将遍历的子节点的值不同的时候，就要翻转一下当前这个节点的左右子树，继续 DFS。递归结束可能有 2 种情况，一种是找出了所有要翻转的节点，另一种情况是没有需要翻转的，即原树先序遍历的结果与 `voyage` 是完全一致的。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func flipMatchVoyage(root *TreeNode, voyage []int) []int {
+	res, index := make([]int, 0, len(voyage)), 0
+	if travelTree(root, &index, voyage, &res) {
+		return res
+	}
+	return []int{-1}
+}
+
+func travelTree(root *TreeNode, index *int, voyage []int, res *[]int) bool {
+	if root == nil {
+		return true
+	}
+	if root.Val != voyage[*index] {
+		return false
+	}
+	*index++
+	if root.Left != nil && root.Left.Val != voyage[*index] {
+		*res = append(*res, root.Val)
+		return travelTree(root.Right, index, voyage, res) && travelTree(root.Left, index, voyage, res)
+	}
+	return travelTree(root.Left, index, voyage, res) && travelTree(root.Right, index, voyage, res)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0970.Powerful-Integers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0973.K-Closest-Points-to-Origin/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0973.K-Closest-Points-to-Origin.md b/website/content/ChapterFour/0900~0999/0973.K-Closest-Points-to-Origin.md
new file mode 100644
index 000000000..8a50e28fc
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0973.K-Closest-Points-to-Origin.md
@@ -0,0 +1,78 @@
+# [973. K Closest Points to Origin](https://leetcode.com/problems/k-closest-points-to-origin/)
+
+## 题目
+
+We have a list of points on the plane.  Find the K closest points to the origin (0, 0).
+
+(Here, the distance between two points on a plane is the Euclidean distance.)
+
+You may return the answer in any order.  The answer is guaranteed to be unique (except for the order that it is in.)
+
+
+**Example 1**:
+
+```
+
+Input: points = [[1,3],[-2,2]], K = 1
+Output: [[-2,2]]
+Explanation: 
+The distance between (1, 3) and the origin is sqrt(10).
+The distance between (-2, 2) and the origin is sqrt(8).
+Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
+We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].
+
+```
+
+**Example 2**:
+
+```
+
+Input: points = [[3,3],[5,-1],[-2,4]], K = 2
+Output: [[3,3],[-2,4]]
+(The answer [[-2,4],[3,3]] would also be accepted.)
+
+```
+
+**Note**:
+
+- 1 <= K <= points.length <= 10000
+- -10000 < points[i][0] < 10000
+- -10000 < points[i][1] < 10000
+
+## 题目大意
+
+找出 K 个距离坐标原点最近的坐标点。
+
+## 解题思路
+
+这题也是排序题，先将所有点距离坐标原点的距离都算出来，然后从小到大排序。取前 K 个即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+// KClosest define
+func KClosest(points [][]int, K int) [][]int {
+	sort.Slice(points, func(i, j int) bool {
+		return points[i][0]*points[i][0]+points[i][1]*points[i][1] <
+			points[j][0]*points[j][0]+points[j][1]*points[j][1]
+	})
+	ans := make([][]int, K)
+	for i := 0; i < K; i++ {
+		ans[i] = points[i]
+	}
+	return ans
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0976.Largest-Perimeter-Triangle/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0976.Largest-Perimeter-Triangle.md b/website/content/ChapterFour/0900~0999/0976.Largest-Perimeter-Triangle.md
new file mode 100644
index 000000000..02ade12a3
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0976.Largest-Perimeter-Triangle.md
@@ -0,0 +1,85 @@
+# [976. Largest Perimeter Triangle](https://leetcode.com/problems/largest-perimeter-triangle/)
+
+## 题目
+
+Given an array A of positive lengths, return the largest perimeter of a triangle with non-zero area, formed from 3 of these lengths.
+
+If it is impossible to form any triangle of non-zero area, return 0.
+
+
+**Example 1**:
+
+```
+
+Input: [2,1,2]
+Output: 5
+
+```
+
+**Example 2**:
+
+```
+
+Input: [1,2,1]
+Output: 0
+
+```
+
+**Example 3**:
+
+```
+
+Input: [3,2,3,4]
+Output: 10
+
+```
+
+**Example 4**:
+
+```
+
+Input: [3,6,2,3]
+Output: 8
+
+```
+
+**Note**:
+
+- 3 <= A.length <= 10000
+- 1 <= A[i] <= 10^6
+
+## 题目大意
+
+找到可以组成三角形三条边的长度，要求输出三条边之和最长的，即三角形周长最长。
+
+## 解题思路
+
+这道题也是排序题，先讲所有的长度进行排序，从大边开始往前找，找到第一个任意两边之和大于第三边(满足能构成三角形的条件)的下标，然后输出这 3 条边之和即可，如果没有找到输出 0 。
+
+## 代码
+
+```go
+
+package leetcode
+
+func largestPerimeter(A []int) int {
+	if len(A) < 3 {
+		return 0
+	}
+	quickSort164(A, 0, len(A)-1)
+	for i := len(A) - 1; i >= 2; i-- {
+		if (A[i]+A[i-1] > A[i-2]) && (A[i]+A[i-2] > A[i-1]) && (A[i-2]+A[i-1] > A[i]) {
+			return A[i] + A[i-1] + A[i-2]
+		}
+	}
+	return 0
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0973.K-Closest-Points-to-Origin/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0977.Squares-of-a-Sorted-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0977.Squares-of-a-Sorted-Array.md b/website/content/ChapterFour/0900~0999/0977.Squares-of-a-Sorted-Array.md
new file mode 100644
index 000000000..3b279c16a
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0977.Squares-of-a-Sorted-Array.md
@@ -0,0 +1,93 @@
+# [977. Squares of a Sorted Array](https://leetcode.com/problems/squares-of-a-sorted-array/)
+
+## 题目
+
+Given an array of integers A sorted in non-decreasing order, return an array of the squares of each number, also in sorted non-decreasing order.
+
+
+
+**Example 1**:
+
+```
+
+Input: [-4,-1,0,3,10]
+Output: [0,1,9,16,100]
+
+```
+
+**Example 2**:
+
+```
+
+Input: [-7,-3,2,3,11]
+Output: [4,9,9,49,121]
+
+```
+
+**Note**:
+
+1. 1 <= A.length <= 10000
+2. -10000 <= A[i] <= 10000
+3. A is sorted in non-decreasing order.
+
+## 题目大意
+
+给一个已经有序的数组，返回的数组也必须是有序的，且数组中的每个元素是由原数组中每个数字的平方得到的。
+
+## 解题思路
+
+这一题由于原数组是有序的，所以要尽量利用这一特点来减少时间复杂度。
+
+最终返回的数组，最后一位，是最大值，这个值应该是由原数组最大值，或者最小值得来的，所以可以从数组的最后一位开始排列最终数组。用 2 个指针分别指向原数组的首尾，分别计算平方值，然后比较两者大小，大的放在最终数组的后面。然后大的一个指针移动。直至两个指针相撞，最终数组就排列完成了。
+
+
+
+
+
+
+
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+// 解法一
+func sortedSquares(A []int) []int {
+	ans := make([]int, len(A))
+	for i, k, j := 0, len(A)-1, len(ans)-1; i <= j; k-- {
+		if A[i]*A[i] > A[j]*A[j] {
+			ans[k] = A[i] * A[i]
+			i++
+		} else {
+			ans[k] = A[j] * A[j]
+			j--
+		}
+	}
+	return ans
+}
+
+// 解法二
+func sortedSquares1(A []int) []int {
+	for i, value := range A {
+		A[i] = value * value
+	}
+	sort.Ints(A)
+	return A
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0976.Largest-Perimeter-Triangle/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0978.Longest-Turbulent-Subarray/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0978.Longest-Turbulent-Subarray.md b/website/content/ChapterFour/0900~0999/0978.Longest-Turbulent-Subarray.md
new file mode 100755
index 000000000..58ff3044d
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0978.Longest-Turbulent-Subarray.md
@@ -0,0 +1,113 @@
+# [978. Longest Turbulent Subarray](https://leetcode.com/problems/longest-turbulent-subarray/)
+
+## 题目
+
+A subarray `A[i], A[i+1], ..., A[j]` of `A` is said to be *turbulent* if and only if:
+
+- For `i <= k < j`, `A[k] > A[k+1]` when `k` is odd, and `A[k] < A[k+1]` when `k` is even;
+- **OR**, for `i <= k < j`, `A[k] > A[k+1]` when `k` is even, and `A[k] < A[k+1]` when `k` is odd.
+
+That is, the subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray.
+
+Return the **length** of a maximum size turbulent subarray of A.
+
+**Example 1**:
+
+    Input: [9,4,2,10,7,8,8,1,9]
+    Output: 5
+    Explanation: (A[1] > A[2] < A[3] > A[4] < A[5])
+
+**Example 2**:
+
+    Input: [4,8,12,16]
+    Output: 2
+
+**Example 3**:
+
+    Input: [100]
+    Output: 1
+
+**Note**:
+
+1. `1 <= A.length <= 40000`
+2. `0 <= A[i] <= 10^9`
+
+
+## 题目大意
+
+
+当 A 的子数组 A[i], A[i+1], ..., A[j] 满足下列条件时，我们称其为湍流子数组：
+
+若 i <= k < j，当 k 为奇数时， A[k] > A[k+1]，且当 k 为偶数时，A[k] < A[k+1]；
+或 若 i <= k < j，当 k 为偶数时，A[k] > A[k+1] ，且当 k 为奇数时， A[k] < A[k+1]。
+也就是说，如果比较符号在子数组中的每个相邻元素对之间翻转，则该子数组是湍流子数组。
+
+返回 A 的最大湍流子数组的长度。
+
+提示：
+
+- 1 <= A.length <= 40000
+- 0 <= A[i] <= 10^9
+
+
+
+## 解题思路
+
+
+- 给出一个数组，要求找出“摆动数组”的最大长度。所谓“摆动数组”的意思是，元素一大一小间隔的。
+- 这一题可以用滑动窗口来解答。用相邻元素差的乘积大于零（a ^ b >= 0 说明a b乘积大于零）来判断是否是湍流， 如果是，那么扩大窗口。否则窗口缩小为0，开始新的一个窗口。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 模拟法
+func maxTurbulenceSize(arr []int) int {
+	inc, dec := 1, 1
+	maxLen := min(1, len(arr))
+	for i := 1; i < len(arr); i++ {
+		if arr[i-1] < arr[i] {
+			inc = dec + 1
+			dec = 1
+		} else if arr[i-1] > arr[i] {
+			dec = inc + 1
+			inc = 1
+		} else {
+			inc = 1
+			dec = 1
+		}
+		maxLen = max(maxLen, max(inc, dec))
+	}
+	return maxLen
+}
+
+// 解法二 滑动窗口
+func maxTurbulenceSize1(arr []int) int {
+	var maxLength int
+	if len(arr) == 2 && arr[0] != arr[1] {
+		maxLength = 2
+	} else {
+		maxLength = 1
+	}
+	left := 0
+	for right := 2; right < len(arr); right++ {
+		if arr[right] == arr[right-1] {
+			left = right
+		} else if (arr[right]-arr[right-1])^(arr[right-1]-arr[right-2]) >= 0 {
+			left = right - 1
+		}
+		maxLength = max(maxLength, right-left+1)
+	}
+	return maxLength
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0977.Squares-of-a-Sorted-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0979.Distribute-Coins-in-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0979.Distribute-Coins-in-Binary-Tree.md b/website/content/ChapterFour/0900~0999/0979.Distribute-Coins-in-Binary-Tree.md
new file mode 100755
index 000000000..c76f68080
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0979.Distribute-Coins-in-Binary-Tree.md
@@ -0,0 +1,103 @@
+# [979. Distribute Coins in Binary Tree](https://leetcode.com/problems/distribute-coins-in-binary-tree/)
+
+
+## 题目
+
+Given the `root` of a binary tree with `N` nodes, each `node` in the tree has `node.val` coins, and there are `N` coins total.
+
+In one move, we may choose two adjacent nodes and move one coin from one node to another. (The move may be from parent to child, or from child to parent.)
+
+Return the number of moves required to make every node have exactly one coin.
+
+**Example 1**:
+
+![](https://assets.leetcode.com/uploads/2019/01/18/tree1.png)
+
+    Input: [3,0,0]
+    Output: 2
+    Explanation: From the root of the tree, we move one coin to its left child, and one coin to its right child.
+
+**Example 2**:
+
+![](https://assets.leetcode.com/uploads/2019/01/18/tree2.png)
+
+    Input: [0,3,0]
+    Output: 3
+    Explanation: From the left child of the root, we move two coins to the root [taking two moves].  Then, we move one coin from the root of the tree to the right child.
+
+**Example 3**:
+
+![](https://assets.leetcode.com/uploads/2019/01/18/tree3.png)
+
+    Input: [1,0,2]
+    Output: 2
+
+**Example 4**:
+
+![](https://assets.leetcode.com/uploads/2019/01/18/tree4.png)
+
+    Input: [1,0,0,null,3]
+    Output: 4
+
+**Note**:
+
+1. `1<= N <= 100`
+2. `0 <= node.val <= N`
+
+## 题目大意
+
+给定一个有 N 个结点的二叉树的根结点 root，树中的每个结点上都对应有 node.val 枚硬币，并且总共有 N 枚硬币。在一次移动中，我们可以选择两个相邻的结点，然后将一枚硬币从其中一个结点移动到另一个结点。(移动可以是从父结点到子结点，或者从子结点移动到父结点。)。返回使每个结点上只有一枚硬币所需的移动次数。
+
+提示：
+
+1. 1<= N <= 100
+2. 0 <= node.val <= N
+
+
+## 解题思路
+
+- 给出一棵树，有 N 个节点。有 N 个硬币分散在这 N 个节点中，问经过多少次移动以后，所有节点都有一枚硬币。
+- 这一题乍一看比较难分析，仔细一想，可以用贪心和分治的思想来解决。一个树的最小单元是一个根节点和两个孩子。在这种情况下，3 个节点谁的硬币多就可以分给没有硬币的那个节点，这种移动方法也能保证移动步数最少。不难证明，硬币由相邻的节点移动过来的步数是最少的。那么一棵树从最下一层开始往上推，逐步从下往上把硬币移动上去，直到最后根节点也都拥有硬币。多余 1 枚的节点记为 `n -1`，没有硬币的节点记为 -1 。例如，下图中左下角的 3 个节点，有 4 枚硬币的节点可以送出 3 枚硬币，叶子节点有 0 枚硬币的节点需要接收 1 枚硬币。根节点有 0 枚硬币，左孩子给了 3 枚，右孩子需要 1 枚，自己本身还要留一枚，所以最终还能剩 1 枚。
+
+![](https://img.halfrost.com/Leetcode/leetcode_979_1.png)
+
+- 所以每个节点移动的步数应该是 `left + right + root.Val - 1`。最后递归求解即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree root.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func distributeCoins(root *TreeNode) int {
+	res := 0
+	distributeCoinsDFS(root, &res)
+	return res
+}
+
+func distributeCoinsDFS(root *TreeNode, res *int) int {
+	if root == nil {
+		return 0
+	}
+	left, right := distributeCoinsDFS(root.Left, res), distributeCoinsDFS(root.Right, res)
+	*res += abs(left) + abs(right)
+	return left + right + root.Val - 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0978.Longest-Turbulent-Subarray/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0980.Unique-Paths-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0980.Unique-Paths-III.md b/website/content/ChapterFour/0900~0999/0980.Unique-Paths-III.md
new file mode 100755
index 000000000..cd74994ff
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0980.Unique-Paths-III.md
@@ -0,0 +1,138 @@
+# [980. Unique Paths III](https://leetcode.com/problems/unique-paths-iii/)
+
+
+## 题目
+
+On a 2-dimensional `grid`, there are 4 types of squares:
+
+- `1` represents the starting square. There is exactly one starting square.
+- `2` represents the ending square. There is exactly one ending square.
+- `0` represents empty squares we can walk over.
+- `-1` represents obstacles that we cannot walk over.
+
+Return the number of 4-directional walks from the starting square to the ending square, that **walk over every non-obstacle square exactly once**.
+
+**Example 1**:
+
+    Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
+    Output: 2
+    Explanation: We have the following two paths: 
+    1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
+    2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)
+
+**Example 2**:
+
+    Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
+    Output: 4
+    Explanation: We have the following four paths: 
+    1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
+    2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
+    3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
+    4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)
+
+**Example 3**:
+
+    Input: [[0,1],[2,0]]
+    Output: 0
+    Explanation: 
+    There is no path that walks over every empty square exactly once.
+    Note that the starting and ending square can be anywhere in the grid.
+
+**Note**:
+
+1. `1 <= grid.length * grid[0].length <= 20`
+
+
+## 题目大意
+
+在二维网格 grid 上，有 4 种类型的方格：
+
+- 1 表示起始方格。且只有一个起始方格。
+- 2 表示结束方格，且只有一个结束方格。
+- 0 表示我们可以走过的空方格。
+- -1 表示我们无法跨越的障碍。
+
+返回在四个方向（上、下、左、右）上行走时，从起始方格到结束方格的不同路径的数目，**每一个无障碍方格都要通过一次**。
+
+
+
+## 解题思路
+
+
+- 这一题也可以按照第 79 题的思路来做。题目要求输出地图中从起点到终点的路径条数。注意路径要求必须走满所有空白的格子。
+- 唯一需要注意的一点是，空白的格子并不是最后走的总步数，`总步数 = 空白格子数 + 1`，因为要走到终点，走到终点也算一步。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func uniquePathsIII(grid [][]int) int {
+	visited := make([][]bool, len(grid))
+	for i := 0; i < len(visited); i++ {
+		visited[i] = make([]bool, len(grid[0]))
+	}
+	res, empty, startx, starty, endx, endy, path := 0, 0, 0, 0, 0, 0, []int{}
+	for i, v := range grid {
+		for j, vv := range v {
+			switch vv {
+			case 0:
+				empty++
+			case 1:
+				startx, starty = i, j
+			case 2:
+				endx, endy = i, j
+			}
+		}
+	}
+	findUniquePathIII(grid, visited, path, empty+1, startx, starty, endx, endy, &res) // 可走的步数要加一，因为终点格子也算一步，不然永远走不到终点！
+	return res
+}
+
+func isInPath(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+
+func findUniquePathIII(board [][]int, visited [][]bool, path []int, empty, startx, starty, endx, endy int, res *int) {
+	if startx == endx && starty == endy {
+		if empty == 0 {
+			*res++
+		}
+		return
+	}
+	if board[startx][starty] >= 0 {
+		visited[startx][starty] = true
+		empty--
+		path = append(path, startx)
+		path = append(path, starty)
+		for i := 0; i < 4; i++ {
+			nx := startx + dir[i][0]
+			ny := starty + dir[i][1]
+			if isInPath(board, nx, ny) && !visited[nx][ny] {
+				findUniquePathIII(board, visited, path, empty, nx, ny, endx, endy, res)
+			}
+		}
+		visited[startx][starty] = false
+		//empty++ 这里虽然可以还原这个变量值，但是赋值没有意义，干脆不写了
+		path = path[:len(path)-2]
+	}
+	return
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0979.Distribute-Coins-in-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0981.Time-Based-Key-Value-Store/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0981.Time-Based-Key-Value-Store.md b/website/content/ChapterFour/0900~0999/0981.Time-Based-Key-Value-Store.md
new file mode 100755
index 000000000..4ef63e1d7
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0981.Time-Based-Key-Value-Store.md
@@ -0,0 +1,131 @@
+# [981. Time Based Key-Value Store](https://leetcode.com/problems/time-based-key-value-store/)
+
+
+## 题目
+
+Create a timebased key-value store class `TimeMap`, that supports two operations.
+
+1. `set(string key, string value, int timestamp)`
+
+- Stores the `key` and `value`, along with the given `timestamp`.
+
+2. `get(string key, int timestamp)`
+
+- Returns a value such that `set(key, value, timestamp_prev)` was called previously, with `timestamp_prev <= timestamp`.
+- If there are multiple such values, it returns the one with the largest `timestamp_prev`.
+- If there are no values, it returns the empty string (`""`).
+
+**Example 1**:
+
+    Input: inputs = ["TimeMap","set","get","get","set","get","get"], inputs = [[],["foo","bar",1],["foo",1],["foo",3],["foo","bar2",4],["foo",4],["foo",5]]
+    Output: [null,null,"bar","bar",null,"bar2","bar2"]
+    Explanation:   
+    TimeMap kv;   
+    kv.set("foo", "bar", 1); // store the key "foo" and value "bar" along with timestamp = 1   
+    kv.get("foo", 1);  // output "bar"   
+    kv.get("foo", 3); // output "bar" since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 ie "bar"   
+    kv.set("foo", "bar2", 4);   
+    kv.get("foo", 4); // output "bar2"   
+    kv.get("foo", 5); //output "bar2"
+
+**Example 2**:
+
+    Input: inputs = ["TimeMap","set","set","get","get","get","get","get"], inputs = [[],["love","high",10],["love","low",20],["love",5],["love",10],["love",15],["love",20],["love",25]]
+    Output: [null,null,null,"","high","high","low","low"]
+
+**Note**:
+
+1. All key/value strings are lowercase.
+2. All key/value strings have length in the range `[1, 100]`
+3. The `timestamps` for all `TimeMap.set` operations are strictly increasing.
+4. `1 <= timestamp <= 10^7`
+5. `TimeMap.set` and `TimeMap.get` functions will be called a total of `120000` times (combined) per test case.
+
+## 题目大意
+
+创建一个基于时间的键值存储类 TimeMap，它支持下面两个操作：
+
+1. set(string key, string value, int timestamp)
+
+- 存储键 key、值 value，以及给定的时间戳 timestamp。
+
+2. get(string key, int timestamp)
+
+- 返回先前调用 set(key, value, timestamp_prev) 所存储的值，其中 timestamp_prev <= timestamp。
+- 如果有多个这样的值，则返回对应最大的  timestamp_prev 的那个值。
+- 如果没有值，则返回空字符串（""）。
+
+提示：
+
+1. 所有的键/值字符串都是小写的。
+2. 所有的键/值字符串长度都在 [1, 100] 范围内。
+3. 所有 TimeMap.set 操作中的时间戳 timestamps 都是严格递增的。
+4. 1 <= timestamp <= 10^7
+5. TimeMap.set 和 TimeMap.get 函数在每个测试用例中将（组合）调用总计 120000 次。
+
+
+## 解题思路
+
+- 要求设计一个基于时间戳的 `kv` 存储。`set()` 操作里面会会包含一个时间戳。get() 操作的时候查找时间戳小于等于 `timestamp` 的 `key` 对应的 `value`，如果有多个解，输出满足条件的最大时间戳对应的 `value` 值。
+- 这一题可以用二分搜索来解答，用 `map` 存储 `kv` 数据，`key` 对应的就是 `key`，`value` 对应一个结构体，里面包含 `value` 和 `timestamp`。执行 `get()` 操作的时候，先取出 `key` 对应的结构体数组，然后在这个数组里面根据 `timestamp` 进行二分搜索。由于题意是要找小于等于 `timestamp` 的最大 `timestamp` ，这会有很多满足条件的解，变换一下，先找 `> timestamp` 的最小解，然后下标减一即是满足题意的最大解。
+- 另外题目中提到“`TimeMap.set` 操作中的 `timestamp` 是严格递增的”。所以在 `map` 中存储 `value` 结构体的时候，不需要排序了，天然有序。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+type data struct {
+	time  int
+	value string
+}
+
+// TimeMap is a timebased key-value store
+// TimeMap define
+type TimeMap map[string][]data
+
+// Constructor981 define
+func Constructor981() TimeMap {
+	return make(map[string][]data, 1024)
+}
+
+// Set define
+func (t TimeMap) Set(key string, value string, timestamp int) {
+	if _, ok := t[key]; !ok {
+		t[key] = make([]data, 1, 1024)
+	}
+	t[key] = append(t[key], data{
+		time:  timestamp,
+		value: value,
+	})
+}
+
+// Get define
+func (t TimeMap) Get(key string, timestamp int) string {
+	d := t[key]
+	i := sort.Search(len(d), func(i int) bool {
+		return timestamp < d[i].time
+	})
+	i--
+	return t[key][i].value
+}
+
+/**
+ * Your TimeMap object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.Set(key,value,timestamp);
+ * param_2 := obj.Get(key,timestamp);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0980.Unique-Paths-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0984.String-Without-AAA-or-BBB/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0984.String-Without-AAA-or-BBB.md b/website/content/ChapterFour/0900~0999/0984.String-Without-AAA-or-BBB.md
new file mode 100755
index 000000000..df90e6f04
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0984.String-Without-AAA-or-BBB.md
@@ -0,0 +1,100 @@
+# [984. String Without AAA or BBB](https://leetcode.com/problems/string-without-aaa-or-bbb/)
+
+
+## 题目
+
+Given two integers `A` and `B`, return **any** string `S` such that:
+
+- `S` has length `A + B` and contains exactly `A` `'a'` letters, and exactly `B` `'b'`letters;
+- The substring `'aaa'` does not occur in `S`;
+- The substring `'bbb'` does not occur in `S`.
+
+**Example 1**:
+
+    Input: A = 1, B = 2
+    Output: "abb"
+    Explanation: "abb", "bab" and "bba" are all correct answers.
+
+**Example 2**:
+
+    Input: A = 4, B = 1
+    Output: "aabaa"
+
+**Note**:
+
+1. `0 <= A <= 100`
+2. `0 <= B <= 100`
+3. It is guaranteed such an `S` exists for the given `A` and `B`.
+
+
+## 题目大意
+
+给定两个整数 A 和 B，返回任意字符串 S，要求满足：
+
+- S 的长度为 A + B，且正好包含 A 个 'a' 字母与 B 个 'b' 字母；
+- 子串 'aaa' 没有出现在 S 中；
+- 子串 'bbb' 没有出现在 S 中。
+
+
+提示：
+
+- 0 <= A <= 100
+- 0 <= B <= 100
+- 对于给定的 A 和 B，保证存在满足要求的 S。
+
+
+## 解题思路
+
+
+- 给出 A 和 B 的个数，要求组合出字符串，不能出现 3 个连续的 A 和 3 个连续的 B。这题由于只可能有 4 种情况，暴力枚举就可以了。假设 B 的个数比 A 多(如果 A 多，就交换一下 A 和 B)，最终可能的情况只可能是这 4 种情况中的一种： `ba`，`bbabb`，`bbabbabb`，`bbabbabbabbabababa`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func strWithout3a3b(A int, B int) string {
+	ans, a, b := "", "a", "b"
+	if B < A {
+		A, B = B, A
+		a, b = b, a
+	}
+	Dif := B - A
+	if A == 1 && B == 1 { // ba
+		ans = b + a
+	} else if A == 1 && B < 5 { // bbabb
+		for i := 0; i < B-2; i++ {
+			ans = ans + b
+		}
+		ans = b + b + a + ans
+	} else if (B-A)/A >= 1 { //bbabbabb
+		for i := 0; i < A; i++ {
+			ans = ans + b + b + a
+			B = B - 2
+		}
+		for i := 0; i < B; i++ {
+			ans = ans + b
+		}
+	} else { //bbabbabbabbabababa
+		for i := 0; i < Dif; i++ {
+			ans = ans + b + b + a
+			B -= 2
+			A--
+		}
+		for i := 0; i < B; i++ {
+			ans = ans + b + a
+		}
+	}
+	return ans
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0981.Time-Based-Key-Value-Store/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0985.Sum-of-Even-Numbers-After-Queries/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0985.Sum-of-Even-Numbers-After-Queries.md b/website/content/ChapterFour/0900~0999/0985.Sum-of-Even-Numbers-After-Queries.md
new file mode 100644
index 000000000..07f6c748e
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0985.Sum-of-Even-Numbers-After-Queries.md
@@ -0,0 +1,79 @@
+# [985. Sum of Even Numbers After Queries](https://leetcode.com/problems/sum-of-even-numbers-after-queries/)
+
+## 题目
+
+We have an array `A` of integers, and an array `queries` of queries.
+
+For the `i`-th query `val = queries[i][0], index = queries[i][1]`, we add val to `A[index]`. Then, the answer to the `i`-th query is the sum of the even values of `A`.
+
+*(Here, the given `index = queries[i][1]` is a 0-based index, and each query permanently modifies the array `A`.)*
+
+Return the answer to all queries. Your `answer` array should have `answer[i]` as the answer to the `i`-th query.
+
+**Example 1**:
+
+```
+Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
+Output: [8,6,2,4]
+Explanation: 
+At the beginning, the array is [1,2,3,4].
+After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
+After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
+After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
+After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
+```
+
+**Note**:
+
+1. `1 <= A.length <= 10000`
+2. `-10000 <= A[i] <= 10000`
+3. `1 <= queries.length <= 10000`
+4. `-10000 <= queries[i][0] <= 10000`
+5. `0 <= queries[i][1] < A.length`
+
+## 题目大意
+
+给出一个整数数组 A 和一个查询数组 queries。
+
+对于第 i 次查询，有 val = queries[i][0], index = queries[i][1]，我们会把 val 加到 A[index] 上。然后，第 i 次查询的答案是 A 中偶数值的和。（此处给定的 index = queries[i][1] 是从 0 开始的索引，每次查询都会永久修改数组 A。）返回所有查询的答案。你的答案应当以数组 answer 给出，answer[i] 为第 i 次查询的答案。
+
+
+## 解题思路
+
+- 给出一个数组 A 和 query 数组。要求每次 query 操作都改变数组 A 中的元素值，并计算此次操作结束数组 A 中偶数值之和。
+- 简单题，先计算 A 中所有偶数之和。再每次 query 操作的时候，动态维护这个偶数之和即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func sumEvenAfterQueries(A []int, queries [][]int) []int {
+	cur, res := 0, []int{}
+	for _, v := range A {
+		if v%2 == 0 {
+			cur += v
+		}
+	}
+	for _, q := range queries {
+		if A[q[1]]%2 == 0 {
+			cur -= A[q[1]]
+		}
+		A[q[1]] += q[0]
+		if A[q[1]]%2 == 0 {
+			cur += A[q[1]]
+		}
+		res = append(res, cur)
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0984.String-Without-AAA-or-BBB/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0986.Interval-List-Intersections/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0986.Interval-List-Intersections.md b/website/content/ChapterFour/0900~0999/0986.Interval-List-Intersections.md
new file mode 100644
index 000000000..304c9ce37
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0986.Interval-List-Intersections.md
@@ -0,0 +1,80 @@
+# [986. Interval List Intersections](https://leetcode.com/problems/interval-list-intersections/)
+
+## 题目
+
+Given two lists of closed intervals, each list of intervals is pairwise disjoint and in sorted order.
+
+Return the intersection of these two interval lists.
+
+(Formally, a closed interval [a, b] (with a <= b) denotes the set of real numbers x with a <= x <= b.  The intersection of two closed intervals is a set of real numbers that is either empty, or can be represented as a closed interval.  For example, the intersection of [1, 3] and [2, 4] is [2, 3].)
+
+
+
+
+**Example 1**:
+
+![](https://assets.leetcode.com/uploads/2019/01/30/interval1.png)
+
+```
+Input: A = [[0,2],[5,10],[13,23],[24,25]], B = [[1,5],[8,12],[15,24],[25,26]]
+Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
+Reminder: The inputs and the desired output are lists of Interval objects, and not arrays or lists.
+```
+
+**Note**:  
+
+- 0 <= A.length < 1000
+- 0 <= B.length < 1000
+- 0 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9
+
+**Note**: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
+
+## 题目大意
+
+这道题考察的是滑动窗口的问题。
+
+给出 2 个数组 A 和数组 B。要求求出这 2 个数组的交集数组。题意见图。
+
+## 解题思路
+
+交集的左边界应该为，`start := max(A[i].Start, B[j].Start)`，右边界为，`end := min(A[i].End, B[j].End)`，如果 `start <= end`，那么这个就是一个满足条件的交集，放入最终数组中。如果 `A[i].End <= B[j].End`，代表 B 数组范围比 A 数组大，A 的游标右移。如果 `A[i].End > B[j].End`，代表 A 数组范围比 B 数组大，B 的游标右移。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for an interval.
+ * type Interval struct {
+ *	   Start int
+ *	   End   int
+ * }
+ */
+func intervalIntersection(A []Interval, B []Interval) []Interval {
+	res := []Interval{}
+	for i, j := 0, 0; i < len(A) && j < len(B); {
+		start := max(A[i].Start, B[j].Start)
+		end := min(A[i].End, B[j].End)
+		if start <= end {
+			res = append(res, Interval{Start: start, End: end})
+		}
+		if A[i].End <= B[j].End {
+			i++
+		} else {
+			j++
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0985.Sum-of-Even-Numbers-After-Queries/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0987.Vertical-Order-Traversal-of-a-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0987.Vertical-Order-Traversal-of-a-Binary-Tree.md b/website/content/ChapterFour/0900~0999/0987.Vertical-Order-Traversal-of-a-Binary-Tree.md
new file mode 100644
index 000000000..aba6488b5
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0987.Vertical-Order-Traversal-of-a-Binary-Tree.md
@@ -0,0 +1,143 @@
+# [987. Vertical Order Traversal of a Binary Tree](https://leetcode.com/problems/vertical-order-traversal-of-a-binary-tree/)
+
+
+## 题目
+
+Given the `root` of a binary tree, calculate the **vertical order traversal** of the binary tree.
+
+For each node at position `(row, col)`, its left and right children will be at positions `(row + 1, col - 1)` and `(row + 1, col + 1)` respectively. The root of the tree is at `(0, 0)`.
+
+The **vertical order traversal** of a binary tree is a list of top-to-bottom orderings for each column index starting from the leftmost column and ending on the rightmost column. There may be multiple nodes in the same row and same column. In such a case, sort these nodes by their values.
+
+Return *the **vertical order traversal** of the binary tree*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/01/29/vtree1.jpg](https://assets.leetcode.com/uploads/2021/01/29/vtree1.jpg)
+
+```
+Input: root = [3,9,20,null,null,15,7]
+Output: [[9],[3,15],[20],[7]]
+Explanation:
+Column -1: Only node 9 is in this column.
+Column 0: Nodes 3 and 15 are in this column in that order from top to bottom.
+Column 1: Only node 20 is in this column.
+Column 2: Only node 7 is in this column.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2021/01/29/vtree2.jpg](https://assets.leetcode.com/uploads/2021/01/29/vtree2.jpg)
+
+```
+Input: root = [1,2,3,4,5,6,7]
+Output: [[4],[2],[1,5,6],[3],[7]]
+Explanation:
+Column -2: Only node 4 is in this column.
+Column -1: Only node 2 is in this column.
+Column 0: Nodes 1, 5, and 6 are in this column.
+          1 is at the top, so it comes first.
+          5 and 6 are at the same position (2, 0), so we order them by their value, 5 before 6.
+Column 1: Only node 3 is in this column.
+Column 2: Only node 7 is in this column.
+
+```
+
+**Example 3:**
+
+![https://assets.leetcode.com/uploads/2021/01/29/vtree3.jpg](https://assets.leetcode.com/uploads/2021/01/29/vtree3.jpg)
+
+```
+Input: root = [1,2,3,4,6,5,7]
+Output: [[4],[2],[1,5,6],[3],[7]]
+Explanation:
+This case is the exact same as example 2, but with nodes 5 and 6 swapped.
+Note that the solution remains the same since 5 and 6 are in the same location and should be ordered by their values.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 1000]`.
+- `0 <= Node.val <= 1000`
+
+## 题目大意
+
+给你二叉树的根结点 root ，请你设计算法计算二叉树的 垂序遍历 序列。
+
+对位于 (row, col) 的每个结点而言，其左右子结点分别位于 (row + 1, col - 1) 和 (row + 1, col + 1) 。树的根结点位于 (0, 0) 。二叉树的 垂序遍历 从最左边的列开始直到最右边的列结束，按列索引每一列上的所有结点，形成一个按出现位置从上到下排序的有序列表。如果同行同列上有多个结点，则按结点的值从小到大进行排序。返回二叉树的 垂序遍历 序列。
+
+## 解题思路
+
+- 题目要求按照一列一列的遍历二叉树。需要解决 2 个问题。第一个问题，二叉树上每个结点的二维坐标如何计算。第二个问题，同一个二维坐标点上摞起来多个结点，需要按照从小到大的顺序排序，如例子二和例子三，同一个二维坐标点 (2，0) 上，摞了 2 个不同的结点。
+- 先解决第一个问题，由于题目要求根结点是 (0，0) ，即根结点是坐标原点，它的左子树的 x 坐标都是负数，它的右子树的 x 坐标都是正数。按照先序遍历，就可以将这些结点的二维坐标计算出来。再进行一次排序，按照 x 坐标从小到大排序，坐标相同的情况对应着结点摞起来的情况，摞起来的结点按照 val 值的大小从小到大排序。这样在 x 轴方向，所有结点就排列好了。排序完成，也顺便解决了第二个问题。
+- 最后一步只需要扫描一遍这个排好序的数组，按照列的顺序，依次将同一列的结点打包至一个一维数组中。最终输出的二维数组即为题目所求。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+	"sort"
+
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+type node struct {
+	x, y, val int
+}
+
+func verticalTraversal(root *TreeNode) [][]int {
+	var dfs func(root *TreeNode, x, y int)
+	var nodes []node
+	dfs = func(root *TreeNode, x, y int) {
+		if root == nil {
+			return
+		}
+		nodes = append(nodes, node{x, y, root.Val})
+		dfs(root.Left, x+1, y-1)
+		dfs(root.Right, x+1, y+1)
+	}
+	dfs(root, 0, 0)
+
+	sort.Slice(nodes, func(i, j int) bool {
+		a, b := nodes[i], nodes[j]
+		return a.y < b.y || a.y == b.y &&
+			(a.x < b.x || a.x == b.x && a.val < b.val)
+	})
+
+	var res [][]int
+	lastY := math.MinInt32
+	for _, node := range nodes {
+		if lastY != node.y {
+			res = append(res, []int{node.val})
+			lastY = node.y
+		} else {
+			res[len(res)-1] = append(res[len(res)-1], node.val)
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0986.Interval-List-Intersections/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0989.Add-to-Array-Form-of-Integer/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0989.Add-to-Array-Form-of-Integer.md b/website/content/ChapterFour/0900~0999/0989.Add-to-Array-Form-of-Integer.md
new file mode 100644
index 000000000..ab9a6d8a2
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0989.Add-to-Array-Form-of-Integer.md
@@ -0,0 +1,91 @@
+# [989. Add to Array-Form of Integer](https://leetcode.com/problems/add-to-array-form-of-integer/)
+
+## 题目
+
+For a non-negative integer `X`, the *array-form of `X`* is an array of its digits in left to right order.  For example, if `X = 1231`, then the array form is `[1,2,3,1]`.
+
+Given the array-form `A` of a non-negative integer `X`, return the array-form of the integer `X+K`.
+
+**Example 1:**
+
+```
+Input: A = [1,2,0,0], K = 34
+Output: [1,2,3,4]
+Explanation: 1200 + 34 = 1234
+```
+
+**Example 2:**
+
+```
+Input: A = [2,7,4], K = 181
+Output: [4,5,5]
+Explanation: 274 + 181 = 455
+```
+
+**Example 3:**
+
+```
+Input: A = [2,1,5], K = 806
+Output: [1,0,2,1]
+Explanation: 215 + 806 = 1021
+```
+
+**Example 4:**
+
+```
+Input: A = [9,9,9,9,9,9,9,9,9,9], K = 1
+Output: [1,0,0,0,0,0,0,0,0,0,0]
+Explanation: 9999999999 + 1 = 10000000000
+```
+
+**Note：**
+
+1. `1 <= A.length <= 10000`
+2. `0 <= A[i] <= 9`
+3. `0 <= K <= 10000`
+4. If `A.length > 1`, then `A[0] != 0`
+
+## 题目大意
+
+对于非负整数 X 而言，X 的数组形式是每位数字按从左到右的顺序形成的数组。例如，如果 X = 1231，那么其数组形式为 [1,2,3,1]。给定非负整数 X 的数组形式 A，返回整数 X+K 的数组形式。
+
+## 解题思路
+
+- 简单题，计算 2 个非负整数的和。累加过程中不断的进位，最终输出到数组中记得需要逆序，即数字的高位排在数组下标较小的位置。
+
+## 代码
+
+```go
+package leetcode
+
+func addToArrayForm(A []int, K int) []int {
+	res := []int{}
+	for i := len(A) - 1; i >= 0; i-- {
+		sum := A[i] + K%10
+		K /= 10
+		if sum >= 10 {
+			K++
+			sum -= 10
+		}
+		res = append(res, sum)
+	}
+	for ; K > 0; K /= 10 {
+		res = append(res, K%10)
+	}
+	reverse(res)
+	return res
+}
+
+func reverse(A []int) {
+	for i, n := 0, len(A); i < n/2; i++ {
+		A[i], A[n-1-i] = A[n-1-i], A[i]
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0987.Vertical-Order-Traversal-of-a-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0990.Satisfiability-of-Equality-Equations/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0990.Satisfiability-of-Equality-Equations.md b/website/content/ChapterFour/0900~0999/0990.Satisfiability-of-Equality-Equations.md
new file mode 100755
index 000000000..bca338729
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0990.Satisfiability-of-Equality-Equations.md
@@ -0,0 +1,106 @@
+# [990. Satisfiability of Equality Equations](https://leetcode.com/problems/satisfiability-of-equality-equations/)
+
+
+## 题目
+
+Given an array equations of strings that represent relationships between variables, each string `equations[i]` has length `4` and takes one of two different forms: `"a==b"` or `"a!=b"`. Here, `a` and `b` are lowercase letters (not necessarily different) that represent one-letter variable names.
+
+Return `true` if and only if it is possible to assign integers to variable names so as to satisfy all the given equations.
+
+**Example 1**:
+
+    Input: ["a==b","b!=a"]
+    Output: false
+    Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second.  There is no way to assign the variables to satisfy both equations.
+
+**Example 2**:
+
+    Input: ["b==a","a==b"]
+    Output: true
+    Explanation: We could assign a = 1 and b = 1 to satisfy both equations.
+
+**Example 3**:
+
+    Input: ["a==b","b==c","a==c"]
+    Output: true
+
+**Example 4**:
+
+    Input: ["a==b","b!=c","c==a"]
+    Output: false
+
+**Example 5**:
+
+    Input: ["c==c","b==d","x!=z"]
+    Output: true
+
+**Note**:
+
+1. `1 <= equations.length <= 500`
+2. `equations[i].length == 4`
+3. `equations[i][0]` and `equations[i][3]` are lowercase letters
+4. `equations[i][1]` is either `'='` or `'!'`
+5. `equations[i][2]` is `'='`
+
+
+
+## 题目大意
+
+给定一个由表示变量之间关系的字符串方程组成的数组，每个字符串方程 equations[i] 的长度为 4，并采用两种不同的形式之一："a==b" 或 "a!=b"。在这里，a 和 b 是小写字母（不一定不同），表示单字母变量名。只有当可以将整数分配给变量名，以便满足所有给定的方程时才返回 true，否则返回 false。 
+
+提示：
+
+1. 1 <= equations.length <= 500
+2. equations[i].length == 4
+3. equations[i][0] 和 equations[i][3] 是小写字母
+4. equations[i][1] 要么是 '='，要么是 '!'
+5. equations[i][2] 是 '='
+
+
+
+## 解题思路
+
+
+- 给出一个字符串数组，数组里面给出的是一些字母的关系，只有 `'=='` 和 `'! ='` 两种关系。问给出的这些关系中是否存在悖论？
+- 这一题是简单的并查集的问题。先将所有 `'=='` 关系的字母 `union()` 起来，然后再一一查看 `'! ='` 关系中是否有 `'=='` 关系的组合，如果有，就返回 `false`，如果遍历完都没有找到，则返回 `true`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func equationsPossible(equations []string) bool {
+	if len(equations) == 0 {
+		return false
+	}
+	uf := template.UnionFind{}
+	uf.Init(26)
+	for _, equ := range equations {
+		if equ[1] == '=' && equ[2] == '=' {
+			uf.Union(int(equ[0]-'a'), int(equ[3]-'a'))
+		}
+	}
+	for _, equ := range equations {
+		if equ[1] == '!' && equ[2] == '=' {
+			if uf.Find(int(equ[0]-'a')) == uf.Find(int(equ[3]-'a')) {
+				return false
+			}
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0989.Add-to-Array-Form-of-Integer/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0991.Broken-Calculator/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0991.Broken-Calculator.md b/website/content/ChapterFour/0900~0999/0991.Broken-Calculator.md
new file mode 100644
index 000000000..e5dfd8dc9
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0991.Broken-Calculator.md
@@ -0,0 +1,90 @@
+# [991. Broken Calculator](https://leetcode.com/problems/broken-calculator/)
+
+
+## 题目
+
+On a broken calculator that has a number showing on its display, we can perform two operations:
+
+- **Double**: Multiply the number on the display by 2, or;
+- **Decrement**: Subtract 1 from the number on the display.
+
+Initially, the calculator is displaying the number `X`.
+
+Return the minimum number of operations needed to display the number `Y`.
+
+**Example 1:**
+
+```
+Input: X = 2, Y = 3
+Output: 2
+Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.
+```
+
+**Example 2:**
+
+```
+Input: X = 5, Y = 8
+Output: 2
+Explanation: Use decrement and then double {5 -> 4 -> 8}.
+```
+
+**Example 3:**
+
+```
+Input: X = 3, Y = 10
+Output: 3
+Explanation:  Use double, decrement and double {3 -> 6 -> 5 -> 10}.
+```
+
+**Example 4:**
+
+```
+Input: X = 1024, Y = 1
+Output: 1023
+Explanation: Use decrement operations 1023 times.
+```
+
+**Note:**
+
+1. `1 <= X <= 10^9`
+2. `1 <= Y <= 10^9`
+
+## 题目大意
+
+在显示着数字的坏计算器上，我们可以执行以下两种操作：
+
+- 双倍（Double）：将显示屏上的数字乘 2；
+- 递减（Decrement）：将显示屏上的数字减 1 。
+
+最初，计算器显示数字 X。返回显示数字 Y 所需的最小操作数。
+
+## 解题思路
+
+- 看到本题的数据规模非常大，`10^9`，算法只能采用 `O(sqrt(n))`、`O(log n)`、`O(1)` 的算法。`O(sqrt(n))` 和 `O(1)` 在本题中是不可能的。所以按照数据规模来估计，本题只能尝试 `O(log n)` 的算法。`O(log n)` 的算法有二分搜索，不过本题不太符合二分搜索算法背景。题目中明显出现乘 2，这很明显是可以达到 `O(log n)` 的。最终确定解题思路是数学方法，循环中会用到乘 2 或者除 2 的计算。
+- 既然出现了乘 2 和减一的操作，很容易考虑到奇偶性上。题目要求最小操作数，贪心思想，应该尽可能多的使用除 2 操作，使得 Y 和 X 大小差不多，最后再利用加一操作微调。只要 Y 比 X 大就执行除法操作。当然这里要考虑一次奇偶性，如果 Y 是奇数，先加一变成偶数再除二；如果 Y 是偶数，直接除二。如此操作直到 Y 不大于 X，最后执行 `X-Y` 次加法操作微调即可。
+
+## 代码
+
+```go
+package leetcode
+
+func brokenCalc(X int, Y int) int {
+	res := 0
+	for Y > X {
+		res++
+		if Y&1 == 1 {
+			Y++
+		} else {
+			Y /= 2
+		}
+	}
+	return res + X - Y
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0990.Satisfiability-of-Equality-Equations/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0992.Subarrays-with-K-Different-Integers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0992.Subarrays-with-K-Different-Integers.md b/website/content/ChapterFour/0900~0999/0992.Subarrays-with-K-Different-Integers.md
new file mode 100644
index 000000000..d9c975512
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0992.Subarrays-with-K-Different-Integers.md
@@ -0,0 +1,120 @@
+# [992. Subarrays with K Different Integers](https://leetcode.com/problems/subarrays-with-k-different-integers/)
+
+## 题目
+
+Given an array A of positive integers, call a (contiguous, not necessarily distinct) subarray of A good if the number of different integers in that subarray is exactly K.
+
+(For example, [1,2,3,1,2] has 3 different integers: 1, 2, and 3.)
+
+Return the number of good subarrays of A.
+
+
+**Example 1**:
+
+```
+
+Input: A = [1,2,1,2,3], K = 2
+Output: 7
+Explanation: Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2].
+
+```
+
+**Example 2**:
+
+```
+
+Input: A = [1,2,1,3,4], K = 3
+Output: 3
+Explanation: Subarrays formed with exactly 3 different integers: [1,2,1,3], [2,1,3], [1,3,4].
+
+```
+
+**Note**:  
+
+- 1 <= A.length <= 20000
+- 1 <= A[i] <= A.length
+- 1 <= K <= A.length
+
+## 题目大意
+
+这道题考察的是滑动窗口的问题。
+
+给出一个数组 和 K，K 代表窗口能能包含的不同数字的个数。K = 2 代表窗口内只能有 2 个不同的数字。求数组中满足条件 K 的窗口个数。
+
+## 解题思路
+
+如果只是单纯的滑动窗口去做，会错过一些解。比如在例子 1 中，滑动窗口可以得到 [1,2], [1,2,1], [1,2,1,2], [2,1,2], [1,2], [2,3], 会少 [2,1] 这个解，原因是右边窗口滑动到最右边了，左边窗口在缩小的过程中，右边窗口不会再跟着动了。有同学可能会说，每次左边窗口移动的时候，右边窗口都再次从左边窗口的位置开始重新滑动。这样做确实可以，但是这样做完会发现超时。因为中间包含大量的重复计算。
+
+这道题就需要第 3 个指针。原有滑动窗口的 2 个指针，右窗口保留这个窗口里面最长的子数组，正好有 K 个元素，左窗口右移的逻辑不变。再多用一个指针用来标识正好有 K - 1 个元素的位置。那么正好有 K 个不同元素的解就等于 ans = atMostK(A, K) - atMostK(A, K - 1)。最多有 K 个元素减去最多有 K - 1 个元素得到的窗口中正好有 K 个元素的解。
+
+以例子 1 为例，先求最多有 K 个元素的窗口个数。
+
+```c
+[1]     
+[1,2], [2]     
+[1,2,1], [2,1], [1]  
+[1,2,1,2], [2,1,2], [1,2], [2]  
+[2,3], [3]  
+```
+
+每当窗口滑动到把 K 消耗为 0 的时候，res = right - left + 1 。为什么要这么计算，right - left + 1 代表的含义是，终点为 right，至多为 K 个元素的窗口有多少个。[left,right], [left + 1,right], [left + 2,right] …… [right,right]。这样算出来的解是包含这道题最终求得的解的，还多出了一部分解。多出来的部分减掉即可，即减掉最多为 K - 1 个元素的解。
+
+最多为 K - 1 个元素的解如下：
+
+```c
+[1]
+[2]
+[1]
+[2]
+[3]
+```
+
+两者相减以后得到的结果就是最终结果：
+
+```c   
+[1,2]    
+[1,2,1], [2,1]  
+[1,2,1,2], [2,1,2], [1,2]  
+[2,3]  
+```
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func subarraysWithKDistinct(A []int, K int) int {
+	return subarraysWithKDistinctSlideWindow(A, K) - subarraysWithKDistinctSlideWindow(A, K-1)
+}
+
+func subarraysWithKDistinctSlideWindow(A []int, K int) int {
+	left, right, counter, res, freq := 0, 0, K, 0, map[int]int{}
+	for right = 0; right < len(A); right++ {
+		if freq[A[right]] == 0 {
+			counter--
+		}
+		freq[A[right]]++
+		for counter < 0 {
+			freq[A[left]]--
+			if freq[A[left]] == 0 {
+				counter++
+			}
+			left++
+		}
+		res += right - left + 1
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0991.Broken-Calculator/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0993.Cousins-in-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0993.Cousins-in-Binary-Tree.md b/website/content/ChapterFour/0900~0999/0993.Cousins-in-Binary-Tree.md
new file mode 100755
index 000000000..8a75f3e20
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0993.Cousins-in-Binary-Tree.md
@@ -0,0 +1,170 @@
+# [993. Cousins in Binary Tree](https://leetcode.com/problems/cousins-in-binary-tree/)
+
+## 题目
+
+In a binary tree, the root node is at depth `0`, and children of each depth `k` node are at depth `k+1`.
+
+Two nodes of a binary tree are *cousins* if they have the same depth, but have **different parents**.
+
+We are given the `root` of a binary tree with unique values, and the values `x` and `y` of two different nodes in the tree.
+
+Return `true` if and only if the nodes corresponding to the values `x` and `y` are cousins.
+
+**Example 1**:
+
+![](https://assets.leetcode.com/uploads/2019/02/12/q1248-01.png)
+
+    Input: root = [1,2,3,4], x = 4, y = 3
+    Output: false
+
+**Example 2**:
+
+![](https://assets.leetcode.com/uploads/2019/02/12/q1248-02.png)
+
+    Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
+    Output: true
+
+**Example 3**:
+
+![](https://assets.leetcode.com/uploads/2019/02/13/q1248-03.png)
+
+    Input: root = [1,2,3,null,4], x = 2, y = 3
+    Output: false
+
+**Note**:
+
+1. The number of nodes in the tree will be between `2` and `100`.
+2. Each node has a unique integer value from `1` to `100`.
+
+
+## 题目大意
+
+在二叉树中，根节点位于深度 0 处，每个深度为 k 的节点的子节点位于深度 k+1 处。如果二叉树的两个节点深度相同，但父节点不同，则它们是一对堂兄弟节点。我们给出了具有唯一值的二叉树的根节点 root，以及树中两个不同节点的值 x 和 y。只有与值 x 和 y 对应的节点是堂兄弟节点时，才返回 true。否则，返回 false。
+
+
+
+## 解题思路
+
+
+- 给出一个二叉树，和 x ，y 两个值，要求判断这两个值是不是兄弟结点。兄弟结点的定义：都位于同一层，并且父结点是同一个结点。
+- 这一题有 3 种解题方法，DFS、BFS、递归。思路都不难。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 递归
+func isCousins(root *TreeNode, x int, y int) bool {
+	if root == nil {
+		return false
+	}
+	levelX, levelY := findLevel(root, x, 1), findLevel(root, y, 1)
+	if levelX != levelY {
+		return false
+	}
+	return !haveSameParents(root, x, y)
+}
+
+func findLevel(root *TreeNode, x, level int) int {
+	if root == nil {
+		return 0
+	}
+	if root.Val != x {
+		leftLevel, rightLevel := findLevel(root.Left, x, level+1), findLevel(root.Right, x, level+1)
+		if leftLevel == 0 {
+			return rightLevel
+		}
+		return leftLevel
+	}
+	return level
+}
+
+func haveSameParents(root *TreeNode, x, y int) bool {
+	if root == nil {
+		return false
+	}
+	if (root.Left != nil && root.Right != nil && root.Left.Val == x && root.Right.Val == y) ||
+		(root.Left != nil && root.Right != nil && root.Left.Val == y && root.Right.Val == x) {
+		return true
+	}
+	return haveSameParents(root.Left, x, y) || haveSameParents(root.Right, x, y)
+}
+
+// 解法二 BFS
+type mark struct {
+	prev  int
+	depth int
+}
+
+func isCousinsBFS(root *TreeNode, x int, y int) bool {
+	if root == nil {
+		return false
+	}
+	queue := []*TreeNode{root}
+	visited := [101]*mark{}
+	visited[root.Val] = &mark{prev: -1, depth: 1}
+
+	for len(queue) > 0 {
+		node := queue[0]
+		queue = queue[1:]
+		depth := visited[node.Val].depth
+		if node.Left != nil {
+			visited[node.Left.Val] = &mark{prev: node.Val, depth: depth + 1}
+			queue = append(queue, node.Left)
+		}
+		if node.Right != nil {
+			visited[node.Right.Val] = &mark{prev: node.Val, depth: depth + 1}
+			queue = append(queue, node.Right)
+		}
+	}
+	if visited[x] == nil || visited[y] == nil {
+		return false
+	}
+	if visited[x].depth == visited[y].depth && visited[x].prev != visited[y].prev {
+		return true
+	}
+	return false
+}
+
+// 解法三 DFS
+func isCousinsDFS(root *TreeNode, x int, y int) bool {
+	var depth1, depth2, parent1, parent2 int
+	dfsCousins(root, x, 0, -1, &parent1, &depth1)
+	dfsCousins(root, y, 0, -1, &parent2, &depth2)
+	return depth1 > 1 && depth1 == depth2 && parent1 != parent2
+}
+
+func dfsCousins(root *TreeNode, val, depth, last int, parent, res *int) {
+	if root == nil {
+		return
+	}
+	if root.Val == val {
+		*res = depth
+		*parent = last
+		return
+	}
+	depth++
+	dfsCousins(root.Left, val, depth, root.Val, parent, res)
+	dfsCousins(root.Right, val, depth, root.Val, parent, res)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0992.Subarrays-with-K-Different-Integers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0995.Minimum-Number-of-K-Consecutive-Bit-Flips/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0995.Minimum-Number-of-K-Consecutive-Bit-Flips.md b/website/content/ChapterFour/0900~0999/0995.Minimum-Number-of-K-Consecutive-Bit-Flips.md
new file mode 100755
index 000000000..31066661c
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0995.Minimum-Number-of-K-Consecutive-Bit-Flips.md
@@ -0,0 +1,104 @@
+# [995. Minimum Number of K Consecutive Bit Flips](https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips/)
+
+
+## 题目
+
+In an array `A` containing only 0s and 1s, a `K`-bit flip consists of choosing a (contiguous) subarray of length `K` and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0.
+
+Return the minimum number of `K`-bit flips required so that there is no 0 in the array. If it is not possible, return `-1`.
+
+**Example 1**:
+
+    Input: A = [0,1,0], K = 1
+    Output: 2
+    Explanation: Flip A[0], then flip A[2].
+
+**Example 2**:
+
+    Input: A = [1,1,0], K = 2
+    Output: -1
+    Explanation: No matter how we flip subarrays of size 2, we can't make the array become [1,1,1].
+
+**Example 3**:
+
+    Input: A = [0,0,0,1,0,1,1,0], K = 3
+    Output: 3
+    Explanation:
+    Flip A[0],A[1],A[2]: A becomes [1,1,1,1,0,1,1,0]
+    Flip A[4],A[5],A[6]: A becomes [1,1,1,1,1,0,0,0]
+    Flip A[5],A[6],A[7]: A becomes [1,1,1,1,1,1,1,1]
+
+**Note**:
+
+1. `1 <= A.length <= 30000`
+2. `1 <= K <= A.length`
+
+
+## 题目大意
+
+在仅包含 0 和 1 的数组 A 中，一次 K 位翻转包括选择一个长度为 K 的（连续）子数组，同时将子数组中的每个 0 更改为 1，而每个 1 更改为 0。返回所需的 K 位翻转的次数，以便数组没有值为 0 的元素。如果不可能，返回 -1。
+
+提示：
+
+1. 1 <= A.length <= 30000
+2. 1 <= K <= A.length
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面的元素只有 0 和 1。给一个长度为 K 的窗口，在这个窗口内的所有元素都会 0-1 翻转。问最后需要翻转几次，使得整个数组都为 1 。如果不能翻转使得最后数组元素都为 1，则输出 -1。
+- 拿到这一题首先想到的是贪心算法。例如第 765 题，这类题的描述都是这样的：在一个数组中或者环形数组中通过交换位置，或者翻转变换，达到最终结果，要求找到最少步数。贪心能保证是最小步数(证明略)。按照贪心的思想，这一题也这样做，从数组 0 下标开始往后扫，依次翻转每个 K 大小的窗口内元素。
+- 由于窗口大小限制了，所以这题滑动窗口只需要一个边界坐标，用左边界就可以判断了。每一个 `A[i]` 是否需要翻转，是由于 `[ i-k+1，i ]`、`[ i-k+2，i+1 ]`、`[ i-k+3，i+2 ]`……`[ i-1，i+k ]` 这一系列的窗口翻转`累积影响`的。那如何之前这些窗口`累积`到 `A[i]` 上翻转的次数呢？可以动态的维护一个翻转次数，当 `i` 摆脱了上一次翻转窗口 `K` 的时候，翻转次数就 `-1` 。举个例子：
+
+        A = [0 0 0 1 0 1 1 0] K = 3
+        
+        A = [2 0 0 1 0 1 1 0] i = 0 flippedTime = 1
+        A = [2 0 0 1 0 1 1 0] i = 1 flippedTime = 1
+        A = [2 0 0 1 0 1 1 0] i = 2 flippedTime = 1
+        A = [2 0 0 1 0 1 1 0] i = 3 flippedTime = 0
+        A = [2 0 0 1 2 1 1 0] i = 4 flippedTime = 1
+        A = [2 0 0 1 2 2 1 0] i = 5 flippedTime = 2
+        A = [2 0 0 1 2 2 1 0] i = 6 flippedTime = 2
+        A = [2 0 0 1 2 2 1 0] i = 7 flippedTime = 1
+
+    当判断 `A[i]` 是否需要翻转的时候，只需要留意每个宽度为 `K` 窗口的左边界。会影响到 A[i] 的窗口的左边界分别是 `i-k+1`、`i-k+2`、`i-k+3`、…… `i-1`，只需要分别看这些窗口有没有翻转就行。这里可以用特殊标记来记录这些窗口的左边界是否被翻转了。如果翻转过，则把窗口左边界的那个数字标记为 2 (为什么要标记为 2 呢？其实设置成什么都可以，只要不是 0 和 1 ，和原有的数字区分开就行)。当 `i≥k` 的时候，代表 `i` 已经脱离了 `i-k` 的这个窗口，因为能影响 `A[i]` 的窗口是从 `i-k+1` 开始的，如果 `A[i-k] == 2` 代表 `i-k` 窗口已经翻转过了，现在既然脱离了它的窗口影响，那么就要把累积的 `flippedTime - 1` 。这样就维护了累积 `flippedTime` 和滑动窗口中累积影响的关系。
+
+- 接下来还需要处理的是 `flippedTime` 与当前 `A[i]` 是否翻转的问题。如果 `flippedTime` 是偶数次，原来的 0 还是 0，就需要再次翻转，如果 `flippedTime` 是奇数次，原来的 0 变成了 1 就不需要翻转了。总结成一条结论就是 `A[i]` 与 `flippedTime` 同奇偶性的时候就要翻转。当 `i + K` 比 `len(A)` 大的时候，代表剩下的这些元素肯定不能在一个窗口里面翻转，则输出 -1 。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func minKBitFlips(A []int, K int) int {
+	flippedTime, count := 0, 0
+	for i := 0; i < len(A); i++ {
+		if i >= K && A[i-K] == 2 {
+			flippedTime--
+		}
+		// 下面这个判断包含了两种情况：
+		// 如果 flippedTime 是奇数，且 A[i] == 1 就需要翻转
+		// 如果 flippedTime 是偶数，且 A[i] == 0 就需要翻转
+		if flippedTime%2 == A[i] {
+			if i+K > len(A) {
+				return -1
+			}
+			A[i] = 2
+			flippedTime++
+			count++
+		}
+	}
+	return count
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0993.Cousins-in-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0996.Number-of-Squareful-Arrays/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0996.Number-of-Squareful-Arrays.md b/website/content/ChapterFour/0900~0999/0996.Number-of-Squareful-Arrays.md
new file mode 100755
index 000000000..c305fa181
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0996.Number-of-Squareful-Arrays.md
@@ -0,0 +1,114 @@
+# [996. Number of Squareful Arrays](https://leetcode.com/problems/number-of-squareful-arrays/)
+
+
+
+## 题目
+
+Given an array `A` of non-negative integers, the array is *squareful* if for every pair of adjacent elements, their sum is a perfect square.
+
+Return the number of permutations of A that are squareful. Two permutations `A1` and `A2` differ if and only if there is some index `i` such that `A1[i] != A2[i]`.
+
+**Example 1**:
+
+    Input: [1,17,8]
+    Output: 2
+    Explanation: 
+    [1,8,17] and [17,8,1] are the valid permutations.
+
+**Example 2**:
+
+    Input: [2,2,2]
+    Output: 1
+
+**Note**:
+
+1. `1 <= A.length <= 12`
+2. `0 <= A[i] <= 1e9`
+
+
+## 题目大意
+
+给定一个非负整数数组 A，如果该数组每对相邻元素之和是一个完全平方数，则称这一数组为正方形数组。
+
+返回 A 的正方形排列的数目。两个排列 A1 和 A2 不同的充要条件是存在某个索引 i，使得 A1[i] != A2[i]。
+
+
+
+## 解题思路
+
+
+- 这一题是第 47 题的加强版。第 47 题要求求出一个数组的所有不重复的排列。这一题要求求出一个数组的所有不重复，且相邻两个数字之和都为完全平方数的排列。
+- 思路和第 47 题完全一致，只不过增加判断相邻两个数字之和为完全平方数的判断，注意在 DFS 的过程中，需要剪枝，否则时间复杂度很高，会超时。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func numSquarefulPerms(A []int) int {
+	if len(A) == 0 {
+		return 0
+	}
+	used, p, res := make([]bool, len(A)), []int{}, [][]int{}
+	sort.Ints(A) // 这里是去重的关键逻辑
+	generatePermutation996(A, 0, p, &res, &used)
+	return len(res)
+}
+
+func generatePermutation996(nums []int, index int, p []int, res *[][]int, used *[]bool) {
+	if index == len(nums) {
+		checkSquareful := true
+		for i := 0; i < len(p)-1; i++ {
+			if !checkSquare(p[i] + p[i+1]) {
+				checkSquareful = false
+				break
+			}
+		}
+		if checkSquareful {
+			temp := make([]int, len(p))
+			copy(temp, p)
+			*res = append(*res, temp)
+		}
+		return
+	}
+	for i := 0; i < len(nums); i++ {
+		if !(*used)[i] {
+			if i > 0 && nums[i] == nums[i-1] && !(*used)[i-1] { // 这里是去重的关键逻辑
+				continue
+			}
+			if len(p) > 0 && !checkSquare(nums[i]+p[len(p)-1]) { // 关键的剪枝条件
+				continue
+			}
+			(*used)[i] = true
+			p = append(p, nums[i])
+			generatePermutation996(nums, index+1, p, res, used)
+			p = p[:len(p)-1]
+			(*used)[i] = false
+		}
+	}
+	return
+}
+
+func checkSquare(num int) bool {
+	tmp := math.Sqrt(float64(num))
+	if int(tmp)*int(tmp) == num {
+		return true
+	}
+	return false
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0995.Minimum-Number-of-K-Consecutive-Bit-Flips/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0997.Find-the-Town-Judge/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0997.Find-the-Town-Judge.md b/website/content/ChapterFour/0900~0999/0997.Find-the-Town-Judge.md
new file mode 100644
index 000000000..495f18cad
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0997.Find-the-Town-Judge.md
@@ -0,0 +1,94 @@
+# [997. Find the Town Judge](https://leetcode.com/problems/find-the-town-judge/)
+
+## 题目
+
+In a town, there are n people labeled from 1 to n. There is a rumor that one of these people is secretly the town judge.
+
+If the town judge exists, then:
+
+- The town judge trusts nobody.
+- Everybody (except for the town judge) trusts the town judge.
+- There is exactly one person that satisfies properties 1 and 2.
+
+You are given an array trust where trust[i] = [ai, bi] representing that the person labeled ai trusts the person labeled bi.
+
+Return the label of the town judge if the town judge exists and can be identified, or return -1 otherwise.
+
+**Example 1**:
+
+    Input: n = 2, trust = [[1,2]]
+    Output: 2
+
+**Example 2**:
+
+    Input: n = 3, trust = [[1,3],[2,3]]
+    Output: 3
+
+**Example 3**:
+
+    Input: n = 3, trust = [[1,3],[2,3],[3,1]]
+    Output: -1
+
+**Constraints:**
+
+- 1 <= n <= 1000
+- 0 <= trust.length <= 10000
+- trust[i].length == 2
+- All the pairs of trust are unique.
+- ai != bi
+- 1 <= ai, bi <= n
+
+## 题目大意
+
+小镇里有 n 个人，按从 1 到 n 的顺序编号。传言称，这些人中有一个暗地里是小镇法官。
+
+如果小镇法官真的存在，那么：
+
+- 小镇法官不会信任任何人。
+- 每个人（除了小镇法官）都信任这位小镇法官。
+- 只有一个人同时满足属性 1 和属性 2 。
+
+给你一个数组 trust ，其中 trust[i] = [ai, bi] 表示编号为 ai 的人信任编号为 bi 的人。
+
+如果小镇法官存在并且可以确定他的身份，请返回该法官的编号；否则，返回 -1 。
+
+## 解题思路
+
+入度和出度统计
+
+- 被人信任定义为入度, 信任别人定义为出度
+- 如果 1-n 之间有数字 x 的入度为 n - 1，出度为 0，则返回 x
+
+## 代码
+
+```go
+package leetcode
+
+func findJudge(n int, trust [][]int) int {
+	if n == 1 && len(trust) == 0 {
+		return 1
+	}
+	judges := make(map[int]int)
+	for _, v := range trust {
+		judges[v[1]] += 1
+	}
+	for _, v := range trust {
+		if _, ok := judges[v[0]]; ok {
+			delete(judges, v[0])
+		}
+	}
+	for k, v := range judges {
+		if v == n-1 {
+			return k
+		}
+	}
+	return -1
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0996.Number-of-Squareful-Arrays/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0999.Available-Captures-for-Rook/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/0999.Available-Captures-for-Rook.md b/website/content/ChapterFour/0900~0999/0999.Available-Captures-for-Rook.md
new file mode 100644
index 000000000..8797ca961
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/0999.Available-Captures-for-Rook.md
@@ -0,0 +1,106 @@
+# [999. Available Captures for Rook](https://leetcode.com/problems/available-captures-for-rook/)
+
+
+## 题目
+
+On an 8 x 8 chessboard, there is one white rook. There also may be empty squares, white bishops, and black pawns. These are given as characters 'R', '.', 'B', and 'p' respectively. Uppercase characters represent white pieces, and lowercase characters represent black pieces.
+
+The rook moves as in the rules of Chess: it chooses one of four cardinal directions (north, east, west, and south), then moves in that direction until it chooses to stop, reaches the edge of the board, or captures an opposite colored pawn by moving to the same square it occupies. Also, rooks cannot move into the same square as other friendly bishops.
+
+Return the number of pawns the rook can capture in one move.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/02/20/1253_example_1_improved.PNG](https://assets.leetcode.com/uploads/2019/02/20/1253_example_1_improved.PNG)
+
+```
+Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
+Output: 3
+Explanation: 
+In this example the rook is able to capture all the pawns.
+```
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/02/19/1253_example_2_improved.PNG](https://assets.leetcode.com/uploads/2019/02/19/1253_example_2_improved.PNG)
+
+```
+Input: [[".",".",".",".",".",".",".","."],[".","p","p","p","p","p",".","."],[".","p","p","B","p","p",".","."],[".","p","B","R","B","p",".","."],[".","p","p","B","p","p",".","."],[".","p","p","p","p","p",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
+Output: 0
+Explanation: 
+Bishops are blocking the rook to capture any pawn.
+```
+
+**Example 3**:
+
+![https://assets.leetcode.com/uploads/2019/02/20/1253_example_3_improved.PNG](https://assets.leetcode.com/uploads/2019/02/20/1253_example_3_improved.PNG)
+
+```
+Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
+Output: 3
+Explanation: 
+The rook can capture the pawns at positions b5, d6 and f5.
+```
+
+**Note**:
+
+1. `board.length == board[i].length == 8`
+2. `board[i][j]` is either `'R'`, `'.'`, `'B'`, or `'p'`
+3. There is exactly one cell with `board[i][j] == 'R'`
+
+## 题目大意
+
+在一个 8 x 8 的棋盘上，有一个白色的车（Rook），用字符 'R' 表示。棋盘上还可能存在空方块，白色的象（Bishop）以及黑色的卒（pawn），分别用字符 '.'，'B' 和 'p' 表示。不难看出，大写字符表示的是白棋，小写字符表示的是黑棋。车按国际象棋中的规则移动。东，西，南，北四个基本方向任选其一，然后一直向选定的方向移动，直到满足下列四个条件之一：
+
+- 棋手选择主动停下来。
+- 棋子因到达棋盘的边缘而停下。
+- 棋子移动到某一方格来捕获位于该方格上敌方（黑色）的卒，停在该方格内。
+- 车不能进入/越过已经放有其他友方棋子（白色的象）的方格，停在友方棋子前。
+
+你现在可以控制车移动一次，请你统计有多少敌方的卒处于你的捕获范围内（即，可以被一步捕获的棋子数）。
+
+## 解题思路
+
+- 按照国际象棋的规则移动车，要求输出只移动一次，有多少个卒在车的捕获范围之内
+- 简单题，按照国际象棋车的移动规则， 4 个方向分别枚举即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func numRookCaptures(board [][]byte) int {
+	num := 0
+	for i := 0; i < len(board); i++ {
+		for j := 0; j < len(board[i]); j++ {
+			if board[i][j] == 'R' {
+				num += caputure(board, i-1, j, -1, 0) // Up
+				num += caputure(board, i+1, j, 1, 0)  // Down
+				num += caputure(board, i, j-1, 0, -1) // Left
+				num += caputure(board, i, j+1, 0, 1)  // Right
+			}
+		}
+	}
+	return num
+}
+
+func caputure(board [][]byte, x, y int, bx, by int) int {
+	for x >= 0 && x < len(board) && y >= 0 && y < len(board[x]) && board[x][y] != 'B' {
+		if board[x][y] == 'p' {
+			return 1
+		}
+		x += bx
+		y += by
+	}
+	return 0
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0997.Find-the-Town-Judge/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1002.Find-Common-Characters/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/0900~0999/_index.md b/website/content/ChapterFour/0900~0999/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/0900~0999/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/1000~1099/1002.Find-Common-Characters.md b/website/content/ChapterFour/1000~1099/1002.Find-Common-Characters.md
new file mode 100755
index 000000000..2e1f62b4d
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1002.Find-Common-Characters.md
@@ -0,0 +1,81 @@
+# [1002. Find Common Characters](https://leetcode.com/problems/find-common-characters/)
+
+
+## 题目
+
+Given an array `A` of strings made only from lowercase letters, return a list of all characters that show up in all strings within the list **(including duplicates)**. For example, if a character occurs 3 times in all strings but not 4 times, you need to include that character three times in the final answer.
+
+You may return the answer in any order.
+
+**Example 1**:
+
+    Input: ["bella","label","roller"]
+    Output: ["e","l","l"]
+
+**Example 2**:
+
+    Input: ["cool","lock","cook"]
+    Output: ["c","o"]
+
+**Note**:
+
+1. `1 <= A.length <= 100`
+2. `1 <= A[i].length <= 100`
+3. `A[i][j]` is a lowercase letter
+
+## 题目大意
+
+给定仅有小写字母组成的字符串数组 A，返回列表中的每个字符串中都显示的全部字符（包括重复字符）组成的列表。例如，如果一个字符在每个字符串中出现 3 次，但不是 4 次，则需要在最终答案中包含该字符 3 次。你可以按任意顺序返回答案。
+
+
+## 解题思路
+
+- 简单题。给出一个字符串数组 A，要求找出这个数组中每个字符串都包含字符，如果字符出现多次，在最终结果中也需要出现多次。这一题可以用 map 来统计每个字符串的频次，但是如果用数组统计会更快。题目中说了只有小写字母，那么用 2 个 26 位长度的数组就可以统计出来了。遍历字符串数组的过程中，不过的缩小每个字符在每个字符串中出现的频次(因为需要找所有字符串公共的字符，公共的频次肯定就是最小的频次)，得到了最终公共字符的频次数组以后，按顺序输出就可以了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func commonChars(A []string) []string {
+	cnt := [26]int{}
+	for i := range cnt {
+		cnt[i] = math.MaxUint16
+	}
+	cntInWord := [26]int{}
+	for _, word := range A {
+		for _, char := range []byte(word) { // compiler trick - here we will not allocate new memory
+			cntInWord[char-'a']++
+		}
+		for i := 0; i < 26; i++ {
+			// 缩小频次，使得统计的公共频次更加准确
+			if cntInWord[i] < cnt[i] {
+				cnt[i] = cntInWord[i]
+			}
+		}
+		// 重置状态
+		for i := range cntInWord {
+			cntInWord[i] = 0
+		}
+	}
+	result := make([]string, 0)
+	for i := 0; i < 26; i++ {
+		for j := 0; j < cnt[i]; j++ {
+			result = append(result, string(i+'a'))
+		}
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0900~0999/0999.Available-Captures-for-Rook/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1003.Check-If-Word-Is-Valid-After-Substitutions/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1003.Check-If-Word-Is-Valid-After-Substitutions.md b/website/content/ChapterFour/1000~1099/1003.Check-If-Word-Is-Valid-After-Substitutions.md
new file mode 100644
index 000000000..69a4c6786
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1003.Check-If-Word-Is-Valid-After-Substitutions.md
@@ -0,0 +1,114 @@
+# [1003. Check If Word Is Valid After Substitutions](https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/)
+
+## 题目
+
+We are given that the string "abc" is valid.
+
+From any valid string V, we may split V into two pieces X and Y such that X + Y (X concatenated with Y) is equal to V.  (X or Y may be empty.)  Then, X + "abc" + Y is also valid.
+
+If for example S = "abc", then examples of valid strings are: "abc", "aabcbc", "abcabc", "abcabcababcc".  Examples of invalid strings are: "abccba", "ab", "cababc", "bac".
+
+Return true if and only if the given string S is valid.
+
+ 
+
+**Example 1**:
+
+```
+
+Input: "aabcbc"
+Output: true
+Explanation: 
+We start with the valid string "abc".
+Then we can insert another "abc" between "a" and "bc", resulting in "a" + "abc" + "bc" which is "aabcbc".
+
+```
+
+**Example 2**:
+
+```
+
+Input: "abcabcababcc"
+Output: true
+Explanation: 
+"abcabcabc" is valid after consecutive insertings of "abc".
+Then we can insert "abc" before the last letter, resulting in "abcabcab" + "abc" + "c" which is "abcabcababcc".
+
+```
+
+**Example 3**:
+
+```
+
+Input: "abccba"
+Output: false
+
+```
+
+**Example 4**:
+
+```
+
+Input: "cababc"
+Output: false
+
+```
+
+**Note**:
+
+1. 1 <= S.length <= 20000
+2. S[i] is 'a', 'b', or 'c'
+
+## 题目大意
+
+假设 abc 是有效的字符串，对于任何 字符串 V，如果用 abc 把字符串 V 切成 2 半，X 和 Y，组成 X + abc + Y 的字符串，X + abc + Y 的这个字符串依旧是有效的。X 和 Y 可以是空字符串。
+
+例如，"abc"( "" + "abc" + ""), "aabcbc"( "a" + "abc" + "bc"), "abcabc"( "" + "abc" + "abc"), "abcabcababcc"( "abc" + "abc" + "ababcc"，其中 "ababcc" 也是有效的，"ab" + "abc" + "c") 都是有效的字符串。
+
+"abccba"( "" + "abc" + "cba"，"cba" 不是有效的字符串), "ab"("ab" 也不是有效字符串), "cababc"("c" + "abc" + "bc"，"c"，"bc" 都不是有效字符串), "bac" ("bac" 也不是有效字符串)这些都不是有效的字符串。
+
+任意给一个字符串 S ，要求判断它是否有效，如果有效则输出 true。
+
+## 解题思路
+
+这一题可以类似括号匹配问题，因为 "abc" 这样的组合就代表是有效的，类似于括号匹配，遇到 "a" 就入栈，当遇到 "b" 字符的时候判断栈顶是不是 "a"，当遇到 "c" 字符的时候需要判断栈顶是不是 "a" 和 "b"。最后如果栈都清空了，就输出 true。
+
+## 代码
+
+```go
+
+package leetcode
+
+func isValid1003(S string) bool {
+	if len(S) < 3 {
+		return false
+	}
+	stack := []byte{}
+	for i := 0; i < len(S); i++ {
+		if S[i] == 'a' {
+			stack = append(stack, S[i])
+		} else if S[i] == 'b' {
+			if len(stack) > 0 && stack[len(stack)-1] == 'a' {
+				stack = append(stack, S[i])
+			} else {
+				return false
+			}
+		} else {
+			if len(stack) > 1 && stack[len(stack)-1] == 'b' && stack[len(stack)-2] == 'a' {
+				stack = stack[:len(stack)-2]
+			} else {
+				return false
+			}
+		}
+	}
+	return len(stack) == 0
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1002.Find-Common-Characters/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1004.Max-Consecutive-Ones-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1004.Max-Consecutive-Ones-III.md b/website/content/ChapterFour/1000~1099/1004.Max-Consecutive-Ones-III.md
new file mode 100644
index 000000000..8731490ef
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1004.Max-Consecutive-Ones-III.md
@@ -0,0 +1,94 @@
+# [1004. Max Consecutive Ones III](https://leetcode.com/problems/max-consecutive-ones-iii/)
+
+## 题目
+
+Given an array A of 0s and 1s, we may change up to K values from 0 to 1.
+
+Return the length of the longest (contiguous) subarray that contains only 1s. 
+
+
+**Example 1**:
+
+```
+
+Input: A = [1,1,1,0,0,0,1,1,1,1,0], K = 2
+Output: 6
+Explanation: 
+[1,1,1,0,0,1,1,1,1,1,1]
+Bolded numbers were flipped from 0 to 1.  The longest subarray is underlined.
+
+```
+
+**Example 2**:
+
+```
+
+Input: A = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], K = 3
+Output: 10
+Explanation: 
+[0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
+Bolded numbers were flipped from 0 to 1.  The longest subarray is underlined.
+
+```
+
+
+**Note**:
+
+- 1 <= A.length <= 20000
+- 0 <= K <= A.length
+- A[i] is 0 or 1 
+
+
+## 题目大意
+
+这道题考察的是滑动窗口的问题。
+
+给出一个数组，数组中元素只包含 0 和 1 。再给一个 K，代表能将 0 变成 1 的次数。要求出经过变换以后，1 连续的最长长度。
+
+## 解题思路
+
+按照滑动窗口的思路处理即可，不断的更新和维护最大长度。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func longestOnes(A []int, K int) int {
+	res, left, right := 0, 0, 0
+	for left < len(A) {
+		if right < len(A) && ((A[right] == 0 && K > 0) || A[right] == 1) {
+			if A[right] == 0 {
+				K--
+			}
+			right++
+		} else {
+			if K == 0 || (right == len(A) && K > 0) {
+				res = max(res, right-left)
+			}
+			if A[left] == 0 {
+				K++
+			}
+			left++
+		}
+	}
+	return res
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1003.Check-If-Word-Is-Valid-After-Substitutions/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1005.Maximize-Sum-Of-Array-After-K-Negations/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1005.Maximize-Sum-Of-Array-After-K-Negations.md b/website/content/ChapterFour/1000~1099/1005.Maximize-Sum-Of-Array-After-K-Negations.md
new file mode 100644
index 000000000..7cb38aaab
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1005.Maximize-Sum-Of-Array-After-K-Negations.md
@@ -0,0 +1,90 @@
+# [1005. Maximize Sum Of Array After K Negations](https://leetcode.com/problems/maximize-sum-of-array-after-k-negations/)
+
+## 题目
+
+Given an array A of integers, we must modify the array in the following way: we choose an i and replace A[i] with -A[i], and we repeat this process K times in total.  (We may choose the same index i multiple times.)
+
+Return the largest possible sum of the array after modifying it in this way.
+
+
+**Example 1**:
+
+```
+
+Input: A = [4,2,3], K = 1
+Output: 5
+Explanation: Choose indices (1,) and A becomes [4,-2,3].
+
+```
+
+**Example 2**:
+
+```
+
+Input: A = [3,-1,0,2], K = 3
+Output: 6
+Explanation: Choose indices (1, 2, 2) and A becomes [3,1,0,2].
+
+```
+
+**Example 3**:
+
+```
+
+Input: A = [2,-3,-1,5,-4], K = 2
+Output: 13
+Explanation: Choose indices (1, 4) and A becomes [2,3,-1,5,4].
+
+```
+
+**Note**:
+
+- 1 <= A.length <= 10000
+- 1 <= K <= 10000
+- -100 <= A[i] <= 100
+
+## 题目大意
+
+将数组中的元素变成它的相反数，这种操作执行 K 次之后，求出数组中所有元素的总和最大。
+
+## 解题思路
+
+这一题可以用最小堆来做，构建最小堆，每次将最小的元素变成它的相反数。然后最小堆调整，再将新的最小元素变成它的相反数。执行 K 次以后求数组中所有的值之和就是最大值。
+
+这道题也可以用排序来实现。排序一次，从最小值开始往后扫，依次将最小值变为相反数。这里需要注意一点，负数都改变成正数以后，接着不是再改变这些变成正数的负数，而是接着改变顺序的正数。因为这些正数是比较小的正数。负数越小，变成正数以后值越大。正数越小，变成负数以后对总和影响最小。具体实现见代码。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+func largestSumAfterKNegations(A []int, K int) int {
+	sort.Ints(A)
+	minIdx := 0
+	for i := 0; i < K; i++ {
+		A[minIdx] = -A[minIdx]
+		if A[minIdx+1] < A[minIdx] {
+			minIdx++
+		}
+	}
+	sum := 0
+	for _, a := range A {
+		sum += a
+	}
+	return sum
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1004.Max-Consecutive-Ones-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1006.Clumsy-Factorial/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1006.Clumsy-Factorial.md b/website/content/ChapterFour/1000~1099/1006.Clumsy-Factorial.md
new file mode 100644
index 000000000..737ec2205
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1006.Clumsy-Factorial.md
@@ -0,0 +1,82 @@
+# [1006. Clumsy Factorial](https://leetcode.com/problems/clumsy-factorial/)
+
+
+## 题目
+
+Normally, the factorial of a positive integer `n` is the product of all positive integers less than or equal to `n`.  For example, `factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1`.
+
+We instead make a *clumsy factorial:* using the integers in decreasing order, we swap out the multiply operations for a fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in this order.
+
+For example, `clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1`.  However, these operations are still applied using the usual order of operations of arithmetic: we do all multiplication and division steps before any addition or subtraction steps, and multiplication and division steps are processed left to right.
+
+Additionally, the division that we use is *floor division* such that `10 * 9 / 8` equals `11`.  This guarantees the result is an integer.
+
+`Implement the clumsy` function as defined above: given an integer `N`, it returns the clumsy factorial of `N`.
+
+**Example 1:**
+
+```
+Input:4
+Output: 7
+Explanation: 7 = 4 * 3 / 2 + 1
+```
+
+**Example 2:**
+
+```
+Input:10
+Output:12
+Explanation:12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
+```
+
+**Note:**
+
+1. `1 <= N <= 10000`
+2. `2^31 <= answer <= 2^31 - 1` (The answer is guaranteed to fit within a 32-bit integer.)
+
+## 题目大意
+
+通常，正整数 n 的阶乘是所有小于或等于 n 的正整数的乘积。例如，factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1。相反，我们设计了一个笨阶乘 clumsy：在整数的递减序列中，我们以一个固定顺序的操作符序列来依次替换原有的乘法操作符：乘法(*)，除法(/)，加法(+)和减法(-)。例如，clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1。然而，这些运算仍然使用通常的算术运算顺序：我们在任何加、减步骤之前执行所有的乘法和除法步骤，并且按从左到右处理乘法和除法步骤。另外，我们使用的除法是地板除法（floor division），所以 10 * 9 / 8 等于 11。这保证结果是一个整数。实现上面定义的笨函数：给定一个整数 N，它返回 N 的笨阶乘。
+
+## 解题思路
+
+- 按照题意，由于本题没有括号，所以先乘除后加减。4 个操作一组，先算乘法，再算除法，再算加法，最后算减法。减法也可以看成是加法，只是带负号的加法。
+
+## 代码
+
+```go
+package leetcode
+
+func clumsy(N int) int {
+	res, count, tmp, flag := 0, 1, N, false
+	for i := N - 1; i > 0; i-- {
+		count = count % 4
+		switch count {
+		case 1:
+			tmp = tmp * i
+		case 2:
+			tmp = tmp / i
+		case 3:
+			res = res + tmp
+			flag = true
+			tmp = -1
+			res = res + i
+		case 0:
+			flag = false
+			tmp = tmp * (i)
+		}
+		count++
+	}
+	if !flag {
+		res = res + tmp
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1005.Maximize-Sum-Of-Array-After-K-Negations/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1009.Complement-of-Base-10-Integer/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1009.Complement-of-Base-10-Integer.md b/website/content/ChapterFour/1000~1099/1009.Complement-of-Base-10-Integer.md
new file mode 100644
index 000000000..b448d6c2b
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1009.Complement-of-Base-10-Integer.md
@@ -0,0 +1,74 @@
+# [1009. Complement of Base 10 Integer](https://leetcode.com/problems/complement-of-base-10-integer/)
+
+
+## 题目
+
+The **complement** of an integer is the integer you get when you flip all the `0`'s to `1`'s and all the `1`'s to `0`'s in its binary representation.
+
+- For example, The integer `5` is `"101"` in binary and its **complement** is `"010"` which is the integer `2`.
+
+Given an integer `n`, return *its complement*.
+
+**Example 1:**
+
+```
+Input: n = 5
+Output: 2
+Explanation: 5 is "101" in binary, with complement "010" in binary, which is 2 in base-10.
+
+```
+
+**Example 2:**
+
+```
+Input: n = 7
+Output: 0
+Explanation: 7 is "111" in binary, with complement "000" in binary, which is 0 in base-10.
+
+```
+
+**Example 3:**
+
+```
+Input: n = 10
+Output: 5
+Explanation: 10 is "1010" in binary, with complement "0101" in binary, which is 5 in base-10.
+
+```
+
+**Constraints:**
+
+- `0 <= n < 109`
+
+## 题目大意
+
+每个非负整数 N 都有其二进制表示。例如， 5 可以被表示为二进制 "101"，11 可以用二进制 "1011" 表示，依此类推。注意，除 N = 0 外，任何二进制表示中都不含前导零。
+
+二进制的反码表示是将每个 1 改为 0 且每个 0 变为 1。例如，二进制数 "101" 的二进制反码为 "010"。
+
+给你一个十进制数 N，请你返回其二进制表示的反码所对应的十进制整数。
+
+## 解题思路
+
+- 简单题。求一个十进制数的反码，只需要让该数和全 1 的数进行异或计算即可。所以本题重点在如何构造 mask 上。
+
+## 代码
+
+```go
+package leetcode
+
+func bitwiseComplement(n int) int {
+    mask := 1 
+    for mask < n {
+        mask = (mask << 1) + 1
+    }
+    return mask ^ n
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1006.Clumsy-Factorial/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60.md b/website/content/ChapterFour/1000~1099/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60.md
new file mode 100644
index 000000000..b8edb0ebb
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60.md
@@ -0,0 +1,70 @@
+# [1010. Pairs of Songs With Total Durations Divisible by 60](https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/)
+
+
+## 题目
+
+You are given a list of songs where the ith song has a duration of `time[i]` seconds.
+
+Return *the number of pairs of songs for which their total duration in seconds is divisible by* `60`. Formally, we want the number of indices `i`, `j` such that `i < j` with `(time[i] + time[j]) % 60 == 0`.
+
+**Example 1:**
+
+```
+Input: time = [30,20,150,100,40]
+Output: 3
+Explanation: Three pairs have a total duration divisible by 60:
+(time[0] = 30, time[2] = 150): total duration 180
+(time[1] = 20, time[3] = 100): total duration 120
+(time[1] = 20, time[4] = 40): total duration 60
+
+```
+
+**Example 2:**
+
+```
+Input: time = [60,60,60]
+Output: 3
+Explanation: All three pairs have a total duration of 120, which is divisible by 60.
+
+```
+
+**Constraints:**
+
+- `1 <= time.length <= 6 * 104`
+- `1 <= time[i] <= 500`
+
+## 题目大意
+
+在歌曲列表中，第 i 首歌曲的持续时间为 time[i] 秒。
+
+返回其总持续时间（以秒为单位）可被 60 整除的歌曲对的数量。形式上，我们希望下标数字 i 和 j 满足  i < j 且有 (time[i] + time[j]) % 60 == 0。
+
+## 解题思路
+
+- 简单题。先将数组每个元素对 60 取余，将它们都转换到 [0,59] 之间。然后在数组中找两两元素之和等于 60 的数对。可以在 0-30 之内对半查找符合条件的数对。对 0 和 30 单独计算。因为多个 0 相加，余数还为 0 。2 个 30 相加之和为 60。
+
+## 代码
+
+```go
+func numPairsDivisibleBy60(time []int) int {
+	counts := make([]int, 60)
+	for _, v := range time {
+		v %= 60
+		counts[v]++
+	}
+	res := 0
+	for i := 1; i < len(counts)/2; i++ {
+		res += counts[i] * counts[60-i]
+	}
+	res += (counts[0] * (counts[0] - 1)) / 2
+	res += (counts[30] * (counts[30] - 1)) / 2
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1009.Complement-of-Base-10-Integer/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1011.Capacity-To-Ship-Packages-Within-D-Days/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1011.Capacity-To-Ship-Packages-Within-D-Days.md b/website/content/ChapterFour/1000~1099/1011.Capacity-To-Ship-Packages-Within-D-Days.md
new file mode 100755
index 000000000..c812630ac
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1011.Capacity-To-Ship-Packages-Within-D-Days.md
@@ -0,0 +1,122 @@
+# [1011. Capacity To Ship Packages Within D Days](https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/)
+
+
+## 题目
+
+A conveyor belt has packages that must be shipped from one port to another within `D` days.
+
+The `i`-th package on the conveyor belt has a weight of `weights[i]`. Each day, we load the ship with packages on the conveyor belt (in the order given by `weights`). We may not load more weight than the maximum weight capacity of the ship.
+
+Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within `D` days.
+
+**Example 1**:
+
+    Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5
+    Output: 15
+    Explanation: 
+    A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
+    1st day: 1, 2, 3, 4, 5
+    2nd day: 6, 7
+    3rd day: 8
+    4th day: 9
+    5th day: 10
+    
+    Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed.
+
+**Example 2**:
+
+    Input: weights = [3,2,2,4,1,4], D = 3
+    Output: 6
+    Explanation: 
+    A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
+    1st day: 3, 2
+    2nd day: 2, 4
+    3rd day: 1, 4
+
+**Example 3**:
+
+    Input: weights = [1,2,3,1,1], D = 4
+    Output: 3
+    Explanation: 
+    1st day: 1
+    2nd day: 2
+    3rd day: 3
+    4th day: 1, 1
+
+**Note**:
+
+1. `1 <= D <= weights.length <= 50000`
+2. `1 <= weights[i] <= 500`
+
+
+## 题目大意
+
+传送带上的包裹必须在 D 天内从一个港口运送到另一个港口。
+
+传送带上的第 i 个包裹的重量为 weights[i]。每一天，我们都会按给出重量的顺序往传送带上装载包裹。我们装载的重量不会超过船的最大运载重量。
+
+返回能在 D 天内将传送带上的所有包裹送达的船的最低运载能力。
+
+提示：
+
+- 1 <= D <= weights.length <= 50000
+- 1 <= weights[i] <= 500
+
+
+## 解题思路
+
+- 给出一个数组和天数 D，要求正好在 D 天把数组中的货物都运完。求传输带上能承受的最小货物重量是多少。
+- 这一题和第 410 题完全一样，只不过换了一个题面。代码完全不变。思路解析见第 410 题。
+
+
+## 代码
+
+```go
+
+func shipWithinDays(weights []int, D int) int {
+	maxNum, sum := 0, 0
+	for _, num := range weights {
+		sum += num
+		if num > maxNum {
+			maxNum = num
+		}
+	}
+	if D == 1 {
+		return sum
+	}
+	low, high := maxNum, sum
+	for low < high {
+		mid := low + (high-low)>>1
+		if calSum(mid, D, weights) {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+func calSum(mid, m int, nums []int) bool {
+	sum, count := 0, 0
+	for _, v := range nums {
+		sum += v
+		if sum > mid {
+			sum = v
+			count++
+			// 分成 m 块，只需要插桩 m -1 个
+			if count > m-1 {
+				return false
+			}
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1017.Convert-to-Base-2/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1017.Convert-to-Base-2.md b/website/content/ChapterFour/1000~1099/1017.Convert-to-Base-2.md
new file mode 100755
index 000000000..d6aaa9ade
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1017.Convert-to-Base-2.md
@@ -0,0 +1,82 @@
+# [1017. Convert to Base -2](https://leetcode.com/problems/convert-to-base-2/)
+
+
+## 题目
+
+Given a number `N`, return a string consisting of `"0"`s and `"1"`s that represents its value in base **`-2`** (negative two).
+
+The returned string must have no leading zeroes, unless the string is `"0"`.
+
+**Example 1**:
+
+    Input: 2
+    Output: "110"
+    Explantion: (-2) ^ 2 + (-2) ^ 1 = 2
+
+**Example 2**:
+
+    Input: 3
+    Output: "111"
+    Explantion: (-2) ^ 2 + (-2) ^ 1 + (-2) ^ 0 = 3
+
+**Example 3**:
+
+    Input: 4
+    Output: "100"
+    Explantion: (-2) ^ 2 = 4
+
+**Note**:
+
+1. `0 <= N <= 10^9`
+
+
+## 题目大意
+
+给出数字 N，返回由若干 "0" 和 "1"组成的字符串，该字符串为 N 的负二进制（base -2）表示。除非字符串就是 "0"，否则返回的字符串中不能含有前导零。
+
+提示：
+
+- 0 <= N <= 10^9
+
+
+
+## 解题思路
+
+- 给出一个十进制的数，要求转换成 -2 进制的数
+- 这一题仿造十进制转二进制的思路，短除法即可。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strconv"
+
+func baseNeg2(N int) string {
+	if N == 0 {
+		return "0"
+	}
+	res := ""
+	for N != 0 {
+		remainder := N % (-2)
+		N = N / (-2)
+		if remainder < 0 {
+			remainder += 2
+			N++
+		}
+		res = strconv.Itoa(remainder) + res
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1011.Capacity-To-Ship-Packages-Within-D-Days/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1018.Binary-Prefix-Divisible-By-5/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1018.Binary-Prefix-Divisible-By-5.md b/website/content/ChapterFour/1000~1099/1018.Binary-Prefix-Divisible-By-5.md
new file mode 100644
index 000000000..5afbdaa2a
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1018.Binary-Prefix-Divisible-By-5.md
@@ -0,0 +1,77 @@
+# [1018. Binary Prefix Divisible By 5](https://leetcode.com/problems/binary-prefix-divisible-by-5/)
+
+
+## 题目
+
+Given an array `A` of `0`s and `1`s, consider `N_i`: the i-th subarray from `A[0]` to `A[i]` interpreted as a binary number (from most-significant-bit to least-significant-bit.)
+
+Return a list of booleans `answer`, where `answer[i]` is `true` if and only if `N_i` is divisible by 5.
+
+**Example 1**:
+
+```
+Input: [0,1,1]
+Output: [true,false,false]
+Explanation: 
+The input numbers in binary are 0, 01, 011; which are 0, 1, and 3 in base-10.  Only the first number is divisible by 5, so answer[0] is true.
+
+```
+
+**Example 2**:
+
+```
+Input: [1,1,1]
+Output: [false,false,false]
+
+```
+
+**Example 3**:
+
+```
+Input: [0,1,1,1,1,1]
+Output: [true,false,false,false,true,false]
+
+```
+
+**Example 4**:
+
+```
+Input: [1,1,1,0,1]
+Output: [false,false,false,false,false]
+
+```
+
+**Note**:
+
+1. `1 <= A.length <= 30000`
+2. `A[i]` is `0` or `1`
+
+## 题目大意
+
+给定由若干 0 和 1 组成的数组 A。我们定义 N_i：从 A[0] 到 A[i] 的第 i 个子数组被解释为一个二进制数（从最高有效位到最低有效位）。返回布尔值列表 answer，只有当 N_i 可以被 5 整除时，答案 answer[i] 为 true，否则为 false。
+
+## 解题思路
+
+- 简单题。每扫描数组中的一个数字，累计转换成二进制数对 5 取余，如果余数为 0，则存入 true，否则存入 false。
+
+## 代码
+
+```go
+package leetcode
+
+func prefixesDivBy5(a []int) []bool {
+	res, num := make([]bool, len(a)), 0
+	for i, v := range a {
+		num = (num<<1 | v) % 5
+		res[i] = num == 0
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1017.Convert-to-Base-2/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1019.Next-Greater-Node-In-Linked-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1019.Next-Greater-Node-In-Linked-List.md b/website/content/ChapterFour/1000~1099/1019.Next-Greater-Node-In-Linked-List.md
new file mode 100644
index 000000000..a4f1ec424
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1019.Next-Greater-Node-In-Linked-List.md
@@ -0,0 +1,101 @@
+# [1019. Next Greater Node In Linked List](https://leetcode.com/problems/next-greater-node-in-linked-list/)
+
+## 题目
+
+We are given a linked list with head as the first node.  Let's number the nodes in the list: node\_1, node\_2, node\_3, ... etc.
+
+Each node may have a next larger value: for node_i, next\_larger(node\_i) is the node\_j.val such that j > i, node\_j.val > node\_i.val, and j is the smallest possible choice.  If such a j does not exist, the next larger value is 0.
+
+Return an array of integers answer, where answer[i] = next\_larger(node\_{i+1}).
+
+Note that in the example inputs (not outputs) below, arrays such as [2,1,5] represent the serialization of a linked list with a head node value of 2, second node value of 1, and third node value of 5.
+
+ 
+
+**Example 1**:
+
+```
+
+Input: [2,1,5]
+Output: [5,5,0]
+
+```
+
+**Example 2**:
+
+```
+
+Input: [2,7,4,3,5]
+Output: [7,0,5,5,0]
+
+```
+
+**Example 3**:
+
+```
+
+Input: [1,7,5,1,9,2,5,1]
+Output: [7,9,9,9,0,5,0,0]
+
+```
+
+**Note**:
+
+- 1 <= node.val <= 10^9 for each node in the linked list.
+- The given list has length in the range [0, 10000].
+
+
+## 题目大意
+
+给出一个链表，要求找出每个结点后面比该结点值大的第一个结点，如果找不到这个结点，则输出 0 。
+
+
+## 解题思路
+
+这一题和第 739 题、第 496 题、第 503 题类似。也有 2 种解题方法。先把链表中的数字存到数组中，整道题的思路就和第 739 题完全一致了。普通做法就是 2 层循环。优化的做法就是用单调栈，维护一个单调递减的栈即可。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// 解法一 单调栈
+func nextLargerNodes(head *ListNode) []int {
+	type node struct {
+		index, val int
+	}
+	var monoStack []node
+	var res []int
+	for head != nil {
+		for len(monoStack) > 0 && monoStack[len(monoStack)-1].val < head.Val {
+			res[monoStack[len(monoStack)-1].index] = head.Val
+			monoStack = monoStack[:len(monoStack)-1]
+		}
+		monoStack = append(monoStack, node{len(res), head.Val})
+		res = append(res, 0)
+		head = head.Next
+	}
+	return res
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1018.Binary-Prefix-Divisible-By-5/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1020.Number-of-Enclaves/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1020.Number-of-Enclaves.md b/website/content/ChapterFour/1000~1099/1020.Number-of-Enclaves.md
new file mode 100644
index 000000000..85dcb5ff9
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1020.Number-of-Enclaves.md
@@ -0,0 +1,100 @@
+# [1020. Number of Enclaves](https://leetcode.com/problems/number-of-enclaves/)
+
+
+
+## 题目
+
+Given a 2D array `A`, each cell is 0 (representing sea) or 1 (representing land)
+
+A move consists of walking from one land square 4-directionally to another land square, or off the boundary of the grid.
+
+Return the number of land squares in the grid for which we **cannot** walk off the boundary of the grid in any number of moves.
+
+**Example 1**:
+
+```
+Input: [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
+Output: 3
+Explanation: 
+There are three 1s that are enclosed by 0s, and one 1 that isn't enclosed because its on the boundary.
+```
+
+**Example 2**:
+
+```
+Input: [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
+Output: 0
+Explanation: 
+All 1s are either on the boundary or can reach the boundary.
+```
+
+**Note**:
+
+1. `1 <= A.length <= 500`
+2. `1 <= A[i].length <= 500`
+3. `0 <= A[i][j] <= 1`
+4. All rows have the same size.
+
+## 题目大意
+
+给出一个二维数组 A，每个单元格为 0（代表海）或 1（代表陆地）。移动是指在陆地上从一个地方走到另一个地方（朝四个方向之一）或离开网格的边界。返回网格中无法在任意次数的移动中离开网格边界的陆地单元格的数量。
+
+提示：
+
+- 1 <= A.length <= 500
+- 1 <= A[i].length <= 500
+- 0 <= A[i][j] <= 1
+- 所有行的大小都相同
+
+
+## 解题思路
+
+- 给出一个地图，要求输出不和边界连通的 1 的个数。
+- 这一题可以用 DFS 也可以用并查集解答。DFS 的思路是深搜的过程中把和边界连通的点都覆盖成 0，最后遍历一遍地图，输出 1 的个数即可。并查集的思路就比较直接了，把能和边界连通的放在一个集合中，剩下的就是不能和边界连通的都在另外一个集合中，输出这个集合里面元素的个数即可。
+- 这一题和第 200 题，第 1254 题，第 695 题类似。可以放在一起练习。
+
+## 代码
+
+```go
+func numEnclaves(A [][]int) int {
+	m, n := len(A), len(A[0])
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			if i == 0 || i == m-1 || j == 0 || j == n-1 {
+				if A[i][j] == 1 {
+					dfsNumEnclaves(A, i, j)
+				}
+			}
+		}
+	}
+	count := 0
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			if A[i][j] == 1 {
+				count++
+			}
+		}
+	}
+	return count
+}
+
+func dfsNumEnclaves(A [][]int, x, y int) {
+	if !isInGrid(A, x, y) || A[x][y] == 0 {
+		return
+	}
+	A[x][y] = 0
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		dfsNumEnclaves(A, nx, ny)
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1019.Next-Greater-Node-In-Linked-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1021.Remove-Outermost-Parentheses/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1021.Remove-Outermost-Parentheses.md b/website/content/ChapterFour/1000~1099/1021.Remove-Outermost-Parentheses.md
new file mode 100644
index 000000000..7742aec2c
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1021.Remove-Outermost-Parentheses.md
@@ -0,0 +1,130 @@
+# [1021. Remove Outermost Parentheses](https://leetcode.com/problems/remove-outermost-parentheses/)
+
+## 题目
+
+A valid parentheses string is either empty (""), "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation.  For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.
+
+A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.
+
+Given a valid parentheses string S, consider its primitive decomposition: S = P\_1 + P\_2 + ... + P\_k, where P\_i are primitive valid parentheses strings.
+
+Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.
+
+
+**Example 1**:
+
+```
+
+Input: "(()())(())"
+Output: "()()()"
+Explanation: 
+The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
+After removing outer parentheses of each part, this is "()()" + "()" = "()()()".
+
+```
+
+**Example 2**:
+
+```
+
+Input: "(()())(())(()(()))"
+Output: "()()()()(())"
+Explanation: 
+The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
+After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".
+
+```
+
+**Example 3**:
+
+```
+
+Input: "()()"
+Output: ""
+Explanation: 
+The input string is "()()", with primitive decomposition "()" + "()".
+After removing outer parentheses of each part, this is "" + "" = "".
+
+```
+
+**Note**:
+
+- S.length <= 10000
+- S[i] is "(" or ")"
+- S is a valid parentheses string
+ 
+
+## 题目大意
+
+题目要求去掉最外层的括号。
+
+## 解题思路
+
+用栈模拟即可。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func removeOuterParentheses(S string) string {
+	now, current, ans := 0, "", ""
+	for _, char := range S {
+		if string(char) == "(" {
+			now++
+		} else if string(char) == ")" {
+			now--
+		}
+		current += string(char)
+		if now == 0 {
+			ans += current[1 : len(current)-1]
+			current = ""
+		}
+	}
+	return ans
+}
+
+// 解法二
+func removeOuterParentheses1(S string) string {
+	stack, res, counter := []byte{}, "", 0
+	for i := 0; i < len(S); i++ {
+		if counter == 0 && len(stack) == 1 && S[i] == ')' {
+			stack = stack[1:]
+			continue
+		}
+		if len(stack) == 0 && S[i] == '(' {
+			stack = append(stack, S[i])
+			continue
+		}
+		if len(stack) > 0 {
+			switch S[i] {
+			case '(':
+				{
+					counter++
+					res += "("
+				}
+			case ')':
+				{
+					counter--
+					res += ")"
+				}
+			}
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1020.Number-of-Enclaves/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1022.Sum-of-Root-To-Leaf-Binary-Numbers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1022.Sum-of-Root-To-Leaf-Binary-Numbers.md b/website/content/ChapterFour/1000~1099/1022.Sum-of-Root-To-Leaf-Binary-Numbers.md
new file mode 100644
index 000000000..299a22219
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1022.Sum-of-Root-To-Leaf-Binary-Numbers.md
@@ -0,0 +1,61 @@
+# [1022. Sum of Root To Leaf Binary Numbers](https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers/)
+
+## 题目
+
+You are given the root of a binary tree where each node has a value 0 or 1. Each root-to-leaf path represents a binary number starting with the most significant bit.
+
+For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is 13.
+For all leaves in the tree, consider the numbers represented by the path from the root to that leaf. Return the sum of these numbers.
+
+The test cases are generated so that the answer fits in a 32-bits integer.
+
+**Example 1:**
+
+```c
+Input: root = [1,0,1,0,1,0,1]
+Output: 22
+Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22
+```
+
+**Example 2:**
+
+```c
+Input: root = [0]
+Output: 0
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 1000]`.
+
+- `Node.val` is `0` or `1`.
+
+
+## 题目大意
+
+给定一棵结点值都是`0`或`1`的二叉树，每条从根结点到叶结点的路径都代表一个从最高有效位开始的二进制数。
+
+返回从根节点到所有叶结点的路径所表示的数字之和。
+
+
+## 解题思路
+
+采用递归的方式对根结点`root`进行后序遍历（左子树-右子树-根结点）。
+
+**递归函数的返回值**：
+
+递归遍历每个结点时，计算从根结点到当前访问结点的所表示数值`sum`都用到了上次的计算结果，所以递归函数的返回值是当前访问结点的计算结果值。
+
+**递归函数的逻辑**：
+
+- 当前遍历结点为`nil`，表示本层递归结束了，直接`return 0`。
+
+- 如果当前访问结点是叶结点，则返回从根结点到该结点所表示的数值`sum`。
+- 如果当前访问结点不是叶结点，则返回左子树和右子树所对应的结果之和。
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1021.Remove-Outermost-Parentheses/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1025.Divisor-Game/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1025.Divisor-Game.md b/website/content/ChapterFour/1000~1099/1025.Divisor-Game.md
new file mode 100755
index 000000000..868900959
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1025.Divisor-Game.md
@@ -0,0 +1,69 @@
+# [1025. Divisor Game](https://leetcode.com/problems/divisor-game/)
+
+
+## 题目
+
+Alice and Bob take turns playing a game, with Alice starting first.
+
+Initially, there is a number `N` on the chalkboard. On each player's turn, that player makes a *move* consisting of:
+
+- Choosing any `x` with `0 < x < N` and `N % x == 0`.
+- Replacing the number `N` on the chalkboard with `N - x`.
+
+Also, if a player cannot make a move, they lose the game.
+
+Return `True` if and only if Alice wins the game, assuming both players play optimally.
+
+**Example 1**:
+
+    Input: 2
+    Output: true
+    Explanation: Alice chooses 1, and Bob has no more moves.
+
+**Example 2**:
+
+    Input: 3
+    Output: false
+    Explanation: Alice chooses 1, Bob chooses 1, and Alice has no more moves.
+
+**Note**:
+
+1. `1 <= N <= 1000`
+
+
+## 题目大意
+
+
+爱丽丝和鲍勃一起玩游戏，他们轮流行动。爱丽丝先手开局。最初，黑板上有一个数字 N 。在每个玩家的回合，玩家需要执行以下操作：
+
+- 选出任一 x，满足 0 < x < N 且 N % x == 0 。
+- 用 N - x 替换黑板上的数字 N 。
+
+如果玩家无法执行这些操作，就会输掉游戏。只有在爱丽丝在游戏中取得胜利时才返回 True，否则返回 false。假设两个玩家都以最佳状态参与游戏。
+
+
+## 解题思路
+
+
+- 两人相互玩一个游戏，游戏初始有一个数 N，开始游戏的时候，任一方选择一个数 x，满足 `0 < x < N` 并且 `N % x == 0` 的条件，然后 `N-x` 为下一轮开始的数。此轮结束，该另外一个人继续选择数字，两人相互轮流选择。直到某一方再也没法选择数字的时候，输掉游戏。问如果你先手开始游戏，给出 N 的时候，能否直到这局你是否会必胜或者必输？
+- 这一题当 `N = 1` 的时候，那一轮的人必输。因为没法找到一个数字能满足 `0 < x < N` 并且 `N % x == 0` 的条件了。必胜策略就是把对方逼至 `N = 1` 的情况。题目中假设了对手也是一个很有头脑的人。初始如果 `N 为偶数`，我就选择 x = 1，对手拿到的数字就是奇数。只要最终能让对手拿到奇数，他就会输。初始如果 `N 为奇数`，N = 1 的时候直接输了，N 为其他奇数的时候，我们也只能选择一个奇数 x，(因为 `N % x == 0` ，N 为奇数，x 一定不会是偶数，因为偶数就能被 2 整除了)，对手由于是一个很有头脑的人，当我们选完 N - x 是偶数的时候，他就选择 1，那么轮到我们拿到的数字又是奇数。对手只要一直保证我们拿到奇数，最终肯定会逼着我们拿到 1，最终他就会获得胜利。所以经过分析可得，初始数字如果是偶数，有必胜策略，如果初始数字是奇数，有必输的策略。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func divisorGame(N int) bool {
+	return N%2 == 0
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1022.Sum-of-Root-To-Leaf-Binary-Numbers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1026.Maximum-Difference-Between-Node-and-Ancestor/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1026.Maximum-Difference-Between-Node-and-Ancestor.md b/website/content/ChapterFour/1000~1099/1026.Maximum-Difference-Between-Node-and-Ancestor.md
new file mode 100644
index 000000000..325536b0d
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1026.Maximum-Difference-Between-Node-and-Ancestor.md
@@ -0,0 +1,81 @@
+# [1026. Maximum Difference Between Node and Ancestor](https://leetcode.com/problems/maximum-difference-between-node-and-ancestor/)
+
+
+
+## 题目
+
+Given the `root` of a binary tree, find the maximum value `V` for which there exists **different** nodes `A` and `B` where `V = |A.val - B.val|` and `A` is an ancestor of `B`.
+
+(A node A is an ancestor of B if either: any child of A is equal to B, or any child of A is an ancestor of B.)
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/09/09/2whqcep.jpg](https://assets.leetcode.com/uploads/2019/09/09/2whqcep.jpg)
+
+```
+Input: [8,3,10,1,6,null,14,null,null,4,7,13]
+Output: 7
+Explanation: 
+We have various ancestor-node differences, some of which are given below :
+|8 - 3| = 5
+|3 - 7| = 4
+|8 - 1| = 7
+|10 - 13| = 3
+Among all possible differences, the maximum value of 7 is obtained by |8 - 1| = 7.
+```
+
+**Note**:
+
+1. The number of nodes in the tree is between `2` and `5000`.
+2. Each node will have value between `0` and `100000`.
+
+## 题目大意
+
+给定二叉树的根节点 root，找出存在于不同节点 A 和 B 之间的最大值 V，其中 V = |A.val - B.val|，且 A 是 B 的祖先。（如果 A 的任何子节点之一为 B，或者 A 的任何子节点是 B 的祖先，那么我们认为 A 是 B 的祖先）
+
+提示：
+
+- 树中的节点数在 2 到 5000 之间。
+- 每个节点的值介于 0 到 100000 之间。
+
+
+
+## 解题思路
+
+- 给出一颗树，要求找出祖先和孩子的最大差值。
+- DPS 深搜即可。每个节点和其所有孩子的`最大值`来自于 3 个值，节点本身，递归遍历左子树的最大值，递归遍历右子树的最大值；每个节点和其所有孩子的`最小值`来自于 3 个值，节点本身，递归遍历左子树的最小值，递归遍历右子树的最小值。依次求出自身节点和其所有孩子节点的最大差值，深搜的过程中动态维护最大差值即可。
+
+## 代码
+
+```go
+func maxAncestorDiff(root *TreeNode) int {
+	res := 0
+	dfsAncestorDiff(root, &res)
+	return res
+}
+
+func dfsAncestorDiff(root *TreeNode, res *int) (int, int) {
+	if root == nil {
+		return -1, -1
+	}
+	leftMax, leftMin := dfsAncestorDiff(root.Left, res)
+	if leftMax == -1 && leftMin == -1 {
+		leftMax = root.Val
+		leftMin = root.Val
+	}
+	rightMax, rightMin := dfsAncestorDiff(root.Right, res)
+	if rightMax == -1 && rightMin == -1 {
+		rightMax = root.Val
+		rightMin = root.Val
+	}
+	*res = max(*res, max(abs(root.Val-min(leftMin, rightMin)), abs(root.Val-max(leftMax, rightMax))))
+	return max(leftMax, max(rightMax, root.Val)), min(leftMin, min(rightMin, root.Val))
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1025.Divisor-Game/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1028.Recover-a-Tree-From-Preorder-Traversal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1028.Recover-a-Tree-From-Preorder-Traversal.md b/website/content/ChapterFour/1000~1099/1028.Recover-a-Tree-From-Preorder-Traversal.md
new file mode 100755
index 000000000..a7195a543
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1028.Recover-a-Tree-From-Preorder-Traversal.md
@@ -0,0 +1,142 @@
+# [1028. Recover a Tree From Preorder Traversal](https://leetcode.com/problems/recover-a-tree-from-preorder-traversal/)
+
+
+## 题目
+
+We run a preorder depth first search on the `root` of a binary tree.
+
+At each node in this traversal, we output `D` dashes (where `D` is the *depth* of this node), then we output the value of this node. *(If the depth of a node is `D`, the depth of its immediate child is `D+1`. The depth of the root node is `0`.)*
+
+If a node has only one child, that child is guaranteed to be the left child.
+
+Given the output `S` of this traversal, recover the tree and return its `root`.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/04/08/recover-a-tree-from-preorder-traversal.png](https://assets.leetcode.com/uploads/2019/04/08/recover-a-tree-from-preorder-traversal.png)
+
+    Input: "1-2--3--4-5--6--7"
+    Output: [1,2,5,3,4,6,7]
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/04/11/screen-shot-2019-04-10-at-114101-pm.png](https://assets.leetcode.com/uploads/2019/04/11/screen-shot-2019-04-10-at-114101-pm.png)
+
+    Input: "1-2--3---4-5--6---7"
+    Output: [1,2,5,3,null,6,null,4,null,7]
+
+**Example 3**:
+
+![https://assets.leetcode.com/uploads/2019/04/11/screen-shot-2019-04-10-at-114955-pm.png](https://assets.leetcode.com/uploads/2019/04/11/screen-shot-2019-04-10-at-114955-pm.png)
+
+    Input: "1-401--349---90--88"
+    Output: [1,401,null,349,88,90]
+
+**Note**:
+
+- The number of nodes in the original tree is between `1` and `1000`.
+- Each node will have a value between `1` and `10^9`.
+
+## 题目大意
+
+我们从二叉树的根节点 root 开始进行深度优先搜索。
+
+在遍历中的每个节点处，我们输出 D 条短划线（其中 D 是该节点的深度），然后输出该节点的值。（如果节点的深度为 D，则其直接子节点的深度为 D + 1。根节点的深度为 0）。如果节点只有一个子节点，那么保证该子节点为左子节点。给出遍历输出 S，还原树并返回其根节点 root。
+
+
+提示：
+
+- 原始树中的节点数介于 1 和 1000 之间。
+- 每个节点的值介于 1 和 10 ^ 9 之间。
+
+
+## 解题思路
+
+- 给出一个字符串，字符串是一个树的先根遍历的结果，其中破折号的个数代表层数。请根据这个字符串生成对应的树。
+- 这一题解题思路比较明确，用 DFS 就可以解题。边深搜字符串，边根据破折号的个数判断当前节点是否属于本层。如果不属于本层，回溯到之前的根节点，添加叶子节点以后再继续深搜。需要注意的是每次深搜时，扫描字符串的 index 需要一直保留，回溯也需要用到这个 index。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+)
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func recoverFromPreorder(S string) *TreeNode {
+	if len(S) == 0 {
+		return &TreeNode{}
+	}
+	root, index, level := &TreeNode{}, 0, 0
+	cur := root
+	dfsBuildPreorderTree(S, &index, &level, cur)
+	return root.Right
+}
+
+func dfsBuildPreorderTree(S string, index, level *int, cur *TreeNode) (newIndex *int) {
+	if *index == len(S) {
+		return index
+	}
+	if *index == 0 && *level == 0 {
+		i := 0
+		for i = *index; i < len(S); i++ {
+			if !isDigital(S[i]) {
+				break
+			}
+		}
+		num, _ := strconv.Atoi(S[*index:i])
+		tmp := &TreeNode{Val: num, Left: nil, Right: nil}
+		cur.Right = tmp
+		nLevel := *level + 1
+		index = dfsBuildPreorderTree(S, &i, &nLevel, tmp)
+		index = dfsBuildPreorderTree(S, index, &nLevel, tmp)
+	}
+	i := 0
+	for i = *index; i < len(S); i++ {
+		if isDigital(S[i]) {
+			break
+		}
+	}
+	if *level == i-*index {
+		j := 0
+		for j = i; j < len(S); j++ {
+			if !isDigital(S[j]) {
+				break
+			}
+		}
+		num, _ := strconv.Atoi(S[i:j])
+		tmp := &TreeNode{Val: num, Left: nil, Right: nil}
+		if cur.Left == nil {
+			cur.Left = tmp
+			nLevel := *level + 1
+			index = dfsBuildPreorderTree(S, &j, &nLevel, tmp)
+			index = dfsBuildPreorderTree(S, index, level, cur)
+		} else if cur.Right == nil {
+			cur.Right = tmp
+			nLevel := *level + 1
+			index = dfsBuildPreorderTree(S, &j, &nLevel, tmp)
+			index = dfsBuildPreorderTree(S, index, level, cur)
+		}
+	}
+	return index
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1026.Maximum-Difference-Between-Node-and-Ancestor/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1030.Matrix-Cells-in-Distance-Order/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1030.Matrix-Cells-in-Distance-Order.md b/website/content/ChapterFour/1000~1099/1030.Matrix-Cells-in-Distance-Order.md
new file mode 100755
index 000000000..a54d2fddd
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1030.Matrix-Cells-in-Distance-Order.md
@@ -0,0 +1,91 @@
+# [1030. Matrix Cells in Distance Order](https://leetcode.com/problems/matrix-cells-in-distance-order/)
+
+
+## 题目
+
+We are given a matrix with `R` rows and `C` columns has cells with integer coordinates `(r, c)`, where `0 <= r < R` and `0 <= c < C`.
+
+Additionally, we are given a cell in that matrix with coordinates `(r0, c0)`.
+
+Return the coordinates of all cells in the matrix, sorted by their distance from `(r0, c0)` from smallest distance to largest distance. Here, the distance between two cells `(r1, c1)` and `(r2, c2)` is the Manhattan distance, `|r1 - r2| + |c1 - c2|`. (You may return the answer in any order that satisfies this condition.)
+
+**Example 1**:
+
+    Input: R = 1, C = 2, r0 = 0, c0 = 0
+    Output: [[0,0],[0,1]]
+    Explanation: The distances from (r0, c0) to other cells are: [0,1]
+
+**Example 2**:
+
+    Input: R = 2, C = 2, r0 = 0, c0 = 1
+    Output: [[0,1],[0,0],[1,1],[1,0]]
+    Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2]
+    The answer [[0,1],[1,1],[0,0],[1,0]] would also be accepted as correct.
+
+**Example 3**:
+
+    Input: R = 2, C = 3, r0 = 1, c0 = 2
+    Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]
+    Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2,2,3]
+    There are other answers that would also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]].
+
+**Note**:
+
+1. `1 <= R <= 100`
+2. `1 <= C <= 100`
+3. `0 <= r0 < R`
+4. `0 <= c0 < C`
+
+
+
+## 题目大意
+
+
+给出 R 行 C 列的矩阵，其中的单元格的整数坐标为 (r, c)，满足 0 <= r < R 且 0 <= c < C。另外，我们在该矩阵中给出了一个坐标为 (r0, c0) 的单元格。
+
+返回矩阵中的所有单元格的坐标，并按到 (r0, c0) 的距离从最小到最大的顺序排，其中，两单元格(r1, c1) 和 (r2, c2) 之间的距离是曼哈顿距离，|r1 - r2| + |c1 - c2|。（你可以按任何满足此条件的顺序返回答案。）
+
+
+## 解题思路
+
+
+- 按照题意计算矩阵内给定点到其他每个点的距离即可
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func allCellsDistOrder(R int, C int, r0 int, c0 int) [][]int {
+	longRow, longCol, result := max(abs(r0-0), abs(R-r0)), max(abs(c0-0), abs(C-c0)), make([][]int, 0)
+	maxDistance := longRow + longCol
+	bucket := make([][][]int, maxDistance+1)
+	for i := 0; i <= maxDistance; i++ {
+		bucket[i] = make([][]int, 0)
+	}
+	for r := 0; r < R; r++ {
+		for c := 0; c < C; c++ {
+			distance := abs(r-r0) + abs(c-c0)
+			tmp := []int{r, c}
+			bucket[distance] = append(bucket[distance], tmp)
+		}
+	}
+	for i := 0; i <= maxDistance; i++ {
+		for _, buk := range bucket[i] {
+			result = append(result, buk)
+		}
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1028.Recover-a-Tree-From-Preorder-Traversal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1034.Coloring-A-Border/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1034.Coloring-A-Border.md b/website/content/ChapterFour/1000~1099/1034.Coloring-A-Border.md
new file mode 100644
index 000000000..cfcb67675
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1034.Coloring-A-Border.md
@@ -0,0 +1,115 @@
+# [1034. Coloring A Border](https://leetcode.com/problems/coloring-a-border/)
+
+## 题目
+
+You are given an m x n integer matrix grid, and three integers row, col, and color. Each value in the grid represents the color of the grid square at that location.
+
+Two squares belong to the same connected component if they have the same color and are next to each other in any of the 4 directions.
+
+The border of a connected component is all the squares in the connected component that are either 4-directionally adjacent to a square not in the component, or on the boundary of the grid (the first or last row or column).
+
+You should color the border of the connected component that contains the square grid[row][col] with color.
+
+Return the final grid.
+
+**Example 1**:
+
+    Input: grid = [[1,1],[1,2]], row = 0, col = 0, color = 3
+    Output: [[3,3],[3,2]]
+
+**Example 2**:
+
+    Input: grid = [[1,2,2],[2,3,2]], row = 0, col = 1, color = 3
+    Output: [[1,3,3],[2,3,3]]
+
+**Example 3**:
+
+    Input: grid = [[1,1,1],[1,1,1],[1,1,1]], row = 1, col = 1, color = 2
+    Output: [[2,2,2],[2,1,2],[2,2,2]]
+
+**Constraints:**
+
+- m == grid.length
+- n == grid[i].length
+- 1 <= m, n <= 50
+- 1 <= grid[i][j], color <= 1000
+- 0 <= row < m
+- 0 <= col < n
+
+## 题目大意
+
+给你一个大小为 m x n 的整数矩阵 grid ，表示一个网格。另给你三个整数 row、col 和 color 。网格中的每个值表示该位置处的网格块的颜色。
+
+当两个网格块的颜色相同，而且在四个方向中任意一个方向上相邻时，它们属于同一连通分量
+
+边界：在连通分量的块中（前提）并且满足以下条件之一：
+（1）要么上下左右存在一个块不在连通分量里面
+（2）要么这个块的位置在整个grid的边框上
+
+请你使用指定颜色 color 为所有包含网格块 grid[row][col] 的连通分量的边界进行着色，并返回最终的网格 grid 。
+
+## 解题思路
+
+- 用 bfs 进行遍历选出边界，使用 color 给边界着色
+
+## 代码
+
+```go
+package leetcode
+
+type point struct {
+	x int
+	y int
+}
+
+type gridInfo struct {
+	m             int
+	n             int
+	grid          [][]int
+	originalColor int
+}
+
+func colorBorder(grid [][]int, row, col, color int) [][]int {
+	m, n := len(grid), len(grid[0])
+	dirs := []point{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}
+	vis := make([][]bool, m)
+	for i := range vis {
+		vis[i] = make([]bool, n)
+	}
+	var borders []point
+	gInfo := gridInfo{
+		m:             m,
+		n:             n,
+		grid:          grid,
+		originalColor: grid[row][col],
+	}
+	dfs(row, col, gInfo, dirs, vis, &borders)
+	for _, p := range borders {
+		grid[p.x][p.y] = color
+	}
+	return grid
+}
+
+func dfs(x, y int, gInfo gridInfo, dirs []point, vis [][]bool, borders *[]point) {
+	vis[x][y] = true
+	isBorder := false
+	for _, dir := range dirs {
+		nx, ny := x+dir.x, y+dir.y
+		if !(0 <= nx && nx < gInfo.m && 0 <= ny && ny < gInfo.n && gInfo.grid[nx][ny] == gInfo.originalColor) {
+			isBorder = true
+		} else if !vis[nx][ny] {
+			dfs(nx, ny, gInfo, dirs, vis, borders)
+		}
+	}
+	if isBorder {
+		*borders = append(*borders, point{x, y})
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1030.Matrix-Cells-in-Distance-Order/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1037.Valid-Boomerang/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1037.Valid-Boomerang.md b/website/content/ChapterFour/1000~1099/1037.Valid-Boomerang.md
new file mode 100644
index 000000000..d01462887
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1037.Valid-Boomerang.md
@@ -0,0 +1,56 @@
+# [1037. Valid Boomerang](https://leetcode.com/problems/valid-boomerang/)
+
+
+## 题目
+
+A *boomerang* is a set of 3 points that are all distinct and **not** in a straight line.
+
+Given a list of three points in the plane, return whether these points are a boomerang.
+
+**Example 1**:
+
+```
+Input: [[1,1],[2,3],[3,2]]
+Output: true
+```
+
+**Example 2**:
+
+```
+Input: [[1,1],[2,2],[3,3]]
+Output: false
+```
+
+**Note**:
+
+1. `points.length == 3`
+2. `points[i].length == 2`
+3. `0 <= points[i][j] <= 100`
+
+## 题目大意
+
+回旋镖定义为一组三个点，这些点各不相同且不在一条直线上。给出平面上三个点组成的列表，判断这些点是否可以构成回旋镖。
+
+## 解题思路
+
+- 判断给出的 3 组点能否满足回旋镖。
+- 简单题。判断 3 个点组成的 2 条直线的斜率是否相等。由于斜率的计算是除法，还可能遇到分母为 0 的情况，那么可以转换成乘法，交叉相乘再判断是否相等，就可以省去判断分母为 0 的情况了，代码也简洁成一行了。
+
+## 代码
+
+```go
+
+package leetcode
+
+func isBoomerang(points [][]int) bool {
+	return (points[0][0]-points[1][0])*(points[0][1]-points[2][1]) != (points[0][0]-points[2][0])*(points[0][1]-points[1][1])
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1034.Coloring-A-Border/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1038.Binary-Search-Tree-to-Greater-Sum-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1038.Binary-Search-Tree-to-Greater-Sum-Tree.md b/website/content/ChapterFour/1000~1099/1038.Binary-Search-Tree-to-Greater-Sum-Tree.md
new file mode 100644
index 000000000..31c0a5142
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1038.Binary-Search-Tree-to-Greater-Sum-Tree.md
@@ -0,0 +1,114 @@
+# [1038. Binary Search Tree to Greater Sum Tree](https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/)
+
+
+## 题目
+
+Given the `root` of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.
+
+As a reminder, a *binary search tree* is a tree that satisfies these constraints:
+
+- The left subtree of a node contains only nodes with keys **less than** the node's key.
+- The right subtree of a node contains only nodes with keys **greater than** the node's key.
+- Both the left and right subtrees must also be binary search trees.
+
+**Note:** This question is the same as 538: [https://leetcode.com/problems/convert-bst-to-greater-tree/](https://leetcode.com/problems/convert-bst-to-greater-tree/)
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/05/02/tree.png](https://assets.leetcode.com/uploads/2019/05/02/tree.png)
+
+```
+Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
+Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]
+```
+
+**Example 2:**
+
+```
+Input: root = [0,null,1]
+Output: [1,null,1]
+```
+
+**Example 3:**
+
+```
+Input: root = [1,0,2]
+Output: [3,3,2]
+```
+
+**Example 4:**
+
+```
+Input: root = [3,2,4,1]
+Output: [7,9,4,10]
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[1, 100]`.
+- `0 <= Node.val <= 100`
+- All the values in the tree are **unique**.
+- `root` is guaranteed to be a valid binary search tree.
+
+## 题目大意
+
+给出二叉 搜索 树的根节点，该树的节点值各不相同，请你将其转换为累加树（Greater Sum Tree），使每个节点 node 的新值等于原树中大于或等于 node.val 的值之和。
+
+提醒一下，二叉搜索树满足下列约束条件：
+
+- 节点的左子树仅包含键 小于 节点键的节点。
+- 节点的右子树仅包含键 大于 节点键的节点。
+- 左右子树也必须是二叉搜索树。
+
+## 解题思路
+
+- 根据二叉搜索树的有序性，想要将其转换为累加树，只需按照 右节点 - 根节点 - 左节点的顺序遍历，并累加和即可。
+- 此题同第 538 题。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func bstToGst(root *TreeNode) *TreeNode {
+	if root == nil {
+		return root
+	}
+	sum := 0
+	dfs1038(root, &sum)
+	return root
+}
+
+func dfs1038(root *TreeNode, sum *int) {
+	if root == nil {
+		return
+	}
+	dfs1038(root.Right, sum)
+	root.Val += *sum
+	*sum = root.Val
+	dfs1038(root.Left, sum)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1037.Valid-Boomerang/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1040.Moving-Stones-Until-Consecutive-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1040.Moving-Stones-Until-Consecutive-II.md b/website/content/ChapterFour/1000~1099/1040.Moving-Stones-Until-Consecutive-II.md
new file mode 100755
index 000000000..9d03de77e
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1040.Moving-Stones-Until-Consecutive-II.md
@@ -0,0 +1,128 @@
+# [1040. Moving Stones Until Consecutive II](https://leetcode.com/problems/moving-stones-until-consecutive-ii/)
+
+
+## 题目
+
+On an **infinite** number line, the position of the i-th stone is given by `stones[i]`. Call a stone an *endpoint stone* if it has the smallest or largest position.
+
+Each turn, you pick up an endpoint stone and move it to an unoccupied position so that it is no longer an endpoint stone.
+
+In particular, if the stones are at say, `stones = [1,2,5]`, you **cannot** move the endpoint stone at position 5, since moving it to any position (such as 0, or 3) will still keep that stone as an endpoint stone.
+
+The game ends when you cannot make any more moves, ie. the stones are in consecutive positions.
+
+When the game ends, what is the minimum and maximum number of moves that you could have made? Return the answer as an length 2 array: `answer = [minimum_moves, maximum_moves]`
+
+**Example 1**:
+
+    Input: [7,4,9]
+    Output: [1,2]
+    Explanation: 
+    We can move 4 -> 8 for one move to finish the game.
+    Or, we can move 9 -> 5, 4 -> 6 for two moves to finish the game.
+
+**Example 2**:
+
+    Input: [6,5,4,3,10]
+    Output: [2,3]
+    We can move 3 -> 8 then 10 -> 7 to finish the game.
+    Or, we can move 3 -> 7, 4 -> 8, 5 -> 9 to finish the game.
+    Notice we cannot move 10 -> 2 to finish the game, because that would be an illegal move.
+
+**Example 3**:
+
+    Input: [100,101,104,102,103]
+    Output: [0,0]
+
+**Note**:
+
+1. `3 <= stones.length <= 10^4`
+2. `1 <= stones[i] <= 10^9`
+3. `stones[i]` have distinct values.
+
+
+## 题目大意
+
+在一个长度无限的数轴上，第 i 颗石子的位置为 stones[i]。如果一颗石子的位置最小/最大，那么该石子被称作端点石子。每个回合，你可以将一颗端点石子拿起并移动到一个未占用的位置，使得该石子不再是一颗端点石子。值得注意的是，如果石子像 stones = [1,2,5] 这样，你将无法移动位于位置 5 的端点石子，因为无论将它移动到任何位置（例如 0 或 3），该石子都仍然会是端点石子。当你无法进行任何移动时，即，这些石子的位置连续时，游戏结束。
+
+要使游戏结束，你可以执行的最小和最大移动次数分别是多少？ 以长度为 2 的数组形式返回答案：answer = [minimum\_moves, maximum\_moves] 。
+
+提示：
+
+1. 3 <= stones.length <= 10^4
+2. 1 <= stones[i] <= 10^9
+3. stones[i] 的值各不相同。
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面代表的是石头的坐标。要求移动石头，最终使得这些石头的坐标是一个连续的自然数列。但是规定，当一个石头是端点的时候，是不能移动的，例如 [1,2,5]，5 是端点，不能把 5 移到 3 或者 0 的位置，因为移动之后，这个石头仍然是端点。最终输出将所有石头排成连续的自然数列所需的最小步数和最大步数。
+- 这道题的关键就是如何保证端点石头不能再次移动到端点的限制。例如，[5,6,8,9,20]，20 是端点，但是 20 就可以移动到 7 的位置，最终形成 [5,6,7,8,9] 的连续序列。但是 [5,6,7,8,20]，这种情况 20 就不能移动到 9 了，只能让 8 移动到 9，20 再移动到 8 的位置，最终还是形成了 [5,6,7,8,9]，但是步数需要 2 步。经过上述分析，可以得到，端点石头只能往中间空挡的位置移动，如果中间没有空挡，那么需要借助一个石头先制造一个空挡，然后端点石头再插入到中间，这样最少是需要 2 步。
+- 再来考虑极值的情况。先看最大步数，最大步数肯定慢慢移动，一次移动一格，并且移动的格数最多。这里有两个极端情况，把数组里面的数全部都移动到最左端点，把数组里面的数全部都移动到最右端点。每次只移动一格。例如，全部都移到最右端点：
+
+        [3,4,5,6,10] // 初始状态，连续的情况
+        [4,5,6,7,10] // 第一步，把 3 挪到右边第一个可以插入的位置，即 7
+        [5,6,7,8,10] // 第二步，把 4 挪到右边第一个可以插入的位置，即 8
+        [6,7,8,9,10] // 第三步，把 5 挪到右边第一个可以插入的位置，即 9
+        
+        
+        [1,3,5,7,10] // 初始状态，不连续的情况
+        [3,4,5,7,10] // 第一步，把 1 挪到右边第一个可以插入的位置，即 4
+        [4,5,6,7,10] // 第二步，把 3 挪到右边第一个可以插入的位置，即 6
+        [5,6,7,8,10] // 第三步，把 4 挪到右边第一个可以插入的位置，即 8
+        [6,7,8,9,10] // 第四步，把 5 挪到右边第一个可以插入的位置，即 9
+
+    挪动的过程类似翻滚，最左边的石头挪到右边第一个可以放下的地方。然后不断的往右翻滚。把数组中的数全部都移动到最左边也同理。对比这两种情况的最大值，即是移动的最大步数。
+
+- 再看最小步数。这里就涉及到了滑动窗口了。由于最终是要形成连续的自然数列，所以滑动窗口的大小已经固定成 n 了，从数组的 0 下标可以往右滑动窗口，这个窗口中能包含的数字越多，代表窗口外的数字越少，那么把这些数字放进窗口内的步数也最小。于是可以求得最小步数。这里有一个比较坑的地方就是题目中的那个`“端点不能移动以后还是端点”`的限制。针对这种情况，需要额外的判断。如果当前窗口内有 n-1 个元素了，即只有一个端点在窗口外，并且窗口右边界的值减去左边界的值也等于 n-1，代表这个窗口内已经都是连续数字了。这种情况端点想融合到这个连续数列中，最少需要 2 步(上文已经分析过了)。
+- 注意一些边界情况。如果窗口从左往右滑动，窗口右边界滑到最右边了，但是窗口右边界的数字减去左边界的数字还是小于窗口大小 n，代表已经滑到头了，可以直接 break 出去。为什么滑到头了呢？由于数组经过从小到大排序以后，数字越往右边越大，当前数字是小值，已经满足了 `stones[right]-stones[left] < n`，左边界继续往右移动只会使得 `stones[left]` 更大，就更加小于 n 了。而我们需要寻找的是 `stones[right]-stones[left] >= n` 的边界点，肯定再也找不到了。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func numMovesStonesII(stones []int) []int {
+	if len(stones) == 0 {
+		return []int{0, 0}
+	}
+	sort.Ints(stones)
+	n := len(stones)
+	maxStep, minStep, left, right := max(stones[n-1]-stones[1]-n+2, stones[n-2]-stones[0]-n+2), math.MaxInt64, 0, 0
+	for left < n {
+		if right+1 < n && stones[right]-stones[left] < n {
+			right++
+		} else {
+			if stones[right]-stones[left] >= n {
+				right--
+			}
+			if right-left+1 == n-1 && stones[right]-stones[left]+1 == n-1 {
+				minStep = min(minStep, 2)
+			} else {
+				minStep = min(minStep, n-(right-left+1))
+			}
+			if right == n-1 && stones[right]-stones[left] < n {
+				break
+			}
+			left++
+		}
+	}
+	return []int{minStep, maxStep}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1038.Binary-Search-Tree-to-Greater-Sum-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1047.Remove-All-Adjacent-Duplicates-In-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1047.Remove-All-Adjacent-Duplicates-In-String.md b/website/content/ChapterFour/1000~1099/1047.Remove-All-Adjacent-Duplicates-In-String.md
new file mode 100644
index 000000000..5508dc6ff
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1047.Remove-All-Adjacent-Duplicates-In-String.md
@@ -0,0 +1,64 @@
+# [1047. Remove All Adjacent Duplicates In String](https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string/)
+
+## 题目
+
+Given a string S of lowercase letters, a duplicate removal consists of choosing two adjacent and equal letters, and removing them.
+
+We repeatedly make duplicate removals on S until we no longer can.
+
+Return the final string after all such duplicate removals have been made.  It is guaranteed the answer is unique.
+
+ 
+
+**Example 1**:
+
+```
+
+Input: "abbaca"
+Output: "ca"
+Explanation: 
+For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move.  The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca".
+
+```
+
+**Note**:
+
+1. 1 <= S.length <= 20000
+2. S consists only of English lowercase letters.
+
+
+## 题目大意
+
+给出由小写字母组成的字符串 S，重复项删除操作会选择两个相邻且相同的字母，并删除它们。在 S 上反复执行重复项删除操作，直到无法继续删除。在完成所有重复项删除操作后返回最终的字符串。答案保证唯一。
+
+
+## 解题思路
+
+用栈模拟，类似“对对碰”，一旦新来的字符和栈顶的字符一样的话，就弹出栈顶字符，直至扫完整个字符串。栈中剩下的字符串就是最终要输出的结果。
+
+## 代码
+
+```go
+
+package leetcode
+
+func removeDuplicates1047(S string) string {
+	stack := []rune{}
+	for _, s := range S {
+		if len(stack) == 0 || len(stack) > 0 && stack[len(stack)-1] != s {
+			stack = append(stack, s)
+		} else {
+			stack = stack[:len(stack)-1]
+		}
+	}
+	return string(stack)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1040.Moving-Stones-Until-Consecutive-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1048.Longest-String-Chain/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1048.Longest-String-Chain.md b/website/content/ChapterFour/1000~1099/1048.Longest-String-Chain.md
new file mode 100644
index 000000000..703bf3794
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1048.Longest-String-Chain.md
@@ -0,0 +1,103 @@
+# [1048. Longest String Chain](https://leetcode.com/problems/longest-string-chain/)
+
+
+## 题目
+
+Given a list of words, each word consists of English lowercase letters.
+
+Let's say `word1` is a predecessor of `word2` if and only if we can add exactly one letter anywhere in `word1` to make it equal to `word2`. For example, `"abc"` is a predecessor of `"abac"`.
+
+A *word chain* is a sequence of words `[word_1, word_2, ..., word_k]` with `k >= 1`, where `word_1` is a predecessor of `word_2`, `word_2` is a predecessor of `word_3`, and so on.
+
+Return the longest possible length of a word chain with words chosen from the given list of `words`.
+
+**Example 1:**
+
+```
+Input: words = ["a","b","ba","bca","bda","bdca"]
+Output: 4
+Explanation: One of the longest word chain is "a","ba","bda","bdca".
+```
+
+**Example 2:**
+
+```
+Input: words = ["xbc","pcxbcf","xb","cxbc","pcxbc"]
+Output: 5
+```
+
+**Constraints:**
+
+- `1 <= words.length <= 1000`
+- `1 <= words[i].length <= 16`
+- `words[i]` only consists of English lowercase letters.
+
+## 题目大意
+
+给出一个单词列表，其中每个单词都由小写英文字母组成。如果我们可以在 word1 的任何地方添加一个字母使其变成 word2，那么我们认为 word1 是 word2 的前身。例如，"abc" 是 "abac" 的前身。词链是单词 [word_1, word_2, ..., word_k] 组成的序列，k >= 1，其中 word_1 是 word_2 的前身，word_2 是 word_3 的前身，依此类推。从给定单词列表 words 中选择单词组成词链，返回词链的最长可能长度。
+
+## 解题思路
+
+- 从这题的数据规模上分析，可以猜出此题是 DFS 或者 DP 的题。简单暴力的方法是以每个字符串为链条的起点开始枚举之后的字符串，两两判断能否构成满足题意的前身字符串。这种做法包含很多重叠子问题，例如 a 和 b 能构成前身字符串，以 c 为起点的字符串链条可能用到 a 和 b，以 d 为起点的字符串链条也可能用到 a 和 b。顺其自然，考虑用 DP 的思路解题。
+- 先将 words 字符串数组排序，然后用 poss 数组记录下每种长度字符串的在排序数组中的起始下标。然后逆序往前递推。因为初始条件只能得到以最长字符串为起始的字符串链长度为 1 。每选择一个起始字符串，从它的长度 + 1 的每个字符串 j 开始比较，是否能为其前身字符串。如果能构成前身字符串，那么 dp[i] = max(dp[i], 1+dp[j])。最终递推到下标为 0 的字符串。最终输出整个递推过程中的最大长度即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func longestStrChain(words []string) int {
+	sort.Slice(words, func(i, j int) bool { return len(words[i]) < len(words[j]) })
+	poss, res := make([]int, 16+2), 0
+	for i, w := range words {
+		if poss[len(w)] == 0 {
+			poss[len(w)] = i
+		}
+	}
+	dp := make([]int, len(words))
+	for i := len(words) - 1; i >= 0; i-- {
+		dp[i] = 1
+		for j := poss[len(words[i])+1]; j < len(words) && len(words[j]) == len(words[i])+1; j++ {
+			if isPredecessor(words[j], words[i]) {
+				dp[i] = max(dp[i], 1+dp[j])
+			}
+		}
+		res = max(res, dp[i])
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func isPredecessor(long, short string) bool {
+	i, j := 0, 0
+	wasMismatch := false
+	for j < len(short) {
+		if long[i] != short[j] {
+			if wasMismatch {
+				return false
+			}
+			wasMismatch = true
+			i++
+			continue
+		}
+		i++
+		j++
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1047.Remove-All-Adjacent-Duplicates-In-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1049.Last-Stone-Weight-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1049.Last-Stone-Weight-II.md b/website/content/ChapterFour/1000~1099/1049.Last-Stone-Weight-II.md
new file mode 100755
index 000000000..4645224f1
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1049.Last-Stone-Weight-II.md
@@ -0,0 +1,86 @@
+# [1049. Last Stone Weight II](https://leetcode.com/problems/last-stone-weight-ii/)
+
+## 题目
+
+We have a collection of rocks, each rock has a positive integer weight.
+
+Each turn, we choose **any two rocks** and smash them together. Suppose the stones have weights `x` and `y` with `x <= y`. The result of this smash is:
+
+- If `x == y`, both stones are totally destroyed;
+- If `x != y`, the stone of weight `x` is totally destroyed, and the stone of weight `y`has new weight `y-x`.
+
+At the end, there is at most 1 stone left. Return the **smallest possible** weight of this stone (the weight is 0 if there are no stones left.)
+
+**Example 1**:
+
+    Input: [2,7,4,1,8,1]
+    Output: 1
+    Explanation: 
+    We can combine 2 and 4 to get 2 so the array converts to [2,7,1,8,1] then,
+    we can combine 7 and 8 to get 1 so the array converts to [2,1,1,1] then,
+    we can combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
+    we can combine 1 and 1 to get 0 so the array converts to [1] then that's the optimal value.
+
+**Note**:
+
+1. `1 <= stones.length <= 30`
+2. `1 <= stones[i] <= 100`
+
+
+
+## 题目大意
+
+有一堆石头，每块石头的重量都是正整数。每一回合，从中选出任意两块石头，然后将它们一起粉碎。假设石头的重量分别为 x 和 y，且 x <= y。那么粉碎的可能结果如下：
+
+如果 x == y，那么两块石头都会被完全粉碎；
+如果 x != y，那么重量为 x 的石头将会完全粉碎，而重量为 y 的石头新重量为 y-x。
+最后，最多只会剩下一块石头。返回此石头最小的可能重量。如果没有石头剩下，就返回 0。
+
+提示：
+
+1. 1 <= stones.length <= 30
+2. 1 <= stones[i] <= 1000
+
+
+## 解题思路
+
+
+- 给出一个数组，数组里面的元素代表的是石头的重量。现在要求两个石头对碰，如果重量相同，两个石头都消失，如果一个重一个轻，剩下的石头是两者的差值。问经过这样的多次碰撞以后，能剩下的石头的重量最轻是多少？
+- 由于两两石头要发生碰撞，所以可以将整个数组可以分为两部分，如果这两部分的石头重量总和相差不大，那么经过若干次碰撞以后，剩下的石头重量一定是最小的。现在就需要找到这样两堆总重量差不多的两堆石头。这个问题就可以转化为 01 背包问题。从数组中找到 `sum/2` 重量的石头集合，如果一半能尽量达到 `sum/2`，那么另外一半和 `sum/2` 的差是最小的，最好的情况就是两堆石头的重量都是 `sum/2`，那么两两石头对碰以后最后都能消失。01 背包的经典模板可以参考第 416 题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func lastStoneWeightII(stones []int) int {
+	sum := 0
+	for _, v := range stones {
+		sum += v
+	}
+	n, C, dp := len(stones), sum/2, make([]int, sum/2+1)
+	for i := 0; i <= C; i++ {
+		if stones[0] <= i {
+			dp[i] = stones[0]
+		} else {
+			dp[i] = 0
+		}
+	}
+	for i := 1; i < n; i++ {
+		for j := C; j >= stones[i]; j-- {
+			dp[j] = max(dp[j], dp[j-stones[i]]+stones[i])
+		}
+	}
+	return sum - 2*dp[C]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1048.Longest-String-Chain/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1051.Height-Checker/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1051.Height-Checker.md b/website/content/ChapterFour/1000~1099/1051.Height-Checker.md
new file mode 100644
index 000000000..cb5879d46
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1051.Height-Checker.md
@@ -0,0 +1,79 @@
+# [1051. Height Checker](https://leetcode.com/problems/height-checker/)
+
+
+## 题目
+
+Students are asked to stand in non-decreasing order of heights for an annual photo.
+
+Return the minimum number of students that must move in order for all students to be standing in non-decreasing order of height.
+
+Notice that when a group of students is selected they can reorder in any possible way between themselves and the non selected students remain on their seats.
+
+**Example 1**:
+
+```
+Input: heights = [1,1,4,2,1,3]
+Output: 3
+Explanation: 
+Current array : [1,1,4,2,1,3]
+Target array  : [1,1,1,2,3,4]
+On index 2 (0-based) we have 4 vs 1 so we have to move this student.
+On index 4 (0-based) we have 1 vs 3 so we have to move this student.
+On index 5 (0-based) we have 3 vs 4 so we have to move this student.
+```
+
+**Example 2**:
+
+```
+Input: heights = [5,1,2,3,4]
+Output: 5
+```
+
+**Example 3**:
+
+```
+Input: heights = [1,2,3,4,5]
+Output: 0
+```
+
+**Constraints**:
+
+- `1 <= heights.length <= 100`
+- `1 <= heights[i] <= 100`
+
+## 题目大意
+
+学校在拍年度纪念照时，一般要求学生按照 非递减 的高度顺序排列。请你返回能让所有学生以 非递减 高度排列的最小必要移动人数。注意，当一组学生被选中时，他们之间可以以任何可能的方式重新排序，而未被选中的学生应该保持不动。
+
+
+## 解题思路
+
+- 给定一个高度数组，要求输出把这个数组按照非递减高度排列所需移动的最少次数。
+- 简单题，最少次数意味着每次移动，一步到位，一步就移动到它所在的最终位置。那么用一个辅助排好序的数组，一一比对计数即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func heightChecker(heights []int) int {
+	result, checker := 0, []int{}
+	checker = append(checker, heights...)
+	sort.Ints(checker)
+	for i := 0; i < len(heights); i++ {
+		if heights[i] != checker[i] {
+			result++
+		}
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1049.Last-Stone-Weight-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1052.Grumpy-Bookstore-Owner/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1052.Grumpy-Bookstore-Owner.md b/website/content/ChapterFour/1000~1099/1052.Grumpy-Bookstore-Owner.md
new file mode 100755
index 000000000..4e48b5af8
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1052.Grumpy-Bookstore-Owner.md
@@ -0,0 +1,113 @@
+# [1052. Grumpy Bookstore Owner](https://leetcode.com/problems/grumpy-bookstore-owner/)
+
+
+## 题目
+
+Today, the bookstore owner has a store open for `customers.length`minutes. Every minute, some number of customers (`customers[i]`) enter the store, and all those customers leave after the end of that minute.
+
+On some minutes, the bookstore owner is grumpy. If the bookstore owner is grumpy on the i-th minute, `grumpy[i] = 1`, otherwise `grumpy[i] = 0`. When the bookstore owner is grumpy, the customers of that minute are not satisfied, otherwise they are satisfied.
+
+The bookstore owner knows a secret technique to keep themselves not grumpy for `X` minutes straight, but can only use it once.
+
+Return the maximum number of customers that can be satisfied throughout the day.
+
+**Example 1**:
+
+    Input: customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], X = 3
+    Output: 16
+    Explanation: The bookstore owner keeps themselves not grumpy for the last 3 minutes. 
+    The maximum number of customers that can be satisfied = 1 + 1 + 1 + 1 + 7 + 5 = 16.
+
+**Note**:
+
+- `1 <= X <= customers.length == grumpy.length <= 20000`
+- `0 <= customers[i] <= 1000`
+- `0 <= grumpy[i] <= 1`
+
+
+## 题目大意
+
+今天，书店老板有一家店打算试营业 customers.length 分钟。每分钟都有一些顾客（customers[i]）会进入书店，所有这些顾客都会在那一分钟结束后离开。在某些时候，书店老板会生气。 如果书店老板在第 i 分钟生气，那么 grumpy[i] = 1，否则 grumpy[i] = 0。 当书店老板生气时，那一分钟的顾客就会不满意，不生气则他们是满意的。书店老板知道一个秘密技巧，能抑制自己的情绪，可以让自己连续 X 分钟不生气，但却只能使用一次。请你返回这一天营业下来，最多有多少客户能够感到满意的数量。
+
+提示：
+
+1. 1 <= X <= customers.length == grumpy.length <= 20000
+2. 0 <= customers[i] <= 1000
+3. 0 <= grumpy[i] <= 1
+
+
+
+## 解题思路
+
+
+- 给出一个顾客入店时间表和书店老板发脾气的时间表。两个数组的时间是一一对应的，即相同下标对应的相同的时间。书店老板可以控制自己在 X 分钟内不发火，但是只能控制一次。问有多少顾客能在书店老板不发火的时候在书店里看书。抽象一下，给出一个价值数组和一个装着 0 和 1 的数组，当价值数组的下标对应另外一个数组相同下标的值是 0 的时候，那么这个价值可以累加，当对应是 1 的时候，就不能加上这个价值。现在可以让装着 0 和 1 的数组中连续 X 个数都变成 0，问最终价值最大是多少？
+- 这道题是典型的滑动窗口的题目。最暴力的解法是滑动窗口右边界，当与左边界的距离等于 X 的时候，计算此刻对应的数组的总价值。当整个宽度为 X 的窗口滑过整个数组以后，输出维护的最大值即可。这个方法耗时比较长。因为每次计算数组总价值的时候都要遍历整个数组。这里是可以优化的地方。
+- 每次计算数组总价值的时候，其实目的是为了找到宽度为 X 的窗口对应里面为 1 的数累加和最大，因为如果把这个窗口里面的 1 都变成 0 以后，那么对最终价值的影响也最大。所以用一个变量 `customer0` 专门记录脾气数组中为 0 的对应的价值，累加起来。因为不管怎么改变，为 0 的永远为 0，唯一变化的是 1 变成 0 。用 `customer1` 专门记录脾气数组中为 1 的对应的价值。在窗口滑动过程中找到 `customer1` 的最大值。最终要求的最大值就是 `customer0 + maxCustomer1`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 滑动窗口优化版
+func maxSatisfied(customers []int, grumpy []int, X int) int {
+	customer0, customer1, maxCustomer1, left, right := 0, 0, 0, 0, 0
+	for ; right < len(customers); right++ {
+		if grumpy[right] == 0 {
+			customer0 += customers[right]
+		} else {
+			customer1 += customers[right]
+			for right-left+1 > X {
+				if grumpy[left] == 1 {
+					customer1 -= customers[left]
+				}
+				left++
+			}
+			if customer1 > maxCustomer1 {
+				maxCustomer1 = customer1
+			}
+		}
+	}
+	return maxCustomer1 + customer0
+}
+
+// 解法二 滑动窗口暴力版
+func maxSatisfied1(customers []int, grumpy []int, X int) int {
+	left, right, res := 0, -1, 0
+	for left < len(customers) {
+		if right+1 < len(customers) && right-left < X-1 {
+			right++
+		} else {
+			if right-left+1 == X {
+				res = max(res, sumSatisfied(customers, grumpy, left, right))
+			}
+			left++
+		}
+	}
+	return res
+}
+
+func sumSatisfied(customers []int, grumpy []int, start, end int) int {
+	sum := 0
+	for i := 0; i < len(customers); i++ {
+		if i < start || i > end {
+			if grumpy[i] == 0 {
+				sum += customers[i]
+			}
+		} else {
+			sum += customers[i]
+		}
+	}
+	return sum
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1051.Height-Checker/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1054.Distant-Barcodes/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1054.Distant-Barcodes.md b/website/content/ChapterFour/1000~1099/1054.Distant-Barcodes.md
new file mode 100755
index 000000000..ea71fd103
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1054.Distant-Barcodes.md
@@ -0,0 +1,99 @@
+# [1054. Distant Barcodes](https://leetcode.com/problems/distant-barcodes/)
+
+
+## 题目
+
+In a warehouse, there is a row of barcodes, where the `i`-th barcode is `barcodes[i]`.
+
+Rearrange the barcodes so that no two adjacent barcodes are equal. You may return any answer, and it is guaranteed an answer exists.
+
+**Example 1**:
+
+    Input: [1,1,1,2,2,2]
+    Output: [2,1,2,1,2,1]
+
+**Example 2**:
+
+    Input: [1,1,1,1,2,2,3,3]
+    Output: [1,3,1,3,2,1,2,1]
+
+**Note**:
+
+1. `1 <= barcodes.length <= 10000`
+2. `1 <= barcodes[i] <= 10000`
+
+
+## 题目大意
+
+在一个仓库里，有一排条形码，其中第 i 个条形码为 barcodes[i]。请你重新排列这些条形码，使其中两个相邻的条形码 不能 相等。 你可以返回任何满足该要求的答案，此题保证存在答案。
+
+
+
+## 解题思路
+
+
+- 这一题和第 767 题原理是完全一样的。第 767 题是 Google 的面试题。
+- 解题思路比较简单，先按照每个数字的频次从高到低进行排序，注意会有频次相同的数字。排序以后，分别从第 0 号位和中间的位置开始往后取数，取完以后即为最终解。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+func rearrangeBarcodes(barcodes []int) []int {
+	bfs := barcodesFrequencySort(barcodes)
+	if len(bfs) == 0 {
+		return []int{}
+	}
+	res := []int{}
+	j := (len(bfs)-1)/2 + 1
+	for i := 0; i <= (len(bfs)-1)/2; i++ {
+		res = append(res, bfs[i])
+		if j < len(bfs) {
+			res = append(res, bfs[j])
+		}
+		j++
+	}
+	return res
+}
+
+func barcodesFrequencySort(s []int) []int {
+	if len(s) == 0 {
+		return []int{}
+	}
+	sMap := map[int]int{}   // 统计每个数字出现的频次
+	cMap := map[int][]int{} // 按照频次作为 key 排序
+	for _, b := range s {
+		sMap[b]++
+	}
+	for key, value := range sMap {
+		cMap[value] = append(cMap[value], key)
+	}
+	var keys []int
+	for k := range cMap {
+		keys = append(keys, k)
+	}
+	sort.Sort(sort.Reverse(sort.IntSlice(keys)))
+	res := make([]int, 0)
+	for _, k := range keys {
+		for i := 0; i < len(cMap[k]); i++ {
+			for j := 0; j < k; j++ {
+				res = append(res, cMap[k][i])
+			}
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1052.Grumpy-Bookstore-Owner/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1073.Adding-Two-Negabinary-Numbers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1073.Adding-Two-Negabinary-Numbers.md b/website/content/ChapterFour/1000~1099/1073.Adding-Two-Negabinary-Numbers.md
new file mode 100755
index 000000000..ef540b112
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1073.Adding-Two-Negabinary-Numbers.md
@@ -0,0 +1,153 @@
+# [1073. Adding Two Negabinary Numbers](https://leetcode.com/problems/adding-two-negabinary-numbers/)
+
+
+## 题目
+
+Given two numbers `arr1` and `arr2` in base **-2**, return the result of adding them together.
+
+Each number is given in *array format*: as an array of 0s and 1s, from most significant bit to least significant bit. For example, `arr = [1,1,0,1]`represents the number `(-2)^3 + (-2)^2 + (-2)^0 = -3`. A number `arr` in *array format* is also guaranteed to have no leading zeros: either `arr == [0]` or `arr[0] == 1`.
+
+Return the result of adding `arr1` and `arr2` in the same format: as an array of 0s and 1s with no leading zeros.
+
+**Example 1**:
+
+    Input: arr1 = [1,1,1,1,1], arr2 = [1,0,1]
+    Output: [1,0,0,0,0]
+    Explanation: arr1 represents 11, arr2 represents 5, the output represents 16.
+
+**Note**:
+
+1. `1 <= arr1.length <= 1000`
+2. `1 <= arr2.length <= 1000`
+3. `arr1` and `arr2` have no leading zeros
+4. `arr1[i]` is `0` or `1`
+5. `arr2[i]` is `0` or `1`
+
+
+## 题目大意
+
+给出基数为 -2 的两个数 arr1 和 arr2，返回两数相加的结果。数字以 数组形式 给出：数组由若干 0 和 1 组成，按最高有效位到最低有效位的顺序排列。例如，arr = [1,1,0,1] 表示数字 (-2)^3 + (-2)^2 + (-2)^0 = -3。数组形式 的数字也同样不含前导零：以 arr 为例，这意味着要么 arr == [0]，要么 arr[0] == 1。
+
+返回相同表示形式的 arr1 和 arr2 相加的结果。两数的表示形式为：不含前导零、由若干 0 和 1 组成的数组。
+
+提示：
+
+- 1 <= arr1.length <= 1000
+- 1 <= arr2.length <= 1000
+- arr1 和 arr2 都不含前导零
+- arr1[i] 为 0 或 1
+- arr2[i] 为 0 或 1
+
+
+
+## 解题思路
+
+- 给出两个 -2 进制的数，要求计算出这两个数的和，最终表示形式还是 -2 进制。
+- 这一题最先想到的思路是先把两个 -2 进制的数转成 10 进制以后做加法，然后把结果表示成 -2 进制。这个思路可行，但是在提交以后会发现数据溢出 int64 了。在第 257 / 267 组测试数据会出现 WA。测试数据见 test 文件。另外换成 big.Add 也不是很方便。所以考虑换一个思路。
+- 这道题实际上就是求两个 -2 进制数的加法，为什么还要先转到 10 进制再换回 -2 进制呢？为何不直接进行 -2 进制的加法。所以开始尝试直接进行加法运算。加法是从低位到高位依次累加，遇到进位要从低往高位进位。所以从两个数组的末尾往前扫，模拟低位相加的过程。关键的是进位问题。进位分 3 种情况，依次来讨论：
+
+1. 进位到高位 k ，高位 k 上的两个数数字分别是 0 和 0 。这种情况最终 k 位为 1 。
+```c
+        证明：由于进位是由 k - 1 位进过来的，所以 k - 1 位是 2 个 1 。现在 k 位是 2 个 0，
+        所以加起来的和是 2 * (-2)^(k - 1)。
+        当 k 为奇数的时候，2 * (-2)^(k - 1) = (-1)^(k - 1)* 2 * 2^(k - 1) = 2^k
+        当 k 为偶数的时候，2 * (-2)^(k - 1) = (-1)^(k - 1)* 2 * 2^(k - 1) = -2^k
+        综合起来就是 (-2)^k，所以最终 k 位上有一个 1
+```
+2. 进位到高位 k ，高位 k 上的两个数数字分别是 0 和 1 。这种情况最终 k 位为 0 。
+```c
+        证明：由于进位是由 k - 1 位进过来的，所以 k - 1 位是 2 个 1。现在 k 位是 1 个 0 和 1 个 1,
+        所以加起来的和是 (-2)^k + 2 * (-2)^(k - 1)。
+        当 k 为奇数的时候，(-2)^k + 2 * (-2)^(k - 1) = -2^k + 2^k = 0
+        当 k 为偶数的时候，(-2)^k + 2 * (-2)^(k - 1) = 2^k - 2^k = 0
+        综合起来就是 0，所以最终 k 位上有一个 0
+```
+3. 进位到高位 k ，高位 k 上的两个数数字分别是 1 和 1 。这种情况最终 k 位为 1 。
+```c
+        证明：由于进位是由 k - 1 位进过来的，所以 k - 1 位是 2 个 1 。现在 k 位是 2 个 1，
+        所以加起来的和是 2 * (-2)^k + 2 * (-2)^(k - 1)。
+        当 k 为奇数的时候，2 * (-2)^k + 2 * (-2)^(k - 1) = -2^(k + 1) + 2^k = 2^k*(1 - 2) = -2^k
+        当 k 为偶数的时候，2 * (-2)^k + 2 * (-2)^(k - 1) = 2^(k + 1) - 2^k = 2^k*(2 - 1) = 2^k
+        综合起来就是 (-2)^k，所以最终 k 位上有一个 1
+```
+
+- 所以综上所属，-2 进制的进位和 2 进制的进位原理是完全一致的，只不过 -2 进制的进位是 -1，而 2 进制的进位是 1 。由于进位可能在 -2 进制上出现前导 0 ，所以最终结果需要再去除前导 0 。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 模拟进位
+func addNegabinary(arr1 []int, arr2 []int) []int {
+	carry, ans := 0, []int{}
+	for i, j := len(arr1)-1, len(arr2)-1; i >= 0 || j >= 0 || carry != 0; {
+		if i >= 0 {
+			carry += arr1[i]
+			i--
+		}
+		if j >= 0 {
+			carry += arr2[j]
+			j--
+		}
+		ans = append([]int{carry & 1}, ans...)
+		carry = -(carry >> 1)
+	}
+	for idx, num := range ans { // 去掉前导 0
+		if num != 0 {
+			return ans[idx:]
+		}
+	}
+	return []int{0}
+}
+
+// 解法二 标准的模拟，但是这个方法不能 AC，因为测试数据超过了 64 位，普通数据类型无法存储
+func addNegabinary1(arr1 []int, arr2 []int) []int {
+	return intToNegabinary(negabinaryToInt(arr1) + negabinaryToInt(arr2))
+}
+
+func negabinaryToInt(arr []int) int {
+	if len(arr) == 0 {
+		return 0
+	}
+	res := 0
+	for i := 0; i < len(arr)-1; i++ {
+		if res == 0 {
+			res += (-2) * arr[i]
+		} else {
+			res = res * (-2)
+			res += (-2) * arr[i]
+		}
+	}
+	return res + 1*arr[len(arr)-1]
+}
+
+func intToNegabinary(num int) []int {
+	if num == 0 {
+		return []int{0}
+	}
+	res := []int{}
+
+	for num != 0 {
+		remainder := num % (-2)
+		num = num / (-2)
+		if remainder < 0 {
+			remainder += 2
+			num++
+		}
+		res = append([]int{remainder}, res...)
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1054.Distant-Barcodes/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1074.Number-of-Submatrices-That-Sum-to-Target/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1074.Number-of-Submatrices-That-Sum-to-Target.md b/website/content/ChapterFour/1000~1099/1074.Number-of-Submatrices-That-Sum-to-Target.md
new file mode 100755
index 000000000..82fa07ba8
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1074.Number-of-Submatrices-That-Sum-to-Target.md
@@ -0,0 +1,153 @@
+# [1074. Number of Submatrices That Sum to Target](https://leetcode.com/problems/number-of-submatrices-that-sum-to-target/)
+
+
+## 题目
+
+Given a `matrix`, and a `target`, return the number of non-empty submatrices that sum to target.
+
+A submatrix `x1, y1, x2, y2` is the set of all cells `matrix[y]` with `x1 <= x <= x2` and `y1 <= y <= y2`.
+
+Two submatrices `(x1, y1, x2, y2)` and `(x1', y1', x2', y2')` are different if they have some coordinate that is different: for example, if `x1 != x1'`.
+
+**Example 1**:
+
+![](https://assets.leetcode.com/uploads/2020/09/02/mate1.jpg)
+
+    Input: matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
+    Output: 4
+    Explanation: The four 1x1 submatrices that only contain 0.
+
+**Example 2**:
+
+    Input: matrix = [[1,-1],[-1,1]], target = 0
+    Output: 5
+    Explanation: The two 1x2 submatrices, plus the two 2x1 submatrices, plus the 2x2 submatrix.
+
+**Note**:
+
+1. `1 <= matrix.length <= 300`
+2. `1 <= matrix[0].length <= 300`
+3. `-1000 <= matrix[i] <= 1000`
+4. `-10^8 <= target <= 10^8`
+
+
+## 题目大意
+
+给出矩阵 matrix 和目标值 target，返回元素总和等于目标值的非空子矩阵的数量。
+
+子矩阵 x1, y1, x2, y2 是满足 x1 <= x <= x2 且 y1 <= y <= y2 的所有单元 matrix[x][y] 的集合。
+
+如果 (x1, y1, x2, y2) 和 (x1', y1', x2', y2') 两个子矩阵中部分坐标不同（如：x1 != x1'），那么这两个子矩阵也不同。
+
+
+提示：
+
+1. 1 <= matrix.length <= 300
+2. 1 <= matrix[0].length <= 300
+3. -1000 <= matrix[i] <= 1000
+4. -10^8 <= target <= 10^8
+
+
+
+
+## 解题思路
+
+- 给出一个矩阵，要求在这个矩阵中找出子矩阵的和等于 target 的矩阵个数。
+- 这一题读完题感觉是滑动窗口的二维版本。如果把它拍扁，在一维数组中，求连续的子数组和为 target，这样就很好做。如果这题不降维，纯暴力解是 O(n^6)。如何优化降低时间复杂度呢？
+- 联想到第 1 题 Two Sum 问题，可以把 2 个数求和的问题优化到 O(n)。这里也用类似的思想，用一个 map 来保存行方向上曾经出现过的累加和，相减就可以得到本行的和。这里可能读者会有疑惑，为什么不能每一行都单独保存呢？为什么一定要用累加和相减的方式来获取每一行的和呢？因为这一题要求子矩阵所有解，如果只单独保存每一行的和，只能求得小的子矩阵，子矩阵和子矩阵组成的大矩阵的情况会漏掉(当然再循环一遍，把子矩阵累加起来也可以，但是这样就多了一层循环了)，例如子矩阵是 1*4 的，但是 2 个这样的子矩阵摞在一起形成 2 * 4 也能满足条件，如果不用累加和的办法，只单独存每一行的和，最终还要有组合的步骤。经过这样的优化，可以从 O(n^6) 优化到 O(n^4)，能 AC 这道题，但是时间复杂度太高了。如何优化？
+- 首先，子矩阵需要上下左右 4 个边界，4 个变量控制循环就需要 O(n^4)，行和列的区间累加还需要 O(n^2)。行和列的区间累加可以通过 preSum 来解决。例如 `sum[i,j] = sum[j] - sum[i - 1]`，其中 sum[k] 中存的是从 0 到 K 的累加和： {{< katex display >}} 
+\sum_{0}^{k} matrix[i]
+{{< /katex >}}
+    那么一个区间内的累加和可以由这个区间的右边界减去区间左边界左边的那个累加和得到(由于是闭区间，所需要取左边界左边的和)。经过这样的处理，列方向的维度就被我们拍扁了。
+
+- 再来看看行方向的和，现在每一列的和都可以通过区间相减的方法得到。那么这道题就变成了第 1 题 Two Sum 的问题了。Two Sum 问题只需要 O(n) 的时间复杂度求解，这一题由于是二维的，所以两个列的边界还需要循环，所以最终优化下来的时间复杂度是 O(n^3)。计算 presum 可以直接用原数组，所以空间复杂度只有一个 O(n) 的字典。
+- 类似思路的题目有第 560 题，第 304 题。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func numSubmatrixSumTarget(matrix [][]int, target int) int {
+	m, n, res := len(matrix), len(matrix[0]), 0
+	for row := range matrix {
+		for col := 1; col < len(matrix[row]); col++ {
+			matrix[row][col] += matrix[row][col-1]
+		}
+	}
+	for i := 0; i < n; i++ {
+		for j := i; j < n; j++ {
+			counterMap, sum := make(map[int]int, m), 0
+			counterMap[0] = 1 // 题目保证一定有解，所以这里初始化是 1
+			for row := 0; row < m; row++ {
+				if i > 0 {
+					sum += matrix[row][j] - matrix[row][i-1]
+				} else {
+					sum += matrix[row][j]
+				}
+				res += counterMap[sum-target]
+				counterMap[sum]++
+			}
+		}
+	}
+	return res
+}
+
+// 暴力解法 O(n^4)
+func numSubmatrixSumTarget1(matrix [][]int, target int) int {
+	m, n, res, sum := len(matrix), len(matrix[0]), 0, 0
+	for i := 0; i < n; i++ {
+		for j := i; j < n; j++ {
+			counterMap := map[int]int{}
+			counterMap[0] = 1 // 题目保证一定有解，所以这里初始化是 1
+			sum = 0
+			for row := 0; row < m; row++ {
+				for k := i; k <= j; k++ {
+					sum += matrix[row][k]
+				}
+				res += counterMap[sum-target]
+				counterMap[sum]++
+			}
+		}
+	}
+	return res
+}
+
+// 暴力解法超时！ O(n^6)
+func numSubmatrixSumTarget2(matrix [][]int, target int) int {
+	res := 0
+	for startx := 0; startx < len(matrix); startx++ {
+		for starty := 0; starty < len(matrix[startx]); starty++ {
+			for endx := startx; endx < len(matrix); endx++ {
+				for endy := starty; endy < len(matrix[startx]); endy++ {
+					if sumSubmatrix(matrix, startx, starty, endx, endy) == target {
+						//fmt.Printf("startx = %v, starty = %v, endx = %v, endy = %v\n", startx, starty, endx, endy)
+						res++
+					}
+				}
+			}
+		}
+	}
+	return res
+}
+
+func sumSubmatrix(matrix [][]int, startx, starty, endx, endy int) int {
+	sum := 0
+	for i := startx; i <= endx; i++ {
+		for j := starty; j <= endy; j++ {
+			sum += matrix[i][j]
+		}
+	}
+	return sum
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1073.Adding-Two-Negabinary-Numbers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1078.Occurrences-After-Bigram/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1078.Occurrences-After-Bigram.md b/website/content/ChapterFour/1000~1099/1078.Occurrences-After-Bigram.md
new file mode 100755
index 000000000..877eeff77
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1078.Occurrences-After-Bigram.md
@@ -0,0 +1,71 @@
+# [1078. Occurrences After Bigram](https://leetcode.com/problems/occurrences-after-bigram/)
+
+
+## 题目
+
+Given words `first` and `second`, consider occurrences in some `text` of the form "`first second third`", where `second` comes immediately after `first`, and `third`comes immediately after `second`.
+
+For each such occurrence, add "`third`" to the answer, and return the answer.
+
+**Example 1**:
+
+    Input: text = "alice is a good girl she is a good student", first = "a", second = "good"
+    Output: ["girl","student"]
+
+**Example 2**:
+
+    Input: text = "we will we will rock you", first = "we", second = "will"
+    Output: ["we","rock"]
+
+**Note**:
+
+1. `1 <= text.length <= 1000`
+2. `text` consists of space separated words, where each word consists of lowercase English letters.
+3. `1 <= first.length, second.length <= 10`
+4. `first` and `second` consist of lowercase English letters.
+
+
+## 题目大意
+
+
+给出第一个词 first 和第二个词 second，考虑在某些文本 text 中可能以 "first second third" 形式出现的情况，其中 second 紧随 first 出现，third 紧随 second 出现。对于每种这样的情况，将第三个词 "third" 添加到答案中，并返回答案。
+
+
+
+
+## 解题思路
+
+
+- 简单题。给出一个 text，要求找出紧接在 first 和 second 后面的那个字符串，有多个就输出多个。解法很简单，先分解出 words 每个字符串，然后依次遍历进行字符串匹配。匹配到 first 和 second 以后，输出之后的那个字符串。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func findOcurrences(text string, first string, second string) []string {
+	var res []string
+	words := strings.Split(text, " ")
+	if len(words) < 3 {
+		return []string{}
+	}
+	for i := 2; i < len(words); i++ {
+		if words[i-2] == first && words[i-1] == second {
+			res = append(res, words[i])
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1074.Number-of-Submatrices-That-Sum-to-Target/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1079.Letter-Tile-Possibilities/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1079.Letter-Tile-Possibilities.md b/website/content/ChapterFour/1000~1099/1079.Letter-Tile-Possibilities.md
new file mode 100755
index 000000000..1227f74ca
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1079.Letter-Tile-Possibilities.md
@@ -0,0 +1,114 @@
+# [1079. Letter Tile Possibilities](https://leetcode.com/problems/letter-tile-possibilities/)
+
+
+## 题目
+
+You have a set of `tiles`, where each tile has one letter `tiles[i]` printed on it. Return the number of possible non-empty sequences of letters you can make.
+
+**Example 1**:
+
+    Input: "AAB"
+    Output: 8
+    Explanation: The possible sequences are "A", "B", "AA", "AB", "BA", "AAB", "ABA", "BAA".
+
+**Example 2**:
+
+    Input: "AAABBC"
+    Output: 188
+
+**Note**:
+
+1. `1 <= tiles.length <= 7`
+2. `tiles` consists of uppercase English letters.
+
+## 题目大意
+
+你有一套活字字模 tiles，其中每个字模上都刻有一个字母 tiles[i]。返回你可以印出的非空字母序列的数目。提示：  
+
+- 1 <= tiles.length <= 7
+- tiles 由大写英文字母组成
+
+## 解题思路
+
+- 题目要求输出所有非空字母序列的数目。这一题是排列和组合的结合题目。组合是可以选择一个字母，二个字母，…… n 个字母。每个组合内是排列问题。比如选择 2 个字母，字母之间相互排序不同是影响最终结果的，不同的排列顺序是不同的解。
+- 这道题目由于不需要输出所有解，所以解法可以优化，例如我们在递归计算解的时候，不需要真的遍历原字符串，只需要累加一些字母的频次就可以。当然如果要输出所有解，就需要真实遍历原字符串了(见解法二)。简单的做法是每次递归按照频次累加。因为每次增加一个字母一定是 26 个大写字母中的一个。这里需要注意的是，增加的只能是 26 个字母里面还能取出“机会”的字母，例如递归到到第 3 轮了，A 用完了，这个时候只能取频次还不为 0 的字母拼上去。
+
+![](https://img.halfrost.com/Leetcode/leetcode_1079_0.png)
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 DFS
+func numTilePossibilities(tiles string) int {
+	m := make(map[byte]int)
+	for i := range tiles {
+		m[tiles[i]]++
+	}
+	arr := make([]int, 0)
+	for _, v := range m {
+		arr = append(arr, v)
+	}
+	return numTileDFS(arr)
+}
+
+func numTileDFS(arr []int) (r int) {
+	for i := 0; i < len(arr); i++ {
+		if arr[i] == 0 {
+			continue
+		}
+		r++
+		arr[i]--
+		r += numTileDFS(arr)
+		arr[i]++
+	}
+	return
+}
+
+// 解法二 DFS 暴力解法
+func numTilePossibilities1(tiles string) int {
+	res, tmp, tMap, used := 0, []byte{}, make(map[string]string, 0), make([]bool, len(tiles))
+	findTile([]byte(tiles), tmp, &used, 0, &res, tMap)
+	return res
+}
+
+func findTile(tiles, tmp []byte, used *[]bool, index int, res *int, tMap map[string]string) {
+	flag := true
+	for _, v := range *used {
+		if v == false {
+			flag = false
+			break
+		}
+	}
+	if flag {
+		return
+	}
+	for i := 0; i < len(tiles); i++ {
+		if (*used)[i] == true {
+			continue
+		}
+		tmp = append(tmp, tiles[i])
+		(*used)[i] = true
+		if _, ok := tMap[string(tmp)]; !ok {
+			//fmt.Printf("i = %v tiles = %v 找到了结果 = %v\n", i, string(tiles), string(tmp))
+			*res++
+		}
+		tMap[string(tmp)] = string(tmp)
+
+		findTile([]byte(tiles), tmp, used, i+1, res, tMap)
+		tmp = tmp[:len(tmp)-1]
+		(*used)[i] = false
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1078.Occurrences-After-Bigram/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1089.Duplicate-Zeros/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1089.Duplicate-Zeros.md b/website/content/ChapterFour/1000~1099/1089.Duplicate-Zeros.md
new file mode 100644
index 000000000..1bc3b7757
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1089.Duplicate-Zeros.md
@@ -0,0 +1,65 @@
+# [1089. Duplicate Zeros](https://leetcode.com/problems/duplicate-zeros/)
+
+
+## 题目
+
+Given a fixed length array `arr` of integers, duplicate each occurrence of zero, shifting the remaining elements to the right.
+
+Note that elements beyond the length of the original array are not written.
+
+Do the above modifications to the input array **in place**, do not return anything from your function.
+
+**Example 1**:
+
+```
+Input: [1,0,2,3,0,4,5,0]
+Output: null
+Explanation: After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]
+```
+
+**Example 2**:
+
+```
+Input: [1,2,3]
+Output: null
+Explanation: After calling your function, the input array is modified to: [1,2,3]
+```
+
+**Note**:
+
+1. `1 <= arr.length <= 10000`
+2. `0 <= arr[i] <= 9`
+
+## 题目大意
+
+给你一个长度固定的整数数组 arr，请你将该数组中出现的每个零都复写一遍，并将其余的元素向右平移。注意：请不要在超过该数组长度的位置写入元素。要求：请对输入的数组 就地 进行上述修改，不要从函数返回任何东西。
+
+
+## 解题思路
+
+- 给一个固定长度的数组，把数组元素为 0 的元素都往后复制一遍，后面的元素往后移，超出数组长度的部分删除。
+- 简单题，按照题意，用 append 和 slice 操作即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func duplicateZeros(arr []int) {
+	for i := 0; i < len(arr); i++ {
+		if arr[i] == 0 && i+1 < len(arr) {
+			arr = append(arr[:i+1], arr[i:len(arr)-1]...)
+			i++
+		}
+	}
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1079.Letter-Tile-Possibilities/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1091.Shortest-Path-in-Binary-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1091.Shortest-Path-in-Binary-Matrix.md b/website/content/ChapterFour/1000~1099/1091.Shortest-Path-in-Binary-Matrix.md
new file mode 100644
index 000000000..849bb15e4
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1091.Shortest-Path-in-Binary-Matrix.md
@@ -0,0 +1,121 @@
+# [1091. Shortest Path in Binary Matrix](https://leetcode.com/problems/shortest-path-in-binary-matrix/)
+
+
+## 题目
+
+In an N by N square grid, each cell is either empty (0) or blocked (1).
+
+A *clear path from top-left to bottom-right* has length `k` if and only if it is composed of cells `C_1, C_2, ..., C_k` such that:
+
+- Adjacent cells `C_i` and `C_{i+1}` are connected 8-directionally (ie., they are different and share an edge or corner)
+- `C_1` is at location `(0, 0)` (ie. has value `grid[0][0]`)
+- `C_k` is at location `(N-1, N-1)` (ie. has value `grid[N-1][N-1]`)
+- If `C_i` is located at `(r, c)`, then `grid[r][c]` is empty (ie. `grid[r][c] == 0`).
+
+Return the length of the shortest such clear path from top-left to bottom-right.  If such a path does not exist, return -1.
+
+**Example 1:**
+
+```
+Input: [[0,1],[1,0]]
+Output: 2
+```
+
+![https://assets.leetcode.com/uploads/2019/08/04/example1_1.png](https://assets.leetcode.com/uploads/2019/08/04/example1_1.png)
+
+![https://assets.leetcode.com/uploads/2019/08/04/example1_2.png](https://assets.leetcode.com/uploads/2019/08/04/example1_2.png)
+
+**Example 2:**
+
+```
+Input: [[0,0,0],[1,1,0],[1,1,0]]
+Output: 4
+```
+
+![https://assets.leetcode.com/uploads/2019/08/04/example2_1.png](https://assets.leetcode.com/uploads/2019/08/04/example2_1.png)
+
+![https://assets.leetcode.com/uploads/2019/08/04/example2_2.png](https://assets.leetcode.com/uploads/2019/08/04/example2_2.png)
+
+**Note:**
+
+1. `1 <= grid.length == grid[0].length <= 100`
+2. `grid[r][c]` is `0` or `1`
+
+## 题目大意
+
+在一个 N × N 的方形网格中，每个单元格有两种状态：空（0）或者阻塞（1）。一条从左上角到右下角、长度为 k 的畅通路径，由满足下述条件的单元格 C_1, C_2, ..., C_k 组成：
+
+- 相邻单元格 C_i 和 C_{i+1} 在八个方向之一上连通（此时，C_i 和 C_{i+1} 不同且共享边或角）
+- C_1 位于 (0, 0)（即，值为 grid[0][0]）
+- C_k 位于 (N-1, N-1)（即，值为 grid[N-1][N-1]）
+- 如果 C_i 位于 (r, c)，则 grid[r][c] 为空（即，grid[r][c] == 0）
+
+返回这条从左上角到右下角的最短畅通路径的长度。如果不存在这样的路径，返回 -1 。
+
+## 解题思路
+
+- 这一题是简单的找最短路径。利用 BFS 从左上角逐步扩展到右下角，便可以很容易求解。注意每轮扩展需要考虑 8 个方向。
+
+## 代码
+
+```go
+var dir = [][]int{
+	{-1, -1},
+	{-1, 0},
+	{-1, 1},
+	{0, 1},
+	{0, -1},
+	{1, -1},
+	{1, 0},
+	{1, 1},
+}
+
+func shortestPathBinaryMatrix(grid [][]int) int {
+	visited := make([][]bool, 0)
+	for range make([]int, len(grid)) {
+		visited = append(visited, make([]bool, len(grid[0])))
+	}
+	dis := make([][]int, 0)
+	for range make([]int, len(grid)) {
+		dis = append(dis, make([]int, len(grid[0])))
+	}
+	if grid[0][0] == 1 {
+		return -1
+	}
+	if len(grid) == 1 && len(grid[0]) == 1 {
+		return 1
+	}
+
+	queue := []int{0}
+	visited[0][0], dis[0][0] = true, 1
+	for len(queue) > 0 {
+		cur := queue[0]
+		queue = queue[1:]
+		curx, cury := cur/len(grid[0]), cur%len(grid[0])
+		for d := 0; d < 8; d++ {
+			nextx := curx + dir[d][0]
+			nexty := cury + dir[d][1]
+			if isInBoard(grid, nextx, nexty) && !visited[nextx][nexty] && grid[nextx][nexty] == 0 {
+				queue = append(queue, nextx*len(grid[0])+nexty)
+				visited[nextx][nexty] = true
+				dis[nextx][nexty] = dis[curx][cury] + 1
+				if nextx == len(grid)-1 && nexty == len(grid[0])-1 {
+					return dis[nextx][nexty]
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func isInBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1089.Duplicate-Zeros/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1093.Statistics-from-a-Large-Sample/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/1093.Statistics-from-a-Large-Sample.md b/website/content/ChapterFour/1000~1099/1093.Statistics-from-a-Large-Sample.md
new file mode 100755
index 000000000..39779662f
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1093.Statistics-from-a-Large-Sample.md
@@ -0,0 +1,105 @@
+# [1093. Statistics from a Large Sample](https://leetcode.com/problems/statistics-from-a-large-sample/)
+
+
+## 题目
+
+We sampled integers between `0` and `255`, and stored the results in an array `count`: `count[k]` is the number of integers we sampled equal to `k`.
+
+Return the minimum, maximum, mean, median, and mode of the sample respectively, as an array of **floating point numbers**. The mode is guaranteed to be unique.
+
+*(Recall that the median of a sample is:*
+
+- *The middle element, if the elements of the sample were sorted and the number of elements is odd;*
+- *The average of the middle two elements, if the elements of the sample were sorted and the number of elements is even.)*
+
+**Example 1**:
+
+    Input: count = [0,1,3,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
+    Output: [1.00000,3.00000,2.37500,2.50000,3.00000]
+
+**Example 2**:
+
+    Input: count = [0,4,3,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
+    Output: [1.00000,4.00000,2.18182,2.00000,1.00000]
+
+**Constraints**:
+
+1. `count.length == 256`
+2. `1 <= sum(count) <= 10^9`
+3. The mode of the sample that count represents is unique.
+4. Answers within `10^-5` of the true value will be accepted as correct.
+
+
+## 题目大意
+
+我们对 0 到 255 之间的整数进行采样，并将结果存储在数组 count 中：count[k] 就是整数 k 的采样个数。
+
+我们以 浮点数 数组的形式，分别返回样本的最小值、最大值、平均值、中位数和众数。其中，众数是保证唯一的。我们先来回顾一下中位数的知识：
+
+- 如果样本中的元素有序，并且元素数量为奇数时，中位数为最中间的那个元素；
+- 如果样本中的元素有序，并且元素数量为偶数时，中位数为中间的两个元素的平均值。
+
+
+
+## 解题思路
+
+
+- 这个问题的关键需要理解题目的意思，什么是采样？`count[k]` 就是整数 `k` 的采样个数。
+- 题目要求返回样本的最小值、最大值、平均值、中位数和众数。最大值和最小值就很好处理，只需要遍历 count 判断最小的非 0 的 index 就是最小值，最大的非 0 的 index 就是最大值。平均值也非常好处理，对于所有非 0 的 count，我们通过累加 count[k] * index 得到所有数的和，然后除上所有非 0 的 count 的和。
+  	{{< katex display >}} 
+    \sum_{n=0}^{256}count[n],count[n]!=0
+	{{< /katex>}}
+
+- 众数也非常容易，只需统计 count 值最大时的 index 即可。
+- 中位数的处理相对麻烦一些，因为需要分非 0 的 count 之和是奇数还是偶数两种情况。先假设非 0 的 count 和为 cnt，那么如果 cnt 是奇数的话，只需要找到 cnt/2 的位置即可，通过不断累加 count 的值，直到累加和超过 ≥ cnt/2。如果 cnt 是偶数的话，需要找到 cnt/2 + 1 和 cnt/2 的位置，找法和奇数情况相同，不过需要找两次(可以放到一个循环中做两次判断)。
+
+## 代码
+
+```go
+
+package leetcode
+
+func sampleStats(count []int) []float64 {
+	res := make([]float64, 5)
+	res[0] = 255
+	sum := 0
+	for _, val := range count {
+		sum += val
+	}
+	left, right := sum/2, sum/2
+	if (sum % 2) == 0 {
+		right++
+	}
+	pre, mode := 0, 0
+	for i, val := range count {
+		if val > 0 {
+			if i < int(res[0]) {
+				res[0] = float64(i)
+			}
+			res[1] = float64(i)
+		}
+		res[2] += float64(i*val) / float64(sum)
+		if pre < left && pre+val >= left {
+			res[3] += float64(i) / 2.0
+		}
+		if pre < right && pre+val >= right {
+			res[3] += float64(i) / 2.0
+		}
+		pre += val
+
+		if val > mode {
+			mode = val
+			res[4] = float64(i)
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1091.Shortest-Path-in-Binary-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1104.Path-In-Zigzag-Labelled-Binary-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1000~1099/_index.md b/website/content/ChapterFour/1000~1099/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/1100~1199/1104.Path-In-Zigzag-Labelled-Binary-Tree.md b/website/content/ChapterFour/1100~1199/1104.Path-In-Zigzag-Labelled-Binary-Tree.md
new file mode 100644
index 000000000..05076ee8d
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1104.Path-In-Zigzag-Labelled-Binary-Tree.md
@@ -0,0 +1,99 @@
+# [1104. Path In Zigzag Labelled Binary Tree](https://leetcode.com/problems/path-in-zigzag-labelled-binary-tree/)
+
+
+## 题目
+
+In an infinite binary tree where every node has two children, the nodes are labelled in row order.
+
+In the odd numbered rows (ie., the first, third, fifth,...), the labelling is left to right, while in the even numbered rows (second, fourth, sixth,...), the labelling is right to left.
+
+![https://assets.leetcode.com/uploads/2019/06/24/tree.png](https://assets.leetcode.com/uploads/2019/06/24/tree.png)
+
+Given the `label` of a node in this tree, return the labels in the path from the root of the tree to the node with that `label`.
+
+**Example 1:**
+
+```
+Input: label = 14
+Output: [1,3,4,14]
+
+```
+
+**Example 2:**
+
+```
+Input: label = 26
+Output: [1,2,6,10,26]
+
+```
+
+**Constraints:**
+
+- `1 <= label <= 10^6`
+
+## 题目大意
+
+在一棵无限的二叉树上，每个节点都有两个子节点，树中的节点 逐行 依次按 “之” 字形进行标记。如下图所示，在奇数行（即，第一行、第三行、第五行……）中，按从左到右的顺序进行标记；而偶数行（即，第二行、第四行、第六行……）中，按从右到左的顺序进行标记。
+
+给你树上某一个节点的标号 label，请你返回从根节点到该标号为 label 节点的路径，该路径是由途经的节点标号所组成的。
+
+## 解题思路
+
+- 计算出 label 所在的 level 和 index。
+- 根据 index 和 level 计算出父节点的 index 和 value。
+- level 减一，循环计算出对应的父节点直到根节点。
+
+## 代码
+
+```go
+package leetcode
+
+func pathInZigZagTree(label int) []int {
+	level := getLevel(label)
+	ans := []int{label}
+	curIndex := label - (1 << level)
+	parent := 0
+	for level >= 1 {
+		parent, curIndex = getParent(curIndex, level)
+		ans = append(ans, parent)
+		level--
+	}
+	ans = reverse(ans)
+	return ans
+}
+
+func getLevel(label int) int {
+	level := 0
+	nums := 0
+	for {
+		nums += 1 << level
+		if nums >= label {
+			return level
+		}
+		level++
+	}
+}
+
+func getParent(index int, level int) (parent int, parentIndex int) {
+	parentIndex = 1<<(level-1) - 1 + (index/2)*(-1)
+	parent = 1<<(level-1) + parentIndex
+	return
+}
+
+func reverse(nums []int) []int {
+	left, right := 0, len(nums)-1
+	for left < right {
+		nums[left], nums[right] = nums[right], nums[left]
+		left++
+		right--
+	}
+	return nums
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1000~1099/1093.Statistics-from-a-Large-Sample/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1105.Filling-Bookcase-Shelves/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1105.Filling-Bookcase-Shelves.md b/website/content/ChapterFour/1100~1199/1105.Filling-Bookcase-Shelves.md
new file mode 100755
index 000000000..8a914c8d0
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1105.Filling-Bookcase-Shelves.md
@@ -0,0 +1,75 @@
+# [1105. Filling Bookcase Shelves](https://leetcode.com/problems/filling-bookcase-shelves/)
+
+
+## 题目
+
+We have a sequence of `books`: the `i`-th book has thickness `books[i][0]`and height `books[i][1]`.
+
+We want to place these books **in order** onto bookcase shelves that have total width `shelf_width`.
+
+We choose some of the books to place on this shelf (such that the sum of their thickness is `<= shelf_width`), then build another level of shelf of the bookcase so that the total height of the bookcase has increased by the maximum height of the books we just put down. We repeat this process until there are no more books to place.
+
+Note again that at each step of the above process, the order of the books we place is the same order as the given sequence of books. For example, if we have an ordered list of 5 books, we might place the first and second book onto the first shelf, the third book on the second shelf, and the fourth and fifth book on the last shelf.
+
+Return the minimum possible height that the total bookshelf can be after placing shelves in this manner.
+
+**Example 1**:
+
+![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/06/28/shelves.png)
+
+    Input: books = [[1,1],[2,3],[2,3],[1,1],[1,1],[1,1],[1,2]], shelf_width = 4
+    Output: 6
+    Explanation:
+    The sum of the heights of the 3 shelves are 1 + 3 + 2 = 6.
+    Notice that book number 2 does not have to be on the first shelf.
+
+**Constraints**:
+
+- `1 <= books.length <= 1000`
+- `1 <= books[i][0] <= shelf_width <= 1000`
+- `1 <= books[i][1] <= 1000`
+
+## 题目大意
+
+附近的家居城促销，你买回了一直心仪的可调节书架，打算把自己的书都整理到新的书架上。你把要摆放的书 books 都整理好，叠成一摞：从上往下，第 i 本书的厚度为 books[i][0]，高度为 books[i][1]。按顺序将这些书摆放到总宽度为 shelf\_width 的书架上。
+
+先选几本书放在书架上（它们的厚度之和小于等于书架的宽度 shelf_width），然后再建一层书架。重复这个过程，直到把所有的书都放在书架上。
+
+需要注意的是，在上述过程的每个步骤中，摆放书的顺序与你整理好的顺序相同。 例如，如果这里有 5 本书，那么可能的一种摆放情况是：第一和第二本书放在第一层书架上，第三本书放在第二层书架上，第四和第五本书放在最后一层书架上。每一层所摆放的书的最大高度就是这一层书架的层高，书架整体的高度为各层高之和。以这种方式布置书架，返回书架整体可能的最小高度。
+
+
+
+## 解题思路
+
+- 给出一个数组，数组里面每个元素代表的是一个本书的宽度和高度。要求按照书籍的顺序，把书摆到宽度为 `shelf_width` 的书架上。问最终放下所有书籍以后，书架的最小高度。
+- 这一题的解题思路是动态规划。`dp[i]` 代表放置前 `i` 本书所需要的书架最小高度。初始值 dp[0] = 0，其他为最大值 1000*1000。遍历每一本书，把当前这本书作为书架最后一层的最后一本书，将这本书之前的书向后调整，看看是否可以减少之前的书架高度。状态转移方程为 `dp[i] = min(dp[i] , dp[j - 1] + h)`，其中 `j` 表示最后一层所能容下书籍的索引，`h` 表示最后一层最大高度。`j` 调整完一遍以后就能找出书架最小高度值了。时间复杂度 O(n^2)。
+
+## 代码
+
+```go
+
+package leetcode
+
+func minHeightShelves(books [][]int, shelfWidth int) int {
+	dp := make([]int, len(books)+1)
+	dp[0] = 0
+	for i := 1; i <= len(books); i++ {
+		width, height := books[i-1][0], books[i-1][1]
+		dp[i] = dp[i-1] + height
+		for j := i - 1; j > 0 && width+books[j-1][0] <= shelfWidth; j-- {
+			height = max(height, books[j-1][1])
+			width += books[j-1][0]
+			dp[i] = min(dp[i], dp[j-1]+height)
+		}
+	}
+	return dp[len(books)]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1104.Path-In-Zigzag-Labelled-Binary-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1108.Defanging-an-IP-Address/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1108.Defanging-an-IP-Address.md b/website/content/ChapterFour/1100~1199/1108.Defanging-an-IP-Address.md
new file mode 100755
index 000000000..90724672f
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1108.Defanging-an-IP-Address.md
@@ -0,0 +1,61 @@
+# [1108. Defanging an IP Address](https://leetcode.com/problems/defanging-an-ip-address/)
+
+
+## 题目
+
+Given a valid (IPv4) IP `address`, return a defanged version of that IP address.
+
+A *defanged IP address* replaces every period `"."` with `"[.]"`.
+
+**Example 1**:
+
+    Input: address = "1.1.1.1"
+    Output: "1[.]1[.]1[.]1"
+
+**Example 2**:
+
+    Input: address = "255.100.50.0"
+    Output: "255[.]100[.]50[.]0"
+
+**Constraints**:
+
+- The given `address` is a valid IPv4 address.
+
+## 题目大意
+
+
+给你一个有效的 IPv4 地址 address，返回这个 IP 地址的无效化版本。所谓无效化 IP 地址，其实就是用 "[.]" 代替了每个 "."。
+
+
+提示：
+
+- 给出的 address 是一个有效的 IPv4 地址
+
+
+
+## 解题思路
+
+- 给出一个 IP 地址，要求把点替换成 `[.]`。
+- 简单题，按照题意替换即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func defangIPaddr(address string) string {
+	return strings.Replace(address, ".", "[.]", -1)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1105.Filling-Bookcase-Shelves/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1110.Delete-Nodes-And-Return-Forest/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1110.Delete-Nodes-And-Return-Forest.md b/website/content/ChapterFour/1100~1199/1110.Delete-Nodes-And-Return-Forest.md
new file mode 100644
index 000000000..7b317cc4f
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1110.Delete-Nodes-And-Return-Forest.md
@@ -0,0 +1,89 @@
+# [1110. Delete Nodes And Return Forest](https://leetcode.com/problems/delete-nodes-and-return-forest/)
+
+
+## 题目
+
+Given the `root` of a binary tree, each node in the tree has a distinct value.
+
+After deleting all nodes with a value in `to_delete`, we are left with a forest (a disjoint union of trees).
+
+Return the roots of the trees in the remaining forest. You may return the result in any order.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/07/01/screen-shot-2019-07-01-at-53836-pm.png](https://assets.leetcode.com/uploads/2019/07/01/screen-shot-2019-07-01-at-53836-pm.png)
+
+```
+Input: root = [1,2,3,4,5,6,7], to_delete = [3,5]
+Output: [[1,2,null,4],[6],[7]]
+```
+
+**Constraints**:
+
+- The number of nodes in the given tree is at most `1000`.
+- Each node has a distinct value between `1` and `1000`.
+- `to_delete.length <= 1000`
+- `to_delete` contains distinct values between `1` and `1000`.
+
+
+## 题目大意
+
+给出二叉树的根节点 root，树上每个节点都有一个不同的值。如果节点值在 to_delete 中出现，我们就把该节点从树上删去，最后得到一个森林（一些不相交的树构成的集合）。返回森林中的每棵树。你可以按任意顺序组织答案。
+
+
+提示：
+
+- 树中的节点数最大为 1000。
+- 每个节点都有一个介于 1 到 1000 之间的值，且各不相同。
+- to_delete.length <= 1000
+- to_delete 包含一些从 1 到 1000、各不相同的值。
+
+
+
+## 解题思路
+
+- 给出一棵树，再给出一个数组，要求删除数组中相同元素值的节点。输出最终删除以后的森林。
+- 简单题。边遍历树，边删除数组中的元素。这里可以先把数组里面的元素放入 map 中，加速查找。遇到相同的元素就删除节点。这里需要特殊判断的是当前删除的节点是否是根节点，如果是根节点需要根据条件置空它的左节点或者右节点。
+
+## 代码
+
+```go
+func delNodes(root *TreeNode, toDelete []int) []*TreeNode {
+	if root == nil {
+		return nil
+	}
+	res, deleteMap := []*TreeNode{}, map[int]bool{}
+	for _, v := range toDelete {
+		deleteMap[v] = true
+	}
+	dfsDelNodes(root, deleteMap, true, &res)
+	return res
+}
+
+func dfsDelNodes(root *TreeNode, toDel map[int]bool, isRoot bool, res *[]*TreeNode) bool {
+	if root == nil {
+		return false
+	}
+	if isRoot && !toDel[root.Val] {
+		*res = append(*res, root)
+	}
+	isRoot = false
+	if toDel[root.Val] {
+		isRoot = true
+	}
+	if dfsDelNodes(root.Left, toDel, isRoot, res) {
+		root.Left = nil
+	}
+	if dfsDelNodes(root.Right, toDel, isRoot, res) {
+		root.Right = nil
+	}
+	return isRoot
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1108.Defanging-an-IP-Address/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings.md b/website/content/ChapterFour/1100~1199/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings.md
new file mode 100755
index 000000000..4289e982c
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings.md
@@ -0,0 +1,139 @@
+# [1111. Maximum Nesting Depth of Two Valid Parentheses Strings](https://leetcode.com/problems/maximum-nesting-depth-of-two-valid-parentheses-strings/)
+
+
+## 题目
+
+A string is a *valid parentheses string* (denoted VPS) if and only if it consists of `"("` and `")"` characters only, and:
+
+- It is the empty string, or
+- It can be written as `AB` (`A` concatenated with `B`), where `A` and `B` are VPS's, or
+- It can be written as `(A)`, where `A` is a VPS.
+
+We can similarly define the *nesting depth* `depth(S)` of any VPS `S` as follows:
+
+- `depth("") = 0`
+- `depth(A + B) = max(depth(A), depth(B))`, where `A` and `B` are VPS's
+- `depth("(" + A + ")") = 1 + depth(A)`, where `A` is a VPS.
+
+For example, `""`, `"()()"`, and `"()(()())"` are VPS's (with nesting depths 0, 1, and 2), and `")("` and `"(()"` are not VPS's.
+
+Given a VPS seq, split it into two disjoint subsequences `A` and `B`, such that `A` and `B` are VPS's (and `A.length + B.length = seq.length`).
+
+Now choose **any** such `A` and `B` such that `max(depth(A), depth(B))` is the minimum possible value.
+
+Return an `answer` array (of length `seq.length`) that encodes such a choice of `A` and `B`: `answer[i] = 0` if `seq[i]` is part of `A`, else `answer[i] = 1`. Note that even though multiple answers may exist, you may return any of them.
+
+**Example 1**:
+
+    Input: seq = "(()())"
+    Output: [0,1,1,1,1,0]
+
+**Example 2**:
+
+    Input: seq = "()(())()"
+    Output: [0,0,0,1,1,0,1,1]
+
+**Constraints**:
+
+- `1 <= seq.size <= 10000`
+
+
+## 题目大意
+
+
+有效括号字符串 仅由 "(" 和 ")" 构成，并符合下述几个条件之一：
+
+- 空字符串
+- 连接，可以记作 AB（A 与 B 连接），其中 A 和 B 都是有效括号字符串
+- 嵌套，可以记作 (A)，其中 A 是有效括号字符串
+
+类似地，我们可以定义任意有效括号字符串 s 的 嵌套深度 depth(S)：
+
+- s 为空时，depth("") = 0
+- s 为 A 与 B 连接时，depth(A + B) = max(depth(A), depth(B))，其中 A 和 B 都是有效括号字符串
+- s 为嵌套情况，depth("(" + A + ")") = 1 + depth(A)，其中 A 是有效括号字符串
+
+
+例如：""，"()()"，和 "()(()())" 都是有效括号字符串，嵌套深度分别为 0，1，2，而 ")(" 和 "(()" 都不是有效括号字符串。
+
+ 
+
+给你一个有效括号字符串 seq，将其分成两个不相交的子序列 A 和 B，且 A 和 B 满足有效括号字符串的定义（注意：A.length + B.length = seq.length）。
+
+现在，你需要从中选出 任意 一组有效括号字符串 A 和 B，使 max(depth(A), depth(B)) 的可能取值最小。
+
+返回长度为 seq.length 答案数组 answer ，选择 A 还是 B 的编码规则是：如果 seq[i] 是 A 的一部分，那么 answer[i] = 0。否则，answer[i] = 1。即便有多个满足要求的答案存在，你也只需返回 一个。
+
+
+
+## 解题思路
+
+- 给出一个括号字符串。选出 A 部分和 B 部分，使得 `max(depth(A), depth(B))` 值最小。在最终的数组中输出 0 和 1，0 标识是 A 部分，1 标识是 B 部分。
+- 这一题想要 `max(depth(A), depth(B))` 值最小，可以使用贪心思想。如果 A 部分和 B 部分都尽快括号匹配，不深层次嵌套，那么总的层次就会变小。只要让嵌套的括号中属于 A 的和属于 B 的间隔排列即可。例如：“`(((())))`”，上面的字符串的嵌套深度是 4，按照上述的贪心思想，则标记为 0101 1010。
+- 这一题也可以用二分的思想来解答。把深度平分给 A 部分和 B 部分。
+    - 第一次遍历，先计算最大深度
+    - 第二次遍历，把深度小于等于最大深度一半的括号标记为 0(给 A 部分)，否则标记为 1(给 B 部分)
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 二分思想
+func maxDepthAfterSplit(seq string) []int {
+	stack, maxDepth, res := 0, 0, []int{}
+	for _, v := range seq {
+		if v == '(' {
+			stack++
+			maxDepth = max(stack, maxDepth)
+		} else {
+			stack--
+		}
+	}
+	stack = 0
+	for i := 0; i < len(seq); i++ {
+		if seq[i] == '(' {
+			stack++
+			if stack <= maxDepth/2 {
+				res = append(res, 0)
+			} else {
+				res = append(res, 1)
+			}
+		} else {
+			if stack <= maxDepth/2 {
+				res = append(res, 0)
+			} else {
+				res = append(res, 1)
+			}
+			stack--
+		}
+	}
+	return res
+}
+
+// 解法二 模拟
+func maxDepthAfterSplit1(seq string) []int {
+	stack, top, res := make([]int, len(seq)), -1, make([]int, len(seq))
+	for i, r := range seq {
+		if r == ')' {
+			res[i] = res[stack[top]]
+			top--
+			continue
+		}
+		top++
+		stack[top] = i
+		res[i] = top % 2
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1110.Delete-Nodes-And-Return-Forest/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1122.Relative-Sort-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1122.Relative-Sort-Array.md b/website/content/ChapterFour/1100~1199/1122.Relative-Sort-Array.md
new file mode 100755
index 000000000..c36c2f10e
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1122.Relative-Sort-Array.md
@@ -0,0 +1,108 @@
+# [1122. Relative Sort Array](https://leetcode.com/problems/relative-sort-array/)
+
+
+## 题目
+
+Given two arrays `arr1` and `arr2`, the elements of `arr2` are distinct, and all elements in `arr2` are also in `arr1`.
+
+Sort the elements of `arr1` such that the relative ordering of items in `arr1` are the same as in `arr2`. Elements that don't appear in `arr2` should be placed at the end of `arr1` in **ascending** order.
+
+**Example 1**:
+
+    Input: arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
+    Output: [2,2,2,1,4,3,3,9,6,7,19]
+
+**Constraints**:
+
+- `arr1.length, arr2.length <= 1000`
+- `0 <= arr1[i], arr2[i] <= 1000`
+- Each `arr2[i]` is distinct.
+- Each `arr2[i]` is in `arr1`.
+
+
+## 题目大意
+
+
+给你两个数组，arr1 和 arr2，
+
+- arr2 中的元素各不相同
+- arr2 中的每个元素都出现在 arr1 中
+
+对 arr1 中的元素进行排序，使 arr1 中项的相对顺序和 arr2 中的相对顺序相同。未在 arr2 中出现过的元素需要按照升序放在 arr1 的末尾。
+
+提示：
+
+- arr1.length, arr2.length <= 1000
+- 0 <= arr1[i], arr2[i] <= 1000
+- arr2 中的元素 arr2[i] 各不相同
+- arr2 中的每个元素 arr2[i] 都出现在 arr1 中
+
+
+
+## 解题思路
+
+- 给出 2 个数组 A 和 B，A 中包含 B 中的所有元素。要求 A 按照 B 的元素顺序排序，B 中没有的元素再接着排在后面，从小到大排序。
+- 这一题有多种解法。一种暴力的解法就是依照题意，先把 A 中的元素都统计频次放在 map 中，然后 按照 B 的顺序输出，接着再把剩下的元素排序接在后面。还有一种桶排序的思想，由于题目限定了 A 的大小是 1000，这个数量级很小，所以可以用 1001 个桶装所有的数，把数都放在桶里，这样默认就已经排好序了。接下来的做法和前面暴力解法差不多，按照频次输出。B 中以外的元素就按照桶的顺序依次输出即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+// 解法一 桶排序，时间复杂度 O(n^2)
+func relativeSortArray(A, B []int) []int {
+	count := [1001]int{}
+	for _, a := range A {
+		count[a]++
+	}
+	res := make([]int, 0, len(A))
+	for _, b := range B {
+		for count[b] > 0 {
+			res = append(res, b)
+			count[b]--
+		}
+	}
+	for i := 0; i < 1001; i++ {
+		for count[i] > 0 {
+			res = append(res, i)
+			count[i]--
+		}
+	}
+	return res
+}
+
+// 解法二 模拟，时间复杂度 O(n^2)
+func relativeSortArray1(arr1 []int, arr2 []int) []int {
+	leftover, m, res := []int{}, make(map[int]int), []int{}
+	for _, v := range arr1 {
+		m[v]++
+	}
+	for _, s := range arr2 {
+		count := m[s]
+		for i := 0; i < count; i++ {
+			res = append(res, s)
+		}
+		m[s] = 0
+	}
+	for v, count := range m {
+		for i := 0; i < count; i++ {
+			leftover = append(leftover, v)
+		}
+	}
+	sort.Ints(leftover)
+	res = append(res, leftover...)
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1123.Lowest-Common-Ancestor-of-Deepest-Leaves/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1123.Lowest-Common-Ancestor-of-Deepest-Leaves.md b/website/content/ChapterFour/1100~1199/1123.Lowest-Common-Ancestor-of-Deepest-Leaves.md
new file mode 100755
index 000000000..af07abcaf
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1123.Lowest-Common-Ancestor-of-Deepest-Leaves.md
@@ -0,0 +1,108 @@
+# [1123. Lowest Common Ancestor of Deepest Leaves](https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/)
+
+
+## 题目
+
+Given a rooted binary tree, return the lowest common ancestor of its deepest leaves.
+
+Recall that:
+
+- The node of a binary tree is a *leaf* if and only if it has no children
+- The *depth* of the root of the tree is 0, and if the depth of a node is `d`, the depth of each of its children is `d+1`.
+- The *lowest common ancestor* of a set `S` of nodes is the node `A` with the largest depth such that every node in S is in the subtree with root `A`.
+
+**Example 1**:
+
+    Input: root = [1,2,3]
+    Output: [1,2,3]
+    Explanation: 
+    The deepest leaves are the nodes with values 2 and 3.
+    The lowest common ancestor of these leaves is the node with value 1.
+    The answer returned is a TreeNode object (not an array) with serialization "[1,2,3]".
+
+**Example 2**:
+
+    Input: root = [1,2,3,4]
+    Output: [4]
+
+**Example 3**:
+
+    Input: root = [1,2,3,4,5]
+    Output: [2,4,5]
+
+**Constraints**:
+
+- The given tree will have between 1 and 1000 nodes.
+- Each node of the tree will have a distinct value between 1 and 1000.
+
+
+## 题目大意
+
+
+给你一个有根节点的二叉树，找到它最深的叶节点的最近公共祖先。
+
+回想一下：
+
+- 叶节点 是二叉树中没有子节点的节点
+- 树的根节点的 深度 为 0，如果某一节点的深度为 d，那它的子节点的深度就是 d+1
+- 如果我们假定 A 是一组节点 S 的 最近公共祖先，S 中的每个节点都在以 A 为根节点的子树中，且 A 的深度达到此条件下可能的最大值。
+ 
+
+提示：
+
+- 给你的树中将有 1 到 1000 个节点。
+- 树中每个节点的值都在 1 到 1000 之间。
+
+
+## 解题思路
+
+
+- 给出一颗树，找出最深的叶子节点的最近公共祖先 LCA。
+- 这一题思路比较直接。先遍历找到最深的叶子节点，如果左右子树的最深的叶子节点深度相同，那么当前节点就是它们的最近公共祖先。如果左右子树的最深的深度不等，那么需要继续递归往下找符合题意的 LCA。如果最深的叶子节点没有兄弟，那么公共父节点就是叶子本身，否则返回它的 LCA。
+- 有几个特殊的测试用例，见测试文件。特殊的点就是最深的叶子节点没有兄弟节点的情况。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func lcaDeepestLeaves(root *TreeNode) *TreeNode {
+	if root == nil {
+		return nil
+	}
+	lca, maxLevel := &TreeNode{}, 0
+	lcaDeepestLeavesDFS(&lca, &maxLevel, 0, root)
+	return lca
+}
+
+func lcaDeepestLeavesDFS(lca **TreeNode, maxLevel *int, depth int, root *TreeNode) int {
+	*maxLevel = max(*maxLevel, depth)
+	if root == nil {
+		return depth
+	}
+	depthLeft := lcaDeepestLeavesDFS(lca, maxLevel, depth+1, root.Left)
+	depthRight := lcaDeepestLeavesDFS(lca, maxLevel, depth+1, root.Right)
+	if depthLeft == *maxLevel && depthRight == *maxLevel {
+		*lca = root
+	}
+	return max(depthLeft, depthRight)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1122.Relative-Sort-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1128.Number-of-Equivalent-Domino-Pairs/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1128.Number-of-Equivalent-Domino-Pairs.md b/website/content/ChapterFour/1100~1199/1128.Number-of-Equivalent-Domino-Pairs.md
new file mode 100755
index 000000000..396e61e1e
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1128.Number-of-Equivalent-Domino-Pairs.md
@@ -0,0 +1,75 @@
+# [1128. Number of Equivalent Domino Pairs](https://leetcode.com/problems/number-of-equivalent-domino-pairs/)
+
+
+## 题目
+
+Given a list of `dominoes`, `dominoes[i] = [a, b]` is *equivalent* to `dominoes[j] = [c, d]` if and only if either (`a==c` and `b==d`), or (`a==d` and `b==c`) - that is, one domino can be rotated to be equal to another domino.
+
+Return the number of pairs `(i, j)` for which `0 <= i < j < dominoes.length`, and `dominoes[i]` is equivalent to `dominoes[j]`.
+
+**Example 1**:
+
+    Input: dominoes = [[1,2],[2,1],[3,4],[5,6]]
+    Output: 1
+
+**Constraints**:
+
+- `1 <= dominoes.length <= 40000`
+- `1 <= dominoes[i][j] <= 9`
+
+
+## 题目大意
+
+给你一个由一些多米诺骨牌组成的列表 dominoes。如果其中某一张多米诺骨牌可以通过旋转 0 度或 180 度得到另一张多米诺骨牌，我们就认为这两张牌是等价的。形式上，dominoes[i] = [a, b] 和 dominoes[j] = [c, d] 等价的前提是 a==c 且 b==d，或是 a==d 且 b==c。
+
+在 0 <= i < j < dominoes.length 的前提下，找出满足 dominoes[i] 和 dominoes[j] 等价的骨牌对 (i, j) 的数量。
+
+提示：
+
+- 1 <= dominoes.length <= 40000
+- 1 <= dominoes[i][j] <= 9
+
+
+
+## 解题思路
+
+- 给出一组多米诺骨牌，求出这组牌中相同牌的个数。牌相同的定义是：牌的 2 个数字相同(正序或者逆序相同都算相同)
+- 简单题。由于牌是 2 个数，所以将牌的 2 个数 hash 成一个 2 位数，比较大小即可，正序和逆序都 hash 成 2 位数，然后在桶中比较是否已经存在，如果不存在，跳过，如果存在，计数。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func numEquivDominoPairs(dominoes [][]int) int {
+	if dominoes == nil || len(dominoes) == 0 {
+		return 0
+	}
+	result, buckets := 0, [100]int{}
+	for _, dominoe := range dominoes {
+		key, rotatedKey := dominoe[0]*10+dominoe[1], dominoe[1]*10+dominoe[0]
+		if dominoe[0] != dominoe[1] {
+			if buckets[rotatedKey] > 0 {
+				result += buckets[rotatedKey]
+			}
+		}
+		if buckets[key] > 0 {
+			result += buckets[key]
+			buckets[key]++
+		} else {
+			buckets[key]++
+		}
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1123.Lowest-Common-Ancestor-of-Deepest-Leaves/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1137.N-th-Tribonacci-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1137.N-th-Tribonacci-Number.md b/website/content/ChapterFour/1100~1199/1137.N-th-Tribonacci-Number.md
new file mode 100755
index 000000000..74c1b0ed9
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1137.N-th-Tribonacci-Number.md
@@ -0,0 +1,78 @@
+# [1137. N-th Tribonacci Number](https://leetcode.com/problems/n-th-tribonacci-number/)
+
+
+## 题目
+
+The Tribonacci sequence Tn is defined as follows:
+
+T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0.
+
+Given `n`, return the value of Tn.
+
+**Example 1**:
+
+    Input: n = 4
+    Output: 4
+    Explanation:
+    T_3 = 0 + 1 + 1 = 2
+    T_4 = 1 + 1 + 2 = 4
+
+**Example 2**:
+
+    Input: n = 25
+    Output: 1389537
+
+**Constraints**:
+
+- `0 <= n <= 37`
+- The answer is guaranteed to fit within a 32-bit integer, ie. `answer <= 2^31 - 1`.
+
+
+## 题目大意
+
+
+泰波那契序列 Tn 定义如下： 
+
+T0 = 0, T1 = 1, T2 = 1, 且在 n >= 0 的条件下 Tn+3 = Tn + Tn+1 + Tn+2
+
+给你整数 n，请返回第 n 个泰波那契数 Tn 的值。
+
+提示：
+
+- 0 <= n <= 37
+- 答案保证是一个 32 位整数，即 answer <= 2^31 - 1。
+
+
+
+## 解题思路
+
+- 求泰波那契数列中的第 n 个数。
+- 简单题，按照题意定义计算即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func tribonacci(n int) int {
+	if n < 2 {
+		return n
+	}
+	trib, prev, prev2 := 1, 1, 0
+	for n > 2 {
+		trib, prev, prev2 = trib+prev+prev2, trib, prev
+		n--
+	}
+	return trib
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1128.Number-of-Equivalent-Domino-Pairs/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1143.Longest-Common-Subsequence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1143.Longest-Common-Subsequence.md b/website/content/ChapterFour/1100~1199/1143.Longest-Common-Subsequence.md
new file mode 100644
index 000000000..01e8b2fb8
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1143.Longest-Common-Subsequence.md
@@ -0,0 +1,95 @@
+# [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)
+
+## 题目
+
+Given two strings `text1` and `text2`, return *the length of their longest **common subsequence**.* If there is no **common subsequence**, return `0`.
+
+A **subsequence** of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.
+
+- For example, `"ace"` is a subsequence of `"abcde"`.
+
+A **common subsequence** of two strings is a subsequence that is common to both strings.
+
+**Example 1:**
+
+```
+Input: text1 = "abcde", text2 = "ace"
+Output: 3
+Explanation: The longest common subsequence is "ace" and its length is 3.
+```
+
+**Example 2:**
+
+```
+Input: text1 = "abc", text2 = "abc"
+Output: 3
+Explanation: The longest common subsequence is "abc" and its length is 3.
+```
+
+**Example 3:**
+
+```
+Input: text1 = "abc", text2 = "def"
+Output: 0
+Explanation: There is no such common subsequence, so the result is 0.
+```
+
+**Constraints:**
+
+- `1 <= text1.length, text2.length <= 1000`
+- `text1` and `text2` consist of only lowercase English characters.
+
+## 题目大意
+
+给定两个字符串 text1 和 text2，返回这两个字符串的最长 公共子序列 的长度。如果不存在 公共子序列 ，返回 0 。一个字符串的 子序列 是指这样一个新的字符串：它是由原字符串在不改变字符的相对顺序的情况下删除某些字符（也可以不删除任何字符）后组成的新字符串。例如，"ace" 是 "abcde" 的子序列，但 "aec" 不是 "abcde" 的子序列。两个字符串的 公共子序列 是这两个字符串所共同拥有的子序列。
+
+## 解题思路
+
+- 这一题是经典的最长公共子序列的问题。解题思路是二维动态规划。假设字符串 `text1` 和 `text2` 的长度分别为 `m` 和 `n`，创建 `m+1` 行 `n+1` 列的二维数组 `dp`，定义 `dp[i][j]` 表示长度为 i 的 `text1[0:i-1]` 和长度为 j 的 `text2[0:j-1]` 的最长公共子序列的长度。先考虑边界条件。当 `i = 0` 时，`text1[]` 为空字符串，它与任何字符串的最长公共子序列的长度都是 `0`，所以 `dp[0][j] = 0`。同理当 `j = 0` 时，`text2[]` 为空字符串，它与任何字符串的最长公共子序列的长度都是 `0`，所以 `dp[i][0] = 0`。由于二维数组的大小特意增加了 `1`，即 `m+1` 和 `n+1`，并且默认值是 `0`，所以不需要再初始化赋值了。
+- 当 `text1[i−1] = text2[j−1]` 时，将这两个相同的字符称为公共字符，考虑 `text1[0:i−1]` 和 `text2[0:j−1]` 的最长公共子序列，再增加一个字符（即公共字符）即可得到 `text1[0:i]` 和 `text2[0:j]` 的最长公共子序列，所以 `dp[i][j]=dp[i−1][j−1]+1`。当 `text1[i−1] != text2[j−1]` 时，最长公共子序列一定在 `text[0:i-1], text2[0:j]` 和 `text[0:i], text2[0:j-1]` 中取得。即 `dp[i][j] = max(dp[i-1][j], dp[i][j-1])`。所以状态转移方程如下：
+
+	{{< katex display >}}
+    dp[i][j] = \left\{\begin{matrix}dp[i-1][j-1]+1 &,text1[i-1]=text2[j-1]\\max(dp[i-1][j],dp[i][j-1])&,text1[i-1]\neq text2[j-1]\end{matrix}\right.
+	{{< /katex >}}
+
+- 最终结果存储在 `dp[len(text1)][len(text2)]` 中。时间复杂度 `O(mn)`，空间复杂度 `O(mn)`，其中 `m` 和 `n` 分别是 `text1` 和 `text2` 的长度。
+
+## 代码
+
+```go
+package leetcode
+
+func longestCommonSubsequence(text1 string, text2 string) int {
+	if len(text1) == 0 || len(text2) == 0 {
+		return 0
+	}
+	dp := make([][]int, len(text1)+1)
+	for i := range dp {
+		dp[i] = make([]int, len(text2)+1)
+	}
+	for i := 1; i < len(text1)+1; i++ {
+		for j := 1; j < len(text2)+1; j++ {
+			if text1[i-1] == text2[j-1] {
+				dp[i][j] = dp[i-1][j-1] + 1
+			} else {
+				dp[i][j] = max(dp[i][j-1], dp[i-1][j])
+			}
+		}
+	}
+	return dp[len(text1)][len(text2)]
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1137.N-th-Tribonacci-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1145.Binary-Tree-Coloring-Game/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1145.Binary-Tree-Coloring-Game.md b/website/content/ChapterFour/1100~1199/1145.Binary-Tree-Coloring-Game.md
new file mode 100644
index 000000000..f04cc161c
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1145.Binary-Tree-Coloring-Game.md
@@ -0,0 +1,89 @@
+# [1145. Binary Tree Coloring Game](https://leetcode.com/problems/binary-tree-coloring-game/)
+
+
+
+## 题目
+
+Two players play a turn based game on a binary tree. We are given the `root` of this binary tree, and the number of nodes `n` in the tree. `n` is odd, and each node has a distinct value from `1` to `n`.
+
+Initially, the first player names a value `x` with `1 <= x <= n`, and the second player names a value `y` with `1 <= y <= n` and `y != x`. The first player colors the node with value `x` red, and the second player colors the node with value `y` blue.
+
+Then, the players take turns starting with the first player. In each turn, that player chooses a node of their color (red if player 1, blue if player 2) and colors an **uncolored** neighbor of the chosen node (either the left child, right child, or parent of the chosen node.)
+
+If (and only if) a player cannot choose such a node in this way, they must pass their turn. If both players pass their turn, the game ends, and the winner is the player that colored more nodes.
+
+You are the second player. If it is possible to choose such a `y` to ensure you win the game, return `true`. If it is not possible, return `false`.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/08/01/1480-binary-tree-coloring-game.png](https://assets.leetcode.com/uploads/2019/08/01/1480-binary-tree-coloring-game.png)
+
+```
+Input: root = [1,2,3,4,5,6,7,8,9,10,11], n = 11, x = 3
+Output: true
+Explanation: The second player can choose the node with value 2.
+```
+
+**Constraints**:
+
+- `root` is the root of a binary tree with `n` nodes and distinct node values from `1` to `n`.
+- `n` is odd.
+- `1 <= x <= n <= 100`
+
+## 题目大意
+
+有两位极客玩家参与了一场「二叉树着色」的游戏。游戏中，给出二叉树的根节点 root，树上总共有 n 个节点，且 n 为奇数，其中每个节点上的值从 1 到 n 各不相同。游戏从「一号」玩家开始（「一号」玩家为红色，「二号」玩家为蓝色），最开始时，
+
+- 「一号」玩家从 [1, n] 中取一个值 x（1 <= x <= n）；
+- 「二号」玩家也从 [1, n] 中取一个值 y（1 <= y <= n）且 y != x。
+- 「一号」玩家给值为 x 的节点染上红色，而「二号」玩家给值为 y 的节点染上蓝色。
+
+之后两位玩家轮流进行操作，每一回合，玩家选择一个他之前涂好颜色的节点，将所选节点一个 未着色 的邻节点（即左右子节点、或父节点）进行染色。如果当前玩家无法找到这样的节点来染色时，他的回合就会被跳过。若两个玩家都没有可以染色的节点时，游戏结束。着色节点最多的那位玩家获得胜利 ✌️。现在，假设你是「二号」玩家，根据所给出的输入，假如存在一个 y 值可以确保你赢得这场游戏，则返回 true；若无法获胜，就请返回 false。
+
+
+提示：
+
+- 二叉树的根节点为 root，树上由 n 个节点，节点上的值从 1 到 n 各不相同。
+- n 为奇数。
+- 1 <= x <= n <= 100
+
+## 解题思路
+
+- 2 个人参加二叉树着色游戏。二叉树节点数为奇数。1 号玩家和 2 号玩家分别在二叉树上选项一个点着色。每一回合，玩家选择一个他之前涂好颜色的节点，将所选节点一个 未着色 的邻节点（即左右子节点、或父节点）进行染色。当有人不能选点着色的时候，他的那个回合会被跳过。双方都没法继续着色的时候游戏结束。着色多的人获胜。问二号玩家是否存在必胜策略？
+
+![](https://img.halfrost.com/Leetcode/leetcode_1145.png)
+
+- 如图所示，当一号玩家选择了一个红色的结点，可能会将二叉树切割为 3 个部分（连通分量），如果选择的是根结点，则可能是 2 个部分或 1 个部分，如果选择叶结点，则是 1 个部分。不过无论哪种情况都无关紧要，我们都可以当成 3 个部分来对待，例如一号玩家选择了一个叶结点，我们也可以把叶结点的左右两个空指针看成大小为 0 的两个部分。
+- 那么二号玩家怎样选择蓝色结点才是最优呢？答案是：选择离红色结点最近，且所属连通分量规模最大的那个点。也就是示例图中的 1 号结点。如果我们选择了 1 号结点为蓝色结点，那么可以染成红色的点就只剩下 6 号点和 7 号点了，而蓝色可以把根结点和其左子树全部占据。
+- 如何确定蓝色是否有必胜策略，就可以转换为，被红色点切割的三个连通分量中，是否存在一个连通分量，大小大于所有结点数目的一半。统计三个连通分量大小的过程，可以用深度优先搜索（DFS）来实现。当遍历到某一结点，其结点值等于选定的红色结点时，我们统计这个结点的左子树 `red_left` 和右子树 `red_right` 的大小，那么我们就已经找到两个连通分量的大小了，最后一个父结点连通分量的大小，可以用结点总数减去这两个连通分量大小，再减去红色所占结点，即 `parent = n - red_left - red_right - 1`。
+
+## 代码
+
+```go
+
+func btreeGameWinningMove(root *TreeNode, n int, x int) bool {
+	var left, right int
+	dfsBtreeGameWinningMove(root, &left, &right, x)
+	up := n - left - right - 1
+	n /= 2
+	return left > n || right > n || up > n
+}
+
+func dfsBtreeGameWinningMove(node *TreeNode, left, right *int, x int) int {
+	if node == nil {
+		return 0
+	}
+	l, r := dfsBtreeGameWinningMove(node.Left, left, right, x), dfsBtreeGameWinningMove(node.Right, left, right, x)
+	if node.Val == x {
+		*left, *right = l, r
+	}
+	return l + r + 1
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1143.Longest-Common-Subsequence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1154.Day-of-the-Year/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1154.Day-of-the-Year.md b/website/content/ChapterFour/1100~1199/1154.Day-of-the-Year.md
new file mode 100755
index 000000000..98e392d01
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1154.Day-of-the-Year.md
@@ -0,0 +1,88 @@
+# [1154. Day of the Year](https://leetcode.com/problems/day-of-the-year/)
+
+
+## 题目
+
+Given a string `date` representing a [Gregorian calendar](https://en.wikipedia.org/wiki/Gregorian_calendar) date formatted as `YYYY-MM-DD`, return the day number of the year.
+
+**Example 1**:
+
+    Input: date = "2019-01-09"
+    Output: 9
+    Explanation: Given date is the 9th day of the year in 2019.
+
+**Example 2**:
+
+    Input: date = "2019-02-10"
+    Output: 41
+
+**Example 3**:
+
+    Input: date = "2003-03-01"
+    Output: 60
+
+**Example 4**:
+
+    Input: date = "2004-03-01"
+    Output: 61
+
+**Constraints**:
+
+- `date.length == 10`
+- `date[4] == date[7] == '-'`, and all other `date[i]`'s are digits
+- `date` represents a calendar date between Jan 1st, 1900 and Dec 31, 2019.
+
+## 题目大意
+
+
+实现一个 MajorityChecker 的类，它应该具有下述几个 API：
+
+- MajorityChecker(int[] arr) 会用给定的数组 arr 来构造一个 MajorityChecker 的实例。
+- int query(int left, int right, int threshold) 有这么几个参数：  
+	- 0 <= left <= right < arr.length 表示数组 arr 的子数组的长度。
+	- 2 * threshold > right - left + 1，也就是说阈值 threshold 始终比子序列长度的一半还要大。
+
+每次查询 query(...) 会返回在 arr[left], arr[left+1], ..., arr[right] 中至少出现阈值次数 threshold 的元素，如果不存在这样的元素，就返回 -1。
+
+提示：
+
+- 1 <= arr.length <= 20000
+- 1 <= arr[i] <= 20000
+- 对于每次查询，0 <= left <= right < len(arr)
+- 对于每次查询，2 * threshold > right - left + 1
+- 查询次数最多为 10000
+
+
+
+
+
+## 解题思路
+
+- 给出一个时间字符串，求出这一天是这一年当中的第几天。
+- 简单题。依照题意处理即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "time"
+
+func dayOfYear(date string) int {
+	first := date[:4] + "-01-01"
+	firstDay, _ := time.Parse("2006-01-02", first)
+	dateDay, _ := time.Parse("2006-01-02", date)
+	duration := dateDay.Sub(firstDay)
+	return int(duration.Hours())/24 + 1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1145.Binary-Tree-Coloring-Game/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1157.Online-Majority-Element-In-Subarray/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1157.Online-Majority-Element-In-Subarray.md b/website/content/ChapterFour/1100~1199/1157.Online-Majority-Element-In-Subarray.md
new file mode 100755
index 000000000..0fd147e49
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1157.Online-Majority-Element-In-Subarray.md
@@ -0,0 +1,203 @@
+# [1157. Online Majority Element In Subarray](https://leetcode.com/problems/online-majority-element-in-subarray/)
+
+
+## 题目
+
+Implementing the class `MajorityChecker`, which has the following API:
+
+- `MajorityChecker(int[] arr)` constructs an instance of MajorityChecker with the given array `arr`;
+- `int query(int left, int right, int threshold)` has arguments such that:
+    - `0 <= left <= right < arr.length` representing a subarray of `arr`;
+    - `2 * threshold > right - left + 1`, ie. the threshold is always a strict majority of the length of the subarray
+
+Each `query(...)` returns the element in `arr[left], arr[left+1], ..., arr[right]` that occurs at least `threshold` times, or `-1` if no such element exists.
+
+**Example**:
+
+    MajorityChecker majorityChecker = new MajorityChecker([1,1,2,2,1,1]);
+    majorityChecker.query(0,5,4); // returns 1
+    majorityChecker.query(0,3,3); // returns -1
+    majorityChecker.query(2,3,2); // returns 2
+
+**Constraints**:
+
+- `1 <= arr.length <= 20000`
+- `1 <= arr[i] <= 20000`
+- For each query, `0 <= left <= right < len(arr)`
+- For each query, `2 * threshold > right - left + 1`
+- The number of queries is at most `10000`
+
+
+## 题目大意
+
+实现一个 MajorityChecker 的类，它应该具有下述几个 API：
+
+- MajorityChecker(int[] arr) 会用给定的数组 arr 来构造一个 MajorityChecker 的实例。
+- int query(int left, int right, int threshold) 有这么几个参数：
+- 0 <= left <= right < arr.length 表示数组 arr 的子数组的长度。
+- 2 * threshold > right - left + 1，也就是说阈值 threshold 始终比子序列长度的一半还要大。
+
+每次查询 query(...) 会返回在 arr[left], arr[left+1], ..., arr[right] 中至少出现阈值次数 threshold 的元素，如果不存在这样的元素，就返回 -1。
+
+提示：
+
+- 1 <= arr.length <= 20000
+- 1 <= arr[i] <= 20000
+- 对于每次查询，0 <= left <= right < len(arr)
+- 对于每次查询，2 * threshold > right - left + 1
+- 查询次数最多为 10000
+
+
+
+
+## 解题思路
+
+
+- 设计一个数据结构，能在任意的一个区间内，查找是否存在众数，众数的定义是：该数字出现的次数大于区间的一半。如果存在众数，一定唯一。如果在给定的区间内找不到众数，则输出 -1 。
+- 这一题有一个很显眼的“暗示”，`2 * threshold > right - left + 1`，这个条件就是摩尔投票算法的前提条件。摩尔投票的思想可以见第 169 题。这一题又要在区间内查询，所以选用线段树这个数据结构来实现。经过分析，可以确定此题的解题思路，摩尔投票 + 线段树。
+- 摩尔投票的思想是用两个变量，candidate 和 count，用来记录待被投票投出去的元素，和候选人累积没被投出去的轮数。如果候选人累积没有被投出去的轮数越多，那么最终成为众数的可能越大。从左往右扫描整个数组，先去第一个元素为 candidate，如果遇到相同的元素就累加轮数，如果遇到不同的元素，就把 candidate 和不同的元素一起投出去。当轮数变成 0 了，再选下一个元素作为 candidate。从左扫到右，就能找到众数了。那怎么和线段树结合起来呢？
+- 线段树是把一个大的区间拆分成很多个小区间，那么考虑这样一个问题。每个小区间内使用摩尔投票，最终把所有小区间合并起来再用一次摩尔投票，得到的结果和对整个区间使用一次摩尔投票的结果是一样的么？答案是一样的。可以这样想，众数总会在一个区间内被选出来，那么其他区间的摩尔投票都是起“中和”作用的，即两两元素一起出局。这个问题想通以后，说明摩尔投票具有可加的性质。既然满足可加，就可以和线段树结合，因为线段树每个线段就是加起来，最终合并成大区间的。
+- 举个例子，arr = [1,1,2,2,1,1]，先构造线段树，如下左图。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_1157_0.png)
+
+    现在每个线段树的节点不是只存一个 int 数字了，而是存 candidate 和 count。每个节点的 candidate 和 count 分别代表的是该区间内摩尔投票的结果。初始化的时候，先把每个叶子都填满，candidate 是自己，count = 1 。即右图绿色节点。然后在 pushUp 的时候，进行摩尔投票：
+
+        mc.merge = func(i, j segmentItem) segmentItem {
+        		if i.candidate == j.candidate {
+        			return segmentItem{candidate: i.candidate, count: i.count + j.count}
+        		}
+        		if i.count > j.count {
+        			return segmentItem{candidate: i.candidate, count: i.count - j.count}
+        		}
+        		return segmentItem{candidate: j.candidate, count: j.count - i.count}
+        	}
+
+    直到根节点的 candidate 和 count 都填满。**注意，这里的 count 并不是元素出现的总次数，而是摩尔投票中坚持没有被投出去的轮数**。当线段树构建完成以后，就可以开始查询任意区间内的众数了，candidate 即为众数。接下来还要确定众数是否满足 `threshold` 的条件。
+
+- 用一个字典记录每个元素在数组中出现位置的下标，例如上述这个例子，用 map 记录下标：count = map[1:[0 1 4 5] 2:[2 3]]。由于下标在记录过程中是递增的，所以满足二分查找的条件。利用这个字典就可以查出在任意区间内，指定元素出现的次数。例如这里要查找 1 在 [0,5] 区间内出现的个数，那么利用 2 次二分查找，分别找到 `lowerBound` 和 `upperBound`，在 [lowerBound，upperBound) 区间内，都是元素 1 ，那么区间长度即是该元素重复出现的次数，和 `threshold` 比较，如果 ≥ `threshold` 说明找到了答案，否则没有找到就输出 -1 。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+)
+
+type segmentItem struct {
+	candidate int
+	count     int
+}
+
+// MajorityChecker define
+type MajorityChecker struct {
+	segmentTree []segmentItem
+	data        []int
+	merge       func(i, j segmentItem) segmentItem
+	count       map[int][]int
+}
+
+// Constructor1157 define
+func Constructor1157(arr []int) MajorityChecker {
+	data, tree, mc, count := make([]int, len(arr)), make([]segmentItem, 4*len(arr)), MajorityChecker{}, make(map[int][]int)
+	// 这个 merge 函数就是摩尔投票算法
+	mc.merge = func(i, j segmentItem) segmentItem {
+		if i.candidate == j.candidate {
+			return segmentItem{candidate: i.candidate, count: i.count + j.count}
+		}
+		if i.count > j.count {
+			return segmentItem{candidate: i.candidate, count: i.count - j.count}
+		}
+		return segmentItem{candidate: j.candidate, count: j.count - i.count}
+	}
+	for i := 0; i < len(arr); i++ {
+		data[i] = arr[i]
+	}
+	for i := 0; i < len(arr); i++ {
+		if _, ok := count[arr[i]]; !ok {
+			count[arr[i]] = []int{}
+		}
+		count[arr[i]] = append(count[arr[i]], i)
+	}
+	mc.data, mc.segmentTree, mc.count = data, tree, count
+	if len(arr) > 0 {
+		mc.buildSegmentTree(0, 0, len(arr)-1)
+	}
+	return mc
+}
+
+func (mc *MajorityChecker) buildSegmentTree(treeIndex, left, right int) {
+	if left == right {
+		mc.segmentTree[treeIndex] = segmentItem{candidate: mc.data[left], count: 1}
+		return
+	}
+	leftTreeIndex, rightTreeIndex := mc.leftChild(treeIndex), mc.rightChild(treeIndex)
+	midTreeIndex := left + (right-left)>>1
+	mc.buildSegmentTree(leftTreeIndex, left, midTreeIndex)
+	mc.buildSegmentTree(rightTreeIndex, midTreeIndex+1, right)
+	mc.segmentTree[treeIndex] = mc.merge(mc.segmentTree[leftTreeIndex], mc.segmentTree[rightTreeIndex])
+}
+
+func (mc *MajorityChecker) leftChild(index int) int {
+	return 2*index + 1
+}
+
+func (mc *MajorityChecker) rightChild(index int) int {
+	return 2*index + 2
+}
+
+// Query define
+func (mc *MajorityChecker) query(left, right int) segmentItem {
+	if len(mc.data) > 0 {
+		return mc.queryInTree(0, 0, len(mc.data)-1, left, right)
+	}
+	return segmentItem{candidate: -1, count: -1}
+}
+
+func (mc *MajorityChecker) queryInTree(treeIndex, left, right, queryLeft, queryRight int) segmentItem {
+	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, mc.leftChild(treeIndex), mc.rightChild(treeIndex)
+	if queryLeft <= left && queryRight >= right { // segment completely inside range
+		return mc.segmentTree[treeIndex]
+	}
+	if queryLeft > midTreeIndex {
+		return mc.queryInTree(rightTreeIndex, midTreeIndex+1, right, queryLeft, queryRight)
+	} else if queryRight <= midTreeIndex {
+		return mc.queryInTree(leftTreeIndex, left, midTreeIndex, queryLeft, queryRight)
+	}
+	// merge query results
+	return mc.merge(mc.queryInTree(leftTreeIndex, left, midTreeIndex, queryLeft, midTreeIndex),
+		mc.queryInTree(rightTreeIndex, midTreeIndex+1, right, midTreeIndex+1, queryRight))
+}
+
+// Query define
+func (mc *MajorityChecker) Query(left int, right int, threshold int) int {
+	res := mc.query(left, right)
+	if _, ok := mc.count[res.candidate]; !ok {
+		return -1
+	}
+	start := sort.Search(len(mc.count[res.candidate]), func(i int) bool { return left <= mc.count[res.candidate][i] })
+	end := sort.Search(len(mc.count[res.candidate]), func(i int) bool { return right < mc.count[res.candidate][i] }) - 1
+	if (end - start + 1) >= threshold {
+		return res.candidate
+	}
+	return -1
+}
+
+/**
+ * Your MajorityChecker object will be instantiated and called as such:
+ * obj := Constructor(arr);
+ * param_1 := obj.Query(left,right,threshold);
+ */
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1154.Day-of-the-Year/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1160.Find-Words-That-Can-Be-Formed-by-Characters/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1160.Find-Words-That-Can-Be-Formed-by-Characters.md b/website/content/ChapterFour/1100~1199/1160.Find-Words-That-Can-Be-Formed-by-Characters.md
new file mode 100755
index 000000000..a0e7603d9
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1160.Find-Words-That-Can-Be-Formed-by-Characters.md
@@ -0,0 +1,88 @@
+# [1160. Find Words That Can Be Formed by Characters](https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/)
+
+
+## 题目
+
+You are given an array of strings `words` and a string `chars`.
+
+A string is *good* if it can be formed by characters from `chars` (each character can only be used once).
+
+Return the sum of lengths of all good strings in `words`.
+
+**Example 1**:
+
+    Input: words = ["cat","bt","hat","tree"], chars = "atach"
+    Output: 6
+    Explanation: 
+    The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.
+
+**Example 2**:
+
+    Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr"
+    Output: 10
+    Explanation: 
+    The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10.
+
+**Note**:
+
+1. `1 <= words.length <= 1000`
+2. `1 <= words[i].length, chars.length <= 100`
+3. All strings contain lowercase English letters only.
+
+
+## 题目大意
+
+
+给你一份『词汇表』（字符串数组） words 和一张『字母表』（字符串） chars。假如你可以用 chars 中的『字母』（字符）拼写出 words 中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。注意：每次拼写时，chars 中的每个字母都只能用一次。返回词汇表 words 中你掌握的所有单词的 长度之和。
+
+提示：
+
+1. 1 <= words.length <= 1000
+2. 1 <= words[i].length, chars.length <= 100
+3. 所有字符串中都仅包含小写英文字母
+
+
+
+## 解题思路
+
+- 给出一个字符串数组 `words` 和一个字符串 `chars`，要求输出 `chars` 中能构成 `words` 字符串的字符数总数。
+- 简单题。先分别统计 `words` 和 `chars` 里面字符的频次。然后针对 `words` 中每个 `word` 判断能够能由 `chars` 构成，如果能构成，最终结果加上这个 `word` 的长度。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func countCharacters(words []string, chars string) int {
+	count, res := make([]int, 26), 0
+	for i := 0; i < len(chars); i++ {
+		count[chars[i]-'a']++
+	}
+	for _, w := range words {
+		if canBeFormed(w, count) {
+			res += len(w)
+		}
+	}
+	return res
+}
+func canBeFormed(w string, c []int) bool {
+	count := make([]int, 26)
+	for i := 0; i < len(w); i++ {
+		count[w[i]-'a']++
+		if count[w[i]-'a'] > c[w[i]-'a'] {
+			return false
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1157.Online-Majority-Element-In-Subarray/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character.md b/website/content/ChapterFour/1100~1199/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character.md
new file mode 100755
index 000000000..987a39a81
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character.md
@@ -0,0 +1,93 @@
+# [1170. Compare Strings by Frequency of the Smallest Character](https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/)
+
+## 题目
+
+Let's define a function `f(s)` over a non-empty string `s`, which calculates the frequency of the smallest character in `s`. For example, if `s = "dcce"` then `f(s) = 2` because the smallest character is `"c"` and its frequency is 2.
+
+Now, given string arrays `queries` and `words`, return an integer array `answer`, where each `answer[i]` is the number of words such that `f(queries[i])` < `f(W)`, where `W` is a word in `words`.
+
+**Example 1**:
+
+    Input: queries = ["cbd"], words = ["zaaaz"]
+    Output: [1]
+    Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").
+
+**Example 2**:
+
+    Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
+    Output: [1,2]
+    Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").
+
+**Constraints**:
+
+- `1 <= queries.length <= 2000`
+- `1 <= words.length <= 2000`
+- `1 <= queries[i].length, words[i].length <= 10`
+- `queries[i][j]`, `words[i][j]` are English lowercase letters.
+
+
+## 题目大意
+
+
+我们来定义一个函数 f(s)，其中传入参数 s 是一个非空字符串；该函数的功能是统计 s  中（按字典序比较）最小字母的出现频次。
+
+例如，若 s = "dcce"，那么 f(s) = 2，因为最小的字母是 "c"，它出现了 2 次。
+
+现在，给你两个字符串数组待查表 queries 和词汇表 words，请你返回一个整数数组 answer 作为答案，其中每个 answer[i] 是满足 f(queries[i]) < f(W) 的词的数目，W 是词汇表 words 中的词。
+
+提示：
+
+- 1 <= queries.length <= 2000
+- 1 <= words.length <= 2000
+- 1 <= queries[i].length, words[i].length <= 10
+- queries[i][j], words[i][j] 都是小写英文字母
+
+
+
+
+## 解题思路
+
+- 给出 2 个数组，`queries` 和 `words`，针对每一个 `queries[i]` 统计在 `words[j]` 中满足 `f(queries[i]) < f(words[j])` 条件的 `words[j]` 的个数。`f(string)` 的定义是 `string` 中字典序最小的字母的频次。
+- 先依照题意，构造出 `f()` 函数，算出每个 `words[j]` 的 `f()` 值，然后排序。再依次计算 `queries[i]` 的 `f()` 值。针对每个 `f()` 值，在 `words[j]` 的 `f()` 值中二分搜索，查找比它大的值的下标 `k`，`n-k` 即是比 `queries[i]` 的 `f()` 值大的元素个数。依次输出到结果数组中即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+func numSmallerByFrequency(queries []string, words []string) []int {
+	ws, res := make([]int, len(words)), make([]int, len(queries))
+	for i, w := range words {
+		ws[i] = countFunc(w)
+	}
+	sort.Ints(ws)
+	for i, q := range queries {
+		fq := countFunc(q)
+		res[i] = len(words) - sort.Search(len(words), func(i int) bool { return fq < ws[i] })
+	}
+	return res
+}
+
+func countFunc(s string) int {
+	count, i := [26]int{}, 0
+	for _, b := range s {
+		count[b-'a']++
+	}
+	for count[i] == 0 {
+		i++
+	}
+	return count[i]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1160.Find-Words-That-Can-Be-Formed-by-Characters/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List.md b/website/content/ChapterFour/1100~1199/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List.md
new file mode 100755
index 000000000..49e9862b9
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List.md
@@ -0,0 +1,147 @@
+# [1171. Remove Zero Sum Consecutive Nodes from Linked List](https://leetcode.com/problems/remove-zero-sum-consecutive-nodes-from-linked-list/)
+
+
+## 题目
+
+Given the `head` of a linked list, we repeatedly delete consecutive sequences of nodes that sum to `0` until there are no such sequences.
+
+After doing so, return the head of the final linked list. You may return any such answer.
+
+(Note that in the examples below, all sequences are serializations of `ListNode` objects.)
+
+**Example 1**:
+
+    Input: head = [1,2,-3,3,1]
+    Output: [3,1]
+    **Note**: The answer [1,2,1] would also be accepted.
+
+**Example 2**:
+
+    Input: head = [1,2,3,-3,4]
+    Output: [1,2,4]
+
+**Example 3**:
+
+    Input: head = [1,2,3,-3,-2]
+    Output: [1]
+
+**Constraints**:
+
+- The given linked list will contain between `1` and `1000` nodes.
+- Each node in the linked list has `-1000 <= node.val <= 1000`.
+
+
+## 题目大意
+
+
+给你一个链表的头节点 head，请你编写代码，反复删去链表中由 总和 值为 0 的连续节点组成的序列，直到不存在这样的序列为止。删除完毕后，请你返回最终结果链表的头节点。你可以返回任何满足题目要求的答案。
+
+（注意，下面示例中的所有序列，都是对 ListNode 对象序列化的表示。）
+
+提示：
+
+- 给你的链表中可能有 1 到 1000 个节点。
+- 对于链表中的每个节点，节点的值：-1000 <= node.val <= 1000.
+
+
+
+## 解题思路
+
+- 给出一个链表，要求把链表中和为 0 的结点都移除。
+- 由于链表的特性，不能随机访问。所以从链表的头开始往后扫，把累加和存到字典中。当再次出现相同的累加和的时候，代表这中间的一段和是 0，于是要删除这一段。删除这一段的过程中，也要删除这一段在字典中存过的累加和。有一个特殊情况需要处理，即整个链表的总和是 0，那么最终结果是空链表。针对这个特殊情况，字典中先预存入 0 值。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// 解法一
+func removeZeroSumSublists(head *ListNode) *ListNode {
+	// 计算累加和，和作为 key 存在 map 中，value 存那个节点的指针。如果字典中出现了重复的和，代表出现了和为 0 的段。
+	sum, sumMap, cur := 0, make(map[int]*ListNode), head
+	// 字典中增加 0 这个特殊值，是为了防止最终链表全部消除完
+	sumMap[0] = nil
+	for cur != nil {
+		sum = sum + cur.Val
+		if ptr, ok := sumMap[sum]; ok {
+			// 在字典中找到了重复的和，代表 [ptr, tmp] 中间的是和为 0 的段，要删除的就是这一段。
+			// 同时删除 map 中中间这一段的和
+			if ptr != nil {
+				iter := ptr.Next
+				tmpSum := sum + iter.Val
+				for tmpSum != sum {
+					// 删除中间为 0 的那一段，tmpSum 不断的累加删除 map 中的和
+					delete(sumMap, tmpSum)
+					iter = iter.Next
+					tmpSum = tmpSum + iter.Val
+				}
+				ptr.Next = cur.Next
+			} else {
+				head = cur.Next
+				sumMap = make(map[int]*ListNode)
+				sumMap[0] = nil
+			}
+		} else {
+			sumMap[sum] = cur
+		}
+		cur = cur.Next
+	}
+	return head
+}
+
+// 解法二 暴力解法
+func removeZeroSumSublists1(head *ListNode) *ListNode {
+	if head == nil {
+		return nil
+	}
+	h, prefixSumMap, sum, counter, lastNode := head, map[int]int{}, 0, 0, &ListNode{Val: 1010}
+	for h != nil {
+		for h != nil {
+			sum += h.Val
+			counter++
+			if v, ok := prefixSumMap[sum]; ok {
+				lastNode, counter = h, v
+				break
+			}
+			if sum == 0 {
+				head = h.Next
+				break
+			}
+			prefixSumMap[sum] = counter
+			h = h.Next
+		}
+		if lastNode.Val != 1010 {
+			h = head
+			for counter > 1 {
+				counter--
+				h = h.Next
+			}
+			h.Next = lastNode.Next
+		}
+		if h == nil {
+			break
+		} else {
+			h, prefixSumMap, sum, counter, lastNode = head, map[int]int{}, 0, 0, &ListNode{Val: 1010}
+		}
+	}
+	return head
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1175.Prime-Arrangements/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1175.Prime-Arrangements.md b/website/content/ChapterFour/1100~1199/1175.Prime-Arrangements.md
new file mode 100755
index 000000000..a147da041
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1175.Prime-Arrangements.md
@@ -0,0 +1,72 @@
+# [1175. Prime Arrangements](https://leetcode.com/problems/prime-arrangements/)
+
+
+## 题目
+
+Return the number of permutations of 1 to `n` so that prime numbers are at prime indices (1-indexed.)
+
+*(Recall that an integer is prime if and only if it is greater than 1, and cannot be written as a product of two positive integers both smaller than it.)*
+
+Since the answer may be large, return the answer **modulo `10^9 + 7`**.
+
+**Example 1**:
+
+    Input: n = 5
+    Output: 12
+    Explanation: For example [1,2,5,4,3] is a valid permutation, but [5,2,3,4,1] is not because the prime number 5 is at index 1.
+
+**Example 2**:
+
+    Input: n = 100
+    Output: 682289015
+
+**Constraints**:
+
+- `1 <= n <= 100`
+
+
+## 题目大意
+
+
+请你帮忙给从 1 到 n 的数设计排列方案，使得所有的「质数」都应该被放在「质数索引」（索引从 1 开始）上；你需要返回可能的方案总数。让我们一起来回顾一下「质数」：质数一定是大于 1 的，并且不能用两个小于它的正整数的乘积来表示。由于答案可能会很大，所以请你返回答案 模 mod 10^9 + 7 之后的结果即可。
+
+提示：
+
+- 1 <= n <= 100
+
+## 解题思路
+
+- 给出一个数 n，要求在 1-n 这 n 个数中，素数在素数索引下标位置上的全排列个数。
+- 由于这一题的 `n` 小于 100，所以可以用打表法。先把小于 100 个素数都打表打出来。然后对小于 n 的素数进行全排列，即 n！，然后再对剩下来的非素数进行全排列，即 (n-c)！。两个的乘积即为最终答案。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+var primes = []int{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
+
+func numPrimeArrangements(n int) int {
+	primeCount := sort.Search(25, func(i int) bool { return primes[i] > n })
+	return factorial(primeCount) * factorial(n-primeCount) % 1000000007
+}
+
+func factorial(n int) int {
+	if n == 1 || n == 0 {
+		return 1
+	}
+	return n * factorial(n-1) % 1000000007
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1178.Number-of-Valid-Words-for-Each-Puzzle/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1178.Number-of-Valid-Words-for-Each-Puzzle.md b/website/content/ChapterFour/1100~1199/1178.Number-of-Valid-Words-for-Each-Puzzle.md
new file mode 100644
index 000000000..bf708bb7d
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1178.Number-of-Valid-Words-for-Each-Puzzle.md
@@ -0,0 +1,126 @@
+# [1178. Number of Valid Words for Each Puzzle](https://leetcode.com/problems/number-of-valid-words-for-each-puzzle/)
+
+
+## 题目
+
+With respect to a given `puzzle` string, a `word` is *valid* if both the following conditions are satisfied:
+
+- `word` contains the first letter of `puzzle`.
+- For each letter in `word`, that letter is in `puzzle`.For example, if the puzzle is "abcdefg", then valid words are "faced", "cabbage", and "baggage"; while invalid words are "beefed" (doesn't include "a") and "based" (includes "s" which isn't in the puzzle).
+
+Return an array `answer`, where `answer[i]` is the number of words in the given word list `words` that are valid with respect to the puzzle `puzzles[i]`.
+
+**Example :**
+
+```
+Input: 
+words = ["aaaa","asas","able","ability","actt","actor","access"], 
+puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
+Output: [1,1,3,2,4,0]
+Explanation:
+1 valid word for "aboveyz" : "aaaa" 
+1 valid word for "abrodyz" : "aaaa"
+3 valid words for "abslute" : "aaaa", "asas", "able"
+2 valid words for "absoryz" : "aaaa", "asas"
+4 valid words for "actresz" : "aaaa", "asas", "actt", "access"
+There're no valid words for "gaswxyz" cause none of the words in the list contains letter 'g'.
+
+```
+
+**Constraints:**
+
+- `1 <= words.length <= 10^5`
+- `4 <= words[i].length <= 50`
+- `1 <= puzzles.length <= 10^4`
+- `puzzles[i].length == 7`
+- `words[i][j]`, `puzzles[i][j]` are English lowercase letters.
+- Each `puzzles[i]` doesn't contain repeated characters.
+
+## 题目大意
+
+外国友人仿照中国字谜设计了一个英文版猜字谜小游戏，请你来猜猜看吧。
+
+字谜的迷面 puzzle 按字符串形式给出，如果一个单词 word 符合下面两个条件，那么它就可以算作谜底：
+
+- 单词 word 中包含谜面 puzzle 的第一个字母。
+- 单词 word 中的每一个字母都可以在谜面 puzzle 中找到。
+例如，如果字谜的谜面是 "abcdefg"，那么可以作为谜底的单词有 "faced", "cabbage", 和 "baggage"；而 "beefed"（不含字母 "a"）以及 "based"（其中的 "s" 没有出现在谜面中）都不能作为谜底。
+
+返回一个答案数组 answer，数组中的每个元素 answer[i] 是在给出的单词列表 words 中可以作为字谜迷面 puzzles[i] 所对应的谜底的单词数目。
+
+提示：
+
+- 1 <= words.length <= 10^5
+- 4 <= words[i].length <= 50
+- 1 <= puzzles.length <= 10^4
+- puzzles[i].length == 7
+- words[i][j], puzzles[i][j] 都是小写英文字母。
+- 每个 puzzles[i] 所包含的字符都不重复。
+
+## 解题思路
+
+- 首先题目中两个限制条件非常关键：**puzzles[i].length == 7**，**每个 puzzles[i] 所包含的字符都不重复**。也就是说穷举每个puzzle的子串的搜索空间就是2^7=128，而且不用考虑去重问题。
+- 因为谜底的判断只跟字符是否出现有关，跟字符的个数无关，另外都是小写的英文字母，所以可以用 `bitmap` 来表示单词(word)。
+- 利用 `map` 记录不同状态的单词(word)的个数。
+- 根据题意，如果某个单词(word)是某个字谜(puzzle)的谜底，那么 `word` 的 `bitmap` 肯定对应于 `puzzle` 某个子串的 `bitmap` 表示，且 `bitmap` 中包含 `puzzle` 的第一个字母的 `bit` 占用。
+- 问题就转换为：求每一个 `puzzle` 的每一个子串，然后求和这个子串具有相同 `bitmap` 表示且 `word` 中包含 `puzzle` 的第一个字母的 `word` 的个数。
+
+## 代码
+
+```go
+
+package leetcode
+
+/*
+	匹配跟单词中的字母顺序，字母个数都无关，可以用 bitmap 压缩
+	1. 记录 word 中 利用 map 记录各种 bit 标示的个数
+	2. puzzles 中各个字母都不相同! 记录 bitmap，然后搜索子空间中各种 bit 标识的个数的和
+	   因为 puzzles 长度最长是7，所以搜索空间 2^7
+*/
+func findNumOfValidWords(words []string, puzzles []string) []int {
+	wordBitStatusMap, res := make(map[uint32]int, 0), []int{}
+	for _, w := range words {
+		wordBitStatusMap[toBitMap([]byte(w))]++
+	}
+	for _, p := range puzzles {
+		var bitMap uint32
+		var totalNum int
+		bitMap |= (1 << (p[0] - 'a')) //work 中要包含 p 的第一个字母 所以这个 bit 位上必须是 1
+		findNum([]byte(p)[1:], bitMap, &totalNum, wordBitStatusMap)
+		res = append(res, totalNum)
+	}
+	return res
+}
+
+func toBitMap(word []byte) uint32 {
+	var res uint32
+	for _, b := range word {
+		res |= (1 << (b - 'a'))
+	}
+	return res
+}
+
+//利用 dfs 搜索 puzzles 的子空间
+func findNum(puzzles []byte, bitMap uint32, totalNum *int, m map[uint32]int) {
+	if len(puzzles) == 0 {
+		*totalNum = *totalNum + m[bitMap]
+		return
+	}
+	//不包含 puzzles[0],即 puzzles[0] 对应 bit 是 0
+	findNum(puzzles[1:], bitMap, totalNum, m)
+	//包含 puzzles[0],即 puzzles[0] 对应 bit 是 1
+	bitMap |= (1 << (puzzles[0] - 'a'))
+	findNum(puzzles[1:], bitMap, totalNum, m)
+	bitMap ^= (1 << (puzzles[0] - 'a')) //异或 清零
+	return
+}
+
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1175.Prime-Arrangements/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1184.Distance-Between-Bus-Stops/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1184.Distance-Between-Bus-Stops.md b/website/content/ChapterFour/1100~1199/1184.Distance-Between-Bus-Stops.md
new file mode 100755
index 000000000..6dbf09a27
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1184.Distance-Between-Bus-Stops.md
@@ -0,0 +1,89 @@
+# [1184. Distance Between Bus Stops](https://leetcode.com/problems/distance-between-bus-stops/)
+
+## 题目
+
+A bus has `n` stops numbered from `0` to `n - 1` that form a circle. We know the distance between all pairs of neighboring stops where `distance[i]` is the distance between the stops number `i` and `(i + 1) % n`.
+
+The bus goes along both directions i.e. clockwise and counterclockwise.
+
+Return the shortest distance between the given `start` and `destination` stops.
+
+**Example 1**:
+
+![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1.jpg)
+
+    Input: distance = [1,2,3,4], start = 0, destination = 1
+    Output: 1
+    Explanation: Distance between 0 and 1 is 1 or 9, minimum is 1.
+
+**Example 2**:
+
+![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1-1.jpg)
+
+    Input: distance = [1,2,3,4], start = 0, destination = 2
+    Output: 3
+    Explanation: Distance between 0 and 2 is 3 or 7, minimum is 3.
+
+**Example 3**:
+
+![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1-2.jpg)
+
+    Input: distance = [1,2,3,4], start = 0, destination = 3
+    Output: 4
+    Explanation: Distance between 0 and 3 is 6 or 4, minimum is 4.
+
+**Constraints**:
+
+- `1 <= n <= 10^4`
+- `distance.length == n`
+- `0 <= start, destination < n`
+- `0 <= distance[i] <= 10^4`
+
+
+## 题目大意
+
+环形公交路线上有 n 个站，按次序从 0 到 n - 1 进行编号。我们已知每一对相邻公交站之间的距离，distance[i] 表示编号为 i 的车站和编号为 (i + 1) % n 的车站之间的距离。环线上的公交车都可以按顺时针和逆时针的方向行驶。返回乘客从出发点 start 到目的地 destination 之间的最短距离。
+
+提示：
+
+- 1 <= n <= 10^4
+- distance.length == n
+- 0 <= start, destination < n
+- 0 <= distance[i] <= 10^4
+
+
+## 解题思路
+
+
+- 给出一个数组，代表的是公交车站每站直接的距离。距离是按照数组下标的顺序给出的，公交车可以按照顺时针行驶，也可以按照逆时针行驶。问行驶的最短距离是多少。
+- 按照题意，分别算出顺时针和逆时针的行驶距离，比较两者距离，取出小值就是结果。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func distanceBetweenBusStops(distance []int, start int, destination int) int {
+	clockwiseDis, counterclockwiseDis, n := 0, 0, len(distance)
+	for i := start; i != destination; i = (i + 1) % n {
+		clockwiseDis += distance[i]
+	}
+	for i := destination; i != start; i = (i + 1) % n {
+		counterclockwiseDis += distance[i]
+	}
+	if clockwiseDis < counterclockwiseDis {
+		return clockwiseDis
+	}
+	return counterclockwiseDis
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1178.Number-of-Valid-Words-for-Each-Puzzle/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1185.Day-of-the-Week/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1185.Day-of-the-Week.md b/website/content/ChapterFour/1100~1199/1185.Day-of-the-Week.md
new file mode 100755
index 000000000..e7489d101
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1185.Day-of-the-Week.md
@@ -0,0 +1,68 @@
+# [1185. Day of the Week](https://leetcode.com/problems/day-of-the-week/)
+
+
+## 题目
+
+Given a date, return the corresponding day of the week for that date.
+
+The input is given as three integers representing the `day`, `month` and `year` respectively.
+
+Return the answer as one of the following values `{"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}`.
+
+**Example 1**:
+
+    Input: day = 31, month = 8, year = 2019
+    Output: "Saturday"
+
+**Example 2**:
+
+    Input: day = 18, month = 7, year = 1999
+    Output: "Sunday"
+
+**Example 3**:
+
+    Input: day = 15, month = 8, year = 1993
+    Output: "Sunday"
+
+**Constraints**:
+
+- The given dates are valid dates between the years `1971` and `2100`.
+
+
+## 题目大意
+
+给你一个日期，请你设计一个算法来判断它是对应一周中的哪一天。输入为三个整数：day、month 和 year，分别表示日、月、年。
+
+您返回的结果必须是这几个值中的一个 {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}。
+
+提示：
+
+- 给出的日期一定是在 1971 到 2100 年之间的有效日期。
+
+## 解题思路
+
+
+- 给出一个日期，要求算出这一天是星期几。
+- 简单题，按照常识计算即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "time"
+
+func dayOfTheWeek(day int, month int, year int) string {
+	return time.Date(year, time.Month(month), day, 0, 0, 0, 0, time.Local).Weekday().String()
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1184.Distance-Between-Bus-Stops/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1189.Maximum-Number-of-Balloons/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1189.Maximum-Number-of-Balloons.md b/website/content/ChapterFour/1100~1199/1189.Maximum-Number-of-Balloons.md
new file mode 100755
index 000000000..ed9b096d8
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1189.Maximum-Number-of-Balloons.md
@@ -0,0 +1,77 @@
+# [1189. Maximum Number of Balloons](https://leetcode.com/problems/maximum-number-of-balloons/)
+
+
+## 题目
+
+Given a string `text`, you want to use the characters of `text` to form as many instances of the word **"balloon"** as possible.
+
+You can use each character in `text` **at most once**. Return the maximum number of instances that can be formed.
+
+**Example 1**:
+
+![](https://assets.leetcode.com/uploads/2019/09/05/1536_ex1_upd.JPG)
+
+    Input: text = "nlaebolko"
+    Output: 1
+
+**Example 2**:
+
+![](https://assets.leetcode.com/uploads/2019/09/05/1536_ex2_upd.JPG)
+
+    Input: text = "loonbalxballpoon"
+    Output: 2
+
+**Example 3**:
+
+    Input: text = "leetcode"
+    Output: 0
+
+**Constraints**:
+
+- `1 <= text.length <= 10^4`
+- `text` consists of lower case English letters only.
+
+
+## 题目大意
+
+给你一个字符串 text，你需要使用 text 中的字母来拼凑尽可能多的单词 "balloon"（气球）。字符串 text 中的每个字母最多只能被使用一次。请你返回最多可以拼凑出多少个单词 "balloon"。
+
+提示：
+
+- 1 <= text.length <= 10^4
+- text 全部由小写英文字母组成
+
+## 解题思路
+
+
+- 给出一个字符串，问这个字符串里面的数组能组成多少个 **balloon** 这个单词。
+- 简单题，先统计 26 个字母每个字母的频次，然后取出 balloon 这 5 个字母出现频次最小的值就是结果。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func maxNumberOfBalloons(text string) int {
+	fre := make([]int, 26)
+	for _, t := range text {
+		fre[t-'a']++
+	}
+	// 字符 b 的频次是数组下标 1 对应的元素值
+	// 字符 a 的频次是数组下标 0 对应的元素值
+	// 字符 l 的频次是数组下标 11 对应的元素值，这里有 2 个 l，所以元素值需要除以 2
+	// 字符 o 的频次是数组下标 14 对应的元素值，这里有 2 个 o，所以元素值需要除以 2
+	// 字符 n 的频次是数组下标 13 对应的元素值
+	return min(fre[1], min(fre[0], min(fre[11]/2, min(fre[14]/2, fre[13]))))
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1185.Day-of-the-Week/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses.md b/website/content/ChapterFour/1100~1199/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses.md
new file mode 100644
index 000000000..0e9fdafc6
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses.md
@@ -0,0 +1,91 @@
+# [1190. Reverse Substrings Between Each Pair of Parentheses](https://leetcode.com/problems/reverse-substrings-between-each-pair-of-parentheses/)
+
+
+## 题目
+
+You are given a string `s` that consists of lower case English letters and brackets.
+
+Reverse the strings in each pair of matching parentheses, starting from the innermost one.
+
+Your result should **not** contain any brackets.
+
+**Example 1:**
+
+```
+Input: s = "(abcd)"
+Output: "dcba"
+```
+
+**Example 2:**
+
+```
+Input: s = "(u(love)i)"
+Output: "iloveu"
+Explanation: The substring "love" is reversed first, then the whole string is reversed.
+```
+
+**Example 3:**
+
+```
+Input: s = "(ed(et(oc))el)"
+Output: "leetcode"
+Explanation: First, we reverse the substring "oc", then "etco", and finally, the whole string.
+```
+
+**Example 4:**
+
+```
+Input: s = "a(bcdefghijkl(mno)p)q"
+Output: "apmnolkjihgfedcbq"
+```
+
+**Constraints:**
+
+- `0 <= s.length <= 2000`
+- `s` only contains lower case English characters and parentheses.
+- It's guaranteed that all parentheses are balanced.
+
+## 题目大意
+
+给出一个字符串 s（仅含有小写英文字母和括号）。请你按照从括号内到外的顺序，逐层反转每对匹配括号中的字符串，并返回最终的结果。注意，您的结果中 不应 包含任何括号。
+
+## 解题思路
+
+- 本题最容易想到的思路是利用栈将每对括号里面的字符串入栈，当遇到 ")" 括号时出栈并逆序。由于用到了栈的数据结构，多层括号嵌套的问题也不用担心。这种边入栈出栈，逆序字符串的方法，时间复杂度是 O(n^2)，有没有可能进一步降低时间复杂度呢？
+- 上述解法中，存在重复遍历的情况。扫描原字符串的时候，入栈出栈已经扫描了一次，在 ")" 括号出栈时，逆序又会扫一遍已经入栈的字符串。这部分重复遍历的过程可以优化掉。第一次循环先标记出逆序区间。例如遇到 "(" 的时候，入栈并记录下它的下标，当遇到 ")" 的时候，意味着这一对括号匹配上了，所以将 ")" 的下标和之前入栈 "(" 的下标交换。此次遍历将逆序区间标记出来了。再遍历一次，根据逆序区间逆序字符串。不在逆序区间的字符串正常 append。如果在逆序区间内的，逆序遍历，添加到最终结果字符串中。这样做，时间复杂度仅为 O(n)。具体实现见下面代码。
+
+## 代码
+
+```go
+package leetcode
+
+func reverseParentheses(s string) string {
+	pair, stack := make([]int, len(s)), []int{}
+	for i, b := range s {
+		if b == '(' {
+			stack = append(stack, i)
+		} else if b == ')' {
+			j := stack[len(stack)-1]
+			stack = stack[:len(stack)-1]
+			pair[i], pair[j] = j, i
+		}
+	}
+	res := []byte{}
+	for i, step := 0, 1; i < len(s); i += step {
+		if s[i] == '(' || s[i] == ')' {
+			i = pair[i]
+			step = -step
+		} else {
+			res = append(res, s[i])
+		}
+	}
+	return string(res)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1189.Maximum-Number-of-Balloons/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1200.Minimum-Absolute-Difference/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1100~1199/_index.md b/website/content/ChapterFour/1100~1199/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/1100~1199/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/1200~1299/1200.Minimum-Absolute-Difference.md b/website/content/ChapterFour/1200~1299/1200.Minimum-Absolute-Difference.md
new file mode 100755
index 000000000..cfc42996d
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1200.Minimum-Absolute-Difference.md
@@ -0,0 +1,84 @@
+# [1200. Minimum Absolute Difference](https://leetcode.com/problems/minimum-absolute-difference/)
+
+
+## 题目
+
+Given an array of **distinct** integers `arr`, find all pairs of elements with the minimum absolute difference of any two elements.
+
+Return a list of pairs in ascending order(with respect to pairs), each pair `[a, b]` follows
+
+- `a, b` are from `arr`
+- `a < b`
+- `b - a` equals to the minimum absolute difference of any two elements in `arr`
+
+**Example 1**:
+
+    Input: arr = [4,2,1,3]
+    Output: [[1,2],[2,3],[3,4]]
+    Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order.
+
+**Example 2**:
+
+    Input: arr = [1,3,6,10,15]
+    Output: [[1,3]]
+
+**Example 3**:
+
+    Input: arr = [3,8,-10,23,19,-4,-14,27]
+    Output: [[-14,-10],[19,23],[23,27]]
+
+**Constraints**:
+
+- `2 <= arr.length <= 10^5`
+- `-10^6 <= arr[i] <= 10^6`
+
+
+## 题目大意
+
+给出一个数组，要求找出所有满足条件的数值对 [a,b]：`a<b` 并且 `b-a` 的差值是数组中所有两个元素差值的最小值。
+
+## 解题思路
+
+
+- 给出一个数组，要求找出所有满足条件的数值对 [a,b]：`a<b` 并且 `b-a` 的差值是数组中所有两个元素差值的最小值。
+- 简单题，按照题意先排序，然后依次求出两个相邻元素的差值，求出最小的差值。最后遍历一遍数组，把所有等于最小差值的数值对都输出。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func minimumAbsDifference(arr []int) [][]int {
+	minDiff, res := math.MaxInt32, [][]int{}
+	sort.Ints(arr)
+	for i := 1; i < len(arr); i++ {
+		if arr[i]-arr[i-1] < minDiff {
+			minDiff = arr[i] - arr[i-1]
+		}
+		if minDiff == 1 {
+			break
+		}
+	}
+	for i := 1; i < len(arr); i++ {
+		if arr[i]-arr[i-1] == minDiff {
+			res = append(res, []int{arr[i-1], arr[i]})
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1100~1199/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1201.Ugly-Number-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1201.Ugly-Number-III.md b/website/content/ChapterFour/1200~1299/1201.Ugly-Number-III.md
new file mode 100755
index 000000000..a4a3299d5
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1201.Ugly-Number-III.md
@@ -0,0 +1,99 @@
+# [1201. Ugly Number III](https://leetcode.com/problems/ugly-number-iii/)
+
+
+## 题目
+
+Write a program to find the `n`-th ugly number.
+
+Ugly numbers are **positive integers** which are divisible by `a` **or** `b` **or** `c`.
+
+**Example 1**:
+
+    Input: n = 3, a = 2, b = 3, c = 5
+    Output: 4
+    Explanation: The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3rd is 4.
+
+**Example 2**:
+
+    Input: n = 4, a = 2, b = 3, c = 4
+    Output: 6
+    Explanation: The ugly numbers are 2, 3, 4, 6, 8, 9, 10, 12... The 4th is 6.
+
+**Example 3**:
+
+    Input: n = 5, a = 2, b = 11, c = 13
+    Output: 10
+    Explanation: The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5th is 10.
+
+**Example 4**:
+
+    Input: n = 1000000000, a = 2, b = 217983653, c = 336916467
+    Output: 1999999984
+
+**Constraints**:
+
+- `1 <= n, a, b, c <= 10^9`
+- `1 <= a * b * c <= 10^18`
+- It's guaranteed that the result will be in range `[1, 2 * 10^9]`
+
+
+## 题目大意
+
+
+请你帮忙设计一个程序，用来找出第 n 个丑数。丑数是可以被 a 或 b 或 c 整除的 正整数。
+
+
+提示：
+
+- 1 <= n, a, b, c <= 10^9
+- 1 <= a * b * c <= 10^18
+- 本题结果在 [1, 2 * 10^9] 的范围内
+
+## 解题思路
+
+
+- 给出 4 个数字，a，b，c，n。要求输出可以整除 a 或者整除 b 或者整除 c 的第 n 个数。
+- 这一题限定了解的范围， `[1, 2 * 10^9]`，所以直接二分搜索来求解。逐步二分逼近 low 值，直到找到能满足条件的 low 的最小值，即为最终答案。
+- 这一题的关键在如何判断一个数是第几个数。一个数能整除 a，能整除 b，能整除 c，那么它应该是第 `num/a + num/b + num/c - num/ab - num/bc - num/ac + num/abc` 个数。这个就是韦恩图。需要注意的是，求 `ab`、`bc`、`ac`、`abc` 的时候需要再除以各自的最大公约数 `gcd()`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func nthUglyNumber(n int, a int, b int, c int) int {
+	low, high := int64(0), int64(2*1e9)
+	for low < high {
+		mid := low + (high-low)>>1
+		if calNthCount(mid, int64(a), int64(b), int64(c)) < int64(n) {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return int(low)
+}
+
+func calNthCount(num, a, b, c int64) int64 {
+	ab, bc, ac := a*b/gcd(a, b), b*c/gcd(b, c), a*c/gcd(a, c)
+	abc := a * bc / gcd(a, bc)
+	return num/a + num/b + num/c - num/ab - num/bc - num/ac + num/abc
+}
+
+func gcd(a, b int64) int64 {
+	for b != 0 {
+		a, b = b, a%b
+	}
+	return a
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1200.Minimum-Absolute-Difference/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1202.Smallest-String-With-Swaps/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1202.Smallest-String-With-Swaps.md b/website/content/ChapterFour/1200~1299/1202.Smallest-String-With-Swaps.md
new file mode 100755
index 000000000..da9f6cc3b
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1202.Smallest-String-With-Swaps.md
@@ -0,0 +1,109 @@
+# [1202. Smallest String With Swaps](https://leetcode.com/problems/smallest-string-with-swaps/)
+
+
+## 题目
+
+You are given a string `s`, and an array of pairs of indices in the string `pairs` where `pairs[i] = [a, b]` indicates 2 indices(0-indexed) of the string.
+
+You can swap the characters at any pair of indices in the given `pairs` **any number of times**.
+
+Return the lexicographically smallest string that `s` can be changed to after using the swaps.
+
+**Example 1**:
+
+    Input: s = "dcab", pairs = [[0,3],[1,2]]
+    Output: "bacd"
+    Explaination: 
+    Swap s[0] and s[3], s = "bcad"
+    Swap s[1] and s[2], s = "bacd"
+
+**Example 2**:
+
+    Input: s = "dcab", pairs = [[0,3],[1,2],[0,2]]
+    Output: "abcd"
+    Explaination: 
+    Swap s[0] and s[3], s = "bcad"
+    Swap s[0] and s[2], s = "acbd"
+    Swap s[1] and s[2], s = "abcd"
+
+**Example 3**:
+
+    Input: s = "cba", pairs = [[0,1],[1,2]]
+    Output: "abc"
+    Explaination: 
+    Swap s[0] and s[1], s = "bca"
+    Swap s[1] and s[2], s = "bac"
+    Swap s[0] and s[1], s = "abc"
+
+**Constraints**:
+
+- `1 <= s.length <= 10^5`
+- `0 <= pairs.length <= 10^5`
+- `0 <= pairs[i][0], pairs[i][1] < s.length`
+- `s` only contains lower case English letters.
+
+
+## 题目大意
+
+给你一个字符串 s，以及该字符串中的一些「索引对」数组 pairs，其中 pairs[i] = [a, b] 表示字符串中的两个索引（编号从 0 开始）。你可以 任意多次交换 在 pairs 中任意一对索引处的字符。返回在经过若干次交换后，s 可以变成的按字典序最小的字符串。
+
+提示：
+
+- 1 <= s.length <= 10^5
+- 0 <= pairs.length <= 10^5
+- 0 <= pairs[i][0], pairs[i][1] < s.length
+- s 中只含有小写英文字母
+
+
+
+## 解题思路
+
+
+- 给出一个字符串和一个字符串里可交换的下标。要求交换以后字典序最小的字符串。
+- 这一题可以用并查集来解题，先把可交换下标都 `Union()` 起来，每个集合内，按照字典序从小到大排列。最后扫描原有字符串，从左到右依次找到各自对应的集合里面最小的字符进行替换，每次替换完以后，删除集合中该字符(防止下次重复替换)。最终得到的字符就是最小字典序的字符。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func smallestStringWithSwaps(s string, pairs [][]int) string {
+	uf, res, sMap := template.UnionFind{}, []byte(s), map[int][]byte{}
+	uf.Init(len(s))
+	for _, pair := range pairs {
+		uf.Union(pair[0], pair[1])
+	}
+	for i := 0; i < len(s); i++ {
+		r := uf.Find(i)
+		sMap[r] = append(sMap[r], s[i])
+	}
+	for _, v := range sMap {
+		sort.Slice(v, func(i, j int) bool {
+			return v[i] < v[j]
+		})
+	}
+	for i := 0; i < len(s); i++ {
+		r := uf.Find(i)
+		bytes := sMap[r]
+		res[i] = bytes[0]
+		sMap[r] = bytes[1:len(bytes)]
+	}
+	return string(res)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1201.Ugly-Number-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1203.Sort-Items-by-Groups-Respecting-Dependencies/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1203.Sort-Items-by-Groups-Respecting-Dependencies.md b/website/content/ChapterFour/1200~1299/1203.Sort-Items-by-Groups-Respecting-Dependencies.md
new file mode 100644
index 000000000..6cbdc8ced
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1203.Sort-Items-by-Groups-Respecting-Dependencies.md
@@ -0,0 +1,198 @@
+# [1203. Sort Items by Groups Respecting Dependencies](https://leetcode.com/problems/sort-items-by-groups-respecting-dependencies/)
+
+
+## 题目
+
+There are `n` items each belonging to zero or one of `m` groups where `group[i]` is the group that the `i`-th item belongs to and it's equal to `-1` if the `i`-th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it.
+
+Return a sorted list of the items such that:
+
+- The items that belong to the same group are next to each other in the sorted list.
+- There are some relations between these items where `beforeItems[i]` is a list containing all the items that should come before the `i`th item in the sorted array (to the left of the `i`th item).
+
+Return any solution if there is more than one solution and return an **empty list** if there is no solution.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2019/09/11/1359_ex1.png)
+
+```
+Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]]
+Output: [6,3,4,1,5,2,0,7]
+
+```
+
+**Example 2:**
+
+```
+Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]]
+Output: []
+Explanation: This is the same as example 1 except that 4 needs to be before 6 in the sorted list.
+
+```
+
+**Constraints:**
+
+- `1 <= m <= n <= 3 * 104`
+- `group.length == beforeItems.length == n`
+- `1 <= group[i] <= m - 1`
+- `0 <= beforeItems[i].length <= n - 1`
+- `0 <= beforeItems[i][j] <= n - 1`
+- `i != beforeItems[i][j]`
+- `beforeItems[i]` does not contain duplicates elements.
+
+## 题目大意
+
+有 n 个项目，每个项目或者不属于任何小组，或者属于 m 个小组之一。group[i] 表示第 i 个小组所属的小组，如果第 i 个项目不属于任何小组，则 group[i] 等于 -1。项目和小组都是从零开始编号的。可能存在小组不负责任何项目，即没有任何项目属于这个小组。
+
+请你帮忙按要求安排这些项目的进度，并返回排序后的项目列表：
+
+- 同一小组的项目，排序后在列表中彼此相邻。
+- 项目之间存在一定的依赖关系，我们用一个列表 beforeItems 来表示，其中 beforeItems[i] 表示在进行第 i 个项目前（位于第 i 个项目左侧）应该完成的所有项目。
+
+如果存在多个解决方案，只需要返回其中任意一个即可。如果没有合适的解决方案，就请返回一个 空列表 。
+
+## 解题思路
+
+- 读完题能确定这一题是拓扑排序。但是和单纯的拓扑排序有区别的是，同一小组内的项目需要彼此相邻。用 2 次拓扑排序即可解决。第一次拓扑排序排出组间的顺序，第二次拓扑排序排出组内的顺序。为了实现方便，用 map 给虚拟分组标记编号。如下图，将 3，4，6 三个任务打包到 0 号分组里面，将 2，5 两个任务打包到 1 号分组里面，其他任务单独各自为一组。组间的依赖是 6 号任务依赖 1 号任务。由于 6 号任务封装在 0 号分组里，所以 3 号分组依赖 0 号分组。先组间排序，确定分组顺序，再组内拓扑排序，排出最终顺序。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_1203_1.png)
+
+- 上面的解法可以 AC，但是时间太慢了。因为做了一些不必要的操作。有没有可能只用一次拓扑排序呢？将必须要在一起的结点统一依赖一个虚拟结点，例如下图中的虚拟结点 8 和 9 。3，4，6 都依赖 8 号任务，2 和 5 都依赖 9 号任务。1 号任务本来依赖 6 号任务，由于 6 由依赖 8 ，所以添加 1 依赖 8 的边。通过增加虚拟结点，增加了需要打包在一起结点的入度。构建出以上关系以后，按照入度为 0 的原则，依次进行 DFS。8 号和 9 号两个虚拟结点的入度都为 0 ，对它们进行 DFS，必定会使得与它关联的节点都被安排在一起，这样就满足了题意：同一小组的项目，排序后在列表中彼此相邻。一遍扫完，满足题意的顺序就排出来了。这个解法 beat 100%！
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_1203_2.png)
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 拓扑排序版的 DFS
+func sortItems(n int, m int, group []int, beforeItems [][]int) []int {
+	groups, inDegrees := make([][]int, n+m), make([]int, n+m)
+	for i, g := range group {
+		if g > -1 {
+			g += n
+			groups[g] = append(groups[g], i)
+			inDegrees[i]++
+		}
+	}
+	for i, ancestors := range beforeItems {
+		gi := group[i]
+		if gi == -1 {
+			gi = i
+		} else {
+			gi += n
+		}
+		for _, ancestor := range ancestors {
+			ga := group[ancestor]
+			if ga == -1 {
+				ga = ancestor
+			} else {
+				ga += n
+			}
+			if gi == ga {
+				groups[ancestor] = append(groups[ancestor], i)
+				inDegrees[i]++
+			} else {
+				groups[ga] = append(groups[ga], gi)
+				inDegrees[gi]++
+			}
+		}
+	}
+	res := []int{}
+	for i, d := range inDegrees {
+		if d == 0 {
+			sortItemsDFS(i, n, &res, &inDegrees, &groups)
+		}
+	}
+	if len(res) != n {
+		return nil
+	}
+	return res
+}
+
+func sortItemsDFS(i, n int, res, inDegrees *[]int, groups *[][]int) {
+	if i < n {
+		*res = append(*res, i)
+	}
+	(*inDegrees)[i] = -1
+	for _, ch := range (*groups)[i] {
+		if (*inDegrees)[ch]--; (*inDegrees)[ch] == 0 {
+			sortItemsDFS(ch, n, res, inDegrees, groups)
+		}
+	}
+}
+
+// 解法二 二维拓扑排序 时间复杂度 O(m+n)，空间复杂度 O(m+n)
+func sortItems1(n int, m int, group []int, beforeItems [][]int) []int {
+	groupItems, res := map[int][]int{}, []int{}
+	for i := 0; i < len(group); i++ {
+		if group[i] == -1 {
+			group[i] = m + i
+		}
+		groupItems[group[i]] = append(groupItems[group[i]], i)
+	}
+	groupGraph, groupDegree, itemGraph, itemDegree := make([][]int, m+n), make([]int, m+n), make([][]int, n), make([]int, n)
+	for i := 0; i < len(beforeItems); i++ {
+		for j := 0; j < len(beforeItems[i]); j++ {
+			if group[beforeItems[i][j]] != group[i] {
+				// 不同组项目，确定组间依赖关系
+				groupGraph[group[beforeItems[i][j]]] = append(groupGraph[group[beforeItems[i][j]]], group[i])
+				groupDegree[group[i]]++
+			} else {
+				// 同组项目，确定组内依赖关系
+				itemGraph[beforeItems[i][j]] = append(itemGraph[beforeItems[i][j]], i)
+				itemDegree[i]++
+			}
+		}
+	}
+	items := []int{}
+	for i := 0; i < m+n; i++ {
+		items = append(items, i)
+	}
+	// 组间拓扑
+	groupOrders := topSort(groupGraph, groupDegree, items)
+	if len(groupOrders) < len(items) {
+		return nil
+	}
+	for i := 0; i < len(groupOrders); i++ {
+		items := groupItems[groupOrders[i]]
+		// 组内拓扑
+		orders := topSort(itemGraph, itemDegree, items)
+		if len(orders) < len(items) {
+			return nil
+		}
+		res = append(res, orders...)
+	}
+	return res
+}
+
+func topSort(graph [][]int, deg, items []int) (orders []int) {
+	q := []int{}
+	for _, i := range items {
+		if deg[i] == 0 {
+			q = append(q, i)
+		}
+	}
+	for len(q) > 0 {
+		from := q[0]
+		q = q[1:]
+		orders = append(orders, from)
+		for _, to := range graph[from] {
+			deg[to]--
+			if deg[to] == 0 {
+				q = append(q, to)
+			}
+		}
+	}
+	return
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1202.Smallest-String-With-Swaps/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1207.Unique-Number-of-Occurrences/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1207.Unique-Number-of-Occurrences.md b/website/content/ChapterFour/1200~1299/1207.Unique-Number-of-Occurrences.md
new file mode 100755
index 000000000..acd4c99da
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1207.Unique-Number-of-Occurrences.md
@@ -0,0 +1,75 @@
+# [1207. Unique Number of Occurrences](https://leetcode.com/problems/unique-number-of-occurrences/)
+
+
+## 题目
+
+Given an array of integers `arr`, write a function that returns `true` if and only if the number of occurrences of each value in the array is unique.
+
+**Example 1**:
+
+    Input: arr = [1,2,2,1,1,3]
+    Output: true
+    Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences.
+
+**Example 2**:
+
+    Input: arr = [1,2]
+    Output: false
+
+**Example 3**:
+
+    Input: arr = [-3,0,1,-3,1,1,1,-3,10,0]
+    Output: true
+
+**Constraints**:
+
+- `1 <= arr.length <= 1000`
+- `-1000 <= arr[i] <= 1000`
+
+
+
+## 题目大意
+
+给你一个整数数组 arr，请你帮忙统计数组中每个数的出现次数。如果每个数的出现次数都是独一无二的，就返回 true；否则返回 false。
+
+提示：
+
+- 1 <= arr.length <= 1000
+- -1000 <= arr[i] <= 1000
+
+## 解题思路
+
+
+- 给出一个数组，先统计每个数字出现的频次，判断在这个数组中是否存在相同的频次。
+- 简单题，先统计数组中每个数字的频次，然后用一个 map 判断频次是否重复。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func uniqueOccurrences(arr []int) bool {
+	freq, m := map[int]int{}, map[int]bool{}
+	for _, v := range arr {
+		freq[v]++
+	}
+	for _, v := range freq {
+		if _, ok := m[v]; !ok {
+			m[v] = true
+		} else {
+			return false
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1203.Sort-Items-by-Groups-Respecting-Dependencies/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1208.Get-Equal-Substrings-Within-Budget/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1208.Get-Equal-Substrings-Within-Budget.md b/website/content/ChapterFour/1200~1299/1208.Get-Equal-Substrings-Within-Budget.md
new file mode 100755
index 000000000..24435f65d
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1208.Get-Equal-Substrings-Within-Budget.md
@@ -0,0 +1,84 @@
+# [1208. Get Equal Substrings Within Budget](https://leetcode.com/problems/get-equal-substrings-within-budget/)
+
+
+## 题目
+
+You are given two strings `s` and `t` of the same length. You want to change `s` to `t`. Changing the `i`-th character of `s` to `i`-th character of `t` costs `|s[i] - t[i]|` that is, the absolute difference between the ASCII values of the characters.
+
+You are also given an integer `maxCost`.
+
+Return the maximum length of a substring of `s` that can be changed to be the same as the corresponding substring of `t`with a cost less than or equal to `maxCost`.
+
+If there is no substring from `s` that can be changed to its corresponding substring from `t`, return `0`.
+
+**Example 1**:
+
+    Input: s = "abcd", t = "bcdf", maxCost = 3
+    Output: 3
+    Explanation: "abc" of s can change to "bcd". That costs 3, so the maximum length is 3.
+
+**Example 2**:
+
+    Input: s = "abcd", t = "cdef", maxCost = 3
+    Output: 1
+    Explanation: Each character in s costs 2 to change to charactor in t, so the maximum length is 1.
+
+**Example 3**:
+
+    Input: s = "abcd", t = "acde", maxCost = 0
+    Output: 1
+    Explanation: You can't make any change, so the maximum length is 1.
+
+**Constraints**:
+
+- `1 <= s.length, t.length <= 10^5`
+- `0 <= maxCost <= 10^6`
+- `s` and `t` only contain lower case English letters.
+
+## 题目大意
+
+给你两个长度相同的字符串，s 和 t。将 s 中的第 i 个字符变到 t 中的第 i 个字符需要 |s[i] - t[i]| 的开销（开销可能为 0），也就是两个字符的 ASCII 码值的差的绝对值。
+
+用于变更字符串的最大预算是 maxCost。在转化字符串时，总开销应当小于等于该预算，这也意味着字符串的转化可能是不完全的。如果你可以将 s 的子字符串转化为它在 t 中对应的子字符串，则返回可以转化的最大长度。如果 s 中没有子字符串可以转化成 t 中对应的子字符串，则返回 0。
+
+提示：
+
+- 1 <= s.length, t.length <= 10^5
+- 0 <= maxCost <= 10^6
+- s 和 t 都只含小写英文字母。
+
+## 解题思路
+
+- 给出 2 个字符串 `s` 和 `t` 和一个“预算”，要求把“预算”尽可能的花完，`s` 中最多连续有几个字母能变成 `t` 中的字母。“预算”的定义是：|s[i] - t[i]| 。
+- 这一题是滑动窗口的题目，滑动窗口右边界每移动一格，就减少一定的预算，直到预算不能减少，再移动滑动窗口的左边界，这个时候注意要把预算还原回去。当整个窗口把字符 `s` 或 `t` 都滑动完了的时候，取出滑动过程中窗口的最大值即为结果。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func equalSubstring(s string, t string, maxCost int) int {
+	left, right, res := 0, -1, 0
+	for left < len(s) {
+		if right+1 < len(s) && maxCost-abs(int(s[right+1]-'a')-int(t[right+1]-'a')) >= 0 {
+			right++
+			maxCost -= abs(int(s[right]-'a') - int(t[right]-'a'))
+		} else {
+			res = max(res, right-left+1)
+			maxCost += abs(int(s[left]-'a') - int(t[left]-'a'))
+			left++
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1207.Unique-Number-of-Occurrences/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1209.Remove-All-Adjacent-Duplicates-in-String-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1209.Remove-All-Adjacent-Duplicates-in-String-II.md b/website/content/ChapterFour/1200~1299/1209.Remove-All-Adjacent-Duplicates-in-String-II.md
new file mode 100644
index 000000000..6a59f20e9
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1209.Remove-All-Adjacent-Duplicates-in-String-II.md
@@ -0,0 +1,113 @@
+# [1209. Remove All Adjacent Duplicates in String II](https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/)
+
+
+## 题目
+
+Given a string `s`, a *k* *duplicate removal* consists of choosing `k` adjacent and equal letters from `s` and removing them causing the left and the right side of the deleted substring to concatenate together.
+
+We repeatedly make `k` duplicate removals on `s` until we no longer can.
+
+Return the final string after all such duplicate removals have been made.
+
+It is guaranteed that the answer is unique.
+
+**Example 1:**
+
+```
+Input: s = "abcd", k = 2
+Output: "abcd"
+Explanation:There's nothing to delete.
+```
+
+**Example 2:**
+
+```
+Input: s = "deeedbbcccbdaa", k = 3
+Output: "aa"
+Explanation:
+First delete "eee" and "ccc", get "ddbbbdaa"
+Then delete "bbb", get "dddaa"
+Finally delete "ddd", get "aa"
+```
+
+**Example 3:**
+
+```
+Input: s = "pbbcggttciiippooaais", k = 2
+Output: "ps"
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 10^5`
+- `2 <= k <= 10^4`
+- `s` only contains lower case English letters.
+
+## 题目大意
+
+给你一个字符串 s，「k 倍重复项删除操作」将会从 s 中选择 k 个相邻且相等的字母，并删除它们，使被删去的字符串的左侧和右侧连在一起。你需要对 s 重复进行无限次这样的删除操作，直到无法继续为止。在执行完所有删除操作后，返回最终得到的字符串。本题答案保证唯一。
+
+## 解题思路
+
+- 暴力解法。每增加一个字符，就往前扫描 `k` 位，判断是否存在 `k` 个连续相同的字符。消除了 `k` 个相同字符后，重新组成的字符串还可能再次产出 `k` 位相同的字符，（类似消消乐，`k` 个相同的字符碰到一起就“消除”），还需要继续消除。最差情况要再次扫描一次字符串。时间复杂度 O(n^2)，空间复杂度 O(n)。
+- 暴力解法的低效在于重复统计字符频次，如果每个字符的频次统计一次就好了。按照这个思路，利用 stack ，每个栈元素存 2 个值，一个是字符，一个是该字符对应的频次。有了栈顶字符频次信息，就不需要重复往前扫描了。只要栈顶字符频次到达了 `k`，就弹出该字符。如此反复，最终剩下的字符串为所求。时间复杂度 O(n)，空间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 stack
+func removeDuplicates(s string, k int) string {
+	stack, arr := [][2]int{}, []byte{}
+	for _, c := range s {
+		i := int(c - 'a')
+		if len(stack) > 0 && stack[len(stack)-1][0] == i {
+			stack[len(stack)-1][1]++
+			if stack[len(stack)-1][1] == k {
+				stack = stack[:len(stack)-1]
+			}
+		} else {
+			stack = append(stack, [2]int{i, 1})
+		}
+	}
+	for _, pair := range stack {
+		c := byte(pair[0] + 'a')
+		for i := 0; i < pair[1]; i++ {
+			arr = append(arr, c)
+		}
+	}
+	return string(arr)
+}
+
+// 解法二 暴力
+func removeDuplicates1(s string, k int) string {
+	arr, count, tmp := []rune{}, 0, '#'
+	for _, v := range s {
+		arr = append(arr, v)
+		for len(arr) > 0 {
+			count = 0
+			tmp = arr[len(arr)-1]
+			for i := len(arr) - 1; i >= 0; i-- {
+				if arr[i] != tmp {
+					break
+				}
+				count++
+			}
+			if count == k {
+				arr = arr[:len(arr)-k]
+			} else {
+				break
+			}
+		}
+	}
+	return string(arr)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1208.Get-Equal-Substrings-Within-Budget/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position.md b/website/content/ChapterFour/1200~1299/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position.md
new file mode 100755
index 000000000..feea53ba6
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position.md
@@ -0,0 +1,83 @@
+# [1217. Minimum Cost to Move Chips to The Same Position](https://leetcode.com/problems/minimum-cost-to-move-chips-to-the-same-position/)
+
+
+## 题目
+
+There are some chips, and the i-th chip is at position `chips[i]`.
+
+You can perform any of the two following types of moves **any number of times** (possibly zero) **on any chip**:
+
+- Move the `i`-th chip by 2 units to the left or to the right with a cost of **0**.
+- Move the `i`-th chip by 1 unit to the left or to the right with a cost of **1**.
+
+There can be two or more chips at the same position initially.
+
+Return the minimum cost needed to move all the chips to the same position (any position).
+
+**Example 1**:
+
+    Input: chips = [1,2,3]
+    Output: 1
+    Explanation: Second chip will be moved to positon 3 with cost 1. First chip will be moved to position 3 with cost 0. Total cost is 1.
+
+**Example 2**:
+
+    Input: chips = [2,2,2,3,3]
+    Output: 2
+    Explanation: Both fourth and fifth chip will be moved to position two with cost 1. Total minimum cost will be 2.
+
+**Constraints**:
+
+- `1 <= chips.length <= 100`
+- `1 <= chips[i] <= 10^9`
+
+
+## 题目大意
+
+
+数轴上放置了一些筹码，每个筹码的位置存在数组 chips 当中。你可以对 任何筹码 执行下面两种操作之一（不限操作次数，0 次也可以）：
+
+- 将第 i 个筹码向左或者右移动 2 个单位，代价为 0。
+- 将第 i 个筹码向左或者右移动 1 个单位，代价为 1。
+
+最开始的时候，同一位置上也可能放着两个或者更多的筹码。返回将所有筹码移动到同一位置（任意位置）上所需要的最小代价。
+
+
+提示：
+
+- 1 <= chips.length <= 100
+- 1 <= chips[i] <= 10^9
+
+
+## 解题思路
+
+- 给出一个数组，数组的下标代表的是数轴上的坐标点，数组的元素代表的是砝码大小。砝码移动规则，左右移动 2 格，没有代价，左右移动 1 个，代价是 1 。问最终把砝码都移动到一个格子上，最小代价是多少。
+- 先解读砝码移动规则：偶数位置的到偶数位置的没有代价，奇数到奇数位置的没有代价。利用这个规则，我们可以把所有的砝码**无代价**的摞在一个奇数的位置上和一个偶数的位置上。这样我们只用关心这两个位置了。并且这两个位置可以连续在一起。最后一步即将相邻的这两摞砝码合并到一起。由于左右移动一个代价是 1，所以最小代价的操作是移动最少砝码的那一边。奇数位置上砝码少就移动奇数位置上的，偶数位置上砝码少就移动偶数位置上的。所以这道题解法变的异常简单，遍历一次数组，找到其中有多少个奇数和偶数位置的砝码，取其中比较少的，就是最终答案。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func minCostToMoveChips(chips []int) int {
+	odd, even := 0, 0
+	for _, c := range chips {
+		if c%2 == 0 {
+			even++
+		} else {
+			odd++
+		}
+	}
+	return min(odd, even)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1209.Remove-All-Adjacent-Duplicates-in-String-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1221.Split-a-String-in-Balanced-Strings/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1221.Split-a-String-in-Balanced-Strings.md b/website/content/ChapterFour/1200~1299/1221.Split-a-String-in-Balanced-Strings.md
new file mode 100755
index 000000000..d17358657
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1221.Split-a-String-in-Balanced-Strings.md
@@ -0,0 +1,80 @@
+# [1221. Split a String in Balanced Strings](https://leetcode.com/problems/split-a-string-in-balanced-strings/)
+
+
+## 题目
+
+Balanced strings are those who have equal quantity of 'L' and 'R' characters.
+
+Given a balanced string `s` split it in the maximum amount of balanced strings.
+
+Return the maximum amount of splitted balanced strings.
+
+**Example 1**:
+
+    Input: s = "RLRRLLRLRL"
+    Output: 4
+    Explanation: s can be split into "RL", "RRLL", "RL", "RL", each substring contains same number of 'L' and 'R'.
+
+**Example 2**:
+
+    Input: s = "RLLLLRRRLR"
+    Output: 3
+    Explanation: s can be split into "RL", "LLLRRR", "LR", each substring contains same number of 'L' and 'R'.
+
+**Example 3**:
+
+    Input: s = "LLLLRRRR"
+    Output: 1
+    Explanation: s can be split into "LLLLRRRR".
+
+**Constraints**:
+
+- `1 <= s.length <= 1000`
+- `s[i] = 'L' or 'R'`
+
+## 题目大意
+
+
+在一个「平衡字符串」中，'L' 和 'R' 字符的数量是相同的。给出一个平衡字符串 s，请你将它分割成尽可能多的平衡字符串。返回可以通过分割得到的平衡字符串的最大数量。
+
+提示：
+
+- 1 <= s.length <= 1000
+- s[i] = 'L' 或 'R'
+
+
+## 解题思路
+
+- 给出一个字符串，要求把这个字符串切成一些子串，这些子串中 R 和 L 的字符数是相等的。问能切成多少个满足条件的子串。
+- 这道题是简单题，按照题意模拟即可。从左往右扫，遇到 `R` 就加一，遇到 `L` 就减一，当计数是 `0` 的时候就是平衡的时候，就切割。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func balancedStringSplit(s string) int {
+	count, res := 0, 0
+	for _, r := range s {
+		if r == 'R' {
+			count++
+		} else {
+			count--
+		}
+		if count == 0 {
+			res++
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1232.Check-If-It-Is-a-Straight-Line/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1232.Check-If-It-Is-a-Straight-Line.md b/website/content/ChapterFour/1200~1299/1232.Check-If-It-Is-a-Straight-Line.md
new file mode 100755
index 000000000..07ad32d47
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1232.Check-If-It-Is-a-Straight-Line.md
@@ -0,0 +1,78 @@
+# [1232. Check If It Is a Straight Line](https://leetcode.com/problems/check-if-it-is-a-straight-line/)
+
+
+## 题目
+
+You are given an array `coordinates`, `coordinates[i] = [x, y]`, where `[x, y]` represents the coordinate of a point. Check if these points make a straight line in the XY plane.
+
+**Example 1**:
+
+![](https://img.halfrost.com/Leetcode/leetcode_1232_1.png)
+
+    Input: coordinates = [[1,2],[2,3],[3,4],[4,5],[5,6],[6,7]]
+    Output: true
+
+**Example 2**:
+
+
+![](https://img.halfrost.com/Leetcode/leetcode_1232_2.png)
+
+    Input: coordinates = [[1,1],[2,2],[3,4],[4,5],[5,6],[7,7]]
+    Output: false
+
+**Constraints**:
+
+- `2 <= coordinates.length <= 1000`
+- `coordinates[i].length == 2`
+- `-10^4 <= coordinates[i][0], coordinates[i][1] <= 10^4`
+- `coordinates` contains no duplicate point.
+
+## 题目大意
+
+
+在一个 XY 坐标系中有一些点，我们用数组 coordinates 来分别记录它们的坐标，其中 coordinates[i] = [x, y] 表示横坐标为 x、纵坐标为 y 的点。
+
+请你来判断，这些点是否在该坐标系中属于同一条直线上，是则返回 true，否则请返回 false。
+
+提示：
+
+- 2 <= coordinates.length <= 1000
+- coordinates[i].length == 2
+- -10^4 <= coordinates[i][0], coordinates[i][1] <= 10^4
+- coordinates 中不含重复的点
+
+
+
+## 解题思路
+
+- 给出一组坐标点，要求判断这些点是否在同一直线上。
+- 按照几何原理，依次计算这些点的斜率是否相等即可。斜率需要做除法，这里采用一个技巧是换成乘法。例如 `a/b = c/d` 换成乘法是 `a*d  = c*d`  。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func checkStraightLine(coordinates [][]int) bool {
+	dx0 := coordinates[1][0] - coordinates[0][0]
+	dy0 := coordinates[1][1] - coordinates[0][1]
+	for i := 1; i < len(coordinates)-1; i++ {
+		dx := coordinates[i+1][0] - coordinates[i][0]
+		dy := coordinates[i+1][1] - coordinates[i][1]
+		if dy*dx0 != dy0*dx { // check cross product
+			return false
+		}
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1221.Split-a-String-in-Balanced-Strings/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1234.Replace-the-Substring-for-Balanced-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1234.Replace-the-Substring-for-Balanced-String.md b/website/content/ChapterFour/1200~1299/1234.Replace-the-Substring-for-Balanced-String.md
new file mode 100755
index 000000000..50847fe1e
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1234.Replace-the-Substring-for-Balanced-String.md
@@ -0,0 +1,100 @@
+# [1234. Replace the Substring for Balanced String](https://leetcode.com/problems/replace-the-substring-for-balanced-string/)
+
+
+## 题目
+
+You are given a string containing only 4 kinds of characters `'Q',` `'W', 'E'` and `'R'`.
+
+A string is said to be **balanced** **if each of its characters appears `n/4` times where `n` is the length of the string.
+
+Return the minimum length of the substring that can be replaced with **any** other string of the same length to make the original string `s` **balanced**.
+
+Return 0 if the string is already **balanced**.
+
+**Example 1**:
+
+    Input: s = "QWER"
+    Output: 0
+    Explanation: s is already balanced.
+
+**Example 2**:
+
+    Input: s = "QQWE"
+    Output: 1
+    Explanation: We need to replace a 'Q' to 'R', so that "RQWE" (or "QRWE") is balanced.
+
+**Example 3**:
+
+    Input: s = "QQQW"
+    Output: 2
+    Explanation: We can replace the first "QQ" to "ER".
+
+**Example 4**:
+
+    Input: s = "QQQQ"
+    Output: 3
+    Explanation: We can replace the last 3 'Q' to make s = "QWER".
+
+**Constraints**:
+
+- `1 <= s.length <= 10^5`
+- `s.length` is a multiple of `4`
+- `s` contains only `'Q'`, `'W'`, `'E'` and `'R'`.
+
+## 题目大意
+
+
+有一个只含有 'Q', 'W', 'E', 'R' 四种字符，且长度为 n 的字符串。假如在该字符串中，这四个字符都恰好出现 n/4 次，那么它就是一个「平衡字符串」。给你一个这样的字符串 s，请通过「替换一个子串」的方式，使原字符串 s 变成一个「平衡字符串」。你可以用和「待替换子串」长度相同的 任何 其他字符串来完成替换。请返回待替换子串的最小可能长度。如果原字符串自身就是一个平衡字符串，则返回 0。
+
+提示：
+
+- 1 <= s.length <= 10^5
+- s.length 是 4 的倍数
+- s 中只含有 'Q', 'W', 'E', 'R' 四种字符
+
+
+
+## 解题思路
+
+- 给出一个字符串，要求输出把这个字符串变成“平衡字符串”的最小替换字符串的长度(替换只能替换一串，不能单个字母替换)。“平衡字符串”的定义是：字符串中，`‘Q’`，`‘W’`，`‘E’`，`‘R’`，出现的次数当且仅当只有 `len(s)/4` 次。
+- 这一题是滑动窗口的题目。先统计 4 个字母的频次并计算出 `k = len(s)/4` 。滑动窗口向右滑动一次，对应右窗口的那个字母频次减 1，直到滑到所有字母的频次都 `≤ k`  的地方停止。此时，窗口外的字母的频次都 `≤ k` 了。这是只要变换窗口内字符串即可。但是这个窗口内还可能包含本来频次就小于 `k`  的字母，如果能把它们剔除掉，窗口可以进一步的减少。所以继续移动左边界，试探移动完左边界以后，是否所有字母频次都 `≤ k`。在所有窗口移动过程中取出最小值，即为最终答案。
+- 举个例子：`"WQWRQQQW"`。`w` 有 3 个，`Q` 有 4 个，`R` 有 1 个，`E` 有 0 个。最后平衡状态是每个字母 2 个，那么我们需要拿出 1 个 `W` 和 2 个 `Q` 替换掉。即要找到一个最短的字符串包含 1 个 `W` 和 2 个 `Q`。滑动窗口正好可以解决这个问题。向右滑到 `"WQWRQ"` 停止，这时窗口外的所有字母频次都  `≤ k` 了。这个窗口包含了多余的 1 个 `W`，和 1 个 `R`。`W` 可以踢除掉，那么要替换的字符串是 `"QWRQ"`。`R` 不能踢除了(因为要找包含 1 个 `W` 和 2 个 `Q` 的字符串) 。窗口不断的滑动，直到结束。这个例子中最小的字符串其实位于末尾，`"QQW"`。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func balancedString(s string) int {
+	count, k := make([]int, 128), len(s)/4
+	for _, v := range s {
+		count[int(v)]++
+	}
+	left, right, res := 0, -1, len(s)
+	for left < len(s) {
+		if count['Q'] > k || count['W'] > k || count['E'] > k || count['R'] > k {
+			if right+1 < len(s) {
+				right++
+				count[s[right]]--
+			} else {
+				break
+			}
+		} else {
+			res = min(res, right-left+1)
+			count[s[left]]++
+			left++
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1232.Check-If-It-Is-a-Straight-Line/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1235.Maximum-Profit-in-Job-Scheduling/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1235.Maximum-Profit-in-Job-Scheduling.md b/website/content/ChapterFour/1200~1299/1235.Maximum-Profit-in-Job-Scheduling.md
new file mode 100755
index 000000000..ca4826f49
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1235.Maximum-Profit-in-Job-Scheduling.md
@@ -0,0 +1,125 @@
+# [1235. Maximum Profit in Job Scheduling](https://leetcode.com/problems/maximum-profit-in-job-scheduling/)
+
+
+## 题目
+
+We have `n` jobs, where every job is scheduled to be done from `startTime[i]` to `endTime[i]`, obtaining a profit of `profit[i]`.
+
+You're given the `startTime` , `endTime` and `profit` arrays, you need to output the maximum profit you can take such that there are no 2 jobs in the subset with overlapping time range.
+
+If you choose a job that ends at time `X` you will be able to start another job that starts at time `X`.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png](https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png)
+
+    Input: startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
+    Output: 120
+    Explanation: The subset chosen is the first and fourth job. 
+    Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png](https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png)
+
+    Input: startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
+    Output: 150
+    Explanation: The subset chosen is the first, fourth and fifth job. 
+    Profit obtained 150 = 20 + 70 + 60.
+
+**Example 3**:
+
+![https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png](https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png)
+
+    Input: startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
+    Output: 6
+
+**Constraints**:
+
+- `1 <= startTime.length == endTime.length == profit.length <= 5 * 10^4`
+- `1 <= startTime[i] < endTime[i] <= 10^9`
+- `1 <= profit[i] <= 10^4`
+
+## 题目大意
+
+
+你打算利用空闲时间来做兼职工作赚些零花钱。这里有 n 份兼职工作，每份工作预计从 startTime[i] 开始到 endTime[i] 结束，报酬为 profit[i]。给你一份兼职工作表，包含开始时间 startTime，结束时间 endTime 和预计报酬 profit 三个数组，请你计算并返回可以获得的最大报酬。注意，时间上出现重叠的 2 份工作不能同时进行。如果你选择的工作在时间 X 结束，那么你可以立刻进行在时间 X 开始的下一份工作。
+
+
+提示：
+
+- 1 <= startTime.length == endTime.length == profit.length <= 5 * 10^4
+- 1 <= startTime[i] < endTime[i] <= 10^9
+- 1 <= profit[i] <= 10^4
+
+
+
+## 解题思路
+
+- 给出一组任务，任务有开始时间，结束时间，和任务收益。一个任务开始还没有结束，中间就不能再安排其他任务。问如何安排任务，能使得最后收益最大？
+- 一般任务的题目，区间的题目，都会考虑是否能排序。这一题可以先按照任务的结束时间从小到大排序，如果结束时间相同，则按照收益从小到大排序。`dp[i]` 代表前 `i` 份工作能获得的最大收益。初始值，`dp[0] = job[1].profit` 。对于任意一个任务 `i` ，看能否找到满足 `jobs[j].enTime <= jobs[j].startTime && j < i` 条件的 `j`，即查找 `upper_bound` 。由于 `jobs` 被我们排序了，所以这里可以使用二分搜索来查找。如果能找到满足条件的任务 j，那么状态转移方程是：`dp[i] = max(dp[i-1], jobs[i].profit)`。如果能找到满足条件的任务 j，那么状态转移方程是：`dp[i] = max(dp[i-1], dp[low]+jobs[i].profit)`。最终求得的解在 `dp[len(startTime)-1]` 中。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "sort"
+
+type job struct {
+	startTime int
+	endTime   int
+	profit    int
+}
+
+func jobScheduling(startTime []int, endTime []int, profit []int) int {
+	jobs, dp := []job{}, make([]int, len(startTime))
+	for i := 0; i < len(startTime); i++ {
+		jobs = append(jobs, job{startTime: startTime[i], endTime: endTime[i], profit: profit[i]})
+	}
+	sort.Sort(sortJobs(jobs))
+	dp[0] = jobs[0].profit
+	for i := 1; i < len(jobs); i++ {
+		low, high := 0, i-1
+		for low < high {
+			mid := low + (high-low)>>1
+			if jobs[mid+1].endTime <= jobs[i].startTime {
+				low = mid + 1
+			} else {
+				high = mid
+			}
+		}
+		if jobs[low].endTime <= jobs[i].startTime {
+			dp[i] = max(dp[i-1], dp[low]+jobs[i].profit)
+		} else {
+			dp[i] = max(dp[i-1], jobs[i].profit)
+		}
+	}
+	return dp[len(startTime)-1]
+}
+
+type sortJobs []job
+
+func (s sortJobs) Len() int {
+	return len(s)
+}
+func (s sortJobs) Less(i, j int) bool {
+	if s[i].endTime == s[j].endTime {
+		return s[i].profit < s[j].profit
+	}
+	return s[i].endTime < s[j].endTime
+}
+func (s sortJobs) Swap(i, j int) {
+	s[i], s[j] = s[j], s[i]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1234.Replace-the-Substring-for-Balanced-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters.md b/website/content/ChapterFour/1200~1299/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters.md
new file mode 100644
index 000000000..86d530b44
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters.md
@@ -0,0 +1,95 @@
+# [1239. Maximum Length of a Concatenated String with Unique Characters](https://leetcode.com/problems/maximum-length-of-a-concatenated-string-with-unique-characters/)
+
+## 题目
+
+Given an array of strings `arr`. String `s` is a concatenation of a sub-sequence of `arr` which have **unique characters**.
+
+Return *the maximum possible length* of `s`.
+
+**Example 1:**
+
+```
+Input: arr = ["un","iq","ue"]
+Output: 4
+Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique".
+Maximum length is 4.
+```
+
+**Example 2:**
+
+```
+Input: arr = ["cha","r","act","ers"]
+Output: 6
+Explanation: Possible solutions are "chaers" and "acters".
+```
+
+**Example 3:**
+
+```
+Input: arr = ["abcdefghijklmnopqrstuvwxyz"]
+Output: 26
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 16`
+- `1 <= arr[i].length <= 26`
+- `arr[i]` contains only lower case English letters.
+
+## 题目大意
+
+给定一个字符串数组 arr，字符串 s 是将 arr 某一子序列字符串连接所得的字符串，如果 s 中的每一个字符都只出现过一次，那么它就是一个可行解。请返回所有可行解 s 中最长长度。
+
+## 解题思路
+
+- 每个字符串数组可以想象为 26 位的 0101 二进制串。出现的字符对应的位上标记为 1，没有出现的字符对应的位上标记为 0 。如果一个字符串中包含重复的字符，那么它所有 1 的个数一定不等于字符串的长度。如果 2 个字符串每个字母都只出现了一次，那么它们俩对应的二进制串 mask 相互与运算的结果一定为 0 ，即 0，1 互补了。利用这个特点，深搜所有解，保存出最长可行解的长度即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math/bits"
+)
+
+func maxLength(arr []string) int {
+	c, res := []uint32{}, 0
+	for _, s := range arr {
+		var mask uint32
+		for _, c := range s {
+			mask = mask | 1<<(c-'a')
+		}
+		if len(s) != bits.OnesCount32(mask) { // 如果字符串本身带有重复的字符，需要排除
+			continue
+		}
+		c = append(c, mask)
+	}
+	dfs(c, 0, 0, &res)
+	return res
+}
+
+func dfs(c []uint32, index int, mask uint32, res *int) {
+	*res = max(*res, bits.OnesCount32(mask))
+	for i := index; i < len(c); i++ {
+		if mask&c[i] == 0 {
+			dfs(c, i+1, mask|c[i], res)
+		}
+	}
+	return
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1235.Maximum-Profit-in-Job-Scheduling/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1249.Minimum-Remove-to-Make-Valid-Parentheses/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1249.Minimum-Remove-to-Make-Valid-Parentheses.md b/website/content/ChapterFour/1200~1299/1249.Minimum-Remove-to-Make-Valid-Parentheses.md
new file mode 100644
index 000000000..c1936b889
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1249.Minimum-Remove-to-Make-Valid-Parentheses.md
@@ -0,0 +1,102 @@
+# [1249. Minimum Remove to Make Valid Parentheses](https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/)
+
+
+## 题目
+
+Given a string s of `'('` , `')'` and lowercase English characters.
+
+Your task is to remove the minimum number of parentheses ( `'('` or `')'`, in any positions ) so that the resulting *parentheses string* is valid and return **any** valid string.
+
+Formally, a *parentheses string* is valid if and only if:
+
+- It is the empty string, contains only lowercase characters, or
+- It can be written as `AB` (`A` concatenated with `B`), where `A` and `B` are valid strings, or
+- It can be written as `(A)`, where `A` is a valid string.
+
+**Example 1:**
+
+```
+Input: s = "lee(t(c)o)de)"
+Output: "lee(t(c)o)de"
+Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "a)b(c)d"
+Output: "ab(c)d"
+
+```
+
+**Example 3:**
+
+```
+Input: s = "))(("
+Output: ""
+Explanation: An empty string is also valid.
+
+```
+
+**Example 4:**
+
+```
+Input: s = "(a(b(c)d)"
+Output: "a(b(c)d)"
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 10^5`
+- `s[i]` is one of `'('` , `')'` and lowercase English letters`.`
+
+## 题目大意
+
+给你一个由 '('、')' 和小写字母组成的字符串 s。你需要从字符串中删除最少数目的 '(' 或者 ')' （可以删除任意位置的括号)，使得剩下的「括号字符串」有效。请返回任意一个合法字符串。有效「括号字符串」应当符合以下 任意一条 要求：
+
+- 空字符串或只包含小写字母的字符串
+- 可以被写作 AB（A 连接 B）的字符串，其中 A 和 B 都是有效「括号字符串」
+- 可以被写作 (A) 的字符串，其中 A 是一个有效的「括号字符串」
+
+## 解题思路
+
+- 最容易想到的思路是利用栈判断括号匹配是否有效。这个思路可行，时间复杂度也只是 O(n)。
+- 不用栈，可以 2 次循环遍历，正向遍历一次，标记出多余的 `'('` ，逆向遍历一次，再标记出多余的 `')'`，最后将所有这些标记多余的字符删掉即可。这种解法写出来的代码也很简洁，时间复杂度也是 O(n)。
+- 针对上面的解法再改进一点。正向遍历的时候不仅标记出多余的 `'('`，还可以顺手把多余的 `')'` 删除。这样只用循环一次。最后再删除掉多余的 `'('` 即可。时间复杂度还是 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+func minRemoveToMakeValid(s string) string {
+	res, opens := []byte{}, 0
+	for i := 0; i < len(s); i++ {
+		if s[i] == '(' {
+			opens++
+		} else if s[i] == ')' {
+			if opens == 0 {
+				continue
+			}
+			opens--
+		}
+		res = append(res, s[i])
+	}
+	for i := len(res) - 1; i >= 0; i-- {
+		if res[i] == '(' && opens > 0 {
+			opens--
+			res = append(res[:i], res[i+1:]...)
+		}
+	}
+	return string(res)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1252.Cells-with-Odd-Values-in-a-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1252.Cells-with-Odd-Values-in-a-Matrix.md b/website/content/ChapterFour/1200~1299/1252.Cells-with-Odd-Values-in-a-Matrix.md
new file mode 100755
index 000000000..61913d739
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1252.Cells-with-Odd-Values-in-a-Matrix.md
@@ -0,0 +1,110 @@
+# [1252. Cells with Odd Values in a Matrix](https://leetcode.com/problems/cells-with-odd-values-in-a-matrix/)
+
+
+## 题目
+
+Given `n` and `m` which are the dimensions of a matrix initialized by zeros and given an array `indices` where `indices[i] = [ri, ci]`. For each pair of `[ri, ci]` you have to increment all cells in row `ri` and column `ci` by 1.
+
+Return *the number of cells with odd values* in the matrix after applying the increment to all `indices`.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/10/30/e1.png](https://assets.leetcode.com/uploads/2019/10/30/e1.png)
+
+    Input: n = 2, m = 3, indices = [[0,1],[1,1]]
+    Output: 6
+    Explanation: Initial matrix = [[0,0,0],[0,0,0]].
+    After applying first increment it becomes [[1,2,1],[0,1,0]].
+    The final matrix will be [[1,3,1],[1,3,1]] which contains 6 odd numbers.
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/10/30/e2.png](https://assets.leetcode.com/uploads/2019/10/30/e2.png)
+
+    Input: n = 2, m = 2, indices = [[1,1],[0,0]]
+    Output: 0
+    Explanation: Final matrix = [[2,2],[2,2]]. There is no odd number in the final matrix.
+
+**Constraints**:
+
+- `1 <= n <= 50`
+- `1 <= m <= 50`
+- `1 <= indices.length <= 100`
+- `0 <= indices[i][0] < n`
+- `0 <= indices[i][1] < m`
+
+## 题目大意
+
+
+给你一个 n 行 m 列的矩阵，最开始的时候，每个单元格中的值都是 0。另有一个索引数组 indices，indices[i] = [ri, ci] 中的 ri 和 ci 分别表示指定的行和列（从 0 开始编号）。你需要将每对 [ri, ci] 指定的行和列上的所有单元格的值加 1。请你在执行完所有 indices 指定的增量操作后，返回矩阵中 「奇数值单元格」 的数目。
+
+提示：
+
+- 1 <= n <= 50
+- 1 <= m <= 50
+- 1 <= indices.length <= 100
+- 0 <= indices[i][0] < n
+- 0 <= indices[i][1] < m
+
+
+## 解题思路
+
+- 给出一个 n * m 的矩阵，和一个数组，数组里面包含一些行列坐标，并在指定坐标上 + 1，问最后 n * m 的矩阵中奇数的总数。
+- 暴力方法按照题意模拟即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 暴力法
+func oddCells(n int, m int, indices [][]int) int {
+	matrix, res := make([][]int, n), 0
+	for i := range matrix {
+		matrix[i] = make([]int, m)
+	}
+	for _, indice := range indices {
+		for i := 0; i < m; i++ {
+			matrix[indice[0]][i]++
+		}
+		for j := 0; j < n; j++ {
+			matrix[j][indice[1]]++
+		}
+	}
+	for _, m := range matrix {
+		for _, v := range m {
+			if v&1 == 1 {
+				res++
+			}
+		}
+	}
+	return res
+}
+
+// 解法二 暴力法
+func oddCells1(n int, m int, indices [][]int) int {
+	rows, cols, count := make([]int, n), make([]int, m), 0
+	for _, pair := range indices {
+		rows[pair[0]]++
+		cols[pair[1]]++
+	}
+	for i := 0; i < n; i++ {
+		for j := 0; j < m; j++ {
+			if (rows[i]+cols[j])%2 == 1 {
+				count++
+			}
+		}
+	}
+	return count
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1249.Minimum-Remove-to-Make-Valid-Parentheses/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1254.Number-of-Closed-Islands/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1254.Number-of-Closed-Islands.md b/website/content/ChapterFour/1200~1299/1254.Number-of-Closed-Islands.md
new file mode 100755
index 000000000..96e175c9a
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1254.Number-of-Closed-Islands.md
@@ -0,0 +1,117 @@
+# [1254. Number of Closed Islands](https://leetcode.com/problems/number-of-closed-islands/)
+
+
+## 题目
+
+Given a 2D `grid` consists of `0s` (land) and `1s` (water). An *island* is a maximal 4-directionally connected group of `0s` and a *closed island* is an island **totally** (all left, top, right, bottom) surrounded by `1s.`
+
+Return the number of *closed islands*.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/10/31/sample_3_1610.png](https://assets.leetcode.com/uploads/2019/10/31/sample_3_1610.png)
+
+    Input: grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
+	Output: 2
+	Explanation: 
+	Islands in gray are closed because they are completely surrounded by water (group of 1s).
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/10/31/sample_4_1610.png](https://assets.leetcode.com/uploads/2019/10/31/sample_4_1610.png)
+
+    Input: grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
+	Output: 1
+
+**Example 3**:
+
+    Input: grid = [[1,1,1,1,1,1,1],
+               [1,0,0,0,0,0,1],
+               [1,0,1,1,1,0,1],
+               [1,0,1,0,1,0,1],
+               [1,0,1,1,1,0,1],
+               [1,0,0,0,0,0,1],
+               [1,1,1,1,1,1,1]]
+	Output: 2
+
+**Constraints**:
+
+- `1 <= grid.length, grid[0].length <= 100`
+- `0 <= grid[i][j] <=1`
+
+## 题目大意
+
+有一个二维矩阵 grid ，每个位置要么是陆地（记号为 0 ）要么是水域（记号为 1 ）。我们从一块陆地出发，每次可以往上下左右 4 个方向相邻区域走，能走到的所有陆地区域，我们将其称为一座「岛屿」。如果一座岛屿 完全 由水域包围，即陆地边缘上下左右所有相邻区域都是水域，那么我们将其称为 「封闭岛屿」。请返回封闭岛屿的数目。
+
+提示：
+
+- 1 <= grid.length, grid[0].length <= 100
+- 0 <= grid[i][j] <=1
+
+
+## 解题思路
+
+- 给出一个地图，1 代表海水，0 代表陆地。要求找出四周都是海水的陆地的总个数。
+- 这一题和第 200 题解题思路完全一致。只不过这一题要求必须四周都是海水，第 200 题的陆地可以是靠着地图边缘的。在此题中，靠着地图边缘的陆地不能最终计数到结果中。
+
+## 代码
+
+```go
+
+package leetcode
+
+func closedIsland(grid [][]int) int {
+	m := len(grid)
+	if m == 0 {
+		return 0
+	}
+	n := len(grid[0])
+	if n == 0 {
+		return 0
+	}
+	res, visited := 0, make([][]bool, m)
+	for i := 0; i < m; i++ {
+		visited[i] = make([]bool, n)
+	}
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			isEdge := false
+			if grid[i][j] == 0 && !visited[i][j] {
+				checkIslands(grid, &visited, i, j, &isEdge)
+				if !isEdge {
+					res++
+				}
+
+			}
+		}
+	}
+	return res
+}
+
+func checkIslands(grid [][]int, visited *[][]bool, x, y int, isEdge *bool) {
+	if (x == 0 || x == len(grid)-1 || y == 0 || y == len(grid[0])-1) && grid[x][y] == 0 {
+		*isEdge = true
+	}
+	(*visited)[x][y] = true
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		if isIntInBoard(grid, nx, ny) && !(*visited)[nx][ny] && grid[nx][ny] == 0 {
+			checkIslands(grid, visited, nx, ny, isEdge)
+		}
+	}
+	*isEdge = *isEdge || false
+}
+
+func isIntInBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1252.Cells-with-Odd-Values-in-a-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1260.Shift-2D-Grid/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1260.Shift-2D-Grid.md b/website/content/ChapterFour/1200~1299/1260.Shift-2D-Grid.md
new file mode 100644
index 000000000..83fcdef1d
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1260.Shift-2D-Grid.md
@@ -0,0 +1,97 @@
+# [1260. Shift 2D Grid](https://leetcode.com/problems/shift-2d-grid/)
+
+
+## 题目
+
+Given a 2D `grid` of size `m x n` and an integer `k`. You need to shift the `grid` `k` times.
+
+In one shift operation:
+
+- Element at `grid[i][j]` moves to `grid[i][j + 1]`.
+- Element at `grid[i][n - 1]` moves to `grid[i + 1][0]`.
+- Element at `grid[m - 1][n - 1]` moves to `grid[0][0]`.
+
+Return the *2D grid* after applying shift operation `k` times.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/11/05/e1.png](https://assets.leetcode.com/uploads/2019/11/05/e1.png)
+
+```
+Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1
+Output: [[9,1,2],[3,4,5],[6,7,8]]
+```
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2019/11/05/e2.png](https://assets.leetcode.com/uploads/2019/11/05/e2.png)
+
+```
+Input: grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4
+Output: [[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]
+```
+
+**Example 3**:
+
+```
+Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9
+Output: [[1,2,3],[4,5,6],[7,8,9]]
+```
+
+**Constraints**:
+
+- `m == grid.length`
+- `n == grid[i].length`
+- `1 <= m <= 50`
+- `1 <= n <= 50`
+- `-1000 <= grid[i][j] <= 1000`
+- `0 <= k <= 100`
+
+## 题目大意
+
+给你一个 m 行 n 列的二维网格 grid 和一个整数 k。你需要将 grid 迁移 k 次。每次「迁移」操作将会引发下述活动：
+
+- 位于 grid[i][j] 的元素将会移动到 grid[i][j + 1]。
+- 位于 grid[i][n - 1] 的元素将会移动到 grid[i + 1][0]。
+- 位于 grid[m - 1][n - 1] 的元素将会移动到 grid[0][0]。
+
+请你返回 k 次迁移操作后最终得到的 二维网格。
+
+
+## 解题思路
+
+- 给一个矩阵和一个移动步数 k，要求把矩阵每个元素往后移动 k 步，最后的元素移动头部，循环移动，最后输出移动结束的矩阵。
+- 简单题，按照题意循环移动即可，注意判断边界情况。
+
+## 代码
+
+```go
+
+package leetcode
+
+func shiftGrid(grid [][]int, k int) [][]int {
+	x, y := len(grid[0]), len(grid)
+	newGrid := make([][]int, y)
+	for i := 0; i < y; i++ {
+		newGrid[i] = make([]int, x)
+	}
+	for i := 0; i < y; i++ {
+		for j := 0; j < x; j++ {
+			ny := (k / x) + i
+			if (j + (k % x)) >= x {
+				ny++
+			}
+			newGrid[ny%y][(j+(k%x))%x] = grid[i][j]
+		}
+	}
+	return newGrid
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1254.Number-of-Closed-Islands/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1266.Minimum-Time-Visiting-All-Points/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1266.Minimum-Time-Visiting-All-Points.md b/website/content/ChapterFour/1200~1299/1266.Minimum-Time-Visiting-All-Points.md
new file mode 100755
index 000000000..5b20c53a1
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1266.Minimum-Time-Visiting-All-Points.md
@@ -0,0 +1,82 @@
+# [1266. Minimum Time Visiting All Points](https://leetcode.com/problems/minimum-time-visiting-all-points/)
+
+
+## 题目
+
+On a plane there are `n` points with integer coordinates `points[i] = [xi, yi]`. Your task is to find the minimum time in seconds to visit all points.
+
+You can move according to the next rules:
+
+- In one second always you can either move vertically, horizontally by one unit or diagonally (it means to move one unit vertically and one unit horizontally in one second).
+- You have to visit the points in the same order as they appear in the array.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/11/14/1626_example_1.PNG](https://assets.leetcode.com/uploads/2019/11/14/1626_example_1.PNG)
+
+    Input: points = [[1,1],[3,4],[-1,0]]
+    Output: 7
+    Explanation: One optimal path is [1,1] -> [2,2] -> [3,3] -> [3,4] -> [2,3] -> [1,2] -> [0,1] -> [-1,0]   
+    Time from [1,1] to [3,4] = 3 seconds 
+    Time from [3,4] to [-1,0] = 4 seconds
+    Total time = 7 seconds
+
+**Example 2**:
+
+    Input: points = [[3,2],[-2,2]]
+    Output: 5
+
+**Constraints**:
+
+- `points.length == n`
+- `1 <= n <= 100`
+- `points[i].length == 2`
+- `-1000 <= points[i][0], points[i][1] <= 1000`
+
+## 题目大意
+
+
+平面上有 n 个点，点的位置用整数坐标表示 points[i] = [xi, yi]。请你计算访问所有这些点需要的最小时间（以秒为单位）。你可以按照下面的规则在平面上移动：
+
+- 每一秒沿水平或者竖直方向移动一个单位长度，或者跨过对角线（可以看作在一秒内向水平和竖直方向各移动一个单位长度）。
+- 必须按照数组中出现的顺序来访问这些点。
+
+提示：
+
+- points.length == n
+- 1 <= n <= 100
+- points[i].length == 2
+- -1000 <= points[i][0], points[i][1] <= 1000
+
+
+
+
+
+## 解题思路
+
+- 在直角坐标系上给出一个数组，数组里面的点是飞机飞行经过的点。飞机飞行只能沿着水平方向、垂直方向、45°方向飞行。问飞机经过所有点的最短时间。
+- 简单的数学问题。依次遍历数组，分别计算 x 轴和 y 轴上的差值，取最大值即是这两点之间飞行的最短时间。最后累加每次计算的最大值就是最短时间。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func minTimeToVisitAllPoints(points [][]int) int {
+	res := 0
+	for i := 1; i < len(points); i++ {
+		res += max(abs(points[i][0]-points[i-1][0]), abs(points[i][1]-points[i-1][1]))
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1260.Shift-2D-Grid/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1268.Search-Suggestions-System/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1268.Search-Suggestions-System.md b/website/content/ChapterFour/1200~1299/1268.Search-Suggestions-System.md
new file mode 100644
index 000000000..759f58221
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1268.Search-Suggestions-System.md
@@ -0,0 +1,96 @@
+# [1268. Search Suggestions System](https://leetcode.com/problems/search-suggestions-system/)
+
+## 题目
+
+Given an array of strings `products` and a string `searchWord`. We want to design a system that suggests at most three product names from `products` after each character of `searchWord` is typed. Suggested products should have common prefix with the searchWord. If there are more than three products with a common prefix return the three lexicographically minimums products.
+
+Return *list of lists* of the suggested `products` after each character of `searchWord` is typed.
+
+**Example 1:**
+
+```
+Input: products = ["mobile","mouse","moneypot","monitor","mousepad"], searchWord = "mouse"
+Output: [
+["mobile","moneypot","monitor"],
+["mobile","moneypot","monitor"],
+["mouse","mousepad"],
+["mouse","mousepad"],
+["mouse","mousepad"]
+]
+Explanation: products sorted lexicographically = ["mobile","moneypot","monitor","mouse","mousepad"]
+After typing m and mo all products match and we show user ["mobile","moneypot","monitor"]
+After typing mou, mous and mouse the system suggests ["mouse","mousepad"]
+
+```
+
+**Example 2:**
+
+```
+Input: products = ["havana"], searchWord = "havana"
+Output: [["havana"],["havana"],["havana"],["havana"],["havana"],["havana"]]
+```
+
+**Example 3:**
+
+```
+Input: products = ["bags","baggage","banner","box","cloths"], searchWord = "bags"
+Output: [["baggage","bags","banner"],["baggage","bags","banner"],["baggage","bags"],["bags"]]
+```
+
+**Example 4:**
+
+```
+Input: products = ["havana"], searchWord = "tatiana"
+Output: [[],[],[],[],[],[],[]]
+```
+
+**Constraints:**
+
+- `1 <= products.length <= 1000`
+- There are no repeated elements in `products`.
+- `1 <= Σ products[i].length <= 2 * 10^4`
+- All characters of `products[i]` are lower-case English letters.
+- `1 <= searchWord.length <= 1000`
+- All characters of `searchWord` are lower-case English letters.
+
+## 题目大意
+
+给你一个产品数组 products 和一个字符串 searchWord ，products  数组中每个产品都是一个字符串。请你设计一个推荐系统，在依次输入单词 searchWord 的每一个字母后，推荐 products 数组中前缀与 searchWord 相同的最多三个产品。如果前缀相同的可推荐产品超过三个，请按字典序返回最小的三个。请你以二维列表的形式，返回在输入 searchWord 每个字母后相应的推荐产品的列表。
+
+## 解题思路
+
+- 由于题目要求返回的答案要按照字典序输出，所以先排序。有序字符串又满足了二分搜索的条件，于是可以用二分搜索。sort.SearchStrings 返回的是满足搜索条件的第一个起始下标。末尾不满足条件的字符串要切掉。所以要搜 2 次，第一次二分搜索先将不满足目标串前缀的字符串筛掉。第二次二分搜索再搜索出最终满足题意的字符串。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func suggestedProducts(products []string, searchWord string) [][]string {
+	sort.Strings(products)
+	searchWordBytes, result := []byte(searchWord), make([][]string, 0, len(searchWord))
+	for i := 1; i <= len(searchWord); i++ {
+		searchWordBytes[i-1]++
+		products = products[:sort.SearchStrings(products, string(searchWordBytes[:i]))]
+		searchWordBytes[i-1]--
+		products = products[sort.SearchStrings(products, searchWord[:i]):]
+		if len(products) > 3 {
+			result = append(result, products[:3])
+		} else {
+			result = append(result, products)
+		}
+	}
+	return result
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1266.Minimum-Time-Visiting-All-Points/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1275.Find-Winner-on-a-Tic-Tac-Toe-Game.md b/website/content/ChapterFour/1200~1299/1275.Find-Winner-on-a-Tic-Tac-Toe-Game.md
new file mode 100644
index 000000000..eda51cd07
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1275.Find-Winner-on-a-Tic-Tac-Toe-Game.md
@@ -0,0 +1,161 @@
+# [1275. Find Winner on a Tic Tac Toe Game](https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/)
+
+
+## 题目
+
+Tic-tac-toe is played by two players *A* and *B* on a 3 x 3 grid.
+
+Here are the rules of Tic-Tac-Toe:
+
+- Players take turns placing characters into empty squares (" ").
+- The first player *A* always places "X" characters, while the second player *B* always places "O" characters.
+- "X" and "O" characters are always placed into empty squares, never on filled ones.
+- The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
+- The game also ends if all squares are non-empty.
+- No more moves can be played if the game is over.
+
+Given an array `moves` where each element is another array of size 2 corresponding to the row and column of the grid where they mark their respective character in the order in which *A* and *B* play.
+
+Return the winner of the game if it exists (*A* or *B*), in case the game ends in a draw return "Draw", if there are still movements to play return "Pending".
+
+You can assume that `moves` is **valid** (It follows the rules of Tic-Tac-Toe), the grid is initially empty and *A* will play **first**.
+
+**Example 1**:
+
+```
+Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
+Output: "A"
+Explanation: "A" wins, he always plays first.
+"X  "    "X  "    "X  "    "X  "    "X  "
+"   " -> "   " -> " X " -> " X " -> " X "
+"   "    "O  "    "O  "    "OO "    "OOX"
+
+```
+
+**Example 2**:
+
+```
+Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
+Output: "B"
+Explanation: "B" wins.
+"X  "    "X  "    "XX "    "XXO"    "XXO"    "XXO"
+"   " -> " O " -> " O " -> " O " -> "XO " -> "XO " 
+"   "    "   "    "   "    "   "    "   "    "O  "
+
+```
+
+**Example 3**:
+
+```
+Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
+Output: "Draw"
+Explanation: The game ends in a draw since there are no moves to make.
+"XXO"
+"OOX"
+"XOX"
+
+```
+
+**Example 4**:
+
+```
+Input: moves = [[0,0],[1,1]]
+Output: "Pending"
+Explanation: The game has not finished yet.
+"X  "
+" O "
+"   "
+
+```
+
+**Constraints**:
+
+- `1 <= moves.length <= 9`
+- `moves[i].length == 2`
+- `0 <= moves[i][j] <= 2`
+- There are no repeated elements on `moves`.
+- `moves` follow the rules of tic tac toe.
+
+
+## 题目大意
+
+A 和 B 在一个 3 x 3 的网格上玩井字棋。井字棋游戏的规则如下：
+
+- 玩家轮流将棋子放在空方格 (" ") 上。
+- 第一个玩家 A 总是用 "X" 作为棋子，而第二个玩家 B 总是用 "O" 作为棋子。
+- "X" 和 "O" 只能放在空方格中，而不能放在已经被占用的方格上。
+- 只要有 3 个相同的（非空）棋子排成一条直线（行、列、对角线）时，游戏结束。
+- 如果所有方块都放满棋子（不为空），游戏也会结束。
+- 游戏结束后，棋子无法再进行任何移动。
+
+给你一个数组 moves，其中每个元素是大小为 2 的另一个数组（元素分别对应网格的行和列），它按照 A 和 B 的行动顺序（先 A 后 B）记录了两人各自的棋子位置。如果游戏存在获胜者（A 或 B），就返回该游戏的获胜者；如果游戏以平局结束，则返回 "Draw"；如果仍会有行动（游戏未结束），则返回 "Pending"。你可以假设 moves 都 有效（遵循井字棋规则），网格最初是空的，A 将先行动。
+
+提示：
+
+- 1 <= moves.length <= 9
+- moves[i].length == 2
+- 0 <= moves[i][j] <= 2
+- moves 里没有重复的元素。
+- moves 遵循井字棋的规则。
+
+
+## 解题思路
+
+- 两人玩 3*3 井字棋，A 先走，B 再走。谁能获胜就输出谁，如果平局输出 “Draw”，如果游戏还未结束，输出 “Pending”。游戏规则：谁能先占满行、列或者对角线任意一条线，谁就赢。
+- 简单题。题目给定 move 数组最多 3 步，而要赢得比赛，必须走满 3 步，所以可以先模拟，按照给的步数数组把 A 和 B 的步数都放在棋盘上。然后依次判断行、列，对角线的三种情况。如果都判完了，剩下的情况就是平局和死局的情况。
+
+## 代码
+
+```go
+
+package leetcode
+
+func tictactoe(moves [][]int) string {
+	board := [3][3]byte{}
+	for i := 0; i < len(moves); i++ {
+		if i%2 == 0 {
+			board[moves[i][0]][moves[i][1]] = 'X'
+		} else {
+			board[moves[i][0]][moves[i][1]] = 'O'
+		}
+	}
+	for i := 0; i < 3; i++ {
+		if board[i][0] == 'X' && board[i][1] == 'X' && board[i][2] == 'X' {
+			return "A"
+		}
+		if board[i][0] == 'O' && board[i][1] == 'O' && board[i][2] == 'O' {
+			return "B"
+		}
+		if board[0][i] == 'X' && board[1][i] == 'X' && board[2][i] == 'X' {
+			return "A"
+		}
+		if board[0][i] == 'O' && board[1][i] == 'O' && board[2][i] == 'O' {
+			return "B"
+		}
+	}
+	if board[0][0] == 'X' && board[1][1] == 'X' && board[2][2] == 'X' {
+		return "A"
+	}
+	if board[0][0] == 'O' && board[1][1] == 'O' && board[2][2] == 'O' {
+		return "B"
+	}
+	if board[0][2] == 'X' && board[1][1] == 'X' && board[2][0] == 'X' {
+		return "A"
+	}
+	if board[0][2] == 'O' && board[1][1] == 'O' && board[2][0] == 'O' {
+		return "B"
+	}
+	if len(moves) < 9 {
+		return "Pending"
+	}
+	return "Draw"
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1268.Search-Suggestions-System/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer.md b/website/content/ChapterFour/1200~1299/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer.md
new file mode 100644
index 000000000..f512f8476
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer.md
@@ -0,0 +1,66 @@
+# [1281. Subtract the Product and Sum of Digits of an Integer](https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/)
+
+
+
+## 题目
+
+Given an integer number `n`, return the difference between the product of its digits and the sum of its digits.
+
+**Example 1**:
+
+```
+Input: n = 234
+Output: 15 
+Explanation: 
+Product of digits = 2 * 3 * 4 = 24 
+Sum of digits = 2 + 3 + 4 = 9 
+Result = 24 - 9 = 15
+```
+
+**Example 2**:
+
+```
+Input: n = 4421
+Output: 21
+Explanation: 
+Product of digits = 4 * 4 * 2 * 1 = 32 
+Sum of digits = 4 + 4 + 2 + 1 = 11 
+Result = 32 - 11 = 21
+```
+
+**Constraints**:
+
+- `1 <= n <= 10^5`
+
+## 题目大意
+
+给你一个整数 n，请你帮忙计算并返回该整数「各位数字之积」与「各位数字之和」的差。
+
+提示：
+
+- 1 <= n <= 10^5
+
+## 解题思路
+
+- 给出一个数，计算这个数每位数字乘积减去每位数字累加的差值。
+- 简单题，按照题意输入输出即可。
+
+## 代码
+
+```go
+func subtractProductAndSum(n int) int {
+	sum, product := 0, 1
+	for ; n > 0; n /= 10 {
+		sum += n % 10
+		product *= n % 10
+	}
+	return product - sum
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1283.Find-the-Smallest-Divisor-Given-a-Threshold/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1283.Find-the-Smallest-Divisor-Given-a-Threshold.md b/website/content/ChapterFour/1200~1299/1283.Find-the-Smallest-Divisor-Given-a-Threshold.md
new file mode 100644
index 000000000..044a91c7b
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1283.Find-the-Smallest-Divisor-Given-a-Threshold.md
@@ -0,0 +1,94 @@
+# [1283. Find the Smallest Divisor Given a Threshold](https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/)
+
+
+
+## 题目
+
+Given an array of integers `nums` and an integer `threshold`, we will choose a positive integer divisor and divide all the array by it and sum the result of the division. Find the **smallest** divisor such that the result mentioned above is less than or equal to `threshold`.
+
+Each result of division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).
+
+It is guaranteed that there will be an answer.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,5,9], threshold = 6
+Output: 5
+Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1. 
+If the divisor is 4 we can get a sum to 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).
+```
+
+**Example 2**:
+
+```
+Input: nums = [2,3,5,7,11], threshold = 11
+Output: 3
+```
+
+**Example 3**:
+
+```
+Input: nums = [19], threshold = 5
+Output: 4
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 5 * 10^4`
+- `1 <= nums[i] <= 10^6`
+- `nums.length <= threshold <= 10^6`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个正整数 threshold  ，你需要选择一个正整数作为除数，然后将数组里每个数都除以它，并对除法结果求和。请你找出能够使上述结果小于等于阈值 threshold 的除数中 最小 的那个。每个数除以除数后都向上取整，比方说 7/3 = 3 ， 10/2 = 5 。题目保证一定有解。
+
+提示：
+
+- 1 <= nums.length <= 5 * 10^4
+- 1 <= nums[i] <= 10^6
+- nums.length <= threshold <= 10^6
+
+## 解题思路
+
+- 给出一个数组和一个阈值，要求找到一个除数，使得数组里面每个数和这个除数的商之和不超过这个阈值。求除数的最小值。
+- 这一题是典型的二分搜索的题目。根据题意，在 [1, 1000000] 区间内搜索除数，针对每次 `mid`，计算一次商的累加和。如果和比 `threshold` 小，说明除数太大，所以缩小右区间；如果和比 `threshold` 大，说明除数太小，所以缩小左区间。最终找到的 `low` 值就是最求的最小除数。
+
+## 代码
+
+```go
+func smallestDivisor(nums []int, threshold int) int {
+	low, high := 1, 1000000
+	for low < high {
+		mid := low + (high-low)>>1
+		if calDivisor(nums, mid, threshold) {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+func calDivisor(nums []int, mid, threshold int) bool {
+	sum := 0
+	for i := range nums {
+		if nums[i]%mid != 0 {
+			sum += nums[i]/mid + 1
+		} else {
+			sum += nums[i] / mid
+		}
+	}
+	if sum <= threshold {
+		return true
+	}
+	return false
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1287.Element-Appearing-More-Than-25-In-Sorted-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1287.Element-Appearing-More-Than-25-In-Sorted-Array.md b/website/content/ChapterFour/1200~1299/1287.Element-Appearing-More-Than-25-In-Sorted-Array.md
new file mode 100644
index 000000000..03294b899
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1287.Element-Appearing-More-Than-25-In-Sorted-Array.md
@@ -0,0 +1,56 @@
+# [1287. Element Appearing More Than 25% In Sorted Array](https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/)
+
+
+
+## 题目
+
+Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time.
+
+Return that integer.
+
+**Example 1**:
+
+```
+Input: arr = [1,2,2,6,6,6,6,7,10]
+Output: 6
+```
+
+**Constraints**:
+
+- `1 <= arr.length <= 10^4`
+- `0 <= arr[i] <= 10^5`
+
+## 题目大意
+
+给你一个非递减的 有序 整数数组，已知这个数组中恰好有一个整数，它的出现次数超过数组元素总数的 25%。请你找到并返回这个整数。
+
+提示：
+
+- 1 <= arr.length <= 10^4
+- 0 <= arr[i] <= 10^5
+
+## 解题思路
+
+- 给出一个非递减的有序数组，要求输出出现次数超过数组元素总数 25% 的元素。
+- 简单题，由于已经非递减有序了，所以只需要判断 `arr[i] == arr[i+n/4]` 是否相等即可。
+
+## 代码
+
+```go
+func findSpecialInteger(arr []int) int {
+	n := len(arr)
+	for i := 0; i < n-n/4; i++ {
+		if arr[i] == arr[i+n/4] {
+			return arr[i]
+		}
+	}
+	return -1
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1283.Find-the-Smallest-Divisor-Given-a-Threshold/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer.md b/website/content/ChapterFour/1200~1299/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer.md
new file mode 100644
index 000000000..af3a2ca0a
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer.md
@@ -0,0 +1,88 @@
+# [1290. Convert Binary Number in a Linked List to Integer](https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/)
+
+
+
+## 题目
+
+Given `head` which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary representation of a number.
+
+Return the *decimal value* of the number in the linked list.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/12/05/graph-1.png](https://assets.leetcode.com/uploads/2019/12/05/graph-1.png)
+
+```
+Input: head = [1,0,1]
+Output: 5
+Explanation: (101) in base 2 = (5) in base 10
+```
+
+**Example 2**:
+
+```
+Input: head = [0]
+Output: 0
+```
+
+**Example 3**:
+
+```
+Input: head = [1]
+Output: 1
+```
+
+**Example 4**:
+
+```
+Input: head = [1,0,0,1,0,0,1,1,1,0,0,0,0,0,0]
+Output: 18880
+```
+
+**Example 5**:
+
+```
+Input: head = [0,0]
+Output: 0
+```
+
+**Constraints**:
+
+- The Linked List is not empty.
+- Number of nodes will not exceed `30`.
+- Each node's value is either `0` or `1`.
+
+## 题目大意
+
+给你一个单链表的引用结点 head。链表中每个结点的值不是 0 就是 1。已知此链表是一个整数数字的二进制表示形式。请你返回该链表所表示数字的 十进制值 。
+
+提示：
+
+- 链表不为空。
+- 链表的结点总数不超过 30。
+- 每个结点的值不是 0 就是 1。
+
+## 解题思路
+
+- 给出一个链表，链表从头到尾表示的数是一个整数的二进制形式，要求输出这个整数的十进制。
+- 简单题，从头到尾遍历一次链表，边遍历边累加二进制位。
+
+## 代码
+
+```go
+func getDecimalValue(head *ListNode) int {
+	sum := 0
+	for head != nil {
+		sum = sum*2 + head.Val
+		head = head.Next
+	}
+	return sum
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1287.Element-Appearing-More-Than-25-In-Sorted-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination.md b/website/content/ChapterFour/1200~1299/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination.md
new file mode 100644
index 000000000..f4cebe1fa
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination.md
@@ -0,0 +1,138 @@
+# [1293. Shortest Path in a Grid with Obstacles Elimination](https://leetcode.com/problems/shortest-path-in-a-grid-with-obstacles-elimination/)
+
+
+
+## 题目
+
+You are given an m x n integer matrix grid where each cell is either 0 (empty) or 1 (obstacle). You can move up, down, left, or right from and to an empty cell in one step.
+
+Return the minimum number of steps to walk from the upper left corner (0, 0) to the lower right corner (m - 1, n - 1) given that you can eliminate at most k obstacles. If it is not possible to find such walk return -1.
+
+
+
+Example 1:
+
+
+![](https://assets.leetcode.com/uploads/2021/09/30/short1-grid.jpg)
+ 
+
+```
+Input: grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1
+Output: 6
+Explanation: 
+The shortest path without eliminating any obstacle is 10.
+The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).
+```
+
+Example 2:
+
+![](https://assets.leetcode.com/uploads/2021/09/30/short2-grid.jpg)
+
+```
+Input: grid = [[0,1,1],[1,1,1],[1,0,0]], k = 1
+Output: -1
+Explanation: We need to eliminate at least two obstacles to find such a walk.
+```
+
+Constraints:
+
+- m == grid.length
+- n == grid[i].length
+- 1 <= m, n <= 40
+- 1 <= k <= m * n
+- grid[i][j] is either 0 or 1.
+- grid[0][0] == grid[m - 1][n - 1] == 0
+
+
+
+## 题目大意
+
+给你一个 m * n 的网格，其中每个单元格不是 0（空）就是 1（障碍物）。每一步，您都可以在空白单元格中上、下、左、右移动。
+
+如果您 最多 可以消除 k 个障碍物，请找出从左上角 (0, 0) 到右下角 (m-1, n-1) 的最短路径，并返回通过该路径所需的步数。如果找不到这样的路径，则返回 -1 。
+
+
+## 解题思路
+
+使用 BFS 遍历棋盘。这题比普通可达性问题多了一个障碍物的限制。这个也不难。每个点往周边四个方向扩展的时候，如果遇到障碍物，先算上这个障碍物，障碍物累积总个数小于 K 的时候，从障碍物的这个格子继续开始遍历。如果没有遇到障碍物，判断当前累积障碍物个数是否已经小于 K 个，如果小于 K 便继续遍历。如果大于 K，便终止此轮遍历。
+
+## 代码
+
+```go
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+type pos struct {
+	x, y     int
+	obstacle int
+	step     int
+}
+
+func shortestPath(grid [][]int, k int) int {
+	queue, m, n := []pos{}, len(grid), len(grid[0])
+	visitor := make([][][]int, m)
+	if len(grid) == 1 && len(grid[0]) == 1 {
+		return 0
+	}
+	for i := 0; i < m; i++ {
+		visitor[i] = make([][]int, n)
+		for j := 0; j < n; j++ {
+			visitor[i][j] = make([]int, k+1)
+		}
+	}
+	visitor[0][0][0] = 1
+	queue = append(queue, pos{x: 0, y: 0, obstacle: 0, step: 0})
+	for len(queue) > 0 {
+		size := len(queue)
+		for size > 0 {
+			size--
+			node := queue[0]
+			queue = queue[1:]
+			for i := 0; i < len(dir); i++ {
+				newX := node.x + dir[i][0]
+				newY := node.y + dir[i][1]
+				if newX == m-1 && newY == n-1 {
+					if node.obstacle != 0 {
+						if node.obstacle <= k {
+							return node.step + 1
+						} else {
+							continue
+						}
+					}
+					return node.step + 1
+				}
+				if isInBoard(grid, newX, newY) {
+					if grid[newX][newY] == 1 {
+						if node.obstacle+1 <= k && visitor[newX][newY][node.obstacle+1] != 1 {
+							queue = append(queue, pos{x: newX, y: newY, obstacle: node.obstacle + 1, step: node.step + 1})
+							visitor[newX][newY][node.obstacle+1] = 1
+						}
+					} else {
+						if node.obstacle <= k && visitor[newX][newY][node.obstacle] != 1 {
+							queue = append(queue, pos{x: newX, y: newY, obstacle: node.obstacle, step: node.step + 1})
+							visitor[newX][newY][node.obstacle] = 1
+						}
+					}
+
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func isInBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md b/website/content/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md
new file mode 100644
index 000000000..afb613611
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md
@@ -0,0 +1,70 @@
+# [1295. Find Numbers with Even Number of Digits](https://leetcode.com/problems/find-numbers-with-even-number-of-digits/)
+
+
+
+## 题目
+
+Given an array `nums` of integers, return how many of them contain an **even number** of digits.
+
+**Example 1**:
+
+```
+Input: nums = [12,345,2,6,7896]
+Output: 2
+Explanation: 
+12 contains 2 digits (even number of digits). 
+345 contains 3 digits (odd number of digits). 
+2 contains 1 digit (odd number of digits). 
+6 contains 1 digit (odd number of digits). 
+7896 contains 4 digits (even number of digits). 
+Therefore only 12 and 7896 contain an even number of digits.
+```
+
+**Example 2**:
+
+```
+Input: nums = [555,901,482,1771]
+Output: 1 
+Explanation: 
+Only 1771 contains an even number of digits.
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 500`
+- `1 <= nums[i] <= 10^5`
+
+## 题目大意
+
+给你一个整数数组 nums，请你返回其中位数为 偶数 的数字的个数。
+
+提示：
+
+- 1 <= nums.length <= 500
+- 1 <= nums[i] <= 10^5
+
+
+
+## 解题思路
+
+- 给你一个整数数组，要求输出位数为偶数的数字的个数。
+- 简单题，把每个数字转换为字符串判断长度是否是偶数即可。
+
+## 代码
+
+```go
+func findNumbers(nums []int) int {
+	res := 0
+	for _, n := range nums {
+		res += 1 - len(strconv.Itoa(n))%2
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers.md b/website/content/ChapterFour/1200~1299/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers.md
new file mode 100644
index 000000000..ea345986c
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers.md
@@ -0,0 +1,79 @@
+# [1296. Divide Array in Sets of K Consecutive Numbers](https://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/)
+
+## 题目
+
+Given an array of integers nums and a positive integer k, check whether it is possible to divide this array into sets of k consecutive numbers.
+
+Return true if it is possible. Otherwise, return false.
+
+**Example 1**:
+
+    Input: nums = [1,2,3,3,4,4,5,6], k = 4
+    Output: true
+    Explanation: Array can be divided into [1,2,3,4] and [3,4,5,6].
+
+**Example 2**:
+
+    Input: nums = [3,2,1,2,3,4,3,4,5,9,10,11], k = 3
+    Output: true
+    Explanation: Array can be divided into [1,2,3] , [2,3,4] , [3,4,5] and [9,10,11].
+
+**Example 3**:
+
+    Input: nums = [1,2,3,4], k = 3
+    Output: false
+    Explanation: Each array should be divided in subarrays of size 3.
+
+**Constraints:**
+
+- 1 <= k <= nums.length <= 100000
+- 1 <= nums[i] <= 1000000000
+
+## 题目大意
+
+给你一个整数数组 nums 和一个正整数 k，请你判断是否可以把这个数组划分成一些由 k 个连续数字组成的集合。
+如果可以，请返回 true；否则，返回 false。
+
+## 解题思路
+
+贪心算法
+
+- 对nums升序排序
+- 对nums内数字进行哈希计数（key:数字，value:数量）
+- 遍历nums中的数字，以数量大于1的数字作为连续数字开头，寻找连续数字后续元素，若无法找到 k 个连续数字则返回false
+- 所有数字都能找到 k 个连续数字返回true
+
+##代码
+
+```go
+package leetcode
+
+import "sort"
+
+func isPossibleDivide(nums []int, k int) bool {
+	mp := make(map[int]int)
+	for _, v := range nums {
+		mp[v] += 1
+	}
+	sort.Ints(nums)
+	for _, num := range nums {
+		if mp[num] == 0 {
+			continue
+		}
+		for diff := 0; diff < k; diff++ {
+			if mp[num+diff] == 0 {
+				return false
+			}
+			mp[num+diff] -= 1
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1299.Replace-Elements-with-Greatest-Element-on-Right-Side/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/1299.Replace-Elements-with-Greatest-Element-on-Right-Side.md b/website/content/ChapterFour/1200~1299/1299.Replace-Elements-with-Greatest-Element-on-Right-Side.md
new file mode 100644
index 000000000..ea964e728
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/1299.Replace-Elements-with-Greatest-Element-on-Right-Side.md
@@ -0,0 +1,58 @@
+# [1299. Replace Elements with Greatest Element on Right Side](https://leetcode.com/problems/replace-elements-with-greatest-element-on-right-side/)
+
+
+
+## 题目
+
+Given an array `arr`, replace every element in that array with the greatest element among the elements to its right, and replace the last element with `-1`.
+
+After doing so, return the array.
+
+**Example 1**:
+
+```
+Input: arr = [17,18,5,4,6,1]
+Output: [18,6,6,6,1,-1]
+```
+
+**Constraints**:
+
+- `1 <= arr.length <= 10^4`
+- `1 <= arr[i] <= 10^5`
+
+
+## 题目大意
+
+给你一个数组 arr ，请你将每个元素用它右边最大的元素替换，如果是最后一个元素，用 -1 替换。完成所有替换操作后，请你返回这个数组。
+
+提示：
+
+- 1 <= arr.length <= 10^4
+- 1 <= arr[i] <= 10^5
+
+
+## 解题思路
+
+- 给出一个数组，要求把所有元素都替换成自己右边最大的元素，最后一位补上 -1 。最后输出变化以后的数组。
+- 简单题，按照题意操作即可。
+
+## 代码
+
+```go
+func replaceElements(arr []int) []int {
+	j, temp := -1, 0
+	for i := len(arr) - 1; i >= 0; i-- {
+		temp = arr[i]
+		arr[i] = j
+		j = max(j, temp)
+	}
+	return arr
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1300.Sum-of-Mutated-Array-Closest-to-Target/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1200~1299/_index.md b/website/content/ChapterFour/1200~1299/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/1200~1299/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/1300~1399/1300.Sum-of-Mutated-Array-Closest-to-Target.md b/website/content/ChapterFour/1300~1399/1300.Sum-of-Mutated-Array-Closest-to-Target.md
new file mode 100644
index 000000000..3e13fe77b
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1300.Sum-of-Mutated-Array-Closest-to-Target.md
@@ -0,0 +1,110 @@
+# [1300. Sum of Mutated Array Closest to Target](https://leetcode.com/problems/sum-of-mutated-array-closest-to-target/)
+
+
+
+## 题目
+
+Given an integer array `arr` and a target value `target`, return the integer `value` such that when we change all the integers larger than `value` in the given array to be equal to `value`, the sum of the array gets as close as possible (in absolute difference) to `target`.
+
+In case of a tie, return the minimum such integer.
+
+Notice that the answer is not neccesarilly a number from `arr`.
+
+**Example 1**:
+
+```
+Input: arr = [4,9,3], target = 10
+Output: 3
+Explanation: When using 3 arr converts to [3, 3, 3] which sums 9 and that's the optimal answer.
+```
+
+**Example 2**:
+
+```
+Input: arr = [2,3,5], target = 10
+Output: 5
+```
+
+**Example 3**:
+
+```
+Input: arr = [60864,25176,27249,21296,20204], target = 56803
+Output: 11361
+```
+
+**Constraints**:
+
+- `1 <= arr.length <= 10^4`
+- `1 <= arr[i], target <= 10^5`
+
+
+## 题目大意
+
+给你一个整数数组 arr 和一个目标值 target ，请你返回一个整数 value ，使得将数组中所有大于 value 的值变成 value 后，数组的和最接近  target （最接近表示两者之差的绝对值最小）。如果有多种使得和最接近 target 的方案，请你返回这些整数中的最小值。请注意，答案不一定是 arr 中的数字。
+
+提示：
+
+- 1 <= arr.length <= 10^4
+- 1 <= arr[i], target <= 10^5
+
+
+
+## 解题思路
+
+- 给出一个数组 arr 和 target。能否找到一个 value 值，使得将数组中所有大于 value 的值变成 value 后，数组的和最接近 target。如果有多种方法，输出 value 值最小的。
+- 这一题可以用二分搜索来求解。最后输出的唯一解有 2 个限制条件，一个是变化后的数组和最接近 target 。另一个是输出的 value 是所有可能方法中最小值。二分搜索最终的 value 值。mid 就是尝试的 value 值，每选择一次 mid，就算一次总和，和 target 比较。由于数组里面每个数和 mid 差距各不相同，所以每次调整 mid 有可能出现 mid 选小了以后，距离 target 反而更大了；mid 选大了以后，距离 target 反而更小了。这里的解决办法是，把 value 上下方可能的值都拿出来比较一下。
+
+## 代码
+
+```go
+
+func findBestValue(arr []int, target int) int {
+	low, high := 0, 100000
+	for low < high {
+		mid := low + (high-low)>>1
+		if calculateSum(arr, mid) < target {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	if high == 100000 {
+		res := 0
+		for _, num := range arr {
+			if res < num {
+				res = num
+			}
+		}
+		return res
+	}
+	// 比较阈值线分别定在 left - 1 和 left 的时候与 target 的接近程度
+	sum1, sum2 := calculateSum(arr, low-1), calculateSum(arr, low)
+	if target-sum1 <= sum2-target {
+		return low - 1
+	}
+	return low
+}
+
+func calculateSum(arr []int, mid int) int {
+	sum := 0
+	for _, num := range arr {
+		sum += min(num, mid)
+	}
+	return sum
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1200~1299/1299.Replace-Elements-with-Greatest-Element-on-Right-Side/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1302.Deepest-Leaves-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1302.Deepest-Leaves-Sum.md b/website/content/ChapterFour/1300~1399/1302.Deepest-Leaves-Sum.md
new file mode 100644
index 000000000..6037e30c5
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1302.Deepest-Leaves-Sum.md
@@ -0,0 +1,65 @@
+# [1302. Deepest Leaves Sum](https://leetcode.com/problems/deepest-leaves-sum/)
+
+
+
+## 题目
+
+Given a binary tree, return the sum of values of its deepest leaves.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/07/31/1483_ex1.png](https://assets.leetcode.com/uploads/2019/07/31/1483_ex1.png)
+
+```
+Input: root = [1,2,3,4,5,null,6,7,null,null,null,null,8]
+Output: 15
+```
+
+**Constraints**:
+
+- The number of nodes in the tree is between `1` and `10^4`.
+- The value of nodes is between `1` and `100`.
+
+## 题目大意
+
+给你一棵二叉树，请你返回层数最深的叶子节点的和。
+
+提示：
+
+- 树中节点数目在 1 到 10^4 之间。
+- 每个节点的值在 1 到 100 之间。
+
+## 解题思路
+
+- 给你一棵二叉树，请你返回层数最深的叶子节点的和。
+- 这一题不难，DFS 遍历把最底层的叶子节点和都加起来即可。
+
+## 代码
+
+```go
+func deepestLeavesSum(root *TreeNode) int {
+	maxLevel, sum := 0, 0
+	dfsDeepestLeavesSum(root, 0, &maxLevel, &sum)
+	return sum
+}
+
+func dfsDeepestLeavesSum(root *TreeNode, level int, maxLevel, sum *int) {
+	if root == nil {
+		return
+	}
+	if level > *maxLevel {
+		*maxLevel, *sum = level, root.Val
+	} else if level == *maxLevel {
+		*sum += root.Val
+	}
+	dfsDeepestLeavesSum(root.Left, level+1, maxLevel, sum)
+	dfsDeepestLeavesSum(root.Right, level+1, maxLevel, sum)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1300.Sum-of-Mutated-Array-Closest-to-Target/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1304.Find-N-Unique-Integers-Sum-up-to-Zero/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1304.Find-N-Unique-Integers-Sum-up-to-Zero.md b/website/content/ChapterFour/1300~1399/1304.Find-N-Unique-Integers-Sum-up-to-Zero.md
new file mode 100644
index 000000000..35a82147a
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1304.Find-N-Unique-Integers-Sum-up-to-Zero.md
@@ -0,0 +1,69 @@
+# [1304. Find N Unique Integers Sum up to Zero](https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/)
+
+
+
+## 题目
+
+Given an integer `n`, return **any** array containing `n` **unique** integers such that they add up to 0.
+
+**Example 1**:
+
+```
+Input: n = 5
+Output: [-7,-1,1,3,4]
+Explanation: These arrays also are accepted [-5,-1,1,2,3] , [-3,-1,2,-2,4].
+```
+
+**Example 2**:
+
+```
+Input: n = 3
+Output: [-1,0,1]
+```
+
+**Example 3**:
+
+```
+Input: n = 1
+Output: [0]
+```
+
+**Constraints**:
+
+- `1 <= n <= 1000`
+
+## 题目大意
+
+给你一个整数 n，请你返回 任意 一个由 n 个 各不相同 的整数组成的数组，并且这 n 个数相加和为 0 。
+
+提示：
+
+- 1 <= n <= 1000
+
+## 解题思路
+
+- 给出一个数 n，输出一个有 n 个数的数组，里面元素之和为 0 。
+- 简单题，简单循环即可。
+
+## 代码
+
+```go
+func sumZero(n int) []int {
+	res, left, right, start := make([]int, n), 0, n-1, 1
+	for left < right {
+		res[left] = start
+		res[right] = -start
+		start++
+		left = left + 1
+		right = right - 1
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1302.Deepest-Leaves-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1305.All-Elements-in-Two-Binary-Search-Trees/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1305.All-Elements-in-Two-Binary-Search-Trees.md b/website/content/ChapterFour/1300~1399/1305.All-Elements-in-Two-Binary-Search-Trees.md
new file mode 100644
index 000000000..a29b324c2
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1305.All-Elements-in-Two-Binary-Search-Trees.md
@@ -0,0 +1,97 @@
+# [1305. All Elements in Two Binary Search Trees](https://leetcode.com/problems/all-elements-in-two-binary-search-trees/)
+
+
+
+## 题目
+
+Given two binary search trees `root1` and `root2`.
+
+Return a list containing *all the integers* from *both trees* sorted in **ascending** order.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2019/12/18/q2-e1.png](https://assets.leetcode.com/uploads/2019/12/18/q2-e1.png)
+
+```
+Input: root1 = [2,1,4], root2 = [1,0,3]
+Output: [0,1,1,2,3,4]
+```
+
+**Example 2**:
+
+```
+Input: root1 = [0,-10,10], root2 = [5,1,7,0,2]
+Output: [-10,0,0,1,2,5,7,10]
+```
+
+**Example 3**:
+
+```
+Input: root1 = [], root2 = [5,1,7,0,2]
+Output: [0,1,2,5,7]
+```
+
+**Example 4**:
+
+```
+Input: root1 = [0,-10,10], root2 = []
+Output: [-10,0,10]
+```
+
+**Example 5**:
+
+![https://assets.leetcode.com/uploads/2019/12/18/q2-e5-.png](https://assets.leetcode.com/uploads/2019/12/18/q2-e5-.png)
+
+```
+Input: root1 = [1,null,8], root2 = [8,1]
+Output: [1,1,8,8]
+```
+
+**Constraints**:
+
+- Each tree has at most `5000` nodes.
+- Each node's value is between `[-10^5, 10^5]`.
+
+## 题目大意
+
+给你 root1 和 root2 这两棵二叉搜索树。请你返回一个列表，其中包含 两棵树 中的所有整数并按 升序 排序。.
+
+提示：
+
+- 每棵树最多有 5000 个节点。
+- 每个节点的值在 [-10^5, 10^5] 之间。
+
+
+## 解题思路
+
+- 给出 2 棵二叉排序树，要求将 2 棵树所有节点的值按照升序排序。
+- 这一题最暴力简单的方法就是把 2 棵树的节点都遍历出来，然后放在一个数组里面从小到大排序即可。这样做虽然能 AC，但是时间复杂度高。因为题目中给的二叉排序树这一条件没有用上。由于树中节点本来已经有序了，所以题目实质想要我们合并 2 个有序数组。利用中根遍历，把 2 个二叉排序树的所有节点值都遍历出来，遍历出来以后就是有序的。接下来再合并这两个有序数组即可。合并 2 个有序数组是第 88 题。
+
+## 代码
+
+```go
+// 解法一 合并排序
+func getAllElements(root1 *TreeNode, root2 *TreeNode) []int {
+	arr1 := inorderTraversal(root1)
+	arr2 := inorderTraversal(root2)
+	arr1 = append(arr1, make([]int, len(arr2))...)
+	merge(arr1, len(arr1)-len(arr2), arr2, len(arr2))
+	return arr1
+}
+
+// 解法二 暴力遍历排序，时间复杂度高
+func getAllElements1(root1 *TreeNode, root2 *TreeNode) []int {
+	arr := []int{}
+	arr = append(arr, preorderTraversal(root1)...)
+	arr = append(arr, preorderTraversal(root2)...)
+	sort.Ints(arr)
+	return arr
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1304.Find-N-Unique-Integers-Sum-up-to-Zero/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1306.Jump-Game-III/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1306.Jump-Game-III.md b/website/content/ChapterFour/1300~1399/1306.Jump-Game-III.md
new file mode 100644
index 000000000..e34767f0e
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1306.Jump-Game-III.md
@@ -0,0 +1,87 @@
+# [1306. Jump Game III](https://leetcode.com/problems/jump-game-iii/)
+
+
+## 题目
+
+Given an array of non-negative integers `arr`, you are initially positioned at `start` index of the array. When you are at index `i`, you can jump to `i + arr[i]` or `i - arr[i]`, check if you can reach to **any** index with value 0.
+
+Notice that you can not jump outside of the array at any time.
+
+**Example 1**:
+
+```
+Input: arr = [4,2,3,0,3,1,2], start = 5
+Output: true
+Explanation: 
+All possible ways to reach at index 3 with value 0 are: 
+index 5 -> index 4 -> index 1 -> index 3 
+index 5 -> index 6 -> index 4 -> index 1 -> index 3
+```
+
+**Example 2**:
+
+```
+Input: arr = [4,2,3,0,3,1,2], start = 0
+Output: true 
+Explanation: 
+One possible way to reach at index 3 with value 0 is: 
+index 0 -> index 4 -> index 1 -> index 3
+```
+
+**Example 3**:
+
+```
+Input: arr = [3,0,2,1,2], start = 2
+Output: false
+Explanation: There is no way to reach at index 1 with value 0.
+```
+
+**Constraints**:
+
+- `1 <= arr.length <= 5 * 10^4`
+- `0 <= arr[i] < arr.length`
+- `0 <= start < arr.length`
+
+
+## 题目大意
+
+这里有一个非负整数数组 arr，你最开始位于该数组的起始下标 start 处。当你位于下标 i 处时，你可以跳到 i + arr[i] 或者 i - arr[i]。请你判断自己是否能够跳到对应元素值为 0 的 任一 下标处。注意，不管是什么情况下，你都无法跳到数组之外。
+
+提示：
+
+- 1 <= arr.length <= 5 * 10^4
+- 0 <= arr[i] < arr.length
+- 0 <= start < arr.length
+
+
+## 解题思路
+
+- 给出一个非负数组和一个起始下标 `start`。站在 `start`，每次可以跳到 `i + arr[i]` 或者 `i - arr[i]` 。要求判断能否跳到元素值为 0 的下标处。
+- 这一题考察的是递归。每一步都需要判断 3 种可能：
+    1. 当前是否站在元素值为 0 的目标点上。
+    2. 往前跳 arr[start]，是否能站在元素值为 0 的目标点上。
+    3. 往后跳 arr[start]，是否能站在元素值为 0 的目标点上。
+
+    第 2 种可能和第 3 种可能递归即可，每一步都判断这 3 种可能是否有一种能跳到元素值为 0 的下标处。
+
+- `arr[start] += len(arr)`  这一步仅仅只是为了标记此下标已经用过了，下次判断的时候该下标会被 `if` 条件筛选掉。
+
+## 代码
+
+```go
+func canReach(arr []int, start int) bool {
+	if start >= 0 && start < len(arr) && arr[start] < len(arr) {
+		jump := arr[start]
+		arr[start] += len(arr)
+		return jump == 0 || canReach(arr, start+jump) || canReach(arr, start-jump)
+	}
+	return false
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1305.All-Elements-in-Two-Binary-Search-Trees/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1310.XOR-Queries-of-a-Subarray/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1310.XOR-Queries-of-a-Subarray.md b/website/content/ChapterFour/1300~1399/1310.XOR-Queries-of-a-Subarray.md
new file mode 100644
index 000000000..4c90917b3
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1310.XOR-Queries-of-a-Subarray.md
@@ -0,0 +1,84 @@
+# [1310. XOR Queries of a Subarray](https://leetcode.com/problems/xor-queries-of-a-subarray/)
+
+
+## 题目
+
+Given the array `arr` of positive integers and the array `queries` where `queries[i] = [Li,Ri]`, for each query `i` compute the **XOR** of elements from `Li` to `Ri` (that is, `arr[Li]xor arr[Li+1]xor ...xor arr[Ri]`). Return an array containing the result for the given `queries`.
+
+**Example 1:**
+
+```
+Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
+Output: [2,7,14,8]
+Explanation:
+The binary representation of the elements in the array are:
+1 = 0001
+3 = 0011
+4 = 0100
+8 = 1000
+The XOR values for queries are:
+[0,1] = 1 xor 3 = 2
+[1,2] = 3 xor 4 = 7
+[0,3] = 1 xor 3 xor 4 xor 8 = 14
+[3,3] = 8
+
+```
+
+**Example 2:**
+
+```
+Input: arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
+Output: [8,0,4,4]
+
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 3 * 10^4`
+- `1 <= arr[i] <= 10^9`
+- `1 <= queries.length <= 3 * 10^4`
+- `queries[i].length == 2`
+- `0 <= queries[i][0] <= queries[i][1] < arr.length`
+
+## 题目大意
+
+有一个正整数数组 arr，现给你一个对应的查询数组 queries，其中 queries[i] = [Li, Ri]。对于每个查询 i，请你计算从 Li 到 Ri 的 XOR 值（即 arr[Li] xor arr[Li+1] xor ... xor arr[Ri]）作为本次查询的结果。并返回一个包含给定查询 queries 所有结果的数组。
+
+## 解题思路
+
+- 此题求区间异或，很容易让人联想到区间求和。区间求和利用前缀和，可以使得 query 从 O(n) 降为 O(1)。区间异或能否也用类似前缀和的思想呢？答案是肯定的。利用异或的两个性质，x ^ x = 0，x ^ 0 = x。那么有：（由于 LaTeX 中异或符号 ^ 是特殊字符，笔者用 {{< katex >}} \oplus {{< /katex >}}  代替异或）
+
+	{{< katex display >}}
+    \begin{aligned}Query(left,right) &=arr[left] \oplus \cdots  \oplus arr[right]\\&=(arr[0] \oplus \cdots  \oplus arr[left-1]) \oplus (arr[0] \oplus \cdots  \oplus arr[left-1]) \oplus (arr[left] \oplus \cdots  \oplus arr[right])\\ &=(arr[0] \oplus \cdots  \oplus arr[left-1]) \oplus (arr[0] \oplus \cdots  \oplus arr[right])\\ &=xors[left] \oplus xors[right+1]\\ \end{aligned}
+	{{< /katex >}}
+
+    按照这个思路解题，便可以将 query 从 O(n) 降为 O(1)，总的时间复杂度为 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+func xorQueries(arr []int, queries [][]int) []int {
+	xors := make([]int, len(arr))
+	xors[0] = arr[0]
+	for i := 1; i < len(arr); i++ {
+		xors[i] = arr[i] ^ xors[i-1]
+	}
+	res := make([]int, len(queries))
+	for i, q := range queries {
+		res[i] = xors[q[1]]
+		if q[0] > 0 {
+			res[i] ^= xors[q[0]-1]
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1306.Jump-Game-III/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1313.Decompress-Run-Length-Encoded-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1313.Decompress-Run-Length-Encoded-List.md b/website/content/ChapterFour/1300~1399/1313.Decompress-Run-Length-Encoded-List.md
new file mode 100644
index 000000000..4d56280fb
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1313.Decompress-Run-Length-Encoded-List.md
@@ -0,0 +1,67 @@
+# [1313. Decompress Run-Length Encoded List](https://leetcode.com/problems/decompress-run-length-encoded-list/)
+
+
+## 题目
+
+We are given a list `nums` of integers representing a list compressed with run-length encoding.
+
+Consider each adjacent pair of elements `[freq, val] = [nums[2*i], nums[2*i+1]]` (with `i >= 0`). For each such pair, there are `freq` elements with value `val` concatenated in a sublist. Concatenate all the sublists from left to right to generate the decompressed list.
+
+Return the decompressed list.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,4]
+Output: [2,4,4,4]
+Explanation: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2].
+The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4].
+At the end the concatenation [2] + [4,4,4] is [2,4,4,4].
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,1,2,3]
+Output: [1,3,3]
+```
+
+**Constraints**:
+
+- `2 <= nums.length <= 100`
+- `nums.length % 2 == 0`
+- `1 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个以行程长度编码压缩的整数列表 nums 。考虑每对相邻的两个元素 [freq, val] = [nums[2*i], nums[2*i+1]] （其中 i >= 0 ），每一对都表示解压后子列表中有 freq 个值为 val 的元素，你需要从左到右连接所有子列表以生成解压后的列表。请你返回解压后的列表。
+
+## 解题思路
+
+- 给定一个带编码长度的数组，要求解压这个数组。
+- 简单题。按照题目要求，下标从 0 开始，奇数位下标为前一个下标对应元素重复次数，那么就把这个元素 append 几次。最终输出解压后的数组即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func decompressRLElist(nums []int) []int {
+	res := []int{}
+	for i := 0; i < len(nums); i += 2 {
+		for j := 0; j < nums[i]; j++ {
+			res = append(res, nums[i+1])
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1310.XOR-Queries-of-a-Subarray/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers.md b/website/content/ChapterFour/1300~1399/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers.md
new file mode 100644
index 000000000..889849408
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers.md
@@ -0,0 +1,103 @@
+# [1317. Convert Integer to the Sum of Two No-Zero Integers](https://leetcode.com/problems/convert-integer-to-the-sum-of-two-no-zero-integers/)
+
+
+## 题目
+
+Given an integer `n`. No-Zero integer is a positive integer which **doesn't contain any 0** in its decimal representation.
+
+Return *a list of two integers* `[A, B]` where:
+
+- `A` and `B` are No-Zero integers.
+- `A + B = n`
+
+It's guarateed that there is at least one valid solution. If there are many valid solutions you can return any of them.
+
+**Example 1**:
+
+```
+Input: n = 2
+Output: [1,1]
+Explanation: A = 1, B = 1. A + B = n and both A and B don't contain any 0 in their decimal representation.
+```
+
+**Example 2**:
+
+```
+Input: n = 11
+Output: [2,9]
+```
+
+**Example 3**:
+
+```
+Input: n = 10000
+Output: [1,9999]
+```
+
+**Example 4**:
+
+```
+Input: n = 69
+Output: [1,68]
+```
+
+**Example 5**:
+
+```
+Input: n = 1010
+Output: [11,999]
+```
+
+**Constraints**:
+
+- `2 <= n <= 10^4`
+
+## 题目大意
+
+「无零整数」是十进制表示中 不含任何 0 的正整数。给你一个整数 n，请你返回一个 由两个整数组成的列表 [A, B]，满足：
+
+- A 和 B 都是无零整数
+- A + B = n
+
+题目数据保证至少有一个有效的解决方案。如果存在多个有效解决方案，你可以返回其中任意一个。
+
+## 解题思路
+
+- 给定一个整数 n，要求把它分解为 2 个十进制位中不含 0 的正整数且这两个正整数之和为 n。
+- 简单题。在 [1, n/2] 区间内搜索，只要有一组满足条件的解就 break。题目保证了至少有一组解，并且多组解返回任意一组即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func getNoZeroIntegers(n int) []int {
+	noZeroPair := []int{}
+	for i := 1; i <= n/2; i++ {
+		if isNoZero(i) && isNoZero(n-i) {
+			noZeroPair = append(noZeroPair, []int{i, n - i}...)
+			break
+		}
+	}
+	return noZeroPair
+}
+
+func isNoZero(n int) bool {
+	for n != 0 {
+		if n%10 == 0 {
+			return false
+		}
+		n /= 10
+	}
+	return true
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1313.Decompress-Run-Length-Encoded-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1319.Number-of-Operations-to-Make-Network-Connected/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1319.Number-of-Operations-to-Make-Network-Connected.md b/website/content/ChapterFour/1300~1399/1319.Number-of-Operations-to-Make-Network-Connected.md
new file mode 100644
index 000000000..2adf943ce
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1319.Number-of-Operations-to-Make-Network-Connected.md
@@ -0,0 +1,97 @@
+# [1319. Number of Operations to Make Network Connected](https://leetcode.com/problems/number-of-operations-to-make-network-connected/)
+
+
+## 题目
+
+There are `n` computers numbered from `0` to `n-1` connected by ethernet cables `connections` forming a network where `connections[i] = [a, b]` represents a connection between computers `a` and `b`. Any computer can reach any other computer directly or indirectly through the network.
+
+Given an initial computer network `connections`. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected. Return the *minimum number of times* you need to do this in order to make all the computers connected. If it's not possible, return -1.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/01/02/sample_1_1677.png](https://assets.leetcode.com/uploads/2020/01/02/sample_1_1677.png)
+
+```
+Input: n = 4, connections = [[0,1],[0,2],[1,2]]
+Output: 1
+Explanation: Remove cable between computer 1 and 2 and place between computers 1 and 3.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/01/02/sample_2_1677.png](https://assets.leetcode.com/uploads/2020/01/02/sample_2_1677.png)
+
+```
+Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
+Output: 2
+```
+
+**Example 3:**
+
+```
+Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
+Output: -1
+Explanation: There are not enough cables.
+```
+
+**Example 4:**
+
+```
+Input: n = 5, connections = [[0,1],[0,2],[3,4],[2,3]]
+Output: 0
+```
+
+**Constraints:**
+
+- `1 <= n <= 10^5`
+- `1 <= connections.length <= min(n*(n-1)/2, 10^5)`
+- `connections[i].length == 2`
+- `0 <= connections[i][0], connections[i][1] < n`
+- `connections[i][0] != connections[i][1]`
+- There are no repeated connections.
+- No two computers are connected by more than one cable.
+
+## 题目大意
+
+用以太网线缆将 n 台计算机连接成一个网络，计算机的编号从 0 到 n-1。线缆用 connections 表示，其中 connections[i] = [a, b] 连接了计算机 a 和 b。网络中的任何一台计算机都可以通过网络直接或者间接访问同一个网络中其他任意一台计算机。给你这个计算机网络的初始布线 connections，你可以拔开任意两台直连计算机之间的线缆，并用它连接一对未直连的计算机。请你计算并返回使所有计算机都连通所需的最少操作次数。如果不可能，则返回 -1 。
+
+## 解题思路
+
+- 很明显这题的解题思路是并查集。先将每个 connections 构建出并查集。构建中需要累加冗余的连接。例如 2 个节点已经连通，再连接这个集合中的任意 2 个节点就算冗余连接。冗余连接的线都可以移动，去连接还没有连通的节点。计算出冗余连接数，再根据并查集的集合总数，即可得出答案。
+- 这一题答案有 3 种可能。第一种，所有点都在一个集合内，即全部连通，这时输出 0 。第二种，冗余的连接不够串起所有点，这时输出 -1 。第三种情况是可以连通的情况。 m 个集合需要连通，最少需要 m - 1 条线。如果冗余连接数大于 m - 1，则输出 m - 1 即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func makeConnected(n int, connections [][]int) int {
+	if n-1 > len(connections) {
+		return -1
+	}
+	uf, redundance := template.UnionFind{}, 0
+	uf.Init(n)
+	for _, connection := range connections {
+		if uf.Find(connection[0]) == uf.Find(connection[1]) {
+			redundance++
+		} else {
+			uf.Union(connection[0], connection[1])
+		}
+	}
+	if uf.TotalCount() == 1 || redundance < uf.TotalCount()-1 {
+		return 0
+	}
+	return uf.TotalCount() - 1
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1329.Sort-the-Matrix-Diagonally/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1329.Sort-the-Matrix-Diagonally.md b/website/content/ChapterFour/1300~1399/1329.Sort-the-Matrix-Diagonally.md
new file mode 100644
index 000000000..7389fe3a5
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1329.Sort-the-Matrix-Diagonally.md
@@ -0,0 +1,64 @@
+# [1329. Sort the Matrix Diagonally](https://leetcode.com/problems/sort-the-matrix-diagonally/)
+
+
+## 题目
+
+A **matrix diagonal** is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix's end. For example, the **matrix diagonal** starting from `mat[2][0]`, where `mat` is a `6 x 3` matrix, includes cells `mat[2][0]`, `mat[3][1]`, and `mat[4][2]`.
+
+Given an `m x n` matrix `mat` of integers, sort each **matrix diagonal** in ascending order and return *the resulting matrix*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/01/21/1482_example_1_2.png](https://assets.leetcode.com/uploads/2020/01/21/1482_example_1_2.png)
+
+```
+Input: mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]
+Output: [[1,1,1,1],[1,2,2,2],[1,2,3,3]]
+```
+
+**Constraints:**
+
+- `m == mat.length`
+- `n == mat[i].length`
+- `1 <= m, n <= 100`
+- `1 <= mat[i][j] <= 100`
+
+## 题目大意
+
+给你一个 m * n 的整数矩阵 mat ，请你将同一条对角线上的元素（从左上到右下）按升序排序后，返回排好序的矩阵。
+
+## 解题思路
+
+- 这道题思路很简单。按照对角线，把每条对角线的元素读取出来放在数组中。这里可以利用 map 保存这些数组。再将这些数组排序。最后按照对角线还原矩阵即可。
+
+## 代码
+
+```go
+package leetcode
+
+func diagonalSort(mat [][]int) [][]int {
+	m, n, diagonalsMap := len(mat), len(mat[0]), make(map[int][]int)
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			diagonalsMap[i-j] = append(diagonalsMap[i-j], mat[i][j])
+		}
+	}
+	for _, v := range diagonalsMap {
+		sort.Ints(v)
+	}
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			mat[i][j] = diagonalsMap[i-j][0]
+			diagonalsMap[i-j] = diagonalsMap[i-j][1:]
+		}
+	}
+	return mat
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1319.Number-of-Operations-to-Make-Network-Connected/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1332.Remove-Palindromic-Subsequences/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1332.Remove-Palindromic-Subsequences.md b/website/content/ChapterFour/1300~1399/1332.Remove-Palindromic-Subsequences.md
new file mode 100644
index 000000000..7814cbb69
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1332.Remove-Palindromic-Subsequences.md
@@ -0,0 +1,88 @@
+# [1332. Remove Palindromic Subsequences](https://leetcode.com/problems/remove-palindromic-subsequences/)
+
+
+## 题目
+
+Given a string `s` consisting only of letters `'a'` and `'b'`. In a single step you can remove one palindromic **subsequence** from `s`.
+
+Return the minimum number of steps to make the given string empty.
+
+A string is a subsequence of a given string, if it is generated by deleting some characters of a given string without changing its order.
+
+A string is called palindrome if is one that reads the same backward as well as forward.
+
+**Example 1:**
+
+```
+Input: s = "ababa"
+Output: 1
+Explanation: String is already palindrome
+```
+
+**Example 2:**
+
+```
+Input: s = "abb"
+Output: 2
+Explanation: "abb" -> "bb" -> "". 
+Remove palindromic subsequence "a" then "bb".
+```
+
+**Example 3:**
+
+```
+Input: s = "baabb"
+Output: 2
+Explanation: "baabb" -> "b" -> "". 
+Remove palindromic subsequence "baab" then "b".
+```
+
+**Example 4:**
+
+```
+Input: s = ""
+Output: 0
+```
+
+**Constraints:**
+
+- `0 <= s.length <= 1000`
+- `s` only consists of letters 'a' and 'b'
+
+## 题目大意
+
+给你一个字符串 s，它仅由字母 'a' 和 'b' 组成。每一次删除操作都可以从 s 中删除一个回文 子序列。返回删除给定字符串中所有字符（字符串为空）的最小删除次数。
+
+「子序列」定义：如果一个字符串可以通过删除原字符串某些字符而不改变原字符顺序得到，那么这个字符串就是原字符串的一个子序列。
+
+「回文」定义：如果一个字符串向后和向前读是一致的，那么这个字符串就是一个回文。
+
+## 解题思路
+
+- 笔者读完题以为是第 5 题的加强版。在字符串中每次都找到最长的回文子串删除，一直删除到找不到回文子串结束，删除的总次数 + 剩余字母数 = 最小删除次数。提交以后 `wrong answer` 了，在 `bbaabaaa` 这组测试用例出错了。如果按照找最长回文字符串的思路，先找到最长回文子串 `aabaa`，剩余 `bba`，还需要再删除 2 次，`bb` 和 `a`。总共删除次数是 3 。为什么出错误了呢？仔细再读题，题目中说的是子序列，这不是连续的，再加上这道题是 `easy` 难度，其实很简单。
+- 这道题的答案只可能是 0，1，2 。空串对应的 0 。如果有一个字母，单个字母可以构成回文，所以是 1，如果字符串长度大于等于 2，即 `a` 和 `b` 都有，第一步先删除所有 `a`，因为所有的 `a` 构成了回文子序列。第二步删除所有的 `b`，因为所有的 `b` 构成了回文子序列。经过这样两步，一定能删除所有字符。
+
+## 代码
+
+```go
+package leetcode
+
+func removePalindromeSub(s string) int {
+	if len(s) == 0 {
+		return 0
+	}
+	for i := 0; i < len(s)/2; i++ {
+		if s[i] != s[len(s)-1-i] {
+			return 2
+		}
+	}
+	return 1
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1329.Sort-the-Matrix-Diagonally/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1337.The-K-Weakest-Rows-in-a-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1337.The-K-Weakest-Rows-in-a-Matrix.md b/website/content/ChapterFour/1300~1399/1337.The-K-Weakest-Rows-in-a-Matrix.md
new file mode 100644
index 000000000..a00d8d9b0
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1337.The-K-Weakest-Rows-in-a-Matrix.md
@@ -0,0 +1,97 @@
+# [1337. The K Weakest Rows in a Matrix](https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/)
+
+
+## 题目
+
+Given a `m * n` matrix `mat` of *ones* (representing soldiers) and *zeros* (representing civilians), return the indexes of the `k` weakest rows in the matrix ordered from the weakest to the strongest.
+
+A row ***i*** is weaker than row ***j***, if the number of soldiers in row ***i*** is less than the number of soldiers in row ***j***, or they have the same number of soldiers but ***i*** is less than ***j***. Soldiers are **always** stand in the frontier of a row, that is, always *ones* may appear first and then *zeros*.
+
+**Example 1:**
+
+```
+Input: mat = 
+[[1,1,0,0,0],
+ [1,1,1,1,0],
+ [1,0,0,0,0],
+ [1,1,0,0,0],
+ [1,1,1,1,1]], 
+k = 3
+Output: [2,0,3]
+Explanation: 
+The number of soldiers for each row is: 
+row 0 -> 2 
+row 1 -> 4 
+row 2 -> 1 
+row 3 -> 2 
+row 4 -> 5 
+Rows ordered from the weakest to the strongest are [2,0,3,1,4]
+
+```
+
+**Example 2:**
+
+```
+Input: mat = 
+[[1,0,0,0],
+ [1,1,1,1],
+ [1,0,0,0],
+ [1,0,0,0]], 
+k = 2
+Output: [0,2]
+Explanation: 
+The number of soldiers for each row is: 
+row 0 -> 1 
+row 1 -> 4 
+row 2 -> 1 
+row 3 -> 1 
+Rows ordered from the weakest to the strongest are [0,2,3,1]
+
+```
+
+**Constraints:**
+
+- `m == mat.length`
+- `n == mat[i].length`
+- `2 <= n, m <= 100`
+- `1 <= k <= m`
+- `matrix[i][j]` is either 0 **or** 1.
+
+## 题目大意
+
+给你一个大小为 m * n 的矩阵 mat，矩阵由若干军人和平民组成，分别用 1 和 0 表示。请你返回矩阵中战斗力最弱的 k 行的索引，按从最弱到最强排序。如果第 i 行的军人数量少于第 j 行，或者两行军人数量相同但 i 小于 j，那么我们认为第 i 行的战斗力比第 j 行弱。军人 总是 排在一行中的靠前位置，也就是说 1 总是出现在 0 之前。
+
+## 解题思路
+
+- 简单题。第一个能想到的解题思路是，先统计每一行 1 的个数，然后将结果进行排序，按照 1 的个数从小到大排序，如果 1 的个数相同，再按照行号从小到大排序。排好序的数组取出前 K 位即为答案。
+- 此题还有第二种解法。在第一种解法中，并没有用到题目中“军人 总是 排在一行中的靠前位置，也就是说 1 总是出现在 0 之前。”这一条件。由于有了这个条件，使得如果按照列去遍历，最先出现 0 的行，则是最弱的行。行号小的先被遍历到，所以相同数量 1 的行，行号小的会排在前面。最后记得再添加上全 1 的行。同样，最终输出取出前 K 位即为答案。此题解法二才是最优雅最高效的解法。
+
+## 代码
+
+```go
+package leetcode
+
+func kWeakestRows(mat [][]int, k int) []int {
+	res := []int{}
+	for j := 0; j < len(mat[0]); j++ {
+		for i := 0; i < len(mat); i++ {
+			if mat[i][j] == 0 && ((j == 0) || (mat[i][j-1] != 0)) {
+				res = append(res, i)
+			}
+		}
+	}
+	for i := 0; i < len(mat); i++ {
+		if mat[i][len(mat[0])-1] == 1 {
+			res = append(res, i)
+		}
+	}
+	return res[:k]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1332.Remove-Palindromic-Subsequences/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1353.Maximum-Number-of-Events-That-Can-Be-Attended/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1353.Maximum-Number-of-Events-That-Can-Be-Attended.md b/website/content/ChapterFour/1300~1399/1353.Maximum-Number-of-Events-That-Can-Be-Attended.md
new file mode 100644
index 000000000..c0929cfa3
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1353.Maximum-Number-of-Events-That-Can-Be-Attended.md
@@ -0,0 +1,120 @@
+# [1353. Maximum Number of Events That Can Be Attended](https://leetcode.com/problems/maximum-number-of-events-that-can-be-attended/)
+
+
+## 题目
+
+Given an array of `events` where `events[i] = [startDayi, endDayi]`. Every event `i` starts at `startDayi` and ends at `endDayi`.
+
+You can attend an event `i` at any day `d` where `startTimei <= d <= endTimei`. Notice that you can only attend one event at any time `d`.
+
+Return *the maximum number of events* you can attend.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/02/05/e1.png](https://assets.leetcode.com/uploads/2020/02/05/e1.png)
+
+```
+Input: events = [[1,2],[2,3],[3,4]]
+Output: 3
+Explanation: You can attend all the three events.
+One way to attend them all is as shown.
+Attend the first event on day 1.
+Attend the second event on day 2.
+Attend the third event on day 3.
+
+```
+
+**Example 2:**
+
+```
+Input: events= [[1,2],[2,3],[3,4],[1,2]]
+Output: 4
+
+```
+
+**Example 3:**
+
+```
+Input: events = [[1,4],[4,4],[2,2],[3,4],[1,1]]
+Output: 4
+
+```
+
+**Example 4:**
+
+```
+Input: events = [[1,100000]]
+Output: 1
+
+```
+
+**Example 5:**
+
+```
+Input: events = [[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7]]
+Output: 7
+
+```
+
+**Constraints:**
+
+- `1 <= events.length <= 10^5`
+- `events[i].length == 2`
+- `1 <= startDayi <= endDayi <= 10^5`
+
+## 题目大意
+
+给你一个数组 events，其中 events[i] = [startDayi, endDayi] ，表示会议 i 开始于 startDayi ，结束于 endDayi 。你可以在满足 startDayi <= d <= endDayi 中的任意一天 d 参加会议 i 。注意，一天只能参加一个会议。请你返回你可以参加的 最大 会议数目。
+
+## 解题思路
+
+- 关于会议安排，活动安排这类题，第一直觉是贪心问题。先按照会议开始时间从小到大排序，如果开始时间相同，再按照结束时间从小到大排序。贪心策略是，优先选择参加早结束的会议。因为一个结束时间晚的会议，代表这个会议持续时间长，先参加马上要结束的会议，这样可以参加更多的会议。
+- 注意题目给的数据代表的是天数。比较大小的时候最好转换成坐标轴上的坐标点。例如 [1,2] 代表这个会议持续 2 天，如果在坐标轴上表示，是 [0,2]，0-1 表示第一天，1-2 表示第二天。所以比较会议时需要把开始时间减一。选定了这个会议以后记得要把这一天排除，例如选择了第二天，那么下次对比起始时间需要从坐标 2 开始，因为第二天的时间范围是 1-2，所以下一轮比较会议前需要把开始时间加一。从左往右依次扫描各个会议时间段，选择结束时间大于起始时间的会议，不断累加次数，扫描完所有会议，最终结果即为可参加的最大会议数。
+- 测试数据中有一组很恶心的数据，见 test 文件中最后一组数据。这组数据在同一天叠加了多个会议，并且起始时间完全一致。这种特殊情况需要加判断条件排除，见下面代码 continue 条件。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func maxEvents(events [][]int) int {
+	sort.Slice(events, func(i, j int) bool {
+		if events[i][0] == events[j][0] {
+			return events[i][1] < events[j][1]
+		}
+		return events[i][0] < events[j][0]
+	})
+	attended, current := 1, events[0]
+	for i := 1; i < len(events); i++ {
+		prev, event := events[i-1], events[i]
+		if event[0] == prev[0] && event[1] == prev[1] && event[1] == event[0] {
+			continue
+		}
+		start, end := max(current[0], event[0]-1), max(current[1], event[1])
+		if end-start > 0 {
+			current[0] = start + 1
+			current[1] = end
+			attended++
+		}
+	}
+	return attended
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1337.The-K-Weakest-Rows-in-a-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1380.Lucky-Numbers-in-a-Matrix/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1380.Lucky-Numbers-in-a-Matrix.md b/website/content/ChapterFour/1300~1399/1380.Lucky-Numbers-in-a-Matrix.md
new file mode 100644
index 000000000..e17321058
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1380.Lucky-Numbers-in-a-Matrix.md
@@ -0,0 +1,94 @@
+# [1380. Lucky Numbers in a Matrix](https://leetcode.com/problems/lucky-numbers-in-a-matrix/)
+
+
+## 题目
+
+Given a `m * n` matrix of **distinct** numbers, return all lucky numbers in the matrix in **any** order.
+
+A lucky number is an element of the matrix such that it is the minimum element in its row and maximum in its column.
+
+**Example 1**:
+
+```
+Input: matrix = [[3,7,8],[9,11,13],[15,16,17]]
+Output: [15]
+Explanation: 15 is the only lucky number since it is the minimum in its row and the maximum in its column
+```
+
+**Example 2**:
+
+```
+Input: matrix = [[1,10,4,2],[9,3,8,7],[15,16,17,12]]
+Output: [12]
+Explanation: 12 is the only lucky number since it is the minimum in its row and the maximum in its column.
+```
+
+**Example 3**:
+
+```
+Input: matrix = [[7,8],[1,2]]
+Output: [7]
+```
+
+**Constraints**:
+
+- `m == mat.length`
+- `n == mat[i].length`
+- `1 <= n, m <= 50`
+- `1 <= matrix[i][j] <= 10^5`.
+- All elements in the matrix are distinct.
+
+## 题目大意
+
+给你一个 m * n 的矩阵，矩阵中的数字 各不相同 。请你按 任意 顺序返回矩阵中的所有幸运数。幸运数是指矩阵中满足同时下列两个条件的元素：
+
+- 在同一行的所有元素中最小
+- 在同一列的所有元素中最大
+
+
+
+## 解题思路
+
+- 找出矩阵中的幸运数。幸运数的定义：同时满足 2 个条件，在同一行的所有元素中最小并且在同一列的所有元素中最大。
+- 简单题。按照题意遍历矩阵，找到同时满足 2 个条件的数输出即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func luckyNumbers(matrix [][]int) []int {
+	t, r, res := make([]int, len(matrix[0])), make([]int, len(matrix[0])), []int{}
+	for _, val := range matrix {
+		m, k := val[0], 0
+		for j := 0; j < len(matrix[0]); j++ {
+			if val[j] < m {
+				m = val[j]
+				k = j
+			}
+			if t[j] < val[j] {
+				t[j] = val[j]
+			}
+		}
+
+		if t[k] == m {
+			r[k] = m
+		}
+	}
+	for k, v := range r {
+		if v > 0 && v == t[k] {
+			res = append(res, v)
+		}
+	}
+	return res
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1353.Maximum-Number-of-Events-That-Can-Be-Attended/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1383.Maximum-Performance-of-a-Team/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1383.Maximum-Performance-of-a-Team.md b/website/content/ChapterFour/1300~1399/1383.Maximum-Performance-of-a-Team.md
new file mode 100644
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--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1383.Maximum-Performance-of-a-Team.md
@@ -0,0 +1,112 @@
+# [1383. Maximum Performance of a Team](https://leetcode.com/problems/maximum-performance-of-a-team/)
+
+## 题目
+
+You are given two integers `n` and `k` and two integer arrays `speed` and `efficiency` both of length `n`. There are `n` engineers numbered from `1` to `n`. `speed[i]` and `efficiency[i]` represent the speed and efficiency of the `ith` engineer respectively.
+
+Choose **at most** `k` different engineers out of the `n` engineers to form a team with the maximum **performance**.
+
+The performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers.
+
+Return *the maximum performance of this team*. Since the answer can be a huge number, return it **modulo** `109 + 7`.
+
+**Example 1:**
+
+```
+Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
+Output: 60
+Explanation:
+We have the maximum performance of the team by selecting engineer 2 (with speed=10 and efficiency=4) and engineer 5 (with speed=5 and efficiency=7). That is, performance = (10 + 5) * min(4, 7) = 60.
+```
+
+**Example 2:**
+
+```
+Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
+Output: 68
+Explanation:
+This is the same example as the first but k = 3. We can select engineer 1, engineer 2 and engineer 5 to get the maximum performance of the team. That is, performance = (2 + 10 + 5) * min(5, 4, 7) = 68.
+```
+
+**Example 3:**
+
+```
+Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4
+Output: 72
+```
+
+**Constraints:**
+
+- `1 <= <= k <= n <= 105`
+- `speed.length == n`
+- `efficiency.length == n`
+- `1 <= speed[i] <= 105`
+- `1 <= efficiency[i] <= 108`
+
+## 题目大意
+
+公司有编号为 1 到 n 的 n 个工程师，给你两个数组 speed 和 efficiency ，其中 speed[i] 和 efficiency[i] 分别代表第 i 位工程师的速度和效率。请你返回由最多 k 个工程师组成的 最大团队表现值 ，由于答案可能很大，请你返回结果对 10^9 + 7 取余后的结果。团队表现值 的定义为：一个团队中「所有工程师速度的和」乘以他们「效率值中的最小值」。
+
+## 解题思路
+
+- 题目要求返回最大团队表现值，表现值需要考虑速度的累加和，和效率的最小值。即使速度快，效率的最小值很小，总的表现值还是很小。先将效率从大到小排序。从效率高的工程师开始选起，遍历过程中维护一个大小为 k 的速度最小堆。每次遍历都计算一次团队最大表现值。扫描完成，最大团队表现值也筛选出来了。具体实现见下面的代码。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+func maxPerformance(n int, speed []int, efficiency []int, k int) int {
+	indexes := make([]int, n)
+	for i := range indexes {
+		indexes[i] = i
+	}
+	sort.Slice(indexes, func(i, j int) bool {
+		return efficiency[indexes[i]] > efficiency[indexes[j]]
+	})
+	ph := speedHeap{}
+	heap.Init(&ph)
+	speedSum := 0
+	var max int64
+	for _, index := range indexes {
+		if ph.Len() == k {
+			speedSum -= heap.Pop(&ph).(int)
+		}
+		speedSum += speed[index]
+		heap.Push(&ph, speed[index])
+		max = Max(max, int64(speedSum)*int64(efficiency[index]))
+	}
+	return int(max % (1e9 + 7))
+}
+
+type speedHeap []int
+
+func (h speedHeap) Less(i, j int) bool  { return h[i] < h[j] }
+func (h speedHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
+func (h speedHeap) Len() int            { return len(h) }
+func (h *speedHeap) Push(x interface{}) { *h = append(*h, x.(int)) }
+func (h *speedHeap) Pop() interface{} {
+	res := (*h)[len(*h)-1]
+	*h = (*h)[:h.Len()-1]
+	return res
+}
+
+func Max(a, b int64) int64 {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1380.Lucky-Numbers-in-a-Matrix/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1385.Find-the-Distance-Value-Between-Two-Arrays/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1385.Find-the-Distance-Value-Between-Two-Arrays.md b/website/content/ChapterFour/1300~1399/1385.Find-the-Distance-Value-Between-Two-Arrays.md
new file mode 100644
index 000000000..02541558a
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1385.Find-the-Distance-Value-Between-Two-Arrays.md
@@ -0,0 +1,105 @@
+# [1385. Find the Distance Value Between Two Arrays](https://leetcode.com/problems/find-the-distance-value-between-two-arrays/)
+
+
+## 题目
+
+Given two integer arrays `arr1` and `arr2`, and the integer `d`, *return the distance value between the two arrays*.
+
+The distance value is defined as the number of elements `arr1[i]` such that there is not any element `arr2[j]` where `|arr1[i]-arr2[j]| <= d`.
+
+**Example 1**:
+
+```
+Input: arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2
+Output: 2
+Explanation: 
+For arr1[0]=4 we have: 
+|4-10|=6 > d=2 
+|4-9|=5 > d=2 
+|4-1|=3 > d=2 
+|4-8|=4 > d=2 
+For arr1[1]=5 we have: 
+|5-10|=5 > d=2 
+|5-9|=4 > d=2 
+|5-1|=4 > d=2 
+|5-8|=3 > d=2
+For arr1[2]=8 we have:
+|8-10|=2 <= d=2
+|8-9|=1 <= d=2
+|8-1|=7 > d=2
+|8-8|=0 <= d=2
+```
+
+**Example 2**:
+
+```
+Input: arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3
+Output: 2
+```
+
+**Example 3**:
+
+```
+Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6
+Output: 1
+```
+
+**Constraints**:
+
+- `1 <= arr1.length, arr2.length <= 500`
+- `-10^3 <= arr1[i], arr2[j] <= 10^3`
+- `0 <= d <= 100`
+
+
+## 题目大意
+
+给你两个整数数组 arr1 ， arr2 和一个整数 d ，请你返回两个数组之间的 距离值 。「距离值」 定义为符合此距离要求的元素数目：对于元素 arr1[i] ，不存在任何元素 arr2[j] 满足 |arr1[i]-arr2[j]| <= d 。
+
+提示：
+
+- 1 <= arr1.length, arr2.length <= 500
+- -10^3 <= arr1[i], arr2[j] <= 10^3
+- 0 <= d <= 100
+
+
+## 解题思路
+
+- 计算两个数组之间的距离，距离值的定义：满足对于元素 arr1[i] ，不存在任何元素 arr2[j] 满足 |arr1[i]-arr2[j]| <= d 这一条件的元素数目。
+- 简单题，按照距离值的定义，双层循环计数即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func findTheDistanceValue(arr1 []int, arr2 []int, d int) int {
+	res := 0
+	for i := range arr1 {
+		for j := range arr2 {
+			if abs(arr1[i]-arr2[j]) <= d {
+				break
+			}
+			if j == len(arr2)-1 {
+				res++
+			}
+		}
+	}
+	return res
+}
+
+func abs(a int) int {
+	if a < 0 {
+		return -1 * a
+	}
+	return a
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1383.Maximum-Performance-of-a-Team/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1389.Create-Target-Array-in-the-Given-Order/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1389.Create-Target-Array-in-the-Given-Order.md b/website/content/ChapterFour/1300~1399/1389.Create-Target-Array-in-the-Given-Order.md
new file mode 100644
index 000000000..678e8f053
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1389.Create-Target-Array-in-the-Given-Order.md
@@ -0,0 +1,97 @@
+# [1389. Create Target Array in the Given Order](https://leetcode.com/problems/create-target-array-in-the-given-order/)
+
+## 题目
+
+Given two arrays of integers `nums` and `index`. Your task is to create *target* array under the following rules:
+
+- Initially *target* array is empty.
+- From left to right read nums[i] and index[i], insert at index `index[i]` the value `nums[i]` in *target* array.
+- Repeat the previous step until there are no elements to read in `nums` and `index.`
+
+Return the *target* array.
+
+It is guaranteed that the insertion operations will be valid.
+
+**Example 1**:
+
+```
+Input: nums = [0,1,2,3,4], index = [0,1,2,2,1]
+Output: [0,4,1,3,2]
+Explanation:
+nums       index     target
+0            0        [0]
+1            1        [0,1]
+2            2        [0,1,2]
+3            2        [0,1,3,2]
+4            1        [0,4,1,3,2]
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,2,3,4,0], index = [0,1,2,3,0]
+Output: [0,1,2,3,4]
+Explanation:
+nums       index     target
+1            0        [1]
+2            1        [1,2]
+3            2        [1,2,3]
+4            3        [1,2,3,4]
+0            0        [0,1,2,3,4]
+```
+
+**Example 3**:
+
+```
+Input: nums = [1], index = [0]
+Output: [1]
+```
+
+**Constraints**:
+
+- `1 <= nums.length, index.length <= 100`
+- `nums.length == index.length`
+- `0 <= nums[i] <= 100`
+- `0 <= index[i] <= i`
+
+## 题目大意
+
+给你两个整数数组 nums 和 index。你需要按照以下规则创建目标数组：
+
+- 目标数组 target 最初为空。
+- 按从左到右的顺序依次读取 nums[i] 和 index[i]，在 target 数组中的下标 index[i] 处插入值 nums[i] 。
+- 重复上一步，直到在 nums 和 index 中都没有要读取的元素。
+
+请你返回目标数组。题目保证数字插入位置总是存在。
+
+
+
+
+## 解题思路
+
+- 给定 2 个数组，分别装的是待插入的元素和待插入的位置。最后输出操作完成的数组。
+- 简单题，按照题意插入元素即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func createTargetArray(nums []int, index []int) []int {
+	result := make([]int, len(nums))
+	for i, pos := range index {
+		copy(result[pos+1:i+1], result[pos:i])
+		result[pos] = nums[i]
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1385.Find-the-Distance-Value-Between-Two-Arrays/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1396.Design-Underground-System/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/1396.Design-Underground-System.md b/website/content/ChapterFour/1300~1399/1396.Design-Underground-System.md
new file mode 100644
index 000000000..883a83945
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/1396.Design-Underground-System.md
@@ -0,0 +1,162 @@
+# [1396. Design Underground System](https://leetcode.com/problems/design-underground-system/)
+
+
+## 题目
+
+Implement the `UndergroundSystem` class:
+
+- `void checkIn(int id, string stationName, int t)`
+    - A customer with a card id equal to `id`, gets in the station `stationName` at time `t`.
+    - A customer can only be checked into one place at a time.
+- `void checkOut(int id, string stationName, int t)`
+    - A customer with a card id equal to `id`, gets out from the station `stationName` at time `t`.
+- `double getAverageTime(string startStation, string endStation)`
+    - Returns the average time to travel between the `startStation` and the `endStation`.
+    - The average time is computed from all the previous traveling from `startStation` to `endStation` that happened **directly**.
+    - Call to `getAverageTime` is always valid.
+
+You can assume all calls to `checkIn` and `checkOut` methods are consistent. If a customer gets in at time **t1** at some station, they get out at time **t2** with **t2 > t1**. All events happen in chronological order.
+
+**Example 1:**
+
+```
+Input
+["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"]
+[[],[45,"Leyton",3],[32,"Paradise",8],[27,"Leyton",10],[45,"Waterloo",15],[27,"Waterloo",20],[32,"Cambridge",22],["Paradise","Cambridge"],["Leyton","Waterloo"],[10,"Leyton",24],["Leyton","Waterloo"],[10,"Waterloo",38],["Leyton","Waterloo"]]
+
+Output
+[null,null,null,null,null,null,null,14.00000,11.00000,null,11.00000,null,12.00000]
+
+Explanation
+UndergroundSystem undergroundSystem = new UndergroundSystem();
+undergroundSystem.checkIn(45, "Leyton", 3);
+undergroundSystem.checkIn(32, "Paradise", 8);
+undergroundSystem.checkIn(27, "Leyton", 10);
+undergroundSystem.checkOut(45, "Waterloo", 15);
+undergroundSystem.checkOut(27, "Waterloo", 20);
+undergroundSystem.checkOut(32, "Cambridge", 22);
+undergroundSystem.getAverageTime("Paradise", "Cambridge");       // return 14.00000. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
+undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.00000. There were two travels from "Leyton" to "Waterloo", a customer with id=45 from time=3 to time=15 and a customer with id=27 from time=10 to time=20. So the average time is ( (15-3) + (20-10) ) / 2 = 11.00000
+undergroundSystem.checkIn(10, "Leyton", 24);
+undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.00000
+undergroundSystem.checkOut(10, "Waterloo", 38);
+undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 12.00000
+```
+
+**Example 2:**
+
+```
+Input
+["UndergroundSystem","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime"]
+[[],[10,"Leyton",3],[10,"Paradise",8],["Leyton","Paradise"],[5,"Leyton",10],[5,"Paradise",16],["Leyton","Paradise"],[2,"Leyton",21],[2,"Paradise",30],["Leyton","Paradise"]]
+
+Output
+[null,null,null,5.00000,null,null,5.50000,null,null,6.66667]
+
+Explanation
+UndergroundSystem undergroundSystem = new UndergroundSystem();
+undergroundSystem.checkIn(10, "Leyton", 3);
+undergroundSystem.checkOut(10, "Paradise", 8);
+undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000
+undergroundSystem.checkIn(5, "Leyton", 10);
+undergroundSystem.checkOut(5, "Paradise", 16);
+undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.50000
+undergroundSystem.checkIn(2, "Leyton", 21);
+undergroundSystem.checkOut(2, "Paradise", 30);
+undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 6.66667
+```
+
+**Constraints:**
+
+- There will be at most `20000` operations.
+- `1 <= id, t <= 106`
+- All strings consist of uppercase and lowercase English letters, and digits.
+- `1 <= stationName.length <= 10`
+- Answers within `105` of the actual value will be accepted as correct.
+
+## 题目大意
+
+请你实现一个类 UndergroundSystem ，它支持以下 3 种方法：
+
+- 1. checkIn(int id, string stationName, int t)
+    - 编号为 id 的乘客在 t 时刻进入地铁站 stationName 。
+    - 一个乘客在同一时间只能在一个地铁站进入或者离开。
+- 2. checkOut(int id, string stationName, int t)
+    - 编号为 id 的乘客在 t 时刻离开地铁站 stationName 。
+- 3. getAverageTime(string startStation, string endStation)
+    - 返回从地铁站 startStation 到地铁站 endStation 的平均花费时间。
+    - 平均时间计算的行程包括当前为止所有从 startStation 直接到达 endStation 的行程。
+    - 调用 getAverageTime 时，询问的路线至少包含一趟行程。
+
+你可以假设所有对 checkIn 和 checkOut 的调用都是符合逻辑的。也就是说，如果一个顾客在 t1 时刻到达某个地铁站，那么他离开的时间 t2 一定满足 t2 > t1 。所有的事件都按时间顺序给出。
+
+## 解题思路
+
+- 维护 2 个 `map`。一个 `mapA` 内部存储的是乘客 `id` 与（入站时间，站名）的对应关系。另外一个 `mapB` 存储的是起点站与终点站花费总时间与人数总数的关系。每当有人 `checkin()`，就更新 `mapA` 中的信息。每当有人 `checkout()`，就更新 `mapB` 中的信息，并删除 `mapA` 对应乘客 `id` 的键值对。最后调用 `getAverageTime()` 函数的时候根据 `mapB` 中存储的信息计算即可。
+
+## 代码
+
+```go
+package leetcode
+
+type checkin struct {
+	station string
+	time    int
+}
+
+type stationTime struct {
+	sum, count float64
+}
+
+type UndergroundSystem struct {
+	checkins     map[int]*checkin
+	stationTimes map[string]map[string]*stationTime
+}
+
+func Constructor() UndergroundSystem {
+	return UndergroundSystem{
+		make(map[int]*checkin),
+		make(map[string]map[string]*stationTime),
+	}
+}
+
+func (s *UndergroundSystem) CheckIn(id int, stationName string, t int) {
+	s.checkins[id] = &checkin{stationName, t}
+}
+
+func (s *UndergroundSystem) CheckOut(id int, stationName string, t int) {
+	checkin := s.checkins[id]
+	destination := s.stationTimes[checkin.station]
+	if destination == nil {
+		s.stationTimes[checkin.station] = make(map[string]*stationTime)
+	}
+	st := s.stationTimes[checkin.station][stationName]
+	if st == nil {
+		st = new(stationTime)
+		s.stationTimes[checkin.station][stationName] = st
+	}
+	st.sum += float64(t - checkin.time)
+	st.count++
+	delete(s.checkins, id)
+}
+
+func (s *UndergroundSystem) GetAverageTime(startStation string, endStation string) float64 {
+	st := s.stationTimes[startStation][endStation]
+	return st.sum / st.count
+}
+
+/**
+ * Your UndergroundSystem object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.CheckIn(id,stationName,t);
+ * obj.CheckOut(id,stationName,t);
+ * param_3 := obj.GetAverageTime(startStation,endStation);
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1389.Create-Target-Array-in-the-Given-Order/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1423.Maximum-Points-You-Can-Obtain-from-Cards/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1300~1399/_index.md b/website/content/ChapterFour/1300~1399/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/1300~1399/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/1400~1499/1423.Maximum-Points-You-Can-Obtain-from-Cards.md b/website/content/ChapterFour/1400~1499/1423.Maximum-Points-You-Can-Obtain-from-Cards.md
new file mode 100644
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@@ -0,0 +1,97 @@
+# [1423. Maximum Points You Can Obtain from Cards](https://leetcode.com/problems/maximum-points-you-can-obtain-from-cards/)
+
+
+## 题目
+
+There are several cards **arranged in a row**, and each card has an associated number of points The points are given in the integer array `cardPoints`.
+
+In one step, you can take one card from the beginning or from the end of the row. You have to take exactly `k` cards.
+
+Your score is the sum of the points of the cards you have taken.
+
+Given the integer array `cardPoints` and the integer `k`, return the *maximum score* you can obtain.
+
+**Example 1:**
+
+```
+Input: cardPoints = [1,2,3,4,5,6,1], k = 3
+Output: 12
+Explanation: After the first step, your score will always be 1. However, choosing the rightmost card first will maximize your total score. The optimal strategy is to take the three cards on the right, giving a final score of 1 + 6 + 5 = 12.
+```
+
+**Example 2:**
+
+```
+Input: cardPoints = [2,2,2], k = 2
+Output: 4
+Explanation: Regardless of which two cards you take, your score will always be 4.
+```
+
+**Example 3:**
+
+```
+Input: cardPoints = [9,7,7,9,7,7,9], k = 7
+Output: 55
+Explanation: You have to take all the cards. Your score is the sum of points of all cards.
+```
+
+**Example 4:**
+
+```
+Input: cardPoints = [1,1000,1], k = 1
+Output: 1
+Explanation: You cannot take the card in the middle. Your best score is 1. 
+```
+
+**Example 5:**
+
+```
+Input: cardPoints = [1,79,80,1,1,1,200,1], k = 3
+Output: 202
+```
+
+**Constraints:**
+
+- `1 <= cardPoints.length <= 10^5`
+- `1 <= cardPoints[i] <= 10^4`
+- `1 <= k <= cardPoints.length`
+
+## 题目大意
+
+几张卡牌 排成一行，每张卡牌都有一个对应的点数。点数由整数数组 cardPoints 给出。每次行动，你可以从行的开头或者末尾拿一张卡牌，最终你必须正好拿 k 张卡牌。你的点数就是你拿到手中的所有卡牌的点数之和。给你一个整数数组 cardPoints 和整数 k，请你返回可以获得的最大点数。
+
+## 解题思路
+
+- 这一题是滑动窗口题的简化题。从卡牌两边取 K 张牌，可以转换成在中间连续取 n-K 张牌。从两边取牌的点数最大，意味着剩下来中间牌的点数最小。扫描一遍数组，在每一个窗口大小为 n-K 的窗口内计算累加和，记录下最小的累加和。题目最终求的最大点数等于牌的总和减去中间最小的累加和。
+
+## 代码
+
+```go
+package leetcode
+
+func maxScore(cardPoints []int, k int) int {
+	windowSize, sum := len(cardPoints)-k, 0
+	for _, val := range cardPoints[:windowSize] {
+		sum += val
+	}
+	minSum := sum
+	for i := windowSize; i < len(cardPoints); i++ {
+		sum += cardPoints[i] - cardPoints[i-windowSize]
+		if sum < minSum {
+			minSum = sum
+		}
+	}
+	total := 0
+	for _, pt := range cardPoints {
+		total += pt
+	}
+	return total - minSum
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1300~1399/1396.Design-Underground-System/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away.md b/website/content/ChapterFour/1400~1499/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away.md
new file mode 100644
index 000000000..430eb24bc
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+++ b/website/content/ChapterFour/1400~1499/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away.md
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+# [1437. Check If All 1's Are at Least Length K Places Away](https://leetcode.com/problems/check-if-all-1s-are-at-least-length-k-places-away/)
+
+
+## 题目
+
+Given an array `nums` of 0s and 1s and an integer `k`, return `True` if all 1's are at least `k` places away from each other, otherwise return `False`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/04/15/sample_1_1791.png](https://assets.leetcode.com/uploads/2020/04/15/sample_1_1791.png)
+
+```
+Input: nums = [1,0,0,0,1,0,0,1], k = 2
+Output: true
+Explanation: Each of the 1s are at least 2 places away from each other.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/04/15/sample_2_1791.png](https://assets.leetcode.com/uploads/2020/04/15/sample_2_1791.png)
+
+```
+Input: nums = [1,0,0,1,0,1], k = 2
+Output: false
+Explanation: The second 1 and third 1 are only one apart from each other.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,1,1,1,1], k = 0
+Output: true
+```
+
+**Example 4:**
+
+```
+Input: nums = [0,1,0,1], k = 1
+Output: true
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 10^5`
+- `0 <= k <= nums.length`
+- `nums[i]` is `0` or `1`
+
+## 题目大意
+
+给你一个由若干 0 和 1 组成的数组 nums 以及整数 k。如果所有 1 都至少相隔 k 个元素，则返回 True ；否则，返回 False 。
+
+## 解题思路
+
+- 简单题。扫描一遍数组，遇到 1 的时候比较前一个 1 的下标索引，如果相隔小于 k 则返回 false。如果大于等于 k 就更新下标索引，继续循环。循环结束输出 true 即可。
+
+## 代码
+
+```go
+package leetcode
+
+func kLengthApart(nums []int, k int) bool {
+	prevIndex := -1
+	for i, num := range nums {
+		if num == 1 {
+			if prevIndex != -1 && i-prevIndex-1 < k {
+				return false
+			}
+			prevIndex = i
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1423.Maximum-Points-You-Can-Obtain-from-Cards/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit.md b/website/content/ChapterFour/1400~1499/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit.md
new file mode 100644
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+++ b/website/content/ChapterFour/1400~1499/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit.md
@@ -0,0 +1,96 @@
+# [1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit](https://leetcode.com/problems/longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit/)
+
+
+## 题目
+
+Given an array of integers `nums` and an integer `limit`, return the size of the longest **non-empty** subarray such that the absolute difference between any two elements of this subarray is less than or equal to `limit`*.*
+
+**Example 1:**
+
+```
+Input: nums = [8,2,4,7], limit = 4
+Output: 2 
+Explanation: All subarrays are: 
+[8] with maximum absolute diff |8-8| = 0 <= 4.
+[8,2] with maximum absolute diff |8-2| = 6 > 4. 
+[8,2,4] with maximum absolute diff |8-2| = 6 > 4.
+[8,2,4,7] with maximum absolute diff |8-2| = 6 > 4.
+[2] with maximum absolute diff |2-2| = 0 <= 4.
+[2,4] with maximum absolute diff |2-4| = 2 <= 4.
+[2,4,7] with maximum absolute diff |2-7| = 5 > 4.
+[4] with maximum absolute diff |4-4| = 0 <= 4.
+[4,7] with maximum absolute diff |4-7| = 3 <= 4.
+[7] with maximum absolute diff |7-7| = 0 <= 4. 
+Therefore, the size of the longest subarray is 2.
+```
+
+**Example 2:**
+
+```
+Input: nums = [10,1,2,4,7,2], limit = 5
+Output: 4 
+Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5.
+```
+
+**Example 3:**
+
+```
+Input: nums = [4,2,2,2,4,4,2,2], limit = 0
+Output: 3
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 10^5`
+- `1 <= nums[i] <= 10^9`
+- `0 <= limit <= 10^9`
+
+## 题目大意
+
+给你一个整数数组 nums ，和一个表示限制的整数 limit，请你返回最长连续子数组的长度，该子数组中的任意两个元素之间的绝对差必须小于或者等于 limit 。如果不存在满足条件的子数组，则返回 0 。
+
+## 解题思路
+
+- 最开始想到的思路是利用滑动窗口遍历一遍数组，每个窗口内排序，取出最大最小值。滑动窗口遍历一次的时间复杂度是 O(n)，所以此题时间复杂度是否高效落在了排序算法上了。由于前后 2 个窗口数据是有关联的，仅仅只变动了 2 个数据（左窗口移出的数据和右窗口移进的数据），所以排序没有必要每次都重新排序。这里利用二叉排序树来排序，添加和删除元素时间复杂度是 O(log n)，这种方法总的时间复杂度是 O(n log n)。空间复杂度 O(n)。
+- 二叉排序树的思路是否还有再优化的空间？答案是有。二叉排序树内维护了所有结点的有序关系，但是这个关系是多余的。此题只需要找到最大值和最小值，并不需要除此以外节点的有序信息。所以用二叉排序树是大材小用了。可以换成 2 个单调队列，一个维护窗口内的最大值，另一个维护窗口内的最小值。这样优化以后，时间复杂度降低到 O(n)，空间复杂度 O(n)。具体实现见代码。
+- 单调栈的题还有第 42 题，第 84 题，第 496 题，第 503 题，第 739 题，第 856 题，第 901 题，第 907 题，第 1130 题，第 1425 题，第 1673 题。
+
+## 代码
+
+```go
+package leetcode
+
+func longestSubarray(nums []int, limit int) int {
+	minStack, maxStack, left, res := []int{}, []int{}, 0, 0
+	for right, num := range nums {
+		for len(minStack) > 0 && nums[minStack[len(minStack)-1]] > num {
+			minStack = minStack[:len(minStack)-1]
+		}
+		minStack = append(minStack, right)
+		for len(maxStack) > 0 && nums[maxStack[len(maxStack)-1]] < num {
+			maxStack = maxStack[:len(maxStack)-1]
+		}
+		maxStack = append(maxStack, right)
+		if len(minStack) > 0 && len(maxStack) > 0 && nums[maxStack[0]]-nums[minStack[0]] > limit {
+			if left == minStack[0] {
+				minStack = minStack[1:]
+			}
+			if left == maxStack[0] {
+				maxStack = maxStack[1:]
+			}
+			left++
+		}
+		if right-left+1 > res {
+			res = right - left + 1
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows.md b/website/content/ChapterFour/1400~1499/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows.md
new file mode 100644
index 000000000..574a8f3a3
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows.md
@@ -0,0 +1,151 @@
+# [1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows](https://leetcode.com/problems/find-the-kth-smallest-sum-of-a-matrix-with-sorted-rows/)
+
+
+## 题目
+
+You are given an `m * n` matrix, `mat`, and an integer `k`, which has its rows sorted in non-decreasing order.
+
+You are allowed to choose exactly 1 element from each row to form an array. Return the Kth **smallest** array sum among all possible arrays.
+
+**Example 1:**
+
+```
+Input: mat = [[1,3,11],[2,4,6]], k = 5
+Output: 7
+Explanation: Choosing one element from each row, the first k smallest sum are:
+[1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.  
+```
+
+**Example 2:**
+
+```
+Input: mat = [[1,3,11],[2,4,6]], k = 9
+Output: 17
+```
+
+**Example 3:**
+
+```
+Input: mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7
+Output: 9
+Explanation: Choosing one element from each row, the first k smallest sum are:
+[1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.  
+```
+
+**Example 4:**
+
+```
+Input: mat = [[1,1,10],[2,2,9]], k = 7
+Output: 12
+```
+
+**Constraints:**
+
+- `m == mat.length`
+- `n == mat.length[i]`
+- `1 <= m, n <= 40`
+- `1 <= k <= min(200, n ^ m)`
+- `1 <= mat[i][j] <= 5000`
+- `mat[i]` is a non decreasing array.
+
+## 题目大意
+
+给你一个 m * n 的矩阵 mat，以及一个整数 k ，矩阵中的每一行都以非递减的顺序排列。你可以从每一行中选出 1 个元素形成一个数组。返回所有可能数组中的第 k 个 最小数组和。
+
+## 解题思路
+
+- 这一题是第 373 题的升级版。在第 373 题中，给定 2 个有序数组，要求分别从这 2 个数组中选出一个数组成一个数对，最终输出和最小的 K 组。这一题中给出的是 m*n 的矩阵。其实是将第 373 题的 2 个数组升级为了 m 个数组。无非外层多了一层循环。这层循环依次从每一行中选出一个数，先从第 0 行和第 1 行取数，找到前 K 小的组合以后，再从第 2 行取数，以此类推。其他做法和第 373 题一致。维护一个长度为 k 的最小堆。每次从堆中 pop 出最小的数组和 sum 和对应的下标 index，然后依次将下标向后移动一位，生成新的 sum，加入堆中。
+
+## 代码
+
+```go
+package leetcode
+
+import "container/heap"
+
+func kthSmallest(mat [][]int, k int) int {
+	if len(mat) == 0 || len(mat[0]) == 0 || k == 0 {
+		return 0
+	}
+	prev := mat[0]
+	for i := 1; i < len(mat); i++ {
+		prev = kSmallestPairs(prev, mat[i], k)
+	}
+	if k < len(prev) {
+		return -1
+	}
+	return prev[k-1]
+}
+
+func kSmallestPairs(nums1 []int, nums2 []int, k int) []int {
+	res := []int{}
+	if len(nums2) == 0 {
+		return res
+	}
+	pq := newPriorityQueue()
+	for i := 0; i < len(nums1) && i < k; i++ {
+		heap.Push(pq, &pddata{
+			n1:    nums1[i],
+			n2:    nums2[0],
+			n2Idx: 0,
+		})
+	}
+	for pq.Len() > 0 {
+		i := heap.Pop(pq)
+		data := i.(*pddata)
+		res = append(res, data.n1+data.n2)
+		k--
+		if k <= 0 {
+			break
+		}
+		idx := data.n2Idx
+		idx++
+		if idx >= len(nums2) {
+			continue
+		}
+		heap.Push(pq, &pddata{
+			n1:    data.n1,
+			n2:    nums2[idx],
+			n2Idx: idx,
+		})
+	}
+	return res
+}
+
+type pddata struct {
+	n1    int
+	n2    int
+	n2Idx int
+}
+
+type priorityQueue []*pddata
+
+func newPriorityQueue() *priorityQueue {
+	pq := priorityQueue([]*pddata{})
+	heap.Init(&pq)
+	return &pq
+}
+
+func (pq priorityQueue) Len() int           { return len(pq) }
+func (pq priorityQueue) Swap(i, j int)      { pq[i], pq[j] = pq[j], pq[i] }
+func (pq priorityQueue) Less(i, j int) bool { return pq[i].n1+pq[i].n2 < pq[j].n1+pq[j].n2 }
+func (pq *priorityQueue) Pop() interface{} {
+	old := *pq
+	val := old[len(old)-1]
+	old[len(old)-1] = nil
+	*pq = old[0 : len(old)-1]
+	return val
+}
+
+func (pq *priorityQueue) Push(i interface{}) {
+	val := i.(*pddata)
+	*pq = append(*pq, val)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR.md b/website/content/ChapterFour/1400~1499/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR.md
new file mode 100644
index 000000000..e8fcfcce1
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR.md
@@ -0,0 +1,104 @@
+# [1442. Count Triplets That Can Form Two Arrays of Equal XOR](https://leetcode.com/problems/count-triplets-that-can-form-two-arrays-of-equal-xor/)
+
+
+## 题目
+
+Given an array of integers `arr`.
+
+We want to select three indices `i`, `j` and `k` where `(0 <= i < j <= k < arr.length)`.
+
+Let's define `a` and `b` as follows:
+
+- `a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]`
+- `b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]`
+
+Note that **^** denotes the **bitwise-xor** operation.
+
+Return *the number of triplets* (`i`, `j` and `k`) Where `a == b`.
+
+**Example 1:**
+
+```
+Input: arr = [2,3,1,6,7]
+Output: 4
+Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)
+```
+
+**Example 2:**
+
+```
+Input: arr = [1,1,1,1,1]
+Output: 10
+```
+
+**Example 3:**
+
+```
+Input: arr = [2,3]
+Output: 0
+```
+
+**Example 4:**
+
+```
+Input: arr = [1,3,5,7,9]
+Output: 3
+```
+
+**Example 5:**
+
+```
+Input: arr = [7,11,12,9,5,2,7,17,22]
+Output: 8
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 300`
+- `1 <= arr[i] <= 10^8`
+
+## 题目大意
+
+给你一个整数数组 arr 。现需要从数组中取三个下标 i、j 和 k ，其中 (0 <= i < j <= k < arr.length) 。a 和 b 定义如下：
+
+- a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
+- b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]
+
+注意：^ 表示 按位异或 操作。请返回能够令 a == b 成立的三元组 (i, j , k) 的数目。
+
+## 解题思路
+
+- 这一题需要用到 `x^x = 0` 这个异或特性。题目要求 `a == b`，可以等效转化为 `arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1] ^ arr[j] ^ arr[j + 1] ^ ... ^ arr[k] = 0`，这样 j 相当于可以“忽略”，专注找到所有元素异或结果为 0 的区间 [i,k] 即为答案。利用前缀和的思想，只不过此题非累加和，而是异或。又由 `x^x = 0` 这个异或特性，相同部分异或相当于消除，于是有 `prefix[i,k] = prefix[0,k] ^ prefix[0,i-1]`，找到每一个 `prefix[i,k] = 0` 的 i，k 组合，i < j <= k，那么满足条件的三元组 (i,j,k) 的个数完全取决于 j 的取值范围，(因为 i 和 k 已经固定了)，j 的取值范围为 k-i，所以累加所有满足条件的 k-i，输出即为最终答案。
+
+## 代码
+
+```go
+package leetcode
+
+func countTriplets(arr []int) int {
+	prefix, num, count, total := make([]int, len(arr)), 0, 0, 0
+	for i, v := range arr {
+		num ^= v
+		prefix[i] = num
+	}
+	for i := 0; i < len(prefix)-1; i++ {
+		for k := i + 1; k < len(prefix); k++ {
+			total = prefix[k]
+			if i > 0 {
+				total ^= prefix[i-1]
+			}
+			if total == 0 {
+				count += k - i
+			}
+		}
+	}
+	return count
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1446.Consecutive-Characters/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1446.Consecutive-Characters.md b/website/content/ChapterFour/1400~1499/1446.Consecutive-Characters.md
new file mode 100644
index 000000000..1d1beb7ff
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1446.Consecutive-Characters.md
@@ -0,0 +1,82 @@
+# [1446. Consecutive Characters](https://leetcode.com/problems/consecutive-characters/)
+
+## 题目
+
+The power of the string is the maximum length of a non-empty substring that contains only one unique character.
+
+Given a string s, return the power of s.
+
+**Example 1**:
+
+    Input: s = "leetcode"
+    Output: 2
+    Explanation: The substring "ee" is of length 2 with the character 'e' only.
+
+**Example 2**:
+
+    Input: s = "abbcccddddeeeeedcba"
+    Output: 5
+    Explanation: The substring "eeeee" is of length 5 with the character 'e' only.
+
+**Example 3**:
+
+    Input: s = "triplepillooooow"
+    Output: 5
+
+**Example 4**:
+
+    Input: s = "hooraaaaaaaaaaay"
+    Output: 11
+
+**Example 5**:
+
+    Input: s = "tourist"
+    Output: 1
+
+**Constraints:**
+
+- 1 <= s.length <= 500
+- s consists of only lowercase English letters.
+
+## 题目大意
+
+给你一个字符串 s ，字符串的「能量」定义为：只包含一种字符的最长非空子字符串的长度。
+
+请你返回字符串的能量。
+
+## 解题思路
+
+- 顺序遍历进行统计
+
+## 代码
+
+```go
+package leetcode
+
+func maxPower(s string) int {
+	cur, cnt, ans := s[0], 1, 1
+	for i := 1; i < len(s); i++ {
+		if cur == s[i] {
+			cnt++
+		} else {
+			if cnt > ans {
+				ans = cnt
+			}
+			cur = s[i]
+			cnt = 1
+		}
+	}
+	if cnt > ans {
+		ans = cnt
+	}
+	return ans
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence.md b/website/content/ChapterFour/1400~1499/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence.md
new file mode 100644
index 000000000..3bba6e43b
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence.md
@@ -0,0 +1,105 @@
+# [1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence](https://leetcode.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/)
+
+
+## 题目
+
+Given a `sentence` that consists of some words separated by a **single space**, and a `searchWord`.
+
+You have to check if `searchWord` is a prefix of any word in `sentence`.
+
+Return *the index of the word* in `sentence` where `searchWord` is a prefix of this word (**1-indexed**).
+
+If `searchWord` is a prefix of more than one word, return the index of the first word **(minimum index)**. If there is no such word return **-1**.
+
+A **prefix** of a string `S` is any leading contiguous substring of `S`.
+
+**Example 1**:
+
+```
+Input: sentence = "i love eating burger", searchWord = "burg"
+Output: 4
+Explanation: "burg" is prefix of "burger" which is the 4th word in the sentence.
+
+```
+
+**Example 2**:
+
+```
+Input: sentence = "this problem is an easy problem", searchWord = "pro"
+Output: 2
+Explanation: "pro" is prefix of "problem" which is the 2nd and the 6th word in the sentence, but we return 2 as it's the minimal index.
+
+```
+
+**Example 3**:
+
+```
+Input: sentence = "i am tired", searchWord = "you"
+Output: -1
+Explanation: "you" is not a prefix of any word in the sentence.
+
+```
+
+**Example 4**:
+
+```
+Input: sentence = "i use triple pillow", searchWord = "pill"
+Output: 4
+
+```
+
+**Example 5**:
+
+```
+Input: sentence = "hello from the other side", searchWord = "they"
+Output: -1
+
+```
+
+**Constraints**:
+
+- `1 <= sentence.length <= 100`
+- `1 <= searchWord.length <= 10`
+- `sentence` consists of lowercase English letters and spaces.
+- `searchWord` consists of lowercase English letters.
+
+## 题目大意
+
+给你一个字符串 sentence 作为句子并指定检索词为 searchWord ，其中句子由若干用 单个空格 分隔的单词组成。请你检查检索词 searchWord 是否为句子 sentence 中任意单词的前缀。
+
+- 如果 searchWord 是某一个单词的前缀，则返回句子 sentence 中该单词所对应的下标（下标从 1 开始）。
+- 如果 searchWord 是多个单词的前缀，则返回匹配的第一个单词的下标（最小下标）。
+- 如果 searchWord 不是任何单词的前缀，则返回 -1 。
+
+字符串 S 的 「前缀」是 S 的任何前导连续子字符串。
+
+## 解题思路
+
+- 给出 2 个字符串，一个是匹配串，另外一个是句子。在句子里面查找带匹配串前缀的单词，并返回第一个匹配单词的下标。
+- 简单题。按照题意，扫描一遍句子，一次匹配即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func isPrefixOfWord(sentence string, searchWord string) int {
+	for i, v := range strings.Split(sentence, " ") {
+		if strings.HasPrefix(v, searchWord) {
+			return i + 1
+		}
+	}
+	return -1
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1446.Consecutive-Characters/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K.md b/website/content/ChapterFour/1400~1499/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K.md
new file mode 100644
index 000000000..606ccb6dd
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K.md
@@ -0,0 +1,93 @@
+# [1461. Check If a String Contains All Binary Codes of Size K](https://leetcode.com/problems/check-if-a-string-contains-all-binary-codes-of-size-k/)
+
+
+## 题目
+
+Given a binary string `s` and an integer `k`.
+
+Return *True* if every binary code of length `k` is a substring of `s`. Otherwise, return *False*.
+
+**Example 1:**
+
+```
+Input: s = "00110110", k = 2
+Output: true
+Explanation: The binary codes of length 2 are "00", "01", "10" and "11". They can be all found as substrings at indicies 0, 1, 3 and 2 respectively.
+```
+
+**Example 2:**
+
+```
+Input: s = "00110", k = 2
+Output: true
+```
+
+**Example 3:**
+
+```
+Input: s = "0110", k = 1
+Output: true
+Explanation: The binary codes of length 1 are "0" and "1", it is clear that both exist as a substring. 
+```
+
+**Example 4:**
+
+```
+Input: s = "0110", k = 2
+Output: false
+Explanation: The binary code "00" is of length 2 and doesn't exist in the array.
+```
+
+**Example 5:**
+
+```
+Input: s = "0000000001011100", k = 4
+Output: false
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 5 * 10^5`
+- `s` consists of 0's and 1's only.
+- `1 <= k <= 20`
+
+## 题目大意
+
+给你一个二进制字符串 `s` 和一个整数 `k` 。如果所有长度为 `k` 的二进制字符串都是 `s` 的子串，请返回 `True` ，否则请返回 `False` 。
+
+## 解题思路
+
+- 构造一个 `mask` 遮罩，依次划过整个二进制字符串，每次滑动即取出遮罩遮住的 `k` 位二进制字符。可以用 `map` 存储不同的二进制转换成的十进制数，最后判断 `len(map)` 是否等于 `k` 即可。但是用 `map` 存比较慢，此处换成 `bool` 数组。先构造一个长度为 `k` 的数组，然后每次通过 `mask` 更新这个 `bool` 数组对应十进制的 `bool` 值，并且记录剩余还缺几个二进制数。等剩余的等于 0 的时候，说明所有二进制字符串都出现了，直接输出 `true`，否则循环完以后输出 `false`。
+
+## 代码
+
+```go
+package leetcode
+
+import "math"
+
+func hasAllCodes(s string, k int) bool {
+	need := int(math.Pow(2.0, float64(k)))
+	visited, mask, curr := make([]bool, need), (1<<k)-1, 0
+	for i := 0; i < len(s); i++ {
+		curr = ((curr << 1) | int(s[i]-'0')) & mask
+		if i >= k-1 { // mask 有效位达到了 k 位
+			if !visited[curr] {
+				need--
+				visited[curr] = true
+				if need == 0 {
+					return true
+				}
+			}
+		}
+	}
+	return false
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1463.Cherry-Pickup-II/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1463.Cherry-Pickup-II.md b/website/content/ChapterFour/1400~1499/1463.Cherry-Pickup-II.md
new file mode 100644
index 000000000..bf04d0caf
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1463.Cherry-Pickup-II.md
@@ -0,0 +1,156 @@
+# [1463. Cherry Pickup II](https://leetcode.com/problems/cherry-pickup-ii/)
+
+## 题目
+
+Given a `rows x cols` matrix `grid` representing a field of cherries. Each cell in `grid` represents the number of cherries that you can collect.
+
+You have two robots that can collect cherries for you, Robot #1 is located at the top-left corner (0,0) , and Robot #2 is located at the top-right corner (0, cols-1) of the grid.
+
+Return the maximum number of cherries collection using both robots  by following the rules below:
+
+- From a cell (i,j), robots can move to cell (i+1, j-1) , (i+1, j) or (i+1, j+1).
+- When any robot is passing through a cell, It picks it up all cherries, and the cell becomes an empty cell (0).
+- When both robots stay on the same cell, only one of them takes the cherries.
+- Both robots cannot move outside of the grid at any moment.
+- Both robots should reach the bottom row in the `grid`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/04/29/sample_1_1802.png](https://assets.leetcode.com/uploads/2020/04/29/sample_1_1802.png)
+
+```
+Input: grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
+Output: 24
+Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
+Cherries taken by Robot #1, (3 + 2 + 5 + 2) = 12.
+Cherries taken by Robot #2, (1 + 5 + 5 + 1) = 12.
+Total of cherries: 12 + 12 = 24.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/04/23/sample_2_1802.png](https://assets.leetcode.com/uploads/2020/04/23/sample_2_1802.png)
+
+```
+Input: grid = [[1,0,0,0,0,0,1],[2,0,0,0,0,3,0],[2,0,9,0,0,0,0],[0,3,0,5,4,0,0],[1,0,2,3,0,0,6]]
+Output: 28
+Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
+Cherries taken by Robot #1, (1 + 9 + 5 + 2) = 17.
+Cherries taken by Robot #2, (1 + 3 + 4 + 3) = 11.
+Total of cherries: 17 + 11 = 28.
+```
+
+**Example 3:**
+
+```
+Input: grid = [[1,0,0,3],[0,0,0,3],[0,0,3,3],[9,0,3,3]]
+Output: 22
+```
+
+**Example 4:**
+
+```
+Input: grid = [[1,1],[1,1]]
+Output: 4
+```
+
+**Constraints:**
+
+- `rows == grid.length`
+- `cols == grid[i].length`
+- `2 <= rows, cols <= 70`
+- `0 <= grid[i][j] <= 100`
+
+## 题目大意
+
+给你一个 rows x cols 的矩阵 grid 来表示一块樱桃地。 grid 中每个格子的数字表示你能获得的樱桃数目。你有两个机器人帮你收集樱桃，机器人 1 从左上角格子 (0,0) 出发，机器人 2 从右上角格子 (0, cols-1) 出发。请你按照如下规则，返回两个机器人能收集的最多樱桃数目：
+
+- 从格子 (i,j) 出发，机器人可以移动到格子 (i+1, j-1)，(i+1, j) 或者 (i+1, j+1) 。
+- 当一个机器人经过某个格子时，它会把该格子内所有的樱桃都摘走，然后这个位置会变成空格子，即没有樱桃的格子。
+- 当两个机器人同时到达同一个格子时，它们中只有一个可以摘到樱桃。
+- 两个机器人在任意时刻都不能移动到 grid 外面。
+- 两个机器人最后都要到达 grid 最底下一行。
+
+## 解题思路
+
+- 如果没有思路可以先用暴力解法 DFS 尝试。读完题可以分析出求最多樱桃数目，里面包含了很多重叠子问题，于是乎自然而然思路是用动态规划。数据规模上看，100 的数据规模最多能保证 O(n^3) 时间复杂度的算法不超时。
+- 这一题的变量有 2 个，一个是行号，另外一个是机器人所在的列。具体来说，机器人每走一步的移动范围只能往下走，不能往上走，所以 2 个机器人所在行号一定相同。两个机器人的列号不同。综上，变量有 3 个，1 个行号和2 个列号。定义 `dp[i][j][k]` 代表第一个机器人从 (0,0) 走到 (i,k) 坐标，第二个机器人从 (0,n-1) 走到 (i,k) 坐标，两者最多能收集樱桃的数目。状态转移方程为  ：
+	
+	{{< katex display >}} 
+    dp[i][j][k] = max \begin{pmatrix}\begin{array}{lr} dp[i-1][f(j_1))][f(j_2)] + grid[i][j_1] + grid[i][j_2], j_1\neq j_2  \\ dp[i-1][f(j_1))][f(j_2)] + grid[i][j_1], j_1 = j_2 \end{array} \end{pmatrix}
+    {{< /katex>}} 
+
+    其中：
+	
+	{{< katex display >}} 
+    \left\{\begin{matrix}f(j_1) \in [0,n), f(j_1) - j_1 \in [-1,0,1]\\ f(j_2) \in [0,n), f(j_2) - j_2 \in [-1,0,1]\end{matrix}\right.
+    {{< /katex>}}
+
+    即状态转移过程中需要在 `[j1 - 1, j1, j1 + 1]` 中枚举 `j1`，同理，在 在 `[j2 - 1, j2, j2 + 1]` 中枚举 `j2`，每个状态转移需要枚举这 3*3 = 9 种状态。
+
+- 边界条件 `dp[i][0][n-1] = grid[0][0] + grid[0][n-1]`，最终答案存储在 `dp[m-1]` 行中，循环找出 `dp[m-1][j1][j2]` 中的最大值，到此该题得解。
+
+## 代码
+
+```go
+package leetcode
+
+func cherryPickup(grid [][]int) int {
+	rows, cols := len(grid), len(grid[0])
+	dp := make([][][]int, rows)
+	for i := 0; i < rows; i++ {
+		dp[i] = make([][]int, cols)
+		for j := 0; j < cols; j++ {
+			dp[i][j] = make([]int, cols)
+		}
+	}
+	for i := 0; i < rows; i++ {
+		for j := 0; j <= i && j < cols; j++ {
+			for k := cols - 1; k >= cols-1-i && k >= 0; k-- {
+				max := 0
+				for a := j - 1; a <= j+1; a++ {
+					for b := k - 1; b <= k+1; b++ {
+						sum := isInBoard(dp, i-1, a, b)
+						if a == b && i > 0 && a >= 0 && a < cols {
+							sum -= grid[i-1][a]
+						}
+						if sum > max {
+							max = sum
+						}
+					}
+				}
+				if j == k {
+					max += grid[i][j]
+				} else {
+					max += grid[i][j] + grid[i][k]
+				}
+				dp[i][j][k] = max
+			}
+		}
+	}
+	count := 0
+	for j := 0; j < cols && j < rows; j++ {
+		for k := cols - 1; k >= 0 && k >= cols-rows; k-- {
+			if dp[rows-1][j][k] > count {
+				count = dp[rows-1][j][k]
+			}
+		}
+	}
+	return count
+}
+
+func isInBoard(dp [][][]int, i, j, k int) int {
+	if i < 0 || j < 0 || j >= len(dp[0]) || k < 0 || k >= len(dp[0]) {
+		return 0
+	}
+	return dp[i][j][k]
+}
+```
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1464.Maximum-Product-of-Two-Elements-in-an-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1464.Maximum-Product-of-Two-Elements-in-an-Array.md b/website/content/ChapterFour/1400~1499/1464.Maximum-Product-of-Two-Elements-in-an-Array.md
new file mode 100644
index 000000000..d318eb25b
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1464.Maximum-Product-of-Two-Elements-in-an-Array.md
@@ -0,0 +1,73 @@
+# [1464. Maximum Product of Two Elements in an Array](https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/)
+
+
+## 题目
+
+Given the array of integers `nums`, you will choose two different indices `i` and `j` of that array. Return the maximum value of `(nums[i]-1)*(nums[j]-1)`.
+
+**Example 1**:
+
+```
+Input: nums = [3,4,5,2]
+Output: 12 
+Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12. 
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,5,4,5]
+Output: 16
+Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16.
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [3,7]
+Output: 12
+
+```
+
+**Constraints**:
+
+- `2 <= nums.length <= 500`
+- `1 <= nums[i] <= 10^3`
+
+## 题目大意
+
+给你一个整数数组 nums，请你选择数组的两个不同下标 i 和 j，使 (nums[i]-1)*(nums[j]-1) 取得最大值。请你计算并返回该式的最大值。
+
+## 解题思路
+
+- 简单题。循环一次，按照题意动态维护 2 个最大值，从而也使得 `(nums[i]-1)*(nums[j]-1)` 能取到最大值。
+
+## 代码
+
+```go
+
+package leetcode
+
+func maxProduct(nums []int) int {
+	max1, max2 := 0, 0
+	for _, num := range nums {
+		if num >= max1 {
+			max2 = max1
+			max1 = num
+		} else if num <= max1 && num >= max2 {
+			max2 = num
+		}
+	}
+	return (max1 - 1) * (max2 - 1)
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1463.Cherry-Pickup-II/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts.md b/website/content/ChapterFour/1400~1499/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts.md
new file mode 100644
index 000000000..9a4a4bf64
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts.md
@@ -0,0 +1,96 @@
+# [1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts](https://leetcode.com/problems/maximum-area-of-a-piece-of-cake-after-horizontal-and-vertical-cuts/)
+
+
+## 题目
+
+Given a rectangular cake with height `h` and width `w`, and two arrays of integers `horizontalCuts` and `verticalCuts` where `horizontalCuts[i]` is the distance from the top of the rectangular cake to the `ith` horizontal cut and similarly, `verticalCuts[j]` is the distance from the left of the rectangular cake to the `jth` vertical cut.
+
+*Return the maximum area of a piece of cake after you cut at each horizontal and vertical position provided in the arrays `horizontalCuts` and `verticalCuts`.* Since the answer can be a huge number, return this modulo 10^9 + 7.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/05/14/leetcode_max_area_2.png](https://assets.leetcode.com/uploads/2020/05/14/leetcode_max_area_2.png)
+
+```
+Input: h = 5, w = 4, horizontalCuts = [1,2,4], verticalCuts = [1,3]
+Output: 4
+Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. After you cut the cake, the green piece of cake has the maximum area.
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/05/14/leetcode_max_area_3.png](https://assets.leetcode.com/uploads/2020/05/14/leetcode_max_area_3.png)
+
+```
+Input: h = 5, w = 4, horizontalCuts = [3,1], verticalCuts = [1]
+Output: 6
+Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. After you cut the cake, the green and yellow pieces of cake have the maximum area.
+
+```
+
+**Example 3:**
+
+```
+Input: h = 5, w = 4, horizontalCuts = [3], verticalCuts = [3]
+Output: 9
+
+```
+
+**Constraints:**
+
+- `2 <= h, w <= 10^9`
+- `1 <= horizontalCuts.length < min(h, 10^5)`
+- `1 <= verticalCuts.length < min(w, 10^5)`
+- `1 <= horizontalCuts[i] < h`
+- `1 <= verticalCuts[i] < w`
+- It is guaranteed that all elements in `horizontalCuts` are distinct.
+- It is guaranteed that all elements in `verticalCuts` are distinct.
+
+## 题目大意
+
+矩形蛋糕的高度为 h 且宽度为 w，给你两个整数数组 horizontalCuts 和 verticalCuts，其中 horizontalCuts[i] 是从矩形蛋糕顶部到第  i 个水平切口的距离，类似地， verticalCuts[j] 是从矩形蛋糕的左侧到第 j 个竖直切口的距离。请你按数组 horizontalCuts 和 verticalCuts 中提供的水平和竖直位置切割后，请你找出 面积最大 的那份蛋糕，并返回其 面积 。由于答案可能是一个很大的数字，因此需要将结果对 10^9 + 7 取余后返回。
+
+## 解题思路
+
+- 读完题比较容易想到解题思路。找到水平切口最大的差值和竖直切口最大的差值，这 4 条边构成的矩形即为最大矩形。不过有特殊情况需要判断，切口除了题目给的切口坐标以外，默认还有 4 个切口，即蛋糕原始的 4 条边。如下图二，最大的矩形其实在切口之外。所以找水平切口最大差值和竖直切口最大差值需要考虑到蛋糕原始的 4 条边。
+
+![https://img.halfrost.com/Leetcode/leetcode_1465.png](https://img.halfrost.com/Leetcode/leetcode_1465.png)
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func maxArea(h int, w int, hcuts []int, vcuts []int) int {
+	sort.Ints(hcuts)
+	sort.Ints(vcuts)
+	maxw, maxl := hcuts[0], vcuts[0]
+	for i, c := range hcuts[1:] {
+		if c-hcuts[i] > maxw {
+			maxw = c - hcuts[i]
+		}
+	}
+	if h-hcuts[len(hcuts)-1] > maxw {
+		maxw = h - hcuts[len(hcuts)-1]
+	}
+	for i, c := range vcuts[1:] {
+		if c-vcuts[i] > maxl {
+			maxl = c - vcuts[i]
+		}
+	}
+	if w-vcuts[len(vcuts)-1] > maxl {
+		maxl = w - vcuts[len(vcuts)-1]
+	}
+	return (maxw * maxl) % (1000000007)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1464.Maximum-Product-of-Two-Elements-in-an-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1470.Shuffle-the-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1470.Shuffle-the-Array.md b/website/content/ChapterFour/1400~1499/1470.Shuffle-the-Array.md
new file mode 100644
index 000000000..fc653d637
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1470.Shuffle-the-Array.md
@@ -0,0 +1,71 @@
+# [1470. Shuffle the Array](https://leetcode.com/problems/shuffle-the-array/)
+
+## 题目
+
+Given the array `nums` consisting of `2n` elements in the form `[x1,x2,...,xn,y1,y2,...,yn]`.
+
+*Return the array in the form* `[x1,y1,x2,y2,...,xn,yn]`.
+
+**Example 1**:
+
+```
+Input: nums = [2,5,1,3,4,7], n = 3
+Output: [2,3,5,4,1,7] 
+Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7].
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,2,3,4,4,3,2,1], n = 4
+Output: [1,4,2,3,3,2,4,1]
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [1,1,2,2], n = 2
+Output: [1,2,1,2]
+
+```
+
+**Constraints**:
+
+- `1 <= n <= 500`
+- `nums.length == 2n`
+- `1 <= nums[i] <= 10^3`
+
+## 题目大意
+
+给你一个数组 nums ，数组中有 2n 个元素，按 [x1,x2,...,xn,y1,y2,...,yn] 的格式排列。请你将数组按 [x1,y1,x2,y2,...,xn,yn] 格式重新排列，返回重排后的数组。
+
+## 解题思路
+
+- 给定一个 2n 的数组，把后 n 个元素插空放到前 n 个元素里面。输出最终完成的数组。
+- 简单题，按照题意插空即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+func shuffle(nums []int, n int) []int {
+	result := make([]int, 0)
+	for i := 0; i < n; i++ {
+		result = append(result, nums[i])
+		result = append(result, nums[n+i])
+	}
+	return result
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1480.Running-Sum-of-1d-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1480.Running-Sum-of-1d-Array.md b/website/content/ChapterFour/1400~1499/1480.Running-Sum-of-1d-Array.md
new file mode 100644
index 000000000..6398e56a0
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1480.Running-Sum-of-1d-Array.md
@@ -0,0 +1,71 @@
+# [1480. Running Sum of 1d Array](https://leetcode.com/problems/running-sum-of-1d-array/)
+
+## 题目
+
+Given an array `nums`. We define a running sum of an array as `runningSum[i] = sum(nums[0]…nums[i])`.
+
+Return the running sum of `nums`.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,4]
+Output: [1,3,6,10]
+Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,1,1,1,1]
+Output: [1,2,3,4,5]
+Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [3,1,2,10,1]
+Output: [3,4,6,16,17]
+
+```
+
+
+**Constraints**:
+
+- `1 <= nums.length <= 1000`
+- `-10^6 <= nums[i] <= 10^6`
+
+## 题目大意
+
+给你一个数组 nums 。数组「动态和」的计算公式为：runningSum[i] = sum(nums[0]…nums[i]) 。请返回 nums 的动态和。
+
+
+## 解题思路
+
+- 简单题，按照题意依次循环计算前缀和即可。
+
+## 代码
+
+```go
+package leetcode
+
+func runningSum(nums []int) []int {
+	dp := make([]int, len(nums)+1)
+	dp[0] = 0
+	for i := 1; i <= len(nums); i++ {
+		dp[i] = dp[i-1] + nums[i-1]
+	}
+	return dp[1:]
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1470.Shuffle-the-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1482.Minimum-Number-of-Days-to-Make-m-Bouquets/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1482.Minimum-Number-of-Days-to-Make-m-Bouquets.md b/website/content/ChapterFour/1400~1499/1482.Minimum-Number-of-Days-to-Make-m-Bouquets.md
new file mode 100644
index 000000000..92c3ba07c
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1482.Minimum-Number-of-Days-to-Make-m-Bouquets.md
@@ -0,0 +1,117 @@
+# [1482. Minimum Number of Days to Make m Bouquets](https://leetcode.com/problems/minimum-number-of-days-to-make-m-bouquets/)
+
+## 题目
+
+Given an integer array `bloomDay`, an integer `m` and an integer `k`.
+
+We need to make `m` bouquets. To make a bouquet, you need to use `k` **adjacent flowers** from the garden.
+
+The garden consists of `n` flowers, the `ith` flower will bloom in the `bloomDay[i]` and then can be used in **exactly one** bouquet.
+
+Return *the minimum number of days* you need to wait to be able to make `m` bouquets from the garden. If it is impossible to make `m` bouquets return **-1**.
+
+**Example 1:**
+
+```
+Input: bloomDay = [1,10,3,10,2], m = 3, k = 1
+Output: 3
+Explanation: Let's see what happened in the first three days. x means flower bloomed and _ means flower didn't bloom in the garden.
+We need 3 bouquets each should contain 1 flower.
+After day 1: [x, _, _, _, _]   // we can only make one bouquet.
+After day 2: [x, _, _, _, x]   // we can only make two bouquets.
+After day 3: [x, _, x, _, x]   // we can make 3 bouquets. The answer is 3.
+```
+
+**Example 2:**
+
+```
+Input: bloomDay = [1,10,3,10,2], m = 3, k = 2
+Output: -1
+Explanation: We need 3 bouquets each has 2 flowers, that means we need 6 flowers. We only have 5 flowers so it is impossible to get the needed bouquets and we return -1.
+```
+
+**Example 3:**
+
+```
+Input: bloomDay = [7,7,7,7,12,7,7], m = 2, k = 3
+Output: 12
+Explanation: We need 2 bouquets each should have 3 flowers.
+Here's the garden after the 7 and 12 days:
+After day 7: [x, x, x, x, _, x, x]
+We can make one bouquet of the first three flowers that bloomed. We cannot make another bouquet from the last three flowers that bloomed because they are not adjacent.
+After day 12: [x, x, x, x, x, x, x]
+It is obvious that we can make two bouquets in different ways.
+```
+
+**Example 4:**
+
+```
+Input: bloomDay = [1000000000,1000000000], m = 1, k = 1
+Output: 1000000000
+Explanation: You need to wait 1000000000 days to have a flower ready for a bouquet.
+```
+
+**Example 5:**
+
+```
+Input: bloomDay = [1,10,2,9,3,8,4,7,5,6], m = 4, k = 2
+Output: 9
+```
+
+**Constraints:**
+
+- `bloomDay.length == n`
+- `1 <= n <= 10^5`
+- `1 <= bloomDay[i] <= 10^9`
+- `1 <= m <= 10^6`
+- `1 <= k <= n`
+
+## 题目大意
+
+给你一个整数数组 bloomDay，以及两个整数 m 和 k 。现需要制作 m 束花。制作花束时，需要使用花园中 相邻的 k 朵花 。花园中有 n 朵花，第 i 朵花会在 bloomDay[i] 时盛开，恰好 可以用于 一束 花中。请你返回从花园中摘 m 束花需要等待的最少的天数。如果不能摘到 m 束花则返回 -1 。
+
+## 解题思路
+
+- 本题是二分搜索提醒。题目解空间固定，答案区间一定在 [0, maxDay] 中。这是单调增且有序区间，所以可以在这个解空间内使用二分搜索。在区间 [0, maxDay] 中找到第一个能满足 m 束花的解。二分搜索判断是否为 true 的条件为：从左往右遍历数组，依次统计当前日期下，花是否开了，如果连续开花 k 朵，便为 1 束，数组遍历结束如果花束总数 ≥ k 即为答案。二分搜索会返回最小的下标，即对应满足题意的最少天数。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func minDays(bloomDay []int, m int, k int) int {
+	if m*k > len(bloomDay) {
+		return -1
+	}
+	maxDay := 0
+	for _, day := range bloomDay {
+		if day > maxDay {
+			maxDay = day
+		}
+	}
+	return sort.Search(maxDay, func(days int) bool {
+		flowers, bouquets := 0, 0
+		for _, d := range bloomDay {
+			if d > days {
+				flowers = 0
+			} else {
+				flowers++
+				if flowers == k {
+					bouquets++
+					flowers = 0
+				}
+			}
+		}
+		return bouquets >= m
+	})
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1480.Running-Sum-of-1d-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1486.XOR-Operation-in-an-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/1486.XOR-Operation-in-an-Array.md b/website/content/ChapterFour/1400~1499/1486.XOR-Operation-in-an-Array.md
new file mode 100644
index 000000000..5536755fb
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/1486.XOR-Operation-in-an-Array.md
@@ -0,0 +1,76 @@
+# [1486. XOR Operation in an Array](https://leetcode.com/problems/xor-operation-in-an-array/)
+
+
+## 题目
+
+Given an integer `n` and an integer `start`.
+
+Define an array `nums` where `nums[i] = start + 2*i` (0-indexed) and `n == nums.length`.
+
+Return the bitwise XOR of all elements of `nums`.
+
+**Example 1:**
+
+```
+Input: n = 5, start = 0
+Output: 8
+Explanation:Array nums is equal to [0, 2, 4, 6, 8] where (0 ^ 2 ^ 4 ^ 6 ^ 8) = 8.
+Where "^" corresponds to bitwise XOR operator.
+```
+
+**Example 2:**
+
+```
+Input: n = 4, start = 3
+Output: 8
+Explanation:Array nums is equal to [3, 5, 7, 9] where (3 ^ 5 ^ 7 ^ 9) = 8.
+```
+
+**Example 3:**
+
+```
+Input: n = 1, start = 7
+Output: 7
+```
+
+**Example 4:**
+
+```
+Input: n = 10, start = 5
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= n <= 1000`
+- `0 <= start <= 1000`
+- `n == nums.length`
+
+## 题目大意
+
+给你两个整数，n 和 start 。数组 nums 定义为：nums[i] = start + 2*i（下标从 0 开始）且 n == nums.length 。请返回 nums 中所有元素按位异或（XOR）后得到的结果。
+
+## 解题思路
+
+- 简单题。按照题意，一层循环依次累积异或数组中每个元素。
+
+## 代码
+
+```go
+package leetcode
+
+func xorOperation(n int, start int) int {
+	res := 0
+	for i := 0; i < n; i++ {
+		res ^= start + 2*i
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1482.Minimum-Number-of-Days-to-Make-m-Bouquets/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1512.Number-of-Good-Pairs/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1400~1499/_index.md b/website/content/ChapterFour/1400~1499/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/1400~1499/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/1500~1599/1512.Number-of-Good-Pairs.md b/website/content/ChapterFour/1500~1599/1512.Number-of-Good-Pairs.md
new file mode 100644
index 000000000..009b74b35
--- /dev/null
+++ b/website/content/ChapterFour/1500~1599/1512.Number-of-Good-Pairs.md
@@ -0,0 +1,74 @@
+# [1512. Number of Good Pairs](https://leetcode.com/problems/number-of-good-pairs/)
+
+## 题目
+
+Given an array of integers `nums`.
+
+A pair `(i,j)` is called good if `nums[i] == nums[j]` and `i < j`.
+
+Return the number of good pairs.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,1,1,3]
+Output: 4
+Explanation: There are 4 good pairs (0,3), (0,4), (3,4), (2,5) 0-indexed.
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,1,1,1]
+Output: 6
+Explanation: Each pair in the array are good.
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [1,2,3]
+Output: 0
+
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 1000`
+- `1 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个整数数组 nums。如果一组数字 (i,j) 满足 nums[i] == nums[j] 且 i < j ，就可以认为这是一组好数对。返回好数对的数目。
+
+## 解题思路
+
+- 简单题，按照题目中好数对的定义，循环遍历判断两数是否相等，累加计数即可。
+
+## 代码
+
+```go
+package leetcode
+
+func numIdenticalPairs(nums []int) int {
+	total := 0
+	for x := 0; x < len(nums); x++ {
+		for y := x + 1; y < len(nums); y++ {
+			if nums[x] == nums[y] {
+				total++
+			}
+		}
+	}
+	return total
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1400~1499/1486.XOR-Operation-in-an-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1518.Water-Bottles/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1500~1599/1518.Water-Bottles.md b/website/content/ChapterFour/1500~1599/1518.Water-Bottles.md
new file mode 100644
index 000000000..e7817708c
--- /dev/null
+++ b/website/content/ChapterFour/1500~1599/1518.Water-Bottles.md
@@ -0,0 +1,81 @@
+# [1518. Water Bottles](https://leetcode.com/problems/water-bottles/)
+
+## 题目
+
+Given numBottles full water bottles, you can exchange numExchange empty water bottles for one full water bottle.
+
+The operation of drinking a full water bottle turns it into an empty bottle.
+
+Return the maximum number of water bottles you can drink.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2020/07/01/sample_1_1875.png](https://assets.leetcode.com/uploads/2020/07/01/sample_1_1875.png)
+
+    Input: numBottles = 9, numExchange = 3
+    Output: 13
+    Explanation: You can exchange 3 empty bottles to get 1 full water bottle.
+    Number of water bottles you can drink: 9 + 3 + 1 = 13.
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2020/07/01/sample_2_1875.png](https://assets.leetcode.com/uploads/2020/07/01/sample_2_1875.png)
+
+    Input: numBottles = 15, numExchange = 4
+    Output: 19
+    Explanation: You can exchange 4 empty bottles to get 1 full water bottle.
+    Number of water bottles you can drink: 15 + 3 + 1 = 19.
+
+**Example 3**:
+
+    Input: numBottles = 5, numExchange = 5
+    Output: 6
+
+**Example 4**:
+
+    Input: numBottles = 2, numExchange = 3
+    Output: 2
+
+**Constraints:**
+
+- 1 <= numBottles <= 100
+- 2 <= numExchange <= 100
+
+## 题目大意
+
+小区便利店正在促销，用 numExchange 个空酒瓶可以兑换一瓶新酒。你购入了 numBottles 瓶酒。
+
+如果喝掉了酒瓶中的酒，那么酒瓶就会变成空的。
+
+请你计算 最多 能喝到多少瓶酒。
+
+## 解题思路
+
+- 模拟。首先我们一定可以喝到 numBottles 瓶酒，剩下 numBottles 个空瓶。接下来我们可以拿空瓶子换酒，每次拿出 numExchange 个瓶子换一瓶酒，然后再喝完这瓶酒，得到一个空瓶。这样模拟下去，直到所有的空瓶子小于numExchange结束。
+
+## 代码
+
+```go
+package leetcode
+
+func numWaterBottles(numBottles int, numExchange int) int {
+	if numBottles < numExchange {
+		return numBottles
+	}
+	quotient := numBottles / numExchange
+	reminder := numBottles % numExchange
+	ans := numBottles + quotient
+	for quotient+reminder >= numExchange {
+		quotient, reminder = (quotient+reminder)/numExchange, (quotient+reminder)%numExchange
+		ans += quotient
+	}
+	return ans
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1512.Number-of-Good-Pairs/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1539.Kth-Missing-Positive-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1500~1599/1539.Kth-Missing-Positive-Number.md b/website/content/ChapterFour/1500~1599/1539.Kth-Missing-Positive-Number.md
new file mode 100644
index 000000000..c22e07e0c
--- /dev/null
+++ b/website/content/ChapterFour/1500~1599/1539.Kth-Missing-Positive-Number.md
@@ -0,0 +1,70 @@
+# [1539. Kth Missing Positive Number](https://leetcode.com/problems/kth-missing-positive-number/)
+
+## 题目
+
+Given an array `arr` of positive integers sorted in a **strictly increasing order**, and an integer `k`.
+
+*Find the* `kth` *positive integer that is missing from this array.*
+
+**Example 1**:
+
+```
+Input: arr = [2,3,4,7,11], k = 5
+Output: 9
+Explanation: The missing positive integers are [1,5,6,8,9,10,12,13,...]. The 5th missing positive integer is 9.
+```
+
+**Example 2**:
+
+```
+Input: arr = [1,2,3,4], k = 2
+Output: 6
+Explanation: The missing positive integers are [5,6,7,...]. The 2nd missing positive integer is 6.
+```
+
+**Constraints**:
+
+- `1 <= arr.length <= 1000`
+- `1 <= arr[i] <= 1000`
+- `1 <= k <= 1000`
+- `arr[i] < arr[j]` for `1 <= i < j <= arr.length`
+
+## 题目大意
+
+给你一个 **严格升序排列** 的正整数数组 `arr` 和一个整数 `k` 。请你找到这个数组里第 `k` 个缺失的正整数。
+
+## 解题思路
+
+- 简单题。用一个变量从 1 开始累加，依次比对数组中是否存在，不存在的话就把 k - -，直到 k 为 0 的时候即是要输出的值。特殊情况，missing positive 都在数组之外，如例子 2 。
+
+## 代码
+
+```go
+package leetcode
+
+func findKthPositive(arr []int, k int) int {
+	positive, index := 1, 0
+	for index < len(arr) {
+		if arr[index] != positive {
+			k--
+		} else {
+			index++
+		}
+		if k == 0 {
+			break
+		}
+		positive++
+	}
+	if k != 0 {
+		positive += k - 1
+	}
+	return positive
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1518.Water-Bottles/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1551.Minimum-Operations-to-Make-Array-Equal/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1500~1599/1551.Minimum-Operations-to-Make-Array-Equal.md b/website/content/ChapterFour/1500~1599/1551.Minimum-Operations-to-Make-Array-Equal.md
new file mode 100644
index 000000000..74ef1db9e
--- /dev/null
+++ b/website/content/ChapterFour/1500~1599/1551.Minimum-Operations-to-Make-Array-Equal.md
@@ -0,0 +1,69 @@
+# [1551. Minimum Operations to Make Array Equal](https://leetcode.com/problems/minimum-operations-to-make-array-equal/)
+
+
+## 题目
+
+You have an array `arr` of length `n` where `arr[i] = (2 * i) + 1` for all valid values of `i` (i.e. `0 <= i < n`).
+
+In one operation, you can select two indices `x` and `y` where `0 <= x, y < n` and subtract `1` from `arr[x]` and add `1` to `arr[y]` (i.e. perform `arr[x] -=1` and `arr[y] += 1`). The goal is to make all the elements of the array **equal**. It is **guaranteed** that all the elements of the array can be made equal using some operations.
+
+Given an integer `n`, the length of the array. Return *the minimum number of operations* needed to make all the elements of arr equal.
+
+**Example 1:**
+
+```
+Input: n = 3
+Output: 2
+Explanation: arr = [1, 3, 5]
+First operation choose x = 2 and y = 0, this leads arr to be [2, 3, 4]
+In the second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].
+```
+
+**Example 2:**
+
+```
+Input: n = 6
+Output: 9
+```
+
+**Constraints:**
+
+- `1 <= n <= 10^4`
+
+## 题目大意
+
+存在一个长度为 n 的数组 arr ，其中 arr[i] = (2 * i) + 1 （ 0 <= i < n ）。一次操作中，你可以选出两个下标，记作 x 和 y （ 0 <= x, y < n ）并使 arr[x] 减去 1 、arr[y] 加上 1 （即 arr[x] -=1 且 arr[y] += 1 ）。最终的目标是使数组中的所有元素都 相等 。题目测试用例将会 保证 ：在执行若干步操作后，数组中的所有元素最终可以全部相等。给你一个整数 n，即数组的长度。请你返回使数组 arr 中所有元素相等所需的 最小操作数 。
+
+## 解题思路
+
+- 这一题是数学题。题目给定的操作并不会使数组中所有元素之和变化，最终让所有元素相等，那么数组中所有元素的平均值即为最后数组中每一个元素的值。最少操作数的策略应该是以平均数为中心，中心右边的数减小，对称的中心左边的数增大。由于原数组是等差数列，两两元素之间相差 2，利用数学方法可以算出操作数。
+- 数组长度分为奇数和偶数分别讨论。如果数组长度为奇数，所需要的操作数是：
+
+    {{< katex display >}}
+    \begin{aligned} &\quad 2 + 4 + \cdots + 2\cdot\left\lfloor\frac{n}{2}\right\rfloor \\ &= \frac{1}{2}\left\lfloor\frac{n}{2}\right\rfloor\left(2\cdot\left\lfloor\frac{n}{2}\right\rfloor + 2\right) \\ &= \left\lfloor\frac{n}{2}\right\rfloor \left(\left\lfloor\frac{n}{2}\right\rfloor + 1\right) \\ &= \frac{n-1}{2}\left(\frac{n-1}{2} + 1\right) \\ &= \frac{n-1}{2}\cdot\frac{n+1}{2} \\ &= \frac{n^2-1}{4} \\ &= \left\lfloor\frac{n^2}{4}\right\rfloor \end{aligned}
+    {{< /katex >}}
+
+    数组长度是偶数，所需要的操作数是：
+
+    {{< katex display >}}
+    \begin{aligned} &\quad 1 + 3 + \cdots + \left(2\cdot\left\lfloor\frac{n}{2}\right\rfloor - 1\right) \\ &= \frac{1}{2}\left\lfloor\frac{n}{2}\right\rfloor\left(2\cdot\left\lfloor\frac{n}{2}\right\rfloor - 1 + 1\right)\\ &= \left(\left\lfloor\frac{n}{2}\right\rfloor\right)^2 \\ &= \frac{n^2}{4} \end{aligned}
+    {{< /katex >}}
+
+    综上所述，最小操作数是 n^2/4
+
+## 代码
+
+```go
+package leetcode
+
+func minOperations(n int) int {
+	return n * n / 4
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1539.Kth-Missing-Positive-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1572.Matrix-Diagonal-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1500~1599/1572.Matrix-Diagonal-Sum.md b/website/content/ChapterFour/1500~1599/1572.Matrix-Diagonal-Sum.md
new file mode 100644
index 000000000..c0fe7ee41
--- /dev/null
+++ b/website/content/ChapterFour/1500~1599/1572.Matrix-Diagonal-Sum.md
@@ -0,0 +1,84 @@
+# [1572. Matrix Diagonal Sum](https://leetcode.com/problems/matrix-diagonal-sum/)
+
+
+## 题目
+
+Given a square matrix `mat`, return the sum of the matrix diagonals.
+
+Only include the sum of all the elements on the primary diagonal and all the elements on the secondary diagonal that are not part of the primary diagonal.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/08/14/sample_1911.png](https://assets.leetcode.com/uploads/2020/08/14/sample_1911.png)
+
+```
+Input: mat = [[1,2,3],
+              [4,5,6],
+              [7,8,9]]
+Output: 25
+Explanation:Diagonals sum: 1 + 5 + 9 + 3 + 7 = 25
+Notice that element mat[1][1] = 5 is counted only once.
+
+```
+
+**Example 2:**
+
+```
+Input: mat = [[1,1,1,1],
+              [1,1,1,1],
+              [1,1,1,1],
+              [1,1,1,1]]
+Output: 8
+
+```
+
+**Example 3:**
+
+```
+Input: mat = [[5]]
+Output: 5
+
+```
+
+**Constraints:**
+
+- `n == mat.length == mat[i].length`
+- `1 <= n <= 100`
+- `1 <= mat[i][j] <= 100`
+
+## 题目大意
+
+给你一个正方形矩阵 mat，请你返回矩阵对角线元素的和。请你返回在矩阵主对角线上的元素和副对角线上且不在主对角线上元素的和。
+
+## 解题思路
+
+- 简单题。根据题意，把主对角线和副对角线上的元素相加。
+- 如果正方形矩阵的长度 n 为奇数，相加的结果需要减去 mat[n/2][n/2]。
+
+## 代码
+
+```go
+package leetcode
+
+func diagonalSum(mat [][]int) int {
+	n := len(mat)
+	ans := 0
+	for pi := 0; pi < n; pi++ {
+		ans += mat[pi][pi]
+	}
+	for si, sj := n-1, 0; sj < n; si, sj = si-1, sj+1 {
+		ans += mat[si][sj]
+	}
+	if n%2 == 0 {
+		return ans
+	}
+	return ans - mat[n/2][n/2]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1551.Minimum-Operations-to-Make-Array-Equal/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1573.Number-of-Ways-to-Split-a-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1500~1599/1573.Number-of-Ways-to-Split-a-String.md b/website/content/ChapterFour/1500~1599/1573.Number-of-Ways-to-Split-a-String.md
new file mode 100644
index 000000000..916d942a9
--- /dev/null
+++ b/website/content/ChapterFour/1500~1599/1573.Number-of-Ways-to-Split-a-String.md
@@ -0,0 +1,115 @@
+# [1573. Number of Ways to Split a String](https://leetcode.com/problems/number-of-ways-to-split-a-string/)
+
+
+## 题目
+
+Given a binary string `s` (a string consisting only of '0's and '1's), we can split `s` into 3 **non-empty** strings s1, s2, s3 (s1+ s2+ s3 = s).
+
+Return the number of ways `s` can be split such that the number of characters '1' is the same in s1, s2, and s3.
+
+Since the answer may be too large, return it modulo 10^9 + 7.
+
+**Example 1**:
+
+```
+Input: s = "10101"
+Output: 4
+Explanation: There are four ways to split s in 3 parts where each part contain the same number of letters '1'.
+"1|010|1"
+"1|01|01"
+"10|10|1"
+"10|1|01"
+
+```
+
+**Example 2**:
+
+```
+Input: s = "1001"
+Output: 0
+
+```
+
+**Example 3**:
+
+```
+Input: s = "0000"
+Output: 3
+Explanation: There are three ways to split s in 3 parts.
+"0|0|00"
+"0|00|0"
+"00|0|0"
+
+```
+
+**Example 4**:
+
+```
+Input: s = "100100010100110"
+Output: 12
+
+```
+
+**Constraints**:
+
+- `3 <= s.length <= 10^5`
+- `s[i]` is `'0'` or `'1'`.
+
+## 题目大意
+
+给你一个二进制串 s  （一个只包含 0 和 1 的字符串），我们可以将 s 分割成 3 个 非空 字符串 s1, s2, s3 （s1 + s2 + s3 = s）。请你返回分割 s 的方案数，满足 s1，s2 和 s3 中字符 '1' 的数目相同。由于答案可能很大，请将它对 10^9 + 7 取余后返回。
+
+## 解题思路
+
+- 这一题是考察的排列组合的知识。根据题意，如果 1 的个数不是 3 的倍数，直接返回 -1。如果字符串里面没有 1，那么切分的方案就是组合，在 n-1 个字母里面选出 2 个位置。利用组合的计算方法，组合数是 (n-1) * (n-2) / 2 。
+- 剩下的是 3 的倍数的情况。在字符串中选 2 个位置隔成 3 段。从第一段最后一个 1 到第二段第一个 1 之间的 0 的个数为 m1，从第二段最后一个 1 到第三段第一个 1 之间的 0 的个数为 m2。利用乘法原理，方案数为 m1 * m2。
+
+## 代码
+
+```go
+package leetcode
+
+func numWays(s string) int {
+	ones := 0
+	for _, c := range s {
+		if c == '1' {
+			ones++
+		}
+	}
+	if ones%3 != 0 {
+		return 0
+	}
+	if ones == 0 {
+		return (len(s) - 1) * (len(s) - 2) / 2 % 1000000007
+	}
+	N, a, b, c, d, count := ones/3, 0, 0, 0, 0, 0
+	for i, letter := range s {
+		if letter == '0' {
+			continue
+		}
+		if letter == '1' {
+			count++
+		}
+		if count == N {
+			a = i
+		}
+		if count == N+1 {
+			b = i
+		}
+		if count == 2*N {
+			c = i
+		}
+		if count == 2*N+1 {
+			d = i
+		}
+	}
+	return (b - a) * (d - c) % 1000000007
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1572.Matrix-Diagonal-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1500~1599/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters.md b/website/content/ChapterFour/1500~1599/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters.md
new file mode 100644
index 000000000..7d57d2b0a
--- /dev/null
+++ b/website/content/ChapterFour/1500~1599/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters.md
@@ -0,0 +1,70 @@
+# [1576. Replace All ?'s to Avoid Consecutive Repeating Characters](https://leetcode.com/problems/replace-all-s-to-avoid-consecutive-repeating-characters/)
+
+## 题目
+
+Given a string `s` containing only lowercase English letters and the `'?'` character, convert **all** the `'?'` characters into lowercase letters such that the final string does not contain any **consecutive repeating** characters. You **cannot** modify the non `'?'` characters.
+
+It is **guaranteed** that there are no consecutive repeating characters in the given string **except** for `'?'`.
+
+Return *the final string after all the conversions (possibly zero) have been made*. If there is more than one solution, return **any of them**. It can be shown that an answer is always possible with the given constraints.
+
+**Example 1:**
+
+```
+Input: s = "?zs"
+Output: "azs"
+Explanation: There are 25 solutions for this problem. From "azs" to "yzs", all are valid. Only "z" is an invalid modification as the string will consist of consecutive repeating characters in "zzs".
+
+```
+
+**Example 2:**
+
+```
+Input: s = "ubv?w"
+Output: "ubvaw"
+Explanation: There are 24 solutions for this problem. Only "v" and "w" are invalid modifications as the strings will consist of consecutive repeating characters in "ubvvw" and "ubvww".
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 100`
+- `s` consist of lowercase English letters and `'?'`.
+
+## 题目大意
+
+给你一个仅包含小写英文字母和 '?' 字符的字符串 s，请你将所有的 '?' 转换为若干小写字母，使最终的字符串不包含任何 连续重复 的字符。注意：你 不能 修改非 '?' 字符。
+
+题目测试用例保证 除 '?' 字符 之外，不存在连续重复的字符。在完成所有转换（可能无需转换）后返回最终的字符串。如果有多个解决方案，请返回其中任何一个。可以证明，在给定的约束条件下，答案总是存在的。
+
+## 解题思路
+
+- 简单题。找到源字符串中 ‘?’ 字符的位置，然后依次用 a ~ z 字符去替换，替换进去的字符不能和前后字符相同即可。
+
+## 代码
+
+```go
+package leetcode
+
+func modifyString(s string) string {
+	res := []byte(s)
+	for i, ch := range res {
+		if ch == '?' {
+			for b := byte('a'); b <= 'z'; b++ {
+				if !(i > 0 && res[i-1] == b || i < len(res)-1 && res[i+1] == b) {
+					res[i] = b
+					break
+				}
+			}
+		}
+	}
+	return string(res)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1573.Number-of-Ways-to-Split-a-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1500~1599/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable.md b/website/content/ChapterFour/1500~1599/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable.md
new file mode 100644
index 000000000..8806ed95d
--- /dev/null
+++ b/website/content/ChapterFour/1500~1599/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable.md
@@ -0,0 +1,110 @@
+# [1579. Remove Max Number of Edges to Keep Graph Fully Traversable](https://leetcode.com/problems/remove-max-number-of-edges-to-keep-graph-fully-traversable/)
+
+
+## 题目
+
+Alice and Bob have an undirected graph of `n` nodes and 3 types of edges:
+
+- Type 1: Can be traversed by Alice only.
+- Type 2: Can be traversed by Bob only.
+- Type 3: Can by traversed by both Alice and Bob.
+
+Given an array `edges` where `edges[i] = [typei, ui, vi]` represents a bidirectional edge of type `typei` between nodes `ui` and `vi`, find the maximum number of edges you can remove so that after removing the edges, the graph can still be fully traversed by both Alice and Bob. The graph is fully traversed by Alice and Bob if starting from any node, they can reach all other nodes.
+
+Return *the maximum number of edges you can remove, or return* `-1` *if it's impossible for the graph to be fully traversed by Alice and Bob.*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/08/19/ex1.png](https://assets.leetcode.com/uploads/2020/08/19/ex1.png)
+
+```
+Input: n = 4, edges = [[3,1,2],[3,2,3],[1,1,3],[1,2,4],[1,1,2],[2,3,4]]
+Output: 2
+Explanation: If we remove the 2 edges [1,1,2] and [1,1,3]. The graph will still be fully traversable by Alice and Bob. Removing any additional edge will not make it so. So the maximum number of edges we can remove is 2.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/08/19/ex2.png](https://assets.leetcode.com/uploads/2020/08/19/ex2.png)
+
+```
+Input: n = 4, edges = [[3,1,2],[3,2,3],[1,1,4],[2,1,4]]
+Output: 0
+Explanation: Notice that removing any edge will not make the graph fully traversable by Alice and Bob.
+```
+
+**Example 3:**
+
+![https://assets.leetcode.com/uploads/2020/08/19/ex3.png](https://assets.leetcode.com/uploads/2020/08/19/ex3.png)
+
+```
+Input: n = 4, edges = [[3,2,3],[1,1,2],[2,3,4]]
+Output: -1
+Explanation: In the current graph, Alice cannot reach node 4 from the other nodes. Likewise, Bob cannot reach 1. Therefore it's impossible to make the graph fully traversable.
+```
+
+**Constraints:**
+
+- `1 <= n <= 10^5`
+- `1 <= edges.length <= min(10^5, 3 * n * (n-1) / 2)`
+- `edges[i].length == 3`
+- `1 <= edges[i][0] <= 3`
+- `1 <= edges[i][1] < edges[i][2] <= n`
+- All tuples `(typei, ui, vi)` are distinct.
+
+## 题目大意
+
+Alice 和 Bob 共有一个无向图，其中包含 n 个节点和 3  种类型的边：
+
+- 类型 1：只能由 Alice 遍历。
+- 类型 2：只能由 Bob 遍历。
+- 类型 3：Alice 和 Bob 都可以遍历。
+
+给你一个数组 edges ，其中 edges[i] = [typei, ui, vi] 表示节点 ui 和 vi 之间存在类型为 typei 的双向边。请你在保证图仍能够被 Alice和 Bob 完全遍历的前提下，找出可以删除的最大边数。如果从任何节点开始，Alice 和 Bob 都可以到达所有其他节点，则认为图是可以完全遍历的。返回可以删除的最大边数，如果 Alice 和 Bob 无法完全遍历图，则返回 -1 。
+
+## 解题思路
+
+- 本题是第 1319 题的加强版。在第 1319 题中只有一个人，同样也是判断在保证图可连通的基础上，删掉最多边的条数。这一题只不过变成了 2 个人。解题思路依旧是并查集。
+- 初始化 2 个并查集，分别表示 Alice 和 Bob。先合并公共边，每合并一条边，可删除的最大总边数便减少 1 。再合并 2 人各自的单独的边，同样是每合并一条边，每合并一条边，可删除的最大总边数便减少 1 。合并完所有的边，2 人的并查集内部集合数仍大于 1，那么则代表 2 人无法完全遍历图，则输出 -1。如果 2 人的并查集内部集合都是 1，代表整个图都连通了。输出可以删除的最大边数。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/template"
+)
+
+func maxNumEdgesToRemove(n int, edges [][]int) int {
+	alice, bob, res := template.UnionFind{}, template.UnionFind{}, len(edges)
+	alice.Init(n)
+	bob.Init(n)
+	for _, e := range edges {
+		x, y := e[1]-1, e[2]-1
+		if e[0] == 3 && (!(alice.Find(x) == alice.Find(y)) || !(bob.Find(x) == bob.Find(y))) {
+			alice.Union(x, y)
+			bob.Union(x, y)
+			res--
+		}
+	}
+	ufs := [2]*template.UnionFind{&alice, &bob}
+	for _, e := range edges {
+		if tp := e[0]; tp < 3 && !(ufs[tp-1].Find(e[1]-1) == ufs[tp-1].Find(e[2]-1)) {
+			ufs[tp-1].Union(e[1]-1, e[2]-1)
+			res--
+		}
+	}
+	if alice.TotalCount() > 1 || bob.TotalCount() > 1 {
+		return -1
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1600.Throne-Inheritance/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1500~1599/_index.md b/website/content/ChapterFour/1500~1599/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/1500~1599/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/1600~1699/1600.Throne-Inheritance.md b/website/content/ChapterFour/1600~1699/1600.Throne-Inheritance.md
new file mode 100644
index 000000000..c20245511
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1600.Throne-Inheritance.md
@@ -0,0 +1,127 @@
+# [1600. Throne Inheritance](https://leetcode.com/problems/throne-inheritance/)
+
+
+## 题目
+
+A kingdom consists of a king, his children, his grandchildren, and so on. Every once in a while, someone in the family dies or a child is born.
+
+The kingdom has a well-defined order of inheritance that consists of the king as the first member. Let's define the recursive function `Successor(x, curOrder)`, which given a person `x` and the inheritance order so far, returns who should be the next person after `x` in the order of inheritance.
+
+```
+Successor(x, curOrder):
+    if x has no children or all of x's children are in curOrder:
+        if x is the king return null
+        else return Successor(x's parent, curOrder)
+    else return x's oldest child who's not in curOrder
+```
+
+For example, assume we have a kingdom that consists of the king, his children Alice and Bob (Alice is older than Bob), and finally Alice's son Jack.
+
+1. In the beginning, `curOrder` will be `["king"]`.
+2. Calling `Successor(king, curOrder)` will return Alice, so we append to `curOrder` to get `["king", "Alice"]`.
+3. Calling `Successor(Alice, curOrder)` will return Jack, so we append to `curOrder` to get `["king", "Alice", "Jack"]`.
+4. Calling `Successor(Jack, curOrder)` will return Bob, so we append to `curOrder` to get `["king", "Alice", "Jack", "Bob"]`.
+5. Calling `Successor(Bob, curOrder)` will return `null`. Thus the order of inheritance will be `["king", "Alice", "Jack", "Bob"]`.
+
+Using the above function, we can always obtain a unique order of inheritance.
+
+Implement the `ThroneInheritance` class:
+
+- `ThroneInheritance(string kingName)` Initializes an object of the `ThroneInheritance` class. The name of the king is given as part of the constructor.
+- `void birth(string parentName, string childName)` Indicates that `parentName` gave birth to `childName`.
+- `void death(string name)` Indicates the death of `name`. The death of the person doesn't affect the `Successor` function nor the current inheritance order. You can treat it as just marking the person as dead.
+- `string[] getInheritanceOrder()` Returns a list representing the current order of inheritance **excluding** dead people.
+
+**Example 1:**
+
+```
+Input
+["ThroneInheritance", "birth", "birth", "birth", "birth", "birth", "birth", "getInheritanceOrder", "death", "getInheritanceOrder"]
+[["king"], ["king", "andy"], ["king", "bob"], ["king", "catherine"], ["andy", "matthew"], ["bob", "alex"], ["bob", "asha"], [null], ["bob"], [null]]
+Output
+[null, null, null, null, null, null, null, ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"], null, ["king", "andy", "matthew", "alex", "asha", "catherine"]]
+
+Explanation
+ThroneInheritance t= new ThroneInheritance("king"); // order:king
+t.birth("king", "andy"); // order: king >andy
+t.birth("king", "bob"); // order: king > andy >bob
+t.birth("king", "catherine"); // order: king > andy > bob >catherine
+t.birth("andy", "matthew"); // order: king > andy >matthew > bob > catherine
+t.birth("bob", "alex"); // order: king > andy > matthew > bob >alex > catherine
+t.birth("bob", "asha"); // order: king > andy > matthew > bob > alex >asha > catherine
+t.getInheritanceOrder(); // return ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"]
+t.death("bob"); // order: king > andy > matthew >bob > alex > asha > catherine
+t.getInheritanceOrder(); // return ["king", "andy", "matthew", "alex", "asha", "catherine"]
+
+```
+
+**Constraints:**
+
+- `1 <= kingName.length, parentName.length, childName.length, name.length <= 15`
+- `kingName`, `parentName`, `childName`, and `name` consist of lowercase English letters only.
+- All arguments `childName` and `kingName` are **distinct**.
+- All `name` arguments of `death` will be passed to either the constructor or as `childName` to `birth` first.
+- For each call to `birth(parentName, childName)`, it is guaranteed that `parentName` is alive.
+- At most `105` calls will be made to `birth` and `death`.
+- At most `10` calls will be made to `getInheritanceOrder`.
+
+## 题目大意
+
+一个王国里住着国王、他的孩子们、他的孙子们等等。每一个时间点，这个家庭里有人出生也有人死亡。这个王国有一个明确规定的皇位继承顺序，第一继承人总是国王自己。我们定义递归函数 Successor(x, curOrder) ，给定一个人 x 和当前的继承顺序，该函数返回 x 的下一继承人。
+
+## 解题思路
+
+- 这道题思路不难。先将国王每个孩子按照顺序存在一个 map 中，然后每个国王的孩子还存在父子关系，同理也按顺序存在 map 中。执行 GetInheritanceOrder() 函数时，将国王的孩子按顺序遍历，如果每个孩子还有孩子，递归遍历到底。如果把继承关系看成一棵树，此题便是多叉树的先根遍历的问题。
+
+## 代码
+
+```go
+package leetcode
+
+type ThroneInheritance struct {
+	king  string
+	edges map[string][]string
+	dead  map[string]bool
+}
+
+func Constructor(kingName string) (t ThroneInheritance) {
+	return ThroneInheritance{kingName, map[string][]string{}, map[string]bool{}}
+}
+
+func (t *ThroneInheritance) Birth(parentName, childName string) {
+	t.edges[parentName] = append(t.edges[parentName], childName)
+}
+
+func (t *ThroneInheritance) Death(name string) {
+	t.dead[name] = true
+}
+
+func (t *ThroneInheritance) GetInheritanceOrder() (res []string) {
+	var preorder func(string)
+	preorder = func(name string) {
+		if !t.dead[name] {
+			res = append(res, name)
+		}
+		for _, childName := range t.edges[name] {
+			preorder(childName)
+		}
+	}
+	preorder(t.king)
+	return
+}
+
+/**
+ * Your ThroneInheritance object will be instantiated and called as such:
+ * obj := Constructor(kingName);
+ * obj.Birth(parentName,childName);
+ * obj.Death(name);
+ * param_3 := obj.GetInheritanceOrder();
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1500~1599/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1603.Design-Parking-System/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1603.Design-Parking-System.md b/website/content/ChapterFour/1600~1699/1603.Design-Parking-System.md
new file mode 100644
index 000000000..274894e60
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1603.Design-Parking-System.md
@@ -0,0 +1,110 @@
+# [1603. Design Parking System](https://leetcode.com/problems/design-parking-system/)
+
+
+## 题目
+
+Design a parking system for a parking lot. The parking lot has three kinds of parking spaces: big, medium, and small, with a fixed number of slots for each size.
+
+Implement the `ParkingSystem` class:
+
+- `ParkingSystem(int big, int medium, int small)` Initializes object of the `ParkingSystem` class. The number of slots for each parking space are given as part of the constructor.
+- `bool addCar(int carType)` Checks whether there is a parking space of `carType` for the car that wants to get into the parking lot. `carType` can be of three kinds: big, medium, or small, which are represented by `1`, `2`, and `3` respectively. **A car can only park in a parking space of its** `carType`. If there is no space available, return `false`, else park the car in that size space and return `true`.
+
+**Example 1:**
+
+```
+Input
+["ParkingSystem", "addCar", "addCar", "addCar", "addCar"]
+[[1, 1, 0], [1], [2], [3], [1]]
+Output
+[null, true, true, false, false]
+
+Explanation
+ParkingSystem parkingSystem = new ParkingSystem(1, 1, 0);
+parkingSystem.addCar(1); // return true because there is 1 available slot for a big car
+parkingSystem.addCar(2); // return true because there is 1 available slot for a medium car
+parkingSystem.addCar(3); // return false because there is no available slot for a small car
+parkingSystem.addCar(1); // return false because there is no available slot for a big car. It is already occupied.
+```
+
+**Constraints:**
+
+- `0 <= big, medium, small <= 1000`
+- `carType` is `1`, `2`, or `3`
+- At most `1000` calls will be made to `addCar`
+
+## 题目大意
+
+请你给一个停车场设计一个停车系统。停车场总共有三种不同大小的车位：大，中和小，每种尺寸分别有固定数目的车位。
+
+请你实现 ParkingSystem 类：
+
+- ParkingSystem(int big, int medium, int small) 初始化 ParkingSystem 类，三个参数分别对应每种停车位的数目。
+- bool addCar(int carType) 检查是否有 carType 对应的停车位。 carType 有三种类型：大，中，小，分别用数字 1， 2 和 3 表示。一辆车只能停在  carType 对应尺寸的停车位中。如果没有空车位，请返回 false ，否则将该车停入车位并返回 true 。
+
+## 解题思路
+
+- 简单题。分别用 3 个变量表示大，中和小车位。`addCar()` 判断这 3 个变量是否还有空车位即可。
+
+## 代码
+
+```go
+package leetcode
+
+type ParkingSystem struct {
+	Big    int
+	Medium int
+	Small  int
+}
+
+func Constructor(big int, medium int, small int) ParkingSystem {
+	return ParkingSystem{
+		Big:    big,
+		Medium: medium,
+		Small:  small,
+	}
+}
+
+func (this *ParkingSystem) AddCar(carType int) bool {
+	switch carType {
+	case 1:
+		{
+			if this.Big > 0 {
+				this.Big--
+				return true
+			}
+			return false
+		}
+	case 2:
+		{
+			if this.Medium > 0 {
+				this.Medium--
+				return true
+			}
+			return false
+		}
+	case 3:
+		{
+			if this.Small > 0 {
+				this.Small--
+				return true
+			}
+			return false
+		}
+	}
+	return false
+}
+
+/**
+ * Your ParkingSystem object will be instantiated and called as such:
+ * obj := Constructor(big, medium, small);
+ * param_1 := obj.AddCar(carType);
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1600.Throne-Inheritance/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X.md b/website/content/ChapterFour/1600~1699/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X.md
new file mode 100644
index 000000000..447f61cc5
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X.md
@@ -0,0 +1,87 @@
+# [1608. Special Array With X Elements Greater Than or Equal X](https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/)
+
+## 题目
+
+You are given an array `nums` of non-negative integers. `nums` is considered **special** if there exists a number `x` such that there are **exactly** `x` numbers in `nums` that are **greater than or equal to** `x`.
+
+Notice that `x` **does not** have to be an element in `nums`.
+
+Return `x` *if the array is **special**, otherwise, return* `-1`. It can be proven that if `nums` is special, the value for `x` is **unique**.
+
+**Example 1:**
+
+```
+Input: nums = [3,5]
+Output: 2
+Explanation: There are 2 values (3 and 5) that are greater than or equal to 2.
+```
+
+**Example 2:**
+
+```
+Input: nums = [0,0]
+Output: -1
+Explanation: No numbers fit the criteria for x.
+If x = 0, there should be 0 numbers >= x, but there are 2.
+If x = 1, there should be 1 number >= x, but there are 0.
+If x = 2, there should be 2 numbers >= x, but there are 0.
+x cannot be greater since there are only 2 numbers in nums.
+```
+
+**Example 3:**
+
+```
+Input: nums = [0,4,3,0,4]
+Output: 3
+Explanation: There are 3 values that are greater than or equal to 3.
+```
+
+**Example 4:**
+
+```
+Input: nums = [3,6,7,7,0]
+Output: -1
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `0 <= nums[i] <= 1000`
+
+## 题目大意
+
+给你一个非负整数数组 nums 。如果存在一个数 x ，使得 nums 中恰好有 x 个元素 大于或者等于 x ，那么就称 nums 是一个 特殊数组 ，而 x 是该数组的 特征值 。（注意： x 不必 是 nums 的中的元素。）如果数组 nums 是一个 特殊数组 ，请返回它的特征值 x 。否则，返回 -1 。可以证明的是，如果 nums 是特殊数组，那么其特征值 x 是 唯一的 。
+
+## 解题思路
+
+- 简单题。抓住题干中给的证明，特征值是唯一的。先将数组从小到大排序，下标的含义与特征值就等价了。下标 `i` 代表大于等于 `nums[i]` 的元素有 `len(nums) - i` 个，那么从第 0 个下标的元素开始遍历，如果这个元素都大于 `len(nums)`，那么后面 `len(nums)` 个元素也都大于等于它，特征值就找到了。如果特征值减一以后，仍然满足 `nums[i] >= x`，说明满足条件的值有多个，这一点不满足特征值唯一性，可以直接返回 -1 了。下标继续右移，特征值继续减一。如果最终循环结束依旧找不到特征值，返回 -1 。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func specialArray(nums []int) int {
+	sort.Ints(nums)
+	x := len(nums)
+	for _, num := range nums {
+		if num >= x {
+			return x
+		}
+		x--
+		if num >= x {
+			return -1
+		}
+	}
+	return -1
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1603.Design-Parking-System/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1609.Even-Odd-Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1609.Even-Odd-Tree.md b/website/content/ChapterFour/1600~1699/1609.Even-Odd-Tree.md
new file mode 100644
index 000000000..66be442a9
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1609.Even-Odd-Tree.md
@@ -0,0 +1,157 @@
+# [1609. Even Odd Tree](https://leetcode.com/problems/even-odd-tree/)
+
+## 题目
+
+A binary tree is named Even-Odd if it meets the following conditions:
+
+- The root of the binary tree is at level index 0, its children are at level index 1, their children are at level index 2, etc.
+- For every even-indexed level, all nodes at the level have odd integer values in strictly increasing order (from left to right).
+- For every odd-indexed level, all nodes at the level have even integer values in strictly decreasing order (from left to right).
+
+Given the root of a binary tree, return true if the binary tree is Even-Odd, otherwise return false.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2020/09/15/sample_1_1966.png](https://assets.leetcode.com/uploads/2020/09/15/sample_1_1966.png)
+
+    Input: root = [1,10,4,3,null,7,9,12,8,6,null,null,2]
+    Output: true
+    Explanation: The node values on each level are:
+    Level 0: [1]
+    Level 1: [10,4]
+    Level 2: [3,7,9]
+    Level 3: [12,8,6,2]
+    Since levels 0 and 2 are all odd and increasing and levels 1 and 3 are all even and decreasing, the tree is Even-Odd.
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2020/09/15/sample_2_1966.png](https://assets.leetcode.com/uploads/2020/09/15/sample_2_1966.png)
+
+    Input: root = [5,4,2,3,3,7]
+    Output: false
+    Explanation: The node values on each level are:
+    Level 0: [5]
+    Level 1: [4,2]
+    Level 2: [3,3,7]
+    Node values in level 2 must be in strictly increasing order, so the tree is not Even-Odd.
+
+**Example 3**:
+
+![https://assets.leetcode.com/uploads/2020/09/22/sample_1_333_1966.png](https://assets.leetcode.com/uploads/2020/09/22/sample_1_333_1966.png)
+
+    Input: root = [5,9,1,3,5,7]
+    Output: false
+    Explanation: Node values in the level 1 should be even integers.
+
+**Example 4**:
+
+    Input: root = [1]
+    Output: true
+
+**Example 5**:
+
+    Input: root = [11,8,6,1,3,9,11,30,20,18,16,12,10,4,2,17]
+    Output: True
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range [1, 100000]. 
+- 1 <= Node.val <= 1000000
+
+## 题目大意
+
+如果一棵二叉树满足下述几个条件，则可以称为 奇偶树 ：
+
+- 二叉树根节点所在层下标为 0 ，根的子节点所在层下标为 1 ，根的孙节点所在层下标为 2 ，依此类推。
+- 偶数下标 层上的所有节点的值都是 奇 整数，从左到右按顺序 严格递增
+- 奇数下标 层上的所有节点的值都是 偶 整数，从左到右按顺序 严格递减
+
+给你二叉树的根节点，如果二叉树为 奇偶树 ，则返回 true ，否则返回 false 。
+
+## 解题思路
+
+- 广度优先遍历(分别判断奇数层和偶数层)
+
+## 代码
+
+```go
+package leetcode
+
+type TreeNode struct {
+	Val int
+	Left *TreeNode
+	Right *TreeNode
+}
+
+func isEvenOddTree(root *TreeNode) bool {
+	level := 0
+	queue := []*TreeNode{root}
+	for len(queue) != 0 {
+		length := len(queue)
+		var nums []int
+		for i := 0; i < length; i++ {
+			node := queue[i]
+			if node.Left != nil {
+				queue = append(queue, node.Left)
+			}
+			if node.Right != nil {
+				queue = append(queue, node.Right)
+			}
+			nums = append(nums, node.Val)
+		}
+		if level%2 == 0 {
+			if !even(nums) {
+				return false
+			}
+		} else {
+			if !odd(nums) {
+				return false
+			}
+		}
+		queue = queue[length:]
+		level++
+	}
+	return true
+}
+
+func odd(nums []int) bool {
+	cur := nums[0]
+	if cur%2 != 0 {
+		return false
+	}
+	for _, num := range nums[1:] {
+		if num >= cur {
+			return false
+		}
+		if num%2 != 0 {
+			return false
+		}
+		cur = num
+	}
+	return true
+}
+
+func even(nums []int) bool {
+	cur := nums[0]
+	if cur%2 == 0 {
+		return false
+	}
+	for _, num := range nums[1:] {
+		if num <= cur {
+			return false
+		}
+		if num%2 == 0 {
+			return false
+		}
+		cur = num
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1614.Maximum-Nesting-Depth-of-the-Parentheses/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1614.Maximum-Nesting-Depth-of-the-Parentheses.md b/website/content/ChapterFour/1600~1699/1614.Maximum-Nesting-Depth-of-the-Parentheses.md
new file mode 100644
index 000000000..6800959bc
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1614.Maximum-Nesting-Depth-of-the-Parentheses.md
@@ -0,0 +1,109 @@
+# [1614. Maximum Nesting Depth of the Parentheses](https://leetcode.com/problems/maximum-nesting-depth-of-the-parentheses/)
+
+## 题目
+
+A string is a **valid parentheses string** (denoted **VPS**) if it meets one of the following:
+
+- It is an empty string `""`, or a single character not equal to `"("` or `")"`,
+- It can be written as `AB` (`A` concatenated with `B`), where `A` and `B` are **VPS**'s, or
+- It can be written as `(A)`, where `A` is a **VPS**.
+
+We can similarly define the **nesting depth** `depth(S)` of any VPS `S` as follows:
+
+- `depth("") = 0`
+- `depth(C) = 0`, where `C` is a string with a single character not equal to `"("` or `")"`.
+- `depth(A + B) = max(depth(A), depth(B))`, where `A` and `B` are **VPS**'s.
+- `depth("(" + A + ")") = 1 + depth(A)`, where `A` is a **VPS**.
+
+For example, `""`, `"()()"`, and `"()(()())"` are **VPS**'s (with nesting depths 0, 1, and 2), and `")("` and `"(()"` are not **VPS**'s.
+
+Given a **VPS** represented as string `s`, return *the **nesting depth** of* `s`.
+
+**Example 1:**
+
+```
+Input: s = "(1+(2*3)+((8)/4))+1"
+Output: 3
+Explanation: Digit 8 is inside of 3 nested parentheses in the string.
+```
+
+**Example 2:**
+
+```
+Input: s = "(1)+((2))+(((3)))"
+Output: 3
+```
+
+**Example 3:**
+
+```
+Input: s = "1+(2*3)/(2-1)"
+Output: 1
+```
+
+**Example 4:**
+
+```
+Input: s = "1"
+Output: 0
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 100`
+- `s` consists of digits `0-9` and characters `'+'`, `'-'`, `'*'`, `'/'`, `'('`, and `')'`.
+- It is guaranteed that parentheses expression `s` is a **VPS**.
+
+## 题目大意
+
+如果字符串满足以下条件之一，则可以称之为 有效括号字符串（valid parentheses string，可以简写为 VPS）：
+
+- 字符串是一个空字符串 ""，或者是一个不为 "(" 或 ")" 的单字符。
+- 字符串可以写为 AB（A 与 B 字符串连接），其中 A 和 B 都是 有效括号字符串 。
+- 字符串可以写为 (A)，其中 A 是一个 有效括号字符串 。
+
+类似地，可以定义任何有效括号字符串 S 的 嵌套深度 depth(S)：
+
+- depth("") = 0
+- depth(C) = 0，其中 C 是单个字符的字符串，且该字符不是 "(" 或者 ")"
+- depth(A + B) = max(depth(A), depth(B))，其中 A 和 B 都是 有效括号字符串
+- depth("(" + A + ")") = 1 + depth(A)，其中 A 是一个 有效括号字符串
+
+例如：""、"()()"、"()(()())" 都是 有效括号字符串（嵌套深度分别为 0、1、2），而 ")(" 、"(()" 都不是 有效括号字符串 。给你一个 有效括号字符串 s，返回该字符串的 s 嵌套深度 。
+
+## 解题思路
+
+- 简单题。求一个括号字符串的嵌套深度。题目给的括号字符串都是有效的，所以不需要考虑非法的情况。扫描一遍括号字符串，遇到 `(` 可以无脑 ++，并动态维护最大值，遇到 `)` 可以无脑 - - 。最后输出最大值即可。
+
+## 代码
+
+```go
+package leetcode
+
+func maxDepth(s string) int {
+	res, cur := 0, 0
+	for _, c := range s {
+		if c == '(' {
+			cur++
+			res = max(res, cur)
+		} else if c == ')' {
+			cur--
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1609.Even-Odd-Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1619.Mean-of-Array-After-Removing-Some-Elements/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1619.Mean-of-Array-After-Removing-Some-Elements.md b/website/content/ChapterFour/1600~1699/1619.Mean-of-Array-After-Removing-Some-Elements.md
new file mode 100644
index 000000000..27feadef2
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1619.Mean-of-Array-After-Removing-Some-Elements.md
@@ -0,0 +1,81 @@
+# [1619. Mean of Array After Removing Some Elements](https://leetcode.com/problems/mean-of-array-after-removing-some-elements/)
+
+## 题目
+
+Given an integer array `arr`, return *the mean of the remaining integers after removing the smallest `5%` and the largest `5%` of the elements.*
+
+Answers within `10-5` of the **actual answer** will be considered accepted.
+
+**Example 1:**
+
+```
+Input: arr = [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3]
+Output: 2.00000
+Explanation: After erasing the minimum and the maximum values of this array, all elements are equal to 2, so the mean is 2.
+```
+
+**Example 2:**
+
+```
+Input: arr = [6,2,7,5,1,2,0,3,10,2,5,0,5,5,0,8,7,6,8,0]
+Output: 4.00000
+```
+
+**Example 3:**
+
+```
+Input: arr = [6,0,7,0,7,5,7,8,3,4,0,7,8,1,6,8,1,1,2,4,8,1,9,5,4,3,8,5,10,8,6,6,1,0,6,10,8,2,3,4]
+Output: 4.77778
+```
+
+**Example 4:**
+
+```
+Input: arr = [9,7,8,7,7,8,4,4,6,8,8,7,6,8,8,9,2,6,0,0,1,10,8,6,3,3,5,1,10,9,0,7,10,0,10,4,1,10,6,9,3,6,0,0,2,7,0,6,7,2,9,7,7,3,0,1,6,1,10,3]
+Output: 5.27778
+```
+
+**Example 5:**
+
+```
+Input: arr = [4,8,4,10,0,7,1,3,7,8,8,3,4,1,6,2,1,1,8,0,9,8,0,3,9,10,3,10,1,10,7,3,2,1,4,9,10,7,6,4,0,8,5,1,2,1,6,2,5,0,7,10,9,10,3,7,10,5,8,5,7,6,7,6,10,9,5,10,5,5,7,2,10,7,7,8,2,0,1,1]
+Output: 5.29167
+```
+
+**Constraints:**
+
+- `20 <= arr.length <= 1000`
+- `arr.length` **is a multiple** of `20`.
+- `0 <= arr[i] <= 10^5`
+
+## 题目大意
+
+给你一个整数数组 `arr` ，请你删除最小 `5%` 的数字和最大 `5%` 的数字后，剩余数字的平均值。与 **标准答案** 误差在 `10-5` 的结果都被视为正确结果。
+
+## 解题思路
+
+- 简单题。先将数组排序，然后累加中间 90% 的元素，最后计算平均值。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func trimMean(arr []int) float64 {
+	sort.Ints(arr)
+	n, sum := len(arr), 0
+	for i := n / 20; i < n-(n/20); i++ {
+		sum += arr[i]
+	}
+	return float64(sum) / float64((n - (n / 10)))
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1614.Maximum-Nesting-Depth-of-the-Parentheses/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1624.Largest-Substring-Between-Two-Equal-Characters/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1624.Largest-Substring-Between-Two-Equal-Characters.md b/website/content/ChapterFour/1600~1699/1624.Largest-Substring-Between-Two-Equal-Characters.md
new file mode 100644
index 000000000..0699f7d7a
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1624.Largest-Substring-Between-Two-Equal-Characters.md
@@ -0,0 +1,80 @@
+# [1624. Largest Substring Between Two Equal Characters](https://leetcode.com/problems/largest-substring-between-two-equal-characters/)
+
+## 题目
+
+Given a string `s`, return *the length of the longest substring between two equal characters, excluding the two characters.* If there is no such substring return `-1`.
+
+A **substring** is a contiguous sequence of characters within a string.
+
+**Example 1:**
+
+```
+Input: s = "aa"
+Output: 0
+Explanation: The optimal substring here is an empty substring between the two 'a's.
+```
+
+**Example 2:**
+
+```
+Input: s = "abca"
+Output: 2
+Explanation: The optimal substring here is "bc".
+```
+
+**Example 3:**
+
+```
+Input: s = "cbzxy"
+Output: -1
+Explanation: There are no characters that appear twice in s.
+```
+
+**Example 4:**
+
+```
+Input: s = "cabbac"
+Output: 4
+Explanation: The optimal substring here is "abba". Other non-optimal substrings include "bb" and "".
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 300`
+- `s` contains only lowercase English letters.
+
+## 题目大意
+
+给你一个字符串 s，请你返回 两个相同字符之间的最长子字符串的长度 ，计算长度时不含这两个字符。如果不存在这样的子字符串，返回 -1 。子字符串 是字符串中的一个连续字符序列。
+
+## 解题思路
+
+- 简单题。取每个字符，扫描一次字符串，如果在字符串中还能找到相同的字符，则返回最末尾的那个字符，计算这两个字符之间的距离。取最末尾的字符是为了让两个相同的字符距离最长。扫描字符串过程中动态维护最长长度。如果字符串中不存在 2 个相同的字符，则返回 -1 。
+
+## 代码
+
+```go
+package leetcode
+
+import "strings"
+
+func maxLengthBetweenEqualCharacters(s string) int {
+	res := -1
+	for k, v := range s {
+		tmp := strings.LastIndex(s, string(v))
+		if tmp > 0 {
+			if res < tmp-k-1 {
+				res = tmp - k - 1
+			}
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1619.Mean-of-Array-After-Removing-Some-Elements/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1629.Slowest-Key/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1629.Slowest-Key.md b/website/content/ChapterFour/1600~1699/1629.Slowest-Key.md
new file mode 100644
index 000000000..3134c791a
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1629.Slowest-Key.md
@@ -0,0 +1,90 @@
+# [1629. Slowest Key](https://leetcode.com/problems/slowest-key/)
+
+
+## 题目
+
+A newly designed keypad was tested, where a tester pressed a sequence of `n` keys, one at a time.
+
+You are given a string `keysPressed` of length `n`, where `keysPressed[i]` was the `ith` key pressed in the testing sequence, and a sorted list `releaseTimes`, where `releaseTimes[i]` was the time the `ith` key was released. Both arrays are **0-indexed**. The `0th` key was pressed at the time `0`, and every subsequent key was pressed at the **exact** time the previous key was released.
+
+The tester wants to know the key of the keypress that had the **longest duration**. The `ith` keypress had a **duration** of `releaseTimes[i] - releaseTimes[i - 1]`, and the `0th` keypress had a duration of `releaseTimes[0]`.
+
+Note that the same key could have been pressed multiple times during the test, and these multiple presses of the same key **may not** have had the same **duration**.
+
+*Return the key of the keypress that had the **longest duration**. If there are multiple such keypresses, return the lexicographically largest key of the keypresses.*
+
+**Example 1:**
+
+```
+Input: releaseTimes = [9,29,49,50], keysPressed = "cbcd"
+Output: "c"
+Explanation: The keypresses were as follows:
+Keypress for 'c' had a duration of 9 (pressed at time 0 and released at time 9).
+Keypress for 'b' had a duration of 29 - 9 = 20 (pressed at time 9 right after the release of the previous character and released at time 29).
+Keypress for 'c' had a duration of 49 - 29 = 20 (pressed at time 29 right after the release of the previous character and released at time 49).
+Keypress for 'd' had a duration of 50 - 49 = 1 (pressed at time 49 right after the release of the previous character and released at time 50).
+The longest of these was the keypress for 'b' and the second keypress for 'c', both with duration 20.
+'c' is lexicographically larger than 'b', so the answer is 'c'.
+```
+
+**Example 2:**
+
+```
+Input: releaseTimes = [12,23,36,46,62], keysPressed = "spuda"
+Output: "a"
+Explanation: The keypresses were as follows:
+Keypress for 's' had a duration of 12.
+Keypress for 'p' had a duration of 23 - 12 = 11.
+Keypress for 'u' had a duration of 36 - 23 = 13.
+Keypress for 'd' had a duration of 46 - 36 = 10.
+Keypress for 'a' had a duration of 62 - 46 = 16.
+The longest of these was the keypress for 'a' with duration 16.
+```
+
+**Constraints:**
+
+- `releaseTimes.length == n`
+- `keysPressed.length == n`
+- `2 <= n <= 1000`
+- `1 <= releaseTimes[i] <= 109`
+- `releaseTimes[i] < releaseTimes[i+1]`
+- `keysPressed` contains only lowercase English letters.
+
+## 题目大意
+
+LeetCode 设计了一款新式键盘，正在测试其可用性。测试人员将会点击一系列键（总计 n 个），每次一个。
+
+给你一个长度为 n 的字符串 keysPressed ，其中 keysPressed[i] 表示测试序列中第 i 个被按下的键。releaseTimes 是一个升序排列的列表，其中 releaseTimes[i] 表示松开第 i 个键的时间。字符串和数组的 下标都从 0 开始 。第 0 个键在时间为 0 时被按下，接下来每个键都 恰好 在前一个键松开时被按下。测试人员想要找出按键 持续时间最长 的键。第 i 次按键的持续时间为 releaseTimes[i] - releaseTimes[i - 1] ，第 0 次按键的持续时间为 releaseTimes[0] 。
+
+注意，测试期间，同一个键可以在不同时刻被多次按下，而每次的持续时间都可能不同。请返回按键 持续时间最长 的键，如果有多个这样的键，则返回 按字母顺序排列最大 的那个键。
+
+## 解题思路
+
+- 题干很长，不过还是简单题。循环扫描一遍数组，计算出每个按键的持续时间。动态更新按键最长时间的键。如果持续时间最长的有多个键，需要返回字母序最大的那一个键。
+
+## 代码
+
+```go
+package leetcode
+
+func slowestKey(releaseTimes []int, keysPressed string) byte {
+	longestDuration, key := releaseTimes[0], keysPressed[0]
+	for i := 1; i < len(releaseTimes); i++ {
+		duration := releaseTimes[i] - releaseTimes[i-1]
+		if duration > longestDuration {
+			longestDuration = duration
+			key = keysPressed[i]
+		} else if duration == longestDuration && keysPressed[i] > key {
+			key = keysPressed[i]
+		}
+	}
+	return key
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1624.Largest-Substring-Between-Two-Equal-Characters/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1631.Path-With-Minimum-Effort/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1631.Path-With-Minimum-Effort.md b/website/content/ChapterFour/1600~1699/1631.Path-With-Minimum-Effort.md
new file mode 100644
index 000000000..e219d8c62
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1631.Path-With-Minimum-Effort.md
@@ -0,0 +1,177 @@
+# [1631. Path With Minimum Effort](https://leetcode.com/problems/path-with-minimum-effort/)
+
+## 题目
+
+You are a hiker preparing for an upcoming hike. You are given `heights`, a 2D array of size `rows x columns`, where `heights[row][col]` represents the height of cell `(row, col)`. You are situated in the top-left cell, `(0, 0)`, and you hope to travel to the bottom-right cell, `(rows-1, columns-1)` (i.e., **0-indexed**). You can move **up**, **down**, **left**, or **right**, and you wish to find a route that requires the minimum **effort**.
+
+A route's **effort** is the **maximum absolute difference** in heights between two consecutive cells of the route.
+
+Return *the minimum **effort** required to travel from the top-left cell to the bottom-right cell.*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/10/04/ex1.png](https://assets.leetcode.com/uploads/2020/10/04/ex1.png)
+
+```
+Input: heights = [[1,2,2],[3,8,2],[5,3,5]]
+Output: 2
+Explanation: The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
+This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2020/10/04/ex2.png](https://assets.leetcode.com/uploads/2020/10/04/ex2.png)
+
+```
+Input: heights = [[1,2,3],[3,8,4],[5,3,5]]
+Output: 1
+Explanation: The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].
+```
+
+**Example 3:**
+
+![https://assets.leetcode.com/uploads/2020/10/04/ex3.png](https://assets.leetcode.com/uploads/2020/10/04/ex3.png)
+
+```
+Input: heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
+Output: 0
+Explanation: This route does not require any effort.
+```
+
+**Constraints:**
+
+- `rows == heights.length`
+- `columns == heights[i].length`
+- `1 <= rows, columns <= 100`
+- `1 <= heights[i][j] <= 10^6`
+
+## 题目大意
+
+你准备参加一场远足活动。给你一个二维 `rows x columns` 的地图 `heights` ，其中 `heights[row][col]` 表示格子 `(row, col)` 的高度。一开始你在最左上角的格子 `(0, 0)` ，且你希望去最右下角的格子 `(rows-1, columns-1)` （注意下标从 0 开始编号）。你每次可以往 上，下，左，右 四个方向之一移动，你想要找到耗费 体力 最小的一条路径。一条路径耗费的 体力值 是路径上相邻格子之间 高度差绝对值 的 最大值 决定的。请你返回从左上角走到右下角的最小 体力消耗值 。
+
+## 解题思路
+
+- 此题和第 778 题解题思路完全一致。在第 778 题中求的是最短连通时间。此题求的是连通路径下的最小体力值。都是求的最小值，只是 2 个值的意义不同罢了。
+- 按照第 778 题的思路，本题也有多种解法。第一种解法是 DFS + 二分。先将题目变换一个等价问法。题目要求找到最小体力消耗值，也相当于问是否存在一个体力消耗值 x，只要大于等于 x，一定能连通。利用二分搜索来找到这个临界值。体力消耗值是有序的，此处满足二分搜索的条件。题目给定柱子高度是 [1,10^6]，所以体力值一定在 [0,10^6-1] 这个区间内。判断是否取中值的条件是用 DFS 或者 BFS 搜索 (0,0) 点和 (N-1, N-1) 点之间是否连通。时间复杂度：O(mnlogC)，其中 m 和 n 分别是地图的行数和列数，C 是格子的最大高度。C 最大为 10^6，所以 logC 常数也很小。空间复杂度 O(mn)。
+- 第二种解法是并查集。将图中所有边按照权值从小到大进行排序，并依次加入并查集中。直到加入一条权值为 x 的边以后，左上角到右下角连通了。最小体力消耗值也就找到了。注意加入边的时候，只加入 `i-1` 和 `i` ，`j-1` 和 `j` 这 2 类相邻的边。因为最小体力消耗意味着不走回头路。上下左右四个方向到达一个节点，只可能从上边和左边走过来。从下边和右边走过来肯定是浪费体力了。时间复杂度：O(mnlog(mn))，其中 m 和 n 分别是地图的行数和列数，图中的边数为 O(mn)。空间复杂度 O(mn)，即为存储所有边以及并查集需要的空间。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/leetcode-go/template"
+)
+
+var dir = [4][2]int{
+	{0, 1},
+	{1, 0},
+	{0, -1},
+	{-1, 0},
+}
+
+// 解法一 DFS + 二分
+func minimumEffortPath(heights [][]int) int {
+	n, m := len(heights), len(heights[0])
+	visited := make([][]bool, n)
+	for i := range visited {
+		visited[i] = make([]bool, m)
+	}
+	low, high := 0, 1000000
+	for low < high {
+		threshold := low + (high-low)>>1
+		if !hasPath(heights, visited, 0, 0, threshold) {
+			low = threshold + 1
+		} else {
+			high = threshold
+		}
+		for i := range visited {
+			for j := range visited[i] {
+				visited[i][j] = false
+			}
+		}
+	}
+	return low
+}
+
+func hasPath(heights [][]int, visited [][]bool, i, j, threshold int) bool {
+	n, m := len(heights), len(heights[0])
+	if i == n-1 && j == m-1 {
+		return true
+	}
+	visited[i][j] = true
+	res := false
+	for _, d := range dir {
+		ni, nj := i+d[0], j+d[1]
+		if ni < 0 || ni >= n || nj < 0 || nj >= m || visited[ni][nj] || res {
+			continue
+		}
+		diff := abs(heights[i][j] - heights[ni][nj])
+		if diff <= threshold && hasPath(heights, visited, ni, nj, threshold) {
+			res = true
+		}
+	}
+	return res
+}
+
+func abs(a int) int {
+	if a < 0 {
+		a = -a
+	}
+	return a
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+func max(a, b int) int {
+	if a < b {
+		return b
+	}
+	return a
+}
+
+// 解法二 并查集
+func minimumEffortPath1(heights [][]int) int {
+	n, m, edges, uf := len(heights), len(heights[0]), []edge{}, template.UnionFind{}
+	uf.Init(n * m)
+	for i, row := range heights {
+		for j, h := range row {
+			id := i*m + j
+			if i > 0 {
+				edges = append(edges, edge{id - m, id, abs(h - heights[i-1][j])})
+			}
+			if j > 0 {
+				edges = append(edges, edge{id - 1, id, abs(h - heights[i][j-1])})
+			}
+		}
+	}
+	sort.Slice(edges, func(i, j int) bool { return edges[i].diff < edges[j].diff })
+	for _, e := range edges {
+		uf.Union(e.v, e.w)
+		if uf.Find(0) == uf.Find(n*m-1) {
+			return e.diff
+		}
+	}
+	return 0
+}
+
+type edge struct {
+	v, w, diff int
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1629.Slowest-Key/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1636.Sort-Array-by-Increasing-Frequency/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1636.Sort-Array-by-Increasing-Frequency.md b/website/content/ChapterFour/1600~1699/1636.Sort-Array-by-Increasing-Frequency.md
new file mode 100644
index 000000000..517e4a895
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1636.Sort-Array-by-Increasing-Frequency.md
@@ -0,0 +1,73 @@
+# [1636. Sort Array by Increasing Frequency](https://leetcode.com/problems/sort-array-by-increasing-frequency/)
+
+
+## 题目
+
+Given an array of integers `nums`, sort the array in **increasing** order based on the frequency of the values. If multiple values have the same frequency, sort them in **decreasing** order.
+
+Return the *sorted array*.
+
+**Example 1:**
+
+```
+Input: nums = [1,1,2,2,2,3]
+Output: [3,1,1,2,2,2]
+Explanation: '3' has a frequency of 1, '1' has a frequency of 2, and '2' has a frequency of 3.
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,3,1,3,2]
+Output: [1,3,3,2,2]
+Explanation: '2' and '3' both have a frequency of 2, so they are sorted in decreasing order.
+```
+
+**Example 3:**
+
+```
+Input: nums = [-1,1,-6,4,5,-6,1,4,1]
+Output: [5,-1,4,4,-6,-6,1,1,1]
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `100 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个整数数组 `nums` ，请你将数组按照每个值的频率 **升序** 排序。如果有多个值的频率相同，请你按照数值本身将它们 **降序** 排序。请你返回排序后的数组。
+
+## 解题思路
+
+- 简单题。先统计每个值的频率，然后按照频率从小到大排序，相同频率的按照值的大小，从大到小排序。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func frequencySort(nums []int) []int {
+	freq := map[int]int{}
+	for _, v := range nums {
+		freq[v]++
+	}
+	sort.Slice(nums, func(i, j int) bool {
+		if freq[nums[i]] == freq[nums[j]] {
+			return nums[j] < nums[i]
+		}
+		return freq[nums[i]] < freq[nums[j]]
+	})
+	return nums
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1631.Path-With-Minimum-Effort/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1640.Check-Array-Formation-Through-Concatenation/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1640.Check-Array-Formation-Through-Concatenation.md b/website/content/ChapterFour/1600~1699/1640.Check-Array-Formation-Through-Concatenation.md
new file mode 100644
index 000000000..6b85c7d2f
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1640.Check-Array-Formation-Through-Concatenation.md
@@ -0,0 +1,102 @@
+# [1640. Check Array Formation Through Concatenation](https://leetcode.com/problems/check-array-formation-through-concatenation/)
+
+
+## 题目
+
+You are given an array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
+
+Return `true` *if it is possible to form the array* `arr` *from* `pieces`. Otherwise, return `false`.
+
+**Example 1**:
+
+```
+Input: arr = [85], pieces = [[85]]
+Output: true
+```
+
+**Example 2**:
+
+```
+Input: arr = [15,88], pieces = [[88],[15]]
+Output: true
+Explanation: Concatenate [15] then [88]
+```
+
+**Example 3**:
+
+```
+Input: arr = [49,18,16], pieces = [[16,18,49]]
+Output: false
+Explanation: Even though the numbers match, we cannot reorder pieces[0].
+```
+
+**Example 4**:
+
+```
+Input: arr = [91,4,64,78], pieces = [[78],[4,64],[91]]
+Output: true
+Explanation: Concatenate [91] then [4,64] then [78]
+```
+
+**Example 5**:
+
+```
+Input: arr = [1,3,5,7], pieces = [[2,4,6,8]]
+Output: false
+
+```
+
+**Constraints**:
+
+- `1 <= pieces.length <= arr.length <= 100`
+- `sum(pieces[i].length) == arr.length`
+- `1 <= pieces[i].length <= arr.length`
+- `1 <= arr[i], pieces[i][j] <= 100`
+- The integers in `arr` are **distinct**.
+- The integers in `pieces` are **distinct** (i.e., If we flatten pieces in a 1D array, all the integers in this array are distinct).
+
+## 题目大意
+
+给你一个整数数组 arr ，数组中的每个整数 互不相同 。另有一个由整数数组构成的数组 pieces，其中的整数也 互不相同 。请你以 任意顺序 连接 pieces 中的数组以形成 arr 。但是，不允许 对每个数组 pieces[i] 中的整数重新排序。如果可以连接 pieces 中的数组形成 arr ，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 简单题。题目保证了 arr 中的元素唯一，所以可以用 map 把每个元素的 index 存起来，方便查找。遍历 pieces 数组，在每个一维数组中判断元素顺序是否和原 arr 元素相对顺序一致。这个时候就用 map 查找，如果顺序是一一相连的，那么就是正确的。有一个顺序不是一一相连，或者出现了 arr 不存在的元素，都返回 false。
+
+## 代码
+
+```go
+package leetcode
+
+func canFormArray(arr []int, pieces [][]int) bool {
+	arrMap := map[int]int{}
+	for i, v := range arr {
+		arrMap[v] = i
+	}
+	for i := 0; i < len(pieces); i++ {
+		order := -1
+		for j := 0; j < len(pieces[i]); j++ {
+			if _, ok := arrMap[pieces[i][j]]; !ok {
+				return false
+			}
+			if order == -1 {
+				order = arrMap[pieces[i][j]]
+			} else {
+				if arrMap[pieces[i][j]] == order+1 {
+					order = arrMap[pieces[i][j]]
+				} else {
+					return false
+				}
+			}
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1636.Sort-Array-by-Increasing-Frequency/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1641.Count-Sorted-Vowel-Strings/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1641.Count-Sorted-Vowel-Strings.md b/website/content/ChapterFour/1600~1699/1641.Count-Sorted-Vowel-Strings.md
new file mode 100644
index 000000000..3967392c4
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1641.Count-Sorted-Vowel-Strings.md
@@ -0,0 +1,97 @@
+# [1641. Count Sorted Vowel Strings](https://leetcode.com/problems/count-sorted-vowel-strings/)
+
+
+## 题目
+
+Given an integer `n`, return *the number of strings of length* `n` *that consist only of vowels (*`a`*,* `e`*,* `i`*,* `o`*,* `u`*) and are **lexicographically sorted**.*
+
+A string `s` is **lexicographically sorted** if for all valid `i`, `s[i]` is the same as or comes before `s[i+1]` in the alphabet.
+
+**Example 1:**
+
+```
+Input: n = 1
+Output: 5
+Explanation: The 5 sorted strings that consist of vowels only are ["a","e","i","o","u"].
+```
+
+**Example 2:**
+
+```
+Input: n = 2
+Output: 15
+Explanation: The 15 sorted strings that consist of vowels only are
+["aa","ae","ai","ao","au","ee","ei","eo","eu","ii","io","iu","oo","ou","uu"].
+Note that "ea" is not a valid string since 'e' comes after 'a' in the alphabet.
+
+```
+
+**Example 3:**
+
+```
+Input: n = 33
+Output: 66045
+
+```
+
+**Constraints:**
+
+- `1 <= n <= 50`
+
+## 题目大意
+
+给你一个整数 n，请返回长度为 n 、仅由元音 (a, e, i, o, u) 组成且按 字典序排列 的字符串数量。
+
+字符串 s 按 字典序排列 需要满足：对于所有有效的 i，s[i] 在字母表中的位置总是与 s[i+1] 相同或在 s[i+1] 之前。
+
+## 解题思路
+
+- 题目给的数据量并不大，第一个思路是利用 DFS 遍历打表法。时间复杂度 O(1)，空间复杂度 O(1)。
+- 第二个思路是利用数学中的组合公式计算结果。题目等价于假设现在有 n 个字母，要求取 4 次球（可以选择不取）将字母分为 5 堆，问有几种取法。确定了取法以后，`a`，`e`，`i`，`o`，`u`，每个字母的个数就确定了，据题意要求按照字母序排序，那么最终字符串也就确定了。现在关注解决这个组合问题就可以了。把问题再转化一次，等价于，有 n+4 个字母，取 4 次，问有几种取法。+4 代表 4 个空操作，取走它们意味着不取。根据组合的数学定义，答案为 C(n+4,4)。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 打表
+func countVowelStrings(n int) int {
+	res := []int{1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315, 8855, 10626, 12650, 14950, 17550, 20475, 23751, 27405, 31465, 35960, 40920, 46376, 52360, 58905, 66045, 73815, 82251, 91390, 101270, 111930, 123410, 135751, 148995, 163185, 178365, 194580, 211876, 230300, 249900, 270725, 292825, 316251}
+	return res[n]
+}
+
+func makeTable() []int {
+	res, array := 0, []int{}
+	for i := 0; i < 51; i++ {
+		countVowelStringsDFS(i, 0, []string{}, []string{"a", "e", "i", "o", "u"}, &res)
+		array = append(array, res)
+		res = 0
+	}
+	return array
+}
+
+func countVowelStringsDFS(n, index int, cur []string, vowels []string, res *int) {
+	vowels = vowels[index:]
+	if len(cur) == n {
+		(*res)++
+		return
+	}
+	for i := 0; i < len(vowels); i++ {
+		cur = append(cur, vowels[i])
+		countVowelStringsDFS(n, i, cur, vowels, res)
+		cur = cur[:len(cur)-1]
+	}
+}
+
+// 解法二 数学方法 —— 组合
+func countVowelStrings1(n int) int {
+	return (n + 1) * (n + 2) * (n + 3) * (n + 4) / 24
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1640.Check-Array-Formation-Through-Concatenation/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1642.Furthest-Building-You-Can-Reach/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1642.Furthest-Building-You-Can-Reach.md b/website/content/ChapterFour/1600~1699/1642.Furthest-Building-You-Can-Reach.md
new file mode 100644
index 000000000..25a330745
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1642.Furthest-Building-You-Can-Reach.md
@@ -0,0 +1,120 @@
+# [1642. Furthest Building You Can Reach](https://leetcode.com/problems/furthest-building-you-can-reach/)
+
+
+## 题目
+
+You are given an integer array `heights` representing the heights of buildings, some `bricks`, and some `ladders`.
+
+You start your journey from building `0` and move to the next building by possibly using bricks or ladders.
+
+While moving from building `i` to building `i+1` (**0-indexed**),
+
+- If the current building's height is **greater than or equal** to the next building's height, you do **not** need a ladder or bricks.
+- If the current building's height is **less than** the next building's height, you can either use **one ladder** or `(h[i+1] - h[i])` **bricks**.
+
+*Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/10/27/q4.gif](https://assets.leetcode.com/uploads/2020/10/27/q4.gif)
+
+```
+Input: heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1
+Output: 4
+Explanation: Starting at building 0, you can follow these steps:
+- Go to building 1 without using ladders nor bricks since 4 >= 2.
+- Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 < 7.
+- Go to building 3 without using ladders nor bricks since 7 >= 6.
+- Go to building 4 using your only ladder. You must use either bricks or ladders because 6 < 9.
+It is impossible to go beyond building 4 because you do not have any more bricks or ladders.
+
+```
+
+**Example 2:**
+
+```
+Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
+Output: 7
+
+```
+
+**Example 3:**
+
+```
+Input: heights = [14,3,19,3], bricks = 17, ladders = 0
+Output: 3
+
+```
+
+**Constraints:**
+
+- `1 <= heights.length <= 10^5`
+- `1 <= heights[i] <= 10^6`
+- `0 <= bricks <= 10^9`
+- `0 <= ladders <= heights.length`
+
+## 题目大意
+
+给你一个整数数组 heights ，表示建筑物的高度。另有一些砖块 bricks 和梯子 ladders 。你从建筑物 0 开始旅程，不断向后面的建筑物移动，期间可能会用到砖块或梯子。当从建筑物 i 移动到建筑物 i+1（下标 从 0 开始 ）时：
+
+- 如果当前建筑物的高度 大于或等于 下一建筑物的高度，则不需要梯子或砖块。
+- 如果当前建筑的高度 小于 下一个建筑的高度，您可以使用 一架梯子 或 (h[i+1] - h[i]) 个砖块
+
+如果以最佳方式使用给定的梯子和砖块，返回你可以到达的最远建筑物的下标（下标 从 0 开始 ）。
+
+## 解题思路
+
+- 这一题可能会想到贪心算法。梯子很厉害，可以无限长，所以梯子用来跨越最高的楼。遇到非最高的距离差，先用砖头。这样贪心的话不正确。例如，[1, 5, 1, 2, 3, 4, 10000] 这组数据，梯子有 1 个，4 块砖头。最大的差距在 10000 和 4 之间，贪心选择在此处用梯子。但是砖头不足以让我们走到最后两栋楼。贪心得到的结果是 3，正确的结果是 5，先用梯子，再用砖头走过 3，4，5 号楼。
+- 上面的贪心解法错误在于没有“动态”的贪心，使用梯子应该选择能爬过楼里面最高的 2 个。于是顺理成章的想到了优先队列。维护一个长度为梯子个数的最小堆，当队列中元素超过梯子个数，便将队首最小值出队，出队的这个楼与楼的差距用砖头填补。所有砖头用完了，即是可以到达的最远楼号。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+)
+
+func furthestBuilding(heights []int, bricks int, ladder int) int {
+	usedLadder := &heightDiffPQ{}
+	for i := 1; i < len(heights); i++ {
+		needbricks := heights[i] - heights[i-1]
+		if needbricks < 0 {
+			continue
+		}
+		if ladder > 0 {
+			heap.Push(usedLadder, needbricks)
+			ladder--
+		} else {
+			if len(*usedLadder) > 0 && needbricks > (*usedLadder)[0] {
+				needbricks, (*usedLadder)[0] = (*usedLadder)[0], needbricks
+				heap.Fix(usedLadder, 0)
+			}
+			if bricks -= needbricks; bricks < 0 {
+				return i - 1
+			}
+		}
+	}
+	return len(heights) - 1
+}
+
+type heightDiffPQ []int
+
+func (pq heightDiffPQ) Len() int            { return len(pq) }
+func (pq heightDiffPQ) Less(i, j int) bool  { return pq[i] < pq[j] }
+func (pq heightDiffPQ) Swap(i, j int)       { pq[i], pq[j] = pq[j], pq[i] }
+func (pq *heightDiffPQ) Push(x interface{}) { *pq = append(*pq, x.(int)) }
+func (pq *heightDiffPQ) Pop() interface{} {
+	x := (*pq)[len(*pq)-1]
+	*pq = (*pq)[:len(*pq)-1]
+	return x
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1641.Count-Sorted-Vowel-Strings/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1646.Get-Maximum-in-Generated-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1646.Get-Maximum-in-Generated-Array.md b/website/content/ChapterFour/1600~1699/1646.Get-Maximum-in-Generated-Array.md
new file mode 100644
index 000000000..05c63906c
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1646.Get-Maximum-in-Generated-Array.md
@@ -0,0 +1,103 @@
+# [1646. Get Maximum in Generated Array](https://leetcode.com/problems/get-maximum-in-generated-array/)
+
+
+## 题目
+
+You are given an integer `n`. An array `nums` of length `n + 1` is generated in the following way:
+
+- `nums[0] = 0`
+- `nums[1] = 1`
+- `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
+- `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
+
+Return *****the **maximum** integer in the array* `nums`.
+
+**Example 1**:
+
+```
+Input: n = 7
+Output: 3
+Explanation: According to the given rules:
+  nums[0] = 0
+  nums[1] = 1
+  nums[(1 * 2) = 2] = nums[1] = 1
+  nums[(1 * 2) + 1 = 3] = nums[1] + nums[2] = 1 + 1 = 2
+  nums[(2 * 2) = 4] = nums[2] = 1
+  nums[(2 * 2) + 1 = 5] = nums[2] + nums[3] = 1 + 2 = 3
+  nums[(3 * 2) = 6] = nums[3] = 2
+  nums[(3 * 2) + 1 = 7] = nums[3] + nums[4] = 2 + 1 = 3
+Hence, nums = [0,1,1,2,1,3,2,3], and the maximum is 3.
+
+```
+
+**Example 2**:
+
+```
+Input: n = 2
+Output: 1
+Explanation: According to the given rules, the maximum between nums[0], nums[1], and nums[2] is 1.
+
+```
+
+**Example 3**:
+
+```
+Input: n = 3
+Output: 2
+Explanation: According to the given rules, the maximum between nums[0], nums[1], nums[2], and nums[3] is 2.
+
+```
+
+**Constraints**:
+
+- `0 <= n <= 100`
+
+## 题目大意
+
+给你一个整数 n 。按下述规则生成一个长度为 n + 1 的数组 nums ：
+
+- nums[0] = 0
+- nums[1] = 1
+- 当 2 <= 2 * i <= n 时，nums[2 * i] = nums[i]
+- 当 2 <= 2 * i + 1 <= n 时，nums[2 * i + 1] = nums[i] + nums[i + 1]
+
+返回生成数组 nums 中的 最大值。
+
+## 解题思路
+
+- 给出一个 n + 1 的数组，并按照生成规则生成这个数组，求出这个数组中的最大值。
+- 简单题，按照题意生成数组，边生成边记录和更新最大值即可。
+- 注意边界条件，当 n 为 0 的时候，数组里面只有一个元素 0 。
+
+## 代码
+
+```go
+package leetcode
+
+func getMaximumGenerated(n int) int {
+	if n == 0 {
+		return 0
+	}
+	nums, max := make([]int, n+1), 0
+	nums[0], nums[1] = 0, 1
+	for i := 0; i <= n; i++ {
+		if nums[i] > max {
+			max = nums[i]
+		}
+		if 2*i >= 2 && 2*i <= n {
+			nums[2*i] = nums[i]
+		}
+		if 2*i+1 >= 2 && 2*i+1 <= n {
+			nums[2*i+1] = nums[i] + nums[i+1]
+		}
+	}
+	return max
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1642.Furthest-Building-You-Can-Reach/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique.md b/website/content/ChapterFour/1600~1699/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique.md
new file mode 100644
index 000000000..738cb9175
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique.md
@@ -0,0 +1,96 @@
+# [1647. Minimum Deletions to Make Character Frequencies Unique](https://leetcode.com/problems/minimum-deletions-to-make-character-frequencies-unique/)
+
+
+## 题目
+
+A string `s` is called **good** if there are no two different characters in `s` that have the same **frequency**.
+
+Given a string `s`, return *the **minimum** number of characters you need to delete to make* `s` ***good**.*
+
+The **frequency** of a character in a string is the number of times it appears in the string. For example, in the string `"aab"`, the **frequency** of `'a'` is `2`, while the **frequency** of `'b'` is `1`.
+
+**Example 1**:
+
+```
+Input: s = "aab"
+Output: 0
+Explanation: s is already good.
+
+```
+
+**Example 2**:
+
+```
+Input: s = "aaabbbcc"
+Output: 2
+Explanation: You can delete two 'b's resulting in the good string "aaabcc".
+Another way it to delete one 'b' and one 'c' resulting in the good string "aaabbc".
+```
+
+**Example 3**:
+
+```
+Input: s = "ceabaacb"
+Output: 2
+Explanation: You can delete both 'c's resulting in the good string "eabaab".
+Note that we only care about characters that are still in the string at the end (i.e. frequency of 0 is ignored).
+
+```
+
+**Constraints**:
+
+- `1 <= s.length <= 105`
+- `s` contains only lowercase English letters.
+
+## 题目大意
+
+如果字符串 s 中 不存在 两个不同字符 频次 相同的情况，就称 s 是 优质字符串 。
+
+给你一个字符串 s，返回使 s 成为优质字符串需要删除的最小字符数。
+
+字符串中字符的 频次 是该字符在字符串中的出现次数。例如，在字符串 "aab" 中，'a' 的频次是 2，而 'b' 的频次是 1 。
+
+**提示：**
+
+- `1 <= s.length <= 105`
+- `s` 仅含小写英文字母
+
+## 解题思路
+
+- 给出一个字符串 s，要求输出使 s 变成“优质字符串”需要删除的最小字符数。“优质字符串”的定义是：字符串 s 中不存在频次相同的两个不同字符。
+- 首先将 26 个字母在字符串中的频次分别统计出来，然后把频次从大到小排列，从频次大的开始，依次调整：例如，假设前一个和后一个频次相等，就把前一个字符删除一个，频次减一，再次排序，如果频次还相等，继续调整，如果频次不同了，游标往后移，继续调整后面的频次。直到所有的频次都不同了，就可以输出最终结果了。
+- 这里需要注意频次为 0 的情况，即字母都被删光了。频次为 0 以后，就不需要再比较了。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func minDeletions(s string) int {
+	frequency, res := make([]int, 26), 0
+	for i := 0; i < len(s); i++ {
+		frequency[s[i]-'a']++
+	}
+	sort.Sort(sort.Reverse(sort.IntSlice(frequency)))
+	for i := 1; i <= 25; i++ {
+		if frequency[i] == frequency[i-1] && frequency[i] != 0 {
+			res++
+			frequency[i]--
+			sort.Sort(sort.Reverse(sort.IntSlice(frequency)))
+			i--
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1646.Get-Maximum-in-Generated-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1648.Sell-Diminishing-Valued-Colored-Balls/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1648.Sell-Diminishing-Valued-Colored-Balls.md b/website/content/ChapterFour/1600~1699/1648.Sell-Diminishing-Valued-Colored-Balls.md
new file mode 100644
index 000000000..a341c7c9e
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1648.Sell-Diminishing-Valued-Colored-Balls.md
@@ -0,0 +1,138 @@
+# [1648. Sell Diminishing-Valued Colored Balls](https://leetcode.com/problems/sell-diminishing-valued-colored-balls/)
+
+
+## 题目
+
+You have an `inventory` of different colored balls, and there is a customer that wants `orders` balls of **any** color.
+
+The customer weirdly values the colored balls. Each colored ball's value is the number of balls **of that color** you currently have in your `inventory`. For example, if you own `6` yellow balls, the customer would pay `6` for the first yellow ball. After the transaction, there are only `5` yellow balls left, so the next yellow ball is then valued at `5` (i.e., the value of the balls decreases as you sell more to the customer).
+
+You are given an integer array, `inventory`, where `inventory[i]` represents the number of balls of the `ith` color that you initially own. You are also given an integer `orders`, which represents the total number of balls that the customer wants. You can sell the balls **in any order**.
+
+Return *the **maximum** total value that you can attain after selling* `orders` *colored balls*. As the answer may be too large, return it **modulo** `109 + 7`.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2020/11/05/jj.gif](https://assets.leetcode.com/uploads/2020/11/05/jj.gif)
+
+```
+Input: inventory = [2,5], orders = 4
+Output: 14
+Explanation: Sell the 1st color 1 time (2) and the 2nd color 3 times (5 + 4 + 3).
+The maximum total value is 2 + 5 + 4 + 3 = 14.
+
+```
+
+**Example 2**:
+
+```
+Input: inventory = [3,5], orders = 6
+Output: 19
+Explanation: Sell the 1st color 2 times (3 + 2) and the 2nd color 4 times (5 + 4 + 3 + 2).
+The maximum total value is 3 + 2 + 5 + 4 + 3 + 2 = 19.
+```
+
+**Example 3**:
+
+```
+Input: inventory = [2,8,4,10,6], orders = 20
+Output: 110
+```
+
+**Example 4**:
+
+```
+Input: inventory = [1000000000], orders = 1000000000
+Output: 21
+Explanation: Sell the 1st color 1000000000 times for a total value of 500000000500000000. 500000000500000000 modulo 109 + 7 = 21.
+```
+
+**Constraints**:
+
+- `1 <= inventory.length <= 10^5`
+- `1 <= inventory[i] <= 10^9`
+- `1 <= orders <= min(sum(inventory[i]), 10^9)`
+
+## 题目大意
+
+你有一些球的库存 inventory ，里面包含着不同颜色的球。一个顾客想要 任意颜色 总数为 orders 的球。这位顾客有一种特殊的方式衡量球的价值：每个球的价值是目前剩下的 同色球 的数目。比方说还剩下 6 个黄球，那么顾客买第一个黄球的时候该黄球的价值为 6 。这笔交易以后，只剩下 5 个黄球了，所以下一个黄球的价值为 5 （也就是球的价值随着顾客购买同色球是递减的）
+
+给你整数数组 inventory ，其中 inventory[i] 表示第 i 种颜色球一开始的数目。同时给你整数 orders ，表示顾客总共想买的球数目。你可以按照 任意顺序 卖球。请你返回卖了 orders 个球以后 最大 总价值之和。由于答案可能会很大，请你返回答案对 109 + 7 取余数 的结果。
+
+提示：
+
+- 1 <= inventory.length <= 10^5
+- 1 <= inventory[i] <= 10^9
+- 1 <= orders <= min(sum(inventory[i]), 10^9)
+
+## 解题思路
+
+- 给出一个 `inventory` 数组和 `orders` 次操作，要求输出数组中前 `orders` 大个元素累加和。需要注意的是，每累加一个元素 `inventory[i]`，这个元素都会减一，下次再累加的时候，需要选取更新以后的数组的最大值。
+- 拿到这个题目以后很容易想到优先队列，建立大根堆以后，`pop` 出当前最大值 `maxItem`，累加，以后把 `maxItem` 减一再 `push` 回去。循环执行 `orders` 次以后即是最终结果。题目是这个意思，但是我们不能这么写代码，因为题目条件里面给出了 `orders` 的数据大小。orders 最大为 10^9。按照优先队列的这个方法一定会超时，时间复杂度为 O(orders⋅logn)。那就换一个思路。优先队列这个思路中，重复操作了 `orders` 次，其实在这些操作中，有一些是没有必要的废操作。这些大量的“废”操作导致了超时。试想，在 `orders` 次操作中，能否合并 `n` 个 `pop` 操作，一口气先 `pop` 掉 `n` 个前 `n` 大的数呢？这个是可行的，因为每次 `pop` 出去，元素都只会减一，这个是非常有规律的。
+- 为了接下来的描述更加清晰易懂，还需要再定义 1 个值， `thresholdValue` 为操作 `n` 次以后，当前  `inventory` 数组的最大值。关于 `thresholdValue` 的理解，这里要说明一下。 `thresholdValue` 的来源有 2 种，一种是本来数组里面就有这个值，还有一种来源是 `inventory[i]` 元素减少到了 `thresholdValue` 这个值。举个例子：原始数组是 [2,3,3,4,5]，`orders` = 4，取 4 次以后，剩下的数组是 [2,2,3,3,3]。3 个 3 里面其中一个 3 就来自于 `4-1=3`，或者 `5-2=3`。
+- 用二分搜索在 [0，max(`inventory`)] 区间内找到这个 `thresholdValue` 值，能满足下列不等式的最小 `thresholdValue` 值：
+	{{< katex display >}} 
+    \sum_{inventory[i]\geqslant thresholdValue}^{} \left ( inventory[i] - thresholdValue \right )\leqslant orders
+	{{< /katex >}} 
+    `thresholdValue` 越小，不等式左边的值越大，随着 `thresholdValue` 的增大，不等式左边的值越来越小，直到刚刚能小于等于 `orders`。求出了 `thresholdValue` 值以后，还需要再判断有多少值等于 `thresholdValue - 1` 值了。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_1648.png)
+
+- 还是举上面的例子，原始数组是 [2,3,3,4,5]，`orders` = 4，我们可以求得 `thresholdValue` = 3 。`inventory[i]` > `thresholdValue` 的那部分 100% 的要取走，`thresholdValue` 就像一个水平面，突出水平面的那些都要拿走，每列的值按照等差数列求和公式计算即可。但是 `orders` - `thresholdValue` = 1，说明水平面以下还要拿走一个，即 `thresholdValue` 线下的虚线框里面的那 4 个球，还需要任意取走一个。最后总的结果是这 2 部分的总和，( ( 5 + 4 ) + 4 ) + 3 = 16 。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+)
+
+// 解法一 贪心 + 二分搜索
+func maxProfit(inventory []int, orders int) int {
+	maxItem, thresholdValue, count, res, mod := 0, -1, 0, 0, 1000000007
+	for i := 0; i < len(inventory); i++ {
+		if inventory[i] > maxItem {
+			maxItem = inventory[i]
+		}
+	}
+	low, high := 0, maxItem
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		for i := 0; i < len(inventory); i++ {
+			count += max(inventory[i]-mid, 0)
+		}
+		if count <= orders {
+			thresholdValue = mid
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+		count = 0
+	}
+	count = 0
+	for i := 0; i < len(inventory); i++ {
+		count += max(inventory[i]-thresholdValue, 0)
+	}
+	count = orders - count
+	for i := 0; i < len(inventory); i++ {
+		if inventory[i] >= thresholdValue {
+			if count > 0 {
+				res += (thresholdValue + inventory[i]) * (inventory[i] - thresholdValue + 1) / 2
+				count--
+			} else {
+				res += (thresholdValue + 1 + inventory[i]) * (inventory[i] - thresholdValue) / 2
+			}
+		}
+	}
+	return res % mod
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1649.Create-Sorted-Array-through-Instructions/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1649.Create-Sorted-Array-through-Instructions.md b/website/content/ChapterFour/1600~1699/1649.Create-Sorted-Array-through-Instructions.md
new file mode 100644
index 000000000..cdd804681
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1649.Create-Sorted-Array-through-Instructions.md
@@ -0,0 +1,156 @@
+# [1649. Create Sorted Array through Instructions](https://leetcode.com/problems/create-sorted-array-through-instructions/)
+
+## 题目
+
+Given an integer array `instructions`, you are asked to create a sorted array from the elements in `instructions`. You start with an empty container `nums`. For each element from **left to right** in `instructions`, insert it into `nums`. The **cost** of each insertion is the **minimum** of the following:
+
+- The number of elements currently in `nums` that are **strictly less than** `instructions[i]`.
+- The number of elements currently in `nums` that are **strictly greater than** `instructions[i]`.
+
+For example, if inserting element `3` into `nums = [1,2,3,5]`, the **cost** of insertion is `min(2, 1)` (elements `1` and `2` are less than `3`, element `5` is greater than `3`) and `nums` will become `[1,2,3,3,5]`.
+
+Return *the **total cost** to insert all elements from* `instructions` *into* `nums`. Since the answer may be large, return it **modulo** `10^9 + 7`
+
+**Example 1**:
+
+```
+Input: instructions = [1,5,6,2]
+Output: 1
+Explanation: Begin with nums = [].
+Insert 1 with cost min(0, 0) = 0, now nums = [1].
+Insert 5 with cost min(1, 0) = 0, now nums = [1,5].
+Insert 6 with cost min(2, 0) = 0, now nums = [1,5,6].
+Insert 2 with cost min(1, 2) = 1, now nums = [1,2,5,6].
+The total cost is 0 + 0 + 0 + 1 = 1.
+```
+
+**Example 2**:
+
+```
+Input: instructions = [1,2,3,6,5,4]
+Output: 3
+Explanation: Begin with nums = [].
+Insert 1 with cost min(0, 0) = 0, now nums = [1].
+Insert 2 with cost min(1, 0) = 0, now nums = [1,2].
+Insert 3 with cost min(2, 0) = 0, now nums = [1,2,3].
+Insert 6 with cost min(3, 0) = 0, now nums = [1,2,3,6].
+Insert 5 with cost min(3, 1) = 1, now nums = [1,2,3,5,6].
+Insert 4 with cost min(3, 2) = 2, now nums = [1,2,3,4,5,6].
+The total cost is 0 + 0 + 0 + 0 + 1 + 2 = 3.
+```
+
+**Example 3**:
+
+```
+Input: instructions = [1,3,3,3,2,4,2,1,2]
+Output: 4
+Explanation: Begin with nums = [].
+Insert 1 with cost min(0, 0) = 0, now nums = [1].
+Insert 3 with cost min(1, 0) = 0, now nums = [1,3].
+Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3].
+Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3,3].
+Insert 2 with cost min(1, 3) = 1, now nums = [1,2,3,3,3].
+Insert 4 with cost min(5, 0) = 0, now nums = [1,2,3,3,3,4].
+Insert 2 with cost min(1, 4) = 1, now nums = [1,2,2,3,3,3,4].
+Insert 1 with cost min(0, 6) = 0, now nums = [1,1,2,2,3,3,3,4].
+Insert 2 with cost min(2, 4) = 2, now nums = [1,1,2,2,2,3,3,3,4].
+The total cost is 0 + 0 + 0 + 0 + 1 + 0 + 1 + 0 + 2 = 4.
+```
+
+**Constraints**:
+
+- `1 <= instructions.length <= 105`
+- `1 <= instructions[i] <= 105`
+
+## 题目大意
+
+给你一个整数数组 instructions ，你需要根据 instructions 中的元素创建一个有序数组。一开始你有一个空的数组 nums ，你需要 从左到右 遍历 instructions 中的元素，将它们依次插入 nums 数组中。每一次插入操作的 代价 是以下两者的 较小值 ：
+
+- nums 中 严格小于 instructions[i] 的数字数目。
+- nums 中 严格大于 instructions[i] 的数字数目。
+
+比方说，如果要将 3 插入到 nums = [1,2,3,5] ，那么插入操作的 代价 为 min(2, 1) (元素 1 和 2 小于 3 ，元素 5 大于 3 ），插入后 nums 变成 [1,2,3,3,5] 。请你返回将 instructions 中所有元素依次插入 nums 后的 总最小代价 。由于答案会很大，请将它对 10^9 + 7 取余 后返回。
+
+## 解题思路
+
+- 给出一个数组，要求将其中的元素从头开始往另外一个空数组中插入，每次插入前，累加代价值 cost = min(**strictly less than**, **strictly greater than**)。最后输出累加值。
+- 这一题虽然是 Hard 题，但是读完题以后就可以判定这是模板题了。可以用线段树和树状数组来解决。这里简单说说线段树的思路吧，先将待插入的数组排序，获得总的区间。每次循环做 4 步：2 次 `query` 分别得到 `strictlyLessThan` 和 `strictlyGreaterThan` ，再比较出两者中的最小值累加，最后一步就是 `update`。
+- 由于题目给的数据比较大，所以建立线段树之前记得要先离散化。这一题核心代码不超过 10 行，其他的都是模板代码。具体实现见代码。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+
+	"github.com/halfrost/leetcode-go/template"
+)
+
+// 解法一 树状数组 Binary Indexed Tree
+func createSortedArray(instructions []int) int {
+	bit, res := template.BinaryIndexedTree{}, 0
+	bit.Init(100001)
+	for i, v := range instructions {
+		less := bit.Query(v - 1)
+		greater := i - bit.Query(v)
+		res = (res + min(less, greater)) % (1e9 + 7)
+		bit.Add(v, 1)
+	}
+	return res
+}
+
+// 解法二 线段树 SegmentTree
+func createSortedArray1(instructions []int) int {
+	if len(instructions) == 0 {
+		return 0
+	}
+	st, res, mod := template.SegmentCountTree{}, 0, 1000000007
+	numsMap, numsArray, tmpArray := discretization1649(instructions)
+	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
+	st.Init(tmpArray, func(i, j int) int {
+		return 0
+	})
+	for i := 0; i < len(instructions); i++ {
+		strictlyLessThan := st.Query(0, numsMap[instructions[i]]-1)
+		strictlyGreaterThan := st.Query(numsMap[instructions[i]]+1, numsArray[len(numsArray)-1])
+		res = (res + min(strictlyLessThan, strictlyGreaterThan)) % mod
+		st.UpdateCount(numsMap[instructions[i]])
+	}
+	return res
+}
+
+func discretization1649(instructions []int) (map[int]int, []int, []int) {
+	tmpArray, numsArray, numsMap := []int{}, []int{}, map[int]int{}
+	for i := 0; i < len(instructions); i++ {
+		numsMap[instructions[i]] = instructions[i]
+	}
+	for _, v := range numsMap {
+		numsArray = append(numsArray, v)
+	}
+	sort.Ints(numsArray)
+	for i, num := range numsArray {
+		numsMap[num] = i
+	}
+	for i := range numsArray {
+		tmpArray = append(tmpArray, i)
+	}
+	return numsMap, numsArray, tmpArray
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1648.Sell-Diminishing-Valued-Colored-Balls/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1652.Defuse-the-Bomb/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1652.Defuse-the-Bomb.md b/website/content/ChapterFour/1600~1699/1652.Defuse-the-Bomb.md
new file mode 100644
index 000000000..8f5dc35f7
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1652.Defuse-the-Bomb.md
@@ -0,0 +1,132 @@
+# [1652. Defuse the Bomb](https://leetcode.com/problems/defuse-the-bomb/)
+
+
+## 题目
+
+You have a bomb to defuse, and your time is running out! Your informer will provide you with a **circular** array `code` of length of `n` and a key `k`.
+
+To decrypt the code, you must replace every number. All the numbers are replaced **simultaneously**.
+
+- If `k > 0`, replace the `ith` number with the sum of the **next** `k` numbers.
+- If `k < 0`, replace the `ith` number with the sum of the **previous** `k` numbers.
+- If `k == 0`, replace the `ith` number with `0`.
+
+As `code` is circular, the next element of `code[n-1]` is `code[0]`, and the previous element of `code[0]` is `code[n-1]`.
+
+Given the **circular** array `code` and an integer key `k`, return *the decrypted code to defuse the bomb*!
+
+**Example 1**:
+
+```
+Input: code = [5,7,1,4], k = 3
+Output: [12,10,16,13]
+Explanation: Each number is replaced by the sum of the next 3 numbers. The decrypted code is [7+1+4, 1+4+5, 4+5+7, 5+7+1]. Notice that the numbers wrap around.
+```
+
+**Example 2**:
+
+```
+Input: code = [1,2,3,4], k = 0
+Output: [0,0,0,0]
+Explanation: When k is zero, the numbers are replaced by 0. 
+```
+
+**Example 3**:
+
+```
+Input: code = [2,4,9,3], k = -2
+Output: [12,5,6,13]
+Explanation: The decrypted code is [3+9, 2+3, 4+2, 9+4]. Notice that the numbers wrap around again. If k is negative, the sum is of the previous numbers.
+```
+
+**Constraints**:
+
+- `n == code.length`
+- `1 <= n <= 100`
+- `1 <= code[i] <= 100`
+- `(n - 1) <= k <= n - 1`
+
+## 题目大意
+
+你有一个炸弹需要拆除，时间紧迫！你的情报员会给你一个长度为 n 的 循环 数组 code 以及一个密钥 k 。为了获得正确的密码，你需要替换掉每一个数字。所有数字会 同时 被替换。
+
+- 如果 k > 0 ，将第 i 个数字用 接下来 k 个数字之和替换。
+- 如果 k < 0 ，将第 i 个数字用 之前 k 个数字之和替换。
+- 如果 k == 0 ，将第 i 个数字用 0 替换。
+
+由于 code 是循环的， code[n-1] 下一个元素是 code[0] ，且 code[0] 前一个元素是 code[n-1] 。
+
+给你 循环 数组 code 和整数密钥 k ，请你返回解密后的结果来拆除炸弹！
+
+## 解题思路
+
+- 给出一个 code 数组，要求按照规则替换每个字母。
+- 简单题，按照题意描述循环即可。
+
+## 代码
+
+```go
+package leetcode
+
+func decrypt(code []int, k int) []int {
+	if k == 0 {
+		for i := 0; i < len(code); i++ {
+			code[i] = 0
+		}
+		return code
+	}
+	count, sum, res := k, 0, make([]int, len(code))
+	if k > 0 {
+		for i := 0; i < len(code); i++ {
+			for j := i + 1; j < len(code); j++ {
+				if count == 0 {
+					break
+				}
+				sum += code[j]
+				count--
+			}
+			if count > 0 {
+				for j := 0; j < len(code); j++ {
+					if count == 0 {
+						break
+					}
+					sum += code[j]
+					count--
+				}
+			}
+			res[i] = sum
+			sum, count = 0, k
+		}
+	}
+	if k < 0 {
+		for i := 0; i < len(code); i++ {
+			for j := i - 1; j >= 0; j-- {
+				if count == 0 {
+					break
+				}
+				sum += code[j]
+				count++
+			}
+			if count < 0 {
+				for j := len(code) - 1; j >= 0; j-- {
+					if count == 0 {
+						break
+					}
+					sum += code[j]
+					count++
+				}
+			}
+			res[i] = sum
+			sum, count = 0, k
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1649.Create-Sorted-Array-through-Instructions/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1653.Minimum-Deletions-to-Make-String-Balanced/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1653.Minimum-Deletions-to-Make-String-Balanced.md b/website/content/ChapterFour/1600~1699/1653.Minimum-Deletions-to-Make-String-Balanced.md
new file mode 100644
index 000000000..15d8427ea
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1653.Minimum-Deletions-to-Make-String-Balanced.md
@@ -0,0 +1,97 @@
+# [1653. Minimum Deletions to Make String Balanced](https://leetcode.com/problems/minimum-deletions-to-make-string-balanced/)
+
+
+## 题目
+
+You are given a string `s` consisting only of characters `'a'` and `'b'`.
+
+You can delete any number of characters in `s` to make `s` **balanced**. `s` is **balanced** if there is no pair of indices `(i,j)` such that `i < j` and `s[i] = 'b'` and `s[j]= 'a'`.
+
+Return *the **minimum** number of deletions needed to make* `s` ***balanced***.
+
+**Example 1**:
+
+```
+Input: s = "aababbab"
+Output: 2
+Explanation: You can either:
+Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
+Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").
+```
+
+**Example 2**:
+
+```
+Input: s = "bbaaaaabb"
+Output: 2
+Explanation: The only solution is to delete the first two characters.
+```
+
+**Constraints**:
+
+- `1 <= s.length <= 105`
+- `s[i]` is `'a'` or `'b'`.
+
+## 题目大意
+
+给你一个字符串 s ，它仅包含字符 'a' 和 'b' 。你可以删除 s 中任意数目的字符，使得 s 平衡 。我们称 s 平衡的 当不存在下标对 (i,j) 满足 i < j 且 s[i] = 'b' 同时 s[j]= 'a' 。请你返回使 s 平衡 的 最少 删除次数。
+
+## 解题思路
+
+- 给定一个字符串，要求删除最少次数，使得字母 a 都排在字母 b 的前面。
+- 很容易想到的一个解题思路是 DP。定义 `dp[i]` 为字符串下标 [ 0, i ] 这个区间内使得字符串平衡的最少删除次数。当 `s[i] == 'a'` 的时候，有 2 种情况，一种是 `s[i]` 前面全是 `[aa……aa]` 的情况，这个时候只需要把其中的所有的字母 `b` 删除即可。还有一种情况是 `s[i]` 前面有字母 `a` 也有字母 `b`，即 `[aaa……abb……b]`，这种情况就需要考虑 `dp[i-1]` 了。当前字母是 `a`，那么肯定要删除字母 `a`，来维持前面有一段字母 `b` 的情况。当 `s[i] == 'b'` 的时候，不管是 `[aa……aa]` 这种情况，还是 `[aaa……abb……b]` 这种情况，当前字母 `b` 都可以直接附加在后面，也能保证整个字符串是平衡的。所以状态转移方程为 `dp[i+1] = min(dp[i] + 1, bCount), s[i] == 'a'`，`dp[i+1] = dp[i], s[i] == 'b'`。最终答案存在 `dp[n]` 中。由于前后项的递推关系中只用到一次前一项，所以我们还可以优化一下空间，用一个变量保存前一项的结果。优化以后的代码见解法一。
+- 这一题还有一个模拟的思路。题目要求找到最小删除字数，那么就是要找到一个“临界点”，在这个临界点的左边删除所有的字母 b，在这个临界点的右边删除所有的字母 a。在所有的“临界点”中找到删除最少的次数。代码实现见解法二。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 DP
+func minimumDeletions(s string) int {
+	prev, res, bCount := 0, 0, 0
+	for _, c := range s {
+		if c == 'a' {
+			res = min(prev+1, bCount)
+			prev = res
+		} else {
+			bCount++
+		}
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+// 解法二 模拟
+func minimumDeletions1(s string) int {
+	aCount, bCount, res := 0, 0, 0
+	for i := 0; i < len(s); i++ {
+		if s[i] == 'a' {
+			aCount++
+		}
+	}
+	res = aCount
+	for i := 0; i < len(s); i++ {
+		if s[i] == 'a' {
+			aCount--
+		} else {
+			bCount++
+		}
+		res = min(res, aCount+bCount)
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1652.Defuse-the-Bomb/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1654.Minimum-Jumps-to-Reach-Home/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1654.Minimum-Jumps-to-Reach-Home.md b/website/content/ChapterFour/1600~1699/1654.Minimum-Jumps-to-Reach-Home.md
new file mode 100644
index 000000000..bc5f9a413
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1654.Minimum-Jumps-to-Reach-Home.md
@@ -0,0 +1,108 @@
+# [1654. Minimum Jumps to Reach Home](https://leetcode.com/problems/minimum-jumps-to-reach-home/)
+
+
+## 题目
+
+A certain bug's home is on the x-axis at position `x`. Help them get there from position `0`.
+
+The bug jumps according to the following rules:
+
+- It can jump exactly `a` positions **forward** (to the right).
+- It can jump exactly `b` positions **backward** (to the left).
+- It cannot jump backward twice in a row.
+- It cannot jump to any `forbidden` positions.
+
+The bug may jump forward **beyond** its home, but it **cannot jump** to positions numbered with **negative** integers.
+
+Given an array of integers `forbidden`, where `forbidden[i]` means that the bug cannot jump to the position `forbidden[i]`, and integers `a`, `b`, and `x`, return *the minimum number of jumps needed for the bug to reach its home*. If there is no possible sequence of jumps that lands the bug on position `x`, return `1.`
+
+**Example 1**:
+
+```
+Input: forbidden = [14,4,18,1,15], a = 3, b = 15, x = 9
+Output: 3
+Explanation: 3 jumps forward (0 -> 3 -> 6 -> 9) will get the bug home.
+```
+
+**Example 2**:
+
+```
+Input: forbidden = [8,3,16,6,12,20], a = 15, b = 13, x = 11
+Output: -1
+```
+
+**Example 3**:
+
+```
+Input: forbidden = [1,6,2,14,5,17,4], a = 16, b = 9, x = 7
+Output: 2
+Explanation: One jump forward (0 -> 16) then one jump backward (16 -> 7) will get the bug home.
+
+```
+
+**Constraints**:
+
+- `1 <= forbidden.length <= 1000`
+- `1 <= a, b, forbidden[i] <= 2000`
+- `0 <= x <= 2000`
+- All the elements in `forbidden` are distinct.
+- Position `x` is not forbidden.
+
+## 题目大意
+
+有一只跳蚤的家在数轴上的位置 x 处。请你帮助它从位置 0 出发，到达它的家。
+
+跳蚤跳跃的规则如下：
+
+- 它可以 往前 跳恰好 a 个位置（即往右跳）。
+- 它可以 往后 跳恰好 b 个位置（即往左跳）。
+- 它不能 连续 往后跳 2 次。
+- 它不能跳到任何 forbidden 数组中的位置。
+
+跳蚤可以往前跳 超过 它的家的位置，但是它 不能跳到负整数 的位置。给你一个整数数组 forbidden ，其中 forbidden[i] 是跳蚤不能跳到的位置，同时给你整数 a， b 和 x ，请你返回跳蚤到家的最少跳跃次数。如果没有恰好到达 x 的可行方案，请你返回 -1 。
+
+## 解题思路
+
+- 给出坐标 x ，可以往前跳的步长 a，往后跳的步长 b。要求输出能跳回家的最少跳跃次数。
+- 求最少跳跃次数，思路用 BFS 求解，最先到达坐标 x 的方案即是最少跳跃次数。对`forbidden` 的处理是把记忆化数组里面把他们标记为 true。禁止连续往后跳 2 次的限制，要求我们在 BFS 入队的时候再记录一下跳跃方向，每次往后跳的时候判断前一跳是否是往后跳，如果是往后跳，此次就不能往后跳了。
+
+## 代码
+
+```go
+package leetcode
+
+func minimumJumps(forbidden []int, a int, b int, x int) int {
+	visited := make([]bool, 6000)
+	for i := range forbidden {
+		visited[forbidden[i]] = true
+	}
+	queue, res := [][2]int{{0, 0}}, -1
+	for len(queue) > 0 {
+		length := len(queue)
+		res++
+		for i := 0; i < length; i++ {
+			cur, isBack := queue[i][0], queue[i][1]
+			if cur == x {
+				return res
+			}
+			if isBack == 0 && cur-b > 0 && !visited[cur-b] {
+				visited[cur-b] = true
+				queue = append(queue, [2]int{cur - b, 1})
+			}
+			if cur+a < len(visited) && !visited[cur+a] {
+				visited[cur+a] = true
+				queue = append(queue, [2]int{cur + a, 0})
+			}
+		}
+		queue = queue[length:]
+	}
+	return -1
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1653.Minimum-Deletions-to-Make-String-Balanced/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1655.Distribute-Repeating-Integers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1655.Distribute-Repeating-Integers.md b/website/content/ChapterFour/1600~1699/1655.Distribute-Repeating-Integers.md
new file mode 100644
index 000000000..ca4122a86
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1655.Distribute-Repeating-Integers.md
@@ -0,0 +1,119 @@
+# [1655. Distribute Repeating Integers](https://leetcode.com/problems/distribute-repeating-integers/)
+
+
+## 题目
+
+You are given an array of `n` integers, `nums`, where there are at most `50` unique values in the array. You are also given an array of `m` customer order quantities, `quantity`, where `quantity[i]` is the amount of integers the `ith` customer ordered. Determine if it is possible to distribute `nums` such that:
+
+- The `ith` customer gets **exactly** `quantity[i]` integers,
+- The integers the `ith` customer gets are **all equal**, and
+- Every customer is satisfied.
+
+Return `true` *if it is possible to distribute* `nums` *according to the above conditions*.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,4], quantity = [2]
+Output: false
+Explanation: The 0th customer cannot be given two different integers.
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,2,3,3], quantity = [2]
+Output: true
+Explanation: The 0th customer is given [3,3]. The integers [1,2] are not used.
+```
+
+**Example 3**:
+
+```
+Input: nums = [1,1,2,2], quantity = [2,2]
+Output: true
+Explanation: The 0th customer is given [1,1], and the 1st customer is given [2,2].
+```
+
+**Example 4**:
+
+```
+Input: nums = [1,1,2,3], quantity = [2,2]
+Output: false
+Explanation: Although the 0th customer could be given [1,1], the 1st customer cannot be satisfied.
+```
+
+**Example 5**:
+
+```
+Input: nums = [1,1,1,1,1], quantity = [2,3]
+Output: true
+Explanation: The 0th customer is given [1,1], and the 1st customer is given [1,1,1].
+```
+
+**Constraints**:
+
+- `n == nums.length`
+- `1 <= n <= 105`
+- `1 <= nums[i] <= 1000`
+- `m == quantity.length`
+- `1 <= m <= 10`
+- `1 <= quantity[i] <= 105`
+- There are at most `50` unique values in `nums`.
+
+## 题目大意
+
+给你一个长度为 n 的整数数组 nums ，这个数组中至多有 50 个不同的值。同时你有 m 个顾客的订单 quantity ，其中，整数 quantity[i] 是第 i 位顾客订单的数目。请你判断是否能将 nums 中的整数分配给这些顾客，且满足：
+
+- 第 i 位顾客 恰好 有 quantity[i] 个整数。
+- 第 i 位顾客拿到的整数都是 相同的 。
+- 每位顾客都满足上述两个要求。
+
+如果你可以分配 nums 中的整数满足上面的要求，那么请返回 true ，否则返回 false 。
+
+## 解题思路
+
+- 给定一个数组 nums，订单数组 quantity，要求按照订单满足顾客的需求。如果能满足输出 true，不能满足输出 false。
+- 用 DFS 记忆化暴力搜索。代码实现不难。（不知道此题为什么是 Hard）
+
+## 代码
+
+```go
+package leetcode
+
+func canDistribute(nums []int, quantity []int) bool {
+	freq := make(map[int]int)
+	for _, n := range nums {
+		freq[n]++
+	}
+	return dfs(freq, quantity)
+}
+
+func dfs(freq map[int]int, quantity []int) bool {
+	if len(quantity) == 0 {
+		return true
+	}
+	visited := make(map[int]bool)
+	for i := range freq {
+		if visited[freq[i]] {
+			continue
+		}
+		visited[freq[i]] = true
+		if freq[i] >= quantity[0] {
+			freq[i] -= quantity[0]
+			if dfs(freq, quantity[1:]) {
+				return true
+			}
+			freq[i] += quantity[0]
+		}
+	}
+	return false
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1654.Minimum-Jumps-to-Reach-Home/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1656.Design-an-Ordered-Stream/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1656.Design-an-Ordered-Stream.md b/website/content/ChapterFour/1600~1699/1656.Design-an-Ordered-Stream.md
new file mode 100644
index 000000000..d71767f07
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1656.Design-an-Ordered-Stream.md
@@ -0,0 +1,113 @@
+# [1656. Design an Ordered Stream](https://leetcode.com/problems/design-an-ordered-stream/)
+
+## 题目
+
+There is a stream of `n` `(id, value)` pairs arriving in an **arbitrary** order, where `id` is an integer between `1` and `n` and `value` is a string. No two pairs have the same `id`.
+
+Design a stream that returns the values in **increasing order of their IDs** by returning a **chunk** (list) of values after each insertion. The concatenation of all the **chunks** should result in a list of the sorted values.
+
+Implement the `OrderedStream` class:
+
+- `OrderedStream(int n)` Constructs the stream to take `n` values.
+- `String[] insert(int id, String value)` Inserts the pair `(id, value)` into the stream, then returns the **largest possible chunk** of currently inserted values that appear next in the order.
+
+**Example**:
+
+![https://assets.leetcode.com/uploads/2020/11/10/q1.gif](https://assets.leetcode.com/uploads/2020/11/10/q1.gif)
+
+```
+Input
+["OrderedStream", "insert", "insert", "insert", "insert", "insert"]
+[[5], [3, "ccccc"], [1, "aaaaa"], [2, "bbbbb"], [5, "eeeee"], [4, "ddddd"]]
+Output
+[null, [], ["aaaaa"], ["bbbbb", "ccccc"], [], ["ddddd", "eeeee"]]
+
+Explanation
+// Note that the values ordered by ID is ["aaaaa", "bbbbb", "ccccc", "ddddd", "eeeee"].
+OrderedStream os = new OrderedStream(5);
+os.insert(3, "ccccc"); // Inserts (3, "ccccc"), returns [].
+os.insert(1, "aaaaa"); // Inserts (1, "aaaaa"), returns ["aaaaa"].
+os.insert(2, "bbbbb"); // Inserts (2, "bbbbb"), returns ["bbbbb", "ccccc"].
+os.insert(5, "eeeee"); // Inserts (5, "eeeee"), returns [].
+os.insert(4, "ddddd"); // Inserts (4, "ddddd"), returns ["ddddd", "eeeee"].
+// Concatentating all the chunks returned:
+// [] + ["aaaaa"] + ["bbbbb", "ccccc"] + [] + ["ddddd", "eeeee"] = ["aaaaa", "bbbbb", "ccccc", "ddddd", "eeeee"]
+// The resulting order is the same as the order above.
+
+```
+
+**Constraints**:
+
+- `1 <= n <= 1000`
+- `1 <= id <= n`
+- `value.length == 5`
+- `value` consists only of lowercase letters.
+- Each call to `insert` will have a unique `id.`
+- Exactly `n` calls will be made to `insert`.
+
+## 题目大意
+
+有 n 个 (id, value) 对，其中 id 是 1 到 n 之间的一个整数，value 是一个字符串。不存在 id 相同的两个 (id, value) 对。
+
+设计一个流，以 任意 顺序获取 n 个 (id, value) 对，并在多次调用时 按 id 递增的顺序 返回一些值。
+
+实现 OrderedStream 类：
+
+- OrderedStream(int n) 构造一个能接收 n 个值的流，并将当前指针 ptr 设为 1 。
+- String[] insert(int id, String value) 向流中存储新的 (id, value) 对。存储后：
+如果流存储有 id = ptr 的 (id, value) 对，则找出从 id = ptr 开始的 最长 id 连续递增序列 ，并 按顺序 返回与这些 id 关联的值的列表。然后，将 ptr 更新为最后那个 id + 1 。
+否则，返回一个空列表。
+
+## 解题思路
+
+- 设计一个具有插入操作的 Ordered Stream。insert 操作先在指定位置插入 value，然后返回当前指针 ptr 到最近一个空位置的最长连续递增字符串。如果字符串不为空，ptr 移动到非空 value 的后一个下标位置处。
+- 简单题。按照题目描述模拟即可。注意控制好 ptr 的位置。
+
+## 代码
+
+```go
+package leetcode
+
+type OrderedStream struct {
+	ptr    int
+	stream []string
+}
+
+func Constructor(n int) OrderedStream {
+	ptr, stream := 1, make([]string, n+1)
+	return OrderedStream{ptr: ptr, stream: stream}
+}
+
+func (this *OrderedStream) Insert(id int, value string) []string {
+	this.stream[id] = value
+	res := []string{}
+	if this.ptr == id || this.stream[this.ptr] != "" {
+		res = append(res, this.stream[this.ptr])
+		for i := id + 1; i < len(this.stream); i++ {
+			if this.stream[i] != "" {
+				res = append(res, this.stream[i])
+			} else {
+				this.ptr = i
+				return res
+			}
+		}
+	}
+	if len(res) > 0 {
+		return res
+	}
+	return []string{}
+}
+
+/**
+ * Your OrderedStream object will be instantiated and called as such:
+ * obj := Constructor(n);
+ * param_1 := obj.Insert(id,value);
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1655.Distribute-Repeating-Integers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1657.Determine-if-Two-Strings-Are-Close/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1657.Determine-if-Two-Strings-Are-Close.md b/website/content/ChapterFour/1600~1699/1657.Determine-if-Two-Strings-Are-Close.md
new file mode 100644
index 000000000..c36f7fa20
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1657.Determine-if-Two-Strings-Are-Close.md
@@ -0,0 +1,121 @@
+# [1657. Determine if Two Strings Are Close](https://leetcode.com/problems/determine-if-two-strings-are-close/)
+
+
+## 题目
+
+Two strings are considered **close** if you can attain one from the other using the following operations:
+
+- Operation 1: Swap any two **existing** characters.
+    - For example, `abcde -> aecdb`
+- Operation 2: Transform **every** occurrence of one **existing** character into another **existing** character, and do the same with the other character.
+    - For example, `aacabb -> bbcbaa` (all `a`'s turn into `b`'s, and all `b`'s turn into `a`'s)
+
+You can use the operations on either string as many times as necessary.
+
+Given two strings, `word1` and `word2`, return `true` *if* `word1` *and* `word2` *are **close**, and* `false` *otherwise.*
+
+**Example 1**:
+
+```
+Input: word1 = "abc", word2 = "bca"
+Output: true
+Explanation: You can attain word2 from word1 in 2 operations.
+Apply Operation 1: "abc" -> "acb"
+Apply Operation 1: "acb" -> "bca"
+
+```
+
+**Example 2**:
+
+```
+Input: word1 = "a", word2 = "aa"
+Output: false
+Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.
+
+```
+
+**Example 3**:
+
+```
+Input: word1 = "cabbba", word2 = "abbccc"
+Output: true
+Explanation: You can attain word2 from word1 in 3 operations.
+Apply Operation 1: "cabbba" -> "caabbb"
+Apply Operation 2: "caabbb" -> "baaccc"
+Apply Operation 2: "baaccc" -> "abbccc"
+
+```
+
+**Example 4**:
+
+```
+Input: word1 = "cabbba", word2 = "aabbss"
+Output: false
+Explanation: It is impossible to attain word2 from word1, or vice versa, in any amount of operations.
+
+```
+
+**Constraints**:
+
+- `1 <= word1.length, word2.length <= 105`
+- `word1` and `word2` contain only lowercase English letters.
+
+## 题目大意
+
+如果可以使用以下操作从一个字符串得到另一个字符串，则认为两个字符串 接近 ：
+
+- 操作 1：交换任意两个 现有 字符。例如，abcde -> aecdb
+- 操作 2：将一个 现有 字符的每次出现转换为另一个 现有 字符，并对另一个字符执行相同的操作。例如，aacabb -> bbcbaa（所有 a 转化为 b ，而所有的 b 转换为 a ）
+
+你可以根据需要对任意一个字符串多次使用这两种操作。给你两个字符串，word1 和 word2 。如果 word1 和 word2 接近 ，就返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 判断 2 个字符串是否“接近”。“接近”的定义是能否通过交换 2 个字符或者 2 个字母互换，从一个字符串变换成另外一个字符串，如果存在这样的变换，即是“接近”。
+- 先统计 2 个字符串的 26 个字母的频次，如果频次有不相同的，直接返回 false。在频次相同的情况下，再从小到大排序，再次扫描判断频次是否相同。
+- 注意几种特殊情况：频次相同，再判断字母交换是否合法存在，如果字母不存在，输出 false。例如测试文件中的 case 5 。出现频次个数相同，但是频次不同。例如测试文件中的 case 6 。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func closeStrings(word1 string, word2 string) bool {
+	if len(word1) != len(word2) {
+		return false
+	}
+	freqCount1, freqCount2 := make([]int, 26), make([]int, 26)
+	for _, c := range word1 {
+		freqCount1[c-97]++
+	}
+	for _, c := range word2 {
+		freqCount2[c-97]++
+	}
+	for i := 0; i < 26; i++ {
+		if (freqCount1[i] == freqCount2[i]) ||
+			(freqCount1[i] > 0 && freqCount2[i] > 0) {
+			continue
+		}
+		return false
+	}
+	sort.Ints(freqCount1)
+	sort.Ints(freqCount2)
+	for i := 0; i < 26; i++ {
+		if freqCount1[i] != freqCount2[i] {
+			return false
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1656.Design-an-Ordered-Stream/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1658.Minimum-Operations-to-Reduce-X-to-Zero/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1658.Minimum-Operations-to-Reduce-X-to-Zero.md b/website/content/ChapterFour/1600~1699/1658.Minimum-Operations-to-Reduce-X-to-Zero.md
new file mode 100644
index 000000000..021824d22
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1658.Minimum-Operations-to-Reduce-X-to-Zero.md
@@ -0,0 +1,103 @@
+# [1658. Minimum Operations to Reduce X to Zero](https://leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/)
+
+
+## 题目
+
+You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
+
+Return *the **minimum number** of operations to reduce* `x` *to **exactly*** `0` *if it's possible, otherwise, return* `1`.
+
+**Example 1**:
+
+```
+Input: nums = [1,1,4,2,3], x = 5
+Output: 2
+Explanation: The optimal solution is to remove the last two elements to reduce x to zero.
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [5,6,7,8,9], x = 4
+Output: -1
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [3,2,20,1,1,3], x = 10
+Output: 5
+Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.
+
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 105`
+- `1 <= nums[i] <= 104`
+- `1 <= x <= 109`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个整数 x 。每一次操作时，你应当移除数组 nums 最左边或最右边的元素，然后从 x 中减去该元素的值。请注意，需要 修改 数组以供接下来的操作使用。如果可以将 x 恰好 减到 0 ，返回 最小操作数 ；否则，返回 -1 。
+
+## 解题思路
+
+- 给定一个数组 nums 和一个整数 x，要求从数组两端分别移除一些数，使得这些数加起来正好等于整数 x，要求输出最小操作数。
+- 要求输出最小操作数，即数组两头的数字个数最少，并且加起来和正好等于整数 x。由于在数组的两头，用 2 个指针分别操作不太方便。我当时解题的时候的思路是把它变成循环数组，这样两边的指针就在一个区间内了。利用滑动窗口找到一个最小的窗口，使得窗口内的累加和等于整数 k。这个方法可行，但是代码挺多的。
+- 有没有更优美的方法呢？有的。要想两头的长度最少，也就是中间这段的长度最大。这样就转换成直接在数组上使用滑动窗口求解，累加和等于一个固定值的连续最长的子数组。
+- 和这道题类似思路的题目，209，1040(循环数组)，325。强烈推荐这 3 题。
+
+## 代码
+
+```go
+package leetcode
+
+func minOperations(nums []int, x int) int {
+	total := 0
+	for _, n := range nums {
+		total += n
+	}
+	target := total - x
+	if target < 0 {
+		return -1
+	}
+	if target == 0 {
+		return len(nums)
+	}
+	left, right, sum, res := 0, 0, 0, -1
+	for right < len(nums) {
+		if sum < target {
+			sum += nums[right]
+			right++
+		}
+		for sum >= target {
+			if sum == target {
+				res = max(res, right-left)
+			}
+			sum -= nums[left]
+			left++
+		}
+	}
+	if res == -1 {
+		return -1
+	}
+	return len(nums) - res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1657.Determine-if-Two-Strings-Are-Close/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1659.Maximize-Grid-Happiness/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1659.Maximize-Grid-Happiness.md b/website/content/ChapterFour/1600~1699/1659.Maximize-Grid-Happiness.md
new file mode 100644
index 000000000..049e5fcf1
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1659.Maximize-Grid-Happiness.md
@@ -0,0 +1,192 @@
+# [1659. Maximize Grid Happiness](https://leetcode.com/problems/maximize-grid-happiness/)
+
+## 题目
+
+You are given four integers, `m`, `n`, `introvertsCount`, and `extrovertsCount`. You have an `m x n` grid, and there are two types of people: introverts and extroverts. There are `introvertsCount` introverts and `extrovertsCount` extroverts.
+
+You should decide how many people you want to live in the grid and assign each of them one grid cell. Note that you **do not** have to have all the people living in the grid.
+
+The **happiness** of each person is calculated as follows:
+
+- Introverts **start** with `120` happiness and **lose** `30` happiness for each neighbor (introvert or extrovert).
+- Extroverts **start** with `40` happiness and **gain** `20` happiness for each neighbor (introvert or extrovert).
+
+Neighbors live in the directly adjacent cells north, east, south, and west of a person's cell.
+
+The **grid happiness** is the **sum** of each person's happiness. Return *the **maximum possible grid happiness**.*
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2020/11/05/grid_happiness.png](https://assets.leetcode.com/uploads/2020/11/05/grid_happiness.png)
+
+```
+Input: m = 2, n = 3, introvertsCount = 1, extrovertsCount = 2
+Output: 240
+Explanation: Assume the grid is 1-indexed with coordinates (row, column).
+We can put the introvert in cell (1,1) and put the extroverts in cells (1,3) and (2,3).
+- Introvert at (1,1) happiness: 120 (starting happiness) - (0 * 30) (0 neighbors) = 120
+- Extrovert at (1,3) happiness: 40 (starting happiness) + (1 * 20) (1 neighbor) = 60
+- Extrovert at (2,3) happiness: 40 (starting happiness) + (1 * 20) (1 neighbor) = 60
+The grid happiness is 120 + 60 + 60 = 240.
+The above figure shows the grid in this example with each person's happiness. The introvert stays in the light green cell while the extroverts live on the light purple cells.
+```
+
+**Example 2**:
+
+```
+Input: m = 3, n = 1, introvertsCount = 2, extrovertsCount = 1
+Output: 260
+Explanation: Place the two introverts in (1,1) and (3,1) and the extrovert at (2,1).
+- Introvert at (1,1) happiness: 120 (starting happiness) - (1 * 30) (1 neighbor) = 90
+- Extrovert at (2,1) happiness: 40 (starting happiness) + (2 * 20) (2 neighbors) = 80
+- Introvert at (3,1) happiness: 120 (starting happiness) - (1 * 30) (1 neighbor) = 90
+The grid happiness is 90 + 80 + 90 = 260.
+```
+
+**Example 3**:
+
+```
+Input: m = 2, n = 2, introvertsCount = 4, extrovertsCount = 0
+Output: 240
+```
+
+**Constraints**:
+
+- `1 <= m, n <= 5`
+- `0 <= introvertsCount, extrovertsCount <= min(m * n, 6)`
+
+## 题目大意
+
+给你四个整数 m、n、introvertsCount 和 extrovertsCount 。有一个 m x n 网格，和两种类型的人：内向的人和外向的人。总共有 introvertsCount 个内向的人和 extrovertsCount 个外向的人。请你决定网格中应当居住多少人，并为每个人分配一个网格单元。 注意，不必 让所有人都生活在网格中。每个人的 幸福感 计算如下：
+
+- 内向的人 开始 时有 120 个幸福感，但每存在一个邻居（内向的或外向的）他都会 失去 30 个幸福感。
+- 外向的人 开始 时有 40 个幸福感，每存在一个邻居（内向的或外向的）他都会 得到 20 个幸福感。
+
+邻居是指居住在一个人所在单元的上、下、左、右四个直接相邻的单元中的其他人。网格幸福感 是每个人幸福感的 总和 。 返回 最大可能的网格幸福感 。
+
+## 解题思路
+
+- 给出 `m` x `n` 网格和两种人，要求如何安排这两种人能使得网格的得分最大。两种人有各自的初始分，相邻可能会加分也有可能减分。
+- 这一题状态很多。首先每个格子有 3 种状态，那么每一行有 3^6 = 729 种不同的状态。每行行内分数变化值可能是 -60(两个内向)，+40(两个外向)，-10(一个内向一个外向)。两行行间分数变化值可能是 -60(两个内向)，+40(两个外向)，-10(一个内向一个外向)。那么我们可以把每行的状态压缩成一个三进制，那么网格就变成了一维，每两个三进制之间的关系是行间关系，每个三进制内部还需要根据内向和外向的人数决定行内最终分数。定义 `dp[lineStatusLast][row][introvertsCount][extrovertsCount]` 代表在上一行 `row - 1` 的状态是 `lineStatusLast` 的情况下，当前枚举到了第 `row` 行，内向还有 `introvertsCount` 个人，外向还有 `extrovertsCount` 个人能获得的最大分数。状态转移方程是 `dp[lineStatusLast(row-1)][row][introvertsCount][extrovertsCount] = max{dp[lineStatusLast(row)][row+1][introvertsCount - countIC(lineStatusLast(row)) ][extrovertsCount - countEC(lineStatusLast(row)) ] + scoreInner(lineStatusLast(row)) + scoreOuter(lineStatusLast(row-1),lineStatusLast(row))}` ，这里有 2 个统计函数，`countIC` 是统计当前行状态三进制里面有多少个内向人。`countEC` 是统计当前行状态三进制里面有多少个外向人。`scoreInner` 是计算当前行状态三进制的行内分数。`scoreOuter` 是计算 `row -1` 行和 `row` 行之间的行间分数。
+- 由于这个状态转移方程的计算量是巨大的。所以需要预先初始化一些计算结果。比如把 729 中行状态分别对应的行内、行间的分数都计算好，在动态规划状态转移的时候，直接查表获取分数即可。这样我们在深搜的时候，利用 dp 的记忆化，可以大幅减少时间复杂度。
+- 题目中还提到，人数可以不用完。如果 `introvertsCount = 0`, `extrovertsCount = 0` ，即人数都用完了的情况，这时候 `dp = 0`。如果 `row = m`，即已经枚举完了所有行，那么不管剩下多少人，这一行的 `dp = 0` 。
+- 初始化的时候，注意，特殊处理 0 的情况，0 行 0 列都初始化为 -1 。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+)
+
+func getMaxGridHappiness(m int, n int, introvertsCount int, extrovertsCount int) int {
+	// lineStatus 					将每一行中 3 种状态进行编码，空白 - 0，内向人 - 1，外向人 - 2，每行状态用三进制表示
+	// lineStatusList[729][6] 		每一行的三进制表示
+	// introvertsCountInner[729] 	每一个 lineStatus 包含的内向人数
+	// extrovertsCountInner[729] 	每一个 lineStatus 包含的外向人数
+	// scoreInner[729] 			 	每一个 lineStatus 包含的行内得分（只统计 lineStatus 本身的得分，不包括它与上一行的）
+	// scoreOuter[729][729] 	 	每一个 lineStatus 包含的行外得分
+	// dp[上一行的 lineStatus][当前处理到的行][剩余的内向人数][剩余的外向人数]
+	n3, lineStatus, introvertsCountInner, extrovertsCountInner, scoreInner, scoreOuter, lineStatusList, dp := math.Pow(3.0, float64(n)), 0, [729]int{}, [729]int{}, [729]int{}, [729][729]int{}, [729][6]int{}, [729][6][7][7]int{}
+	for i := 0; i < 729; i++ {
+		lineStatusList[i] = [6]int{}
+	}
+	for i := 0; i < 729; i++ {
+		dp[i] = [6][7][7]int{}
+		for j := 0; j < 6; j++ {
+			dp[i][j] = [7][7]int{}
+			for k := 0; k < 7; k++ {
+				dp[i][j][k] = [7]int{-1, -1, -1, -1, -1, -1, -1}
+			}
+		}
+	}
+	// 预处理
+	for lineStatus = 0; lineStatus < int(n3); lineStatus++ {
+		tmp := lineStatus
+		for i := 0; i < n; i++ {
+			lineStatusList[lineStatus][i] = tmp % 3
+			tmp /= 3
+		}
+		introvertsCountInner[lineStatus], extrovertsCountInner[lineStatus], scoreInner[lineStatus] = 0, 0, 0
+		for i := 0; i < n; i++ {
+			if lineStatusList[lineStatus][i] != 0 {
+				// 个人分数
+				if lineStatusList[lineStatus][i] == 1 {
+					introvertsCountInner[lineStatus]++
+					scoreInner[lineStatus] += 120
+				} else if lineStatusList[lineStatus][i] == 2 {
+					extrovertsCountInner[lineStatus]++
+					scoreInner[lineStatus] += 40
+				}
+				// 行内分数
+				if i-1 >= 0 {
+					scoreInner[lineStatus] += closeScore(lineStatusList[lineStatus][i], lineStatusList[lineStatus][i-1])
+				}
+			}
+		}
+	}
+	// 行外分数
+	for lineStatus0 := 0; lineStatus0 < int(n3); lineStatus0++ {
+		for lineStatus1 := 0; lineStatus1 < int(n3); lineStatus1++ {
+			scoreOuter[lineStatus0][lineStatus1] = 0
+			for i := 0; i < n; i++ {
+				scoreOuter[lineStatus0][lineStatus1] += closeScore(lineStatusList[lineStatus0][i], lineStatusList[lineStatus1][i])
+			}
+		}
+	}
+	return dfs(0, 0, introvertsCount, extrovertsCount, m, int(n3), &dp, &introvertsCountInner, &extrovertsCountInner, &scoreInner, &scoreOuter)
+}
+
+// 如果 x 和 y 相邻，需要加上的分数
+func closeScore(x, y int) int {
+	if x == 0 || y == 0 {
+		return 0
+	}
+	// 两个内向的人，每个人要 -30，一共 -60
+	if x == 1 && y == 1 {
+		return -60
+	}
+	if x == 2 && y == 2 {
+		return 40
+	}
+	return -10
+}
+
+// dfs(上一行的 lineStatus，当前处理到的行，剩余的内向人数，剩余的外向人数）
+func dfs(lineStatusLast, row, introvertsCount, extrovertsCount, m, n3 int, dp *[729][6][7][7]int, introvertsCountInner, extrovertsCountInner, scoreInner *[729]int, scoreOuter *[729][729]int) int {
+	// 边界条件：如果已经处理完，或者没有人了
+	if row == m || introvertsCount+extrovertsCount == 0 {
+		return 0
+	}
+	// 记忆化
+	if dp[lineStatusLast][row][introvertsCount][extrovertsCount] != -1 {
+		return dp[lineStatusLast][row][introvertsCount][extrovertsCount]
+	}
+	best := 0
+	for lineStatus := 0; lineStatus < n3; lineStatus++ {
+		if introvertsCountInner[lineStatus] > introvertsCount || extrovertsCountInner[lineStatus] > extrovertsCount {
+			continue
+		}
+		score := scoreInner[lineStatus] + scoreOuter[lineStatus][lineStatusLast]
+		best = max(best, score+dfs(lineStatus, row+1, introvertsCount-introvertsCountInner[lineStatus], extrovertsCount-extrovertsCountInner[lineStatus], m, n3, dp, introvertsCountInner, extrovertsCountInner, scoreInner, scoreOuter))
+	}
+	dp[lineStatusLast][row][introvertsCount][extrovertsCount] = best
+	return best
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1658.Minimum-Operations-to-Reduce-X-to-Zero/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1662.Check-If-Two-String-Arrays-are-Equivalent/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1662.Check-If-Two-String-Arrays-are-Equivalent.md b/website/content/ChapterFour/1600~1699/1662.Check-If-Two-String-Arrays-are-Equivalent.md
new file mode 100644
index 000000000..4b99dc3eb
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1662.Check-If-Two-String-Arrays-are-Equivalent.md
@@ -0,0 +1,72 @@
+# [1662. Check If Two String Arrays are Equivalent](https://leetcode.com/problems/check-if-two-string-arrays-are-equivalent/)
+
+
+## 题目
+
+Given two string arrays `word1` and `word2`, return **`true` *if the two arrays **represent** the same string, and* `false` *otherwise.*
+
+A string is **represented** by an array if the array elements concatenated **in order** forms the string.
+
+**Example 1**:
+
+```
+Input: word1 = ["ab", "c"], word2 = ["a", "bc"]
+Output: true
+Explanation:
+word1 represents string "ab" + "c" -> "abc"
+word2 represents string "a" + "bc" -> "abc"
+The strings are the same, so return true.
+```
+
+**Example 2**:
+
+```
+Input: word1 = ["a", "cb"], word2 = ["ab", "c"]
+Output: false
+```
+
+**Example 3**:
+
+```
+Input: word1  = ["abc", "d", "defg"], word2 = ["abcddefg"]
+Output: true
+```
+
+**Constraints**:
+
+- `1 <= word1.length, word2.length <= 103`
+- `1 <= word1[i].length, word2[i].length <= 103`
+- `1 <= sum(word1[i].length), sum(word2[i].length) <= 103`
+- `word1[i]` and `word2[i]` consist of lowercase letters.
+
+## 题目大意
+
+给你两个字符串数组 word1 和 word2 。如果两个数组表示的字符串相同，返回 true ；否则，返回 false 。数组表示的字符串 是由数组中的所有元素 按顺序 连接形成的字符串。
+
+## 解题思路
+
+- 简单题，依次拼接 2 个数组内的字符串，然后比较 str1 和 str2 是否相同即可。
+
+## 代码
+
+```go
+package leetcode
+
+func arrayStringsAreEqual(word1 []string, word2 []string) bool {
+	str1, str2 := "", ""
+	for i := 0; i < len(word1); i++ {
+		str1 += word1[i]
+	}
+	for i := 0; i < len(word2); i++ {
+		str2 += word2[i]
+	}
+	return str1 == str2
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1659.Maximize-Grid-Happiness/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1663.Smallest-String-With-A-Given-Numeric-Value/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1663.Smallest-String-With-A-Given-Numeric-Value.md b/website/content/ChapterFour/1600~1699/1663.Smallest-String-With-A-Given-Numeric-Value.md
new file mode 100644
index 000000000..b373c410f
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1663.Smallest-String-With-A-Given-Numeric-Value.md
@@ -0,0 +1,109 @@
+# [1663. Smallest String With A Given Numeric Value](https://leetcode.com/problems/smallest-string-with-a-given-numeric-value/)
+
+## 题目
+
+The **numeric value** of a **lowercase character** is defined as its position `(1-indexed)` in the alphabet, so the numeric value of `a` is `1`, the numeric value of `b` is `2`, the numeric value of `c` is `3`, and so on.
+
+The **numeric value** of a **string** consisting of lowercase characters is defined as the sum of its characters' numeric values. For example, the numeric value of the string `"abe"` is equal to `1 + 2 + 5 = 8`.
+
+You are given two integers `n` and `k`. Return *the **lexicographically smallest string** with **length** equal to `n` and **numeric value** equal to `k`.*
+
+Note that a string `x` is lexicographically smaller than string `y` if `x` comes before `y` in dictionary order, that is, either `x` is a prefix of `y`, or if `i` is the first position such that `x[i] != y[i]`, then `x[i]` comes before `y[i]` in alphabetic order.
+
+**Example 1**:
+
+```
+Input: n = 3, k = 27
+Output: "aay"
+Explanation: The numeric value of the string is 1 + 1 + 25 = 27, and it is the smallest string with such a value and length equal to 3.
+```
+
+**Example 2**:
+
+```
+Input: n = 5, k = 73
+Output: "aaszz"
+```
+
+**Constraints**:
+
+- `1 <= n <= 105`
+- `n <= k <= 26 * n`
+
+## 题目大意
+
+小写字符 的 数值 是它在字母表中的位置（从 1 开始），因此 a 的数值为 1 ，b 的数值为 2 ，c 的数值为 3 ，以此类推。字符串由若干小写字符组成，字符串的数值 为各字符的数值之和。例如，字符串 "abe" 的数值等于 1 + 2 + 5 = 8 。给你两个整数 n 和 k 。返回 长度 等于 n 且 数值 等于 k 的 字典序最小 的字符串。注意，如果字符串 x 在字典排序中位于 y 之前，就认为 x 字典序比 y 小，有以下两种情况：
+
+- x 是 y 的一个前缀；
+- 如果 i 是 x[i] != y[i] 的第一个位置，且 x[i] 在字母表中的位置比 y[i] 靠前。
+
+## 解题思路
+
+- 给出 n 和 k，要求找到字符串长度为 n，字母在字母表内位置总和为 k 的最小字典序字符串。
+- 这一题笔者读完题，比赛的时候直接用 DFS 撸了一版。赛后看了时间复杂度马马虎虎，感觉还有优化的空间。DFS 会遍历出所有的解，实际上这一题只要求最小字典序，所以 DFS 剪枝的时候要加上判断字典序的判断，如果新添加进来的字母比已经保存的字符串的相应位置上的字母字典序大，那么就直接 return，这个答案一定不会是最小字典序。代码见解法二
+- 想到这里，其实 DFS 不必要，直接用 for 循环就可找到最小字典序的字符串。代码见解法一。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 贪心
+func getSmallestString(n int, k int) string {
+    str, i, j := make([]byte, n), 0, 0
+    for i = n-1; i <= k-26; i, k = i-1, k-26 {
+        str[i] = 'z'
+    }
+    if i >= 0 {
+        str[i] = byte('a' + k-1-i)
+        for ; j < i; j++ {
+            str[j] = 'a'
+        }
+    }
+    return string(str)
+}
+
+// 解法二 DFS
+func getSmallestString1(n int, k int) string {
+	if n == 0 {
+		return ""
+	}
+	res, c := "", []byte{}
+	findSmallestString(0, n, k, 0, c, &res)
+	return res
+}
+
+func findSmallestString(value int, length, k, index int, str []byte, res *string) {
+	if len(str) == length && value == k {
+		tmp := string(str)
+		if (*res) == "" {
+			*res = tmp
+		}
+		if tmp < *res && *res != "" {
+			*res = tmp
+		}
+		return
+	}
+	if len(str) >= index && (*res) != "" && str[index-1] > (*res)[index-1] {
+		return
+	}
+	for j := 0; j < 26; j++ {
+		if k-value > (length-len(str))*26 || value > k {
+			return
+		}
+		str = append(str, byte(int('a')+j))
+		value += j + 1
+		findSmallestString(value, length, k, index+1, str, res)
+		str = str[:len(str)-1]
+		value -= j + 1
+
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1662.Check-If-Two-String-Arrays-are-Equivalent/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1664.Ways-to-Make-a-Fair-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1664.Ways-to-Make-a-Fair-Array.md b/website/content/ChapterFour/1600~1699/1664.Ways-to-Make-a-Fair-Array.md
new file mode 100644
index 000000000..1c6076177
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1664.Ways-to-Make-a-Fair-Array.md
@@ -0,0 +1,123 @@
+# [1664. Ways to Make a Fair Array](https://leetcode.com/problems/ways-to-make-a-fair-array/)
+
+
+## 题目
+
+You are given an integer array `nums`. You can choose **exactly one** index (**0-indexed**) and remove the element. Notice that the index of the elements may change after the removal.
+
+For example, if `nums = [6,1,7,4,1]`:
+
+- Choosing to remove index `1` results in `nums = [6,7,4,1]`.
+- Choosing to remove index `2` results in `nums = [6,1,4,1]`.
+- Choosing to remove index `4` results in `nums = [6,1,7,4]`.
+
+An array is **fair** if the sum of the odd-indexed values equals the sum of the even-indexed values.
+
+Return the ***number** of indices that you could choose such that after the removal,* `nums` *is **fair**.*
+
+**Example 1**:
+
+```
+Input: nums = [2,1,6,4]
+Output: 1
+Explanation:
+Remove index 0: [1,6,4] -> Even sum: 1 + 4 = 5. Odd sum: 6. Not fair.
+Remove index 1: [2,6,4] -> Even sum: 2 + 4 = 6. Odd sum: 6. Fair.
+Remove index 2: [2,1,4] -> Even sum: 2 + 4 = 6. Odd sum: 1. Not fair.
+Remove index 3: [2,1,6] -> Even sum: 2 + 6 = 8. Odd sum: 1. Not fair.
+There is 1 index that you can remove to make nums fair.
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,1,1]
+Output: 3
+Explanation: You can remove any index and the remaining array is fair.
+```
+
+**Example 3**:
+
+```
+Input: nums = [1,2,3]
+Output: 0
+Explanation: You cannot make a fair array after removing any index.
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 105`
+- `1 <= nums[i] <= 104`
+
+## 题目大意
+
+给你一个整数数组 nums 。你需要选择 恰好 一个下标（下标从 0 开始）并删除对应的元素。请注意剩下元素的下标可能会因为删除操作而发生改变。
+
+比方说，如果 nums = [6,1,7,4,1] ，那么：
+
+- 选择删除下标 1 ，剩下的数组为 nums = [6,7,4,1] 。
+- 选择删除下标 2 ，剩下的数组为 nums = [6,1,4,1] 。
+- 选择删除下标 4 ，剩下的数组为 nums = [6,1,7,4] 。
+
+如果一个数组满足奇数下标元素的和与偶数下标元素的和相等，该数组就是一个 平衡数组 。请你返回删除操作后，剩下的数组 nums 是 平衡数组 的 方案数 。
+
+## 解题思路
+
+- 给定一个数组 nums，要求输出仅删除一个元素以后能使得整个数组平衡的方案数。平衡的定义是奇数下标元素总和等于偶数下标元素总和。
+- 这一题如果暴力解答，会超时。原因是每次删除元素以后，都重新计算奇偶数位总和比较耗时。应该利用前面计算过的累加和，推导出此次删除元素以后的情况。这样修改以后就不超时了。具体的，如果删除的是元素是奇数位，这个下标的前缀和不变，要变化的是后面的。删除元素后面，原来偶数位的总和变成了奇数位了，原来奇数位的总和变成偶数位了。删除元素后面这半段的总和可以用前缀和计算出来，奇数位的总和减去删除元素的前缀和，就得到了删除元素后面的后缀和。通过这个办法就可以得到删除元素后面的，奇数位总和，偶数位总和。注意这个后缀和是包含了删除元素的。所以最后需要判断删除元素是奇数位还是偶数位，如果是奇数位，那么在计算出来的偶数和上再减去这个删除元素；如果是偶数位，就在计算出来的奇数和上再减去这个删除元素。代码见解法二。
+- 这一题还有一种更简洁的写法，就是解法一了。通过了解法二的思考，我们可以知道，每次变换以后的操作可以抽象出来，即三步，减去一个数，判断是否相等，再加上一个数。只不过这三步在解法二中都去判断了奇偶性。如果我们不判断奇偶性，那么代码就可以写成解法一的样子。为什么可以不用管奇偶性呢？因为每次删除一个元素以后，下次再删除，奇偶就发生颠倒了，上次的奇数和到了下次就是偶数和了。想通这一点就可以把代码写成解法一的样子。
+
+## 代码
+
+```go
+// 解法一 超简洁写法
+func waysToMakeFair(nums []int) int {
+	sum, res := [2]int{}, 0
+	for i := 0; i < len(nums); i++ {
+		sum[i%2] += nums[i]
+	}
+	for i := 0; i < len(nums); i++ {
+		sum[i%2] -= nums[i]
+		if sum[i%2] == sum[1-(i%2)] {
+			res++
+		}
+		sum[1-(i%2)] += nums[i]
+	}
+	return res
+}
+
+// 解法二 前缀和，后缀和
+func waysToMakeFair1(nums []int) int {
+	evenPrefix, oddPrefix, evenSuffix, oddSuffix, res := 0, 0, 0, 0, 0
+	for i := 0; i < len(nums); i++ {
+		if i%2 == 0 {
+			evenSuffix += nums[i]
+		} else {
+			oddSuffix += nums[i]
+		}
+	}
+	for i := 0; i < len(nums); i++ {
+		if i%2 == 0 {
+			evenSuffix -= nums[i]
+		} else {
+			oddSuffix -= nums[i]
+		}
+		if (evenPrefix + oddSuffix) == (oddPrefix + evenSuffix) {
+			res++
+		}
+		if i%2 == 0 {
+			evenPrefix += nums[i]
+		} else {
+			oddPrefix += nums[i]
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1663.Smallest-String-With-A-Given-Numeric-Value/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1665.Minimum-Initial-Energy-to-Finish-Tasks/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1665.Minimum-Initial-Energy-to-Finish-Tasks.md b/website/content/ChapterFour/1600~1699/1665.Minimum-Initial-Energy-to-Finish-Tasks.md
new file mode 100644
index 000000000..e201a20d5
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1665.Minimum-Initial-Energy-to-Finish-Tasks.md
@@ -0,0 +1,135 @@
+# [1665. Minimum Initial Energy to Finish Tasks](https://leetcode.com/problems/minimum-initial-energy-to-finish-tasks/)
+
+## 题目
+
+You are given an array `tasks` where `tasks[i] = [actuali, minimumi]`:
+
+- `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
+- `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
+
+For example, if the task is `[10, 12]` and your current energy is `11`, you cannot start this task. However, if your current energy is `13`, you can complete this task, and your energy will be `3` after finishing it.
+
+You can finish the tasks in **any order** you like.
+
+Return *the **minimum** initial amount of energy you will need* *to finish all the tasks*.
+
+**Example 1**:
+
+```
+Input: tasks = [[1,2],[2,4],[4,8]]
+Output: 8
+Explanation:
+Starting with 8 energy, we finish the tasks in the following order:
+    - 3rd task. Now energy = 8 - 4 = 4.
+    - 2nd task. Now energy = 4 - 2 = 2.
+    - 1st task. Now energy = 2 - 1 = 1.
+Notice that even though we have leftover energy, starting with 7 energy does not work because we cannot do the 3rd task.
+```
+
+**Example 2**:
+
+```
+Input: tasks = [[1,3],[2,4],[10,11],[10,12],[8,9]]
+Output: 32
+Explanation:
+Starting with 32 energy, we finish the tasks in the following order:
+    - 1st task. Now energy = 32 - 1 = 31.
+    - 2nd task. Now energy = 31 - 2 = 29.
+    - 3rd task. Now energy = 29 - 10 = 19.
+    - 4th task. Now energy = 19 - 10 = 9.
+    - 5th task. Now energy = 9 - 8 = 1.
+```
+
+**Example 3**:
+
+```
+Input: tasks = [[1,7],[2,8],[3,9],[4,10],[5,11],[6,12]]
+Output: 27
+Explanation:
+Starting with 27 energy, we finish the tasks in the following order:
+    - 5th task. Now energy = 27 - 5 = 22.
+    - 2nd task. Now energy = 22 - 2 = 20.
+    - 3rd task. Now energy = 20 - 3 = 17.
+    - 1st task. Now energy = 17 - 1 = 16.
+    - 4th task. Now energy = 16 - 4 = 12.
+    - 6th task. Now energy = 12 - 6 = 6.
+
+```
+
+**Constraints**:
+
+- `1 <= tasks.length <= 105`
+- `1 <= actuali <= minimumi <= 104`
+
+## 题目大意
+
+给你一个任务数组 tasks ，其中 tasks[i] = [actuali, minimumi] ：
+
+- actual i 是完成第 i 个任务 需要耗费 的实际能量。
+- minimum i 是开始第 i 个任务前需要达到的最低能量。
+
+比方说，如果任务为 [10, 12] 且你当前的能量为 11 ，那么你不能开始这个任务。如果你当前的能量为 13 ，你可以完成这个任务，且完成它后剩余能量为 3 。你可以按照 任意顺序 完成任务。请你返回完成所有任务的 最少 初始能量。
+
+## 解题思路
+
+- 给出一个 task 数组，每个元素代表一个任务，每个任务有实际消费能量值和开始这个任务需要的最低能量。要求输出能完成所有任务的最少初始能量。
+- 这一题直觉是贪心。先将任务按照 `minimum - actual` 进行排序。先完成差值大的任务，那么接下来的能量能最大限度的满足接下来的任务。这样可能完成所有任务的可能性越大。循环任务数组的时候，保存当前能量在 `cur` 中，如果当前能量不够开启下一个任务，那么这个差值就是需要弥补的，这些能量就是最少初始能量中的，所以加上这些差值能量。如果当前能量可以开启下一个任务，那么就更新当前能量，减去实际消耗的能量以后，再继续循环。循环结束就能得到最少初始能量了。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func minimumEffort(tasks [][]int) int {
+	sort.Sort(Task(tasks))
+	res, cur := 0, 0
+	for _, t := range tasks {
+		if t[1] > cur {
+			res += t[1] - cur
+			cur = t[1] - t[0]
+		} else {
+			cur -= t[0]
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// Task define
+type Task [][]int
+
+func (task Task) Len() int {
+	return len(task)
+}
+
+func (task Task) Less(i, j int) bool {
+	t1, t2 := task[i][1]-task[i][0], task[j][1]-task[j][0]
+	if t1 != t2 {
+		return t2 < t1
+	}
+	return task[j][1] < task[i][1]
+}
+
+func (task Task) Swap(i, j int) {
+	t := task[i]
+	task[i] = task[j]
+	task[j] = t
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1664.Ways-to-Make-a-Fair-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1668.Maximum-Repeating-Substring/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1668.Maximum-Repeating-Substring.md b/website/content/ChapterFour/1600~1699/1668.Maximum-Repeating-Substring.md
new file mode 100644
index 000000000..f86e6bc16
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1668.Maximum-Repeating-Substring.md
@@ -0,0 +1,76 @@
+# [1668. Maximum Repeating Substring](https://leetcode.com/problems/maximum-repeating-substring/)
+
+
+## 题目
+
+For a string `sequence`, a string `word` is **`k`-repeating** if `word` concatenated `k` times is a substring of `sequence`. The `word`'s **maximum `k`-repeating value** is the highest value `k` where `word` is `k`-repeating in `sequence`. If `word` is not a substring of `sequence`, `word`'s maximum `k`-repeating value is `0`.
+
+Given strings `sequence` and `word`, return *the **maximum `k`-repeating value** of `word` in `sequence`*.
+
+**Example 1**:
+
+```
+Input: sequence = "ababc", word = "ab"
+Output: 2
+Explanation: "abab" is a substring in "ababc".
+```
+
+**Example 2**:
+
+```
+Input: sequence = "ababc", word = "ba"
+Output: 1
+Explanation: "ba" is a substring in "ababc". "baba" is not a substring in "ababc".
+```
+
+**Example 3**:
+
+```
+Input: sequence = "ababc", word = "ac"
+Output: 0
+Explanation: "ac" is not a substring in "ababc". 
+```
+
+**Constraints**:
+
+- `1 <= sequence.length <= 100`
+- `1 <= word.length <= 100`
+- `sequence` and `word` contains only lowercase English letters.
+
+## 题目大意
+
+给你一个字符串 sequence ，如果字符串 word 连续重复 k 次形成的字符串是 sequence 的一个子字符串，那么单词 word 的 重复值为 k 。单词 word 的 最大重复值 是单词 word 在 sequence 中最大的重复值。如果 word 不是 sequence 的子串，那么重复值 k 为 0 。给你一个字符串 sequence 和 word ，请你返回 最大重复值 k 。
+
+## 解题思路
+
+- 循环叠加构造 `word`，每次构造出新的 `word` 都在 `sequence` 查找一次，如果找到就输出叠加次数，否则继续叠加构造，直到字符串长度和 `sequence` 一样长，最终都没有找到则输出 0 。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"strings"
+)
+
+func maxRepeating(sequence string, word string) int {
+	for i := len(sequence) / len(word); i >= 0; i-- {
+		tmp := ""
+		for j := 0; j < i; j++ {
+			tmp += word
+		}
+		if strings.Contains(sequence, tmp) {
+			return i
+		}
+	}
+	return 0
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1665.Minimum-Initial-Energy-to-Finish-Tasks/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1669.Merge-In-Between-Linked-Lists/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1669.Merge-In-Between-Linked-Lists.md b/website/content/ChapterFour/1600~1699/1669.Merge-In-Between-Linked-Lists.md
new file mode 100644
index 000000000..05e4afc49
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1669.Merge-In-Between-Linked-Lists.md
@@ -0,0 +1,82 @@
+# [1669. Merge In Between Linked Lists](https://leetcode.com/problems/merge-in-between-linked-lists/)
+
+
+## 题目
+
+You are given two linked lists: `list1` and `list2` of sizes `n` and `m` respectively.
+
+Remove `list1`'s nodes from the `ath` node to the `bth` node, and put `list2` in their place.
+
+The blue edges and nodes in the following figure incidate the result:
+
+![https://assets.leetcode.com/uploads/2020/11/05/fig1.png](https://assets.leetcode.com/uploads/2020/11/05/fig1.png)
+
+*Build the result list and return its head.*
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2020/11/05/merge_linked_list_ex1.png](https://assets.leetcode.com/uploads/2020/11/05/merge_linked_list_ex1.png)
+
+```
+Input: list1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
+Output: [0,1,2,1000000,1000001,1000002,5]
+Explanation: We remove the nodes 3 and 4 and put the entire list2 in their place. The blue edges and nodes in the above figure indicate the result.
+
+```
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2020/11/05/merge_linked_list_ex2.png](https://assets.leetcode.com/uploads/2020/11/05/merge_linked_list_ex2.png)
+
+```
+Input: list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
+Output: [0,1,1000000,1000001,1000002,1000003,1000004,6]
+Explanation: The blue edges and nodes in the above figure indicate the result.
+
+```
+
+**Constraints**:
+
+- `3 <= list1.length <= 104`
+- `1 <= a <= b < list1.length - 1`
+- `1 <= list2.length <= 104`
+
+## 题目大意
+
+给你两个链表 list1 和 list2 ，它们包含的元素分别为 n 个和 m 个。请你将 list1 中第 a 个节点到第 b 个节点删除，并将list2 接在被删除节点的位置。
+
+## 解题思路
+
+- 简单题，考查链表的基本操作。此题注意 a == b 的情况。
+
+## 代码
+
+```go
+func mergeInBetween(list1 *ListNode, a int, b int, list2 *ListNode) *ListNode {
+	n := list1
+	var startRef, endRef *ListNode
+	for i := 0; i <= b; i++ {
+		if i == a-1 {
+			startRef = n
+		}
+		if i == b {
+			endRef = n
+		}
+		n = n.Next
+	}
+	startRef.Next = list2
+	n = list2
+	for n.Next != nil {
+		n = n.Next
+	}
+	n.Next = endRef.Next
+	return list1
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1668.Maximum-Repeating-Substring/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1670.Design-Front-Middle-Back-Queue/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1670.Design-Front-Middle-Back-Queue.md b/website/content/ChapterFour/1600~1699/1670.Design-Front-Middle-Back-Queue.md
new file mode 100644
index 000000000..db98ab366
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1670.Design-Front-Middle-Back-Queue.md
@@ -0,0 +1,179 @@
+# [1670. Design Front Middle Back Queue](https://leetcode.com/problems/design-front-middle-back-queue/)
+
+
+## 题目
+
+Design a queue that supports `push` and `pop` operations in the front, middle, and back.
+
+Implement the `FrontMiddleBack` class:
+
+- `FrontMiddleBack()` Initializes the queue.
+- `void pushFront(int val)` Adds `val` to the **front** of the queue.
+- `void pushMiddle(int val)` Adds `val` to the **middle** of the queue.
+- `void pushBack(int val)` Adds `val` to the **back** of the queue.
+- `int popFront()` Removes the **front** element of the queue and returns it. If the queue is empty, return `1`.
+- `int popMiddle()` Removes the **middle** element of the queue and returns it. If the queue is empty, return `1`.
+- `int popBack()` Removes the **back** element of the queue and returns it. If the queue is empty, return `1`.
+
+**Notice** that when there are **two** middle position choices, the operation is performed on the **frontmost** middle position choice. For example:
+
+- Pushing `6` into the middle of `[1, 2, 3, 4, 5]` results in `[1, 2, 6, 3, 4, 5]`.
+- Popping the middle from `[1, 2, 3, 4, 5, 6]` returns `3` and results in `[1, 2, 4, 5, 6]`.
+
+**Example 1**:
+
+```
+Input:
+["FrontMiddleBackQueue", "pushFront", "pushBack", "pushMiddle", "pushMiddle", "popFront", "popMiddle", "popMiddle", "popBack", "popFront"]
+[[], [1], [2], [3], [4], [], [], [], [], []]
+Output:
+[null, null, null, null, null, 1, 3, 4, 2, -1]
+
+Explanation:
+FrontMiddleBackQueue q = new FrontMiddleBackQueue();
+q.pushFront(1);   // [1]
+q.pushBack(2);    // [1, 2]
+q.pushMiddle(3);  // [1, 3, 2]
+q.pushMiddle(4);  // [1, 4, 3, 2]
+q.popFront();     // return 1 -> [4, 3, 2]
+q.popMiddle();    // return 3 -> [4, 2]
+q.popMiddle();    // return 4 -> [2]
+q.popBack();      // return 2 -> []
+q.popFront();     // return -1 -> [] (The queue is empty)
+
+```
+
+**Constraints**:
+
+- `1 <= val <= 109`
+- At most `1000` calls will be made to `pushFront`, `pushMiddle`, `pushBack`, `popFront`, `popMiddle`, and `popBack`.
+
+## 题目大意
+
+请你设计一个队列，支持在前，中，后三个位置的 push 和 pop 操作。
+
+请你完成 FrontMiddleBack 类：
+
+- FrontMiddleBack() 初始化队列。
+- void pushFront(int val) 将 val 添加到队列的 最前面 。
+- void pushMiddle(int val) 将 val 添加到队列的 正中间 。
+- void pushBack(int val) 将 val 添加到队里的 最后面 。
+- int popFront() 将 最前面 的元素从队列中删除并返回值，如果删除之前队列为空，那么返回 -1 。
+- int popMiddle() 将 正中间 的元素从队列中删除并返回值，如果删除之前队列为空，那么返回 -1 。
+- int popBack() 将 最后面 的元素从队列中删除并返回值，如果删除之前队列为空，那么返回 -1 。
+
+请注意当有 两个 中间位置的时候，选择靠前面的位置进行操作。比方说：
+
+- 将 6 添加到 [1, 2, 3, 4, 5] 的中间位置，结果数组为 [1, 2, 6, 3, 4, 5] 。
+- 从 [1, 2, 3, 4, 5, 6] 的中间位置弹出元素，返回 3 ，数组变为 [1, 2, 4, 5, 6] 。
+
+## 解题思路
+
+- 简单题，利用 go 原生的双向队列 list 的实现，可以轻松实现这个“前中后队列”。
+- 具体实现见代码，几组特殊测试用例见测试文件。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/list"
+)
+
+type FrontMiddleBackQueue struct {
+	list   *list.List
+	middle *list.Element
+}
+
+func Constructor() FrontMiddleBackQueue {
+	return FrontMiddleBackQueue{list: list.New()}
+}
+
+func (this *FrontMiddleBackQueue) PushFront(val int) {
+	e := this.list.PushFront(val)
+	if this.middle == nil {
+		this.middle = e
+	} else if this.list.Len()%2 == 0 && this.middle.Prev() != nil {
+		this.middle = this.middle.Prev()
+	}
+}
+
+func (this *FrontMiddleBackQueue) PushMiddle(val int) {
+	if this.middle == nil {
+		this.PushFront(val)
+	} else {
+		if this.list.Len()%2 != 0 {
+			this.middle = this.list.InsertBefore(val, this.middle)
+		} else {
+			this.middle = this.list.InsertAfter(val, this.middle)
+		}
+	}
+}
+
+func (this *FrontMiddleBackQueue) PushBack(val int) {
+	e := this.list.PushBack(val)
+	if this.middle == nil {
+		this.middle = e
+	} else if this.list.Len()%2 != 0 && this.middle.Next() != nil {
+		this.middle = this.middle.Next()
+	}
+}
+
+func (this *FrontMiddleBackQueue) PopFront() int {
+	if this.list.Len() == 0 {
+		return -1
+	}
+	e := this.list.Front()
+	if this.list.Len() == 1 {
+		this.middle = nil
+	} else if this.list.Len()%2 == 0 && this.middle.Next() != nil {
+		this.middle = this.middle.Next()
+	}
+	return this.list.Remove(e).(int)
+}
+
+func (this *FrontMiddleBackQueue) PopMiddle() int {
+	if this.middle == nil {
+		return -1
+	}
+	e := this.middle
+	if this.list.Len()%2 != 0 {
+		this.middle = e.Prev()
+	} else {
+		this.middle = e.Next()
+	}
+	return this.list.Remove(e).(int)
+}
+
+func (this *FrontMiddleBackQueue) PopBack() int {
+	if this.list.Len() == 0 {
+		return -1
+	}
+	e := this.list.Back()
+	if this.list.Len() == 1 {
+		this.middle = nil
+	} else if this.list.Len()%2 != 0 && this.middle.Prev() != nil {
+		this.middle = this.middle.Prev()
+	}
+	return this.list.Remove(e).(int)
+}
+
+/**
+ * Your FrontMiddleBackQueue object will be instantiated and called as such:
+ * obj := Constructor();
+ * obj.PushFront(val);
+ * obj.PushMiddle(val);
+ * obj.PushBack(val);
+ * param_4 := obj.PopFront();
+ * param_5 := obj.PopMiddle();
+ * param_6 := obj.PopBack();
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1669.Merge-In-Between-Linked-Lists/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1672.Richest-Customer-Wealth/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1672.Richest-Customer-Wealth.md b/website/content/ChapterFour/1600~1699/1672.Richest-Customer-Wealth.md
new file mode 100644
index 000000000..72e287b64
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1672.Richest-Customer-Wealth.md
@@ -0,0 +1,79 @@
+# [1672. Richest Customer Wealth](https://leetcode.com/problems/richest-customer-wealth/)
+
+
+## 题目
+
+You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return *the **wealth** that the richest customer has.*
+
+A customer's **wealth** is the amount of money they have in all their bank accounts. The richest customer is the customer that has the maximum **wealth**.
+
+**Example 1**:
+
+```
+Input: accounts = [[1,2,3],[3,2,1]]
+Output: 6
+Explanation:1st customer has wealth = 1 + 2 + 3 = 6
+2nd customer has wealth = 3 + 2 + 1 = 6
+Both customers are considered the richest with a wealth of 6 each, so return 6.
+```
+
+**Example 2**:
+
+```
+Input: accounts = [[1,5],[7,3],[3,5]]
+Output: 10
+Explanation: 
+1st customer has wealth = 6
+2nd customer has wealth = 10 
+3rd customer has wealth = 8
+The 2nd customer is the richest with a wealth of 10.
+```
+
+**Example 3**:
+
+```
+Input: accounts = [[2,8,7],[7,1,3],[1,9,5]]
+Output: 17
+```
+
+**Constraints**:
+
+- `m == accounts.length`
+- `n == accounts[i].length`
+- `1 <= m, n <= 50`
+- `1 <= accounts[i][j] <= 100`
+
+## 题目大意
+
+给你一个 m x n 的整数网格 accounts ，其中 accounts[i][j] 是第 i 位客户在第 j 家银行托管的资产数量。返回最富有客户所拥有的 资产总量 。客户的 资产总量 就是他们在各家银行托管的资产数量之和。最富有客户就是 资产总量 最大的客户。
+
+## 解题思路
+
+- 简单题。计算二维数组中每个一位数组的元素总和，然后动态维护这些一位数组和的最大值即可。
+
+## 代码
+
+```go
+package leetcode
+
+func maximumWealth(accounts [][]int) int {
+	res := 0
+	for _, banks := range accounts {
+		sAmount := 0
+		for _, amount := range banks {
+			sAmount += amount
+		}
+		if sAmount > res {
+			res = sAmount
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1670.Design-Front-Middle-Back-Queue/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1673.Find-the-Most-Competitive-Subsequence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1673.Find-the-Most-Competitive-Subsequence.md b/website/content/ChapterFour/1600~1699/1673.Find-the-Most-Competitive-Subsequence.md
new file mode 100644
index 000000000..02292dd4f
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1673.Find-the-Most-Competitive-Subsequence.md
@@ -0,0 +1,69 @@
+# [1673. Find the Most Competitive Subsequence](https://leetcode.com/problems/find-the-most-competitive-subsequence/)
+
+
+## 题目
+
+Given an integer array `nums` and a positive integer `k`, return *the most **competitive** subsequence of* `nums` *of size* `k`.
+
+An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.
+
+We define that a subsequence `a` is more **competitive** than a subsequence `b` (of the same length) if in the first position where `a` and `b` differ, subsequence `a` has a number **less** than the corresponding number in `b`. For example, `[1,3,4]` is more competitive than `[1,3,5]` because the first position they differ is at the final number, and `4` is less than `5`.
+
+**Example 1**:
+
+```
+Input: nums = [3,5,2,6], k = 2
+Output: [2,6]
+Explanation: Among the set of every possible subsequence: {[3,5], [3,2], [3,6], [5,2], [5,6], [2,6]}, [2,6] is the most competitive.
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [2,4,3,3,5,4,9,6], k = 4
+Output: [2,3,3,4]
+
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 105`
+- `0 <= nums[i] <= 109`
+- `1 <= k <= nums.length`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个正整数 k ，返回长度为 k 且最具 竞争力 的 nums 子序列。数组的子序列是从数组中删除一些元素（可能不删除元素）得到的序列。
+
+在子序列 a 和子序列 b 第一个不相同的位置上，如果 a 中的数字小于 b 中对应的数字，那么我们称子序列 a 比子序列 b（相同长度下）更具 竞争力 。 例如，[1,3,4] 比 [1,3,5] 更具竞争力，在第一个不相同的位置，也就是最后一个位置上， 4 小于 5 。
+
+## 解题思路
+
+- 这一题是单调栈的典型题型。利用单调栈，可以保证原数组中元素相对位置不变，这满足题意中删除元素但不移动元素的要求。单调栈又能保证每次进栈，元素是最小的。
+- 类似的题目还有第 42 题，第 84 题，第 496 题，第 503 题，第 856 题，第 901 题，第 907 题，第 1130 题，第 1425 题，第 1673 题。
+
+## 代码
+
+```go
+package leetcode
+
+// 单调栈
+func mostCompetitive(nums []int, k int) []int {
+	stack := make([]int, 0, len(nums))
+	for i := 0; i < len(nums); i++ {
+		for len(stack)+len(nums)-i > k && len(stack) > 0 && nums[i] < stack[len(stack)-1] {
+			stack = stack[:len(stack)-1]
+		}
+		stack = append(stack, nums[i])
+	}
+	return stack[:k]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1672.Richest-Customer-Wealth/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1674.Minimum-Moves-to-Make-Array-Complementary/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1674.Minimum-Moves-to-Make-Array-Complementary.md b/website/content/ChapterFour/1600~1699/1674.Minimum-Moves-to-Make-Array-Complementary.md
new file mode 100644
index 000000000..32997f177
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1674.Minimum-Moves-to-Make-Array-Complementary.md
@@ -0,0 +1,104 @@
+# [1674. Minimum Moves to Make Array Complementary](https://leetcode.com/problems/minimum-moves-to-make-array-complementary/)
+
+## 题目
+
+You are given an integer array `nums` of **even** length `n` and an integer `limit`. In one move, you can replace any integer from `nums` with another integer between `1` and `limit`, inclusive.
+
+The array `nums` is **complementary** if for all indices `i` (**0-indexed**), `nums[i] + nums[n - 1 - i]` equals the same number. For example, the array `[1,2,3,4]` is complementary because for all indices `i`, `nums[i] + nums[n - 1 - i] = 5`.
+
+Return the ***minimum** number of moves required to make* `nums` ***complementary***.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,4,3], limit = 4
+Output: 1
+Explanation: In 1 move, you can change nums to [1,2,2,3] (underlined elements are changed).
+nums[0] + nums[3] = 1 + 3 = 4.
+nums[1] + nums[2] = 2 + 2 = 4.
+nums[2] + nums[1] = 2 + 2 = 4.
+nums[3] + nums[0] = 3 + 1 = 4.
+Therefore, nums[i] + nums[n-1-i] = 4 for every i, so nums is complementary.
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,2,2,1], limit = 2
+Output: 2
+Explanation: In 2 moves, you can change nums to [2,2,2,2]. You cannot change any number to 3 since 3 > limit.
+```
+
+**Example 3**:
+
+```
+Input: nums = [1,2,1,2], limit = 2
+Output: 0
+Explanation: nums is already complementary.
+```
+
+**Constraints**:
+
+- `n == nums.length`
+- `2 <= n <= 105`
+- `1 <= nums[i] <= limit <= 105`
+- `n` is even.
+
+## 题目大意
+
+给你一个长度为 偶数 n 的整数数组 nums 和一个整数 limit 。每一次操作，你可以将 nums 中的任何整数替换为 1 到 limit 之间的另一个整数。
+
+如果对于所有下标 i（下标从 0 开始），nums[i] + nums[n - 1 - i] 都等于同一个数，则数组 nums 是 互补的 。例如，数组 [1,2,3,4] 是互补的，因为对于所有下标 i ，nums[i] + nums[n - 1 - i] = 5 。
+
+返回使数组 互补 的 最少 操作次数。
+
+## 解题思路
+
+- 这一题考察的是差分数组。通过分析题意，可以得出，针对每一个 `sum` 的取值范围是 `[2, 2* limt]`，定义 `a = min(nums[i], nums[n - i - 1])`，`b = max(nums[i], nums[n - i - 1])`，在这个区间内，又可以细分成 5 个区间，`[2, a + 1)`，`[a + 1, a + b)`，`[a + b + 1, a + b + 1)`，`[a + b + 1, b + limit + 1)`，`[b + limit + 1, 2 * limit)`，在这 5 个区间内使得数组互补的最小操作次数分别是 `2(减少 a, 减少 b)`，`1(减少 b)`，`0(不用操作)`，`1(增大 a)`，`+2(增大 a, 增大 b)`，换个表达方式，按照扫描线从左往右扫描，在这 5 个区间内使得数组互补的最小操作次数叠加变化分别是 `+2(减少 a, 减少 b)`，`-1(减少 a)`，`-1(不用操作)`，`+1(增大 a)`，`+1(增大 a, 增大 b)`，利用这前后两个区间的关系，就可以构造一个差分数组。差分数组反应的是前后两者的关系。如果想求得 0 ~ n 的总关系，只需要求一次前缀和即可。
+- 这道题要求输出最少的操作次数，所以利用差分数组 + 前缀和，累加前缀和的同时维护最小值。从左往右扫描完一遍以后，输出最小值即可。
+
+## 代码
+
+```go
+package leetcode
+
+func minMoves(nums []int, limit int) int {
+	diff := make([]int, limit*2+2) // nums[i] <= limit, b+limit+1 is maximum limit+limit+1
+	for j := 0; j < len(nums)/2; j++ {
+		a, b := min(nums[j], nums[len(nums)-j-1]), max(nums[j], nums[len(nums)-j-1])
+		// using prefix sum: most interesting point, and is the key to reduce complexity
+		diff[2] += 2
+		diff[a+1]--
+		diff[a+b]--
+		diff[a+b+1]++
+		diff[b+limit+1]++
+	}
+	cur, res := 0, len(nums)
+	for i := 2; i <= 2*limit; i++ {
+		cur += diff[i]
+		res = min(res, cur)
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1673.Find-the-Most-Competitive-Subsequence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1675.Minimize-Deviation-in-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1675.Minimize-Deviation-in-Array.md b/website/content/ChapterFour/1600~1699/1675.Minimize-Deviation-in-Array.md
new file mode 100644
index 000000000..974e444a2
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1675.Minimize-Deviation-in-Array.md
@@ -0,0 +1,112 @@
+# [1675. Minimize Deviation in Array](https://leetcode.com/problems/minimize-deviation-in-array/)
+
+## 题目
+
+You are given an array `nums` of `n` positive integers.
+
+You can perform two types of operations on any element of the array any number of times:
+
+- If the element is **even**, **divide** it by `2`.
+    - For example, if the array is `[1,2,3,4]`, then you can do this operation on the last element, and the array will be `[1,2,3,2].`
+- If the element is **odd**, **multiply** it by `2`.
+    - For example, if the array is `[1,2,3,4]`, then you can do this operation on the first element, and the array will be `[2,2,3,4].`
+
+The **deviation** of the array is the **maximum difference** between any two elements in the array.
+
+Return *the **minimum deviation** the array can have after performing some number of operations.*
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3,4]
+Output: 1
+Explanation: You can transform the array to [1,2,3,2], then to [2,2,3,2], then the deviation will be 3 - 2 = 1.
+```
+
+**Example 2:**
+
+```
+Input: nums = [4,1,5,20,3]
+Output: 3
+Explanation: You can transform the array after two operations to [4,2,5,5,3], then the deviation will be 5 - 2 = 3.
+```
+
+**Example 3:**
+
+```
+Input: nums = [2,10,8]
+Output: 3
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `2 <= n <= 105`
+- `1 <= nums[i] <= 10^9`
+
+## 题目大意
+
+给你一个由 n 个正整数组成的数组 nums 。你可以对数组的任意元素执行任意次数的两类操作：
+
+- 如果元素是 偶数 ，除以 2。例如，如果数组是 [1,2,3,4] ，那么你可以对最后一个元素执行此操作，使其变成 [1,2,3,2]
+- 如果元素是 奇数 ，乘上 2。例如，如果数组是 [1,2,3,4] ，那么你可以对第一个元素执行此操作，使其变成 [2,2,3,4]
+数组的 偏移量 是数组中任意两个元素之间的 最大差值 。
+
+返回数组在执行某些操作之后可以拥有的 最小偏移量 。
+
+## 解题思路
+
+- 要找到最小偏移量，即需要令最大值变小，最小值变大。要想达到这个要求，需要对奇数偶数做乘法和除法。这里特殊的是，奇数一旦乘以 2 以后，就变成偶数了。偶数除以 2 以后可能是奇数也可能是偶数。所以可以先将所有的奇数都乘以 2 统一变成偶数。
+- 第二轮不断的将最大值除 2，直到最大值为奇数，不能再操作了。每轮循环中把比 min 值大的偶数也都除以 2 。这里除以 2 有 2 个目的，一个目的是将第一步奇数乘 2 还原回去，另一个目的是将本来的偶数除以 2 。可能有人有疑问，为什么只把最大值变小，没有将最小值变大呢？如果最小值是奇数，那么它一定是由上一个偶数除以 2 变过来的，我们在上一个状态已经计算过这个偶数了，因此没必要扩大它；如果最小值是偶数，那么它一定会在某一轮的除 2 操作中，不操作，即它不会满足 `min <= nums[i]/2`  这个条件。每次循环都更新该次循环的最大值和最小值，并记录偏移量。不断的循环，直到最大值为奇数，退出循环。最终输出最小偏移量。
+
+## 代码
+
+```go
+package leetcode
+
+func minimumDeviation(nums []int) int {
+	min, max := 0, 0
+	for i := range nums {
+		if nums[i]%2 == 1 {
+			nums[i] *= 2
+		}
+		if i == 0 {
+			min = nums[i]
+			max = nums[i]
+		} else if nums[i] < min {
+			min = nums[i]
+		} else if max < nums[i] {
+			max = nums[i]
+		}
+	}
+	res := max - min
+	for max%2 == 0 {
+		tmax, tmin := 0, 0
+		for i := range nums {
+			if nums[i] == max || (nums[i]%2 == 0 && min <= nums[i]/2) {
+				nums[i] /= 2
+			}
+			if i == 0 {
+				tmin = nums[i]
+				tmax = nums[i]
+			} else if nums[i] < tmin {
+				tmin = nums[i]
+			} else if tmax < nums[i] {
+				tmax = nums[i]
+			}
+		}
+		if tmax-tmin < res {
+			res = tmax - tmin
+		}
+		min, max = tmin, tmax
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1674.Minimum-Moves-to-Make-Array-Complementary/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1678.Goal-Parser-Interpretation/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1678.Goal-Parser-Interpretation.md b/website/content/ChapterFour/1600~1699/1678.Goal-Parser-Interpretation.md
new file mode 100644
index 000000000..54cd358bd
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1678.Goal-Parser-Interpretation.md
@@ -0,0 +1,80 @@
+# [1678. Goal Parser Interpretation](https://leetcode.com/problems/goal-parser-interpretation/)
+
+## 题目
+
+You own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G"`, `"()"` and/or `"(al)"` in some order. The Goal Parser will interpret `"G"` as the string `"G"`, `"()"` as the string `"o"`, and `"(al)"` as the string `"al"`. The interpreted strings are then concatenated in the original order.
+
+Given the string `command`, return *the **Goal Parser**'s interpretation of* `command`.
+
+**Example 1**:
+
+```
+Input: command = "G()(al)"
+Output: "Goal"
+Explanation: The Goal Parser interprets the command as follows:
+G -> G
+() -> o
+(al) -> al
+The final concatenated result is "Goal".
+```
+
+**Example 2**:
+
+```
+Input: command = "G()()()()(al)"
+Output: "Gooooal"
+```
+
+**Example 3**:
+
+```
+Input: command = "(al)G(al)()()G"
+Output: "alGalooG"
+```
+
+**Constraints**:
+
+- `1 <= command.length <= 100`
+- `command` consists of `"G"`, `"()"`, and/or `"(al)"` in some order.
+
+## 题目大意
+
+请你设计一个可以解释字符串 command 的 Goal 解析器 。command 由 "G"、"()" 和/或 "(al)" 按某种顺序组成。Goal 解析器会将 "G" 解释为字符串 "G"、"()" 解释为字符串 "o" ，"(al)" 解释为字符串 "al" 。然后，按原顺序将经解释得到的字符串连接成一个字符串。给你字符串 command ，返回 Goal 解析器 对 command 的解释结果。
+
+## 解题思路
+
+- 简单题，按照题意修改字符串即可。由于是简单题，这一题也不用考虑嵌套的情况。
+
+## 代码
+
+```go
+package leetcode
+
+func interpret(command string) string {
+	if command == "" {
+		return ""
+	}
+	res := ""
+	for i := 0; i < len(command); i++ {
+		if command[i] == 'G' {
+			res += "G"
+		} else {
+			if command[i] == '(' && command[i+1] == 'a' {
+				res += "al"
+				i += 3
+			} else {
+				res += "o"
+				i ++
+			}
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1675.Minimize-Deviation-in-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1679.Max-Number-of-K-Sum-Pairs/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1679.Max-Number-of-K-Sum-Pairs.md b/website/content/ChapterFour/1600~1699/1679.Max-Number-of-K-Sum-Pairs.md
new file mode 100644
index 000000000..f95055aa9
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1679.Max-Number-of-K-Sum-Pairs.md
@@ -0,0 +1,105 @@
+# [1679. Max Number of K-Sum Pairs](https://leetcode.com/problems/max-number-of-k-sum-pairs/)
+
+
+## 题目
+
+You are given an integer array `nums` and an integer `k`.
+
+In one operation, you can pick two numbers from the array whose sum equals `k` and remove them from the array.
+
+Return *the maximum number of operations you can perform on the array*.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,4], k = 5
+Output: 2
+Explanation: Starting with nums = [1,2,3,4]:
+- Remove numbers 1 and 4, then nums = [2,3]
+- Remove numbers 2 and 3, then nums = []
+There are no more pairs that sum up to 5, hence a total of 2 operations.
+```
+
+**Example 2**:
+
+```
+Input: nums = [3,1,3,4,3], k = 6
+Output: 1
+Explanation: Starting with nums = [3,1,3,4,3]:
+- Remove the first two 3's, then nums = [1,4,3]
+There are no more pairs that sum up to 6, hence a total of 1 operation.
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 105`
+- `1 <= nums[i] <= 109`
+- `1 <= k <= 109`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个整数 k 。每一步操作中，你需要从数组中选出和为 k 的两个整数，并将它们移出数组。返回你可以对数组执行的最大操作数。
+
+## 解题思路
+
+- 读完题第一感觉这道题是 TWO SUM 题目的加强版。需要找到所有满足和是 k 的数对。先考虑能不能找到两个数都是 k/2 ，如果能找到多个这样的数，可以先移除他们。其次在利用 TWO SUM 的思路，找出和为 k 的数对。利用 TWO SUM 里面 map 的做法，时间复杂度 O(n)。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 优化版
+func maxOperations(nums []int, k int) int {
+	counter, res := make(map[int]int), 0
+	for _, n := range nums {
+		counter[n]++
+	}
+	if (k & 1) == 0 {
+		res += counter[k>>1] >> 1
+		// 能够由 2 个相同的数构成 k 的组合已经都排除出去了，剩下的一个单独的也不能组成 k 了
+		// 所以这里要把它的频次置为 0 。如果这里不置为 0，下面代码判断逻辑还需要考虑重复使用数字的情况
+		counter[k>>1] = 0
+	}
+	for num, freq := range counter {
+		if num <= k/2 {
+			remain := k - num
+			if counter[remain] < freq {
+				res += counter[remain]
+			} else {
+				res += freq
+			}
+		}
+	}
+	return res
+}
+
+// 解法二
+func maxOperations_(nums []int, k int) int {
+	counter, res := make(map[int]int), 0
+	for _, num := range nums {
+		counter[num]++
+		remain := k - num
+		if num == remain {
+			if counter[num] >= 2 {
+				res++
+				counter[num] -= 2
+			}
+		} else {
+			if counter[remain] > 0 {
+				res++
+				counter[remain]--
+				counter[num]--
+			}
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1678.Goal-Parser-Interpretation/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1680.Concatenation-of-Consecutive-Binary-Numbers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1680.Concatenation-of-Consecutive-Binary-Numbers.md b/website/content/ChapterFour/1600~1699/1680.Concatenation-of-Consecutive-Binary-Numbers.md
new file mode 100644
index 000000000..085376b2c
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1680.Concatenation-of-Consecutive-Binary-Numbers.md
@@ -0,0 +1,102 @@
+# [1680. Concatenation of Consecutive Binary Numbers](https://leetcode.com/problems/concatenation-of-consecutive-binary-numbers/)
+
+
+## 题目
+
+Given an integer `n`, return *the **decimal value** of the binary string formed by concatenating the binary representations of* `1` *to* `n` *in order, **modulo*** `109 + 7`.
+
+**Example 1**:
+
+```
+Input: n = 1
+Output: 1
+Explanation: "1" in binary corresponds to the decimal value 1. 
+```
+
+**Example 2**:
+
+```
+Input: n = 3
+Output: 27
+Explanation: In binary, 1, 2, and 3 corresponds to "1", "10", and "11".
+After concatenating them, we have "11011", which corresponds to the decimal value 27.
+```
+
+**Example 3**:
+
+```
+Input: n = 12
+Output: 505379714
+Explanation: The concatenation results in "1101110010111011110001001101010111100".
+The decimal value of that is 118505380540.
+After modulo 109 + 7, the result is 505379714.
+```
+
+**Constraints**:
+
+- `1 <= n <= 10^5`
+
+## 题目大意
+
+给你一个整数 n ，请你将 1 到 n 的二进制表示连接起来，并返回连接结果对应的 十进制 数字对 10^9 + 7 取余的结果。
+
+## 解题思路
+
+- 理解题意以后，先找到如何拼接最终二进制数的规律。假设 `f(n)` 为最终变换以后的十进制数。那么根据题意，`f(n) = f(n-1) << shift + n` 这是一个递推公式。`shift` 左移的位数就是 `n` 的二进制对应的长度。`shift` 的值是随着 `n` 变化而变化的。由二进制进位规律可以知道，2 的整数次幂的时候，对应的二进制长度会增加 1 位。这里可以利用位运算来判断是否是 2 的整数次幂。
+- 这道题另外一个需要处理的是模运算的法则。此题需要用到模运算的加法法则。
+
+    ```go
+    模运算与基本四则运算有些相似，但是除法例外。
+    (a + b) % p = (a % p + b % p) % p （1）
+    (a - b) % p = (a % p - b % p) % p （2）
+    (a * b) % p = (a % p * b % p) % p （3）
+    a ^ b % p = ((a % p)^b) % p （4）
+    结合律：
+    ((a+b) % p + c) % p = (a + (b+c) % p) % p （5）
+    ((a*b) % p * c)% p = (a * (b*c) % p) % p （6）
+    交换律：
+    (a + b) % p = (b+a) % p （7）
+    (a * b) % p = (b * a) % p （8）
+    分配律：
+    ((a +b)% p * c) % p = ((a * c) % p + (b * c) % p) % p （9）
+    ```
+
+    这一题需要用到模运算的加法运算法则。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math/bits"
+)
+
+// 解法一 模拟
+func concatenatedBinary(n int) int {
+	res, mod, shift := 0, 1000000007, 0
+	for i := 1; i <= n; i++ {
+		if (i & (i - 1)) == 0 {
+			shift++
+		}
+		res = ((res << shift) + i) % mod
+	}
+	return res
+}
+
+// 解法二 位运算
+func concatenatedBinary1(n int) int {
+	res := 0
+	for i := 1; i <= n; i++ {
+		res = (res<<bits.Len(uint(i)) | i) % (1e9 + 7)
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1679.Max-Number-of-K-Sum-Pairs/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1681.Minimum-Incompatibility/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1681.Minimum-Incompatibility.md b/website/content/ChapterFour/1600~1699/1681.Minimum-Incompatibility.md
new file mode 100644
index 000000000..e8a267e6b
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1681.Minimum-Incompatibility.md
@@ -0,0 +1,129 @@
+# [1681. Minimum Incompatibility](https://leetcode.com/problems/minimum-incompatibility/)
+
+
+## 题目
+
+You are given an integer array `nums` and an integer `k`. You are asked to distribute this array into `k` subsets of **equal size** such that there are no two equal elements in the same subset.
+
+A subset's **incompatibility** is the difference between the maximum and minimum elements in that array.
+
+Return *the **minimum possible sum of incompatibilities** of the* `k` *subsets after distributing the array optimally, or return* `-1` *if it is not possible.*
+
+A subset is a group integers that appear in the array with no particular order.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,1,4], k = 2
+Output: 4
+Explanation: The optimal distribution of subsets is [1,2] and [1,4].
+The incompatibility is (2-1) + (4-1) = 4.
+Note that [1,1] and [2,4] would result in a smaller sum, but the first subset contains 2 equal elements.
+```
+
+**Example 2**:
+
+```
+Input: nums = [6,3,8,1,3,1,2,2], k = 4
+Output: 6
+Explanation: The optimal distribution of subsets is [1,2], [2,3], [6,8], and [1,3].
+The incompatibility is (2-1) + (3-2) + (8-6) + (3-1) = 6.
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [5,3,3,6,3,3], k = 3
+Output: -1
+Explanation: It is impossible to distribute nums into 3 subsets where no two elements are equal in the same subset.
+
+```
+
+**Constraints**:
+
+- `1 <= k <= nums.length <= 16`
+- `nums.length` is divisible by `k`
+- `1 <= nums[i] <= nums.length`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个整数 k 。你需要将这个数组划分到 k 个相同大小的子集中，使得同一个子集里面没有两个相同的元素。一个子集的 不兼容性 是该子集里面最大值和最小值的差。
+
+请你返回将数组分成 k 个子集后，各子集 **不兼容性** 的 **和** 的 **最小值** ，如果无法分成分成 k 个子集，返回 -1 。子集的定义是数组中一些数字的集合，对数字顺序没有要求。
+
+## 解题思路
+
+- 读完题最直白的思路就是 DFS。做法类似第 77 题。这里就不赘述了。可以见第 77 题题解。
+- 这一题还需要用到贪心的思想。每次取数都取最小的数。这样可以不会让最大数和最小数在一个集合中。由于每次取数都是取最小的，那么能保证不兼容性每次都尽量最小。于是在 order 数组中定义取数的顺序。然后再把数组从小到大排列。这样每次按照 order 顺序取数，都是取的最小值。
+- 正常的 DFS 写完提交，耗时是很长的。大概是 1532ms。如何优化到极致呢？这里需要加上 2 个剪枝条件。第一个剪枝条件比较简单，如果累计 sum 比之前存储的 res 大，那么直接 return，不需要继续递归了。第二个剪枝条件就非常重要了，可以一下子减少很多次递归。每次取数产生新的集合的时候，要从第一个最小数开始取，一旦取了，后面就不需要再循环递归了。举个例子，[1,2,3,4]，第一个数如果取 2，集合可以是 [[2,3],[1,4]] 或 [[2,4], [1,3]], 这个集合和[[1,3],[2,4]]、[[1,4], [2,3]] 情况一样。可以看到如果取出第一个最小值以后，后面的循环是不必要的了。所以在取下标为 0 的数的时候，递归到底层以后，返回就可以直接 break，不用接下去的循环了，接下去的循环和递归是不必要的。每组组内的顺序我们并不关心，只要最大值和最小值在分组内即可。另外组间顺序我们也不关心。所以可以把排列问题 O(n!) 时间复杂度降低到组合问题 O(2^n)。加了这 2 个剪枝条件以后，耗时就变成了 0ms 了。beats 100%
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func minimumIncompatibility(nums []int, k int) int {
+	sort.Ints(nums)
+	eachSize, counts := len(nums)/k, make([]int, len(nums)+1)
+	for i := range nums {
+		counts[nums[i]]++
+		if counts[nums[i]] > k {
+			return -1
+		}
+	}
+	orders := []int{}
+	for i := range counts {
+		orders = append(orders, i)
+	}
+	sort.Ints(orders)
+	res := math.MaxInt32
+	generatePermutation1681(nums, counts, orders, 0, 0, eachSize, &res, []int{})
+	if res == math.MaxInt32 {
+		return -1
+	}
+	return res
+}
+
+func generatePermutation1681(nums, counts, order []int, index, sum, eachSize int, res *int, current []int) {
+	if len(current) > 0 && len(current)%eachSize == 0 {
+		sum += current[len(current)-1] - current[len(current)-eachSize]
+		index = 0
+	}
+	if sum >= *res {
+		return
+	}
+	if len(current) == len(nums) {
+		if sum < *res {
+			*res = sum
+		}
+		return
+	}
+	for i := index; i < len(counts); i++ {
+		if counts[order[i]] == 0 {
+			continue
+		}
+		counts[order[i]]--
+		current = append(current, order[i])
+		generatePermutation1681(nums, counts, order, i+1, sum, eachSize, res, current)
+		current = current[:len(current)-1]
+		counts[order[i]]++
+		// 这里是关键的剪枝
+		if index == 0 {
+			break
+		}
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1680.Concatenation-of-Consecutive-Binary-Numbers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1684.Count-the-Number-of-Consistent-Strings/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1684.Count-the-Number-of-Consistent-Strings.md b/website/content/ChapterFour/1600~1699/1684.Count-the-Number-of-Consistent-Strings.md
new file mode 100644
index 000000000..afe034b5c
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1684.Count-the-Number-of-Consistent-Strings.md
@@ -0,0 +1,86 @@
+# [1684. Count the Number of Consistent Strings](https://leetcode.com/problems/count-the-number-of-consistent-strings/)
+
+
+## 题目
+
+You are given a string `allowed` consisting of **distinct** characters and an array of strings `words`. A string is **consistent** if all characters in the string appear in the string `allowed`.
+
+Return *the number of **consistent** strings in the array* `words`.
+
+**Example 1**:
+
+```
+Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
+Output: 2
+Explanation: Strings "aaab" and "baa" are consistent since they only contain characters 'a' and 'b'.
+
+```
+
+**Example 2**:
+
+```
+Input: allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
+Output: 7
+Explanation: All strings are consistent.
+
+```
+
+**Example 3**:
+
+```
+Input: allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
+Output: 4
+Explanation: Strings "cc", "acd", "ac", and "d" are consistent.
+
+```
+
+**Constraints**:
+
+- `1 <= words.length <= 104`
+- `1 <= allowed.length <= 26`
+- `1 <= words[i].length <= 10`
+- The characters in `allowed` are **distinct**.
+- `words[i]` and `allowed` contain only lowercase English letters.
+
+## 题目大意
+
+给你一个由不同字符组成的字符串 `allowed` 和一个字符串数组 `words` 。如果一个字符串的每一个字符都在 `allowed` 中，就称这个字符串是 一致字符串 。
+
+请你返回 `words` 数组中 一致字符串 的数目。
+
+## 解题思路
+
+- 简单题。先将 `allowed` 转化成 map。将 `words` 数组中每个单词的字符都在 map 中查找一遍，如果都存在就累加 res。如果有不存在的字母，不累加。最终输出 res 即可。
+
+## 代码
+
+```go
+package leetcode
+
+func countConsistentStrings(allowed string, words []string) int {
+	allowedMap, res, flag := map[rune]int{}, 0, true
+	for _, str := range allowed {
+		allowedMap[str]++
+	}
+	for i := 0; i < len(words); i++ {
+		flag = true
+		for j := 0; j < len(words[i]); j++ {
+			if _, ok := allowedMap[rune(words[i][j])]; !ok {
+				flag = false
+				break
+			}
+		}
+		if flag {
+			res++
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1681.Minimum-Incompatibility/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array.md b/website/content/ChapterFour/1600~1699/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array.md
new file mode 100644
index 000000000..4ab16dc75
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array.md
@@ -0,0 +1,95 @@
+# [1685. Sum of Absolute Differences in a Sorted Array](https://leetcode.com/problems/sum-of-absolute-differences-in-a-sorted-array/)
+
+
+## 题目
+
+You are given an integer array `nums` sorted in **non-decreasing** order.
+
+Build and return *an integer array* `result` *with the same length as* `nums` *such that* `result[i]` *is equal to the **summation of absolute differences** between* `nums[i]` *and all the other elements in the array.*
+
+In other words, `result[i]` is equal to `sum(|nums[i]-nums[j]|)` where `0 <= j < nums.length` and `j != i` (**0-indexed**).
+
+**Example 1**:
+
+```
+Input: nums = [2,3,5]
+Output: [4,3,5]
+Explanation: Assuming the arrays are 0-indexed, then
+result[0] = |2-2| + |2-3| + |2-5| = 0 + 1 + 3 = 4,
+result[1] = |3-2| + |3-3| + |3-5| = 1 + 0 + 2 = 3,
+result[2] = |5-2| + |5-3| + |5-5| = 3 + 2 + 0 = 5.
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,4,6,8,10]
+Output: [24,15,13,15,21]
+```
+
+**Constraints**:
+
+- `2 <= nums.length <= 105`
+- `1 <= nums[i] <= nums[i + 1] <= 104`
+
+## 题目大意
+
+给你一个 非递减 有序整数数组 `nums` 。请你建立并返回一个整数数组 `result`，它跟 `nums` 长度相同，且`result[i]` 等于 `nums[i]` 与数组中所有其他元素差的绝对值之和。换句话说， `result[i]` 等于 `sum(|nums[i]-nums[j]|)` ，其中 `0 <= j < nums.length` 且 `j != i` （下标从 0 开始）。
+
+## 解题思路
+
+- 利用前缀和思路解题。题目中说明了是有序数组，所以在计算绝对值的时候可以拆开绝对值符号。假设要计算当前 `result[i]`，以 `i` 为界，把原数组 `nums` 分成了 3 段。`nums[0 ~ i-1]` 和 `nums[i+1 ~ n]`，前面一段 `nums[0 ~ i-1]` 中的每个元素都比 `nums[i]` 小，拆掉绝对值以后，`sum(|nums[i]-nums[j]|) = nums[i] * i - prefixSum[0 ~ i-1]`，后面一段 `nums[i+1 ~ n]` 中的每个元素都比 `nums[i]` 大，拆掉绝对值以后，`sum(|nums[i]-nums[j]|) = prefixSum[i+1 ~ n] - nums[i] * (n - 1 - i)`。特殊的情况，`i = 0` 和 `i = n` 的情况特殊处理一下就行。
+
+## 代码
+
+```go
+package leetcode
+
+//解法一 优化版 prefixSum + sufixSum
+func getSumAbsoluteDifferences(nums []int) []int {
+	size := len(nums)
+	sufixSum := make([]int, size)
+	sufixSum[size-1] = nums[size-1]
+	for i := size - 2; i >= 0; i-- {
+		sufixSum[i] = sufixSum[i+1] + nums[i]
+	}
+	ans, preSum := make([]int, size), 0
+	for i := 0; i < size; i++ {
+		// 后面可以加到的值
+		res, sum := 0, sufixSum[i]-nums[i]
+		res += (sum - (size-i-1)*nums[i])
+		// 前面可以加到的值
+		res += (i*nums[i] - preSum)
+		ans[i] = res
+		preSum += nums[i]
+	}
+	return ans
+}
+
+// 解法二 prefixSum
+func getSumAbsoluteDifferences1(nums []int) []int {
+	preSum, res, sum := []int{}, []int{}, nums[0]
+	preSum = append(preSum, nums[0])
+	for i := 1; i < len(nums); i++ {
+		sum += nums[i]
+		preSum = append(preSum, sum)
+	}
+	for i := 0; i < len(nums); i++ {
+		if i == 0 {
+			res = append(res, preSum[len(nums)-1]-preSum[0]-nums[i]*(len(nums)-1))
+		} else if i > 0 && i < len(nums)-1 {
+			res = append(res, preSum[len(nums)-1]-preSum[i]-preSum[i-1]+nums[i]*i-nums[i]*(len(nums)-1-i))
+		} else {
+			res = append(res, nums[i]*len(nums)-preSum[len(nums)-1])
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1684.Count-the-Number-of-Consistent-Strings/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1688.Count-of-Matches-in-Tournament/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1688.Count-of-Matches-in-Tournament.md b/website/content/ChapterFour/1600~1699/1688.Count-of-Matches-in-Tournament.md
new file mode 100644
index 000000000..e8a8c3921
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1688.Count-of-Matches-in-Tournament.md
@@ -0,0 +1,87 @@
+# [1688. Count of Matches in Tournament](https://leetcode.com/problems/count-of-matches-in-tournament/)
+
+
+## 题目
+
+You are given an integer `n`, the number of teams in a tournament that has strange rules:
+
+- If the current number of teams is **even**, each team gets paired with another team. A total of `n / 2` matches are played, and `n / 2` teams advance to the next round.
+- If the current number of teams is **odd**, one team randomly advances in the tournament, and the rest gets paired. A total of `(n - 1) / 2` matches are played, and `(n - 1) / 2 + 1` teams advance to the next round.
+
+Return *the number of matches played in the tournament until a winner is decided.*
+
+**Example 1**:
+
+```
+Input: n = 7
+Output: 6
+Explanation: Details of the tournament: 
+- 1st Round: Teams = 7, Matches = 3, and 4 teams advance.
+- 2nd Round: Teams = 4, Matches = 2, and 2 teams advance.
+- 3rd Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
+Total number of matches = 3 + 2 + 1 = 6.
+```
+
+**Example 2**:
+
+```
+Input: n = 14
+Output: 13
+Explanation: Details of the tournament:
+- 1st Round: Teams = 14, Matches = 7, and 7 teams advance.
+- 2nd Round: Teams = 7, Matches = 3, and 4 teams advance.
+- 3rd Round: Teams = 4, Matches = 2, and 2 teams advance.
+- 4th Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
+Total number of matches = 7 + 3 + 2 + 1 = 13.
+```
+
+**Constraints**:
+
+- `1 <= n <= 200`
+
+## 题目大意
+
+给你一个整数 n ，表示比赛中的队伍数。比赛遵循一种独特的赛制：
+
+- 如果当前队伍数是 偶数 ，那么每支队伍都会与另一支队伍配对。总共进行 n / 2 场比赛，且产生 n / 2 支队伍进入下一轮。
+- 如果当前队伍数为 奇数 ，那么将会随机轮空并晋级一支队伍，其余的队伍配对。总共进行 (n - 1) / 2 场比赛，且产生 (n - 1) / 2 + 1 支队伍进入下一轮。
+
+返回在比赛中进行的配对次数，直到决出获胜队伍为止。
+
+## 解题思路
+
+- 简单题，按照题目的规则模拟。
+- 这一题还有更加简洁的代码，见解法一。n 个队伍，一个冠军，需要淘汰 n-1 个队伍。每一场比赛淘汰一个队伍，因此进行了 n-1 场比赛。所以共有 n-1 个配对。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一
+func numberOfMatches(n int) int {
+	return n - 1
+}
+
+// 解法二 模拟
+func numberOfMatches1(n int) int {
+	sum := 0
+	for n != 1 {
+		if n&1 == 0 {
+			sum += n / 2
+			n = n / 2
+		} else {
+			sum += (n - 1) / 2
+			n = (n-1)/2 + 1
+		}
+	}
+	return sum
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers.md b/website/content/ChapterFour/1600~1699/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers.md
new file mode 100644
index 000000000..f3494ed47
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers.md
@@ -0,0 +1,67 @@
+# [1689. Partitioning Into Minimum Number Of Deci-Binary Numbers](https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/)
+
+## 题目
+
+A decimal number is called **deci-binary** if each of its digits is either `0` or `1` without any leading zeros. For example, `101` and `1100` are **deci-binary**, while `112` and `3001` are not.
+
+Given a string `n` that represents a positive decimal integer, return *the **minimum** number of positive **deci-binary** numbers needed so that they sum up to* `n`*.*
+
+**Example 1**:
+
+```
+Input: n = "32"
+Output: 3
+Explanation: 10 + 11 + 11 = 32
+```
+
+**Example 2**:
+
+```
+Input: n = "82734"
+Output: 8
+```
+
+**Example 3**:
+
+```
+Input: n = "27346209830709182346"
+Output: 9
+```
+
+**Constraints**:
+
+- `1 <= n.length <= 105`
+- `n` consists of only digits.
+- `n` does not contain any leading zeros and represents a positive integer.
+
+## 题目大意
+
+如果一个十进制数字不含任何前导零，且每一位上的数字不是 0 就是 1 ，那么该数字就是一个 十-二进制数 。例如，101 和 1100 都是 十-二进制数，而 112 和 3001 不是。给你一个表示十进制整数的字符串 n ，返回和为 n 的 十-二进制数 的最少数目。
+
+## 解题思路
+
+- 这一题也算是简单题，相通了以后，代码就 3 行。
+- 要想由 01 组成的十进制数组成 n，只需要在 n 这个数的各个数位上依次排上 0 和 1 即可。例如 n = 23423723，这是一个 8 位数。最大数字是 7，所以至少需要 7 个数累加能得到这个 n。这 7 个数的百位都为 1，其他数位按需求取 0 和 1 即可。例如万位是 2，那么这 7 个数中任找 2 个数的万位是 1 ，其他 5 个数的万位是 0 即可。
+
+## 代码
+
+```go
+package leetcode
+
+func minPartitions(n string) int {
+	res := 0
+	for i := 0; i < len(n); i++ {
+		if int(n[i]-'0') > res {
+			res = int(n[i] - '0')
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1688.Count-of-Matches-in-Tournament/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1690.Stone-Game-VII/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1690.Stone-Game-VII.md b/website/content/ChapterFour/1600~1699/1690.Stone-Game-VII.md
new file mode 100644
index 000000000..c81e0619f
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1690.Stone-Game-VII.md
@@ -0,0 +1,137 @@
+# [1690. Stone Game VII](https://leetcode.com/problems/stone-game-vii/)
+
+## 题目
+
+Alice and Bob take turns playing a game, with **Alice starting first**.
+
+There are `n` stones arranged in a row. On each player's turn, they can **remove** either the leftmost stone or the rightmost stone from the row and receive points equal to the **sum** of the remaining stones' values in the row. The winner is the one with the higher score when there are no stones left to remove.
+
+Bob found that he will always lose this game (poor Bob, he always loses), so he decided to **minimize the score's difference**. Alice's goal is to **maximize the difference** in the score.
+
+Given an array of integers `stones` where `stones[i]` represents the value of the `ith` stone **from the left**, return *the **difference** in Alice and Bob's score if they both play **optimally**.*
+
+**Example 1**:
+
+```
+Input: stones = [5,3,1,4,2]
+Output: 6
+Explanation: 
+- Alice removes 2 and gets 5 + 3 + 1 + 4 = 13 points. Alice = 13, Bob = 0, stones = [5,3,1,4].
+- Bob removes 5 and gets 3 + 1 + 4 = 8 points. Alice = 13, Bob = 8, stones = [3,1,4].
+- Alice removes 3 and gets 1 + 4 = 5 points. Alice = 18, Bob = 8, stones = [1,4].
+- Bob removes 1 and gets 4 points. Alice = 18, Bob = 12, stones = [4].
+- Alice removes 4 and gets 0 points. Alice = 18, Bob = 12, stones = [].
+The score difference is 18 - 12 = 6.
+```
+
+**Example 2**:
+
+```
+Input: stones = [7,90,5,1,100,10,10,2]
+Output: 122
+```
+
+**Constraints**:
+
+- `n == stones.length`
+- `2 <= n <= 1000`
+- `1 <= stones[i] <= 1000`
+
+## 题目大意
+
+石子游戏中，爱丽丝和鲍勃轮流进行自己的回合，爱丽丝先开始 。有 n 块石子排成一排。每个玩家的回合中，可以从行中 移除 最左边的石头或最右边的石头，并获得与该行中剩余石头值之 和 相等的得分。当没有石头可移除时，得分较高者获胜。鲍勃发现他总是输掉游戏（可怜的鲍勃，他总是输），所以他决定尽力 减小得分的差值 。爱丽丝的目标是最大限度地 扩大得分的差值 。
+
+给你一个整数数组 stones ，其中 stones[i] 表示 从左边开始 的第 i 个石头的值，如果爱丽丝和鲍勃都 发挥出最佳水平 ，请返回他们 得分的差值 。
+
+## 解题思路
+
+- 首先考虑 Bob 缩小分值差距意味着什么，意味着他想让他和 Alice 相对分数最小。Bob 已经明确肯定是输，所以他的分数一定比 Alice 小，那么 Bob - Alice 分数相减一定是负数。相对分数越小，意味着差值越大。负数越大，差值越小。-50 和 -10，-10 数值大，相差小。所以 Bob 的操作是让相对分数越大。Alice 的目的也是这样，要让 Alice - Bob 的相对分数越大，这里是正数的越大。综上，两者的目的相同，都是让相对分数最大化。
+- 定义 `dp[i][j]` 代表在当前 `stone[i ~ j]` 区间内能获得的最大分差。状态转移方程为：
+
+    ```go
+    dp[i][j] = max(
+                    sum(i + 1, j) - dp[i + 1][j],   // 这一局取走 `stone[i]`，获得 sum(i + 1, j) 分数，再减去剩下对手能获得的分数，即是此局能获得的最大分差。
+                    sum(i, j - 1) - dp[i][j - 1]    // 这一局取走 `stone[j]`，获得 sum(i, j - 1) 分数，再减去剩下对手能获得的分数，即是此局能获得的最大分差。
+                  )
+    ```
+
+    计算 `sum(i + 1, j) = stone[i + 1] + stone[i + 2] + …… + stone[j]` 利用前缀和计算区间和。
+
+- 解法二是正常思路解答出来的代码。解法一是压缩了 DP 数组，在 DP 状态转移的时候，生成下一个 `dp[j]` 实际上是有规律的。利用这个规律可以少存一维数据，压缩空间。解法一的代码直接写出来，比较难想。先写出解法二的代码，然后找到递推规律，优化空间压缩一维，再写出解法一的代码。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 优化空间版 DP
+func stoneGameVII(stones []int) int {
+	n := len(stones)
+	sum := make([]int, n)
+	dp := make([]int, n)
+	for i, d := range stones {
+		sum[i] = d
+	}
+	for i := 1; i < n; i++ {
+		for j := 0; j+i < n; j++ {
+			if (n-i)%2 == 1 {
+				d0 := dp[j] + sum[j]
+				d1 := dp[j+1] + sum[j+1]
+				if d0 > d1 {
+					dp[j] = d0
+				} else {
+					dp[j] = d1
+				}
+			} else {
+				d0 := dp[j] - sum[j]
+				d1 := dp[j+1] - sum[j+1]
+				if d0 < d1 {
+					dp[j] = d0
+				} else {
+					dp[j] = d1
+				}
+			}
+			sum[j] = sum[j] + stones[i+j]
+		}
+	}
+	return dp[0]
+}
+
+// 解法二 常规 DP
+func stoneGameVII1(stones []int) int {
+	prefixSum := make([]int, len(stones))
+	for i := 0; i < len(stones); i++ {
+		if i == 0 {
+			prefixSum[i] = stones[i]
+		} else {
+			prefixSum[i] = prefixSum[i-1] + stones[i]
+		}
+	}
+	dp := make([][]int, len(stones))
+	for i := range dp {
+		dp[i] = make([]int, len(stones))
+		dp[i][i] = 0
+	}
+	n := len(stones)
+	for l := 2; l <= n; l++ {
+		for i := 0; i+l <= n; i++ {
+			dp[i][i+l-1] = max(prefixSum[i+l-1]-prefixSum[i+1]+stones[i+1]-dp[i+1][i+l-1], prefixSum[i+l-2]-prefixSum[i]+stones[i]-dp[i][i+l-2])
+		}
+	}
+	return dp[0][n-1]
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1691.Maximum-Height-by-Stacking-Cuboids/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1691.Maximum-Height-by-Stacking-Cuboids.md b/website/content/ChapterFour/1600~1699/1691.Maximum-Height-by-Stacking-Cuboids.md
new file mode 100644
index 000000000..623ab35ec
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1691.Maximum-Height-by-Stacking-Cuboids.md
@@ -0,0 +1,113 @@
+# [1691. Maximum Height by Stacking Cuboids](https://leetcode.com/problems/maximum-height-by-stacking-cuboids/)
+
+## 题目
+
+Given `n` `cuboids` where the dimensions of the `ith` cuboid is `cuboids[i] = [widthi, lengthi, heighti]` (**0-indexed**). Choose a **subset** of `cuboids` and place them on each other.
+
+You can place cuboid `i` on cuboid `j` if `widthi <= widthj` and `lengthi <= lengthj` and `heighti <= heightj`. You can rearrange any cuboid's dimensions by rotating it to put it on another cuboid.
+
+Return *the **maximum height** of the stacked* `cuboids`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2019/10/21/image.jpg](https://assets.leetcode.com/uploads/2019/10/21/image.jpg)
+
+```
+Input: cuboids = [[50,45,20],[95,37,53],[45,23,12]]
+Output: 190
+Explanation:
+Cuboid 1 is placed on the bottom with the 53x37 side facing down with height 95.
+Cuboid 0 is placed next with the 45x20 side facing down with height 50.
+Cuboid 2 is placed next with the 23x12 side facing down with height 45.
+The total height is 95 + 50 + 45 = 190.
+```
+
+**Example 2:**
+
+```
+Input: cuboids = [[38,25,45],[76,35,3]]
+Output: 76
+Explanation:
+You can't place any of the cuboids on the other.
+We choose cuboid 1 and rotate it so that the 35x3 side is facing down and its height is 76.
+```
+
+**Example 3:**
+
+```
+Input: cuboids = [[7,11,17],[7,17,11],[11,7,17],[11,17,7],[17,7,11],[17,11,7]]
+Output: 102
+Explanation:
+After rearranging the cuboids, you can see that all cuboids have the same dimension.
+You can place the 11x7 side down on all cuboids so their heights are 17.
+The maximum height of stacked cuboids is 6 * 17 = 102.
+```
+
+**Constraints:**
+
+- `n == cuboids.length`
+- `1 <= n <= 100`
+- `1 <= widthi, lengthi, heighti <= 100`
+
+## 题目大意
+
+给你 n 个长方体 cuboids ，其中第 i 个长方体的长宽高表示为 cuboids[i] = [widthi, lengthi, heighti]（下标从 0 开始）。请你从 cuboids 选出一个 子集 ，并将它们堆叠起来。如果 widthi <= widthj 且 lengthi <= lengthj 且 heighti <= heightj ，你就可以将长方体 i 堆叠在长方体 j 上。你可以通过旋转把长方体的长宽高重新排列，以将它放在另一个长方体上。返回 堆叠长方体 cuboids 可以得到的 最大高度 。
+
+## 解题思路
+
+- 这一题是 LIS 最长递增子序列系列问题的延续。一维 LIS 问题是第 300 题。二维 LIS 问题是 354 题。这一题是三维的 LIS 问题。
+- 题目要求最终摞起来的长方体尽可能的高，那么把长宽高中最大的值旋转为高。这是针对单个方块的排序。多个方块间还要排序，因为他们摞起来有要求，大的方块必须放在下面。所以针对多个方块，按照长，宽，高的顺序进行排序。两次排序完成以后，可以用动态规划找出最大值了。定义 `dp[i]` 为以 `i` 为最后一块砖块所能堆叠的最高高度。由于长和宽已经排好序了。所以只需要在 [0, i - 1] 这个区间内动态更新 dp 最大值。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func maxHeight(cuboids [][]int) int {
+	n := len(cuboids)
+	for i := 0; i < n; i++ {
+		sort.Ints(cuboids[i]) // 立方体三边内部排序
+	}
+	// 立方体排序，先按最短边，再到最长边
+	sort.Slice(cuboids, func(i, j int) bool {
+		if cuboids[i][0] != cuboids[j][0] {
+			return cuboids[i][0] < cuboids[j][0]
+		}
+		if cuboids[i][1] != cuboids[j][1] {
+			return cuboids[i][1] < cuboids[j][1]
+		}
+		return cuboids[i][2] < cuboids[j][2]
+	})
+	res := 0
+	dp := make([]int, n)
+	for i := 0; i < n; i++ {
+		dp[i] = cuboids[i][2]
+		res = max(res, dp[i])
+	}
+	for i := 1; i < n; i++ {
+		for j := 0; j < i; j++ {
+			if cuboids[j][0] <= cuboids[i][0] && cuboids[j][1] <= cuboids[i][1] && cuboids[j][2] <= cuboids[i][2] {
+				dp[i] = max(dp[i], dp[j]+cuboids[i][2])
+			}
+		}
+		res = max(res, dp[i])
+	}
+	return res
+}
+
+func max(x, y int) int {
+	if x > y {
+		return x
+	}
+	return y
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1690.Stone-Game-VII/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1694.Reformat-Phone-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1694.Reformat-Phone-Number.md b/website/content/ChapterFour/1600~1699/1694.Reformat-Phone-Number.md
new file mode 100644
index 000000000..d2705b8a9
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1694.Reformat-Phone-Number.md
@@ -0,0 +1,142 @@
+# [1694. Reformat Phone Number](https://leetcode.com/problems/reformat-phone-number/)
+
+
+## 题目
+
+You are given a phone number as a string `number`. `number` consists of digits, spaces `' '`, and/or dashes `'-'`.
+
+You would like to reformat the phone number in a certain manner. Firstly, **remove** all spaces and dashes. Then, **group** the digits from left to right into blocks of length 3 **until** there are 4 or fewer digits. The final digits are then grouped as follows:
+
+- 2 digits: A single block of length 2.
+- 3 digits: A single block of length 3.
+- 4 digits: Two blocks of length 2 each.
+
+The blocks are then joined by dashes. Notice that the reformatting process should **never** produce any blocks of length 1 and produce **at most** two blocks of length 2.
+
+Return *the phone number after formatting.*
+
+**Example 1**:
+
+```
+Input: number = "1-23-45 6"
+Output: "123-456"
+Explanation: The digits are "123456".
+Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".
+Step 2: There are 3 digits remaining, so put them in a single block of length 3. The 2nd block is "456".
+Joining the blocks gives "123-456".
+
+```
+
+**Example 2**:
+
+```
+Input: number = "123 4-567"
+Output: "123-45-67"
+Explanation: The digits are "1234567".
+Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".
+Step 2: There are 4 digits left, so split them into two blocks of length 2. The blocks are "45" and "67".
+Joining the blocks gives "123-45-67".
+
+```
+
+**Example 3**:
+
+```
+Input: number = "123 4-5678"
+Output: "123-456-78"
+Explanation: The digits are "12345678".
+Step 1: The 1st block is "123".
+Step 2: The 2nd block is "456".
+Step 3: There are 2 digits left, so put them in a single block of length 2. The 3rd block is "78".
+Joining the blocks gives "123-456-78".
+
+```
+
+**Example 4**:
+
+```
+Input: number = "12"
+Output: "12"
+
+```
+
+**Example 5**:
+
+```
+Input: number = "--17-5 229 35-39475 "
+Output: "175-229-353-94-75"
+
+```
+
+**Constraints**:
+
+- `2 <= number.length <= 100`
+- `number` consists of digits and the characters `'-'` and `' '`.
+- There are at least **two** digits in `number`.
+
+## 题目大意
+
+给你一个字符串形式的电话号码 number 。number 由数字、空格 ' '、和破折号 '-' 组成。
+
+请你按下述方式重新格式化电话号码。
+
+- 首先，删除 所有的空格和破折号。
+- 其次，将数组从左到右 每 3 个一组 分块，直到 剩下 4 个或更少数字。剩下的数字将按下述规定再分块：
+    - 2 个数字：单个含 2 个数字的块。
+    - 3 个数字：单个含 3 个数字的块。
+    - 4 个数字：两个分别含 2 个数字的块。
+
+最后用破折号将这些块连接起来。注意，重新格式化过程中 不应该 生成仅含 1 个数字的块，并且 最多 生成两个含 2 个数字的块。返回格式化后的电话号码。
+
+## 解题思路
+
+- 简单题。先判断号码是不是 2 位和 4 位，如果是，单独输出这 2 种情况。剩下的都是 3 位以上了，取余，判断剩余的数字是 2 个还是 4 个。这时不可能存在剩 1 位数的情况。除 3 余 1，即剩 4 位的情况，末尾 4 位需要 2 个一组输出。除 3 余 2，即剩  2 位的情况。处理好末尾，再逆序每 3 个一组连接 "-" 即可。可能需要注意的 case 见 test 文件。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"strings"
+)
+
+func reformatNumber(number string) string {
+	parts, nums := []string{}, []rune{}
+	for _, r := range number {
+		if r != '-' && r != ' ' {
+			nums = append(nums, r)
+		}
+	}
+	threeDigits, twoDigits := len(nums)/3, 0
+	switch len(nums) % 3 {
+	case 1:
+		threeDigits--
+		twoDigits = 2
+	case 2:
+		twoDigits = 1
+	default:
+		twoDigits = 0
+	}
+	for i := 0; i < threeDigits; i++ {
+		s := ""
+		s += string(nums[0:3])
+		nums = nums[3:]
+		parts = append(parts, s)
+	}
+	for i := 0; i < twoDigits; i++ {
+		s := ""
+		s += string(nums[0:2])
+		nums = nums[2:]
+		parts = append(parts, s)
+	}
+	return strings.Join(parts, "-")
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1691.Maximum-Height-by-Stacking-Cuboids/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1695.Maximum-Erasure-Value/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1695.Maximum-Erasure-Value.md b/website/content/ChapterFour/1600~1699/1695.Maximum-Erasure-Value.md
new file mode 100644
index 000000000..7ca5b8842
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1695.Maximum-Erasure-Value.md
@@ -0,0 +1,82 @@
+# [1695. Maximum Erasure Value](https://leetcode.com/problems/maximum-erasure-value/)
+
+
+## 题目
+
+You are given an array of positive integers `nums` and want to erase a subarray containing **unique elements**. The **score** you get by erasing the subarray is equal to the **sum** of its elements.
+
+Return *the **maximum score** you can get by erasing **exactly one** subarray.*
+
+An array `b` is called to be a subarray of `a` if it forms a contiguous subsequence of `a`, that is, if it is equal to `a[l],a[l+1],...,a[r]` for some `(l,r)`.
+
+**Example 1**:
+
+```
+Input: nums = [4,2,4,5,6]
+Output: 17
+Explanation: The optimal subarray here is [2,4,5,6].
+```
+
+**Example 2**:
+
+```
+Input: nums = [5,2,1,2,5,2,1,2,5]
+Output: 8
+Explanation: The optimal subarray here is [5,2,1] or [1,2,5].
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 105`
+- `1 <= nums[i] <= 104`
+
+## 题目大意
+
+给你一个正整数数组 nums ，请你从中删除一个含有 若干不同元素 的子数组。删除子数组的 得分 就是子数组各元素之 和 。返回 只删除一个 子数组可获得的 最大得分 。如果数组 b 是数组 a 的一个连续子序列，即如果它等于 a[l],a[l+1],...,a[r] ，那么它就是 a 的一个子数组。
+
+## 解题思路
+
+- 读完题立马能识别出这是经典的滑动窗口题。利用滑动窗口从左往右滑动窗口，滑动过程中统计频次，如果是不同元素，右边界窗口又移，否则左边窗口缩小。每次移动更新 max 值。最终扫完一遍以后，max 值即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+func maximumUniqueSubarray(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	result, left, right, freq := 0, 0, -1, map[int]int{}
+	for left < len(nums) {
+		if right+1 < len(nums) && freq[nums[right+1]] == 0 {
+			freq[nums[right+1]]++
+			right++
+		} else {
+			freq[nums[left]]--
+			left++
+		}
+		sum := 0
+		for i := left; i <= right; i++ {
+			sum += nums[i]
+		}
+		result = max(result, sum)
+	}
+	return result
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1694.Reformat-Phone-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1696.Jump-Game-VI/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/1696.Jump-Game-VI.md b/website/content/ChapterFour/1600~1699/1696.Jump-Game-VI.md
new file mode 100644
index 000000000..155af11ea
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/1696.Jump-Game-VI.md
@@ -0,0 +1,118 @@
+# [1696. Jump Game VI](https://leetcode.com/problems/jump-game-vi/)
+
+## 题目
+
+You are given a **0-indexed** integer array `nums` and an integer `k`.
+
+You are initially standing at index `0`. In one move, you can jump at most `k` steps forward without going outside the boundaries of the array. That is, you can jump from index `i` to any index in the range `[i + 1, min(n - 1, i + k)]` **inclusive**.
+
+You want to reach the last index of the array (index `n - 1`). Your **score** is the **sum** of all `nums[j]` for each index `j` you visited in the array.
+
+Return *the **maximum score** you can get*.
+
+**Example 1:**
+
+```
+Input: nums = [1,-1,-2,4,-7,3], k = 2
+Output: 7
+Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [10,-5,-2,4,0,3], k = 3
+Output: 17
+Explanation: You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17.
+
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
+Output: 0
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length, k <= 10^5`
+- `10^4 <= nums[i] <= 10^4`
+
+## 题目大意
+
+给你一个下标从 0 开始的整数数组 nums 和一个整数 k 。一开始你在下标 0 处。每一步，你最多可以往前跳 k 步，但你不能跳出数组的边界。也就是说，你可以从下标 i 跳到 [i + 1， min(n - 1, i + k)] 包含 两个端点的任意位置。你的目标是到达数组最后一个位置（下标为 n - 1 ），你的 得分 为经过的所有数字之和。请你返回你能得到的 最大得分 。
+
+## 解题思路
+
+- 首先能想到的解题思路是动态规划。定义 `dp[i]` 为跳到第 `i` 个位子能获得的最大分数。题目要求的是 `dp[n-1]`，状态转移方程是：`dp[i] = nums[i] + max(dp[j]), max(0, i - k ) <= j < i`，这里需要注意 `j` 的下界，题目中说到不能跳到负数区间，所以左边界下界为 0 。求 `max(dp[j])` 需要遍历一次求得最大值，所以这个解法整体时间复杂度是 O((n - k) * k)，但是提交以后提示超时了。
+- 分析一下超时原因。每次都要在 `[max(0, i - k ), i)` 区间内扫描找到最大值，下一轮的区间是 `[max(0, i - k + 1), i + 1)`，前后这两轮扫描的区间存在大量重合部分  `[max(0, i - k + 1), i)`，正是这部分反反复复的扫描导致算法低效。如何高效的在一个区间内找到最大值是本题的关键。利用单调队列可以完成此题。单调队列里面存一个区间内最大值的下标。这里单调队列有 2 个性质。性质一，队列的队首永远都是最大值，队列从大到小降序排列。如果来了一个比队首更大的值的下标，需要将单调队列清空，只存这个新的最大值的下标。性质二，队列的长度为 k。从队尾插入新值，并把队头的最大值“挤”出队首。拥有了这个单调队列以后，再进行 DP 状态转移，效率就很高了。每次只需取出队首的最大值即可。具体代码见下面。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+)
+
+// 单调队列
+func maxResult(nums []int, k int) int {
+	dp := make([]int, len(nums))
+	dp[0] = nums[0]
+	for i := 1; i < len(dp); i++ {
+		dp[i] = math.MinInt32
+	}
+	window := make([]int, k)
+	for i := 1; i < len(nums); i++ {
+		dp[i] = nums[i] + dp[window[0]]
+		for len(window) > 0 && dp[window[len(window)-1]] <= dp[i] {
+			window = window[:len(window)-1]
+		}
+		for len(window) > 0 && i-k >= window[0] {
+			window = window[1:]
+		}
+		window = append(window, i)
+	}
+	return dp[len(nums)-1]
+}
+
+// 超时
+func maxResult1(nums []int, k int) int {
+	dp := make([]int, len(nums))
+	if k > len(nums) {
+		k = len(nums)
+	}
+	dp[0] = nums[0]
+	for i := 1; i < len(dp); i++ {
+		dp[i] = math.MinInt32
+	}
+	for i := 1; i < len(nums); i++ {
+		left, tmp := max(0, i-k), math.MinInt32
+		for j := left; j < i; j++ {
+			tmp = max(tmp, dp[j])
+		}
+		dp[i] = nums[i] + tmp
+	}
+	return dp[len(nums)-1]
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1695.Maximum-Erasure-Value/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1700.Number-of-Students-Unable-to-Eat-Lunch/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1600~1699/_index.md b/website/content/ChapterFour/1600~1699/_index.md
new file mode 100644
index 000000000..d2021683f
--- /dev/null
+++ b/website/content/ChapterFour/1600~1699/_index.md
@@ -0,0 +1,5 @@
+---
+bookCollapseSection: true
+weight: 20
+---
+
diff --git a/website/content/ChapterFour/1700~1799/1700.Number-of-Students-Unable-to-Eat-Lunch.md b/website/content/ChapterFour/1700~1799/1700.Number-of-Students-Unable-to-Eat-Lunch.md
new file mode 100644
index 000000000..8ba364f7c
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1700.Number-of-Students-Unable-to-Eat-Lunch.md
@@ -0,0 +1,86 @@
+# [1700. Number of Students Unable to Eat Lunch](https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/)
+
+
+## 题目
+
+The school cafeteria offers circular and square sandwiches at lunch break, referred to by numbers `0` and `1` respectively. All students stand in a queue. Each student either prefers square or circular sandwiches.
+
+The number of sandwiches in the cafeteria is equal to the number of students. The sandwiches are placed in a **stack**. At each step:
+
+- If the student at the front of the queue **prefers** the sandwich on the top of the stack, they will **take it** and leave the queue.
+- Otherwise, they will **leave it** and go to the queue's end.
+
+This continues until none of the queue students want to take the top sandwich and are thus unable to eat.
+
+You are given two integer arrays `students` and `sandwiches` where `sandwiches[i]` is the type of the `ith` sandwich in the stack (`i = 0` is the top of the stack) and `students[j]` is the preference of the `jth` student in the initial queue (`j = 0` is the front of the queue). Return *the number of students that are unable to eat.*
+
+**Example 1:**
+
+```
+Input: students = [1,1,0,0], sandwiches = [0,1,0,1]
+Output: 0 
+Explanation:
+- Front student leaves the top sandwich and returns to the end of the line making students = [1,0,0,1].
+- Front student leaves the top sandwich and returns to the end of the line making students = [0,0,1,1].
+- Front student takes the top sandwich and leaves the line making students = [0,1,1] and sandwiches = [1,0,1].
+- Front student leaves the top sandwich and returns to the end of the line making students = [1,1,0].
+- Front student takes the top sandwich and leaves the line making students = [1,0] and sandwiches = [0,1].
+- Front student leaves the top sandwich and returns to the end of the line making students = [0,1].
+- Front student takes the top sandwich and leaves the line making students = [1] and sandwiches = [1].
+- Front student takes the top sandwich and leaves the line making students = [] and sandwiches = [].
+Hence all students are able to eat.
+```
+
+**Example 2:**
+
+```
+Input: students = [1,1,1,0,0,1], sandwiches = [1,0,0,0,1,1]
+Output: 3
+```
+
+**Constraints:**
+
+- `1 <= students.length, sandwiches.length <= 100`
+- `students.length == sandwiches.length`
+- `sandwiches[i]` is `0` or `1`.
+- `students[i]` is `0` or `1`.
+
+## 题目大意
+
+学校的自助午餐提供圆形和方形的三明治，分别用数字 0 和 1 表示。所有学生站在一个队列里，每个学生要么喜欢圆形的要么喜欢方形的。
+餐厅里三明治的数量与学生的数量相同。所有三明治都放在一个 栈 里，每一轮：
+
+- 如果队列最前面的学生 喜欢 栈顶的三明治，那么会 拿走它 并离开队列。
+- 否则，这名学生会 放弃这个三明治 并回到队列的尾部。
+这个过程会一直持续到队列里所有学生都不喜欢栈顶的三明治为止。
+
+给你两个整数数组 students 和 sandwiches ，其中 sandwiches[i] 是栈里面第 i 个三明治的类型（i = 0 是栈的顶部）， students[j] 是初始队列里第 j 名学生对三明治的喜好（j = 0 是队列的最开始位置）。请你返回无法吃午餐的学生数量。
+
+## 解题思路
+
+- 简单题。按照题意，学生不管怎么轮流领三明治，如果数量够，经过多轮循环，总能领到。问题可以等价为，学生依次到队列前面领取三明治。2 个种类的三明治都摆好放在那里了。最终领不到三明治的学生都是因为喜欢的三明治不够发放了。按照这个思路，先统计 2 种三明治的总个数，然后减去学生对三明治的需求总数，剩下的学生即都是无法满足的。
+
+## 代码
+
+```go
+package leetcode
+
+func countStudents(students []int, sandwiches []int) int {
+	tmp, n, i := [2]int{}, len(students), 0
+	for _, v := range students {
+		tmp[v]++
+	}
+	for i < n && tmp[sandwiches[i]] > 0 {
+		tmp[sandwiches[i]]--
+		i++
+	}
+	return n - i
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1696.Jump-Game-VI/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1704.Determine-if-String-Halves-Are-Alike/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1704.Determine-if-String-Halves-Are-Alike.md b/website/content/ChapterFour/1700~1799/1704.Determine-if-String-Halves-Are-Alike.md
new file mode 100644
index 000000000..e3e454298
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1704.Determine-if-String-Halves-Are-Alike.md
@@ -0,0 +1,82 @@
+# [1704. Determine if String Halves Are Alike](https://leetcode.com/problems/determine-if-string-halves-are-alike/)
+
+## 题目
+
+You are given a string `s` of even length. Split this string into two halves of equal lengths, and let `a` be the first half and `b` be the second half.
+
+Two strings are **alike** if they have the same number of vowels (`'a'`, `'e'`, `'i'`, `'o'`, `'u'`, `'A'`, `'E'`, `'I'`, `'O'`, `'U'`). Notice that `s` contains uppercase and lowercase letters.
+
+Return `true` *if* `a` *and* `b` *are **alike***. Otherwise, return `false`.
+
+**Example 1:**
+
+```
+Input: s = "book"
+Output: true
+Explanation: a = "bo" and b = "ok". a has 1 vowel and b has 1 vowel. Therefore, they are alike.
+```
+
+**Example 2:**
+
+```
+Input: s = "textbook"
+Output: false
+Explanation: a = "text" and b = "book". a has 1 vowel whereas b has 2. Therefore, they are not alike.
+Notice that the vowel o is counted twice.
+```
+
+**Example 3:**
+
+```
+Input: s = "MerryChristmas"
+Output: false
+```
+
+**Example 4:**
+
+```
+Input: s = "AbCdEfGh"
+Output: true
+```
+
+**Constraints:**
+
+- `2 <= s.length <= 1000`
+- `s.length` is even.
+- `s` consists of **uppercase and lowercase** letters.
+
+## 题目大意
+
+给你一个偶数长度的字符串 s 。将其拆分成长度相同的两半，前一半为 a ，后一半为 b 。两个字符串 相似 的前提是它们都含有相同数目的元音（'a'，'e'，'i'，'o'，'u'，'A'，'E'，'I'，'O'，'U'）。注意，s 可能同时含有大写和小写字母。如果 a 和 b 相似，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+- 简单题。依题意，分别统计前半段元音字母的个数和后半段元音字母的个数，个数相同则输出 true，不同就输出 false。
+
+## 代码
+
+```go
+package leetcode
+
+func halvesAreAlike(s string) bool {
+	return numVowels(s[len(s)/2:]) == numVowels(s[:len(s)/2])
+}
+
+func numVowels(x string) int {
+	res := 0
+	for _, c := range x {
+		switch c {
+		case 'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U':
+			res++
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1700.Number-of-Students-Unable-to-Eat-Lunch/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1705.Maximum-Number-of-Eaten-Apples/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1705.Maximum-Number-of-Eaten-Apples.md b/website/content/ChapterFour/1700~1799/1705.Maximum-Number-of-Eaten-Apples.md
new file mode 100644
index 000000000..fb4094c33
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1705.Maximum-Number-of-Eaten-Apples.md
@@ -0,0 +1,131 @@
+# [1705. Maximum Number of Eaten Apples](https://leetcode.com/problems/maximum-number-of-eaten-apples/)
+
+## 题目
+
+There is a special kind of apple tree that grows apples every day for n days. On the ith day, the tree grows apples[i] apples that will rot after days[i] days, that is on day i + days[i] the apples will be rotten and cannot be eaten. On some days, the apple tree does not grow any apples, which are denoted by apples[i] == 0 and days[i] == 0.
+
+You decided to eat at most one apple a day (to keep the doctors away). Note that you can keep eating after the first n days.
+
+Given two integer arrays days and apples of length n, return the maximum number of apples you can eat.
+
+**Example 1**:
+
+    Input: apples = [1,2,3,5,2], days = [3,2,1,4,2]
+    Output: 7
+    Explanation: You can eat 7 apples:
+    - On the first day, you eat an apple that grew on the first day.
+    - On the second day, you eat an apple that grew on the second day.
+    - On the third day, you eat an apple that grew on the second day. After this day, the apples that grew on the third day rot.
+    - On the fourth to the seventh days, you eat apples that grew on the fourth day.
+
+**Example 2**:
+
+    Input: apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
+    Output: 5
+    Explanation: You can eat 5 apples:
+    - On the first to the third day you eat apples that grew on the first day.
+    - Do nothing on the fouth and fifth days.
+    - On the sixth and seventh days you eat apples that grew on the sixth day.
+
+**Constraints:**
+
+- apples.length == n
+- days.length == n
+- 1 <= n <= 2 * 10000
+- 0 <= apples[i], days[i] <= 2 * 10000
+- days[i] = 0 if and only if apples[i] = 0.
+
+## 题目大意
+
+有一棵特殊的苹果树，一连 n 天，每天都可以长出若干个苹果。在第 i 天，树上会长出 apples[i] 个苹果，这些苹果将会在 days[i] 天后（也就是说，第 i + days[i] 天时）腐烂，变得无法食用。也可能有那么几天，树上不会长出新的苹果，此时用 apples[i] == 0 且 days[i] == 0 表示。
+
+你打算每天 最多 吃一个苹果来保证营养均衡。注意，你可以在这 n 天之后继续吃苹果。
+
+给你两个长度为 n 的整数数组 days 和 apples ，返回你可以吃掉的苹果的最大数目。
+
+## 解题思路
+
+贪心算法和最小堆
+
+  - data中的end表示腐烂的日期，left表示拥有的苹果数量
+  - 贪心:每天吃掉end最小但没有腐烂的苹果
+  - 最小堆:构造类型为数组(数组中元素的类型为data)的最小堆
+
+## 代码
+
+```go
+package leetcode
+
+import "container/heap"
+
+func eatenApples(apples []int, days []int) int {
+	h := hp{}
+	i := 0
+	var ans int
+	for ; i < len(apples); i++ {
+		for len(h) > 0 && h[0].end <= i {
+			heap.Pop(&h)
+		}
+		if apples[i] > 0 {
+			heap.Push(&h, data{apples[i], i + days[i]})
+		}
+		if len(h) > 0 {
+			minData := heap.Pop(&h).(data)
+			ans++
+			if minData.left > 1 {
+				heap.Push(&h, data{minData.left - 1, minData.end})
+			}
+		}
+	}
+	for len(h) > 0 {
+		for len(h) > 0 && h[0].end <= i {
+			heap.Pop(&h)
+		}
+		if len(h) == 0 {
+			break
+		}
+		minData := heap.Pop(&h).(data)
+		nums := min(minData.left, minData.end-i)
+		ans += nums
+		i += nums
+	}
+	return ans
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+type data struct {
+	left int
+	end  int
+}
+
+type hp []data
+
+func (h hp) Len() int           { return len(h) }
+func (h hp) Less(i, j int) bool { return h[i].end < h[j].end }
+func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
+
+func (h *hp) Push(x interface{}) {
+	*h = append(*h, x.(data))
+}
+
+func (h *hp) Pop() interface{} {
+	old := *h
+	n := len(old)
+	x := old[n-1]
+	*h = old[0 : n-1]
+	return x
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1704.Determine-if-String-Halves-Are-Alike/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1710.Maximum-Units-on-a-Truck/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1710.Maximum-Units-on-a-Truck.md b/website/content/ChapterFour/1700~1799/1710.Maximum-Units-on-a-Truck.md
new file mode 100644
index 000000000..2b311d0c0
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1710.Maximum-Units-on-a-Truck.md
@@ -0,0 +1,84 @@
+# [1710. Maximum Units on a Truck](https://leetcode.com/problems/maximum-units-on-a-truck/)
+
+
+## 题目
+
+You are assigned to put some amount of boxes onto **one truck**. You are given a 2D array `boxTypes`, where `boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]`:
+
+- `numberOfBoxesi` is the number of boxes of type `i`.
+- `numberOfUnitsPerBoxi`is the number of units in each box of the type `i`.
+
+You are also given an integer `truckSize`, which is the **maximum** number of **boxes** that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed `truckSize`.
+
+Return *the **maximum** total number of **units** that can be put on the truck.*
+
+**Example 1:**
+
+```
+Input: boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4
+Output: 8
+Explanation: There are:
+- 1 box of the first type that contains 3 units.
+- 2 boxes of the second type that contain 2 units each.
+- 3 boxes of the third type that contain 1 unit each.
+You can take all the boxes of the first and second types, and one box of the third type.
+The total number of units will be = (1 * 3) + (2 * 2) + (1 * 1) = 8.
+```
+
+**Example 2:**
+
+```
+Input: boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10
+Output: 91
+```
+
+**Constraints:**
+
+- `1 <= boxTypes.length <= 1000`
+- `1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000`
+- `1 <= truckSize <= 106`
+
+## 题目大意
+
+请你将一些箱子装在 一辆卡车 上。给你一个二维数组 boxTypes ，其中 boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi] ：
+
+- numberOfBoxesi 是类型 i 的箱子的数量。-
+- numberOfUnitsPerBoxi 是类型 i 每个箱子可以装载的单元数量。
+
+整数 truckSize 表示卡车上可以装载 箱子 的 最大数量 。只要箱子数量不超过 truckSize ，你就可以选择任意箱子装到卡车上。返回卡车可以装载 单元 的 最大 总数。
+
+## 解题思路
+
+- 简单题。先将箱子按照单元数量从大到小排序。要想卡车装载单元数最大，那么需要尽量装单元数多的箱子。所以排序以后从单元数量多的箱子开始取。一直取至 truckSize 没有空间。累积的单元数即最大总数。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func maximumUnits(boxTypes [][]int, truckSize int) int {
+	sort.Slice(boxTypes, func(i, j int) bool {
+		return boxTypes[i][1] > boxTypes[j][1]
+	})
+	res := 0
+	for i := 0; truckSize > 0 && i < len(boxTypes); i++ {
+		if truckSize >= boxTypes[i][0] {
+			truckSize -= boxTypes[i][0]
+			res += (boxTypes[i][1] * boxTypes[i][0])
+		} else {
+			res += (truckSize * boxTypes[i][1])
+			truckSize = 0
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1705.Maximum-Number-of-Eaten-Apples/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1716.Calculate-Money-in-Leetcode-Bank/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1716.Calculate-Money-in-Leetcode-Bank.md b/website/content/ChapterFour/1700~1799/1716.Calculate-Money-in-Leetcode-Bank.md
new file mode 100644
index 000000000..48256ac74
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1716.Calculate-Money-in-Leetcode-Bank.md
@@ -0,0 +1,71 @@
+# [1716. Calculate Money in Leetcode Bank](https://leetcode.com/problems/calculate-money-in-leetcode-bank/)
+
+
+## 题目
+
+Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
+
+He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
+
+Given `n`, return *the total amount of money he will have in the Leetcode bank at the end of the* `nth` *day.*
+
+**Example 1:**
+
+```
+Input: n = 4
+Output: 10
+Explanation: After the 4th day, the total is 1 + 2 + 3 + 4 = 10.
+```
+
+**Example 2:**
+
+```
+Input: n = 10
+Output: 37
+Explanation: After the 10th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4) = 37. Notice that on the 2nd Monday, Hercy only puts in $2.
+```
+
+**Example 3:**
+
+```
+Input: n = 20
+Output: 96
+Explanation: After the 20th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4 + 5 + 6 + 7 + 8) + (3 + 4 + 5 + 6 + 7 + 8) = 96.
+```
+
+**Constraints:**
+
+- `1 <= n <= 1000`
+
+## 题目大意
+
+Hercy 想要为购买第一辆车存钱。他 每天 都往力扣银行里存钱。最开始，他在周一的时候存入 1 块钱。从周二到周日，他每天都比前一天多存入 1 块钱。在接下来每一个周一，他都会比 前一个周一 多存入 1 块钱。给你 n ，请你返回在第 n 天结束的时候他在力扣银行总共存了多少块钱。
+
+## 解题思路
+
+- 简单题。按照题意写 2 层循环即可。
+
+## 代码
+
+```go
+package leetcode
+
+func totalMoney(n int) int {
+	res := 0
+	for tmp, count := 1, 7; n > 0; tmp, count = tmp+1, 7 {
+		for m := tmp; n > 0 && count > 0; m++ {
+			res += m
+			n--
+			count--
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1710.Maximum-Units-on-a-Truck/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1720.Decode-XORed-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1720.Decode-XORed-Array.md b/website/content/ChapterFour/1700~1799/1720.Decode-XORed-Array.md
new file mode 100644
index 000000000..130eb3d4f
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1720.Decode-XORed-Array.md
@@ -0,0 +1,66 @@
+# [1720. Decode XORed Array](https://leetcode.com/problems/decode-xored-array/)
+
+
+## 题目
+
+There is a **hidden** integer array `arr` that consists of `n` non-negative integers.
+
+It was encoded into another integer array `encoded` of length `n - 1`, such that `encoded[i] = arr[i] XOR arr[i + 1]`. For example, if `arr = [1,0,2,1]`, then `encoded = [1,2,3]`.
+
+You are given the `encoded` array. You are also given an integer `first`, that is the first element of `arr`, i.e. `arr[0]`.
+
+Return *the original array* `arr`. It can be proved that the answer exists and is unique.
+
+**Example 1:**
+
+```
+Input: encoded = [1,2,3], first = 1
+Output: [1,0,2,1]
+Explanation: If arr = [1,0,2,1], then first = 1 and encoded = [1 XOR 0, 0 XOR 2, 2 XOR 1] = [1,2,3]
+
+```
+
+**Example 2:**
+
+```
+Input: encoded = [6,2,7,3], first = 4
+Output: [4,2,0,7,4]
+
+```
+
+**Constraints:**
+
+- `2 <= n <= 104`
+- `encoded.length == n - 1`
+- `0 <= encoded[i] <= 105`
+- `0 <= first <= 10^5`
+
+## 题目大意
+
+未知 整数数组 arr 由 n 个非负整数组成。经编码后变为长度为 n - 1 的另一个整数数组 encoded ，其中 encoded[i] = arr[i] XOR arr[i + 1] 。例如，arr = [1,0,2,1] 经编码后得到 encoded = [1,2,3] 。给你编码后的数组 encoded 和原数组 arr 的第一个元素 first（arr[0]）。请解码返回原数组 arr 。可以证明答案存在并且是唯一的。
+
+## 解题思路
+
+- 简单题。按照题意，求返回解码以后的原数组，即将编码后的数组前后两两元素依次做异或 `XOR` 运算。
+
+## 代码
+
+```go
+package leetcode
+
+func decode(encoded []int, first int) []int {
+	arr := make([]int, len(encoded)+1)
+	arr[0] = first
+	for i, val := range encoded {
+		arr[i+1] = arr[i] ^ val
+	}
+	return arr
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1716.Calculate-Money-in-Leetcode-Bank/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1721.Swapping-Nodes-in-a-Linked-List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1721.Swapping-Nodes-in-a-Linked-List.md b/website/content/ChapterFour/1700~1799/1721.Swapping-Nodes-in-a-Linked-List.md
new file mode 100644
index 000000000..70e7b2412
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1721.Swapping-Nodes-in-a-Linked-List.md
@@ -0,0 +1,107 @@
+# [1721. Swapping Nodes in a Linked List](https://leetcode.com/problems/swapping-nodes-in-a-linked-list/)
+
+
+## 题目
+
+You are given the `head` of a linked list, and an integer `k`.
+
+Return *the head of the linked list after **swapping** the values of the* `kth` *node from the beginning and the* `kth` *node from the end (the list is **1-indexed**).*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/09/21/linked1.jpg](https://assets.leetcode.com/uploads/2020/09/21/linked1.jpg)
+
+```
+Input: head = [1,2,3,4,5], k = 2
+Output: [1,4,3,2,5]
+```
+
+**Example 2:**
+
+```
+Input: head = [7,9,6,6,7,8,3,0,9,5], k = 5
+Output: [7,9,6,6,8,7,3,0,9,5]
+```
+
+**Example 3:**
+
+```
+Input: head = [1], k = 1
+Output: [1]
+```
+
+**Example 4:**
+
+```
+Input: head = [1,2], k = 1
+Output: [2,1]
+```
+
+**Example 5:**
+
+```
+Input: head = [1,2,3], k = 2
+Output: [1,2,3]
+```
+
+**Constraints:**
+
+- The number of nodes in the list is `n`.
+- `1 <= k <= n <= 10^5`
+- `0 <= Node.val <= 100`
+
+## 题目大意
+
+给你链表的头节点 `head` 和一个整数 `k` 。**交换** 链表正数第 `k` 个节点和倒数第 `k` 个节点的值后，返回链表的头节点（链表 **从 1 开始索引**）。
+
+## 解题思路
+
+- 这道题虽然是 medium，但是实际非常简单。题目要求链表中 2 个节点的值，无非是先找到这 2 个节点，然后再交换即可。链表查询节点需要 O(n)，2 次循环找到对应的 2 个节点，交换值即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func swapNodes(head *ListNode, k int) *ListNode {
+	count := 1
+	var a, b *ListNode
+	for node := head; node != nil; node = node.Next {
+		if count == k {
+			a = node
+		}
+		count++
+	}
+	length := count
+	count = 1
+	for node := head; node != nil; node = node.Next {
+		if count == length-k {
+			b = node
+		}
+		count++
+	}
+	a.Val, b.Val = b.Val, a.Val
+	return head
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1720.Decode-XORed-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square.md b/website/content/ChapterFour/1700~1799/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square.md
new file mode 100644
index 000000000..05bb9b25d
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square.md
@@ -0,0 +1,75 @@
+# [1725. Number Of Rectangles That Can Form The Largest Square](https://leetcode.com/problems/number-of-rectangles-that-can-form-the-largest-square/)
+
+
+## 题目
+
+You are given an array `rectangles` where `rectangles[i] = [li, wi]` represents the `ith` rectangle of length `li` and width `wi`.
+
+You can cut the `ith` rectangle to form a square with a side length of `k` if both `k <= li` and `k <= wi`. For example, if you have a rectangle `[4,6]`, you can cut it to get a square with a side length of at most `4`.
+
+Let `maxLen` be the side length of the **largest** square you can obtain from any of the given rectangles.
+
+Return *the **number** of rectangles that can make a square with a side length of* `maxLen`.
+
+**Example 1:**
+
+```
+Input: rectangles = [[5,8],[3,9],[5,12],[16,5]]
+Output: 3
+Explanation: The largest squares you can get from each rectangle are of lengths [5,3,5,5].
+The largest possible square is of length 5, and you can get it out of 3 rectangles.
+```
+
+**Example 2:**
+
+```
+Input: rectangles = [[2,3],[3,7],[4,3],[3,7]]
+Output: 3
+```
+
+**Constraints:**
+
+- `1 <= rectangles.length <= 1000`
+- `rectangles[i].length == 2`
+- `1 <= li, wi <= 10^9`
+- `li != wi`
+
+## 题目大意
+
+给你一个数组 rectangles ，其中 rectangles[i] = [li, wi] 表示第 i 个矩形的长度为 li 、宽度为 wi 。如果存在 k 同时满足 k <= li 和 k <= wi ，就可以将第 i 个矩形切成边长为 k 的正方形。例如，矩形 [4,6] 可以切成边长最大为 4 的正方形。设 maxLen 为可以从矩形数组 rectangles 切分得到的 最大正方形 的边长。返回可以切出边长为 maxLen 的正方形的矩形 数目 。
+
+## 解题思路
+
+- 简单题。扫描数组中的每一个矩形，先找到边长较小的那条边，作为正方形的边长。扫描过程中动态更新最大的正方形边长，并累加计数。循环一遍结束，输出最终计数值即可。
+
+## 代码
+
+```go
+package leetcode
+
+func countGoodRectangles(rectangles [][]int) int {
+	minLength, count := 0, 0
+	for i, _ := range rectangles {
+		minSide := 0
+		if rectangles[i][0] <= rectangles[i][1] {
+			minSide = rectangles[i][0]
+		} else {
+			minSide = rectangles[i][1]
+		}
+		if minSide > minLength {
+			minLength = minSide
+			count = 1
+		} else if minSide == minLength {
+			count++
+		}
+	}
+	return count
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1721.Swapping-Nodes-in-a-Linked-List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1732.Find-the-Highest-Altitude/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1732.Find-the-Highest-Altitude.md b/website/content/ChapterFour/1700~1799/1732.Find-the-Highest-Altitude.md
new file mode 100644
index 000000000..7ff314990
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1732.Find-the-Highest-Altitude.md
@@ -0,0 +1,62 @@
+# [1732. Find the Highest Altitude](https://leetcode.com/problems/find-the-highest-altitude/)
+
+
+## 题目
+
+There is a biker going on a road trip. The road trip consists of `n + 1` points at different altitudes. The biker starts his trip on point `0` with altitude equal `0`.
+
+You are given an integer array `gain` of length `n` where `gain[i]` is the **net gain in altitude** between points `i` and `i + 1` for all (`0 <= i < n)`. Return *the **highest altitude** of a point.*
+
+**Example 1:**
+
+```
+Input: gain = [-5,1,5,0,-7]
+Output: 1
+Explanation: The altitudes are [0,-5,-4,1,1,-6]. The highest is 1.
+```
+
+**Example 2:**
+
+```
+Input: gain = [-4,-3,-2,-1,4,3,2]
+Output: 0
+Explanation: The altitudes are [0,-4,-7,-9,-10,-6,-3,-1]. The highest is 0.
+```
+
+**Constraints:**
+
+- `n == gain.length`
+- `1 <= n <= 100`
+- `100 <= gain[i] <= 100`
+
+## 题目大意
+
+有一个自行车手打算进行一场公路骑行，这条路线总共由 n + 1 个不同海拔的点组成。自行车手从海拔为 0 的点 0 开始骑行。给你一个长度为 n 的整数数组 gain ，其中 gain[i] 是点 i 和点 i + 1 的 净海拔高度差（0 <= i < n）。请你返回 最高点的海拔 。
+
+## 解题思路
+
+- 简单题。循环数组依次从第一个海拔点开始还原每个海拔点，动态记录最大高度。循环结束输出最大高度即可。
+
+## 代码
+
+```go
+package leetcode
+
+func largestAltitude(gain []int) int {
+	max, height := 0, 0
+	for _, g := range gain {
+		height += g
+		if height > max {
+			max = height
+		}
+	}
+	return max
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1734.Decode-XORed-Permutation/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1734.Decode-XORed-Permutation.md b/website/content/ChapterFour/1700~1799/1734.Decode-XORed-Permutation.md
new file mode 100644
index 000000000..eef845b91
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1734.Decode-XORed-Permutation.md
@@ -0,0 +1,82 @@
+# [1734. Decode XORed Permutation](https://leetcode.com/problems/decode-xored-permutation/)
+
+
+## 题目
+
+There is an integer array `perm` that is a permutation of the first `n` positive integers, where `n` is always **odd**.
+
+It was encoded into another integer array `encoded` of length `n - 1`, such that `encoded[i] = perm[i] XOR perm[i + 1]`. For example, if `perm = [1,3,2]`, then `encoded = [2,1]`.
+
+Given the `encoded` array, return *the original array* `perm`. It is guaranteed that the answer exists and is unique.
+
+**Example 1:**
+
+```
+Input: encoded = [3,1]
+Output: [1,2,3]
+Explanation: If perm = [1,2,3], then encoded = [1 XOR 2,2 XOR 3] = [3,1]
+
+```
+
+**Example 2:**
+
+```
+Input: encoded = [6,5,4,6]
+Output: [2,4,1,5,3]
+
+```
+
+**Constraints:**
+
+- `3 <= n < 10^5`
+- `n` is odd.
+- `encoded.length == n - 1`
+
+## 题目大意
+
+给你一个整数数组 perm ，它是前 n 个正整数的排列，且 n 是个奇数 。它被加密成另一个长度为 n - 1 的整数数组 encoded ，满足 encoded[i] = perm[i] XOR perm[i + 1] 。比方说，如果 perm = [1,3,2] ，那么 encoded = [2,1] 。给你 encoded 数组，请你返回原始数组 perm 。题目保证答案存在且唯一。
+
+## 解题思路
+
+- 这一题与第 136 题和第 137 题思路类似，借用 `x ^ x = 0` 这个性质解题。依题意，原数组 perm 是 n 个正整数，即取值在 `[1,n+1]` 区间内，但是排列顺序未知。可以考虑先将 `[1,n+1]` 区间内的所有数异或得到 total。再将 encoded 数组中奇数下标的元素异或得到 odd：
+
+	{{< katex display >}}
+    \begin{aligned}odd &= encoded[1] + encoded[3] + ... + encoded[n-1]\\&= (perm[1] \,\, XOR \,\, perm[2]) + (perm[3] \,\,  XOR  \,\, perm[4]) + ... + (perm[n-1]  \,\, XOR \,\, perm[n])\end{aligned}
+	{{< /katex >}}
+
+    total 是 n 个正整数异或全集，odd 是 `n-1` 个正整数异或集。两者异或 `total ^ odd` 得到的值必定是 perm[0]，因为 `x ^ x = 0`，那么重复出现的元素被异或以后消失了。算出 perm[0] 就好办了。
+
+	{{< katex display >}}
+    \begin{aligned}encoded[0] &= perm[0] \,\, XOR \,\, perm[1]\\perm[0] \,\, XOR \,\, encoded[0] &= perm[0] \,\, XOR \,\, perm[0] \,\, XOR \,\, perm[1] = perm[1]\\perm[1] \,\, XOR \,\, encoded[1] &= perm[1] \,\, XOR \,\, perm[1] \,\, XOR \,\, perm[2] = perm[2]\\...\\perm[n-1] \,\, XOR \,\, encoded[n-1] &= perm[n-1] \,\, XOR \,\, perm[n-1] \,\, XOR \,\, perm[n] = perm[n]\\\end{aligned}
+	{{< /katex >}}
+
+    依次类推，便可以推出原数组 perm 中的所有数。
+
+## 代码
+
+```go
+package leetcode
+
+func decode(encoded []int) []int {
+	n, total, odd := len(encoded), 0, 0
+	for i := 1; i <= n+1; i++ {
+		total ^= i
+	}
+	for i := 1; i < n; i += 2 {
+		odd ^= encoded[i]
+	}
+	perm := make([]int, n+1)
+	perm[0] = total ^ odd
+	for i, v := range encoded {
+		perm[i+1] = perm[i] ^ v
+	}
+	return perm
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1732.Find-the-Highest-Altitude/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1736.Latest-Time-by-Replacing-Hidden-Digits/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1736.Latest-Time-by-Replacing-Hidden-Digits.md b/website/content/ChapterFour/1700~1799/1736.Latest-Time-by-Replacing-Hidden-Digits.md
new file mode 100644
index 000000000..c8d3cf920
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1736.Latest-Time-by-Replacing-Hidden-Digits.md
@@ -0,0 +1,82 @@
+# [1736. Latest Time by Replacing Hidden Digits](https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/)
+
+
+## 题目
+
+You are given a string `time` in the form of `hh:mm`, where some of the digits in the string are hidden (represented by `?`).
+
+The valid times are those inclusively between `00:00` and `23:59`.
+
+Return *the latest valid time you can get from* `time` *by replacing the hidden* *digits*.
+
+**Example 1:**
+
+```
+Input: time = "2?:?0"
+Output: "23:50"
+Explanation: The latest hour beginning with the digit '2' is 23 and the latest minute ending with the digit '0' is 50.
+```
+
+**Example 2:**
+
+```
+Input: time = "0?:3?"
+Output: "09:39"
+```
+
+**Example 3:**
+
+```
+Input: time = "1?:22"
+Output: "19:22"
+```
+
+**Constraints:**
+
+- `time` is in the format `hh:mm`.
+- It is guaranteed that you can produce a valid time from the given string.
+
+## 题目大意
+
+给你一个字符串 time ，格式为 hh:mm（小时：分钟），其中某几位数字被隐藏（用 ? 表示）。有效的时间为 00:00 到 23:59 之间的所有时间，包括 00:00 和 23:59 。替换 time 中隐藏的数字，返回你可以得到的最晚有效时间。
+
+## 解题思路
+
+- 简单题。根据题意，需要找到最晚的有效时间。枚举时间 4 个位置即可。如果第 3 个位置是 ？，那么它最晚时间是 5；如果第 4 个位置是 ？，那么它最晚时间是 9；如果第 2 个位置是 ？，那么它最晚时间是 9；如果第 1 个位置是 ？，根据第 2 个位置判断，如果第 2 个位置是大于 3 的数，那么第一个位置最晚时间是 1，如果第 2 个位置是小于 3 的数那么第一个位置最晚时间是 2 。按照上述规则即可还原最晚时间。
+
+## 代码
+
+```go
+package leetcode
+
+func maximumTime(time string) string {
+	timeb := []byte(time)
+	if timeb[3] == '?' {
+		timeb[3] = '5'
+	}
+	if timeb[4] == '?' {
+		timeb[4] = '9'
+	}
+	if timeb[0] == '?' {
+		if int(timeb[1]-'0') > 3 && int(timeb[1]-'0') < 10 {
+			timeb[0] = '1'
+		} else {
+			timeb[0] = '2'
+		}
+	}
+	if timeb[1] == '?' {
+		timeb[1] = '9'
+	}
+	if timeb[0] == '2' && timeb[1] == '9' {
+		timeb[1] = '3'
+	}
+	return string(timeb)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1734.Decode-XORed-Permutation/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1738.Find-Kth-Largest-XOR-Coordinate-Value/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1738.Find-Kth-Largest-XOR-Coordinate-Value.md b/website/content/ChapterFour/1700~1799/1738.Find-Kth-Largest-XOR-Coordinate-Value.md
new file mode 100644
index 000000000..c5d0cc20b
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1738.Find-Kth-Largest-XOR-Coordinate-Value.md
@@ -0,0 +1,118 @@
+# [1738. Find Kth Largest XOR Coordinate Value](https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/)
+
+
+## 题目
+
+You are given a 2D `matrix` of size `m x n`, consisting of non-negative integers. You are also given an integer `k`.
+
+The **value** of coordinate `(a, b)` of the matrix is the XOR of all `matrix[i][j]` where `0 <= i <= a < m` and `0 <= j <= b < n` **(0-indexed)**.
+
+Find the `kth` largest value **(1-indexed)** of all the coordinates of `matrix`.
+
+**Example 1:**
+
+```
+Input: matrix = [[5,2],[1,6]], k = 1
+Output: 7
+Explanation: The value of coordinate (0,1) is 5 XOR 2 = 7, which is the largest value.
+```
+
+**Example 2:**
+
+```
+Input: matrix = [[5,2],[1,6]], k = 2
+Output: 5
+Explanation:The value of coordinate (0,0) is 5 = 5, which is the 2nd largest value.
+```
+
+**Example 3:**
+
+```
+Input: matrix = [[5,2],[1,6]], k = 3
+Output: 4
+Explanation: The value of coordinate (1,0) is 5 XOR 1 = 4, which is the 3rd largest value.
+```
+
+**Example 4:**
+
+```
+Input: matrix = [[5,2],[1,6]], k = 4
+Output: 0
+Explanation: The value of coordinate (1,1) is 5 XOR 2 XOR 1 XOR 6 = 0, which is the 4th largest value.
+```
+
+**Constraints:**
+
+- `m == matrix.length`
+- `n == matrix[i].length`
+- `1 <= m, n <= 1000`
+- `0 <= matrix[i][j] <= 10^6`
+- `1 <= k <= m * n`
+
+## 题目大意
+
+给你一个二维矩阵 matrix 和一个整数 k ，矩阵大小为 m x n 由非负整数组成。矩阵中坐标 (a, b) 的 值 可由对所有满足 0 <= i <= a < m 且 0 <= j <= b < n 的元素 matrix[i][j]（下标从 0 开始计数）执行异或运算得到。请你找出 matrix 的所有坐标中第 k 大的值（k 的值从 1 开始计数）。
+
+## 解题思路
+
+- 区间异或结果类比于区间二维前缀和。只不过需要注意 x^x = 0 这一性质。举例：
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_1738_0_.png)
+
+    通过简单推理，可以得出区间二维前缀和 preSum 的递推式。具体代码见解法二。
+
+- 上面的解法中，preSum 用二维数组计算的。能否再优化空间复杂度，降低成 O(n)？答案是可以的。通过观察可以发现。preSum 可以按照一行一行来生成。先生成 preSum 前一行，下一行生成过程中会用到前一行的信息，异或计算以后，可以覆盖原数据(前一行的信息)，对之后的计算没有影响。这个优化空间复杂度的方法和优化 DP 空间复杂度是完全一样的思路和方法。
+
+    ![](https://img.halfrost.com/Leetcode/leetcode_1738_1_.png)
+
+    具体代码见解法一。
+
+- 计算出了 preSum，还需要考虑如何输出第 k 大的值。有 3 种做法，第一种是排序，第二种是优先队列，第三种是第 215 题中的 O(n) 的 partition 方法。时间复杂度最低的当然是 O(n)。但是经过实际测试，runtime 最优的是排序的方法。所以笔者以下两种方法均采用了排序的方法。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+// 解法一 压缩版的前缀和
+func kthLargestValue(matrix [][]int, k int) int {
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return 0
+	}
+	res, prefixSum := make([]int, 0, len(matrix)*len(matrix[0])), make([]int, len(matrix[0]))
+	for i := range matrix {
+		line := 0
+		for j, v := range matrix[i] {
+			line ^= v
+			prefixSum[j] ^= line
+			res = append(res, prefixSum[j])
+		}
+	}
+	sort.Ints(res)
+	return res[len(res)-k]
+}
+
+// 解法二 前缀和
+func kthLargestValue1(matrix [][]int, k int) int {
+	nums, prefixSum := []int{}, make([][]int, len(matrix)+1)
+	prefixSum[0] = make([]int, len(matrix[0])+1)
+	for i, row := range matrix {
+		prefixSum[i+1] = make([]int, len(matrix[0])+1)
+		for j, val := range row {
+			prefixSum[i+1][j+1] = prefixSum[i+1][j] ^ prefixSum[i][j+1] ^ prefixSum[i][j] ^ val
+			nums = append(nums, prefixSum[i+1][j+1])
+		}
+	}
+	sort.Ints(nums)
+	return nums[len(nums)-k]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1736.Latest-Time-by-Replacing-Hidden-Digits/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1742.Maximum-Number-of-Balls-in-a-Box/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1742.Maximum-Number-of-Balls-in-a-Box.md b/website/content/ChapterFour/1700~1799/1742.Maximum-Number-of-Balls-in-a-Box.md
new file mode 100644
index 000000000..188f72959
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1742.Maximum-Number-of-Balls-in-a-Box.md
@@ -0,0 +1,88 @@
+# [1742. Maximum Number of Balls in a Box](https://leetcode.com/problems/maximum-number-of-balls-in-a-box/)
+
+
+## 题目
+
+You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
+
+Your job at this factory is to put each ball in the box with a number equal to the sum of digits of the ball's number. For example, the ball number `321` will be put in the box number `3 + 2 + 1 = 6` and the ball number `10` will be put in the box number `1 + 0 = 1`.
+
+Given two integers `lowLimit` and `highLimit`, return *the number of balls in the box with the most balls.*
+
+**Example 1:**
+
+```
+Input: lowLimit = 1, highLimit = 10
+Output: 2
+Explanation:
+Box Number:  1 2 3 4 5 6 7 8 9 10 11 ...
+Ball Count:  2 1 1 1 1 1 1 1 1 0  0  ...
+Box 1 has the most number of balls with 2 balls.
+```
+
+**Example 2:**
+
+```
+Input: lowLimit = 5, highLimit = 15
+Output: 2
+Explanation:
+Box Number:  1 2 3 4 5 6 7 8 9 10 11 ...
+Ball Count:  1 1 1 1 2 2 1 1 1 0  0  ...
+Boxes 5 and 6 have the most number of balls with 2 balls in each.
+
+```
+
+**Example 3:**
+
+```
+Input: lowLimit = 19, highLimit = 28
+Output: 2
+Explanation:
+Box Number:  1 2 3 4 5 6 7 8 9 10 11 12 ...
+Ball Count:  0 1 1 1 1 1 1 1 1 2  0  0  ...
+Box 10 has the most number of balls with 2 balls.
+
+```
+
+**Constraints:**
+
+- `1 <= lowLimit <= highLimit <= 10^5`
+
+## 题目大意
+
+你在一家生产小球的玩具厂工作，有 n 个小球，编号从 lowLimit 开始，到 highLimit 结束（包括 lowLimit 和 highLimit ，即 n == highLimit - lowLimit + 1）。另有无限数量的盒子，编号从 1 到 infinity 。你的工作是将每个小球放入盒子中，其中盒子的编号应当等于小球编号上每位数字的和。例如，编号 321 的小球应当放入编号 3 + 2 + 1 = 6 的盒子，而编号 10 的小球应当放入编号 1 + 0 = 1 的盒子。
+
+给你两个整数 lowLimit 和 highLimit ，返回放有最多小球的盒子中的小球数量。如果有多个盒子都满足放有最多小球，只需返回其中任一盒子的小球数量。
+
+## 解题思路
+
+- 简单题。循环遍历一遍数组，依次计算出所有小球的编号各位数字累加和，并且动态维护放有小球最多的数目。循环结束，输出最多小球个数即可。
+
+## 代码
+
+```go
+package leetcode
+
+func countBalls(lowLimit int, highLimit int) int {
+	buckets, maxBall := [46]int{}, 0
+	for i := lowLimit; i <= highLimit; i++ {
+		t := 0
+		for j := i; j > 0; {
+			t += j % 10
+			j = j / 10
+		}
+		buckets[t]++
+		if buckets[t] > maxBall {
+			maxBall = buckets[t]
+		}
+	}
+	return maxBall
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1738.Find-Kth-Largest-XOR-Coordinate-Value/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day.md b/website/content/ChapterFour/1700~1799/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day.md
new file mode 100644
index 000000000..1be8f7864
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day.md
@@ -0,0 +1,95 @@
+# [1744. Can You Eat Your Favorite Candy on Your Favorite Day?](https://leetcode.com/problems/can-you-eat-your-favorite-candy-on-your-favorite-day/)
+
+## 题目
+
+You are given a **(0-indexed)** array of positive integers `candiesCount` where `candiesCount[i]` represents the number of candies of the `ith` type you have. You are also given a 2D array `queries` where `queries[i] = [favoriteTypei, favoriteDayi, dailyCapi]`.
+
+You play a game with the following rules:
+
+- You start eating candies on day **`0`**.
+- You **cannot** eat **any** candy of type `i` unless you have eaten **all** candies of type `i - 1`.
+- You must eat **at least** **one** candy per day until you have eaten all the candies.
+
+Construct a boolean array `answer` such that `answer.length == queries.length` and `answer[i]` is `true` if you can eat a candy of type `favoriteTypei` on day `favoriteDayi` without eating **more than** `dailyCapi` candies on **any** day, and `false` otherwise. Note that you can eat different types of candy on the same day, provided that you follow rule 2.
+
+Return *the constructed array* `answer`.
+
+**Example 1:**
+
+```
+Input: candiesCount = [7,4,5,3,8], queries = [[0,2,2],[4,2,4],[2,13,1000000000]]
+Output: [true,false,true]
+Explanation:
+1- If you eat 2 candies (type 0) on day 0 and 2 candies (type 0) on day 1, you will eat a candy of type 0 on day 2.
+2- You can eat at most 4 candies each day.
+   If you eat 4 candies every day, you will eat 4 candies (type 0) on day 0 and 4 candies (type 0 and type 1) on day 1.
+   On day 2, you can only eat 4 candies (type 1 and type 2), so you cannot eat a candy of type 4 on day 2.
+3- If you eat 1 candy each day, you will eat a candy of type 2 on day 13.
+```
+
+**Example 2:**
+
+```
+Input: candiesCount = [5,2,6,4,1], queries = [[3,1,2],[4,10,3],[3,10,100],[4,100,30],[1,3,1]]
+Output: [false,true,true,false,false]
+```
+
+**Constraints:**
+
+- `1 <= candiesCount.length <= 105`
+- `1 <= candiesCount[i] <= 105`
+- `1 <= queries.length <= 105`
+- `queries[i].length == 3`
+- `0 <= favoriteTypei < candiesCount.length`
+- `0 <= favoriteDayi <= 109`
+- `1 <= dailyCapi <= 109`
+
+## 题目大意
+
+给你一个下标从 0 开始的正整数数组 candiesCount ，其中 candiesCount[i] 表示你拥有的第 i 类糖果的数目。同时给你一个二维数组 queries ，其中 queries[i] = [favoriteTypei, favoriteDayi, dailyCapi] 。你按照如下规则进行一场游戏：
+
+- 你从第 0 天开始吃糖果。
+- 你在吃完 所有 第 i - 1 类糖果之前，不能 吃任何一颗第 i 类糖果。
+- 在吃完所有糖果之前，你必须每天 至少 吃 一颗 糖果。
+
+请你构建一个布尔型数组 answer ，满足 answer.length == queries.length 。answer[i] 为 true 的条件是：在每天吃 不超过 dailyCapi 颗糖果的前提下，你可以在第 favoriteDayi 天吃到第 favoriteTypei 类糖果；否则 answer[i] 为 false 。注意，只要满足上面 3 条规则中的第二条规则，你就可以在同一天吃不同类型的糖果。请你返回得到的数组 answer 。
+
+## 解题思路
+
+- 每天吃糖个数的下限是 1 颗，上限是 dailyCap。针对每一个 query 查询在第 i 天能否吃到 i 类型的糖果，要想吃到 i 类型的糖果，必须吃完 i-1 类型的糖果。意味着在 [favoriteDayi + 1, (favoriteDayi+1)×dailyCapi] 区间内能否包含一颗第 favoriteTypei 类型的糖果。如果能包含则输出 true，不能包含则输出 false。吃的糖果数是累积的，所以这里利用前缀和计算出累积吃糖果数所在区间 [sum[favoriteTypei−1]+1, sum[favoriteTypei]]。最后判断 query 区间和累积吃糖果数的区间是否有重叠即可。如果重叠即输出 true。
+- 判断两个区间是否重合，情况有好几种：内包含，全包含，半包含等等。没有交集的情况比较少，所以可以用排除法。对于区间 [x1, y1] 以及 [x2, y2]，它们没有交集当且仅当 x1 > y2 或者 y1 < x2。
+
+## 代码
+
+```go
+package leetcode
+
+func canEat(candiesCount []int, queries [][]int) []bool {
+	n := len(candiesCount)
+	prefixSum := make([]int, n)
+	prefixSum[0] = candiesCount[0]
+	for i := 1; i < n; i++ {
+		prefixSum[i] = prefixSum[i-1] + candiesCount[i]
+	}
+	res := make([]bool, len(queries))
+	for i, q := range queries {
+		favoriteType, favoriteDay, dailyCap := q[0], q[1], q[2]
+		x1 := favoriteDay + 1
+		y1 := (favoriteDay + 1) * dailyCap
+		x2 := 1
+		if favoriteType > 0 {
+			x2 = prefixSum[favoriteType-1] + 1
+		}
+		y2 := prefixSum[favoriteType]
+		res[i] = !(x1 > y2 || y1 < x2)
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1742.Maximum-Number-of-Balls-in-a-Box/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1748.Sum-of-Unique-Elements/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1748.Sum-of-Unique-Elements.md b/website/content/ChapterFour/1700~1799/1748.Sum-of-Unique-Elements.md
new file mode 100644
index 000000000..27fade95e
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1748.Sum-of-Unique-Elements.md
@@ -0,0 +1,74 @@
+# [1748. Sum of Unique Elements](https://leetcode.com/problems/sum-of-unique-elements/)
+
+
+## 题目
+
+You are given an integer array `nums`. The unique elements of an array are the elements that appear **exactly once** in the array.
+
+Return *the **sum** of all the unique elements of* `nums`.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3,2]
+Output: 4
+Explanation: The unique elements are [1,3], and the sum is 4.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,1,1,1,1]
+Output: 0
+Explanation: There are no unique elements, and the sum is 0.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,2,3,4,5]
+Output: 15
+Explanation: The unique elements are [1,2,3,4,5], and the sum is 15.
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `1 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个整数数组 `nums` 。数组中唯一元素是那些只出现 **恰好一次** 的元素。请你返回 `nums` 中唯一元素的 **和** 。
+
+## 解题思路
+
+- 简单题。利用 map 统计出每个元素出现的频次。再累加所有频次为 1 的元素，最后输出累加和即可。
+
+## 代码
+
+```go
+package leetcode
+
+func sumOfUnique(nums []int) int {
+	freq, res := make(map[int]int), 0
+	for _, v := range nums {
+		if _, ok := freq[v]; !ok {
+			freq[v] = 0
+		}
+		freq[v]++
+	}
+	for k, v := range freq {
+		if v == 1 {
+			res += k
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1752.Check-if-Array-Is-Sorted-and-Rotated/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1752.Check-if-Array-Is-Sorted-and-Rotated.md b/website/content/ChapterFour/1700~1799/1752.Check-if-Array-Is-Sorted-and-Rotated.md
new file mode 100644
index 000000000..b72e4ff6e
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1752.Check-if-Array-Is-Sorted-and-Rotated.md
@@ -0,0 +1,93 @@
+# [1752. Check if Array Is Sorted and Rotated](https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/)
+
+
+## 题目
+
+Given an array `nums`, return `true` *if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)*. Otherwise, return `false`.
+
+There may be **duplicates** in the original array.
+
+**Note:** An array `A` rotated by `x` positions results in an array `B` of the same length such that `A[i] == B[(i+x) % A.length]`, where `%` is the modulo operation.
+
+**Example 1:**
+
+```
+Input: nums = [3,4,5,1,2]
+Output: true
+Explanation: [1,2,3,4,5] is the original sorted array.
+You can rotate the array by x = 3 positions to begin on the the element of value 3: [3,4,5,1,2].
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,1,3,4]
+Output: false
+Explanation: There is no sorted array once rotated that can make nums.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,2,3]
+Output: true
+Explanation: [1,2,3] is the original sorted array.
+You can rotate the array by x = 0 positions (i.e. no rotation) to make nums.
+```
+
+**Example 4:**
+
+```
+Input: nums = [1,1,1]
+Output: true
+Explanation: [1,1,1] is the original sorted array.
+You can rotate any number of positions to make nums.
+```
+
+**Example 5:**
+
+```
+Input: nums = [2,1]
+Output: true
+Explanation: [1,2] is the original sorted array.
+You can rotate the array by x = 5 positions to begin on the element of value 2: [2,1].
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `1 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个数组 nums 。nums 的源数组中，所有元素与 nums 相同，但按非递减顺序排列。如果 nums 能够由源数组轮转若干位置（包括 0 个位置）得到，则返回 true ；否则，返回 false 。源数组中可能存在 重复项 。
+
+## 解题思路
+
+- 简单题。从头扫描一遍数组，找出相邻两个元素递减的数对。如果递减的数对只有 1 个，则有可能是轮转得来的，超过 1 个，则返回 false。题干里面还提到可能有多个重复元素，针对这一情况还需要判断一下 `nums[0]` 和 `nums[len(nums)-1]` 。如果是相同元素，`nums[0] < nums[len(nums)-1]`，并且数组中间还存在一对递减的数对，这时候也是 false。判断好上述这 2 种情况，本题得解。
+
+## 代码
+
+```go
+package leetcode
+
+func check(nums []int) bool {
+	count := 0
+	for i := 0; i < len(nums)-1; i++ {
+		if nums[i] > nums[i+1] {
+			count++
+			if count > 1 || nums[0] < nums[len(nums)-1] {
+				return false
+			}
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1748.Sum-of-Unique-Elements/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1758.Minimum-Changes-To-Make-Alternating-Binary-String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1758.Minimum-Changes-To-Make-Alternating-Binary-String.md b/website/content/ChapterFour/1700~1799/1758.Minimum-Changes-To-Make-Alternating-Binary-String.md
new file mode 100644
index 000000000..7e06a4478
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1758.Minimum-Changes-To-Make-Alternating-Binary-String.md
@@ -0,0 +1,77 @@
+# [1758. Minimum Changes To Make Alternating Binary String](https://leetcode.com/problems/minimum-changes-to-make-alternating-binary-string/)
+
+
+## 题目
+
+You are given a string `s` consisting only of the characters `'0'` and `'1'`. In one operation, you can change any `'0'` to `'1'` or vice versa.
+
+The string is called alternating if no two adjacent characters are equal. For example, the string `"010"` is alternating, while the string `"0100"` is not.
+
+Return *the **minimum** number of operations needed to make* `s` *alternating*.
+
+**Example 1:**
+
+```
+Input: s = "0100"
+Output: 1
+Explanation: If you change the last character to '1', s will be "0101", which is alternating.
+```
+
+**Example 2:**
+
+```
+Input: s = "10"
+Output: 0
+Explanation: s is already alternating.
+```
+
+**Example 3:**
+
+```
+Input: s = "1111"
+Output: 2
+Explanation: You need two operations to reach "0101" or "1010".
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 104`
+- `s[i]` is either `'0'` or `'1'`.
+
+## 题目大意
+
+你将得到一个仅包含字符“ 0”和“ 1”的字符串 `s`。 在一项操作中，你可以将任何 `'0'` 更改为 `'1'`，反之亦然。 如果两个相邻字符都不相等，则该字符串称为交替字符串。 例如，字符串“ 010”是交替的，而字符串“ 0100”则不是。 返回使 `s` 交替所需的最小操作数。
+
+## 解题思路
+
+- 简单题。利用数组下标奇偶交替性来判断交替字符串。交替字符串有 2 种，一个是 `'01010101……'` 还有一个是 `'1010101010……'`，这两个只需要计算出一个即可，另外一个利用 `len(s) - res` 就是答案。
+
+## 代码
+
+```go
+package leetcode
+
+func minOperations(s string) int {
+	res := 0
+	for i := 0; i < len(s); i++ {
+		if int(s[i]-'0') != i%2 {
+			res++
+		}
+	}
+	return min(res, len(s)-res)
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1752.Check-if-Array-Is-Sorted-and-Rotated/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1763.Longest-Nice-Substring/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1763.Longest-Nice-Substring.md b/website/content/ChapterFour/1700~1799/1763.Longest-Nice-Substring.md
new file mode 100644
index 000000000..4b574b199
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1763.Longest-Nice-Substring.md
@@ -0,0 +1,147 @@
+# [1763. Longest Nice Substring](https://leetcode.com/problems/longest-nice-substring/)
+
+
+## 题目
+
+A string `s` is **nice** if, for every letter of the alphabet that `s` contains, it appears **both** in uppercase and lowercase. For example, `"abABB"` is nice because `'A'` and `'a'` appear, and `'B'` and `'b'` appear. However, `"abA"` is not because `'b'` appears, but `'B'` does not.
+
+Given a string `s`, return *the longest **substring** of `s` that is **nice**. If there are multiple, return the substring of the **earliest** occurrence. If there are none, return an empty string*.
+
+**Example 1:**
+
+```
+Input: s = "YazaAay"
+Output: "aAa"
+Explanation:"aAa" is a nice string because 'A/a' is the only letter of the alphabet in s, and both 'A' and 'a' appear.
+"aAa" is the longest nice substring.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "Bb"
+Output: "Bb"
+Explanation: "Bb" is a nice string because both 'B' and 'b' appear. The whole string is a substring.
+
+```
+
+**Example 3:**
+
+```
+Input: s = "c"
+Output: ""
+Explanation: There are no nice substrings.
+
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 100`
+- `s` consists of uppercase and lowercase English letters.
+
+## 题目大意
+
+当一个字符串 s 包含的每一种字母的大写和小写形式 同时 出现在 s 中，就称这个字符串 s 是 美好 字符串。比方说，"abABB" 是美好字符串，因为 'A' 和 'a' 同时出现了，且 'B' 和 'b' 也同时出现了。然而，"abA" 不是美好字符串因为 'b' 出现了，而 'B' 没有出现。
+
+给你一个字符串 s ，请你返回 s 最长的 美好子字符串 。如果有多个答案，请你返回 最早 出现的一个。如果不存在美好子字符串，请你返回一个空字符串。
+
+## 解题思路
+
+- 解法一，暴力解法。枚举每一段字符串，判断这个子字符串内是否满足美好字符串的定义，即字母的大小写是否同时出现。
+- 解法二，这个解法是解法一的小幅优化版，利用二进制记录状态。先构造二进制状态串，再利用直接比较这个二进制串。
+- 解法三，分治。以 `i` 为分割点依次切开字符串。左右两个字符串分别判断是否满足美好字符串的定义。左右分开的字符串还可以继续划分。直至分到一个字母为止。在这个过程中记录最早出现的字符串。
+
+## 代码
+
+```go
+package leetcode
+
+import "unicode"
+
+// 解法一 分治，时间复杂度 O(n)
+func longestNiceSubstring(s string) string {
+	if len(s) < 2 {
+		return ""
+	}
+
+	chars := map[rune]int{}
+	for _, r := range s {
+		chars[r]++
+	}
+
+	for i := 0; i < len(s); i++ {
+		r := rune(s[i])
+		_, u := chars[unicode.ToUpper(r)]
+		_, l := chars[unicode.ToLower(r)]
+		if u && l {
+			continue
+		}
+		left := longestNiceSubstring(s[:i])
+		right := longestNiceSubstring(s[i+1:])
+		if len(left) >= len(right) {
+			return left
+		} else {
+			return right
+		}
+	}
+	return s
+}
+
+// 解法二 用二进制表示状态
+func longestNiceSubstring1(s string) (ans string) {
+	for i := range s {
+		lower, upper := 0, 0
+		for j := i; j < len(s); j++ {
+			if unicode.IsLower(rune(s[j])) {
+				lower |= 1 << (s[j] - 'a')
+			} else {
+				upper |= 1 << (s[j] - 'A')
+			}
+			if lower == upper && j-i+1 > len(ans) {
+				ans = s[i : j+1]
+			}
+		}
+	}
+	return
+}
+
+// 解法三 暴力枚举，时间复杂度 O(n^2)
+func longestNiceSubstring2(s string) string {
+	res := ""
+	for i := 0; i < len(s); i++ {
+		m := map[byte]int{}
+		m[s[i]]++
+		for j := i + 1; j < len(s); j++ {
+			m[s[j]]++
+			if checkNiceString(m) && (j-i+1 > len(res)) {
+				res = s[i : j+1]
+			}
+		}
+	}
+	return res
+}
+
+func checkNiceString(m map[byte]int) bool {
+	for k := range m {
+		if k >= 97 && k <= 122 {
+			if _, ok := m[k-32]; !ok {
+				return false
+			}
+		}
+		if k >= 65 && k <= 90 {
+			if _, ok := m[k+32]; !ok {
+				return false
+			}
+		}
+	}
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1758.Minimum-Changes-To-Make-Alternating-Binary-String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1791.Find-Center-of-Star-Graph/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/1791.Find-Center-of-Star-Graph.md b/website/content/ChapterFour/1700~1799/1791.Find-Center-of-Star-Graph.md
new file mode 100644
index 000000000..6e91c8abc
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/1791.Find-Center-of-Star-Graph.md
@@ -0,0 +1,59 @@
+# [1791.Find Center of Star Graph](https://leetcode.com/problems/find-center-of-star-graph/)
+
+## 题目
+
+There is an undirected star graph consisting of n nodes labeled from 1 to n. A star graph is a graph where there is one center node and exactly n - 1 edges that connect the center node with every other node.
+
+You are given a 2D integer array edges where each edges[i] = [ui, vi] indicates that there is an edge between the nodes ui and vi. Return the center of the given star graph.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/02/24/star_graph.png:w](https://assets.leetcode.com/uploads/2021/02/24/star_graph.png)
+
+    Input: edges = [[1,2],[2,3],[4,2]]
+    Output: 2
+    Explanation: As shown in the figure above, node 2 is connected to every other node, so 2 is the center.
+
+**Example 2:**
+
+    Input: edges = [[1,2],[5,1],[1,3],[1,4]]
+    Output: 1
+
+**Constraints:**
+
+- 3 <= n <= 100000
+- edges.length == n - 1
+- edges[i].length == 2
+- 1 <= ui, vi <= n
+- ui != vi
+- The given edges represent a valid star graph.
+
+## 题目大意
+
+有一个无向的 星型 图，由 n 个编号从 1 到 n 的节点组成。星型图有一个 中心 节点，并且恰有 n - 1 条边将中心节点与其他每个节点连接起来。
+
+给你一个二维整数数组 edges ，其中 edges[i] = [ui, vi] 表示在节点 ui 和 vi 之间存在一条边。请你找出并返回 edges 所表示星型图的中心节点。
+
+## 解题思路
+
+- 求出edges中前两个元素的共同值，即是中心节点
+
+## 代码
+
+```go
+package leetcode
+
+func findCenter(edges [][]int) int {
+	if edges[0][0] == edges[1][0] || edges[0][0] == edges[1][1] {
+		return edges[0][0]
+	}
+	return edges[0][1]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1763.Longest-Nice-Substring/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1800~1899/1816.Truncate-Sentence/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1700~1799/_index.md b/website/content/ChapterFour/1700~1799/_index.md
new file mode 100644
index 000000000..e954f0877
--- /dev/null
+++ b/website/content/ChapterFour/1700~1799/_index.md
@@ -0,0 +1,4 @@
+---
+bookCollapseSection: true
+weight: 20
+---
diff --git a/website/content/ChapterFour/1800~1899/1816.Truncate-Sentence.md b/website/content/ChapterFour/1800~1899/1816.Truncate-Sentence.md
new file mode 100644
index 000000000..74b52d890
--- /dev/null
+++ b/website/content/ChapterFour/1800~1899/1816.Truncate-Sentence.md
@@ -0,0 +1,83 @@
+# [1816. Truncate Sentence](https://leetcode.com/problems/truncate-sentence/)
+
+## 题目
+
+A sentence is a list of words that are separated by a single space with no leading or trailing spaces. Each of the words consists of only uppercase and lowercase English letters (no punctuation).
+
+- For example, "Hello World", "HELLO", and "hello world hello world" are all sentences.
+
+You are given a sentence s and an integer k. You want to truncate s such that it contains only the first k words. Return s after truncating it.
+
+**Example 1**:
+
+    Input: s = "Hello how are you Contestant", k = 4
+    Output: "Hello how are you"
+    Explanation:
+    The words in s are ["Hello", "how" "are", "you", "Contestant"].
+    The first 4 words are ["Hello", "how", "are", "you"].
+    Hence, you should return "Hello how are you".
+
+**Example 2**:
+
+    Input: s = "What is the solution to this problem", k = 4
+    Output: "What is the solution"
+    Explanation:
+    The words in s are ["What", "is" "the", "solution", "to", "this", "problem"].
+    The first 4 words are ["What", "is", "the", "solution"].
+    Hence, you should return "What is the solution".
+
+**Example 3**:
+
+    Input: s = "chopper is not a tanuki", k = 5
+    Output: "chopper is not a tanuki"
+
+**Constraints:**
+
+- 1 <= s.length <= 500
+- k is in the range [1, the number of words in s].
+- s consist of only lowercase and uppercase English letters and spaces.
+- The words in s are separated by a single space.
+- There are no leading or trailing spaces.
+
+## 题目大意
+
+句子 是一个单词列表，列表中的单词之间用单个空格隔开，且不存在前导或尾随空格。每个单词仅由大小写英文字母组成（不含标点符号）。
+
+- 例如，"Hello World"、"HELLO" 和 "hello world hello world" 都是句子。
+
+给你一个句子 s 和一个整数 k ，请你将 s 截断使截断后的句子仅含前 k 个单词。返回截断 s 后得到的句子。
+
+## 解题思路
+
+- 遍历字符串 s，找到最后一个空格的下标 end
+- 如果 end 为 0，直接返回 s,否则返回 s[:end]
+
+## 代码
+
+```go
+package leetcode
+
+func truncateSentence(s string, k int) string {
+	end := 0
+	for i := range s {
+		if k > 0 && s[i] == ' ' {
+			k--
+		}
+		if k == 0 {
+			end = i
+			break
+		}
+	}
+	if end == 0 {
+		return s
+	}
+	return s[:end]
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1700~1799/1791.Find-Center-of-Star-Graph/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1800~1899/1818.Minimum-Absolute-Sum-Difference/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1800~1899/1818.Minimum-Absolute-Sum-Difference.md b/website/content/ChapterFour/1800~1899/1818.Minimum-Absolute-Sum-Difference.md
new file mode 100644
index 000000000..204e09c1a
--- /dev/null
+++ b/website/content/ChapterFour/1800~1899/1818.Minimum-Absolute-Sum-Difference.md
@@ -0,0 +1,115 @@
+# [1818. Minimum Absolute Sum Difference](https://leetcode.com/problems/minimum-absolute-sum-difference/)
+
+## 题目
+
+You are given two positive integer arrays `nums1` and `nums2`, both of length `n`.
+
+The **absolute sum difference** of arrays `nums1` and `nums2` is defined as the **sum** of `|nums1[i] - nums2[i]|` for each `0 <= i < n` (**0-indexed**).
+
+You can replace **at most one** element of `nums1` with **any** other element in `nums1` to **minimize** the absolute sum difference.
+
+Return the *minimum absolute sum difference **after** replacing at most one ****element in the array `nums1`.* Since the answer may be large, return it **modulo** `109 + 7`.
+
+`|x|` is defined as:
+
+- `x` if `x >= 0`, or
+- `x` if `x < 0`.
+
+**Example 1:**
+
+```
+Input: nums1 = [1,7,5], nums2 = [2,3,5]
+Output: 3
+Explanation:There are two possible optimal solutions:
+- Replace the second element with the first: [1,7,5] => [1,1,5], or
+- Replace the second element with the third: [1,7,5] => [1,5,5].
+Both will yield an absolute sum difference of|1-2| + (|1-3| or |5-3|) + |5-5| =3.
+
+```
+
+**Example 2:**
+
+```
+Input: nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10]
+Output: 0
+Explanation:nums1 is equal to nums2 so no replacement is needed. This will result in an
+absolute sum difference of 0.
+
+```
+
+**Example 3:**
+
+```
+Input: nums1 = [1,10,4,4,2,7], nums2 = [9,3,5,1,7,4]
+Output: 20
+Explanation:Replace the first element with the second: [1,10,4,4,2,7] => [10,10,4,4,2,7].
+This yields an absolute sum difference of|10-9| + |10-3| + |4-5| + |4-1| + |2-7| + |7-4| = 20
+```
+
+**Constraints:**
+
+- `n == nums1.length`
+- `n == nums2.length`
+- `1 <= n <= 10^5`
+- `1 <= nums1[i], nums2[i] <= 10^5`
+
+## 题目大意
+
+给你两个正整数数组 nums1 和 nums2 ，数组的长度都是 n 。数组 nums1 和 nums2 的 绝对差值和 定义为所有 |nums1[i] - nums2[i]|（0 <= i < n）的 总和（下标从 0 开始）。你可以选用 nums1 中的 任意一个 元素来替换 nums1 中的 至多 一个元素，以 最小化 绝对差值和。在替换数组 nums1 中最多一个元素 之后 ，返回最小绝对差值和。因为答案可能很大，所以需要对 10^9 + 7 取余 后返回。
+
+## 解题思路
+
+- 如果不改变任何元素，绝对差值和为 {{< katex >}} \sum \left | nums1[i] - nums2[i] \right | {{< /katex >}}。如果改变一个元素后，那么绝对差值和为 
+  {{< katex display>}} \begin{aligned}&\sum \left | nums1[i] - nums2[i] \right | - \left ( \left | nums1[i] - nums2[i] \right | - \left | nums1[j] - nums2[i] \right |\right )\\= &\sum \left | nums1[i] - nums2[i] \right | - \Delta \end{aligned} {{< /katex >}}
+  
+  题目要求返回最小绝对差值和，即求 {{< katex >}}\Delta {{< /katex >}} 的最大值。暴力枚举 nums1 和 nums2 中两两差值，找到 maxdiff，即 {{< katex >}}\Delta {{< /katex >}} 的最大值，此题得到解。
+
+## 代码
+
+```go
+package leetcode
+
+func minAbsoluteSumDiff(nums1 []int, nums2 []int) int {
+	diff := 0
+	maxDiff := 0
+	for i, n2 := range nums2 {
+		d := abs(nums1[i] - n2)
+		diff += d
+		if maxDiff < d {
+			t := 100001
+			for _, n1 := range nums1 {
+				maxDiff = max(maxDiff, d-min(t, abs(n1-n2)))
+			}
+		}
+	}
+	return (diff - maxDiff) % (1e9 + 7)
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func abs(a int) int {
+	if a > 0 {
+		return a
+	}
+	return -a
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1800~1899/1816.Truncate-Sentence/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1800~1899/1846.Maximum-Element-After-Decreasing-and-Rearranging/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1800~1899/1846.Maximum-Element-After-Decreasing-and-Rearranging.md b/website/content/ChapterFour/1800~1899/1846.Maximum-Element-After-Decreasing-and-Rearranging.md
new file mode 100644
index 000000000..8b68e57ac
--- /dev/null
+++ b/website/content/ChapterFour/1800~1899/1846.Maximum-Element-After-Decreasing-and-Rearranging.md
@@ -0,0 +1,110 @@
+# [1846. Maximum Element After Decreasing and Rearranging](https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/)
+
+
+## 题目
+
+You are given an array of positive integers `arr`. Perform some operations (possibly none) on `arr` so that it satisfies these conditions:
+
+- The value of the **first** element in `arr` must be `1`.
+- The absolute difference between any 2 adjacent elements must be **less than or equal to** `1`. In other words, `abs(arr[i] - arr[i - 1]) <= 1` for each `i` where `1 <= i < arr.length` (**0-indexed**). `abs(x)` is the absolute value of `x`.
+
+There are 2 types of operations that you can perform any number of times:
+
+- **Decrease** the value of any element of `arr` to a **smaller positive integer**.
+- **Rearrange** the elements of `arr` to be in any order.
+
+Return *the **maximum** possible value of an element in* `arr` *after performing the operations to satisfy the conditions*.
+
+**Example 1:**
+
+```
+Input: arr = [2,2,1,2,1]
+Output: 2
+Explanation:
+We can satisfy the conditions by rearrangingarr so it becomes[1,2,2,2,1].
+The largest element inarr is 2.
+
+```
+
+**Example 2:**
+
+```
+Input: arr = [100,1,1000]
+Output: 3
+Explanation:
+One possible way to satisfy the conditions is by doing the following:
+1. Rearrangearr so it becomes[1,100,1000].
+2. Decrease the value of the second element to 2.
+3. Decrease the value of the third element to 3.
+Nowarr = [1,2,3], whichsatisfies the conditions.
+The largest element inarr is 3.
+```
+
+**Example 3:**
+
+```
+Input: arr = [1,2,3,4,5]
+Output: 5
+Explanation: The array already satisfies the conditions, and the largest element is 5.
+
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 10^5`
+- `1 <= arr[i] <= 10^9`
+
+## 题目大意
+
+给你一个正整数数组 arr 。请你对 arr 执行一些操作（也可以不进行任何操作），使得数组满足以下条件：
+
+- arr 中 第一个 元素必须为 1 。
+- 任意相邻两个元素的差的绝对值 小于等于 1 ，也就是说，对于任意的 1 <= i < arr.length （数组下标从 0 开始），都满足 abs(arr[i] - arr[i - 1]) <= 1 。abs(x) 为 x 的绝对值。
+
+你可以执行以下 2 种操作任意次：
+
+- 减小 arr 中任意元素的值，使其变为一个 更小的正整数 。
+- 重新排列 arr 中的元素，你可以以任意顺序重新排列。
+
+请你返回执行以上操作后，在满足前文所述的条件下，arr 中可能的 最大值 。
+
+## 解题思路
+
+- 正整数数组 arr 第一个元素必须为 1，且两两元素绝对值小于等于 1，那么 arr 最大值肯定不大于 n。采用贪心的策略，先统计所有元素出现的次数，大于 n 的元素出现次数都累加到 n 上。然后从 1 扫描到 n，遇到“空隙”（出现次数为 0 的元素），便将最近一个出现次数大于 1 的元素“挪”过来填补“空隙”。题目所求最大值出现在，“填补空隙”之后，数组从左往右连续的最右端。
+
+## 代码
+
+```go
+package leetcode
+
+func maximumElementAfterDecrementingAndRearranging(arr []int) int {
+	n := len(arr)
+	count := make([]int, n+1)
+	for _, v := range arr {
+		count[min(v, n)]++
+	}
+	miss := 0
+	for _, c := range count[1:] {
+		if c == 0 {
+			miss++
+		} else {
+			miss -= min(c-1, miss)
+		}
+	}
+	return n - miss
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1800~1899/1818.Minimum-Absolute-Sum-Difference/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1800~1899/1877.Minimize-Maximum-Pair-Sum-in-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1800~1899/1877.Minimize-Maximum-Pair-Sum-in-Array.md b/website/content/ChapterFour/1800~1899/1877.Minimize-Maximum-Pair-Sum-in-Array.md
new file mode 100644
index 000000000..26bc64c66
--- /dev/null
+++ b/website/content/ChapterFour/1800~1899/1877.Minimize-Maximum-Pair-Sum-in-Array.md
@@ -0,0 +1,87 @@
+# [1877. Minimize Maximum Pair Sum in Array](https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/)
+
+
+## 题目
+
+The **pair sum** of a pair `(a,b)` is equal to `a + b`. The **maximum pair sum** is the largest **pair sum** in a list of pairs.
+
+- For example, if we have pairs `(1,5)`, `(2,3)`, and `(4,4)`, the **maximum pair sum** would be `max(1+5, 2+3, 4+4) = max(6, 5, 8) = 8`.
+
+Given an array `nums` of **even** length `n`, pair up the elements of `nums` into `n / 2` pairs such that:
+
+- Each element of `nums` is in **exactly one** pair, and
+- The **maximum pair sum** is **minimized**.
+
+Return *the minimized **maximum pair sum** after optimally pairing up the elements*.
+
+**Example 1:**
+
+```
+Input: nums = [3,5,2,3]
+Output: 7
+Explanation: The elements can be paired up into pairs (3,3) and (5,2).
+The maximum pair sum is max(3+3, 5+2) = max(6, 7) = 7.
+```
+
+**Example 2:**
+
+```
+Input: nums = [3,5,4,2,4,6]
+Output: 8
+Explanation: The elements can be paired up into pairs (3,5), (4,4), and (6,2).
+The maximum pair sum is max(3+5, 4+4, 6+2) = max(8, 8, 8) = 8.
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `2 <= n <= 105`
+- `n` is **even**.
+- `1 <= nums[i] <= 105`
+
+## 题目大意
+
+一个数对 (a,b) 的 **数对和** 等于 a + b 。**最大数对和** 是一个数对数组中最大的 数对和 。
+
+- 比方说，如果我们有数对 (1,5) ，(2,3) 和 (4,4)，**最大数对和** 为 max(1+5, 2+3, 4+4) = max(6, 5, 8) = 8 。
+
+给你一个长度为 **偶数** n 的数组 nums ，请你将 nums 中的元素分成 n / 2 个数对，使得：
+
+- nums 中每个元素 **恰好** 在 一个 数对中，且
+- **最大数对和** 的值 **最小** 。
+
+请你在最优数对划分的方案下，返回最小的 最大数对和 。
+
+## 解题思路
+
+- 要想最大数对和最小，那么最大的元素一定只能和最小的元素组合在一起，不然一定不是最小。当最大元素和最小元素组合在一起了，剩下的次最大元素也应该和次最小元素组合在一起。按照这个思路，先将数组从小到大排序，然后依次取出首尾元素，两两组合在一起。输出这些数对的最大值即为所求。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func minPairSum(nums []int) int {
+	sort.Ints(nums)
+	n, res := len(nums), 0
+	for i, val := range nums[:n/2] {
+		res = max(res, val+nums[n-1-i])
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1800~1899/1846.Maximum-Element-After-Decreasing-and-Rearranging/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1900~1999/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1800~1899/_index.md b/website/content/ChapterFour/1800~1899/_index.md
new file mode 100644
index 000000000..e954f0877
--- /dev/null
+++ b/website/content/ChapterFour/1800~1899/_index.md
@@ -0,0 +1,4 @@
+---
+bookCollapseSection: true
+weight: 20
+---
diff --git a/website/content/ChapterFour/1900~1999/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores.md b/website/content/ChapterFour/1900~1999/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores.md
new file mode 100644
index 000000000..31ce16d63
--- /dev/null
+++ b/website/content/ChapterFour/1900~1999/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores.md
@@ -0,0 +1,78 @@
+# [1984. Minimum Difference Between Highest and Lowest of K Scores](https://leetcode.com/problems/minimum-difference-between-highest-and-lowest-of-k-scores/)
+
+## 题目
+
+You are given a 0-indexed integer array nums, where nums[i] represents the score of the ith student. You are also given an integer k.
+
+Pick the scores of any k students from the array so that the difference between the highest and the lowest of the k scores is minimized.
+
+Return the minimum possible difference.
+
+**Example 1:**
+
+    Input: nums = [90], k = 1
+    Output: 0
+    Explanation: There is one way to pick score(s) of one student:
+    - [90]. The difference between the highest and lowest score is 90 - 90 = 0.
+    The minimum possible difference is 0.
+
+**Example 2:**
+
+    Input: nums = [9,4,1,7], k = 2
+    Output: 2
+    Explanation: There are six ways to pick score(s) of two students:
+    - [9,4,1,7]. The difference between the highest and lowest score is 9 - 4 = 5.
+    - [9,4,1,7]. The difference between the highest and lowest score is 9 - 1 = 8.
+    - [9,4,1,7]. The difference between the highest and lowest score is 9 - 7 = 2.
+    - [9,4,1,7]. The difference between the highest and lowest score is 4 - 1 = 3.
+    - [9,4,1,7]. The difference between the highest and lowest score is 7 - 4 = 3.
+    - [9,4,1,7]. The difference between the highest and lowest score is 7 - 1 = 6.
+    The minimum possible difference is 2.
+
+**Constraints:**
+
+- 1 <= k <= nums.length <= 1000
+- 0 <= nums[i] <= 100000
+
+## 题目大意
+
+给你一个下标从 0 开始的整数数组 nums ，其中 nums[i] 表示第 i 名学生的分数。另给你一个整数 k 。
+
+从数组中选出任意 k 名学生的分数，使这 k 个分数间最高分和最低分的差值达到最小化 。
+
+返回可能的最小差值 。
+
+## 解题思路
+
+- nums 排序
+- 求出nums[i+k-1] - nums[i]中的最小差值
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func minimumDifference(nums []int, k int) int {
+	sort.Ints(nums)
+	minDiff := 100000 + 1
+	for i := 0; i < len(nums); i++ {
+		if i+k-1 >= len(nums) {
+			break
+		}
+		diff := nums[i+k-1] - nums[i]
+		if diff < minDiff {
+			minDiff = diff
+		}
+	}
+	return minDiff
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1800~1899/1877.Minimize-Maximum-Pair-Sum-in-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2021.Brightest-Position-on-Street/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/1900~1999/_index.md b/website/content/ChapterFour/1900~1999/_index.md
new file mode 100644
index 000000000..e954f0877
--- /dev/null
+++ b/website/content/ChapterFour/1900~1999/_index.md
@@ -0,0 +1,4 @@
+---
+bookCollapseSection: true
+weight: 20
+---
diff --git a/website/content/ChapterFour/2000~2099/2021.Brightest-Position-on-Street.md b/website/content/ChapterFour/2000~2099/2021.Brightest-Position-on-Street.md
new file mode 100644
index 000000000..d223af086
--- /dev/null
+++ b/website/content/ChapterFour/2000~2099/2021.Brightest-Position-on-Street.md
@@ -0,0 +1,116 @@
+# [2021. Brightest Position on Street](https://leetcode.com/problems/brightest-position-on-street/)
+
+
+## 题目
+
+A perfectly straight street is represented by a number line. The street has street lamp(s) on it and is represented by a 2D integer array `lights`. Each `lights[i] = [positioni, rangei]` indicates that there is a street lamp at position `positioni` that lights up the area from `[positioni - rangei, positioni + rangei]` (**inclusive**).
+
+The **brightness** of a position `p` is defined as the number of street lamp that light up the position `p`.
+
+Given `lights`, return *the **brightest** position on the street. If there are multiple brightest positions, return the **smallest** one.*
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/09/28/image-20210928155140-1.png](https://assets.leetcode.com/uploads/2021/09/28/image-20210928155140-1.png)
+
+```
+Input: lights = [[-3,2],[1,2],[3,3]]
+Output: -1
+Explanation:
+The first street lamp lights up the area from [(-3) - 2, (-3) + 2] = [-5, -1].
+The second street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3].
+The third street lamp lights up the area from [3 - 3, 3 + 3] = [0, 6].
+
+Position -1 has a brightness of 2, illuminated by the first and second street light.
+Positions 0, 1, 2, and 3 have a brightness of 2, illuminated by the second and third street light.
+Out of all these positions, -1 is the smallest, so return it.
+
+```
+
+**Example 2:**
+
+```
+Input: lights = [[1,0],[0,1]]
+Output: 1
+Explanation:
+The first street lamp lights up the area from [1 - 0, 1 + 0] = [1, 1].
+The second street lamp lights up the area from [0 - 1, 0 + 1] = [-1, 1].
+
+Position 1 has a brightness of 2, illuminated by the first and second street light.
+Return 1 because it is the brightest position on the street.
+
+```
+
+**Example 3:**
+
+```
+Input: lights = [[1,2]]
+Output: -1
+Explanation:
+The first street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3].
+
+Positions -1, 0, 1, 2, and 3 have a brightness of 1, illuminated by the first street light.
+Out of all these positions, -1 is the smallest, so return it.
+
+```
+
+**Constraints:**
+
+- `1 <= lights.length <= 105`
+- `lights[i].length == 2`
+- `108 <= positioni <= 108`
+- `0 <= rangei <= 108`
+
+## 题目大意
+
+一条完全笔直的街道由一条数字线表示。街道上有路灯，由二维数据表示。每个 `lights[i] = [positioni, rangei]` 表示位置 `i` 处有一盏路灯，灯可以照亮从 `[positioni - rangei, positioni + rangei]` （含）的区域。 位置 `p` 的亮度定义为点亮位置 `p` 的路灯数量。 给定路灯，返回街道上最亮的位置。如果有多个最亮的位置，则返回最小的一个。
+
+## 解题思路
+
+- 先将每个路灯的起始和终点位置计算出来。这样我们得到了一堆坐标点。假设灯照亮的范围是 [A, B]，那么在坐标轴上 A 坐标点处 + 1， B + 1 坐标点处 -1 。这样处理的含义是：坐标点 A 可以被一盏灯照亮，所以它照亮次数加一，坐标点 B + 1 出了灯照亮的范围了，所以照亮次数减一。那么从坐标轴坐标开始扫一遍，每次遇到 + 1 的时候就 + 1，遇到 - 1 的地方就 - 1。如此可以算出某个坐标点处，可以被灯照亮的总次数。
+- 需要注意的点是，题目给的测试数据可能会有单点照亮的情况，即某一盏灯只照亮一个坐标点，灯照范围为 0。同一个坐标点也可能是多个灯的起点。用一个 map 去重坐标点即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+type lightItem struct {
+	index int
+	sign  int
+}
+
+func brightestPosition(lights [][]int) int {
+	lightMap, lightItems := map[int]int{}, []lightItem{}
+	for _, light := range lights {
+		lightMap[light[0]-light[1]] += 1
+		lightMap[light[0]+light[1]+1] -= 1
+	}
+	for k, v := range lightMap {
+		lightItems = append(lightItems, lightItem{index: k, sign: v})
+	}
+	sort.SliceStable(lightItems, func(i, j int) bool {
+		return lightItems[i].index < lightItems[j].index
+	})
+	res, border, tmp := 0, 0, 0
+	for _, v := range lightItems {
+		tmp += v.sign
+		if border < tmp {
+			res = v.index
+			border = tmp
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/1900~1999/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2022.Convert-1D-Array-Into-2D-Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2000~2099/2022.Convert-1D-Array-Into-2D-Array.md b/website/content/ChapterFour/2000~2099/2022.Convert-1D-Array-Into-2D-Array.md
new file mode 100644
index 000000000..29646b4a0
--- /dev/null
+++ b/website/content/ChapterFour/2000~2099/2022.Convert-1D-Array-Into-2D-Array.md
@@ -0,0 +1,84 @@
+# [2022. Convert 1D Array Into 2D Array](https://leetcode.com/problems/convert-1d-array-into-2d-array/)
+
+## 题目
+
+You are given a **0-indexed** 1-dimensional (1D) integer array `original`, and two integers, `m` and `n`. You are tasked with creating a 2-dimensional (2D) array with `m` rows and `n` columns using **all** the elements from `original`.
+
+The elements from indices `0` to `n - 1` (**inclusive**) of `original` should form the first row of the constructed 2D array, the elements from indices `n` to `2 * n - 1` (**inclusive**) should form the second row of the constructed 2D array, and so on.
+
+Return *an* `m x n` *2D array constructed according to the above procedure, or an empty 2D array if it is impossible*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/08/26/image-20210826114243-1.png](https://assets.leetcode.com/uploads/2021/08/26/image-20210826114243-1.png)
+
+```
+Input: original = [1,2,3,4], m = 2, n = 2
+Output: [[1,2],[3,4]]
+Explanation: The constructed 2D array should contain 2 rows and 2 columns.
+The first group of n=2 elements in original, [1,2], becomes the first row in the constructed 2D array.
+The second group of n=2 elements in original, [3,4], becomes the second row in the constructed 2D array.
+
+```
+
+**Example 2:**
+
+```
+Input: original = [1,2,3], m = 1, n = 3
+Output: [[1,2,3]]
+Explanation: The constructed 2D array should contain 1 row and 3 columns.
+Put all three elements in original into the first row of the constructed 2D array.
+
+```
+
+**Example 3:**
+
+```
+Input: original = [1,2], m = 1, n = 1
+Output: []
+Explanation: There are 2 elements in original.
+It is impossible to fit 2 elements in a 1x1 2D array, so return an empty 2D array.
+
+```
+
+**Constraints:**
+
+- `1 <= original.length <= 5 * 104`
+- `1 <= original[i] <= 105`
+- `1 <= m, n <= 4 * 104`
+
+## 题目大意
+
+给你一个下标从 0 开始的一维整数数组 original 和两个整数 m 和  n 。你需要使用 original 中 所有 元素创建一个 m 行 n 列的二维数组。
+
+original 中下标从 0 到 n - 1 （都 包含 ）的元素构成二维数组的第一行，下标从 n 到 2 * n - 1 （都 包含 ）的元素构成二维数组的第二行，依此类推。
+
+请你根据上述过程返回一个 m x n 的二维数组。如果无法构成这样的二维数组，请你返回一个空的二维数组。
+
+## 解题思路
+
+- 简单题。从一维数组 original 中依次取出每行 n 个元素，顺序放到 m 行中。此题中，如果 m*n 大于或者小于 original 的长度，都输出空数组。
+
+## 代码
+
+```go
+package leetcode
+
+func construct2DArray(original []int, m int, n int) [][]int {
+	if m*n != len(original) {
+		return [][]int{}
+	}
+	res := make([][]int, m)
+	for i := 0; i < m; i++ {
+		res[i] = original[n*i : n*(i+1)]
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2021.Brightest-Position-on-Street/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2037.Minimum-Number-of-Moves-to-Seat-Everyone/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2000~2099/2037.Minimum-Number-of-Moves-to-Seat-Everyone.md b/website/content/ChapterFour/2000~2099/2037.Minimum-Number-of-Moves-to-Seat-Everyone.md
new file mode 100644
index 000000000..8f7878c44
--- /dev/null
+++ b/website/content/ChapterFour/2000~2099/2037.Minimum-Number-of-Moves-to-Seat-Everyone.md
@@ -0,0 +1,101 @@
+# [2037. Minimum Number of Moves to Seat Everyone](https://leetcode.com/problems/minimum-number-of-moves-to-seat-everyone/)
+
+## 题目
+
+There are n seats and n students in a room. You are given an array seats of length n, where seats[i] is the position of the ith seat. You are also given the array students of length n, where students[j] is the position of the jth student.
+
+You may perform the following move any number of times:
+
+- Increase or decrease the position of the ith student by 1 (i.e., moving the ith student from position x to x + 1 or x - 1)
+
+Return the minimum number of moves required to move each student to a seat such that no two students are in the same seat.
+
+Note that there may be multiple seats or students in the same position at the beginning.
+
+**Example 1:**
+
+    Input: seats = [3,1,5], students = [2,7,4]
+    Output: 4
+    Explanation: The students are moved as follows:
+    - The first student is moved from from position 2 to position 1 using 1 move.
+    - The second student is moved from from position 7 to position 5 using 2 moves.
+    - The third student is moved from from position 4 to position 3 using 1 move.
+    In total, 1 + 2 + 1 = 4 moves were used.
+
+**Example 2:**
+
+    Input: seats = [4,1,5,9], students = [1,3,2,6]
+    Output: 7
+    Explanation: The students are moved as follows:
+    - The first student is not moved.
+    - The second student is moved from from position 3 to position 4 using 1 move.
+    - The third student is moved from from position 2 to position 5 using 3 moves.
+    - The fourth student is moved from from position 6 to position 9 using 3 moves.
+    In total, 0 + 1 + 3 + 3 = 7 moves were used.
+
+**Example 3:**
+
+    Input: seats = [2,2,6,6], students = [1,3,2,6]
+    Output: 4
+    Explanation: Note that there are two seats at position 2 and two seats at position 6.
+    The students are moved as follows:
+    - The first student is moved from from position 1 to position 2 using 1 move.
+    - The second student is moved from from position 3 to position 6 using 3 moves.
+    - The third student is not moved.
+    - The fourth student is not moved.
+    In total, 1 + 3 + 0 + 0 = 4 moves were used.
+
+**Constraints:**
+
+- n == seats.length == students.length
+- 1 <= n <= 100
+- 1 <= seats[i], students[j] <= 100
+
+## 题目大意
+
+一个房间里有 n 个座位和 n 名学生，房间用一个数轴表示。给你一个长度为 n 的数组 seats，其中 seats[i] 是第 i 个座位的位置。同时给你一个长度为 n 的数组 students ，其中 students[j] 是第 j 位学生的位置。
+
+你可以执行以下操作任意次：
+
+增加或者减少第 i 位学生的位置，每次变化量为 1（也就是将第 i 位学生从位置 x 移动到 x + 1或者 x - 1）
+
+请你返回使所有学生都有座位坐的最少移动次数，并确保没有两位学生的座位相同。
+
+请注意，初始时有可能有多个座位或者多位学生在 同一位置。
+
+## 解题思路
+
+- 排序+模拟计算
+
+# 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func minMovesToSeat(seats []int, students []int) int {
+	sort.Ints(seats)
+	sort.Ints(students)
+	n := len(students)
+	moves := 0
+	for i := 0; i < n; i++ {
+		moves += abs(seats[i], students[i])
+	}
+	return moves
+}
+
+func abs(a, b int) int {
+	if a > b {
+		return a - b
+	}
+	return b - a
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2022.Convert-1D-Array-Into-2D-Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2000~2099/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color.md b/website/content/ChapterFour/2000~2099/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color.md
new file mode 100644
index 000000000..09091fcc8
--- /dev/null
+++ b/website/content/ChapterFour/2000~2099/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color.md
@@ -0,0 +1,112 @@
+# [2038. Remove Colored Pieces if Both Neighbors are the Same Color](https://leetcode.com/problems/remove-colored-pieces-if-both-neighbors-are-the-same-color/)
+
+## 题目
+
+There are n pieces arranged in a line, and each piece is colored either by 'A' or by 'B'. You are given a string colors of length n where colors[i] is the color of the ith piece.
+
+Alice and Bob are playing a game where they take alternating turns removing pieces from the line. In this game, Alice moves first.
+
+- Alice is only allowed to remove a piece colored 'A' if both its neighbors are also colored 'A'. She is not allowed to remove pieces that are colored 'B'.
+- Bob is only allowed to remove a piece colored 'B' if both its neighbors are also colored 'B'. He is not allowed to remove pieces that are colored 'A'.
+- Alice and Bob cannot remove pieces from the edge of the line.
+- If a player cannot make a move on their turn, that player loses and the other player wins.
+
+Assuming Alice and Bob play optimally, return true if Alice wins, or return false if Bob wins.
+
+**Example 1:**
+
+    Input: colors = "AAABABB"
+    Output: true
+    Explanation:
+    AAABABB -> AABABB
+    Alice moves first.
+    She removes the second 'A' from the left since that is the only 'A' whose neighbors are both 'A'.
+
+    Now it's Bob's turn.
+    Bob cannot make a move on his turn since there are no 'B's whose neighbors are both 'B'.
+    Thus, Alice wins, so return true.
+
+**Example 2:**
+
+    Input: colors = "AA"
+    Output: false
+    Explanation:
+    Alice has her turn first.
+    There are only two 'A's and both are on the edge of the line, so she cannot move on her turn.
+    Thus, Bob wins, so return false.
+
+**Example 3:**
+
+    Input: colors = "ABBBBBBBAAA"
+    Output: false
+    Explanation:
+    ABBBBBBBAAA -> ABBBBBBBAA
+    Alice moves first.
+    Her only option is to remove the second to last 'A' from the right.
+
+    ABBBBBBBAA -> ABBBBBBAA
+    Next is Bob's turn.
+    He has many options for which 'B' piece to remove. He can pick any.
+
+    On Alice's second turn, she has no more pieces that she can remove.
+    Thus, Bob wins, so return false.
+
+**Constraints:**
+
+- 1 <= colors.length <= 100000
+- colors consists of only the letters 'A' and 'B'
+
+## 题目大意
+
+总共有 n 个颜色片段排成一列，每个颜色片段要么是 'A' 要么是 'B' 。给你一个长度为 n 的字符串 colors ，其中 colors[i] 表示第 i 个颜色片段的颜色。
+
+Alice 和 Bob 在玩一个游戏，他们轮流从这个字符串中删除颜色。Alice 先手。
+
+- 如果一个颜色片段为 'A' 且相邻两个颜色都是颜色 'A'，那么 Alice 可以删除该颜色片段。Alice不可以删除任何颜色 'B' 片段。
+- 如果一个颜色片段为 'B'且相邻两个颜色都是颜色 'B' ，那么 Bob 可以删除该颜色片段。Bob 不可以删除任何颜色 'A' 片段。
+- Alice 和 Bob 不能从字符串两端删除颜色片段。
+- 如果其中一人无法继续操作，则该玩家 输掉游戏且另一玩家 获胜。
+
+假设 Alice 和 Bob 都采用最优策略，如果 Alice 获胜，请返回true，否则 Bob 获胜，返回false。
+
+## 解题思路
+
+- 统计 Alice 和 Bob 分别可以操作的次数记为 As，Bs 
+- 因为 Alice 先手，所以只要 As 大于 Bs，Alice 获胜返回 true，否则 Bob 获胜返回 false
+
+# 代码
+
+```go
+package leetcode
+
+func winnerOfGame(colors string) bool {
+	As, Bs := 0, 0
+	Acont, Bcont := 0, 0
+	for _, color := range colors {
+		if color == 'A' {
+			Acont += 1
+			Bcont = 0
+		} else {
+			Bcont += 1
+			Acont = 0
+		}
+		if Acont >= 3 {
+			As += Acont - 2
+		}
+		if Bcont >= 3 {
+			Bs += Bcont - 2
+		}
+	}
+	if As > Bs {
+		return true
+	}
+	return false
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2037.Minimum-Number-of-Moves-to-Seat-Everyone/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2043.Simple-Bank-System/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2000~2099/2043.Simple-Bank-System.md b/website/content/ChapterFour/2000~2099/2043.Simple-Bank-System.md
new file mode 100644
index 000000000..b756eb8e0
--- /dev/null
+++ b/website/content/ChapterFour/2000~2099/2043.Simple-Bank-System.md
@@ -0,0 +1,120 @@
+# [2043. Simple Bank System](https://leetcode.com/problems/simple-bank-system/)
+
+## 题目
+
+You have been tasked with writing a program for a popular bank that will automate all its incoming transactions (transfer, deposit, and withdraw). The bank has n accounts numbered from 1 to n. The initial balance of each account is stored in a 0-indexed integer array balance, with the (i + 1)th account having an initial balance of balance[i].
+
+Execute all the valid transactions. A transaction is valid if:
+
+- The given account number(s) are between 1 and n, and
+- The amount of money withdrawn or transferred from is less than or equal to the balance of the account.
+
+Implement the Bank class:
+
+- Bank(long[] balance) Initializes the object with the 0-indexed integer array balance.
+- boolean transfer(int account1, int account2, long money) Transfers money dollars from the account numbered account1 to the account numbered account2. Return true if the transaction was successful, false otherwise.
+- boolean deposit(int account, long money) Deposit money dollars into the account numbered account. Return true if the transaction was successful, false otherwise.
+- boolean withdraw(int account, long money) Withdraw money dollars from the account numbered account. Return true if the transaction was successful, false otherwise.
+
+**Example 1:**
+
+    Input
+    ["Bank", "withdraw", "transfer", "deposit", "transfer", "withdraw"]
+    [[[10, 100, 20, 50, 30]], [3, 10], [5, 1, 20], [5, 20], [3, 4, 15], [10, 50]]
+    Output
+    [null, true, true, true, false, false]
+
+    Explanation
+    Bank bank = new Bank([10, 100, 20, 50, 30]);
+    bank.withdraw(3, 10);    // return true, account 3 has a balance of $20, so it is valid to withdraw $10.
+    // Account 3 has $20 - $10 = $10.
+    bank.transfer(5, 1, 20); // return true, account 5 has a balance of $30, so it is valid to transfer $20.
+    // Account 5 has $30 - $20 = $10, and account 1 has $10 + $20 = $30.
+    bank.deposit(5, 20);     // return true, it is valid to deposit $20 to account 5.
+    // Account 5 has $10 + $20 = $30.
+    bank.transfer(3, 4, 15); // return false, the current balance of account 3 is $10,
+    // so it is invalid to transfer $15 from it.
+    bank.withdraw(10, 50);   // return false, it is invalid because account 10 does not exist.
+
+**Constraints:**
+
+- n == balance.length
+- 1 <= n, account, account1, account2 <= 100000
+- 0 <= balance[i], money <= 1000000000000
+- At most 104 calls will be made to each function transfer, deposit, withdraw.
+
+## 题目大意
+
+你的任务是为一个很受欢迎的银行设计一款程序，以自动化执行所有传入的交易（转账，存款和取款）。银行共有 n 个账户，编号从 1 到 n 。每个账号的初始余额存储在一个下标从 0 开始的整数数组 balance 中，其中第 (i + 1) 个账户的初始余额是 balance[i] 。
+
+请你执行所有 有效的 交易。如果满足下面全部条件，则交易 有效 ：
+
+- 指定的账户数量在 1 和 n 之间，且
+- 取款或者转账需要的钱的总数 小于或者等于 账户余额。
+
+实现 Bank 类：
+
+- Bank(long[] balance) 使用下标从 0 开始的整数数组 balance 初始化该对象。
+- boolean transfer(int account1, int account2, long money) 从编号为 account1 的账户向编号为 account2 的账户转帐 money 美元。如果交易成功，返回 true ，否则，返回 false 。
+- boolean deposit(int account, long money) 向编号为 account 的账户存款 money 美元。如果交易成功，返回 true ；否则，返回 false 。
+- boolean withdraw(int account, long money) 从编号为 account 的账户取款 money 美元。如果交易成功，返回 true ；否则，返回 false 。
+
+## 解题思路
+
+    根据题意进行简单模拟
+
+# 代码
+
+```go
+package leetcode
+
+type Bank struct {
+	accounts []int64
+	n        int
+}
+
+func Constructor(balance []int64) Bank {
+	return Bank{
+		accounts: balance,
+		n:        len(balance),
+	}
+}
+
+func (this *Bank) Transfer(account1 int, account2 int, money int64) bool {
+	if account1 > this.n || account2 > this.n {
+		return false
+	}
+	if this.accounts[account1-1] < money {
+		return false
+	}
+	this.accounts[account1-1] -= money
+	this.accounts[account2-1] += money
+	return true
+}
+
+func (this *Bank) Deposit(account int, money int64) bool {
+	if account > this.n {
+		return false
+	}
+	this.accounts[account-1] += money
+	return true
+}
+
+func (this *Bank) Withdraw(account int, money int64) bool {
+	if account > this.n {
+		return false
+	}
+	if this.accounts[account-1] < money {
+		return false
+	}
+	this.accounts[account-1] -= money
+	return true
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2000~2099/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another.md b/website/content/ChapterFour/2000~2099/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another.md
new file mode 100644
index 000000000..65dcf54fb
--- /dev/null
+++ b/website/content/ChapterFour/2000~2099/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another.md
@@ -0,0 +1,152 @@
+# [2096. Step-By-Step Directions From a Binary Tree Node to Another](https://leetcode.com/problems/step-by-step-directions-from-a-binary-tree-node-to-another/)
+
+
+## 题目
+
+You are given the `root` of a **binary tree** with `n` nodes. Each node is uniquely assigned a value from `1` to `n`. You are also given an integer `startValue` representing the value of the start node `s`, and a different integer `destValue` representing the value of the destination node `t`.
+
+Find the **shortest path** starting from node `s` and ending at node `t`. Generate step-by-step directions of such path as a string consisting of only the **uppercase** letters `'L'`, `'R'`, and `'U'`. Each letter indicates a specific direction:
+
+- `'L'` means to go from a node to its **left child** node.
+- `'R'` means to go from a node to its **right child** node.
+- `'U'` means to go from a node to its **parent** node.
+
+Return *the step-by-step directions of the **shortest path** from node* `s` *to node* `t`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/11/15/eg1.png](https://assets.leetcode.com/uploads/2021/11/15/eg1.png)
+
+```
+Input: root = [5,1,2,3,null,6,4], startValue = 3, destValue = 6
+Output: "UURL"
+Explanation: The shortest path is: 3 → 1 → 5 → 2 → 6.
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2021/11/15/eg2.png](https://assets.leetcode.com/uploads/2021/11/15/eg2.png)
+
+```
+Input: root = [2,1], startValue = 2, destValue = 1
+Output: "L"
+Explanation: The shortest path is: 2 → 1.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is `n`.
+- `2 <= n <= 105`
+- `1 <= Node.val <= n`
+- All the values in the tree are **unique**.
+- `1 <= startValue, destValue <= n`
+- `startValue != destValue`
+
+## 题目大意
+
+给你一棵 二叉树 的根节点 root ，这棵二叉树总共有 n 个节点。每个节点的值为 1 到 n 中的一个整数，且互不相同。给你一个整数 startValue ，表示起点节点 s 的值，和另一个不同的整数 destValue ，表示终点节点 t 的值。
+
+请找到从节点 s 到节点 t 的 最短路径 ，并以字符串的形式返回每一步的方向。每一步用 大写 字母 'L' ，'R' 和 'U' 分别表示一种方向：
+
+- 'L' 表示从一个节点前往它的 左孩子 节点。
+- 'R' 表示从一个节点前往它的 右孩子 节点。
+- 'U' 表示从一个节点前往它的 父 节点。
+
+请你返回从 s 到 t 最短路径 每一步的方向。
+
+## 解题思路
+
+- 二叉树中一个节点到另一个节点的最短路径一定可以分为两个部分（可能为空）：从起点节点向上到两个节点的**最近公共祖先**，再从最近公共祖先向下到达终点节点。
+- 首先需要找到起点 s 与公共祖先的节点之间的 path1，公共祖先节点与终点 t 的 path2。再删掉 2 个 path 的公共前缀。如果起点 s 和终点 t 在不同的分支上，不存在公共前缀。如果他们在相同的分支上，那么最终答案要去掉这个公共前缀。
+- 删除掉公共前缀以后，需要再整理一下最终答案的输出格式。由于题目要求，起点到公共祖先节点需要输出 U，所以把这段 path1 全部改成 U，然后再拼接上 path2 字符串，即可得到的字符串即为待求 ss 到 tt 每一步的最短路径。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// TreeNode define
+type TreeNode = structures.TreeNode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+func getDirections(root *TreeNode, startValue int, destValue int) string {
+	sPath, dPath := make([]byte, 0), make([]byte, 0)
+	findPath(root, startValue, &sPath)
+	findPath(root, destValue, &dPath)
+	size, i := min(len(sPath), len(dPath)), 0
+	for i < size {
+		if sPath[len(sPath)-1-i] == dPath[len(dPath)-1-i] {
+			i++
+		} else {
+			break
+		}
+	}
+	sPath = sPath[:len(sPath)-i]
+	replace(sPath)
+	dPath = dPath[:len(dPath)-i]
+	reverse(dPath)
+	sPath = append(sPath, dPath...)
+	return string(sPath)
+}
+
+func findPath(root *TreeNode, value int, path *[]byte) bool {
+	if root.Val == value {
+		return true
+	}
+
+	if root.Left != nil && findPath(root.Left, value, path) {
+		*path = append(*path, 'L')
+		return true
+	}
+
+	if root.Right != nil && findPath(root.Right, value, path) {
+		*path = append(*path, 'R')
+		return true
+	}
+
+	return false
+}
+
+func reverse(path []byte) {
+	left, right := 0, len(path)-1
+	for left < right {
+		path[left], path[right] = path[right], path[left]
+		left++
+		right--
+	}
+}
+
+func replace(path []byte) {
+	for i := 0; i < len(path); i++ {
+		path[i] = 'U'
+	}
+}
+
+func min(i, j int) int {
+	if i < j {
+		return i
+	}
+	return j
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2043.Simple-Bank-System/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2164.Sort-Even-and-Odd-Indices-Independently/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2000~2099/_index.md b/website/content/ChapterFour/2000~2099/_index.md
new file mode 100644
index 000000000..e954f0877
--- /dev/null
+++ b/website/content/ChapterFour/2000~2099/_index.md
@@ -0,0 +1,4 @@
+---
+bookCollapseSection: true
+weight: 20
+---
diff --git a/website/content/ChapterFour/2100~2199/2164.Sort-Even-and-Odd-Indices-Independently.md b/website/content/ChapterFour/2100~2199/2164.Sort-Even-and-Odd-Indices-Independently.md
new file mode 100644
index 000000000..bb3a1e28c
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/2164.Sort-Even-and-Odd-Indices-Independently.md
@@ -0,0 +1,100 @@
+# [2164. Sort Even and Odd Indices Independently](https://leetcode.com/problems/sort-even-and-odd-indices-independently/)
+
+
+## 题目
+
+You are given a **0-indexed** integer array `nums`. Rearrange the values of `nums` according to the following rules:
+
+1. Sort the values at **odd indices** of `nums` in **non-increasing** order.
+    - For example, if `nums = [4,**1**,2,**3**]` before this step, it becomes `[4,**3**,2,**1**]` after. The values at odd indices `1` and `3` are sorted in non-increasing order.
+2. Sort the values at **even indices** of `nums` in **non-decreasing** order.
+    - For example, if `nums = [**4**,1,**2**,3]` before this step, it becomes `[**2**,1,**4**,3]` after. The values at even indices `0` and `2` are sorted in non-decreasing order.
+
+Return *the array formed after rearranging the values of* `nums`.
+
+**Example 1:**
+
+```
+Input: nums = [4,1,2,3]
+Output: [2,3,4,1]
+Explanation:
+First, we sort the values present at odd indices (1 and 3) in non-increasing order.
+So, nums changes from [4,1,2,3] to [4,3,2,1].
+Next, we sort the values present at even indices (0 and 2) in non-decreasing order.
+So, nums changes from [4,1,2,3] to [2,3,4,1].
+Thus, the array formed after rearranging the values is [2,3,4,1].
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,1]
+Output: [2,1]
+Explanation:
+Since there is exactly one odd index and one even index, no rearrangement of values takes place.
+The resultant array formed is [2,1], which is the same as the initial array.
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `1 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个下标从 0 开始的整数数组 nums 。根据下述规则重排 nums 中的值：
+
+1. 按 非递增 顺序排列 nums 奇数下标 上的所有值。
+举个例子，如果排序前 nums = [4,1,2,3] ，对奇数下标的值排序后变为 [4,3,2,1] 。奇数下标 1 和 3 的值按照非递增顺序重排。
+2. 按 非递减 顺序排列 nums 偶数下标 上的所有值。
+举个例子，如果排序前 nums = [4,1,2,3] ，对偶数下标的值排序后变为 [2,1,4,3] 。偶数下标 0 和 2 的值按照非递减顺序重排。
+
+返回重排 nums 的值之后形成的数组。
+
+## 解题思路
+
+- 简单题。分别将奇数和偶数位上的数字排序，奇数位的数从大到小，偶数位的数从小到大。最后将他们组合成一个数组。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func sortEvenOdd(nums []int) []int {
+	odd, even, res := []int{}, []int{}, []int{}
+	for index, v := range nums {
+		if index%2 == 0 {
+			even = append(even, v)
+		} else {
+			odd = append(odd, v)
+		}
+	}
+	sort.Ints(even)
+	sort.Sort(sort.Reverse(sort.IntSlice(odd)))
+
+	indexO, indexE := 0, 0
+	for i := 0; i < len(nums); i++ {
+		if i%2 == 0 {
+			res = append(res, even[indexE])
+			indexE++
+		} else {
+			res = append(res, odd[indexO])
+			indexO++
+		}
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2000~2099/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2165.Smallest-Value-of-the-Rearranged-Number/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2100~2199/2165.Smallest-Value-of-the-Rearranged-Number.md b/website/content/ChapterFour/2100~2199/2165.Smallest-Value-of-the-Rearranged-Number.md
new file mode 100644
index 000000000..d2ce438b0
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/2165.Smallest-Value-of-the-Rearranged-Number.md
@@ -0,0 +1,108 @@
+# [2165. Smallest Value of the Rearranged Number](https://leetcode.com/problems/smallest-value-of-the-rearranged-number/)
+
+
+## 题目
+
+You are given an integer `num.` **Rearrange** the digits of `num` such that its value is **minimized** and it does not contain **any** leading zeros.
+
+Return *the rearranged number with minimal value*.
+
+Note that the sign of the number does not change after rearranging the digits.
+
+**Example 1:**
+
+```
+Input: num = 310
+Output: 103
+Explanation: The possible arrangements for the digits of 310 are 013, 031, 103, 130, 301, 310.
+The arrangement with the smallest value that does not contain any leading zeros is 103.
+
+```
+
+**Example 2:**
+
+```
+Input: num = -7605
+Output: -7650
+Explanation: Some possible arrangements for the digits of -7605 are -7650, -6705, -5076, -0567.
+The arrangement with the smallest value that does not contain any leading zeros is -7650.
+
+```
+
+**Constraints:**
+
+- `10^15 <= num <= 10^15`
+
+## 题目大意
+
+给你一个整数 num 。重排 num 中的各位数字，使其值 最小化 且不含 任何 前导零。
+
+返回不含前导零且值最小的重排数字。注意，重排各位数字后，num 的符号不会改变。
+
+## 解题思路
+
+- 先将每个数字出现次数统计出来。然后将数字大小从小到大排序。如果原数是正数，当出现有数字 0 的情况的时候，需先将第二小的数字排列到第一个，再把 0 排列完。再继续排列第二小，第三小。。。
+- 如果原数是负数。那么就逆序排列，即先排列最大的数字，然后次大的数字，直到排列最小的数字。因为数字越大，对应的这个数的负数就越小。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func smallestNumber(num int64) int64 {
+	pos := true
+	if num < 0 {
+		pos = false
+		num *= -1
+	}
+	nums, m, res := []int{}, map[int]int{}, 0
+	for num != 0 {
+		tmp := int(num % 10)
+		m[tmp]++
+		num = num / 10
+	}
+
+	for k := range m {
+		nums = append(nums, k)
+	}
+	if pos {
+		sort.Ints(nums)
+	} else {
+		sort.Sort(sort.Reverse(sort.IntSlice(nums)))
+	}
+
+	if nums[0] == 0 && len(nums) > 1 {
+		res += nums[1]
+		m[nums[1]]--
+	}
+
+	for _, v := range nums {
+		if res != 0 {
+			for j := m[v]; j > 0; j-- {
+				res = res * 10
+				res += v
+			}
+		} else {
+			res += v
+			tmp := m[v] - 1
+			for j := tmp; j > 0; j-- {
+				res = res * 10
+				res += v
+			}
+		}
+	}
+	if !pos {
+		return -1 * int64(res)
+	}
+	return int64(res)
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2164.Sort-Even-and-Odd-Indices-Independently/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2166.Design-Bitset/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2100~2199/2166.Design-Bitset.md b/website/content/ChapterFour/2100~2199/2166.Design-Bitset.md
new file mode 100644
index 000000000..8fbc2a786
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/2166.Design-Bitset.md
@@ -0,0 +1,163 @@
+# [2166. Design Bitset](https://leetcode.com/problems/design-bitset/)
+
+
+## 题目
+
+A **Bitset** is a data structure that compactly stores bits.
+
+Implement the `Bitset` class:
+
+- `Bitset(int size)` Initializes the Bitset with `size` bits, all of which are `0`.
+- `void fix(int idx)` Updates the value of the bit at the index `idx` to `1`. If the value was already `1`, no change occurs.
+- `void unfix(int idx)` Updates the value of the bit at the index `idx` to `0`. If the value was already `0`, no change occurs.
+- `void flip()` Flips the values of each bit in the Bitset. In other words, all bits with value `0` will now have value `1` and vice versa.
+- `boolean all()` Checks if the value of **each** bit in the Bitset is `1`. Returns `true` if it satisfies the condition, `false` otherwise.
+- `boolean one()` Checks if there is **at least one** bit in the Bitset with value `1`. Returns `true` if it satisfies the condition, `false` otherwise.
+- `int count()` Returns the **total number** of bits in the Bitset which have value `1`.
+- `String toString()` Returns the current composition of the Bitset. Note that in the resultant string, the character at the `ith` index should coincide with the value at the `ith` bit of the Bitset.
+
+**Example 1:**
+
+```
+Input
+["Bitset", "fix", "fix", "flip", "all", "unfix", "flip", "one", "unfix", "count", "toString"]
+[[5], [3], [1], [], [], [0], [], [], [0], [], []]
+Output
+[null, null, null, null, false, null, null, true, null, 2, "01010"]
+
+Explanation
+Bitset bs = new Bitset(5); // bitset = "00000".
+bs.fix(3);     // the value at idx = 3 is updated to 1, so bitset = "00010".
+bs.fix(1);     // the value at idx = 1 is updated to 1, so bitset = "01010".
+bs.flip();     // the value of each bit is flipped, so bitset = "10101".
+bs.all();      // return False, as not all values of the bitset are 1.
+bs.unfix(0);   // the value at idx = 0 is updated to 0, so bitset = "00101".
+bs.flip();     // the value of each bit is flipped, so bitset = "11010".
+bs.one();      // return True, as there is at least 1 index with value 1.
+bs.unfix(0);   // the value at idx = 0 is updated to 0, so bitset = "01010".
+bs.count();    // return 2, as there are 2 bits with value 1.
+bs.toString(); // return "01010", which is the composition of bitset.
+
+```
+
+**Constraints:**
+
+- `1 <= size <= 10^5`
+- `0 <= idx <= size - 1`
+- At most `10^5` calls will be made **in total** to `fix`, `unfix`, `flip`, `all`, `one`, `count`, and `toString`.
+- At least one call will be made to `all`, `one`, `count`, or `toString`.
+- At most `5` calls will be made to `toString`.
+
+## 题目大意
+
+位集 Bitset 是一种能以紧凑形式存储位的数据结构。
+
+请你实现 Bitset 类。
+
+- Bitset(int size) 用 size 个位初始化 Bitset ，所有位都是 0 。
+- void fix(int idx) 将下标为 idx 的位上的值更新为 1 。如果值已经是 1 ，则不会发生任何改变。
+- void unfix(int idx) 将下标为 idx 的位上的值更新为 0 。如果值已经是 0 ，则不会发生任何改变。
+- void flip() 翻转 Bitset 中每一位上的值。换句话说，所有值为 0 的位将会变成 1 ，反之亦然。
+- boolean all() 检查 Bitset 中 每一位 的值是否都是 1 。如果满足此条件，返回 true ；否则，返回 false 。
+- boolean one() 检查 Bitset 中 是否 至少一位 的值是 1 。如果满足此条件，返回 true ；否则，返回 false 。
+- int count() 返回 Bitset 中值为 1 的位的 总数 。
+- String toString() 返回 Bitset 的当前组成情况。注意，在结果字符串中，第 i 个下标处的字符应该与 Bitset 中的第 i 位一致。
+
+提示：
+
+- 1 <= size <= 10^5
+- 0 <= idx <= size - 1
+- 至多调用 fix、unfix、flip、all、one、count 和 toString 方法 总共 10^5 次
+- 至少调用 all、one、count 或 toString 方法一次
+- 至多调用 toString 方法 5 次
+
+## 解题思路
+
+- 题目中给出了 size 大小，10^5 位二进制。所以不能用 int64 数据类型。
+- 用数组模拟二进制位的一系列操作。flip 操作并不需要每次去翻转，偶数次翻转等于没有翻转，奇数次翻转记下标记，同时更新 1 的个数。这次懒操作在调用 fix 和 unfix 时，更新到原来数组中。
+- fix 和 unfix 根据懒数组中的标记对应更新二进制位。同时更新 1 的个数。
+- all，one，count 都是判断 1 的个数。toString 输出即可。
+
+## 代码
+
+```go
+package leetcode
+
+type Bitset struct {
+	set      []byte
+	flipped  []byte
+	oneCount int
+	size     int
+}
+
+func Constructor(size int) Bitset {
+	set := make([]byte, size)
+	flipped := make([]byte, size)
+	for i := 0; i < size; i++ {
+		set[i] = byte('0')
+		flipped[i] = byte('1')
+	}
+	return Bitset{
+		set:      set,
+		flipped:  flipped,
+		oneCount: 0,
+		size:     size,
+	}
+}
+
+func (this *Bitset) Fix(idx int) {
+	if this.set[idx] == byte('0') {
+		this.set[idx] = byte('1')
+		this.flipped[idx] = byte('0')
+		this.oneCount++
+	}
+}
+
+func (this *Bitset) Unfix(idx int) {
+	if this.set[idx] == byte('1') {
+		this.set[idx] = byte('0')
+		this.flipped[idx] = byte('1')
+		this.oneCount--
+	}
+}
+
+func (this *Bitset) Flip() {
+	this.set, this.flipped = this.flipped, this.set
+	this.oneCount = this.size - this.oneCount
+}
+
+func (this *Bitset) All() bool {
+	return this.oneCount == this.size
+}
+
+func (this *Bitset) One() bool {
+	return this.oneCount != 0
+}
+
+func (this *Bitset) Count() int {
+	return this.oneCount
+}
+
+func (this *Bitset) ToString() string {
+	return string(this.set)
+}
+
+/**
+ * Your Bitset object will be instantiated and called as such:
+ * obj := Constructor(size);
+ * obj.Fix(idx);
+ * obj.Unfix(idx);
+ * obj.Flip();
+ * param_4 := obj.All();
+ * param_5 := obj.One();
+ * param_6 := obj.Count();
+ * param_7 := obj.ToString();
+ */
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2165.Smallest-Value-of-the-Rearranged-Number/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2100~2199/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods.md b/website/content/ChapterFour/2100~2199/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods.md
new file mode 100644
index 000000000..867f7fc8e
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods.md
@@ -0,0 +1,156 @@
+# [2167. Minimum Time to Remove All Cars Containing Illegal Goods](https://leetcode.com/problems/minimum-time-to-remove-all-cars-containing-illegal-goods/)
+
+
+## 题目
+
+You are given a **0-indexed** binary string `s` which represents a sequence of train cars. `s[i] = '0'` denotes that the `ith` car does **not** contain illegal goods and `s[i] = '1'` denotes that the `ith` car does contain illegal goods.
+
+As the train conductor, you would like to get rid of all the cars containing illegal goods. You can do any of the following three operations **any** number of times:
+
+1. Remove a train car from the **left** end (i.e., remove `s[0]`) which takes 1 unit of time.
+2. Remove a train car from the **right** end (i.e., remove `s[s.length - 1]`) which takes 1 unit of time.
+3. Remove a train car from **anywhere** in the sequence which takes 2 units of time.
+
+Return *the **minimum** time to remove all the cars containing illegal goods*.
+
+Note that an empty sequence of cars is considered to have no cars containing illegal goods.
+
+**Example 1:**
+
+```
+Input: s = "1100101"
+Output: 5
+Explanation:
+One way to remove all the cars containing illegal goods from the sequence is to
+- remove a car from the left end 2 times. Time taken is 2 * 1 = 2.
+- remove a car from the right end. Time taken is 1.
+- remove the car containing illegal goods found in the middle. Time taken is 2.
+This obtains a total time of 2 + 1 + 2 = 5.
+
+An alternative way is to
+- remove a car from the left end 2 times. Time taken is 2 * 1 = 2.
+- remove a car from the right end 3 times. Time taken is 3 * 1 = 3.
+This also obtains a total time of 2 + 3 = 5.
+
+5 is the minimum time taken to remove all the cars containing illegal goods.
+There are no other ways to remove them with less time.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "0010"
+Output: 2
+Explanation:
+One way to remove all the cars containing illegal goods from the sequence is to
+- remove a car from the left end 3 times. Time taken is 3 * 1 = 3.
+This obtains a total time of 3.
+
+Another way to remove all the cars containing illegal goods from the sequence is to
+- remove the car containing illegal goods found in the middle. Time taken is 2.
+This obtains a total time of 2.
+
+Another way to remove all the cars containing illegal goods from the sequence is to
+- remove a car from the right end 2 times. Time taken is 2 * 1 = 2.
+This obtains a total time of 2.
+
+2 is the minimum time taken to remove all the cars containing illegal goods.
+There are no other ways to remove them with less time.
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 2 * 10^5`
+- `s[i]` is either `'0'` or `'1'`.
+
+## 题目大意
+
+给你一个下标从 0 开始的二进制字符串 s ，表示一个列车车厢序列。s[i] = '0' 表示第 i 节车厢 不 含违禁货物，而 s[i] = '1' 表示第 i 节车厢含违禁货物。
+
+作为列车长，你需要清理掉所有载有违禁货物的车厢。你可以不限次数执行下述三种操作中的任意一个：
+
+1. 从列车 左 端移除一节车厢（即移除 s[0]），用去 1 单位时间。
+2. 从列车 右 端移除一节车厢（即移除 s[s.length - 1]），用去 1 单位时间。
+3. 从列车车厢序列的 任意位置 移除一节车厢，用去 2 单位时间。
+
+返回移除所有载有违禁货物车厢所需要的 最少 单位时间数。注意，空的列车车厢序列视为没有车厢含违禁货物。
+
+## 解题思路
+
+- 这道题求最少单位时间数，最少时间数一定是尽量少使用 2 个单位时间的操作，多用 1 个时间的操作。从列车两头移除车厢，只需要移除和旁边车厢的金属连接处即可。由于列车位于两边，所以与其他车厢的金属连接处只有 1 个，故只需要 1 个单位时间；当车厢在中间，该车厢与两边的车厢有 2 个金属连接处，移除它需要断开与两边车厢的连接。所以需要 2 个单位时间。
+- 断开中间一节车厢以后，列车会被断成 2 部分。2 部分列车分别有 2 个头 2 个尾。举例：`1100111101`，如果把它从第 5 节开始断开，剩下的列车为 `11001 (1)` 和 `1101`。剩下的 1 都位于 2 边，移除他们都只需要 1 个单位时间。那么移除所有违禁品最少时间是 2 * 1 + 1 * 6 = 8。
+- 左半部分，定义 prefixSum[i] 表示移除前 i 节车厢所花费的最少时间。状态转移方程为：
+	
+	{{< katex display >}} 
+    prefixSum[i] =\left\{\begin{matrix}prefixSum[i-1],s[i]=0\\ min(prefixSum[i-1]+2, i+1), s[i]=1\end{matrix}\right.
+	{{< /katex >}} 
+    
+- 同理，右半部分定义 suffixSum[i] 表示移除后 i 节车厢所花费的最少时间。状态转移方程为：
+    
+	{{< katex display >}} 
+    suffixSum[i] =\left\{\begin{matrix} suffixSum[i+1],s[i]=0\\ min(suffixSum[i+1]+2, n-i), s[i]=1\end{matrix}\right.
+	{{< /katex >}} 
+    
+- 最后一层循环枚举 prefixSum[i] + suffixSum[i+1] 的最小值即为答案。
+- 这一题在解法一的基础上还可以再简化。当 s[i] = 1 时，prefixSum 和 suffixSum 是两种计算方法。我们可以假设中间断开的部分在 prefixSum 中。于是可以合并上面两个状态转移方程。简化以后的代码见解法二。
+
+## 代码
+
+```go
+package leetcode
+
+import "runtime/debug"
+
+// 解法一 DP
+func minimumTime(s string) int {
+	suffixSum, prefixSum, res := make([]int, len(s)+1), make([]int, len(s)+1), 0
+	for i := len(s) - 1; i >= 0; i-- {
+		if s[i] == '0' {
+			suffixSum[i] = suffixSum[i+1]
+		} else {
+			suffixSum[i] = min(suffixSum[i+1]+2, len(s)-i)
+		}
+	}
+	res = suffixSum[0]
+	if s[0] == '1' {
+		prefixSum[0] = 1
+	}
+	for i := 1; i < len(s); i++ {
+		if s[i] == '0' {
+			prefixSum[i] = prefixSum[i-1]
+		} else {
+			prefixSum[i] = min(prefixSum[i-1]+2, i+1)
+		}
+		res = min(res, prefixSum[i]+suffixSum[i+1])
+	}
+	return res
+}
+
+func init() { debug.SetGCPercent(-1) }
+
+// 解法二 小幅优化时间和空间复杂度
+func minimumTime1(s string) int {
+	res, count := len(s), 0
+	for i := 0; i < len(s); i++ {
+		count = min(count+int(s[i]-'0')*2, i+1)
+		res = min(res, count+len(s)-i-1)
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	} else {
+		return b
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2166.Design-Bitset/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2169.Count-Operations-to-Obtain-Zero/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2100~2199/2169.Count-Operations-to-Obtain-Zero.md b/website/content/ChapterFour/2100~2199/2169.Count-Operations-to-Obtain-Zero.md
new file mode 100644
index 000000000..588a18794
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/2169.Count-Operations-to-Obtain-Zero.md
@@ -0,0 +1,80 @@
+# [2169. Count Operations to Obtain Zero](https://leetcode.com/problems/count-operations-to-obtain-zero/)
+
+
+## 题目
+
+You are given two **non-negative** integers `num1` and `num2`.
+
+In one **operation**, if `num1 >= num2`, you must subtract `num2` from `num1`, otherwise subtract `num1` from `num2`.
+
+- For example, if `num1 = 5` and `num2 = 4`, subtract `num2` from `num1`, thus obtaining `num1 = 1` and `num2 = 4`. However, if `num1 = 4` and `num2 = 5`, after one operation, `num1 = 4` and `num2 = 1`.
+
+Return *the **number of operations** required to make either* `num1 = 0` *or* `num2 = 0`.
+
+**Example 1:**
+
+```
+Input: num1 = 2, num2 = 3
+Output: 3
+Explanation:
+- Operation 1: num1 = 2, num2 = 3. Since num1 < num2, we subtract num1 from num2 and get num1 = 2, num2 = 3 - 2 = 1.
+- Operation 2: num1 = 2, num2 = 1. Since num1 > num2, we subtract num2 from num1.
+- Operation 3: num1 = 1, num2 = 1. Since num1 == num2, we subtract num2 from num1.
+Now num1 = 0 and num2 = 1. Since num1 == 0, we do not need to perform any further operations.
+So the total number of operations required is 3.
+
+```
+
+**Example 2:**
+
+```
+Input: num1 = 10, num2 = 10
+Output: 1
+Explanation:
+- Operation 1: num1 = 10, num2 = 10. Since num1 == num2, we subtract num2 from num1 and get num1 = 10 - 10 = 0.
+Now num1 = 0 and num2 = 10. Since num1 == 0, we are done.
+So the total number of operations required is 1.
+
+```
+
+**Constraints:**
+
+- `0 <= num1, num2 <= 10^5`
+
+## 题目大意
+
+给你两个 非负 整数 num1 和 num2 。每一步 操作 中，如果 num1 >= num2 ，你必须用 num1 减 num2 ；否则，你必须用 num2 减 num1 。
+
+- 例如，num1 = 5 且 num2 = 4 ，应该用 num1 减 num2 ，因此，得到 num1 = 1 和 num2 = 4 。然而，如果 num1 = 4且 num2 = 5 ，一步操作后，得到 num1 = 4 和 num2 = 1 。
+
+返回使 num1 = 0 或 num2 = 0 的 操作数 。
+
+## 解题思路
+
+- 简单题，按照题意模拟，每次两个数字相减，便累加操作次数。当某个数字变为 0 时，输出操作次数。
+
+## 代码
+
+```go
+package leetcode
+
+func countOperations(num1 int, num2 int) int {
+	res := 0
+	for num1 != 0 && num2 != 0 {
+		if num1 >= num2 {
+			num1 -= num2
+		} else {
+			num2 -= num1
+		}
+		res++
+	}
+	return res
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2170.Minimum-Operations-to-Make-the-Array-Alternating/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2100~2199/2170.Minimum-Operations-to-Make-the-Array-Alternating.md b/website/content/ChapterFour/2100~2199/2170.Minimum-Operations-to-Make-the-Array-Alternating.md
new file mode 100644
index 000000000..e41c64ec8
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/2170.Minimum-Operations-to-Make-the-Array-Alternating.md
@@ -0,0 +1,137 @@
+# [2170. Minimum Operations to Make the Array Alternating](https://leetcode.com/problems/minimum-operations-to-make-the-array-alternating/)
+
+
+## 题目
+
+You are given a **0-indexed** array `nums` consisting of `n` positive integers.
+
+The array `nums` is called **alternating** if:
+
+- `nums[i - 2] == nums[i]`, where `2 <= i <= n - 1`.
+- `nums[i - 1] != nums[i]`, where `1 <= i <= n - 1`.
+
+In one **operation**, you can choose an index `i` and **change** `nums[i]` into **any** positive integer.
+
+Return *the **minimum number of operations** required to make the array alternating*.
+
+**Example 1:**
+
+```
+Input: nums = [3,1,3,2,4,3]
+Output: 3
+Explanation:
+One way to make the array alternating is by converting it to [3,1,3,1,3,1].
+The number of operations required in this case is 3.
+It can be proven that it is not possible to make the array alternating in less than 3 operations.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,2,2,2]
+Output: 2
+Explanation:
+One way to make the array alternating is by converting it to [1,2,1,2,1].
+The number of operations required in this case is 2.
+Note that the array cannot be converted to [2,2,2,2,2] because in this case nums[0] == nums[1] which violates the conditions of an alternating array.
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 10^5`
+- `1 <= nums[i] <= 10^5`
+
+## 题目大意
+
+给你一个下标从 0 开始的数组 nums ，该数组由 n 个正整数组成。
+
+如果满足下述条件，则数组 nums 是一个 交替数组 ：
+
+- nums[i - 2] == nums[i] ，其中 2 <= i <= n - 1 。
+- nums[i - 1] != nums[i] ，其中 1 <= i <= n - 1 。
+
+在一步 操作 中，你可以选择下标 i 并将 nums[i] 更改 为 任一 正整数。返回使数组变成交替数组的 最少操作数 。
+
+**提示：**
+
+- `1 <= nums.length <= 10^5`
+- `1 <= nums[i] <= 10^5`
+
+## 解题思路
+
+- 题目要求最少操作数，即留下出现频次最多的数字，剩下的数字都替换成这个数字。先将每个数字出现的频次统计出来，然后按照频次从大到小排序。优先选择出现频次高的数字。
+- 有几种“特殊”情况需要处理：当奇数下标的数字频次最大的数字和偶数下标的数字频次最大的数字相同（数字相同，频次不同），这时应选取频次大的数字留下；当数字相同，频次也相同，这时要看奇数下标和偶数下标的数字分别有几个。 如果其中一个只有一种数字，那么另外一组数字则需都变成该组频次第二大的数字，例如奇数下标的数字全是 1，频次是 3，偶数下标的数字是 1，最大频次是 2。第二频次的数字是 9，频次是 1 。那么这种情况下，选择奇数下标的数字 1，和偶数下标数字 9 。将偶数下标不是 9 的数字改变成 9 ；更近一步，如果奇数下标和偶数下标都只有一个数字，频次相同，那么只能改变奇数下标或者偶数下标的所有数字。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+type node struct {
+	value int
+	count int
+}
+
+func minimumOperations(nums []int) int {
+	if len(nums) == 1 {
+		return 0
+	}
+	res, odd, even, oddMap, evenMap := 0, []node{}, []node{}, map[int]int{}, map[int]int{}
+
+	for i := 0; i < len(nums); i += 2 {
+		evenMap[nums[i]]++
+	}
+	for k, v := range evenMap {
+		even = append(even, node{value: k, count: v})
+	}
+	sort.Slice(even, func(i, j int) bool {
+		return even[i].count > even[j].count
+	})
+
+	for i := 1; i < len(nums); i += 2 {
+		oddMap[nums[i]]++
+	}
+	for k, v := range oddMap {
+		odd = append(odd, node{value: k, count: v})
+	}
+	sort.Slice(odd, func(i, j int) bool {
+		return odd[i].count > odd[j].count
+	})
+
+	if even[0].value == odd[0].value {
+		if len(even) == 1 && len(odd) != 1 {
+			res = len(nums) - even[0].count - odd[1].count
+		} else if len(odd) == 1 && len(even) != 1 {
+			res = len(nums) - odd[0].count - even[1].count
+		} else if len(odd) == 1 && len(even) == 1 {
+			res = len(nums) / 2
+		} else {
+			// both != 1
+			res = min(len(nums)-odd[0].count-even[1].count, len(nums)-odd[1].count-even[0].count)
+		}
+	} else {
+		res = len(nums) - even[0].count - odd[0].count
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2169.Count-Operations-to-Obtain-Zero/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2171.Removing-Minimum-Number-of-Magic-Beans/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2100~2199/2171.Removing-Minimum-Number-of-Magic-Beans.md b/website/content/ChapterFour/2100~2199/2171.Removing-Minimum-Number-of-Magic-Beans.md
new file mode 100644
index 000000000..410c13c7c
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/2171.Removing-Minimum-Number-of-Magic-Beans.md
@@ -0,0 +1,97 @@
+# [2171. Removing Minimum Number of Magic Beans](https://leetcode.com/problems/removing-minimum-number-of-magic-beans/)
+
+
+## 题目
+
+You are given an array of **positive** integers `beans`, where each integer represents the number of magic beans found in a particular magic bag.
+
+**Remove** any number of beans (**possibly none**) from each bag such that the number of beans in each remaining **non-empty** bag (still containing **at least one** bean) is **equal**. Once a bean has been removed from a bag, you are **not** allowed to return it to any of the bags.
+
+Return *the **minimum** number of magic beans that you have to remove*.
+
+**Example 1:**
+
+```
+Input: beans = [4,1,6,5]
+Output: 4
+Explanation:
+- We remove 1 bean from the bag with only 1 bean.
+  This results in the remaining bags: [4,0,6,5]
+- Then we remove 2 beans from the bag with 6 beans.
+  This results in the remaining bags: [4,0,4,5]
+- Then we remove 1 bean from the bag with 5 beans.
+  This results in the remaining bags: [4,0,4,4]
+We removed a total of 1 + 2 + 1 = 4 beans to make the remaining non-empty bags have an equal number of beans.
+There are no other solutions that remove 4 beans or fewer.
+
+```
+
+**Example 2:**
+
+```
+Input: beans = [2,10,3,2]
+Output: 7
+Explanation:
+- We remove 2 beans from one of the bags with 2 beans.
+  This results in the remaining bags: [0,10,3,2]
+- Then we remove 2 beans from the other bag with 2 beans.
+  This results in the remaining bags: [0,10,3,0]
+- Then we remove 3 beans from the bag with 3 beans.
+  This results in the remaining bags: [0,10,0,0]
+We removed a total of 2 + 2 + 3 = 7 beans to make the remaining non-empty bags have an equal number of beans.
+There are no other solutions that removes 7 beans or fewer.
+
+```
+
+**Constraints:**
+
+- `1 <= beans.length <= 10^5`
+- `1 <= beans[i] <= 10^5`
+
+## 题目大意
+
+给你一个 正 整数数组 beans ，其中每个整数表示一个袋子里装的魔法豆的数目。
+
+请你从每个袋子中 拿出 一些豆子（也可以 不拿出），使得剩下的 非空 袋子中（即 至少 还有 一颗 魔法豆的袋子）魔法豆的数目 相等 。一旦魔法豆从袋子中取出，你不能将它放到任何其他的袋子中。请你返回你需要拿出魔法豆的 最少数目。
+
+**提示：**
+
+- `1 <= beans.length <= 10^5`
+- `1 <= beans[i] <= 10^5`
+
+## 解题思路
+
+- 这一题没有特别巧妙的方法。最初思路来源于暴力解法。从第一个袋子开始，依次以每个袋子中的豆子为基准，改变其他袋子里面的豆子数，使得其他袋子里面的豆子都和基准袋子中豆子一样多。
+- 如果从下标为 0 扫到下标 n-1 ，这中间会有大量重复计算。有些计算区间和的操作，反复计算了很多遍，导致算法不高效。由于移除豆子数量多少和基准袋豆子数量强相关，所以先排序。如果袋子内豆子数目小于基准袋的豆子，`0 ≤ j < i`，那么这些袋子内的豆子数量会归零。需要移除 `beans[0] + beans[1] + ... + beans[i-1]` 个豆子；如果袋子内豆子数目大于等于基准袋的豆子，`j ≥ i` ，那么这些袋子内的豆子需要调整为 `beans[i]` 个。需要移除 `(beans[i] - beans[i]) + (beans[i+1] - beans[i]) + (beans[i+2] - beans[i]) + ... + (beans[n-1] - beans[i]) = beans[i]+ ... + beans[n-1] - (n-i) * beans[i]` 个豆子。将这 2 种情况综合起来，那么总共需要移除 `sum(beans) - (N - i) * beans[i]` 个豆子。综上，先排序，然后从小到大扫一遍数组，动态维护最少移除豆子的个数即可。
+
+## 代码
+
+```go
+package leetcode
+
+import "sort"
+
+func minimumRemoval(beans []int) int64 {
+	sort.Ints(beans)
+	sum, mx := 0, 0
+	for i, v := range beans {
+		sum += v
+		mx = max(mx, (len(beans)-i)*v)
+	}
+	return int64(sum - mx)
+}
+
+func max(a, b int) int {
+	if b > a {
+		return b
+	}
+	return a
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2170.Minimum-Operations-to-Make-the-Array-Alternating/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2180.Count-Integers-With-Even-Digit-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2100~2199/2180.Count-Integers-With-Even-Digit-Sum.md b/website/content/ChapterFour/2100~2199/2180.Count-Integers-With-Even-Digit-Sum.md
new file mode 100644
index 000000000..4685d7d37
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/2180.Count-Integers-With-Even-Digit-Sum.md
@@ -0,0 +1,76 @@
+# [2180. Count Integers With Even Digit Sum](https://leetcode.com/problems/count-integers-with-even-digit-sum/)
+
+
+## 题目
+
+Given a positive integer `num`, return *the number of positive integers **less than or equal to*** `num` *whose digit sums are **even***.
+
+The **digit sum** of a positive integer is the sum of all its digits.
+
+**Example 1:**
+
+```
+Input: num = 4
+Output: 2
+Explanation:
+The only integers less than or equal to 4 whose digit sums are even are 2 and 4.
+
+```
+
+**Example 2:**
+
+```
+Input: num = 30
+Output: 14
+Explanation:
+The 14 integers less than or equal to 30 whose digit sums are even are
+2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22, 24, 26, and 28.
+
+```
+
+**Constraints:**
+
+- `1 <= num <= 1000`
+
+## 题目大意
+
+给你一个正整数 num ，请你统计并返回 小于或等于 num 且各位数字之和为 偶数 的正整数的数目。
+
+正整数的 各位数字之和 是其所有位上的对应数字相加的结果。
+
+## 解题思路
+
+- 简单题。依照题意，计算每个数的各位数字之和，如何和为偶数，则统计结果加一。最后输出统计结果即可。
+
+## 代码
+
+```go
+package leetcode
+
+func countEven(num int) int {
+	count := 0
+	for i := 1; i <= num; i++ {
+		if addSum(i)%2 == 0 {
+			count++
+		}
+	}
+	return count
+}
+
+func addSum(num int) int {
+	sum := 0
+	tmp := num
+	for tmp != 0 {
+		sum += tmp % 10
+		tmp = tmp / 10
+	}
+	return sum
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2171.Removing-Minimum-Number-of-Magic-Beans/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2181.Merge-Nodes-in-Between-Zeros/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2100~2199/2181.Merge-Nodes-in-Between-Zeros.md b/website/content/ChapterFour/2100~2199/2181.Merge-Nodes-in-Between-Zeros.md
new file mode 100644
index 000000000..35ef63ea0
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/2181.Merge-Nodes-in-Between-Zeros.md
@@ -0,0 +1,103 @@
+# [2181. Merge Nodes in Between Zeros](https://leetcode.com/problems/merge-nodes-in-between-zeros/)
+
+## 题目
+
+You are given the `head` of a linked list, which contains a series of integers **separated** by `0`'s. The **beginning** and **end** of the linked list will have `Node.val == 0`.
+
+For **every** two consecutive `0`'s, **merge** all the nodes lying in between them into a single node whose value is the **sum** of all the merged nodes. The modified list should not contain any `0`'s.
+
+Return *the* `head` *of the modified linked list*.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2022/02/02/ex1-1.png](https://assets.leetcode.com/uploads/2022/02/02/ex1-1.png)
+
+```
+Input: head = [0,3,1,0,4,5,2,0]
+Output: [4,11]
+Explanation:
+The above figure represents the given linked list. The modified list contains
+- The sum of the nodes marked in green: 3 + 1 = 4.
+- The sum of the nodes marked in red: 4 + 5 + 2 = 11.
+
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2022/02/02/ex2-1.png](https://assets.leetcode.com/uploads/2022/02/02/ex2-1.png)
+
+```
+Input: head = [0,1,0,3,0,2,2,0]
+Output: [1,3,4]
+Explanation:
+The above figure represents the given linked list. The modified list contains
+- The sum of the nodes marked in green: 1 = 1.
+- The sum of the nodes marked in red: 3 = 3.
+- The sum of the nodes marked in yellow: 2 + 2 = 4.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the list is in the range `[3, 2 * 10^5]`.
+- `0 <= Node.val <= 1000`
+- There are **no** two consecutive nodes with `Node.val == 0`.
+- The **beginning** and **end** of the linked list have `Node.val == 0`.
+
+## 题目大意
+
+给你一个链表的头节点 head ，该链表包含由 0 分隔开的一连串整数。链表的 开端 和 末尾 的节点都满足 Node.val == 0 。对于每两个相邻的 0 ，请你将它们之间的所有节点合并成一个节点，其值是所有已合并节点的值之和。然后将所有 0 移除，修改后的链表不应该含有任何 0 。
+
+返回修改后链表的头节点 head 。
+
+## 解题思路
+
+- 简单题。合并链表中两个值为 0 的节点。从头开始遍历链表，遇到节点值不为 0 的节点便累加；遇到节点值为 0 的节点，将累加值转换成结果链表要输出的节点值，然后继续遍历。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"github.com/halfrost/leetcode-go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func mergeNodes(head *ListNode) *ListNode {
+	res := &ListNode{}
+	h := res
+	if head.Next == nil {
+		return &structures.ListNode{}
+	}
+	cur := head
+	sum := 0
+	for cur.Next != nil {
+		if cur.Next.Val != 0 {
+			sum += cur.Next.Val
+		} else {
+			h.Next = &ListNode{Val: sum, Next: nil}
+			h = h.Next
+			sum = 0
+		}
+		cur = cur.Next
+	}
+	return res.Next
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2180.Count-Integers-With-Even-Digit-Sum/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2182.Construct-String-With-Repeat-Limit/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2100~2199/2182.Construct-String-With-Repeat-Limit.md b/website/content/ChapterFour/2100~2199/2182.Construct-String-With-Repeat-Limit.md
new file mode 100644
index 000000000..0507b90ee
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/2182.Construct-String-With-Repeat-Limit.md
@@ -0,0 +1,105 @@
+# [2182. Construct String With Repeat Limit](https://leetcode.com/problems/construct-string-with-repeat-limit/)
+
+
+## 题目
+
+You are given a string `s` and an integer `repeatLimit`. Construct a new string `repeatLimitedString` using the characters of `s` such that no letter appears **more than** `repeatLimit` times **in a row**. You do **not** have to use all characters from `s`.
+
+Return *the **lexicographically largest*** `repeatLimitedString` *possible*.
+
+A string `a` is **lexicographically larger** than a string `b` if in the first position where `a` and `b` differ, string `a` has a letter that appears later in the alphabet than the corresponding letter in `b`. If the first `min(a.length, b.length)` characters do not differ, then the longer string is the lexicographically larger one.
+
+**Example 1:**
+
+```
+Input: s = "cczazcc", repeatLimit = 3
+Output: "zzcccac"
+Explanation: We use all of the characters from s to construct the repeatLimitedString "zzcccac".
+The letter 'a' appears at most 1 time in a row.
+The letter 'c' appears at most 3 times in a row.
+The letter 'z' appears at most 2 times in a row.
+Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
+The string is the lexicographically largest repeatLimitedString possible so we return "zzcccac".
+Note that the string "zzcccca" is lexicographically larger but the letter 'c' appears more than 3 times in a row, so it is not a valid repeatLimitedString.
+
+```
+
+**Example 2:**
+
+```
+Input: s = "aababab", repeatLimit = 2
+Output: "bbabaa"
+Explanation: We use only some of the characters from s to construct the repeatLimitedString "bbabaa".
+The letter 'a' appears at most 2 times in a row.
+The letter 'b' appears at most 2 times in a row.
+Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
+The string is the lexicographically largest repeatLimitedString possible so we return "bbabaa".
+Note that the string "bbabaaa" is lexicographically larger but the letter 'a' appears more than 2 times in a row, so it is not a valid repeatLimitedString.
+
+```
+
+**Constraints:**
+
+- `1 <= repeatLimit <= s.length <= 10^5`
+- `s` consists of lowercase English letters.
+
+## 题目大意
+
+给你一个字符串 s 和一个整数 repeatLimit ，用 s 中的字符构造一个新字符串 repeatLimitedString ，使任何字母 连续 出现的次数都不超过 repeatLimit 次。你不必使用 s 中的全部字符。
+
+返回 字典序最大的 repeatLimitedString 。
+
+如果在字符串 a 和 b 不同的第一个位置，字符串 a 中的字母在字母表中出现时间比字符串 b 对应的字母晚，则认为字符串 a 比字符串 b 字典序更大 。如果字符串中前 min(a.length, b.length) 个字符都相同，那么较长的字符串字典序更大。
+
+## 解题思路
+
+- 利用贪心的思想，由于题意要求返回字典序最大的字符串，所以先从字典序最大的字母开始选起。然后选择当前字典序最大的字母个数和 limit 的最小值。如果当前字典序最大的字母比较多，多于 limit，不能一直选择它。选完 limit 个以后，需要选一个字典序次大的字母，选完这个字母以后再次选择字典序最大的字母。因为 limit 限制字母不能连续多于 limit 个。如此循环，直到所有的字母都选完。这样的策略排列出来的字母串为最大字典序。
+
+## 代码
+
+```go
+package leetcode
+
+func repeatLimitedString(s string, repeatLimit int) string {
+	cnt := make([]int, 26)
+	for _, c := range s {
+		cnt[int(c-'a')]++
+	}
+	var ns []byte
+	for i := 25; i >= 0; {
+		k := i - 1
+		for cnt[i] > 0 {
+			for j := 0; j < min(cnt[i], repeatLimit); j++ {
+				ns = append(ns, byte(i)+'a')
+			}
+			cnt[i] -= repeatLimit
+			if cnt[i] > 0 {
+				for ; k >= 0 && cnt[k] == 0; k-- {
+				}
+				if k < 0 {
+					break
+				} else {
+					ns = append(ns, byte(k)+'a')
+					cnt[k]--
+				}
+			}
+		}
+		i = k
+	}
+	return string(ns)
+}
+func min(a, b int) int {
+	if a < b {
+		return a
+	} else {
+		return b
+	}
+}
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2181.Merge-Nodes-in-Between-Zeros/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2183.Count-Array-Pairs-Divisible-by-K/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/2100~2199/2183.Count-Array-Pairs-Divisible-by-K.md b/website/content/ChapterFour/2100~2199/2183.Count-Array-Pairs-Divisible-by-K.md
new file mode 100644
index 000000000..71d925113
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/2183.Count-Array-Pairs-Divisible-by-K.md
@@ -0,0 +1,90 @@
+# [2183. Count Array Pairs Divisible by K](https://leetcode.com/problems/count-array-pairs-divisible-by-k/)
+
+
+## 题目
+
+Given a **0-indexed** integer array `nums` of length `n` and an integer `k`, return *the **number of pairs*** `(i, j)` *such that:*
+
+- `0 <= i < j <= n - 1` *and*
+- `nums[i] * nums[j]` *is divisible by* `k`.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3,4,5], k = 2
+Output: 7
+Explanation:
+The 7 pairs of indices whose corresponding products are divisible by 2 are
+(0, 1), (0, 3), (1, 2), (1, 3), (1, 4), (2, 3), and (3, 4).
+Their products are 2, 4, 6, 8, 10, 12, and 20 respectively.
+Other pairs such as (0, 2) and (2, 4) have products 3 and 15 respectively, which are not divisible by 2.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3,4], k = 5
+Output: 0
+Explanation: There does not exist any pair of indices whose corresponding product is divisible by 5.
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 10^5`
+- `1 <= nums[i], k <= 10^5`
+
+## 题目大意
+
+给你一个下标从 0 开始、长度为 n 的整数数组 nums 和一个整数 k ，返回满足下述条件的下标对 (i, j) 的数目：
+
+- 0 <= i < j <= n - 1 且
+- nums[i] * nums[j] 能被 k 整除。
+
+## 解题思路
+
+- 先找出 num 中每个元素与 k 的最大公约数。并统计这些公约数出现的频次，将数据保存在 map 中。在计算过程中，循环可以只需算到 {{< katex >}}{O(\sqrt {k})}{{< /katex >}} , 因为每一个 gcd[i] 一定是 k 的因数，而它出现的频次不会超过 {{< katex >}}{O(\sqrt {k})}{{< /katex >}}。简单证明一下：假设因子 v 和 k/v 这两个因数为 k 的因子。v 和 k/v 必至少有 1 个小于等于 {{< katex >}}\sqrt {k}{{< /katex >}}。所以 k 的因子也不会超过 2 * {{< katex >}}\sqrt {k}{{< /katex >}} = {{< katex >}}{O(\sqrt {k})}{{< /katex >}} 个。
+- 算出上述的 map 以后，2 层循环暴力遍历 key 值，如果 a * b 能被 k 整除，并且 a 和 b 不相同，那么 a 和 b 对应的 value 值相乘即为满足条件的下标对数；如果 a 和 b 相同，那么下标对数为 {{< katex >}}C_{n}^{2}{{< /katex >}}。最后累加结果即可。
+
+## 代码
+
+```go
+package leetcode
+
+import "math"
+
+func countPairs(nums []int, k int) int64 {
+	n := int(math.Sqrt(float64(k)))
+	gcds, res := make(map[int]int, n), 0
+	for _, num := range nums {
+		gcds[gcd(num, k)]++
+	}
+
+	for a, n1 := range gcds {
+		for b, n2 := range gcds {
+			if a > b || (a*b)%k != 0 {
+				continue
+			}
+			if a != b {
+				res += n1 * n2
+			} else { // a == b
+				res += n1 * (n1 - 1) / 2
+			}
+		}
+	}
+	return int64(res)
+}
+
+func gcd(a, b int) int {
+	for a%b != 0 {
+		a, b = b, a%b
+	}
+	return b
+}
+```
+
+
+----------------------------------------------
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/2100~2199/2182.Construct-String-With-Repeat-Limit/">⬅️上一页</a></p>
+
diff --git a/website/content/ChapterFour/2100~2199/_index.md b/website/content/ChapterFour/2100~2199/_index.md
new file mode 100644
index 000000000..e954f0877
--- /dev/null
+++ b/website/content/ChapterFour/2100~2199/_index.md
@@ -0,0 +1,4 @@
+---
+bookCollapseSection: true
+weight: 20
+---
diff --git a/website/content/ChapterFour/2200~2299/_index.md b/website/content/ChapterFour/2200~2299/_index.md
new file mode 100644
index 000000000..e954f0877
--- /dev/null
+++ b/website/content/ChapterFour/2200~2299/_index.md
@@ -0,0 +1,4 @@
+---
+bookCollapseSection: true
+weight: 20
+---
diff --git a/website/content/ChapterFour/_index.md b/website/content/ChapterFour/_index.md
new file mode 100644
index 000000000..b3eaba6b8
--- /dev/null
+++ b/website/content/ChapterFour/_index.md
@@ -0,0 +1,25 @@
+---
+title: 第四章 LeetCode 题解
+type: docs
+weight: 4
+---
+
+# 第四章 LeetCode 题解
+
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/GOPHERCON_.png'>
+</p>
+
+
+这一章就是 LeetCode 的题解了。笔者目前只刷到 608 题，题解这里有 520 题，都已经 runtime beats 100% 了。相差的 88 题是还没有 beats 100% 的，笔者还需要继续优化~
+
+题解慢慢更新中，欢迎大家提出更好的解法。点击页面下方的 edit，会跳转到 github 对应的页面 markdown 中，可以提交你的最优解 PR。
+
+让我们在题解的太空遨游吧~
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterThree/LFUCache/">⬅️上一章</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0001.Two-Sum/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterFour/pytool/AddTOC.py b/website/content/ChapterFour/pytool/AddTOC.py
new file mode 100644
index 000000000..356882662
--- /dev/null
+++ b/website/content/ChapterFour/pytool/AddTOC.py
@@ -0,0 +1,25 @@
+import os
+from os.path import join
+from shutil import move
+
+toc_string = """\n
+
+---
+bookToc: false
+---
+
+\n"""
+
+current_working_dir = os.getcwd()
+# print(f"current_working_dir: {current_working_dir}")
+
+dir_names = os.listdir(current_working_dir)
+# print(dir_names)
+print(len(dir_names))
+for file_name in dir_names:
+        if(file_name.endswith('.md')):
+                with open(file_name, "r") as myfile:
+                        codeContent = toc_string + myfile.read()
+                with open(file_name, "w") as myfile:
+                        myfile.write(codeContent)
+print("Finished")
\ No newline at end of file
diff --git a/website/content/ChapterFour/pytool/CutContent.py b/website/content/ChapterFour/pytool/CutContent.py
new file mode 100644
index 000000000..28dbc720d
--- /dev/null
+++ b/website/content/ChapterFour/pytool/CutContent.py
@@ -0,0 +1,24 @@
+import os
+import re
+import glob
+
+# file_name = 'Array.md'
+reg = "## 题目大意"
+
+current_working_dir = os.getcwd()
+# print(f"current_working_dir: {current_working_dir}")
+
+dir_names = glob.glob("*.md")
+dir_names.sort()
+print(len(dir_names))
+
+for file_name in dir_names:
+    # print(file_name)
+    with open(file_name, "r") as myfile:
+        codeContent = myfile.read()
+        findIndex = codeContent.find(reg)
+        # print(findIndex)
+        content = codeContent[findIndex:]
+    with open(file_name, "w") as myfile:
+        myfile.write(content)
+print("Finished")
\ No newline at end of file
diff --git a/website/content/ChapterFour/pytool/GenerateIndex.py b/website/content/ChapterFour/pytool/GenerateIndex.py
new file mode 100644
index 000000000..e4dd33f70
--- /dev/null
+++ b/website/content/ChapterFour/pytool/GenerateIndex.py
@@ -0,0 +1,21 @@
+import os
+import glob
+
+content = []
+indexFile = 'index.txt'
+current_working_dir = os.getcwd()
+# print(f"current_working_dir: {current_working_dir}")
+
+dir_names = glob.glob("*.md")
+dir_names.sort()
+# print(dir_names)
+print(len(dir_names))
+for file_name in dir_names:
+        # - [0001. Two Sum]({{< relref "/ChapterFour/withouttoc/0001. Two Sum.md" >}})
+        # content = '- [' + file_name[:-3] + ']' + '({{< relref "/ChapterFour/withouttoc/' + file_name + '" >}})'
+        content.append('- [{}]({{{{< relref "/ChapterFour/{}" >}}}})'.format(file_name[:-3],file_name))
+
+with open(indexFile, "w") as myfile:
+        myfile.write('\n'.join(content))
+
+print("Finished")
\ No newline at end of file
diff --git a/website/content/ChapterFour/pytool/GenerateOne.py b/website/content/ChapterFour/pytool/GenerateOne.py
new file mode 100644
index 000000000..233171b52
--- /dev/null
+++ b/website/content/ChapterFour/pytool/GenerateOne.py
@@ -0,0 +1,24 @@
+import os
+from os.path import join
+from shutil import move
+import glob
+
+content = []
+oneFile = 'ChapterTwo_OnePDF.md'
+current_working_dir = os.getcwd()
+# print(f"current_working_dir: {current_working_dir}")
+
+dir_names = glob.glob("*.md")
+dir_names.sort()
+# print(dir_names)
+print(len(dir_names))
+for file_name in dir_names:
+        # print(file_name)
+        with open(file_name, "r") as myfile:
+                content = myfile.read()
+        with open(oneFile, "a") as myfile:
+                myfile.write("\n")
+                myfile.write(content)
+                myfile.write("\n")
+
+print("Finished")
\ No newline at end of file
diff --git a/website/content/ChapterFour/pytool/GetFile.py b/website/content/ChapterFour/pytool/GetFile.py
new file mode 100644
index 000000000..154f4397f
--- /dev/null
+++ b/website/content/ChapterFour/pytool/GetFile.py
@@ -0,0 +1,56 @@
+import os
+from os.path import join
+from shutil import move
+
+toc_string = """\n
+
+---
+bookToc: false
+---
+
+\n"""
+
+pre_string = """
+
+## 代码
+
+```go
+
+"""
+end_string = "\n```"
+codeContent = ""
+current_working_dir = os.getcwd()
+# print(f"current_working_dir: {current_working_dir}")
+
+# dir_names = os.listdir(current_working_dir)
+
+# 重命名目录。 os.rename(dir_name, new_dir_name)
+
+# 遍历目录
+# dir_path是当前遍历到的目录。dir_names是dir_path下的文件夹列表。file_names是是dir_path下的文件列表
+# 如果想实现目录白名单，将白名单目录从dir_names中去除即可
+for (dir_path, dir_names, file_names) in os.walk(current_working_dir):
+    # print(f"当前遍历到的目录: {dir_path}")
+    os.chdir(dir_path)
+    files = dir_path.split("/")
+    new_file_name = files[len(files) - 1] + '.md'
+    
+    print(f"当前所在的文件夹: {os.getcwd()}")
+    for file_name in file_names:
+        if(file_name.endswith('.go') and not file_name.endswith('_test.go')):
+            print(f"当前所在文件: {file_name}")
+            with open(file_name, "r") as myfile:
+                codeContent = myfile.read()
+            break
+            # print(codeContent)
+    for file_name in file_names:
+        if(file_name.endswith('.md')):
+            print(f"当前所在文件: {file_name}")
+            with open(file_name, "a") as myfile:
+                myfile.write(pre_string)
+                myfile.write(codeContent)
+                myfile.write(end_string)
+            os.rename(file_name, new_file_name)
+            move(join(dir_path, new_file_name), current_working_dir)
+            break
+    
\ No newline at end of file
diff --git a/website/content/ChapterFour/pytool/WordCount.py b/website/content/ChapterFour/pytool/WordCount.py
new file mode 100644
index 000000000..3bae213e9
--- /dev/null
+++ b/website/content/ChapterFour/pytool/WordCount.py
@@ -0,0 +1,29 @@
+from collections import defaultdict
+import glob
+import os
+
+
+def str_count2(str):
+    count_zh = count_dg = 0
+    for s in str:
+        # 中文字符范围
+        if '\u4e00' <= s <= '\u9fff':
+            count_zh += 1
+        if s.isdigit():
+            count_dg += 1
+    # print(count_zh + count_dg)
+    return count_zh + count_dg
+
+current_working_dir = os.getcwd()
+# print(f"current_working_dir: {current_working_dir}")
+
+dir_names = glob.glob("*.md")
+dir_names.sort()
+
+word_count = 0
+for file_name in dir_names:
+    with open(file_name, "r") as myfile:
+        codeContent = myfile.read()
+        print("当前读取文件: {}, 字数统计: {}".format(file_name, str_count2(codeContent)))
+        word_count += str_count2(codeContent)
+print(word_count)
\ No newline at end of file
diff --git a/website/content/ChapterOne/Algorithm.md b/website/content/ChapterOne/Algorithm.md
new file mode 100644
index 000000000..45b4f74ff
--- /dev/null
+++ b/website/content/ChapterOne/Algorithm.md
@@ -0,0 +1,34 @@
+---
+title: 1.2 算法知识
+type: docs
+weight: 2
+---
+
+# 算法知识
+
+以下是笔者整理的算法相关的知识。希望能把常见的算法都枚举穷尽。如有遗漏，欢迎大家赐教，提 PR。相关题目还在慢慢整理中，讲解的文章还在创作中。
+
+> 刷题只是提升算法能力的手段，最终目的应该是提升自我的思维能力，知识需要凝结成块，那么就把这些总结在第一章这两节中，让它得到升华吧~希望读者在刷完题之后再回过头来看这个表格，能很清晰的梳理自己的知识体系，查缺补漏，尽早完善。
+
+| 算法 | 具体类型 | 相关题目 | 讲解文章 | 
+|:-------:|:-------|:------|:------|
+|排序算法|1. 冒泡排序<br>2. 插入排序<br>3. 选择排序<br>4. 希尔 Shell 排序<br>5. 快速排序<br>6. 归并排序<br>7. 堆排序<br>8. 线性排序算法<br>9. 自省排序<br>10. 间接排序<br>11. 计数排序<br>12. 基数排序<br>13. 桶排序<br>14. 外部排序 - k 路归并败者树<br>15. 外部排序 - 最佳归并树|||
+|递归与分治||1. 二分搜索/查找<br>2. 大整数的乘法<br>3. Strassen 矩阵乘法<br>4. 棋盘覆盖<br>5. 合并排序<br>6. 快速排序<br>7. 线性时间选择<br>8. 最接近点对问题<br>9. 循环赛日程表<br>||
+|动态规划||1. 矩阵连乘问题<br>2. 最长公共子序列<br>3. 最大子段和<br>4. 凸多边形最优三角剖分<br>5. 多边形游戏<br>6. 图像压缩<br>7. 电路布线<br>8. 流水作业调度<br>9. 0-1 背包问题/背包九讲<br>10. 最优二叉搜索树<br>11. 动态规划加速原理<br>12. 树型 DP<br>||
+|贪心||1. 活动安排问题<br>2. 最优装载<br>3. 哈夫曼编码<br>4. 单源最短路径<br>5. 最小生成树<br>6. 多机调度问题<br>||
+|回溯法||1. 装载问题<br>2. 批处理作业调度<br>3. 符号三角形问题<br>4. n 后问题<br>5. 0-1 背包问题<br>6. 最大团问题<br>7. 图的 m 着色问题<br>8. 旅行售货员问题<br>9. 圆排列问题<br>10. 电路板排列问题<br>11. 连续邮资问题<br>||
+|搜索|1. 枚举<br>2. DFS<br>3. BFS<br>4. 启发式搜索<br>|||
+|随机化|1. 随机数<br>2. 数值随机化算法<br>3. Sherwood 舍伍德算法<br>4. Las Vegas 拉斯维加斯算法<br>5. Monte Carlo 蒙特卡罗算法<br>|1. 计算 π 值<br>2. 计算定积分<br>3. 解非线性方程组<br>4. 线性时间选择算法<br>5. 跳跃表<br>6. n 后问题<br>7. 整数因子分解<br>8. 主元素问题<br>9. 素数测试<br>||
+|图论|1. 遍历 DFS / BFS<br>2. AOV / AOE 网络<br>3. Kruskal 算法(最小生成树)<br>4. Prim 算法(最小生成树)<br>5. Boruvka 算法(最小生成树)<br>6. Dijkstra 算法(单源最短路径)<br>7. Bellman-Ford 算法(单源最短路径)<br>8. SPFA 算法(单源最短路径)<br>9. Floyd 算法(多源最短路径)<br>10. Johnson 算法(多源最短路径)<br>11. Fleury 算法(欧拉回路)<br>12. Ford-Fulkerson 算法(最大网络流增广路)<br>13. Edmonds-Karp 算法(最大网络流)<br>14. Dinic 算法(最大网络流)<br>15. 一般预流推进算法<br>16. 最高标号预流推进 HLPP 算法<br>17. Primal-Dual 原始对偶算法(最小费用流)18. Kosaraju 算法(有向图强连通分量)<br>19. Tarjan 算法(有向图强连通分量)<br>20. Gabow 算法(有向图强连通分量)<br>21. 匈牙利算法(二分图匹配)<br>22. Hopcroft－Karp 算法(二分图匹配)<br>23. kuhn munkras 算法(二分图最佳匹配)<br>24. Edmonds’ Blossom-Contraction 算法(一般图匹配)<br>|1. 图遍历<br>2. 有向图和无向图的强弱连通性<br>3. 割点/割边<br>3. AOV 网络和拓扑排序<br>4. AOE 网络和关键路径<br>5. 最小代价生成树/次小生成树<br>6. 最短路径问题/第 K 短路问题<br>7. 最大网络流问题<br>8. 最小费用流问题<br>9. 图着色问题<br>10. 差分约束系统<br>11. 欧拉回路<br>12. 中国邮递员问题<br>13. 汉密尔顿回路<br>14. 最佳边割集/最佳点割集/最小边割集/最小点割集/最小路径覆盖/最小点集覆盖 <br>15. 边覆盖集<br>16. 二分图完美匹配和最大匹配问题<br>17. 仙人掌图<br>18. 弦图<br>19. 稳定婚姻问题<br>20. 最大团问题<br>||
+|数论||1. 最大公约数<br> 2. 最小公倍数<br>3. 分解质因数<br>4. 素数判定<br>5. 进制转换<br>6. 高精度计算<br>7. 整除问题<br>8. 同余问题<br>9. 欧拉函数<br>10. 扩展欧几里得<br>11. 置换群<br>12. 母函数<br>13. 离散变换<br>14. 康托展开<br>15. 矩阵<br>16. 向量<br>17. 线性方程组<br>18. 线性规划<br> ||
+|几何||1. 凸包 - Gift wrapping<br>2. 凸包 - Graham scan<br>3. 线段问题<br> 4. 多边形和多面体相关问题<br>||
+|NP 完全|1. 计算模型<br>2. P 类与 NP 类问题<br>3. NP 完全问题<br>4. NP 完全问题的近似算法<br>|1. 随机存取机 RAM<br>2. 随机存取存储程序机 RASP<br>3. 图灵机<br>4. 非确定性图灵机<br>5. P 类与 NP 类语言<br>6. 多项式时间验证<br>7. 多项式时间变换<br>8. Cook定理<br>9. 合取范式的可满足性问题 CNF-SAT<br>10. 3 元合取范式的可满足性问题 3-SAT<br>11. 团问题 CLIQUE<br>12. 顶点覆盖问题 VERTEX-COVER<br>13. 子集和问题 SUBSET-SUM<br>14. 哈密顿回路问题 HAM-CYCLE<br>15. 旅行售货员问题 TSP<br>16. 顶点覆盖问题的近似算法<br>17. 旅行售货员问题近似算法<br>18. 具有三角不等式性质的旅行售货员问题<br>19. 一般的旅行售货员问题<br>20. 集合覆盖问题的近似算法<br>21. 子集和问题的近似算法<br>22. 子集和问题的指数时间算法<br>23. 子集和问题的多项式时间近似格式<br>||
+|位运算| 位操作包括：<br> 1. 取反（NOT）<br> 2. 按位或（OR） <br> 3. 按位异或（XOR） <br> 4. 按位与（AND） <br> 5. 移位: 是一个二元运算符，用来将一个二进制数中的每一位全部都向一个方向移动指定位，溢出的部分将被舍弃，而空缺的部分填入一定的值。<br> | 1.数字范围按位与<br> 2.UTF-8 编码验证<br> 3.数字转换为十六进制数<br> 4.找出最长的超赞子字符串<br> 5.数组异或操作<br> 6.幂集<br> 7.位1的个数<br> 8.二进制表示中质数个计算置位<br> 9.子数组异或查询<br>| [力扣：位运算](https://leetcode-cn.com/tag/bit-manipulation/)|
+|------------|------------------------------------------------------------------|-----------------------------------------------------------------|--------------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterOne/Data_Structure/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterOne/Time_Complexity/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterOne/Data_Structure.md b/website/content/ChapterOne/Data_Structure.md
new file mode 100644
index 000000000..7ccb955e7
--- /dev/null
+++ b/website/content/ChapterOne/Data_Structure.md
@@ -0,0 +1,32 @@
+---
+title: 1.1 数据结构知识
+type: docs
+weight: 1
+---
+
+# 数据结构知识
+
+以下是笔者整理的数据结构相关的知识。希望能把常见的数据结构都枚举穷尽。如有遗漏，欢迎大家赐教，提 PR。相关题目还在慢慢整理中，讲解的文章还在创作中。
+
+> 刷题只是提升算法能力的手段，最终目的应该是提升自我的思维能力，知识需要凝结成块，那么就把这些总结在第一章这两节中，让它得到升华吧~希望读者在刷完题之后再回过头来看这个表格，能很清晰的梳理自己的知识体系，查缺补漏，尽早完善。
+
+| 数据结构 | 变种 | 相关题目 | 讲解文章 | 
+|:-------:|:-------|:------|:------|
+|顺序线性表：向量 <br>Vector||||
+|单链表<br>Singly Linked List|1. 双向链表 Double Linked Lists<br>2. 静态链表 Static List<br>3. 对称矩阵 Symmetric Matrix<br>4. 稀疏矩阵 Sparse Matrix|||
+|哈希表<br>Hash Table|1. 散列函数 Hash Function<br>2. 解决碰撞/填充因子 Collision Resolution<br>|||
+|栈和队列<br>Stack & Queue|1. 广义表 Generalized List/GList<br>2. 双端队列 Deque<br>|||
+|队列<br>Queue|1. 链表实现 Linked List Implementation<br>2. 循环数组实现 ArrayQueue<br>3. 双端队列 Deque<br>4. 优先队列 Priority Queue<br>5. 循环队列 Circular Queue|||
+|字符串<br>String|1. KMP 算法<br>2. 有限状态自动机<br>3. 模式匹配有限状态自动机<br>4. BM 模式匹配算法<br>5. BM-KMP 算法<br>6. BF 算法|||
+|树<br>Tree|1. 二叉树 Binary Tree<br>2. 并查集 Union-Find<br>3. Huffman 树|||
+|数组实现的堆<br>Heap|1. 极大堆和极小堆 Max Heap and Min Heap<br>2. 极大极小堆<br>3. 双端堆 Deap<br>4. d 叉堆|||
+|树实现的堆<br>Heap|1. 左堆 Leftist Tree/Leftist Heap<br>2. 扁堆<br>3. 二项式堆<br>4. 斐波那契堆 Fibonacco Heap<br>5. 配对堆 Pairing Heap|||
+|查找<br>Search|1. 哈希表 Hash<br>2. 跳跃表 Skip List<br>3. 排序二叉树 Binary Sort Tree<br>4. AVL 树<br>5. B 树 / B+ 树 / B* 树<br>6. AA 树<br>7. 红黑树 Red Black Tree<br>8. 排序二叉堆 Binary Heap<br>9. Splay 树<br>10. 双链树 Double Chained Tree<br>11. Trie 树<br>12. R 树|||
+|--------------------------------------------|--------------------------------------------------------------------------------------------|---------------------------|-----------------------------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterOne/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterOne/Algorithm/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterOne/Time_Complexity.md b/website/content/ChapterOne/Time_Complexity.md
new file mode 100644
index 000000000..b68708bb2
--- /dev/null
+++ b/website/content/ChapterOne/Time_Complexity.md
@@ -0,0 +1,139 @@
+---
+title: 1.3 时间复杂度
+type: docs
+weight: 3
+---
+
+# 时间复杂度和空间复杂度
+
+
+## 一. 时间复杂度数据规模
+
+1s 内能解决问题的数据规模：10^6 ~ 10^7
+
+- O(n^2) 算法可以处理 10^4 级别的数据规模(保守估计，处理 1000 级别的问题肯定没问题)
+- O(n) 算法可以处理 10^8 级别的数据规模(保守估计，处理 10^7 级别的问题肯定没问题)
+- O(nlog n) 算法可以处理 10^7 级别的数据规模(保守估计，处理 10^6 级别的问题肯定没问题)
+
+| | 数据规模|时间复杂度 | 算法举例|
+|:------:|:------:|:------:|:------:|
+|1|10|O(n!)|permutation 排列|
+|2|20~30|O(2^n)|combination 组合|
+|3|50|O(n^4)|DFS 搜索、DP 动态规划|
+|4|100|O(n^3)|任意两点最短路径、DP 动态规划|
+|5|1000|O(n^2)|稠密图、DP 动态规划|
+|6|10^6|O(nlog n)|排序，堆，递归与分治|
+|7|10^7|O(n)|DP 动态规划、图遍历、拓扑排序、树遍历|
+|8|10^9|O(sqrt(n))|筛素数、求平方根|
+|9|10^10|O(log n)|二分搜索|
+|10|+∞|O(1)|数学相关算法|
+|------------------------------|------------------------------|------------------------------------------------------------------|------------------------------------------------------------------|
+
+
+一些具有迷惑性的例子：
+
+```c
+void hello (int n){
+    for( int sz = 1 ; sz < n ; sz += sz )
+        for( int i = 1 ; i < n ; i ++ )
+            cout << "Hello" << endl;
+}
+```
+
+上面这段代码的时间复杂度是 O(nlog n) 而不是 O(n^2)
+
+```c
+bool isPrime (int n){
+    if (num <= 1) return false;
+    for( int x = 2 ; x * x <= n ; x ++ )
+        if( n % x == 0 )
+            return false;
+    return true;
+}
+```
+
+上面这段代码的时间复杂度是 O(sqrt(n)) 而不是 O(n)。
+
+再举一个例子，有一个字符串数组，将数组中的每一个字符串按照字母序排序，之后再将整个字符串数组按照字典序排序。两步操作的整体时间复杂度是多少呢？
+
+如果回答是 O(n*nlog n + nlog n) = O(n^2log n)，这个答案是错误的。字符串的长度和数组的长度是没有关系的，所以这两个变量应该单独计算。假设最长的字符串长度为 s，数组中有 n 个字符串。对每个字符串排序的时间复杂度是 O(slog s)，将数组中每个字符串都按照字母序排序的时间复杂度是 O(n * slog s)。
+
+将整个字符串数组按照字典序排序的时间复杂度是 O(s * nlog n)。排序算法中的 O(nlog n) 是比较的次数，由于比较的是整型数字，所以每次比较是 O(1)。但是字符串按照字典序比较，时间复杂度是 O(s)。所以字符串数组按照字典序排序的时间复杂度是 O(s * nlog n)。所以整体复杂度是 O(n * slog s) + O(s * nlog n) = O(n\*slog s + s\*nlogn) = O(n\*s\*(log s + log n)) = O(n\*s\*log(n\*s)）。
+
+## 二. 空间复杂度
+
+递归调用是有空间代价的，递归算法需要保存递归栈信息，所以花费的空间复杂度会比非递归算法要高。
+
+```c
+int sum( int n ){
+    assert( n >= 0 )
+    int ret = 0;
+    for ( int i = 0 ; i <= n ; i ++ )
+        ret += i;
+    return ret;
+}
+```
+
+上面算法的时间复杂度为 O(n)，空间复杂度 O(1)。
+
+```c
+int sum( int n ){
+    assert( n >= 0 )
+    if ( n == 0 )
+        return 0;
+    return n + sum( n - 1 );
+}
+```
+
+上面算法的时间复杂度为 O(n)，空间复杂度 O(n)。
+
+## 三. 递归的时间复杂度
+
+### 1. 只有一次递归调用
+
+如果递归函数中，只进行了一次递归调用，且递归深度为 depth，在每个递归函数中，时间复杂度为 T，那么总体的时间复杂度为 O(T * depth)
+
+举个例子：
+
+```c
+int binarySearch(int arr[], int l, int r, int target){
+	if( l > r )
+	    return -1;
+    int mid = l + ( r - l ) / 2; // 防溢出
+    if(arr[mid] == target)
+        return mid;
+    else if (arr[mid] > target)
+        return binarySearch(arr,l,mid-1,target);
+    else
+        return binarySearch(arr,mid+1,r,target);
+}
+
+```
+
+在二分查找的递归实现中，只递归调用了自身。递归深度是 log n ，每次递归里面的复杂度是 O(1) 的，所以二分查找的递归实现的时间复杂度为 O(log n) 的。
+
+
+### 2. 只有多次递归调用
+
+针对多次递归调用的情况，就需要看它的计算调用的次数了。通常可以画一颗递归树来看。举例：
+
+```c
+int f(int n){
+    assert( n >= 0 );
+    if( n == 0 )
+        return 1;
+    return f( n - 1 ) + f ( n - 1 );
+}
+```
+
+上述这次递归调用的次数为 2^0^ + 2^1^ + 2^2^ + …… + 2^n^ = 2^n+1^ - 1 = O(2^n)
+
+
+> 关于更加复杂的递归的复杂度分析，请参考主定理。主定理中针对各种复杂情况都给出了正确的结论。
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterOne/Algorithm/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/">下一章➡️</a></p>
+</div>
diff --git a/website/content/ChapterOne/_index.md b/website/content/ChapterOne/_index.md
new file mode 100644
index 000000000..6d535294a
--- /dev/null
+++ b/website/content/ChapterOne/_index.md
@@ -0,0 +1,95 @@
+---
+title: 第一章 序章
+type: docs
+weight: 1
+---
+
+# 第一章 序章
+
+{{< columns >}}
+## 关于 LeetCode
+
+说到 LeetCode，作为一个程序员来说，应该不陌生，近几年参加面试都会提到它。国内外的程序员用它刷题主要是为了面试。据历史记载，这个网站 2011 年就成立了，马上就要到自己 10 周年的生日了。每周举行周赛，双周赛，月赛，在有限时间内编码，确实非常能考验人的算法能力。一些大公司赞助冠名的比赛获得前几名除了有奖品，还能直接拿到内推的机会。
+
+<--->
+
+## 什么是 Cookbook
+
+直译的话就是烹饪书，教你做各种食谱美食的书。经常看 O'Reilly 技术书的同学对这个名词会很熟悉。一般动手操作，实践类的书都会有这个名字。
+
+{{< /columns >}}
+
+<img src="https://books.halfrost.com/leetcode/logo.png" alt="logo" height="600" align="right" style="padding-left: 30px;"/>
+
+## 为什么会写这个开源书
+
+笔者刷题刷了一年了，想和大家分享分享一些做题心得，解题方法。想和有相同爱好的人交个朋友，一起交流学习。对于自己来说，写题解也是一种提高。把一道深奥的题目讲给一点都没有头绪的人，并能让他完全听懂，很能锻炼人的表达能力。在讲解中很可能还会遇到听者的一些提问，这些问题可能是自己的知识漏洞，强迫自己去弥补。笔者在公司做过相关的分享，感受很深，双方受益都还不错。
+
+> 另外，在大学期间，笔者做题的时候最讨厌写题解，感觉是浪费时间，用更多的时间去做更多的题。现在不知道算不算是“出来混的，总是要还的”。
+
+
+## 关于书的封面
+
+常看 O'Reilly 动物书的同学一看这个封面就知道是向他们致敬。确实是这个目的。O'Reilly 的封面动物都是稀缺动物，并且画风都是黑白素描风。这些动物都有版权了，所以只能在网上找没有版权的黑白素描风的图片。常见的能找到 40 张这种风格的图片。不过用的人太多了，笔者费劲的找了其他几张这种图片，这张孔雀开屏是其中一张。孔雀开屏的意义是希望大家刷完 LeetCode 以后，提高了自身的算法能力，在人生的舞台上开出自己的“屏”。全书配色也都是绿色，因为这是 AC 的颜色。
+
+
+## 关于作者
+
+笔者是一个刚刚入行一年半的 gopher 新人，还请各位大佬多多指点小弟我。大学参加了 3 年 ACM-ICPC，但是由于资质不高，没有拿到一块金牌。所以在算法方面，我对自己的评价算是新手吧。参加 ACM-ICPC 最大的收获是训练了思维能力，这种能力也会运用到生活中。其次是认识了很多国内很聪明的选手，看到了自己和他们的差距。最后，就是那 200 多页，有些自己都没有完全理解的，打印的密密麻麻的[算法模板](https://github.com/halfrost/leetcode-go/releases/tag/Special)。知识学会了，终身都是自己的，没有学会，那些知识都是身外之物。
+
+笔者从 2019 年 3 月 25 号开始刷题，到 2020 年 3 月 25 号，整整一年的时间。原计划是每天一题。实际上每天有时候不止一题，最终完成了 600+：
+
+![](https://img.halfrost.com/Blog/ArticleImage/2019_leetcode.png)
+
+> 一个温馨提示：笔者本以为每天做一题，会让这个 submissions 图全绿，但是我发现我错了。如果你也想坚持，让这个图全绿，一定要注意以下的问题：LeetCode 服务器是在 +0 时区的，这个图也是按照这个时区计算的。也就是说，中国每天早上 8 点之前，是算前一天的！也是因为时区的问题，导致我空白了这 22 个格子。比如有一道 Hard 题很难，当天工作也很多，晚上下班回家想出来了就到第二天凌晨了。于是再做一题当做第二天的量。结果会发现这 2 题都算前一天的。有时候笔者早上 6 点起床刷题，提交以后也都是前一天的。
+> 
+> (当然这些都是过去了，不重要了，全当是奋斗路上的一些小插曲)
+
+2020 年笔者肯定还会继续刷题，因为还没有达到自己的一些目标。可能会朝着 1000 题奋进，也有可能刷到 800 题的时候回头开始二刷，三刷。(不达目的不罢休吧~)
+
+## 关于书中的代码
+
+代码都放在 [github repo](https://github.com/halfrost/leetcode-go/tree/master/leetcode) 中，按题号可以搜索到题目。
+本书题目的代码都已经 beats 100% 了。没有 beats 100% 题解就没有放到本书中了。那些题目笔者会继续优化到 100% 再放进来。
+
+有可能读者会问，为何要追求 beats 100%。笔者认为优化到 beats 100% 才算是把这题做出感觉了。有好几道 Hard 题，笔者都用暴力解法 AC 了，然后只 beats 了 5%。这题就如同没做一样。而且面试中如果给了这样的答案，面试官也不会满意，“还有没有更优解？”。如果通过自己的思考能给出更优解，面试官会更满意一些。
+
+LeetCode 统计代码运行时长会有波动的，相同的代码提交 10 次可能就会 beats 100% 了。笔者开始没有发现这个问题，很多题用正确的代码连续交了很多次，一年提交 3400+ 次，导致我的正确率也变的奇高。😢
+
+当然，如果还有其他更优美的解法，也能 beats 100% 的，欢迎提交 PR，笔者和大家一起学习。
+
+## 目标读者
+
+想通过 LeetCode 提高算法能力的编程爱好者。
+
+
+## 编程语言
+
+本书的算法全部用 Go 语言实现。
+
+
+## 使用说明
+
+- 本电子书的左上角有搜索栏，可以迅速帮你找到你想看的章节和题号。
+- 本电子书每页都接入了 Gitalk，每一页的最下方都有评论框可以评论，如果没有显示出来，请检查自己的网络。
+- 关于题解，笔者建议这样使用：先自己读题，思考如何解题。如果 15 分钟还没有思路，那么先看笔者的解题思路，但是不要看代码。有思路以后自己用代码实现一遍。如果完全不会写，那就看笔者提供的代码，找出自己到底哪里不会写，找出问题记下来，这就是自己要弥补的知识漏洞。如果自己实现出来了，提交以后有错误，自己先 debug。AC 以后没有到 100% 也先自己思考如何优化。如果每道题自己都能优化到 100% 了，那么一段时间以后进步会很大。所以总的来说，实在没思路，看解题思路；实在优化不到 100%，看看代码。
+
+## 互动与勘误
+
+如果书中文章有所遗漏，欢迎点击所在页面下边的 edit 按钮进行评论和互动，感谢您的支持与帮助。
+
+## 最后
+
+一起开始刷题吧~
+
+![](https://img.halfrost.com/Blog/ArticleImage/hello_leetcode.png)
+
+本作品采用 [知识署名-非商业性使用-禁止演绎 （BY-NC-ND） 4.0 国际许可协议](https://creativecommons.org/licenses/by-nc-nd/4.0/legalcode.zh-Hans) 进行许可。
+
+
+题解里面的所有题目版权均归 [LeetCode](https://leetcode.com/) 和 [力扣中国](https://leetcode-cn.com/) 所有
+
+
+----------------------------------------------
+<p align = "right"><a href="https://books.halfrost.com/leetcode/ChapterOne/Data_Structure/">下一页➡️</a></p>
+
diff --git a/website/content/ChapterThree/Binary_Indexed_Tree.md b/website/content/ChapterThree/Binary_Indexed_Tree.md
new file mode 100644
index 000000000..1d58fdeed
--- /dev/null
+++ b/website/content/ChapterThree/Binary_Indexed_Tree.md
@@ -0,0 +1,637 @@
+---
+title: 3.5 Binary Indexed Tree 
+type: docs
+weight: 5
+---
+
+# 树状数组 Binary Indexed Tree (二叉索引树)
+
+树状数组或二叉索引树（Binary Indexed Tree），又以其发明者命名为 Fenwick 树，最早由 Peter M. Fenwick 于 1994 年以 A New Data Structure for Cumulative Frequency Tables 为题发表在 SOFTWARE PRACTICE AND EXPERIENCE 上。其初衷是解决数据压缩里的累积频率（Cumulative Frequency）的计算问题，现多用于高效计算数列的前缀和，区间和。针对区间问题，除了常见的线段树解法，还可以考虑树状数组。它可以以 O(log n) 的时间得到任意前缀和{{< katex >}} \sum_{i=1}^{j}A[i],1<=j<=N {{< /katex >}}，并同时支持在 O(log n)时间内支持动态单点值的修改(增加或者减少)。空间复杂度 O(n)。
+
+> 利用数组实现前缀和，查询本来是 O(1)，但是对于频繁更新的数组，每次重新计算前缀和，时间复杂度 O(n)。此时树状数组的优势便立即显现。
+
+## 一. 一维树状数组概念
+
+
+![](https://img.halfrost.com/Blog/ArticleImage/152_0.png)
+
+树状数组名字虽然又有树，又有数组，但是它实际上物理形式还是数组，不过每个节点的含义是树的关系，如上图。树状数组中父子节点下标关系是 {{< katex >}}parent = son + 2^{k}{{< /katex >}}，其中 k 是子节点下标对应二进制末尾 0 的个数。
+
+例如上图中 A 和 B 都是数组。A 数组正常存储数据，B 数组是树状数组。B4，B6，B7 是 B8 的子节点。4 的二进制是 100，4 + {{< katex >}}2^{2}{{< /katex >}} = 8，所以 8 是 4 的父节点。同理，7 的二进制 111，7 + {{< katex >}}2^{0}{{< /katex >}} = 8，8 也是 7 的父节点。
+
+
+### 1. 节点意义
+
+在树状数组中，所有的奇数下标的节点的含义是叶子节点，表示单点，它存的值是原数组相同下标存的值。例如上图中 B1，B3，B5，B7 分别存的值是 A1，A3，A5，A7。所有的偶数下标的节点均是父节点。父节点内存的是区间和。例如 B4 内存的是 B1 + B2 + B3 + A4 = A1 + A2 + A3 + A4。这个区间的左边界是该父节点最左边叶子节点对应的下标，右边界就是自己的下标。例如 B8 表示的区间左边界是 B1，右边界是 B8，所以它表示的区间和是 A1 + A2 + …… + A8。
+
+{{< katex display >}}
+\begin{aligned}
+B_{1} &= A_{1} \\
+B_{2} &= B_{1} + A_{2} = A_{1} + A_{2} \\
+B_{3} &= A_{3} \\
+B_{4} &= B_{2} + B_{3} + A_{4} = A_{1} + A_{2} + A_{3} + A_{4} \\
+B_{5} &= A_{5} \\
+B_{6} &= B_{5} + A_{6} = A_{5} + A_{6} \\
+B_{7} &= A_{7} \\
+B_{8} &= B_{4} + B_{6} + B_{7} + A_{8} = A_{1} + A_{2} + A_{3} + A_{4} + A_{5} + A_{6} + A_{7} + A_{8} \\
+\end{aligned}
+{{< /katex >}}
+
+
+由数学归纳法可以得出，左边界的下标一定是 {{< katex >}}i - 2^{k} + 1{{< /katex >}}，其中 i 为父节点的下标，k 为 i 的二进制中末尾 0 的个数。用数学方式表达偶数节点的区间和：
+
+{{< katex display >}}
+B_{i} = \sum_{j = i - 2^{k} + 1}^{i} A_{j}
+{{< /katex >}}
+
+初始化树状数组的代码如下：
+
+```go
+// BinaryIndexedTree define
+type BinaryIndexedTree struct {
+	tree     []int
+	capacity int
+}
+
+// Init define
+func (bit *BinaryIndexedTree) Init(nums []int) {
+	bit.tree, bit.capacity = make([]int, len(nums)+1), len(nums)+1
+	for i := 1; i <= len(nums); i++ {
+		bit.tree[i] += nums[i-1]
+		for j := i - 2; j >= i-lowbit(i); j-- {
+			bit.tree[i] += nums[j]
+		}
+	}
+}
+```
+
+lowbit(i) 函数返回 i 转换成二进制以后，末尾最后一个 1 代表的数值，即 {{< katex >}}2^{k}{{< /katex >}}，k 为 i 末尾 0 的个数。我们都知道，在计算机系统中，数值一律用补码来表示和存储。原因在于，使用补码，可以将符号位和数值域统一处理；同时，加法和减法也可以统一处理。利用补码，可以 O(1) 算出 lowbit(i)。负数的补码等于正数的原码每位取反再 + 1，加一会使得负数的补码末尾的 0 和正数原码末尾的 0 一样。这两个数进行 & 运算以后，结果即为 lowbit(i)：
+
+```go
+func lowbit(x int) int {
+	return x & -x
+}
+```
+
+如果还想不通的读者，可以看这个例子，34 的二进制是 {{< katex >}}(0010 0010)_{2} {{< /katex >}}，它的补码是 {{< katex >}}(1101 1110)_{2} {{< /katex >}}。
+
+{{< katex display >}}
+(0010 0010)_{2} \& (1101 1110)_{2} = (0000 0010)_{2}
+{{< /katex >}}
+
+lowbit(34) 结果是 {{< katex >}}2^{k} = 2^{1} = 2 {{< /katex >}}
+
+### 2. 插入操作
+
+树状数组上的父子的下标满足 {{< katex >}}parent = son + 2^{k}{{< /katex >}} 关系，所以可以通过这个公式从叶子结点不断往上递归，直到访问到最大节点值为止，祖先结点最多为 logn 个。插入操作可以实现节点值的增加或者减少，代码实现如下：
+
+```go
+// Add define
+func (bit *BinaryIndexedTree) Add(index int, val int) {
+	for index <= bit.capacity {
+		bit.tree[index] += val
+		index += lowbit(index)
+	}
+}
+```
+
+
+
+
+### 3. 查询操作
+
+	
+树状数组中查询 [1, i] 区间内的和。按照节点的含义，可以得出下面的关系：
+
+{{< katex display >}}
+\begin{aligned}
+Query(i) &= A_{1} + A_{2} + ...... + A_{i} \\
+&= A_{1} + A_{2} + A_{i-2^{k}} + A_{i-2^{k}+1} + ...... + A_{i} \\
+&= A_{1} + A_{2} + A_{i-2^{k}} + B_{i} \\
+&= Query(i-2^{k}) + B_{i} \\
+&= Query(i-lowbit(i)) + B_{i} \\
+\end{aligned}
+{{< /katex >}}
+
+{{< katex >}}B_{i}{{< /katex >}} 是树状数组存的值。Query 操作实际是一个递归的过程。lowbit(i) 表示 {{< katex >}}2^{k}{{< /katex >}}，其中 k 是 i 的二进制表示中末尾 0 的个数。i - lowbit(i) 将 i 的二进制中末尾的 1 去掉，最多有 {{< katex >}}log(i){{< /katex >}} 个 1，所以查询操作最坏的时间复杂度是 O(log n)。查询操作实现代码如下：
+
+```go
+// Query define
+func (bit *BinaryIndexedTree) Query(index int) int {
+	sum := 0
+	for index >= 1 {
+		sum += bit.tree[index]
+		index -= lowbit(index)
+	}
+	return sum
+}
+```
+
+## 二. 不同场景下树状数组的功能
+
+根据节点维护的数据含义不同，树状数组可以提供不同的功能来满足各种各样的区间场景。下面我们先以上例中讲述的区间和为例，进而引出 RMQ 的使用场景。
+
+### 1. 单点增减 + 区间求和
+
+这种场景是树状数组最经典的场景。单点增减分别调用 add(i,v) 和 add(i,-v)。区间求和，利用前缀和的思想，求 [m,n] 区间和，即 query(n) - query(m-1)。query(n) 代表 [1,n] 区间内的和，query(m-1) 代表 [1,m-1] 区间内的和，两者相减，即 [m,n] 区间内的和。
+
+> LeetCode 对应题目是 [307. Range Sum Query - Mutable](https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0307.Range-Sum-Query-Mutable/)、[327. Count of Range Sum](https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0327.Count-of-Range-Sum/)
+
+### 2. 区间增减 + 单点查询
+
+这种情况需要做一下转化。定义差分数组 {{< katex >}}C_{i}{{< /katex >}} 代表 {{< katex >}}C_{i} = A_{i} - A_{i-1}{{< /katex >}}。那么：
+
+{{< katex display >}}
+\begin{aligned}
+C_{0} &= A_{0} \\
+C_{1} &= A_{1} - A_{0}\\
+C_{2} &= A_{2} - A_{1}\\
+......\\
+C_{n} &= A_{n} - A_{n-1}\\
+\sum_{j=1}^{n}C_{j} &= A_{n}\\
+\end{aligned}
+{{< /katex >}}
+
+区间增减：在 [m,n] 区间内每一个数都增加 v，只影响 2 个单点的值：
+
+{{< katex display >}}
+\begin{aligned}
+C_{m} &= (A_{m} + v) - A_{m-1}\\
+C_{m+1} &= (A_{m+1} + v) - (A_{m} + v)\\
+C_{m+2} &= (A_{m+2} + v) - (A_{m+1} + v)\\
+......\\
+C_{n} &= (A_{n} + v) - (A_{n-1} + v)\\
+C_{n+1} &= A_{n+1} - (A_{n} + v)\\
+\end{aligned}
+{{< /katex >}}
+
+
+可以观察看，{{< katex >}}C_{m+1}, C_{m+2}, ......, C_{n}{{< /katex >}} 值都不变，变化的是 {{< katex >}}C_{m}, C_{n+1}{{< /katex >}}。所以在这种情况下，区间增加只需要执行 add(m,v) 和 add(n+1,-v) 即可。
+
+单点查询这时就是求前缀和了，{{< katex >}}A_{n} = \sum_{j=1}^{n}C_{j}{{< /katex >}}，即 query(n)。
+
+### 3. 区间增减 + 区间求和
+
+这种情况是上面一种情况的增强版。区间增减的做法和上面做法一致，构造差分数组。这里主要说明区间查询怎么做。先来看 [1,n] 区间和如何求：
+
+
+
+{{< katex display >}}
+A_{1} + A_{2} + A_{3} + ...... + A_{n}\\
+\begin{aligned}
+ &= (C_{1}) + (C_{1} + C_{2}) + (C_{1} + C_{2} + C_{3}) + ...... + \sum_{1}^{n}C_{n}\\
+&= n * C_{1} + (n-1) * C_{2} + ...... + C_{n}\\
+&= n * (C_{1} + C_{2} + C_{3} + ...... + C_{n}) - (0 * C_{1} + 1 * C_{2} + 2 * C_{3} + ...... + (n - 1) * C_{n})\\
+&= n * \sum_{1}^{n}C_{n} - (D_{1} + D_{2} + D_{3} + ...... + D_{n})\\
+&= n * \sum_{1}^{n}C_{n} - \sum_{1}^{n}D_{n}\\
+\end{aligned}
+{{< /katex >}}
+
+其中 {{< katex >}}D_{n} = (n - 1) * C_{n}{{< /katex >}}
+
+所以求区间和，只需要再构造一个 {{< katex >}}D_{n}{{< /katex >}} 即可。
+
+{{< katex display >}}
+\begin{aligned}
+\sum_{1}^{n}A_{n} &= A_{1} + A_{2} + A_{3} + ...... + A_{n} \\
+&= n * \sum_{1}^{n}C_{n} - \sum_{1}^{n}D_{n}\\
+\end{aligned}
+{{< /katex >}}
+
+以此类推，推到更一般的情况：
+
+{{< katex display >}}
+\begin{aligned}
+\sum_{m}^{n}A_{n} &= A_{m} + A_{m+1} + A_{m+2} + ...... + A_{n} \\
+&= \sum_{1}^{n}A_{n} - \sum_{1}^{m-1}A_{n}\\
+&= (n * \sum_{1}^{n}C_{n} - \sum_{1}^{n}D_{n}) - ((m-1) * \sum_{1}^{m-1}C_{m-1} - \sum_{1}^{m-1}D_{m-1})\\
+\end{aligned}
+{{< /katex >}}
+
+至此区间查询问题得解。
+
+### 4. 单点增减 + 区间最值
+
+线段树最基础的运用是区间求和，但是将 sum 操作换成 max 操作以后，也可以求区间最值，并且时间复杂度完全没有变。那树状数组呢？也可以实现相同的功能么？答案是可以的，不过时间复杂度会下降一点。
+
+线段树求区间和，把每个小区间的和计算好，然后依次 pushUp，往上更新。把 sum 换成 max 操作，含义完全相同：取出小区间的最大值，然后依次 pushUp 得到整个区间的最大值。
+
+树状数组求区间和，是将单点增减的增量影响更新到固定区间 {{< katex >}}[i-2^{k}+1, i]{{< /katex >}}。但是把 sum 换成 max 操作，含义就变了。此时单点的增量和区间 max 值并无直接联系。暴力的方式是将该点与区间内所有值比较大小，取出最大值，时间复杂度 O(n * log n)。仔细观察树状数组的结构，可以发现不必枚举所有区间。例如更新 {{< katex >}}A_{i}{{< /katex >}} 的值，那么受到影响的树状数组下标为 {{< katex >}}i-2^{0}, i-2^{1}, i-2^{2}, i-2^{3}, ......, i-2^{k}{{< /katex >}}，其中 {{< katex >}}2^{k} < lowbit(i) \leqslant 2^{k+1}{{< /katex >}}。需要更新至多 k 个下标，外层循环由 O(n) 降为了 O(log n)。区间内部每次都需要重新比较，需要 O(log n) 的复杂度，总的时间复杂度为 {{< katex >}}(O(log n))^2 {{< /katex >}}。
+
+```go
+func (bit *BinaryIndexedTree) Add(index int, val int) {
+	for index <= bit.capacity {
+		bit.tree[index] = val
+		for i := 1; i < lowbit(index); i = i << 1 {
+			bit.tree[index] = max(bit.tree[index], bit.tree[index-i])
+		}
+		index += lowbit(index)
+	}
+}
+```
+
+上面解决了单点更新的问题，再来看区间最值。线段树划分区间是均分，对半分，而树状数组不是均分。在树状数组中 {{< katex >}}B_{i} {{< /katex >}} 表示的区间是 {{< katex >}}[i-2^{k}+1, i]{{< /katex >}}，据此划分“不规则区间”。对于树状数组求 [m,n] 区间内最值，
+
+- 如果 {{< katex >}} m < n - 2^{k} {{< /katex >}}，那么 {{< katex >}} query(m,n) = max(query(m,n-2^{k}), B_{n}){{< /katex >}}
+- 如果 {{< katex >}} m >= n - 2^{k} {{< /katex >}}，那么 {{< katex >}} query(m,n) = max(query(m,n-1), A_{n}){{< /katex >}}
+
+
+```go
+func (bit *BinaryIndexedTree) Query(m, n int) int {
+	res := 0
+	for n >= m {
+		res = max(nums[n], res)
+		n--
+		for ; n-lowbit(n) >= m; n -= lowbit(n) {
+			res = max(bit.tree[n], res)
+		}
+	}
+	return res
+}
+```
+
+n 最多经过 {{< katex >}}(O(log n))^2 {{< /katex >}} 变化，最终 n < m。时间复杂度为 {{< katex >}}(O(log n))^2 {{< /katex >}}。
+
+针对这类问题放一道经典例题[《HDU 1754 I Hate It》](http://acm.hdu.edu.cn/showproblem.php?pid=1754)：
+
+Problem Description  
+很多学校流行一种比较的习惯。老师们很喜欢询问，从某某到某某当中，分数最高的是多少。这让很多学生很反感。不管你喜不喜欢，现在需要你做的是，就是按照老师的要求，写一个程序，模拟老师的询问。当然，老师有时候需要更新某位同学的成绩。
+ 
+
+Input  
+本题目包含多组测试，请处理到文件结束。
+在每个测试的第一行，有两个正整数 N 和 M ( 0<N<=200000,0<M<5000 )，分别代表学生的数目和操作的数目。学生 ID 编号分别从 1 编到 N。第二行包含 N 个整数，代表这 N 个学生的初始成绩，其中第 i 个数代表 ID 为 i 的学生的成绩。接下来有 M 行。每一行有一个字符 C (只取'Q'或'U') ，和两个正整数 A，B。当 C 为 'Q' 的时候，表示这是一条询问操作，它询问 ID 从 A 到 B(包括 A,B)的学生当中，成绩最高的是多少。当 C 为 'U' 的时候，表示这是一条更新操作，要求把 ID 为 A 的学生的成绩更改为 B。
+ 
+
+Output  
+对于每一次询问操作，在一行里面输出最高成绩。
+
+
+	Sample Input
+	5 6
+	1 2 3 4 5
+	Q 1 5
+	U 3 6
+	Q 3 4
+	Q 4 5
+	U 2 9
+	Q 1 5
+ 
+
+	Sample Output
+	5
+	6
+	5
+	9
+
+读完题可以很快反应是单点增减 + 区间最大值的题。利用上面讲解的思想写出代码：
+
+> 由于 OJ 不支持 Go，所以此处用 C 代码实现。这里还有一个 Hint，对于超大量的输入，scanf() 的性能明显优于 cin。
+
+```c
+#include <iostream>
+#include <stdio.h>
+#include <stdlib.h>
+using namespace std;
+ 
+const int MAXN = 3e5;
+int a[MAXN], h[MAXN];
+int n, m;
+ 
+int lowbit(int x)
+{
+	return x & (-x);
+}
+void updata(int x)
+{
+	int lx, i;
+	while (x <= n)
+	{
+		h[x] = a[x];
+		lx = lowbit(x);
+		for (i=1; i<lx; i<<=1)
+			h[x] = max(h[x], h[x-i]);
+		x += lowbit(x);
+	}		
+}
+int query(int x, int y)
+{
+	int ans = 0;
+	while (y >= x)
+	{
+		ans = max(a[y], ans);
+		y --;
+		for (; y-lowbit(y) >= x; y -= lowbit(y))
+			ans = max(h[y], ans);
+	}
+	return ans;
+}
+int main()
+{
+	int i, j, x, y, ans;
+	char c;
+	while (scanf("%d%d",&n,&m)!=EOF)
+	{
+		for (i=1; i<=n; i++)
+			h[i] = 0;
+		for (i=1; i<=n; i++)
+		{
+			scanf("%d",&a[i]);
+			updata(i);
+		}
+		for (i=1; i<=m; i++)
+		{
+			scanf("%c",&c);
+			scanf("%c",&c);
+			if (c == 'Q')
+			{
+				scanf("%d%d",&x,&y);
+				ans = query(x, y);
+				printf("%d\n",ans);
+			}
+			else if (c == 'U')
+			{
+				scanf("%d%d",&x,&y);
+				a[x] = y;
+				updata(x);
+			}
+		}
+	}
+	return 0;
+}
+```
+
+上述代码已 AC。感兴趣的读者可以自己做一做这道 ACM 的简单题。
+
+### 5. 区间叠加 + 单点最值
+
+看到这里可能有细心的读者疑惑，这一类题不就是第二类“区间增减 + 单点查询”类似么？可以考虑用第二类题的思路解决这一类题。不过麻烦点在于，区间叠加以后，每个单点的更新不是直接告诉增减变化，而是需要我们自己维护一个最值。例如在 [5,7] 区间当前值是 7，接下来区间 [1,9] 区间内增加了一个 2 的值。正确的做法是把 [1,4] 区间内增加 2，[8,9] 区间增加 2，[5,7] 区间维持不变，因为 7 > 2。这仅仅是 2 个区间叠加的情况，如果区间叠加的越多，需要拆分的区间也越多了。看到这里有些读者可能会考虑线段树的解法了。线段树确实是解决区间叠加问题的利器。笔者这里只讨论树状数组的解法。
+
+![](https://img.halfrost.com/Leetcode/leetcode_218_0.png)
+
+当前 LeetCode 有 1836 题，Binary Indexed Tree tag 下面只有 7 题，[218. The Skyline Problem](https://leetcode.com/problems/the-skyline-problem/) 这一题算是 7 道 BIT 里面最“难”的。这道天际线的题就属于区间叠加 + 单点最值的题。笔者以这道题为例，讲讲此类题的常用解法。
+
+![](https://img.halfrost.com/Leetcode/leetcode_218_1.png)
+
+要求天际线，即找到楼与楼重叠区间外边缘的线，说白了是维护各个区间内的最值。这有 2 个需要解决的问题。
+
+1. 如何维护最值。当一个高楼的右边界消失，剩下的各个小楼间还需要选出最大值作为天际线。剩下重重叠叠的小楼很多，树状数组如何维护区间最值是解决此类题的关键。
+2. 如何维护天际线的转折点。有些楼与楼并非完全重叠，重叠一半的情况导致天际线出现转折点。如上图中标记的红色转折点。树状数组如何维护这些点呢？
+
+
+先解决第一个问题(维护最值)。树状数组只有 2 个操作，一个是 Add() 一个是 Query()。从上面关于这 2 个操作的讲解中可以知道这 2 个操作都不能满足我们的需求。Add() 操作可以改成维护区间内 max() 的操作。但是 max() 容易获得却很难“去除”。如上图 [3,7] 这个区间内的最大值是 15。根据树状数组的定义，[3,12] 这个区间内最值还是 15。观察上图可以看到 [5,12] 区间内最值其实是 12。树状数组如何维护这种最值呢？最大值既然难以“去除”，那么需要考虑如何让最大值“来的晚一点”。解决办法是将 Query() 操作含义从前缀含义改成后缀含义。Query(i) 查询区间是 [1,i]，现在查询区间变成 {{< katex >}}[i,+\infty){{< /katex >}}。例如：[i,j] 区间内最值是 {{< katex >}}max_{i...j}{{< /katex >}}，Query(j+1) 的结果不会包含 {{< katex >}}max_{i...j}{{< /katex >}}，因为它查询的区间是 {{< katex >}}[j+1,+\infty){{< /katex >}}。这样更改以后，可以有效避免前驱高楼对后面楼的累积 max() 最值的影响。
+
+具体做法，将 x 轴上的各个区间排序，按照 x 值大小从小到大排序。从左往右依次遍历各个区间。Add() 操作含义是加入每个区间右边界代表后缀区间的最值。这样不需要考虑“移除”最值的问题了。细心的读者可能又有疑问了：能否从右往左遍历区间，Query() 的含义继续延续前缀区间？这样做是可行的，解决第一个问题(维护最值)是可以的。但是这种处理办法解决第二个问题(维护转折点)会遇到麻烦。
+
+再解决第二个问题(维护转折点)。如果用前缀含义的 Query()，在单点 i 上除了考虑以这个点为结束点的区间，还需要考虑以这个单点 i 为起点的区间。如果是后缀含义的 Query() 就没有这个问题了，{{< katex >}}[i+1,+\infty){{< /katex >}} 这个区间内不用考虑以单点 i 为结束点的区间。此题用树状数组代码实现如下：
+
+
+```go
+const LEFTSIDE = 1
+const RIGHTSIDE = 2
+
+type Point struct {
+	xAxis int
+	side  int
+	index int
+}
+
+func getSkyline3(buildings [][]int) [][]int {
+	res := [][]int{}
+	if len(buildings) == 0 {
+		return res
+	}
+	allPoints, bit := make([]Point, 0), BinaryIndexedTree{}
+	// [x-axis (value), [1 (left) | 2 (right)], index (building number)]
+	for i, b := range buildings {
+		allPoints = append(allPoints, Point{xAxis: b[0], side: LEFTSIDE, index: i})
+		allPoints = append(allPoints, Point{xAxis: b[1], side: RIGHTSIDE, index: i})
+	}
+	sort.Slice(allPoints, func(i, j int) bool {
+		if allPoints[i].xAxis == allPoints[j].xAxis {
+			return allPoints[i].side < allPoints[j].side
+		}
+		return allPoints[i].xAxis < allPoints[j].xAxis
+	})
+	bit.Init(len(allPoints))
+	kth := make(map[Point]int)
+	for i := 0; i < len(allPoints); i++ {
+		kth[allPoints[i]] = i
+	}
+	for i := 0; i < len(allPoints); i++ {
+		pt := allPoints[i]
+		if pt.side == LEFTSIDE {
+			bit.Add(kth[Point{xAxis: buildings[pt.index][1], side: RIGHTSIDE, index: pt.index}], buildings[pt.index][2])
+		}
+		currHeight := bit.Query(kth[pt] + 1)
+		if len(res) == 0 || res[len(res)-1][1] != currHeight {
+			if len(res) > 0 && res[len(res)-1][0] == pt.xAxis {
+				res[len(res)-1][1] = currHeight
+			} else {
+				res = append(res, []int{pt.xAxis, currHeight})
+			}
+		}
+	}
+	return res
+}
+
+type BinaryIndexedTree struct {
+	tree     []int
+	capacity int
+}
+
+// Init define
+func (bit *BinaryIndexedTree) Init(capacity int) {
+	bit.tree, bit.capacity = make([]int, capacity+1), capacity
+}
+
+// Add define
+func (bit *BinaryIndexedTree) Add(index int, val int) {
+	for ; index > 0; index -= index & -index {
+		bit.tree[index] = max(bit.tree[index], val)
+	}
+}
+
+// Query define
+func (bit *BinaryIndexedTree) Query(index int) int {
+	sum := 0
+	for ; index <= bit.capacity; index += index & -index {
+		sum = max(sum, bit.tree[index])
+	}
+	return sum
+}
+
+```
+
+> 此题还可以用线段树和扫描线解答。扫描线和树状数组解答此题，非常快。线段树稍微慢一些。
+
+
+## 三. 常见应用
+
+这一章节来谈谈树状数组的常见应用。
+
+### 1. 求逆序对
+
+给定 {{< katex >}} n {{< /katex >}} 个数 {{< katex >}} A[n] \in [1,n]  {{< /katex >}} 的排列 P，求满足 {{< katex >}}i < j {{< /katex >}} 且 {{< katex >}} A[i] > A[j] {{< /katex >}} 的数对 {{< katex >}} (i,j) {{< /katex >}} 的个数。
+
+
+这个问题就是经典的逆序数问题，如果采用朴素算法，就是枚举 i 和 j，并且判断   A[i] 和 A[j] 的值进行数值统计，如果 A[i] > A[j] 则计数器加一，统计完后计数器的值就是答案。时间复杂度为 {{< katex >}} O(n^{2}) {{< /katex >}}，这个时间复杂度太高，是否存在 {{< katex >}} O(log n) {{< /katex >}} 的解法呢？
+
+> 如果题目换成 {{< katex >}} A[n] \in [1,10^{10}]  {{< /katex >}}，解题思路不变，只不过一开始再多加一步，离散化的操作。
+
+假设第一步需要离散化。先把数列中的数按大小顺序转化成 1 到 n 的整数，将重复的数据编相同的号，将空缺的数据编上连续的号。使得原数列映射成为一个 1,2,...,n 的数组 B。注意，数组 B 中存的元素也是乱序的，是根据原数组映射而来的。例如原数组是 int[9,8,5,4,6,2,3,8,7,0]，数组中 8 是重复的，且少了数字 1，将这个数组映射到 [1,9] 区间内，调整后的数组 B 为 int[9,8,5,4,6,2,3,8,7,1]。
+
+再创建一个树状数组，用来记录这样一个数组 C（下标从1算起）的前缀和：若 [1, N] 这个排列中的数 i 当前已经出现，则 C[i] 的值为 1 ，否则为 0。初始时数组 C 的值均为 0。从数组 B 第一个元素开始遍历，对树状数组执行修改数组 C 的第 B[j] 个数值加 1 的操作。再在树状数组中查询有多少个数小于等于当前的数 B[j]（即用树状数组查询数组 C 中的 [1,B[j]] 区间前缀和），当前插入总数 i 减去小于等于 B[j] 元素总数，差值即为大于 B[j] 元素的个数，并加入计数器。
+
+```go
+func reversePairs(nums []int) int {
+	if len(nums) <= 1 {
+		return 0
+	}
+	arr, newPermutation, bit, res := make([]Element, len(nums)), make([]int, len(nums)), template.BinaryIndexedTree{}, 0
+	for i := 0; i < len(nums); i++ {
+		arr[i].data = nums[i]
+		arr[i].pos = i
+	}
+	sort.Slice(arr, func(i, j int) bool {
+		if arr[i].data == arr[j].data {
+			if arr[i].pos < arr[j].pos {
+				return true
+			} else {
+				return false
+			}
+		}
+		return arr[i].data < arr[j].data
+	})
+	id := 1
+	newPermutation[arr[0].pos] = 1
+	for i := 1; i < len(arr); i++ {
+		if arr[i].data == arr[i-1].data {
+			newPermutation[arr[i].pos] = id
+		} else {
+			id++
+			newPermutation[arr[i].pos] = id
+		}
+	}
+	bit.Init(id)
+	for i := 0; i < len(newPermutation); i++ {
+		bit.Add(newPermutation[i], 1)
+		res += (i + 1) - bit.Query(newPermutation[i])
+	}
+	return res
+}
+```
+
+上述代码中的 newPermutation 就是映射调整后的数组 B。遍历数组 B，按顺序把元素插入到树状数组中。例如数组 B 是 int[9,8,5,4,6,2,3,8,7,1]，现在往树状数组中插入 6，代表 6 这个元素出现了。query() 查询 [1,6] 区间内是否有元素出现，区间前缀和代表区间内元素出现次数和。如果有 k 个元素出现，且当前插入了 5 个元素，那么 5-k 的差值即是逆序的元素个数，这些元素一定比 6 大。这种方法是正序构造树状数组。
+
+
+还有一种方法是倒序构造树状数组。例如下面代码：
+
+```go
+	for i := len(s) - 1; i > 0; i-- {
+		bit.Add(newPermutation[i], 1)
+		res += bit.Query(newPermutation[i] - 1)
+	}
+```
+
+由于是倒序插入，每次 Query 之前的元素下标一定比当前 i 要大。下标比 i 大，元素值比 A[i] 小，这样的元素和 i 可以构成逆序对。Query 查找 [1, B[j]] 区间内元素总个数，即为逆序对的总数。
+
+> 注意，计算逆序对的时候不要算重复了。比如，计算当前 j 下标前面比 B[j] 值大的数，又算上 j 下标后面比 B[j] 值小的数。这样计算出现了很多重复。因为 j 下标前面的下标 k，也会寻找 k 下标后面比 B[k] 值小的数，重复计算了。那么统一找比自己下标小，但是值大的元素，那么统一找比自己下标大，但是值小的元素。切勿交叉计算。
+
+
+> LeetCode 对应题目是 [315. Count of Smaller Numbers After Self](https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0315.Count-of-Smaller-Numbers-After-Self/)、[493. Reverse Pairs](https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0493.Reverse-Pairs/)、[1649. Create Sorted Array through Instructions](https://books.halfrost.com/leetcode/ChapterFour/1600~1699/1649.Create-Sorted-Array-through-Instructions/)
+
+
+
+### 2. 求区间逆序对
+
+给定 {{< katex >}} n {{< /katex >}} 个数的序列 {{< katex >}} A[n] \in [1,2^{31}-1] {{< /katex >}}，然后给出 {{< katex >}} n \in [1,10^{5}] {{< /katex >}} 次询问 {{< katex >}} [L,R] {{< /katex >}}，每次询问区间 {{< katex >}} [L,R] {{< /katex >}} 中满足 {{< katex >}} L \leqslant i < j \leqslant R {{< /katex >}} 且 {{< katex >}} A[i] > A[j] {{< /katex >}} 的下标 {{< katex >}} (i,j) {{< /katex >}} 的对数。
+
+这个问题比上一题多了一个区间限制。这个区间的限制影响对逆序对的选择。例如：[1,3,5,2,1,1,8,9,8,6,5,3,7,7,2]，求在 [3,7] 区间内的逆序数。元素 2 在区间内，比元素 2 大的元素只有 2 个。元素 3 和 5 在区间外，所以 3 和 5 不能参与逆序数的统计。比元素 2 小的元素也只有 2 个，黄色标识的 3 个 1 都比 2 小，但是第一个 1 不能算在内，因为它在区间外。
+
+先将所有查询区间按照右端点单调不减排序，如下图所示。
+
+> 这里也可以按照查询区间左端点单调不增排序。如果这样排序，下面构建树状数组需要倒序插入。并且查找的是下标靠后但是元素值小的逆序对。两者方法都可以实现，这里讲解选其中一种。
+
+![](https://img.halfrost.com/Blog/ArticleImage/152_1_0.png)
+
+总的区间覆盖的范围决定了树状数组待插入数字的范围。如上图，总的区间位于 [1,12]，那么下标为 0，13，14 的元素不需要理会，它们不会被用到，所以也不用插入到树状数组中。
+
+
+求区间逆序对的过程中还需要利用到一个辅助数组 C[k]，这个数组的含义是下标为 k 的元素，在插入到树状数组之前，比 A[k] 值小的元素有几个。举个例子，例如下标为  7 的元素值为 9 。C[7] = 6，因为当前比 9 小的元素是 3，5，2，1，1，8。这个辅助数组 C[k] 的意义是找到下标比它小，且元素值也比它小的元素个数。
+
+由于这里选择区间右区间排序，所以构造树状数组插入是顺序插入。这样区间从左有右的查询可以依次得到结果。如上图中最下一行的图示，假设当前查询到了第 4 个区间。第 4 个区间包含元素值 1,8,9,8,6,5 。当前从左往右插入构造树状数组，已经插入了下标为 [1,10] 区间的元素值，即如图显示插入的数值。现在遍历查询区间内所有元素，Query(A[i] - 1) - C[i] 即为下标 i 在当前查询区间内的逆序对总个数。例如元素 9：
+
+
+{{< katex display >}}
+\begin{aligned}
+Query(A[i] - 1) - C[i] &=  Query(A[7] - 1) - C[7]  \\
+&= Query(9 - 1) - C[7] = Query(8) - C[7]\\
+&= 9 - 6 = 3\\
+\end{aligned}
+{{< /katex >}}
+
+插入 A[i] 元素构造树状数组在先，Query() 查询针对当前全局情况，即查询下标 [1,10] 区间内所有比元素 9 小的元素总数，不难发现所有元素都比元素 9 小，那么 Query(A[i] - 1) 得到的结果是 9。C[7] 是元素 9 插入到树状数组之前比元素 9 小的元素总数，是 6。两者相减，最终结果是 9 - 6 = 3。看上图也很容易看出来结果是正确的，在区间内比 9 下标值大且元素值比 9 小的只有 3 个，分别对应的下标是 8，9，10，对应的元素值是 8，6，5。
+
+总结：
+
+1. 离散化数组 A[i]
+2. 对所有区间按照右端点单调不减排序
+3. 按照区间排序后的结果，从左往右依次遍历每个区间。依照从左往右的区间覆盖元素范围，从左往右将 A[i] 插入至树状数组中，每个元素插入之前计算辅助数组 C[i]。
+4. 依次遍历每个区间内的所有元素，对每个元素计算 Query(A[i] - 1) - C[i]，累加逆序对的结果即是这个区间所有逆序对的总数。
+
+### 3. 求树上逆序对
+
+给定 {{< katex >}} n \in [0,10^{5}] {{< /katex >}} 个结点的树，求每个结点的子树中结点编号比它小的数的个数。
+
+
+树上逆序对的问题可以通过树的先序遍历可以将树转换成数组，令树上的某个结点 i，先序遍历到的顺序为 pre[i]，i 的子结点个数为 a[i]，则转换成数组后 i 管理的区间为 [pre[i], pre[i] + a[i] - 1]，然后就可以转换成区间逆序对问题进行求解了。
+
+
+## 四. 二维树状数组
+
+树状数组可以扩展到二维、三维或者更高维。二维树状数组可以解决离散平面上的统计问题。
+
+```go
+// BinaryIndexedTree2D define
+type BinaryIndexedTree2D struct {
+	tree [][]int
+	row  int
+	col  int
+}
+
+// Add define
+func (bit2 *BinaryIndexedTree2D) Add(i, j int, val int) {
+	for i <= bit2.row {
+		k := j
+		for k <= bit2.col {
+			bit2.tree[i][k] += val
+			k += lowbit(k)
+		}
+		i += lowbit(i)
+	}
+}
+
+// Query define
+func (bit2 *BinaryIndexedTree2D) Query(i, j int) int {
+	sum := 0
+	for i >= 1 {
+		k := j
+		for k >= 1 {
+			sum += bit2.tree[i][k]
+			k -= lowbit(k)
+		}
+		i -= lowbit(i)
+	}
+	return sum
+}
+```
+
+如果把一维树状数组维护的是数轴上的统计问题，
+
+![](https://img.halfrost.com/Blog/ArticleImage/152_2.png)
+
+
+那么二维数组维护的是二维坐标系下的统计问题。X 和 Y 分别都满足一维树状数组的性质。
+
+![](https://img.halfrost.com/Blog/ArticleImage/152_3.png)
diff --git a/website/content/ChapterThree/LFUCache.md b/website/content/ChapterThree/LFUCache.md
new file mode 100644
index 000000000..95dfb6a75
--- /dev/null
+++ b/website/content/ChapterThree/LFUCache.md
@@ -0,0 +1,368 @@
+---
+title: 3.4 LFUCache
+type: docs
+weight: 4
+---
+
+# 最不经常最少使用 LFUCache
+
+![](https://img.halfrost.com/Blog/ArticleImage/146_1_.png)
+
+LFU 是 Least Frequently Used 的缩写，即最不经常最少使用，也是一种常用的页面置换算法，选择访问计数器最小的页面予以淘汰。如下图，缓存中每个页面带一个访问计数器。
+
+
+![](https://img.halfrost.com/Blog/ArticleImage/146_3.png)
+
+根据 LFU 的策略，每访问一次都要更新访问计数器。当插入 B 的时候，发现缓存中有 B，所以增加访问计数器的计数，并把 B 移动到访问计数器从大到小排序的地方。再插入 D，同理先更新计数器，再移动到它排序以后的位置。当插入 F 的时候，缓存中不存在 F，所以淘汰计数器最小的页面的页面，所以淘汰 A 页面。此时 F 排在最下面，计数为 1。
+
+![](https://img.halfrost.com/Blog/ArticleImage/146_8_.png)
+
+这里有一个比 LRU 特别的地方。如果淘汰的页面访问次数有多个相同的访问次数，选择最靠尾部的。如上图中，A、B、C 三者的访问次数相同，都是 1 次。要插入 F，F 不在缓存中，此时要淘汰 A 页面。F 是新插入的页面，访问次数为 1，排在 C 的前面。也就是说相同的访问次数，按照新旧顺序排列，淘汰掉最旧的页面。这一点是和 LRU 最大的不同的地方。
+
+可以发现，**LFU 更新和插入新页面可以发生在链表中任意位置，删除页面都发生在表尾**。
+
+
+## 解法一 Get O(1) / Put O(1)
+
+LFU 同样要求查询尽量高效，O(1) 内查询。依旧选用 map 查询。修改和删除也需要 O(1) 完成，依旧选用双向链表，继续复用 container 包中的 list 数据结构。LFU 需要记录访问次数，所以每个结点除了存储 key，value，需要再多存储 frequency 访问次数。
+
+还有 1 个问题需要考虑，一个是如何按频次排序？相同频次，按照先后顺序排序。如果你开始考虑排序算法的话，思考方向就偏离最佳答案了。排序至少 O(nlogn)。重新回看 LFU 的工作原理，会发现它只关心最小频次。其他频次之间的顺序并不关心。所以不需要排序。用一个 min 变量保存最小频次，淘汰时读取这个最小值能找到要删除的结点。相同频次按照先后顺序排列，这个需求还是用双向链表实现，双向链表插入的顺序体现了结点的先后顺序。相同频次对应一个双向链表，可能有多个相同频次，所以可能有多个双向链表。用一个 map 维护访问频次和双向链表的对应关系。删除最小频次时，通过 min 找到最小频次，然后再这个 map 中找到这个频次对应的双向链表，在双向链表中找到最旧的那个结点删除。这就解决了 LFU 删除操作。
+
+LFU 的更新操作和 LRU 类似，也需要用一个 map 保存 key 和双向链表结点的映射关系。这个双向链表结点中存储的是 key-value-frequency 三个元素的元组。这样通过结点中的 key 和 frequency 可以反过来删除 map 中的 key。
+
+定义 LFUCache 的数据结构如下：
+
+```go
+
+import "container/list"
+
+type LFUCache struct {
+	nodes    map[int]*list.Element
+	lists    map[int]*list.List
+	capacity int
+	min      int
+}
+
+type node struct {
+	key       int
+	value     int
+	frequency int
+}
+
+func Constructor(capacity int) LFUCache {
+	return LFUCache{nodes: make(map[int]*list.Element),
+		lists:    make(map[int]*list.List),
+		capacity: capacity,
+		min:      0,
+	}
+}
+
+```
+
+LFUCache 的 Get 操作涉及更新 frequency 值和 2 个 map。在 nodes map 中通过 key 获取到结点信息。在 lists 删除结点当前 frequency 结点。删完以后 frequency ++。新的 frequency 如果在 lists 中存在，添加到双向链表表首，如果不存在，需要新建一个双向链表并把当前结点加到表首。再更新双向链表结点作为 value 的 map。最后更新 min 值，判断老的 frequency 对应的双向链表中是否已经为空，如果空了，min++。
+
+```go
+func (this *LFUCache) Get(key int) int {
+	value, ok := this.nodes[key]
+	if !ok {
+		return -1
+	}
+	currentNode := value.Value.(*node)
+	this.lists[currentNode.frequency].Remove(value)
+	currentNode.frequency++
+	if _, ok := this.lists[currentNode.frequency]; !ok {
+		this.lists[currentNode.frequency] = list.New()
+	}
+	newList := this.lists[currentNode.frequency]
+	newNode := newList.PushFront(currentNode)
+	this.nodes[key] = newNode
+	if currentNode.frequency-1 == this.min && this.lists[currentNode.frequency-1].Len() == 0 {
+		this.min++
+	}
+	return currentNode.value
+}
+
+```
+
+LFU 的 Put 操作逻辑稍微多一点。先在 nodes map 中查询 key 是否存在，如果存在，获取这个结点，更新它的 value 值，然后手动调用一次 Get 操作，因为下面的更新逻辑和 Get 操作一致。如果 map 中不存在，接下来进行插入或者删除操作。判断 capacity 是否装满，如果装满，执行删除操作。在 min 对应的双向链表中删除表尾的结点，对应的也要删除 nodes map 中的键值。
+
+由于新插入的页面访问次数一定为 1，所以 min 此时置为 1。新建结点，插入到 2 个 map 中。
+
+```go
+
+func (this *LFUCache) Put(key int, value int) {
+	if this.capacity == 0 {
+		return
+	}
+	// 如果存在，更新访问次数
+	if currentValue, ok := this.nodes[key]; ok {
+		currentNode := currentValue.Value.(*node)
+		currentNode.value = value
+		this.Get(key)
+		return
+	}
+	// 如果不存在且缓存满了，需要删除
+	if this.capacity == len(this.nodes) {
+		currentList := this.lists[this.min]
+		backNode := currentList.Back()
+		delete(this.nodes, backNode.Value.(*node).key)
+		currentList.Remove(backNode)
+	}
+	// 新建结点，插入到 2 个 map 中
+	this.min = 1
+	currentNode := &node{
+		key:       key,
+		value:     value,
+		frequency: 1,
+	}
+	if _, ok := this.lists[1]; !ok {
+		this.lists[1] = list.New()
+	}
+	newList := this.lists[1]
+	newNode := newList.PushFront(currentNode)
+	this.nodes[key] = newNode
+}
+
+```
+
+总结，LFU 是由两个 map 和一个 min 指针组成的数据结构。一个 map 中 key 存的是访问次数，对应的 value 是一个个的双向链表，此处双向链表的作用是在相同频次的情况下，淘汰表尾最旧的那个页面。另一个 map 中 key 对应的 value 是双向链表的结点，结点中比 LRU 多存储了一个访问次数的值，即结点中存储 key-value-frequency 的元组。此处双向链表的作用和 LRU 是类似的，可以根据 map 中的 key 更新双向链表结点中的 value 和 frequency 的值，也可以根据双向链表结点中的 key 和 frequency 反向更新 map 中的对应关系。如下图：
+
+![](https://img.halfrost.com/Blog/ArticleImage/146_10_0.png)
+
+提交代码以后，成功通过所有测试用例。
+
+
+![](https://img.halfrost.com/Blog/ArticleImage/146_5.png)
+
+
+## 解法二 Get O(capacity) / Put O(capacity)
+
+LFU 的另外一个思路是利用 [Index Priority Queue](https://algs4.cs.princeton.edu/24pq/) 这个数据结构。别被名字吓到，Index Priority Queue = map + Priority Queue，仅此而已。
+
+利用 Priority Queue 维护一个最小堆，堆顶是访问次数最小的元素。map 中的 value 存储的是优先队列中结点。
+
+```go
+import "container/heap"
+
+type LFUCache struct {
+	capacity int
+	pq       PriorityQueue
+	hash     map[int]*Item
+	counter  int
+}
+
+func Constructor(capacity int) LFUCache {
+	lfu := LFUCache{
+		pq:       PriorityQueue{},
+		hash:     make(map[int]*Item, capacity),
+		capacity: capacity,
+	}
+	return lfu
+}
+
+```
+
+Get 和 Put 操作要尽量的快，有 2 个问题需要解决。当访问次数相同时，如何删除掉最久的元素？当元素的访问次数发生变化时，如何快速调整堆？为了解决这 2 个问题，定义如下的数据结构：
+
+```go
+// An Item is something we manage in a priority queue.
+type Item struct {
+	value     int // The value of the item; arbitrary.
+	key       int
+	frequency int // The priority of the item in the queue.
+	count     int // use for evicting the oldest element
+	// The index is needed by update and is maintained by the heap.Interface methods.
+	index int // The index of the item in the heap.
+}
+
+```
+
+堆中的结点存储这 5 个值。count 值用来决定哪个是最老的元素，类似一个操作时间戳。index 值用来 re-heapify 调整堆的。接下来实现 PriorityQueue 的方法。
+
+```go
+// A PriorityQueue implements heap.Interface and holds Items.
+type PriorityQueue []*Item
+
+func (pq PriorityQueue) Len() int { return len(pq) }
+
+func (pq PriorityQueue) Less(i, j int) bool {
+	// We want Pop to give us the highest, not lowest, priority so we use greater than here.
+	if pq[i].frequency == pq[j].frequency {
+		return pq[i].count < pq[j].count
+	}
+	return pq[i].frequency < pq[j].frequency
+}
+
+func (pq PriorityQueue) Swap(i, j int) {
+	pq[i], pq[j] = pq[j], pq[i]
+	pq[i].index = i
+	pq[j].index = j
+}
+
+func (pq *PriorityQueue) Push(x interface{}) {
+	n := len(*pq)
+	item := x.(*Item)
+	item.index = n
+	*pq = append(*pq, item)
+}
+
+func (pq *PriorityQueue) Pop() interface{} {
+	old := *pq
+	n := len(old)
+	item := old[n-1]
+	old[n-1] = nil  // avoid memory leak
+	item.index = -1 // for safety
+	*pq = old[0 : n-1]
+	return item
+}
+
+// update modifies the priority and value of an Item in the queue.
+func (pq *PriorityQueue) update(item *Item, value int, frequency int, count int) {
+	item.value = value
+	item.count = count
+	item.frequency = frequency
+	heap.Fix(pq, item.index)
+}
+```
+
+在 Less() 方法中，frequency 从小到大排序，frequency 相同的，按 count 从小到大排序。按照优先队列建堆规则，可以得到，frequency 最小的在堆顶，相同的 frequency，count 最小的越靠近堆顶。
+
+在 Swap() 方法中，记得要更新 index 值。在 Push() 方法中，插入时队列的长度即是该元素的 index 值，此处也要记得更新 index 值。update() 方法调用 Fix() 函数。Fix() 函数比先 Remove() 再 Push() 一个新的值，花销要小。所以此处调用 Fix() 函数，这个操作的时间复杂度是 O(log n)。
+
+这样就维护了最小 Index Priority Queue。Get 操作非常简单：
+
+```go
+func (this *LFUCache) Get(key int) int {
+	if this.capacity == 0 {
+		return -1
+	}
+	if item, ok := this.hash[key]; ok {
+		this.counter++
+		this.pq.update(item, item.value, item.frequency+1, this.counter)
+		return item.value
+	}
+	return -1
+}
+
+```
+
+在 hashmap 中查询 key，如果存在，counter 时间戳累加，调用 Priority Queue 的 update 方法，调整堆。
+
+```go
+func (this *LFUCache) Put(key int, value int) {
+	if this.capacity == 0 {
+		return
+	}
+	this.counter++
+	// 如果存在，增加 frequency，再调整堆
+	if item, ok := this.hash[key]; ok {
+		this.pq.update(item, value, item.frequency+1, this.counter)
+		return
+	}
+	// 如果不存在且缓存满了，需要删除。在 hashmap 和 pq 中删除。
+	if len(this.pq) == this.capacity {
+		item := heap.Pop(&this.pq).(*Item)
+		delete(this.hash, item.key)
+	}
+	// 新建结点，在 hashmap 和 pq 中添加。
+	item := &Item{
+		value: value,
+		key:   key,
+		count: this.counter,
+	}
+	heap.Push(&this.pq, item)
+	this.hash[key] = item
+}
+```
+
+
+用最小堆实现的 LFU，Put 时间复杂度是 O(capacity)，Get 时间复杂度是 O(capacity)，不及 2 个 map 实现的版本。巧的是最小堆的版本居然打败了 100%。
+
+![](https://img.halfrost.com/Blog/ArticleImage/146_7.png)
+
+
+## 模板
+
+
+```go
+import "container/list"
+
+type LFUCache struct {
+	nodes    map[int]*list.Element
+	lists    map[int]*list.List
+	capacity int
+	min      int
+}
+
+type node struct {
+	key       int
+	value     int
+	frequency int
+}
+
+func Constructor(capacity int) LFUCache {
+	return LFUCache{nodes: make(map[int]*list.Element),
+		lists:    make(map[int]*list.List),
+		capacity: capacity,
+		min:      0,
+	}
+}
+
+func (this *LFUCache) Get(key int) int {
+	value, ok := this.nodes[key]
+	if !ok {
+		return -1
+	}
+	currentNode := value.Value.(*node)
+	this.lists[currentNode.frequency].Remove(value)
+	currentNode.frequency++
+	if _, ok := this.lists[currentNode.frequency]; !ok {
+		this.lists[currentNode.frequency] = list.New()
+	}
+	newList := this.lists[currentNode.frequency]
+	newNode := newList.PushBack(currentNode)
+	this.nodes[key] = newNode
+	if currentNode.frequency-1 == this.min && this.lists[currentNode.frequency-1].Len() == 0 {
+		this.min++
+	}
+	return currentNode.value
+}
+
+func (this *LFUCache) Put(key int, value int) {
+	if this.capacity == 0 {
+		return
+	}
+	if currentValue, ok := this.nodes[key]; ok {
+		currentNode := currentValue.Value.(*node)
+		currentNode.value = value
+		this.Get(key)
+		return
+	}
+	if this.capacity == len(this.nodes) {
+		currentList := this.lists[this.min]
+		frontNode := currentList.Front()
+		delete(this.nodes, frontNode.Value.(*node).key)
+		currentList.Remove(frontNode)
+	}
+	this.min = 1
+	currentNode := &node{
+		key:       key,
+		value:     value,
+		frequency: 1,
+	}
+	if _, ok := this.lists[1]; !ok {
+		this.lists[1] = list.New()
+	}
+	newList := this.lists[1]
+	newNode := newList.PushBack(currentNode)
+	this.nodes[key] = newNode
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterThree/LRUCache/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterFour/">下一章➡️</a></p>
+</div>
diff --git a/website/content/ChapterThree/LRUCache.md b/website/content/ChapterThree/LRUCache.md
new file mode 100644
index 000000000..272748971
--- /dev/null
+++ b/website/content/ChapterThree/LRUCache.md
@@ -0,0 +1,280 @@
+---
+title: 3.3 LRUCache
+type: docs
+weight: 3
+---
+
+# 最近最少使用 LRUCache
+
+![](https://img.halfrost.com/Blog/ArticleImage/146_1_.png)
+
+LRU 是 Least Recently Used 的缩写，即最近最少使用，是一种常用的页面置换算法，选择最近最久未使用的页面予以淘汰。如上图，要插入 F 的时候，此时需要淘汰掉原来的一个页面。
+
+![](https://img.halfrost.com/Blog/ArticleImage/146_2_0.png)
+
+根据 LRU 的策略，每次都淘汰最近最久未使用的页面，所以先淘汰 A 页面。再插入 C 的时候，发现缓存中有 C 页面，这个时候需要把 C 页面放到首位，因为它被使用了。以此类推，插入 G 页面，G 页面是新页面，不在缓存中，所以淘汰掉 B 页面。插入 H 页面，H 页面是新页面，不在缓存中，所以淘汰掉 D 页面。插入 E 的时候，发现缓存中有 E 页面，这个时候需要把 E 页面放到首位。插入 I 页面，I 页面是新页面，不在缓存中，所以淘汰掉 F 页面。
+
+可以发现，**LRU 更新和插入新页面都发生在链表首，删除页面都发生在链表尾**。
+
+## 解法一 Get O(1) / Put O(1)
+
+LRU 要求查询尽量高效，O(1) 内查询。那肯定选用 map 查询。修改，删除也要尽量 O(1) 完成。搜寻常见的数据结构，链表，栈，队列，树，图。树和图排除，栈和队列无法任意查询中间的元素，也排除。所以选用链表来实现。但是如果选用单链表，删除这个结点，需要 O(n) 遍历一遍找到前驱结点。所以选用双向链表，在删除的时候也能 O(1) 完成。
+
+由于 Go 的 container 包中的 list 底层实现是双向链表，所以可以直接复用这个数据结构。定义 LRUCache 的数据结构如下：
+
+```go
+import "container/list"
+
+type LRUCache struct {
+    Cap  int
+    Keys map[int]*list.Element
+    List *list.List
+}
+
+type pair struct {
+    K, V int
+}
+
+func Constructor(capacity int) LRUCache {
+    return LRUCache{
+        Cap: capacity,
+        Keys: make(map[int]*list.Element),
+        List: list.New(),
+    }
+}
+
+```
+
+这里需要解释 2 个问题，list 中的值存的是什么？pair 这个结构体有什么用？
+
+```go
+type Element struct {
+	// Next and previous pointers in the doubly-linked list of elements.
+	// To simplify the implementation, internally a list l is implemented
+	// as a ring, such that &l.root is both the next element of the last
+	// list element (l.Back()) and the previous element of the first list
+	// element (l.Front()).
+	next, prev *Element
+
+	// The list to which this element belongs.
+	list *List
+
+	// The value stored with this element.
+	Value interface{}
+}
+```
+
+在 container/list 中，这个双向链表的每个结点的类型是 Element。Element 中存了 4 个值，前驱和后继结点，双向链表的头结点，value 值。这里的 value 是 interface 类型。笔者在这个 value 里面存了 pair 这个结构体。这就解释了 list 里面存的是什么数据。
+
+为什么要存 pair 呢？单单只存 v 不行么，为什么还要存一份 key ？原因是在 LRUCache 执行删除操作的时候，需要维护 2 个数据结构，一个是 map，一个是双向链表。在双向链表中删除淘汰出去的 value，在 map 中删除淘汰出去 value 对应的 key。如果在双向链表的 value 中不存储 key，那么再删除 map 中的 key 的时候有点麻烦。如果硬要实现，需要先获取到双向链表这个结点 Element 的地址。然后遍历 map，在 map 中找到存有这个 Element 元素地址对应的 key，再删除。这样做时间复杂度是 O(n)，做不到 O(1)。所以双向链表中的 Value 需要存储这个 pair。
+
+LRUCache 的 Get 操作很简单，在 map 中直接读取双向链表的结点。如果 map 中存在，将它移动到双向链表的表头，并返回它的 value 值，如果 map 中不存在，返回 -1。
+
+```go 
+func (c *LRUCache) Get(key int) int {
+	if el, ok := c.Keys[key]; ok {
+		c.List.MoveToFront(el)
+		return el.Value.(pair).V
+	}
+	return -1
+}
+```
+
+LRUCache 的 Put 操作也不难。先查询 map 中是否存在 key，如果存在，更新它的 value，并且把该结点移到双向链表的表头。如果 map 中不存在，新建这个结点加入到双向链表和 map 中。最后别忘记还需要维护双向链表的 cap，如果超过 cap，需要淘汰最后一个结点，双向链表中删除这个结点，map 中删掉这个结点对应的 key。
+
+```go
+func (c *LRUCache) Put(key int, value int) {
+	if el, ok := c.Keys[key]; ok {
+		el.Value = pair{K: key, V: value}
+		c.List.MoveToFront(el)
+	} else {
+		el := c.List.PushFront(pair{K: key, V: value})
+		c.Keys[key] = el
+	}
+	if c.List.Len() > c.Cap {
+		el := c.List.Back()
+		c.List.Remove(el)
+		delete(c.Keys, el.Value.(pair).K)
+	}
+}
+
+```
+
+总结，LRU 是由一个 map 和一个双向链表组成的数据结构。map 中 key 对应的 value 是双向链表的结点。双向链表中存储 key-value 的 pair。双向链表表首更新缓存，表尾淘汰缓存。如下图：
+
+![](https://img.halfrost.com/Blog/ArticleImage/146_9.png)
+
+提交代码以后，成功通过所有测试用例。
+
+![](https://img.halfrost.com/Blog/ArticleImage/146_4_.png)
+
+
+## 解法二 Get O(1) / Put O(1)
+
+数据结构上想不到其他解法了，但从打败的百分比上，看似还有常数的优化空间。笔者反复思考，觉得可能导致运行时间变长的地方是在 interface{} 类型推断，其他地方已无优化的空间。手写一个双向链表提交试试，代码如下：
+
+```go
+
+type LRUCache struct {
+	head, tail *Node
+	keys       map[int]*Node
+	capacity   int
+}
+
+type Node struct {
+	key, val   int
+	prev, next *Node
+}
+
+func ConstructorLRU(capacity int) LRUCache {
+	return LRUCache{keys: make(map[int]*Node), capacity: capacity}
+}
+
+func (this *LRUCache) Get(key int) int {
+	if node, ok := this.keys[key]; ok {
+		this.Remove(node)
+		this.Add(node)
+		return node.val
+	}
+	return -1
+}
+
+func (this *LRUCache) Put(key int, value int) {
+	if node, ok := this.keys[key]; ok {
+		node.val = value
+		this.Remove(node)
+		this.Add(node)
+		return
+	} else {
+		node = &Node{key: key, val: value}
+		this.keys[key] = node
+		this.Add(node)
+	}
+	if len(this.keys) > this.capacity {
+		delete(this.keys, this.tail.key)
+		this.Remove(this.tail)
+	}
+}
+
+func (this *LRUCache) Add(node *Node) {
+	node.prev = nil
+	node.next = this.head
+	if this.head != nil {
+		this.head.prev = node
+	}
+	this.head = node
+	if this.tail == nil {
+		this.tail = node
+		this.tail.next = nil
+	}
+}
+
+func (this *LRUCache) Remove(node *Node) {
+	if node == this.head {
+		this.head = node.next
+		if node.next != nil {
+			node.next.prev = nil
+		}
+		node.next = nil
+		return
+	}
+	if node == this.tail {
+		this.tail = node.prev
+		node.prev.next = nil
+		node.prev = nil
+		return
+	}
+	node.prev.next = node.next
+	node.next.prev = node.prev
+}
+
+```
+
+提交以后还真的 100% 了。
+
+![](https://img.halfrost.com/Blog/ArticleImage/146_6.png)
+
+上述代码实现的 LRU 本质并没有优化，只是换了一个写法，没有用 container 包而已。
+
+
+## 模板
+
+```go
+type LRUCache struct {
+	head, tail *Node
+	Keys       map[int]*Node
+	Cap        int
+}
+
+type Node struct {
+	Key, Val   int
+	Prev, Next *Node
+}
+
+func Constructor(capacity int) LRUCache {
+	return LRUCache{Keys: make(map[int]*Node), Cap: capacity}
+}
+
+func (this *LRUCache) Get(key int) int {
+	if node, ok := this.Keys[key]; ok {
+		this.Remove(node)
+		this.Add(node)
+		return node.Val
+	}
+	return -1
+}
+
+func (this *LRUCache) Put(key int, value int) {
+	if node, ok := this.Keys[key]; ok {
+		node.Val = value
+		this.Remove(node)
+		this.Add(node)
+		return
+	} else {
+		node = &Node{Key: key, Val: value}
+		this.Keys[key] = node
+		this.Add(node)
+	}
+	if len(this.Keys) > this.Cap {
+		delete(this.Keys, this.tail.Key)
+		this.Remove(this.tail)
+	}
+}
+
+func (this *LRUCache) Add(node *Node) {
+	node.Prev = nil
+	node.Next = this.head
+	if this.head != nil {
+		this.head.Prev = node
+	}
+	this.head = node
+	if this.tail == nil {
+		this.tail = node
+		this.tail.Next = nil
+	}
+}
+
+func (this *LRUCache) Remove(node *Node) {
+	if node == this.head {
+		this.head = node.Next
+		node.Next = nil
+		return
+	}
+	if node == this.tail {
+		this.tail = node.Prev
+		node.Prev.Next = nil
+		node.Prev = nil
+		return
+	}
+	node.Prev.Next = node.Next
+	node.Next.Prev = node.Prev
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterThree/UnionFind/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterThree/LFUCache/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterThree/Segment_Tree.md b/website/content/ChapterThree/Segment_Tree.md
new file mode 100644
index 000000000..162d7e6b6
--- /dev/null
+++ b/website/content/ChapterThree/Segment_Tree.md
@@ -0,0 +1,502 @@
+---
+title: 3.1 Segment Tree
+type: docs
+weight: 1
+---
+
+# 线段树 Segment Tree
+
+线段树 Segment tree 是一种二叉树形数据结构，1977年由 Jon Louis Bentley 发明，用以存储区间或线段，并且允许快速查询结构内包含某一点的所有区间。
+
+一个包含 {{< katex >}}n {{< /katex >}}个区间的线段树，空间复杂度为 {{< katex >}} O(n) {{< /katex >}}，查询的时间复杂度则为{{< katex >}}O(log n+k) {{< /katex >}}，其中 {{< katex >}} k {{< /katex >}} 是符合条件的区间数量。线段树的数据结构也可推广到高维度。
+
+## 一. 什么是线段树
+
+以一维的线段树为例。
+
+![](https://img.halfrost.com/Blog/ArticleImage/153_1.png)
+
+
+令 S 是一维线段的集合。将这些线段的端点坐标由小到大排序，令其为{{< katex >}}x_{1},x_{2},\cdots ,x_{m} {{< /katex >}}。我们将被这些端点切分的每一个区间称为“单位区间”（每个端点所在的位置会单独成为一个单位区间），从左到右包含：
+
+{{< katex display>}} 
+(-\infty ,x_{1}),[x_{1},x_{1}],(x_{1},x_{2}),[x_{2},x_{2}],...,(x_{m-1},x_{m}),[x_{m},x_{m}],(x_{m},+\infty )
+{{< /katex >}}
+
+线段树的结构为一个二叉树，每个节点都代表一个坐标区间，节点 N 所代表的区间记为 Int(N)，则其需符合以下条件：
+
+- 其每一个叶节点，从左到右代表每个单位区间。
+- 其内部节点代表的区间是其两个儿子代表的区间之并集。
+- 每个节点（包含叶子）中有一个存储线段的数据结构。若一个线段 S 的坐标区间包含 Int(N) 但不包含 Int(parent(N))，则节点 N 中会存储线段 S。
+
+
+![](https://img.halfrost.com/Blog/ArticleImage/153_2.png)
+
+
+线段树是二叉树，其中每个节点代表一个区间。通常，一个节点将存储一个或多个合并的区间的数据，以便可以执行查询操作。
+
+
+## 二. 为什么需要这种数据结构
+
+许多问题要求我们基于对可用数据范围或区间的查询来给出结果。这可能是一个繁琐而缓慢的过程，尤其是在查询数量众多且重复的情况下。线段树让我们以对数时间复杂度有效地处理此类查询。
+
+线段树可用于计算几何和[地理信息系统领域](https://en.wikipedia.org/wiki/Geographic_information_systems)。例如，距中心参考点/原点一定距离的空间中可能会有大量点。假设我们要查找距原点一定距离范围内的点。一个普通的查找表将需要对所有可能的点或所有可能的距离进行线性扫描（假设是散列图）。线段树使我们能够以对数时间实现这一需求，而所需空间却少得多。这样的问题称为[平面范围搜索](https://en.wikipedia.org/wiki/Range_searching)。有效地解决此类问题至关重要，尤其是在处理动态数据且瞬息万变的情况下（例如，用于空中交通的雷达系统）。下文会以线段树解决 Range Sum Query problem 为例。
+
+![](https://img.halfrost.com/Blog/ArticleImage/153_3.png)
+
+
+上图即作为范围查询的线段树。
+
+## 三. 构造线段树
+
+
+假设数据存在 size 为 n 的 arr[] 中。
+
+1. 线段树的根通常代表整个数据区间。这里是 arr[0：n-1]。
+2. 树的每个叶子代表一个范围，其中仅包含一个元素。 因此，叶子代表 arr[0]，arr[1] 等等，直到 arr[n-1]。
+3. 树的内部节点将代表其子节点的合并或并集结果。
+4. 每个子节点可代表其父节点所代表范围的大约一半。(二分的思想)
+
+使用大小为 {{< katex >}}\approx 4 \ast n {{< /katex >}} 的数组可以轻松表示 n 个元素范围的线段树。（[Stack Overflow](http://stackoverflow.com/q/28470692/2844164) 对原因进行了很好的讨论。如果你还不确定，请不要担心。本文将在稍后进行讨论。）
+
+下标为 i 的节点有两个节点，下标分别为 {{< katex >}}(2 \ast i + 1) {{< /katex >}}和 {{< katex >}}(2 \ast i + 2){{< /katex >}}。
+
+![](https://img.halfrost.com/Blog/ArticleImage/153_4.png)
+
+线段树看上去很直观并且非常适合递归构造。
+
+我们将使用数组 tree[] 来存储线段树的节点（初始化为全零）。 下标从 0 开始。
+
+- 树的节点在下标 0 处。因此 tree[0] 是树的根。
+- tree[i] 的孩子存在 tree[2 * i + 1] 和 tree[2 * i + 2] 中。
+- 用额外的 0 或 null 值填充 arr[]，使得 {{< katex >}}n = 2^{k} {{< /katex >}}（其中 n 是 arr[] 的总长度，而 k 是非负整数。）
+- 叶子节点的下标取值范围在 {{< katex >}} \in [2^{k}-1, 2^{k+1}-2]{{< /katex >}}
+
+![](https://img.halfrost.com/Blog/ArticleImage/153_5.png)
+
+构造线段树的代码如下：
+
+
+```go
+// SegmentTree define
+type SegmentTree struct {
+	data, tree, lazy []int
+	left, right      int
+	merge            func(i, j int) int
+}
+
+// Init define
+func (st *SegmentTree) Init(nums []int, oper func(i, j int) int) {
+	st.merge = oper
+	data, tree, lazy := make([]int, len(nums)), make([]int, 4*len(nums)), make([]int, 4*len(nums))
+	for i := 0; i < len(nums); i++ {
+		data[i] = nums[i]
+	}
+	st.data, st.tree, st.lazy = data, tree, lazy
+	if len(nums) > 0 {
+		st.buildSegmentTree(0, 0, len(nums)-1)
+	}
+}
+
+// 在 treeIndex 的位置创建 [left....right] 区间的线段树
+func (st *SegmentTree) buildSegmentTree(treeIndex, left, right int) {
+	if left == right {
+		st.tree[treeIndex] = st.data[left]
+		return
+	}
+	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, st.leftChild(treeIndex), st.rightChild(treeIndex)
+	st.buildSegmentTree(leftTreeIndex, left, midTreeIndex)
+	st.buildSegmentTree(rightTreeIndex, midTreeIndex+1, right)
+	st.tree[treeIndex] = st.merge(st.tree[leftTreeIndex], st.tree[rightTreeIndex])
+}
+
+func (st *SegmentTree) leftChild(index int) int {
+	return 2*index + 1
+}
+
+func (st *SegmentTree) rightChild(index int) int {
+	return 2*index + 2
+}
+```
+
+笔者将线段树合并的操作变成了一个函数。合并操作根据题意变化，常见的有加法，取 max，min 等等。
+
+我们以 arr[] = [18, 17, 13, 19, 15, 11, 20, 12, 33, 25 ] 为例构造线段树：
+
+![](https://img.halfrost.com/Blog/ArticleImage/153_6.png)
+
+线段树构造好以后，数组里面的数据是：
+
+```c
+tree[] = [ 183, 82, 101, 48, 34, 43, 58, 35, 13, 19, 15, 31, 12, 33, 25, 18, 17, 0, 0, 0, 0, 0, 0, 11, 20, 0, 0, 0, 0, 0, 0 ]
+```
+
+线段树用 0 填充到 4*n 个元素。
+
+
+> LeetCode 对应题目是 [218. The Skyline Problem](https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0218.The-Skyline-Problem/)、[303. Range Sum Query - Immutable](https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0303.Range-Sum-Query-Immutable/)、[307. Range Sum Query - Mutable](https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0307.Range-Sum-Query-Mutable/)、[699. Falling Squares](https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0699.Falling-Squares/)
+
+## 四. 线段树的查询
+
+线段树的查询方法有两种，一种是直接查询，另外一种是懒查询。
+
+### 1. 直接查询
+
+当查询范围与当前节点表示的范围完全匹配时，该方法返回结果。否则，它会更深入地遍历线段树树，以找到与节点的一部分完全匹配的节点。
+
+
+```go
+// 查询 [left....right] 区间内的值
+
+// Query define
+func (st *SegmentTree) Query(left, right int) int {
+	if len(st.data) > 0 {
+		return st.queryInTree(0, 0, len(st.data)-1, left, right)
+	}
+	return 0
+}
+
+// 在以 treeIndex 为根的线段树中 [left...right] 的范围里，搜索区间 [queryLeft...queryRight] 的值
+func (st *SegmentTree) queryInTree(treeIndex, left, right, queryLeft, queryRight int) int {
+	if left == queryLeft && right == queryRight {
+		return st.tree[treeIndex]
+	}
+	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, st.leftChild(treeIndex), st.rightChild(treeIndex)
+	if queryLeft > midTreeIndex {
+		return st.queryInTree(rightTreeIndex, midTreeIndex+1, right, queryLeft, queryRight)
+	} else if queryRight <= midTreeIndex {
+		return st.queryInTree(leftTreeIndex, left, midTreeIndex, queryLeft, queryRight)
+	}
+	return st.merge(st.queryInTree(leftTreeIndex, left, midTreeIndex, queryLeft, midTreeIndex),
+		st.queryInTree(rightTreeIndex, midTreeIndex+1, right, midTreeIndex+1, queryRight))
+}
+```
+
+
+![](https://img.halfrost.com/Blog/ArticleImage/153_7.png)
+
+
+在上面的示例中，查询的区间范围为[2，8] 的元素之和。没有任何线段可以完全代表[2，8] 范围。但是可以观察到，可以使用范围 [2，2]，[3，4]，[5，7]，[8，8] 这 4 个区间构成 [8，8]。快速验证 [2,8] 处的输入元素之和为 13 + 19 + 15 + 11 + 20 + 12 + 33 = 123。[2，2]，[3，4]，[5，7] 和 [8，8] 的节点总和是 13 + 34 + 43 + 33 = 123。答案正确。
+
+
+
+### 2. 懒查询
+
+懒查询对应懒更新，两者是配套操作。在区间更新时，并不直接更新区间内所有节点，而是把区间内节点增减变化的值存在 lazy 数组中。等到下次查询的时候再把增减应用到具体的节点上。这样做也是为了分摊时间复杂度，保证查询和更新的时间复杂度在 O(log n) 级别，不会退化成 O(n) 级别。
+
+懒查询节点的步骤：
+
+1. 先判断当前节点是否是懒节点。通过查询 lazy[i] 是否为 0 判断。如果是懒节点，将它的增减变化应用到该节点上。并且更新它的孩子节点。这一步和更新操作的第一步完全一样。
+2. 递归查询子节点，以找到适合的查询节点。
+
+具体代码如下：
+
+```go
+// 查询 [left....right] 区间内的值
+
+// QueryLazy define
+func (st *SegmentTree) QueryLazy(left, right int) int {
+	if len(st.data) > 0 {
+		return st.queryLazyInTree(0, 0, len(st.data)-1, left, right)
+	}
+	return 0
+}
+
+func (st *SegmentTree) queryLazyInTree(treeIndex, left, right, queryLeft, queryRight int) int {
+	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, st.leftChild(treeIndex), st.rightChild(treeIndex)
+	if left > queryRight || right < queryLeft { // segment completely outside range
+		return 0 // represents a null node
+	}
+	if st.lazy[treeIndex] != 0 { // this node is lazy
+		for i := 0; i < right-left+1; i++ {
+			st.tree[treeIndex] = st.merge(st.tree[treeIndex], st.lazy[treeIndex])
+			// st.tree[treeIndex] += (right - left + 1) * st.lazy[treeIndex] // normalize current node by removing lazinesss
+		}
+		if left != right { // update lazy[] for children nodes
+			st.lazy[leftTreeIndex] = st.merge(st.lazy[leftTreeIndex], st.lazy[treeIndex])
+			st.lazy[rightTreeIndex] = st.merge(st.lazy[rightTreeIndex], st.lazy[treeIndex])
+			// st.lazy[leftTreeIndex] += st.lazy[treeIndex]
+			// st.lazy[rightTreeIndex] += st.lazy[treeIndex]
+		}
+		st.lazy[treeIndex] = 0 // current node processed. No longer lazy
+	}
+	if queryLeft <= left && queryRight >= right { // segment completely inside range
+		return st.tree[treeIndex]
+	}
+	if queryLeft > midTreeIndex {
+		return st.queryLazyInTree(rightTreeIndex, midTreeIndex+1, right, queryLeft, queryRight)
+	} else if queryRight <= midTreeIndex {
+		return st.queryLazyInTree(leftTreeIndex, left, midTreeIndex, queryLeft, queryRight)
+	}
+	// merge query results
+	return st.merge(st.queryLazyInTree(leftTreeIndex, left, midTreeIndex, queryLeft, midTreeIndex),
+		st.queryLazyInTree(rightTreeIndex, midTreeIndex+1, right, midTreeIndex+1, queryRight))
+}
+```
+
+
+## 五. 线段树的更新
+
+### 1. 单点更新
+
+单点更新类似于 `buildSegTree`。更新树的叶子节点的值，该值与更新后的元素相对应。这些更新的值会通过树的上层节点把影响传播到根。
+
+
+```go
+// 更新 index 位置的值
+
+// Update define
+func (st *SegmentTree) Update(index, val int) {
+	if len(st.data) > 0 {
+		st.updateInTree(0, 0, len(st.data)-1, index, val)
+	}
+}
+
+// 以 treeIndex 为根，更新 index 位置上的值为 val
+func (st *SegmentTree) updateInTree(treeIndex, left, right, index, val int) {
+	if left == right {
+		st.tree[treeIndex] = val
+		return
+	}
+	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, st.leftChild(treeIndex), st.rightChild(treeIndex)
+	if index > midTreeIndex {
+		st.updateInTree(rightTreeIndex, midTreeIndex+1, right, index, val)
+	} else {
+		st.updateInTree(leftTreeIndex, left, midTreeIndex, index, val)
+	}
+	st.tree[treeIndex] = st.merge(st.tree[leftTreeIndex], st.tree[rightTreeIndex])
+}
+```
+
+![](https://img.halfrost.com/Blog/ArticleImage/153_8.png)
+
+在此示例中，下标为（在原始输入数据中）1、3 和 6 处的元素分别增加了 +3，-1 和 +2。可以看到更改如何沿树传播，一直到根。
+
+
+### 2. 区间更新
+
+
+线段树仅更新单个元素，非常有效，时间复杂度 O(log n)。 但是，如果我们要更新一系列元素怎么办？按照当前的方法，每个元素都必须独立更新，每个元素都会花费一些时间。分别更新每一个叶子节点意味着要多次处理它们的共同祖先。祖先节点可能被更新多次。如果想要减少这种重复计算，该怎么办？
+
+
+![](https://img.halfrost.com/Blog/ArticleImage/153_11.png)
+
+在上面的示例中，根节点被更新了三次，而编号为 82 的节点被更新了两次。这是因为更新叶子节点对上层父亲节点有影响。最差的情况，查询的区间内不包含频繁更新的元素，于是需要花费很多时间更新不怎么访问的节点。增加额外的 lazy 数组，可以减少不必要的计算，并且能按需处理节点。
+
+使用另一个数组 lazy[]，它的大小与我们的线段树 array tree[] 完全相同，代表一个惰性节点。当访问或查询该节点时，lazy[i] 中保留需要增加或者减少该节点 tree[i] 的数量。 当 lazy[i] 为 0 时，表示 tree[i] 该节点不是惰性的，并且没有缓存的更新。
+
+更新区间内节点的步骤：
+
+1. 先判断当前节点是否是懒节点。通过查询 lazy[i] 是否为 0 判断。如果是懒节点，将它的增减变化应用到该节点上。并且更新它的孩子节点。
+2. 如果当前节点代表的区间位于更新范围内，则将当前更新操作应用于当前节点。
+3. 递归更新子节点。
+
+具体代码如下：
+
+```go
+
+// 更新 [updateLeft....updateRight] 位置的值
+// 注意这里的更新值是在原来值的基础上增加或者减少，而不是把这个区间内的值都赋值为 x，区间更新和单点更新不同
+// 这里的区间更新关注的是变化，单点更新关注的是定值
+// 当然区间更新也可以都更新成定值，如果只区间更新成定值，那么 lazy 更新策略需要变化，merge 策略也需要变化，这里暂不详细讨论
+
+// UpdateLazy define
+func (st *SegmentTree) UpdateLazy(updateLeft, updateRight, val int) {
+	if len(st.data) > 0 {
+		st.updateLazyInTree(0, 0, len(st.data)-1, updateLeft, updateRight, val)
+	}
+}
+
+func (st *SegmentTree) updateLazyInTree(treeIndex, left, right, updateLeft, updateRight, val int) {
+	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, st.leftChild(treeIndex), st.rightChild(treeIndex)
+	if st.lazy[treeIndex] != 0 { // this node is lazy
+		for i := 0; i < right-left+1; i++ {
+			st.tree[treeIndex] = st.merge(st.tree[treeIndex], st.lazy[treeIndex])
+			//st.tree[treeIndex] += (right - left + 1) * st.lazy[treeIndex] // normalize current node by removing laziness
+		}
+		if left != right { // update lazy[] for children nodes
+			st.lazy[leftTreeIndex] = st.merge(st.lazy[leftTreeIndex], st.lazy[treeIndex])
+			st.lazy[rightTreeIndex] = st.merge(st.lazy[rightTreeIndex], st.lazy[treeIndex])
+			// st.lazy[leftTreeIndex] += st.lazy[treeIndex]
+			// st.lazy[rightTreeIndex] += st.lazy[treeIndex]
+		}
+		st.lazy[treeIndex] = 0 // current node processed. No longer lazy
+	}
+
+	if left > right || left > updateRight || right < updateLeft {
+		return // out of range. escape.
+	}
+
+	if updateLeft <= left && right <= updateRight { // segment is fully within update range
+		for i := 0; i < right-left+1; i++ {
+			st.tree[treeIndex] = st.merge(st.tree[treeIndex], val)
+			//st.tree[treeIndex] += (right - left + 1) * val // update segment
+		}
+		if left != right { // update lazy[] for children
+			st.lazy[leftTreeIndex] = st.merge(st.lazy[leftTreeIndex], val)
+			st.lazy[rightTreeIndex] = st.merge(st.lazy[rightTreeIndex], val)
+			// st.lazy[leftTreeIndex] += val
+			// st.lazy[rightTreeIndex] += val
+		}
+		return
+	}
+	st.updateLazyInTree(leftTreeIndex, left, midTreeIndex, updateLeft, updateRight, val)
+	st.updateLazyInTree(rightTreeIndex, midTreeIndex+1, right, updateLeft, updateRight, val)
+	// merge updates
+	st.tree[treeIndex] = st.merge(st.tree[leftTreeIndex], st.tree[rightTreeIndex])
+}
+
+```
+
+> LeetCode 对应题目是 [218. The Skyline Problem](https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0218.The-Skyline-Problem/)、[699. Falling Squares](https://books.halfrost.com/leetcode/ChapterFour/0600~0699/0699.Falling-Squares/)
+
+## 六. 时间复杂度分析
+
+让我们看一下构建过程。我们访问了线段树的每个叶子（对应于数组 arr[] 中的每个元素）。因此，我们处理大约 2 * n 个节点。这使构建过程时间复杂度为 O(n)。对于每个递归更新的过程都将丢弃区间范围的一半，以到达树中的叶子节点。这类似于二分搜索，只需要对数时间。更新叶子后，将更新树的每个级别上的直接祖先。这花费时间与树的高度成线性关系。
+
+![](https://img.halfrost.com/Blog/ArticleImage/153_9.png)
+
+
+4\*n 个节点可以确保将线段树构建为完整的二叉树，从而树的高度为 log(4\*n + 1) 向上取整。线段树读取和更新的时间复杂度都为 O(log n)。
+
+## 七. 常见题型
+
+
+### 1. Range Sum Queries
+
+![](https://img.halfrost.com/Blog/ArticleImage/153_10.png)
+
+
+Range Sum Queries 是 [Range Queries](https://en.wikipedia.org/wiki/Range_query_(data_structures)) 问题的子集。给定一个数据元素数组或序列，需要处理由元素范围组成的读取和更新查询。线段树 Segment Tree 和树状数组 Binary Indexed Tree (a.k.a. Fenwick Tree)) 都能很快的解决这类问题。
+
+Range Sum Query 问题专门处理查询范围内的元素总和。这个问题存在许多变体，包括[不可变数据](https://leetcode.com/problems/range-sum-query-immutable/)，[可变数据](https://leetcode.com/problems/range-sum-query-mutable/)，[多次更新，单次查询](https://leetcode.com/problems/range-addition/) 和 [多次更新，多次查询](https://leetcode.com/problems/range-sum-query-2d-mutable/)。
+
+
+
+### 2. 单点更新    
+- [HDU 1166 敌兵布阵](http://acm.hdu.edu.cn/showproblem.php?pid=1166) update:单点增减 query:区间求和  
+- [HDU 1754 I Hate It](http://acm.hdu.edu.cn/showproblem.php?pid=1754) update:单点替换 query:区间最值  
+- [HDU 1394 Minimum Inversion Number](http://acm.hdu.edu.cn/showproblem.php?pid=1394) update:单点增减 query:区间求和  
+- [HDU 2795 Billboard](http://acm.hdu.edu.cn/showproblem.php?pid=2795) query:区间求最大值的位子(直接把update的操作在query里做了)
+
+### 3. 区间更新  
+
+- [HDU 1698 Just a Hook](http://acm.hdu.edu.cn/showproblem.php?pid=1698) update:成段替换 (由于只query一次总区间,所以可以直接输出 1 结点的信息)  
+- [POJ 3468 A Simple Problem with Integers](http://poj.org/problem?id=3468) update:成段增减 query:区间求和  
+- [POJ 2528 Mayor’s posters](http://poj.org/problem?id=2528) 离散化 + update:成段替换 query:简单hash  
+- [POJ 3225 Help with Intervals](http://poj.org/problem?id=3225) update:成段替换,区间异或 query:简单hash
+
+### 4. 区间合并
+
+这类题目会询问区间中满足条件的连续最长区间,所以PushUp的时候需要对左右儿子的区间进行合并
+
+- [POJ 3667 Hotel](http://poj.org/problem?id=3667) update:区间替换 query:询问满足条件的最左端点
+
+### 5. 扫描线
+
+这类题目需要将一些操作排序,然后从左到右用一根扫描线扫过去最典型的就是矩形面积并,周长并等题
+
+- [HDU 1542 Atlantis](http://acm.hdu.edu.cn/showproblem.php?pid=1542) update:区间增减 query:直接取根节点的值  
+- [HDU 1828 Picture](http://acm.hdu.edu.cn/showproblem.php?pid=1828) update:区间增减 query:直接取根节点的值
+
+
+### 6. 计数问题
+
+在 LeetCode 中还有一类问题涉及到计数的。[315. Count of Smaller Numbers After Self](https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0315.Count-of-Smaller-Numbers-After-Self/)，[327. Count of Range Sum](https://books.halfrost.com/leetcode/ChapterFour/0300~0399/0327.Count-of-Range-Sum/)，[493. Reverse Pairs](https://books.halfrost.com/leetcode/ChapterFour/0400~0499/0493.Reverse-Pairs/) 这类问题可以用下面的套路解决。线段树的每个节点存的是区间计数。
+
+
+
+```go
+// SegmentCountTree define
+type SegmentCountTree struct {
+	data, tree  []int
+	left, right int
+	merge       func(i, j int) int
+}
+
+// Init define
+func (st *SegmentCountTree) Init(nums []int, oper func(i, j int) int) {
+	st.merge = oper
+
+	data, tree := make([]int, len(nums)), make([]int, 4*len(nums))
+	for i := 0; i < len(nums); i++ {
+		data[i] = nums[i]
+	}
+	st.data, st.tree = data, tree
+}
+
+// 在 treeIndex 的位置创建 [left....right] 区间的线段树
+func (st *SegmentCountTree) buildSegmentTree(treeIndex, left, right int) {
+	if left == right {
+		st.tree[treeIndex] = st.data[left]
+		return
+	}
+	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, st.leftChild(treeIndex), st.rightChild(treeIndex)
+	st.buildSegmentTree(leftTreeIndex, left, midTreeIndex)
+	st.buildSegmentTree(rightTreeIndex, midTreeIndex+1, right)
+	st.tree[treeIndex] = st.merge(st.tree[leftTreeIndex], st.tree[rightTreeIndex])
+}
+
+func (st *SegmentCountTree) leftChild(index int) int {
+	return 2*index + 1
+}
+
+func (st *SegmentCountTree) rightChild(index int) int {
+	return 2*index + 2
+}
+
+// 查询 [left....right] 区间内的值
+
+// Query define
+func (st *SegmentCountTree) Query(left, right int) int {
+	if len(st.data) > 0 {
+		return st.queryInTree(0, 0, len(st.data)-1, left, right)
+	}
+	return 0
+}
+
+// 在以 treeIndex 为根的线段树中 [left...right] 的范围里，搜索区间 [queryLeft...queryRight] 的值，值是计数值
+func (st *SegmentCountTree) queryInTree(treeIndex, left, right, queryLeft, queryRight int) int {
+	if queryRight < st.data[left] || queryLeft > st.data[right] {
+		return 0
+	}
+	if queryLeft <= st.data[left] && queryRight >= st.data[right] || left == right {
+		return st.tree[treeIndex]
+	}
+	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, st.leftChild(treeIndex), st.rightChild(treeIndex)
+	return st.queryInTree(rightTreeIndex, midTreeIndex+1, right, queryLeft, queryRight) +
+		st.queryInTree(leftTreeIndex, left, midTreeIndex, queryLeft, queryRight)
+}
+
+// 更新计数
+
+// UpdateCount define
+func (st *SegmentCountTree) UpdateCount(val int) {
+	if len(st.data) > 0 {
+		st.updateCountInTree(0, 0, len(st.data)-1, val)
+	}
+}
+
+// 以 treeIndex 为根，更新 [left...right] 区间内的计数
+func (st *SegmentCountTree) updateCountInTree(treeIndex, left, right, val int) {
+	if val >= st.data[left] && val <= st.data[right] {
+		st.tree[treeIndex]++
+		if left == right {
+			return
+		}
+		midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, st.leftChild(treeIndex), st.rightChild(treeIndex)
+		st.updateCountInTree(rightTreeIndex, midTreeIndex+1, right, val)
+		st.updateCountInTree(leftTreeIndex, left, midTreeIndex, val)
+	}
+}
+
+```
+
+
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterThree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterThree/UnionFind/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterThree/UnionFind.md b/website/content/ChapterThree/UnionFind.md
new file mode 100644
index 000000000..c1e2009b0
--- /dev/null
+++ b/website/content/ChapterThree/UnionFind.md
@@ -0,0 +1,152 @@
+---
+title: 3.2 UnionFind
+type: docs
+weight: 2
+---
+
+# 并查集 UnionFind
+
+```go
+package template
+
+// UnionFind defind
+// 路径压缩 + 秩优化
+type UnionFind struct {
+	parent, rank []int
+	count        int
+}
+
+// Init define
+func (uf *UnionFind) Init(n int) {
+	uf.count = n
+	uf.parent = make([]int, n)
+	uf.rank = make([]int, n)
+	for i := range uf.parent {
+		uf.parent[i] = i
+	}
+}
+
+// Find define
+func (uf *UnionFind) Find(p int) int {
+	root := p
+	for root != uf.parent[root] {
+		root = uf.parent[root]
+	}
+	// compress path
+	for p != uf.parent[p] {
+		tmp := uf.parent[p]
+		uf.parent[p] = root
+		p = tmp
+	}
+	return root
+}
+
+// Union define
+func (uf *UnionFind) Union(p, q int) {
+	proot := uf.Find(p)
+	qroot := uf.Find(q)
+	if proot == qroot {
+		return
+	}
+	if uf.rank[qroot] > uf.rank[proot] {
+		uf.parent[proot] = qroot
+	} else {
+		uf.parent[qroot] = proot
+		if uf.rank[proot] == uf.rank[qroot] {
+			uf.rank[proot]++
+		}
+	}
+	uf.count--
+}
+
+// TotalCount define
+func (uf *UnionFind) TotalCount() int {
+	return uf.count
+}
+
+// UnionFindCount define
+// 计算每个集合中元素的个数 + 最大集合元素个数
+type UnionFindCount struct {
+	parent, count []int
+	maxUnionCount int
+}
+
+// Init define
+func (uf *UnionFindCount) Init(n int) {
+	uf.parent = make([]int, n)
+	uf.count = make([]int, n)
+	for i := range uf.parent {
+		uf.parent[i] = i
+		uf.count[i] = 1
+	}
+}
+
+// Find define
+func (uf *UnionFindCount) Find(p int) int {
+	root := p
+	for root != uf.parent[root] {
+		root = uf.parent[root]
+	}
+	return root
+}
+
+// 不进行秩压缩，时间复杂度爆炸，太高了
+// func (uf *UnionFindCount) union(p, q int) {
+// 	proot := uf.find(p)
+// 	qroot := uf.find(q)
+// 	if proot == qroot {
+// 		return
+// 	}
+// 	if proot != qroot {
+// 		uf.parent[proot] = qroot
+// 		uf.count[qroot] += uf.count[proot]
+// 	}
+// }
+
+// Union define
+func (uf *UnionFindCount) Union(p, q int) {
+	proot := uf.Find(p)
+	qroot := uf.Find(q)
+	if proot == qroot {
+		return
+	}
+	if proot == len(uf.parent)-1 {
+		//proot is root
+	} else if qroot == len(uf.parent)-1 {
+		// qroot is root, always attach to root
+		proot, qroot = qroot, proot
+	} else if uf.count[qroot] > uf.count[proot] {
+		proot, qroot = qroot, proot
+	}
+
+	//set relation[0] as parent
+	uf.maxUnionCount = max(uf.maxUnionCount, (uf.count[proot] + uf.count[qroot]))
+	uf.parent[qroot] = proot
+	uf.count[proot] += uf.count[qroot]
+}
+
+// Count define
+func (uf *UnionFindCount) Count() []int {
+	return uf.count
+}
+
+// MaxUnionCount define
+func (uf *UnionFindCount) MaxUnionCount() int {
+	return uf.maxUnionCount
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+```
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterThree/Segment_Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterThree/LRUCache/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterThree/_index.md b/website/content/ChapterThree/_index.md
new file mode 100644
index 000000000..b078b1d66
--- /dev/null
+++ b/website/content/ChapterThree/_index.md
@@ -0,0 +1,21 @@
+---
+title: 第三章 一些模板
+type: docs
+weight: 3
+---
+
+# 第三章 一些模板
+
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/GOPHER_DAD_.png'>
+</p>
+
+
+这一章会罗列一些整理好的模板。一起来看看吧。
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Binary_Indexed_Tree/">⬅️上一章</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterThree/Segment_Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Array.md b/website/content/ChapterTwo/Array.md
new file mode 100644
index 000000000..1ef86606b
--- /dev/null
+++ b/website/content/ChapterTwo/Array.md
@@ -0,0 +1,434 @@
+---
+title: 2.01 Array
+type: docs
+weight: 1
+---
+
+# Array
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0001|Two Sum|[Go]({{< relref "/ChapterFour/0001~0099/0001.Two-Sum.md" >}})|Easy| O(n)| O(n)||49.7%|
+|0004|Median of Two Sorted Arrays|[Go]({{< relref "/ChapterFour/0001~0099/0004.Median-of-Two-Sorted-Arrays.md" >}})|Hard||||36.2%|
+|0011|Container With Most Water|[Go]({{< relref "/ChapterFour/0001~0099/0011.Container-With-Most-Water.md" >}})|Medium| O(n)| O(1)||54.0%|
+|0015|3Sum|[Go]({{< relref "/ChapterFour/0001~0099/0015.3Sum.md" >}})|Medium| O(n^2)| O(n)|❤️|32.6%|
+|0016|3Sum Closest|[Go]({{< relref "/ChapterFour/0001~0099/0016.3Sum-Closest.md" >}})|Medium| O(n^2)| O(1)|❤️|45.7%|
+|0018|4Sum|[Go]({{< relref "/ChapterFour/0001~0099/0018.4Sum.md" >}})|Medium| O(n^3)| O(n^2)|❤️|35.9%|
+|0026|Remove Duplicates from Sorted Array|[Go]({{< relref "/ChapterFour/0001~0099/0026.Remove-Duplicates-from-Sorted-Array.md" >}})|Easy| O(n)| O(1)||51.6%|
+|0027|Remove Element|[Go]({{< relref "/ChapterFour/0001~0099/0027.Remove-Element.md" >}})|Easy| O(n)| O(1)||53.0%|
+|0031|Next Permutation|[Go]({{< relref "/ChapterFour/0001~0099/0031.Next-Permutation.md" >}})|Medium||||37.6%|
+|0033|Search in Rotated Sorted Array|[Go]({{< relref "/ChapterFour/0001~0099/0033.Search-in-Rotated-Sorted-Array.md" >}})|Medium||||39.0%|
+|0034|Find First and Last Position of Element in Sorted Array|[Go]({{< relref "/ChapterFour/0001~0099/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array.md" >}})|Medium||||41.9%|
+|0035|Search Insert Position|[Go]({{< relref "/ChapterFour/0001~0099/0035.Search-Insert-Position.md" >}})|Easy||||43.4%|
+|0036|Valid Sudoku|[Go]({{< relref "/ChapterFour/0001~0099/0036.Valid-Sudoku.md" >}})|Medium||||58.1%|
+|0037|Sudoku Solver|[Go]({{< relref "/ChapterFour/0001~0099/0037.Sudoku-Solver.md" >}})|Hard||||57.7%|
+|0039|Combination Sum|[Go]({{< relref "/ChapterFour/0001~0099/0039.Combination-Sum.md" >}})|Medium| O(n log n)| O(n)||68.6%|
+|0040|Combination Sum II|[Go]({{< relref "/ChapterFour/0001~0099/0040.Combination-Sum-II.md" >}})|Medium| O(n log n)| O(n)||53.4%|
+|0041|First Missing Positive|[Go]({{< relref "/ChapterFour/0001~0099/0041.First-Missing-Positive.md" >}})|Hard| O(n)| O(n)||36.7%|
+|0042|Trapping Rain Water|[Go]({{< relref "/ChapterFour/0001~0099/0042.Trapping-Rain-Water.md" >}})|Hard| O(n)| O(1)|❤️|59.3%|
+|0045|Jump Game II|[Go]({{< relref "/ChapterFour/0001~0099/0045.Jump-Game-II.md" >}})|Medium||||39.8%|
+|0046|Permutations|[Go]({{< relref "/ChapterFour/0001~0099/0046.Permutations.md" >}})|Medium||||75.7%|
+|0047|Permutations II|[Go]({{< relref "/ChapterFour/0001~0099/0047.Permutations-II.md" >}})|Medium||||57.4%|
+|0048|Rotate Image|[Go]({{< relref "/ChapterFour/0001~0099/0048.Rotate-Image.md" >}})|Medium| O(n)| O(1)||71.0%|
+|0049|Group Anagrams|[Go]({{< relref "/ChapterFour/0001~0099/0049.Group-Anagrams.md" >}})|Medium||||66.8%|
+|0051|N-Queens|[Go]({{< relref "/ChapterFour/0001~0099/0051.N-Queens.md" >}})|Hard||||64.2%|
+|0053|Maximum Subarray|[Go]({{< relref "/ChapterFour/0001~0099/0053.Maximum-Subarray.md" >}})|Medium| O(n)| O(n)||50.2%|
+|0054|Spiral Matrix|[Go]({{< relref "/ChapterFour/0001~0099/0054.Spiral-Matrix.md" >}})|Medium| O(n)| O(n^2)||45.0%|
+|0055|Jump Game|[Go]({{< relref "/ChapterFour/0001~0099/0055.Jump-Game.md" >}})|Medium||||38.9%|
+|0056|Merge Intervals|[Go]({{< relref "/ChapterFour/0001~0099/0056.Merge-Intervals.md" >}})|Medium| O(n log n)| O(1)||46.2%|
+|0057|Insert Interval|[Go]({{< relref "/ChapterFour/0001~0099/0057.Insert-Interval.md" >}})|Medium| O(n)| O(1)||39.0%|
+|0059|Spiral Matrix II|[Go]({{< relref "/ChapterFour/0001~0099/0059.Spiral-Matrix-II.md" >}})|Medium| O(n)| O(n^2)||67.4%|
+|0063|Unique Paths II|[Go]({{< relref "/ChapterFour/0001~0099/0063.Unique-Paths-II.md" >}})|Medium| O(n^2)| O(n^2)||39.4%|
+|0064|Minimum Path Sum|[Go]({{< relref "/ChapterFour/0001~0099/0064.Minimum-Path-Sum.md" >}})|Medium| O(n^2)| O(n^2)||62.0%|
+|0066|Plus One|[Go]({{< relref "/ChapterFour/0001~0099/0066.Plus-One.md" >}})|Easy||||43.7%|
+|0073|Set Matrix Zeroes|[Go]({{< relref "/ChapterFour/0001~0099/0073.Set-Matrix-Zeroes.md" >}})|Medium||||51.3%|
+|0074|Search a 2D Matrix|[Go]({{< relref "/ChapterFour/0001~0099/0074.Search-a-2D-Matrix.md" >}})|Medium||||47.7%|
+|0075|Sort Colors|[Go]({{< relref "/ChapterFour/0001~0099/0075.Sort-Colors.md" >}})|Medium| O(n)| O(1)|❤️|58.6%|
+|0078|Subsets|[Go]({{< relref "/ChapterFour/0001~0099/0078.Subsets.md" >}})|Medium| O(n^2)| O(n)|❤️|74.9%|
+|0079|Word Search|[Go]({{< relref "/ChapterFour/0001~0099/0079.Word-Search.md" >}})|Medium| O(n^2)| O(n^2)|❤️|40.2%|
+|0080|Remove Duplicates from Sorted Array II|[Go]({{< relref "/ChapterFour/0001~0099/0080.Remove-Duplicates-from-Sorted-Array-II.md" >}})|Medium| O(n)| O(1||52.3%|
+|0081|Search in Rotated Sorted Array II|[Go]({{< relref "/ChapterFour/0001~0099/0081.Search-in-Rotated-Sorted-Array-II.md" >}})|Medium||||35.7%|
+|0084|Largest Rectangle in Histogram|[Go]({{< relref "/ChapterFour/0001~0099/0084.Largest-Rectangle-in-Histogram.md" >}})|Hard| O(n)| O(n)|❤️|42.6%|
+|0088|Merge Sorted Array|[Go]({{< relref "/ChapterFour/0001~0099/0088.Merge-Sorted-Array.md" >}})|Easy| O(n)| O(1)|❤️|46.6%|
+|0090|Subsets II|[Go]({{< relref "/ChapterFour/0001~0099/0090.Subsets-II.md" >}})|Medium| O(n^2)| O(n)|❤️|55.9%|
+|0105|Construct Binary Tree from Preorder and Inorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal.md" >}})|Medium||||61.6%|
+|0106|Construct Binary Tree from Inorder and Postorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.md" >}})|Medium||||60.0%|
+|0108|Convert Sorted Array to Binary Search Tree|[Go]({{< relref "/ChapterFour/0100~0199/0108.Convert-Sorted-Array-to-Binary-Search-Tree.md" >}})|Easy||||69.9%|
+|0118|Pascal's Triangle|[Go]({{< relref "/ChapterFour/0100~0199/0118.Pascals-Triangle.md" >}})|Easy||||70.8%|
+|0119|Pascal's Triangle II|[Go]({{< relref "/ChapterFour/0100~0199/0119.Pascals-Triangle-II.md" >}})|Easy||||60.8%|
+|0120|Triangle|[Go]({{< relref "/ChapterFour/0100~0199/0120.Triangle.md" >}})|Medium| O(n^2)| O(n)||54.5%|
+|0121|Best Time to Buy and Sell Stock|[Go]({{< relref "/ChapterFour/0100~0199/0121.Best-Time-to-Buy-and-Sell-Stock.md" >}})|Easy| O(n)| O(1)||54.3%|
+|0122|Best Time to Buy and Sell Stock II|[Go]({{< relref "/ChapterFour/0100~0199/0122.Best-Time-to-Buy-and-Sell-Stock-II.md" >}})|Medium| O(n)| O(1)||63.9%|
+|0128|Longest Consecutive Sequence|[Go]({{< relref "/ChapterFour/0100~0199/0128.Longest-Consecutive-Sequence.md" >}})|Medium||||48.5%|
+|0130|Surrounded Regions|[Go]({{< relref "/ChapterFour/0100~0199/0130.Surrounded-Regions.md" >}})|Medium||||36.8%|
+|0135|Candy|[Go]({{< relref "/ChapterFour/0100~0199/0135.Candy.md" >}})|Hard||||41.0%|
+|0136|Single Number|[Go]({{< relref "/ChapterFour/0100~0199/0136.Single-Number.md" >}})|Easy||||70.7%|
+|0137|Single Number II|[Go]({{< relref "/ChapterFour/0100~0199/0137.Single-Number-II.md" >}})|Medium||||58.5%|
+|0150|Evaluate Reverse Polish Notation|[Go]({{< relref "/ChapterFour/0100~0199/0150.Evaluate-Reverse-Polish-Notation.md" >}})|Medium||||45.8%|
+|0152|Maximum Product Subarray|[Go]({{< relref "/ChapterFour/0100~0199/0152.Maximum-Product-Subarray.md" >}})|Medium| O(n)| O(1)||34.9%|
+|0153|Find Minimum in Rotated Sorted Array|[Go]({{< relref "/ChapterFour/0100~0199/0153.Find-Minimum-in-Rotated-Sorted-Array.md" >}})|Medium||||48.9%|
+|0154|Find Minimum in Rotated Sorted Array II|[Go]({{< relref "/ChapterFour/0100~0199/0154.Find-Minimum-in-Rotated-Sorted-Array-II.md" >}})|Hard||||43.5%|
+|0162|Find Peak Element|[Go]({{< relref "/ChapterFour/0100~0199/0162.Find-Peak-Element.md" >}})|Medium||||46.0%|
+|0164|Maximum Gap|[Go]({{< relref "/ChapterFour/0100~0199/0164.Maximum-Gap.md" >}})|Hard||||43.4%|
+|0167|Two Sum II - Input Array Is Sorted|[Go]({{< relref "/ChapterFour/0100~0199/0167.Two-Sum-II-Input-Array-Is-Sorted.md" >}})|Medium| O(n)| O(1)||60.0%|
+|0169|Majority Element|[Go]({{< relref "/ChapterFour/0100~0199/0169.Majority-Element.md" >}})|Easy||||63.9%|
+|0174|Dungeon Game|[Go]({{< relref "/ChapterFour/0100~0199/0174.Dungeon-Game.md" >}})|Hard||||37.5%|
+|0179|Largest Number|[Go]({{< relref "/ChapterFour/0100~0199/0179.Largest-Number.md" >}})|Medium||||34.6%|
+|0189|Rotate Array|[Go]({{< relref "/ChapterFour/0100~0199/0189.Rotate-Array.md" >}})|Medium||||39.4%|
+|0198|House Robber|[Go]({{< relref "/ChapterFour/0100~0199/0198.House-Robber.md" >}})|Medium||||49.4%|
+|0200|Number of Islands|[Go]({{< relref "/ChapterFour/0200~0299/0200.Number-of-Islands.md" >}})|Medium||||57.0%|
+|0204|Count Primes|[Go]({{< relref "/ChapterFour/0200~0299/0204.Count-Primes.md" >}})|Medium||||33.1%|
+|0209|Minimum Size Subarray Sum|[Go]({{< relref "/ChapterFour/0200~0299/0209.Minimum-Size-Subarray-Sum.md" >}})|Medium| O(n)| O(1)||45.0%|
+|0212|Word Search II|[Go]({{< relref "/ChapterFour/0200~0299/0212.Word-Search-II.md" >}})|Hard||||36.4%|
+|0213|House Robber II|[Go]({{< relref "/ChapterFour/0200~0299/0213.House-Robber-II.md" >}})|Medium||||41.0%|
+|0215|Kth Largest Element in an Array|[Go]({{< relref "/ChapterFour/0200~0299/0215.Kth-Largest-Element-in-an-Array.md" >}})|Medium||||66.2%|
+|0216|Combination Sum III|[Go]({{< relref "/ChapterFour/0200~0299/0216.Combination-Sum-III.md" >}})|Medium| O(n)| O(1)|❤️|67.6%|
+|0217|Contains Duplicate|[Go]({{< relref "/ChapterFour/0200~0299/0217.Contains-Duplicate.md" >}})|Easy| O(n)| O(n)||61.4%|
+|0218|The Skyline Problem|[Go]({{< relref "/ChapterFour/0200~0299/0218.The-Skyline-Problem.md" >}})|Hard||||41.9%|
+|0219|Contains Duplicate II|[Go]({{< relref "/ChapterFour/0200~0299/0219.Contains-Duplicate-II.md" >}})|Easy| O(n)| O(n)||42.6%|
+|0220|Contains Duplicate III|[Go]({{< relref "/ChapterFour/0200~0299/0220.Contains-Duplicate-III.md" >}})|Hard||||22.1%|
+|0228|Summary Ranges|[Go]({{< relref "/ChapterFour/0200~0299/0228.Summary-Ranges.md" >}})|Easy||||47.2%|
+|0229|Majority Element II|[Go]({{< relref "/ChapterFour/0200~0299/0229.Majority-Element-II.md" >}})|Medium||||45.1%|
+|0239|Sliding Window Maximum|[Go]({{< relref "/ChapterFour/0200~0299/0239.Sliding-Window-Maximum.md" >}})|Hard||||46.3%|
+|0240|Search a 2D Matrix II|[Go]({{< relref "/ChapterFour/0200~0299/0240.Search-a-2D-Matrix-II.md" >}})|Medium||||51.0%|
+|0260|Single Number III|[Go]({{< relref "/ChapterFour/0200~0299/0260.Single-Number-III.md" >}})|Medium||||67.7%|
+|0268|Missing Number|[Go]({{< relref "/ChapterFour/0200~0299/0268.Missing-Number.md" >}})|Easy||||62.6%|
+|0274|H-Index|[Go]({{< relref "/ChapterFour/0200~0299/0274.H-Index.md" >}})|Medium||||38.3%|
+|0275|H-Index II|[Go]({{< relref "/ChapterFour/0200~0299/0275.H-Index-II.md" >}})|Medium||||37.5%|
+|0283|Move Zeroes|[Go]({{< relref "/ChapterFour/0200~0299/0283.Move-Zeroes.md" >}})|Easy| O(n)| O(1)||61.4%|
+|0284|Peeking Iterator|[Go]({{< relref "/ChapterFour/0200~0299/0284.Peeking-Iterator.md" >}})|Medium||||58.7%|
+|0287|Find the Duplicate Number|[Go]({{< relref "/ChapterFour/0200~0299/0287.Find-the-Duplicate-Number.md" >}})|Medium| O(n)| O(1)|❤️|59.1%|
+|0300|Longest Increasing Subsequence|[Go]({{< relref "/ChapterFour/0300~0399/0300.Longest-Increasing-Subsequence.md" >}})|Medium||||52.2%|
+|0303|Range Sum Query - Immutable|[Go]({{< relref "/ChapterFour/0300~0399/0303.Range-Sum-Query-Immutable.md" >}})|Easy||||59.5%|
+|0304|Range Sum Query 2D - Immutable|[Go]({{< relref "/ChapterFour/0300~0399/0304.Range-Sum-Query-2D-Immutable.md" >}})|Medium||||52.9%|
+|0307|Range Sum Query - Mutable|[Go]({{< relref "/ChapterFour/0300~0399/0307.Range-Sum-Query-Mutable.md" >}})|Medium||||40.7%|
+|0309|Best Time to Buy and Sell Stock with Cooldown|[Go]({{< relref "/ChapterFour/0300~0399/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown.md" >}})|Medium||||56.2%|
+|0315|Count of Smaller Numbers After Self|[Go]({{< relref "/ChapterFour/0300~0399/0315.Count-of-Smaller-Numbers-After-Self.md" >}})|Hard||||42.6%|
+|0318|Maximum Product of Word Lengths|[Go]({{< relref "/ChapterFour/0300~0399/0318.Maximum-Product-of-Word-Lengths.md" >}})|Medium||||59.9%|
+|0322|Coin Change|[Go]({{< relref "/ChapterFour/0300~0399/0322.Coin-Change.md" >}})|Medium||||42.1%|
+|0324|Wiggle Sort II|[Go]({{< relref "/ChapterFour/0300~0399/0324.Wiggle-Sort-II.md" >}})|Medium||||33.3%|
+|0327|Count of Range Sum|[Go]({{< relref "/ChapterFour/0300~0399/0327.Count-of-Range-Sum.md" >}})|Hard||||35.8%|
+|0329|Longest Increasing Path in a Matrix|[Go]({{< relref "/ChapterFour/0300~0399/0329.Longest-Increasing-Path-in-a-Matrix.md" >}})|Hard||||52.4%|
+|0347|Top K Frequent Elements|[Go]({{< relref "/ChapterFour/0300~0399/0347.Top-K-Frequent-Elements.md" >}})|Medium||||64.2%|
+|0349|Intersection of Two Arrays|[Go]({{< relref "/ChapterFour/0300~0399/0349.Intersection-of-Two-Arrays.md" >}})|Easy||||70.9%|
+|0350|Intersection of Two Arrays II|[Go]({{< relref "/ChapterFour/0300~0399/0350.Intersection-of-Two-Arrays-II.md" >}})|Easy||||56.0%|
+|0354|Russian Doll Envelopes|[Go]({{< relref "/ChapterFour/0300~0399/0354.Russian-Doll-Envelopes.md" >}})|Hard||||37.9%|
+|0368|Largest Divisible Subset|[Go]({{< relref "/ChapterFour/0300~0399/0368.Largest-Divisible-Subset.md" >}})|Medium||||41.6%|
+|0373|Find K Pairs with Smallest Sums|[Go]({{< relref "/ChapterFour/0300~0399/0373.Find-K-Pairs-with-Smallest-Sums.md" >}})|Medium||||38.3%|
+|0376|Wiggle Subsequence|[Go]({{< relref "/ChapterFour/0300~0399/0376.Wiggle-Subsequence.md" >}})|Medium||||48.3%|
+|0377|Combination Sum IV|[Go]({{< relref "/ChapterFour/0300~0399/0377.Combination-Sum-IV.md" >}})|Medium||||52.2%|
+|0378|Kth Smallest Element in a Sorted Matrix|[Go]({{< relref "/ChapterFour/0300~0399/0378.Kth-Smallest-Element-in-a-Sorted-Matrix.md" >}})|Medium||||61.8%|
+|0384|Shuffle an Array|[Go]({{< relref "/ChapterFour/0300~0399/0384.Shuffle-an-Array.md" >}})|Medium||||57.8%|
+|0391|Perfect Rectangle|[Go]({{< relref "/ChapterFour/0300~0399/0391.Perfect-Rectangle.md" >}})|Hard||||32.8%|
+|0393|UTF-8 Validation|[Go]({{< relref "/ChapterFour/0300~0399/0393.UTF-8-Validation.md" >}})|Medium||||45.1%|
+|0396|Rotate Function|[Go]({{< relref "/ChapterFour/0300~0399/0396.Rotate-Function.md" >}})|Medium||||41.1%|
+|0399|Evaluate Division|[Go]({{< relref "/ChapterFour/0300~0399/0399.Evaluate-Division.md" >}})|Medium||||59.7%|
+|0410|Split Array Largest Sum|[Go]({{< relref "/ChapterFour/0400~0499/0410.Split-Array-Largest-Sum.md" >}})|Hard||||53.5%|
+|0413|Arithmetic Slices|[Go]({{< relref "/ChapterFour/0400~0499/0413.Arithmetic-Slices.md" >}})|Medium||||65.1%|
+|0414|Third Maximum Number|[Go]({{< relref "/ChapterFour/0400~0499/0414.Third-Maximum-Number.md" >}})|Easy||||33.2%|
+|0416|Partition Equal Subset Sum|[Go]({{< relref "/ChapterFour/0400~0499/0416.Partition-Equal-Subset-Sum.md" >}})|Medium||||46.3%|
+|0417|Pacific Atlantic Water Flow|[Go]({{< relref "/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow.md" >}})|Medium||||54.4%|
+|0419|Battleships in a Board|[Go]({{< relref "/ChapterFour/0400~0499/0419.Battleships-in-a-Board.md" >}})|Medium||||74.8%|
+|0421|Maximum XOR of Two Numbers in an Array|[Go]({{< relref "/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array.md" >}})|Medium||||54.0%|
+|0435|Non-overlapping Intervals|[Go]({{< relref "/ChapterFour/0400~0499/0435.Non-overlapping-Intervals.md" >}})|Medium||||50.3%|
+|0436|Find Right Interval|[Go]({{< relref "/ChapterFour/0400~0499/0436.Find-Right-Interval.md" >}})|Medium||||50.8%|
+|0447|Number of Boomerangs|[Go]({{< relref "/ChapterFour/0400~0499/0447.Number-of-Boomerangs.md" >}})|Medium||||54.9%|
+|0448|Find All Numbers Disappeared in an Array|[Go]({{< relref "/ChapterFour/0400~0499/0448.Find-All-Numbers-Disappeared-in-an-Array.md" >}})|Easy||||59.9%|
+|0453|Minimum Moves to Equal Array Elements|[Go]({{< relref "/ChapterFour/0400~0499/0453.Minimum-Moves-to-Equal-Array-Elements.md" >}})|Medium||||56.0%|
+|0454|4Sum II|[Go]({{< relref "/ChapterFour/0400~0499/0454.4Sum-II.md" >}})|Medium||||57.2%|
+|0455|Assign Cookies|[Go]({{< relref "/ChapterFour/0400~0499/0455.Assign-Cookies.md" >}})|Easy||||49.9%|
+|0456|132 Pattern|[Go]({{< relref "/ChapterFour/0400~0499/0456.132-Pattern.md" >}})|Medium||||32.4%|
+|0457|Circular Array Loop|[Go]({{< relref "/ChapterFour/0400~0499/0457.Circular-Array-Loop.md" >}})|Medium||||32.6%|
+|0462|Minimum Moves to Equal Array Elements II|[Go]({{< relref "/ChapterFour/0400~0499/0462.Minimum-Moves-to-Equal-Array-Elements-II.md" >}})|Medium||||60.0%|
+|0463|Island Perimeter|[Go]({{< relref "/ChapterFour/0400~0499/0463.Island-Perimeter.md" >}})|Easy||||69.7%|
+|0473|Matchsticks to Square|[Go]({{< relref "/ChapterFour/0400~0499/0473.Matchsticks-to-Square.md" >}})|Medium||||40.2%|
+|0474|Ones and Zeroes|[Go]({{< relref "/ChapterFour/0400~0499/0474.Ones-and-Zeroes.md" >}})|Medium||||46.8%|
+|0475|Heaters|[Go]({{< relref "/ChapterFour/0400~0499/0475.Heaters.md" >}})|Medium||||36.5%|
+|0477|Total Hamming Distance|[Go]({{< relref "/ChapterFour/0400~0499/0477.Total-Hamming-Distance.md" >}})|Medium||||52.2%|
+|0480|Sliding Window Median|[Go]({{< relref "/ChapterFour/0400~0499/0480.Sliding-Window-Median.md" >}})|Hard||||41.1%|
+|0485|Max Consecutive Ones|[Go]({{< relref "/ChapterFour/0400~0499/0485.Max-Consecutive-Ones.md" >}})|Easy||||56.6%|
+|0491|Non-decreasing Subsequences|[Go]({{< relref "/ChapterFour/0400~0499/0491.Non-decreasing-Subsequences.md" >}})|Medium||||60.2%|
+|0493|Reverse Pairs|[Go]({{< relref "/ChapterFour/0400~0499/0493.Reverse-Pairs.md" >}})|Hard||||30.9%|
+|0494|Target Sum|[Go]({{< relref "/ChapterFour/0400~0499/0494.Target-Sum.md" >}})|Medium||||45.7%|
+|0495|Teemo Attacking|[Go]({{< relref "/ChapterFour/0400~0499/0495.Teemo-Attacking.md" >}})|Easy||||56.8%|
+|0496|Next Greater Element I|[Go]({{< relref "/ChapterFour/0400~0499/0496.Next-Greater-Element-I.md" >}})|Easy||||71.4%|
+|0497|Random Point in Non-overlapping Rectangles|[Go]({{< relref "/ChapterFour/0400~0499/0497.Random-Point-in-Non-overlapping-Rectangles.md" >}})|Medium||||39.4%|
+|0498|Diagonal Traverse|[Go]({{< relref "/ChapterFour/0400~0499/0498.Diagonal-Traverse.md" >}})|Medium||||58.3%|
+|0500|Keyboard Row|[Go]({{< relref "/ChapterFour/0500~0599/0500.Keyboard-Row.md" >}})|Easy||||69.6%|
+|0503|Next Greater Element II|[Go]({{< relref "/ChapterFour/0500~0599/0503.Next-Greater-Element-II.md" >}})|Medium||||63.2%|
+|0506|Relative Ranks|[Go]({{< relref "/ChapterFour/0500~0599/0506.Relative-Ranks.md" >}})|Easy||||60.6%|
+|0518|Coin Change II|[Go]({{< relref "/ChapterFour/0500~0599/0518.Coin-Change-II.md" >}})|Medium||||60.6%|
+|0523|Continuous Subarray Sum|[Go]({{< relref "/ChapterFour/0500~0599/0523.Continuous-Subarray-Sum.md" >}})|Medium||||28.5%|
+|0524|Longest Word in Dictionary through Deleting|[Go]({{< relref "/ChapterFour/0500~0599/0524.Longest-Word-in-Dictionary-through-Deleting.md" >}})|Medium||||51.0%|
+|0525|Contiguous Array|[Go]({{< relref "/ChapterFour/0500~0599/0525.Contiguous-Array.md" >}})|Medium||||46.8%|
+|0526|Beautiful Arrangement|[Go]({{< relref "/ChapterFour/0500~0599/0526.Beautiful-Arrangement.md" >}})|Medium||||64.4%|
+|0528|Random Pick with Weight|[Go]({{< relref "/ChapterFour/0500~0599/0528.Random-Pick-with-Weight.md" >}})|Medium||||46.1%|
+|0529|Minesweeper|[Go]({{< relref "/ChapterFour/0500~0599/0529.Minesweeper.md" >}})|Medium||||65.7%|
+|0532|K-diff Pairs in an Array|[Go]({{< relref "/ChapterFour/0500~0599/0532.K-diff-Pairs-in-an-Array.md" >}})|Medium| O(n)| O(n)||41.2%|
+|0540|Single Element in a Sorted Array|[Go]({{< relref "/ChapterFour/0500~0599/0540.Single-Element-in-a-Sorted-Array.md" >}})|Medium||||59.1%|
+|0542|01 Matrix|[Go]({{< relref "/ChapterFour/0500~0599/0542.01-Matrix.md" >}})|Medium||||44.8%|
+|0554|Brick Wall|[Go]({{< relref "/ChapterFour/0500~0599/0554.Brick-Wall.md" >}})|Medium||||53.6%|
+|0560|Subarray Sum Equals K|[Go]({{< relref "/ChapterFour/0500~0599/0560.Subarray-Sum-Equals-K.md" >}})|Medium||||43.7%|
+|0561|Array Partition|[Go]({{< relref "/ChapterFour/0500~0599/0561.Array-Partition.md" >}})|Easy||||77.2%|
+|0566|Reshape the Matrix|[Go]({{< relref "/ChapterFour/0500~0599/0566.Reshape-the-Matrix.md" >}})|Easy| O(n^2)| O(n^2)||62.9%|
+|0575|Distribute Candies|[Go]({{< relref "/ChapterFour/0500~0599/0575.Distribute-Candies.md" >}})|Easy||||66.5%|
+|0581|Shortest Unsorted Continuous Subarray|[Go]({{< relref "/ChapterFour/0500~0599/0581.Shortest-Unsorted-Continuous-Subarray.md" >}})|Medium||||36.4%|
+|0594|Longest Harmonious Subsequence|[Go]({{< relref "/ChapterFour/0500~0599/0594.Longest-Harmonious-Subsequence.md" >}})|Easy||||53.5%|
+|0598|Range Addition II|[Go]({{< relref "/ChapterFour/0500~0599/0598.Range-Addition-II.md" >}})|Easy||||55.3%|
+|0599|Minimum Index Sum of Two Lists|[Go]({{< relref "/ChapterFour/0500~0599/0599.Minimum-Index-Sum-of-Two-Lists.md" >}})|Easy||||53.4%|
+|0605|Can Place Flowers|[Go]({{< relref "/ChapterFour/0600~0699/0605.Can-Place-Flowers.md" >}})|Easy||||32.8%|
+|0609|Find Duplicate File in System|[Go]({{< relref "/ChapterFour/0600~0699/0609.Find-Duplicate-File-in-System.md" >}})|Medium||||67.7%|
+|0611|Valid Triangle Number|[Go]({{< relref "/ChapterFour/0600~0699/0611.Valid-Triangle-Number.md" >}})|Medium||||50.6%|
+|0622|Design Circular Queue|[Go]({{< relref "/ChapterFour/0600~0699/0622.Design-Circular-Queue.md" >}})|Medium||||51.5%|
+|0628|Maximum Product of Three Numbers|[Go]({{< relref "/ChapterFour/0600~0699/0628.Maximum-Product-of-Three-Numbers.md" >}})|Easy| O(n)| O(1)||45.9%|
+|0630|Course Schedule III|[Go]({{< relref "/ChapterFour/0600~0699/0630.Course-Schedule-III.md" >}})|Hard||||40.1%|
+|0632|Smallest Range Covering Elements from K Lists|[Go]({{< relref "/ChapterFour/0600~0699/0632.Smallest-Range-Covering-Elements-from-K-Lists.md" >}})|Hard||||61.0%|
+|0636|Exclusive Time of Functions|[Go]({{< relref "/ChapterFour/0600~0699/0636.Exclusive-Time-of-Functions.md" >}})|Medium||||61.2%|
+|0638|Shopping Offers|[Go]({{< relref "/ChapterFour/0600~0699/0638.Shopping-Offers.md" >}})|Medium||||53.3%|
+|0643|Maximum Average Subarray I|[Go]({{< relref "/ChapterFour/0600~0699/0643.Maximum-Average-Subarray-I.md" >}})|Easy||||43.7%|
+|0645|Set Mismatch|[Go]({{< relref "/ChapterFour/0600~0699/0645.Set-Mismatch.md" >}})|Easy||||42.7%|
+|0648|Replace Words|[Go]({{< relref "/ChapterFour/0600~0699/0648.Replace-Words.md" >}})|Medium||||62.7%|
+|0658|Find K Closest Elements|[Go]({{< relref "/ChapterFour/0600~0699/0658.Find-K-Closest-Elements.md" >}})|Medium||||46.8%|
+|0661|Image Smoother|[Go]({{< relref "/ChapterFour/0600~0699/0661.Image-Smoother.md" >}})|Easy||||55.4%|
+|0665|Non-decreasing Array|[Go]({{< relref "/ChapterFour/0600~0699/0665.Non-decreasing-Array.md" >}})|Medium||||24.3%|
+|0667|Beautiful Arrangement II|[Go]({{< relref "/ChapterFour/0600~0699/0667.Beautiful-Arrangement-II.md" >}})|Medium||||59.8%|
+|0674|Longest Continuous Increasing Subsequence|[Go]({{< relref "/ChapterFour/0600~0699/0674.Longest-Continuous-Increasing-Subsequence.md" >}})|Easy||||49.3%|
+|0682|Baseball Game|[Go]({{< relref "/ChapterFour/0600~0699/0682.Baseball-Game.md" >}})|Easy||||74.3%|
+|0695|Max Area of Island|[Go]({{< relref "/ChapterFour/0600~0699/0695.Max-Area-of-Island.md" >}})|Medium||||71.8%|
+|0697|Degree of an Array|[Go]({{< relref "/ChapterFour/0600~0699/0697.Degree-of-an-Array.md" >}})|Easy||||56.0%|
+|0699|Falling Squares|[Go]({{< relref "/ChapterFour/0600~0699/0699.Falling-Squares.md" >}})|Hard||||44.7%|
+|0704|Binary Search|[Go]({{< relref "/ChapterFour/0700~0799/0704.Binary-Search.md" >}})|Easy||||56.1%|
+|0705|Design HashSet|[Go]({{< relref "/ChapterFour/0700~0799/0705.Design-HashSet.md" >}})|Easy||||65.6%|
+|0706|Design HashMap|[Go]({{< relref "/ChapterFour/0700~0799/0706.Design-HashMap.md" >}})|Easy||||64.7%|
+|0710|Random Pick with Blacklist|[Go]({{< relref "/ChapterFour/0700~0799/0710.Random-Pick-with-Blacklist.md" >}})|Hard||||33.5%|
+|0713|Subarray Product Less Than K|[Go]({{< relref "/ChapterFour/0700~0799/0713.Subarray-Product-Less-Than-K.md" >}})|Medium| O(n)| O(1)||45.8%|
+|0714|Best Time to Buy and Sell Stock with Transaction Fee|[Go]({{< relref "/ChapterFour/0700~0799/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee.md" >}})|Medium| O(n)| O(1)||65.2%|
+|0717|1-bit and 2-bit Characters|[Go]({{< relref "/ChapterFour/0700~0799/0717.1-bit-and-2-bit-Characters.md" >}})|Easy||||45.7%|
+|0718|Maximum Length of Repeated Subarray|[Go]({{< relref "/ChapterFour/0700~0799/0718.Maximum-Length-of-Repeated-Subarray.md" >}})|Medium||||51.3%|
+|0719|Find K-th Smallest Pair Distance|[Go]({{< relref "/ChapterFour/0700~0799/0719.Find-K-th-Smallest-Pair-Distance.md" >}})|Hard||||36.7%|
+|0720|Longest Word in Dictionary|[Go]({{< relref "/ChapterFour/0700~0799/0720.Longest-Word-in-Dictionary.md" >}})|Medium||||52.0%|
+|0721|Accounts Merge|[Go]({{< relref "/ChapterFour/0700~0799/0721.Accounts-Merge.md" >}})|Medium||||56.3%|
+|0724|Find Pivot Index|[Go]({{< relref "/ChapterFour/0700~0799/0724.Find-Pivot-Index.md" >}})|Easy||||54.7%|
+|0733|Flood Fill|[Go]({{< relref "/ChapterFour/0700~0799/0733.Flood-Fill.md" >}})|Easy||||62.0%|
+|0735|Asteroid Collision|[Go]({{< relref "/ChapterFour/0700~0799/0735.Asteroid-Collision.md" >}})|Medium||||44.4%|
+|0739|Daily Temperatures|[Go]({{< relref "/ChapterFour/0700~0799/0739.Daily-Temperatures.md" >}})|Medium||||66.3%|
+|0744|Find Smallest Letter Greater Than Target|[Go]({{< relref "/ChapterFour/0700~0799/0744.Find-Smallest-Letter-Greater-Than-Target.md" >}})|Easy||||45.8%|
+|0746|Min Cost Climbing Stairs|[Go]({{< relref "/ChapterFour/0700~0799/0746.Min-Cost-Climbing-Stairs.md" >}})|Easy| O(n)| O(1)||63.2%|
+|0747|Largest Number At Least Twice of Others|[Go]({{< relref "/ChapterFour/0700~0799/0747.Largest-Number-At-Least-Twice-of-Others.md" >}})|Easy||||47.1%|
+|0748|Shortest Completing Word|[Go]({{< relref "/ChapterFour/0700~0799/0748.Shortest-Completing-Word.md" >}})|Easy||||59.3%|
+|0752|Open the Lock|[Go]({{< relref "/ChapterFour/0700~0799/0752.Open-the-Lock.md" >}})|Medium||||55.6%|
+|0766|Toeplitz Matrix|[Go]({{< relref "/ChapterFour/0700~0799/0766.Toeplitz-Matrix.md" >}})|Easy| O(n)| O(1)||68.6%|
+|0775|Global and Local Inversions|[Go]({{< relref "/ChapterFour/0700~0799/0775.Global-and-Local-Inversions.md" >}})|Medium||||43.3%|
+|0778|Swim in Rising Water|[Go]({{< relref "/ChapterFour/0700~0799/0778.Swim-in-Rising-Water.md" >}})|Hard||||59.8%|
+|0781|Rabbits in Forest|[Go]({{< relref "/ChapterFour/0700~0799/0781.Rabbits-in-Forest.md" >}})|Medium||||54.7%|
+|0786|K-th Smallest Prime Fraction|[Go]({{< relref "/ChapterFour/0700~0799/0786.K-th-Smallest-Prime-Fraction.md" >}})|Medium||||51.7%|
+|0794|Valid Tic-Tac-Toe State|[Go]({{< relref "/ChapterFour/0700~0799/0794.Valid-Tic-Tac-Toe-State.md" >}})|Medium||||35.1%|
+|0795|Number of Subarrays with Bounded Maximum|[Go]({{< relref "/ChapterFour/0700~0799/0795.Number-of-Subarrays-with-Bounded-Maximum.md" >}})|Medium||||52.8%|
+|0803|Bricks Falling When Hit|[Go]({{< relref "/ChapterFour/0800~0899/0803.Bricks-Falling-When-Hit.md" >}})|Hard||||34.4%|
+|0807|Max Increase to Keep City Skyline|[Go]({{< relref "/ChapterFour/0800~0899/0807.Max-Increase-to-Keep-City-Skyline.md" >}})|Medium||||85.9%|
+|0810|Chalkboard XOR Game|[Go]({{< relref "/ChapterFour/0800~0899/0810.Chalkboard-XOR-Game.md" >}})|Hard||||55.8%|
+|0811|Subdomain Visit Count|[Go]({{< relref "/ChapterFour/0800~0899/0811.Subdomain-Visit-Count.md" >}})|Medium||||75.5%|
+|0812|Largest Triangle Area|[Go]({{< relref "/ChapterFour/0800~0899/0812.Largest-Triangle-Area.md" >}})|Easy||||59.9%|
+|0815|Bus Routes|[Go]({{< relref "/ChapterFour/0800~0899/0815.Bus-Routes.md" >}})|Hard||||45.6%|
+|0817|Linked List Components|[Go]({{< relref "/ChapterFour/0800~0899/0817.Linked-List-Components.md" >}})|Medium||||57.7%|
+|0820|Short Encoding of Words|[Go]({{< relref "/ChapterFour/0800~0899/0820.Short-Encoding-of-Words.md" >}})|Medium||||60.6%|
+|0821|Shortest Distance to a Character|[Go]({{< relref "/ChapterFour/0800~0899/0821.Shortest-Distance-to-a-Character.md" >}})|Easy||||71.3%|
+|0823|Binary Trees With Factors|[Go]({{< relref "/ChapterFour/0800~0899/0823.Binary-Trees-With-Factors.md" >}})|Medium||||49.7%|
+|0825|Friends Of Appropriate Ages|[Go]({{< relref "/ChapterFour/0800~0899/0825.Friends-Of-Appropriate-Ages.md" >}})|Medium||||46.3%|
+|0826|Most Profit Assigning Work|[Go]({{< relref "/ChapterFour/0800~0899/0826.Most-Profit-Assigning-Work.md" >}})|Medium||||44.9%|
+|0832|Flipping an Image|[Go]({{< relref "/ChapterFour/0800~0899/0832.Flipping-an-Image.md" >}})|Easy||||80.8%|
+|0839|Similar String Groups|[Go]({{< relref "/ChapterFour/0800~0899/0839.Similar-String-Groups.md" >}})|Hard||||48.0%|
+|0845|Longest Mountain in Array|[Go]({{< relref "/ChapterFour/0800~0899/0845.Longest-Mountain-in-Array.md" >}})|Medium||||40.2%|
+|0846|Hand of Straights|[Go]({{< relref "/ChapterFour/0800~0899/0846.Hand-of-Straights.md" >}})|Medium||||56.2%|
+|0850|Rectangle Area II|[Go]({{< relref "/ChapterFour/0800~0899/0850.Rectangle-Area-II.md" >}})|Hard||||53.9%|
+|0851|Loud and Rich|[Go]({{< relref "/ChapterFour/0800~0899/0851.Loud-and-Rich.md" >}})|Medium||||58.4%|
+|0852|Peak Index in a Mountain Array|[Go]({{< relref "/ChapterFour/0800~0899/0852.Peak-Index-in-a-Mountain-Array.md" >}})|Medium||||69.0%|
+|0853|Car Fleet|[Go]({{< relref "/ChapterFour/0800~0899/0853.Car-Fleet.md" >}})|Medium||||50.3%|
+|0862|Shortest Subarray with Sum at Least K|[Go]({{< relref "/ChapterFour/0800~0899/0862.Shortest-Subarray-with-Sum-at-Least-K.md" >}})|Hard||||26.0%|
+|0864|Shortest Path to Get All Keys|[Go]({{< relref "/ChapterFour/0800~0899/0864.Shortest-Path-to-Get-All-Keys.md" >}})|Hard||||45.6%|
+|0867|Transpose Matrix|[Go]({{< relref "/ChapterFour/0800~0899/0867.Transpose-Matrix.md" >}})|Easy| O(n)| O(1)||64.3%|
+|0870|Advantage Shuffle|[Go]({{< relref "/ChapterFour/0800~0899/0870.Advantage-Shuffle.md" >}})|Medium||||51.9%|
+|0874|Walking Robot Simulation|[Go]({{< relref "/ChapterFour/0800~0899/0874.Walking-Robot-Simulation.md" >}})|Medium||||39.0%|
+|0875|Koko Eating Bananas|[Go]({{< relref "/ChapterFour/0800~0899/0875.Koko-Eating-Bananas.md" >}})|Medium||||52.1%|
+|0877|Stone Game|[Go]({{< relref "/ChapterFour/0800~0899/0877.Stone-Game.md" >}})|Medium||||69.7%|
+|0881|Boats to Save People|[Go]({{< relref "/ChapterFour/0800~0899/0881.Boats-to-Save-People.md" >}})|Medium||||55.8%|
+|0885|Spiral Matrix III|[Go]({{< relref "/ChapterFour/0800~0899/0885.Spiral-Matrix-III.md" >}})|Medium||||73.5%|
+|0888|Fair Candy Swap|[Go]({{< relref "/ChapterFour/0800~0899/0888.Fair-Candy-Swap.md" >}})|Easy||||60.7%|
+|0890|Find and Replace Pattern|[Go]({{< relref "/ChapterFour/0800~0899/0890.Find-and-Replace-Pattern.md" >}})|Medium||||77.6%|
+|0891|Sum of Subsequence Widths|[Go]({{< relref "/ChapterFour/0800~0899/0891.Sum-of-Subsequence-Widths.md" >}})|Hard| O(n log n)| O(1)||36.7%|
+|0892|Surface Area of 3D Shapes|[Go]({{< relref "/ChapterFour/0800~0899/0892.Surface-Area-of-3D-Shapes.md" >}})|Easy||||64.0%|
+|0896|Monotonic Array|[Go]({{< relref "/ChapterFour/0800~0899/0896.Monotonic-Array.md" >}})|Easy||||58.4%|
+|0898|Bitwise ORs of Subarrays|[Go]({{< relref "/ChapterFour/0800~0899/0898.Bitwise-ORs-of-Subarrays.md" >}})|Medium||||37.2%|
+|0904|Fruit Into Baskets|[Go]({{< relref "/ChapterFour/0900~0999/0904.Fruit-Into-Baskets.md" >}})|Medium||||43.7%|
+|0907|Sum of Subarray Minimums|[Go]({{< relref "/ChapterFour/0900~0999/0907.Sum-of-Subarray-Minimums.md" >}})|Medium| O(n)| O(n)|❤️|35.8%|
+|0909|Snakes and Ladders|[Go]({{< relref "/ChapterFour/0900~0999/0909.Snakes-and-Ladders.md" >}})|Medium||||45.1%|
+|0910|Smallest Range II|[Go]({{< relref "/ChapterFour/0900~0999/0910.Smallest-Range-II.md" >}})|Medium||||35.2%|
+|0911|Online Election|[Go]({{< relref "/ChapterFour/0900~0999/0911.Online-Election.md" >}})|Medium||||52.2%|
+|0914|X of a Kind in a Deck of Cards|[Go]({{< relref "/ChapterFour/0900~0999/0914.X-of-a-Kind-in-a-Deck-of-Cards.md" >}})|Easy||||31.2%|
+|0916|Word Subsets|[Go]({{< relref "/ChapterFour/0900~0999/0916.Word-Subsets.md" >}})|Medium||||53.7%|
+|0918|Maximum Sum Circular Subarray|[Go]({{< relref "/ChapterFour/0900~0999/0918.Maximum-Sum-Circular-Subarray.md" >}})|Medium||||43.0%|
+|0922|Sort Array By Parity II|[Go]({{< relref "/ChapterFour/0900~0999/0922.Sort-Array-By-Parity-II.md" >}})|Easy| O(n)| O(1)||70.7%|
+|0923|3Sum With Multiplicity|[Go]({{< relref "/ChapterFour/0900~0999/0923.3Sum-With-Multiplicity.md" >}})|Medium||||45.3%|
+|0924|Minimize Malware Spread|[Go]({{< relref "/ChapterFour/0900~0999/0924.Minimize-Malware-Spread.md" >}})|Hard||||42.1%|
+|0927|Three Equal Parts|[Go]({{< relref "/ChapterFour/0900~0999/0927.Three-Equal-Parts.md" >}})|Hard||||39.6%|
+|0928|Minimize Malware Spread II|[Go]({{< relref "/ChapterFour/0900~0999/0928.Minimize-Malware-Spread-II.md" >}})|Hard||||42.8%|
+|0930|Binary Subarrays With Sum|[Go]({{< relref "/ChapterFour/0900~0999/0930.Binary-Subarrays-With-Sum.md" >}})|Medium||||52.2%|
+|0942|DI String Match|[Go]({{< relref "/ChapterFour/0900~0999/0942.DI-String-Match.md" >}})|Easy||||77.3%|
+|0946|Validate Stack Sequences|[Go]({{< relref "/ChapterFour/0900~0999/0946.Validate-Stack-Sequences.md" >}})|Medium||||67.7%|
+|0952|Largest Component Size by Common Factor|[Go]({{< relref "/ChapterFour/0900~0999/0952.Largest-Component-Size-by-Common-Factor.md" >}})|Hard||||40.0%|
+|0953|Verifying an Alien Dictionary|[Go]({{< relref "/ChapterFour/0900~0999/0953.Verifying-an-Alien-Dictionary.md" >}})|Easy||||54.5%|
+|0961|N-Repeated Element in Size 2N Array|[Go]({{< relref "/ChapterFour/0900~0999/0961.N-Repeated-Element-in-Size-2N-Array.md" >}})|Easy||||76.1%|
+|0966|Vowel Spellchecker|[Go]({{< relref "/ChapterFour/0900~0999/0966.Vowel-Spellchecker.md" >}})|Medium||||51.4%|
+|0969|Pancake Sorting|[Go]({{< relref "/ChapterFour/0900~0999/0969.Pancake-Sorting.md" >}})|Medium| O(n)| O(1)|❤️|70.1%|
+|0973|K Closest Points to Origin|[Go]({{< relref "/ChapterFour/0900~0999/0973.K-Closest-Points-to-Origin.md" >}})|Medium||||65.7%|
+|0976|Largest Perimeter Triangle|[Go]({{< relref "/ChapterFour/0900~0999/0976.Largest-Perimeter-Triangle.md" >}})|Easy||||54.7%|
+|0977|Squares of a Sorted Array|[Go]({{< relref "/ChapterFour/0900~0999/0977.Squares-of-a-Sorted-Array.md" >}})|Easy| O(n)| O(1)||71.9%|
+|0978|Longest Turbulent Subarray|[Go]({{< relref "/ChapterFour/0900~0999/0978.Longest-Turbulent-Subarray.md" >}})|Medium||||47.2%|
+|0980|Unique Paths III|[Go]({{< relref "/ChapterFour/0900~0999/0980.Unique-Paths-III.md" >}})|Hard||||81.7%|
+|0985|Sum of Even Numbers After Queries|[Go]({{< relref "/ChapterFour/0900~0999/0985.Sum-of-Even-Numbers-After-Queries.md" >}})|Medium||||68.1%|
+|0986|Interval List Intersections|[Go]({{< relref "/ChapterFour/0900~0999/0986.Interval-List-Intersections.md" >}})|Medium||||71.3%|
+|0989|Add to Array-Form of Integer|[Go]({{< relref "/ChapterFour/0900~0999/0989.Add-to-Array-Form-of-Integer.md" >}})|Easy||||47.1%|
+|0990|Satisfiability of Equality Equations|[Go]({{< relref "/ChapterFour/0900~0999/0990.Satisfiability-of-Equality-Equations.md" >}})|Medium||||50.5%|
+|0992|Subarrays with K Different Integers|[Go]({{< relref "/ChapterFour/0900~0999/0992.Subarrays-with-K-Different-Integers.md" >}})|Hard||||54.6%|
+|0995|Minimum Number of K Consecutive Bit Flips|[Go]({{< relref "/ChapterFour/0900~0999/0995.Minimum-Number-of-K-Consecutive-Bit-Flips.md" >}})|Hard||||51.2%|
+|0996|Number of Squareful Arrays|[Go]({{< relref "/ChapterFour/0900~0999/0996.Number-of-Squareful-Arrays.md" >}})|Hard||||49.2%|
+|0997|Find the Town Judge|[Go]({{< relref "/ChapterFour/0900~0999/0997.Find-the-Town-Judge.md" >}})|Easy||||49.5%|
+|0999|Available Captures for Rook|[Go]({{< relref "/ChapterFour/0900~0999/0999.Available-Captures-for-Rook.md" >}})|Easy||||68.2%|
+|1002|Find Common Characters|[Go]({{< relref "/ChapterFour/1000~1099/1002.Find-Common-Characters.md" >}})|Easy||||68.5%|
+|1004|Max Consecutive Ones III|[Go]({{< relref "/ChapterFour/1000~1099/1004.Max-Consecutive-Ones-III.md" >}})|Medium||||63.2%|
+|1005|Maximize Sum Of Array After K Negations|[Go]({{< relref "/ChapterFour/1000~1099/1005.Maximize-Sum-Of-Array-After-K-Negations.md" >}})|Easy||||50.8%|
+|1010|Pairs of Songs With Total Durations Divisible by 60|[Go]({{< relref "/ChapterFour/1000~1099/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60.md" >}})|Medium||||52.8%|
+|1011|Capacity To Ship Packages Within D Days|[Go]({{< relref "/ChapterFour/1000~1099/1011.Capacity-To-Ship-Packages-Within-D-Days.md" >}})|Medium||||67.7%|
+|1018|Binary Prefix Divisible By 5|[Go]({{< relref "/ChapterFour/1000~1099/1018.Binary-Prefix-Divisible-By-5.md" >}})|Easy||||46.9%|
+|1019|Next Greater Node In Linked List|[Go]({{< relref "/ChapterFour/1000~1099/1019.Next-Greater-Node-In-Linked-List.md" >}})|Medium||||59.9%|
+|1020|Number of Enclaves|[Go]({{< relref "/ChapterFour/1000~1099/1020.Number-of-Enclaves.md" >}})|Medium||||65.6%|
+|1030|Matrix Cells in Distance Order|[Go]({{< relref "/ChapterFour/1000~1099/1030.Matrix-Cells-in-Distance-Order.md" >}})|Easy||||69.7%|
+|1034|Coloring A Border|[Go]({{< relref "/ChapterFour/1000~1099/1034.Coloring-A-Border.md" >}})|Medium||||49.2%|
+|1037|Valid Boomerang|[Go]({{< relref "/ChapterFour/1000~1099/1037.Valid-Boomerang.md" >}})|Easy||||37.0%|
+|1040|Moving Stones Until Consecutive II|[Go]({{< relref "/ChapterFour/1000~1099/1040.Moving-Stones-Until-Consecutive-II.md" >}})|Medium||||55.9%|
+|1048|Longest String Chain|[Go]({{< relref "/ChapterFour/1000~1099/1048.Longest-String-Chain.md" >}})|Medium||||59.3%|
+|1049|Last Stone Weight II|[Go]({{< relref "/ChapterFour/1000~1099/1049.Last-Stone-Weight-II.md" >}})|Medium||||53.2%|
+|1051|Height Checker|[Go]({{< relref "/ChapterFour/1000~1099/1051.Height-Checker.md" >}})|Easy||||75.6%|
+|1052|Grumpy Bookstore Owner|[Go]({{< relref "/ChapterFour/1000~1099/1052.Grumpy-Bookstore-Owner.md" >}})|Medium||||57.1%|
+|1054|Distant Barcodes|[Go]({{< relref "/ChapterFour/1000~1099/1054.Distant-Barcodes.md" >}})|Medium||||45.9%|
+|1073|Adding Two Negabinary Numbers|[Go]({{< relref "/ChapterFour/1000~1099/1073.Adding-Two-Negabinary-Numbers.md" >}})|Medium||||36.5%|
+|1074|Number of Submatrices That Sum to Target|[Go]({{< relref "/ChapterFour/1000~1099/1074.Number-of-Submatrices-That-Sum-to-Target.md" >}})|Hard||||69.5%|
+|1089|Duplicate Zeros|[Go]({{< relref "/ChapterFour/1000~1099/1089.Duplicate-Zeros.md" >}})|Easy||||51.5%|
+|1091|Shortest Path in Binary Matrix|[Go]({{< relref "/ChapterFour/1000~1099/1091.Shortest-Path-in-Binary-Matrix.md" >}})|Medium||||44.7%|
+|1093|Statistics from a Large Sample|[Go]({{< relref "/ChapterFour/1000~1099/1093.Statistics-from-a-Large-Sample.md" >}})|Medium||||43.5%|
+|1105|Filling Bookcase Shelves|[Go]({{< relref "/ChapterFour/1100~1199/1105.Filling-Bookcase-Shelves.md" >}})|Medium||||59.3%|
+|1122|Relative Sort Array|[Go]({{< relref "/ChapterFour/1100~1199/1122.Relative-Sort-Array.md" >}})|Easy||||68.6%|
+|1128|Number of Equivalent Domino Pairs|[Go]({{< relref "/ChapterFour/1100~1199/1128.Number-of-Equivalent-Domino-Pairs.md" >}})|Easy||||47.1%|
+|1157|Online Majority Element In Subarray|[Go]({{< relref "/ChapterFour/1100~1199/1157.Online-Majority-Element-In-Subarray.md" >}})|Hard||||41.8%|
+|1160|Find Words That Can Be Formed by Characters|[Go]({{< relref "/ChapterFour/1100~1199/1160.Find-Words-That-Can-Be-Formed-by-Characters.md" >}})|Easy||||67.5%|
+|1170|Compare Strings by Frequency of the Smallest Character|[Go]({{< relref "/ChapterFour/1100~1199/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character.md" >}})|Medium||||61.5%|
+|1178|Number of Valid Words for Each Puzzle|[Go]({{< relref "/ChapterFour/1100~1199/1178.Number-of-Valid-Words-for-Each-Puzzle.md" >}})|Hard||||46.3%|
+|1184|Distance Between Bus Stops|[Go]({{< relref "/ChapterFour/1100~1199/1184.Distance-Between-Bus-Stops.md" >}})|Easy||||54.0%|
+|1200|Minimum Absolute Difference|[Go]({{< relref "/ChapterFour/1200~1299/1200.Minimum-Absolute-Difference.md" >}})|Easy||||69.6%|
+|1207|Unique Number of Occurrences|[Go]({{< relref "/ChapterFour/1200~1299/1207.Unique-Number-of-Occurrences.md" >}})|Easy||||73.5%|
+|1217|Minimum Cost to Move Chips to The Same Position|[Go]({{< relref "/ChapterFour/1200~1299/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position.md" >}})|Easy||||71.9%|
+|1232|Check If It Is a Straight Line|[Go]({{< relref "/ChapterFour/1200~1299/1232.Check-If-It-Is-a-Straight-Line.md" >}})|Easy||||40.3%|
+|1235|Maximum Profit in Job Scheduling|[Go]({{< relref "/ChapterFour/1200~1299/1235.Maximum-Profit-in-Job-Scheduling.md" >}})|Hard||||53.4%|
+|1239|Maximum Length of a Concatenated String with Unique Characters|[Go]({{< relref "/ChapterFour/1200~1299/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters.md" >}})|Medium||||52.2%|
+|1252|Cells with Odd Values in a Matrix|[Go]({{< relref "/ChapterFour/1200~1299/1252.Cells-with-Odd-Values-in-a-Matrix.md" >}})|Easy||||78.5%|
+|1254|Number of Closed Islands|[Go]({{< relref "/ChapterFour/1200~1299/1254.Number-of-Closed-Islands.md" >}})|Medium||||66.9%|
+|1260|Shift 2D Grid|[Go]({{< relref "/ChapterFour/1200~1299/1260.Shift-2D-Grid.md" >}})|Easy||||67.8%|
+|1266|Minimum Time Visiting All Points|[Go]({{< relref "/ChapterFour/1200~1299/1266.Minimum-Time-Visiting-All-Points.md" >}})|Easy||||79.1%|
+|1268|Search Suggestions System|[Go]({{< relref "/ChapterFour/1200~1299/1268.Search-Suggestions-System.md" >}})|Medium||||66.2%|
+|1275|Find Winner on a Tic Tac Toe Game|[Go]({{< relref "/ChapterFour/1200~1299/1275.Find-Winner-on-a-Tic-Tac-Toe-Game.md" >}})|Easy||||54.2%|
+|1283|Find the Smallest Divisor Given a Threshold|[Go]({{< relref "/ChapterFour/1200~1299/1283.Find-the-Smallest-Divisor-Given-a-Threshold.md" >}})|Medium||||56.2%|
+|1287|Element Appearing More Than 25% In Sorted Array|[Go]({{< relref "/ChapterFour/1200~1299/1287.Element-Appearing-More-Than-25-In-Sorted-Array.md" >}})|Easy||||59.4%|
+|1293|Shortest Path in a Grid with Obstacles Elimination|[Go]({{< relref "/ChapterFour/1200~1299/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination.md" >}})|Hard||||45.3%|
+|1295|Find Numbers with Even Number of Digits|[Go]({{< relref "/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md" >}})|Easy||||77.0%|
+|1296|Divide Array in Sets of K Consecutive Numbers|[Go]({{< relref "/ChapterFour/1200~1299/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers.md" >}})|Medium||||56.5%|
+|1299|Replace Elements with Greatest Element on Right Side|[Go]({{< relref "/ChapterFour/1200~1299/1299.Replace-Elements-with-Greatest-Element-on-Right-Side.md" >}})|Easy||||73.3%|
+|1300|Sum of Mutated Array Closest to Target|[Go]({{< relref "/ChapterFour/1300~1399/1300.Sum-of-Mutated-Array-Closest-to-Target.md" >}})|Medium||||43.6%|
+|1304|Find N Unique Integers Sum up to Zero|[Go]({{< relref "/ChapterFour/1300~1399/1304.Find-N-Unique-Integers-Sum-up-to-Zero.md" >}})|Easy||||76.9%|
+|1306|Jump Game III|[Go]({{< relref "/ChapterFour/1300~1399/1306.Jump-Game-III.md" >}})|Medium||||63.5%|
+|1310|XOR Queries of a Subarray|[Go]({{< relref "/ChapterFour/1300~1399/1310.XOR-Queries-of-a-Subarray.md" >}})|Medium||||72.3%|
+|1313|Decompress Run-Length Encoded List|[Go]({{< relref "/ChapterFour/1300~1399/1313.Decompress-Run-Length-Encoded-List.md" >}})|Easy||||85.8%|
+|1329|Sort the Matrix Diagonally|[Go]({{< relref "/ChapterFour/1300~1399/1329.Sort-the-Matrix-Diagonally.md" >}})|Medium||||83.3%|
+|1337|The K Weakest Rows in a Matrix|[Go]({{< relref "/ChapterFour/1300~1399/1337.The-K-Weakest-Rows-in-a-Matrix.md" >}})|Easy||||72.1%|
+|1353|Maximum Number of Events That Can Be Attended|[Go]({{< relref "/ChapterFour/1300~1399/1353.Maximum-Number-of-Events-That-Can-Be-Attended.md" >}})|Medium||||32.5%|
+|1380|Lucky Numbers in a Matrix|[Go]({{< relref "/ChapterFour/1300~1399/1380.Lucky-Numbers-in-a-Matrix.md" >}})|Easy||||70.7%|
+|1383|Maximum Performance of a Team|[Go]({{< relref "/ChapterFour/1300~1399/1383.Maximum-Performance-of-a-Team.md" >}})|Hard||||48.5%|
+|1385|Find the Distance Value Between Two Arrays|[Go]({{< relref "/ChapterFour/1300~1399/1385.Find-the-Distance-Value-Between-Two-Arrays.md" >}})|Easy||||66.6%|
+|1389|Create Target Array in the Given Order|[Go]({{< relref "/ChapterFour/1300~1399/1389.Create-Target-Array-in-the-Given-Order.md" >}})|Easy||||85.8%|
+|1423|Maximum Points You Can Obtain from Cards|[Go]({{< relref "/ChapterFour/1400~1499/1423.Maximum-Points-You-Can-Obtain-from-Cards.md" >}})|Medium||||52.2%|
+|1437|Check If All 1's Are at Least Length K Places Away|[Go]({{< relref "/ChapterFour/1400~1499/1437.Check-If-All-1s-Are-at-Least-Length-K-Places-Away.md" >}})|Easy||||58.7%|
+|1438|Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit|[Go]({{< relref "/ChapterFour/1400~1499/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit.md" >}})|Medium||||48.3%|
+|1439|Find the Kth Smallest Sum of a Matrix With Sorted Rows|[Go]({{< relref "/ChapterFour/1400~1499/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows.md" >}})|Hard||||61.4%|
+|1442|Count Triplets That Can Form Two Arrays of Equal XOR|[Go]({{< relref "/ChapterFour/1400~1499/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR.md" >}})|Medium||||76.1%|
+|1463|Cherry Pickup II|[Go]({{< relref "/ChapterFour/1400~1499/1463.Cherry-Pickup-II.md" >}})|Hard||||69.5%|
+|1464|Maximum Product of Two Elements in an Array|[Go]({{< relref "/ChapterFour/1400~1499/1464.Maximum-Product-of-Two-Elements-in-an-Array.md" >}})|Easy||||80.0%|
+|1465|Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts|[Go]({{< relref "/ChapterFour/1400~1499/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts.md" >}})|Medium||||40.9%|
+|1470|Shuffle the Array|[Go]({{< relref "/ChapterFour/1400~1499/1470.Shuffle-the-Array.md" >}})|Easy||||89.0%|
+|1480|Running Sum of 1d Array|[Go]({{< relref "/ChapterFour/1400~1499/1480.Running-Sum-of-1d-Array.md" >}})|Easy||||87.4%|
+|1482|Minimum Number of Days to Make m Bouquets|[Go]({{< relref "/ChapterFour/1400~1499/1482.Minimum-Number-of-Days-to-Make-m-Bouquets.md" >}})|Medium||||54.0%|
+|1512|Number of Good Pairs|[Go]({{< relref "/ChapterFour/1500~1599/1512.Number-of-Good-Pairs.md" >}})|Easy||||88.2%|
+|1539|Kth Missing Positive Number|[Go]({{< relref "/ChapterFour/1500~1599/1539.Kth-Missing-Positive-Number.md" >}})|Easy||||58.6%|
+|1572|Matrix Diagonal Sum|[Go]({{< relref "/ChapterFour/1500~1599/1572.Matrix-Diagonal-Sum.md" >}})|Easy||||80.3%|
+|1608|Special Array With X Elements Greater Than or Equal X|[Go]({{< relref "/ChapterFour/1600~1699/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X.md" >}})|Easy||||60.5%|
+|1619|Mean of Array After Removing Some Elements|[Go]({{< relref "/ChapterFour/1600~1699/1619.Mean-of-Array-After-Removing-Some-Elements.md" >}})|Easy||||65.8%|
+|1629|Slowest Key|[Go]({{< relref "/ChapterFour/1600~1699/1629.Slowest-Key.md" >}})|Easy||||59.2%|
+|1631|Path With Minimum Effort|[Go]({{< relref "/ChapterFour/1600~1699/1631.Path-With-Minimum-Effort.md" >}})|Medium||||55.7%|
+|1636|Sort Array by Increasing Frequency|[Go]({{< relref "/ChapterFour/1600~1699/1636.Sort-Array-by-Increasing-Frequency.md" >}})|Easy||||69.5%|
+|1640|Check Array Formation Through Concatenation|[Go]({{< relref "/ChapterFour/1600~1699/1640.Check-Array-Formation-Through-Concatenation.md" >}})|Easy||||56.2%|
+|1642|Furthest Building You Can Reach|[Go]({{< relref "/ChapterFour/1600~1699/1642.Furthest-Building-You-Can-Reach.md" >}})|Medium||||48.3%|
+|1646|Get Maximum in Generated Array|[Go]({{< relref "/ChapterFour/1600~1699/1646.Get-Maximum-in-Generated-Array.md" >}})|Easy||||50.2%|
+|1648|Sell Diminishing-Valued Colored Balls|[Go]({{< relref "/ChapterFour/1600~1699/1648.Sell-Diminishing-Valued-Colored-Balls.md" >}})|Medium||||30.4%|
+|1649|Create Sorted Array through Instructions|[Go]({{< relref "/ChapterFour/1600~1699/1649.Create-Sorted-Array-through-Instructions.md" >}})|Hard||||37.5%|
+|1652|Defuse the Bomb|[Go]({{< relref "/ChapterFour/1600~1699/1652.Defuse-the-Bomb.md" >}})|Easy||||62.4%|
+|1654|Minimum Jumps to Reach Home|[Go]({{< relref "/ChapterFour/1600~1699/1654.Minimum-Jumps-to-Reach-Home.md" >}})|Medium||||29.1%|
+|1655|Distribute Repeating Integers|[Go]({{< relref "/ChapterFour/1600~1699/1655.Distribute-Repeating-Integers.md" >}})|Hard||||39.3%|
+|1656|Design an Ordered Stream|[Go]({{< relref "/ChapterFour/1600~1699/1656.Design-an-Ordered-Stream.md" >}})|Easy||||85.2%|
+|1658|Minimum Operations to Reduce X to Zero|[Go]({{< relref "/ChapterFour/1600~1699/1658.Minimum-Operations-to-Reduce-X-to-Zero.md" >}})|Medium||||37.6%|
+|1662|Check If Two String Arrays are Equivalent|[Go]({{< relref "/ChapterFour/1600~1699/1662.Check-If-Two-String-Arrays-are-Equivalent.md" >}})|Easy||||83.5%|
+|1664|Ways to Make a Fair Array|[Go]({{< relref "/ChapterFour/1600~1699/1664.Ways-to-Make-a-Fair-Array.md" >}})|Medium||||63.3%|
+|1665|Minimum Initial Energy to Finish Tasks|[Go]({{< relref "/ChapterFour/1600~1699/1665.Minimum-Initial-Energy-to-Finish-Tasks.md" >}})|Hard||||56.4%|
+|1670|Design Front Middle Back Queue|[Go]({{< relref "/ChapterFour/1600~1699/1670.Design-Front-Middle-Back-Queue.md" >}})|Medium||||57.2%|
+|1672|Richest Customer Wealth|[Go]({{< relref "/ChapterFour/1600~1699/1672.Richest-Customer-Wealth.md" >}})|Easy||||87.9%|
+|1673|Find the Most Competitive Subsequence|[Go]({{< relref "/ChapterFour/1600~1699/1673.Find-the-Most-Competitive-Subsequence.md" >}})|Medium||||49.3%|
+|1674|Minimum Moves to Make Array Complementary|[Go]({{< relref "/ChapterFour/1600~1699/1674.Minimum-Moves-to-Make-Array-Complementary.md" >}})|Medium||||38.7%|
+|1675|Minimize Deviation in Array|[Go]({{< relref "/ChapterFour/1600~1699/1675.Minimize-Deviation-in-Array.md" >}})|Hard||||54.6%|
+|1679|Max Number of K-Sum Pairs|[Go]({{< relref "/ChapterFour/1600~1699/1679.Max-Number-of-K-Sum-Pairs.md" >}})|Medium||||57.3%|
+|1681|Minimum Incompatibility|[Go]({{< relref "/ChapterFour/1600~1699/1681.Minimum-Incompatibility.md" >}})|Hard||||37.8%|
+|1684|Count the Number of Consistent Strings|[Go]({{< relref "/ChapterFour/1600~1699/1684.Count-the-Number-of-Consistent-Strings.md" >}})|Easy||||82.3%|
+|1685|Sum of Absolute Differences in a Sorted Array|[Go]({{< relref "/ChapterFour/1600~1699/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array.md" >}})|Medium||||63.5%|
+|1690|Stone Game VII|[Go]({{< relref "/ChapterFour/1600~1699/1690.Stone-Game-VII.md" >}})|Medium||||58.1%|
+|1691|Maximum Height by Stacking Cuboids|[Go]({{< relref "/ChapterFour/1600~1699/1691.Maximum-Height-by-Stacking-Cuboids.md" >}})|Hard||||54.6%|
+|1695|Maximum Erasure Value|[Go]({{< relref "/ChapterFour/1600~1699/1695.Maximum-Erasure-Value.md" >}})|Medium||||57.6%|
+|1696|Jump Game VI|[Go]({{< relref "/ChapterFour/1600~1699/1696.Jump-Game-VI.md" >}})|Medium||||46.1%|
+|1700|Number of Students Unable to Eat Lunch|[Go]({{< relref "/ChapterFour/1700~1799/1700.Number-of-Students-Unable-to-Eat-Lunch.md" >}})|Easy||||68.8%|
+|1705|Maximum Number of Eaten Apples|[Go]({{< relref "/ChapterFour/1700~1799/1705.Maximum-Number-of-Eaten-Apples.md" >}})|Medium||||38.0%|
+|1710|Maximum Units on a Truck|[Go]({{< relref "/ChapterFour/1700~1799/1710.Maximum-Units-on-a-Truck.md" >}})|Easy||||73.8%|
+|1720|Decode XORed Array|[Go]({{< relref "/ChapterFour/1700~1799/1720.Decode-XORed-Array.md" >}})|Easy||||85.8%|
+|1725|Number Of Rectangles That Can Form The Largest Square|[Go]({{< relref "/ChapterFour/1700~1799/1725.Number-Of-Rectangles-That-Can-Form-The-Largest-Square.md" >}})|Easy||||78.6%|
+|1732|Find the Highest Altitude|[Go]({{< relref "/ChapterFour/1700~1799/1732.Find-the-Highest-Altitude.md" >}})|Easy||||78.9%|
+|1734|Decode XORed Permutation|[Go]({{< relref "/ChapterFour/1700~1799/1734.Decode-XORed-Permutation.md" >}})|Medium||||63.0%|
+|1738|Find Kth Largest XOR Coordinate Value|[Go]({{< relref "/ChapterFour/1700~1799/1738.Find-Kth-Largest-XOR-Coordinate-Value.md" >}})|Medium||||61.0%|
+|1744|Can You Eat Your Favorite Candy on Your Favorite Day?|[Go]({{< relref "/ChapterFour/1700~1799/1744.Can-You-Eat-Your-Favorite-Candy-on-Your-Favorite-Day.md" >}})|Medium||||33.1%|
+|1748|Sum of Unique Elements|[Go]({{< relref "/ChapterFour/1700~1799/1748.Sum-of-Unique-Elements.md" >}})|Easy||||76.3%|
+|1752|Check if Array Is Sorted and Rotated|[Go]({{< relref "/ChapterFour/1700~1799/1752.Check-if-Array-Is-Sorted-and-Rotated.md" >}})|Easy||||50.1%|
+|1816|Truncate Sentence|[Go]({{< relref "/ChapterFour/1800~1899/1816.Truncate-Sentence.md" >}})|Easy||||83.1%|
+|1818|Minimum Absolute Sum Difference|[Go]({{< relref "/ChapterFour/1800~1899/1818.Minimum-Absolute-Sum-Difference.md" >}})|Medium||||30.4%|
+|1846|Maximum Element After Decreasing and Rearranging|[Go]({{< relref "/ChapterFour/1800~1899/1846.Maximum-Element-After-Decreasing-and-Rearranging.md" >}})|Medium||||58.9%|
+|1877|Minimize Maximum Pair Sum in Array|[Go]({{< relref "/ChapterFour/1800~1899/1877.Minimize-Maximum-Pair-Sum-in-Array.md" >}})|Medium||||79.9%|
+|1984|Minimum Difference Between Highest and Lowest of K Scores|[Go]({{< relref "/ChapterFour/1900~1999/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores.md" >}})|Easy||||54.5%|
+|2021|Brightest Position on Street|[Go]({{< relref "/ChapterFour/2000~2099/2021.Brightest-Position-on-Street.md" >}})|Medium||||62.1%|
+|2022|Convert 1D Array Into 2D Array|[Go]({{< relref "/ChapterFour/2000~2099/2022.Convert-1D-Array-Into-2D-Array.md" >}})|Easy||||59.1%|
+|2037|Minimum Number of Moves to Seat Everyone|[Go]({{< relref "/ChapterFour/2000~2099/2037.Minimum-Number-of-Moves-to-Seat-Everyone.md" >}})|Easy||||82.1%|
+|2043|Simple Bank System|[Go]({{< relref "/ChapterFour/2000~2099/2043.Simple-Bank-System.md" >}})|Medium||||65.2%|
+|2164|Sort Even and Odd Indices Independently|[Go]({{< relref "/ChapterFour/2100~2199/2164.Sort-Even-and-Odd-Indices-Independently.md" >}})|Easy||||64.9%|
+|2166|Design Bitset|[Go]({{< relref "/ChapterFour/2100~2199/2166.Design-Bitset.md" >}})|Medium||||31.8%|
+|2170|Minimum Operations to Make the Array Alternating|[Go]({{< relref "/ChapterFour/2100~2199/2170.Minimum-Operations-to-Make-the-Array-Alternating.md" >}})|Medium||||33.2%|
+|2171|Removing Minimum Number of Magic Beans|[Go]({{< relref "/ChapterFour/2100~2199/2171.Removing-Minimum-Number-of-Magic-Beans.md" >}})|Medium||||42.1%|
+|2183|Count Array Pairs Divisible by K|[Go]({{< relref "/ChapterFour/2100~2199/2183.Count-Array-Pairs-Divisible-by-K.md" >}})|Hard||||28.3%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/String/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Backtracking.md b/website/content/ChapterTwo/Backtracking.md
new file mode 100644
index 000000000..fd1b61831
--- /dev/null
+++ b/website/content/ChapterTwo/Backtracking.md
@@ -0,0 +1,149 @@
+---
+title: 2.08 ✅ Backtracking
+type: docs
+weight: 8
+---
+
+# Backtracking
+
+![](https://img.halfrost.com/Leetcode/Backtracking.png)
+
+- 排列问题 Permutations。第 46 题，第 47 题。第 60 题，第 526 题，第 996 题。
+- 组合问题 Combination。第 39 题，第 40 题，第 77 题，第 216 题。
+- 排列和组合杂交问题。第 1079 题。
+- N 皇后终极解法(二进制解法)。第 51 题，第 52 题。
+- 数独问题。第 37 题。
+- 四个方向搜索。第 79 题，第 212 题，第 980 题。
+- 子集合问题。第 78 题，第 90 题。
+- Trie。第 208 题，第 211 题。
+- BFS 优化。第 126 题，第 127 题。
+- DFS 模板。(只是一个例子，不对应任何题)
+
+```go
+func combinationSum2(candidates []int, target int) [][]int {
+	if len(candidates) == 0 {
+		return [][]int{}
+	}
+	c, res := []int{}, [][]int{}
+	sort.Ints(candidates)
+	findcombinationSum2(candidates, target, 0, c, &res)
+	return res
+}
+
+func findcombinationSum2(nums []int, target, index int, c []int, res *[][]int) {
+	if target == 0 {
+		b := make([]int, len(c))
+		copy(b, c)
+		*res = append(*res, b)
+		return
+	}
+	for i := index; i < len(nums); i++ {
+		if i > index && nums[i] == nums[i-1] { // 这里是去重的关键逻辑
+			continue
+		}
+		if target >= nums[i] {
+			c = append(c, nums[i])
+			findcombinationSum2(nums, target-nums[i], i+1, c, res)
+			c = c[:len(c)-1]
+		}
+	}
+}
+```
+- BFS 模板。(只是一个例子，不对应任何题)
+
+```go
+func updateMatrix_BFS(matrix [][]int) [][]int {
+	res := make([][]int, len(matrix))
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return res
+	}
+	queue := make([][]int, 0)
+	for i, _ := range matrix {
+		res[i] = make([]int, len(matrix[0]))
+		for j, _ := range res[i] {
+			if matrix[i][j] == 0 {
+				res[i][j] = -1
+				queue = append(queue, []int{i, j})
+			}
+		}
+	}
+	level := 1
+	for len(queue) > 0 {
+		size := len(queue)
+		for size > 0 {
+			size -= 1
+			node := queue[0]
+			queue = queue[1:]
+			i, j := node[0], node[1]
+			for _, direction := range [][]int{{-1, 0}, {1, 0}, {0, 1}, {0, -1}} {
+				x := i + direction[0]
+				y := j + direction[1]
+				if x < 0 || x >= len(matrix) || y < 0 || y >= len(matrix[0]) || res[x][y] < 0 || res[x][y] > 0 {
+					continue
+				}
+				res[x][y] = level
+				queue = append(queue, []int{x, y})
+			}
+		}
+		level++
+	}
+	for i, row := range res {
+		for j, cell := range row {
+			if cell == -1 {
+				res[i][j] = 0
+			}
+		}
+	}
+	return res
+}
+```
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0017|Letter Combinations of a Phone Number|[Go]({{< relref "/ChapterFour/0001~0099/0017.Letter-Combinations-of-a-Phone-Number.md" >}})|Medium| O(log n)| O(1)||56.6%|
+|0022|Generate Parentheses|[Go]({{< relref "/ChapterFour/0001~0099/0022.Generate-Parentheses.md" >}})|Medium| O(log n)| O(1)||72.5%|
+|0037|Sudoku Solver|[Go]({{< relref "/ChapterFour/0001~0099/0037.Sudoku-Solver.md" >}})|Hard| O(n^2)| O(n^2)|❤️|57.7%|
+|0039|Combination Sum|[Go]({{< relref "/ChapterFour/0001~0099/0039.Combination-Sum.md" >}})|Medium| O(n log n)| O(n)||68.6%|
+|0040|Combination Sum II|[Go]({{< relref "/ChapterFour/0001~0099/0040.Combination-Sum-II.md" >}})|Medium| O(n log n)| O(n)||53.4%|
+|0046|Permutations|[Go]({{< relref "/ChapterFour/0001~0099/0046.Permutations.md" >}})|Medium| O(n)| O(n)|❤️|75.7%|
+|0047|Permutations II|[Go]({{< relref "/ChapterFour/0001~0099/0047.Permutations-II.md" >}})|Medium| O(n^2)| O(n)|❤️|57.4%|
+|0051|N-Queens|[Go]({{< relref "/ChapterFour/0001~0099/0051.N-Queens.md" >}})|Hard| O(n!)| O(n)|❤️|64.2%|
+|0052|N-Queens II|[Go]({{< relref "/ChapterFour/0001~0099/0052.N-Queens-II.md" >}})|Hard| O(n!)| O(n)|❤️|71.6%|
+|0077|Combinations|[Go]({{< relref "/ChapterFour/0001~0099/0077.Combinations.md" >}})|Medium| O(n)| O(n)|❤️|67.0%|
+|0078|Subsets|[Go]({{< relref "/ChapterFour/0001~0099/0078.Subsets.md" >}})|Medium| O(n^2)| O(n)|❤️|74.9%|
+|0079|Word Search|[Go]({{< relref "/ChapterFour/0001~0099/0079.Word-Search.md" >}})|Medium| O(n^2)| O(n^2)|❤️|40.2%|
+|0089|Gray Code|[Go]({{< relref "/ChapterFour/0001~0099/0089.Gray-Code.md" >}})|Medium| O(n)| O(1)||57.2%|
+|0090|Subsets II|[Go]({{< relref "/ChapterFour/0001~0099/0090.Subsets-II.md" >}})|Medium| O(n^2)| O(n)|❤️|55.9%|
+|0093|Restore IP Addresses|[Go]({{< relref "/ChapterFour/0001~0099/0093.Restore-IP-Addresses.md" >}})|Medium| O(n)| O(n)|❤️|47.4%|
+|0095|Unique Binary Search Trees II|[Go]({{< relref "/ChapterFour/0001~0099/0095.Unique-Binary-Search-Trees-II.md" >}})|Medium||||52.4%|
+|0113|Path Sum II|[Go]({{< relref "/ChapterFour/0100~0199/0113.Path-Sum-II.md" >}})|Medium||||57.1%|
+|0126|Word Ladder II|[Go]({{< relref "/ChapterFour/0100~0199/0126.Word-Ladder-II.md" >}})|Hard| O(n)| O(n^2)|❤️|27.5%|
+|0131|Palindrome Partitioning|[Go]({{< relref "/ChapterFour/0100~0199/0131.Palindrome-Partitioning.md" >}})|Medium| O(n)| O(n^2)|❤️|64.9%|
+|0212|Word Search II|[Go]({{< relref "/ChapterFour/0200~0299/0212.Word-Search-II.md" >}})|Hard| O(n^2)| O(n^2)|❤️|36.4%|
+|0216|Combination Sum III|[Go]({{< relref "/ChapterFour/0200~0299/0216.Combination-Sum-III.md" >}})|Medium| O(n)| O(1)|❤️|67.6%|
+|0257|Binary Tree Paths|[Go]({{< relref "/ChapterFour/0200~0299/0257.Binary-Tree-Paths.md" >}})|Easy||||61.4%|
+|0301|Remove Invalid Parentheses|[Go]({{< relref "/ChapterFour/0300~0399/0301.Remove-Invalid-Parentheses.md" >}})|Hard||||47.2%|
+|0306|Additive Number|[Go]({{< relref "/ChapterFour/0300~0399/0306.Additive-Number.md" >}})|Medium| O(n^2)| O(1)|❤️|31.1%|
+|0357|Count Numbers with Unique Digits|[Go]({{< relref "/ChapterFour/0300~0399/0357.Count-Numbers-with-Unique-Digits.md" >}})|Medium| O(1)| O(1)||51.9%|
+|0401|Binary Watch|[Go]({{< relref "/ChapterFour/0400~0499/0401.Binary-Watch.md" >}})|Easy| O(1)| O(1)||52.3%|
+|0473|Matchsticks to Square|[Go]({{< relref "/ChapterFour/0400~0499/0473.Matchsticks-to-Square.md" >}})|Medium||||40.2%|
+|0491|Non-decreasing Subsequences|[Go]({{< relref "/ChapterFour/0400~0499/0491.Non-decreasing-Subsequences.md" >}})|Medium||||60.2%|
+|0494|Target Sum|[Go]({{< relref "/ChapterFour/0400~0499/0494.Target-Sum.md" >}})|Medium||||45.7%|
+|0526|Beautiful Arrangement|[Go]({{< relref "/ChapterFour/0500~0599/0526.Beautiful-Arrangement.md" >}})|Medium| O(n^2)| O(1)|❤️|64.4%|
+|0638|Shopping Offers|[Go]({{< relref "/ChapterFour/0600~0699/0638.Shopping-Offers.md" >}})|Medium||||53.3%|
+|0784|Letter Case Permutation|[Go]({{< relref "/ChapterFour/0700~0799/0784.Letter-Case-Permutation.md" >}})|Medium| O(n)| O(n)||73.8%|
+|0816|Ambiguous Coordinates|[Go]({{< relref "/ChapterFour/0800~0899/0816.Ambiguous-Coordinates.md" >}})|Medium||||56.4%|
+|0842|Split Array into Fibonacci Sequence|[Go]({{< relref "/ChapterFour/0800~0899/0842.Split-Array-into-Fibonacci-Sequence.md" >}})|Medium| O(n^2)| O(1)|❤️|38.4%|
+|0980|Unique Paths III|[Go]({{< relref "/ChapterFour/0900~0999/0980.Unique-Paths-III.md" >}})|Hard| O(n log n)| O(n)||81.7%|
+|0996|Number of Squareful Arrays|[Go]({{< relref "/ChapterFour/0900~0999/0996.Number-of-Squareful-Arrays.md" >}})|Hard| O(n log n)| O(n) ||49.2%|
+|1079|Letter Tile Possibilities|[Go]({{< relref "/ChapterFour/1000~1099/1079.Letter-Tile-Possibilities.md" >}})|Medium| O(n^2)| O(1)|❤️|76.0%|
+|1239|Maximum Length of a Concatenated String with Unique Characters|[Go]({{< relref "/ChapterFour/1200~1299/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters.md" >}})|Medium||||52.2%|
+|1655|Distribute Repeating Integers|[Go]({{< relref "/ChapterFour/1600~1699/1655.Distribute-Repeating-Integers.md" >}})|Hard||||39.3%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Dynamic_Programming/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Depth_First_Search/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Binary_Indexed_Tree.md b/website/content/ChapterTwo/Binary_Indexed_Tree.md
new file mode 100644
index 000000000..6a988987e
--- /dev/null
+++ b/website/content/ChapterTwo/Binary_Indexed_Tree.md
@@ -0,0 +1,27 @@
+---
+title: 2.19 ✅ Binary Indexed Tree
+type: docs
+weight: 19
+---
+
+# Binary Indexed Tree
+
+![](https://img.halfrost.com/Leetcode/Binary_Indexed_Tree.png)
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0218|The Skyline Problem|[Go]({{< relref "/ChapterFour/0200~0299/0218.The-Skyline-Problem.md" >}})|Hard||||41.9%|
+|0307|Range Sum Query - Mutable|[Go]({{< relref "/ChapterFour/0300~0399/0307.Range-Sum-Query-Mutable.md" >}})|Medium||||40.7%|
+|0315|Count of Smaller Numbers After Self|[Go]({{< relref "/ChapterFour/0300~0399/0315.Count-of-Smaller-Numbers-After-Self.md" >}})|Hard||||42.6%|
+|0327|Count of Range Sum|[Go]({{< relref "/ChapterFour/0300~0399/0327.Count-of-Range-Sum.md" >}})|Hard||||35.8%|
+|0493|Reverse Pairs|[Go]({{< relref "/ChapterFour/0400~0499/0493.Reverse-Pairs.md" >}})|Hard||||30.9%|
+|1157|Online Majority Element In Subarray|[Go]({{< relref "/ChapterFour/1100~1199/1157.Online-Majority-Element-In-Subarray.md" >}})|Hard||||41.8%|
+|1649|Create Sorted Array through Instructions|[Go]({{< relref "/ChapterFour/1600~1699/1649.Create-Sorted-Array-through-Instructions.md" >}})|Hard||||37.5%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Segment_Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterThree/">下一章➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Binary_Search.md b/website/content/ChapterTwo/Binary_Search.md
new file mode 100644
index 000000000..bbfa80272
--- /dev/null
+++ b/website/content/ChapterTwo/Binary_Search.md
@@ -0,0 +1,225 @@
+---
+title: 2.11 Binary Search
+type: docs
+weight: 11
+---
+
+# Binary Search
+
+- 二分搜索的经典写法。需要注意的三点：
+	1. 循环退出条件，注意是 low <= high，而不是 low < high。
+	2. mid 的取值，mid := low + (high-low)>>1
+	3. low 和 high 的更新。low = mid + 1，high = mid - 1。
+
+```go
+func binarySearchMatrix(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[mid] == target {
+			return mid
+		} else if nums[mid] > target {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+```
+
+- 二分搜索的变种写法。有 4 个基本变种:
+	1. 查找第一个与 target 相等的元素，时间复杂度 O(logn)
+	2. 查找最后一个与 target 相等的元素，时间复杂度 O(logn)
+	3. 查找第一个大于等于 target 的元素，时间复杂度 O(logn)
+	4. 查找最后一个小于等于 target 的元素，时间复杂度 O(logn)
+
+```go
+// 二分查找第一个与 target 相等的元素，时间复杂度 O(logn)
+func searchFirstEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == 0) || (nums[mid-1] != target) { // 找到第一个与 target 相等的元素
+				return mid
+			}
+			high = mid - 1
+		}
+	}
+	return -1
+}
+
+// 二分查找最后一个与 target 相等的元素，时间复杂度 O(logn)
+func searchLastEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == len(nums)-1) || (nums[mid+1] != target) { // 找到最后一个与 target 相等的元素
+				return mid
+			}
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// 二分查找第一个大于等于 target 的元素，时间复杂度 O(logn)
+func searchFirstGreaterElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] >= target {
+			if (mid == 0) || (nums[mid-1] < target) { // 找到第一个大于等于 target 的元素
+				return mid
+			}
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// 二分查找最后一个小于等于 target 的元素，时间复杂度 O(logn)
+func searchLastLessElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] <= target {
+			if (mid == len(nums)-1) || (nums[mid+1] > target) { // 找到最后一个小于等于 target 的元素
+				return mid
+			}
+			low = mid + 1
+		} else {
+			high = mid - 1
+		}
+	}
+	return -1
+}
+```
+
+- 在基本有序的数组中用二分搜索。经典解法可以解，变种写法也可以写，常见的题型，在山峰数组中找山峰，在旋转有序数组中找分界点。第 33 题，第 81 题，第 153 题，第 154 题，第 162 题，第 852 题
+
+```go
+func peakIndexInMountainArray(A []int) int {
+	low, high := 0, len(A)-1
+	for low < high {
+		mid := low + (high-low)>>1
+		// 如果 mid 较大，则左侧存在峰值，high = m，如果 mid + 1 较大，则右侧存在峰值，low = mid + 1
+		if A[mid] > A[mid+1] {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+```
+
+- max-min 最大值最小化问题。求在最小满足条件的情况下的最大值。第 410 题，第 875 题，第 1011 题，第 1283 题。
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0004|Median of Two Sorted Arrays|[Go]({{< relref "/ChapterFour/0001~0099/0004.Median-of-Two-Sorted-Arrays.md" >}})|Hard||||36.2%|
+|0033|Search in Rotated Sorted Array|[Go]({{< relref "/ChapterFour/0001~0099/0033.Search-in-Rotated-Sorted-Array.md" >}})|Medium||||39.0%|
+|0034|Find First and Last Position of Element in Sorted Array|[Go]({{< relref "/ChapterFour/0001~0099/0034.Find-First-and-Last-Position-of-Element-in-Sorted-Array.md" >}})|Medium||||41.9%|
+|0035|Search Insert Position|[Go]({{< relref "/ChapterFour/0001~0099/0035.Search-Insert-Position.md" >}})|Easy||||43.4%|
+|0069|Sqrt(x)|[Go]({{< relref "/ChapterFour/0001~0099/0069.Sqrtx.md" >}})|Easy| O(log n)| O(1)||37.4%|
+|0074|Search a 2D Matrix|[Go]({{< relref "/ChapterFour/0001~0099/0074.Search-a-2D-Matrix.md" >}})|Medium||||47.7%|
+|0081|Search in Rotated Sorted Array II|[Go]({{< relref "/ChapterFour/0001~0099/0081.Search-in-Rotated-Sorted-Array-II.md" >}})|Medium||||35.7%|
+|0153|Find Minimum in Rotated Sorted Array|[Go]({{< relref "/ChapterFour/0100~0199/0153.Find-Minimum-in-Rotated-Sorted-Array.md" >}})|Medium||||48.9%|
+|0154|Find Minimum in Rotated Sorted Array II|[Go]({{< relref "/ChapterFour/0100~0199/0154.Find-Minimum-in-Rotated-Sorted-Array-II.md" >}})|Hard||||43.5%|
+|0162|Find Peak Element|[Go]({{< relref "/ChapterFour/0100~0199/0162.Find-Peak-Element.md" >}})|Medium||||46.0%|
+|0167|Two Sum II - Input Array Is Sorted|[Go]({{< relref "/ChapterFour/0100~0199/0167.Two-Sum-II-Input-Array-Is-Sorted.md" >}})|Medium| O(n)| O(1)||60.0%|
+|0209|Minimum Size Subarray Sum|[Go]({{< relref "/ChapterFour/0200~0299/0209.Minimum-Size-Subarray-Sum.md" >}})|Medium| O(n)| O(1)||45.0%|
+|0222|Count Complete Tree Nodes|[Go]({{< relref "/ChapterFour/0200~0299/0222.Count-Complete-Tree-Nodes.md" >}})|Medium| O(n)| O(1)||60.6%|
+|0240|Search a 2D Matrix II|[Go]({{< relref "/ChapterFour/0200~0299/0240.Search-a-2D-Matrix-II.md" >}})|Medium||||51.0%|
+|0268|Missing Number|[Go]({{< relref "/ChapterFour/0200~0299/0268.Missing-Number.md" >}})|Easy||||62.6%|
+|0275|H-Index II|[Go]({{< relref "/ChapterFour/0200~0299/0275.H-Index-II.md" >}})|Medium||||37.5%|
+|0278|First Bad Version|[Go]({{< relref "/ChapterFour/0200~0299/0278.First-Bad-Version.md" >}})|Easy||||43.3%|
+|0287|Find the Duplicate Number|[Go]({{< relref "/ChapterFour/0200~0299/0287.Find-the-Duplicate-Number.md" >}})|Medium| O(n)| O(1)|❤️|59.1%|
+|0300|Longest Increasing Subsequence|[Go]({{< relref "/ChapterFour/0300~0399/0300.Longest-Increasing-Subsequence.md" >}})|Medium| O(n log n)| O(n)||52.2%|
+|0315|Count of Smaller Numbers After Self|[Go]({{< relref "/ChapterFour/0300~0399/0315.Count-of-Smaller-Numbers-After-Self.md" >}})|Hard||||42.6%|
+|0327|Count of Range Sum|[Go]({{< relref "/ChapterFour/0300~0399/0327.Count-of-Range-Sum.md" >}})|Hard||||35.8%|
+|0349|Intersection of Two Arrays|[Go]({{< relref "/ChapterFour/0300~0399/0349.Intersection-of-Two-Arrays.md" >}})|Easy| O(n)| O(n) ||70.9%|
+|0350|Intersection of Two Arrays II|[Go]({{< relref "/ChapterFour/0300~0399/0350.Intersection-of-Two-Arrays-II.md" >}})|Easy| O(n)| O(n) ||56.0%|
+|0352|Data Stream as Disjoint Intervals|[Go]({{< relref "/ChapterFour/0300~0399/0352.Data-Stream-as-Disjoint-Intervals.md" >}})|Hard||||59.7%|
+|0354|Russian Doll Envelopes|[Go]({{< relref "/ChapterFour/0300~0399/0354.Russian-Doll-Envelopes.md" >}})|Hard||||37.9%|
+|0367|Valid Perfect Square|[Go]({{< relref "/ChapterFour/0300~0399/0367.Valid-Perfect-Square.md" >}})|Easy||||43.3%|
+|0374|Guess Number Higher or Lower|[Go]({{< relref "/ChapterFour/0300~0399/0374.Guess-Number-Higher-or-Lower.md" >}})|Easy||||51.9%|
+|0378|Kth Smallest Element in a Sorted Matrix|[Go]({{< relref "/ChapterFour/0300~0399/0378.Kth-Smallest-Element-in-a-Sorted-Matrix.md" >}})|Medium||||61.8%|
+|0400|Nth Digit|[Go]({{< relref "/ChapterFour/0400~0499/0400.Nth-Digit.md" >}})|Medium||||34.1%|
+|0410|Split Array Largest Sum|[Go]({{< relref "/ChapterFour/0400~0499/0410.Split-Array-Largest-Sum.md" >}})|Hard||||53.5%|
+|0436|Find Right Interval|[Go]({{< relref "/ChapterFour/0400~0499/0436.Find-Right-Interval.md" >}})|Medium||||50.8%|
+|0441|Arranging Coins|[Go]({{< relref "/ChapterFour/0400~0499/0441.Arranging-Coins.md" >}})|Easy||||46.2%|
+|0456|132 Pattern|[Go]({{< relref "/ChapterFour/0400~0499/0456.132-Pattern.md" >}})|Medium||||32.4%|
+|0475|Heaters|[Go]({{< relref "/ChapterFour/0400~0499/0475.Heaters.md" >}})|Medium||||36.5%|
+|0483|Smallest Good Base|[Go]({{< relref "/ChapterFour/0400~0499/0483.Smallest-Good-Base.md" >}})|Hard||||38.8%|
+|0493|Reverse Pairs|[Go]({{< relref "/ChapterFour/0400~0499/0493.Reverse-Pairs.md" >}})|Hard||||30.9%|
+|0497|Random Point in Non-overlapping Rectangles|[Go]({{< relref "/ChapterFour/0400~0499/0497.Random-Point-in-Non-overlapping-Rectangles.md" >}})|Medium||||39.4%|
+|0528|Random Pick with Weight|[Go]({{< relref "/ChapterFour/0500~0599/0528.Random-Pick-with-Weight.md" >}})|Medium||||46.1%|
+|0532|K-diff Pairs in an Array|[Go]({{< relref "/ChapterFour/0500~0599/0532.K-diff-Pairs-in-an-Array.md" >}})|Medium||||41.2%|
+|0540|Single Element in a Sorted Array|[Go]({{< relref "/ChapterFour/0500~0599/0540.Single-Element-in-a-Sorted-Array.md" >}})|Medium||||59.1%|
+|0611|Valid Triangle Number|[Go]({{< relref "/ChapterFour/0600~0699/0611.Valid-Triangle-Number.md" >}})|Medium||||50.6%|
+|0633|Sum of Square Numbers|[Go]({{< relref "/ChapterFour/0600~0699/0633.Sum-of-Square-Numbers.md" >}})|Medium||||34.4%|
+|0658|Find K Closest Elements|[Go]({{< relref "/ChapterFour/0600~0699/0658.Find-K-Closest-Elements.md" >}})|Medium||||46.8%|
+|0668|Kth Smallest Number in Multiplication Table|[Go]({{< relref "/ChapterFour/0600~0699/0668.Kth-Smallest-Number-in-Multiplication-Table.md" >}})|Hard||||51.4%|
+|0704|Binary Search|[Go]({{< relref "/ChapterFour/0700~0799/0704.Binary-Search.md" >}})|Easy||||56.1%|
+|0710|Random Pick with Blacklist|[Go]({{< relref "/ChapterFour/0700~0799/0710.Random-Pick-with-Blacklist.md" >}})|Hard| O(n)| O(n)  ||33.5%|
+|0718|Maximum Length of Repeated Subarray|[Go]({{< relref "/ChapterFour/0700~0799/0718.Maximum-Length-of-Repeated-Subarray.md" >}})|Medium||||51.3%|
+|0719|Find K-th Smallest Pair Distance|[Go]({{< relref "/ChapterFour/0700~0799/0719.Find-K-th-Smallest-Pair-Distance.md" >}})|Hard||||36.7%|
+|0729|My Calendar I|[Go]({{< relref "/ChapterFour/0700~0799/0729.My-Calendar-I.md" >}})|Medium||||56.8%|
+|0732|My Calendar III|[Go]({{< relref "/ChapterFour/0700~0799/0732.My-Calendar-III.md" >}})|Hard||||71.5%|
+|0744|Find Smallest Letter Greater Than Target|[Go]({{< relref "/ChapterFour/0700~0799/0744.Find-Smallest-Letter-Greater-Than-Target.md" >}})|Easy||||45.8%|
+|0778|Swim in Rising Water|[Go]({{< relref "/ChapterFour/0700~0799/0778.Swim-in-Rising-Water.md" >}})|Hard||||59.8%|
+|0786|K-th Smallest Prime Fraction|[Go]({{< relref "/ChapterFour/0700~0799/0786.K-th-Smallest-Prime-Fraction.md" >}})|Medium||||51.7%|
+|0793|Preimage Size of Factorial Zeroes Function|[Go]({{< relref "/ChapterFour/0700~0799/0793.Preimage-Size-of-Factorial-Zeroes-Function.md" >}})|Hard||||43.2%|
+|0825|Friends Of Appropriate Ages|[Go]({{< relref "/ChapterFour/0800~0899/0825.Friends-Of-Appropriate-Ages.md" >}})|Medium||||46.3%|
+|0826|Most Profit Assigning Work|[Go]({{< relref "/ChapterFour/0800~0899/0826.Most-Profit-Assigning-Work.md" >}})|Medium||||44.9%|
+|0852|Peak Index in a Mountain Array|[Go]({{< relref "/ChapterFour/0800~0899/0852.Peak-Index-in-a-Mountain-Array.md" >}})|Medium||||69.0%|
+|0862|Shortest Subarray with Sum at Least K|[Go]({{< relref "/ChapterFour/0800~0899/0862.Shortest-Subarray-with-Sum-at-Least-K.md" >}})|Hard||||26.0%|
+|0875|Koko Eating Bananas|[Go]({{< relref "/ChapterFour/0800~0899/0875.Koko-Eating-Bananas.md" >}})|Medium||||52.1%|
+|0878|Nth Magical Number|[Go]({{< relref "/ChapterFour/0800~0899/0878.Nth-Magical-Number.md" >}})|Hard||||35.4%|
+|0887|Super Egg Drop|[Go]({{< relref "/ChapterFour/0800~0899/0887.Super-Egg-Drop.md" >}})|Hard||||27.1%|
+|0888|Fair Candy Swap|[Go]({{< relref "/ChapterFour/0800~0899/0888.Fair-Candy-Swap.md" >}})|Easy||||60.7%|
+|0911|Online Election|[Go]({{< relref "/ChapterFour/0900~0999/0911.Online-Election.md" >}})|Medium||||52.2%|
+|0981|Time Based Key-Value Store|[Go]({{< relref "/ChapterFour/0900~0999/0981.Time-Based-Key-Value-Store.md" >}})|Medium||||52.2%|
+|1004|Max Consecutive Ones III|[Go]({{< relref "/ChapterFour/1000~1099/1004.Max-Consecutive-Ones-III.md" >}})|Medium||||63.2%|
+|1011|Capacity To Ship Packages Within D Days|[Go]({{< relref "/ChapterFour/1000~1099/1011.Capacity-To-Ship-Packages-Within-D-Days.md" >}})|Medium||||67.7%|
+|1157|Online Majority Element In Subarray|[Go]({{< relref "/ChapterFour/1100~1199/1157.Online-Majority-Element-In-Subarray.md" >}})|Hard||||41.8%|
+|1170|Compare Strings by Frequency of the Smallest Character|[Go]({{< relref "/ChapterFour/1100~1199/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character.md" >}})|Medium||||61.5%|
+|1201|Ugly Number III|[Go]({{< relref "/ChapterFour/1200~1299/1201.Ugly-Number-III.md" >}})|Medium||||28.9%|
+|1208|Get Equal Substrings Within Budget|[Go]({{< relref "/ChapterFour/1200~1299/1208.Get-Equal-Substrings-Within-Budget.md" >}})|Medium||||48.6%|
+|1235|Maximum Profit in Job Scheduling|[Go]({{< relref "/ChapterFour/1200~1299/1235.Maximum-Profit-in-Job-Scheduling.md" >}})|Hard||||53.4%|
+|1283|Find the Smallest Divisor Given a Threshold|[Go]({{< relref "/ChapterFour/1200~1299/1283.Find-the-Smallest-Divisor-Given-a-Threshold.md" >}})|Medium||||56.2%|
+|1300|Sum of Mutated Array Closest to Target|[Go]({{< relref "/ChapterFour/1300~1399/1300.Sum-of-Mutated-Array-Closest-to-Target.md" >}})|Medium||||43.6%|
+|1337|The K Weakest Rows in a Matrix|[Go]({{< relref "/ChapterFour/1300~1399/1337.The-K-Weakest-Rows-in-a-Matrix.md" >}})|Easy||||72.1%|
+|1385|Find the Distance Value Between Two Arrays|[Go]({{< relref "/ChapterFour/1300~1399/1385.Find-the-Distance-Value-Between-Two-Arrays.md" >}})|Easy||||66.6%|
+|1439|Find the Kth Smallest Sum of a Matrix With Sorted Rows|[Go]({{< relref "/ChapterFour/1400~1499/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows.md" >}})|Hard||||61.4%|
+|1482|Minimum Number of Days to Make m Bouquets|[Go]({{< relref "/ChapterFour/1400~1499/1482.Minimum-Number-of-Days-to-Make-m-Bouquets.md" >}})|Medium||||54.0%|
+|1539|Kth Missing Positive Number|[Go]({{< relref "/ChapterFour/1500~1599/1539.Kth-Missing-Positive-Number.md" >}})|Easy||||58.6%|
+|1608|Special Array With X Elements Greater Than or Equal X|[Go]({{< relref "/ChapterFour/1600~1699/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X.md" >}})|Easy||||60.5%|
+|1631|Path With Minimum Effort|[Go]({{< relref "/ChapterFour/1600~1699/1631.Path-With-Minimum-Effort.md" >}})|Medium||||55.7%|
+|1648|Sell Diminishing-Valued Colored Balls|[Go]({{< relref "/ChapterFour/1600~1699/1648.Sell-Diminishing-Valued-Colored-Balls.md" >}})|Medium||||30.4%|
+|1649|Create Sorted Array through Instructions|[Go]({{< relref "/ChapterFour/1600~1699/1649.Create-Sorted-Array-through-Instructions.md" >}})|Hard||||37.5%|
+|1658|Minimum Operations to Reduce X to Zero|[Go]({{< relref "/ChapterFour/1600~1699/1658.Minimum-Operations-to-Reduce-X-to-Zero.md" >}})|Medium||||37.6%|
+|1818|Minimum Absolute Sum Difference|[Go]({{< relref "/ChapterFour/1800~1899/1818.Minimum-Absolute-Sum-Difference.md" >}})|Medium||||30.4%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Breadth_First_Search/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Math/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Bit_Manipulation.md b/website/content/ChapterTwo/Bit_Manipulation.md
new file mode 100644
index 000000000..e1c9ac95d
--- /dev/null
+++ b/website/content/ChapterTwo/Bit_Manipulation.md
@@ -0,0 +1,113 @@
+---
+title: 2.15 ✅ Bit Manipulation
+type: docs
+weight: 15
+---
+
+# Bit Manipulation
+
+![](https://img.halfrost.com/Leetcode/Bit_Manipulation.png)
+
+- 异或的特性。第 136 题，第 268 题，第 389 题，第 421 题，
+
+```go
+x ^ 0 = x
+x ^ 11111……1111 = ~x
+x ^ (~x) = 11111……1111
+x ^ x = 0
+a ^ b = c  => a ^ c = b  => b ^ c = a (交换律)
+a ^ b ^ c = a ^ (b ^ c) = (a ^ b）^ c (结合律)
+```
+
+- 构造特殊 Mask，将特殊位置放 0 或 1。
+
+```go
+将 x 最右边的 n 位清零， x & ( ~0 << n )
+获取 x 的第 n 位值(0 或者 1)，(x >> n) & 1
+获取 x 的第 n 位的幂值，x & (1 << (n - 1))
+仅将第 n 位置为 1，x | (1 << n)
+仅将第 n 位置为 0，x & (~(1 << n))
+将 x 最高位至第 n 位(含)清零，x & ((1 << n) - 1)
+将第 n 位至第 0 位(含)清零，x & (~((1 << (n + 1)) - 1)）
+```
+
+- 有特殊意义的 & 位操作运算。第 260 题，第 201 题，第 318 题，第 371 题，第 397 题，第 461 题，第 693 题，
+
+```go
+X & 1 == 1 判断是否是奇数(偶数)
+X & = (X - 1) 将最低位(LSB)的 1 清零
+X & -X 得到最低位(LSB)的 1
+X & ~X = 0
+```
+
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0029|Divide Two Integers|[Go]({{< relref "/ChapterFour/0001~0099/0029.Divide-Two-Integers.md" >}})|Medium||||17.2%|
+|0067|Add Binary|[Go]({{< relref "/ChapterFour/0001~0099/0067.Add-Binary.md" >}})|Easy||||52.4%|
+|0078|Subsets|[Go]({{< relref "/ChapterFour/0001~0099/0078.Subsets.md" >}})|Medium| O(n^2)| O(n)|❤️|74.9%|
+|0089|Gray Code|[Go]({{< relref "/ChapterFour/0001~0099/0089.Gray-Code.md" >}})|Medium||||57.2%|
+|0090|Subsets II|[Go]({{< relref "/ChapterFour/0001~0099/0090.Subsets-II.md" >}})|Medium||||55.9%|
+|0136|Single Number|[Go]({{< relref "/ChapterFour/0100~0199/0136.Single-Number.md" >}})|Easy| O(n)| O(1)||70.7%|
+|0137|Single Number II|[Go]({{< relref "/ChapterFour/0100~0199/0137.Single-Number-II.md" >}})|Medium| O(n)| O(1)|❤️|58.5%|
+|0187|Repeated DNA Sequences|[Go]({{< relref "/ChapterFour/0100~0199/0187.Repeated-DNA-Sequences.md" >}})|Medium| O(n)| O(1)||47.0%|
+|0190|Reverse Bits|[Go]({{< relref "/ChapterFour/0100~0199/0190.Reverse-Bits.md" >}})|Easy| O(n)| O(1)|❤️|54.0%|
+|0191|Number of 1 Bits|[Go]({{< relref "/ChapterFour/0100~0199/0191.Number-of-1-Bits.md" >}})|Easy| O(n)| O(1)||66.6%|
+|0201|Bitwise AND of Numbers Range|[Go]({{< relref "/ChapterFour/0200~0299/0201.Bitwise-AND-of-Numbers-Range.md" >}})|Medium| O(n)| O(1)|❤️|42.5%|
+|0231|Power of Two|[Go]({{< relref "/ChapterFour/0200~0299/0231.Power-of-Two.md" >}})|Easy| O(1)| O(1)||46.0%|
+|0260|Single Number III|[Go]({{< relref "/ChapterFour/0200~0299/0260.Single-Number-III.md" >}})|Medium| O(n)| O(1)|❤️|67.7%|
+|0268|Missing Number|[Go]({{< relref "/ChapterFour/0200~0299/0268.Missing-Number.md" >}})|Easy| O(n)| O(1)||62.6%|
+|0287|Find the Duplicate Number|[Go]({{< relref "/ChapterFour/0200~0299/0287.Find-the-Duplicate-Number.md" >}})|Medium||||59.1%|
+|0318|Maximum Product of Word Lengths|[Go]({{< relref "/ChapterFour/0300~0399/0318.Maximum-Product-of-Word-Lengths.md" >}})|Medium| O(n)| O(1)||59.9%|
+|0338|Counting Bits|[Go]({{< relref "/ChapterFour/0300~0399/0338.Counting-Bits.md" >}})|Easy| O(n)| O(n)||75.8%|
+|0342|Power of Four|[Go]({{< relref "/ChapterFour/0300~0399/0342.Power-of-Four.md" >}})|Easy| O(n)| O(1)||46.2%|
+|0371|Sum of Two Integers|[Go]({{< relref "/ChapterFour/0300~0399/0371.Sum-of-Two-Integers.md" >}})|Medium| O(n)| O(1)||50.7%|
+|0389|Find the Difference|[Go]({{< relref "/ChapterFour/0300~0399/0389.Find-the-Difference.md" >}})|Easy| O(n)| O(1)||59.9%|
+|0393|UTF-8 Validation|[Go]({{< relref "/ChapterFour/0300~0399/0393.UTF-8-Validation.md" >}})|Medium| O(n)| O(1)||45.1%|
+|0397|Integer Replacement|[Go]({{< relref "/ChapterFour/0300~0399/0397.Integer-Replacement.md" >}})|Medium| O(n)| O(1)||35.2%|
+|0401|Binary Watch|[Go]({{< relref "/ChapterFour/0400~0499/0401.Binary-Watch.md" >}})|Easy| O(1)| O(1)||52.3%|
+|0405|Convert a Number to Hexadecimal|[Go]({{< relref "/ChapterFour/0400~0499/0405.Convert-a-Number-to-Hexadecimal.md" >}})|Easy| O(n)| O(1)||46.8%|
+|0421|Maximum XOR of Two Numbers in an Array|[Go]({{< relref "/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array.md" >}})|Medium| O(n)| O(1)|❤️|54.0%|
+|0461|Hamming Distance|[Go]({{< relref "/ChapterFour/0400~0499/0461.Hamming-Distance.md" >}})|Easy| O(n)| O(1)||75.0%|
+|0473|Matchsticks to Square|[Go]({{< relref "/ChapterFour/0400~0499/0473.Matchsticks-to-Square.md" >}})|Medium||||40.2%|
+|0476|Number Complement|[Go]({{< relref "/ChapterFour/0400~0499/0476.Number-Complement.md" >}})|Easy| O(n)| O(1)||67.4%|
+|0477|Total Hamming Distance|[Go]({{< relref "/ChapterFour/0400~0499/0477.Total-Hamming-Distance.md" >}})|Medium| O(n)| O(1)||52.2%|
+|0491|Non-decreasing Subsequences|[Go]({{< relref "/ChapterFour/0400~0499/0491.Non-decreasing-Subsequences.md" >}})|Medium||||60.2%|
+|0526|Beautiful Arrangement|[Go]({{< relref "/ChapterFour/0500~0599/0526.Beautiful-Arrangement.md" >}})|Medium||||64.4%|
+|0638|Shopping Offers|[Go]({{< relref "/ChapterFour/0600~0699/0638.Shopping-Offers.md" >}})|Medium||||53.3%|
+|0645|Set Mismatch|[Go]({{< relref "/ChapterFour/0600~0699/0645.Set-Mismatch.md" >}})|Easy||||42.7%|
+|0693|Binary Number with Alternating Bits|[Go]({{< relref "/ChapterFour/0600~0699/0693.Binary-Number-with-Alternating-Bits.md" >}})|Easy| O(n)| O(1)|❤️|61.6%|
+|0756|Pyramid Transition Matrix|[Go]({{< relref "/ChapterFour/0700~0799/0756.Pyramid-Transition-Matrix.md" >}})|Medium| O(n log n)| O(n)||52.7%|
+|0762|Prime Number of Set Bits in Binary Representation|[Go]({{< relref "/ChapterFour/0700~0799/0762.Prime-Number-of-Set-Bits-in-Binary-Representation.md" >}})|Easy| O(n)| O(1)||68.0%|
+|0784|Letter Case Permutation|[Go]({{< relref "/ChapterFour/0700~0799/0784.Letter-Case-Permutation.md" >}})|Medium| O(n)| O(1)||73.8%|
+|0810|Chalkboard XOR Game|[Go]({{< relref "/ChapterFour/0800~0899/0810.Chalkboard-XOR-Game.md" >}})|Hard||||55.8%|
+|0864|Shortest Path to Get All Keys|[Go]({{< relref "/ChapterFour/0800~0899/0864.Shortest-Path-to-Get-All-Keys.md" >}})|Hard||||45.6%|
+|0898|Bitwise ORs of Subarrays|[Go]({{< relref "/ChapterFour/0800~0899/0898.Bitwise-ORs-of-Subarrays.md" >}})|Medium| O(n)| O(1)||37.2%|
+|0980|Unique Paths III|[Go]({{< relref "/ChapterFour/0900~0999/0980.Unique-Paths-III.md" >}})|Hard||||81.7%|
+|0995|Minimum Number of K Consecutive Bit Flips|[Go]({{< relref "/ChapterFour/0900~0999/0995.Minimum-Number-of-K-Consecutive-Bit-Flips.md" >}})|Hard||||51.2%|
+|0996|Number of Squareful Arrays|[Go]({{< relref "/ChapterFour/0900~0999/0996.Number-of-Squareful-Arrays.md" >}})|Hard||||49.2%|
+|1009|Complement of Base 10 Integer|[Go]({{< relref "/ChapterFour/1000~1099/1009.Complement-of-Base-10-Integer.md" >}})|Easy||||61.5%|
+|1178|Number of Valid Words for Each Puzzle|[Go]({{< relref "/ChapterFour/1100~1199/1178.Number-of-Valid-Words-for-Each-Puzzle.md" >}})|Hard||||46.3%|
+|1239|Maximum Length of a Concatenated String with Unique Characters|[Go]({{< relref "/ChapterFour/1200~1299/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters.md" >}})|Medium||||52.2%|
+|1310|XOR Queries of a Subarray|[Go]({{< relref "/ChapterFour/1300~1399/1310.XOR-Queries-of-a-Subarray.md" >}})|Medium||||72.3%|
+|1442|Count Triplets That Can Form Two Arrays of Equal XOR|[Go]({{< relref "/ChapterFour/1400~1499/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR.md" >}})|Medium||||76.1%|
+|1461|Check If a String Contains All Binary Codes of Size K|[Go]({{< relref "/ChapterFour/1400~1499/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K.md" >}})|Medium||||56.6%|
+|1486|XOR Operation in an Array|[Go]({{< relref "/ChapterFour/1400~1499/1486.XOR-Operation-in-an-Array.md" >}})|Easy||||84.6%|
+|1655|Distribute Repeating Integers|[Go]({{< relref "/ChapterFour/1600~1699/1655.Distribute-Repeating-Integers.md" >}})|Hard||||39.3%|
+|1659|Maximize Grid Happiness|[Go]({{< relref "/ChapterFour/1600~1699/1659.Maximize-Grid-Happiness.md" >}})|Hard||||38.8%|
+|1680|Concatenation of Consecutive Binary Numbers|[Go]({{< relref "/ChapterFour/1600~1699/1680.Concatenation-of-Consecutive-Binary-Numbers.md" >}})|Medium||||57.0%|
+|1681|Minimum Incompatibility|[Go]({{< relref "/ChapterFour/1600~1699/1681.Minimum-Incompatibility.md" >}})|Hard||||37.8%|
+|1684|Count the Number of Consistent Strings|[Go]({{< relref "/ChapterFour/1600~1699/1684.Count-the-Number-of-Consistent-Strings.md" >}})|Easy||||82.3%|
+|1720|Decode XORed Array|[Go]({{< relref "/ChapterFour/1700~1799/1720.Decode-XORed-Array.md" >}})|Easy||||85.8%|
+|1734|Decode XORed Permutation|[Go]({{< relref "/ChapterFour/1700~1799/1734.Decode-XORed-Permutation.md" >}})|Medium||||63.0%|
+|1738|Find Kth Largest XOR Coordinate Value|[Go]({{< relref "/ChapterFour/1700~1799/1738.Find-Kth-Largest-XOR-Coordinate-Value.md" >}})|Medium||||61.0%|
+|1763|Longest Nice Substring|[Go]({{< relref "/ChapterFour/1700~1799/1763.Longest-Nice-Substring.md" >}})|Easy||||61.5%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Sorting/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Union_Find/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Breadth_First_Search.md b/website/content/ChapterTwo/Breadth_First_Search.md
new file mode 100644
index 000000000..631b6dff9
--- /dev/null
+++ b/website/content/ChapterTwo/Breadth_First_Search.md
@@ -0,0 +1,99 @@
+---
+title: 2.10 Breadth First Search
+type: docs
+weight: 10
+---
+
+# Breadth First Search
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0100|Same Tree|[Go]({{< relref "/ChapterFour/0100~0199/0100.Same-Tree.md" >}})|Easy||||58.2%|
+|0101|Symmetric Tree|[Go]({{< relref "/ChapterFour/0100~0199/0101.Symmetric-Tree.md" >}})|Easy| O(n)| O(1)||54.3%|
+|0102|Binary Tree Level Order Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0102.Binary-Tree-Level-Order-Traversal.md" >}})|Medium| O(n)| O(1)||64.4%|
+|0103|Binary Tree Zigzag Level Order Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0103.Binary-Tree-Zigzag-Level-Order-Traversal.md" >}})|Medium| O(n)| O(n)||56.9%|
+|0104|Maximum Depth of Binary Tree|[Go]({{< relref "/ChapterFour/0100~0199/0104.Maximum-Depth-of-Binary-Tree.md" >}})|Easy||||73.9%|
+|0107|Binary Tree Level Order Traversal II|[Go]({{< relref "/ChapterFour/0100~0199/0107.Binary-Tree-Level-Order-Traversal-II.md" >}})|Medium| O(n)| O(1)||61.2%|
+|0111|Minimum Depth of Binary Tree|[Go]({{< relref "/ChapterFour/0100~0199/0111.Minimum-Depth-of-Binary-Tree.md" >}})|Easy| O(n)| O(1)||44.5%|
+|0112|Path Sum|[Go]({{< relref "/ChapterFour/0100~0199/0112.Path-Sum.md" >}})|Easy||||48.3%|
+|0116|Populating Next Right Pointers in Each Node|[Go]({{< relref "/ChapterFour/0100~0199/0116.Populating-Next-Right-Pointers-in-Each-Node.md" >}})|Medium||||60.4%|
+|0126|Word Ladder II|[Go]({{< relref "/ChapterFour/0100~0199/0126.Word-Ladder-II.md" >}})|Hard| O(n)| O(n^2)|❤️|27.5%|
+|0127|Word Ladder|[Go]({{< relref "/ChapterFour/0100~0199/0127.Word-Ladder.md" >}})|Hard| O(n)| O(n)||37.2%|
+|0130|Surrounded Regions|[Go]({{< relref "/ChapterFour/0100~0199/0130.Surrounded-Regions.md" >}})|Medium||||36.8%|
+|0199|Binary Tree Right Side View|[Go]({{< relref "/ChapterFour/0100~0199/0199.Binary-Tree-Right-Side-View.md" >}})|Medium| O(n)| O(1)||61.6%|
+|0200|Number of Islands|[Go]({{< relref "/ChapterFour/0200~0299/0200.Number-of-Islands.md" >}})|Medium| O(n^2)| O(n^2)||57.0%|
+|0207|Course Schedule|[Go]({{< relref "/ChapterFour/0200~0299/0207.Course-Schedule.md" >}})|Medium| O(n^2)| O(n^2)||45.4%|
+|0210|Course Schedule II|[Go]({{< relref "/ChapterFour/0200~0299/0210.Course-Schedule-II.md" >}})|Medium| O(n^2)| O(n^2)||48.5%|
+|0226|Invert Binary Tree|[Go]({{< relref "/ChapterFour/0200~0299/0226.Invert-Binary-Tree.md" >}})|Easy||||74.7%|
+|0279|Perfect Squares|[Go]({{< relref "/ChapterFour/0200~0299/0279.Perfect-Squares.md" >}})|Medium||||52.7%|
+|0297|Serialize and Deserialize Binary Tree|[Go]({{< relref "/ChapterFour/0200~0299/0297.Serialize-and-Deserialize-Binary-Tree.md" >}})|Hard||||55.4%|
+|0301|Remove Invalid Parentheses|[Go]({{< relref "/ChapterFour/0300~0399/0301.Remove-Invalid-Parentheses.md" >}})|Hard||||47.2%|
+|0322|Coin Change|[Go]({{< relref "/ChapterFour/0300~0399/0322.Coin-Change.md" >}})|Medium||||42.1%|
+|0329|Longest Increasing Path in a Matrix|[Go]({{< relref "/ChapterFour/0300~0399/0329.Longest-Increasing-Path-in-a-Matrix.md" >}})|Hard||||52.4%|
+|0399|Evaluate Division|[Go]({{< relref "/ChapterFour/0300~0399/0399.Evaluate-Division.md" >}})|Medium||||59.7%|
+|0404|Sum of Left Leaves|[Go]({{< relref "/ChapterFour/0400~0499/0404.Sum-of-Left-Leaves.md" >}})|Easy||||56.7%|
+|0417|Pacific Atlantic Water Flow|[Go]({{< relref "/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow.md" >}})|Medium||||54.4%|
+|0429|N-ary Tree Level Order Traversal|[Go]({{< relref "/ChapterFour/0400~0499/0429.N-ary-Tree-Level-Order-Traversal.md" >}})|Medium||||70.7%|
+|0433|Minimum Genetic Mutation|[Go]({{< relref "/ChapterFour/0400~0499/0433.Minimum-Genetic-Mutation.md" >}})|Medium||||52.4%|
+|0463|Island Perimeter|[Go]({{< relref "/ChapterFour/0400~0499/0463.Island-Perimeter.md" >}})|Easy||||69.7%|
+|0488|Zuma Game|[Go]({{< relref "/ChapterFour/0400~0499/0488.Zuma-Game.md" >}})|Hard||||33.9%|
+|0513|Find Bottom Left Tree Value|[Go]({{< relref "/ChapterFour/0500~0599/0513.Find-Bottom-Left-Tree-Value.md" >}})|Medium||||66.9%|
+|0515|Find Largest Value in Each Tree Row|[Go]({{< relref "/ChapterFour/0500~0599/0515.Find-Largest-Value-in-Each-Tree-Row.md" >}})|Medium| O(n)| O(n)||64.6%|
+|0529|Minesweeper|[Go]({{< relref "/ChapterFour/0500~0599/0529.Minesweeper.md" >}})|Medium||||65.7%|
+|0530|Minimum Absolute Difference in BST|[Go]({{< relref "/ChapterFour/0500~0599/0530.Minimum-Absolute-Difference-in-BST.md" >}})|Easy||||57.3%|
+|0542|01 Matrix|[Go]({{< relref "/ChapterFour/0500~0599/0542.01-Matrix.md" >}})|Medium| O(n)| O(1)||44.8%|
+|0547|Number of Provinces|[Go]({{< relref "/ChapterFour/0500~0599/0547.Number-of-Provinces.md" >}})|Medium||||63.8%|
+|0559|Maximum Depth of N-ary Tree|[Go]({{< relref "/ChapterFour/0500~0599/0559.Maximum-Depth-of-N-ary-Tree.md" >}})|Easy||||71.7%|
+|0617|Merge Two Binary Trees|[Go]({{< relref "/ChapterFour/0600~0699/0617.Merge-Two-Binary-Trees.md" >}})|Easy||||78.7%|
+|0623|Add One Row to Tree|[Go]({{< relref "/ChapterFour/0600~0699/0623.Add-One-Row-to-Tree.md" >}})|Medium||||59.5%|
+|0637|Average of Levels in Binary Tree|[Go]({{< relref "/ChapterFour/0600~0699/0637.Average-of-Levels-in-Binary-Tree.md" >}})|Easy||||71.7%|
+|0653|Two Sum IV - Input is a BST|[Go]({{< relref "/ChapterFour/0600~0699/0653.Two-Sum-IV-Input-is-a-BST.md" >}})|Easy||||61.0%|
+|0662|Maximum Width of Binary Tree|[Go]({{< relref "/ChapterFour/0600~0699/0662.Maximum-Width-of-Binary-Tree.md" >}})|Medium||||40.7%|
+|0684|Redundant Connection|[Go]({{< relref "/ChapterFour/0600~0699/0684.Redundant-Connection.md" >}})|Medium||||62.2%|
+|0685|Redundant Connection II|[Go]({{< relref "/ChapterFour/0600~0699/0685.Redundant-Connection-II.md" >}})|Hard||||34.1%|
+|0690|Employee Importance|[Go]({{< relref "/ChapterFour/0600~0699/0690.Employee-Importance.md" >}})|Medium||||65.6%|
+|0695|Max Area of Island|[Go]({{< relref "/ChapterFour/0600~0699/0695.Max-Area-of-Island.md" >}})|Medium||||71.8%|
+|0721|Accounts Merge|[Go]({{< relref "/ChapterFour/0700~0799/0721.Accounts-Merge.md" >}})|Medium||||56.3%|
+|0733|Flood Fill|[Go]({{< relref "/ChapterFour/0700~0799/0733.Flood-Fill.md" >}})|Easy||||62.0%|
+|0752|Open the Lock|[Go]({{< relref "/ChapterFour/0700~0799/0752.Open-the-Lock.md" >}})|Medium||||55.6%|
+|0756|Pyramid Transition Matrix|[Go]({{< relref "/ChapterFour/0700~0799/0756.Pyramid-Transition-Matrix.md" >}})|Medium||||52.7%|
+|0765|Couples Holding Hands|[Go]({{< relref "/ChapterFour/0700~0799/0765.Couples-Holding-Hands.md" >}})|Hard||||56.6%|
+|0778|Swim in Rising Water|[Go]({{< relref "/ChapterFour/0700~0799/0778.Swim-in-Rising-Water.md" >}})|Hard||||59.8%|
+|0783|Minimum Distance Between BST Nodes|[Go]({{< relref "/ChapterFour/0700~0799/0783.Minimum-Distance-Between-BST-Nodes.md" >}})|Easy||||59.3%|
+|0785|Is Graph Bipartite?|[Go]({{< relref "/ChapterFour/0700~0799/0785.Is-Graph-Bipartite.md" >}})|Medium||||53.1%|
+|0802|Find Eventual Safe States|[Go]({{< relref "/ChapterFour/0800~0899/0802.Find-Eventual-Safe-States.md" >}})|Medium||||56.6%|
+|0815|Bus Routes|[Go]({{< relref "/ChapterFour/0800~0899/0815.Bus-Routes.md" >}})|Hard||||45.6%|
+|0839|Similar String Groups|[Go]({{< relref "/ChapterFour/0800~0899/0839.Similar-String-Groups.md" >}})|Hard||||48.0%|
+|0841|Keys and Rooms|[Go]({{< relref "/ChapterFour/0800~0899/0841.Keys-and-Rooms.md" >}})|Medium||||71.5%|
+|0863|All Nodes Distance K in Binary Tree|[Go]({{< relref "/ChapterFour/0800~0899/0863.All-Nodes-Distance-K-in-Binary-Tree.md" >}})|Medium||||62.3%|
+|0864|Shortest Path to Get All Keys|[Go]({{< relref "/ChapterFour/0800~0899/0864.Shortest-Path-to-Get-All-Keys.md" >}})|Hard||||45.6%|
+|0909|Snakes and Ladders|[Go]({{< relref "/ChapterFour/0900~0999/0909.Snakes-and-Ladders.md" >}})|Medium||||45.1%|
+|0924|Minimize Malware Spread|[Go]({{< relref "/ChapterFour/0900~0999/0924.Minimize-Malware-Spread.md" >}})|Hard||||42.1%|
+|0928|Minimize Malware Spread II|[Go]({{< relref "/ChapterFour/0900~0999/0928.Minimize-Malware-Spread-II.md" >}})|Hard||||42.8%|
+|0958|Check Completeness of a Binary Tree|[Go]({{< relref "/ChapterFour/0900~0999/0958.Check-Completeness-of-a-Binary-Tree.md" >}})|Medium||||56.2%|
+|0959|Regions Cut By Slashes|[Go]({{< relref "/ChapterFour/0900~0999/0959.Regions-Cut-By-Slashes.md" >}})|Medium||||69.1%|
+|0987|Vertical Order Traversal of a Binary Tree|[Go]({{< relref "/ChapterFour/0900~0999/0987.Vertical-Order-Traversal-of-a-Binary-Tree.md" >}})|Hard||||45.1%|
+|0993|Cousins in Binary Tree|[Go]({{< relref "/ChapterFour/0900~0999/0993.Cousins-in-Binary-Tree.md" >}})|Easy| O(n)| O(1)||54.6%|
+|1020|Number of Enclaves|[Go]({{< relref "/ChapterFour/1000~1099/1020.Number-of-Enclaves.md" >}})|Medium||||65.6%|
+|1034|Coloring A Border|[Go]({{< relref "/ChapterFour/1000~1099/1034.Coloring-A-Border.md" >}})|Medium||||49.2%|
+|1091|Shortest Path in Binary Matrix|[Go]({{< relref "/ChapterFour/1000~1099/1091.Shortest-Path-in-Binary-Matrix.md" >}})|Medium||||44.7%|
+|1123|Lowest Common Ancestor of Deepest Leaves|[Go]({{< relref "/ChapterFour/1100~1199/1123.Lowest-Common-Ancestor-of-Deepest-Leaves.md" >}})|Medium||||70.9%|
+|1202|Smallest String With Swaps|[Go]({{< relref "/ChapterFour/1200~1299/1202.Smallest-String-With-Swaps.md" >}})|Medium||||57.7%|
+|1203|Sort Items by Groups Respecting Dependencies|[Go]({{< relref "/ChapterFour/1200~1299/1203.Sort-Items-by-Groups-Respecting-Dependencies.md" >}})|Hard||||51.2%|
+|1254|Number of Closed Islands|[Go]({{< relref "/ChapterFour/1200~1299/1254.Number-of-Closed-Islands.md" >}})|Medium||||66.9%|
+|1293|Shortest Path in a Grid with Obstacles Elimination|[Go]({{< relref "/ChapterFour/1200~1299/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination.md" >}})|Hard||||45.3%|
+|1302|Deepest Leaves Sum|[Go]({{< relref "/ChapterFour/1300~1399/1302.Deepest-Leaves-Sum.md" >}})|Medium||||86.6%|
+|1306|Jump Game III|[Go]({{< relref "/ChapterFour/1300~1399/1306.Jump-Game-III.md" >}})|Medium||||63.5%|
+|1319|Number of Operations to Make Network Connected|[Go]({{< relref "/ChapterFour/1300~1399/1319.Number-of-Operations-to-Make-Network-Connected.md" >}})|Medium||||62.1%|
+|1609|Even Odd Tree|[Go]({{< relref "/ChapterFour/1600~1699/1609.Even-Odd-Tree.md" >}})|Medium||||54.4%|
+|1631|Path With Minimum Effort|[Go]({{< relref "/ChapterFour/1600~1699/1631.Path-With-Minimum-Effort.md" >}})|Medium||||55.7%|
+|1654|Minimum Jumps to Reach Home|[Go]({{< relref "/ChapterFour/1600~1699/1654.Minimum-Jumps-to-Reach-Home.md" >}})|Medium||||29.1%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Depth_First_Search/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Binary_Search/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Depth_First_Search.md b/website/content/ChapterTwo/Depth_First_Search.md
new file mode 100644
index 000000000..ac15634e1
--- /dev/null
+++ b/website/content/ChapterTwo/Depth_First_Search.md
@@ -0,0 +1,126 @@
+---
+title: 2.09 Depth First Search
+type: docs
+weight: 9
+---
+
+# Depth First Search
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0094|Binary Tree Inorder Traversal|[Go]({{< relref "/ChapterFour/0001~0099/0094.Binary-Tree-Inorder-Traversal.md" >}})|Easy||||73.8%|
+|0098|Validate Binary Search Tree|[Go]({{< relref "/ChapterFour/0001~0099/0098.Validate-Binary-Search-Tree.md" >}})|Medium| O(n)| O(1)||32.0%|
+|0099|Recover Binary Search Tree|[Go]({{< relref "/ChapterFour/0001~0099/0099.Recover-Binary-Search-Tree.md" >}})|Medium| O(n)| O(1)||51.0%|
+|0100|Same Tree|[Go]({{< relref "/ChapterFour/0100~0199/0100.Same-Tree.md" >}})|Easy| O(n)| O(1)||58.2%|
+|0101|Symmetric Tree|[Go]({{< relref "/ChapterFour/0100~0199/0101.Symmetric-Tree.md" >}})|Easy| O(n)| O(1)||54.3%|
+|0104|Maximum Depth of Binary Tree|[Go]({{< relref "/ChapterFour/0100~0199/0104.Maximum-Depth-of-Binary-Tree.md" >}})|Easy| O(n)| O(1)||73.9%|
+|0110|Balanced Binary Tree|[Go]({{< relref "/ChapterFour/0100~0199/0110.Balanced-Binary-Tree.md" >}})|Easy| O(n)| O(1)||49.1%|
+|0111|Minimum Depth of Binary Tree|[Go]({{< relref "/ChapterFour/0100~0199/0111.Minimum-Depth-of-Binary-Tree.md" >}})|Easy| O(n)| O(1)||44.5%|
+|0112|Path Sum|[Go]({{< relref "/ChapterFour/0100~0199/0112.Path-Sum.md" >}})|Easy| O(n)| O(1)||48.3%|
+|0113|Path Sum II|[Go]({{< relref "/ChapterFour/0100~0199/0113.Path-Sum-II.md" >}})|Medium| O(n)| O(1)||57.1%|
+|0114|Flatten Binary Tree to Linked List|[Go]({{< relref "/ChapterFour/0100~0199/0114.Flatten-Binary-Tree-to-Linked-List.md" >}})|Medium| O(n)| O(1)||61.8%|
+|0116|Populating Next Right Pointers in Each Node|[Go]({{< relref "/ChapterFour/0100~0199/0116.Populating-Next-Right-Pointers-in-Each-Node.md" >}})|Medium||||60.4%|
+|0124|Binary Tree Maximum Path Sum|[Go]({{< relref "/ChapterFour/0100~0199/0124.Binary-Tree-Maximum-Path-Sum.md" >}})|Hard| O(n)| O(1)||39.2%|
+|0129|Sum Root to Leaf Numbers|[Go]({{< relref "/ChapterFour/0100~0199/0129.Sum-Root-to-Leaf-Numbers.md" >}})|Medium| O(n)| O(1)||61.0%|
+|0130|Surrounded Regions|[Go]({{< relref "/ChapterFour/0100~0199/0130.Surrounded-Regions.md" >}})|Medium||||36.8%|
+|0144|Binary Tree Preorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0144.Binary-Tree-Preorder-Traversal.md" >}})|Easy||||66.9%|
+|0145|Binary Tree Postorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0145.Binary-Tree-Postorder-Traversal.md" >}})|Easy||||68.0%|
+|0199|Binary Tree Right Side View|[Go]({{< relref "/ChapterFour/0100~0199/0199.Binary-Tree-Right-Side-View.md" >}})|Medium| O(n)| O(1)||61.6%|
+|0200|Number of Islands|[Go]({{< relref "/ChapterFour/0200~0299/0200.Number-of-Islands.md" >}})|Medium| O(n^2)| O(n^2)||57.0%|
+|0207|Course Schedule|[Go]({{< relref "/ChapterFour/0200~0299/0207.Course-Schedule.md" >}})|Medium| O(n^2)| O(n^2)||45.4%|
+|0210|Course Schedule II|[Go]({{< relref "/ChapterFour/0200~0299/0210.Course-Schedule-II.md" >}})|Medium| O(n^2)| O(n^2)||48.5%|
+|0211|Design Add and Search Words Data Structure|[Go]({{< relref "/ChapterFour/0200~0299/0211.Design-Add-and-Search-Words-Data-Structure.md" >}})|Medium||||44.0%|
+|0222|Count Complete Tree Nodes|[Go]({{< relref "/ChapterFour/0200~0299/0222.Count-Complete-Tree-Nodes.md" >}})|Medium||||60.6%|
+|0226|Invert Binary Tree|[Go]({{< relref "/ChapterFour/0200~0299/0226.Invert-Binary-Tree.md" >}})|Easy||||74.7%|
+|0230|Kth Smallest Element in a BST|[Go]({{< relref "/ChapterFour/0200~0299/0230.Kth-Smallest-Element-in-a-BST.md" >}})|Medium||||70.2%|
+|0235|Lowest Common Ancestor of a Binary Search Tree|[Go]({{< relref "/ChapterFour/0200~0299/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree.md" >}})|Medium||||61.6%|
+|0236|Lowest Common Ancestor of a Binary Tree|[Go]({{< relref "/ChapterFour/0200~0299/0236.Lowest-Common-Ancestor-of-a-Binary-Tree.md" >}})|Medium||||58.8%|
+|0257|Binary Tree Paths|[Go]({{< relref "/ChapterFour/0200~0299/0257.Binary-Tree-Paths.md" >}})|Easy| O(n)| O(1)||61.4%|
+|0297|Serialize and Deserialize Binary Tree|[Go]({{< relref "/ChapterFour/0200~0299/0297.Serialize-and-Deserialize-Binary-Tree.md" >}})|Hard||||55.4%|
+|0329|Longest Increasing Path in a Matrix|[Go]({{< relref "/ChapterFour/0300~0399/0329.Longest-Increasing-Path-in-a-Matrix.md" >}})|Hard||||52.4%|
+|0337|House Robber III|[Go]({{< relref "/ChapterFour/0300~0399/0337.House-Robber-III.md" >}})|Medium||||53.9%|
+|0341|Flatten Nested List Iterator|[Go]({{< relref "/ChapterFour/0300~0399/0341.Flatten-Nested-List-Iterator.md" >}})|Medium||||61.8%|
+|0385|Mini Parser|[Go]({{< relref "/ChapterFour/0300~0399/0385.Mini-Parser.md" >}})|Medium||||36.9%|
+|0386|Lexicographical Numbers|[Go]({{< relref "/ChapterFour/0300~0399/0386.Lexicographical-Numbers.md" >}})|Medium||||61.6%|
+|0399|Evaluate Division|[Go]({{< relref "/ChapterFour/0300~0399/0399.Evaluate-Division.md" >}})|Medium||||59.7%|
+|0404|Sum of Left Leaves|[Go]({{< relref "/ChapterFour/0400~0499/0404.Sum-of-Left-Leaves.md" >}})|Easy||||56.7%|
+|0417|Pacific Atlantic Water Flow|[Go]({{< relref "/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow.md" >}})|Medium||||54.4%|
+|0419|Battleships in a Board|[Go]({{< relref "/ChapterFour/0400~0499/0419.Battleships-in-a-Board.md" >}})|Medium||||74.8%|
+|0437|Path Sum III|[Go]({{< relref "/ChapterFour/0400~0499/0437.Path-Sum-III.md" >}})|Medium||||48.0%|
+|0463|Island Perimeter|[Go]({{< relref "/ChapterFour/0400~0499/0463.Island-Perimeter.md" >}})|Easy||||69.7%|
+|0508|Most Frequent Subtree Sum|[Go]({{< relref "/ChapterFour/0500~0599/0508.Most-Frequent-Subtree-Sum.md" >}})|Medium||||64.9%|
+|0513|Find Bottom Left Tree Value|[Go]({{< relref "/ChapterFour/0500~0599/0513.Find-Bottom-Left-Tree-Value.md" >}})|Medium||||66.9%|
+|0515|Find Largest Value in Each Tree Row|[Go]({{< relref "/ChapterFour/0500~0599/0515.Find-Largest-Value-in-Each-Tree-Row.md" >}})|Medium| O(n)| O(n)||64.6%|
+|0529|Minesweeper|[Go]({{< relref "/ChapterFour/0500~0599/0529.Minesweeper.md" >}})|Medium||||65.7%|
+|0530|Minimum Absolute Difference in BST|[Go]({{< relref "/ChapterFour/0500~0599/0530.Minimum-Absolute-Difference-in-BST.md" >}})|Easy||||57.3%|
+|0538|Convert BST to Greater Tree|[Go]({{< relref "/ChapterFour/0500~0599/0538.Convert-BST-to-Greater-Tree.md" >}})|Medium||||67.8%|
+|0543|Diameter of Binary Tree|[Go]({{< relref "/ChapterFour/0500~0599/0543.Diameter-of-Binary-Tree.md" >}})|Easy||||56.8%|
+|0547|Number of Provinces|[Go]({{< relref "/ChapterFour/0500~0599/0547.Number-of-Provinces.md" >}})|Medium||||63.8%|
+|0559|Maximum Depth of N-ary Tree|[Go]({{< relref "/ChapterFour/0500~0599/0559.Maximum-Depth-of-N-ary-Tree.md" >}})|Easy||||71.7%|
+|0563|Binary Tree Tilt|[Go]({{< relref "/ChapterFour/0500~0599/0563.Binary-Tree-Tilt.md" >}})|Easy||||60.1%|
+|0572|Subtree of Another Tree|[Go]({{< relref "/ChapterFour/0500~0599/0572.Subtree-of-Another-Tree.md" >}})|Easy||||46.4%|
+|0589|N-ary Tree Preorder Traversal|[Go]({{< relref "/ChapterFour/0500~0599/0589.N-ary-Tree-Preorder-Traversal.md" >}})|Easy||||75.8%|
+|0617|Merge Two Binary Trees|[Go]({{< relref "/ChapterFour/0600~0699/0617.Merge-Two-Binary-Trees.md" >}})|Easy||||78.7%|
+|0623|Add One Row to Tree|[Go]({{< relref "/ChapterFour/0600~0699/0623.Add-One-Row-to-Tree.md" >}})|Medium||||59.5%|
+|0637|Average of Levels in Binary Tree|[Go]({{< relref "/ChapterFour/0600~0699/0637.Average-of-Levels-in-Binary-Tree.md" >}})|Easy||||71.7%|
+|0653|Two Sum IV - Input is a BST|[Go]({{< relref "/ChapterFour/0600~0699/0653.Two-Sum-IV-Input-is-a-BST.md" >}})|Easy||||61.0%|
+|0662|Maximum Width of Binary Tree|[Go]({{< relref "/ChapterFour/0600~0699/0662.Maximum-Width-of-Binary-Tree.md" >}})|Medium||||40.7%|
+|0669|Trim a Binary Search Tree|[Go]({{< relref "/ChapterFour/0600~0699/0669.Trim-a-Binary-Search-Tree.md" >}})|Medium||||66.4%|
+|0684|Redundant Connection|[Go]({{< relref "/ChapterFour/0600~0699/0684.Redundant-Connection.md" >}})|Medium||||62.2%|
+|0685|Redundant Connection II|[Go]({{< relref "/ChapterFour/0600~0699/0685.Redundant-Connection-II.md" >}})|Hard||||34.1%|
+|0690|Employee Importance|[Go]({{< relref "/ChapterFour/0600~0699/0690.Employee-Importance.md" >}})|Medium||||65.6%|
+|0695|Max Area of Island|[Go]({{< relref "/ChapterFour/0600~0699/0695.Max-Area-of-Island.md" >}})|Medium||||71.8%|
+|0721|Accounts Merge|[Go]({{< relref "/ChapterFour/0700~0799/0721.Accounts-Merge.md" >}})|Medium||||56.3%|
+|0733|Flood Fill|[Go]({{< relref "/ChapterFour/0700~0799/0733.Flood-Fill.md" >}})|Easy||||62.0%|
+|0753|Cracking the Safe|[Go]({{< relref "/ChapterFour/0700~0799/0753.Cracking-the-Safe.md" >}})|Hard||||55.8%|
+|0756|Pyramid Transition Matrix|[Go]({{< relref "/ChapterFour/0700~0799/0756.Pyramid-Transition-Matrix.md" >}})|Medium||||52.7%|
+|0765|Couples Holding Hands|[Go]({{< relref "/ChapterFour/0700~0799/0765.Couples-Holding-Hands.md" >}})|Hard||||56.6%|
+|0778|Swim in Rising Water|[Go]({{< relref "/ChapterFour/0700~0799/0778.Swim-in-Rising-Water.md" >}})|Hard||||59.8%|
+|0783|Minimum Distance Between BST Nodes|[Go]({{< relref "/ChapterFour/0700~0799/0783.Minimum-Distance-Between-BST-Nodes.md" >}})|Easy||||59.3%|
+|0785|Is Graph Bipartite?|[Go]({{< relref "/ChapterFour/0700~0799/0785.Is-Graph-Bipartite.md" >}})|Medium||||53.1%|
+|0802|Find Eventual Safe States|[Go]({{< relref "/ChapterFour/0800~0899/0802.Find-Eventual-Safe-States.md" >}})|Medium||||56.6%|
+|0834|Sum of Distances in Tree|[Go]({{< relref "/ChapterFour/0800~0899/0834.Sum-of-Distances-in-Tree.md" >}})|Hard||||59.1%|
+|0839|Similar String Groups|[Go]({{< relref "/ChapterFour/0800~0899/0839.Similar-String-Groups.md" >}})|Hard||||48.0%|
+|0841|Keys and Rooms|[Go]({{< relref "/ChapterFour/0800~0899/0841.Keys-and-Rooms.md" >}})|Medium||||71.5%|
+|0851|Loud and Rich|[Go]({{< relref "/ChapterFour/0800~0899/0851.Loud-and-Rich.md" >}})|Medium||||58.4%|
+|0863|All Nodes Distance K in Binary Tree|[Go]({{< relref "/ChapterFour/0800~0899/0863.All-Nodes-Distance-K-in-Binary-Tree.md" >}})|Medium||||62.3%|
+|0872|Leaf-Similar Trees|[Go]({{< relref "/ChapterFour/0800~0899/0872.Leaf-Similar-Trees.md" >}})|Easy||||67.6%|
+|0897|Increasing Order Search Tree|[Go]({{< relref "/ChapterFour/0800~0899/0897.Increasing-Order-Search-Tree.md" >}})|Easy||||78.4%|
+|0924|Minimize Malware Spread|[Go]({{< relref "/ChapterFour/0900~0999/0924.Minimize-Malware-Spread.md" >}})|Hard||||42.1%|
+|0928|Minimize Malware Spread II|[Go]({{< relref "/ChapterFour/0900~0999/0928.Minimize-Malware-Spread-II.md" >}})|Hard||||42.8%|
+|0938|Range Sum of BST|[Go]({{< relref "/ChapterFour/0900~0999/0938.Range-Sum-of-BST.md" >}})|Easy||||85.9%|
+|0947|Most Stones Removed with Same Row or Column|[Go]({{< relref "/ChapterFour/0900~0999/0947.Most-Stones-Removed-with-Same-Row-or-Column.md" >}})|Medium||||58.9%|
+|0959|Regions Cut By Slashes|[Go]({{< relref "/ChapterFour/0900~0999/0959.Regions-Cut-By-Slashes.md" >}})|Medium||||69.1%|
+|0968|Binary Tree Cameras|[Go]({{< relref "/ChapterFour/0900~0999/0968.Binary-Tree-Cameras.md" >}})|Hard||||46.6%|
+|0971|Flip Binary Tree To Match Preorder Traversal|[Go]({{< relref "/ChapterFour/0900~0999/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal.md" >}})|Medium||||50.0%|
+|0979|Distribute Coins in Binary Tree|[Go]({{< relref "/ChapterFour/0900~0999/0979.Distribute-Coins-in-Binary-Tree.md" >}})|Medium||||72.2%|
+|0987|Vertical Order Traversal of a Binary Tree|[Go]({{< relref "/ChapterFour/0900~0999/0987.Vertical-Order-Traversal-of-a-Binary-Tree.md" >}})|Hard||||45.1%|
+|0993|Cousins in Binary Tree|[Go]({{< relref "/ChapterFour/0900~0999/0993.Cousins-in-Binary-Tree.md" >}})|Easy||||54.6%|
+|1020|Number of Enclaves|[Go]({{< relref "/ChapterFour/1000~1099/1020.Number-of-Enclaves.md" >}})|Medium||||65.6%|
+|1022|Sum of Root To Leaf Binary Numbers|[Go]({{< relref "/ChapterFour/1000~1099/1022.Sum-of-Root-To-Leaf-Binary-Numbers.md" >}})|Easy||||73.5%|
+|1026|Maximum Difference Between Node and Ancestor|[Go]({{< relref "/ChapterFour/1000~1099/1026.Maximum-Difference-Between-Node-and-Ancestor.md" >}})|Medium||||75.8%|
+|1028|Recover a Tree From Preorder Traversal|[Go]({{< relref "/ChapterFour/1000~1099/1028.Recover-a-Tree-From-Preorder-Traversal.md" >}})|Hard||||73.3%|
+|1034|Coloring A Border|[Go]({{< relref "/ChapterFour/1000~1099/1034.Coloring-A-Border.md" >}})|Medium||||49.2%|
+|1038|Binary Search Tree to Greater Sum Tree|[Go]({{< relref "/ChapterFour/1000~1099/1038.Binary-Search-Tree-to-Greater-Sum-Tree.md" >}})|Medium||||85.5%|
+|1110|Delete Nodes And Return Forest|[Go]({{< relref "/ChapterFour/1100~1199/1110.Delete-Nodes-And-Return-Forest.md" >}})|Medium||||69.3%|
+|1123|Lowest Common Ancestor of Deepest Leaves|[Go]({{< relref "/ChapterFour/1100~1199/1123.Lowest-Common-Ancestor-of-Deepest-Leaves.md" >}})|Medium||||70.9%|
+|1145|Binary Tree Coloring Game|[Go]({{< relref "/ChapterFour/1100~1199/1145.Binary-Tree-Coloring-Game.md" >}})|Medium||||51.7%|
+|1202|Smallest String With Swaps|[Go]({{< relref "/ChapterFour/1200~1299/1202.Smallest-String-With-Swaps.md" >}})|Medium||||57.7%|
+|1203|Sort Items by Groups Respecting Dependencies|[Go]({{< relref "/ChapterFour/1200~1299/1203.Sort-Items-by-Groups-Respecting-Dependencies.md" >}})|Hard||||51.2%|
+|1254|Number of Closed Islands|[Go]({{< relref "/ChapterFour/1200~1299/1254.Number-of-Closed-Islands.md" >}})|Medium||||66.9%|
+|1302|Deepest Leaves Sum|[Go]({{< relref "/ChapterFour/1300~1399/1302.Deepest-Leaves-Sum.md" >}})|Medium||||86.6%|
+|1305|All Elements in Two Binary Search Trees|[Go]({{< relref "/ChapterFour/1300~1399/1305.All-Elements-in-Two-Binary-Search-Trees.md" >}})|Medium||||79.8%|
+|1306|Jump Game III|[Go]({{< relref "/ChapterFour/1300~1399/1306.Jump-Game-III.md" >}})|Medium||||63.5%|
+|1319|Number of Operations to Make Network Connected|[Go]({{< relref "/ChapterFour/1300~1399/1319.Number-of-Operations-to-Make-Network-Connected.md" >}})|Medium||||62.1%|
+|1600|Throne Inheritance|[Go]({{< relref "/ChapterFour/1600~1699/1600.Throne-Inheritance.md" >}})|Medium||||63.6%|
+|1631|Path With Minimum Effort|[Go]({{< relref "/ChapterFour/1600~1699/1631.Path-With-Minimum-Effort.md" >}})|Medium||||55.7%|
+|2096|Step-By-Step Directions From a Binary Tree Node to Another|[Go]({{< relref "/ChapterFour/2000~2099/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another.md" >}})|Medium||||48.5%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Backtracking/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Breadth_First_Search/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Dynamic_Programming.md b/website/content/ChapterTwo/Dynamic_Programming.md
new file mode 100644
index 000000000..79c094207
--- /dev/null
+++ b/website/content/ChapterTwo/Dynamic_Programming.md
@@ -0,0 +1,117 @@
+---
+title: 2.07 Dynamic Programming
+type: docs
+weight: 7
+---
+
+# Dynamic Programming
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0005|Longest Palindromic Substring|[Go]({{< relref "/ChapterFour/0001~0099/0005.Longest-Palindromic-Substring.md" >}})|Medium||||32.4%|
+|0022|Generate Parentheses|[Go]({{< relref "/ChapterFour/0001~0099/0022.Generate-Parentheses.md" >}})|Medium||||72.5%|
+|0032|Longest Valid Parentheses|[Go]({{< relref "/ChapterFour/0001~0099/0032.Longest-Valid-Parentheses.md" >}})|Hard||||32.8%|
+|0042|Trapping Rain Water|[Go]({{< relref "/ChapterFour/0001~0099/0042.Trapping-Rain-Water.md" >}})|Hard||||59.3%|
+|0045|Jump Game II|[Go]({{< relref "/ChapterFour/0001~0099/0045.Jump-Game-II.md" >}})|Medium||||39.8%|
+|0053|Maximum Subarray|[Go]({{< relref "/ChapterFour/0001~0099/0053.Maximum-Subarray.md" >}})|Medium| O(n)| O(n)||50.2%|
+|0055|Jump Game|[Go]({{< relref "/ChapterFour/0001~0099/0055.Jump-Game.md" >}})|Medium||||38.9%|
+|0062|Unique Paths|[Go]({{< relref "/ChapterFour/0001~0099/0062.Unique-Paths.md" >}})|Medium| O(n^2)| O(n^2)||62.7%|
+|0063|Unique Paths II|[Go]({{< relref "/ChapterFour/0001~0099/0063.Unique-Paths-II.md" >}})|Medium| O(n^2)| O(n^2)||39.4%|
+|0064|Minimum Path Sum|[Go]({{< relref "/ChapterFour/0001~0099/0064.Minimum-Path-Sum.md" >}})|Medium| O(n^2)| O(n^2)||62.0%|
+|0070|Climbing Stairs|[Go]({{< relref "/ChapterFour/0001~0099/0070.Climbing-Stairs.md" >}})|Easy| O(n)| O(n)||52.2%|
+|0091|Decode Ways|[Go]({{< relref "/ChapterFour/0001~0099/0091.Decode-Ways.md" >}})|Medium| O(n)| O(n)||32.7%|
+|0095|Unique Binary Search Trees II|[Go]({{< relref "/ChapterFour/0001~0099/0095.Unique-Binary-Search-Trees-II.md" >}})|Medium||||52.4%|
+|0096|Unique Binary Search Trees|[Go]({{< relref "/ChapterFour/0001~0099/0096.Unique-Binary-Search-Trees.md" >}})|Medium| O(n)| O(n)||59.7%|
+|0097|Interleaving String|[Go]({{< relref "/ChapterFour/0001~0099/0097.Interleaving-String.md" >}})|Medium||||37.3%|
+|0115|Distinct Subsequences|[Go]({{< relref "/ChapterFour/0100~0199/0115.Distinct-Subsequences.md" >}})|Hard||||44.5%|
+|0118|Pascal's Triangle|[Go]({{< relref "/ChapterFour/0100~0199/0118.Pascals-Triangle.md" >}})|Easy||||70.8%|
+|0119|Pascal's Triangle II|[Go]({{< relref "/ChapterFour/0100~0199/0119.Pascals-Triangle-II.md" >}})|Easy||||60.8%|
+|0120|Triangle|[Go]({{< relref "/ChapterFour/0100~0199/0120.Triangle.md" >}})|Medium| O(n^2)| O(n)||54.5%|
+|0121|Best Time to Buy and Sell Stock|[Go]({{< relref "/ChapterFour/0100~0199/0121.Best-Time-to-Buy-and-Sell-Stock.md" >}})|Easy| O(n)| O(1)||54.3%|
+|0122|Best Time to Buy and Sell Stock II|[Go]({{< relref "/ChapterFour/0100~0199/0122.Best-Time-to-Buy-and-Sell-Stock-II.md" >}})|Medium||||63.9%|
+|0124|Binary Tree Maximum Path Sum|[Go]({{< relref "/ChapterFour/0100~0199/0124.Binary-Tree-Maximum-Path-Sum.md" >}})|Hard||||39.2%|
+|0131|Palindrome Partitioning|[Go]({{< relref "/ChapterFour/0100~0199/0131.Palindrome-Partitioning.md" >}})|Medium||||64.9%|
+|0152|Maximum Product Subarray|[Go]({{< relref "/ChapterFour/0100~0199/0152.Maximum-Product-Subarray.md" >}})|Medium| O(n)| O(1)||34.9%|
+|0174|Dungeon Game|[Go]({{< relref "/ChapterFour/0100~0199/0174.Dungeon-Game.md" >}})|Hard||||37.5%|
+|0198|House Robber|[Go]({{< relref "/ChapterFour/0100~0199/0198.House-Robber.md" >}})|Medium| O(n)| O(n)||49.4%|
+|0213|House Robber II|[Go]({{< relref "/ChapterFour/0200~0299/0213.House-Robber-II.md" >}})|Medium| O(n)| O(n)||41.0%|
+|0264|Ugly Number II|[Go]({{< relref "/ChapterFour/0200~0299/0264.Ugly-Number-II.md" >}})|Medium||||46.2%|
+|0279|Perfect Squares|[Go]({{< relref "/ChapterFour/0200~0299/0279.Perfect-Squares.md" >}})|Medium||||52.7%|
+|0300|Longest Increasing Subsequence|[Go]({{< relref "/ChapterFour/0300~0399/0300.Longest-Increasing-Subsequence.md" >}})|Medium| O(n log n)| O(n)||52.2%|
+|0309|Best Time to Buy and Sell Stock with Cooldown|[Go]({{< relref "/ChapterFour/0300~0399/0309.Best-Time-to-Buy-and-Sell-Stock-with-Cooldown.md" >}})|Medium| O(n)| O(n)||56.2%|
+|0322|Coin Change|[Go]({{< relref "/ChapterFour/0300~0399/0322.Coin-Change.md" >}})|Medium| O(n)| O(n)||42.1%|
+|0329|Longest Increasing Path in a Matrix|[Go]({{< relref "/ChapterFour/0300~0399/0329.Longest-Increasing-Path-in-a-Matrix.md" >}})|Hard||||52.4%|
+|0337|House Robber III|[Go]({{< relref "/ChapterFour/0300~0399/0337.House-Robber-III.md" >}})|Medium||||53.9%|
+|0338|Counting Bits|[Go]({{< relref "/ChapterFour/0300~0399/0338.Counting-Bits.md" >}})|Easy| O(n)| O(n)||75.8%|
+|0343|Integer Break|[Go]({{< relref "/ChapterFour/0300~0399/0343.Integer-Break.md" >}})|Medium| O(n^2)| O(n)||56.1%|
+|0354|Russian Doll Envelopes|[Go]({{< relref "/ChapterFour/0300~0399/0354.Russian-Doll-Envelopes.md" >}})|Hard||||37.9%|
+|0357|Count Numbers with Unique Digits|[Go]({{< relref "/ChapterFour/0300~0399/0357.Count-Numbers-with-Unique-Digits.md" >}})|Medium| O(1)| O(1)||51.9%|
+|0368|Largest Divisible Subset|[Go]({{< relref "/ChapterFour/0300~0399/0368.Largest-Divisible-Subset.md" >}})|Medium||||41.6%|
+|0376|Wiggle Subsequence|[Go]({{< relref "/ChapterFour/0300~0399/0376.Wiggle-Subsequence.md" >}})|Medium||||48.3%|
+|0377|Combination Sum IV|[Go]({{< relref "/ChapterFour/0300~0399/0377.Combination-Sum-IV.md" >}})|Medium||||52.2%|
+|0392|Is Subsequence|[Go]({{< relref "/ChapterFour/0300~0399/0392.Is-Subsequence.md" >}})|Easy| O(n)| O(1)||47.5%|
+|0396|Rotate Function|[Go]({{< relref "/ChapterFour/0300~0399/0396.Rotate-Function.md" >}})|Medium||||41.1%|
+|0397|Integer Replacement|[Go]({{< relref "/ChapterFour/0300~0399/0397.Integer-Replacement.md" >}})|Medium||||35.2%|
+|0410|Split Array Largest Sum|[Go]({{< relref "/ChapterFour/0400~0499/0410.Split-Array-Largest-Sum.md" >}})|Hard||||53.5%|
+|0413|Arithmetic Slices|[Go]({{< relref "/ChapterFour/0400~0499/0413.Arithmetic-Slices.md" >}})|Medium||||65.1%|
+|0416|Partition Equal Subset Sum|[Go]({{< relref "/ChapterFour/0400~0499/0416.Partition-Equal-Subset-Sum.md" >}})|Medium| O(n^2)| O(n)||46.3%|
+|0435|Non-overlapping Intervals|[Go]({{< relref "/ChapterFour/0400~0499/0435.Non-overlapping-Intervals.md" >}})|Medium||||50.3%|
+|0458|Poor Pigs|[Go]({{< relref "/ChapterFour/0400~0499/0458.Poor-Pigs.md" >}})|Hard||||62.8%|
+|0473|Matchsticks to Square|[Go]({{< relref "/ChapterFour/0400~0499/0473.Matchsticks-to-Square.md" >}})|Medium||||40.2%|
+|0474|Ones and Zeroes|[Go]({{< relref "/ChapterFour/0400~0499/0474.Ones-and-Zeroes.md" >}})|Medium||||46.8%|
+|0488|Zuma Game|[Go]({{< relref "/ChapterFour/0400~0499/0488.Zuma-Game.md" >}})|Hard||||33.9%|
+|0494|Target Sum|[Go]({{< relref "/ChapterFour/0400~0499/0494.Target-Sum.md" >}})|Medium||||45.7%|
+|0509|Fibonacci Number|[Go]({{< relref "/ChapterFour/0500~0599/0509.Fibonacci-Number.md" >}})|Easy||||69.8%|
+|0518|Coin Change II|[Go]({{< relref "/ChapterFour/0500~0599/0518.Coin-Change-II.md" >}})|Medium||||60.6%|
+|0526|Beautiful Arrangement|[Go]({{< relref "/ChapterFour/0500~0599/0526.Beautiful-Arrangement.md" >}})|Medium||||64.4%|
+|0542|01 Matrix|[Go]({{< relref "/ChapterFour/0500~0599/0542.01-Matrix.md" >}})|Medium||||44.8%|
+|0576|Out of Boundary Paths|[Go]({{< relref "/ChapterFour/0500~0599/0576.Out-of-Boundary-Paths.md" >}})|Medium||||44.3%|
+|0583|Delete Operation for Two Strings|[Go]({{< relref "/ChapterFour/0500~0599/0583.Delete-Operation-for-Two-Strings.md" >}})|Medium||||59.8%|
+|0638|Shopping Offers|[Go]({{< relref "/ChapterFour/0600~0699/0638.Shopping-Offers.md" >}})|Medium||||53.3%|
+|0647|Palindromic Substrings|[Go]({{< relref "/ChapterFour/0600~0699/0647.Palindromic-Substrings.md" >}})|Medium||||66.9%|
+|0714|Best Time to Buy and Sell Stock with Transaction Fee|[Go]({{< relref "/ChapterFour/0700~0799/0714.Best-Time-to-Buy-and-Sell-Stock-with-Transaction-Fee.md" >}})|Medium| O(n)| O(1)||65.2%|
+|0718|Maximum Length of Repeated Subarray|[Go]({{< relref "/ChapterFour/0700~0799/0718.Maximum-Length-of-Repeated-Subarray.md" >}})|Medium||||51.3%|
+|0746|Min Cost Climbing Stairs|[Go]({{< relref "/ChapterFour/0700~0799/0746.Min-Cost-Climbing-Stairs.md" >}})|Easy| O(n)| O(1)||63.2%|
+|0823|Binary Trees With Factors|[Go]({{< relref "/ChapterFour/0800~0899/0823.Binary-Trees-With-Factors.md" >}})|Medium||||49.7%|
+|0828|Count Unique Characters of All Substrings of a Given String|[Go]({{< relref "/ChapterFour/0800~0899/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String.md" >}})|Hard||||51.6%|
+|0834|Sum of Distances in Tree|[Go]({{< relref "/ChapterFour/0800~0899/0834.Sum-of-Distances-in-Tree.md" >}})|Hard||||59.1%|
+|0838|Push Dominoes|[Go]({{< relref "/ChapterFour/0800~0899/0838.Push-Dominoes.md" >}})|Medium| O(n)| O(n)||57.0%|
+|0845|Longest Mountain in Array|[Go]({{< relref "/ChapterFour/0800~0899/0845.Longest-Mountain-in-Array.md" >}})|Medium||||40.2%|
+|0877|Stone Game|[Go]({{< relref "/ChapterFour/0800~0899/0877.Stone-Game.md" >}})|Medium||||69.7%|
+|0887|Super Egg Drop|[Go]({{< relref "/ChapterFour/0800~0899/0887.Super-Egg-Drop.md" >}})|Hard||||27.1%|
+|0898|Bitwise ORs of Subarrays|[Go]({{< relref "/ChapterFour/0800~0899/0898.Bitwise-ORs-of-Subarrays.md" >}})|Medium||||37.2%|
+|0907|Sum of Subarray Minimums|[Go]({{< relref "/ChapterFour/0900~0999/0907.Sum-of-Subarray-Minimums.md" >}})|Medium||||35.8%|
+|0918|Maximum Sum Circular Subarray|[Go]({{< relref "/ChapterFour/0900~0999/0918.Maximum-Sum-Circular-Subarray.md" >}})|Medium||||43.0%|
+|0920|Number of Music Playlists|[Go]({{< relref "/ChapterFour/0900~0999/0920.Number-of-Music-Playlists.md" >}})|Hard||||50.7%|
+|0968|Binary Tree Cameras|[Go]({{< relref "/ChapterFour/0900~0999/0968.Binary-Tree-Cameras.md" >}})|Hard||||46.6%|
+|0978|Longest Turbulent Subarray|[Go]({{< relref "/ChapterFour/0900~0999/0978.Longest-Turbulent-Subarray.md" >}})|Medium||||47.2%|
+|0996|Number of Squareful Arrays|[Go]({{< relref "/ChapterFour/0900~0999/0996.Number-of-Squareful-Arrays.md" >}})|Hard||||49.2%|
+|1025|Divisor Game|[Go]({{< relref "/ChapterFour/1000~1099/1025.Divisor-Game.md" >}})|Easy| O(1)| O(1)||67.6%|
+|1048|Longest String Chain|[Go]({{< relref "/ChapterFour/1000~1099/1048.Longest-String-Chain.md" >}})|Medium||||59.3%|
+|1049|Last Stone Weight II|[Go]({{< relref "/ChapterFour/1000~1099/1049.Last-Stone-Weight-II.md" >}})|Medium||||53.2%|
+|1105|Filling Bookcase Shelves|[Go]({{< relref "/ChapterFour/1100~1199/1105.Filling-Bookcase-Shelves.md" >}})|Medium||||59.3%|
+|1137|N-th Tribonacci Number|[Go]({{< relref "/ChapterFour/1100~1199/1137.N-th-Tribonacci-Number.md" >}})|Easy||||63.7%|
+|1143|Longest Common Subsequence|[Go]({{< relref "/ChapterFour/1100~1199/1143.Longest-Common-Subsequence.md" >}})|Medium||||58.4%|
+|1235|Maximum Profit in Job Scheduling|[Go]({{< relref "/ChapterFour/1200~1299/1235.Maximum-Profit-in-Job-Scheduling.md" >}})|Hard||||53.4%|
+|1463|Cherry Pickup II|[Go]({{< relref "/ChapterFour/1400~1499/1463.Cherry-Pickup-II.md" >}})|Hard||||69.5%|
+|1641|Count Sorted Vowel Strings|[Go]({{< relref "/ChapterFour/1600~1699/1641.Count-Sorted-Vowel-Strings.md" >}})|Medium||||77.4%|
+|1646|Get Maximum in Generated Array|[Go]({{< relref "/ChapterFour/1600~1699/1646.Get-Maximum-in-Generated-Array.md" >}})|Easy||||50.2%|
+|1653|Minimum Deletions to Make String Balanced|[Go]({{< relref "/ChapterFour/1600~1699/1653.Minimum-Deletions-to-Make-String-Balanced.md" >}})|Medium||||58.9%|
+|1654|Minimum Jumps to Reach Home|[Go]({{< relref "/ChapterFour/1600~1699/1654.Minimum-Jumps-to-Reach-Home.md" >}})|Medium||||29.1%|
+|1655|Distribute Repeating Integers|[Go]({{< relref "/ChapterFour/1600~1699/1655.Distribute-Repeating-Integers.md" >}})|Hard||||39.3%|
+|1659|Maximize Grid Happiness|[Go]({{< relref "/ChapterFour/1600~1699/1659.Maximize-Grid-Happiness.md" >}})|Hard||||38.8%|
+|1664|Ways to Make a Fair Array|[Go]({{< relref "/ChapterFour/1600~1699/1664.Ways-to-Make-a-Fair-Array.md" >}})|Medium||||63.3%|
+|1681|Minimum Incompatibility|[Go]({{< relref "/ChapterFour/1600~1699/1681.Minimum-Incompatibility.md" >}})|Hard||||37.8%|
+|1690|Stone Game VII|[Go]({{< relref "/ChapterFour/1600~1699/1690.Stone-Game-VII.md" >}})|Medium||||58.1%|
+|1691|Maximum Height by Stacking Cuboids|[Go]({{< relref "/ChapterFour/1600~1699/1691.Maximum-Height-by-Stacking-Cuboids.md" >}})|Hard||||54.6%|
+|1696|Jump Game VI|[Go]({{< relref "/ChapterFour/1600~1699/1696.Jump-Game-VI.md" >}})|Medium||||46.1%|
+|2167|Minimum Time to Remove All Cars Containing Illegal Goods|[Go]({{< relref "/ChapterFour/2100~2199/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods.md" >}})|Hard||||40.8%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Tree/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Backtracking/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Hash_Table.md b/website/content/ChapterTwo/Hash_Table.md
new file mode 100644
index 000000000..d0dc53572
--- /dev/null
+++ b/website/content/ChapterTwo/Hash_Table.md
@@ -0,0 +1,180 @@
+---
+title: 2.13 Hash Table
+type: docs
+weight: 13
+---
+
+# Hash Table
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0001|Two Sum|[Go]({{< relref "/ChapterFour/0001~0099/0001.Two-Sum.md" >}})|Easy| O(n)| O(n)||49.7%|
+|0003|Longest Substring Without Repeating Characters|[Go]({{< relref "/ChapterFour/0001~0099/0003.Longest-Substring-Without-Repeating-Characters.md" >}})|Medium| O(n)| O(1)|❤️|33.8%|
+|0012|Integer to Roman|[Go]({{< relref "/ChapterFour/0001~0099/0012.Integer-to-Roman.md" >}})|Medium||||62.0%|
+|0013|Roman to Integer|[Go]({{< relref "/ChapterFour/0001~0099/0013.Roman-to-Integer.md" >}})|Easy||||58.6%|
+|0017|Letter Combinations of a Phone Number|[Go]({{< relref "/ChapterFour/0001~0099/0017.Letter-Combinations-of-a-Phone-Number.md" >}})|Medium||||56.6%|
+|0030|Substring with Concatenation of All Words|[Go]({{< relref "/ChapterFour/0001~0099/0030.Substring-with-Concatenation-of-All-Words.md" >}})|Hard| O(n)| O(n)|❤️|31.2%|
+|0036|Valid Sudoku|[Go]({{< relref "/ChapterFour/0001~0099/0036.Valid-Sudoku.md" >}})|Medium| O(n^2)| O(n^2)||58.1%|
+|0037|Sudoku Solver|[Go]({{< relref "/ChapterFour/0001~0099/0037.Sudoku-Solver.md" >}})|Hard| O(n^2)| O(n^2)|❤️|57.7%|
+|0041|First Missing Positive|[Go]({{< relref "/ChapterFour/0001~0099/0041.First-Missing-Positive.md" >}})|Hard||||36.7%|
+|0049|Group Anagrams|[Go]({{< relref "/ChapterFour/0001~0099/0049.Group-Anagrams.md" >}})|Medium| O(n log n)| O(n)||66.8%|
+|0073|Set Matrix Zeroes|[Go]({{< relref "/ChapterFour/0001~0099/0073.Set-Matrix-Zeroes.md" >}})|Medium||||51.3%|
+|0076|Minimum Window Substring|[Go]({{< relref "/ChapterFour/0001~0099/0076.Minimum-Window-Substring.md" >}})|Hard| O(n)| O(n)|❤️|40.9%|
+|0105|Construct Binary Tree from Preorder and Inorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal.md" >}})|Medium||||61.6%|
+|0106|Construct Binary Tree from Inorder and Postorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.md" >}})|Medium||||60.0%|
+|0126|Word Ladder II|[Go]({{< relref "/ChapterFour/0100~0199/0126.Word-Ladder-II.md" >}})|Hard||||27.5%|
+|0127|Word Ladder|[Go]({{< relref "/ChapterFour/0100~0199/0127.Word-Ladder.md" >}})|Hard||||37.2%|
+|0128|Longest Consecutive Sequence|[Go]({{< relref "/ChapterFour/0100~0199/0128.Longest-Consecutive-Sequence.md" >}})|Medium||||48.5%|
+|0138|Copy List with Random Pointer|[Go]({{< relref "/ChapterFour/0100~0199/0138.Copy-List-with-Random-Pointer.md" >}})|Medium| O(n)| O(1)||51.4%|
+|0141|Linked List Cycle|[Go]({{< relref "/ChapterFour/0100~0199/0141.Linked-List-Cycle.md" >}})|Easy||||47.5%|
+|0142|Linked List Cycle II|[Go]({{< relref "/ChapterFour/0100~0199/0142.Linked-List-Cycle-II.md" >}})|Medium||||48.8%|
+|0146|LRU Cache|[Go]({{< relref "/ChapterFour/0100~0199/0146.LRU-Cache.md" >}})|Medium||||40.7%|
+|0160|Intersection of Two Linked Lists|[Go]({{< relref "/ChapterFour/0100~0199/0160.Intersection-of-Two-Linked-Lists.md" >}})|Easy||||54.4%|
+|0169|Majority Element|[Go]({{< relref "/ChapterFour/0100~0199/0169.Majority-Element.md" >}})|Easy||||63.9%|
+|0187|Repeated DNA Sequences|[Go]({{< relref "/ChapterFour/0100~0199/0187.Repeated-DNA-Sequences.md" >}})|Medium||||47.0%|
+|0202|Happy Number|[Go]({{< relref "/ChapterFour/0200~0299/0202.Happy-Number.md" >}})|Easy| O(log n)| O(1)||54.8%|
+|0205|Isomorphic Strings|[Go]({{< relref "/ChapterFour/0200~0299/0205.Isomorphic-Strings.md" >}})|Easy| O(log n)| O(n)||42.9%|
+|0208|Implement Trie (Prefix Tree)|[Go]({{< relref "/ChapterFour/0200~0299/0208.Implement-Trie-Prefix-Tree.md" >}})|Medium||||62.8%|
+|0217|Contains Duplicate|[Go]({{< relref "/ChapterFour/0200~0299/0217.Contains-Duplicate.md" >}})|Easy| O(n)| O(n)||61.4%|
+|0219|Contains Duplicate II|[Go]({{< relref "/ChapterFour/0200~0299/0219.Contains-Duplicate-II.md" >}})|Easy| O(n)| O(n)||42.6%|
+|0229|Majority Element II|[Go]({{< relref "/ChapterFour/0200~0299/0229.Majority-Element-II.md" >}})|Medium||||45.1%|
+|0242|Valid Anagram|[Go]({{< relref "/ChapterFour/0200~0299/0242.Valid-Anagram.md" >}})|Easy| O(n)| O(n) ||63.1%|
+|0264|Ugly Number II|[Go]({{< relref "/ChapterFour/0200~0299/0264.Ugly-Number-II.md" >}})|Medium||||46.2%|
+|0268|Missing Number|[Go]({{< relref "/ChapterFour/0200~0299/0268.Missing-Number.md" >}})|Easy||||62.6%|
+|0290|Word Pattern|[Go]({{< relref "/ChapterFour/0200~0299/0290.Word-Pattern.md" >}})|Easy|  O(n)| O(n) ||41.7%|
+|0299|Bulls and Cows|[Go]({{< relref "/ChapterFour/0200~0299/0299.Bulls-and-Cows.md" >}})|Medium||||49.4%|
+|0347|Top K Frequent Elements|[Go]({{< relref "/ChapterFour/0300~0399/0347.Top-K-Frequent-Elements.md" >}})|Medium|  O(n)| O(n) ||64.2%|
+|0349|Intersection of Two Arrays|[Go]({{< relref "/ChapterFour/0300~0399/0349.Intersection-of-Two-Arrays.md" >}})|Easy| O(n)| O(n) ||70.9%|
+|0350|Intersection of Two Arrays II|[Go]({{< relref "/ChapterFour/0300~0399/0350.Intersection-of-Two-Arrays-II.md" >}})|Easy| O(n)| O(n) ||56.0%|
+|0383|Ransom Note|[Go]({{< relref "/ChapterFour/0300~0399/0383.Ransom-Note.md" >}})|Easy||||58.3%|
+|0387|First Unique Character in a String|[Go]({{< relref "/ChapterFour/0300~0399/0387.First-Unique-Character-in-a-String.md" >}})|Easy||||59.6%|
+|0389|Find the Difference|[Go]({{< relref "/ChapterFour/0300~0399/0389.Find-the-Difference.md" >}})|Easy||||59.9%|
+|0395|Longest Substring with At Least K Repeating Characters|[Go]({{< relref "/ChapterFour/0300~0399/0395.Longest-Substring-with-At-Least-K-Repeating-Characters.md" >}})|Medium||||44.8%|
+|0409|Longest Palindrome|[Go]({{< relref "/ChapterFour/0400~0499/0409.Longest-Palindrome.md" >}})|Easy||||54.2%|
+|0421|Maximum XOR of Two Numbers in an Array|[Go]({{< relref "/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array.md" >}})|Medium||||54.0%|
+|0423|Reconstruct Original Digits from English|[Go]({{< relref "/ChapterFour/0400~0499/0423.Reconstruct-Original-Digits-from-English.md" >}})|Medium||||51.2%|
+|0424|Longest Repeating Character Replacement|[Go]({{< relref "/ChapterFour/0400~0499/0424.Longest-Repeating-Character-Replacement.md" >}})|Medium||||52.0%|
+|0433|Minimum Genetic Mutation|[Go]({{< relref "/ChapterFour/0400~0499/0433.Minimum-Genetic-Mutation.md" >}})|Medium||||52.4%|
+|0438|Find All Anagrams in a String|[Go]({{< relref "/ChapterFour/0400~0499/0438.Find-All-Anagrams-in-a-String.md" >}})|Medium| O(n)| O(1) ||50.2%|
+|0447|Number of Boomerangs|[Go]({{< relref "/ChapterFour/0400~0499/0447.Number-of-Boomerangs.md" >}})|Medium| O(n)| O(1) ||54.9%|
+|0448|Find All Numbers Disappeared in an Array|[Go]({{< relref "/ChapterFour/0400~0499/0448.Find-All-Numbers-Disappeared-in-an-Array.md" >}})|Easy||||59.9%|
+|0451|Sort Characters By Frequency|[Go]({{< relref "/ChapterFour/0400~0499/0451.Sort-Characters-By-Frequency.md" >}})|Medium| O(n log n)| O(1) ||70.1%|
+|0454|4Sum II|[Go]({{< relref "/ChapterFour/0400~0499/0454.4Sum-II.md" >}})|Medium| O(n^2)| O(n) ||57.2%|
+|0457|Circular Array Loop|[Go]({{< relref "/ChapterFour/0400~0499/0457.Circular-Array-Loop.md" >}})|Medium||||32.6%|
+|0460|LFU Cache|[Go]({{< relref "/ChapterFour/0400~0499/0460.LFU-Cache.md" >}})|Hard||||43.0%|
+|0480|Sliding Window Median|[Go]({{< relref "/ChapterFour/0400~0499/0480.Sliding-Window-Median.md" >}})|Hard||||41.1%|
+|0491|Non-decreasing Subsequences|[Go]({{< relref "/ChapterFour/0400~0499/0491.Non-decreasing-Subsequences.md" >}})|Medium||||60.2%|
+|0496|Next Greater Element I|[Go]({{< relref "/ChapterFour/0400~0499/0496.Next-Greater-Element-I.md" >}})|Easy||||71.4%|
+|0500|Keyboard Row|[Go]({{< relref "/ChapterFour/0500~0599/0500.Keyboard-Row.md" >}})|Easy||||69.6%|
+|0508|Most Frequent Subtree Sum|[Go]({{< relref "/ChapterFour/0500~0599/0508.Most-Frequent-Subtree-Sum.md" >}})|Medium||||64.9%|
+|0519|Random Flip Matrix|[Go]({{< relref "/ChapterFour/0500~0599/0519.Random-Flip-Matrix.md" >}})|Medium||||40.0%|
+|0523|Continuous Subarray Sum|[Go]({{< relref "/ChapterFour/0500~0599/0523.Continuous-Subarray-Sum.md" >}})|Medium||||28.5%|
+|0525|Contiguous Array|[Go]({{< relref "/ChapterFour/0500~0599/0525.Contiguous-Array.md" >}})|Medium||||46.8%|
+|0532|K-diff Pairs in an Array|[Go]({{< relref "/ChapterFour/0500~0599/0532.K-diff-Pairs-in-an-Array.md" >}})|Medium||||41.2%|
+|0535|Encode and Decode TinyURL|[Go]({{< relref "/ChapterFour/0500~0599/0535.Encode-and-Decode-TinyURL.md" >}})|Medium||||86.0%|
+|0554|Brick Wall|[Go]({{< relref "/ChapterFour/0500~0599/0554.Brick-Wall.md" >}})|Medium||||53.6%|
+|0560|Subarray Sum Equals K|[Go]({{< relref "/ChapterFour/0500~0599/0560.Subarray-Sum-Equals-K.md" >}})|Medium||||43.7%|
+|0567|Permutation in String|[Go]({{< relref "/ChapterFour/0500~0599/0567.Permutation-in-String.md" >}})|Medium||||44.3%|
+|0575|Distribute Candies|[Go]({{< relref "/ChapterFour/0500~0599/0575.Distribute-Candies.md" >}})|Easy||||66.5%|
+|0594|Longest Harmonious Subsequence|[Go]({{< relref "/ChapterFour/0500~0599/0594.Longest-Harmonious-Subsequence.md" >}})|Easy||||53.5%|
+|0599|Minimum Index Sum of Two Lists|[Go]({{< relref "/ChapterFour/0500~0599/0599.Minimum-Index-Sum-of-Two-Lists.md" >}})|Easy||||53.4%|
+|0609|Find Duplicate File in System|[Go]({{< relref "/ChapterFour/0600~0699/0609.Find-Duplicate-File-in-System.md" >}})|Medium||||67.7%|
+|0632|Smallest Range Covering Elements from K Lists|[Go]({{< relref "/ChapterFour/0600~0699/0632.Smallest-Range-Covering-Elements-from-K-Lists.md" >}})|Hard||||61.0%|
+|0645|Set Mismatch|[Go]({{< relref "/ChapterFour/0600~0699/0645.Set-Mismatch.md" >}})|Easy||||42.7%|
+|0648|Replace Words|[Go]({{< relref "/ChapterFour/0600~0699/0648.Replace-Words.md" >}})|Medium| O(n)| O(n) ||62.7%|
+|0653|Two Sum IV - Input is a BST|[Go]({{< relref "/ChapterFour/0600~0699/0653.Two-Sum-IV-Input-is-a-BST.md" >}})|Easy||||61.0%|
+|0676|Implement Magic Dictionary|[Go]({{< relref "/ChapterFour/0600~0699/0676.Implement-Magic-Dictionary.md" >}})|Medium| O(n)| O(n) ||56.9%|
+|0677|Map Sum Pairs|[Go]({{< relref "/ChapterFour/0600~0699/0677.Map-Sum-Pairs.md" >}})|Medium||||56.8%|
+|0690|Employee Importance|[Go]({{< relref "/ChapterFour/0600~0699/0690.Employee-Importance.md" >}})|Medium||||65.6%|
+|0692|Top K Frequent Words|[Go]({{< relref "/ChapterFour/0600~0699/0692.Top-K-Frequent-Words.md" >}})|Medium||||57.2%|
+|0697|Degree of an Array|[Go]({{< relref "/ChapterFour/0600~0699/0697.Degree-of-an-Array.md" >}})|Easy||||56.0%|
+|0705|Design HashSet|[Go]({{< relref "/ChapterFour/0700~0799/0705.Design-HashSet.md" >}})|Easy||||65.6%|
+|0706|Design HashMap|[Go]({{< relref "/ChapterFour/0700~0799/0706.Design-HashMap.md" >}})|Easy||||64.7%|
+|0710|Random Pick with Blacklist|[Go]({{< relref "/ChapterFour/0700~0799/0710.Random-Pick-with-Blacklist.md" >}})|Hard| O(n)| O(n)  ||33.5%|
+|0720|Longest Word in Dictionary|[Go]({{< relref "/ChapterFour/0700~0799/0720.Longest-Word-in-Dictionary.md" >}})|Medium| O(n)| O(n) ||52.0%|
+|0726|Number of Atoms|[Go]({{< relref "/ChapterFour/0700~0799/0726.Number-of-Atoms.md" >}})|Hard| O(n)| O(n) |❤️|52.2%|
+|0745|Prefix and Suffix Search|[Go]({{< relref "/ChapterFour/0700~0799/0745.Prefix-and-Suffix-Search.md" >}})|Hard||||41.2%|
+|0748|Shortest Completing Word|[Go]({{< relref "/ChapterFour/0700~0799/0748.Shortest-Completing-Word.md" >}})|Easy||||59.3%|
+|0752|Open the Lock|[Go]({{< relref "/ChapterFour/0700~0799/0752.Open-the-Lock.md" >}})|Medium||||55.6%|
+|0763|Partition Labels|[Go]({{< relref "/ChapterFour/0700~0799/0763.Partition-Labels.md" >}})|Medium||||79.7%|
+|0767|Reorganize String|[Go]({{< relref "/ChapterFour/0700~0799/0767.Reorganize-String.md" >}})|Medium||||52.9%|
+|0771|Jewels and Stones|[Go]({{< relref "/ChapterFour/0700~0799/0771.Jewels-and-Stones.md" >}})|Easy||||88.2%|
+|0781|Rabbits in Forest|[Go]({{< relref "/ChapterFour/0700~0799/0781.Rabbits-in-Forest.md" >}})|Medium||||54.7%|
+|0791|Custom Sort String|[Go]({{< relref "/ChapterFour/0700~0799/0791.Custom-Sort-String.md" >}})|Medium||||69.1%|
+|0792|Number of Matching Subsequences|[Go]({{< relref "/ChapterFour/0700~0799/0792.Number-of-Matching-Subsequences.md" >}})|Medium||||51.6%|
+|0811|Subdomain Visit Count|[Go]({{< relref "/ChapterFour/0800~0899/0811.Subdomain-Visit-Count.md" >}})|Medium||||75.5%|
+|0815|Bus Routes|[Go]({{< relref "/ChapterFour/0800~0899/0815.Bus-Routes.md" >}})|Hard||||45.6%|
+|0817|Linked List Components|[Go]({{< relref "/ChapterFour/0800~0899/0817.Linked-List-Components.md" >}})|Medium||||57.7%|
+|0819|Most Common Word|[Go]({{< relref "/ChapterFour/0800~0899/0819.Most-Common-Word.md" >}})|Easy||||44.7%|
+|0820|Short Encoding of Words|[Go]({{< relref "/ChapterFour/0800~0899/0820.Short-Encoding-of-Words.md" >}})|Medium||||60.6%|
+|0823|Binary Trees With Factors|[Go]({{< relref "/ChapterFour/0800~0899/0823.Binary-Trees-With-Factors.md" >}})|Medium||||49.7%|
+|0828|Count Unique Characters of All Substrings of a Given String|[Go]({{< relref "/ChapterFour/0800~0899/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String.md" >}})|Hard||||51.6%|
+|0846|Hand of Straights|[Go]({{< relref "/ChapterFour/0800~0899/0846.Hand-of-Straights.md" >}})|Medium||||56.2%|
+|0859|Buddy Strings|[Go]({{< relref "/ChapterFour/0800~0899/0859.Buddy-Strings.md" >}})|Easy||||29.2%|
+|0884|Uncommon Words from Two Sentences|[Go]({{< relref "/ChapterFour/0800~0899/0884.Uncommon-Words-from-Two-Sentences.md" >}})|Easy||||66.4%|
+|0888|Fair Candy Swap|[Go]({{< relref "/ChapterFour/0800~0899/0888.Fair-Candy-Swap.md" >}})|Easy||||60.7%|
+|0890|Find and Replace Pattern|[Go]({{< relref "/ChapterFour/0800~0899/0890.Find-and-Replace-Pattern.md" >}})|Medium||||77.6%|
+|0895|Maximum Frequency Stack|[Go]({{< relref "/ChapterFour/0800~0899/0895.Maximum-Frequency-Stack.md" >}})|Hard| O(n)| O(n)  ||66.6%|
+|0904|Fruit Into Baskets|[Go]({{< relref "/ChapterFour/0900~0999/0904.Fruit-Into-Baskets.md" >}})|Medium||||43.7%|
+|0911|Online Election|[Go]({{< relref "/ChapterFour/0900~0999/0911.Online-Election.md" >}})|Medium||||52.2%|
+|0914|X of a Kind in a Deck of Cards|[Go]({{< relref "/ChapterFour/0900~0999/0914.X-of-a-Kind-in-a-Deck-of-Cards.md" >}})|Easy||||31.2%|
+|0916|Word Subsets|[Go]({{< relref "/ChapterFour/0900~0999/0916.Word-Subsets.md" >}})|Medium||||53.7%|
+|0923|3Sum With Multiplicity|[Go]({{< relref "/ChapterFour/0900~0999/0923.3Sum-With-Multiplicity.md" >}})|Medium||||45.3%|
+|0930|Binary Subarrays With Sum|[Go]({{< relref "/ChapterFour/0900~0999/0930.Binary-Subarrays-With-Sum.md" >}})|Medium| O(n)| O(n)  |❤️|52.2%|
+|0953|Verifying an Alien Dictionary|[Go]({{< relref "/ChapterFour/0900~0999/0953.Verifying-an-Alien-Dictionary.md" >}})|Easy||||54.5%|
+|0961|N-Repeated Element in Size 2N Array|[Go]({{< relref "/ChapterFour/0900~0999/0961.N-Repeated-Element-in-Size-2N-Array.md" >}})|Easy||||76.1%|
+|0966|Vowel Spellchecker|[Go]({{< relref "/ChapterFour/0900~0999/0966.Vowel-Spellchecker.md" >}})|Medium||||51.4%|
+|0970|Powerful Integers|[Go]({{< relref "/ChapterFour/0900~0999/0970.Powerful-Integers.md" >}})|Medium||||43.6%|
+|0981|Time Based Key-Value Store|[Go]({{< relref "/ChapterFour/0900~0999/0981.Time-Based-Key-Value-Store.md" >}})|Medium||||52.2%|
+|0987|Vertical Order Traversal of a Binary Tree|[Go]({{< relref "/ChapterFour/0900~0999/0987.Vertical-Order-Traversal-of-a-Binary-Tree.md" >}})|Hard||||45.1%|
+|0992|Subarrays with K Different Integers|[Go]({{< relref "/ChapterFour/0900~0999/0992.Subarrays-with-K-Different-Integers.md" >}})|Hard| O(n)| O(n)  |❤️|54.6%|
+|0997|Find the Town Judge|[Go]({{< relref "/ChapterFour/0900~0999/0997.Find-the-Town-Judge.md" >}})|Easy||||49.5%|
+|1002|Find Common Characters|[Go]({{< relref "/ChapterFour/1000~1099/1002.Find-Common-Characters.md" >}})|Easy||||68.5%|
+|1010|Pairs of Songs With Total Durations Divisible by 60|[Go]({{< relref "/ChapterFour/1000~1099/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60.md" >}})|Medium||||52.8%|
+|1048|Longest String Chain|[Go]({{< relref "/ChapterFour/1000~1099/1048.Longest-String-Chain.md" >}})|Medium||||59.3%|
+|1054|Distant Barcodes|[Go]({{< relref "/ChapterFour/1000~1099/1054.Distant-Barcodes.md" >}})|Medium||||45.9%|
+|1074|Number of Submatrices That Sum to Target|[Go]({{< relref "/ChapterFour/1000~1099/1074.Number-of-Submatrices-That-Sum-to-Target.md" >}})|Hard||||69.5%|
+|1079|Letter Tile Possibilities|[Go]({{< relref "/ChapterFour/1000~1099/1079.Letter-Tile-Possibilities.md" >}})|Medium||||76.0%|
+|1122|Relative Sort Array|[Go]({{< relref "/ChapterFour/1100~1199/1122.Relative-Sort-Array.md" >}})|Easy||||68.6%|
+|1123|Lowest Common Ancestor of Deepest Leaves|[Go]({{< relref "/ChapterFour/1100~1199/1123.Lowest-Common-Ancestor-of-Deepest-Leaves.md" >}})|Medium||||70.9%|
+|1128|Number of Equivalent Domino Pairs|[Go]({{< relref "/ChapterFour/1100~1199/1128.Number-of-Equivalent-Domino-Pairs.md" >}})|Easy||||47.1%|
+|1160|Find Words That Can Be Formed by Characters|[Go]({{< relref "/ChapterFour/1100~1199/1160.Find-Words-That-Can-Be-Formed-by-Characters.md" >}})|Easy||||67.5%|
+|1170|Compare Strings by Frequency of the Smallest Character|[Go]({{< relref "/ChapterFour/1100~1199/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character.md" >}})|Medium||||61.5%|
+|1171|Remove Zero Sum Consecutive Nodes from Linked List|[Go]({{< relref "/ChapterFour/1100~1199/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List.md" >}})|Medium||||43.2%|
+|1178|Number of Valid Words for Each Puzzle|[Go]({{< relref "/ChapterFour/1100~1199/1178.Number-of-Valid-Words-for-Each-Puzzle.md" >}})|Hard||||46.3%|
+|1189|Maximum Number of Balloons|[Go]({{< relref "/ChapterFour/1100~1199/1189.Maximum-Number-of-Balloons.md" >}})|Easy||||61.0%|
+|1202|Smallest String With Swaps|[Go]({{< relref "/ChapterFour/1200~1299/1202.Smallest-String-With-Swaps.md" >}})|Medium||||57.7%|
+|1207|Unique Number of Occurrences|[Go]({{< relref "/ChapterFour/1200~1299/1207.Unique-Number-of-Occurrences.md" >}})|Easy||||73.5%|
+|1275|Find Winner on a Tic Tac Toe Game|[Go]({{< relref "/ChapterFour/1200~1299/1275.Find-Winner-on-a-Tic-Tac-Toe-Game.md" >}})|Easy||||54.2%|
+|1296|Divide Array in Sets of K Consecutive Numbers|[Go]({{< relref "/ChapterFour/1200~1299/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers.md" >}})|Medium||||56.5%|
+|1396|Design Underground System|[Go]({{< relref "/ChapterFour/1300~1399/1396.Design-Underground-System.md" >}})|Medium||||73.6%|
+|1442|Count Triplets That Can Form Two Arrays of Equal XOR|[Go]({{< relref "/ChapterFour/1400~1499/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR.md" >}})|Medium||||76.1%|
+|1461|Check If a String Contains All Binary Codes of Size K|[Go]({{< relref "/ChapterFour/1400~1499/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K.md" >}})|Medium||||56.6%|
+|1512|Number of Good Pairs|[Go]({{< relref "/ChapterFour/1500~1599/1512.Number-of-Good-Pairs.md" >}})|Easy||||88.2%|
+|1600|Throne Inheritance|[Go]({{< relref "/ChapterFour/1600~1699/1600.Throne-Inheritance.md" >}})|Medium||||63.6%|
+|1624|Largest Substring Between Two Equal Characters|[Go]({{< relref "/ChapterFour/1600~1699/1624.Largest-Substring-Between-Two-Equal-Characters.md" >}})|Easy||||59.1%|
+|1636|Sort Array by Increasing Frequency|[Go]({{< relref "/ChapterFour/1600~1699/1636.Sort-Array-by-Increasing-Frequency.md" >}})|Easy||||69.5%|
+|1640|Check Array Formation Through Concatenation|[Go]({{< relref "/ChapterFour/1600~1699/1640.Check-Array-Formation-Through-Concatenation.md" >}})|Easy||||56.2%|
+|1647|Minimum Deletions to Make Character Frequencies Unique|[Go]({{< relref "/ChapterFour/1600~1699/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique.md" >}})|Medium||||59.1%|
+|1656|Design an Ordered Stream|[Go]({{< relref "/ChapterFour/1600~1699/1656.Design-an-Ordered-Stream.md" >}})|Easy||||85.2%|
+|1657|Determine if Two Strings Are Close|[Go]({{< relref "/ChapterFour/1600~1699/1657.Determine-if-Two-Strings-Are-Close.md" >}})|Medium||||56.3%|
+|1658|Minimum Operations to Reduce X to Zero|[Go]({{< relref "/ChapterFour/1600~1699/1658.Minimum-Operations-to-Reduce-X-to-Zero.md" >}})|Medium||||37.6%|
+|1674|Minimum Moves to Make Array Complementary|[Go]({{< relref "/ChapterFour/1600~1699/1674.Minimum-Moves-to-Make-Array-Complementary.md" >}})|Medium||||38.7%|
+|1679|Max Number of K-Sum Pairs|[Go]({{< relref "/ChapterFour/1600~1699/1679.Max-Number-of-K-Sum-Pairs.md" >}})|Medium||||57.3%|
+|1684|Count the Number of Consistent Strings|[Go]({{< relref "/ChapterFour/1600~1699/1684.Count-the-Number-of-Consistent-Strings.md" >}})|Easy||||82.3%|
+|1695|Maximum Erasure Value|[Go]({{< relref "/ChapterFour/1600~1699/1695.Maximum-Erasure-Value.md" >}})|Medium||||57.6%|
+|1742|Maximum Number of Balls in a Box|[Go]({{< relref "/ChapterFour/1700~1799/1742.Maximum-Number-of-Balls-in-a-Box.md" >}})|Easy||||73.6%|
+|1748|Sum of Unique Elements|[Go]({{< relref "/ChapterFour/1700~1799/1748.Sum-of-Unique-Elements.md" >}})|Easy||||76.3%|
+|1763|Longest Nice Substring|[Go]({{< relref "/ChapterFour/1700~1799/1763.Longest-Nice-Substring.md" >}})|Easy||||61.5%|
+|2043|Simple Bank System|[Go]({{< relref "/ChapterFour/2000~2099/2043.Simple-Bank-System.md" >}})|Medium||||65.2%|
+|2166|Design Bitset|[Go]({{< relref "/ChapterFour/2100~2199/2166.Design-Bitset.md" >}})|Medium||||31.8%|
+|2170|Minimum Operations to Make the Array Alternating|[Go]({{< relref "/ChapterFour/2100~2199/2170.Minimum-Operations-to-Make-the-Array-Alternating.md" >}})|Medium||||33.2%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Math/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Sorting/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Linked_List.md b/website/content/ChapterTwo/Linked_List.md
new file mode 100644
index 000000000..aa72c67ec
--- /dev/null
+++ b/website/content/ChapterTwo/Linked_List.md
@@ -0,0 +1,77 @@
+---
+title: 2.04 ✅ Linked List
+type: docs
+weight: 4
+---
+
+# Linked List
+
+![](https://img.halfrost.com/Leetcode/Linked_List.png)
+
+
+- 巧妙的构造虚拟头结点。可以使遍历处理逻辑更加统一。
+- 灵活使用递归。构造递归条件，使用递归可以巧妙的解题。不过需要注意有些题目不能使用递归，因为递归深度太深会导致超时和栈溢出。
+- 链表区间逆序。第 92 题。
+- 链表寻找中间节点。第 876 题。链表寻找倒数第 n 个节点。第 19 题。只需要一次遍历就可以得到答案。
+- 合并 K 个有序链表。第 21 题，第 23 题。
+- 链表归类。第 86 题，第 328 题。
+- 链表排序，时间复杂度要求 O(n * log n)，空间复杂度 O(1)。只有一种做法，归并排序，至顶向下归并。第 148 题。
+- 判断链表是否存在环，如果有环，输出环的交叉点的下标；判断 2 个链表是否有交叉点，如果有交叉点，输出交叉点。第 141 题，第 142 题，第 160 题。
+
+
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0002|Add Two Numbers|[Go]({{< relref "/ChapterFour/0001~0099/0002.Add-Two-Numbers.md" >}})|Medium| O(n)| O(1)||40.4%|
+|0019|Remove Nth Node From End of List|[Go]({{< relref "/ChapterFour/0001~0099/0019.Remove-Nth-Node-From-End-of-List.md" >}})|Medium| O(n)| O(1)||41.1%|
+|0021|Merge Two Sorted Lists|[Go]({{< relref "/ChapterFour/0001~0099/0021.Merge-Two-Sorted-Lists.md" >}})|Easy| O(log n)| O(1)||62.5%|
+|0023|Merge k Sorted Lists|[Go]({{< relref "/ChapterFour/0001~0099/0023.Merge-k-Sorted-Lists.md" >}})|Hard| O(log n)| O(1)|❤️|49.8%|
+|0024|Swap Nodes in Pairs|[Go]({{< relref "/ChapterFour/0001~0099/0024.Swap-Nodes-in-Pairs.md" >}})|Medium| O(n)| O(1)||61.3%|
+|0025|Reverse Nodes in k-Group|[Go]({{< relref "/ChapterFour/0001~0099/0025.Reverse-Nodes-in-k-Group.md" >}})|Hard| O(log n)| O(1)|❤️|54.7%|
+|0061|Rotate List|[Go]({{< relref "/ChapterFour/0001~0099/0061.Rotate-List.md" >}})|Medium| O(n)| O(1)||36.1%|
+|0082|Remove Duplicates from Sorted List II|[Go]({{< relref "/ChapterFour/0001~0099/0082.Remove-Duplicates-from-Sorted-List-II.md" >}})|Medium| O(n)| O(1)||45.9%|
+|0083|Remove Duplicates from Sorted List|[Go]({{< relref "/ChapterFour/0001~0099/0083.Remove-Duplicates-from-Sorted-List.md" >}})|Easy| O(n)| O(1)||50.6%|
+|0086|Partition List|[Go]({{< relref "/ChapterFour/0001~0099/0086.Partition-List.md" >}})|Medium| O(n)| O(1)|❤️|52.0%|
+|0092|Reverse Linked List II|[Go]({{< relref "/ChapterFour/0001~0099/0092.Reverse-Linked-List-II.md" >}})|Medium| O(n)| O(1)|❤️|45.4%|
+|0109|Convert Sorted List to Binary Search Tree|[Go]({{< relref "/ChapterFour/0100~0199/0109.Convert-Sorted-List-to-Binary-Search-Tree.md" >}})|Medium| O(log n)| O(n)||60.2%|
+|0114|Flatten Binary Tree to Linked List|[Go]({{< relref "/ChapterFour/0100~0199/0114.Flatten-Binary-Tree-to-Linked-List.md" >}})|Medium||||61.8%|
+|0116|Populating Next Right Pointers in Each Node|[Go]({{< relref "/ChapterFour/0100~0199/0116.Populating-Next-Right-Pointers-in-Each-Node.md" >}})|Medium||||60.4%|
+|0138|Copy List with Random Pointer|[Go]({{< relref "/ChapterFour/0100~0199/0138.Copy-List-with-Random-Pointer.md" >}})|Medium||||51.4%|
+|0141|Linked List Cycle|[Go]({{< relref "/ChapterFour/0100~0199/0141.Linked-List-Cycle.md" >}})|Easy| O(n)| O(1)|❤️|47.5%|
+|0142|Linked List Cycle II|[Go]({{< relref "/ChapterFour/0100~0199/0142.Linked-List-Cycle-II.md" >}})|Medium| O(n)| O(1)|❤️|48.8%|
+|0143|Reorder List|[Go]({{< relref "/ChapterFour/0100~0199/0143.Reorder-List.md" >}})|Medium| O(n)| O(1)|❤️|52.6%|
+|0146|LRU Cache|[Go]({{< relref "/ChapterFour/0100~0199/0146.LRU-Cache.md" >}})|Medium||||40.7%|
+|0147|Insertion Sort List|[Go]({{< relref "/ChapterFour/0100~0199/0147.Insertion-Sort-List.md" >}})|Medium| O(n)| O(1)|❤️|51.1%|
+|0148|Sort List|[Go]({{< relref "/ChapterFour/0100~0199/0148.Sort-List.md" >}})|Medium| O(n log n)| O(n)|❤️|55.1%|
+|0160|Intersection of Two Linked Lists|[Go]({{< relref "/ChapterFour/0100~0199/0160.Intersection-of-Two-Linked-Lists.md" >}})|Easy| O(n)| O(1)|❤️|54.4%|
+|0203|Remove Linked List Elements|[Go]({{< relref "/ChapterFour/0200~0299/0203.Remove-Linked-List-Elements.md" >}})|Easy| O(n)| O(1)||46.0%|
+|0206|Reverse Linked List|[Go]({{< relref "/ChapterFour/0200~0299/0206.Reverse-Linked-List.md" >}})|Easy| O(n)| O(1)||73.6%|
+|0234|Palindrome Linked List|[Go]({{< relref "/ChapterFour/0200~0299/0234.Palindrome-Linked-List.md" >}})|Easy| O(n)| O(1)||50.2%|
+|0237|Delete Node in a Linked List|[Go]({{< relref "/ChapterFour/0200~0299/0237.Delete-Node-in-a-Linked-List.md" >}})|Medium| O(n)| O(1)||76.0%|
+|0328|Odd Even Linked List|[Go]({{< relref "/ChapterFour/0300~0399/0328.Odd-Even-Linked-List.md" >}})|Medium| O(n)| O(1)||61.3%|
+|0382|Linked List Random Node|[Go]({{< relref "/ChapterFour/0300~0399/0382.Linked-List-Random-Node.md" >}})|Medium||||62.8%|
+|0445|Add Two Numbers II|[Go]({{< relref "/ChapterFour/0400~0499/0445.Add-Two-Numbers-II.md" >}})|Medium| O(n)| O(n)||59.6%|
+|0460|LFU Cache|[Go]({{< relref "/ChapterFour/0400~0499/0460.LFU-Cache.md" >}})|Hard||||43.0%|
+|0622|Design Circular Queue|[Go]({{< relref "/ChapterFour/0600~0699/0622.Design-Circular-Queue.md" >}})|Medium||||51.5%|
+|0705|Design HashSet|[Go]({{< relref "/ChapterFour/0700~0799/0705.Design-HashSet.md" >}})|Easy||||65.6%|
+|0706|Design HashMap|[Go]({{< relref "/ChapterFour/0700~0799/0706.Design-HashMap.md" >}})|Easy||||64.7%|
+|0707|Design Linked List|[Go]({{< relref "/ChapterFour/0700~0799/0707.Design-Linked-List.md" >}})|Medium| O(n)| O(1)||27.7%|
+|0725|Split Linked List in Parts|[Go]({{< relref "/ChapterFour/0700~0799/0725.Split-Linked-List-in-Parts.md" >}})|Medium| O(n)| O(1)||57.2%|
+|0817|Linked List Components|[Go]({{< relref "/ChapterFour/0800~0899/0817.Linked-List-Components.md" >}})|Medium| O(n)| O(1)||57.7%|
+|0876|Middle of the Linked List|[Go]({{< relref "/ChapterFour/0800~0899/0876.Middle-of-the-Linked-List.md" >}})|Easy| O(n)| O(1)|❤️|75.7%|
+|1019|Next Greater Node In Linked List|[Go]({{< relref "/ChapterFour/1000~1099/1019.Next-Greater-Node-In-Linked-List.md" >}})|Medium| O(n)| O(1)||59.9%|
+|1171|Remove Zero Sum Consecutive Nodes from Linked List|[Go]({{< relref "/ChapterFour/1100~1199/1171.Remove-Zero-Sum-Consecutive-Nodes-from-Linked-List.md" >}})|Medium||||43.2%|
+|1290|Convert Binary Number in a Linked List to Integer|[Go]({{< relref "/ChapterFour/1200~1299/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer.md" >}})|Easy||||82.1%|
+|1669|Merge In Between Linked Lists|[Go]({{< relref "/ChapterFour/1600~1699/1669.Merge-In-Between-Linked-Lists.md" >}})|Medium||||73.7%|
+|1670|Design Front Middle Back Queue|[Go]({{< relref "/ChapterFour/1600~1699/1670.Design-Front-Middle-Back-Queue.md" >}})|Medium||||57.2%|
+|1721|Swapping Nodes in a Linked List|[Go]({{< relref "/ChapterFour/1700~1799/1721.Swapping-Nodes-in-a-Linked-List.md" >}})|Medium||||67.1%|
+|2181|Merge Nodes in Between Zeros|[Go]({{< relref "/ChapterFour/2100~2199/2181.Merge-Nodes-in-Between-Zeros.md" >}})|Medium||||86.3%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Two_Pointers/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Stack/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Math.md b/website/content/ChapterTwo/Math.md
new file mode 100644
index 000000000..20e6ba342
--- /dev/null
+++ b/website/content/ChapterTwo/Math.md
@@ -0,0 +1,160 @@
+---
+title: 2.12 Math
+type: docs
+weight: 12
+---
+
+# Math
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0002|Add Two Numbers|[Go]({{< relref "/ChapterFour/0001~0099/0002.Add-Two-Numbers.md" >}})|Medium| O(n)| O(1)||40.4%|
+|0007|Reverse Integer|[Go]({{< relref "/ChapterFour/0001~0099/0007.Reverse-Integer.md" >}})|Medium||||27.5%|
+|0009|Palindrome Number|[Go]({{< relref "/ChapterFour/0001~0099/0009.Palindrome-Number.md" >}})|Easy||||53.5%|
+|0012|Integer to Roman|[Go]({{< relref "/ChapterFour/0001~0099/0012.Integer-to-Roman.md" >}})|Medium||||62.0%|
+|0013|Roman to Integer|[Go]({{< relref "/ChapterFour/0001~0099/0013.Roman-to-Integer.md" >}})|Easy||||58.6%|
+|0029|Divide Two Integers|[Go]({{< relref "/ChapterFour/0001~0099/0029.Divide-Two-Integers.md" >}})|Medium||||17.2%|
+|0043|Multiply Strings|[Go]({{< relref "/ChapterFour/0001~0099/0043.Multiply-Strings.md" >}})|Medium||||39.2%|
+|0048|Rotate Image|[Go]({{< relref "/ChapterFour/0001~0099/0048.Rotate-Image.md" >}})|Medium||||71.0%|
+|0050|Pow(x, n)|[Go]({{< relref "/ChapterFour/0001~0099/0050.Powx-n.md" >}})|Medium| O(log n)| O(1)||33.0%|
+|0060|Permutation Sequence|[Go]({{< relref "/ChapterFour/0001~0099/0060.Permutation-Sequence.md" >}})|Hard| O(n log n)| O(1)||44.4%|
+|0062|Unique Paths|[Go]({{< relref "/ChapterFour/0001~0099/0062.Unique-Paths.md" >}})|Medium||||62.7%|
+|0066|Plus One|[Go]({{< relref "/ChapterFour/0001~0099/0066.Plus-One.md" >}})|Easy||||43.7%|
+|0067|Add Binary|[Go]({{< relref "/ChapterFour/0001~0099/0067.Add-Binary.md" >}})|Easy||||52.4%|
+|0069|Sqrt(x)|[Go]({{< relref "/ChapterFour/0001~0099/0069.Sqrtx.md" >}})|Easy| O(log n)| O(1)||37.4%|
+|0070|Climbing Stairs|[Go]({{< relref "/ChapterFour/0001~0099/0070.Climbing-Stairs.md" >}})|Easy||||52.2%|
+|0089|Gray Code|[Go]({{< relref "/ChapterFour/0001~0099/0089.Gray-Code.md" >}})|Medium||||57.2%|
+|0096|Unique Binary Search Trees|[Go]({{< relref "/ChapterFour/0001~0099/0096.Unique-Binary-Search-Trees.md" >}})|Medium||||59.7%|
+|0150|Evaluate Reverse Polish Notation|[Go]({{< relref "/ChapterFour/0100~0199/0150.Evaluate-Reverse-Polish-Notation.md" >}})|Medium||||45.8%|
+|0168|Excel Sheet Column Title|[Go]({{< relref "/ChapterFour/0100~0199/0168.Excel-Sheet-Column-Title.md" >}})|Easy||||35.5%|
+|0171|Excel Sheet Column Number|[Go]({{< relref "/ChapterFour/0100~0199/0171.Excel-Sheet-Column-Number.md" >}})|Easy||||62.1%|
+|0172|Factorial Trailing Zeroes|[Go]({{< relref "/ChapterFour/0100~0199/0172.Factorial-Trailing-Zeroes.md" >}})|Medium||||42.2%|
+|0189|Rotate Array|[Go]({{< relref "/ChapterFour/0100~0199/0189.Rotate-Array.md" >}})|Medium||||39.4%|
+|0202|Happy Number|[Go]({{< relref "/ChapterFour/0200~0299/0202.Happy-Number.md" >}})|Easy| O(log n)| O(1)||54.8%|
+|0204|Count Primes|[Go]({{< relref "/ChapterFour/0200~0299/0204.Count-Primes.md" >}})|Medium||||33.1%|
+|0223|Rectangle Area|[Go]({{< relref "/ChapterFour/0200~0299/0223.Rectangle-Area.md" >}})|Medium||||45.1%|
+|0224|Basic Calculator|[Go]({{< relref "/ChapterFour/0200~0299/0224.Basic-Calculator.md" >}})|Hard| O(n)| O(n)||42.4%|
+|0227|Basic Calculator II|[Go]({{< relref "/ChapterFour/0200~0299/0227.Basic-Calculator-II.md" >}})|Medium||||42.4%|
+|0231|Power of Two|[Go]({{< relref "/ChapterFour/0200~0299/0231.Power-of-Two.md" >}})|Easy| O(1)| O(1)||46.0%|
+|0258|Add Digits|[Go]({{< relref "/ChapterFour/0200~0299/0258.Add-Digits.md" >}})|Easy||||64.0%|
+|0263|Ugly Number|[Go]({{< relref "/ChapterFour/0200~0299/0263.Ugly-Number.md" >}})|Easy| O(log n)| O(1)||42.3%|
+|0264|Ugly Number II|[Go]({{< relref "/ChapterFour/0200~0299/0264.Ugly-Number-II.md" >}})|Medium||||46.2%|
+|0268|Missing Number|[Go]({{< relref "/ChapterFour/0200~0299/0268.Missing-Number.md" >}})|Easy||||62.6%|
+|0279|Perfect Squares|[Go]({{< relref "/ChapterFour/0200~0299/0279.Perfect-Squares.md" >}})|Medium||||52.7%|
+|0319|Bulb Switcher|[Go]({{< relref "/ChapterFour/0300~0399/0319.Bulb-Switcher.md" >}})|Medium||||48.3%|
+|0326|Power of Three|[Go]({{< relref "/ChapterFour/0300~0399/0326.Power-of-Three.md" >}})|Easy| O(1)| O(1)||45.5%|
+|0342|Power of Four|[Go]({{< relref "/ChapterFour/0300~0399/0342.Power-of-Four.md" >}})|Easy||||46.2%|
+|0343|Integer Break|[Go]({{< relref "/ChapterFour/0300~0399/0343.Integer-Break.md" >}})|Medium| O(n^2)| O(n)||56.1%|
+|0357|Count Numbers with Unique Digits|[Go]({{< relref "/ChapterFour/0300~0399/0357.Count-Numbers-with-Unique-Digits.md" >}})|Medium| O(1)| O(1)||51.9%|
+|0367|Valid Perfect Square|[Go]({{< relref "/ChapterFour/0300~0399/0367.Valid-Perfect-Square.md" >}})|Easy||||43.3%|
+|0368|Largest Divisible Subset|[Go]({{< relref "/ChapterFour/0300~0399/0368.Largest-Divisible-Subset.md" >}})|Medium||||41.6%|
+|0371|Sum of Two Integers|[Go]({{< relref "/ChapterFour/0300~0399/0371.Sum-of-Two-Integers.md" >}})|Medium||||50.7%|
+|0372|Super Pow|[Go]({{< relref "/ChapterFour/0300~0399/0372.Super-Pow.md" >}})|Medium||||36.3%|
+|0382|Linked List Random Node|[Go]({{< relref "/ChapterFour/0300~0399/0382.Linked-List-Random-Node.md" >}})|Medium||||62.8%|
+|0384|Shuffle an Array|[Go]({{< relref "/ChapterFour/0300~0399/0384.Shuffle-an-Array.md" >}})|Medium||||57.8%|
+|0390|Elimination Game|[Go]({{< relref "/ChapterFour/0300~0399/0390.Elimination-Game.md" >}})|Medium||||46.1%|
+|0396|Rotate Function|[Go]({{< relref "/ChapterFour/0300~0399/0396.Rotate-Function.md" >}})|Medium||||41.1%|
+|0400|Nth Digit|[Go]({{< relref "/ChapterFour/0400~0499/0400.Nth-Digit.md" >}})|Medium||||34.1%|
+|0405|Convert a Number to Hexadecimal|[Go]({{< relref "/ChapterFour/0400~0499/0405.Convert-a-Number-to-Hexadecimal.md" >}})|Easy||||46.8%|
+|0412|Fizz Buzz|[Go]({{< relref "/ChapterFour/0400~0499/0412.Fizz-Buzz.md" >}})|Easy||||70.0%|
+|0423|Reconstruct Original Digits from English|[Go]({{< relref "/ChapterFour/0400~0499/0423.Reconstruct-Original-Digits-from-English.md" >}})|Medium||||51.2%|
+|0441|Arranging Coins|[Go]({{< relref "/ChapterFour/0400~0499/0441.Arranging-Coins.md" >}})|Easy||||46.2%|
+|0445|Add Two Numbers II|[Go]({{< relref "/ChapterFour/0400~0499/0445.Add-Two-Numbers-II.md" >}})|Medium||||59.6%|
+|0447|Number of Boomerangs|[Go]({{< relref "/ChapterFour/0400~0499/0447.Number-of-Boomerangs.md" >}})|Medium||||54.9%|
+|0453|Minimum Moves to Equal Array Elements|[Go]({{< relref "/ChapterFour/0400~0499/0453.Minimum-Moves-to-Equal-Array-Elements.md" >}})|Medium||||56.0%|
+|0458|Poor Pigs|[Go]({{< relref "/ChapterFour/0400~0499/0458.Poor-Pigs.md" >}})|Hard||||62.8%|
+|0462|Minimum Moves to Equal Array Elements II|[Go]({{< relref "/ChapterFour/0400~0499/0462.Minimum-Moves-to-Equal-Array-Elements-II.md" >}})|Medium||||60.0%|
+|0470|Implement Rand10() Using Rand7()|[Go]({{< relref "/ChapterFour/0400~0499/0470.Implement-Rand10-Using-Rand7.md" >}})|Medium||||46.4%|
+|0477|Total Hamming Distance|[Go]({{< relref "/ChapterFour/0400~0499/0477.Total-Hamming-Distance.md" >}})|Medium||||52.2%|
+|0478|Generate Random Point in a Circle|[Go]({{< relref "/ChapterFour/0400~0499/0478.Generate-Random-Point-in-a-Circle.md" >}})|Medium||||39.6%|
+|0483|Smallest Good Base|[Go]({{< relref "/ChapterFour/0400~0499/0483.Smallest-Good-Base.md" >}})|Hard||||38.8%|
+|0492|Construct the Rectangle|[Go]({{< relref "/ChapterFour/0400~0499/0492.Construct-the-Rectangle.md" >}})|Easy||||54.8%|
+|0497|Random Point in Non-overlapping Rectangles|[Go]({{< relref "/ChapterFour/0400~0499/0497.Random-Point-in-Non-overlapping-Rectangles.md" >}})|Medium||||39.4%|
+|0504|Base 7|[Go]({{< relref "/ChapterFour/0500~0599/0504.Base-7.md" >}})|Easy||||48.5%|
+|0507|Perfect Number|[Go]({{< relref "/ChapterFour/0500~0599/0507.Perfect-Number.md" >}})|Easy||||37.7%|
+|0509|Fibonacci Number|[Go]({{< relref "/ChapterFour/0500~0599/0509.Fibonacci-Number.md" >}})|Easy||||69.8%|
+|0519|Random Flip Matrix|[Go]({{< relref "/ChapterFour/0500~0599/0519.Random-Flip-Matrix.md" >}})|Medium||||40.0%|
+|0523|Continuous Subarray Sum|[Go]({{< relref "/ChapterFour/0500~0599/0523.Continuous-Subarray-Sum.md" >}})|Medium||||28.5%|
+|0528|Random Pick with Weight|[Go]({{< relref "/ChapterFour/0500~0599/0528.Random-Pick-with-Weight.md" >}})|Medium||||46.1%|
+|0537|Complex Number Multiplication|[Go]({{< relref "/ChapterFour/0500~0599/0537.Complex-Number-Multiplication.md" >}})|Medium||||71.4%|
+|0598|Range Addition II|[Go]({{< relref "/ChapterFour/0500~0599/0598.Range-Addition-II.md" >}})|Easy||||55.3%|
+|0628|Maximum Product of Three Numbers|[Go]({{< relref "/ChapterFour/0600~0699/0628.Maximum-Product-of-Three-Numbers.md" >}})|Easy| O(n)| O(1)||45.9%|
+|0633|Sum of Square Numbers|[Go]({{< relref "/ChapterFour/0600~0699/0633.Sum-of-Square-Numbers.md" >}})|Medium||||34.4%|
+|0667|Beautiful Arrangement II|[Go]({{< relref "/ChapterFour/0600~0699/0667.Beautiful-Arrangement-II.md" >}})|Medium||||59.8%|
+|0668|Kth Smallest Number in Multiplication Table|[Go]({{< relref "/ChapterFour/0600~0699/0668.Kth-Smallest-Number-in-Multiplication-Table.md" >}})|Hard||||51.4%|
+|0710|Random Pick with Blacklist|[Go]({{< relref "/ChapterFour/0700~0799/0710.Random-Pick-with-Blacklist.md" >}})|Hard||||33.5%|
+|0728|Self Dividing Numbers|[Go]({{< relref "/ChapterFour/0700~0799/0728.Self-Dividing-Numbers.md" >}})|Easy||||77.9%|
+|0762|Prime Number of Set Bits in Binary Representation|[Go]({{< relref "/ChapterFour/0700~0799/0762.Prime-Number-of-Set-Bits-in-Binary-Representation.md" >}})|Easy||||68.0%|
+|0775|Global and Local Inversions|[Go]({{< relref "/ChapterFour/0700~0799/0775.Global-and-Local-Inversions.md" >}})|Medium||||43.3%|
+|0781|Rabbits in Forest|[Go]({{< relref "/ChapterFour/0700~0799/0781.Rabbits-in-Forest.md" >}})|Medium||||54.7%|
+|0793|Preimage Size of Factorial Zeroes Function|[Go]({{< relref "/ChapterFour/0700~0799/0793.Preimage-Size-of-Factorial-Zeroes-Function.md" >}})|Hard||||43.2%|
+|0810|Chalkboard XOR Game|[Go]({{< relref "/ChapterFour/0800~0899/0810.Chalkboard-XOR-Game.md" >}})|Hard||||55.8%|
+|0812|Largest Triangle Area|[Go]({{< relref "/ChapterFour/0800~0899/0812.Largest-Triangle-Area.md" >}})|Easy||||59.9%|
+|0836|Rectangle Overlap|[Go]({{< relref "/ChapterFour/0800~0899/0836.Rectangle-Overlap.md" >}})|Easy||||43.9%|
+|0869|Reordered Power of 2|[Go]({{< relref "/ChapterFour/0800~0899/0869.Reordered-Power-of-2.md" >}})|Medium||||63.5%|
+|0877|Stone Game|[Go]({{< relref "/ChapterFour/0800~0899/0877.Stone-Game.md" >}})|Medium||||69.7%|
+|0878|Nth Magical Number|[Go]({{< relref "/ChapterFour/0800~0899/0878.Nth-Magical-Number.md" >}})|Hard||||35.4%|
+|0887|Super Egg Drop|[Go]({{< relref "/ChapterFour/0800~0899/0887.Super-Egg-Drop.md" >}})|Hard||||27.1%|
+|0891|Sum of Subsequence Widths|[Go]({{< relref "/ChapterFour/0800~0899/0891.Sum-of-Subsequence-Widths.md" >}})|Hard| O(n log n)| O(1)||36.7%|
+|0892|Surface Area of 3D Shapes|[Go]({{< relref "/ChapterFour/0800~0899/0892.Surface-Area-of-3D-Shapes.md" >}})|Easy||||64.0%|
+|0910|Smallest Range II|[Go]({{< relref "/ChapterFour/0900~0999/0910.Smallest-Range-II.md" >}})|Medium||||35.2%|
+|0914|X of a Kind in a Deck of Cards|[Go]({{< relref "/ChapterFour/0900~0999/0914.X-of-a-Kind-in-a-Deck-of-Cards.md" >}})|Easy||||31.2%|
+|0920|Number of Music Playlists|[Go]({{< relref "/ChapterFour/0900~0999/0920.Number-of-Music-Playlists.md" >}})|Hard||||50.7%|
+|0927|Three Equal Parts|[Go]({{< relref "/ChapterFour/0900~0999/0927.Three-Equal-Parts.md" >}})|Hard||||39.6%|
+|0952|Largest Component Size by Common Factor|[Go]({{< relref "/ChapterFour/0900~0999/0952.Largest-Component-Size-by-Common-Factor.md" >}})|Hard||||40.0%|
+|0970|Powerful Integers|[Go]({{< relref "/ChapterFour/0900~0999/0970.Powerful-Integers.md" >}})|Medium||||43.6%|
+|0973|K Closest Points to Origin|[Go]({{< relref "/ChapterFour/0900~0999/0973.K-Closest-Points-to-Origin.md" >}})|Medium||||65.7%|
+|0976|Largest Perimeter Triangle|[Go]({{< relref "/ChapterFour/0900~0999/0976.Largest-Perimeter-Triangle.md" >}})|Easy| O(n log n)| O(log n) ||54.7%|
+|0989|Add to Array-Form of Integer|[Go]({{< relref "/ChapterFour/0900~0999/0989.Add-to-Array-Form-of-Integer.md" >}})|Easy||||47.1%|
+|0991|Broken Calculator|[Go]({{< relref "/ChapterFour/0900~0999/0991.Broken-Calculator.md" >}})|Medium||||54.1%|
+|0996|Number of Squareful Arrays|[Go]({{< relref "/ChapterFour/0900~0999/0996.Number-of-Squareful-Arrays.md" >}})|Hard| O(n log n)| O(n) ||49.2%|
+|1006|Clumsy Factorial|[Go]({{< relref "/ChapterFour/1000~1099/1006.Clumsy-Factorial.md" >}})|Medium||||55.5%|
+|1017|Convert to Base -2|[Go]({{< relref "/ChapterFour/1000~1099/1017.Convert-to-Base-2.md" >}})|Medium||||60.9%|
+|1025|Divisor Game|[Go]({{< relref "/ChapterFour/1000~1099/1025.Divisor-Game.md" >}})|Easy| O(1)| O(1)||67.6%|
+|1030|Matrix Cells in Distance Order|[Go]({{< relref "/ChapterFour/1000~1099/1030.Matrix-Cells-in-Distance-Order.md" >}})|Easy||||69.7%|
+|1037|Valid Boomerang|[Go]({{< relref "/ChapterFour/1000~1099/1037.Valid-Boomerang.md" >}})|Easy||||37.0%|
+|1040|Moving Stones Until Consecutive II|[Go]({{< relref "/ChapterFour/1000~1099/1040.Moving-Stones-Until-Consecutive-II.md" >}})|Medium||||55.9%|
+|1073|Adding Two Negabinary Numbers|[Go]({{< relref "/ChapterFour/1000~1099/1073.Adding-Two-Negabinary-Numbers.md" >}})|Medium||||36.5%|
+|1093|Statistics from a Large Sample|[Go]({{< relref "/ChapterFour/1000~1099/1093.Statistics-from-a-Large-Sample.md" >}})|Medium||||43.5%|
+|1104|Path In Zigzag Labelled Binary Tree|[Go]({{< relref "/ChapterFour/1100~1199/1104.Path-In-Zigzag-Labelled-Binary-Tree.md" >}})|Medium||||75.1%|
+|1137|N-th Tribonacci Number|[Go]({{< relref "/ChapterFour/1100~1199/1137.N-th-Tribonacci-Number.md" >}})|Easy||||63.7%|
+|1154|Day of the Year|[Go]({{< relref "/ChapterFour/1100~1199/1154.Day-of-the-Year.md" >}})|Easy||||49.6%|
+|1175|Prime Arrangements|[Go]({{< relref "/ChapterFour/1100~1199/1175.Prime-Arrangements.md" >}})|Easy||||54.7%|
+|1185|Day of the Week|[Go]({{< relref "/ChapterFour/1100~1199/1185.Day-of-the-Week.md" >}})|Easy||||57.4%|
+|1201|Ugly Number III|[Go]({{< relref "/ChapterFour/1200~1299/1201.Ugly-Number-III.md" >}})|Medium||||28.9%|
+|1217|Minimum Cost to Move Chips to The Same Position|[Go]({{< relref "/ChapterFour/1200~1299/1217.Minimum-Cost-to-Move-Chips-to-The-Same-Position.md" >}})|Easy||||71.9%|
+|1232|Check If It Is a Straight Line|[Go]({{< relref "/ChapterFour/1200~1299/1232.Check-If-It-Is-a-Straight-Line.md" >}})|Easy||||40.3%|
+|1252|Cells with Odd Values in a Matrix|[Go]({{< relref "/ChapterFour/1200~1299/1252.Cells-with-Odd-Values-in-a-Matrix.md" >}})|Easy||||78.5%|
+|1266|Minimum Time Visiting All Points|[Go]({{< relref "/ChapterFour/1200~1299/1266.Minimum-Time-Visiting-All-Points.md" >}})|Easy||||79.1%|
+|1281|Subtract the Product and Sum of Digits of an Integer|[Go]({{< relref "/ChapterFour/1200~1299/1281.Subtract-the-Product-and-Sum-of-Digits-of-an-Integer.md" >}})|Easy||||86.6%|
+|1290|Convert Binary Number in a Linked List to Integer|[Go]({{< relref "/ChapterFour/1200~1299/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer.md" >}})|Easy||||82.1%|
+|1304|Find N Unique Integers Sum up to Zero|[Go]({{< relref "/ChapterFour/1300~1399/1304.Find-N-Unique-Integers-Sum-up-to-Zero.md" >}})|Easy||||76.9%|
+|1317|Convert Integer to the Sum of Two No-Zero Integers|[Go]({{< relref "/ChapterFour/1300~1399/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers.md" >}})|Easy||||55.4%|
+|1442|Count Triplets That Can Form Two Arrays of Equal XOR|[Go]({{< relref "/ChapterFour/1400~1499/1442.Count-Triplets-That-Can-Form-Two-Arrays-of-Equal-XOR.md" >}})|Medium||||76.1%|
+|1486|XOR Operation in an Array|[Go]({{< relref "/ChapterFour/1400~1499/1486.XOR-Operation-in-an-Array.md" >}})|Easy||||84.6%|
+|1512|Number of Good Pairs|[Go]({{< relref "/ChapterFour/1500~1599/1512.Number-of-Good-Pairs.md" >}})|Easy||||88.2%|
+|1518|Water Bottles|[Go]({{< relref "/ChapterFour/1500~1599/1518.Water-Bottles.md" >}})|Easy||||60.5%|
+|1551|Minimum Operations to Make Array Equal|[Go]({{< relref "/ChapterFour/1500~1599/1551.Minimum-Operations-to-Make-Array-Equal.md" >}})|Medium||||81.5%|
+|1573|Number of Ways to Split a String|[Go]({{< relref "/ChapterFour/1500~1599/1573.Number-of-Ways-to-Split-a-String.md" >}})|Medium||||32.5%|
+|1641|Count Sorted Vowel Strings|[Go]({{< relref "/ChapterFour/1600~1699/1641.Count-Sorted-Vowel-Strings.md" >}})|Medium||||77.4%|
+|1648|Sell Diminishing-Valued Colored Balls|[Go]({{< relref "/ChapterFour/1600~1699/1648.Sell-Diminishing-Valued-Colored-Balls.md" >}})|Medium||||30.4%|
+|1680|Concatenation of Consecutive Binary Numbers|[Go]({{< relref "/ChapterFour/1600~1699/1680.Concatenation-of-Consecutive-Binary-Numbers.md" >}})|Medium||||57.0%|
+|1685|Sum of Absolute Differences in a Sorted Array|[Go]({{< relref "/ChapterFour/1600~1699/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array.md" >}})|Medium||||63.5%|
+|1688|Count of Matches in Tournament|[Go]({{< relref "/ChapterFour/1600~1699/1688.Count-of-Matches-in-Tournament.md" >}})|Easy||||83.1%|
+|1690|Stone Game VII|[Go]({{< relref "/ChapterFour/1600~1699/1690.Stone-Game-VII.md" >}})|Medium||||58.1%|
+|1716|Calculate Money in Leetcode Bank|[Go]({{< relref "/ChapterFour/1700~1799/1716.Calculate-Money-in-Leetcode-Bank.md" >}})|Easy||||66.1%|
+|1742|Maximum Number of Balls in a Box|[Go]({{< relref "/ChapterFour/1700~1799/1742.Maximum-Number-of-Balls-in-a-Box.md" >}})|Easy||||73.6%|
+|2038|Remove Colored Pieces if Both Neighbors are the Same Color|[Go]({{< relref "/ChapterFour/2000~2099/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color.md" >}})|Medium||||57.9%|
+|2165|Smallest Value of the Rearranged Number|[Go]({{< relref "/ChapterFour/2100~2199/2165.Smallest-Value-of-the-Rearranged-Number.md" >}})|Medium||||51.4%|
+|2169|Count Operations to Obtain Zero|[Go]({{< relref "/ChapterFour/2100~2199/2169.Count-Operations-to-Obtain-Zero.md" >}})|Easy||||75.2%|
+|2180|Count Integers With Even Digit Sum|[Go]({{< relref "/ChapterFour/2100~2199/2180.Count-Integers-With-Even-Digit-Sum.md" >}})|Easy||||65.5%|
+|2183|Count Array Pairs Divisible by K|[Go]({{< relref "/ChapterFour/2100~2199/2183.Count-Array-Pairs-Divisible-by-K.md" >}})|Hard||||28.3%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Binary_Search/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Hash_Table/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Segment_Tree.md b/website/content/ChapterTwo/Segment_Tree.md
new file mode 100644
index 000000000..85d41ea58
--- /dev/null
+++ b/website/content/ChapterTwo/Segment_Tree.md
@@ -0,0 +1,59 @@
+---
+title: 2.18 ✅ Segment Tree
+type: docs
+weight: 18
+---
+
+# Segment Tree
+
+![](https://img.halfrost.com/Leetcode/Segment_Tree.png)
+
+- 线段树的经典数组实现写法。将合并两个节点 pushUp 逻辑抽象出来了，可以实现任意操作(常见的操作有：加法，取 max，min 等等)。第 218 题，第 303 题，第 307 题，第 699 题。
+- 计数线段树的经典写法。第 315 题，第 327 题，第 493 题。
+- 线段树的树的实现写法。第 715 题，第 732 题。
+- 区间懒惰更新。第 218 题，第 699 题。
+- 离散化。离散化需要注意一个特殊情况：假如三个区间为 [1,10] [1,4] [6,10]，离散化后 x[1]=1,x[2]=4,x[3]=6,x[4]=10。第一个区间为 [1,4]，第二个区间为 [1,2]，第三个区间为 [3,4]，这样一来，区间一 = 区间二 + 区间三，这和离散前的模型不符，离散前，很明显，区间一 > 区间二 + 区间三。正确的做法是：在相差大于 1 的数间加一个数，例如在上面 1 4 6 10 中间加 5，即可 x[1]=1,x[2]=4,x[3]=5,x[4]=6,x[5]=10。这样处理之后，区间一是 1-5 ，区间二是 1-2 ，区间三是 4-5 。
+- 灵活构建线段树。线段树节点可以存储多条信息，合并两个节点的 pushUp 操作也可以是多样的。第 850 题，第 1157 题。
+
+
+线段树[题型](https://blog.csdn.net/xuechelingxiao/article/details/38313105)从简单到困难:
+
+1. 单点更新:  
+	[HDU 1166 敌兵布阵](http://acm.hdu.edu.cn/showproblem.php?pid=1166) update:单点增减 query:区间求和  
+	[HDU 1754 I Hate It](http://acm.hdu.edu.cn/showproblem.php?pid=1754) update:单点替换 query:区间最值  
+	[HDU 1394 Minimum Inversion Number](http://acm.hdu.edu.cn/showproblem.php?pid=1394) update:单点增减 query:区间求和  
+	[HDU 2795 Billboard](http://acm.hdu.edu.cn/showproblem.php?pid=2795) query:区间求最大值的位子(直接把update的操作在query里做了)
+2. 区间更新:  
+	[HDU 1698 Just a Hook](http://acm.hdu.edu.cn/showproblem.php?pid=1698) update:成段替换 (由于只query一次总区间,所以可以直接输出 1 结点的信息)  
+	[POJ 3468 A Simple Problem with Integers](http://poj.org/problem?id=3468) update:成段增减 query:区间求和  
+	[POJ 2528 Mayor’s posters](http://poj.org/problem?id=2528) 离散化 + update:成段替换 query:简单hash  
+	[POJ 3225 Help with Intervals](http://poj.org/problem?id=3225) update:成段替换,区间异或 query:简单hash
+3. 区间合并(这类题目会询问区间中满足条件的连续最长区间,所以PushUp的时候需要对左右儿子的区间进行合并):  
+	[POJ 3667 Hotel](http://poj.org/problem?id=3667) update:区间替换 query:询问满足条件的最左端点
+4. 扫描线(这类题目需要将一些操作排序,然后从左到右用一根扫描线扫过去最典型的就是矩形面积并,周长并等题):  
+	[HDU 1542 Atlantis](http://acm.hdu.edu.cn/showproblem.php?pid=1542) update:区间增减 query:直接取根节点的值  
+	[HDU 1828 Picture](http://acm.hdu.edu.cn/showproblem.php?pid=1828) update:区间增减 query:直接取根节点的值
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0218|The Skyline Problem|[Go]({{< relref "/ChapterFour/0200~0299/0218.The-Skyline-Problem.md" >}})|Hard| O(n log n)| O(n)|❤️|41.9%|
+|0307|Range Sum Query - Mutable|[Go]({{< relref "/ChapterFour/0300~0399/0307.Range-Sum-Query-Mutable.md" >}})|Medium| O(1)| O(n)||40.7%|
+|0315|Count of Smaller Numbers After Self|[Go]({{< relref "/ChapterFour/0300~0399/0315.Count-of-Smaller-Numbers-After-Self.md" >}})|Hard| O(n log n)| O(n)||42.6%|
+|0327|Count of Range Sum|[Go]({{< relref "/ChapterFour/0300~0399/0327.Count-of-Range-Sum.md" >}})|Hard| O(n log n)| O(n)|❤️|35.8%|
+|0493|Reverse Pairs|[Go]({{< relref "/ChapterFour/0400~0499/0493.Reverse-Pairs.md" >}})|Hard| O(n log n)| O(n)||30.9%|
+|0699|Falling Squares|[Go]({{< relref "/ChapterFour/0600~0699/0699.Falling-Squares.md" >}})|Hard| O(n log n)| O(n)|❤️|44.7%|
+|0715|Range Module|[Go]({{< relref "/ChapterFour/0700~0799/0715.Range-Module.md" >}})|Hard| O(log n)| O(n)|❤️|44.6%|
+|0729|My Calendar I|[Go]({{< relref "/ChapterFour/0700~0799/0729.My-Calendar-I.md" >}})|Medium||||56.8%|
+|0732|My Calendar III|[Go]({{< relref "/ChapterFour/0700~0799/0732.My-Calendar-III.md" >}})|Hard| O(log n)| O(n)|❤️|71.5%|
+|0850|Rectangle Area II|[Go]({{< relref "/ChapterFour/0800~0899/0850.Rectangle-Area-II.md" >}})|Hard| O(n log n)| O(n)|❤️|53.9%|
+|1157|Online Majority Element In Subarray|[Go]({{< relref "/ChapterFour/1100~1199/1157.Online-Majority-Element-In-Subarray.md" >}})|Hard| O(log n)| O(n)|❤️|41.8%|
+|1649|Create Sorted Array through Instructions|[Go]({{< relref "/ChapterFour/1600~1699/1649.Create-Sorted-Array-through-Instructions.md" >}})|Hard||||37.5%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Sliding_Window/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Binary_Indexed_Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Sliding_Window.md b/website/content/ChapterTwo/Sliding_Window.md
new file mode 100644
index 000000000..f0e75b163
--- /dev/null
+++ b/website/content/ChapterTwo/Sliding_Window.md
@@ -0,0 +1,74 @@
+---
+title: 2.17 ✅ Sliding Window
+type: docs
+weight: 17
+---
+
+# Sliding Window
+
+![](https://img.halfrost.com/Leetcode/Sliding_Window.png)
+
+- 双指针滑动窗口的经典写法。右指针不断往右移，移动到不能往右移动为止(具体条件根据题目而定)。当右指针到最右边以后，开始挪动左指针，释放窗口左边界。第 3 题，第 76 题，第 209 题，第 424 题，第 438 题，第 567 题，第 713 题，第 763 题，第 845 题，第 881 题，第 904 题，第 978 题，第 992 题，第 1004 题，第 1040 题，第 1052 题。
+
+```c
+	left, right := 0, -1
+
+	for left < len(s) {
+		if right+1 < len(s) && freq[s[right+1]-'a'] == 0 {
+			freq[s[right+1]-'a']++
+			right++
+		} else {
+			freq[s[left]-'a']--
+			left++
+		}
+		result = max(result, right-left+1)
+	}
+```
+- 滑动窗口经典题。第 239 题，第 480 题。
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0003|Longest Substring Without Repeating Characters|[Go]({{< relref "/ChapterFour/0001~0099/0003.Longest-Substring-Without-Repeating-Characters.md" >}})|Medium| O(n)| O(1)|❤️|33.8%|
+|0030|Substring with Concatenation of All Words|[Go]({{< relref "/ChapterFour/0001~0099/0030.Substring-with-Concatenation-of-All-Words.md" >}})|Hard||||31.2%|
+|0076|Minimum Window Substring|[Go]({{< relref "/ChapterFour/0001~0099/0076.Minimum-Window-Substring.md" >}})|Hard| O(n)| O(n)|❤️|40.9%|
+|0187|Repeated DNA Sequences|[Go]({{< relref "/ChapterFour/0100~0199/0187.Repeated-DNA-Sequences.md" >}})|Medium||||47.0%|
+|0209|Minimum Size Subarray Sum|[Go]({{< relref "/ChapterFour/0200~0299/0209.Minimum-Size-Subarray-Sum.md" >}})|Medium||||45.0%|
+|0219|Contains Duplicate II|[Go]({{< relref "/ChapterFour/0200~0299/0219.Contains-Duplicate-II.md" >}})|Easy||||42.6%|
+|0220|Contains Duplicate III|[Go]({{< relref "/ChapterFour/0200~0299/0220.Contains-Duplicate-III.md" >}})|Hard||||22.1%|
+|0239|Sliding Window Maximum|[Go]({{< relref "/ChapterFour/0200~0299/0239.Sliding-Window-Maximum.md" >}})|Hard| O(n * k)| O(n)|❤️|46.3%|
+|0395|Longest Substring with At Least K Repeating Characters|[Go]({{< relref "/ChapterFour/0300~0399/0395.Longest-Substring-with-At-Least-K-Repeating-Characters.md" >}})|Medium||||44.8%|
+|0424|Longest Repeating Character Replacement|[Go]({{< relref "/ChapterFour/0400~0499/0424.Longest-Repeating-Character-Replacement.md" >}})|Medium| O(n)| O(1) ||52.0%|
+|0438|Find All Anagrams in a String|[Go]({{< relref "/ChapterFour/0400~0499/0438.Find-All-Anagrams-in-a-String.md" >}})|Medium||||50.2%|
+|0480|Sliding Window Median|[Go]({{< relref "/ChapterFour/0400~0499/0480.Sliding-Window-Median.md" >}})|Hard| O(n * log k)| O(k)|❤️|41.1%|
+|0567|Permutation in String|[Go]({{< relref "/ChapterFour/0500~0599/0567.Permutation-in-String.md" >}})|Medium| O(n)| O(1)|❤️|44.3%|
+|0632|Smallest Range Covering Elements from K Lists|[Go]({{< relref "/ChapterFour/0600~0699/0632.Smallest-Range-Covering-Elements-from-K-Lists.md" >}})|Hard||||61.0%|
+|0643|Maximum Average Subarray I|[Go]({{< relref "/ChapterFour/0600~0699/0643.Maximum-Average-Subarray-I.md" >}})|Easy||||43.7%|
+|0658|Find K Closest Elements|[Go]({{< relref "/ChapterFour/0600~0699/0658.Find-K-Closest-Elements.md" >}})|Medium||||46.8%|
+|0713|Subarray Product Less Than K|[Go]({{< relref "/ChapterFour/0700~0799/0713.Subarray-Product-Less-Than-K.md" >}})|Medium||||45.8%|
+|0718|Maximum Length of Repeated Subarray|[Go]({{< relref "/ChapterFour/0700~0799/0718.Maximum-Length-of-Repeated-Subarray.md" >}})|Medium||||51.3%|
+|0862|Shortest Subarray with Sum at Least K|[Go]({{< relref "/ChapterFour/0800~0899/0862.Shortest-Subarray-with-Sum-at-Least-K.md" >}})|Hard||||26.0%|
+|0904|Fruit Into Baskets|[Go]({{< relref "/ChapterFour/0900~0999/0904.Fruit-Into-Baskets.md" >}})|Medium||||43.7%|
+|0930|Binary Subarrays With Sum|[Go]({{< relref "/ChapterFour/0900~0999/0930.Binary-Subarrays-With-Sum.md" >}})|Medium||||52.2%|
+|0978|Longest Turbulent Subarray|[Go]({{< relref "/ChapterFour/0900~0999/0978.Longest-Turbulent-Subarray.md" >}})|Medium| O(n)| O(1)|❤️|47.2%|
+|0992|Subarrays with K Different Integers|[Go]({{< relref "/ChapterFour/0900~0999/0992.Subarrays-with-K-Different-Integers.md" >}})|Hard| O(n)| O(n)|❤️|54.6%|
+|0995|Minimum Number of K Consecutive Bit Flips|[Go]({{< relref "/ChapterFour/0900~0999/0995.Minimum-Number-of-K-Consecutive-Bit-Flips.md" >}})|Hard| O(n)| O(1)|❤️|51.2%|
+|1004|Max Consecutive Ones III|[Go]({{< relref "/ChapterFour/1000~1099/1004.Max-Consecutive-Ones-III.md" >}})|Medium| O(n)| O(1) ||63.2%|
+|1052|Grumpy Bookstore Owner|[Go]({{< relref "/ChapterFour/1000~1099/1052.Grumpy-Bookstore-Owner.md" >}})|Medium| O(n log n)| O(1) ||57.1%|
+|1208|Get Equal Substrings Within Budget|[Go]({{< relref "/ChapterFour/1200~1299/1208.Get-Equal-Substrings-Within-Budget.md" >}})|Medium||||48.6%|
+|1234|Replace the Substring for Balanced String|[Go]({{< relref "/ChapterFour/1200~1299/1234.Replace-the-Substring-for-Balanced-String.md" >}})|Medium||||37.2%|
+|1423|Maximum Points You Can Obtain from Cards|[Go]({{< relref "/ChapterFour/1400~1499/1423.Maximum-Points-You-Can-Obtain-from-Cards.md" >}})|Medium||||52.2%|
+|1438|Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit|[Go]({{< relref "/ChapterFour/1400~1499/1438.Longest-Continuous-Subarray-With-Absolute-Diff-Less-Than-or-Equal-to-Limit.md" >}})|Medium||||48.3%|
+|1658|Minimum Operations to Reduce X to Zero|[Go]({{< relref "/ChapterFour/1600~1699/1658.Minimum-Operations-to-Reduce-X-to-Zero.md" >}})|Medium||||37.6%|
+|1695|Maximum Erasure Value|[Go]({{< relref "/ChapterFour/1600~1699/1695.Maximum-Erasure-Value.md" >}})|Medium||||57.6%|
+|1696|Jump Game VI|[Go]({{< relref "/ChapterFour/1600~1699/1696.Jump-Game-VI.md" >}})|Medium||||46.1%|
+|1763|Longest Nice Substring|[Go]({{< relref "/ChapterFour/1700~1799/1763.Longest-Nice-Substring.md" >}})|Easy||||61.5%|
+|1984|Minimum Difference Between Highest and Lowest of K Scores|[Go]({{< relref "/ChapterFour/1900~1999/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores.md" >}})|Easy||||54.5%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Union_Find/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Segment_Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Sorting.md b/website/content/ChapterTwo/Sorting.md
new file mode 100755
index 000000000..f50ce2687
--- /dev/null
+++ b/website/content/ChapterTwo/Sorting.md
@@ -0,0 +1,136 @@
+---
+title: 2.14 ✅ Sorting
+type: docs
+weight: 14
+---
+
+# Sorting
+
+![](https://img.halfrost.com/Leetcode/Sort.png)
+
+- 深刻的理解多路快排。第 75 题。
+- 链表的排序，插入排序(第 147 题)和归并排序(第 148 题)
+- 桶排序和基数排序。第 164 题。
+- "摆动排序"。第 324 题。
+- 两两不相邻的排序。第 767 题，第 1054 题。
+- "饼子排序"。第 969 题。
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0015|3Sum|[Go]({{< relref "/ChapterFour/0001~0099/0015.3Sum.md" >}})|Medium||||32.6%|
+|0016|3Sum Closest|[Go]({{< relref "/ChapterFour/0001~0099/0016.3Sum-Closest.md" >}})|Medium||||45.8%|
+|0018|4Sum|[Go]({{< relref "/ChapterFour/0001~0099/0018.4Sum.md" >}})|Medium||||35.9%|
+|0049|Group Anagrams|[Go]({{< relref "/ChapterFour/0001~0099/0049.Group-Anagrams.md" >}})|Medium||||66.7%|
+|0056|Merge Intervals|[Go]({{< relref "/ChapterFour/0001~0099/0056.Merge-Intervals.md" >}})|Medium| O(n log n)| O(log n)||46.2%|
+|0075|Sort Colors|[Go]({{< relref "/ChapterFour/0001~0099/0075.Sort-Colors.md" >}})|Medium| O(n)| O(1)|❤️|58.5%|
+|0088|Merge Sorted Array|[Go]({{< relref "/ChapterFour/0001~0099/0088.Merge-Sorted-Array.md" >}})|Easy||||46.6%|
+|0147|Insertion Sort List|[Go]({{< relref "/ChapterFour/0100~0199/0147.Insertion-Sort-List.md" >}})|Medium| O(n^2)| O(1) |❤️|51.0%|
+|0148|Sort List|[Go]({{< relref "/ChapterFour/0100~0199/0148.Sort-List.md" >}})|Medium|O(n log n)| O(log n)|❤️|55.1%|
+|0164|Maximum Gap|[Go]({{< relref "/ChapterFour/0100~0199/0164.Maximum-Gap.md" >}})|Hard| O(n log n)| O(log n) |❤️|43.3%|
+|0169|Majority Element|[Go]({{< relref "/ChapterFour/0100~0199/0169.Majority-Element.md" >}})|Easy||||63.9%|
+|0179|Largest Number|[Go]({{< relref "/ChapterFour/0100~0199/0179.Largest-Number.md" >}})|Medium| O(n log n)| O(log n) |❤️|34.5%|
+|0215|Kth Largest Element in an Array|[Go]({{< relref "/ChapterFour/0200~0299/0215.Kth-Largest-Element-in-an-Array.md" >}})|Medium||||66.1%|
+|0217|Contains Duplicate|[Go]({{< relref "/ChapterFour/0200~0299/0217.Contains-Duplicate.md" >}})|Easy||||61.4%|
+|0220|Contains Duplicate III|[Go]({{< relref "/ChapterFour/0200~0299/0220.Contains-Duplicate-III.md" >}})|Hard| O(n log n)| O(1) |❤️|22.1%|
+|0229|Majority Element II|[Go]({{< relref "/ChapterFour/0200~0299/0229.Majority-Element-II.md" >}})|Medium||||45.0%|
+|0242|Valid Anagram|[Go]({{< relref "/ChapterFour/0200~0299/0242.Valid-Anagram.md" >}})|Easy| O(n)| O(n) ||63.0%|
+|0268|Missing Number|[Go]({{< relref "/ChapterFour/0200~0299/0268.Missing-Number.md" >}})|Easy||||62.5%|
+|0274|H-Index|[Go]({{< relref "/ChapterFour/0200~0299/0274.H-Index.md" >}})|Medium|  O(n)| O(n) ||38.3%|
+|0324|Wiggle Sort II|[Go]({{< relref "/ChapterFour/0300~0399/0324.Wiggle-Sort-II.md" >}})|Medium| O(n)| O(n)|❤️|33.3%|
+|0347|Top K Frequent Elements|[Go]({{< relref "/ChapterFour/0300~0399/0347.Top-K-Frequent-Elements.md" >}})|Medium||||64.2%|
+|0349|Intersection of Two Arrays|[Go]({{< relref "/ChapterFour/0300~0399/0349.Intersection-of-Two-Arrays.md" >}})|Easy| O(n)| O(n) ||70.9%|
+|0350|Intersection of Two Arrays II|[Go]({{< relref "/ChapterFour/0300~0399/0350.Intersection-of-Two-Arrays-II.md" >}})|Easy| O(n)| O(n) ||56.0%|
+|0354|Russian Doll Envelopes|[Go]({{< relref "/ChapterFour/0300~0399/0354.Russian-Doll-Envelopes.md" >}})|Hard||||38.0%|
+|0368|Largest Divisible Subset|[Go]({{< relref "/ChapterFour/0300~0399/0368.Largest-Divisible-Subset.md" >}})|Medium||||41.5%|
+|0378|Kth Smallest Element in a Sorted Matrix|[Go]({{< relref "/ChapterFour/0300~0399/0378.Kth-Smallest-Element-in-a-Sorted-Matrix.md" >}})|Medium||||61.7%|
+|0389|Find the Difference|[Go]({{< relref "/ChapterFour/0300~0399/0389.Find-the-Difference.md" >}})|Easy||||59.9%|
+|0414|Third Maximum Number|[Go]({{< relref "/ChapterFour/0400~0499/0414.Third-Maximum-Number.md" >}})|Easy||||33.2%|
+|0435|Non-overlapping Intervals|[Go]({{< relref "/ChapterFour/0400~0499/0435.Non-overlapping-Intervals.md" >}})|Medium||||50.3%|
+|0436|Find Right Interval|[Go]({{< relref "/ChapterFour/0400~0499/0436.Find-Right-Interval.md" >}})|Medium||||50.8%|
+|0451|Sort Characters By Frequency|[Go]({{< relref "/ChapterFour/0400~0499/0451.Sort-Characters-By-Frequency.md" >}})|Medium||||70.1%|
+|0455|Assign Cookies|[Go]({{< relref "/ChapterFour/0400~0499/0455.Assign-Cookies.md" >}})|Easy||||49.9%|
+|0462|Minimum Moves to Equal Array Elements II|[Go]({{< relref "/ChapterFour/0400~0499/0462.Minimum-Moves-to-Equal-Array-Elements-II.md" >}})|Medium||||60.0%|
+|0475|Heaters|[Go]({{< relref "/ChapterFour/0400~0499/0475.Heaters.md" >}})|Medium||||36.5%|
+|0506|Relative Ranks|[Go]({{< relref "/ChapterFour/0500~0599/0506.Relative-Ranks.md" >}})|Easy||||60.5%|
+|0524|Longest Word in Dictionary through Deleting|[Go]({{< relref "/ChapterFour/0500~0599/0524.Longest-Word-in-Dictionary-through-Deleting.md" >}})|Medium| O(n)| O(1) ||51.0%|
+|0532|K-diff Pairs in an Array|[Go]({{< relref "/ChapterFour/0500~0599/0532.K-diff-Pairs-in-an-Array.md" >}})|Medium||||41.2%|
+|0561|Array Partition|[Go]({{< relref "/ChapterFour/0500~0599/0561.Array-Partition.md" >}})|Easy||||77.2%|
+|0581|Shortest Unsorted Continuous Subarray|[Go]({{< relref "/ChapterFour/0500~0599/0581.Shortest-Unsorted-Continuous-Subarray.md" >}})|Medium||||36.4%|
+|0594|Longest Harmonious Subsequence|[Go]({{< relref "/ChapterFour/0500~0599/0594.Longest-Harmonious-Subsequence.md" >}})|Easy||||53.5%|
+|0611|Valid Triangle Number|[Go]({{< relref "/ChapterFour/0600~0699/0611.Valid-Triangle-Number.md" >}})|Medium||||50.5%|
+|0628|Maximum Product of Three Numbers|[Go]({{< relref "/ChapterFour/0600~0699/0628.Maximum-Product-of-Three-Numbers.md" >}})|Easy||||45.9%|
+|0632|Smallest Range Covering Elements from K Lists|[Go]({{< relref "/ChapterFour/0600~0699/0632.Smallest-Range-Covering-Elements-from-K-Lists.md" >}})|Hard||||61.0%|
+|0645|Set Mismatch|[Go]({{< relref "/ChapterFour/0600~0699/0645.Set-Mismatch.md" >}})|Easy||||42.7%|
+|0658|Find K Closest Elements|[Go]({{< relref "/ChapterFour/0600~0699/0658.Find-K-Closest-Elements.md" >}})|Medium||||46.8%|
+|0692|Top K Frequent Words|[Go]({{< relref "/ChapterFour/0600~0699/0692.Top-K-Frequent-Words.md" >}})|Medium||||57.2%|
+|0710|Random Pick with Blacklist|[Go]({{< relref "/ChapterFour/0700~0799/0710.Random-Pick-with-Blacklist.md" >}})|Hard| O(n)| O(n)  ||33.5%|
+|0719|Find K-th Smallest Pair Distance|[Go]({{< relref "/ChapterFour/0700~0799/0719.Find-K-th-Smallest-Pair-Distance.md" >}})|Hard||||36.7%|
+|0720|Longest Word in Dictionary|[Go]({{< relref "/ChapterFour/0700~0799/0720.Longest-Word-in-Dictionary.md" >}})|Medium||||52.0%|
+|0726|Number of Atoms|[Go]({{< relref "/ChapterFour/0700~0799/0726.Number-of-Atoms.md" >}})|Hard||||52.1%|
+|0747|Largest Number At Least Twice of Others|[Go]({{< relref "/ChapterFour/0700~0799/0747.Largest-Number-At-Least-Twice-of-Others.md" >}})|Easy||||47.1%|
+|0767|Reorganize String|[Go]({{< relref "/ChapterFour/0700~0799/0767.Reorganize-String.md" >}})|Medium| O(n log n)| O(log n)  |❤️|52.9%|
+|0786|K-th Smallest Prime Fraction|[Go]({{< relref "/ChapterFour/0700~0799/0786.K-th-Smallest-Prime-Fraction.md" >}})|Medium||||51.6%|
+|0791|Custom Sort String|[Go]({{< relref "/ChapterFour/0700~0799/0791.Custom-Sort-String.md" >}})|Medium||||69.1%|
+|0792|Number of Matching Subsequences|[Go]({{< relref "/ChapterFour/0700~0799/0792.Number-of-Matching-Subsequences.md" >}})|Medium||||51.6%|
+|0825|Friends Of Appropriate Ages|[Go]({{< relref "/ChapterFour/0800~0899/0825.Friends-Of-Appropriate-Ages.md" >}})|Medium||||46.3%|
+|0826|Most Profit Assigning Work|[Go]({{< relref "/ChapterFour/0800~0899/0826.Most-Profit-Assigning-Work.md" >}})|Medium||||44.9%|
+|0846|Hand of Straights|[Go]({{< relref "/ChapterFour/0800~0899/0846.Hand-of-Straights.md" >}})|Medium||||56.2%|
+|0853|Car Fleet|[Go]({{< relref "/ChapterFour/0800~0899/0853.Car-Fleet.md" >}})|Medium| O(n log n)| O(log n)  ||50.3%|
+|0869|Reordered Power of 2|[Go]({{< relref "/ChapterFour/0800~0899/0869.Reordered-Power-of-2.md" >}})|Medium||||63.5%|
+|0870|Advantage Shuffle|[Go]({{< relref "/ChapterFour/0800~0899/0870.Advantage-Shuffle.md" >}})|Medium||||51.8%|
+|0881|Boats to Save People|[Go]({{< relref "/ChapterFour/0800~0899/0881.Boats-to-Save-People.md" >}})|Medium||||53.1%|
+|0888|Fair Candy Swap|[Go]({{< relref "/ChapterFour/0800~0899/0888.Fair-Candy-Swap.md" >}})|Easy||||60.7%|
+|0891|Sum of Subsequence Widths|[Go]({{< relref "/ChapterFour/0800~0899/0891.Sum-of-Subsequence-Widths.md" >}})|Hard||||36.6%|
+|0910|Smallest Range II|[Go]({{< relref "/ChapterFour/0900~0999/0910.Smallest-Range-II.md" >}})|Medium||||35.1%|
+|0922|Sort Array By Parity II|[Go]({{< relref "/ChapterFour/0900~0999/0922.Sort-Array-By-Parity-II.md" >}})|Easy| O(n)| O(1) ||70.7%|
+|0923|3Sum With Multiplicity|[Go]({{< relref "/ChapterFour/0900~0999/0923.3Sum-With-Multiplicity.md" >}})|Medium||||45.3%|
+|0969|Pancake Sorting|[Go]({{< relref "/ChapterFour/0900~0999/0969.Pancake-Sorting.md" >}})|Medium| O(n log n)| O(log n) |❤️|70.1%|
+|0973|K Closest Points to Origin|[Go]({{< relref "/ChapterFour/0900~0999/0973.K-Closest-Points-to-Origin.md" >}})|Medium| O(n log n)| O(log n) ||65.7%|
+|0976|Largest Perimeter Triangle|[Go]({{< relref "/ChapterFour/0900~0999/0976.Largest-Perimeter-Triangle.md" >}})|Easy| O(n log n)| O(log n) ||54.6%|
+|0977|Squares of a Sorted Array|[Go]({{< relref "/ChapterFour/0900~0999/0977.Squares-of-a-Sorted-Array.md" >}})|Easy||||71.9%|
+|1005|Maximize Sum Of Array After K Negations|[Go]({{< relref "/ChapterFour/1000~1099/1005.Maximize-Sum-Of-Array-After-K-Negations.md" >}})|Easy||||50.9%|
+|1030|Matrix Cells in Distance Order|[Go]({{< relref "/ChapterFour/1000~1099/1030.Matrix-Cells-in-Distance-Order.md" >}})|Easy| O(n^2)| O(1) ||69.7%|
+|1040|Moving Stones Until Consecutive II|[Go]({{< relref "/ChapterFour/1000~1099/1040.Moving-Stones-Until-Consecutive-II.md" >}})|Medium||||55.9%|
+|1051|Height Checker|[Go]({{< relref "/ChapterFour/1000~1099/1051.Height-Checker.md" >}})|Easy||||75.6%|
+|1054|Distant Barcodes|[Go]({{< relref "/ChapterFour/1000~1099/1054.Distant-Barcodes.md" >}})|Medium| O(n log n)| O(log n) |❤️|45.8%|
+|1122|Relative Sort Array|[Go]({{< relref "/ChapterFour/1100~1199/1122.Relative-Sort-Array.md" >}})|Easy||||68.6%|
+|1170|Compare Strings by Frequency of the Smallest Character|[Go]({{< relref "/ChapterFour/1100~1199/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character.md" >}})|Medium||||61.5%|
+|1200|Minimum Absolute Difference|[Go]({{< relref "/ChapterFour/1200~1299/1200.Minimum-Absolute-Difference.md" >}})|Easy||||69.6%|
+|1235|Maximum Profit in Job Scheduling|[Go]({{< relref "/ChapterFour/1200~1299/1235.Maximum-Profit-in-Job-Scheduling.md" >}})|Hard||||53.4%|
+|1296|Divide Array in Sets of K Consecutive Numbers|[Go]({{< relref "/ChapterFour/1200~1299/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers.md" >}})|Medium||||56.5%|
+|1300|Sum of Mutated Array Closest to Target|[Go]({{< relref "/ChapterFour/1300~1399/1300.Sum-of-Mutated-Array-Closest-to-Target.md" >}})|Medium||||43.6%|
+|1305|All Elements in Two Binary Search Trees|[Go]({{< relref "/ChapterFour/1300~1399/1305.All-Elements-in-Two-Binary-Search-Trees.md" >}})|Medium||||79.8%|
+|1329|Sort the Matrix Diagonally|[Go]({{< relref "/ChapterFour/1300~1399/1329.Sort-the-Matrix-Diagonally.md" >}})|Medium||||83.3%|
+|1337|The K Weakest Rows in a Matrix|[Go]({{< relref "/ChapterFour/1300~1399/1337.The-K-Weakest-Rows-in-a-Matrix.md" >}})|Easy||||72.1%|
+|1353|Maximum Number of Events That Can Be Attended|[Go]({{< relref "/ChapterFour/1300~1399/1353.Maximum-Number-of-Events-That-Can-Be-Attended.md" >}})|Medium||||32.5%|
+|1383|Maximum Performance of a Team|[Go]({{< relref "/ChapterFour/1300~1399/1383.Maximum-Performance-of-a-Team.md" >}})|Hard||||48.5%|
+|1385|Find the Distance Value Between Two Arrays|[Go]({{< relref "/ChapterFour/1300~1399/1385.Find-the-Distance-Value-Between-Two-Arrays.md" >}})|Easy||||66.5%|
+|1464|Maximum Product of Two Elements in an Array|[Go]({{< relref "/ChapterFour/1400~1499/1464.Maximum-Product-of-Two-Elements-in-an-Array.md" >}})|Easy||||79.9%|
+|1465|Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts|[Go]({{< relref "/ChapterFour/1400~1499/1465.Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts.md" >}})|Medium||||40.9%|
+|1608|Special Array With X Elements Greater Than or Equal X|[Go]({{< relref "/ChapterFour/1600~1699/1608.Special-Array-With-X-Elements-Greater-Than-or-Equal-X.md" >}})|Easy||||60.5%|
+|1619|Mean of Array After Removing Some Elements|[Go]({{< relref "/ChapterFour/1600~1699/1619.Mean-of-Array-After-Removing-Some-Elements.md" >}})|Easy||||65.7%|
+|1636|Sort Array by Increasing Frequency|[Go]({{< relref "/ChapterFour/1600~1699/1636.Sort-Array-by-Increasing-Frequency.md" >}})|Easy||||69.5%|
+|1647|Minimum Deletions to Make Character Frequencies Unique|[Go]({{< relref "/ChapterFour/1600~1699/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique.md" >}})|Medium||||59.1%|
+|1648|Sell Diminishing-Valued Colored Balls|[Go]({{< relref "/ChapterFour/1600~1699/1648.Sell-Diminishing-Valued-Colored-Balls.md" >}})|Medium||||30.5%|
+|1657|Determine if Two Strings Are Close|[Go]({{< relref "/ChapterFour/1600~1699/1657.Determine-if-Two-Strings-Are-Close.md" >}})|Medium||||56.3%|
+|1665|Minimum Initial Energy to Finish Tasks|[Go]({{< relref "/ChapterFour/1600~1699/1665.Minimum-Initial-Energy-to-Finish-Tasks.md" >}})|Hard||||56.3%|
+|1679|Max Number of K-Sum Pairs|[Go]({{< relref "/ChapterFour/1600~1699/1679.Max-Number-of-K-Sum-Pairs.md" >}})|Medium||||57.3%|
+|1691|Maximum Height by Stacking Cuboids|[Go]({{< relref "/ChapterFour/1600~1699/1691.Maximum-Height-by-Stacking-Cuboids.md" >}})|Hard||||54.4%|
+|1710|Maximum Units on a Truck|[Go]({{< relref "/ChapterFour/1700~1799/1710.Maximum-Units-on-a-Truck.md" >}})|Easy||||73.8%|
+|1818|Minimum Absolute Sum Difference|[Go]({{< relref "/ChapterFour/1800~1899/1818.Minimum-Absolute-Sum-Difference.md" >}})|Medium||||30.4%|
+|1846|Maximum Element After Decreasing and Rearranging|[Go]({{< relref "/ChapterFour/1800~1899/1846.Maximum-Element-After-Decreasing-and-Rearranging.md" >}})|Medium||||58.9%|
+|1877|Minimize Maximum Pair Sum in Array|[Go]({{< relref "/ChapterFour/1800~1899/1877.Minimize-Maximum-Pair-Sum-in-Array.md" >}})|Medium||||79.9%|
+|1984|Minimum Difference Between Highest and Lowest of K Scores|[Go]({{< relref "/ChapterFour/1900~1999/1984.Minimum-Difference-Between-Highest-and-Lowest-of-K-Scores.md" >}})|Easy||||54.4%|
+|2037|Minimum Number of Moves to Seat Everyone|[Go]({{< relref "/ChapterFour/2000~2099/2037.Minimum-Number-of-Moves-to-Seat-Everyone.md" >}})|Easy||||82.1%|
+|2164|Sort Even and Odd Indices Independently|[Go]({{< relref "/ChapterFour/2100~2199/2164.Sort-Even-and-Odd-Indices-Independently.md" >}})|Easy||||65.0%|
+|2165|Smallest Value of the Rearranged Number|[Go]({{< relref "/ChapterFour/2100~2199/2165.Smallest-Value-of-the-Rearranged-Number.md" >}})|Medium||||51.4%|
+|2171|Removing Minimum Number of Magic Beans|[Go]({{< relref "/ChapterFour/2100~2199/2171.Removing-Minimum-Number-of-Magic-Beans.md" >}})|Medium||||42.1%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Hash_Table/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Bit_Manipulation/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Stack.md b/website/content/ChapterTwo/Stack.md
new file mode 100644
index 000000000..5b5c28b9a
--- /dev/null
+++ b/website/content/ChapterTwo/Stack.md
@@ -0,0 +1,84 @@
+---
+title: 2.05 ✅ Stack
+type: docs
+weight: 5
+---
+
+# Stack
+
+![](https://img.halfrost.com/Leetcode/Stack.png)
+
+- 括号匹配问题及类似问题。第 20 题，第 921 题，第 1021 题。
+- 栈的基本 pop 和 push 操作。第 71 题，第 150 题，第 155 题，第 224 题，第 225 题，第 232 题，第 946 题，第 1047 题。
+- 利用栈进行编码问题。第 394 题，第 682 题，第 856 题，第 880 题。
+- **单调栈**。**利用栈维护一个单调递增或者递减的下标数组**。第 84 题，第 456 题，第 496 题，第 503 题，第 739 题，第 901 题，第 907 题，第 1019 题。
+
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0020|Valid Parentheses|[Go]({{< relref "/ChapterFour/0001~0099/0020.Valid-Parentheses.md" >}})|Easy| O(log n)| O(1)||40.2%|
+|0032|Longest Valid Parentheses|[Go]({{< relref "/ChapterFour/0001~0099/0032.Longest-Valid-Parentheses.md" >}})|Hard||||32.8%|
+|0042|Trapping Rain Water|[Go]({{< relref "/ChapterFour/0001~0099/0042.Trapping-Rain-Water.md" >}})|Hard| O(n)| O(1)|❤️|59.2%|
+|0071|Simplify Path|[Go]({{< relref "/ChapterFour/0001~0099/0071.Simplify-Path.md" >}})|Medium| O(n)| O(n)|❤️|39.3%|
+|0084|Largest Rectangle in Histogram|[Go]({{< relref "/ChapterFour/0001~0099/0084.Largest-Rectangle-in-Histogram.md" >}})|Hard| O(n)| O(n)|❤️|42.6%|
+|0094|Binary Tree Inorder Traversal|[Go]({{< relref "/ChapterFour/0001~0099/0094.Binary-Tree-Inorder-Traversal.md" >}})|Easy| O(n)| O(1)||73.8%|
+|0114|Flatten Binary Tree to Linked List|[Go]({{< relref "/ChapterFour/0100~0199/0114.Flatten-Binary-Tree-to-Linked-List.md" >}})|Medium||||61.8%|
+|0143|Reorder List|[Go]({{< relref "/ChapterFour/0100~0199/0143.Reorder-List.md" >}})|Medium||||52.5%|
+|0144|Binary Tree Preorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0144.Binary-Tree-Preorder-Traversal.md" >}})|Easy| O(n)| O(1)||66.8%|
+|0145|Binary Tree Postorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0145.Binary-Tree-Postorder-Traversal.md" >}})|Easy| O(n)| O(1)||67.9%|
+|0150|Evaluate Reverse Polish Notation|[Go]({{< relref "/ChapterFour/0100~0199/0150.Evaluate-Reverse-Polish-Notation.md" >}})|Medium| O(n)| O(1)||45.7%|
+|0155|Min Stack|[Go]({{< relref "/ChapterFour/0100~0199/0155.Min-Stack.md" >}})|Medium| O(n)| O(n)||52.3%|
+|0173|Binary Search Tree Iterator|[Go]({{< relref "/ChapterFour/0100~0199/0173.Binary-Search-Tree-Iterator.md" >}})|Medium| O(n)| O(1)||69.7%|
+|0224|Basic Calculator|[Go]({{< relref "/ChapterFour/0200~0299/0224.Basic-Calculator.md" >}})|Hard| O(n)| O(n)||42.4%|
+|0225|Implement Stack using Queues|[Go]({{< relref "/ChapterFour/0200~0299/0225.Implement-Stack-using-Queues.md" >}})|Easy| O(n)| O(n)||58.6%|
+|0227|Basic Calculator II|[Go]({{< relref "/ChapterFour/0200~0299/0227.Basic-Calculator-II.md" >}})|Medium||||42.4%|
+|0232|Implement Queue using Stacks|[Go]({{< relref "/ChapterFour/0200~0299/0232.Implement-Queue-using-Stacks.md" >}})|Easy| O(n)| O(n)||63.2%|
+|0234|Palindrome Linked List|[Go]({{< relref "/ChapterFour/0200~0299/0234.Palindrome-Linked-List.md" >}})|Easy||||50.2%|
+|0331|Verify Preorder Serialization of a Binary Tree|[Go]({{< relref "/ChapterFour/0300~0399/0331.Verify-Preorder-Serialization-of-a-Binary-Tree.md" >}})|Medium| O(n)| O(1)||44.6%|
+|0341|Flatten Nested List Iterator|[Go]({{< relref "/ChapterFour/0300~0399/0341.Flatten-Nested-List-Iterator.md" >}})|Medium||||61.8%|
+|0385|Mini Parser|[Go]({{< relref "/ChapterFour/0300~0399/0385.Mini-Parser.md" >}})|Medium||||36.9%|
+|0394|Decode String|[Go]({{< relref "/ChapterFour/0300~0399/0394.Decode-String.md" >}})|Medium| O(n)| O(n)||57.9%|
+|0402|Remove K Digits|[Go]({{< relref "/ChapterFour/0400~0499/0402.Remove-K-Digits.md" >}})|Medium| O(n)| O(1)||30.6%|
+|0445|Add Two Numbers II|[Go]({{< relref "/ChapterFour/0400~0499/0445.Add-Two-Numbers-II.md" >}})|Medium||||59.6%|
+|0456|132 Pattern|[Go]({{< relref "/ChapterFour/0400~0499/0456.132-Pattern.md" >}})|Medium| O(n)| O(n)||32.4%|
+|0496|Next Greater Element I|[Go]({{< relref "/ChapterFour/0400~0499/0496.Next-Greater-Element-I.md" >}})|Easy| O(n)| O(n)||71.4%|
+|0503|Next Greater Element II|[Go]({{< relref "/ChapterFour/0500~0599/0503.Next-Greater-Element-II.md" >}})|Medium| O(n)| O(n)||63.2%|
+|0581|Shortest Unsorted Continuous Subarray|[Go]({{< relref "/ChapterFour/0500~0599/0581.Shortest-Unsorted-Continuous-Subarray.md" >}})|Medium||||36.4%|
+|0589|N-ary Tree Preorder Traversal|[Go]({{< relref "/ChapterFour/0500~0599/0589.N-ary-Tree-Preorder-Traversal.md" >}})|Easy||||75.9%|
+|0636|Exclusive Time of Functions|[Go]({{< relref "/ChapterFour/0600~0699/0636.Exclusive-Time-of-Functions.md" >}})|Medium| O(n)| O(n)||61.2%|
+|0682|Baseball Game|[Go]({{< relref "/ChapterFour/0600~0699/0682.Baseball-Game.md" >}})|Easy| O(n)| O(n)||74.3%|
+|0726|Number of Atoms|[Go]({{< relref "/ChapterFour/0700~0799/0726.Number-of-Atoms.md" >}})|Hard| O(n)| O(n) |❤️|52.1%|
+|0735|Asteroid Collision|[Go]({{< relref "/ChapterFour/0700~0799/0735.Asteroid-Collision.md" >}})|Medium| O(n)| O(n) ||44.4%|
+|0739|Daily Temperatures|[Go]({{< relref "/ChapterFour/0700~0799/0739.Daily-Temperatures.md" >}})|Medium| O(n)| O(n) ||66.3%|
+|0844|Backspace String Compare|[Go]({{< relref "/ChapterFour/0800~0899/0844.Backspace-String-Compare.md" >}})|Easy| O(n)| O(n) ||48.1%|
+|0853|Car Fleet|[Go]({{< relref "/ChapterFour/0800~0899/0853.Car-Fleet.md" >}})|Medium||||50.3%|
+|0856|Score of Parentheses|[Go]({{< relref "/ChapterFour/0800~0899/0856.Score-of-Parentheses.md" >}})|Medium| O(n)| O(n)||64.8%|
+|0880|Decoded String at Index|[Go]({{< relref "/ChapterFour/0800~0899/0880.Decoded-String-at-Index.md" >}})|Medium| O(n)| O(n)||28.3%|
+|0895|Maximum Frequency Stack|[Go]({{< relref "/ChapterFour/0800~0899/0895.Maximum-Frequency-Stack.md" >}})|Hard| O(n)| O(n)  ||66.6%|
+|0897|Increasing Order Search Tree|[Go]({{< relref "/ChapterFour/0800~0899/0897.Increasing-Order-Search-Tree.md" >}})|Easy||||78.4%|
+|0901|Online Stock Span|[Go]({{< relref "/ChapterFour/0900~0999/0901.Online-Stock-Span.md" >}})|Medium| O(n)| O(n)  ||65.2%|
+|0907|Sum of Subarray Minimums|[Go]({{< relref "/ChapterFour/0900~0999/0907.Sum-of-Subarray-Minimums.md" >}})|Medium| O(n)| O(n)|❤️|35.8%|
+|0921|Minimum Add to Make Parentheses Valid|[Go]({{< relref "/ChapterFour/0900~0999/0921.Minimum-Add-to-Make-Parentheses-Valid.md" >}})|Medium| O(n)| O(n)||75.8%|
+|0946|Validate Stack Sequences|[Go]({{< relref "/ChapterFour/0900~0999/0946.Validate-Stack-Sequences.md" >}})|Medium| O(n)| O(n)||67.7%|
+|1003|Check If Word Is Valid After Substitutions|[Go]({{< relref "/ChapterFour/1000~1099/1003.Check-If-Word-Is-Valid-After-Substitutions.md" >}})|Medium| O(n)| O(1)||58.2%|
+|1006|Clumsy Factorial|[Go]({{< relref "/ChapterFour/1000~1099/1006.Clumsy-Factorial.md" >}})|Medium||||55.4%|
+|1019|Next Greater Node In Linked List|[Go]({{< relref "/ChapterFour/1000~1099/1019.Next-Greater-Node-In-Linked-List.md" >}})|Medium| O(n)| O(1)||59.9%|
+|1021|Remove Outermost Parentheses|[Go]({{< relref "/ChapterFour/1000~1099/1021.Remove-Outermost-Parentheses.md" >}})|Easy| O(n)| O(1)||80.6%|
+|1047|Remove All Adjacent Duplicates In String|[Go]({{< relref "/ChapterFour/1000~1099/1047.Remove-All-Adjacent-Duplicates-In-String.md" >}})|Easy| O(n)| O(1)||69.7%|
+|1111|Maximum Nesting Depth of Two Valid Parentheses Strings|[Go]({{< relref "/ChapterFour/1100~1199/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings.md" >}})|Medium||||73.0%|
+|1190|Reverse Substrings Between Each Pair of Parentheses|[Go]({{< relref "/ChapterFour/1100~1199/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses.md" >}})|Medium||||65.9%|
+|1209|Remove All Adjacent Duplicates in String II|[Go]({{< relref "/ChapterFour/1200~1299/1209.Remove-All-Adjacent-Duplicates-in-String-II.md" >}})|Medium||||56.2%|
+|1249|Minimum Remove to Make Valid Parentheses|[Go]({{< relref "/ChapterFour/1200~1299/1249.Minimum-Remove-to-Make-Valid-Parentheses.md" >}})|Medium||||65.8%|
+|1614|Maximum Nesting Depth of the Parentheses|[Go]({{< relref "/ChapterFour/1600~1699/1614.Maximum-Nesting-Depth-of-the-Parentheses.md" >}})|Easy||||82.3%|
+|1653|Minimum Deletions to Make String Balanced|[Go]({{< relref "/ChapterFour/1600~1699/1653.Minimum-Deletions-to-Make-String-Balanced.md" >}})|Medium||||58.9%|
+|1673|Find the Most Competitive Subsequence|[Go]({{< relref "/ChapterFour/1600~1699/1673.Find-the-Most-Competitive-Subsequence.md" >}})|Medium||||49.3%|
+|1700|Number of Students Unable to Eat Lunch|[Go]({{< relref "/ChapterFour/1700~1799/1700.Number-of-Students-Unable-to-Eat-Lunch.md" >}})|Easy||||68.7%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Linked_List/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Tree/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/String.md b/website/content/ChapterTwo/String.md
new file mode 100644
index 000000000..240b6b1ac
--- /dev/null
+++ b/website/content/ChapterTwo/String.md
@@ -0,0 +1,199 @@
+---
+title: 2.02 String
+type: docs
+weight: 2
+---
+
+# String
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0003|Longest Substring Without Repeating Characters|[Go]({{< relref "/ChapterFour/0001~0099/0003.Longest-Substring-Without-Repeating-Characters.md" >}})|Medium| O(n)| O(1)|❤️|33.8%|
+|0005|Longest Palindromic Substring|[Go]({{< relref "/ChapterFour/0001~0099/0005.Longest-Palindromic-Substring.md" >}})|Medium||||32.4%|
+|0006|Zigzag Conversion|[Go]({{< relref "/ChapterFour/0001~0099/0006.Zigzag-Conversion.md" >}})|Medium||||44.8%|
+|0008|String to Integer (atoi)|[Go]({{< relref "/ChapterFour/0001~0099/0008.String-to-Integer-atoi.md" >}})|Medium||||16.6%|
+|0012|Integer to Roman|[Go]({{< relref "/ChapterFour/0001~0099/0012.Integer-to-Roman.md" >}})|Medium||||62.0%|
+|0013|Roman to Integer|[Go]({{< relref "/ChapterFour/0001~0099/0013.Roman-to-Integer.md" >}})|Easy||||58.5%|
+|0014|Longest Common Prefix|[Go]({{< relref "/ChapterFour/0001~0099/0014.Longest-Common-Prefix.md" >}})|Easy||||40.9%|
+|0017|Letter Combinations of a Phone Number|[Go]({{< relref "/ChapterFour/0001~0099/0017.Letter-Combinations-of-a-Phone-Number.md" >}})|Medium| O(log n)| O(1)||56.5%|
+|0020|Valid Parentheses|[Go]({{< relref "/ChapterFour/0001~0099/0020.Valid-Parentheses.md" >}})|Easy| O(log n)| O(1)||40.2%|
+|0022|Generate Parentheses|[Go]({{< relref "/ChapterFour/0001~0099/0022.Generate-Parentheses.md" >}})|Medium| O(log n)| O(1)||72.5%|
+|0028|Find the Index of the First Occurrence in a String|[Go]({{< relref "/ChapterFour/0001~0099/0028.Find-the-Index-of-the-First-Occurrence-in-a-String.md" >}})|Easy| O(n)| O(1)||39.0%|
+|0030|Substring with Concatenation of All Words|[Go]({{< relref "/ChapterFour/0001~0099/0030.Substring-with-Concatenation-of-All-Words.md" >}})|Hard| O(n)| O(n)|❤️|31.2%|
+|0032|Longest Valid Parentheses|[Go]({{< relref "/ChapterFour/0001~0099/0032.Longest-Valid-Parentheses.md" >}})|Hard||||32.8%|
+|0043|Multiply Strings|[Go]({{< relref "/ChapterFour/0001~0099/0043.Multiply-Strings.md" >}})|Medium||||39.2%|
+|0049|Group Anagrams|[Go]({{< relref "/ChapterFour/0001~0099/0049.Group-Anagrams.md" >}})|Medium| O(n log n)| O(n)||66.7%|
+|0058|Length of Last Word|[Go]({{< relref "/ChapterFour/0001~0099/0058.Length-of-Last-Word.md" >}})|Easy||||42.8%|
+|0065|Valid Number|[Go]({{< relref "/ChapterFour/0001~0099/0065.Valid-Number.md" >}})|Hard||||18.7%|
+|0067|Add Binary|[Go]({{< relref "/ChapterFour/0001~0099/0067.Add-Binary.md" >}})|Easy||||52.4%|
+|0071|Simplify Path|[Go]({{< relref "/ChapterFour/0001~0099/0071.Simplify-Path.md" >}})|Medium| O(n)| O(n)||39.3%|
+|0076|Minimum Window Substring|[Go]({{< relref "/ChapterFour/0001~0099/0076.Minimum-Window-Substring.md" >}})|Hard| O(n)| O(n)|❤️|40.9%|
+|0091|Decode Ways|[Go]({{< relref "/ChapterFour/0001~0099/0091.Decode-Ways.md" >}})|Medium| O(n)| O(n)||32.7%|
+|0093|Restore IP Addresses|[Go]({{< relref "/ChapterFour/0001~0099/0093.Restore-IP-Addresses.md" >}})|Medium| O(n)| O(n)|❤️|47.4%|
+|0097|Interleaving String|[Go]({{< relref "/ChapterFour/0001~0099/0097.Interleaving-String.md" >}})|Medium||||37.3%|
+|0115|Distinct Subsequences|[Go]({{< relref "/ChapterFour/0100~0199/0115.Distinct-Subsequences.md" >}})|Hard||||44.4%|
+|0125|Valid Palindrome|[Go]({{< relref "/ChapterFour/0100~0199/0125.Valid-Palindrome.md" >}})|Easy| O(n)| O(1)||44.3%|
+|0126|Word Ladder II|[Go]({{< relref "/ChapterFour/0100~0199/0126.Word-Ladder-II.md" >}})|Hard| O(n)| O(n^2)|❤️|27.5%|
+|0127|Word Ladder|[Go]({{< relref "/ChapterFour/0100~0199/0127.Word-Ladder.md" >}})|Hard||||37.1%|
+|0131|Palindrome Partitioning|[Go]({{< relref "/ChapterFour/0100~0199/0131.Palindrome-Partitioning.md" >}})|Medium||||64.8%|
+|0151|Reverse Words in a String|[Go]({{< relref "/ChapterFour/0100~0199/0151.Reverse-Words-in-a-String.md" >}})|Medium||||32.7%|
+|0168|Excel Sheet Column Title|[Go]({{< relref "/ChapterFour/0100~0199/0168.Excel-Sheet-Column-Title.md" >}})|Easy||||35.5%|
+|0171|Excel Sheet Column Number|[Go]({{< relref "/ChapterFour/0100~0199/0171.Excel-Sheet-Column-Number.md" >}})|Easy||||62.0%|
+|0179|Largest Number|[Go]({{< relref "/ChapterFour/0100~0199/0179.Largest-Number.md" >}})|Medium||||34.5%|
+|0187|Repeated DNA Sequences|[Go]({{< relref "/ChapterFour/0100~0199/0187.Repeated-DNA-Sequences.md" >}})|Medium||||46.9%|
+|0205|Isomorphic Strings|[Go]({{< relref "/ChapterFour/0200~0299/0205.Isomorphic-Strings.md" >}})|Easy||||42.9%|
+|0208|Implement Trie (Prefix Tree)|[Go]({{< relref "/ChapterFour/0200~0299/0208.Implement-Trie-Prefix-Tree.md" >}})|Medium||||62.7%|
+|0211|Design Add and Search Words Data Structure|[Go]({{< relref "/ChapterFour/0200~0299/0211.Design-Add-and-Search-Words-Data-Structure.md" >}})|Medium||||44.0%|
+|0212|Word Search II|[Go]({{< relref "/ChapterFour/0200~0299/0212.Word-Search-II.md" >}})|Hard||||36.4%|
+|0224|Basic Calculator|[Go]({{< relref "/ChapterFour/0200~0299/0224.Basic-Calculator.md" >}})|Hard||||42.4%|
+|0227|Basic Calculator II|[Go]({{< relref "/ChapterFour/0200~0299/0227.Basic-Calculator-II.md" >}})|Medium||||42.4%|
+|0242|Valid Anagram|[Go]({{< relref "/ChapterFour/0200~0299/0242.Valid-Anagram.md" >}})|Easy||||63.0%|
+|0257|Binary Tree Paths|[Go]({{< relref "/ChapterFour/0200~0299/0257.Binary-Tree-Paths.md" >}})|Easy||||61.4%|
+|0290|Word Pattern|[Go]({{< relref "/ChapterFour/0200~0299/0290.Word-Pattern.md" >}})|Easy||||41.7%|
+|0297|Serialize and Deserialize Binary Tree|[Go]({{< relref "/ChapterFour/0200~0299/0297.Serialize-and-Deserialize-Binary-Tree.md" >}})|Hard||||55.4%|
+|0299|Bulls and Cows|[Go]({{< relref "/ChapterFour/0200~0299/0299.Bulls-and-Cows.md" >}})|Medium||||49.4%|
+|0301|Remove Invalid Parentheses|[Go]({{< relref "/ChapterFour/0300~0399/0301.Remove-Invalid-Parentheses.md" >}})|Hard||||47.2%|
+|0306|Additive Number|[Go]({{< relref "/ChapterFour/0300~0399/0306.Additive-Number.md" >}})|Medium||||31.1%|
+|0318|Maximum Product of Word Lengths|[Go]({{< relref "/ChapterFour/0300~0399/0318.Maximum-Product-of-Word-Lengths.md" >}})|Medium||||59.9%|
+|0331|Verify Preorder Serialization of a Binary Tree|[Go]({{< relref "/ChapterFour/0300~0399/0331.Verify-Preorder-Serialization-of-a-Binary-Tree.md" >}})|Medium||||44.6%|
+|0344|Reverse String|[Go]({{< relref "/ChapterFour/0300~0399/0344.Reverse-String.md" >}})|Easy| O(n)| O(1)||76.7%|
+|0345|Reverse Vowels of a String|[Go]({{< relref "/ChapterFour/0300~0399/0345.Reverse-Vowels-of-a-String.md" >}})|Easy| O(n)| O(1)||50.1%|
+|0383|Ransom Note|[Go]({{< relref "/ChapterFour/0300~0399/0383.Ransom-Note.md" >}})|Easy||||58.2%|
+|0385|Mini Parser|[Go]({{< relref "/ChapterFour/0300~0399/0385.Mini-Parser.md" >}})|Medium||||36.9%|
+|0387|First Unique Character in a String|[Go]({{< relref "/ChapterFour/0300~0399/0387.First-Unique-Character-in-a-String.md" >}})|Easy||||59.6%|
+|0389|Find the Difference|[Go]({{< relref "/ChapterFour/0300~0399/0389.Find-the-Difference.md" >}})|Easy||||59.9%|
+|0392|Is Subsequence|[Go]({{< relref "/ChapterFour/0300~0399/0392.Is-Subsequence.md" >}})|Easy||||47.6%|
+|0394|Decode String|[Go]({{< relref "/ChapterFour/0300~0399/0394.Decode-String.md" >}})|Medium||||57.9%|
+|0395|Longest Substring with At Least K Repeating Characters|[Go]({{< relref "/ChapterFour/0300~0399/0395.Longest-Substring-with-At-Least-K-Repeating-Characters.md" >}})|Medium||||44.8%|
+|0402|Remove K Digits|[Go]({{< relref "/ChapterFour/0400~0499/0402.Remove-K-Digits.md" >}})|Medium||||30.6%|
+|0409|Longest Palindrome|[Go]({{< relref "/ChapterFour/0400~0499/0409.Longest-Palindrome.md" >}})|Easy||||54.2%|
+|0412|Fizz Buzz|[Go]({{< relref "/ChapterFour/0400~0499/0412.Fizz-Buzz.md" >}})|Easy||||69.9%|
+|0423|Reconstruct Original Digits from English|[Go]({{< relref "/ChapterFour/0400~0499/0423.Reconstruct-Original-Digits-from-English.md" >}})|Medium||||51.3%|
+|0424|Longest Repeating Character Replacement|[Go]({{< relref "/ChapterFour/0400~0499/0424.Longest-Repeating-Character-Replacement.md" >}})|Medium||||51.9%|
+|0433|Minimum Genetic Mutation|[Go]({{< relref "/ChapterFour/0400~0499/0433.Minimum-Genetic-Mutation.md" >}})|Medium||||52.3%|
+|0434|Number of Segments in a String|[Go]({{< relref "/ChapterFour/0400~0499/0434.Number-of-Segments-in-a-String.md" >}})|Easy||||37.2%|
+|0438|Find All Anagrams in a String|[Go]({{< relref "/ChapterFour/0400~0499/0438.Find-All-Anagrams-in-a-String.md" >}})|Medium||||50.2%|
+|0451|Sort Characters By Frequency|[Go]({{< relref "/ChapterFour/0400~0499/0451.Sort-Characters-By-Frequency.md" >}})|Medium||||70.1%|
+|0474|Ones and Zeroes|[Go]({{< relref "/ChapterFour/0400~0499/0474.Ones-and-Zeroes.md" >}})|Medium||||46.8%|
+|0488|Zuma Game|[Go]({{< relref "/ChapterFour/0400~0499/0488.Zuma-Game.md" >}})|Hard||||33.9%|
+|0500|Keyboard Row|[Go]({{< relref "/ChapterFour/0500~0599/0500.Keyboard-Row.md" >}})|Easy||||69.5%|
+|0520|Detect Capital|[Go]({{< relref "/ChapterFour/0500~0599/0520.Detect-Capital.md" >}})|Easy||||57.0%|
+|0524|Longest Word in Dictionary through Deleting|[Go]({{< relref "/ChapterFour/0500~0599/0524.Longest-Word-in-Dictionary-through-Deleting.md" >}})|Medium||||51.0%|
+|0535|Encode and Decode TinyURL|[Go]({{< relref "/ChapterFour/0500~0599/0535.Encode-and-Decode-TinyURL.md" >}})|Medium||||85.9%|
+|0537|Complex Number Multiplication|[Go]({{< relref "/ChapterFour/0500~0599/0537.Complex-Number-Multiplication.md" >}})|Medium||||71.4%|
+|0541|Reverse String II|[Go]({{< relref "/ChapterFour/0500~0599/0541.Reverse-String-II.md" >}})|Easy||||50.5%|
+|0551|Student Attendance Record I|[Go]({{< relref "/ChapterFour/0500~0599/0551.Student-Attendance-Record-I.md" >}})|Easy||||48.2%|
+|0557|Reverse Words in a String III|[Go]({{< relref "/ChapterFour/0500~0599/0557.Reverse-Words-in-a-String-III.md" >}})|Easy||||81.9%|
+|0567|Permutation in String|[Go]({{< relref "/ChapterFour/0500~0599/0567.Permutation-in-String.md" >}})|Medium||||44.3%|
+|0583|Delete Operation for Two Strings|[Go]({{< relref "/ChapterFour/0500~0599/0583.Delete-Operation-for-Two-Strings.md" >}})|Medium||||59.8%|
+|0599|Minimum Index Sum of Two Lists|[Go]({{< relref "/ChapterFour/0500~0599/0599.Minimum-Index-Sum-of-Two-Lists.md" >}})|Easy||||53.4%|
+|0609|Find Duplicate File in System|[Go]({{< relref "/ChapterFour/0600~0699/0609.Find-Duplicate-File-in-System.md" >}})|Medium||||67.7%|
+|0647|Palindromic Substrings|[Go]({{< relref "/ChapterFour/0600~0699/0647.Palindromic-Substrings.md" >}})|Medium||||66.8%|
+|0648|Replace Words|[Go]({{< relref "/ChapterFour/0600~0699/0648.Replace-Words.md" >}})|Medium||||62.7%|
+|0676|Implement Magic Dictionary|[Go]({{< relref "/ChapterFour/0600~0699/0676.Implement-Magic-Dictionary.md" >}})|Medium||||57.0%|
+|0677|Map Sum Pairs|[Go]({{< relref "/ChapterFour/0600~0699/0677.Map-Sum-Pairs.md" >}})|Medium||||56.8%|
+|0692|Top K Frequent Words|[Go]({{< relref "/ChapterFour/0600~0699/0692.Top-K-Frequent-Words.md" >}})|Medium||||57.2%|
+|0696|Count Binary Substrings|[Go]({{< relref "/ChapterFour/0600~0699/0696.Count-Binary-Substrings.md" >}})|Easy||||65.5%|
+|0709|To Lower Case|[Go]({{< relref "/ChapterFour/0700~0799/0709.To-Lower-Case.md" >}})|Easy||||82.4%|
+|0720|Longest Word in Dictionary|[Go]({{< relref "/ChapterFour/0700~0799/0720.Longest-Word-in-Dictionary.md" >}})|Medium||||52.0%|
+|0721|Accounts Merge|[Go]({{< relref "/ChapterFour/0700~0799/0721.Accounts-Merge.md" >}})|Medium||||56.3%|
+|0726|Number of Atoms|[Go]({{< relref "/ChapterFour/0700~0799/0726.Number-of-Atoms.md" >}})|Hard||||52.1%|
+|0745|Prefix and Suffix Search|[Go]({{< relref "/ChapterFour/0700~0799/0745.Prefix-and-Suffix-Search.md" >}})|Hard||||41.2%|
+|0748|Shortest Completing Word|[Go]({{< relref "/ChapterFour/0700~0799/0748.Shortest-Completing-Word.md" >}})|Easy||||59.3%|
+|0752|Open the Lock|[Go]({{< relref "/ChapterFour/0700~0799/0752.Open-the-Lock.md" >}})|Medium||||55.6%|
+|0763|Partition Labels|[Go]({{< relref "/ChapterFour/0700~0799/0763.Partition-Labels.md" >}})|Medium||||79.7%|
+|0767|Reorganize String|[Go]({{< relref "/ChapterFour/0700~0799/0767.Reorganize-String.md" >}})|Medium| O(n log n)| O(log n)  |❤️|52.9%|
+|0771|Jewels and Stones|[Go]({{< relref "/ChapterFour/0700~0799/0771.Jewels-and-Stones.md" >}})|Easy||||88.2%|
+|0784|Letter Case Permutation|[Go]({{< relref "/ChapterFour/0700~0799/0784.Letter-Case-Permutation.md" >}})|Medium||||73.8%|
+|0791|Custom Sort String|[Go]({{< relref "/ChapterFour/0700~0799/0791.Custom-Sort-String.md" >}})|Medium||||69.1%|
+|0792|Number of Matching Subsequences|[Go]({{< relref "/ChapterFour/0700~0799/0792.Number-of-Matching-Subsequences.md" >}})|Medium||||51.6%|
+|0794|Valid Tic-Tac-Toe State|[Go]({{< relref "/ChapterFour/0700~0799/0794.Valid-Tic-Tac-Toe-State.md" >}})|Medium||||35.1%|
+|0811|Subdomain Visit Count|[Go]({{< relref "/ChapterFour/0800~0899/0811.Subdomain-Visit-Count.md" >}})|Medium||||75.5%|
+|0816|Ambiguous Coordinates|[Go]({{< relref "/ChapterFour/0800~0899/0816.Ambiguous-Coordinates.md" >}})|Medium||||56.3%|
+|0819|Most Common Word|[Go]({{< relref "/ChapterFour/0800~0899/0819.Most-Common-Word.md" >}})|Easy||||44.8%|
+|0820|Short Encoding of Words|[Go]({{< relref "/ChapterFour/0800~0899/0820.Short-Encoding-of-Words.md" >}})|Medium||||60.6%|
+|0821|Shortest Distance to a Character|[Go]({{< relref "/ChapterFour/0800~0899/0821.Shortest-Distance-to-a-Character.md" >}})|Easy||||71.3%|
+|0828|Count Unique Characters of All Substrings of a Given String|[Go]({{< relref "/ChapterFour/0800~0899/0828.Count-Unique-Characters-of-All-Substrings-of-a-Given-String.md" >}})|Hard||||51.6%|
+|0830|Positions of Large Groups|[Go]({{< relref "/ChapterFour/0800~0899/0830.Positions-of-Large-Groups.md" >}})|Easy||||51.8%|
+|0838|Push Dominoes|[Go]({{< relref "/ChapterFour/0800~0899/0838.Push-Dominoes.md" >}})|Medium||||57.0%|
+|0839|Similar String Groups|[Go]({{< relref "/ChapterFour/0800~0899/0839.Similar-String-Groups.md" >}})|Hard||||48.0%|
+|0842|Split Array into Fibonacci Sequence|[Go]({{< relref "/ChapterFour/0800~0899/0842.Split-Array-into-Fibonacci-Sequence.md" >}})|Medium| O(n^2)| O(1)|❤️|38.4%|
+|0844|Backspace String Compare|[Go]({{< relref "/ChapterFour/0800~0899/0844.Backspace-String-Compare.md" >}})|Easy||||48.1%|
+|0856|Score of Parentheses|[Go]({{< relref "/ChapterFour/0800~0899/0856.Score-of-Parentheses.md" >}})|Medium| O(n)| O(n)||64.8%|
+|0859|Buddy Strings|[Go]({{< relref "/ChapterFour/0800~0899/0859.Buddy-Strings.md" >}})|Easy||||29.2%|
+|0880|Decoded String at Index|[Go]({{< relref "/ChapterFour/0800~0899/0880.Decoded-String-at-Index.md" >}})|Medium||||28.3%|
+|0884|Uncommon Words from Two Sentences|[Go]({{< relref "/ChapterFour/0800~0899/0884.Uncommon-Words-from-Two-Sentences.md" >}})|Easy||||66.3%|
+|0890|Find and Replace Pattern|[Go]({{< relref "/ChapterFour/0800~0899/0890.Find-and-Replace-Pattern.md" >}})|Medium||||77.6%|
+|0916|Word Subsets|[Go]({{< relref "/ChapterFour/0900~0999/0916.Word-Subsets.md" >}})|Medium||||53.7%|
+|0921|Minimum Add to Make Parentheses Valid|[Go]({{< relref "/ChapterFour/0900~0999/0921.Minimum-Add-to-Make-Parentheses-Valid.md" >}})|Medium||||75.8%|
+|0925|Long Pressed Name|[Go]({{< relref "/ChapterFour/0900~0999/0925.Long-Pressed-Name.md" >}})|Easy| O(n)| O(1)||33.1%|
+|0942|DI String Match|[Go]({{< relref "/ChapterFour/0900~0999/0942.DI-String-Match.md" >}})|Easy||||77.3%|
+|0949|Largest Time for Given Digits|[Go]({{< relref "/ChapterFour/0900~0999/0949.Largest-Time-for-Given-Digits.md" >}})|Medium||||35.2%|
+|0953|Verifying an Alien Dictionary|[Go]({{< relref "/ChapterFour/0900~0999/0953.Verifying-an-Alien-Dictionary.md" >}})|Easy||||54.5%|
+|0966|Vowel Spellchecker|[Go]({{< relref "/ChapterFour/0900~0999/0966.Vowel-Spellchecker.md" >}})|Medium||||51.4%|
+|0981|Time Based Key-Value Store|[Go]({{< relref "/ChapterFour/0900~0999/0981.Time-Based-Key-Value-Store.md" >}})|Medium||||52.3%|
+|0984|String Without AAA or BBB|[Go]({{< relref "/ChapterFour/0900~0999/0984.String-Without-AAA-or-BBB.md" >}})|Medium||||43.1%|
+|0990|Satisfiability of Equality Equations|[Go]({{< relref "/ChapterFour/0900~0999/0990.Satisfiability-of-Equality-Equations.md" >}})|Medium||||50.5%|
+|1002|Find Common Characters|[Go]({{< relref "/ChapterFour/1000~1099/1002.Find-Common-Characters.md" >}})|Easy||||68.5%|
+|1003|Check If Word Is Valid After Substitutions|[Go]({{< relref "/ChapterFour/1000~1099/1003.Check-If-Word-Is-Valid-After-Substitutions.md" >}})|Medium| O(n)| O(1)||58.2%|
+|1021|Remove Outermost Parentheses|[Go]({{< relref "/ChapterFour/1000~1099/1021.Remove-Outermost-Parentheses.md" >}})|Easy||||80.6%|
+|1028|Recover a Tree From Preorder Traversal|[Go]({{< relref "/ChapterFour/1000~1099/1028.Recover-a-Tree-From-Preorder-Traversal.md" >}})|Hard||||73.3%|
+|1047|Remove All Adjacent Duplicates In String|[Go]({{< relref "/ChapterFour/1000~1099/1047.Remove-All-Adjacent-Duplicates-In-String.md" >}})|Easy||||69.7%|
+|1048|Longest String Chain|[Go]({{< relref "/ChapterFour/1000~1099/1048.Longest-String-Chain.md" >}})|Medium||||59.2%|
+|1078|Occurrences After Bigram|[Go]({{< relref "/ChapterFour/1000~1099/1078.Occurrences-After-Bigram.md" >}})|Easy||||63.6%|
+|1079|Letter Tile Possibilities|[Go]({{< relref "/ChapterFour/1000~1099/1079.Letter-Tile-Possibilities.md" >}})|Medium||||76.0%|
+|1108|Defanging an IP Address|[Go]({{< relref "/ChapterFour/1100~1199/1108.Defanging-an-IP-Address.md" >}})|Easy||||89.1%|
+|1111|Maximum Nesting Depth of Two Valid Parentheses Strings|[Go]({{< relref "/ChapterFour/1100~1199/1111.Maximum-Nesting-Depth-of-Two-Valid-Parentheses-Strings.md" >}})|Medium||||73.0%|
+|1143|Longest Common Subsequence|[Go]({{< relref "/ChapterFour/1100~1199/1143.Longest-Common-Subsequence.md" >}})|Medium||||58.4%|
+|1154|Day of the Year|[Go]({{< relref "/ChapterFour/1100~1199/1154.Day-of-the-Year.md" >}})|Easy||||49.6%|
+|1160|Find Words That Can Be Formed by Characters|[Go]({{< relref "/ChapterFour/1100~1199/1160.Find-Words-That-Can-Be-Formed-by-Characters.md" >}})|Easy||||67.5%|
+|1170|Compare Strings by Frequency of the Smallest Character|[Go]({{< relref "/ChapterFour/1100~1199/1170.Compare-Strings-by-Frequency-of-the-Smallest-Character.md" >}})|Medium||||61.5%|
+|1178|Number of Valid Words for Each Puzzle|[Go]({{< relref "/ChapterFour/1100~1199/1178.Number-of-Valid-Words-for-Each-Puzzle.md" >}})|Hard||||46.3%|
+|1189|Maximum Number of Balloons|[Go]({{< relref "/ChapterFour/1100~1199/1189.Maximum-Number-of-Balloons.md" >}})|Easy||||61.0%|
+|1190|Reverse Substrings Between Each Pair of Parentheses|[Go]({{< relref "/ChapterFour/1100~1199/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses.md" >}})|Medium||||65.9%|
+|1202|Smallest String With Swaps|[Go]({{< relref "/ChapterFour/1200~1299/1202.Smallest-String-With-Swaps.md" >}})|Medium||||57.7%|
+|1208|Get Equal Substrings Within Budget|[Go]({{< relref "/ChapterFour/1200~1299/1208.Get-Equal-Substrings-Within-Budget.md" >}})|Medium||||48.5%|
+|1209|Remove All Adjacent Duplicates in String II|[Go]({{< relref "/ChapterFour/1200~1299/1209.Remove-All-Adjacent-Duplicates-in-String-II.md" >}})|Medium||||56.2%|
+|1221|Split a String in Balanced Strings|[Go]({{< relref "/ChapterFour/1200~1299/1221.Split-a-String-in-Balanced-Strings.md" >}})|Easy||||85.1%|
+|1234|Replace the Substring for Balanced String|[Go]({{< relref "/ChapterFour/1200~1299/1234.Replace-the-Substring-for-Balanced-String.md" >}})|Medium||||37.2%|
+|1239|Maximum Length of a Concatenated String with Unique Characters|[Go]({{< relref "/ChapterFour/1200~1299/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters.md" >}})|Medium||||52.2%|
+|1249|Minimum Remove to Make Valid Parentheses|[Go]({{< relref "/ChapterFour/1200~1299/1249.Minimum-Remove-to-Make-Valid-Parentheses.md" >}})|Medium||||65.8%|
+|1268|Search Suggestions System|[Go]({{< relref "/ChapterFour/1200~1299/1268.Search-Suggestions-System.md" >}})|Medium||||66.2%|
+|1332|Remove Palindromic Subsequences|[Go]({{< relref "/ChapterFour/1300~1399/1332.Remove-Palindromic-Subsequences.md" >}})|Easy||||76.2%|
+|1396|Design Underground System|[Go]({{< relref "/ChapterFour/1300~1399/1396.Design-Underground-System.md" >}})|Medium||||73.6%|
+|1446|Consecutive Characters|[Go]({{< relref "/ChapterFour/1400~1499/1446.Consecutive-Characters.md" >}})|Easy||||61.2%|
+|1455|Check If a Word Occurs As a Prefix of Any Word in a Sentence|[Go]({{< relref "/ChapterFour/1400~1499/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence.md" >}})|Easy||||64.2%|
+|1461|Check If a String Contains All Binary Codes of Size K|[Go]({{< relref "/ChapterFour/1400~1499/1461.Check-If-a-String-Contains-All-Binary-Codes-of-Size-K.md" >}})|Medium||||56.6%|
+|1573|Number of Ways to Split a String|[Go]({{< relref "/ChapterFour/1500~1599/1573.Number-of-Ways-to-Split-a-String.md" >}})|Medium||||32.5%|
+|1576|Replace All ?'s to Avoid Consecutive Repeating Characters|[Go]({{< relref "/ChapterFour/1500~1599/1576.Replace-All-s-to-Avoid-Consecutive-Repeating-Characters.md" >}})|Easy||||48.3%|
+|1614|Maximum Nesting Depth of the Parentheses|[Go]({{< relref "/ChapterFour/1600~1699/1614.Maximum-Nesting-Depth-of-the-Parentheses.md" >}})|Easy||||82.3%|
+|1624|Largest Substring Between Two Equal Characters|[Go]({{< relref "/ChapterFour/1600~1699/1624.Largest-Substring-Between-Two-Equal-Characters.md" >}})|Easy||||59.1%|
+|1629|Slowest Key|[Go]({{< relref "/ChapterFour/1600~1699/1629.Slowest-Key.md" >}})|Easy||||59.2%|
+|1647|Minimum Deletions to Make Character Frequencies Unique|[Go]({{< relref "/ChapterFour/1600~1699/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique.md" >}})|Medium||||59.1%|
+|1653|Minimum Deletions to Make String Balanced|[Go]({{< relref "/ChapterFour/1600~1699/1653.Minimum-Deletions-to-Make-String-Balanced.md" >}})|Medium||||58.9%|
+|1657|Determine if Two Strings Are Close|[Go]({{< relref "/ChapterFour/1600~1699/1657.Determine-if-Two-Strings-Are-Close.md" >}})|Medium||||56.3%|
+|1662|Check If Two String Arrays are Equivalent|[Go]({{< relref "/ChapterFour/1600~1699/1662.Check-If-Two-String-Arrays-are-Equivalent.md" >}})|Easy||||83.5%|
+|1663|Smallest String With A Given Numeric Value|[Go]({{< relref "/ChapterFour/1600~1699/1663.Smallest-String-With-A-Given-Numeric-Value.md" >}})|Medium||||66.8%|
+|1668|Maximum Repeating Substring|[Go]({{< relref "/ChapterFour/1600~1699/1668.Maximum-Repeating-Substring.md" >}})|Easy||||39.5%|
+|1678|Goal Parser Interpretation|[Go]({{< relref "/ChapterFour/1600~1699/1678.Goal-Parser-Interpretation.md" >}})|Easy||||86.5%|
+|1684|Count the Number of Consistent Strings|[Go]({{< relref "/ChapterFour/1600~1699/1684.Count-the-Number-of-Consistent-Strings.md" >}})|Easy||||82.2%|
+|1689|Partitioning Into Minimum Number Of Deci-Binary Numbers|[Go]({{< relref "/ChapterFour/1600~1699/1689.Partitioning-Into-Minimum-Number-Of-Deci-Binary-Numbers.md" >}})|Medium||||89.2%|
+|1694|Reformat Phone Number|[Go]({{< relref "/ChapterFour/1600~1699/1694.Reformat-Phone-Number.md" >}})|Easy||||65.1%|
+|1704|Determine if String Halves Are Alike|[Go]({{< relref "/ChapterFour/1700~1799/1704.Determine-if-String-Halves-Are-Alike.md" >}})|Easy||||77.7%|
+|1736|Latest Time by Replacing Hidden Digits|[Go]({{< relref "/ChapterFour/1700~1799/1736.Latest-Time-by-Replacing-Hidden-Digits.md" >}})|Easy||||42.4%|
+|1758|Minimum Changes To Make Alternating Binary String|[Go]({{< relref "/ChapterFour/1700~1799/1758.Minimum-Changes-To-Make-Alternating-Binary-String.md" >}})|Easy||||58.3%|
+|1763|Longest Nice Substring|[Go]({{< relref "/ChapterFour/1700~1799/1763.Longest-Nice-Substring.md" >}})|Easy||||61.6%|
+|1816|Truncate Sentence|[Go]({{< relref "/ChapterFour/1800~1899/1816.Truncate-Sentence.md" >}})|Easy||||83.0%|
+|2038|Remove Colored Pieces if Both Neighbors are the Same Color|[Go]({{< relref "/ChapterFour/2000~2099/2038.Remove-Colored-Pieces-if-Both-Neighbors-are-the-Same-Color.md" >}})|Medium||||57.9%|
+|2096|Step-By-Step Directions From a Binary Tree Node to Another|[Go]({{< relref "/ChapterFour/2000~2099/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another.md" >}})|Medium||||48.4%|
+|2167|Minimum Time to Remove All Cars Containing Illegal Goods|[Go]({{< relref "/ChapterFour/2100~2199/2167.Minimum-Time-to-Remove-All-Cars-Containing-Illegal-Goods.md" >}})|Hard||||40.7%|
+|2182|Construct String With Repeat Limit|[Go]({{< relref "/ChapterFour/2100~2199/2182.Construct-String-With-Repeat-Limit.md" >}})|Medium||||52.2%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Array/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Two_Pointers/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Tree.md b/website/content/ChapterTwo/Tree.md
new file mode 100644
index 000000000..ebd6b5b9b
--- /dev/null
+++ b/website/content/ChapterTwo/Tree.md
@@ -0,0 +1,103 @@
+---
+title: 2.06 Tree
+type: docs
+weight: 6
+---
+
+# Tree
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0094|Binary Tree Inorder Traversal|[Go]({{< relref "/ChapterFour/0001~0099/0094.Binary-Tree-Inorder-Traversal.md" >}})|Easy| O(n)| O(1)||73.8%|
+|0095|Unique Binary Search Trees II|[Go]({{< relref "/ChapterFour/0001~0099/0095.Unique-Binary-Search-Trees-II.md" >}})|Medium||||52.3%|
+|0096|Unique Binary Search Trees|[Go]({{< relref "/ChapterFour/0001~0099/0096.Unique-Binary-Search-Trees.md" >}})|Medium| O(n^2)| O(n)||59.6%|
+|0098|Validate Binary Search Tree|[Go]({{< relref "/ChapterFour/0001~0099/0098.Validate-Binary-Search-Tree.md" >}})|Medium| O(n)| O(1)||32.0%|
+|0099|Recover Binary Search Tree|[Go]({{< relref "/ChapterFour/0001~0099/0099.Recover-Binary-Search-Tree.md" >}})|Medium| O(n)| O(1)||51.0%|
+|0100|Same Tree|[Go]({{< relref "/ChapterFour/0100~0199/0100.Same-Tree.md" >}})|Easy| O(n)| O(1)||58.1%|
+|0101|Symmetric Tree|[Go]({{< relref "/ChapterFour/0100~0199/0101.Symmetric-Tree.md" >}})|Easy| O(n)| O(1)||54.3%|
+|0102|Binary Tree Level Order Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0102.Binary-Tree-Level-Order-Traversal.md" >}})|Medium| O(n)| O(1)||64.3%|
+|0103|Binary Tree Zigzag Level Order Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0103.Binary-Tree-Zigzag-Level-Order-Traversal.md" >}})|Medium| O(n)| O(n)||56.9%|
+|0104|Maximum Depth of Binary Tree|[Go]({{< relref "/ChapterFour/0100~0199/0104.Maximum-Depth-of-Binary-Tree.md" >}})|Easy| O(n)| O(1)||73.9%|
+|0105|Construct Binary Tree from Preorder and Inorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal.md" >}})|Medium||||61.5%|
+|0106|Construct Binary Tree from Inorder and Postorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.md" >}})|Medium||||60.0%|
+|0107|Binary Tree Level Order Traversal II|[Go]({{< relref "/ChapterFour/0100~0199/0107.Binary-Tree-Level-Order-Traversal-II.md" >}})|Medium| O(n)| O(1)||61.1%|
+|0108|Convert Sorted Array to Binary Search Tree|[Go]({{< relref "/ChapterFour/0100~0199/0108.Convert-Sorted-Array-to-Binary-Search-Tree.md" >}})|Easy| O(n)| O(1)||69.8%|
+|0109|Convert Sorted List to Binary Search Tree|[Go]({{< relref "/ChapterFour/0100~0199/0109.Convert-Sorted-List-to-Binary-Search-Tree.md" >}})|Medium||||60.2%|
+|0110|Balanced Binary Tree|[Go]({{< relref "/ChapterFour/0100~0199/0110.Balanced-Binary-Tree.md" >}})|Easy| O(n)| O(1)||49.0%|
+|0111|Minimum Depth of Binary Tree|[Go]({{< relref "/ChapterFour/0100~0199/0111.Minimum-Depth-of-Binary-Tree.md" >}})|Easy| O(n)| O(1)||44.4%|
+|0112|Path Sum|[Go]({{< relref "/ChapterFour/0100~0199/0112.Path-Sum.md" >}})|Easy| O(n)| O(1)||48.2%|
+|0113|Path Sum II|[Go]({{< relref "/ChapterFour/0100~0199/0113.Path-Sum-II.md" >}})|Medium| O(n)| O(1)||57.1%|
+|0114|Flatten Binary Tree to Linked List|[Go]({{< relref "/ChapterFour/0100~0199/0114.Flatten-Binary-Tree-to-Linked-List.md" >}})|Medium| O(n)| O(1)||61.8%|
+|0116|Populating Next Right Pointers in Each Node|[Go]({{< relref "/ChapterFour/0100~0199/0116.Populating-Next-Right-Pointers-in-Each-Node.md" >}})|Medium||||60.4%|
+|0124|Binary Tree Maximum Path Sum|[Go]({{< relref "/ChapterFour/0100~0199/0124.Binary-Tree-Maximum-Path-Sum.md" >}})|Hard| O(n)| O(1)||39.2%|
+|0129|Sum Root to Leaf Numbers|[Go]({{< relref "/ChapterFour/0100~0199/0129.Sum-Root-to-Leaf-Numbers.md" >}})|Medium| O(n)| O(1)||61.0%|
+|0144|Binary Tree Preorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0144.Binary-Tree-Preorder-Traversal.md" >}})|Easy| O(n)| O(1)||66.8%|
+|0145|Binary Tree Postorder Traversal|[Go]({{< relref "/ChapterFour/0100~0199/0145.Binary-Tree-Postorder-Traversal.md" >}})|Easy| O(n)| O(1)||67.9%|
+|0173|Binary Search Tree Iterator|[Go]({{< relref "/ChapterFour/0100~0199/0173.Binary-Search-Tree-Iterator.md" >}})|Medium| O(n)| O(1)||69.7%|
+|0199|Binary Tree Right Side View|[Go]({{< relref "/ChapterFour/0100~0199/0199.Binary-Tree-Right-Side-View.md" >}})|Medium| O(n)| O(1)||61.6%|
+|0222|Count Complete Tree Nodes|[Go]({{< relref "/ChapterFour/0200~0299/0222.Count-Complete-Tree-Nodes.md" >}})|Medium| O(n)| O(1)||60.5%|
+|0226|Invert Binary Tree|[Go]({{< relref "/ChapterFour/0200~0299/0226.Invert-Binary-Tree.md" >}})|Easy| O(n)| O(1)||74.7%|
+|0230|Kth Smallest Element in a BST|[Go]({{< relref "/ChapterFour/0200~0299/0230.Kth-Smallest-Element-in-a-BST.md" >}})|Medium| O(n)| O(1)||70.1%|
+|0235|Lowest Common Ancestor of a Binary Search Tree|[Go]({{< relref "/ChapterFour/0200~0299/0235.Lowest-Common-Ancestor-of-a-Binary-Search-Tree.md" >}})|Medium| O(n)| O(1)||61.5%|
+|0236|Lowest Common Ancestor of a Binary Tree|[Go]({{< relref "/ChapterFour/0200~0299/0236.Lowest-Common-Ancestor-of-a-Binary-Tree.md" >}})|Medium| O(n)| O(1)||58.8%|
+|0257|Binary Tree Paths|[Go]({{< relref "/ChapterFour/0200~0299/0257.Binary-Tree-Paths.md" >}})|Easy| O(n)| O(1)||61.4%|
+|0297|Serialize and Deserialize Binary Tree|[Go]({{< relref "/ChapterFour/0200~0299/0297.Serialize-and-Deserialize-Binary-Tree.md" >}})|Hard||||55.4%|
+|0331|Verify Preorder Serialization of a Binary Tree|[Go]({{< relref "/ChapterFour/0300~0399/0331.Verify-Preorder-Serialization-of-a-Binary-Tree.md" >}})|Medium||||44.6%|
+|0337|House Robber III|[Go]({{< relref "/ChapterFour/0300~0399/0337.House-Robber-III.md" >}})|Medium||||53.9%|
+|0341|Flatten Nested List Iterator|[Go]({{< relref "/ChapterFour/0300~0399/0341.Flatten-Nested-List-Iterator.md" >}})|Medium||||61.8%|
+|0404|Sum of Left Leaves|[Go]({{< relref "/ChapterFour/0400~0499/0404.Sum-of-Left-Leaves.md" >}})|Easy| O(n)| O(1)||56.7%|
+|0429|N-ary Tree Level Order Traversal|[Go]({{< relref "/ChapterFour/0400~0499/0429.N-ary-Tree-Level-Order-Traversal.md" >}})|Medium||||70.7%|
+|0437|Path Sum III|[Go]({{< relref "/ChapterFour/0400~0499/0437.Path-Sum-III.md" >}})|Medium| O(n)| O(1)||48.0%|
+|0508|Most Frequent Subtree Sum|[Go]({{< relref "/ChapterFour/0500~0599/0508.Most-Frequent-Subtree-Sum.md" >}})|Medium||||64.9%|
+|0513|Find Bottom Left Tree Value|[Go]({{< relref "/ChapterFour/0500~0599/0513.Find-Bottom-Left-Tree-Value.md" >}})|Medium||||66.9%|
+|0515|Find Largest Value in Each Tree Row|[Go]({{< relref "/ChapterFour/0500~0599/0515.Find-Largest-Value-in-Each-Tree-Row.md" >}})|Medium| O(n)| O(n)||64.6%|
+|0530|Minimum Absolute Difference in BST|[Go]({{< relref "/ChapterFour/0500~0599/0530.Minimum-Absolute-Difference-in-BST.md" >}})|Easy||||57.3%|
+|0538|Convert BST to Greater Tree|[Go]({{< relref "/ChapterFour/0500~0599/0538.Convert-BST-to-Greater-Tree.md" >}})|Medium||||67.8%|
+|0543|Diameter of Binary Tree|[Go]({{< relref "/ChapterFour/0500~0599/0543.Diameter-of-Binary-Tree.md" >}})|Easy||||56.8%|
+|0559|Maximum Depth of N-ary Tree|[Go]({{< relref "/ChapterFour/0500~0599/0559.Maximum-Depth-of-N-ary-Tree.md" >}})|Easy||||71.7%|
+|0563|Binary Tree Tilt|[Go]({{< relref "/ChapterFour/0500~0599/0563.Binary-Tree-Tilt.md" >}})|Easy||||60.0%|
+|0572|Subtree of Another Tree|[Go]({{< relref "/ChapterFour/0500~0599/0572.Subtree-of-Another-Tree.md" >}})|Easy||||46.4%|
+|0589|N-ary Tree Preorder Traversal|[Go]({{< relref "/ChapterFour/0500~0599/0589.N-ary-Tree-Preorder-Traversal.md" >}})|Easy||||75.9%|
+|0617|Merge Two Binary Trees|[Go]({{< relref "/ChapterFour/0600~0699/0617.Merge-Two-Binary-Trees.md" >}})|Easy||||78.6%|
+|0623|Add One Row to Tree|[Go]({{< relref "/ChapterFour/0600~0699/0623.Add-One-Row-to-Tree.md" >}})|Medium||||59.5%|
+|0637|Average of Levels in Binary Tree|[Go]({{< relref "/ChapterFour/0600~0699/0637.Average-of-Levels-in-Binary-Tree.md" >}})|Easy| O(n)| O(n)||71.8%|
+|0653|Two Sum IV - Input is a BST|[Go]({{< relref "/ChapterFour/0600~0699/0653.Two-Sum-IV-Input-is-a-BST.md" >}})|Easy||||61.0%|
+|0662|Maximum Width of Binary Tree|[Go]({{< relref "/ChapterFour/0600~0699/0662.Maximum-Width-of-Binary-Tree.md" >}})|Medium||||40.7%|
+|0669|Trim a Binary Search Tree|[Go]({{< relref "/ChapterFour/0600~0699/0669.Trim-a-Binary-Search-Tree.md" >}})|Medium||||66.4%|
+|0700|Search in a Binary Search Tree|[Go]({{< relref "/ChapterFour/0700~0799/0700.Search-in-a-Binary-Search-Tree.md" >}})|Easy||||77.7%|
+|0701|Insert into a Binary Search Tree|[Go]({{< relref "/ChapterFour/0700~0799/0701.Insert-into-a-Binary-Search-Tree.md" >}})|Medium||||74.3%|
+|0703|Kth Largest Element in a Stream|[Go]({{< relref "/ChapterFour/0700~0799/0703.Kth-Largest-Element-in-a-Stream.md" >}})|Easy||||55.5%|
+|0783|Minimum Distance Between BST Nodes|[Go]({{< relref "/ChapterFour/0700~0799/0783.Minimum-Distance-Between-BST-Nodes.md" >}})|Easy||||59.3%|
+|0834|Sum of Distances in Tree|[Go]({{< relref "/ChapterFour/0800~0899/0834.Sum-of-Distances-in-Tree.md" >}})|Hard||||59.1%|
+|0863|All Nodes Distance K in Binary Tree|[Go]({{< relref "/ChapterFour/0800~0899/0863.All-Nodes-Distance-K-in-Binary-Tree.md" >}})|Medium||||62.2%|
+|0872|Leaf-Similar Trees|[Go]({{< relref "/ChapterFour/0800~0899/0872.Leaf-Similar-Trees.md" >}})|Easy||||67.6%|
+|0897|Increasing Order Search Tree|[Go]({{< relref "/ChapterFour/0800~0899/0897.Increasing-Order-Search-Tree.md" >}})|Easy||||78.4%|
+|0938|Range Sum of BST|[Go]({{< relref "/ChapterFour/0900~0999/0938.Range-Sum-of-BST.md" >}})|Easy||||85.9%|
+|0958|Check Completeness of a Binary Tree|[Go]({{< relref "/ChapterFour/0900~0999/0958.Check-Completeness-of-a-Binary-Tree.md" >}})|Medium||||56.2%|
+|0968|Binary Tree Cameras|[Go]({{< relref "/ChapterFour/0900~0999/0968.Binary-Tree-Cameras.md" >}})|Hard||||46.6%|
+|0971|Flip Binary Tree To Match Preorder Traversal|[Go]({{< relref "/ChapterFour/0900~0999/0971.Flip-Binary-Tree-To-Match-Preorder-Traversal.md" >}})|Medium||||50.0%|
+|0979|Distribute Coins in Binary Tree|[Go]({{< relref "/ChapterFour/0900~0999/0979.Distribute-Coins-in-Binary-Tree.md" >}})|Medium||||72.2%|
+|0987|Vertical Order Traversal of a Binary Tree|[Go]({{< relref "/ChapterFour/0900~0999/0987.Vertical-Order-Traversal-of-a-Binary-Tree.md" >}})|Hard||||45.1%|
+|0993|Cousins in Binary Tree|[Go]({{< relref "/ChapterFour/0900~0999/0993.Cousins-in-Binary-Tree.md" >}})|Easy| O(n)| O(1)||54.6%|
+|1022|Sum of Root To Leaf Binary Numbers|[Go]({{< relref "/ChapterFour/1000~1099/1022.Sum-of-Root-To-Leaf-Binary-Numbers.md" >}})|Easy||||73.6%|
+|1026|Maximum Difference Between Node and Ancestor|[Go]({{< relref "/ChapterFour/1000~1099/1026.Maximum-Difference-Between-Node-and-Ancestor.md" >}})|Medium||||75.8%|
+|1028|Recover a Tree From Preorder Traversal|[Go]({{< relref "/ChapterFour/1000~1099/1028.Recover-a-Tree-From-Preorder-Traversal.md" >}})|Hard||||73.3%|
+|1038|Binary Search Tree to Greater Sum Tree|[Go]({{< relref "/ChapterFour/1000~1099/1038.Binary-Search-Tree-to-Greater-Sum-Tree.md" >}})|Medium||||85.5%|
+|1104|Path In Zigzag Labelled Binary Tree|[Go]({{< relref "/ChapterFour/1100~1199/1104.Path-In-Zigzag-Labelled-Binary-Tree.md" >}})|Medium||||75.1%|
+|1110|Delete Nodes And Return Forest|[Go]({{< relref "/ChapterFour/1100~1199/1110.Delete-Nodes-And-Return-Forest.md" >}})|Medium||||69.3%|
+|1123|Lowest Common Ancestor of Deepest Leaves|[Go]({{< relref "/ChapterFour/1100~1199/1123.Lowest-Common-Ancestor-of-Deepest-Leaves.md" >}})|Medium||||70.9%|
+|1145|Binary Tree Coloring Game|[Go]({{< relref "/ChapterFour/1100~1199/1145.Binary-Tree-Coloring-Game.md" >}})|Medium||||51.7%|
+|1302|Deepest Leaves Sum|[Go]({{< relref "/ChapterFour/1300~1399/1302.Deepest-Leaves-Sum.md" >}})|Medium||||86.7%|
+|1305|All Elements in Two Binary Search Trees|[Go]({{< relref "/ChapterFour/1300~1399/1305.All-Elements-in-Two-Binary-Search-Trees.md" >}})|Medium||||79.8%|
+|1600|Throne Inheritance|[Go]({{< relref "/ChapterFour/1600~1699/1600.Throne-Inheritance.md" >}})|Medium||||63.6%|
+|1609|Even Odd Tree|[Go]({{< relref "/ChapterFour/1600~1699/1609.Even-Odd-Tree.md" >}})|Medium||||54.3%|
+|2096|Step-By-Step Directions From a Binary Tree Node to Another|[Go]({{< relref "/ChapterFour/2000~2099/2096.Step-By-Step-Directions-From-a-Binary-Tree-Node-to-Another.md" >}})|Medium||||48.4%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Stack/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Dynamic_Programming/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Two_Pointers.md b/website/content/ChapterTwo/Two_Pointers.md
new file mode 100644
index 000000000..6959d5ff8
--- /dev/null
+++ b/website/content/ChapterTwo/Two_Pointers.md
@@ -0,0 +1,118 @@
+---
+title: 2.03 ✅ Two Pointers
+type: docs
+weight: 3
+---
+
+# Two Pointers
+
+![](https://img.halfrost.com/Leetcode/Two_pointers.png)
+
+- 双指针滑动窗口的经典写法。右指针不断往右移，移动到不能往右移动为止(具体条件根据题目而定)。当右指针到最右边以后，开始挪动左指针，释放窗口左边界。第 3 题，第 76 题，第 209 题，第 424 题，第 438 题，第 567 题，第 713 题，第 763 题，第 845 题，第 881 题，第 904 题，第 978 题，第 992 题，第 1004 题，第 1040 题，第 1052 题。
+
+```c
+	left, right := 0, -1
+
+	for left < len(s) {
+		if right+1 < len(s) && freq[s[right+1]-'a'] == 0 {
+			freq[s[right+1]-'a']++
+			right++
+		} else {
+			freq[s[left]-'a']--
+			left++
+		}
+		result = max(result, right-left+1)
+	}
+```
+
+- 快慢指针可以查找重复数字，时间复杂度 O(n)，第 287 题。
+- 替换字母以后，相同字母能出现连续最长的长度。第 424 题。
+- SUM 问题集。第 1 题，第 15 题，第 16 题，第 18 题，第 167 题，第 923 题，第 1074 题。
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0011|Container With Most Water|[Go]({{< relref "/ChapterFour/0001~0099/0011.Container-With-Most-Water.md" >}})|Medium| O(n)| O(1)||54.0%|
+|0015|3Sum|[Go]({{< relref "/ChapterFour/0001~0099/0015.3Sum.md" >}})|Medium| O(n^2)| O(n)|❤️|32.6%|
+|0016|3Sum Closest|[Go]({{< relref "/ChapterFour/0001~0099/0016.3Sum-Closest.md" >}})|Medium| O(n^2)| O(1)|❤️|45.8%|
+|0018|4Sum|[Go]({{< relref "/ChapterFour/0001~0099/0018.4Sum.md" >}})|Medium| O(n^3)| O(n^2)|❤️|35.9%|
+|0019|Remove Nth Node From End of List|[Go]({{< relref "/ChapterFour/0001~0099/0019.Remove-Nth-Node-From-End-of-List.md" >}})|Medium| O(n)| O(1)||41.0%|
+|0026|Remove Duplicates from Sorted Array|[Go]({{< relref "/ChapterFour/0001~0099/0026.Remove-Duplicates-from-Sorted-Array.md" >}})|Easy| O(n)| O(1)||51.5%|
+|0027|Remove Element|[Go]({{< relref "/ChapterFour/0001~0099/0027.Remove-Element.md" >}})|Easy| O(n)| O(1)||53.0%|
+|0028|Find the Index of the First Occurrence in a String|[Go]({{< relref "/ChapterFour/0001~0099/0028.Find-the-Index-of-the-First-Occurrence-in-a-String.md" >}})|Easy| O(n)| O(1)||39.0%|
+|0031|Next Permutation|[Go]({{< relref "/ChapterFour/0001~0099/0031.Next-Permutation.md" >}})|Medium||||37.5%|
+|0042|Trapping Rain Water|[Go]({{< relref "/ChapterFour/0001~0099/0042.Trapping-Rain-Water.md" >}})|Hard| O(n)| O(1)|❤️|59.2%|
+|0061|Rotate List|[Go]({{< relref "/ChapterFour/0001~0099/0061.Rotate-List.md" >}})|Medium| O(n)| O(1)||36.1%|
+|0075|Sort Colors|[Go]({{< relref "/ChapterFour/0001~0099/0075.Sort-Colors.md" >}})|Medium| O(n)| O(1)|❤️|58.5%|
+|0080|Remove Duplicates from Sorted Array II|[Go]({{< relref "/ChapterFour/0001~0099/0080.Remove-Duplicates-from-Sorted-Array-II.md" >}})|Medium| O(n)| O(1||52.3%|
+|0082|Remove Duplicates from Sorted List II|[Go]({{< relref "/ChapterFour/0001~0099/0082.Remove-Duplicates-from-Sorted-List-II.md" >}})|Medium||||45.9%|
+|0086|Partition List|[Go]({{< relref "/ChapterFour/0001~0099/0086.Partition-List.md" >}})|Medium| O(n)| O(1)|❤️|52.0%|
+|0088|Merge Sorted Array|[Go]({{< relref "/ChapterFour/0001~0099/0088.Merge-Sorted-Array.md" >}})|Easy| O(n)| O(1)|❤️|46.6%|
+|0125|Valid Palindrome|[Go]({{< relref "/ChapterFour/0100~0199/0125.Valid-Palindrome.md" >}})|Easy| O(n)| O(1)||44.3%|
+|0141|Linked List Cycle|[Go]({{< relref "/ChapterFour/0100~0199/0141.Linked-List-Cycle.md" >}})|Easy| O(n)| O(1)|❤️|47.4%|
+|0142|Linked List Cycle II|[Go]({{< relref "/ChapterFour/0100~0199/0142.Linked-List-Cycle-II.md" >}})|Medium| O(n)| O(1)|❤️|48.7%|
+|0143|Reorder List|[Go]({{< relref "/ChapterFour/0100~0199/0143.Reorder-List.md" >}})|Medium||||52.5%|
+|0148|Sort List|[Go]({{< relref "/ChapterFour/0100~0199/0148.Sort-List.md" >}})|Medium||||55.1%|
+|0151|Reverse Words in a String|[Go]({{< relref "/ChapterFour/0100~0199/0151.Reverse-Words-in-a-String.md" >}})|Medium||||32.7%|
+|0160|Intersection of Two Linked Lists|[Go]({{< relref "/ChapterFour/0100~0199/0160.Intersection-of-Two-Linked-Lists.md" >}})|Easy||||54.3%|
+|0167|Two Sum II - Input Array Is Sorted|[Go]({{< relref "/ChapterFour/0100~0199/0167.Two-Sum-II-Input-Array-Is-Sorted.md" >}})|Medium| O(n)| O(1)||60.0%|
+|0189|Rotate Array|[Go]({{< relref "/ChapterFour/0100~0199/0189.Rotate-Array.md" >}})|Medium||||39.4%|
+|0202|Happy Number|[Go]({{< relref "/ChapterFour/0200~0299/0202.Happy-Number.md" >}})|Easy||||54.8%|
+|0234|Palindrome Linked List|[Go]({{< relref "/ChapterFour/0200~0299/0234.Palindrome-Linked-List.md" >}})|Easy| O(n)| O(1)||50.2%|
+|0283|Move Zeroes|[Go]({{< relref "/ChapterFour/0200~0299/0283.Move-Zeroes.md" >}})|Easy| O(n)| O(1)||61.4%|
+|0287|Find the Duplicate Number|[Go]({{< relref "/ChapterFour/0200~0299/0287.Find-the-Duplicate-Number.md" >}})|Medium| O(n)| O(1)|❤️|59.1%|
+|0344|Reverse String|[Go]({{< relref "/ChapterFour/0300~0399/0344.Reverse-String.md" >}})|Easy| O(n)| O(1)||76.7%|
+|0345|Reverse Vowels of a String|[Go]({{< relref "/ChapterFour/0300~0399/0345.Reverse-Vowels-of-a-String.md" >}})|Easy| O(n)| O(1)||50.1%|
+|0349|Intersection of Two Arrays|[Go]({{< relref "/ChapterFour/0300~0399/0349.Intersection-of-Two-Arrays.md" >}})|Easy| O(n)| O(n) ||70.9%|
+|0350|Intersection of Two Arrays II|[Go]({{< relref "/ChapterFour/0300~0399/0350.Intersection-of-Two-Arrays-II.md" >}})|Easy| O(n)| O(n) ||56.0%|
+|0392|Is Subsequence|[Go]({{< relref "/ChapterFour/0300~0399/0392.Is-Subsequence.md" >}})|Easy||||47.6%|
+|0455|Assign Cookies|[Go]({{< relref "/ChapterFour/0400~0499/0455.Assign-Cookies.md" >}})|Easy||||49.9%|
+|0457|Circular Array Loop|[Go]({{< relref "/ChapterFour/0400~0499/0457.Circular-Array-Loop.md" >}})|Medium||||32.6%|
+|0475|Heaters|[Go]({{< relref "/ChapterFour/0400~0499/0475.Heaters.md" >}})|Medium||||36.5%|
+|0524|Longest Word in Dictionary through Deleting|[Go]({{< relref "/ChapterFour/0500~0599/0524.Longest-Word-in-Dictionary-through-Deleting.md" >}})|Medium| O(n)| O(1) ||51.0%|
+|0532|K-diff Pairs in an Array|[Go]({{< relref "/ChapterFour/0500~0599/0532.K-diff-Pairs-in-an-Array.md" >}})|Medium| O(n)| O(n)||41.2%|
+|0541|Reverse String II|[Go]({{< relref "/ChapterFour/0500~0599/0541.Reverse-String-II.md" >}})|Easy||||50.5%|
+|0557|Reverse Words in a String III|[Go]({{< relref "/ChapterFour/0500~0599/0557.Reverse-Words-in-a-String-III.md" >}})|Easy||||81.9%|
+|0567|Permutation in String|[Go]({{< relref "/ChapterFour/0500~0599/0567.Permutation-in-String.md" >}})|Medium| O(n)| O(1)|❤️|44.3%|
+|0581|Shortest Unsorted Continuous Subarray|[Go]({{< relref "/ChapterFour/0500~0599/0581.Shortest-Unsorted-Continuous-Subarray.md" >}})|Medium||||36.4%|
+|0611|Valid Triangle Number|[Go]({{< relref "/ChapterFour/0600~0699/0611.Valid-Triangle-Number.md" >}})|Medium||||50.5%|
+|0633|Sum of Square Numbers|[Go]({{< relref "/ChapterFour/0600~0699/0633.Sum-of-Square-Numbers.md" >}})|Medium||||34.4%|
+|0653|Two Sum IV - Input is a BST|[Go]({{< relref "/ChapterFour/0600~0699/0653.Two-Sum-IV-Input-is-a-BST.md" >}})|Easy||||61.0%|
+|0658|Find K Closest Elements|[Go]({{< relref "/ChapterFour/0600~0699/0658.Find-K-Closest-Elements.md" >}})|Medium||||46.8%|
+|0696|Count Binary Substrings|[Go]({{< relref "/ChapterFour/0600~0699/0696.Count-Binary-Substrings.md" >}})|Easy||||65.5%|
+|0719|Find K-th Smallest Pair Distance|[Go]({{< relref "/ChapterFour/0700~0799/0719.Find-K-th-Smallest-Pair-Distance.md" >}})|Hard||||36.7%|
+|0763|Partition Labels|[Go]({{< relref "/ChapterFour/0700~0799/0763.Partition-Labels.md" >}})|Medium| O(n)| O(1)|❤️|79.7%|
+|0795|Number of Subarrays with Bounded Maximum|[Go]({{< relref "/ChapterFour/0700~0799/0795.Number-of-Subarrays-with-Bounded-Maximum.md" >}})|Medium||||52.8%|
+|0821|Shortest Distance to a Character|[Go]({{< relref "/ChapterFour/0800~0899/0821.Shortest-Distance-to-a-Character.md" >}})|Easy||||71.3%|
+|0825|Friends Of Appropriate Ages|[Go]({{< relref "/ChapterFour/0800~0899/0825.Friends-Of-Appropriate-Ages.md" >}})|Medium||||46.3%|
+|0826|Most Profit Assigning Work|[Go]({{< relref "/ChapterFour/0800~0899/0826.Most-Profit-Assigning-Work.md" >}})|Medium| O(n log n)| O(n)||44.9%|
+|0832|Flipping an Image|[Go]({{< relref "/ChapterFour/0800~0899/0832.Flipping-an-Image.md" >}})|Easy||||80.8%|
+|0838|Push Dominoes|[Go]({{< relref "/ChapterFour/0800~0899/0838.Push-Dominoes.md" >}})|Medium| O(n)| O(n)||57.0%|
+|0844|Backspace String Compare|[Go]({{< relref "/ChapterFour/0800~0899/0844.Backspace-String-Compare.md" >}})|Easy| O(n)| O(n) ||48.1%|
+|0845|Longest Mountain in Array|[Go]({{< relref "/ChapterFour/0800~0899/0845.Longest-Mountain-in-Array.md" >}})|Medium| O(n)| O(1) ||40.2%|
+|0870|Advantage Shuffle|[Go]({{< relref "/ChapterFour/0800~0899/0870.Advantage-Shuffle.md" >}})|Medium||||51.8%|
+|0876|Middle of the Linked List|[Go]({{< relref "/ChapterFour/0800~0899/0876.Middle-of-the-Linked-List.md" >}})|Easy||||75.6%|
+|0881|Boats to Save People|[Go]({{< relref "/ChapterFour/0800~0899/0881.Boats-to-Save-People.md" >}})|Medium| O(n log n)| O(1) ||53.1%|
+|0922|Sort Array By Parity II|[Go]({{< relref "/ChapterFour/0900~0999/0922.Sort-Array-By-Parity-II.md" >}})|Easy||||70.7%|
+|0923|3Sum With Multiplicity|[Go]({{< relref "/ChapterFour/0900~0999/0923.3Sum-With-Multiplicity.md" >}})|Medium| O(n^2)| O(n) ||45.3%|
+|0925|Long Pressed Name|[Go]({{< relref "/ChapterFour/0900~0999/0925.Long-Pressed-Name.md" >}})|Easy| O(n)| O(1)||33.1%|
+|0942|DI String Match|[Go]({{< relref "/ChapterFour/0900~0999/0942.DI-String-Match.md" >}})|Easy||||77.3%|
+|0969|Pancake Sorting|[Go]({{< relref "/ChapterFour/0900~0999/0969.Pancake-Sorting.md" >}})|Medium||||70.1%|
+|0977|Squares of a Sorted Array|[Go]({{< relref "/ChapterFour/0900~0999/0977.Squares-of-a-Sorted-Array.md" >}})|Easy| O(n)| O(1)||71.9%|
+|0986|Interval List Intersections|[Go]({{< relref "/ChapterFour/0900~0999/0986.Interval-List-Intersections.md" >}})|Medium| O(n)| O(1)||71.3%|
+|1040|Moving Stones Until Consecutive II|[Go]({{< relref "/ChapterFour/1000~1099/1040.Moving-Stones-Until-Consecutive-II.md" >}})|Medium||||55.9%|
+|1048|Longest String Chain|[Go]({{< relref "/ChapterFour/1000~1099/1048.Longest-String-Chain.md" >}})|Medium||||59.2%|
+|1089|Duplicate Zeros|[Go]({{< relref "/ChapterFour/1000~1099/1089.Duplicate-Zeros.md" >}})|Easy||||51.5%|
+|1332|Remove Palindromic Subsequences|[Go]({{< relref "/ChapterFour/1300~1399/1332.Remove-Palindromic-Subsequences.md" >}})|Easy||||76.2%|
+|1385|Find the Distance Value Between Two Arrays|[Go]({{< relref "/ChapterFour/1300~1399/1385.Find-the-Distance-Value-Between-Two-Arrays.md" >}})|Easy||||66.5%|
+|1679|Max Number of K-Sum Pairs|[Go]({{< relref "/ChapterFour/1600~1699/1679.Max-Number-of-K-Sum-Pairs.md" >}})|Medium||||57.3%|
+|1721|Swapping Nodes in a Linked List|[Go]({{< relref "/ChapterFour/1700~1799/1721.Swapping-Nodes-in-a-Linked-List.md" >}})|Medium||||67.2%|
+|1877|Minimize Maximum Pair Sum in Array|[Go]({{< relref "/ChapterFour/1800~1899/1877.Minimize-Maximum-Pair-Sum-in-Array.md" >}})|Medium||||79.9%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/String/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Linked_List/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/Union_Find.md b/website/content/ChapterTwo/Union_Find.md
new file mode 100644
index 000000000..391edabe5
--- /dev/null
+++ b/website/content/ChapterTwo/Union_Find.md
@@ -0,0 +1,55 @@
+---
+title: 2.16 ✅ Union Find
+type: docs
+weight: 16
+---
+
+# Union Find
+
+![](https://img.halfrost.com/Leetcode/Union_Find.png)
+
+- 灵活使用并查集的思想，熟练掌握并查集的[模板]({{< relref "/ChapterThree/UnionFind.md" >}})，模板中有两种并查集的实现方式，一种是路径压缩 + 秩优化的版本，另外一种是计算每个集合中元素的个数 + 最大集合元素个数的版本，这两种版本都有各自使用的地方。能使用第一类并查集模板的题目有：第 128 题，第 130 题，第 547 题，第 684 题，第 721 题，第 765 题，第 778 题，第 839 题，第 924 题，第 928 题，第 947 题，第 952 题，第 959 题，第 990 题。能使用第二类并查集模板的题目有：第 803 题，第 952 题。第 803 题秩优化和统计集合个数这些地方会卡时间，如果不优化，会 TLE。
+- 并查集是一种思想，有些题需要灵活使用这种思想，而不是死套模板，如第 399 题，这一题是 stringUnionFind，利用并查集思想实现的。这里每个节点是基于字符串和 map 的，而不是单纯的用 int 节点编号实现的。
+- 有些题死套模板反而做不出来，比如第 685 题，这一题不能路径压缩和秩优化，因为题目中涉及到有向图，需要知道节点的前驱节点，如果路径压缩了，这一题就没法做了。这一题不需要路径压缩和秩优化。
+- 灵活的抽象题目给的信息，将给定的信息合理的编号，使用并查集解题，并用 map 降低时间复杂度，如第 721 题，第 959 题。
+- 关于地图，砖块，网格的题目，可以新建一个特殊节点，将四周边缘的砖块或者网格都 union() 到这个特殊节点上。第 130 题，第 803 题。
+- 能用并查集的题目，一般也可以用 DFS 和 BFS 解答，只不过时间复杂度会高一点。
+
+
+
+| No.      | Title | Solution | Difficulty | TimeComplexity | SpaceComplexity |Favorite| Acceptance |
+|:--------:|:------- | :--------: | :----------: | :----: | :-----: | :-----: |:-----: |
+|0128|Longest Consecutive Sequence|[Go]({{< relref "/ChapterFour/0100~0199/0128.Longest-Consecutive-Sequence.md" >}})|Medium| O(n)| O(n)|❤️|48.5%|
+|0130|Surrounded Regions|[Go]({{< relref "/ChapterFour/0100~0199/0130.Surrounded-Regions.md" >}})|Medium| O(m\*n)| O(m\*n)||36.7%|
+|0200|Number of Islands|[Go]({{< relref "/ChapterFour/0200~0299/0200.Number-of-Islands.md" >}})|Medium| O(m\*n)| O(m\*n)||57.0%|
+|0399|Evaluate Division|[Go]({{< relref "/ChapterFour/0300~0399/0399.Evaluate-Division.md" >}})|Medium| O(n)| O(n)||59.6%|
+|0547|Number of Provinces|[Go]({{< relref "/ChapterFour/0500~0599/0547.Number-of-Provinces.md" >}})|Medium| O(n^2)| O(n)||63.7%|
+|0684|Redundant Connection|[Go]({{< relref "/ChapterFour/0600~0699/0684.Redundant-Connection.md" >}})|Medium| O(n)| O(n)||62.2%|
+|0685|Redundant Connection II|[Go]({{< relref "/ChapterFour/0600~0699/0685.Redundant-Connection-II.md" >}})|Hard| O(n)| O(n)||34.1%|
+|0695|Max Area of Island|[Go]({{< relref "/ChapterFour/0600~0699/0695.Max-Area-of-Island.md" >}})|Medium||||71.8%|
+|0721|Accounts Merge|[Go]({{< relref "/ChapterFour/0700~0799/0721.Accounts-Merge.md" >}})|Medium| O(n)| O(n)|❤️|56.3%|
+|0765|Couples Holding Hands|[Go]({{< relref "/ChapterFour/0700~0799/0765.Couples-Holding-Hands.md" >}})|Hard| O(n)| O(n)|❤️|56.6%|
+|0778|Swim in Rising Water|[Go]({{< relref "/ChapterFour/0700~0799/0778.Swim-in-Rising-Water.md" >}})|Hard| O(n^2)| O(n)|❤️|59.8%|
+|0785|Is Graph Bipartite?|[Go]({{< relref "/ChapterFour/0700~0799/0785.Is-Graph-Bipartite.md" >}})|Medium||||53.1%|
+|0803|Bricks Falling When Hit|[Go]({{< relref "/ChapterFour/0800~0899/0803.Bricks-Falling-When-Hit.md" >}})|Hard| O(n^2)| O(n)|❤️|34.4%|
+|0839|Similar String Groups|[Go]({{< relref "/ChapterFour/0800~0899/0839.Similar-String-Groups.md" >}})|Hard| O(n^2)| O(n)||48.0%|
+|0924|Minimize Malware Spread|[Go]({{< relref "/ChapterFour/0900~0999/0924.Minimize-Malware-Spread.md" >}})|Hard| O(m\*n)| O(n)||42.1%|
+|0928|Minimize Malware Spread II|[Go]({{< relref "/ChapterFour/0900~0999/0928.Minimize-Malware-Spread-II.md" >}})|Hard| O(m\*n)| O(n)|❤️|42.7%|
+|0947|Most Stones Removed with Same Row or Column|[Go]({{< relref "/ChapterFour/0900~0999/0947.Most-Stones-Removed-with-Same-Row-or-Column.md" >}})|Medium| O(n)| O(n)||58.9%|
+|0952|Largest Component Size by Common Factor|[Go]({{< relref "/ChapterFour/0900~0999/0952.Largest-Component-Size-by-Common-Factor.md" >}})|Hard| O(n)| O(n)|❤️|40.0%|
+|0959|Regions Cut By Slashes|[Go]({{< relref "/ChapterFour/0900~0999/0959.Regions-Cut-By-Slashes.md" >}})|Medium| O(n^2)| O(n^2)|❤️|69.1%|
+|0990|Satisfiability of Equality Equations|[Go]({{< relref "/ChapterFour/0900~0999/0990.Satisfiability-of-Equality-Equations.md" >}})|Medium| O(n)| O(n)||50.5%|
+|1020|Number of Enclaves|[Go]({{< relref "/ChapterFour/1000~1099/1020.Number-of-Enclaves.md" >}})|Medium||||65.5%|
+|1202|Smallest String With Swaps|[Go]({{< relref "/ChapterFour/1200~1299/1202.Smallest-String-With-Swaps.md" >}})|Medium||||57.7%|
+|1254|Number of Closed Islands|[Go]({{< relref "/ChapterFour/1200~1299/1254.Number-of-Closed-Islands.md" >}})|Medium||||64.1%|
+|1319|Number of Operations to Make Network Connected|[Go]({{< relref "/ChapterFour/1300~1399/1319.Number-of-Operations-to-Make-Network-Connected.md" >}})|Medium||||62.1%|
+|1579|Remove Max Number of Edges to Keep Graph Fully Traversable|[Go]({{< relref "/ChapterFour/1500~1599/1579.Remove-Max-Number-of-Edges-to-Keep-Graph-Fully-Traversable.md" >}})|Hard||||53.2%|
+|1631|Path With Minimum Effort|[Go]({{< relref "/ChapterFour/1600~1699/1631.Path-With-Minimum-Effort.md" >}})|Medium||||55.7%|
+|------------|-------------------------------------------------------|-------| ----------------| ---------------|-------------|-------------|-------------|
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Bit_Manipulation/">⬅️上一页</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Sliding_Window/">下一页➡️</a></p>
+</div>
diff --git a/website/content/ChapterTwo/_index.md b/website/content/ChapterTwo/_index.md
new file mode 100644
index 000000000..6aa84542e
--- /dev/null
+++ b/website/content/ChapterTwo/_index.md
@@ -0,0 +1,29 @@
+---
+title: 第二章 算法专题
+type: docs
+weight: 2
+---
+
+# 第二章 算法专题
+
+<p align='center'>
+<img src='https://img.halfrost.com/Leetcode/GO_LEARN_.png'>
+</p>
+
+
+
+
+本来天真的认为，把 LeetCode 所有题都完整刷一遍，就可以完整这本书了。经过事实证明，确实是天真了。因为 LeetCode 每天都会增加新题，有时候工作忙了，刷题进度就完全追不上题目更新的速度了。而且以我当前的刷题速度，一年才完成 500+，一年 LeetCode 也会更新 400+ 多题，要起码 5~10 年才能把所有的题目刷完。时间太长了。所以先给自己定了一个小目标，500 题就先把书写出来，总结这个阶段的刷题心得，和大家一起交流。要想把 LeetCode 所有题目都刷完，看来这本书要迭代 5 ~ 10 个版本了(一年迭代一版)。
+
+那么这一章就把已经刷完了的专题都整理一遍。有相似套路的题目都放在一起，如果想快速面试的话，其实相同的题目刷 2，3 道就可以了。相同类型的题目非常熟练的情况下，再多刷几道也是做无用功。
+
+做到目前为止，笔者认为动态规划是最灵活的类型，这类题目没有一个模板可以给你套用，它也是算法之优雅的地方。笔者认为称它为算法的艺术不为过。动态规划这类型，笔者也还没有刷完，只刷了一部分，还在学习中。
+
+那么就分享一下笔者目前刷过的题，和有相似点的题目吧。
+
+
+----------------------------------------------
+<div style="display: flex;justify-content: space-between;align-items: center;">
+<p><a href="https://books.halfrost.com/leetcode/ChapterOne/Time_Complexity/">⬅️上一章</a></p>
+<p><a href="https://books.halfrost.com/leetcode/ChapterTwo/Array/">下一页➡️</a></p>
+</div>
diff --git a/website/content/_index.md b/website/content/_index.md
new file mode 100644
index 000000000..50500e868
--- /dev/null
+++ b/website/content/_index.md
@@ -0,0 +1,88 @@
+---
+title: 序
+type: docs
+---
+
+# 序
+
+{{< columns >}}
+## 关于 LeetCode
+
+说到 LeetCode，作为一个程序员来说，应该不陌生，近几年参加面试都会提到它。国内外的程序员用它刷题主要是为了面试。据历史记载，这个网站 2011 年就成立了，马上就要到自己 10 周年的生日了。每周举行周赛，双周赛，月赛，在有限时间内编码，确实非常能考验人的算法能力。一些大公司赞助冠名的比赛获得前几名除了有奖品，还能直接拿到内推的机会。
+
+<--->
+
+## 什么是 Cookbook
+
+直译的话就是烹饪书，教你做各种食谱美食的书。经常看 O'Reilly 技术书的同学对这个名词会很熟悉。一般动手操作，实践类的书都会有这个名字。
+
+{{< /columns >}}
+
+<img src="https://books.halfrost.com/leetcode/logo.png" alt="logo" height="600" align="right" style="padding-left: 30px;"/>
+
+## 为什么会写这个开源书
+
+笔者刷题刷了一年了，想和大家分享分享一些做题心得，解题方法。想和有相同爱好的人交个朋友，一起交流学习。对于自己来说，写题解也是一种提高。把一道深奥的题目讲给一点都没有头绪的人，并能让他完全听懂，很能锻炼人的表达能力。在讲解中很可能还会遇到听者的一些提问，这些问题可能是自己的知识漏洞，强迫自己去弥补。笔者在公司做过相关的分享，感受很深，双方受益都还不错。
+
+> 另外，在大学期间，笔者做题的时候最讨厌写题解，感觉是浪费时间，用更多的时间去做更多的题。现在不知道算不算是“出来混的，总是要还的”。
+
+
+## 关于书的封面
+
+常看 O'Reilly 动物书的同学一看这个封面就知道是向他们致敬。确实是这个目的。O'Reilly 的封面动物都是稀缺动物，并且画风都是黑白素描风。这些动物都有版权了，所以只能在网上找没有版权的黑白素描风的图片。常见的能找到 40 张这种风格的图片。不过用的人太多了，笔者费劲的找了其他几张这种图片，这张孔雀开屏是其中一张。孔雀开屏的意义是希望大家刷完 LeetCode 以后，提高了自身的算法能力，在人生的舞台上开出自己的“屏”。全书配色也都是绿色，因为这是 AC 的颜色。
+
+
+## 关于作者
+
+笔者是一个刚刚入行一年半的 gopher 新人，还请各位大佬多多指点小弟我。大学参加了 3 年 ACM-ICPC，但是由于资质不高，没有拿到一块金牌。所以在算法方面，我对自己的评价算是新手吧。参加 ACM-ICPC 最大的收获是训练了思维能力，这种能力也会运用到生活中。其次是认识了很多国内很聪明的选手，看到了自己和他们的差距。最后，就是那 200 多页，有些自己都没有完全理解的，打印的密密麻麻的[算法模板](https://github.com/halfrost/leetcode-go/releases/tag/Special)。知识学会了，终身都是自己的，没有学会，那些知识都是身外之物。
+
+笔者从 2019 年 3 月 25 号开始刷题，到 2020 年 3 月 25 号，整整一年的时间。原计划是每天一题。实际上每天有时候不止一题，最终完成了 600+：
+
+![](https://img.halfrost.com/Blog/ArticleImage/2019_leetcode.png)
+
+> 一个温馨提示：笔者本以为每天做一题，会让这个 submissions 图全绿，但是我发现我错了。如果你也想坚持，让这个图全绿，一定要注意以下的问题：LeetCode 服务器是在 +0 时区的，这个图也是按照这个时区计算的。也就是说，中国每天早上 8 点之前，是算前一天的！也是因为时区的问题，导致我空白了这 22 个格子。比如有一道 Hard 题很难，当天工作也很多，晚上下班回家想出来了就到第二天凌晨了。于是再做一题当做第二天的量。结果会发现这 2 题都算前一天的。有时候笔者早上 6 点起床刷题，提交以后也都是前一天的。
+> 
+> (当然这些都是过去了，不重要了，全当是奋斗路上的一些小插曲)
+
+2020 年笔者肯定还会继续刷题，因为还没有达到自己的一些目标。可能会朝着 1000 题奋进，也有可能刷到 800 题的时候回头开始二刷，三刷。(不达目的不罢休吧~)
+
+## 关于书中的代码
+
+代码都放在 [github repo](https://github.com/halfrost/leetcode-go/tree/master/leetcode) 中，按题号可以搜索到题目。
+本书题目的代码都已经 beats 100% 了。没有 beats 100% 题解就没有放到本书中了。那些题目笔者会继续优化到 100% 再放进来。
+
+有可能读者会问，为何要追求 beats 100%。笔者认为优化到 beats 100% 才算是把这题做出感觉了。有好几道 Hard 题，笔者都用暴力解法 AC 了，然后只 beats 了 5%。这题就如同没做一样。而且面试中如果给了这样的答案，面试官也不会满意，“还有没有更优解？”。如果通过自己的思考能给出更优解，面试官会更满意一些。
+
+LeetCode 统计代码运行时长会有波动的，相同的代码提交 10 次可能就会 beats 100% 了。笔者开始没有发现这个问题，很多题用正确的代码连续交了很多次，一年提交 3400+ 次，导致我的正确率也变的奇高。😢
+
+当然，如果还有其他更优美的解法，也能 beats 100% 的，欢迎提交 PR，笔者和大家一起学习。
+
+## 目标读者
+
+想通过 LeetCode 提高算法能力的编程爱好者。
+
+
+## 编程语言
+
+本书的算法全部用 Go 语言实现。
+
+## 使用说明
+
+- 本电子书的左上角有搜索栏，可以迅速帮你找到你想看的章节和题号。
+- 本电子书每页都接入了 Gitalk，每一页的最下方都有评论框可以评论，如果没有显示出来，请检查自己的网络。
+- 关于题解，笔者建议这样使用：先自己读题，思考如何解题。如果 15 分钟还没有思路，那么先看笔者的解题思路，但是不要看代码。有思路以后自己用代码实现一遍。如果完全不会写，那就看笔者提供的代码，找出自己到底哪里不会写，找出问题记下来，这就是自己要弥补的知识漏洞。如果自己实现出来了，提交以后有错误，自己先 debug。AC 以后没有到 100% 也先自己思考如何优化。如果每道题自己都能优化到 100% 了，那么一段时间以后进步会很大。所以总的来说，实在没思路，看解题思路；实在优化不到 100%，看看代码。
+
+## 互动与勘误
+
+如果书中文章有所遗漏，欢迎点击所在页面下边的 edit 按钮进行评论和互动，感谢您的支持与帮助。
+
+## 最后
+
+一起开始刷题吧~
+
+![](https://img.halfrost.com/Blog/ArticleImage/hello_leetcode.png)
+
+本作品采用 [知识署名-非商业性使用-禁止演绎 （BY-NC-ND） 4.0 国际许可协议](https://creativecommons.org/licenses/by-nc-nd/4.0/legalcode.zh-Hans) 进行许可。
+
+题解里面的所有题目版权均归 [LeetCode](https://leetcode.com/) 和 [力扣中国](https://leetcode-cn.com/) 所有
+ 
\ No newline at end of file
diff --git a/website/resources/_gen/assets/scss/book.scss_50fc8c04e12a2f59027287995557ceff.content b/website/resources/_gen/assets/scss/book.scss_50fc8c04e12a2f59027287995557ceff.content
new file mode 100644
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+++ b/website/resources/_gen/assets/scss/book.scss_50fc8c04e12a2f59027287995557ceff.content
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diff --git a/website/resources/_gen/assets/scss/leetcode/book.scss_50fc8c04e12a2f59027287995557ceff.json b/website/resources/_gen/assets/scss/leetcode/book.scss_50fc8c04e12a2f59027287995557ceff.json
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+ * @author ocodia
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+     font-weight: bold;
+ }
+ .token.italic {
+     font-style: italic;
+ }
+ 
+ .token.entity {
+     cursor: help;
+ }
+ 
+ div.prism-show-language {
+     position: relative;
+ }
+ 
+ div.prism-show-language > div.prism-show-language-label[data-language] {
+   content: attr(data-language);
+     color: black;
+     background-color: #CFCFCF;
+     display: inline-block;
+     position: absolute;
+     top: 0;
+     right: 0;
+     font-size: 0.9em;
+     border-radius: 0 0 0 5px;
+     padding: 0 0.5em;
+     text-shadow: none;
+ }
\ No newline at end of file
diff --git a/website/static/prism.js b/website/static/prism.js
new file mode 100755
index 000000000..647d69ca8
--- /dev/null
+++ b/website/static/prism.js
@@ -0,0 +1,78 @@
+/* http://prismjs.com/download.html?themes=prism&languages=markup+css+clike+javascript+bash+css-extras+git+handlebars+http+makefile+markdown+nginx+php+sass+scss&plugins=line-highlight+line-numbers+autolinker+file-highlight+show-language */
+var _self="undefined"!=typeof window?window:"undefined"!=typeof WorkerGlobalScope&&self instanceof WorkerGlobalScope?self:{},Prism=function(){var e=/\blang(?:uage)?-(\w+)\b/i,t=0,n=_self.Prism={util:{encode:function(e){return e instanceof a?new a(e.type,n.util.encode(e.content),e.alias):"Array"===n.util.type(e)?e.map(n.util.encode):e.replace(/&/g,"&amp;").replace(/</g,"&lt;").replace(/\u00a0/g," ")},type:function(e){return Object.prototype.toString.call(e).match(/\[object (\w+)\]/)[1]},objId:function(e){return e.__id||Object.defineProperty(e,"__id",{value:++t}),e.__id},clone:function(e){var t=n.util.type(e);switch(t){case"Object":var a={};for(var r in e)e.hasOwnProperty(r)&&(a[r]=n.util.clone(e[r]));return a;case"Array":return e.map&&e.map(function(e){return n.util.clone(e)})}return e}},languages:{extend:function(e,t){var a=n.util.clone(n.languages[e]);for(var r in t)a[r]=t[r];return a},insertBefore:function(e,t,a,r){r=r||n.languages;var i=r[e];if(2==arguments.length){a=arguments[1];for(var l in a)a.hasOwnProperty(l)&&(i[l]=a[l]);return i}var o={};for(var s in i)if(i.hasOwnProperty(s)){if(s==t)for(var l in a)a.hasOwnProperty(l)&&(o[l]=a[l]);o[s]=i[s]}return n.languages.DFS(n.languages,function(t,n){n===r[e]&&t!=e&&(this[t]=o)}),r[e]=o},DFS:function(e,t,a,r){r=r||{};for(var i in e)e.hasOwnProperty(i)&&(t.call(e,i,e[i],a||i),"Object"!==n.util.type(e[i])||r[n.util.objId(e[i])]?"Array"!==n.util.type(e[i])||r[n.util.objId(e[i])]||(r[n.util.objId(e[i])]=!0,n.languages.DFS(e[i],t,i,r)):(r[n.util.objId(e[i])]=!0,n.languages.DFS(e[i],t,null,r)))}},plugins:{},highlightAll:function(e,t){var a={callback:t,selector:'code[class*="language-"], [class*="language-"] code, code[class*="lang-"], [class*="lang-"] code'};n.hooks.run("before-highlightall",a);for(var r,i=a.elements||document.querySelectorAll(a.selector),l=0;r=i[l++];)n.highlightElement(r,e===!0,a.callback)},highlightElement:function(t,a,r){for(var i,l,o=t;o&&!e.test(o.className);)o=o.parentNode;o&&(i=(o.className.match(e)||[,""])[1].toLowerCase(),l=n.languages[i]),t.className=t.className.replace(e,"").replace(/\s+/g," ")+" language-"+i,o=t.parentNode,/pre/i.test(o.nodeName)&&(o.className=o.className.replace(e,"").replace(/\s+/g," ")+" language-"+i);var s=t.textContent,u={element:t,language:i,grammar:l,code:s};if(n.hooks.run("before-sanity-check",u),!u.code||!u.grammar)return n.hooks.run("complete",u),void 0;if(n.hooks.run("before-highlight",u),a&&_self.Worker){var c=new Worker(n.filename);c.onmessage=function(e){u.highlightedCode=e.data,n.hooks.run("before-insert",u),u.element.innerHTML=u.highlightedCode,r&&r.call(u.element),n.hooks.run("after-highlight",u),n.hooks.run("complete",u)},c.postMessage(JSON.stringify({language:u.language,code:u.code,immediateClose:!0}))}else u.highlightedCode=n.highlight(u.code,u.grammar,u.language),n.hooks.run("before-insert",u),u.element.innerHTML=u.highlightedCode,r&&r.call(t),n.hooks.run("after-highlight",u),n.hooks.run("complete",u)},highlight:function(e,t,r){var i=n.tokenize(e,t);return a.stringify(n.util.encode(i),r)},tokenize:function(e,t){var a=n.Token,r=[e],i=t.rest;if(i){for(var l in i)t[l]=i[l];delete t.rest}e:for(var l in t)if(t.hasOwnProperty(l)&&t[l]){var o=t[l];o="Array"===n.util.type(o)?o:[o];for(var s=0;s<o.length;++s){var u=o[s],c=u.inside,g=!!u.lookbehind,h=!!u.greedy,f=0,d=u.alias;if(h&&!u.pattern.global){var p=u.pattern.toString().match(/[imuy]*$/)[0];u.pattern=RegExp(u.pattern.source,p+"g")}u=u.pattern||u;for(var m=0,y=0;m<r.length;y+=(r[m].matchedStr||r[m]).length,++m){var v=r[m];if(r.length>e.length)break e;if(!(v instanceof a)){u.lastIndex=0;var b=u.exec(v),k=1;if(!b&&h&&m!=r.length-1){if(u.lastIndex=y,b=u.exec(e),!b)break;for(var w=b.index+(g?b[1].length:0),_=b.index+b[0].length,A=m,S=y,P=r.length;P>A&&_>S;++A)S+=(r[A].matchedStr||r[A]).length,w>=S&&(++m,y=S);if(r[m]instanceof a||r[A-1].greedy)continue;k=A-m,v=e.slice(y,S),b.index-=y}if(b){g&&(f=b[1].length);var w=b.index+f,b=b[0].slice(f),_=w+b.length,x=v.slice(0,w),O=v.slice(_),j=[m,k];x&&j.push(x);var N=new a(l,c?n.tokenize(b,c):b,d,b,h);j.push(N),O&&j.push(O),Array.prototype.splice.apply(r,j)}}}}}return r},hooks:{all:{},add:function(e,t){var a=n.hooks.all;a[e]=a[e]||[],a[e].push(t)},run:function(e,t){var a=n.hooks.all[e];if(a&&a.length)for(var r,i=0;r=a[i++];)r(t)}}},a=n.Token=function(e,t,n,a,r){this.type=e,this.content=t,this.alias=n,this.matchedStr=a||null,this.greedy=!!r};if(a.stringify=function(e,t,r){if("string"==typeof e)return e;if("Array"===n.util.type(e))return e.map(function(n){return a.stringify(n,t,e)}).join("");var i={type:e.type,content:a.stringify(e.content,t,r),tag:"span",classes:["token",e.type],attributes:{},language:t,parent:r};if("comment"==i.type&&(i.attributes.spellcheck="true"),e.alias){var l="Array"===n.util.type(e.alias)?e.alias:[e.alias];Array.prototype.push.apply(i.classes,l)}n.hooks.run("wrap",i);var o="";for(var s in i.attributes)o+=(o?" ":"")+s+'="'+(i.attributes[s]||"")+'"';return"<"+i.tag+' class="'+i.classes.join(" ")+'"'+(o?" "+o:"")+">"+i.content+"</"+i.tag+">"},!_self.document)return _self.addEventListener?(_self.addEventListener("message",function(e){var t=JSON.parse(e.data),a=t.language,r=t.code,i=t.immediateClose;_self.postMessage(n.highlight(r,n.languages[a],a)),i&&_self.close()},!1),_self.Prism):_self.Prism;var r=document.currentScript||[].slice.call(document.getElementsByTagName("script")).pop();return r&&(n.filename=r.src,document.addEventListener&&!r.hasAttribute("data-manual")&&("loading"!==document.readyState?window.requestAnimationFrame?window.requestAnimationFrame(n.highlightAll):window.setTimeout(n.highlightAll,16):document.addEventListener("DOMContentLoaded",n.highlightAll))),_self.Prism}();"undefined"!=typeof module&&module.exports&&(module.exports=Prism),"undefined"!=typeof global&&(global.Prism=Prism);
+
+Prism.languages.markup={comment:/<!--[\w\W]*?-->/,prolog:/<\?[\w\W]+?\?>/,doctype:/<!DOCTYPE[\w\W]+?>/,cdata:/<!\[CDATA\[[\w\W]*?]]>/i,tag:{pattern:/<\/?(?!\d)[^\s>\/=$<]+(?:\s+[^\s>\/=]+(?:=(?:("|')(?:\\\1|\\?(?!\1)[\w\W])*\1|[^\s'">=]+))?)*\s*\/?>/i,inside:{tag:{pattern:/^<\/?[^\s>\/]+/i,inside:{punctuation:/^<\/?/,namespace:/^[^\s>\/:]+:/}},"attr-value":{pattern:/=(?:('|")[\w\W]*?(\1)|[^\s>]+)/i,inside:{punctuation:/[=>"']/}},punctuation:/\/?>/,"attr-name":{pattern:/[^\s>\/]+/,inside:{namespace:/^[^\s>\/:]+:/}}}},entity:/&#?[\da-z]{1,8};/i},Prism.hooks.add("wrap",function(a){"entity"===a.type&&(a.attributes.title=a.content.replace(/&amp;/,"&"))}),Prism.languages.xml=Prism.languages.markup,Prism.languages.html=Prism.languages.markup,Prism.languages.mathml=Prism.languages.markup,Prism.languages.svg=Prism.languages.markup;
+Prism.languages.css={comment:/\/\*[\w\W]*?\*\//,atrule:{pattern:/@[\w-]+?.*?(;|(?=\s*\{))/i,inside:{rule:/@[\w-]+/}},url:/url\((?:(["'])(\\(?:\r\n|[\w\W])|(?!\1)[^\\\r\n])*\1|.*?)\)/i,selector:/[^\{\}\s][^\{\};]*?(?=\s*\{)/,string:/("|')(\\(?:\r\n|[\w\W])|(?!\1)[^\\\r\n])*\1/,property:/(\b|\B)[\w-]+(?=\s*:)/i,important:/\B!important\b/i,"function":/[-a-z0-9]+(?=\()/i,punctuation:/[(){};:]/},Prism.languages.css.atrule.inside.rest=Prism.util.clone(Prism.languages.css),Prism.languages.markup&&(Prism.languages.insertBefore("markup","tag",{style:{pattern:/(<style[\w\W]*?>)[\w\W]*?(?=<\/style>)/i,lookbehind:!0,inside:Prism.languages.css,alias:"language-css"}}),Prism.languages.insertBefore("inside","attr-value",{"style-attr":{pattern:/\s*style=("|').*?\1/i,inside:{"attr-name":{pattern:/^\s*style/i,inside:Prism.languages.markup.tag.inside},punctuation:/^\s*=\s*['"]|['"]\s*$/,"attr-value":{pattern:/.+/i,inside:Prism.languages.css}},alias:"language-css"}},Prism.languages.markup.tag));
+Prism.languages.clike={comment:[{pattern:/(^|[^\\])\/\*[\w\W]*?\*\//,lookbehind:!0},{pattern:/(^|[^\\:])\/\/.*/,lookbehind:!0}],string:{pattern:/(["'])(\\(?:\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1/,greedy:!0},"class-name":{pattern:/((?:\b(?:class|interface|extends|implements|trait|instanceof|new)\s+)|(?:catch\s+\())[a-z0-9_\.\\]+/i,lookbehind:!0,inside:{punctuation:/(\.|\\)/}},keyword:/\b(if|else|while|do|for|return|in|instanceof|function|new|try|throw|catch|finally|null|break|continue)\b/,"boolean":/\b(true|false)\b/,"function":/[a-z0-9_]+(?=\()/i,number:/\b-?(?:0x[\da-f]+|\d*\.?\d+(?:e[+-]?\d+)?)\b/i,operator:/--?|\+\+?|!=?=?|<=?|>=?|==?=?|&&?|\|\|?|\?|\*|\/|~|\^|%/,punctuation:/[{}[\];(),.:]/};
+Prism.languages.javascript=Prism.languages.extend("clike",{keyword:/\b(as|async|await|break|case|catch|class|const|continue|debugger|default|delete|do|else|enum|export|extends|finally|for|from|function|get|if|implements|import|in|instanceof|interface|let|new|null|of|package|private|protected|public|return|set|static|super|switch|this|throw|try|typeof|var|void|while|with|yield)\b/,number:/\b-?(0x[\dA-Fa-f]+|0b[01]+|0o[0-7]+|\d*\.?\d+([Ee][+-]?\d+)?|NaN|Infinity)\b/,"function":/[_$a-zA-Z\xA0-\uFFFF][_$a-zA-Z0-9\xA0-\uFFFF]*(?=\()/i,operator:/--?|\+\+?|!=?=?|<=?|>=?|==?=?|&&?|\|\|?|\?|\*\*?|\/|~|\^|%|\.{3}/}),Prism.languages.insertBefore("javascript","keyword",{regex:{pattern:/(^|[^\/])\/(?!\/)(\[.+?]|\\.|[^\/\\\r\n])+\/[gimyu]{0,5}(?=\s*($|[\r\n,.;})]))/,lookbehind:!0,greedy:!0}}),Prism.languages.insertBefore("javascript","string",{"template-string":{pattern:/`(?:\\\\|\\?[^\\])*?`/,greedy:!0,inside:{interpolation:{pattern:/\$\{[^}]+\}/,inside:{"interpolation-punctuation":{pattern:/^\$\{|\}$/,alias:"punctuation"},rest:Prism.languages.javascript}},string:/[\s\S]+/}}}),Prism.languages.markup&&Prism.languages.insertBefore("markup","tag",{script:{pattern:/(<script[\w\W]*?>)[\w\W]*?(?=<\/script>)/i,lookbehind:!0,inside:Prism.languages.javascript,alias:"language-javascript"}}),Prism.languages.js=Prism.languages.javascript;
+Prism.languages.actionscript=Prism.languages.extend("javascript",{keyword:/\b(?:as|break|case|catch|class|const|default|delete|do|else|extends|finally|for|function|if|implements|import|in|instanceof|interface|internal|is|native|new|null|package|private|protected|public|return|super|switch|this|throw|try|typeof|use|var|void|while|with|dynamic|each|final|get|include|namespace|native|override|set|static)\b/,operator:/\+\+|--|(?:[+\-*\/%^]|&&?|\|\|?|<<?|>>?>?|[!=]=?)=?|[~?@]/}),Prism.languages.actionscript["class-name"].alias="function",Prism.languages.markup&&Prism.languages.insertBefore("actionscript","string",{xml:{pattern:/(^|[^.])<\/?\w+(?:\s+[^\s>\/=]+=("|')(?:\\\1|\\?(?!\1)[\w\W])*\2)*\s*\/?>/,lookbehind:!0,inside:{rest:Prism.languages.markup}}});
+Prism.languages.applescript={comment:[/\(\*(?:\(\*[\w\W]*?\*\)|[\w\W])*?\*\)/,/--.+/,/#.+/],string:/"(?:\\?.)*?"/,number:/\b-?\d*\.?\d+([Ee]-?\d+)?\b/,operator:[/[&=≠≤≥*+\-\/÷^]|[<>]=?/,/\b(?:(?:start|begin|end)s? with|(?:(?:does not|doesn't) contain|contains?)|(?:is|isn't|is not) (?:in|contained by)|(?:(?:is|isn't|is not) )?(?:greater|less) than(?: or equal)?(?: to)?|(?:(?:does not|doesn't) come|comes) (?:before|after)|(?:is|isn't|is not) equal(?: to)?|(?:(?:does not|doesn't) equal|equals|equal to|isn't|is not)|(?:a )?(?:ref(?: to)?|reference to)|(?:and|or|div|mod|as|not))\b/],keyword:/\b(?:about|above|after|against|apart from|around|aside from|at|back|before|beginning|behind|below|beneath|beside|between|but|by|considering|continue|copy|does|eighth|else|end|equal|error|every|exit|false|fifth|first|for|fourth|from|front|get|given|global|if|ignoring|in|instead of|into|is|it|its|last|local|me|middle|my|ninth|of|on|onto|out of|over|prop|property|put|repeat|return|returning|second|set|seventh|since|sixth|some|tell|tenth|that|the|then|third|through|thru|timeout|times|to|transaction|true|try|until|where|while|whose|with|without)\b/,"class":{pattern:/\b(?:alias|application|boolean|class|constant|date|file|integer|list|number|POSIX file|real|record|reference|RGB color|script|text|centimetres|centimeters|feet|inches|kilometres|kilometers|metres|meters|miles|yards|square feet|square kilometres|square kilometers|square metres|square meters|square miles|square yards|cubic centimetres|cubic centimeters|cubic feet|cubic inches|cubic metres|cubic meters|cubic yards|gallons|litres|liters|quarts|grams|kilograms|ounces|pounds|degrees Celsius|degrees Fahrenheit|degrees Kelvin)\b/,alias:"builtin"},punctuation:/[{}():,¬«»《》]/};
+Prism.languages.aspnet=Prism.languages.extend("markup",{"page-directive tag":{pattern:/<%\s*@.*%>/i,inside:{"page-directive tag":/<%\s*@\s*(?:Assembly|Control|Implements|Import|Master(?:Type)?|OutputCache|Page|PreviousPageType|Reference|Register)?|%>/i,rest:Prism.languages.markup.tag.inside}},"directive tag":{pattern:/<%.*%>/i,inside:{"directive tag":/<%\s*?[$=%#:]{0,2}|%>/i,rest:Prism.languages.csharp}}}),Prism.languages.aspnet.tag.pattern=/<(?!%)\/?[^\s>\/]+(?:\s+[^\s>\/=]+(?:=(?:("|')(?:\\\1|\\?(?!\1)[\w\W])*\1|[^\s'">=]+))?)*\s*\/?>/i,Prism.languages.insertBefore("inside","punctuation",{"directive tag":Prism.languages.aspnet["directive tag"]},Prism.languages.aspnet.tag.inside["attr-value"]),Prism.languages.insertBefore("aspnet","comment",{"asp comment":/<%--[\w\W]*?--%>/}),Prism.languages.insertBefore("aspnet",Prism.languages.javascript?"script":"tag",{"asp script":{pattern:/(<script(?=.*runat=['"]?server['"]?)[\w\W]*?>)[\w\W]*?(?=<\/script>)/i,lookbehind:!0,inside:Prism.languages.csharp||{}}});
+!function(e){var t={variable:[{pattern:/\$?\(\([\w\W]+?\)\)/,inside:{variable:[{pattern:/(^\$\(\([\w\W]+)\)\)/,lookbehind:!0},/^\$\(\(/],number:/\b-?(?:0x[\dA-Fa-f]+|\d*\.?\d+(?:[Ee]-?\d+)?)\b/,operator:/--?|-=|\+\+?|\+=|!=?|~|\*\*?|\*=|\/=?|%=?|<<=?|>>=?|<=?|>=?|==?|&&?|&=|\^=?|\|\|?|\|=|\?|:/,punctuation:/\(\(?|\)\)?|,|;/}},{pattern:/\$\([^)]+\)|`[^`]+`/,inside:{variable:/^\$\(|^`|\)$|`$/}},/\$(?:[a-z0-9_#\?\*!@]+|\{[^}]+\})/i]};e.languages.bash={shebang:{pattern:/^#!\s*\/bin\/bash|^#!\s*\/bin\/sh/,alias:"important"},comment:{pattern:/(^|[^"{\\])#.*/,lookbehind:!0},string:[{pattern:/((?:^|[^<])<<\s*)(?:"|')?(\w+?)(?:"|')?\s*\r?\n(?:[\s\S])*?\r?\n\2/g,lookbehind:!0,greedy:!0,inside:t},{pattern:/(["'])(?:\\\\|\\?[^\\])*?\1/g,greedy:!0,inside:t}],variable:t.variable,"function":{pattern:/(^|\s|;|\||&)(?:alias|apropos|apt-get|aptitude|aspell|awk|basename|bash|bc|bg|builtin|bzip2|cal|cat|cd|cfdisk|chgrp|chmod|chown|chroot|chkconfig|cksum|clear|cmp|comm|command|cp|cron|crontab|csplit|cut|date|dc|dd|ddrescue|df|diff|diff3|dig|dir|dircolors|dirname|dirs|dmesg|du|egrep|eject|enable|env|ethtool|eval|exec|expand|expect|export|expr|fdformat|fdisk|fg|fgrep|file|find|fmt|fold|format|free|fsck|ftp|fuser|gawk|getopts|git|grep|groupadd|groupdel|groupmod|groups|gzip|hash|head|help|hg|history|hostname|htop|iconv|id|ifconfig|ifdown|ifup|import|install|jobs|join|kill|killall|less|link|ln|locate|logname|logout|look|lpc|lpr|lprint|lprintd|lprintq|lprm|ls|lsof|make|man|mkdir|mkfifo|mkisofs|mknod|more|most|mount|mtools|mtr|mv|mmv|nano|netstat|nice|nl|nohup|notify-send|npm|nslookup|open|op|passwd|paste|pathchk|ping|pkill|popd|pr|printcap|printenv|printf|ps|pushd|pv|pwd|quota|quotacheck|quotactl|ram|rar|rcp|read|readarray|readonly|reboot|rename|renice|remsync|rev|rm|rmdir|rsync|screen|scp|sdiff|sed|seq|service|sftp|shift|shopt|shutdown|sleep|slocate|sort|source|split|ssh|stat|strace|su|sudo|sum|suspend|sync|tail|tar|tee|test|time|timeout|times|touch|top|traceroute|trap|tr|tsort|tty|type|ulimit|umask|umount|unalias|uname|unexpand|uniq|units|unrar|unshar|uptime|useradd|userdel|usermod|users|uuencode|uudecode|v|vdir|vi|vmstat|wait|watch|wc|wget|whereis|which|who|whoami|write|xargs|xdg-open|yes|zip)(?=$|\s|;|\||&)/,lookbehind:!0},keyword:{pattern:/(^|\s|;|\||&)(?:let|:|\.|if|then|else|elif|fi|for|break|continue|while|in|case|function|select|do|done|until|echo|exit|return|set|declare)(?=$|\s|;|\||&)/,lookbehind:!0},"boolean":{pattern:/(^|\s|;|\||&)(?:true|false)(?=$|\s|;|\||&)/,lookbehind:!0},operator:/&&?|\|\|?|==?|!=?|<<<?|>>|<=?|>=?|=~/,punctuation:/\$?\(\(?|\)\)?|\.\.|[{}[\];]/};var a=t.variable[1].inside;a["function"]=e.languages.bash["function"],a.keyword=e.languages.bash.keyword,a.boolean=e.languages.bash.boolean,a.operator=e.languages.bash.operator,a.punctuation=e.languages.bash.punctuation}(Prism);
+Prism.languages.basic={string:/"(?:""|[!#$%&'()*,\/:;<=>?^_ +\-.A-Z\d])*"/i,comment:{pattern:/(?:!|REM\b).+/i,inside:{keyword:/^REM/i}},number:/(?:\b|\B[.-])(?:\d+\.?\d*)(?:E[+-]?\d+)?/i,keyword:/\b(?:AS|BEEP|BLOAD|BSAVE|CALL(?: ABSOLUTE)?|CASE|CHAIN|CHDIR|CLEAR|CLOSE|CLS|COM|COMMON|CONST|DATA|DECLARE|DEF(?: FN| SEG|DBL|INT|LNG|SNG|STR)|DIM|DO|DOUBLE|ELSE|ELSEIF|END|ENVIRON|ERASE|ERROR|EXIT|FIELD|FILES|FOR|FUNCTION|GET|GOSUB|GOTO|IF|INPUT|INTEGER|IOCTL|KEY|KILL|LINE INPUT|LOCATE|LOCK|LONG|LOOP|LSET|MKDIR|NAME|NEXT|OFF|ON(?: COM| ERROR| KEY| TIMER)?|OPEN|OPTION BASE|OUT|POKE|PUT|READ|REDIM|REM|RESTORE|RESUME|RETURN|RMDIR|RSET|RUN|SHARED|SINGLE|SELECT CASE|SHELL|SLEEP|STATIC|STEP|STOP|STRING|SUB|SWAP|SYSTEM|THEN|TIMER|TO|TROFF|TRON|TYPE|UNLOCK|UNTIL|USING|VIEW PRINT|WAIT|WEND|WHILE|WRITE)(?:\$|\b)/i,"function":/\b(?:ABS|ACCESS|ACOS|ANGLE|AREA|ARITHMETIC|ARRAY|ASIN|ASK|AT|ATN|BASE|BEGIN|BREAK|CAUSE|CEIL|CHR|CLIP|COLLATE|COLOR|CON|COS|COSH|COT|CSC|DATE|DATUM|DEBUG|DECIMAL|DEF|DEG|DEGREES|DELETE|DET|DEVICE|DISPLAY|DOT|ELAPSED|EPS|ERASABLE|EXLINE|EXP|EXTERNAL|EXTYPE|FILETYPE|FIXED|FP|GO|GRAPH|HANDLER|IDN|IMAGE|IN|INT|INTERNAL|IP|IS|KEYED|LBOUND|LCASE|LEFT|LEN|LENGTH|LET|LINE|LINES|LOG|LOG10|LOG2|LTRIM|MARGIN|MAT|MAX|MAXNUM|MID|MIN|MISSING|MOD|NATIVE|NUL|NUMERIC|OF|OPTION|ORD|ORGANIZATION|OUTIN|OUTPUT|PI|POINT|POINTER|POINTS|POS|PRINT|PROGRAM|PROMPT|RAD|RADIANS|RANDOMIZE|RECORD|RECSIZE|RECTYPE|RELATIVE|REMAINDER|REPEAT|REST|RETRY|REWRITE|RIGHT|RND|ROUND|RTRIM|SAME|SEC|SELECT|SEQUENTIAL|SET|SETTER|SGN|SIN|SINH|SIZE|SKIP|SQR|STANDARD|STATUS|STR|STREAM|STYLE|TAB|TAN|TANH|TEMPLATE|TEXT|THERE|TIME|TIMEOUT|TRACE|TRANSFORM|TRUNCATE|UBOUND|UCASE|USE|VAL|VARIABLE|VIEWPORT|WHEN|WINDOW|WITH|ZER|ZONEWIDTH)(?:\$|\b)/i,operator:/<[=>]?|>=?|[+\-*\/^=&]|\b(?:AND|EQV|IMP|NOT|OR|XOR)\b/i,punctuation:/[,;:()]/};
+Prism.languages.c=Prism.languages.extend("clike",{keyword:/\b(asm|typeof|inline|auto|break|case|char|const|continue|default|do|double|else|enum|extern|float|for|goto|if|int|long|register|return|short|signed|sizeof|static|struct|switch|typedef|union|unsigned|void|volatile|while)\b/,operator:/\-[>-]?|\+\+?|!=?|<<?=?|>>?=?|==?|&&?|\|?\||[~^%?*\/]/,number:/\b-?(?:0x[\da-f]+|\d*\.?\d+(?:e[+-]?\d+)?)[ful]*\b/i}),Prism.languages.insertBefore("c","string",{macro:{pattern:/(^\s*)#\s*[a-z]+([^\r\n\\]|\\.|\\(?:\r\n?|\n))*/im,lookbehind:!0,alias:"property",inside:{string:{pattern:/(#\s*include\s*)(<.+?>|("|')(\\?.)+?\3)/,lookbehind:!0},directive:{pattern:/(#\s*)\b(define|elif|else|endif|error|ifdef|ifndef|if|import|include|line|pragma|undef|using)\b/,lookbehind:!0,alias:"keyword"}}},constant:/\b(__FILE__|__LINE__|__DATE__|__TIME__|__TIMESTAMP__|__func__|EOF|NULL|stdin|stdout|stderr)\b/}),delete Prism.languages.c["class-name"],delete Prism.languages.c["boolean"];
+Prism.languages.bison=Prism.languages.extend("c",{}),Prism.languages.insertBefore("bison","comment",{bison:{pattern:/^[\s\S]*?%%[\s\S]*?%%/,inside:{c:{pattern:/%\{[\s\S]*?%\}|\{(?:\{[^}]*\}|[^{}])*\}/,inside:{delimiter:{pattern:/^%?\{|%?\}$/,alias:"punctuation"},"bison-variable":{pattern:/[$@](?:<[^\s>]+>)?[\w$]+/,alias:"variable",inside:{punctuation:/<|>/}},rest:Prism.languages.c}},comment:Prism.languages.c.comment,string:Prism.languages.c.string,property:/\S+(?=:)/,keyword:/%\w+/,number:{pattern:/(^|[^@])\b(?:0x[\da-f]+|\d+)/i,lookbehind:!0},punctuation:/%[%?]|[|:;\[\]<>]/}}});
+Prism.languages.csharp=Prism.languages.extend("clike",{keyword:/\b(abstract|as|async|await|base|bool|break|byte|case|catch|char|checked|class|const|continue|decimal|default|delegate|do|double|else|enum|event|explicit|extern|false|finally|fixed|float|for|foreach|goto|if|implicit|in|int|interface|internal|is|lock|long|namespace|new|null|object|operator|out|override|params|private|protected|public|readonly|ref|return|sbyte|sealed|short|sizeof|stackalloc|static|string|struct|switch|this|throw|true|try|typeof|uint|ulong|unchecked|unsafe|ushort|using|virtual|void|volatile|while|add|alias|ascending|async|await|descending|dynamic|from|get|global|group|into|join|let|orderby|partial|remove|select|set|value|var|where|yield)\b/,string:[/@("|')(\1\1|\\\1|\\?(?!\1)[\s\S])*\1/,/("|')(\\?.)*?\1/],number:/\b-?(0x[\da-f]+|\d*\.?\d+f?)\b/i}),Prism.languages.insertBefore("csharp","keyword",{"generic-method":{pattern:/[a-z0-9_]+\s*<[^>\r\n]+?>\s*(?=\()/i,alias:"function",inside:{keyword:Prism.languages.csharp.keyword,punctuation:/[<>(),.:]/}},preprocessor:{pattern:/(^\s*)#.*/m,lookbehind:!0,alias:"property",inside:{directive:{pattern:/(\s*#)\b(define|elif|else|endif|endregion|error|if|line|pragma|region|undef|warning)\b/,lookbehind:!0,alias:"keyword"}}}});
+Prism.languages.cpp=Prism.languages.extend("c",{keyword:/\b(alignas|alignof|asm|auto|bool|break|case|catch|char|char16_t|char32_t|class|compl|const|constexpr|const_cast|continue|decltype|default|delete|do|double|dynamic_cast|else|enum|explicit|export|extern|float|for|friend|goto|if|inline|int|long|mutable|namespace|new|noexcept|nullptr|operator|private|protected|public|register|reinterpret_cast|return|short|signed|sizeof|static|static_assert|static_cast|struct|switch|template|this|thread_local|throw|try|typedef|typeid|typename|union|unsigned|using|virtual|void|volatile|wchar_t|while)\b/,"boolean":/\b(true|false)\b/,operator:/[-+]{1,2}|!=?|<{1,2}=?|>{1,2}=?|\->|:{1,2}|={1,2}|\^|~|%|&{1,2}|\|?\||\?|\*|\/|\b(and|and_eq|bitand|bitor|not|not_eq|or|or_eq|xor|xor_eq)\b/}),Prism.languages.insertBefore("cpp","keyword",{"class-name":{pattern:/(class\s+)[a-z0-9_]+/i,lookbehind:!0}});
+!function(e){var t=/#(?!\{).+/,n={pattern:/#\{[^}]+\}/,alias:"variable"};e.languages.coffeescript=e.languages.extend("javascript",{comment:t,string:[{pattern:/'(?:\\?[^\\])*?'/,greedy:!0},{pattern:/"(?:\\?[^\\])*?"/,greedy:!0,inside:{interpolation:n}}],keyword:/\b(and|break|by|catch|class|continue|debugger|delete|do|each|else|extend|extends|false|finally|for|if|in|instanceof|is|isnt|let|loop|namespace|new|no|not|null|of|off|on|or|own|return|super|switch|then|this|throw|true|try|typeof|undefined|unless|until|when|while|window|with|yes|yield)\b/,"class-member":{pattern:/@(?!\d)\w+/,alias:"variable"}}),e.languages.insertBefore("coffeescript","comment",{"multiline-comment":{pattern:/###[\s\S]+?###/,alias:"comment"},"block-regex":{pattern:/\/{3}[\s\S]*?\/{3}/,alias:"regex",inside:{comment:t,interpolation:n}}}),e.languages.insertBefore("coffeescript","string",{"inline-javascript":{pattern:/`(?:\\?[\s\S])*?`/,inside:{delimiter:{pattern:/^`|`$/,alias:"punctuation"},rest:e.languages.javascript}},"multiline-string":[{pattern:/'''[\s\S]*?'''/,greedy:!0,alias:"string"},{pattern:/"""[\s\S]*?"""/,greedy:!0,alias:"string",inside:{interpolation:n}}]}),e.languages.insertBefore("coffeescript","keyword",{property:/(?!\d)\w+(?=\s*:(?!:))/}),delete e.languages.coffeescript["template-string"]}(Prism);
+!function(e){e.languages.ruby=e.languages.extend("clike",{comment:/#(?!\{[^\r\n]*?\}).*/,keyword:/\b(alias|and|BEGIN|begin|break|case|class|def|define_method|defined|do|each|else|elsif|END|end|ensure|false|for|if|in|module|new|next|nil|not|or|raise|redo|require|rescue|retry|return|self|super|then|throw|true|undef|unless|until|when|while|yield)\b/});var n={pattern:/#\{[^}]+\}/,inside:{delimiter:{pattern:/^#\{|\}$/,alias:"tag"},rest:e.util.clone(e.languages.ruby)}};e.languages.insertBefore("ruby","keyword",{regex:[{pattern:/%r([^a-zA-Z0-9\s\{\(\[<])(?:[^\\]|\\[\s\S])*?\1[gim]{0,3}/,inside:{interpolation:n}},{pattern:/%r\((?:[^()\\]|\\[\s\S])*\)[gim]{0,3}/,inside:{interpolation:n}},{pattern:/%r\{(?:[^#{}\\]|#(?:\{[^}]+\})?|\\[\s\S])*\}[gim]{0,3}/,inside:{interpolation:n}},{pattern:/%r\[(?:[^\[\]\\]|\\[\s\S])*\][gim]{0,3}/,inside:{interpolation:n}},{pattern:/%r<(?:[^<>\\]|\\[\s\S])*>[gim]{0,3}/,inside:{interpolation:n}},{pattern:/(^|[^\/])\/(?!\/)(\[.+?]|\\.|[^\/\r\n])+\/[gim]{0,3}(?=\s*($|[\r\n,.;})]))/,lookbehind:!0}],variable:/[@$]+[a-zA-Z_][a-zA-Z_0-9]*(?:[?!]|\b)/,symbol:/:[a-zA-Z_][a-zA-Z_0-9]*(?:[?!]|\b)/}),e.languages.insertBefore("ruby","number",{builtin:/\b(Array|Bignum|Binding|Class|Continuation|Dir|Exception|FalseClass|File|Stat|File|Fixnum|Fload|Hash|Integer|IO|MatchData|Method|Module|NilClass|Numeric|Object|Proc|Range|Regexp|String|Struct|TMS|Symbol|ThreadGroup|Thread|Time|TrueClass)\b/,constant:/\b[A-Z][a-zA-Z_0-9]*(?:[?!]|\b)/}),e.languages.ruby.string=[{pattern:/%[qQiIwWxs]?([^a-zA-Z0-9\s\{\(\[<])(?:[^\\]|\\[\s\S])*?\1/,inside:{interpolation:n}},{pattern:/%[qQiIwWxs]?\((?:[^()\\]|\\[\s\S])*\)/,inside:{interpolation:n}},{pattern:/%[qQiIwWxs]?\{(?:[^#{}\\]|#(?:\{[^}]+\})?|\\[\s\S])*\}/,inside:{interpolation:n}},{pattern:/%[qQiIwWxs]?\[(?:[^\[\]\\]|\\[\s\S])*\]/,inside:{interpolation:n}},{pattern:/%[qQiIwWxs]?<(?:[^<>\\]|\\[\s\S])*>/,inside:{interpolation:n}},{pattern:/("|')(#\{[^}]+\}|\\(?:\r?\n|\r)|\\?.)*?\1/,inside:{interpolation:n}}]}(Prism);
+Prism.languages.css.selector={pattern:/[^\{\}\s][^\{\}]*(?=\s*\{)/,inside:{"pseudo-element":/:(?:after|before|first-letter|first-line|selection)|::[-\w]+/,"pseudo-class":/:[-\w]+(?:\(.*\))?/,"class":/\.[-:\.\w]+/,id:/#[-:\.\w]+/,attribute:/\[[^\]]+\]/}},Prism.languages.insertBefore("css","function",{hexcode:/#[\da-f]{3,6}/i,entity:/\\[\da-f]{1,8}/i,number:/[\d%\.]+/});
+Prism.languages.diff={coord:[/^(?:\*{3}|-{3}|\+{3}).*$/m,/^@@.*@@$/m,/^\d+.*$/m],deleted:/^[-<].*$/m,inserted:/^[+>].*$/m,diff:{pattern:/^!(?!!).+$/m,alias:"important"}};
+Prism.languages.docker={keyword:{pattern:/(^\s*)(?:ONBUILD|FROM|MAINTAINER|RUN|EXPOSE|ENV|ADD|COPY|VOLUME|USER|WORKDIR|CMD|LABEL|ENTRYPOINT)(?=\s)/im,lookbehind:!0},string:/("|')(?:(?!\1)[^\\\r\n]|\\(?:\r\n|[\s\S]))*?\1/,comment:/#.*/,punctuation:/---|\.\.\.|[:[\]{}\-,|>?]/};
+Prism.languages.erlang={comment:/%.+/,string:/"(?:\\?.)*?"/,"quoted-function":{pattern:/'(?:\\.|[^'\\])+'(?=\()/,alias:"function"},"quoted-atom":{pattern:/'(?:\\.|[^'\\])+'/,alias:"atom"},"boolean":/\b(?:true|false)\b/,keyword:/\b(?:fun|when|case|of|end|if|receive|after|try|catch)\b/,number:[/\$\\?./,/\d+#[a-z0-9]+/i,/(?:\b|-)\d*\.?\d+([Ee][+-]?\d+)?\b/],"function":/\b[a-z][\w@]*(?=\()/,variable:{pattern:/(^|[^@])(?:\b|\?)[A-Z_][\w@]*/,lookbehind:!0},operator:[/[=\/<>:]=|=[:\/]=|\+\+?|--?|[=*\/!]|\b(?:bnot|div|rem|band|bor|bxor|bsl|bsr|not|and|or|xor|orelse|andalso)\b/,{pattern:/(^|[^<])<(?!<)/,lookbehind:!0},{pattern:/(^|[^>])>(?!>)/,lookbehind:!0}],atom:/\b[a-z][\w@]*/,punctuation:/[()[\]{}:;,.#|]|<<|>>/};
+Prism.languages.fsharp=Prism.languages.extend("clike",{comment:[{pattern:/(^|[^\\])\(\*[\w\W]*?\*\)/,lookbehind:!0},{pattern:/(^|[^\\:])\/\/.*/,lookbehind:!0}],keyword:/\b(?:let|return|use|yield)(?:!\B|\b)|\b(abstract|and|as|assert|base|begin|class|default|delegate|do|done|downcast|downto|elif|else|end|exception|extern|false|finally|for|fun|function|global|if|in|inherit|inline|interface|internal|lazy|match|member|module|mutable|namespace|new|not|null|of|open|or|override|private|public|rec|select|static|struct|then|to|true|try|type|upcast|val|void|when|while|with|asr|land|lor|lsl|lsr|lxor|mod|sig|atomic|break|checked|component|const|constraint|constructor|continue|eager|event|external|fixed|functor|include|method|mixin|object|parallel|process|protected|pure|sealed|tailcall|trait|virtual|volatile)\b/,string:/(?:"""[\s\S]*?"""|@"(?:""|[^"])*"|("|')(?:\\\1|\\?(?!\1)[\s\S])*\1)B?/,number:[/\b-?0x[\da-fA-F]+(un|lf|LF)?\b/,/\b-?0b[01]+(y|uy)?\b/,/\b-?(\d*\.?\d+|\d+\.)([fFmM]|[eE][+-]?\d+)?\b/,/\b-?\d+(y|uy|s|us|l|u|ul|L|UL|I)?\b/]}),Prism.languages.insertBefore("fsharp","keyword",{preprocessor:{pattern:/^[^\r\n\S]*#.*/m,alias:"property",inside:{directive:{pattern:/(\s*#)\b(else|endif|if|light|line|nowarn)\b/,lookbehind:!0,alias:"keyword"}}}});
+Prism.languages.fortran={"quoted-number":{pattern:/[BOZ](['"])[A-F0-9]+\1/i,alias:"number"},string:{pattern:/(?:\w+_)?(['"])(?:\1\1|&(?:\r\n?|\n)(?:\s*!.+(?:\r\n?|\n))?|(?!\1).)*(?:\1|&)/,inside:{comment:{pattern:/(&(?:\r\n?|\n)\s*)!.*/,lookbehind:!0}}},comment:/!.*/,"boolean":/\.(?:TRUE|FALSE)\.(?:_\w+)?/i,number:/(?:\b|[+-])(?:\d+(?:\.\d*)?|\.\d+)(?:[ED][+-]?\d+)?(?:_\w+)?/i,keyword:[/\b(?:INTEGER|REAL|DOUBLE ?PRECISION|COMPLEX|CHARACTER|LOGICAL)\b/i,/\b(?:END ?)?(?:BLOCK ?DATA|DO|FILE|FORALL|FUNCTION|IF|INTERFACE|MODULE(?! PROCEDURE)|PROGRAM|SELECT|SUBROUTINE|TYPE|WHERE)\b/i,/\b(?:ALLOCATABLE|ALLOCATE|BACKSPACE|CALL|CASE|CLOSE|COMMON|CONTAINS|CONTINUE|CYCLE|DATA|DEALLOCATE|DIMENSION|DO|END|EQUIVALENCE|EXIT|EXTERNAL|FORMAT|GO ?TO|IMPLICIT(?: NONE)?|INQUIRE|INTENT|INTRINSIC|MODULE PROCEDURE|NAMELIST|NULLIFY|OPEN|OPTIONAL|PARAMETER|POINTER|PRINT|PRIVATE|PUBLIC|READ|RETURN|REWIND|SAVE|SELECT|STOP|TARGET|WHILE|WRITE)\b/i,/\b(?:ASSIGNMENT|DEFAULT|ELEMENTAL|ELSE|ELSEWHERE|ELSEIF|ENTRY|IN|INCLUDE|INOUT|KIND|NULL|ONLY|OPERATOR|OUT|PURE|RECURSIVE|RESULT|SEQUENCE|STAT|THEN|USE)\b/i],operator:[/\*\*|\/\/|=>|[=\/]=|[<>]=?|::|[+\-*=%]|\.(?:EQ|NE|LT|LE|GT|GE|NOT|AND|OR|EQV|NEQV)\.|\.[A-Z]+\./i,{pattern:/(^|(?!\().)\/(?!\))/,lookbehind:!0}],punctuation:/\(\/|\/\)|[(),;:&]/};
+Prism.languages.git={comment:/^#.*/m,deleted:/^[-–].*/m,inserted:/^\+.*/m,string:/("|')(\\?.)*?\1/m,command:{pattern:/^.*\$ git .*$/m,inside:{parameter:/\s(--|-)\w+/m}},coord:/^@@.*@@$/m,commit_sha1:/^commit \w{40}$/m};
+Prism.languages.go=Prism.languages.extend("clike",{keyword:/\b(break|case|chan|const|continue|default|defer|else|fallthrough|for|func|go(to)?|if|import|interface|map|package|range|return|select|struct|switch|type|var)\b/,builtin:/\b(bool|byte|complex(64|128)|error|float(32|64)|rune|string|u?int(8|16|32|64|)|uintptr|append|cap|close|complex|copy|delete|imag|len|make|new|panic|print(ln)?|real|recover)\b/,"boolean":/\b(_|iota|nil|true|false)\b/,operator:/[*\/%^!=]=?|\+[=+]?|-[=-]?|\|[=|]?|&(?:=|&|\^=?)?|>(?:>=?|=)?|<(?:<=?|=|-)?|:=|\.\.\./,number:/\b(-?(0x[a-f\d]+|(\d+\.?\d*|\.\d+)(e[-+]?\d+)?)i?)\b/i,string:/("|'|`)(\\?.|\r|\n)*?\1/}),delete Prism.languages.go["class-name"];
+Prism.languages.haskell={comment:{pattern:/(^|[^-!#$%*+=?&@|~.:<>^\\\/])(--[^-!#$%*+=?&@|~.:<>^\\\/].*|{-[\w\W]*?-})/m,lookbehind:!0},"char":/'([^\\']|\\([abfnrtv\\"'&]|\^[A-Z@[\]\^_]|NUL|SOH|STX|ETX|EOT|ENQ|ACK|BEL|BS|HT|LF|VT|FF|CR|SO|SI|DLE|DC1|DC2|DC3|DC4|NAK|SYN|ETB|CAN|EM|SUB|ESC|FS|GS|RS|US|SP|DEL|\d+|o[0-7]+|x[0-9a-fA-F]+))'/,string:{pattern:/"([^\\"]|\\([abfnrtv\\"'&]|\^[A-Z@[\]\^_]|NUL|SOH|STX|ETX|EOT|ENQ|ACK|BEL|BS|HT|LF|VT|FF|CR|SO|SI|DLE|DC1|DC2|DC3|DC4|NAK|SYN|ETB|CAN|EM|SUB|ESC|FS|GS|RS|US|SP|DEL|\d+|o[0-7]+|x[0-9a-fA-F]+)|\\\s+\\)*"/,greedy:!0},keyword:/\b(case|class|data|deriving|do|else|if|in|infixl|infixr|instance|let|module|newtype|of|primitive|then|type|where)\b/,import_statement:{pattern:/(\r?\n|\r|^)\s*import\s+(qualified\s+)?([A-Z][_a-zA-Z0-9']*)(\.[A-Z][_a-zA-Z0-9']*)*(\s+as\s+([A-Z][_a-zA-Z0-9']*)(\.[A-Z][_a-zA-Z0-9']*)*)?(\s+hiding\b)?/m,inside:{keyword:/\b(import|qualified|as|hiding)\b/}},builtin:/\b(abs|acos|acosh|all|and|any|appendFile|approxRational|asTypeOf|asin|asinh|atan|atan2|atanh|basicIORun|break|catch|ceiling|chr|compare|concat|concatMap|const|cos|cosh|curry|cycle|decodeFloat|denominator|digitToInt|div|divMod|drop|dropWhile|either|elem|encodeFloat|enumFrom|enumFromThen|enumFromThenTo|enumFromTo|error|even|exp|exponent|fail|filter|flip|floatDigits|floatRadix|floatRange|floor|fmap|foldl|foldl1|foldr|foldr1|fromDouble|fromEnum|fromInt|fromInteger|fromIntegral|fromRational|fst|gcd|getChar|getContents|getLine|group|head|id|inRange|index|init|intToDigit|interact|ioError|isAlpha|isAlphaNum|isAscii|isControl|isDenormalized|isDigit|isHexDigit|isIEEE|isInfinite|isLower|isNaN|isNegativeZero|isOctDigit|isPrint|isSpace|isUpper|iterate|last|lcm|length|lex|lexDigits|lexLitChar|lines|log|logBase|lookup|map|mapM|mapM_|max|maxBound|maximum|maybe|min|minBound|minimum|mod|negate|not|notElem|null|numerator|odd|or|ord|otherwise|pack|pi|pred|primExitWith|print|product|properFraction|putChar|putStr|putStrLn|quot|quotRem|range|rangeSize|read|readDec|readFile|readFloat|readHex|readIO|readInt|readList|readLitChar|readLn|readOct|readParen|readSigned|reads|readsPrec|realToFrac|recip|rem|repeat|replicate|return|reverse|round|scaleFloat|scanl|scanl1|scanr|scanr1|seq|sequence|sequence_|show|showChar|showInt|showList|showLitChar|showParen|showSigned|showString|shows|showsPrec|significand|signum|sin|sinh|snd|sort|span|splitAt|sqrt|subtract|succ|sum|tail|take|takeWhile|tan|tanh|threadToIOResult|toEnum|toInt|toInteger|toLower|toRational|toUpper|truncate|uncurry|undefined|unlines|until|unwords|unzip|unzip3|userError|words|writeFile|zip|zip3|zipWith|zipWith3)\b/,number:/\b(\d+(\.\d+)?(e[+-]?\d+)?|0o[0-7]+|0x[0-9a-f]+)\b/i,operator:/\s\.\s|[-!#$%*+=?&@|~.:<>^\\\/]*\.[-!#$%*+=?&@|~.:<>^\\\/]+|[-!#$%*+=?&@|~.:<>^\\\/]+\.[-!#$%*+=?&@|~.:<>^\\\/]*|[-!#$%*+=?&@|~:<>^\\\/]+|`([A-Z][_a-zA-Z0-9']*\.)*[_a-z][_a-zA-Z0-9']*`/,hvariable:/\b([A-Z][_a-zA-Z0-9']*\.)*[_a-z][_a-zA-Z0-9']*\b/,constant:/\b([A-Z][_a-zA-Z0-9']*\.)*[A-Z][_a-zA-Z0-9']*\b/,punctuation:/[{}[\];(),.:]/};
+Prism.languages.http={"request-line":{pattern:/^(POST|GET|PUT|DELETE|OPTIONS|PATCH|TRACE|CONNECT)\b\shttps?:\/\/\S+\sHTTP\/[0-9.]+/m,inside:{property:/^(POST|GET|PUT|DELETE|OPTIONS|PATCH|TRACE|CONNECT)\b/,"attr-name":/:\w+/}},"response-status":{pattern:/^HTTP\/1.[01] [0-9]+.*/m,inside:{property:{pattern:/(^HTTP\/1.[01] )[0-9]+.*/i,lookbehind:!0}}},"header-name":{pattern:/^[\w-]+:(?=.)/m,alias:"keyword"}};var httpLanguages={"application/json":Prism.languages.javascript,"application/xml":Prism.languages.markup,"text/xml":Prism.languages.markup,"text/html":Prism.languages.markup};for(var contentType in httpLanguages)if(httpLanguages[contentType]){var options={};options[contentType]={pattern:new RegExp("(content-type:\\s*"+contentType+"[\\w\\W]*?)(?:\\r?\\n|\\r){2}[\\w\\W]*","i"),lookbehind:!0,inside:{rest:httpLanguages[contentType]}},Prism.languages.insertBefore("http","header-name",options)};
+Prism.languages.icon={comment:/#.*/,string:/(["'])(?:(?!\1)[^\\\r\n]|\\.|_(?:\r?\n|\r))*\1/,number:/\b(?:\d+r[a-z\d]+|\d+(?:\.\d+)?(?:e[+-]?\d+)?)\b|\.\d+\b/i,"builtin-keyword":{pattern:/&(?:allocated|ascii|clock|collections|cset|current|date|dateline|digits|dump|e|error(?:number|text|value)?|errout|fail|features|file|host|input|lcase|letters|level|line|main|null|output|phi|pi|pos|progname|random|regions|source|storage|subject|time|trace|ucase|version)\b/,alias:"variable"},directive:{pattern:/\$\w+/,alias:"builtin"},keyword:/\b(?:break|by|case|create|default|do|else|end|every|fail|global|if|initial|invocable|link|local|next|not|of|procedure|record|repeat|return|static|suspend|then|to|until|while)\b/,"function":/(?!\d)\w+(?=\s*[({]|\s*!\s*\[)/,operator:/[+-]:(?!=)|(?:[\/?@^%&]|\+\+?|--?|==?=?|~==?=?|\*\*?|\|\|\|?|<(?:->?|<?=?)|>>?=?)(?::=)?|:(?:=:?)?|[!.\\|~]/,punctuation:/[\[\](){},;]/};
+Prism.languages.java=Prism.languages.extend("clike",{keyword:/\b(abstract|continue|for|new|switch|assert|default|goto|package|synchronized|boolean|do|if|private|this|break|double|implements|protected|throw|byte|else|import|public|throws|case|enum|instanceof|return|transient|catch|extends|int|short|try|char|final|interface|static|void|class|finally|long|strictfp|volatile|const|float|native|super|while)\b/,number:/\b0b[01]+\b|\b0x[\da-f]*\.?[\da-fp\-]+\b|\b\d*\.?\d+(?:e[+-]?\d+)?[df]?\b/i,operator:{pattern:/(^|[^.])(?:\+[+=]?|-[-=]?|!=?|<<?=?|>>?>?=?|==?|&[&=]?|\|[|=]?|\*=?|\/=?|%=?|\^=?|[?:~])/m,lookbehind:!0}}),Prism.languages.insertBefore("java","function",{annotation:{alias:"punctuation",pattern:/(^|[^.])@\w+/,lookbehind:!0}});
+Prism.languages.json={property:/".*?"(?=\s*:)/gi,string:/"(?!:)(\\?[^"])*?"(?!:)/g,number:/\b-?(0x[\dA-Fa-f]+|\d*\.?\d+([Ee]-?\d+)?)\b/g,punctuation:/[{}[\]);,]/g,operator:/:/g,"boolean":/\b(true|false)\b/gi,"null":/\bnull\b/gi},Prism.languages.jsonp=Prism.languages.json;
+!function(a){var e=/\\([^a-z()[\]]|[a-z\*]+)/i,n={"equation-command":{pattern:e,alias:"regex"}};a.languages.latex={comment:/%.*/m,cdata:{pattern:/(\\begin\{((?:verbatim|lstlisting)\*?)\})([\w\W]*?)(?=\\end\{\2\})/,lookbehind:!0},equation:[{pattern:/\$(?:\\?[\w\W])*?\$|\\\((?:\\?[\w\W])*?\\\)|\\\[(?:\\?[\w\W])*?\\\]/,inside:n,alias:"string"},{pattern:/(\\begin\{((?:equation|math|eqnarray|align|multline|gather)\*?)\})([\w\W]*?)(?=\\end\{\2\})/,lookbehind:!0,inside:n,alias:"string"}],keyword:{pattern:/(\\(?:begin|end|ref|cite|label|usepackage|documentclass)(?:\[[^\]]+\])?\{)[^}]+(?=\})/,lookbehind:!0},url:{pattern:/(\\url\{)[^}]+(?=\})/,lookbehind:!0},headline:{pattern:/(\\(?:part|chapter|section|subsection|frametitle|subsubsection|paragraph|subparagraph|subsubparagraph|subsubsubparagraph)\*?(?:\[[^\]]+\])?\{)[^}]+(?=\}(?:\[[^\]]+\])?)/,lookbehind:!0,alias:"class-name"},"function":{pattern:e,alias:"selector"},punctuation:/[[\]{}&]/}}(Prism);
+Prism.languages.lua={comment:/^#!.+|--(?:\[(=*)\[[\s\S]*?\]\1\]|.*)/m,string:{pattern:/(["'])(?:(?!\1)[^\\\r\n]|\\z(?:\r\n|\s)|\\(?:\r\n|[\s\S]))*\1|\[(=*)\[[\s\S]*?\]\2\]/,greedy:!0},number:/\b0x[a-f\d]+\.?[a-f\d]*(?:p[+-]?\d+)?\b|\b\d+(?:\.\B|\.?\d*(?:e[+-]?\d+)?\b)|\B\.\d+(?:e[+-]?\d+)?\b/i,keyword:/\b(?:and|break|do|else|elseif|end|false|for|function|goto|if|in|local|nil|not|or|repeat|return|then|true|until|while)\b/,"function":/(?!\d)\w+(?=\s*(?:[({]))/,operator:[/[-+*%^&|#]|\/\/?|<[<=]?|>[>=]?|[=~]=?/,{pattern:/(^|[^.])\.\.(?!\.)/,lookbehind:!0}],punctuation:/[\[\](){},;]|\.+|:+/};
+Prism.languages.makefile={comment:{pattern:/(^|[^\\])#(?:\\(?:\r\n|[\s\S])|.)*/,lookbehind:!0},string:/(["'])(?:\\(?:\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1/,builtin:/\.[A-Z][^:#=\s]+(?=\s*:(?!=))/,symbol:{pattern:/^[^:=\r\n]+(?=\s*:(?!=))/m,inside:{variable:/\$+(?:[^(){}:#=\s]+|(?=[({]))/}},variable:/\$+(?:[^(){}:#=\s]+|\([@*%<^+?][DF]\)|(?=[({]))/,keyword:[/-include\b|\b(?:define|else|endef|endif|export|ifn?def|ifn?eq|include|override|private|sinclude|undefine|unexport|vpath)\b/,{pattern:/(\()(?:addsuffix|abspath|and|basename|call|dir|error|eval|file|filter(?:-out)?|findstring|firstword|flavor|foreach|guile|if|info|join|lastword|load|notdir|or|origin|patsubst|realpath|shell|sort|strip|subst|suffix|value|warning|wildcard|word(?:s|list)?)(?=[ \t])/,lookbehind:!0}],operator:/(?:::|[?:+!])?=|[|@]/,punctuation:/[:;(){}]/};
+Prism.languages.markdown=Prism.languages.extend("markup",{}),Prism.languages.insertBefore("markdown","prolog",{blockquote:{pattern:/^>(?:[\t ]*>)*/m,alias:"punctuation"},code:[{pattern:/^(?: {4}|\t).+/m,alias:"keyword"},{pattern:/``.+?``|`[^`\n]+`/,alias:"keyword"}],title:[{pattern:/\w+.*(?:\r?\n|\r)(?:==+|--+)/,alias:"important",inside:{punctuation:/==+$|--+$/}},{pattern:/(^\s*)#+.+/m,lookbehind:!0,alias:"important",inside:{punctuation:/^#+|#+$/}}],hr:{pattern:/(^\s*)([*-])([\t ]*\2){2,}(?=\s*$)/m,lookbehind:!0,alias:"punctuation"},list:{pattern:/(^\s*)(?:[*+-]|\d+\.)(?=[\t ].)/m,lookbehind:!0,alias:"punctuation"},"url-reference":{pattern:/!?\[[^\]]+\]:[\t ]+(?:\S+|<(?:\\.|[^>\\])+>)(?:[\t ]+(?:"(?:\\.|[^"\\])*"|'(?:\\.|[^'\\])*'|\((?:\\.|[^)\\])*\)))?/,inside:{variable:{pattern:/^(!?\[)[^\]]+/,lookbehind:!0},string:/(?:"(?:\\.|[^"\\])*"|'(?:\\.|[^'\\])*'|\((?:\\.|[^)\\])*\))$/,punctuation:/^[\[\]!:]|[<>]/},alias:"url"},bold:{pattern:/(^|[^\\])(\*\*|__)(?:(?:\r?\n|\r)(?!\r?\n|\r)|.)+?\2/,lookbehind:!0,inside:{punctuation:/^\*\*|^__|\*\*$|__$/}},italic:{pattern:/(^|[^\\])([*_])(?:(?:\r?\n|\r)(?!\r?\n|\r)|.)+?\2/,lookbehind:!0,inside:{punctuation:/^[*_]|[*_]$/}},url:{pattern:/!?\[[^\]]+\](?:\([^\s)]+(?:[\t ]+"(?:\\.|[^"\\])*")?\)| ?\[[^\]\n]*\])/,inside:{variable:{pattern:/(!?\[)[^\]]+(?=\]$)/,lookbehind:!0},string:{pattern:/"(?:\\.|[^"\\])*"(?=\)$)/}}}}),Prism.languages.markdown.bold.inside.url=Prism.util.clone(Prism.languages.markdown.url),Prism.languages.markdown.italic.inside.url=Prism.util.clone(Prism.languages.markdown.url),Prism.languages.markdown.bold.inside.italic=Prism.util.clone(Prism.languages.markdown.italic),Prism.languages.markdown.italic.inside.bold=Prism.util.clone(Prism.languages.markdown.bold);
+Prism.languages.matlab={string:/\B'(?:''|[^'\n])*'/,comment:[/%\{[\s\S]*?\}%/,/%.+/],number:/\b-?(?:\d*\.?\d+(?:[eE][+-]?\d+)?(?:[ij])?|[ij])\b/,keyword:/\b(?:break|case|catch|continue|else|elseif|end|for|function|if|inf|NaN|otherwise|parfor|pause|pi|return|switch|try|while)\b/,"function":/(?!\d)\w+(?=\s*\()/,operator:/\.?[*^\/\\']|[+\-:@]|[<>=~]=?|&&?|\|\|?/,punctuation:/\.{3}|[.,;\[\](){}!]/};
+Prism.languages.nasm={comment:/;.*$/m,string:/("|'|`)(\\?.)*?\1/m,label:{pattern:/(^\s*)[A-Za-z._?$][\w.?$@~#]*:/m,lookbehind:!0,alias:"function"},keyword:[/\[?BITS (16|32|64)\]?/m,{pattern:/(^\s*)section\s*[a-zA-Z\.]+:?/im,lookbehind:!0},/(?:extern|global)[^;\r\n]*/im,/(?:CPU|FLOAT|DEFAULT).*$/m],register:{pattern:/\b(?:st\d|[xyz]mm\d\d?|[cdt]r\d|r\d\d?[bwd]?|[er]?[abcd]x|[abcd][hl]|[er]?(bp|sp|si|di)|[cdefgs]s)\b/i,alias:"variable"},number:/(\b|-|(?=\$))(0[hx][\da-f]*\.?[\da-f]+(p[+-]?\d+)?|\d[\da-f]+[hx]|\$\d[\da-f]*|0[oq][0-7]+|[0-7]+[oq]|0[by][01]+|[01]+[by]|0[dt]\d+|\d*\.?\d+(\.?e[+-]?\d+)?[dt]?)\b/i,operator:/[\[\]*+\-\/%<>=&|$!]/};
+Prism.languages.nginx=Prism.languages.extend("clike",{comment:{pattern:/(^|[^"{\\])#.*/,lookbehind:!0},keyword:/\b(?:CONTENT_|DOCUMENT_|GATEWAY_|HTTP_|HTTPS|if_not_empty|PATH_|QUERY_|REDIRECT_|REMOTE_|REQUEST_|SCGI|SCRIPT_|SERVER_|http|server|events|location|include|accept_mutex|accept_mutex_delay|access_log|add_after_body|add_before_body|add_header|addition_types|aio|alias|allow|ancient_browser|ancient_browser_value|auth|auth_basic|auth_basic_user_file|auth_http|auth_http_header|auth_http_timeout|autoindex|autoindex_exact_size|autoindex_localtime|break|charset|charset_map|charset_types|chunked_transfer_encoding|client_body_buffer_size|client_body_in_file_only|client_body_in_single_buffer|client_body_temp_path|client_body_timeout|client_header_buffer_size|client_header_timeout|client_max_body_size|connection_pool_size|create_full_put_path|daemon|dav_access|dav_methods|debug_connection|debug_points|default_type|deny|devpoll_changes|devpoll_events|directio|directio_alignment|disable_symlinks|empty_gif|env|epoll_events|error_log|error_page|expires|fastcgi_buffer_size|fastcgi_buffers|fastcgi_busy_buffers_size|fastcgi_cache|fastcgi_cache_bypass|fastcgi_cache_key|fastcgi_cache_lock|fastcgi_cache_lock_timeout|fastcgi_cache_methods|fastcgi_cache_min_uses|fastcgi_cache_path|fastcgi_cache_purge|fastcgi_cache_use_stale|fastcgi_cache_valid|fastcgi_connect_timeout|fastcgi_hide_header|fastcgi_ignore_client_abort|fastcgi_ignore_headers|fastcgi_index|fastcgi_intercept_errors|fastcgi_keep_conn|fastcgi_max_temp_file_size|fastcgi_next_upstream|fastcgi_no_cache|fastcgi_param|fastcgi_pass|fastcgi_pass_header|fastcgi_read_timeout|fastcgi_redirect_errors|fastcgi_send_timeout|fastcgi_split_path_info|fastcgi_store|fastcgi_store_access|fastcgi_temp_file_write_size|fastcgi_temp_path|flv|geo|geoip_city|geoip_country|google_perftools_profiles|gzip|gzip_buffers|gzip_comp_level|gzip_disable|gzip_http_version|gzip_min_length|gzip_proxied|gzip_static|gzip_types|gzip_vary|if|if_modified_since|ignore_invalid_headers|image_filter|image_filter_buffer|image_filter_jpeg_quality|image_filter_sharpen|image_filter_transparency|imap_capabilities|imap_client_buffer|include|index|internal|ip_hash|keepalive|keepalive_disable|keepalive_requests|keepalive_timeout|kqueue_changes|kqueue_events|large_client_header_buffers|limit_conn|limit_conn_log_level|limit_conn_zone|limit_except|limit_rate|limit_rate_after|limit_req|limit_req_log_level|limit_req_zone|limit_zone|lingering_close|lingering_time|lingering_timeout|listen|location|lock_file|log_format|log_format_combined|log_not_found|log_subrequest|map|map_hash_bucket_size|map_hash_max_size|master_process|max_ranges|memcached_buffer_size|memcached_connect_timeout|memcached_next_upstream|memcached_pass|memcached_read_timeout|memcached_send_timeout|merge_slashes|min_delete_depth|modern_browser|modern_browser_value|mp4|mp4_buffer_size|mp4_max_buffer_size|msie_padding|msie_refresh|multi_accept|open_file_cache|open_file_cache_errors|open_file_cache_min_uses|open_file_cache_valid|open_log_file_cache|optimize_server_names|override_charset|pcre_jit|perl|perl_modules|perl_require|perl_set|pid|pop3_auth|pop3_capabilities|port_in_redirect|post_action|postpone_output|protocol|proxy|proxy_buffer|proxy_buffer_size|proxy_buffering|proxy_buffers|proxy_busy_buffers_size|proxy_cache|proxy_cache_bypass|proxy_cache_key|proxy_cache_lock|proxy_cache_lock_timeout|proxy_cache_methods|proxy_cache_min_uses|proxy_cache_path|proxy_cache_use_stale|proxy_cache_valid|proxy_connect_timeout|proxy_cookie_domain|proxy_cookie_path|proxy_headers_hash_bucket_size|proxy_headers_hash_max_size|proxy_hide_header|proxy_http_version|proxy_ignore_client_abort|proxy_ignore_headers|proxy_intercept_errors|proxy_max_temp_file_size|proxy_method|proxy_next_upstream|proxy_no_cache|proxy_pass|proxy_pass_error_message|proxy_pass_header|proxy_pass_request_body|proxy_pass_request_headers|proxy_read_timeout|proxy_redirect|proxy_redirect_errors|proxy_send_lowat|proxy_send_timeout|proxy_set_body|proxy_set_header|proxy_ssl_session_reuse|proxy_store|proxy_store_access|proxy_temp_file_write_size|proxy_temp_path|proxy_timeout|proxy_upstream_fail_timeout|proxy_upstream_max_fails|random_index|read_ahead|real_ip_header|recursive_error_pages|request_pool_size|reset_timedout_connection|resolver|resolver_timeout|return|rewrite|root|rtsig_overflow_events|rtsig_overflow_test|rtsig_overflow_threshold|rtsig_signo|satisfy|satisfy_any|secure_link_secret|send_lowat|send_timeout|sendfile|sendfile_max_chunk|server|server_name|server_name_in_redirect|server_names_hash_bucket_size|server_names_hash_max_size|server_tokens|set|set_real_ip_from|smtp_auth|smtp_capabilities|so_keepalive|source_charset|split_clients|ssi|ssi_silent_errors|ssi_types|ssi_value_length|ssl|ssl_certificate|ssl_certificate_key|ssl_ciphers|ssl_client_certificate|ssl_crl|ssl_dhparam|ssl_engine|ssl_prefer_server_ciphers|ssl_protocols|ssl_session_cache|ssl_session_timeout|ssl_verify_client|ssl_verify_depth|starttls|stub_status|sub_filter|sub_filter_once|sub_filter_types|tcp_nodelay|tcp_nopush|timeout|timer_resolution|try_files|types|types_hash_bucket_size|types_hash_max_size|underscores_in_headers|uninitialized_variable_warn|upstream|use|user|userid|userid_domain|userid_expires|userid_name|userid_p3p|userid_path|userid_service|valid_referers|variables_hash_bucket_size|variables_hash_max_size|worker_connections|worker_cpu_affinity|worker_priority|worker_processes|worker_rlimit_core|worker_rlimit_nofile|worker_rlimit_sigpending|working_directory|xclient|xml_entities|xslt_entities|xslt_stylesheet|xslt_types)\b/i}),Prism.languages.insertBefore("nginx","keyword",{variable:/\$[a-z_]+/i});
+Prism.languages.objectivec=Prism.languages.extend("c",{keyword:/\b(asm|typeof|inline|auto|break|case|char|const|continue|default|do|double|else|enum|extern|float|for|goto|if|int|long|register|return|short|signed|sizeof|static|struct|switch|typedef|union|unsigned|void|volatile|while|in|self|super)\b|(@interface|@end|@implementation|@protocol|@class|@public|@protected|@private|@property|@try|@catch|@finally|@throw|@synthesize|@dynamic|@selector)\b/,string:/("|')(\\(?:\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1|@"(\\(?:\r\n|[\s\S])|[^"\\\r\n])*"/,operator:/-[->]?|\+\+?|!=?|<<?=?|>>?=?|==?|&&?|\|\|?|[~^%?*\/@]/});
+Prism.languages.pascal={comment:[/\(\*[\s\S]+?\*\)/,/\{[\s\S]+?\}/,/\/\/.*/],string:/(?:'(?:''|[^'\r\n])*'|#[&$%]?[a-f\d]+)+|\^[a-z]/i,keyword:[{pattern:/(^|[^&])\b(?:absolute|array|asm|begin|case|const|constructor|destructor|do|downto|else|end|file|for|function|goto|if|implementation|inherited|inline|interface|label|nil|object|of|operator|packed|procedure|program|record|reintroduce|repeat|self|set|string|then|to|type|unit|until|uses|var|while|with)\b/i,lookbehind:!0},{pattern:/(^|[^&])\b(?:dispose|exit|false|new|true)\b/i,lookbehind:!0},{pattern:/(^|[^&])\b(?:class|dispinterface|except|exports|finalization|finally|initialization|inline|library|on|out|packed|property|raise|resourcestring|threadvar|try)\b/i,lookbehind:!0},{pattern:/(^|[^&])\b(?:absolute|abstract|alias|assembler|bitpacked|break|cdecl|continue|cppdecl|cvar|default|deprecated|dynamic|enumerator|experimental|export|external|far|far16|forward|generic|helper|implements|index|interrupt|iochecks|local|message|name|near|nodefault|noreturn|nostackframe|oldfpccall|otherwise|overload|override|pascal|platform|private|protected|public|published|read|register|reintroduce|result|safecall|saveregisters|softfloat|specialize|static|stdcall|stored|strict|unaligned|unimplemented|varargs|virtual|write)\b/i,lookbehind:!0}],number:[/[+-]?(?:[&%]\d+|\$[a-f\d]+)/i,/([+-]|\b)\d+(?:\.\d+)?(?:e[+-]?\d+)?/i],operator:[/\.\.|\*\*|:=|<[<=>]?|>[>=]?|[+\-*\/]=?|[@^=]/i,{pattern:/(^|[^&])\b(?:and|as|div|exclude|in|include|is|mod|not|or|shl|shr|xor)\b/,lookbehind:!0}],punctuation:/\(\.|\.\)|[()\[\]:;,.]/};
+Prism.languages.perl={comment:[{pattern:/(^\s*)=\w+[\s\S]*?=cut.*/m,lookbehind:!0},{pattern:/(^|[^\\$])#.*/,lookbehind:!0}],string:[/\b(?:q|qq|qx|qw)\s*([^a-zA-Z0-9\s\{\(\[<])(?:[^\\]|\\[\s\S])*?\1/,/\b(?:q|qq|qx|qw)\s+([a-zA-Z0-9])(?:[^\\]|\\[\s\S])*?\1/,/\b(?:q|qq|qx|qw)\s*\((?:[^()\\]|\\[\s\S])*\)/,/\b(?:q|qq|qx|qw)\s*\{(?:[^{}\\]|\\[\s\S])*\}/,/\b(?:q|qq|qx|qw)\s*\[(?:[^[\]\\]|\\[\s\S])*\]/,/\b(?:q|qq|qx|qw)\s*<(?:[^<>\\]|\\[\s\S])*>/,/("|`)(?:[^\\]|\\[\s\S])*?\1/,/'(?:[^'\\\r\n]|\\.)*'/],regex:[/\b(?:m|qr)\s*([^a-zA-Z0-9\s\{\(\[<])(?:[^\\]|\\[\s\S])*?\1[msixpodualngc]*/,/\b(?:m|qr)\s+([a-zA-Z0-9])(?:[^\\]|\\.)*?\1[msixpodualngc]*/,/\b(?:m|qr)\s*\((?:[^()\\]|\\[\s\S])*\)[msixpodualngc]*/,/\b(?:m|qr)\s*\{(?:[^{}\\]|\\[\s\S])*\}[msixpodualngc]*/,/\b(?:m|qr)\s*\[(?:[^[\]\\]|\\[\s\S])*\][msixpodualngc]*/,/\b(?:m|qr)\s*<(?:[^<>\\]|\\[\s\S])*>[msixpodualngc]*/,{pattern:/(^|[^-]\b)(?:s|tr|y)\s*([^a-zA-Z0-9\s\{\(\[<])(?:[^\\]|\\[\s\S])*?\2(?:[^\\]|\\[\s\S])*?\2[msixpodualngcer]*/,lookbehind:!0},{pattern:/(^|[^-]\b)(?:s|tr|y)\s+([a-zA-Z0-9])(?:[^\\]|\\[\s\S])*?\2(?:[^\\]|\\[\s\S])*?\2[msixpodualngcer]*/,lookbehind:!0},{pattern:/(^|[^-]\b)(?:s|tr|y)\s*\((?:[^()\\]|\\[\s\S])*\)\s*\((?:[^()\\]|\\[\s\S])*\)[msixpodualngcer]*/,lookbehind:!0},{pattern:/(^|[^-]\b)(?:s|tr|y)\s*\{(?:[^{}\\]|\\[\s\S])*\}\s*\{(?:[^{}\\]|\\[\s\S])*\}[msixpodualngcer]*/,lookbehind:!0},{pattern:/(^|[^-]\b)(?:s|tr|y)\s*\[(?:[^[\]\\]|\\[\s\S])*\]\s*\[(?:[^[\]\\]|\\[\s\S])*\][msixpodualngcer]*/,lookbehind:!0},{pattern:/(^|[^-]\b)(?:s|tr|y)\s*<(?:[^<>\\]|\\[\s\S])*>\s*<(?:[^<>\\]|\\[\s\S])*>[msixpodualngcer]*/,lookbehind:!0},/\/(?:[^\/\\\r\n]|\\.)*\/[msixpodualngc]*(?=\s*(?:$|[\r\n,.;})&|\-+*~<>!?^]|(lt|gt|le|ge|eq|ne|cmp|not|and|or|xor|x)\b))/],variable:[/[&*$@%]\{\^[A-Z]+\}/,/[&*$@%]\^[A-Z_]/,/[&*$@%]#?(?=\{)/,/[&*$@%]#?((::)*'?(?!\d)[\w$]+)+(::)*/i,/[&*$@%]\d+/,/(?!%=)[$@%][!"#$%&'()*+,\-.\/:;<=>?@[\\\]^_`{|}~]/],filehandle:{pattern:/<(?![<=])\S*>|\b_\b/,alias:"symbol"},vstring:{pattern:/v\d+(\.\d+)*|\d+(\.\d+){2,}/,alias:"string"},"function":{pattern:/sub [a-z0-9_]+/i,inside:{keyword:/sub/}},keyword:/\b(any|break|continue|default|delete|die|do|else|elsif|eval|for|foreach|given|goto|if|last|local|my|next|our|package|print|redo|require|say|state|sub|switch|undef|unless|until|use|when|while)\b/,number:/\b-?(0x[\dA-Fa-f](_?[\dA-Fa-f])*|0b[01](_?[01])*|(\d(_?\d)*)?\.?\d(_?\d)*([Ee][+-]?\d+)?)\b/,operator:/-[rwxoRWXOezsfdlpSbctugkTBMAC]\b|\+[+=]?|-[-=>]?|\*\*?=?|\/\/?=?|=[=~>]?|~[~=]?|\|\|?=?|&&?=?|<(?:=>?|<=?)?|>>?=?|![~=]?|[%^]=?|\.(?:=|\.\.?)?|[\\?]|\bx(?:=|\b)|\b(lt|gt|le|ge|eq|ne|cmp|not|and|or|xor)\b/,punctuation:/[{}[\];(),:]/};
+Prism.languages.php=Prism.languages.extend("clike",{keyword:/\b(and|or|xor|array|as|break|case|cfunction|class|const|continue|declare|default|die|do|else|elseif|enddeclare|endfor|endforeach|endif|endswitch|endwhile|extends|for|foreach|function|include|include_once|global|if|new|return|static|switch|use|require|require_once|var|while|abstract|interface|public|implements|private|protected|parent|throw|null|echo|print|trait|namespace|final|yield|goto|instanceof|finally|try|catch)\b/i,constant:/\b[A-Z0-9_]{2,}\b/,comment:{pattern:/(^|[^\\])(?:\/\*[\w\W]*?\*\/|\/\/.*)/,lookbehind:!0,greedy:!0}}),Prism.languages.insertBefore("php","class-name",{"shell-comment":{pattern:/(^|[^\\])#.*/,lookbehind:!0,alias:"comment"}}),Prism.languages.insertBefore("php","keyword",{delimiter:/\?>|<\?(?:php)?/i,variable:/\$\w+\b/i,"package":{pattern:/(\\|namespace\s+|use\s+)[\w\\]+/,lookbehind:!0,inside:{punctuation:/\\/}}}),Prism.languages.insertBefore("php","operator",{property:{pattern:/(->)[\w]+/,lookbehind:!0}}),Prism.languages.markup&&(Prism.hooks.add("before-highlight",function(e){"php"===e.language&&(e.tokenStack=[],e.code=e.code.replace(/(?:<\?php|<\?)[\w\W]*?(?:\?>)/gi,function(n){return e.tokenStack.push(n),"{{{PHP"+e.tokenStack.length+"}}}"}))}),Prism.hooks.add("after-highlight",function(e){if("php"===e.language){for(var n,a=0;n=e.tokenStack[a];a++)e.highlightedCode=e.highlightedCode.replace("{{{PHP"+(a+1)+"}}}",Prism.highlight(n,e.grammar,"php").replace(/\$/g,"$$$$"));e.element.innerHTML=e.highlightedCode}}),Prism.hooks.add("wrap",function(e){"php"===e.language&&"markup"===e.type&&(e.content=e.content.replace(/(\{\{\{PHP[0-9]+\}\}\})/g,'<span class="token php">$1</span>'))}),Prism.languages.insertBefore("php","comment",{markup:{pattern:/<[^?]\/?(.*?)>/,inside:Prism.languages.markup},php:/\{\{\{PHP[0-9]+\}\}\}/}));
+Prism.languages.insertBefore("php","variable",{"this":/\$this\b/,global:/\$(?:_(?:SERVER|GET|POST|FILES|REQUEST|SESSION|ENV|COOKIE)|GLOBALS|HTTP_RAW_POST_DATA|argc|argv|php_errormsg|http_response_header)/,scope:{pattern:/\b[\w\\]+::/,inside:{keyword:/(static|self|parent)/,punctuation:/(::|\\)/}}});
+Prism.languages.properties={comment:/^[ \t]*[#!].*$/m,"attr-value":{pattern:/(^[ \t]*(?:\\(?:\r\n|[\s\S])|[^\\\s:=])+?(?: *[=:] *| ))(?:\\(?:\r\n|[\s\S])|.)+/m,lookbehind:!0},"attr-name":/^[ \t]*(?:\\(?:\r\n|[\s\S])|[^\\\s:=])+?(?= *[ =:]| )/m,punctuation:/[=:]/};
+Prism.languages.python={"triple-quoted-string":{pattern:/"""[\s\S]+?"""|'''[\s\S]+?'''/,alias:"string"},comment:{pattern:/(^|[^\\])#.*/,lookbehind:!0},string:{pattern:/("|')(?:\\\\|\\?[^\\\r\n])*?\1/,greedy:!0},"function":{pattern:/((?:^|\s)def[ \t]+)[a-zA-Z_][a-zA-Z0-9_]*(?=\()/g,lookbehind:!0},"class-name":{pattern:/(\bclass\s+)[a-z0-9_]+/i,lookbehind:!0},keyword:/\b(?:as|assert|async|await|break|class|continue|def|del|elif|else|except|exec|finally|for|from|global|if|import|in|is|lambda|pass|print|raise|return|try|while|with|yield)\b/,"boolean":/\b(?:True|False)\b/,number:/\b-?(?:0[bo])?(?:(?:\d|0x[\da-f])[\da-f]*\.?\d*|\.\d+)(?:e[+-]?\d+)?j?\b/i,operator:/[-+%=]=?|!=|\*\*?=?|\/\/?=?|<[<=>]?|>[=>]?|[&|^~]|\b(?:or|and|not)\b/,punctuation:/[{}[\];(),.:]/};
+Prism.languages.r={comment:/#.*/,string:/(['"])(?:\\?.)*?\1/,"percent-operator":{pattern:/%[^%\s]*%/,alias:"operator"},"boolean":/\b(?:TRUE|FALSE)\b/,ellipsis:/\.\.(?:\.|\d+)/,number:[/\b(?:NaN|Inf)\b/,/\b(?:0x[\dA-Fa-f]+(?:\.\d*)?|\d*\.?\d+)(?:[EePp][+-]?\d+)?[iL]?\b/],keyword:/\b(?:if|else|repeat|while|function|for|in|next|break|NULL|NA|NA_integer_|NA_real_|NA_complex_|NA_character_)\b/,operator:/->?>?|<(?:=|<?-)?|[>=!]=?|::?|&&?|\|\|?|[+*\/^$@~]/,punctuation:/[(){}\[\],;]/};
+!function(a){var e=a.util.clone(a.languages.javascript);a.languages.jsx=a.languages.extend("markup",e),a.languages.jsx.tag.pattern=/<\/?[\w\.:-]+\s*(?:\s+[\w\.:-]+(?:=(?:("|')(\\?[\w\W])*?\1|[^\s'">=]+|(\{[\w\W]*?\})))?\s*)*\/?>/i,a.languages.jsx.tag.inside["attr-value"].pattern=/=[^\{](?:('|")[\w\W]*?(\1)|[^\s>]+)/i;var s=a.util.clone(a.languages.jsx);delete s.punctuation,s=a.languages.insertBefore("jsx","operator",{punctuation:/=(?={)|[{}[\];(),.:]/},{jsx:s}),a.languages.insertBefore("inside","attr-value",{script:{pattern:/=(\{(?:\{[^}]*\}|[^}])+\})/i,inside:s,alias:"language-javascript"}},a.languages.jsx.tag)}(Prism);
+Prism.languages.rust={comment:[{pattern:/(^|[^\\])\/\*[\w\W]*?\*\//,lookbehind:!0},{pattern:/(^|[^\\:])\/\/.*/,lookbehind:!0}],string:[/b?r(#*)"(?:\\?.)*?"\1/,/b?("|')(?:\\?.)*?\1/],keyword:/\b(?:abstract|alignof|as|be|box|break|const|continue|crate|do|else|enum|extern|false|final|fn|for|if|impl|in|let|loop|match|mod|move|mut|offsetof|once|override|priv|pub|pure|ref|return|sizeof|static|self|struct|super|true|trait|type|typeof|unsafe|unsized|use|virtual|where|while|yield)\b/,attribute:{pattern:/#!?\[.+?\]/,alias:"attr-name"},"function":[/[a-z0-9_]+(?=\s*\()/i,/[a-z0-9_]+!(?=\s*\(|\[)/i],"macro-rules":{pattern:/[a-z0-9_]+!/i,alias:"function"},number:/\b-?(?:0x[\dA-Fa-f](?:_?[\dA-Fa-f])*|0o[0-7](?:_?[0-7])*|0b[01](?:_?[01])*|(\d(_?\d)*)?\.?\d(_?\d)*([Ee][+-]?\d+)?)(?:_?(?:[iu](?:8|16|32|64)?|f32|f64))?\b/,"closure-params":{pattern:/\|[^|]*\|(?=\s*[{-])/,inside:{punctuation:/[\|:,]/,operator:/[&*]/}},punctuation:/[{}[\];(),:]|\.+|->/,operator:/[-+*\/%!^=]=?|@|&[&=]?|\|[|=]?|<<?=?|>>?=?/};
+Prism.languages.scala=Prism.languages.extend("java",{keyword:/<-|=>|\b(?:abstract|case|catch|class|def|do|else|extends|final|finally|for|forSome|if|implicit|import|lazy|match|new|null|object|override|package|private|protected|return|sealed|self|super|this|throw|trait|try|type|val|var|while|with|yield)\b/,string:[{pattern:/"""[\W\w]*?"""/,greedy:!0},{pattern:/("|')(?:\\\\|\\?[^\\\r\n])*?\1/,greedy:!0}],builtin:/\b(?:String|Int|Long|Short|Byte|Boolean|Double|Float|Char|Any|AnyRef|AnyVal|Unit|Nothing)\b/,number:/\b(?:0x[\da-f]*\.?[\da-f]+|\d*\.?\d+e?\d*[dfl]?)\b/i,symbol:/'[^\d\s\\]\w*/}),delete Prism.languages.scala["class-name"],delete Prism.languages.scala["function"];
+Prism.languages.scheme={comment:/;.*/,string:/"(?:[^"\\\r\n]|\\.)*?"|'[^('\s]*/,keyword:{pattern:/(\()(?:define(?:-syntax|-library|-values)?|(?:case-)?lambda|let(?:\*|rec)?(?:-values)?|else|if|cond|begin|delay(?:-force)?|parameterize|guard|set!|(?:quasi-)?quote|syntax-rules)/,lookbehind:!0},builtin:{pattern:/(\()(?:(?:cons|car|cdr|list|call-with-current-continuation|call\/cc|append|abs|apply|eval)\b|null\?|pair\?|boolean\?|eof-object\?|char\?|procedure\?|number\?|port\?|string\?|vector\?|symbol\?|bytevector\?)/,lookbehind:!0},number:{pattern:/(\s|\))[-+]?[0-9]*\.?[0-9]+(?:\s*[-+]\s*[0-9]*\.?[0-9]+i)?\b/,lookbehind:!0},"boolean":/#[tf]/,operator:{pattern:/(\()(?:[-+*%\/]|[<>]=?|=>?)/,lookbehind:!0},"function":{pattern:/(\()[^\s()]*(?=\s)/,lookbehind:!0},punctuation:/[()]/};
+Prism.languages.smalltalk={comment:/"(?:""|[^"])+"/,string:/'(?:''|[^'])+'/,symbol:/#[\da-z]+|#(?:-|([+\/\\*~<>=@%|&?!])\1?)|#(?=\()/i,"block-arguments":{pattern:/(\[\s*):[^\[|]*?\|/,lookbehind:!0,inside:{variable:/:[\da-z]+/i,punctuation:/\|/}},"temporary-variables":{pattern:/\|[^|]+\|/,inside:{variable:/[\da-z]+/i,punctuation:/\|/}},keyword:/\b(?:nil|true|false|self|super|new)\b/,character:{pattern:/\$./,alias:"string"},number:[/\d+r-?[\dA-Z]+(?:\.[\dA-Z]+)?(?:e-?\d+)?/,/(?:\B-|\b)\d+(?:\.\d+)?(?:e-?\d+)?/],operator:/[<=]=?|:=|~[~=]|\/\/?|\\\\|>[>=]?|[!^+\-*&|,@]/,punctuation:/[.;:?\[\](){}]/};
+Prism.languages.sql={comment:{pattern:/(^|[^\\])(?:\/\*[\w\W]*?\*\/|(?:--|\/\/|#).*)/,lookbehind:!0},string:{pattern:/(^|[^@\\])("|')(?:\\?[\s\S])*?\2/,lookbehind:!0},variable:/@[\w.$]+|@("|'|`)(?:\\?[\s\S])+?\1/,"function":/\b(?:COUNT|SUM|AVG|MIN|MAX|FIRST|LAST|UCASE|LCASE|MID|LEN|ROUND|NOW|FORMAT)(?=\s*\()/i,keyword:/\b(?:ACTION|ADD|AFTER|ALGORITHM|ALL|ALTER|ANALYZE|ANY|APPLY|AS|ASC|AUTHORIZATION|AUTO_INCREMENT|BACKUP|BDB|BEGIN|BERKELEYDB|BIGINT|BINARY|BIT|BLOB|BOOL|BOOLEAN|BREAK|BROWSE|BTREE|BULK|BY|CALL|CASCADED?|CASE|CHAIN|CHAR VARYING|CHARACTER (?:SET|VARYING)|CHARSET|CHECK|CHECKPOINT|CLOSE|CLUSTERED|COALESCE|COLLATE|COLUMN|COLUMNS|COMMENT|COMMIT|COMMITTED|COMPUTE|CONNECT|CONSISTENT|CONSTRAINT|CONTAINS|CONTAINSTABLE|CONTINUE|CONVERT|CREATE|CROSS|CURRENT(?:_DATE|_TIME|_TIMESTAMP|_USER)?|CURSOR|DATA(?:BASES?)?|DATE(?:TIME)?|DBCC|DEALLOCATE|DEC|DECIMAL|DECLARE|DEFAULT|DEFINER|DELAYED|DELETE|DELIMITER(?:S)?|DENY|DESC|DESCRIBE|DETERMINISTIC|DISABLE|DISCARD|DISK|DISTINCT|DISTINCTROW|DISTRIBUTED|DO|DOUBLE(?: PRECISION)?|DROP|DUMMY|DUMP(?:FILE)?|DUPLICATE KEY|ELSE|ENABLE|ENCLOSED BY|END|ENGINE|ENUM|ERRLVL|ERRORS|ESCAPE(?:D BY)?|EXCEPT|EXEC(?:UTE)?|EXISTS|EXIT|EXPLAIN|EXTENDED|FETCH|FIELDS|FILE|FILLFACTOR|FIRST|FIXED|FLOAT|FOLLOWING|FOR(?: EACH ROW)?|FORCE|FOREIGN|FREETEXT(?:TABLE)?|FROM|FULL|FUNCTION|GEOMETRY(?:COLLECTION)?|GLOBAL|GOTO|GRANT|GROUP|HANDLER|HASH|HAVING|HOLDLOCK|IDENTITY(?:_INSERT|COL)?|IF|IGNORE|IMPORT|INDEX|INFILE|INNER|INNODB|INOUT|INSERT|INT|INTEGER|INTERSECT|INTO|INVOKER|ISOLATION LEVEL|JOIN|KEYS?|KILL|LANGUAGE SQL|LAST|LEFT|LIMIT|LINENO|LINES|LINESTRING|LOAD|LOCAL|LOCK|LONG(?:BLOB|TEXT)|MATCH(?:ED)?|MEDIUM(?:BLOB|INT|TEXT)|MERGE|MIDDLEINT|MODIFIES SQL DATA|MODIFY|MULTI(?:LINESTRING|POINT|POLYGON)|NATIONAL(?: CHAR VARYING| CHARACTER(?: VARYING)?| VARCHAR)?|NATURAL|NCHAR(?: VARCHAR)?|NEXT|NO(?: SQL|CHECK|CYCLE)?|NONCLUSTERED|NULLIF|NUMERIC|OFF?|OFFSETS?|ON|OPEN(?:DATASOURCE|QUERY|ROWSET)?|OPTIMIZE|OPTION(?:ALLY)?|ORDER|OUT(?:ER|FILE)?|OVER|PARTIAL|PARTITION|PERCENT|PIVOT|PLAN|POINT|POLYGON|PRECEDING|PRECISION|PREV|PRIMARY|PRINT|PRIVILEGES|PROC(?:EDURE)?|PUBLIC|PURGE|QUICK|RAISERROR|READ(?:S SQL DATA|TEXT)?|REAL|RECONFIGURE|REFERENCES|RELEASE|RENAME|REPEATABLE|REPLICATION|REQUIRE|RESTORE|RESTRICT|RETURNS?|REVOKE|RIGHT|ROLLBACK|ROUTINE|ROW(?:COUNT|GUIDCOL|S)?|RTREE|RULE|SAVE(?:POINT)?|SCHEMA|SELECT|SERIAL(?:IZABLE)?|SESSION(?:_USER)?|SET(?:USER)?|SHARE MODE|SHOW|SHUTDOWN|SIMPLE|SMALLINT|SNAPSHOT|SOME|SONAME|START(?:ING BY)?|STATISTICS|STATUS|STRIPED|SYSTEM_USER|TABLES?|TABLESPACE|TEMP(?:ORARY|TABLE)?|TERMINATED BY|TEXT(?:SIZE)?|THEN|TIMESTAMP|TINY(?:BLOB|INT|TEXT)|TOP?|TRAN(?:SACTIONS?)?|TRIGGER|TRUNCATE|TSEQUAL|TYPES?|UNBOUNDED|UNCOMMITTED|UNDEFINED|UNION|UNIQUE|UNPIVOT|UPDATE(?:TEXT)?|USAGE|USE|USER|USING|VALUES?|VAR(?:BINARY|CHAR|CHARACTER|YING)|VIEW|WAITFOR|WARNINGS|WHEN|WHERE|WHILE|WITH(?: ROLLUP|IN)?|WORK|WRITE(?:TEXT)?)\b/i,"boolean":/\b(?:TRUE|FALSE|NULL)\b/i,number:/\b-?(?:0x)?\d*\.?[\da-f]+\b/,operator:/[-+*\/=%^~]|&&?|\|?\||!=?|<(?:=>?|<|>)?|>[>=]?|\b(?:AND|BETWEEN|IN|LIKE|NOT|OR|IS|DIV|REGEXP|RLIKE|SOUNDS LIKE|XOR)\b/i,punctuation:/[;[\]()`,.]/};
+Prism.languages.swift=Prism.languages.extend("clike",{string:{pattern:/("|')(\\(?:\((?:[^()]|\([^)]+\))+\)|\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1/,greedy:!0,inside:{interpolation:{pattern:/\\\((?:[^()]|\([^)]+\))+\)/,inside:{delimiter:{pattern:/^\\\(|\)$/,alias:"variable"}}}}},keyword:/\b(as|associativity|break|case|catch|class|continue|convenience|default|defer|deinit|didSet|do|dynamic(?:Type)?|else|enum|extension|fallthrough|final|for|func|get|guard|if|import|in|infix|init|inout|internal|is|lazy|left|let|mutating|new|none|nonmutating|operator|optional|override|postfix|precedence|prefix|private|Protocol|public|repeat|required|rethrows|return|right|safe|self|Self|set|static|struct|subscript|super|switch|throws?|try|Type|typealias|unowned|unsafe|var|weak|where|while|willSet|__(?:COLUMN__|FILE__|FUNCTION__|LINE__))\b/,number:/\b([\d_]+(\.[\de_]+)?|0x[a-f0-9_]+(\.[a-f0-9p_]+)?|0b[01_]+|0o[0-7_]+)\b/i,constant:/\b(nil|[A-Z_]{2,}|k[A-Z][A-Za-z_]+)\b/,atrule:/@\b(IB(?:Outlet|Designable|Action|Inspectable)|class_protocol|exported|noreturn|NS(?:Copying|Managed)|objc|UIApplicationMain|auto_closure)\b/,builtin:/\b([A-Z]\S+|abs|advance|alignof(?:Value)?|assert|contains|count(?:Elements)?|debugPrint(?:ln)?|distance|drop(?:First|Last)|dump|enumerate|equal|filter|find|first|getVaList|indices|isEmpty|join|last|lexicographicalCompare|map|max(?:Element)?|min(?:Element)?|numericCast|overlaps|partition|print(?:ln)?|reduce|reflect|reverse|sizeof(?:Value)?|sort(?:ed)?|split|startsWith|stride(?:of(?:Value)?)?|suffix|swap|toDebugString|toString|transcode|underestimateCount|unsafeBitCast|with(?:ExtendedLifetime|Unsafe(?:MutablePointers?|Pointers?)|VaList))\b/}),Prism.languages.swift.string.inside.interpolation.inside.rest=Prism.util.clone(Prism.languages.swift);
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+Prism.languages.css={comment:/\/\*[\w\W]*?\*\//,atrule:{pattern:/@[\w-]+?.*?(;|(?=\s*\{))/i,inside:{rule:/@[\w-]+/}},url:/url\((?:(["'])(\\(?:\r\n|[\w\W])|(?!\1)[^\\\r\n])*\1|.*?)\)/i,selector:/[^\{\}\s][^\{\};]*?(?=\s*\{)/,string:/("|')(\\(?:\r\n|[\w\W])|(?!\1)[^\\\r\n])*\1/,property:/(\b|\B)[\w-]+(?=\s*:)/i,important:/\B!important\b/i,"function":/[-a-z0-9]+(?=\()/i,punctuation:/[(){};:]/},Prism.languages.css.atrule.inside.rest=Prism.util.clone(Prism.languages.css),Prism.languages.markup&&(Prism.languages.insertBefore("markup","tag",{style:{pattern:/(<style[\w\W]*?>)[\w\W]*?(?=<\/style>)/i,lookbehind:!0,inside:Prism.languages.css,alias:"language-css"}}),Prism.languages.insertBefore("inside","attr-value",{"style-attr":{pattern:/\s*style=("|').*?\1/i,inside:{"attr-name":{pattern:/^\s*style/i,inside:Prism.languages.markup.tag.inside},punctuation:/^\s*=\s*['"]|['"]\s*$/,"attr-value":{pattern:/.+/i,inside:Prism.languages.css}},alias:"language-css"}},Prism.languages.markup.tag));
+Prism.languages.clike={comment:[{pattern:/(^|[^\\])\/\*[\w\W]*?\*\//,lookbehind:!0},{pattern:/(^|[^\\:])\/\/.*/,lookbehind:!0}],string:/(["'])(\\(?:\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1/,"class-name":{pattern:/((?:\b(?:class|interface|extends|implements|trait|instanceof|new)\s+)|(?:catch\s+\())[a-z0-9_\.\\]+/i,lookbehind:!0,inside:{punctuation:/(\.|\\)/}},keyword:/\b(if|else|while|do|for|return|in|instanceof|function|new|try|throw|catch|finally|null|break|continue)\b/,"boolean":/\b(true|false)\b/,"function":/[a-z0-9_]+(?=\()/i,number:/\b-?(?:0x[\da-f]+|\d*\.?\d+(?:e[+-]?\d+)?)\b/i,operator:/--?|\+\+?|!=?=?|<=?|>=?|==?=?|&&?|\|\|?|\?|\*|\/|~|\^|%/,punctuation:/[{}[\];(),.:]/};
+Prism.languages.javascript=Prism.languages.extend("clike",{keyword:/\b(as|async|await|break|case|catch|class|const|continue|debugger|default|delete|do|else|enum|export|extends|finally|for|from|function|get|if|implements|import|in|instanceof|interface|let|new|null|of|package|private|protected|public|return|set|static|super|switch|this|throw|try|typeof|var|void|while|with|yield)\b/,number:/\b-?(0x[\dA-Fa-f]+|0b[01]+|0o[0-7]+|\d*\.?\d+([Ee][+-]?\d+)?|NaN|Infinity)\b/,"function":/[_$a-zA-Z\xA0-\uFFFF][_$a-zA-Z0-9\xA0-\uFFFF]*(?=\()/i}),Prism.languages.insertBefore("javascript","keyword",{regex:{pattern:/(^|[^\/])\/(?!\/)(\[.+?]|\\.|[^\/\\\r\n])+\/[gimyu]{0,5}(?=\s*($|[\r\n,.;})]))/,lookbehind:!0}}),Prism.languages.insertBefore("javascript","class-name",{"template-string":{pattern:/`(?:\\`|\\?[^`])*`/,inside:{interpolation:{pattern:/\$\{[^}]+\}/,inside:{"interpolation-punctuation":{pattern:/^\$\{|\}$/,alias:"punctuation"},rest:Prism.languages.javascript}},string:/[\s\S]+/}}}),Prism.languages.markup&&Prism.languages.insertBefore("markup","tag",{script:{pattern:/(<script[\w\W]*?>)[\w\W]*?(?=<\/script>)/i,lookbehind:!0,inside:Prism.languages.javascript,alias:"language-javascript"}}),Prism.languages.js=Prism.languages.javascript;
+!function(e){var t={variable:[{pattern:/\$?\(\([\w\W]+?\)\)/,inside:{variable:[{pattern:/(^\$\(\([\w\W]+)\)\)/,lookbehind:!0},/^\$\(\(/],number:/\b-?(?:0x[\dA-Fa-f]+|\d*\.?\d+(?:[Ee]-?\d+)?)\b/,operator:/--?|-=|\+\+?|\+=|!=?|~|\*\*?|\*=|\/=?|%=?|<<=?|>>=?|<=?|>=?|==?|&&?|&=|\^=?|\|\|?|\|=|\?|:/,punctuation:/\(\(?|\)\)?|,|;/}},{pattern:/\$\([^)]+\)|`[^`]+`/,inside:{variable:/^\$\(|^`|\)$|`$/}},/\$(?:[a-z0-9_#\?\*!@]+|\{[^}]+\})/i]};e.languages.bash={shebang:{pattern:/^#!\s*\/bin\/bash|^#!\s*\/bin\/sh/,alias:"important"},comment:{pattern:/(^|[^"{\\])#.*/,lookbehind:!0},string:[{pattern:/((?:^|[^<])<<\s*)(?:"|')?(\w+?)(?:"|')?\s*\r?\n(?:[\s\S])*?\r?\n\2/g,lookbehind:!0,inside:t},{pattern:/("|')(?:\\?[\s\S])*?\1/g,inside:t}],variable:t.variable,"function":{pattern:/(^|\s|;|\||&)(?:alias|apropos|apt-get|aptitude|aspell|awk|basename|bash|bc|bg|builtin|bzip2|cal|cat|cd|cfdisk|chgrp|chmod|chown|chroot|chkconfig|cksum|clear|cmp|comm|command|cp|cron|crontab|csplit|cut|date|dc|dd|ddrescue|df|diff|diff3|dig|dir|dircolors|dirname|dirs|dmesg|du|egrep|eject|enable|env|ethtool|eval|exec|expand|expect|export|expr|fdformat|fdisk|fg|fgrep|file|find|fmt|fold|format|free|fsck|ftp|fuser|gawk|getopts|git|grep|groupadd|groupdel|groupmod|groups|gzip|hash|head|help|hg|history|hostname|htop|iconv|id|ifconfig|ifdown|ifup|import|install|jobs|join|kill|killall|less|link|ln|locate|logname|logout|look|lpc|lpr|lprint|lprintd|lprintq|lprm|ls|lsof|make|man|mkdir|mkfifo|mkisofs|mknod|more|most|mount|mtools|mtr|mv|mmv|nano|netstat|nice|nl|nohup|notify-send|nslookup|open|op|passwd|paste|pathchk|ping|pkill|popd|pr|printcap|printenv|printf|ps|pushd|pv|pwd|quota|quotacheck|quotactl|ram|rar|rcp|read|readarray|readonly|reboot|rename|renice|remsync|rev|rm|rmdir|rsync|screen|scp|sdiff|sed|seq|service|sftp|shift|shopt|shutdown|sleep|slocate|sort|source|split|ssh|stat|strace|su|sudo|sum|suspend|sync|tail|tar|tee|test|time|timeout|times|touch|top|traceroute|trap|tr|tsort|tty|type|ulimit|umask|umount|unalias|uname|unexpand|uniq|units|unrar|unshar|uptime|useradd|userdel|usermod|users|uuencode|uudecode|v|vdir|vi|vmstat|wait|watch|wc|wget|whereis|which|who|whoami|write|xargs|xdg-open|yes|zip)(?=$|\s|;|\||&)/,lookbehind:!0},keyword:{pattern:/(^|\s|;|\||&)(?:let|:|\.|if|then|else|elif|fi|for|break|continue|while|in|case|function|select|do|done|until|echo|exit|return|set|declare)(?=$|\s|;|\||&)/,lookbehind:!0},"boolean":{pattern:/(^|\s|;|\||&)(?:true|false)(?=$|\s|;|\||&)/,lookbehind:!0},operator:/&&?|\|\|?|==?|!=?|<<<?|>>|<=?|>=?|=~/,punctuation:/\$?\(\(?|\)\)?|\.\.|[{}[\];]/};var a=t.variable[1].inside;a["function"]=e.languages.bash["function"],a.keyword=e.languages.bash.keyword,a.boolean=e.languages.bash.boolean,a.operator=e.languages.bash.operator,a.punctuation=e.languages.bash.punctuation}(Prism);
+Prism.languages.css.selector={pattern:/[^\{\}\s][^\{\}]*(?=\s*\{)/,inside:{"pseudo-element":/:(?:after|before|first-letter|first-line|selection)|::[-\w]+/,"pseudo-class":/:[-\w]+(?:\(.*\))?/,"class":/\.[-:\.\w]+/,id:/#[-:\.\w]+/}},Prism.languages.insertBefore("css","function",{hexcode:/#[\da-f]{3,6}/i,entity:/\\[\da-f]{1,8}/i,number:/[\d%\.]+/});
+Prism.languages.git={comment:/^#.*/m,deleted:/^[-–].*/m,inserted:/^\+.*/m,string:/("|')(\\?.)*?\1/m,command:{pattern:/^.*\$ git .*$/m,inside:{parameter:/\s(--|-)\w+/m}},coord:/^@@.*@@$/m,commit_sha1:/^commit \w{40}$/m};
+!function(e){var a=/\{\{\{[\w\W]+?\}\}\}|\{\{[\w\W]+?\}\}/g;e.languages.handlebars=e.languages.extend("markup",{handlebars:{pattern:a,inside:{delimiter:{pattern:/^\{\{\{?|\}\}\}?$/i,alias:"punctuation"},string:/(["'])(\\?.)*?\1/,number:/\b-?(0x[\dA-Fa-f]+|\d*\.?\d+([Ee][+-]?\d+)?)\b/,"boolean":/\b(true|false)\b/,block:{pattern:/^(\s*~?\s*)[#\/]\S+?(?=\s*~?\s*$|\s)/i,lookbehind:!0,alias:"keyword"},brackets:{pattern:/\[[^\]]+\]/,inside:{punctuation:/\[|\]/,variable:/[\w\W]+/}},punctuation:/[!"#%&'()*+,.\/;<=>@\[\\\]^`{|}~]/,variable:/[^!"#%&'()*+,.\/;<=>@\[\\\]^`{|}~\s]+/}}}),e.languages.insertBefore("handlebars","tag",{"handlebars-comment":{pattern:/\{\{![\w\W]*?\}\}/,alias:["handlebars","comment"]}}),e.hooks.add("before-highlight",function(e){"handlebars"===e.language&&(e.tokenStack=[],e.backupCode=e.code,e.code=e.code.replace(a,function(a){return e.tokenStack.push(a),"___HANDLEBARS"+e.tokenStack.length+"___"}))}),e.hooks.add("before-insert",function(e){"handlebars"===e.language&&(e.code=e.backupCode,delete e.backupCode)}),e.hooks.add("after-highlight",function(a){if("handlebars"===a.language){for(var n,t=0;n=a.tokenStack[t];t++)a.highlightedCode=a.highlightedCode.replace("___HANDLEBARS"+(t+1)+"___",e.highlight(n,a.grammar,"handlebars").replace(/\$/g,"$$$$"));a.element.innerHTML=a.highlightedCode}})}(Prism);
+Prism.languages.http={"request-line":{pattern:/^(POST|GET|PUT|DELETE|OPTIONS|PATCH|TRACE|CONNECT)\b\shttps?:\/\/\S+\sHTTP\/[0-9.]+/m,inside:{property:/^(POST|GET|PUT|DELETE|OPTIONS|PATCH|TRACE|CONNECT)\b/,"attr-name":/:\w+/}},"response-status":{pattern:/^HTTP\/1.[01] [0-9]+.*/m,inside:{property:{pattern:/(^HTTP\/1.[01] )[0-9]+.*/i,lookbehind:!0}}},"header-name":{pattern:/^[\w-]+:(?=.)/m,alias:"keyword"}};var httpLanguages={"application/json":Prism.languages.javascript,"application/xml":Prism.languages.markup,"text/xml":Prism.languages.markup,"text/html":Prism.languages.markup};for(var contentType in httpLanguages)if(httpLanguages[contentType]){var options={};options[contentType]={pattern:new RegExp("(content-type:\\s*"+contentType+"[\\w\\W]*?)(?:\\r?\\n|\\r){2}[\\w\\W]*","i"),lookbehind:!0,inside:{rest:httpLanguages[contentType]}},Prism.languages.insertBefore("http","header-name",options)};
+Prism.languages.makefile={comment:{pattern:/(^|[^\\])#(?:\\(?:\r\n|[\s\S])|.)*/,lookbehind:!0},string:/(["'])(?:\\(?:\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1/,builtin:/\.[A-Z][^:#=\s]+(?=\s*:(?!=))/,symbol:{pattern:/^[^:=\r\n]+(?=\s*:(?!=))/m,inside:{variable:/\$+(?:[^(){}:#=\s]+|(?=[({]))/}},variable:/\$+(?:[^(){}:#=\s]+|\([@*%<^+?][DF]\)|(?=[({]))/,keyword:[/-include\b|\b(?:define|else|endef|endif|export|ifn?def|ifn?eq|include|override|private|sinclude|undefine|unexport|vpath)\b/,{pattern:/(\()(?:addsuffix|abspath|and|basename|call|dir|error|eval|file|filter(?:-out)?|findstring|firstword|flavor|foreach|guile|if|info|join|lastword|load|notdir|or|origin|patsubst|realpath|shell|sort|strip|subst|suffix|value|warning|wildcard|word(?:s|list)?)(?=[ \t])/,lookbehind:!0}],operator:/(?:::|[?:+!])?=|[|@]/,punctuation:/[:;(){}]/};
+Prism.languages.markdown=Prism.languages.extend("markup",{}),Prism.languages.insertBefore("markdown","prolog",{blockquote:{pattern:/^>(?:[\t ]*>)*/m,alias:"punctuation"},code:[{pattern:/^(?: {4}|\t).+/m,alias:"keyword"},{pattern:/``.+?``|`[^`\n]+`/,alias:"keyword"}],title:[{pattern:/\w+.*(?:\r?\n|\r)(?:==+|--+)/,alias:"important",inside:{punctuation:/==+$|--+$/}},{pattern:/(^\s*)#+.+/m,lookbehind:!0,alias:"important",inside:{punctuation:/^#+|#+$/}}],hr:{pattern:/(^\s*)([*-])([\t ]*\2){2,}(?=\s*$)/m,lookbehind:!0,alias:"punctuation"},list:{pattern:/(^\s*)(?:[*+-]|\d+\.)(?=[\t ].)/m,lookbehind:!0,alias:"punctuation"},"url-reference":{pattern:/!?\[[^\]]+\]:[\t ]+(?:\S+|<(?:\\.|[^>\\])+>)(?:[\t ]+(?:"(?:\\.|[^"\\])*"|'(?:\\.|[^'\\])*'|\((?:\\.|[^)\\])*\)))?/,inside:{variable:{pattern:/^(!?\[)[^\]]+/,lookbehind:!0},string:/(?:"(?:\\.|[^"\\])*"|'(?:\\.|[^'\\])*'|\((?:\\.|[^)\\])*\))$/,punctuation:/^[\[\]!:]|[<>]/},alias:"url"},bold:{pattern:/(^|[^\\])(\*\*|__)(?:(?:\r?\n|\r)(?!\r?\n|\r)|.)+?\2/,lookbehind:!0,inside:{punctuation:/^\*\*|^__|\*\*$|__$/}},italic:{pattern:/(^|[^\\])([*_])(?:(?:\r?\n|\r)(?!\r?\n|\r)|.)+?\2/,lookbehind:!0,inside:{punctuation:/^[*_]|[*_]$/}},url:{pattern:/!?\[[^\]]+\](?:\([^\s)]+(?:[\t ]+"(?:\\.|[^"\\])*")?\)| ?\[[^\]\n]*\])/,inside:{variable:{pattern:/(!?\[)[^\]]+(?=\]$)/,lookbehind:!0},string:{pattern:/"(?:\\.|[^"\\])*"(?=\)$)/}}}}),Prism.languages.markdown.bold.inside.url=Prism.util.clone(Prism.languages.markdown.url),Prism.languages.markdown.italic.inside.url=Prism.util.clone(Prism.languages.markdown.url),Prism.languages.markdown.bold.inside.italic=Prism.util.clone(Prism.languages.markdown.italic),Prism.languages.markdown.italic.inside.bold=Prism.util.clone(Prism.languages.markdown.bold);
+Prism.languages.nginx=Prism.languages.extend("clike",{comment:{pattern:/(^|[^"{\\])#.*/,lookbehind:!0},keyword:/\b(?:CONTENT_|DOCUMENT_|GATEWAY_|HTTP_|HTTPS|if_not_empty|PATH_|QUERY_|REDIRECT_|REMOTE_|REQUEST_|SCGI|SCRIPT_|SERVER_|http|server|events|location|include|accept_mutex|accept_mutex_delay|access_log|add_after_body|add_before_body|add_header|addition_types|aio|alias|allow|ancient_browser|ancient_browser_value|auth|auth_basic|auth_basic_user_file|auth_http|auth_http_header|auth_http_timeout|autoindex|autoindex_exact_size|autoindex_localtime|break|charset|charset_map|charset_types|chunked_transfer_encoding|client_body_buffer_size|client_body_in_file_only|client_body_in_single_buffer|client_body_temp_path|client_body_timeout|client_header_buffer_size|client_header_timeout|client_max_body_size|connection_pool_size|create_full_put_path|daemon|dav_access|dav_methods|debug_connection|debug_points|default_type|deny|devpoll_changes|devpoll_events|directio|directio_alignment|disable_symlinks|empty_gif|env|epoll_events|error_log|error_page|expires|fastcgi_buffer_size|fastcgi_buffers|fastcgi_busy_buffers_size|fastcgi_cache|fastcgi_cache_bypass|fastcgi_cache_key|fastcgi_cache_lock|fastcgi_cache_lock_timeout|fastcgi_cache_methods|fastcgi_cache_min_uses|fastcgi_cache_path|fastcgi_cache_purge|fastcgi_cache_use_stale|fastcgi_cache_valid|fastcgi_connect_timeout|fastcgi_hide_header|fastcgi_ignore_client_abort|fastcgi_ignore_headers|fastcgi_index|fastcgi_intercept_errors|fastcgi_keep_conn|fastcgi_max_temp_file_size|fastcgi_next_upstream|fastcgi_no_cache|fastcgi_param|fastcgi_pass|fastcgi_pass_header|fastcgi_read_timeout|fastcgi_redirect_errors|fastcgi_send_timeout|fastcgi_split_path_info|fastcgi_store|fastcgi_store_access|fastcgi_temp_file_write_size|fastcgi_temp_path|flv|geo|geoip_city|geoip_country|google_perftools_profiles|gzip|gzip_buffers|gzip_comp_level|gzip_disable|gzip_http_version|gzip_min_length|gzip_proxied|gzip_static|gzip_types|gzip_vary|if|if_modified_since|ignore_invalid_headers|image_filter|image_filter_buffer|image_filter_jpeg_quality|image_filter_sharpen|image_filter_transparency|imap_capabilities|imap_client_buffer|include|index|internal|ip_hash|keepalive|keepalive_disable|keepalive_requests|keepalive_timeout|kqueue_changes|kqueue_events|large_client_header_buffers|limit_conn|limit_conn_log_level|limit_conn_zone|limit_except|limit_rate|limit_rate_after|limit_req|limit_req_log_level|limit_req_zone|limit_zone|lingering_close|lingering_time|lingering_timeout|listen|location|lock_file|log_format|log_format_combined|log_not_found|log_subrequest|map|map_hash_bucket_size|map_hash_max_size|master_process|max_ranges|memcached_buffer_size|memcached_connect_timeout|memcached_next_upstream|memcached_pass|memcached_read_timeout|memcached_send_timeout|merge_slashes|min_delete_depth|modern_browser|modern_browser_value|mp4|mp4_buffer_size|mp4_max_buffer_size|msie_padding|msie_refresh|multi_accept|open_file_cache|open_file_cache_errors|open_file_cache_min_uses|open_file_cache_valid|open_log_file_cache|optimize_server_names|override_charset|pcre_jit|perl|perl_modules|perl_require|perl_set|pid|pop3_auth|pop3_capabilities|port_in_redirect|post_action|postpone_output|protocol|proxy|proxy_buffer|proxy_buffer_size|proxy_buffering|proxy_buffers|proxy_busy_buffers_size|proxy_cache|proxy_cache_bypass|proxy_cache_key|proxy_cache_lock|proxy_cache_lock_timeout|proxy_cache_methods|proxy_cache_min_uses|proxy_cache_path|proxy_cache_use_stale|proxy_cache_valid|proxy_connect_timeout|proxy_cookie_domain|proxy_cookie_path|proxy_headers_hash_bucket_size|proxy_headers_hash_max_size|proxy_hide_header|proxy_http_version|proxy_ignore_client_abort|proxy_ignore_headers|proxy_intercept_errors|proxy_max_temp_file_size|proxy_method|proxy_next_upstream|proxy_no_cache|proxy_pass|proxy_pass_error_message|proxy_pass_header|proxy_pass_request_body|proxy_pass_request_headers|proxy_read_timeout|proxy_redirect|proxy_redirect_errors|proxy_send_lowat|proxy_send_timeout|proxy_set_body|proxy_set_header|proxy_ssl_session_reuse|proxy_store|proxy_store_access|proxy_temp_file_write_size|proxy_temp_path|proxy_timeout|proxy_upstream_fail_timeout|proxy_upstream_max_fails|random_index|read_ahead|real_ip_header|recursive_error_pages|request_pool_size|reset_timedout_connection|resolver|resolver_timeout|return|rewrite|root|rtsig_overflow_events|rtsig_overflow_test|rtsig_overflow_threshold|rtsig_signo|satisfy|satisfy_any|secure_link_secret|send_lowat|send_timeout|sendfile|sendfile_max_chunk|server|server_name|server_name_in_redirect|server_names_hash_bucket_size|server_names_hash_max_size|server_tokens|set|set_real_ip_from|smtp_auth|smtp_capabilities|so_keepalive|source_charset|split_clients|ssi|ssi_silent_errors|ssi_types|ssi_value_length|ssl|ssl_certificate|ssl_certificate_key|ssl_ciphers|ssl_client_certificate|ssl_crl|ssl_dhparam|ssl_engine|ssl_prefer_server_ciphers|ssl_protocols|ssl_session_cache|ssl_session_timeout|ssl_verify_client|ssl_verify_depth|starttls|stub_status|sub_filter|sub_filter_once|sub_filter_types|tcp_nodelay|tcp_nopush|timeout|timer_resolution|try_files|types|types_hash_bucket_size|types_hash_max_size|underscores_in_headers|uninitialized_variable_warn|upstream|use|user|userid|userid_domain|userid_expires|userid_name|userid_p3p|userid_path|userid_service|valid_referers|variables_hash_bucket_size|variables_hash_max_size|worker_connections|worker_cpu_affinity|worker_priority|worker_processes|worker_rlimit_core|worker_rlimit_nofile|worker_rlimit_sigpending|working_directory|xclient|xml_entities|xslt_entities|xslt_stylesheet|xslt_types)\b/i}),Prism.languages.insertBefore("nginx","keyword",{variable:/\$[a-z_]+/i});
+Prism.languages.php=Prism.languages.extend("clike",{keyword:/\b(and|or|xor|array|as|break|case|cfunction|class|const|continue|declare|default|die|do|else|elseif|enddeclare|endfor|endforeach|endif|endswitch|endwhile|extends|for|foreach|function|include|include_once|global|if|new|return|static|switch|use|require|require_once|var|while|abstract|interface|public|implements|private|protected|parent|throw|null|echo|print|trait|namespace|final|yield|goto|instanceof|finally|try|catch)\b/i,constant:/\b[A-Z0-9_]{2,}\b/,comment:{pattern:/(^|[^\\])(?:\/\*[\w\W]*?\*\/|\/\/.*)/,lookbehind:!0}}),Prism.languages.insertBefore("php","class-name",{"shell-comment":{pattern:/(^|[^\\])#.*/,lookbehind:!0,alias:"comment"}}),Prism.languages.insertBefore("php","keyword",{delimiter:/\?>|<\?(?:php)?/i,variable:/\$\w+\b/i,"package":{pattern:/(\\|namespace\s+|use\s+)[\w\\]+/,lookbehind:!0,inside:{punctuation:/\\/}}}),Prism.languages.insertBefore("php","operator",{property:{pattern:/(->)[\w]+/,lookbehind:!0}}),Prism.languages.markup&&(Prism.hooks.add("before-highlight",function(e){"php"===e.language&&(e.tokenStack=[],e.backupCode=e.code,e.code=e.code.replace(/(?:<\?php|<\?)[\w\W]*?(?:\?>)/gi,function(a){return e.tokenStack.push(a),"{{{PHP"+e.tokenStack.length+"}}}"}))}),Prism.hooks.add("before-insert",function(e){"php"===e.language&&(e.code=e.backupCode,delete e.backupCode)}),Prism.hooks.add("after-highlight",function(e){if("php"===e.language){for(var a,n=0;a=e.tokenStack[n];n++)e.highlightedCode=e.highlightedCode.replace("{{{PHP"+(n+1)+"}}}",Prism.highlight(a,e.grammar,"php").replace(/\$/g,"$$$$"));e.element.innerHTML=e.highlightedCode}}),Prism.hooks.add("wrap",function(e){"php"===e.language&&"markup"===e.type&&(e.content=e.content.replace(/(\{\{\{PHP[0-9]+\}\}\})/g,'<span class="token php">$1</span>'))}),Prism.languages.insertBefore("php","comment",{markup:{pattern:/<[^?]\/?(.*?)>/,inside:Prism.languages.markup},php:/\{\{\{PHP[0-9]+\}\}\}/}));
+!function(e){e.languages.sass=e.languages.extend("css",{comment:{pattern:/^([ \t]*)\/[\/*].*(?:(?:\r?\n|\r)\1[ \t]+.+)*/m,lookbehind:!0}}),e.languages.insertBefore("sass","atrule",{"atrule-line":{pattern:/^(?:[ \t]*)[@+=].+/m,inside:{atrule:/(?:@[\w-]+|[+=])/m}}}),delete e.languages.sass.atrule;var a=/((\$[-_\w]+)|(#\{\$[-_\w]+\}))/i,t=[/[+*\/%]|[=!]=|<=?|>=?|\b(?:and|or|not)\b/,{pattern:/(\s+)-(?=\s)/,lookbehind:!0}];e.languages.insertBefore("sass","property",{"variable-line":{pattern:/^[ \t]*\$.+/m,inside:{punctuation:/:/,variable:a,operator:t}},"property-line":{pattern:/^[ \t]*(?:[^:\s]+ *:.*|:[^:\s]+.*)/m,inside:{property:[/[^:\s]+(?=\s*:)/,{pattern:/(:)[^:\s]+/,lookbehind:!0}],punctuation:/:/,variable:a,operator:t,important:e.languages.sass.important}}}),delete e.languages.sass.property,delete e.languages.sass.important,delete e.languages.sass.selector,e.languages.insertBefore("sass","punctuation",{selector:{pattern:/([ \t]*)\S(?:,?[^,\r\n]+)*(?:,(?:\r?\n|\r)\1[ \t]+\S(?:,?[^,\r\n]+)*)*/,lookbehind:!0}})}(Prism);
+Prism.languages.scss=Prism.languages.extend("css",{comment:{pattern:/(^|[^\\])(?:\/\*[\w\W]*?\*\/|\/\/.*)/,lookbehind:!0},atrule:{pattern:/@[\w-]+(?:\([^()]+\)|[^(])*?(?=\s+[{;])/,inside:{rule:/@[\w-]+/}},url:/(?:[-a-z]+-)*url(?=\()/i,selector:{pattern:/(?=\S)[^@;\{\}\(\)]?([^@;\{\}\(\)]|&|#\{\$[-_\w]+\})+(?=\s*\{(\}|\s|[^\}]+(:|\{)[^\}]+))/m,inside:{placeholder:/%[-_\w]+/}}}),Prism.languages.insertBefore("scss","atrule",{keyword:[/@(?:if|else(?: if)?|for|each|while|import|extend|debug|warn|mixin|include|function|return|content)/i,{pattern:/( +)(?:from|through)(?= )/,lookbehind:!0}]}),Prism.languages.insertBefore("scss","property",{variable:/\$[-_\w]+|#\{\$[-_\w]+\}/}),Prism.languages.insertBefore("scss","function",{placeholder:{pattern:/%[-_\w]+/,alias:"selector"},statement:/\B!(?:default|optional)\b/i,"boolean":/\b(?:true|false)\b/,"null":/\bnull\b/,operator:{pattern:/(\s)(?:[-+*\/%]|[=!]=|<=?|>=?|and|or|not)(?=\s)/,lookbehind:!0}}),Prism.languages.scss.atrule.inside.rest=Prism.util.clone(Prism.languages.scss);
+!function(){function e(e,t){return Array.prototype.slice.call((t||document).querySelectorAll(e))}function t(e,t){return t=" "+t+" ",(" "+e.className+" ").replace(/[\n\t]/g," ").indexOf(t)>-1}function n(e,n,i){for(var o,a=n.replace(/\s+/g,"").split(","),l=+e.getAttribute("data-line-offset")||0,d=r()?parseInt:parseFloat,c=d(getComputedStyle(e).lineHeight),s=0;o=a[s++];){o=o.split("-");var u=+o[0],m=+o[1]||u,h=document.createElement("div");h.textContent=Array(m-u+2).join(" \n"),h.className=(i||"")+" line-highlight",t(e,"line-numbers")||(h.setAttribute("data-start",u),m>u&&h.setAttribute("data-end",m)),h.style.top=(u-l-1)*c+"px",t(e,"line-numbers")?e.appendChild(h):(e.querySelector("code")||e).appendChild(h)}}function i(){var t=location.hash.slice(1);e(".temporary.line-highlight").forEach(function(e){e.parentNode.removeChild(e)});var i=(t.match(/\.([\d,-]+)$/)||[,""])[1];if(i&&!document.getElementById(t)){var r=t.slice(0,t.lastIndexOf(".")),o=document.getElementById(r);o&&(o.hasAttribute("data-line")||o.setAttribute("data-line",""),n(o,i,"temporary "),document.querySelector(".temporary.line-highlight").scrollIntoView())}}if("undefined"!=typeof self&&self.Prism&&self.document&&document.querySelector){var r=function(){var e;return function(){if("undefined"==typeof e){var t=document.createElement("div");t.style.fontSize="13px",t.style.lineHeight="1.5",t.style.padding=0,t.style.border=0,t.innerHTML="&nbsp;<br />&nbsp;",document.body.appendChild(t),e=38===t.offsetHeight,document.body.removeChild(t)}return e}}(),o=0;Prism.hooks.add("complete",function(t){var r=t.element.parentNode,a=r&&r.getAttribute("data-line");r&&a&&/pre/i.test(r.nodeName)&&(clearTimeout(o),e(".line-highlight",r).forEach(function(e){e.parentNode.removeChild(e)}),n(r,a),o=setTimeout(i,1))}),window.addEventListener&&window.addEventListener("hashchange",i)}}();
+
+
+!function(){"undefined"!=typeof self&&self.Prism&&self.document&&Prism.hooks.add("complete",function(e){if(e.code){var t=e.element.parentNode,s=/\s*\bline-numbers\b\s*/;if(t&&/pre/i.test(t.nodeName)&&(s.test(t.className)||s.test(e.element.className))&&!e.element.querySelector(".line-numbers-rows")){s.test(e.element.className)&&(e.element.className=e.element.className.replace(s,"")),s.test(t.className)||(t.className+=" line-numbers");var n,a=e.code.match(/\n(?!$)/g),l=a?a.length+1:1,m=new Array(l+1);m=m.join("<span></span>"),n=document.createElement("span"),n.className="line-numbers-rows",n.innerHTML=m,t.hasAttribute("data-start")&&(t.style.counterReset="linenumber "+(parseInt(t.getAttribute("data-start"),10)-1)),e.element.appendChild(n)}}})}();
+!function(){if(("undefined"==typeof self||self.Prism)&&("undefined"==typeof global||global.Prism)){var i=/\b([a-z]{3,7}:\/\/|tel:)[\w\-+%~\/.:#=?&amp;]+/,n=/\b\S+@[\w.]+[a-z]{2}/,e=/\[([^\]]+)]\(([^)]+)\)/,t=["comment","url","attr-value","string"];Prism.hooks.add("before-highlight",function(a){a.grammar&&!a.grammar["url-link"]&&(Prism.languages.DFS(a.grammar,function(a,r,l){t.indexOf(l)>-1&&"Array"!==Prism.util.type(r)&&(r.pattern||(r=this[a]={pattern:r}),r.inside=r.inside||{},"comment"==l&&(r.inside["md-link"]=e),"attr-value"==l?Prism.languages.insertBefore("inside","punctuation",{"url-link":i},r):r.inside["url-link"]=i,r.inside["email-link"]=n)}),a.grammar["url-link"]=i,a.grammar["email-link"]=n)}),Prism.hooks.add("wrap",function(i){if(/-link$/.test(i.type)){i.tag="a";var n=i.content;if("email-link"==i.type&&0!=n.indexOf("mailto:"))n="mailto:"+n;else if("md-link"==i.type){var t=i.content.match(e);n=t[2],i.content=t[1]}i.attributes.href=n}})}}();
+!function(){"undefined"!=typeof self&&self.Prism&&self.document&&document.querySelector&&(self.Prism.fileHighlight=function(){var e={js:"javascript",html:"markup",svg:"markup",xml:"markup",py:"python",rb:"ruby",ps1:"powershell",psm1:"powershell"};Array.prototype.forEach&&Array.prototype.slice.call(document.querySelectorAll("pre[data-src]")).forEach(function(t){for(var r,a=t.getAttribute("data-src"),s=t,n=/\blang(?:uage)?-(?!\*)(\w+)\b/i;s&&!n.test(s.className);)s=s.parentNode;if(s&&(r=(t.className.match(n)||[,""])[1]),!r){var o=(a.match(/\.(\w+)$/)||[,""])[1];r=e[o]||o}var l=document.createElement("code");l.className="language-"+r,t.textContent="",l.textContent="Loading…",t.appendChild(l);var i=new XMLHttpRequest;i.open("GET",a,!0),i.onreadystatechange=function(){4==i.readyState&&(i.status<400&&i.responseText?(l.textContent=i.responseText,Prism.highlightElement(l)):l.textContent=i.status>=400?"✖ Error "+i.status+" while fetching file: "+i.statusText:"✖ Error: File does not exist or is empty")},i.send(null)})},self.Prism.fileHighlight())}();
+!function(){if("undefined"!=typeof self&&self.Prism&&self.document){var e={css:"CSS",clike:"C-like",javascript:"JavaScript",abap:"ABAP",actionscript:"ActionScript",apacheconf:"Apache Configuration",apl:"APL",applescript:"AppleScript",asciidoc:"AsciiDoc",aspnet:"ASP.NET (C#)",autoit:"AutoIt",autohotkey:"AutoHotkey",basic:"BASIC",csharp:"C#",cpp:"C++",coffeescript:"CoffeeScript","css-extras":"CSS Extras",fsharp:"F#",glsl:"GLSL",http:"HTTP",inform7:"Inform 7",latex:"LaTeX",lolcode:"LOLCODE",matlab:"MATLAB",mel:"MEL",nasm:"NASM",nginx:"nginx",nsis:"NSIS",objectivec:"Objective-C",ocaml:"OCaml",parigp:"PARI/GP",php:"PHP","php-extras":"PHP Extras",powershell:"PowerShell",jsx:"React JSX",rest:"reST (reStructuredText)",sas:"SAS",sass:"Sass (Sass)",scss:"Sass (Scss)",sql:"SQL",typescript:"TypeScript",vhdl:"VHDL",vim:"vim",wiki:"Wiki markup",yaml:"YAML"};Prism.hooks.add("before-highlight",function(a){var s=a.element.parentNode;if(s&&/pre/i.test(s.nodeName)){var t=e[a.language]||a.language.substring(0,1).toUpperCase()+a.language.substring(1);s.setAttribute("data-language",t);var i,r,l=s.previousSibling;l&&/\s*\bprism-show-language\b\s*/.test(l.className)&&l.firstChild&&/\s*\bprism-show-language-label\b\s*/.test(l.firstChild.className)?(r=l.firstChild,r.getAttribute("data-language")!==t&&(r.setAttribute("data-language",t),r.innerHTML=t)):(i=document.createElement("div"),r=document.createElement("div"),r.className="prism-show-language-label",r.setAttribute("data-language",t),r.innerHTML=t,i.className="prism-show-language",i.appendChild(r),s.parentNode.insertBefore(i,s))}})}}();
diff --git a/website/static/wechat-qr-code.png b/website/static/wechat-qr-code.png
new file mode 100644
index 000000000..c604e7331
Binary files /dev/null and b/website/static/wechat-qr-code.png differ
diff --git a/website/themes/book/.github/workflows/main.yml b/website/themes/book/.github/workflows/main.yml
new file mode 100644
index 000000000..19f30e9cf
--- /dev/null
+++ b/website/themes/book/.github/workflows/main.yml
@@ -0,0 +1,33 @@
+name: Build with Hugo
+
+on: [push]
+
+jobs:
+  hugo-latest:
+    runs-on: ubuntu-latest
+    steps:
+      - uses: actions/checkout@master
+
+      - name: Install Hugo
+        run: |
+          LATEST_VERSION=`curl --silent "https://api.github.com/repos/gohugoio/hugo/releases/latest" | grep '"tag_name":' | sed -E 's/.*"([^"]+)".*/\1/'`
+          VERSION_NO_PREFIX=`echo $LATEST_VERSION | cut -c 2-`
+          wget "https://github.com/gohugoio/hugo/releases/download/$LATEST_VERSION/hugo_extended_${VERSION_NO_PREFIX}_Linux-64bit.deb" -O /tmp/hugo.deb
+          sudo dpkg -i /tmp/hugo.deb
+
+      - name: Run Hugo
+        working-directory: exampleSite
+        run: hugo --themesDir ../..
+  hugo-minimum:
+    runs-on: ubuntu-latest
+    steps:
+      - uses: actions/checkout@master
+
+      - name: Install Hugo
+        run: |
+          wget "https://github.com/gohugoio/hugo/releases/download/v0.68.0/hugo_extended_0.68.0_Linux-64bit.deb" -O /tmp/hugo.deb
+          sudo dpkg -i /tmp/hugo.deb
+
+      - name: Run Hugo
+        working-directory: exampleSite
+        run: hugo --themesDir ../..
diff --git a/website/themes/book/.gitignore b/website/themes/book/.gitignore
new file mode 100644
index 000000000..e52eb5292
--- /dev/null
+++ b/website/themes/book/.gitignore
@@ -0,0 +1,3 @@
+public/
+exampleSite/public/
+.DS_Store
diff --git a/website/themes/book/LICENSE b/website/themes/book/LICENSE
new file mode 100644
index 000000000..e7a669ab2
--- /dev/null
+++ b/website/themes/book/LICENSE
@@ -0,0 +1,20 @@
+The MIT License (MIT)
+
+Copyright (c) 2018 Alex Shpak
+
+Permission is hereby granted, free of charge, to any person obtaining a copy of
+this software and associated documentation files (the "Software"), to deal in
+the Software without restriction, including without limitation the rights to
+use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of
+the Software, and to permit persons to whom the Software is furnished to do so,
+subject to the following conditions:
+
+The above copyright notice and this permission notice shall be included in all
+copies or substantial portions of the Software.
+
+THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
+IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS
+FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR
+COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER
+IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
+CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
diff --git a/website/themes/book/README.md b/website/themes/book/README.md
new file mode 100644
index 000000000..052730c5d
--- /dev/null
+++ b/website/themes/book/README.md
@@ -0,0 +1,321 @@
+# Hugo Book Theme
+
+[![Hugo](https://img.shields.io/badge/hugo-0.68-blue.svg)](https://gohugo.io)
+[![License: MIT](https://img.shields.io/badge/License-MIT-blue.svg)](LICENSE)
+![Build with Hugo](https://github.com/alex-shpak/hugo-book/workflows/Build%20with%20Hugo/badge.svg)
+
+### [Hugo](https://gohugo.io) documentation theme as simple as plain book
+
+![Screenshot](https://github.com/alex-shpak/hugo-book/blob/master/images/screenshot.png)
+
+- [Features](#features)
+- [Requirements](#requirements)
+- [Installation](#installation)
+- [Menu](#menu)
+- [Blog](#blog)
+- [Configuration](#configuration)
+- [Shortcodes](#shortcodes)
+- [Versioning](#versioning)
+- [Contributing](#contributing)
+
+## Features
+
+- Clean simple design
+- Light and Mobile-Friendly
+- Multi-language support
+- Customisable
+- Zero initial configuration
+- Handy shortcodes
+- Comments support
+- Simple blog and taxonomy
+- Primary features work without JavaScript
+- Dark Mode
+
+## Requirements
+
+- Hugo 0.68 or higher
+- Hugo extended version, read more [here](https://gohugo.io/news/0.48-relnotes/)
+
+## Installation
+
+Navigate to your hugo project root and run:
+
+```
+git submodule add https://github.com/alex-shpak/hugo-book themes/book
+```
+
+Then run hugo (or set `theme = "book"`/`theme: book` in configuration file)
+
+```
+hugo server --minify --theme book
+```
+
+### Creating site from scratch
+
+Below is an example on how to create a new site from scratch:
+
+```sh
+hugo new site mydocs; cd mydocs
+git init
+git submodule add https://github.com/alex-shpak/hugo-book themes/book
+cp -R themes/book/exampleSite/content .
+```
+
+```sh
+hugo server --minify --theme book
+```
+
+## Menu
+
+### File tree menu (default)
+
+By default, the theme will render pages from the `content/docs` section as a menu in a tree structure.  
+You can set `title` and `weight` in the front matter of pages to adjust the order and titles in the menu.
+
+### Leaf bundle menu
+
+You can also use leaf bundle and the content of its `index.md` file as menu.  
+Given you have the following file structure:
+
+```
+├── content
+│   ├── docs
+│   │   ├── page-one.md
+│   │   └── page-two.md
+│   └── posts
+│       ├── post-one.md
+│       └── post-two.md
+```
+
+Create a file `content/menu/index.md` with the content:
+
+```md
++++
+headless = true
++++
+
+- [Book Example]({{< relref "/docs/" >}})
+  - [Page One]({{< relref "/docs/page-one" >}})
+  - [Page Two]({{< relref "/docs/page-two" >}})
+- [Blog]({{< relref "/posts" >}})
+```
+
+And Enable it by setting `BookMenuBundle: /menu` in Site configuration.
+
+- [Example menu](https://github.com/alex-shpak/hugo-book/blob/master/exampleSite/content/menu/index.md)
+- [Example config file](https://github.com/alex-shpak/hugo-book/blob/master/exampleSite/config.yaml)
+- [Leaf bundles](https://gohugo.io/content-management/page-bundles/)
+
+## Blog
+
+A simple blog is supported in the section `posts`.  
+A blog is not the primary usecase of this theme, so it has only minimal features.
+
+## Configuration
+
+### Site Configuration
+
+There are a few configuration options that you can add to your `config.toml` file.  
+You can also see the `yaml` example [here](https://github.com/alex-shpak/hugo-book/blob/master/exampleSite/config.yaml).
+
+```toml
+# (Optional) Set Google Analytics if you use it to track your website.
+# Always put it on the top of the configuration file, otherwise it won't work
+googleAnalytics = "UA-XXXXXXXXX-X"
+
+# (Optional) If you provide a Disqus shortname, comments will be enabled on
+# all pages.
+disqusShortname = "my-site"
+
+# (Optional) Set this to true if you use capital letters in file names
+disablePathToLower = true
+
+# (Optional) Set this to true to enable 'Last Modified by' date and git author
+#  information on 'doc' type pages.
+enableGitInfo = true
+
+# (Optional) Theme is intended for documentation use, therefore it doesn't render taxonomy.
+# You can remove related files with config below
+disableKinds = ['taxonomy', 'taxonomyTerm']
+
+[params]
+  # (Optional, default light) Sets color theme: light, dark or auto.
+  # Theme 'auto' switches between dark and light modes based on browser/os preferences
+  BookTheme = 'light'
+
+  # (Optional, default true) Controls table of contents visibility on right side of pages.
+  # Start and end levels can be controlled with markup.tableOfContents setting.
+  # You can also specify this parameter per page in front matter.
+  BookToC = true
+
+  # (Optional, default none) Set the path to a logo for the book. If the logo is
+  # /static/logo.png then the path would be 'logo.png'
+  BookLogo = 'logo.png'
+
+  # (Optional, default none) Set leaf bundle to render as side menu
+  # When not specified file structure and weights will be used
+  BookMenuBundle = '/menu'
+
+  # (Optional, default docs) Specify section of content to render as menu
+  # You can also set value to "*" to render all sections to menu
+  BookSection = 'docs'
+
+  # Set source repository location.
+  # Used for 'Last Modified' and 'Edit this page' links.
+  BookRepo = 'https://github.com/alex-shpak/hugo-book'
+  
+  # Specifies commit portion of the link to the page's last modified commit hash for 'doc' page
+  # type.
+  # Required if 'BookRepo' param is set.
+  # Value used to construct a URL consisting of BookRepo/BookCommitPath/<commit-hash>
+  # Github uses 'commit', Bitbucket uses 'commits'
+  BookCommitPath = 'commit'
+
+  # Enable 'Edit this page' links for 'doc' page type.
+  # Disabled by default. Uncomment to enable. Requires 'BookRepo' param.
+  # Path must point to the site directory.
+  BookEditPath = 'edit/master/exampleSite'
+
+  # (Optional, default January 2, 2006) Configure the date format used on the pages
+  # - In git information
+  # - In blog posts
+  BookDateFormat = 'Jan 2, 2006'
+
+  # (Optional, default true) Enables search function with flexsearch,
+  # Index is built on fly, therefore it might slowdown your website.
+  # Configuration for indexing can be adjusted in i18n folder per language.
+  BookSearch = true
+
+  # (Optional, default true) Enables comments template on pages
+  # By default partials/docs/comments.html includes Disqus template
+  # See https://gohugo.io/content-management/comments/#configure-disqus
+  # Can be overwritten by same param in page frontmatter
+  BookComments = true
+
+  # /!\ This is an experimental feature, might be removed or changed at any time
+  # (Optional, experimental, default false) Enables portable links and link checks in markdown pages.
+  # Portable links meant to work with text editors and let you write markdown without {{< relref >}} shortcode
+  # Theme will print warning if page referenced in markdown does not exists.
+  BookPortableLinks = true
+
+  # /!\ This is an experimental feature, might be removed or changed at any time
+  # (Optional, experimental, default false) Enables service worker that caches visited pages and resources for offline use.
+  BookServiceWorker = true
+```
+
+### Multi-Language Support
+
+Theme supports Hugo's [multilingual mode](https://gohugo.io/content-management/multilingual/), just follow configuration guide there. You can also tweak search indexing configuration per language in `i18n` folder.
+
+### Page Configuration
+
+You can specify additional params in the front matter of individual pages:
+
+```toml
+# Set type to 'docs' if you want to render page outside of configured section or if you render section other than 'docs'
+type = 'docs'
+
+# Set page weight to re-arrange items in file-tree menu (if BookMenuBundle not set)
+weight = 10
+
+# (Optional) Set to 'true' to mark page as flat section in file-tree menu (if BookMenuBundle not set)
+bookFlatSection = false
+
+# (Optional) Set to hide nested sections or pages at that level. Works only with file-tree menu mode
+bookCollapseSection = true
+
+# (Optional) Set true to hide page or section from side menu (if BookMenuBundle not set)
+bookHidden = false
+
+# (Optional) Set 'false' to hide ToC from page
+bookToC = true
+
+# (Optional) If you have enabled BookComments for the site, you can disable it for specific pages.
+bookComments = true
+
+# (Optional) Set to 'false' to exclude page from search index.
+bookSearchExclude = true
+```
+
+### Partials
+
+There are few empty partials you can override in `layouts/partials/`
+
+| Partial                                            | Placement                                   |
+| -------------------------------------------------- | ------------------------------------------- |
+| `layouts/partials/docs/inject/head.html`           | Before closing `<head>` tag                 |
+| `layouts/partials/docs/inject/body.html`           | Before closing `<body>` tag                 |
+| `layouts/partials/docs/inject/footer.html`         | After page footer content                   |
+| `layouts/partials/docs/inject/menu-before.html`    | At the beginning of `<nav>` menu block      |
+| `layouts/partials/docs/inject/menu-after.html`     | At the end of `<nav>` menu block            |
+| `layouts/partials/docs/inject/content-before.html` | Before page content                         |
+| `layouts/partials/docs/inject/content-after.html`  | After page content                          |
+| `layouts/partials/docs/inject/toc-before.html`     | At the beginning of table of contents block |
+| `layouts/partials/docs/inject/toc-after.html`      | At the end of table of contents block       |
+
+### Extra Customisation
+
+| File                     | Description                                                                           |
+| ------------------------ | ------------------------------------------------------------------------------------- |
+| `static/favicon.png`     | Override default favicon                                                              |
+| `assets/_custom.scss`    | Customise or override scss styles                                                     |
+| `assets/_variables.scss` | Override default SCSS variables                                                       |
+| `assets/_fonts.scss`     | Replace default font with custom fonts (e.g. local files or remote like google fonts) |
+| `assets/mermaid.json`    | Replace Mermaid initialization config                                                 |
+
+### Plugins
+
+There are a few features implemented as plugable `scss` styles. Usually these are features that don't make it to the core but can still be useful.
+
+| Plugin                            | Description                                                 |
+| --------------------------------- | ----------------------------------------------------------- |
+| `assets/plugins/_numbered.scss`   | Makes headings in markdown numbered, e.g. `1.1`, `1.2`      |
+| `assets/plugins/_scrollbars.scss` | Overrides scrollbar styles to look similar across platforms |
+
+To enable plugins, add `@import "plugins/{name}";` to `assets/_custom.scss` in your website root.
+
+### Hugo Internal Templates
+
+There are a few hugo templates inserted in `<head>`
+
+- [Google Analytics](https://gohugo.io/templates/internal/#google-analytics)
+- [Open Graph](https://gohugo.io/templates/internal/#open-graph)
+
+## Shortcodes
+
+- [Buttons](https://themes.gohugo.io/theme/hugo-book/docs/shortcodes/buttons/)
+- [Columns](https://themes.gohugo.io/theme/hugo-book/docs/shortcodes/columns/)
+- [Expand](https://themes.gohugo.io/theme/hugo-book/docs/shortcodes/expand/)
+- [Hints](https://themes.gohugo.io/theme/hugo-book/docs/shortcodes/hints/)
+- [KaTeX](https://themes.gohugo.io/theme/hugo-book/docs/shortcodes/katex/)
+- [Mermaid](https://themes.gohugo.io/theme/hugo-book/docs/shortcodes/mermaid/)
+- [Tabs](https://themes.gohugo.io/theme/hugo-book/docs/shortcodes/tabs/)
+
+By default, Goldmark trims unsafe outputs which might prevent some shortcodes from rendering. It is recommended to set `markup.goldmark.renderer.unsafe=true` if you encounter problems.
+
+```toml
+[markup.goldmark.renderer]
+  unsafe = true
+```
+
+If you are using `config.yaml` or `config.json`, consult the [configuration markup](https://gohugo.io/getting-started/configuration-markup/)
+
+## Versioning
+
+This theme follows a simple incremental versioning. e.g. `v1`, `v2` and so on. There might be breaking changes between versions.
+
+If you want lower maintenance, use one of the released versions. If you want to live on the bleeding edge of changes, you can use the `master` branch and update your website when needed.
+
+## Contributing
+
+### [Extra credits to contributors](https://github.com/alex-shpak/hugo-book/graphs/contributors)
+
+Contributions are welcome and I will review and consider pull requests.  
+Primary goals are:
+
+- Keep it simple.
+- Keep minimal (or zero) default configuration.
+- Avoid interference with user-defined layouts.
+- Avoid using JS if it can be solved by CSS.
+
+Feel free to open issues if you find missing configuration or customisation options.
diff --git a/website/themes/book/archetypes/docs.md b/website/themes/book/archetypes/docs.md
new file mode 100644
index 000000000..7e5e24b57
--- /dev/null
+++ b/website/themes/book/archetypes/docs.md
@@ -0,0 +1,9 @@
+---
+title: "{{ .Name | humanize | title }}"
+weight: 1
+# bookFlatSection: false
+# bookToc: true
+# bookHidden: false
+# bookCollapseSection: false
+# bookComments: true
+---
diff --git a/website/themes/book/archetypes/posts.md b/website/themes/book/archetypes/posts.md
new file mode 100644
index 000000000..fdccff8ae
--- /dev/null
+++ b/website/themes/book/archetypes/posts.md
@@ -0,0 +1,4 @@
+---
+title: "{{ replace .Name "-" " " | title }}"
+date: {{ .Date }}
+---
diff --git a/website/themes/book/assets/_custom.scss b/website/themes/book/assets/_custom.scss
new file mode 100644
index 000000000..eedd64fde
--- /dev/null
+++ b/website/themes/book/assets/_custom.scss
@@ -0,0 +1,158 @@
+/* You can add custom styles here. */
+
+// @import "plugins/numbered";
+/*
+ * Dark Mode Toggle
+ * Copyright (c) 2015 instructure-react
+ * Forked from Dan Abramov's personal blog
+ * https://github.com/gaearon/overreacted.io/blob/master/src/components/Toggle.css
+ * MIT License
+**/
+
+.js-toggle-wrapper {
+    display: table;
+    margin: 1rem 0 0 5px;
+    flex: 1;
+}
+
+.js-toggle {
+    touch-action: pan-x;
+    display: inline-block;
+    position: relative;
+    cursor: pointer;
+    background-color: transparent;
+    border: 0;
+    padding: 0;
+    -webkit-touch-callout: none;
+    user-select: none;
+    -webkit-tap-highlight-color: rgba(0, 0, 0, 0);
+    -webkit-tap-highlight-color: transparent;
+  }
+  
+  .js-toggle-screenreader-only {
+    border: 0;
+    clip: rect(0 0 0 0);
+    height: 1px;
+    margin: -1px;
+    overflow: hidden;
+    padding: 0;
+    position: absolute;
+    width: 1px;
+  }
+  
+  .js-toggle-track {
+    width: 60px;
+    height: 34px;
+    padding: 0;
+    // border-radius: 30px;
+    // background-color: hsl(222, 14%, 7%);
+    // background-color: var(--body-background);
+    transition: all 0.2s ease;
+  }
+  
+  .js-toggle-track-check {
+    position: absolute;
+    width: 17px;
+    height: 17px;
+    left: 5px;
+    top: 0px;
+    bottom: 0px;
+    margin-top: auto;
+    margin-bottom: auto;
+    line-height: 0;
+    opacity: 0;
+    transition: opacity 0.25s ease;
+  }
+  
+  .js-toggle--checked .js-toggle-track-check {
+    opacity: 1;
+    transition: opacity 0.25s ease;
+  }
+  
+  .js-toggle-track-x {
+    position: absolute;
+    width: 17px;
+    height: 17px;
+    right: 5px;
+    top: 0px;
+    bottom: 0px;
+    margin-top: auto;
+    margin-bottom: auto;
+    line-height: 0;
+    opacity: 1;
+    transition: opacity 0.25s ease;
+  }
+  
+  .js-toggle--checked .js-toggle-track-x {
+    opacity: 0;
+  }
+  
+  .js-toggle-thumb {
+    position: absolute;
+    top: 1px;
+    left: 1px;
+    width: 30px;
+    height: 32px;
+    // border-radius: 50%;
+    // background-color: var(--gray-100);
+    box-sizing: border-box;
+    transition: all 0.5s cubic-bezier(0.23, 1, 0.32, 1) 0ms;
+    transform: translateX(0);
+  }
+  
+  .js-toggle--checked .js-toggle-thumb {
+    transform: translateX(26px);
+    border-color: #19ab27;
+  }
+  
+  // .js-toggle--focus .js-toggle-thumb {
+  //   // box-shadow: 0px 0px 2px 3px rgb(255, 167, 196);
+  //   box-shadow: 0px 0px 2px 3px #64CEAA;
+  // }
+  
+  // .js-toggle:active .js-toggle-thumb {
+  //   box-shadow: 0px 0px 5px 5px #64CEAA;
+  //   // box-shadow: 0px 0px 5px 5px rgb(255, 167, 196);
+  // }
+
+  .magic {
+      display: flex;
+  }
+
+  body.dark-mode, 
+  body.dark-mode main * {
+      background: rgb(45, 45, 45);
+      color: #f5f5f5;
+  }
+
+$theme-light: (
+    gray-100: #f8f9fa,
+    gray-200: #e9ecef,
+    body-background: white,
+    body-font-color: black,
+    color-link: #05b,
+    color-visited-link: #8440f1,
+    icon-filter: none,
+);
+
+$theme-dark: (
+    gray-100: rgba(255, 255, 255, 0.1),
+    gray-200: rgba(255, 255, 255, 0.2),
+    body-background: #343a40,
+    body-font-color: #e9ecef,
+    color-link: #84b2ff,
+    color-visited-link: #b88dff,
+    icon-filter: brightness(0) invert(1),
+);
+
+[data-theme="dark"] {
+    @each $name, $value in $theme-dark {
+        --#{$name}: #{$value};
+    }
+}
+
+[data-theme="light"] {
+    @each $name, $value in $theme-light {
+        --#{$name}: #{$value};
+    }
+}
\ No newline at end of file
diff --git a/website/themes/book/assets/_defaults.scss b/website/themes/book/assets/_defaults.scss
new file mode 100644
index 000000000..4c8669736
--- /dev/null
+++ b/website/themes/book/assets/_defaults.scss
@@ -0,0 +1,66 @@
+// Used in layout
+$padding-1: 1px !default;
+$padding-4: 0.25rem !default;
+$padding-8: 0.5rem !default;
+$padding-16: 1rem !default;
+
+$font-size-base: 16px !default;
+$font-size-12: 0.75rem !default;
+$font-size-14: 0.875rem !default;
+$font-size-16: 1rem !default;
+
+$border-radius: $padding-4 !default;
+
+$body-font-weight: normal !default;
+
+$body-min-width: 20rem !default;
+$container-max-width: 80rem !default;
+
+$header-height: 3.5rem !default;
+$menu-width: 16rem !default;
+$toc-width: 16rem !default;
+
+$mobile-breakpoint: $menu-width + $body-min-width * 1.2 + $toc-width !default;
+
+$hint-colors: (
+  info: #6bf,
+  warning: #fd6,
+  danger: #f66,
+) !default;
+
+// Themes
+@mixin theme-light {
+  --gray-100: #f8f9fa;
+  --gray-200: #e9ecef;
+  --gray-500: #adb5bd;
+
+  --color-link: #0055bb;
+  --color-visited-link: #8440f1;
+
+  --body-background: white;
+  --body-font-color: black;
+
+  --icon-filter: none;
+
+  --hint-color-info: #6bf;
+  --hint-color-warning: #fd6;
+  --hint-color-danger: #f66;
+}
+
+@mixin theme-dark {
+  --gray-100: rgba(255, 255, 255, 0.1);
+  --gray-200: rgba(255, 255, 255, 0.2);
+  --gray-500: rgba(255, 255, 255, 0.5);
+
+  --color-link: #84b2ff;
+  --color-visited-link: #b88dff;
+
+  --body-background: #343a40;
+  --body-font-color: #e9ecef;
+
+  --icon-filter: brightness(0) invert(1);
+
+  --hint-color-info: #6bf;
+  --hint-color-warning: #fd6;
+  --hint-color-danger: #f66;
+}
diff --git a/website/themes/book/assets/_fonts.scss b/website/themes/book/assets/_fonts.scss
new file mode 100644
index 000000000..1cf735ad3
--- /dev/null
+++ b/website/themes/book/assets/_fonts.scss
@@ -0,0 +1,49 @@
+/* roboto-300italic - latin */
+@font-face {
+  font-family: 'Roboto';
+  font-style: italic;
+  font-weight: 300;
+  font-display: swap;
+  src: local('Roboto Light Italic'), local('Roboto-LightItalic'),
+       url('fonts/roboto-v19-latin-300italic.woff2') format('woff2'), /* Chrome 26+, Opera 23+, Firefox 39+ */
+       url('fonts/roboto-v19-latin-300italic.woff') format('woff'); /* Chrome 6+, Firefox 3.6+, IE 9+, Safari 5.1+ */
+}
+/* roboto-regular - latin */
+@font-face {
+  font-family: 'Roboto';
+  font-style: normal;
+  font-weight: 400;
+  font-display: swap;
+  src: local('Roboto'), local('Roboto-Regular'),
+       url('fonts/roboto-v19-latin-regular.woff2') format('woff2'), /* Chrome 26+, Opera 23+, Firefox 39+ */
+       url('fonts/roboto-v19-latin-regular.woff') format('woff'); /* Chrome 6+, Firefox 3.6+, IE 9+, Safari 5.1+ */
+}
+/* roboto-700 - latin */
+@font-face {
+  font-family: 'Roboto';
+  font-style: normal;
+  font-weight: 700;
+  font-display: swap;
+  src: local('Roboto Bold'), local('Roboto-Bold'),
+       url('fonts/roboto-v19-latin-700.woff2') format('woff2'), /* Chrome 26+, Opera 23+, Firefox 39+ */
+       url('fonts/roboto-v19-latin-700.woff') format('woff'); /* Chrome 6+, Firefox 3.6+, IE 9+, Safari 5.1+ */
+}
+
+/* roboto-mono-regular - latin */
+@font-face {
+  font-family: 'Roboto Mono';
+  font-style: normal;
+  font-weight: 400;
+  font-display: swap;
+  src: local('Roboto Mono'), local('RobotoMono-Regular'),
+       url('fonts/roboto-mono-v6-latin-regular.woff2') format('woff2'), /* Chrome 26+, Opera 23+, Firefox 39+ */
+       url('fonts/roboto-mono-v6-latin-regular.woff') format('woff'); /* Chrome 6+, Firefox 3.6+, IE 9+, Safari 5.1+ */
+}
+
+body {
+  font-family: 'Roboto', sans-serif;
+}
+
+code {
+  font-family: 'Roboto Mono', monospace;
+}
diff --git a/website/themes/book/assets/_main.scss b/website/themes/book/assets/_main.scss
new file mode 100644
index 000000000..7085e3648
--- /dev/null
+++ b/website/themes/book/assets/_main.scss
@@ -0,0 +1,392 @@
+html {
+  font-size: $font-size-base;
+  scroll-behavior: smooth;
+  touch-action: manipulation;
+}
+
+body {
+  min-width: $body-min-width;
+  color: var(--body-font-color);
+  background: var(--body-background);
+
+  letter-spacing: 0.33px;
+  font-weight: $body-font-weight;
+  text-rendering: optimizeLegibility;
+  -webkit-font-smoothing: antialiased;
+  -moz-osx-font-smoothing: grayscale;
+
+  box-sizing: border-box;
+  * {
+    box-sizing: inherit;
+  }
+}
+
+h1,
+h2,
+h3,
+h4,
+h5 {
+  font-weight: $body-font-weight;
+}
+
+a {
+  text-decoration: none;
+  color: var(--color-link);
+}
+
+img {
+  vertical-align: baseline;
+}
+
+:focus {
+  @include outline;
+}
+
+aside nav ul {
+  padding: 0;
+  margin: 0;
+  list-style: none;
+
+  li {
+    margin: 1em 0;
+    position: relative;
+  }
+
+  a {
+    display: block;
+  }
+
+  a:hover {
+    opacity: 0.5;
+  }
+
+  ul {
+    padding-inline-start: $padding-16;
+  }
+}
+
+ul.pagination {
+  display: flex;
+  justify-content: center;
+  list-style-type: none;
+
+  .page-item a {
+    padding: $padding-16;
+  }
+}
+
+.container {
+  max-width: $container-max-width;
+  margin: 0 auto;
+}
+
+.book-icon {
+  filter: var(--icon-filter);
+}
+
+.book-brand {
+  margin-top: 0;
+
+  img {
+    height: 1.5em;
+    width: auto;
+    vertical-align: middle;
+    margin-inline-end: $padding-8;
+  }
+}
+
+.book-menu {
+  flex: 0 0 $menu-width;
+  font-size: $font-size-14;
+
+  .book-menu-content {
+    width: $menu-width;
+    padding: $padding-16;
+    background: var(--body-background);
+
+    @include fixed;
+  }
+
+  a,
+  label {
+    color: inherit;
+    cursor: pointer;
+    word-wrap: break-word;
+  }
+
+  a.active {
+    color: var(--color-link);
+  }
+
+  input.toggle + label + ul {
+    display: none;
+  }
+
+  input.toggle:checked + label + ul {
+    display: block;
+  }
+
+  input.toggle + label::after {
+    content: "▸";
+  }
+
+  input.toggle:checked + label::after {
+    content: "▾";
+  }
+}
+
+.book-section-flat {
+  margin-bottom: $padding-16 * 2;
+
+  &:not(:first-child) {
+    margin-top: $padding-16 * 2;
+  }
+
+  > a,
+  > span,
+  > label {
+    font-weight: bolder;
+  }
+
+  > ul {
+    padding-inline-start: 0;
+  }
+}
+
+.book-page {
+  min-width: $body-min-width;
+  flex-grow: 1;
+  padding: $padding-16;
+}
+
+.book-post {
+  margin-bottom: $padding-16 * 3;
+}
+
+.book-header {
+  display: none;
+  margin-bottom: $padding-16;
+
+  label {
+    line-height: 0;
+  }
+}
+
+.book-search {
+  position: relative;
+  margin: $padding-16 0;
+  border-bottom: 1px solid transparent;
+
+  input {
+    width: 100%;
+    padding: $padding-8;
+
+    border: 0;
+    border-radius: $border-radius;
+
+    background: var(--gray-100);
+    color: var(--body-font-color);
+
+    &:required + .book-search-spinner {
+      display: block;
+    }
+  }
+
+  .book-search-spinner {
+    position: absolute;
+    top: 0;
+    margin: $padding-8;
+    margin-inline-start: calc(100% - #{$padding-16 + $padding-8});
+
+    width: $padding-16;
+    height: $padding-16;
+
+    border: $padding-1 solid transparent;
+    border-top-color: var(--body-font-color);
+    border-radius: 50%;
+
+    @include spin(1s);
+  }
+
+  small {
+    opacity: 0.5;
+  }
+}
+
+.book-toc {
+  // flex: 0 0 $toc-width; // 使 toc 靠右贴边
+  font-size: $font-size-12;
+
+  .book-toc-content {
+    width: $toc-width;
+    padding: $padding-16;
+
+    @include fixed;
+  }
+
+  img {
+    height: 1em;
+  }
+
+  nav > ul > li:first-child {
+    margin-top: 0;
+  }
+}
+
+.book-footer {
+  padding-top: $padding-16;
+  font-size: $font-size-14;
+
+  img {
+    height: 1em;
+    margin-inline-end: $padding-8;
+  }
+}
+
+.book-comments {
+  margin-top: $padding-16;
+}
+
+.book-languages {
+  position: relative;
+  overflow: visible;
+
+  padding: $padding-16;
+  margin: -$padding-16;
+
+  ul {
+    margin: 0;
+    padding: 0;
+    list-style: none;
+
+    li {
+      white-space: nowrap;
+      cursor: pointer;
+    }
+  }
+
+  &:hover,
+  &:focus,
+  &:focus-within {
+    .book-languages-list {
+      display: block;
+    }
+  }
+
+  .book-languages-list {
+    display: none;
+
+    position: absolute;
+    bottom: 100%;
+    left: 0;
+    padding: $padding-8 0;
+
+    background: var(--body-background);
+    box-shadow: 0 0 $padding-4 rgba(0, 0, 0, 0.1);
+
+    li img {
+      opacity: 0.25;
+    }
+
+    li.active img,
+    li:hover img {
+      opacity: initial;
+    }
+
+    a {
+      color: inherit;
+      padding: $padding-8 $padding-16;
+    }
+  }
+}
+
+.book-home {
+  padding: $padding-16;
+}
+
+// Responsive styles
+.book-menu-content,
+.book-toc-content,
+.book-page,
+.book-header aside,
+.markdown {
+  transition: 0.2s ease-in-out;
+  transition-property: transform, margin, opacity, visibility;
+  will-change: transform, margin, opacity;
+}
+
+@media screen and (max-width: $mobile-breakpoint) {
+  #menu-control,
+  #toc-control {
+    display: inline;
+  }
+
+  .book-menu {
+    visibility: hidden;
+    margin-inline-start: -$menu-width;
+    font-size: $font-size-base;
+    z-index: 1;
+  }
+
+  .book-toc {
+    display: none;
+  }
+
+  .book-header {
+    display: block;
+  }
+
+  #menu-control:focus ~ main label[for="menu-control"] {
+    @include outline;
+  }
+
+  #menu-control:checked ~ main {
+    .book-menu {
+      visibility: initial;
+    }
+
+    .book-menu .book-menu-content {
+      transform: translateX($menu-width);
+      box-shadow: 0 0 $padding-8 rgba(0, 0, 0, 0.1);
+    }
+
+    .book-page {
+      opacity: 0.25;
+    }
+
+    .book-menu-overlay {
+      display: block;
+      position: absolute;
+      top: 0;
+      bottom: 0;
+      left: 0;
+      right: 0;
+    }
+  }
+
+  #toc-control:focus ~ main label[for="toc-control"] {
+    @include outline;
+  }
+
+  #toc-control:checked ~ main {
+    .book-header aside {
+      display: block;
+    }
+  }
+
+  //for RTL support
+  body[dir="rtl"] #menu-control:checked + main {
+    .book-menu .book-menu-content {
+      transform: translateX(-$menu-width);
+    }
+  }
+}
+
+// Extra space for big screens
+@media screen and (min-width: $container-max-width) {
+  .book-page,
+  .book-menu .book-menu-content,
+  .book-toc .book-toc-content {
+    padding: $padding-16 * 2 $padding-16;
+  }
+}
diff --git a/website/themes/book/assets/_markdown.scss b/website/themes/book/assets/_markdown.scss
new file mode 100644
index 000000000..e71de431d
--- /dev/null
+++ b/website/themes/book/assets/_markdown.scss
@@ -0,0 +1,191 @@
+@import "variables";
+
+.markdown {
+  line-height: 1.6;
+
+  // remove padding at the beginning of page
+  > :first-child {
+    margin-top: 0;
+  }
+
+  h1,
+  h2,
+  h3,
+  h4,
+  h5,
+  h6 {
+    font-weight: normal;
+    line-height: 1;
+    margin-top: 1.5em;
+    margin-bottom: $padding-16;
+
+    a.anchor {
+      opacity: 0;
+      font-size: 0.75em;
+      vertical-align: middle;
+      text-decoration: none;
+    }
+
+    &:hover a.anchor,
+    a.anchor:focus {
+      opacity: initial;
+    }
+  }
+
+  h4,
+  h5,
+  h6 {
+    font-weight: bolder;
+  }
+
+  h5 {
+    font-size: 0.875em;
+  }
+
+  h6 {
+    font-size: 0.75em;
+  }
+
+  b,
+  optgroup,
+  strong {
+    font-weight: bolder;
+  }
+
+  a {
+    text-decoration: none;
+
+    &:hover {
+      text-decoration: underline;
+    }
+    &:visited {
+      color: var(--color-visited-link);
+    }
+  }
+
+  img {
+    max-width: 100%;
+  }
+
+  code {
+    padding: 0 $padding-4;
+    background: var(--gray-200);
+    border-radius: $border-radius;
+    font-size: 0.875em;
+  }
+
+  pre {
+    padding: $padding-16;
+    background: var(--gray-100);
+    border-radius: $border-radius;
+    overflow-x: auto;
+
+    code {
+      padding: 0;
+      background: none;
+    }
+  }
+
+  blockquote {
+    margin: $padding-16 0;
+    padding: $padding-8 $padding-16 $padding-8 ($padding-16 - $padding-4); //to keep total left space 16dp
+
+    border-inline-start: $padding-4 solid var(--gray-200);
+    border-radius: $border-radius;
+
+    :first-child {
+      margin-top: 0;
+    }
+    :last-child {
+      margin-bottom: 0;
+    }
+  }
+
+  table {
+    overflow: auto;
+    display: block;
+    border-spacing: 0;
+    border-collapse: collapse;
+    margin-top: $padding-16;
+    margin-bottom: $padding-16;
+
+    tr th,
+    tr td {
+      padding: $padding-8 $padding-16;
+      border: $padding-1 solid var(--gray-200);
+    }
+
+    tr:nth-child(2n) {
+      background: var(--gray-100);
+    }
+  }
+
+  hr {
+    height: $padding-1;
+    border: none;
+    background: var(--gray-200);
+  }
+
+  ul,
+  ol {
+    padding-inline-start: $padding-16 * 2;
+  }
+
+  dl {
+    dt {
+      font-weight: bolder;
+      margin-top: $padding-16;
+    }
+
+    dd {
+      margin-inline-start: $padding-16;
+      margin-bottom: $padding-16;
+    }
+  }
+
+  // Special case for highlighted code with line numbers
+  .highlight table tr {
+    td:nth-child(1) pre {
+      margin: 0;
+      padding-inline-end: 0;
+    }
+    td:nth-child(2) pre {
+      margin: 0;
+      padding-inline-start: 0;
+    }
+  }
+
+  details {
+    padding: $padding-16;
+    border: $padding-1 solid var(--gray-200);
+    border-radius: $border-radius;
+
+    summary {
+      line-height: 1;
+      padding: $padding-16;
+      margin: -$padding-16;
+      cursor: pointer;
+    }
+
+    &[open] summary {
+      margin-bottom: 0;
+    }
+  }
+
+  figure {
+    margin: $padding-16 0;
+    figcaption p {
+      margin-top: 0;
+    }
+  }
+}
+
+.markdown-inner {
+  // Util class to remove extra margin in nested markdown content
+  > :first-child {
+    margin-top: 0;
+  }
+  > :last-child {
+    margin-bottom: 0;
+  }
+}
diff --git a/website/themes/book/assets/_print.scss b/website/themes/book/assets/_print.scss
new file mode 100644
index 000000000..8ae290121
--- /dev/null
+++ b/website/themes/book/assets/_print.scss
@@ -0,0 +1,17 @@
+@media print {
+  .book-menu,
+  .book-footer,
+  .book-toc {
+    display: none;
+  }
+
+  .book-header,
+  .book-header aside {
+    display: block;
+  }
+
+  main {
+    // Fix for https://bugzilla.mozilla.org/show_bug.cgi?id=939897
+    display: block !important;
+  }
+}
diff --git a/website/themes/book/assets/_shortcodes.scss b/website/themes/book/assets/_shortcodes.scss
new file mode 100644
index 000000000..714de2ae3
--- /dev/null
+++ b/website/themes/book/assets/_shortcodes.scss
@@ -0,0 +1,104 @@
+.markdown {
+  // {{< expand "Label" "icon" >}}
+  .book-expand {
+    margin-top: $padding-16;
+    margin-bottom: $padding-16;
+
+    border: $padding-1 solid var(--gray-200);
+    border-radius: $border-radius;
+
+    overflow: hidden;
+
+    .book-expand-head {
+      background: var(--gray-100);
+      padding: $padding-8 $padding-16;
+      cursor: pointer;
+    }
+
+    .book-expand-content {
+      display: none;
+      padding: $padding-16;
+    }
+
+    input[type="checkbox"]:checked + .book-expand-content {
+      display: block;
+    }
+  }
+
+  // {{< tabs >}}
+  .book-tabs {
+    margin-top: $padding-16;
+    margin-bottom: $padding-16;
+
+    border: $padding-1 solid var(--gray-200);
+    border-radius: $border-radius;
+
+    overflow: hidden;
+
+    display: flex;
+    flex-wrap: wrap;
+
+    label {
+      display: inline-block;
+      padding: $padding-8 $padding-16;
+      border-bottom: $padding-1 transparent;
+      cursor: pointer;
+    }
+
+    .book-tabs-content {
+      order: 999; //Move content blocks to the end
+      width: 100%;
+      border-top: $padding-1 solid var(--gray-100);
+      padding: $padding-16;
+      display: none;
+    }
+
+    input[type="radio"]:checked + label {
+      border-bottom: $padding-1 solid var(--color-link);
+    }
+    input[type="radio"]:checked + label + .book-tabs-content {
+      display: block;
+    }
+    input[type="radio"]:focus + label {
+      @include outline;
+    }
+  }
+
+  // {{< columns >}}
+  .book-columns {
+    margin-left: -$padding-16;
+    margin-right: -$padding-16;
+
+    > div {
+      margin: $padding-16 0;
+      min-width: $body-min-width / 2;
+      padding: 0 $padding-16;
+    }
+  }
+
+  // {{< button >}}
+  a.book-btn {
+    display: inline-block;
+    font-size: $font-size-14;
+    color: var(--color-link);
+    line-height: $padding-16 * 2;
+    padding: 0 $padding-16;
+    border: $padding-1 solid var(--color-link);
+    border-radius: $border-radius;
+    cursor: pointer;
+
+    &:hover {
+      text-decoration: none;
+    }
+  }
+
+  // {{< hint >}}
+  .book-hint {
+    @each $name, $color in $hint-colors {
+      &.#{$name} {
+        border-color: $color;
+        background-color: rgba($color, 0.1);
+      }
+    }
+  }
+}
diff --git a/website/themes/book/assets/_utils.scss b/website/themes/book/assets/_utils.scss
new file mode 100644
index 000000000..29ef1caab
--- /dev/null
+++ b/website/themes/book/assets/_utils.scss
@@ -0,0 +1,92 @@
+.flex {
+  display: flex;
+}
+
+.flex-auto {
+  flex: 1 1 auto;
+}
+
+.flex-even {
+  flex: 1 1;
+}
+
+.flex-wrap {
+  flex-wrap: wrap;
+}
+
+.justify-start {
+  justify-content: flex-start;
+}
+
+.justify-end {
+  justify-content: flex-end;
+}
+
+.justify-center {
+  justify-content: center;
+}
+
+.justify-between {
+  justify-content: space-between;
+}
+
+.align-center {
+  align-items: center;
+}
+
+.mx-auto {
+  margin: 0 auto;
+}
+
+.text-center {
+  text-align: center;
+}
+
+.text-left {
+  text-align: left;
+}
+
+.text-right {
+  text-align: right;
+}
+
+.hidden {
+  display: none;
+}
+
+input.toggle {
+  height: 0;
+  width: 0;
+  overflow: hidden;
+  opacity: 0;
+  position: absolute;
+}
+
+.clearfix::after {
+  content: "";
+  display: table;
+  clear: both;
+}
+
+@mixin spin($duration) {
+  animation: spin $duration ease infinite;
+  @keyframes spin {
+    100% {
+      transform: rotate(360deg);
+    }
+  }
+}
+
+@mixin fixed {
+  position: fixed;
+  top: 0;
+  bottom: 0;
+  overflow-x: hidden;
+  overflow-y: auto;
+}
+
+@mixin outline {
+  outline-style: auto;
+  outline-color: currentColor;
+  outline-color: -webkit-focus-ring-color;
+}
diff --git a/website/themes/book/assets/_variables.scss b/website/themes/book/assets/_variables.scss
new file mode 100644
index 000000000..6e34d16b8
--- /dev/null
+++ b/website/themes/book/assets/_variables.scss
@@ -0,0 +1,3 @@
+/* You can override SASS variables here. */
+
+// @import "plugins/dark";
diff --git a/website/themes/book/assets/book.scss b/website/themes/book/assets/book.scss
new file mode 100644
index 000000000..59369fabd
--- /dev/null
+++ b/website/themes/book/assets/book.scss
@@ -0,0 +1,15 @@
+@import "defaults";
+@import "variables";
+@import "themes/{{ default "light" .Site.Params.BookTheme }}";
+
+@import "normalize";
+@import "utils";
+@import "main";
+@import "fonts";
+@import "print";
+
+@import "markdown";
+@import "shortcodes";
+
+// Custom defined styles
+@import "custom";
diff --git a/website/themes/book/assets/manifest.json b/website/themes/book/assets/manifest.json
new file mode 100644
index 000000000..982a4ebab
--- /dev/null
+++ b/website/themes/book/assets/manifest.json
@@ -0,0 +1,54 @@
+{
+  "name": "{{ .Site.Title }}",
+  "short_name": "Cookbook",
+  "start_url": "{{ "/" | relURL }}",
+  "scope": "{{ "/" | relURL }}",
+  "display": "standalone",
+  "background_color": "#FFFFFF",
+  "theme_color": "#FFFFFF",
+  "icons": [
+    {
+      "src": "{{ "/apple-touch-icon-180x180.png" | relURL }}",
+      "sizes": "180x180",
+      "type": "image/png",
+      "density": "3.0",
+      "purpose": "any maskable"
+    },
+    {
+      "src": "{{ "/apple-touch-icon-1024x1024.png" | relURL }}",
+      "sizes": "1024x1024",
+      "type": "image/png",
+      "purpose": "any maskable"
+    },
+    {
+      "src": "{{ "/apple-touch-icon-60x60.png" | relURL }}",
+      "sizes": "60x60",
+      "type": "image/png",
+      "purpose": "any maskable"
+    },
+    {
+      "src": "{{ "/apple-touch-icon-76x76.png" | relURL }}",
+      "sizes": "76x76",
+      "type": "image/png",
+      "purpose": "any maskable"
+    },
+    {
+      "src": "{{ "/apple-touch-icon-120x120.png" | relURL }}",
+      "sizes": "120x120",
+      "type": "image/png",
+      "purpose": "any maskable"
+    },
+    {
+      "src": "{{ "/apple-touch-icon-152x152.png" | relURL }}",
+      "sizes": "152x152",
+      "type": "image/png",
+      "purpose": "any maskable"
+    },
+    {
+      "src": "{{ "/apple-touch-icon-512x512.png" | relURL }}",
+      "sizes": "512x512",
+      "type": "image/png",
+      "purpose": "any maskable"
+    }
+  ]
+}
diff --git a/website/themes/book/assets/menu-reset.js b/website/themes/book/assets/menu-reset.js
new file mode 100644
index 000000000..f3b90c54e
--- /dev/null
+++ b/website/themes/book/assets/menu-reset.js
@@ -0,0 +1,7 @@
+(function() {
+  var menu = document.querySelector("aside.book-menu nav");
+  addEventListener("beforeunload", function(event) {
+      localStorage.setItem("menu.scrollTop", menu.scrollTop);
+  });
+  menu.scrollTop = localStorage.getItem("menu.scrollTop");
+})();
diff --git a/website/themes/book/assets/mermaid.json b/website/themes/book/assets/mermaid.json
new file mode 100644
index 000000000..0a3f4fbad
--- /dev/null
+++ b/website/themes/book/assets/mermaid.json
@@ -0,0 +1,6 @@
+{
+  "flowchart": {
+    "useMaxWidth":true
+  },
+  "theme": "default"
+}
diff --git a/website/themes/book/assets/normalize.css b/website/themes/book/assets/normalize.css
new file mode 100644
index 000000000..192eb9ce4
--- /dev/null
+++ b/website/themes/book/assets/normalize.css
@@ -0,0 +1,349 @@
+/*! normalize.css v8.0.1 | MIT License | github.com/necolas/normalize.css */
+
+/* Document
+   ========================================================================== */
+
+/**
+ * 1. Correct the line height in all browsers.
+ * 2. Prevent adjustments of font size after orientation changes in iOS.
+ */
+
+html {
+  line-height: 1.15; /* 1 */
+  -webkit-text-size-adjust: 100%; /* 2 */
+}
+
+/* Sections
+   ========================================================================== */
+
+/**
+ * Remove the margin in all browsers.
+ */
+
+body {
+  margin: 0;
+}
+
+/**
+ * Render the `main` element consistently in IE.
+ */
+
+main {
+  display: block;
+}
+
+/**
+ * Correct the font size and margin on `h1` elements within `section` and
+ * `article` contexts in Chrome, Firefox, and Safari.
+ */
+
+h1 {
+  font-size: 2em;
+  margin: 0.67em 0;
+}
+
+/* Grouping content
+   ========================================================================== */
+
+/**
+ * 1. Add the correct box sizing in Firefox.
+ * 2. Show the overflow in Edge and IE.
+ */
+
+hr {
+  box-sizing: content-box; /* 1 */
+  height: 0; /* 1 */
+  overflow: visible; /* 2 */
+}
+
+/**
+ * 1. Correct the inheritance and scaling of font size in all browsers.
+ * 2. Correct the odd `em` font sizing in all browsers.
+ */
+
+pre {
+  font-family: monospace, monospace; /* 1 */
+  font-size: 1em; /* 2 */
+}
+
+/* Text-level semantics
+   ========================================================================== */
+
+/**
+ * Remove the gray background on active links in IE 10.
+ */
+
+a {
+  background-color: transparent;
+}
+
+/**
+ * 1. Remove the bottom border in Chrome 57-
+ * 2. Add the correct text decoration in Chrome, Edge, IE, Opera, and Safari.
+ */
+
+abbr[title] {
+  border-bottom: none; /* 1 */
+  text-decoration: underline; /* 2 */
+  text-decoration: underline dotted; /* 2 */
+}
+
+/**
+ * Add the correct font weight in Chrome, Edge, and Safari.
+ */
+
+b,
+strong {
+  font-weight: bolder;
+}
+
+/**
+ * 1. Correct the inheritance and scaling of font size in all browsers.
+ * 2. Correct the odd `em` font sizing in all browsers.
+ */
+
+code,
+kbd,
+samp {
+  font-family: monospace, monospace; /* 1 */
+  font-size: 1em; /* 2 */
+}
+
+/**
+ * Add the correct font size in all browsers.
+ */
+
+small {
+  font-size: 80%;
+}
+
+/**
+ * Prevent `sub` and `sup` elements from affecting the line height in
+ * all browsers.
+ */
+
+sub,
+sup {
+  font-size: 75%;
+  line-height: 0;
+  position: relative;
+  vertical-align: baseline;
+}
+
+sub {
+  bottom: -0.25em;
+}
+
+sup {
+  top: -0.5em;
+}
+
+/* Embedded content
+   ========================================================================== */
+
+/**
+ * Remove the border on images inside links in IE 10.
+ */
+
+img {
+  border-style: none;
+}
+
+/* Forms
+   ========================================================================== */
+
+/**
+ * 1. Change the font styles in all browsers.
+ * 2. Remove the margin in Firefox and Safari.
+ */
+
+button,
+input,
+optgroup,
+select,
+textarea {
+  font-family: inherit; /* 1 */
+  font-size: 100%; /* 1 */
+  line-height: 1.15; /* 1 */
+  margin: 0; /* 2 */
+}
+
+/**
+ * Show the overflow in IE.
+ * 1. Show the overflow in Edge.
+ */
+
+button,
+input { /* 1 */
+  overflow: visible;
+}
+
+/**
+ * Remove the inheritance of text transform in Edge, Firefox, and IE.
+ * 1. Remove the inheritance of text transform in Firefox.
+ */
+
+button,
+select { /* 1 */
+  text-transform: none;
+}
+
+/**
+ * Correct the inability to style clickable types in iOS and Safari.
+ */
+
+button,
+[type="button"],
+[type="reset"],
+[type="submit"] {
+  -webkit-appearance: button;
+}
+
+/**
+ * Remove the inner border and padding in Firefox.
+ */
+
+button::-moz-focus-inner,
+[type="button"]::-moz-focus-inner,
+[type="reset"]::-moz-focus-inner,
+[type="submit"]::-moz-focus-inner {
+  border-style: none;
+  padding: 0;
+}
+
+/**
+ * Restore the focus styles unset by the previous rule.
+ */
+
+button:-moz-focusring,
+[type="button"]:-moz-focusring,
+[type="reset"]:-moz-focusring,
+[type="submit"]:-moz-focusring {
+  outline: 1px dotted ButtonText;
+}
+
+/**
+ * Correct the padding in Firefox.
+ */
+
+fieldset {
+  padding: 0.35em 0.75em 0.625em;
+}
+
+/**
+ * 1. Correct the text wrapping in Edge and IE.
+ * 2. Correct the color inheritance from `fieldset` elements in IE.
+ * 3. Remove the padding so developers are not caught out when they zero out
+ *    `fieldset` elements in all browsers.
+ */
+
+legend {
+  box-sizing: border-box; /* 1 */
+  color: inherit; /* 2 */
+  display: table; /* 1 */
+  max-width: 100%; /* 1 */
+  padding: 0; /* 3 */
+  white-space: normal; /* 1 */
+}
+
+/**
+ * Add the correct vertical alignment in Chrome, Firefox, and Opera.
+ */
+
+progress {
+  vertical-align: baseline;
+}
+
+/**
+ * Remove the default vertical scrollbar in IE 10+.
+ */
+
+textarea {
+  overflow: auto;
+}
+
+/**
+ * 1. Add the correct box sizing in IE 10.
+ * 2. Remove the padding in IE 10.
+ */
+
+[type="checkbox"],
+[type="radio"] {
+  box-sizing: border-box; /* 1 */
+  padding: 0; /* 2 */
+}
+
+/**
+ * Correct the cursor style of increment and decrement buttons in Chrome.
+ */
+
+[type="number"]::-webkit-inner-spin-button,
+[type="number"]::-webkit-outer-spin-button {
+  height: auto;
+}
+
+/**
+ * 1. Correct the odd appearance in Chrome and Safari.
+ * 2. Correct the outline style in Safari.
+ */
+
+[type="search"] {
+  -webkit-appearance: textfield; /* 1 */
+  outline-offset: -2px; /* 2 */
+}
+
+/**
+ * Remove the inner padding in Chrome and Safari on macOS.
+ */
+
+[type="search"]::-webkit-search-decoration {
+  -webkit-appearance: none;
+}
+
+/**
+ * 1. Correct the inability to style clickable types in iOS and Safari.
+ * 2. Change font properties to `inherit` in Safari.
+ */
+
+::-webkit-file-upload-button {
+  -webkit-appearance: button; /* 1 */
+  font: inherit; /* 2 */
+}
+
+/* Interactive
+   ========================================================================== */
+
+/*
+ * Add the correct display in Edge, IE 10+, and Firefox.
+ */
+
+details {
+  display: block;
+}
+
+/*
+ * Add the correct display in all browsers.
+ */
+
+summary {
+  display: list-item;
+}
+
+/* Misc
+   ========================================================================== */
+
+/**
+ * Add the correct display in IE 10+.
+ */
+
+template {
+  display: none;
+}
+
+/**
+ * Add the correct display in IE 10.
+ */
+
+[hidden] {
+  display: none;
+}
diff --git a/website/themes/book/assets/plugins/_dark.scss b/website/themes/book/assets/plugins/_dark.scss
new file mode 100644
index 000000000..5796a898d
--- /dev/null
+++ b/website/themes/book/assets/plugins/_dark.scss
@@ -0,0 +1,10 @@
+$gray-100: rgba(255, 255, 255, 0.1);
+$gray-200: rgba(255, 255, 255, 0.2);
+
+$body-background: #343a40;
+$body-font-color: #e9ecef;
+
+$color-link: #84b2ff;
+$color-visited-link: #b88dff;
+
+$icon-filter: brightness(0) invert(1);
diff --git a/website/themes/book/assets/plugins/_numbered.scss b/website/themes/book/assets/plugins/_numbered.scss
new file mode 100644
index 000000000..56cda5ac5
--- /dev/null
+++ b/website/themes/book/assets/plugins/_numbered.scss
@@ -0,0 +1,36 @@
+$startLevel: 1;
+$endLevel: 6;
+
+.book-page .markdown {
+  @for $currentLevel from $startLevel through $endLevel {
+    > h#{$currentLevel} {
+      counter-increment: h#{$currentLevel};
+      counter-reset: h#{$currentLevel + 1};
+
+      $content: "";
+      @for $n from $startLevel through $currentLevel {
+        $content: $content + 'counter(h#{$n})"."';
+      }
+
+      &::before {
+        content: unquote($content) " ";
+      }
+    }
+  }
+}
+
+.book-toc nav ul {
+  li {
+    counter-increment: item;
+
+    &:first-child {
+      counter-reset: item;
+    }
+
+    &:before {
+      content: counters(item, ".") ". ";
+      float: left;
+      margin-inline-end: $padding-4;
+    }
+  }
+}
diff --git a/website/themes/book/assets/plugins/_scrollbars.scss b/website/themes/book/assets/plugins/_scrollbars.scss
new file mode 100644
index 000000000..00625827b
--- /dev/null
+++ b/website/themes/book/assets/plugins/_scrollbars.scss
@@ -0,0 +1,26 @@
+@import "defaults";
+@import "variables";
+
+// Webkit
+::-webkit-scrollbar {
+  width: $padding-8;
+}
+
+::-webkit-scrollbar-thumb {
+  background: transparent;
+  border-radius: $padding-8;
+}
+
+:hover::-webkit-scrollbar-thumb {
+  background: var(--gray-500);
+}
+
+// MS
+body {
+  -ms-overflow-style: -ms-autohiding-scrollbar;
+}
+
+// Future
+.book-menu nav {
+  scrollbar-color: transparent var(--gray-500);
+}
diff --git a/website/themes/book/assets/search-data.js b/website/themes/book/assets/search-data.js
new file mode 100644
index 000000000..a93664ec0
--- /dev/null
+++ b/website/themes/book/assets/search-data.js
@@ -0,0 +1,32 @@
+'use strict';
+
+(function () {
+  const indexCfg = {{ with i18n "bookSearchConfig" }}
+    {{ . }};
+  {{ else }}
+   {};
+  {{ end }}
+
+  indexCfg.doc = {
+    id: 'id',
+    field: ['title', 'content'],
+    store: ['title', 'href', 'section'],
+  };
+
+  const index = FlexSearch.create('balance', indexCfg);
+  window.bookSearchIndex = index;
+
+  {{- $pages := where .Site.Pages "Kind" "in" (slice "page" "section") -}}
+  {{- $pages = where $pages "Params.booksearchexclude" "!=" true -}}
+  {{- $pages = where $pages "Content" "not in" (slice nil "") -}}
+
+  {{ range $index, $page := $pages }}
+  index.add({
+    'id': {{ $index }},
+    'href': '{{ $page.RelPermalink }}',
+    'title': {{ (partial "docs/title" $page) | jsonify }},
+    'section': {{ (partial "docs/title" $page.Parent) | jsonify }},
+    'content': {{ $page.Plain | jsonify }}
+  });
+  {{- end -}}
+})();
diff --git a/website/themes/book/assets/search.js b/website/themes/book/assets/search.js
new file mode 100644
index 000000000..3635a1fd3
--- /dev/null
+++ b/website/themes/book/assets/search.js
@@ -0,0 +1,101 @@
+'use strict';
+
+{{ $searchDataFile := printf "%s.search-data.js" .Language.Lang }}
+{{ $searchData := resources.Get "search-data.js" | resources.ExecuteAsTemplate $searchDataFile . | resources.Minify | resources.Fingerprint }}
+
+(function () {
+  const input = document.querySelector('#book-search-input');
+  const results = document.querySelector('#book-search-results');
+
+  if (!input) {
+    return
+  }
+
+  input.addEventListener('focus', init);
+  input.addEventListener('keyup', search);
+
+  document.addEventListener('keypress', focusSearchFieldOnKeyPress);
+
+  /**
+   * @param {Event} event
+   */
+  function focusSearchFieldOnKeyPress(event) {
+    if (input === document.activeElement) {
+      return;
+    }
+
+    const characterPressed = String.fromCharCode(event.charCode);
+    if (!isHotkey(characterPressed)) {
+      return;
+    }
+
+    input.focus();
+    event.preventDefault();
+  }
+
+  /**
+   * @param {String} character
+   * @returns {Boolean} 
+   */
+  function isHotkey(character) {
+    const dataHotkeys = input.getAttribute('data-hotkeys') || '';
+    return dataHotkeys.indexOf(character) >= 0;
+  }
+
+  function init() {
+    input.removeEventListener('focus', init); // init once
+    input.required = true;
+
+    loadScript('{{ "flexsearch.min.js" | relURL }}');
+    loadScript('{{ $searchData.RelPermalink }}', function () {
+      input.required = false;
+      search();
+    });
+  }
+
+  function search() {
+    while (results.firstChild) {
+      results.removeChild(results.firstChild);
+    }
+
+    if (!input.value) {
+      return;
+    }
+
+    const searchHits = window.bookSearchIndex.search(input.value, 10);
+    searchHits.forEach(function (page) {
+      const li = element('<li><a href></a><small></small></li>');
+      const a = li.querySelector('a'), small = li.querySelector('small');
+
+      a.href = page.href;
+      a.textContent = page.title;
+      small.textContent = page.section;
+
+      results.appendChild(li);
+    });
+  }
+
+  /**
+   * @param {String} src 
+   * @param {Function} callback 
+   */
+  function loadScript(src, callback) {
+    const script = document.createElement('script');
+    script.defer = true;
+    script.async = false;
+    script.src = src;
+    script.onload = callback;
+
+    document.head.appendChild(script);
+  }
+
+  /**
+   * @param {String} content
+   * @returns {Node}
+   */
+  function element(content) {
+    const div = document.createElement('div');
+    div.innerHTML = content;
+    return div.firstChild;
+  }
+})();
diff --git a/website/themes/book/assets/serviceworker-v1.js b/website/themes/book/assets/serviceworker-v1.js
new file mode 100644
index 000000000..dd7ef6600
--- /dev/null
+++ b/website/themes/book/assets/serviceworker-v1.js
@@ -0,0 +1,86 @@
+'use strict';
+(function () {
+  'use strict';
+    /**
+    * Service Worker Toolbox caching
+    */
+    var cacheVersion = '-toolbox-v1';
+    var dynamicVendorCacheName = 'dynamic-vendor' + cacheVersion;
+    var staticVendorCacheName = 'static-vendor' + cacheVersion;
+    var staticAssetsCacheName = 'static-assets' + cacheVersion;
+    var contentCacheName = 'content' + cacheVersion;
+    var maxEntries = 50;
+    self.importScripts("/leetcode/js/sw-toolbox.js");
+    self.toolbox.options.debug = false;
+
+    // Cache own static assets
+    self.toolbox.router.get('/(.*)', self.toolbox.cacheFirst, {
+        cache: {
+          name: staticAssetsCacheName,
+          maxEntries: maxEntries
+        }
+    });
+    // cache dynamic vendor assets, things which have no other update mechanism like filename change/version hash
+    self.toolbox.router.get('/css', self.toolbox.fastest, {
+        origin: /fonts\.googleapis\.com/,
+            cache: {
+              name: dynamicVendorCacheName,
+              maxEntries: maxEntries
+            }
+    });
+    // Do not cache disqus
+    self.toolbox.router.get('/(.*)', self.toolbox.networkOnly, {
+        origin: /disqus\.com/
+    });
+    self.toolbox.router.get('/(.*)', self.toolbox.networkOnly, {
+        origin: /disquscdn\.com/
+    });
+    // Cache all static vendor assets, e.g. fonts whose version is bind to the according url
+    self.toolbox.router.get('/(.*)', self.toolbox.cacheFirst, {
+        origin: /(fonts\.gstatic\.com|www\.google-analytics\.com|www\.googletagmanager\.com|cdnjs\.cloudflare\.com)/,
+        cache: {
+          name: staticVendorCacheName,
+          maxEntries: maxEntries
+        }
+    });
+    self.toolbox.router.get('/(.*)', self.toolbox.cacheFirst, {
+        origin: /(img\.halfrost\.com)/,
+        cache: {
+          name: staticAssetsCacheName,
+          maxEntries: maxEntries
+        }
+    });
+    self.toolbox.router.get('/content/(.*)', self.toolbox.fastest, {
+        cache: {
+          name: contentCacheName,
+          maxEntries: maxEntries
+        }
+    });
+    self.toolbox.router.get('/*', function (request, values, options) {
+        if (!request.url.match(/(\/ghost\/|\/page\/)/) && request.headers.get('accept').includes('text/html')) {
+            return self.toolbox.fastest(request, values, options);
+        } else {
+          // DevTools opening will trigger these o-i-c requests, which this SW can't handle.
+          // There's probaly more going on here, but I'd rather just ignore this problem. :)
+          // https://github.com/paulirish/caltrainschedule.io/issues/49
+          if (request.cache === 'only-if-cached' && request.mode !== 'same-origin'){
+            return;
+          } else {
+            return self.toolbox.networkOnly(request, values, options);
+          }
+        }
+        }, {
+        cache: {
+            name: contentCacheName,
+            maxEntries: maxEntries
+        }
+    });
+    // immediately activate this serviceworker
+    self.addEventListener('install', function (event) {
+        return event.waitUntil(self.skipWaiting());
+    });
+    self.addEventListener('activate', function (event) {
+        return event.waitUntil(self.clients.claim());
+    });
+})();
+//# sourceMappingURL=serviceworker-v1.js.map
diff --git a/website/themes/book/assets/sw-register.js b/website/themes/book/assets/sw-register.js
new file mode 100644
index 000000000..b67dbfdea
--- /dev/null
+++ b/website/themes/book/assets/sw-register.js
@@ -0,0 +1,7 @@
+{{- $swJS := resources.Get "sw.js" | resources.ExecuteAsTemplate "sw.js" . -}}
+if (navigator.serviceWorker) {
+  navigator.serviceWorker.register(
+    "{{ $swJS.RelPermalink }}", 
+    { scope: "{{ "/" | relURL }}" }
+  );
+}
diff --git a/website/themes/book/assets/sw.js b/website/themes/book/assets/sw.js
new file mode 100644
index 000000000..2ff11fc7f
--- /dev/null
+++ b/website/themes/book/assets/sw.js
@@ -0,0 +1,55 @@
+const cacheName = self.location.pathname
+const pages = [
+{{ if eq .Site.Params.BookServiceWorker "precache" }}
+  {{ range .Site.AllPages -}}
+  "{{ .RelPermalink }}",
+  {{ end -}}
+{{ end }}
+];
+
+self.addEventListener("install", function (event) {
+  self.skipWaiting();
+
+  caches.open(cacheName).then((cache) => {
+    return cache.addAll(pages);
+  });
+});
+
+self.addEventListener("fetch", (event) => {
+  const request = event.request;
+  if (request.method !== "GET") {
+    return;
+  }
+
+  /**
+   * @param {Response} response
+   * @returns {Promise<Response>}
+   */
+  function saveToCache(response) {
+    if (cacheable(response)) {
+      return caches
+        .open(cacheName)
+        .then((cache) => cache.put(request, response.clone()))
+        .then(() => response);
+    } else {
+      return response;
+    }
+  }
+
+  /**
+   * @param {Error} error
+   */
+  function serveFromCache(error) {
+    return caches.open(cacheName).then((cache) => cache.match(request.url));
+  }
+
+  /**
+   * @param {Response} response
+   * @returns {Boolean}
+   */
+  function cacheable(response) {
+    return response.type === "basic" && response.ok && !response.headers.has("Content-Disposition")
+  }
+
+  event.respondWith(fetch(request).then(saveToCache).catch(serveFromCache));
+});
diff --git a/website/themes/book/assets/themes/_auto.scss b/website/themes/book/assets/themes/_auto.scss
new file mode 100644
index 000000000..31d7f9ac4
--- /dev/null
+++ b/website/themes/book/assets/themes/_auto.scss
@@ -0,0 +1,9 @@
+:root {
+  @include theme-light;
+}
+
+@media (prefers-color-scheme: dark) {
+  :root {
+    @include theme-dark;
+  }
+}
diff --git a/website/themes/book/assets/themes/_dark.scss b/website/themes/book/assets/themes/_dark.scss
new file mode 100644
index 000000000..e00e38ef6
--- /dev/null
+++ b/website/themes/book/assets/themes/_dark.scss
@@ -0,0 +1,3 @@
+:root {
+  @include theme-dark;
+}
diff --git a/website/themes/book/assets/themes/_light.scss b/website/themes/book/assets/themes/_light.scss
new file mode 100644
index 000000000..8c0e34653
--- /dev/null
+++ b/website/themes/book/assets/themes/_light.scss
@@ -0,0 +1,3 @@
+:root {
+  @include theme-light;
+}
diff --git a/website/themes/book/exampleSite/assets/_custom.scss b/website/themes/book/exampleSite/assets/_custom.scss
new file mode 100644
index 000000000..9be7a1d21
--- /dev/null
+++ b/website/themes/book/exampleSite/assets/_custom.scss
@@ -0,0 +1,4 @@
+/* You can add custom styles here. */
+
+// @import "plugins/numbered";
+// @import "plugins/scrollbars";
diff --git a/website/themes/book/exampleSite/assets/_variables.scss b/website/themes/book/exampleSite/assets/_variables.scss
new file mode 100644
index 000000000..6e34d16b8
--- /dev/null
+++ b/website/themes/book/exampleSite/assets/_variables.scss
@@ -0,0 +1,3 @@
+/* You can override SASS variables here. */
+
+// @import "plugins/dark";
diff --git a/website/themes/book/exampleSite/config.toml b/website/themes/book/exampleSite/config.toml
new file mode 100644
index 000000000..dfd9a31cc
--- /dev/null
+++ b/website/themes/book/exampleSite/config.toml
@@ -0,0 +1,118 @@
+# hugo server --minify --themesDir ... --baseURL=http://0.0.0.0:1313/theme/hugo-book/
+
+baseURL = 'https://example.com/'
+title = 'Hugo Book'
+theme = 'hugo-book'
+
+# Book configuration
+disablePathToLower = true
+enableGitInfo = true
+
+# Needed for mermaid/katex shortcodes
+[markup]
+[markup.goldmark.renderer]
+  unsafe = true
+
+[markup.tableOfContents]
+  startLevel = 1
+
+# Multi-lingual mode config
+# There are different options to translate files
+# See https://gohugo.io/content-management/multilingual/#translation-by-filename
+# And https://gohugo.io/content-management/multilingual/#translation-by-content-directory
+[languages]
+[languages.en]
+  languageName = 'English'
+  contentDir = 'content'
+  weight = 1
+
+[languages.ru]
+  languageName = 'Russian'
+  contentDir = 'content.ru'
+  weight = 2
+
+[languages.zh]
+  languageName = 'Chinese'
+  contentDir = 'content.zh'
+  weight = 3
+
+[menu]
+# [[menu.before]]
+[[menu.after]]
+  name = "Github"
+  url = "https://github.com/alex-shpak/hugo-book"
+  weight = 10
+
+[[menu.after]]
+  name = "Hugo Themes"
+  url = "https://themes.gohugo.io/hugo-book/"
+  weight = 20
+
+[params]
+  # (Optional, default light) Sets color theme: light, dark or auto.
+  # Theme 'auto' switches between dark and light modes based on browser/os preferences
+  BookTheme = 'light'
+
+  # (Optional, default true) Controls table of contents visibility on right side of pages.
+  # Start and end levels can be controlled with markup.tableOfContents setting.
+  # You can also specify this parameter per page in front matter.
+  BookToC = true
+
+  # (Optional, default none) Set the path to a logo for the book. If the logo is
+  # /static/logo.png then the path would be logo.png
+  # BookLogo = 'logo.png'
+
+  # (Optional, default none) Set leaf bundle to render as side menu
+  # When not specified file structure and weights will be used
+  # BookMenuBundle = '/menu'
+
+  # (Optional, default docs) Specify root page to render child pages as menu.
+  # Page is resoled by .GetPage function: https://gohugo.io/functions/getpage/
+  # For backward compatibility you can set '*' to render all sections to menu. Acts same as '/'
+  BookSection = 'docs'
+
+  # Set source repository location.
+  # Used for 'Last Modified' and 'Edit this page' links.
+  BookRepo = 'https://github.com/alex-shpak/hugo-book'
+
+  # (Optional, default 'commit') Specifies commit portion of the link to the page's last modified
+  # commit hash for 'doc' page type.
+  # Requires 'BookRepo' param.
+  # Value used to construct a URL consisting of BookRepo/BookCommitPath/<commit-hash>
+  # Github uses 'commit', Bitbucket uses 'commits'
+  # BookCommitPath = 'commit'
+
+  # Enable "Edit this page" links for 'doc' page type.
+  # Disabled by default. Uncomment to enable. Requires 'BookRepo' param.
+  # Edit path must point to root directory of repo.
+  BookEditPath = 'edit/main/exampleSite'
+
+  # Configure the date format used on the pages
+  # - In git information
+  # - In blog posts
+  BookDateFormat = 'January 2, 2006'
+
+  # (Optional, default true) Enables search function with flexsearch,
+  # Index is built on fly, therefore it might slowdown your website.
+  # Configuration for indexing can be adjusted in i18n folder per language.
+  BookSearch = true
+
+  # (Optional, default true) Enables comments template on pages
+  # By default partals/docs/comments.html includes Disqus template
+  # See https://gohugo.io/content-management/comments/#configure-disqus
+  # Can be overwritten by same param in page frontmatter
+  BookComments = true
+
+  # /!\ This is an experimental feature, might be removed or changed at any time
+  # (Optional, experimental, default false) Enables portable links and link checks in markdown pages.
+  # Portable links meant to work with text editors and let you write markdown without {{< relref >}} shortcode
+  # Theme will print warning if page referenced in markdown does not exists.
+  BookPortableLinks = true
+
+  # /!\ This is an experimental feature, might be removed or changed at any time
+  # (Optional, experimental, default false) Enables service worker that caches visited pages and resources for offline use.
+  BookServiceWorker = true
+
+  # /!\ This is an experimental feature, might be removed or changed at any time
+  # (Optional, experimental, default false) Enables a drop-down menu for translations only if a translation is present.
+  BookTranslatedOnly = false
diff --git a/website/themes/book/exampleSite/config.yaml b/website/themes/book/exampleSite/config.yaml
new file mode 100644
index 000000000..4f439c689
--- /dev/null
+++ b/website/themes/book/exampleSite/config.yaml
@@ -0,0 +1,114 @@
+# hugo server --minify --themesDir ... --baseURL=http://0.0.0.0:1313/theme/hugo-book/
+
+baseURL: https://example.com/
+title: Hugo Book
+theme: hugo-book
+
+# Book configuration
+disablePathToLower: true
+enableGitInfo: true
+
+# Needed for mermaid/katex shortcodes
+markup:
+  goldmark:
+    renderer:
+      unsafe: true
+  tableOfContents:
+    startLevel: 1
+
+# Multi-lingual mode config
+# There are different options to translate files
+# See https://gohugo.io/content-management/multilingual/#translation-by-filename
+# And https://gohugo.io/content-management/multilingual/#translation-by-content-directory
+languages:
+  en:
+    languageName: English
+    contentDir: content
+    weight: 1
+  ru:
+    languageName: Russian
+    contentDir: content.ru
+    weight: 2
+  zh:
+    languageName: Chinese
+    contentDir: content.zh
+    weight: 3
+
+menu:
+  # before: []
+  after:
+    - name: "Github"
+      url: "https://github.com/alex-shpak/hugo-book"
+      weight: 10
+    - name: "Hugo Themes"
+      url: "https://themes.gohugo.io/hugo-book/"
+      weight: 20
+
+params:
+  # (Optional, default light) Sets color theme: light, dark or auto.
+  # Theme 'auto' switches between dark and light modes based on browser/os preferences
+  BookTheme: "light"
+
+  # (Optional, default true) Controls table of contents visibility on right side of pages.
+  # Start and end levels can be controlled with markup.tableOfContents setting.
+  # You can also specify this parameter per page in front matter.
+  BookToC: true
+
+  # (Optional, default none) Set the path to a logo for the book. If the logo is
+  # /static/logo.png then the path would be logo.png
+  # BookLogo: /logo.png
+
+  # (Optional, default none) Set leaf bundle to render as side menu
+  # When not specified file structure and weights will be used
+  # BookMenuBundle: /menu
+
+  # (Optional, default docs) Specify root page to render child pages as menu.
+  # Page is resoled by .GetPage function: https://gohugo.io/functions/getpage/
+  # For backward compatibility you can set '*' to render all sections to menu. Acts same as '/'
+  BookSection: docs
+
+  # Set source repository location.
+  # Used for 'Last Modified' and 'Edit this page' links.
+  BookRepo: https://github.com/alex-shpak/hugo-book
+
+  # (Optional, default 'commit') Specifies commit portion of the link to the page's last modified
+  # commit hash for 'doc' page type.
+  # Requires 'BookRepo' param.
+  # Value used to construct a URL consisting of BookRepo/BookCommitPath/<commit-hash>
+  # Github uses 'commit', Bitbucket uses 'commits'
+  # BookCommitPath: commit
+
+  # Enable "Edit this page" links for 'doc' page type.
+  # Disabled by default. Uncomment to enable. Requires 'BookRepo' param.
+  # Edit path must point to root directory of repo.
+  BookEditPath: edit/main/exampleSite
+
+  # Configure the date format used on the pages
+  # - In git information
+  # - In blog posts
+  BookDateFormat: "January 2, 2006"
+
+  # (Optional, default true) Enables search function with flexsearch,
+  # Index is built on fly, therefore it might slowdown your website.
+  # Configuration for indexing can be adjusted in i18n folder per language.
+  BookSearch: true
+
+  # (Optional, default true) Enables comments template on pages
+  # By default partals/docs/comments.html includes Disqus template
+  # See https://gohugo.io/content-management/comments/#configure-disqus
+  # Can be overwritten by same param in page frontmatter
+  BookComments: true
+
+  # /!\ This is an experimental feature, might be removed or changed at any time
+  # (Optional, experimental, default false) Enables portable links and link checks in markdown pages.
+  # Portable links meant to work with text editors and let you write markdown without {{< relref >}} shortcode
+  # Theme will print warning if page referenced in markdown does not exists.
+  BookPortableLinks: true
+
+  # /!\ This is an experimental feature, might be removed or changed at any time
+  # (Optional, experimental, default false) Enables service worker that caches visited pages and resources for offline use.
+  BookServiceWorker: true
+
+  # /!\ This is an experimental feature, might be removed or changed at any time
+  # (Optional, experimental, default false) Enables a drop-down menu for translations only if a translation is present.
+  BookTranslatedOnly: false
diff --git a/website/themes/book/exampleSite/content.bn/_index.md b/website/themes/book/exampleSite/content.bn/_index.md
new file mode 100644
index 000000000..85c03e006
--- /dev/null
+++ b/website/themes/book/exampleSite/content.bn/_index.md
@@ -0,0 +1,79 @@
+---
+title: ভূমিকা
+type: docs
+---
+
+# বাংলা ভাষায় শুরু করুন
+
+{{< columns >}}
+## অস্ট্রিস চিপসে ফুর্তিভা 
+
+Est in vagis et Pittheus tu arge accipiter regia iram vocatur nurus. Omnes ut
+olivae sensit **arma sorori** deducit, inesset **crudus**, ego vetuere aliis,
+modo arsit? Utinam rapta fiducia valuere litora _adicit cursu_, ad facies
+
+<--->
+
+## সুইস কোটা ভোটে
+
+Ea _furtique_ risere fratres edidit terrae magis. Colla tam mihi tenebat:
+miseram excita suadent es pecudes iam. Concilio _quam_ velatus posset ait quod
+nunc! Fragosis suae dextra geruntur functus vulgata.
+{{< /columns >}}
+
+
+## টেম্পোরার নিশি
+
+Lorem **markdownum** emicat gestu. Cannis sol pressit ducta. **Est** Idaei,
+tremens ausim se tutaeque, illi ulnis hausit, sed, lumina cutem. Quae avis
+sequens!
+
+    var panel = ram_design;
+    if (backup + system) {
+        file.readPoint = network_native;
+        sidebar_engine_device(cell_tftp_raster,
+                dual_login_paper.adf_vci.application_reader_design(
+                graphicsNvramCdma, lpi_footer_snmp, integer_model));
+    }
+    public_keyboard_docking += error.controller_gibibyte_plug.ip(4,
+            asciiPetaflops, software(supercomputer_compatible_status + 4));
+    dynamic_disk.indexModeLaptop = bufferTftpReality;
+    var export_vlog_sequence = trinitron_flowchart + supercomputer_cluster_rj(
+            -1, toolbar_powerpoint_query, -2 / multiprocessing_impression);
+
+## Locis suis novi cum suoque decidit eadem
+
+Idmoniae ripis, at aves, ali missa adest, ut _et autem_, et ab? Venit spes
+versus finis sermonibus patefecit murum nec est sine oculis. _Ille_ inmota
+macies domoque caelestia cadit tantummodo scelus procul, corde!
+
+1. Dolentem capi parte rostro alvum habentem pudor
+2. Fulgentia sanguine paret
+3. E punior consurgit lentus
+4. Vox hasta eras micantes
+
+## Facibus pharetrae indetonsusque indulsit sic incurrite foliis
+
+Nefandam et prisci palmas! Blandita cutis flectitur montis macies, te _nati_
+Latiis; turbaque inferias. Virginis tibi peracta avidusque facies caper nec, e
+at ademptae, mira.
+
+    direct *= font(inputScareware(sliHome), crossplatform.byte(
+            ppl_encryption.excel_e_rte(integratedModelModifier), timeVirtual,
+            floating_speakers.media_printer(us, yahoo, primaryPhp)));
+    friendly_metal_flatbed(cd, isoPrimaryStorage(reader), dmaMirrored);
+    if (parse_flash_cron.metalGif(1, adServiceDevice, utility)) {
+        adf -= operation_cdma_samba;
+        imapGif.switch += torrent;
+    } else {
+        pmu.disk_captcha = digital_ppp_pci + recursionTransistor(5, dram);
+        ajax_service += grayscalePythonLock;
+        google_scroll_capacity = ftp + engine_dslam_sidebar / tape - 1;
+    }
+    drive_rw = zipTftp;
+    var suffix = software_router_extension.dimm_ddr(-5,
+            kernel_digital_minisite);
+
+Vocavit toto; alas **mitis** maestus in liquidarum ab legi finitimosque dominam
+tibi subitus; Orionis vertitur nota. Currere alti etiam seroque cernitis
+innumeris miraturus amplectique collo sustinet quemque! Litora ante turba?
diff --git a/website/themes/book/exampleSite/content.ru/_index.md b/website/themes/book/exampleSite/content.ru/_index.md
new file mode 100644
index 000000000..8fa610a41
--- /dev/null
+++ b/website/themes/book/exampleSite/content.ru/_index.md
@@ -0,0 +1,79 @@
+---
+title: Введение
+type: docs
+---
+
+# Стартовая страница на русском языке
+
+{{< columns >}}
+## Astris ipse furtiva
+
+Est in vagis et Pittheus tu arge accipiter regia iram vocatur nurus. Omnes ut
+olivae sensit **arma sorori** deducit, inesset **crudus**, ego vetuere aliis,
+modo arsit? Utinam rapta fiducia valuere litora _adicit cursu_, ad facies
+
+<--->
+
+## Suis quot vota
+
+Ea _furtique_ risere fratres edidit terrae magis. Colla tam mihi tenebat:
+miseram excita suadent es pecudes iam. Concilio _quam_ velatus posset ait quod
+nunc! Fragosis suae dextra geruntur functus vulgata.
+{{< /columns >}}
+
+
+## Tempora nisi nunc
+
+Lorem **markdownum** emicat gestu. Cannis sol pressit ducta. **Est** Idaei,
+tremens ausim se tutaeque, illi ulnis hausit, sed, lumina cutem. Quae avis
+sequens!
+
+    var panel = ram_design;
+    if (backup + system) {
+        file.readPoint = network_native;
+        sidebar_engine_device(cell_tftp_raster,
+                dual_login_paper.adf_vci.application_reader_design(
+                graphicsNvramCdma, lpi_footer_snmp, integer_model));
+    }
+    public_keyboard_docking += error.controller_gibibyte_plug.ip(4,
+            asciiPetaflops, software(supercomputer_compatible_status + 4));
+    dynamic_disk.indexModeLaptop = bufferTftpReality;
+    var export_vlog_sequence = trinitron_flowchart + supercomputer_cluster_rj(
+            -1, toolbar_powerpoint_query, -2 / multiprocessing_impression);
+
+## Locis suis novi cum suoque decidit eadem
+
+Idmoniae ripis, at aves, ali missa adest, ut _et autem_, et ab? Venit spes
+versus finis sermonibus patefecit murum nec est sine oculis. _Ille_ inmota
+macies domoque caelestia cadit tantummodo scelus procul, corde!
+
+1. Dolentem capi parte rostro alvum habentem pudor
+2. Fulgentia sanguine paret
+3. E punior consurgit lentus
+4. Vox hasta eras micantes
+
+## Facibus pharetrae indetonsusque indulsit sic incurrite foliis
+
+Nefandam et prisci palmas! Blandita cutis flectitur montis macies, te _nati_
+Latiis; turbaque inferias. Virginis tibi peracta avidusque facies caper nec, e
+at ademptae, mira.
+
+    direct *= font(inputScareware(sliHome), crossplatform.byte(
+            ppl_encryption.excel_e_rte(integratedModelModifier), timeVirtual,
+            floating_speakers.media_printer(us, yahoo, primaryPhp)));
+    friendly_metal_flatbed(cd, isoPrimaryStorage(reader), dmaMirrored);
+    if (parse_flash_cron.metalGif(1, adServiceDevice, utility)) {
+        adf -= operation_cdma_samba;
+        imapGif.switch += torrent;
+    } else {
+        pmu.disk_captcha = digital_ppp_pci + recursionTransistor(5, dram);
+        ajax_service += grayscalePythonLock;
+        google_scroll_capacity = ftp + engine_dslam_sidebar / tape - 1;
+    }
+    drive_rw = zipTftp;
+    var suffix = software_router_extension.dimm_ddr(-5,
+            kernel_digital_minisite);
+
+Vocavit toto; alas **mitis** maestus in liquidarum ab legi finitimosque dominam
+tibi subitus; Orionis vertitur nota. Currere alti etiam seroque cernitis
+innumeris miraturus amplectique collo sustinet quemque! Litora ante turba?
diff --git a/website/themes/book/exampleSite/content.zh/_index.md b/website/themes/book/exampleSite/content.zh/_index.md
new file mode 100644
index 000000000..3e9982357
--- /dev/null
+++ b/website/themes/book/exampleSite/content.zh/_index.md
@@ -0,0 +1,79 @@
+---
+title: 介绍
+type: docs
+---
+
+# 中文索引页
+
+{{< columns >}}
+## Astris ipse furtiva
+
+Est in vagis et Pittheus tu arge accipiter regia iram vocatur nurus. Omnes ut
+olivae sensit **arma sorori** deducit, inesset **crudus**, ego vetuere aliis,
+modo arsit? Utinam rapta fiducia valuere litora _adicit cursu_, ad facies
+
+<--->
+
+## Suis quot vota
+
+Ea _furtique_ risere fratres edidit terrae magis. Colla tam mihi tenebat:
+miseram excita suadent es pecudes iam. Concilio _quam_ velatus posset ait quod
+nunc! Fragosis suae dextra geruntur functus vulgata.
+{{< /columns >}}
+
+
+## Tempora nisi nunc
+
+Lorem **markdownum** emicat gestu. Cannis sol pressit ducta. **Est** Idaei,
+tremens ausim se tutaeque, illi ulnis hausit, sed, lumina cutem. Quae avis
+sequens!
+
+    var panel = ram_design;
+    if (backup + system) {
+        file.readPoint = network_native;
+        sidebar_engine_device(cell_tftp_raster,
+                dual_login_paper.adf_vci.application_reader_design(
+                graphicsNvramCdma, lpi_footer_snmp, integer_model));
+    }
+    public_keyboard_docking += error.controller_gibibyte_plug.ip(4,
+            asciiPetaflops, software(supercomputer_compatible_status + 4));
+    dynamic_disk.indexModeLaptop = bufferTftpReality;
+    var export_vlog_sequence = trinitron_flowchart + supercomputer_cluster_rj(
+            -1, toolbar_powerpoint_query, -2 / multiprocessing_impression);
+
+## Locis suis novi cum suoque decidit eadem
+
+Idmoniae ripis, at aves, ali missa adest, ut _et autem_, et ab? Venit spes
+versus finis sermonibus patefecit murum nec est sine oculis. _Ille_ inmota
+macies domoque caelestia cadit tantummodo scelus procul, corde!
+
+1. Dolentem capi parte rostro alvum habentem pudor
+2. Fulgentia sanguine paret
+3. E punior consurgit lentus
+4. Vox hasta eras micantes
+
+## Facibus pharetrae indetonsusque indulsit sic incurrite foliis
+
+Nefandam et prisci palmas! Blandita cutis flectitur montis macies, te _nati_
+Latiis; turbaque inferias. Virginis tibi peracta avidusque facies caper nec, e
+at ademptae, mira.
+
+    direct *= font(inputScareware(sliHome), crossplatform.byte(
+            ppl_encryption.excel_e_rte(integratedModelModifier), timeVirtual,
+            floating_speakers.media_printer(us, yahoo, primaryPhp)));
+    friendly_metal_flatbed(cd, isoPrimaryStorage(reader), dmaMirrored);
+    if (parse_flash_cron.metalGif(1, adServiceDevice, utility)) {
+        adf -= operation_cdma_samba;
+        imapGif.switch += torrent;
+    } else {
+        pmu.disk_captcha = digital_ppp_pci + recursionTransistor(5, dram);
+        ajax_service += grayscalePythonLock;
+        google_scroll_capacity = ftp + engine_dslam_sidebar / tape - 1;
+    }
+    drive_rw = zipTftp;
+    var suffix = software_router_extension.dimm_ddr(-5,
+            kernel_digital_minisite);
+
+Vocavit toto; alas **mitis** maestus in liquidarum ab legi finitimosque dominam
+tibi subitus; Orionis vertitur nota. Currere alti etiam seroque cernitis
+innumeris miraturus amplectique collo sustinet quemque! Litora ante turba?
diff --git a/website/themes/book/exampleSite/content/_index.md b/website/themes/book/exampleSite/content/_index.md
new file mode 100644
index 000000000..6123c2829
--- /dev/null
+++ b/website/themes/book/exampleSite/content/_index.md
@@ -0,0 +1,41 @@
+---
+title: Introduction
+type: docs
+---
+
+# Acerbo datus maxime
+
+{{< columns >}}
+## Astris ipse furtiva
+
+Est in vagis et Pittheus tu arge accipiter regia iram vocatur nurus. Omnes ut
+olivae sensit **arma sorori** deducit, inesset **crudus**, ego vetuere aliis,
+modo arsit? Utinam rapta fiducia valuere litora _adicit cursu_, ad facies
+
+<--->
+
+## Suis quot vota
+
+Ea _furtique_ risere fratres edidit terrae magis. Colla tam mihi tenebat:
+miseram excita suadent es pecudes iam. Concilio _quam_ velatus posset ait quod
+nunc! Fragosis suae dextra geruntur functus vulgata.
+{{< /columns >}}
+
+
+## Tempora nisi nunc
+
+Lorem **markdownum** emicat gestu. Cannis sol pressit ducta. **Est** Idaei,
+tremens ausim se tutaeque, illi ulnis hausit, sed, lumina cutem. Quae avis
+sequens!
+
+    var panel = ram_design;
+    if (backup + system) {
+        file.readPoint = network_native;
+        sidebar_engine_device(cell_tftp_raster,
+                dual_login_paper.adf_vci.application_reader_design(
+                graphicsNvramCdma, lpi_footer_snmp, integer_model));
+    }
+
+## Locis suis novi cum suoque decidit eadem
+
+Idmoniae ripis, at aves, ali missa adest, ut _et autem_, et ab?
diff --git a/website/themes/book/exampleSite/content/docs/example/_index.md b/website/themes/book/exampleSite/content/docs/example/_index.md
new file mode 100644
index 000000000..4835b7ca0
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/example/_index.md
@@ -0,0 +1,71 @@
+---
+weight: 1
+bookFlatSection: true
+title: "Example Site"
+---
+
+# Introduction
+
+## Ferre hinnitibus erat accipitrem dixi Troiae tollens
+
+Lorem markdownum, a quoque nutu est *quodcumque mandasset* veluti. Passim
+inportuna totidemque nympha fert; repetens pendent, poenarum guttura sed vacet
+non, mortali undas. Omnis pharetramque gramen portentificisque membris servatum
+novabis fallit de nubibus atque silvas mihi. **Dixit repetitaque Quid**; verrit
+longa; sententia [mandat](http://pastor-ad.io/questussilvas) quascumque nescio
+solebat [litore](http://lacrimas-ab.net/); noctes. *Hostem haerentem* circuit
+[plenaque tamen](http://www.sine.io/in).
+
+- Pedum ne indigenae finire invergens carpebat
+- Velit posses summoque
+- De fumos illa foret
+
+## Est simul fameque tauri qua ad
+
+Locum nullus nisi vomentes. Ab Persea sermone vela, miratur aratro; eandem
+Argolicas gener.
+
+## Me sol
+
+Nec dis certa fuit socer, Nonacria **dies** manet tacitaque sibi? Sucis est
+iactata Castrumque iudex, et iactato quoque terraeque es tandem et maternos
+vittis. Lumina litus bene poenamque animos callem ne tuas in leones illam dea
+cadunt genus, et pleno nunc in quod. Anumque crescentesque sanguinis
+[progenies](http://www.late.net/alimentavirides) nuribus rustica tinguet. Pater
+omnes liquido creditis noctem.
+
+    if (mirrored(icmp_dvd_pim, 3, smbMirroredHard) != lion(clickImportQueue,
+            viralItunesBalancing, bankruptcy_file_pptp)) {
+        file += ip_cybercrime_suffix;
+    }
+    if (runtimeSmartRom == netMarketingWord) {
+        virusBalancingWin *= scriptPromptBespoke + raster(post_drive,
+                windowsSli);
+        cd = address_hertz_trojan;
+        soap_ccd.pcbServerGigahertz(asp_hardware_isa, offlinePeopleware, nui);
+    } else {
+        megabyte.api = modem_flowchart - web + syntaxHalftoneAddress;
+    }
+    if (3 < mebibyteNetworkAnimated) {
+        pharming_regular_error *= jsp_ribbon + algorithm * recycleMediaKindle(
+                dvrSyntax, cdma);
+        adf_sla *= hoverCropDrive;
+        templateNtfs = -1 - vertical;
+    } else {
+        expressionCompressionVariable.bootMulti = white_eup_javascript(
+                table_suffix);
+        guidPpiPram.tracerouteLinux += rtfTerabyteQuicktime(1,
+                managementRosetta(webcamActivex), 740874);
+    }
+    var virusTweetSsl = nullGigo;
+
+## Trepident sitimque
+
+Sentiet et ferali errorem fessam, coercet superbus, Ascaniumque in pennis
+mediis; dolor? Vidit imi **Aeacon** perfida propositos adde, tua Somni Fluctibus
+errante lustrat non.
+
+Tamen inde, vos videt e flammis Scythica parantem rupisque pectora umbras. Haec
+ficta canistris repercusso simul ego aris Dixit! Esse Fama trepidare hunc
+crescendo vigor ululasse vertice *exspatiantur* celer tepidique petita aversata
+oculis iussa est me ferro.
diff --git a/website/themes/book/exampleSite/content/docs/example/collapsed/3rd-level/4th-level.md b/website/themes/book/exampleSite/content/docs/example/collapsed/3rd-level/4th-level.md
new file mode 100644
index 000000000..aa451f19e
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/example/collapsed/3rd-level/4th-level.md
@@ -0,0 +1,12 @@
+# 4th Level of Menu
+
+## Caesorum illa tu sentit micat vestes papyriferi
+
+Inde aderam facti; Theseus vis de tauri illa peream. Oculos **uberaque** non
+regisque vobis cursuque, opus venit quam vulnera. Et maiora necemque, lege modo;
+gestanda nitidi, vero? Dum ne pectoraque testantur.
+
+Venasque repulsa Samos qui, exspectatum eram animosque hinc, [aut
+manes](http://www.creveratnon.net/apricaaetheriis), Assyrii. Cupiens auctoribus
+pariter rubet, profana magni super nocens. Vos ius sibilat inpar turba visae
+iusto! Sedes ante dum superest **extrema**.
diff --git a/website/themes/book/exampleSite/content/docs/example/collapsed/3rd-level/_index.md b/website/themes/book/exampleSite/content/docs/example/collapsed/3rd-level/_index.md
new file mode 100644
index 000000000..cc0100f3c
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/example/collapsed/3rd-level/_index.md
@@ -0,0 +1,26 @@
+# 3rd Level of Menu
+
+Nefas discordemque domino montes numen tum humili nexilibusque exit, Iove. Quae
+miror esse, scelerisque Melaneus viribus. Miseri laurus. Hoc est proposita me
+ante aliquid, aura inponere candidioribus quidque accendit bella, sumpta.
+Intravit quam erat figentem hunc, motus de fontes parvo tempestate.
+
+    iscsi_virus = pitch(json_in_on(eupViral),
+            northbridge_services_troubleshooting, personal(
+            firmware_rw.trash_rw_crm.device(interactive_gopher_personal,
+            software, -1), megabit, ergonomicsSoftware(cmyk_usb_panel,
+            mips_whitelist_duplex, cpa)));
+    if (5) {
+        managementNetwork += dma - boolean;
+        kilohertz_token = 2;
+        honeypot_affiliate_ergonomics = fiber;
+    }
+    mouseNorthbridge = byte(nybble_xmp_modem.horse_subnet(
+            analogThroughputService * graphicPoint, drop(daw_bit, dnsIntranet),
+            gateway_ospf), repository.domain_key.mouse(serverData(fileNetwork,
+            trim_duplex_file), cellTapeDirect, token_tooltip_mashup(
+            ripcordingMashup)));
+    module_it = honeypot_driver(client_cold_dvr(593902, ripping_frequency) +
+            coreLog.joystick(componentUdpLink), windows_expansion_touchscreen);
+    bashGigabit.external.reality(2, server_hardware_codec.flops.ebookSampling(
+            ciscNavigationBacklink, table + cleanDriver), indexProtocolIsp);
diff --git a/website/themes/book/exampleSite/content/docs/example/collapsed/_index.md b/website/themes/book/exampleSite/content/docs/example/collapsed/_index.md
new file mode 100644
index 000000000..e954f0877
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/example/collapsed/_index.md
@@ -0,0 +1,4 @@
+---
+bookCollapseSection: true
+weight: 20
+---
diff --git a/website/themes/book/exampleSite/content/docs/example/hidden.md b/website/themes/book/exampleSite/content/docs/example/hidden.md
new file mode 100644
index 000000000..df7cb9ebb
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/example/hidden.md
@@ -0,0 +1,52 @@
+---
+bookHidden: true
+---
+
+# This page is hidden in menu
+
+# Quondam non pater est dignior ille Eurotas
+
+## Latent te facies
+
+Lorem markdownum arma ignoscas vocavit quoque ille texit mandata mentis ultimus,
+frementes, qui in vel. Hippotades Peleus [pennas
+conscia](http://gratia.net/tot-qua.php) cuiquam Caeneus quas.
+
+- Pater demittere evincitque reddunt
+- Maxime adhuc pressit huc Danaas quid freta
+- Soror ego
+- Luctus linguam saxa ultroque prior Tatiumque inquit
+- Saepe liquitur subita superata dederat Anius sudor
+
+## Cum honorum Latona
+
+O fallor [in sustinui
+iussorum](http://www.spectataharundine.org/aquas-relinquit.html) equidem.
+Nymphae operi oris alii fronde parens dumque, in auro ait mox ingenti proxima
+iamdudum maius?
+
+    reality(burnDocking(apache_nanometer),
+            pad.property_data_programming.sectorBrowserPpga(dataMask, 37,
+            recycleRup));
+    intellectualVaporwareUser += -5 * 4;
+    traceroute_key_upnp /= lag_optical(android.smb(thyristorTftp));
+    surge_host_golden = mca_compact_device(dual_dpi_opengl, 33,
+            commerce_add_ppc);
+    if (lun_ipv) {
+        verticalExtranet(1, thumbnail_ttl, 3);
+        bar_graphics_jpeg(chipset - sector_xmp_beta);
+    }
+
+## Fronde cetera dextrae sequens pennis voce muneris
+
+Acta cretus diem restet utque; move integer, oscula non inspirat, noctisque
+scelus! Nantemque in suas vobis quamvis, et labori!
+
+    var runtimeDiskCompiler = home - array_ad_software;
+    if (internic > disk) {
+        emoticonLockCron += 37 + bps - 4;
+        wan_ansi_honeypot.cardGigaflops = artificialStorageCgi;
+        simplex -= downloadAccess;
+    }
+    var volumeHardeningAndroid = pixel + tftp + onProcessorUnmount;
+    sector(memory(firewire + interlaced, wired));
\ No newline at end of file
diff --git a/website/themes/book/exampleSite/content/docs/example/table-of-contents/_index.md b/website/themes/book/exampleSite/content/docs/example/table-of-contents/_index.md
new file mode 100644
index 000000000..c7ee0d872
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/example/table-of-contents/_index.md
@@ -0,0 +1,85 @@
+---
+weight: 10
+---
+
+# Ubi loqui
+
+## Mentem genus facietque salire tempus bracchia
+
+Lorem markdownum partu paterno Achillem. Habent amne generosi aderant ad pellem
+nec erat sustinet merces columque haec et, dixit minus nutrit accipiam subibis
+subdidit. Temeraria servatum agros qui sed fulva facta. Primum ultima, dedit,
+suo quisque linguae medentes fixo: tum petis.
+
+## Rapit vocant si hunc siste adspice
+
+Ora precari Patraeque Neptunia, dixit Danae [Cithaeron
+armaque](http://mersis-an.org/litoristum) maxima in **nati Coniugis** templis
+fluidove. Effugit usus nec ingreditur agmen *ac manus* conlato. Nullis vagis
+nequiquam vultibus aliquos altera *suum venis* teneas fretum. Armos [remotis
+hoc](http://tutum.io/me) sine ferrea iuncta quam!
+
+## Locus fuit caecis
+
+Nefas discordemque domino montes numen tum humili nexilibusque exit, Iove. Quae
+miror esse, scelerisque Melaneus viribus. Miseri laurus. Hoc est proposita me
+ante aliquid, aura inponere candidioribus quidque accendit bella, sumpta.
+Intravit quam erat figentem hunc, motus de fontes parvo tempestate.
+
+    iscsi_virus = pitch(json_in_on(eupViral),
+            northbridge_services_troubleshooting, personal(
+            firmware_rw.trash_rw_crm.device(interactive_gopher_personal,
+            software, -1), megabit, ergonomicsSoftware(cmyk_usb_panel,
+            mips_whitelist_duplex, cpa)));
+    if (5) {
+        managementNetwork += dma - boolean;
+        kilohertz_token = 2;
+        honeypot_affiliate_ergonomics = fiber;
+    }
+    mouseNorthbridge = byte(nybble_xmp_modem.horse_subnet(
+            analogThroughputService * graphicPoint, drop(daw_bit, dnsIntranet),
+            gateway_ospf), repository.domain_key.mouse(serverData(fileNetwork,
+            trim_duplex_file), cellTapeDirect, token_tooltip_mashup(
+            ripcordingMashup)));
+    module_it = honeypot_driver(client_cold_dvr(593902, ripping_frequency) +
+            coreLog.joystick(componentUdpLink), windows_expansion_touchscreen);
+    bashGigabit.external.reality(2, server_hardware_codec.flops.ebookSampling(
+            ciscNavigationBacklink, table + cleanDriver), indexProtocolIsp);
+
+## Placabilis coactis nega ingemuit ignoscat nimia non
+
+Frontis turba. Oculi gravis est Delphice; *inque praedaque* sanguine manu non.
+
+    if (ad_api) {
+        zif += usb.tiffAvatarRate(subnet, digital_rt) + exploitDrive;
+        gigaflops(2 - bluetooth, edi_asp_memory.gopher(queryCursor, laptop),
+                panel_point_firmware);
+        spyware_bash.statePopApplet = express_netbios_digital(
+                insertion_troubleshooting.brouter(recordFolderUs), 65);
+    }
+    recursionCoreRay = -5;
+    if (hub == non) {
+        portBoxVirus = soundWeb(recursive_card(rwTechnologyLeopard),
+                font_radcab, guidCmsScalable + reciprocalMatrixPim);
+        left.bug = screenshot;
+    } else {
+        tooltipOpacity = raw_process_permalink(webcamFontUser, -1);
+        executable_router += tape;
+    }
+    if (tft) {
+        bandwidthWeb *= social_page;
+    } else {
+        regular += 611883;
+        thumbnail /= system_lag_keyboard;
+    }
+
+## Caesorum illa tu sentit micat vestes papyriferi
+
+Inde aderam facti; Theseus vis de tauri illa peream. Oculos **uberaque** non
+regisque vobis cursuque, opus venit quam vulnera. Et maiora necemque, lege modo;
+gestanda nitidi, vero? Dum ne pectoraque testantur.
+
+Venasque repulsa Samos qui, exspectatum eram animosque hinc, [aut
+manes](http://www.creveratnon.net/apricaaetheriis), Assyrii. Cupiens auctoribus
+pariter rubet, profana magni super nocens. Vos ius sibilat inpar turba visae
+iusto! Sedes ante dum superest **extrema**.
diff --git a/website/themes/book/exampleSite/content/docs/example/table-of-contents/with-toc.md b/website/themes/book/exampleSite/content/docs/example/table-of-contents/with-toc.md
new file mode 100644
index 000000000..5345c668a
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/example/table-of-contents/with-toc.md
@@ -0,0 +1,64 @@
+---
+title: With ToC
+weight: 1
+---
+# Caput vino delphine in tamen vias
+
+## Cognita laeva illo fracta
+
+Lorem markdownum pavent auras, surgit nunc cingentibus libet **Laomedonque que**
+est. Pastor [An](http://est.org/ire.aspx) arbor filia foedat, ne [fugit
+aliter](http://www.indiciumturbam.org/moramquid.php), per. Helicona illas et
+callida neptem est *Oresitrophos* caput, dentibus est venit. Tenet reddite
+[famuli](http://www.antro-et.net/) praesentem fortibus, quaeque vis foret si
+frondes *gelidos* gravidae circumtulit [inpulit armenta
+nativum](http://incurvasustulit.io/illi-virtute.html).
+
+1. Te at cruciabere vides rubentis manebo
+2. Maturuit in praetemptat ruborem ignara postquam habitasse
+3. Subitarum supplevit quoque fontesque venabula spretis modo
+4. Montis tot est mali quasque gravis
+5. Quinquennem domus arsit ipse
+6. Pellem turis pugnabant locavit
+
+## Natus quaerere
+
+Pectora et sine mulcere, coniuge dum tincta incurvae. Quis iam; est dextra
+Peneosque, metuis a verba, primo. Illa sed colloque suis: magno: gramen, aera
+excutiunt concipit.
+
+> Phrygiae petendo suisque extimuit, super, pars quod audet! Turba negarem.
+> Fuerat attonitus; et dextra retinet sidera ulnas undas instimulat vacuae
+> generis? *Agnus* dabat et ignotis dextera, sic tibi pacis **feriente at mora**
+> euhoeque *comites hostem* vestras Phineus. Vultuque sanguine dominoque [metuit
+> risi](http://iuvat.org/eundem.php) fama vergit summaque meus clarissimus
+> artesque tinguebat successor nominis cervice caelicolae.
+
+## Limitibus misere sit
+
+Aurea non fata repertis praerupit feruntur simul, meae hosti lentaque *citius
+levibus*, cum sede dixit, Phaethon texta. *Albentibus summos* multifidasque
+iungitur loquendi an pectore, mihi ursaque omnia adfata, aeno parvumque in animi
+perlucentes. Epytus agis ait vixque clamat ornum adversam spondet, quid sceptra
+ipsum **est**. Reseret nec; saeva suo passu debentia linguam terga et aures et
+cervix [de](http://www.amnem.io/pervenit.aspx) ubera. Coercet gelidumque manus,
+doluit volvitur induta?
+
+## Enim sua
+
+Iuvenilior filia inlustre templa quidem herbis permittat trahens huic. In
+cruribus proceres sole crescitque *fata*, quos quos; merui maris se non tamen
+in, mea.
+
+## Germana aves pignus tecta
+
+Mortalia rudibusque caelum cognosceret tantum aquis redito felicior texit, nec,
+aris parvo acre. Me parum contulerant multi tenentem, gratissime suis; vultum tu
+occupat deficeret corpora, sonum. E Actaea inplevit Phinea concepit nomenque
+potest sanguine captam nulla et, in duxisses campis non; mercede. Dicere cur
+Leucothoen obitum?
+
+Postibus mittam est *nubibus principium pluma*, exsecratur facta et. Iunge
+Mnemonidas pallamque pars; vere restitit alis flumina quae **quoque**, est
+ignara infestus Pyrrha. Di ducis terris maculatum At sede praemia manes
+nullaque!
diff --git a/website/themes/book/exampleSite/content/docs/example/table-of-contents/without-toc.md b/website/themes/book/exampleSite/content/docs/example/table-of-contents/without-toc.md
new file mode 100644
index 000000000..9b1631883
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/example/table-of-contents/without-toc.md
@@ -0,0 +1,59 @@
+---
+title: Without ToC
+weight: 2
+bookToc: false
+---
+
+# At me ipso nepotibus nunc celebratior genus
+
+## Tanto oblite
+
+Lorem markdownum pectora novis patenti igne sua opus aurae feras materiaque
+illic demersit imago et aristas questaque posset. Vomit quoque suo inhaesuro
+clara. Esse cumque, per referri triste. Ut exponit solisque communis in tendens
+vincetis agisque iamque huic bene ante vetat omina Thebae rates. Aeacus servat
+admonitu concidit, ad resimas vultus et rugas vultu **dignamque** Siphnon.
+
+Quam iugulum regia simulacra, plus meruit humo pecorumque haesit, ab discedunt
+dixit: ritu pharetramque. Exul Laurenti orantem modo, per densum missisque labor
+manibus non colla unum, obiectat. Tu pervia collo, fessus quae Cretenque Myconon
+crate! Tegumenque quae invisi sudore per vocari quaque plus ventis fluidos. Nodo
+perque, fugisse pectora sorores.
+
+## Summe promissa supple vadit lenius
+
+Quibus largis latebris aethera versato est, ait sentiat faciemque. Aequata alis
+nec Caeneus exululat inclite corpus est, ire **tibi** ostendens et tibi. Rigent
+et vires dique possent lumina; **eadem** dixit poma funeribus paret et felix
+reddebant ventis utile lignum.
+
+1. Remansit notam Stygia feroxque
+2. Et dabit materna
+3. Vipereas Phrygiaeque umbram sollicito cruore conlucere suus
+4. Quarum Elis corniger
+5. Nec ieiunia dixit
+
+Vertitur mos ortu ramosam contudit dumque; placabat ac lumen. Coniunx Amoris
+spatium poenamque cavernis Thebae Pleiadasque ponunt, rapiare cum quae parum
+nimium rima.
+
+## Quidem resupinus inducto solebat una facinus quae
+
+Credulitas iniqua praepetibus paruit prospexit, voce poena, sub rupit sinuatur,
+quin suum ventorumque arcadiae priori. Soporiferam erat formamque, fecit,
+invergens, nymphae mutat fessas ait finge.
+
+1. Baculum mandataque ne addere capiti violentior
+2. Altera duas quam hoc ille tenues inquit
+3. Sicula sidereus latrantis domoque ratae polluit comites
+4. Possit oro clausura namque se nunc iuvenisque
+5. Faciem posuit
+6. Quodque cum ponunt novercae nata vestrae aratra
+
+Ite extrema Phrygiis, patre dentibus, tonso perculit, enim blanda, manibus fide
+quos caput armis, posse! Nocendo fas Alcyonae lacertis structa ferarum manus
+fulmen dubius, saxa caelum effuge extremis fixum tumor adfecit **bella**,
+potentes? Dum nec insidiosa tempora tegit
+[spirarunt](http://mihiferre.net/iuvenes-peto.html). Per lupi pars foliis,
+porreximus humum negant sunt subposuere Sidone steterant auro. Memoraverit sine:
+ferrum idem Orion caelum heres gerebat fixis?
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/_index.md b/website/themes/book/exampleSite/content/docs/shortcodes/_index.md
new file mode 100644
index 000000000..9bb0430cb
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/_index.md
@@ -0,0 +1,3 @@
+---
+bookFlatSection: true
+---
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/buttons.md b/website/themes/book/exampleSite/content/docs/shortcodes/buttons.md
new file mode 100644
index 000000000..c2ef1e755
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/buttons.md
@@ -0,0 +1,13 @@
+# Buttons
+
+Buttons are styled links that can lead to local page or external link.
+
+## Example
+
+```tpl
+{{</* button relref="/" [class="..."] */>}}Get Home{{</* /button */>}}
+{{</* button href="https://github.com/alex-shpak/hugo-book" */>}}Contribute{{</* /button */>}}
+```
+
+{{< button relref="/" >}}Get Home{{< /button >}}
+{{< button href="https://github.com/alex-shpak/hugo-book" >}}Contribute{{< /button >}}
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/columns.md b/website/themes/book/exampleSite/content/docs/shortcodes/columns.md
new file mode 100644
index 000000000..0b8fde836
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/columns.md
@@ -0,0 +1,45 @@
+# Columns
+
+Columns help organize shorter pieces of content horizontally for readability.
+
+
+```html
+{{</* columns */>}} <!-- begin columns block -->
+# Left Content
+Lorem markdownum insigne...
+
+<---> <!-- magic separator, between columns -->
+
+# Mid Content
+Lorem markdownum insigne...
+
+<---> <!-- magic separator, between columns -->
+
+# Right Content
+Lorem markdownum insigne...
+{{</* /columns */>}}
+```
+
+## Example
+
+{{< columns >}}
+## Left Content
+Lorem markdownum insigne. Olympo signis Delphis! Retexi Nereius nova develat
+stringit, frustra Saturnius uteroque inter! Oculis non ritibus Telethusa
+protulit, sed sed aere valvis inhaesuro Pallas animam: qui _quid_, ignes.
+Miseratus fonte Ditis conubia.
+
+<--->
+
+## Mid Content
+Lorem markdownum insigne. Olympo signis Delphis! Retexi Nereius nova develat
+stringit, frustra Saturnius uteroque inter!
+
+<--->
+
+## Right Content
+Lorem markdownum insigne. Olympo signis Delphis! Retexi Nereius nova develat
+stringit, frustra Saturnius uteroque inter! Oculis non ritibus Telethusa
+protulit, sed sed aere valvis inhaesuro Pallas animam: qui _quid_, ignes.
+Miseratus fonte Ditis conubia.
+{{< /columns >}}
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/details.md b/website/themes/book/exampleSite/content/docs/shortcodes/details.md
new file mode 100644
index 000000000..248bafd97
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/details.md
@@ -0,0 +1,22 @@
+# Details
+
+Details shortcode is a helper for `details` html5 element. It is going to replace `expand` shortcode.
+
+## Example
+```tpl
+{{</* details "Title" [open] */>}}
+## Markdown content
+Lorem markdownum insigne...
+{{</* /details */>}}
+```
+```tpl
+{{</* details title="Title" open=true */>}}
+## Markdown content
+Lorem markdownum insigne...
+{{</* /details */>}}
+```
+
+{{< details "Title" open >}}
+## Markdown content
+Lorem markdownum insigne...
+{{< /details >}}
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/expand.md b/website/themes/book/exampleSite/content/docs/shortcodes/expand.md
new file mode 100644
index 000000000..c62520f3f
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/expand.md
@@ -0,0 +1,35 @@
+# Expand
+
+Expand shortcode can help to decrease clutter on screen by hiding part of text. Expand content by clicking on it.
+
+## Example
+### Default
+
+```tpl
+{{</* expand */>}}
+## Markdown content
+Lorem markdownum insigne...
+{{</* /expand */>}}
+```
+
+{{< expand >}}
+## Markdown content
+Lorem markdownum insigne...
+{{< /expand >}}
+
+### With Custom Label
+
+```tpl
+{{</* expand "Custom Label" "..." */>}}
+## Markdown content
+Lorem markdownum insigne...
+{{</* /expand */>}}
+```
+
+{{< expand "Custom Label" "..." >}}
+## Markdown content
+Lorem markdownum insigne. Olympo signis Delphis! Retexi Nereius nova develat
+stringit, frustra Saturnius uteroque inter! Oculis non ritibus Telethusa
+protulit, sed sed aere valvis inhaesuro Pallas animam: qui _quid_, ignes.
+Miseratus fonte Ditis conubia.
+{{< /expand >}}
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/hints.md b/website/themes/book/exampleSite/content/docs/shortcodes/hints.md
new file mode 100644
index 000000000..3477113de
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/hints.md
@@ -0,0 +1,32 @@
+# Hints
+
+Hint shortcode can be used as hint/alerts/notification block.  
+There are 3 colors to choose: `info`, `warning` and `danger`.
+
+```tpl
+{{</* hint [info|warning|danger] */>}}
+**Markdown content**  
+Lorem markdownum insigne. Olympo signis Delphis! Retexi Nereius nova develat
+stringit, frustra Saturnius uteroque inter! Oculis non ritibus Telethusa
+{{</* /hint */>}}
+```
+
+## Example
+
+{{< hint info >}}
+**Markdown content**  
+Lorem markdownum insigne. Olympo signis Delphis! Retexi Nereius nova develat
+stringit, frustra Saturnius uteroque inter! Oculis non ritibus Telethusa
+{{< /hint >}}
+
+{{< hint warning >}}
+**Markdown content**  
+Lorem markdownum insigne. Olympo signis Delphis! Retexi Nereius nova develat
+stringit, frustra Saturnius uteroque inter! Oculis non ritibus Telethusa
+{{< /hint >}}
+
+{{< hint danger >}}
+**Markdown content**  
+Lorem markdownum insigne. Olympo signis Delphis! Retexi Nereius nova develat
+stringit, frustra Saturnius uteroque inter! Oculis non ritibus Telethusa
+{{< /hint >}}
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/katex.md b/website/themes/book/exampleSite/content/docs/shortcodes/katex.md
new file mode 100644
index 000000000..4beeaa004
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/katex.md
@@ -0,0 +1,28 @@
+# KaTeX
+
+KaTeX shortcode let you render math typesetting in markdown document. See [KaTeX](https://katex.org/)
+
+## Example
+{{< columns >}}
+
+```latex
+{{</* katex [display] [class="text-center"]  */>}}
+f(x) = \int_{-\infty}^\infty\hat f(\xi)\,e^{2 \pi i \xi x}\,d\xi
+{{</* /katex */>}}
+```
+
+<--->
+
+{{< katex display >}}
+f(x) = \int_{-\infty}^\infty\hat f(\xi)\,e^{2 \pi i \xi x}\,d\xi
+{{< /katex >}}
+
+{{< /columns >}}
+
+## Display Mode Example
+
+Here is some inline example: {{< katex >}}\pi(x){{< /katex >}}, rendered in the same line. And below is `display` example, having `display: block`
+{{< katex display >}}
+f(x) = \int_{-\infty}^\infty\hat f(\xi)\,e^{2 \pi i \xi x}\,d\xi
+{{< /katex >}}
+Text continues here.
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/mermaid.md b/website/themes/book/exampleSite/content/docs/shortcodes/mermaid.md
new file mode 100644
index 000000000..c37e6b20e
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/mermaid.md
@@ -0,0 +1,46 @@
+# Mermaid Chart
+
+[Mermaid](https://mermaidjs.github.io/) is library for generating svg charts and diagrams from text.
+
+{{< hint info >}}
+**Override Mermaid Initialization Config**
+
+To override the [initialization config](https://mermaid-js.github.io/mermaid/#/Setup) for Mermaid,
+create a `mermaid.json` file in your `assets` folder!
+{{< /hint >}}
+
+## Example
+
+{{< columns >}}
+```tpl
+{{</* mermaid [class="text-center"]*/>}}
+sequenceDiagram
+    Alice->>Bob: Hello Bob, how are you?
+    alt is sick
+        Bob->>Alice: Not so good :(
+    else is well
+        Bob->>Alice: Feeling fresh like a daisy
+    end
+    opt Extra response
+        Bob->>Alice: Thanks for asking
+    end
+{{</* /mermaid */>}}
+```
+
+<--->
+
+{{< mermaid >}}
+sequenceDiagram
+    Alice->>Bob: Hello Bob, how are you?
+    alt is sick
+        Bob->>Alice: Not so good :(
+    else is well
+        Bob->>Alice: Feeling fresh like a daisy
+    end
+    opt Extra response
+        Bob->>Alice: Thanks for asking
+    end
+{{< /mermaid >}}
+
+{{< /columns >}}
+
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/section/_index.md b/website/themes/book/exampleSite/content/docs/shortcodes/section/_index.md
new file mode 100644
index 000000000..bd5db38b3
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/section/_index.md
@@ -0,0 +1,15 @@
+---
+bookCollapseSection: true
+---
+
+# Section
+
+Section renders pages in section as definition list, using title and description.
+
+## Example
+
+```tpl
+{{</* section */>}}
+```
+
+{{<section>}}
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/section/page1.md b/website/themes/book/exampleSite/content/docs/shortcodes/section/page1.md
new file mode 100644
index 000000000..960800143
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/section/page1.md
@@ -0,0 +1 @@
+# Page 1
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/section/page2.md b/website/themes/book/exampleSite/content/docs/shortcodes/section/page2.md
new file mode 100644
index 000000000..f310be332
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/section/page2.md
@@ -0,0 +1 @@
+# Page 2
diff --git a/website/themes/book/exampleSite/content/docs/shortcodes/tabs.md b/website/themes/book/exampleSite/content/docs/shortcodes/tabs.md
new file mode 100644
index 000000000..096892c67
--- /dev/null
+++ b/website/themes/book/exampleSite/content/docs/shortcodes/tabs.md
@@ -0,0 +1,50 @@
+# Tabs
+
+Tabs let you organize content by context, for example installation instructions for each supported platform.
+
+```tpl
+{{</* tabs "uniqueid" */>}}
+{{</* tab "MacOS" */>}} # MacOS Content {{</* /tab */>}}
+{{</* tab "Linux" */>}} # Linux Content {{</* /tab */>}}
+{{</* tab "Windows" */>}} # Windows Content {{</* /tab */>}}
+{{</* /tabs */>}}
+```
+
+## Example
+
+{{< tabs "uniqueid" >}}
+{{< tab "MacOS" >}}
+# MacOS
+
+This is tab **MacOS** content.
+
+Lorem markdownum insigne. Olympo signis Delphis! Retexi Nereius nova develat
+stringit, frustra Saturnius uteroque inter! Oculis non ritibus Telethusa
+protulit, sed sed aere valvis inhaesuro Pallas animam: qui _quid_, ignes.
+Miseratus fonte Ditis conubia.
+{{< /tab >}}
+
+{{< tab "Linux" >}}
+
+# Linux
+
+This is tab **Linux** content.
+
+Lorem markdownum insigne. Olympo signis Delphis! Retexi Nereius nova develat
+stringit, frustra Saturnius uteroque inter! Oculis non ritibus Telethusa
+protulit, sed sed aere valvis inhaesuro Pallas animam: qui _quid_, ignes.
+Miseratus fonte Ditis conubia.
+{{< /tab >}}
+
+{{< tab "Windows" >}}
+
+# Windows
+
+This is tab **Windows** content.
+
+Lorem markdownum insigne. Olympo signis Delphis! Retexi Nereius nova develat
+stringit, frustra Saturnius uteroque inter! Oculis non ritibus Telethusa
+protulit, sed sed aere valvis inhaesuro Pallas animam: qui _quid_, ignes.
+Miseratus fonte Ditis conubia.
+{{< /tab >}}
+{{< /tabs >}}
diff --git a/website/themes/book/exampleSite/content/menu/index.md b/website/themes/book/exampleSite/content/menu/index.md
new file mode 100644
index 000000000..810bcfd1c
--- /dev/null
+++ b/website/themes/book/exampleSite/content/menu/index.md
@@ -0,0 +1,22 @@
+---
+headless: true
+---
+
+- [**Example Site**]({{< relref "/docs/example" >}})
+- [Table of Contents]({{< relref "/docs/example/table-of-contents" >}})
+  - [With ToC]({{< relref "/docs/example/table-of-contents/with-toc" >}})
+  - [Without ToC]({{< relref "/docs/example/table-of-contents/without-toc" >}})
+- [Collapsed]({{< relref "/docs/example/collapsed" >}})
+  - [3rd]({{< relref "/docs/example/collapsed/3rd-level" >}})
+    - [4th]({{< relref "/docs/example/collapsed/3rd-level/4th-level" >}})
+<br />
+
+- **Shortcodes**
+- [Buttons]({{< relref "/docs/shortcodes/buttons" >}})
+- [Columns]({{< relref "/docs/shortcodes/columns" >}})
+- [Expand]({{< relref "/docs/shortcodes/expand" >}})
+- [Hints]({{< relref "/docs/shortcodes/hints" >}})
+- [Katex]({{< relref "/docs/shortcodes/katex" >}})
+- [Mermaid]({{< relref "/docs/shortcodes/mermaid" >}})
+- [Tabs]({{< relref "/docs/shortcodes/tabs" >}})
+<br />
diff --git a/website/themes/book/exampleSite/content/posts/_index.md b/website/themes/book/exampleSite/content/posts/_index.md
new file mode 100644
index 000000000..001ae24e0
--- /dev/null
+++ b/website/themes/book/exampleSite/content/posts/_index.md
@@ -0,0 +1,7 @@
+---
+menu:
+  after:
+    name: blog
+    weight: 5
+title: Blog
+---
diff --git a/website/themes/book/exampleSite/content/posts/creating-a-new-theme.md b/website/themes/book/exampleSite/content/posts/creating-a-new-theme.md
new file mode 100644
index 000000000..f8230a170
--- /dev/null
+++ b/website/themes/book/exampleSite/content/posts/creating-a-new-theme.md
@@ -0,0 +1,1150 @@
+---
+author: "Michael Henderson"
+date: 2014-09-28
+linktitle: Creating a New Theme
+menu:
+  main:
+    parent: tutorials
+next: /tutorials/github-pages-blog
+prev: /tutorials/automated-deployments
+title: Creating a New Theme
+weight: 10
+---
+
+
+## Introduction
+
+This tutorial will show you how to create a simple theme in Hugo. I assume that you are familiar with HTML, the bash command line, and that you are comfortable using Markdown to format content. I'll explain how Hugo uses templates and how you can organize your templates to create a theme. I won't cover using CSS to style your theme.
+
+We'll start with creating a new site with a very basic template. Then we'll add in a few pages and posts. With small variations on that, you will be able to create many different types of web sites.
+
+In this tutorial, commands that you enter will start with the "$" prompt. The output will follow. Lines that start with "#" are comments that I've added to explain a point. When I show updates to a file, the ":wq" on the last line means to save the file.
+
+Here's an example:
+
+```
+## this is a comment
+$ echo this is a command
+this is a command
+
+## edit the file
+$ vi foo.md
++++
+date = "2014-09-28"
+title = "creating a new theme"
++++
+
+bah and humbug
+:wq
+
+## show it
+$ cat foo.md
++++
+date = "2014-09-28"
+title = "creating a new theme"
++++
+
+bah and humbug
+$
+```
+
+
+## Some Definitions
+
+There are a few concepts that you need to understand before creating a theme.
+
+### Skins
+
+Skins are the files responsible for the look and feel of your site. It’s the CSS that controls colors and fonts, it’s the Javascript that determines actions and reactions. It’s also the rules that Hugo uses to transform your content into the HTML that the site will serve to visitors.
+
+You have two ways to create a skin. The simplest way is to create it in the ```layouts/``` directory. If you do, then you don’t have to worry about configuring Hugo to recognize it. The first place that Hugo will look for rules and files is in the ```layouts/``` directory so it will always find the skin.
+
+Your second choice is to create it in a sub-directory of the ```themes/``` directory. If you do, then you must always tell Hugo where to search for the skin. It’s extra work, though, so why bother with it?
+
+The difference between creating a skin in ```layouts/``` and creating it in ```themes/``` is very subtle. A skin in ```layouts/``` can’t be customized without updating the templates and static files that it is built from. A skin created in ```themes/```, on the other hand, can be and that makes it easier for other people to use it.
+
+The rest of this tutorial will call a skin created in the ```themes/``` directory a theme.
+
+Note that you can use this tutorial to create a skin in the ```layouts/``` directory if you wish to. The main difference will be that you won’t need to update the site’s configuration file to use a theme.
+
+### The Home Page
+
+The home page, or landing page, is the first page that many visitors to a site see. It is the index.html file in the root directory of the web site. Since Hugo writes files to the public/ directory, our home page is public/index.html.
+
+### Site Configuration File
+
+When Hugo runs, it looks for a configuration file that contains settings that override default values for the entire site. The file can use TOML, YAML, or JSON. I prefer to use TOML for my configuration files. If you prefer to use JSON or YAML, you’ll need to translate my examples. You’ll also need to change the name of the file since Hugo uses the extension to determine how to process it.
+
+Hugo translates Markdown files into HTML. By default, Hugo expects to find Markdown files in your ```content/``` directory and template files in your ```themes/``` directory. It will create HTML files in your ```public/``` directory. You can change this by specifying alternate locations in the configuration file.
+
+### Content
+
+Content is stored in text files that contain two sections. The first section is the “front matter,” which is the meta-information on the content. The second section contains Markdown that will be converted to HTML.
+
+#### Front Matter
+
+The front matter is information about the content. Like the configuration file, it can be written in TOML, YAML, or JSON. Unlike the configuration file, Hugo doesn’t use the file’s extension to know the format. It looks for markers to signal the type. TOML is surrounded by “`+++`”, YAML by “`---`”, and JSON is enclosed in curly braces. I prefer to use TOML, so you’ll need to translate my examples if you prefer YAML or JSON.
+
+The information in the front matter is passed into the template before the content is rendered into HTML.
+
+#### Markdown
+
+Content is written in Markdown which makes it easier to create the content. Hugo runs the content through a Markdown engine to create the HTML which will be written to the output file.
+
+### Template Files
+
+Hugo uses template files to render content into HTML. Template files are a bridge between the content and presentation. Rules in the template define what content is published, where it's published to, and how it will rendered to the HTML file. The template guides the presentation by specifying the style to use.
+
+There are three types of templates: single, list, and partial. Each type takes a bit of content as input and transforms it based on the commands in the template.
+
+Hugo uses its knowledge of the content to find the template file used to render the content. If it can’t find a template that is an exact match for the content, it will shift up a level and search from there. It will continue to do so until it finds a matching template or runs out of templates to try. If it can’t find a template, it will use the default template for the site.
+
+Please note that you can use the front matter to influence Hugo’s choice of templates.
+
+#### Single Template
+
+A single template is used to render a single piece of content. For example, an article or post would be a single piece of content and use a single template.
+
+#### List Template
+
+A list template renders a group of related content. That could be a summary of recent postings or all articles in a category. List templates can contain multiple groups.
+
+The homepage template is a special type of list template. Hugo assumes that the home page of your site will act as the portal for the rest of the content in the site.
+
+#### Partial Template
+
+A partial template is a template that can be included in other templates. Partial templates must be called using the “partial” template command. They are very handy for rolling up common behavior. For example, your site may have a banner that all pages use. Instead of copying the text of the banner into every single and list template, you could create a partial with the banner in it. That way if you decide to change the banner, you only have to change the partial template.
+
+## Create a New Site
+
+Let's use Hugo to create a new web site. I'm a Mac user, so I'll create mine in my home directory, in the Sites folder. If you're using Linux, you might have to create the folder first.
+
+The "new site" command will create a skeleton of a site. It will give you the basic directory structure and a useable configuration file.
+
+```
+$ hugo new site ~/Sites/zafta
+$ cd ~/Sites/zafta
+$ ls -l
+total 8
+drwxr-xr-x  7 quoha  staff  238 Sep 29 16:49 .
+drwxr-xr-x  3 quoha  staff  102 Sep 29 16:49 ..
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 archetypes
+-rw-r--r--  1 quoha  staff   82 Sep 29 16:49 config.toml
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 content
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 layouts
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 static
+$
+```
+
+Take a look in the content/ directory to confirm that it is empty.
+
+The other directories (archetypes/, layouts/, and static/) are used when customizing a theme. That's a topic for a different tutorial, so please ignore them for now.
+
+### Generate the HTML For the New Site
+
+Running the `hugo` command with no options will read all the available content and generate the HTML files. It will also copy all static files (that's everything that's not content). Since we have an empty site, it won't do much, but it will do it very quickly.
+
+```
+$ hugo --verbose
+INFO: 2014/09/29 Using config file: config.toml
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/static/ to /Users/quoha/Sites/zafta/public/
+WARN: 2014/09/29 Unable to locate layout: [index.html _default/list.html _default/single.html]
+WARN: 2014/09/29 Unable to locate layout: [404.html]
+0 draft content 
+0 future content 
+0 pages created 
+0 tags created
+0 categories created
+in 2 ms
+$ 
+```
+
+The "`--verbose`" flag gives extra information that will be helpful when we build the template. Every line of the output that starts with "INFO:" or "WARN:" is present because we used that flag. The lines that start with "WARN:" are warning messages. We'll go over them later.
+
+We can verify that the command worked by looking at the directory again.
+
+```
+$ ls -l
+total 8
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 archetypes
+-rw-r--r--  1 quoha  staff   82 Sep 29 16:49 config.toml
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 content
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 layouts
+drwxr-xr-x  4 quoha  staff  136 Sep 29 17:02 public
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 static
+$
+```
+
+See that new public/ directory? Hugo placed all generated content there. When you're ready to publish your web site, that's the place to start. For now, though, let's just confirm that we have what we'd expect from a site with no content.
+
+```
+$ ls -l public
+total 16
+-rw-r--r--  1 quoha  staff  416 Sep 29 17:02 index.xml
+-rw-r--r--  1 quoha  staff  262 Sep 29 17:02 sitemap.xml
+$ 
+```
+
+Hugo created two XML files, which is standard, but there are no HTML files.
+
+
+
+### Test the New Site
+
+Verify that you can run the built-in web server. It will dramatically shorten your development cycle if you do. Start it by running the "server" command. If it is successful, you will see output similar to the following:
+
+```
+$ hugo server --verbose
+INFO: 2014/09/29 Using config file: /Users/quoha/Sites/zafta/config.toml
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/static/ to /Users/quoha/Sites/zafta/public/
+WARN: 2014/09/29 Unable to locate layout: [index.html _default/list.html _default/single.html]
+WARN: 2014/09/29 Unable to locate layout: [404.html]
+0 draft content 
+0 future content 
+0 pages created 
+0 tags created
+0 categories created
+in 2 ms
+Serving pages from /Users/quoha/Sites/zafta/public
+Web Server is available at http://localhost:1313
+Press Ctrl+C to stop
+```
+
+Connect to the listed URL (it's on the line that starts with "Web Server"). If everything is working correctly, you should get a page that shows the following:
+
+```
+index.xml
+sitemap.xml
+```
+
+That's a listing of your public/ directory. Hugo didn't create a home page because our site has no content. When there's no index.html file in a directory, the server lists the files in the directory, which is what you should see in your browser.
+
+Let’s go back and look at those warnings again.
+
+```
+WARN: 2014/09/29 Unable to locate layout: [index.html _default/list.html _default/single.html]
+WARN: 2014/09/29 Unable to locate layout: [404.html]
+```
+
+That second warning is easier to explain. We haven’t created a template to be used to generate “page not found errors.” The 404 message is a topic for a separate tutorial.
+
+Now for the first warning. It is for the home page. You can tell because the first layout that it looked for was “index.html.” That’s only used by the home page.
+
+I like that the verbose flag causes Hugo to list the files that it's searching for. For the home page, they are index.html, _default/list.html, and _default/single.html. There are some rules that we'll cover later that explain the names and paths. For now, just remember that Hugo couldn't find a template for the home page and it told you so.
+
+At this point, you've got a working installation and site that we can build upon. All that’s left is to add some content and a theme to display it.
+
+## Create a New Theme
+
+Hugo doesn't ship with a default theme. There are a few available (I counted a dozen when I first installed Hugo) and Hugo comes with a command to create new themes.
+
+We're going to create a new theme called "zafta." Since the goal of this tutorial is to show you how to fill out the files to pull in your content, the theme will not contain any CSS. In other words, ugly but functional.
+
+All themes have opinions on content and layout. For example, Zafta uses "post" over "blog". Strong opinions make for simpler templates but differing opinions make it tougher to use themes. When you build a theme, consider using the terms that other themes do.
+
+
+### Create a Skeleton
+
+Use the hugo "new" command to create the skeleton of a theme. This creates the directory structure and places empty files for you to fill out.
+
+```
+$ hugo new theme zafta
+
+$ ls -l
+total 8
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 archetypes
+-rw-r--r--  1 quoha  staff   82 Sep 29 16:49 config.toml
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 content
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 layouts
+drwxr-xr-x  4 quoha  staff  136 Sep 29 17:02 public
+drwxr-xr-x  2 quoha  staff   68 Sep 29 16:49 static
+drwxr-xr-x  3 quoha  staff  102 Sep 29 17:31 themes
+
+$ find themes -type f | xargs ls -l
+-rw-r--r--  1 quoha  staff  1081 Sep 29 17:31 themes/zafta/LICENSE.md
+-rw-r--r--  1 quoha  staff     0 Sep 29 17:31 themes/zafta/archetypes/default.md
+-rw-r--r--  1 quoha  staff     0 Sep 29 17:31 themes/zafta/layouts/_default/list.html
+-rw-r--r--  1 quoha  staff     0 Sep 29 17:31 themes/zafta/layouts/_default/single.html
+-rw-r--r--  1 quoha  staff     0 Sep 29 17:31 themes/zafta/layouts/index.html
+-rw-r--r--  1 quoha  staff     0 Sep 29 17:31 themes/zafta/layouts/partials/footer.html
+-rw-r--r--  1 quoha  staff     0 Sep 29 17:31 themes/zafta/layouts/partials/header.html
+-rw-r--r--  1 quoha  staff    93 Sep 29 17:31 themes/zafta/theme.toml
+$ 
+```
+
+The skeleton includes templates (the files ending in .html), license file, a description of your theme (the theme.toml file), and an empty archetype.
+
+Please take a minute to fill out the theme.toml and LICENSE.md files. They're optional, but if you're going to be distributing your theme, it tells the world who to praise (or blame). It's also nice to declare the license so that people will know how they can use the theme.
+
+```
+$ vi themes/zafta/theme.toml
+author = "michael d henderson"
+description = "a minimal working template"
+license = "MIT"
+name = "zafta"
+source_repo = ""
+tags = ["tags", "categories"]
+:wq
+
+## also edit themes/zafta/LICENSE.md and change
+## the bit that says "YOUR_NAME_HERE"
+```
+
+Note that the the skeleton's template files are empty. Don't worry, we'll be changing that shortly.
+
+```
+$ find themes/zafta -name '*.html' | xargs ls -l
+-rw-r--r--  1 quoha  staff  0 Sep 29 17:31 themes/zafta/layouts/_default/list.html
+-rw-r--r--  1 quoha  staff  0 Sep 29 17:31 themes/zafta/layouts/_default/single.html
+-rw-r--r--  1 quoha  staff  0 Sep 29 17:31 themes/zafta/layouts/index.html
+-rw-r--r--  1 quoha  staff  0 Sep 29 17:31 themes/zafta/layouts/partials/footer.html
+-rw-r--r--  1 quoha  staff  0 Sep 29 17:31 themes/zafta/layouts/partials/header.html
+$
+```
+
+
+
+### Update the Configuration File to Use the Theme
+
+Now that we've got a theme to work with, it's a good idea to add the theme name to the configuration file. This is optional, because you can always add "-t zafta" on all your commands. I like to put it the configuration file because I like shorter command lines. If you don't put it in the configuration file or specify it on the command line, you won't use the template that you're expecting to.
+
+Edit the file to add the theme, add a title for the site, and specify that all of our content will use the TOML format.
+
+```
+$ vi config.toml
+theme = "zafta"
+baseurl = ""
+languageCode = "en-us"
+title = "zafta - totally refreshing"
+MetaDataFormat = "toml"
+:wq
+
+$
+```
+
+### Generate the Site
+
+Now that we have an empty theme, let's generate the site again.
+
+```
+$ hugo --verbose
+INFO: 2014/09/29 Using config file: /Users/quoha/Sites/zafta/config.toml
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/themes/zafta/static/ to /Users/quoha/Sites/zafta/public/
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/static/ to /Users/quoha/Sites/zafta/public/
+WARN: 2014/09/29 Unable to locate layout: [404.html theme/404.html]
+0 draft content 
+0 future content 
+0 pages created 
+0 tags created
+0 categories created
+in 2 ms
+$
+```
+
+Did you notice that the output is different? The warning message for the home page has disappeared and we have an additional information line saying that Hugo is syncing from the theme's directory.
+
+Let's check the public/ directory to see what Hugo's created.
+
+```
+$ ls -l public
+total 16
+drwxr-xr-x  2 quoha  staff   68 Sep 29 17:56 css
+-rw-r--r--  1 quoha  staff    0 Sep 29 17:56 index.html
+-rw-r--r--  1 quoha  staff  407 Sep 29 17:56 index.xml
+drwxr-xr-x  2 quoha  staff   68 Sep 29 17:56 js
+-rw-r--r--  1 quoha  staff  243 Sep 29 17:56 sitemap.xml
+$
+```
+
+Notice four things:
+
+1. Hugo created a home page. This is the file public/index.html.
+2. Hugo created a css/ directory.
+3. Hugo created a js/ directory.
+4. Hugo claimed that it created 0 pages. It created a file and copied over static files, but didn't create any pages. That's because it considers a "page" to be a file created directly from a content file. It doesn't count things like the index.html files that it creates automatically.
+
+#### The Home Page
+
+Hugo supports many different types of templates. The home page is special because it gets its own type of template and its own template file. The file, layouts/index.html, is used to generate the HTML for the home page. The Hugo documentation says that this is the only required template, but that depends. Hugo's warning message shows that it looks for three different templates:
+
+```
+WARN: 2014/09/29 Unable to locate layout: [index.html _default/list.html _default/single.html]
+```
+
+If it can't find any of these, it completely skips creating the home page. We noticed that when we built the site without having a theme installed.
+
+When Hugo created our theme, it created an empty home page template. Now, when we build the site, Hugo finds the template and uses it to generate the HTML for the home page. Since the template file is empty, the HTML file is empty, too. If the template had any rules in it, then Hugo would have used them to generate the home page.
+
+```
+$ find . -name index.html | xargs ls -l
+-rw-r--r--  1 quoha  staff  0 Sep 29 20:21 ./public/index.html
+-rw-r--r--  1 quoha  staff  0 Sep 29 17:31 ./themes/zafta/layouts/index.html
+$ 
+```
+
+#### The Magic of Static
+
+Hugo does two things when generating the site. It uses templates to transform content into HTML and it copies static files into the site. Unlike content, static files are not transformed. They are copied exactly as they are.
+
+Hugo assumes that your site will use both CSS and JavaScript, so it creates directories in your theme to hold them. Remember opinions? Well, Hugo's opinion is that you'll store your CSS in a directory named css/ and your JavaScript in a directory named js/. If you don't like that, you can change the directory names in your theme directory or even delete them completely. Hugo's nice enough to offer its opinion, then behave nicely if you disagree.
+
+```
+$ find themes/zafta -type d | xargs ls -ld
+drwxr-xr-x  7 quoha  staff  238 Sep 29 17:38 themes/zafta
+drwxr-xr-x  3 quoha  staff  102 Sep 29 17:31 themes/zafta/archetypes
+drwxr-xr-x  5 quoha  staff  170 Sep 29 17:31 themes/zafta/layouts
+drwxr-xr-x  4 quoha  staff  136 Sep 29 17:31 themes/zafta/layouts/_default
+drwxr-xr-x  4 quoha  staff  136 Sep 29 17:31 themes/zafta/layouts/partials
+drwxr-xr-x  4 quoha  staff  136 Sep 29 17:31 themes/zafta/static
+drwxr-xr-x  2 quoha  staff   68 Sep 29 17:31 themes/zafta/static/css
+drwxr-xr-x  2 quoha  staff   68 Sep 29 17:31 themes/zafta/static/js
+$ 
+```
+
+## The Theme Development Cycle
+
+When you're working on a theme, you will make changes in the theme's directory, rebuild the site, and check your changes in the browser. Hugo makes this very easy:
+
+1. Purge the public/ directory.
+2. Run the built in web server in watch mode.
+3. Open your site in a browser.
+4. Update the theme.
+5. Glance at your browser window to see changes.
+6. Return to step 4.
+
+I’ll throw in one more opinion: never work on a theme on a live site. Always work on a copy of your site. Make changes to your theme, test them, then copy them up to your site. For added safety, use a tool like Git to keep a revision history of your content and your theme. Believe me when I say that it is too easy to lose both your mind and your changes.
+
+Check the main Hugo site for information on using Git with Hugo.
+
+### Purge the public/ Directory
+
+When generating the site, Hugo will create new files and update existing ones in the ```public/``` directory. It will not delete files that are no longer used. For example, files that were created in the wrong directory or with the wrong title will remain. If you leave them, you might get confused by them later. I recommend cleaning out your site prior to generating it.
+
+Note: If you're building on an SSD, you should ignore this. Churning on a SSD can be costly.
+
+### Hugo's Watch Option
+
+Hugo's "`--watch`" option will monitor the content/ and your theme directories for changes and rebuild the site automatically.
+
+### Live Reload
+
+Hugo's built in web server supports live reload. As pages are saved on the server, the browser is told to refresh the page. Usually, this happens faster than you can say, "Wow, that's totally amazing."
+
+### Development Commands
+
+Use the following commands as the basis for your workflow.
+
+```
+## purge old files. hugo will recreate the public directory.
+##
+$ rm -rf public
+##
+## run hugo in watch mode
+##
+$ hugo server --watch --verbose
+```
+
+Here's sample output showing Hugo detecting a change to the template for the home page. Once generated, the web browser automatically reloaded the page. I've said this before, it's amazing.
+
+
+```
+$ rm -rf public
+$ hugo server --watch --verbose
+INFO: 2014/09/29 Using config file: /Users/quoha/Sites/zafta/config.toml
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/themes/zafta/static/ to /Users/quoha/Sites/zafta/public/
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/static/ to /Users/quoha/Sites/zafta/public/
+WARN: 2014/09/29 Unable to locate layout: [404.html theme/404.html]
+0 draft content 
+0 future content 
+0 pages created 
+0 tags created
+0 categories created
+in 2 ms
+Watching for changes in /Users/quoha/Sites/zafta/content
+Serving pages from /Users/quoha/Sites/zafta/public
+Web Server is available at http://localhost:1313
+Press Ctrl+C to stop
+INFO: 2014/09/29 File System Event: ["/Users/quoha/Sites/zafta/themes/zafta/layouts/index.html": MODIFY|ATTRIB]
+Change detected, rebuilding site
+
+WARN: 2014/09/29 Unable to locate layout: [404.html theme/404.html]
+0 draft content 
+0 future content 
+0 pages created 
+0 tags created
+0 categories created
+in 1 ms
+```
+
+## Update the Home Page Template
+
+The home page is one of a few special pages that Hugo creates automatically. As mentioned earlier, it looks for one of three files in the theme's layout/ directory:
+
+1. index.html
+2. _default/list.html
+3. _default/single.html
+
+We could update one of the default templates, but a good design decision is to update the most specific template available. That's not a hard and fast rule (in fact, we'll break it a few times in this tutorial), but it is a good generalization.
+
+### Make a Static Home Page
+
+Right now, that page is empty because we don't have any content and we don't have any logic in the template. Let's change that by adding some text to the template.
+
+```
+$ vi themes/zafta/layouts/index.html
+<!DOCTYPE html> 
+<html> 
+<body> 
+  <p>hugo says hello!</p> 
+</body> 
+</html> 
+:wq
+
+$
+```
+
+Build the web site and then verify the results.
+
+```
+$ hugo --verbose
+INFO: 2014/09/29 Using config file: /Users/quoha/Sites/zafta/config.toml
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/themes/zafta/static/ to /Users/quoha/Sites/zafta/public/
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/static/ to /Users/quoha/Sites/zafta/public/
+WARN: 2014/09/29 Unable to locate layout: [404.html theme/404.html]
+0 draft content 
+0 future content 
+0 pages created 
+0 tags created
+0 categories created
+in 2 ms
+
+$ find public -type f -name '*.html' | xargs ls -l
+-rw-r--r--  1 quoha  staff  78 Sep 29 21:26 public/index.html
+
+$ cat public/index.html 
+<!DOCTYPE html> 
+<html> 
+<body> 
+  <p>hugo says hello!</p> 
+</html>
+```
+
+#### Live Reload
+
+Note: If you're running the server with the `--watch` option, you'll see different content in the file:
+
+```
+$ cat public/index.html 
+<!DOCTYPE html> 
+<html> 
+<body> 
+  <p>hugo says hello!</p> 
+<script>document.write('<script src="http://' 
+        + (location.host || 'localhost').split(':')[0] 
+    + ':1313/livereload.js?mindelay=10"></' 
+        + 'script>')</script></body> 
+</html>
+```
+
+When you use `--watch`, the Live Reload script is added by Hugo. Look for live reload in the documentation to see what it does and how to disable it.
+
+### Build a "Dynamic" Home Page
+
+"Dynamic home page?" Hugo's a static web site generator, so this seems an odd thing to say. I mean let's have the home page automatically reflect the content in the site every time Hugo builds it. We'll use iteration in the template to do that.
+
+#### Create New Posts
+
+Now that we have the home page generating static content, let's add some content to the site. We'll display these posts as a list on the home page and on their own page, too.
+
+Hugo has a command to generate a skeleton post, just like it does for sites and themes.
+
+```
+$ hugo --verbose new post/first.md
+INFO: 2014/09/29 Using config file: /Users/quoha/Sites/zafta/config.toml
+INFO: 2014/09/29 attempting to create  post/first.md of post
+INFO: 2014/09/29 curpath: /Users/quoha/Sites/zafta/themes/zafta/archetypes/default.md
+ERROR: 2014/09/29 Unable to Cast <nil> to map[string]interface{}
+
+$ 
+```
+
+That wasn't very nice, was it?
+
+The "new" command uses an archetype to create the post file. Hugo created an empty default archetype file, but that causes an error when there's a theme. For me, the workaround was to create an archetypes file specifically for the post type.
+
+```
+$ vi themes/zafta/archetypes/post.md
++++
+Description = ""
+Tags = []
+Categories = []
++++
+:wq
+
+$ find themes/zafta/archetypes -type f | xargs ls -l
+-rw-r--r--  1 quoha  staff   0 Sep 29 21:53 themes/zafta/archetypes/default.md
+-rw-r--r--  1 quoha  staff  51 Sep 29 21:54 themes/zafta/archetypes/post.md
+
+$ hugo --verbose new post/first.md
+INFO: 2014/09/29 Using config file: /Users/quoha/Sites/zafta/config.toml
+INFO: 2014/09/29 attempting to create  post/first.md of post
+INFO: 2014/09/29 curpath: /Users/quoha/Sites/zafta/themes/zafta/archetypes/post.md
+INFO: 2014/09/29 creating /Users/quoha/Sites/zafta/content/post/first.md
+/Users/quoha/Sites/zafta/content/post/first.md created
+
+$ hugo --verbose new post/second.md
+INFO: 2014/09/29 Using config file: /Users/quoha/Sites/zafta/config.toml
+INFO: 2014/09/29 attempting to create  post/second.md of post
+INFO: 2014/09/29 curpath: /Users/quoha/Sites/zafta/themes/zafta/archetypes/post.md
+INFO: 2014/09/29 creating /Users/quoha/Sites/zafta/content/post/second.md
+/Users/quoha/Sites/zafta/content/post/second.md created
+
+$ ls -l content/post
+total 16
+-rw-r--r--  1 quoha  staff  104 Sep 29 21:54 first.md
+-rw-r--r--  1 quoha  staff  105 Sep 29 21:57 second.md
+
+$ cat content/post/first.md 
++++
+Categories = []
+Description = ""
+Tags = []
+date = "2014-09-29T21:54:53-05:00"
+title = "first"
+
++++
+my first post
+
+$ cat content/post/second.md 
++++
+Categories = []
+Description = ""
+Tags = []
+date = "2014-09-29T21:57:09-05:00"
+title = "second"
+
++++
+my second post
+
+$ 
+```
+
+Build the web site and then verify the results.
+
+```
+$ rm -rf public
+$ hugo --verbose
+INFO: 2014/09/29 Using config file: /Users/quoha/Sites/zafta/config.toml
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/themes/zafta/static/ to /Users/quoha/Sites/zafta/public/
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/static/ to /Users/quoha/Sites/zafta/public/
+INFO: 2014/09/29 found taxonomies: map[string]string{"category":"categories", "tag":"tags"}
+WARN: 2014/09/29 Unable to locate layout: [404.html theme/404.html]
+0 draft content 
+0 future content 
+2 pages created 
+0 tags created
+0 categories created
+in 4 ms
+$
+```
+
+The output says that it created 2 pages. Those are our new posts:
+
+```
+$ find public -type f -name '*.html' | xargs ls -l
+-rw-r--r--  1 quoha  staff  78 Sep 29 22:13 public/index.html
+-rw-r--r--  1 quoha  staff   0 Sep 29 22:13 public/post/first/index.html
+-rw-r--r--  1 quoha  staff   0 Sep 29 22:13 public/post/index.html
+-rw-r--r--  1 quoha  staff   0 Sep 29 22:13 public/post/second/index.html
+$
+```
+
+The new files are empty because because the templates used to generate the content are empty. The homepage doesn't show the new content, either. We have to update the templates to add the posts.
+
+### List and Single Templates
+
+In Hugo, we have three major kinds of templates. There's the home page template that we updated previously. It is used only by the home page. We also have "single" templates which are used to generate output for a single content file. We also have "list" templates that are used to group multiple pieces of content before generating output.
+
+Generally speaking, list templates are named "list.html" and single templates are named "single.html."
+
+There are three other types of templates: partials, content views, and terms. We will not go into much detail on these.
+
+### Add Content to the Homepage
+
+The home page will contain a list of posts. Let's update its template to add the posts that we just created. The logic in the template will run every time we build the site.
+
+```
+$ vi themes/zafta/layouts/index.html 
+<!DOCTYPE html>
+<html>
+<body>
+  {{ range first 10 .Data.Pages }}
+    <h1>{{ .Title }}</h1>
+  {{ end }}
+</body>
+</html>
+:wq
+
+$
+```
+
+Hugo uses the Go template engine. That engine scans the template files for commands which are enclosed between "{{" and "}}". In our template, the commands are:
+
+1. range
+2. .Title
+3. end
+
+The "range" command is an iterator. We're going to use it to go through the first ten pages. Every HTML file that Hugo creates is treated as a page, so looping through the list of pages will look at every file that will be created.
+
+The ".Title" command prints the value of the "title" variable. Hugo pulls it from the front matter in the Markdown file.
+
+The "end" command signals the end of the range iterator. The engine loops back to the top of the iteration when it finds "end." Everything between the "range" and "end" is evaluated every time the engine goes through the iteration. In this file, that would cause the title from the first ten pages to be output as heading level one.
+
+It's helpful to remember that some variables, like .Data, are created before any output files. Hugo loads every content file into the variable and then gives the template a chance to process before creating the HTML files.
+
+Build the web site and then verify the results.
+
+```
+$ rm -rf public
+$ hugo --verbose
+INFO: 2014/09/29 Using config file: /Users/quoha/Sites/zafta/config.toml
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/themes/zafta/static/ to /Users/quoha/Sites/zafta/public/
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/static/ to /Users/quoha/Sites/zafta/public/
+INFO: 2014/09/29 found taxonomies: map[string]string{"tag":"tags", "category":"categories"}
+WARN: 2014/09/29 Unable to locate layout: [404.html theme/404.html]
+0 draft content 
+0 future content 
+2 pages created 
+0 tags created
+0 categories created
+in 4 ms
+$ find public -type f -name '*.html' | xargs ls -l 
+-rw-r--r--  1 quoha  staff  94 Sep 29 22:23 public/index.html
+-rw-r--r--  1 quoha  staff   0 Sep 29 22:23 public/post/first/index.html
+-rw-r--r--  1 quoha  staff   0 Sep 29 22:23 public/post/index.html
+-rw-r--r--  1 quoha  staff   0 Sep 29 22:23 public/post/second/index.html
+$ cat public/index.html 
+<!DOCTYPE html>
+<html>
+<body>
+  
+    <h1>second</h1>
+  
+    <h1>first</h1>
+  
+</body>
+</html>
+$
+```
+
+Congratulations, the home page shows the title of the two posts. The posts themselves are still empty, but let's take a moment to appreciate what we've done. Your template now generates output dynamically. Believe it or not, by inserting the range command inside of those curly braces, you've learned everything you need to know to build a theme. All that's really left is understanding which template will be used to generate each content file and becoming familiar with the commands for the template engine.
+
+And, if that were entirely true, this tutorial would be much shorter. There are a few things to know that will make creating a new template much easier. Don't worry, though, that's all to come.
+
+### Add Content to the Posts
+
+We're working with posts, which are in the content/post/ directory. That means that their section is "post" (and if we don't do something weird, their type is also "post").
+
+Hugo uses the section and type to find the template file for every piece of content. Hugo will first look for a template file that matches the section or type name. If it can't find one, then it will look in the _default/ directory. There are some twists that we'll cover when we get to categories and tags, but for now we can assume that Hugo will try post/single.html, then _default/single.html.
+
+Now that we know the search rule, let's see what we actually have available:
+
+```
+$ find themes/zafta -name single.html | xargs ls -l
+-rw-r--r--  1 quoha  staff  132 Sep 29 17:31 themes/zafta/layouts/_default/single.html
+```
+
+We could create a new template, post/single.html, or change the default. Since we don't know of any other content types, let's start with updating the default.
+
+Remember, any content that we haven't created a template for will end up using this template. That can be good or bad. Bad because I know that we're going to be adding different types of content and we're going to end up undoing some of the changes we've made. It's good because we'll be able to see immediate results. It's also good to start here because we can start to build the basic layout for the site. As we add more content types, we'll refactor this file and move logic around. Hugo makes that fairly painless, so we'll accept the cost and proceed.
+
+Please see the Hugo documentation on template rendering for all the details on determining which template to use. And, as the docs mention, if you're building a single page application (SPA) web site, you can delete all of the other templates and work with just the default single page. That's a refreshing amount of joy right there.
+
+#### Update the Template File
+
+```
+$ vi themes/zafta/layouts/_default/single.html 
+<!DOCTYPE html>
+<html>
+<head>
+  <title>{{ .Title }}</title>
+</head>
+<body>
+  <h1>{{ .Title }}</h1>
+  {{ .Content }}
+</body>
+</html>
+:wq
+
+$
+```
+
+Build the web site and verify the results.
+
+```
+$ rm -rf public
+$ hugo --verbose
+INFO: 2014/09/29 Using config file: /Users/quoha/Sites/zafta/config.toml
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/themes/zafta/static/ to /Users/quoha/Sites/zafta/public/
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/static/ to /Users/quoha/Sites/zafta/public/
+INFO: 2014/09/29 found taxonomies: map[string]string{"tag":"tags", "category":"categories"}
+WARN: 2014/09/29 Unable to locate layout: [404.html theme/404.html]
+0 draft content 
+0 future content 
+2 pages created 
+0 tags created
+0 categories created
+in 4 ms
+
+$ find public -type f -name '*.html' | xargs ls -l
+-rw-r--r--  1 quoha  staff   94 Sep 29 22:40 public/index.html
+-rw-r--r--  1 quoha  staff  125 Sep 29 22:40 public/post/first/index.html
+-rw-r--r--  1 quoha  staff    0 Sep 29 22:40 public/post/index.html
+-rw-r--r--  1 quoha  staff  128 Sep 29 22:40 public/post/second/index.html
+
+$ cat public/post/first/index.html 
+<!DOCTYPE html>
+<html>
+<head>
+  <title>first</title>
+</head>
+<body>
+  <h1>first</h1>
+  <p>my first post</p>
+
+</body>
+</html>
+
+$ cat public/post/second/index.html 
+<!DOCTYPE html>
+<html>
+<head>
+  <title>second</title>
+</head>
+<body>
+  <h1>second</h1>
+  <p>my second post</p>
+
+</body>
+</html>
+$
+```
+
+Notice that the posts now have content. You can go to localhost:1313/post/first to verify.
+
+### Linking to Content
+
+The posts are on the home page. Let's add a link from there to the post. Since this is the home page, we'll update its template.
+
+```
+$ vi themes/zafta/layouts/index.html
+<!DOCTYPE html>
+<html>
+<body>
+  {{ range first 10 .Data.Pages }}
+    <h1><a href="{{ .Permalink }}">{{ .Title }}</a></h1>
+  {{ end }}
+</body>
+</html>
+```
+
+Build the web site and verify the results.
+
+```
+$ rm -rf public
+$ hugo --verbose
+INFO: 2014/09/29 Using config file: /Users/quoha/Sites/zafta/config.toml
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/themes/zafta/static/ to /Users/quoha/Sites/zafta/public/
+INFO: 2014/09/29 syncing from /Users/quoha/Sites/zafta/static/ to /Users/quoha/Sites/zafta/public/
+INFO: 2014/09/29 found taxonomies: map[string]string{"tag":"tags", "category":"categories"}
+WARN: 2014/09/29 Unable to locate layout: [404.html theme/404.html]
+0 draft content 
+0 future content 
+2 pages created 
+0 tags created
+0 categories created
+in 4 ms
+
+$ find public -type f -name '*.html' | xargs ls -l
+-rw-r--r--  1 quoha  staff  149 Sep 29 22:44 public/index.html
+-rw-r--r--  1 quoha  staff  125 Sep 29 22:44 public/post/first/index.html
+-rw-r--r--  1 quoha  staff    0 Sep 29 22:44 public/post/index.html
+-rw-r--r--  1 quoha  staff  128 Sep 29 22:44 public/post/second/index.html
+
+$ cat public/index.html 
+<!DOCTYPE html>
+<html>
+<body>
+  
+    <h1><a href="/post/second/">second</a></h1>
+  
+    <h1><a href="/post/first/">first</a></h1>
+  
+</body>
+</html>
+
+$
+```
+
+### Create a Post Listing
+
+We have the posts displaying on the home page and on their own page. We also have a file public/post/index.html that is empty. Let's make it show a list of all posts (not just the first ten).
+
+We need to decide which template to update. This will be a listing, so it should be a list template. Let's take a quick look and see which list templates are available.
+
+```
+$ find themes/zafta -name list.html | xargs ls -l
+-rw-r--r--  1 quoha  staff  0 Sep 29 17:31 themes/zafta/layouts/_default/list.html
+```
+
+As with the single post, we have to decide to update _default/list.html or create post/list.html. We still don't have multiple content types, so let's stay consistent and update the default list template.
+
+## Creating Top Level Pages
+
+Let's add an "about" page and display it at the top level (as opposed to a sub-level like we did with posts).
+
+The default in Hugo is to use the directory structure of the content/ directory to guide the location of the generated html in the public/ directory. Let's verify that by creating an "about" page at the top level:
+
+```
+$ vi content/about.md 
++++
+title = "about"
+description = "about this site"
+date = "2014-09-27"
+slug = "about time"
++++
+
+## about us
+
+i'm speechless
+:wq
+```
+
+Generate the web site and verify the results.
+
+```
+$ find public -name '*.html' | xargs ls -l
+-rw-rw-r--  1 mdhender  staff   334 Sep 27 15:08 public/about-time/index.html
+-rw-rw-r--  1 mdhender  staff   527 Sep 27 15:08 public/index.html
+-rw-rw-r--  1 mdhender  staff   358 Sep 27 15:08 public/post/first-post/index.html
+-rw-rw-r--  1 mdhender  staff     0 Sep 27 15:08 public/post/index.html
+-rw-rw-r--  1 mdhender  staff   342 Sep 27 15:08 public/post/second-post/index.html
+```
+
+Notice that the page wasn't created at the top level. It was created in a sub-directory named 'about-time/'. That name came from our slug. Hugo will use the slug to name the generated content. It's a reasonable default, by the way, but we can learn a few things by fighting it for this file.
+
+One other thing. Take a look at the home page.
+
+```
+$ cat public/index.html
+<!DOCTYPE html>
+<html>
+<body>
+    <h1><a href="http://localhost:1313/post/theme/">creating a new theme</a></h1>
+    <h1><a href="http://localhost:1313/about-time/">about</a></h1>
+    <h1><a href="http://localhost:1313/post/second-post/">second</a></h1>
+    <h1><a href="http://localhost:1313/post/first-post/">first</a></h1>
+<script>document.write('<script src="http://'
+        + (location.host || 'localhost').split(':')[0]
+		+ ':1313/livereload.js?mindelay=10"></'
+        + 'script>')</script></body>
+</html>
+```
+
+Notice that the "about" link is listed with the posts? That's not desirable, so let's change that first.
+
+```
+$ vi themes/zafta/layouts/index.html
+<!DOCTYPE html>
+<html>
+<body>
+  <h1>posts</h1>
+  {{ range first 10 .Data.Pages }}
+    {{ if eq .Type "post"}}
+      <h2><a href="{{ .Permalink }}">{{ .Title }}</a></h2>
+    {{ end }}
+  {{ end }}
+
+  <h1>pages</h1>
+  {{ range .Data.Pages }}
+    {{ if eq .Type "page" }}
+      <h2><a href="{{ .Permalink }}">{{ .Title }}</a></h2>
+    {{ end }}
+  {{ end }}
+</body>
+</html>
+:wq
+```
+
+Generate the web site and verify the results. The home page has two sections, posts and pages, and each section has the right set of headings and links in it.
+
+But, that about page still renders to about-time/index.html.
+
+```
+$ find public -name '*.html' | xargs ls -l
+-rw-rw-r--  1 mdhender  staff    334 Sep 27 15:33 public/about-time/index.html
+-rw-rw-r--  1 mdhender  staff    645 Sep 27 15:33 public/index.html
+-rw-rw-r--  1 mdhender  staff    358 Sep 27 15:33 public/post/first-post/index.html
+-rw-rw-r--  1 mdhender  staff      0 Sep 27 15:33 public/post/index.html
+-rw-rw-r--  1 mdhender  staff    342 Sep 27 15:33 public/post/second-post/index.html
+```
+
+Knowing that hugo is using the slug to generate the file name, the simplest solution is to change the slug. Let's do it the hard way and change the permalink in the configuration file.
+
+```
+$ vi config.toml
+[permalinks]
+	page = "/:title/"
+	about = "/:filename/"
+```
+
+Generate the web site and verify that this didn't work. Hugo lets "slug" or "URL" override the permalinks setting in the configuration file. Go ahead and comment out the slug in content/about.md, then generate the web site to get it to be created in the right place.
+
+## Sharing Templates
+
+If you've been following along, you probably noticed that posts have titles in the browser and the home page doesn't. That's because we didn't put the title in the home page's template (layouts/index.html). That's an easy thing to do, but let's look at a different option.
+
+We can put the common bits into a shared template that's stored in the themes/zafta/layouts/partials/ directory.
+
+### Create the Header and Footer Partials
+
+In Hugo, a partial is a sugar-coated template. Normally a template reference has a path specified. Partials are different. Hugo searches for them along a TODO defined search path. This makes it easier for end-users to override the theme's presentation.
+
+```
+$ vi themes/zafta/layouts/partials/header.html
+<!DOCTYPE html>
+<html>
+<head>
+	<title>{{ .Title }}</title>
+</head>
+<body>
+:wq
+
+$ vi themes/zafta/layouts/partials/footer.html
+</body>
+</html>
+:wq
+```
+
+### Update the Home Page Template to Use the Partials
+
+The most noticeable difference between a template call and a partials call is the lack of path:
+
+```
+{{ template "theme/partials/header.html" . }}
+```
+versus
+```
+{{ partial "header.html" . }}
+```
+Both pass in the context.
+
+Let's change the home page template to use these new partials.
+
+```
+$ vi themes/zafta/layouts/index.html
+{{ partial "header.html" . }}
+
+  <h1>posts</h1>
+  {{ range first 10 .Data.Pages }}
+    {{ if eq .Type "post"}}
+      <h2><a href="{{ .Permalink }}">{{ .Title }}</a></h2>
+    {{ end }}
+  {{ end }}
+
+  <h1>pages</h1>
+  {{ range .Data.Pages }}
+    {{ if or (eq .Type "page") (eq .Type "about") }}
+      <h2><a href="{{ .Permalink }}">{{ .Type }} - {{ .Title }} - {{ .RelPermalink }}</a></h2>
+    {{ end }}
+  {{ end }}
+
+{{ partial "footer.html" . }}
+:wq
+```
+
+Generate the web site and verify the results. The title on the home page is now "your title here", which comes from the "title" variable in the config.toml file.
+
+### Update the Default Single Template to Use the Partials
+
+```
+$ vi themes/zafta/layouts/_default/single.html
+{{ partial "header.html" . }}
+
+  <h1>{{ .Title }}</h1>
+  {{ .Content }}
+
+{{ partial "footer.html" . }}
+:wq
+```
+
+Generate the web site and verify the results. The title on the posts and the about page should both reflect the value in the markdown file.
+
+## Add “Date Published” to Posts
+
+It's common to have posts display the date that they were written or published, so let's add that. The front matter of our posts has a variable named "date." It's usually the date the content was created, but let's pretend that's the value we want to display.
+
+### Add “Date Published” to the Template
+
+We'll start by updating the template used to render the posts. The template code will look like:
+
+```
+{{ .Date.Format "Mon, Jan 2, 2006" }}
+```
+
+Posts use the default single template, so we'll change that file.
+
+```
+$ vi themes/zafta/layouts/_default/single.html
+{{ partial "header.html" . }}
+
+  <h1>{{ .Title }}</h1>
+  <h2>{{ .Date.Format "Mon, Jan 2, 2006" }}</h2>
+  {{ .Content }}
+
+{{ partial "footer.html" . }}
+:wq
+```
+
+Generate the web site and verify the results. The posts now have the date displayed in them. There's a problem, though. The "about" page also has the date displayed.
+
+As usual, there are a couple of ways to make the date display only on posts. We could do an "if" statement like we did on the home page. Another way would be to create a separate template for posts.
+
+The "if" solution works for sites that have just a couple of content types. It aligns with the principle of "code for today," too.
+
+Let's assume, though, that we've made our site so complex that we feel we have to create a new template type. In Hugo-speak, we're going to create a section template.
+
+Let's restore the default single template before we forget.
+
+```
+$ mkdir themes/zafta/layouts/post
+$ vi themes/zafta/layouts/_default/single.html
+{{ partial "header.html" . }}
+
+  <h1>{{ .Title }}</h1>
+  {{ .Content }}
+
+{{ partial "footer.html" . }}
+:wq
+```
+
+Now we'll update the post's version of the single template. If you remember Hugo's rules, the template engine will use this version over the default.
+
+```
+$ vi themes/zafta/layouts/post/single.html
+{{ partial "header.html" . }}
+
+  <h1>{{ .Title }}</h1>
+  <h2>{{ .Date.Format "Mon, Jan 2, 2006" }}</h2>
+  {{ .Content }}
+
+{{ partial "footer.html" . }}
+:wq
+
+```
+
+Note that we removed the date logic from the default template and put it in the post template. Generate the web site and verify the results. Posts have dates and the about page doesn't.
+
+### Don't Repeat Yourself
+
+DRY is a good design goal and Hugo does a great job supporting it. Part of the art of a good template is knowing when to add a new template and when to update an existing one. While you're figuring that out, accept that you'll be doing some refactoring. Hugo makes that easy and fast, so it's okay to delay splitting up a template.
diff --git a/website/themes/book/exampleSite/content/posts/goisforlovers.md b/website/themes/book/exampleSite/content/posts/goisforlovers.md
new file mode 100644
index 000000000..df125d820
--- /dev/null
+++ b/website/themes/book/exampleSite/content/posts/goisforlovers.md
@@ -0,0 +1,344 @@
++++
+title = "(Hu)go Template Primer"
+description = ""
+tags = [
+    "go",
+    "golang",
+    "templates",
+    "themes",
+    "development",
+]
+date = "2014-04-02"
+categories = [
+    "Development",
+    "golang",
+]
+menu = "main"
++++
+
+Hugo uses the excellent [Go][] [html/template][gohtmltemplate] library for
+its template engine. It is an extremely lightweight engine that provides a very
+small amount of logic. In our experience that it is just the right amount of
+logic to be able to create a good static website. If you have used other
+template systems from different languages or frameworks you will find a lot of
+similarities in Go templates.
+
+This document is a brief primer on using Go templates. The [Go docs][gohtmltemplate]
+provide more details.
+
+## Introduction to Go Templates
+
+Go templates provide an extremely simple template language. It adheres to the
+belief that only the most basic of logic belongs in the template or view layer.
+One consequence of this simplicity is that Go templates parse very quickly.
+
+A unique characteristic of Go templates is they are content aware. Variables and
+content will be sanitized depending on the context of where they are used. More
+details can be found in the [Go docs][gohtmltemplate].
+
+## Basic Syntax
+
+Golang templates are HTML files with the addition of variables and
+functions.
+
+**Go variables and functions are accessible within {{ }}**
+
+Accessing a predefined variable "foo":
+
+    {{ foo }}
+
+**Parameters are separated using spaces**
+
+Calling the add function with input of 1, 2:
+
+    {{ add 1 2 }}
+
+**Methods and fields are accessed via dot notation**
+
+Accessing the Page Parameter "bar"
+
+    {{ .Params.bar }}
+
+**Parentheses can be used to group items together**
+
+    {{ if or (isset .Params "alt") (isset .Params "caption") }} Caption {{ end }}
+
+
+## Variables
+
+Each Go template has a struct (object) made available to it. In hugo each
+template is passed either a page or a node struct depending on which type of
+page you are rendering. More details are available on the
+[variables](/layout/variables) page.
+
+A variable is accessed by referencing the variable name.
+
+    <title>{{ .Title }}</title>
+
+Variables can also be defined and referenced.
+
+    {{ $address := "123 Main St."}}
+    {{ $address }}
+
+
+## Functions
+
+Go template ship with a few functions which provide basic functionality. The Go
+template system also provides a mechanism for applications to extend the
+available functions with their own. [Hugo template
+functions](/layout/functions) provide some additional functionality we believe
+are useful for building websites. Functions are called by using their name
+followed by the required parameters separated by spaces. Template
+functions cannot be added without recompiling hugo.
+
+**Example:**
+
+    {{ add 1 2 }}
+
+## Includes
+
+When including another template you will pass to it the data it will be
+able to access. To pass along the current context please remember to
+include a trailing dot. The templates location will always be starting at
+the /layout/ directory within Hugo.
+
+**Example:**
+
+    {{ template "chrome/header.html" . }}
+
+
+## Logic
+
+Go templates provide the most basic iteration and conditional logic.
+
+### Iteration
+
+Just like in Go, the Go templates make heavy use of range to iterate over
+a map, array or slice. The following are different examples of how to use
+range.
+
+**Example 1: Using Context**
+
+    {{ range array }}
+        {{ . }}
+    {{ end }}
+
+**Example 2: Declaring value variable name**
+
+    {{range $element := array}}
+        {{ $element }}
+    {{ end }}
+
+**Example 2: Declaring key and value variable name**
+
+    {{range $index, $element := array}}
+        {{ $index }}
+        {{ $element }}
+    {{ end }}
+
+### Conditionals
+
+If, else, with, or, & and provide the framework for handling conditional
+logic in Go Templates. Like range, each statement is closed with `end`.
+
+
+Go Templates treat the following values as false:
+
+* false
+* 0
+* any array, slice, map, or string of length zero
+
+**Example 1: If**
+
+    {{ if isset .Params "title" }}<h4>{{ index .Params "title" }}</h4>{{ end }}
+
+**Example 2: If -> Else**
+
+    {{ if isset .Params "alt" }}
+        {{ index .Params "alt" }}
+    {{else}}
+        {{ index .Params "caption" }}
+    {{ end }}
+
+**Example 3: And & Or**
+
+    {{ if and (or (isset .Params "title") (isset .Params "caption")) (isset .Params "attr")}}
+
+**Example 4: With**
+
+An alternative way of writing "if" and then referencing the same value
+is to use "with" instead. With rebinds the context `.` within its scope,
+and skips the block if the variable is absent.
+
+The first example above could be simplified as:
+
+    {{ with .Params.title }}<h4>{{ . }}</h4>{{ end }}
+
+**Example 5: If -> Else If**
+
+    {{ if isset .Params "alt" }}
+        {{ index .Params "alt" }}
+    {{ else if isset .Params "caption" }}
+        {{ index .Params "caption" }}
+    {{ end }}
+
+## Pipes
+
+One of the most powerful components of Go templates is the ability to
+stack actions one after another. This is done by using pipes. Borrowed
+from unix pipes, the concept is simple, each pipeline's output becomes the
+input of the following pipe.
+
+Because of the very simple syntax of Go templates, the pipe is essential
+to being able to chain together function calls. One limitation of the
+pipes is that they only can work with a single value and that value
+becomes the last parameter of the next pipeline.
+
+A few simple examples should help convey how to use the pipe.
+
+**Example 1 :**
+
+    {{ if eq 1 1 }} Same {{ end }}
+
+is the same as
+
+    {{ eq 1 1 | if }} Same {{ end }}
+
+It does look odd to place the if at the end, but it does provide a good
+illustration of how to use the pipes.
+
+**Example 2 :**
+
+    {{ index .Params "disqus_url" | html }}
+
+Access the page parameter called "disqus_url" and escape the HTML.
+
+**Example 3 :**
+
+    {{ if or (or (isset .Params "title") (isset .Params "caption")) (isset .Params "attr")}}
+    Stuff Here
+    {{ end }}
+
+Could be rewritten as
+
+    {{  isset .Params "caption" | or isset .Params "title" | or isset .Params "attr" | if }}
+    Stuff Here
+    {{ end }}
+
+
+## Context (aka. the dot)
+
+The most easily overlooked concept to understand about Go templates is that {{ . }}
+always refers to the current context. In the top level of your template this
+will be the data set made available to it. Inside of a iteration it will have
+the value of the current item. When inside of a loop the context has changed. .
+will no longer refer to the data available to the entire page. If you need to
+access this from within the loop you will likely want to set it to a variable
+instead of depending on the context.
+
+**Example:**
+
+      {{ $title := .Site.Title }}
+      {{ range .Params.tags }}
+        <li> <a href="{{ $baseurl }}/tags/{{ . | urlize }}">{{ . }}</a> - {{ $title }} </li>
+      {{ end }}
+
+Notice how once we have entered the loop the value of {{ . }} has changed. We
+have defined a variable outside of the loop so we have access to it from within
+the loop.
+
+# Hugo Parameters
+
+Hugo provides the option of passing values to the template language
+through the site configuration (for sitewide values), or through the meta
+data of each specific piece of content. You can define any values of any
+type (supported by your front matter/config format) and use them however
+you want to inside of your templates.
+
+
+## Using Content (page) Parameters
+
+In each piece of content you can provide variables to be used by the
+templates. This happens in the [front matter](/content/front-matter).
+
+An example of this is used in this documentation site. Most of the pages
+benefit from having the table of contents provided. Sometimes the TOC just
+doesn't make a lot of sense. We've defined a variable in our front matter
+of some pages to turn off the TOC from being displayed.
+
+Here is the example front matter:
+
+```
+---
+title: "Permalinks"
+date: "2013-11-18"
+aliases:
+  - "/doc/permalinks/"
+groups: ["extras"]
+groups_weight: 30
+notoc: true
+---
+```
+
+Here is the corresponding code inside of the template:
+
+      {{ if not .Params.notoc }}
+        <div id="toc" class="well col-md-4 col-sm-6">
+        {{ .TableOfContents }}
+        </div>
+      {{ end }}
+
+
+
+## Using Site (config) Parameters
+In your top-level configuration file (eg, `config.yaml`) you can define site
+parameters, which are values which will be available to you in chrome.
+
+For instance, you might declare:
+
+```yaml
+params:
+  CopyrightHTML: "Copyright &#xA9; 2013 John Doe. All Rights Reserved."
+  TwitterUser: "spf13"
+  SidebarRecentLimit: 5
+```
+
+Within a footer layout, you might then declare a `<footer>` which is only
+provided if the `CopyrightHTML` parameter is provided, and if it is given,
+you would declare it to be HTML-safe, so that the HTML entity is not escaped
+again.  This would let you easily update just your top-level config file each
+January 1st, instead of hunting through your templates.
+
+```
+{{if .Site.Params.CopyrightHTML}}<footer>
+<div class="text-center">{{.Site.Params.CopyrightHTML | safeHtml}}</div>
+</footer>{{end}}
+```
+
+An alternative way of writing the "if" and then referencing the same value
+is to use "with" instead. With rebinds the context `.` within its scope,
+and skips the block if the variable is absent:
+
+```
+{{with .Site.Params.TwitterUser}}<span class="twitter">
+<a href="https://twitter.com/{{.}}" rel="author">
+<img src="/images/twitter.png" width="48" height="48" title="Twitter: {{.}}"
+ alt="Twitter"></a>
+</span>{{end}}
+```
+
+Finally, if you want to pull "magic constants" out of your layouts, you can do
+so, such as in this example:
+
+```
+<nav class="recent">
+  <h1>Recent Posts</h1>
+  <ul>{{range first .Site.Params.SidebarRecentLimit .Site.Recent}}
+    <li><a href="{{.RelPermalink}}">{{.Title}}</a></li>
+  {{end}}</ul>
+</nav>
+```
+
+
+[go]: https://golang.org/
+[gohtmltemplate]: https://golang.org/pkg/html/template/
diff --git a/website/themes/book/exampleSite/content/posts/hugoisforlovers.md b/website/themes/book/exampleSite/content/posts/hugoisforlovers.md
new file mode 100644
index 000000000..02bb5091c
--- /dev/null
+++ b/website/themes/book/exampleSite/content/posts/hugoisforlovers.md
@@ -0,0 +1,89 @@
++++
+title = "Getting Started with Hugo"
+description = ""
+tags = [
+    "go",
+    "golang",
+    "hugo",
+    "development",
+]
+date = "2014-04-02"
+categories = [
+    "Development",
+    "golang",
+]
+menu = "main"
++++
+
+## Step 1. Install Hugo
+
+Go to [Hugo releases](https://github.com/spf13/hugo/releases) and download the
+appropriate version for your OS and architecture.
+
+Save it somewhere specific as we will be using it in the next step.
+
+More complete instructions are available at [Install Hugo](https://gohugo.io/getting-started/installing/)
+
+## Step 2. Build the Docs
+
+Hugo has its own example site which happens to also be the documentation site
+you are reading right now.
+
+Follow the following steps:
+
+ 1. Clone the [Hugo repository](http://github.com/spf13/hugo)
+ 2. Go into the repo
+ 3. Run hugo in server mode and build the docs
+ 4. Open your browser to http://localhost:1313
+
+Corresponding pseudo commands:
+
+    git clone https://github.com/spf13/hugo
+    cd hugo
+    /path/to/where/you/installed/hugo server --source=./docs
+    > 29 pages created
+    > 0 tags index created
+    > in 27 ms
+    > Web Server is available at http://localhost:1313
+    > Press ctrl+c to stop
+
+Once you've gotten here, follow along the rest of this page on your local build.
+
+## Step 3. Change the docs site
+
+Stop the Hugo process by hitting Ctrl+C.
+
+Now we are going to run hugo again, but this time with hugo in watch mode.
+
+    /path/to/hugo/from/step/1/hugo server --source=./docs --watch
+    > 29 pages created
+    > 0 tags index created
+    > in 27 ms
+    > Web Server is available at http://localhost:1313
+    > Watching for changes in /Users/spf13/Code/hugo/docs/content
+    > Press ctrl+c to stop
+
+
+Open your [favorite editor](http://vim.spf13.com) and change one of the source
+content pages. How about changing this very file to *fix the typo*. How about changing this very file to *fix the typo*.
+
+Content files are found in `docs/content/`. Unless otherwise specified, files
+are located at the same relative location as the url, in our case
+`docs/content/overview/quickstart.md`.
+
+Change and save this file.. Notice what happened in your terminal.
+
+    > Change detected, rebuilding site
+
+    > 29 pages created
+    > 0 tags index created
+    > in 26 ms
+
+Refresh the browser and observe that the typo is now fixed.
+
+Notice how quick that was. Try to refresh the site before it's finished building. I double dare you.
+Having nearly instant feedback enables you to have your creativity flow without waiting for long builds.
+
+## Step 4. Have fun
+
+The best way to learn something is to play with it.
diff --git a/website/themes/book/exampleSite/content/posts/migrate-from-jekyll.md b/website/themes/book/exampleSite/content/posts/migrate-from-jekyll.md
new file mode 100644
index 000000000..75f955854
--- /dev/null
+++ b/website/themes/book/exampleSite/content/posts/migrate-from-jekyll.md
@@ -0,0 +1,156 @@
+---
+date: 2014-03-10
+linktitle: Migrating from Jekyll
+menu:
+  main:
+    parent: tutorials
+prev: /tutorials/mathjax
+title: Migrate to Hugo from Jekyll
+weight: 10
+---
+
+## Move static content to `static`
+Jekyll has a rule that any directory not starting with `_` will be copied as-is to the `_site` output. Hugo keeps all static content under `static`. You should therefore move it all there.
+With Jekyll, something that looked like
+
+    ▾ <root>/
+        ▾ images/
+            logo.png
+
+should become
+
+    ▾ <root>/
+        ▾ static/
+            ▾ images/
+                logo.png
+
+Additionally, you'll want any files that should reside at the root (such as `CNAME`) to be moved to `static`.
+
+## Create your Hugo configuration file
+Hugo can read your configuration as JSON, YAML or TOML. Hugo supports parameters custom configuration too. Refer to the [Hugo configuration documentation](/overview/configuration/) for details.
+
+## Set your configuration publish folder to `_site`
+The default is for Jekyll to publish to `_site` and for Hugo to publish to `public`. If, like me, you have [`_site` mapped to a git submodule on the `gh-pages` branch](http://blog.blindgaenger.net/generate_github_pages_in_a_submodule.html), you'll want to do one of two alternatives:
+
+1. Change your submodule to point to map `gh-pages` to public instead of `_site` (recommended).
+
+        git submodule deinit _site
+        git rm _site
+        git submodule add -b gh-pages git@github.com:your-username/your-repo.git public
+
+2. Or, change the Hugo configuration to use `_site` instead of `public`.
+
+        {
+            ..
+            "publishdir": "_site",
+            ..
+        }
+
+## Convert Jekyll templates to Hugo templates
+That's the bulk of the work right here. The documentation is your friend. You should refer to [Jekyll's template documentation](http://jekyllrb.com/docs/templates/) if you need to refresh your memory on how you built your blog and [Hugo's template](/layout/templates/) to learn Hugo's way.
+
+As a single reference data point, converting my templates for [heyitsalex.net](http://heyitsalex.net/) took me no more than a few hours.
+
+## Convert Jekyll plugins to Hugo shortcodes
+Jekyll has [plugins](http://jekyllrb.com/docs/plugins/); Hugo has [shortcodes](/doc/shortcodes/). It's fairly trivial to do a port.
+
+### Implementation
+As an example, I was using a custom [`image_tag`](https://github.com/alexandre-normand/alexandre-normand/blob/74bb12036a71334fdb7dba84e073382fc06908ec/_plugins/image_tag.rb) plugin to generate figures with caption when running Jekyll. As I read about shortcodes, I found Hugo had a nice built-in shortcode that does exactly the same thing.
+
+Jekyll's plugin:
+
+    module Jekyll
+      class ImageTag < Liquid::Tag
+        @url = nil
+        @caption = nil
+        @class = nil
+        @link = nil
+        // Patterns
+        IMAGE_URL_WITH_CLASS_AND_CAPTION =
+        IMAGE_URL_WITH_CLASS_AND_CAPTION_AND_LINK = /(\w+)(\s+)((https?:\/\/|\/)(\S+))(\s+)"(.*?)"(\s+)->((https?:\/\/|\/)(\S+))(\s*)/i
+        IMAGE_URL_WITH_CAPTION = /((https?:\/\/|\/)(\S+))(\s+)"(.*?)"/i
+        IMAGE_URL_WITH_CLASS = /(\w+)(\s+)((https?:\/\/|\/)(\S+))/i
+        IMAGE_URL = /((https?:\/\/|\/)(\S+))/i
+        def initialize(tag_name, markup, tokens)
+          super
+          if markup =~ IMAGE_URL_WITH_CLASS_AND_CAPTION_AND_LINK
+            @class   = $1
+            @url     = $3
+            @caption = $7
+            @link = $9
+          elsif markup =~ IMAGE_URL_WITH_CLASS_AND_CAPTION
+            @class   = $1
+            @url     = $3
+            @caption = $7
+          elsif markup =~ IMAGE_URL_WITH_CAPTION
+            @url     = $1
+            @caption = $5
+          elsif markup =~ IMAGE_URL_WITH_CLASS
+            @class = $1
+            @url   = $3
+          elsif markup =~ IMAGE_URL
+            @url = $1
+          end
+        end
+        def render(context)
+          if @class
+            source = "<figure class='#{@class}'>"
+          else
+            source = "<figure>"
+          end
+          if @link
+            source += "<a href=\"#{@link}\">"
+          end
+          source += "<img src=\"#{@url}\">"
+          if @link
+            source += "</a>"
+          end
+          source += "<figcaption>#{@caption}</figcaption>" if @caption
+          source += "</figure>"
+          source
+        end
+      end
+    end
+    Liquid::Template.register_tag('image', Jekyll::ImageTag)
+
+is written as this Hugo shortcode:
+
+    <!-- image -->
+    <figure {{ with .Get "class" }}class="{{.}}"{{ end }}>
+        {{ with .Get "link"}}<a href="{{.}}">{{ end }}
+            <img src="{{ .Get "src" }}" {{ if or (.Get "alt") (.Get "caption") }}alt="{{ with .Get "alt"}}{{.}}{{else}}{{ .Get "caption" }}{{ end }}"{{ end }} />
+        {{ if .Get "link"}}</a>{{ end }}
+        {{ if or (or (.Get "title") (.Get "caption")) (.Get "attr")}}
+        <figcaption>{{ if isset .Params "title" }}
+            {{ .Get "title" }}{{ end }}
+            {{ if or (.Get "caption") (.Get "attr")}}<p>
+            {{ .Get "caption" }}
+            {{ with .Get "attrlink"}}<a href="{{.}}"> {{ end }}
+                {{ .Get "attr" }}
+            {{ if .Get "attrlink"}}</a> {{ end }}
+            </p> {{ end }}
+        </figcaption>
+        {{ end }}
+    </figure>
+    <!-- image -->
+
+### Usage
+I simply changed:
+
+    {% image full http://farm5.staticflickr.com/4136/4829260124_57712e570a_o_d.jpg "One of my favorite touristy-type photos. I secretly waited for the good light while we were "having fun" and took this. Only regret: a stupid pole in the top-left corner of the frame I had to clumsily get rid of at post-processing." ->http://www.flickr.com/photos/alexnormand/4829260124/in/set-72157624547713078/ %}
+
+to this (this example uses a slightly extended version named `fig`, different than the built-in `figure`):
+
+    {{%/* fig class="full" src="http://farm5.staticflickr.com/4136/4829260124_57712e570a_o_d.jpg" title="One of my favorite touristy-type photos. I secretly waited for the good light while we were having fun and took this. Only regret: a stupid pole in the top-left corner of the frame I had to clumsily get rid of at post-processing." link="http://www.flickr.com/photos/alexnormand/4829260124/in/set-72157624547713078/" */%}}
+
+As a bonus, the shortcode named parameters are, arguably, more readable.
+
+## Finishing touches
+### Fix content
+Depending on the amount of customization that was done with each post with Jekyll, this step will require more or less effort. There are no hard and fast rules here except that `hugo server --watch` is your friend. Test your changes and fix errors as needed.
+
+### Clean up
+You'll want to remove the Jekyll configuration at this point. If you have anything else that isn't used, delete it.
+
+## A practical example in a diff
+[Hey, it's Alex](http://heyitsalex.net/) was migrated in less than a _father-with-kids day_ from Jekyll to Hugo. You can see all the changes (and screw-ups) by looking at this [diff](https://github.com/alexandre-normand/alexandre-normand/compare/869d69435bd2665c3fbf5b5c78d4c22759d7613a...b7f6605b1265e83b4b81495423294208cc74d610).
diff --git a/website/themes/book/exampleSite/resources/_gen/assets/scss/book.scss_50fc8c04e12a2f59027287995557ceff.content b/website/themes/book/exampleSite/resources/_gen/assets/scss/book.scss_50fc8c04e12a2f59027287995557ceff.content
new file mode 100644
index 000000000..d6ea2eecc
--- /dev/null
+++ b/website/themes/book/exampleSite/resources/_gen/assets/scss/book.scss_50fc8c04e12a2f59027287995557ceff.content
@@ -0,0 +1 @@
+@charset "UTF-8";:root{--gray-100:#f8f9fa;--gray-200:#e9ecef;--gray-500:#adb5bd;--color-link:#0055bb;--color-visited-link:#8440f1;--body-background:white;--body-font-color:black;--icon-filter:none;--hint-color-info:#6bf;--hint-color-warning:#fd6;--hint-color-danger:#f66}/*!normalize.css v8.0.1 | MIT License | github.com/necolas/normalize.css*/html{line-height:1.15;-webkit-text-size-adjust:100%}body{margin:0}main{display:block}h1{font-size:2em;margin:.67em 0}hr{box-sizing:content-box;height:0;overflow:visible}pre{font-family:monospace,monospace;font-size:1em}a{background-color:initial}abbr[title]{border-bottom:none;text-decoration:underline;text-decoration:underline dotted}b,strong{font-weight:bolder}code,kbd,samp{font-family:monospace,monospace;font-size:1em}small{font-size:80%}sub,sup{font-size:75%;line-height:0;position:relative;vertical-align:baseline}sub{bottom:-.25em}sup{top:-.5em}img{border-style:none}button,input,optgroup,select,textarea{font-family:inherit;font-size:100%;line-height:1.15;margin:0}button,input{overflow:visible}button,select{text-transform:none}button,[type=button],[type=reset],[type=submit]{-webkit-appearance:button}button::-moz-focus-inner,[type=button]::-moz-focus-inner,[type=reset]::-moz-focus-inner,[type=submit]::-moz-focus-inner{border-style:none;padding:0}button:-moz-focusring,[type=button]:-moz-focusring,[type=reset]:-moz-focusring,[type=submit]:-moz-focusring{outline:1px dotted ButtonText}fieldset{padding:.35em .75em .625em}legend{box-sizing:border-box;color:inherit;display:table;max-width:100%;padding:0;white-space:normal}progress{vertical-align:baseline}textarea{overflow:auto}[type=checkbox],[type=radio]{box-sizing:border-box;padding:0}[type=number]::-webkit-inner-spin-button,[type=number]::-webkit-outer-spin-button{height:auto}[type=search]{-webkit-appearance:textfield;outline-offset:-2px}[type=search]::-webkit-search-decoration{-webkit-appearance:none}::-webkit-file-upload-button{-webkit-appearance:button;font:inherit}details{display:block}summary{display:list-item}template{display:none}[hidden]{display:none}.flex{display:flex}.flex-auto{flex:auto}.flex-even{flex:1 1}.flex-wrap{flex-wrap:wrap}.justify-start{justify-content:flex-start}.justify-end{justify-content:flex-end}.justify-center{justify-content:center}.justify-between{justify-content:space-between}.align-center{align-items:center}.mx-auto{margin:0 auto}.text-center{text-align:center}.text-left{text-align:left}.text-right{text-align:right}.hidden{display:none}input.toggle{height:0;width:0;overflow:hidden;opacity:0;position:absolute}.clearfix::after{content:"";display:table;clear:both}html{font-size:16px;scroll-behavior:smooth;touch-action:manipulation}body{min-width:20rem;color:var(--body-font-color);background:var(--body-background);letter-spacing:.33px;font-weight:400;text-rendering:optimizeLegibility;-webkit-font-smoothing:antialiased;-moz-osx-font-smoothing:grayscale;box-sizing:border-box}body *{box-sizing:inherit}h1,h2,h3,h4,h5{font-weight:400}a{text-decoration:none;color:var(--color-link)}img{vertical-align:baseline}:focus{outline-style:auto;outline-color:currentColor;outline-color:-webkit-focus-ring-color}aside nav ul{padding:0;margin:0;list-style:none}aside nav ul li{margin:1em 0;position:relative}aside nav ul a{display:block}aside nav ul a:hover{opacity:.5}aside nav ul ul{padding-inline-start:1rem}ul.pagination{display:flex;justify-content:center;list-style-type:none}ul.pagination .page-item a{padding:1rem}.container{max-width:80rem;margin:0 auto}.book-icon{filter:var(--icon-filter)}.book-brand{margin-top:0}.book-brand img{height:1.5em;width:auto;vertical-align:middle;margin-inline-end:.5rem}.book-menu{flex:0 0 16rem;font-size:.875rem}.book-menu .book-menu-content{width:16rem;padding:1rem;background:var(--body-background);position:fixed;top:0;bottom:0;overflow-x:hidden;overflow-y:auto}.book-menu a,.book-menu label{color:inherit;cursor:pointer;word-wrap:break-word}.book-menu a.active{color:var(--color-link)}.book-menu input.toggle+label+ul{display:none}.book-menu input.toggle:checked+label+ul{display:block}.book-menu input.toggle+label::after{content:"▸"}.book-menu input.toggle:checked+label::after{content:"▾"}.book-section-flat{margin-bottom:2rem}.book-section-flat:not(:first-child){margin-top:2rem}.book-section-flat>a,.book-section-flat>span,.book-section-flat>label{font-weight:bolder}.book-section-flat>ul{padding-inline-start:0}.book-page{min-width:20rem;flex-grow:1;padding:1rem}.book-post{margin-bottom:3rem}.book-header{display:none;margin-bottom:1rem}.book-header label{line-height:0}.book-search{position:relative;margin:1rem 0;border-bottom:1px solid transparent}.book-search input{width:100%;padding:.5rem;border:0;border-radius:.25rem;background:var(--gray-100);color:var(--body-font-color)}.book-search input:required+.book-search-spinner{display:block}.book-search .book-search-spinner{position:absolute;top:0;margin:.5rem;margin-inline-start:calc(100% - 1.5rem);width:1rem;height:1rem;border:1px solid transparent;border-top-color:var(--body-font-color);border-radius:50%;animation:spin 1s ease infinite}@keyframes spin{100%{transform:rotate(360deg)}}.book-search small{opacity:.5}.book-toc{flex:0 0 16rem;font-size:.75rem}.book-toc .book-toc-content{width:16rem;padding:1rem;position:fixed;top:0;bottom:0;overflow-x:hidden;overflow-y:auto}.book-toc img{height:1em}.book-toc nav>ul>li:first-child{margin-top:0}.book-footer{padding-top:1rem;font-size:.875rem}.book-footer img{height:1em;margin-inline-end:.5rem}.book-comments{margin-top:1rem}.book-languages{position:relative;overflow:visible;padding:1rem;margin:-1rem}.book-languages ul{margin:0;padding:0;list-style:none}.book-languages ul li{white-space:nowrap;cursor:pointer}.book-languages:hover .book-languages-list,.book-languages:focus .book-languages-list,.book-languages:focus-within .book-languages-list{display:block}.book-languages .book-languages-list{display:none;position:absolute;bottom:100%;left:0;padding:.5rem 0;background:var(--body-background);box-shadow:0 0 .25rem rgba(0,0,0,.1)}.book-languages .book-languages-list li img{opacity:.25}.book-languages .book-languages-list li.active img,.book-languages .book-languages-list li:hover img{opacity:initial}.book-languages .book-languages-list a{color:inherit;padding:.5rem 1rem}.book-home{padding:1rem}.book-menu-content,.book-toc-content,.book-page,.book-header aside,.markdown{transition:.2s ease-in-out;transition-property:transform,margin,opacity,visibility;will-change:transform,margin,opacity}@media screen and (max-width:56rem){#menu-control,#toc-control{display:inline}.book-menu{visibility:hidden;margin-inline-start:-16rem;font-size:16px;z-index:1}.book-toc{display:none}.book-header{display:block}#menu-control:focus~main label[for=menu-control]{outline-style:auto;outline-color:currentColor;outline-color:-webkit-focus-ring-color}#menu-control:checked~main .book-menu{visibility:initial}#menu-control:checked~main .book-menu .book-menu-content{transform:translateX(16rem);box-shadow:0 0 .5rem rgba(0,0,0,.1)}#menu-control:checked~main .book-page{opacity:.25}#menu-control:checked~main .book-menu-overlay{display:block;position:absolute;top:0;bottom:0;left:0;right:0}#toc-control:focus~main label[for=toc-control]{outline-style:auto;outline-color:currentColor;outline-color:-webkit-focus-ring-color}#toc-control:checked~main .book-header aside{display:block}body[dir=rtl] #menu-control:checked+main .book-menu .book-menu-content{transform:translateX(-16rem)}}@media screen and (min-width:80rem){.book-page,.book-menu .book-menu-content,.book-toc .book-toc-content{padding:2rem 1rem}}@font-face{font-family:roboto;font-style:italic;font-weight:300;font-display:swap;src:local("Roboto Light Italic"),local("Roboto-LightItalic"),url(fonts/roboto-v19-latin-300italic.woff2)format("woff2"),url(fonts/roboto-v19-latin-300italic.woff)format("woff")}@font-face{font-family:roboto;font-style:normal;font-weight:400;font-display:swap;src:local("Roboto"),local("Roboto-Regular"),url(fonts/roboto-v19-latin-regular.woff2)format("woff2"),url(fonts/roboto-v19-latin-regular.woff)format("woff")}@font-face{font-family:roboto;font-style:normal;font-weight:700;font-display:swap;src:local("Roboto Bold"),local("Roboto-Bold"),url(fonts/roboto-v19-latin-700.woff2)format("woff2"),url(fonts/roboto-v19-latin-700.woff)format("woff")}@font-face{font-family:roboto mono;font-style:normal;font-weight:400;font-display:swap;src:local("Roboto Mono"),local("RobotoMono-Regular"),url(fonts/roboto-mono-v6-latin-regular.woff2)format("woff2"),url(fonts/roboto-mono-v6-latin-regular.woff)format("woff")}body{font-family:roboto,sans-serif}code{font-family:roboto mono,monospace}@media print{.book-menu,.book-footer,.book-toc{display:none}.book-header,.book-header aside{display:block}main{display:block!important}}.markdown{line-height:1.6}.markdown>:first-child{margin-top:0}.markdown h1,.markdown h2,.markdown h3,.markdown h4,.markdown h5,.markdown h6{font-weight:400;line-height:1;margin-top:1.5em;margin-bottom:1rem}.markdown h1 a.anchor,.markdown h2 a.anchor,.markdown h3 a.anchor,.markdown h4 a.anchor,.markdown h5 a.anchor,.markdown h6 a.anchor{opacity:0;font-size:.75em;vertical-align:middle;text-decoration:none}.markdown h1:hover a.anchor,.markdown h1 a.anchor:focus,.markdown h2:hover a.anchor,.markdown h2 a.anchor:focus,.markdown h3:hover a.anchor,.markdown h3 a.anchor:focus,.markdown h4:hover a.anchor,.markdown h4 a.anchor:focus,.markdown h5:hover a.anchor,.markdown h5 a.anchor:focus,.markdown h6:hover a.anchor,.markdown h6 a.anchor:focus{opacity:initial}.markdown h4,.markdown h5,.markdown h6{font-weight:bolder}.markdown h5{font-size:.875em}.markdown h6{font-size:.75em}.markdown b,.markdown optgroup,.markdown strong{font-weight:bolder}.markdown a{text-decoration:none}.markdown a:hover{text-decoration:underline}.markdown a:visited{color:var(--color-visited-link)}.markdown img{max-width:100%}.markdown code{padding:0 .25rem;background:var(--gray-200);border-radius:.25rem;font-size:.875em}.markdown pre{padding:1rem;background:var(--gray-100);border-radius:.25rem;overflow-x:auto}.markdown pre code{padding:0;background:0 0}.markdown blockquote{margin:1rem 0;padding:.5rem 1rem .5rem .75rem;border-inline-start:.25rem solid var(--gray-200);border-radius:.25rem}.markdown blockquote :first-child{margin-top:0}.markdown blockquote :last-child{margin-bottom:0}.markdown table{overflow:auto;display:block;border-spacing:0;border-collapse:collapse;margin-top:1rem;margin-bottom:1rem}.markdown table tr th,.markdown table tr td{padding:.5rem 1rem;border:1px solid var(--gray-200)}.markdown table tr:nth-child(2n){background:var(--gray-100)}.markdown hr{height:1px;border:none;background:var(--gray-200)}.markdown ul,.markdown ol{padding-inline-start:2rem}.markdown dl dt{font-weight:bolder;margin-top:1rem}.markdown dl dd{margin-inline-start:1rem;margin-bottom:1rem}.markdown .highlight table tr td:nth-child(1) pre{margin:0;padding-inline-end:0}.markdown .highlight table tr td:nth-child(2) pre{margin:0;padding-inline-start:0}.markdown details{padding:1rem;border:1px solid var(--gray-200);border-radius:.25rem}.markdown details summary{line-height:1;padding:1rem;margin:-1rem;cursor:pointer}.markdown details[open] summary{margin-bottom:0}.markdown figure{margin:1rem 0}.markdown figure figcaption p{margin-top:0}.markdown-inner>:first-child{margin-top:0}.markdown-inner>:last-child{margin-bottom:0}.markdown .book-expand{margin-top:1rem;margin-bottom:1rem;border:1px solid var(--gray-200);border-radius:.25rem;overflow:hidden}.markdown .book-expand .book-expand-head{background:var(--gray-100);padding:.5rem 1rem;cursor:pointer}.markdown .book-expand .book-expand-content{display:none;padding:1rem}.markdown .book-expand input[type=checkbox]:checked+.book-expand-content{display:block}.markdown .book-tabs{margin-top:1rem;margin-bottom:1rem;border:1px solid var(--gray-200);border-radius:.25rem;overflow:hidden;display:flex;flex-wrap:wrap}.markdown .book-tabs label{display:inline-block;padding:.5rem 1rem;border-bottom:1px transparent;cursor:pointer}.markdown .book-tabs .book-tabs-content{order:999;width:100%;border-top:1px solid var(--gray-100);padding:1rem;display:none}.markdown .book-tabs input[type=radio]:checked+label{border-bottom:1px solid var(--color-link)}.markdown .book-tabs input[type=radio]:checked+label+.book-tabs-content{display:block}.markdown .book-tabs input[type=radio]:focus+label{outline-style:auto;outline-color:currentColor;outline-color:-webkit-focus-ring-color}.markdown .book-columns{margin-left:-1rem;margin-right:-1rem}.markdown .book-columns>div{margin:1rem 0;min-width:10rem;padding:0 1rem}.markdown a.book-btn{display:inline-block;font-size:.875rem;color:var(--color-link);line-height:2rem;padding:0 1rem;border:1px solid var(--color-link);border-radius:.25rem;cursor:pointer}.markdown a.book-btn:hover{text-decoration:none}.markdown .book-hint.info{border-color:#6bf;background-color:rgba(102,187,255,.1)}.markdown .book-hint.warning{border-color:#fd6;background-color:rgba(255,221,102,.1)}.markdown .book-hint.danger{border-color:#f66;background-color:rgba(255,102,102,.1)}
\ No newline at end of file
diff --git a/website/themes/book/exampleSite/resources/_gen/assets/scss/book.scss_50fc8c04e12a2f59027287995557ceff.json b/website/themes/book/exampleSite/resources/_gen/assets/scss/book.scss_50fc8c04e12a2f59027287995557ceff.json
new file mode 100644
index 000000000..f22fdaa76
--- /dev/null
+++ b/website/themes/book/exampleSite/resources/_gen/assets/scss/book.scss_50fc8c04e12a2f59027287995557ceff.json
@@ -0,0 +1 @@
+{"Target":"book.min.5ac6c2989f0943405962be6800b442aef429ef26ade26545ecf0617a21d1197a.css","MediaType":"text/css","Data":{"Integrity":"sha256-WsbCmJ8JQ0BZYr5oALRCrvQp7yat4mVF7PBheiHRGXo="}}
\ No newline at end of file
diff --git a/website/themes/book/i18n/bn.yaml b/website/themes/book/i18n/bn.yaml
new file mode 100644
index 000000000..853b72745
--- /dev/null
+++ b/website/themes/book/i18n/bn.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: অনুসন্ধান
+
+- id: Edit this page
+  translation: এই পৃষ্ঠাটি সম্পাদনা করুন
+
+- id: Last modified by
+  translation: সর্বশেষ সম্পাদনা করেছেন
+
+- id: Expand
+  translation: বিস্তৃত করা
+
+- id: bookSearchConfig
+  translation: '{ cache: true }'
diff --git a/website/themes/book/i18n/cn.yaml b/website/themes/book/i18n/cn.yaml
new file mode 100644
index 000000000..ba91f48bb
--- /dev/null
+++ b/website/themes/book/i18n/cn.yaml
@@ -0,0 +1,21 @@
+# This should be removed in future, 'cn' is moved to `zh'
+- id: Search
+  translation: 搜索
+
+- id: Edit this page
+  translation: 编辑本页
+
+- id: Last modified by
+  translation: 最后修改者
+
+- id: Expand
+  translation: 展开
+
+- id: bookSearchConfig
+  translation: |
+    {
+      encode: false,
+      tokenize: function(str) {
+        return str.replace(/[\x00-\x7F]/g, '').split('');
+      }
+    }
diff --git a/website/themes/book/i18n/cs.yaml b/website/themes/book/i18n/cs.yaml
new file mode 100644
index 000000000..b93e63f2e
--- /dev/null
+++ b/website/themes/book/i18n/cs.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: Vyhledávat
+
+- id: Edit this page
+  translation: Upravit tuto stránku
+
+- id: Last modified by
+  translation: Autor poslední změny
+
+- id: Expand
+  translation: Rozbalit
+
+- id: bookSearchConfig
+  translation: '{ cache: true }'
diff --git a/website/themes/book/i18n/de.yaml b/website/themes/book/i18n/de.yaml
new file mode 100644
index 000000000..620160868
--- /dev/null
+++ b/website/themes/book/i18n/de.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: Suche
+
+- id: Edit this page
+  translation: Seite bearbeiten
+
+- id: Last modified by
+  translation: Zuletzt geändert von
+
+- id: Expand
+  translation: Erweitern
+
+- id: bookSearchConfig
+  translation: '{ cache: true }'
diff --git a/website/themes/book/i18n/en.yaml b/website/themes/book/i18n/en.yaml
new file mode 100644
index 000000000..94b9c7751
--- /dev/null
+++ b/website/themes/book/i18n/en.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: Search
+
+- id: Edit this page
+  translation: Edit this page
+
+- id: Last modified by
+  translation: Last modified by
+
+- id: Expand
+  translation: Expand
+
+- id: bookSearchConfig
+  translation: '{ cache: true }'
diff --git a/website/themes/book/i18n/es.yaml b/website/themes/book/i18n/es.yaml
new file mode 100644
index 000000000..db4da6016
--- /dev/null
+++ b/website/themes/book/i18n/es.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: Buscar
+
+- id: Edit this page
+  translation: Editar esta página
+
+- id: Last modified by
+  translation: Última modificación por
+
+- id: Expand
+  translation: Expand
+
+- id: bookSearchConfig
+  translation: '{ cache: true }'
diff --git a/website/themes/book/i18n/fr.yaml b/website/themes/book/i18n/fr.yaml
new file mode 100644
index 000000000..3397f9a8c
--- /dev/null
+++ b/website/themes/book/i18n/fr.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: Rechercher
+
+- id: Edit this page
+  translation: Modifier cette page
+
+- id: Last modified by
+  translation: Dernière modification par
+
+- id: Expand
+  translation: Développer
+
+- id: bookSearchConfig
+  translation: '{ cache: true }'
diff --git a/website/themes/book/i18n/ja.yaml b/website/themes/book/i18n/ja.yaml
new file mode 100644
index 000000000..77f87becf
--- /dev/null
+++ b/website/themes/book/i18n/ja.yaml
@@ -0,0 +1,20 @@
+- id: Search
+  translation: 検索
+
+- id: Edit this page
+  translation: このページを編集する
+
+- id: Last modified by
+  translation: 最終更新者
+
+- id: Expand
+  translation: 展開
+
+- id: bookSearchConfig
+  translation: |
+    {
+      encode: false,
+      tokenize: function(str) {
+        return str.replace(/[\x00-\x7F]/g, '').split('');
+      }
+    }
diff --git a/website/themes/book/i18n/jp.yaml b/website/themes/book/i18n/jp.yaml
new file mode 100644
index 000000000..c42d20ed3
--- /dev/null
+++ b/website/themes/book/i18n/jp.yaml
@@ -0,0 +1,21 @@
+# This should be removed in future, 'jp' is moved to `ja'
+- id: Search
+  translation: 検索
+
+- id: Edit this page
+  translation: このページを編集する
+
+- id: Last modified by
+  translation: 最終更新者
+
+- id: Expand
+  translation: 展開
+
+- id: bookSearchConfig
+  translation: |
+    {
+      encode: false,
+      tokenize: function(str) {
+        return str.replace(/[\x00-\x7F]/g, '').split('');
+      }
+    }
diff --git a/website/themes/book/i18n/ko.yaml b/website/themes/book/i18n/ko.yaml
new file mode 100644
index 000000000..5c1fdb07d
--- /dev/null
+++ b/website/themes/book/i18n/ko.yaml
@@ -0,0 +1,20 @@
+- id: Search
+  translation: Search
+
+- id: Edit this page
+  translation: Edit this page
+
+- id: Last modified by
+  translation: Last modified by
+
+- id: Expand
+  translation: Expand
+
+- id: bookSearchConfig
+  translation: |
+    {
+      encode: false,
+      tokenize: function(str) {
+        return str.replace(/[\x00-\x7F]/g, '').split('');
+      }
+    }
diff --git a/website/themes/book/i18n/nb.yaml b/website/themes/book/i18n/nb.yaml
new file mode 100644
index 000000000..4b8537865
--- /dev/null
+++ b/website/themes/book/i18n/nb.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: Søk
+
+- id: Edit this page
+  translation: Rediger denne siden
+
+- id: Last modified by
+  translation: Sist endret av
+
+- id: Expand
+  translation: Utvid
+
+- id: bookSearchConfig
+  translation: '{ cache: true }'
diff --git a/website/themes/book/i18n/pt.yaml b/website/themes/book/i18n/pt.yaml
new file mode 100644
index 000000000..96216a0cb
--- /dev/null
+++ b/website/themes/book/i18n/pt.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: Buscar
+
+- id: Edit this page
+  translation: Editar página
+
+- id: Last modified by
+  translation: Última modificação por
+
+- id: Expand
+  translation: Expandir
+
+- id: bookSearchConfig
+  translation: '{ cache: true }'
diff --git a/website/themes/book/i18n/ru.yaml b/website/themes/book/i18n/ru.yaml
new file mode 100644
index 000000000..b4fd64555
--- /dev/null
+++ b/website/themes/book/i18n/ru.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: Поиск
+
+- id: Edit this page
+  translation: Редактировать эту страницу
+
+- id: Last modified by
+  translation: Последнее изменение от
+
+- id: Expand
+  translation: Развернуть
+
+- id: bookSearchConfig
+  translation: '{ split: /[^a-zа-яё0-9\w]/gi }'
diff --git a/website/themes/book/i18n/sv.yaml b/website/themes/book/i18n/sv.yaml
new file mode 100644
index 000000000..c2b9d4e0b
--- /dev/null
+++ b/website/themes/book/i18n/sv.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: Sök
+
+- id: Edit this page
+  translation: Redigera denna sida
+
+- id: Last modified by
+  translation: Senast modifierad av
+
+- id: Expand
+  translation: Expandera
+
+- id: bookSearchConfig
+  translation: '{ cache: true }'
diff --git a/website/themes/book/i18n/tr.yaml b/website/themes/book/i18n/tr.yaml
new file mode 100644
index 000000000..518c84190
--- /dev/null
+++ b/website/themes/book/i18n/tr.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: Arama
+
+- id: Edit this page
+  translation: Bu sayfayı düzenle
+
+- id: Last modified by
+  translation: Son düzenleyen
+
+- id: Expand
+  translation: Genişlet
+
+- id: bookSearchConfig
+  translation: '{ cache: true }'
diff --git a/website/themes/book/i18n/uk.yaml b/website/themes/book/i18n/uk.yaml
new file mode 100644
index 000000000..03e21f773
--- /dev/null
+++ b/website/themes/book/i18n/uk.yaml
@@ -0,0 +1,14 @@
+- id: Search
+  translation: Пошук
+
+- id: Edit this page
+  translation: Редагувати цю сторінку
+
+- id: Last modified by
+  translation: Остання зміна від
+
+- id: Expand
+  translation: Розгорнути
+
+- id: bookSearchConfig
+  translation: '{ split: /[^a-zа-яё0-9\w]/gi }'
diff --git a/website/themes/book/i18n/zh-TW.yaml b/website/themes/book/i18n/zh-TW.yaml
new file mode 100644
index 000000000..7ed7f3af5
--- /dev/null
+++ b/website/themes/book/i18n/zh-TW.yaml
@@ -0,0 +1,20 @@
+- id: Search
+  translation: 搜索
+
+- id: Edit this page
+  translation: 編輯頁面
+
+- id: Last modified by
+  translation: 最後修改者
+
+- id: Expand
+  translation: 展開
+
+- id: bookSearchConfig
+  translation: |
+    {
+      encode: false,
+      tokenize: function(str) {
+        return str.replace(/[\x00-\x7F]/g, '').split('');
+      }
+    }
diff --git a/website/themes/book/i18n/zh.yaml b/website/themes/book/i18n/zh.yaml
new file mode 100644
index 000000000..6f326d132
--- /dev/null
+++ b/website/themes/book/i18n/zh.yaml
@@ -0,0 +1,20 @@
+- id: Search
+  translation: 搜索
+
+- id: Edit this page
+  translation: 编辑本页
+
+- id: Last modified by
+  translation: 最后修改者
+
+- id: Expand
+  translation: 展开
+
+- id: bookSearchConfig
+  translation: |
+    {
+      encode: false,
+      tokenize: function(str) {
+        return str.replace(/[\x00-\x7F]/g, '').split('');
+      }
+    }
diff --git a/website/themes/book/images/screenshot.png b/website/themes/book/images/screenshot.png
new file mode 100644
index 000000000..e7da28903
Binary files /dev/null and b/website/themes/book/images/screenshot.png differ
diff --git a/website/themes/book/images/tn.png b/website/themes/book/images/tn.png
new file mode 100644
index 000000000..129ea9aca
Binary files /dev/null and b/website/themes/book/images/tn.png differ
diff --git a/website/themes/book/layouts/404.html b/website/themes/book/layouts/404.html
new file mode 100644
index 000000000..3f76ed472
--- /dev/null
+++ b/website/themes/book/layouts/404.html
@@ -0,0 +1,35 @@
+<!DOCTYPE html>
+<html lang="{{ .Site.Language.Lang }}">
+
+  <head>
+    {{ partial "docs/html-head" . }}
+    {{ partial "docs/inject/head" . }}
+
+    <style>
+      .not-found {
+        text-align: center;
+      }
+      .not-found h1 {
+        margin: .25em 0 0 0;
+        opacity: .25;
+        font-size: 40vmin;
+      }
+    </style>
+  </head>
+
+  <body>
+    <main class="flex justify-center not-found">
+      <div>
+        <h1>404</h1>
+        <h2>Page Not Found</h2>
+        <h3>
+          <a href="{{ .Site.Home.RelPermalink }}">{{ .Site.Title }}</a>
+        </h3>
+      </div>
+    </main>
+
+    {{ partial "docs/inject/body" . }}
+    {{ template "_internal/google_analytics_async.html" . }}
+  </body>
+
+  </html>
diff --git a/website/themes/book/layouts/_default/_markup/render-heading.html b/website/themes/book/layouts/_default/_markup/render-heading.html
new file mode 100644
index 000000000..5439d2000
--- /dev/null
+++ b/website/themes/book/layouts/_default/_markup/render-heading.html
@@ -0,0 +1,4 @@
+<h{{ .Level }} id="{{ .Anchor | safeURL }}">
+  {{ .Text | safeHTML }}
+  <a class="anchor" href="#{{ .Anchor | safeURL }}">#</a>
+</h{{ .Level }}>
diff --git a/website/themes/book/layouts/_default/_markup/render-image.html b/website/themes/book/layouts/_default/_markup/render-image.html
new file mode 100644
index 000000000..148cbaf3e
--- /dev/null
+++ b/website/themes/book/layouts/_default/_markup/render-image.html
@@ -0,0 +1,19 @@
+{{- if .Page.Site.Params.BookPortableLinks -}}
+  {{- template "portable-image" . -}}
+{{- else -}}
+  <img src="{{ .Destination | safeURL }}" alt="{{ .Text }}" {{ with .Title }}title="{{ . }}"{{ end }}/>
+{{- end -}}
+
+{{- define "portable-image" -}}
+  {{- $isRemote := or (in .Destination "://") (strings.HasPrefix .Destination "//") }}
+  {{- if not $isRemote }}
+    {{- $path := print .Page.File.Dir .Destination }}
+    {{- if strings.HasPrefix .Destination "/" }}
+      {{- $path = print "/static" .Destination }}
+    {{- end }}
+    {{- if not (fileExists $path) }}
+      {{- warnf "Image '%s' not found in '%s'" .Destination .Page.File }}
+    {{- end }}
+  {{- end }}
+  <img src="{{ .Destination | safeURL }}" alt="{{ .Text }}" {{ with .Title }}title="{{ . }}"{{ end }}/>
+{{- end -}}
diff --git a/website/themes/book/layouts/_default/_markup/render-link.html b/website/themes/book/layouts/_default/_markup/render-link.html
new file mode 100644
index 000000000..5c5d67851
--- /dev/null
+++ b/website/themes/book/layouts/_default/_markup/render-link.html
@@ -0,0 +1,28 @@
+{{- if .Page.Site.Params.BookPortableLinks -}}
+  {{- template "portable-link" . -}}
+{{- else -}}
+  <a href="{{ .Destination | safeURL }}"{{ with .Title}} title="{{ . }}"{{ end }}{{ if strings.HasPrefix .Destination "http" }} target="_blank" rel="noreferrer noopener"{{ end }}>{{ .Text | safeHTML }}</a>
+{{- end -}}
+
+{{- define "portable-link" -}}
+  {{- $destination := .Destination }}
+  {{- $isRemote := or (in .Destination ":") (strings.HasPrefix .Destination "//") }}
+  {{- if not $isRemote }}
+    {{- $url := urls.Parse .Destination }}
+    {{- $path := strings.TrimSuffix "/_index.md" $url.Path }}
+    {{- $path = strings.TrimSuffix "/_index" $path }}
+    {{- $path = strings.TrimSuffix ".md" $path }}
+    {{- $page := .Page.GetPage $path }}
+    {{- if $page }}
+      {{- $destination = $page.RelPermalink }}
+      {{- if $url.Fragment }}
+        {{- $destination = print $destination "#" $url.Fragment }}
+      {{- end }}
+    {{- else if fileExists (print .Page.File.Dir .Destination) }}
+      <!-- Nothing -->
+    {{- else -}}
+      {{- warnf "Page '%s' not found in '%s'" .Destination .Page.File }}
+    {{- end }}
+  {{- end }}
+  <a href="{{ $destination | safeURL }}"{{ with .Title}} title="{{ . }}"{{ end }}>{{ .Text | safeHTML }}</a>
+{{- end -}}
diff --git a/website/themes/book/layouts/_default/baseof.html b/website/themes/book/layouts/_default/baseof.html
new file mode 100644
index 000000000..b30c0bee4
--- /dev/null
+++ b/website/themes/book/layouts/_default/baseof.html
@@ -0,0 +1,90 @@
+<!DOCTYPE html>
+<html lang="{{ .Site.Language.Lang }}" dir="{{ .Site.Language.LanguageDirection | default "ltr" }}">
+
+<head>
+  {{ hugo.Generator }}
+  {{ partial "docs/html-head" . }}
+  {{ partial "docs/inject/head" . }}
+</head>
+
+<body dir="{{ .Site.Language.LanguageDirection | default "ltr" }}">
+  <input type="checkbox" class="hidden toggle" id="menu-control" />
+  <input type="checkbox" class="hidden toggle" id="toc-control" />
+
+  <a href="https://github.com/halfrost/LeetCode-Go" class="github-corner" aria-label="View source on GitHub"><svg width="80" height="80" viewBox="0 0 250 250" style="fill:#64CEAA; color:#fff; position: absolute; top: 0; border: 0; right: 0;z-index: 100;" aria-hidden="true"><path d="M0,0 L115,115 L130,115 L142,142 L250,250 L250,0 Z"></path><path d="M128.3,109.0 C113.8,99.7 119.0,89.6 119.0,89.6 C122.0,82.7 120.5,78.6 120.5,78.6 C119.2,72.0 123.4,76.3 123.4,76.3 C127.3,80.9 125.5,87.3 125.5,87.3 C122.9,97.6 130.6,101.9 134.4,103.2" fill="currentColor" style="transform-origin: 130px 106px;" class="octo-arm"></path><path d="M115.0,115.0 C114.9,115.1 118.7,116.5 119.8,115.4 L133.7,101.6 C136.9,99.2 139.9,98.4 142.2,98.6 C133.8,88.0 127.5,74.4 143.8,58.0 C148.5,53.4 154.0,51.2 159.7,51.0 C160.3,49.4 163.2,43.6 171.4,40.1 C171.4,40.1 176.1,42.5 178.8,56.2 C183.1,58.6 187.2,61.8 190.9,65.4 C194.5,69.0 197.7,73.2 200.1,77.6 C213.8,80.2 216.3,84.9 216.3,84.9 C212.7,93.1 206.9,96.0 205.4,96.6 C205.1,102.4 203.0,107.8 198.3,112.5 C181.9,128.9 168.3,122.5 157.7,114.1 C157.9,116.9 156.7,120.9 152.7,124.9 L141.0,136.5 C139.8,137.7 141.6,141.9 141.8,141.8 Z" fill="currentColor" class="octo-body"></path></svg></a><style>.github-corner:hover .octo-arm{animation:octocat-wave 560ms ease-in-out}@keyframes octocat-wave{0%,100%{transform:rotate(0)}20%,60%{transform:rotate(-25deg)}40%,80%{transform:rotate(10deg)}}@media (max-width:500px){.github-corner:hover .octo-arm{animation:none}.github-corner .octo-arm{animation:octocat-wave 560ms ease-in-out}}</style>
+
+  <main class="container flex">
+    <aside class="book-menu">
+      <div class="book-menu-content">
+        {{ template "menu" . }} <!-- Left menu Content -->
+      </div>
+    </aside>
+
+    <div class="book-page">
+      <header class="book-header">
+        {{ template "header" . }} <!-- Mobile layout header -->
+      </header>
+
+      {{ partial "docs/inject/content-before" . }}
+      {{ template "main" . }} <!-- Page Content -->
+      {{ partial "docs/inject/content-after" . }}
+
+      <footer class="book-footer">
+        {{ template "footer" . }} <!-- Footer under page content -->
+        {{ partial "docs/inject/footer" . }}
+      </footer>
+
+      {{ template "comments" . }} <!-- Comments block -->
+
+      <label for="menu-control" class="hidden book-menu-overlay"></label>
+    </div>
+
+    {{ if default true (default .Site.Params.BookToC .Params.BookToC) }}
+    <aside class="book-toc">
+      <div class="book-toc-content">
+        {{ template "toc" . }} <!-- Table of Contents -->
+      </div>
+    </aside>
+    {{ end }}
+  </main>
+
+  {{ partial "docs/inject/body" . }}
+</body>
+
+</html>
+
+{{ define "menu" }}
+  {{ partial "docs/menu" . }}
+{{ end }}
+
+{{ define "header" }}
+  {{ partial "docs/header" . }}
+
+  {{ if default true (default .Site.Params.BookToC .Params.BookToC) }}
+  <aside class="hidden clearfix">
+    {{ template "toc" . }}
+  </aside>
+  {{ end }}
+{{ end }}
+
+{{ define "footer" }}
+  {{ partial "docs/footer" . }}
+{{ end }}
+
+{{ define "comments" }}
+  {{ if and .Content (default true (default .Site.Params.BookComments .Params.BookComments)) }}
+  <div class="book-comments">
+    {{- partial "docs/comments" . -}}
+  </div>
+  {{ end }}
+{{ end }}
+
+{{ define "main" }}
+  <article class="markdown">
+    {{- .Content -}}
+  </article>
+{{ end }}
+
+{{ define "toc" }}
+  {{ partial "docs/toc" . }}
+{{ end }}
diff --git a/website/themes/book/layouts/_default/list.html b/website/themes/book/layouts/_default/list.html
new file mode 100644
index 000000000..0dc8b68d4
--- /dev/null
+++ b/website/themes/book/layouts/_default/list.html
@@ -0,0 +1 @@
+{{ define "dummy" }}{{ end }}
diff --git a/website/themes/book/layouts/_default/single.html b/website/themes/book/layouts/_default/single.html
new file mode 100644
index 000000000..0dc8b68d4
--- /dev/null
+++ b/website/themes/book/layouts/_default/single.html
@@ -0,0 +1 @@
+{{ define "dummy" }}{{ end }}
diff --git a/website/themes/book/layouts/partials/docs/brand.html b/website/themes/book/layouts/partials/docs/brand.html
new file mode 100644
index 000000000..f8171df60
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/brand.html
@@ -0,0 +1,588 @@
+
+<a href="{{ .Site.BaseURL | relLangURL }}">
+    {{- with .Site.Params.BookLogo -}}
+    <img src="{{ . | relURL }}" alt="Logo" />
+    {{- end -}}
+    <svg version="1.1" id="Layer_1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" x="0px" y="0px" width="1075px" height="342px" viewBox="0 0 1075 342" enable-background="new 0 0 1075 342" xml:space="preserve" style="width: 224px;height:70px;">  <image id="image0" width="1075" height="342" x="0" y="0"
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diff --git a/website/themes/book/layouts/partials/docs/comments.html b/website/themes/book/layouts/partials/docs/comments.html
new file mode 100644
index 000000000..59c5f2293
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/comments.html
@@ -0,0 +1,2 @@
+<!-- This partial can be replaced to support other commenting engines -->
+{{ template "_internal/disqus.html" . }}
diff --git a/website/themes/book/layouts/partials/docs/date.html b/website/themes/book/layouts/partials/docs/date.html
new file mode 100644
index 000000000..73d69a332
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/date.html
@@ -0,0 +1,6 @@
+<!--
+  Returns formatted date.
+  Usage: partial "docs/date" (dict "Date" .Date "Format" .Site.Params.BookDateFormat)
+-->
+{{- $format := default "January 2, 2006" .Format -}}
+{{- return (.Date.Format $format) -}}
diff --git a/website/themes/book/layouts/partials/docs/footer.html b/website/themes/book/layouts/partials/docs/footer.html
new file mode 100644
index 000000000..b506f18c5
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/footer.html
@@ -0,0 +1,28 @@
+<div class="flex flex-wrap justify-between">
+{{ if .Site.IsMultiLingual }}
+  {{ partial "docs/languages" . }}
+{{ end }}
+
+{{ if and .GitInfo .Site.Params.BookRepo }}
+  <div>
+    {{- $date := partial "docs/date" (dict "Date" .GitInfo.AuthorDate.Local "Format" .Site.Params.BookDateFormat) -}}
+    {{- $commitPath := default "commit" .Site.Params.BookCommitPath -}}
+    <a class="flex align-center" href="{{ .Site.Params.BookRepo }}/{{ $commitPath }}/{{ .GitInfo.Hash }}" title='{{ i18n "Last modified by" }} {{ .GitInfo.AuthorName }} | {{ $date }}' target="_blank" rel="noopener">
+      <img src="{{ "svg/calendar.svg" | relURL }}" class="book-icon" alt="Calendar" />
+      <span>{{ $date }}</span>
+    </a>
+  </div>
+{{ end }}
+
+{{ if and .File .Site.Params.BookRepo .Site.Params.BookEditPath }}
+  <div>
+    <a class="flex align-center" href="{{ .Site.Params.BookRepo }}/{{ .Site.Params.BookEditPath }}/{{ .Site.Params.contentDir | default "content" }}/{{ replace .File.Path "\\" "/" }}" target="_blank" rel="noopener">
+      <img src="{{ "svg/edit.svg" | relURL }}" class="book-icon" alt="Edit" />
+      <span>{{ i18n "Edit this page" }}</span>
+    </a>
+  </div>
+{{ end }}
+{{ partial "docs/gitalk.html" . }}
+<span id="busuanzi_container_site_pv">本站总访问量: <span id="busuanzi_value_site_pv"></span> &nbsp;次</span>
+<span id="busuanzi_container_site_uv">您是本站第 <span id="busuanzi_value_site_uv"></span> &nbsp;位访问者</span>
+</div>
diff --git a/website/themes/book/layouts/partials/docs/gitalk.html b/website/themes/book/layouts/partials/docs/gitalk.html
new file mode 100644
index 000000000..0bd80aa0c
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/gitalk.html
@@ -0,0 +1,26 @@
+{{ if .Site.Params.enableGitalk }}
+<div id="gitalk-container" style="width : 100%;"></div>
+<link rel="stylesheet" href="https://unpkg.com/gitalk/dist/gitalk.css">
+<script src="https://unpkg.com/gitalk/dist/gitalk.min.js"></script>
+<script src="https://rawgit.com/qhh0205/78e9e0b1f3114db6737f3ed8cdd51d3a/raw/3894c5be5aa2378336b1f5ee0f296fa0b22d06e9/md5.min.js"></script>
+<script>
+  const gitalk = new Gitalk({
+    clientID: '{{ .Site.Params.Gitalk.clientID }}',
+    clientSecret: '{{ .Site.Params.Gitalk.clientSecret }}',
+    repo: '{{ .Site.Params.Gitalk.repo }}',
+    owner: '{{ .Site.Params.Gitalk.owner }}',
+    admin: ['{{ .Site.Params.Gitalk.owner }}'],
+    id: md5(location.pathname), // Ensure uniqueness and length less than 50
+    distractionFreeMode: false, // Facebook-like distraction free mode
+    body: location.href, // GitHub issue body.
+    proxy: '{{ .Site.Params.Gitalk.proxy }}' // Gitalke proxy
+  });
+  (function() {
+    if (["localhost", "127.0.0.1"].indexOf(window.location.hostname) != -1) {
+      document.getElementById('gitalk-container').innerHTML = 'Gitalk comments not available by default when the website is previewed locally.';
+      return;
+    }
+    gitalk.render('gitalk-container');
+  })();
+</script>
+{{ end }}
diff --git a/website/themes/book/layouts/partials/docs/header.html b/website/themes/book/layouts/partials/docs/header.html
new file mode 100644
index 000000000..089859cc0
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/header.html
@@ -0,0 +1,13 @@
+<div class="flex align-center justify-between">
+  <label for="menu-control">
+    <img src="{{ "svg/menu.svg" | relURL }}" class="book-icon" alt="Menu" />
+  </label>
+
+  <strong>{{ partial "docs/title" . }}</strong>
+
+  <label for="toc-control">
+    {{ if default true (default .Site.Params.BookToC .Params.BookToC) }}
+    <img src="{{ "svg/toc.svg" | relURL }}" class="book-icon" alt="Table of Contents" />
+    {{ end }}
+  </label>
+</div>
diff --git a/website/themes/book/layouts/partials/docs/html-head.html b/website/themes/book/layouts/partials/docs/html-head.html
new file mode 100644
index 000000000..cc915d0ef
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/html-head.html
@@ -0,0 +1,60 @@
+<meta charset="UTF-8">
+<meta name="viewport" content="width=device-width, initial-scale=1.0">
+<meta name="description" content="{{ default .Summary .Description }}">
+<meta name="theme-color" content="#FFFFFF">
+
+{{- template "_internal/opengraph.html" . -}}
+
+<title>{{ partial "docs/title" . }} | {{ .Site.Title -}}</title>
+
+{{- $manifest := resources.Get "manifest.json" | resources.ExecuteAsTemplate "manifest.json" . }}
+<link rel="manifest" href="{{ $manifest.RelPermalink }}">
+<link rel="icon" href="{{ "favicon.png" | relURL }}" type="image/x-icon">
+
+{{- range .Translations }}
+<link rel="alternate" hreflang="{{ .Language.Lang }}" href="{{ .Permalink }}" title="{{ partial "docs/title" . }}">
+{{ end -}}
+
+<!-- Theme stylesheet, you can customize scss by creating `assets/custom.scss` in your website -->
+{{- $styles := resources.Get "book.scss" | resources.ExecuteAsTemplate "book.scss" . | resources.ToCSS | resources.Minify | resources.Fingerprint }}
+<link rel="stylesheet" href="{{ $styles.RelPermalink }}" integrity="{{ $styles.Data.Integrity }}">
+
+{{- if default true .Site.Params.BookSearch }}
+{{- $searchJSFile := printf "%s.search.js" .Language.Lang }}
+{{- $searchJS := resources.Get "search.js" | resources.ExecuteAsTemplate $searchJSFile . | resources.Minify | resources.Fingerprint }}
+<script defer src="{{ $searchJS.RelPermalink }}" integrity="{{ $searchJS.Data.Integrity }}"></script>
+{{ end -}}
+
+{{- if .Site.Params.BookServiceWorker }}
+<!-- {{- $swJS := resources.Get "sw-register.js" | resources.ExecuteAsTemplate "sw.js" . | resources.Minify | resources.Fingerprint }}
+<script defer src="{{ $swJS.RelPermalink }}" integrity="{{ $swJS.Data.Integrity }}"></script> -->
+{{- $swJS := resources.Get "serviceworker-v1.js" | resources.Minify | resources.Fingerprint }}
+<script async>
+    if ('serviceWorker' in navigator) {
+      navigator.serviceWorker.register("{{ $swJS.RelPermalink }}").then(function() {
+          if (navigator.serviceWorker.controller) {
+            console.log('Assets cached by the controlling service worker.');
+          } else {
+            console.log('Please reload this page to allow the service worker to handle network operations.');
+          }
+      }).catch(function(error) {
+          console.log('ERROR: ' + error);
+        });
+  } else {
+      console.log('Service workers are not supported in the current browser.');
+  }
+</script>
+{{ end -}}
+
+<script async src="//busuanzi.ibruce.info/busuanzi/2.3/busuanzi.pure.mini.js"></script>
+{{- template "_internal/google_analytics_async.html" . -}}
+
+<!-- RSS -->
+{{- with .OutputFormats.Get "rss" -}}
+  {{ printf `<link rel="%s" type="%s" href="%s" title="%s" />` .Rel .MediaType.Type .Permalink $.Site.Title | safeHTML }}
+{{ end -}}
+
+{{ "<!--" | safeHTML }}
+Made with Book Theme
+https://github.com/alex-shpak/hugo-book
+{{ "-->" | safeHTML }}
diff --git a/website/themes/book/layouts/partials/docs/inject/body.html b/website/themes/book/layouts/partials/docs/inject/body.html
new file mode 100644
index 000000000..9b4a1fc14
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/inject/body.html
@@ -0,0 +1,26 @@
+<script src="/leetcode/prism.js"></script>
+
+<script async="async">
+    var html = document.documentElement;
+    var switcher = document.getElementsByClassName('js-toggle')[0];
+    if(typeof(Storage) !== 'undefined') {
+        if (localStorage.getItem('themeMode') == 'dark' || document.documentElement.getAttribute('data-theme') == 'dark') {
+            switcher.classList.add('js-toggle--checked');
+        }
+        switcher.classList.add('js-toggle--focus');
+    }
+    //Click on dark mode icon. Add dark mode classes and wrappers. Store user preference through sessions
+    switcher.addEventListener("click", function() {
+            this.classList.toggle('js-toggle--checked');
+            this.classList.add('js-toggle--focus');
+            //If dark mode is selected
+            if (this.classList.contains('js-toggle--checked')) {
+                html.setAttribute("data-theme","dark");
+                //Save user preference in storage
+                localStorage.setItem('themeMode', 'dark');
+            } else {
+                html.setAttribute("data-theme","light");
+                localStorage.setItem('themeMode', 'light');
+            }
+        })
+</script>
\ No newline at end of file
diff --git a/website/themes/book/layouts/partials/docs/inject/content-after.html b/website/themes/book/layouts/partials/docs/inject/content-after.html
new file mode 100644
index 000000000..e69de29bb
diff --git a/website/themes/book/layouts/partials/docs/inject/content-before.html b/website/themes/book/layouts/partials/docs/inject/content-before.html
new file mode 100644
index 000000000..e69de29bb
diff --git a/website/themes/book/layouts/partials/docs/inject/footer.html b/website/themes/book/layouts/partials/docs/inject/footer.html
new file mode 100644
index 000000000..e69de29bb
diff --git a/website/themes/book/layouts/partials/docs/inject/head.html b/website/themes/book/layouts/partials/docs/inject/head.html
new file mode 100644
index 000000000..83617b7fc
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/inject/head.html
@@ -0,0 +1,19 @@
+<link rel="stylesheet" href="/leetcode/prism.css"/>
+<link href="{{ "apple-touch-icon-60x60.png" | relURL }}" rel="apple-touch-icon" sizes="60x60" />
+<link href="{{ "apple-touch-icon-76x76.png" | relURL }}" rel="apple-touch-icon" sizes="76x76" />
+<link href="{{ "apple-touch-icon-120x120.png" | relURL }}" rel="apple-touch-icon" sizes="120x120" />
+<link href="{{ "apple-touch-icon-152x152.png" | relURL }}" rel="apple-touch-icon" sizes="152x152" />
+<link href="{{ "apple-touch-icon-180x180.png" | relURL }}" rel="apple-touch-icon" sizes="180x180"/>
+<link href="{{ "apple-touch-icon-512x512.png" | relURL }}" rel="apple-touch-icon" sizes="512x512" />
+<link href="{{ "apple-touch-icon-1024x1024.png" | relURL }}" rel="apple-touch-icon" sizes="1024x1024" />
+
+<script>
+    var html = document.documentElement;  
+    //Check Storage. Keep user preference on page reload
+    if (localStorage.getItem('themeMode') == 'dark') {
+      html.setAttribute("data-theme","dark");
+    }
+    if (localStorage.getItem('themeMode') == 'light') {
+      html.setAttribute("data-theme","light");
+    }
+  </script>
\ No newline at end of file
diff --git a/website/themes/book/layouts/partials/docs/inject/menu-after.html b/website/themes/book/layouts/partials/docs/inject/menu-after.html
new file mode 100644
index 000000000..e69de29bb
diff --git a/website/themes/book/layouts/partials/docs/inject/menu-before.html b/website/themes/book/layouts/partials/docs/inject/menu-before.html
new file mode 100644
index 000000000..e69de29bb
diff --git a/website/themes/book/layouts/partials/docs/inject/toc-after.html b/website/themes/book/layouts/partials/docs/inject/toc-after.html
new file mode 100644
index 000000000..e69de29bb
diff --git a/website/themes/book/layouts/partials/docs/inject/toc-before.html b/website/themes/book/layouts/partials/docs/inject/toc-before.html
new file mode 100644
index 000000000..e69de29bb
diff --git a/website/themes/book/layouts/partials/docs/languages.html b/website/themes/book/layouts/partials/docs/languages.html
new file mode 100644
index 000000000..402027566
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/languages.html
@@ -0,0 +1,31 @@
+<!-- Merge home and current page translations -->
+{{ $bookTranslatedOnly := default false .Site.Params.BookTranslatedOnly }}
+{{ $translations := dict }}
+{{ if (eq $bookTranslatedOnly false ) }}
+  {{ range .Site.Home.AllTranslations }}
+    {{ $translations = merge $translations (dict .Language.Lang .) }}
+  {{ end }}
+{{ end }}
+{{ range .Translations }}
+  {{ $translations = merge $translations (dict .Language.Lang .) }}
+{{ end }}
+
+<div class="book-languages" tabindex="0" aria-haspopup="true">
+  <ul>
+    <li class="flex align-center">
+      <img src="{{ "svg/translate.svg" | relURL }}" class="book-icon" alt="Languages" />
+      {{ $.Site.Language.LanguageName }}
+    </li> 
+  </ul>
+
+  <ul class="book-languages-list">
+    {{ range .Site.Languages }}{{ with index $translations .Lang }}
+    <li class="{{ if (eq $.Site.Language .Language) }}active{{ end }}">
+      <a href="{{ .Permalink }}" class="flex align-center">
+        <img src="{{ "svg/translate.svg" | relURL }}" class="book-icon" alt="Languages" />
+        {{ .Language.LanguageName }}
+      </a>
+    </li>
+    {{ end }}{{ end }}
+  </ul>
+</div>
diff --git a/website/themes/book/layouts/partials/docs/menu-bundle.html b/website/themes/book/layouts/partials/docs/menu-bundle.html
new file mode 100644
index 000000000..ddaac5230
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/menu-bundle.html
@@ -0,0 +1,4 @@
+{{ with .Site.GetPage .Site.Params.BookMenuBundle }}
+  {{- $href := printf "href=\"%s\"" $.RelPermalink -}}
+  {{- replace .Content $href (print $href "class=active") | safeHTML -}}
+{{ end }}
diff --git a/website/themes/book/layouts/partials/docs/menu-filetree.html b/website/themes/book/layouts/partials/docs/menu-filetree.html
new file mode 100644
index 000000000..a4835b214
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/menu-filetree.html
@@ -0,0 +1,45 @@
+{{ $bookSection := default "docs" .Site.Params.BookSection  }}
+{{ if eq $bookSection "*" }}
+  {{ $bookSection = "/" }}{{/* Backward compatibility */}}
+{{ end }}
+
+{{ with .Site.GetPage $bookSection }}
+  {{ template "book-section-children" (dict "Section" . "CurrentPage" $) }}
+{{ end }}
+
+{{ define "book-section-children" }}{{/* (dict "Section" .Section "CurrentPage" .CurrentPage) */}}
+  <ul>
+    {{ range (where .Section.Pages "Params.bookhidden" "ne" true) }}
+      {{ if .IsSection }}
+        <li {{- if .Params.BookFlatSection }} class="book-section-flat" {{ end -}}>
+          {{ template "book-page-link" (dict "Page" . "CurrentPage" $.CurrentPage) }}
+          {{ template "book-section-children" (dict "Section" . "CurrentPage" $.CurrentPage) }}
+        </li>
+      {{ else if and .IsPage .Content }}
+        <li>
+          {{ template "book-page-link" (dict "Page" . "CurrentPage" $.CurrentPage) }}
+        </li>
+      {{ end }}
+    {{ end }}
+  </ul>
+{{ end }}
+
+{{ define "book-page-link" }}{{/* (dict "Page" .Page "CurrentPage" .CurrentPage) */}}
+  {{ $current := eq .CurrentPage .Page }}
+  {{ $ancestor := .Page.IsAncestor .CurrentPage }}
+
+  {{ if .Page.Params.bookCollapseSection }}
+    <input type="checkbox" id="section-{{ md5 .Page }}" class="toggle" {{ if or $current $ancestor }}checked{{ end }} />
+    <label for="section-{{ md5 .Page }}" class="flex justify-between">
+      <a {{ if .Page.Content }}href="{{ .Page.Permalink }}"{{ end }} class="{{ if $current }}active{{ end }}">
+        {{- partial "docs/title" .Page -}}
+      </a>
+    </label>
+  {{ else if .Page.Content }}
+    <a href="{{ .Page.Permalink }}" class="{{ if $current }} active{{ end }}">
+      {{- partial "docs/title" .Page -}}
+    </a>
+  {{ else }}
+    <span>{{- partial "docs/title" .Page -}}</span>
+  {{ end }}
+{{ end }}
diff --git a/website/themes/book/layouts/partials/docs/menu-hugo.html b/website/themes/book/layouts/partials/docs/menu-hugo.html
new file mode 100644
index 000000000..02391ae83
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/menu-hugo.html
@@ -0,0 +1,28 @@
+<!--
+  This is template for hugo menus, accepts MenuEntity as context
+  https://gohugo.io/variables/menus/
+-->
+{{ if . }}
+  {{ template "book-menu-hugo" . }}
+{{ end }}
+
+{{ define "book-menu-hugo" }}
+<ul>
+  {{ range . }}
+  <li>
+    <a href="{{ .URL }}" {{ if not .Page }}target="_blank" rel="noopener"{{ end }}>
+      {{- .Pre -}}
+      {{ with .Page }}
+        {{ partial "docs/title" .Page }}
+      {{ else }}
+        {{ .Name }}
+      {{ end }}
+      {{- .Post -}}
+    </a>
+    {{- with .Children }}
+      {{ template "book-menu-hugo" . }}
+    {{- end }}
+  </li>
+  {{ end }}
+</ul>
+{{ end }}
diff --git a/website/themes/book/layouts/partials/docs/menu.html b/website/themes/book/layouts/partials/docs/menu.html
new file mode 100644
index 000000000..ee76e8739
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/menu.html
@@ -0,0 +1,24 @@
+<nav>
+{{ partial "docs/brand" . }}
+{{ partial "docs/search" . }}
+
+{{ partial "docs/inject/menu-before" . }}
+{{ partial "docs/menu-hugo" .Site.Menus.before }}
+
+<hr>
+
+{{ if .Site.Params.BookMenuBundle }}
+  {{ partial "docs/menu-bundle" . }}
+{{ else }}
+  {{ partial "docs/menu-filetree" . }}
+{{ end }}
+
+{{ partial "docs/menu-hugo" .Site.Menus.after }}
+{{ partial "docs/inject/menu-after" . }}
+</nav>
+
+<!-- Restore menu position as soon as possible to avoid flickering -->
+{{ $script := resources.Get "menu-reset.js" | resources.Minify }}
+{{ with $script.Content }}
+  <script>{{ . | safeJS }}</script>
+{{ end }}
diff --git a/website/themes/book/layouts/partials/docs/post-meta.html b/website/themes/book/layouts/partials/docs/post-meta.html
new file mode 100644
index 000000000..01adf8b7c
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/post-meta.html
@@ -0,0 +1,23 @@
+{{ with .Date }}
+  <h5>{{ partial "docs/date" (dict "Date" . "Format" $.Site.Params.BookDateFormat) }}</h5>
+{{ end }}
+
+{{ range $taxonomy, $_ := .Site.Taxonomies }}
+  {{ with $terms := $.GetTerms $taxonomy }}
+  <div>
+    {{ range $n, $term := $terms }}{{ if $n }}, {{ end }}
+      <a href="{{ $term.RelPermalink }}">{{ $term.Title }}</a>
+    {{- end }}
+  </div>
+  {{ end }}
+{{ end }}
+
+{{ if .Params.image }}
+<p>
+  {{ with .Resources.GetMatch .Params.image }}
+    <img src={{ .RelPermalink }} />
+  {{ else }}
+    <img src={{ .Params.image | relURL }} />
+  {{ end }}
+</p>
+{{ end }}
diff --git a/website/themes/book/layouts/partials/docs/search.html b/website/themes/book/layouts/partials/docs/search.html
new file mode 100644
index 000000000..aac63de02
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/search.html
@@ -0,0 +1,25 @@
+{{ if default true .Site.Params.BookSearch }}
+<div class="magic">
+    <div class="book-search">
+      <input type="text" id="book-search-input" placeholder="{{ i18n "Search" }}" aria-label="{{ i18n "Search" }}" maxlength="64" data-hotkeys="s/" />
+      <div class="book-search-spinner hidden"></div>
+      <ul id="book-search-results"></ul>
+    </div>
+    <div class="js-toggle-wrapper">
+      <div class="js-toggle">
+          <div class="js-toggle-track">
+              <div class="js-toggle-track-check">
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src="data:image/png;base64,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" role="presentation" style="pointer-events: none;" width="32" height="32">
+          </div>
+          <input class="js-toggle-screenreader-only" type="checkbox" aria-label="Switch between Dark and Light mode">
+      </div>
+    </div>
+</div>
+{{ end }}
diff --git a/website/themes/book/layouts/partials/docs/taxonomy.html b/website/themes/book/layouts/partials/docs/taxonomy.html
new file mode 100644
index 000000000..63ef315de
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/taxonomy.html
@@ -0,0 +1,19 @@
+<nav>
+  <ul>
+  {{ range $term, $_ := .Site.Taxonomies }}
+    {{ with $.Site.GetPage (printf "/%s" $term | urlize) }}
+    <li class="book-section-flat">
+      <strong>{{ .Title | title }}</strong>
+      <ul>
+      {{ range .Pages }}
+        <li class="flex justify-between">
+          <a href="{{ .RelPermalink }}">{{ .Title }}</a>
+          <span>{{ len .Pages }}</span>
+        </li>
+      {{ end }}
+      </ul>
+    </li>
+    {{ end }}
+  {{ end }}
+  </ul>
+</nav>
diff --git a/website/themes/book/layouts/partials/docs/title.html b/website/themes/book/layouts/partials/docs/title.html
new file mode 100644
index 000000000..a37ac7a27
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/title.html
@@ -0,0 +1,15 @@
+<!-- 
+  Partial to generate page name from Title or File name.
+  Accepts Page as context
+-->
+{{ $title := "" }}
+
+{{ if .Title }}
+  {{ $title = .Title }}
+{{ else if and .IsSection .File }}
+  {{ $title = path.Base .File.Dir | humanize | title }}
+{{ else if and .IsPage .File }}
+  {{ $title = .File.BaseFileName | humanize | title }}
+{{ end }}
+
+{{ return $title }}
diff --git a/website/themes/book/layouts/partials/docs/toc.html b/website/themes/book/layouts/partials/docs/toc.html
new file mode 100644
index 000000000..64697a470
--- /dev/null
+++ b/website/themes/book/layouts/partials/docs/toc.html
@@ -0,0 +1,3 @@
+{{ partial "docs/inject/toc-before" . }}
+{{ .TableOfContents }}
+{{ partial "docs/inject/toc-after" . }}
diff --git a/website/themes/book/layouts/posts/list.html b/website/themes/book/layouts/posts/list.html
new file mode 100644
index 000000000..badf0f618
--- /dev/null
+++ b/website/themes/book/layouts/posts/list.html
@@ -0,0 +1,22 @@
+{{ define "main" }}
+  {{ range sort .Paginator.Pages }}
+  <article class="markdown book-post">
+    <h2>
+      <a href="{{ .RelPermalink }}">{{ partial "docs/title.html" . }}</a>
+    </h2>
+    {{ partial "docs/post-meta" . }}
+    <p>
+      {{- .Summary -}}
+      {{ if .Truncated }}
+        <a href="{{ .RelPermalink }}">...</a>
+      {{ end }}
+    </p>
+  </article>
+  {{ end }}
+
+  {{ template "_internal/pagination.html" . }}
+{{ end }}
+
+{{ define "toc" }}
+  {{ partial "docs/taxonomy" . }}
+{{ end }}
diff --git a/website/themes/book/layouts/posts/single.html b/website/themes/book/layouts/posts/single.html
new file mode 100644
index 000000000..301ca1eee
--- /dev/null
+++ b/website/themes/book/layouts/posts/single.html
@@ -0,0 +1,13 @@
+{{ define "main" }}
+<article class="markdown">
+  <h1>
+    <a href="{{ .RelPermalink }}">{{ partial "docs/title.html" . }}</a>
+  </h1>
+  {{ partial "docs/post-meta" . }}
+  {{- .Content -}}
+</article>
+{{ end }}
+
+{{ define "toc" }}
+  {{ partial "docs/toc" . }}
+{{ end }}
diff --git a/website/themes/book/layouts/shortcodes/button.html b/website/themes/book/layouts/shortcodes/button.html
new file mode 100644
index 000000000..337a50f42
--- /dev/null
+++ b/website/themes/book/layouts/shortcodes/button.html
@@ -0,0 +1,12 @@
+{{ $ref := "" }}
+{{ $target := "" }}
+{{ with .Get "href" }}
+  {{ $ref = . }}
+  {{ $target = "_blank" }}
+{{ end }}
+{{ with .Get "relref" }}
+  {{ $ref = relref $ . }}
+{{ end }}
+<a {{ with $ref }} href="{{.}}" {{ end }} {{ with $target }} target="{{.}}" {{ end }} class="book-btn{{ with .Get "class" }} {{ . }}{{ end }}">
+  {{ $.Inner | markdownify }}
+</a>
diff --git a/website/themes/book/layouts/shortcodes/columns.html b/website/themes/book/layouts/shortcodes/columns.html
new file mode 100644
index 000000000..163f16ddf
--- /dev/null
+++ b/website/themes/book/layouts/shortcodes/columns.html
@@ -0,0 +1,7 @@
+<div class="book-columns flex flex-wrap">
+  {{ range split .Inner "<--->" }}
+  <div class="flex-even markdown-inner">
+    {{ . | markdownify }}
+  </div>
+  {{ end }}
+</div>
diff --git a/website/themes/book/layouts/shortcodes/details.html b/website/themes/book/layouts/shortcodes/details.html
new file mode 100644
index 000000000..f6fff3893
--- /dev/null
+++ b/website/themes/book/layouts/shortcodes/details.html
@@ -0,0 +1,6 @@
+<details {{ if or (.Get "open") (in .Params "open") }}open{{ end }}>
+  <summary>{{ cond .IsNamedParams (.Get "title") (.Get 0) }}</summary>
+  <div class="markdown-inner">
+    {{ .Inner | markdownify }}
+  </div>
+</details>
diff --git a/website/themes/book/layouts/shortcodes/expand.html b/website/themes/book/layouts/shortcodes/expand.html
new file mode 100644
index 000000000..f254944c6
--- /dev/null
+++ b/website/themes/book/layouts/shortcodes/expand.html
@@ -0,0 +1,13 @@
+{{ warnf "Expand shortcode is deprecated. Use 'details' instead." }}
+<div class="book-expand">
+  <label>
+    <div class="book-expand-head flex justify-between">
+      <span>{{ default (i18n "Expand") (.Get 0) }}</span>
+      <span>{{ default "↕" (.Get 1) }}</span>
+    </div>
+    <input type="checkbox" class="hidden" />
+    <div class="book-expand-content markdown-inner">
+      {{ .Inner | markdownify }}
+    </div>
+  </label>
+</div>
diff --git a/website/themes/book/layouts/shortcodes/hint.html b/website/themes/book/layouts/shortcodes/hint.html
new file mode 100644
index 000000000..066aa68cf
--- /dev/null
+++ b/website/themes/book/layouts/shortcodes/hint.html
@@ -0,0 +1,3 @@
+<blockquote class="book-hint {{ .Get 0 }}">
+  {{ .Inner | markdownify }}
+</blockquote>
diff --git a/website/themes/book/layouts/shortcodes/katex.html b/website/themes/book/layouts/shortcodes/katex.html
new file mode 100644
index 000000000..b255cb80b
--- /dev/null
+++ b/website/themes/book/layouts/shortcodes/katex.html
@@ -0,0 +1,13 @@
+{{- if not (.Page.Scratch.Get "katex") -}}
+<!-- Include katext only first time -->
+<link rel="stylesheet" href="{{ "katex/katex.min.css" | relURL }}" />
+<script defer src="{{ "katex/katex.min.js" | relURL }}"></script>
+<script defer src="{{ "katex/auto-render.min.js" | relURL }}" onload="renderMathInElement(document.body);"></script>
+{{- .Page.Scratch.Set "katex" true -}}
+{{- end -}}
+
+<span {{- with .Get "class" }} class="{{ . }}"{{ end }}>
+  {{ cond (in .Params "display") "\\[" "\\(" -}}
+  {{- trim .Inner "\n" -}}
+  {{- cond (in .Params "display") "\\]" "\\)" }}
+</span>
diff --git a/website/themes/book/layouts/shortcodes/mermaid.html b/website/themes/book/layouts/shortcodes/mermaid.html
new file mode 100644
index 000000000..ff9033503
--- /dev/null
+++ b/website/themes/book/layouts/shortcodes/mermaid.html
@@ -0,0 +1,12 @@
+{{ if not (.Page.Scratch.Get "mermaid") }}
+<!-- Include mermaid only first time -->
+<script src="{{ "mermaid.min.js" | relURL }}"></script>
+{{ with resources.Get "mermaid.json" }}
+  <script>mermaid.initialize({{ .Content | safeJS }})</script>
+{{ end }}
+{{ .Page.Scratch.Set "mermaid" true }}
+{{ end }}
+
+<p class="mermaid{{ with .Get "class" }} {{ . }}{{ end }}">
+  {{- .Inner -}}
+</p>
diff --git a/website/themes/book/layouts/shortcodes/section.html b/website/themes/book/layouts/shortcodes/section.html
new file mode 100644
index 000000000..21d2e7541
--- /dev/null
+++ b/website/themes/book/layouts/shortcodes/section.html
@@ -0,0 +1,10 @@
+<dl>
+{{ range .Page.Pages }}
+  <dt>
+    <a href="{{ .RelPermalink }}">{{ partial "docs/title" . }}</a>
+  </dt>
+  <dd>
+    {{ default .Summary .Description }}
+  </dd>
+{{ end }}
+</dl>
diff --git a/website/themes/book/layouts/shortcodes/tab.html b/website/themes/book/layouts/shortcodes/tab.html
new file mode 100644
index 000000000..e2a207ca9
--- /dev/null
+++ b/website/themes/book/layouts/shortcodes/tab.html
@@ -0,0 +1,12 @@
+{{ if .Parent }}
+  {{ $name := .Get 0 }}
+  {{ $group := printf "tabs-%s" (.Parent.Get 0) }}
+
+  {{ if not (.Parent.Scratch.Get $group) }}
+    {{ .Parent.Scratch.Set $group slice }}
+  {{ end }}
+
+  {{ .Parent.Scratch.Add $group (dict "Name" $name "Content" .Inner) }}
+{{ else }}
+  {{ errorf "%q: 'tab' shortcode must be inside 'tabs' shortcode" .Page.Path }}
+{{ end}}
diff --git a/website/themes/book/layouts/shortcodes/tabs.html b/website/themes/book/layouts/shortcodes/tabs.html
new file mode 100644
index 000000000..01c1a63bf
--- /dev/null
+++ b/website/themes/book/layouts/shortcodes/tabs.html
@@ -0,0 +1,15 @@
+{{ if .Inner }}{{ end }}
+{{ $id := .Get 0 }}
+{{ $group := printf "tabs-%s" $id }}
+
+<div class="book-tabs">
+{{- range $index, $tab := .Scratch.Get $group -}}
+  <input type="radio" class="toggle" name="{{ $group }}" id="{{ printf "%s-%d" $group $index }}" {{ if not $index }}checked="checked"{{ end }} />
+  <label for="{{ printf "%s-%d" $group $index }}">
+    {{- $tab.Name -}}
+  </label>
+  <div class="book-tabs-content markdown-inner">
+    {{- .Content | markdownify -}}
+  </div>
+{{- end -}}
+</div>
diff --git a/website/themes/book/layouts/taxonomy/list.html b/website/themes/book/layouts/taxonomy/list.html
new file mode 100644
index 000000000..31ecd9035
--- /dev/null
+++ b/website/themes/book/layouts/taxonomy/list.html
@@ -0,0 +1,13 @@
+{{ define "main" }}
+<article class="markdown">
+  <h1>{{ .Title | title }}</h1>
+  {{ $taxonomies := index .Site.Taxonomies .Page.Type }}
+  {{ range $taxonomies }}
+    <div><a href="{{ .Page.Permalink }}">{{ .Page.Title }} <sup>{{ .Count }}</sup></a></div>
+  {{ end }}
+</article>
+{{ end }}
+
+{{ define "toc" }}
+  {{ partial "docs/taxonomy" . }}
+{{ end }}
diff --git a/website/themes/book/layouts/taxonomy/taxonomy.html b/website/themes/book/layouts/taxonomy/taxonomy.html
new file mode 100644
index 000000000..badf0f618
--- /dev/null
+++ b/website/themes/book/layouts/taxonomy/taxonomy.html
@@ -0,0 +1,22 @@
+{{ define "main" }}
+  {{ range sort .Paginator.Pages }}
+  <article class="markdown book-post">
+    <h2>
+      <a href="{{ .RelPermalink }}">{{ partial "docs/title.html" . }}</a>
+    </h2>
+    {{ partial "docs/post-meta" . }}
+    <p>
+      {{- .Summary -}}
+      {{ if .Truncated }}
+        <a href="{{ .RelPermalink }}">...</a>
+      {{ end }}
+    </p>
+  </article>
+  {{ end }}
+
+  {{ template "_internal/pagination.html" . }}
+{{ end }}
+
+{{ define "toc" }}
+  {{ partial "docs/taxonomy" . }}
+{{ end }}
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diff --git a/website/themes/book/static/LeetCode_logo2048.png b/website/themes/book/static/LeetCode_logo2048.png
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diff --git a/website/themes/book/static/apple-touch-icon-180x180.png b/website/themes/book/static/apple-touch-icon-180x180.png
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diff --git a/website/themes/book/static/apple-touch-icon-512x512.png b/website/themes/book/static/apple-touch-icon-512x512.png
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diff --git a/website/themes/book/static/apple-touch-icon-76x76.png b/website/themes/book/static/apple-touch-icon-76x76.png
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diff --git a/website/themes/book/static/flexsearch.min.js b/website/themes/book/static/flexsearch.min.js
new file mode 100644
index 000000000..984d8c6e6
--- /dev/null
+++ b/website/themes/book/static/flexsearch.min.js
@@ -0,0 +1,42 @@
+/*
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+ Copyright 2019 Nextapps GmbH
+ Author: Thomas Wilkerling
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+ https://github.com/nextapps-de/flexsearch
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diff --git a/website/themes/book/static/js/sw-toolbox.js b/website/themes/book/static/js/sw-toolbox.js
new file mode 100644
index 000000000..dacb8553a
--- /dev/null
+++ b/website/themes/book/static/js/sw-toolbox.js
@@ -0,0 +1,16 @@
+/*
+ Copyright 2016 Google Inc. All Rights Reserved.
+
+ Licensed under the Apache License, Version 2.0 (the "License");
+ you may not use this file except in compliance with the License.
+ You may obtain a copy of the License at
+
+     http://www.apache.org/licenses/LICENSE-2.0
+
+ Unless required by applicable law or agreed to in writing, software
+ distributed under the License is distributed on an "AS IS" BASIS,
+ WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
+ See the License for the specific language governing permissions and
+ limitations under the License.
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diff --git a/website/themes/book/static/katex/auto-render.min.js b/website/themes/book/static/katex/auto-render.min.js
new file mode 100644
index 000000000..3a6d66390
--- /dev/null
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new file mode 100644
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t(t,e){this.mode=void 0,this.gullet=void 0,this.settings=void 0,this.leftrightDepth=void 0,this.nextToken=void 0,this.mode="math",this.gullet=new wa(t,e,this.mode),this.settings=e,this.leftrightDepth=0}var e=t.prototype;return e.expect=function(t,e){if(void 0===e&&(e=!0),this.fetch().text!==t)throw new o("Expected '"+t+"', got '"+this.fetch().text+"'",this.fetch());e&&this.consume()},e.consume=function(){this.nextToken=null},e.fetch=function(){return null==this.nextToken&&(this.nextToken=this.gullet.expandNextToken()),this.nextToken},e.switchMode=function(t){this.mode=t,this.gullet.switchMode(t)},e.parse=function(){this.gullet.beginGroup(),this.settings.colorIsTextColor&&this.gullet.macros.set("\\color","\\textcolor");var t=this.parseExpression(!1);return this.expect("EOF"),this.gullet.endGroup(),t},e.parseExpression=function(e,r){for(var a=[];;){"math"===this.mode&&this.consumeSpaces();var n=this.fetch();if(-1!==t.endOfExpression.indexOf(n.text))break;if(r&&n.text===r)break;if(e&&ia[n.text]&&ia[n.text].infix)break;var i=this.parseAtom(r);if(!i)break;a.push(i)}return"text"===this.mode&&this.formLigatures(a),this.handleInfixNodes(a)},e.handleInfixNodes=function(t){for(var e,r=-1,a=0;a<t.length;a++){var n=Vt(t[a],"infix");if(n){if(-1!==r)throw new o("only one infix operator per group",n.token);r=a,e=n.replaceWith}}if(-1!==r&&e){var i,s,h=t.slice(0,r),l=t.slice(r+1);return i=1===h.length&&"ordgroup"===h[0].type?h[0]:{type:"ordgroup",mode:this.mode,body:h},s=1===l.length&&"ordgroup"===l[0].type?l[0]:{type:"ordgroup",mode:this.mode,body:l},["\\\\abovefrac"===e?this.callFunction(e,[i,t[r],s],[]):this.callFunction(e,[i,s],[])]}return t},e.handleSupSubscript=function(e){var r=this.fetch(),a=r.text;this.consume();var n=this.parseGroup(e,!1,t.SUPSUB_GREEDINESS,void 0,void 0,!0);if(!n)throw new o("Expected group after '"+a+"'",r);return n},e.formatUnsupportedCmd=function(t){for(var e=[],r=0;r<t.length;r++)e.push({type:"textord",mode:"text",text:t[r]});var a={type:"text",mode:this.mode,body:e};return{type:"color",mode:this.mode,color:this.settings.errorColor,body:[a]}},e.parseAtom=function(t){var e,r,a=this.parseGroup("atom",!1,null,t);if("text"===this.mode)return a;for(;;){this.consumeSpaces();var n=this.fetch();if("\\limits"===n.text||"\\nolimits"===n.text){var i=Vt(a,"op");if(i){var s="\\limits"===n.text;i.limits=s,i.alwaysHandleSupSub=!0}else{if(!(i=Vt(a,"operatorname"))||!i.alwaysHandleSupSub)throw new o("Limit controls must follow a math operator",n);var h="\\limits"===n.text;i.limits=h}this.consume()}else if("^"===n.text){if(e)throw new o("Double superscript",n);e=this.handleSupSubscript("superscript")}else if("_"===n.text){if(r)throw new o("Double subscript",n);r=this.handleSupSubscript("subscript")}else{if("'"!==n.text)break;if(e)throw new o("Double superscript",n);var l={type:"textord",mode:this.mode,text:"\\prime"},m=[l];for(this.consume();"'"===this.fetch().text;)m.push(l),this.consume();"^"===this.fetch().text&&m.push(this.handleSupSubscript("superscript")),e={type:"ordgroup",mode:this.mode,body:m}}}return e||r?{type:"supsub",mode:this.mode,base:a,sup:e,sub:r}:a},e.parseFunction=function(t,e,r){var a=this.fetch(),n=a.text,i=ia[n];if(!i)return null;if(this.consume(),null!=r&&i.greediness<=r)throw new o("Got function '"+n+"' with no arguments"+(e?" as "+e:""),a);if("text"===this.mode&&!i.allowedInText)throw new o("Can't use function '"+n+"' in text mode",a);if("math"===this.mode&&!1===i.allowedInMath)throw new o("Can't use function '"+n+"' in math mode",a);var s=this.parseArguments(n,i),h=s.args,l=s.optArgs;return this.callFunction(n,h,l,a,t)},e.callFunction=function(t,e,r,a,n){var i={funcName:t,parser:this,token:a,breakOnTokenText:n},s=ia[t];if(s&&s.handler)return s.handler(i,e,r);throw new o("No function handler for 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n=new Error(t);throw n.hash=e,n}this.trace(t)},parse:function(t){var e=this,n=[0],r=[],i=[null],a=[],o=this.table,s="",c=0,u=0,l=0,h=2,f=1,d=a.slice.call(arguments,1),p=Object.create(this.lexer),g={yy:{}};for(var y in this.yy)Object.prototype.hasOwnProperty.call(this.yy,y)&&(g.yy[y]=this.yy[y]);p.setInput(t,g.yy),g.yy.lexer=p,g.yy.parser=this,void 0===p.yylloc&&(p.yylloc={});var v=p.yylloc;a.push(v);var m=p.options&&p.options.ranges;function b(){var t;return"number"!=typeof(t=r.pop()||p.lex()||f)&&(t instanceof Array&&(t=(r=t).pop()),t=e.symbols_[t]||t),t}"function"==typeof g.yy.parseError?this.parseError=g.yy.parseError:this.parseError=Object.getPrototypeOf(this).parseError;for(var x,_,k,w,E,T,C,A,S,M={};;){if(k=n[n.length-1],this.defaultActions[k]?w=this.defaultActions[k]:(null==x&&(x=b()),w=o[k]&&o[k][x]),void 0===w||!w.length||!w[0]){var O="";for(T in S=[],o[k])this.terminals_[T]&&T>h&&S.push("'"+this.terminals_[T]+"'");O=p.showPosition?"Parse error on line 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E;C&&(n=n.slice(0,-1*C*2),i=i.slice(0,-1*C),a=a.slice(0,-1*C)),n.push(this.productions_[w[1]][0]),i.push(M.$),a.push(M._$),A=o[n[n.length-2]][n[n.length-1]],n.push(A);break;case 3:return!0}}return!0}},M={EOF:1,parseError:function(t,e){if(!this.yy.parser)throw new Error(t);this.yy.parser.parseError(t,e)},setInput:function(t,e){return this.yy=e||this.yy||{},this._input=t,this._more=this._backtrack=this.done=!1,this.yylineno=this.yyleng=0,this.yytext=this.matched=this.match="",this.conditionStack=["INITIAL"],this.yylloc={first_line:1,first_column:0,last_line:1,last_column:0},this.options.ranges&&(this.yylloc.range=[0,0]),this.offset=0,this},input:function(){var t=this._input[0];return this.yytext+=t,this.yyleng++,this.offset++,this.match+=t,this.matched+=t,t.match(/(?:\r\n?|\n).*/g)?(this.yylineno++,this.yylloc.last_line++):this.yylloc.last_column++,this.options.ranges&&this.yylloc.range[1]++,this._input=this._input.slice(1),t},unput:function(t){var e=t.length,n=t.split(/(?:\r\n?|\n)/g);this._input=t+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-e),this.offset-=e;var r=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),n.length-1&&(this.yylineno-=n.length-1);var i=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:n?(n.length===r.length?this.yylloc.first_column:0)+r[r.length-n.length].length-n[0].length:this.yylloc.first_column-e},this.options.ranges&&(this.yylloc.range=[i[0],i[0]+this.yyleng-e]),this.yyleng=this.yytext.length,this},more:function(){return this._more=!0,this},reject:function(){return this.options.backtrack_lexer?(this._backtrack=!0,this):this.parseError("Lexical error on line "+(this.yylineno+1)+". You can only invoke reject() in the lexer when the lexer is of the backtracking persuasion (options.backtrack_lexer = true).\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},less:function(t){this.unput(this.match.slice(t))},pastInput:function(){var t=this.matched.substr(0,this.matched.length-this.match.length);return(t.length>20?"...":"")+t.substr(-20).replace(/\n/g,"")},upcomingInput:function(){var t=this.match;return t.length<20&&(t+=this._input.substr(0,20-t.length)),(t.substr(0,20)+(t.length>20?"...":"")).replace(/\n/g,"")},showPosition:function(){var t=this.pastInput(),e=new Array(t.length+1).join("-");return t+this.upcomingInput()+"\n"+e+"^"},test_match:function(t,e){var n,r,i;if(this.options.backtrack_lexer&&(i={yylineno:this.yylineno,yylloc:{first_line:this.yylloc.first_line,last_line:this.last_line,first_column:this.yylloc.first_column,last_column:this.yylloc.last_column},yytext:this.yytext,match:this.match,matches:this.matches,matched:this.matched,yyleng:this.yyleng,offset:this.offset,_more:this._more,_input:this._input,yy:this.yy,conditionStack:this.conditionStack.slice(0),done:this.done},this.options.ranges&&(i.yylloc.range=this.yylloc.range.slice(0))),(r=t[0].match(/(?:\r\n?|\n).*/g))&&(this.yylineno+=r.length),this.yylloc={first_line:this.yylloc.last_line,last_line:this.yylineno+1,first_column:this.yylloc.last_column,last_column:r?r[r.length-1].length-r[r.length-1].match(/\r?\n?/)[0].length:this.yylloc.last_column+t[0].length},this.yytext+=t[0],this.match+=t[0],this.matches=t,this.yyleng=this.yytext.length,this.options.ranges&&(this.yylloc.range=[this.offset,this.offset+=this.yyleng]),this._more=!1,this._backtrack=!1,this._input=this._input.slice(t[0].length),this.matched+=t[0],n=this.performAction.call(this,this.yy,this,e,this.conditionStack[this.conditionStack.length-1]),this.done&&this._input&&(this.done=!1),n)return n;if(this._backtrack){for(var a in i)this[a]=i[a];return!1}return!1},next:function(){if(this.done)return this.EOF;var t,e,n,r;this._input||(this.done=!0),this._more||(this.yytext="",this.match="");for(var i=this._currentRules(),a=0;a<i.length;a++)if((n=this._input.match(this.rules[i[a]]))&&(!e||n[0].length>e[0].length)){if(e=n,r=a,this.options.backtrack_lexer){if(!1!==(t=this.test_match(n,i[a])))return t;if(this._backtrack){e=!1;continue}return!1}if(!this.options.flex)break}return e?!1!==(t=this.test_match(e,i[r]))&&t:""===this._input?this.EOF:this.parseError("Lexical error on line "+(this.yylineno+1)+". Unrecognized text.\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},lex:function(){var t=this.next();return t||this.lex()},begin:function(t){this.conditionStack.push(t)},popState:function(){return this.conditionStack.length-1>0?this.conditionStack.pop():this.conditionStack[0]},_currentRules:function(){return this.conditionStack.length&&this.conditionStack[this.conditionStack.length-1]?this.conditions[this.conditionStack[this.conditionStack.length-1]].rules:this.conditions.INITIAL.rules},topState:function(t){return(t=this.conditionStack.length-1-Math.abs(t||0))>=0?this.conditionStack[t]:"INITIAL"},pushState:function(t){this.begin(t)},stateStackSize:function(){return this.conditionStack.length},options:{"case-insensitive":!0},performAction:function(t,e,n,r){switch(n){case 0:return this.begin("open_directive"),58;case 1:return this.begin("type_directive"),59;case 2:return this.popState(),this.begin("arg_directive"),14;case 3:return this.popState(),this.popState(),61;case 4:return 60;case 5:return 5;case 6:case 7:case 8:case 9:case 10:break;case 11:return this.begin("ID"),16;case 12:return e.yytext=e.yytext.trim(),this.begin("ALIAS"),48;case 13:return this.popState(),this.popState(),this.begin("LINE"),18;case 14:return this.popState(),this.popState(),5;case 15:return this.begin("LINE"),27;case 16:return this.begin("LINE"),29;case 17:return this.begin("LINE"),30;case 18:return this.begin("LINE"),31;case 19:return this.begin("LINE"),36;case 20:return this.begin("LINE"),33;case 21:return this.begin("LINE"),35;case 22:return this.popState(),19;case 23:return 28;case 24:return 43;case 25:return 44;case 26:return 39;case 27:return 37;case 28:return this.begin("ID"),22;case 29:return this.begin("ID"),23;case 30:return 25;case 31:return 7;case 32:return 21;case 33:return 42;case 34:return 5;case 35:return e.yytext=e.yytext.trim(),48;case 36:return 51;case 37:return 52;case 38:return 49;case 39:return 50;case 40:return 53;case 41:return 54;case 42:return 55;case 43:return 56;case 44:return 57;case 45:return 46;case 46:return 47;case 47:return 5;case 48:return"INVALID"}},rules:[/^(?:%%\{)/i,/^(?:((?:(?!\}%%)[^:.])*))/i,/^(?::)/i,/^(?:\}%%)/i,/^(?:((?:(?!\}%%).|\n)*))/i,/^(?:[\n]+)/i,/^(?:\s+)/i,/^(?:((?!\n)\s)+)/i,/^(?:#[^\n]*)/i,/^(?:%(?!\{)[^\n]*)/i,/^(?:[^\}]%%[^\n]*)/i,/^(?:participant\b)/i,/^(?:[^\->:\n,;]+?(?=((?!\n)\s)+as(?!\n)\s|[#\n;]|$))/i,/^(?:as\b)/i,/^(?:(?:))/i,/^(?:loop\b)/i,/^(?:rect\b)/i,/^(?:opt\b)/i,/^(?:alt\b)/i,/^(?:else\b)/i,/^(?:par\b)/i,/^(?:and\b)/i,/^(?:(?:[:]?(?:no)?wrap)?[^#\n;]*)/i,/^(?:end\b)/i,/^(?:left of\b)/i,/^(?:right of\b)/i,/^(?:over\b)/i,/^(?:note\b)/i,/^(?:activate\b)/i,/^(?:deactivate\b)/i,/^(?:title\b)/i,/^(?:sequenceDiagram\b)/i,/^(?:autonumber\b)/i,/^(?:,)/i,/^(?:;)/i,/^(?:[^\+\->:\n,;]+((?!(-x|--x|-\)|--\)))[\-]*[^\+\->:\n,;]+)*)/i,/^(?:->>)/i,/^(?:-->>)/i,/^(?:->)/i,/^(?:-->)/i,/^(?:-[x])/i,/^(?:--[x])/i,/^(?:-[\)])/i,/^(?:--[\)])/i,/^(?::(?:(?:no)?wrap)?[^#\n;]+)/i,/^(?:\+)/i,/^(?:-)/i,/^(?:$)/i,/^(?:.)/i],conditions:{open_directive:{rules:[1,8],inclusive:!1},type_directive:{rules:[2,3,8],inclusive:!1},arg_directive:{rules:[3,4,8],inclusive:!1},ID:{rules:[7,8,12],inclusive:!1},ALIAS:{rules:[7,8,13,14],inclusive:!1},LINE:{rules:[7,8,22],inclusive:!1},INITIAL:{rules:[0,5,6,8,9,10,11,15,16,17,18,19,20,21,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48],inclusive:!0}}};function O(){this.yy={}}return S.lexer=M,O.prototype=S,S.Parser=O,new O}();e.parser=i,e.Parser=i.Parser,e.parse=function(){return i.parse.apply(i,arguments)},e.main=function(r){r[1]||(console.log("Usage: "+r[0]+" FILE"),t.exit(1));var i=n(19).readFileSync(n(20).normalize(r[1]),"utf8");return e.parser.parse(i)},n.c[n.s]===r&&e.main(t.argv.slice(1))}).call(this,n(14),n(7)(t))},function(t,e,n){var r=n(198);t.exports={Graph:r.Graph,json:n(301),alg:n(302),version:r.version}},function(t,e,n){var r;try{r={cloneDeep:n(313),constant:n(86),defaults:n(154),each:n(87),filter:n(128),find:n(314),flatten:n(156),forEach:n(126),forIn:n(319),has:n(93),isUndefined:n(139),last:n(320),map:n(140),mapValues:n(321),max:n(322),merge:n(324),min:n(329),minBy:n(330),now:n(331),pick:n(161),range:n(162),reduce:n(142),sortBy:n(338),uniqueId:n(163),values:n(147),zipObject:n(343)}}catch(t){}r||(r=window._),t.exports=r},function(t,e){var n=Array.isArray;t.exports=n},function(t,e,n){
+/**
+ * @license
+ * Copyright (c) 2012-2013 Chris Pettitt
+ *
+ * Permission is hereby granted, free of charge, to any person obtaining a copy
+ * of this software and associated documentation files (the "Software"), to deal
+ * in the Software without restriction, including without limitation the rights
+ * to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
+ * copies of the Software, and to permit persons to whom the Software is
+ * furnished to do so, subject to the following conditions:
+ *
+ * The above copyright notice and this permission notice shall be included in
+ * all copies or substantial portions of the Software.
+ *
+ * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
+ * IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
+ * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
+ * AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
+ * LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
+ * OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
+ * THE SOFTWARE.
+ */
+t.exports={graphlib:n(311),dagre:n(153),intersect:n(368),render:n(370),util:n(12),version:n(382)}},function(t,e){t.exports=function(t){return t.webpackPolyfill||(t.deprecate=function(){},t.paths=[],t.children||(t.children=[]),Object.defineProperty(t,"loaded",{enumerable:!0,get:function(){return t.l}}),Object.defineProperty(t,"id",{enumerable:!0,get:function(){return t.i}}),t.webpackPolyfill=1),t}},function(t,e,n){"use strict";var r=n(4),i=n(17).Graph;function a(t,e,n,i){var a;do{a=r.uniqueId(i)}while(t.hasNode(a));return n.dummy=e,t.setNode(a,n),a}function o(t){return r.max(r.map(t.nodes(),(function(e){var n=t.node(e).rank;if(!r.isUndefined(n))return n})))}t.exports={addDummyNode:a,simplify:function(t){var e=(new i).setGraph(t.graph());return r.forEach(t.nodes(),(function(n){e.setNode(n,t.node(n))})),r.forEach(t.edges(),(function(n){var r=e.edge(n.v,n.w)||{weight:0,minlen:1},i=t.edge(n);e.setEdge(n.v,n.w,{weight:r.weight+i.weight,minlen:Math.max(r.minlen,i.minlen)})})),e},asNonCompoundGraph:function(t){var e=new i({multigraph:t.isMultigraph()}).setGraph(t.graph());return r.forEach(t.nodes(),(function(n){t.children(n).length||e.setNode(n,t.node(n))})),r.forEach(t.edges(),(function(n){e.setEdge(n,t.edge(n))})),e},successorWeights:function(t){var e=r.map(t.nodes(),(function(e){var n={};return r.forEach(t.outEdges(e),(function(e){n[e.w]=(n[e.w]||0)+t.edge(e).weight})),n}));return r.zipObject(t.nodes(),e)},predecessorWeights:function(t){var e=r.map(t.nodes(),(function(e){var n={};return r.forEach(t.inEdges(e),(function(e){n[e.v]=(n[e.v]||0)+t.edge(e).weight})),n}));return r.zipObject(t.nodes(),e)},intersectRect:function(t,e){var n,r,i=t.x,a=t.y,o=e.x-i,s=e.y-a,c=t.width/2,u=t.height/2;if(!o&&!s)throw new Error("Not possible to find intersection inside of the rectangle");Math.abs(s)*c>Math.abs(o)*u?(s<0&&(u=-u),n=u*o/s,r=u):(o<0&&(c=-c),n=c,r=c*s/o);return{x:i+n,y:a+r}},buildLayerMatrix:function(t){var e=r.map(r.range(o(t)+1),(function(){return[]}));return r.forEach(t.nodes(),(function(n){var i=t.node(n),a=i.rank;r.isUndefined(a)||(e[a][i.order]=n)})),e},normalizeRanks:function(t){var e=r.min(r.map(t.nodes(),(function(e){return t.node(e).rank})));r.forEach(t.nodes(),(function(n){var i=t.node(n);r.has(i,"rank")&&(i.rank-=e)}))},removeEmptyRanks:function(t){var e=r.min(r.map(t.nodes(),(function(e){return t.node(e).rank}))),n=[];r.forEach(t.nodes(),(function(r){var i=t.node(r).rank-e;n[i]||(n[i]=[]),n[i].push(r)}));var i=0,a=t.graph().nodeRankFactor;r.forEach(n,(function(e,n){r.isUndefined(e)&&n%a!=0?--i:i&&r.forEach(e,(function(e){t.node(e).rank+=i}))}))},addBorderNode:function(t,e,n,r){var i={width:0,height:0};arguments.length>=4&&(i.rank=n,i.order=r);return a(t,"border",i,e)},maxRank:o,partition:function(t,e){var n={lhs:[],rhs:[]};return r.forEach(t,(function(t){e(t)?n.lhs.push(t):n.rhs.push(t)})),n},time:function(t,e){var n=r.now();try{return e()}finally{console.log(t+" time: "+(r.now()-n)+"ms")}},notime:function(t,e){return e()}}},function(t,e,n){"use strict";Object.defineProperty(e,"__esModule",{value:!0});var r=n(173),i=n(174),a=n(175),o={channel:r.default,lang:i.default,unit:a.default};e.default=o},function(t,e,n){var r;try{r={clone:n(199),constant:n(86),each:n(87),filter:n(128),has:n(93),isArray:n(5),isEmpty:n(276),isFunction:n(37),isUndefined:n(139),keys:n(30),map:n(140),reduce:n(142),size:n(279),transform:n(285),union:n(286),values:n(147)}}catch(t){}r||(r=window._),t.exports=r},function(t,e){t.exports=function(t){var e=typeof t;return null!=t&&("object"==e||"function"==e)}},function(t,e,n){var r=n(43);t.exports={isSubgraph:function(t,e){return!!t.children(e).length},edgeToId:function(t){return 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37:this.$={id1:a[s-2],id2:a[s],relation:a[s-1],relationTitle1:"none",relationTitle2:"none"};break;case 38:this.$={id1:a[s-3],id2:a[s],relation:a[s-1],relationTitle1:a[s-2],relationTitle2:"none"};break;case 39:this.$={id1:a[s-3],id2:a[s],relation:a[s-2],relationTitle1:"none",relationTitle2:a[s-1]};break;case 40:this.$={id1:a[s-4],id2:a[s],relation:a[s-2],relationTitle1:a[s-3],relationTitle2:a[s-1]};break;case 41:this.$={type1:a[s-2],type2:a[s],lineType:a[s-1]};break;case 42:this.$={type1:"none",type2:a[s],lineType:a[s-1]};break;case 43:this.$={type1:a[s-1],type2:"none",lineType:a[s]};break;case 44:this.$={type1:"none",type2:"none",lineType:a[s]};break;case 45:this.$=r.relationType.AGGREGATION;break;case 46:this.$=r.relationType.EXTENSION;break;case 47:this.$=r.relationType.COMPOSITION;break;case 48:this.$=r.relationType.DEPENDENCY;break;case 49:this.$=r.lineType.LINE;break;case 50:this.$=r.lineType.DOTTED_LINE;break;case 51:case 57:this.$=a[s-2],r.setClickEvent(a[s-1],a[s]);break;case 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n=new Error(t);throw n.hash=e,n}this.trace(t)},parse:function(t){var e=this,n=[0],r=[],i=[null],a=[],o=this.table,s="",c=0,u=0,l=0,h=2,f=1,d=a.slice.call(arguments,1),p=Object.create(this.lexer),g={yy:{}};for(var y in this.yy)Object.prototype.hasOwnProperty.call(this.yy,y)&&(g.yy[y]=this.yy[y]);p.setInput(t,g.yy),g.yy.lexer=p,g.yy.parser=this,void 0===p.yylloc&&(p.yylloc={});var v=p.yylloc;a.push(v);var m=p.options&&p.options.ranges;function b(){var t;return"number"!=typeof(t=r.pop()||p.lex()||f)&&(t instanceof Array&&(t=(r=t).pop()),t=e.symbols_[t]||t),t}"function"==typeof g.yy.parseError?this.parseError=g.yy.parseError:this.parseError=Object.getPrototypeOf(this).parseError;for(var x,_,k,w,E,T,C,A,S,M={};;){if(k=n[n.length-1],this.defaultActions[k]?w=this.defaultActions[k]:(null==x&&(x=b()),w=o[k]&&o[k][x]),void 0===w||!w.length||!w[0]){var O="";for(T in S=[],o[k])this.terminals_[T]&&T>h&&S.push("'"+this.terminals_[T]+"'");O=p.showPosition?"Parse error on line 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e=t.length,n=t.split(/(?:\r\n?|\n)/g);this._input=t+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-e),this.offset-=e;var r=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),n.length-1&&(this.yylineno-=n.length-1);var i=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:n?(n.length===r.length?this.yylloc.first_column:0)+r[r.length-n.length].length-n[0].length:this.yylloc.first_column-e},this.options.ranges&&(this.yylloc.range=[i[0],i[0]+this.yyleng-e]),this.yyleng=this.yytext.length,this},more:function(){return this._more=!0,this},reject:function(){return this.options.backtrack_lexer?(this._backtrack=!0,this):this.parseError("Lexical error on line "+(this.yylineno+1)+". 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19:this.$={stmt:"state",id:a[s-3],type:"default",description:a[s-5],doc:a[s-1]};break;case 20:this.$={stmt:"state",id:a[s],type:"fork"};break;case 21:this.$={stmt:"state",id:a[s],type:"join"};break;case 22:this.$={stmt:"state",id:r.getDividerId(),type:"divider"};break;case 23:this.$={stmt:"state",id:a[s-1].trim(),note:{position:a[s-2].trim(),text:a[s].trim()}};break;case 30:case 31:this.$=a[s];break;case 34:r.parseDirective("%%{","open_directive");break;case 35:r.parseDirective(a[s],"type_directive");break;case 36:a[s]=a[s].trim().replace(/'/g,'"'),r.parseDirective(a[s],"arg_directive");break;case 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n=new Error(t);throw n.hash=e,n}this.trace(t)},parse:function(t){var e=this,n=[0],r=[],i=[null],a=[],o=this.table,s="",c=0,u=0,l=0,h=2,f=1,d=a.slice.call(arguments,1),p=Object.create(this.lexer),g={yy:{}};for(var y in this.yy)Object.prototype.hasOwnProperty.call(this.yy,y)&&(g.yy[y]=this.yy[y]);p.setInput(t,g.yy),g.yy.lexer=p,g.yy.parser=this,void 0===p.yylloc&&(p.yylloc={});var v=p.yylloc;a.push(v);var m=p.options&&p.options.ranges;function b(){var t;return"number"!=typeof(t=r.pop()||p.lex()||f)&&(t instanceof Array&&(t=(r=t).pop()),t=e.symbols_[t]||t),t}"function"==typeof g.yy.parseError?this.parseError=g.yy.parseError:this.parseError=Object.getPrototypeOf(this).parseError;for(var x,_,k,w,E,T,C,A,S,M={};;){if(k=n[n.length-1],this.defaultActions[k]?w=this.defaultActions[k]:(null==x&&(x=b()),w=o[k]&&o[k][x]),void 0===w||!w.length||!w[0]){var O="";for(T in S=[],o[k])this.terminals_[T]&&T>h&&S.push("'"+this.terminals_[T]+"'");O=p.showPosition?"Parse error on line 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e=t.length,n=t.split(/(?:\r\n?|\n)/g);this._input=t+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-e),this.offset-=e;var r=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),n.length-1&&(this.yylineno-=n.length-1);var i=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:n?(n.length===r.length?this.yylloc.first_column:0)+r[r.length-n.length].length-n[0].length:this.yylloc.first_column-e},this.options.ranges&&(this.yylloc.range=[i[0],i[0]+this.yyleng-e]),this.yyleng=this.yytext.length,this},more:function(){return this._more=!0,this},reject:function(){return this.options.backtrack_lexer?(this._backtrack=!0,this):this.parseError("Lexical error on line "+(this.yylineno+1)+". 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4:return 41;case 5:break;case 6:console.log("Crap after close");break;case 7:return 5;case 8:case 9:case 10:case 11:break;case 12:return this.pushState("SCALE"),15;case 13:return 16;case 14:this.popState();break;case 15:this.pushState("STATE");break;case 16:return this.popState(),e.yytext=e.yytext.slice(0,-8).trim(),23;case 17:return this.popState(),e.yytext=e.yytext.slice(0,-8).trim(),24;case 18:return this.popState(),e.yytext=e.yytext.slice(0,-8).trim(),23;case 19:return this.popState(),e.yytext=e.yytext.slice(0,-8).trim(),24;case 20:this.begin("STATE_STRING");break;case 21:return this.popState(),this.pushState("STATE_ID"),"AS";case 22:return this.popState(),"ID";case 23:this.popState();break;case 24:return"STATE_DESCR";case 25:return 17;case 26:this.popState();break;case 27:return this.popState(),this.pushState("struct"),18;case 28:return this.popState(),19;case 29:break;case 30:return this.begin("NOTE"),26;case 31:return this.popState(),this.pushState("NOTE_ID"),37;case 32:return this.popState(),this.pushState("NOTE_ID"),38;case 33:this.popState(),this.pushState("FLOATING_NOTE");break;case 34:return this.popState(),this.pushState("FLOATING_NOTE_ID"),"AS";case 35:break;case 36:return"NOTE_TEXT";case 37:return this.popState(),"ID";case 38:return this.popState(),this.pushState("NOTE_TEXT"),22;case 39:return this.popState(),e.yytext=e.yytext.substr(2).trim(),28;case 40:return this.popState(),e.yytext=e.yytext.slice(0,-8).trim(),28;case 41:case 42:return 7;case 43:return 14;case 44:return 36;case 45:return 22;case 46:return e.yytext=e.yytext.trim(),12;case 47:return 13;case 48:return 25;case 49:return 5;case 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64:this.$=a[s-3],r.addVertex(a[s-3],a[s-1],"trapezoid");break;case 65:this.$=a[s-3],r.addVertex(a[s-3],a[s-1],"inv_trapezoid");break;case 66:this.$=a[s-3],r.addVertex(a[s-3],a[s-1],"lean_right");break;case 67:this.$=a[s-3],r.addVertex(a[s-3],a[s-1],"lean_left");break;case 68:this.$=a[s],r.addVertex(a[s]);break;case 69:a[s-1].text=a[s],this.$=a[s-1];break;case 70:case 71:a[s-2].text=a[s-1],this.$=a[s-2];break;case 72:this.$=a[s];break;case 73:var c=r.destructLink(a[s],a[s-2]);this.$={type:c.type,stroke:c.stroke,length:c.length,text:a[s-1]};break;case 74:c=r.destructLink(a[s]);this.$={type:c.type,stroke:c.stroke,length:c.length};break;case 75:this.$=a[s-1];break;case 77:case 91:case 147:this.$=a[s-1]+""+a[s];break;case 92:case 93:this.$=a[s-4],r.addClass(a[s-2],a[s]);break;case 94:this.$=a[s-4],r.setClass(a[s-2],a[s]);break;case 95:case 103:this.$=a[s-1],r.setClickEvent(a[s-1],a[s]);break;case 96:case 104:this.$=a[s-3],r.setClickEvent(a[s-3],a[s-2]),r.setTooltip(a[s-3],a[s]);break;case 97:this.$=a[s-2],r.setClickEvent(a[s-2],a[s-1],a[s]);break;case 98:this.$=a[s-4],r.setClickEvent(a[s-4],a[s-3],a[s-2]),r.setTooltip(a[s-4],a[s]);break;case 99:case 105:this.$=a[s-1],r.setLink(a[s-1],a[s]);break;case 100:case 106:this.$=a[s-3],r.setLink(a[s-3],a[s-2]),r.setTooltip(a[s-3],a[s]);break;case 101:case 107:this.$=a[s-3],r.setLink(a[s-3],a[s-2],a[s]);break;case 102:case 108:this.$=a[s-5],r.setLink(a[s-5],a[s-4],a[s]),r.setTooltip(a[s-5],a[s-2]);break;case 109:this.$=a[s-4],r.addVertex(a[s-2],void 0,void 0,a[s]);break;case 110:case 112:this.$=a[s-4],r.updateLink(a[s-2],a[s]);break;case 111:this.$=a[s-4],r.updateLink([a[s-2]],a[s]);break;case 113:this.$=a[s-8],r.updateLinkInterpolate([a[s-6]],a[s-2]),r.updateLink([a[s-6]],a[s]);break;case 114:this.$=a[s-8],r.updateLinkInterpolate(a[s-6],a[s-2]),r.updateLink(a[s-6],a[s]);break;case 115:this.$=a[s-6],r.updateLinkInterpolate([a[s-4]],a[s]);break;case 116:this.$=a[s-6],r.updateLinkInterpolate(a[s-4],a[s]);break;case 117:case 119:this.$=[a[s]];break;case 118:case 120:a[s-2].push(a[s]),this.$=a[s-2];break;case 122:this.$=a[s-1]+a[s];break;case 144:this.$=a[s];break;case 145:this.$=a[s-1]+""+a[s];break;case 150:this.$="v";break;case 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"+(c+1)+":\n"+p.showPosition()+"\nExpecting "+S.join(", ")+", got '"+(this.terminals_[x]||x)+"'":"Parse error on line "+(c+1)+": Unexpected "+(x==f?"end of input":"'"+(this.terminals_[x]||x)+"'"),this.parseError(O,{text:p.match,token:this.terminals_[x]||x,line:p.yylineno,loc:v,expected:S})}if(w[0]instanceof Array&&w.length>1)throw new Error("Parse Error: multiple actions possible at state: "+k+", token: "+x);switch(w[0]){case 1:n.push(x),i.push(p.yytext),a.push(p.yylloc),n.push(w[1]),x=null,_?(x=_,_=null):(u=p.yyleng,s=p.yytext,c=p.yylineno,v=p.yylloc,l>0&&l--);break;case 2:if(C=this.productions_[w[1]][1],M.$=i[i.length-C],M._$={first_line:a[a.length-(C||1)].first_line,last_line:a[a.length-1].last_line,first_column:a[a.length-(C||1)].first_column,last_column:a[a.length-1].last_column},m&&(M._$.range=[a[a.length-(C||1)].range[0],a[a.length-1].range[1]]),void 0!==(E=this.performAction.apply(M,[s,u,c,g.yy,w[1],i,a].concat(d))))return E;C&&(n=n.slice(0,-1*C*2),i=i.slice(0,-1*C),a=a.slice(0,-1*C)),n.push(this.productions_[w[1]][0]),i.push(M.$),a.push(M._$),A=o[n[n.length-2]][n[n.length-1]],n.push(A);break;case 3:return!0}}return!0}},qt={EOF:1,parseError:function(t,e){if(!this.yy.parser)throw new Error(t);this.yy.parser.parseError(t,e)},setInput:function(t,e){return this.yy=e||this.yy||{},this._input=t,this._more=this._backtrack=this.done=!1,this.yylineno=this.yyleng=0,this.yytext=this.matched=this.match="",this.conditionStack=["INITIAL"],this.yylloc={first_line:1,first_column:0,last_line:1,last_column:0},this.options.ranges&&(this.yylloc.range=[0,0]),this.offset=0,this},input:function(){var t=this._input[0];return this.yytext+=t,this.yyleng++,this.offset++,this.match+=t,this.matched+=t,t.match(/(?:\r\n?|\n).*/g)?(this.yylineno++,this.yylloc.last_line++):this.yylloc.last_column++,this.options.ranges&&this.yylloc.range[1]++,this._input=this._input.slice(1),t},unput:function(t){var e=t.length,n=t.split(/(?:\r\n?|\n)/g);this._input=t+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-e),this.offset-=e;var r=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),n.length-1&&(this.yylineno-=n.length-1);var i=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:n?(n.length===r.length?this.yylloc.first_column:0)+r[r.length-n.length].length-n[0].length:this.yylloc.first_column-e},this.options.ranges&&(this.yylloc.range=[i[0],i[0]+this.yyleng-e]),this.yyleng=this.yytext.length,this},more:function(){return this._more=!0,this},reject:function(){return this.options.backtrack_lexer?(this._backtrack=!0,this):this.parseError("Lexical error on line "+(this.yylineno+1)+". You can only invoke reject() in the lexer when the lexer is of the backtracking persuasion (options.backtrack_lexer = true).\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},less:function(t){this.unput(this.match.slice(t))},pastInput:function(){var t=this.matched.substr(0,this.matched.length-this.match.length);return(t.length>20?"...":"")+t.substr(-20).replace(/\n/g,"")},upcomingInput:function(){var t=this.match;return t.length<20&&(t+=this._input.substr(0,20-t.length)),(t.substr(0,20)+(t.length>20?"...":"")).replace(/\n/g,"")},showPosition:function(){var t=this.pastInput(),e=new Array(t.length+1).join("-");return t+this.upcomingInput()+"\n"+e+"^"},test_match:function(t,e){var n,r,i;if(this.options.backtrack_lexer&&(i={yylineno:this.yylineno,yylloc:{first_line:this.yylloc.first_line,last_line:this.last_line,first_column:this.yylloc.first_column,last_column:this.yylloc.last_column},yytext:this.yytext,match:this.match,matches:this.matches,matched:this.matched,yyleng:this.yyleng,offset:this.offset,_more:this._more,_input:this._input,yy:this.yy,conditionStack:this.conditionStack.slice(0),done:this.done},this.options.ranges&&(i.yylloc.range=this.yylloc.range.slice(0))),(r=t[0].match(/(?:\r\n?|\n).*/g))&&(this.yylineno+=r.length),this.yylloc={first_line:this.yylloc.last_line,last_line:this.yylineno+1,first_column:this.yylloc.last_column,last_column:r?r[r.length-1].length-r[r.length-1].match(/\r?\n?/)[0].length:this.yylloc.last_column+t[0].length},this.yytext+=t[0],this.match+=t[0],this.matches=t,this.yyleng=this.yytext.length,this.options.ranges&&(this.yylloc.range=[this.offset,this.offset+=this.yyleng]),this._more=!1,this._backtrack=!1,this._input=this._input.slice(t[0].length),this.matched+=t[0],n=this.performAction.call(this,this.yy,this,e,this.conditionStack[this.conditionStack.length-1]),this.done&&this._input&&(this.done=!1),n)return n;if(this._backtrack){for(var a in i)this[a]=i[a];return!1}return!1},next:function(){if(this.done)return this.EOF;var t,e,n,r;this._input||(this.done=!0),this._more||(this.yytext="",this.match="");for(var i=this._currentRules(),a=0;a<i.length;a++)if((n=this._input.match(this.rules[i[a]]))&&(!e||n[0].length>e[0].length)){if(e=n,r=a,this.options.backtrack_lexer){if(!1!==(t=this.test_match(n,i[a])))return t;if(this._backtrack){e=!1;continue}return!1}if(!this.options.flex)break}return e?!1!==(t=this.test_match(e,i[r]))&&t:""===this._input?this.EOF:this.parseError("Lexical error on line "+(this.yylineno+1)+". Unrecognized text.\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},lex:function(){var t=this.next();return t||this.lex()},begin:function(t){this.conditionStack.push(t)},popState:function(){return this.conditionStack.length-1>0?this.conditionStack.pop():this.conditionStack[0]},_currentRules:function(){return this.conditionStack.length&&this.conditionStack[this.conditionStack.length-1]?this.conditions[this.conditionStack[this.conditionStack.length-1]].rules:this.conditions.INITIAL.rules},topState:function(t){return(t=this.conditionStack.length-1-Math.abs(t||0))>=0?this.conditionStack[t]:"INITIAL"},pushState:function(t){this.begin(t)},stateStackSize:function(){return this.conditionStack.length},options:{},performAction:function(t,e,n,r){switch(n){case 0:return this.begin("open_directive"),12;case 1:return this.begin("type_directive"),13;case 2:return this.popState(),this.begin("arg_directive"),10;case 3:return this.popState(),this.popState(),15;case 4:return 14;case 5:case 6:break;case 7:this.begin("string");break;case 8:this.popState();break;case 9:return"STR";case 10:return 75;case 11:return 84;case 12:return 76;case 13:return 93;case 14:return 77;case 15:return 78;case 16:this.begin("href");break;case 17:this.popState();break;case 18:return 89;case 19:this.begin("callbackname");break;case 20:this.popState();break;case 21:this.popState(),this.begin("callbackargs");break;case 22:return 87;case 23:this.popState();break;case 24:return 88;case 25:this.begin("click");break;case 26:this.popState();break;case 27:return 79;case 28:case 29:return t.lex.firstGraph()&&this.begin("dir"),24;case 30:return 38;case 31:return 42;case 32:case 33:case 34:case 35:return 90;case 36:return this.popState(),25;case 37:case 38:case 39:case 40:case 41:case 42:case 43:case 44:case 45:case 46:return this.popState(),26;case 47:return 94;case 48:return 102;case 49:return 47;case 50:return 99;case 51:return 46;case 52:return 20;case 53:return 95;case 54:return 113;case 55:case 56:case 57:return 70;case 58:case 59:case 60:return 69;case 61:return 51;case 62:return 52;case 63:return 53;case 64:return 54;case 65:return 55;case 66:return 56;case 67:return 57;case 68:return 58;case 69:return 100;case 70:return 103;case 71:return 114;case 72:return 111;case 73:return 104;case 74:case 75:return 112;case 76:return 105;case 77:return 61;case 78:return 81;case 79:return"SEP";case 80:return 80;case 81:return 98;case 82:return 63;case 83:return 62;case 84:return 65;case 85:return 64;case 86:return 109;case 87:return 110;case 88:return 71;case 89:return 49;case 90:return 50;case 91:return 40;case 92:return 41;case 93:return 59;case 94:return 60;case 95:return 120;case 96:return 21;case 97:return 22;case 98:return 23}},rules:[/^(?:%%\{)/,/^(?:((?:(?!\}%%)[^:.])*))/,/^(?::)/,/^(?:\}%%)/,/^(?:((?:(?!\}%%).|\n)*))/,/^(?:%%(?!\{)[^\n]*)/,/^(?:[^\}]%%[^\n]*)/,/^(?:["])/,/^(?:["])/,/^(?:[^"]*)/,/^(?:style\b)/,/^(?:default\b)/,/^(?:linkStyle\b)/,/^(?:interpolate\b)/,/^(?:classDef\b)/,/^(?:class\b)/,/^(?:href[\s]+["])/,/^(?:["])/,/^(?:[^"]*)/,/^(?:call[\s]+)/,/^(?:\([\s]*\))/,/^(?:\()/,/^(?:[^(]*)/,/^(?:\))/,/^(?:[^)]*)/,/^(?:click[\s]+)/,/^(?:[\s\n])/,/^(?:[^\s\n]*)/,/^(?:graph\b)/,/^(?:flowchart\b)/,/^(?:subgraph\b)/,/^(?:end\b\s*)/,/^(?:_self\b)/,/^(?:_blank\b)/,/^(?:_parent\b)/,/^(?:_top\b)/,/^(?:(\r?\n)*\s*\n)/,/^(?:\s*LR\b)/,/^(?:\s*RL\b)/,/^(?:\s*TB\b)/,/^(?:\s*BT\b)/,/^(?:\s*TD\b)/,/^(?:\s*BR\b)/,/^(?:\s*<)/,/^(?:\s*>)/,/^(?:\s*\^)/,/^(?:\s*v\b)/,/^(?:[0-9]+)/,/^(?:#)/,/^(?::::)/,/^(?::)/,/^(?:&)/,/^(?:;)/,/^(?:,)/,/^(?:\*)/,/^(?:\s*[xo<]?--+[-xo>]\s*)/,/^(?:\s*[xo<]?==+[=xo>]\s*)/,/^(?:\s*[xo<]?-?\.+-[xo>]?\s*)/,/^(?:\s*[xo<]?--\s*)/,/^(?:\s*[xo<]?==\s*)/,/^(?:\s*[xo<]?-\.\s*)/,/^(?:\(-)/,/^(?:-\))/,/^(?:\(\[)/,/^(?:\]\))/,/^(?:\[\[)/,/^(?:\]\])/,/^(?:\[\()/,/^(?:\)\])/,/^(?:-)/,/^(?:\.)/,/^(?:[\_])/,/^(?:\+)/,/^(?:%)/,/^(?:=)/,/^(?:=)/,/^(?:<)/,/^(?:>)/,/^(?:\^)/,/^(?:\\\|)/,/^(?:v\b)/,/^(?:[A-Za-z]+)/,/^(?:\\\])/,/^(?:\[\/)/,/^(?:\/\])/,/^(?:\[\\)/,/^(?:[!"#$%&'*+,-.`?\\_/])/,/^(?:[\u00AA\u00B5\u00BA\u00C0-\u00D6\u00D8-\u00F6]|[\u00F8-\u02C1\u02C6-\u02D1\u02E0-\u02E4\u02EC\u02EE\u0370-\u0374\u0376\u0377]|[\u037A-\u037D\u0386\u0388-\u038A\u038C\u038E-\u03A1\u03A3-\u03F5]|[\u03F7-\u0481\u048A-\u0527\u0531-\u0556\u0559\u0561-\u0587\u05D0-\u05EA]|[\u05F0-\u05F2\u0620-\u064A\u066E\u066F\u0671-\u06D3\u06D5\u06E5\u06E6\u06EE]|[\u06EF\u06FA-\u06FC\u06FF\u0710\u0712-\u072F\u074D-\u07A5\u07B1\u07CA-\u07EA]|[\u07F4\u07F5\u07FA\u0800-\u0815\u081A\u0824\u0828\u0840-\u0858\u08A0]|[\u08A2-\u08AC\u0904-\u0939\u093D\u0950\u0958-\u0961\u0971-\u0977]|[\u0979-\u097F\u0985-\u098C\u098F\u0990\u0993-\u09A8\u09AA-\u09B0\u09B2]|[\u09B6-\u09B9\u09BD\u09CE\u09DC\u09DD\u09DF-\u09E1\u09F0\u09F1\u0A05-\u0A0A]|[\u0A0F\u0A10\u0A13-\u0A28\u0A2A-\u0A30\u0A32\u0A33\u0A35\u0A36\u0A38\u0A39]|[\u0A59-\u0A5C\u0A5E\u0A72-\u0A74\u0A85-\u0A8D\u0A8F-\u0A91\u0A93-\u0AA8]|[\u0AAA-\u0AB0\u0AB2\u0AB3\u0AB5-\u0AB9\u0ABD\u0AD0\u0AE0\u0AE1\u0B05-\u0B0C]|[\u0B0F\u0B10\u0B13-\u0B28\u0B2A-\u0B30\u0B32\u0B33\u0B35-\u0B39\u0B3D\u0B5C]|[\u0B5D\u0B5F-\u0B61\u0B71\u0B83\u0B85-\u0B8A\u0B8E-\u0B90\u0B92-\u0B95\u0B99]|[\u0B9A\u0B9C\u0B9E\u0B9F\u0BA3\u0BA4\u0BA8-\u0BAA\u0BAE-\u0BB9\u0BD0]|[\u0C05-\u0C0C\u0C0E-\u0C10\u0C12-\u0C28\u0C2A-\u0C33\u0C35-\u0C39\u0C3D]|[\u0C58\u0C59\u0C60\u0C61\u0C85-\u0C8C\u0C8E-\u0C90\u0C92-\u0CA8\u0CAA-\u0CB3]|[\u0CB5-\u0CB9\u0CBD\u0CDE\u0CE0\u0CE1\u0CF1\u0CF2\u0D05-\u0D0C\u0D0E-\u0D10]|[\u0D12-\u0D3A\u0D3D\u0D4E\u0D60\u0D61\u0D7A-\u0D7F\u0D85-\u0D96\u0D9A-\u0DB1]|[\u0DB3-\u0DBB\u0DBD\u0DC0-\u0DC6\u0E01-\u0E30\u0E32\u0E33\u0E40-\u0E46\u0E81]|[\u0E82\u0E84\u0E87\u0E88\u0E8A\u0E8D\u0E94-\u0E97\u0E99-\u0E9F\u0EA1-\u0EA3]|[\u0EA5\u0EA7\u0EAA\u0EAB\u0EAD-\u0EB0\u0EB2\u0EB3\u0EBD\u0EC0-\u0EC4\u0EC6]|[\u0EDC-\u0EDF\u0F00\u0F40-\u0F47\u0F49-\u0F6C\u0F88-\u0F8C\u1000-\u102A]|[\u103F\u1050-\u1055\u105A-\u105D\u1061\u1065\u1066\u106E-\u1070\u1075-\u1081]|[\u108E\u10A0-\u10C5\u10C7\u10CD\u10D0-\u10FA\u10FC-\u1248\u124A-\u124D]|[\u1250-\u1256\u1258\u125A-\u125D\u1260-\u1288\u128A-\u128D\u1290-\u12B0]|[\u12B2-\u12B5\u12B8-\u12BE\u12C0\u12C2-\u12C5\u12C8-\u12D6\u12D8-\u1310]|[\u1312-\u1315\u1318-\u135A\u1380-\u138F\u13A0-\u13F4\u1401-\u166C]|[\u166F-\u167F\u1681-\u169A\u16A0-\u16EA\u1700-\u170C\u170E-\u1711]|[\u1720-\u1731\u1740-\u1751\u1760-\u176C\u176E-\u1770\u1780-\u17B3\u17D7]|[\u17DC\u1820-\u1877\u1880-\u18A8\u18AA\u18B0-\u18F5\u1900-\u191C]|[\u1950-\u196D\u1970-\u1974\u1980-\u19AB\u19C1-\u19C7\u1A00-\u1A16]|[\u1A20-\u1A54\u1AA7\u1B05-\u1B33\u1B45-\u1B4B\u1B83-\u1BA0\u1BAE\u1BAF]|[\u1BBA-\u1BE5\u1C00-\u1C23\u1C4D-\u1C4F\u1C5A-\u1C7D\u1CE9-\u1CEC]|[\u1CEE-\u1CF1\u1CF5\u1CF6\u1D00-\u1DBF\u1E00-\u1F15\u1F18-\u1F1D]|[\u1F20-\u1F45\u1F48-\u1F4D\u1F50-\u1F57\u1F59\u1F5B\u1F5D\u1F5F-\u1F7D]|[\u1F80-\u1FB4\u1FB6-\u1FBC\u1FBE\u1FC2-\u1FC4\u1FC6-\u1FCC\u1FD0-\u1FD3]|[\u1FD6-\u1FDB\u1FE0-\u1FEC\u1FF2-\u1FF4\u1FF6-\u1FFC\u2071\u207F]|[\u2090-\u209C\u2102\u2107\u210A-\u2113\u2115\u2119-\u211D\u2124\u2126\u2128]|[\u212A-\u212D\u212F-\u2139\u213C-\u213F\u2145-\u2149\u214E\u2183\u2184]|[\u2C00-\u2C2E\u2C30-\u2C5E\u2C60-\u2CE4\u2CEB-\u2CEE\u2CF2\u2CF3]|[\u2D00-\u2D25\u2D27\u2D2D\u2D30-\u2D67\u2D6F\u2D80-\u2D96\u2DA0-\u2DA6]|[\u2DA8-\u2DAE\u2DB0-\u2DB6\u2DB8-\u2DBE\u2DC0-\u2DC6\u2DC8-\u2DCE]|[\u2DD0-\u2DD6\u2DD8-\u2DDE\u2E2F\u3005\u3006\u3031-\u3035\u303B\u303C]|[\u3041-\u3096\u309D-\u309F\u30A1-\u30FA\u30FC-\u30FF\u3105-\u312D]|[\u3131-\u318E\u31A0-\u31BA\u31F0-\u31FF\u3400-\u4DB5\u4E00-\u9FCC]|[\uA000-\uA48C\uA4D0-\uA4FD\uA500-\uA60C\uA610-\uA61F\uA62A\uA62B]|[\uA640-\uA66E\uA67F-\uA697\uA6A0-\uA6E5\uA717-\uA71F\uA722-\uA788]|[\uA78B-\uA78E\uA790-\uA793\uA7A0-\uA7AA\uA7F8-\uA801\uA803-\uA805]|[\uA807-\uA80A\uA80C-\uA822\uA840-\uA873\uA882-\uA8B3\uA8F2-\uA8F7\uA8FB]|[\uA90A-\uA925\uA930-\uA946\uA960-\uA97C\uA984-\uA9B2\uA9CF\uAA00-\uAA28]|[\uAA40-\uAA42\uAA44-\uAA4B\uAA60-\uAA76\uAA7A\uAA80-\uAAAF\uAAB1\uAAB5]|[\uAAB6\uAAB9-\uAABD\uAAC0\uAAC2\uAADB-\uAADD\uAAE0-\uAAEA\uAAF2-\uAAF4]|[\uAB01-\uAB06\uAB09-\uAB0E\uAB11-\uAB16\uAB20-\uAB26\uAB28-\uAB2E]|[\uABC0-\uABE2\uAC00-\uD7A3\uD7B0-\uD7C6\uD7CB-\uD7FB\uF900-\uFA6D]|[\uFA70-\uFAD9\uFB00-\uFB06\uFB13-\uFB17\uFB1D\uFB1F-\uFB28\uFB2A-\uFB36]|[\uFB38-\uFB3C\uFB3E\uFB40\uFB41\uFB43\uFB44\uFB46-\uFBB1\uFBD3-\uFD3D]|[\uFD50-\uFD8F\uFD92-\uFDC7\uFDF0-\uFDFB\uFE70-\uFE74\uFE76-\uFEFC]|[\uFF21-\uFF3A\uFF41-\uFF5A\uFF66-\uFFBE\uFFC2-\uFFC7\uFFCA-\uFFCF]|[\uFFD2-\uFFD7\uFFDA-\uFFDC])/,/^(?:\|)/,/^(?:\()/,/^(?:\))/,/^(?:\[)/,/^(?:\])/,/^(?:\{)/,/^(?:\})/,/^(?:")/,/^(?:(\r?\n)+)/,/^(?:\s)/,/^(?:$)/],conditions:{close_directive:{rules:[],inclusive:!1},arg_directive:{rules:[3,4],inclusive:!1},type_directive:{rules:[2,3],inclusive:!1},open_directive:{rules:[1],inclusive:!1},callbackargs:{rules:[23,24],inclusive:!1},callbackname:{rules:[20,21,22],inclusive:!1},href:{rules:[17,18],inclusive:!1},click:{rules:[26,27],inclusive:!1},vertex:{rules:[],inclusive:!1},dir:{rules:[36,37,38,39,40,41,42,43,44,45,46],inclusive:!1},string:{rules:[8,9],inclusive:!1},INITIAL:{rules:[0,5,6,7,10,11,12,13,14,15,16,19,25,28,29,30,31,32,33,34,35,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98],inclusive:!0}}};function Xt(){this.yy={}}return Gt.lexer=qt,Xt.prototype=Gt,Gt.Parser=Xt,new Xt}();e.parser=i,e.Parser=i.Parser,e.parse=function(){return i.parse.apply(i,arguments)},e.main=function(r){r[1]||(console.log("Usage: "+r[0]+" FILE"),t.exit(1));var i=n(19).readFileSync(n(20).normalize(r[1]),"utf8");return e.parser.parse(i)},n.c[n.s]===r&&e.main(t.argv.slice(1))}).call(this,n(14),n(7)(t))},function(t,e,n){(function(t,r){var i=function(){var t=function(t,e,n,r){for(n=n||{},r=t.length;r--;n[t[r]]=e);return n},e=[1,3],n=[1,5],r=[7,9,11,12,13,14,15,16,17,18,20,27,32],i=[1,15],a=[1,16],o=[1,17],s=[1,18],c=[1,19],u=[1,20],l=[1,21],h=[1,23],f=[1,25],d=[1,28],p=[5,7,9,11,12,13,14,15,16,17,18,20,27,32],g={trace:function(){},yy:{},symbols_:{error:2,start:3,directive:4,gantt:5,document:6,EOF:7,line:8,SPACE:9,statement:10,NL:11,dateFormat:12,inclusiveEndDates:13,axisFormat:14,excludes:15,todayMarker:16,title:17,section:18,clickStatement:19,taskTxt:20,taskData:21,openDirective:22,typeDirective:23,closeDirective:24,":":25,argDirective:26,click:27,callbackname:28,callbackargs:29,href:30,clickStatementDebug:31,open_directive:32,type_directive:33,arg_directive:34,close_directive:35,$accept:0,$end:1},terminals_:{2:"error",5:"gantt",7:"EOF",9:"SPACE",11:"NL",12:"dateFormat",13:"inclusiveEndDates",14:"axisFormat",15:"excludes",16:"todayMarker",17:"title",18:"section",20:"taskTxt",21:"taskData",25:":",27:"click",28:"callbackname",29:"callbackargs",30:"href",32:"open_directive",33:"type_directive",34:"arg_directive",35:"close_directive"},productions_:[0,[3,2],[3,3],[6,0],[6,2],[8,2],[8,1],[8,1],[8,1],[10,1],[10,1],[10,1],[10,1],[10,1],[10,1],[10,1],[10,1],[10,2],[10,1],[4,4],[4,6],[19,2],[19,3],[19,3],[19,4],[19,3],[19,4],[19,2],[31,2],[31,3],[31,3],[31,4],[31,3],[31,4],[31,2],[22,1],[23,1],[26,1],[24,1]],performAction:function(t,e,n,r,i,a,o){var s=a.length-1;switch(i){case 2:return a[s-1];case 3:this.$=[];break;case 4:a[s-1].push(a[s]),this.$=a[s-1];break;case 5:case 6:this.$=a[s];break;case 7:case 8:this.$=[];break;case 9:r.setDateFormat(a[s].substr(11)),this.$=a[s].substr(11);break;case 10:r.enableInclusiveEndDates(),this.$=a[s].substr(18);break;case 11:r.setAxisFormat(a[s].substr(11)),this.$=a[s].substr(11);break;case 12:r.setExcludes(a[s].substr(9)),this.$=a[s].substr(9);break;case 13:r.setTodayMarker(a[s].substr(12)),this.$=a[s].substr(12);break;case 14:r.setTitle(a[s].substr(6)),this.$=a[s].substr(6);break;case 15:r.addSection(a[s].substr(8)),this.$=a[s].substr(8);break;case 17:r.addTask(a[s-1],a[s]),this.$="task";break;case 21:this.$=a[s-1],r.setClickEvent(a[s-1],a[s],null);break;case 22:this.$=a[s-2],r.setClickEvent(a[s-2],a[s-1],a[s]);break;case 23:this.$=a[s-2],r.setClickEvent(a[s-2],a[s-1],null),r.setLink(a[s-2],a[s]);break;case 24:this.$=a[s-3],r.setClickEvent(a[s-3],a[s-2],a[s-1]),r.setLink(a[s-3],a[s]);break;case 25:this.$=a[s-2],r.setClickEvent(a[s-2],a[s],null),r.setLink(a[s-2],a[s-1]);break;case 26:this.$=a[s-3],r.setClickEvent(a[s-3],a[s-1],a[s]),r.setLink(a[s-3],a[s-2]);break;case 27:this.$=a[s-1],r.setLink(a[s-1],a[s]);break;case 28:case 34:this.$=a[s-1]+" "+a[s];break;case 29:case 30:case 32:this.$=a[s-2]+" "+a[s-1]+" "+a[s];break;case 31:case 33:this.$=a[s-3]+" "+a[s-2]+" "+a[s-1]+" "+a[s];break;case 35:r.parseDirective("%%{","open_directive");break;case 36:r.parseDirective(a[s],"type_directive");break;case 37:a[s]=a[s].trim().replace(/'/g,'"'),r.parseDirective(a[s],"arg_directive");break;case 38:r.parseDirective("}%%","close_directive","gantt")}},table:[{3:1,4:2,5:e,22:4,32:n},{1:[3]},{3:6,4:2,5:e,22:4,32:n},t(r,[2,3],{6:7}),{23:8,33:[1,9]},{33:[2,35]},{1:[2,1]},{4:24,7:[1,10],8:11,9:[1,12],10:13,11:[1,14],12:i,13:a,14:o,15:s,16:c,17:u,18:l,19:22,20:h,22:4,27:f,32:n},{24:26,25:[1,27],35:d},t([25,35],[2,36]),t(r,[2,8],{1:[2,2]}),t(r,[2,4]),{4:24,10:29,12:i,13:a,14:o,15:s,16:c,17:u,18:l,19:22,20:h,22:4,27:f,32:n},t(r,[2,6]),t(r,[2,7]),t(r,[2,9]),t(r,[2,10]),t(r,[2,11]),t(r,[2,12]),t(r,[2,13]),t(r,[2,14]),t(r,[2,15]),t(r,[2,16]),{21:[1,30]},t(r,[2,18]),{28:[1,31],30:[1,32]},{11:[1,33]},{26:34,34:[1,35]},{11:[2,38]},t(r,[2,5]),t(r,[2,17]),t(r,[2,21],{29:[1,36],30:[1,37]}),t(r,[2,27],{28:[1,38]}),t(p,[2,19]),{24:39,35:d},{35:[2,37]},t(r,[2,22],{30:[1,40]}),t(r,[2,23]),t(r,[2,25],{29:[1,41]}),{11:[1,42]},t(r,[2,24]),t(r,[2,26]),t(p,[2,20])],defaultActions:{5:[2,35],6:[2,1],28:[2,38],35:[2,37]},parseError:function(t,e){if(!e.recoverable){var n=new Error(t);throw n.hash=e,n}this.trace(t)},parse:function(t){var e=this,n=[0],r=[],i=[null],a=[],o=this.table,s="",c=0,u=0,l=0,h=2,f=1,d=a.slice.call(arguments,1),p=Object.create(this.lexer),g={yy:{}};for(var y in this.yy)Object.prototype.hasOwnProperty.call(this.yy,y)&&(g.yy[y]=this.yy[y]);p.setInput(t,g.yy),g.yy.lexer=p,g.yy.parser=this,void 0===p.yylloc&&(p.yylloc={});var v=p.yylloc;a.push(v);var m=p.options&&p.options.ranges;function b(){var t;return"number"!=typeof(t=r.pop()||p.lex()||f)&&(t instanceof Array&&(t=(r=t).pop()),t=e.symbols_[t]||t),t}"function"==typeof g.yy.parseError?this.parseError=g.yy.parseError:this.parseError=Object.getPrototypeOf(this).parseError;for(var x,_,k,w,E,T,C,A,S,M={};;){if(k=n[n.length-1],this.defaultActions[k]?w=this.defaultActions[k]:(null==x&&(x=b()),w=o[k]&&o[k][x]),void 0===w||!w.length||!w[0]){var O="";for(T in S=[],o[k])this.terminals_[T]&&T>h&&S.push("'"+this.terminals_[T]+"'");O=p.showPosition?"Parse error on line "+(c+1)+":\n"+p.showPosition()+"\nExpecting "+S.join(", ")+", got '"+(this.terminals_[x]||x)+"'":"Parse error on line "+(c+1)+": Unexpected "+(x==f?"end of input":"'"+(this.terminals_[x]||x)+"'"),this.parseError(O,{text:p.match,token:this.terminals_[x]||x,line:p.yylineno,loc:v,expected:S})}if(w[0]instanceof Array&&w.length>1)throw new Error("Parse Error: multiple actions possible at state: "+k+", token: "+x);switch(w[0]){case 1:n.push(x),i.push(p.yytext),a.push(p.yylloc),n.push(w[1]),x=null,_?(x=_,_=null):(u=p.yyleng,s=p.yytext,c=p.yylineno,v=p.yylloc,l>0&&l--);break;case 2:if(C=this.productions_[w[1]][1],M.$=i[i.length-C],M._$={first_line:a[a.length-(C||1)].first_line,last_line:a[a.length-1].last_line,first_column:a[a.length-(C||1)].first_column,last_column:a[a.length-1].last_column},m&&(M._$.range=[a[a.length-(C||1)].range[0],a[a.length-1].range[1]]),void 0!==(E=this.performAction.apply(M,[s,u,c,g.yy,w[1],i,a].concat(d))))return E;C&&(n=n.slice(0,-1*C*2),i=i.slice(0,-1*C),a=a.slice(0,-1*C)),n.push(this.productions_[w[1]][0]),i.push(M.$),a.push(M._$),A=o[n[n.length-2]][n[n.length-1]],n.push(A);break;case 3:return!0}}return!0}},y={EOF:1,parseError:function(t,e){if(!this.yy.parser)throw new Error(t);this.yy.parser.parseError(t,e)},setInput:function(t,e){return this.yy=e||this.yy||{},this._input=t,this._more=this._backtrack=this.done=!1,this.yylineno=this.yyleng=0,this.yytext=this.matched=this.match="",this.conditionStack=["INITIAL"],this.yylloc={first_line:1,first_column:0,last_line:1,last_column:0},this.options.ranges&&(this.yylloc.range=[0,0]),this.offset=0,this},input:function(){var t=this._input[0];return this.yytext+=t,this.yyleng++,this.offset++,this.match+=t,this.matched+=t,t.match(/(?:\r\n?|\n).*/g)?(this.yylineno++,this.yylloc.last_line++):this.yylloc.last_column++,this.options.ranges&&this.yylloc.range[1]++,this._input=this._input.slice(1),t},unput:function(t){var e=t.length,n=t.split(/(?:\r\n?|\n)/g);this._input=t+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-e),this.offset-=e;var r=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),n.length-1&&(this.yylineno-=n.length-1);var i=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:n?(n.length===r.length?this.yylloc.first_column:0)+r[r.length-n.length].length-n[0].length:this.yylloc.first_column-e},this.options.ranges&&(this.yylloc.range=[i[0],i[0]+this.yyleng-e]),this.yyleng=this.yytext.length,this},more:function(){return this._more=!0,this},reject:function(){return this.options.backtrack_lexer?(this._backtrack=!0,this):this.parseError("Lexical error on line "+(this.yylineno+1)+". You can only invoke reject() in the lexer when the lexer is of the backtracking persuasion (options.backtrack_lexer = true).\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},less:function(t){this.unput(this.match.slice(t))},pastInput:function(){var t=this.matched.substr(0,this.matched.length-this.match.length);return(t.length>20?"...":"")+t.substr(-20).replace(/\n/g,"")},upcomingInput:function(){var t=this.match;return t.length<20&&(t+=this._input.substr(0,20-t.length)),(t.substr(0,20)+(t.length>20?"...":"")).replace(/\n/g,"")},showPosition:function(){var t=this.pastInput(),e=new Array(t.length+1).join("-");return t+this.upcomingInput()+"\n"+e+"^"},test_match:function(t,e){var n,r,i;if(this.options.backtrack_lexer&&(i={yylineno:this.yylineno,yylloc:{first_line:this.yylloc.first_line,last_line:this.last_line,first_column:this.yylloc.first_column,last_column:this.yylloc.last_column},yytext:this.yytext,match:this.match,matches:this.matches,matched:this.matched,yyleng:this.yyleng,offset:this.offset,_more:this._more,_input:this._input,yy:this.yy,conditionStack:this.conditionStack.slice(0),done:this.done},this.options.ranges&&(i.yylloc.range=this.yylloc.range.slice(0))),(r=t[0].match(/(?:\r\n?|\n).*/g))&&(this.yylineno+=r.length),this.yylloc={first_line:this.yylloc.last_line,last_line:this.yylineno+1,first_column:this.yylloc.last_column,last_column:r?r[r.length-1].length-r[r.length-1].match(/\r?\n?/)[0].length:this.yylloc.last_column+t[0].length},this.yytext+=t[0],this.match+=t[0],this.matches=t,this.yyleng=this.yytext.length,this.options.ranges&&(this.yylloc.range=[this.offset,this.offset+=this.yyleng]),this._more=!1,this._backtrack=!1,this._input=this._input.slice(t[0].length),this.matched+=t[0],n=this.performAction.call(this,this.yy,this,e,this.conditionStack[this.conditionStack.length-1]),this.done&&this._input&&(this.done=!1),n)return n;if(this._backtrack){for(var a in i)this[a]=i[a];return!1}return!1},next:function(){if(this.done)return this.EOF;var t,e,n,r;this._input||(this.done=!0),this._more||(this.yytext="",this.match="");for(var i=this._currentRules(),a=0;a<i.length;a++)if((n=this._input.match(this.rules[i[a]]))&&(!e||n[0].length>e[0].length)){if(e=n,r=a,this.options.backtrack_lexer){if(!1!==(t=this.test_match(n,i[a])))return t;if(this._backtrack){e=!1;continue}return!1}if(!this.options.flex)break}return e?!1!==(t=this.test_match(e,i[r]))&&t:""===this._input?this.EOF:this.parseError("Lexical error on line "+(this.yylineno+1)+". Unrecognized text.\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},lex:function(){var t=this.next();return t||this.lex()},begin:function(t){this.conditionStack.push(t)},popState:function(){return this.conditionStack.length-1>0?this.conditionStack.pop():this.conditionStack[0]},_currentRules:function(){return this.conditionStack.length&&this.conditionStack[this.conditionStack.length-1]?this.conditions[this.conditionStack[this.conditionStack.length-1]].rules:this.conditions.INITIAL.rules},topState:function(t){return(t=this.conditionStack.length-1-Math.abs(t||0))>=0?this.conditionStack[t]:"INITIAL"},pushState:function(t){this.begin(t)},stateStackSize:function(){return this.conditionStack.length},options:{"case-insensitive":!0},performAction:function(t,e,n,r){switch(n){case 0:return this.begin("open_directive"),32;case 1:return this.begin("type_directive"),33;case 2:return this.popState(),this.begin("arg_directive"),25;case 3:return this.popState(),this.popState(),35;case 4:return 34;case 5:case 6:case 7:break;case 8:return 11;case 9:case 10:case 11:break;case 12:this.begin("href");break;case 13:this.popState();break;case 14:return 30;case 15:this.begin("callbackname");break;case 16:this.popState();break;case 17:this.popState(),this.begin("callbackargs");break;case 18:return 28;case 19:this.popState();break;case 20:return 29;case 21:this.begin("click");break;case 22:this.popState();break;case 23:return 27;case 24:return 5;case 25:return 12;case 26:return 13;case 27:return 14;case 28:return 15;case 29:return 16;case 30:return"date";case 31:return 17;case 32:return 18;case 33:return 20;case 34:return 21;case 35:return 25;case 36:return 7;case 37:return"INVALID"}},rules:[/^(?:%%\{)/i,/^(?:((?:(?!\}%%)[^:.])*))/i,/^(?::)/i,/^(?:\}%%)/i,/^(?:((?:(?!\}%%).|\n)*))/i,/^(?:%%(?!\{)*[^\n]*)/i,/^(?:[^\}]%%*[^\n]*)/i,/^(?:%%*[^\n]*[\n]*)/i,/^(?:[\n]+)/i,/^(?:\s+)/i,/^(?:#[^\n]*)/i,/^(?:%[^\n]*)/i,/^(?:href[\s]+["])/i,/^(?:["])/i,/^(?:[^"]*)/i,/^(?:call[\s]+)/i,/^(?:\([\s]*\))/i,/^(?:\()/i,/^(?:[^(]*)/i,/^(?:\))/i,/^(?:[^)]*)/i,/^(?:click[\s]+)/i,/^(?:[\s\n])/i,/^(?:[^\s\n]*)/i,/^(?:gantt\b)/i,/^(?:dateFormat\s[^#\n;]+)/i,/^(?:inclusiveEndDates\b)/i,/^(?:axisFormat\s[^#\n;]+)/i,/^(?:excludes\s[^#\n;]+)/i,/^(?:todayMarker\s[^\n;]+)/i,/^(?:\d\d\d\d-\d\d-\d\d\b)/i,/^(?:title\s[^#\n;]+)/i,/^(?:section\s[^#:\n;]+)/i,/^(?:[^#:\n;]+)/i,/^(?::[^#\n;]+)/i,/^(?::)/i,/^(?:$)/i,/^(?:.)/i],conditions:{close_directive:{rules:[],inclusive:!1},arg_directive:{rules:[3,4],inclusive:!1},type_directive:{rules:[2,3],inclusive:!1},open_directive:{rules:[1],inclusive:!1},callbackargs:{rules:[19,20],inclusive:!1},callbackname:{rules:[16,17,18],inclusive:!1},href:{rules:[13,14],inclusive:!1},click:{rules:[22,23],inclusive:!1},INITIAL:{rules:[0,5,6,7,8,9,10,11,12,15,21,24,25,26,27,28,29,30,31,32,33,34,35,36,37],inclusive:!0}}};function v(){this.yy={}}return g.lexer=y,v.prototype=g,g.Parser=v,new v}();e.parser=i,e.Parser=i.Parser,e.parse=function(){return i.parse.apply(i,arguments)},e.main=function(r){r[1]||(console.log("Usage: "+r[0]+" FILE"),t.exit(1));var i=n(19).readFileSync(n(20).normalize(r[1]),"utf8");return e.parser.parse(i)},n.c[n.s]===r&&e.main(t.argv.slice(1))}).call(this,n(14),n(7)(t))},function(t,e,n){(function(t,r){var i=function(){var t=function(t,e,n,r){for(n=n||{},r=t.length;r--;n[t[r]]=e);return 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t=function(t,e,n,r){for(n=n||{},r=t.length;r--;n[t[r]]=e);return n},e=[2,3],n=[1,7],r=[7,12,15,17,19,20,21],i=[7,11,12,15,17,19,20,21],a=[2,20],o=[1,32],s={trace:function(){},yy:{},symbols_:{error:2,start:3,GG:4,":":5,document:6,EOF:7,DIR:8,options:9,body:10,OPT:11,NL:12,line:13,statement:14,COMMIT:15,commit_arg:16,BRANCH:17,ID:18,CHECKOUT:19,MERGE:20,RESET:21,reset_arg:22,STR:23,HEAD:24,reset_parents:25,CARET:26,$accept:0,$end:1},terminals_:{2:"error",4:"GG",5:":",7:"EOF",8:"DIR",11:"OPT",12:"NL",15:"COMMIT",17:"BRANCH",18:"ID",19:"CHECKOUT",20:"MERGE",21:"RESET",23:"STR",24:"HEAD",26:"CARET"},productions_:[0,[3,4],[3,5],[6,0],[6,2],[9,2],[9,1],[10,0],[10,2],[13,2],[13,1],[14,2],[14,2],[14,2],[14,2],[14,2],[16,0],[16,1],[22,2],[22,2],[25,0],[25,2]],performAction:function(t,e,n,r,i,a,o){var s=a.length-1;switch(i){case 1:return a[s-1];case 2:return r.setDirection(a[s-3]),a[s-1];case 4:r.setOptions(a[s-1]),this.$=a[s];break;case 5:a[s-1]+=a[s],this.$=a[s-1];break;case 7:this.$=[];break;case 8:a[s-1].push(a[s]),this.$=a[s-1];break;case 9:this.$=a[s-1];break;case 11:r.commit(a[s]);break;case 12:r.branch(a[s]);break;case 13:r.checkout(a[s]);break;case 14:r.merge(a[s]);break;case 15:r.reset(a[s]);break;case 16:this.$="";break;case 17:this.$=a[s];break;case 18:this.$=a[s-1]+":"+a[s];break;case 19:this.$=a[s-1]+":"+r.count,r.count=0;break;case 20:r.count=0;break;case 21:r.count+=1}},table:[{3:1,4:[1,2]},{1:[3]},{5:[1,3],8:[1,4]},{6:5,7:e,9:6,12:n},{5:[1,8]},{7:[1,9]},t(r,[2,7],{10:10,11:[1,11]}),t(i,[2,6]),{6:12,7:e,9:6,12:n},{1:[2,1]},{7:[2,4],12:[1,15],13:13,14:14,15:[1,16],17:[1,17],19:[1,18],20:[1,19],21:[1,20]},t(i,[2,5]),{7:[1,21]},t(r,[2,8]),{12:[1,22]},t(r,[2,10]),{12:[2,16],16:23,23:[1,24]},{18:[1,25]},{18:[1,26]},{18:[1,27]},{18:[1,30],22:28,24:[1,29]},{1:[2,2]},t(r,[2,9]),{12:[2,11]},{12:[2,17]},{12:[2,12]},{12:[2,13]},{12:[2,14]},{12:[2,15]},{12:a,25:31,26:o},{12:a,25:33,26:o},{12:[2,18]},{12:a,25:34,26:o},{12:[2,19]},{12:[2,21]}],defaultActions:{9:[2,1],21:[2,2],23:[2,11],24:[2,17],25:[2,12],26:[2,13],27:[2,14],28:[2,15],31:[2,18],33:[2,19],34:[2,21]},parseError:function(t,e){if(!e.recoverable){var n=new Error(t);throw n.hash=e,n}this.trace(t)},parse:function(t){var e=this,n=[0],r=[],i=[null],a=[],o=this.table,s="",c=0,u=0,l=0,h=2,f=1,d=a.slice.call(arguments,1),p=Object.create(this.lexer),g={yy:{}};for(var y in this.yy)Object.prototype.hasOwnProperty.call(this.yy,y)&&(g.yy[y]=this.yy[y]);p.setInput(t,g.yy),g.yy.lexer=p,g.yy.parser=this,void 0===p.yylloc&&(p.yylloc={});var v=p.yylloc;a.push(v);var m=p.options&&p.options.ranges;function b(){var t;return"number"!=typeof(t=r.pop()||p.lex()||f)&&(t instanceof Array&&(t=(r=t).pop()),t=e.symbols_[t]||t),t}"function"==typeof g.yy.parseError?this.parseError=g.yy.parseError:this.parseError=Object.getPrototypeOf(this).parseError;for(var x,_,k,w,E,T,C,A,S,M={};;){if(k=n[n.length-1],this.defaultActions[k]?w=this.defaultActions[k]:(null==x&&(x=b()),w=o[k]&&o[k][x]),void 0===w||!w.length||!w[0]){var O="";for(T in S=[],o[k])this.terminals_[T]&&T>h&&S.push("'"+this.terminals_[T]+"'");O=p.showPosition?"Parse error on line "+(c+1)+":\n"+p.showPosition()+"\nExpecting "+S.join(", ")+", got '"+(this.terminals_[x]||x)+"'":"Parse error on line "+(c+1)+": Unexpected "+(x==f?"end of 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3:return!0}}return!0}},c={EOF:1,parseError:function(t,e){if(!this.yy.parser)throw new Error(t);this.yy.parser.parseError(t,e)},setInput:function(t,e){return this.yy=e||this.yy||{},this._input=t,this._more=this._backtrack=this.done=!1,this.yylineno=this.yyleng=0,this.yytext=this.matched=this.match="",this.conditionStack=["INITIAL"],this.yylloc={first_line:1,first_column:0,last_line:1,last_column:0},this.options.ranges&&(this.yylloc.range=[0,0]),this.offset=0,this},input:function(){var t=this._input[0];return this.yytext+=t,this.yyleng++,this.offset++,this.match+=t,this.matched+=t,t.match(/(?:\r\n?|\n).*/g)?(this.yylineno++,this.yylloc.last_line++):this.yylloc.last_column++,this.options.ranges&&this.yylloc.range[1]++,this._input=this._input.slice(1),t},unput:function(t){var e=t.length,n=t.split(/(?:\r\n?|\n)/g);this._input=t+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-e),this.offset-=e;var r=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),n.length-1&&(this.yylineno-=n.length-1);var i=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:n?(n.length===r.length?this.yylloc.first_column:0)+r[r.length-n.length].length-n[0].length:this.yylloc.first_column-e},this.options.ranges&&(this.yylloc.range=[i[0],i[0]+this.yyleng-e]),this.yyleng=this.yytext.length,this},more:function(){return this._more=!0,this},reject:function(){return this.options.backtrack_lexer?(this._backtrack=!0,this):this.parseError("Lexical error on line "+(this.yylineno+1)+". You can only invoke reject() in the lexer when the lexer is of the backtracking persuasion (options.backtrack_lexer = true).\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},less:function(t){this.unput(this.match.slice(t))},pastInput:function(){var t=this.matched.substr(0,this.matched.length-this.match.length);return(t.length>20?"...":"")+t.substr(-20).replace(/\n/g,"")},upcomingInput:function(){var t=this.match;return t.length<20&&(t+=this._input.substr(0,20-t.length)),(t.substr(0,20)+(t.length>20?"...":"")).replace(/\n/g,"")},showPosition:function(){var t=this.pastInput(),e=new Array(t.length+1).join("-");return t+this.upcomingInput()+"\n"+e+"^"},test_match:function(t,e){var n,r,i;if(this.options.backtrack_lexer&&(i={yylineno:this.yylineno,yylloc:{first_line:this.yylloc.first_line,last_line:this.last_line,first_column:this.yylloc.first_column,last_column:this.yylloc.last_column},yytext:this.yytext,match:this.match,matches:this.matches,matched:this.matched,yyleng:this.yyleng,offset:this.offset,_more:this._more,_input:this._input,yy:this.yy,conditionStack:this.conditionStack.slice(0),done:this.done},this.options.ranges&&(i.yylloc.range=this.yylloc.range.slice(0))),(r=t[0].match(/(?:\r\n?|\n).*/g))&&(this.yylineno+=r.length),this.yylloc={first_line:this.yylloc.last_line,last_line:this.yylineno+1,first_column:this.yylloc.last_column,last_column:r?r[r.length-1].length-r[r.length-1].match(/\r?\n?/)[0].length:this.yylloc.last_column+t[0].length},this.yytext+=t[0],this.match+=t[0],this.matches=t,this.yyleng=this.yytext.length,this.options.ranges&&(this.yylloc.range=[this.offset,this.offset+=this.yyleng]),this._more=!1,this._backtrack=!1,this._input=this._input.slice(t[0].length),this.matched+=t[0],n=this.performAction.call(this,this.yy,this,e,this.conditionStack[this.conditionStack.length-1]),this.done&&this._input&&(this.done=!1),n)return n;if(this._backtrack){for(var a in i)this[a]=i[a];return!1}return!1},next:function(){if(this.done)return this.EOF;var t,e,n,r;this._input||(this.done=!0),this._more||(this.yytext="",this.match="");for(var i=this._currentRules(),a=0;a<i.length;a++)if((n=this._input.match(this.rules[i[a]]))&&(!e||n[0].length>e[0].length)){if(e=n,r=a,this.options.backtrack_lexer){if(!1!==(t=this.test_match(n,i[a])))return t;if(this._backtrack){e=!1;continue}return!1}if(!this.options.flex)break}return e?!1!==(t=this.test_match(e,i[r]))&&t:""===this._input?this.EOF:this.parseError("Lexical error on line "+(this.yylineno+1)+". Unrecognized text.\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},lex:function(){var t=this.next();return t||this.lex()},begin:function(t){this.conditionStack.push(t)},popState:function(){return this.conditionStack.length-1>0?this.conditionStack.pop():this.conditionStack[0]},_currentRules:function(){return this.conditionStack.length&&this.conditionStack[this.conditionStack.length-1]?this.conditions[this.conditionStack[this.conditionStack.length-1]].rules:this.conditions.INITIAL.rules},topState:function(t){return(t=this.conditionStack.length-1-Math.abs(t||0))>=0?this.conditionStack[t]:"INITIAL"},pushState:function(t){this.begin(t)},stateStackSize:function(){return this.conditionStack.length},options:{"case-insensitive":!0},performAction:function(t,e,n,r){switch(n){case 0:return 12;case 1:case 2:case 3:break;case 4:return 4;case 5:return 15;case 6:return 17;case 7:return 20;case 8:return 21;case 9:return 19;case 10:case 11:return 8;case 12:return 5;case 13:return 26;case 14:this.begin("options");break;case 15:this.popState();break;case 16:return 11;case 17:this.begin("string");break;case 18:this.popState();break;case 19:return 23;case 20:return 18;case 21:return 7}},rules:[/^(?:(\r?\n)+)/i,/^(?:\s+)/i,/^(?:#[^\n]*)/i,/^(?:%[^\n]*)/i,/^(?:gitGraph\b)/i,/^(?:commit\b)/i,/^(?:branch\b)/i,/^(?:merge\b)/i,/^(?:reset\b)/i,/^(?:checkout\b)/i,/^(?:LR\b)/i,/^(?:BT\b)/i,/^(?::)/i,/^(?:\^)/i,/^(?:options\r?\n)/i,/^(?:end\r?\n)/i,/^(?:[^\n]+\r?\n)/i,/^(?:["])/i,/^(?:["])/i,/^(?:[^"]*)/i,/^(?:[a-zA-Z][-_\.a-zA-Z0-9]*[-_a-zA-Z0-9])/i,/^(?:$)/i],conditions:{options:{rules:[15,16],inclusive:!1},string:{rules:[18,19],inclusive:!1},INITIAL:{rules:[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,17,20,21],inclusive:!0}}};function u(){this.yy={}}return s.lexer=c,u.prototype=s,s.Parser=u,new u}();e.parser=i,e.Parser=i.Parser,e.parse=function(){return i.parse.apply(i,arguments)},e.main=function(r){r[1]||(console.log("Usage: "+r[0]+" FILE"),t.exit(1));var i=n(19).readFileSync(n(20).normalize(r[1]),"utf8");return e.parser.parse(i)},n.c[n.s]===r&&e.main(t.argv.slice(1))}).call(this,n(14),n(7)(t))},function(t,e,n){(function(t,r){var i=function(){var t=function(t,e,n,r){for(n=n||{},r=t.length;r--;n[t[r]]=e);return n},e=[6,9,10],n={trace:function(){},yy:{},symbols_:{error:2,start:3,info:4,document:5,EOF:6,line:7,statement:8,NL:9,showInfo:10,$accept:0,$end:1},terminals_:{2:"error",4:"info",6:"EOF",9:"NL",10:"showInfo"},productions_:[0,[3,3],[5,0],[5,2],[7,1],[7,1],[8,1]],performAction:function(t,e,n,r,i,a,o){a.length;switch(i){case 1:return r;case 4:break;case 6:r.setInfo(!0)}},table:[{3:1,4:[1,2]},{1:[3]},t(e,[2,2],{5:3}),{6:[1,4],7:5,8:6,9:[1,7],10:[1,8]},{1:[2,1]},t(e,[2,3]),t(e,[2,4]),t(e,[2,5]),t(e,[2,6])],defaultActions:{4:[2,1]},parseError:function(t,e){if(!e.recoverable){var n=new Error(t);throw n.hash=e,n}this.trace(t)},parse:function(t){var e=this,n=[0],r=[],i=[null],a=[],o=this.table,s="",c=0,u=0,l=0,h=2,f=1,d=a.slice.call(arguments,1),p=Object.create(this.lexer),g={yy:{}};for(var y in this.yy)Object.prototype.hasOwnProperty.call(this.yy,y)&&(g.yy[y]=this.yy[y]);p.setInput(t,g.yy),g.yy.lexer=p,g.yy.parser=this,void 0===p.yylloc&&(p.yylloc={});var v=p.yylloc;a.push(v);var m=p.options&&p.options.ranges;function b(){var t;return"number"!=typeof(t=r.pop()||p.lex()||f)&&(t instanceof Array&&(t=(r=t).pop()),t=e.symbols_[t]||t),t}"function"==typeof g.yy.parseError?this.parseError=g.yy.parseError:this.parseError=Object.getPrototypeOf(this).parseError;for(var x,_,k,w,E,T,C,A,S,M={};;){if(k=n[n.length-1],this.defaultActions[k]?w=this.defaultActions[k]:(null==x&&(x=b()),w=o[k]&&o[k][x]),void 0===w||!w.length||!w[0]){var O="";for(T in S=[],o[k])this.terminals_[T]&&T>h&&S.push("'"+this.terminals_[T]+"'");O=p.showPosition?"Parse error on line "+(c+1)+":\n"+p.showPosition()+"\nExpecting "+S.join(", ")+", got '"+(this.terminals_[x]||x)+"'":"Parse error on line "+(c+1)+": Unexpected "+(x==f?"end of input":"'"+(this.terminals_[x]||x)+"'"),this.parseError(O,{text:p.match,token:this.terminals_[x]||x,line:p.yylineno,loc:v,expected:S})}if(w[0]instanceof Array&&w.length>1)throw new Error("Parse Error: multiple actions possible at state: "+k+", token: "+x);switch(w[0]){case 1:n.push(x),i.push(p.yytext),a.push(p.yylloc),n.push(w[1]),x=null,_?(x=_,_=null):(u=p.yyleng,s=p.yytext,c=p.yylineno,v=p.yylloc,l>0&&l--);break;case 2:if(C=this.productions_[w[1]][1],M.$=i[i.length-C],M._$={first_line:a[a.length-(C||1)].first_line,last_line:a[a.length-1].last_line,first_column:a[a.length-(C||1)].first_column,last_column:a[a.length-1].last_column},m&&(M._$.range=[a[a.length-(C||1)].range[0],a[a.length-1].range[1]]),void 0!==(E=this.performAction.apply(M,[s,u,c,g.yy,w[1],i,a].concat(d))))return E;C&&(n=n.slice(0,-1*C*2),i=i.slice(0,-1*C),a=a.slice(0,-1*C)),n.push(this.productions_[w[1]][0]),i.push(M.$),a.push(M._$),A=o[n[n.length-2]][n[n.length-1]],n.push(A);break;case 3:return!0}}return!0}},r={EOF:1,parseError:function(t,e){if(!this.yy.parser)throw new Error(t);this.yy.parser.parseError(t,e)},setInput:function(t,e){return this.yy=e||this.yy||{},this._input=t,this._more=this._backtrack=this.done=!1,this.yylineno=this.yyleng=0,this.yytext=this.matched=this.match="",this.conditionStack=["INITIAL"],this.yylloc={first_line:1,first_column:0,last_line:1,last_column:0},this.options.ranges&&(this.yylloc.range=[0,0]),this.offset=0,this},input:function(){var t=this._input[0];return this.yytext+=t,this.yyleng++,this.offset++,this.match+=t,this.matched+=t,t.match(/(?:\r\n?|\n).*/g)?(this.yylineno++,this.yylloc.last_line++):this.yylloc.last_column++,this.options.ranges&&this.yylloc.range[1]++,this._input=this._input.slice(1),t},unput:function(t){var e=t.length,n=t.split(/(?:\r\n?|\n)/g);this._input=t+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-e),this.offset-=e;var r=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),n.length-1&&(this.yylineno-=n.length-1);var i=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:n?(n.length===r.length?this.yylloc.first_column:0)+r[r.length-n.length].length-n[0].length:this.yylloc.first_column-e},this.options.ranges&&(this.yylloc.range=[i[0],i[0]+this.yyleng-e]),this.yyleng=this.yytext.length,this},more:function(){return this._more=!0,this},reject:function(){return this.options.backtrack_lexer?(this._backtrack=!0,this):this.parseError("Lexical error on line "+(this.yylineno+1)+". You can only invoke reject() in the lexer when the lexer is of the backtracking persuasion (options.backtrack_lexer = true).\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},less:function(t){this.unput(this.match.slice(t))},pastInput:function(){var t=this.matched.substr(0,this.matched.length-this.match.length);return(t.length>20?"...":"")+t.substr(-20).replace(/\n/g,"")},upcomingInput:function(){var t=this.match;return t.length<20&&(t+=this._input.substr(0,20-t.length)),(t.substr(0,20)+(t.length>20?"...":"")).replace(/\n/g,"")},showPosition:function(){var t=this.pastInput(),e=new Array(t.length+1).join("-");return t+this.upcomingInput()+"\n"+e+"^"},test_match:function(t,e){var n,r,i;if(this.options.backtrack_lexer&&(i={yylineno:this.yylineno,yylloc:{first_line:this.yylloc.first_line,last_line:this.last_line,first_column:this.yylloc.first_column,last_column:this.yylloc.last_column},yytext:this.yytext,match:this.match,matches:this.matches,matched:this.matched,yyleng:this.yyleng,offset:this.offset,_more:this._more,_input:this._input,yy:this.yy,conditionStack:this.conditionStack.slice(0),done:this.done},this.options.ranges&&(i.yylloc.range=this.yylloc.range.slice(0))),(r=t[0].match(/(?:\r\n?|\n).*/g))&&(this.yylineno+=r.length),this.yylloc={first_line:this.yylloc.last_line,last_line:this.yylineno+1,first_column:this.yylloc.last_column,last_column:r?r[r.length-1].length-r[r.length-1].match(/\r?\n?/)[0].length:this.yylloc.last_column+t[0].length},this.yytext+=t[0],this.match+=t[0],this.matches=t,this.yyleng=this.yytext.length,this.options.ranges&&(this.yylloc.range=[this.offset,this.offset+=this.yyleng]),this._more=!1,this._backtrack=!1,this._input=this._input.slice(t[0].length),this.matched+=t[0],n=this.performAction.call(this,this.yy,this,e,this.conditionStack[this.conditionStack.length-1]),this.done&&this._input&&(this.done=!1),n)return n;if(this._backtrack){for(var a in i)this[a]=i[a];return!1}return!1},next:function(){if(this.done)return this.EOF;var t,e,n,r;this._input||(this.done=!0),this._more||(this.yytext="",this.match="");for(var i=this._currentRules(),a=0;a<i.length;a++)if((n=this._input.match(this.rules[i[a]]))&&(!e||n[0].length>e[0].length)){if(e=n,r=a,this.options.backtrack_lexer){if(!1!==(t=this.test_match(n,i[a])))return t;if(this._backtrack){e=!1;continue}return!1}if(!this.options.flex)break}return e?!1!==(t=this.test_match(e,i[r]))&&t:""===this._input?this.EOF:this.parseError("Lexical error on line "+(this.yylineno+1)+". Unrecognized text.\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},lex:function(){var t=this.next();return t||this.lex()},begin:function(t){this.conditionStack.push(t)},popState:function(){return this.conditionStack.length-1>0?this.conditionStack.pop():this.conditionStack[0]},_currentRules:function(){return this.conditionStack.length&&this.conditionStack[this.conditionStack.length-1]?this.conditions[this.conditionStack[this.conditionStack.length-1]].rules:this.conditions.INITIAL.rules},topState:function(t){return(t=this.conditionStack.length-1-Math.abs(t||0))>=0?this.conditionStack[t]:"INITIAL"},pushState:function(t){this.begin(t)},stateStackSize:function(){return this.conditionStack.length},options:{"case-insensitive":!0},performAction:function(t,e,n,r){switch(n){case 0:return 4;case 1:return 9;case 2:return"space";case 3:return 10;case 4:return 6;case 5:return"TXT"}},rules:[/^(?:info\b)/i,/^(?:[\s\n\r]+)/i,/^(?:[\s]+)/i,/^(?:showInfo\b)/i,/^(?:$)/i,/^(?:.)/i],conditions:{INITIAL:{rules:[0,1,2,3,4,5],inclusive:!0}}};function i(){this.yy={}}return n.lexer=r,i.prototype=n,n.Parser=i,new i}();e.parser=i,e.Parser=i.Parser,e.parse=function(){return i.parse.apply(i,arguments)},e.main=function(r){r[1]||(console.log("Usage: "+r[0]+" FILE"),t.exit(1));var i=n(19).readFileSync(n(20).normalize(r[1]),"utf8");return e.parser.parse(i)},n.c[n.s]===r&&e.main(t.argv.slice(1))}).call(this,n(14),n(7)(t))},function(t,e,n){(function(t,r){var i=function(){var t=function(t,e,n,r){for(n=n||{},r=t.length;r--;n[t[r]]=e);return n},e=[1,4],n=[1,5],r=[1,6],i=[1,7],a=[1,9],o=[1,10,12,19,20,21,22],s=[1,6,10,12,19,20,21,22],c=[19,20,21],u=[1,22],l=[6,19,20,21,22],h={trace:function(){},yy:{},symbols_:{error:2,start:3,eol:4,directive:5,PIE:6,document:7,line:8,statement:9,txt:10,value:11,title:12,title_value:13,openDirective:14,typeDirective:15,closeDirective:16,":":17,argDirective:18,NEWLINE:19,";":20,EOF:21,open_directive:22,type_directive:23,arg_directive:24,close_directive:25,$accept:0,$end:1},terminals_:{2:"error",6:"PIE",10:"txt",11:"value",12:"title",13:"title_value",17:":",19:"NEWLINE",20:";",21:"EOF",22:"open_directive",23:"type_directive",24:"arg_directive",25:"close_directive"},productions_:[0,[3,2],[3,2],[3,2],[7,0],[7,2],[8,2],[9,0],[9,2],[9,2],[9,1],[5,3],[5,5],[4,1],[4,1],[4,1],[14,1],[15,1],[18,1],[16,1]],performAction:function(t,e,n,r,i,a,o){var s=a.length-1;switch(i){case 6:this.$=a[s-1];break;case 8:r.addSection(a[s-1],r.cleanupValue(a[s]));break;case 9:this.$=a[s].trim(),r.setTitle(this.$);break;case 16:r.parseDirective("%%{","open_directive");break;case 17:r.parseDirective(a[s],"type_directive");break;case 18:a[s]=a[s].trim().replace(/'/g,'"'),r.parseDirective(a[s],"arg_directive");break;case 19:r.parseDirective("}%%","close_directive","pie")}},table:[{3:1,4:2,5:3,6:e,14:8,19:n,20:r,21:i,22:a},{1:[3]},{3:10,4:2,5:3,6:e,14:8,19:n,20:r,21:i,22:a},{3:11,4:2,5:3,6:e,14:8,19:n,20:r,21:i,22:a},t(o,[2,4],{7:12}),t(s,[2,13]),t(s,[2,14]),t(s,[2,15]),{15:13,23:[1,14]},{23:[2,16]},{1:[2,1]},{1:[2,2]},t(c,[2,7],{14:8,8:15,9:16,5:19,1:[2,3],10:[1,17],12:[1,18],22:a}),{16:20,17:[1,21],25:u},t([17,25],[2,17]),t(o,[2,5]),{4:23,19:n,20:r,21:i},{11:[1,24]},{13:[1,25]},t(c,[2,10]),t(l,[2,11]),{18:26,24:[1,27]},t(l,[2,19]),t(o,[2,6]),t(c,[2,8]),t(c,[2,9]),{16:28,25:u},{25:[2,18]},t(l,[2,12])],defaultActions:{9:[2,16],10:[2,1],11:[2,2],27:[2,18]},parseError:function(t,e){if(!e.recoverable){var n=new Error(t);throw n.hash=e,n}this.trace(t)},parse:function(t){var e=this,n=[0],r=[],i=[null],a=[],o=this.table,s="",c=0,u=0,l=0,h=2,f=1,d=a.slice.call(arguments,1),p=Object.create(this.lexer),g={yy:{}};for(var y in this.yy)Object.prototype.hasOwnProperty.call(this.yy,y)&&(g.yy[y]=this.yy[y]);p.setInput(t,g.yy),g.yy.lexer=p,g.yy.parser=this,void 0===p.yylloc&&(p.yylloc={});var v=p.yylloc;a.push(v);var m=p.options&&p.options.ranges;function b(){var t;return"number"!=typeof(t=r.pop()||p.lex()||f)&&(t instanceof Array&&(t=(r=t).pop()),t=e.symbols_[t]||t),t}"function"==typeof g.yy.parseError?this.parseError=g.yy.parseError:this.parseError=Object.getPrototypeOf(this).parseError;for(var x,_,k,w,E,T,C,A,S,M={};;){if(k=n[n.length-1],this.defaultActions[k]?w=this.defaultActions[k]:(null==x&&(x=b()),w=o[k]&&o[k][x]),void 0===w||!w.length||!w[0]){var O="";for(T in S=[],o[k])this.terminals_[T]&&T>h&&S.push("'"+this.terminals_[T]+"'");O=p.showPosition?"Parse error on line "+(c+1)+":\n"+p.showPosition()+"\nExpecting "+S.join(", ")+", got '"+(this.terminals_[x]||x)+"'":"Parse error on line "+(c+1)+": Unexpected "+(x==f?"end of input":"'"+(this.terminals_[x]||x)+"'"),this.parseError(O,{text:p.match,token:this.terminals_[x]||x,line:p.yylineno,loc:v,expected:S})}if(w[0]instanceof Array&&w.length>1)throw new Error("Parse Error: multiple actions possible at state: "+k+", token: "+x);switch(w[0]){case 1:n.push(x),i.push(p.yytext),a.push(p.yylloc),n.push(w[1]),x=null,_?(x=_,_=null):(u=p.yyleng,s=p.yytext,c=p.yylineno,v=p.yylloc,l>0&&l--);break;case 2:if(C=this.productions_[w[1]][1],M.$=i[i.length-C],M._$={first_line:a[a.length-(C||1)].first_line,last_line:a[a.length-1].last_line,first_column:a[a.length-(C||1)].first_column,last_column:a[a.length-1].last_column},m&&(M._$.range=[a[a.length-(C||1)].range[0],a[a.length-1].range[1]]),void 0!==(E=this.performAction.apply(M,[s,u,c,g.yy,w[1],i,a].concat(d))))return E;C&&(n=n.slice(0,-1*C*2),i=i.slice(0,-1*C),a=a.slice(0,-1*C)),n.push(this.productions_[w[1]][0]),i.push(M.$),a.push(M._$),A=o[n[n.length-2]][n[n.length-1]],n.push(A);break;case 3:return!0}}return!0}},f={EOF:1,parseError:function(t,e){if(!this.yy.parser)throw new Error(t);this.yy.parser.parseError(t,e)},setInput:function(t,e){return this.yy=e||this.yy||{},this._input=t,this._more=this._backtrack=this.done=!1,this.yylineno=this.yyleng=0,this.yytext=this.matched=this.match="",this.conditionStack=["INITIAL"],this.yylloc={first_line:1,first_column:0,last_line:1,last_column:0},this.options.ranges&&(this.yylloc.range=[0,0]),this.offset=0,this},input:function(){var t=this._input[0];return this.yytext+=t,this.yyleng++,this.offset++,this.match+=t,this.matched+=t,t.match(/(?:\r\n?|\n).*/g)?(this.yylineno++,this.yylloc.last_line++):this.yylloc.last_column++,this.options.ranges&&this.yylloc.range[1]++,this._input=this._input.slice(1),t},unput:function(t){var e=t.length,n=t.split(/(?:\r\n?|\n)/g);this._input=t+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-e),this.offset-=e;var r=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),n.length-1&&(this.yylineno-=n.length-1);var i=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:n?(n.length===r.length?this.yylloc.first_column:0)+r[r.length-n.length].length-n[0].length:this.yylloc.first_column-e},this.options.ranges&&(this.yylloc.range=[i[0],i[0]+this.yyleng-e]),this.yyleng=this.yytext.length,this},more:function(){return this._more=!0,this},reject:function(){return this.options.backtrack_lexer?(this._backtrack=!0,this):this.parseError("Lexical error on line "+(this.yylineno+1)+". You can only invoke reject() in the lexer when the lexer is of the backtracking persuasion (options.backtrack_lexer = true).\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},less:function(t){this.unput(this.match.slice(t))},pastInput:function(){var t=this.matched.substr(0,this.matched.length-this.match.length);return(t.length>20?"...":"")+t.substr(-20).replace(/\n/g,"")},upcomingInput:function(){var t=this.match;return t.length<20&&(t+=this._input.substr(0,20-t.length)),(t.substr(0,20)+(t.length>20?"...":"")).replace(/\n/g,"")},showPosition:function(){var t=this.pastInput(),e=new Array(t.length+1).join("-");return t+this.upcomingInput()+"\n"+e+"^"},test_match:function(t,e){var n,r,i;if(this.options.backtrack_lexer&&(i={yylineno:this.yylineno,yylloc:{first_line:this.yylloc.first_line,last_line:this.last_line,first_column:this.yylloc.first_column,last_column:this.yylloc.last_column},yytext:this.yytext,match:this.match,matches:this.matches,matched:this.matched,yyleng:this.yyleng,offset:this.offset,_more:this._more,_input:this._input,yy:this.yy,conditionStack:this.conditionStack.slice(0),done:this.done},this.options.ranges&&(i.yylloc.range=this.yylloc.range.slice(0))),(r=t[0].match(/(?:\r\n?|\n).*/g))&&(this.yylineno+=r.length),this.yylloc={first_line:this.yylloc.last_line,last_line:this.yylineno+1,first_column:this.yylloc.last_column,last_column:r?r[r.length-1].length-r[r.length-1].match(/\r?\n?/)[0].length:this.yylloc.last_column+t[0].length},this.yytext+=t[0],this.match+=t[0],this.matches=t,this.yyleng=this.yytext.length,this.options.ranges&&(this.yylloc.range=[this.offset,this.offset+=this.yyleng]),this._more=!1,this._backtrack=!1,this._input=this._input.slice(t[0].length),this.matched+=t[0],n=this.performAction.call(this,this.yy,this,e,this.conditionStack[this.conditionStack.length-1]),this.done&&this._input&&(this.done=!1),n)return n;if(this._backtrack){for(var a in i)this[a]=i[a];return!1}return!1},next:function(){if(this.done)return this.EOF;var t,e,n,r;this._input||(this.done=!0),this._more||(this.yytext="",this.match="");for(var i=this._currentRules(),a=0;a<i.length;a++)if((n=this._input.match(this.rules[i[a]]))&&(!e||n[0].length>e[0].length)){if(e=n,r=a,this.options.backtrack_lexer){if(!1!==(t=this.test_match(n,i[a])))return t;if(this._backtrack){e=!1;continue}return!1}if(!this.options.flex)break}return e?!1!==(t=this.test_match(e,i[r]))&&t:""===this._input?this.EOF:this.parseError("Lexical error on line "+(this.yylineno+1)+". Unrecognized text.\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},lex:function(){var t=this.next();return t||this.lex()},begin:function(t){this.conditionStack.push(t)},popState:function(){return this.conditionStack.length-1>0?this.conditionStack.pop():this.conditionStack[0]},_currentRules:function(){return this.conditionStack.length&&this.conditionStack[this.conditionStack.length-1]?this.conditions[this.conditionStack[this.conditionStack.length-1]].rules:this.conditions.INITIAL.rules},topState:function(t){return(t=this.conditionStack.length-1-Math.abs(t||0))>=0?this.conditionStack[t]:"INITIAL"},pushState:function(t){this.begin(t)},stateStackSize:function(){return this.conditionStack.length},options:{"case-insensitive":!0},performAction:function(t,e,n,r){switch(n){case 0:return this.begin("open_directive"),22;case 1:return this.begin("type_directive"),23;case 2:return this.popState(),this.begin("arg_directive"),17;case 3:return this.popState(),this.popState(),25;case 4:return 24;case 5:case 6:break;case 7:return 19;case 8:case 9:break;case 10:return this.begin("title"),12;case 11:return this.popState(),"title_value";case 12:this.begin("string");break;case 13:this.popState();break;case 14:return"txt";case 15:return 6;case 16:return"value";case 17:return 21}},rules:[/^(?:%%\{)/i,/^(?:((?:(?!\}%%)[^:.])*))/i,/^(?::)/i,/^(?:\}%%)/i,/^(?:((?:(?!\}%%).|\n)*))/i,/^(?:%%(?!\{)[^\n]*)/i,/^(?:[^\}]%%[^\n]*)/i,/^(?:[\n\r]+)/i,/^(?:%%[^\n]*)/i,/^(?:[\s]+)/i,/^(?:title\b)/i,/^(?:(?!\n||)*[^\n]*)/i,/^(?:["])/i,/^(?:["])/i,/^(?:[^"]*)/i,/^(?:pie\b)/i,/^(?::[\s]*[\d]+(?:\.[\d]+)?)/i,/^(?:$)/i],conditions:{close_directive:{rules:[],inclusive:!1},arg_directive:{rules:[3,4],inclusive:!1},type_directive:{rules:[2,3],inclusive:!1},open_directive:{rules:[1],inclusive:!1},title:{rules:[11],inclusive:!1},string:{rules:[13,14],inclusive:!1},INITIAL:{rules:[0,5,6,7,8,9,10,12,15,16,17],inclusive:!0}}};function d(){this.yy={}}return h.lexer=f,d.prototype=h,h.Parser=d,new d}();e.parser=i,e.Parser=i.Parser,e.parse=function(){return i.parse.apply(i,arguments)},e.main=function(r){r[1]||(console.log("Usage: "+r[0]+" FILE"),t.exit(1));var i=n(19).readFileSync(n(20).normalize(r[1]),"utf8");return e.parser.parse(i)},n.c[n.s]===r&&e.main(t.argv.slice(1))}).call(this,n(14),n(7)(t))},function(t,e,n){(function(t,r){var i=function(){var t=function(t,e,n,r){for(n=n||{},r=t.length;r--;n[t[r]]=e);return n},e=[1,2],n=[1,5],r=[6,9,11,23,37],i=[1,17],a=[1,20],o=[1,25],s=[1,26],c=[1,27],u=[1,28],l=[1,37],h=[23,34,35],f=[4,6,9,11,23,37],d=[30,31,32,33],p=[22,27],g={trace:function(){},yy:{},symbols_:{error:2,start:3,ER_DIAGRAM:4,document:5,EOF:6,directive:7,line:8,SPACE:9,statement:10,NEWLINE:11,openDirective:12,typeDirective:13,closeDirective:14,":":15,argDirective:16,entityName:17,relSpec:18,role:19,BLOCK_START:20,attributes:21,BLOCK_STOP:22,ALPHANUM:23,attribute:24,attributeType:25,attributeName:26,ATTRIBUTE_WORD:27,cardinality:28,relType:29,ZERO_OR_ONE:30,ZERO_OR_MORE:31,ONE_OR_MORE:32,ONLY_ONE:33,NON_IDENTIFYING:34,IDENTIFYING:35,WORD:36,open_directive:37,type_directive:38,arg_directive:39,close_directive:40,$accept:0,$end:1},terminals_:{2:"error",4:"ER_DIAGRAM",6:"EOF",9:"SPACE",11:"NEWLINE",15:":",20:"BLOCK_START",22:"BLOCK_STOP",23:"ALPHANUM",27:"ATTRIBUTE_WORD",30:"ZERO_OR_ONE",31:"ZERO_OR_MORE",32:"ONE_OR_MORE",33:"ONLY_ONE",34:"NON_IDENTIFYING",35:"IDENTIFYING",36:"WORD",37:"open_directive",38:"type_directive",39:"arg_directive",40:"close_directive"},productions_:[0,[3,3],[3,2],[5,0],[5,2],[8,2],[8,1],[8,1],[8,1],[7,4],[7,6],[10,1],[10,5],[10,4],[10,3],[10,1],[17,1],[21,1],[21,2],[24,2],[25,1],[26,1],[18,3],[28,1],[28,1],[28,1],[28,1],[29,1],[29,1],[19,1],[19,1],[12,1],[13,1],[16,1],[14,1]],performAction:function(t,e,n,r,i,a,o){var s=a.length-1;switch(i){case 1:break;case 3:this.$=[];break;case 4:a[s-1].push(a[s]),this.$=a[s-1];break;case 5:case 6:this.$=a[s];break;case 7:case 8:this.$=[];break;case 12:r.addEntity(a[s-4]),r.addEntity(a[s-2]),r.addRelationship(a[s-4],a[s],a[s-2],a[s-3]);break;case 13:r.addEntity(a[s-3]),r.addAttributes(a[s-3],a[s-1]);break;case 14:r.addEntity(a[s-2]);break;case 15:r.addEntity(a[s]);break;case 16:this.$=a[s];break;case 17:this.$=[a[s]];break;case 18:a[s].push(a[s-1]),this.$=a[s];break;case 19:this.$={attributeType:a[s-1],attributeName:a[s]};break;case 20:case 21:this.$=a[s];break;case 22:this.$={cardA:a[s],relType:a[s-1],cardB:a[s-2]};break;case 23:this.$=r.Cardinality.ZERO_OR_ONE;break;case 24:this.$=r.Cardinality.ZERO_OR_MORE;break;case 25:this.$=r.Cardinality.ONE_OR_MORE;break;case 26:this.$=r.Cardinality.ONLY_ONE;break;case 27:this.$=r.Identification.NON_IDENTIFYING;break;case 28:this.$=r.Identification.IDENTIFYING;break;case 29:this.$=a[s].replace(/"/g,"");break;case 30:this.$=a[s];break;case 31:r.parseDirective("%%{","open_directive");break;case 32:r.parseDirective(a[s],"type_directive");break;case 33:a[s]=a[s].trim().replace(/'/g,'"'),r.parseDirective(a[s],"arg_directive");break;case 34:r.parseDirective("}%%","close_directive","er")}},table:[{3:1,4:e,7:3,12:4,37:n},{1:[3]},t(r,[2,3],{5:6}),{3:7,4:e,7:3,12:4,37:n},{13:8,38:[1,9]},{38:[2,31]},{6:[1,10],7:15,8:11,9:[1,12],10:13,11:[1,14],12:4,17:16,23:i,37:n},{1:[2,2]},{14:18,15:[1,19],40:a},t([15,40],[2,32]),t(r,[2,8],{1:[2,1]}),t(r,[2,4]),{7:15,10:21,12:4,17:16,23:i,37:n},t(r,[2,6]),t(r,[2,7]),t(r,[2,11]),t(r,[2,15],{18:22,28:24,20:[1,23],30:o,31:s,32:c,33:u}),t([6,9,11,15,20,23,30,31,32,33,37],[2,16]),{11:[1,29]},{16:30,39:[1,31]},{11:[2,34]},t(r,[2,5]),{17:32,23:i},{21:33,22:[1,34],24:35,25:36,27:l},{29:38,34:[1,39],35:[1,40]},t(h,[2,23]),t(h,[2,24]),t(h,[2,25]),t(h,[2,26]),t(f,[2,9]),{14:41,40:a},{40:[2,33]},{15:[1,42]},{22:[1,43]},t(r,[2,14]),{21:44,22:[2,17],24:35,25:36,27:l},{26:45,27:[1,46]},{27:[2,20]},{28:47,30:o,31:s,32:c,33:u},t(d,[2,27]),t(d,[2,28]),{11:[1,48]},{19:49,23:[1,51],36:[1,50]},t(r,[2,13]),{22:[2,18]},t(p,[2,19]),t(p,[2,21]),{23:[2,22]},t(f,[2,10]),t(r,[2,12]),t(r,[2,29]),t(r,[2,30])],defaultActions:{5:[2,31],7:[2,2],20:[2,34],31:[2,33],37:[2,20],44:[2,18],47:[2,22]},parseError:function(t,e){if(!e.recoverable){var n=new Error(t);throw n.hash=e,n}this.trace(t)},parse:function(t){var e=this,n=[0],r=[],i=[null],a=[],o=this.table,s="",c=0,u=0,l=0,h=2,f=1,d=a.slice.call(arguments,1),p=Object.create(this.lexer),g={yy:{}};for(var y in this.yy)Object.prototype.hasOwnProperty.call(this.yy,y)&&(g.yy[y]=this.yy[y]);p.setInput(t,g.yy),g.yy.lexer=p,g.yy.parser=this,void 0===p.yylloc&&(p.yylloc={});var v=p.yylloc;a.push(v);var m=p.options&&p.options.ranges;function b(){var t;return"number"!=typeof(t=r.pop()||p.lex()||f)&&(t instanceof Array&&(t=(r=t).pop()),t=e.symbols_[t]||t),t}"function"==typeof g.yy.parseError?this.parseError=g.yy.parseError:this.parseError=Object.getPrototypeOf(this).parseError;for(var x,_,k,w,E,T,C,A,S,M={};;){if(k=n[n.length-1],this.defaultActions[k]?w=this.defaultActions[k]:(null==x&&(x=b()),w=o[k]&&o[k][x]),void 0===w||!w.length||!w[0]){var O="";for(T in S=[],o[k])this.terminals_[T]&&T>h&&S.push("'"+this.terminals_[T]+"'");O=p.showPosition?"Parse error on line "+(c+1)+":\n"+p.showPosition()+"\nExpecting "+S.join(", ")+", got '"+(this.terminals_[x]||x)+"'":"Parse error on line "+(c+1)+": Unexpected "+(x==f?"end of input":"'"+(this.terminals_[x]||x)+"'"),this.parseError(O,{text:p.match,token:this.terminals_[x]||x,line:p.yylineno,loc:v,expected:S})}if(w[0]instanceof Array&&w.length>1)throw new Error("Parse Error: multiple actions possible at state: "+k+", token: "+x);switch(w[0]){case 1:n.push(x),i.push(p.yytext),a.push(p.yylloc),n.push(w[1]),x=null,_?(x=_,_=null):(u=p.yyleng,s=p.yytext,c=p.yylineno,v=p.yylloc,l>0&&l--);break;case 2:if(C=this.productions_[w[1]][1],M.$=i[i.length-C],M._$={first_line:a[a.length-(C||1)].first_line,last_line:a[a.length-1].last_line,first_column:a[a.length-(C||1)].first_column,last_column:a[a.length-1].last_column},m&&(M._$.range=[a[a.length-(C||1)].range[0],a[a.length-1].range[1]]),void 0!==(E=this.performAction.apply(M,[s,u,c,g.yy,w[1],i,a].concat(d))))return E;C&&(n=n.slice(0,-1*C*2),i=i.slice(0,-1*C),a=a.slice(0,-1*C)),n.push(this.productions_[w[1]][0]),i.push(M.$),a.push(M._$),A=o[n[n.length-2]][n[n.length-1]],n.push(A);break;case 3:return!0}}return!0}},y={EOF:1,parseError:function(t,e){if(!this.yy.parser)throw new Error(t);this.yy.parser.parseError(t,e)},setInput:function(t,e){return this.yy=e||this.yy||{},this._input=t,this._more=this._backtrack=this.done=!1,this.yylineno=this.yyleng=0,this.yytext=this.matched=this.match="",this.conditionStack=["INITIAL"],this.yylloc={first_line:1,first_column:0,last_line:1,last_column:0},this.options.ranges&&(this.yylloc.range=[0,0]),this.offset=0,this},input:function(){var t=this._input[0];return this.yytext+=t,this.yyleng++,this.offset++,this.match+=t,this.matched+=t,t.match(/(?:\r\n?|\n).*/g)?(this.yylineno++,this.yylloc.last_line++):this.yylloc.last_column++,this.options.ranges&&this.yylloc.range[1]++,this._input=this._input.slice(1),t},unput:function(t){var e=t.length,n=t.split(/(?:\r\n?|\n)/g);this._input=t+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-e),this.offset-=e;var r=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),n.length-1&&(this.yylineno-=n.length-1);var i=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:n?(n.length===r.length?this.yylloc.first_column:0)+r[r.length-n.length].length-n[0].length:this.yylloc.first_column-e},this.options.ranges&&(this.yylloc.range=[i[0],i[0]+this.yyleng-e]),this.yyleng=this.yytext.length,this},more:function(){return this._more=!0,this},reject:function(){return this.options.backtrack_lexer?(this._backtrack=!0,this):this.parseError("Lexical error on line "+(this.yylineno+1)+". You can only invoke reject() in the lexer when the lexer is of the backtracking persuasion (options.backtrack_lexer = true).\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},less:function(t){this.unput(this.match.slice(t))},pastInput:function(){var t=this.matched.substr(0,this.matched.length-this.match.length);return(t.length>20?"...":"")+t.substr(-20).replace(/\n/g,"")},upcomingInput:function(){var t=this.match;return t.length<20&&(t+=this._input.substr(0,20-t.length)),(t.substr(0,20)+(t.length>20?"...":"")).replace(/\n/g,"")},showPosition:function(){var t=this.pastInput(),e=new Array(t.length+1).join("-");return t+this.upcomingInput()+"\n"+e+"^"},test_match:function(t,e){var n,r,i;if(this.options.backtrack_lexer&&(i={yylineno:this.yylineno,yylloc:{first_line:this.yylloc.first_line,last_line:this.last_line,first_column:this.yylloc.first_column,last_column:this.yylloc.last_column},yytext:this.yytext,match:this.match,matches:this.matches,matched:this.matched,yyleng:this.yyleng,offset:this.offset,_more:this._more,_input:this._input,yy:this.yy,conditionStack:this.conditionStack.slice(0),done:this.done},this.options.ranges&&(i.yylloc.range=this.yylloc.range.slice(0))),(r=t[0].match(/(?:\r\n?|\n).*/g))&&(this.yylineno+=r.length),this.yylloc={first_line:this.yylloc.last_line,last_line:this.yylineno+1,first_column:this.yylloc.last_column,last_column:r?r[r.length-1].length-r[r.length-1].match(/\r?\n?/)[0].length:this.yylloc.last_column+t[0].length},this.yytext+=t[0],this.match+=t[0],this.matches=t,this.yyleng=this.yytext.length,this.options.ranges&&(this.yylloc.range=[this.offset,this.offset+=this.yyleng]),this._more=!1,this._backtrack=!1,this._input=this._input.slice(t[0].length),this.matched+=t[0],n=this.performAction.call(this,this.yy,this,e,this.conditionStack[this.conditionStack.length-1]),this.done&&this._input&&(this.done=!1),n)return n;if(this._backtrack){for(var a in i)this[a]=i[a];return!1}return!1},next:function(){if(this.done)return this.EOF;var t,e,n,r;this._input||(this.done=!0),this._more||(this.yytext="",this.match="");for(var i=this._currentRules(),a=0;a<i.length;a++)if((n=this._input.match(this.rules[i[a]]))&&(!e||n[0].length>e[0].length)){if(e=n,r=a,this.options.backtrack_lexer){if(!1!==(t=this.test_match(n,i[a])))return t;if(this._backtrack){e=!1;continue}return!1}if(!this.options.flex)break}return e?!1!==(t=this.test_match(e,i[r]))&&t:""===this._input?this.EOF:this.parseError("Lexical error on line "+(this.yylineno+1)+". Unrecognized text.\n"+this.showPosition(),{text:"",token:null,line:this.yylineno})},lex:function(){var t=this.next();return t||this.lex()},begin:function(t){this.conditionStack.push(t)},popState:function(){return this.conditionStack.length-1>0?this.conditionStack.pop():this.conditionStack[0]},_currentRules:function(){return this.conditionStack.length&&this.conditionStack[this.conditionStack.length-1]?this.conditions[this.conditionStack[this.conditionStack.length-1]].rules:this.conditions.INITIAL.rules},topState:function(t){return(t=this.conditionStack.length-1-Math.abs(t||0))>=0?this.conditionStack[t]:"INITIAL"},pushState:function(t){this.begin(t)},stateStackSize:function(){return this.conditionStack.length},options:{"case-insensitive":!0},performAction:function(t,e,n,r){switch(n){case 0:return this.begin("open_directive"),37;case 1:return this.begin("type_directive"),38;case 2:return this.popState(),this.begin("arg_directive"),15;case 3:return this.popState(),this.popState(),40;case 4:return 39;case 5:case 6:break;case 7:return 11;case 8:break;case 9:return 9;case 10:return 36;case 11:return 4;case 12:return this.begin("block"),20;case 13:break;case 14:return 27;case 15:break;case 16:return this.popState(),22;case 17:return e.yytext[0];case 18:return 30;case 19:return 31;case 20:return 32;case 21:return 33;case 22:return 30;case 23:return 31;case 24:return 32;case 25:return 34;case 26:return 35;case 27:case 28:return 34;case 29:return 23;case 30:return e.yytext[0];case 31:return 6}},rules:[/^(?:%%\{)/i,/^(?:((?:(?!\}%%)[^:.])*))/i,/^(?::)/i,/^(?:\}%%)/i,/^(?:((?:(?!\}%%).|\n)*))/i,/^(?:%(?!\{)[^\n]*)/i,/^(?:[^\}]%%[^\n]*)/i,/^(?:[\n]+)/i,/^(?:\s+)/i,/^(?:[\s]+)/i,/^(?:"[^"]*")/i,/^(?:erDiagram\b)/i,/^(?:\{)/i,/^(?:\s+)/i,/^(?:[A-Za-z][A-Za-z0-9\-_]*)/i,/^(?:[\n]+)/i,/^(?:\})/i,/^(?:.)/i,/^(?:\|o\b)/i,/^(?:\}o\b)/i,/^(?:\}\|)/i,/^(?:\|\|)/i,/^(?:o\|)/i,/^(?:o\{)/i,/^(?:\|\{)/i,/^(?:\.\.)/i,/^(?:--)/i,/^(?:\.-)/i,/^(?:-\.)/i,/^(?:[A-Za-z][A-Za-z0-9\-_]*)/i,/^(?:.)/i,/^(?:$)/i],conditions:{open_directive:{rules:[1],inclusive:!1},type_directive:{rules:[2,3],inclusive:!1},arg_directive:{rules:[3,4],inclusive:!1},block:{rules:[13,14,15,16,17],inclusive:!1},INITIAL:{rules:[0,5,6,7,8,9,10,11,12,18,19,20,21,22,23,24,25,26,27,28,29,30,31],inclusive:!0}}};function v(){this.yy={}}return g.lexer=y,v.prototype=g,g.Parser=v,new v}();e.parser=i,e.Parser=i.Parser,e.parse=function(){return i.parse.apply(i,arguments)},e.main=function(r){r[1]||(console.log("Usage: "+r[0]+" FILE"),t.exit(1));var i=n(19).readFileSync(n(20).normalize(r[1]),"utf8");return e.parser.parse(i)},n.c[n.s]===r&&e.main(t.argv.slice(1))}).call(this,n(14),n(7)(t))},function(t,e,n){"use strict";var r;Object.defineProperty(e,"__esModule",{value:!0}),function(t){t[t.ALL=0]="ALL",t[t.RGB=1]="RGB",t[t.HSL=2]="HSL"}(r||(r={})),e.TYPE=r},function(t,e,n){"use strict";var r=n(10);t.exports=i;function i(t){this._isDirected=!r.has(t,"directed")||t.directed,this._isMultigraph=!!r.has(t,"multigraph")&&t.multigraph,this._isCompound=!!r.has(t,"compound")&&t.compound,this._label=void 0,this._defaultNodeLabelFn=r.constant(void 0),this._defaultEdgeLabelFn=r.constant(void 0),this._nodes={},this._isCompound&&(this._parent={},this._children={},this._children["\0"]={}),this._in={},this._preds={},this._out={},this._sucs={},this._edgeObjs={},this._edgeLabels={}}function a(t,e){t[e]?t[e]++:t[e]=1}function o(t,e){--t[e]||delete t[e]}function s(t,e,n,i){var a=""+e,o=""+n;if(!t&&a>o){var s=a;a=o,o=s}return a+"
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e=t,(n=[{key:"updateColors",value:function(){this.secondBkg=Object(q.lighten)(this.mainBkg,16),this.lineColor=this.mainContrastColor,this.arrowheadColor=this.mainContrastColor,this.nodeBkg=this.mainBkg,this.nodeBorder=this.border1,this.clusterBkg=this.secondBkg,this.clusterBorder=this.border2,this.defaultLinkColor=this.lineColor,this.edgeLabelBackground=Object(q.lighten)(this.labelBackground,25),this.actorBorder=this.border1,this.actorBkg=this.mainBkg,this.actorTextColor=this.mainContrastColor,this.actorLineColor=this.mainContrastColor,this.signalColor=this.mainContrastColor,this.signalTextColor=this.mainContrastColor,this.labelBoxBkgColor=this.actorBkg,this.labelBoxBorderColor=this.actorBorder,this.labelTextColor=this.mainContrastColor,this.loopTextColor=this.mainContrastColor,this.noteBorderColor=this.border2,this.noteTextColor=this.mainBkg,this.activationBorderColor=this.border1,this.activationBkgColor=this.secondBkg,this.altSectionBkgColor=this.background,this.taskBkgColor=Object(q.lighten)(this.mainBkg,23),this.taskTextColor=this.darkTextColor,this.taskTextLightColor=this.mainContrastColor,this.taskTextOutsideColor=this.taskTextLightColor,this.gridColor=this.mainContrastColor,this.doneTaskBkgColor=this.mainContrastColor,this.taskTextDarkColor=this.darkTextColor,this.labelColor=this.textColor,this.altBackground=Object(q.lighten)(this.background,20),this.fillType0=this.primaryColor,this.fillType1=this.secondaryColor,this.fillType2=Object(q.adjust)(this.primaryColor,{h:64}),this.fillType3=Object(q.adjust)(this.secondaryColor,{h:64}),this.fillType4=Object(q.adjust)(this.primaryColor,{h:-64}),this.fillType5=Object(q.adjust)(this.secondaryColor,{h:-64}),this.fillType6=Object(q.adjust)(this.primaryColor,{h:128}),this.fillType7=Object(q.adjust)(this.secondaryColor,{h:128}),this.classText=this.primaryTextColor}},{key:"calculate",value:function(t){var e=this;if("object"===Q(t)){var n=Object.keys(t);n.forEach((function(n){e[n]=t[n]})),this.updateColors(),n.forEach((function(n){e[n]=t[n]}))}else this.updateColors()}}])&&tt(e.prototype,n),r&&tt(e,r),t}();function nt(t){return(nt="function"==typeof Symbol&&"symbol"==typeof Symbol.iterator?function(t){return typeof t}:function(t){return t&&"function"==typeof Symbol&&t.constructor===Symbol&&t!==Symbol.prototype?"symbol":typeof t})(t)}function rt(t,e){for(var n=0;n<e.length;n++){var r=e[n];r.enumerable=r.enumerable||!1,r.configurable=!0,"value"in r&&(r.writable=!0),Object.defineProperty(t,r.key,r)}}var it=function(){function t(){!function(t,e){if(!(t instanceof e))throw new TypeError("Cannot call a class as a function")}(this,t),this.background="#f4f4f4",this.primaryColor="#ECECFF",this.secondaryColor=Object(q.adjust)(this.primaryColor,{h:120}),this.secondaryColor="#ffffde",this.tertiaryColor=Object(q.adjust)(this.primaryColor,{h:-160}),this.primaryBorderColor=X(this.primaryColor,this.darkMode),this.secondaryBorderColor=X(this.secondaryColor,this.darkMode),this.tertiaryBorderColor=X(this.tertiaryColor,this.darkMode),this.primaryTextColor=Object(q.invert)(this.primaryColor),this.secondaryTextColor=Object(q.invert)(this.secondaryColor),this.tertiaryTextColor=Object(q.invert)(this.tertiaryColor),this.lineColor=Object(q.invert)(this.background),this.textColor=Object(q.invert)(this.background),this.background="white",this.mainBkg="#ECECFF",this.secondBkg="#ffffde",this.lineColor="#333333",this.border1="#9370DB",this.border2="#aaaa33",this.arrowheadColor="#333333",this.fontFamily='"trebuchet ms", verdana, arial, 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e,n,r;return e=t,(n=[{key:"updateColors",value:function(){this.nodeBkg=this.mainBkg,this.nodeBorder=this.border1,this.clusterBkg=this.secondBkg,this.clusterBorder=this.border2,this.defaultLinkColor=this.lineColor,this.titleColor=this.textColor,this.edgeLabelBackground=this.labelBackground,this.actorBorder=Object(q.lighten)(this.border1,23),this.actorBkg=this.mainBkg,this.labelBoxBkgColor=this.actorBkg,this.signalColor=this.textColor,this.signalTextColor=this.textColor,this.labelBoxBorderColor=this.actorBorder,this.labelTextColor=this.actorTextColor,this.loopTextColor=this.actorTextColor,this.noteBorderColor=this.border2,this.noteTextColor=this.actorTextColor,this.taskTextColor=this.taskTextLightColor,this.taskTextOutsideColor=this.taskTextDarkColor,this.classText=this.primaryTextColor,this.fillType0=this.primaryColor,this.fillType1=this.secondaryColor,this.fillType2=Object(q.adjust)(this.primaryColor,{h:64}),this.fillType3=Object(q.adjust)(this.secondaryColor,{h:64}),this.fillType4=Object(q.adjust)(this.primaryColor,{h:-64}),this.fillType5=Object(q.adjust)(this.secondaryColor,{h:-64}),this.fillType6=Object(q.adjust)(this.primaryColor,{h:128}),this.fillType7=Object(q.adjust)(this.secondaryColor,{h:128})}},{key:"calculate",value:function(t){var e=this;if("object"===nt(t)){var n=Object.keys(t);n.forEach((function(n){e[n]=t[n]})),this.updateColors(),n.forEach((function(n){e[n]=t[n]}))}else this.updateColors()}}])&&rt(e.prototype,n),r&&rt(e,r),t}();function at(t){return(at="function"==typeof Symbol&&"symbol"==typeof Symbol.iterator?function(t){return typeof t}:function(t){return t&&"function"==typeof Symbol&&t.constructor===Symbol&&t!==Symbol.prototype?"symbol":typeof t})(t)}function ot(t,e){for(var n=0;n<e.length;n++){var r=e[n];r.enumerable=r.enumerable||!1,r.configurable=!0,"value"in r&&(r.writable=!0),Object.defineProperty(t,r.key,r)}}var st=function(){function t(){!function(t,e){if(!(t instanceof e))throw new TypeError("Cannot call a class as a function")}(this,t),this.background="#f4f4f4",this.primaryColor="#cde498",this.secondaryColor="#cdffb2",this.background="white",this.mainBkg="#cde498",this.secondBkg="#cdffb2",this.lineColor="green",this.border1="#13540c",this.border2="#6eaa49",this.arrowheadColor="green",this.fontFamily='"trebuchet ms", verdana, arial, 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e,n,r;return e=t,(n=[{key:"updateColors",value:function(){this.nodeBkg=this.mainBkg,this.nodeBorder=this.border1,this.clusterBkg=this.secondBkg,this.clusterBorder=this.border2,this.defaultLinkColor=this.lineColor,this.actorBorder=Object(q.darken)(this.mainBkg,20),this.actorBkg=this.mainBkg,this.labelBoxBkgColor=this.actorBkg,this.labelTextColor=this.actorTextColor,this.loopTextColor=this.actorTextColor,this.noteBorderColor=this.border2,this.noteTextColor=this.actorTextColor,this.taskBorderColor=this.border1,this.taskTextColor=this.taskTextLightColor,this.taskTextOutsideColor=this.taskTextDarkColor,this.activeTaskBorderColor=this.taskBorderColor,this.activeTaskBkgColor=this.mainBkg,this.classText=this.primaryTextColor,this.fillType0=this.primaryColor,this.fillType1=this.secondaryColor,this.fillType2=Object(q.adjust)(this.primaryColor,{h:64}),this.fillType3=Object(q.adjust)(this.secondaryColor,{h:64}),this.fillType4=Object(q.adjust)(this.primaryColor,{h:-64}),this.fillType5=Object(q.adjust)(this.secondaryColor,{h:-64}),this.fillType6=Object(q.adjust)(this.primaryColor,{h:128}),this.fillType7=Object(q.adjust)(this.secondaryColor,{h:128})}},{key:"calculate",value:function(t){var e=this;if("object"===at(t)){var n=Object.keys(t);n.forEach((function(n){e[n]=t[n]})),this.updateColors(),n.forEach((function(n){e[n]=t[n]}))}else this.updateColors()}}])&&ot(e.prototype,n),r&&ot(e,r),t}();function ct(t){return(ct="function"==typeof Symbol&&"symbol"==typeof Symbol.iterator?function(t){return typeof t}:function(t){return t&&"function"==typeof Symbol&&t.constructor===Symbol&&t!==Symbol.prototype?"symbol":typeof t})(t)}function ut(t,e){for(var n=0;n<e.length;n++){var r=e[n];r.enumerable=r.enumerable||!1,r.configurable=!0,"value"in r&&(r.writable=!0),Object.defineProperty(t,r.key,r)}}var lt=function(){function t(){!function(t,e){if(!(t instanceof e))throw new TypeError("Cannot call a class as a function")}(this,t),this.primaryColor="#eee",this.contrast="#26a",this.secondaryColor=Object(q.lighten)(this.contrast,55),this.background="#ffffff",this.tertiaryColor=Object(q.adjust)(this.primaryColor,{h:-160}),this.primaryBorderColor=X(this.primaryColor,this.darkMode),this.secondaryBorderColor=X(this.secondaryColor,this.darkMode),this.tertiaryBorderColor=X(this.tertiaryColor,this.darkMode),this.primaryTextColor=Object(q.invert)(this.primaryColor),this.secondaryTextColor=Object(q.invert)(this.secondaryColor),this.tertiaryTextColor=Object(q.invert)(this.tertiaryColor),this.lineColor=Object(q.invert)(this.background),this.textColor=Object(q.invert)(this.background),this.altBackground=Object(q.lighten)(this.contrast,55),this.mainBkg="#eee",this.secondBkg="calculated",this.lineColor="#666",this.border1="#999",this.border2="calculated",this.note="#ffa",this.text="#333",this.critical="#d42",this.done="#bbb",this.arrowheadColor="#333333",this.fontFamily='"trebuchet ms", verdana, arial, sans-serif',this.fontSize="16px",this.nodeBkg="calculated",this.nodeBorder="calculated",this.clusterBkg="calculated",this.clusterBorder="calculated",this.defaultLinkColor="calculated",this.titleColor="calculated",this.edgeLabelBackground="white",this.actorBorder="calculated",this.actorBkg="calculated",this.actorTextColor="calculated",this.actorLineColor="calculated",this.signalColor="calculated",this.signalTextColor="calculated",this.labelBoxBkgColor="calculated",this.labelBoxBorderColor="calculated",this.labelTextColor="calculated",this.loopTextColor="calculated",this.noteBorderColor="calculated",this.noteBkgColor="calculated",this.noteTextColor="calculated",this.activationBorderColor="#666",this.activationBkgColor="#f4f4f4",this.sequenceNumberColor="white",this.sectionBkgColor="calculated",this.altSectionBkgColor="white",this.sectionBkgColor2="calculated",this.taskBorderColor="calculated",this.taskBkgColor="calculated",this.taskTextLightColor="white",this.taskTextColor="calculated",this.taskTextDarkColor="calculated",this.taskTextOutsideColor="calculated",this.taskTextClickableColor="#003163",this.activeTaskBorderColor="calculated",this.activeTaskBkgColor="calculated",this.gridColor="calculated",this.doneTaskBkgColor="calculated",this.doneTaskBorderColor="calculated",this.critBkgColor="calculated",this.critBorderColor="calculated",this.todayLineColor="calculated",this.labelColor="black",this.errorBkgColor="#552222",this.errorTextColor="#552222"}var e,n,r;return e=t,(n=[{key:"updateColors",value:function(){this.secondBkg=Object(q.lighten)(this.contrast,55),this.border2=this.contrast,this.nodeBkg=this.mainBkg,this.nodeBorder=this.border1,this.clusterBkg=this.secondBkg,this.clusterBorder=this.border2,this.defaultLinkColor=this.lineColor,this.titleColor=this.text,this.actorBorder=Object(q.lighten)(this.border1,23),this.actorBkg=this.mainBkg,this.actorTextColor=this.text,this.actorLineColor=this.lineColor,this.signalColor=this.text,this.signalTextColor=this.text,this.labelBoxBkgColor=this.actorBkg,this.labelBoxBorderColor=this.actorBorder,this.labelTextColor=this.text,this.loopTextColor=this.text,this.noteBorderColor=Object(q.darken)(this.note,60),this.noteBkgColor=this.note,this.noteTextColor=this.actorTextColor,this.sectionBkgColor=Object(q.lighten)(this.contrast,30),this.sectionBkgColor2=Object(q.lighten)(this.contrast,30),this.taskBorderColor=Object(q.darken)(this.contrast,10),this.taskBkgColor=this.contrast,this.taskTextColor=this.taskTextLightColor,this.taskTextDarkColor=this.text,this.taskTextOutsideColor=this.taskTextDarkColor,this.activeTaskBorderColor=this.taskBorderColor,this.activeTaskBkgColor=this.mainBkg,this.gridColor=Object(q.lighten)(this.border1,30),this.doneTaskBkgColor=this.done,this.doneTaskBorderColor=this.lineColor,this.critBkgColor=this.critical,this.critBorderColor=Object(q.darken)(this.critBkgColor,10),this.todayLineColor=this.critBkgColor,this.classText=this.primaryTextColor,this.fillType0=this.primaryColor,this.fillType1=this.secondaryColor,this.fillType2=Object(q.adjust)(this.primaryColor,{h:64}),this.fillType3=Object(q.adjust)(this.secondaryColor,{h:64}),this.fillType4=Object(q.adjust)(this.primaryColor,{h:-64}),this.fillType5=Object(q.adjust)(this.secondaryColor,{h:-64}),this.fillType6=Object(q.adjust)(this.primaryColor,{h:128}),this.fillType7=Object(q.adjust)(this.secondaryColor,{h:128})}},{key:"calculate",value:function(t){var e=this;if("object"===ct(t)){var n=Object.keys(t);n.forEach((function(n){e[n]=t[n]})),this.updateColors(),n.forEach((function(n){e[n]=t[n]}))}else this.updateColors()}}])&&ut(e.prototype,n),r&&ut(e,r),t}(),ht={base:{getThemeVariables:function(t){var e=new K;return e.calculate(t),e}},dark:{getThemeVariables:function(t){var e=new et;return e.calculate(t),e}},default:{getThemeVariables:function(t){var e=new it;return e.calculate(t),e}},forest:{getThemeVariables:function(t){var e=new st;return e.calculate(t),e}},neutral:{getThemeVariables:function(t){var e=new lt;return e.calculate(t),e}}},ft={theme:"default",themeVariables:ht.default.getThemeVariables(),themeCSS:void 0,maxTextSize:5e4,fontFamily:'"trebuchet ms", verdana, arial, sans-serif;',logLevel:5,securityLevel:"strict",startOnLoad:!0,arrowMarkerAbsolute:!1,secure:["secure","securityLevel","startOnLoad","maxTextSize"],deterministicIds:!1,deterministicIDSeed:void 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0,useMaxWidth:!0},state:{dividerMargin:10,sizeUnit:5,padding:8,textHeight:10,titleShift:-15,noteMargin:10,forkWidth:70,forkHeight:7,miniPadding:2,fontSizeFactor:5.02,fontSize:24,labelHeight:16,edgeLengthFactor:"20",compositTitleSize:35,radius:5,useMaxWidth:!0},er:{diagramPadding:20,layoutDirection:"TB",minEntityWidth:100,minEntityHeight:75,entityPadding:15,stroke:"gray",fill:"honeydew",fontSize:12,useMaxWidth:!0},pie:{useWidth:void 0,useMaxWidth:!0}};ft.class.arrowMarkerAbsolute=ft.arrowMarkerAbsolute,ft.git.arrowMarkerAbsolute=ft.arrowMarkerAbsolute;var dt,pt=ft,gt=Object.freeze(pt),yt=F({},gt),vt=[],mt=F({},gt),bt=function(t,e){for(var n=F({},t),r={},i=0;i<e.length;i++){var a=e[i];kt(a),r=F(r,a)}if(n=F(n,r),r.theme){var o=F(dt.themeVariables||{},r.themeVariables);n.themeVariables=ht[n.theme].getThemeVariables(o)}return mt=n,n},xt=function(){return F({},yt)},_t=function(){return F({},mt)},kt=function(t){Object.keys(yt.secure).forEach((function(e){void 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i=this,a=0;function o(o){return function(s){a++;var c=i.sequenceItems.length-a+1;i.updateVal(s,"starty",e-c*cr.boxMargin,Math.min),i.updateVal(s,"stopy",r+c*cr.boxMargin,Math.max),i.updateVal(ur.data,"startx",t-c*cr.boxMargin,Math.min),i.updateVal(ur.data,"stopx",n+c*cr.boxMargin,Math.max),"activation"!==o&&(i.updateVal(s,"startx",t-c*cr.boxMargin,Math.min),i.updateVal(s,"stopx",n+c*cr.boxMargin,Math.max),i.updateVal(ur.data,"starty",e-c*cr.boxMargin,Math.min),i.updateVal(ur.data,"stopy",r+c*cr.boxMargin,Math.max))}}this.sequenceItems.forEach(o()),this.activations.forEach(o("activation"))},insert:function(t,e,n,r){var i=Math.min(t,n),a=Math.max(t,n),o=Math.min(e,r),s=Math.max(e,r);this.updateVal(ur.data,"startx",i,Math.min),this.updateVal(ur.data,"starty",o,Math.min),this.updateVal(ur.data,"stopx",a,Math.max),this.updateVal(ur.data,"stopy",s,Math.max),this.updateBounds(i,o,a,s)},newActivation:function(t,e,n){var r=n[t.from.actor],i=gr(t.from.actor).length||0,a=r.x+r.width/2+(i-1)*cr.activationWidth/2;this.activations.push({startx:a,starty:this.verticalPos+2,stopx:a+cr.activationWidth,stopy:void 0,actor:t.from.actor,anchored:$n.anchorElement(e)})},endActivation:function(t){var e=this.activations.map((function(t){return t.actor})).lastIndexOf(t.from.actor);return this.activations.splice(e,1)[0]},createLoop:function(){var t=arguments.length>0&&void 0!==arguments[0]?arguments[0]:{message:void 0,wrap:!1,width:void 0},e=arguments.length>1?arguments[1]:void 0;return{startx:void 0,starty:this.verticalPos,stopx:void 0,stopy:void 0,title:t.message,wrap:t.wrap,width:t.width,height:0,fill:e}},newLoop:function(){var t=arguments.length>0&&void 0!==arguments[0]?arguments[0]:{message:void 0,wrap:!1,width:void 0},e=arguments.length>1?arguments[1]:void 0;this.sequenceItems.push(this.createLoop(t,e))},endLoop:function(){return this.sequenceItems.pop()},addSectionToLoop:function(t){var e=this.sequenceItems.pop();e.sections=e.sections||[],e.sectionTitles=e.sectionTitles||[],e.sections.push({y:ur.getVerticalPos(),height:0}),e.sectionTitles.push(t),this.sequenceItems.push(e)},bumpVerticalPos:function(t){this.verticalPos=this.verticalPos+t,this.data.stopy=this.verticalPos},getVerticalPos:function(){return this.verticalPos},getBounds:function(){return{bounds:this.data,models:this.models}}},lr=function(t){return{fontFamily:t.messageFontFamily,fontSize:t.messageFontSize,fontWeight:t.messageFontWeight}},hr=function(t){return{fontFamily:t.noteFontFamily,fontSize:t.noteFontSize,fontWeight:t.noteFontWeight}},fr=function(t){return{fontFamily:t.actorFontFamily,fontSize:t.actorFontSize,fontWeight:t.actorFontWeight}},dr=function(t,e,n,r){for(var i=0,a=0,o=0;o<n.length;o++){var s=e[n[o]];s.width=s.width||cr.width,s.height=Math.max(s.height||cr.height,cr.height),s.margin=s.margin||cr.actorMargin,s.x=i+a,s.y=r,$n.drawActor(t,s,cr),ur.insert(s.x,r,s.x+s.width,s.height),i+=s.width,a+=s.margin,ur.models.addActor(s)}ur.bumpVerticalPos(cr.height)},pr=function(t){F(cr,t),t.fontFamily&&(cr.actorFontFamily=cr.noteFontFamily=cr.messageFontFamily=t.fontFamily),t.fontSize&&(cr.actorFontSize=cr.noteFontSize=cr.messageFontSize=t.fontSize),t.fontWeight&&(cr.actorFontWeight=cr.noteFontWeight=cr.messageFontWeight=t.fontWeight)},gr=function(t){return ur.activations.filter((function(e){return e.actor===t}))},yr=function(t,e){var n=e[t],r=gr(t);return[r.reduce((function(t,e){return Math.min(t,e.startx)}),n.x+n.width/2),r.reduce((function(t,e){return Math.max(t,e.stopx)}),n.x+n.width/2)]};function vr(t,e,n,r,i){ur.bumpVerticalPos(n);var a=r;if(e.id&&e.message&&t[e.id]){var o=t[e.id].width,s=lr(cr);e.message=H.wrapLabel("[".concat(e.message,"]"),o-2*cr.wrapPadding,s),e.width=o,e.wrap=!0;var u=H.calculateTextDimensions(e.message,s),l=Math.max(u.height,cr.labelBoxHeight);a=r+l,c.debug("".concat(l," - ").concat(e.message))}i(e),ur.bumpVerticalPos(a)}var mr=function(t,e){var n={};return e.forEach((function(e){if(t[e.to]&&t[e.from]){var r=t[e.to];if(e.placement===Wn.parser.yy.PLACEMENT.LEFTOF&&!r.prevActor)return;if(e.placement===Wn.parser.yy.PLACEMENT.RIGHTOF&&!r.nextActor)return;var i=void 0!==e.placement,a=!i,o=i?hr(cr):lr(cr),s=e.wrap?H.wrapLabel(e.message,cr.width-2*cr.wrapPadding,o):e.message,c=H.calculateTextDimensions(s,o).width+2*cr.wrapPadding;a&&e.from===r.nextActor?n[e.to]=Math.max(n[e.to]||0,c):a&&e.from===r.prevActor?n[e.from]=Math.max(n[e.from]||0,c):a&&e.from===e.to?(n[e.from]=Math.max(n[e.from]||0,c/2),n[e.to]=Math.max(n[e.to]||0,c/2)):e.placement===Wn.parser.yy.PLACEMENT.RIGHTOF?n[e.from]=Math.max(n[e.from]||0,c):e.placement===Wn.parser.yy.PLACEMENT.LEFTOF?n[r.prevActor]=Math.max(n[r.prevActor]||0,c):e.placement===Wn.parser.yy.PLACEMENT.OVER&&(r.prevActor&&(n[r.prevActor]=Math.max(n[r.prevActor]||0,c/2)),r.nextActor&&(n[e.from]=Math.max(n[e.from]||0,c/2)))}})),c.debug("maxMessageWidthPerActor:",n),n},br=function(t,e){var n=0;for(var r in Object.keys(t).forEach((function(e){var r=t[e];r.wrap&&(r.description=H.wrapLabel(r.description,cr.width-2*cr.wrapPadding,fr(cr)));var i=H.calculateTextDimensions(r.description,fr(cr));r.width=r.wrap?cr.width:Math.max(cr.width,i.width+2*cr.wrapPadding),r.height=r.wrap?Math.max(i.height,cr.height):cr.height,n=Math.max(n,r.height)})),e){var i=t[r];if(i){var a=t[i.nextActor];if(a){var o=e[r]+cr.actorMargin-i.width/2-a.width/2;i.margin=Math.max(o,cr.actorMargin)}}}return Math.max(n,cr.height)},xr=function(t,e){var n,r,i,a={},o=[];return t.forEach((function(t){switch(t.id=H.random({length:10}),t.type){case Wn.parser.yy.LINETYPE.LOOP_START:case Wn.parser.yy.LINETYPE.ALT_START:case Wn.parser.yy.LINETYPE.OPT_START:case Wn.parser.yy.LINETYPE.PAR_START:o.push({id:t.id,msg:t.message,from:Number.MAX_SAFE_INTEGER,to:Number.MIN_SAFE_INTEGER,width:0});break;case Wn.parser.yy.LINETYPE.ALT_ELSE:case Wn.parser.yy.LINETYPE.PAR_AND:t.message&&(n=o.pop(),a[n.id]=n,a[t.id]=n,o.push(n));break;case Wn.parser.yy.LINETYPE.LOOP_END:case Wn.parser.yy.LINETYPE.ALT_END:case Wn.parser.yy.LINETYPE.OPT_END:case Wn.parser.yy.LINETYPE.PAR_END:n=o.pop(),a[n.id]=n;break;case Wn.parser.yy.LINETYPE.ACTIVE_START:var s=e[t.from?t.from.actor:t.to.actor],u=gr(t.from?t.from.actor:t.to.actor).length,l=s.x+s.width/2+(u-1)*cr.activationWidth/2,h={startx:l,stopx:l+cr.activationWidth,actor:t.from.actor,enabled:!0};ur.activations.push(h);break;case Wn.parser.yy.LINETYPE.ACTIVE_END:var f=ur.activations.map((function(t){return t.actor})).lastIndexOf(t.from.actor);delete ur.activations.splice(f,1)[0]}void 0!==t.placement?(r=function(t,e){var n=e[t.from].x,r=e[t.to].x,i=t.wrap&&t.message,a=H.calculateTextDimensions(i?H.wrapLabel(t.message,cr.width,hr(cr)):t.message,hr(cr)),o={width:i?cr.width:Math.max(cr.width,a.width+2*cr.noteMargin),height:0,startx:e[t.from].x,stopx:0,starty:0,stopy:0,message:t.message};return t.placement===Wn.parser.yy.PLACEMENT.RIGHTOF?(o.width=i?Math.max(cr.width,a.width):Math.max(e[t.from].width/2+e[t.to].width/2,a.width+2*cr.noteMargin),o.startx=n+(e[t.from].width+cr.actorMargin)/2):t.placement===Wn.parser.yy.PLACEMENT.LEFTOF?(o.width=i?Math.max(cr.width,a.width+2*cr.noteMargin):Math.max(e[t.from].width/2+e[t.to].width/2,a.width+2*cr.noteMargin),o.startx=n-o.width+(e[t.from].width-cr.actorMargin)/2):t.to===t.from?(a=H.calculateTextDimensions(i?H.wrapLabel(t.message,Math.max(cr.width,e[t.from].width),hr(cr)):t.message,hr(cr)),o.width=i?Math.max(cr.width,e[t.from].width):Math.max(e[t.from].width,cr.width,a.width+2*cr.noteMargin),o.startx=n+(e[t.from].width-o.width)/2):(o.width=Math.abs(n+e[t.from].width/2-(r+e[t.to].width/2))+cr.actorMargin,o.startx=n<r?n+e[t.from].width/2-cr.actorMargin/2:r+e[t.to].width/2-cr.actorMargin/2),i&&(o.message=H.wrapLabel(t.message,o.width-2*cr.wrapPadding,hr(cr))),c.debug("NM:[".concat(o.startx,",").concat(o.stopx,",").concat(o.starty,",").concat(o.stopy,":").concat(o.width,",").concat(o.height,"=").concat(t.message,"]")),o}(t,e),t.noteModel=r,o.forEach((function(t){(n=t).from=Math.min(n.from,r.startx),n.to=Math.max(n.to,r.startx+r.width),n.width=Math.max(n.width,Math.abs(n.from-n.to))-cr.labelBoxWidth}))):(i=function(t,e){var n=!1;if([Wn.parser.yy.LINETYPE.SOLID_OPEN,Wn.parser.yy.LINETYPE.DOTTED_OPEN,Wn.parser.yy.LINETYPE.SOLID,Wn.parser.yy.LINETYPE.DOTTED,Wn.parser.yy.LINETYPE.SOLID_CROSS,Wn.parser.yy.LINETYPE.DOTTED_CROSS,Wn.parser.yy.LINETYPE.SOLID_POINT,Wn.parser.yy.LINETYPE.DOTTED_POINT].includes(t.type)&&(n=!0),!n)return{};var r=yr(t.from,e),i=yr(t.to,e),a=r[0]<=i[0]?1:0,o=r[0]<i[0]?0:1,s=r.concat(i),c=Math.abs(i[o]-r[a]);t.wrap&&t.message&&(t.message=H.wrapLabel(t.message,Math.max(c+2*cr.wrapPadding,cr.width),lr(cr)));var u=H.calculateTextDimensions(t.message,lr(cr));return{width:Math.max(t.wrap?0:u.width+2*cr.wrapPadding,c+2*cr.wrapPadding,cr.width),height:0,startx:r[a],stopx:i[o],starty:0,stopy:0,message:t.message,type:t.type,wrap:t.wrap,fromBounds:Math.min.apply(null,s),toBounds:Math.max.apply(null,s)}}(t,e),t.msgModel=i,i.startx&&i.stopx&&o.length>0&&o.forEach((function(r){if(n=r,i.startx===i.stopx){var a=e[t.from],o=e[t.to];n.from=Math.min(a.x-i.width/2,a.x-a.width/2,n.from),n.to=Math.max(o.x+i.width/2,o.x+a.width/2,n.to),n.width=Math.max(n.width,Math.abs(n.to-n.from))-cr.labelBoxWidth}else n.from=Math.min(i.startx,n.from),n.to=Math.max(i.stopx,n.to),n.width=Math.max(n.width,i.width)-cr.labelBoxWidth})))})),ur.activations=[],c.debug("Loop type widths:",a),a},_r={bounds:ur,drawActors:dr,setConf:pr,draw:function(t,e){cr=_t().sequence,Wn.parser.yy.clear(),Wn.parser.yy.setWrap(cr.wrap),Wn.parser.parse(t+"\n"),ur.init(),c.debug("C:".concat(JSON.stringify(cr,null,2)));var n=Object(d.select)('[id="'.concat(e,'"]')),r=Wn.parser.yy.getActors(),i=Wn.parser.yy.getActorKeys(),a=Wn.parser.yy.getMessages(),o=Wn.parser.yy.getTitle(),s=mr(r,a);cr.height=br(r,s),dr(n,r,i,0);var u=xr(a,r,s);$n.insertArrowHead(n),$n.insertArrowCrossHead(n),$n.insertArrowFilledHead(n),$n.insertSequenceNumber(n);var l=1;a.forEach((function(t){var e,i,a;switch(t.type){case Wn.parser.yy.LINETYPE.NOTE:i=t.noteModel,function(t,e){ur.bumpVerticalPos(cr.boxMargin),e.height=cr.boxMargin,e.starty=ur.getVerticalPos();var n=$n.getNoteRect();n.x=e.startx,n.y=e.starty,n.width=e.width||cr.width,n.class="note";var r=t.append("g"),i=$n.drawRect(r,n),a=$n.getTextObj();a.x=e.startx,a.y=e.starty,a.width=n.width,a.dy="1em",a.text=e.message,a.class="noteText",a.fontFamily=cr.noteFontFamily,a.fontSize=cr.noteFontSize,a.fontWeight=cr.noteFontWeight,a.anchor=cr.noteAlign,a.textMargin=cr.noteMargin,a.valign=cr.noteAlign;var o=Fn(r,a),s=Math.round(o.map((function(t){return(t._groups||t)[0][0].getBBox().height})).reduce((function(t,e){return t+e})));i.attr("height",s+2*cr.noteMargin),e.height+=s+2*cr.noteMargin,ur.bumpVerticalPos(s+2*cr.noteMargin),e.stopy=e.starty+s+2*cr.noteMargin,e.stopx=e.startx+n.width,ur.insert(e.startx,e.starty,e.stopx,e.stopy),ur.models.addNote(e)}(n,i);break;case Wn.parser.yy.LINETYPE.ACTIVE_START:ur.newActivation(t,n,r);break;case Wn.parser.yy.LINETYPE.ACTIVE_END:!function(t,e){var r=ur.endActivation(t);r.starty+18>e&&(r.starty=e-6,e+=12),$n.drawActivation(n,r,e,cr,gr(t.from.actor).length),ur.insert(r.startx,e-10,r.stopx,e)}(t,ur.getVerticalPos());break;case Wn.parser.yy.LINETYPE.LOOP_START:vr(u,t,cr.boxMargin,cr.boxMargin+cr.boxTextMargin,(function(t){return ur.newLoop(t)}));break;case Wn.parser.yy.LINETYPE.LOOP_END:e=ur.endLoop(),$n.drawLoop(n,e,"loop",cr),ur.bumpVerticalPos(e.stopy-ur.getVerticalPos()),ur.models.addLoop(e);break;case Wn.parser.yy.LINETYPE.RECT_START:vr(u,t,cr.boxMargin,cr.boxMargin,(function(t){return ur.newLoop(void 0,t.message)}));break;case Wn.parser.yy.LINETYPE.RECT_END:e=ur.endLoop(),$n.drawBackgroundRect(n,e),ur.models.addLoop(e),ur.bumpVerticalPos(e.stopy-ur.getVerticalPos());break;case Wn.parser.yy.LINETYPE.OPT_START:vr(u,t,cr.boxMargin,cr.boxMargin+cr.boxTextMargin,(function(t){return ur.newLoop(t)}));break;case Wn.parser.yy.LINETYPE.OPT_END:e=ur.endLoop(),$n.drawLoop(n,e,"opt",cr),ur.bumpVerticalPos(e.stopy-ur.getVerticalPos()),ur.models.addLoop(e);break;case Wn.parser.yy.LINETYPE.ALT_START:vr(u,t,cr.boxMargin,cr.boxMargin+cr.boxTextMargin,(function(t){return ur.newLoop(t)}));break;case Wn.parser.yy.LINETYPE.ALT_ELSE:vr(u,t,cr.boxMargin+cr.boxTextMargin,cr.boxMargin,(function(t){return ur.addSectionToLoop(t)}));break;case Wn.parser.yy.LINETYPE.ALT_END:e=ur.endLoop(),$n.drawLoop(n,e,"alt",cr),ur.bumpVerticalPos(e.stopy-ur.getVerticalPos()),ur.models.addLoop(e);break;case Wn.parser.yy.LINETYPE.PAR_START:vr(u,t,cr.boxMargin,cr.boxMargin+cr.boxTextMargin,(function(t){return ur.newLoop(t)}));break;case Wn.parser.yy.LINETYPE.PAR_AND:vr(u,t,cr.boxMargin+cr.boxTextMargin,cr.boxMargin,(function(t){return ur.addSectionToLoop(t)}));break;case Wn.parser.yy.LINETYPE.PAR_END:e=ur.endLoop(),$n.drawLoop(n,e,"par",cr),ur.bumpVerticalPos(e.stopy-ur.getVerticalPos()),ur.models.addLoop(e);break;default:try{(a=t.msgModel).starty=ur.getVerticalPos(),a.sequenceIndex=l,function(t,e){ur.bumpVerticalPos(10);var n=e.startx,r=e.stopx,i=e.starty,a=e.message,o=e.type,s=e.sequenceIndex,c=x.splitBreaks(a).length,u=H.calculateTextDimensions(a,lr(cr)),l=u.height/c;e.height+=l,ur.bumpVerticalPos(l);var h=$n.getTextObj();h.x=n,h.y=i+10,h.width=r-n,h.class="messageText",h.dy="1em",h.text=a,h.fontFamily=cr.messageFontFamily,h.fontSize=cr.messageFontSize,h.fontWeight=cr.messageFontWeight,h.anchor=cr.messageAlign,h.valign=cr.messageAlign,h.textMargin=cr.wrapPadding,h.tspan=!1,Fn(t,h);var f,d,p=u.height-10,g=u.width;if(n===r){d=ur.getVerticalPos()+p,cr.rightAngles?f=t.append("path").attr("d","M  ".concat(n,",").concat(d," H ").concat(n+Math.max(cr.width/2,g/2)," V ").concat(d+25," H ").concat(n)):(p+=cr.boxMargin,d=ur.getVerticalPos()+p,f=t.append("path").attr("d","M "+n+","+d+" C "+(n+60)+","+(d-10)+" "+(n+60)+","+(d+30)+" "+n+","+(d+20))),p+=30;var y=Math.max(g/2,cr.width/2);ur.insert(n-y,ur.getVerticalPos()-10+p,r+y,ur.getVerticalPos()+30+p)}else p+=cr.boxMargin,d=ur.getVerticalPos()+p,(f=t.append("line")).attr("x1",n),f.attr("y1",d),f.attr("x2",r),f.attr("y2",d),ur.insert(n,d-10,r,d);o===Wn.parser.yy.LINETYPE.DOTTED||o===Wn.parser.yy.LINETYPE.DOTTED_CROSS||o===Wn.parser.yy.LINETYPE.DOTTED_POINT||o===Wn.parser.yy.LINETYPE.DOTTED_OPEN?(f.style("stroke-dasharray","3, 3"),f.attr("class","messageLine1")):f.attr("class","messageLine0");var v="";cr.arrowMarkerAbsolute&&(v=(v=(v=window.location.protocol+"//"+window.location.host+window.location.pathname+window.location.search).replace(/\(/g,"\\(")).replace(/\)/g,"\\)")),f.attr("stroke-width",2),f.attr("stroke","none"),f.style("fill","none"),o!==Wn.parser.yy.LINETYPE.SOLID&&o!==Wn.parser.yy.LINETYPE.DOTTED||f.attr("marker-end","url("+v+"#arrowhead)"),o!==Wn.parser.yy.LINETYPE.SOLID_POINT&&o!==Wn.parser.yy.LINETYPE.DOTTED_POINT||f.attr("marker-end","url("+v+"#filled-head)"),o!==Wn.parser.yy.LINETYPE.SOLID_CROSS&&o!==Wn.parser.yy.LINETYPE.DOTTED_CROSS||f.attr("marker-end","url("+v+"#crosshead)"),(sr.showSequenceNumbers()||cr.showSequenceNumbers)&&(f.attr("marker-start","url("+v+"#sequencenumber)"),t.append("text").attr("x",n).attr("y",d+4).attr("font-family","sans-serif").attr("font-size","12px").attr("text-anchor","middle").attr("textLength","16px").attr("class","sequenceNumber").text(s)),ur.bumpVerticalPos(p),e.height+=p,e.stopy=e.starty+e.height,ur.insert(e.fromBounds,e.starty,e.toBounds,e.stopy)}(n,a),ur.models.addMessage(a)}catch(t){c.error("error while drawing message",t)}}[Wn.parser.yy.LINETYPE.SOLID_OPEN,Wn.parser.yy.LINETYPE.DOTTED_OPEN,Wn.parser.yy.LINETYPE.SOLID,Wn.parser.yy.LINETYPE.DOTTED,Wn.parser.yy.LINETYPE.SOLID_CROSS,Wn.parser.yy.LINETYPE.DOTTED_CROSS,Wn.parser.yy.LINETYPE.SOLID_POINT,Wn.parser.yy.LINETYPE.DOTTED_POINT].includes(t.type)&&l++})),cr.mirrorActors&&(ur.bumpVerticalPos(2*cr.boxMargin),dr(n,r,i,ur.getVerticalPos()));var h=ur.getBounds().bounds;c.debug("For line height fix Querying: #"+e+" .actor-line"),Object(d.selectAll)("#"+e+" .actor-line").attr("y2",h.stopy);var f=h.stopy-h.starty+2*cr.diagramMarginY;cr.mirrorActors&&(f=f-cr.boxMargin+cr.bottomMarginAdj);var p=h.stopx-h.startx+2*cr.diagramMarginX;o&&n.append("text").text(o).attr("x",(h.stopx-h.startx)/2-2*cr.diagramMarginX).attr("y",-25),W(n,f,p,cr.useMaxWidth);var g=o?40:0;n.attr("viewBox",h.startx-cr.diagramMarginX+" -"+(cr.diagramMarginY+g)+" "+p+" "+(f+g)),c.debug("models:",ur.models)}},kr=n(27),wr=n.n(kr);function Er(t){return 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p=o.height+2*r,g="attributeBoxOdd";s.forEach((function(e){var n=p+r+Math.max(e.tn.node().getBBox().height,e.nn.node().getBBox().height)/2;e.tn.attr("transform","translate("+i+","+n+")");var a=t.insert("rect","#"+e.tn.node().id).attr("class","er ".concat(g)).attr("fill",Qa.fill).attr("fill-opacity","100%").attr("stroke",Qa.stroke).attr("x",0).attr("y",p).attr("width",c+2*i+d/2).attr("height",e.tn.node().getBBox().height+2*r);e.nn.attr("transform","translate("+(parseFloat(a.attr("width"))+i)+","+n+")"),t.insert("rect","#"+e.nn.node().id).attr("class","er ".concat(g)).attr("fill",Qa.fill).attr("fill-opacity","100%").attr("stroke",Qa.stroke).attr("x","".concat(a.attr("x")+a.attr("width"))).attr("y",p).attr("width",u+2*i+d/2).attr("height",e.nn.node().getBBox().height+2*r),p+=Math.max(e.tn.node().getBBox().height,e.nn.node().getBBox().height)+2*r,g="attributeBoxOdd"==g?"attributeBoxEven":"attributeBoxOdd"}))}else 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a,o,s=to(i,Ga.getEntities(),r),u=function(t,e){return t.forEach((function(t){e.setEdge(t.entityA,t.entityB,{relationship:t},eo(t))})),t}(Ga.getRelationships(),r);ke.a.layout(r),a=i,(o=r).nodes().forEach((function(t){void 0!==t&&void 0!==o.node(t)&&a.select("#"+t).attr("transform","translate("+(o.node(t).x-o.node(t).width/2)+","+(o.node(t).y-o.node(t).height/2)+" )")})),u.forEach((function(t){!function(t,e,n,r){no++;var i=n.edge(e.entityA,e.entityB,eo(e)),a=Object(d.line)().x((function(t){return t.x})).y((function(t){return t.y})).curve(d.curveBasis),o=t.insert("path","#"+r).attr("class","er relationshipLine").attr("d",a(i.points)).attr("stroke",Qa.stroke).attr("fill","none");e.relSpec.relType===Ga.Identification.NON_IDENTIFYING&&o.attr("stroke-dasharray","8,8");var s="";switch(Qa.arrowMarkerAbsolute&&(s=(s=(s=window.location.protocol+"//"+window.location.host+window.location.pathname+window.location.search).replace(/\(/g,"\\(")).replace(/\)/g,"\\)")),e.relSpec.cardA){case 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e=Object(d.arc)().startAngle(3*Math.PI/2).endAngle(Math.PI/2*5).innerRadius(7.5).outerRadius(15/2.2);t.append("path").attr("class","mouth").attr("d",e).attr("transform","translate("+o.cx+","+(o.cy+7)+")")}(s):function(t){t.append("line").attr("class","mouth").attr("stroke",2).attr("x1",o.cx-5).attr("y1",o.cy+7).attr("x2",o.cx+5).attr("y2",o.cy+7).attr("class","mouth").attr("stroke-width","1px").attr("stroke","#666")}(s);var c=xo();c.x=e.x,c.y=e.y,c.fill=e.fill,c.width=n.width,c.height=n.height,c.class="task task-type-"+e.num,c.rx=3,c.ry=3,yo(i,c);var u=e.x+14;e.people.forEach((function(t){var n=e.actors[t],r={cx:u,cy:e.y,r:7,fill:n,stroke:"#000",title:t};vo(i,r),u+=10})),_o(n)(e.task,i,c.x,c.y,c.width,c.height,{class:"task"},n,e.colour)},Co=function(t){t.append("defs").append("marker").attr("id","arrowhead").attr("refX",5).attr("refY",2).attr("markerWidth",6).attr("markerHeight",4).attr("orient","auto").append("path").attr("d","M 0,0 V 4 L6,2 Z")};ao.parser.yy=go;var Ao={leftMargin:150,diagramMarginX:50,diagramMarginY:20,taskMargin:50,width:150,height:50,taskFontSize:14,taskFontFamily:'"Open-Sans", "sans-serif"',boxMargin:10,boxTextMargin:5,noteMargin:10,messageMargin:35,messageAlign:"center",bottomMarginAdj:1,activationWidth:10,textPlacement:"fo",actorColours:["#8FBC8F","#7CFC00","#00FFFF","#20B2AA","#B0E0E6","#FFFFE0"],sectionFills:["#191970","#8B008B","#4B0082","#2F4F4F","#800000","#8B4513","#00008B"],sectionColours:["#fff"]},So={};var Mo=Ao.leftMargin,Oo={data:{startx:void 0,stopx:void 0,starty:void 0,stopy:void 0},verticalPos:0,sequenceItems:[],init:function(){this.sequenceItems=[],this.data={startx:void 0,stopx:void 0,starty:void 0,stopy:void 0},this.verticalPos=0},updateVal:function(t,e,n,r){void 0===t[e]?t[e]=n:t[e]=r(n,t[e])},updateBounds:function(t,e,n,r){var i,a=this,o=0;this.sequenceItems.forEach((function(s){o++;var 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this.data}},Do=Ao.sectionFills,No=Ao.sectionColours,Bo=function(t,e,n){for(var r="",i=n+(2*Ao.height+Ao.diagramMarginY),a=0,o="#CCC",s="black",c=0,u=0;u<e.length;u++){var l=e[u];if(r!==l.section){o=Do[a%Do.length],c=a%Do.length,s=No[a%No.length];var h={x:u*Ao.taskMargin+u*Ao.width+Mo,y:50,text:l.section,fill:o,num:c,colour:s};wo(t,h,Ao),r=l.section,a++}var f=l.people.reduce((function(t,e){return So[e]&&(t[e]=So[e]),t}),{});l.x=u*Ao.taskMargin+u*Ao.width+Mo,l.y=i,l.width=Ao.diagramMarginX,l.height=Ao.diagramMarginY,l.colour=s,l.fill=o,l.num=c,l.actors=f,To(t,l,Ao),Oo.insert(l.x,l.y,l.x+l.width+Ao.taskMargin,450)}},Lo=function(t){Object.keys(t).forEach((function(e){Ao[e]=t[e]}))},Po=function(t,e){ao.parser.yy.clear(),ao.parser.parse(t+"\n"),Oo.init();var n=Object(d.select)("#"+e);n.attr("xmlns:xlink","http://www.w3.org/1999/xlink"),Co(n);var r=ao.parser.yy.getTasks(),i=ao.parser.yy.getTitle(),a=ao.parser.yy.getActors();for(var o in So)delete So[o];var s=0;a.forEach((function(t){So[t]=Ao.actorColours[s%Ao.actorColours.length],s++})),function(t){var e=60;Object.keys(So).forEach((function(n){var r=So[n];ko(t,{cx:20,cy:e,r:7,fill:r,stroke:"#000"});var i={x:40,y:e+7,fill:"#666",text:n,textMargin:5|Ao.boxTextMargin};Eo(t,i),e+=20}))}(n),Oo.insert(0,0,Mo,50*Object.keys(So).length),Bo(n,r,0);var c=Oo.getBounds();i&&n.append("text").text(i).attr("x",Mo).attr("font-size","4ex").attr("font-weight","bold").attr("y",25);var u=c.stopy-c.starty+2*Ao.diagramMarginY,l=Mo+c.stopx+2*Ao.diagramMarginX;W(n,u,l,Ao.useMaxWidth),n.append("line").attr("x1",Mo).attr("y1",4*Ao.height).attr("x2",l-Mo-4).attr("y2",4*Ao.height).attr("stroke-width",4).attr("stroke","black").attr("marker-end","url(#arrowhead)");var h=i?70:0;n.attr("viewBox","".concat(c.startx," -25 ").concat(l," ").concat(u+h)),n.attr("preserveAspectRatio","xMinYMin meet")},Io=function(t){return"g.classGroup text {\n  fill: ".concat(t.nodeBorder,";\n  fill: ").concat(t.classText,";\n  stroke: 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fill: ").concat(t.lineColor," !important;\n  stroke: ").concat(t.lineColor," !important;\n  stroke-width: 1;\n}\n\n#compositionEnd, .composition {\n  fill: ").concat(t.lineColor," !important;\n  stroke: ").concat(t.lineColor," !important;\n  stroke-width: 1;\n}\n\n#dependencyStart, .dependency {\n  fill: ").concat(t.lineColor," !important;\n  stroke: ").concat(t.lineColor," !important;\n  stroke-width: 1;\n}\n\n#dependencyStart, .dependency {\n  fill: ").concat(t.lineColor," !important;\n  stroke: ").concat(t.lineColor," !important;\n  stroke-width: 1;\n}\n\n#extensionStart, .extension {\n  fill: ").concat(t.lineColor," !important;\n  stroke: ").concat(t.lineColor," !important;\n  stroke-width: 1;\n}\n\n#extensionEnd, .extension {\n  fill: ").concat(t.lineColor," !important;\n  stroke: ").concat(t.lineColor," !important;\n  stroke-width: 1;\n}\n\n#aggregationStart, .aggregation {\n  fill: ").concat(t.mainBkg," !important;\n  stroke: ").concat(t.lineColor," !important;\n  stroke-width: 1;\n}\n\n#aggregationEnd, .aggregation {\n  fill: ").concat(t.mainBkg," !important;\n  stroke: ").concat(t.lineColor," !important;\n  stroke-width: 1;\n}\n\n.edgeTerminals {\n  font-size: 11px;\n}\n\n")},Fo=function(t){return".label {\n    font-family: ".concat(t.fontFamily,";\n    color: ").concat(t.nodeTextColor||t.textColor,";\n  }\n\n  .label text {\n    fill: ").concat(t.nodeTextColor||t.textColor,";\n  }\n\n  .node rect,\n  .node circle,\n  .node ellipse,\n  .node polygon,\n  .node path {\n    fill: ").concat(t.mainBkg,";\n    stroke: ").concat(t.nodeBorder,";\n    stroke-width: 1px;\n  }\n\n  .node .label {\n    text-align: center;\n  }\n  .node.clickable {\n    cursor: pointer;\n  }\n\n  .arrowheadPath {\n    fill: ").concat(t.arrowheadColor,";\n  }\n\n  .edgePath .path {\n    stroke: ").concat(t.lineColor,";\n    stroke-width: 1.5px;\n  }\n\n  .flowchart-link {\n    stroke: ").concat(t.lineColor,";\n    fill: none;\n  }\n\n  .edgeLabel {\n    background-color: ").concat(t.edgeLabelBackground,";\n    rect {\n      opacity: 0.5;\n      background-color: ").concat(t.edgeLabelBackground,";\n      fill: ").concat(t.edgeLabelBackground,";\n    }\n    text-align: center;\n  }\n\n  .cluster rect {\n    fill: ").concat(t.clusterBkg,";\n    stroke: ").concat(t.clusterBorder,";\n    stroke-width: 1px;\n  }\n\n  .cluster text {\n    fill: ").concat(t.titleColor,";\n  }\n\n  div.mermaidTooltip {\n    position: absolute;\n    text-align: center;\n    max-width: 200px;\n    padding: 2px;\n    font-family: ").concat(t.fontFamily,";\n    font-size: 12px;\n    background: ").concat(t.tertiaryColor,";\n    border: 1px solid ").concat(t.border2,";\n    border-radius: 2px;\n    pointer-events: none;\n    z-index: 100;\n  }\n")},jo=function(t){return"g.stateGroup text {\n  fill: ".concat(t.nodeBorder,";\n  stroke: none;\n  font-size: 10px;\n}\ng.stateGroup text {\n  fill: ").concat(t.textColor,";\n  stroke: none;\n  font-size: 10px;\n\n}\ng.stateGroup .state-title {\n  font-weight: bolder;\n  fill: ").concat(t.labelColor,";\n}\n\ng.stateGroup rect {\n  fill: ").concat(t.mainBkg,";\n  stroke: ").concat(t.nodeBorder,";\n}\n\ng.stateGroup line {\n  stroke: ").concat(t.lineColor,";\n  stroke-width: 1;\n}\n\n.transition {\n  stroke: ").concat(t.lineColor,";\n  stroke-width: 1;\n  fill: none;\n}\n\n.stateGroup .composit {\n  fill: ").concat(t.background,";\n  border-bottom: 1px\n}\n\n.stateGroup .alt-composit {\n  fill: #e0e0e0;\n  border-bottom: 1px\n}\n\n.state-note {\n  stroke: ").concat(t.noteBorderColor,";\n  fill: ").concat(t.noteBkgColor,";\n\n  text {\n    fill: black;\n    stroke: none;\n    font-size: 10px;\n  }\n}\n\n.stateLabel .box {\n  stroke: none;\n  stroke-width: 0;\n  fill: ").concat(t.mainBkg,";\n  opacity: 0.5;\n}\n\n.edgeLabel .label rect {\n  fill: ").concat(t.tertiaryColor,";\n  opacity: 0.5;\n}\n.edgeLabel .label text {\n  fill: ").concat(t.tertiaryTextColor,";\n}\n.label div .edgeLabel {\n  color: ").concat(t.tertiaryTextColor,";\n}\n\n.stateLabel text {\n  fill: ").concat(t.labelColor,";\n  font-size: 10px;\n  font-weight: bold;\n}\n\n.node circle.state-start {\n  fill: ").concat(t.lineColor,";\n  stroke: black;\n}\n.node circle.state-end {\n  fill: ").concat(t.primaryBorderColor,";\n  stroke: ").concat(t.background,";\n  stroke-width: 1.5\n}\n.end-state-inner {\n  fill: ").concat(t.background,";\n  // stroke: ").concat(t.background,";\n  stroke-width: 1.5\n}\n\n.node rect {\n  fill: ").concat(t.mainBkg,";\n  stroke: ").concat(t.nodeBorder,";\n  stroke-width: 1px;\n}\n#statediagram-barbEnd {\n  fill: ").concat(t.lineColor,";\n}\n\n.statediagram-cluster rect {\n  fill: ").concat(t.mainBkg,";\n  stroke: ").concat(t.nodeBorder,";\n  stroke-width: 1px;\n}\n\n.cluster-label, .nodeLabel {\n  color: ").concat(t.textColor,";\n}\n\n.statediagram-cluster rect.outer {\n  rx: 5px;\n  ry: 5px;\n}\n.statediagram-state .divider {\n  stroke: ").concat(t.nodeBorder,";\n}\n\n.statediagram-state .title-state {\n  rx: 5px;\n  ry: 5px;\n}\n.statediagram-cluster.statediagram-cluster .inner {\n  fill: ").concat(t.background,";\n}\n.statediagram-cluster.statediagram-cluster-alt .inner {\n  fill: #e0e0e0;\n}\n\n.statediagram-cluster .inner {\n  rx:0;\n  ry:0;\n}\n\n.statediagram-state rect.basic {\n  rx: 5px;\n  ry: 5px;\n}\n.statediagram-state rect.divider {\n  stroke-dasharray: 10,10;\n  fill: ").concat(t.altBackground?t.altBackground:"#efefef",";\n}\n\n.note-edge {\n  stroke-dasharray: 5;\n}\n\n.statediagram-note rect {\n  fill: ").concat(t.noteBkgColor,";\n  stroke: ").concat(t.noteBorderColor,";\n  stroke-width: 1px;\n  rx: 0;\n  ry: 0;\n}\n.statediagram-note rect {\n  fill: ").concat(t.noteBkgColor,";\n  stroke: ").concat(t.noteBorderColor,";\n  stroke-width: 1px;\n  rx: 0;\n  ry: 0;\n}\n\n.statediagram-note text {\n  fill: ").concat(t.noteTextColor,";\n}\n\n.statediagram-note .nodeLabel {\n  color: ").concat(t.noteTextColor,";\n}\n\n#dependencyStart, #dependencyEnd {\n  fill: ").concat(t.lineColor,";\n  stroke: ").concat(t.lineColor,";\n  stroke-width: 1;\n}\n")},Ro={flowchart:Fo,"flowchart-v2":Fo,sequence:function(t){return".actor {\n    stroke: ".concat(t.actorBorder,";\n    fill: ").concat(t.actorBkg,";\n  }\n\n  text.actor > tspan {\n    fill: ").concat(t.actorTextColor,";\n    stroke: none;\n  }\n\n  .actor-line {\n    stroke: ").concat(t.actorLineColor,";\n  }\n\n  .messageLine0 {\n    stroke-width: 1.5;\n    stroke-dasharray: none;\n    stroke: ").concat(t.signalColor,";\n  }\n\n  .messageLine1 {\n    stroke-width: 1.5;\n    stroke-dasharray: 2, 2;\n    stroke: ").concat(t.signalColor,";\n  }\n\n  #arrowhead path {\n    fill: ").concat(t.signalColor,";\n    stroke: ").concat(t.signalColor,";\n  }\n\n  .sequenceNumber {\n    fill: ").concat(t.sequenceNumberColor,";\n  }\n\n  #sequencenumber {\n    fill: ").concat(t.signalColor,";\n  }\n\n  #crosshead path {\n    fill: ").concat(t.signalColor,";\n    stroke: ").concat(t.signalColor,";\n  }\n\n  .messageText {\n    fill: ").concat(t.signalTextColor,";\n    stroke: ").concat(t.signalTextColor,";\n  }\n\n  .labelBox {\n    stroke: ").concat(t.labelBoxBorderColor,";\n    fill: ").concat(t.labelBoxBkgColor,";\n  }\n\n  .labelText, .labelText > tspan {\n    fill: ").concat(t.labelTextColor,";\n    stroke: none;\n  }\n\n  .loopText, .loopText > tspan {\n    fill: ").concat(t.loopTextColor,";\n    stroke: none;\n  }\n\n  .loopLine {\n    stroke-width: 2px;\n    stroke-dasharray: 2, 2;\n    stroke: ").concat(t.labelBoxBorderColor,";\n    fill: ").concat(t.labelBoxBorderColor,";\n  }\n\n  .note {\n    //stroke: #decc93;\n    stroke: ").concat(t.noteBorderColor,";\n    fill: ").concat(t.noteBkgColor,";\n  }\n\n  .noteText, .noteText > tspan {\n    fill: ").concat(t.noteTextColor,";\n    stroke: none;\n  }\n\n  .activation0 {\n    fill: ").concat(t.activationBkgColor,";\n    stroke: ").concat(t.activationBorderColor,";\n  }\n\n  .activation1 {\n    fill: ").concat(t.activationBkgColor,";\n    stroke: ").concat(t.activationBorderColor,";\n  }\n\n  .activation2 {\n    fill: ").concat(t.activationBkgColor,";\n    stroke: ").concat(t.activationBorderColor,";\n  }\n")},gantt:function(t){return'\n  .mermaid-main-font {\n    font-family: "trebuchet ms", verdana, arial, sans-serif;\n    font-family: var(--mermaid-font-family);\n  }\n\n  .section {\n    stroke: none;\n    opacity: 0.2;\n  }\n\n  .section0 {\n    fill: '.concat(t.sectionBkgColor,";\n  }\n\n  .section2 {\n    fill: ").concat(t.sectionBkgColor2,";\n  }\n\n  .section1,\n  .section3 {\n    fill: ").concat(t.altSectionBkgColor,";\n    opacity: 0.2;\n  }\n\n  .sectionTitle0 {\n    fill: ").concat(t.titleColor,";\n  }\n\n  .sectionTitle1 {\n    fill: ").concat(t.titleColor,";\n  }\n\n  .sectionTitle2 {\n    fill: ").concat(t.titleColor,";\n  }\n\n  .sectionTitle3 {\n    fill: ").concat(t.titleColor,";\n  }\n\n  .sectionTitle {\n    text-anchor: start;\n    font-size: 11px;\n    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f.a)("#".concat(t),Yo(h,v,o.themeVariables)),_=document.createElement("style");_.innerHTML=x,g.insertBefore(_,y);try{switch(h){case"git":o.flowchart.arrowMarkerAbsolute=o.arrowMarkerAbsolute,ka(o.git),wa(i,t,!1);break;case"flowchart":o.flowchart.arrowMarkerAbsolute=o.arrowMarkerAbsolute,me(o.flowchart),xe(i,t,!1);break;case"flowchart-v2":o.flowchart.arrowMarkerAbsolute=o.arrowMarkerAbsolute,Ln(o.flowchart),Pn(i,t,!1);break;case"sequence":o.sequence.arrowMarkerAbsolute=o.arrowMarkerAbsolute,o.sequenceDiagram?(_r.setConf(Object.assign(o.sequence,o.sequenceDiagram)),console.error("`mermaid config.sequenceDiagram` has been renamed to `config.sequence`. Please update your mermaid config.")):_r.setConf(o.sequence),_r.draw(i,t);break;case"gantt":o.gantt.arrowMarkerAbsolute=o.arrowMarkerAbsolute,ri(o.gantt),ii(i,t);break;case"class":o.class.arrowMarkerAbsolute=o.arrowMarkerAbsolute,li(o.class),hi(i,t);break;case"classDiagram":o.class.arrowMarkerAbsolute=o.arrowMarkerAbsolute,di(o.class),pi(i,t);break;case"state":o.class.arrowMarkerAbsolute=o.arrowMarkerAbsolute,zi(o.state),Ui(i,t);break;case"stateDiagram":o.class.arrowMarkerAbsolute=o.arrowMarkerAbsolute,qi(o.state),Xi(i,t);break;case"info":o.class.arrowMarkerAbsolute=o.arrowMarkerAbsolute,Oa(o.class),Da(i,t,p.version);break;case"pie":o.class.arrowMarkerAbsolute=o.arrowMarkerAbsolute,za(o.pie),Ua(i,t,p.version);break;case"er":ro(o.er),io(i,t,p.version);break;case"journey":Lo(o.journey),Po(i,t,p.version)}}catch(e){throw La(t,p.version),e}Object(d.select)('[id="'.concat(t,'"]')).selectAll("foreignobject > *").attr("xmlns","http://www.w3.org/1999/xhtml");var k=Object(d.select)("#d"+t).node().innerHTML;if(c.debug("cnf.arrowMarkerAbsolute",o.arrowMarkerAbsolute),o.arrowMarkerAbsolute&&"false"!==o.arrowMarkerAbsolute||(k=k.replace(/marker-end="url\(.*?#/g,'marker-end="url(#',"g")),k=(k=function(t){var e=t;return e=(e=(e=e.replace(/fl°°/g,(function(){return"&#"}))).replace(/fl°/g,(function(){return"&"}))).replace(/¶ß/g,(function(){return";"}))}(k)).replace(/<br>/g,"<br/>"),void 0!==n)switch(h){case"flowchart":case"flowchart-v2":n(k,Xt.bindFunctions);break;case"gantt":n(k,Qr.bindFunctions);break;case"class":case"classDiagram":n(k,cn.bindFunctions);break;default:n(k)}else c.debug("CB = undefined!");var w=Object(d.select)("#d"+t).node();return null!==w&&"function"==typeof w.remove&&Object(d.select)("#d"+t).node().remove(),k},parse:function(t){var e=H.detectInit(t);e&&c.debug("reinit ",e);var n,r=H.detectType(t);switch(c.debug("Type "+r),r){case"git":(n=ha.a).parser.yy=ua;break;case"flowchart":case"flowchart-v2":Xt.clear(),(n=Jt.a).parser.yy=Xt;break;case"sequence":(n=Hn.a).parser.yy=sr;break;case"gantt":(n=wr.a).parser.yy=Qr;break;case"class":case"classDiagram":(n=oi.a).parser.yy=cn;break;case"state":case"stateDiagram":(n=Di.a).parser.yy=Mi;break;case"info":c.debug("info info info"),(n=Sa.a).parser.yy=Ca;break;case"pie":c.debug("pie"),(n=Ra.a).parser.yy=Fa;break;case"er":c.debug("er"),(n=Xa.a).parser.yy=Ga;break;case"journey":c.debug("Journey"),(n=oo.a).parser.yy=go}return n.parser.yy.graphType=r,n.parser.yy.parseError=function(t,e){throw{str:t,hash:e}},n.parse(t),n},parseDirective:function(t,e,n,r){try{if(void 0!==e)switch(e=e.trim(),n){case"open_directive":Uo={};break;case"type_directive":Uo.type=e.toLowerCase();break;case"arg_directive":Uo.args=JSON.parse(e);break;case"close_directive":$o(t,Uo,r),Uo=null}}catch(t){c.error("Error while rendering sequenceDiagram directive: ".concat(e," jison context: ").concat(n)),c.error(t.message)}},initialize:function(t){t&&t.fontFamily&&(t.themeVariables&&t.themeVariables.fontFamily||(t.themeVariables={fontFamily:t.fontFamily})),dt=F({},t),t&&t.theme&&ht[t.theme]?t.themeVariables=ht[t.theme].getThemeVariables(t.themeVariables):t&&(t.themeVariables=ht.default.getThemeVariables(t.themeVariables));var e="object"===zo(t)?function(t){return yt=F({},gt),yt=F(yt,t),t.theme&&(yt.themeVariables=ht[t.theme].getThemeVariables(t.themeVariables)),mt=bt(yt,vt),yt}(t):xt();Wo(e),u(e.logLevel)},reinitialize:Ho,getConfig:_t,setConfig:function(t){return F(mt,t),_t()},getSiteConfig:xt,updateSiteConfig:function(t){return yt=F(yt,t),bt(yt,vt),yt},reset:function(){Et()},globalReset:function(){Et(),Wo(_t())},defaultConfig:gt});u(_t().logLevel),Et(_t());var Go=Vo,qo=function(){Xo.startOnLoad?Go.getConfig().startOnLoad&&Xo.init():void 0===Xo.startOnLoad&&(c.debug("In start, no config"),Go.getConfig().startOnLoad&&Xo.init())};"undefined"!=typeof document&&
+/*!
+   * Wait for document loaded before starting the execution
+   */
+window.addEventListener("load",(function(){qo()}),!1);var Xo={startOnLoad:!0,htmlLabels:!0,mermaidAPI:Go,parse:Go.parse,render:Go.render,init:function(){var t,e,n=this,r=Go.getConfig();arguments.length>=2?(
+/*! sequence config was passed as #1 */
+void 0!==arguments[0]&&(Xo.sequenceConfig=arguments[0]),t=arguments[1]):t=arguments[0],"function"==typeof arguments[arguments.length-1]?(e=arguments[arguments.length-1],c.debug("Callback function found")):void 0!==r.mermaid&&("function"==typeof r.mermaid.callback?(e=r.mermaid.callback,c.debug("Callback function found")):c.debug("No Callback function found")),t=void 0===t?document.querySelectorAll(".mermaid"):"string"==typeof t?document.querySelectorAll(t):t instanceof window.Node?[t]:t,c.debug("Start On Load before: "+Xo.startOnLoad),void 0!==Xo.startOnLoad&&(c.debug("Start On Load inner: "+Xo.startOnLoad),Go.updateSiteConfig({startOnLoad:Xo.startOnLoad})),void 0!==Xo.ganttConfig&&Go.updateSiteConfig({gantt:Xo.ganttConfig});for(var a,o=H.initIdGeneratior(r.deterministicIds,r.deterministicIDSeed).next,s=function(r){var s=t[r];
+/*! Check if previously processed */if(s.getAttribute("data-processed"))return"continue";s.setAttribute("data-processed",!0);var u="mermaid-".concat(o());a=i(a=s.innerHTML).trim().replace(/<br\s*\/?>/gi,"<br/>");var l=H.detectInit(a);l&&c.debug("Detected early reinit: ",l);try{Go.render(u,a,(function(t,n){s.innerHTML=t,void 0!==e&&e(u),n&&n(s)}),s)}catch(t){c.warn("Syntax Error rendering"),c.warn(t),n.parseError&&n.parseError(t)}},u=0;u<t.length;u++)s(u)},initialize:function(t){void 0!==t.mermaid&&(void 0!==t.mermaid.startOnLoad&&(Xo.startOnLoad=t.mermaid.startOnLoad),void 0!==t.mermaid.htmlLabels&&(Xo.htmlLabels=t.mermaid.htmlLabels)),Go.initialize(t)},contentLoaded:qo};e.default=Xo}]).default}));
+//# sourceMappingURL=mermaid.min.js.map
\ No newline at end of file
diff --git a/website/themes/book/static/prism.css b/website/themes/book/static/prism.css
new file mode 100644
index 000000000..6d58efa47
--- /dev/null
+++ b/website/themes/book/static/prism.css
@@ -0,0 +1,140 @@
+/* http://prismjs.com/download.html?themes=prism-okaidia&languages=markup+css+clike+javascript+actionscript+applescript+aspnet+bash+basic+c+bison+csharp+cpp+coffeescript+ruby+css-extras+diff+docker+erlang+fsharp+fortran+git+go+haskell+http+icon+java+json+latex+lua+makefile+markdown+matlab+nasm+nginx+objectivec+pascal+perl+php+php-extras+properties+python+r+jsx+rust+scala+scheme+smalltalk+sql+swift+vim+wiki */
+/**
+ * okaidia theme for JavaScript, CSS and HTML
+ * Loosely based on Monokai textmate theme by http://www.monokai.nl/
+ * @author ocodia
+ */
+
+ code[class*="language-"],
+ pre[class*="language-"] {
+     color: #f8f8f2;
+     background: none;
+     text-shadow: 0 1px rgba(0, 0, 0, 0.3);
+     font-family: Consolas, Monaco, 'Andale Mono', 'Ubuntu Mono', monospace;
+     text-align: left;
+     white-space: pre;
+     word-spacing: normal;
+     word-break: normal;
+     word-wrap: normal;
+     line-height: 1.5;
+ 
+     -moz-tab-size: 4;
+     -o-tab-size: 4;
+     tab-size: 4;
+ 
+     -webkit-hyphens: none;
+     -moz-hyphens: none;
+     -ms-hyphens: none;
+     hyphens: none;
+ }
+ 
+ /* Code blocks */
+ pre[class*="language-"] {
+     padding: 1em;
+     margin: .5em 0;
+     overflow: auto;
+     border-radius: 0.3em;
+ }
+ 
+ :not(pre) > code[class*="language-"],
+ pre[class*="language-"] {
+     background: #272822;
+ }
+ 
+ /* Inline code */
+ :not(pre) > code[class*="language-"] {
+     padding: .1em;
+     border-radius: .3em;
+     white-space: normal;
+ }
+ 
+ .token.comment,
+ .token.prolog,
+ .token.doctype,
+ .token.cdata {
+     color: slategray;
+ }
+ 
+ .token.punctuation {
+     color: #f8f8f2;
+ }
+ 
+ .namespace {
+     opacity: .7;
+ }
+ 
+ .token.property,
+ .token.tag,
+ .token.constant,
+ .token.symbol,
+ .token.deleted {
+     color: #f92672;
+ }
+ 
+ .token.boolean,
+ .token.number {
+     color: #ae81ff;
+ }
+ 
+ .token.selector,
+ .token.attr-name,
+ .token.string,
+ .token.char,
+ .token.builtin,
+ .token.inserted {
+     color: #a6e22e;
+ }
+ 
+ .token.operator,
+ .token.entity,
+ .token.url,
+ .language-css .token.string,
+ .style .token.string,
+ .token.variable {
+     color: #f8f8f2;
+ }
+ 
+ .token.atrule,
+ .token.attr-value,
+ .token.function {
+     color: #e6db74;
+ }
+ 
+ .token.keyword {
+     color: #66d9ef;
+ }
+ 
+ .token.regex,
+ .token.important {
+     color: #fd971f;
+ }
+ 
+ .token.important,
+ .token.bold {
+     font-weight: bold;
+ }
+ .token.italic {
+     font-style: italic;
+ }
+ 
+ .token.entity {
+     cursor: help;
+ }
+ 
+ div.prism-show-language {
+     position: relative;
+ }
+ 
+ div.prism-show-language > div.prism-show-language-label[data-language] {
+   content: attr(data-language);
+     color: black;
+     background-color: #CFCFCF;
+     display: inline-block;
+     position: absolute;
+     top: 0;
+     right: 0;
+     font-size: 0.9em;
+     border-radius: 0 0 0 5px;
+     padding: 0 0.5em;
+     text-shadow: none;
+ }
\ No newline at end of file
diff --git a/website/themes/book/static/prism.js b/website/themes/book/static/prism.js
new file mode 100755
index 000000000..647d69ca8
--- /dev/null
+++ b/website/themes/book/static/prism.js
@@ -0,0 +1,78 @@
+/* http://prismjs.com/download.html?themes=prism&languages=markup+css+clike+javascript+bash+css-extras+git+handlebars+http+makefile+markdown+nginx+php+sass+scss&plugins=line-highlight+line-numbers+autolinker+file-highlight+show-language */
+var _self="undefined"!=typeof window?window:"undefined"!=typeof WorkerGlobalScope&&self instanceof WorkerGlobalScope?self:{},Prism=function(){var e=/\blang(?:uage)?-(\w+)\b/i,t=0,n=_self.Prism={util:{encode:function(e){return e instanceof a?new a(e.type,n.util.encode(e.content),e.alias):"Array"===n.util.type(e)?e.map(n.util.encode):e.replace(/&/g,"&amp;").replace(/</g,"&lt;").replace(/\u00a0/g," ")},type:function(e){return Object.prototype.toString.call(e).match(/\[object (\w+)\]/)[1]},objId:function(e){return e.__id||Object.defineProperty(e,"__id",{value:++t}),e.__id},clone:function(e){var t=n.util.type(e);switch(t){case"Object":var a={};for(var r in e)e.hasOwnProperty(r)&&(a[r]=n.util.clone(e[r]));return a;case"Array":return e.map&&e.map(function(e){return n.util.clone(e)})}return e}},languages:{extend:function(e,t){var a=n.util.clone(n.languages[e]);for(var r in t)a[r]=t[r];return a},insertBefore:function(e,t,a,r){r=r||n.languages;var i=r[e];if(2==arguments.length){a=arguments[1];for(var l in a)a.hasOwnProperty(l)&&(i[l]=a[l]);return i}var o={};for(var s in i)if(i.hasOwnProperty(s)){if(s==t)for(var l in a)a.hasOwnProperty(l)&&(o[l]=a[l]);o[s]=i[s]}return n.languages.DFS(n.languages,function(t,n){n===r[e]&&t!=e&&(this[t]=o)}),r[e]=o},DFS:function(e,t,a,r){r=r||{};for(var i in e)e.hasOwnProperty(i)&&(t.call(e,i,e[i],a||i),"Object"!==n.util.type(e[i])||r[n.util.objId(e[i])]?"Array"!==n.util.type(e[i])||r[n.util.objId(e[i])]||(r[n.util.objId(e[i])]=!0,n.languages.DFS(e[i],t,i,r)):(r[n.util.objId(e[i])]=!0,n.languages.DFS(e[i],t,null,r)))}},plugins:{},highlightAll:function(e,t){var a={callback:t,selector:'code[class*="language-"], [class*="language-"] code, code[class*="lang-"], [class*="lang-"] code'};n.hooks.run("before-highlightall",a);for(var r,i=a.elements||document.querySelectorAll(a.selector),l=0;r=i[l++];)n.highlightElement(r,e===!0,a.callback)},highlightElement:function(t,a,r){for(var i,l,o=t;o&&!e.test(o.className);)o=o.parentNode;o&&(i=(o.className.match(e)||[,""])[1].toLowerCase(),l=n.languages[i]),t.className=t.className.replace(e,"").replace(/\s+/g," ")+" language-"+i,o=t.parentNode,/pre/i.test(o.nodeName)&&(o.className=o.className.replace(e,"").replace(/\s+/g," ")+" language-"+i);var s=t.textContent,u={element:t,language:i,grammar:l,code:s};if(n.hooks.run("before-sanity-check",u),!u.code||!u.grammar)return n.hooks.run("complete",u),void 0;if(n.hooks.run("before-highlight",u),a&&_self.Worker){var c=new Worker(n.filename);c.onmessage=function(e){u.highlightedCode=e.data,n.hooks.run("before-insert",u),u.element.innerHTML=u.highlightedCode,r&&r.call(u.element),n.hooks.run("after-highlight",u),n.hooks.run("complete",u)},c.postMessage(JSON.stringify({language:u.language,code:u.code,immediateClose:!0}))}else u.highlightedCode=n.highlight(u.code,u.grammar,u.language),n.hooks.run("before-insert",u),u.element.innerHTML=u.highlightedCode,r&&r.call(t),n.hooks.run("after-highlight",u),n.hooks.run("complete",u)},highlight:function(e,t,r){var i=n.tokenize(e,t);return a.stringify(n.util.encode(i),r)},tokenize:function(e,t){var a=n.Token,r=[e],i=t.rest;if(i){for(var l in i)t[l]=i[l];delete t.rest}e:for(var l in t)if(t.hasOwnProperty(l)&&t[l]){var o=t[l];o="Array"===n.util.type(o)?o:[o];for(var s=0;s<o.length;++s){var u=o[s],c=u.inside,g=!!u.lookbehind,h=!!u.greedy,f=0,d=u.alias;if(h&&!u.pattern.global){var p=u.pattern.toString().match(/[imuy]*$/)[0];u.pattern=RegExp(u.pattern.source,p+"g")}u=u.pattern||u;for(var m=0,y=0;m<r.length;y+=(r[m].matchedStr||r[m]).length,++m){var v=r[m];if(r.length>e.length)break e;if(!(v instanceof a)){u.lastIndex=0;var b=u.exec(v),k=1;if(!b&&h&&m!=r.length-1){if(u.lastIndex=y,b=u.exec(e),!b)break;for(var w=b.index+(g?b[1].length:0),_=b.index+b[0].length,A=m,S=y,P=r.length;P>A&&_>S;++A)S+=(r[A].matchedStr||r[A]).length,w>=S&&(++m,y=S);if(r[m]instanceof a||r[A-1].greedy)continue;k=A-m,v=e.slice(y,S),b.index-=y}if(b){g&&(f=b[1].length);var w=b.index+f,b=b[0].slice(f),_=w+b.length,x=v.slice(0,w),O=v.slice(_),j=[m,k];x&&j.push(x);var N=new a(l,c?n.tokenize(b,c):b,d,b,h);j.push(N),O&&j.push(O),Array.prototype.splice.apply(r,j)}}}}}return r},hooks:{all:{},add:function(e,t){var a=n.hooks.all;a[e]=a[e]||[],a[e].push(t)},run:function(e,t){var a=n.hooks.all[e];if(a&&a.length)for(var r,i=0;r=a[i++];)r(t)}}},a=n.Token=function(e,t,n,a,r){this.type=e,this.content=t,this.alias=n,this.matchedStr=a||null,this.greedy=!!r};if(a.stringify=function(e,t,r){if("string"==typeof e)return e;if("Array"===n.util.type(e))return e.map(function(n){return a.stringify(n,t,e)}).join("");var i={type:e.type,content:a.stringify(e.content,t,r),tag:"span",classes:["token",e.type],attributes:{},language:t,parent:r};if("comment"==i.type&&(i.attributes.spellcheck="true"),e.alias){var l="Array"===n.util.type(e.alias)?e.alias:[e.alias];Array.prototype.push.apply(i.classes,l)}n.hooks.run("wrap",i);var o="";for(var s in i.attributes)o+=(o?" ":"")+s+'="'+(i.attributes[s]||"")+'"';return"<"+i.tag+' class="'+i.classes.join(" ")+'"'+(o?" "+o:"")+">"+i.content+"</"+i.tag+">"},!_self.document)return _self.addEventListener?(_self.addEventListener("message",function(e){var t=JSON.parse(e.data),a=t.language,r=t.code,i=t.immediateClose;_self.postMessage(n.highlight(r,n.languages[a],a)),i&&_self.close()},!1),_self.Prism):_self.Prism;var r=document.currentScript||[].slice.call(document.getElementsByTagName("script")).pop();return r&&(n.filename=r.src,document.addEventListener&&!r.hasAttribute("data-manual")&&("loading"!==document.readyState?window.requestAnimationFrame?window.requestAnimationFrame(n.highlightAll):window.setTimeout(n.highlightAll,16):document.addEventListener("DOMContentLoaded",n.highlightAll))),_self.Prism}();"undefined"!=typeof module&&module.exports&&(module.exports=Prism),"undefined"!=typeof global&&(global.Prism=Prism);
+
+Prism.languages.markup={comment:/<!--[\w\W]*?-->/,prolog:/<\?[\w\W]+?\?>/,doctype:/<!DOCTYPE[\w\W]+?>/,cdata:/<!\[CDATA\[[\w\W]*?]]>/i,tag:{pattern:/<\/?(?!\d)[^\s>\/=$<]+(?:\s+[^\s>\/=]+(?:=(?:("|')(?:\\\1|\\?(?!\1)[\w\W])*\1|[^\s'">=]+))?)*\s*\/?>/i,inside:{tag:{pattern:/^<\/?[^\s>\/]+/i,inside:{punctuation:/^<\/?/,namespace:/^[^\s>\/:]+:/}},"attr-value":{pattern:/=(?:('|")[\w\W]*?(\1)|[^\s>]+)/i,inside:{punctuation:/[=>"']/}},punctuation:/\/?>/,"attr-name":{pattern:/[^\s>\/]+/,inside:{namespace:/^[^\s>\/:]+:/}}}},entity:/&#?[\da-z]{1,8};/i},Prism.hooks.add("wrap",function(a){"entity"===a.type&&(a.attributes.title=a.content.replace(/&amp;/,"&"))}),Prism.languages.xml=Prism.languages.markup,Prism.languages.html=Prism.languages.markup,Prism.languages.mathml=Prism.languages.markup,Prism.languages.svg=Prism.languages.markup;
+Prism.languages.css={comment:/\/\*[\w\W]*?\*\//,atrule:{pattern:/@[\w-]+?.*?(;|(?=\s*\{))/i,inside:{rule:/@[\w-]+/}},url:/url\((?:(["'])(\\(?:\r\n|[\w\W])|(?!\1)[^\\\r\n])*\1|.*?)\)/i,selector:/[^\{\}\s][^\{\};]*?(?=\s*\{)/,string:/("|')(\\(?:\r\n|[\w\W])|(?!\1)[^\\\r\n])*\1/,property:/(\b|\B)[\w-]+(?=\s*:)/i,important:/\B!important\b/i,"function":/[-a-z0-9]+(?=\()/i,punctuation:/[(){};:]/},Prism.languages.css.atrule.inside.rest=Prism.util.clone(Prism.languages.css),Prism.languages.markup&&(Prism.languages.insertBefore("markup","tag",{style:{pattern:/(<style[\w\W]*?>)[\w\W]*?(?=<\/style>)/i,lookbehind:!0,inside:Prism.languages.css,alias:"language-css"}}),Prism.languages.insertBefore("inside","attr-value",{"style-attr":{pattern:/\s*style=("|').*?\1/i,inside:{"attr-name":{pattern:/^\s*style/i,inside:Prism.languages.markup.tag.inside},punctuation:/^\s*=\s*['"]|['"]\s*$/,"attr-value":{pattern:/.+/i,inside:Prism.languages.css}},alias:"language-css"}},Prism.languages.markup.tag));
+Prism.languages.clike={comment:[{pattern:/(^|[^\\])\/\*[\w\W]*?\*\//,lookbehind:!0},{pattern:/(^|[^\\:])\/\/.*/,lookbehind:!0}],string:{pattern:/(["'])(\\(?:\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1/,greedy:!0},"class-name":{pattern:/((?:\b(?:class|interface|extends|implements|trait|instanceof|new)\s+)|(?:catch\s+\())[a-z0-9_\.\\]+/i,lookbehind:!0,inside:{punctuation:/(\.|\\)/}},keyword:/\b(if|else|while|do|for|return|in|instanceof|function|new|try|throw|catch|finally|null|break|continue)\b/,"boolean":/\b(true|false)\b/,"function":/[a-z0-9_]+(?=\()/i,number:/\b-?(?:0x[\da-f]+|\d*\.?\d+(?:e[+-]?\d+)?)\b/i,operator:/--?|\+\+?|!=?=?|<=?|>=?|==?=?|&&?|\|\|?|\?|\*|\/|~|\^|%/,punctuation:/[{}[\];(),.:]/};
+Prism.languages.javascript=Prism.languages.extend("clike",{keyword:/\b(as|async|await|break|case|catch|class|const|continue|debugger|default|delete|do|else|enum|export|extends|finally|for|from|function|get|if|implements|import|in|instanceof|interface|let|new|null|of|package|private|protected|public|return|set|static|super|switch|this|throw|try|typeof|var|void|while|with|yield)\b/,number:/\b-?(0x[\dA-Fa-f]+|0b[01]+|0o[0-7]+|\d*\.?\d+([Ee][+-]?\d+)?|NaN|Infinity)\b/,"function":/[_$a-zA-Z\xA0-\uFFFF][_$a-zA-Z0-9\xA0-\uFFFF]*(?=\()/i,operator:/--?|\+\+?|!=?=?|<=?|>=?|==?=?|&&?|\|\|?|\?|\*\*?|\/|~|\^|%|\.{3}/}),Prism.languages.insertBefore("javascript","keyword",{regex:{pattern:/(^|[^\/])\/(?!\/)(\[.+?]|\\.|[^\/\\\r\n])+\/[gimyu]{0,5}(?=\s*($|[\r\n,.;})]))/,lookbehind:!0,greedy:!0}}),Prism.languages.insertBefore("javascript","string",{"template-string":{pattern:/`(?:\\\\|\\?[^\\])*?`/,greedy:!0,inside:{interpolation:{pattern:/\$\{[^}]+\}/,inside:{"interpolation-punctuation":{pattern:/^\$\{|\}$/,alias:"punctuation"},rest:Prism.languages.javascript}},string:/[\s\S]+/}}}),Prism.languages.markup&&Prism.languages.insertBefore("markup","tag",{script:{pattern:/(<script[\w\W]*?>)[\w\W]*?(?=<\/script>)/i,lookbehind:!0,inside:Prism.languages.javascript,alias:"language-javascript"}}),Prism.languages.js=Prism.languages.javascript;
+Prism.languages.actionscript=Prism.languages.extend("javascript",{keyword:/\b(?:as|break|case|catch|class|const|default|delete|do|else|extends|finally|for|function|if|implements|import|in|instanceof|interface|internal|is|native|new|null|package|private|protected|public|return|super|switch|this|throw|try|typeof|use|var|void|while|with|dynamic|each|final|get|include|namespace|native|override|set|static)\b/,operator:/\+\+|--|(?:[+\-*\/%^]|&&?|\|\|?|<<?|>>?>?|[!=]=?)=?|[~?@]/}),Prism.languages.actionscript["class-name"].alias="function",Prism.languages.markup&&Prism.languages.insertBefore("actionscript","string",{xml:{pattern:/(^|[^.])<\/?\w+(?:\s+[^\s>\/=]+=("|')(?:\\\1|\\?(?!\1)[\w\W])*\2)*\s*\/?>/,lookbehind:!0,inside:{rest:Prism.languages.markup}}});
+Prism.languages.applescript={comment:[/\(\*(?:\(\*[\w\W]*?\*\)|[\w\W])*?\*\)/,/--.+/,/#.+/],string:/"(?:\\?.)*?"/,number:/\b-?\d*\.?\d+([Ee]-?\d+)?\b/,operator:[/[&=≠≤≥*+\-\/÷^]|[<>]=?/,/\b(?:(?:start|begin|end)s? with|(?:(?:does not|doesn't) contain|contains?)|(?:is|isn't|is not) (?:in|contained by)|(?:(?:is|isn't|is not) )?(?:greater|less) than(?: or equal)?(?: to)?|(?:(?:does not|doesn't) come|comes) (?:before|after)|(?:is|isn't|is not) equal(?: to)?|(?:(?:does not|doesn't) equal|equals|equal to|isn't|is not)|(?:a )?(?:ref(?: to)?|reference to)|(?:and|or|div|mod|as|not))\b/],keyword:/\b(?:about|above|after|against|apart from|around|aside from|at|back|before|beginning|behind|below|beneath|beside|between|but|by|considering|continue|copy|does|eighth|else|end|equal|error|every|exit|false|fifth|first|for|fourth|from|front|get|given|global|if|ignoring|in|instead of|into|is|it|its|last|local|me|middle|my|ninth|of|on|onto|out of|over|prop|property|put|repeat|return|returning|second|set|seventh|since|sixth|some|tell|tenth|that|the|then|third|through|thru|timeout|times|to|transaction|true|try|until|where|while|whose|with|without)\b/,"class":{pattern:/\b(?:alias|application|boolean|class|constant|date|file|integer|list|number|POSIX file|real|record|reference|RGB color|script|text|centimetres|centimeters|feet|inches|kilometres|kilometers|metres|meters|miles|yards|square feet|square kilometres|square kilometers|square metres|square meters|square miles|square yards|cubic centimetres|cubic centimeters|cubic feet|cubic inches|cubic metres|cubic meters|cubic yards|gallons|litres|liters|quarts|grams|kilograms|ounces|pounds|degrees Celsius|degrees Fahrenheit|degrees Kelvin)\b/,alias:"builtin"},punctuation:/[{}():,¬«»《》]/};
+Prism.languages.aspnet=Prism.languages.extend("markup",{"page-directive tag":{pattern:/<%\s*@.*%>/i,inside:{"page-directive tag":/<%\s*@\s*(?:Assembly|Control|Implements|Import|Master(?:Type)?|OutputCache|Page|PreviousPageType|Reference|Register)?|%>/i,rest:Prism.languages.markup.tag.inside}},"directive tag":{pattern:/<%.*%>/i,inside:{"directive tag":/<%\s*?[$=%#:]{0,2}|%>/i,rest:Prism.languages.csharp}}}),Prism.languages.aspnet.tag.pattern=/<(?!%)\/?[^\s>\/]+(?:\s+[^\s>\/=]+(?:=(?:("|')(?:\\\1|\\?(?!\1)[\w\W])*\1|[^\s'">=]+))?)*\s*\/?>/i,Prism.languages.insertBefore("inside","punctuation",{"directive tag":Prism.languages.aspnet["directive tag"]},Prism.languages.aspnet.tag.inside["attr-value"]),Prism.languages.insertBefore("aspnet","comment",{"asp comment":/<%--[\w\W]*?--%>/}),Prism.languages.insertBefore("aspnet",Prism.languages.javascript?"script":"tag",{"asp script":{pattern:/(<script(?=.*runat=['"]?server['"]?)[\w\W]*?>)[\w\W]*?(?=<\/script>)/i,lookbehind:!0,inside:Prism.languages.csharp||{}}});
+!function(e){var t={variable:[{pattern:/\$?\(\([\w\W]+?\)\)/,inside:{variable:[{pattern:/(^\$\(\([\w\W]+)\)\)/,lookbehind:!0},/^\$\(\(/],number:/\b-?(?:0x[\dA-Fa-f]+|\d*\.?\d+(?:[Ee]-?\d+)?)\b/,operator:/--?|-=|\+\+?|\+=|!=?|~|\*\*?|\*=|\/=?|%=?|<<=?|>>=?|<=?|>=?|==?|&&?|&=|\^=?|\|\|?|\|=|\?|:/,punctuation:/\(\(?|\)\)?|,|;/}},{pattern:/\$\([^)]+\)|`[^`]+`/,inside:{variable:/^\$\(|^`|\)$|`$/}},/\$(?:[a-z0-9_#\?\*!@]+|\{[^}]+\})/i]};e.languages.bash={shebang:{pattern:/^#!\s*\/bin\/bash|^#!\s*\/bin\/sh/,alias:"important"},comment:{pattern:/(^|[^"{\\])#.*/,lookbehind:!0},string:[{pattern:/((?:^|[^<])<<\s*)(?:"|')?(\w+?)(?:"|')?\s*\r?\n(?:[\s\S])*?\r?\n\2/g,lookbehind:!0,greedy:!0,inside:t},{pattern:/(["'])(?:\\\\|\\?[^\\])*?\1/g,greedy:!0,inside:t}],variable:t.variable,"function":{pattern:/(^|\s|;|\||&)(?:alias|apropos|apt-get|aptitude|aspell|awk|basename|bash|bc|bg|builtin|bzip2|cal|cat|cd|cfdisk|chgrp|chmod|chown|chroot|chkconfig|cksum|clear|cmp|comm|command|cp|cron|crontab|csplit|cut|date|dc|dd|ddrescue|df|diff|diff3|dig|dir|dircolors|dirname|dirs|dmesg|du|egrep|eject|enable|env|ethtool|eval|exec|expand|expect|export|expr|fdformat|fdisk|fg|fgrep|file|find|fmt|fold|format|free|fsck|ftp|fuser|gawk|getopts|git|grep|groupadd|groupdel|groupmod|groups|gzip|hash|head|help|hg|history|hostname|htop|iconv|id|ifconfig|ifdown|ifup|import|install|jobs|join|kill|killall|less|link|ln|locate|logname|logout|look|lpc|lpr|lprint|lprintd|lprintq|lprm|ls|lsof|make|man|mkdir|mkfifo|mkisofs|mknod|more|most|mount|mtools|mtr|mv|mmv|nano|netstat|nice|nl|nohup|notify-send|npm|nslookup|open|op|passwd|paste|pathchk|ping|pkill|popd|pr|printcap|printenv|printf|ps|pushd|pv|pwd|quota|quotacheck|quotactl|ram|rar|rcp|read|readarray|readonly|reboot|rename|renice|remsync|rev|rm|rmdir|rsync|screen|scp|sdiff|sed|seq|service|sftp|shift|shopt|shutdown|sleep|slocate|sort|source|split|ssh|stat|strace|su|sudo|sum|suspend|sync|tail|tar|tee|test|time|timeout|times|touch|top|traceroute|trap|tr|tsort|tty|type|ulimit|umask|umount|unalias|uname|unexpand|uniq|units|unrar|unshar|uptime|useradd|userdel|usermod|users|uuencode|uudecode|v|vdir|vi|vmstat|wait|watch|wc|wget|whereis|which|who|whoami|write|xargs|xdg-open|yes|zip)(?=$|\s|;|\||&)/,lookbehind:!0},keyword:{pattern:/(^|\s|;|\||&)(?:let|:|\.|if|then|else|elif|fi|for|break|continue|while|in|case|function|select|do|done|until|echo|exit|return|set|declare)(?=$|\s|;|\||&)/,lookbehind:!0},"boolean":{pattern:/(^|\s|;|\||&)(?:true|false)(?=$|\s|;|\||&)/,lookbehind:!0},operator:/&&?|\|\|?|==?|!=?|<<<?|>>|<=?|>=?|=~/,punctuation:/\$?\(\(?|\)\)?|\.\.|[{}[\];]/};var a=t.variable[1].inside;a["function"]=e.languages.bash["function"],a.keyword=e.languages.bash.keyword,a.boolean=e.languages.bash.boolean,a.operator=e.languages.bash.operator,a.punctuation=e.languages.bash.punctuation}(Prism);
+Prism.languages.basic={string:/"(?:""|[!#$%&'()*,\/:;<=>?^_ +\-.A-Z\d])*"/i,comment:{pattern:/(?:!|REM\b).+/i,inside:{keyword:/^REM/i}},number:/(?:\b|\B[.-])(?:\d+\.?\d*)(?:E[+-]?\d+)?/i,keyword:/\b(?:AS|BEEP|BLOAD|BSAVE|CALL(?: ABSOLUTE)?|CASE|CHAIN|CHDIR|CLEAR|CLOSE|CLS|COM|COMMON|CONST|DATA|DECLARE|DEF(?: FN| SEG|DBL|INT|LNG|SNG|STR)|DIM|DO|DOUBLE|ELSE|ELSEIF|END|ENVIRON|ERASE|ERROR|EXIT|FIELD|FILES|FOR|FUNCTION|GET|GOSUB|GOTO|IF|INPUT|INTEGER|IOCTL|KEY|KILL|LINE INPUT|LOCATE|LOCK|LONG|LOOP|LSET|MKDIR|NAME|NEXT|OFF|ON(?: COM| ERROR| KEY| TIMER)?|OPEN|OPTION BASE|OUT|POKE|PUT|READ|REDIM|REM|RESTORE|RESUME|RETURN|RMDIR|RSET|RUN|SHARED|SINGLE|SELECT CASE|SHELL|SLEEP|STATIC|STEP|STOP|STRING|SUB|SWAP|SYSTEM|THEN|TIMER|TO|TROFF|TRON|TYPE|UNLOCK|UNTIL|USING|VIEW PRINT|WAIT|WEND|WHILE|WRITE)(?:\$|\b)/i,"function":/\b(?:ABS|ACCESS|ACOS|ANGLE|AREA|ARITHMETIC|ARRAY|ASIN|ASK|AT|ATN|BASE|BEGIN|BREAK|CAUSE|CEIL|CHR|CLIP|COLLATE|COLOR|CON|COS|COSH|COT|CSC|DATE|DATUM|DEBUG|DECIMAL|DEF|DEG|DEGREES|DELETE|DET|DEVICE|DISPLAY|DOT|ELAPSED|EPS|ERASABLE|EXLINE|EXP|EXTERNAL|EXTYPE|FILETYPE|FIXED|FP|GO|GRAPH|HANDLER|IDN|IMAGE|IN|INT|INTERNAL|IP|IS|KEYED|LBOUND|LCASE|LEFT|LEN|LENGTH|LET|LINE|LINES|LOG|LOG10|LOG2|LTRIM|MARGIN|MAT|MAX|MAXNUM|MID|MIN|MISSING|MOD|NATIVE|NUL|NUMERIC|OF|OPTION|ORD|ORGANIZATION|OUTIN|OUTPUT|PI|POINT|POINTER|POINTS|POS|PRINT|PROGRAM|PROMPT|RAD|RADIANS|RANDOMIZE|RECORD|RECSIZE|RECTYPE|RELATIVE|REMAINDER|REPEAT|REST|RETRY|REWRITE|RIGHT|RND|ROUND|RTRIM|SAME|SEC|SELECT|SEQUENTIAL|SET|SETTER|SGN|SIN|SINH|SIZE|SKIP|SQR|STANDARD|STATUS|STR|STREAM|STYLE|TAB|TAN|TANH|TEMPLATE|TEXT|THERE|TIME|TIMEOUT|TRACE|TRANSFORM|TRUNCATE|UBOUND|UCASE|USE|VAL|VARIABLE|VIEWPORT|WHEN|WINDOW|WITH|ZER|ZONEWIDTH)(?:\$|\b)/i,operator:/<[=>]?|>=?|[+\-*\/^=&]|\b(?:AND|EQV|IMP|NOT|OR|XOR)\b/i,punctuation:/[,;:()]/};
+Prism.languages.c=Prism.languages.extend("clike",{keyword:/\b(asm|typeof|inline|auto|break|case|char|const|continue|default|do|double|else|enum|extern|float|for|goto|if|int|long|register|return|short|signed|sizeof|static|struct|switch|typedef|union|unsigned|void|volatile|while)\b/,operator:/\-[>-]?|\+\+?|!=?|<<?=?|>>?=?|==?|&&?|\|?\||[~^%?*\/]/,number:/\b-?(?:0x[\da-f]+|\d*\.?\d+(?:e[+-]?\d+)?)[ful]*\b/i}),Prism.languages.insertBefore("c","string",{macro:{pattern:/(^\s*)#\s*[a-z]+([^\r\n\\]|\\.|\\(?:\r\n?|\n))*/im,lookbehind:!0,alias:"property",inside:{string:{pattern:/(#\s*include\s*)(<.+?>|("|')(\\?.)+?\3)/,lookbehind:!0},directive:{pattern:/(#\s*)\b(define|elif|else|endif|error|ifdef|ifndef|if|import|include|line|pragma|undef|using)\b/,lookbehind:!0,alias:"keyword"}}},constant:/\b(__FILE__|__LINE__|__DATE__|__TIME__|__TIMESTAMP__|__func__|EOF|NULL|stdin|stdout|stderr)\b/}),delete Prism.languages.c["class-name"],delete Prism.languages.c["boolean"];
+Prism.languages.bison=Prism.languages.extend("c",{}),Prism.languages.insertBefore("bison","comment",{bison:{pattern:/^[\s\S]*?%%[\s\S]*?%%/,inside:{c:{pattern:/%\{[\s\S]*?%\}|\{(?:\{[^}]*\}|[^{}])*\}/,inside:{delimiter:{pattern:/^%?\{|%?\}$/,alias:"punctuation"},"bison-variable":{pattern:/[$@](?:<[^\s>]+>)?[\w$]+/,alias:"variable",inside:{punctuation:/<|>/}},rest:Prism.languages.c}},comment:Prism.languages.c.comment,string:Prism.languages.c.string,property:/\S+(?=:)/,keyword:/%\w+/,number:{pattern:/(^|[^@])\b(?:0x[\da-f]+|\d+)/i,lookbehind:!0},punctuation:/%[%?]|[|:;\[\]<>]/}}});
+Prism.languages.csharp=Prism.languages.extend("clike",{keyword:/\b(abstract|as|async|await|base|bool|break|byte|case|catch|char|checked|class|const|continue|decimal|default|delegate|do|double|else|enum|event|explicit|extern|false|finally|fixed|float|for|foreach|goto|if|implicit|in|int|interface|internal|is|lock|long|namespace|new|null|object|operator|out|override|params|private|protected|public|readonly|ref|return|sbyte|sealed|short|sizeof|stackalloc|static|string|struct|switch|this|throw|true|try|typeof|uint|ulong|unchecked|unsafe|ushort|using|virtual|void|volatile|while|add|alias|ascending|async|await|descending|dynamic|from|get|global|group|into|join|let|orderby|partial|remove|select|set|value|var|where|yield)\b/,string:[/@("|')(\1\1|\\\1|\\?(?!\1)[\s\S])*\1/,/("|')(\\?.)*?\1/],number:/\b-?(0x[\da-f]+|\d*\.?\d+f?)\b/i}),Prism.languages.insertBefore("csharp","keyword",{"generic-method":{pattern:/[a-z0-9_]+\s*<[^>\r\n]+?>\s*(?=\()/i,alias:"function",inside:{keyword:Prism.languages.csharp.keyword,punctuation:/[<>(),.:]/}},preprocessor:{pattern:/(^\s*)#.*/m,lookbehind:!0,alias:"property",inside:{directive:{pattern:/(\s*#)\b(define|elif|else|endif|endregion|error|if|line|pragma|region|undef|warning)\b/,lookbehind:!0,alias:"keyword"}}}});
+Prism.languages.cpp=Prism.languages.extend("c",{keyword:/\b(alignas|alignof|asm|auto|bool|break|case|catch|char|char16_t|char32_t|class|compl|const|constexpr|const_cast|continue|decltype|default|delete|do|double|dynamic_cast|else|enum|explicit|export|extern|float|for|friend|goto|if|inline|int|long|mutable|namespace|new|noexcept|nullptr|operator|private|protected|public|register|reinterpret_cast|return|short|signed|sizeof|static|static_assert|static_cast|struct|switch|template|this|thread_local|throw|try|typedef|typeid|typename|union|unsigned|using|virtual|void|volatile|wchar_t|while)\b/,"boolean":/\b(true|false)\b/,operator:/[-+]{1,2}|!=?|<{1,2}=?|>{1,2}=?|\->|:{1,2}|={1,2}|\^|~|%|&{1,2}|\|?\||\?|\*|\/|\b(and|and_eq|bitand|bitor|not|not_eq|or|or_eq|xor|xor_eq)\b/}),Prism.languages.insertBefore("cpp","keyword",{"class-name":{pattern:/(class\s+)[a-z0-9_]+/i,lookbehind:!0}});
+!function(e){var t=/#(?!\{).+/,n={pattern:/#\{[^}]+\}/,alias:"variable"};e.languages.coffeescript=e.languages.extend("javascript",{comment:t,string:[{pattern:/'(?:\\?[^\\])*?'/,greedy:!0},{pattern:/"(?:\\?[^\\])*?"/,greedy:!0,inside:{interpolation:n}}],keyword:/\b(and|break|by|catch|class|continue|debugger|delete|do|each|else|extend|extends|false|finally|for|if|in|instanceof|is|isnt|let|loop|namespace|new|no|not|null|of|off|on|or|own|return|super|switch|then|this|throw|true|try|typeof|undefined|unless|until|when|while|window|with|yes|yield)\b/,"class-member":{pattern:/@(?!\d)\w+/,alias:"variable"}}),e.languages.insertBefore("coffeescript","comment",{"multiline-comment":{pattern:/###[\s\S]+?###/,alias:"comment"},"block-regex":{pattern:/\/{3}[\s\S]*?\/{3}/,alias:"regex",inside:{comment:t,interpolation:n}}}),e.languages.insertBefore("coffeescript","string",{"inline-javascript":{pattern:/`(?:\\?[\s\S])*?`/,inside:{delimiter:{pattern:/^`|`$/,alias:"punctuation"},rest:e.languages.javascript}},"multiline-string":[{pattern:/'''[\s\S]*?'''/,greedy:!0,alias:"string"},{pattern:/"""[\s\S]*?"""/,greedy:!0,alias:"string",inside:{interpolation:n}}]}),e.languages.insertBefore("coffeescript","keyword",{property:/(?!\d)\w+(?=\s*:(?!:))/}),delete e.languages.coffeescript["template-string"]}(Prism);
+!function(e){e.languages.ruby=e.languages.extend("clike",{comment:/#(?!\{[^\r\n]*?\}).*/,keyword:/\b(alias|and|BEGIN|begin|break|case|class|def|define_method|defined|do|each|else|elsif|END|end|ensure|false|for|if|in|module|new|next|nil|not|or|raise|redo|require|rescue|retry|return|self|super|then|throw|true|undef|unless|until|when|while|yield)\b/});var n={pattern:/#\{[^}]+\}/,inside:{delimiter:{pattern:/^#\{|\}$/,alias:"tag"},rest:e.util.clone(e.languages.ruby)}};e.languages.insertBefore("ruby","keyword",{regex:[{pattern:/%r([^a-zA-Z0-9\s\{\(\[<])(?:[^\\]|\\[\s\S])*?\1[gim]{0,3}/,inside:{interpolation:n}},{pattern:/%r\((?:[^()\\]|\\[\s\S])*\)[gim]{0,3}/,inside:{interpolation:n}},{pattern:/%r\{(?:[^#{}\\]|#(?:\{[^}]+\})?|\\[\s\S])*\}[gim]{0,3}/,inside:{interpolation:n}},{pattern:/%r\[(?:[^\[\]\\]|\\[\s\S])*\][gim]{0,3}/,inside:{interpolation:n}},{pattern:/%r<(?:[^<>\\]|\\[\s\S])*>[gim]{0,3}/,inside:{interpolation:n}},{pattern:/(^|[^\/])\/(?!\/)(\[.+?]|\\.|[^\/\r\n])+\/[gim]{0,3}(?=\s*($|[\r\n,.;})]))/,lookbehind:!0}],variable:/[@$]+[a-zA-Z_][a-zA-Z_0-9]*(?:[?!]|\b)/,symbol:/:[a-zA-Z_][a-zA-Z_0-9]*(?:[?!]|\b)/}),e.languages.insertBefore("ruby","number",{builtin:/\b(Array|Bignum|Binding|Class|Continuation|Dir|Exception|FalseClass|File|Stat|File|Fixnum|Fload|Hash|Integer|IO|MatchData|Method|Module|NilClass|Numeric|Object|Proc|Range|Regexp|String|Struct|TMS|Symbol|ThreadGroup|Thread|Time|TrueClass)\b/,constant:/\b[A-Z][a-zA-Z_0-9]*(?:[?!]|\b)/}),e.languages.ruby.string=[{pattern:/%[qQiIwWxs]?([^a-zA-Z0-9\s\{\(\[<])(?:[^\\]|\\[\s\S])*?\1/,inside:{interpolation:n}},{pattern:/%[qQiIwWxs]?\((?:[^()\\]|\\[\s\S])*\)/,inside:{interpolation:n}},{pattern:/%[qQiIwWxs]?\{(?:[^#{}\\]|#(?:\{[^}]+\})?|\\[\s\S])*\}/,inside:{interpolation:n}},{pattern:/%[qQiIwWxs]?\[(?:[^\[\]\\]|\\[\s\S])*\]/,inside:{interpolation:n}},{pattern:/%[qQiIwWxs]?<(?:[^<>\\]|\\[\s\S])*>/,inside:{interpolation:n}},{pattern:/("|')(#\{[^}]+\}|\\(?:\r?\n|\r)|\\?.)*?\1/,inside:{interpolation:n}}]}(Prism);
+Prism.languages.css.selector={pattern:/[^\{\}\s][^\{\}]*(?=\s*\{)/,inside:{"pseudo-element":/:(?:after|before|first-letter|first-line|selection)|::[-\w]+/,"pseudo-class":/:[-\w]+(?:\(.*\))?/,"class":/\.[-:\.\w]+/,id:/#[-:\.\w]+/,attribute:/\[[^\]]+\]/}},Prism.languages.insertBefore("css","function",{hexcode:/#[\da-f]{3,6}/i,entity:/\\[\da-f]{1,8}/i,number:/[\d%\.]+/});
+Prism.languages.diff={coord:[/^(?:\*{3}|-{3}|\+{3}).*$/m,/^@@.*@@$/m,/^\d+.*$/m],deleted:/^[-<].*$/m,inserted:/^[+>].*$/m,diff:{pattern:/^!(?!!).+$/m,alias:"important"}};
+Prism.languages.docker={keyword:{pattern:/(^\s*)(?:ONBUILD|FROM|MAINTAINER|RUN|EXPOSE|ENV|ADD|COPY|VOLUME|USER|WORKDIR|CMD|LABEL|ENTRYPOINT)(?=\s)/im,lookbehind:!0},string:/("|')(?:(?!\1)[^\\\r\n]|\\(?:\r\n|[\s\S]))*?\1/,comment:/#.*/,punctuation:/---|\.\.\.|[:[\]{}\-,|>?]/};
+Prism.languages.erlang={comment:/%.+/,string:/"(?:\\?.)*?"/,"quoted-function":{pattern:/'(?:\\.|[^'\\])+'(?=\()/,alias:"function"},"quoted-atom":{pattern:/'(?:\\.|[^'\\])+'/,alias:"atom"},"boolean":/\b(?:true|false)\b/,keyword:/\b(?:fun|when|case|of|end|if|receive|after|try|catch)\b/,number:[/\$\\?./,/\d+#[a-z0-9]+/i,/(?:\b|-)\d*\.?\d+([Ee][+-]?\d+)?\b/],"function":/\b[a-z][\w@]*(?=\()/,variable:{pattern:/(^|[^@])(?:\b|\?)[A-Z_][\w@]*/,lookbehind:!0},operator:[/[=\/<>:]=|=[:\/]=|\+\+?|--?|[=*\/!]|\b(?:bnot|div|rem|band|bor|bxor|bsl|bsr|not|and|or|xor|orelse|andalso)\b/,{pattern:/(^|[^<])<(?!<)/,lookbehind:!0},{pattern:/(^|[^>])>(?!>)/,lookbehind:!0}],atom:/\b[a-z][\w@]*/,punctuation:/[()[\]{}:;,.#|]|<<|>>/};
+Prism.languages.fsharp=Prism.languages.extend("clike",{comment:[{pattern:/(^|[^\\])\(\*[\w\W]*?\*\)/,lookbehind:!0},{pattern:/(^|[^\\:])\/\/.*/,lookbehind:!0}],keyword:/\b(?:let|return|use|yield)(?:!\B|\b)|\b(abstract|and|as|assert|base|begin|class|default|delegate|do|done|downcast|downto|elif|else|end|exception|extern|false|finally|for|fun|function|global|if|in|inherit|inline|interface|internal|lazy|match|member|module|mutable|namespace|new|not|null|of|open|or|override|private|public|rec|select|static|struct|then|to|true|try|type|upcast|val|void|when|while|with|asr|land|lor|lsl|lsr|lxor|mod|sig|atomic|break|checked|component|const|constraint|constructor|continue|eager|event|external|fixed|functor|include|method|mixin|object|parallel|process|protected|pure|sealed|tailcall|trait|virtual|volatile)\b/,string:/(?:"""[\s\S]*?"""|@"(?:""|[^"])*"|("|')(?:\\\1|\\?(?!\1)[\s\S])*\1)B?/,number:[/\b-?0x[\da-fA-F]+(un|lf|LF)?\b/,/\b-?0b[01]+(y|uy)?\b/,/\b-?(\d*\.?\d+|\d+\.)([fFmM]|[eE][+-]?\d+)?\b/,/\b-?\d+(y|uy|s|us|l|u|ul|L|UL|I)?\b/]}),Prism.languages.insertBefore("fsharp","keyword",{preprocessor:{pattern:/^[^\r\n\S]*#.*/m,alias:"property",inside:{directive:{pattern:/(\s*#)\b(else|endif|if|light|line|nowarn)\b/,lookbehind:!0,alias:"keyword"}}}});
+Prism.languages.fortran={"quoted-number":{pattern:/[BOZ](['"])[A-F0-9]+\1/i,alias:"number"},string:{pattern:/(?:\w+_)?(['"])(?:\1\1|&(?:\r\n?|\n)(?:\s*!.+(?:\r\n?|\n))?|(?!\1).)*(?:\1|&)/,inside:{comment:{pattern:/(&(?:\r\n?|\n)\s*)!.*/,lookbehind:!0}}},comment:/!.*/,"boolean":/\.(?:TRUE|FALSE)\.(?:_\w+)?/i,number:/(?:\b|[+-])(?:\d+(?:\.\d*)?|\.\d+)(?:[ED][+-]?\d+)?(?:_\w+)?/i,keyword:[/\b(?:INTEGER|REAL|DOUBLE ?PRECISION|COMPLEX|CHARACTER|LOGICAL)\b/i,/\b(?:END ?)?(?:BLOCK ?DATA|DO|FILE|FORALL|FUNCTION|IF|INTERFACE|MODULE(?! PROCEDURE)|PROGRAM|SELECT|SUBROUTINE|TYPE|WHERE)\b/i,/\b(?:ALLOCATABLE|ALLOCATE|BACKSPACE|CALL|CASE|CLOSE|COMMON|CONTAINS|CONTINUE|CYCLE|DATA|DEALLOCATE|DIMENSION|DO|END|EQUIVALENCE|EXIT|EXTERNAL|FORMAT|GO ?TO|IMPLICIT(?: NONE)?|INQUIRE|INTENT|INTRINSIC|MODULE PROCEDURE|NAMELIST|NULLIFY|OPEN|OPTIONAL|PARAMETER|POINTER|PRINT|PRIVATE|PUBLIC|READ|RETURN|REWIND|SAVE|SELECT|STOP|TARGET|WHILE|WRITE)\b/i,/\b(?:ASSIGNMENT|DEFAULT|ELEMENTAL|ELSE|ELSEWHERE|ELSEIF|ENTRY|IN|INCLUDE|INOUT|KIND|NULL|ONLY|OPERATOR|OUT|PURE|RECURSIVE|RESULT|SEQUENCE|STAT|THEN|USE)\b/i],operator:[/\*\*|\/\/|=>|[=\/]=|[<>]=?|::|[+\-*=%]|\.(?:EQ|NE|LT|LE|GT|GE|NOT|AND|OR|EQV|NEQV)\.|\.[A-Z]+\./i,{pattern:/(^|(?!\().)\/(?!\))/,lookbehind:!0}],punctuation:/\(\/|\/\)|[(),;:&]/};
+Prism.languages.git={comment:/^#.*/m,deleted:/^[-–].*/m,inserted:/^\+.*/m,string:/("|')(\\?.)*?\1/m,command:{pattern:/^.*\$ git .*$/m,inside:{parameter:/\s(--|-)\w+/m}},coord:/^@@.*@@$/m,commit_sha1:/^commit \w{40}$/m};
+Prism.languages.go=Prism.languages.extend("clike",{keyword:/\b(break|case|chan|const|continue|default|defer|else|fallthrough|for|func|go(to)?|if|import|interface|map|package|range|return|select|struct|switch|type|var)\b/,builtin:/\b(bool|byte|complex(64|128)|error|float(32|64)|rune|string|u?int(8|16|32|64|)|uintptr|append|cap|close|complex|copy|delete|imag|len|make|new|panic|print(ln)?|real|recover)\b/,"boolean":/\b(_|iota|nil|true|false)\b/,operator:/[*\/%^!=]=?|\+[=+]?|-[=-]?|\|[=|]?|&(?:=|&|\^=?)?|>(?:>=?|=)?|<(?:<=?|=|-)?|:=|\.\.\./,number:/\b(-?(0x[a-f\d]+|(\d+\.?\d*|\.\d+)(e[-+]?\d+)?)i?)\b/i,string:/("|'|`)(\\?.|\r|\n)*?\1/}),delete Prism.languages.go["class-name"];
+Prism.languages.haskell={comment:{pattern:/(^|[^-!#$%*+=?&@|~.:<>^\\\/])(--[^-!#$%*+=?&@|~.:<>^\\\/].*|{-[\w\W]*?-})/m,lookbehind:!0},"char":/'([^\\']|\\([abfnrtv\\"'&]|\^[A-Z@[\]\^_]|NUL|SOH|STX|ETX|EOT|ENQ|ACK|BEL|BS|HT|LF|VT|FF|CR|SO|SI|DLE|DC1|DC2|DC3|DC4|NAK|SYN|ETB|CAN|EM|SUB|ESC|FS|GS|RS|US|SP|DEL|\d+|o[0-7]+|x[0-9a-fA-F]+))'/,string:{pattern:/"([^\\"]|\\([abfnrtv\\"'&]|\^[A-Z@[\]\^_]|NUL|SOH|STX|ETX|EOT|ENQ|ACK|BEL|BS|HT|LF|VT|FF|CR|SO|SI|DLE|DC1|DC2|DC3|DC4|NAK|SYN|ETB|CAN|EM|SUB|ESC|FS|GS|RS|US|SP|DEL|\d+|o[0-7]+|x[0-9a-fA-F]+)|\\\s+\\)*"/,greedy:!0},keyword:/\b(case|class|data|deriving|do|else|if|in|infixl|infixr|instance|let|module|newtype|of|primitive|then|type|where)\b/,import_statement:{pattern:/(\r?\n|\r|^)\s*import\s+(qualified\s+)?([A-Z][_a-zA-Z0-9']*)(\.[A-Z][_a-zA-Z0-9']*)*(\s+as\s+([A-Z][_a-zA-Z0-9']*)(\.[A-Z][_a-zA-Z0-9']*)*)?(\s+hiding\b)?/m,inside:{keyword:/\b(import|qualified|as|hiding)\b/}},builtin:/\b(abs|acos|acosh|all|and|any|appendFile|approxRational|asTypeOf|asin|asinh|atan|atan2|atanh|basicIORun|break|catch|ceiling|chr|compare|concat|concatMap|const|cos|cosh|curry|cycle|decodeFloat|denominator|digitToInt|div|divMod|drop|dropWhile|either|elem|encodeFloat|enumFrom|enumFromThen|enumFromThenTo|enumFromTo|error|even|exp|exponent|fail|filter|flip|floatDigits|floatRadix|floatRange|floor|fmap|foldl|foldl1|foldr|foldr1|fromDouble|fromEnum|fromInt|fromInteger|fromIntegral|fromRational|fst|gcd|getChar|getContents|getLine|group|head|id|inRange|index|init|intToDigit|interact|ioError|isAlpha|isAlphaNum|isAscii|isControl|isDenormalized|isDigit|isHexDigit|isIEEE|isInfinite|isLower|isNaN|isNegativeZero|isOctDigit|isPrint|isSpace|isUpper|iterate|last|lcm|length|lex|lexDigits|lexLitChar|lines|log|logBase|lookup|map|mapM|mapM_|max|maxBound|maximum|maybe|min|minBound|minimum|mod|negate|not|notElem|null|numerator|odd|or|ord|otherwise|pack|pi|pred|primExitWith|print|product|properFraction|putChar|putStr|putStrLn|quot|quotRem|range|rangeSize|read|readDec|readFile|readFloat|readHex|readIO|readInt|readList|readLitChar|readLn|readOct|readParen|readSigned|reads|readsPrec|realToFrac|recip|rem|repeat|replicate|return|reverse|round|scaleFloat|scanl|scanl1|scanr|scanr1|seq|sequence|sequence_|show|showChar|showInt|showList|showLitChar|showParen|showSigned|showString|shows|showsPrec|significand|signum|sin|sinh|snd|sort|span|splitAt|sqrt|subtract|succ|sum|tail|take|takeWhile|tan|tanh|threadToIOResult|toEnum|toInt|toInteger|toLower|toRational|toUpper|truncate|uncurry|undefined|unlines|until|unwords|unzip|unzip3|userError|words|writeFile|zip|zip3|zipWith|zipWith3)\b/,number:/\b(\d+(\.\d+)?(e[+-]?\d+)?|0o[0-7]+|0x[0-9a-f]+)\b/i,operator:/\s\.\s|[-!#$%*+=?&@|~.:<>^\\\/]*\.[-!#$%*+=?&@|~.:<>^\\\/]+|[-!#$%*+=?&@|~.:<>^\\\/]+\.[-!#$%*+=?&@|~.:<>^\\\/]*|[-!#$%*+=?&@|~:<>^\\\/]+|`([A-Z][_a-zA-Z0-9']*\.)*[_a-z][_a-zA-Z0-9']*`/,hvariable:/\b([A-Z][_a-zA-Z0-9']*\.)*[_a-z][_a-zA-Z0-9']*\b/,constant:/\b([A-Z][_a-zA-Z0-9']*\.)*[A-Z][_a-zA-Z0-9']*\b/,punctuation:/[{}[\];(),.:]/};
+Prism.languages.http={"request-line":{pattern:/^(POST|GET|PUT|DELETE|OPTIONS|PATCH|TRACE|CONNECT)\b\shttps?:\/\/\S+\sHTTP\/[0-9.]+/m,inside:{property:/^(POST|GET|PUT|DELETE|OPTIONS|PATCH|TRACE|CONNECT)\b/,"attr-name":/:\w+/}},"response-status":{pattern:/^HTTP\/1.[01] [0-9]+.*/m,inside:{property:{pattern:/(^HTTP\/1.[01] )[0-9]+.*/i,lookbehind:!0}}},"header-name":{pattern:/^[\w-]+:(?=.)/m,alias:"keyword"}};var httpLanguages={"application/json":Prism.languages.javascript,"application/xml":Prism.languages.markup,"text/xml":Prism.languages.markup,"text/html":Prism.languages.markup};for(var contentType in httpLanguages)if(httpLanguages[contentType]){var options={};options[contentType]={pattern:new RegExp("(content-type:\\s*"+contentType+"[\\w\\W]*?)(?:\\r?\\n|\\r){2}[\\w\\W]*","i"),lookbehind:!0,inside:{rest:httpLanguages[contentType]}},Prism.languages.insertBefore("http","header-name",options)};
+Prism.languages.icon={comment:/#.*/,string:/(["'])(?:(?!\1)[^\\\r\n]|\\.|_(?:\r?\n|\r))*\1/,number:/\b(?:\d+r[a-z\d]+|\d+(?:\.\d+)?(?:e[+-]?\d+)?)\b|\.\d+\b/i,"builtin-keyword":{pattern:/&(?:allocated|ascii|clock|collections|cset|current|date|dateline|digits|dump|e|error(?:number|text|value)?|errout|fail|features|file|host|input|lcase|letters|level|line|main|null|output|phi|pi|pos|progname|random|regions|source|storage|subject|time|trace|ucase|version)\b/,alias:"variable"},directive:{pattern:/\$\w+/,alias:"builtin"},keyword:/\b(?:break|by|case|create|default|do|else|end|every|fail|global|if|initial|invocable|link|local|next|not|of|procedure|record|repeat|return|static|suspend|then|to|until|while)\b/,"function":/(?!\d)\w+(?=\s*[({]|\s*!\s*\[)/,operator:/[+-]:(?!=)|(?:[\/?@^%&]|\+\+?|--?|==?=?|~==?=?|\*\*?|\|\|\|?|<(?:->?|<?=?)|>>?=?)(?::=)?|:(?:=:?)?|[!.\\|~]/,punctuation:/[\[\](){},;]/};
+Prism.languages.java=Prism.languages.extend("clike",{keyword:/\b(abstract|continue|for|new|switch|assert|default|goto|package|synchronized|boolean|do|if|private|this|break|double|implements|protected|throw|byte|else|import|public|throws|case|enum|instanceof|return|transient|catch|extends|int|short|try|char|final|interface|static|void|class|finally|long|strictfp|volatile|const|float|native|super|while)\b/,number:/\b0b[01]+\b|\b0x[\da-f]*\.?[\da-fp\-]+\b|\b\d*\.?\d+(?:e[+-]?\d+)?[df]?\b/i,operator:{pattern:/(^|[^.])(?:\+[+=]?|-[-=]?|!=?|<<?=?|>>?>?=?|==?|&[&=]?|\|[|=]?|\*=?|\/=?|%=?|\^=?|[?:~])/m,lookbehind:!0}}),Prism.languages.insertBefore("java","function",{annotation:{alias:"punctuation",pattern:/(^|[^.])@\w+/,lookbehind:!0}});
+Prism.languages.json={property:/".*?"(?=\s*:)/gi,string:/"(?!:)(\\?[^"])*?"(?!:)/g,number:/\b-?(0x[\dA-Fa-f]+|\d*\.?\d+([Ee]-?\d+)?)\b/g,punctuation:/[{}[\]);,]/g,operator:/:/g,"boolean":/\b(true|false)\b/gi,"null":/\bnull\b/gi},Prism.languages.jsonp=Prism.languages.json;
+!function(a){var e=/\\([^a-z()[\]]|[a-z\*]+)/i,n={"equation-command":{pattern:e,alias:"regex"}};a.languages.latex={comment:/%.*/m,cdata:{pattern:/(\\begin\{((?:verbatim|lstlisting)\*?)\})([\w\W]*?)(?=\\end\{\2\})/,lookbehind:!0},equation:[{pattern:/\$(?:\\?[\w\W])*?\$|\\\((?:\\?[\w\W])*?\\\)|\\\[(?:\\?[\w\W])*?\\\]/,inside:n,alias:"string"},{pattern:/(\\begin\{((?:equation|math|eqnarray|align|multline|gather)\*?)\})([\w\W]*?)(?=\\end\{\2\})/,lookbehind:!0,inside:n,alias:"string"}],keyword:{pattern:/(\\(?:begin|end|ref|cite|label|usepackage|documentclass)(?:\[[^\]]+\])?\{)[^}]+(?=\})/,lookbehind:!0},url:{pattern:/(\\url\{)[^}]+(?=\})/,lookbehind:!0},headline:{pattern:/(\\(?:part|chapter|section|subsection|frametitle|subsubsection|paragraph|subparagraph|subsubparagraph|subsubsubparagraph)\*?(?:\[[^\]]+\])?\{)[^}]+(?=\}(?:\[[^\]]+\])?)/,lookbehind:!0,alias:"class-name"},"function":{pattern:e,alias:"selector"},punctuation:/[[\]{}&]/}}(Prism);
+Prism.languages.lua={comment:/^#!.+|--(?:\[(=*)\[[\s\S]*?\]\1\]|.*)/m,string:{pattern:/(["'])(?:(?!\1)[^\\\r\n]|\\z(?:\r\n|\s)|\\(?:\r\n|[\s\S]))*\1|\[(=*)\[[\s\S]*?\]\2\]/,greedy:!0},number:/\b0x[a-f\d]+\.?[a-f\d]*(?:p[+-]?\d+)?\b|\b\d+(?:\.\B|\.?\d*(?:e[+-]?\d+)?\b)|\B\.\d+(?:e[+-]?\d+)?\b/i,keyword:/\b(?:and|break|do|else|elseif|end|false|for|function|goto|if|in|local|nil|not|or|repeat|return|then|true|until|while)\b/,"function":/(?!\d)\w+(?=\s*(?:[({]))/,operator:[/[-+*%^&|#]|\/\/?|<[<=]?|>[>=]?|[=~]=?/,{pattern:/(^|[^.])\.\.(?!\.)/,lookbehind:!0}],punctuation:/[\[\](){},;]|\.+|:+/};
+Prism.languages.makefile={comment:{pattern:/(^|[^\\])#(?:\\(?:\r\n|[\s\S])|.)*/,lookbehind:!0},string:/(["'])(?:\\(?:\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1/,builtin:/\.[A-Z][^:#=\s]+(?=\s*:(?!=))/,symbol:{pattern:/^[^:=\r\n]+(?=\s*:(?!=))/m,inside:{variable:/\$+(?:[^(){}:#=\s]+|(?=[({]))/}},variable:/\$+(?:[^(){}:#=\s]+|\([@*%<^+?][DF]\)|(?=[({]))/,keyword:[/-include\b|\b(?:define|else|endef|endif|export|ifn?def|ifn?eq|include|override|private|sinclude|undefine|unexport|vpath)\b/,{pattern:/(\()(?:addsuffix|abspath|and|basename|call|dir|error|eval|file|filter(?:-out)?|findstring|firstword|flavor|foreach|guile|if|info|join|lastword|load|notdir|or|origin|patsubst|realpath|shell|sort|strip|subst|suffix|value|warning|wildcard|word(?:s|list)?)(?=[ \t])/,lookbehind:!0}],operator:/(?:::|[?:+!])?=|[|@]/,punctuation:/[:;(){}]/};
+Prism.languages.markdown=Prism.languages.extend("markup",{}),Prism.languages.insertBefore("markdown","prolog",{blockquote:{pattern:/^>(?:[\t ]*>)*/m,alias:"punctuation"},code:[{pattern:/^(?: {4}|\t).+/m,alias:"keyword"},{pattern:/``.+?``|`[^`\n]+`/,alias:"keyword"}],title:[{pattern:/\w+.*(?:\r?\n|\r)(?:==+|--+)/,alias:"important",inside:{punctuation:/==+$|--+$/}},{pattern:/(^\s*)#+.+/m,lookbehind:!0,alias:"important",inside:{punctuation:/^#+|#+$/}}],hr:{pattern:/(^\s*)([*-])([\t ]*\2){2,}(?=\s*$)/m,lookbehind:!0,alias:"punctuation"},list:{pattern:/(^\s*)(?:[*+-]|\d+\.)(?=[\t ].)/m,lookbehind:!0,alias:"punctuation"},"url-reference":{pattern:/!?\[[^\]]+\]:[\t ]+(?:\S+|<(?:\\.|[^>\\])+>)(?:[\t ]+(?:"(?:\\.|[^"\\])*"|'(?:\\.|[^'\\])*'|\((?:\\.|[^)\\])*\)))?/,inside:{variable:{pattern:/^(!?\[)[^\]]+/,lookbehind:!0},string:/(?:"(?:\\.|[^"\\])*"|'(?:\\.|[^'\\])*'|\((?:\\.|[^)\\])*\))$/,punctuation:/^[\[\]!:]|[<>]/},alias:"url"},bold:{pattern:/(^|[^\\])(\*\*|__)(?:(?:\r?\n|\r)(?!\r?\n|\r)|.)+?\2/,lookbehind:!0,inside:{punctuation:/^\*\*|^__|\*\*$|__$/}},italic:{pattern:/(^|[^\\])([*_])(?:(?:\r?\n|\r)(?!\r?\n|\r)|.)+?\2/,lookbehind:!0,inside:{punctuation:/^[*_]|[*_]$/}},url:{pattern:/!?\[[^\]]+\](?:\([^\s)]+(?:[\t ]+"(?:\\.|[^"\\])*")?\)| ?\[[^\]\n]*\])/,inside:{variable:{pattern:/(!?\[)[^\]]+(?=\]$)/,lookbehind:!0},string:{pattern:/"(?:\\.|[^"\\])*"(?=\)$)/}}}}),Prism.languages.markdown.bold.inside.url=Prism.util.clone(Prism.languages.markdown.url),Prism.languages.markdown.italic.inside.url=Prism.util.clone(Prism.languages.markdown.url),Prism.languages.markdown.bold.inside.italic=Prism.util.clone(Prism.languages.markdown.italic),Prism.languages.markdown.italic.inside.bold=Prism.util.clone(Prism.languages.markdown.bold);
+Prism.languages.matlab={string:/\B'(?:''|[^'\n])*'/,comment:[/%\{[\s\S]*?\}%/,/%.+/],number:/\b-?(?:\d*\.?\d+(?:[eE][+-]?\d+)?(?:[ij])?|[ij])\b/,keyword:/\b(?:break|case|catch|continue|else|elseif|end|for|function|if|inf|NaN|otherwise|parfor|pause|pi|return|switch|try|while)\b/,"function":/(?!\d)\w+(?=\s*\()/,operator:/\.?[*^\/\\']|[+\-:@]|[<>=~]=?|&&?|\|\|?/,punctuation:/\.{3}|[.,;\[\](){}!]/};
+Prism.languages.nasm={comment:/;.*$/m,string:/("|'|`)(\\?.)*?\1/m,label:{pattern:/(^\s*)[A-Za-z._?$][\w.?$@~#]*:/m,lookbehind:!0,alias:"function"},keyword:[/\[?BITS (16|32|64)\]?/m,{pattern:/(^\s*)section\s*[a-zA-Z\.]+:?/im,lookbehind:!0},/(?:extern|global)[^;\r\n]*/im,/(?:CPU|FLOAT|DEFAULT).*$/m],register:{pattern:/\b(?:st\d|[xyz]mm\d\d?|[cdt]r\d|r\d\d?[bwd]?|[er]?[abcd]x|[abcd][hl]|[er]?(bp|sp|si|di)|[cdefgs]s)\b/i,alias:"variable"},number:/(\b|-|(?=\$))(0[hx][\da-f]*\.?[\da-f]+(p[+-]?\d+)?|\d[\da-f]+[hx]|\$\d[\da-f]*|0[oq][0-7]+|[0-7]+[oq]|0[by][01]+|[01]+[by]|0[dt]\d+|\d*\.?\d+(\.?e[+-]?\d+)?[dt]?)\b/i,operator:/[\[\]*+\-\/%<>=&|$!]/};
+Prism.languages.nginx=Prism.languages.extend("clike",{comment:{pattern:/(^|[^"{\\])#.*/,lookbehind:!0},keyword:/\b(?:CONTENT_|DOCUMENT_|GATEWAY_|HTTP_|HTTPS|if_not_empty|PATH_|QUERY_|REDIRECT_|REMOTE_|REQUEST_|SCGI|SCRIPT_|SERVER_|http|server|events|location|include|accept_mutex|accept_mutex_delay|access_log|add_after_body|add_before_body|add_header|addition_types|aio|alias|allow|ancient_browser|ancient_browser_value|auth|auth_basic|auth_basic_user_file|auth_http|auth_http_header|auth_http_timeout|autoindex|autoindex_exact_size|autoindex_localtime|break|charset|charset_map|charset_types|chunked_transfer_encoding|client_body_buffer_size|client_body_in_file_only|client_body_in_single_buffer|client_body_temp_path|client_body_timeout|client_header_buffer_size|client_header_timeout|client_max_body_size|connection_pool_size|create_full_put_path|daemon|dav_access|dav_methods|debug_connection|debug_points|default_type|deny|devpoll_changes|devpoll_events|directio|directio_alignment|disable_symlinks|empty_gif|env|epoll_events|error_log|error_page|expires|fastcgi_buffer_size|fastcgi_buffers|fastcgi_busy_buffers_size|fastcgi_cache|fastcgi_cache_bypass|fastcgi_cache_key|fastcgi_cache_lock|fastcgi_cache_lock_timeout|fastcgi_cache_methods|fastcgi_cache_min_uses|fastcgi_cache_path|fastcgi_cache_purge|fastcgi_cache_use_stale|fastcgi_cache_valid|fastcgi_connect_timeout|fastcgi_hide_header|fastcgi_ignore_client_abort|fastcgi_ignore_headers|fastcgi_index|fastcgi_intercept_errors|fastcgi_keep_conn|fastcgi_max_temp_file_size|fastcgi_next_upstream|fastcgi_no_cache|fastcgi_param|fastcgi_pass|fastcgi_pass_header|fastcgi_read_timeout|fastcgi_redirect_errors|fastcgi_send_timeout|fastcgi_split_path_info|fastcgi_store|fastcgi_store_access|fastcgi_temp_file_write_size|fastcgi_temp_path|flv|geo|geoip_city|geoip_country|google_perftools_profiles|gzip|gzip_buffers|gzip_comp_level|gzip_disable|gzip_http_version|gzip_min_length|gzip_proxied|gzip_static|gzip_types|gzip_vary|if|if_modified_since|ignore_invalid_headers|image_filter|image_filter_buffer|image_filter_jpeg_quality|image_filter_sharpen|image_filter_transparency|imap_capabilities|imap_client_buffer|include|index|internal|ip_hash|keepalive|keepalive_disable|keepalive_requests|keepalive_timeout|kqueue_changes|kqueue_events|large_client_header_buffers|limit_conn|limit_conn_log_level|limit_conn_zone|limit_except|limit_rate|limit_rate_after|limit_req|limit_req_log_level|limit_req_zone|limit_zone|lingering_close|lingering_time|lingering_timeout|listen|location|lock_file|log_format|log_format_combined|log_not_found|log_subrequest|map|map_hash_bucket_size|map_hash_max_size|master_process|max_ranges|memcached_buffer_size|memcached_connect_timeout|memcached_next_upstream|memcached_pass|memcached_read_timeout|memcached_send_timeout|merge_slashes|min_delete_depth|modern_browser|modern_browser_value|mp4|mp4_buffer_size|mp4_max_buffer_size|msie_padding|msie_refresh|multi_accept|open_file_cache|open_file_cache_errors|open_file_cache_min_uses|open_file_cache_valid|open_log_file_cache|optimize_server_names|override_charset|pcre_jit|perl|perl_modules|perl_require|perl_set|pid|pop3_auth|pop3_capabilities|port_in_redirect|post_action|postpone_output|protocol|proxy|proxy_buffer|proxy_buffer_size|proxy_buffering|proxy_buffers|proxy_busy_buffers_size|proxy_cache|proxy_cache_bypass|proxy_cache_key|proxy_cache_lock|proxy_cache_lock_timeout|proxy_cache_methods|proxy_cache_min_uses|proxy_cache_path|proxy_cache_use_stale|proxy_cache_valid|proxy_connect_timeout|proxy_cookie_domain|proxy_cookie_path|proxy_headers_hash_bucket_size|proxy_headers_hash_max_size|proxy_hide_header|proxy_http_version|proxy_ignore_client_abort|proxy_ignore_headers|proxy_intercept_errors|proxy_max_temp_file_size|proxy_method|proxy_next_upstream|proxy_no_cache|proxy_pass|proxy_pass_error_message|proxy_pass_header|proxy_pass_request_body|proxy_pass_request_headers|proxy_read_timeout|proxy_redirect|proxy_redirect_errors|proxy_send_lowat|proxy_send_timeout|proxy_set_body|proxy_set_header|proxy_ssl_session_reuse|proxy_store|proxy_store_access|proxy_temp_file_write_size|proxy_temp_path|proxy_timeout|proxy_upstream_fail_timeout|proxy_upstream_max_fails|random_index|read_ahead|real_ip_header|recursive_error_pages|request_pool_size|reset_timedout_connection|resolver|resolver_timeout|return|rewrite|root|rtsig_overflow_events|rtsig_overflow_test|rtsig_overflow_threshold|rtsig_signo|satisfy|satisfy_any|secure_link_secret|send_lowat|send_timeout|sendfile|sendfile_max_chunk|server|server_name|server_name_in_redirect|server_names_hash_bucket_size|server_names_hash_max_size|server_tokens|set|set_real_ip_from|smtp_auth|smtp_capabilities|so_keepalive|source_charset|split_clients|ssi|ssi_silent_errors|ssi_types|ssi_value_length|ssl|ssl_certificate|ssl_certificate_key|ssl_ciphers|ssl_client_certificate|ssl_crl|ssl_dhparam|ssl_engine|ssl_prefer_server_ciphers|ssl_protocols|ssl_session_cache|ssl_session_timeout|ssl_verify_client|ssl_verify_depth|starttls|stub_status|sub_filter|sub_filter_once|sub_filter_types|tcp_nodelay|tcp_nopush|timeout|timer_resolution|try_files|types|types_hash_bucket_size|types_hash_max_size|underscores_in_headers|uninitialized_variable_warn|upstream|use|user|userid|userid_domain|userid_expires|userid_name|userid_p3p|userid_path|userid_service|valid_referers|variables_hash_bucket_size|variables_hash_max_size|worker_connections|worker_cpu_affinity|worker_priority|worker_processes|worker_rlimit_core|worker_rlimit_nofile|worker_rlimit_sigpending|working_directory|xclient|xml_entities|xslt_entities|xslt_stylesheet|xslt_types)\b/i}),Prism.languages.insertBefore("nginx","keyword",{variable:/\$[a-z_]+/i});
+Prism.languages.objectivec=Prism.languages.extend("c",{keyword:/\b(asm|typeof|inline|auto|break|case|char|const|continue|default|do|double|else|enum|extern|float|for|goto|if|int|long|register|return|short|signed|sizeof|static|struct|switch|typedef|union|unsigned|void|volatile|while|in|self|super)\b|(@interface|@end|@implementation|@protocol|@class|@public|@protected|@private|@property|@try|@catch|@finally|@throw|@synthesize|@dynamic|@selector)\b/,string:/("|')(\\(?:\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1|@"(\\(?:\r\n|[\s\S])|[^"\\\r\n])*"/,operator:/-[->]?|\+\+?|!=?|<<?=?|>>?=?|==?|&&?|\|\|?|[~^%?*\/@]/});
+Prism.languages.pascal={comment:[/\(\*[\s\S]+?\*\)/,/\{[\s\S]+?\}/,/\/\/.*/],string:/(?:'(?:''|[^'\r\n])*'|#[&$%]?[a-f\d]+)+|\^[a-z]/i,keyword:[{pattern:/(^|[^&])\b(?:absolute|array|asm|begin|case|const|constructor|destructor|do|downto|else|end|file|for|function|goto|if|implementation|inherited|inline|interface|label|nil|object|of|operator|packed|procedure|program|record|reintroduce|repeat|self|set|string|then|to|type|unit|until|uses|var|while|with)\b/i,lookbehind:!0},{pattern:/(^|[^&])\b(?:dispose|exit|false|new|true)\b/i,lookbehind:!0},{pattern:/(^|[^&])\b(?:class|dispinterface|except|exports|finalization|finally|initialization|inline|library|on|out|packed|property|raise|resourcestring|threadvar|try)\b/i,lookbehind:!0},{pattern:/(^|[^&])\b(?:absolute|abstract|alias|assembler|bitpacked|break|cdecl|continue|cppdecl|cvar|default|deprecated|dynamic|enumerator|experimental|export|external|far|far16|forward|generic|helper|implements|index|interrupt|iochecks|local|message|name|near|nodefault|noreturn|nostackframe|oldfpccall|otherwise|overload|override|pascal|platform|private|protected|public|published|read|register|reintroduce|result|safecall|saveregisters|softfloat|specialize|static|stdcall|stored|strict|unaligned|unimplemented|varargs|virtual|write)\b/i,lookbehind:!0}],number:[/[+-]?(?:[&%]\d+|\$[a-f\d]+)/i,/([+-]|\b)\d+(?:\.\d+)?(?:e[+-]?\d+)?/i],operator:[/\.\.|\*\*|:=|<[<=>]?|>[>=]?|[+\-*\/]=?|[@^=]/i,{pattern:/(^|[^&])\b(?:and|as|div|exclude|in|include|is|mod|not|or|shl|shr|xor)\b/,lookbehind:!0}],punctuation:/\(\.|\.\)|[()\[\]:;,.]/};
+Prism.languages.perl={comment:[{pattern:/(^\s*)=\w+[\s\S]*?=cut.*/m,lookbehind:!0},{pattern:/(^|[^\\$])#.*/,lookbehind:!0}],string:[/\b(?:q|qq|qx|qw)\s*([^a-zA-Z0-9\s\{\(\[<])(?:[^\\]|\\[\s\S])*?\1/,/\b(?:q|qq|qx|qw)\s+([a-zA-Z0-9])(?:[^\\]|\\[\s\S])*?\1/,/\b(?:q|qq|qx|qw)\s*\((?:[^()\\]|\\[\s\S])*\)/,/\b(?:q|qq|qx|qw)\s*\{(?:[^{}\\]|\\[\s\S])*\}/,/\b(?:q|qq|qx|qw)\s*\[(?:[^[\]\\]|\\[\s\S])*\]/,/\b(?:q|qq|qx|qw)\s*<(?:[^<>\\]|\\[\s\S])*>/,/("|`)(?:[^\\]|\\[\s\S])*?\1/,/'(?:[^'\\\r\n]|\\.)*'/],regex:[/\b(?:m|qr)\s*([^a-zA-Z0-9\s\{\(\[<])(?:[^\\]|\\[\s\S])*?\1[msixpodualngc]*/,/\b(?:m|qr)\s+([a-zA-Z0-9])(?:[^\\]|\\.)*?\1[msixpodualngc]*/,/\b(?:m|qr)\s*\((?:[^()\\]|\\[\s\S])*\)[msixpodualngc]*/,/\b(?:m|qr)\s*\{(?:[^{}\\]|\\[\s\S])*\}[msixpodualngc]*/,/\b(?:m|qr)\s*\[(?:[^[\]\\]|\\[\s\S])*\][msixpodualngc]*/,/\b(?:m|qr)\s*<(?:[^<>\\]|\\[\s\S])*>[msixpodualngc]*/,{pattern:/(^|[^-]\b)(?:s|tr|y)\s*([^a-zA-Z0-9\s\{\(\[<])(?:[^\\]|\\[\s\S])*?\2(?:[^\\]|\\[\s\S])*?\2[msixpodualngcer]*/,lookbehind:!0},{pattern:/(^|[^-]\b)(?:s|tr|y)\s+([a-zA-Z0-9])(?:[^\\]|\\[\s\S])*?\2(?:[^\\]|\\[\s\S])*?\2[msixpodualngcer]*/,lookbehind:!0},{pattern:/(^|[^-]\b)(?:s|tr|y)\s*\((?:[^()\\]|\\[\s\S])*\)\s*\((?:[^()\\]|\\[\s\S])*\)[msixpodualngcer]*/,lookbehind:!0},{pattern:/(^|[^-]\b)(?:s|tr|y)\s*\{(?:[^{}\\]|\\[\s\S])*\}\s*\{(?:[^{}\\]|\\[\s\S])*\}[msixpodualngcer]*/,lookbehind:!0},{pattern:/(^|[^-]\b)(?:s|tr|y)\s*\[(?:[^[\]\\]|\\[\s\S])*\]\s*\[(?:[^[\]\\]|\\[\s\S])*\][msixpodualngcer]*/,lookbehind:!0},{pattern:/(^|[^-]\b)(?:s|tr|y)\s*<(?:[^<>\\]|\\[\s\S])*>\s*<(?:[^<>\\]|\\[\s\S])*>[msixpodualngcer]*/,lookbehind:!0},/\/(?:[^\/\\\r\n]|\\.)*\/[msixpodualngc]*(?=\s*(?:$|[\r\n,.;})&|\-+*~<>!?^]|(lt|gt|le|ge|eq|ne|cmp|not|and|or|xor|x)\b))/],variable:[/[&*$@%]\{\^[A-Z]+\}/,/[&*$@%]\^[A-Z_]/,/[&*$@%]#?(?=\{)/,/[&*$@%]#?((::)*'?(?!\d)[\w$]+)+(::)*/i,/[&*$@%]\d+/,/(?!%=)[$@%][!"#$%&'()*+,\-.\/:;<=>?@[\\\]^_`{|}~]/],filehandle:{pattern:/<(?![<=])\S*>|\b_\b/,alias:"symbol"},vstring:{pattern:/v\d+(\.\d+)*|\d+(\.\d+){2,}/,alias:"string"},"function":{pattern:/sub [a-z0-9_]+/i,inside:{keyword:/sub/}},keyword:/\b(any|break|continue|default|delete|die|do|else|elsif|eval|for|foreach|given|goto|if|last|local|my|next|our|package|print|redo|require|say|state|sub|switch|undef|unless|until|use|when|while)\b/,number:/\b-?(0x[\dA-Fa-f](_?[\dA-Fa-f])*|0b[01](_?[01])*|(\d(_?\d)*)?\.?\d(_?\d)*([Ee][+-]?\d+)?)\b/,operator:/-[rwxoRWXOezsfdlpSbctugkTBMAC]\b|\+[+=]?|-[-=>]?|\*\*?=?|\/\/?=?|=[=~>]?|~[~=]?|\|\|?=?|&&?=?|<(?:=>?|<=?)?|>>?=?|![~=]?|[%^]=?|\.(?:=|\.\.?)?|[\\?]|\bx(?:=|\b)|\b(lt|gt|le|ge|eq|ne|cmp|not|and|or|xor)\b/,punctuation:/[{}[\];(),:]/};
+Prism.languages.php=Prism.languages.extend("clike",{keyword:/\b(and|or|xor|array|as|break|case|cfunction|class|const|continue|declare|default|die|do|else|elseif|enddeclare|endfor|endforeach|endif|endswitch|endwhile|extends|for|foreach|function|include|include_once|global|if|new|return|static|switch|use|require|require_once|var|while|abstract|interface|public|implements|private|protected|parent|throw|null|echo|print|trait|namespace|final|yield|goto|instanceof|finally|try|catch)\b/i,constant:/\b[A-Z0-9_]{2,}\b/,comment:{pattern:/(^|[^\\])(?:\/\*[\w\W]*?\*\/|\/\/.*)/,lookbehind:!0,greedy:!0}}),Prism.languages.insertBefore("php","class-name",{"shell-comment":{pattern:/(^|[^\\])#.*/,lookbehind:!0,alias:"comment"}}),Prism.languages.insertBefore("php","keyword",{delimiter:/\?>|<\?(?:php)?/i,variable:/\$\w+\b/i,"package":{pattern:/(\\|namespace\s+|use\s+)[\w\\]+/,lookbehind:!0,inside:{punctuation:/\\/}}}),Prism.languages.insertBefore("php","operator",{property:{pattern:/(->)[\w]+/,lookbehind:!0}}),Prism.languages.markup&&(Prism.hooks.add("before-highlight",function(e){"php"===e.language&&(e.tokenStack=[],e.code=e.code.replace(/(?:<\?php|<\?)[\w\W]*?(?:\?>)/gi,function(n){return e.tokenStack.push(n),"{{{PHP"+e.tokenStack.length+"}}}"}))}),Prism.hooks.add("after-highlight",function(e){if("php"===e.language){for(var n,a=0;n=e.tokenStack[a];a++)e.highlightedCode=e.highlightedCode.replace("{{{PHP"+(a+1)+"}}}",Prism.highlight(n,e.grammar,"php").replace(/\$/g,"$$$$"));e.element.innerHTML=e.highlightedCode}}),Prism.hooks.add("wrap",function(e){"php"===e.language&&"markup"===e.type&&(e.content=e.content.replace(/(\{\{\{PHP[0-9]+\}\}\})/g,'<span class="token php">$1</span>'))}),Prism.languages.insertBefore("php","comment",{markup:{pattern:/<[^?]\/?(.*?)>/,inside:Prism.languages.markup},php:/\{\{\{PHP[0-9]+\}\}\}/}));
+Prism.languages.insertBefore("php","variable",{"this":/\$this\b/,global:/\$(?:_(?:SERVER|GET|POST|FILES|REQUEST|SESSION|ENV|COOKIE)|GLOBALS|HTTP_RAW_POST_DATA|argc|argv|php_errormsg|http_response_header)/,scope:{pattern:/\b[\w\\]+::/,inside:{keyword:/(static|self|parent)/,punctuation:/(::|\\)/}}});
+Prism.languages.properties={comment:/^[ \t]*[#!].*$/m,"attr-value":{pattern:/(^[ \t]*(?:\\(?:\r\n|[\s\S])|[^\\\s:=])+?(?: *[=:] *| ))(?:\\(?:\r\n|[\s\S])|.)+/m,lookbehind:!0},"attr-name":/^[ \t]*(?:\\(?:\r\n|[\s\S])|[^\\\s:=])+?(?= *[ =:]| )/m,punctuation:/[=:]/};
+Prism.languages.python={"triple-quoted-string":{pattern:/"""[\s\S]+?"""|'''[\s\S]+?'''/,alias:"string"},comment:{pattern:/(^|[^\\])#.*/,lookbehind:!0},string:{pattern:/("|')(?:\\\\|\\?[^\\\r\n])*?\1/,greedy:!0},"function":{pattern:/((?:^|\s)def[ \t]+)[a-zA-Z_][a-zA-Z0-9_]*(?=\()/g,lookbehind:!0},"class-name":{pattern:/(\bclass\s+)[a-z0-9_]+/i,lookbehind:!0},keyword:/\b(?:as|assert|async|await|break|class|continue|def|del|elif|else|except|exec|finally|for|from|global|if|import|in|is|lambda|pass|print|raise|return|try|while|with|yield)\b/,"boolean":/\b(?:True|False)\b/,number:/\b-?(?:0[bo])?(?:(?:\d|0x[\da-f])[\da-f]*\.?\d*|\.\d+)(?:e[+-]?\d+)?j?\b/i,operator:/[-+%=]=?|!=|\*\*?=?|\/\/?=?|<[<=>]?|>[=>]?|[&|^~]|\b(?:or|and|not)\b/,punctuation:/[{}[\];(),.:]/};
+Prism.languages.r={comment:/#.*/,string:/(['"])(?:\\?.)*?\1/,"percent-operator":{pattern:/%[^%\s]*%/,alias:"operator"},"boolean":/\b(?:TRUE|FALSE)\b/,ellipsis:/\.\.(?:\.|\d+)/,number:[/\b(?:NaN|Inf)\b/,/\b(?:0x[\dA-Fa-f]+(?:\.\d*)?|\d*\.?\d+)(?:[EePp][+-]?\d+)?[iL]?\b/],keyword:/\b(?:if|else|repeat|while|function|for|in|next|break|NULL|NA|NA_integer_|NA_real_|NA_complex_|NA_character_)\b/,operator:/->?>?|<(?:=|<?-)?|[>=!]=?|::?|&&?|\|\|?|[+*\/^$@~]/,punctuation:/[(){}\[\],;]/};
+!function(a){var e=a.util.clone(a.languages.javascript);a.languages.jsx=a.languages.extend("markup",e),a.languages.jsx.tag.pattern=/<\/?[\w\.:-]+\s*(?:\s+[\w\.:-]+(?:=(?:("|')(\\?[\w\W])*?\1|[^\s'">=]+|(\{[\w\W]*?\})))?\s*)*\/?>/i,a.languages.jsx.tag.inside["attr-value"].pattern=/=[^\{](?:('|")[\w\W]*?(\1)|[^\s>]+)/i;var s=a.util.clone(a.languages.jsx);delete s.punctuation,s=a.languages.insertBefore("jsx","operator",{punctuation:/=(?={)|[{}[\];(),.:]/},{jsx:s}),a.languages.insertBefore("inside","attr-value",{script:{pattern:/=(\{(?:\{[^}]*\}|[^}])+\})/i,inside:s,alias:"language-javascript"}},a.languages.jsx.tag)}(Prism);
+Prism.languages.rust={comment:[{pattern:/(^|[^\\])\/\*[\w\W]*?\*\//,lookbehind:!0},{pattern:/(^|[^\\:])\/\/.*/,lookbehind:!0}],string:[/b?r(#*)"(?:\\?.)*?"\1/,/b?("|')(?:\\?.)*?\1/],keyword:/\b(?:abstract|alignof|as|be|box|break|const|continue|crate|do|else|enum|extern|false|final|fn|for|if|impl|in|let|loop|match|mod|move|mut|offsetof|once|override|priv|pub|pure|ref|return|sizeof|static|self|struct|super|true|trait|type|typeof|unsafe|unsized|use|virtual|where|while|yield)\b/,attribute:{pattern:/#!?\[.+?\]/,alias:"attr-name"},"function":[/[a-z0-9_]+(?=\s*\()/i,/[a-z0-9_]+!(?=\s*\(|\[)/i],"macro-rules":{pattern:/[a-z0-9_]+!/i,alias:"function"},number:/\b-?(?:0x[\dA-Fa-f](?:_?[\dA-Fa-f])*|0o[0-7](?:_?[0-7])*|0b[01](?:_?[01])*|(\d(_?\d)*)?\.?\d(_?\d)*([Ee][+-]?\d+)?)(?:_?(?:[iu](?:8|16|32|64)?|f32|f64))?\b/,"closure-params":{pattern:/\|[^|]*\|(?=\s*[{-])/,inside:{punctuation:/[\|:,]/,operator:/[&*]/}},punctuation:/[{}[\];(),:]|\.+|->/,operator:/[-+*\/%!^=]=?|@|&[&=]?|\|[|=]?|<<?=?|>>?=?/};
+Prism.languages.scala=Prism.languages.extend("java",{keyword:/<-|=>|\b(?:abstract|case|catch|class|def|do|else|extends|final|finally|for|forSome|if|implicit|import|lazy|match|new|null|object|override|package|private|protected|return|sealed|self|super|this|throw|trait|try|type|val|var|while|with|yield)\b/,string:[{pattern:/"""[\W\w]*?"""/,greedy:!0},{pattern:/("|')(?:\\\\|\\?[^\\\r\n])*?\1/,greedy:!0}],builtin:/\b(?:String|Int|Long|Short|Byte|Boolean|Double|Float|Char|Any|AnyRef|AnyVal|Unit|Nothing)\b/,number:/\b(?:0x[\da-f]*\.?[\da-f]+|\d*\.?\d+e?\d*[dfl]?)\b/i,symbol:/'[^\d\s\\]\w*/}),delete Prism.languages.scala["class-name"],delete Prism.languages.scala["function"];
+Prism.languages.scheme={comment:/;.*/,string:/"(?:[^"\\\r\n]|\\.)*?"|'[^('\s]*/,keyword:{pattern:/(\()(?:define(?:-syntax|-library|-values)?|(?:case-)?lambda|let(?:\*|rec)?(?:-values)?|else|if|cond|begin|delay(?:-force)?|parameterize|guard|set!|(?:quasi-)?quote|syntax-rules)/,lookbehind:!0},builtin:{pattern:/(\()(?:(?:cons|car|cdr|list|call-with-current-continuation|call\/cc|append|abs|apply|eval)\b|null\?|pair\?|boolean\?|eof-object\?|char\?|procedure\?|number\?|port\?|string\?|vector\?|symbol\?|bytevector\?)/,lookbehind:!0},number:{pattern:/(\s|\))[-+]?[0-9]*\.?[0-9]+(?:\s*[-+]\s*[0-9]*\.?[0-9]+i)?\b/,lookbehind:!0},"boolean":/#[tf]/,operator:{pattern:/(\()(?:[-+*%\/]|[<>]=?|=>?)/,lookbehind:!0},"function":{pattern:/(\()[^\s()]*(?=\s)/,lookbehind:!0},punctuation:/[()]/};
+Prism.languages.smalltalk={comment:/"(?:""|[^"])+"/,string:/'(?:''|[^'])+'/,symbol:/#[\da-z]+|#(?:-|([+\/\\*~<>=@%|&?!])\1?)|#(?=\()/i,"block-arguments":{pattern:/(\[\s*):[^\[|]*?\|/,lookbehind:!0,inside:{variable:/:[\da-z]+/i,punctuation:/\|/}},"temporary-variables":{pattern:/\|[^|]+\|/,inside:{variable:/[\da-z]+/i,punctuation:/\|/}},keyword:/\b(?:nil|true|false|self|super|new)\b/,character:{pattern:/\$./,alias:"string"},number:[/\d+r-?[\dA-Z]+(?:\.[\dA-Z]+)?(?:e-?\d+)?/,/(?:\B-|\b)\d+(?:\.\d+)?(?:e-?\d+)?/],operator:/[<=]=?|:=|~[~=]|\/\/?|\\\\|>[>=]?|[!^+\-*&|,@]/,punctuation:/[.;:?\[\](){}]/};
+Prism.languages.sql={comment:{pattern:/(^|[^\\])(?:\/\*[\w\W]*?\*\/|(?:--|\/\/|#).*)/,lookbehind:!0},string:{pattern:/(^|[^@\\])("|')(?:\\?[\s\S])*?\2/,lookbehind:!0},variable:/@[\w.$]+|@("|'|`)(?:\\?[\s\S])+?\1/,"function":/\b(?:COUNT|SUM|AVG|MIN|MAX|FIRST|LAST|UCASE|LCASE|MID|LEN|ROUND|NOW|FORMAT)(?=\s*\()/i,keyword:/\b(?:ACTION|ADD|AFTER|ALGORITHM|ALL|ALTER|ANALYZE|ANY|APPLY|AS|ASC|AUTHORIZATION|AUTO_INCREMENT|BACKUP|BDB|BEGIN|BERKELEYDB|BIGINT|BINARY|BIT|BLOB|BOOL|BOOLEAN|BREAK|BROWSE|BTREE|BULK|BY|CALL|CASCADED?|CASE|CHAIN|CHAR VARYING|CHARACTER (?:SET|VARYING)|CHARSET|CHECK|CHECKPOINT|CLOSE|CLUSTERED|COALESCE|COLLATE|COLUMN|COLUMNS|COMMENT|COMMIT|COMMITTED|COMPUTE|CONNECT|CONSISTENT|CONSTRAINT|CONTAINS|CONTAINSTABLE|CONTINUE|CONVERT|CREATE|CROSS|CURRENT(?:_DATE|_TIME|_TIMESTAMP|_USER)?|CURSOR|DATA(?:BASES?)?|DATE(?:TIME)?|DBCC|DEALLOCATE|DEC|DECIMAL|DECLARE|DEFAULT|DEFINER|DELAYED|DELETE|DELIMITER(?:S)?|DENY|DESC|DESCRIBE|DETERMINISTIC|DISABLE|DISCARD|DISK|DISTINCT|DISTINCTROW|DISTRIBUTED|DO|DOUBLE(?: PRECISION)?|DROP|DUMMY|DUMP(?:FILE)?|DUPLICATE KEY|ELSE|ENABLE|ENCLOSED BY|END|ENGINE|ENUM|ERRLVL|ERRORS|ESCAPE(?:D BY)?|EXCEPT|EXEC(?:UTE)?|EXISTS|EXIT|EXPLAIN|EXTENDED|FETCH|FIELDS|FILE|FILLFACTOR|FIRST|FIXED|FLOAT|FOLLOWING|FOR(?: EACH ROW)?|FORCE|FOREIGN|FREETEXT(?:TABLE)?|FROM|FULL|FUNCTION|GEOMETRY(?:COLLECTION)?|GLOBAL|GOTO|GRANT|GROUP|HANDLER|HASH|HAVING|HOLDLOCK|IDENTITY(?:_INSERT|COL)?|IF|IGNORE|IMPORT|INDEX|INFILE|INNER|INNODB|INOUT|INSERT|INT|INTEGER|INTERSECT|INTO|INVOKER|ISOLATION LEVEL|JOIN|KEYS?|KILL|LANGUAGE SQL|LAST|LEFT|LIMIT|LINENO|LINES|LINESTRING|LOAD|LOCAL|LOCK|LONG(?:BLOB|TEXT)|MATCH(?:ED)?|MEDIUM(?:BLOB|INT|TEXT)|MERGE|MIDDLEINT|MODIFIES SQL DATA|MODIFY|MULTI(?:LINESTRING|POINT|POLYGON)|NATIONAL(?: CHAR VARYING| CHARACTER(?: VARYING)?| VARCHAR)?|NATURAL|NCHAR(?: VARCHAR)?|NEXT|NO(?: SQL|CHECK|CYCLE)?|NONCLUSTERED|NULLIF|NUMERIC|OFF?|OFFSETS?|ON|OPEN(?:DATASOURCE|QUERY|ROWSET)?|OPTIMIZE|OPTION(?:ALLY)?|ORDER|OUT(?:ER|FILE)?|OVER|PARTIAL|PARTITION|PERCENT|PIVOT|PLAN|POINT|POLYGON|PRECEDING|PRECISION|PREV|PRIMARY|PRINT|PRIVILEGES|PROC(?:EDURE)?|PUBLIC|PURGE|QUICK|RAISERROR|READ(?:S SQL DATA|TEXT)?|REAL|RECONFIGURE|REFERENCES|RELEASE|RENAME|REPEATABLE|REPLICATION|REQUIRE|RESTORE|RESTRICT|RETURNS?|REVOKE|RIGHT|ROLLBACK|ROUTINE|ROW(?:COUNT|GUIDCOL|S)?|RTREE|RULE|SAVE(?:POINT)?|SCHEMA|SELECT|SERIAL(?:IZABLE)?|SESSION(?:_USER)?|SET(?:USER)?|SHARE MODE|SHOW|SHUTDOWN|SIMPLE|SMALLINT|SNAPSHOT|SOME|SONAME|START(?:ING BY)?|STATISTICS|STATUS|STRIPED|SYSTEM_USER|TABLES?|TABLESPACE|TEMP(?:ORARY|TABLE)?|TERMINATED BY|TEXT(?:SIZE)?|THEN|TIMESTAMP|TINY(?:BLOB|INT|TEXT)|TOP?|TRAN(?:SACTIONS?)?|TRIGGER|TRUNCATE|TSEQUAL|TYPES?|UNBOUNDED|UNCOMMITTED|UNDEFINED|UNION|UNIQUE|UNPIVOT|UPDATE(?:TEXT)?|USAGE|USE|USER|USING|VALUES?|VAR(?:BINARY|CHAR|CHARACTER|YING)|VIEW|WAITFOR|WARNINGS|WHEN|WHERE|WHILE|WITH(?: ROLLUP|IN)?|WORK|WRITE(?:TEXT)?)\b/i,"boolean":/\b(?:TRUE|FALSE|NULL)\b/i,number:/\b-?(?:0x)?\d*\.?[\da-f]+\b/,operator:/[-+*\/=%^~]|&&?|\|?\||!=?|<(?:=>?|<|>)?|>[>=]?|\b(?:AND|BETWEEN|IN|LIKE|NOT|OR|IS|DIV|REGEXP|RLIKE|SOUNDS LIKE|XOR)\b/i,punctuation:/[;[\]()`,.]/};
+Prism.languages.swift=Prism.languages.extend("clike",{string:{pattern:/("|')(\\(?:\((?:[^()]|\([^)]+\))+\)|\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1/,greedy:!0,inside:{interpolation:{pattern:/\\\((?:[^()]|\([^)]+\))+\)/,inside:{delimiter:{pattern:/^\\\(|\)$/,alias:"variable"}}}}},keyword:/\b(as|associativity|break|case|catch|class|continue|convenience|default|defer|deinit|didSet|do|dynamic(?:Type)?|else|enum|extension|fallthrough|final|for|func|get|guard|if|import|in|infix|init|inout|internal|is|lazy|left|let|mutating|new|none|nonmutating|operator|optional|override|postfix|precedence|prefix|private|Protocol|public|repeat|required|rethrows|return|right|safe|self|Self|set|static|struct|subscript|super|switch|throws?|try|Type|typealias|unowned|unsafe|var|weak|where|while|willSet|__(?:COLUMN__|FILE__|FUNCTION__|LINE__))\b/,number:/\b([\d_]+(\.[\de_]+)?|0x[a-f0-9_]+(\.[a-f0-9p_]+)?|0b[01_]+|0o[0-7_]+)\b/i,constant:/\b(nil|[A-Z_]{2,}|k[A-Z][A-Za-z_]+)\b/,atrule:/@\b(IB(?:Outlet|Designable|Action|Inspectable)|class_protocol|exported|noreturn|NS(?:Copying|Managed)|objc|UIApplicationMain|auto_closure)\b/,builtin:/\b([A-Z]\S+|abs|advance|alignof(?:Value)?|assert|contains|count(?:Elements)?|debugPrint(?:ln)?|distance|drop(?:First|Last)|dump|enumerate|equal|filter|find|first|getVaList|indices|isEmpty|join|last|lexicographicalCompare|map|max(?:Element)?|min(?:Element)?|numericCast|overlaps|partition|print(?:ln)?|reduce|reflect|reverse|sizeof(?:Value)?|sort(?:ed)?|split|startsWith|stride(?:of(?:Value)?)?|suffix|swap|toDebugString|toString|transcode|underestimateCount|unsafeBitCast|with(?:ExtendedLifetime|Unsafe(?:MutablePointers?|Pointers?)|VaList))\b/}),Prism.languages.swift.string.inside.interpolation.inside.rest=Prism.util.clone(Prism.languages.swift);
+Prism.languages.vim={string:/"(?:[^"\\\r\n]|\\.)*"|'(?:[^'\r\n]|'')*'/,comment:/".*/,"function":/\w+(?=\()/,keyword:/\b(?:ab|abbreviate|abc|abclear|abo|aboveleft|al|all|arga|argadd|argd|argdelete|argdo|arge|argedit|argg|argglobal|argl|arglocal|ar|args|argu|argument|as|ascii|bad|badd|ba|ball|bd|bdelete|be|bel|belowright|bf|bfirst|bl|blast|bm|bmodified|bn|bnext|bN|bNext|bo|botright|bp|bprevious|brea|break|breaka|breakadd|breakd|breakdel|breakl|breaklist|br|brewind|bro|browse|bufdo|b|buffer|buffers|bun|bunload|bw|bwipeout|ca|cabbrev|cabc|cabclear|caddb|caddbuffer|cad|caddexpr|caddf|caddfile|cal|call|cat|catch|cb|cbuffer|cc|ccl|cclose|cd|ce|center|cex|cexpr|cf|cfile|cfir|cfirst|cgetb|cgetbuffer|cgete|cgetexpr|cg|cgetfile|c|change|changes|chd|chdir|che|checkpath|checkt|checktime|cla|clast|cl|clist|clo|close|cmapc|cmapclear|cnew|cnewer|cn|cnext|cN|cNext|cnf|cnfile|cNfcNfile|cnorea|cnoreabbrev|col|colder|colo|colorscheme|comc|comclear|comp|compiler|conf|confirm|con|continue|cope|copen|co|copy|cpf|cpfile|cp|cprevious|cq|cquit|cr|crewind|cuna|cunabbrev|cu|cunmap|cw|cwindow|debugg|debuggreedy|delc|delcommand|d|delete|delf|delfunction|delm|delmarks|diffg|diffget|diffoff|diffpatch|diffpu|diffput|diffsplit|diffthis|diffu|diffupdate|dig|digraphs|di|display|dj|djump|dl|dlist|dr|drop|ds|dsearch|dsp|dsplit|earlier|echoe|echoerr|echom|echomsg|echon|e|edit|el|else|elsei|elseif|em|emenu|endfo|endfor|endf|endfunction|endfun|en|endif|endt|endtry|endw|endwhile|ene|enew|ex|exi|exit|exu|exusage|f|file|files|filetype|fina|finally|fin|find|fini|finish|fir|first|fix|fixdel|fo|fold|foldc|foldclose|folddoc|folddoclosed|foldd|folddoopen|foldo|foldopen|for|fu|fun|function|go|goto|gr|grep|grepa|grepadd|ha|hardcopy|h|help|helpf|helpfind|helpg|helpgrep|helpt|helptags|hid|hide|his|history|ia|iabbrev|iabc|iabclear|if|ij|ijump|il|ilist|imapc|imapclear|in|inorea|inoreabbrev|isearch|isp|isplit|iuna|iunabbrev|iu|iunmap|j|join|ju|jumps|k|keepalt|keepj|keepjumps|kee|keepmarks|laddb|laddbuffer|lad|laddexpr|laddf|laddfile|lan|language|la|last|later|lb|lbuffer|lc|lcd|lch|lchdir|lcl|lclose|let|left|lefta|leftabove|lex|lexpr|lf|lfile|lfir|lfirst|lgetb|lgetbuffer|lgete|lgetexpr|lg|lgetfile|lgr|lgrep|lgrepa|lgrepadd|lh|lhelpgrep|l|list|ll|lla|llast|lli|llist|lmak|lmake|lm|lmap|lmapc|lmapclear|lnew|lnewer|lne|lnext|lN|lNext|lnf|lnfile|lNf|lNfile|ln|lnoremap|lo|loadview|loc|lockmarks|lockv|lockvar|lol|lolder|lop|lopen|lpf|lpfile|lp|lprevious|lr|lrewind|ls|lt|ltag|lu|lunmap|lv|lvimgrep|lvimgrepa|lvimgrepadd|lw|lwindow|mak|make|ma|mark|marks|mat|match|menut|menutranslate|mk|mkexrc|mks|mksession|mksp|mkspell|mkvie|mkview|mkv|mkvimrc|mod|mode|m|move|mzf|mzfile|mz|mzscheme|nbkey|new|n|next|N|Next|nmapc|nmapclear|noh|nohlsearch|norea|noreabbrev|nu|number|nun|nunmap|omapc|omapclear|on|only|o|open|opt|options|ou|ounmap|pc|pclose|ped|pedit|pe|perl|perld|perldo|po|pop|popu|popu|popup|pp|ppop|pre|preserve|prev|previous|p|print|P|Print|profd|profdel|prof|profile|promptf|promptfind|promptr|promptrepl|ps|psearch|pta|ptag|ptf|ptfirst|ptj|ptjump|ptl|ptlast|ptn|ptnext|ptN|ptNext|ptp|ptprevious|ptr|ptrewind|pts|ptselect|pu|put|pw|pwd|pyf|pyfile|py|python|qa|qall|q|quit|quita|quitall|r|read|rec|recover|redi|redir|red|redo|redr|redraw|redraws|redrawstatus|reg|registers|res|resize|ret|retab|retu|return|rew|rewind|ri|right|rightb|rightbelow|rub|ruby|rubyd|rubydo|rubyf|rubyfile|ru|runtime|rv|rviminfo|sal|sall|san|sandbox|sa|sargument|sav|saveas|sba|sball|sbf|sbfirst|sbl|sblast|sbm|sbmodified|sbn|sbnext|sbN|sbNext|sbp|sbprevious|sbr|sbrewind|sb|sbuffer|scripte|scriptencoding|scrip|scriptnames|se|set|setf|setfiletype|setg|setglobal|setl|setlocal|sf|sfind|sfir|sfirst|sh|shell|sign|sil|silent|sim|simalt|sla|slast|sl|sleep|sm|smagic|sm|smap|smapc|smapclear|sme|smenu|sn|snext|sN|sNext|sni|sniff|sno|snomagic|snor|snoremap|snoreme|snoremenu|sor|sort|so|source|spelld|spelldump|spe|spellgood|spelli|spellinfo|spellr|spellrepall|spellu|spellundo|spellw|spellwrong|sp|split|spr|sprevious|sre|srewind|sta|stag|startg|startgreplace|star|startinsert|startr|startreplace|stj|stjump|st|stop|stopi|stopinsert|sts|stselect|sun|sunhide|sunm|sunmap|sus|suspend|sv|sview|syncbind|t|tab|tabc|tabclose|tabd|tabdo|tabe|tabedit|tabf|tabfind|tabfir|tabfirst|tabl|tablast|tabm|tabmove|tabnew|tabn|tabnext|tabN|tabNext|tabo|tabonly|tabp|tabprevious|tabr|tabrewind|tabs|ta|tag|tags|tc|tcl|tcld|tcldo|tclf|tclfile|te|tearoff|tf|tfirst|th|throw|tj|tjump|tl|tlast|tm|tm|tmenu|tn|tnext|tN|tNext|to|topleft|tp|tprevious|tr|trewind|try|ts|tselect|tu|tu|tunmenu|una|unabbreviate|u|undo|undoj|undojoin|undol|undolist|unh|unhide|unlet|unlo|unlockvar|unm|unmap|up|update|verb|verbose|ve|version|vert|vertical|vie|view|vim|vimgrep|vimgrepa|vimgrepadd|vi|visual|viu|viusage|vmapc|vmapclear|vne|vnew|vs|vsplit|vu|vunmap|wa|wall|wh|while|winc|wincmd|windo|winp|winpos|win|winsize|wn|wnext|wN|wNext|wp|wprevious|wq|wqa|wqall|w|write|ws|wsverb|wv|wviminfo|X|xa|xall|x|xit|xm|xmap|xmapc|xmapclear|xme|xmenu|XMLent|XMLns|xn|xnoremap|xnoreme|xnoremen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+Prism.languages.wiki=Prism.languages.extend("markup",{"block-comment":{pattern:/(^|[^\\])\/\*[\w\W]*?\*\//,lookbehind:!0,alias:"comment"},heading:{pattern:/^(=+).+?\1/m,inside:{punctuation:/^=+|=+$/,important:/.+/}},emphasis:{pattern:/('{2,5}).+?\1/,inside:{"bold italic":{pattern:/(''''').+?(?=\1)/,lookbehind:!0},bold:{pattern:/(''')[^'](?:.*?[^'])?(?=\1)/,lookbehind:!0},italic:{pattern:/('')[^'](?:.*?[^'])?(?=\1)/,lookbehind:!0},punctuation:/^''+|''+$/}},hr:{pattern:/^-{4,}/m,alias:"punctuation"},url:[/ISBN +(?:97[89][ -]?)?(?:\d[ -]?){9}[\dx]\b|(?:RFC|PMID) +\d+/i,/\[\[.+?\]\]|\[.+?\]/],variable:[/__[A-Z]+__/,/\{{3}.+?\}{3}/,/\{\{.+?}}/],symbol:[/^#redirect/im,/~{3,5}/],"table-tag":{pattern:/((?:^|[|!])[|!])[^|\r\n]+\|(?!\|)/m,lookbehind:!0,inside:{"table-bar":{pattern:/\|$/,alias:"punctuation"},rest:Prism.languages.markup.tag.inside}},punctuation:/^(?:\{\||\|\}|\|-|[*#:;!|])|\|\||!!/m}),Prism.languages.insertBefore("wiki","tag",{nowiki:{pattern:/<(nowiki|pre|source)\b[\w\W]*?>[\w\W]*?<\/\1>/i,inside:{tag:{pattern:/<(?:nowiki|pre|source)\b[\w\W]*?>|<\/(?:nowiki|pre|source)>/i,inside:Prism.languages.markup.tag.inside}}}});
+
+Prism.languages.markup={comment:/<!--[\w\W]*?-->/,prolog:/<\?[\w\W]+?\?>/,doctype:/<!DOCTYPE[\w\W]+?>/,cdata:/<!\[CDATA\[[\w\W]*?]]>/i,tag:{pattern:/<\/?(?!\d)[^\s>\/=.$<]+(?:\s+[^\s>\/=]+(?:=(?:("|')(?:\\\1|\\?(?!\1)[\w\W])*\1|[^\s'">=]+))?)*\s*\/?>/i,inside:{tag:{pattern:/^<\/?[^\s>\/]+/i,inside:{punctuation:/^<\/?/,namespace:/^[^\s>\/:]+:/}},"attr-value":{pattern:/=(?:('|")[\w\W]*?(\1)|[^\s>]+)/i,inside:{punctuation:/[=>"']/}},punctuation:/\/?>/,"attr-name":{pattern:/[^\s>\/]+/,inside:{namespace:/^[^\s>\/:]+:/}}}},entity:/&#?[\da-z]{1,8};/i},Prism.hooks.add("wrap",function(a){"entity"===a.type&&(a.attributes.title=a.content.replace(/&amp;/,"&"))}),Prism.languages.xml=Prism.languages.markup,Prism.languages.html=Prism.languages.markup,Prism.languages.mathml=Prism.languages.markup,Prism.languages.svg=Prism.languages.markup;
+Prism.languages.css={comment:/\/\*[\w\W]*?\*\//,atrule:{pattern:/@[\w-]+?.*?(;|(?=\s*\{))/i,inside:{rule:/@[\w-]+/}},url:/url\((?:(["'])(\\(?:\r\n|[\w\W])|(?!\1)[^\\\r\n])*\1|.*?)\)/i,selector:/[^\{\}\s][^\{\};]*?(?=\s*\{)/,string:/("|')(\\(?:\r\n|[\w\W])|(?!\1)[^\\\r\n])*\1/,property:/(\b|\B)[\w-]+(?=\s*:)/i,important:/\B!important\b/i,"function":/[-a-z0-9]+(?=\()/i,punctuation:/[(){};:]/},Prism.languages.css.atrule.inside.rest=Prism.util.clone(Prism.languages.css),Prism.languages.markup&&(Prism.languages.insertBefore("markup","tag",{style:{pattern:/(<style[\w\W]*?>)[\w\W]*?(?=<\/style>)/i,lookbehind:!0,inside:Prism.languages.css,alias:"language-css"}}),Prism.languages.insertBefore("inside","attr-value",{"style-attr":{pattern:/\s*style=("|').*?\1/i,inside:{"attr-name":{pattern:/^\s*style/i,inside:Prism.languages.markup.tag.inside},punctuation:/^\s*=\s*['"]|['"]\s*$/,"attr-value":{pattern:/.+/i,inside:Prism.languages.css}},alias:"language-css"}},Prism.languages.markup.tag));
+Prism.languages.clike={comment:[{pattern:/(^|[^\\])\/\*[\w\W]*?\*\//,lookbehind:!0},{pattern:/(^|[^\\:])\/\/.*/,lookbehind:!0}],string:/(["'])(\\(?:\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1/,"class-name":{pattern:/((?:\b(?:class|interface|extends|implements|trait|instanceof|new)\s+)|(?:catch\s+\())[a-z0-9_\.\\]+/i,lookbehind:!0,inside:{punctuation:/(\.|\\)/}},keyword:/\b(if|else|while|do|for|return|in|instanceof|function|new|try|throw|catch|finally|null|break|continue)\b/,"boolean":/\b(true|false)\b/,"function":/[a-z0-9_]+(?=\()/i,number:/\b-?(?:0x[\da-f]+|\d*\.?\d+(?:e[+-]?\d+)?)\b/i,operator:/--?|\+\+?|!=?=?|<=?|>=?|==?=?|&&?|\|\|?|\?|\*|\/|~|\^|%/,punctuation:/[{}[\];(),.:]/};
+Prism.languages.javascript=Prism.languages.extend("clike",{keyword:/\b(as|async|await|break|case|catch|class|const|continue|debugger|default|delete|do|else|enum|export|extends|finally|for|from|function|get|if|implements|import|in|instanceof|interface|let|new|null|of|package|private|protected|public|return|set|static|super|switch|this|throw|try|typeof|var|void|while|with|yield)\b/,number:/\b-?(0x[\dA-Fa-f]+|0b[01]+|0o[0-7]+|\d*\.?\d+([Ee][+-]?\d+)?|NaN|Infinity)\b/,"function":/[_$a-zA-Z\xA0-\uFFFF][_$a-zA-Z0-9\xA0-\uFFFF]*(?=\()/i}),Prism.languages.insertBefore("javascript","keyword",{regex:{pattern:/(^|[^\/])\/(?!\/)(\[.+?]|\\.|[^\/\\\r\n])+\/[gimyu]{0,5}(?=\s*($|[\r\n,.;})]))/,lookbehind:!0}}),Prism.languages.insertBefore("javascript","class-name",{"template-string":{pattern:/`(?:\\`|\\?[^`])*`/,inside:{interpolation:{pattern:/\$\{[^}]+\}/,inside:{"interpolation-punctuation":{pattern:/^\$\{|\}$/,alias:"punctuation"},rest:Prism.languages.javascript}},string:/[\s\S]+/}}}),Prism.languages.markup&&Prism.languages.insertBefore("markup","tag",{script:{pattern:/(<script[\w\W]*?>)[\w\W]*?(?=<\/script>)/i,lookbehind:!0,inside:Prism.languages.javascript,alias:"language-javascript"}}),Prism.languages.js=Prism.languages.javascript;
+!function(e){var t={variable:[{pattern:/\$?\(\([\w\W]+?\)\)/,inside:{variable:[{pattern:/(^\$\(\([\w\W]+)\)\)/,lookbehind:!0},/^\$\(\(/],number:/\b-?(?:0x[\dA-Fa-f]+|\d*\.?\d+(?:[Ee]-?\d+)?)\b/,operator:/--?|-=|\+\+?|\+=|!=?|~|\*\*?|\*=|\/=?|%=?|<<=?|>>=?|<=?|>=?|==?|&&?|&=|\^=?|\|\|?|\|=|\?|:/,punctuation:/\(\(?|\)\)?|,|;/}},{pattern:/\$\([^)]+\)|`[^`]+`/,inside:{variable:/^\$\(|^`|\)$|`$/}},/\$(?:[a-z0-9_#\?\*!@]+|\{[^}]+\})/i]};e.languages.bash={shebang:{pattern:/^#!\s*\/bin\/bash|^#!\s*\/bin\/sh/,alias:"important"},comment:{pattern:/(^|[^"{\\])#.*/,lookbehind:!0},string:[{pattern:/((?:^|[^<])<<\s*)(?:"|')?(\w+?)(?:"|')?\s*\r?\n(?:[\s\S])*?\r?\n\2/g,lookbehind:!0,inside:t},{pattern:/("|')(?:\\?[\s\S])*?\1/g,inside:t}],variable:t.variable,"function":{pattern:/(^|\s|;|\||&)(?:alias|apropos|apt-get|aptitude|aspell|awk|basename|bash|bc|bg|builtin|bzip2|cal|cat|cd|cfdisk|chgrp|chmod|chown|chroot|chkconfig|cksum|clear|cmp|comm|command|cp|cron|crontab|csplit|cut|date|dc|dd|ddrescue|df|diff|diff3|dig|dir|dircolors|dirname|dirs|dmesg|du|egrep|eject|enable|env|ethtool|eval|exec|expand|expect|export|expr|fdformat|fdisk|fg|fgrep|file|find|fmt|fold|format|free|fsck|ftp|fuser|gawk|getopts|git|grep|groupadd|groupdel|groupmod|groups|gzip|hash|head|help|hg|history|hostname|htop|iconv|id|ifconfig|ifdown|ifup|import|install|jobs|join|kill|killall|less|link|ln|locate|logname|logout|look|lpc|lpr|lprint|lprintd|lprintq|lprm|ls|lsof|make|man|mkdir|mkfifo|mkisofs|mknod|more|most|mount|mtools|mtr|mv|mmv|nano|netstat|nice|nl|nohup|notify-send|nslookup|open|op|passwd|paste|pathchk|ping|pkill|popd|pr|printcap|printenv|printf|ps|pushd|pv|pwd|quota|quotacheck|quotactl|ram|rar|rcp|read|readarray|readonly|reboot|rename|renice|remsync|rev|rm|rmdir|rsync|screen|scp|sdiff|sed|seq|service|sftp|shift|shopt|shutdown|sleep|slocate|sort|source|split|ssh|stat|strace|su|sudo|sum|suspend|sync|tail|tar|tee|test|time|timeout|times|touch|top|traceroute|trap|tr|tsort|tty|type|ulimit|umask|umount|unalias|uname|unexpand|uniq|units|unrar|unshar|uptime|useradd|userdel|usermod|users|uuencode|uudecode|v|vdir|vi|vmstat|wait|watch|wc|wget|whereis|which|who|whoami|write|xargs|xdg-open|yes|zip)(?=$|\s|;|\||&)/,lookbehind:!0},keyword:{pattern:/(^|\s|;|\||&)(?:let|:|\.|if|then|else|elif|fi|for|break|continue|while|in|case|function|select|do|done|until|echo|exit|return|set|declare)(?=$|\s|;|\||&)/,lookbehind:!0},"boolean":{pattern:/(^|\s|;|\||&)(?:true|false)(?=$|\s|;|\||&)/,lookbehind:!0},operator:/&&?|\|\|?|==?|!=?|<<<?|>>|<=?|>=?|=~/,punctuation:/\$?\(\(?|\)\)?|\.\.|[{}[\];]/};var a=t.variable[1].inside;a["function"]=e.languages.bash["function"],a.keyword=e.languages.bash.keyword,a.boolean=e.languages.bash.boolean,a.operator=e.languages.bash.operator,a.punctuation=e.languages.bash.punctuation}(Prism);
+Prism.languages.css.selector={pattern:/[^\{\}\s][^\{\}]*(?=\s*\{)/,inside:{"pseudo-element":/:(?:after|before|first-letter|first-line|selection)|::[-\w]+/,"pseudo-class":/:[-\w]+(?:\(.*\))?/,"class":/\.[-:\.\w]+/,id:/#[-:\.\w]+/}},Prism.languages.insertBefore("css","function",{hexcode:/#[\da-f]{3,6}/i,entity:/\\[\da-f]{1,8}/i,number:/[\d%\.]+/});
+Prism.languages.git={comment:/^#.*/m,deleted:/^[-–].*/m,inserted:/^\+.*/m,string:/("|')(\\?.)*?\1/m,command:{pattern:/^.*\$ git .*$/m,inside:{parameter:/\s(--|-)\w+/m}},coord:/^@@.*@@$/m,commit_sha1:/^commit \w{40}$/m};
+!function(e){var a=/\{\{\{[\w\W]+?\}\}\}|\{\{[\w\W]+?\}\}/g;e.languages.handlebars=e.languages.extend("markup",{handlebars:{pattern:a,inside:{delimiter:{pattern:/^\{\{\{?|\}\}\}?$/i,alias:"punctuation"},string:/(["'])(\\?.)*?\1/,number:/\b-?(0x[\dA-Fa-f]+|\d*\.?\d+([Ee][+-]?\d+)?)\b/,"boolean":/\b(true|false)\b/,block:{pattern:/^(\s*~?\s*)[#\/]\S+?(?=\s*~?\s*$|\s)/i,lookbehind:!0,alias:"keyword"},brackets:{pattern:/\[[^\]]+\]/,inside:{punctuation:/\[|\]/,variable:/[\w\W]+/}},punctuation:/[!"#%&'()*+,.\/;<=>@\[\\\]^`{|}~]/,variable:/[^!"#%&'()*+,.\/;<=>@\[\\\]^`{|}~\s]+/}}}),e.languages.insertBefore("handlebars","tag",{"handlebars-comment":{pattern:/\{\{![\w\W]*?\}\}/,alias:["handlebars","comment"]}}),e.hooks.add("before-highlight",function(e){"handlebars"===e.language&&(e.tokenStack=[],e.backupCode=e.code,e.code=e.code.replace(a,function(a){return e.tokenStack.push(a),"___HANDLEBARS"+e.tokenStack.length+"___"}))}),e.hooks.add("before-insert",function(e){"handlebars"===e.language&&(e.code=e.backupCode,delete e.backupCode)}),e.hooks.add("after-highlight",function(a){if("handlebars"===a.language){for(var n,t=0;n=a.tokenStack[t];t++)a.highlightedCode=a.highlightedCode.replace("___HANDLEBARS"+(t+1)+"___",e.highlight(n,a.grammar,"handlebars").replace(/\$/g,"$$$$"));a.element.innerHTML=a.highlightedCode}})}(Prism);
+Prism.languages.http={"request-line":{pattern:/^(POST|GET|PUT|DELETE|OPTIONS|PATCH|TRACE|CONNECT)\b\shttps?:\/\/\S+\sHTTP\/[0-9.]+/m,inside:{property:/^(POST|GET|PUT|DELETE|OPTIONS|PATCH|TRACE|CONNECT)\b/,"attr-name":/:\w+/}},"response-status":{pattern:/^HTTP\/1.[01] [0-9]+.*/m,inside:{property:{pattern:/(^HTTP\/1.[01] )[0-9]+.*/i,lookbehind:!0}}},"header-name":{pattern:/^[\w-]+:(?=.)/m,alias:"keyword"}};var httpLanguages={"application/json":Prism.languages.javascript,"application/xml":Prism.languages.markup,"text/xml":Prism.languages.markup,"text/html":Prism.languages.markup};for(var contentType in httpLanguages)if(httpLanguages[contentType]){var options={};options[contentType]={pattern:new RegExp("(content-type:\\s*"+contentType+"[\\w\\W]*?)(?:\\r?\\n|\\r){2}[\\w\\W]*","i"),lookbehind:!0,inside:{rest:httpLanguages[contentType]}},Prism.languages.insertBefore("http","header-name",options)};
+Prism.languages.makefile={comment:{pattern:/(^|[^\\])#(?:\\(?:\r\n|[\s\S])|.)*/,lookbehind:!0},string:/(["'])(?:\\(?:\r\n|[\s\S])|(?!\1)[^\\\r\n])*\1/,builtin:/\.[A-Z][^:#=\s]+(?=\s*:(?!=))/,symbol:{pattern:/^[^:=\r\n]+(?=\s*:(?!=))/m,inside:{variable:/\$+(?:[^(){}:#=\s]+|(?=[({]))/}},variable:/\$+(?:[^(){}:#=\s]+|\([@*%<^+?][DF]\)|(?=[({]))/,keyword:[/-include\b|\b(?:define|else|endef|endif|export|ifn?def|ifn?eq|include|override|private|sinclude|undefine|unexport|vpath)\b/,{pattern:/(\()(?:addsuffix|abspath|and|basename|call|dir|error|eval|file|filter(?:-out)?|findstring|firstword|flavor|foreach|guile|if|info|join|lastword|load|notdir|or|origin|patsubst|realpath|shell|sort|strip|subst|suffix|value|warning|wildcard|word(?:s|list)?)(?=[ \t])/,lookbehind:!0}],operator:/(?:::|[?:+!])?=|[|@]/,punctuation:/[:;(){}]/};
+Prism.languages.markdown=Prism.languages.extend("markup",{}),Prism.languages.insertBefore("markdown","prolog",{blockquote:{pattern:/^>(?:[\t ]*>)*/m,alias:"punctuation"},code:[{pattern:/^(?: {4}|\t).+/m,alias:"keyword"},{pattern:/``.+?``|`[^`\n]+`/,alias:"keyword"}],title:[{pattern:/\w+.*(?:\r?\n|\r)(?:==+|--+)/,alias:"important",inside:{punctuation:/==+$|--+$/}},{pattern:/(^\s*)#+.+/m,lookbehind:!0,alias:"important",inside:{punctuation:/^#+|#+$/}}],hr:{pattern:/(^\s*)([*-])([\t ]*\2){2,}(?=\s*$)/m,lookbehind:!0,alias:"punctuation"},list:{pattern:/(^\s*)(?:[*+-]|\d+\.)(?=[\t ].)/m,lookbehind:!0,alias:"punctuation"},"url-reference":{pattern:/!?\[[^\]]+\]:[\t ]+(?:\S+|<(?:\\.|[^>\\])+>)(?:[\t ]+(?:"(?:\\.|[^"\\])*"|'(?:\\.|[^'\\])*'|\((?:\\.|[^)\\])*\)))?/,inside:{variable:{pattern:/^(!?\[)[^\]]+/,lookbehind:!0},string:/(?:"(?:\\.|[^"\\])*"|'(?:\\.|[^'\\])*'|\((?:\\.|[^)\\])*\))$/,punctuation:/^[\[\]!:]|[<>]/},alias:"url"},bold:{pattern:/(^|[^\\])(\*\*|__)(?:(?:\r?\n|\r)(?!\r?\n|\r)|.)+?\2/,lookbehind:!0,inside:{punctuation:/^\*\*|^__|\*\*$|__$/}},italic:{pattern:/(^|[^\\])([*_])(?:(?:\r?\n|\r)(?!\r?\n|\r)|.)+?\2/,lookbehind:!0,inside:{punctuation:/^[*_]|[*_]$/}},url:{pattern:/!?\[[^\]]+\](?:\([^\s)]+(?:[\t ]+"(?:\\.|[^"\\])*")?\)| ?\[[^\]\n]*\])/,inside:{variable:{pattern:/(!?\[)[^\]]+(?=\]$)/,lookbehind:!0},string:{pattern:/"(?:\\.|[^"\\])*"(?=\)$)/}}}}),Prism.languages.markdown.bold.inside.url=Prism.util.clone(Prism.languages.markdown.url),Prism.languages.markdown.italic.inside.url=Prism.util.clone(Prism.languages.markdown.url),Prism.languages.markdown.bold.inside.italic=Prism.util.clone(Prism.languages.markdown.italic),Prism.languages.markdown.italic.inside.bold=Prism.util.clone(Prism.languages.markdown.bold);
+Prism.languages.nginx=Prism.languages.extend("clike",{comment:{pattern:/(^|[^"{\\])#.*/,lookbehind:!0},keyword:/\b(?:CONTENT_|DOCUMENT_|GATEWAY_|HTTP_|HTTPS|if_not_empty|PATH_|QUERY_|REDIRECT_|REMOTE_|REQUEST_|SCGI|SCRIPT_|SERVER_|http|server|events|location|include|accept_mutex|accept_mutex_delay|access_log|add_after_body|add_before_body|add_header|addition_types|aio|alias|allow|ancient_browser|ancient_browser_value|auth|auth_basic|auth_basic_user_file|auth_http|auth_http_header|auth_http_timeout|autoindex|autoindex_exact_size|autoindex_localtime|break|charset|charset_map|charset_types|chunked_transfer_encoding|client_body_buffer_size|client_body_in_file_only|client_body_in_single_buffer|client_body_temp_path|client_body_timeout|client_header_buffer_size|client_header_timeout|client_max_body_size|connection_pool_size|create_full_put_path|daemon|dav_access|dav_methods|debug_connection|debug_points|default_type|deny|devpoll_changes|devpoll_events|directio|directio_alignment|disable_symlinks|empty_gif|env|epoll_events|error_log|error_page|expires|fastcgi_buffer_size|fastcgi_buffers|fastcgi_busy_buffers_size|fastcgi_cache|fastcgi_cache_bypass|fastcgi_cache_key|fastcgi_cache_lock|fastcgi_cache_lock_timeout|fastcgi_cache_methods|fastcgi_cache_min_uses|fastcgi_cache_path|fastcgi_cache_purge|fastcgi_cache_use_stale|fastcgi_cache_valid|fastcgi_connect_timeout|fastcgi_hide_header|fastcgi_ignore_client_abort|fastcgi_ignore_headers|fastcgi_index|fastcgi_intercept_errors|fastcgi_keep_conn|fastcgi_max_temp_file_size|fastcgi_next_upstream|fastcgi_no_cache|fastcgi_param|fastcgi_pass|fastcgi_pass_header|fastcgi_read_timeout|fastcgi_redirect_errors|fastcgi_send_timeout|fastcgi_split_path_info|fastcgi_store|fastcgi_store_access|fastcgi_temp_file_write_size|fastcgi_temp_path|flv|geo|geoip_city|geoip_country|google_perftools_profiles|gzip|gzip_buffers|gzip_comp_level|gzip_disable|gzip_http_version|gzip_min_length|gzip_proxied|gzip_static|gzip_types|gzip_vary|if|if_modified_since|ignore_invalid_headers|image_filter|image_filter_buffer|image_filter_jpeg_quality|image_filter_sharpen|image_filter_transparency|imap_capabilities|imap_client_buffer|include|index|internal|ip_hash|keepalive|keepalive_disable|keepalive_requests|keepalive_timeout|kqueue_changes|kqueue_events|large_client_header_buffers|limit_conn|limit_conn_log_level|limit_conn_zone|limit_except|limit_rate|limit_rate_after|limit_req|limit_req_log_level|limit_req_zone|limit_zone|lingering_close|lingering_time|lingering_timeout|listen|location|lock_file|log_format|log_format_combined|log_not_found|log_subrequest|map|map_hash_bucket_size|map_hash_max_size|master_process|max_ranges|memcached_buffer_size|memcached_connect_timeout|memcached_next_upstream|memcached_pass|memcached_read_timeout|memcached_send_timeout|merge_slashes|min_delete_depth|modern_browser|modern_browser_value|mp4|mp4_buffer_size|mp4_max_buffer_size|msie_padding|msie_refresh|multi_accept|open_file_cache|open_file_cache_errors|open_file_cache_min_uses|open_file_cache_valid|open_log_file_cache|optimize_server_names|override_charset|pcre_jit|perl|perl_modules|perl_require|perl_set|pid|pop3_auth|pop3_capabilities|port_in_redirect|post_action|postpone_output|protocol|proxy|proxy_buffer|proxy_buffer_size|proxy_buffering|proxy_buffers|proxy_busy_buffers_size|proxy_cache|proxy_cache_bypass|proxy_cache_key|proxy_cache_lock|proxy_cache_lock_timeout|proxy_cache_methods|proxy_cache_min_uses|proxy_cache_path|proxy_cache_use_stale|proxy_cache_valid|proxy_connect_timeout|proxy_cookie_domain|proxy_cookie_path|proxy_headers_hash_bucket_size|proxy_headers_hash_max_size|proxy_hide_header|proxy_http_version|proxy_ignore_client_abort|proxy_ignore_headers|proxy_intercept_errors|proxy_max_temp_file_size|proxy_method|proxy_next_upstream|proxy_no_cache|proxy_pass|proxy_pass_error_message|proxy_pass_header|proxy_pass_request_body|proxy_pass_request_headers|proxy_read_timeout|proxy_redirect|proxy_redirect_errors|proxy_send_lowat|proxy_send_timeout|proxy_set_body|proxy_set_header|proxy_ssl_session_reuse|proxy_store|proxy_store_access|proxy_temp_file_write_size|proxy_temp_path|proxy_timeout|proxy_upstream_fail_timeout|proxy_upstream_max_fails|random_index|read_ahead|real_ip_header|recursive_error_pages|request_pool_size|reset_timedout_connection|resolver|resolver_timeout|return|rewrite|root|rtsig_overflow_events|rtsig_overflow_test|rtsig_overflow_threshold|rtsig_signo|satisfy|satisfy_any|secure_link_secret|send_lowat|send_timeout|sendfile|sendfile_max_chunk|server|server_name|server_name_in_redirect|server_names_hash_bucket_size|server_names_hash_max_size|server_tokens|set|set_real_ip_from|smtp_auth|smtp_capabilities|so_keepalive|source_charset|split_clients|ssi|ssi_silent_errors|ssi_types|ssi_value_length|ssl|ssl_certificate|ssl_certificate_key|ssl_ciphers|ssl_client_certificate|ssl_crl|ssl_dhparam|ssl_engine|ssl_prefer_server_ciphers|ssl_protocols|ssl_session_cache|ssl_session_timeout|ssl_verify_client|ssl_verify_depth|starttls|stub_status|sub_filter|sub_filter_once|sub_filter_types|tcp_nodelay|tcp_nopush|timeout|timer_resolution|try_files|types|types_hash_bucket_size|types_hash_max_size|underscores_in_headers|uninitialized_variable_warn|upstream|use|user|userid|userid_domain|userid_expires|userid_name|userid_p3p|userid_path|userid_service|valid_referers|variables_hash_bucket_size|variables_hash_max_size|worker_connections|worker_cpu_affinity|worker_priority|worker_processes|worker_rlimit_core|worker_rlimit_nofile|worker_rlimit_sigpending|working_directory|xclient|xml_entities|xslt_entities|xslt_stylesheet|xslt_types)\b/i}),Prism.languages.insertBefore("nginx","keyword",{variable:/\$[a-z_]+/i});
+Prism.languages.php=Prism.languages.extend("clike",{keyword:/\b(and|or|xor|array|as|break|case|cfunction|class|const|continue|declare|default|die|do|else|elseif|enddeclare|endfor|endforeach|endif|endswitch|endwhile|extends|for|foreach|function|include|include_once|global|if|new|return|static|switch|use|require|require_once|var|while|abstract|interface|public|implements|private|protected|parent|throw|null|echo|print|trait|namespace|final|yield|goto|instanceof|finally|try|catch)\b/i,constant:/\b[A-Z0-9_]{2,}\b/,comment:{pattern:/(^|[^\\])(?:\/\*[\w\W]*?\*\/|\/\/.*)/,lookbehind:!0}}),Prism.languages.insertBefore("php","class-name",{"shell-comment":{pattern:/(^|[^\\])#.*/,lookbehind:!0,alias:"comment"}}),Prism.languages.insertBefore("php","keyword",{delimiter:/\?>|<\?(?:php)?/i,variable:/\$\w+\b/i,"package":{pattern:/(\\|namespace\s+|use\s+)[\w\\]+/,lookbehind:!0,inside:{punctuation:/\\/}}}),Prism.languages.insertBefore("php","operator",{property:{pattern:/(->)[\w]+/,lookbehind:!0}}),Prism.languages.markup&&(Prism.hooks.add("before-highlight",function(e){"php"===e.language&&(e.tokenStack=[],e.backupCode=e.code,e.code=e.code.replace(/(?:<\?php|<\?)[\w\W]*?(?:\?>)/gi,function(a){return e.tokenStack.push(a),"{{{PHP"+e.tokenStack.length+"}}}"}))}),Prism.hooks.add("before-insert",function(e){"php"===e.language&&(e.code=e.backupCode,delete e.backupCode)}),Prism.hooks.add("after-highlight",function(e){if("php"===e.language){for(var a,n=0;a=e.tokenStack[n];n++)e.highlightedCode=e.highlightedCode.replace("{{{PHP"+(n+1)+"}}}",Prism.highlight(a,e.grammar,"php").replace(/\$/g,"$$$$"));e.element.innerHTML=e.highlightedCode}}),Prism.hooks.add("wrap",function(e){"php"===e.language&&"markup"===e.type&&(e.content=e.content.replace(/(\{\{\{PHP[0-9]+\}\}\})/g,'<span class="token php">$1</span>'))}),Prism.languages.insertBefore("php","comment",{markup:{pattern:/<[^?]\/?(.*?)>/,inside:Prism.languages.markup},php:/\{\{\{PHP[0-9]+\}\}\}/}));
+!function(e){e.languages.sass=e.languages.extend("css",{comment:{pattern:/^([ \t]*)\/[\/*].*(?:(?:\r?\n|\r)\1[ \t]+.+)*/m,lookbehind:!0}}),e.languages.insertBefore("sass","atrule",{"atrule-line":{pattern:/^(?:[ \t]*)[@+=].+/m,inside:{atrule:/(?:@[\w-]+|[+=])/m}}}),delete e.languages.sass.atrule;var a=/((\$[-_\w]+)|(#\{\$[-_\w]+\}))/i,t=[/[+*\/%]|[=!]=|<=?|>=?|\b(?:and|or|not)\b/,{pattern:/(\s+)-(?=\s)/,lookbehind:!0}];e.languages.insertBefore("sass","property",{"variable-line":{pattern:/^[ \t]*\$.+/m,inside:{punctuation:/:/,variable:a,operator:t}},"property-line":{pattern:/^[ \t]*(?:[^:\s]+ *:.*|:[^:\s]+.*)/m,inside:{property:[/[^:\s]+(?=\s*:)/,{pattern:/(:)[^:\s]+/,lookbehind:!0}],punctuation:/:/,variable:a,operator:t,important:e.languages.sass.important}}}),delete e.languages.sass.property,delete e.languages.sass.important,delete e.languages.sass.selector,e.languages.insertBefore("sass","punctuation",{selector:{pattern:/([ \t]*)\S(?:,?[^,\r\n]+)*(?:,(?:\r?\n|\r)\1[ \t]+\S(?:,?[^,\r\n]+)*)*/,lookbehind:!0}})}(Prism);
+Prism.languages.scss=Prism.languages.extend("css",{comment:{pattern:/(^|[^\\])(?:\/\*[\w\W]*?\*\/|\/\/.*)/,lookbehind:!0},atrule:{pattern:/@[\w-]+(?:\([^()]+\)|[^(])*?(?=\s+[{;])/,inside:{rule:/@[\w-]+/}},url:/(?:[-a-z]+-)*url(?=\()/i,selector:{pattern:/(?=\S)[^@;\{\}\(\)]?([^@;\{\}\(\)]|&|#\{\$[-_\w]+\})+(?=\s*\{(\}|\s|[^\}]+(:|\{)[^\}]+))/m,inside:{placeholder:/%[-_\w]+/}}}),Prism.languages.insertBefore("scss","atrule",{keyword:[/@(?:if|else(?: if)?|for|each|while|import|extend|debug|warn|mixin|include|function|return|content)/i,{pattern:/( +)(?:from|through)(?= )/,lookbehind:!0}]}),Prism.languages.insertBefore("scss","property",{variable:/\$[-_\w]+|#\{\$[-_\w]+\}/}),Prism.languages.insertBefore("scss","function",{placeholder:{pattern:/%[-_\w]+/,alias:"selector"},statement:/\B!(?:default|optional)\b/i,"boolean":/\b(?:true|false)\b/,"null":/\bnull\b/,operator:{pattern:/(\s)(?:[-+*\/%]|[=!]=|<=?|>=?|and|or|not)(?=\s)/,lookbehind:!0}}),Prism.languages.scss.atrule.inside.rest=Prism.util.clone(Prism.languages.scss);
+!function(){function e(e,t){return Array.prototype.slice.call((t||document).querySelectorAll(e))}function t(e,t){return t=" "+t+" ",(" "+e.className+" ").replace(/[\n\t]/g," ").indexOf(t)>-1}function n(e,n,i){for(var o,a=n.replace(/\s+/g,"").split(","),l=+e.getAttribute("data-line-offset")||0,d=r()?parseInt:parseFloat,c=d(getComputedStyle(e).lineHeight),s=0;o=a[s++];){o=o.split("-");var u=+o[0],m=+o[1]||u,h=document.createElement("div");h.textContent=Array(m-u+2).join(" \n"),h.className=(i||"")+" line-highlight",t(e,"line-numbers")||(h.setAttribute("data-start",u),m>u&&h.setAttribute("data-end",m)),h.style.top=(u-l-1)*c+"px",t(e,"line-numbers")?e.appendChild(h):(e.querySelector("code")||e).appendChild(h)}}function i(){var t=location.hash.slice(1);e(".temporary.line-highlight").forEach(function(e){e.parentNode.removeChild(e)});var i=(t.match(/\.([\d,-]+)$/)||[,""])[1];if(i&&!document.getElementById(t)){var r=t.slice(0,t.lastIndexOf(".")),o=document.getElementById(r);o&&(o.hasAttribute("data-line")||o.setAttribute("data-line",""),n(o,i,"temporary "),document.querySelector(".temporary.line-highlight").scrollIntoView())}}if("undefined"!=typeof self&&self.Prism&&self.document&&document.querySelector){var r=function(){var e;return function(){if("undefined"==typeof e){var t=document.createElement("div");t.style.fontSize="13px",t.style.lineHeight="1.5",t.style.padding=0,t.style.border=0,t.innerHTML="&nbsp;<br />&nbsp;",document.body.appendChild(t),e=38===t.offsetHeight,document.body.removeChild(t)}return e}}(),o=0;Prism.hooks.add("complete",function(t){var r=t.element.parentNode,a=r&&r.getAttribute("data-line");r&&a&&/pre/i.test(r.nodeName)&&(clearTimeout(o),e(".line-highlight",r).forEach(function(e){e.parentNode.removeChild(e)}),n(r,a),o=setTimeout(i,1))}),window.addEventListener&&window.addEventListener("hashchange",i)}}();
+
+
+!function(){"undefined"!=typeof self&&self.Prism&&self.document&&Prism.hooks.add("complete",function(e){if(e.code){var t=e.element.parentNode,s=/\s*\bline-numbers\b\s*/;if(t&&/pre/i.test(t.nodeName)&&(s.test(t.className)||s.test(e.element.className))&&!e.element.querySelector(".line-numbers-rows")){s.test(e.element.className)&&(e.element.className=e.element.className.replace(s,"")),s.test(t.className)||(t.className+=" line-numbers");var n,a=e.code.match(/\n(?!$)/g),l=a?a.length+1:1,m=new Array(l+1);m=m.join("<span></span>"),n=document.createElement("span"),n.className="line-numbers-rows",n.innerHTML=m,t.hasAttribute("data-start")&&(t.style.counterReset="linenumber "+(parseInt(t.getAttribute("data-start"),10)-1)),e.element.appendChild(n)}}})}();
+!function(){if(("undefined"==typeof self||self.Prism)&&("undefined"==typeof global||global.Prism)){var i=/\b([a-z]{3,7}:\/\/|tel:)[\w\-+%~\/.:#=?&amp;]+/,n=/\b\S+@[\w.]+[a-z]{2}/,e=/\[([^\]]+)]\(([^)]+)\)/,t=["comment","url","attr-value","string"];Prism.hooks.add("before-highlight",function(a){a.grammar&&!a.grammar["url-link"]&&(Prism.languages.DFS(a.grammar,function(a,r,l){t.indexOf(l)>-1&&"Array"!==Prism.util.type(r)&&(r.pattern||(r=this[a]={pattern:r}),r.inside=r.inside||{},"comment"==l&&(r.inside["md-link"]=e),"attr-value"==l?Prism.languages.insertBefore("inside","punctuation",{"url-link":i},r):r.inside["url-link"]=i,r.inside["email-link"]=n)}),a.grammar["url-link"]=i,a.grammar["email-link"]=n)}),Prism.hooks.add("wrap",function(i){if(/-link$/.test(i.type)){i.tag="a";var n=i.content;if("email-link"==i.type&&0!=n.indexOf("mailto:"))n="mailto:"+n;else if("md-link"==i.type){var t=i.content.match(e);n=t[2],i.content=t[1]}i.attributes.href=n}})}}();
+!function(){"undefined"!=typeof self&&self.Prism&&self.document&&document.querySelector&&(self.Prism.fileHighlight=function(){var e={js:"javascript",html:"markup",svg:"markup",xml:"markup",py:"python",rb:"ruby",ps1:"powershell",psm1:"powershell"};Array.prototype.forEach&&Array.prototype.slice.call(document.querySelectorAll("pre[data-src]")).forEach(function(t){for(var r,a=t.getAttribute("data-src"),s=t,n=/\blang(?:uage)?-(?!\*)(\w+)\b/i;s&&!n.test(s.className);)s=s.parentNode;if(s&&(r=(t.className.match(n)||[,""])[1]),!r){var o=(a.match(/\.(\w+)$/)||[,""])[1];r=e[o]||o}var l=document.createElement("code");l.className="language-"+r,t.textContent="",l.textContent="Loading…",t.appendChild(l);var i=new XMLHttpRequest;i.open("GET",a,!0),i.onreadystatechange=function(){4==i.readyState&&(i.status<400&&i.responseText?(l.textContent=i.responseText,Prism.highlightElement(l)):l.textContent=i.status>=400?"✖ Error "+i.status+" while fetching file: "+i.statusText:"✖ Error: File does not exist or is empty")},i.send(null)})},self.Prism.fileHighlight())}();
+!function(){if("undefined"!=typeof self&&self.Prism&&self.document){var e={css:"CSS",clike:"C-like",javascript:"JavaScript",abap:"ABAP",actionscript:"ActionScript",apacheconf:"Apache Configuration",apl:"APL",applescript:"AppleScript",asciidoc:"AsciiDoc",aspnet:"ASP.NET (C#)",autoit:"AutoIt",autohotkey:"AutoHotkey",basic:"BASIC",csharp:"C#",cpp:"C++",coffeescript:"CoffeeScript","css-extras":"CSS Extras",fsharp:"F#",glsl:"GLSL",http:"HTTP",inform7:"Inform 7",latex:"LaTeX",lolcode:"LOLCODE",matlab:"MATLAB",mel:"MEL",nasm:"NASM",nginx:"nginx",nsis:"NSIS",objectivec:"Objective-C",ocaml:"OCaml",parigp:"PARI/GP",php:"PHP","php-extras":"PHP Extras",powershell:"PowerShell",jsx:"React JSX",rest:"reST (reStructuredText)",sas:"SAS",sass:"Sass (Sass)",scss:"Sass (Scss)",sql:"SQL",typescript:"TypeScript",vhdl:"VHDL",vim:"vim",wiki:"Wiki markup",yaml:"YAML"};Prism.hooks.add("before-highlight",function(a){var s=a.element.parentNode;if(s&&/pre/i.test(s.nodeName)){var t=e[a.language]||a.language.substring(0,1).toUpperCase()+a.language.substring(1);s.setAttribute("data-language",t);var i,r,l=s.previousSibling;l&&/\s*\bprism-show-language\b\s*/.test(l.className)&&l.firstChild&&/\s*\bprism-show-language-label\b\s*/.test(l.firstChild.className)?(r=l.firstChild,r.getAttribute("data-language")!==t&&(r.setAttribute("data-language",t),r.innerHTML=t)):(i=document.createElement("div"),r=document.createElement("div"),r.className="prism-show-language-label",r.setAttribute("data-language",t),r.innerHTML=t,i.className="prism-show-language",i.appendChild(r),s.parentNode.insertBefore(i,s))}})}}();
diff --git a/website/themes/book/static/svg/calendar.svg b/website/themes/book/static/svg/calendar.svg
new file mode 100644
index 000000000..f84811207
--- /dev/null
+++ b/website/themes/book/static/svg/calendar.svg
@@ -0,0 +1 @@
+<svg xmlns="http://www.w3.org/2000/svg" width="24" height="24" viewBox="0 0 24 24"><path d="M20 3h-1V1h-2v2H7V1H5v2H4c-1.1 0-2 .9-2 2v16c0 1.1.9 2 2 2h16c1.1 0 2-.9 2-2V5c0-1.1-.9-2-2-2zm0 18H4V8h16v13z"/><path fill="none" d="M0 0h24v24H0z"/></svg>
\ No newline at end of file
diff --git a/website/themes/book/static/svg/edit.svg b/website/themes/book/static/svg/edit.svg
new file mode 100644
index 000000000..5b54e693b
--- /dev/null
+++ b/website/themes/book/static/svg/edit.svg
@@ -0,0 +1 @@
+<svg xmlns="http://www.w3.org/2000/svg" width="24" height="24" viewBox="0 0 24 24"><path d="M3 17.25V21h3.75L17.81 9.94l-3.75-3.75L3 17.25zM20.71 7.04c.39-.39.39-1.02 0-1.41l-2.34-2.34c-.39-.39-1.02-.39-1.41 0l-1.83 1.83 3.75 3.75 1.83-1.83z"/><path d="M0 0h24v24H0z" fill="none"/></svg>
\ No newline at end of file
diff --git a/website/themes/book/static/svg/menu.svg b/website/themes/book/static/svg/menu.svg
new file mode 100644
index 000000000..770b19236
--- /dev/null
+++ b/website/themes/book/static/svg/menu.svg
@@ -0,0 +1 @@
+<svg xmlns="http://www.w3.org/2000/svg" width="24" height="24" viewBox="0 0 24 24"><path fill="none" d="M0 0h24v24H0V0z"/><path d="M3 18h18v-2H3v2zm0-5h18v-2H3v2zm0-7v2h18V6H3z"/></svg>
\ No newline at end of file
diff --git a/website/themes/book/static/svg/toc.svg b/website/themes/book/static/svg/toc.svg
new file mode 100644
index 000000000..1889904ec
--- /dev/null
+++ b/website/themes/book/static/svg/toc.svg
@@ -0,0 +1 @@
+<svg xmlns="http://www.w3.org/2000/svg" width="24" height="24" viewBox="0 0 24 24"><path d="M3 9h14V7H3v2zm0 4h14v-2H3v2zm0 4h14v-2H3v2zm16 0h2v-2h-2v2zm0-10v2h2V7h-2zm0 6h2v-2h-2v2z"/><path d="M0 0h24v24H0z" fill="none"/></svg>
\ No newline at end of file
diff --git a/website/themes/book/static/svg/translate.svg b/website/themes/book/static/svg/translate.svg
new file mode 100644
index 000000000..a1bbe1666
--- /dev/null
+++ b/website/themes/book/static/svg/translate.svg
@@ -0,0 +1 @@
+<svg xmlns="http://www.w3.org/2000/svg" width="24" height="24" viewBox="0 0 24 24"><path d="M0 0h24v24H0z" fill="none"/><path d="M12.87 15.07l-2.54-2.51.03-.03c1.74-1.94 2.98-4.17 3.71-6.53H17V4h-7V2H8v2H1v1.99h11.17C11.5 7.92 10.44 9.75 9 11.35 8.07 10.32 7.3 9.19 6.69 8h-2c.73 1.63 1.73 3.17 2.98 4.56l-5.09 5.02L4 19l5-5 3.11 3.11.76-2.04zM18.5 10h-2L12 22h2l1.12-3h4.75L21 22h2l-4.5-12zm-2.62 7l1.62-4.33L19.12 17h-3.24z"/></svg>
\ No newline at end of file
diff --git a/website/themes/book/theme.toml b/website/themes/book/theme.toml
new file mode 100644
index 000000000..7f3adec55
--- /dev/null
+++ b/website/themes/book/theme.toml
@@ -0,0 +1,15 @@
+# theme.toml template for a Hugo theme
+# See https://github.com/gohugoio/hugoThemes#themetoml for an example
+
+name = "Book"
+license = "MIT"
+licenselink = "https://github.com/alex-shpak/hugo-book/blob/master/LICENSE"
+description = "Hugo documentation theme as simple as plain book"
+homepage = "https://github.com/alex-shpak/hugo-book"
+tags = ["responsive", "clean", "documentation", "docs", "flexbox", "search", "mobile", "multilingual", "disqus"]
+features = []
+min_version = "0.68"
+
+[author]
+  name = "Alex Shpak"
+  homepage = "https://github.com/alex-shpak/"