Merge pull request halfrost#37 from halfrost/add_hugo

Add hugo

Author: halfrost

leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/448. Find All Numbers Disappeared in an Array.go

@@ -0,0 +1,19 @@
+package leetcode
+
+func findDisappearedNumbers(nums []int) []int {
+	res := []int{}
+	for _, v := range nums {
+		if v < 0 {
+			v = -v
+		}
+		if nums[v-1] > 0 {
+			nums[v-1] = -nums[v-1]
+		}
+	}
+	for i, v := range nums {
+		if v > 0 {
+			res = append(res, i+1)
+		}
+	}
+	return res
+}

leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/448. Find All Numbers Disappeared in an Array_test.go

@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question448 struct {
+	para448
+	ans448
+}
+
+// para 是参数
+// one 代表第一个参数
+type para448 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans448 struct {
+	one []int
+}
+
+func Test_Problem448(t *testing.T) {
+
+	qs := []question448{
+
+		question448{
+			para448{[]int{4, 3, 2, 7, 8, 2, 3, 1}},
+			ans448{[]int{5, 6}},
+		},
+
+		question448{
+			para448{[]int{4, 3, 2, 10, 9, 2, 3, 1, 1, 1, 1}},
+			ans448{[]int{5, 6, 7, 8, 11}},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 448------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans448, q.para448
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findDisappearedNumbers(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0448.Find-All-Numbers-Disappeared-in-an-Array/README.md

@@ -0,0 +1,53 @@
+# [448. Find All Numbers Disappeared in an Array](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/)
+
+
+## 题目
+
+Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
+
+Find all the elements of [1, n] inclusive that do not appear in this array.
+
+Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.
+
+**Example**:
+
+```
+Input:
+[4,3,2,7,8,2,3,1]
+
+Output:
+[5,6]
+```
+
+## 题目大意
+
+给定一个范围在  1 ≤ a[i] ≤ n ( n = 数组大小 ) 的 整型数组，数组中的元素一些出现了两次，另一些只出现一次。找到所有在 [1, n] 范围之间没有出现在数组中的数字。你能在不使用额外空间且时间复杂度为 O(n) 的情况下完成这个任务吗? 你可以假定返回的数组不算在额外空间内。
+
+
+
+## 解题思路
+
+- 找出 [1,n] 范围内没有出现在数组中的数字。要求不使用额外空间，并且时间复杂度为 O(n)。
+- 要求不能使用额外的空间，那么只能想办法在原有数组上进行修改，并且这个修改是可还原的。时间复杂度也只能允许我们一层循环。只要循环一次能标记出已经出现过的数字，这道题就可以按要求解答出来。这里笔者的标记方法是把 |nums[i]|-1 索引位置的元素标记为负数。即 nums[| nums[i] |- 1] * -1。这里需要注意的是，nums[i] 需要加绝对值，因为它可能被之前的数置为负数了，需要还原一下。最后再遍历一次数组，若当前数组元素 nums[i] 为负数，说明我们在数组中存在数字 i+1。把结果输出到最终数组里即可。
+
+## 代码
+
+```go
+func findDisappearedNumbers(nums []int) []int {
+	res := []int{}
+	for _, v := range nums {
+		if v < 0 {
+			v = -v
+		}
+		if nums[v-1] > 0 {
+			nums[v-1] = -nums[v-1]
+		}
+	}
+	for i, v := range nums {
+		if v > 0 {
+			res = append(res, i+1)
+		}
+	}
+	return res
+}
+```

leetcode/0485.Max-Consecutive-Ones/485. Max Consecutive Ones.go

@@ -0,0 +1,16 @@
+package leetcode
+
+func findMaxConsecutiveOnes(nums []int) int {
+	maxCount, currentCount := 0, 0
+	for _, v := range nums {
+		if v == 1 {
+			currentCount++
+		} else {
+			currentCount = 0
+		}
+		if currentCount > maxCount {
+			maxCount = currentCount
+		}
+	}
+	return maxCount
+}

leetcode/0485.Max-Consecutive-Ones/485. Max Consecutive Ones_test.go

@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question485 struct {
+	para485
+	ans485
+}
+
+// para 是参数
+// one 代表第一个参数
+type para485 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans485 struct {
+	one int
+}
+
+func Test_Problem485(t *testing.T) {
+
+	qs := []question485{
+
+		question485{
+			para485{[]int{1, 1, 0, 1, 1, 1}},
+			ans485{3},
+		},
+
+		question485{
+			para485{[]int{1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1}},
+			ans485{4},
+		},
+		// 如需多个测试，可以复制上方元素。
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 485------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans485, q.para485
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findMaxConsecutiveOnes(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0485.Max-Consecutive-Ones/README.md

@@ -0,0 +1,55 @@
+# [485. Max Consecutive Ones](https://leetcode.com/problems/max-consecutive-ones/)
+
+
+## 题目
+
+Given a binary array, find the maximum number of consecutive 1s in this array.
+
+**Example 1**:
+
+```
+Input: [1,1,0,1,1,1]
+Output: 3
+Explanation: The first two digits or the last three digits are consecutive 1s.
+    The maximum number of consecutive 1s is 3.
+```
+
+**Note**:
+
+- The input array will only contain `0` and `1`.
+- The length of input array is a positive integer and will not exceed 10,000
+
+
+## 题目大意
+
+给定一个二进制数组， 计算其中最大连续1的个数。
+
+注意：
+
+- 输入的数组只包含 0 和 1。
+- 输入数组的长度是正整数，且不超过 10,000。
+
+
+## 解题思路
+
+- 给定一个二进制数组， 计算其中最大连续1的个数。
+- 简单题。扫一遍数组，累计 1 的个数，动态维护最大的计数，最终输出即可。
+
+## 代码
+
+```go
+func findMaxConsecutiveOnes(nums []int) int {
+	maxCount, currentCount := 0, 0
+	for _, v := range nums {
+		if v == 1 {
+			currentCount++
+		} else {
+			currentCount = 0
+		}
+		if currentCount > maxCount {
+			maxCount = currentCount
+		}
+	}
+	return maxCount
+}
+```

leetcode/0661.Image-Smoother/661. Image Smoother.go

@@ -0,0 +1,60 @@
+package leetcode
+
+func imageSmoother(M [][]int) [][]int {
+	res := make([][]int, len(M))
+	for i := range M {
+		res[i] = make([]int, len(M[0]))
+	}
+	for y := 0; y < len(M); y++ {
+		for x := 0; x < len(M[0]); x++ {
+			res[y][x] = smooth(x, y, M)
+		}
+	}
+	return res
+}
+
+func smooth(x, y int, M [][]int) int {
+	count, sum := 1, M[y][x]
+	// Check bottom
+	if y+1 < len(M) {
+		sum += M[y+1][x]
+		count++
+	}
+	// Check Top
+	if y-1 >= 0 {
+		sum += M[y-1][x]
+		count++
+	}
+	// Check left
+	if x-1 >= 0 {
+		sum += M[y][x-1]
+		count++
+	}
+	// Check Right
+	if x+1 < len(M[y]) {
+		sum += M[y][x+1]
+		count++
+	}
+	// Check Coners
+	// Top Left
+	if y-1 >= 0 && x-1 >= 0 {
+		sum += M[y-1][x-1]
+		count++
+	}
+	// Top Right
+	if y-1 >= 0 && x+1 < len(M[0]) {
+		sum += M[y-1][x+1]
+		count++
+	}
+	// Bottom Left
+	if y+1 < len(M) && x-1 >= 0 {
+		sum += M[y+1][x-1]
+		count++
+	}
+	//Bottom Right
+	if y+1 < len(M) && x+1 < len(M[0]) {
+		sum += M[y+1][x+1]
+		count++
+	}
+	return sum / count
+}

leetcode/0661.Image-Smoother/661. Image Smoother_test.go

@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question661 struct {
+	para661
+	ans661
+}
+
+// para 是参数
+// one 代表第一个参数
+type para661 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans661 struct {
+	one [][]int
+}
+
+func Test_Problem661(t *testing.T) {
+
+	qs := []question661{
+
+		question661{
+			para661{[][]int{[]int{1, 1, 1}, []int{1, 1, 2}}},
+			ans661{[][]int{[]int{1, 1, 1}, []int{1, 1, 1}}},
+		},
+
+		question661{
+			para661{[][]int{[]int{1, 1, 1}, []int{1, 1, 2}, []int{1, 1, 1}}},
+			ans661{[][]int{[]int{1, 1, 1}, []int{1, 1, 1}, []int{1, 1, 1}}},
+		},
+
+		question661{
+			para661{[][]int{[]int{1, 1, 1}, []int{1, 0, 1}, []int{1, 1, 1}}},
+			ans661{[][]int{[]int{0, 0, 0}, []int{0, 0, 0}, []int{0, 0, 0}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 661------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans661, q.para661
+		fmt.Printf("【input】:%v       【output】:%v\n", p, imageSmoother(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}