feat: 修改azl397985856#209 azl397985856#279， 增加todo

README.md

@@ -118,9 +118,9 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 - [199.binary-tree-right-side-view](./problems/199.binary-tree-right-side-view.md)
 - [201.bitwise-and-of-numbers-range](./problems/201.bitwise-and-of-numbers-range.md)
 - 🆕 [208.implement-trie-prefix-tree](./problems/208.implement-trie-prefix-tree.md)
-- [209.minimum-size-subarray-sum](./problems/209.minimum-size-subarray-sum.md)
+- 🖊 [209.minimum-size-subarray-sum](./problems/209.minimum-size-subarray-sum.md)
 - [240.search-a-2-d-matrix-ii](./problems/240.search-a-2-d-matrix-ii.md)
-- [279.perfect-squares](./problems/279.perfect-squares.md)
+- 🖊 [279.perfect-squares](./problems/279.perfect-squares.md)
 - [322.coin-change](./problems/322.coin-change.md)
 - [328.odd-even-linked-list](./problems/328.odd-even-linked-list.md)
 - [416.partition-equal-subset-sum](./problems/416.partition-equal-subset-sum.md)
@@ -169,9 +169,9 @@ anki - 文件 - 导入 - 下拉格式选择“打包的 anki集合”，然后
 
 - [365.water-and-jug-problem](./todo/365.water-and-jug-problem.js)
 
-- anki 卡片 完善
+- [anki 卡片 完善](./assets/anki/)
 
-- 字符串类问题汇总
+- [字符串类问题汇总](./todo/str/)
 
 ## 交流群
 

assets/drawio/209.minimum-size-subarray-sum.drawio

@@ -0,0 +1 @@
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problems/209.minimum-size-subarray-sum.md

@@ -21,11 +21,13 @@ If you have figured out the O(n) solution, try coding another solution of which
 
 用滑动窗口来记录序列， 每当滑动窗口中的 sum 超过 s， 就去更新最小值，并根据先进先出的原则更新滑动窗口，直至 sum 刚好小于 s
 
+![209.minimum-size-subarray-sum](../assets/problems/209.minimum-size-subarray-sum.png)
+
 > 这道题目和 leetcode 3 号题目有点像，都可以用滑动窗口的思路来解决
 
 ## 关键点
 
-- 滑动窗口简化操作
+- 滑动窗口简化操作(滑窗口适合用于求解这种要求`连续`的题目)
 
 ## 代码
 

problems/279.perfect-squares.md

@@ -1,16 +1,16 @@
 ## 题目地址
+
 https://leetcode.com/problems/perfect-squares/description/
 
 ## 题目描述
 
-
 ```
 Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
 
 Example 1:
 
 Input: n = 12
-Output: 3 
+Output: 3
 Explanation: 12 = 4 + 4 + 4.
 Example 2:
 
@@ -23,19 +23,66 @@ Explanation: 13 = 4 + 9.
 ## 思路
 
 直接递归处理即可，但是这种暴力的解法很容易超时。如果你把递归的过程化成一棵树的话（其实就是递归树），
-可以看出中间有很多重复的计算。 
+可以看出中间有很多重复的计算。
 
 如果能将重复的计算缓存下来，说不定能够解决时间复杂度太高的问题。
 
 > 递归对内存的要求也很高， 如果数字非常大，也会面临爆栈的风险，将递归转化为循环可以解决。
 
-如果使用DP，其实本质上和递归 + 缓存 差不多。
+递归 + 缓存的方式代码如下：
+
+```js
+const mapper = {};
+
+function d(n, level) {
+  if (n === 0) return level;
+
+  let i = 1;
+  const arr = [];
+
+  while (n - i * i >= 0) {
+    const hit = mapper[n - i * i];
+    if (hit) {
+      arr.push(hit + level);
+    } else {
+      const depth = d(n - i * i, level + 1) - level;
+      mapper[n - i * i] = depth;
+      arr.push(depth + level);
+    }
+    i++;
+  }
+
+  return Math.min(...arr);
+}
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var numSquares = function(n) {
+  return d(n, 0);
+};
+```
+
+如果使用 DP，其实本质上和递归 + 缓存 差不多。
+
+DP 的代码见代码区。
 
 ## 关键点解析
 
 - 如果用递归 + 缓存， 缓存的设计很重要
-我的做法是key就是n，value是以n为起点，到达底端的深度。
-下次取出缓存的时候用当前的level + 存的深度 就是我们想要的level.
+  我的做法是 key 就是 n，value 是以 n 为起点，到达底端的深度。
+  下次取出缓存的时候用当前的 level + 存的深度 就是我们想要的 level.
+
+- 使用递归的核心点还是选和不选的问题
+
+```js
+for (let i = 1; i <= n; i++) {
+  for (let j = 1; j * j <= i; j++) {
+    // 选（dp[i]） 还是  不选（dp[i - j * j]）
+    dp[i] = Math.min(dp[i], dp[i - j * j] + 1);
+  }
+}
+```
 
 ## 代码
 
@@ -74,13 +121,12 @@ Explanation: 13 = 4 + 9.
 const mapper = {};
 
 function d(n, level) {
-
-  if (n  === 0) return level;
+  if (n === 0) return level;
 
   let i = 1;
   const arr = [];
 
-  while(n - i * i >= 0) {
+  while (n - i * i >= 0) {
     const hit = mapper[n - i * i];
     if (hit) {
       arr.push(hit + level);
@@ -99,6 +145,19 @@ function d(n, level) {
  * @return {number}
  */
 var numSquares = function(n) {
-  return d(n, 0);
+  if (n <= 0) {
+    return 0;
+  }
+
+  const dp = Array(n + 1).fill(Number.MAX_VALUE);
+  dp[0] = 0;
+  for (let i = 1; i <= n; i++) {
+    for (let j = 1; j * j <= i; j++) {
+      // 选（dp[i]） 还是  不选（dp[i - j * j]）
+      dp[i] = Math.min(dp[i], dp[i - j * j] + 1);
+    }
+  }
+
+  return dp[n];
 };
 ```

todo/slideWindow/239.sliding-window-maximum.js

@@ -0,0 +1,62 @@
+/*
+ * @lc app=leetcode id=239 lang=javascript
+ *
+ * [239] Sliding Window Maximum
+ *
+ * https://leetcode.com/problems/sliding-window-maximum/description/
+ *
+ * algorithms
+ * Hard (37.22%)
+ * Total Accepted:    150.8K
+ * Total Submissions: 399.5K
+ * Testcase Example:  '[1,3,-1,-3,5,3,6,7]\n3'
+ *
+ * Given an array nums, there is a sliding window of size k which is moving
+ * from the very left of the array to the very right. You can only see the k
+ * numbers in the window. Each time the sliding window moves right by one
+ * position. Return the max sliding window.
+ *
+ * Example:
+ *
+ *
+ * Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
+ * Output: [3,3,5,5,6,7]
+ * Explanation:
+ *
+ * Window position                Max
+ * ---------------               -----
+ * [1  3  -1] -3  5  3  6  7       3
+ * ⁠1 [3  -1  -3] 5  3  6  7       3
+ * ⁠1  3 [-1  -3  5] 3  6  7       5
+ * ⁠1  3  -1 [-3  5  3] 6  7       5
+ * ⁠1  3  -1  -3 [5  3  6] 7       6
+ * ⁠1  3  -1  -3  5 [3  6  7]      7
+ *
+ *
+ * Note:
+ * You may assume k is always valid, 1 ≤ k ≤ input array's size for non-empty
+ * array.
+ *
+ * Follow up:
+ * Could you solve it in linear time?
+ */
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+var maxSlidingWindow = function(nums, k) {
+  // bad 时间复杂度O(n^2)
+  if (nums.length === 0 || k === 0) return [];
+  let slideWindow = [];
+  const ret = [];
+  for (let i = 0; i < nums.length - k + 1; i++) {
+    for (let j = 0; j < k; j++) {
+      slideWindow.push(nums[i + j]);
+    }
+    ret.push(Math.max(...slideWindow));
+    slideWindow = [];
+  }
+  return ret;
+  // 双端队列优化时间复杂度
+};

todo/str/214.shortest-palindrome.js

@@ -0,0 +1,38 @@
+/*
+ * @lc app=leetcode id=214 lang=javascript
+ *
+ * [214] Shortest Palindrome
+ *
+ * https://leetcode.com/problems/shortest-palindrome/description/
+ *
+ * algorithms
+ * Hard (27.15%)
+ * Total Accepted:    73.1K
+ * Total Submissions: 266.3K
+ * Testcase Example:  '"aacecaaa"'
+ *
+ * Given a string s, you are allowed to convert it to a palindrome by adding
+ * characters in front of it. Find and return the shortest palindrome you can
+ * find by performing this transformation.
+ * 
+ * Example 1:
+ * 
+ * 
+ * Input: "aacecaaa"
+ * Output: "aaacecaaa"
+ * 
+ * 
+ * Example 2:
+ * 
+ * 
+ * Input: "abcd"
+ * Output: "dcbabcd"
+ */
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var shortestPalindrome = function(s) {
+
+};
+

todo/str/5.longest-palindromic-substring.js

@@ -0,0 +1,40 @@
+/*
+ * @lc app=leetcode id=5 lang=javascript
+ *
+ * [5] Longest Palindromic Substring
+ *
+ * https://leetcode.com/problems/longest-palindromic-substring/description/
+ *
+ * algorithms
+ * Medium (26.70%)
+ * Total Accepted:    525K
+ * Total Submissions: 1.9M
+ * Testcase Example:  '"babad"'
+ *
+ * Given a string s, find the longest palindromic substring in s. You may
+ * assume that the maximum length of s is 1000.
+ * 
+ * Example 1:
+ * 
+ * 
+ * Input: "babad"
+ * Output: "bab"
+ * Note: "aba" is also a valid answer.
+ * 
+ * 
+ * Example 2:
+ * 
+ * 
+ * Input: "cbbd"
+ * Output: "bb"
+ * 
+ * 
+ */
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var longestPalindrome = function(s) {
+
+};
+