feat: azl397985856#239

README.md

@@ -136,6 +136,7 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 - 🆕 [128.longest-consecutive-sequence](./problems/128.longest-consecutive-sequence.md)
 - [145.binary-tree-postorder-traversal](./problems/145.binary-tree-postorder-traversal.md)
 - [146.lru-cache](./problems/146.lru-cache.md)
+- 🆕 [239.sliding-window-maximum](./problems/239.sliding-window-maximum.md)
 - 🆕 [295.find-median-from-data-stream.md](./problems/295.find-median-from-data-stream.md)
 - [301.remove-invalid-parentheses](./problems/301.remove-invalid-parentheses.md)
 

assets/drawio/239.sliding-window-maximum.drawio

@@ -0,0 +1 @@
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problems/239.sliding-window-maximum.md

@@ -0,0 +1,160 @@
+## 题目地址
+
+https://leetcode.com/problems/sliding-window-maximum/description/
+
+## 题目描述
+
+```
+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?
+```
+
+## 思路
+
+符合直觉的想法是直接遍历 nums, 然后然后用一个变量 slideWindow 去承载 k 个元素，
+然后对 slideWindow 求最大值，这是可以的，时间复杂度是 O(n \* k).代码如下：
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+var maxSlidingWindow = function(nums, k) {
+  // bad 时间复杂度O(n * k)
+  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;
+};
+```
+
+但是如果真的是这样，这道题也不会是 hard 吧？这道题有一个 follow up，要求你用线性的时间去完成。
+我们可以用双端队列来完成，思路是用一个双端队列来保存`接下来的滑动窗口可能成为最大值的数`。具体做法：
+
+
+- 入队列
+
+- 移除失效元素，失效元素有两种
+
+1. 一种是已经超出窗口范围了，比如我遍历到第4个元素，k = 3，那么i = 0的元素就不应该出现在双端队列中了
+具体就是`索引大于 i - k + 1的元素都应该被清除`
+
+2. 小于当前元素都没有利用价值了，具体就是`从后往前遍历（双端队列是一个递减队列）双端队列，如果小于当前元素就出队列`
+
+
+如果你仔细观察的话，发现双端队列其实是一个递减的一个队列。因此队首的元素一定是最大的。用图来表示就是：
+
+![239.sliding-window-maximum](../assets/problems/239.sliding-window-maximum.png)
+
+## 关键点解析
+
+- 递归简化操作
+- 如果树很高，建议使用栈来代替递归
+- 这道题目对顺序没要求的，因此队列数组操作都是一样的，无任何区别
+
+## 代码
+
+```js
+/*
+ * @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) {
+  // 双端队列优化时间复杂度, 时间复杂度O(n)
+  const deque = []; // 存放在接下来的滑动窗口可能成为最大值的数
+  const ret = [];
+  for (let i = 0; i < nums.length; i++) {
+    // 清空失效元素
+    while (deque[0] < i - k + 1) {
+      deque.shift();
+    }
+
+    while (nums[deque[deque.length - 1]] < nums[i]) {
+      deque.pop();
+    }
+
+    deque.push(i);
+
+    if (i >= k - 1) {
+      ret.push(nums[deque[0]]);
+    }
+  }
+  return ret;
+};
+```
+
+## 扩展
+
+### 为什么用双端队列
+因为删除无效元素的时候，会清除队首的元素（索引太小了
+）或者队尾(元素太小了)的元素。 因此需要同时对队首和队尾进行操作，使用双端队列是一种合乎情理的做法。