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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：

/**
 * @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

关键点解析

  • 双端队列简化时间复杂度

  • 滑动窗口

代码

/*
 * @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。