leetcode239：Sliding Window Maximum（单调队列）

最新推荐文章于 2021-11-04 14:36:47 发布

原创最新推荐文章于 2021-11-04 14:36:47 发布 · 302 阅读

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本文介绍了一种解决滑动窗口最大值问题的高效算法，使用双端队列维护一个单调递减的队列，实现O(n)的时间复杂度。通过实例解析，详细展示了算法的实现过程。

为了避免毕业后失业，还是要好好刷题，上次面试的时候问leetcode和剑指offer的比较多，去这里练习一下。

239. Sliding Window Maximum

Hard

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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.

题意：给一个数组，维护一个长度为K的滑动窗口，输出滑动过程窗口内的各个最大值。

思路：比较简单，用双端队列维护一个单调递减的单调队列，复杂度O(n)。

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        int len = nums.size();
        vector<int>ans;
        deque<pair<int,int>>q;
        for(int i=0; i<len; ++i){
            while(!q.empty() && i-q.front().first >= k) q.pop_front();
            while(!q.empty() && nums[i] >= q.back().second) q.pop_back();
            q.push_back(make_pair(i,nums[i]));
            if(i >= k-1) ans.push_back(q.front().second);
        }
        return ans;
    }
};