Leetcode 239. Sliding Window Maximum

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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.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

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

Therefore, return the max sliding window as [3,3,5,5,6,7].

输入：数组，滑动窗口大小
输出：滑动窗口中最大的数

思路

维护优先队列：维护一个优先队列，在窗口滑动时，剔除堆顶的无效元素，对于无效元素的定义是其下标不属于滑动窗口中；将新的元素加入堆中，同时输出堆顶元素。复杂度应该是O(n2)

代码

class Solution {public:    vector<int> maxSlidingWindow(vector<int>& nums, int k) {        vector<int> res;        if(nums.empty()||nums.size()<k) return res;        priority_queue<pair<int,int> > pq;        //Init - insert k elements to heap        for(int i =0;i<k;i++){            pq.push(pair<int, int>(nums[i],i));        }        res.push_back(pq.top().first);        //Check the heap top, pop, then insert        for(int i=k;i<nums.size();i++){            //invalid            while(!pq.empty()&&pq.top().second<=i-k){                pq.pop();            }            pq.push(pair<int,int>(nums[i],i));            res.push_back(pq.top().first);        }        return res;    }};