239.Sliding Window Maximum

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.

Example1

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?

难度系数

Hard

解法一：通过STL中的multiset解决，对元素自动排序，且允许有重复值

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> res;
        //允许有重复的元素，如若删除某个元素，则所有相同值的元素均被删除
        multiset<int> st;
        for (int i = 0; i < nums.size(); ++i) {
            //从开始查找，找到则返回位置，并删除
            if (i >= k) st.erase(st.find(nums[i - k]));
            st.insert(nums[i]);
            if (i >= k - 1) res.push_back(*st.rbegin());
        }
        return res;
    }
};

解法二：通过优先队列实现，将数值和下标均放到优先队列中，优先队列会对按照数值大小堆排，只需要保证数值在窗口中即可。

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> res;
        priority_queue<pair<int, int>> q;//大顶堆
        for (int i = 0; i < nums.size(); ++i) {
            while (!q.empty() && q.top().second <= i - k) q.pop();
            q.push({nums[i], i});
            if (i >= k - 1) res.push_back(q.top().first);
        }
        return res;
    }
};

解法三：双端队列添加下标，下标对应的元素是由front到back递减的

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> res;
        //双向队列
        deque<int> q;
        for (int i = 0; i < nums.size(); ++i) {
            //说明此时队列中的元素长度大于k，或者是这个元素，已经不在窗口中了，故而删除
            if (!q.empty() && q.front() == i - k) q.pop_front();
            //队列中的的下标对应nums中的值由front到back是递减的
            while (!q.empty() && nums[q.back()] < nums[i]) q.pop_back();
            q.push_back(i);
            if (i >= k - 1) res.push_back(nums[q.front()]);
        }
        return res;
    }
};