LeetCode239. Sliding Window Maximum

一、题目

You are given an array of integers 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 1:

Input: nums = [1,3,-1,-3,5,3,6,7], 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
Example 2:

Input: nums = [1], k = 1
Output: [1]

Constraints:

1 <= nums.length <= 105
-104 <= nums[i] <= 104
1 <= k <= nums.length

二、题解

方法一：优先级队列法

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        int n = nums.size();
        priority_queue<pair<int,int>> q;
        for(int i = 0;i < k;i++){
            q.emplace(nums[i],i);
        }
        vector<int> res = {q.top().first};
        for(int i = k;i < n;i++){
            q.emplace(nums[i],i);
            //若最大值不在窗口中，不断移除元素
            while(q.top().second <= i-k) q.pop();
            res.push_back(q.top().first);
        }
        return res;
    }
};

方法二、单调队列法

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        int n = nums.size();
        deque<int> q;
        for(int i = 0;i < k;i++){
            while(!q.empty() && nums[i] > nums[q.back()]) q.pop_back();
            q.push_back(i);
        }
        vector<int> res = {nums[q.front()]};
        for(int i = k;i < n;i++){
            while(!q.empty() && nums[i] > nums[q.back()]) q.pop_back();
            q.push_back(i);
            while(!q.empty() && q.front() <= i - k) q.pop_front();
            res.push_back(nums[q.front()]);
        }
        return res;
    }
};