# Sliding Window Maximum

## Sliding window maximum

[原题链接](https://leetcode.com/problems/sliding-window-maximum/)

### 描述

> 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*.

### 例子

> 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

### 解法一 Time Limit Exceeded O(kN)

```cpp
class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> maxSW;
        for(int i=0; i<nums.size()-k+1; i++){
            int maxV=INT_MIN;
            for(int j=0; j<k; j++){
                maxV=max(maxV, nums[i+j]);
            }
            maxSW.push_back(maxV);
        }
        return maxSW;
    }
};
```

### 解法二 use dequeue O(N)

```cpp
class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector <int> ans;
        int n=nums.size();
        deque <int> q; //only indexes are stored
        for(int i=0;i<n;i++)
        {   
            while(!q.empty() && i-q.front()>=k)
                q.pop_front(); //only window size of k is allowed
            while(!q.empty() && nums[q.back()]<nums[i])
                q.pop_back();
            q.push_back(i);
            if(i>=k-1)ans.push_back(nums[q.front()]); //our max value in O(1)
        }
        return ans;
    }
};
```

better one

```cpp
class Solution {
public:
   vector<int> maxSlidingWindow(const vector<int> &A, int B) {
       vector<int> ans;
       deque<int> dq;
       for(int i = 0; i < A.size(); i++){
           while(!dq.empty() && dq.back() < A[i]) dq.pop_back();
           dq.push_back(A[i]);
           if(i >= B - 1){
               ans.push_back(dq.front());
               if(A[i - B + 1] == dq.front()) dq.pop_front();
           }
       }
       return ans;
   }
};
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://celia-qian.gitbook.io/leetcode-notebook-2020-2021/others/slide-window/sliding-window-maximum.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.