Sliding Window Maximum

hard 原题链接：https://leetcode.com/problems/sliding-window-maximum/

Sliding window maximum

原题链接

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)

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)

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

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;
   }
};