239. Sliding Window Maximum

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].

Note: 
You may assume k is always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.

解题思路：本题主要是运用了双端队列deque来维护以当前节点i为末位置节点时，i-(k-1)这段区间内的最大值。另外 如果nums[x]<nums[i],且x<i，则无论如何在此区间内都取不到nums[x]。

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