239. Sliding Window Maximum ----------H

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]

Example 3:

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

Example 4:

Input: nums = [9,11], k = 2
Output: [11]

Example 5:

Input: nums = [4,-2], k = 2
Output: [4]

 

Constraints:

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

A:

首先，用了一个monotonous decreasing vector ， say V, 来记录每次windows里从大到小的值。

每次加入一个新值，则V 最后的一定要比前面的都小，否则，如果大，就可以pop出来

然后，每次需要删除前面的值。 而被删除的值， 只有等于 V的第一个值的时候，才可能是存在V里的。（那个时候，为了不删除，我们直接前进一个index即可）

**************第二遍，用std::deque ， 方便前后删*******

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> res;
        deque<int> DQ;
        for(int i =0;i<nums.size();i++){
            if(i>=k && nums[i-k]>=DQ.front()){
                DQ.pop_front();
            }
            while(not DQ.empty() && DQ.back() < nums[i]){
                DQ.pop_back();
            }
            DQ.push_back(nums[i]);
            if(i>=k-1){
                res.push_back(DQ.front());
            }
        }
        return res;
    }
};

Amortized analysis 复杂度是： O(n). 

虽然感觉while loop.然后，每个数字，最多被push 一次， 也最多被pop一次。