leetcode239 - Sliding Window Maximum - hard

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. Return the max sliding window.

Follow up:
Could you solve it in linear time?

Example:

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

 

Constraints:

• 1 <= nums.length <= 10^5
• -10^4 <= nums[i] <= 10^4
• 1 <= k <= nums.length

 

总共n-k+1个window。

维护一个长度为k的deque代表window:

1. 用index维护

2. push新的数之前用pop_back把比它小的清掉

3. 长度到k了：dq.front() == i-k，用pop_front把之前的踢出去

4. window内单调递减，每个window的max都在front，从i >= k-1开始

Time O(n)

 

实现：

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++){
            while (!dq.empty() && nums[i] >= nums[dq.back()])
                dq.pop_back();
            dq.push_back(i);
            if (dq.front() == i-k)
                dq.pop_front();
            if (i >= k-1)
                res.push_back(nums[dq.front()]);
        }
        
        return res;
        
    }
};