239. Sliding Window Maximum

Problem:

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

思路：

Solution (C++):

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

性能：

Runtime: 60 ms  Memory Usage: 10.5 MB

思路：

Solution (C++):

性能：

Runtime: ms  Memory Usage: MB

思路：

Solution (C++):

性能：

Runtime: ms  Memory Usage: MB