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  • 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

    相关链接如下:

    知乎:littledy

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  • 原文地址:https://www.cnblogs.com/dysjtu1995/p/12760344.html
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