Leetcode:239. Sliding Window Maximum

最新推荐文章于 2020-09-20 08:59:22 发布

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本文介绍了一种解决滑动窗口最大值问题的高效算法。通过使用双端队列维护当前窗口内元素的最大值，算法能在O(n)的时间复杂度内求解。示例输入为数组[1,3,-1,-3,5,3,6,7]及窗口大小3，输出为[3,3,5,5,6,7]。

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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.

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

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

public int[] maxSlidingWindow(int[] nums, int k) {
        if(nums==null || nums.length ==0 || k==0) {
            return new int[0];
        }
        Deque<Integer> window = new LinkedList<Integer>();
        int[] result = new int[nums.length > k ? nums.length - k + 1 : 1];
        int h = nums.length > k ? k : nums.length;
        int i = 0;
        for(int j=0;j<nums.length;j++){
            while(!window.isEmpty() && nums[window.getLast()]<nums[j]) {
                window.removeLast();
            }
            window.addLast(j);
            if(j-window.getFirst()>=k){
                window.removeFirst();
            }
            if(j>=h-1) {
                result[i++] = nums[window.getFirst()];
            }
        }
        return result;
    }