Leetcode 239. Sliding Window Maximum-优快云博客

本文介绍了一种求解滑动窗口中最大值的高效算法。对于给定数组和窗口大小，通过双端队列记录当前窗口内的元素索引，确保队列头部始终为最大值的索引，从而实现线性时间内获取所有滑动窗口的最大值。

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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.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

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

Therefore, return the max sliding window as [3,3,5,5,6,7].

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

Follow up:
Could you solve it in linear time?

    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || k == 0) return new int[0];
        int len = nums.length;
        int[] res = new int[len - k + 1];
        int index = 0;
        Deque<Integer> q = new LinkedList<>();
        for (int i = 0; i < len; i++) {
            while (!q.isEmpty() && i - q.peekFirst() + 1 > k) q.pollFirst();
            while (!q.isEmpty() && nums[i] > nums[q.peekLast()]) q.pollLast();
            q.offer(i);
            if(i + 1>= k) res[index++] = nums[q.peek()]; 
        }
        return res;
    }