LeetCode - 239. Sliding Window Maximum

最新推荐文章于 2022-04-20 21:57:46 发布

转载最新推荐文章于 2022-04-20 21:57:46 发布 · 92 阅读

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原文链接：http://www.cnblogs.com/wxisme/p/10629126.html

本文介绍了一种使用双端队列高效求解滑动窗口中最大值的方法。通过维护一个只包含当前滑动窗口内元素下标的双端队列，确保队列头部始终是窗口内的最大值，从而实现快速查找。

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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

滑动窗口中的最大值，使用双端队列保存下标，窗口内左边比右边第一个小的不用考虑。

class Solution {
    private Deque<Integer> deque = new ArrayDeque<>();
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || nums.length <= 0)
            return new int[0];
        int[] res = new int[nums.length - k + 1];
        for (int i=0; i<nums.length; i++) {
            if (!deque.isEmpty() && deque.peekFirst() == i-k)
                deque.pollFirst();
            while (!deque.isEmpty() && nums[deque.peekLast()] < nums[i])
                deque.pollLast();
            deque.offerLast(i);
            if (i >= k - 1)
                res[i-k+1] = nums[deque.peekFirst()];
        }
        return res;
    }
}

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

转载于:https://www.cnblogs.com/wxisme/p/10629126.html