LeetCode239 滑动窗口的最大数

最新推荐文章于 2025-07-27 18:14:06 发布

原创最新推荐文章于 2025-07-27 18:14:06 发布 · 85 阅读

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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窗口内的元素，确保每次移动窗口时能快速找到最大值。算法在遍历数组的同时更新队列，保证了线性时间复杂度。

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.

Follow up:
Could you solve it in linear time?

public int[] maxSlidingWindow(int[] a, int k) {
        int n = a.length;
        if(n == 0) return new int[0];
		int[] r = new int[n-k+1];
        Deque<Integer> q = new ArrayDeque<>();
        int ri = 0;
        for(int i = 0; i < n; i++){
            while(!q.isEmpty() && q.peekFirst() < i - k + 1){ //去除索引过小而超出范围的
                q.pollFirst();
            }
            while (!q.isEmpty() && a[q.peekLast()] < a[i]) {  //去除索引比当前新索引小而其值也小的
				q.pollLast();
			}
			// q contains index... r contains content
			q.offerLast(i);
			if (i >= k - 1) {
				r[ri++] = a[q.peek()];
			}
		}
		return r;
    }