栏目分类:
子分类:
返回
名师互学网用户登录
快速导航关闭
当前搜索
当前分类
子分类
实用工具
热门搜索
名师互学网 > IT > 软件开发 > 后端开发 > Java

LeetCode-239. Sliding Window Maximum [C++][Java]

Java 更新时间: 发布时间: IT归档 最新发布 模块sitemap 名妆网 法律咨询 聚返吧 英语巴士网 伯小乐 网商动力

LeetCode-239. Sliding Window Maximum [C++][Java]

LeetCode-239. Sliding Window Maximumhttps://leetcode.com/problems/sliding-window-maximum/

You are given an array of integers 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 1:

Input: nums = [1,3,-1,-3,5,3,6,7], 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

Example 2:

Input: nums = [1], k = 1
Output: [1]

Constraints:

  • 1 <= nums.length <= 10^5
  • -10^4 <= nums[i] <= 10^4
  • 1 <= k <= nums.length

【C++】
class Solution {
public:
    vector maxSlidingWindow(vector& nums, int k) {
        deque dq;
        vector ans;
        for (int i = 0; i < nums.size(); ++i) {
            if (!dq.empty() && dq.front() == i - k) {
                dq.pop_front();
            }
            while (!dq.empty() && nums[dq.back()] < nums[i]) {
                dq.pop_back();
            }
            dq.push_back(i);
            if (i >= k - 1) {
                ans.push_back(nums[dq.front()]);
            }
        }
        return ans;
    }
};

【Java】
class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        int n = nums.length;
        Deque dq = new LinkedList<>();
        int[] ans = new int[n-k+1];
        for (int i = 0; i < n; ++i) {
            if (!dq.isEmpty() && dq.peekFirst() == i - k) {
                dq.pollFirst();
            }
            while (!dq.isEmpty() && nums[dq.peekLast()] < nums[i]) {
                dq.pollLast();
            }
            dq.offerLast(i);
            if (i >= k - 1) {
                ans[i-k+1] = nums[dq.peekFirst()];
            }
        }
        return ans;
    }
}

参考文献

【1】【Java】Java双端队列Deque使用详解_devnn的博客-CSDN博客_deque java

转载请注明:文章转载自 www.mshxw.com
本文地址:https://www.mshxw.com/it/821759.html
我们一直用心在做
关于我们 文章归档 网站地图 联系我们

版权所有 (c)2021-2022 MSHXW.COM

ICP备案号:晋ICP备2021003244-6号