滑动窗口(sliding window)方法是一种在序列数据(如字符串、数组等)中找到满足特定条件的 连续子序列的算法。滑动窗口方法的核心思想是使用两个指针(通常称为左指针和右指针)表示当前的窗口。在遍历过程中,…
The key condition r - l > k + max is used to determine when the window is too large to meet the problem's requirement. This condition checks whether the number of characters in the current window (r - l) minus the number of occurrences of the most frequent character (max) exceeds k...
Minimum Window Substring Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). Example: Input: S = "ADOBECODEBANC", T = "ABC" Output: "BANC" Note: If there is no such window in S that covers all characters...
在LeetCode写题目的时候评论区看到一个方法,一开始没看懂,后来查了一些资料整理了一下。原题见文中例3 什么是滑动窗口算法? The Sliding Problem contains a sliding window which is a sub – list that runs over a Large Array which is an underlying collection of elements. 滑动窗口算法可以用以解决数组/...
Can you solve this real interview question? 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 win
LeetCode 题目详细解析 关注博客注册登录 赞1收藏1 分享 阅读2.3k发布于2019-04-17 小鹿 82声望16粉丝 « 上一篇 LeetCode 之 JavaScript 解答第641题 —— 设计双端队列(Design Circular Deque) 下一篇 » LeetCode 之 JavaScript 解答第69题 —— X 的平方根(Squrt(x)) ...
Sliding Window(RMQ) Sliding Window Time Limit: 12000MS Memory Limit: 65536K Total Submissions: 35974 Accepted: 10648 Case Time Limit: 5000MS Description An array of size n ...Sliding Window(单调队列 ) 附上题目链接:点击这里 题意: 给你一个长度为 n 的数组 , 再给你一个只能看见 k 个...
/sliding-window-maximum/ 英文版:https://leetcode.com/problems/sliding-window-maximum/给定一个数组nums,有一个大小为k的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口k内的数字。滑动窗口每次只向右移动一位。返回滑动窗口最大值。 示例: 输入:nums= [1,3,-1,-3,5,3,6,7], 和...
https://leetcode-cn.com/problems/sliding-window-median/ 文章目录 分析 解法 `addNum` `eraseNum` 代码 分析 这道题是295.数据流的中位数这题的进阶版,此时数据流变成了一个固定大小的窗口,每次新加入一个数字就会有一个旧的数字被移除,因此,这相当于在295题的基础上引入了删除操作。 295题的解法是用...
Store the max window length possible and return this value. int longestOnes(vector<int>& A, int K) { int ans = 0; int l = 0; int r = 0; int n = A.size(); int k = K; while(r < n) { if (!A[r]) { if (!k) { ans = max(ans, r-l); while(!k) { if(!A[l...