刚开始刷力扣 LeetCode吃力是非常正常的一件事情，各位同学可千万别被刚开始的一点小困难吓退了。俗话说...
如果任一值在数组中出现至少两次 ,返回true ;如果数组中每个元素互不相同,返回false 。 来源:力扣(LeetCode) 链接: 力扣leetcode.cn/problems/contains-duplicate/?envType=study-plan&id=shu-ju-jie-gou-ru-men&plan=data-structures&plan_progress=b584641 题目中给出的class Solution定义: class ...
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https://leetcode-cn.com/problems/falling-squares/solution/cong-xian-duan-shu-dao-tu-lun-jian-mo-by-wotxdx/ 线段树 https://cp-algorithms.com/data_structures/segment_tree.html ZKW线段树 https://www.cnblogs.com/Judge/p/9514862.html 初版16ms namespaceSegTree{//描述方块以及高度classNode {publi...
LeetCode链接:https://leetcode-cn.com/problems/word-break/ 题目: 给定一个非空字符串s和一个包含非空单词列表的字典wordDict,判定s是否可以被空格拆分为一个或多个在字典中出现的单词; 注: 1、拆分时可以重复使用字典中的单词; 2、可以假设字典中没有重复的单词; ...
About DSA problems leetcode.com/AyanOrNot/ Topics sql algorithms cpp data-structures problem-solving dsa Activity Stars 1 star Watchers 1 watching Forks 0 forks Report repository Releases No releases published Packages No packages published Languages C++ 99.8% Python 0.2% Footer...
Save this event: Practice Solving Data Structures and Algorithms Problems Practice Solving Data Structures and Algorithms Problems Fri, Dec 27 • 8:00 AM GMT+8 Free Software SchoolSave this event: Practice Solving Data Structures and Algorithms Problems Practice Solving Data Structures and Algorithm...
These problems may require you to implement a given interface of a class, and may involve using one or more data structures. These are great exercises to improve your data structure skills. We recommend: Serialize and Deserialize Binary Tree and Insert Delete GetRandom O(1). ...
These problems may require you to implement a given interface of a class, and may involve using one or more data structures. These are great exercises to improve your data structure skills. We recommend: Shuffle an Array and Min Stack. ...
方法一:// https://leetcode.cn/problems/trim-a-binary-search-tree/description/ // 669. 修剪二叉搜索树 #include<iostream> #include<vector> using namespace std; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode() : val(...