Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b" or "a!=b".  Here, a and b are lowercase letters (not necessarily dif…

Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b" or "a!=b".  Here, a and b are lowercase letters (not necessarily dif…

题目如下: Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b" or "a!=b".  Here, a and bare lowercase letters (not necessaril…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS 并查集 日期 题目地址:https://leetcode.com/problems/satisfiability-of-equality-equations/ 题目描述 Given an array equations of strings that represent relationships between variables, eac…

原题链接在这里:https://leetcode.com/problems/satisfiability-of-equality-equations/ 题目: Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b"…

目录 题目链接 注意点 解法 小结 题目链接 Satisfiability of Equality Equations - LeetCode 注意点 必须要初始化pre 解法 解法一:典型的并查集算法应用.先遍历所有等式,将等号两边的字母加入同一分类,每类中的字母都是相等的.然后遍历不等式,如果不等号两边的字母属于同一类则返回false.时间复杂度O(nm) class Solution { public: map<char,char> pre; char Find(char x) { cha…

Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4and takes one of two different forms: "a==b" or "a!=b".  Here, a and b are lowercase letters (not necessarily diff…

最长连续序列 题目[128]:链接. 解题思路 节点本身的值作为节点的标号,两节点相邻,即允许合并(x, y)的条件为x == y+1 . 因为数组中可能会出现值为 -1 的节点,因此不能把 root[x] == -1 作为根节点的特征,所以采取 root[x] == x 作为判断是否为根节点的条件.默认较小的节点作为连通分量的根. 此外,使用 map<int, int> counter 记录节点所在连通分量的节点个数(也是merge 的返回值). class Solution { publi…

并查集 这部分主要是学习了 labuladong 公众号中对于并查集的讲解,文章链接如下: Union-Find 并查集算法详解 Union-Find 算法怎么应用? 概述 并查集用于解决图论中「动态连通性」问题: 主要实现以下几个API class UF { /* 将 p 和 q 连接 */ public void union(int p, int q); /* 判断 p 和 q 是否连通 */ public boolean connected(int p, int q); /* 返回图中有多…

PS:做到第四题才发现 2,3题的路径压缩等于没写 How Many Tables Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14542    Accepted Submission(s): 7132 Problem Description Today is Ignatius' birthday. He invites a lot…