广州网站建设高端网百度推广联系人

学习要点

1. 理解关系之间的相互传递

题目链接

        990. 等式方程的可满足性 - 力扣（LeetCode）

题目描述

解法：图联通

class Solution {
public:char my_find_root(unordered_map<char,int>& tu_map_char,char ch){while(tu_map_char[ch] >= 0){ch = tu_map_char[ch];}return ch;}bool equationsPossible(vector<string>& equations) {int n = equations.size();// if(equations.size() == 1) return true;vector<string> v1;vector<string> v2;for(auto& i: equations){if(i[1] == '='){v1.push_back(i);}else{v2.push_back(i);}}unordered_map<char,int> tu_map_char;for(char ch = 'a';ch<='z';ch++){tu_map_char[ch] = -1;}for(int i = 0;i<v1.size();i++){if(v1[i][0] == v1[i][3])continue;if(my_find_root(tu_map_char,v1[i][0]) != my_find_root(tu_map_char,v1[i][3])){tu_map_char[my_find_root(tu_map_char,v1[i][3])] = v1[i][0];}}bool flag = true;for(int i = 0;i<v2.size();i++){if(v2[i][0] == v2[i][3]){flag  = false;break;}if(my_find_root(tu_map_char,v2[i][0]) == my_find_root(tu_map_char,v2[i][3])){flag  = false;break;}}return flag;}
};