题目大意: 同一个昵称可能不是同一个人,也可能是同一个人,但不同的昵称肯定不是同一个人.按照题意将同一个人的邮箱地址链接起来

 

并查集:

　　对对应的邮箱地址进行并查集操作,如果存在有交集的邮箱地址,则两个列表肯定归属于同一个人,将他们连接起来.

 1 class Solution {
 2 public:
 3     int find(int idr, vector<int>& u) {
 4         while (idr != u[idr]) idr = u[idr];
 5         return idr;
 6     }
 7     void join(int idr1, int idr2, vector<int>& u) {
 8         int f1 = find(idr1, u);
 9         int f2 = find(idr2, u);
10         u[f2] = f1;
11     }
12     map<string, int> hashmap1;
13     map<int, vector<vector<string>>> hashmap2;
14     vector<vector<string>> res; //返回向量
15     vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
16         vector<int> u(accounts.size()); 
17         for (int i = 0; i < u.size(); i++) u[i] = i; //初始化并查集
18         for (int k = 0; k < accounts.size(); k++) { //查找,逻辑连接
19             for (int i = 1;i<accounts[k].size();i++) {
20                 auto addr = accounts[k][i];
21                 auto iter = hashmap1.find(addr);
22                 if (iter == hashmap1.end()) hashmap1[addr] = k;
23                 else join(hashmap1[addr], k, u);
24             }
25         }
26 
27         //打印并查集情况
28         //for (int i = 0; i < u.size(); i++) {
29         //    cout << u[i] << " ";
30         //    for (auto val : accounts[i]) cout << val << " ";
31         //    cout << endl;
32         //}
33 
34         for (int i = 0; i < accounts.size(); i++) { //物理连接
35             int f = find(u[i], u); //对顶层节点进行分类
36             auto iter = hashmap2.find(f);
37             if (iter == hashmap2.end()) { //如果还不存在先创建一个
38                 hashmap2[f] = {};
39             }
40             hashmap2[f].push_back(accounts[i]);
41         }
42         for (auto iter = hashmap2.begin(); iter != hashmap2.end(); iter++) {
43             set<string> map;
44             for (auto list : iter->second) {
45                 for (int i = 1;i<list.size();i++) {
46                     string val = list[i];
47                     auto itr = map.find(val);
48                     if (itr == map.end()) map.insert(val);
49                 }
50             }
51             if (map.empty()) continue;
52             vector<string> temp(map.begin(), map.end());
53             sort(temp.begin(), temp.end());
54             temp.insert(temp.begin(), accounts[iter->first][0]);
55             res.push_back(temp);
56         }
57         return res;
58     }
59 };