721. consolidating the accounts of (disjoint-set / search deep / wide search)

Subject to the effect: the same nickname may not be the same person, it could be the same person, but a different nickname is certainly not the same person that Italy will follow the same personal email address linked.

 

Disjoint-set:

　　E-mail address of the corresponding disjoint-set operations, e-mail address if there is an intersection exists, the two lists is certainly attributable to the same person, to connect them together.

 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;18initialize disjoint-set//
         for ( int K = 0 ; K <accounts.size (); K ++) { // find the logical connection 
. 19              for ( int I = . 1 ; I <Accounts [K] .size (); I ++ ) {
 20 is                  Auto addr = Accounts [K] [I];
 21 is                  Auto ITER = hashmap1.find (addr);
 22 is                  IF (ITER hashmap1.end == ()) hashmap1 [addr] = K;
 23 is                  the else the Join (hashmap1 [addr], K, U );
 24              }
 25          }
 26 is  
27          // print disjoint-set case
 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 ()) {                  hashmap2 [F] =38 isif it does not exist to create a//
 {};
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 };