721. Accounts Merge

 1 class Solution {
 2     public List<List<String>> accountsMerge(List<List<String>> accounts) {
 3         List<List<String>> res = new ArrayList<>();
 4         if(accounts == null) return res;
 5         HashMap<String, String> map = new HashMap<>();
 6         HashMap<String, Set<String>> graph = new HashMap<>();
 7         for(int j = 0; j < accounts.size(); j++){
 8             List<String> account = accounts.get(j);
 9             String name = account.get(0);
10             for(int i = 1; i < account.size(); i++){
11                 if(!graph.containsKey(account.get(i))){
12                     graph.put(account.get(i), new HashSet<>());
13                 }
14                 map.put(account.get(i), name);
15                 
16                 if(i != 1){
17                     graph.get(account.get(i)).add(account.get(i-1));
18                     graph.get(account.get(i-1)).add(account.get(i));
19                 }
20             }
21         }
22         
23         Set<String> visited = new HashSet<>();
24         for(String email : map.keySet()){
25             List<String> list = new ArrayList<>();
26             String name = map.get(email);
27             if(!visited.contains(email)){  //判断是否visited之后， 没有的话要加进去
28                 visited.add(email);
29                 dfs(graph, visited, list, email);
30                 Collections.sort(list);
31                 list.add(0, name);
32                 res.add(list);
33             }   
34         }
35         return res;    
36     }
37     
38     public void dfs(HashMap<String, Set<String>> graph, Set<String> visited, List<String> list, String email){
39         list.add(email);
40         for(String neighbor : graph.get(email)){
41             if(!visited.contains(neighbor)){
42                 visited.add(neighbor);
43                 dfs(graph, visited, list, neighbor);
44             }
45         }
46     }
47 }