[Leetcode] Alien Dictionary, Solution

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

Example 1:
Given the following words in dictionary,

[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

The correct order is: "wertf".

Example 2:
Given the following words in dictionary,

[
  "z",
  "x"
]

The correct order is: "zx".

Example 3:
Given the following words in dictionary,

[
  "z",
  "x",
  "z"
]

The order is invalid, so return "".

Note:

1. You may assume all letters are in lowercase.
2. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary.
3. If the order is invalid, return an empty string.
4. There may be multiple valid order of letters, return any one of them is fine.

[Thoughts]
有向图的拓扑排序。进来的边为入度，出去的边为出度。生成图的结构，然后按照如下规则排序：
1. 找出当前入度为0的节点
2. 所有和该节点相连接的节点入度减1
3. go to #1

看下图示例：

[Code]

1:  string alienOrder(vector<string>& words) {  
2:       unordered_map<char, set<char>> outbound;  
3:       unordered_map<char, set<char>> inbound;  
4:    set<char> no_pre;  
5:      
6:    string s = "";  
7:    for(int i =0; i< words.size(); i++) {  
8:      string t = words[i];  
9:      no_pre.insert(t.begin(), t.end());  
10:      for(int j = 0; j< min(s.size(), t.size()); j++) {  
11:        if(t[j] != s[j]) {  
12:          inbound[t[j]].insert(s[j]);  
13:          outbound[s[j]].insert(t[j]);  
14:          break;  
15:        }  
16:      }  
17:      s = t;  
18:    }  
19:      
20:    // get the nodes which has 0 inbound edge  
21:    int char_count = no_pre.size();  
22:    for(auto p : inbound) {  
23:      no_pre.erase(p.first);  
24:    }  
25:      
26:    string result = "";  
27:    while(no_pre.size() >0) {  
28:      auto it = no_pre.begin();  
29:      char c = *it;  
30:      no_pre.erase(c);  
31:      result+=c;  
32:        
33:      for(char su : outbound[c]) {  
34:        inbound[su].erase(c);  
35:        if(inbound[su].size() == 0) {  
36:          no_pre.insert(su);  
37:        }  
38:      }  
39:        
40:    }   
41:       return result.size() == char_count? result : "";  
42:  }