还是先抄一下题目：

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:

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

Output: "wertf"

Example 2:

Input:
[
  "z",
  "x"
]

Output: "zx"

Example 3:

Input:
[
  "z",
  "x",
  "z"
] 

Output: "" 

Explanation: The order is invalid, so return "".

Note:

• You may assume all letters are in lowercase.
• If the order is invalid, return an empty string.
• There may be multiple valid order of letters, return any one of them is fine.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/alien-dictionary

简单说一下思路

今天刷的第二道题目，本质上是一道拓扑排序的题目，只不过需要注意的小细节比较多。（不然怎么是Hard难度）我认为这题的hard难度主要就在与图中的edge需要自己找，也可能会重复，然后节点是用分散的字母而不是连续的数字表示的。

先贴一下代码：

class Solution {
public:
    string alienOrder(vector<string>& words) {
        unordered_map<char, unordered_set<char>> graph;
        unordered_map<char, int> indegree;
        for(auto word:words){
            for(auto letter:word){
                if(indegree.find(letter)==indegree.end()){
                    indegree[letter]=0;
                    graph[letter]=unordered_set<char>();
                }
            }
        }
        //build graph
        for(int i=0;i+1<words.size();i++){
            int j=0;
            while(j<words[i].size()&&j<words[i+1].size()&&words[i][j]==words[i+1][j]){
                j++;
            }

            if(j<words[i].size()&&j<words[i+1].size()){
                char a = words[i][j];
                char b = words[i+1][j];
                if(graph[a].count(b)==0){
                    graph[a].insert(b);
                    indegree[b]++;
                }                
            }
            else{
                if(j<words[i].size())return "";
            }
        }
        queue<char> q;
        unordered_map<char, int>::iterator iter;
        for(iter=indegree.begin();iter!=indegree.end();iter++){
            if(iter->second==0) {
                q.push(iter->first);
            }
        }
        string ans="";
        while(!q.empty()){
            char front=q.front();
            q.pop();
            ans+=front;
            for(auto i:graph[front]){
                indegree[i]--;
                if(indegree[i]==0)q.push(i);
            }
            
        }
        for(auto kv :indegree){
            if(kv.second!=0)return "";
        }
        return ans;

    }
};

这题其实可以和 //www.greatytc.com/p/dc3a1c9cdd3a 题号207的那题做一下对比，题目本质是一样的，只不过这题不太直观，要自己提炼出来才知道啊这原来是一道拓扑排序的题。
此题和207那题的区别是：在这题里，因为node都有char表示，将它重新标号其实有点麻烦也不划算，所以可以直接把图存在unordered_map<char, unordered_set>里，对应的每个node的入度也存在字典里。
这种方法还要注意为了让graph和indegree的size与节点数目相同，还要在开头做一个初始化操作。