拓扑排序关键：

1. 无环有向图Directed Acyclic Graph (DAG)
2. 用backtracking或stack从后往前
3. 记录访问过的节点避免重复访问

269.Alien Dictionary

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list ofnon-emptywords from the dictionary, wherewords 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.

S： Topological Sort

1. 根据前后两个单词第一个不同的字母构建有向边，从而组成图。注意无序节点也要加入图
2. 对图进行拓扑排序，用visited记录访问过的节点，用ancestor记录当前路径的前后关系判断是否成环

class Solution {
public:
    string alienOrder(vector<string>& words) {
        if (words.size() == 1) {
            return words[0];
        }
        unordered_map<char, unordered_set<char>> map;
        for (int i = 1; i < words.size(); ++i) {
            makeGraph(map, words[i - 1], words[i]);
        }
        vector<bool> visited(26);
        vector<bool> ancestor(26);
        string result = "";
        string test = "";
        for (const auto& i : map) {
            test += i.first;
            if (!dfs(map, i.first, result, visited, ancestor)){
                return "";
            }
        }
        return result;
    }

private:
    void makeGraph(unordered_map<char, unordered_set<char>>& map, string word1, string word2) {
        int idx = 0;
        while (idx < word1.size() && idx < word2.size() && word1[idx] == word2[idx]) {
            addChar(map, word1[idx]);
            idx++;
        }
        if (idx < word1.size() && idx < word2.size()) {
            map[word1[idx]].insert(word2[idx]);
            addChar(map, word2[idx]);
            idx++;
        }
        for (int i = idx; i < word1.size(); ++i) {
            addChar(map, word1[idx]);
        }
        for (int i = idx; i < word2.size(); ++i) {
            addChar(map, word2[idx]);
        }
    }
    void addChar(unordered_map<char, unordered_set<char>>& map, char a) {
        if (map.find(a) == map.end()) {
            map[a] = unordered_set<char>{};
        }
    }
    bool dfs(unordered_map<char, unordered_set<char>>& map, char letter, string& result, 
             vector<bool>& visited, vector<bool>& ancestor) {
        if (ancestor[letter - 'a']) {
            return false;
        }
        if (visited[letter - 'a']) {
            return true;
        }
        visited[letter - 'a'] = true;
        ancestor[letter - 'a'] = true;
        for (char i : map[letter]) {
            if (!dfs(map, i, result, visited, ancestor)) {
                return false;
            }
        }
        ancestor[letter - 'a'] = false;
        result = letter + result;
        return true;
    }
};