[Leetcode] 269. Alien Dictionary 解题报告

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本文介绍了一种基于拓扑排序的算法，用于确定一种未知外星语言中字母的正确顺序。通过分析字典中按该语言规则排序的单词列表，可以推导出字母间的先后关系。

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:
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:

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

思路：

使用拓扑排序是这一道题目的正解。我们可以利用这道题目总结一下拓扑排序的实现框架：

1）定义两个哈希表，一个用来表示每次出现过的字符的入度，另一个来存一个字符指向的字符集合。

2）每次找出一个入度为0的字符，并且更新这个字符指向的字符集入度减1。如果在没有遍历完所有字符的情况下找不到入度为0的字符，那么说明没有合法的拓扑排序，返回“”。如果在一次遍历过程中出现了多个入度为0的字符，说明拓扑排序的解不唯一，任意选择其中一个就可以了。

代码：

class Solution {
public:
    string alienOrder(vector<string>& words) {
        if(words.size() == 0) {
            return "";  
        }
        unordered_map<char, int> indegree;  
        unordered_map<char, multiset<char>> hash;  
        for(auto word: words) {  
            for(auto ch: word) {
                indegree[ch] = 0; 
            }
        }
        for(int i = 1; i < words.size(); i++) {  
            int k = 0, len1 = words[i-1].size(), len2 = words[i].size();  
            while(words[i-1][k] == words[i][k]) {
                ++k;
            }  
            if(k >= min(len1, len2)) {
                continue;
            }
            ++indegree[words[i][k]];
            hash[words[i - 1][k]].insert(words[i][k]);  
        }  
        string ret;  
        for(int i = 0; i < indegree.size(); i++) {  
            char ch = ' ';  
            for(auto val: indegree) {
                if(!val.second) { 
                    ch = val.first; 
                    break; 
                } 
            }
            if(ch == ' ') {
                return ""; 
            }
            ret += ch;
            --indegree[ch];  
            for(auto val: hash[ch]) {
                --indegree[val];
            }
        }  
        return ret;  
    }
};