LeetCode #269 - Alien Dictionary_leetcode269-优快云博客

本文介绍了一种算法，用于根据字典中的单词顺序推导出一种未知的外星语言的字母排序规则。通过构建字符有向图并进行层序遍历，实现了对外星语言字母顺序的有效推断。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述：

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:

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.

利用两个单词的字典序大小，构造字符的有向图，从入度为0的点开始层序遍历，入度越小说明字典序越大。

class Solution {
public:
    string alienOrder(vector<string>& words) {
        vector<int> indegree(26,-1);
        set<char> s;
        for(int i=0;i<words.size();i++)
        {
            for(int j=0;j<words[i].size();j++)
            {
                s.insert(words[i][j]);
                indegree[words[i][j]-'a']=0;
            }
        }
        
        vector<vector<int>> v(26);
        for(int i=1;i<words.size();i++)
        {
            string x=words[i-1];
            string y=words[i];
            for(int j=0;j<min(x.size(),y.size());j++)
            {
                if(x[j]!=y[j])
                {
                    v[x[j]-'a'].push_back(y[j]-'a');
                    indegree[y[j]-'a']++;
                    break;
                }
            }
        }
        
        queue<int> q;
        for(int i=0;i<26;i++) if(indegree[i]==0) q.push(i);
        string result;
        while(!q.empty())
        {
            int i=q.front();
            q.pop();
            result+=(i+'a');
            for(int j=0;j<v[i].size();j++)
            {
                indegree[v[i][j]]--;
                if(indegree[v[i][j]]==0) q.push(v[i][j]);
            }
        }
        if(result.size()!=s.size()) return "";
        return result;
    }
};