[leetcode] 127. Word Ladder

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是念 2022-08-12 07:30:48 博主文章分类：C++ ©著作权

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Description

Given two words (beginWord and endWord), and a dictionary’s word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time.
Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.
You may assume no duplicates in the word list.
You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

Input:

beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output:

Explanation:

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Example 2:

Input:

beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output:

Explanation:

The endWord "cog" is not in wordList, therefore no possible 
transformation.

分析

题目的意思是：给定开始单词和结束单词，还有一个单词词典，求从开始单词到结束单词的最小路径。

本题用了一个队列，先把start放进去，然后每次从队列里面取字符串，计算字符串与词典之间的距离，如果距离为1，我们就把字符串加入队列，然后长度+1，这样一直遍历到队列为空为止。

代码

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string> dict(wordList.begin(),wordList.end());
        unordered_map<string,int> m;
        queue<string> q;
        m[beginWord]=1;
        q.push(beginWord);
        while(!q.empty()){
            string word=q.front();
            q.pop();
            for(int i=0;i<word.size();i++){
                string newWord=word;
                for(char ch='a';ch<='z';ch++){
                    newWord[i]=ch;
                    if(dict.count(newWord)&&newWord==endWord){
                        return m[word]+1;
                    }
                    if(dict.count(newWord)&&!m.count(newWord)){
                        q.push(newWord);
                        m[newWord]=m[word]+1;
                    }
                }
            }
        }
        return 0;
    }
};