## 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:

1. Only one letter can be changed at a time.
2. 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:

5

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:

0

Explanation:

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

## 分析

• 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;
}
};