Only one letter can be changed at a time
Each intermediate word must exist in the word list
For example,
Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.
Note:
Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.
给定两个单词和一个单词集合，从一个单词变换到另一个单词，每次只能改变一个字母，并且变换后的单词都要在单词集合中，找出最少的变换次数。我们从beginWord开始，每个字符都可能有26种变换,我们用一个set来记录在wordlist中已经被访问过的单词，用path来记录走过的长度，如果变换后的单词与endword相同就返回path+1，如果不相同，看是否在wordlist中并且检查是否被访问过，然后更新beginSet，直到找到一条路，或者beginSet为空，返回0。代码如下：
public class Solution {
public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
Set<String> beginSet = new HashSet<String>();
Set<String> isVisited = new HashSet<String>();
beginSet.add(beginWord);
int path = 1;
while(!beginSet.isEmpty()) {
Set<String> tem = new HashSet<String>();
for(String word : beginSet) {
char[] c = word.toCharArray();
for(int i = 0; i < c.length; i++) {
char oldChar = c[i];
for(char j = 'a'; j <= 'z'; j++) {
c[i] = j;
String s = new String(c);
if(s.equals(endWord))
return path + 1;
if(wordList.contains(s) && !isVisited.contains(s)) {
tem.add(s);
isVisited.add(s);
}
}
c[i] = oldChar;
}
}
beginSet = tem;
path ++;
}
return 0;
}
}