题目：

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 intermediate word must exist in the word listFor 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.

由于只需要求最短的距离，所以直接BFS就行了。这题OJ的数据卡得比较严，裸BFS会TLE。没有必要写双向的BFS，注意下剪枝就可以了。直接放代码。

code:

import java.util.*;public class WordLadder {    public int ladderLength(String beginWord, String endWord, Set
    
      wordList) {        HashMap
     
       map = new HashMap
      
       ();        LinkedList
       
         queue = new LinkedList
        
         ();                queue.offer(beginWord);        map.put(beginWord, 1);        boolean flag = false;        while(!queue.isEmpty()){            String curr = queue.poll();            for(String next : getNextWords(curr, wordList)){                if(next.equals(endWord)){                    flag = true;                    map.put(next, map.get(curr) + 1);                    System.out.println(next + " " + map.get(next));                    return map.get(next);                }                if(wordList.contains(next)){                    if(!map.containsKey(next)){                        map.put(next, map.get(curr) + 1);                        System.out.println(next + " " + map.get(next));                        queue.offer(next);                    }                }            }        }        if(flag) return map.get(endWord);        else return 0;    }        public List
         
           getNextWords(String word, Set
          
            wordList){ List
           
             nextWords = new ArrayList
            
             (); for(int i = 0; i < word.length(); i++){ for(char c = 'a'; c <= 'z'; c++){ if(c == word.charAt(i)) continue; char[] arr = word.toCharArray(); arr[i] = c; String next = new String(arr); if(wordList.contains(next)) nextWords.add(next); } } return nextWords; } public static void main(String[] args){ WordLadder WL = new WordLadder(); Set
             
               wordList = new HashSet
              
               (); wordList.add("a"); wordList.add("b"); wordList.add("c"); String beginWord = "a"; String endWord = "c"; int result = WL.ladderLength(beginWord, endWord, wordList); System.out.println(result); }}