Word Ladder

1. Title

Word Ladder

2.   Http address

https://leetcode.com/problems/word-ladder/

3. The question

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 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.

4. My code (AC)

import java.util.*;
public class WordLadder {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

    }
    // Accepted
       public int ladderLength(String beginWord, String endWord, Set<String> wordDict) {
            
           if ( beginWord == null || beginWord.equals(endWord))
                   return 1;
           Queue<String> currQue = new LinkedList<String>();
           HashSet<String> used = new HashSet<String>();
           currQue.add(beginWord);
           used.add(beginWord);
           int count = 1;
           while( !currQue.isEmpty())
           {
               Queue<String> nextQue = new LinkedList<String>();
               count++;
                while( !currQue.isEmpty())
                {
                    String word = currQue.poll();
                    for(String transform : getTransform(word, used, wordDict,endWord))
                    {
                        if (transform.equals(endWord))
                            return count;
                        nextQue.add(transform);
                    }
                    
                }
                
                currQue = nextQue;
           }
           return 0;
        
       }
           
       public List<String> getTransform(String word, HashSet<String> used, Set<String> dict, String endWord)
       {
           List<String> res = new ArrayList<String>();
           char [] letters = word.toCharArray();
           for( int i = 0 ; i < letters.length; i++)
           {
               char original = letters[i];
               for( char c = 'a' ; c <= 'z' ; c++)
               {
                   if ( c != original)
                   {
                       letters[i] = c;
                       String trans = new String(letters);
                       if ( trans.endsWith(endWord) || (!used.contains(trans) && dict.contains(trans)) ){
                           res.add(trans);
                           used.add(trans);
                       }
                   }
               }
               
               letters[i] = original;
           }
           return res;
       }
}