Java Word Ladder(字梯)

问题：

Given two words (start and end), and a dictionary, find the length of shortest transformation 
sequence from start to end, such that:
Only one letter can be changed at a time
Each intermediate word must exist in the dictionary
For example,

Given:
start = "hit"
end = "cog"
dict = ["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.

答案：

import java.util.HashSet;
import java.util.LinkedList;

public class WordLadderTest1 {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		String start = "hit";
		String end = "cog";
		HashSet<String> dict = new HashSet<String>();
		dict.add("hot");
		dict.add("dot");
		dict.add("dog");
		dict.add("lot");
		dict.add("cog");
		System.out.println(ladderLength(start, end, dict));
	}

	 public static int ladderLength(String start, String end, HashSet<String> dict) {
		 
	        if (dict.size() == 0)  
	            return 0; 
	 
	        LinkedList<String> wordQueue = new LinkedList<String>();
	        LinkedList<Integer> distanceQueue = new LinkedList<Integer>();
	 
	        wordQueue.add(start);
	        distanceQueue.add(1);
	 
	        while(!wordQueue.isEmpty()){
	            String currWord = wordQueue.pop();
	            Integer currDistance = distanceQueue.pop();
	            if(currWord.equals(end)){
	                return currDistance;
	            }
	            for(int i=0; i<currWord.length(); i++){
	                char[] currCharArr = currWord.toCharArray();
	                for(char c='a'; c<='z'; c++){
	                    currCharArr[i] = c;
	 
	                    String newWord = new String(currCharArr);
	                    if(dict.contains(newWord)){
	                        wordQueue.add(newWord);
	                        distanceQueue.add(currDistance+1);
	                        dict.remove(newWord);
	                    }
	                }
	            }
	        }
	 	        return 0;
	    }
}