[LeetCode] Word Ladder

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.

[Solution].java

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

public class Solution {
    public int ladderLength(String start, String end, HashSet<String> dict) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(start.equals(end)) return 0;
        HashSet<String> hasVisited = new HashSet<String>();
        Queue<Path> q = new LinkedList<Path>();
        Path begin = new Path(start, 1);
        q.offer(begin);
        hasVisited.add(start);
        while(!q.isEmpty()){
          Path path = q.poll();
        	char[] chars = path.word.toCharArray();
    		for(int i = 0; i < chars.length; i++){
    			for(int j = 97; j <= 122; j++){
    				char odd = chars[i];
    				if(chars[i] != (char) j){
    					chars[i] = (char) j;
    					String newWord = String.valueOf(chars);
    					if(newWord.equals(end)) return path.level + 1;
    					if(dict.contains(newWord) && !hasVisited.contains(newWord)){
    						q.offer(new Path(newWord, path.level + 1));
    						hasVisited.add(newWord);
    					}
    				}
    				chars[i] = odd;
    			}
    		}
        }
        return 0;
    }

	class Path{
		String word;
		int level;
		Path(String word, int level){
			this.word = word;
			this.level = level;
		}
	}
}