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:

1. Only one letter can be changed at a time
2. 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

 

public class Solution {
   	/**
	 * This is a typical bfs problem.
	 * @param start
	 * @param end
	 * @param dict
	 * @return
	 */
    public int ladderLength(String start, String end, Set<String> dict) {
        if(dict.size() == 0) return 0;
        if(start.equals(end)) return 1;
		
		Queue<String> iteratedQueue = new LinkedList<String>();
		Queue<Integer> depth = new LinkedList<Integer>();
		
		iteratedQueue.add(start);
		depth.add(1);
		
		
		while(iteratedQueue.peek() != null) {
			String currentWord = iteratedQueue.poll();
			int currentDepth = depth.poll();

			for(int i = 0; i < currentWord.length(); ++i){
				char[] carray = currentWord.toCharArray();
				for(char c = 'a'; c <= 'z'; ++c){
					carray[i] = c;
					String possibleWord = new String(carray);
					if(possibleWord.equals(end))
					    return currentDepth + 1;
					if(dict.contains(possibleWord)) {
						iteratedQueue.add(possibleWord);
						depth.add(currentDepth + 1);
						dict.remove(possibleWord);
					}
				}
			}
		}
		return 0;
    }
}