LeetCode Word Ladder

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.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.

题意：一个字符串每次改变一个字符，问是否可以通过这种改变变成另一个字符串，要求每次改变后的字符串都必须在wordList里，求最少的方法。

思路：最短路BFS，每次尝试换成另外25个字符，答案是走的步数，也就是每新一轮BFS入队后的都出队后，那么我们才需要更新一次答案。

public class Solution {
    public int ladderLength(String beginWord, String endWord, Set wordList) {
        if (beginWord.length() != endWord.length()) return 0;
    	if (beginWord.length() == 0 || endWord.length() == 0) return 0;
    	
    	LinkedList queue = new LinkedList();
    	queue.add(beginWord);
    	wordList.remove(beginWord);
    	int ans = 1;
    	int count = 1;
    	while (wordList.size() > 0 && !queue.isEmpty()) {
    		String current = queue.poll();
    		--count;
    		for (int i = 0; i < current.length(); i++) {
    			char[] tmp = current.toCharArray();
    			for (char j = 'a'; j <= 'z'; j++) {
    				if (tmp[i] == j) continue;
    				tmp[i] = j;
    				String tmpStr = new String(tmp);
    				if (tmpStr.equals(endWord)) return ans + 1;
    				if (wordList.contains(tmpStr)) {
    					queue.add(tmpStr);
    					wordList.remove(tmpStr);
    				}
    			}
    		}
    		
    		if (count == 0) {
    			count = queue.size();
    			++ans;
    		}
    	}
    	
    	return 0;
    }
}