Leetcode - Word Ladder

Given two words (beginWord and endWord), and a dictionary, find the length of shortest transformation sequence from beginWord to endWord, 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.
[分析] 使用bfs思路解决：维护一个层级遍历队列，考察当前层的每个单词，通过修改一个字母得到的新单词若出现在dict中则将新单词添加到考察队列的下一层中，且将该新单词从dict中删除，以避免出现hot-dot-hot的死循环，一层遍历完后考察下一层。bfs保证第一次找到endWord肯定是最短路径，因为bfs是层次遍历，同一层每个单词到beginWord的变换距离是相等的。参考 http://www.cnblogs.com/TenosDoIt/p/3443512.html

public class Solution {
    public int ladderLength(String beginWord, String endWord, Set<String> wordDict) {
        if (beginWord == null || endWord == null || wordDict == null)
            return -1;
        LinkedList<String> queue = new LinkedList<String>();
        queue.offer(beginWord);
        int len = 0;
        int currLevel = 0, nextLevel = 1;
        int L = beginWord.length();
        while (!queue.isEmpty()) {
            if (currLevel == 0) {
                currLevel = nextLevel;
                nextLevel = 0;
                len++;
            }
            char[] curr = queue.poll().toCharArray();
            for (int i = 0; i < L; i++) {
                char oldChar = curr[i];
                for (char c = 'a'; c <= 'z'; c++) {
                    if (c == oldChar) continue;
                    curr[i] = c;
                    String newWord = new String(curr);
                    if (newWord.equals(endWord))
                        return len + 1;
                    else if (wordDict.contains(newWord)) {
                        nextLevel++;
                        queue.offer(newWord);
                        wordDict.remove(newWord);
                    }
                }
                curr[i] = oldChar;
            }
            currLevel--;
        }
        return 0;
    }
}