[Hard BFS] Word Ladders

Description

给出两个单词（start和end）和一个字典，找出从start到end的最短转换序列，输出最短序列的长度。

变换规则如下：

每次只能改变一个字母。
变换过程中的中间单词必须在字典中出现。(起始单词和结束单词不需要出现在字典中)

class Solution:
    """
    @param: start: a string
    @param: end: a string
    @param: dict: a set of string
    @return: An integer
    """
    def ladderLength(self, start, end, dict):
        dict.add(end)
        queue = collections.deque([start])
        visited = set([start])
        
        distance = 0
        while queue:
            distance += 1
            for i in range(len(queue)):
                word = queue.popleft()
                if word == end:
                    return distance
                
                for next_word in self.get_next_words(word):
                    if next_word not in dict or next_word in visited:
                        continue
                    queue.append(next_word)
                    visited.add(next_word) 

        return 0
        
    # O(26 * L^2)
    # L is the length of word
    def get_next_words(self, word):
        words = []
        for i in range(len(word)):
            left, right = word[:i], word[i + 1:]
            for char in 'abcdefghijklmnopqrstuvwxyz':
                if word[i] == char:
                    continue
                words.append(left + char + right)
        return words