LeetCode 127. 单词接龙

Posted 2020-12-17 机器狗mo

给定两个单词（beginWord?和 endWord）和一个字典，找到从?beginWord 到?endWord 的最短转换序列的长度。转换需遵循如下规则：

每次转换只能改变一个字母。
转换过程中的中间单词必须是字典中的单词。
说明:

如果不存在这样的转换序列，返回 0。
所有单词具有相同的长度。
所有单词只由小写字母组成。
字典中不存在重复的单词。
你可以假设 beginWord 和 endWord 是非空的，且二者不相同。
示例?1:

输入:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

输出: 5

解释: 一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog",
     返回它的长度 5。
示例 2:

输入:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

输出:?0

解释:?endWord "cog" 不在字典中，所以无法进行转换。


class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        w_len = len(wordList[0])
        pattern_dict = {}
        used_dict = {}
        for word in wordList:
            used_dict[word] = False
            for i in range(w_len):
                tmp_list = list(word)
                tmp_list[i] = ‘*‘
                pattern = ‘‘.join(tmp_list)
                if pattern not in pattern_dict:
                    pattern_dict[pattern] = []
                pattern_dict[pattern].append(word)
        #print(pattern_dict)

        queue = [(beginWord,1)]
        if beginWord in used_dict:
            used_dict[beginWord] = True

        while len(queue)>0:
            #print(queue)
            queue_size = len(queue)
            for q_i in range(queue_size):
                cur_word,step = queue.pop(0)
                match_flag = False
                for s_i in range(w_len):
                    tmp_list = list(cur_word)
                    tmp_list[s_i] = ‘*‘
                    cur_pattern = ‘‘.join(tmp_list)
                    if cur_pattern in pattern_dict:
                        for match_word in pattern_dict[cur_pattern]:
                            if match_word == endWord:
                                #print((cur_word,step),(match_word,step+1))
                                return step+1
                            if used_dict[match_word] is False:
                                match_flag = True
                                used_dict[match_word] = True
                                queue.append((match_word,step+1))

        return 0