每日一道Leetcode - 面试题 17.22. 单词转换

class Solution:
    def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[str]:
        def change_only_first_element(first, second):
            if len(first) != len(second):
                return False
            count = 0
            for i in range(len(first)):
                if first[i] != second[i]:
                    count += 1
            return count == 1
        """
        # 超时了...
        ans = []
        # 如果目标词不再字典中，就不能转换
        if endWord not in wordList:
            return ans
        # 先存放第一个词
        queue = [beginWord]
        # 判断是否找到了
        flag = False
        # 存 与当前字符只变换一个的情况
        dic = collections.OrderedDict()
        # 标记单词是否被访问过
        visited = [False] * len(wordList)

        while queue:
            # 取第一个元素
            queueHead = queue.pop(0)
            # 如果这个元素就是目标endWord元素，则直接返回答案
            if queueHead == endWord:
                flag = True
                break
            # 在字典中寻找可能的转换情况，加入队列，同时将能够从当前只变换一个字符的情况存起来
            for i in range(len(wordList)):
                # 该词未被访问的情况，且能够转换
                if not visited[i] and change_only_first_element(wordList[i], queueHead)==True:
                    queue.append(wordList[i])
                    visited[i] = True
                    dic[wordList[i]] = queueHead
        # print(dic)

        if not flag:
            return ans
        key = endWord
        while dic[key] != beginWord:
            ans.append(key)
            key = dic[key]
        ans.append(key)
        ans.append(dic[key])
        return ans[::-1]
        """
       
        def dfs(curWord,endWord,wordList):
            if curWord == endWord:
                return True
            n = len(wordList)
            for i in range(n):
                if self.visited[i]==True or change_only_first_element(curWord,wordList[i])==False:
                    continue
                self.visited[i] = True
                self.path.append(wordList[i])
                if dfs(wordList[i],endWord,wordList)==True:
                    return True
                self.path.pop(-1)
            return False
        
        n = len(wordList)
        self.visited = [False]*n
        self.path = [beginWord]
        if dfs(beginWord,endWord,wordList)==True:
            return self.path[:]
        else:
            return []