【LeetCode】127. Word Ladder 解题报告（Python）

作者：负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目地址: https://leetcode.com/problems/word-ladder/description/

题目描述：

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:

Only one letter can be changed at a time.
Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.
You may assume no duplicates in the word list.
You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Example 2:

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
Output: 0

Explanation: The endWord “cog” is not in wordList, therefore no possible transformation.

题目大意

这个题名字是词语梯子，简单理解就是从begin开始，每次只能替换已经转化了的单词的其中一个字符，看最终能不能得到end。有个要求就是，每次变化不是任意的，是必须变成wordList中的其中一个才行。

解题方法

拿到这个题没有什么思路，看了别人解答之后，才猛然发现这个题是走迷宫问题的变形！也就是说，我们每次变化有26个方向，如果变化之后的位置在wordList中，我们认为这个走法是合规的，最后问能不能走到endWord？

很显然这个问题是BFS的问题，只是把走迷宫问题的4个方向转变成了26个方向，直接BFS会超时，所以我使用了个visited来保存已经遍历了的字符串，代表已经走过了的位置。代码总体思路很简单，就是利用队列保存每个遍历的有效的字符串，然后对队列中的每个字符串再次遍历，保存每次遍历的长度即可。

时间复杂度是O(NL)，空间复杂度是O(N).其中N是wordList中的单词个数，L是其实字符串的长度。

class Solution(object):
    def ladderLength(self, beginWord, endWord, wordList):
        """
        :type beginWord: str
        :type endWord: str
        :type wordList: List[str]
        :rtype: int
        """
        wordset = set(wordList)
        if endWord not in wordset:
            return 0
        visited = set([beginWord])
        chrs = [chr(ord('a') + i) for i in range(26)]
        bfs = collections.deque([beginWord])
        res = 1
        while bfs:
            len_bfs = len(bfs)
            for _ in range(len_bfs):
                origin = bfs.popleft()
                for i in range(len(origin)):
                    originlist = list(origin)
                    for c in chrs:
                        originlist[i] = c
                        transword = "".join(originlist)
                        if transword not in visited:
                            if transword == endWord:
                                return res + 1
                            elif transword in wordset:
                                bfs.append(transword)
                                visited.add(transword)
            res += 1
        return 0

显然上面的这个做法还是可以变短一点的，想起之前的二叉树的BFS的时候，会在每个节点入队列的时候同时保存了这个节点的深度，这样就少了一层对bfs当前长度的循环，可以使得代码变短。同时，学会了一个技巧，直接把已经遍历过的位置从wordList中删除，这样就相当于我上面的那个visited数组。下面这个代码很经典了，可以记住。

class Solution(object):
    def ladderLength(self, beginWord, endWord, wordList):
        """
        :type beginWord: str
        :type endWord: str
        :type wordList: List[str]
        :rtype: int
        """
        wordset = set(wordList)
        bfs = collections.deque()
        bfs.append((beginWord, 1))
        while bfs:
            word, length = bfs.popleft()
            if word == endWord:
                return length
            for i in range(len(word)):
                for c in "abcdefghijklmnopqrstuvwxyz":
                    newWord = word[:i] + c + word[i + 1:]
                    if newWord in wordset and newWord != word:
                        wordset.remove(newWord)
                        bfs.append((newWord, length + 1))
        return 0

参考资料：

http://www.cnblogs.com/grandyang/p/4539768.html

日期

2018 年 9 月 29 日 —— 国庆9天长假第一天！