word Ladder2

[LeetCode]Word Ladder 
作者是 在线疯狂 发布于 2015年8月17日 在 LeetCode.

题目描述：

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.

题目大意：

给定两个单词（beginWord 和 endWord），以及一个字典，寻找从 beginWord 到 endWord 的最短转换序列的长度，满足约束条件：

每次只能改变一个字母

每一个得到的单词必须存在于字典中

例如，给定：

start = "hit"end = "cog"dict = ["hot","dot","dog","lot","log"]

最短的转换序列为："hit" -> "hot" -> "dot" -> "dog" -> "cog"，

返回其长度5。

注意：

如果不存在这样的转换序列，返回0。

所有的单词都是等长的。

所有的单词都只包含小写字母。

解题思路：

方法I：BFS（广度优先搜索）

将wordDict中单词word的每一个字母使用下划线“_”代替，得到wordKey

可以构造wordKey -> word的映射，记为neighbors

for word in wordDict:    for x in range(len(word)):        token = word[:x] + '_' + word[x+1:]        neighbors[token] += word,
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在BFS的执行过程中，将当前单词word的每一位以下划线“_”代替，

在neighbors字典中查找其“相邻”单词（只改变一个字母得到的单词）

Python代码：

class Solution:    # @param {string} beginWord    # @param {string} endWord    # @param {set<string>} wordDict    # @return {integer}    def ladderLength(self, beginWord, endWord, wordDict):        from collections import defaultdict, deque        queue = deque( [ [beginWord, 1] ] )        visited = set( [ beginWord ] )        neighbors = defaultdict(list)        for word in wordDict:            for x in range(len(word)):                token = word[:x] + '_' + word[x+1:]                neighbors[token] += word,        while queue:            word, length = queue.popleft()            if self.wordDist(word, endWord) <= 1:                return length + 1            for x in range(len(word)):                token = word[:x] + '_' + word[x+1:]                for ladder in neighbors[token]:                    if ladder not in visited:                        visited.add(ladder)                        queue += [ladder, length + 1],        return 0    def wordDist(self, wordA, wordB):        return sum([wordA[x] != wordB[x] for x in range(len(wordA))])
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方法II：双向BFS

从起点beginWord和终点endWord同时进行广度优先搜索，执行速度较方法I更快

Python代码：

class Solution:    # @param {string} beginWord    # @param {string} endWord    # @param {set<string>} wordDict    # @return {integer}    def ladderLength(self, beginWord, endWord, wordDict):        from collections import defaultdict, deque        qf = deque( [ [beginWord, 0] ] )        qe = deque( [ [endWord, 0] ] )        sf = { }        se = { }        neighbors = defaultdict(list)        for word in wordDict:            for x in range(len(word)):                token = word[:x] + '_' + word[x+1:]                neighbors[token] += word,        while qf or qe:            if qf:                word, length = qf.popleft()                if word in se :                    return length + se[word] + 1                for x in range(len(word)):                    token = word[:x] + '_' + word[x+1:]                    for ladder in neighbors[token]:                        if ladder not in sf:                            sf[ladder] = length + 1                            qf += [ladder, length + 1],            if qe:                word, length = qe.popleft()                if word in sf :                    return length + sf[word] + 1                for x in range(len(word)):                    token = word[:x] + '_' + word[x+1:]                    for ladder in neighbors[token]:                        if ladder not in se:                            se[ladder] = length + 1                            qe += [ladder, length + 1],        return 0
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