[leetcode126]word ladder 2

word ladder 2

Given two words (beginWord and endWord), and a dictionary’s word list, find all shortest transformation sequence(s) from beginWord to endWord, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the word list
For example,

Given:
beginWord = “hit”
endWord = “cog”
wordList = [“hot”,”dot”,”dog”,”lot”,”log”]
Return
[
[“hit”,”hot”,”dot”,”dog”,”cog”],
[“hit”,”hot”,”lot”,”log”,”cog”]
]
Note:
All words have the same length.
All words contain only lowercase alphabetic characters.

算法：
同样是最短路径问题，应该使用BFS。
但是，结果要求得到所有最短路径。
这时仅仅有BFS是不够的，我们需要构建从起点到终点的倒序图。
这就需要BFS以父层，子层遍历的形式进行搜索。
为了得到最后的路径，要利用生成的图从终点进行回溯，直到得到所有最短路径。

from collections import defaultdictclass Solution(object):    def findLadders(self, start, end, wordDict):        """        :type start: str        :type end: str        :type dict: Set[str]        :rtype: List[List[str]]        """        # first build the child - father tree. tree is implemented by dict        # second get the path from the tree         tree = defaultdict(list)        wordDict.add(end)        curLevel = set()        nextLevel = set()        visited = set()        curLevel.add(start)        find = False        while curLevel and not find:            for node in curLevel:                visited.add(node)            for node in curLevel:                for scss in self.successor(node, wordDict):                    if scss == end:                        find = True                    if scss not in visited:                        nextLevel.add(scss)                        tree[scss].append(node)            curLevel = nextLevel            nextLevel = set()        result = []        print('tree is', tree)        if find:            path = []            self.gener_path(tree, start, end, path, result)        return result    def gener_path(self, tree, start, word, path, result):        path = path + [word]        print(word, path, result)        if start == word:            result.append(list(reversed(path)))        else:            for father in tree[word]:                self.gener_path(tree, start, father, path, result)    def successor(self, curStr, wordDict):        result = []        curList = list(curStr)        for idx in range(len(curStr)):            for ele in 'abcdefghijklmnopqrstuvwxyz':                temp = curList[idx]                if ele != temp:                    curList[idx] = ele                    newStr = ''.join(curList)                    curList[idx] = temp                    if newStr in wordDict:                        result.append(newStr)        return resultsolution =  Solution()wordSet = set(["hot","cog","dog","tot","hog","hop","pot","dot"])start = 'hot'end = 'dog'print(solution.findLadders(start,end,wordSet))('tree is', defaultdict(<type 'list'>, {'hog': ['hot'], 'hop': ['hot'], 'pot': ['hot'], 'dog': ['hog', 'dot'], 'tot': ['hot'], 'cog': ['hog'], 'dot': ['hot']}))('dog', ['dog'], [])('hog', ['dog', 'hog'], [])('hot', ['dog', 'hog', 'hot'], [])('dot', ['dog', 'dot'], [['hot', 'hog', 'dog']])('hot', ['dog', 'dot', 'hot'], [['hot', 'hog', 'dog']])[['hot', 'hog', 'dog'], ['hot', 'dot', 'dog']]