leetcode126. Word Ladder II

Posted 2021-01-03 seyjs

题目如下：

解题思路：DFS或者BFS都行。本题的关键在于减少重复计算。我采用了两种方法：一是用字典dic_ladderlist记录每一个单词可以ladder的单词列表；另外是用dp数组记录从startword开始到wordlist每一个word的最小转换次数，这一点非常重要，可以过滤很多无效的运算。

代码如下：

class Solution(object):
    def getLadderList(self, w,d):
        l = []
        r = []
        for i in xrange(26):
            l.append(chr(i + ord(\'a\')))
        for i in range(len(w)):
            for j in l:
                if w[i] != j:
                    v = w[:i] + j + w[i+1:]
                    if v in d:
                        r.append(v)
        return r

    def findLadders(self, beginWord, endWord, wordList):
        #print len(wordList)
        dic = {}
        for i,v in enumerate(wordList):
            dic[v] = i

        if endWord not in dic:
            return []

        queue = [(beginWord,0,beginWord)]
        res = []
        shortest = len(wordList) + 1
        dp = [shortest for i in wordList]

        dic_ladderlist = {}
        while len(queue) > 0:
            word,count,path = queue.pop(0)
            if count > shortest:
                continue
            if word in dic_ladderlist:
                wl = dic_ladderlist[word]
            else:
                wl = self.getLadderList(word,dic)
                dic_ladderlist[word] = wl

            for i in wl:
                if dp[dic[i]] >= count+1 :
                    if i == endWord:
                        #path = path + \' \' + i
                        pl = path.split(\' \')
                        pl.append(endWord)
                        if len(pl) < shortest:
                            res = []
                            res.append(pl)
                            shortest = len(pl)
                        elif len(pl) == shortest:
                            res.append(pl)
                            shortest = len(pl)
                        continue
                    queue.append((i,count + 1,path + \' \' + i))
                    dp[dic[i]] = count + 1
        return res