双向BFS及其应用

127. 单词接龙

相关题目
打开转盘锁 – BFS

关于双向BFS介绍：双向BFS

# 127. 单词接龙

class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        self.wordId = dict()
        self.edge = collections.defaultdict(list)
        self.nodeNum = 0

        for word in wordList:
            self.addEdge(word)

        self.addEdge(beginWord)
        if endWord not in self.wordId:
            return 0
        
        disBegin = [float('inf')] * self.nodeNum
        beginId = self.wordId[beginWord]
        disBegin[beginId] = 0
        queBegin = collections.deque([beginId])

        disEnd = [float("inf")] * self.nodeNum
        endId = self.wordId[endWord]
        disEnd[endId] = 0
        queEnd = collections.deque([endId])

        while queBegin or queEnd:
            queBeginSize = len(queBegin)
            for _ in range(queBeginSize):
                nodeBegin = queBegin.popleft()
                if disEnd[nodeBegin] != float('inf'):
                # 因为添加了虚拟节点，所以我们得到的距离为实际最短路径长度的两倍。
                    return (disBegin[nodeBegin] + disEnd[nodeBegin]) // 2 + 1

                for it in self.edge[nodeBegin]:
                    if disBegin[it] == float('inf'):
                        disBegin[it] = disBegin[nodeBegin] + 1
                        queBegin.append(it)

            queEndSize = len(queEnd)
            for _ in range(queEndSize):
                nodeEnd = queEnd.popleft()
                if disBegin[nodeEnd] != float('inf'):
                    return (disBegin[nodeEnd] + disEnd[nodeEnd]) // 2 + 1

                for it in self.edge[nodeEnd]:
                    if disEnd[it] == float('inf'):
                        disEnd[it] = disEnd[nodeEnd] + 1
                        queEnd.append(it)

        return 0


    
    def addWord(self, word: str):
        if word not in self.wordId:
            self.wordId[word] = self.nodeNum
            self.nodeNum += 1

    def addEdge(self, word: str):
        self.addWord(word)
        id1 = self.wordId[word]
        chars = list(word)
		# 创建虚拟节点。对于单词 hit，我们创建三个虚拟节点 *it、h*t、hi*，
		# 并让 hit 向这三个虚拟节点分别连一条边即可。
        for i in range(len(chars)):
            tmp = chars[i]
            chars[i] = '*'
            newWord = ''.join(chars)
            self.addWord(newWord)
            id2 = self.wordId[newWord]
            self.edge[id1].append(id2)
            self.edge[id2].append(id1)
            chars[i] = tmp