给定两个单词(beginWord 和 endWord)和一个字典,找到从 beginWord 到 endWord 的最短转换序列的长度。
找最短路径,用bfs(广度优先搜索,按照声波的方式,一层一层的扩散;或者生活中找人)
遍历的时候要把访问过的节点保存下来
每个位置枚举(一位一位来),看单词在不在集合中;边枚举边建立图(无向图,双向转换)
hit 枚举第一位:ait,bit.... 枚举第二位 hat,hbt,...枚举第三位 hia,hib,...
共26*len(word)种情况 O(26*wordLen) :常数的时间复杂度
class Solution: def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int: word_set = set(wordList) if len(word_set) == 0 oor endWord not in word_set: return 0 if beginWord in word_set: word_set.remove(beginWord) queue = [beginWord] word_len = len(beginWord) step = 1 while queue: current_size = len(queue) for i in range(current_size): word = queue.pop(0) word_list = list(word) for j in range(len(word_list)): origin_char = word_list[j] for k in range(26): word_list[j] = chr(ord('a') + k) next_word = ''.join(word_list) if next_word in word_set: if next_word == endWord: return step + 1 queue.append(next_word) word_set.remove(next_word) word_list[j] = origin_char step += 1 return 0