Lyft VO 面试真题解析：Word Ladder（最短单词转换序列）

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

• Every adjacent pair of words differs by a single letter.
• Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
• sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> "cog", which is 5 words long.

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: 0
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

Constraints:

• 1 <= beginWord.length <= 10
• endWord.length == beginWord.length

这道题本质上是在单词图中求从 beginWord 到 endWord 的最短路径：如果两个单词只差一个字母，就可以连边，而 wordList 就是允许经过的节点集合。最常见的做法是用 BFS 按层扩展，因为题目要求“最短转换序列”，BFS 天然适合求最少步数；如果直接暴力枚举每个单词的所有替换，会在词表较大时变慢，因此通常会结合哈希集合快速判断单词是否存在，并在搜索过程中及时删除已访问单词避免重复。示例 1 中可以走出 hit -> hot -> dot -> dog -> cog，共 5 个单词；示例 2 里 endWord 不在词表中，因此直接返回 0。