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Lyft 技术面试：单词转换问题深度解析（Word Ladder）

2025-09-12

Lyft 的技术面试以注重算法与逻辑推理能力而闻名，其中常包含复杂的图论、搜索类问题。本文将通过一道真实的 Lyft 面试题目，还原候选人解题的全过程，展现如何借助 csvosupport 的实时辅助，让候选人不仅能够清晰阐述解题思路，还能自信应对深度追问与优化讨论

📋 面试题目：单词转换（Word Ladder

题目描述（英文原题）

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

  1. Every adjacent pair of words differs by a single letter
  2. Every si for 1 <= i <= k is in wordList
  3. beginWord does not need to be in wordList
  4. 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.

示例

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: "hit" -> "hot" -> "dot" -> "dog" -> "cog"

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: 0
Explanation: endWord "cog" is not in wordList

🎯 核心考点分析

这道题是经典LeetCode 127. Word Ladder，主要考察

  1. BFS（广度优先搜索） - 寻找最短路
  2. 图的建模能力 - 将单词转换问题转化为图遍
  3. *优化技 - 如何高效生成相邻单词
  4. 边界处理 - 处理无解情况

💡 解题思路（csvosupport 实战指导

方法一：BFS + 哈希集合

*csvosupport 建议 这道题的本质是在图中寻找最短路径，BFS 是最优选择

from collections import deque

def ladderLength(beginWord, endWord, wordList):
    # 边界检查：endWord 必须wordList 
    word_set = set(wordList)
    if endWord not in word_set:
        return 0
    
    # BFS 初始
    queue = deque([(beginWord, 1)])  # (当前单词, 路径长度)
    visited = {beginWord}
    
    while queue:
        current_word, length = queue.popleft()
        
        # 找到目标单词
        if current_word == endWord:
            return length
        
        # 生成所有可能的下一个单
        for i in range(len(current_word)):
            for c in 'abcdefghijklmnopqrstuvwxyz':
                next_word = current_word[:i] + c + current_word[i+1:]
                
                # 如果下一个单词在字典中且未访问过
                if next_word in word_set and next_word not in visited:
                    visited.add(next_word)
                    queue.append((next_word, length + 1))
    
    return 0

# 测试
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]
print(ladderLength(beginWord, endWord, wordList))  # 输出: 5

时间复杂度： O(M² × N)

空间复杂度： O(N)

方法二：双向 BFS（优化版

*csvosupport 高级指导 对于大规模数据，双向 BFS 可以显著提升效率

def ladderLength_bidirectional(beginWord, endWord, wordList):
    word_set = set(wordList)
    if endWord not in word_set:
        return 0
    
    # 双向 BFS
    begin_set = {beginWord}
    end_set = {endWord}
    visited = set()
    length = 1
    
    while begin_set and end_set:
        # 优化：总是从较小的集合开始扩
        if len(begin_set) > len(end_set):
            begin_set, end_set = end_set, begin_set
        
        next_set = set()
        for word in begin_set:
            for i in range(len(word)):
                for c in 'abcdefghijklmnopqrstuvwxyz':
                    next_word = word[:i] + c + word[i+1:]
                    
                    # 如果两个方向相遇
                    if next_word in end_set:
                        return length + 1
                    
                    if next_word in word_set and next_word not in visited:
                        visited.add(next_word)
                        next_set.add(next_word)
        
        begin_set = next_set
        length += 1
    
    return 0

时间复杂度： O(M² × N/2) - 双向搜索减少一半搜索空 空间复杂度： O(N)

🚀 面试中的深度追问

Q1: 如何优化字符替换的过程？

*csvosupport 建议

from collections import defaultdict

def build_pattern_dict(wordList):
    pattern_dict = defaultdict(list)
    for word in wordList:
        for i in range(len(word)):
            pattern = word[:i] + '*' + word[i+1:]
            pattern_dict[pattern].append(word)
    return pattern_dict

Q2: 如果需要返回所有最短路径怎么办？

*csvosupport 建议

Q3: 如何处理超大规模字典

*csvosupport 建议

📊 复杂度对

方法 时间复杂 空间复杂 适用场景
单向 BFS O(M² × N) O(N) 通用场景
双向 BFS O(M² × N/2) O(N) 大规模数
预处+ BFS O(M × N) O(M × N) 多次查询

💼 csvosupport 如何助力 Lyft 面试

Lyft 面试中，csvosupport 提供

算法选择指导 - 快速识别最优算 代码实现辅助 - 确保代码逻辑清晰 优化方案讨论 - 展现工程实践能力 边界条件提醒 - 避免遗漏特殊情况 追问应对策略 - 自信回答深度追问

想在 Lyft、Uber、DoorDash 等公司的技术面试中脱颖而出

联系 csvosupport，我们提供专业的 VO 面试辅助服务，助你顺利拿Offer

*标签 #Lyft #WordLadder #BFS #图论 #最短路#VO辅助 #面试辅助 #算法面试 #一亩三分地

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