字梯
“Word Ladder”问题的详细指南和 Python 实现。
1. 学习
“字阶梯”问题是图表部分的一个关键挑战。
此实现侧重于 Python 中的简单级逻辑。
在我们提供的解决方案中,我们优先考虑技术准确性和代码可读性。
2. Real-World Applications
3. Visual Intuition
可视化字梯形图的逻辑流程。
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
仔细阅读 Word Ladder 的问题陈述。
2. Formulate brute force
起草一个简单的迭代解决方案。
3. Identify inefficiency
寻找冗余计算。
4. Optimize search path
使用散列或排序来加速该过程。
5. Final Implementation
清理生产标准代码。
问题陈述
使用字典 wordList 从单词 beginWord 到单词 endWord 的转换序列是单词 beginWord -> s1 -> s2 -> ... -> sk 的序列,使得:
- 每对相邻的单词都有一个字母不同。
- 1 <= i <= k 的每个 si 都在 wordList 中。请注意,beginWord 不需要位于 wordList 中。
- sk == endWord。
给定两个单词 beginWord 和 endWord 以及字典 wordList,返回从 beginWord 到 endWord 的最短转换序列中的单词数,如果不存在这样的序列则返回 0。
编写一个函数 ladderLength(beginWord: str, endWord: str, wordList: List[str]) -> int。
- •1 <= len(beginWord) <= 10
- •endWord.length == beginWord.length
- •1 <= len(wordList) <= 5000
- •wordList[i].length == beginWord.length
- •beginWord, endWord, and wordList[i] consist of lowercase English letters
- •All the words in wordList are unique
示例
beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
5
One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> "cog", which is 5 words long.
beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
0
The endWord "cog" is not in wordList, so there is no valid transformation sequence.
Need a Hint?
Edge Cases to Watch
- 空输入结构
- 单元素输入
- 大数值范围
准备好解决了吗?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
面试见解和变化
复杂性分析分解
为什么时间: Directly evaluates all possibilities.
为什么选择太空: Uses standard local memory.
为什么时间: Optimized paths reduce total operations.
为什么选择太空: May trade memory for speed.
优化解决方案Python代码
优化解决方案Python代码
from collections import deque, defaultdict
def ladder_length_opt(beginWord, endWord, wordList):
if endWord not in wordList: return 0
nei = defaultdict(list)
wordList.append(beginWord)
for word in wordList:
for j in range(len(word)):
pattern = word[:j] + "*" + word[j + 1 :]
nei[pattern].append(word)
visit = {beginWord}
q = deque([beginWord])
res = 1
while q:
for i in range(len(q)):
word = q.popleft()
if word == endWord: return res
for j in range(len(word)):
pattern = word[:j] + "*" + word[j + 1 :]
for neighbor in nei[pattern]:
if neighbor not in visit:
visit.add(neighbor)
q.append(neighbor)
res += 1
return 0暴力破解代码(剧透保护)
暴力破解代码(剧透保护)
def ladder_length_brute(beginWord, endWord, wordList):
if endWord not in wordList: return 0
def is_diff_one(w1, w2):
diff = 0
for i in range(len(w1)):
if w1[i] != w2[i]: diff += 1
return diff == 1
q = [(beginWord, 1)]
visit = {beginWord}
while q:
word, dist = q.pop(0)
if word == endWord: return dist
for w in wordList:
if w not in visit and is_diff_one(word, w):
visit.add(w)
q.append((w, dist + 1))
return 0Algorithm Pattern Checklist
When dealing with Graphs data patterns.
- Are constraints clear?
- Is there a linear or logarithmic optimization possible?
PyRun is built and maintained by an independent solo developer. If this helped your interview prep, consider buying a coffee!
推荐的 Python 资源
通过相关的交互式教程、备忘单和代码比较来扩展您的知识。
Python 循环:For 和 While 循环解释
了解如何使用 Python 循环来迭代数据。通过交互式示例掌握 for 循环、while 循环、break、continue 和循环最佳实践。
如何在 Python 中对列表进行排序(升序和降序)
了解如何在 Python 中使用 sort() 方法和sorted() 函数对列表进行排序。发现自定义键排序和逆序示例。
Python 字符串方法备忘单
Python 字符串操作的完整参考指南。掌握格式化、搜索、拆分、替换和检查字符串属性。
Python 与 JavaScript:哪种编程语言最好?
Python 和 JavaScript 的全面比较。探索语法差异、性能、用例(后端与前端)和编码示例。