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| 1 | +""" |
| 2 | +Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that: |
| 3 | + |
| 4 | +Only one letter can be changed at a time. |
| 5 | +Each transformed word must exist in the word list. Note that beginWord is not a transformed word. |
| 6 | +Note: |
| 7 | + |
| 8 | +Return 0 if there is no such transformation sequence. |
| 9 | +All words have the same length. |
| 10 | +All words contain only lowercase alphabetic characters. |
| 11 | +You may assume no duplicates in the word list. |
| 12 | +You may assume beginWord and endWord are non-empty and are not the same. |
| 13 | +Example 1: |
| 14 | + |
| 15 | +Input: |
| 16 | +beginWord = "hit", |
| 17 | +endWord = "cog", |
| 18 | +wordList = ["hot","dot","dog","lot","log","cog"] |
| 19 | + |
| 20 | +Output: 5 |
| 21 | + |
| 22 | +Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", |
| 23 | +return its length 5. |
| 24 | +Example 2: |
| 25 | + |
| 26 | +Input: |
| 27 | +beginWord = "hit" |
| 28 | +endWord = "cog" |
| 29 | +wordList = ["hot","dot","dog","lot","log"] |
| 30 | + |
| 31 | +Output: 0 |
| 32 | + |
| 33 | +Explanation: The endWord "cog" is not in wordList, therefore no possible transformation. |
| 34 | + |
| 35 | + |
| 36 | +这个BFS有点... |
| 37 | + |
| 38 | +一开始的思路就是 BFS,写法不同。 |
| 39 | + |
| 40 | +一开始的写法是每次都从 wordList 里找单词,结果是有一个case跑不通。 |
| 41 | + |
| 42 | +后来实在没思路学习 Discuss 里的内容,Discuss 里的 BFS 是每一个都生成26个新的单词,然后判断是否在 wordList 中。 |
| 43 | + |
| 44 | +这种方法可以全部跑通... |
| 45 | + |
| 46 | +可优化的点在于如何高效的找到存在于 _wordList 中可变换的值。 |
| 47 | + |
| 48 | +beat 66% |
| 49 | + |
| 50 | +测试地址: |
| 51 | +https://leetcode.com/problems/word-ladder/description/ |
| 52 | + |
| 53 | +""" |
| 54 | +from collections import deque |
| 55 | +class Solution(object): |
| 56 | + def ladderLength(self, beginWord, endWord, wordList): |
| 57 | + """ |
| 58 | + :type beginWord: str |
| 59 | + :type endWord: str |
| 60 | + :type wordList: List[str] |
| 61 | + :rtype: int |
| 62 | + """ |
| 63 | + |
| 64 | + if len(beginWord) == 1: |
| 65 | + return 2 |
| 66 | + |
| 67 | + _wordList = set(wordList) |
| 68 | + |
| 69 | + result = deque([[beginWord, 1]]) |
| 70 | + |
| 71 | + while result: |
| 72 | + word, length = result.popleft() |
| 73 | + |
| 74 | + if word == endWord: |
| 75 | + return length |
| 76 | + |
| 77 | + _length = length + 1 |
| 78 | + |
| 79 | + for i in range(len(word)): |
| 80 | + for c in 'qwertyuiopasdfghjklzxcvbnm': |
| 81 | + new = word[:i]+c+word[i+1:] |
| 82 | + |
| 83 | + if new in _wordList: |
| 84 | + _wordList.remove(new) |
| 85 | + result.append([new, _length]) |
| 86 | + |
| 87 | + return 0 |
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