word ladder medium

BFS/WordLadder.py

@@ -0,0 +1,87 @@
+"""
+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:
+
+Only one letter can be changed at a time.
+Each transformed word must exist in the word list. Note that beginWord is not a transformed word.
+Note:
+
+Return 0 if there is no such transformation sequence.
+All words have the same length.
+All words contain only lowercase alphabetic characters.
+You may assume no duplicates in the word list.
+You may assume beginWord and endWord are non-empty and are not the same.
+Example 1:
+
+Input:
+beginWord = "hit",
+endWord = "cog",
+wordList = ["hot","dot","dog","lot","log","cog"]
+
+Output: 5
+
+Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
+return its length 5.
+Example 2:
+
+Input:
+beginWord = "hit"
+endWord = "cog"
+wordList = ["hot","dot","dog","lot","log"]
+
+Output: 0
+
+Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
+
+
+这个BFS有点...
+
+一开始的思路就是 BFS，写法不同。
+
+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。
+
+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。
+
+这种方法可以全部跑通... 
+
+可优化的点在于如何高效的找到存在于 _wordList 中可变换的值。
+
+beat 66%
+
+测试地址：
+https://leetcode.com/problems/word-ladder/description/
+
+"""
+from collections import deque
+class Solution(object):
+    def ladderLength(self, beginWord, endWord, wordList):
+        """
+        :type beginWord: str
+        :type endWord: str
+        :type wordList: List[str]
+        :rtype: int
+        """
+
+        if len(beginWord) == 1:
+            return 2
+
+        _wordList = set(wordList)
+
+        result = deque([[beginWord, 1]])
+
+        while result:
+            word, length = result.popleft()
+
+            if word == endWord:
+                return length 
+
+            _length = length + 1
+
+            for i in range(len(word)):
+                for c in 'qwertyuiopasdfghjklzxcvbnm':
+                    new = word[:i]+c+word[i+1:]
+
+                    if new in _wordList:
+                        _wordList.remove(new)
+                        result.append([new, _length])
+
+        return 0