Description

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:

  1. Only one letter can be changed at a time.
  2. 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.

Solutions

1. BFS

-w1210

class Solution(object):
    def ladderLength(self, beginWord, endWord, wordList):
        """
        :type beginWord: str
        :type endWord: str
        :type wordList: List[str]
        :rtype: int
        """
        wordList = set(wordList)
        visited = set()
        queue = collections.deque([(beginWord, 1)])
        alpha = string.ascii_lowercase # 'abcd...z'
        while queue:
            word, length = queue.popleft()
            
            if word == endWord:
                return length
            n = len(word)
            for i in range(n):
                for ch in alpha:
                    newWord = word[:i] + ch + word[i+1:]
                    if newWord in wordList and newWord not in visited:
                        queue.append((newWord, length+1))
                        visited.add(newWord)
        return 0
# Runtime: 420 ms, faster than 37.62% of Python online submissions for Word Ladder.
# Memory Usage: 12.9 MB, less than 75.68% of Python online submissions for Word Ladder.

2. Bidirectional BFS

-w1091 -w1215

3. DFS

  第一个想法总是用 DFS 回溯,其实也可以做,但是这里不合适,因为要求是求最短的路径,如果是回溯的话会优先往深了找,这样会超时。

References

  1. 127. Word Ladder
  2. huahua
  3. Python (BFS) - 公瑾™