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127. Word Ladder

{leetcode}/problems/word-ladder/[LeetCode - Word Ladder^]

通过 {leetcode}/problems/word-ladder/solution/[Word Ladder - LeetCode^] 讲解,竟然可以抽象成无向无权图,然后通过 Queue 将其串联起来,实在好精巧。

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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.

思路分析

BFS 解题思路,与 752. Open the Lock 的套路是一样的。

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link:{sourcedir}/_0127_WordLadder.java[role=include]
link:{sourcedir}/_0127_WordLadder_2.java[role=include]