Class Solution
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- All Implemented Interfaces:
public final class Solution127 - Word Ladder\.
Hard
A transformation sequence from word
beginWordto wordendWordusing a dictionarywordListis a sequence of words <code>beginWord -> s<sub>1</sub> -> s<sub>2</sub> -> ... -> s<sub>k</sub></code> such that:Every adjacent pair of words differs by a single letter.
Every <code>s<sub>i</sub></code> for
1 <= i <= kis inwordList. Note thatbeginWorddoes not need to be inwordList.<code>s<sub>k</sub> == endWord</code>
Given two words,
beginWordandendWord, and a dictionarywordList, return the number of words in the shortest transformation sequence frombeginWordtoendWord, or0if no such sequence exists.Example 1:
Input: beginWord = "hit", endWord = "cog", wordList = "hot","dot","dog","lot","log","cog"
Output: 5
Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.
Example 2:
Input: beginWord = "hit", endWord = "cog", wordList = "hot","dot","dog","lot","log"
Output: 0
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.
Constraints:
1 <= beginWord.length <= 10endWord.length == beginWord.length1 <= wordList.length <= 5000wordList[i].length == beginWord.lengthbeginWord,endWord, andwordList[i]consist of lowercase English letters.beginWord != endWordAll the words in
wordListare unique.
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Constructor Summary
Constructors Constructor Description Solution()
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