LeetCode 솔루션 분류

[8/14] 126. Word Ladder II

컨텐츠 정보

본문

Hard
4650587Add to ListShare

transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

  • Every adjacent pair of words differs by a single letter.
  • Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
  • sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s1, s2, ..., sk].

 

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: [["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
Explanation: There are 2 shortest transformation sequences:
"hit" -> "hot" -> "dot" -> "dog" -> "cog"
"hit" -> "hot" -> "lot" -> "log" -> "cog"

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: []
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

 

Constraints:

  • 1 <= beginWord.length <= 5
  • endWord.length == beginWord.length
  • 1 <= wordList.length <= 500
  • wordList[i].length == beginWord.length
  • beginWordendWord, and wordList[i] consist of lowercase English letters.
  • beginWord != endWord
  • All the words in wordList are unique.
Accepted
326,866
Submissions
1,183,724

관련자료

댓글 1

학부유학생님의 댓글

  • 익명
  • 작성일
class Solution:
    def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:
        wordList = set(wordList)
        if endWord not in wordList:
            return []
        # BFS visit
        curr_level = {beginWord}
        parents = collections.defaultdict(list)
        while curr_level:
            wordList -= curr_level
            next_level = set()
            for word in curr_level:
                for i in range(len(word)):
                    for c in 'abcdefghijklmnopqrstuvwxyz':
                        new_word = word[:i] + c + word[i+1:]
                        if new_word in wordList:
                            next_level.add(new_word)
                            parents[new_word].append(word)
            if endWord in next_level:
                break
            curr_level = next_level
        # DFS reconstruction
        res = []
        def dfs(word, path):
            if word == beginWord:
                path.append(word)
                res.append(path[::-1])
            else:
                for p_word in parents[word]:
                    dfs(p_word, path + [word])
        dfs(endWord, [])
        return res
LeetCode 솔루션 357 / 9 페이지
번호
제목
이름

최근글


인기글


새댓글


Stats


알림 0