leetcode - 127. Word Ladder

Description

A 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 the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if 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 <= 10
endWord.length == beginWord.length
1 <= wordList.length <= 5000
wordList[i].length == beginWord.length
beginWord, endWord, and wordList[i] consist of lowercase English letters.
beginWord != endWord
All the words in wordList are unique.

Solution

BFS, start with endWord, every time change one character to decide if we want to add this to the queue.

Time complexity: o ( n ∗ n ∗ w o r d . l e n + n ) o(n*n*word.len + n) o(n∗n∗word.len+n), where n is the length of wordList, word.len is the length of each word in wordList

Space complexity: o ( n ) o(n) o(n)

Code

python3 复制代码

class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        queue = collections.deque([(endWord, 1)])
        visited = set()
        wordList = set(wordList)
        if endWord not in wordList:
            return 0
        while queue:
            cur_word, step = queue.popleft()
            if cur_word in visited:
                continue
            visited.add(cur_word)
            if cur_word == beginWord:
                return step
            for i in range(len(cur_word)):
                for new_char in 'abcdefghijklmnopqrstuvwxyz':
                    if new_char == cur_word[i]:
                        continue
                    new_word = f'{cur_word[:i]}{new_char}{cur_word[i+1:]}'
                    if new_word in wordList or new_word == beginWord:
                        queue.append((new_word, step + 1))
        return 0