140. 单词拆分 II

2021/04/05 leetcode 共 811 字,约 3 分钟

title

problem link

给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict,在字符串中增加空格来构建一个句子,使得句子中所有的单词都在词典中。返回所有这些可能的句子。

说明:

  • 分隔时可以重复使用字典中的单词。
  • 你可以假设字典中没有重复的单词。

示例 1:

输入:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
输出:
[
  "cats and dog",
  "cat sand dog"
]

示例 2:

输入:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
输出:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
解释: 注意你可以重复使用字典中的单词。

示例 3:

输入:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
输出:
[]

solution

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
        n = len(s)
        wordSet = set(wordDict)

        import functools
        @functools.lru_cache(None)
        def dfs(i):
            if i == n:
                return []
            ans = []
            if s[i:] in wordSet:
                ans.append([s[i:]])
            for j in range(i, n - 1):
                if s[i:j + 1] in wordSet:
                    for v in dfs(j + 1):
                        ans.append([s[i:j + 1]] + v)
            return ans

        d = dfs(0)
        ans = []
        for l in d:
            ans.append(" ".join(l))
        return ans

Search

    Table of Contents