140-word-break-ii.md

140. Word Break II - 单词拆分 II

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:
[]

---

题目标签：Dynamic Programming / Backtracking

题目链接：LeetCode / LeetCode中国

题解

Language	Runtime	Memory
python3	52 ms	14.3 MB

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
        cache = {}
        dic = set(wordDict)
        
        def func(s):
            if not s:
                return []

            if s in cache:
                return cache[s]

            res = []
            for i in range(1, len(s)):
                if s[:i] in dic:
                    for r in func(s[i:]):
                        res.append(s[:i] + ' ' + r)
            if s in dic:
                res.append(s)
            cache[s] = res
            return res

        return func(s)