DFS——单词分割，原来还是要使用cached dp才能避免不超时

582. 单词拆分II

中文

English

给一字串s和单词的字典dict,在字串中增加空格来构建一个句子，并且所有单词都来自字典。
返回所有有可能的句子。

样例

样例 1:

输入："lintcode"，["de","ding","co","code","lint"]
输出：["lint code", "lint co de"]
解释：
插入一个空格是"lint code"，插入两个空格是 "lint co de"

样例 2:

输入："a"，[]
输出：[]
解释：字典为空

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class Solution:     """     @param: s: A string     @param: wordDict: A set of words.     @return: All possible sentences.     """     def wordBreak(self, s, wordDict):         # write your code here         if not s:             return []                       cache = {}         def dfs(s):             if not s:                 return [[]]                               if s in cache:                 return cache[s]                               ans = []             for i in range(len(s)):                 word = s[0:i+1]                 if word in wordDict:                     words = dfs(s[i+1:])                     for p in words:                         ans.append([word] + p)                           cache[s] = ans                           return ans                   ans = [" ".join(l) for l in dfs(s)]                   return ans

 超时的例子：

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class Solution:     """     @param: s: A string     @param: wordDict: A set of words.     @return: All possible sentences.     """     def wordBreak(self, s, wordDict):         # write your code here         def get_words_matrix(s):             n = len(s)             ans = collections.defaultdict(list)             for i in range(n):                 for j in range(i, n):                     if s[i:j+1] in wordDict:                         ans[i].append(j)                                       return ans                   if not s:             return []                       ans = []         dp = get_words_matrix(s)                   def dfs(s, start, path):             if start == len(s):                 ans.append(" ".join(path))                 return                           for pos in dp[start]:                 path.append(s[start:pos+1])                 dfs(s, pos+1, path)                 path.pop()                                 dfs(s, start=0, path=[])                   return ans

 

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输入   "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]