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

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：

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

 

输入

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