断字Leetcode＃139Python解决方案超时-备忘录不起作用吗？

问题描述

我正在解决leetcode #139，由于某种原因，我超出了时间限制。我使用备忘录的方式不正确吗？

class Solution:
    
    memo = set()
    
    def wordBreak(self,s: str,wordDict: List[str]) -> bool:
        self.memo = set(wordDict)
        
        return self.word_break(s)
    
    def word_break(self,s):
        if s in self.memo:
            return True

        for i in range(1,len(s)):
            head = s[:i]
            tail = s[i:]
            
            head_possible = self.word_break(head)
            tail_possible = self.word_break(tail)
            
            if head_possible:
                self.memo.add(head)
            if tail_possible:
                self.memo.add(tail)
            
            if head_possible and tail_possible:
                return True
        
        return False     

谢谢！

解决方法

• 如果我们仅启用lru_cache()，则您的解决方案无需TLE即可正常工作：

class Solution:
    
    memo = set()
    
    def wordBreak(self,s: str,wordDict: List[str]) -> bool:
        self.memo = set(wordDict)
        
        return self.word_break(s)
    @lru_cache(None)
    def word_break(self,s):
        if s in self.memo:
            return True

        for i in range(1,len(s)):
            head = s[:i]
            tail = s[i:]
            
            head_possible = self.word_break(head)
            tail_possible = self.word_break(tail)
            
            if head_possible:
                self.memo.add(head)
            if tail_possible:
                self.memo.add(tail)
            
            if head_possible and tail_possible:
                return True
        
        return False

• 这也会在没有超过时间限制的情况下通过：

class Solution:
    def wordBreak(self,s,words):
        dp = [False] * len(s)
        for i in range(len(s)):
            for word in words:
                k = i - len(word)
                if word == s[k + 1:i + 1] and (dp[k] or k == -1):
                    dp[i] = True
        return dp[-1]