Leetcode 139. Word Break （python+cpp）

题目

解法1：recursion (TLE)

最直观的解法就是用recursion将string不断分成小的部分，但是这样的方法会超时，代码如下：

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        # recurrsion 解法,TLE
        def dfs(start):
            if start == len(s):
                return True
            
            for i in range(start+1,len(s)+1):
                if s[start:i] in wordDict and dfs(i):
                    return True
            return False
    
        return dfs(0)

解法2：recursion + memorization(bottom up)

其实上面recursion的解法会对string的某个位置重复计算很多遍，所以可以用一个字典来记录已经做过的位置，避免重复计算，只需要加几行代码即可。
python代码如下：

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        def helper(curr):
            if curr == len(s):
                return True
            if curr in memo:
                return memo[curr]
            memo[curr] = False
            for i in range(curr,len(s)+1):
                if s[curr:i] in wordDict and helper(i):
                    memo[curr] = True
                    break
            return memo[curr]
        
        memo = {
   }
        return helper(0)

解法3：recursion + memorization(top down)

class Solution:
    def wordBreak(self, s: str, wordDict: List[str