题解 | 单词拆分(一)

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param s string字符串 
# @param dic string字符串一维数组 
# @return bool布尔型
#
class Solution:
    def wordDiv(self , s: str, dic: List[str]) -> bool:
        # write code here 动态规划解题
        n = len(s)
        dp = [False]*(n+1)
        dp[0] = True
        for i in range(1,n+1):
            for j in range(i):
                if dp[j] and s[j:i] in dic:
                    dp[i] = True
                    break
        return dp[-1]
        # write code here 递归回溯解题
        '''n = len(s)
        def dfs(idx):
            if idx>=n:
                return True
            for i in range(idx,n):
                c = s[idx:i+1]
                if c in dic and dfs(i+1):
                    return True
            return False
        return dfs(0)'''