leetcode 139. Word Break

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题目链接：leetcode

思路：
用递归做会超时，发现可以分解为更小的问题，用动态规划

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class Solution(object):
    def wordBreak(self, s, wordDict):
        """
        :type s: str
        :type wordDict: List[str]
        :rtype: bool
        """
        d = {}
        dp = [False]*len(s)
        for word in wordDict:
            if word not in d:
                d[word] = 1
        dp[0] = True if s[0] in d else False
        for i in range(1,len(s)):
            if s[:i+1] in d:
                dp[i] = True
                continue
            for j in range(i):
                if dp[j] and s[j+1:i+1] in d:
                    dp[i] = True
                    break
        return dp[-1]