Word Break

问题链接[https://leetcode.com/problems/word-break/]
给定：
1. 字符串 s = "leetcode",
2. 字典 dict = ["leet", "code"].
判断字符串是否能分解成一个或多个字典里的单词。
Return true because "leetcode" can be segmented as "leet code".

经典的动态规划问题，一直想写下来，一直都没时间写，现在先记下来，以后有空补上

class Solution:
    # @param s, a string
    # @param wordDict, a set<string>
    # @return a boolean
    def wordBreak(self, s, wordDict):
        s_len=len(s)
        DP=[False]*(s_len+1)
        DP[0]=True
        for i in range(s_len+1):
            for j in range(i):
                if(s[j:i] in wordDict and DP[j]):
                    DP[i]=True
        return DP[s_len]

DP[i]==True 表示到第i个字符为止，该字符串能由字典里的单词组成