leetcode || 139、Word Break

problem：

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given

s = "leetcode",

dict = ["leet", "code"].

Return true because "leetcode" can
be segmented as "leet code".

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Dynamic Programming

thinking：

从s的第一个字母向后匹配，如果i前面的前缀可以匹配，就看s字符串i以后的后缀是否匹配

code：

class Solution{
    public:
bool wordBreakHelper(string s, unordered_set<string> &dict,set<string> &unmatch) {
        if(s.length() < 1) return true;
		bool flag = false;
		for(int i = 1; i <= s.length(); i++)
		{
			string prefixstr = s.substr(0,i);
			unordered_set<string>::iterator it = dict.find(prefixstr);
			if(it != dict.end())
			{
				string suffixstr = s.substr(i);
				set<string>::iterator its = unmatch.find(suffixstr);
				if(its != unmatch.end())continue;
				else{
					flag = wordBreakHelper(suffixstr,dict,unmatch);
					if(flag) return true;
					else unmatch.insert(suffixstr);
				}
			}
		}
		return false;
    }
	bool wordBreak(string s, unordered_set<string> &dict) {
        int len = s.length();
		if(len < 1) return true;
		set<string> unmatch;
		return wordBreakHelper(s,dict,unmatch);
    }
};