leetcode || 139、Word Break

标签：leetcode   dp   

problem：

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);
    }
};

leetcode || 139、Word Break

标签：leetcode   dp   

原文地址：http://blog.csdn.net/hustyangju/article/details/45476583