[LeetCode]Word Break II

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

参考：http://blog.csdn.net/doc_sgl/article/details/13064739

class Solution {
public:
    void wordBreakHelper(string s,unordered_set<string> &dict,set<string> &unmatched,
		int mn,int mx, vector<string> &path, vector<string> &res) {
		int i = mx < s.length() ? mx : s.length();
		for(; i >= mn ; i--)
		{
			string preffixstr = s.substr(0,i);
			if(dict.find(preffixstr) != dict.end()){
				path.push_back(preffixstr);
				if(preffixstr.size() == s.size())
				{
					string tmp = path[0];
					for(int i=1; i<path.size(); i++)
						tmp+= " "+path[i];
					res.push_back(tmp);
				}
				string suffixstr = s.substr(i);
				if(unmatched.find(suffixstr) == unmatched.end())
				{
					int oldsz = res.size();
					wordBreakHelper(suffixstr,dict,unmatched,mn,mx,path,res);
					if(res.size() == oldsz)
						unmatched.insert(suffixstr);
				}
				path.pop_back();
			}
		}
    }
	vector<string> wordBreak(string s, unordered_set<string> &dict) {
        // Note: The Solution object is instantiated only once.
        vector<string> res;
		if(s.size() < 1 || dict.empty()) return res;
		unordered_set<string>::iterator it = dict.begin();
		int maxlen=(*it).length(), minlen=(*it).length();
		for(it++; it != dict.end(); it++)
			if((*it).length() > maxlen)
				maxlen = (*it).length();
			else if((*it).length() < minlen)
				minlen = (*it).length();
        set<string> unmatched;
		vector<string> path;
		wordBreakHelper(s,dict,unmatched,minlen,maxlen,path,res);
		return res;
    }
};