[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;

    }

};