word break II

DP+剪枝
class Solution {
public:
    
    bool dfs(string s, unordered_set<string> &dict, unordered_set<string> &unmatch, int maxLen, vector<string> &result,string path){
        
        if (s.size() == 0){
            path.pop_back();
            result.push_back(path);
            return true;
        }
        
        bool flag = false;
        int size = s.size();
        for (int i = 1; i <= min(maxLen,size); i++){
            
            if (dict.find(s.substr(0,i)) != dict.end()){
                
                if (unmatch.find(s.substr(i) )== unmatch.end()){
                    
                    path+=s.substr(0,i);
                    path.push_back(' ');
                    if (dfs(s.substr(i), dict, unmatch, maxLen, result, path)) flag = true;
                    else unmatch.insert(s.substr(i));
                    for (int j = 0; j < i+1; j++) path.pop_back();
                }
            }
        }
        return flag;
    }
    
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<string> result;
        if (s.size() == 0) return result;
        int maxLen = 0;
        unordered_set<string> :: iterator it;
        for (it = dict.begin(); it != dict.end(); it++){
            if (it->size() > maxLen) maxLen = it->size();
        }
        
        unordered_set<string> unmatch;
        string path;
        bool hasAnswer = dfs(s,dict,unmatch,maxLen,result,path);
        return result;
    }
};