Restore IP Addresses

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

For example:
Given "25525511135",

return ["255.255.11.135", "255.255.111.35"]. (Order does not matter)

1.我的答案 回溯

class Solution {
public:
   void split(vector<string>&res, string s, vector<string>& vec, int num){
    int len = s.size();
    if(num == 0 && len > 0 && len < 4){
        if(s.size()>1 && s[0] == '0')
        return;
        else{
        int temp = atoi(s.c_str());
        if(temp <=255 && temp >=0){
            string sres = vec[0]+"."+vec[1]+"."+vec[2]+"."+s;
            res.push_back(sres);
            return;
            }
        }
    }
    int i = 1;
    while(i <=3 && i < len){
        string stemp = s.substr(0,i);
        if(stemp.size()>1 && stemp[0] == '0'){//用来判断是否有“010”这种情况出现，出现就删掉
            ++i;
            continue;
        }
        else{
        int sint = atoi(stemp.c_str());
        if(sint <= 255 && sint >=0){
            vec.push_back(stemp);
            string str2 = s.substr(i,len-i);
            split(res,str2,vec,num-1);
            vec.pop_back();
        }
        }
       ++i; 
    }
    return;
}

vector<string> restoreIpAddresses(string s) {
    vector<string>res,vec;
    int num = 3;
    int len = s.size();
    if(len < 4 || len > 12)
        return res;
    split(res,s,vec,num);
    return res;
}
};

2.别人的答案，时间一样，就是比我的要整洁

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> result;
        string ip;
        dfs(s,0,0,ip,result); //paras:string s,start index of s,step(from0-3),intermediate ip,final result
        return result;
    }
    void dfs(string s,int start,int step,string ip,vector<string>& result){
        if(start==s.size()&&step==4){
            ip.erase(ip.end()-1); //remove the last '.' from the last decimal number
            result.push_back(ip);
            return;
        }
        if(s.size()-start>(4-step)*3) return;
        if(s.size()-start<(4-step)) return;
        int num=0;
        for(int i=start;i<start+3;i++){
            num=num*10+(s[i]-'0');
            if(num<=255){
                ip+=s[i];
                dfs(s,i+1,step+1,ip+'.',result);
            }
            if(num==0) break;
        }
    }
};