leetcode Question 83: Restore IP Addresses

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)

Analysis:
This problem can be viewed as a DP problem. There needed 3 dots to divide the string, and make sure the IP address is valid:  less than or equal to 255, greater or equal to 0, and note that, "0X" or "00X" is not valid.
For the DP, the length of each part is from 1 to 3. We use a vector<string> to store each part, and cut the string every time. Details see the code.

Note that "atoi" is for c-string, <string> need to convert to cstring by str.c_str();

Code(Updated 201309): 

class Solution {
public :
    bool valid(string s){
        if (s.size()==3 && ( atoi (s.c_str())>255 || atoi (s.c_str())==0)){ return false ;}
        if (s.size()==3 && s[0]== '0' ){ return false ;}
        if (s.size()==2 && atoi (s.c_str())==0){ return false ;}
        if (s.size()==2 && s[0]== '0' ){ return false ;}
        return true ;
    }

    void getRes(string s, string r, vector<string> &res, int k){
        if (k==0){
            if (s.empty()){res.push_back(r);}
            return ;
        } else {
            for ( int i=1;i<=3;i++){
                if (s.size()>=i && valid(s.substr(0,i))){
                    if (k==1){getRes(s.substr(i),r+s.substr(0,i),res,k-1);}
                    else {getRes(s.substr(i),r+s.substr(0,i)+ "." ,res,k-1);}
                }
            }
        }
    }

    vector<string> restoreIpAddresses(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<string> res;
        getRes(s, "" ,res,4);
        return res;
    }
};

Code(old version):

class Solution {
public :

    void dp(string s,vector<string> &cur ,vector<string> &res){
        if (cur.size()==3){ // if there are 4 parts in the original string
            cur.push_back(s); //all 4 parts and check if valid
            bool r = true ;
            for ( int i=0;i<4;i++){
                if ( atoi (cur[i].c_str())>255){  //check value
                    r = false ;
                    break ;
                }
                if ((cur[i].size()>1 && cur[i][0]== '0' )){ //check "0X" "00X" and "0XX" cases
                    r = false ;
                    break ;
                }
            }
            if (r){
                res.push_back(cur[0]+ "." +cur[1]+ "." +cur[2]+ "." +cur[3]);
            }
            cur.pop_back();

        } else {
            for ( int i=0;i<3;i++){
                if (s.size()>i+1){
                    cur.push_back(s.substr(0,i+1));
                    dp(s.substr(i+1,(s.size()-i-1)),cur,res);
                    cur.pop_back();
                }
            }
        }

    }


    vector<string> restoreIpAddresses(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<string> res,cur;
        if (s.size()>12 || s.size()<4 ){ return res;}
        dp(s,cur,res); // cur stores the current separation
        return res;
    }
};