LeetCode: Restore IP Address

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)

---

We can apply depth first search to this problem. But we need to pay attention to every corner case. If current char is 0, it can only appear in a single segment. For example, if the string is '11011', it cannot be converted to '1.1.01.1'.

C++ Code:

/*
 * func: restore_ip_address_helper
 * goal: DFS helper function to find possible ip address
 * @param candidate: current candidate ip address
 * @param section: which section of ip is dealt with now
 * @param s: input string s
 * @param result: final result set
 * return:
 */
void restore_ip_address_helper(string &candidate, int section, string s, vector<string> &result){
    //If the remaining string is longer than the maximum, then return
    if(s.length() == 0 || s.length() > (4-section+1) * 3){
        return;
    }
    
    //If current section is the last one
    if(section == 4){
        //If there is only one bit remained
        if(s.length() == 1){
            candidate += s;
            result.emplace_back(candidate);
        }
        //If there are more than one bits remained and the first one is not 0
        else if(s.length() > 1 && s.length() <= 3 && s[0] != '0' && atoi(s.c_str()) <= 255){
            candidate += s;
            result.emplace_back(candidate);
        }
        return;
    }
    //If the first one is 0, we have to put it into a single section
    if(s[0] == '0'){
        candidate += '0';
        candidate += '.';
        restore_ip_address_helper(candidate, section+1, s.substr(1), result);
    }else{
        string cpy = candidate;
        for(int i = 1; i <= std::min(3, static_cast<int>(s.length())); i++){
            string tmp = s.substr(0, i);
            if(atoi(tmp.c_str()) <= 255){
                candidate += tmp;
                candidate += '.';
                restore_ip_address_helper(candidate, section+1, s.substr(i), result);
                candidate = cpy;
            }
        }
    }
}

/*
 * func: restore_ip_address
 * goal: restore ip address from a string
 * @param s: a input string
 * return: all possible ips
 */
vector<string> restore_ip_address(string s){
    vector<string> result;
    
    string candidate = "";
    restore_ip_address_helper(candidate, 1, s, result);
    
    return result;
}

Python Code:

# func: restore ip address from a string
# @param s: input string
# @return: all possible ips
def restore_ip_address(s):
    result = []

    def restore_ip_address_helper(candidate, section, str):
        if len(str) == 0 or len(str) > (5-section) * 3:
            return
        if section == 4:
            if len(str) == 1:
                candidate += str
                result.append(candidate)
            elif 1 < len(str) <= 3 and str[0] != '0' and int(str) <= 255:
                candidate += str
                result.append(candidate)
            return

        if str[0] == '0':
            candidate += '0.'
            restore_ip_address_helper(candidate[:], section+1, str[1:])
        else:
            tmp = candidate[:]
            for i in xrange(1, min(4, len(str)+1)):
                if int(str[0:i]) <= 255:
                    candidate += str[0:i] + '.'
                    restore_ip_address_helper(candidate[:], section+1, str[i:])
                    candidate = tmp[:]

    restore_ip_address_helper("", 1, s)
    return result