Addictive Number — C++

// Time:  O(n^3)
// Space: O(n)

class Solution {
public:
    bool isAdditiveNumber(string num) {
        for (int i = 1; i < num.length(); ++i) {
            for (int j = i + 1; j < num.length(); ++j) {
                string s1 = num.substr(0, i), s2 = num.substr(i, j - i);
                if ((s1.length() > 1 && s1[0] == '0') ||
                    (s2.length() > 1 && s2[0] == '0')) {
                    continue;
                }
                
                string next = add(s1, s2);
                string cur = s1 + s2 + next;
                while (cur.length() < num.length()) {
                    s1 = s2;
                    s2 = next;
                    next = add(s1, s2);
                    cur += next;
                }
                if (cur == num) {
                    return true;
                }
            }
        }
        return false;
    }

private:
        string add(const string& m, const string& n) {
            string res;
            int res_length = max(m.length(), n.length()) ;
    
            int carry = 0;
            for (int i = 0; i < res_length; ++i) {
                int m_digit_i = i < m.length() ? m[m.length() - 1 - i] - '0' : 0;
                int n_digit_i = i < n.length() ? n[n.length() - 1 - i] - '0' : 0;
                int sum = carry + m_digit_i + n_digit_i;
                carry = sum / 10;
                sum %= 10;
                res.push_back('0' + sum);
            }
            if (carry) {
                res.push_back('0' + carry);
            }
            reverse(res.begin(), res.end());
    
            return res;
        }
};