LeetCode 306. Additive Number

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I first thought about the commented out method. However, that one is hard to split the numbers which start with '0'.

Second Method is from leetcode discuss forum.

// try backtracking?//bool isAdditiveNumber(string nums, int start, int end) {//  if(end == nums.size()) return true;//  if(end - start < 2) return false;//  for(int i = start; i < nums.size() - 2; ++i) {//    string num_1 = nums.substr(start, i - start + 1);//    for(int j = i  + 1; j < nums.size() - 1; ++j) {//      string num_2 = nums.substr(i + 1, j - i);//      for(int k = j  + 1; k < nums.size(); ++k) {//        string num_3 = nums.substr(j + 1, k - j);//        int a = stoi(num_1);//        int b = stoi(num_2);//        int c = stoi(num_3);//        if(a + b == c) {//          return isAdditiveNumber(nums, i + 1, k + 1);//        }//      }//    }//  }//  return false;//}////bool isAdditiveNumber(string num) {//  int start = 0;//  int end = num.size() - 1;//  return isAdditiveNumber(num, start, end);//}

bool ok(string& num, int idx, unsigned long a, unsigned long b) {  if(idx == num.size()) return true;  unsigned long c = 0;  for(int i = idx; i < num.size(); ++i) {    c = c * 10 + num[i] - '0';    if(a + b == c) {      if(ok(num, i + 1, b, c)) return true;    } else if(a + b < c) break;    if(num[idx] == '0') break;  }  return false;}bool isAdditiveNumber(string num) {  unsigned long a = 0;   // pay attention to the unsigned long type.  for(int i = 0; i < num.size(); ++i) {    a = a * 10 + num[i] - '0';    unsigned long b = 0;    for(int j = i + 1; j < num.size(); ++j) {      b = b * 10 + num[j] - '0';      if(j + 1 < num.size() && ok(num, j + 1, a, b)) return true;      if(num[i+1] == '0') break;    }    if(num[0] == '0') break;  }  return false;}int main(void) {  cout << isAdditiveNumber("12358") << endl;  cout << isAdditiveNumber("211738") << endl;  cout << isAdditiveNumber("199100199") << endl;  cout << isAdditiveNumber("198019823962") << endl;}

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