Multiply Strings LeetCode solution in C++

1
class Solution {
2
  void compute(string &num, int dig, int ind, string &ans) {
3
    int c = 0; // carry digit..
4
    int i = num.size() - 1;
5
    //  again travarsing the string in reverse
6
    while (i >= 0) {
7
      int mul = dig * (num[i] - '0') + c; // the curr digit's multiplication
8
      c = mul / 10;                       // carry update..
9
      mul %= 10;                          // mul update in  a single digit..
10
      if (ind >= ans.length()) {          // here if the index where we'll put the value
11
                                          // is out of bounds...
12
        ans.push_back('0' + mul);
13
      } else {
14
        //  here adding the val with the previous digit and further computing
15
        //  the carry...
16
        mul += (ans[ind] - '0');
17
        c += mul / 10;
18
        mul %= 10;
19
        ans[ind] = ('0' + mul);
20
      }
21
      i--;
22
      ind++; // increment the index where we'll put the val;
23
    }
24

25
    if (c > 0) { //   if carry is non-zero...
26
      ans.push_back('0' + c);
27
    }
28
  }
29

30
 public:
31
  string multiply(string num1, string num2) {
32
    string ans = "";                            //  the string which we have to return as answer..
33
    if (num1 == "0" || num2 == "0") return "0"; // base case..
34
    int ind = 0; // the index from which we'll add the multiplied value to the
35
                 // ans string
36
    for (int i = num1.size() - 1; i >= 0; ind++, i--) {
37
      // travarsing in reverse dir. on num1 and increasing ind bcz for every
38
      // digit
39
      //  of num1 we'll add the multiplication in a index greater than the
40
      //  previous iteration
41
      int dig = num1[i] - '0';      // the digit with which we'll multiply by num2..
42
      compute(num2, dig, ind, ans); // function call for every digit and num2
43
    }
44
    //   we have calculated the ans in a reverse way such that we can easily add
45
    //   a leading digit & now reversing it...
46
    reverse(ans.begin(), ans.end());
47
    return ans;
48
  }
49
};