Basic Calculator LeetCode solution in C++

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class Solution {
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 public:
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  int calculate(string s) {
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    stack<pair<int, int>> st; // pair(prev_calc_value , sign before next bracket () )
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    long long int sum = 0;
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    int sign = +1;
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    for (int i = 0; i < s.size(); ++i) {
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      char ch = s[i];
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      if (isdigit(ch)) {
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        long long int num = 0;
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        while (i < s.size() and isdigit(s[i])) {
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          num = (num * 10) + s[i] - '0';
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          i++;
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        }
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        i--; // as for loop also increase i , so if we don't decrease i here a
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             // sign will be skipped
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        sum += (num * sign);
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        sign = +1; // reseting sign
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      } else if (ch == '(') {
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        // Saving current state of (sum , sign) in stack
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        st.push(make_pair(sum, sign));
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        // Reseting sum and sign for inner bracket calculation
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        sum = 0;
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        sign = +1;
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      } else if (ch == ')') {
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        sum = st.top().first + (st.top().second * sum);
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        st.pop();
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      } else if (ch == '-') {
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        // toggle sign
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        sign = (-1 * sign);
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      }
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    }
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    return sum;
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  }
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};