Basic calculator with +,-,(,) and unary operator

/*
Logic:
We will keep a variable sum and add in sum one by one eg 1+21+(3+ -4)
add 1 with 21 then with (3+ -4) ...

sign will contain the sign of next upcoming integer(1,-1) eg. sign of full bracket is +
but sign of 4 is -
eg +- 9 so,sign of 9 is -

Iterate through whole string:
case 1:if currnt element is integer
   extract full integer eg. 789
   sign will have the sign for next integer ie. 789 so multiply element with sign
   now sign again will be +,so make sign =1.
case 2:
   push current sum in stack and then sign of bracket also so that after finding sum of
   bracket we will multiply bracket sum with sign and add with sum.
   new sing for bracket will be positive so sign=1

case 3:
   if closing bracket then we have solution of expression inside bracket in sum,
   so multiply sign of bracket which is top of stack with sum and add with
   actual sum which is in stack after sign.

case 4:
   if current char = '-' means it will reverse sign in all cases
   '+','-' ='-'
   '-', '-'= '+'

   '+' sign will not impact previous sum.
*/
class Solution {
public:
    int calculate(string s) {
        int sign=1;
        int sum=0;
        stack <int> st;
        for(int i=0;i<s.size();i++){
            if(isdigit(s[i])){
                int num=0;
                while(i<s.size() && isdigit(s[i])){
                    num=num*10+(s[i]-'0');
                    i++;
                }
                i--;
                num*=sign;
                sum+=num;
                continue;
            }
            else if(s[i]=='('){
                st.push(sum);
                st.push(sign);
                sum=0;
                sign=1;
            }
            else if(s[i]==')'){
                sum=sum*st.top();
                st.pop();
                sum+=st.top();
                st.pop();
            }
            else if(s[i]=='-') sign=-1;
            else if(s[i]=='+') sign=1;
        }
        return sum;
    }
};