Basic Calculator III in C++

Suppose we have a simple expression string and we have to Implement a basic calculator to evaluate that expression. The expression string may contain opening and closing parentheses, the plus + or minus sign -, non-negative integers and empty spaces. The expression string contains only non-negative integers, +, -, *, / operators, opening and closing parentheses and empty spaces. The integer division should truncate toward zero.

So, if the input is like "6-4 / 2", then the output will be 4

To solve this, we will follow these steps −

• l1 := 0, l2 := 1

• o1 := 1, o2 := 1

• Define one stack st

• n := size of s

• for initialize i := 0, when i < n, update (increase i by 1), do −

  • x := s[i]

  • if x >= '0' and x <= '9', then −

    • num := x - '0'

    • while (i + 1 < n and s[i + 1] >= '0' and s[i + 1] <= '9'), do −

      • (increase i by 1)

      • num := (num * 10) + (s[i] - '0')

    • l2 := (if o2 is same as 1, then l2 * num, otherwise l2 / num)

  • otherwise when x is same as '(', then −

    • insert l1 into st, insert o1 into st

    • insert l2 into st, insert o2 into st

    • l1 := 0, o2 := 1

    • o1 := 1, l2 := 1

  • otherwise when x is same as ')', then −

    • temp := l1 + o1 * l2

    • o2 := top element of st

    • delete element from st

    • l2 := top element of st

    • delete element from st

    • o1 := top element of st

    • delete element from st

    • l1 := top element of st

    • delete element from st

    • l2 := (if o2 is same as 1, then l2 * temp, otherwise l2 / temp)

  • otherwise when x is same as '*' or x is same as '/', then −

    • o2 := (if x is same as '*', then 1, otherwise -1)

  • otherwise when x is same as '+' or x is same as '-', then −

    • if x is same as '-' and (i is same as 0 or (i - 1 >= 0 and s[i - 1] is same as '(')), then −

      • o1 := -1

      • Ignore following part, skip to the next iteration

    • l1 := l1 + o1 * l2

    • o1 := (if x is same as '+', then 1, otherwise -1)

    • l2 := 1, o2 := 1

• return l1 + o1 * l2

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
   public:
   int calculate(string s) {
      lli l1 = 0;
      lli l2 = 1;
      lli o1 = 1;
      lli o2 = 1;
      stack<lli> st;
      lli n = s.size();
      for (lli i = 0; i < n; i++) {
         char x = s[i];
         if (x >= '0' && x <= '9') {
            lli num = x - '0';
            while (i + 1 < n && s[i + 1] >= '0' && s[i + 1] <= '9') {
               i++;
               num = (num * 10) + (s[i] - '0');
            }
            l2 = (o2 == 1) ? l2 * num : l2 / num;
         }
         else if (x == '(') {
            st.push(l1);
            st.push(o1);
            st.push(l2);
            st.push(o2);
            l1 = 0;
            o2 = 1;
            o1 = 1;
            l2 = 1;
         }
         else if (x == ')') {
            lli temp = l1 + o1 * l2;
            o2 = st.top();
            st.pop();
            l2 = st.top();
            st.pop();
            o1 = st.top();
            st.pop();
            l1 = st.top();
            st.pop();
            l2 = (o2 == 1) ? l2 * temp : l2 / temp;
         }
         else if (x == '*' || x == '/') {
            o2 = (x == '*') ? 1 : -1;
         }
         else if (x == '+' || x == '-') {
            if (x == '-' && (i == 0 || (i - 1 >= 0 && s[i - 1] == '('))) {
               o1 = -1;
               continue;
            }
            l1 += o1 * l2;
            o1 = (x == '+') ? 1 : -1;
            l2 = 1;
            o2 = 1;
         }
      }
      return l1 + o1 * l2;
   }
};
main(){
   Solution ob;
   cout << (ob.calculate("(5+9*3)/8"));
}

Input

"(5+9*3)/8"

Output

4