The algorithm for evaluating any postfix expression is fairly straightforward:

While there are input tokens left

• • Read the next token from input.
  • If the token is a value
    • Push it onto the stack.
  • Otherwise, the token is an operator (operator here includes both operators and functions).
    • It is known  that the operator takes n arguments.
    • If there are fewer than n values on the stack
      • (Error) The user has not input sufficient values in the expression.
    • Else, Pop the top n values from the stack.
    • Evaluate the operator, with the values as arguments.
    • Push the returned results, if any, back onto the stack.
• If there is only one value in the stack
  • That value is the result of the calculation.
• Otherwise, there are more values in the stack
  • (Error) The user input has too many values.

1 class Solution { 2 public: 3     int evalRPN(vector
     
       &tokens) { 4         if (tokens.empty()) return 0; 5          6         stack
      
        st; 7          8         for (int i = 0; i < tokens.size(); ++i) { 9             if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") {10                 if (st.empty()) return 0;11                 int n1 = st.top();12                 st.pop();13                 if (st.empty()) return 0;14                 int n2 = st.top();15                 st.pop();16                 if (tokens[i] == "+") st.push(n1 + n2);17                 else if (tokens[i] == "-") st.push(n2 - n1);18                 else if (tokens[i] == "*") st.push(n2 * n1);19                 else if (n1 == 0) return 0;20                 else st.push(n2 / n1);21             } else {22                 st.push(atoi(tokens[i].c_str()));23             }24         }25         26         return st.top();27     }28 };