150. Evaluate Reverse Polish Notation

https://leetcode.com/problems/evaluate-reverse-polish-notation/

Problem

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Note:

• Division between two integers should truncate toward zero.

• The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.

Example 1:

Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9

Example 2:

Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6

Example 3:

Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation: 
  ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

Solution

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<int> nums;
        for (const auto &t: tokens) {
            if (isInt(t)) {
                nums.push(parseInt(t));
            } else {
                const int op1 = nums.top(); nums.pop();
                const int op2 = nums.top(); nums.pop();
                switch(t[0]) {
                    case '+':
                        nums.push(op1 + op2);
                        break;
                    case '-':
                        nums.push(op2 - op1);
                        break;
                    case '*':
                        nums.push(op1 * op2);
                        break;
                    default:
                        nums.push(op2 / op1);
                }
            }
        }
        return nums.top();
    }
    inline bool isInt(const string &token) {
        if (token.size() > 1) return true;
        return isdigit(token[0]);
    }
    
    inline int parseInt(const string &token) {
        int i = (isdigit(token[0])) ? 0 : 1;
        int pos = (i == 0) ? 1 : -1;
        int num = 0;
        for (; i < token.size(); ++i) num = num * 10 + (token[i] - '0');
        return num * pos;
    }
};

• #stack