LeetCode - Evaluate Reverse Polish Notation

You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation.

Evaluate the expression. Return an integer that represents the value of the expression.

Note that:

• The valid operators are '+', '-', '*', and '/'.
• Each operand may be an integer or another expression.
• The division between two integers always truncates toward zero.
• There will not be any division by zero.
• The input represents a valid arithmetic expression in a reverse polish notation.
• The answer and all the intermediate calculations can be represented in a 32-bit integer.

 

Example 1:

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

Example 2:

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

Example 3:

Input: tokens = ["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

 

Constraints:

• 1 <= tokens.length <= 104
• tokens[i] is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].

 

My Solution:

import math


class Solution(object):
    def evalRPN(self, tokens):
        """
        :type tokens: List[str]
        :rtype: int
        """

        if len(tokens) == 1:
            return int(tokens[0])

        operators = ['+', '-', '*', '/']
        stack = []
        for t in tokens:
            # t is an operator
            if t in operators:
                right = stack.pop()
                left = stack.pop()
                stack.append(self.calc(left, t, right))
            # t is a number
            else:
                stack.append(int(t))
        
        return stack[0]


    def calc(self, left, operator, right):
        a = int(left)
        b = int(right)
        if operator == '+':
            return a + b
        elif operator == '-':
            return a - b
        elif operator == '*':
            return a * b
        elif operator == '/':
            # in some python versions math.trunc() does not truncate toward zero
            if a * b >= 0:
                return a // b 
            elif a < 0:
                return -(-1 * a // right)
            elif b < 0:
                return -(a // (-1 * b))