
'''
150. Evaluate Reverse Polish Notation
Medium

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

'''

class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        stack=[]
        for si in tokens:
            if si == '+':
                b = int(stack.pop())
                a = int(stack.pop())
                stack+=[a+b]
            elif si == '-':
                b = int(stack.pop())
                a = int(stack.pop())
                stack+=[a-b]
            elif si == '*':
                b = int(stack.pop())
                a = int(stack.pop())
                stack+=[a*b]
            elif si == '/':
                b = int(stack.pop())
                a = int(stack.pop())
                stack+=[a//b if (a >= 0 and b>= 0 or a < 0 and b < 0) else -(-a//b)]
            else:
                stack+=[si]
        return stack[-1] if stack else None
        
#using stack, 
#when meet operator "+-*/" , pop twice,
#when operand, push