[LeetCode] 150. Evaluate Reverse Polish Notation_Medium tag: Stack

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

这个题目就是用stack，如果看到{‘+‘, ‘-‘,‘*‘,‘/‘}, 将stack pop两次，然后进行相应的计算，注意 1// -32 == -1， 而 int(1/ -32) == 0;

T: O(n)   S: O(n)

Code

class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        if not tokens: return 0
        stack, d = [], {‘+‘, ‘-‘,‘*‘,‘/‘}
        for token in tokens:
            if token in d:
                num2 = stack.pop()
                num1 = stack.pop()
                if token == ‘+‘:
                    num = num1 + num2
                elif token == ‘-‘:
                    num = num1 - num2
                elif token == ‘*‘:
                    num = num1 * num2
                else:
                    num = int(num1/num2)
                stack.append(num)
            else:
                stack.append(int(token))
        return stack[0]