動機

原來要用一般除法

Problem

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

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

Note that division between two integers should truncate toward zero.

It is guaranteed that the given RPN expression is always valid. That means the expression would always evaluate to a result, and there will not be any division by zero operation.

 

Example 1:

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

Example 2:

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

Example 3:

Input: tokens = [10,6,9,3,+,-11,*,/,*,17,+,5,+]Output: 22Explanation: ((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].

Sol

最初用整數除法處理，就吃了一個奇怪的WA，最後才想到一般除法

class Solution:
    def evalRPN(self, ts: List[str]) -> int:
        stk = []
        for t in ts:
            if t == "+":
                stk.append(stk.pop()+stk.pop())
            elif t == "*":
                stk.append(stk.pop()*stk.pop())
            elif t == "-":
                b,a = stk.pop(),stk.pop()
                stk.append(a-b)
            elif t == "/":
                b,a = stk.pop(),stk.pop()
                stk.append(int(a/b))
            else:
                stk.append(int(t))
            #print(t,stk)
        return stk[-1]