原題

對(duì)于一個(gè)逆波蘭數(shù)學(xué)表達(dá)式求值

["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

解題思路

• 使用stack，遍歷逆波蘭表達(dá)式
• 遇到+-*/則從stack中pop出兩個(gè)數(shù)做相應(yīng)的運(yùn)算，結(jié)果push入棧
• 遇到數(shù)字直接入棧

完整代碼

class Solution(object):
    def evalRPN(self, tokens):
        """
        :type tokens: List[str]
        :rtype: int
        """
        stack = []
        operators = "+-*/"
        for str in tokens:
            if str not in operators:
                stack.append(int(str))
            else:
                a = stack.pop()
                b = stack.pop()
                if str == "+":
                    stack.append(a + b)
                elif str == "-":
                    stack.append(b - a)
                elif str == "*":
                    stack.append(a * b)
                elif str == "/":
                    if a * b < 0 and b % a != 0:
                        stack.append(b / a + 1)
                    else:
                        stack.append(b / a)
        if not stack:
            return 0
        return stack.pop()