leetcode150 Evaluate Reverse Polish Notation

"""
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
"""
"""
此题思路很简单，用栈即可实现，自己AC，但遇到三个麻烦
第一个是加减乘除操作要将字符型转为整型
第二个是被减数和减数的问题。先弹出的是减数，后弹出的是被减数。除法也需这样考虑
第三个是python向下取整的问题 -6 // 132 == -1
"""
class Solution1:
    def evalRPN(self, tokens):
        if not tokens:
            return 0
        stack = []
        for i in range(len(tokens)):
            stack.append(tokens[i])
            if tokens[i] == '+':
                stack.pop()
                x = int(stack.pop())
                y = int(stack.pop())
                stack.append(y+x)
            if tokens[i] == '-':
                stack.pop()
                x = int(stack.pop())
                y = int(stack.pop())
                stack.append(y-x) #减法注意先出栈的是减数,同除法
            if tokens[i] == '*':
                stack.pop()
                x = int(stack.pop())
                y = int(stack.pop())
                stack.append(y*x)
            if tokens[i] == '/':
                stack.pop()
                x = int(stack.pop())
                y = int(stack.pop())
                if x > 0 and y < 0 or x < 0 and y > 0:
                    stack.append(-(abs(y)//abs(x)))
                else:                    #6//-132 == -1,python的向下取整,所以加了这条语句
                    stack.append(y//x)
        return stack.pop()

if __name__ == '__main__':
    ans = Solution1()
    x = ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
    s = ans.evalRPN(x)
    print(s)

print(6//-132)