Description
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
描述
根据逆波兰表示法,求表达式的值。
有效的运算符包括 +, -, *, / 。每个运算对象可以是整数,也可以是另一个逆波兰表达式。
说明:
整数除法只保留整数部分。
给定逆波兰表达式总是有效的。换句话说,表达式总会得出有效数值且不存在除数为 0 的情况。
思路
- 使用两个栈来实现逆波兰表达式.
- 先将表达式中的所有元素压入到栈1,接下来我们pop出栈,如果是数字则压入栈2,如果是符号我们从栈2中弹出两个数组,用刚才的符号对其进行操作,再将操作的结果压入栈2.最后的结果是栈2中剩下的一个元素.
# -*- coding: utf-8 -*- # @Author: 何睿 # @Create Date: 2019-01-14 17:15:18 # @Last Modified by: 何睿 # @Last Modified time: 2019-01-14 17:50:02 class Solution: def evalRPN(self, tokens): """ :type tokens: List[str] :rtype: int """ stack1, stack2 = [], [] for i in range(len(tokens) - 1, -1, -1): stack1.append(tokens[i]) operateset = set(["+", "-", "*", "/"]) while stack1: top = stack1.pop() if top in operateset: num1, num2 = stack2.pop(), stack2.pop() stack2.append(self.operate(top, num1, num2)) else: stack2.append(int(top)) return stack2[-1] def operate(self, op, num1, num2): if op == "+": return num2 + num1 if op == "-": return num2 - num1 if op == "*": return num2 * num1 if op == "/": # 这里可以不用做判断,题目保证了除数不会是0 return int(num2 / num1) if num1 != 0 else 0
源代码文件在这里.