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","-11","*","17","+"] 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]
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。