0150. Evaluate Reverse Polish Notation

# 150. Evaluate Reverse Polish Notation#

## 题目 #

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

``````

## 代码 #

``````
package leetcode

import (
"strconv"
)

func evalRPN(tokens []string) int {
stack := make([]int, 0, len(tokens))
for _, token := range tokens {
v, err := strconv.Atoi(token)
if err == nil {
stack = append(stack, v)
} else {
num1, num2 := stack[len(stack)-2], stack[len(stack)-1]
stack = stack[:len(stack)-2]
switch token {
case "+":
stack = append(stack, num1+num2)
case "-":
stack = append(stack, num1-num2)
case "*":
stack = append(stack, num1*num2)
case "/":
stack = append(stack, num1/num2)
}
}
}
return stack
}

`````` Oct 9, 2021 Edit this page