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

题目大意 #

计算逆波兰表达式。

解题思路 #

这道题就是经典的考察栈的知识的题目。

代码 #


package leetcode

import (
	"strconv"
)

func evalRPN(tokens []string) int {
	if len(tokens) == 1 {
		i, _ := strconv.Atoi(tokens[0])
		return i
	}
	stack, top := []int{}, 0
	for _, v := range tokens {
		switch v {
		case "+":
			{
				sum := stack[top-2] + stack[top-1]
				stack = stack[:top-2]
				stack = append(stack, sum)
				top--
			}
		case "-":
			{
				sub := stack[top-2] - stack[top-1]
				stack = stack[:top-2]
				stack = append(stack, sub)
				top--
			}
		case "*":
			{
				mul := stack[top-2] * stack[top-1]
				stack = stack[:top-2]
				stack = append(stack, mul)
				top--
			}
		case "/":
			{
				div := stack[top-2] / stack[top-1]
				stack = stack[:top-2]
				stack = append(stack, div)
				top--
			}
		default:
			{
				i, _ := strconv.Atoi(v)
				stack = append(stack, i)
				top++
			}
		}
	}
	return stack[0]
}


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