《leetcode-go》逆波兰式计算

Evaluate the value of an arithmetic expression in Reverse Polish Notation. 

Valid operators are+,-,*,/. Each operand may be an integer or another expression. 

Some examples: 

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
  ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

在做这个题目之前，我们先了解一下什么是逆波兰式？

我们平常人类常用的计算方式(1+2)*(3+4)-(5*6)是中缀表达式，这种表达式对我们来说很简单，但是对计算机来说却很复杂。

所以转化成计算机喜欢的逆波兰式，上面的例子转化为逆波兰式则为

->"(1+2)*(3+4)","(5*6)","-"

->"(1+2)","(3+4)","*","5","6","*","-"

->"1","2","+","3","4","+","*","5","6","*","-"

这种算式需要采用栈式结构进行计算，先进后出。

package main

import (
	"fmt"
	"strconv"
)

/**
 *
 * @param tokens string字符串一维数组
 * @return int整型
 */
type Stack struct {
	i       int
	numList []int
}

func (stack *Stack) Push(num int) {
	stack.numList = append(stack.numList, num)
	stack.i++
}
func (stack *Stack) Pop() (num int) {
	num = stack.numList[stack.i-1]
	stack.numList = stack.numList[0 : stack.i-1]
	stack.i--
	return
}
func main() {
	var tokens = []string{"2", "1", "+", "3", "*"}
	res := evalRPN(tokens)
	fmt.Println(res)

}
func evalRPN(tokens []string) int {
	// write code here
	stack := new(Stack)
	for _, str := range tokens {
		if str == "+" || str == "-" || str == "*" || str == "/" {
			num1 := stack.Pop()
			num2 := stack.Pop()
			var res int
			switch str {
			case "+":
				res = num2 + num1
				break
			case "-":
				res = num2 - num1
				break
			case "*":
				res = num2 * num1
				break
			case "/":
				res = num2 / num1
				break
			}
			stack.Push(res)
			fmt.Println("ha", stack)
		} else {
			intNum, _ := strconv.Atoi(str)
			stack.Push(intNum)
			fmt.Println("hei", stack)
		}
	}
	return stack.Pop()
}