添加 0150.逆波兰表达式求值.md Scala版本

This commit is contained in:
ZongqinWang
2022-05-17 17:08:11 +08:00
parent 8b44527e94
commit 8c394c9f85

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@ -325,6 +325,33 @@ func evalRPN(_ tokens: [String]) -> Int {
return stack.last! return stack.last!
} }
``` ```
Scala:
```scala
object Solution {
import scala.collection.mutable
def evalRPN(tokens: Array[String]): Int = {
val stack = mutable.Stack[Int]() // 定义栈
// 抽取运算操作需要传递x,y和一个函数
def operator(x: Int, y: Int, f: (Int, Int) => Int): Int = f(x, y)
for (token <- tokens) {
// 模式匹配,匹配不同的操作符做什么样的运算
token match {
// 最后一个参数 _+_代表x+y遵循Scala的函数至简原则以下运算同理
case "+" => stack.push(operator(stack.pop(), stack.pop(), _ + _))
case "-" => stack.push(operator(stack.pop(), stack.pop(), -_ + _))
case "*" => stack.push(operator(stack.pop(), stack.pop(), _ * _))
case "/" => {
var pop1 = stack.pop()
var pop2 = stack.pop()
stack.push(operator(pop2, pop1, _ / _))
}
case _ => stack.push(token.toInt) // 不是运算符就入栈
}
}
// 最后返回栈顶不需要加return关键字
stack.pop()
}
}
```
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