Question

我有这个递归函数：

 def series(n: Int): Double = 
 {
   if(n == 1)  
     return 3 
   if (n == 2)
     return 5
   Math.pow(-1, n) * 5 * series(n - 1) + Math.pow(-1, n-1) * 3 * series(n - 2)
 }

我必须把它变成尾递归的。我不知道如何执行此操作谢谢

Answer 1

这个系列的尾递归版本是可能的。诀窍是将current和previous结果保留在函数参数中。

fooSeries遵循自下而上的方法，首先将n = 1和n = 2结果作为输入，然后根据较小的n计算较大的n。

注意helper将counter的初始值设为3，初始current为5，previous为3

def fooSeries(n: Int): Double = {

    @scala.annotation.tailrec
    def helper(counter: Int, current: Double, previous: Double): Double = {
      n match {
        case 1 => previous
        case 2 => current
        case _ =>
          if (counter > n) current
          else
            helper(counter + 1, Math.pow(-1, counter) * 5 * current + Math.pow(-1, counter - 1) * 3 * previous, current)
      }
    }

    helper(3, 5, 3)
  }

Scala REPL

scala> fooSeries(1)
res46: Double = 3.0

scala> series(1)
res47: Double = 3.0

scala> fooSeries(2)
res48: Double = 5.0

scala> series(2)
res49: Double = 5.0

scala> fooSeries(3)
res50: Double = -16.0

scala> series(3)
res51: Double = -16.0

scala> fooSeries(4)
res52: Double = -95.0

scala> series(4)
res53: Double = -95.0

scala> fooSeries(5)
res54: Double = 427.0

scala> series(5)
res55: Double = 427.0

scala> fooSeries(6)
res56: Double = 2420.0

scala> series(6)
res57: Double = 2420.0

scala> fooSeries(11)
res58: Double = -6955321.0

scala> series(11)
res59: Double = -6955321.0

在Scala中将递归函数转换为尾递归函数

1 个答案: