scala - 如何使此代码更具功能性？

我是函数式编程的新手。我只是尝试解决以下问题：

[ a rough specification ]

e.g.1:
dividend : {3,5,9}
divisor : {2,2}
radix = 10
ans (remainder) : {7}

Procedure :
dividend = 3*10^2+5*10^1+9*10^0 = 359
similarly, divisor = 22
so 359 % 22 = 7

e.g.2:
dividend : {555,555,555,555,555,555,555,555,555,555}
divisor: {112,112,112,112,112,112,112,112,112,112}
radix = 1000
ans (remainder) : {107,107,107,107,107,107,107,107,107,107}

我对这个问题的解决方案是：

object Tornedo {
  def main(args: Array[String]) {
    val radix: BigInt = 1000
    def buildNum(segs: BigInt*) = (BigInt(0) /: segs.toList) { _ * radix + _ }
    val dividend = buildNum(555,555,555,555,555,555,555,555,555,555)
    val divisor = buildNum(112,112,112,112,112,112,112,112,112,112)
    var remainder = dividend % divisor
    var rem = List[BigInt]()
    while(remainder > 0) {
      rem = (remainder % radix) :: rem
      remainder /= radix
    }
    println(rem)
  }
}

尽管我对此代码感到非常满意，但我想知道如何消除while循环和两个可变变量并使此代码更具功能。

任何帮助将不胜感激。

谢谢。 :)

最佳答案

这个尾部递归函数删除了两个可变的var和循环：

object Tornedo {
  def main(args: Array[String]) {
    val radix: BigInt = 1000
    def buildNum(segs: BigInt*) = (BigInt(0) /: segs.toList) { _ * radix + _ }
    val dividend = buildNum(555,555,555,555,555,555,555,555,555,555)
    val divisor = buildNum(112,112,112,112,112,112,112,112,112,112)
    def breakup(n: BigInt, segs: List[BigInt]): List[BigInt] =
      if (n == 0) segs else breakup(n / radix, n % radix :: segs)
    println(breakup(dividend % divisor, Nil))
  }
}