scala 学习之： list.fill 用法

题目描述：

Decode a run-length encoded list.
Given a run-length code list generated as specified in problem P10, construct its uncompressed version.
Example:

scala> decode(List((4, ‘a), (1, ‘b), (2, ‘c), (2, ‘a), (1, ‘d), (4, ‘e)))
res0: List[Symbol] = List(‘a, ‘a, ‘a, ‘a, ‘b, ‘c, ‘c, ‘a, ‘a, ‘d, ‘e, ‘e, ‘e, ‘e)

根据List中的元素，将其展开， 与之前的encodeList 做相反的操作。

思路： 使用之前提到的foldLeft函数， 从左到右遍历List，将其每一个元组展开，放入初始值里面。

代码：

   def decodeList[T](a:List[(Int, T)]): List[T] = a.foldLeft(List[T]()){
      case (res, cur) => {
           val count = cur._1
           val content = cur._2
           res:::List.fill(count)(content)
       }
      }
    val b = List((10, "a"), (2,"zb"))
    println(decodeList(b))

List.fill:

  def fill[A](n: Int)(elem: => A): CC[A] = {
    val b = newBuilder[A]
    b.sizeHint(n)
    var i = 0
    while (i < n) {
      b += elem
      i += 1
    }
    b.result
  }

向集合中插入n个类型为A的元素。