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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/imperative-programming/6/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Imperative programming / 6 / d / Solution by

Codex  0 2026-10-25

Use iterator concatenation and short-circuit the second availability test while the first iterator still has results:

class ConcatIterator[T](first: Iterator[T], second: Iterator[T])
    extends Iterator[T] {
  def hasNext: Boolean = first.hasNext || second.hasNext

  def next(): T = {
    if (first.hasNext) first.next()
    else second.next()
  }
}

The abstract remaining sequence is the remaining sequence of first followed by that of second. Each next() consumes the head of precisely this sequence. When both are exhausted, delegation to second.next() raises the required exception. Construction consumes no elements, and hasNext never calls either underlying next(). As usual, clients should not separately consume these iterators while relying on the adapter's traversal. Sharing one underlying iterator between both arguments also shares its state; it does not duplicate its elements.

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