How could Scala do a merge sort?

Merge sort is a classical “divide and conquer” sorting algorithm. You should have to never write one because you’d be silly to do that when a standard library class already will already do it for you. But, it is useful to demonstrate a few characteristics of programming techniques in Scala. Firstly a quick recap on the merge sort. It is a divide and conquer algorithm. A list of elements is split up into smaller and smaller lists. When a list has one element it is considered sorted. It is then merged with the list beside it. When there are no more lists to merged the original data set is considered sorted. Now let’s take a look how to do that using an imperative approach in Java.

public void sort(int[] values) {
   int[] numbers = values;
   int[] auxillaryNumbers = new int[values.length];
   mergesort(numbers, auxillaryNumbers, 0, values.length - 1);

private void mergesort(int [] numbers, int [] auxillaryNumbers, int low, int high) {
    // Check if low is smaller then high, if not then the array is sorted
    if (low < high) {
        // Get the index of the element which is in the middle
        int middle = low + (high - low) / 2;

        // Sort the left side of the array
        mergesort(numbers, auxillaryNumbers, low, middle);
        // Sort the right side of the array
        mergesort(numbers, auxillaryNumbers, middle + 1, high);

        // Combine them both
        // Alex: the first time we hit this when there is min difference between high and low.
        merge(numbers, auxillaryNumbers, low, middle, high);

 * Merges a[low .. middle] with a[middle..high].
 * This method assumes a[low .. middle] and a[middle..high] are sorted. It returns
 * a[low..high] as an sorted array. 
private void merge(int [] a, int[] aux, int low, int middle, int high) {
    // Copy both parts into the aux array
    for (int k = low; k <= high; k++) {
        aux[k] = a[k];

    int i = low, j = middle + 1;
    for (int k = low; k <= high; k++) {
        if (i > middle)                      a[k] = aux[j++];
        else if (j > high)                   a[k] = aux[i++];
        else if (aux[j] < aux[i])            a[k] = aux[j++];
        else                                 a[k] = aux[i++];

public static void main(String args[]){
     ms.sort(new int[] {5, 3, 1, 17, 2, 8, 19, 11});



  1. An auxillary array is used to achieve the sort. Elements to be sorted are copied into it and then once sorted copied back. It is important this array is only created once otherwise there can be a performance hit from extensive array created. The merge method does not have to create an auxiliary array however since it changes an object it means the merge method has side effects.
  2. Merge sort big(O) performance is N log N.

Now let’s have a go at a Scala solution.

def mergeSort(xs: List[Int]): List[Int] = {
    val n = xs.length / 2
    if (n == 0) xs
    else {
      def merge(xs: List[Int], ys: List[Int]): List[Int] =
          (xs, ys) match {
          case(Nil, ys) => ys
          case(xs, Nil) => xs
          case(x :: xs1, y :: ys1) =>
            if (x < y) x::merge(xs1, ys)
            else y :: merge(xs, ys1)
      val (left, right) = xs splitAt(n)
      merge(mergeSort(left), mergeSort(right))

Key discussion points:

  1. It is the same divide and conquer idea.
  2. The splitAt function is used to divide up the data up each time into a tuple. For every recursion this will new a new tuple.
  3. The local function merge is then used to perform the merging. Local functions are a useful feature as they help promote encapsulation and prevent code bloat.
  4. Neiher the mergeSort() or merge() functions have any side effects. They don’t change any object. They create (and throw away) objects.
  5. Because the data is not been passed across iterations of the merging, there is no need to pass beginning and ending pointers which can get very buggy.
  6. This merge recursion uses pattern matching to great effect here. Not only is there matching for data lists but when a match happens the data lists are assigned to variables:
    • x meaning the top element in the left list
    • xs1 the rest of the left list
    • y meaning the top element in the right list
    • ys1 meaning the rest of the data in the right list

This makes it very easy to compare the top elements and to pass around the rest of the date to compare. Would the iterative approach be possible in Java? Of course. But it would be much more complex. You don’t have any pattern matching and you don’t get a nudge to declare objects as immutable as Scala does with making you make something val or var. In Java, it would always be easier to read the code for this problem if it was done in an imperative style where objects are being changed across iterations of a loop. But Scala a functional recursive approach can be quite neat. So here we see an example of how Scala makes it easier to achieve good, clean, concise recursion and a make a functional approach much more possible.

Reference: How could Scala do a merge sort? from our JCG partner Alex Staveley at the Dublin’s Tech Blog blog.
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