Sorting a linked list

Question

I have written a basic linked list class in C#. It has a Node object, which (obviously) represents every node in the list.

The code does not use IEnumerable, however

Answer 1

Some people (including me) may have wanted to sort LinkedList from .net library.

The easy way is to use Linq, sort and finally create a new linked list. but this creates garbage. for small collections it would not be a problem, but for large collections it may not be so efficient.

for people who want some degree of optimization, here is generic In-place quick sort implementation for .net doubly linked list.

this implementation does not split/merge, instead it checks nodes for boundaries of each recursion.

/// 

/// in place linked list sort using quick sort.
/// 
public static void QuickSort(this LinkedList linkedList, IComparer comparer)
{
    if (linkedList == null || linkedList.Count <= 1) return; // there is nothing to sort
    SortImpl(linkedList.First, linkedList.Last, comparer);
}

private static void SortImpl(LinkedListNode head, LinkedListNode tail, IComparer comparer)
{
    if (head == tail) return; // there is nothing to sort

    void Swap(LinkedListNode a, LinkedListNode b)
    {
        var tmp = a.Value;
        a.Value = b.Value;
        b.Value = tmp;
    }

    var pivot = tail;
    var node = head;
    while (node.Next != pivot)
    {
        if (comparer.Compare(node.Value, pivot.Value) > 0)
        {
            Swap(pivot, pivot.Previous);
            Swap(node, pivot);
            pivot = pivot.Previous; // move pivot backward
        }
        else node = node.Next; // move node forward
    }
    if (comparer.Compare(node.Value, pivot.Value) > 0)
    {
        Swap(node, pivot);
        pivot = node;
    }

    // pivot location is found, now sort nodes below and above pivot.
    // if head or tail is equal to pivot we reached boundaries and we should stop recursion.
    if (head != pivot) SortImpl(head, pivot.Previous, comparer);
    if (tail != pivot) SortImpl(pivot.Next, tail, comparer);
}