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, can I implement a sorting function? The language I am using is C#. Is there an example of this in C#?

I am working from this sample:

Thanks

Answer 1

Here is an implementation based on Insertion sort. It should be fast enough for small linked lists and/or if the list is almost already sorted. For large lists, check M.kazem Akhgary answer (which use Quicksort).

static void Sort<T>(this LinkedList<T> list, IComparer<T> comparer)
{
    var node = list.First;
    while (node != null)
    {
        var next = node.Next;
        
        var min = node;
        for (var comp = node.Previous; 
             comp != null && comparer.Compare(node.Value, comp.Value) < 0; 
             comp = comp.Previous)
        {
            min = comp;
        }
        
        if (node != min) 
        {               
            list.Remove(node);
            list.AddBefore(min, node);
        }
        
        node = next;
    }
}

Answer 2

Some people (including me) may have wanted to sort LinkedList<T> 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.

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

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

    void Swap(LinkedListNode<T> a, LinkedListNode<T> 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);
}

Answer 3

public LinkedListNode<int> Sort2(LinkedListNode<int> head, int count)
    {
        var cur = head;
        var prev = cur;
        var min = cur;
        var minprev = min;

        LinkedListNode<int> newHead = null;
        LinkedListNode<int> newTail = newHead;

        for (int i = 0; i < count; i++)
        {
            cur = head;
            min = cur;
            minprev = min;

            while (cur != null)
            {
                if (cur.Value < min.Value)
                {
                    min = cur;
                    minprev = prev;
                }
                prev = cur;
                cur = cur.Next;
            }

            if (min == head)
                head = head.Next;
            else if (min.Next == null)
                minprev.Next = null;
            else
                minprev.Next = minprev.Next.Next;

            if (newHead != null)
            {
                newTail.Next = min;
                newTail = newTail.Next;
            }
            else
            {
                newHead = min;
                newTail = newHead;
            }
        }
        return newHead;
    }

Answer 4

If you want to really utilize the fact that you're using a linked list, as opposed to working around it, I would suggest insertion sort.

Normally an insertion sort is not very efficient - O(n^2) in the worst case, but with linked list this can be improved to O(n log n)

pseudo:

for i in range (1,n):
    item = arr[i]
    location = binary_search(l=root, r=i, value=item.val)  // O(log i)
    insert_item_before(arr[location], item) // O(1)

In the regular algorithm the insert part takes O(i) because we may need to shift all the items that are larger then item. because a linked list enables us to do the insertion without shifting, we can search for the location with a binary search and so the insertion part takes only O(log i)

note: usually an insertion sort has O(n) performance in the best case (if the array is sorted). Unfortunately this isn't the case here because binary search takes always O(log n) steps.

Answer 5

for(int i=0; i<counter;i++)
{
  while(current.next!=null)
  {
    if(current.elemen>current.next.element)
    {
      Swap(current.elment,current.next.element);
    }
    current=current.next;
  }
}

increment counter when you add or insert elements to your linked lists

Answer 6

This might not be the best solution, but it's as simple as I can come up with. If anyone can think of something simpler but still fast, I'd love to hear it.
SORRY THAT IT'S C++ it should translate.

List * SortList(List * p_list)
{
     if(p_list == NULL || p_list->next == NULL) 
          return p_list;

     List left, right;
     List * piviot = p_list;
     List * piviotEnd = piviot;
     List * next = p_list->next;
     do
     {
              p_list = next;
              next = next->next;
              //FIGURE OUT WHICH SIDE I'M ADDING TO.
              if(p_list->data > piviot->data ) 
                  right.InsertNode(p_list);
              else if(p_list->data < piviot->data)
                  left.InsertNode(p_list);
              else
              {   //we put this in it's own list because it doesn't need to be sorted
                  piviotEnd->next = p_list;
                  piviotEnd= p_list;
              }  
     }while(next);

     //now left contains values < piviot and more contains all the values more
     left.next = SortList(left.next);
     right.next = SortList(right.next);

     //add the piviot to the list then add the rigth list to the full list
     left.GetTail()->next = piviot;
     piviotEnd->next = right.next;

     return left.next;
}

Answer 7

Functional Quicksort and Mergesort

Here's a linked list with quicksort and mergesort methods written in a functional style:

class List
{
    public int item;
    public List rest;

    public List(int item, List rest)
    {
        this.item = item;
        this.rest = rest;
    }

    // helper methods for quicksort

    public static List Append(List xs, List ys)
    {
        if (xs == null) return ys;
        else return new List(xs.item, Append(xs.rest, ys));
    }

    public static List Filter(Func<int,bool> p, List xs)
    {
        if (xs == null) return null;
        else if (p(xs.item)) return new List(xs.item, Filter(p, xs.rest));
        else return Filter(p, xs.rest);
    }

    public static List QSort(List xs)
    {
        if (xs == null) return null;
        else
        {
            int pivot = xs.item;
            List less = QSort(Filter(x => x <= pivot, xs.rest));
            List more = QSort(Filter(x => x > pivot, xs.rest));
            return Append(less, new List(pivot, more));
        }
    }

    // Helper methods for mergesort

    public static int Length(List xs)
    {
        if (xs == null) return 0;
        else return 1 + Length(xs.rest);
    }

    public static List Take(int n, List xs)
    {
        if (n == 0) return null;
        else return new List(xs.item, Take(n - 1, xs.rest));
    }

    public static List Drop(int n, List xs)
    {
        if (n == 0) return xs;
        else return Drop(n - 1, xs.rest);
    }

    public static List Merge(List xs, List ys)
    {
        if (xs == null) return ys;
        else if (ys == null) return xs;
        else if (xs.item <= ys.item) return new List(xs.item, Merge(xs.rest, ys));
        else return new List(ys.item, Merge(xs, ys.rest));
    }

    public static List MSort(List xs)
    {
        if (Length(xs) <= 1) return xs;
        else
        {
            int len = Length(xs) / 2;
            List left  = MSort(Take(len, xs));
            List right = MSort(Drop(len, xs));
            return Merge(left, right);
        }
    }

    public static string Show(List xs)
    {
        if(xs == null) return "";
        else return xs.item.ToString() + " " + Show(xs.rest);
    }
}

Functional heapsort using a Pairing Heap

Bonus: heapsort (using functional pairing heap).

class List
{
    // ...

    public static Heap List2Heap(List xs)
    {
        if (xs == null) return null;
        else return Heap.Merge(new Heap(null, xs.item, null), List2Heap(xs.rest));
    }

    public static List HSort(List xs)
    {
        return Heap.Heap2List(List2Heap(xs));
    }
}

class Heap
{
    Heap left;
    int min;
    Heap right;

    public Heap(Heap left, int min, Heap right)
    {
        this.left = left;
        this.min = min;
        this.right = right;
    }

    public static Heap Merge(Heap a, Heap b)
    {
        if (a == null) return b;
        if (b == null) return a;

        Heap smaller = a.min <= b.min ? a : b;
        Heap larger = a.min <= b.min ? b : a;
        return new Heap(smaller.left, smaller.min, Merge(smaller.right, larger));
    }

    public static Heap DeleteMin(Heap a)
    {
        return Merge(a.left, a.right);
    }

    public static List Heap2List(Heap a)
    {
        if (a == null) return null;
        else return new List(a.min, Heap2List(DeleteMin(a)));
    }
}

For actual use you want to rewrite the helper methods without using recursion, and maybe use a mutable list like the built-in one.

How to use:

List xs = new List(4, new List(2, new List(3, new List(1, null))));
Console.WriteLine(List.Show(List.QSort(xs)));
Console.WriteLine(List.Show(List.MSort(xs)));
Console.WriteLine(List.Show(List.HSort(xs)));

Imperative In-place Quicksort for linked lists

An in-place version was requested. Here's a very quick implementation. I wrote this code top to bottom without looking for opportunities to make the code better, i.e. every line is the first line that came to mind. It's extremely ugly because I used null as the empty list :) The indentation is inconsistent, etc.

Additionally I tested this code on only one example:

        MList ys = new MList(4, new MList(2, new MList(3, new MList(1, null))));
        MList.QSortInPlace(ref ys);
        Console.WriteLine(MList.Show(ys));

Magically it worked the first time! I'm pretty sure that this code contains bugs though. Don't hold me accountable.

class MList
{
    public int item;
    public MList rest;

    public MList(int item, MList rest)
    {
        this.item = item;
        this.rest = rest;
    }

    public static void QSortInPlace(ref MList xs)
    {
        if (xs == null) return;

        int pivot = xs.item;
        MList pivotNode = xs;
        xs = xs.rest;
        pivotNode.rest = null;
        // partition the list into two parts
        MList smaller = null; // items smaller than pivot
        MList larger = null; // items larger than pivot
        while (xs != null)
        {
            var rest = xs.rest;
            if (xs.item < pivot) {
                xs.rest = smaller;
                smaller = xs;
            } else {
                xs.rest = larger;
                larger = xs;
            }
            xs = rest;
        }

        // sort the smaller and larger lists
        QSortInPlace(ref smaller);
        QSortInPlace(ref larger);

        // append smaller + [pivot] + larger
        AppendInPlace(ref pivotNode, larger);
        AppendInPlace(ref smaller, pivotNode);
        xs = smaller;
    }

    public static void AppendInPlace(ref MList xs, MList ys)
    {
        if(xs == null){
            xs = ys;
            return;
        }

        // find the last node in xs
        MList last = xs;
        while (last.rest != null)
        {
            last = last.rest;
        }
        last.rest = ys;
    }

    public static string Show(MList xs)
    {
        if (xs == null) return "";
        else return xs.item.ToString() + " " + Show(xs.rest);
    }
}

Answer 8

The simplest option is probably to extract the data, sort it in a mechanism that already supports sorting (arrays, List<T> or IEnumerable<T> via LINQ), and re-build the linked list with the sorted data.

If you want to write your own sort algorithm, then you may find Comparer<T>.Default useful (assuming you are using generics). This should allow you to compare any items that are IComparable<T> or IComparable.

As an aside - note that .NET already includes LinkedList<T> etc; if this is just for learning etc, then fine ;-p

Answer 9

Of course you can implement a sorting function using a plain linked list. Merge sort might be a suitable algorithm to try, it's fairly simple.