QuickSort on Singly Linked List

Given a linked list, apply the Quick sort algorithm to sort the linked list. The important things about implementation are, that it changes pointers rather than swapping data.

Example:

Follow the given steps to solve the problem:

• Call partition function to get a pivot node placed at its correct position
  • In the partition function, the last element is considered a pivot
  • Then traverse the current list and if a node has a value greater than the pivot, then move it after the tail. If the node has a smaller value, then keep it at its current position. 
  • Return pivot node
• Find the tail node of the list which is on the left side of the pivot and recur for the left list
• Similarly, after the left side, recur for the list on the right side of the pivot
• Now return the head of the linked list after joining the left and the right list, as the whole linked list is now sorted

Below is the implementation of the above approach:

C

#include <stdio.h>
#include <stdlib.h>

// Creating  structure
struct Node {
    int data;
    struct Node* next;
};

// Add new node at end of linked list
void insert(struct Node** head, int value)
{
    // Create dynamic node
    struct Node* node
        = (struct Node*)malloc(sizeof(struct Node));
    if (node == NULL) {
        // checking memory overflow
        printf("Memory overflow\n");
    }
    else {
        node->data = value;
        node->next = NULL;
        if (*head == NULL) {
            *head = node;
        }
        else {
            struct Node* temp = *head;

            // finding last node
            while (temp->next != NULL) {
                temp = temp->next;
            }

            // adding node at last position
            temp->next = node;
        }
    }
}

// Displaying linked list element
void display(struct Node* head)
{
    if (head == NULL) {
        printf("Empty linked list");
        return;
    }
    struct Node* temp = head;
    printf("\n Linked List :");
    while (temp != NULL) {
        printf("  %d", temp->data);
        temp = temp->next;
    }
}

// Finding last node of linked list
struct Node* last_node(struct Node* head)
{
    struct Node* temp = head;
    while (temp != NULL && temp->next != NULL) {
        temp = temp->next;
    }
    return temp;
}

// We are Setting the given last node position to its proper
// position
struct Node* partition(struct Node* first, struct Node* last)
{
    // Get first node of given linked list
    struct Node* pivot = first;
    struct Node* front = first;
    int temp = 0;
    while (front != NULL && front != last) {
        if (front->data < last->data) {
            pivot = first;

            // Swapping  node values
            temp = first->data;
            first->data = front->data;
            front->data = temp;

            // Visiting the next node
            first = first->next;
        }

        // Visiting the next node
        front = front->next;
    }

    // Change last node value to current node
    temp = first->data;
    first->data = last->data;
    last->data = temp;
    return pivot;
}

// Performing quick sort in  the given linked list
void quick_sort(struct Node* first, struct Node* last)
{
    if (first == last) {
        return;
    }
    struct Node* pivot = partition(first, last);

    if (pivot != NULL && pivot->next != NULL) {
        quick_sort(pivot->next, last);
    }

    if (pivot != NULL && first != pivot) {
        quick_sort(first, pivot);
    }
}

// Driver's code
int main()
{
    struct Node* head = NULL;

    // Create linked list
    insert(&head, 41);
    insert(&head, 5);
    insert(&head, 7);
    insert(&head, 22);
    insert(&head, 28);
    insert(&head, 63);
    insert(&head, 4);
    insert(&head, 8);
    insert(&head, 2);
    insert(&head, 11);
    printf("\n Before Sort ");
    display(head);

    // Function call
    quick_sort(head, last_node(head));
    printf("\n After Sort ");
    display(head);
    return 0;
}

C++

// C++ program for Quick Sort on Singly Linked List

#include <cstdio>
#include <iostream>
using namespace std;

/* a node of the singly linked list */
struct Node {
    int data;
    struct Node* next;
};

/* A utility function to insert a node at the beginning of
 * linked list */
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node = new Node;

    /* put in the data */
    new_node->data = new_data;

    /* link the old list of the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

/* A utility function to print linked list */
void printList(struct Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
    printf("\n");
}

// Returns the last node of the list
struct Node* getTail(struct Node* cur)
{
    while (cur != NULL && cur->next != NULL)
        cur = cur->next;
    return cur;
}

// Partitions the list taking the last element as the pivot
struct Node* partition(struct Node* head, struct Node* end,
                       struct Node** newHead,
                       struct Node** newEnd)
{
    struct Node* pivot = end;
    struct Node *prev = NULL, *cur = head, *tail = pivot;

    // During partition, both the head and end of the list
    // might change which is updated in the newHead and
    // newEnd variables
    while (cur != pivot) {
        if (cur->data < pivot->data) {
            // First node that has a value less than the
            // pivot - becomes the new head
            if ((*newHead) == NULL)
                (*newHead) = cur;

            prev = cur;
            cur = cur->next;
        }
        else // If cur node is greater than pivot
        {
            // Move cur node to next of tail, and change
            // tail
            if (prev)
                prev->next = cur->next;
            struct Node* tmp = cur->next;
            cur->next = NULL;
            tail->next = cur;
            tail = cur;
            cur = tmp;
        }
    }

    // If the pivot data is the smallest element in the
    // current list, pivot becomes the head
    if ((*newHead) == NULL)
        (*newHead) = pivot;

    // Update newEnd to the current last node
    (*newEnd) = tail;

    // Return the pivot node
    return pivot;
}

// here the sorting happens exclusive of the end node
struct Node* quickSortRecur(struct Node* head,
                            struct Node* end)
{
    // base condition
    if (!head || head == end)
        return head;

    Node *newHead = NULL, *newEnd = NULL;

    // Partition the list, newHead and newEnd will be
    // updated by the partition function
    struct Node* pivot
        = partition(head, end, &newHead, &newEnd);

    // If pivot is the smallest element - no need to recur
    // for the left part.
    if (newHead != pivot) {
        // Set the node before the pivot node as NULL
        struct Node* tmp = newHead;
        while (tmp->next != pivot)
            tmp = tmp->next;
        tmp->next = NULL;

        // Recur for the list before pivot
        newHead = quickSortRecur(newHead, tmp);

        // Change next of last node of the left half to
        // pivot
        tmp = getTail(newHead);
        tmp->next = pivot;
    }

    // Recur for the list after the pivot element
    pivot->next = quickSortRecur(pivot->next, newEnd);

    return newHead;
}

// The main function for quick sort. This is a wrapper over
// recursive function quickSortRecur()
void quickSort(struct Node** headRef)
{
    (*headRef)
        = quickSortRecur(*headRef, getTail(*headRef));
    return;
}

// Driver's code
int main()
{
    struct Node* a = NULL;
    push(&a, 5);
    push(&a, 20);
    push(&a, 4);
    push(&a, 3);
    push(&a, 30);

    cout << "Linked List before sorting \n";
    printList(a);

    // Function call
    quickSort(&a);

    cout << "Linked List after sorting \n";
    printList(a);

    return 0;
}

Java

// Java program for Quick Sort on Singly Linked List

/*Sort a linked list using quick sort*/
public class QuickSortLinkedList {
    static class Node {

        int data;
        Node next;

        Node(int d)
        {
            this.data = d;
            this.next = null;
        }
    }

    Node head;

    void addNode(int data)
    {
        if (head == null) {
            head = new Node(data);
            return;
        }

        Node curr = head;
        while (curr.next != null)
            curr = curr.next;

        Node newNode = new Node(data);
        curr.next = newNode;
    }

    void printList(Node n)
    {
        while (n != null) {
            System.out.print(n.data);
            System.out.print(" ");
            n = n.next;
        }
    }

    // Takes first and last node,
    // but do not break any links in
    // the whole linked list
    Node partitionLast(Node start, Node end)
    {
        if (start == end || start == null || end == null)
            return start;

        Node pivot_prev = start;
        Node curr = start;
        int pivot = end.data;

        // iterate till one before the end,
        // no need to iterate till the end
        // because end is pivot
        while (start != end) {
            if (start.data < pivot) {

                // keep tracks of last modified item
                pivot_prev = curr;
                int temp = curr.data;
                curr.data = start.data;
                start.data = temp;
                curr = curr.next;
            }
            start = start.next;
        }

        // Swap the position of curr i.e.
        // next suitable index and pivot
        int temp = curr.data;
        curr.data = pivot;
        end.data = temp;

        // Return one previous to current
        // because current is now pointing to pivot
        return pivot_prev;
    }

    void sort(Node start, Node end)
    {
        if (start == null || start == end
            || start == end.next)
            return;

        // Split list and partition recurse
        Node pivot_prev = partitionLast(start, end);
        sort(start, pivot_prev);

        // If pivot is picked and moved to the start,
        // that means start and pivot is same
        // so pick from next of pivot
        if (pivot_prev != null && pivot_prev == start)
            sort(pivot_prev.next, end);

        // If pivot is in between of the list,
        // start from next of pivot,
        // since we have pivot_prev, so we move two nodes
        else if (pivot_prev != null
                 && pivot_prev.next != null)
            sort(pivot_prev.next.next, end);
    }

    // Driver's Code
    public static void main(String[] args)
    {
        QuickSortLinkedList list
            = new QuickSortLinkedList();
        list.addNode(30);
        list.addNode(3);
        list.addNode(4);
        list.addNode(20);
        list.addNode(5);

        Node n = list.head;
        while (n.next != null)
            n = n.next;

        System.out.println("Linked List before sorting");
        list.printList(list.head);

        // Function call
        list.sort(list.head, n);

        System.out.println("\nLinked List after sorting");
        list.printList(list.head);
    }
}

// This code is contributed by trinadumca

Python3

'''

sort a linked list using quick sort

'''


class Node:
    def __init__(self, val):
        self.data = val
        self.next = None


class QuickSortLinkedList:

    def __init__(self):
        self.head = None

    def addNode(self, data):
        if (self.head == None):
            self.head = Node(data)
            return

        curr = self.head
        while (curr.next != None):
            curr = curr.next

        newNode = Node(data)
        curr.next = newNode

    def printList(self, n):
        while (n != None):
            print(n.data, end=" ")
            n = n.next

    ''' takes first and last node,but do not
    break any links in    the whole linked list'''

    def partitionLast(self, start, end):
        if (start == end or start == None or end == None):
            return start

        pivot_prev = start
        curr = start
        pivot = end.data

        '''iterate till one before the end,
        no need to iterate till the end because end is pivot'''

        while (start != end):
            if (start.data < pivot):

                # keep tracks of last modified item
                pivot_prev = curr
                temp = curr.data
                curr.data = start.data
                start.data = temp
                curr = curr.next
            start = start.next

        '''swap the position of curr i.e.
        next suitable index and pivot'''

        temp = curr.data
        curr.data = pivot
        end.data = temp

        ''' return one previous to current because
        current is now pointing to pivot '''
        return pivot_prev

    def sort(self, start, end):
        if(start == None or start == end or start == end.next):
            return

        # split list and partition recurse
        pivot_prev = self.partitionLast(start, end)
        self.sort(start, pivot_prev)

        '''
        if pivot is picked and moved to the start,
        that means start and pivot is same
        so pick from next of pivot
        '''
        if(pivot_prev != None and pivot_prev == start):
            self.sort(pivot_prev.next, end)

        # if pivot is in between of the list,start from next of pivot,
        # since we have pivot_prev, so we move two nodes
        elif (pivot_prev != None and pivot_prev.next != None):
            self.sort(pivot_prev.next.next, end)


if __name__ == "__main__":
    ll = QuickSortLinkedList()
    ll.addNode(30)
    ll.addNode(3)
    ll.addNode(4)
    ll.addNode(20)
    ll.addNode(5)

    N = ll.head
    while (N.next != None):
        N = N.next

    print("\nLinked List before sorting")
    ll.printList(ll.head)

    # Function call
    ll.sort(ll.head, N)

    print("\nLinked List after sorting")
    ll.printList(ll.head)

    # This code is contributed by humpheykibet.

C#

// C# program for Quick Sort on
// Singly Linked List
using System;

/*Sort a linked list using quick sort*/
class GFG {
    public class Node {
        public int data;
        public Node next;

        public Node(int d)
        {
            this.data = d;
            this.next = null;
        }
    }

    Node head;

    void addNode(int data)
    {
        if (head == null) {
            head = new Node(data);
            return;
        }

        Node curr = head;
        while (curr.next != null)
            curr = curr.next;

        Node newNode = new Node(data);
        curr.next = newNode;
    }

    void printList(Node n)
    {
        while (n != null) {
            Console.Write(n.data);
            Console.Write(" ");
            n = n.next;
        }
    }

    // takes first and last node,
    // but do not break any links in
    // the whole linked list
    Node partitionLast(Node start, Node end)
    {
        if (start == end || start == null || end == null)
            return start;

        Node pivot_prev = start;
        Node curr = start;
        int pivot = end.data;

        // iterate till one before the end,
        // no need to iterate till the end
        // because end is pivot
        int temp;
        while (start != end) {

            if (start.data < pivot) {
                // keep tracks of last modified item
                pivot_prev = curr;
                temp = curr.data;
                curr.data = start.data;
                start.data = temp;
                curr = curr.next;
            }
            start = start.next;
        }

        // swap the position of curr i.e.
        // next suitable index and pivot
        temp = curr.data;
        curr.data = pivot;
        end.data = temp;

        // return one previous to current
        // because current is now pointing to pivot
        return pivot_prev;
    }

    void sort(Node start, Node end)
    {
        if (start == end)
            return;

        // split list and partition recurse
        Node pivot_prev = partitionLast(start, end);
        sort(start, pivot_prev);

        // if pivot is picked and moved to the start,
        // that means start and pivot is same
        // so pick from next of pivot
        if (pivot_prev != null && pivot_prev == start)
            sort(pivot_prev.next, end);

        // if pivot is in between of the list,
        // start from next of pivot,
        // since we have pivot_prev, so we move two nodes
        else if (pivot_prev != null
                 && pivot_prev.next != null)
            sort(pivot_prev.next.next, end);
    }

    // Driver Code
    public static void Main(String[] args)
    {
        GFG list = new GFG();
        list.addNode(30);
        list.addNode(3);
        list.addNode(4);
        list.addNode(20);
        list.addNode(5);

        Node N = list.head;
        while (N.next != null)
            N = N.next;

        Console.WriteLine("Linked List before sorting");
        list.printList(list.head);

        // Function call
        list.sort(list.head, N);

        Console.WriteLine("\nLinked List after sorting");
        list.printList(list.head);
    }
}

// This code is contributed by 29AjayKumar

Javascript

// javascript program for Quick Sort on Singly Linked List

/*sort a linked list using quick sort*/

     class Node {
        constructor(val) {
            this.data = val;
            this.next = null;
        }
    }

    var head;

    function addNode(data) {
        if (head == null) {
            head = new Node(data);
            return;
        }

        var curr = head;
        while (curr.next != null)
            curr = curr.next;

        var newNode = new Node(data);
        curr.next = newNode;
    }

    function printList( n) {
        while (n != null) {
            document.write(n.data);
            document.write(" ");
            n = n.next;
        }
    }

    // takes first and last node,
    // but do not break any links in
    // the whole linked list
    function partitionLast( start,  end) {
        if (start == end || start == null || end == null)
            return start;

        var pivot_prev = start;
        var curr = start;
        var pivot = end.data;

        // iterate till one before the end,
        // no need to iterate till the end
        // because end is pivot
        while (start != end) {
            if (start.data < pivot) {
                // keep tracks of last modified item
                pivot_prev = curr;
                var temp = curr.data;
                curr.data = start.data;
                start.data = temp;
                curr = curr.next;
            }
            start = start.next;
        }

        // swap the position of curr i.e.
        // next suitable index and pivot
        var temp = curr.data;
        curr.data = pivot;
        end.data = temp;

        // return one previous to current
        // because current is now pointing to pivot
        return pivot_prev;
    }

    function sort( start,  end) {
        if (start == null || start == end || start == end.next)
            return;

        // split list and partition recurse
        var pivot_prev = partitionLast(start, end);
        sort(start, pivot_prev);

        // if pivot is picked and moved to the start,
        // that means start and pivot is same
        // so pick from next of pivot
        if (pivot_prev != null && pivot_prev == start)
            sort(pivot_prev.next, end);

        // if pivot is in between of the list,
        // start from next of pivot,
        // since we have pivot_prev, so we move two nodes
        else if (pivot_prev != null && pivot_prev.next != null)
            sort(pivot_prev.next.next, end);
    }

    // Driver Code

        addNode(30);
        addNode(3);
        addNode(4);
        addNode(20);
        addNode(5);

        var n = head;
        while (n.next != null)
            n = n.next;

        document.write("Linked List before sorting<br/>");
        printList(head);

        sort(head, n);

        document.write("<br/>Linked List after sorting<br/>");
        printList(head);

// This code contributed by umadevi9616

Output

Linked List before sorting 
30 3 4 20 5 
Linked List after sorting 
3 4 5 20 30 

Time Complexity: O(N * log N), It takes O(N²) time in the worst case and O(N log N) in the average or best case.
Auxiliary Space: O(N), As extra space is used in the recursion call stack.