QuickSort on Singly Linked List

QuickSort on Doubly Linked List is discussed here. QuickSort on Singly linked list was given as an exercise. Following is C++ implementation for same. The important things about implementation are, it changes pointers rather swapping data and time complexity is same as the implementation for Doubly Linked List.
sorting image

In partition(), we consider last element as pivot. We traverse through the current list and if a node has value greater than pivot, we move it after tail. If the node has smaller value, we keep it at its current position.
In QuickSortRecur(), we first call partition() which places pivot at correct position and returns pivot. After pivot is placed at correct position, we find tail node of left side (list before pivot) and recur for left list. Finally, we recur for right list.

C++

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// C++ program for Quick Sort on Singly Linled List
#include <iostream>
#include <cstdio>
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 off 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 program to test above functions
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);
  
    quickSort(&a);
  
    cout << "Linked List after sorting \n";
    printList(a);
  
    return 0;
}

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Java

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// Java program for Quick Sort on Singly Linled 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 paritionLast(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 == end )
        return;
          
    // split list and partion recurse
    Node pivot_prev = paritionLast(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)
{
    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);
  
    list.sort(list.head , n);
  
    System.out.println("\nLinked List after sorting");
    list.printList(list.head);
}
}
  
// This code is contributed by trinadumca

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C#

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// C# program for Quick Sort on 
// Singly Linled 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 paritionLast(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 partion recurse
    Node pivot_prev = paritionLast(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);
  
    list.sort(list.head , n);
  
    Console.WriteLine("\nLinked List after sorting");
    list.printList(list.head);
}
}
  
// This code is contributed by 29AjayKumar

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Output:



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

This article is contributed by Balasubramanian.N . Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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Improved By : TrinaD, 29AjayKumar