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C# Program For Pairwise Swapping Elements Of A Given Linked List

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  • Last Updated : 15 Dec, 2021
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Given a singly linked list, write a function to swap elements pairwise.

Input: 1->2->3->4->5->6->NULL 
Output: 2->1->4->3->6->5->NULL Input: 1->2->3->4->5->NULL 
Output: 2->1->4->3->5->NULL Input: 1->NULL 
Output: 1->NULL

For example, if the linked list is 1->2->3->4->5 then the function should change it to 2->1->4->3->5, and if the linked list is then the function should change it to.

METHOD 1 (Iterative): 
Start from the head node and traverse the list. While traversing swap data of each node with its next node’s data. 
Below is the implementation of the above approach:

C#




// C# program to pairwise swap elements 
// of a linked list
using System;
class LinkedList 
{
    // Head of list
    Node head; 
  
    // Linked list Node
    public class Node 
    {
        public int data;
        public Node next;
        public Node(int d)
        {
            data = d;
            next = null;
        }
    }
  
    void pairWiseSwap()
    {
        Node temp = head;
  
        /* Traverse only till there are 
           atleast 2 nodes left */
        while (temp != null && 
               temp.next != null
        {
            // Swap the data 
            int k = temp.data;
            temp.data = temp.next.data;
            temp.next.data = k;
            temp = temp.next.next;
        }
    }
  
    // Utility functions 
    /* Inserts a new Node at front 
       of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node & 
                  Put in the data*/
        Node new_node = new Node(new_data);
  
        // 3. Make next of new Node as head 
        new_node.next = head;
  
        /* 4. Move the head to point to 
              new Node */
        head = new_node;
    }
  
    // Function to print linked list 
    void printList()
    {
        Node temp = head;
        while (temp != null
        {
            Console.Write(temp.data + " ");
            temp = temp.next;
        }
  
        Console.WriteLine();
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        LinkedList llist = new LinkedList();
  
        /* Created Linked List 
           1->2->3->4->5 */
        llist.push(5);
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);
  
        Console.WriteLine(
                "Linked List before calling pairWiseSwap() ");
        llist.printList();
  
        llist.pairWiseSwap();
  
        Console.WriteLine(
                "Linked List after calling pairWiseSwap() ");
        llist.printList();
    }
}
// This code is contributed by Arnab Kundu


Output:

Linked list before calling pairWiseSwap()
1 2 3 4 5 
Linked list after calling pairWiseSwap()
2 1 4 3 5 

Time complexity: O(n)

METHOD 2 (Recursive): 
If there are 2 or more than 2 nodes in Linked List then swap the first two nodes and recursively call for rest of the list.
Below image is a dry run of the above approach:

Below is the implementation of the above approach:

C#




/* Recursive function to pairwise swap 
   elements of a linked list */
static void pairWiseSwap(node head)
{
    /* There must be at-least two nodes 
       in the list */
    if (head != null &&
        head.next != null
    {
  
        /* Swap the node's data with data 
           of next node */
        swap(head.data, head.next.data);
  
        /* Call pairWiseSwap() for rest 
           of the list */
        pairWiseSwap(head.next.next);
    }
}
// This code is contributed by aashish1995


Time complexity: O(n)
The solution provided there swaps data of nodes. If data contains many fields, there will be many swap operations. See this for an implementation that changes links rather than swapping data.

Please refer complete article on Pairwise swap elements of a given linked list for more details!


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