Open In App

C# Program For Moving Last Element To Front Of A Given Linked List

Improve
Improve
Like Article
Like
Save
Share
Report

Write a function that moves the last element to the front in a given Singly Linked List. For example, if the given Linked List is 1->2->3->4->5, then the function should change the list to 5->1->2->3->4. Algorithm: Traverse the list till the last node. Use two pointers: one to store the address of the last node and the other for the address of the second last node. After the end of the loop do the following operations.

  1. Make second last as last (secLast->next = NULL).
  2. Set next of last as head (last->next = *head_ref).
  3. Make last as head ( *head_ref = last).

C#




/* C# Program to move last element to
   front in a given 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 moveToFront()
    {
        /* If linked list is empty or
           it contains only one node
           then simply return. */
        if(head == null ||  
           head.next == null)
            return;
 
        /* Initialize second last and
           last pointers */
        Node secLast = null;
        Node last = head;
 
        /* After this loop secLast contains
           address of second last node and
           last contains address of last node
           in Linked List */
        while (last.next != null)
        {
        secLast = last;
        last = last.next;
        }
 
        // Set the next of second last as null
        secLast.next = null;
 
        // Set the next of last as head
        last.next = head;
 
        // Change head to point to last node.
        head = last;
    }                
 
                     
    // 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();
 
        /* Constructed Linked List is
           1->2->3->4->5->null */
        llist.push(5);
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);
         
        Console.WriteLine(
                "Linked List before moving last to front ");
        llist.printList();
         
        llist.moveToFront();
         
        Console.WriteLine(
                "Linked List after moving last to front ");
        llist.printList();
    }
}
// This code is contributed by Arnab Kundu


Output:

Linked list before moving last to front 
1 2 3 4 5 
Linked list after removing last to front 
5 1 2 3 4

Time Complexity: O(n) where n is the number of nodes in the given Linked List.

Auxiliary space: O(1) as it is using constant space

Please refer complete article on Move last element to front of a given Linked List for more details!



Last Updated : 15 Aug, 2022
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads