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.
- Make second last as last (secLast->next = NULL).
- Set next of last as head (last->next = *head_ref).
- Make last as head ( *head_ref = last).
/* 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!