C# Program For Moving Last Element To Front Of A Given Linked List Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like 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. 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# /* 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! Comment More infoAdvertise with us Next Article C++ Program For Deleting A Linked List Node At A Given Position K kartik Follow Improve Article Tags : C# Linked Lists Similar Reads C++ Program For Finding The Middle Element Of A Given Linked List Given a singly linked list, find the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the output should be 3. If there are even nodes, then there would be two middle nodes, we need to print the second middle element. For example, if given linked list 6 min read Program for Nth node from the end of a Linked List Given a Linked List of M nodes and a number N, find the value at the Nth node from the end of the Linked List. 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