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

• Last Updated : 14 Dec, 2021

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).

## Java

 `/* Java Program to move last element to  ` `   ``front in a given linked list */` `class` `LinkedList ` `{ ` `    ``// Head of list ` `    ``Node head;   ` `  `  `    ``// Linked list Node ` `    ``class` `Node ` `    ``{ ` `        ``int` `data; ` `        ``Node next; ` `        ``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``) ` `        ``{ ` `           ``System.out.print(temp.data + ``" "``); ` `           ``temp = temp.next; ` `        ``}   ` `        ``System.out.println(); ` `    ``} ` ` `  `    ``// 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``); ` `         `  `        ``System.out.println( ` `               ``"Linked List before moving last to front "``); ` `        ``llist.printList(); ` `         `  `        ``llist.moveToFront(); ` `         `  `        ``System.out.println( ` `               ``"Linked List after moving last to front "``); ` `        ``llist.printList(); ` `    ``} ` `}  ` `// This code is contributed by Rajat Mishra  `

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.

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

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