# Move last element to front of a given Linked List

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 last node. Use two pointers: one to store the address of last node and other for address of second last node. After the end of loop do following operations.
i) Make second last as last (secLast->next = NULL).

## C++

 `/* CPP Program to move last element  ` `to front in a given linked list */` `#include ` `using` `namespace` `std; ` ` `  `/* A linked list node */` `class` `Node  ` `{  ` `    ``public``: ` `    ``int` `data;  ` `    ``Node *next;  ` `};  ` ` `  `/* We are using a double pointer ` `head_ref here because we change  ` `head of the linked list inside  ` `this function.*/` `void` `moveToFront(Node **head_ref)  ` `{  ` `    ``/* If linked list is empty, or  ` `    ``it contains only one node,  ` `    ``then nothing needs to be done, ` `    ``simply return */` `    ``if` `(*head_ref == NULL || (*head_ref)->next == NULL)  ` `        ``return``;  ` ` `  `    ``/* Initialize second last ` `    ``and last pointers */` `    ``Node *secLast = NULL;  ` `    ``Node *last = *head_ref;  ` ` `  `    ``/*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 next of last as head node */` `    ``last->next = *head_ref;  ` ` `  `    ``/* Change the head pointer ` `    ``to point to last node now */` `    ``*head_ref = last;  ` `}  ` ` `  `/* UTILITY FUNCTIONS */` `/* Function to add a node  ` `at the beginning of Linked List */` `void` `push(Node** head_ref, ``int` `new_data)  ` `{  ` `    ``/* allocate node */` `    ``Node* new_node = ``new` `Node(); ` ` `  `    ``/* put in the data */` `    ``new_node->data = new_data;  ` ` `  `    ``/* link the old list off the new node */` `    ``new_node->next = (*head_ref);  ` ` `  `    ``/* move the head to point to the new node */` `    ``(*head_ref) = new_node;  ` `}  ` ` `  ` `  `/* Function to print nodes in a given linked list */` `void` `printList(Node *node)  ` `{  ` `    ``while``(node != NULL)  ` `    ``{  ` `        ``cout << node->data << ``" "``;  ` `        ``node = node->next;  ` `    ``}  ` `}  ` ` `  `/* Driver code */` `int` `main()  ` `{  ` `    ``Node *start = NULL;  ` ` `  `    ``/* The constructed linked list is:  ` `    ``1->2->3->4->5 */` `    ``push(&start, 5);  ` `    ``push(&start, 4);  ` `    ``push(&start, 3);  ` `    ``push(&start, 2);  ` `    ``push(&start, 1);  ` ` `  `    ``cout<<``"Linked list before moving last to front\n"``;  ` `    ``printList(start);  ` ` `  `    ``moveToFront(&start);  ` ` `  `    ``cout<<``"\nLinked list after removing last to front\n"``;  ` `    ``printList(start);  ` ` `  `    ``return` `0;  ` `}  ` ` `  `// This code is contributed by rathbhupendra `

## C

 `/* C Program to move last element to front in a given linked list */` `#include ` `#include ` ` `  `/* A linked list node */` `struct` `Node ` `{ ` `    ``int` `data; ` `    ``struct` `Node *next; ` `}; ` ` `  `/* We are using a double pointer head_ref here because we change ` `   ``head of the linked list inside this function.*/` `void` `moveToFront(``struct` `Node **head_ref) ` `{ ` `    ``/* If linked list is empty, or it contains only one node, ` `      ``then nothing needs to be done, simply return */` `    ``if` `(*head_ref == NULL || (*head_ref)->next == NULL) ` `        ``return``; ` ` `  `    ``/* Initialize second last and last pointers */` `    ``struct` `Node *secLast = NULL; ` `    ``struct` `Node *last = *head_ref; ` ` `  `    ``/*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 next of last as head node */` `    ``last->next = *head_ref; ` ` `  `    ``/* Change the head pointer to point to last node now */` `    ``*head_ref = last; ` `} ` ` `  `/* UTILITY FUNCTIONS */` `/* Function to add a node at the beginning of Linked List */` `void` `push(``struct` `Node** head_ref, ``int` `new_data) ` `{ ` `    ``/* allocate node */` `    ``struct` `Node* new_node = ` `        ``(``struct` `Node*) ``malloc``(``sizeof``(``struct` `Node)); ` ` `  `    ``/* put in the data  */` `    ``new_node->data  = new_data; ` ` `  `    ``/* link the old list off the new node */` `    ``new_node->next = (*head_ref); ` ` `  `    ``/* move the head to point to the new node */` `    ``(*head_ref)    = new_node; ` `} ` ` `  ` `  `/* Function to print nodes in a given linked list */` `void` `printList(``struct` `Node *node) ` `{ ` `    ``while``(node != NULL) ` `    ``{ ` `        ``printf``(``"%d "``, node->data); ` `        ``node = node->next; ` `    ``} ` `} ` ` `  `/* Driver program to test above function */` `int` `main() ` `{ ` `    ``struct` `Node *start = NULL; ` ` `  `    ``/* The constructed linked list is: ` `     ``1->2->3->4->5 */` `    ``push(&start, 5); ` `    ``push(&start, 4); ` `    ``push(&start, 3); ` `    ``push(&start, 2); ` `    ``push(&start, 1); ` ` `  `    ``printf``(``"\n Linked list before moving last to front\n"``); ` `    ``printList(start); ` ` `  `    ``moveToFront(&start); ` ` `  `    ``printf``(``"\n Linked list after removing last to front\n"``); ` `    ``printList(start); ` ` `  `    ``return` `0; ` `} `

## Java

 `/* Java Program to move last element to front in a given linked list */` `class` `LinkedList ` `{ ` `    ``Node head;  ``// head of list ` `  `  `    ``/* 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 program to test above functions */` `    ``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 */`

## Python3

 `# Python3 code to move the last item to front ` `class` `Node: ` `    ``def` `__init__(``self``, data): ` `        ``self``.data ``=` `data ` `        ``self``.``next` `=` `None` ` `  `class` `LinkedList: ` `    ``def` `__init__(``self``): ` `        ``self``.head ``=` `None` ` `  `    ``# Function to add a node  ` `    ``# at the beginning of Linked List ` `    ``def` `push(``self``, data): ` `        ``new_node ``=` `Node(data) ` `        ``new_node.``next` `=` `self``.head ` `        ``self``.head ``=` `new_node ` `         `  `    ``# Function to print nodes in a ` `    ``# given linked list ` `    ``def` `printList(``self``): ` `        ``tmp ``=` `self``.head ` `        ``while` `tmp ``is` `not` `None``: ` `            ``print``(tmp.data, end``=``", "``) ` `            ``tmp ``=` `tmp.``next` `        ``print``() ` ` `  `    ``# Function to bring the last node to the front ` `    ``def` `moveToFront(``self``): ` `        ``tmp ``=` `self``.head ` `        ``sec_last ``=` `None` `# To maintain the track of ` `                        ``# the second last node ` ` `  `    ``# To check whether we have not received  ` `    ``# the empty list or list with a single node ` `        ``if` `not` `tmp ``or` `not` `tmp.``next``:  ` `            ``return` ` `  `        ``# Iterate till the end to get ` `        ``# the last and second last node  ` `        ``while` `tmp ``and` `tmp.``next` `: ` `            ``sec_last ``=` `tmp ` `            ``tmp ``=` `tmp.``next` ` `  `        ``# point the next of the second ` `        ``# last node to None ` `        ``sec_last.``next` `=` `None` ` `  `        ``# Make the last node as the first Node ` `        ``tmp.``next` `=` `self``.head ` `        ``self``.head ``=` `tmp ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``llist ``=` `LinkedList() ` `     `  `    ``# swap the 2 nodes ` `    ``llist.push(``5``) ` `    ``llist.push(``4``) ` `    ``llist.push(``3``) ` `    ``llist.push(``2``) ` `    ``llist.push(``1``) ` `    ``print` `(``"Linked List before moving last to front "``) ` `    ``llist.printList() ` `    ``llist.moveToFront() ` `    ``print` `(``"Linked List after moving last to front "``) ` `    ``llist.printList() `

## C#

 `/* C# Program to move last element to front in a given linked list */` `using` `System; ` `class` `LinkedList  ` `{  ` `    ``Node head; ``// head of list  ` ` `  `    ``/* 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 program to test above functions */` `    ``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.

Please write comments if you find any bug in the above code/algorithm, or find other ways to solve the same problem.

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