Move last element to front of a given Linked List

Write a function that moves the last node to the front in a given Singly Linked List.

Examples:

Input: 1->2->3->4->5
Output: 5->1->2->3->4

Input: 3->8->1->5->7->12
Output: 12->3->8->1->5->7  

Approach: To solve the problem follow the below idea:

 Traverse the list till the last node. Use two pointers: one to store the address of the last node and other for the address of the second last node. After the end of loop, make the second last node as the last node and the last node as the head node

Follow the given steps to solve the problem using the above approach:

• Traverse the linked list till the last node and Initialize two pointers to store the address of the last and the second last node
• Then follow these three steps to move the last node to the front
  • 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)

Below is the implementation of the above approach:

C

/* C Program to move last element to front
 * in a given linked list */
#include <stdio.h>
#include <stdlib.h>

/* 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 of 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's code
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("Linked list before moving last to front\n");
    printList(start);

      // Function call
    moveToFront(&start);

    printf("\nLinked list after removing last to front\n");
    printList(start);

    return 0;
}

C++

/* CPP Program to move last element
to front in a given linked list */

#include <bits/stdc++.h>
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 of 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's 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);

      // Function call
    moveToFront(&start);

    cout << "\nLinked list after removing last to front\n";
    printList(start);

    return 0;
}

// This code is contributed by rathbhupendra

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's 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();

          // Function call
        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's 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()

    # Function call
    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's 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();

          // Function call
        llist.moveToFront();

        Console.WriteLine(
            "Linked List after moving last to front ");
        llist.printList();
    }
}
// This code is contributed by Arnab Kundu

Javascript

/* javascript Program to move last element to front in a given linked list */

    /* Linked list Node */
     class Node {
            constructor(val) {
                this.data = val;
                this.next = null;
            }
        }
    var head; // head of list

    function 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 */
    var secLast = null;
    var 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. */
     function push(new_data) {
        /*
         * 1 & 2: Allocate the Node & Put in the data
         */
    var 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 */
    function printList() {
    var temp = head;
        while (temp != null) {
            document.write(temp.data + " ");
            temp = temp.next;
        }
        document.write();
    }

    /* Driver program to test above functions */

        /* Constructed Linked List is 1->2->3->4->5->null */
        push(5);
        push(4);
        push(3);
        push(2);
        push(1);

        document.write("Linked List before moving last to front<br/> ");
        printList();

        moveToFront();

        document.write("<br/>Linked List after moving last to front <br/>");
        printList();

// This code is contributed by umadevi9616

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), As we need to traverse the list once.
Auxiliary Space: O(1), As constant extra space is used.

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