Given a Linked List move the last element to the front

Introduction

The linked list is one of the most important concepts and data structures to learn while preparing for interviews. Having a good grasp of Linked Lists can be a huge plus point in a coding interview.

Problem Statement

In this problem, we are given a linked list, and we have to move the last node of the linked list to the very first and make it head or root.

Input:

Output (After moving the last element to the front of the linked list):

Problem Statement Understanding

Let’s try to understand the problem by taking an example.
Initial Linked list = 7 → 5 → 3 → 1

  • We will start with traversing the linked list from the head until we have reached the last node of the linked list.
  • Once we have reached the last node of the linked list, we will make the previous node of the last node point to NULL and the next of the last node will point to the head node.
  • Finally, we will make the node which we inserted at the front of the linked list the new head of the linked list.

After moving the last node to the front of the linked list, our linked list will look like:
Updated Linked list = 1 → 7 → 5 → 3.

Well, this is a very basic problem, and furthermore we will continue with the approach to solve the problem.

Approach and Algorithm

The most naive idea is to traverse the list till the last node or end of the LinkedList. Use two node pointers last and secLast - last used to store the address of the last node and the secLast is used to store the address of the second last node. After the end of the loop, do the following operations.

  • We have to make the second last as last (secLast->next = NULL).
  • Next step is to change next of last as head (_last->next = *headref).
  • The final step is to make last as the head (_*headref = last).

Dry Run

Code Implementation

#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 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;
}

#include <bits stdc++.h="">
using namespace std;
class Node
{
    public:
    int data;
    Node *next;
};
void moveToFront(Node **head_ref)
{
    if (*head_ref == NULL || *head_ref->next == NULL)
        return;
    Node *secLast = NULL;
    Node *last = *head_ref;
    while (last->next != NULL)
    {
        secLast = last;
        last = last->next;
    }
    secLast->next = NULL;
    last->next = *head_ref;
    *head_ref = last;
}
void push(Node** head_ref, int new_data)
{
    Node* new_node = new Node();
    new_node->data = new_data;
    new_node->next = *head_ref;
    *head_ref = new_node;
}
void printList(Node *node)
{
    while(node != NULL)
    {
        cout << node->data << " ";
        node = node->next;
    }
}
int main()
{
    Node *start = NULL;
    push(&start, 1);
    push(&start, 3);
    push(&start, 5);
    push(&start, 7);
    moveToFront(&start);
    printList(start);
    return 0;
}

class MoveLast
{
    Node head; 

    class Node
    {
        int data;
        Node next;
        Node(int d) {data = d; next = null; }
    }

    void moveToFront()
    {
        if(head == null || head.next == null)
            return;

        /* Initialize second last and last pointers */
        Node secLast = null;
        Node last = head;

        while (last.next != null)
        {
        secLast = last;
        last = last.next;
        }

        secLast.next = null;
        last.next = head;
        head = last;
    }                
    void push(int new_data)
    {
        Node new_node = new Node(new_data);
        new_node.next = head;
        head = new_node;
    }
    void printList()
    {
        Node temp = head;
        while(temp != null)
        {
        System.out.print(temp.data+" ");
        temp = temp.next;
        }
        System.out.println();
    }
    public static void main(String args[])
    {
        MoveLast llist = new MoveLast();
    
        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();
    }
}
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def push(self, data):
        new_node = Node(data)
        new_node.next = self.head
        self.head = new_node
        
    def printList(self):
        tmp = self.head
        while tmp is not None:
            print(tmp.data, end=" ")
            tmp = tmp.next
        print()

    def moveToFront(self):
        tmp = self.head
        sec_last = None

        if not tmp or not tmp.next:
            return

        while tmp and tmp.next :
            sec_last = tmp
            tmp = tmp.next

        sec_last.next = None

        tmp.next = self.head
        self.head = tmp

if __name__ == '__main__':
    llist = LinkedList()
    
    llist.push(1)
    llist.push(3)
    llist.push(5)
    llist.push(7)
    llist.moveToFront()
    llist.printList()

Output

1 7 5 3

Time Complexity: O(n), where n is the number of nodes in the given LinkedList.

Space complexity: O(1), constant space complexity, as no extra space is used.

This blog tried to discuss how you can move the last element of the linked list to the front of the list by changing the links and not just by copying the data. If you want to practice more questions on linked lists, feel free to solve them at Linked List.

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