Move last element to front of a given Linked List

Last Updated : 26 Aug, 2024

Given a singly linked list. The task is to move the last node to the front in a given List.

Examples:

Input: 2->5->6->2->1
Output: 1->2->5->6->2
Explanation : Node 1 moved to front.

Input: 1->2->3->4->5
Output: 5->1->2->3->4
Explanation : Node 5 moved to front.

[Expected Approach] Traverse Till Last Node – O(n) Time and O(1) space:

Traverse the list till the last node. Use two pointers to store the reference of the last node and 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.

Below is the implementation of the above approach:

C++

// C++ Program to move last element // to front in a given linked list  #include <iostream> using namespace std;  class Node {   public:     int data;     Node *next;     Node(int x) {         data = x;         next = nullptr;     } };  // Function to move the last node to the // front of the linked list Node *moveToFront(Node *head) {        // If the list is empty or has only one node,     // no need to move     if (head == NULL || head->next == NULL) {         return head;     }      // To keep track of the second last node     Node *secLast = NULL;      // To traverse to the last node     Node *last = head;      // Traverse the list to find the last and     // second last nodes     while (last->next != NULL) {         secLast = last;         last = last->next;     }      // Change the next of second last node to NULL     secLast->next = NULL;      // Make the last node as the new head     last->next = head;     head = last;      return head; }  void printList(Node *node) {     while (node != NULL) {         cout << node->data << " ";         node = node->next;     }     cout << endl; }  int main() {        // Create a linked list 1->2->3->4->5     Node *head = new Node(1);     head->next = new Node(2);     head->next->next = new Node(3);     head->next->next->next = new Node(4);     head->next->next->next->next = new Node(5);      cout << "Linked list before: " << endl;     printList(head);      head = moveToFront(head);      cout << "Linked list after: " << endl;     printList(head);      return 0; }

// C Program to move last element // to front in a given linked list #include <stdio.h> #include <stdlib.h>  struct Node {     int data;     struct Node *next; };   // Function to move the last node to the front of the linked list struct Node *moveToFront(struct Node *head) {        // If the list is empty or has only one node, no need to move     if (head == NULL || head->next == NULL) {         return head;     }      // To keep track of the second last node     struct Node *secLast = NULL;      // To traverse to the last node     struct Node *last = head;      // Traverse the list to find the last and second last nodes     while (last->next != NULL) {         secLast = last;         last = last->next;     }      // Change the next of second last node to NULL     secLast->next = NULL;      // Make the last node as the new head     last->next = head;     head = last;      return head; }  void printList(struct Node *node) {     while (node != NULL) {         printf("%d ", node->data);         node = node->next;     }     printf("\n"); }   struct Node *createNode(int data) {     struct Node *newNode =        (struct Node *)malloc(sizeof(struct Node));     newNode->data = data;     newNode->next = NULL;     return newNode; }  int main() {     // Create a linked list 1->2->3->4->5     struct Node *head = createNode(1);     head->next = createNode(2);     head->next->next = createNode(3);     head->next->next->next = createNode(4);     head->next->next->next->next = createNode(5);      printf("Linked list before:\n");     printList(head);      head = moveToFront(head);      printf("Linked list after:\n");     printList(head);      return 0; }

Java

// Java Program to move last element // to front in a given linked list  class Node {     int data;     Node next;          Node(int x) {         data = x;         next = null;     } }  class GfG {          // Function to move the last node to the      // front of the linked list     static Node moveToFront(Node head) {                // If the list is empty or has only one node,          // no need to move         if (head == null || head.next == null) {             return head;         }                // To keep track of the second last node         Node secLast = null;                   // To traverse to the last node         Node last = head;              // Traverse the list to find the last and          // second last nodes         while (last.next != null) {             secLast = last;             last = last.next;         }          // Change the next of second last node to null         secLast.next = null;          // Make the last node as the new head         last.next = head;         head = last;          return head;     }      static void printList(Node node) {         while (node != null) {             System.out.print(node.data + " ");             node = node.next;         }         System.out.println();     }      public static void main(String[] args) {                // Create a linked list 1->2->3->4->5         Node head = new Node(1);         head.next = new Node(2);         head.next.next = new Node(3);         head.next.next.next = new Node(4);         head.next.next.next.next = new Node(5);          System.out.println("Linked list before: ");         printList(head);          head = moveToFront(head);          System.out.println("Linked list after: ");         printList(head);     } }

Python

# Python code to move the last item to front  class Node:     def __init__(self, data):         self.data = data         self.next = None  # Function to move the last node  # to the front of the linked list def move_to_front(head):        # If the list is empty or has      # only one node, no need to move     if head is None or head.next is None:         return head        	# To keep track of the second last node     sec_last = None            # To traverse to the last node     last = head            # Traverse the list to find the      # last and second last nodes     while last.next is not None:         sec_last = last         last = last.next      # Change the next of second last node to None     sec_last.next = None      # Make the last node as the new head     last.next = head     head = last      return head  def print_list(node):     while node is not None:         print(node.data, end=" ")         node = node.next     print()  # Create a linked list 1->2->3->4->5 head = Node(1) head.next = Node(2) head.next.next = Node(3) head.next.next.next = Node(4) head.next.next.next.next = Node(5)  print("Linked list before:") print_list(head)  head = move_to_front(head)  print("Linked list after:") print_list(head)

// C# Program to move last element // to front in a given linked list  class Node {     public int data;     public Node next;          public Node(int x) {         data = x;         next = null;     } }  class GfG {          // Function to move the last node to the      // front of the linked list     static Node MoveToFront(Node head) {                // If the list is empty or has only one node,          // no need to move         if (head == null || head.next == null) {             return head;         }                // To keep track of the second last node         Node secLast = null;                   // To traverse to the last node         Node last = head;              // Traverse the list to find the last and          // second last nodes         while (last.next != null) {             secLast = last;             last = last.next;         }          // Change the next of second last node to null         secLast.next = null;          // Make the last node as the new head         last.next = head;         head = last;          return head;     }      static void PrintList(Node node) {         while (node != null) {             System.Console.Write(node.data + " ");             node = node.next;         }         System.Console.WriteLine();     }      public static void Main() {                // Create a linked list 1->2->3->4->5         Node head = new Node(1);         head.next = new Node(2);         head.next.next = new Node(3);         head.next.next.next = new Node(4);         head.next.next.next.next = new Node(5);          System.Console.WriteLine("Linked list before:");         PrintList(head);          head = MoveToFront(head);          System.Console.WriteLine("Linked list after:");         PrintList(head);     } }

JavaScript

// Javascript Program to move last element to // front in a given linked list  class Node {     constructor(data) {         this.data = data;         this.next = null;     } }  // Function to move the last  // node to the front of the linked list function moveToFront(head) {      // If the list is empty or has      // only one node, no need to move     if (head === null || head.next === null) {         return head;     } 	     // To keep track of the second last node     let secLast = null;      // To traverse to the last node     let last = head;          // Traverse the list to find the last      // and second last nodes     while (last.next !== null) {         secLast = last;         last = last.next;     }      // Change the next of second last node to null     secLast.next = null;      // Make the last node as the new head     last.next = head;     head = last;      return head; }  function printList(node) {     while (node !== null) {         process.stdout.write(node.data + " ");         node = node.next;     }     console.log(); }  let head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5);  console.log("Linked list before:"); printList(head);  head = moveToFront(head);  console.log("Linked list after:"); printList(head);

Output

Linked list before:  1 2 3 4 5  Linked list after:  5 1 2 3 4

Time Complexity: O(n), As we need to traverse the list once.
Auxiliary Space: O(1)

Move the Kth Element in a Doubly Linked List to the End

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Linked List

Move last element to front of a given Linked List

[Expected Approach] Traverse Till Last Node – O(n) Time and O(1) space:

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