Append the last M nodes to the beginning of the given linked list.
Last Updated :
22 Mar, 2023
Given a linked list and an integer M, the task is to append the last M nodes of the linked list to the front.
Examples:
Input: List = 4 -> 5 -> 6 -> 1 -> 2 -> 3 -> NULL, M = 3
Output: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL
Input: List = 8 -> 7 -> 0 -> 4 -> 1 -> NULL, M = 2
Output: 4 -> 1 -> 8 -> 7 -> 0 -> NULL
Approach: Find the first node of the last M nodes in the list, this node will be the new head node so make the next pointer of the previous node as NULL and point the last node of the original list to the head of the original list. Finally, print the updated list.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Class for a node of
// the linked list
struct Node {
// Data and the pointer
// to the next node
int data;
Node* next;
Node(int data)
{
this->data = data;
this->next = NULL;
}
};
// Function to print the linked list
void printList(Node* node)
{
while (node != NULL) {
cout << (node->data) << " -> ";
node = node->next;
}
cout << "NULL";
}
// Recursive function to return the
// count of nodes in the linked list
int cntNodes(Node* node)
{
if (node == NULL)
return 0;
return (1 + cntNodes(node->next));
}
// Function to update and print
// the updated list nodes
void updateList(Node* head, int m)
{
// Total nodes in the list
int cnt = cntNodes(head);
if (cnt != m && m < cnt) {
// Count of nodes to be skipped
// from the beginning
int skip = cnt - m;
Node* prev = NULL;
Node* curr = head;
// Skip the nodes
while (skip > 0) {
prev = curr;
curr = curr->next;
skip--;
}
// Change the pointers
prev->next = NULL;
Node* tempHead = head;
head = curr;
// Find the last node
while (curr->next != NULL)
curr = curr->next;
// Connect it to the head
// of the sub list
curr->next = tempHead;
}
// Print the updated list
printList(head);
}
// Driver code
int main()
{
// Create the list
Node* head = new Node(4);
head->next = new Node(5);
head->next->next = new Node(6);
head->next->next->next = new Node(1);
head->next->next->next->next = new Node(2);
head->next->next->next->next->next = new Node(3);
int m = 3;
updateList(head, m);
return 0;
}
// This code is contributed by rutvik_56
// Java implementation of the approach
class GFG {
// Class for a node of
// the linked list
static class Node {
// Data and the pointer
// to the next node
int data;
Node next;
Node(int data)
{
this.data = data;
this.next = null;
}
}
// Function to print the linked list
static void printList(Node node)
{
while (node != null) {
System.out.print(node.data + " -> ");
node = node.next;
}
System.out.print("NULL");
}
// Recursive function to return the
// count of nodes in the linked list
static int cntNodes(Node node)
{
if (node == null)
return 0;
return (1 + cntNodes(node.next));
}
// Function to update and print
// the updated list nodes
static void updateList(Node head, int m)
{
// Total nodes in the list
int cnt = cntNodes(head);
if (cnt != m && m < cnt) {
// Count of nodes to be skipped
// from the beginning
int skip = cnt - m;
Node prev = null;
Node curr = head;
// Skip the nodes
while (skip > 0) {
prev = curr;
curr = curr.next;
skip--;
}
// Change the pointers
prev.next = null;
Node tempHead = head;
head = curr;
// Find the last node
while (curr.next != null)
curr = curr.next;
// Connect it to the head
// of the sub list
curr.next = tempHead;
}
// Print the updated list
printList(head);
}
// Driver code
public static void main(String[] args)
{
// Create the list
Node head = new Node(4);
head.next = new Node(5);
head.next.next = new Node(6);
head.next.next.next = new Node(1);
head.next.next.next.next = new Node(2);
head.next.next.next.next.next = new Node(3);
int m = 3;
updateList(head, m);
}
}
# Python3 implementation of the approach
# Class for a node of
# the linked list
class newNode:
# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None
# Function to print the linked list
def printList(node):
while (node != None):
print(node.data, "->", end=" ")
node = node.next
print("NULL")
# Recursive function to return the
# count of nodes in the linked list
def cntNodes(node):
if (node == None):
return 0
return (1 + cntNodes(node.next))
# Function to update and print
# the updated list nodes
def updateList(head, m):
# Total nodes in the list
cnt = cntNodes(head)
if (cnt != m and m < cnt):
# Count of nodes to be skipped
# from the beginning
skip = cnt - m
prev = None
curr = head
# Skip the nodes
while (skip > 0):
prev = curr
curr = curr.next
skip -= 1
# Change the pointers
prev.next = None
tempHead = head
head = curr
# Find the last node
while (curr.next != None):
curr = curr.next
# Connect it to the head
# of the sub list
curr.next = tempHead
# Print the updated list
printList(head)
# Driver code
# Create the list
head = newNode(4)
head.next = newNode(5)
head.next.next = newNode(6)
head.next.next.next = newNode(1)
head.next.next.next.next = newNode(2)
head.next.next.next.next.next = newNode(3)
m = 3
updateList(head, m)
# This code is contributed by shubhamsingh10
// C# implementation of the approach
using System;
class GFG {
// Class for a node of
// the linked list
class Node {
// Data and the pointer
// to the next node
public int data;
public Node next;
public Node(int data)
{
this.data = data;
this.next = null;
}
}
// Function to print the linked list
static void printList(Node node)
{
while (node != null) {
Console.Write(node.data + " -> ");
node = node.next;
}
Console.Write("NULL");
}
// Recursive function to return the
// count of nodes in the linked list
static int cntNodes(Node node)
{
if (node == null)
return 0;
return (1 + cntNodes(node.next));
}
// Function to update and print
// the updated list nodes
static void updateList(Node head, int m)
{
// Total nodes in the list
int cnt = cntNodes(head);
if (cnt != m && m < cnt) {
// Count of nodes to be skipped
// from the beginning
int skip = cnt - m;
Node prev = null;
Node curr = head;
// Skip the nodes
while (skip > 0) {
prev = curr;
curr = curr.next;
skip--;
}
// Change the pointers
prev.next = null;
Node tempHead = head;
head = curr;
// Find the last node
while (curr.next != null)
curr = curr.next;
// Connect it to the head
// of the sub list
curr.next = tempHead;
}
// Print the updated list
printList(head);
}
// Driver code
public static void Main(String[] args)
{
// Create the list
Node head = new Node(4);
head.next = new Node(5);
head.next.next = new Node(6);
head.next.next.next = new Node(1);
head.next.next.next.next = new Node(2);
head.next.next.next.next.next = new Node(3);
int m = 3;
updateList(head, m);
}
}
// This code is contributed by PrinciRaj1992
<script>
// JavaScript implementation of the approach
// Class for a node of
// the linked list
class Node {
// Data and the pointer
// to the next node
constructor(data) {
this.data = data;
this.next = null;
}
}
// Function to print the linked list
function printList(node) {
while (node != null) {
document.write(node.data + " -> ");
node = node.next;
}
document.write("NULL");
}
// Recursive function to return the
// count of nodes in the linked list
function cntNodes(node) {
if (node == null) return 0;
return 1 + cntNodes(node.next);
}
// Function to update and print
// the updated list nodes
function updateList(head, m) {
// Total nodes in the list
var cnt = cntNodes(head);
if (cnt != m && m < cnt) {
// Count of nodes to be skipped
// from the beginning
var skip = cnt - m;
var prev = null;
var curr = head;
// Skip the nodes
while (skip > 0) {
prev = curr;
curr = curr.next;
skip--;
}
// Change the pointers
prev.next = null;
var tempHead = head;
head = curr;
// Find the last node
while (curr.next != null) curr = curr.next;
// Connect it to the head
// of the sub list
curr.next = tempHead;
}
// Print the updated list
printList(head);
}
// Driver code
// Create the list
var head = new Node(4);
head.next = new Node(5);
head.next.next = new Node(6);
head.next.next.next = new Node(1);
head.next.next.next.next = new Node(2);
head.next.next.next.next.next = new Node(3);
var m = 3;
updateList(head, m);
// This code is contributed by rdtank.
</script>
Output1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL
METHOD 2:
We Will use modification of runner's technique :-
1. find the kth node from end using runner technique and do the following modifications
2. now we have to update our pointers as
a) fast->next will be pointing to head,
b)slow->next will be new head,
c)last node will be the slow->next hence it should point to null
C++
#include <iostream>
using namespace std;
struct node {
int data;
node* next;
node(int x)
{
data = x;
next = NULL;
}
};
void insertAtTail(node*& head, int x)
{
if (head == NULL) {
head = new node(x);
return;
}
node* curr = head;
while (curr->next != NULL) {
curr = curr->next;
}
node* t = new node(x);
curr->next = t;
}
void print(node* head)
{
node* curr = head;
while (curr != NULL) {
cout << curr->data << " -> ";
curr = curr->next;
}
cout << "NULL\n";
}
node* appendK(node* head, int k)
{
node* fast = head;
node* slow = head;
for (int i = 0; i < k; i++) {
fast = fast->next;
}
while (fast->next != NULL) {
slow = slow->next;
fast = fast->next;
}
// cout<<"data"<<" "<<slow->data<<" "<<fast->data<<endl;
fast->next = head;
head = slow->next;
slow->next = NULL;
return head;
}
int main()
{
node* head = NULL;
int n;
n = 6;
insertAtTail(head, 4);
insertAtTail(head, 5);
insertAtTail(head, 6);
insertAtTail(head, 1);
insertAtTail(head, 2);
insertAtTail(head, 3);
int k;
k = 3;
head = appendK(head, k % n);
print(head);
return 0;
}
// Java code to the above approach
import java.io.*;
class GFG {
class Node {
int data;
Node next;
Node(int x)
{
data = x;
next = null;
}
}
public Node insertAtTail(Node head, int x)
{
if (head == null) {
head = new Node(x);
return head;
}
Node curr = head;
while (curr.next != null) {
curr = curr.next;
}
Node t = new Node(x);
curr.next = t;
return head;
}
public void print(Node head)
{
Node curr = head;
while (curr != null) {
System.out.print(curr.data + " -> ");
curr = curr.next;
}
System.out.print("NULL\n");
}
public Node appendK(Node head, int k)
{
Node fast = head;
Node slow = head;
for (int i = 0; i < k; i++) {
fast = fast.next;
}
while (fast.next != null) {
slow = slow.next;
fast = fast.next;
}
fast.next = head;
head = slow.next;
slow.next = null;
return head;
}
public static void main(String[] args)
{
GFG l = new GFG();
Node head = null;
int n = 6;
head = l.insertAtTail(head, 4);
head = l.insertAtTail(head, 5);
head = l.insertAtTail(head, 6);
head = l.insertAtTail(head, 1);
head = l.insertAtTail(head, 2);
head = l.insertAtTail(head, 3);
int k = 3;
head = l.appendK(head, k % n);
l.print(head);
}
}
// This code is contributed by lokesh (lokeshmvs21).
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def insertAtTail(self, head, x):
if head is None:
head = Node(x)
return head
curr = head
while curr.next is not None:
curr = curr.next
t = Node(x)
curr.next = t
return head
def printList(self, head):
curr = head
while curr is not None:
print(curr.data, end=" -> ")
curr = curr.next
print("NULL")
def appendK(self, head, k):
fast = head
slow = head
for i in range(k):
fast = fast.next
while fast.next is not None:
slow = slow.next
fast = fast.next
fast.next = head
head = slow.next
slow.next = None
return head
if __name__ == "__main__":
l = LinkedList()
head = None
n = 6
head = l.insertAtTail(head, 4)
head = l.insertAtTail(head, 5)
head = l.insertAtTail(head, 6)
head = l.insertAtTail(head, 1)
head = l.insertAtTail(head, 2)
head = l.insertAtTail(head, 3)
k = 3
head = l.appendK(head, k % n)
l.printList(head)
// JavaScript code to the above approach
class Node {
constructor(x){
this.data = x;
this.next = null;
}
}
function insertAtTail( head, x)
{
if (head == null) {
head = new Node(x);
return head;
}
let curr = head;
while (curr.next != null) {
curr = curr.next;
}
let t = new Node(x);
curr.next = t;
return head;
}
function print(head)
{
let curr = head;
while (curr != null) {
process.stdout.write(curr.data + " -> ");
curr = curr.next;
}
console.log("NULL");
}
function appendK(head, k)
{
let fast = head;
let slow = head;
for (let i = 0; i < k; i++) {
fast = fast.next;
}
while (fast.next != null) {
slow = slow.next;
fast = fast.next;
}
fast.next = head;
head = slow.next;
slow.next = null;
return head;
}
let head = null;
let n = 6;
head = insertAtTail(head, 4);
head = insertAtTail(head, 5);
head = insertAtTail(head, 6);
head = insertAtTail(head, 1);
head = insertAtTail(head, 2);
head = insertAtTail(head, 3);
let k = 3;
head = appendK(head, k % n);
print(head);
// This code is contributed by Nidhi goel.
// C# code to the above approach
using System;
public class GFG {
class Node {
public int data;
public Node next;
public Node(int x)
{
data = x;
next = null;
}
}
Node insertAtTail(Node head, int x)
{
if (head == null) {
head = new Node(x);
return head;
}
Node curr = head;
while (curr.next != null) {
curr = curr.next;
}
Node t = new Node(x);
curr.next = t;
return head;
}
void print(Node head)
{
Node curr = head;
while (curr != null) {
Console.Write(curr.data + " -> ");
curr = curr.next;
}
Console.Write("NULL\n");
}
Node appendK(Node head, int k)
{
Node fast = head;
Node slow = head;
for (int i = 0; i < k; i++) {
fast = fast.next;
}
while (fast.next != null) {
slow = slow.next;
fast = fast.next;
}
fast.next = head;
head = slow.next;
slow.next = null;
return head;
}
static public void Main()
{
GFG l = new GFG();
Node head = null;
int n = 6;
head = l.insertAtTail(head, 4);
head = l.insertAtTail(head, 5);
head = l.insertAtTail(head, 6);
head = l.insertAtTail(head, 1);
head = l.insertAtTail(head, 2);
head = l.insertAtTail(head, 3);
int k = 3;
head = l.appendK(head, k % n);
l.print(head);
}
}
// This code is contributed by lokesh (lokeshmvs21).
Output1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL
Time complexity: O(n) where n is size of given linked list
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