Append the last M nodes to the beginning of the given linked list
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 liststruct 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 listvoid 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 listint cntNodes(Node* node){ if (node == NULL) return 0; return (1 + cntNodes(node->next));}// Function to update and print// the updated list nodesvoid 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 codeint 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 |
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Java
// Java implementation of the approachclass 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); }} |
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Python3
# Python3 implementation of the approach# Class for a node of# the linked listclass newNode: # Constructor to initialize the node object def __init__(self, data): self.data = data self.next = None# Function to print the linked listdef 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 listdef cntNodes(node): if (node == None): return 0 return (1 + cntNodes(node.next)) # Function to update and print# the updated list nodesdef 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 listhead = 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 = 3updateList(head, m)# This code is contributed by shubhamsingh10 |
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C#
// C# implementation of the approachusing 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 |
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Output:
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL
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