Given a linked list with some two adjacent repeating nodes before a zero, task is to double the first and make next 0. After this, append all the zeros to tail.
Prerequisite: Basics of implementation of Singly Linked List
Examples :
Input : 4 -> 4 -> 0 -> 2 -> 3 -> 4 ->
3 -> 3 -> 0 -> 4 ->
Output : 8-> 2-> 3-> 4-> 6-> 4-> 0->
0-> 0-> 0->
Explanation :
First, after doubling the first element and making
second element 0 before all zeros.
8 -> 0 -> 0 -> 2 -> 3 -> 4 -> 6 -> 0
-> 0 -> 4 ->
Next :
8 -> 6 -> 5 -> 6 -> 0 -> 0 -> 0 ->
0 -> 0 -> 0 -> 0 ->
Input : 0 -> 4 -> 4 -> 0 -> 3 -> 3 -> 0
-> 5 -> 0 -> 0 -> 6 ->
Output : 8 -> 6 -> 5 -> 6 -> 0 -> 0 -> 0
-> 0 -> 0 -> 0 -> 0 ->
Traverse through the linked list, and wherever there are two adjacent same data of nodes before a 0 (e.g. 4 -> 4 -> 0), then, double first element and make another as 0 (e.g. 8 -> 0 -> 0 ->). Finally, traverse the linked list and linearly point all the zeros to tail.
Implementation:
Java
// Java code to modify linked listimport java.util.*;// Linked List Nodeclass Node { int data; Node next; // Constructor public Node(int data) { this.data = data; next = null; }}// Class to perform operations// on linked listclass GfG { // Recursive function to double the one of two // elements and make next one as 0, // which are equal before 0 public static void changeTwoBefore0(Node head) { // there should be atleast three elements // to perform required operation if (head == null || head.next == null || head.next.next == null) return; // when two continuous elements // are same if ((head.data == head.next.data) && (head.next.next.data == 0)) { int temp = head.data; head.data = 2 * temp; head.next.data = 0; if (head.next.next.next != null) head = head.next.next.next; else return; } else head = head.next; // recursive call to changeTwoBefore0 // for next element changeTwoBefore0(head); } // function to append zeros at tail public static Node appendZero(Node head) { if (head == null || head.next == null) return head; // Find tail node Node tail = head; while (tail.next != null) tail = tail.next; Node origTail = tail; // Case when starting nodes have 0 values // we need to change head in this case. Node curr = head; while (curr.next != null && curr.data == 0) { tail.next = curr; tail = curr; curr = curr.next; } head = curr; // Now moving other 0s to end Node prev = curr; curr = curr.next; // We check until original tail while (curr != origTail) { // If current data is 0, append // after tail and update tail. if (curr.data == 0) { tail.next = curr; tail = curr; prev.next = curr.next; } else prev = curr; // We always move current curr = curr.next; } // Finally making sure that linked // list is null terminated. tail.next = null; return head; } public static Node doubleAndAppend0(Node head) { // Change two same nodes before 0 changeTwoBefore0(head); // Move all 0s to end return appendZero(head); } // function to display the nodes public static void display(Node head) { while (head != null) { System.out.print(head.data + " -> "); head = head.next; } } // Driver code public static void main(String[] args) { Node head = new Node(4); head.next = new Node(4); head.next.next = new Node(0); head.next.next.next = new Node(2); head.next.next.next.next = new Node(3); head.next.next.next.next.next = new Node(4); head.next.next.next.next.next.next = new Node(3); head.next.next.next.next.next.next.next = new Node(3); head.next.next.next.next.next.next.next.next = new Node(0); head.next.next.next.next.next.next.next.next.next = new Node(4); System.out.println("Original linked list :"); display(head); head = doubleAndAppend0(head); System.out.println("\nModified linked list :"); display(head); }} |
C++
// C++ code to modify linked list#include <bits/stdc++.h>using namespace std;// Linked List Nodeclass Node {public: int data; Node* next; // Constructorpublic: Node(int dat) { data = dat; next = NULL; }};// Recursive function to double the one of two// elements and make next one as 0,// which are equal before 0void changeTwoBefore0(Node* head){ // there should be atleast three elements // to perform required operation if (head == NULL || head->next == NULL || head->next->next == NULL) return; // when two continuous elements // are same if ((head->data == head->next->data) && (head->next->next->data == 0)) { int temp = head->data; head->data = 2 * temp; head->next->data = 0; if (head->next->next->next != NULL) head = head->next->next->next; else return; } else head = head->next; // recursive call to changeTwoBefore0 // for next element changeTwoBefore0(head);}// function to append zeros at tailNode* appendZero(Node* head){ if (head == NULL || head->next == NULL) return head; // Find tail node Node* tail = head; while (tail->next != NULL) tail = tail->next; Node* origTail = tail; // Case when starting nodes have 0 values // we need to change head in this case. Node* curr = head; while (curr->next != NULL && curr->data == 0) { tail->next = curr; tail = curr; curr = curr->next; } head = curr; // Now moving other 0s to end Node* prev = curr; curr = curr->next; // We check until original tail while (curr != origTail) { // If current data is 0, append // after tail and update tail. if (curr->data == 0) { tail->next = curr; tail = curr; prev->next = curr->next; } else prev = curr; // We always move current curr = curr->next; } // Finally making sure that linked // list is NULL terminated. tail->next = NULL; return head;}Node* doubleAndAppend0(Node* head){ // Change two same nodes before 0 changeTwoBefore0(head); // Move all 0s to end return appendZero(head);}// function to display the nodesvoid display(Node* head){ while (head != NULL) { cout << head->data << " -> "; head = head->next; }}// Driver Codeint main(){ Node* head = new Node(4); head->next = new Node(4); head->next->next = new Node(0); head->next->next->next = new Node(2); head->next->next->next->next = new Node(3); head->next->next->next->next->next = new Node(4); head->next->next->next->next->next->next = new Node(3); head->next->next->next->next->next->next->next = new Node(3); head->next->next->next->next->next->next->next->next = new Node(0); head->next->next->next->next->next->next->next->next ->next = new Node(4); cout << "Original linked list :"; display(head); head = doubleAndAppend0(head); cout << "\nModified linked list :"; display(head); return 0;}// contributed by ajaykr00kj |
Python3
# Python3 code to modify linked list# Linked List Nodeclass Node: # Constructor def __init__(self, data): self.data = data self.next = None # Recursive function to double the one of two# elements and make next one as 0,# which are equal before 0def changeTwoBefore0 (head): # there should be atleast three elements # to perform required operation if (head == None or head.next == None or head.next.next == None): return # when two continuous elements # are same if ((head.data == head.next.data) and (head.next.next.data == 0)): temp = head.data head.data = 2*temp head.next.data = 0 if (head.next.next.next != None): head = head.next.next.next else: return else: head = head.next # recursive call to changeTwoBefore0 # for next element changeTwoBefore0(head) # function to append zeros at taildef appendZero( head): if (head == None or head.next == None): return head # Find tail node tail = head while (tail.next != None): tail = tail.next origTail = tail # Case when starting nodes have 0 values # we need to change head in this case. curr = head while (curr.next != None and curr.data == 0): tail.next = curr tail = curr curr = curr.next head = curr # Now moving other 0s to end prev = curr curr = curr.next # We check until original tail while (curr != origTail): # If current data is 0, append # after tail and update tail. if (curr.data == 0): tail.next = curr tail = curr prev.next = curr.next else: prev = curr # We always move current curr = curr.next # Finally making sure that linked # list is None terminated. tail.next = None return head def doubleAndAppend0(head): # Change two same nodes before 0 changeTwoBefore0(head) # Move all 0s to end return appendZero(head) # function to display the nodesdef display( head): while (head != None): print(head.data ,end = " -> ") head = head.next# Driver codehead = Node(4)head.next = Node(4)head.next.next = Node(0)head.next.next.next = Node(2)head.next.next.next.next = Node(3)head.next.next.next.next.next = Node(4)head.next.next.next.next.next.next = Node(3)head.next.next.next.next.next.next.next = Node(3)head.next.next.next.next.next.next.next.next = Node(0)head.next.next.next.next.next.next.next.next.next = Node(4)print("Original linked list :")display(head) head = doubleAndAppend0(head)print("\nModified linked list :")display(head) # This code is contributed by Arnab Kundu |
C#
// C# code to modify linked list using System;// Linked List Node public class Node { public int data; public Node next; // Constructor public Node(int data) { this.data = data; next = null; } } // Class to perform operations // on linked list public class GfG { // Recursive function to double the one of two // elements and make next one as 0, // which are equal before 0 public static void changeTwoBefore0(Node head) { // there should be atleast three elements // to perform required operation if (head == null || head.next == null || head.next.next == null) return; // when two continuous elements // are same if ((head.data == head.next.data) && (head.next.next.data == 0)) { int temp = head.data; head.data = 2*temp; head.next.data = 0; if (head.next.next.next != null) head = head.next.next.next; else return; } else head = head.next; // recursive call to changeTwoBefore0 // for next element changeTwoBefore0(head); } // function to append zeros at tail public static Node appendZero(Node head) { if (head == null || head.next == null) return head; // Find tail node Node tail = head; while (tail.next != null) tail = tail.next; Node origTail = tail; // Case when starting nodes have 0 values // we need to change head in this case. Node curr = head; while (curr.next != null && curr.data == 0) { tail.next = curr; tail = curr; curr = curr.next; } head = curr; // Now moving other 0s to end Node prev = curr; curr = curr.next; // We check until original tail while (curr != origTail) { // If current data is 0, append // after tail and update tail. if (curr.data == 0) { tail.next = curr; tail = curr; prev.next = curr.next; } else prev = curr; // We always move current curr = curr.next; } // Finally making sure that linked // list is null terminated. tail.next = null; return head; } public static Node doubleAndAppend0(Node head) { // Change two same nodes before 0 changeTwoBefore0(head); // Move all 0s to end return appendZero(head); } // function to display the nodes public static void display(Node head) { while (head != null) { Console.Write(head.data + " -> "); head = head.next; } } // Driver code public static void Main() { Node head = new Node(4); head.next = new Node(4); head.next.next = new Node(0); head.next.next.next = new Node(2); head.next.next.next.next = new Node(3); head.next.next.next.next.next = new Node(4); head.next.next.next.next.next.next = new Node(3); head.next.next.next.next.next.next.next = new Node(3); head.next.next.next.next.next.next.next.next = new Node(0); head.next.next.next.next.next.next.next.next.next = new Node(4); Console.Write("Original linked list :\n"); display(head); head = doubleAndAppend0(head); Console.WriteLine("\nModified linked list :"); display(head); } } /* This code contributed by PrinciRaj1992 */ |
Javascript
<script>// JavaScript code to modify linked list // Linked List Node class Node { constructor(data) { this.data = data; this.next = null; }} // Class to perform operations // on linked list // Recursive function to double the one of two // elements and make next one as 0, // which are equal before 0 function changeTwoBefore0(head) { // there should be atleast three elements // to perform required operation if (head == null || head.next == null || head.next.next == null) return; // when two continuous elements // are same if ((head.data == head.next.data) && (head.next.next.data == 0)) { var temp = head.data; head.data = 2*temp; head.next.data = 0; if (head.next.next.next != null) head = head.next.next.next; else return; } else head = head.next; // recursive call to changeTwoBefore0 // for next element changeTwoBefore0(head); } // function to append zeros at tail function appendZero(head) { if (head == null || head.next == null) return head; // Find tail node var tail = head; while (tail.next != null) tail = tail.next; var origTail = tail; // Case when starting nodes have 0 values // we need to change head in this case. var curr = head; while (curr.next != null && curr.data == 0) { tail.next = curr; tail = curr; curr = curr.next; } head = curr; // Now moving other 0s to end var prev = curr; curr = curr.next; // We check until original tail while (curr != origTail) { // If current data is 0, append // after tail and update tail. if (curr.data == 0) { tail.next = curr; tail = curr; prev.next = curr.next; } else prev = curr; // We always move current curr = curr.next; } // Finally making sure that linked // list is null terminated. tail.next = null; return head; } function doubleAndAppend0(head) { // Change two same nodes before 0 changeTwoBefore0(head); // Move all 0s to end return appendZero(head); } // function to display the nodes function display( head) { while (head != null) { document.write(head.data + " -> "); head = head.next; } } // Driver code var head = new Node(4); head.next = new Node(4); head.next.next = new Node(0); head.next.next.next = new Node(2); head.next.next.next.next = new Node(3); head.next.next.next.next.next = new Node(4); head.next.next.next.next.next.next = new Node(3); head.next.next.next.next.next.next.next = new Node(3); head.next.next.next.next.next.next.next.next = new Node(0); head.next.next.next.next.next.next.next.next.next = new Node(4); document.write("Original linked list :<br>"); display(head); head = doubleAndAppend0(head); document.write("<br>Modified linked list :<br>"); display(head); </script> |
Output
Original linked list : 4 -> 4 -> 0 -> 2 -> 3 -> 4 -> 3 -> 3 -> 0 -> 4 -> Modified linked list : 8 -> 2 -> 3 -> 4 -> 6 -> 4 -> 0 -> 0 -> 0 -> 0 ->
Time complexity: O(n), where n is the number of nodes of linked list.
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