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.
Java
// Java code to modify linked list import java.util.*; // Linked List Node class Node { int data; Node next; // Constructor public Node(int data) { this.data = data; next = null; } } // Class ro perform operations // on linked list 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) { 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); } } |
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Python3
# Python3 code to modify linked list # Linked List Node class 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 0 def 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 tail def 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 nodes def display( head): while (head != None): print(head.data ,end = " -> ") head = head.next # Driver code head = 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 |
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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 ro 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 */ |
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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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