Print the Elements of a Linked List-hackerrank-algorithm-solution

If you're new to linked lists, this is a great exercise for learning about them. Given a pointer to the head node of a linked list, print its elements in order, one element per line. If the head pointer is null (indicating the list is empty), don’t print anything.

Input Format

The void Print(Node* head) method takes the head node of a linked list as a parameter. Each struct Node has a data field (which stores integer data) and a next field (which points to the next element in the list).

Note: Do not read any input from stdin/console. Each test case calls the Print method individually and passes it the head of a list.

Output Format

Print the integer data for each element of the linked list to stdout/console (e.g.: using printf, cout, etc.). There should be one element per line.

Sample Input

This example uses the following two linked lists:

NULL 

1->2->3->NULL

 and  are the two head nodes passed as arguments to Print(Node* head).

Note: In linked list diagrams, -> describes a pointer to the next node in the list.

Sample Output

1

2

3

Explanation

Test Case 0: NULL. An empty list is passed to the method, so nothing is printed.
Test Case 1: 1->2->3->NULL. This is a non-empty list so we loop through each element, printing each element's data field on its own line.

c++ program:

/*

  Print elements of a linked list on console 

  head pointer input could be NULL as well for empty list

  Node is defined as 

  struct Node

  {

     int data;

     struct Node *next;

  }

*/

void Print(Node *head)

{

  

    if(head!=NULL)

        {

        cout<<head->data<<"\n";

        Print(head->next);

    }

    // This is a "method-only" submission. 

  // You only need to complete this method. 

}