How to print the node of the given linked list

Linklist Interview Question

How to print the node of the given linked list

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 and printing each element's data field on its own line.

Here i am only providing the main function you have to create the link list first then only you can print its nodes.

/*
  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)
{
  // This is a "method-only" submission. 
  // You only need to complete this method. 
    while(head!=NULL)
        {
        cout<<head->data<<endl;
        head = head->next;
    }
}

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