Print the Elements of a Linked List

Hackerrank Problem

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node’s data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print.

Function Description

Complete the printLinkedList function in the editor below.

printLinkedList has the following parameter(s):

• SinglyLinkedListNode head: a reference to the head of the list

Print

• For each node, print its data value on a new line (console.log in Javascript).

Input Format

The first line of input contains n, the number of elements in the linked list.
The next n lines contain one element each, the data values for each node.

Note: Do not read any input from stdin/console. Complete the printLinkedList function in the editor below.

Constraints

• 1 ≤ n ≤ 1000
• 1 ≤ list[i] ≤ 1000, where list[i] is the ith element of the linked list.

Sample Input

2
16
13

Sample Output

16
13

Explanation

There are two elements in the linked list. They are represented as 16 -> 13 -> NULL. So, the printLinkedList function should print 16 and 13 each on a new line.

void printLinkedList(SinglyLinkedListNode* head) {
     while(head!= NULL){
          cout<<head->data<<endl;
          head = head->next;
     }
}

Happy Coding 😁😁