Print the Elements of a Linked List


Problem Statement :


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):
      1.SinglyLinkedListNode head: a reference to the head of the list

Print:
     1. For each node, print its  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. 1<=n<=1000
    2. 1<=list[i]<=1000



Solution :


                        In C:
//In hacker rank as the solution the program is too big 
//all u need is to code this function.

void printLinkedList(SinglyLinkedListNode* head) {

SinglyLinkedListNode* ptr=head;
    while(ptr!=NULL)
    {
        printf("%d\n",ptr->data);
        ptr=ptr->next;
    }

}







In C++:
//The rest of the code is already coded.
//As the code is too big all you need is complete the following part.

void printLinkedList(SinglyLinkedListNode* head) {

     while(head!=NULL)
     {
         cout << head->data << "\n";
         head = head->next;
     }
}









In Java:
//The following part is all you to complete, the rest is
//already present in hackerrank 

 static void printLinkedList(SinglyLinkedListNode head) {
        while(head != null) { System.out.println(head.data); head = head.next;}

    }






In Python 3:

# The following part of code is all you need to 
# complete the challenge in hacker rank

def printLinkedList(head):
    temp = head
    while (temp.next != None):
        print(temp.data)
        temp = temp.next
    print(temp.data)