**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 :

` ````
Solution in C :
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)
```

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