# Hackerrank Print in Reverse solution

## Problem Statement

This challenge is part of a tutorial track by MyCodeSchool and is accompanied by a video lesson.

Youâ€™re given the pointer to the head node of a linked list and you need to print all its elements in reverse order from tail to head, one element per line. The head pointer may be null meaning that the list is empty - in that case, donâ€™t print anything!

## Soruce code:

/*
Print elements of a linked list in reverse order as standard output
head pointer could be NULL as well for empty list
Node is defined as
struct Node
{
int data;
struct Node *next;
}
*/
void ReversePrint(Node *head)
{
// This is a "method-only" submission.
// You only need to complete this method.
int a[100],i=0;
while(head!=NULL){
a[i] = head->data;
i++;
head = head->next;
}
for(int j = i-1;j>=0;j--){
cout<<a[j]<<endl;
}
}
** The above solution is my own code and it may not be the optimal solution or optimal way to approach the problem but it passes all the testcases in Hackerrank. So if you have any optimal approaches feel free to paste the code as the comment below..... :) :) :)

#### 8 comments:

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5. Many printers have a feature that reverses the output. The problem is there doesn’t seem to be a consistent name for this feature.Some names are given like flip horizontal
,mirror image,reverse or emulsion down.The given source code is very helpful in creating a back print.

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7. u did well but linkedlist means we should allocate memory in dynamic location but u used arrays ,what happen if give more than 100 numbers??