Print in Reverse
Problem Statement :
Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything.
Example
head* refers to the linked list with data values 1->2->3->Null
Print the following:
3
2
1
Function Description:
Complete the reversePrint function in the editor below.
reversePrint has the following parameters:
SinglyLinkedListNode pointer head: a reference to the head of the list
Prints
The data values of each node in the reversed list.
Input Format:
The first line of input contains t, the number of test cases.
The input of each test case is as follows:
1. The first line contains an integer n, the number of elements in the list.
2. Each of the next n lines contains a data element for a list node.
Constraints:
1. 1<=n<=1000
2. 1<=list[i]<=1000
Solution :
Solution in C :
In C:
//the following fuction is all that is needed to complete the
//challenge in hackerrank platform.
void reversePrint(SinglyLinkedListNode* head) {
SinglyLinkedListNode* current=head,*next,*prev=NULL;
while(current!=NULL)
{ next=current->next;
current->next=prev;
prev=current;
current=next;
}
head=prev;
SinglyLinkedListNode* temp=head;
while(temp!=NULL)
{
printf("%d\n",temp->data);
temp=temp->next;
}
}
In C++:
//the following fuction is all that is needed to complete the
//challenge in hackerrank platform.
void ReversePrint(Node *head)
{
// This is a "method-only" submission.
// You only need to complete this method.
if(head!=NULL)
{
ReversePrint(head->next);
cout<<head->data<<"\n";
}
}
In Java:
//the following method is all that is needed to complete the
//challenge in hackerrank platform.
static void reversePrint(SinglyLinkedListNode head) {
if(head.next != null) {
reversePrint(head.next);
}
System.out.println(head.data);
}
In Python 3 :
//the following method is all that is needed to complete the
//challenge in hackerrank platform.
def ReversePrint(head):
if head is None:
return
ReversePrint(head.next)
print(head.data)
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