April 18, 2015
GRK
Problem Statement
You'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, don'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;
}
}