Q1: The function below reverses a linked list in place. Call it on one
of the lists, compare the resulting list to the other list. Then call
it again to put the list back in its original order.
list Reverse(list head)
{
list T, prv, nxt;
prv = head;
for(T = head->next; T; T = nxt)
{
nxt = T->next;
T->next = prv;
prv = T;
T = nxt;
}
head->next = 0;
return prv;
}
Q2:
delete(node *d)
{
if (d->next)
{
node nxt = d->next;
d->value = nxt->value;
d->next = nxt->next;
free nxt;
}
else
{
for(node p = head; p; p = p->next)
if (p->next == d)
{
p->next = 0;
free d;
}
}
}
On Aug 10, 1:14 pm, Piyush Kapoor <[email protected]> wrote:
> Q1)Two linked Lists are given,i.e,their head pointers are given,and the
> problem is to check if the second one is reverse of the first one.Give the
> most efficient algo for it.
> Q2)A linked list is given,and one of its nodes is given.The problem is to
> delete the given node from the linked list.(The head node is not given).
> (In both of the above cases,the linked lists are singly linked lists.)
> --
> *Regards,*
> *Piyush Kapoor,*
> *2nd year,CSE
> IT-BHU*
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