@mohit : +1
On Sat, Aug 13, 2011 at 9:20 AM, Raghavan its...@gmail.com wrote:
First question:
- .Read the data from the first list and put it in a stack
- Traverse the next list and compare by reading elements from the stack
- This would solve it
Second question:
- Take an
folks ..
for second problem .. the mentioned algo doesn't work for last node.. I mean
copying data of next node and aalso making nxt-nxt as nxt node.. and
deleting the next node..
in case of last node.. how can we delete the last node being on that node?
even though if we use free (node) ... the
yes you are right
Thank you,
Siddharam
On Thu, Aug 11, 2011 at 10:21 AM, Abhishek gupta
mailatabhishekgu...@gmail.com wrote:
Q2). i think for second question it will be enough to just swap the data of
current node to next node,
and delete the next node.
it will be like,
//for swap
int
Q2). i think for second question it will be enough to just swap the data of
current node to next node,
and delete the next node.
it will be like,
//for swap
int temp=current-data;
current-data=current-next-data;
current-next-data=temp;
//for delete
struct node *temp;
temp=current-next;
First question:
- .Read the data from the first list and put it in a stack
- Traverse the next list and compare by reading elements from the stack
- This would solve it
Second question:
- Take an list like 1-2--3-4
- If you are given with 2, juz copy the value and reference of
hey guys , can't it be like this without reversing list-
int rec_iterate(Node head1,Node *head2)
{
if(head1 ==NULL ) return 1;
if(rec_iterate(head1-n,head2) == 0) return 0;
if (head1-value == (*head2)-value)
{ *head2=(*head2)-next;
return 1;
}
else
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)
{
