we can do that too.
But if I return cur then I can myself print cur, cur->next, cur->next->prev.
for verification.
On 18 August 2011 11:56, Vijay Kansal <vijaykans...@gmail.com> wrote:
> We should return curr->next in the last statement of ur code
>
> On Aug 18, 7:08 am, Dipankar Patro <dip10c...@gmail.com> wrote:
> > A slight change in above code:
> > make it
> > while(cur && cur->next)
> > ^^ other wise the code will crash at last element in a prefect list, with
> no
> > loop.
> >
> > On 18 August 2011 07:36, Dipankar Patro <dip10c...@gmail.com> wrote:
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > > I have come up with this:
> > > - Use only one pointer, NODE *cur
> > > - initialize cur to headref
> >
> > > - The main loop:
> > > while (cur)
> > > {
> > > if(cur->next->prev != cur)
> > > break;
> > > cur=cur->next;
> > > }
> > > return cur;
> >
> > > ^^ I think the code is self explanatory. It just uses the fact that at
> > > loop, the prev of next to current won't be current.
> > > e.g. A<->B<->C<->D<->E<->F->C
> > > Though F is pointing to C, C won't be pointing back to F as the prev of
> C
> > > is pointing to B.
> > > Complexity: O(n)
> >
> > > On 18 August 2011 04:45, payal gupta <gpt.pa...@gmail.com> wrote:
> >
> > >> ys...i guess i misinterpreted..the question..
> > >> ma fault...
> >
> > >> On Thu, Aug 18, 2011 at 4:27 AM, Brijesh Upadhyay <
> > >> brijeshupadhyay...@gmail.com> wrote:
> >
> > >>> At the node from where the loop just started.. anyway we could not
> use
> > >>> that logic , coz it isnt circular linked list!
> >
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