@terence....i cant get..can u eleborate....thnx for the sol..but plz
elaborate...

On Fri, Mar 9, 2012 at 5:59 PM, Terence <technic....@gmail.com> wrote:

>  @ rahul sharma:
> the linked list is a combination of a list a->b->...->p->q and a cycle
> q->r->...->z->q. (z != p).
> noting that the start of cycle q is the only node with 2 predecessor: p
> and z.
> if 2 pointers meet at some node x, different from q, in last step they
> must have met at x', the predecessor of x.
> the above logic holds for all nodes in cycle except q.
>
> @ sanjiv yadav:
> They will meet at the start of loop.
> ex.  a->b->c->d->e->c->d->e...
> First round:
> A: a->b->c->d
> B: a->c->e->d
> meet at d.
> Second round:
> A: a->b->c
> B: d->e->c
> meet at c.
>
>
> On 2012-3-9 18:39, sanjiv yadav wrote:
>
> No They will not meet at the start in a case containing 5 nods and having
> loop at the third node. once check this
>
> On Fri, Mar 9, 2012 at 3:48 PM, rahul sharma <rahul23111...@gmail.com>wrote:
>
>> i have 2 pointers fast and slow.....now if tehy meet there is a loop...
>>
>>  now keep one ptr at meeting point and take other one to the begining of
>> list....move both at speed of one..they will meet at start of loop....how
>> this happens???why they meet at start..plz tell logic behind this???thnx in
>> advance
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>
>
>  --
> Regards....
>
>  Sanjiv Yadav
>
>  MobNo.-  8050142693
>
>  Email Id-  sanjiv2009...@gmail.com
>
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