suppose linked list is a->b->c->d->e
and suppose loop starts from 'c' according to u let one pointer be at 'c' say *q and another be at 'a' say *p. Now if we move both at the speed of one then After first pass p will be at b q will be at d After second pass p will be at c q will be at e After third pass p will be at d q will be at c After fourth pass p will be at e q will be at d After fifth pass p will be at c q will be at e and so on. Correct me if i am wrong. On Fri, Mar 9, 2012 at 7:28 PM, rahul sharma <rahul23111...@gmail.com>wrote: > @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 >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To post to this group, send email to algogeeks@googlegroups.com. >>> To unsubscribe from this group, send email to >>> algogeeks+unsubscr...@googlegroups.com. >>> For more options, visit this group at >>> http://groups.google.com/group/algogeeks?hl=en. >>> >> >> >> >> -- >> Regards.... >> >> Sanjiv Yadav >> >> MobNo.- 8050142693 >> >> Email Id- sanjiv2009...@gmail.com >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algogeeks@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@googlegroups.com. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algogeeks@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@googlegroups.com. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Regards.... Sanjiv Yadav MobNo.- 8050142693 Email Id- sanjiv2009...@gmail.com -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
