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 <[email protected]> wrote:
> ys...i guess i misinterpreted..the question..
> ma fault...
>
>
> On Thu, Aug 18, 2011 at 4:27 AM, Brijesh Upadhyay <
> [email protected]> 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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