get the size  of both the linked list say d1,d2 ;
get their diff like df=abs(d1-d2);

now take 2 pointers both poiting to the start of both the linked list .
go to bigger list now move this pointer until both the distance becomes
equal .
at this point we have one pointer from smaller list pointing to the start
of the list and
another pointer in bigger list pointing to node where distance came equal ,
move these 2 pointers until both point to the same node .. and we are done .

On Wed, Jul 4, 2012 at 10:44 PM, Amit Chauhan <amitchauhan....@gmail.com>wrote:

> If both the linked list are ordered one then you can solve this problem in
> linear time and with constant space.
>
>
>
> On Wed, Jul 4, 2012 at 10:41 PM, Abhi <abhi120...@gmail.com> wrote:
>
>> Any efficient algorithm to find intersection of two linked lists.Example:
>> Linked List 1)  1 -> 2 -> 3 -> 4 -> 5 -> 6
>> Linked List 2)  3 -> 4 -> 5
>>
>> Intersection 4 -> 5 -> 6
>>
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