@Raj

What do you mean by identical? You are just concerned about the
presence of one element in another LL or you are concerned about the
equality of number of elements too?

On 6/2/10, Raj N <rajn...@gmail.com> wrote:
> @sharad kumar: But don't you think this'll consume a lot of space. Merge
> sort will give O(nlogn) complexity when a separate LL is used to store the
> sorted elements but if we disbuild the same LL it'll be >O(n2). And also wat
> do u mean by combining LL can u explain
>
>
> On Wed, Jun 2, 2010 at 6:48 AM, sharad kumar <aryansmit3...@gmail.com>wrote:
>
>> @Raj:sorting can be done in O(nlogn)..sort both and compare both.or use a
>> hash map to store all elements of one LL and then compare it with
>> other....or combine both the  LL perform merge sort and start deleting
>> adjacent elements.if adjacent elements in equal count then LL are equal
>> and
>> at end of process we get an empty list.....
>>
>>
>> On Tue, Jun 1, 2010 at 11:55 PM, Raj N <rajn...@gmail.com> wrote:
>>
>>> Hi all,
>>> Can someone suggest an efficient algorithm to check if 2 linked lists
>>> are identical. If 2 lists have to be sorted then there'll be a lot of
>>> space consumption for 2 separate linked lists. Can the whole process
>>> be done < O(n2)
>>>
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>>
>>
>> --
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Luv,
S.Antony Vincent Pandian

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