@ligerdave
m nt getting ur algo..can u explain with an example

On 10/8/10, snehal jain <[email protected]> wrote:
>
> @neeraj
> ur worst case complexity will be O(mn)
>
>
>  On 10/8/10, snehal jain <[email protected]> wrote:
>>
>> @tech
>> the ouput will be abhgrtsghgrthswert as no suffix of 1st matches with
>> prefix of 2nd
>>
>>
>>  On 10/7/10, ligerdave <[email protected]> wrote:
>>>
>>> use pointer. shift to left if one more leading char has been found.
>>> any unmatched char resets the pointer to first char
>>>
>>> once you went through the entire list(first one), the pointer on the
>>> second list tells you where to concatenate
>>>
>>> that gives you O(n) where n is the length of first list
>>>
>>> On Oct 7, 3:52 am, snehal jain <[email protected]> wrote:
>>> > There are two linked list, both containing a character in each node.
>>> >
>>> > If one linked list contain characters  o x e n c and second contain
>>> > characters e n c a r t a then the final linked list should contain o x
>>> > e n c a r t a    i.e. if the end of one list is same as the start of
>>> > second then those characters should come only once.
>>> >
>>> > can we do it in O(n+m) where n and m are the length of list. both are
>>> > singly link list.
>>>
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>>
>

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