Update on 2nd line

2.    if( ptr2=ptr1->next->next.......(5 or 10 times) ) goto 3.       else
make linear search till NULL encounter and exit with the solution

On Mon, Jul 18, 2011 at 7:41 PM, sagar pareek <[email protected]> wrote:

> i have one approach :-
>
> first compare root->data  and k
> if k is very much greater than root->data then set next=5or10 ur choice
>
> else set next 2or3  ur choice
> take two pointers ptr1 and ptr2
>
> now lets take k is much greater than root->data
> then
> 1. set ptr1=root //initially
> 2. if( ptr2=ptr1->next->next.......(5 or 10 times) ) else make linear
> search till NULL encounter
> 3. if ptr2->data==k return its position
> 4. else if (ptr2->data>k) set ptr1=ptr2 goto 2
> 5. else traverse the ptr1 upto ptr2, if found return its position else
> return fail
>
> if anyone has more efficient solution then pls tell  :)
>
> On Mon, Jul 18, 2011 at 6:53 PM, Dumanshu <[email protected]> wrote:
>
>> You have a sorted linked list of integers of some length you don't
>> know and it keeps on increasing. You are given a number k. Print the
>> position of the number k.
>> Basically, you have to search for number k in the ever growing sorted
>> list and print its position.
>>
>> Please write the complexity of whatever solution you propose.
>>
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>>
>
>
> --
> **Regards
> SAGAR PAREEK
> COMPUTER SCIENCE AND ENGINEERING
> NIT ALLAHABAD
>
>


-- 
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD

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