Take two pointers ptr1 and ptr2,

ptr1 should move 4 nodes at a time
ptr2 should move 3 nodes at a time,
thats it when the ptr1 reaches the end the ptr2 will be pointing to 3/4th,
same way for m/n th node. But this works best for even number of nodes in
list.

For odd numbers we need to compromise like finding the middle node :)

Cheers,
Mani

On Tue, Jul 19, 2011 at 4:20 PM, Yogesh Yadav <[email protected]> wrote:

> Part 1:
> Take 2 pointers moving at different speed...
>
> 1st with head->next
> 2nd with head->next->next
>
> when 2nd will reach the end 1st will be pointing middle... so one result is
> obtained
>
> Part 2:
> when 1st result is obtained then start with the node pointed by 1st as it
> will be on 1/2th part of the list.... Start 2 pointers again
>
> 3rd with head->next
> 4th with head->next->next
>
> when 4th will reach end point then 3rd will be pointing the node present at
> 3/4th of the list....
>
>
>
>
>
>
>
>
> On Tue, Jul 19, 2011 at 3:54 PM, surender sanke <[email protected]>wrote:
>
>> take two ptrs ptr1 and ptr2 pointing to head
>> move ptr1 until 1/4th of size of list.
>> move ptr1 and ptr2 until ptr1=null
>> ptr2 is pointing at 3/4th
>>
>> surender
>>
>> On Tue, Jul 19, 2011 at 3:42 PM, SAMMM <[email protected]> wrote:
>>
>>> Yaa this will work , you need to handle the case for odd number of
>>> nodes .
>>> For even number of node it will serve the purpose .
>>>
>>> Yaa for the second part also you can use the ratio concept
>>>
>>> fast pointer : slow pointer = 4:3
>>>
>>> slow = ptr->next->next->next
>>> fast = ptr->next->next->next->next
>>>
>>>
>>> Here also we need to handle the case if the number of node is not a
>>> multiple of 4 .
>>>
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Mani
http://www.sanidapa.com - The music Search engine

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