But suppose a random number generate a value 5 and your linked list has
only four elements. In that case what would be the answer ???
On Thu, Dec 27, 2012 at 4:03 PM, Prem Krishna Chettri <hprem...@gmail.com>wrote:
> Well my algo will be Something like this
>
> 1> Get a Random number. Perhaps You can have the function like Randon(List
> *head, int Randomnumber)
>
> 2> Use the function argument Randomnumber to loop the list.
> i.e. for(int count=0;count<=Randomnumber;count++ ){
> head = head -> next;
> }
>
> 3> print (head->value);
>
> 4> return ;
>
> Now as we are using byvalue when we return the value of head remains the
> same old head value. So everytime we call we are traversing the same old
> list.
>
> The Random variable can be taken inside the function itself if the user
> is not taking the random value.
> i.e. int Randomnumber = random(); and now the user can calll Simple
> Random(head);
>
>
>
> On Thu, Dec 27, 2012 at 3:31 PM, naveen shukla <
> naveenshuklasweetdrea...@gmail.com> wrote:
>
>> random node
>
>
> --
>
>
>
--
With Best Wishes
From:
Naveen Shukla
IIIT Allahabad
B.Tech IT 4th year
Mob No: 07860896972
E-mail naveenshuklasweetdrea...@gmail.com
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