i wish u had read the question... it is simple.. push to new stack and then
pop back... number of elements count need to be there

On Mon, Aug 22, 2011 at 7:44 PM, muthu raj <[email protected]> wrote:

> No need to count the number of nodes. Since its implemented as a linked
> list traverse the list with two two pointers one incremented one node next
> and other incremented two nodes next simultaneously.
>
> void delete_MiddleStack(node **h)
> {
>
>      if(*h==NULL)
>      return;
>      node *p,*q;
>     p=*h;
>   q=*h;
>  while(q->next!=NULL)
> {
>     p=p->next;
>     if(q->next==NULL)
>     q=q->next;
>   else q=q->next->next;
> }
> p->ele=p->next->ele;
> q=p->next;
> p->next=p->next->next;
> free(q);
> }
>
>
>
>
>
>
>
> *Muthuraj R
> IV th Year , ISE
> PESIT , Bangalore*
>
>
>
>
> On Mon, Aug 22, 2011 at 5:08 AM, vikas <[email protected]>wrote:
>
>> why to bother this much...? just count the elements when popping and
>> output the middle one .
>> while(!s.empty()){
>> e= s.pop()
>> count++
>> q.enq(e);
>> }
>>
>> count <<= 2;
>>
>> while(count){
>> e = q.deq();
>> s.push(e);
>> count --;
>> }
>> output s.top()
>>
>> while(!q.empty()){
>> e = q.deq();
>> s.push(e);
>> }
>>
>>
>> On Aug 22, 4:27 pm, Shravan Kumar <[email protected]> wrote:
>> > Pop each element and en-queue it twice and de-queue it once. When stack
>> is
>> > empty the front of the queue will be middle element.
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> > On Mon, Aug 22, 2011 at 4:01 PM, Ankur Garg <[email protected]>
>> wrote:
>> > > Find the middle of the stack..(Time complexity should be minimum)
>> >
>> > > Stack is not implemented as Linked List ...u have normal stack with
>> > > push,pop and top
>> >
>> > > How to do this ??
>> >
>> > > --
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