count is required since its not implemented as LL
On Mon, Aug 22, 2011 at 7:47 PM, shady <sinv...@gmail.com> wrote:
> 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 <muthura...@gmail.com> 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 <vikas.rastogi2...@gmail.com>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 <shrava...@gmail.com> 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 <ankurga...@gmail.com>
>>> 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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