then i guess ...it can be done using stack..with O(n) complexity..
it is similar to finding next greater element ....

http://www.geeksforgeeks.org/archives/8405

element in the stack at the end of the algo...are the element which will
remain in the linked list . if stack is not empty then keep poping elements
and create a SLL.

On Thu, May 31, 2012 at 4:29 PM, Ashish Goel <ashg...@gmail.com> wrote:

> yes
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Thu, May 31, 2012 at 2:34 PM, atul anand <atul.87fri...@gmail.com>wrote:
>
>> @Ashish :  please clarify this ques...
>>
>> delete a node in SLL if it is less than *any* of the succesor node ..
>>
>> 1 2 8 10 3 4 7 12
>>
>> delete 1,2,8,10,3,4,7
>>
>> ouput will be single node i.e 12
>>
>> dats what question asks?
>>
>> On Thu, May 31, 2012 at 2:16 PM, Ashish Goel <ashg...@gmail.com> wrote:
>>
>>> the LL is unsorted, is there any better solution that this?
>>>
>>> struct node* deleteNodes(struct node *pHead, struct node *pPrev)
>>> {
>>>   struct  node *pLL = *pHead;
>>>   if (!pLL) return NULL;
>>>   struct node *pCurr = pLL;
>>>
>>>   struct node *pRest = deleteNodes(pCurr->next, pCurr);
>>>   if (!pRest) return pCurr;
>>>   if (pCurr->data <pRest->data)
>>>   {
>>>     if (pPrev) { pPrev->next = pRest; };
>>>     free(pCurr);
>>>   }
>>>  else
>>>  {
>>>    pCurr->next = pRest;
>>>  }
>>>    return pCurr;
>>> }
>>>
>>>
>>> Best Regards
>>> Ashish Goel
>>> "Think positive and find fuel in failure"
>>> +919985813081
>>> +919966006652
>>>
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