@ashish : please provide link for that page
On Sun, Dec 25, 2011 at 10:36 PM, Ashish Goel <[email protected]> wrote:
> refer stanford page
> 1,2,3,4,5,6
>
> will become
>
> 5,3,1
> 6,4,2
>
> void MoveNode(struct node** destRef, struct node** sourceRef) {
> struct node* newNode = *sourceRef; // the front source node
> assert(newNode != NULL);
> *sourceRef = newNode->next; // Advance the source pointer
> newNode->next = *destRef; // Link the old dest off the new node
> *destRef = newNode; // Move dest to point to the new node
> }
>
> void AlternatingSplit(struct node* source,
> struct node** aRef, struct node** bRef) {
> struct node* a = NULL; // Split the nodes to these 'a' and 'b' lists
> struct node* b = NULL;
> struct node* current = source;
> while (current != NULL) {
> MoveNode(&a, ¤t); // Move a node to 'a'
> if (current != NULL) {
> MoveNode(&b, ¤t); // Move a node to 'b'
> }
> }
> *aRef = a;
> *bRef = b;
> }
>
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
>
> On Sat, Dec 24, 2011 at 11:37 PM, atul anand <[email protected]>wrote:
>
>> because you are doing odd=odd->next; and even=even->next;
>> you will lose head pointers for the two linked list formed once you come
>> out of the loop.
>>
>>
>>
>> void segregate(node* head)
>> {
>> node *temp,*odd,*even;
>> toggle=1;
>> if(head==NULL)
>> {
>> return;
>> }
>> odd=head;
>>
>> if(head->next==NULL)
>> {
>>
>> return;
>> }
>> even=head->next;
>>
>> if(even->next==NULL)
>> {
>> return;
>> }
>> else
>> {
>> temp=even->next;
>> }
>> node *otemp,*etemp;
>> otemp=odd;
>> etemp=even;
>>
>> while(temp!=NULL)
>> {
>>
>> if( toggle == 1)
>> {
>>
>> otemp->next=temp;
>> otemp=otemp->next;
>>
>> temp=temp->next;
>> otemp->next=NULL;
>>
>> toggle=0;
>> }
>> else
>> {
>> etemp->next=temp;
>> etemp=etemp->next;
>>
>> temp=temp->next;
>> etemp->next=NULL;
>>
>> toggle=1;
>>
>> }
>>
>>
>>
>> }
>>
>>
>> }
>>
>> On Sat, Dec 24, 2011 at 10:56 PM, Karthikeyan V.B <[email protected]>wrote:
>>
>>> Segregate even and odd psoitioned nodes in a linked list
>>>
>>> Eg:
>>> 2->3->1->9->7->5
>>> The output should be two separate lists
>>> 2->1->7
>>> 3->9->5
>>>
>>> *My code is:*
>>> void segregate(node* head)
>>> {
>>> int i=1;
>>> node* sec=head->next;
>>> node *odd=head,*even=head->next;
>>> while(even)
>>> {
>>> if(i%2)
>>> {
>>> odd->next=even->next;
>>> even->next=NULL;
>>> odd=odd->next;
>>> if(!odd->next) break;
>>> }
>>> else
>>> {
>>> even->next=odd->next;
>>> odd->next=NULL;
>>> even=even->next;
>>> if(!even->next) break;
>>> }
>>> i++;
>>> }
>>> }
>>>
>>> Pls correct me if i'm wrong or suggest me a better approach
>>>
>>> Regards,
>>> KARTHIKEYAN.V.B
>>> PSGTECH
>>> CBE
>>>
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>>
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>
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