how about this one?
Node* reverseBy2(Node* head){
Node* p1 = head;
if(p1 == NULL)
return NULL;
Node* p2 = p1->next;
if(p2 == NULL)
return head;
Node* nextHead = p2->next;
p2->next = p1;
p1->next = reverseBy2(nextHead);
return p2;
}
[?]
2011/6/1 Shivaji Varma <[email protected]>> Hi, > > Please take a look at this link. > > http://mycsinterviewsexperiences.blogspot.com/ > > > -- > Shivaji > > > On Wed, Jun 1, 2011 at 6:26 AM, Anand <[email protected]> wrote: > >> Given a linked list of the form, 1->2->3->4->5->6, convert it into the >> form 2->1->4->3->6->5. Note that the nodes need to be altered and not the >> data contained in them >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
<<1E3.gif>>
