I figured that out, need just resign pointers. |_1_| ----- |_2_| ---del---
|_3_| ------ next {1}
|--------next------------------|
On Nov 16, 2007 8:26 PM, Dave <[EMAIL PROTECTED]> wrote:
>
> If the list contains at least two nodes, simply copy the data from the
> next node into the current one and delete the next node. It is an O(1)
> operation.
>
> Dave
>
> On Nov 11, 3:34 am, "Lukas alkauskas" <[EMAIL PROTECTED]>
> wrote:
> > Hi, I have this *single circular linked list* structure:
> >
> >
> >
> >
> >
> > > public class ListItem {
> > > int n;
> > > ListItem next;
> >
> > > public ListItem() {
> > > this.n = 0;
> > > this.next = null;
> > > }
> >
> > > public ListItem(int n, ListItem e) {
> > > this.n = n;
> > > this.next = e;
> > > }
> >
> > > public int getValue() { return this.n; }
> > > public ListItem getNext() { return this.next; }
> > > public void setValue(int n) { this.n = n; }
> > > public void setNext(ListItem nextItem) { this.next = nextItem;
> }
> > > }
> >
> > public class List {
> >
> >
> >
> >
> >
> > > ListItem head;
> >
> > > public List() {
> > > this.head = null;
> > > }
> >
> > > public ListItem getHead() { return this.head; }
> >
> > > public void Insert(int n) {
> > > if (this.head == null) {
> > > this.head = new ListItem(n, null);
> > > this.head.next = head;
> > > } else if (this.head.getNext() == null) {
> > > this.head = new ListItem(n, this.head);
> > > head.setNext(head);
> > > } else {
> > > this.head.next = new ListItem(n, this.head.next);
> > > }
> >
> > > }
> >
> > > public void Remove(int Key) {
> > > ListItem curr = this.head;
> >
> > > do {
> > > if ( curr.next.getValue() == Key ) {
> > > ListItem temp = curr.getNext();
> > > curr.setNext(temp.getNext());
> > > } curr = curr.getNext();
> >
> > > if (curr.getNext() == this.head && curr.getValue() ==
> Key)
> > > {
> > > this.head.setNext(null);
> > > curr.setNext(null);
> > > }
> >
> > > } while ( curr != this.head );
> > > }
> > > }
> >
> > My question is, how to *optimize Remove(int Key)* method.
> > Maybe anyone have some docs, about SINGLE circular linked list's.
> >
> > Thanks, for any help.
> >
> > --
> > You can contact me by :
> > Google Talk: [EMAIL PROTECTED]
> > Skype: halfas.online.now
> > IRC: HalFas` (irc2.omnitel.net)
> > Home page:http://www.revidev.com/halfas/
> > One's leisure project:http://rvision.gamedev.lt/RVengine- Hide quoted
> text -
> >
> > - Show quoted text -- Hide quoted text -
> >
> > - Show quoted text -
> >
>
--
You can contact me by :
Google Talk: [EMAIL PROTECTED]
Skype: halfas.online.now
IRC: HalFas` (irc2.omnitel.net)
Home page: http://www.revidev.com/halfas/
One's leisure project: http://rvision.gamedev.lt/RVengine
--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com
To unsubscribe from this group, send email to [EMAIL PROTECTED]
For more options, visit this group at http://groups.google.com/group/algogeeks
-~----------~----~----~----~------~----~------~--~---
