On Feb 16, 11:49 pm, Lawrence Louie <lawrence.lo...@gmail.com> wrote:
> import java.util.Collections;
> import java.util.Iterator;
> import java.util.LinkedList;
> import java.util.List;
>
> public class Question {
>
> /**
> * @param args
> */
> public static void main(String[] args) {
> // TODO Auto-generated method stub
>
> List ll = new LinkedList();
> ll.add(new Integer (5));
> ll.add(new Integer (8));
> ll.add(new Integer (3));
> ll.add(new Integer (4));
>
> int index5 = Collections.binarySearch(ll, new Integer(5));
> int index8 = Collections.binarySearch(ll, new Integer(8));
> int index3 = Collections.binarySearch(ll, new Integer(3));
> int index4 = Collections.binarySearch(ll, new Integer(4));
> System.out.println(ll);
> System.out.println("5 is in index " + index5);
> System.out.println("8 is in index " + index8);
> System.out.println("3 is in index " + index3);
> System.out.println("4 is in index " + index4);
>
> }
>
> }
>
> Output is:
> -------------------------------
> [5, 8, 3, 4]
> 5 is in index 0
> 8 is in index 1
> 3 is in index -1
> 4 is in index -1
>
> Instead of using the LinkedList, I have also tried it with ArrayList,
> same result.
>
> Questions: why is 3 or 4 returns -1 in the index? Thx.
You should sort the collection, that is the LinkedList, before
applying binarySearch on it, because binarySearch operates on a sorted
list, and adding multiple objects to a LinkedList does not preserve
the natural order.
Just add:
Collections.sort(ll);
before the first line of int indexx = Collections.binarySearch(...);
Then you should get the following output:
[3, 4, 5, 8]
5 is in index 2
8 is in index 3
3 is in index 0
4 is in index 1
Michèle Garoche
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