Dear All,
Thanks for the response and the clear clarification.
Lawrence
On Thu, Feb 17, 2011 at 5:44 AM, Mike Trentman <mike.trent...@gmail.com>wrote:
> I believe this problem is arising because in order to use
> Colelctions.binarySearch() you must first sort your list.
> Try modifying as below and compare the results.
>
> 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);
>
> // Sort the linked list
> Collections.sort(ll);
> index5 = Collections.binarySearch(ll, new Integer(5));
> index8 = Collections.binarySearch(ll, new Integer(8));
> index3 = Collections.binarySearch(ll, new Integer(3));
> 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 now 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
> [3, 4, 5, 8]
> 5 is in index 2
> 8 is in index 3
> 3 is in index 0
> 4 is in index 1
>
>
>
> On Thu, Feb 17, 2011 at 1:17 AM, Michèle Garoche <migat...@gmail.com>wrote:
>
>>
>>
>> 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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>
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