Hi
i am here again with another question  
if I have the following array:
static float [] in1 = { 0,0,0,80,0,0,100,0,0,90,0,0,70,60};
first I want  to sort it  so I will have as the following array  {100,90,80,70,60,0,0,0,0,0,0,0,0,0}
Then I want an array of the positions of  the first ordered values in the original array
I mean that ,I must have the following
{7,10,4,13,14,0,0,0,0,0}

How can I
1.sort the array (in deascending order)
2.build an array of the positions of the first 10 ordered value (its position in the original array).



Here are the results you're looking for then (remember that Java gives array positions starting at 0):

Code:

[localhost:~] graham% java Open 100   6 90   9 80   3 70   12 60   13 0   0 0   0 0   0 0   0 0   0 0   0 0   0 0   0 0   0 [localhost:~] graham%

And here's an example of the code to do it;   I've made heavy use of Java Utility classes as I really should be encouraging to use those rather than write your own maths routinse, but you may find a shorter way ...

Code:

import java.util.*; public class Open { public static void main (String [] args) {        int [] in1 = { 0,0,0,80,0,0,100,0,0,90,0,0,70,60}; // Remember Original Positions, giving precedence to earlier occurrence        ArrayList Values = new ArrayList();        HashMap OriginalPosition = new HashMap();        for (int i=in1.length-1; i>=0; i--) {                OriginalPosition.put(new Integer(in1[i]),new Integer(i));                Values.add(new Integer(in1[i]));                } // Sort incoming array        Collections.sort(Values); // List out sorted values in reverse order, with position numbers        for (int i=Values.size()-1; i>=0; i--) {                int val = ((Integer)(Values.get(i))).intValue();                System.out.print("" + val + "   ");                System.out.println(OriginalPosition.get(new Integer(val)));                }        } }

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