Array in Order

Question

im trying to make a method that takes in an array and makes a new array with the values from 0 to the array.length-1 so that the array it returns will corrispond with the values in the original array (if the number at the original element[ n ] is the higest number of all the array its value in the new array would be array.length-1
examples:
{81,-81,10,2,-2,-10} becomes {5,0,4,3,2,1}
{79,-17,2,0,0,-2} becomes {5,0,4,2,3,1}
{80,-80,0,0,0,0} becomes {5,0,1,2,3,4}
i tried something like this:
int[] newOrder = new int[order.length];
for(int i = 0; i < newOrder.length; i++)
newOrder=0;
for(int h = 0; h < order.length;h++)
for(int g = 0; g < order.length;g++)
if(order[h]>order[g])
newOrder[h]++;
cept with that if i send it {80,-80,0,0,0,0} it comes out {5,0,1,1,1,1} which is not what i want, i want each number to be a differnt number between 0 and array.length-1

Answer 1

Try something like this:public int[] order(final int[] array) {
  int[] temp = new int[array.length];
  for(int i = 0; i < array.length; i++)
    temp[i] = array;
java.util.Arrays.sort(temp);
java.util.List list = new java.util.ArrayList();
for(int i = 0; i < array.length; i++)
list.add(new Integer(temp[i]));
temp = new int[list.size()];
for(int i = 0; i < array.length; i++) {
Integer n = new Integer(array[i]);
int index = list.indexOf(n);
temp[i] = index;
list.set(index, null);
return temp;
I didn't comment or test it, so you'll need to see for your self how (and if) it works.
Good luck.