Initial indexes of sorted elements of ArrayList

Question

The following code implements sorting of nfit in ascending order.

public static void main(String[] args) {
    ArrayList<Double> nfit = new ArrayList<Double>();

    nfit.add(2.0);
    nfit.add(5.0);
    nfit.add(1.0);
    nfit.add(8.0);
    nfit.add(3.0);

    // Sort individuals in ascending order
    Collections.sort(nfit);

    System.out.print(nfit);

}

The output is:

[1.0, 2.0, 3.0, 5.0, 8.0]

My question is how to get initial indexes of sorted elements? In this example the answer to my question would be the following:

[2, 0, 4, 1, 3]

How can I get these indexes?

Answer 1

If the values are unique, you can use a Map<Double, Integer> to hold the value and initial order. You just need to sort the map keys and get the correspond value of each key.

Answer 2

Copy the ArrayList and sort, then use indexOf.

ArrayList<Double> nfit = new ArrayList<Double>();
nfit.add(2.0);
nfit.add(5.0);
nfit.add(1.0);
nfit.add(8.0);
nfit.add(3.0);
ArrayList<Double> nstore = new ArrayList<Double>(nfit); // may need to be new ArrayList(nfit)
Collections.sort(nfit);
int[] indexes = new int[nfit.size()];
for (int n = 0; n < nfit.size(); n++){
    indexes[n] = nstore.indexOf(nfit.get(n));
}
System.out.println(Arrays.toString(indexes));

If you wanted the indexes in an ArrayList,

Collections.sort(nstore);
for (int n = 0; n < nfit.size(); n++){
    nstore.add(n, nfit.indexOf(nstore.remove(n)));
}
Collections.sort(nfit);

That will result in one sorted ArrayList nfit, and one ArrayList of indexes nstore.

Edited: In the for loop

for (int n = 0; n < nfit.size(); nfit++){
    nstore.add(n, nfit.indexOf(nstore.remove(n)));
}

The loop count must be iterated over n and not on nfit Find the corrected code:

for (int n = 0; n < nfit.size(); n++){
    nstore.add(n, nfit.indexOf(nstore.remove(n)));
}