Re: Array sort problem.

Warren Tang <>
Wed, 19 Oct 2011 10:59:30 +0800
On 10/18/2011 6:54 AM, Lew wrote:

Alex Mentis wrote:

Warren Tang wrote:

I have an array:

index value
0 33
1 22
2 44
3 11

Now I'd like to sort it, but I also need to preserve the original
index, like this:

newIndex originalIndex sortedValue
0 3 11
1 1 22
2 0 33
3 2 44

How can this be done conveniently in Java?

I imagine you'd have to go through the array of values and turn it into
an array of objects that contain fields for both the original index and
value. Then sort the new array based on the values.


Here's a rough outline of such a (value,index) type (not compiled, untried):

  public class ValueIndex<T extends Comparable>
         implements Comparable<ValueIndex<T>>
    private final T value;
    private final int index;
    public ValueIndex( T val, int idx )
      if (val == null) {throw new IllegalArgumentException("null value");}
      this.value = val;
      this.index = idx;
      assert this.value != null;
    public T getValue() {assert value != null; return value;}
    public int getIndex() {return index;}
    @Override public int compareTo(ValueIndex<T> other)
      return other == null ? 1 : getValue().compareTo( other.getValue() );

I had thought to swap the "keys" and "values", and the use a TreeMap to
sort it as follows:

     int[] scores = {33, 22, 44, 11};
     TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
     for(int i = 0; i < scores.length; i++) {
       map.put(scores[i], i);
     int j = 0;
     for(Entry<Integer, Integer> e : map.entrySet()) {
       System.out.println(String.format("%10d%10d%10d", j++,
e.getValue(), e.getKey()));

But it won't work if "values" have duplicates. So I have to define a new
class after all. Thank you guys for the answer.

Warren Tang

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