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From: Warren Tang <nospam@tangcs.com>
Newsgroups: comp.lang.java.help
Subject: Re: Array sort problem.
Date: Wed, 19 Oct 2011 10:59:30 +0800
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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.
>
> +1
>
> 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.

Regards,
Warren Tang