Sorting the Index of Java ArrayList

Here is the solution. Suppose you have an ArrayList named count.

   ArrayList count=new ArrayList();

Let n be the size of the ArrayList.

   int n=count.size();

The index which will be sorted should be an Integer array. Construct the array, and set the default value. By default, the index should indicate the natural order of the array, namely, natural number starting from zero to the size of the array.

   Integer index[]=new Integer[n];
   for(int i=0;i<n;i++)
       index[i]=new Integer(i);

Let's sort the array. The Arrays.sort(Object[],Comparator) method takes the Integer[] index and a Comparator interface as arguments. The Comparator interface defines the comparison method. Here, the interface is declared as an inner interface, defining compare(Object,Object) method where elements of the ArrayList are compared. The int values of n1 and n2 are simply the int values of the Integer's. Get the element value by calling ArrayList.get(int) method, and return the result of the comparison by calling Integer.compareTo(Integer) method.

  Arrays.sort(index,new Comparator()
    {
	public int compare(Object obj1, Object obj2)
	{
	    int n1=((Integer)obj1).intValue();
	    int n2=((Integer)obj2).intValue();
	    return ((Integer)count.get(n1)).compareTo((Integer)count.get(n2));
	}
    });	

Now, the index array is sorted. It is sorted in the ascending order of the ArrayList count. If n1 and n2 values are replaced in compareTo method, the index will be sorted in the descending order of the ArrayList.