将int数组从最高到最低排序

Question

So I just learned about the Arrays.sort(arrayName); and applied it in a project of mine but I found it sorts from lowest to highest. Is there anyway to have it do the opposite? (I don't think I need to post the code in question but if it is needed I'll post it).

Answer 1

If you use an Integer[] instead of int[] , then you can pass a Comparator as 2nd argument to the sort method. To impose reverse ordering, you can make use of Collections.reverseOrder() method:

Arrays.sort(arr, Collections.reverseOrder());

Answer 2

If you have an int[] array, you can sort it using Arrays.sort and then reverse it :

int [] tab2 = new int[]{1,5,0,-2};
Arrays.sort(tab2);
ArrayUtils.reverse(tab2);
System.out.print(Arrays.toString(tab2));

Output :

[5, 1, 0, -2]

Code of the reverse method (from org.apache.commons.lang.ArrayUtils.reverse(int[]) ) :

public static void reverse(int[] array) {
        if (array == null) {
            return;
        }
        int i = 0;
        int j = array.length - 1;
        int tmp;
        while (j > i) {
            tmp = array[j];
            array[j] = array[i];
            array[i] = tmp;
            j--;
            i++;
        }
}

Answer 3

For some usecases you may come along by just treating your sorted array as "reverse sorted". Eg to iterate from highest to lowest number you may use

int[] foo = ...;
Arrays.sort(foo);
for (int i=foo.length-1; i>=0; i--) { 
  doSomethingWith(foo[i]);
 }

Answer 4

1. Use an Integer[] instead of int[]
2. Make use of Collections.reverseOrder() : Returns a comparator that imposes the reverse of the natural ordering on a collection of objects that implement the Comparable interface

3. If possible use ArrayList<Integer> and Collections.sort(list, Collections.reverseOrder()) for stronger case.

    Integer[] intArr = new Integer[10]; // add some integer Arrays.sort(intArr, Collections.reverseOrder())