Java Array Sort descending?

for a list

Collections.sort(list, Collections.reverseOrder());

for an array

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

You can use this:

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

Collections.reverseOrder() returns a Comparator using the inverse natural order. You can get an inverted version of your own comparator using Collections.reverseOrder(myComparator).

an alternative could be (for numbers!!!)

1. multiply the Array by -1
2. sort
3. multiply once again with -1

Literally spoken:

array = -Arrays.sort(-array)

without explicit comparator:

Collections.sort(list, Collections.reverseOrder());

with explicit comparator:

Collections.sort(list, Collections.reverseOrder(new Comparator()));

It's not directly possible to reverse sort an array of primitives (i.e., int[] arr = {1, 2, 3};) using Arrays.sort() and Collections.reverseOrder() because those methods require reference types (Integer) instead of primitive types (int).

However, we can use Java 8 Stream to first box the array to sort in reverse order:

// an array of ints
int[] arr = {1, 2, 3, 4, 5, 6};

// an array of reverse sorted ints
int[] arrDesc = Arrays.stream(arr).boxed()
    .sorted(Collections.reverseOrder())
    .mapToInt(Integer::intValue)
    .toArray();

System.out.println(Arrays.toString(arrDesc)); // outputs [6, 5, 4, 3, 2, 1]

First you need to sort your array using:

Collections.sort(myArray);

Then you need to reverse the order from ascending to descending using:

Collections.reverse(myArray);

Java 8:

Arrays.sort(list, comparator.reversed());

Update: reversed() reverses the specified comparator. Usually, comparators order ascending, so this changes the order to descending.

For array which contains elements of primitives if there is org.apache.commons.lang(3) at disposal easy way to reverse array (after sorting it) is to use:

ArrayUtils.reverse(array);

When an array is a type of Integer class then you can use below:

Integer[] arr = {7, 10, 4, 3, 20, 15};
Arrays.sort(arr, Collections.reverseOrder());

When an array is a type of int data type then you can use below:

int[] arr = {7, 10, 4, 3, 20, 15};
int[] reverseArr = IntStream.rangeClosed(1, arr.length).map(i -> arr[arr.length-i]).toArray();

For 2D arrays to sort in descending order you can just flip the positions of the parameters

int[][] array= {
    {1, 5},
    {13, 1},
    {12, 100},
    {12, 85} 
};
Arrays.sort(array, (a, b) -> Integer.compare(a[1], b[1])); // for ascending order
Arrays.sort(array, (b, a) -> Integer.compare(a[1], b[1])); // for descending order

Output for descending

12, 100
12, 85
1, 5
13, 1

I don't know what your use case was, however in addition to other answers here another (lazy) option is to still sort in ascending order as you indicate but then iterate in reverse order instead.

For discussions above, here is an easy example to sort the primitive arrays in descending order.

import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        int[] nums = { 5, 4, 1, 2, 9, 7, 3, 8, 6, 0 };
        Arrays.sort(nums);

        // reverse the array, just like dumping the array!
        // swap(1st, 1st-last) <= 1st: 0, 1st-last: nums.length - 1
        // swap(2nd, 2nd-last) <= 2nd: i++,  2nd-last: j--
        // swap(3rd, 3rd-last) <= 3rd: i++,  3rd-last: j--
        //
        for (int i = 0, j = nums.length - 1, tmp; i < j; i++, j--) {
            tmp = nums[i];
            nums[i] = nums[j];
            nums[j] = tmp;
        }

        // dump the array (for Java 4/5/6/7/8/9)
        for (int i = 0; i < nums.length; i++) {
            System.out.println("nums[" + i + "] = " + nums[i]);
        }
    }
}

Output:

nums[0] = 9
nums[1] = 8
nums[2] = 7
nums[3] = 6
nums[4] = 5
nums[5] = 4
nums[6] = 3
nums[7] = 2
nums[8] = 1
nums[9] = 0

Another solution is that if you're making use of the Comparable interface you can switch the output values which you had specified in your compareTo(Object bCompared).

For Example :

public int compareTo(freq arg0) 
{
    int ret=0;
    if(this.magnitude>arg0.magnitude)
        ret= 1;
    else if (this.magnitude==arg0.magnitude)
        ret= 0;
    else if (this.magnitude<arg0.magnitude)
        ret= -1;
    return ret;
}

Where magnitude is an attribute with datatype double in my program. This was sorting my defined class freq in reverse order by it's magnitude. So in order to correct that, you switch the values returned by the < and >. This gives you the following :

public int compareTo(freq arg0) 
{
    int ret=0;
    if(this.magnitude>arg0.magnitude)
        ret= -1;
    else if (this.magnitude==arg0.magnitude)
        ret= 0;
    else if (this.magnitude<arg0.magnitude)
        ret= 1;
    return ret;
}

To make use of this compareTo, we simply call Arrays.sort(mFreq) which will give you the sorted array freq [] mFreq.

The beauty (in my opinion) of this solution is that it can be used to sort user defined classes, and even more than that sort them by a specific attribute. If implementation of a Comparable interface sounds daunting to you, I'd encourage you not to think that way, it actually isn't. This link on how to implement comparable made things much easier for me. Hoping persons can make use of this solution, and that your joy will even be comparable to mine.

Adding my answer in here for a couple of different scenarios For an Array

Arrays.sort(a, Comparator.reverseOrder());

FWIW Lists

Lists.reverse(a);

Any and all Collections

Collections.reverse(a);

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

But Arrays.sort() will not work with primitive objects like int[]. For them it will throw,

error: no suitable method found for sort(int[],Comparator)

Arrays.sort() will work with primitive objects only in increasing order.

Better to convert into a collection and then sort

Collections.sort(Arrays.asList(nums), Collections.reverseOrder())

You could use stream operations (Collections.stream()) with Comparator.reverseOrder().

For example, say you have this collection:

List<String> items = new ArrayList<>();
items.add("item01");
items.add("item02");
items.add("item03");
items.add("item04");
items.add("item04");

To print the items in their "natural" order you could use the sorted() method (or leave it out and get the same result):

items.stream()
     .sorted()
     .forEach(item -> System.out.println(item));

Or to print them in descending (reverse) order, you could use the sorted method that takes a Comparator and reverse the order:

items.stream()
     .sorted(Comparator.reverseOrder())
     .forEach(item -> System.out.println(item));

Note this requires the collection to have implemented Comparable (as do Integer, String, etc.).

There is a lot of mess going on here - people suggest solutions for non-primitive values, try to implement some sorting algos from the ground, give solutions involving additional libraries, showing off some hacky ones etc. The answer to the original question is 50/50. For those who just want to copy/paste:

// our initial int[] array containing primitives
int[] arrOfPrimitives = new int[]{1,2,3,4,5,6};

// we have to convert it into array of Objects, using java's boxing
Integer[] arrOfObjects = new Integer[arrOfPrimitives.length];
for (int i = 0; i < arrOfPrimitives.length; i++) 
    arrOfObjects[i] = new Integer(arrOfPrimitives[i]);

// now when we have an array of Objects we can use that nice built-in method
Arrays.sort(arrOfObjects, Collections.reverseOrder());

arrOfObjects is {6,5,4,3,2,1} now. If you have an array of something other than ints - use the corresponding object instead of Integer.

Simple method to sort an int array descending:

private static int[] descendingArray(int[] array) {
    Arrays.sort(array);
    int[] descArray = new int[array.length];
    for(int i=0; i<array.length; i++) {
        descArray[i] = array[(array.length-1)-i];
    }
    return descArray;
}

NOTE: this is a N log N time complexity but easier to read through and understand how to perform sort in reverse.

credits to Ken for suggestion of this solution.

    // this func sorts in n log n time complexity
    public void sort_reverse(int[] arr) {
        // 1. sort the arr in asc order
        Arrays.sort(arr);
        // 2. now sort all values in descending order
        for (int i = 0, j = arr.length - 1; i < arr.length / 2;i++) {
            int tmp = arr[i];
            arr[i] = arr[j];
            arr[j] = tmp;
            j--;
        }
    }

I know that this is a quite old thread, but here is an updated version for Integers and Java 8:

Arrays.sort(array, (o1, o2) -> o2 - o1);

Note that it is "o1 - o2" for the normal ascending order (or Comparator.comparingInt()).

This also works for any other kinds of Objects. Say:

Arrays.sort(array, (o1, o2) -> o2.getValue() - o1.getValue());

This worked for me:

package doublearraysort;

import java.util.Arrays;
import java.util.Collections;

public class Gpa {


    public static void main(String[] args) {
        // initializing unsorted double array
        Double[] dArr = new Double[] {                 
            new Double(3.2),
            new Double(1.2),
            new Double(4.7),
            new Double(3.3),
            new Double(4.6),
           };
        // print all the elements available in list
        for (double number : dArr) {
            System.out.println("GPA = " + number);
        }

        // sorting the array
        Arrays.sort(dArr, Collections.reverseOrder());

        // print all the elements available in list again
        System.out.println("The sorted GPA Scores are:");
        for (double number : dArr) {
            System.out.println("GPA = " + number);
        }
    }
}

Output:

GPA = 3.2
GPA = 1.2
GPA = 4.7
GPA = 3.3
GPA = 4.6
The sorted GPA Scores are:
GPA = 4.7
GPA = 4.6
GPA = 3.3
GPA = 3.2
GPA = 1.2

public double[] sortArrayAlgorithm(double[] array) { //sort in descending order
    for (int i = 0; i < array.length; i++) {
        for (int j = 0; j < array.length; j++) {
            if (array[i] >= array[j]) {
                double x = array[i];
                array[i] = array[j];
                array[j] = x;
            }
        }
    }
    return array;
}

just use this method to sort an array of type double in descending order, you can use it to sort arrays of any other types(like int, float, and etc) just by changing the "return type", the "argument type" and the variable "x" type to the corresponding type. you can also change ">=" to "<=" in the if condition to make the order ascending.

Another way with Comparator

import java.util.Arrays;
import java.util.Comparator;
...

Integer[] aInt = {6,2,3,4,1,5,7,8,9,10};
Arrays.sort(aInt, Comparator.reverseOrder()  );

It's good sometimes we practice over an example, here is a full one:

sortdesc.java

import java.util.Arrays;
import java.util.Collections;
class sortdesc{
public static void main(String[] args){
       // int Array
       Integer[] intArray=new Integer[]{
                 new Integer(15),
                 new Integer(9),
                 new Integer(16),
                 new Integer(2),
                 new Integer(30)};

       // Sorting int Array in descending order
       Arrays.sort(intArray,Collections.reverseOrder());

       // Displaying elements of int Array
       System.out.println("Int Array Elements in reverse order:");
       for(int i=0;i<intArray.length;i++)
          System.out.println(intArray[i]);

       // String Array
       String[] stringArray=new String[]{"FF","PP","AA","OO","DD"};

       // Sorting String Array in descending order
       Arrays.sort(stringArray,Collections.reverseOrder());

       // Displaying elements of String Array
       System.out.println("String Array Elements in reverse order:");
       for(int i=0;i<stringArray.length;i++)
          System.out.println(stringArray[i]);}}

compiling it...

javac sortdec.java

calling it...

java sortdesc

OUTPUT

Int Array Elements in reverse order:
30
16
15
9
2
String Array Elements in reverse order:
PP
OO
FF
DD
AA

If you want to try an alphanumeric array...

//replace this line:
String[] stringArray=new String[]{"FF","PP","AA","OO","DD"};

//with this:
String[] stringArray=new String[]{"10FF","20AA","50AA"};

you gonna get the OUTPUT as follow:

50AA
20AA
10FF

source

There is a way that might be a little bit longer, but it works fine. This is a method to sort an int array descendingly.

Hope that this will help someone ,,, some day:

public static int[] sortArray (int[] array) {
    int [] sortedArray = new int[array.length];
    for (int i = 0; i < sortedArray.length; i++) {
        sortedArray[i] = array[i];
    }
    
    boolean flag = true;
    int temp;
    while (flag) {
        flag = false;
        for (int i = 0; i < sortedArray.length - 1; i++) {
            if(sortedArray[i] < sortedArray[i+1]) {
                temp = sortedArray[i];
                sortedArray[i] = sortedArray[i+1];
                sortedArray[i+1] = temp;
                flag = true;
            }
        }
    }
    
    return sortedArray;
    
}

I had the below working solution

    public static int[] sortArrayDesc(int[] intArray){
    Arrays.sort(intArray);                      //sort intArray in Asc order
    int[] sortedArray = new int[intArray.length];   //this array will hold the sorted values

    int indexSortedArray = 0;
    for(int i=intArray.length-1 ; i >= 0 ; i--){    //insert to sortedArray in reverse order
        sortedArray[indexSortedArray ++] = intArray [i];
    }
    return sortedArray;
}

This is applicable for primitive types examples for integers. First is to sort the array into descending order then loop through the contents but start the index into the last index of the sorted array.

int[] reversedArr = new int[0];
int[] arrNum = {1, 3, 4, 2, 5};
Arrays.sort(arrNum);
int j=0;
for(int i = (arrNum.length - 1); i >= 0; i-- ) {
    reversedArr[j] = arrNum[i];
    j++;
}
System.out.println(Arrays.toString(reversedArr));

Unfortunatley it's quite complicated to sort in desc. order.

It would be good to have just overloaded method like Arrays.sort(arr, reverseOrder())

For now I see the following option:

int[] sortedArr = Arrays.stream(arr)
                .boxed()
                .sorted(Collections.reverseOrder())
                .mapToInt(Integer::intValue)
                .toArray();

Here is how I sorted a primitive type int array.

int[] intArr = new int[] {9,4,1,7};
Arrays.sort(nums);
Collections.reverse(Arrays.asList(nums));

Result:

[1, 4, 7, 9]