problem:https://leetcode.com/problems/maximum-units-on-a-truck/
I am supposed to sort array of arrays of size 2(eg. [[1,3],[2,2],[3,1]]) in descending order according to 2nd value of the inner element. i.e for 1st element[1,3]according to value 3. , but my code is resulting in error: no suitable method found for sort().Some Help would be appreciated.
here is my code in java
class Solution {
public int maximumUnits(int[][] boxTypes, int truckSize) {
Arrays.sort(boxTypes, new Comparator<int[][]>() {
public int compare(final int[][] entry1, final int[][] entry2) {
if (entry1[0][0] < entry2[0][0])
return 1;
else return -1;
}
}
);
for (int i = 0; i < boxTypes.length; i++)
System.out.println(boxTypes[i]);
return 0;
}
}
As mentioned in comments, you are sorting by inner element, which is int[], so you need Comparator<int[]>.
public class Solution {
public static void main(String[] args) {
int[][] input = new int[][]{new int[]{2, 2}, new int[]{1, 3}, new int[]{3, 1}};
Arrays.sort(input, new Comparator<int[]>() {
#Override
public int compare(int[] o1, int[] o2) {
return Integer.compare(o2[1], o1[1]);
}
});
System.out.println(Arrays.deepToString(input));
}
}
Note return Integer.compare(o2[1], o1[1]);, second parameter is compared to first in order to achieve descending order.
You could also achieve same effect using lambda, to make it shorter and more readable.
public class Solution {
public static void main(String[] args) {
int[][] input = new int[][]{new int[]{2, 2}, new int[]{1, 3}, new int[]{3, 1}};
System.out.println("Initial array - " + Arrays.deepToString(input));
Arrays.sort(input, (o1, o2) -> Integer.compare(o2[1], o1[1]));
System.out.println("Sorted array - " + Arrays.deepToString(input));
}
}
First, you can't use native type in <>, you need to use Integer instead. Then what you need to compare is the inner array Integer[] if I'm not mistaken so your comparator can't work. Here you are just trying to sort 2 arrays of arrays based on the first element of the first array.
Here is what I would do (using stream):
Integer[][] sortedBoxTypes = Arrays.stream(boxTypes).sorted(Comparator.comparing(entry -> entry[1])).toArray(Integer[][]::new);
public static int[] convertListToArray(List<Integer> listResult) {
int[] result = new int[listResult.size()];
int i= 0;
for (int num : listResult) {
result[i++] = num;
}
return result;
}
Is there an efficient way to convert List to array without iterating List explicitly ?
Maybe it is possible by using methods like:
Arrays.copyOf(int [] origin , int newLength );
System.arraycopy(Object src, int srcPos,
Object dest, int destPos,
int length);
I know that there is a solution described here. However, I particularly interested in an efficient way of converting List<Integer> to int[]
Given the need to convert from Integer to int, I don't think you're going to find something more efficient than what you have, if I assume you're talking about runtime efficiency.
You might find converting to Integer[] first and then looping might be more efficient (below), but you might not, too. You'd have to test it in your specific scenario and see.
Here's that example:
int size = listResult.size();
int[] result = new int[size];
Integer[] temp = listResult.toArray(new Integer[size]);
for (int n = 0; n < size; ++n) {
result[n] = temp[n];
}
If efficiency is your primary concern, I think you can use your solution and make it more efficient by using an indexed for loop on the listResult if it is RandomAccess. However this makes the code much less readable, and you'd have to benchmark it for your use cases to see if it is more efficient.
public static int[] convertListToArray(List<Integer> listResult) {
int size = listResult.size();
int[] result = new int[size];
if (listResult instanceof RandomAccess)
{
for (int i = 0; i < size; i++)
{
result[i] = listResult.get(i);
}
}
else
{
int i = 0;
for (int num : listResult) {
result[i++] = num;
}
}
return result;
}
If you use Java 8 and would like to write less code you can use the Streams library.
List<Integer> list = Arrays.asList(1, 2, 3, 4, 5);
int[] array = list.stream().mapToInt(i -> i).toArray();
If you are open to using a third party library, you can Eclipse Collections as follows.
MutableList<Integer> list = Lists.mutable.with(1, 2, 3, 4, 5);
int[] array = list.collectInt(i -> i).toArray();
The following is slightly more code, but is the most efficient solution I could come up with using Eclipse Collections.
MutableList<Integer> list = Lists.mutable.with(1, 2, 3, 4, 5);
int[] array = new int[list.size()];
list.forEachWithIndex((each, index) -> array[index] = each);
If you need to use the java.util.List interface, the ListIterate utility class can be used from Eclipse Collections.
List<Integer> list = Arrays.asList(1, 2, 3, 4, 5);
int[] array = new int[list.size()];
ListIterate.forEachWithIndex(list, (each, index) -> array[index] = each);
The ListIterate utility will use different iteration code for RandomAccess lists and non-RandomAccess lists.
The most efficient thing to do would be to change the List<Integer> to a MutableIntList in Eclipse Collections or another library that has support for primitive collections.
Note: I am a committer for Eclipse Collections.
In Java 8:
int[] anArray = list.stream()
.filter(Objects::nonNull)
.mapToInt(Integer::intValue)
.toArray();
There is efficient way you could do this Java. However, this could open room for someone to create the generic function (depend on demand).
Just like this sample i wrote, I suggest you do the same to the specific knowledge of your program.
// Generic function to convert set to list
public static <T> ArrayList<T> convertSetToList(Set<T> set)
{
// create an empty list
ArrayList<T> list = new ArrayList<>();
// push each element in the set into the list
for (T t : set)
list.add(t);
// return the list
return list;
}
I have a list of integer arrays. I need to find the common elements between those. What I can think of is an extension of what is listed in Common elements in two lists
Example would be
[1,3,5],
[1,6,7,9,3],
[1,3,10,11]
should result in [1,3]
There are no duplicates in the arrays as well.
Is there a straight forward way to do this?
You can transform the lists to sets, and then use Set.retainAll method for intersection between the different sets.
Once you intersect all sets, you are left with the common elements, and you can transform the resulting set back to a list.
You can use Set's intersection method offered by Guava, Here is a little example :
public <T> Set<T> intersection(List<T>... list) {
Set<T> result = Sets.newHashSet(list[0]);
for (List<T> numbers : list) {
result = Sets.intersection(result, Sets.newHashSet(numbers));
}
return result;
}
Hope that could help you
We can use retainAll method of Collections. I initialised my commons arraylist with the first array list and called this for each remaining arraylists.
List<List<Integer>> lists = new ArrayList<List<Integer>>();
lists.add(new ArrayList<Integer>(Arrays.asList(1, 3, 5)));
lists.add(new ArrayList<Integer>(Arrays.asList(1, 6, 7, 9, 3)));
lists.add(new ArrayList<Integer>(Arrays.asList(1, 3, 10, 11)));
List<Integer> commons = new ArrayList<Integer>();
commons.addAll(lists.get(1));
for (ListIterator<List<Integer>> iter = lists.listIterator(1); iter.hasNext(); ) {
commons.retainAll(iter.next());
}
System.out.println(commons);
System.out.println(lists.get(1));
with Java 8
ArrayList retain = list1.stream()
.filter(list2::contains).filter(list3::contains).collect(toList())
If you are looking for a function that returns elements that exist in all lists,
then the straight forward & simple way is building a statistic { < member, occurences > }
The condition here is no duplicates among the same list,
private Set<Integer> getCommonElements(ArrayList<Integer[]> idList)
{
MapList<Integer,Short> stat = new MapList<Integer,Short>();
// Here we count how many times each value occur
for (int i = 0; i < idList.size(); i++)
{
for (int j = 0; j < idList.get(i).size; j++)
{
if (stat.containsKey(idList.get(i)[j]))
{
stat.set(idList.get(i)[j], stat.get(idList.get(i)[j])+1);
}
else
{
stat.add(idList.get(i)[j], 1);
}
}
}
// Here we only keep value that occured in all lists
for (int i = 0; i < stat.size(); i++)
{
if (stat.get(i) < idList.size())
{
stat.remove(i);
i--;
}
}
return stat.keySet();
}
public class ArrayListImpl{
public static void main(String s[]){
ArrayList<Integer> al1=new ArrayList<Integer>();
al1.add(21);al1.add(23);al1.add(25);al1.add(26);
ArrayList<Integer> al2=new ArrayList<Integer>();
al2.add(15);al2.add(16);al2.add(23);al2.add(25);
ArrayList Al3=new ArrayList<Integer>();
al3.addAll(al1);
System.out.println("Al3 Elements :"+al3);
al3.retainAll(al2); //Keeps common elements of (al1 & al2) & removes remaining elements
System.out.println("Common Elements Between Two Array List:"+al3);
}
}
If you are using JAVA 8 streams. Then using stream reduce operation we can achieve the same.
Considering your example: Let's say
a = [1,3,5], b = [1,6,7,9,3] and c = [1,3,10,11]
List<Integer> commonElements = Stream.of(a,b,c)
.reduce((s1,s2) -> {
s1.retainAll(s2);
return s1;
}).orElse(Collections.emptyList());
Keep in mind that after running this operation a will get modified with common values as well. So you will lose the actual value of a.
So elements of a and the result elements of commonElements will be essentially the same after running this operation.
Giving some another alternative code using retainAll capability of Set
public List getCommonItems(List... lists) {
Set<Integer> result = new HashSet<>(lists[0]);
for (List list : lists) {
result.retainAll(new HashSet<>(list));
}
return new ArrayList<>(result);;
}
Usage:
List list1 = [1, 2, 3]
List list2 = [3, 2, 1]
List list3 = [2, 5, 1]
List commonItems = getCommonItems(list1, list2, list3)
System.out.println("Common items: " + result);
Result:
commonItems: [1, 2]
public class commonvalue {
Public static void MyMethod(){
Set<integer> S1 = new set<integer>{1,3,5};
Set<integer> S2 = new set<integer>{1,6,7,9,3};
Set<integer> S3 = new set<integer>{1,3,10,11};
s2.retainall(s1);
s3.retainall(s2);
system.debug(s3);
}
}
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Sort arrays of primitive types in descending order
Java : How to sort an array of floats in reverse order?
How do I reverse an int array in Java?
The following code will sort the array in ascending order :
int a[] = {30,7,9,20};
Arrays.sort(a);
System.out.println(Arrays.toString(a));
I need to sort it in descending order. How do I use Comparator to do this?
Please help.
For primitive array types, you would have to write a reverse sort algorithm:
Alternatively, you can convert your int[] to Integer[] and write a comparator:
public class IntegerComparator implements Comparator<Integer> {
#Override
public int compare(Integer o1, Integer o2) {
return o2.compareTo(o1);
}
}
or use Collections.reverseOrder() since it only works on non-primitive array types.
and finally,
Integer[] a2 = convertPrimitiveArrayToBoxableTypeArray(a1);
Arrays.sort(a2, new IntegerComparator()); // OR
// Arrays.sort(a2, Collections.reverseOrder());
//Unbox the array to primitive type
a1 = convertBoxableTypeArrayToPrimitiveTypeArray(a2);
If it's not a big/long array just mirror it:
for( int i = 0; i < arr.length/2; ++i )
{
temp = arr[i];
arr[i] = arr[arr.length - i - 1];
arr[arr.length - i - 1] = temp;
}
Comparator<Integer> comparator = new Comparator<Integer>() {
#Override
public int compare(Integer o1, Integer o2) {
return o2.compareTo(o1);
}
};
// option 1
Integer[] array = new Integer[] { 1, 24, 4, 4, 345 };
Arrays.sort(array, comparator);
// option 2
int[] array2 = new int[] { 1, 24, 4, 4, 345 };
List<Integer>list = Ints.asList(array2);
Collections.sort(list, comparator);
array2 = Ints.toArray(list);
Guava has a method Ints.asList() for creating a List<Integer> backed by an int[] array. You can use this with Collections.sort to apply the Comparator to the underlying array.
List<Integer> integersList = Ints.asList(arr);
Collections.sort(integersList, Collections.reverseOrder());
Note that the latter is a live list backed by the actual array, so it should be pretty efficient.
Is there any EASY way to sort an array in descending order like how they have a sort in ascending order in the Arrays class?
Or do I have to stop being lazy and do this myself :[
You could use this to sort all kind of Objects
sort(T[] a, Comparator<? super T> c)
Arrays.sort(a, Collections.reverseOrder());
Arrays.sort() cannot be used directly to sort primitive arrays in descending order. If you try to call the Arrays.sort() method by passing reverse Comparator defined by Collections.reverseOrder() , it will throw the error
no suitable method found for sort(int[],comparator)
That will work fine with 'Array of Objects' such as Integer array but will not work with a primitive array such as int array.
The only way to sort a primitive array in descending order is, first sort the array in ascending order and then reverse the array in place. This is also true for two-dimensional primitive arrays.
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!!!)
multiply the Array by -1
sort
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();
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.
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
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;
}
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);
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;
}
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]
I know many answers are here, but still thinks , none of them tried using core java.
And using collection api , you will end up wasting so much memory and reseduals.
here is a try with pure core concepts , and yes this may be better way if you are more concerned about memory footprints.
int[] elements = new int [] {10,999,999,-58,548,145,255,889,1,1,4,5555,0,-1,-52};
//int[] elements = null;
if(elements != null && elements.length >1)
{
int max = 0, index = 0;
for(int i =0;i<elements.length;i++)//find out what is Max
{
if(elements[i] > max)
{
max = elements[i];
index = i;
}
}
elements[index] = elements[0];//Swap the places
elements[0] = max;
for(int i =0;i < elements.length;i++)//loop over element
{
for(int j = i+1;j < elements.length;j++)//loop to compare the elements
{
if(elements[j] > elements[i])
{
max = elements[j];
elements[j] = elements[i];
elements[i] = max;
}
}
}
}//i ended up using three loops and 2 extra variables
System.out.println(Arrays.toString(elements));//if null it will print null
// still love to learn more, please advise if we can do it better.
Love to learn from you too !