How do I reverse an int array in Java?

Question

I am trying to reverse an int array in Java.

This method does not reverse the array.

for(int i = 0; i < validData.length; i++)
{
    int temp = validData[i];
    validData[i] = validData[validData.length - i - 1];
    validData[validData.length - i - 1] = temp;
}

What is wrong with it?

Answer 1

For Java 8+, you can use the following to reverse an int[]:

int[] validData = {1, 2, 3};
validData = Arrays.stream(validData)
        .boxed()
        .sorted(Collections.reverseOrder())
        .mapToInt(Integer::intValue)
        .toArray();

After this, validData will contain {3, 2, 1}.

Answer 2

Use a stream to reverse

There are already a lot of answers here, mostly focused on modifying the array in-place. But for the sake of completeness, here is another approach using Java streams to preserve the original array and create a new reversed array:

int[] a = {8, 6, 7, 5, 3, 0, 9};
int[] b = IntStream.rangeClosed(1, a.length).map(i -> a[a.length-i]).toArray();

Answer 3

Guava

Using the Google Guava library:

Collections.reverse(Ints.asList(array));

Answer 4

Simple way to do this:

    for(int i=queue.length-1;i>=0;i--){
                System.out.print(queue[i] + "  ");
    }

Answer 5

A implementation using generics for arrays of non primitive types.

    //Reverse and get new Array -preferred
    public static final <T> T[] reverse(final T[] array) {
        final int len = array.length;
        final T[] reverse = (T[]) Array.newInstance(array.getClass().getComponentType(), len);
        for (int i = 0; i < len; i++) {
            reverse[i] = array[len-(i+1)];
        }
        return reverse;
    }
    
    //Reverse existing array - don't have to return it
    public static final <T> T[] reverseExisting(final T[] array) {
        final int len = array.length;
        for (int i = 0; i < len/2; i++) {
            final T temp = array[i];
            array[i] = array[len-(i+1)];
            array[len-(i+1)] = temp;
        }
        return array;
    }

Answer 6

This has 2 solution

1. Loop

2. Recursion

   public class _1_ReverseArray {

    public static void main(String[] args) {
        int array[] = {2, 3, 1, 4, 9};
        //reverseArray(array, 0, array.length - 1);
        reverseArrayWhileLoop(array, 0, array.length - 1);
        printArray(array);
    }
   
    private static void printArray(int[] array) {
        for (int a : array) {
            System.out.println(a);
        }
    }
   
    private static void reverseArray(int[] array, int start, int end) {
        if (start > end) {
            return;
        } else {
            int temp;
            temp = array[start];
            array[start] = array[end];
            array[end] = temp;
            reverseArray(array, start + 1, end - 1);
        }
    }
   
    private static void reverseArrayWhileLoop(int[] array, int start, int end) {
        while (start < end) {
            int temp;
            temp = array[start];
            array[start] = array[end];
            array[end] = temp;
            start++;
            end--;
        }
    }
   

   }

Answer 7

Just for the sake of it. People often do only need a 'view' on an array or list in reversed order instead of a completely do not need a reversed array when working with streams and collections but a 'reversed' view on the original array/collection. , it is best to create a toolkit that has a reverse view on a list / array.

So create your Iterator implementation that takes an array or list and provide the input.

/// Reverse Iterator
public class ReverseIterator<T> implements Iterator<T> {
  private int index;
  private final List<T> list;
  public ReverseIterator(List<T> list) {
     this.list = list;
     this.index = list.size() - 1;
  }
  public boolean hasNext() {
    return index >= 0 ? true : false;
  }
  public T next() {
    if(index >= 0) 
      return list.get(index--);
    else 
      throw new NoSuchElementException();
  }
}

An implementation for the array situation is quite similar. Of cause an iterator can be source for a stream or a collection as well.

So not always it is best to create a new array just to provide a reverse view when all you want to do is iterating over the array / list or feed it into a stream or a new collection / array.

Answer 8

Here is a condensed version:

My solution creates a new array reversed With each iteration of i the for loop inserts the last index [array.length - 1] into the current index [i] Then continues the same process by subtracting the current iteration array[(array.length - 1) - i] from the last index and inserting the element into the next index of the reverse array!

private static void reverse(int[] array) {
    int[] reversed = new int[array.length];

    for (int i = 0; i < array.length; i++) {
        reversed[i] = array[(array.length - 1) - i];
    }
    System.out.println(Arrays.toString(reversed));
}

Answer 9

A short way to reverse without additional libraries, imports, or static references.

int[] a = {1,2,3,4,5,6,7,23,9}, b; //compound declaration
var j = a.length;
b = new int[j];
for (var i : a)
    b[--j] = i; //--j so you don't have to subtract 1 from j. Otherwise you would get ArrayIndexOutOfBoundsException;
System.out.println(Arrays.toString(b));

Of course if you actually need a to be the reversed array just use

a = b; //after the loop

Answer 10

a piece of cake. i did it for string but, it's not much different

Answer 11

In case of Java 8 we can also use IntStream to reverse the array of integers as:

int[] sample = new int[]{1,2,3,4,5};
int size = sample.length;
int[] reverseSample = IntStream.range(0,size).map(i -> sample[size-i-1])
                      .toArray(); //Output: [5, 4, 3, 2, 1]

Answer 12

There are some great answers above, but this is how I did it:

public static int[] test(int[] arr) {

    int[] output = arr.clone();
    for (int i = arr.length - 1; i > -1; i--) {
        output[i] = arr[arr.length - i - 1];
    }
    return output;
}

Answer 13

    public static void main(String args[])    {
        int [] arr = {10, 20, 30, 40, 50}; 
        reverse(arr, arr.length);
    }

    private static void reverse(int[] arr,    int length)    {

        for(int i=length;i>0;i--)    { 
            System.out.println(arr[i-1]); 
        }
    }

Answer 14

This is how I would personally solve it. The reason behind creating the parametrized method is to allow any array to be sorted... not just your integers.

I hope you glean something from it.

@Test
public void reverseTest(){
   Integer[] ints = { 1, 2, 3, 4 };
   Integer[] reversedInts = reverse(ints);

   assert ints[0].equals(reversedInts[3]);
   assert ints[1].equals(reversedInts[2]);
   assert ints[2].equals(reversedInts[1]);
   assert ints[3].equals(reversedInts[0]);

   reverseInPlace(reversedInts);
   assert ints[0].equals(reversedInts[0]);
}

@SuppressWarnings("unchecked")
private static <T> T[] reverse(T[] array) {
    if (array == null) {
        return (T[]) new ArrayList<T>().toArray();
    }
    List<T> copyOfArray = Arrays.asList(Arrays.copyOf(array, array.length));
    Collections.reverse(copyOfArray);
    return copyOfArray.toArray(array);
}

private static <T> T[] reverseInPlace(T[] array) {
    if(array == null) {
        // didn't want two unchecked suppressions
        return reverse(array);
    }

    Collections.reverse(Arrays.asList(array));
    return array;
}

Answer 15

 public static int[] reverse(int[] array) {

    int j = array.length-1;
    // swap the values at the left and right indices //////
        for(int i=0; i<=j; i++)
        {
             int temp = array[i];
                array[i] = array[j];
                array[j] = temp;
           j--;
        }

         return array;
    }

      public static void main(String []args){
        int[] data = {1,2,3,4,5,6,7,8,9};
        reverse(data);

    }

Answer 16

2 ways to reverse an Array .

1. Using For loop and swap the elements till the mid point with time complexity of O(n/2).

   private static void reverseArray() {
   int[] array = new int[] { 1, 2, 3, 4, 5, 6 };
   
   for (int i = 0; i < array.length / 2; i++) {
       int temp = array[i];
       int index = array.length - i - 1;
       array[i] = array[index];
       array[index] = temp;
   }
   System.out.println(Arrays.toString(array));
   

   }

2. Using built in function (Collections.reverse())

   private static void reverseArrayUsingBuiltInFun() {
   int[] array = new int[] { 1, 2, 3, 4, 5, 6 };
   
   Collections.reverse(Ints.asList(array));
   System.out.println(Arrays.toString(array));
   

   }

   Output : [6, 5, 4, 3, 2, 1]

Answer 17

static int[] reverseArray(int[] a) {
     int ret[] = new int[a.length];
     for(int i=0, j=a.length-1; i<a.length && j>=0; i++, j--)
         ret[i] = a[j];
     return ret;
}

Answer 18

There are two ways to have a solution for the problem:

1. Reverse an array in space.

Step 1. Swap the elements at the start and the end index.

Step 2. Increment the start index decrement the end index.

Step 3. Iterate Step 1 and Step 2 till start index < end index

For this, the time complexity will be O(n) and the space complexity will be O(1)

Sample code for reversing an array in space is like:

public static int[] reverseAnArrayInSpace(int[] array) {
    int startIndex = 0;
    int endIndex = array.length - 1;
    while(startIndex < endIndex) {
        int temp = array[endIndex];
        array[endIndex] = array[startIndex];
        array[startIndex] = temp;
        startIndex++;
        endIndex--;
    }
    return array;
}

2. Reverse an array using an auxiliary array.

Step 1. Create a new array of size equal to the given array.

Step 2. Insert elements to the new array starting from the start index, from the given array starting from end index.

For this, the time complexity will be O(n) and the space complexity will be O(n)

Sample code for reversing an array with auxiliary array is like:

public static int[] reverseAnArrayWithAuxiliaryArray(int[] array) {
    int[] reversedArray = new int[array.length];
    for(int index = 0; index < array.length; index++) {
        reversedArray[index] = array[array.length - index -1]; 
    }
    return reversedArray;
}

Also, we can use the Collections API from Java to do this.

The Collections API internally uses the same reverse in space approach.

Sample code for using the Collections API is like:

public static Integer[] reverseAnArrayWithCollections(Integer[] array) {
    List<Integer> arrayList = Arrays.asList(array);
    Collections.reverse(arrayList);
    return arrayList.toArray(array);
}

Answer 19

   import java.util.Scanner;
class ReverseArray 
{
    public static void main(String[] args) 
    {
        int[] arra = new int[10];
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter Array Elements : ");
        for(int i = 0 ; i <arra.length;i++)
        {
            arra[i] = sc.nextInt();
        }

        System.out.println("Printing  Array : ");
        for(int i = 0; i <arra.length;i++)
        {
            System.out.print(arra[i] + " ");
        }

        System.out.println();
        System.out.println("Printing  Reverse Array : ");
        for(int i = arra.length-1; i >=0;i--)
        {
            System.out.print(arra[i] + " ");
        }
    }
}

Answer 20

Here is what I've come up with:

// solution 1 - boiler plated 
Integer[] original = {100, 200, 300, 400};
Integer[] reverse = new Integer[original.length];

int lastIdx = original.length -1;
int startIdx = 0;

for (int endIdx = lastIdx; endIdx >= 0; endIdx--, startIdx++)
   reverse[startIdx] = original[endIdx];

System.out.printf("reverse form: %s", Arrays.toString(reverse));

// solution 2 - abstracted 
// convert to list then use Collections static reverse()
List<Integer> l = Arrays.asList(original);
Collections.reverse(l);
System.out.printf("reverse form: %s", l);

Answer 21

Solution with o(n) time complexity and o(1) space complexity.

void reverse(int[] array) {
    int start = 0;
    int end = array.length - 1;
    while (start < end) {
        int temp = array[start];
        array[start] = array[end];
        array[end] = temp;
        start++;
        end--;
    }
}

Answer 22

public void getDSCSort(int[] data){
        for (int left = 0, right = data.length - 1; left < right; left++, right--){
            // swap the values at the left and right indices
            int temp = data[left];
            data[left]  = data[right];
            data[right] = temp;
        }
    }

Answer 23

int[] arrTwo = {5, 8, 18, 6, 20, 50, 6};

    for (int i = arrTwo.length-1; i > 0; i--)
    {
        System.out.print(arrTwo[i] + " ");
    }

Answer 24

As I intended to keep my original Array as it was, I solved this problem in the following manner:

List<Integer> normalArray= new ArrayList<>();
List<Integer> reversedArray = new ArrayList<>();

// Fill up array here

for (int i = 1; i <= normalArray.size(); i++) {
  reversedArray .add(normalArray.get(normalArray.size()-i));
}

So basically loop through the initial array and add all the values in reversed order to the new (reversed) array. The type of the list can be anything. I work my way through this code multiple times, this causes some of the other solutions not to work.

Answer 25

Simple for loop!

for (int start = 0, end = array.length - 1; start <= end; start++, end--) {
    int aux = array[start];
    array[start]=array[end];
    array[end]=aux;
}

Answer 26

public class ArrayHandle {
    public static Object[] reverse(Object[] arr) {
        List<Object> list = Arrays.asList(arr);
        Collections.reverse(list);
        return list.toArray();
    }
}

Answer 27

To reverse an int array, you swap items up until you reach the midpoint, like this:

for(int i = 0; i < validData.length / 2; i++)
{
    int temp = validData[i];
    validData[i] = validData[validData.length - i - 1];
    validData[validData.length - i - 1] = temp;
}

The way you are doing it, you swap each element twice, so the result is the same as the initial list.

Answer 28

Another way to reverse array

public static int []reversing(int[] array){
    int arraysize = array.length;
    int[] reverse = new int [arraysize+1];
    for(int i=1; i <= arraysize ; i++){
        int dec= arraysize -i;
        reverse[i] = array[dec];
    }
    return reverse;
}

Answer 29

Collections.reverse(Arrays.asList(yourArray));

java.util.Collections.reverse() can reverse java.util.Lists and java.util.Arrays.asList() returns a list that wraps the the specific array you pass to it, therefore yourArray is reversed after the invocation of Collections.reverse().

The cost is just the creation of one List-object and no additional libraries are required.

A similar solution has been presented in the answer of Tarik and their commentors, but I think this answer would be more concise and more easily parsable.

Answer 30

Here is a simple implementation, to reverse array of any type, plus full/partial support.

import java.util.logging.Logger;

public final class ArrayReverser {
 private static final Logger LOGGER = Logger.getLogger(ArrayReverser.class.getName());

 private ArrayReverser () {

 }

 public static <T> void reverse(T[] seed) {
    reverse(seed, 0, seed.length);
 }

 public static <T> void reverse(T[] seed, int startIndexInclusive, int endIndexExclusive) {
    if (seed == null || seed.length == 0) {
        LOGGER.warning("Nothing to rotate");
    }
    int start = startIndexInclusive < 0 ? 0 : startIndexInclusive;
    int end = Math.min(seed.length, endIndexExclusive) - 1;
    while (start < end) {
        swap(seed, start, end);
        start++;
        end--;
    }
}

 private static <T> void swap(T[] seed, int start, int end) {
    T temp =  seed[start];
    seed[start] = seed[end];
    seed[end] = temp;
 }  

}

Here is the corresponding Unit Test

import static org.hamcrest.CoreMatchers.is;
import static org.junit.Assert.assertThat;

import org.junit.Before;
import org.junit.Test;

public class ArrayReverserTest {
private Integer[] seed;

@Before
public void doBeforeEachTestCase() {
    this.seed = new Integer[]{1,2,3,4,5,6,7,8};
}

@Test
public void wholeArrayReverse() {
    ArrayReverser.<Integer>reverse(seed);
    assertThat(seed[0], is(8));
}

 @Test
 public void partialArrayReverse() {
    ArrayReverser.<Integer>reverse(seed, 1, 5);
    assertThat(seed[1], is(5));
 }
}