Reversing an Array in Java

Question

If I have an array like this:

1 4 9 16 9 7 4 9 11

What is the best way to reverse the array so that it looks like this:

11 9 4 7 9 16 9 4 1

I have the code below, but I feel it is a little tedious:

public int[] reverse3(int[] nums) {
    return new int[] { nums[8], nums[7], nums[6], nums[5], num[4],
                       nums[3], nums[2], nums[1], nums[0] };
}

Is there a simpler way?

Use a loop to swap each of the elements in the first half with the elements in the second half. — Peter Lawrey, Oct 01 '12 at 18:23
If i used void that means i wont be able to use the return statement in this method right? — PHZE OXIDE, Oct 01 '12 at 18:27
That is true. You can either reserve the exist array which may or may not be returned. Or you can return a copy. — Peter Lawrey, Oct 01 '12 at 18:31
Based on the idea you already had, I would hazard a guess that you are very experienced in development, so I will point out an easy alternative. Just do what you want in a reversed for-loop, instead of actually reversing the array. `for (int i = someArray.length - 1; i > 0; i--) { doStuff(someArray[i]); } reversedArray[j++] = firstArray[i]; }` — Letharion, Oct 01 '12 at 18:41

Vikdor · Answer 1 · 2012-10-07T14:06:32.957

71

Collections.reverse() can do that job for you if you put your numbers in a List of Integers.

List<Integer> list = Arrays.asList(1, 4, 9, 16, 9, 7, 4, 9, 11);
System.out.println(list);
Collections.reverse(list);
System.out.println(list);

Output:

[1, 4, 9, 16, 9, 7, 4, 9, 11]
[11, 9, 4, 7, 9, 16, 9, 4, 1]

edited Oct 07 '12 at 14:06

answered Oct 01 '12 at 18:22

Vikdor

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Can you explain a little more thanks – PHZE OXIDE Oct 01 '12 at 18:24
1

Please see the edit to the answer. – Vikdor Oct 01 '12 at 18:28
1

Why instantiating a new ArrayList? Arrays.asList() already returns a List. – Kirill Rakhman Oct 07 '12 at 13:54
Thanks @cypressious, I was under the impression that Arrays.asList would return an unmodifiable list. Corrected that. – Vikdor Oct 07 '12 at 14:08
Arrays.asList actually returns a Frozen list in the sense that you cannot add or remove any element from that newly created List. However you are allowed to do inplace element modifications. – sunny_dev Jan 02 '20 at 04:48

score 69 · Answer 2 · edited Jan 07 '15 at 08:49

69

If you want to reverse the array in-place:

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

It works since Arrays.asList returns a write-through proxy to the original array.

edited Jan 07 '15 at 08:49

Elazar

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answered Oct 01 '12 at 18:26

Jordan Denison

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Though an old answer it costed me some time to figure out... This does not work for the op as int[] is an object and Arrays.asList(array) will return List and not List (which is impossible anyway) so actually a List with exactly one entry which is reversed the same. This is true for all primitive arrays! – Dennis Ich Aug 26 '15 at 16:19
public int decToBinary(int n) { // array to store binary number int[] binaryNum = new int[1000]; // counter for binary array int i = 0; while (n > 0) { // storing remainder in binary array binaryNum[i] = n % 2; n = n / 2; i++; } return Collections.reverse(Arrays.asList(binaryNum)); } ---------------- Gives an error --> error: incompatible types: void cannot be converted to int return Collections.reverse(Arrays.asList(binaryNum)); – Varun Chandran Jun 12 '19 at 06:21

score 33 · Answer 3 · edited Apr 01 '15 at 05:43

33

If you don't want to use Collections then you can do this:

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

edited Apr 01 '15 at 05:43

karmakaze

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answered Oct 01 '12 at 18:29

kanhai shah

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Isn't it worth to put `array.length / 2` part to outside before the for-loop? It will be recalculated every iteration, I believe. – Jin Kwon Mar 10 '14 at 03:24
@JinKwon I believe that the compiler will enhance it that way, you can try it out with javap – Zarathustra Mar 27 '14 at 18:49
@JinKwon To clarify this, I didn't mean javac, I was talking about the JIT compiler, the optimisation that happens at run time. – Zarathustra Mar 28 '14 at 10:24

Luigi R. Viggiano · Answer 4 · 2013-01-27T14:42:08.893

12

I like to keep the original array and return a copy. This is a generic version:

public static <T> T[] reverse(T[] array) {
    T[] copy = array.clone();
    Collections.reverse(Arrays.asList(copy));
    return copy;
}

without keeping the original array:

public static <T> void reverse(T[] array) {
    Collections.reverse(Arrays.asList(array));
}

edited Jan 27 '13 at 14:42

answered Jan 26 '13 at 02:02

Luigi R. Viggiano

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This was really helpful. I like the fact that you didn't change the original array, instead you returned a new array. – The Room Mar 03 '16 at 04:56
Doesn't work for array of Primitives though – sunny_dev Jan 02 '20 at 04:47

score 12 · Answer 5 · edited Apr 22 '18 at 17:41

12

try this:

public int[] reverse3(int[] nums) {
    int[] reversed = new int[nums.length];
    for (int i=0; i<nums.length; i++) {
        reversed[i] = nums[nums.length - 1 - i];
    }
    return reversed;
}

My input was:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

And the output I got:

12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

edited Apr 22 '18 at 17:41

Mooncrater

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answered Mar 21 '14 at 15:05

The Badak

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score 5 · Answer 6 · answered Mar 21 '14 at 14:37

5

You could use org.apache.commons.lang.ArrayUtils : ArrayUtils.reverse(array)

answered Mar 21 '14 at 14:37

EFalco

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score 3 · Answer 7 · answered Oct 01 '12 at 18:33

Or you could loop through it backeards

int[] firstArray = new int[]{1,2,3,4};
int[] reversedArray = new int[firstArray.length];
int j = 0;
for (int i = firstArray.length -1; i > 0; i--){
    reversedArray[j++] = firstArray[i];
}

(note: I have not compiled this but hopefully it is correct)

score 3 · Answer 8 · answered Nov 08 '14 at 22:17

3

In place reversal with minimum amount of swaps.

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

answered Nov 08 '14 at 22:17

Mathias Bak

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Rodrigo Kossmann · Answer 9 · 2012-10-01T18:35:34.950

2

I would do something like this:

public int[] reverse3(int[] nums) {
  int[] numsReturn = new int[nums.length()]; 
  int count = nums.length()-1;
  for(int num : nums) {
    numsReturn[count] = num;
    count--;
  }
  return numsReturn;
}

edited Oct 01 '12 at 18:35

answered Oct 01 '12 at 18:31

Rodrigo Kossmann

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@RodrigoKossmann.. Or you can just use: - `numsReturn[count--] = num` – Rohit Jain Oct 01 '12 at 18:36

score 2 · Answer 10 · answered Aug 26 '13 at 20:13

2

you messed up

int[] firstArray = new int[]{1,2,3,4};
int[] reversedArray = new int[firstArray.length];
int j = 0;
for (int i = firstArray.length -1; i >= 0; i--){
    reversedArray[j++] = firstArray[i];
}

answered Aug 26 '13 at 20:13

user2719349

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score 2 · Answer 11 · edited Jan 12 '15 at 05:42

2

 public void swap(int[] arr,int a,int b)
 {
    int temp=arr[a];
    arr[a]=arr[b];
    arr[b]=temp;        
}
public int[] reverseArray(int[] arr){
    int size=arr.length-1;

    for(int i=0;i<size;i++){

        swap(arr,i,size--); 

    }

    return arr;
}

edited Jan 12 '15 at 05:42

Dharmesh Porwal

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answered Jan 12 '15 at 04:59

Himanshu Gupta

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score 2 · Answer 12 · answered Jun 30 '15 at 16:17

2

The following will reverse in place the array between indexes i and j (to reverse the whole array call reverse(a, 0, a.length - 1))

    public void reverse(int[] a, int i , int j) {
        int ii =  i;
        int jj = j;

        while (ii < jj) {
            swap(ii, jj);
            ++ii;
            --jj;
        }
    }

answered Jun 30 '15 at 16:17

David Soroko

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I like this answer because it avoids calling the .length method inside the loop and doesn't constantly calculate len - 1 - i; it works for any range of the array and uses swap instead of a temp variable. But there's no reason to use the duplicate variables (ii & jj); you can just use i & j directly. I'd swap(ii++, jj--) and avoid the extra lines. – geowar Jan 22 '17 at 16:41
Note also that it only works if i < j; if you want it to always work you could assign ii = min(i,j) and jj = max(i,j). – geowar Jan 22 '17 at 16:42

score 2 · Answer 13 · answered Aug 05 '15 at 17:29

In case you don't want to use a temporary variable, you can also do like this:

final int len = arr.length;
for (int i=0; i < (len/2); i++) {
    arr[i] += arr[len - 1 - i]; //  a = a+b
    arr[len - 1 - i] = arr[i] - arr[len - 1 - i];   //  b = a-b
    arr[i] -= arr[len - 1 - i]; //  a = a-b
}

score -1 · Answer 14 · edited Dec 04 '14 at 17:43

-1

This code would help:

int [] a={1,2,3,4,5,6,7};
for(int i=a.length-1;i>=0;i--)
  System.out.println(a[i]);

edited Dec 04 '14 at 17:43

1ac0

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answered Dec 04 '14 at 17:24

Jaipal Kumar

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@Begueradj - it looks like an answer to me - not necessarily a good one, but it's an answer. – Krease Dec 04 '14 at 17:56

shayyy7 · Answer 15 · 2015-08-16T11:25:55.033

-3

you can send the original array to a method for example:

after that you create a new array to hold the reversed elements

public static void reverse(int[] a){

int[] reversedArray = new int[a.length];

for(int i = 0 ; i<a.length; i++){
reversedArray[i] = a[a.length -1 -i];


}

edited Aug 16 '15 at 11:25

answered Aug 16 '15 at 11:19

shayyy7

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Reversing an Array in Java

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