Reversing an Array in Java [duplicate]

Answer 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]);

Answer 2

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]

Answer 3

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;
        }
    }

Answer 4

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.

Answer 5

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)

Answer 6

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;
}