repeatedly moving elements in an array by 1 index

Question

I was hoping i could get some help concerning, how to move an element of an array, one index over. so that Array[0] is on Array[1] place and Array[1] is on Array[2] place and so on. The code i have so far does not want to run:

int[] p = {1, 2, 3, 4, 5};

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

Any guidance would be appreciated.

What should happen to the index 4? Should it go to 0? Or should it move to index 5? — Frakcool, Oct 09 '17 at 19:06
@Eliyo Index 4 should never move to index 5. No one can make that happen because that will be `ArrayIndexOutOfBoudsException`. — user3437460, Oct 09 '17 at 20:18

Marcos Vasconcelos · Accepted Answer · 2017-10-10T14:27:56.913

1

The easiest is using a circular counter

int[] p = {1, 2, 3, 4, 5};
int[] r = new int[p.length];
for(int i = 0; i < p.length; i++) {
    r[(i + 1) % p.length] = p[i];
}

Thats ensure that index will be inside the lenght of p.lenght and i+1 will be 0

Also note that I created a array for the result, cause else you would need to keep a reference of the last swaped int at index to set, cause you will replace it value by loop.

edited Oct 10 '17 at 14:27

answered Oct 09 '17 at 19:08

Marcos Vasconcelos

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I think you mean `p.length` in the line inside your for-loop. – Juan Carlos Mendoza Oct 09 '17 at 19:14

score 1 · Answer 2 · answered Oct 09 '17 at 20:13

First, you need to correct the upper bound in your for loop. As you have it, you'll get ArrayIndexOutOfBoundsException, because i + 1 will be over last element.

Second, if you want to move elements from i to i+1, you have to loop backwards. Otherwise, you'll keep overwriting the next element. If you can't see that, try doing in on paper.

Third, you might want to use System.arrayCopy, which is almost certainly faster.

And last, "code does not want to run" is a completely useless information. How do you run it, what exceptions do you get, what did you try?

repeatedly moving elements in an array by 1 index

2 Answers2