Shift elements circularly in array by n

Question

I am trying to shift the elements of an array left by n (some random number) using a method. The difficulty is with swapping the elements at the start of the array with the elements at the end. Here is my code so far.

The other array questions asked were different from this one.

public class Shifty {

    private int[] datas = new int[]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13};

    public Shifty() {
        datas = new int[]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13};
    }

    public int shift(int position) {
        int tmp = 0;
        int[] tmp1 = new int[12];
        for (int i = 0; i <= 12; i++) {
            if (i < position) {

                  tmp1[12-position] = this.datas[i];

            } else {
                this.datas[i] = this.datas[i+1];
                            }
            for  (int j = 12-position; j <= 12; j++){
                this.datas[j] = tmp1[j];
            }
        }

        return position;
    }

    public void display() {
        System.out.println(Arrays.toString(this.datas));
        System.out.println();
    }
}

for example see this: http://stackoverflow.com/questions/18659792/circular-arraylist-extending-arraylist and http://stackoverflow.com/questions/7266042/java-ring-buffer — guillaume girod-vitouchkina, Dec 05 '15 at 23:01
Your `tmp1` is too small - should be int[13] at peast but int[datqs.length] would be more robust. Same for literal use of `12` in your code. — Jan, Dec 05 '15 at 23:04
Your solution could be this nice one-liner `Collections.rotate(Arrays.asList(datas), 3);` if only your array would be `Ineger[]` instead of `int[]`. — Pshemo, Dec 05 '15 at 23:17

score 6 · Accepted Answer · answered Dec 05 '15 at 23:13

6

You could make good use of the modulo operation for this.

public void shift(int n) {
  int[] temp = new int[datas.length];
  for (int i = 0; i < temp.length; i++) {
    temp[i] = datas[(i + n) % temp.length];
  }
  datas = temp;
}

answered Dec 05 '15 at 23:13

Will

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score 1 · Answer 2 · answered Dec 06 '15 at 03:05

Since you apparently don't need to do the shift in place and can use a temporary array, the solution is pretty simple:

public class Shifty {

    private int[] datas = new int[]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13};

    public Shifty() {
        datas = new int[]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13};
    }

    public int shift(int position) {
        final int len = datas.length;
        int[] tmp = Arrays.copyOf(datas, position);
        System.arraycopy(datas, position, datas, 0, len - position);
        System.arraycopy(tmp, 0, datas, len - position, position);
        return position;
    }

    public void display() {
        System.out.println(Arrays.toString(this.datas));
        System.out.println();
    }
}

This only works for calls to shift with argument values that are legal subscripts of datas. If you want to handle negative values or values greater than datas.length, you'll need to reduce position to the appropriate range. Note that since this is a circular shift, you can use modular arithmetic to advantage here, since shift(position) and shift(position * n*datas.length) should yield the same result for any integer value n.

Shift elements circularly in array by n

2 Answers2