Codility-CyclicRotation

CyclicRotation:
Rotate an array to the right by a given number of steps.

An array A consisting of N integers is given. Rotation of the array means that each element is shifted right by one index, and the last element of the array is moved to the first place. For example, the rotation of array A = [3, 8, 9, 7, 6] is [6, 3, 8, 9, 7] (elements are shifted right by one index and 6 is moved to the first place).

The goal is to rotate array A K times; that is, each element of A will be shifted to the right K times.

Write a function:

class Solution { public int[] solution(int[] A, int K); }

that, given an array A consisting of N integers and an integer K, returns the array A rotated K times.

For example, given A = [3, 8, 9, 7, 6] K = 3

the function should return [9, 7, 6, 3, 8]. Three rotations were made: [3, 8, 9, 7, 6] -> [6, 3, 8, 9, 7] [6, 3, 8, 9, 7] -> [7, 6, 3, 8, 9] [7, 6, 3, 8, 9] -> [9, 7, 6, 3, 8]

For another example, given A = [0, 0, 0] K = 1

the function should return [0, 0, 0]

Given A = [1, 2, 3, 4] K = 4

the function should return [1, 2, 3, 4]

Assume that:

• N and K are integers within the range [0..100];
• each element of array A is an integer within the range [−1,000..1,000].

In your solution, focus on correctness. The performance of your solution will not be the focus of the assessment.

這題解滿快的，一開始就想到使用陣列複製，因為要轉動的次數也等於數字移位的了幾次，所以原陣列從轉動的次數開始複製到子陣列上，然後子陣列前幾位空著的位置，就是需要放入轉動的數字，然後再依序賦值給那些位置就完成了。

public static int[] cyclicRotation(int[] a, int k) {
        if (a.length == 0 || k == a.length) {
            return a;
        }
        if (k > a.length) {
            k = k % a.length;
        }
        int[] subArr = new int[a.length];
        System.arraycopy(a, 0, subArr, k, (a.length - k));
        for (int i = 0; i < k; i++) {
            subArr[k - i - 1] = a[a.length - i - 1];
        }
        return subArr;
    }