코딩테스트#006 CyclicRotation

A Kind Dev·2023년 2월 1일
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코딩테스트

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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.

(Copyright 2009–2023 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.)


나의 풀이

풀이 1 --> 87% (성능 이슈)

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

        for (int i = 0; i < K; i++) {
            int last = A[A.length - 1];
            for (int j = A.length - 1; j > 0; j--) {
                A[j] = A[j - 1];
            }
            A[0] = last;
        }

        return A;
    }
}

Check1 : 우측으로 한 칸씩 이동

우측값이 먼저 사라지게 되므로 for문을 우측에서부터 돌려야 함.
마지막값을 tmp변수에 저장해야 함.

풀이 2 --> 100%

class Solution {
    public int[] solution(int[] A, int K) {
        
        int[] result = new int[A.length];

        for (int i = 0; i < A.length; i++) {
            result[(i + K) % A.length] = A[i];
        }

        return result;
    }
}
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