Codility_CyclicRotaion

2-1. CyclicRotaion

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.

받은 배열을 받은 K의 수 만큼 우측으로 이동 시키고 이동된 배열을 리턴해주는 문제

문제풀이

import java.util.*;

class Solution {
    public int[] solution(int[] A, int K) {
        // Implement your solution here
        int AL = A.length;
        if(AL == 0) return A;
        K = K % AL;

        for(int i = 0; i < K; i++){
            int lastNum = A[AL - 1];
            for(int j = AL - 1; j > 0; j--){
                A[j] = A[j - 1];
            }
            A[0] = lastNum;
        }
        return A;
    }
}

처음 제출때는 if(AL == 0) return A; 배열의 길이가 없을 경우의 리턴처리를 안 넣어줘서 100%가 나오지 않았다...

이후 배열에 아무것도 없을 경우 바로 int[] A를 리턴하게 수정 후

수정 후에는 100% 로 변경된 모습

문제 풀어보기 ->
https://app.codility.com/programmers/lessons/2-arrays/cyclic_rotation/