Lesson 2 - CyClic Rotation

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.

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맨 뒤의 숫자를 미리 뽑아놓고, 배열 안의 숫자를 오른쪽으로 쉬프트 시켜놓으면 쉽게 풀 수 있다.

핵심은 배열의 위치를 바꾸는 것에 얼마나 익숙하냐 라는 문제이기 때문에 (처음에 풀때는 뒤로 뽑을 생각을 전혀 안했다.)

비슷한 유형의 문제에 익숙해지는게 좋을 것 같다.

class Solution {
	public int[] solution(int[] A, int K) {
    	if (A.length == 0) return A;
        
        for (int i=0; i<K; i++) {
        	int temp = A[A.length - 1];
            
            for (int j=A.length-1; j>0; j--) {
            	A[j] = A[j-1];
            }
			A[0] = temp;            
        }
        
    	return A;        
    }