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:
function solution(A, 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.
문제는 심플했다..
- A라는 임의의 배열값을 K만큼 배열에 (끝값 => 앞)으로 이동하면 되는 문제라고 생각을 했다.
function solution(A, K) {
for(let i=0; i<K; i++){
A.unshift(A.pop())
}
return A;
}
왜 이런일이 발생했을까?
function solution(A, K) {
for (let i = 0, k = K % A.length || 0; i < k; i++) {
A.unshift(A.pop())
}
return A
}
위에 로직에서 this.solution( [ ] , 1 ) 이렇게 값을 넣었을경우에는 [undefined]가 발생하였고
그래서 k값에 대해서 나머지값 또는 0으로 초기값을 두어서 처리하여 해결을 하였다..
정말 항상 초기값 또는 빈값에 대해서 항상 생각을 해야하는것 같다....
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