일 | 월 | 화 | 수 | 목 | 금 | 토 |
---|---|---|---|---|---|---|
1 | 2 | 3 | 4 | |||
5 | 6 | 7 | 8 | 9 | 10 | 11 |
12 | 13 | 14 | 15 | 16 | 17 | 18 |
19 | 20 | 21 | 22 | 23 | 24 | 25 |
26 | 27 | 28 | 29 | 30 | 31 |
Tags
- 고차함수
- espresso
- Fragment에서 Activity의 함수 사용하기
- 안드로이드
- 안드로이드스튜디오
- searchview
- fragment
- high order function
- 구분선
- 리사이클러뷰
- LayoutManger
- 안드로이드13
- 코딜리티
- 뷰변경 감지
- Fragment 수동 추가
- adapter
- Error:Execution failed for task ':app:mergeDebugResources'
- Android
- 코틀린
- 테마 아이콘
- ActivityTestRule
- viewholder
- 생명주기
- ui test
- 스와이프
- IntentTestRule
- 안드로이드개발레벨업교과서
- 재사용
- recyclerview
- binding adapter
Archives
- Today
- Total
룬아님의 취중코딩
Codility 28번 MinPerimeterRectangle 본문
MinPerimeterRectangle
An integer N is given, representing the area of some rectangle.
The area of a rectangle whose sides are of length A and B is A * B, and the perimeter is 2 * (A + B).
The goal is to find the minimal perimeter of any rectangle whose area equals N. The sides of this rectangle should be only integers.
For example, given integer N = 30, rectangles of area 30 are:
- (1, 30), with a perimeter of 62,
- (2, 15), with a perimeter of 34,
- (3, 10), with a perimeter of 26,
- (5, 6), with a perimeter of 22.
Write a function:
class Solution { public int solution(int N); }
that, given an integer N, returns the minimal perimeter of any rectangle whose area is exactly equal to N.
For example, given an integer N = 30, the function should return 22, as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [1..1,000,000,000].
import java.util.*;
class Solution {
public int solution(int N) {
int cnt = 0;
int min = Integer.MAX_VALUE;
int sq = (int)Math.sqrt(N);
for(int i=1; i<=sq; i++){
if(N%i == 0){
if((i + (N/i)) * 2 < min){
min =(i + (N/i)) * 2;
}
}
}
return min;
}
}
반응형
'개발 > 알고리즘' 카테고리의 다른 글
Codility 31번 CountSemiprimes (0) | 2019.10.19 |
---|---|
Codility 29번 Peaks (0) | 2019.09.30 |
Codility 26번 MaxDoubleSliceSum (0) | 2019.09.25 |
Codility 27번 CountFactors (0) | 2019.09.23 |
Codility 25번 MaxSliceSum (0) | 2019.09.22 |
Comments