[casentino] WEEK 03 solutions #2089
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seungriyou
approved these changes
Nov 30, 2025
| output += 1; | ||
| num = (num - 1) / 2; | ||
| } else { | ||
| num = num / 2; |
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n & n-1로 오른쪽에서부터 1인 비트를 하나씩 지워가며 개수를 세는 방식인 Brian Kernighan's 알고리즘이라는 방법도 있어 공유드립니다~!
| if (lowerStart <= characterCode && lowerEnd >= characterCode) { | ||
| return true; | ||
| } | ||
| if (numericStart <= characterCode && numericEnd >= characterCode) { |
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직관적이고 가독성 좋은 two pointer 풀이인 것 같습니다!
추가로 return true 조건들을 한 번에 or 조건으로 비교한 값 자체를 반환한다면 더 간결하게 작성할 수 있을 것 같아요!
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변수로 가독성 좋게 구현하신 것 같습니다 ㅎㅎ
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|
한 주간 고생하셨습니다! |
yuhyeon99
approved these changes
Nov 30, 2025
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