resolve #137-luck-balance 문제풀이 및 문서추가 #138
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yeonjuan
reviewed
Aug 14, 2019
| return (a[1] > b[1]) ? 1 : -1; | ||
| }); | ||
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| let numImportant = length - [...sorted].findIndex(x => x[1] === 1); |
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findIndex를 쓰려고 사용했습니다
yeonjuan
reviewed
Aug 14, 2019
| const loseImportant = [...sorted].slice(length - numImportant, length - numImportant + limit); | ||
| const winImportant = [...sorted].slice(length - numImportant + limit); | ||
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| const sumLuck = (unImportant.reduce((sum, [luck]) => sum + luck, 0) || 0) + |
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(sum, [luck]) => sum + luck반복되는 부분은 함수로 빼는것이 좋을것 같습니다.
yeonjuan
reviewed
Aug 14, 2019
| return (a[1] > b[1]) ? 1 : -1; | ||
| }); | ||
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| let numImportant = length - [...sorted].findIndex(x => x[1] === 1); |
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numImportant 아래의 unImportant, loseImportant, winImportant 와 이름 형식이 같은데
아랫것들은 배열이고 numImportant 는 number 로 보입니다.
배열의 경우 s 복수형을 쓰는것이 좋을것 같습니다.
yeonjuan
reviewed
Aug 14, 2019
| let numImportant = length - [...sorted].findIndex(x => x[1] === 1); | ||
| numImportant = (numImportant > length) ? 0 : numImportant; | ||
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| const unImportant = [...sorted].slice(0, length - numImportant); |
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length - numImportant 반복되서 사용되는 부분은 상수로 선언하는게 좋을것 같습니다.
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