[5주차][정렬][Sue]학습내용정리 #57
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Yeonlisa
reviewed
Oct 31, 2021
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| const solution = (arr) => { | ||
| return [...new Set(arr)].sort((a, b) => b - a); | ||
| }; |
ganeodolu
reviewed
Oct 31, 2021
| tempArr.sort((a, b) => { | ||
| //국어점수 desc | ||
| if (b[1] > a[1]) return -1; | ||
| else if (b[1 < a[1]]) return 1; |
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else if (b[1 < a[1]]) return 1;
오타 찾았습니당
| .slice(1, n + 1) | ||
| .map((v) => v.split(" ").map((x) => Number(x) || x)); | ||
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| tempArr.sort((a, b) => { |
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| const failureRates = []; | ||
| let failure; | ||
| let completedPlayersNum = stages.length || 0; |
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stages.length 는 항상 0이상이니 || 0은 없어도 되지 않을까요..?
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| failure = | ||
| completedPlayersNum == 0 | ||
| ? 0 | ||
| : (failure = count[i] / completedPlayersNum); |
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Suggested change
| failure = | |
| completedPlayersNum == 0 | |
| ? 0 | |
| : (failure = count[i] / completedPlayersNum); | |
| const failure = | |
| completedPlayersNum === 0 | |
| ? 0 | |
| : count[i] / completedPlayersNum; |
요렇게 하는건 어떨까요?
| let res = []; | ||
| for (let i = 0; i < count.length; i++) { | ||
| for (let j = count[i]; j >= 0; j--) { | ||
| res = [...res, i]; |
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spread operator도 O(N) 이기 때문에, res가 충분히 커진다면 [...res, i]도 꽤 큰 비용일 수 있습니다 🐶
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그렇네요 배열의 길이가 크면 그만큼 다시 밀려서 비용이 커지니 push 낫겠네요. 혹시 더 괜찮은 방법 있으면 알려주세요 !
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low-level 한 연산을 수행할 때는 그냥 push 하는게 공간복잡도, 시간복잡도 측면에서 제일 나은거 같아요 😅
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[5주차][정렬][Sue] 학습한 내용 정리입니다