[HodaeSsi] Week 02 #762
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[HodaeSsi] Week 02 #762
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EgonD3V
approved these changes
Dec 21, 2024
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| answerSet = set() | ||
| nums.sort() |
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정렬은 강력한 조건이기 때문에 set을 사용하지 않고 인덱스를 사용해서 중복을 제거하는 방법도 이용해볼 수 있습니다
| leftIdx = leftIdx + 1 | ||
| rightIdx = rightIdx - 1 | ||
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| return list(answerSet) |
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그러면 여기서 set을 list로 돌리지 않아도 될 것 같습니다
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오 안그래도 시간 제한에 아슬아슬한 풀이라 고민 중이었는데,
반영해서 다시 풀어보겠습니다.
리뷰 감사합니다!
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| dp = [] | ||
| dp.append(0) | ||
| dp.append(1) | ||
| dp.append(2) |
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Suggested change
| dp = [] | |
| dp.append(0) | |
| dp.append(1) | |
| dp.append(2) | |
| dp = [0, 1, 2] |
이게 더 깔끔할 것 같아서 제안드립니다
| for i in range(3, n + 1): | ||
| dp.append(dp[i - 1] + dp[i - 2]) | ||
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| return dp[n] |
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Suggested change
| return dp[n] | |
| return dp[-1] |
파이썬에서는 이렇게 하면 배열 마지막 원소라는 점이 명확해서 가독성에 도움이 됩니다
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| continue | ||
| if s[idx] == '0': |
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이렇게 if-elif-else 구문이 많으면 논리적으로 분리되는 지점에 나눠주시면 가독성에 도움이 됩니다
| tLetterDict = {} | ||
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| for letter in s: | ||
| sLetterDict[letter] = sLetterDict.get(letter, 0) + 1 |
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