[forest000014] Week 2 #719
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TonyKim9401
reviewed
Dec 21, 2024
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+14
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+28
| if (n == 1) { | ||
| return 1; | ||
| } else if (n == 2) { | ||
| return 2; | ||
| } else { | ||
| int prev2 = 1; | ||
| int prev1 = 2; | ||
| int cur = 0; | ||
| for (int i = 3; i <= n; i++) { | ||
| cur = prev2 + prev1; | ||
| prev2 = prev1; | ||
| prev1 = cur; | ||
| } | ||
| return cur; | ||
| } |
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로직은 좋다고 생각합니다! 개인적으로는 else if 문을 줄이는 방향으로 하면 가독성 면에서 좀 더 올라가지 않을까 싶어요!
TonyKim9401
reviewed
Dec 21, 2024
Comment on lines
+18
to
+23
| for (int i = 0; i < s.length(); i++) { | ||
| cnt[s.charAt(i) - 'a']++; | ||
| } | ||
| for (int i = 0; i < t.length(); i++) { | ||
| cnt[t.charAt(i) - 'a']--; | ||
| } |
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위에 If문에서 이미 s와 t의 길이를 비교한 이후라면 둘의 길이가 같을테니, 하나의 for문으로 합칠 수 있지 않을까요? :)
TonyKim9401
approved these changes
Dec 21, 2024
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