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…Altu-Bitu-HwangJaeEun into 06_greedy_algorithm
kimhj010502
approved these changes
Nov 16, 2023
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| // 청소 상태를 나타내는 상수 | ||
| const int CLEAN = 0; | ||
| const int WALL = 1; | ||
| const int CLEANED = 2; |
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| clean(nx, ny, nextDirection); | ||
| return; |
| } | ||
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| //연산 & 출력 | ||
| cout << findSolution(arr, k) << '\n'; |
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연산부를 별도 메소드로 분리해서 구현해주셨네요! 덕분에 코드가 더 깔끔한 것 같아 좋습니다 👍👍
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| using namespace std; | ||
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| int findSolution(const vector<int>& arr, int k) { |
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벡터를 const로 전달해주셨네요! 이런 사소한 부분이 코드 퀄리티를 높여주는 것 같아요 좋습니다 🥰🥰
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| while (right < n) { | ||
| frequency[arr[right]]++; | ||
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| // 현재 구간의 서로 다른 숫자의 개수가 k 초과라면 | ||
| while (frequency[arr[right]] > k) { | ||
| frequency[arr[left]]--; | ||
| left++; | ||
| } | ||
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| // 현재 구간의 길이가 최대인지 확인하고 right 증가 | ||
| maxLength = max(maxLength, right - left + 1); | ||
| right++; | ||
| } |
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투포인터 탐색의 종료 조건이나 각 포인터의 이동 조건을 모두 잘 파악해서 구현해주신 것 같습니다! 👍
| crt_ans++; | ||
| } | ||
| check[sushi[i]]++; | ||
| sushi[n + i] = sushi[i]; |
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P3. sushi 벡터를 처음부터 n+k 크기로 생성하고 초기화해주셨네요. 이 풀이도 매우 좋습니다! 그런데 인덱스를 모듈러 연산하는 방식에도 한 번 익숙해지시면 좋을 것 같아요. 조금이나마 메모리 절약에 도움이 될 수도 있으니까요 :)
| check[sushi[i + k]]++; | ||
| } | ||
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| cout << ans; |
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P3. 한 메소드에서 연산과 출력 모두를 수행하고 있네요. 메소드는 출력까지 수행하는 것보다 담당하고 있는 핵심 로직에만 집중하는 게 클린 코드 측면에서 조금 더 좋을 것 같습니다!
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| ans = max(crt_ans, ans); | ||
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| check[sushi[i]]--; | ||
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| if (check[sushi[i]] == 0) { | ||
| crt_ans--; | ||
| } | ||
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| if (check[sushi[i + k]] == 0) { | ||
| crt_ans++; | ||
| } | ||
| check[sushi[i + k]]++; |
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