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[4주차] 백제완 #49
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[4주차] 백제완 #49
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Co-authored-by: 이예진 <86579541+yeahdy@users.noreply.github.com>
// 별과 별 사이의 거리 계산 | ||
private static double calDistance(double[] A, double[] B) { | ||
return Math.abs(Math.sqrt(Math.pow(B[0] - A[0], 2) + Math.pow(B[1] - A[1], 2))); | ||
} |
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마지막에 절대값으로 반환한 이유가 있을까요?
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그러게여..🐹 필요없는 연산입니다!
수정했습니다 ㅎ 감사합니다
System.out.println(left); | ||
} | ||
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// 해당 값으로 주유소를 지었을 때 m개 이하로 지을 수 있는지 |
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오호.. 다들 휴게소 개수 짱 깔끔하게 구하셨군요..!!👍
System.out.println(n - len); | ||
} | ||
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// 빠른 입력 함수 |
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저는 왜 빠른 입력 함수만 쓰면 에러가 날까요?,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, 요상합니다,,,,,,,,,,,,,,,,,,,,,,,,,ㅋㅎㅋㅎ,, ㅠㅠ !
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나중에 같이 입력함수 이야기해봐요!
int[][] board = new int[n + 1][m + 1]; | ||
// 웅덩이가 있는 위치 표시 | ||
for (int[] puddle : puddles) | ||
board[puddle[1]][puddle[0]] = -1; |
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저는 웅덩이가 있는 곳을 2차원 boolean 배열을 만들어서 저장해줬는데, 그냥 int형 배열에 모두 다 저장해줘도 됐군요,,,,👍
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최대한 하나의 배열에서 끝내려고 -1로 표기했더니 계산도 간단해지고 괜찮더라구요!
// 현재 위치가 웅덩이가 아니라면 | ||
if (board[i][j] != -1) { | ||
int up = Math.max(0, board[i - 1][j]); // 위쪽에서 오는 경우 | ||
int lf = Math.max(0, board[i][j - 1]); // 왼쪽에서 오는 경우 |
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이렇게 최댓값으로 바로 받아와도 되는군요..!! 👍 👍
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최댓값으로 0과 함께 받아오면 체크해둔 -1을 무시할 수 있어요!
l++; | ||
} | ||
r--; | ||
answer++; |
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따로 else문 필요 없이 한번에 증가도 가능하군요...!!
저는 2명 타는 경우랑 1명 타는 경우 이렇게 나눠서 생각 하다보니 else문으로 나눠줬었는데, 불필요한 구현 없이 깔끔한 코드 배워갑니다!👍
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