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k0000k
approved these changes
May 27, 2025
Comment on lines
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| count += diffDistance[i]/mid; | ||
| // 0으로 나누어 떨어진 경우 이미 휴게소가 1개 지어진 경우이므로 -1을 해준다. | ||
| if(diffDistance[i] % mid == 0) { | ||
| count--; | ||
| } |
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엇ㅋㅋㅋ 저는 이거를 안될때까지 뺄셈하도록 했는데 나눗셈이라는 똑똑한 방법이...
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| public static void binary_search(int start, int end) { | ||
| while(start < end) { | ||
| int diff = arr[start] + arr[end]; | ||
| if(min > Math.abs(diff)) { | ||
| min = Math.abs(diff); | ||
| a = arr[start]; | ||
| b = arr[end]; | ||
| } | ||
| //두 용액의 합이 음수인 경우 | ||
| if(diff < 0) { | ||
| start++; | ||
| } | ||
| //두 용액의 합이 양수인 경우 | ||
| else if(diff > 0){ | ||
| end--; | ||
| } | ||
| else { | ||
| break; | ||
| } | ||
| } |
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아 세상에.. 이렇게하면 이분탐색으로 풀리는군요! 이걸 못떠올려서 절댓값 순으로 정렬해놓고 인접한 값들만 전부 비교했습니다... 😢
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| } else { | ||
| isFlag = true; | ||
| break; |
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그냥 여기서 프린트하고 return; 해버리면 플래그를 안써도 깔끔하게 만들수 있습니다 !
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