[2주차] 이혜원 #20
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[2주차] 이혜원 #20
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yeongleej
reviewed
Sep 21, 2024
| if(queue.size() ==1 ) return -1; | ||
| queue.add(queue.poll() + queue.poll() *2); | ||
| answer++; | ||
| } |
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while문에서 큐 사이즈에 대한 조건문 제어로 더 깔끔한 풀인 것 같습니다 👍
baexxbin
reviewed
Sep 21, 2024
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| visited = new boolean[n][n]; | ||
| group = new int[n][n]; | ||
| groupCnt = new int[n * n + 1]; |
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배열로 미리 만들어질 수 있는 그룹에 대해 초기화해서 깔끔한 것 같아요!!
baexxbin
reviewed
Sep 21, 2024
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| public static void main(String[] args) { | ||
| Scanner sc = new Scanner(System.in); |
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엇 위의 HW_1647풀이 처럼, 입력받을때 BufferedReader와 StringTokenizer를 사용하면 더 효율적으로 입력받을 수 있을것 같습니다!
Jewan1120
reviewed
Sep 21, 2024
| max = Math.max(max, land[i-1][k]); | ||
| } | ||
| } | ||
| land[i][j] += max; |
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최댓값을 미리 계산해놓고 더하는 방식이 직관적이라 보기 좋은 것 같아요👍🏻
Comment on lines
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| for(int i=0; i<prices.length; i++){ | ||
| for(int j=i+1; j<prices.length; j++){ | ||
| answer[i]++; | ||
| if(prices[i] > prices[j]) break; | ||
| } | ||
| } |
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완전 탐색으로도.. 풀리는 문제였군요...🫥
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