[01주차 이은지] 백준1105_팔 #5
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jiwonsudo
approved these changes
Oct 14, 2023
| System.out.println(0); | ||
| } else { | ||
| int count = 0; | ||
| for( int i = 0; i < L.length()&&L.charAt(i)==R.charAt(i); i++){ |
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c 스타일의 for문에선 이렇게 높은 수준의 코드를 구현할 일이 없어서 잘 몰랐는데, 당연하지만 for문의 조건 검사절 (중간)에서도 and 연산이 가능하군요!! 잘 생각하지 못했던 부분인데 다음 구현에 참고하겠습니다 🙉
hepheir
approved these changes
Oct 14, 2023
Comment on lines
+15
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+17
| String[] arr = LR.split(" "); | ||
| String L = arr[0]; | ||
| String R = arr[1]; |
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오호, 숫자로 변환하지 않고 바로 문자열로서 처리하셨군요
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넵 카운트만 하면되기도 하고 문자열로 처리하면 배열 처리도 쉬울 것 같아서 문자열로 했슴다!!
pmsu2007
approved these changes
Oct 15, 2023
| System.out.println(0); | ||
| } else { | ||
| int count = 0; | ||
| for( int i = 0; i < L.length()&&L.charAt(i)==R.charAt(i); i++){ |
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charAt에 대해 잘 몰랐는데 덕분에 하나 배워갑니다
hepheir
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Oct 21, 2023
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🔎 문제 소개
🖊️ 풀이
처음에 Scanner를 사용하였지만 입력값 제한이 2,000,000,000으로 상당히 크다고 생각하여 BufferedReader을 사용하였고 다음 스트링 문자열을 다시 char의 리스트로 바꾸기 위해 CharAt 을 사용하였다
문제를 풀면서 주의해야할 점은 같은자리 수가 같은 숫자이고 그 숫자가 8인 조건을 계속 만족할때까지만 반복문에 적용시켜 카운트를 세야하는데 처음엔 같은 자리수 같은 숫자일때까지만 반복문을 넣어두고 뒤에 8은 조건문에 붙혀서 틀렸다 조건제시 부분을 헷갈리지 말아야할 것 같다!