6월 4주차 #42
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6월 4주차 #42
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OHHAKO
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Jun 23, 2020
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| int max=Integer.MIN_VALUE; | ||
| for (int i:dp){ | ||
| if (max<i) max=i; | ||
| } |
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푼제풀이 확인 했습니다:) 고생하셨어요! 프로그램의 첫번째 코드에 max 연산을 넣으면 최대값을 찾는 반복문을 수행하지 않을 수 있어요. 이미 알고 계실거라 생각하지만...ㅎㅎ
추가적으로 이 코드에서
MIN_VALUE는 숫자 클래스의 최소값 상수인데 0이 아닌 최소값 상수를 할당한 이유가 무엇인지 궁금해요!
OHHAKO
reviewed
Jun 23, 2020
Comment on lines
+7
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| 스택 두 개를 이용 | ||
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| - 스택1은 push만, 스택2는 pop과 peek할 때만 사용함 |
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영어문제를 풀어야 하는 일이 많으신가봐요! 저도 이 글 보고 링크 문제 확인해봤는데요.
전 스택1과 배열을 생성해서 풀려고 했네요ㅋㅋ
pop 메서드 호출 시
- 배열 생성
- stack 요소를 배열에 0부터 모두 저장
- 마지막 인덱스의 값 return
이 풀이는 배열해제를 시키지 않으면 가비지 데이터가 많이 생길 것 같고 매 pop마다 배열을 생성하는 번거로움이 있었어요. lee님이 푸신 스택 두개 이용하는게 가장 효율적인 것 같네요!
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