[Stack] bj 17608 막대기 / bj 1935 후위표기식2 #2
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loveitall
requested review from
skylartosf,
wkdyujin and
IslandofDream
and removed request for
a team
IslandofDream
approved these changes
Feb 10, 2023
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| op2 = stack.pop() | ||
| op1 = stack.pop() |
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| for i in postfix : | ||
| if 'A' <= i <= 'Z' : | ||
| # 후위표기식의 값이 알파벳이면, 스택에 값을 넣어준다. str -> int 형태로 | ||
| stack.append(num[ord(i) - ord('A')]) |
wkdyujin
approved these changes
Feb 10, 2023
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| for i in range(1, n) : | ||
| last = stack.pop() | ||
| if last > before : | ||
| count += 1 | ||
| before = last |
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| if 'A' <= i <= 'Z' : | ||
| # 후위표기식의 값이 알파벳이면, 스택에 값을 넣어준다. str -> int 형태로 |
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알파벳 여부를 기준으로 판별했구나 이렇게 하면 바로 계산이 가능하네 코드 너무 깔끔해 좋다...
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📒 문제번호 : bj 17608 막대기, bj 1935 후위표기식2
[bj17608] 막대기
[bj1935] 후위표기식2
☑️ PR Point
어려웠습니다... (후위표기식 ㅎㅎㅎ)
둘 다 stack을 사용해서 풀었고, 1935번의 경우 소수점 2자리 출력 관련해서 새롭게 배울 수 있었습니다! op1, op2 사칙연산 계산이 은근 헷갈려서 삽질했습니다 ㅎㅎ stack배열에 알파벳 값을 int 형태로 넣는 것도 생각하는데 시간이 오래걸렸습니다..이런 문제 많이 풀어봐야할 거 같아요!
⏱️시간복잡도
17608 : O(N)
1935 : O(N)