upload W3 - disk, nw, node, spicy #16
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nohsion
reviewed
Feb 14, 2022
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| while(notK(scoville, K)): | ||
| answer+=1 | ||
| try: | ||
| scoville.append(scoville.pop(0) + (scoville.pop(0)*2)) | ||
| scoville.sort() |
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반복문을 돌면서 pop(0)과 sort()를 매번 실행하는 것 때문에 효율성테스트에 통과되지 못하는거 같아요!
그래서 pop(0)대신 popleft가 가능한 queue를 쓰는 것이 좋은데, 여기서는 항상 가장 작은 값 두 개가 필요하니까 정렬을 시켜줘야 하잖아요? 이렇게 정렬된 상태의 큐를 쓸 때는 파이썬의 heapq를 이용하면 효율성도 잡을 수 있을 겁니닷!
heapq 참고: https://www.daleseo.com/python-heapq/
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| for job, i in zip(jobs[idx:], range(idx,len(jobs))): | ||
| if job[0] > current: # 작업의 요청시간이 현재보다 클 경우 | ||
| break | ||
| if i not in finish: | ||
| vaild.append(job + [i]) |
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zip 사용은 볼때마다 놀랍네요.. 저는 enumerate 보고 신세계였는데 아직 배울게 많네요 ㅎㅎ
| shortest[node] = shortest[now] + 1 | ||
| queue += node | ||
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| answer = shortest.count(max(shortest)) |
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| def get_node_list(n, edge): | ||
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| graph = [[] for i in range(n+1)] |
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반복문에서 인덱스 i가 딱히 쓰이지 않는다면, 언더스코어(_)를 쓰는 게 좋다고 해요! (간지)
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| def recursive(p): | ||
| result = [] | ||
| global network | ||
| global visited |
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경험 상.. bfs나 dfs 함수 만들 때, solution 함수 내에서 선언하면 global 없이도 사용가능한 것 같더라구요? 코테상에서는 이런 방법도 편하게 쓸 수 있을 것 같아서 알려드려요!
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간단 예시
def solution():
network = []
visited = []
def recursive():
# network, visited 사용~
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