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Merge pull request #141 from AlgoLeadMe/6-ljedd2-02
6-ljedd2
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Original file line number | Diff line number | Diff line change |
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# 첫 번째 코드 | ||
def solution(k, dungeons): | ||
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def DFS(k, cnt): | ||
nonlocal answer | ||
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answer = max(answer, cnt) | ||
for i in range(len(dungeons)): | ||
# 던전 = ["최소 필요 피로도", "소모 피로도"] | ||
min_fatigue, use_fatigue = dungeons[i][0], dungeons[i][1] | ||
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# 현재 피로도가 해당 던전을 방문하기 위한 최소 피로도보다 클 때 | ||
if not visited[i] and k >= min_fatigue: | ||
visited[i] = True | ||
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# 백트래킹 : 이전 노드로 다시 back할 때, | ||
# 해당 노드를 방문하기 전의 피로도로 다시 복구 | ||
DFS(k-use_fatigue, cnt+1) | ||
visited[i] = False | ||
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answer = 0 | ||
visited = [False] * len(dungeons) # visited | ||
DFS(k, 0) | ||
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return answer | ||
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# 개선된 코드 | ||
def solution(k, dungeons): | ||
def dfs(k, cnt): | ||
# 함수 내부에서 값을 갱신할 수 있는 비전역 변수로 선언 | ||
nonlocal answer | ||
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answer = max(answer, cnt) | ||
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# enumerate()를 활용하여 각 던전의 인덱스와 값을 동시에 반복 | ||
for i, (min_fatigue, use_fatigue) in enumerate(dungeons): | ||
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if not visited[i] and k >= min_fatigue: | ||
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visited[i] = True | ||
dfs(k - use_fatigue, cnt + 1) | ||
visited[i] = False | ||
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answer = 0 | ||
visited = [False] * len(dungeons) | ||
dfs(k, 0) | ||
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return answer |