[kayden] Week 06 Solutions #477
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obzva
reviewed
Sep 20, 2024
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| if len(s) % 2 != 0: | ||
| return False |
obzva
reviewed
Sep 20, 2024
Comment on lines
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| def dfs(x, y, direction): | ||
| answer.append(matrix[x][y]) | ||
| nx = x + dx[direction] | ||
| ny = y + dy[direction] | ||
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| if 0 <= nx < m and 0 <= ny < n and not visited[nx][ny]: | ||
| visited[nx][ny] = True | ||
| return dfs(nx, ny, direction) | ||
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| direction = (direction+1) % 4 | ||
| nx = x + dx[direction] | ||
| ny = y + dy[direction] | ||
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| if 0 <= nx < m and 0 <= ny < n and not visited[nx][ny]: | ||
| visited[nx][ny] = True | ||
| return dfs(nx, ny, direction) | ||
| else: | ||
| return answer |
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혹시 재귀함수를 사용하신 이유가 있을까요? :)
저는 반복문으로 구현하는 것이 공간 사용 측면에서는 더 효율적일 것 같다는 생각을 했습니다
왜냐하면 위처럼 재귀함수를 사용할 경우 마지막 좌표에 도달할 때까지 재귀함수의 호출 스택만큼의 추가적인 공간을 사용하게 될 것 같거든요
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그래프 탐색이라서 DFS, BFS 라고 생각하고 문제가 한쪽 방향으로 탐색하길래 DFS를 사용하는게 직관적으로 편하다고 생각을 했는데 다시 보니 틀에 박힌 생각이였네요. 단순 반복문을 사용해서 하는게 더 효율적이네요!
좋은 답변 감사합니다! 그래프 탐색은 DFS, BFS라고 틀에박힌 생각을 했네요.
obzva
reviewed
Sep 20, 2024
Sunjae95
approved these changes
Sep 21, 2024
| path = [] | ||
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| for num in nums: | ||
| idx = bisect_left(path, num) |
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bisect_left로 정답배열에 들어갈 위치를 지정할 수 있군요..!
리스트에서 num값이 들어갈 위치를 찾는 방법을 적용하는 아이디어는 참신한 아이디어 같아요 👍
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