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挑戦する必要はないのですが、空間計算量 O(1) で書けますか？ |
oda
reviewed
Apr 5, 2024
| if total_chars >= num_chars: | ||
| return num_cols | ||
| # 斜め右上方向に上に上がる | ||
| for _ in range(num_rows - 2, 0, -1): |
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書けそうです。動かしてないですが以下みたいな感じでいけそうな気がします。
max(0, num_rows - 2)は変数に切り出しても良いと思います。いい名前が思いつかないですが、upper_right_lenとかですかね。
total_chars += max(0, num_rows - 2)
num_cols += max(0, num_rows - 2)
if total_chars >= num_chars:
num_cols -= (total_chars - num_chars)
return num_cols
oda
reviewed
Apr 5, 2024
| zigzag = to_zigzag(s, numRows) | ||
| converted = "" | ||
| for zigzag_row in zigzag: | ||
| converted += "".join(zigzag_row) |
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一応、文字列の追記は再構築されます。ただし、CPython ならば高速化される場合があるようです。
liquo-rice
reviewed
Apr 28, 2024
| converted = "" | ||
| for zigzag_row in zigzag: | ||
| converted += "".join(zigzag_row) | ||
| return converted |
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return "".join(c for c in row for row in rows)みたいな感じで書けませんか？（forループの順番は間違っているかもしれないですが…）
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やっぱり反対でした。"".join(char for row in rows for char in row)
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https://leetcode.com/problems/zigzag-conversion