6. Zigzag Conversion#60
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yakataN
reviewed
Jul 31, 2025
| row_index = 0 | ||
| for i, c in enumerate(s): | ||
| zigzag[row_index].append(c) | ||
| if i != 0 and row_index == 0 or row_index == numRows - 1: |
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i==0かどうかと端にたどり着いたかの判定は分けたほうが読みやすく思います。
if i == 0:
continue
if row_index == 0 or row_index == numRows - 1:
going_down = not going_down
nodchip
reviewed
Jul 31, 2025
| * これを理解した段階で解答のコードなとは見ず、再度考えた。 | ||
| * トータル 25:00 弱くらいで `Solution1` 書いた。 | ||
| * 最初、`numRows <= 1` のケースを考えていなくて WA | ||
| * また2次元リストに対する list comprehension を久しぶりに書いたので |
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list comprehension をネストすると読みにくいため、避けたほうが良いと思います。
fuga-98
reviewed
Jul 31, 2025
| row_index += 1 | ||
| else: | ||
| row_index -= 1 | ||
| return ''.join(c for row in zigzag for c in row) |
skypenguins
reviewed
Aug 2, 2025
| while row_index < numRows - 1: | ||
| yield row_index | ||
| row_index += 1 | ||
| while row_index > 0: |
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好みの範囲かもしれませんが、0 < row_index も直観的で一考の余地があると思います。
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6. Zigzag Conversion
https://leetcode.com/problems/zigzag-conversion/