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oda
reviewed
Oct 11, 2025
| row = is_down ? row + 1 : row - 1 | ||
| end | ||
| end | ||
| chars_by_row.inject("") { |result, chars| result << chars.join } |
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コメントありがとうございます。
確かにそっちの方が簡潔で良いですね。いけました。
potrue
reviewed
Oct 11, 2025
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| strings_by_row = Array.new(num_rows) { "" } | ||
| row = 0 | ||
| is_downward = true |
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これをdirectionというような名前の1と-1を切り替える変数にして、rowの更新をrow + directionで済ませてしまうようなコードもよく見ます。(どちらでもよいと思います)
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コメントありがとうございます！
その発想はなかったです。
potrue
reviewed
Oct 11, 2025
| row = is_downward ? row + 1 : row - 1 | ||
| end | ||
| strings_by_row.join | ||
| end |
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0...num_rowsの範囲でループを回し、その行に来るであろうindexを一回で全て計算してしまいstringに入れていく解法もあると思います。
(自分が解いたときは多分違う配列にアクセスを切り替える頻度が少なくなるおかげでこちらの解法の方が少し実行が速かったです)
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ありがとうございます。
それぞれの行に来る文字のインデックスには規則性があるので、それを利用する感じですよね。
tokuhirat
reviewed
Oct 13, 2025
| end | ||
| strings_by_row.join | ||
| end | ||
| ``` |
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解いた問題
6. Zigzag Conversion
使用言語
Ruby
次に解く問題
Kth Largest Element in a Stream