2. Add Two Numbers#5
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liquo-rice
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Mar 26, 2026
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| node1 = l1 | ||
| node2 = l2 |
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新しい変数に置く必要はなさそうです。変数名がheadとかですと、動かすと、意味と名前が一致しなくなるので、避ける方が望ましいですが、この場合ですと、そのまま動かした方がいいかなと。
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| - 繰り上げは真偽値がよいと思っていたが、３つ以上の数の和を考えると数値が自然 | ||
| - コードも簡潔になる | ||
| - ループ条件に`carry != 0`を使っている解答もあるが、個人的には抜けた後に処理した方が意図がわかりやすいと思う。 |
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２つのリストの足し算だったら、最大でも一桁しか位は上がりませんが、3個以上のリストの足し算とかだと、話が変わってくるので、whileに含めた方がより一般的なケースに対応できそうですね。同じ処理を複数箇所に書かなくて良いですし。
| - コードも簡潔になる | ||
| - ループ条件に`carry != 0`を使っている解答もあるが、個人的には抜けた後に処理した方が意図がわかりやすいと思う。 | ||
| - `divmod`を使えば、商と余りが同時に得られる。 | ||
| - (自分は知らなかったので)可読性が低いかとも思ったが、関数名と変数名から想像できる範囲だと思ったので採用。 |
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標準ライブラリや組み込み関数を正しく使って可読性が下がることはなさそうです。知らない人が多いのと、可読性はまた別の概念かなと。
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| # 2. Add Two Numbers | |||
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| ## step1 | |||
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in-placeでやる方法もありますね。空間計算量O(1)でやってくださいとか指定されたりするかもしれません。
| carry, digit = divmod(value1 + value2 + carry, 10) | ||
| sum_node.next = ListNode(digit) | ||
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| node1 = node1.next if node1 is not None else None |
h-masder
reviewed
Mar 26, 2026
| value2 = node2.val if node2 is not None else 0 | ||
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| sum_value = value1 + value2 + (1 if has_carry else 0) | ||
| sum_node.next = ListNode(sum_value % 10) |
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has_carryを数値で保持して、
sum_value = value1 + value 2 + has_carry
sum_node.next = ListNode(val=sum_value % 10)とすると、3個以上のリストの足し算にも対応できそうです。
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今回の問題
https://leetcode.com/problems/add-two-numbers/description/
次回の問題
https://leetcode.com/problems/valid-parentheses/description/