82. Remove Duplicates from Sorted List II#5
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oda
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Dec 21, 2025
| - rightが重複要素であることの判定 | ||
| - 直前の要素の値 == 今の要素の値 or （直後の要素がNoneか今の要素の値 == 直後の要素の値） | ||
| - 解いている中で境界条件が甘く、先頭をきちんと考慮できていないことに気づく | ||
| （アクセプトされなかったコード） |
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| - https://hackage.haskell.org/package/base-4.19.0.0/docs/Data-List.html#v:group | ||
| - Pythonのgroup→filter→concatの順に見えるので、なんでconcat. filter. groupなのかが疑問だった | ||
| - おそらくHaskellの書き方をもとにしてそうとわかり納得 |
nodchip
reviewed
Dec 22, 2025
| left.next = None | ||
| return head | ||
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| def isDuplicate(self, prev_node_val: int, node: ListNode) -> bool: |
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is(動詞の原形) は、英語句としてやや不自然に感じました。 isDuplicated はいかがでしょうか？
mamo3gr
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Dec 22, 2025
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| - 重複を全部削除する→重複している要素の直前の要素を基準とする必要がありそう | ||
| - 重複している直前の要素を特定してから、重複している要素の直後の要素を特定できればそれらを繋げることで目的を達成できそう | ||
| - その重複している要素の直前の要素、直後の要素はどう特定するのか？ | ||
| - 二つのポインタ直前の要素left、直後の要素rightを持ってそれを動かしていけば良さそう→rightを主役にループを回す |
mamo3gr
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Dec 22, 2025
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| tail = dummy | ||
| node = head |
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tail は処理済みのノードを繋げていく先だということが、名前に情報をもう少し付け足せると親切に思いました（中盤で初めてそうだと分かる。それまでは node とセットで探索につかうポインタなのかな？とも思いました）
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確かにそう見えますね。ご指摘ありがとうございます。
以下のようにすると探索には使わないよということがはっきりしますかね。
result_dummy = ListNode()
result_tail = result_dummy
node = head
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この問題：https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/description/
次に解く問題：https://leetcode.com/problems/add-two-numbers/description/