92. Reversed Linked List 2 #46
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oda
reviewed
Apr 4, 2025
| - current, next, prevなどの多用は、操作上の問題で、あまり意図が見えないので好みではない(と言いつつ自分もちょこちょこ使っているが) | ||
| - odaさんの2番目のコードで書いてみた。区間の左端と、その1個まえのnodeがどこにつながっているかというのだけ変わっているが、実質そんな変わらない気がする | ||
| - まあ左端の直前のnodeが、左端のnodeからどんどん奪っていく、という見方をすればちょっと気持ちは違うか | ||
| - odaさんのコード見たらinsert_afterだけreturn NoneのNoneがないけど、なんか意図があるのかな |
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確かにremove_afterはnodeを返しませんね。(ありがとうございます)
fuga-98
reviewed
Apr 4, 2025
| node = node.next | ||
| return node | ||
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| def remove_after(self, node): |
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remove_afterだと後ろのすべてのnodeを消しそうに感じました。
remove_next、pop_nextとかどうでしょうか。
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これ、確かに〜となりました。(ありがとうございます)
fuga-98
reviewed
Apr 4, 2025
| reversed_head = before_reversed.next | ||
| for i in range(right - left): | ||
| removed = self.remove_after(reversed_head) | ||
| self.insert_after(before_reversed, removed) |
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nodeの入れ替えをよく混乱していまうのですが、このように分けると理解しやすいですね
TORUS0818
reviewed
Apr 5, 2025
| ### Step2 | ||
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| - https://discord.com/channels/1084280443945353267/1196498607977799853/1354841302826619021 | ||
| - torusさんの。なるほど、1 passじゃないとそう解くのか |
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逆に私は1-passの方法がパッと浮かびませんでした。
確かnodeの並び順を逆にする、みたいな問題があったのを覚えていて、それと組み合わせて解こう（その問題のフォローアップみたいな位置付け）という感じで考えました。
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問題文: https://leetcode.com/problems/reverse-linked-list-ii/description/
Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list.