206. Reverse Linked List#7
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tom4649
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Mar 29, 2026
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+14
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| while node is not None: | ||
| next_node = node.next | ||
| node.next = new_head | ||
| new_head = node | ||
| node = next_node |
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多重代入を使えば変数名は減らせそうですが、このままでも良いと思います
node.next, new_head, node = new_head, node, node.next
h-masder
reviewed
Mar 29, 2026
| next_node = node.next | ||
| node.next = new_head | ||
| new_head = node | ||
| node = next_node |
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この解法も再帰の解法も元の入力を書き換えていることは意識しておくとよいかもしれません。
h-masder
reviewed
Mar 29, 2026
| reversed_head, reversed_tail = _reverse_list(node_next) | ||
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| reversed_tail.next = node | ||
| return reversed_head, node |
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21行目から26行目は
reversed_head, reversed_tail = _reverse_list(node.next)
node.next = None
reverse_tail.next = node
return reversed_head, nodeのほうがシンプルかなと思いました。
それとreturn文は
return reversed_head, reversed_tail.nextのほうが読みやすいかなと感じます。
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レビューありがとうございます。
21~26行目については、あらかじめ繋がりを切っておけば_reverse_listで帰って来るリストが純粋にnode以降のリストを反転したものになって分かりやすい、と思って実装しました。
ですが、わざわざ変数を増やしてまでやる事ではなかったかもしれませんね。
return文については、提案していただいた物の方が意図が分かりやすいと感じました。
今後参考にさせていただきます。
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今回の問題
https://leetcode.com/problems/reverse-linked-list/
次回の問題
https://leetcode.com/problems/kth-largest-element-in-a-stream/