206. Reversed Linked List#8
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oda
reviewed
Dec 27, 2025
| return reverse_helper(head)[0] | ||
| ``` | ||
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| 2-1） 実はtailは不要（上でnode.next is Noneの時にnode, nodeと二つ同じものを返しているところに違和感を感じて気づく？） |
| ``` | ||
| - https://github.com/resumit30minutes/leetcode-arai60-practice/pull/8/files#diff-56cce9af784f5c572f0706e03269e28af28f8aabbface21c32e6199d6400c657 | ||
| - https://github.com/quinn-sasha/leetcode/pull/7#discussion_r1948355100 | ||
| - stackを利用する場合でもNodeのまま次へのつながりを切ってstackに積むやり方もあり |
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一応、この問題の出題意図は、LinkedList でお手玉ができますかということかと思うので、これは出題意図には沿っていないだろうと推測します。
mamo3gr
reviewed
Dec 28, 2025
| def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: | ||
| reversed_head = None | ||
| while head is not None: | ||
| reversed_head = ListNode(head.val, reversed_head) |
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ListNode を作ると結構シンプルに書けるんですね。この場合はもとのノードが残る（新しいノードが増える）ので、そこの良し悪しを意識できると良いと思いました。
nodchip
reviewed
Dec 29, 2025
| return None | ||
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| node = head | ||
| stack = deque() |
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スタックとして使うのであれば list で十分だと思います。一般に、同じ目的に使えるデータ構造が複数ある場合、その中で一番実装が軽いものを選ぶことが多いように思います。
| ```Python3 | ||
| class Solution: | ||
| def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: | ||
| original_head = head |
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head はリンクトリストの先頭を表すノードのため、変数名を変えたほうが良いかもしれません。ただ、今回は head の位置のノードを逆順のリンクトリストに繋ぎ変え、 head が残りの部分の先頭を表すノードになるため、そのままでも十分通じるかもしれません。
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この問題：https://leetcode.com/problems/reverse-linked-list
次の問題：https://leetcode.com/problems/kth-largest-element-in-a-stream