142. Linked List Cycle II.md#2
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特に突っ込みたい点はなかったです、良いと思います |
oda
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Jan 6, 2025
| visited = set() | ||
| node = head | ||
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| while node is not None and node.next is not None: |
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あ、たしかにそうですね。何も考えずに書いていました。ありがとうございます
oda
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Jan 6, 2025
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| 振り返ってみると前回の141. Linked List Cycleとあまり変わらない | ||
| おそらく型アノテーションの知識もプラスで問われているのでmediumになったのかな（141. Linked List Cycleはeasy) | ||
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フロイドの循環検出法が想定解だと考えて、だと思います。
LeetCode の難易度はあまり参考にならないのであまり気にしなくていいでしょう。
(本来のゴールは動くコードが書けるかどうかではないのだけれども、そのあたりの感覚がなさそうです。)
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プルリク名は「142. Linked List Cycle II」とすると良いでしょう。 |
liquo-rice
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Jan 6, 2025
| 誰かの発言の引用は「」でおこなう | ||
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| step1 | ||
| ```python |
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バグのあるコードは明示するようにすると、レビュワーにとって親切でしょう。
| 今までの話をまとめると、以下のコードの意味は引数headはListNodeの属性だがNoneも含む、返り値も同様という意味 | ||
| def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]: | ||
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| そのため循環がない場合の返り値は-1ではなくNoneで返すべきだった |
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あまり問題文を読めていない印象を受けました。
問題文には
Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.
とありますね。
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あ、たしかに型アノテーションを見なくても問題文に書いていましたね。問題文の指摘ありがとうございます
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| 振り返ってみると前回の141. Linked List Cycleとあまり変わらない | ||
| おそらく型アノテーションの知識もプラスで問われているのでmediumになったのかな（141. Linked List Cycleはeasy) |
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ありがとうございます。修正いたしました |
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142. Linked List Cycle II
Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.
Do not modify the linked list.
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