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oda
reviewed
Jan 2, 2025
| l1 = l1.next | ||
| l2 = l2.next | ||
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| return sentinental |
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番兵は、端っこの処理をまとめるために形式的に現れるノードを指すので、そのまま返しているということは番兵ではないと思われます。
oda
reviewed
Jan 2, 2025
Comment on lines
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| if l1.next is None and l2.next: | ||
| l1.next = ListNode(0, None) | ||
| elif l1.next and l2.next is None: | ||
| l2.next = ListNode(0, None) |
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ここのところは、入力の ListNode を変更していますね。
これがライブラリーだったとして、他の人が
c = addTwoNumbers(a, b)と呼んで、a, b が変更されたらそれなりに驚くと思うのです。
変更していいかは状況次第ですが、呼び出す人のことを考えてみましょう。(考えた上で決行するのはあり。)
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l1, l2 に代入しているだけならば、呼び出し元に影響はないので、
l1 = l1.next
l2 = l2.nextとまとめるのは一つです。
oda
reviewed
Jan 2, 2025
| l2 = l2.next | ||
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| return sentinental | ||
| ``` |
t0hsumi
reviewed
Jan 3, 2025
| def addTwoNumbers( | ||
| self, l1: Optional[ListNode], l2: Optional[ListNode] | ||
| ) -> Optional[ListNode]: | ||
| sentinental = ListNode(0, None) |
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sentinelですね...。ありがとうございます。
| - If l1 is None else 0という書き方は便利：https://github.com/olsen-blue/Arai60/pull/5/file | ||
| - If l1:~ sum+=, If l2:~ sum+=という書き方が読んでて綺麗だと思ったのでこの方向を目指したい | ||
| - Get_valueでNoneを0に置き換えるのはうまいと思った。条件分けが減って読みやすい。：https://github.com/t0hsumi/leetcode/pull/5/files | ||
| - l1やl2を直で動かすべきではない |
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step 2, 3の様に、ただvalを見ていき、nextで次に進むだけであるなら、書き換えられるわけではないので、直で動かすこと自体に問題はないと思います。
ただ、この問題では「ある桁のノードに格納されている値」に注目してくところがあるので、nodeをおいたほうが自然かなくらいで好みの問題だと思いました。
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次に解く問題：https://leetcode.com/problems/valid-parentheses/description/