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nodchip
reviewed
Jan 8, 2026
| list2 = l2 | ||
| carry = 0 | ||
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| while list1 or list2 or carry: |
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以下のコメントをご参照ください。
Kazuuuuuuu-u/arai60#2 (comment)
h1rosaka/arai60#47 (comment)
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@nodchip
今回の場合だと、
while list1 is not None or list2 is not None or carry is not None:にした方がより良いという認識で合っていますでしょうか？
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carry は int 型ですので、
while list1 is not None or list2 is not None or carry != 0:が良いと思います。
| result = ListNode(0, l1) | ||
| list1 = l1.val | ||
| list2 = l2.val | ||
| res = 0 |
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res が residual の略なのか、 result の略なのか、一瞬判断に迷いました。
以下のコメントをご参照ください。
hemispherium/LeetCode_Arai60#10 (comment)
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参考リンクの提供ありがとうございます！
以降は頻繁に使用されるprefix以外の省略は避けていきます。
| # self.next = next | ||
| class Solution: | ||
| def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: | ||
| dummy = ListNode() |
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dummy に番兵ノードを代入しているにもかかわらず、 dummy を先に進めていっている点に違和感を感じました。 dummy の値は動かさず、別のポインターを動かしていったほうが自然に感じられます。
dummy = ListNode()
node = dummy
...
while l1 or l2 or carry:
...
node.next = ListNode(digit)
node = node.next
return dummy.next
05ryt31
commented
Jan 9, 2026
| # self.next = next | ||
| class Solution: | ||
| def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: | ||
| dummy = ListNode() |
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問題リンク
https://leetcode.com/problems/add-two-numbers/description/
問題文の概要
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
次に解く予定の問題
https://leetcode.com/problems/merge-two-binary-trees/description/