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katsukii
changed the title
| 空間計算量: O(1) | ||
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| - head からノードを一つずつ走査。現在の対象ノードの値が次のノードと同一であればリンクをさらにその次につなぎなおす。そうでない場合に対象ノードを次に移動 | ||
| - 主役のノードは一人しかいないので、シンプルに node と命名した |
plushn
reviewed
Apr 11, 2025
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| return head; | ||
| } | ||
| } |
oda
reviewed
Apr 12, 2025
| 他の方が描いたコードを見て、参考にしてコードを書き直してみる。 | ||
| 参考にしたコードのリンクは貼っておく。 | ||
| 読みやすいことを意識する。 | ||
| 他の解法も考えみる。 |
Owner
Author
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ありがとうございます。二重ループは非効率的だと思い込みがあり直感的に避けていましが、実際には変わらないのですね。
class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode node = head;
while (node != null) {
while (node.next != null && node.val == node.next.val) {
node.next = node.next.next;
}
node = node.next;
}
return head;
}
}
Satorien
reviewed
Apr 16, 2025
| 他の解法も考えみる。 | ||
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| - continue で 処理をスキップする方法。現在ノードを次に進める処理が else の中にある事に違和感があったのでこちらの方が見やすい | ||
| - https://github.com/shintaro1993/arai60/pull/6/files |
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仰る通り、elseだと並列の関係にある処理のように読めてしまうのでこっちの方が良いですね
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問題
https://leetcode.com/problems/remove-duplicates-from-sorted-list/
言語
Java
次に解く問題
Remove Duplicates from Sorted List II