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TORUS0818
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May 9, 2024
| if head.next and head.val == head.next.val: | ||
| head = skip_duplicated_nodes(head) | ||
| else: | ||
| current_node.next = ListNode(head.val) |
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ListNode()で新たにオブジェクトを作らなくてもheadの繋ぎかえでいけると思います。
| head = head.next | ||
| return head | ||
| dummy_head = ListNode(-1) | ||
| current_node = dummy_head |
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個人的にはcurrent_nodeは今注目しているもの（今回であれば重複判定とかするポインタ）の名前にする方が分かりやすいかなと思いました。
本コードではheadが使われてますね。
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問題
82. Remove Duplicates from Sorted List II