Linked List Copy

Given the head of a singly linked list, return a new list with the same values.

1def copy_list(head: Optional[ListNode]) -> Optional[ListNode]:
2    dummy = ListNode(0)
3    tail = dummy
4    cur = head
5    while cur:
6        tail.next = ListNode(cur.val)
7        tail = tail.next
8        cur = cur.next
9    return dummy.next
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copied=[]
Step 1 / 3
Step 1:
Initialize dummy and tail to build the copy.
Pointers: cur=0
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