The trick for this one was tracking duplicates via the unique boolean and a pending variable.

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Solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
        start = None
        curr = None
        pending = None
        unique = True
        while head:
            if not pending:
                pending = head
            elif pending.val != head.val and unique:
                if not start:
                    start = pending
                    curr = start
                else:
                    curr.next = pending
                    curr = curr.next
                pending = head
            elif pending.val != head.val and not unique:
                pending = head
                unique = True
            else:
                unique = False
                if curr:
                    curr.next = None
            head = head.next
        
        if pending and unique:
            if start:
                curr.next = pending
            else:
                start = pending

        return start

Time complexity is O(n) = n and space complexity is O(1).