3217. Delete Nodes From Linked List Present in Array
## 思路
先把nums转成set
扫linked list时, 如果下一个node的值在set里面就跳过去
## Code
```python
class Solution:
def modifiedList(self, nums: List[int], head: Optional[ListNode]) ->
Optional[ListNode]:
nums = set(nums)
dummy = ListNode(-1, head)
curr = dummy
while curr.next:
if curr.next.val not in nums:
curr = curr.next
else:
curr.next = curr.next.next
return dummy.next
```

大师

大师

别卷了