Linked List Cycle II LeetCode solution in Python

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# Definition for singly-linked list.
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# class ListNode:
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#     def __init__(self, x):
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#         self.val = x
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#         self.next = None
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class Solution:
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    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
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        if not head or not head.next:
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            return None
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        slow = fast = entry = head
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        while fast and fast.next:
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            slow = slow.next
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            fast = fast.next.next
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            if slow == fast:
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                while slow != entry:
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                    slow = slow.next
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                    entry = entry.next
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                return entry
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        return None