연결리스트 추가문제

Question: 유효한 팰린드롬 2

입력: head = [3,2,0,-4], pos = 1

출력: true

내 풀이

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        if head is None:
            return
        
        slow = fast = head
        
        # fast 와 slow는 연결리스트가 cycle구조일때 무조건 만나게 되있다.
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                return True
        return False

결과

Runtime: 82 ms, faster than 36.59% of Python3 online submissions for Linked List Cycle.
Memory Usage: 17.7 MB, less than 54.54% of Python3 online submissions for Linked List Cycle.