| 题目链接 | 141. 环形链表 |
|---|---|
| 思路 | 快慢指针-经典问题 |
| 题解链接 | 没想明白？一个视频讲透快慢指针！（Python/Java/C++/Go/JS） |
| 关键点 | while fast and fast.next: ... |
| 时间复杂度 | \(O(n)\) |
| 空间复杂度 | \(O(1)\) |
代码实现：
class Solution:
def hasCycle(self, head: Optional[ListNode]) -> bool:
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if fast is slow:
return True
return False