| 题目链接 | 142. 环形链表 II |
|---|---|
| 思路 | 快慢指针-经典应用：相遇后，移动head及slow直至相遇 |
| 题解链接 | 没想明白？一个视频讲透！含数学推导（Python/Java/C++/Go/JS） |
| 关键点 | while fast and fast.next: ... && while head != slow: ... |
| 时间复杂度 | \(O(n)\) |
| 空间复杂度 | \(O(1)\) |
代码实现：
class Solution:
def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if fast is slow:
while slow is not head:
slow = slow.next
head = head.next
return head
return None