Leetcode 142. 环形链表 II

注意的点：

1、注意里面的判断条件，虽然代码少但是其实逻辑很多。

解法：快慢指针

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:

        # 快慢指针
        slow, fast = head, head

        if not head: return None

        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast: break # 得放在后面
        
        if not fast.next or not fast.next.next: 
            return None

        slow = head
        while slow != fast:
            slow = slow.next
            fast = fast.next

        return fast