https://leetcode.cn/problems/linked-list-cycle-ii/
    需要注意的点
    1. 环头节点(环头)，链表的头节点(链头)，fast 跟slow 相遇的节点(相遇点)
    走n圈，环头到链头为a,slow 到相遇点为b， fast 在环里面走c，
    环的长度是b+c
    fast走过的距离是
    fast = a + n(b+c)+b = a + (n+1)b+nc
    slow走过的距离是
    slow = a+b
    等式
    a + (n+1)b+nc = 2(a+b)
    == a = c + (n-1)(b+c)

    1. # Definition for singly-linked list.
    2. # class ListNode:
    3. #     def __init__(self, x):
    4. #         self.val = x
    5. #         self.next = None
    7. class Solution:
    8.     def detectCycle(self, head: ListNode) -> ListNode:
    10.         '''
    11.         环头节点(环头)，链表的头节点(链头)，fast 跟slow 相遇的节点(相遇点)
    12.         走n圈，环头到链头为a,slow 到相遇点为b， fast 在环里面走c，
    13.         环的长度是b+c
    14.         fast走过的距离是
    15.         fast = a + n(b+c)+b = a + (n+1)b+nc
    16.         slow走过的距离是
    17.         slow = a+b
    18.         等式
    19.         a + (n+1)b+nc = 2(a+b)
    20.         == a = c + (n-1)(b+c)
    21.         '''
    22.         slow = head
    23.         fast = head
    25.         while fast and fast.next:
    27.             fast = fast.next.next
    28.             slow = slow.next
    30.             if fast == slow:
    32.                 a = fast
    33.                 b = head
    35.                 while a != b:
    36.                     a = a.next
    37.                     b = b.next
    39.                 return a
    41.         return None