https://leetcode.cn/problems/linked-list-cycle-ii/ (opens new window) 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)

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

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

        '''
        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)
        '''
        slow = head
        fast = head

        while fast and fast.next:

            fast = fast.next.next
            slow = slow.next

            if fast == slow:

                a = fast
                b = head

                while a != b:
                    a = a.next
                    b = b.next
                
                return a

        return None