题解 | #链表中环的入口结点#

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def EntryNodeOfLoop(self, pHead):
        # write code here
        slow = pHead 
        fast = pHead 
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            # 找到快慢指针相遇的点
            if slow == fast:
                index1 = fast
                index2 = pHead
                # 利用数学推导 当 x = z 时即链表中环的入口节点
                while index1!=index2:
                    index1 = index1.next
                    index2 = index2.next
                return index1
        return None

数学推导，当x = z 时即为环入口