链表中环的入口节点

给一个链表，若其中包含环，请找出该链表的环的入口结点，否则，输出null。

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def EntryNodeOfLoop(self, pHead):
        # write code here
        #遍历链表，环的存在，遍历遇见的第一个重复的即为入口节点
        nodeList = []
        while pHead != None:
            if pHead not in nodeList:
                nodeList.append(pHead)
            else:
                return pHead
            pHead = pHead.next

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def EntryNodeOfLoop(self, pHead):
        # write code here
        if pHead == None or pHead.next == None or pHead.next.next == None:
            return None
        pSlow = pHead.next
        pFast = pHead.next.next
        # 快慢指针找到相遇点
        while pSlow != pFast:
            if pFast.next == None or pFast.next.next == None:
                return None
            pSlow = pSlow.next
            pFast = pFast.next.next
        pSlow = pHead
        # （出发点到入口的距离 = n*圈+相遇点到入口距离）速度相同，重新相遇的地方即为入口
        while pSlow != pFast:
            pSlow = pSlow.next
            pFast = pFast.next
        return pSlow