'''
解题思路:
遍历单链表的每个结点
如果当前结点地址没有出现在set中,则存入set中
否则,出现在set中,则当前结点就是环的入口结点
整个单链表遍历完,若没出现在set中,则不存在环
'''
# -*- 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 not pHead or not pHead.next:
return None
visited = set()
while pHead:
if pHead not in visited:
visited.add(pHead)
pHead = pHead.next
else:
return pHead
return None

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