题目:链表中环的入口结点 描述 给一个长度为n链表,若其中包含环,请找出该链表的环的入口结点,否则,返回null。

数据范围: n≤10000 节点值范围:[1,10000] 要求:空间复杂度 O(1),时间复杂度 O(n)

这个题目我之前有思路,但是只在环的长度大于前面链的情况下成立,看评论知道了方法,记录一下

```# -*- 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):
            return
        fast = pHead
        slow = pHead
        isMeeting = False
        while ((fast != None) & (fast.next != None)):
            if isMeeting:
                fast = fast.next
            else:
                fast = fast.next.next
                if fast == None:
                    return
            slow = slow.next
            if slow == fast:
                if slow == pHead:
                    return slow
                if isMeeting:
                    return slow
                slow = pHead
                isMeeting = True
        return