55.链表中环的入口结点(python)

题目描述

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。

思路

如果slow走了L的长度那么fast走了2L

假设从开始到入口点的长度是s，slow在环里走的长度是d

那么 l = s + d

假设环内slow没走的长度是m，fast走的长度是n*(m+d) + d + s = 2L

带入得n*(m+d) + d + s = 2（s+d）  =>   s = m+(n-1)(m+d)     m+d就是绕环一圈   所以s = m  所以相遇后，让slow和head一起走，相遇点就是入环节点

 1 class Solution:
 2     def EntryNodeOfLoop(self, pHead):
 3         # write code here
 4         if pHead == None:
 5             return None
 6         fastPointer = pHead
 7         slowPointer = pHead
 8         while fastPointer and fastPointer.next:
 9             slowPointer = slowPointer.next
10             fastPointer = fastPointer.next.next
11             if fastPointer == slowPointer:
12                 break
13         if fastPointer == None or fastPointer.next == None:
14             return None
15         fastPointer = pHead
16         while fastPointer!=slowPointer:
17             fastPointer=fastPointer.next
18             slowPointer=slowPointer.next
19         return fastPointer

2019-12-31 22:17:13