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
原文：https://www.cnblogs.com/NPC-assange/p/12127747.html