55. The list entry node in the ring (Python)

Title Description

To a linked list, wherein if the ring comprises, find the entry ring node list, otherwise, outputs null.

Thinking

If the slow walk of length L 2L gone so fast

Assuming that the length from the start point to the entry s, slow down the ring in a length d

Then l = s + d

Suppose the length of the ring is not gone slow m, fast walking is a length of n * (m + d) + d + s = 2L

Obtained into n * (m + d) + d + s = 2 (s + d) => s = m + (n-1) (m + d) m + d is a circle around the ring so it met s = m after so slow and head go together, the meeting point is the ingress point

 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