[Inlet] offer the list to prove safety ring

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

Links: https://www.nowcoder.com/questionTerminal/253d2c59ec3e4bc68da16833f79a38e4
Source: Cattle-off network

 

Suppose x is a distance in front of the loop (from red), ring A is the meeting point of the distance to the inlet (green distance, assuming a clockwise down), c is the length of the loop (blue distance)

Fast setting speed pointer and slow, the fast speed of the pointer is twice slower pointer

When the speed of the pointer met:

At this time, the pointer's journey is slow Sslow = x + m * c + a
fast pointer's journey to + n-X * = Sfast C + A
2 Sslow Sfast =
2 * (m * X + C + A) = (X + n * c + a)
can be deduced:
X = (n-- 2 * m) C * - a
= (n-- 2 * m -1) * C + C - a
i.e. the front ring = number of distance rings the length (as possibly 0) + c - a
what is c - a? This is the meeting point, the distance (from orange) cycloalkyl rear portion and x (red distance) is equal to

class Solution:
    def EntryNodeOfLoop(self, pHead):
        # write code here
        if pHead==None or pHead.next==None or pHead.next.next==None:
            return None
        # 一快一慢
        low=pHead.next
        fast=pHead.next.next
        # 如果有环，两者会相遇
        while low!=fast:
            if fast.next==None or fast.next.next==None:
                return None
            low=low.next
            FAST = fast.next.next
         # position to encounter the distance ring starting from the inlet and the inlet to the same ring 
        FAST = PHEAD
         the while Low =! FAST: 
            Low = low.next 
            FAST = fast.next
         return FAST