The thinking about this question comes from Question 142 and 287 in LeetCode.

We all know that we can use two pointers to decide if there is a circle in a list:

fast: move 2 steps each time

slow: move 1 step each time

If slow meet fast in a certain moment, then this list has a circle.

Now I would like to show how we can find the entry of circle. Here is the demonstration:

First we define some variables:

s: when slow meet fast, the steps slow moves (so fast moves 2s steps)

x: the length from head to circle entry (this is what we  want to know)

c: the length of the circle

a: the length from entry points to meet points

we have: 2s = s + n*c (n is a positive integer)

                 s = x + a

so we have : x + a = n*c

                     x = (n-1)*c + c - a

c - a means the length from meet point to circle entry

we can consider the left side of expression as a new pointer who moves x steps from list head.

the right side: the slow moves n-1 circle plus the length from meet point to the entry.

So we can get the entry when new pointer meet slow.