Why use the fast and slow pointer to find the link of the linked list, and the fast pointer and the slow pointer must meet?

You can think about this problem by mathematical induction . First , Because the linked list is a ring , So the process of meeting can be seen as the process of the fast pointer catching up with the slow pointer from behind . Then do the following thinking ：

1. The difference between the fast pointer and the slow pointer is one step . Now continue to walk back , Slow pointer forward one step , Move the pointer forward two steps , The two meet .
2. There is a difference of two steps between the fast pointer and the slow pointer . Now sigh and walk back , Slow pointer forward one step , Move the pointer forward two steps , The difference between the two is one step , Into the first case .
3. Difference between fast pointer and slow pointer N Step . Now continue to walk back , Slow pointer forward one step , Move the pointer forward two steps , The difference between the two is (N+1-2)-> N-1 Step .

therefore , This question is proved . So the fast pointer must meet the slow pointer . And because the speed of the fast pointer is twice that of the slow pointer , So when we meet, we must only make a circle .