# Short Python solution: 8 lines

• ``````class Solution(object):
while True:
if fast == None or fast.next == None: return None
slow = slow.next; fast = fast.next.next
if slow == fast: break
``````

• I understand the detection part, but
why can you make sure that when node head == node fast, it is the start of the linked list cycle?

• @pjerryhu Hello,

Let's say, the first node is node 0, the cycle starts at node L, and the length of the cycle is C;
Moreover, after t steps, fast catches slow.

Now we know that fast totally traveled 2t nodes, and slow traveled t nodes

Then we have:
2t - t = nC (where n is an positive integer.)
i.e. t=nC

Now, think about that, at step t, if we travels L more steps, where are we?
i.e. if we travel L+t = L + nC steps in total, where are we?

Absolutely, at the start of the cycle, because we have covered the first L nodes once and the entire cycle n times.

So, if we travel L more steps at time t, then we get the start of the cycle.

However, how can we travel exactly L step?
The answer is to use an other pointer to travel from node 0, and when they meet together, it is exactly L steps and both of them are at the start of the cycle.

Hope that help :)

• @LostSummer233 Thank you. This is very helpful

• Some really good explanation on youtube :D

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.