LeetCode 142: Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Follow up:
Can you solve it without using extra space?

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        // Suppose the non-cycle length is X
        // Suppose the cycle length is Y
        // Suppose fast pointer and slow pointer encounter on the Kth node of cycle
        // So when both encountered, slow steps = X + nY + K; fast steps = X + mY + K (Note: m > n)
        // We can deduce X + K = (m-2n)Y
        // Therefore, if slow pointer run X steps from the Kth node, it will return to the beginning of cycle.
        ListNode fast = head, slow = head;
        
        while (fast != null && fast.next != null)
        {
            fast = fast.next.next;
            slow = slow.next;
            
            if (fast == slow)
            {
                fast = head;
                
                while (fast != slow)
                {
                    fast = fast.next;
                    slow = slow.next;
                }
                
                return fast;
            }
        }
        
        return null;
    }
}