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) {
        ListNode fast=head;
        ListNode slow=head;
        while(fast!=null&&fast.next!=null){
            slow=slow.next;
            fast=fast.next.next;
            if(slow==fast)break;
        }
        if(fast==null||fast.next==null){
            return null;
        }
        fast=head;
       while(fast!=slow){
                fast=fast.next;
                slow=slow.next;
            }
        return fast;
    }
        
        
        
        /*ListNode fast=head;
        ListNode slow=head;
        if(head==null||head.next==null)return null;
        while(fast!=null){
            slow=slow.next;
            if(fast.next!=null){
                fast=fast.next.next;
            }else return null;
            if(slow==fast)
            break;
            
        }
        if(slow==fast){
            fast=head;
            while(fast!=slow){
                fast=fast.next;
                slow=slow.next;
            }
            return fast;
            
        }
        return null;
        
        
    }*/
}