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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if (head==null) {
            return null;
        }
        ListNode pFast = head;
        ListNode pSlow = head;
        boolean first = true;
        while (pFast != null) {
            if (pSlow == pFast && !first) {
                //从相遇点找入口
                ListNode p = head;
                while (p!=null) {
                    if(p==pFast) {
                        return p;
                    }
                    p = p.next;
                    pFast = pFast.next;
                }
            }
            pSlow = pSlow.next;
            if (pFast.next !=null) {
                pFast = pFast.next.next;
            } else {
                pFast = null;
            }
            first = false;
        }
        return null;
    }
}