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) {        if(head == null || head.next == null || head.next.next == null){            return null;        }        ListNode p1 = head;        ListNode p2 = head.next.next;        while(p2 != null){            if(p1 == p2){                break;            }            p1 = p1.next;            if(p2.next == null || p2.next.next == null){                return null;            }            p2 = p2.next.next;        }        int circleLen = 0;//链长        do{            circleLen++;            p2 = p2.next;        }while(p1 != p2);                int c = 0;        p1 = head;        p2 = head;        while(c < circleLen){            c++;            p2 = p2.next;        }        while(p1 != p2){            p1 = p1.next;            p2 = p2.next;        }        return p1;    }}

Runtime: 396 ms