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?

解题思路：双指针， 一个指针一次走一步，一个指针一次走两步， 如果有环，定相遇。然后将快指针移到头部，再一次走一步，当两只真再次相遇的时候就是指针的头

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