LeetCode-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?

这道题其实是一个数学题，先检测出是不是有环，在有环的情况下slow再走x步（x是环外的节点个数）就可以到达环的开始点。

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