給定一個連結串列，如果連結串列中存在環，則返回到連結串列中環的起始節點的值，如果沒有環，返回null。
樣例
給出 -21->10->4->5, tail connects to node index 1，返回10
挑戰
不使用額外的空間

Given a linked list, return the node where the cycle begins.
If there is no cycle, return null.
Example
Given -21->10->4->5, tail connects to node index 1，return 10
Challenge
Follow up:
Can you solve it without using extra space?

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @return: The node where the cycle begins. 
     *           if there is no cycle, return null
     */
 

    public ListNode detectCycle(ListNode head) {  
        if(null == head || head.next == null || head.next.next == null) return null;
        ListNode fast = head, slow = head;
        while(true) {
            if(slow == null || fast == null || fast.next == null) {
                return null;
            }
            fast = fast.next.next;
            slow = slow.next;
            if
 
(fast == slow) {
                break;
            }
        }
        slow = head;
        while(true) {
            fast = fast.next;
            slow = slow.next;
            if(fast == slow) {
                return fast;
            }
        }
    }
}