Given a linked list, return the node where the cycle begins. If there is no cycle, return null
.
Note: Do not modify the linked list.
Follow up:Can you solve it without using extra space?

Solution:

思路，用141的思路判断有环后令另一个指针从 head 开始与 slow以1的速度同步行走，他们将正好相会在环入口Node 处。

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