142. Linked List Cycle II（找出链表相交的节点）

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?

头结点到cycle begins的点 距离是A, cycle begins的点 到快慢结点相遇的 点的距离是B

A+B+N = 2*(A+B)

A+B = N

所以 快慢指针相遇后，从头结点开始再跑一个慢指针，直到2个慢的相遇，相遇的点就是cycle begin

 public class Solution {
     public ListNode detectCycle(ListNode head) {
         ListNode faster = head;
         ListNode slower = head;

         while(faster !=null && faster.next != null){
             faster = faster.next.next;
             slower = slower.next;

             if(faster == slower){
                 ListNode slower2 = head;
                 while(slower != slower2){
                     slower = slower.next;
                     slower2 = slower2.next;
                 }
                 return slower;
             }
        }
        return null;

     }
 }