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;
        
    }
}