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?

一刷
方法1： hashMap, 但是空间复杂度为O(n)
方法2：

fast: 2N = D+K+C(i)
slow: N = D+K+C(j)
那么D+K = C(i-2j)

所以先找到fast和slow的相遇点。然后fast从head开始，fast和slow相遇点即为D
因为此时fast走了D + C(i), slow走了C-K + jC

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