题解 | #判断链表中是否有环#

 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if(head== null || head.next == null){
            return false;
        }
        //快慢指针 第二次相逢
        //如果有环 看下快慢指针能否相遇  和求相遇节点类似
        ListNode slow = head;
        ListNode fast = head;
        while(slow != null && fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast){
                return true;
            }
        }
        return false;
    }
}