循环链表：快慢指针

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解决方案1:循环链表遍历放入数组，然后循环看节点是否存在于数组中，比较简单，就不实现了
解决方案2：快慢指针

public class Solution {
    public boolean hasCycle(ListNode head) {
       if (head == null) {
           return  false;
       }
       //定义快慢指针
       ListNode slow = head;
       ListNode fast = head.next;
       //遍历链表： 快指针步长为2，慢指针步长为1
        while (fast != null && fast.next != null) {
            //当且仅当快慢指针重合：有环，操作结束
            if (slow == fast) {
                return true;
            }
            fast = fast.next.next;
            slow = slow.next;
        }

        //快指针为null，或其next指向null，没有环，返回false
        return false;
    }

    class ListNode {
       int val;
       ListNode next;
       ListNode(int x) {
           val = x;
           next = null;
       }
   }
}
//leetcode submit region end(Prohibit modification and deletion)

}