2024年HCIE认证考下来要多少钱？能不能自学啊

文章目录

• 题目介绍
• 题解

题目介绍

题解

结论：若有环，则快慢指针相遇时慢指针还没有走完一圈。

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 (fast == slow) {
                while (slow != head) {
                    slow = slow.next;
                    head = head.next;
                }
                return slow;
            }
        }
        return null;
    }
}