题解 | #牛牛队列成环#

经典快慢指针

不过这次一开始考虑漏了两点

• fast指针一次跳两格，所以判断条件的那里应该fast!=null && fast.next!=null
• 快指针的位置要在慢指针后面

public class Solution {

    public boolean hasCycle (ListNode head) {
        // 快慢指针
        ListNode low = head;
        ListNode fast = head.next;
        while (low != null && fast != null && fast.next != null) {
            low = low.next;
            fast = fast.next.next;
            if (low.val == fast.val) return true;
        }
        return false;
    }
}