链表中环的入口节点 牛客网

 

public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode fast = head;          //fast一次走两步 slow一次走一步
        ListNode slow = head;          //这样效率最高
        if(fast == null) {            //若fast为null  则返回null
            return null;
        }
        while(fast != null && fast.next != null) {  //还未遍历完，fast slow继续走
            fast = fast.next.next;
            slow = slow.next;
            if(slow == fast) {               //slow追上fast 则有环
                break;
            }
        }
        if(fast == null || fast.next == null) {
            return null;
        }
                     //以上判断链表是否有环

        slow = head;   //有环的情况下 让slow=this.head
        while (fast != slow) {  //相遇时既是环的入口点
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}