题解 | #链表中环的入口结点#

/*
 public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}
*/
这个题是典型的龟兔赛跑。利用快慢指针解决问题
public class Solution {

    public ListNode EntryNodeOfLoop(ListNode pHead) {
        if(pHead == null || pHead.next == null)return null;//极端情况
        if(pHead == pHead.next)return pHead;//只有一个节点的环形链表
        ListNode fast = pHead;//快指针
        ListNode slow = pHead;//慢指针
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
            if(fast == slow){
                ListNode slow02 = pHead;
                while(slow != slow02){
                    slow = slow.next;
                    slow02 = slow02.next;
                }
                return slow02;//两节点相遇，返回一个即可
            }
        }
        return null;//fast先走出去了就返回空
    }
}