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;
    }
}
*/
public class Solution {

    public ListNode EntryNodeOfLoop(ListNode pHead) {
        if (pHead == null) {
            return null;
        }
        ListNode fast = pHead;
        ListNode slow = fast;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow ) {
                break;
            }
        }
        // 不存在环
        if (fast == null || fast.next == null) {
            return null;
        }
        // ListNode second = pHead;
        //指针重新回到表头
        ListNode second = pHead;
        //再次相遇即是环入口
        while (second != slow) {
            second = second.next;
            slow = slow.next;
        }
        return slow;
    }
}