题解 | #链表中环的入口结点#

public class Solution {
    public ListNode hasCycle(ListNode head) {
        //先判断链表为空的情况 fast-template
        if (head == null)
            return null;
        //快慢双指针
        ListNode fast = head;
        ListNode slow = head;
        //如果没环快指针会先到链表尾
        while (fast != null && fast.next != null) {
            //快指针移动两步
            fast = fast.next.next;
            //慢指针移动一步
            slow = slow.next;
            //相遇则有环，返回相遇的位置
            if (fast == slow)
                return slow;
        }
        //到末尾说明没有环，返回null
        return null;
    }
    public ListNode EntryNodeOfLoop(ListNode pHead) {
        ListNode slow = hasCycle(pHead);
        //没有环
        if (slow == null)
            return null;
        //快指针回到表头
        ListNode fast = pHead;
        //再次相遇即是环入口
        while (fast != slow) {
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}