offer2-21

给定一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

示例 1：

输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]

示例 2：

输入：head = [1], n = 1
输出：[]

示例 3：

输入：head = [1,2], n = 1
输出：[1]

提示：

• 链表中结点的数目为 sz
• 1 <= sz <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

进阶：能尝试使用一趟扫描实现吗？

//leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null || n <= 0) {
            return head;
        }

        ListNode fast = head;
        ListNode slow = null;
        int cnt = 0;

        while (fast != null) {
            fast = fast.next;
            if (cnt >= n) {
                if (slow == null) {
                    slow = head;
                } else {
                    slow = slow.next;
                }
            }
            cnt++;
        }

        if (slow == null) {
            if (cnt == n) {
                ListNode newHead = head.next;
                head.next = null;
                return newHead;
            } else {
                return head;
            }
        } else {
            ListNode remove = slow.next;
            slow.next = remove.next;
            remove.next = null;
            return head;
        }
    }
}
//leetcode submit region end(Prohibit modification and deletion)