给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

进阶：你能尝试使用一趟扫描实现吗？

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

Solution

• 快慢指针
• 慢指针指向的是待删除节点的前一个节点

// @lc code=start
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(!head) return head;
        ListNode* dummyHead = new ListNode(0, head);
        ListNode* slow=dummyHead, *fast=dummyHead;
        while (n-- && fast != nullptr) {
            fast = fast->next;
        }

        while(fast->next){
            slow = slow->next;
            fast = fast->next;
        }
        slow->next = slow->next->next;
        return dummyHead->next;
    }
};
// @lc code=end

java

/**
 * 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) {
            return head;
        }
        ListNode dummyHead = new ListNode(-1, head);
        ListNode slow = dummyHead;
        ListNode fast = dummyHead;
        for (int i = 0; i < n; ++i) {
            if (fast == null) {
                return null;
            }
            fast = fast.next;
        }
        while (fast.next != null) {
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;
        return dummyHead.next;
    }
}