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+# 删除链表的倒数第 N 个结点
+
+## 题目
+
+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。
+
+ 
+
+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：你能尝试使用一趟扫描实现吗？
+
+
+## 思路
+
+- 双指针
+  - 快指针指向 null
+  - 慢指针指向目标结点
+
+## 代码
+
+```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) {
+        // Init: slow, fast, dummyHead
+        ListNode dummyHead = new ListNode(-1);
+        dummyHead.next = head;
+        ListNode slow = dummyHead;
+        ListNode fast = dummyHead;
+        int cnt = 0;
+
+        while (fast != null && cnt < n) {
+            fast = fast.next;
+            cnt++;
+        }
+
+        if (cnt != n) {
+            return null;
+        }
+
+        while (fast != null) {
+            fast = fast.next;
+            
+            
+            if (fast == null) {
+                slow.next = slow.next.next;
+                return dummyHead.next;
+            }
+
+            slow = slow.next;
+        }
+
+        return dummyHead.next;
+
+    }
+}
+```
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