diff --git a/src/main/java/leetcode/editor/cn/RemoveNthNodeFromEndOfList.java b/src/main/java/leetcode/editor/cn/RemoveNthNodeFromEndOfList.java
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+//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 
+// 
+// Related Topics 链表 双指针 
+// 👍 1398 👎 0
+
+package leetcode.editor.cn;
+
+import com.code.leet.entiy.ListNode;
+
+import java.util.ArrayList;
+import java.util.List;
+
+//19:删除链表的倒数第 N 个结点
+public class RemoveNthNodeFromEndOfList{
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new RemoveNthNodeFromEndOfList().new Solution();
+    }
+    //力扣代码
+    //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) {
+        List<ListNode> list = new ArrayList<>();
+        ListNode temp = head;
+        while (temp != null) {
+            list.add(temp);
+            temp = temp.next;
+        }
+        if (n == list.size() && head != null) {
+            head = head.next;
+        } else if (list.size() != 1) {
+            list.get(list.size() - n - 1).next = list.get(list.size() - n).next;
+        }
+        return head;
+    }
+}
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/RemoveNthNodeFromEndOfList.md b/src/main/java/leetcode/editor/cn/RemoveNthNodeFromEndOfList.md
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+<p>给你一个链表，删除链表的倒数第 <code>n</code><em> </em>个结点，并且返回链表的头结点。</p>
+
+<p><strong>进阶：</strong>你能尝试使用一趟扫描实现吗？</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.jpg" style="width: 542px; height: 222px;" />
+<pre>
+<strong>输入：</strong>head = [1,2,3,4,5], n = 2
+<strong>输出：</strong>[1,2,3,5]
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>head = [1], n = 1
+<strong>输出：</strong>[]
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>head = [1,2], n = 1
+<strong>输出：</strong>[1]
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li>链表中结点的数目为 <code>sz</code></li>
+	<li><code>1 <= sz <= 30</code></li>
+	<li><code>0 <= Node.val <= 100</code></li>
+	<li><code>1 <= n <= sz</code></li>
+</ul>
+<div><div>Related Topics</div><div><li>链表</li><li>双指针</li></div></div>\n<div><li>👍 1398</li><li>👎 0</li></div>
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