diff --git a/src/main/java/leetcode/editor/cn/ArrayPartitionI.java b/src/main/java/leetcode/editor/cn/ArrayPartitionI.java
new file mode 100644
index 0000000..6d24acc
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/ArrayPartitionI.java
@@ -0,0 +1,64 @@
+//给定长度为 2n 的整数数组 nums ，你的任务是将这些数分成 n 对, 例如 (a1, b1), (a2, b2), ..., (an, bn) ，使得
+//从 1 到 n 的 min(ai, bi) 总和最大。 
+//
+// 返回该 最大总和 。 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//输入：nums = [1,4,3,2]
+//输出：4
+//解释：所有可能的分法（忽略元素顺序）为：
+//1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
+//2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
+//3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
+//所以最大总和为 4 
+//
+// 示例 2： 
+//
+// 
+//输入：nums = [6,2,6,5,1,2]
+//输出：9
+//解释：最优的分法为 (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 
+//6 = 9
+// 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 1 <= n <= 10⁴ 
+// nums.length == 2 * n 
+// -10⁴ <= nums[i] <= 10⁴ 
+// 
+// Related Topics 贪心 数组 计数排序 排序 👍 260 👎 0
+
+package leetcode.editor.cn;
+
+import java.util.Arrays;
+
+//561:数组拆分 I
+class ArrayPartitionI {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new ArrayPartitionI().new Solution();
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int arrayPairSum(int[] nums) {
+            Arrays.sort(nums);
+            int sum = 0;
+            for (int i = 0; i < nums.length; i = i + 2) {
+                sum += nums[i];
+            }
+            return sum;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
\ No newline at end of file
diff --git a/src/main/java/leetcode/editor/cn/doc/content/ArrayPartitionI.md b/src/main/java/leetcode/editor/cn/doc/content/ArrayPartitionI.md
new file mode 100644
index 0000000..08791c2
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/content/ArrayPartitionI.md
@@ -0,0 +1,35 @@
+<p>给定长度为 <code>2n</code><strong> </strong>的整数数组 <code>nums</code> ，你的任务是将这些数分成 <code>n</code><strong> </strong>对, 例如 <code>(a<sub>1</sub>, b<sub>1</sub>), (a<sub>2</sub>, b<sub>2</sub>), ..., (a<sub>n</sub>, b<sub>n</sub>)</code> ，使得从 <code>1</code> 到 <code>n</code> 的 <code>min(a<sub>i</sub>, b<sub>i</sub>)</code> 总和最大。</p>
+
+<p>返回该 <strong>最大总和</strong> 。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [1,4,3,2]
+<strong>输出：</strong>4
+<strong>解释：</strong>所有可能的分法（忽略元素顺序）为：
+1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
+2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
+3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
+所以最大总和为 4</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [6,2,6,5,1,2]
+<strong>输出：</strong>9
+<strong>解释：</strong>最优的分法为 (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 <= n <= 10<sup>4</sup></code></li>
+	<li><code>nums.length == 2 * n</code></li>
+	<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>
+</ul>
+<div><div>Related Topics</div><div><li>贪心</li><li>数组</li><li>计数排序</li><li>排序</li></div></div><br><div><li>👍 260</li><li>👎 0</li></div>
\ No newline at end of file