diff --git a/src/main/java/leetcode/editor/cn/NumberOfBoomerangs.java b/src/main/java/leetcode/editor/cn/NumberOfBoomerangs.java
new file mode 100644
index 0000000..7cc6a84
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/NumberOfBoomerangs.java
@@ -0,0 +1,95 @@
+//给定平面上 n 对 互不相同 的点 points ，其中 points[i] = [xi, yi] 。回旋镖 是由点 (i, j, k) 表示的元组 ，其中
+// i 和 j 之间的距离和 i 和 k 之间的距离相等（需要考虑元组的顺序）。
+//
+// 返回平面上所有回旋镖的数量。
+//
+//
+// 示例 1：
+//
+//
+//输入：points = [[0,0],[1,0],[2,0]]
+//输出：2
+//解释：两个回旋镖为 [[1,0],[0,0],[2,0]] 和 [[1,0],[2,0],[0,0]]
+//
+//
+// 示例 2：
+//
+//
+//输入：points = [[1,1],[2,2],[3,3]]
+//输出：2
+//
+//
+// 示例 3：
+//
+//
+//输入：points = [[1,1]]
+//输出：0
+//
+//
+//
+//
+// 提示：
+//
+//
+// n == points.length
+// 1 <= n <= 500
+// points[i].length == 2
+// -10⁴ <= xi, yi <= 10⁴
+// 所有点都 互不相同
+//
+// Related Topics 数组 哈希表 数学 👍 176 👎 0
+
+package leetcode.editor.cn;
+
+import com.code.leet.entiy.TwoArray;
+
+import java.util.Arrays;
+
+//447:回旋镖的数量
+class NumberOfBoomerangs {
+ public static void main(String[] args) {
+ //测试代码
+ Solution solution = new NumberOfBoomerangs().new Solution();
+ TwoArray twoArray = new TwoArray("[[0,0],[1,0],[2,0]]", true);
+ System.out.println(solution.numberOfBoomerangs(twoArray.getArr()));
+ }
+
+ //力扣代码
+ //leetcode submit region begin(Prohibit modification and deletion)
+ class Solution {
+ public int numberOfBoomerangs(int[][] points) {
+ int size = points.length;
+ int[][] lengths = new int[size][size];
+ for (int i = 0; i < size; i++) {
+ for (int j = i + 1; j < size; j++) {
+ int x = points[j][0] - points[i][0];
+ int y = points[j][1] - points[i][1];
+ int sum = x * x + y * y;
+ lengths[i][j] = sum;
+ lengths[j][i] = sum;
+ }
+ }
+ int counts = 0;
+ for (int[] length : lengths) {
+ Arrays.sort(length);
+ int count = 0;
+ int bef = length[0];
+ for (int i = 1; i < size; i++) {
+ if (length[i] == bef) {
+ count++;
+ } else {
+ counts += (count + 1) * count;
+ count = 0;
+ bef = length[i];
+ }
+ }
+ if (count > 0) {
+ counts += (count + 1) * count;
+ }
+ }
+ return counts;
+ }
+ }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
\ No newline at end of file
diff --git a/src/main/java/leetcode/editor/cn/doc/all.json b/src/main/java/leetcode/editor/cn/doc/all.json
index 15cdc66..4db60ec 100644
--- a/src/main/java/leetcode/editor/cn/doc/all.json
+++ b/src/main/java/leetcode/editor/cn/doc/all.json
@@ -1 +1 @@
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\ No newline at end of file
diff --git a/src/main/java/leetcode/editor/cn/doc/content/NumberOfBoomerangs.md b/src/main/java/leetcode/editor/cn/doc/content/NumberOfBoomerangs.md
new file mode 100644
index 0000000..31a491e
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/content/NumberOfBoomerangs.md
@@ -0,0 +1,39 @@
+给定平面上 n 对 互不相同 的点 points ，其中 points[i] = [xi, yi] 。回旋镖 是由点 (i, j, k) 表示的元组 ，其中 i 和 j 之间的距离和 i 和 k 之间的距离相等（需要考虑元组的顺序）。
+
+返回平面上所有回旋镖的数量。
+
+
+示例 1：
+
+
+输入：points = [[0,0],[1,0],[2,0]]
+输出：2
+解释：两个回旋镖为 [[1,0],[0,0],[2,0]] 和 [[1,0],[2,0],[0,0]]
+
+
+示例 2：
+
+
+输入：points = [[1,1],[2,2],[3,3]]
+输出：2
+
+
+示例 3：
+
+
+输入：points = [[1,1]]
+输出：0
+
+
+
+
+提示：
+
+
+ n == points.length1 <= n <= 500points[i].length == 2-104 <= xi, yi <= 104- 所有点都 互不相同
+
+
👍 176👎 0
\ No newline at end of file