diff --git a/README.md b/README.md
index 08fd072..e6603ec 100644
--- a/README.md
+++ b/README.md
@@ -272,10 +272,9 @@
 
 #### 数学问题
 
-- 数组中出现次数超过一半的数字
-- 圆圈中最后剩下的数
-- 从1到n整数中1出现的次数
-
+- [【简单】数组中出现次数超过一半的数字](./算法/剑指/数学/MoreThanHalfNum_Solution.js)
+- [【中等】圆圈中最后剩下的数 约瑟夫问题](./算法/剑指/数学/LastRemaining_Solution.js)
+- [【中等】从1到n整数中1出现的次数](./算法/剑指/数学/NumberOf1Between1AndN_Solution.js)
 #### 位运算
 
 - 二进制中1的个数
diff --git a/算法/剑指/.DS_Store b/算法/剑指/.DS_Store
new file mode 100644
index 0000000..5e00245
Binary files /dev/null and b/算法/剑指/.DS_Store differ
diff --git a/算法/剑指/数学/LastRemaining_Solution.js b/算法/剑指/数学/LastRemaining_Solution.js
new file mode 100644
index 0000000..0d6a431
--- /dev/null
+++ b/算法/剑指/数学/LastRemaining_Solution.js
@@ -0,0 +1,72 @@
+/*
+ * @Description: 【中等】圆圈中最后剩下的数 约瑟夫问题
+ * @Version: Beta1.0
+ * @Author: 【B站&公众号】Rong姐姐好可爱
+ * @Date: 2021-05-05 14:48:28
+ * @LastEditors: 【B站&公众号】Rong姐姐好可爱
+ * @LastEditTime: 2021-05-05 15:11:02
+ */
+
+
+
+// 把n个人的编号改为0~n-1，然后对删除的过程进行分析。
+// 第一个删除的数字是(m-1)%n，几位k，则剩余的编号为(0,1,...,k-1,k+1,...,n-1)，下次开始删除时，顺序为(k+1,...,n-1,0,1,...k-1)。
+// 用f(n,m)表示从(0~n-1)开始删除后的最终结果。
+// 用q(n-1,m)表示从(k+1,...,n-1,0,1,...k-1)开始删除后的最终结果。
+// 则f(n,m)=q(n-1,m)。
+
+// 下面把(k+1,...,n-1,0,1,...k-1)转换为(0~n-2)的形式，即
+// k+1对应0
+// k+2对于1
+// ...
+// k-1对应n-2
+// 转化函数设为p(x)=(x-k-1)%n, p(x)的逆函数为p^(x)=(x+k+1)%n。
+// 则f(n,m)=q(n-1,m)=p^(f(n-1,m))=(f(n-1,m)+k+1)%n，又因为k=(m-1)%n。
+// 取余
+// f(n,m)=(f(n-1,m)+m)%n;
+
+// 最终的递推关系式为
+// f(1,m) = 0;                        (n=1)
+// f(n,m)=(f(n-1,m)+m)%n; （n>1）
+
+function LastRemaining_Solution (n, m) {
+  // write code here
+  // 递推公式： f(0)=-1  f(1)=0 f(i)={f(i-1)+m}%i
+
+  if (n === 0) {
+    return -1;
+  }
+
+  if (n === 1) {
+    return 0;
+  }
+
+  // 递归
+  return (LastRemaining_Solution(n - 1, m) + m) % n;
+
+}
+
+// 非递归实现
+function LastRemaining_Solution01 (n, m) {
+
+  // 当然，这里也可以添加上负数的校验情况
+  if (n === 0) {
+    return -1;
+  }
+
+  if (n === 1) {
+    return 0
+  }
+
+  // 循环处理
+  let result = 0
+  for (let index = 2; index <= n; index++) {
+    // f(n,m)=[f(n-1,m)+m]%n
+    result = (result + m) % index
+  }
+  // 返回
+  return result;
+}
+module.exports = {
+  LastRemaining_Solution: LastRemaining_Solution
+};
\ No newline at end of file
diff --git a/算法/剑指/数学/MoreThanHalfNum_Solution.js b/算法/剑指/数学/MoreThanHalfNum_Solution.js
new file mode 100644
index 0000000..21de050
--- /dev/null
+++ b/算法/剑指/数学/MoreThanHalfNum_Solution.js
@@ -0,0 +1,40 @@
+/*
+ * @Description: 【 简单】数组中出现次数超过一半的数字
+ * @Version: Beta1.0
+ * @Author: 【B站&公众号】Rong姐姐好可爱
+ * @Date: 2021-05-05 14:28:36
+ * @LastEditors: 【B站&公众号】Rong姐姐好可爱
+ * @LastEditTime: 2021-05-05 14:29:22
+ */
+
+// 借助map计数即可
+function MoreThanHalfNum_Solution (numbers) {
+  // write code here
+  let map = new Map();
+
+  numbers.forEach(item => {
+    if (map.has(item)) {
+      map.set(item, map.get(item) + 1)
+    } else {
+      map.set(item, 1)
+    }
+  })
+
+  const arr = [...new Set(numbers)];
+
+  //   console.log(map,arr)
+  let result = 0;
+
+  arr.map(item => {
+    if (2 * map.get(item) > numbers.length) {
+      result = item;
+    }
+  })
+
+
+  return result;
+
+}
+module.exports = {
+  MoreThanHalfNum_Solution: MoreThanHalfNum_Solution
+};
\ No newline at end of file
diff --git a/算法/剑指/数学/JZ43.NumberOf1Between1AndN_Solution.js b/算法/剑指/数学/NumberOf1Between1AndN_Solution.js
similarity index 80%
rename from 算法/剑指/数学/JZ43.NumberOf1Between1AndN_Solution.js
rename to 算法/剑指/数学/NumberOf1Between1AndN_Solution.js
index ce27205..e927a1d 100644
--- a/算法/剑指/数学/JZ43.NumberOf1Between1AndN_Solution.js
+++ b/算法/剑指/数学/NumberOf1Between1AndN_Solution.js
@@ -1,10 +1,10 @@
 /*
- * @Description: 
+ * @Description: 【中等】从1到n整数中1出现的次数
  * @Version: Beta1.0
  * @Author: 【B站&公众号】Rong姐姐好可爱
  * @Date: 2021-04-26 22:23:17
  * @LastEditors: 【B站&公众号】Rong姐姐好可爱
- * @LastEditTime: 2021-04-26 22:27:23
+ * @LastEditTime: 2021-05-05 15:15:56
  */
 
 
@@ -28,7 +28,7 @@ function NumberOf1Between1AndN_Solution(n)
     return count;
 }
 
-console.log(NumberOf1Between1AndN_Solution(13))
+// console.log(NumberOf1Between1AndN_Solution(13))
 module.exports = {
     NumberOf1Between1AndN_Solution : NumberOf1Between1AndN_Solution
 };
\ No newline at end of file