diff --git a/.vscode/settings.json b/.vscode/settings.json
index 2315d28..8167afe 100644
--- a/.vscode/settings.json
+++ b/.vscode/settings.json
@@ -11,6 +11,7 @@
     "lrloseumgh",
     "mincost",
     "nums",
+    "powx",
     "pwwkew",
     "umghlrlose",
     "xuan",
diff --git a/README.md b/README.md
index e07f107..afed3c5 100644
--- a/README.md
+++ b/README.md
@@ -73,18 +73,18 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
 
 - [最后一个单词的长度](src/string/length-of-last-word.js)
 
-  - LeetCode 58. 最后一个单词的长度 https://leetcode-cn.com/problems/length-of-last-word/
-  - LintCode 422. 最后一个单词的长度 https://www.lintcode.com/problem/length-of-last-word/description
+  - LeetCode 58. 最后一个单词的长度 <https://leetcode-cn.com/problems/length-of-last-word/>
+  - LintCode 422. 最后一个单词的长度 <https://www.lintcode.com/problem/length-of-last-word/description>
 
 - [整数转罗马数字](src/string/integer-to-roman.js)
 
-  - LeetCode 12. 整数转罗马数字 https://leetcode-cn.com/problems/integer-to-roman/
-  - LintCode 418. 整数转罗马数字 https://www.lintcode.com/problem/integer-to-roman/description
+  - LeetCode 12. 整数转罗马数字 <https://leetcode-cn.com/problems/integer-to-roman/>
+  - LintCode 418. 整数转罗马数字 <https://www.lintcode.com/problem/integer-to-roman/description>
 
 - [罗马数字转整数](src/string/roman-to-integer.js)
 
-  - LeetCode 13. 罗马数字转整数 https://leetcode-cn.com/problems/roman-to-integer/
-  - LintCode 419. 罗马数字转整数 https://www.lintcode.com/problem/roman-to-integer/description
+  - LeetCode 13. 罗马数字转整数 <https://leetcode-cn.com/problems/roman-to-integer/>
+  - LintCode 419. 罗马数字转整数 <https://www.lintcode.com/problem/roman-to-integer/description>
 
 ## 数组
 
@@ -319,6 +319,11 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 9. 回文数 <https://leetcode-cn.com/problems/palindrome-number/>
   - LintCode 491. 回文数 <https://www.lintcode.com/problem/palindrome-number/>
 
+- [Pow(x, n)](src/math/powx-n.js)
+
+  - LeetCode 50. Pow(x, n) <https://leetcode-cn.com/problems/powx-n/>
+  - LintCode 428. x的n次幂 <https://www.lintcode.com/problem/powx-n/>
+
 ## 堆
 
 - [超级丑数](src/stack/super-ugly-number.js)【未完成】
diff --git a/src/math/powx-n.js b/src/math/powx-n.js
new file mode 100644
index 0000000..e34c610
--- /dev/null
+++ b/src/math/powx-n.js
@@ -0,0 +1,21 @@
+/**
+ * @param {number} x
+ * @param {number} n
+ * @return {number}
+ */
+export const myPow = function (x, n) { // 参考：快速幂 + 迭代 https://leetcode-cn.com/problems/powx-n/solution/powx-n-by-leetcode-solution/
+  let ans = 1.0 // 答案
+
+  if (n < 0) { // 如果 n 为负
+    n = -n // 将 n 变为正
+    x = 1.0 / x // 2^-2 = 1/(2^2) = 1/4 = 0.25
+  }
+
+  while (n > 0) {
+    if (n & 1) ans *= x // 如果n的二进制最后一位为1，则计入贡献
+    x *= x // 将贡献不断平方
+    n >>>= 1 // 无符号右移一位，相当于 n = n / 2 | 0 ，去掉二进制的最后一位，相当于下一次判断倒数第二位
+  }
+
+  return ans.toFixed(5)
+}
diff --git a/test/math/powx-n.test.js b/test/math/powx-n.test.js
new file mode 100644
index 0000000..3f79c7f
--- /dev/null
+++ b/test/math/powx-n.test.js
@@ -0,0 +1,7 @@
+import { myPow } from '../../src/math/powx-n'
+
+test('Pow(x, n)', () => {
+  expect(myPow(2.00000, 10)).toBe('1024.00000')
+  expect(myPow(2.10000, 3)).toBe('9.26100')
+  expect(myPow(2.00000, -2)).toBe('0.25000')
+})