diff --git a/README.md b/README.md
index 50e2dac..7a3a3fb 100644
--- a/README.md
+++ b/README.md
@@ -390,6 +390,11 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 507. 完美数 <https://leetcode-cn.com/problems/perfect-number/>
   - LintCode 1199. 完美的数 <https://www.lintcode.com/problem/perfect-number/description>
 
+- [寻找两个正序数组的中位数](src/math/median-of-two-sorted-arrays.js)
+
+  - LeetCode 4. 寻找两个正序数组的中位数 <https://leetcode-cn.com/problems/median-of-two-sorted-arrays/>
+  - LintCode 65. 两个排序数组的中位数 <https://www.lintcode.com/problem/median-of-two-sorted-arrays/description>
+
 ## 堆
 
 - [超级丑数](src/stack/super-ugly-number.js)【未完成】
diff --git a/src/math/median-of-two-sorted-arrays.js b/src/math/median-of-two-sorted-arrays.js
new file mode 100644
index 0000000..5e56fb1
--- /dev/null
+++ b/src/math/median-of-two-sorted-arrays.js
@@ -0,0 +1,52 @@
+// 中位数：将一个集合划分为两个长度相等的子集，其中一个子集中的元素总是大于另一个子集中的元素。
+// 参考：https://leetcode-cn.com/problems/median-of-two-sorted-arrays/solution/xun-zhao-liang-ge-you-xu-shu-zu-de-zhong-wei-s-114/
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number}
+ */
+export const findMedianSortedArrays = function (nums1, nums2) {
+  if (nums1.length > nums2.length) return findMedianSortedArrays(nums2, nums1)
+
+  const m = nums1.length; const n = nums2.length
+  let left = 0; let right = m
+  let median1 = 0; let median2 = 0
+
+  while (left <= right) {
+    const i = (left + right) >> 1
+    const j = ((m + n + 1) >> 1) - i
+
+    const numsIm1 = (i === 0 ? Number.MIN_SAFE_INTEGER : nums1[i - 1]) // nums1[i-1]
+    const numsI = (i === m ? Number.MAX_SAFE_INTEGER : nums1[i]) // nums1[i]
+    const numsJm1 = (j === 0 ? Number.MIN_SAFE_INTEGER : nums2[j - 1]) // nums2[j-1]
+    const numsJ = (j === n ? Number.MAX_SAFE_INTEGER : nums2[j]) // nums2[j]
+
+    if (numsIm1 <= numsJ) {
+      median1 = Math.max(numsIm1, numsJm1)
+      median2 = Math.min(numsI, numsJ)
+      left = i + 1
+    } else {
+      right = i - 1
+    }
+  }
+
+  return (m + n) % 2 === 0 ? (median1 + median2) / 2.0 : median1
+}
+
+// 常规解法，不满足复杂度
+// /**
+//  * @param {number[]} nums1
+//  * @param {number[]} nums2
+//  * @return {number}
+//  */
+// var findMedianSortedArrays = function(nums1, nums2) {
+//   const newArr = nums1.concat(nums2).sort((a, b) => a - b)
+//   const length = newArr.length
+
+//   if (length % 2 === 0) {
+//     const half = length / 2
+//     return (newArr[half - 1] + newArr[half]) / 2
+//   } else {
+//     return newArr[(length - 1) / 2]
+//   }
+// };
diff --git a/test/math/median-of-two-sorted-arrays.test.js b/test/math/median-of-two-sorted-arrays.test.js
new file mode 100644
index 0000000..485bfe8
--- /dev/null
+++ b/test/math/median-of-two-sorted-arrays.test.js
@@ -0,0 +1,7 @@
+import { findMedianSortedArrays } from '../../src/math/median-of-two-sorted-arrays'
+
+test('寻找两个正序数组的中位数', () => {
+  expect(findMedianSortedArrays([], [1])).toBe(1.0)
+  expect(findMedianSortedArrays([1, 3], [2])).toBe(2.0)
+  expect(findMedianSortedArrays([1, 2], [3, 4])).toBe(2.5)
+})