diff --git a/README.en.md b/README.en.md
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--- a/README.en.md
+++ b/README.en.md
@@ -195,6 +195,8 @@ The data structures mainly includes:
 - [1031.maximum-sum-of-two-non-overlapping-subarrays](./problems/1031.maximum-sum-of-two-non-overlapping-subarrays.md) 🆕
 
 #### Hard
+
+- [0004.median-of-two-sorted-array](./problems/4.median-of-two-sorted-array.md) 🆕
 - [0023.merge-k-sorted-lists](./problems/23.merge-k-sorted-lists.md)
 - [0032.longest-valid-parentheses](./problems/32.longest-valid-parentheses.md) 🆕
 - [0042.trapping-rain-water](./problems/42.trapping-rain-water.md)
diff --git a/README.md b/README.md
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--- a/README.md
+++ b/README.md
@@ -215,6 +215,7 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 
 #### 困难难度
 
+- [0004.median-of-two-sorted-array](./problems/4.median-of-two-sorted-array.md) 🆕
 - [0023.merge-k-sorted-lists](./problems/23.merge-k-sorted-lists.md)
 - [0032.longest-valid-parentheses](./problems/32.longest-valid-parentheses.md) 🆕
 - [0042.trapping-rain-water](./problems/42.trapping-rain-water.md)
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+## 题目地址
+https://leetcode.com/problems/median-of-two-sorted-arrays/
+
+## 题目描述
+```
+There are two sorted arrays nums1 and nums2 of size m and n respectively.
+
+Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
+
+You may assume nums1 and nums2 cannot be both empty.
+
+Example 1:
+
+nums1 = [1, 3]
+nums2 = [2]
+
+The median is 2.0
+Example 2:
+
+nums1 = [1, 2]
+nums2 = [3, 4]
+
+The median is (2 + 3)/2 = 2.5
+```
+
+## 思路
+首先了解一下Median的概念，一个数组中median就是把数组分成左右等分的中位数。
+
+如下图：
+![median](../assets/problems/4.median-of-two-sorted-array-1.jpg)
+
+这道题，很容易想到暴力解法，时间复杂度和空间复杂度都是`O(m+n)`, 不符合题中给出`O(log(m+n))`时间复杂度的要求。
+我们可以从简单的解法入手，试了一下，暴力解法也是可以被Leetcode Accept的. 分析中会给出两种解法，暴力求解和二分解法。
+
+#### 解法一 - 暴力 （Brute Force）
+暴力解主要是要merge两个排序的数组`（A，B）`成一个排序的数组。
+
+用两个`pointer（i，j）`，`i` 从数组`A`起始位置开始，即`i=0`开始，`j` 从数组`B`起始位置， 即`j=0`开始. 
+一一比较 `A[i] 和 B[j]`, 
+1. 如果`A[i] <= B[j]`, 则把`A[i]` 放入新的数组中，i往后移一位，即 `i+1`.
+2. 如果`A[i] > B[j]`, 则把`B[j]` 放入新的数组中，j往后移一位，即 `j+1`.
+3. 重复步骤#1 和 #2，直到`i`移到`A`最后，或者`j`移到`B`最后。
+4. 如果`j`移动到`B`数组最后，那么直接把剩下的所有`A`依次放入新的数组中. 
+5. 如果`i`移动到`A`数组最后，那么直接把剩下的所有`B`依次放入新的数组中.
+
+Merge的过程如下图。
+![merge two sorted array](../assets/problems/4.median-of-two-sorted-array-2.jpg)
+
+
+*时间复杂度： `O(m+n) - m is length of A, n is length of B`*
+
+*空间复杂度： `O(m+n)`*
+
+#### 解法二 - 二分查找 （Binary Search）
+由于题中给出的数组都是排好序的，在排好序的数组中查找很容易想到可以用二分查找（Binary Search), 这里对数组长度小的做二分，
+保证数组A 和 数组B 做partition 之后
+
+`len(Aleft)+len(Bleft)=(m+n+1)/2 - m是数组A的长度， n是数组B的长度`
+
+对数组A的做partition的位置是区间`[0,m]`
+
+如图：
+![partition A,B](../assets/problems/4.median-of-two-sorted-array-3.png)
+
+下图给出几种不同情况的例子（注意但左边或者右边没有元素的时候，左边用`INF_MIN`，右边用`INF_MAX`表示左右的元素：
+![median examples](../assets/problems/4.median-of-two-sorted-array-5.png)
+
+下图给出具体做的partition 解题的例子步骤，
+![median partition example](../assets/problems/4.median-of-two-sorted-array-4.png)
+
+*时间复杂度： `O(log(min(m, n)) - m is length of A, n is length of B`*
+
+*空间复杂度： `O(1)` - 这里没有用额外的空间*
+
+## 关键点分析
+1. 暴力求解，在线性时间内merge两个排好序的数组成一个数组。
+2. 二分查找，关键点在于
+  - 要partition两个排好序的数组成左右两等份，partition需要满足`len(Aleft)+len(Bleft)=(m+n+1)/2 - m是数组A的长度， n是数组B的长度`
+ 
+  - 并且partition后 A左边最大(`maxLeftA`), A右边最小（`minRightA`), B左边最大（`maxLeftB`), B右边最小（`minRightB`) 满足
+`(maxLeftA <= minRightB && maxLeftB <= minRightA)`
+
+有了这两个条件，那么median就在这四个数中，根据奇数或者是偶数，
+```
+奇数：
+median = max(maxLeftA, maxLeftB)
+偶数：
+median = (max(maxLeftA, maxLeftB) + min(minRightA, minRightB)) / 2
+```
+
+## 代码（Java code）
+*解法一 - 暴力解法（Brute force）*
+```java
+class MedianTwoSortedArrayBruteForce {
+    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
+      int[] newArr = mergeTwoSortedArray(nums1, nums2);
+      int n = newArr.length;
+      if (n % 2 == 0) {
+        // even
+        return (double) (newArr[n / 2] + newArr[n / 2 - 1]) / 2;
+      } else {
+        // odd
+        return (double) newArr[n / 2];
+      }
+    }
+    private int[] mergeTwoSortedArray(int[] nums1, int[] nums2) {
+      int m = nums1.length;
+      int n = nums2.length;
+      int[] res = new int[m + n];
+      int i = 0;
+      int j = 0;
+      int idx = 0;
+      while (i < m && j < n) {
+        if (nums1[i] <= nums2[j]) {
+          res[idx++] = nums1[i++];
+        } else {
+          res[idx++] = nums2[j++];
+        }
+      }
+      while (i < m) {
+        res[idx++] = nums1[i++];
+      }
+      while (j < n) {
+        res[idx++] = nums2[j++];
+      }
+      return res;
+    }
+}
+```
+*解法二 - 二分查找（Binary Search*
+```java
+class MedianSortedTwoArrayBinarySearch {
+  public static double findMedianSortedArraysBinarySearch(int[] nums1, int[] nums2) {
+     // do binary search for shorter length array, make sure time complexity log(min(m,n)).
+     if (nums1.length > nums2.length) {
+        return findMedianSortedArraysBinarySearch(nums2, nums1);
+      }
+      int m = nums1.length;
+      int n = nums2.length;
+      int lo = 0;
+      int hi = m;
+      while (lo <= hi) {
+        // partition A position i
+        int i = lo + (hi - lo) / 2;
+        // partition B position j
+        int j = (m + n + 1) / 2 - i;
+        
+        int maxLeftA = i == 0 ? Integer.MIN_VALUE : nums1[i - 1];
+        int minRightA = i == m ? Integer.MAX_VALUE : nums1[i];
+  
+        int maxLeftB = j == 0 ? Integer.MIN_VALUE : nums2[j - 1];
+        int minRightB = j == n ? Integer.MAX_VALUE : nums2[j];
+  
+        if (maxLeftA <= minRightB && maxLeftB <= minRightA) {
+          // total length is even
+          if ((m + n) % 2 == 0) {
+            return (double) (Math.max(maxLeftA, maxLeftB) + Math.min(minRightA, minRightB)) / 2;
+          } else {
+            // total length is odd
+            return (double) Math.max(maxLeftA, maxLeftB);
+          }
+        } else if (maxLeftA > minRightB) {
+          // binary search left half
+          hi = i - 1;
+        } else {
+          // binary search right half
+          lo = i + 1;
+        }
+      }
+      return 0.0;
+    }
+}
+```