diff --git a/README.md b/README.md
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@@ -85,6 +85,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 82 |[Remove Duplicates from Sorted List II](https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/82.%20Remove%20Duplicates%20from%20Sorted%20List%20II.md)|Medium|  |
 | 83 |[Remove Duplicates from Sorted List](https://leetcode.com/problems/remove-duplicates-from-sorted-list)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/83.%20Remove%20Duplicates%20from%20Sorted%20List.md)|Easy|  |
 | 84 |[Largest Rectangle in Histogram](https://leetcode.com/problems/largest-rectangle-in-histogram/)|[C++](solutions/84.%20Largest%20Rectangle%20in%20Histogram.md)|Hard|  |
+| 85 |[Maximal Rectangle](https://leetcode.com/problems/maximal-rectangle/)|[C++](solutions/85.%20Maximal%20Rectangle.md)|Hard|  |
 | 86 |[Partition List](https://leetcode.com/problems/partition-list/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/86.%20Partition%20List.md)|Medium|  |
 | 88 |[Merge Sorted Array](https://leetcode.com/problems/merge-sorted-array)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/88.%20Merge%20Sorted%20Array.md)|Easy|  |
 | 89 |[Gray Code](https://leetcode.com/problems/gray-code/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/89.%20Gray%20Code.md)|Medium|  |
diff --git a/solutions/85. Maximal Rectangle.md b/solutions/85. Maximal Rectangle.md
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+# [85. Maximal Rectangle](https://leetcode.com/problems/maximal-rectangle/)
+
+# 思路
+
+## 思路一
+
+对于二维矩阵的第 i 行，这一行及上面的部分可以看成是一个直方图，而在直方图里求最大矩形即[84. Largest Rectangle in Histogram](https://leetcode.com/problems/largest-rectangle-in-histogram/)，我们可以用单调栈在线性复杂度内求解84题，详见[84题解](84.%20Largest%20Rectangle%20in%20Histogram.md)。
+
+所以对于此题，我们只需要开辟一个表示直方图的一维数组`heights`，然后对于每一行，建立直方图，然后再用84题思路（代码中采用的是[84题解思路二](84.%20Largest%20Rectangle%20in%20Histogram.md)）进行求解。我们可以根据上一行的直方图`heights`在线性时间复杂度内建立当前行的直方图`heights`：
+* 如果`matrix[i][j] == '0'`，则`heights[j] = 0`;
+* 否则，`heights[j] += 1`;
+
+时间复杂度O(MN)，空间复杂度O(N)；其中M和N分别为高和宽。
+
+## 思路二
+
+对于每个点，我们会通过以下步骤计算出一个矩形：
+
+* 不断向上方遍历，直到遇到“0”，以此找到矩形的高度。
+* 向左右两边扩展，直到无法容纳这个高度。
+
+例如，找到黄色点对应的矩形（[图片来源](https://leetcode-cn.com/problems/maximal-rectangle/solution/zui-da-ju-xing-by-leetcode/)）：
+
+<div align=center>
+<img width="500" src="img/85.png" alt="85img"/>
+</div>
+
+如果知道了高height和左右边界left、right，这样就可以计算出面积等于`height * (right - left + 1)`。
+
+求height的方法和思路一是一样的，下面讨论求left和right。
+
+设第`i-1`行的left和right已经求出来了分别为`old_left`和`old_right`，当前点为`matrix[i][j]`为`"1"`，
+1. 求left
+设`left_zero`为`matrix[i][j]`向左看第一个0的列坐标，那么易知`new_left[j] = max(old_left[j], left_zero + 1)`；
+
+2. 求right
+类似的，设`right_zero`为`matrix[i][j]`向右看第一个0的列坐标，那么易知`new_right[j] = min(old_right[j], right_zero - 1)`。
+
+所以我们可以开辟一个一维数组`left`和`right`，然后对于每一行：从左往后更新`left`，从右往左更新`right`（就是动态规划的滚动数组思路）。
+
+时间复杂度O(MN)，空间复杂度O(N)；其中M和N分别为高和宽。
+
+
+# C++
+## 思路一
+``` C++
+class Solution {
+public:
+    int maximalRectangle(vector<vector<char>>& matrix) {
+        int m = matrix.size(); 
+        if(!m) return 0;
+        int n = matrix[0].size();
+        
+        int max_res = 0, res = 0;
+        vector<int>heights(n + 1, 0); // 注意大小是n+1, 见84题解
+        for(int i = 0; i < m; i++){
+            for(int j = 0; j < n; j++){
+                if(matrix[i][j] == '0') heights[j] = 0;
+                else heights[j] += 1;
+            }
+            
+            // 下面几行基本就是84题解
+            stack<int>ascend_stk; res = 0;
+            for(int k = 0; k <= n; k++){
+                while(!ascend_stk.empty() && heights[ascend_stk.top()] >= heights[k]){
+                    int height = heights[ascend_stk.top()]; ascend_stk.pop();
+                    int width = ascend_stk.empty() ? k : k - 1 - ascend_stk.top();
+                    res = max(res, height * width);
+                }
+                ascend_stk.push(k);
+            }
+            
+            max_res = max(res, max_res);
+        }
+        return max_res;
+    }
+};
+```
+
+## 思路二
+``` C++
+class Solution {
+public:
+    int maximalRectangle(vector<vector<char>>& matrix) {
+        int m = matrix.size(); 
+        if(!m) return 0;
+        int n = matrix[0].size();
+        
+        int res = 0, left_zero, right_zero;
+        vector<int>heights(n, 0), left(n, -1), right(n, n);
+        for(int i = 0; i < m; i++){
+            // 更新高度
+            for(int j = 0; j < n; j++)
+                if(matrix[i][j] == '0') heights[j] = 0;
+                else heights[j] += 1;
+            
+            // 更新左边界(可以和上面更新高度合在一个循环内)
+            left_zero = -1;
+            for(int j = 0; j < n; j++){
+                if(matrix[i][j] == '0') {
+                    left_zero = j;
+                    left[j] = -1;
+                }
+                else left[j] = max(left_zero + 1, left[j]);
+            }
+            
+            // 更新右边界
+            right_zero = n;
+            for(int j = n-1; j >= 0; j--){ // 逆向循环!!!
+                if(matrix[i][j] == '0') {
+                    right_zero = j;
+                    right[j] = n;
+                }
+                else right[j] = min(right_zero - 1, right[j]);
+                res = max(res, heights[j] * (right[j] - left[j] + 1));
+            }
+        }
+        
+        return res;
+    }
+};
+```
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