# 85. 最大矩形 (Hard)

## 问题描述
[85. 最大矩形 (Hard)](https://leetcode.cn/problems/maximal-rectangle/)
给定一个仅包含 `0` 和 `1` 、大小为 `rows x cols` 的二维二进
制矩阵，找出只包含 `1` 的最大矩形，并返回其面积。

**示例 1：**

![](https://assets.leetcode.com/uploads/2020/09/14/maximal.j
pg)

```
输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],
["1","1","1","1","1"],["1","0","0","1","0"]]
输出：6
解释：最大矩形如上图所示。

```

**示例 2：**

```
输入：matrix = []
输出：0

```

**示例 3：**

```
输入：matrix = [["0"]]
输出：0

```

**示例 4：**

```
输入：matrix = [["1"]]
输出：1

```

**示例 5：**

```
输入：matrix = [["0","0"]]
输出：0

```

**提示：**

- `rows == matrix.length`
- `cols == matrix[0].length`
- `1 <= row, cols <= 200`
- `matrix[i][j]` 为 `'0'` 或 `'1'`

## 解题思路
其实本题就相当于是 [84. 柱状图中最大的矩形 (Hard)](https://blog.zwyyy456.tech/zh/posts/leet/84.largest-rectangle-in-histogram) 套了一层皮，想到了这一点之后就不难了。

## 代码
```cpp
class Solution {
  public:
    int Count(int row, int n, vector<int> &vec, vector<vector<char>> &matrix) {
        for (int i = 0; i < n; ++i) {
            vec[i] = matrix[row][i] == '0' ? 0 : vec[i] + 1;
        }
        // 单调递增栈
        stack<int> stk;
        int res = 0;
        for (int i = 0; i < n; ++i) {
            while (!stk.empty() && vec[i] < vec[stk.top()]) {
                int idx = stk.top(), h = vec[idx];
                stk.pop();
                int rec = stk.empty() ? h * i : h * (i - stk.top() - 1);
                res = max(rec, res);
            }
            stk.push(i);
        }
        while (!stk.empty()) {
            int idx = stk.top();
            int h = vec[idx];
            stk.pop();
            int rec = stk.empty() ? h * n : h * (n - stk.top() - 1);
            res = max(res, rec);
        }
        return res;
    }
    int maximalRectangle(vector<vector<char>> &matrix) {
        // 二维前缀和？
        // 转化为 84 题，即可用单调栈解决
        int m = matrix.size(), n = matrix[0].size();
        int res = 0;
        vector<int> vec(n);
        for (int i = 0; i < m; ++i) {
            res = max(res, Count(i, n, vec, matrix));
        }
        return res;
    }
};
```