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：

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
• 0 <= row, cols <= 200
• matrix[i][j] 为 '0' 或 '1'

参考答案

class Solution {
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        int m = matrix.size();
        if (m == 0) {
            return 0;
        }
        int n = matrix[0].size();
        vector<vector<int>> left(m, vector<int>(n, 0));

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == '1') {
                    left[i][j] = (j == 0 ? 0: left[i][j - 1]) + 1;
                }
            }
        }

        int ret = 0;
        for (int j = 0; j < n; j++) { // 对于每一列，使用基于柱状图的方法
            vector<int> up(m, 0), down(m, 0);

            stack<int> stk;
            for (int i = 0; i < m; i++) {
                while (!stk.empty() && left[stk.top()][j] >= left[i][j]) {
                    stk.pop();
                }
                up[i] = stk.empty() ? -1 : stk.top();
                stk.push(i);
            }
            stk = stack<int>();
            for (int i = m - 1; i >= 0; i--) {
                while (!stk.empty() && left[stk.top()][j] >= left[i][j]) {
                    stk.pop();
                }
                down[i] = stk.empty() ? m : stk.top();
                stk.push(i);
            }

            for (int i = 0; i < m; i++) {
                int height = down[i] - up[i] - 1;
                int area = height * left[i][j];
                ret = max(ret, area);
            }
        }
        return ret;
    }
};