給定一個僅包含0 和 1 、大小為 rows x cols 的二維二進位制矩陣，找出只包含 1 的最大矩形，並返回其面積。



示例 1：


輸入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
輸出：6
解釋：最大矩形如上圖所示。
示例 2：

輸入：matrix = []
輸出：0
示例 3：

輸入：matrix = [["0"]]
輸出：0
示例 4：

輸入：matrix = [["1"]]
輸出：1
示例 5：

輸入：matrix = [["0","0"]]
輸出：
 
0

來源：力扣（LeetCode）
連結：https://leetcode-cn.com/problems/maximal-rectangle
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class Solution {
    public int maximalRectangle(char[][] matrix) {
        int m = matrix.length;
        if (m == 0) {
            return 0;
        }
        int n = matrix[0].length;
        int
 
[][] left = new int[m][n];

        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++) { // 對於每一列，使用基於柱狀圖的方法
            int[] up = new int[m];
            int[] down = new int[m];

            Deque<Integer> stack = new LinkedList<Integer>();
            for (int i = 0; i < m; i++) {
                while (!stack.isEmpty() && left[stack.peek()][j] >= left[i][j]) {
                    stack.pop();
                }
                up[i] = stack.isEmpty() ? -1 : stack.peek();
                stack.push(i);
            }
            stack.clear();
            for (int i = m - 1; i >= 0; i--) {
                while (!stack.isEmpty() && left[stack.peek()][j] >= left[i][j]) {
                    stack.pop();
                }
                down[i] = stack.isEmpty() ? m : stack.peek();
                stack.push(i);
            }

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