510. Maximal Rectangle

Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

RETURN 6

class Solution {
    public int maximalRectangle(char[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        int rows = matrix.length;
        int cols = matrix[0].length;
        int[][] height = new int[rows][cols];
        for (int j = 0; j < cols; j++) {
            height[0][j] = (matrix[0][j] == '1') ? 1 : 0;
        }
        for (int i = 1 ; i < rows ;i++) {
            for (int j = 0; j < cols; j++) {
                if (matrix[i][j] == '1') {
                    height[i][j] = height[i-1][j] + 1;
                }else{
                    height[i][j]  = 0;
                }

            }
        }
        int ans = 0;
        for (int i = 0; i < rows; i++) {
            int area = maxAreainHisto(height[i]);
            ans = Math.max(ans, area);
        }
        return ans;
    }

    public int maxAreainHisto(int[] height) {
        Stack<Integer> stack = new Stack<>();
        int area = 0;
        for (int i = 0; i <= height.length; i++) {
            if (i <height.length ) {
                 System.out.println(height[i]);
            }

            int curt = (i == height.length) ? -1 : height[i];
            while(!stack.isEmpty() && curt <= height[stack.peek()] ) {
                int h = height[stack.pop()];
                int w = stack.isEmpty() ? i : i - stack.peek() - 1;
                area = Math.max(area, w * h);
            }
            stack.push(i);
        }
        return area;
    }


}