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) {
        int m = matrix.length;
        if(matrix == null || m == 0) 
            return 0;
        int n = matrix[0].length;
        int maxA = 0;
        int[] right = new int[n], left = new int[n], heights = new int[n];  // rectangle with highest height
        Arrays.fill(right,n);

        for(int i=0; i<m; i++) {
            int curleft = 0, curright = n;
            for(int j=0; j<n; j++) {
                if(matrix[i][j] == '1') 
                    heights[j]++;
                else 
                    heights[j] = 0;
            }

            for(int j=0; j<n; j++) {
                if(matrix[i][j] == '1') 
                    left[j] = Math.max(curleft, left[j]);  // each i has own left[], each will renew
                else { 
                    left[j] = 0; 
                    curleft = j + 1; 
                }     // most next positon to zero 
            }

            for(int j=n-1; j>=0; j--) {
                if(matrix[i][j] == '1') 
                    right[j] = Math.min(curright, right[j]);
                else { 
                    right[j] = n;
                    curright = j; 
                }         // remain at last zero position
            }

            for(int j=0; j<n; j++) {
                maxA = Math.max(maxA, (right[j] - left[j]) * heights[j]);
            }
        }
        return