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

方法：應用直方圖最大矩形面積的方法。

public class Solution {
    private Stack<Integer> stack = new Stack<>();
    private int histogram(int[] nums) {
        stack.clear();
        int max = 0;
        for(int j=0; j<nums.length; j++) {
            if (stack.isEmpty() || nums[stack.peek()] <= nums[j]) {
                stack.push(j);
                continue;
            }
            do {
                int p = stack.pop();
                int s = stack.isEmpty()? nums[p] * j: nums[p] * (j-stack.peek()-1);
                max = Math.max(max, s);
            } while (!stack.isEmpty() && nums[stack.peek()] > nums[j]);
            stack.push(j);
        }
        while (!stack.isEmpty()) {
            int p = stack.pop();
            int s = stack.isEmpty()? nums[p] * nums.length: nums[p] * (nums.length - stack.peek() - 1);
            max = Math.max(max, s);
        }
        return max;
    }
    public int maximalRectangle(char[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;
        int max = 0;
        int n = matrix.length;
        int m = matrix[0].length;
        int[][] counts = new int[n][m];
        for(int j=0; j<m; j++) counts[0][j] = matrix[0][j] == '1'? 1: 0;
        for(int i=1; i<n; i++) {
            for(int j=0; j<m; j++) {
                counts[i][j] = matrix[i][j] == '1'? counts[i-1][j]+1:0;
            }
            
        }
        for(int i=0; i<n; i++) {
            int s = histogram(counts[i]);
            max = Math.max(max, s);
        }
        return max;
    }
}