java基础编程题（2）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxDepth(TreeNode root) {
                return root == null ? 0 : Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
    }
}

class Solution {
    public int maximalRectangle(char[][] matrix) {
    if (matrix.length == 0) {
        return 0;
    }
    //保存以当前数字结尾的连续 1 的个数
    int[][] width = new int[matrix.length][matrix[0].length];
    int maxArea = 0;
    //遍历每一行
    for (int row = 0; row < matrix.length; row++) {
        for (int col = 0; col < matrix[0].length; col++) {
            //更新 width
            if (matrix[row][col] == ‘1‘) {
                if (col == 0) {
                    width[row][col] = 1;
                } else {
                    width[row][col] = width[row][col - 1] + 1;
                }
            } else {
                width[row][col] = 0;
            }
            //记录所有行中最小的数
            int minWidth = width[row][col];
            //向上扩展行
            for (int up_row = row; up_row >= 0; up_row--) {
                int height = row - up_row + 1;
                //找最小的数作为矩阵的宽
                minWidth = Math.min(minWidth, width[up_row][col]);
                //更新面积
                maxArea = Math.max(maxArea, height * minWidth);
            }
        }
    }
    return maxArea;
    }
}