给定一个仅包含 0 和 1 的二维二进制矩阵，找出只包含 1 的最大矩形，并返回其面积。

示例:

输入:
[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
输出: 6

和leetcode84没什么区别，就是一行一行更新出“高度”，一行一行的跑一遍单调栈即可。

class Solution {public int leetcode84(int[] heights) {Stack < Integer > stack = new Stack < > ();stack.push(-1);int maxarea = 0;for (int i = 0; i < heights.length; ++i) {while (stack.peek() != -1 && heights[stack.peek()] >= heights[i])maxarea = Math.max(maxarea, heights[stack.pop()] * (i - stack.peek() - 1));stack.push(i);}while (stack.peek() != -1)maxarea = Math.max(maxarea, heights[stack.pop()] * (heights.length - stack.peek() -1));return maxarea;}public int maximalRectangle(char[][] matrix) {if (matrix.length == 0) return 0;int maxarea = 0;int[] dp = new int[matrix[0].length];for(int i = 0; i < matrix.length; i++) {for(int j = 0; j < matrix[0].length; j++) {dp[j] = matrix[i][j] == '1' ? dp[j] + 1 : 0;}maxarea = Math.max(maxarea, leetcode84(dp));} return maxarea;}
}