题目：
Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest rectangle containing all ones and return its area.
题解：
这道题可以应用之前解过的Largetst Rectangle in Histogram一题辅助解决。解决方法是：
按照每一行计算列中有1的个数，作为高度，当遇见0时，这一列高度就为0。然后对每一行计算 Largetst Rectangle in Histogram，最后得到的就是结果。
代码如下：
1 public int maximalRectangle(char[][] matrix) {
2 if(matrix==null || matrix.length==0 || matrix[0].length==0)
3 return 0;
4 int m = matrix.length;
5 int n = matrix[0].length;
6 int max = 0;
7 int[] height = new int[n];//对每一列构造数组
8 for(int i=0;i<m;i++){
9 for(int j=0;j<n;j++){
10 if(matrix[i][j] == ‘0‘)//如果遇见0，这一列的高度就为0了
11 height[j] = 0;
12 else
13 height[j] += 1;
14 }
15 max = Math.max(largestRectangleArea(height),max);
16 }
17 return max;
18 }
19
20 public int largestRectangleArea(int[] height) {
21 Stack<Integer> stack = new Stack<Integer>();
22 int i = 0;
23 int maxArea = 0;
24 int[] h = new int[height.length + 1];
25 h = Arrays.copyOf(height, height.length + 1);
26 while(i < h.length){
27 if(stack.isEmpty() || h[stack.peek()] <= h[i]){
28 stack.push(i);
29 i++;
30 }else {
31 int t = stack.pop();
32 int square = -1;
33 if(stack.isEmpty())
34 square = h[t]*i;
35 else{
36 int x = i-stack.peek()-1;
37 square = h[t]*x;
38 }
39 maxArea = Math.max(maxArea, square);
40 }
41 }
42 return maxArea;
43 }
Maximal Rectangle leetcode java