Maximal Rectangle

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

用动态规划去做，用grid[i]记录matrix[][i]向下有多少个连续的1，就转化为求n个直方图最大面积的问题，求出最大值，计算每行的grid数组时间复杂度为O(n),再求该数组直方图最大面积的时间复杂度为O(n)。共有n行，则总体时间复杂度为O(n^2)。

求矩阵直方图的做法参见Trapping Rain Water。

本题代码如下：

public int maximalRectangle(char[][] matrix) {
        if(matrix==null||matrix.length==0){
            return 0;
        }
        int[] grid= new int[matrix[0].length];
        int max = 0;
        for(int i=matrix.length-1;i>=0;i--){
            for(int j=0;j<matrix[0].length;j++){
                if(i==matrix.length-1){
                    grid[j] = (matrix[i][j]=='1'?1:0);
                }else{
                    grid[j] = (matrix[i][j]=='1'?1+grid[j]:0);
                }
            }
            Stack<Integer> pos = new Stack<>();
            Stack<Integer> high = new Stack<>();
            int m = 0;
            pos.add(0);
            high.add(grid[0]);
            for(int k=1;k<grid.length;k++){
                int p = k;
                while(!high.empty()&&grid[k]<high.peek()){
                    p = pos.pop();
                    int area = (k-p)*high.pop();
                    m = m>area?m:area;
                }
                pos.add(p);
                high.add(grid[k]);
            }
            while(!high.empty()){
                int area = (grid.length-pos.pop())*high.pop();
                m = m>area?m:area;
            }
            max=max>m?max:m;
        }
        return max;
    }