Leetcode -- Maximal Rectangle

https://oj.leetcode.com/problems/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

 public int maximalRectangle(char[][] matrix)

Leetcode 把这一题放在Largest Rectangle这一题后面其实就是这一题最大的提示了。

这一题的做法如果没有思路的话是很难的，基本上你很难在不expensive的情况下画清楚一个范围是否矩形。但上一题就给了一个合适的解法。

实际上这一题做法在懂的上一题之后就变得很简单了。其实就是从第一层扫到最后一层，不停构建供给上一题用的histogram，然后根据上一题的解法一行行的解就可以了。

给出代码如下：

    public int maximalRectangle(char[][] matrix) {        if(matrix.length == 0 || matrix[0].length == 0)            return 0;        int[] cur_histogram = new int[matrix[0].length];        int res = 0;        for(int i = 0; i < matrix.length; i++){            for(int j = 0; j < matrix[0].length; j++){                if(matrix[i][j] == '0')                    cur_histogram[j] = 0;                else                    cur_histogram[j]++;            }            res = Math.max(res, maxRectinHistogram(cur_histogram));        }        return res;    }        public int maxRectinHistogram(int[] height){        Stack<Integer> index_stack = new Stack<Integer>();        int res = 0;        for(int i = 0; i < height.length; i++){            if(index_stack.isEmpty() || height[i] >= height[index_stack.peek()])                index_stack.push(i);            else{                while(!index_stack.isEmpty() && height[index_stack.peek()] > height[i]){                    int cur_bar = index_stack.pop();                    int prev_bar = index_stack.isEmpty() ? -1 : index_stack.peek();                    res = Math.max(res, height[cur_bar] * (i - prev_bar - 1));                }                index_stack.push(i);            }        }        while(!index_stack.isEmpty()){            int cur_bar = index_stack.pop();            int prev_bar = index_stack.isEmpty() ? -1 : index_stack.peek();            res = Math.max(res, height[cur_bar] * (height.length - prev_bar - 1));        }        return res;    }