算法描述：

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

Example:

Input:[  ["1","0","1","0","0"],  ["1","0","1","1","1"],  ["1","1","1","1","1"],  ["1","0","0","1","0"]]Output: 6

解题思路：利用直方图最大面积求解，每一行构建一个直方图。当元素为1时，直方图对应元素加1，否则置为0；

int maximalRectangle(vector
     
      
       >& matrix) {        if(matrix.size()==0 || matrix[0].size()==0) return 0;        int m = matrix.size();        int n= matrix[0].size();        int maxArea = 0;        vector
       
         hist(n,0);        for(int i=0; i < m; i++){            for(int j =0; j < n; j++){                if( matrix[i][j] == '1')                    hist[j] += 1;                else                    hist[j] = 0;            }                        maxArea = max(maxArea,maxAreaOfHist(hist));        }        return maxArea;           }    int maxAreaOfHist(vector
        
         & heights){        if(heights.size()==0) return 0;        stack
         
           stk;        int maxArea = 0;        int i=0;        while(i < heights.size()){            if(stk.empty() || heights[i] >= heights[stk.top()]){                stk.push(i);                i++;            }else{                int index = stk.top();                stk.pop();                maxArea = max(maxArea, heights[index]*(stk.empty()? i: i-stk.top()-1));            }        }        while(!stk.empty()){            int index = stk.top();            stk.pop();            maxArea = max(maxArea, heights[index]*(stk.empty()?i:i-stk.top()-1));        }        return maxArea;    }