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

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Return 6.

思路：這道題單獨拿出來絕對是一道難題，但是有了上一題Largest Rectangle in Histogram的基礎，再思考這個問題就變的簡單了許多，可以考慮遍歷矩陣的每一行，把每一行為1的部分當做直方圖的高度，求得當前行以上構成的各個直方圖的最大矩陣面積即可。

class Solution {
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        int res=0;
        vector<int>height;
        for(int i=0;i<matrix.size();i++){
            height.resize(matrix[i].size());
            for(int j=0;j<matrix[i].size();j++){
                height[j]=matrix[i][j]=='0'
 
?0:(1+height[j]);
            }
            res=max(res,largestRectangleArea(height));
        }
        return res;
    }
    int largestRectangleArea(vector<int>&heights){
        int w,h,area=0;
        stack<int>index;
        heights.push_back(0);
        for(int i=0;i<heights.size();i++){
            while
 
(index.size()&&heights[index.top()]>heights[i]){
                h=heights[index.top()];
                index.pop();
                if(index.size())
                    w=i-1-index.top();
                else
                    w=i;
                area=max(area,w*h);
            }
            index.push(i);
        }
        return area;
    }
};