LeetCode_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.

分析： 对于每一列从右到左看成一个直方图，每个直方图计算最大面积的时间复杂度为O(n) 所以总的时间复杂度是O（n²）

class Solution {
public:
    int maxArea(vector<int> &m)
    {
        int size = m.size();
        stack<int> s;
        int area, maxArea = 0;
        for(int i = 0; i< size; ++i)
        {
            if(s.empty() || m[i] >= m[s.top()] )
            {
                s.push(i);
                continue;
            }            
            int tp = s.top();
            s.pop();
            area = m[tp] * (s.empty() ? i : i -s.top() -1);
            maxArea = maxArea > area ? maxArea : area;
            --i;
        }
        
        while(!s.empty()){
            int    tp = s.top();
            s.pop();
            area = m[tp] * (s.empty() ? size: size - s.top() -1);
            maxArea = maxArea > area ? maxArea : area;
        }
        return maxArea;
    }
    int maximalRectangle(vector<vector<char> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int row = matrix.size();
        if(row < 1) return 0;
        int column = matrix[0].size();
        if(column <1) return 0;
        
        int area , res = 0;
        
        vector<int> m(row,0);
        for(int i = 0; i< column; ++i)
        {
            for(int j = 0; j< row; ++j)
               if(matrix[j][i] == '0')
                        m[j] = 0;
                else
                        m[j]++;
        
            area = maxArea(m);
            res = res > area ? res : area;
        }
        return res;
    }
};