85. 最大矩形

  这道题直接做难度较大，可转换为圆柱图的最大矩形面积的问题，先把这个矩阵转换为

就是每一行对应一个圆柱形图像，然后分别求解最大值即可。

    long long  max(long long  a, long long  b)
{
    return a > b ? a : b;
}
int Histogram_sMax(const vector<char> &data)
{
    stack<int> Stack;
    long long sMax = 0;
    int tmp = 0;
    for (int i = 0; i < data.size(); i++)
    {
        if (Stack.empty() || Stack.top() <= data[i]-'0')
            Stack.push(data[i] - '0');
        else
        {
            int count = 1;
            while (!Stack.empty() && Stack.top()>data[i] - '0')
            {
                tmp = (count++) * Stack.top();
                sMax = max(sMax, tmp);
                Stack.pop();
            }
            while (count-->0)
            {
                Stack.push(data[i] - '0');
            }
        }
    }
//    cout << "size : " << Stack.size() << endl;
    int count = 1;
    while (!Stack.empty())
    {
        sMax = max(sMax, Stack.top()*count++);
        Stack.pop();
    }

    return sMax;
}
int maximalRectangle(vector<vector<char>>& matrix) {
    if (matrix.empty())
        return 0;
    const int rows = matrix.size();
    const int cols = matrix[0].size();
    for (int i = 1; i < rows; i++)
    {
        for (int j = 0; j < cols; j++)
        {
            if (matrix[i][j] == '1' &&  matrix[i - 1][j] > 0)
                matrix[i][j] += matrix[i - 1][j] - '0';
        }
    }
    int rmax = 0;
    for (int i = 0; i < rows; i++)
    {
        rmax = max(rmax,Histogram_sMax(matrix[i]));
    }
    return rmax;
}