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.

两个矩阵，一个存行最大连续1，一个存列最大连续1

然后利用这两个矩阵计算，行矩阵值代表最大长度；找到列最大矩阵，在最大长度范围内找打最小高度，相乘得到面积

class Solution {public:    int maximalRectangle(vector<vector<char> > &matrix) {        if( matrix.size() == 0)            return 0;        const int m = matrix.size();        const int n = matrix[0].size();        vector< vector< int> > row( m + 1, vector< int>( n + 1, 0));        vector< vector< int> > col( m + 1, vector< int>( n + 1, 0));        vector< vector< int> > rectangle( m + 1, vector< int>( n + 1, 0));                for( int i = 1; i <= m; ++i){            for( int j = 1; j <= n; ++j){                if( matrix[i-1][j-1] == '1'){                    row[i][j] = row[i][j-1] + 1;                    col[i][j] = col[i-1][j] + 1;                }                else{                    row[i][j] = 0;                    col[i][j] = 0;                }            }        }        for( int i = 1; i <= m; ++i){            for( int j = 1; j <= n; ++j){                int height = col[i][j];                int length = row[i][j];                int min_height = height;                int area = 0;                for( int k = 1; k <= length; ++k){                    min_height = min( min_height, col[i][j-k+1]);//向前回退                    area = max( area, min_height * k);//这里乘以k                }                rectangle[i][j] = area;            }        }        int area = 0;        for( int i = 0; i < rectangle.size(); ++i)            for( int j = 0; j < rectangle[i].size(); ++j)                area = area > rectangle[i][j] ? area : rectangle[i][j];                return area;    }};