[LintCode] Maximal Rectangle 最大矩形

Given a 2D boolean matrix filled with False and True, find the largest rectangle containing all True and return its area.

Example
Given a matrix:

[
[1, 1, 0, 0, 1],
[0, 1, 0, 0, 1],
[0, 0, 1, 1, 1],
[0, 0, 1, 1, 1],
[0, 0, 0, 0, 1]
]
return 6.

LeetCode上的原题，请参见我之前的博客Maximal Rectangle。

解法一：

class Solution {
public:
    /**
     * @param matrix a boolean 2D matrix
     * @return an integer
     */
    int maximalRectangle(vector<vector<bool> > &matrix) {
        if (matrix.empty() || matrix[].empty()) return ;
        int res = , m = matrix.size(), n = matrix[].size();
        vector<int> h(n + , );
        for (int i = ; i < m; ++i) {
            stack<int> s;
            for (int j = ; j < n + ; ++j) {
                if (j < n) {
                    if (matrix[i][j]) ++h[j];
                    else h[j] = ;
                }
                while (!s.empty() && h[s.top()] >= h[j]) {
                    int cur = s.top(); s.pop();
                    res = max(res, h[cur] * (s.empty() ? j : (j - s.top() - )));
                }
                s.push(j);
            }
        }
        return res;
    }
};

解法二：

class Solution {
public:
    /**
     * @param matrix a boolean 2D matrix
     * @return an integer
     */
    int maximalRectangle(vector<vector<bool> > &matrix) {
        if (matrix.empty() || matrix[].empty()) return ;
        int res = , m = matrix.size(), n = matrix[].size();
        vector<int> h(n, ), left(n, ), right(n, n);
        for (int i = ; i < m; ++i) {
            int cur_left = , cur_right = n;
            for (int j = ; j < n; ++j) {
                h[j] = matrix[i][j] ? h[j] +  : ;
            }
            for (int j = ; j < n; ++j) {
                if (matrix[i][j]) left[j] = max(left[j], cur_left);
                else {left[j] = ; cur_left = j + ;}
            }
            for (int j = n - ; j >= ; --j) {
                if (matrix[i][j]) right[j] = min(right[j], cur_right);
                else {right[j] = n; cur_right = j;}
            }
            for (int j = ; j < n; ++j) {
                res = max(res, (right[j] - left[j]) * h[j]);
            }
        }
        return res;
    }
};