[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[0].empty()) return 0;
        int res = 0, m = matrix.size(), n = matrix[0].size();
        vector<int> h(n + 1, 0);
        for (int i = 0; i < m; ++i) {
            stack<int> s;
            for (int j = 0; j < n + 1; ++j) {
                if (j < n) {
                    if (matrix[i][j]) ++h[j];
                    else h[j] = 0;
                }
                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() - 1)));
                }
                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[0].empty()) return 0;
        int res = 0, m = matrix.size(), n = matrix[0].size();
        vector<int> h(n, 0), left(n, 0), right(n, n);
        for (int i = 0; i < m; ++i) {
            int cur_left = 0, cur_right = n;
            for (int j = 0; j < n; ++j) {
                h[j] = matrix[i][j] ? h[j] + 1 : 0;
            }
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j]) left[j] = max(left[j], cur_left);
                else {left[j] = 0; cur_left = j + 1;}
            }
            for (int j = n - 1; j >= 0; --j) {
                if (matrix[i][j]) right[j] = min(right[j], cur_right);
                else {right[j] = n; cur_right = j;}
            }
            for (int j = 0; j < n; ++j) {
                res = max(res, (right[j] - left[j]) * h[j]);
            }
        }
        return res;
    }
};