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; } };
本文转自博客园Grandyang的博客，原文链接：最大矩形[LintCode] Maximal Rectangle ，如需转载请自行联系原博主。
