Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area.
给定一个矩阵中,只有0和1,求出这个矩阵的一个最大的子矩阵,其中只包含1.
例如
01101
11010
01110
11110
11111
00000
其实这个问题可以转化为Largest Rectangle in Histogram,先将上面的矩阵转化为:
01101
12010
03110
14210
25321
00000
然后对每一行求直方图的最大面积。
class Solution { public: int maxArea(vector<int> &line) { if (line.size() < 1) return 0; stack<int> S; line.push_back(0); int sum = 0; for (int i = 0; i < line.size(); i++) { if (S.empty() || line[i] > line[S.top()]) S.push(i); else { int tmp = S.top(); S.pop(); sum = max(sum, line[tmp]*(S.empty()? i : i-S.top()-1)); i--; } } return sum; } int maximalRectangle(vector<vector<char> > &matrix) { if (matrix.size() < 1) return 0; int n = matrix.size(); if (n == 0) return 0; int m = matrix[0].size(); if (m == 0) return 0; vector<vector<int> > lines(n, vector<int>(m, 0)); for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { if (i == 0) { lines[i][j] = ((matrix[i][j] == '1') ? 1 : 0); } else { lines[i][j] += ((matrix[i][j] == '1') ? lines[i-1][j] + 1 : 0); } } } int maxRec = 0, tmpRec; for (int i = 0; i < n; ++i) { tmpRec = maxArea(lines[i]); maxRec = (maxRec > tmpRec) ? maxRec : tmpRec; } return maxRec; } };