LeetCode OJ Maximal Rectangle
2015-03-22 18:32
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Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area.
参考了求直方图的最大矩形,https://oj.leetcode.com/problems/largest-rectangle-in-histogram/,从上往下,将每一行及以上都看成直方图并求出最大矩形。
参考了求直方图的最大矩形,https://oj.leetcode.com/problems/largest-rectangle-in-histogram/,从上往下,将每一行及以上都看成直方图并求出最大矩形。
class Solution { public: int maximalRectangle(vector<vector<char> > &matrix) { if (matrix.size() == 0) return 0; vector<int> height(matrix[0].size(), 0); // record the height of every row's histogram int tempArea, maxArea = 0; for (int i = 0; i < matrix.size(); i++) { int j; for (j = 0; j < matrix[i].size(); j++) { // calculate the height if (matrix[i][j] == '1') height[j]++; else height[j] = 0; } //calculate the max area of every histogram stack<int> s; for (j = 0; j < matrix[i].size();) { if (s.empty() || height[s.top()] <= height[j]) { s.push(j++); } else { int temp = s.top(); s.pop(); tempArea = height[temp] * (s.empty() ? j : j - s.top() - 1); maxArea = tempArea > maxArea ? tempArea : maxArea; } } while (!s.empty()) { int temp = s.top(); s.pop(); tempArea = height[temp] * (s.empty() ? j : j - s.top() - 1); maxArea = tempArea > maxArea ? tempArea : maxArea; } //calculate the max area of every histogram } return maxArea; } };
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