[LeetCode] Maximal Rectangle-优快云博客

本文分享了一个使用动态规划(DP)算法解决最大矩形问题的有效解决方案，并详细解释了算法思路。通过一个示例矩阵，文章给出了清晰的代码实现过程，帮助读者理解如何通过动态规划方法来确定矩阵中全为1的最大矩形区域。

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This link shares a nice solution with explanation using DP. You will be clear of the algorithm after running it on its suggested example:

matrix = [
[0, 0, 0, 1, 0, 0, 0],
[0, 0, 1, 1, 1, 0, 0],
[0, 1, 1, 1, 1, 1, 0]];

The code is rewritten as follows.

 1 class Solution {
 2 public:
 3     int maximalRectangle(vector<vector<char>>& matrix) {
 4         if (matrix.empty()) return 0;
 5         const int m = matrix.size(), n = matrix[0].size();
 6         int *left = new int[n](), *right = new int[n](), *height = new int[n]();
 7         fill_n(right, n, n);
 8         int area = 0;
 9         for (int i = 0; i < m; i++) {
10             int l = 0, r = n;
11             for (int j = 0; j < n; j++)
12                 height[j] += matrix[i][j] == '1' ? 1 : -height[j];
13             for (int j = 0; j < n; j++) {
14                 if (matrix[i][j] == '1') left[j] = max(left[j], l);
15                 else left[j] = 0, l = j + 1;
16             }
17             for (int j = n - 1; j >= 0; j--) {
18                 if (matrix[i][j] == '1') right[j] = min(right[j], r);
19                 else right[j] = n, r = j;
20             }
21             for (int j = 0; j < n; j++)
22                 area = max(area, (right[j] - left[j]) * height[j]);
23         }
24         return area;
25     }
26 };