leetcode85 Maximal Rectangle

思路：

分別按行拆分後將這一行以前的每列疊加起來，而後使用leetcode84http://www.javashuo.com/article/p-hfawwtcx-gt.html的思路計算。

實現：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 class Solution
 5 {
 6 public:
 7     int maximalRectangle(vector<vector<char>>& matrix)
 8     {
 9         int n = matrix.size(); if (n == 0) return 0;
10         int m = matrix[0].size(); if (m == 0) return 0;
11         int ans = 0;
12         vector<int> h(m, 0), l(m, 0), r(m, 0);
13         for (int i = 0; i < n; i++)
14         {
15             for (int j = 0; j < m; j++)
16             {
17                 h[j] = (matrix[i][j] == '0' ? 0 : h[j] + 1);
18             }
19             stack<int> s;
20             for (int j = 0; j < m; j++)
21             {
22                 while (!s.empty() && h[s.top()] >= h[j]) s.pop();
23                 l[j] = (s.empty() ? 0 : s.top() + 1);
24                 s.push(j);
25             }
26             while (!s.empty()) s.pop();
27             for (int j = m - 1; j >= 0; j--)
28             {
29                 while (!s.empty() && h[s.top()] >= h[j]) s.pop();
30                 r[j] = (s.empty() ? m - 1 : s.top() - 1);
31                 s.push(j);
32             }
33             for (int j = 0; j < m; j++)
34             {
35                 ans = max(ans, (r[j] - l[j] + 1) * h[j]);
36             }
37         }
38         return ans;
39     }
40 };
41 int main()
42 {
43     char a[][5] = {{'1','0','1','0','0'},
44                   {'1','0','1','1','1'},
45                   {'1','1','1','1','1'},
46                   {'1','0','0','1','0'}};
47     // char a[][4] = {{'1','1','1','1'},
48     //               {'1','1','1','1'},
49     //               {'1','1','1','1'}};
50     vector<vector<char>> v;
51     for (int i = 0; i < 4; i++)
52     {
53         v.push_back(vector<char>(a[i], a[i] + 5));
54     }
55     cout << Solution().maximalRectangle(v) << endl;
56     return 0;
57 }