思路:

dp。

实现:

 1 class Solution 
 2 {
 3 public:
 4     int maximalSquare(vector<vector<char>>& matrix) 
 5     {
 6         if (matrix.empty()) return 0;
 7         int m = matrix.size(), n = matrix[0].size();
 8         vector<int> dp(m, 0);
 9         int maxn = 0;
10         for (int i = 0; i < m; i++)
11         {
12             dp[i] = matrix[i][0] - '0';
13             maxn = max(maxn, dp[i]);
14         }
15         for (int i = 1; i < n; i++)
16         {
17             int pre = dp[0];
18             dp[0] = matrix[0][i] - '0';
19             maxn = max(dp[0], maxn);
20             for (int j = 1; j < m; j++)
21             {
22                 if (matrix[j][i] == '1')
23                 {
24                     pre = min(pre, min(dp[j - 1], dp[j])) + 1;
25                     swap(dp[j], pre);
26                     maxn = max(maxn, dp[j]);
27                 }
28                 else dp[j] = 0;
29             }
30         }
31         return maxn * maxn;
32     }
33 };

 

相关文章:

  • 2021-12-17
  • 2021-06-27
  • 2021-04-09
  • 2021-12-07
  • 2021-05-10
猜你喜欢
  • 2022-12-23
  • 2022-02-23
  • 2022-01-12
  • 2021-07-20
  • 2021-11-16
  • 2021-10-03
相关资源
相似解决方案