【题目链接】

最大正方形

【题目描述】

在一个由 '0' 和 '1' 组成的二维矩阵内，找到只包含 '1' 的最大正方形，并返回其面积。

【输入输出样例】

输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出：4

【数据范围】

1 <= m, n <= 300

f[i][j] 取决于 f[i -1][j - 1]，之后判断一下紫色区域是否都为1即可，如果为1，则加1。

 1 const int N = 309;
 2 int f[N][N];
 3 int res = 0;
 4 class Solution {
 5 public:
 6     int maximalSquare(vector<vector<char>>& matrix) {
 7         memset(f,0,sizeof f);
 8         res = 0;
 9         int n = matrix.size(),m = matrix[0].size();
10         for(int i = 0;i < n;++i)
11         {
12             f[i][0] = matrix[i][0] - '0';
13             res = max(res,f[i][0]);
14         }
15         for(int i = 0;i < m;++i)
16         {
17             f[0][i] = matrix[0][i] - '0';
18             res = max(res,f[0][i]);
19         }
20         for(int i = 1;i < n;++i)
21             for(int j = 1;j < m;++j)
22             {
23                 if(matrix[i][j] != '0')
24                 {
25                     for(int k  = 0;k <= f[i - 1][j - 1];++k)　　//当前f[i][j]最大值受限于f[i - 1][j - 1]
26                     {
27                         int x = i - k,y = j - k;
28                         if(x >= 0 && y >= 0 && matrix[i][y] == '1' && matrix[x][j] == '1')　
29                             f[i][j] += 1;
30                         else    break;
31                     }
32                     res = max(res,f[i][j]);
33                 }         
34             }
35         return res * res;
36     }
37 };

滚动数组优化：

 1 const int N = 309;
 2 int f[N];
 3 int res = 0;
 4 class Solution {
 5 public:
 6     int maximalSquare(vector<vector<char>>& matrix) {
 7         memset(f,0,sizeof f);
 8         res = 0;
 9         int n = matrix.size(),m = matrix[0].size();
10         for(int i = 0;i < m;++i)
11         {
12             f[i] = matrix[0][i] - '0';
13             res = max(res,matrix[0][i] - '0');
14         }
15         for(int i = 0;i < n;++i)
16             res = max(res,matrix[i][0] - '0');
17         for(int i = 1;i < n;++i)
18         {
19             
20             for(int j = m - 1;j >= 1;--j)
21             {  
22                 f[j] = 0;
23                 if(matrix[i][j] != '0')
24                 {                 
25                     for(int k  = 0;k <= f[j - 1];++k)
26                     {
27                         int x = i - k,y = j - k;
28                         if(x >= 0 && y >= 0 && matrix[i][y] == '1' && matrix[x][j] == '1')
29                             f[j] += 1;
30                         else    break;
31                     }
32                     res = max(res,f[j]);
33                 } 
34             }
35             f[0] = matrix[i][0] - '0';
36         }
37         
38         return res * res;
39     }
40 };

DP做法：

 1 const int N = 309;
 2 int f[N][N];
 3 int res = 0;
 4 class Solution {
 5 public:
 6     int maximalSquare(vector<vector<char>>& matrix) {
 7         memset(f,0,sizeof f);
 8         res = 0;
 9         int n = matrix.size(),m = matrix[0].size();
10         for(int i = 0;i < n ;++i)
11         {
12             f[i][0] = matrix[i][0] - '0';
13             res = max(res,f[i][0]);
14         }
15         for(int i = 0;i < m;++i)
16         {
17             f[0][i] = matrix[0][i] - '0';
18             res = max(res,f[0][i]);
19         }
20         for(int i = 1;i < n;++i)
21             for(int j = 1;j < m;++j)
22                 if(matrix[i][j] == '0')
23                     f[i][j] = 0;
24                 else
25                 {
26                     f[i][j] = min(f[i][j - 1],min(f[i - 1][j - 1],f[i - 1][j])) + 1;
27                     res = max(res,f[i][j]);
28                 }
29         return res * res;
30     }
31 };

滚动数组优化的DP懒得写了，后面想起来再说吧...