【LeetCode 221】Maximal Square

Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest square containing all 1‘s and return its area.

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 4.

思路：

　　DP，到达当前（非0）结点的最长边长square[i][j] = min(square[i-1][j-1], min(square[i-1][j], square[i][j-1])) + 1，即取左上方3个结点中值最小的一个加1。代码写的看起来好不舒服 - -！

 1 class Solution {
 2 public:
 3     int maximalSquare(vector<vector<char>>& matrix) {
 4         const int rows = matrix.size();
 5         if(rows == 0)
 6             return 0;
 7
 8         vector<vector<char> >::iterator it = matrix.begin();
 9         const int columns = (*it).size();
10
11         int square[rows][columns], ret = 0;
12
13         for(int i = 0; i < rows; i++)
14             for(int j = 0; j < columns; j++)
15                 square[i][j] = 0;
16
17         for(int i = 0; i < rows; i++){
18             if(matrix[i][0] == ‘1‘)
19             {
20                 square[i][0] = 1;
21                 ret = 1;
22             }
23         }
24
25         for(int i = 0; i < columns; i++){
26             if(matrix[0][i] == ‘1‘)
27             {
28                 square[0][i] = 1;
29                 ret = 1;
30             }
31         }
32
33         for(int i = 1; i < rows; i++)
34         {
35             for(int j = 1; j < columns; j++)
36             {
37                 square[i][j] = matrix[i][j] == ‘1‘ ? min(square[i-1][j], min(square[i][j-1], square[i-1][j-1])) + 1 : 0;
38
39                 if(square[i][j] > ret)
40                     ret = square[i][j];
41             }
42         }
43         return ret * ret;
44     }
45 };