【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 1
  1 1
    

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 - -！

 class Solution {
 public:
     int maximalSquare(vector<vector<char>>& matrix) {
         const int rows = matrix.size();
         if(rows == )
             return ;

         vector<vector<char> >::iterator it = matrix.begin();
         const int columns = (*it).size();

         int square[rows][columns], ret = ;

         for(int i = ; i < rows; i++)
             for(int j = ; j < columns; j++)
                 square[i][j] = ;

         for(int i = ; i < rows; i++){
             if(matrix[i][] == '')
             {
                 square[i][] = ;
                 ret = ;
             }
         }

         for(int i = ; i < columns; i++){
             if(matrix[][i] == '')
             {
                 square[][i] = ;
                 ret = ;
             }
         }

         for(int i = ; i < rows; i++)
         {
             for(int j = ; j < columns; j++)
             {
                 square[i][j] = matrix[i][j] == '' ? min(square[i-][j], min(square[i][j-], square[i-][j-])) +  : ;

                 if(square[i][j] > ret)
                     ret = square[i][j];
             }
         }
         return ret * ret;
     }
 };