【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 - -！

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        const int rows = matrix.size();
        if(rows == 0)
            return 0;
        
        vector<vector<char> >::iterator it = matrix.begin();
        const int columns = (*it).size();
        
        int square[rows][columns], ret = 0;
        
        for(int i = 0; i < rows; i++)
            for(int j = 0; j < columns; j++)
                square[i][j] = 0;
        
        for(int i = 0; i < rows; i++){
            if(matrix[i][0] == '1')
            {
                square[i][0] = 1;
                ret = 1;
            }
        }
    
        for(int i = 0; i < columns; i++){
            if(matrix[0][i] == '1')
            {
                square[0][i] = 1;
                ret = 1;
            }
        }
        
        for(int i = 1; i < rows; i++)
        {
            for(int j = 1; j < columns; j++)
            {
                square[i][j] = matrix[i][j] == '1' ? min(square[i-1][j], min(square[i][j-1], square[i-1][j-1])) + 1 : 0;
                
                if(square[i][j] > ret)
                    ret = square[i][j];
            }
        }
        return ret * ret;
    }
};