221. Maximal Square -- 矩阵中1组成的最大正方形

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 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.

class Solution {

public:

    inline int min(int x, int y) {

        return x<y? x:y;

    }

    inline int min(int x, int y, int z) {

        return min(x, min(y, z));

    }

    int maximalSquare(vector<vector<char>>& matrix) {

        int row = matrix.size();

        if (row <=) return ;

        int col = matrix[].size();

        int maxSize = ;

        vector<vector<int>> dp(row, vector<int>(col));

        for (int i=; i<matrix.size(); i++) {

            for (int j=; j<matrix[i].size(); j++){

                //convert the `char` to `int`

                dp[i][j] = matrix[i][j] -'';

                //for the first row and first column, or matrix[i][j], dp[i][j] is ZERO

                //so, it's done during the previous conversion

                // i>0 && j>0 && matrix[i][j]=='1'

                if (i!= && j!= & dp[i][j]!=){

                    dp[i][j] = min(dp[i-][j], dp[i][j-], dp[i-][j-]) + ;

                }

                //tracking the maxSize

                if (dp[i][j] > maxSize ){

                    maxSize = dp[i][j];

                }

            }

        }

        return maxSize*maxSize;

    }

};

*   1) P[0][j] = matrix[0][j] (topmost row);

*   2) P[i][0] = matrix[i][0] (leftmost column);

*   3) For i > 0 and j > 0:

*       3.1) if matrix[i][j] = 0, P[i][j] = 0;

*       3.2) if matrix[i][j] = 1, P[i][j] = min(P[i-1][j], P[i][j-1], P[i-1][j-1]) + 1.