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 <=0) return 0;

int col = matrix[0].size();

int maxSize = 0;

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

for (int i=0; i<matrix.size(); i++) {

for (int j=0; j<matrix[i].size(); j++){

//convert the `char` to `int`

dp[i][j] = matrix[i][j] -'0';

//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!=0 && j!=0 & dp[i][j]!=0){

dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1;

}

//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.