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.

Credits:

Special thanks to @Freezen for adding this problem and creating all test cases.
https://leetcode.com/discuss/38489/easy-solution-with-detailed-explanations-8ms-time-and-space
P[i][j]表示在点(i,j)处，可以得到的最大正方形的边长。

P[0][j] = matrix[0][j]

 (topmost row);

P[i][0] = matrix[i][0]

 (leftmost column);
For 

i > 0

 and 

j
> 0

: if 

matrix[i][j] = 0

, 

P[i][j]
= 0

; if 

matrix[i][j] = 1

, 

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

.

class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
int m = matrix.size();
if(m == 0)
return 0;
int n = matrix[0].size();

vector<vector<int>> P(m,vector<int>(n,0));
int maxSize = 0;
for(int i = 0;i<m;i++)
{
P[i][0] = matrix[i][0]-'0';
maxSize = max(maxSize,P[i][0]);
}
for(int j = 0;j<n;j++)
{
P[0][j] = matrix[0][j]-'0';
maxSize = max(maxSize,P[0][j]);
}

for(int i = 1;i<m;i++)
{
for(int j = 1;j<n;j++)
{
if(matrix[i][j] == '1')
{
P[i][j] = min(P[i-1][j-1],min(P[i-1][j],P[i][j-1]))+1;
maxSize = max(maxSize,P[i][j]);
}
}
}

return maxSize * maxSize;

}
};