思路1 dp

注意，状态方程dp[i][j]表示的是边长，不是面积。

对于最左边一列和最上面一行，dp都是0。

When i > 0 and j > 0, if matrix[i][j] = '0', then dp[i][j] = 0 since no square will be able to contain the '0' at that cell. If matrix[i][j] = '1', we will have dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1, which means that the square will be limited by its left, upper and upper-left neighbors.

if matrix[i][j] == '0': dp[i][j] = 0 else: dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1

是min因为只有左、上、左上三个方向都能组成square，[i][j]才能组成square