题目：
在一个由 '0' 和 '1' 组成的二维矩阵内，找到只包含 '1' 的最大正方形，并返回其面积。
示例 1：
输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出：4示例 2：
输入：matrix = [["0","1"],["1","0"]]
输出：1示例 3：
输入：matrix = [["0"]]
输出：0代码实现：
class Solution {
public int maximalSquare(char[][] matrix) {
int maxSide = 0;
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return maxSide;
}
int rows = matrix.length, columns = matrix[0].length;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
if (matrix[i][j] == '1') {
// 遇到一个 1 作为正方形的左上角
maxSide = Math.max(maxSide, 1);
// 计算可能的最大正方形边长
int currentMaxSide = Math.min(rows - i, columns - j);
for (int k = 1; k < currentMaxSide; k++) {
// 判断新增的一行一列是否均为 1
boolean flag = true;
if (matrix[i + k][j + k] == '0') {
break;
}
for (int m = 0; m < k; m++) {
if (matrix[i + k][j + m] == '0' || matrix[i + m][j + k] == '0') {
flag = false;
break;
}
}
if (flag) {
maxSide = Math.max(maxSide, k + 1);
} else {
break;
}
}
}
}
}
int maxSquare = maxSide * maxSide;
return maxSquare;
}
}