解法

class Solution {
    public int maximalSquare(char[][] matrix) {
        int row = matrix.length;
        int col = matrix[0].length;
        // 以对应坐标为正方形右下角，能得到的边长最大值
        int[][] dp = new int[row][col];
        int maxSide = 0; 
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (matrix[i][j] == '1') {
                    if (i == 0 || j == 0) {
                        dp[i][j] = 1;
                    } else {
                        // 左上角3个位置最小值加1
                        dp[i][j] = Math.min(dp[i - 1][j], Math.min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
                    }
                    maxSide = Math.max(dp[i][j], maxSide);
                }
            }
        }
        return maxSide * maxSide;
    }
}