class Solution {
    // 动态规划： dp[i][j]代表以nums[i][j] 为右下角能得到的正方形的最大值
    // 递推公式： dp[i][j] = Math.min(dp[i-1][j], dp[i][j - 1], dp[i - 1][j - 1])
    // 初始化 对i==0 || j==0的情况进行初始化
    // 遍历顺序 从左到右，从上到下
    public int maximalSquare(char[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        int[][] dp = new int[m][n];
        int maxSide = 0;
        for (int i = 0; i < m; i++) {
            if (matrix[i][0] == '1') {
                dp[i][0] = 1;
                maxSide = 1; 
            }
        }
        for (int j = 0; j < n; j++) {
            if (matrix[0][j] == '1') {
                dp[0][j] = 1;
                maxSide = 1;
                
            }
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (matrix[i][j] == '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(maxSide, dp[i][j]);
                }
            }
        }
        return maxSide*maxSide;
    }
}