Maximal Square

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理解这个转移方程：

dp[i][j] = matrix[i][j] == '1' ? Math.min(dp[i-1][j-1], Math.min(dp[i-1][j], dp[i][j-1])) + 1 : 0;
                

public class Solution {
    public int maximalSquare(char[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        int max = 0, m = matrix.length, n = matrix[0].length;
        int[][] dp = new int[m][n];
        for (int i = 0; i < m; i++) {
            dp[i][0] = matrix[i][0] - '0';
            max = Math.max(dp[i][0], max);
        }
        for (int j = 0; j < n; j++) {
            dp[0][j] = matrix[0][j] - '0';
            max = Math.max(dp[0][j], max);
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = matrix[i][j] == '1' ? Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1 : 0;
                max = Math.max(dp[i][j], max);
            }
        }
        return max * max;
    }
}

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.