221. Maximal Square java solutions

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.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

题目是要找出元素全为1 的最大正方形。使用DP，递推式为：

dp[i][j] = Math.min(dp[i-1][j],Math.min(dp[i][j-1],dp[i-1][j-1])) + 1;

dp[i][j] 表示到该位置出现的正方形的边长，dp[i][j] 要构成一个新的正方形，只能从之前的左边，上边，左上 三个里面取边长最小的加上matrix[i][j] 为1 的情况，即边长+1.

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