二维 4x4 阵列中的邻居

Question

Alright... So I've been stuck with an exercise for school for a little while now and I really can't figure out how to solve it. I think I've come really compared to where I started and I hope you guys could help me out.

The final meaning of the exercise would be that the code will output all the neighbors possible of every single digit in the array. I've done the middle four ones, they work perfectly. The outer digits are a pain for me, I can't find a way to make sure to make the code 'notice' that there are no more digits, for example, above the digit in the upper left corner.

I feel like I know how to do it: with an if statement that doesn't let something happen if the value of the index of the array is higher than 3 or lower than 0. Because it's a 4x4 2D Array that means there are 0, 1, 2, 3 indexes for the X Axis and Y Axis.

I hope someone here is willing to help me out. It will be greatly appreciated! Here's my code for so far!

Thanks in advance

public class P620 {

/**
 * @param args the command line arguments
 */
public static void main(String[] args) {
    int[][] counts = 
        {
            { 1, 0, 3, 4},
            { 3, 5, 6, 4 },
            { 9, 7, 1, 4},
            { 1, 1, 1, 1}
        };

    for(int i = 0; i <= 3; i++) {

        for(int j = 0; j <= 3; j++) { 

            System.out.println("Neighbours van de array: " + i + j + " met waarde: " + counts[i][j]);

                if ((i < 4 && i > -1) && (j < 4 && j > -1)) {

                    System.out.println(counts[i - 1][j]); 
                }

                else if ((i < 4 && i > -1) && (j < 4 && j > -1)) {
                    System.out.println(counts[i - 1][j - 1]);
                }

                else if ((i < 4 && i > -1) && (j < 4 && j > -1)) {
                    System.out.println(counts[i - 1][j + 1]);
                }

                else if ((i < 4 && i > -1) && (j < 4 && j > -1)) {
                    System.out.println(counts[i][j - 1]);
                }

                else if ((i < 4 && i > -1) && (j < 4 && j > -1)) {
                    System.out.println(counts[i + 1][j - 1]);         
                }

                else if ((i < 4 && i > -1) && (j < 4 && j > -1)) {
                    System.out.println(counts[i + 1][j]);
                }

                else if ((i < 4 && i > -1) && (j < 4 && j > -1)) {
                    System.out.println(counts[i + 1][j + 1]);
                }

                else if ((i < 4 && i > -1) && (j < 4 && j > -1)) {
                    System.out.println(counts[i][j + 1]);
                }
                else {

                }

            }
        }
}      

}

Answer 1

Here's a little hint:

for (int i=0; i < 4; ++i) {
    for (int j=0; j<4; ++j) {
        System.out.println("Neighbours van de array: " + i + j + " met waarde: " + counts[i][j]);

        // print upper-left (northwest) neighbor, if there is one
        if (i >= 1 && j >= 1)
            System.out.println(counts[i-1][j-1]);
        // print upper (north) neighbor, if there is one
        if (i >= 1)
            System.out.println(counts[i-1][j]);
                .
                .
                .
        // print lower (south) neighbor, if there is one
        if (i < 3) 
            System.out.println(counts[i+1][j]);
        // print lower-right (southeast) neighbor, if there is one
        if (i < 3 && j < 3)
            System.out.println(counts[i+1][j+1]);
    }       
}