1s和0s的二维数组

Question

I need to create a random maze of 1 s and 0 s, and it does not have to be possible to solve. The program I am writing needs to be able to search through this maze and tell me whether not it is possible to get from the beginning to end. 0 s are fine to go through, but 1 s are like walls. Currently I am just trying to create the random array, and what I have so far compiles, but it returns this: [[I@4ea20232 . So far I have this:

import java.util.Random;

public class SearchMaze{    
    public static void main(String [] args){    
        int n = 8;
        int m = 7;
        int[][] maze = new int[n][m];

        Random rand = new Random();

        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
                maze[i][j] = rand.nextInt(2);           

        System.out.println(maze);
    }
}   

I am trying to get something like this:

1000000
1111100
0000001
1110111
1010101
1010101
0000110
1110000

I can't seem to find why something like this wouldn't work.

Answer 1

The problem is that the toString of an array returns useless information. Using a list will not help much because it will print everything on one line with commas. Try to use nested loops to output the data.

for(int i = 0; i < n; i++) {
    for(int j = 0; j < m; j++) {
        System.out.print(maze[i][j]);
    }
    System.out.print('\n');
}

Answer 2

When you print an object like System.out.println(a); , you're really calling the toString() method of that object. Thus you're really calling System.out.println(a.toString()); .

The toString() method of Arrays isn't particularly useful, so by doing System.out.println(maze) you're not going to see anything informative.

Fortunately, the Arrays class has helper methods that can reduce your problem to a single line . Try using:

System.out.println(Arrays.deepToString(maze));

Given that your maze is 2d, you can use some replaceAll calls to format it like your post:

System.out.println(Arrays.deepToString(maze)
        .replaceAll("],\\s\\[", "\n")
        .replaceAll(",\\s|]|\\[", "")
    );

Answer 3

Instead filling the maze with random values, you need to start will all 1 s and fill it with random paths eg going north or east for a given length.

Just like the loops you have you need to print them this way too.

for(int i = 0; i < n; i++) {
    for(int j = 0; j < m; j++) {
        System.out.print(maze[i][j]);
    }
    System.out.println();
}

Answer 4

Not the answer to your problem, as it has already been answered, but how do you know where to begin and where to end?

Anyway, I came here to steer you towards the A* algorithm - it's a path finding algorithm that would be great for this type of application.

Answer 5

I think a better solution would be to create an object for the maze:

public class Maze {

    public Maze(final int[][] maze) {
        this.maze = maze;
    }

    @Override
    public String toString() {
        final StringBuilder builder = new StringBuilder();
        for (int i = 0; i < maze.length; i++) {
            for (int j = 0; j < maze[i].length; j++) {
                builder.append(maze[i][j]);
            }
            builder.append("\r\n");
        }
        return builder.toString();
    }

    private final int[][] maze;
}