矩阵存储。 我们要去哪里？

Question

We're trying to read a file that contains a sudoko plate with integers and dots for the empty spaces. We have no problem reading the file using scanner and new File , but when we can't print the plate in a 9 x 9 matrix.

We wish to load the file and store it in an 9 x 9 matrix with a while loop. But we get the following error when doing so:

Exception in thread "main" java.util.InputMismatchException
at java.base/java.util.Scanner.throwFor(Scanner.java:939)
at java.base/java.util.Scanner.next(Scanner.java:1594)
at java.base/java.util.Scanner.nextInt(Scanner.java:2258)
at java.base/java.util.Scanner.nextInt(Scanner.java:2212)
at Sodoku.main(Sodoku.java:19)

And were not sure, what we're doing wrong. So far we have the following code:

import java.util.*;
import java.lang.*;
import java.io.*;

public class Sodoku{
    public static void main(String[] args) 
        throws java.io.FileNotFoundException{
        Scanner input = new Scanner (new File("Sudoku.txt"));
        int m = 9;
        int n = 9;
        int[][] a = new int [m][n];
        while (input.next()!=null){
            for (int i=0;i<m;i++){
                for (int j=0;j<n;j++)
                    a[i][j]= input.nextInt();
            }   

        }
        //print the input matrix
        System.out.println("The input sorted matrix is : ");
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++)
                System.out.println(a[i][j]);
        }

    }

}

Answer 1

By using input.next in the while you move the pointer(you consume one of the values), you can use

while (input.hasNext()==true){

This will not move the pointer it will only verify it has a "next" value

Answer 2

The best method i can think of would be to load the file as an string and then parse it to the desired format. Here you have the code. Not pretty, but will do the trick:

import java.util.*;
import java.lang.*;
import java.io.*;
import java.nio.charset.Charset;
import java.nio.file.Files;
import java.nio.file.Paths;

public class Sudoku {

    static String readFile(String path, Charset encoding) 
    throws IOException 
    {
        byte[] encoded = Files.readAllBytes(Paths.get(path));
        return new String(encoded, encoding);
    }

    static int[][] parse(String path) throws IOException {
        String input = readFile(path, Charset.defaultCharset());
        int [][] array = new int[10][10];
        int x = 0;
        int y = 0;
        char[] chars = input.toCharArray();
        for (int i = 0; i < input.length(); i++) {
            if (chars[i] == '\n') {
                x = 0;
                y++;
                continue;
            } else if (chars[i] == '.') {
                array[y][x] = -1;
                x++;
                continue;
            }
            array[y][x] = Character.getNumericValue(chars[i]);
            x++;
        }

        return array;
    }
    public static void main(String[] args) 
        throws IOException {
        int m = 9;
        int n = 9;
        int[][] a = parse("Sudoku.txt");
        //print the input matrix
        System.out.println("The input sorted matrix is : ");
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++) {
                if (j==3 || j==6) System.out.print ("| ");
                if (a[i][j]==-1) {
                    System.out.print(". ");
                } else {
                    System.out.print(a[i][j]+" ");
                }
            }

            System.out.println("");
            if (i==2 || i==5) System.out.println("----------------------");
        }
    }
}

With the file Sudoku.txt

.1.3..8..
5.96..7..
7.4.95.2.
4.....1..
.28.71.63
...2.495.
6.3..9..7
...43.516
.52.8..4.

Outputs the following:

. 1 . | 3 . . | 8 . . 
5 . 9 | 6 . . | 7 . . 
7 . 4 | . 9 5 | . 2 . 
----------------------
4 . . | . . . | 1 . . 
. 2 8 | . 7 1 | . 6 3 
. . . | 2 . 4 | 9 5 . 
----------------------
6 . 3 | . . 9 | . . 7 
. . . | 4 3 . | 5 1 6 
. 5 2 | . 8 . | . 4 . 

Answer 3

Not sure but I think using input.next() as a while statements results in loosing this value later on ( when you are using input.nextInt() it' s next value from a file, not the one from the statement). Try using integer as a temporary container for that value:

int tmp;
while(tmp=input.next()!=null) and later on a[i][j]= tmp;

or you can try using different statement like while( input.hasNext() )