获取数独可能的解决方法

Question

所以这是这个问题的后续问题：

我正在编写一种方法，该方法采用二维数组的行、列组合，扫描行、列和 3x3 子数组，以在该特定位置寻找可能的解决方案。到目前为止，一切看起来都很好。第一个 for 循环扫描行并将已使用的数字（1-9 之间）存储在 alreadyInUse 数组中。第二个对列执行相同的操作。第三个 for 循环使用行、列组合确定子数组的位置，搜索此 3x3 子数组并将已找到的数字存储在同一个 alreadyInUse 数组中。

// Method for calculating all possibilities at specific position
public int[] getPossibilities(int row, int col){
    int [] alreadyInUse;
    int [] unsorted = new int[9];
    int currentIndex = 0;
    int size = 0;
    boolean match;
    if(sudoku[row][col] != 0){
        return  new int[]{sudoku[col][row]};
    }
    else{
        alreadyInUse = new int[26];
        //Go into Row x and store all available numbers in alreadyInUse
        for(int i=0; i<sudoku.length; i++){
            if(sudoku[row][i] !=0){
                alreadyInUse[currentIndex] = sudoku[row][i];
                currentIndex++;
            }
        }
        //Go into Column y and store all available numbers in alreadyInUse
        for(int j=0; j<sudoku.length; j++){
            if(sudoku[j][col] !=0){
                alreadyInUse[currentIndex] = sudoku[j][col];
                currentIndex++;
            }
        }
        //Go into subarray and store all available numbers in alreadyInUse
        int x_left = row - (row%3);
        int y_top = col - (col%3);
        for(int i=0; i<3; i++){
            for(int j=0; j<3; j++){
                if(sudoku[i+x_left][j+y_top] != 0){
                    alreadyInUse[currentIndex] = sudoku[i+x_left][j+y_top];
                }
            }
        }
        //compare 1-9 with the numbers stored in alreadyInUse array
        for(int i=1; i<10; i++){
            match = false;
            //if a pair matches, break;
            for(int j=0; j<alreadyInUse.length; j++){
                if(alreadyInUse[j] == i ){
                    match = true;
                    break;
                }
            }
            //If no pair matches, store the number (i) in an unsorted array
            if(!match){
                size++;
                unsorted[i-1] = i;
            }
        }
        //Re-use alreadyInUse array
        alreadyInUse = new int[size];
        size = 0;
        //Remove the zeros and store the rest of the numbers in alreadyInUse array
        for(int i=0; i<unsorted.length; i++){
            if(unsorted[i] != 0){
                alreadyInUse[size] = unsorted[i];
                size++;
            }
        }
        return alreadyInUse;    
    }   
}

当我这样测试输出时：

 public class SudokuTest {
 public static void main(String[] args){
    Sudoku sudoku;
    String sud = "250030901010004000407000208005200000000098100040003000000360072070000003903000604";
    sudoku = new Sudoku(sud);
    System.out.println(sudoku.getPossibilities(3,4));

我将此作为输出：[I@659e0bfd 尽管调试显示 alreadyInUse 数组包含正确的值。有人可以帮忙吗？

Answer 1

不能直接打印int数组。您需要：

import java.util.Arrays;
...
System.out.println(Arrays.toString(sudoku.getPossibilities(3,4)));

获取数独可能的解决方法

Getting sudoku possible solutions Method

java

arrays

sudoku