获得数独可能的解决方案方法

时间:2015-04-18 14:29:04

标签: java arrays sudoku

所以这是对此问题的跟进问题:

From given row(x),column(y), find 3x3 subarray of a 2D 9x9 array

我正在编写一个方法,该方法采用2D数组的行,列组合,扫描行,列和3x3子阵列,以寻找此特定位置的可能解决方案。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(在1-9之间)存储在alreadyInUse数组中。第二个对列进行相同的操作。第三个for循环使用行,列组合确定子阵列的位置,搜索此3x3子阵列并将已找到的数字存储在相同的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;    
    }   
}

当我像这样测试输出时:

 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));

我得到这个作为输出:[I@659e0bfd即使调试显示我hasInsese数组包含正确的值。有人可以在这帮忙吗?

1 个答案:

答案 0 :(得分:2)

您无法直接打印int数组。你需要:

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