Question

在上一篇文章中，我提到了如何从每一行中删除Sudoku矩阵中的一个元素。现在，我正在考虑从Sudoku矩阵中删除一个元素的方法，但是对于每个行和列。我正在考虑创建一个存储列索引的数组，如果该元素已在行中删除。然后在下一行中，您迭代以删除另一个元素检查，以查看您删除的数字是否在先前存储的列索引中。虽然我不认为它会是一个非常有效的算法。

从行方法中删除元素

public static void remove_Sud (int [][] S){ // Remove sudoku method 
    for(int i = 0; i < S.length; i++){ // For loop iterate through length of matrix array
        int randomPosition = (int) Math.floor(Math.random() * S.length);  //Random number between 0 and 2
        S[i][randomPosition] = 0; // 0 or whatever you use to represent blank
    }

}

Sudoku Matrix

Answer 1

您可以使用此代码。我没有使用完美的数独值制作矩阵，但你可以看到输出。该函数是deleteRandom（）

8 1 5 4 2 5 3 0 2 1 
3 6 0 4 5 5 3 3 5 8 
6 9 4 3 8 2 3 8 0 7 
2 9 9 1 0 5 7 6 9 2 
4 0 6 7 7 9 5 6 6 2 
2 9 1 8 8 7 9 9 8 0 
0 4 6 2 7 3 8 5 8 1 
1 8 5 2 1 8 0 4 8 7 
4 7 5 0 6 6 6 4 3 3 
9 6 3 5 6 0 4 7 1 6

在每一行和每一列中只有一个零，每次程序执行时它都会改变。

import java.util.ArrayList;
import java.util.LinkedHashSet;
import java.util.Random;
import java.util.Set;

public class test {
    public static void main(String[] args) {
        int[][] sodoku = new int[10][10];
        for (int i = 0; i < 10; i++) {
            for (int j = 0; j < 10; j++) {
                Random random = new Random();
                int max = 9;
                int min = 1;
                sodoku[i][j] = random.nextInt(max - min + 1) + min;
            }
        }
        print(sodoku);
        deleteRandom(sodoku);
    }

    private static void deleteRandom(int[][] sodoku) {
        Random r = new Random();
        Set<Integer> rowSet = new LinkedHashSet<>();
        while (rowSet.size() != 10) {
            int answer = r.nextInt(10);
            rowSet.add(answer);
        }
        ArrayList<Integer> rowList = new ArrayList<>();
        rowList.addAll(rowSet);

        Set<Integer> colSet = new LinkedHashSet<>();
        while (colSet.size() != 10) {
            int answer = r.nextInt(10);
            colSet.add(answer);
        }
        ArrayList<Integer> colList = new ArrayList<>();
        colList.addAll(colSet);
        for (int i = 0; i < 10; i++) {
            sodoku[rowList.get(i)][colList.get(i)] = 0;
        }
        System.out.println();
        print(sodoku);
    }

    private static void print(int[][] sodoku) {
        for (int i = 0; i < 10; i++) {
            for (int j = 0; j < 10; j++) {
                System.out.print(sodoku[i][j] + " ");
            }
            System.out.println();
        }
    }
}

Answer 2

为了有效地做，我建议创建两个索引数组（一个用于列，一个用于行），每个9个int长。它比使用列表更有效。然后我们用0-8的整数排列填充它们而不重复，并将它们用作矩阵中哪些元素输入0的映射。

以下是示例代码：

public static void removeSudoku(int[][] sudoku) {
    Random rand = new Random();
    int[] cols = {-1,-1,-1,-1,-1,-1,-1,-1,-1};
    int[] rows = {-1,-1,-1,-1,-1,-1,-1,-1,-1};

    //We need to choose an index for each number 0-8 inclusive.
    for (int i=0;i<9;i++) {

      //get a random index on the column array for i
      int randomInt = rand.nextInt(9);
      //In case this random index is already populated - 
      //rand again until an empty spot is available.
      while (cols[randomInt]!=-1) {
        randomInt = rand.nextInt(9);
      }
      cols[randomInt] = i;

      //Same thing for the rows - get a random index in the 
      //array for i, rand again if needed.
      randomInt = rand.nextInt(9);
      while (rows[randomInt]!=-1) {
        randomInt = rand.nextInt(9);
      }
      rows[randomInt] = i;
    }

    //Now that we have the two arrays filled in with a random
    //permutation of ints 0-8, we can use it to remove the 
    //elements from the sudoku.
    for (int i=0;i<9;i++) {
      sudoku[rows[i]][cols[i]] = 0;
    }
  }

  //Just for printout
  public static void printSoduku(int[][] sudoku) {
    for (int i=0;i<9;i++) {
      for(int j=0;j<9;j++) {
        System.out.print(sudoku[i][j]+" ");
        if (j==2 || j==5) {
          System.out.print("|");
        }
      }
      System.out.println();
      if (i==2 || i==5) {
        System.out.println("-------------------");
      } 
    }
  }

  public static void main(String[] args) throws IOException {

    int[][] soduku = new int[][] {{1,2,3,4,5,6,7,8,9},
    {4,5,6,7,8,9,1,2,3},{7,8,9,1,2,3,4,5,6},
    {2,3,4,5,6,7,8,9,1},{5,6,7,8,9,1,2,3,4},
    {8,9,1,2,3,4,5,6,7},{3,4,5,6,7,8,9,1,2},
    {6,7,8,9,1,2,3,4,5},{9,1,2,3,4,5,6,7,8}};
    printSudoku(sudoku);
    removeSudoku(sudoku);
    System.out.println();
    printSudoku (sudoku);
  }

输出将是原始数独矩阵，然后是删除的数据：

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

1 2 3 |0 5 6 |7 8 9 
4 5 6 |7 8 9 |0 2 3 
0 8 9 |1 2 3 |4 5 6 
-------------------
2 3 0 |5 6 7 |8 9 1 
5 6 7 |8 0 1 |2 3 4 
8 9 1 |2 3 0 |5 6 7 
-------------------
3 0 5 |6 7 8 |9 1 2 
6 7 8 |9 1 2 |3 4 0 
9 1 2 |3 4 5 |6 0 8

从数独矩阵

2 个答案: