Java数独游戏破解工具源代码

一、游戏原理，通过填充空格来完成数独游戏。

数独的解法需遵循如下规则：

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

二、解法思路（递归+回溯）

针对游戏规则，我采用对为空的位置，进行猜测填入除横行竖列以及3x3宫外其他数字。

然后依照上述方式再填入以后的数字。

如果在后续中遇到无数可填情况，证明之前填的数字有错误值，这时我进行回溯，重新填入其他数字，再进行上述方法，知道所有空格都填完。游戏破解

其中需要注意，可以先找出空格周围需要填的集合，进行从小到大排序，这样填错的的可能性小，回溯的次数也会降低，提高程序效率。

三、源代码

1. Game2类（游戏主类）

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class Game2 {

    public static void main(String[] args) {
        // 开始时间
        long sTime = System.nanoTime();
        // 执行时间（1s）


        char[][] board = {{'5', '3', '.', '.', '7', '.', '.', '.', '.'}, {'6', '.', '.', '1', '9', '5', '.', '.', '.'}, {'.', '9', '8', '.', '.', '.', '.', '6', '.'}, {'8', '.', '.', '.', '6', '.', '.', '.', '3'}, {'4', '.', '.', '8', '.', '3', '.', '.', '1'}, {'7', '.', '.', '.', '2', '.', '.', '.', '6'}, {'.', '6', '.', '.', '.', '.', '2', '8', '.'}, {'.', '.', '.', '4', '1', '9', '.', '.', '5'}, {'.', '.', '.', '.', '8', '.', '.', '7', '9'}};
        List ListAll = new ArrayList<>();
        for (int i = 0; i < board.length; i++) {
            //board[i] 表示第i+1个一维数组
            for (int j = 0; j < board[i].length; j++) {
                if (board[i][j] == '.') {
                    //进行填充
                    List chance = haveOther(board, i, j);
                    ListAll.add(new EmptyNumberComparable(chance, i, j));
                    Collections.sort(ListAll);
                }
            }
        }
        // 结束时间
        long eTime1 = System.nanoTime();
        // 计算执行时间
        System.out.printf("排序执行时长：%d 纳秒.", (eTime1 - sTime));
        int resultCode = mainSee(board, ListAll);
        if (resultCode == 0) {
            System.out.println("出现大错误");
        } else {
            System.out.println("完成");
        }
        for (int ii = 0; ii < board.length; ii++) {
            //board[i] 表示第i+1个一维数组
            for (int jj = 0; jj < board[ii].length; jj++) {
                System.out.print(board[ii][jj]);
            }
            System.out.println("");
        }
        // 结束时间
        long eTime = System.nanoTime();
        // 计算执行时间
        System.out.printf("执行时长：%d 纳秒.", (eTime - sTime));
    }

    //主递归程序
    public static int mainSee(char[][] board, List ListAll) {
        for (EmptyNumberComparable emptyNumberComparable : ListAll) {
            int i = emptyNumberComparable.getI();
            int j = emptyNumberComparable.getJ();
            if (board[i][j] == '.') {
                //进行填充
                // 开始时间
                long sTime = System.nanoTime();
                // 执行时间（1s）
                List chance = haveOther(board, i, j);
                // 结束时间
                long eTime = System.nanoTime();
                // 计算执行时间
                System.out.printf("找不同执行时长：%d 纳秒.", (eTime - sTime));
                // 开始时间
                long sTime1 = System.nanoTime();
                // 执行时间（1s）
                //寻找其他可能集合,如果为0证明某步填错,回退到前面
                if (chance.size() == 0) {
                    return 0;
                } else {
                    for (char c : chance) {
                        board[i][j] = c;
                        System.out.println(i + "," + j + "," + c);
                        int resultCode = mainSee(board, ListAll);
                        if (resultCode == 0) {
                            System.out.println("返回" + i + "," + j + "," + c);
                            System.out.println("出现错误" + c);
                        } else {
                            return 1;
                        }
                    }
                }
                // 结束时间
                long eTime1 = System.nanoTime();
                // 计算执行时间
                System.out.printf("递归执行时长：%d 纳秒.", (eTime1 - sTime1));
                System.out.println(i + "," + j + "把他弄成点");
                board[i][j] = '.';
                return 0;
            }
        }
        return 1;
    }

    //找不同
    public static List haveOther(char[][] board, int a, int b) {
        char[] all9 = {'1', '2', '3', '4', '5', '6', '7', '8', '9'};
        char[] other = new char[30];
        //遍历二维数组,找出同行与同列不同的数据
        int m = 0;
        for (int i = 0; i < board.length; i++) {
            if (board[i][b] != '.') {
                other[m] = board[i][b];
                m++;
            }
        }
        for (int j = 0; j < board.length; j++) {
            if (board[a][j] != '.') {
                other[m] = board[a][j];
                m++;
            }
        }
        //找身边九宫中的重复数据
        other = nineOther(board, other, a, b, m);

        return arrContrast(all9, other);
    }

    //九宫中不同值
    public static char[] nineOther(char[][] board, char[] other, int a, int b, int m) {
        int leftUp = 0;
        int leftDown = 0;
        int rightUp = 0;
        int rightDown = 0;
        if (a >= 0 && a <= 2) {
            if (b >= 0 && b <= 2) {
                //第1宫
                leftUp = 0;
                leftDown = 2;
                rightUp = 0;
                rightDown = 2;
                other = searchNine(board, other, leftUp, leftDown, rightUp, rightDown, m);
            }
            if (b >= 3 && b <= 5) {
                //第2宫
                leftUp = 0;
                leftDown = 2;
                rightUp = 3;
                rightDown = 5;
                other = searchNine(board, other, leftUp, leftDown, rightUp, rightDown, m);
            }
            if (b >= 6 && b <= 8) {
                //第3宫
                leftUp = 0;
                leftDown = 2;
                rightUp = 6;
                rightDown = 8;
                other = searchNine(board, other, leftUp, leftDown, rightUp, rightDown, m);
            }
        }
        if (a >= 3 && a <= 5) {
            if (b >= 0 && b <= 2) {
                //第4宫
                leftUp = 3;
                leftDown = 5;
                rightUp = 0;
                rightDown = 2;
                other = searchNine(board, other, leftUp, leftDown, rightUp, rightDown, m);
            }
            if (b >= 3 && b <= 5) {
                //第5宫
                leftUp = 3;
                leftDown = 5;
                rightUp = 3;
                rightDown = 5;
                other = searchNine(board, other, leftUp, leftDown, rightUp, rightDown, m);
            }
            if (b >= 6 && b <= 8) {
                //第6宫
                leftUp = 3;
                leftDown = 5;
                rightUp = 6;
                rightDown = 8;
                other = searchNine(board, other, leftUp, leftDown, rightUp, rightDown, m);
            }
        }
        if (a >= 6 && a <= 8) {
            if (b >= 0 && b <= 2) {
                //第7宫
                leftUp = 6;
                leftDown = 8;
                rightUp = 0;
                rightDown = 2;
                other = searchNine(board, other, leftUp, leftDown, rightUp, rightDown, m);
            }
            if (b >= 3 && b <= 5) {
                //第8宫
                leftUp = 6;
                leftDown = 8;
                rightUp = 3;
                rightDown = 5;
                other = searchNine(board, other, leftUp, leftDown, rightUp, rightDown, m);
            }
            if (b >= 6 && b <= 8) {
                //第9宫
                leftUp = 6;
                leftDown = 8;
                rightUp = 6;
                rightDown = 8;
                other = searchNine(board, other, leftUp, leftDown, rightUp, rightDown, m);
            }
        }
        return other;
    }

    //遍历寻找9宫中为空数值
    public static char[] searchNine(char[][] board, char[] other, int leftUp, int leftDown, int rightUp, int rightDown, int m) {
        for (int i = leftUp; i <= leftDown; i++) {
            for (int j = rightUp; j <= rightDown; j++) {
                if (board[i][j] != '.') {
                    other[m] = board[i][j];
                    m++;
                }
            }
        }
        return other;
    }

    //两数组相减
    public static List arrContrast(char[] a, char[] b) {
        List list = new ArrayList<>();
        //输入的部分你写了我就不管了
        for (int i = 0; i < a.length; i++) {
            //a中的数是否b中也存在
            boolean have = false;
            for (int j = 0; j < b.length; j++) {
                if (a[i] == b[j]) {
                    have = true;
                    break;
                }
            }
            if (!have) {
                list.add(a[i]);
            }
        }
        return list;
    }

    //找身边九宫中的重复数据
}

2.EmptyNumberComparable类（用于集合从小到大排序）

import java.util.List;

class EmptyNumberComparable implements Comparable {
    private List chance; //姓名
    private int i; // i
    private int j; // j

    public EmptyNumberComparable(List chance, int i, int j) {
        this.chance = chance;
        this.i = i;
        this.j = j;
    }

    public List getChance() {
        return chance;
    }

    public void setChance(List chance) {
        this.chance = chance;
    }

    public int getI() {
        return i;
    }

    public void setI(int i) {
        this.i = i;
    }

    public int getJ() {
        return j;
    }

    public void setJ(int j) {
        this.j = j;
    }

    @Override
    public int compareTo(EmptyNumberComparable user) {           //重写Comparable接口的compareTo方法，
        return this.chance.size() - user.getChance().size();
    }
}

3.java源文件（懒人专用）

csdn下载方式

数独源码配合文章中一起使用.rar-Java文档类资源-CSDN下载

百度云下载方式

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