在網上找了好多的數獨生成算法發現一個挺好的算法，然后自己稍微修改了一下，將其改為數獨模板生成工具，用了為我的數獨app生成模板。

本文中用到的數獨生成算法從網上借鑒而來，但是記不清是哪個網址了，如果此算法的原作者看到，請海涵！

public class ShuDu {
?? ?/** 存儲數字的數組 */
?? ?private static int[][] n = new int[9][9];
?? ?/** 生成隨機數字的源數組，隨機數字從該數組中產生 */
?? ?private static int[] num = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };

?? ?public static int[][] generateShuDu(){
?? ??? ?// 生成數字
?? ??? ?for (int i = 0; i < 9; i++) {
?? ??? ??? ?// 嘗試填充的數字次數
?? ??? ??? ?int time = 0;
?? ??? ??? ?// 填充數字
?? ??? ??? ?for (int j = 0; j < 9; j++) {
?? ??? ??? ??? ?// 產生數字
?? ??? ??? ??? ?n[i][j] = generateNum(time);
?? ??? ??? ??? ?// 如果返回值為0，則代表卡住，退回處理
?? ??? ??? ??? ?// 退回處理的原則是：如果不是第一列，則先倒退到前一列，否則倒退到前一行的最后一列
?? ??? ??? ??? ?if (n[i][j] == 0) {
?? ??? ??? ??? ??? ?// 不是第一列，則倒退一列
?? ??? ??? ??? ??? ?if (j > 0) {
?? ??? ??? ??? ??? ??? ?j -= 2;
?? ??? ??? ??? ??? ??? ?continue;
?? ??? ??? ??? ??? ?} else {// 是第一列，則倒退到上一行的最后一列
?? ??? ??? ??? ??? ??? ?i--;
?? ??? ??? ??? ??? ??? ?j = 8;
?? ??? ??? ??? ??? ??? ?continue;
?? ??? ??? ??? ??? ?}
?? ??? ??? ??? ?}
?? ??? ??? ??? ?// 填充成功
?? ??? ??? ??? ?if (isCorret(i, j)) {
?? ??? ??? ??? ??? ?// 初始化time，為下一次填充做準備
?? ??? ??? ??? ??? ?time = 0;
?? ??? ??? ??? ?} else { // 繼續填充
?? ??? ??? ??? ??? ?// 次數增加1
?? ??? ??? ??? ??? ?time++;
?? ??? ??? ??? ??? ?// 繼續填充當前格
?? ??? ??? ??? ??? ?j--;
?? ??? ??? ??? ?}
?? ??? ??? ?}
?? ??? ?}
?? ??? ?return n;
?? ?}

?? ?/**
?? ? * 是否滿足行、列和3X3區域不重復的要求
?? ? *
?? ? * @param row
?? ? *??????????? 行號
?? ? * @param col
?? ? *??????????? 列號
?? ? * @return true代表符合要求
?? ? */
?? ?private static boolean isCorret(int row, int col) {
?? ??? ?return (checkRow(row) & checkLine(col) & checkNine(row, col));
?? ?}

?? ?/**
?? ? * 檢查行是否符合要求
?? ? *
?? ? * @param row
?? ? *??????????? 檢查的行號
?? ? * @return true代表符合要求
?? ? */
?? ?private static boolean checkRow(int row) {
?? ??? ?for (int j = 0; j < 8; j++) {
?? ??? ??? ?if (n[row][j] == 0) {
?? ??? ??? ??? ?continue;
?? ??? ??? ?}
?? ??? ??? ?for (int k = j + 1; k < 9; k++) {
?? ??? ??? ??? ?if (n[row][j] == n[row][k]) {
?? ??? ??? ??? ??? ?return false;
?? ??? ??? ??? ?}
?? ??? ??? ?}
?? ??? ?}
?? ??? ?return true;
?? ?}

?? ?/**
?? ? * 檢查列是否符合要求
?? ? *
?? ? * @param col
?? ? *??????????? 檢查的列號
?? ? * @return true代表符合要求
?? ? */
?? ?private static boolean checkLine(int col) {
?? ??? ?for (int j = 0; j < 8; j++) {
?? ??? ??? ?if (n[j][col] == 0) {
?? ??? ??? ??? ?continue;
?? ??? ??? ?}
?? ??? ??? ?for (int k = j + 1; k < 9; k++) {
?? ??? ??? ??? ?if (n[j][col] == n[k][col]) {
?? ??? ??? ??? ??? ?return false;
?? ??? ??? ??? ?}
?? ??? ??? ?}
?? ??? ?}
?? ??? ?return true;
?? ?}

?? ?/**
?? ? * 檢查3X3區域是否符合要求
?? ? *
?? ? * @param row
?? ? *??????????? 檢查的行號
?? ? * @param col
?? ? *??????????? 檢查的列號
?? ? * @return true代表符合要求
?? ? */
?? ?private static boolean checkNine(int row, int col) {
?? ??? ?// 獲得左上角的坐標
?? ??? ?int j = row / 3 * 3;
?? ??? ?int k = col / 3 * 3;
?? ??? ?// 循環比較
?? ??? ?for (int i = 0; i < 8; i++) {
?? ??? ??? ?if (n[j + i / 3][k + i % 3] == 0) {
?? ??? ??? ??? ?continue;
?? ??? ??? ?}
?? ??? ??? ?for (int m = i + 1; m < 9; m++) {
?? ??? ??? ??? ?if (n[j + i / 3][k + i % 3] == n[j + m / 3][k + m % 3]) {
?? ??? ??? ??? ??? ?return false;
?? ??? ??? ??? ?}
?? ??? ??? ?}
?? ??? ?}
?? ??? ?return true;
?? ?}

?? ?/**
?? ? * 產生1-9之間的隨機數字 規則：生成的隨機數字放置在數組8-time下標的位置，隨著time的增加，已經嘗試過的數字將不會在取到
?? ? * 說明：即第一次次是從所有數字中隨機，第二次時從前八個數字中隨機，依次類推， 這樣既保證隨機，也不會再重復取已經不符合要求的數字，提高程序的效率
?? ? * 這個規則是本算法的核心
?? ? *
?? ? * @param time
?? ? *??????????? 填充的次數，0代表第一次填充
?? ? * @return
?? ? */
?? ?private static Random r=new Random();
?? ?private static int generateNum(int time) {
?? ??? ?// 第一次嘗試時，初始化隨機數字源數組
?? ??? ?if (time == 0) {
?? ??? ??? ?for (int i = 0; i < 9; i++) {
?? ??? ??? ??? ?num[i] = i + 1;
?? ??? ??? ?}
?? ??? ?}
?? ??? ?// 第10次填充，表明該位置已經卡住，則返回0，由主程序處理退回
?? ??? ?if (time == 9) {
?? ??? ??? ?return 0;
?? ??? ?}
?? ??? ?// 不是第一次填充
?? ??? ?// 生成隨機數字，該數字是數組的下標，取數組num中該下標對應的數字為隨機數字
//?? ??? ?int ranNum = (int) (Math.random() * (9 - time));//j2se
?? ??? ?int ranNum=r.nextInt(9 - time);//j2me
?? ??? ?// 把數字放置在數組倒數第time個位置，
?? ??? ?int temp = num[8 - time];
?? ??? ?num[8 - time] = num[ranNum];
?? ??? ?num[ranNum] = temp;
?? ??? ?// 返回數字
?? ??? ?return num[8 - time];
?? ?}
?? ?
?? ?public static void main(String[] args) {
?? ??? ?int[][] shuDu=generateShuDu();
?? ??? ?// 輸出結果
?? ??? ?for (int i = 0; i < 9; i++) {
?? ??? ??? ?System.out.print("{");
?? ??? ??? ?for (int j = 0; j < 9; j++) {
?? ??? ??? ??? ?int t=shuDu[i][j];
?? ??? ??? ??? ?if(t==1){
?? ??? ??? ??? ??? ?System.out.print("\"a\",");
?? ??? ??? ??? ?}else if(t==2){
?? ??? ??? ??? ??? ?System.out.print("\"b\",");
?? ??? ??? ??? ?}else if(t==3){
?? ??? ??? ??? ??? ?System.out.print("\"c\",");
?? ??? ??? ??? ?}else if(t==4){
?? ??? ??? ??? ??? ?System.out.print("\"d\",");
?? ??? ??? ??? ?}else if(t==5){
?? ??? ??? ??? ??? ?System.out.print("\"e\",");
?? ??? ??? ??? ?}else if(t==6){
?? ??? ??? ??? ??? ?System.out.print("\"f\",");
?? ??? ??? ??? ?}else if(t==7){
?? ??? ??? ??? ??? ?System.out.print("\"g\",");
?? ??? ??? ??? ?}else if(t==8){
?? ??? ??? ??? ??? ?System.out.print("\"h\",");
?? ??? ??? ??? ?}else if(t==9){
?? ??? ??? ??? ??? ?System.out.print("\"i\",");
?? ??? ??? ??? ?}
?? ??? ??? ?}
?? ??? ??? ?System.out.println("},");
?? ??? ?}
?? ?}
}

程序運行之后生成一個類似以下數組的數獨模板

{"a","b","i","g","d","e","c","h","f",}, {"c","h","f","b","a","i","e","d","g",}, {"e","d","g","c","f","h","b","i","a",}, {"d","e","b","a","h","c","g","f","i",}, {"h","i","c","f","g","b","d","a","e",}, {"g","f","a","i","e","d","h","c","b",}, {"i","c","h","e","b","f","a","g","d",}, {"b","g","d","h","i","a","f","e","c",}, {"f","a","e","d","c","g","i","b","h",}

此數組可作為數獨生成的模板，以降低 app 中數獨生成算法的難度

總結

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