我正在进行的任务要求我创建一个Sudoku游戏，而不使用任何类，方法，封装等。我是Java的新手。我无法验证用户输入到“fourArray”或“nineArray”中的值不包含重复值。到目前为止，我一直在尝试使用嵌套的for循环来迭代任一数组的列和行。例如，我一直在尝试在程序结束时包含以下代码片段，以确定是否存在任何重复值：
for (int i = 0; i < fourArray.length; i++) {
for (int j = i + 1; j < fourArray.length; j++)
if (fourArray[i] == fourArray[j]) {
System.out.println("No Sudoku");
} else {
System.out.println("Sudoku!);
}
}
然而这不起作用。我想遍历数组以查找任何重复的值，如果没有，则打印出“Sudoku！”如果有任何重复值，那么我想打印出“Sudoku！”我是否需要对数组进行排序？还是有一些我不知道的方法？我收录了我的节目。谢谢你的时间。
import java.util.Scanner;
public class Sudoku {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int boardSize = -1;
int[][] fourArray = { {0,0,0,0}, {0,0,0,0}, {0,0,0,0}, {0,0,0,0} };
int[][] nineArray = { {0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0} };
while (true)
{
Scanner boardsizeOption = new Scanner(System.in);
System.out.println("Please select a board size:" + "\n" + "1) 4x4" + "\n" + "2) 9x9");
boardSize = boardsizeOption.nextInt();
if (boardSize == 1 || boardSize == 2) {
break;
}
}
if (boardSize == 1) { //still need to build complete board
int i, j = 0;
for (i = 0; i < fourArray.length; i++)
{
for (j = 0; j < fourArray.length; j++)
System.out.print(fourArray[i][j] + " ");
System.out.println();
}
} else if (boardSize == 2) {
int i, j = 0;
for (i = 0; i < nineArray.length; i++)
{
for (j = 0; j < nineArray.length; j++)
System.out.print(nineArray[i][j] + " ");
System.out.println();
}
}
int dataSelection = -1;
while (true)
{
Scanner rowColumn = new Scanner(System.in);
System.out.println("Please select which way you would like to enter the values:" + "\n" + "1) row" + "\n" + "2) columnn");
dataSelection = rowColumn.nextInt();
if (dataSelection == 1 || dataSelection == 2) {
break;
}
}
//Entering by ROWS
//This is for a 4x4 board size using rows
if (dataSelection == 1) {
if (boardSize == 1) {
int row = 1;
while (row < 5) {
String row1Values4x4 = "-1";
while (true) {
Scanner firstRow4x4 = new Scanner(System.in);
System.out.println("Please enter four values using commas for row " + row); //this needs to loop
row1Values4x4 = firstRow4x4.next();
row1Values4x4 = row1Values4x4.replaceAll(" ",""); //this is in case user enters numbers with spaces
if (row1Values4x4.length() == 7) {
break;
}
}
String strArray[] = row1Values4x4.split(",");
int arraySidesInteger[] = new int[strArray.length];
for (int i = 0; i < strArray.length; i++) {
arraySidesInteger[i] = Integer.parseInt(strArray[i]);
}
fourArray[row-1] = arraySidesInteger;
for (int i = 0; i < fourArray.length; i++) {
for (int j = 0; j < fourArray.length; j++)
System.out.print(fourArray[i][j] + " ");
System.out.println();
}
row++;
}
//This is for a 9x9 board size using rows
} else {
int row = 1;
while (row < 10) {
String row1Values9x9 = "-1";
while (true) {
Scanner firstRow9x9 = new Scanner(System.in);
System.out.println("Please enter nine values using commas for row " + row); //this needs to loop
row1Values9x9 = firstRow9x9.next();
row1Values9x9 = row1Values9x9.replaceAll(" ",""); //this is in case user enters numbers with spaces
if (row1Values9x9.length() == 17) {
break;
}
}
String strArray[] = row1Values9x9.split(",");
int arraySidesInteger[] = new int[strArray.length];
for (int i = 0; i < strArray.length; i++) {
arraySidesInteger[i] = Integer.parseInt(strArray[i]);
}
nineArray[row-1] = arraySidesInteger;
for (int i = 0; i < nineArray.length; i++) {
for (int j = 0; j < nineArray.length; j++)
System.out.print(nineArray[i][j] + " ");
System.out.println();
}
row++;
}
}
//Entering by COLUMNS
//This is for 4x4 board size using columns
} else {
if (boardSize == 1) {
int column = 1;
while (column < 5) {
String column1Values4x4 = "-1";
while (true) {
Scanner firstColumn4x4 = new Scanner(System.in);
System.out.println("Please enter four values using commas for column " + column);
column1Values4x4 = firstColumn4x4.next();
column1Values4x4 = column1Values4x4.replaceAll(" ","");
if (column1Values4x4.length() == 7) {
break;
}
}
String strArray[] = column1Values4x4.split(",");
int arraySidesInteger[] = new int[strArray.length];
for (int i = 0; i < strArray.length; i++) {
arraySidesInteger[i] = Integer.parseInt(strArray[i]);
}
for (int i = 0; i < arraySidesInteger.length; i++) {
fourArray[i][column-1] = arraySidesInteger[i];
}
for (int i = 0; i < fourArray.length; i++) {
for (int j = 0; j < fourArray.length; j++)
System.out.print(fourArray[i][j] + " ");
System.out.println();
}
column++;
}
//This is for a 9x9 board size using columns
} else {
int column = 1;
while (column < 10) {
String column1Values9x9 = "-1";
while (true) {
Scanner firstColumn9x9 = new Scanner(System.in);
System.out.println("Please enter nine values using commas for column " + column);
column1Values9x9 = firstColumn9x9.next();
column1Values9x9 = column1Values9x9.replaceAll(" ","");
//row1Values4x4 = row1Values4x4.replaceAll(",","");
if (column1Values9x9.length() == 17) {
break;
}
}
String strArray[] = column1Values9x9.split(",");
int arraySidesInteger[] = new int[strArray.length];
for (int i = 0; i < strArray.length; i++) {
arraySidesInteger[i] = Integer.parseInt(strArray[i]);
}
for (int i = 0; i < arraySidesInteger.length; i++) {
nineArray[i][column-1] = arraySidesInteger[i];
}
for (int i = 0; i < nineArray.length; i++) {
for (int j = 0; j < nineArray.length; j++)
System.out.print(nineArray[i][j] + " ");
System.out.println();
}
column++;
}
}
for (int i = 0; i < fourArray.length; i++) {
for(int j = i + 1; j < fourArray.length; j++) {
if(fourArray[i] == fourArray[j]) {
System.out.println("No Sudoku");
} else {
System.out.println("Sudoku!");
}
}
}
}
}
答案 0 :(得分：2)
由于它是家庭作业，我将尽量减少代码，但我认为如果您有关于2D数组的更多信息，你会很好的解决它，其中一些相当棘手：
fourArray是一个数组数组，fourArray[i]指的是一个数组（您可以将其视为二维数组的第i行）。fourArray[i][j]。myArray1 = myArray2就会发生这种情况）。Arrays.equals(myArray1, myArray2)。fourArray.length是一维的大小; fourArray[x].length是其他维度的大小（x无关紧要，只要它在0和fourArray.length - 1之间。 在回复评论时添加：我的理解和假设是您试图避免二维fourArray中包含的任何值之间的任何重复值。有很多解决方案。
可能被称为天真的解决方案是首先使用一对嵌套的for循环来遍历fourArray中的每个值。对于每个值，将其与其他每个值进行比较。您的中间代码可能如下所示：
for (int i = 0; i < fourArray.length; i++) {
for (int j = 0; j < fourArray[i].length; j++) {
// TODO: Compare value at (i,j) to every other point by nesting
// two more for loops with new iterators (called, e.g., m and n)
// TODO: If a duplicate is found, either stop searching, or at
// least mark that a duplicate has been found somehow.
}
}
一方面，这有点低效。另一方面，对于小型2-D阵列来说，它仍然是完全无足轻重的计算，所以如果它对你有意义，那么就去做其他问题。
但是，如果您感兴趣并假设允许的值是顺序集的一部分，我会提交另一个想法供您考虑（即，在典型的数独游戏中，您有3x3个框，其中允许的值是总是1-9，永远不会更高）。如果您有一个数组count[]来跟踪这些已知值发生了多少次，该怎么办？因此，其中的所有值都将初始化为零。当您遍历表中的每个点时（如上面的代码示例所示），您可以使用找到的值 - 将其称为v - 来增加count[v]。 count[]中任何大于1的值都表示重复。
答案 1 :(得分：0)
首先，在您的整个班级代码中，您需要将您的数独检查向下移动1 }（之后应该只有2个，主管和类）。
第二件事就像你想的那样，假设我正确理解了问题，你的双循环是错误的。如果要检查网格中每个其他值的每个值，我就会这样做：
boolean sudoku = true;
for (int i = 0; i < fourArray.length; i++) {
for (int j = 0; j < fourArray[i].length; j++) {
if (fourArray[i] == fourArray[j]) {
sudoku = false;
break;
}
if (!sudoku){
break;
}
}
}
if (sudoku){
System.out.println("Sudoku!");
} else {
System.out.println("No Sudoku!");
}