如何在2D数组中找到常见数字或异常，并使其触发其他内容

Question

我有一个2D数组(10x10的矩阵(，值范围从0到-5。

我希望在数组中找到一个值序列时触发一个方法。

例如，有一个由两个负2组成的序列。我希望它能触发一个事件/方法，得到4分的奖励。只有当有两个-2时才应该发生这种情况，而不是只有一个-2时。

我试着实现这样的目标，但我不知道如何告诉程序只在矩阵中找到值的"n"个数时触发。

public class Test {
static int board[][] = new int[10][10];
public static void Test() {
int i, j;
board[0][0] = -1;
board[0][1] = -1;
board[1][1] = -2;
board[1][2] = -2;
board[1][3] = -2;
board[1][4] = -2;
for (i = 0; i < board.length; i++) {
System.out.println("");
for (j = 0; j < board.length; j++) {
//board[i][j] = 0;
System.out.print(board[i][j]);
}
}
System.out.println();
}
public static void scanBoard() {
int i, j;
for (i = 0; i < board.length; i++) {
for (j = 0; j < board.length; j++) {
if (board[i][j] == -1) {
System.out.println("Hello");
}
}
}
}
public static void main(String[] args) {
Test(); //prints out whole array
scanBoard(); //scans for 
}
}

Answer 1

public class Main {
static final int size = 10;
static int[][] matrix = new int[size][size];
public static void main(String[] args) {
System.out.println("The first matrix.n");
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
if (i == 3 && j > 3) {
matrix[i][j] = -2; //-2
} else {
matrix[i][j] = 1;
}
System.out.print(matrix[i][j]);
}
System.out.println();
}
scanBoard();
System.out.println("nThe second matrix.n");
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
if (i == 9 && j > 5) {
matrix[i][j] = 2; //changed it from -2 to 2
} else {
matrix[i][j] = 1;
}
System.out.print(matrix[i][j]);
}
System.out.println();
}
scanBoard();
}
static void scanBoard() {
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
if (matrix[i][j] == -2 && (j + 3 < size)) {
if (matrix[i][j + 1] == -2 && matrix[i][j + 2] == -2 && matrix[i][j + 3] == -2) {
System.out.println("nThere you go, a special effect!".toUpperCase());
}
}
}
}
}
}

Answer 2

我不确定这是否是您根据请求希望看到的结果。我希望这对你有帮助。我对你的代码做了一些修改，这样会更容易阅读(在我看来，哈哈(。

public class Main {
static final int size = 10;
static int[][] matrix = new int[size][size];
public static void main(String[] args) {
System.out.println("The first matrix.n");
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
if (i == 9 && (j == 0 || j == 1)) {
matrix[i][j] = -2; //-2
} else {
matrix[i][j] = 1;
}
System.out.print(matrix[i][j]);
}
System.out.println();
}
scanBoard();
System.out.println("nThe second matrix.n");
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
if (i == 8 && (j == 5 || j == 6)) {
matrix[i][j] = 2; //changed it from -2 to 2
} else {
matrix[i][j] = 1;
}
System.out.print(matrix[i][j]);
}
System.out.println();
}
scanBoard();
}
static void scanBoard() {
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
if (matrix[i][j] == -2 && (j + 1 < size)) {
if (matrix[i][j + 1] == -2) {
//You can remove the '.toUpperCase()', it's just my personal preference
System.out.println("nThere you go, a special effect!".toUpperCase());
}
}
}
}
}
}

Answer 3

根据我从问题陈述和评论中了解到的情况，您希望您的扫描板表现如下：

public static void scanBoard(int value, int frequency) {
int i, j;
if (value <= 0 && value >= -5 && frequency >= 2 && frequency <= 10) {
for (i = 0; i < board.length; i++) {
int rowFrequency = 0;
for (j = 1; j < board.length; j++) {
if (board[i][j] == value && board[i][j - 1] == value) {
rowFrequency++;
} else {
rowFrequency = 0;
}
if (rowFrequency + 1 >= frequency) {
System.out.println("Hello");
}
}
}
}
}
public static void main(String[] args) {
Test(); //prints out whole array
scanBoard(-2, 4); //prints Hello once
scanBoard(-2, 3); //prints Hello twice
scanBoard(-2, 3); //prints Hello thrice
}