java实现一个3×3的九宫格,将1至9的数字填入方格,使矩阵中心的元素为5,且每行每列及每个对角线的值相加都相同。
时间: 2024-06-11 13:04:43 浏览: 195
要实现一个满足条件的九宫格,可以使用回溯算法来解决。下面是一个Java实现的示例代码:
```java
public class MagicSquare {
private int[][] square;
private boolean[] used;
private int sum;
public MagicSquare() {
square = new int[3][3];
used = new boolean[10];
sum = 15;
}
public void solve() {
if (backtrack(0, 0)) {
printSquare();
} else {
System.out.println("No solution exists.");
}
}
private boolean backtrack(int row, int col) {
if (row == 3) {
return isMagicSquare();
}
for (int num = 1; num <= 9; num++) {
if (!used[num]) {
square[row][col] = num;
used[num] = true;
int nextRow = (col == 2) ? row + 1 : row;
int nextCol = (col + 1) % 3;
if (backtrack(nextRow, nextCol)) {
return true;
}
square[row][col] = 0;
used[num] = false;
}
}
return false;
}
private boolean isMagicSquare() {
for (int i = 0; i < 3; i++) {
int rowSum = square[i][0] + square[i][1] + square[i][2];
int colSum = square[0][i] + square[1][i] + square[2][i];
if (rowSum != sum || colSum != sum) {
return false;
}
}
int diagSum1 = square[0][0] + square[1][1] + square[2][2];
int diagSum2 = square[0][2] + square[1][1] + square[2][0];
return diagSum1 == sum && diagSum2 == sum;
}
private void printSquare() {
System.out.println("Magic Square:");
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
System.out.print(square[i][j] + " ");
}
System.out.println();
}
}
public static void main(String[] args) {
MagicSquare magicSquare = new MagicSquare();
magicSquare.solve();
}
}
```
上述代码使用回溯算法来生成满足条件的九宫格。程序将尝试将数字从1到9填入九宫格中的空格,然后验证是否满足条件。如果满足条件则输出结果,否则回溯并继续尝试其他可能的组合。
运行以上代码,将输出一个满足条件的九宫格。
阅读全文