java - Magic Square 给出 ArrayIndexOutOfBoundException

Question

我一直在研究Magic Square的形成，在阅读了算法之后，我发现在形成MagicSquare时需要遵循一定的规则。

我遵循的算法是:

1. The magic constant will always be equal to n(n^2 + 1)/2, where n is the dimension given.
2. Numbers which magicSquare consists will always be equals 1 to n*n.
3. For the first element that is 1, will always be in the position (n/2, n-1).
4. Other elements will be placed like (i--,j++)

5. The condition to be put through for placing an elements are :

   a) If i < 0, then i = n-1.
   b) If j == n, then j = 0.
   c) This is a special case, if i < 0 and j=n happens at the same time, then i = 0, j = n-2.
   d) If the position is already occupied by some other element, then i++, j = j-2.
   

6. 然后根据条件输入magicSquare内部的元素。

根据上述算法，我写下了一段代码，由于某种原因，我得到了

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 3                                                                  
        at Main.generateMagicSquare(Main.java:25)                                                                                       
        at Main.main(Main.java:58) 

这很奇怪。我已经检查过，感觉使用代码来获得所需的结果是安全的，但我不知道哪里出了问题。

代码

static void generateMagicSquare(int n){
    int[][] magicSquare = new int[n][n];

    //initialising for pos of the elem 1
    int i = n/2, j = n-1;
    magicSquare[i][j] = 1;

    //the element consist by the magic square will always be equal to 1 to n*n
    for(int num=2; num <= n*n; num++){
        //it must go like this, for any other element
        i--; j++;

        // if the element is already present
        if(magicSquare[i][j] != 0){
            i++;
            j -= 2; 
        }else{
            if(i < 0)
                i = n-1;

            if(j == n)
                j = 0;

            if(i < 0 && j == n){
                i = 0;
                j = n-2;
            }
        }

        magicSquare[i][j] = num;
    }

    for(int k=0; k<n; k++){
        for(int l=0; l<n; l++){
            System.out.print(magicSquare[k][l] + " ");
        }

        System.out.println();
    }
}

如有任何帮助，我们将不胜感激。谢谢。因为我可以从互联网上复制并粘贴代码，但我想以我的方式学习它，你的帮助将帮助我实现我想要的。 :)

编辑

读完异常后，我对代码做了一些修改，但仍然有一些结果没有达到要求。

这是我更新的代码 =======>

static void generateMagicSquare(int n){
int[][] magicSquare = new int[n][n];

//initialising for pos of the elem 1
int i = n/2, j = n-1;
magicSquare[i][j] = 1;

//the element consist by the magic square will always be equal to 1 to n*n
for(int num=2; num <= n*n; num++){
    //it must go like this, for any other element
    i--; j++;

    if(i < 0){
       i = n-1;
    }

    if(j == n){
       j = 0;
    }

    if(i < 0 && j == n){
       i = 0;
       j = n-2;
    }

    if(magicSquare[i][j] != 0){
       i++;
       j -= 2;
    }else{
       magicSquare[i][j] = num;
    }
}

for(int k=0; k<n; k++){
    for(int l=0; l<n; l++){
        System.out.print(magicSquare[k][l] + " ");
    }

    System.out.println();
}
}

我得到这个输出:

2 0 6                                                                                                                                   
9 5 1                                                                                                                                   
7 3 0 

仍然不是正确的答案。

Answer 1

此行引发错误:

if(magicSquare[i][j] != 0)

问题是数组 magicSquare 初始化为:

int[][] magicSquare = new int[n][n];

意味着它有 n 列，索引从 0 到 n - 1(索引从零开始) )。
变量j初始化为

j = n-1;

以及后来的这一行:

j++;

使j等于n
因此，当您访问 magicSquare[i][j] 时，您正在尝试访问不存在的 magicSquare[i][n]
。