那么，如果你要枚举这些填充模式的索引，你会得到

0,0

1,0

0,1

2,0

1,1

0,2

2,1

1,2

2,2

所以，你需要遍历两个指标的总和。也就是说，总添加量。如您所见，0,0总计0，1,0和0,1总计1，依此类推。给我们这样的东西：

0 1 2

1 2 3

2 3 4

要以这种对角线模式进行迭代，我们可以执行以下操作：

// set up your matrix, any size and shape (MxN) is fine, but jagged arrays will break

int[][] matrix = {{0,0,0},{0,0,0},{0,0,0}};

// number is the value we will put in each position of the matrix

int number = 1;

// iterate while number is less than or equal to the total number of positions

// in the matrix. So, for a 3x3 matrix, 9. (this is why the code won't work for

// jagged arrays)

for (int i = 0; number <= matrix.length * matrix[0].length; i++) {

// start each diagonal at the top row and from the right

int row = 0;

int col = i;

do {

// make sure row and length are within the bounds of the matrix

if (row < matrix.length && col < matrix[row].length) {

matrix[row][col] = number;

number++;

}

// we decrement col while incrementing row in order to traverse down and left

row++;

col--;

} while (row >= 0);

}

请注意，虽然此实现将适用于所有矩阵大小(和形状)，但它不会尽可能高效。其中n是matrix.length(假设为方形矩阵)，该实现是大O符号中的最优O(n ^ 2)类算法;然而，它有效地执行2×n ^ 2次迭代，而最优解只能执行n ^ 2。