Consider the following code.

int[][] matrix = new int[6][4];

Suppose we want to initialize the matrix to the following rows and columns.

3 2 1 0
0 1 2 3
3 2 1 0
0 1 2 3
3 2 1 0
0 1 2 3

Which of the options below correctly initializes the matrix?

I.
for (int i = 0; i

Question

Consider the following code.

int[][] matrix = new int[6][4];

Suppose we want to initialize the matrix to the following rows and columns.

3 2 1 0
0 1 2 3
3 2 1 0
0 1 2 3
3 2 1 0
0 1 2 3

Which of the options below correctly initializes the matrix?

I.	
for (int i = 0; i < matrix.length - 1; i += 2)
{
  for (int j = 0; j < matrix[i].length; j++)
  {
    matrix[i][j] = j;
    matrix[i + 1][j] = j;
  }
}

II.	
for (int i = 0; i < matrix.length - 1; i += 2) {
  for (int j = 0; j < matrix[i].length; j++) {
    matrix[i + 1][j] = j;
    matrix[i][matrix[i].length - j - 1] = j;
  }
}

III.	
for (int i = 0; i < matrix.length - 1; i += 2)
{
  for (int j = 0; j < matrix[i].length; j++)
  {
    matrix[i + 1][j] = j;
    matrix[i][matrix[i].length - j - 1] = i;
  }
}

BlackTom AI · Accepted Answer

Even-indexed rows must contain descending values 3,2,1,0 and odd-indexed rows ascending 0,1,2,3. Relevant points: a 6x4 int[][] has 6 rows (indices 0..5) and 4 columns (indices 0..3). To fill rows in pairs (i and i+1) a loop stepping i by 2 is appropriate. To invert a column index use matrix[i].length - j - 1.

Step-by-step reasoning and intermediate assignments:
1) Consider option II inner body:
   - matrix[i+1][j] = j; assigns odd row i+1 column j the value j, producing 0,1,2,3 for j=0..3.
   - matrix[i][matrix[i].length - j - 1] = j; with length=4 this becomes matrix[i][3 - j] = j.
     Evaluate j loop:
     - j=0: matrix[i][3] = 0
     - j=1: matrix[i][2] = 1
     - j=2: matrix[i][1] = 2
     - j=3: matrix[i][0] = 3
   After all j, row i (even) holds [3,2,1,0] as required. Repeating for i=0,2,4 fills rows 0..5 correctly.

Why II is correct:
- It explicitly assigns ascending values to odd rows and uses the reversed index (length - j - 1) to place ascending j values into reversed positions on even rows, yielding descending sequences.

Why I is incorrect:
- Option I sets matrix[i][j] = j and matrix[i+1][j] = j, so both the even row i and odd row i+1 become ascending 0..3. Even rows should be descending, so I fails.

Why III is incorrect:
- Option III uses matrix[i][matrix[i].length - j - 1] = i; it stores the row index i (0,2,4) into even-row cells rather than the desired column-derived values (0..3). This yields incorrect numbers (0,2,4) not 3,2,1,0.

Conclusion: Only option II produces the exact pattern shown. Correct answer: II only.

类似问题

Consider the following code segment. int[][] mat = new int[5][3]; for (int j = 0; j < mat.length; j++) { for (int k = 0; k < mat[0].length; k++) { mat[j][k] = 10 * k + j; } } What are the contents of mat after the code segment has been executed?

A project has a required return of 12.6 percent, an initial cash outflow of $42,100, and cash inflows of $16,500 in Year 1, $11,700 in Year 2, and $10,400 in Year 4. What is the net present value?

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