You need a method to find the minimum value in every row of an array. Which of the following methods works as intended?

I.
public static int[] findMinList (int[][] a)
{
int temp [] = new int [a.length];
for (int i = 0; i

Question

You need a method to find the minimum value in every row of an array. Which of the following methods works as intended?

I.	
public static int[] findMinList (int[][] a)
{
 int temp [] = new int [a.length];
 for (int i = 0; i < a.length; i++)
  {
    int min = a[i][0];
    for (int j = 0; j < a[0].length; j++)
    {
      if (a[i][j] < min)
        min = a[0][j];
    }
    temp [i] = min;
  }
  return temp;
}

II.	
public static int[] findMinList (int[][] a)
{
  int temp [] = new int [a.length];
  for (int i = 0; i < a.length; i++)
  {
    int min = a[i][0];
    for (int j = 0; j < a[0].length; j++)
    {
      if (a[i][j] < min)
        min = a[i][j];
    }
    temp [i] = min;
  }
  return temp;
}

III.	
public static int[] findMinList (int[][] a)
{
  int temp [] = new int [a.length];
  for (int i = 0; i < a.length; i++)
  {
    int min = a[i][0];
    for (int j = 0; j < a[0].length; j++)
    {
      if (a[i][j] < a[0][j])
        min = a[i][j];
    }
    temp [i] = min;
  }
  return temp;
}

BlackTom AI · Accepted Answer

The correct implementation is the second method because it consistently compares and assigns values from the same row (a[i][j]) to the running minimum.

1. Relevant knowledge points and concepts
- 2D arrays in Java are accessed as a[i][j], where i is the row and j the column.
- To find the minimum in a row, initialize min to an element in that row (commonly a[i][0]) and compare each element of that row to update min.
- Indexing mistakes (using a[0][j] or assigning from the wrong row) produce incorrect results.

2. Step-by-step reasoning for each method
Method II:
- For each row i, set min = a[i][0].
- For each column j in that row, if a[i][j] < min, set min = a[i][j].
- After scanning all columns of row i, temp[i] = min. This exactly implements the intended algorithm.

Method I:
- Same outer/inner loop, and min initialized correctly.
- Bug: inside the if, it does min = a[0][j]; that assigns a value from row 0 instead of the current row (a[i][j]). So when a smaller element appears in a later row, min is set to the corresponding element from row 0, producing wrong results for rows other than the first.

Method III:
- The if condition compares a[i][j] < a[0][j] (element in current row vs element in row 0), not against the running min.
- Example: with row i initialized min=a[i][0], later a[i][1] might be less than a[0][1] and set min=a[i][1], but a subsequent column a[i][2] might also be less than a[0][2] and then overwrite min with a larger value. Hence final min can be incorrect.

3. Intermediate example (showing failure)
Array: {{5,2,7},{6,1,4}}
- Method II row 1 yields min = 2, row 2 yields min = 1 (correct).
- Method I on row 2: when seeing 1, it sets min = a[0][1] = 2 → final min 2 (wrong).
- Method III on row 2: sees 1 → min=1, then sees 4 < 7 → min=4 → final min 4 (wrong).

4. Why the correct answer is correct
Method II always compares and assigns values from the current row to the running min, so it reliably finds the smallest element in each row.

5. Why other options are incorrect
- I uses min = a[0][j] inside the if, assigning from the first row instead of the current row.
- III compares to a[0][j] (first row) rather than to the running min, allowing later comparisons to overwrite a true smaller value.

6. Conclusion
Only method II works as intended. Final answer: II only.

类似问题

Which statement declares a two-dimensional integer array called myArray with 3 rows and 4 columns?

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