Question

我想知道是否有办法在方法中更改2D数组的列而无需接收整个数组并对其进行修改？

public static void main(String[] args){
  int[][] a = {{1,2,3},{4,5,6},{7,8,9}};
  int[] b = reverse(new int[]{1,2,3});
  System.out.println(Arrays.toString(b));
  System.out.println(Arrays.deepToString(a));
  System.out.println(Arrays.toString(new int[]{a[0][0],a[1][0],a[2][0]}));
  System.out.println(Arrays.deepToString(a));
}
public static int[] reverse(int[] start)
{
    int[] result = new int[start.length];
    int x = 0;
    for(int i= start.length-1; i >= 0; i --)
    {
        result[i] = start[x];
        x ++;
    }
    return result;
}

目前，它会反转输入的数字，但不会修改a[0][0]，a[1][0]和a[2][0]的位置。原始数组保持不变，但创建的新数组已修改。我该如何解决这个问题？

Answer 1

您可以先对内部子数组进行排序，然后对外部父数组进行排序。

    //Array for testing
    int[][] a = {{1,2,3},{4,5,6},{7,8,9}};
    System.out.println(Arrays.deepToString(a));

    //loop through inner array only and reverse items
    for (int i = 0; i < a.length; i++){
        a[i] = reverse(a[i]);
    }
    System.out.println(Arrays.deepToString(a));

    //loop through outer array only and reverse parents.
    //Note that this only needs to goe through half the array that is why we use "a.length / 2".
    for (int i = 0; i < a.length / 2; i++){
        int[] temp = a[i];
        a[i] = a[a.length - i - 1];
        a[a.length - i - 1] = temp;
    }
    System.out.println(Arrays.deepToString(a));

输出：

    [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
    [[3, 2, 1], [6, 5, 4], [9, 8, 7]]
    [[9, 8, 7], [6, 5, 4], [3, 2, 1]]

更改2D数组的列，就好像它是一维数组java一样

1 个答案: