从输入的1d数组返回2d数组 [英] Returning a 2d array from inputed 1d array

问题描述

我想将输入的 1x6 一维数组返回到一个 3x2 的二维数组中.我想我已经弄清楚了，但是在我的代码中，我一直都需要此错误数组，但是找到了 int .我认为这是由于输入数组是1d数组而不是2d，但是我不确定.有人可以帮我解决这个问题吗?

I want to take the input 1d array that is a 1x6 and return it into a 2d array that is a 3x2. I think I have figured it out but in my code I keep getting this error array required, but int found. I think this is due to the input array being a 1d array and not a 2d, but I'm not sure. Could anyone help me to fix this?

示例:

int[] d = {4, 1, 20, 45, 2, 31};

返回2D数组 3x2

输出:

4  1  
20 45  
2   31

错误:

transmo[j][i] = d[i][j];
array required, but int found

我的下面的代码:

public class mypractice {
    public void a_three_by_two(int[] d) {
        int transmo[][] = new int[3][2];

        for (int i = 0; i < 1; i++) {
            for (int j = 0; j < 6; j++) {
                transmo[j][i] = d[i][j];
            }
        }

        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 2; j++) {
                System.out.print(transmo[i][j] + "\t");
            }
            System.out.println();
        }
    }
}

推荐答案

您的第一组循环必须与第二组循环相同，否则 transmo [i] [j] 将失败.

Your first set of loops need to be the same as the second set of loops, otherwise transmo[i][j] will fail.

然后您需要将索引计算到 d 中.

You then need to calculate the index into d.

for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 2; j++) {
        transmo[i][j] = d[i * 2 + j];
    }
}

这篇关于从输入的1d数组返回2d数组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！