Question

如何以相同的顺序从1-d数组中的2-d数组传输元素？

Answer 1

以下是我看到的最可能的情景。

class nikhil {

   public static int[] firstIndex(int[][] arr2d) {
        int rows = arr2d.length;
        if(rows == 0) return new int[0];
        int cols = arr2d[0].length;
        if(cols == 0) return new int[0];

        int[] arr1d = new int[rows * cols];
        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {
                arr1d[i * cols + j] = arr2d[i][j];
            }
        }
        return arr1d;

   }

   public static int[] secondIndex(int[][] arr2d) {
       int rows = arr2d.length;
       if(rows == 0) return new int[0];
       int cols = arr2d[0].length;
       if(cols == 0) return new int[0];

        int[] arr1d = new int[rows * cols];
        for(int i = 0; i < cols; i++) {
            for(int j = 0; j < rows; j++) {
                arr1d[i * rows + j] = arr2d[j][i];
            }
        }
        return arr1d;

   }

   public static int[] sorted(int[][] arr2d) {
       int[] arr1d = firstIndex(arr2d);
       java.util.Arrays.sort(arr1d);
       return arr1d;
   }

   public static int[] scramble(int[][] arr2d) {
       int[] arr1d = firstIndex(arr2d);
       java.util.List<Integer> list = new java.util.ArrayList<Integer>();
       for(int i : arr1d) list.add(i);
       java.util.Collections.shuffle(list);

       for(int i = 0; i < list.size(); i++) arr1d[i] = list.get(i).intValue();
       return arr1d;

   }

   public static void main(String[] args) {
           /*
            5  4  6 10
            3  1  9 12
            8  2  7 0
            */
       final int rows = 3;
       final int cols = 4;
       int[][] arr2d = new int[rows][cols];
           arr2d[0][0] = 5;
           arr2d[0][1] = 4;
           arr2d[0][2] = 6;
           arr2d[0][3] = 10;
           arr2d[1][0] = 3;
           arr2d[1][1] = 1;
           arr2d[1][2] = 9;
           arr2d[1][3] = 12;
           arr2d[2][0] = 8;
           arr2d[2][1] = 2;
           arr2d[2][2] = 7;
           arr2d[2][3] = 0;

        int[] first = firstIndex(arr2d);
        int[] second = secondIndex(arr2d);
        int[] sorted = sorted(arr2d);
        int[] scrambled = scramble(arr2d);

        System.out.println(java.util.Arrays.toString(first));
        System.out.println(java.util.Arrays.toString(second));
        System.out.println(java.util.Arrays.toString(sorted));
        // for oli
        System.out.println(java.util.Arrays.toString(scrambled));

   }

}

Answer 2

虽然已经回答here，但我将在此处发布解决方案示例。

假设您有一个2D数组，例如originalArray [ ][ ]

int a [] = {1,2,6,7,2};
int b [] = {2,44,55,2};
int c [] = {2,44,511,33};

int originalArray [][] = new array[][]{a,b,c};

案例1：

如果以上所有1D数组都具有相同的长度，那么您可以这样继续：

int[] newArray = new int[3 * a.length];
int index = 0;
for (int n = 0; n < a.length; n++) {
    newArray[index++] = a[n];
    newArray[index++] = b[n];
    newArray[index++] = c[n];
}

案例2：

如果上述所有1D数组的长度都不同，那么您可以这样做，

int[] newArray = new int[a.length + b.length + c.length];
System.arrayCopy(a, 0, newArray, 0, a.length);
System.arrayCopy(b, 0, newArray, a.length, b.length);
System.arrayCopy(c, 0, newArray, a.length + b.length, c.length);

如何以相同的顺序从1-d数组中的2-d数组转移元素？

2 个答案: