使用Java 8+流，您可以编写以下函数：

private static String[] concatArrays(final String[]... arrays) {
    return Arrays.stream(arrays)
         .flatMap(Arrays::stream)
         .toArray(String[]::new);
}

ArrayList<String> both = new ArrayList(Arrays.asList(first));
both.addAll(Arrays.asList(second));

both.toArray(new String[0]);

如果您想在解决方案中使用ArrayList，可以尝试以下操作：

public final String [] f(final String [] first, final String [] second) {
    // Assuming non-null for brevity.
    final ArrayList<String> resultList = new ArrayList<String>(Arrays.asList(first));
    resultList.addAll(new ArrayList<String>(Arrays.asList(second)));
    return resultList.toArray(new String [resultList.size()]);
}

非Java 8解决方案：

public static int[] combineArrays(int[] a, int[] b) {
        int[] c = new int[a.length + b.length];

        for (int i = 0; i < a.length; i++) {
            c[i] = a[i];
        }

        for (int j = 0, k = a.length; j < b.length; j++, k++) {
            c[k] = b[j];
        }

        return c;
    }

您可以尝试连接多个数组的方法：

public static <T> T[] concatMultipleArrays(T[]... arrays)
{
   int length = 0;
   for (T[] array : arrays)
   {
      length += array.length;
   }
   T[] result = (T[]) Array.newInstance(arrays.getClass().getComponentType(), length) ;

   length = 0;
   for (int i = 0; i < arrays.length; i++)
   {
      System.arraycopy(arrays[i], 0, result, length, arrays[i].length);
      length += arrays[i].length;
   }

   return result;
}

public String[] concat(String[]... arrays)
{
    int length = 0;
    for (String[] array : arrays) {
        length += array.length;
    }
    String[] result = new String[length];
    int destPos = 0;
    for (String[] array : arrays) {
        System.arraycopy(array, 0, result, destPos, array.length);
        destPos += array.length;
    }
    return result;
}