Given the fact that the problem definition specifies "pointers", python is not a suitable language to implement this. But you can use python variables as "pointers" (or rather references) because that is what they are; A python variable is just a name for or refrence to an object.

But if you want to implement in python, I would use a list ot tuples.

The first one is a list of (name, weight) tuples.

In [1]: data = [("Michael", 275), ("Tom", 150), ("Abe", 200)]

The order in this list doesn't matter. Just append new tuples to this list as they arrive.

Now the easy way to do it would be to make shallow copies (which reference the same tuples), and sort them appropriately just before you print them;

In [2]: namelist = [d for d in data]

In [3]: namelist.sort(key=lambda x: x[0])

In [4]: namelist
Out[4]: [('Abe', 200), ('Michael', 275), ('Tom', 150)]

and

In [5]: weightlist = [d for d in data]

In [6]: weightlist.sort(key=lambda x: x[1])

In [7]: weightlist
Out[7]: [('Tom', 150), ('Abe', 200), ('Michael', 275)]

Printing these in the correct sequence is now trivial.

But this is expressly forbidden in the exercise. So what you have to do is something like this;

• Create a new (name, weight) tuple
• Walk the list of tuples sorted by weight and compare the weight in the new tuple with the weight in the existing tuple (hint: use enumerate so you get the index of the tuple in the list). As soon as you've found a weight that is greater than the weight of the listed tuple, insert the new tuple in the weight-sorted list.
• similar for the name-sorted list, but then using the name as the compare value.

Something like this;

In [10]: newvalue = ("Eric", 225)

In [11]: for index, (name, weight) in enumerate(weightlist):
   ....:     if newvalue[1] < weight:
   ....:         weightlist.insert(index, newvalue)
   ....:         break
   ....:     

In [12]: weightlist
Out[12]: [('Tom', 150), ('Abe', 200), ('Eric', 225), ('Michael', 275)]

Note that this algorithm assumes that weightlist is already in sorted order!

Another solution that is more in line of the assignment would be to use a dictionary for every person;

In [37]: newvalue = {"name": "Eric", "weight": 225, "nextname": None, "nextweight": None}

You will also need the data list to hold all the dictionaries. And you will need two variables startname and startweight to hold the first name and lowest weight respectively.

After you have made a newvalue, you start with comparing newvalue["weight"] to startweight["weight"]. If the new weight is smaller than the startweight, then newvalue becomes the new startweight, and newvalue["nextweight"] should be set to the old startweight. If not, you move to the next item in the list and compare again. Note that if you want to insert in the chain, you have to change two nextweight attributes!

This is a double, singly linked list. Beginning with startweight and startname you can print both in order by walking both chains.