合并排序2列表以查找常见元素

Question

For an assignment we have been asked to write a function that takes as inputs 2 lists and then returns a list containing all of names that were in both 'list 1' and 'list 2'.

It has been asked to be done using a merge sort based algorithm. What I have got so far returns the correct list, however I am making too many comparisons to get this list. VoterList is a specified class given to us so that we don't use normal python lists. Only VoterName objects (what the two input lists are made of) are able to be appended to a VoterList. The lists being passed in are both in lexicographical order.

Any advice on how to reduce my comparisons is welcomed. Thank you.

from classes import VoterList
def fraud_detect_merge(first_booth_voters, second_booth_voters):

    fraud = VoterList()
    length_first = len(first_booth_voters)
    length_second = len(second_booth_voters)
    first_booth_position = 0
    second_booth_position = 0
    comparisons = 0
    while first_booth_position < length_first:

        if second_booth_position == length_second:
            first_booth_position += 1
            second_booth_position = 0

        else:
            name_comparison = first_booth_voters[first_booth_position]
            comparisons += 1

            if second_booth_voters[second_booth_position] == name_comparison:
                fraud.append(second_booth_voters[second_booth_position])
                first_booth_position += 1
                second_booth_position = 0

            else:
                second_booth_position += 1


    return fraud, comparisons

Answer 1

It is not clear what your input is, is it already sorted? You are being given lists. Check out what operations can be performed on lists and you will find the pop() operation. This removes one item from the list and give you its value. As the lists are both in order the following approach can be used:

def fraud_detect_merge(first_booth_voters, second_booth_voters):
    fraud = VoterList()
    comparisons = 0

    first_booth_voters.sort()     # if not already sorted
    second_booth_voters.sort()    # if not already sorted

    first = first_booth_voters[0]
    second = second_booth_voters[0]

    while first_booth_voters and second_booth_voters:
        comparisons += 1

        if first == second:
            fraud.append(first)
            first = first_booth_voters.pop(0)
            second = second_booth_voters.pop(0)
        elif first < second:
            first = first_booth_voters.pop(0)
        else:
            second = second_booth_voters.pop(0)

    return fraud, comparisons

Answer 2

The assignment exactly asks you find a solution, and there is the big hint of merge sort, so I'm not going to spill out an answer for you :) But perhaps I can point out what's happening in your code: with your while loop you've essentially done two nested loops of lengths length_first and length_second in worst case:

for name_comparison in first_booth_voters:
  for second_name in second_booth_voters:
    comparisons += 1
    if second_name == name_comparison:
      fraud.append(second_name)
      break

That results in length_first x length_second comparisons in worst case. That is certainly not optimal given that the input lists are sorted. You need to take advantage of the sortedness. And if the input lists aren't sorted, you should consider replacing your harder-to-read/debug/understand while loop with more readable nested for loops.