插入有序链接列表Python和总和重复项

Question

I am trying to insert item pairs (letter, frequency) into an ordered linked list. So far I am able to create the sorted linked list but I cannot figure out how to update the frequency and rearrange the list if a letter gets comes up twice.

So far I have:

def add(self, letter, frequency):

    temp = Frequency(letter, frequency)

    curr = self.head
    prev = None
    stop = False

    while curr != None and not stop:
        if curr.frequency < temp.frequency:
            stop = True
        else:
            prev = curr
            curr = curr.next

    if prev == None:
        temp.set_next(self.head)
        self.head = temp
    else:
        temp.set_next(curr)
        prev.set_next(temp)

f = SortedFrequencyList()
f.add('a', 3)
f.add('b', 5)
f.add('g' 1)

returns

({b: 5}, {a: 3}, {g: 1})

but if I were to do

f = SortedFrequencyList()
f.add('a', 3)
f.add('b', 5)
f.add('g', 1)
f.add('a', 3)

I get

({b: 5}, {a: 3}, {a: 3}, {g: 1})

instead of

({a: 6}, {b: 5}, {g: 1})

Answer 1

We can't do much in the way of coding help, since you haven't posted your support code: we don't have enough to reproduce the problem. You also haven't given us any coding attempt from which to work.

You need to build out more of your linked-list support. So far, all you have is a method that inserts a new entry into the list. I recommend that you write a find method that locates an existing element by name (letter), and an update that will find it, increase the frequency, and sort it back into the list.

The building-block way to do the update is to write a delete routine as well. Find the item, hold a copy, delete it from the list, and then add a new one with the updated frequency.

The nicer way to update is to do the removal and reinsertion in one routine, backing up the list. This suggests either a recursive memory or a doubly-linked list. An even faster way is to implement some sort of indexed tree structure, so that the searches are faster.

Does this get you moving in the right direction?