仅迭代给定列表 Python 中的零

Question

I have a problem that I have written pseudocode for, and I am having a really hard time translating it into workable Python code. It works like this: the 0's in my list represent available spots that I can insert numbers into, and I want to insert the next number into the next available spot by counting the free spaces, and I increase the number of spots I count by 1 for every loop. I am also trying to write this code to work with list of any given size. My first attempt was to try and index past the size of the list thinking it would loop back around but it didn't work as you cannot index a spot in your list that doesn't exist.

Here is the pseudocode:

Cycle 1: Count 1 space starting from first available space:                         0 1 0 0 0
Cycle 2: Count 2 spaces starting from first available space from last insertion:    0 1 0 0 2
Cycle 3: Count 3 spaces starting from first available space from last insertion:    3 1 0 0 2
Cycle 4: Count 4 spaces starting from first available space from last insertion:    3 1 4 0 2
Cycle 5: Count 5 spaces starting from first available space from last insertion:    3 1 4 5 2

Note: The numbers that are being inserted into the list start at 1 and increase by 1 for every loop.

Here is the code I have setup so far:

#The output for list of size 4 should have the numbers in this order: 2 1 4 3
#The output for list of size 5 should have the numbers in this order: 3 1 4 5 2
results = [4, 5]
print(results)

for i in results:

    myList = [0] * i
    print(myList)

    count = 0

    while count < len(myList):
        
        myList[count] = count+1
        
        print(myList)
        count += 1

My goal is to implement this as simply as possible, and I am having a hard time though I feel like I am missing something very obvious.

Answer 1

Well the most straightforward and easy to comprehend way is to just work with an index pointer with which you just keep iterating the list over and over again and a counter for the amount of spaces you need to skip next. A quick example:

list_sizes = [4, 5]
for list_size in list_sizes:
    your_list = [0] * list_size
    index = 0
    spaces_to_skip = 1
    space_count = 0
    while spaces_to_skip <= len(your_list):
        if your_list[index] == 0:
            # Found a space at the current pointer, determine what to do.
            if space_count == spaces_to_skip:
                # Skipped the correct amount of spaces (entries with 0)
                # Set value in the list
                your_list[index] = spaces_to_skip
                # Set the new amount of spaces to skip
                spaces_to_skip += 1
                # Reset the current space counter
                space_count = 0
            else:
                # Increase the current amount of spaces found
                space_count += 1

        # Move to next entry in list or start from the beginning
        index = (index + 1) % len(your_list)
    
    print(your_list)

Which gives as output:

[2, 1, 4, 3]
[3, 1, 4, 5, 2]

Answer 2

You can define generator function which will return element and it's index in source list:

def list_cycle(lst):
    i = 0
    while True:
        idx = i % len(lst)
        i += 1
        yield idx, lst[idx]

Using list_cycle() we can iterate over cycled list and decrement from current counter 1 every time empty space ( 0 ) occurred and write this counter once we counted enough:

def func(size):
    l = [0] * size
    i = curr = 1
    for idx, el in list_cycle(l):
        if el == 0:  # free space
            if i == 0:  # if counted enough
                l[idx] = curr
                i = curr = curr + 1
                if curr > size:
                    return l
            else:  # not enough
                i -= 1

Usage is simple:

print(func(4))   # => [2, 1, 4, 3]
print(func(5))   # => [3, 1, 4, 5, 2]
print(func(10))  # => [9, 1, 8, 5, 2, 4, 7, 6, 3, 10]