发出生成字符串排列列表的问题

Question

I want to generate a list of all permutations of a string using a backtracking algorithm.

I modified some code from a previous post: https://stackoverflow.com/a/20955291/12021192

def permutations(string, step = 0):

    permTrack = []

    if step == len(string):
        permTrack.append(''.join(string))

    for i in range(step, len(string)):
        string_copy = list(string)
        string_copy[step], string_copy[i] = string_copy[i], string_copy[step]
        permutations(string_copy, step + 1)

    return permTrack

permTrack = permutations('ABC')

I expected permTrack = ['ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA] , but the actual output is permTrack = [ ] .

The idea is to is to append to a list permTrack at the base case, when step == len(string). This works for instance in the original code to print the permutations.

Can someone help me figure this out?

Answer 1

You don't do anything with what is returned by permutations when it is called recursively.

If you add what gets returned to permTrack , you should get what you want.