Python - 递归函数不能与len（）一起使用，尽管返回的值是列表

Question

So I want to create a recursive function that functions in a use-it or lose-it sense, where it uses an unlimited number of coins by a list, and counts the number that are needed.

So say you have: change(48, [1, 5, 10, 25, 50]) it would return 6, because it would use 25x1, 10x2 and 1x3, totalling 6 coins.

def change(value, L):

    if not L:
        return L

    if L[-1] > value:
        return change(value, L[:-1])

    else:
        useIt = [L[-1]] + change(value - L[-1], L)
        return useIt

This returns the list of the coins used, however if I return len(useIt), I get this error:

TypeError: can only concatenate list (not "int") to list

However, this would return the right value:

print(len(change(48, [1, 5, 10, 25, 50])))

How do I return the length of the list without doing this? And no loops please, only recursion, this is review for an exam.

Answer 1

You have done excellent work, just place 1 in place of [L[-1]] in else part and return 0 in the base case and you are done.

def change(value, L):
    if not L:
        return 0

    if L[-1] > value:
        return change(value, L[:-1])
    else:
        useIt = 1 + change(value - L[-1], L)
        return useIt


print(change(48, [1, 5, 10, 25, 50]))