我想消除下面代码中的第三个参数,它只是一个空数组,我认为我应该能够在函数本身内将其创建为局部变量。
作为奖励,我还想将其构建为单个函数,尽管我认为当前的代码递归结构无法实现这一点。
我尝试创建一个空数组作为局部变量(请参阅注释掉的整数列表)
我还尝试创建一个计数变量,以随着找到的每个组合而递增(请参阅注释掉的计数变量)
def count_combinations(number, integers_available, integers):
coin_set = []
# integers = []
# count = 0
if sum(integers) == number:
coin_set.append(integers)
# count += 1
elif sum(integers) > number:
pass
elif integers_available == []:
pass
else:
for c in count_combinations(number, integers_available[:], integers + [integers_available[0]]):
coin_set.append(c)
# count += 1
for c in count_combinations(number, integers_available[1:], integers):
coin_set.append(c)
# count += 1
# return count += 1
return coin_set
def count_total(number, integers_available, integers):
return len(count_combinations(number, integers_available, integers))
# Testing the code
number = 15
integers_available = [1, 5, 10]
print(count_total(number, integers_available, []))
我希望得到相同的结果,但函数中的参数较少,因为其中一个参数将被切换为局部变量。
最佳答案
正如评论中所讨论的,动态编程方法在这里可能更Pythonic。
from collections import Counter
def ways(total, coins=(1,2,5,10,20,50,100)):
counts = [[Counter()]] + [[] for _ in range(total)]
for coin in coins:
for i in range(coin, total + 1):
counts[i] += [c + Counter({coin: 1}) for c in counts[i-coin]]
return counts[total]
演示:
>>> ways(15, coins=(1,5,10))
[Counter({1: 15}),
Counter({1: 10, 5: 1}),
Counter({1: 5, 5: 2}),
Counter({5: 3}),
Counter({1: 5, 10: 1}),
Counter({5: 1, 10: 1})]
>>> len(ways(15, coins=(1,5,10)))
6
关于python - 如何消除该函数中的一个参数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54429721/