这是一个很大的算法问题，但老实说，我认为您的实现不正确，或者很抱歉，我可能不了解您函数的输入/输出。
这是实现的修改版本。
def C(i, coins, cdict = None):
if cdict == None:
cdict = {}
if i <= 0:
cdict[i] = 0
return cdict[i]
elif i in cdict:
return cdict[i]
elif i in coins:
cdict[i] = 1
return cdict[i]
else:
min = 0
for cj in coins:
result = C(i - cj, coins)
if result != 0:
if min == 0 or (result + 1) < min:
min = 1 + result
cdict[i] = min
return cdict[i]
这是我尝试解决类似问题的尝试，但是这次返回了硬币列表。我最初从递归算法开始，该算法接受一个总和和一个硬币列表，如果找不到这种配置，它可能返回一个硬币数量最少的列表，或者返回无。
def get_min_coin_configuration(sum = None, coins = None):
if sum in coins: # if sum in coins, nothing to do but return.
return [sum]
elif max(coins) > sum: # if the largest coin is greater then the sum, there's nothing we can do.
return None
else: # check for each coin, keep track of the minimun configuration, then return it.
min_length = None
min_configuration = None
for coin in coins:
results = get_min_coin_configuration(sum = sum - coin, coins = coins)
if results != None:
if min_length == None or (1 + len(results)) < len(min_configuration):
min_configuration = [coin] + results
min_length = len(min_configuration)
return min_configuration
好的，现在让我们看看是否可以通过使用动态编程（我称之为缓存）来改善它。
def get_min_coin_configuration(sum = None, coins = None, cache = None):
if cache == None: # this is quite crucial if its in the deFinition its presistent ...
cache = {}
if sum in cache:
return cache[sum]
elif sum in coins: # if sum in coins, nothing to do but return.
cache[sum] = [sum]
return cache[sum]
elif max(coins) > sum: # if the largest coin is greater then the sum, there's nothing we can do.
cache[sum] = None
return cache[sum]
else: # check for each coin, keep track of the minimun configuration, then return it.
min_length = None
min_configuration = None
for coin in coins:
results = get_min_coin_configuration(sum = sum - coin, coins = coins, cache = cache)
if results != None:
if min_length == None or (1 + len(results)) < len(min_configuration):
min_configuration = [coin] + results
min_length = len(min_configuration)
cache[sum] = min_configuration
return cache[sum]
现在让我们运行一些测试。
assert all([ get_min_coin_configuration(**test[0]) == test[1] for test in
[({'sum':25, 'coins':[1, 5, 10]}, [5, 10, 10]),
({'sum':153, 'coins':[1, 5, 10, 50]}, [1, 1, 1, 50, 50, 50]),
({'sum':100, 'coins':[1, 5, 10, 25]}, [25, 25, 25, 25]),
({'sum':123, 'coins':[5, 10, 25]}, None),
({'sum':100, 'coins':[1,5,25,100]}, [100])] ])
如果此测试不够强大，您也可以这样做。
import random
random_sum = random.randint(10**3, 10**4)
result = get_min_coin_configuration(sum = random_sum, coins = random.sample(range(10**3), 200))
assert sum(result) == random_sum
没有这样的硬币组合可能等于我们的random_sum，但我相信它的可能性很小…
我肯定那里有更好的实现方式，我试图强调可读性而不是性能。祝好运。
了先前的代码，它有一个小错误，它应该检查最小硬币而不是最大硬币，重新编写符合pep8标准的算法，并[]在找不到组合的情况下返回代替None。
def get_min_coin_configuration(total_sum, coins, cache=None): # shadowing python built-ins is frowned upon.
# assert(all(c > 0 for c in coins)) Assuming all coins are > 0
if cache is None: # initialize cache.
cache = {}
if total_sum in cache: # check cache, for prevIoUsly discovered solution.
return cache[total_sum]
elif total_sum in coins: # check if total_sum is one of the coins.
cache[total_sum] = [total_sum]
return [total_sum]
elif min(coins) > total_sum: # check feasibility, if min(coins) > total_sum
cache[total_sum] = [] # no combination of coins will yield solution as per our assumption (all +).
return []
else:
min_configuration = [] # default solution if none found.
for coin in coins: # iterate over all coins, check which one will yield the smallest combination.
results = get_min_coin_configuration(total_sum - coin, coins, cache=cache) # recursively search.
if results and (not min_configuration or (1 + len(results)) < len(min_configuration)): # check if better.
min_configuration = [coin] + results
cache[total_sum] = min_configuration # save this solution, for future calculations.
return cache[total_sum]
assert all([ get_min_coin_configuration(**test[0]) == test[1] for test in
[({'total_sum':25, 'coins':[1, 5, 10]}, [5, 10, 10]),
({'total_sum':153, 'coins':[1, 5, 10, 50]}, [1, 1, 1, 50, 50, 50]),
({'total_sum':100, 'coins':[1, 5, 10, 25]}, [25, 25, 25, 25]),
({'total_sum':123, 'coins':[5, 10, 25]}, []),
({'total_sum':100, 'coins':[1,5,25,100]}, [100])] ])
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