在Python中递归地实现“最小数量的硬币”

Z时代
2024-01-10
分类：问答

给定一个硬币列表，它们的值（c1，c2，c3，… cj，…）和总和i。

找到总和为i的最小数量的硬币（我们可以根据需要使用任意数量的一种类型的硬币），或者报告无法选择以总和为S的方式选择硬币。

我昨天刚刚被介绍给动态编程，我试图为此编写代码。

# Optimal substructure: C[i] = 1 + min_j(C[i-cj])
cdict = {}
def C(i, coins):
    if i <= 0:
        return 0
    if i in cdict:
        return cdict[i]
    else:
        answer = 1 + min([C(i - cj, coins) for cj in coins])
        cdict[i] = answer
        return answer

在此，C [i]是金额“

i”的最佳解决方案。程序中可用的硬币为{c1，c2，…，cj，…}，我增加了递归限制，以避免最大递归深度超过错误。但是，该程序仅给某人一个正确的答案，当不可能解决时，它并不表示这样做。

回答：

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])] ])