递归实现'；最小硬币数量'；用python

此问题与中的问题相同

给出一个硬币列表，它们的值（c1，c2，c3，…cj，…）和总金额i。找到最小数量的硬币，其总和为i（我们可以使用任意数量的硬币），或者报告不可能选择总和为s的硬币

我昨天刚被介绍了动态规划，我试着为它编写一个代码

# 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

最优子结构：C[i]=1+min_j（C[i-cj]） cdict={} def C（一、硬币）：

---

如果我如注释所述，则在

i<0

时需要返回足够大的值，这样它就不会被

min

选择，如下所示：

cdict = {}
def C(i, coins):

    if i == 0:
        return 0

   if i < 0:
        return 1e100    # Return infinity in ideally

    if i in cdict:
        return cdict[i]
    else:
        answer = 1 + min([C(i - cj, coins) for cj in coins])
        cdict[i] = answer
    return answer

使用方法：

python2 coins.py <amount> <coin1> <coin2> ...

python2 coins.py。。。

这是一个很好的算法问题，但老实说，我不认为您的实现是正确的，或者可能是我不理解您函数的输入/输出，为此我道歉。

这是您的实现的修改版本。

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]

假设这些测试不够健壮，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

有可能没有这样的硬币组合等于我们的随机总和，但我相信这不太可能

我确信有更好的实现，我试图强调可读性而不是性能。 祝你好运

更新 以前的代码有一个小错误，它假设检查最小硬币而不是最大硬币，重新编写符合pep8的算法，并在找不到组合时返回

[]

，而不是

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])] ])

def get_min_coin_配置（total_sum，coins，cache=None）：#不赞成隐藏python内置。
#断言（所有（对于硬币中的c，c>0））假设所有硬币都>0
如果缓存为无：#初始化缓存。
缓存={}
如果缓存中的总和：#检查缓存，查找以前发现的解决方案。
返回缓存[总计]
elif total_sum in coins:#检查total_sum是否为硬币之一。
缓存[总和]=[总和]
返回[总计]
elif min（硬币）>总计：#检查可行性，如果min（硬币）>总计
cache[total_sum]=[]#根据我们的假设，任何硬币组合都不会产生解决方案（全部+）。
返回[]
其他：
min_configuration=[]如果未找到默认解决方案，则为默认解决方案。
对于硬币中的硬币：#迭代所有硬币，检查哪一个会产生最小的组合。
结果=获取\u最小\u硬币\u配置（总计\u总和-硬币，硬币，缓存=缓存）#递归搜索。
如果结果和（不是最小配置或（1+len（结果））

这里是一个递归的、效率非常低的更改算法实现，其中

V

是一个硬币列表和

C

目标金额：

def min_change(V, C):
    def min_coins(i, aC):
        if aC == 0:
            return 0
        elif i == -1 or aC < 0:
            return float('inf')
        else:
            return min(min_coins(i-1, aC), 1 + min_coins(i, aC-V[i]))
    return min_coins(len(V)-1, C)

---

这里有一个有趣的方法。有点小技巧，但这就是为什么它很有趣

    import math

    def find_change(coins, value):
        coins = sorted(coins, reverse=True)
        coin_dict = {}
        for c in coins:
            if value % c == 0:
                coin_dict[c] = value / c
                return coin_dict
            else:
                coin_dict[c] = math.trunc(value/ float(c))
                value -= (c * coin_dict[c])

    coins = [1, 5, 10, 25]
    answer = find_change(coins, 69)
    print answer
    [OUT]: {25: 2, 10: 1, 5: 1, 1: 4}

下面是带边缘保护的注释的相同解决方案

    import math
    def find_change(coins, value):
        '''
        :param coins: List of the value of each coin [25, 10, 5, 1]
        :param value: the value you want to find the change for ie; 69 cents
        :return: a change dictionary where the key is the coin, and the value is how many times it is used in finding the minimum change
        '''
        change_dict = {}  # CREATE OUR CHANGE DICT, THIS IS A DICT OF THE COINS WE ARE RETURNING, A COIN PURSE
        coins = sorted(coins, reverse=True)  # SORT COINS SO WE START LOOP WITH BIGGEST COIN VALUE
        for c in coins:
            for d in coins:     # THIS LOOP WAS ADDED BY A SMART STUDENT:  IE IN THE CASE OF IF THERE IS A 7cent COIN AND YOU ARE LOOKING FOR CHANGE FOR 14 CENTS, WITHOUT THIS D FOR LOOP IT WILL RETURN 10: 1, 1: 4
                if (d != 1) & (value % d == 0):
                    change_dict[d] = value / d
                    return change_dict
            if value % c == 0:      # IF THE VALUE DIVIDED BY THE COIN HAS NO REMAINDER,  # ie, if there is no remainder, all the neccessary change has been given  # PLACE THE NUMBER OF TIMES THAT COIN IS USED IN THE change_dict  # YOU ARE FINISHED NOW RETURN THE change_dict
                change_dict[c] = value / c
                return change_dict
            else:
                change_dict[c] = math.trunc(value/ float(c))        # PLACE THAT NUMBER INTO OUR coin_dict  # DIVIDE THE VALUE BY THE COIN, THEN GET JUST THE WHOLE NUMBER  # IE 69 / 25.0 = 2.76  # math.trunc(2.76) == 2    # AND THAT IS HOW MANY TIMES IT WILL EVENLY GO INTO THE VALUE,
                amount = (c * change_dict[c])  # NOW TAKE THE NUMBER OF COINS YOU HAVE IN YOUR  UPDATE THE VALUE BY SUBTRACTING THE c * TIME NUMBER OF TIMES IT WAS USED    # AMOUNT IS HOW MUCH CHANGE HAS BEEN PUT INTO THE CHANGE DICT ON THIS LOOP  # FOR THE CASE OF 69, YOU GIVE 2 25CENT COINS, SO 2 * 25 = 50, 19 = 69 - 50
                value = value - amount              # NOW, UPDATE YOUR VALUE, SO THE NEXT TIME IT GOES INTO THIS LOOP, IT WILL BE LOOKING FOR THE MIN CHANGE FOR 19 CENTS...

    coins = [1, 5, 10, 25]
    answer = find_change(coins, 69)
    print answer
    [OUT]: {25: 2, 10: 1, 5: 1, 1: 4}
    edge_case_coins = [1, 7, 10, 25]
    edge_case_answer = find_change(coins, 14)
    print edge_case_answer
    [OUT]: {7: 2}

---

这是一个使用while循环的方法。算法非常简单。你首先使用最大的硬币来支付这笔钱。如果你知道你会选择下一个较小的硬币，然后重复，直到钱为0。这段代码的优点是，虽然最坏情况下的运行时间较高（我认为是m*n）（m是列表的大小，n是while中的迭代次数），代码要简单得多

我假设没有价值为0的硬币，并且总是有价值为1的硬币。当没有价值为1时，函数将给出价格下最佳硬币数量的答案

def find_change(coins, money):
    coins = sorted(coins, reverse=True)
    coincount = 0

    for coin in coins:
        while money >= coin:
            money = money - coin
            coincount += 1

    return coincount

---

我试着想一个这样做行不通的极端情况（它会溢出任何具有0值硬币的列表）但是我想不出一个。

如果我<0，你真的想返回0吗？如果你让你的程序从标准美国硬币中赚20美分，你的程序将返回1，因为20-25=-5，所以在递归调用中，程序将返回0，所以你所有递归调用的最小值都将是1，这似乎不正确（除非我误解了这个问题）。此外，cdict是某种回忆录吗？是的，cdict将记住以前的解决方案。例如，cdict[10]将包含10笔钱所需的最少数量的硬币。I“我意识到，如果我在某个点上无如果找不到组合例如，如果我给你一枚一角硬币和一枚五分硬币，请你找到可以产生13的硬币组合，但除了5模的数字之外，没有任何镍和一角硬币的组合可以产生任何东西…无论如何现在重写算法，如果找不到匹配项并修复了错误，它将返回

[]

，尽管我希望我的测试能够找到它，尽管错误可能对性能的影响比任何东西都大……当我通过

count\u change调用它时
def min_change(V, C):
    def min_coins(i, aC):
        if aC == 0:
            return 0
        elif i == -1 or aC < 0:
            return float('inf')
        else:
            return min(min_coins(i-1, aC), 1 + min_coins(i, aC-V[i]))
    return min_coins(len(V)-1, C)

def min_change(V, C):
    m, n = len(V)+1, C+1
    table = [[0] * n for x in xrange(m)]
    for j in xrange(1, n):
        table[0][j] = float('inf')
    for i in xrange(1, m):
        for j in xrange(1, n):
            aC = table[i][j - V[i-1]] if j - V[i-1] >= 0 else float('inf')
            table[i][j] = min(table[i-1][j], 1 + aC)
    return table[m-1][n-1]

    import math

    def find_change(coins, value):
        coins = sorted(coins, reverse=True)
        coin_dict = {}
        for c in coins:
            if value % c == 0:
                coin_dict[c] = value / c
                return coin_dict
            else:
                coin_dict[c] = math.trunc(value/ float(c))
                value -= (c * coin_dict[c])

    coins = [1, 5, 10, 25]
    answer = find_change(coins, 69)
    print answer
    [OUT]: {25: 2, 10: 1, 5: 1, 1: 4}

    import math
    def find_change(coins, value):
        '''
        :param coins: List of the value of each coin [25, 10, 5, 1]
        :param value: the value you want to find the change for ie; 69 cents
        :return: a change dictionary where the key is the coin, and the value is how many times it is used in finding the minimum change
        '''
        change_dict = {}  # CREATE OUR CHANGE DICT, THIS IS A DICT OF THE COINS WE ARE RETURNING, A COIN PURSE
        coins = sorted(coins, reverse=True)  # SORT COINS SO WE START LOOP WITH BIGGEST COIN VALUE
        for c in coins:
            for d in coins:     # THIS LOOP WAS ADDED BY A SMART STUDENT:  IE IN THE CASE OF IF THERE IS A 7cent COIN AND YOU ARE LOOKING FOR CHANGE FOR 14 CENTS, WITHOUT THIS D FOR LOOP IT WILL RETURN 10: 1, 1: 4
                if (d != 1) & (value % d == 0):
                    change_dict[d] = value / d
                    return change_dict
            if value % c == 0:      # IF THE VALUE DIVIDED BY THE COIN HAS NO REMAINDER,  # ie, if there is no remainder, all the neccessary change has been given  # PLACE THE NUMBER OF TIMES THAT COIN IS USED IN THE change_dict  # YOU ARE FINISHED NOW RETURN THE change_dict
                change_dict[c] = value / c
                return change_dict
            else:
                change_dict[c] = math.trunc(value/ float(c))        # PLACE THAT NUMBER INTO OUR coin_dict  # DIVIDE THE VALUE BY THE COIN, THEN GET JUST THE WHOLE NUMBER  # IE 69 / 25.0 = 2.76  # math.trunc(2.76) == 2    # AND THAT IS HOW MANY TIMES IT WILL EVENLY GO INTO THE VALUE,
                amount = (c * change_dict[c])  # NOW TAKE THE NUMBER OF COINS YOU HAVE IN YOUR  UPDATE THE VALUE BY SUBTRACTING THE c * TIME NUMBER OF TIMES IT WAS USED    # AMOUNT IS HOW MUCH CHANGE HAS BEEN PUT INTO THE CHANGE DICT ON THIS LOOP  # FOR THE CASE OF 69, YOU GIVE 2 25CENT COINS, SO 2 * 25 = 50, 19 = 69 - 50
                value = value - amount              # NOW, UPDATE YOUR VALUE, SO THE NEXT TIME IT GOES INTO THIS LOOP, IT WILL BE LOOKING FOR THE MIN CHANGE FOR 19 CENTS...

    coins = [1, 5, 10, 25]
    answer = find_change(coins, 69)
    print answer
    [OUT]: {25: 2, 10: 1, 5: 1, 1: 4}
    edge_case_coins = [1, 7, 10, 25]
    edge_case_answer = find_change(coins, 14)
    print edge_case_answer
    [OUT]: {7: 2}

def find_change(coins, money):
    coins = sorted(coins, reverse=True)
    coincount = 0

    for coin in coins:
        while money >= coin:
            money = money - coin
            coincount += 1

    return coincount