Question

我正在尝试获取目标金额之和的所有硬币。我能够获得所需的硬币数量。我将如何解决它。

您可以无限使用相同的硬币例如。 change([2], 10) => [2, 2, 2, 2, 2]

def change(coins, amount):
    result = [amount+1] * (amount+1)

    result[0] = 0

    for i in range(1, amount+1):
        for coin in coins:
            if i >= coin:
                result[i] = min(result[i], result[i-coin] + 1)

    if result[amount] == amount+1:
        return -1

    return result[amount]

change([1, 2, 5,8], 7) => [5, 2]顺序无关紧要。

Answer 1

如果使用dyanmic programming，则只能获得最佳结果，可以通过使用数组存储contentPadding的中间结果来实现这一点，我已经根据您的dp版本进行了修改：

dynamic programming

测试：

def change(coins, amount):
    result = [amount+1] * (amount+1)
    coins_results = [[] for _ in range(amount+1)]

    result[0] = 0

    for i in range(1, amount+1):
        for coin in coins:
            if i >= coin and result[i - coin] + 1 < result[i]:
                result[i] = result[i-coin] + 1
                coins_results[i] = coins_results[i-coin] + [coin]

    if result[amount] == amount+1:
        return []

    return coins_results[amount]

输出：

print(change([1, 2, 5, 8], 7))
print(change([2], 10))

这是一个通过backtracking输出所有结果的版本：

[5, 2]
[2, 2, 2, 2, 2]

测试：

def change(coins, amount):
    res = []

    def backtrack(end, remain, cur_result):
        if end < 0: return
        if remain == 0:
            res.append(cur_result)
            return
        if remain >= coins[end]:
            backtrack(end, remain - coins[end], cur_result + [coins[end]])
        backtrack(end - 1, remain, cur_result)

    backtrack(len(coins) - 1, amount, [])
    return res

输出：

print(change([1, 2, 5, 8], 7))
print(change([2], 10))

希望对您有所帮助，如果还有其他问题，请发表评论。：）

Answer 2

我相信这是您要找的答案，请提出您的问题，以便我们对需要做的事情有更好的了解：）

def change(coins, amount):
    ret = []    # Here we keep all possible solves
    solves = [] # Here we keep all unique solves, to avoid duplicates (Eg.: [5, 2] and [2, 5] are both a solution to 7)

    for c1 in coins:
        for c2 in coins:
            if c1 + c2 == amount: # Check if the solve is a match
                solve = [c1, c2] 
                if not set(solve) in solves: # Check if the solve is not a duplicate
                    ret.append(solve)
                    solves.append(set(solve))

    return ret # Return a list of solves

Answer 3

如果要获取与目标数量相对应的所有组合，则可以使用以下生成器：

def change(coins, amount):
    for i, coin in enumerate(coins):
        if coin == amount:
            yield (coin,)
        elif coin < amount:
            yield from ((coin,) + x for x in change(coins[i:], amount - coin))

print(list(change([2], 10)))  # [(2, 2, 2, 2, 2)]
print(list(change([1, 2, 5, 8], 7)))  # [(1, 1, 1, 1, 1, 1, 1), (1, 1, 1, 1, 1, 2), (1, 1, 1, 2, 2), (1, 1, 5), (1, 2, 2, 2), (2, 5)]

返回数组的动态编程硬币更改

3 个答案: