将低效的递归硬币变换函数转换为迭代

我有一个低效的递归硬币更改功能，可以计算给定金额的硬币组合数量 . 如果可能的话，我想将其转换为更有效的迭代函数 .

一个问题是我使用回溯来尝试在称为面额的数组中的不同硬币 . 我也在使用memoization但是当数量很大时它不会加快速度 .

这是我的代码：

unsigned long long CalculateCombinations(std::vector<double> &denominations, std::vector<double> change,
    double amount, unsigned int index)
{
    double current = 0.0;
    unsigned long long combinations = 0;

    if (amount == 0.0)
    {
        if (change.size() % 2 == 0)
        {
            combinations = Calculate(change);
        }
        return combinations;
    }

    // If amount is less than 0 then no solution exists
    if (amount < 0.0)
        return 0;

    // If there are no coins and index is greater than 0, then no solution exist
    if (index >= denominations.size())
        return 0;

    std::string str = std::to_string(amount) + "-" + std::to_string(index) + "-" + std::to_string(change.size());

    auto it = Memo.find(str);

    if (it != Memo.end())
    {
        return it->second;
    }

    while (current <= amount)
    {
        double remainder = amount - current;
        combinations += CalculateCombinations(denominations, change, remainder, index + 1);
        current += denominations[index];
        change.push_back(denominations[index]);
    }

    Memo[str] = combinations;
    return combinations;
}

有什么想法可以做到这一点？我知道有适用于硬币更换问题的DP解决方案，但我的解决方案并不容易 . 我可以半便士 .

*更新：我将函数更改为迭代，并且我按比例缩放了2倍以使用整数，位没有相当大的差别 .

这是我的新代码：

unsigned long long CalculateCombinations(std::vector<int> &denominations, std::vector<int> change, int amount, unsigned int index)
{
    unsigned long long combinations = 0;

    if (amount <= 0)
        return combinations;

    std::stack<Param> mystack;
    mystack.push({ change, amount, index });

    while (!mystack.empty())
    {
        int current = 0;
        std::vector<int> current_coins = mystack.top().Coins;
        int current_amount = mystack.top().Amount;
        unsigned int current_index = mystack.top().Index;
        mystack.pop();

        if (current_amount == 0)
        {
            if (current_coins.size() % 2 == 0)
            {
                combinations += Calculate(std::move(current_coins));
            }
        }
        else
        {
            std::string str = std::to_string(current_amount) + "-" + std::to_string(current_index);
            if (Memo.find(str) == Memo.end())
            {
                // If amount is less than 0 then no solution exists
                if (current_amount >= 0 && current_index < denominations.size())
                {
                    while (current <= current_amount)
                    {
                        int remainder = current_amount - current;
                        mystack.push({ current_coins, remainder, current_index + 1 });
                        current += denominations[current_index];
                        current_coins.push_back(denominations[current_index]);
                    }
                }
                else
                {
                    Memo.insert(str);
                }
            }
        }
    }

    return combinations;
}

备注定义为std :: unordered_set .

这可以通过DP解决吗？问题是我对所有组合都不感兴趣 - 只有大小合适的组合 .