贪心算法 - 钱币找零问题

这个问题在我们的日常生活中就更加普遍了。假设1元、2元、5元、10元、20元、50元、100元的纸币分别有c0, c1, c2, c3, c4, c5, c6张。现在要用这些钱来支付K元，至少要用多少张纸币？

int Count[N]={3,0,2,1,0,3,5}; 

int Value[N]={1,2,5,10,20,50,100}; 

 

用贪心算法的思想，很显然，每一步尽可能用面值大的纸币即可。

 class Program
    {

        static void Main(string[] args)
        {
            int[] count = { 3, 0, 2, 1, 0, 3, 5 };
            int[] value = { 1, 2, 5, 10, 20, 50, 100 };
            int k = 47;
            int money = 100;
            int remain = money - 47;
            List moneyList = GetMoney(new List(), remain, count, value);
            //List moneyList = GetMoney_Iteration(remain, count, value);

            foreach (var item in moneyList)
            {
                Console.Write(item + " ");
            }

            Console.ReadKey();
        }
        // 递归法
        private static List GetMoney(List resultList,int remain,int[] count,int[] value)
        {
            if (remain <= 0) return resultList;

            for (int i = count.Length - 1; i >= 0; i--)
            {
                if (remain >= value[i] && count[i] > 0)
                {
                    resultList.Add(value[i]);
                    remain -= value[i];
                    count[i] -= 1;
                    break;
                }
            }

            return GetMoney(resultList,remain ,count,value);
        }

        // 迭代法
        private static List GetMoney_Iteration(int remain,int[] count,int[] value)
        {
            List resultList = new List();
            for (int i = count.Length - 1; i >= 0; i--)
            {
                if (remain >= value[i] && count[i] > 0)
                {
                    int h = remain / value[i];
                    // Console.WriteLine(h);
                    for (int j = 0; j < h; j++)
                    {
                        if (count[i] > 0)
                        { 
                            resultList.Add(value[i]);
                            count[i] -= 1;
                            remain -= value[i];
                        }
                    }
                }
            }
            return resultList;
        }
    }