算法：给定一个值（如10），从N个数据当中找最接近该值的组合形式

Question

要求组合形式之和不能大于该值 假如有6个数据：1、2、3、4、5、6
给定值为10，最优的组合为4和6
给定值为5，最优的组合为5假如有6个数据：8.1、1.3、1.5、7.0、0.3、6.6
给定值为10，最优的组合为8.1、1.5和0.3
给定值为7，最优的组合为7

Answer 1

        static void Main(string[] args)
        {
            List<double> list = Find(new double[] { 8.1, 1.3, 1.5, 7.0, 0.3, 6.6 }, 10);
            foreach (double d in list)
                Console.WriteLine(d);
        }        static List<double> Find(double[] dArray, double d)
        {
            if (dArray.Length < 1)
                return null;
            List<double> list = new List<double>();
            int index = Array.IndexOf(dArray, d);
            if (index > -1)
            {
                list.Add(dArray[index]);
                return list;
            }
            double temp = d;
            Array.Sort(dArray);
            for (int i = dArray.Length - 1; i > -1; i--)
            {
                if (dArray[i] <= temp)
                {
                    list.Add(dArray[i]);
                    temp -= dArray[i];
                }
            }
            return list;
        }

Answer 2

 第三楼的算法有点问题：
List<double> list = Find(new double[] { 8.1, 1.3, 1.58, 7.0, 0.3, 6.6 }, 8.2);
算出的结果是8.1，
而事实上，我要的结果是1.58和6.6。1.58+6.6=8.18,最接近8.2

Answer 3

        static void Main(string[] args)
        {
            Find(new double[] { 8.1, 1.3, 1.58, 7.0, 0.3, 6.6 }, 8.2); 
        }        static void Find(double[] dArray, double d)
        {
            if (dArray.Length < 1)
            {
                Console.WriteLine("数组无效");
                return;
            }
            int index = Array.IndexOf(dArray, d);
            if (index > -1)
            {
                Console.WriteLine(dArray[index]);
                return;
            }
            Array.Sort(dArray);
            List<double> listMinus = new List<double>();
            List<List<double>> listCom = new List<List<double>>();
            double minus = dArray[dArray.Length - 1];
            for (int i = dArray.Length - 1; i > -1; i--)
            {
                double temp = d;
                if (dArray[i] <= temp)
                {
                    List<double> list = new List<double>();
                    list.Add(dArray[i]);
                    temp -= dArray[i];
                    for (int j = i - 1; j > -1; j--)
                    {
                        if (dArray[j] <= temp)
                        {
                            list.Add(dArray[j]);
                            temp -= dArray[j];
                        }
                    }
                    listCom.Add(list);
                    listMinus.Add(temp);
                    if (temp < minus)
                        minus = temp;
                }
            }
            for (int i = 0; i < listMinus.Count; i++)
            {
                if (listMinus[i] == minus)
                {
                    Console.Write("可能的组合有:");
                    foreach (double d1 in listCom[i])
                        Console.Write("{0} ", d1);
                    Console.WriteLine();
                }
            }
        }