using System;
using System.Collections.Generic;
using System.Linq;

namespace ConsoleApplication1
{
    class Program
    {

        static void Main(string[] args)
        {
            var lst = new int[] { 1, 1, 2, 2, 4, 4, 1, 3, 2 };
            foreach (var result in Solve(lst, 5))
            {
                result.ToList().ForEach(x => { Console.Write("{0} ", x); });
                Console.WriteLine();
            }
            Console.ReadKey();
        }

        static IEnumerable<IEnumerable<int>> Solve(int[] lst, int sum)
        {
            foreach (var r in lst.Where(x => x == sum))
                yield return new int[] { r };

            if (lst.Length > 1)
                foreach (var r in Solve(lst.Where((x, index) => index > 0).ToArray(), sum - lst[0]))
                    yield return new int[] { lst[0] }.Concat(r);
        }
    }
}

------解决方案--------------------
针对你的“特殊情况”那好写得很。
C# code
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] data = { 1, 1, 2, 2, 4, 4, 1, 3, 2 };
            int n = 5;
            var list = data.OrderByDescending(x => x).Select(x => new { g = -1, v = x }).ToList();
            int g1 = 0;
            while (list.Any(x => x.g == -1))
            {
                int t = 0;
                g1 = list.Max(x => x.g) + 1;
                while (t != n)
                {
                    var i = list.Where(x => x.g == -1 && x.v <= n - t).First();
                    t += i.v;
                    list.Add(new { g = g1, v = i.v });
                    list.Remove(i);
                }
            }
            foreach (var i in list.GroupBy(x => x.g).Select(x => string.Join(",", x.Select(y => y.v))))
            {
                Console.WriteLine(i);
            }
        }
    }
}

------解决方案--------------------
你们代码都是大量的循环。。。
程序这么实现是最差劲的。。。
无语
Int[] s=[1,1,2,2,4,4,1,3,2] 要和等于5 分析下。最少要二个数相加，最多四个数相加。
定义一个新的int[] newS 让int[]的元素之和等于5可以限制newS的元素个数减少循环
  

------解决方案--------------------
可以用0-1背包问题来解决