Question

我有一个

下的数组

var x = new int[] { 1,2,3 };

并给出了限制

int limit=2;

我必须找到一个组合

1+2 = 3
1+3 = 4
2+3 = 5.

如果数组是

var x = new int[] { 1,2,3,4,5,6};

并给出了限制

int limit=3;

组合应

1+2+3 = 6
 1+2+4 = 7
 1+2+5 = 8
 1+2+6 = 9
 2+3+4  = 9
 2+3+5 = 10
 2+3+6 = 11
...........
............

等等

怎么做？

我的嫌疑人

var x = new int[] {1,2,3};
        int limit = 2;

        var m = from a1 in x
                from a2 in x                
                select new
                    {
                        P1 = a1 ,
                        P2 = a2,                     
                        P3 = a1+a2
                    };

Answer 1

我会使用这个有效的组合学项目，这也值得一读：

Permutations, Combinations, and Variations using C# Generics

然后很简单：

var x = new int[] { 1, 2, 3, 4, 5, 6 };
var combis = new Facet.Combinatorics.Combinations<int>(x, 3, Facet.Combinatorics.GenerateOption.WithoutRepetition);
foreach(var combi in combis)
    Console.WriteLine(String.Join("+", combi) + "=" + combi.Sum());
Console.WriteLine("Total: " + combis.Sum(c => c.Sum())); // 201

输出：

1+2+3=6
1+2+4=7
1+2+5=8
1+2+6=9
1+3+4=8
1+3+5=9
1+3+6=10
1+4+5=10
1+4+6=11
1+5+6=12
2+3+4=9
2+3+5=10
2+3+6=11
2+4+5=11
2+4+6=12
2+5+6=13
3+4+5=12
3+4+6=13
3+5+6=14
4+5+6=15
Total: 210

Answer 2

不使用外部库的解决方案：

public static IEnumerable<IEnumerable<T>> Combinations<T>(IEnumerable<T> elements, int k)
{
  return k == 0 ? new[] { new T[0] } :
    elements.SelectMany((e, i) =>
      Combinations(elements.Skip(i + 1),k - 1).Select(c => (new[] {e}).Concat(c)));
}

public static void Main()
{
    var x = new int[] { 1, 2, 3, 4, 5, 6 };
    var limit = 3;
    IEnumerable<IEnumerable<int>> result = Combinations(x, limit);


    foreach(var combi in result)
        Console.WriteLine(String.Join("+", combi.Select(a=>a.ToString()).ToArray()) + "=" + combi.Sum());
    Console.WriteLine("Total: " + result.Sum(c => c.Sum())); // 201
}

修改：所有组合：

public static IEnumerable<IEnumerable<T>> AllCombinations<T>(IEnumerable<T> elements)
{
    int length = elements.Count();
    for(int k = 1; k<=length; k++){
        var comb = Combinations(elements, k);
        foreach (IEnumerable<T> c in comb)
            yield return c;
    }
}

Answer 3

看起来你只需要一种方法

        /// <summary>
        /// Gets all combinations (of a given size) of a given list, either with or without reptitions.
        /// </summary>
        /// <typeparam name="T">The type of the elements in the list.</typeparam>
        /// <param name="list">The list of which to get combinations.</param>
        /// <param name="action">The action to perform on each combination of the list.</param>
        /// <param name="resultSize">The number of elements in each resulting combination; or <see langword="null"/> to get
        /// premutations of the same size as <paramref name="list"/>.</param>
        /// <param name="withRepetition"><see langword="true"/> to get combinations with reptition of elements;
        /// <see langword="false"/> to get combinations without reptition of elements.</param>
        /// <exception cref="ArgumentNullException"><paramref name="list"/> is <see langword="null"/>. -or-
        /// <paramref name="action"/> is <see langword="null"/>.</exception>
        /// <exception cref="ArgumentException"><paramref name="resultSize"/> is less than zero.</exception>
        /// <remarks>
        /// The algorithm performs combinations in-place. <paramref name="list"/> is however not changed.
        /// </remarks>
        public static void GetCombinations<T>(IList<T> list, Action<IList<T>> action, int? resultSize = null,
            bool withRepetition = false)
        {
            if (list == null)
                throw new ArgumentNullException("list");
            if (action == null)
                throw new ArgumentNullException("action");
            if (resultSize.HasValue && resultSize.Value <= 0)
                throw new ArgumentException("error", "resultSize");

            var result = new T[resultSize.HasValue ? resultSize.Value : list.Count];
            var indices = new int[result.Length];
            for (int i = 0; i < indices.Length; i++)
                indices[i] = withRepetition ? -1 : indices.Length - i - 2;

            int curIndex = 0;
            while (curIndex != -1)
            {
                indices[curIndex]++;
                if (indices[curIndex] == (curIndex == 0 ? list.Count : indices[curIndex - 1] + (withRepetition ? 1 : 0)))
                {
                    indices[curIndex] = withRepetition ? -1 : indices.Length - curIndex - 2;
                    curIndex--;
                }
                else
                {
                    result[curIndex] = list[indices[curIndex]];
                    if (curIndex < indices.Length - 1)
                        curIndex++;
                    else
                        action(result);
                }
            }
        }

用法：

var x = new int[] { 1, 2, 3, 4, 5, 6 };
var combinationsResult = new List<int>();
GetCombinations<int>(x, (list => combinationsResult.Add(list.Sum())), limit, false);

代码来自this beautiful blog

如何根据提供的限制进行组合？

3 个答案: