什么是最好的方式来查找数组中的项目的所有组合？
什么是最好的方式来find在C＃中的数组中的项目的所有组合？
它开着！）
static List<List<int>> comb; static bool[] used; static void GetCombinationSample() { int[] arr = { 10, 50, 3, 1, 2 }; used = new bool[arr.Length]; used.Fill(false); comb = new List<List<int>>(); List<int> c = new List<int>(); GetComb(arr, 0, c); foreach (var item in comb) { foreach (var x in item) { Console.Write(x + ","); } Console.WriteLine(""); } } static void GetComb(int[] arr, int colindex, List<int> c) { if (colindex >= arr.Length) { comb.Add(new List<int>(c)); return; } for (int i = 0; i < arr.Length; i++) { if (!used[i]) { used[i] = true; c.Add(arr[i]); GetComb(arr, colindex + 1, c); c.RemoveAt(c.Count - 1); used[i] = false; } } } 更新
这里有一组通用函数（需要.net 3.5或更高版本）用于不同的场景。 输出为{1,2,3,4}，长度为2的列表。
重复排列
static IEnumerable<IEnumerable<T>> GetPermutationsWithRept<T>(IEnumerable<T> list, int length) { if (length == 1) return list.Select(t => new T[] { t }); return GetPermutationsWithRept(list, length - 1) .SelectMany(t => list, (t1, t2) => t1.Concat(new T[] { t2 })); }
输出：
{1,1} {1,2} {1,3} {1,4} {2,1} {2,2} {2,3} {2,4} {3,1} {3,2} {3,3} {3,4} {4,1} {4,2} {4,3} {4,4}
排列
static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> list, int length) { if (length == 1) return list.Select(t => new T[] { t }); return GetPermutations(list, length - 1) .SelectMany(t => list.Where(o => !t.Contains(o)), (t1, t2) => t1.Concat(new T[] { t2 })); }
输出：
{1,2} {1,3} {1,4} {2,1} {2,3} {2,4} {3,1} {3,2} {3,4} {4,1} {4,2} {4,3}
K组合重复
static IEnumerable<IEnumerable<T>> GetKCombsWithRept<T>(IEnumerable<T> list, int length) where T : IComparable { if (length == 1) return list.Select(t => new T[] { t }); return GetKCombsWithRept(list, length - 1) .SelectMany(t => list.Where(o => o.CompareTo(t.Last()) >= 0), (t1, t2) => t1.Concat(new T[] { t2 })); }
输出：
{1,1} {1,2} {1,3} {1,4} {2,2} {2,3} {2,4} {3,3} {3,4} {4,4}
K-组合
static IEnumerable<IEnumerable<T>> GetKCombs<T>(IEnumerable<T> list, int length) where T : IComparable { if (length == 1) return list.Select(t => new T[] { t }); return GetKCombs(list, length - 1) .SelectMany(t => list.Where(o => o.CompareTo(t.Last()) > 0), (t1, t2) => t1.Concat(new T[] { t2 })); }
输出：
{1,2} {1,3} {1,4} {2,3} {2,4} {3,4}
这就是所谓的排列。
这可以给你任何集合的排列：
public class Permutation { public static IEnumerable<T[]> GetPermutations<T>(T[] items) { int[] work = new int[items.Length]; for (int i = 0; i < work.Length; i++) { work[i] = i; } foreach (int[] index in GetIntPermutations(work, 0, work.Length)) { T[] result = new T[index.Length]; for (int i = 0; i < index.Length; i++) result[i] = items[index[i]]; yield return result; } } public static IEnumerable<int[]> GetIntPermutations(int[] index, int offset, int len) { if (len == 1) { yield return index; } else if (len == 2) { yield return index; Swap(index, offset, offset + 1); yield return index; Swap(index, offset, offset + 1); } else { foreach (int[] result in GetIntPermutations(index, offset + 1, len - 1)) { yield return result; } for (int i = 1; i < len; i++) { Swap(index, offset, offset + i); foreach (int[] result in GetIntPermutations(index, offset + 1, len - 1)) { yield return result; } Swap(index, offset, offset + i); } } } private static void Swap(int[] index, int offset1, int offset2) { int temp = index[offset1]; index[offset1] = index[offset2]; index[offset2] = temp; } }
例：
string[] items = { "one", "two", "three" }; foreach (string[] permutation in Permutation.GetPermutations<string>(items)) { Console.WriteLine(String.Join(", ", permutation)); }
关于彭阳答案：这是我的通用函数，它可以返回T列表中的所有组合：
static IEnumerable<IEnumerable<T>> GetCombinations<T>(IEnumerable<T> list, int length) { if (length == 1) return list.Select(t => new T[] { t }); return GetCombinations(list, length - 1) .SelectMany(t => list, (t1, t2) => t1.Concat(new T[] { t2 })); }
例1：n = 3，k = 2
IEnumerable<IEnumerable<int>> result = GetCombinations(Enumerable.Range(1, 3), 2);
输出 – 整数列表的列表：
{1, 1} {1, 2} {1, 3} {2, 1} {2, 2} {2, 3} {3, 1} {3, 2} {3, 3}
………………………………………….. ………………………
我运行这个例子，我不太确定结果的正确性。
例2：n = 3，k = 3
IEnumerable<IEnumerable<int>> result = GetCombinations(Enumerable.Range(1, 3), 3);
输出 – 整数列表的列表：
{1, 1, 1} {1, 1, 2} {1, 1, 3} {1, 2, 1} {1, 2, 2} {1, 2, 3} {1, 3, 1} {1, 3, 2} {1, 3, 3} {2, 1, 1} {2, 1, 2} {2, 1, 3} {2, 2, 1} {2, 2, 2} {2, 2, 3} {2, 3, 1} {2, 3, 2} {2, 3, 3} {3, 1, 1} {3, 1, 2} {3, 1, 3} {3, 2, 1} {3, 2, 2} {3, 2, 3} {3, 3, 1} {3, 3, 2} {3, 3, 3}
这不应该与组合发生，否则它应该指定它与重复。参见文章http://en.wikipedia.org/wiki/Combinations
有一对非常简单的方法来find用户inputstring的组合。
通过使用LINQ的第一种方法
private static IEnumerable<string> FindPermutations(string set) { var output = new List<string>(); switch (set.Length) { case 1: output.Add(set); break; default: output.AddRange(from c in set let tail = set.Remove(set.IndexOf(c), 1) from tailPerms in FindPermutations(tail) select c + tailPerms); break; } return output; }
使用这个function就好
Console.WriteLine("Enter a sting "); var input = Console.ReadLine(); foreach (var stringCombination in FindPermutations(input)) { Console.WriteLine(stringCombination); } Console.ReadLine();
其他的方法是使用循环
// 1. remove first char // 2. find permutations of the rest of chars // 3. Attach the first char to each of those permutations. // 3.1 for each permutation, move firstChar in all indexes to produce even more permutations. // 4. Return list of possible permutations. public static string[] FindPermutationsSet(string word) { if (word.Length == 2) { var c = word.ToCharArray(); var s = new string(new char[] { c[1], c[0] }); return new string[] { word, s }; } var result = new List<string>(); var subsetPermutations = (string[])FindPermutationsSet(word.Substring(1)); var firstChar = word[0]; foreach (var temp in subsetPermutations.Select(s => firstChar.ToString() + s).Where(temp => temp != null).Where(temp => temp != null)) { result.Add(temp); var chars = temp.ToCharArray(); for (var i = 0; i < temp.Length - 1; i++) { var t = chars[i]; chars[i] = chars[i + 1]; chars[i + 1] = t; var s2 = new string(chars); result.Add(s2); } } return result.ToArray(); }
你可以使用这个function
Console.WriteLine("Enter a sting "); var input = Console.ReadLine(); Console.WriteLine("Here is all the possable combination "); foreach (var stringCombination in FindPermutationsSet(input)) { Console.WriteLine(stringCombination); } Console.ReadLine();
也许kwcombinatorics可以提供一些帮助（请参见主页上的示例）：
KwCombinatorics库有3个类，提供3种不同的方式来生成有序（sorting）的数字组合列表。 这些组合对于软件testing非常有用，可以生成各种可能的input组合。 其他用途包括解决math问题和机会游戏。
有关详细的答案请参阅：唐纳德·克努特（Donald Knuth），计算机编程艺术（又名TAOCP）。 第4A卷，枚举和回溯，第7.2章。 生成所有可能性。 http://www-cs-faculty.stanford.edu/~uno/taocp.html
Gufa给出的另一个解决scheme。 在该类的完整源代码下面：
使用System.Collections.Generic;
命名空间ConsoleApplication1 {公共类Permutation {
public IEnumerable<T[]> GetPermutations<T>(T[] items) { var work = new int[items.Length]; for (var i = 0; i < work.Length; i++) { work[i] = i; } foreach (var index in GetIntPermutations(work, 0, work.Length)) { var result = new T[index.Length]; for (var i = 0; i < index.Length; i++) result[i] = items[index[i]]; yield return result; } } public IEnumerable<int[]> GetIntPermutations(int[] index, int offset, int len) { switch (len) { case 1: yield return index; break; case 2: yield return index; Swap(index, offset, offset + 1); yield return index; Swap(index, offset, offset + 1); break; default: foreach (var result in GetIntPermutations(index, offset + 1, len - 1)) { yield return result; } for (var i = 1; i < len; i++) { Swap(index, offset, offset + i); foreach (var result in GetIntPermutations(index, offset + 1, len - 1)) { yield return result; } Swap(index, offset, offset + i); } break; } } private static void Swap(IList<int> index, int offset1, int offset2) { var temp = index[offset1]; index[offset1] = index[offset2]; index[offset2] = temp; } }
}
这实际上工作，因为它应该为组合。但是不允许selectk的组合…
