计算字符串的所有可能组合，并加以扭曲
我试图允许用户在文本框中输入文本，并让程序生成所有可能的组合，除了最少3个字符，最多6个字符。我不需要像’as’这样无用的单词， ‘a’，’i’，’to’等混乱我的arrays。 我还将根据字典检查每个组合，以确保它是一个真正的单词。
我有完整的字典（精心制作， 这里是一个回复的链接 （警告：巨大的加载时间（对我而言）！）
无论如何，如果用户输入’ABCDEF’（没有特定的顺序），我怎么能生成，例如：
'ABC' 'BAC' 'CAB' ... 'ABD' 'ABE' 'ABF' 等……无论什么顺序，每种可能的组合？ 我知道这些组合有一些荒谬的组合，但它只需要计算一次，所以我不太担心。
我发现代码示例以递归方式找到固定宽度字符串（ABCDEF，ABCDFE … ACDBFE等）的组合（不是排列，我不需要那些）。 他们没有做我需要的事情，而且我对这个项目的起点也没有丝毫的线索。
这不是作业，它开始是我的个人项目，这个项目已经成长为一个简单的问题……我无法相信我无法理解这一点！
听起来像你在描述Power Set
这是我在个人图书馆里面的一个实现：
// Helper method to count set bits in an integer public static int CountBits(int n) { int count = 0; while (n != 0) { count++; n &= (n - 1); } return count; } public static IEnumerable> PowerSet( IEnumerable src, int minSetSize = 0, int maxSetSize = int.MaxValue) { // we want fast random access to the source, so we'll // need to ToArray() it var cached = src.ToArray(); var setSize = Math.Pow(2, cached.Length); for(int i=0; i < setSize; i++) { var subSetSize = CountBits(i); if(subSetSize < minSetSize || subSetSize > maxSetSize) { continue; } T[] set = new T[subSetSize]; var temp = i; var srcIdx = 0; var dstIdx = 0; while(temp > 0) { if((temp & 0x01) == 1) { set[dstIdx++] = cached[srcIdx]; } temp >>= 1; srcIdx++; } yield return set; } yield break; }
还有一个快速测试台：
void Main() { var src = "ABCDEF"; var combos = PowerSet(src, 3, 6); // hairy joins for great prettiness Console.WriteLine( string.Join(" , ", combos.Select(subset => string.Concat("[", string.Join(",", subset) , "]"))) ); }
输出：
[A,B,C] , [A,B,D] , [A,C,D] , [B,C,D] , [A,B,C,D] , [A,B,E] , [A,C,E] , [B,C,E] , [A,B,C,E] , [A,D,E] , [B,D,E] , [A,B,D,E] , [C,D,E] , [A,C,D,E] , [B,C,D,E] , [A,B,C,D,E] , [A,B,F] , [A,C,F] , [B,C,F] , [A,B,C,F] , [A,D,F] , [B,D,F] , [A,B,D,F] , [C,D,F] , [A,C,D,F] , [B,C,D,F] , [A,B,C,D,F] , [A,E,F] , [B,E,F] , [A,B,E,F] , [C,E,F] , [A,C,E,F] , [B,C,E,F] , [A,B,C,E,F] , [D,E,F] , [A,D,E,F] , [B,D,E,F] , [A,B,D,E,F] , [C,D,E,F] , [A,C,D,E,F] , [B,C,D,E,F] , [A,B,C,D,E,F]
执行此操作的最佳方法是使用for循环并将每个字符从int转换为char并将它们连接在一起。
例如：
for(int i = 0; i < 26; i++) { Console.WriteLine((char)i + 'A'); }
假设，你也想要像“AAB”这样的东西，你的字母组的“交叉产品”应该是它。
生成可以像LINQ一样简单：
string myset = "ABCDE"; var All = (from char l1 in myset from char l2 in myset from char l3 in myset select new string(new char[] { l1, l2, l3})).ToList();
注意：构造很多字符串和char数组并不快 。 您可能希望用自定义类替换新字符串和新char []，如下所示：
select new MyCustomClass(l1, l2, l3).ToList();
如果您不想要“AAB”（或“EEL”）之类的东西，那么我会将您指向维基百科以获取“组合”。
要从固定长度到“从3到6的任何长度”连接多个集合，如果限制是动态的，则使用循环。
从此链接 （在麻省理工学院许可下）
using System; using System.Collections.Generic; using System.Diagnostics; // Copyright (c) 2010 Alex Regueiro // Licensed under MIT license, available at . // Published originally at /// Gets all permutations (of a given size) of a given list, either with or without reptitions. /// /// The type of the elements in the list. /// The list of which to get permutations. /// The action to perform on each permutation of the list. /// The number of elements in each resulting permutation; or /// The algorithm performs permutations in-place. public static void GetPermutations(this IList list, Action> 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(errorMessageValueLessThanZero, "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 : i - 1; int curIndex = 0; while (curIndex != -1) { indices[curIndex]++; if (indices[curIndex] == list.Count) { indices[curIndex] = withRepetition ? -1 : curIndex - 1; curIndex--; } else { result[curIndex] = list[indices[curIndex]]; if (curIndex < indices.Length - 1) curIndex++; else action(result); } } } /// /// Gets all combinations (of a given size) of a given list, either with or without reptitions. /// /// The type of the elements in the list. /// The list of which to get combinations. /// The action to perform on each combination of the list. /// The number of elements in each resulting combination; or /// The algorithm performs combinations in-place. public static void GetCombinations (this IList list, Action> 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(errorMessageValueLessThanZero, "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); } } } /// /// Gets a specific permutation of a given list. /// /// The type of the elements in the list. /// The list to permute. /// The indices of the elements in the original list at each index in the permuted list. /// /// The specified permutation of the given list. /// Permute (this IList list, IList indices) { if (list == null) throw new ArgumentNullException("list"); if (indices == null) throw new ArgumentNullException("indices"); if (list.Count != indices.Count) throw new ArgumentException(errorMessagesIndicesListInvalidSize, "indices"); var result = new T[list.Count]; for (int i = 0; i < result.Length; i++) result[i] = list[indices[i]]; return result; } }