非常偷懶的寫法，先產生 C(n, r) 的 Combination 集合 C，再將集合 C 中的子集合，作 Permutation 即可。
產生 Combination 參考【C# - 產生 Combination - C(n, r) 組合問題】；產生 Permutation 參考【C# - 產生 Permutation - P(n) 排列問題】。
程式碼：
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using System;
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using System.Collections.Generic;
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using System.Text;
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namespace PermutationNR
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{
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internal class Program
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{
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// from string to String[]: "abc" => {"a", "b", "c"}
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public static string [ ] stringToStringArray( string source)
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{
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char [ ] temp = source.ToCharArray ( );
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string [ ] target = new string [temp. Length ];
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for ( int i = 0; i < temp.Length; i++)
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{
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target[i] = temp[i].ToString ( );
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}
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return target;
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}
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private static void Main( string [ ] args)
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{
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Console.Write ( "Please input set: " );
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string [ ] sign = Console.ReadLine ( ).Split ( ' ' );
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// Get r value
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Console.Write ( "How many numbers do you select: " );
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int rValue;
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if ( int.TryParse (Console.ReadLine ( ), out rValue) == false )
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{
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rValue = 1;
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}
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Combination c = new Combination (sign. Length, rValue );
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int [ ] indices;
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List<string> combinationSet = new List<string> ( );
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// Generate all combination
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while (c.hasMore ( ) )
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{
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StringBuilder sb = new StringBuilder ( );
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indices = c.getNext ( );
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for ( int i = 0; i < indices.Length; i++)
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{
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sb.Append (sign[indices[i] ] );
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}
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combinationSet.Add (sb.ToString ( ) );
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}
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List<string> output = new List<string> ( );
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foreach ( string s in combinationSet)
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{
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sign = stringToStringArray(s);
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Permutation p = new Permutation (sign. Length );
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// Generate all permutation
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while (p.hasMore ( ) )
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{
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StringBuilder sb = new StringBuilder ( );
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indices = p.getNext ( );
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for ( int i = 0; i < indices.Length; i++)
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{
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sb.Append (sign[indices[i] ] );
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}
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output.Add (sb.ToString ( ) );
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}
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}
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// List all permutation
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Console.Write ( "Permutaion: \n" );
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output.Sort ( );
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foreach ( string s in output)
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{
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Console.WriteLine (s);
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}
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Console.WriteLine ( "\npress any key to exit" );
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Console.Read ( );
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}
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}
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}
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