排列在c# [英] permutation in c#
本文介绍了排列在c#的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
是否有可能产生在C#中收集的所有排列?
的char [] inputSet = {'A', B,C};
置换<&烧焦GT;排列=新的置换和LT;焦炭>(inputSet);
的foreach(IList的<焦炭指p在排列)
{
Console.WriteLine(的String.Format({{{0} {1} {2}}},P [0 ],第[1],第[2]));
}
解决方案
我已经面对的问题,我写了这些简单的方法:
公共静态的IList< T []> GeneratePermutations< T>(T [] OBJ文件,长限制吗?)
{
VAR的结果=新的List< T []>();
长N =阶乘(objs.Length);
N =(limit.HasValue || limit.Value> N!)? N:(limit.Value);
为(长K = 0; K< N; k ++)
{
T [] kperm = GenerateKthPermutation< T>(K,OBJ文件);
result.Add(kperm);
}
返回结果;
}
公共静态T [] GenerateKthPermutation< T>(长K,T [] OBJ文件)
{
T [] permutedObjs =新的T [OBJ文件。长度];
的for(int i = 0; I< objs.Length;我++)
{
permutedObjs [I] = OBJ文件[I]
}
为(INT J = 2; J< objs.Length + 1; J ++)
{
K = K /(J - 1); //整数除法切断其余
长I1 =(K%j)条;
长I2 =的J - 1;
如果
{$ B $(B T)= tmpObj1 permutedObjs [I1](I1 = I2!); $ B $(B T)= tmpObj2 permutedObjs [12];
permutedObjs [I1] = tmpObj2;
permutedObjs [12] = tmpObj1;
}
}
返回permutedObjs;
}
公共静态长因子(INT N)
{
如果(N小于0){抛出新的异常(为阶乘未接受输入); } //错误结果 - 未定义
如果(N> 256){抛出新的异常(输入太大了,阶乘); } //错误结果 - 输入太大
如果(N == 0){返回1; }
//计算阶乘的迭代而不是递归:
长tempResult = 1;
的for(int i = 1; I< = N;我++)
{
tempResult * = I;
}
返回tempResult;
}
用法:
VAR烫发= Utilities.GeneratePermutations<焦炭>(新的char [] {'A','B','C'},NULL);
Is it possible to generate all permutations of a collection in c#?
char[] inputSet = { 'A','B','C' };
Permutations<char> permutations = new Permutations<char>(inputSet);
foreach (IList<char> p in permutations)
{
Console.WriteLine(String.Format("{{{0} {1} {2}}}", p[0], p[1], p[2]));
}
解决方案
I've already faced the problem and I wrote these simple methods:
public static IList<T[]> GeneratePermutations<T>(T[] objs, long? limit)
{
var result = new List<T[]>();
long n = Factorial(objs.Length);
n = (!limit.HasValue || limit.Value > n) ? n : (limit.Value);
for (long k = 0; k < n; k++)
{
T[] kperm = GenerateKthPermutation<T>(k, objs);
result.Add(kperm);
}
return result;
}
public static T[] GenerateKthPermutation<T>(long k, T[] objs)
{
T[] permutedObjs = new T[objs.Length];
for (int i = 0; i < objs.Length; i++)
{
permutedObjs[i] = objs[i];
}
for (int j = 2; j < objs.Length + 1; j++)
{
k = k / (j - 1); // integer division cuts off the remainder
long i1 = (k % j);
long i2 = j - 1;
if (i1 != i2)
{
T tmpObj1 = permutedObjs[i1];
T tmpObj2 = permutedObjs[i2];
permutedObjs[i1] = tmpObj2;
permutedObjs[i2] = tmpObj1;
}
}
return permutedObjs;
}
public static long Factorial(int n)
{
if (n < 0) { throw new Exception("Unaccepted input for factorial"); } //error result - undefined
if (n > 256) { throw new Exception("Input too big for factorial"); } //error result - input is too big
if (n == 0) { return 1; }
// Calculate the factorial iteratively rather than recursively:
long tempResult = 1;
for (int i = 1; i <= n; i++)
{
tempResult *= i;
}
return tempResult;
}
Usage:
var perms = Utilities.GeneratePermutations<char>(new char[]{'A','B','C'}, null);
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