在c＃中查找数组中所有项目组合的最佳方法是什么？
答案 0 :(得分：81)
<强>已更新
以下是针对不同场景的一组通用功能（需要.net 3.5或更高版本）。输出的列表为{1,2,3,4}，长度为2。
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}
<强>排列组合
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}
重复的K组合
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}
<强> 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}
答案 1 :(得分：15)
这就是所谓的排列。
这可以为您提供任何集合的排列：
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;
}
}
示例：
string[] items = { "one", "two", "three" };
foreach (string[] permutation in Permutation.GetPermutations<string>(items)) {
Console.WriteLine(String.Join(", ", permutation));
}
答案 2 :(得分：12)
是O（n！）
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;
}
}
}
答案 3 :(得分：4)
关于彭阳的回答： 这是我的通用函数，它可以从T：
列表中返回所有组合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 }));
}
实施例1：n = 3，k = 2
IEnumerable<IEnumerable<int>> result =
GetCombinations(Enumerable.Range(1, 3), 2);
输出 - 整数列表列表：
{1, 1} {1, 2} {1, 3} {2, 1} {2, 2} {2, 3} {3, 1} {3, 2} {3, 3}
............................................... ..............................
我运行了这个例子，我不太确定结果的正确性。
实施例2：n = 3，k = 3
IEnumerable<IEnumerable<int>> result =
GetCombinations(Enumerable.Range(1, 3), 3);
输出 - 整数列表列表：
{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}
这不应该在组合中发生，否则应该指明它是重复的。参见文章http://en.wikipedia.org/wiki/Combinations
答案 4 :(得分：2)
也许kwcombinatorics可以提供一些帮助（请参阅主页上的示例）：
KwCombinatorics库有3个类，提供3种不同的方式来生成有序（排名）的数字组合列表。这些组合可用于软件测试，允许生成各种类型的可能输入组合。其他用途包括解决数学问题和机会游戏。
答案 5 :(得分：2)
有很多很容易找到用户输入字符串组合的方法。
使用LINQ的第一种方式
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;
}
使用此功能，如
Console.WriteLine("Enter a sting ");
var input = Console.ReadLine();
foreach (var stringCombination in FindPermutations(input))
{
Console.WriteLine(stringCombination);
}
Console.ReadLine();
其他方式是使用循环
// 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();
}
您可以使用此功能，如
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();
答案 6 :(得分：1)
详细答案请参阅：Donald Knuth，计算机编程艺术（又名TAOCP）。第4A卷，枚举和回溯，第7.2章。产生所有可能性。 http://www-cs-faculty.stanford.edu/~uno/taocp.html
答案 7 :(得分：1)
Gufa给出的另一个解决方案版本。在类的完整源代码下面：
using System.Collections.Generic;
namespace ConsoleApplication1
{
public class 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;
}
}
}
这实际上适用于组合。但是不允许在k中选择n的组合......
答案 8 :(得分：1)
如何进行递归？
internal HashSet<string> GetAllPermutations(IEnumerable<int> numbers)
{
HashSet<string> results = new HashSet<string>();
if (numbers.Count() > 0)
results.Add(string.Join(",", new SortedSet<int>(numbers)));
for (int i = 0; i <= numbers.Count() - 1; i++)
{
List<int> newNumbers = new List<int>(numbers);
newNumbers.RemoveAt(i);
results.UnionWith(GetAllPermutations(newNumbers));
}
return results;
}
答案 9 :(得分：0)
我创建了一个方法来获取数组中所有整数元素的唯一组合，如下所示。我使用Tuple来表示一对或数字组合：
private static void CombinationsOfItemsInAnArray()
{
int[] arr = { 10, 50, 3, 1, 2 }; //unique elements
var numberSet = new HashSet<int>();
var combinationList = new List<Tuple<int, int>>();
foreach (var number in arr)
{
if (!numberSet.Contains(number))
{
//create all tuple combinations for the current number against all the existing number in the number set
foreach (var item in numberSet)
combinationList.Add(new Tuple<int, int>(number, item));
numberSet.Add(number);
}
}
foreach (var item in combinationList)
{
Console.WriteLine("{{{0}}} - {{50} - {10}
{3} - {10}
{3} - {50}
{1} - {10}
{1} - {50}
{1} - {3}
{2} - {10}
{2} - {50}
{2} - {3}
{2} - {1}
}",item.Item1,item.Item2);
}
}
当我在控制台应用程序中调用此方法时，我得到以下输出：
const userfetched =[];
fetch('https://jsonplaceholder.typicode.com/users').then(response=> response.json()).then(user=>userfetched.push(user));
console.log(userfetched)