我有一个 ArrayList[] myList 并且我正在尝试创建数组中所有值排列的列表。
示例：（所有值都是字符串）
myList[0] = { "1", "5", "3", "9" };
myList[1] = { "2", "3" };
myList[2] = { "93" };
myList 的计数可以变化，因此事先不知道其长度。
我希望能够生成类似于以下所有排列的列表（但有一些额外的格式）。
1 2 93
1 3 93
5 2 93
5 3 93
3 2 93
3 3 93
9 2 93
9 3 93
这对我想要完成的事情有意义吗？我似乎无法想出一个好的方法来做到这一点，（如果有的话）。
编辑：
我不确定递归是否会干扰我以自己的方式格式化输出的愿望。抱歉，我之前没有提到我的格式是什么。
我想最终构建一个 string[] 数组，其中包含以下格式的所有组合：
对于“1 2 93”排列
我希望输出为“val0=1;val1=2;val2=93;”
我现在将尝试递归。谢谢Jokepu博士
我很惊讶没有人发布 LINQ 解决方案。
from val0 in new []{ "1", "5", "3", "9" }
from val1 in new []{ "2", "3" }
from val2 in new []{ "93" }
select String.Format("val0={0};val1={1};val2={2}", val0, val1, val2)
递归解决方案
static List<string> foo(int a, List<Array> x)
{
List<string> retval= new List<string>();
if (a == x.Count)
{
retval.Add("");
return retval;
}
foreach (Object y in x[a])
{
foreach (string x2 in foo(a + 1, x))
{
retval.Add(y.ToString() + " " + x2.ToString());
}
}
return retval;
}
static void Main(string[] args)
{
List<Array> myList = new List<Array>();
myList.Add(new string[0]);
myList.Add(new string[0]);
myList.Add(new string[0]);
myList[0] = new string[]{ "1", "5", "3", "9" };
myList[1] = new string[] { "2", "3" };
myList[2] = new string[] { "93" };
foreach (string x in foo(0, myList))
{
Console.WriteLine(x);
}
Console.ReadKey();
}
请注意，通过将返回更改为字符串列表的列表并将 retval.add 调用更改为使用列表而不是使用连接来返回列表或数组而不是字符串将非常容易。
这个怎么运作：
这是一个经典的递归算法。基本情况是foo(myList.Count, myList)，它返回一个包含一个元素的列表，即空字符串。n 个字符串数组 s1, s2, ..., sN 的列表的排列等于 sA1 的每个成员都以 n-1 个字符串数组 s2, ..., sN 的排列为前缀。基本情况只是为要连接的 sN 的每个元素提供一些东西。
我最近在我的一个项目中遇到了类似的问题，偶然发现了这个问题。我需要一个可以处理任意对象列表的非递归解决方案。这就是我想出的。基本上我正在为每个子列表形成一个枚举器列表并迭代地递增它们。
public static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<IEnumerable<T>> lists)
{
// Check against an empty list.
if (!lists.Any())
{
yield break;
}
// Create a list of iterators into each of the sub-lists.
List<IEnumerator<T>> iterators = new List<IEnumerator<T>>();
foreach (var list in lists)
{
var it = list.GetEnumerator();
// Ensure empty sub-lists are excluded.
if (!it.MoveNext())
{
continue;
}
iterators.Add(it);
}
bool done = false;
while (!done)
{
// Return the current state of all the iterator, this permutation.
yield return from it in iterators select it.Current;
// Move to the next permutation.
bool recurse = false;
var mainIt = iterators.GetEnumerator();
mainIt.MoveNext(); // Move to the first, succeeds; the main list is not empty.
do
{
recurse = false;
var subIt = mainIt.Current;
if (!subIt.MoveNext())
{
subIt.Reset(); // Note the sub-list must be a reset-able IEnumerable!
subIt.MoveNext(); // Move to the first, succeeds; each sub-list is not empty.
if (!mainIt.MoveNext())
{
done = true;
}
else
{
recurse = true;
}
}
}
while (recurse);
}
}
您可以使用factoradics来生成排列的枚举。试试MSDN 上的这篇文章以在 C# 中实现。
无论您将多少个数组添加到 myList 中，这都会起作用：
static void Main(string[] args)
{
string[][] myList = new string[3][];
myList[0] = new string[] { "1", "5", "3", "9" };
myList[1] = new string[] { "2", "3" };
myList[2] = new string[] { "93" };
List<string> permutations = new List<string>(myList[0]);
for (int i = 1; i < myList.Length; ++i)
{
permutations = RecursiveAppend(permutations, myList[i]);
}
//at this point the permutations variable contains all permutations
}
static List<string> RecursiveAppend(List<string> priorPermutations, string[] additions)
{
List<string> newPermutationsResult = new List<string>();
foreach (string priorPermutation in priorPermutations)
{
foreach (string addition in additions)
{
newPermutationsResult.Add(priorPermutation + ":" + addition);
}
}
return newPermutationsResult;
}
请注意，它并不是真正的递归。可能是一个误导性的函数名称。
这是一个符合您的新要求的版本。请注意我输出到控制台的部分，您可以在此处进行自己的格式化：
static void Main(string[] args)
{
string[][] myList = new string[3][];
myList[0] = new string[] { "1", "5", "3", "9" };
myList[1] = new string[] { "2", "3" };
myList[2] = new string[] { "93" };
List<List<string>> permutations = new List<List<string>>();
foreach (string init in myList[0])
{
List<string> temp = new List<string>();
temp.Add(init);
permutations.Add(temp);
}
for (int i = 1; i < myList.Length; ++i)
{
permutations = RecursiveAppend(permutations, myList[i]);
}
//at this point the permutations variable contains all permutations
foreach (List<string> list in permutations)
{
foreach (string item in list)
{
Console.Write(item + ":");
}
Console.WriteLine();
}
}
static List<List<string>> RecursiveAppend(List<List<string>> priorPermutations, string[] additions)
{
List<List<string>> newPermutationsResult = new List<List<string>>();
foreach (List<string> priorPermutation in priorPermutations)
{
foreach (string addition in additions)
{
List<string> priorWithAddition = new List<string>(priorPermutation);
priorWithAddition.Add(addition);
newPermutationsResult.Add(priorWithAddition);
}
}
return newPermutationsResult;
}
您所要求的称为笛卡尔积。一旦你知道它叫什么，Stack Overflow 上有几个类似的问题。他们似乎最终都指向了一个最终写成博客文章的答案：
http://blogs.msdn.com/b/ericlippert/archive/2010/06/28/computing-a-cartesian-product-with-linq.aspx
非递归解决方案：
foreach (String s1 in array1) {
foreach (String s2 in array2) {
foreach (String s3 in array3) {
String result = s1 + " " + s2 + " " + s3;
//do something with the result
}
}
}
递归解决方案：
private ArrayList<String> permute(ArrayList<ArrayList<String>> ar, int startIndex) {
if (ar.Count == 1) {
foreach(String s in ar.Value(0)) {
ar.Value(0) = "val" + startIndex + "=" + ar.Value(0);
return ar.Value(0);
}
ArrayList<String> ret = new ArrayList<String>();
ArrayList<String> tmp1 ar.Value(0);
ar.remove(0);
ArrayList<String> tmp2 = permute(ar, startIndex+1);
foreach (String s in tmp1) {
foreach (String s2 in tmp2) {
ret.Add("val" + startIndex + "=" + s + " " + s2);
}
}
return ret;
}
class Program
{
static void Main(string[] args)
{
var listofInts = new List<List<int>>(3);
listofInts.Add(new List<int>{1, 2, 3});
listofInts.Add(new List<int> { 4,5,6 });
listofInts.Add(new List<int> { 7,8,9,10 });
var temp = CrossJoinLists(listofInts);
foreach (var l in temp)
{
foreach (var i in l)
Console.Write(i + ",");
Console.WriteLine();
}
}
private static IEnumerable<List<T>> CrossJoinLists<T>(IEnumerable<List<T>> listofObjects)
{
var result = from obj in listofObjects.First()
select new List<T> {obj};
for (var i = 1; i < listofObjects.Count(); i++)
{
var iLocal = i;
result = from obj in result
from obj2 in listofObjects.ElementAt(iLocal)
select new List<T>(obj){ obj2 };
}
return result;
}
}
这是一个使用很少代码的版本，并且完全是声明性的
public static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> collection) where T : IComparable
{
if (!collection.Any())
{
return new List<IEnumerable<T>>() {Enumerable.Empty<T>() };
}
var sequence = collection.OrderBy(s => s).ToArray();
return sequence.SelectMany(s => GetPermutations(sequence.Where(s2 => !s2.Equals(s))).Select(sq => (new T[] {s}).Concat(sq)));
}
当我为大量代码执行此操作时遇到的一个问题是，在给出的示例 brian 中，我实际上内存不足。为了解决这个问题，我使用了以下代码。
static void foo(string s, List<Array> x, int a)
{
if (a == x.Count)
{
// output here
Console.WriteLine(s);
}
else
{
foreach (object y in x[a])
{
foo(s + y.ToString(), x, a + 1);
}
}
}
static void Main(string[] args)
{
List<Array> a = new List<Array>();
a.Add(new string[0]);
a.Add(new string[0]);
a.Add(new string[0]);
a[0] = new string[] { "T", "Z" };
a[1] = new string[] { "N", "Z" };
a[2] = new string[] { "3", "2", "Z" };
foo("", a, 0);
Console.Read();
}
这是我编写的通用递归函数（以及可能方便调用的重载）：
Public Shared Function GetCombinationsFromIEnumerables(ByRef chain() As Object, ByRef IEnumerables As IEnumerable(Of IEnumerable(Of Object))) As List(Of Object())
Dim Combinations As New List(Of Object())
If IEnumerables.Any Then
For Each v In IEnumerables.First
Combinations.AddRange(GetCombinationsFromIEnumerables(chain.Concat(New Object() {v}).ToArray, IEnumerables.Skip(1)).ToArray)
Next
Else
Combinations.Add(chain)
End If
Return Combinations
End Function
Public Shared Function GetCombinationsFromIEnumerables(ByVal ParamArray IEnumerables() As IEnumerable(Of Object)) As List(Of Object())
Return GetCombinationsFromIEnumerables(chain:=New Object() {}, IEnumerables:=IEnumerables.AsEnumerable)
End Function
和 C# 中的等价物：
public static List<object[]> GetCombinationsFromIEnumerables(ref object[] chain, ref IEnumerable<IEnumerable<object>> IEnumerables)
{
List<object[]> Combinations = new List<object[]>();
if (IEnumerables.Any) {
foreach ( v in IEnumerables.First) {
Combinations.AddRange(GetCombinationsFromIEnumerables(chain.Concat(new object[] { v }).ToArray, IEnumerables.Skip(1)).ToArray);
}
} else {
Combinations.Add(chain);
}
return Combinations;
}
public static List<object[]> GetCombinationsFromIEnumerables(params IEnumerable<object>[] IEnumerables)
{
return GetCombinationsFromIEnumerables(chain = new object[], IEnumerables = IEnumerables.AsEnumerable);
}
便于使用：
Dim list1 = New String() {"hello", "bonjour", "hallo", "hola"}
Dim list2 = New String() {"Erwin", "Larry", "Bill"}
Dim list3 = New String() {"!", ".."}
Dim result = MyLib.GetCombinationsFromIEnumerables(list1, list2, list3)
For Each r In result
Debug.Print(String.Join(" "c, r))
Next
或在 C# 中：
object list1 = new string[] {"hello","bonjour","hallo","hola"};
object list2 = new string[] {"Erwin", "Larry", "Bill"};
object list3 = new string[] {"!",".."};
object result = MyLib.GetCombinationsFromIEnumerables(list1, list2, list3);
foreach (r in result) {
Debug.Print(string.Join(' ', r));
}
这是一个非递归、非 Linq 的解决方案。我不禁觉得我可以减少循环并用除法和模数计算位置，但不能完全理解这一点。
static void Main(string[] args)
{
//build test list
List<string[]> myList = new List<string[]>();
myList.Add(new string[0]);
myList.Add(new string[0]);
myList.Add(new string[0]);
myList[0] = new string[] { "1", "2", "3"};
myList[1] = new string[] { "4", "5" };
myList[2] = new string[] { "7", "8", "9" };
object[][] xProds = GetProducts(myList.ToArray());
foreach(object[] os in xProds)
{
foreach(object o in os)
{
Console.Write(o.ToString() + " ");
}
Console.WriteLine();
}
Console.ReadKey();
}
static object[][] GetProducts(object[][] jaggedArray){
int numLists = jaggedArray.Length;
int nProducts = 1;
foreach (object[] oArray in jaggedArray)
{
nProducts *= oArray.Length;
}
object[][] productAry = new object[nProducts][];//holds the results
int[] listIdxArray = new int[numLists];
listIdxArray.Initialize();
int listPtr = 0;//point to current list
for(int rowcounter = 0; rowcounter < nProducts; rowcounter++)
{
//create a result row
object[] prodRow = new object[numLists];
//get values for each column
for(int i=0;i<numLists;i++)
{
prodRow[i] = jaggedArray[i][listIdxArray[i]];
}
productAry[rowcounter] = prodRow;
//move the list pointer
//possible states
// 1) in a list, has room to move down
// 2) at bottom of list, can move to next list
// 3) at bottom of list, no more lists left
//in a list, can move down
if (listIdxArray[listPtr] < (jaggedArray[listPtr].Length - 1))
{
listIdxArray[listPtr]++;
}
else
{
//can move to next column?
//move the pointer over until we find a list, or run out of room
while (listPtr < numLists && listIdxArray[listPtr] >= (jaggedArray[listPtr].Length - 1))
{
listPtr++;
}
if (listPtr < listIdxArray.Length && listIdxArray[listPtr] < (jaggedArray[listPtr].Length - 1))
{
//zero out the previous stuff
for (int k = 0; k < listPtr; k++)
{
listIdxArray[k] = 0;
}
listIdxArray[listPtr]++;
listPtr = 0;
}
}
}
return productAry;
}
private static void GetP(List<List<string>> conditions, List<List<string>> combinations, List<string> conditionCombo, List<string> previousT, int selectCnt)
{
for (int i = 0; i < conditions.Count(); i++)
{
List<string> oneField = conditions[i];
for (int k = 0; k < oneField.Count(); k++)
{
List<string> t = new List<string>(conditionCombo);
t.AddRange(previousT);
t.Add(oneField[k]);
if (selectCnt == t.Count )
{
combinations.Add(t);
continue;
}
GetP(conditions.GetRange(i + 1, conditions.Count - 1 - i), combinations, conditionCombo, t, selectCnt);
}
}
}
List<List<string>> a = new List<List<string>>();
a.Add(new List<string> { "1", "5", "3", "9" });
a.Add(new List<string> { "2", "3" });
a.Add(new List<string> { "93" });
List<List<string>> result = new List<List<string>>();
GetP(a, result, new List<string>(), new List<string>(), a.Count);
另一个递归函数。