听起来你想要这个函数,它来自Eric Lippert的博客文章,是为了回应生成所有可能组合而编写的:
public static IEnumerable<IEnumerable<T>> CartesianProduct<T>(
this IEnumerable<IEnumerable<T>> sequences)
{
IEnumerable<IEnumerable<T>> emptyProduct = new[] { Enumerable.Empty<T>() };
return sequences.Aggregate(
emptyProduct,
(accumulator, sequence) =>
from accseq in accumulator
from item in sequence
select accseq.Concat(new[] {item}));
}
那将使您编写以下代码:
这样做:
int[][] items = {
new[] { 11001, 54010, 60621 },
new[] { 11001, 60621 },
new[] { 60621 }
};
var routes = CartesianProduct(items);
foreach (var route in routes)
Console.WriteLine(string.Join(", ", route));
并获得以下输出:
11001、11001、60621
11001、60621、60621
54010、11001、60621
54010、60621、60621
60621、11001、60621
60621、60621、60621
编辑:这是VB.NET版本(在VS2010中)
Imports System.Runtime.CompilerServices
Module Module1
<Extension()>
Private Function CartesianProduct(Of T)(
ByVal sequences As IEnumerable(Of IEnumerable(Of T))) _
As IEnumerable(Of IEnumerable(Of T))
Dim emptyProduct As IEnumerable(Of IEnumerable(Of T)) =
New IEnumerable(Of T)() {Enumerable.Empty(Of T)()}
Return sequences.Aggregate(
emptyProduct,
Function(accumulator, sequence)
Return (From accseq In accumulator
From item In sequence
Select accseq.Concat(New T() {item}))
End Function)
End Function
Sub Main(ByVal args As String())
Dim items = New Integer()() {New Integer() {11001, 54010, 60621},
New Integer() {11001, 60621},
New Integer() {60621}}
Dim routes = items.CartesianProduct()
Dim route As IEnumerable(Of Integer)
For Each route In routes
Console.WriteLine(String.Join(", ", route))
Next
End Sub
End Module
当然,如果您不想使用任何LINQ,这里有一个完全不含LINQ的递归实现:
public static IEnumerable<IEnumerable<T>> CartesianProduct<T>(
this IEnumerable<IEnumerable<T>> sequences)
{
var accum = new List<T[]>();
var list = sequences.ToList();
if (list.Count > 0)
CartesianRecurse(accum, new Stack<T>(), list, list.Count - 1);
return accum;
}
static void CartesianRecurse<T>(List<T[]> accum, Stack<T> stack,
List<IEnumerable<T>> list, int index)
{
foreach (T item in list[index])
{
stack.Push(item);
if (index == 0)
accum.Add(stack.ToArray());
else
CartesianRecurse(accum, stack, list, index - 1);
stack.Pop();
}
}
它不会按照第一个函数返回项目的相同顺序,但在功能上是相同的。如果你不喜欢LINQ 或者 递归,这里有一个单一的不使用LINQ的方法,可以执行与递归版本相同的操作:
public static IEnumerable<IEnumerable<T>> CartesianProduct<T>(
this IEnumerable<IEnumerable<T>> sequences)
{
var accum = new List<T[]>();
var list = sequences.ToList();
if (list.Count > 0)
{
var enumStack = new Stack<IEnumerator<T>>();
var itemStack = new Stack<T>();
int index = list.Count - 1;
var enumerator = list[index].GetEnumerator();
while (true)
if (enumerator.MoveNext())
{
itemStack.Push(enumerator.Current);
if (index == 0)
{
accum.Add(itemStack.ToArray());
itemStack.Pop();
}
else
{
enumStack.Push(enumerator);
enumerator = list[--index].GetEnumerator();
}
}
else
{
if (++index == list.Count)
break;
itemStack.Pop();
enumerator = enumStack.Pop();
}
}
return accum;
}