笛卡尔积+ N x M动态阵列

Question

我已经看了几个小时的解决方案,没有任何成功.希望有人可以帮助我.

我有M个原始邮政编码的动态N个项目数组.

例如:

项目1:11001,54010,60621项目2:11001,60621项目3:60621

我想创建一个如下所示的新数组:

路线1:11001,11001,60621路线2:11001,60621,60621路线3:54010,11001,60621

等 - 直到6号公路.

建议？

----------------------有没有办法在不使用Linq的情况下完成此任务？VB.net和Linq不一起:)

Answer 1

听起来你想要这个函数来自Eric Lippert的博客文章,以回应Generate all Possible Combinations:

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}));                
}

那会让你写这样的代码:

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));

得到这样的输出:

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,如果你不想要任何LINQ,这里是一个完全无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();
    }
}

它不会以与第一个函数相同的顺序返回项目,但在功能上相同.如果你不喜欢LINQ 或递归,这里是一个单一的LINQ-less方法,它与递归版本做同样的事情:

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;
}

Answer 2

您要做的是生成数组中每个项目的组合。

下面是 N == 3 的硬编码示例：

        var array1 = new[] { 1101, 5410, 60621 };
        var array2 = new[] { 1101, 60621 };
        var array3 = new[] { 60621 };

        foreach (var a in array1)
        {
            foreach (var b in array2)
            {
                foreach (var c in array3)
                {
                    Console.WriteLine("{0},{1},{2}", a, b, c);
                }
            }
        }

看看你是否可以针对N 个案例进行调整。