生成前X个排列 [英] Generating first X number of permutations

问题描述

我已经尝试了很多方法，但我无法弄清楚如何生成，例如给定字符数组时的前10个排列。



我正在寻找的是：



chars ='abcd'

count = 10



输出应为：



a

b

c

d

aa

ab

ac

ad

ba

bb



有没有简单的方法来实现这个目标？ />


我的尝试：



-------- -------------------------------------------------- -----------------------------

I've tried so many methods but I just can't figure out how to generate, for example the first 10 permutations when given a character array.

What I'm looking for is this:

chars = 'abcd'
count = 10

The output should then be:

a
b
c
d
aa
ab
ac
ad
ba
bb

Is there a simple way to achieve this?

What I have tried:

---------------------------------------------------------------------------------------

推荐答案

见这里：< a href =https://stackoverflow.com/questions/756055/listing-all-permutations-of-a-string-integer> c＃ - 列出字符串/整数的所有排列 - Stack Overflow [< a href =https://stackoverflow.com/questions/756055/listing-all-permutations-of-a-string-integertarget =_ blanktitle =New Window> ^ ]

Quote：

是一个简单的方法来实现这个目标吗？

Is there a simple way to achieve this?

是的，但在40个问题之后，你应该知道我们不做你的功课。

Yes, but after 40 questions, you should know that we don't do your homework.

引用：

例如前10个排列

您的示例输出不是排列！

Your sample output is not a permutation!

正如我之前提到的那样（解决方案已被删除 - 在您看来 - 与问题无关），您想要获得的结果不是排列 [ ^ ]。



如果您有兴趣实现产生序列集的方法，

基于解决方案＃1中提供的链接 OriginalGriff [ ^ ]，我已将Linq C＃ GetPermutations 方法转换为VB .NET表单：

As i mentioned it earlier (solution has been deleted as it was - in your opinion - unrelated to the question), a result you want to get is not a permutation[^].

If you are interested to implement method which produces set of sequences,
Based on the link provided in solution#1 by OriginalGriff[^], i've translated Linq C# GetPermutations method into VB.NET form:

Sub Main
	
	Dim chars As String = "ABCD"
	Dim count = 10
	'get 10 permutations of ABCD arranged into 3-part sequences 
	Dim result = GetPermutations(chars.ToArray(), 3).Take(count)

	For Each r In result
		Console.WriteLine("{0}", String.Join("", r))
	Next
	
End Sub

Function GetPermutations(Of T)(list AS IEnumerable(Of T), length As Integer) AS IEnumerable(Of IEnumerable(Of T)) 

    If length = 1 Then Return list.Select(Function(o) New T(){o})

    Return GetPermutations(list, length - 1) _
        .SelectMany(Function(a) list.Where(Function(e) Not a.Contains(e)), Function(t1, t2) t1.Concat(New T(){t2}))
		
End Function

以上代码产生10个结果，例如：

Above code produces 10 results, for example:

ABC
ABD
ACB
ACD
ADB
ADC
BAC
BAD
BCA
BCD

如果你想获得N部分的结果，你需要改进上面的代码：

In case you want to get a result of N-part, you need to improve above code:

For i = 1 To chars.Length
		Dim result = GetPermutations(chars.ToArray(), i).Take(count)
		For Each r In result
			Console.WriteLine("{0}", String.Join("", r))
		Next
	Next

如果你想获得非Linq解决方案，你必须实现自己的交换和 GetPer SO提供的方法。

If you want to get non-Linq solution, you have to implement your own Swap and GetPer methods provided on SO.

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