如何在没有嵌套for循环的情况下列出所有基数n

Question

如前所述here 问题是如何生成长度为 n 的所有可能数字，其中每个数字来自集合 {1,2,3,..., n-1} ？即如何创建所有基数为 n 的数字？ 一种解决方案是使用 for 循环：

int n;
int[] myArray = new int[n];
for (int i1 = 1; i1 < n; i1++)
myArray[0]=i1;
for (int i2 = 1; i2 < n; i2++)
    myArray[1]=i2;
    // and so on....
        for (int in = 1; in < n; in++)
        {
            myArray[n]=in;
            foreach (var item in myArray)
                Console.Write(item.ToString());
                Console.Write(Environment.NewLine);
        }

问题是当 n 改变时循环数会改变！

Answer 1

递归解决方案是：

using System;
namespace ConsoleApplication1
{
class Program
{
    static void Main(string[] args)
    {
        new c().f1(3);
        Console.WriteLine("Done!");
        Console.ReadKey();
    }
}
class c
{
    public void f1(int n)
    {
        f(n-1 , n, "");
    }
    public string f(int n,int depth, string r)
    {
        if(depth == 0)
        {
            Console.WriteLine(r);
            return "";
        }
        for (int i = 1; i <= n ; i++)
        {                
            string r1 =  r + i.ToString();
            f(n, depth - 1, r1);
        }
        return "";
    }        
}
}

说明：

for n= 4

f1(4) =>
f(3,4,"")
    f(3,3,"1")
        f(3,2,"11")
            f(3,1,"111")
                f(3,0,"1111") => prints "1111"
                f(3,0,"1112") => prints "1112"
                f(3,0,"1113") => prints "1113"
                f(3,0,"1114") => prints "1114"
            f(3,1,"112")
                f(3,0,"1121") => prints "1121"
                f(3,0,"1122") => prints "1122"
                f(3,0,"1123") => prints "1123"
                f(3,0,"1124") => prints "1124"
            f(3,1,"113")
                f(3,0,"1131") => prints "1131"
                f(3,0,"1132") => prints "1132"
                f(3,0,"1133") => prints "1133"
                f(3,0,"1134") => prints "1134"
            ....
            ....
            ....