Question

C#编程,利用递归、非递归编程实现输出如下字母金字塔

要从头到尾完整的,谢谢.

Answer 1

非递归：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("输出a......k的字母金字塔为：");
string str="`abcdefghijklmnop";
for (int i = 1; i < 12; i++)
{
for (int j = 1; j < 50 - 2 * i; j++)
{
Console.Write(" ");
}
for (int j = -i+1; j < i; j++)
{
if (j < 0)
{
Console.Write(str[i+j] + " ");
}
else
{
Console.Write(str[i-j] + " ");
}

}
Console.WriteLine();
}
Console.Read();
}
}
}

递归：
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
class Program
{
public static string str="`abcdefghijklmnop";
private static void fun(int i)
{
if (i > 11) return;
for (int j = 1; j < 50 - 2 * i; j++)
{
Console.Write(" ");
}
for (int j = -i + 1; j < i; j++)
{
if (j < 0)
{
Console.Write(str[i + j] + " ");
}
else
{
Console.Write(str[i - j] + " ");
}

}
Console.WriteLine();
fun(i + 1);
}
static void Main(string[] args)
{
Console.WriteLine("输出a......k的字母金字塔为：");
fun(1);
Console.Read();
}

}
}

Answer 2

Console.WriteLine("输出a......k的字母金字塔为：");

int sz ='a';

int szk = 'k';

int pos = (szk - sz) * 2 + 1;

for (int i =0; i <=szk-sz; i++)
{
for (int k = pos-i*2; k > 0; k--)
{
Console.Write(" ");
}

for (int j = 0; j <= i; j++)
{
Console.Write(" {0}",(char)(sz + j));
}

for (int j = i-1; j >= 0; j--)
{
Console.Write(" {0}",(char)(sz+j));
}
Console.WriteLine();
}