c#打印出两个菱形在控制台同一高度位置。（便是一左一右）

c#打印出两个菱形在控制台同一高度位置。（就是一左一右）

功能要求
可自定义菱形长度
可自定义菱形个数
可自定义菱形间的间隔
 

程序结果是这样的：
两个菱形在同一水平位置 即一左一右
跪求程序编写方法 代码实现

[解决办法]

C# code

using System;using System.Collections.Generic;using System.Linq;using System.Text;namespace ConsoleApplication1{    class Program    {        static void Main(string[] args)        {            int l = 8; //边长            int n = 5; //数量            int x = 3; //间距            for (int i = 0; i < n / 2; i++)            {                for (int j = 0; j < l; j++)                {                    string lb = new string(' ', l - j);                    string s1 = j == 0 ? "*" : "*" + new string(' ', j * 2 - 1) + "*";                    string mb = new string(' ', x + l - j - 1) +lb;                    Console.WriteLine("{0}{1}{2}{3}", lb, s1, mb, s1);                }                for (int j = l - 2; j >= 0; j--)                {                    string lb = new string(' ', l - j);                    string s1 = j == 0 ? "*" : "*" + new string(' ', j * 2 - 1) + "*";                    string mb = new string(' ', x + l - j - 1) + lb;                    Console.WriteLine("{0}{1}{2}{3}", lb, s1, mb, s1);                }                if (i != (n / 2 - 1) || n % 2 == 1)                {                    for (int j = 0; j < x; j++)                    {                        Console.WriteLine();                    }                }            }            if (n % 2 == 1)            {                for (int j = 0; j < l; j++)                {                    string lb = new string(' ', l - j);                    string s1 = j == 0 ? "*" : "*" + new string(' ', j * 2 - 1) + "*";                    Console.WriteLine("{0}{1}", lb, s1);                }                for (int j = l - 2; j >= 0; j--)                {                    string lb = new string(' ', l - j);                    string s1 = j == 0 ? "*" : "*" + new string(' ', j * 2 - 1) + "*";                    Console.WriteLine("{0}{1}", lb, s1);                }            }        }    }}
[解决办法]
实心只要把
string s1 = j == 0 ? "*" : "*" + new string(' ', j * 2 - 1) + "*";
写成
string s1 = new string('*', j * 2 + 1);
就可以了