我有这种方法来生成新的数值,但完成该功能需要很长时间。如何在快速approch中生成随机数?
public int GeneratenewID(int[] OptionId)
{
Random ran = new Random();
int SearchId = ran.Next(1, OptionId.Length*2);
if (!OptionId.Contains(SearchId))
{
return SearchId;
}
else
{
return GeneratenewID(OptionId);
}
}
这肯定会起作用所以尝试:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace ConsoleApplication4
{
class Program
{
static void Main(string[] args)
{
int [] OptionId=new int[]
{
0, 1,4,7,3,1,37,9
};
Program p = new Program();
int a= p. GeneratenewID(OptionId);
}
public int GeneratenewID(int[] OptionId)
{
Random ran = new Random(1);
int number = 0;
for (int j = 0; j < OptionId.Length ; j++)
{
number = ran.Next(OptionId.Length);
if (!OptionId.Contains(number))
break;
else
j--;
}
return number;
}
}
}