生成随机的,独特的价值观C# [英] Generating random, unique values C#

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问题描述

我已经搜索了一会儿,一直苦苦寻找这一点,我想生成多个随机的,唯一的编号为C#。我使用System.Random,而我使用的是datetime.now.ticks种子:

 公开随机的A =新的随机(DateTime.Now.Ticks.GetHash code());
私人无效NewNumber()
  {
     为mynumber = a.Next(0,10);
  }
 

我打电话NewNumber()定期,但问题是我经常重复的数字。有些人认为,因为我是随机每次我做了声明,就不会产生一个随机数,所以我把我的声明函数外。任何建议或比使用System.Random更好的方法?谢谢

解决方案
  

我打电话NewNumber()定期,但问题是我经常得到   重复的数字。

Random.Next 不guarntee数是唯一的。此外,您的范围是从0到10,那么你会得到重复的值。可能是你可以设置 INT 的列表,并检查,如果它不包含重复的后面插入随机数在列表中。是这样的:

 公开随机的A =新的随机(); //新的随机替换(DateTime.Now.Ticks.GetHash code());
                                //由于类似code工作是在默认构造函数内部
公开名单< INT> randomList =新的名单,其中,INT>();
INT为mynumber = 0;
私人无效NewNumber()
{
    为mynumber = a.Next(0,10);
    如果(!randomList.Contains(为mynumber))
        randomList.Add(为mynumber);
}
 

I've searched for a while and been struggling to find this, I'm trying to generate several random, unique numbers is C#. I'm using System.Random, and I'm using a datetime.now.ticks seed:

public Random a = new Random(DateTime.Now.Ticks.GetHashCode());
private void NewNumber()
  {
     MyNumber = a.Next(0, 10);
  }

I'm calling NewNumber() regularly, but the problem is I often get repeated numbers. Some people suggested because I was declaring the random every time I did it, it would not produce a random number, so I put the declaration outside my function. Any suggestions or better ways than using System.Random ? Thank you

解决方案

I'm calling NewNumber() regularly, but the problem is I often get repeated numbers.

Random.Next doesn't guarntee the number to be unique. Also your range is from 0 to 10 and chances are you will get duplicate values. May be you can setup a list of int and insert random numbers in the list after checking if it doesn't contain the duplicate. Something like:

public Random a = new Random(); // replace from new Random(DateTime.Now.Ticks.GetHashCode());
                                // Since similar code is done in default constructor internally
public List<int> randomList = new List<int>();
int MyNumber = 0;
private void NewNumber()
{
    MyNumber = a.Next(0, 10);
    if (!randomList.Contains(MyNumber))
        randomList.Add(MyNumber);
}

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