Question

c# 如何生成10位随机数

Answer 1

//方法1
public static int[] GetRandom1(int minValue, int maxValue, int count)
{

Random rnd = new Random();
int length = maxValue - minValue + 1;
byte[] keys = new byte[length];
rnd.NextBytes(keys);
int[] items = new int[length];
for (int i = 0; i < length; i++)
{
items[i] = i + minValue;
}
Array.Sort(keys, items);
int[] result = new int[count];
Array.Copy(items, result, count);
return result;

}

//方法2
public static int[] GetRandom2(int minValue, int maxValue, int count)
{
int[] intList = new int[maxValue];
for (int i = 0; i < maxValue; i++)
{
intList[i] = i + minValue;
}
int[] intRet = new int[count];
int n = maxValue;
Random rand = new Random();
for (int i = 0; i < count; i++)
{
int index = rand.Next(0, n);
intRet[i] = intList[index];
intList[index] = intList[--n];
}

return intRet;
}

}

其实方法有很多，采用递归、HASHTABLE、集合等都可以实现

Answer 2

Random rd = new Random();
int RandomValue = rd.Next(999999999) + 1000000000;

思路是 取0~9位数的随机数，加上十位的最小值

Answer 3

/// <summary>
/// 获取10位随即数(字符串类型)
/// </summary>
/// <returns></returns>
private string GetRandom()
{
Random ran = new Random();
return ran.Next(1000000000, 10000000000).ToString();
}

Answer 4

Random rd=new Random()
rd.Next( min, max) 大于或等于min 小于max的随机数