您需要做的就是检查该号码是否已存在于列表中,如果存在则获取另一个:
static void Main(string[] args)
{
ArrayList r = new ArrayList();
Random ran = new Random();
int num = 0;
for (int i = 0; i < 50; i++)
{
do { num = ran.Next(1, 51); } while (r.Contains(num));
r.Add(num);
}
for (int i = 0; i < 50; i++)
Console.WriteLine(r[i]);
Console.ReadKey();
}
编辑:这将大大提高效率,防止长时间暂停等待非冲突数字:
static void Main(string[] args)
{
List<int> numbers = new List<int>();
Random ran = new Random();
int number = 0;
int min = 1;
int max = 51;
for (int i = 0; i < 50; i++)
{
do
{
number = ran.Next(min, max);
}
while (numbers.Contains(number));
numbers.Add(number);
if (number == min) min++;
if (number == max - 1) max--;
}
for (int i = 0; i < 50; i++)
Console.WriteLine(numbers[i]);
Console.ReadKey();
}