随机取２０个无重复数

[newma]

我想在（20－46）这27个数中随机生成２７个数，要求无重复.

random myRan=new random();
Hashtable hashtable = new Hashtable();
for (int i = 20; i <= 46; i++)
{
int temp = myRan.Next(20, 46);
if (!hashtable.ContainsValue(temp))
{
hashtable.Add(temp, temp);
string s = table.Rows[temp][0].ToString();
param = param + s + ",";
}
}

大家看看这样行不行，会不会影响效率。

[lw065]

int [] a ={20,21,22,23,24,25,26,27,28,29,30};
            List<int> tmp = new List<int>(a);
            List<int> result = new List<int>();
            Random r = new Random();
            for (int i = tmp.Count-1; i >= 0; i--)
            {
                int j= r.Next(tmp.Count);
                result.Add(tmp[j]);
                tmp.RemoveAt(j);
            }

[aioria0815]

嗯...收了...............................

[s32702]

3，4楼的都OK。。。。。。

[西安风影]

[Quote=引用 8 楼 gomoku 的回复:]
这是个稳定的O(n)方法:

C# codeint[] nums=newint[27];for(inti=0; i<nums.Length; i++) nums[i]=20+i;

Random rand=newRandom();for(inti=nums.Length-1; i>0; i--)
{//randomly choose a position to pop out, j belongs to [0,i-1]intj=rand.Next( i );//swap the popped-out to the rear the arrayinttmp=nums[i];
nums[i]=nums[j];
nums[j]=tmp;
}
[/Quote]
非常可行

[yagebu1983]

[Quote=引用 3 楼 ojlovecd 的回复:]
C# code
Random r = new Random();
List<int> list = new List<int>();
while(list.Count<28)
{
int i = r.Next(20,46);
if(!list.Contains(i))
list.Add(i);
}
[/Quote]
同意！！

[yagebu1983]

int i = r.Next(20,46);
注意随机范围！！

[s330481]

        static void Main(string[] args)
        {
            int[] a = new int[] { 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46 };
            
            Random r = new Random();
            for (int i = a.Length - 1,j=0; i > 6;i--,j++ )
            {
                int temp = r.Next(0, i);               
                int delete = a[temp];                
                a[temp] = a[i];
                a[i] = delete;
            }
            for (int i = 0; i < 20; i++)
                Console.WriteLine(a[i]);
            Console.Read();
        }

[s330481]

        static void Main(string[] args)
        {
            int[] a = new int[] { 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46 };
            
            Random r = new Random();
            for (int i = a.Length - 1; i > 6;i-- )
            {
                int temp = r.Next(0, i);               
                int delete = a[temp];                
                a[temp] = a[i];
                a[i] = delete;
            }
            for (int i = 0; i < 20; i++)
                Console.WriteLine(a[i]);
            Console.Read();
        }

[s330481]

呃。。。。才发现前面已经有相似版本了

[BIGBIRDINWOODS]

[Quote=引用 3 楼 ojlovecd 的回复:]
C# code
Random r = new Random();
List<int> list = new List<int>();
while(list.Count<28)
{
int i = r.Next(20,46);
if(!list.Contains(i))
list.Add(i);
}
[/Quote]

[newma]

这样不行。还是有重复。
谁有更好的办法帮个忙。。。。。。。。。急。。。。。。。。。。。。。。。。。

[lw065]

你这个方法很有可能无法生成27个数吧，for 循环才27次，除非每次都是不重复的

[我姓区不姓区]

Random r = new Random();
List<int> list = new List<int>();
while(list.Count<28)
{
    int i = r.Next(20,46);
    if(!list.Contains(i))
       list.Add(i);
}

[ericzhangbo1982111]

List<int> list = new List<int>();
random myRan=new random();
for(;;)
{
if(list.count>=28)
break;
else
{
int temp = myRan.Next(20, 46);

if(!list.Contains(temp))
list.Add(temp);
}
}

[fancystyle]

            Random rand = new Random();
            var randomSeq = Enumerable.Repeat(0,20).Select(i => rand.Next(20,46));

[fuadam]

有重复不会吧，要不你每次都new下random

[fancystyle]

错了，当我没说。。。。

[gomoku]

这是个稳定的O(n)方法:

int[] nums = new int[27];
for(int i=0; i<nums.Length; i++) nums[i] = 20 + i;

Random rand = new Random();
for(int i=nums.Length-1; i > 0; i--)
{
   // randomly choose a position to pop out,  j belongs to [0,i-1]
   int j = rand.Next( i );
   
   // swap the popped-out to the rear the array
   int tmp = nums[i];
   nums[i] = nums[j];
   nums[j] = tmp;
}