没有重复的随机值

Question

Random r = new Random();
int randomvalue = r.Next(1,20);
*
*
if(randomvalue == 1) something
if(randomvalue == 2) something
*
*
if(randomvalue == 19) something

使该随机值不重复的最佳方法是什么？顺便说一句：它的 WPF，而不是控制台。

Answer 1

试试这个：

var rnd = new Random();

var shuffled =
    Enumerable
        .Range(1, 20)
        .OrderBy(x => rnd.Next())
        .ToArray();

Answer 2

// 这会起作用

类程序 { 静态无效主要（字符串 [] 参数） {

        List<int> i = new List<int>(new int[] { 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20});
        List<int> r;

        r = ShuffleList(i);


    }


    private static List<E> ShuffleList<E>(List<E> unshuffledList)
    {
        List<E> sList = new List<E>();

        Random r = new Random();
        int randomIndex = 0;
        while (unshuffledList.Count > 0)
        {
            randomIndex = r.Next(0, unshuffledList.Count); 
            sList.Add(unshuffledList[randomIndex]);
            unshuffledList.RemoveAt(randomIndex); //remove so wont be added a second time
        }

        return sList; //return the new shuffled list
    }
}

Answer 3

不清楚重复是什么意思。如果您不想在一行中看到相同的数字，那么只需保留一个哈希表或Dictionary<int,int>，它会保留出现的 last 整数。然后检查下一个数字是否与最后一个数字相同。如果没有，请从字典中删除最后一个数字并放入当前数字。如果它们相同，则从Random 请求另一个随机整数。

var myDictionary = new Dictionary<int,int>;
var randomvalue = r.Next(1,20);
while(myDictionary.ContainsKey(randomvalue))
{
   randomvalue = r.Next(1,20);
}
myDictionary.Clear();
myDictionary.Add(randomvalue, 123); //123 is just a number. Doesn't matter.

这保证了两个相同的整数永远不会连续出现。

注意：：其他答案很有创意，但“了解您的数据结构”。不要使用列表查找。哈希是我们使用的。

Answer 4

您需要存储随机值是使其唯一的唯一方法。

试试这个：

Random r = new Random(); 
public List<int> randomList = new List<int>();
int randomvalue = 0;
Public void newNumber()
{
    randomvalue = r.Next(0, 20);

    if (!randomList.Contains(randomvalue))
    {
        randomList.Add(randomvalue);
        if(randomvalue == 1) 
          //do something
        if(randomvalue == N)
          // do something
    }
}

Answer 5

using System;
using System.Collections;
using System.Collections.Generic;

namespace SOFAcrobatics
{
    public static class Launcher
    {
        public static void Main ()
        {
            // 1 to 20 without duplicates
            List<Int32> ns = new List<Int32>();
            Random r = new Random();
            Int32 ph = 0;
            while (ns.Count < 20)
            {
                while (ns.Contains (ph = r.Next(1, 21))) {}
                ns.Add(ph);
            }
            // ns is now populated with random unique numbers (1 to 20)
        }
    }
}

Answer 6

试试下面的方法：

随机 randomInstance = new Random();

        List<int> NumList = new List<int>();
        NumList.AddRange(new int[] { 1, 2, 3, 4, 5, 6, 7,
            8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,
            22, 23, 24, 25, 26, 27, 28, 29, 30 });

        int index = randomInstance.Next(0, NumList.Count - 1);
        int randomNumber = NumList[index];
        NumList.RemoveAt(index);