随机数发生器，没有重复 [英] Random number generator with no duplicates

问题描述

基本上我创建一个程序随机生成6个独特的开奖号码所以在同一行没有重复，这里是code我到目前为止...

Basically I'm creating a program to randomly generate 6 unique lottery numbers so there is no duplicates in the same line, here is the code I have so far...

        //Generate 6 random numbers using the randomiser object

        int randomNumber1 = random.Next(1, 49);
        int randomNumber2 = random.Next(1, 49);
        int randomNumber3 = random.Next(1, 49);
        int randomNumber4 = random.Next(1, 49);
        int randomNumber5 = random.Next(1, 49);
        int randomNumber6 = random.Next(1, 49);

        textBox1.Text = randomNumber1.ToString();
        textBox2.Text = randomNumber2.ToString();
        textBox3.Text = randomNumber3.ToString();
        textBox4.Text = randomNumber4.ToString();
        textBox5.Text = randomNumber5.ToString();
        textBox6.Text = randomNumber6.ToString();

    }

我得到的随机数但有时在同一行上相同的号码，如何使每个编号唯一????

I'm getting random numbers but sometimes there is the same number on the same line, how do I make each number unique????

在此先感谢

推荐答案

您需要将它们存储在一个集合中，你选择你需要确保它不是present已经是一个新的号码各一次，否则你需要产生一个新的号码，直到找到一个唯一的编号。

You need to store them in a collection and each time you pick a new number you need to make sure it's not present already, otherwise you need to generate a new number until you find a unique number.

取而代之的是，我会产生 1 之间的顺序和 49 ，打乱他们并挑选6号出序列的，例如：

Instead of this, I would generate a sequence between 1 and 49, shuffle them and pick 6 number out of the sequence, for example:

var rnd = new Random();
var randomNumbers = Enumerable.Range(1,49).OrderBy(x => rnd.Next()).Take(6).ToList();

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