使用java创建一个4位随机数，数字不重复 [英] Creating a 4 digit Random Number using java with no repetition in digits

问题描述

我用java写了一个代码来创建一个没有重复数字的随机4位数字，我写的代码如下:-

I wrote a code using java to create a random 4 digit number with no repetition of digits, the code I wrote is given below :-

Random r = new Random();
d1 = r.nextInt(9);
d2 = r.nextInt(9);
d3 = r.nextInt(9);
d4 = r.nextInt(9);
while(d1==d2||d1==d3||d1==d4||d2==d3||d2==d4||d3==d4)
{
    if(d1==d2||d2==d3||d2==d4)
    {
        d2 = r.nextInt(9);
    }
    if(d1==d3||d2==d3||d3==d4)
    {
        d3 = r.nextInt(9);
    }
    if(d1==d4||d2==d4||d3==d4)
    {
        d4 = r.nextInt(9);
    }
}   
System.out.println(d1+""+d2+""+d3+""+d4);

这里是测试用例(从 System.out.println(R1+""+R2+""+R3+""+R4); 生成)如下:-

here are the test cases(generated from System.out.println(R1+""+R2+""+R3+""+R4);) are as following :-

 0123 |  OK as required
 1234 |  OK as required
 2123 |  not OK because 2 is present more than one time 
 9870 |  OK as required
 0444 |  not OK because 4 is present more than one time

现在我的问题是，如果有更好的方法来做到这一点.如果我可以以某种方式增强它?

Now My question here is, that if there is some better way to do this. If I could enhance it in some way?

推荐答案

创建一个从 0 到 9 的整数列表，将其打乱并提取前 4 个.

Create a list of integers from 0 to 9, shuffle it and extract the first 4.

public static void main(String[] args) {
    List<Integer> numbers = new ArrayList<>();
    for(int i = 0; i < 10; i++){
        numbers.add(i);
    }

    Collections.shuffle(numbers);

    String result = "";
    for(int i = 0; i < 4; i++){
        result += numbers.get(i).toString();
    }
    System.out.println(result);
}

正在进行一些丑陋的字符串到整数的转换，但你明白了.根据您的用例，您可以看到需要什么.

There's some ugly string-to-int conversing going on, but you get the idea. Depending on your use case you can see what is needed.

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