使用java创建一个4位随机数，数字不重复
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Creating a 4 digit Random Number using java with no repetition in digits
提问by kakabali
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 :-
我使用java编写了一个代码来创建一个没有重复数字的随机4位数字，我编写的代码如下：-
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);
here are the test cases(generated from System.out.println(R1+""+R2+""+R3+""+R4);) are as following :-
这里是测试用例（生成自System.out.println(R1+""+R2+""+R3+""+R4);）如下：-
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
Now My question here is, that if there is some better way to do this. If I could enhance it in some way?
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？
采纳答案by Jeroen Vannevel
Create a list of integers from 0 to 9, shuffle it and extract the first 4.
创建一个从 0 到 9 的整数列表，将其打乱并提取前 4 个。
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);
}
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.
有一些丑陋的 string-to-int 对话正在进行，但你明白了。根据您的用例，您可以看到需要什么。
回答by fred02138
Use a Set maybe?
可能使用 Set 吗？
Random r = new Random();
Set<Integer> s = new HashSet<Integer>();
while (s.size() < 4) {
s.add(r.nextInt(9));
}
回答by Rob
A couple of approaches:
几种方法：
Use a Set to hold the digits and keep adding random digits until the set has four values in it.
Create an array with values 0-9 in it. Shuffle the array and take the first four values.
使用 Set 来保存数字并不断添加随机数字，直到该集合中有四个值。
创建一个包含 0-9 值的数组。随机排列数组并取前四个值。
If performance matters, you will want to try a couple of different methods and see which is faster.
如果性能很重要，您将需要尝试几种不同的方法，看看哪种方法更快。
回答by Algiz
Create a list with Integer from 0 to 9 (so 10 items in total)
创建一个从 0 到 9 的整数列表（总共 10 个项目）
List<Integer> l = ...
Collections.shuffle(l);
d1 = l.get(0);
d2 = l.get(1);
d3 = l.get(2);
d4 = l.get(3);
回答by Juvanis
Here is my solution without using any additional data structure, looping on generating random number till it has unique digits.
这是我的解决方案，不使用任何额外的数据结构，循环生成随机数，直到它具有唯一的数字。
int a = 0, b = 0, c = 0, d = 0;
int x = 0;
while (true) {
x = r.nextInt(9000) + 1000;
a = x % 10;
b = (x / 10) % 10;
c = (x / 100) % 10;
d = x / 1000;
if (a == b || a == c || a == d || b == c || b == d || c == d)
continue;
else
break;
}
System.out.println(x);
回答by MZ4Code
Here's my approach, even though it uses a lot of string parsing but no data structure:
这是我的方法，即使它使用了大量字符串解析但没有数据结构：
static int generateNumber(int length){
String result = "";
int random;
while(true){
random = (int) ((Math.random() * (10 )));
if(result.length() == 0 && random == 0){//when parsed this insures that the number doesn't start with 0
random+=1;
result+=random;
}
else if(!result.contains(Integer.toString(random))){//if my result doesn't contain the new generated digit then I add it to the result
result+=Integer.toString(random);
}
if(result.length()>=length){//when i reach the number of digits desired i break out of the loop and return the final result
break;
}
}
return Integer.parseInt(result);
}
回答by Hot Licks
Roughly (not tested):
粗略（未测试）：
int randomNum = r.nextInt(5040);
int firstDigit = randomNum % 10;
randomNum = randomNum / 10;
int secondDigit = randomNum % 9;
randomNum = randomNum / 9;
int thirdDigit = randomNum % 8;
randomNum = randomNum / 8;
int fourthDigit = randomNum % 7;
if (secondDigit == firstDigit) {
secondDigit++;
}
while ((thirdDigit == firstDigit) || (thirdDigit == secondDigit)) {
thirdDigit++:
}
while ((fourthDigit == firstDigit) || (fourthDigit == secondDigit) || (fourthDigit == thirdDigit)) {
fourthDigit++;
}
(After coding this I realized that the increment operations need to be done modulo 10.)
（编码后，我意识到增量操作需要以 10 为模进行。）
