blocks|key|515735|text|对所有100个数字使用Collections.shuffle()，并选择前五个数字，如here所示。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|515736|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/2523492/choosing-random-numbers-efficiently/2524394#2524394^0|B|L|17|4|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

Use <code>Collections.shuffle()</code> on all 100 numbers and select the first five, as shown <a href="https://stackoverflow.com/questions/2523492/choosing-random-numbers-efficiently/2524394#2524394">here</a>.

blocks|key|732244|text|这将生成唯一的随机数................|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732245|import+java.util.HashSet;
import+java.util.Random;

public+class+RandomExample+{

++++public+static+void+main(String[]+args)+{
++++++++Random+rand+=+new+Random();
++++++++int+e;
++++++++int+i;
++++++++int+g+=+10;
++++++++HashSet<Integer>+randomNumbers+=+new+HashSet<Integer>();

++++++++for+(i+=+0;+i+<+g;+i%2B%2B)+{
++++++++++++e+=+rand.nextInt(20);
++++++++++++randomNumbers.add(e);
++++++++++++if+(randomNumbers.size()+<=+10)+{
++++++++++++++++if+(randomNumbers.size()+==+10)+{
++++++++++++++++++++g+=+10;
++++++++++++++++}
++++++++++++++++g%2B%2B;
++++++++++++++++randomNumbers.add(e);
++++++++++++}
++++++++}
++++++++System.out.println("Ten+Unique+random+numbers+from+1+to+20+are++:+"+%2B+randomNumbers);
++++}
}|code-block|syntax|javascript|732246|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This will work to generate unique random numbers................

<pre><code>import java.util.HashSet;
import java.util.Random;

public class RandomExample {

 public static void main(String[] args) {
 Random rand = new Random();
 int e;
 int i;
 int g = 10;
 HashSet&lt;Integer&gt; randomNumbers = new HashSet&lt;Integer&gt;();

 for (i = 0; i &lt; g; i++) {
 e = rand.nextInt(20);
 randomNumbers.add(e);
 if (randomNumbers.size() &lt;= 10) {
 if (randomNumbers.size() == 10) {
 g = 10;
 }
 g++;
 randomNumbers.add(e);
 }
 }
 System.out.println("Ten Unique random numbers from 1 to 20 are : " + randomNumbers);
 }
}
</code></pre>

blocks|key|515809|text|我重构了Anand的答案，不仅利用了Set的唯一属性，还使用了在向set添加失败时由set.add()返回的布尔值false。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|515810|import+java.util.HashSet;
import+java.util.Random;
import+java.util.Set;

public+class+randomUniqueNumberGenerator+{

++++public+static+final+int+SET_SIZE_REQUIRED+=+10;
++++public+static+final+int+NUMBER_RANGE+=+100;

++++public+static+void+main(String[]+args)+{
++++++++Random+random+=+new+Random();

++++++++Set+set+=+new+HashSet<Integer>(SET_SIZE_REQUIRED);

++++++++while(set.size()<+SET_SIZE_REQUIRED)+{
++++++++++++while+(set.add(random.nextInt(NUMBER_RANGE))+!=+true)
++++++++++++++++;
++++++++}
++++++++assert+set.size()+==+SET_SIZE_REQUIRED;
++++++++System.out.println(set);
++++}
}|code-block|syntax|javascript|515811|entityMap^0|16|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

I re-factored Anand's answer to make use not only of the unique properties of a Set but also use the boolean false returned by the <code>set.add()</code> when an add to the set fails.

<pre><code>import java.util.HashSet;
import java.util.Random;
import java.util.Set;

public class randomUniqueNumberGenerator {

 public static final int SET_SIZE_REQUIRED = 10;
 public static final int NUMBER_RANGE = 100;

 public static void main(String[] args) {
 Random random = new Random();

 Set set = new HashSet&lt;Integer&gt;(SET_SIZE_REQUIRED);

 while(set.size()&lt; SET_SIZE_REQUIRED) {
 while (set.add(random.nextInt(NUMBER_RANGE)) != true)
 ;
 }
 assert set.size() == SET_SIZE_REQUIRED;
 System.out.println(set);
 }
}
</code></pre>

blocks|key|515834|text|我觉得这个方法值得一提。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|515835|+++private+static+final+Random+RANDOM+=+new+Random();++++
+++/**
+++++*+Pick+n+numbers+between+0+(inclusive)+and+k+(inclusive)
+++++*+While+there+are+very+deterministic+ways+to+do+this,
+++++*+for+large+k+and+small+n,+this+could+be+easier+than+creating
+++++*+an+large+array+and+sorting,+i.e.+k+=+10,000
+++++*/
++++public+Set<Integer>+pickRandom(int+n,+int+k)+{
++++++++final+Set<Integer>+picked+=+new+HashSet<>();
++++++++while+(picked.size()+<+n)+{
++++++++++++picked.add(RANDOM.nextInt(k+%2B+1));
++++++++}
++++++++return+picked;
++++}|code-block|syntax|javascript|515836|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I feel like this method is worth mentioning.

<pre><code> private static final Random RANDOM = new Random(); 
 /**
 * Pick n numbers between 0 (inclusive) and k (inclusive)
 * While there are very deterministic ways to do this,
 * for large k and small n, this could be easier than creating
 * an large array and sorting, i.e. k = 10,000
 */
 public Set&lt;Integer&gt; pickRandom(int n, int k) {
 final Set&lt;Integer&gt; picked = new HashSet&lt;&gt;();
 while (picked.size() &lt; n) {
 picked.add(RANDOM.nextInt(k + 1));
 }
 return picked;
 }
</code></pre>

blocks|key|515869|text|要做到这一点，一个聪明的方法是使用模数中原始元素的指数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|515870|例如，2是一个原根mod+101，这意味着2+mod+101的幂给你一个不重复的序列，它可以看到从1到100+(包括1和100)的每个数字：|515871|2%5E0+mod+101+=+1
2%5E1+mod+101+=+2
2%5E2+mod+101+=+4
...
2%5E50+mod+101+=+100
2%5E51+mod+101+=+99
2%5E52+mod+101+=+97
...
2%5E100+mod+101+=+1|code-block|syntax|javascript|515872|在Java代码中，您可以这样写：|515873|void+randInts()+{
int+num=1;
for+(int+ii=0;+ii<101;+ii%2B%2B)+{
++++System.out.println(num);
++++num=+(num*2)+%25+101;
++++}
}|515874|寻找特定模数的原根可能很棘手，但Maple的"primroot“函数将为您完成这项工作。|515875|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

One clever way to do this is to use exponents of a primitive element in modulus.

For example, 2 is a primitive root mod 101, meaning that the powers of 2 mod 101 give you a non-repeating sequence that sees every number from 1 to 100 inclusive:

<pre><code>2^0 mod 101 = 1
2^1 mod 101 = 2
2^2 mod 101 = 4
...
2^50 mod 101 = 100
2^51 mod 101 = 99
2^52 mod 101 = 97
...
2^100 mod 101 = 1
</code></pre>

In Java code, you would write:

<pre class="lang-java prettyprint-override"><code>void randInts() {
int num=1;
for (int ii=0; ii&lt;101; ii++) {
 System.out.println(num);
 num= (num*2) % 101;
 }
}
</code></pre>

Finding a primitive root for a specific modulus can be tricky, but Maple's "primroot" function will do this for you.

blocks|key|515891|text|看看这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|515892|public+class+RandomNumbers+{

++++public+static+void+main(String[]+args)+{
++++++++//+TODO+Auto-generated+method+stub
++++++++int+n+=+5;
++++++++int+A[]+=+uniqueRandomArray(n);
++++++++for(int+i+=+0;+i<n;+i%2B%2B){
++++++++++++System.out.println(A[i]);
++++++++}
++++}
++++public+static+int[]+uniqueRandomArray(int+n){
++++++++int+[]+A+=+new+int[n];
++++++++for(int+i+=+0;+i<+A.length;+){
++++++++++++if(i+==+A.length){
++++++++++++++++break;
++++++++++++}
++++++++++++int+b+=+(int)(Math.random()+*n)+%2B+1;
++++++++++++if(f(A,b)+==+false){
++++++++++++++++A[i%2B%2B]+=+b;
++++++++++++}+
++++++++}
++++++++return+A;
++++}
++++public+static+boolean+f(int[]+A,+int+n){
++++++++for(int+i=0;+i<A.length;+i%2B%2B){
++++++++++++if(A[i]+==+n){
++++++++++++++++return+true;
++++++++++++}
++++++++}
++++++++return+false;
++++}
}|code-block|syntax|javascript|515893|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Check this 

<pre><code>public class RandomNumbers {

 public static void main(String[] args) {
 // TODO Auto-generated method stub
 int n = 5;
 int A[] = uniqueRandomArray(n);
 for(int i = 0; i&lt;n; i++){
 System.out.println(A[i]);
 }
 }
 public static int[] uniqueRandomArray(int n){
 int [] A = new int[n];
 for(int i = 0; i&lt; A.length; ){
 if(i == A.length){
 break;
 }
 int b = (int)(Math.random() *n) + 1;
 if(f(A,b) == false){
 A[i++] = b;
 } 
 }
 return A;
 }
 public static boolean f(int[] A, int n){
 for(int i=0; i&lt;A.length; i++){
 if(A[i] == n){
 return true;
 }
 }
 return false;
 }
}
</code></pre>

blocks|key|515979|text|对于Java+8%2B，您可以使用Random的ints方法获取随机值的IntStream，然后使用distinct和limit将流减少为一些惟一的随机值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|515980|ThreadLocalRandom.current().ints(0,+100).distinct().limit(5).forEach(System.out::println);|code-block|syntax|javascript|515981|如果需要的话，Random也有创建LongStream和DoubleStream的方法。|515982|如果您希望某个范围中的所有(或大量)数字按随机顺序排列，则将所有数字添加到列表中、对其进行混洗并获取前n可能更有效，因为上面的示例当前是通过在请求的范围中生成随机数并将它们传递到一个集合(类似于Rob+Kielty's+answer)来实现的，这可能需要生成比传递到限制的数量多得多的数量，因为生成新的唯一数字的概率随着找到的每个数字而减少。下面是另一种方式的示例：|515983|List<Integer>+range+=+IntStream.range(0,+100).boxed()
++++++++.collect(Collectors.toCollection(ArrayList::new));
Collections.shuffle(range);
range.subList(0,+99).forEach(System.out::println);|515984|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/util/Random.html#ints-int-int-|1|https://docs.oracle.com/javase/8/docs/api/java/util/stream/IntStream.html|2|https://docs.oracle.com/javase/8/docs/api/java/util/stream/IntStream.html#distinct--|3|https://docs.oracle.com/javase/8/docs/api/java/util/stream/IntStream.html#limit-long-|4|https://stackoverflow.com/a/11336538/3179759^0|F|6|M|4|Y|9|1C|8|1L|5|M|4|0|Y|9|1|1C|8|2|1L|5|3|0|0|7|6|H|A|S|C|0|2P|J|4|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@$9|1H|A|1I|1|1J]|$9|1K|A|1L|1|1M]|$9|1N|A|1O|1|1P]|$9|1Q|A|1R|1|1S]]|E|$]]|$1|F|3|G|5|H|7|1T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1U|8|@$9|1V|A|1W|B|C]|$9|1X|A|1Y|B|C]|$9|1Z|A|20|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|21|8|@]|D|@$9|22|A|23|1|24]]|E|$]]|$1|O|3|P|5|H|7|25|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|26|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]|10|$5|T|U|V|E|$W|11]]|12|$5|T|U|V|E|$W|13]]|14|$5|T|U|V|E|$W|15]]]]

With Java 8+ you can use the <a href="https://docs.oracle.com/javase/8/docs/api/java/util/Random.html#ints-int-int-" rel="noreferrer"><code>ints</code></a> method of <code>Random</code> to get an <a href="https://docs.oracle.com/javase/8/docs/api/java/util/stream/IntStream.html" rel="noreferrer"><code>IntStream</code></a> of random values then <a href="https://docs.oracle.com/javase/8/docs/api/java/util/stream/IntStream.html#distinct--" rel="noreferrer"><code>distinct</code></a> and <a href="https://docs.oracle.com/javase/8/docs/api/java/util/stream/IntStream.html#limit-long-" rel="noreferrer"><code>limit</code></a> to reduce the stream to a number of unique random values.

<pre><code>ThreadLocalRandom.current().ints(0, 100).distinct().limit(5).forEach(System.out::println);
</code></pre>

<code>Random</code> also has methods which create <code>LongStream</code>s and <code>DoubleStream</code>s if you need those instead.

If you want all (or a large amount) of the numbers in a range in a random order it might be more efficient to add all of the numbers to a list, shuffle it, and take the first n because the above example is currently implemented by generating random numbers in the range requested and passing them through a set (similarly to <a href="https://stackoverflow.com/a/11336538/3179759">Rob Kielty's answer</a>), which may require generating many more than the amount passed to limit because the probability of a generating a new unique number decreases with each one found. Here's an example of the other way:

<pre><code>List&lt;Integer&gt; range = IntStream.range(0, 100).boxed()
 .collect(Collectors.toCollection(ArrayList::new));
Collections.shuffle(range);
range.subList(0, 99).forEach(System.out::println);
</code></pre>

blocks|key|515992|text|试试这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|515993|public+class+RandomValueGenerator+{
++++/**
+++++*+
+++++*/
++++private+volatile+List<Double>+previousGenValues+=+new+ArrayList<Double>();

++++public+void+init()+{
++++++++previousGenValues.add(Double.valueOf(0));
++++}

++++public+String+getNextValue()+{
++++++++Random+random+=+new+Random();
++++++++double+nextValue=0;
++++++++while(previousGenValues.contains(Double.valueOf(nextValue)))+{
++++++++++++nextValue+=+random.nextDouble();
++++++++}
++++++++previousGenValues.add(Double.valueOf(nextValue));
++++++++return+String.valueOf(nextValue);
++++}
}|code-block|syntax|javascript|515994|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

try this out

<pre><code>public class RandomValueGenerator {
 /**
 * 
 */
 private volatile List&lt;Double&gt; previousGenValues = new ArrayList&lt;Double&gt;();

 public void init() {
 previousGenValues.add(Double.valueOf(0));
 }

 public String getNextValue() {
 Random random = new Random();
 double nextValue=0;
 while(previousGenValues.contains(Double.valueOf(nextValue))) {
 nextValue = random.nextDouble();
 }
 previousGenValues.add(Double.valueOf(nextValue));
 return String.valueOf(nextValue);
 }
}
</code></pre>

blocks|key|732498|text|选择从0到m-1的n个唯一随机数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732499|int[]+uniqueRand(int+n,+int+m){
++++Random+rand+=+new+Random();
++++int[]+r+=+new+int[n];
++++int[]+result+=+new+int[n];
++++for(int+i+=+0;+i+<+n;+i%2B%2B){
++++++++r[i]+=+rand.nextInt(m-i);
++++++++result[i]+=+r[i];
++++++++for(int+j+=+i-1;+j+>=+0;+j--){
++++++++++++if(result[i]+>=+r[j])
++++++++++++++++result[i]%2B%2B;
++++++++}
++++}
++++return+result;
}|code-block|syntax|javascript|732500|想象一个包含从0到m-1的数字的列表。要选择第一个数字，我们只需使用rand.nextInt(m)。然后从列表中删除该号码。现在仍然有m-1个数字，所以我们调用rand.nextInt(m-1)。我们得到的数字表示列表中的位置。如果它小于第一个数字，那么它就是第二个数字，因为第一个数字之前的列表部分不会因为删除第一个数字而改变。如果位置大于或等于第一个数，则第二个数为position%2B1。进一步推导，即可得到此算法。|offset|length|style|CODE|732501|解释|732502|该算法的复杂度为O(n%5E2)。因此，它对于从大集合中生成少量的唯一数是很好的。而基于混洗的算法至少需要O(m)来做混洗。|732503|同样，基于混洗的算法需要内存来存储所有可能的结果来进行混洗，这个算法不需要。|732504|entityMap^0|0|0|Y|F|28|H|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

Choose n unique random numbers from 0 to m-1.

<pre><code>int[] uniqueRand(int n, int m){
 Random rand = new Random();
 int[] r = new int[n];
 int[] result = new int[n];
 for(int i = 0; i &lt; n; i++){
 r[i] = rand.nextInt(m-i);
 result[i] = r[i];
 for(int j = i-1; j &gt;= 0; j--){
 if(result[i] &gt;= r[j])
 result[i]++;
 }
 }
 return result;
}
</code></pre>

Imagine a list containing numbers from 0 to m-1. To choose the first number, we simply use <code>rand.nextInt(m)</code>. Then remove the number from the list. Now there remains m-1 numbers, so we call <code>rand.nextInt(m-1)</code>. The number we get represents the position in the list. If it is less than the first number, then it is the second number, since the part of list prior to the first number wasn't changed by the removal of the first number. If the position is greater than or equal to the first number, the second number is position+1. Do some further derivation, you can get this algorithm.

<h2>Explanation</h2>

This algorithm has O(n^2) complexity. So it is good for generating small amount of unique numbers from a large set. While the shuffle based algorithm need at least O(m) to do the shuffle.

Also shuffle based algorithm need memory to store every possible outcome to do the shuffle, this algorithm doesn’t need.

blocks|key|732537|text|这与其他答案没有太大不同，但我最终想要的是整数数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732538|++++Integer[]+indices+=+new+Integer[n];
++++Arrays.setAll(indices,+i+->+i);
++++Collections.shuffle(Arrays.asList(indices));
++++return+Arrays.stream(indices).mapToInt(Integer::intValue).toArray();|code-block|syntax|javascript|732539|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This isn't significantly different from other answers, but I wanted the array of integers in the end:

<pre><code> Integer[] indices = new Integer[n];
 Arrays.setAll(indices, i -&gt; i);
 Collections.shuffle(Arrays.asList(indices));
 return Arrays.stream(indices).mapToInt(Integer::intValue).toArray();
</code></pre>

blocks|key|732587|text|下面是我用来生成唯一数字的一种方法。随机函数生成数字并将其存储在文本文件中，然后下一次它在文件中检查它，比较它并生成新的唯一数字，因此在这种方式中总是有新的唯一数字。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732588|public+int+GenerateRandomNo()
{
++++int+_min+=+0000;
++++int+_max+=+9999;
++++Random+_rdm+=+new+Random();
++++return+_rdm.Next(_min,+_max);
}
public+int+rand_num()
{
++++randnum+=+GenerateRandomNo();
++++string+createText+=+randnum.ToString()+%2B+Environment.NewLine;
++++string+file_path+=+System.IO.Path.GetDirectoryName(System.Windows.Forms.Application.ExecutablePath)+%2B+@"\Invoices\numbers.txt";
++++File.AppendAllText(file_path,+createText);
++++int+number+=+File.ReadLines(file_path).Count();+//count+number+of+lines+in+file
++++System.IO.StreamReader+file+=+new+System.IO.StreamReader(file_path);
++++do
++++{
++++++++randnum+=+GenerateRandomNo();
++++}
++++while+((file.ReadLine())+==+randnum.ToString());
++++file.Close();
++++return+randnum;

}|code-block|syntax|javascript|732589|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Below is a way I used to generate unique number always. Random function generates number and stores it in textfile then next time it checks it in file compares it and generate new unique number hence in this way there is always a new unique number.

<pre><code>public int GenerateRandomNo()
{
 int _min = 0000;
 int _max = 9999;
 Random _rdm = new Random();
 return _rdm.Next(_min, _max);
}
public int rand_num()
{
 randnum = GenerateRandomNo();
 string createText = randnum.ToString() + Environment.NewLine;
 string file_path = System.IO.Path.GetDirectoryName(System.Windows.Forms.Application.ExecutablePath) + @"\Invoices\numbers.txt";
 File.AppendAllText(file_path, createText);
 int number = File.ReadLines(file_path).Count(); //count number of lines in file
 System.IO.StreamReader file = new System.IO.StreamReader(file_path);
 do
 {
 randnum = GenerateRandomNo();
 }
 while ((file.ReadLine()) == randnum.ToString());
 file.Close();
 return randnum;

}
</code></pre>

blocks|key|732619|text|我就是这样做的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732620|++++Random+random+=+new+Random();
++++ArrayList<Integer>+arrayList+=+new+ArrayList<Integer>();

++++while+(arrayList.size()+<+6)+{+//+how+many+numbers+u+need+-+it+will+6
++++++++int+a+=+random.nextInt(49)%2B1;+//+this+will+give+numbers+between+1+and+50.

++++++++if+(!arrayList.contains(a))+{
++++++++++++arrayList.add(a);
++++++++}
++++}|code-block|syntax|javascript|732621|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I have made this like that.

<pre><code> Random random = new Random();
 ArrayList&lt;Integer&gt; arrayList = new ArrayList&lt;Integer&gt;();

 while (arrayList.size() &lt; 6) { // how many numbers u need - it will 6
 int a = random.nextInt(49)+1; // this will give numbers between 1 and 50.

 if (!arrayList.contains(a)) {
 arrayList.add(a);
 }
 }
</code></pre>

blocks|key|732656|text|您可以使用布尔数组来填充true+if值，否则设置通过布尔数组导航以获得如下所示的值|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732657|package+study;

import+java.util.ArrayList;
import+java.util.Arrays;
import+java.util.List;

/*
Created+By+Sachin++Rane+on+Jul+18,+2018
*/
public+class+UniqueRandomNumber+{
++++static+Boolean[]+boolArray;
++++public+static+void+main(String+s[]){
++++++++List<Integer>+integers+=+new+ArrayList<>();


++++++++for+(int+i+=+0;+i+<+10;+i%2B%2B)+{
++++++++++++integers.add(i);
++++++++}


++++++++//get+unique+random+numbers
++++++++boolArray+=+new+Boolean[integers.size()%2B1];
++++++++Arrays.fill(boolArray,+false);
++++++++for+(int+i+=+0;+i+<+10;+i%2B%2B)+{
++++++++++++System.out.print(getUniqueRandomNumber(integers)+%2B+"+");

++++++++}

++++}

++++private+static+int++getUniqueRandomNumber(List<Integer>+integers)+{
++++++++int+randNum+=(int)+(Math.random()*integers.size());
++++++++if(boolArray[randNum]){
++++++++++++while(boolArray[randNum]){
++++++++++++++++randNum%2B%2B;
++++++++++++++++if(randNum>boolArray.length){
++++++++++++++++++++randNum=0;
++++++++++++++++}
++++++++++++}
++++++++++++boolArray[randNum]=true;
++++++++++++return+randNum;
++++++++}else+{
++++++++++++boolArray[randNum]=true;
++++++++++++return+randNum;
++++++++}

++++}

}|code-block|syntax|javascript|732658|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

you can use boolean array to fill the true if value taken else set navigate through boolean array to get value as per given below 

<pre><code>package study;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/*
Created By Sachin Rane on Jul 18, 2018
*/
public class UniqueRandomNumber {
 static Boolean[] boolArray;
 public static void main(String s[]){
 List&lt;Integer&gt; integers = new ArrayList&lt;&gt;();


 for (int i = 0; i &lt; 10; i++) {
 integers.add(i);
 }


 //get unique random numbers
 boolArray = new Boolean[integers.size()+1];
 Arrays.fill(boolArray, false);
 for (int i = 0; i &lt; 10; i++) {
 System.out.print(getUniqueRandomNumber(integers) + " ");

 }

 }

 private static int getUniqueRandomNumber(List&lt;Integer&gt; integers) {
 int randNum =(int) (Math.random()*integers.size());
 if(boolArray[randNum]){
 while(boolArray[randNum]){
 randNum++;
 if(randNum&gt;boolArray.length){
 randNum=0;
 }
 }
 boolArray[randNum]=true;
 return randNum;
 }else {
 boolArray[randNum]=true;
 return randNum;
 }

 }

}
</code></pre>

blocks|key|732700|text|我对这个问题有一个简单的解决方案，我们可以很容易地生成n个唯一的随机数，这只是一个逻辑，任何人都可以在任何语言中使用它。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732701|for(int+i=0;i<4;i%2B%2B)
++++++++{
++++++++++++rn[i]=+GenerateRandomNumber();
++++++++++++for+(int+j=0;j<i;j%2B%2B)
++++++++++++{
++++++++++++++++if+(rn[i]+==+rn[j])
++++++++++++++++{
++++++++++++++++++++i--;
++++++++++++++++}
++++++++++++}
++++++++}|code-block|syntax|javascript|732702|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I have easy solution for this problem, 
With this we can easily generate n number of unique random numbers,
Its just logic anyone can use it in any language.

<pre><code>for(int i=0;i&lt;4;i++)
 {
 rn[i]= GenerateRandomNumber();
 for (int j=0;j&lt;i;j++)
 {
 if (rn[i] == rn[j])
 {
 i--;
 }
 }
 }
</code></pre>

blocks|key|516282|text|您可以使用Collections类。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|516283|称为集合的实用程序类提供了可以在集合上执行的不同操作(例如，搜索元素、查找最大或最小元素、颠倒元素的顺序，等等)。它可以执行的操作之一是对元素进行混洗。混洗将随机地将每个元素移动到列表中的不同位置。它通过使用Random对象来做到这一点。这意味着它是确定性的随机性，但它在大多数情况下都可以。|516284|要混洗ArrayList，请将集合导入添加到程序的顶部，然后使用shuffle静态方法。它将ArrayList作为一个参数进行混洗：|516285|import+java.util.Collections;
import+java.util.ArrayList;
public+class+Lottery+{
public+static+void+main(String[]+args)+{
//define+ArrayList+to+hold+Integer+objects
ArrayList+numbers+=+new+ArrayList();
for(int+i+=+0;+i+<+100;+i%2B%2B)
{
numbers.add(i%2B1);
}
Collections.shuffle(numbers);
System.out.println(numbers);
}
}|code-block|syntax|javascript|516286|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You can use the Collections class.

A utility class called Collections offers different actions that can be performed on a collection like an ArrayList (e.g., search the elements, find the maximum or minimum element, reverse the order of elements, and so on). One of the actions it can perform is to shuffle the elements. The shuffle will randomly move each element to a different position in the list. It does this by using a Random object. This means it's deterministic randomness, but it will do in most situations.

To shuffle the ArrayList, add the Collections import to the top of the program and then use the Shuffle static method. It takes the ArrayList to be shuffled as a parameter:

<pre><code>import java.util.Collections;
import java.util.ArrayList;
public class Lottery {
public static void main(String[] args) {
//define ArrayList to hold Integer objects
ArrayList numbers = new ArrayList();
for(int i = 0; i &lt; 100; i++)
{
numbers.add(i+1);
}
Collections.shuffle(numbers);
System.out.println(numbers);
}
}
</code></pre>

blocks|key|516347|text|在java中，可以生成0到n-1之间的n个唯一的随机数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|516348|public+static+void+RandomGenerate(int+n)
{
+++++Set<Integer>+st=new+HashSet<Integer>();
+++++Random+r=new+Random();
+++++while(st.size()<n)
+++++{
++++++++st.add(r.nextInt(n));
+++++}|code-block|syntax|javascript|516349|}|516350|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can generate n unique random number between 0 to n-1 in java

<pre><code>public static void RandomGenerate(int n)
{
 Set&lt;Integer&gt; st=new HashSet&lt;Integer&gt;();
 Random r=new Random();
 while(st.size()&lt;n)
 {
 st.add(r.nextInt(n));
 }
</code></pre>

}

blocks|key|732834|text|下面的代码将生成5个介于1到100之间的随机数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732835|+public+static+void+main(String[]+args)+{
++++++++Random+random+=+new+Random();
++++++++HashSet<Integer>+set+=+new+HashSet<>();
++++++++while+(set.size()+<+5)+{
++++++++++++int+a+=+random.nextInt(99)%2B1;
++++++++++++if+(!set.contains(a))+{
++++++++++++++++set.add(a);
++++++++++++}
++++++++}
++++++++System.out.println(set);
++++}|code-block|syntax|javascript|732836|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Following code will generate 5 random numbers between 1 to 100
<pre><code> public static void main(String[] args) {
 Random random = new Random();
 HashSet&lt;Integer&gt; set = new HashSet&lt;&gt;();
 while (set.size() &lt; 5) {
 int a = random.nextInt(99)+1;
 if (!set.contains(a)) {
 set.add(a);
 }
 }
 System.out.println(set);
 }
</code></pre>

I'm trying to get random numbers between 0 and 100. But I want them to be unique, not repeated in a sequence. For example if I got 5 numbers, they should be 82,12,53,64,32 and not 82,12,53,12,32
I used this, but it generates same numbers in a sequence.

<pre><code>Random rand = new Random();
selected = rand.nextInt(100);
</code></pre>

Generating Unique Random Numbers in Java

Windows

Java

我正在尝试获取0到100之间的随机数。但我希望它们是唯一的，而不是在一个序列中重复。例如，如果我得到5个数字，它们应该是82,12,53,64,32，而不是82,12,53,12,32我使用了这个，但它在一个序列中生成相同的数字。Random rand = new Random();selected = rand.ne...

问在Java中生成唯一随机数
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