Related
I have to create a program which adds two integers and prints the sum vertically.
For example, I have.
a=323, b=322.
The output should be:
6
4
5
I've created the code for when the integers are up to two digits, but I want it to work for at least three digits.
Below is the best I could think of.
It may be completely wrong, but the only problem I'm facing is the declaration of array.
It says that the array might not be initialized.
If I set it to null then also it won't assign values to it later.
I know maybe I'm making a big mistake here, but I'll really appreciate if anyone could help me out.
Please keep in mind that I must not use any other functions for this code.
Hope I'm clear.
public class Vert
{
public static void main(String args[])
{
int n,i=0,j,a=323,b=322;
int s[];
n=a+b;
while(n>9)
{
s[i]=n%10;
i++;
s[i]=n/10;
if(s[i]>9)
{
n=s[i];
}
}
j=i;
for(j=i;j>=0;j--)
{
System.out.println(+s[j]);
}
}
}
String conversion seems like cheating, so here's a Stack.
int a = 323, b = 322;
java.util.Stack<Integer> stack = new java.util.Stack<>();
int n = a + b;
while (n > 0) {
stack.push(n % 10);
n = n / 10;
}
while (!stack.isEmpty())
System.out.println(stack.pop());
If an array is required, you need two passes over the sum
int a = 323, b = 322;
// Get the size of the array
int n = a + b;
int size = 0;
while (n > 0) {
size++;
n = n / 10;
}
// Build the output
int s[] = new int[size];
n = a + b;
for (int i = size - 1; n > 0; i--) {
s[i] = n % 10;
n = n / 10;
}
// Print
for (int x : s) {
System.out.println(x);
}
To initialize an array, you need to specify the size of your array as next:
int s[] = new int[mySize];
If you don't know the size of your array, you should consider using a List of Integer instead as next:
List<Integer> s = new ArrayList<Integer>();
Here is how it could be done:
// Convert the sum into a String
String result = String.valueOf(a + b);
for (int i=0; i <result.length();i++) {
// Print one character corresponding to a digit here per line
System.out.println(result.charAt(i));
}
I'd do it like this:
int a = 322;
int b = 322;
int sum = a + b;
String s = Integer.toString(sum);
for(int i = 0; i < s.length(); i++) {
System.out.println(s.charAt(i));
}
But your problem looks like an array is required.
The steps are same as in my solution:
Use int values
Sum the int values (operation)
Convert the int value in an array/string
Output the array/string
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.
Random rand = new Random();
selected = rand.nextInt(100);
Add each number in the range sequentially in a list structure.
Shuffle it.
Take the first 'n'.
Here is a simple implementation. This will print 3 unique random numbers from the range 1-10.
import java.util.ArrayList;
import java.util.Collections;
public class UniqueRandomNumbers {
public static void main(String[] args) {
ArrayList<Integer> list = new ArrayList<Integer>();
for (int i=1; i<11; i++) list.add(i);
Collections.shuffle(list);
for (int i=0; i<3; i++) System.out.println(list.get(i));
}
}
The first part of the fix with the original approach, as Mark Byers pointed out in an answer now deleted, is to use only a single Random instance.
That is what is causing the numbers to be identical. A Random instance is seeded by the current time in milliseconds. For a particular seed value, the 'random' instance will return the exact same sequence of pseudo random numbers.
With Java 8+ you can use the ints method of Random to get an IntStream of random values then distinct and limit to reduce the stream to a number of unique random values.
ThreadLocalRandom.current().ints(0, 100).distinct().limit(5).forEach(System.out::println);
Random also has methods which create LongStreams and DoubleStreams 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 Rob Kielty's answer), 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:
List<Integer> range = IntStream.range(0, 100).boxed()
.collect(Collectors.toCollection(ArrayList::new));
Collections.shuffle(range);
range.subList(0, 99).forEach(System.out::println);
Create an array of 100 numbers, then randomize their order.
Devise a pseudo-random number generator that has a range of 100.
Create a boolean array of 100 elements, then set an element true when you pick that number. When you pick the next number check against the array and try again if the array element is set. (You can make an easy-to-clear boolean array with an array of long where you shift and mask to access individual bits.)
Use Collections.shuffle() on all 100 numbers and select the first five, as shown here and below.
Console:
59 9 68 24 82
Code:
private static final Random rnd = new Random();
private static final int N = 100;
private static final int K = 5;
private static final List<Integer> S = new ArrayList<>(N);
public static void main(String[] args) {
for (int i = 0; i < N; i++) {
S.add(i + 1);
}
Collections.shuffle(S, rnd);
for (int i = 0; i < K; i++) {
System.out.print(S.get(i) + " ");
}
System.out.println();
}
I feel like this method is worth mentioning.
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 + 1));
}
return picked;
}
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 set.add() when an add to the set fails.
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);
}
}
I have made this like that.
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)+1; // this will give numbers between 1 and 50.
if (!arrayList.contains(a)) {
arrayList.add(a);
}
}
This will work to generate unique random numbers................
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++) {
e = rand.nextInt(20);
randomNumbers.add(e);
if (randomNumbers.size() <= 10) {
if (randomNumbers.size() == 10) {
g = 10;
}
g++;
randomNumbers.add(e);
}
}
System.out.println("Ten Unique random numbers from 1 to 20 are : " + randomNumbers);
}
}
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:
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
In Java code, you would write:
void randInts() {
int num=1;
for (int ii=0; ii<101; ii++) {
System.out.println(num);
num= (num*2) % 101;
}
}
Finding a primitive root for a specific modulus can be tricky, but Maple's "primroot" function will do this for you.
I have come here from another question, which has been duplicate of this question (Generating unique random number in java)
Store 1 to 100 numbers in an Array.
Generate random number between 1 to 100 as position and return array[position-1] to get the value
Once you use a number in array, mark the value as -1 ( No need to maintain another array to check if this number is already used)
If value in array is -1, get the random number again to fetch new location in array.
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.
for(int i=0;i<4;i++)
{
rn[i]= GenerateRandomNumber();
for (int j=0;j<i;j++)
{
if (rn[i] == rn[j])
{
i--;
}
}
}
Choose n unique random numbers from 0 to m-1.
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++){
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]++;
}
}
return result;
}
Imagine a list containing numbers from 0 to m-1. To choose the first number, we simply use rand.nextInt(m). Then remove the number from the list. Now there remains m-1 numbers, so we call rand.nextInt(m-1). 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.
Explanation
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.
Though it's an old thread, but adding another option might not harm. (JDK 1.8 lambda functions seem to make it easy);
The problem could be broken down into the following steps;
Get a minimum value for the provided list of integers (for which to generate unique random numbers)
Get a maximum value for the provided list of integers
Use ThreadLocalRandom class (from JDK 1.8) to generate random integer values against the previously found min and max integer values and then filter to ensure that the values are indeed contained by the originally provided list. Finally apply distinct to the intstream to ensure that generated numbers are unique.
Here is the function with some description:
/**
* Provided an unsequenced / sequenced list of integers, the function returns unique random IDs as defined by the parameter
* #param numberToGenerate
* #param idList
* #return List of unique random integer values from the provided list
*/
private List<Integer> getUniqueRandomInts(List<Integer> idList, Integer numberToGenerate) {
List<Integer> generatedUniqueIds = new ArrayList<>();
Integer minId = idList.stream().mapToInt (v->v).min().orElseThrow(NoSuchElementException::new);
Integer maxId = idList.stream().mapToInt (v->v).max().orElseThrow(NoSuchElementException::new);
ThreadLocalRandom.current().ints(minId,maxId)
.filter(e->idList.contains(e))
.distinct()
.limit(numberToGenerate)
.forEach(generatedUniqueIds:: add);
return generatedUniqueIds;
}
So that, to get 11 unique random numbers for 'allIntegers' list object, we'll call the function like;
List<Integer> ids = getUniqueRandomInts(allIntegers,11);
The function declares new arrayList 'generatedUniqueIds' and populates with each unique random integer up to the required number before returning.
P.S. ThreadLocalRandom class avoids common seed value in case of concurrent threads.
try this out
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);
}
}
This isn't significantly different from other answers, but I wanted the array of integers in the end:
Integer[] indices = new Integer[n];
Arrays.setAll(indices, i -> i);
Collections.shuffle(Arrays.asList(indices));
return Arrays.stream(indices).mapToInt(Integer::intValue).toArray();
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
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++) {
integers.add(i);
}
//get unique random numbers
boolArray = new Boolean[integers.size()+1];
Arrays.fill(boolArray, false);
for (int i = 0; i < 10; i++) {
System.out.print(getUniqueRandomNumber(integers) + " ");
}
}
private static int getUniqueRandomNumber(List<Integer> integers) {
int randNum =(int) (Math.random()*integers.size());
if(boolArray[randNum]){
while(boolArray[randNum]){
randNum++;
if(randNum>boolArray.length){
randNum=0;
}
}
boolArray[randNum]=true;
return randNum;
}else {
boolArray[randNum]=true;
return randNum;
}
}
}
This is the most simple method to generate unique random values in a range or from an array.
In this example, I will be using a predefined array but you can adapt this method to generate random numbers as well. First, we will create a sample array to retrieve our data from.
Generate a random number and add it to the new array.
Generate another random number and check if it is already stored in the new array.
If not then add it and continue
else reiterate the step.
ArrayList<Integer> sampleList = new ArrayList<>();
sampleList.add(1);
sampleList.add(2);
sampleList.add(3);
sampleList.add(4);
sampleList.add(5);
sampleList.add(6);
sampleList.add(7);
sampleList.add(8);
Now from the sampleList we will produce five random numbers that are unique.
int n;
randomList = new ArrayList<>();
for(int i=0;i<5;i++){
Random random = new Random();
n=random.nextInt(8); //Generate a random index between 0-7
if(!randomList.contains(sampleList.get(n)))
randomList.add(sampleList.get(n));
else
i--; //reiterating the step
}
This is conceptually very simple. If the random value generated already exists then we will reiterate the step. This will continue until all the values generated are unique.
If you found this answer useful then you can vote it up as it is much simple in concept as compared to the other answers.
Check this
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++){
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) + 1;
if(f(A,b) == false){
A[i++] = b;
}
}
return A;
}
public static boolean f(int[] A, int n){
for(int i=0; i<A.length; i++){
if(A[i] == n){
return true;
}
}
return false;
}
}
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.
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;
}
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:
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++)
{
numbers.add(i+1);
}
Collections.shuffle(numbers);
System.out.println(numbers);
}
}
You can generate n unique random number between 0 to n-1 in java
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));
}
}
How would i prevent duplicating numbers from random numbers.
I need to generate 5 numbers between 1 and 9 that are each different.
I would often get same numbers like 23334, how can i prevent that?
Any help would be great!
int num2 = (int) Math.round((Math.random()*9) +1);
int num1 = (int) Math.round((Math.random()*9) +1);
int num5 = (int) Math.round((Math.random()*9) +1);
int num3 = (int) Math.round((Math.random()*9) +1);
int num4 = (int) Math.round((Math.random()*9) +1);
One option is to use shuffle algorithm (e.g. Fisher-Yates shuffle ) to generate random sequence from 1 to 9, then take first 5 numbers of the sequence
Further explanation on StackOverflow: https://stackoverflow.com/a/196065/950427
Set<Integer> set=new HashSet<>();
while (set.size()<5) {
set.add( Math.round((Math.random()*9) +1));
}
After the set is filled you have 5 unique random numbers.
UPDATE: just to illustrate Jared Burrows' comment
Create a List includes the numbers that you want (1 to 9).
Generate random number from 0 to (size of the list minus 1).
Remove one element by index from the above generated random number. And add the removed element to a array which to be returned as a results
public static void main(String[] args) {
int []answers= returnRandomNonRepeatingNumbers(5,0,9);
for(int answer: answers) {
System.out.println(answer);
}
}
public static int[] returnRandomNonRepeatingNumbers(int sizeYouWant, int poolStart, int poolEnd) {
List<Integer> pool=new ArrayList<Integer>();
for(int i=poolStart;i<=poolEnd;i++) {
pool.add(i);
}
int []answers=new int[sizeYouWant];
for(int i=0;i<sizeYouWant;i++) {
//random index to be pick and remove from pool
int randomIndex = (int) Math.round((Math.random()*(pool.size()-1)));
answers[i]=pool.remove(randomIndex);
}
return answers;
}
If the number of possible random values is small, you want to use shuffle.
List<Integer> values = IntStream.range(0, 10).boxed().collect(toList());
Collections.shuffle(values);
values = values.subList(0, 5);
If the number of possible random values is large, you want to test adding them to a Set (or the original list if small enough)
Set<Integer> valueSet = new HashSet<>();
Random rand = new Random();
while(valuesSet.size() < 5) valuesSet.add(rand.nextInt(9) + 1);
List<Integer> values = new ArrayList<>(valueSet);
Collections.shuffle(values, rand);
Note: you need to shuffle the set as it doesn't preserve order. e.g. the numbers 1,2,3 will always come out in that order with HashSet, not 3,2,1.
Floyd's subset selection algorithm is designed to do exactly what you want, and is extremely efficient even for large sets. Selecting m items from a set of n is O(m) average running time, independent of n. Here's a Java implementation.
/*
* Floyd's algorithm to chose a random subset of m integers
* from a set of n, zero-based.
*/
public static HashSet<Integer> generateMfromN(int m, int n) {
HashSet<Integer> s = new HashSet<Integer>();
for (int j = n-m; j < n; ++j) {
if(! s.add((int)((j+1) * Math.random()))) {
s.add(j);
}
}
return s;
}
One possible approach to this problem can be divide & conquer. Step of following describes the approach:
Say m is the minimum & n is the maximum, within what i wanna get x number of randoms
Choose a random p between m & n. Save it to an array of answer. decrease x by 1 as we get one answer to our problem.
Now take a q a random number between m & p-1, another r a random number between p+1 & n. Fill up the answer array with q & r decrease x 1 for q and another 1 for the r.
Now carry on this process recursively, until the lower bound (m) & higher bound (n) becomes equal or x becomes 0.
Benefit: benefit of this approach is that, in worst case, it's runtime will be O(x), where x is the number of random number required. The best case scenarion is also o(x), as i have to find at least n number of random. These two comprise average case to θ(x) complexity.
import java.util.Random;
class GenerateDistinctRandom{
static int alreadyPut = 0;
static Random rand = new Random();
public static int[] generateDistinctRandom(int howMany, int rangeMin, int rangeMax)
{
int randomNumbers[] = new int[howMany];
GenerateDistinctRandom.recursiveRandomGenerator(rangeMin, rangeMax, randomNumbers, howMany);
return randomNumbers;
}
private static void recursiveRandomGenerator(int rangeMin, int rangeMax, int[] storage ,int storageSize)
{
if(rangeMax - rangeMin <= 0 || GenerateDistinctRandom.alreadyPut == storageSize)
{
return ;
}
int randomNumber = GenerateDistinctRandom.rand.nextInt(rangeMax-rangeMin) + rangeMin;
storage[GenerateDistinctRandom.alreadyPut] = randomNumber;
GenerateDistinctRandom.alreadyPut++;
//calling the left side of the recursion
recursiveRandomGenerator(rangeMin, randomNumber - 1, storage, storageSize);
recursiveRandomGenerator(randomNumber + 1, rangeMax, storage, storageSize);
}
public static void main(String []args){
int howMany = 5;
int distinctNumber[] = GenerateDistinctRandom.generateDistinctRandom(howMany 0, 9);
for(int i = 0;i < howMany;i++)
{
System.out.println(distinctNumber[i]);
}
}
}
I suppose you would need to store the ones that have been generated into an array and compare the new random number to the list to ensure it is unique.
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
int[] numbers = new int[5];
int tempNumber = 0;
for(int numberCounter = 0; numberCounter < numbers.length;)
{
tempNumber = (int) Math.round((Math.random()*9) +1);
if(!contains(numbers, tempNumber)){
numbers[numberCounter++] = tempNumber;
}
}
}
public static boolean contains(final int[] numbersArray, final int tempNumber) {
for (final int numberFromArray : numbersArray) {
if (numberFromArray == tempNumber) {
return true;
}
}
return false;
}
I notice you did not use an array in your example, so in case you do not know how to use them yet, you could also make 5 variables.
int randomNumber = 0;
int firstNumber = Math.round((Math.random()*9) +1);
int secondNumber = 0;
while(secondNumber == 0){
randomNumber = Math.round((Math.random()*9) +1)l
if(randomNumber != firstNumber){
secondNumber = randomNumber;
}
}
And you could continue making while statements like that. But if you are supposed to know about arrays, you should definitely be using one to store the numbers.
How about this?
package com.se;
import java.util.ArrayList;
import java.util.List;
import java.util.Random;
public class TestRandom {
List<Integer> comp = new ArrayList<>();
int listSize = 20;
public void doTask() {
Random ran = new Random();
int i = 0;
while(i < listSize){
int randomNumber = ran.nextInt(80) + 1;
if(!comp.contains(randomNumber)){
comp.add(randomNumber);
i++;
}
}
for(Integer num : comp){
System.out.println(num);
}
}
public static void main(String[] args) {
TestRandom testRandom = new TestRandom();
testRandom.doTask();
}
}
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.
Random rand = new Random();
selected = rand.nextInt(100);
Add each number in the range sequentially in a list structure.
Shuffle it.
Take the first 'n'.
Here is a simple implementation. This will print 3 unique random numbers from the range 1-10.
import java.util.ArrayList;
import java.util.Collections;
public class UniqueRandomNumbers {
public static void main(String[] args) {
ArrayList<Integer> list = new ArrayList<Integer>();
for (int i=1; i<11; i++) list.add(i);
Collections.shuffle(list);
for (int i=0; i<3; i++) System.out.println(list.get(i));
}
}
The first part of the fix with the original approach, as Mark Byers pointed out in an answer now deleted, is to use only a single Random instance.
That is what is causing the numbers to be identical. A Random instance is seeded by the current time in milliseconds. For a particular seed value, the 'random' instance will return the exact same sequence of pseudo random numbers.
With Java 8+ you can use the ints method of Random to get an IntStream of random values then distinct and limit to reduce the stream to a number of unique random values.
ThreadLocalRandom.current().ints(0, 100).distinct().limit(5).forEach(System.out::println);
Random also has methods which create LongStreams and DoubleStreams 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 Rob Kielty's answer), 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:
List<Integer> range = IntStream.range(0, 100).boxed()
.collect(Collectors.toCollection(ArrayList::new));
Collections.shuffle(range);
range.subList(0, 99).forEach(System.out::println);
Create an array of 100 numbers, then randomize their order.
Devise a pseudo-random number generator that has a range of 100.
Create a boolean array of 100 elements, then set an element true when you pick that number. When you pick the next number check against the array and try again if the array element is set. (You can make an easy-to-clear boolean array with an array of long where you shift and mask to access individual bits.)
Use Collections.shuffle() on all 100 numbers and select the first five, as shown here and below.
Console:
59 9 68 24 82
Code:
private static final Random rnd = new Random();
private static final int N = 100;
private static final int K = 5;
private static final List<Integer> S = new ArrayList<>(N);
public static void main(String[] args) {
for (int i = 0; i < N; i++) {
S.add(i + 1);
}
Collections.shuffle(S, rnd);
for (int i = 0; i < K; i++) {
System.out.print(S.get(i) + " ");
}
System.out.println();
}
I feel like this method is worth mentioning.
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 + 1));
}
return picked;
}
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 set.add() when an add to the set fails.
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);
}
}
I have made this like that.
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)+1; // this will give numbers between 1 and 50.
if (!arrayList.contains(a)) {
arrayList.add(a);
}
}
This will work to generate unique random numbers................
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++) {
e = rand.nextInt(20);
randomNumbers.add(e);
if (randomNumbers.size() <= 10) {
if (randomNumbers.size() == 10) {
g = 10;
}
g++;
randomNumbers.add(e);
}
}
System.out.println("Ten Unique random numbers from 1 to 20 are : " + randomNumbers);
}
}
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:
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
In Java code, you would write:
void randInts() {
int num=1;
for (int ii=0; ii<101; ii++) {
System.out.println(num);
num= (num*2) % 101;
}
}
Finding a primitive root for a specific modulus can be tricky, but Maple's "primroot" function will do this for you.
I have come here from another question, which has been duplicate of this question (Generating unique random number in java)
Store 1 to 100 numbers in an Array.
Generate random number between 1 to 100 as position and return array[position-1] to get the value
Once you use a number in array, mark the value as -1 ( No need to maintain another array to check if this number is already used)
If value in array is -1, get the random number again to fetch new location in array.
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.
for(int i=0;i<4;i++)
{
rn[i]= GenerateRandomNumber();
for (int j=0;j<i;j++)
{
if (rn[i] == rn[j])
{
i--;
}
}
}
Choose n unique random numbers from 0 to m-1.
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++){
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]++;
}
}
return result;
}
Imagine a list containing numbers from 0 to m-1. To choose the first number, we simply use rand.nextInt(m). Then remove the number from the list. Now there remains m-1 numbers, so we call rand.nextInt(m-1). 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.
Explanation
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.
Though it's an old thread, but adding another option might not harm. (JDK 1.8 lambda functions seem to make it easy);
The problem could be broken down into the following steps;
Get a minimum value for the provided list of integers (for which to generate unique random numbers)
Get a maximum value for the provided list of integers
Use ThreadLocalRandom class (from JDK 1.8) to generate random integer values against the previously found min and max integer values and then filter to ensure that the values are indeed contained by the originally provided list. Finally apply distinct to the intstream to ensure that generated numbers are unique.
Here is the function with some description:
/**
* Provided an unsequenced / sequenced list of integers, the function returns unique random IDs as defined by the parameter
* #param numberToGenerate
* #param idList
* #return List of unique random integer values from the provided list
*/
private List<Integer> getUniqueRandomInts(List<Integer> idList, Integer numberToGenerate) {
List<Integer> generatedUniqueIds = new ArrayList<>();
Integer minId = idList.stream().mapToInt (v->v).min().orElseThrow(NoSuchElementException::new);
Integer maxId = idList.stream().mapToInt (v->v).max().orElseThrow(NoSuchElementException::new);
ThreadLocalRandom.current().ints(minId,maxId)
.filter(e->idList.contains(e))
.distinct()
.limit(numberToGenerate)
.forEach(generatedUniqueIds:: add);
return generatedUniqueIds;
}
So that, to get 11 unique random numbers for 'allIntegers' list object, we'll call the function like;
List<Integer> ids = getUniqueRandomInts(allIntegers,11);
The function declares new arrayList 'generatedUniqueIds' and populates with each unique random integer up to the required number before returning.
P.S. ThreadLocalRandom class avoids common seed value in case of concurrent threads.
try this out
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);
}
}
This isn't significantly different from other answers, but I wanted the array of integers in the end:
Integer[] indices = new Integer[n];
Arrays.setAll(indices, i -> i);
Collections.shuffle(Arrays.asList(indices));
return Arrays.stream(indices).mapToInt(Integer::intValue).toArray();
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
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++) {
integers.add(i);
}
//get unique random numbers
boolArray = new Boolean[integers.size()+1];
Arrays.fill(boolArray, false);
for (int i = 0; i < 10; i++) {
System.out.print(getUniqueRandomNumber(integers) + " ");
}
}
private static int getUniqueRandomNumber(List<Integer> integers) {
int randNum =(int) (Math.random()*integers.size());
if(boolArray[randNum]){
while(boolArray[randNum]){
randNum++;
if(randNum>boolArray.length){
randNum=0;
}
}
boolArray[randNum]=true;
return randNum;
}else {
boolArray[randNum]=true;
return randNum;
}
}
}
This is the most simple method to generate unique random values in a range or from an array.
In this example, I will be using a predefined array but you can adapt this method to generate random numbers as well. First, we will create a sample array to retrieve our data from.
Generate a random number and add it to the new array.
Generate another random number and check if it is already stored in the new array.
If not then add it and continue
else reiterate the step.
ArrayList<Integer> sampleList = new ArrayList<>();
sampleList.add(1);
sampleList.add(2);
sampleList.add(3);
sampleList.add(4);
sampleList.add(5);
sampleList.add(6);
sampleList.add(7);
sampleList.add(8);
Now from the sampleList we will produce five random numbers that are unique.
int n;
randomList = new ArrayList<>();
for(int i=0;i<5;i++){
Random random = new Random();
n=random.nextInt(8); //Generate a random index between 0-7
if(!randomList.contains(sampleList.get(n)))
randomList.add(sampleList.get(n));
else
i--; //reiterating the step
}
This is conceptually very simple. If the random value generated already exists then we will reiterate the step. This will continue until all the values generated are unique.
If you found this answer useful then you can vote it up as it is much simple in concept as compared to the other answers.
Check this
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++){
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) + 1;
if(f(A,b) == false){
A[i++] = b;
}
}
return A;
}
public static boolean f(int[] A, int n){
for(int i=0; i<A.length; i++){
if(A[i] == n){
return true;
}
}
return false;
}
}
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.
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;
}
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:
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++)
{
numbers.add(i+1);
}
Collections.shuffle(numbers);
System.out.println(numbers);
}
}
You can generate n unique random number between 0 to n-1 in java
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));
}
}
I've just been looking at the following piece of code
package test;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class Main {
public static void main(final String[] args) {
final int sizeA = 3;
final int sizeB = 5;
final List<int[]> combos = getAllCombinations(sizeA-1, sizeB);
int counter = 1;
for(final int[] combo : combos) {
System.out.println("Combination " + counter);
System.out.println("--------------");
for(final int value : combo) {
System.out.print(value + " ");
}
System.out.println();
System.out.println();
++counter;
}
}
private static List<int[]> getAllCombinations(final int maxIndex, final int size) {
if(maxIndex >= size)
throw new IllegalArgumentException("The maximum index must be smaller than the array size.");
final List<int[]> result = new ArrayList<int[]>();
if(maxIndex == 0) {
final int[] array = new int[size];
Arrays.fill(array, maxIndex);
result.add(array);
return result;
}
//We'll create one array for every time the maxIndex can occur while allowing
//every other index to appear, then create every variation on that array
//by having every possible head generated recursively
for(int i = 1; i < size - maxIndex + 1; ++i) {
//Generating every possible head for the array
final List<int[]> heads = getAllCombinations(maxIndex - 1, size - i);
//Combining every head with the tail
for(final int[] head : heads) {
final int[] array = new int[size];
System.arraycopy(head, 0, array, 0, head.length);
//Filling the tail of the array with i maxIndex values
for(int j = 1; j <= i; ++j)
array[size - j] = maxIndex;
result.add(array);
}
}
return result;
}
}
I'm wondering, how do I eliminate recursion from this, so that it returns a single random combination, rather than a list of all possible combinations?
Thanks
If I understand your code correctly your task is as follows: give a random combination of numbers '0' .. 'sizeA-1' of length sizeB where
the combination is sorted
each number occurs at least once
i.e. in your example e.g. [0,0,1,2,2].
If you want to have a single combination only I'd suggest another algorithm (pseudo-code):
Randomly choose the step-up positions (e.g. for sequence [0,0,1,1,2] it would be steps (1->2) & (3->4)) - we need sizeA-1 steps randomly chosen at sizeB-1 positions.
Calculate your target combination out of this vector
A quick-and-dirty implementation in java looks like follows
// Generate list 0,1,2,...,sizeB-2 of possible step-positions
List<Integer> steps = new ArrayList<Integer>();
for (int h = 0; h < sizeB-1; h++) {
steps.add(h);
}
// Randomly choose sizeA-1 elements
Collections.shuffle(steps);
steps = steps.subList(0, sizeA - 1);
Collections.sort(steps);
// Build result array
int[] result = new int[sizeB];
for (int h = 0, o = 0; h < sizeB; h++) {
result[h] = o;
if (o < steps.size() && steps.get(o) == h) {
o++;
}
}
Note: this can be optimized further - the first step generates a random permutation and later strips this down to desired size. Therefore it is just for demonstration purpose that the algorithm itself works as desired.
This appears to be homework. Without giving you code, here's an idea. Call getAllCombinations, store the result in a List, and return a value from a random index in that list. As Howard pointed out in his comment to your question, eliminating recursion, and returning a random combination are separate tasks.