I am trying to get all prime factors of a number. The for loop should work until it finds the match and it should break and jump to the next if statement which checks if number is not equal to zero.
public class Factor {
public static ArrayList <Integer> HoldNum = new ArrayList();
public static void main(String[]args){
Factor object = new Factor();
object.Factor(104);
System.out.println(HoldNum.get(0));
}
public static int Factor(int number){
int new_numb = 0;
int n=0;
for( n = 1; n < 9; n++) {
if (number % n == 0) {
HoldNum.add(n);
new_numb = number/n;
break;
}
}
System.out.println(new_numb);
if(new_numb < 0) {
HoldNum.add(new_numb);
return 1;
} else {
return Factor(new_numb);
}
}
}
There are at least three errors :
As okiharaherbst wrote, your counter is not incremented.
you start your loop at 1, so yourval % 1 always equals to 0 and new_numb is always equals to your input val, so you'll loop endlessly on 104.
new_numb will never be lesser than 0.
You asked for a recursive solution. Here you go:
public class Example {
public static void main(String[] args) {
System.out.println(factors(104));
}
public static List<Integer> factors(int number) {
return factors(number, new ArrayList<Integer>());
}
private static List<Integer> factors(int number, List<Integer> primes) {
for (int prim = 2; prim <= number; prim++) {
if (number % prim == 0) {
primes.add(prim);
return factors(number / prim, primes);
}
}
return primes;
}
}
The code is not bullet-proof, it is only a quick-and-dirty example.
Java implementation...
public class PrimeFactor {
public int divisor=2;
void printPrimeFactors(int num)
{
if(num == 1)
return;
if(num%divisor!=0)
{
while(num%divisor!=0)
++divisor;
}
if(num%divisor==0){
System.out.println(divisor);
printPrimeFactors(num/divisor);
}
}
public static void main(String[] args)
{
PrimeFactor obj = new PrimeFactor();
obj.printPrimeFactors(90);
}
}
Related
I want to reverse an int but it doesn't work. For example, 123 should return 321, but the printed number is 356.
public class x {
public static void main(String[] args) {
System.out.println(reverse2(123, 0));
}
static int reverse2(int a, int i) {
if(a == 0) {
return 0;
} else {
i = i*10 + a%10;
System.out.println(i);
return i += reverse2(a/10, i);
}
}
}
Your code should look like this:
public class x {
public static void main(String[] args) {
System.out.println(reverse2(123, 0));
}
static int reverse2(int a, int i) {
if(a == 0) {
return i;
} else {
i = i*10 + a%10;
System.out.println(i);
return reverse2(a/10, i);
}
}
}
You should return i when a is 0.
You shouldn't add i when you call the reverse2 function because you're adding i twice.
You are greatly complicating your recursive function for printing an integer in reverse. For one, there is no good reason for reverse2 to have two integer arguments, as you can achieve your desired results with a single argument. The trick is to access the rightmost digit with the % 10 operation then shift that digit off the number with the / 10 operation. Consider these revisions:
public class x {
public static void main(String[] args) {
System.out.println(reverse2(123));
}
static String reverse2(int number) {
if(number == 0) {
return "";
} else {
return number % 10 + reverse2(number / 10);
}
}
}
You can do it like this. You only need to pass the value you are reversing. The math computation computes 10 to the power of the number of digits in the argument.
public static int reverse(int v) {
int reversed = 0;
if (v > 0) {
int d = (int)Math.pow(10,(int)(Math.log10(v)));
reversed = reverse(v%d) * 10 + v/d;
}
return reversed;
}
Of course, if you can pass a second argument as a scratch pad, then it can be done like so. As you tear down the original value you build up the returned value.
public static int reverse(int v, int reversed) {
if (v > 0) {
return reverse(v / 10, reversed * 10 + v % 10);
}
return reversed;
}
The program works fine for small numbers but as soon as i take a big number like this it doesn't work
here is my code
public class Main {
public static void main(String[] args) {
long no=600851475143L,i;
int result=0;
for(i=(no/2);i>=2;i--){
if(no%i==0){
if(checkPrime(i)){
System.out.println("Longest Prime Factor is: " + i);
break;
}
}
}
}
private static boolean checkPrime(long i){
for(long j=2L;j<=(int)Math.sqrt(i);j++){
if(i%j==0)
return false;
}
return true;
}
}
to assign long variable value we does not require write L at last on value Remove L.
It will take time to display the answer. Just try with small number(1000000 ) almost 10 to 15 min for above code.
Try this
public class Main {
public static void main(String[] args) {
//long no=600851475143L,i;
System.out.println(largestPrimeFactor(600851475143L));
}
public static int largestPrimeFactor(long number) {
int i;
for (i = 2; i <= number; i++) {
if (number % i == 0) {
number /= i;
i--;
}
}
return i;
}
}
[1]https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
I'm working on an eulers problem, where you have to get the longest collatz chain. The problem is, you have to find it in a sequence from 1 to 1 000 000. My code works great up to 100 000, then it throws StackOverFlowError. Can I avoid it, or is my code crap?
public class Collatz {
private static final ArrayList<Integer> previous = new ArrayList<>();
public static void main(String[] args) {
for (int i = 1; i < 100000; i++){
collatzLength(i, 1);
}
Integer i = Collections.max(previous);
System.out.println(i);
}
public static void collatzLength(int n, int count){
if (n == 1){
previous.add(count);
} else if (n%2 == 0){
collatzLength(n/2, count+1);
} else {
collatzLength(3*n+1, count+1);
}
}
}
This question already has answers here:
What is a StackOverflowError?
(16 answers)
Closed 7 years ago.
I am trying to solve a problem which asks to find the smallest prime palindrome, which comes after a given number which means that if the input is 24, the output would be 101 as it is the smallest number after 24 which is both prime and a palindrome.
Now my code works perfectly for small values but the moment I plug in something like 543212 as input, I end up with a StackOverFlowError on line 20, followed by multiple instances of StackOverFlowErrors on line 24. Here is my code :
package nisarg;
import java.util.Scanner;
public class Chef_prime_palindromes {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
long num = input.nextLong();
isPalindrome(num + 1);
}
public static boolean isPrime(long num) {
long i;
for (i = 2; i < num; i++) {
if (num % i == 0) {
return false;
}
}
return true;
}
public static void isPalindrome(long num) {
String word = Long.toString(num);
int i;
for (i = 0; i < word.length() / 2; i++) {
if (word.charAt(i) != word.charAt(word.length() - i - 1)) {
isPalindrome(num + 1);
}
}
if (i == word.length() / 2) {
if (isPrime(num)) {
System.out.println(num);
System.exit(0);
} else {
isPalindrome(num + 1);
}
}
}
}
All shown exiting solutions use recursion and have the problem that at some point they will reach the point where a StackOverflowException will occur.
A better solution which would also be parallelizable would be to change it into a loop.
It could be something like:
package nisarg;
import java.math.BigInteger;
import java.util.Scanner;
import java.util.concurrent.CopyOnWriteArrayList;
public class Chef_prime_palindromes {
private static final CopyOnWriteArrayList<BigInteger> PRIMESCACHE
= new CopyOnWriteArrayList<>();
public static void main(String[] args) {
try (Scanner input = new Scanner(System.in)) {
BigInteger num = new BigInteger(input.nextLine());
initPrimes(num);
for (num = num.add(BigInteger.ONE);
!isPrimePalindrome(num);
num = num.add(BigInteger.ONE));
System.out.println(num.toString());
}
}
private static void initPrimes(BigInteger upTo) {
BigInteger i;
for (i = new BigInteger("2"); i.compareTo(upTo) <= 0 ; i = i.add(BigInteger.ONE)) {
isPrime(i);
}
}
public static boolean isPrimePalindrome(BigInteger num) {
return isPrime(num) && isPalindrome(num);
}
// could be optimized
public static boolean isPrime(BigInteger num) {
for (int idx = PRIMESCACHE.size() - 1; idx >= 0; --idx) {
if (num.mod(PRIMESCACHE.get(idx)).compareTo(BigInteger.ZERO) == 0) {
return false;
}
}
if (!PRIMESCACHE.contains(num)) {
PRIMESCACHE.add(num);
}
return true;
}
public static boolean isPalindrome(BigInteger num) {
String word = num.toString();
int i;
for (i = 0; i < word.length() / 2; i++) {
if (word.charAt(i) != word.charAt(word.length() - i - 1)) {
return false;
}
}
return true;
}
}
A new String object is created in each recursive call and placed onto stack (the place where all variables created in methods are stored until you leave the method), which for a deep enough recursion makes JVM reach the end of allocated stack space.
I changed the locality of the String object by placing it into a separate method, thus reducing its locality and bounding its creation and destruction (freeing of stack space) to one recursive call.
package com.company;
import java.util.Scanner;
public class Chef_prime_palindromes {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
long num = input.nextLong();
isPalindrom(num + 1);
}
public static boolean isPrime(long num) {
long i;
for (i = 2; i < num; i++) {
if (num % i == 0) {
return false;
}
}
return true;
}
private static void isPalindrom(long num) {
for (; ; ) {
if (isPalindrome(num)) {
if (isPrime(num)) {
System.out.println(num);
System.exit(0);
} else {
num++;
}
} else {
num++;
}
}
}
public static boolean isPalindrome(long num) {
String string = String.valueOf(num);
return string.equals(new StringBuilder(string).reverse().toString());
}
}
First thing you should be aware of is the fact that your resources are limited. Even if your implementation was precise and all recursive calls were correct, you may still get the error. The error indicates your JVM stack ran out of space. Try to increase the size of your JVM stack ( see here for details).
Another important thing is to look for the distribution of prime and palindrome numbers. Your code runs by testing every num+1 against palindrome property. This is incorrect. You test for palindrome only when the number is prime. This will make the computation much much easier (and reduce recursive calls). I have edited your code accordingly and got the closest palindrome number after 543212 (1003001) . Here it is:
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
long num = input.nextLong();
//isPalindrome(num+1);
nextPrimePalindrome(num+1);
}
public static void nextPrimePalindrome(long num)
{
boolean flag=true;
while(flag)
{
if(isPrime(num))
if(isPalindrome(num))
{
System.out.println(num);
flag=false;
}
num++;
}
}
public static boolean isPrime(long num){
long i;
for(i=2;i<num;i++){
if(num%i == 0){
return false;
}
}
return true;
}
public static boolean isPalindrome(long num)
{
String word=Long.toString(num);
for(int i=0;i<word.length()/2;i++)
if(word.charAt(i)!=word.charAt(word.length()-i-1))
return false;
return true;
}
}
I am a new member here, and I am also a beginner in JAVA. The thing that seems the most abstract to me is recursion, and so I have some difficulties finishing a program that should have this output if we write 3 for example:
1
12
123
12
1
Or if we write 5 for example it should print out this:
1
12
123
1234
12345
1234
123
12
1
And I can do this program with for loop, but I have to use recursion, and here is what I have done so far:
public class Aufgabe3 {
private static void printSequenz(int n) {
if(n<1){
return;
}
printMany(n);
printSequenz(n-1);
}
private static void printMany(int n){
for(int i=1;i<=n;i++){
System.out.print(i);
}
System.out.println();
}
public static void main(String[] args) {
printSequenz(5);
}
}
I would be really happy if someone would help me :).
You need to implement two recursive functions:
void printLoToHi(int n)
{
if (n < 1)
return;
printLoToHi(n-1);
printMany(n);
}
void printHiToLo(int n)
{
if (n < 1)
return;
printMany(n);
printHiToLo(n-1);
}
Then, you need to call them sequentially:
printSequenz(int n)
{
printLoToHi(n);
printHiToLo(n-1); // -1 in order to avoid printing the highest twice
}
Or in a more "symmetrical manner":
printSequenz(int n)
{
printLoToHi(n-1); // print lowest to second highest
printMany(n); // print the highest
printHiToLo(n-1); // print second highest to lowest
}
You could do it like this:
private static void printSequenz(int n) {
printSequenz(1,n, true);
}
private static void printSequenz(int current, int total, boolean goingUp) {
if(!goingUp && current<1){
return;
}
printMany(current);
if(current+1>total){
goingUp=false;
}
if(goingUp){
printSequenz(current+1,total,goingUp);
} else {
printSequenz(current-1,total,goingUp);
}
}
private static void printMany(int n) {
for (int i = 1; i <= n; i++) {
System.out.print(i);
}
System.out.println();
}
public static void main(String[] args) {
printSequenz(5);
}
public class Test {
public static void main(String args[]) {
int seq = 6;
for(int i=1; i<=seq; i++) {
System.out.println("");
int low =seq-(seq-i);
printIncreasing(i,low);
}
for(int i=seq-1; i>=1; i--) {
System.out.println("");
int low = seq-i;
printDecreasing(i,low);
}
}
public static void printIncreasing(int high, int low) {
for(int i = 1; i<=high;i++ ) {
System.out.print(i);
}
}
public static void printDecreasing(int high, int low) {
for(int i = 1; i<=high;i++ ) {
System.out.print(i);
}
}
}