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how can i check if the number is perfect square by not using math.sqrt

i want to make a program which related to this question:
An integer that can be expressed as the square of another integer is called a perfect square, such as 4,9,16,25, etc. Write a progran that checks if a number is a perfect square.
I did built something goes like:
import java.util.Scanner;
class Q3{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int num = 0;
int a = 0;
System.out.println("Type a number to check if it has square");
num = sc.nextInt();
for(a = 1;a<num;a++){ }
if (a*a == num){
System.out.println("Ok");
break;
}
else if (a*a != num){
System.out.println("Not ok");
}
}
}
So it doesn’t give what i want when i run it. What should i change or add ?

I think your for loop interpretation might be wrong, I made up something that might just work. Give this code a try.. You can make the method return a boolean too if you want.
static void perfectSquare(int number) {
for (int i = 1; i < i * number; ++i) {
// 'i' is the divisor, making sure
// it is equal to the quotient
if ((number % i == 0) && (number / i == i)) {
System.out.println(i);
}
}

If you want to brute force every number then you are on the right track but you should only print "Not ok" if all numbers in the loop have failed otherwise you may have a perfect square but "Ok" will be hidden within many "Not ok" lines. Also there is nothing in your for loop so the if statement always checks if 0*0 == num.
This looks like it may be a homework question so I won't give a full answer for you but think about how you can make this more efficient.
If you have found an integer number that matches do you need to keep going?
Do you need to check every number? (a starting point may be following the principles of a binary search)

I ended up like this:
import java.util.Scanner;
class Q3{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int num = 0;
int a = 0;
int b = 0;
System.out.println("Type a number to check if it has square");
num = sc.nextInt();
for(a = 1;a<num;a++){
if (a*a == num){
b = 1;
break;
}
}
if(b==1){
System.out.println("Perfect Square");
}
else {
System.out.println("Not ok");
}
}
}
Thanks for support !

Trying to test when user inputs code if the code is already sorted

I keep trying different kinds of code and I always come back to this. but it never seems to work. The last if statement is making the i's underlined red but I can't even understand why. The homework was to make a program that takes user input and put it into an array and see if the user input is already sorted. Please Help!
import java.util.Scanner;
public class Sorting
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.println("Enter the array size: ");
int a = input.nextInt();
System.out.println("Enter the numbers using spaces between each number: ");
int[] numbers = new int[a];
for (int i=0; i<numbers.length; i++)
{
numbers[i]=input.nextInt();
if(isSorted(numbers))
{
System.out.println("Sort is already sorted");
}
else
{
System.out.println("Sort is not sorted sorry");
}
}
}
public static boolean isSorted(int[] numbers)
{
for(int i = 0; i<numbers.length-1; i++);
{
if(numbers[i]>numbers[i+1])
{
return false;
}
}
return true;
}
}

Close the for loop before the if statement.

for(int i = 0; i<numbers.length-1; i++); //<===== remove the ';' here
I think you missed place the ; after the for loop and that cause your issue.

Using arrays to store primes

Here is my program which is supposed to create an array and initialize prime numbers to it. The prime numbers should then be printed but the program just keeps running.
import java.util.*;
public class primes
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
System.out.println("Enter number of primes ");
int x = scan.nextInt();
int[] prime = new int[x];
int div=2,hold=2;
int c=0;
while (prime[x-1]==0)
{
for(int a=2; div>a;a++)
{
if(div>a && div%a==0)
a=div;
else if(div==(a-1))
hold=div;
}
if(div==2||hold!=prime[c-1])
{
prime[c]=hold;
c++;
}
div++;
}
for(int f =0; f<x;f++)
System.out.print(" "+prime[f]+" ");
}
}
I tried changing my loops but I just don't know whats wrong

Like the others mentioned your logic is not right, try something like:
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter number of primes ");
int x = scan.nextInt();
List<Integer> primes = getPrimes(x);
Integer[] primeArray = primes.toArray(new Integer[primes.size()]);
for(int i :primes.toArray(primeArray)){ // you could just use for(int i :primes){ if you don't need array
System.out.print(i + " ");
}
}
private static List<Integer> getPrimes(int upperLimit) {
ArrayList primes = new ArrayList();
for (int i = 2; i < upperLimit; i++) {
boolean isPrime = true;
// Is it prime?
for (int j = 2; j < i; j++) {
if (i % j == 0) {
isPrime = false;
break;
}
}
if (isPrime)
primes.add(i);
}
return primes;
}
The above will print out up to the numbers entered so if you type 5 it will print out 2 3 but not 5.
The following is an other example with Java 8, this one will print as many prime numbers based on the input, if you input 5 you will get 2 3 5 7 11
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter number of primes ");
int x = scan.nextInt();
long[] prime = primes(x).toArray();
Arrays.stream(prime).forEach(value -> System.out.print(value + " " ));
}
private static LongStream primes(long max) {
return LongStream.iterate(2, i -> i + 1)
.filter(PrimeNumber::isPrime)
.limit(max);
}
private static boolean isPrime(long x) {
return LongStream.rangeClosed(2, (long)(Math.sqrt(x)))
.allMatch(n -> x % n != 0);
}

Your code is wrong. First correct it, And i think you want to store prime numbers coming in range of 1 to N where N is user provided number. Use arrayList (growable) to store it.

It will keep on running because you have this: while (prime[x-1]==0). Where x is an input from the user. Say 5 for instance, then prime[5-1] initially is going to contain a 0 always, and you are running your while loop on this condition which is always going to turn true, thus never ending. Also, your prime number generation logic is not right!

I ran your code in debugger mode and I found the problem.
I tested your program with x=5.
At the end of the first while loop iteration you have :
prime[0] = 2
div = 3
hold = 2
c = 1
And here's the problem :
if(div==2||hold!=prime[c-1])
{
prime[c]=hold;
c++;
}
This part won't ever be reached anymore because :
div is never decrement, so it will always be superior to 2.
hold is
equal to prime[c-1], and never change value.
So prime will always stick to be : 2 0 0 0 0, and your while loop will never end.

I found what was wrong and rewrote the code, it works now. The program asks the user for the number primes they want to see and it prints them after storing them in a basic integer array.
import java.util.*;
public class Prime
{
public static void main(String [] args)
{
Scanner scan= new Scanner(System.in);
int i=0, hold=2, d=2;
boolean flag = true;
System.out.println("Enter the number of primes.");
int[] prime= new int[scan.nextInt()];
for(;flag;){
for(int a=2;d>a;a++){
if(d==(a)||d%a==0){
break;
}
if((d-1)==a){
hold = d;
}
}
d++;
if(hold==2 || hold!=prime[i-1]){
prime[i] = hold;
i++;
}
if(i==prime.length)
flag= false;
}
for(int x=0;x<prime.length;x++)
System.out.print(prime[x]+" ");
System.out.println("");
}
}

How do I send user inputted integers to a linked list in Java?

I am trying to write code to have users input positive integers and have them sent to a linked list. The users input should end after entering a negative number. Furthermore, I am having issues writing an isSorted boolean method that will return true if the linked list is sorted in increasing order and false otherwise.
Here is the only code I have so far
import java.util.*;
public class List {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Please type positive integers one by one separated by a space.");
System.out.println("When you are done, please type a negative integer.");
int num = input.nextInt();
}
public boolean isSorted(){
if(){
return true;
}
else{
return false;
}
}
}

public void input() {
System.out.println("Please type positive integers one by one separated by a space.");
System.out.println("When you are done, please type a negative integer.");
LinkedList<Integer> ll = new LinkedList<>();
//System.in.available()
Scanner input = new Scanner(System.in);
int num;
while ( input.hasNextInt() ) {
int i = input.nextInt();
if (i >= 0) {
ll.add(i);
}
}
System.out.println(ll+" <-- ll"); //TODO remove debugging code
System.out.println(isSortedAccending(ll)+" <-- isSortedAccending(ll)");//TODO
}
This works by returning false the moment something is found out of order.
public static boolean isSortedAccending(List<Integer> list){
if (list.size() < 2) {
return true;
}
Integer previous = list.get(0);
for (Integer next : list) {
if (previous > next) {
return false;
}
}
return true;
}
Outputs:
Please type positive integers one by one separated by a space.
When you are done, please type a negative integer.
1
2
3
-1
[1, 2, 3] <-- ll
true <-- isSortedAccending(ll)
isSortedDecending() looks exactly the same except it uses <.

import java.util.LinkedList;
import java.util.Scanner;
public class NumberListing {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Please type positive integers one by one separated by a space.");
System.out.println("When you are done, please type a negative integer.");
int num = input.nextInt();
LinkedList<Integer> list = new LinkedList<Integer>();
while(num > 0){
list.add(num);
num = input.nextInt();
}
if(isSorted(list)){
System.out.println("The list is sorted");
} else{
System.out.println("The list is not sorted");
}
}
public static boolean isSorted(LinkedList<Integer> list){
boolean done = false;
boolean sorted = true;
for(int i = 1; i < list.size() && !done ; i++){
if(list.get(i) < list.get(i-1)){
done = true;
sorted = false;
}
}
return sorted;
}
}
Should be fairly simple. While the number you intake is greater than 0, store the number in a list and get the next one from the console.
After that, loop through the list starting with index 1 to check that each current number is greater than the previous number. I added an early exit so the loop quits anytime a number that is less than the previous number. this means it is unordered and we don't care about the rest.
However, if you're trying to create your own linked list that creates a lot more code and I would suggest looking in a text book or something of the sort to help.

You can put your int's in a container(LinkedList) by using the wrapper class Integer for int. The normal int is a standard primitive type and can not be used by containers, as containers can only work with objects of a class.
LinkedList<Integer> list = new LinkedList();
As for the isSorted method, you could iterate through the list and check if the current value is higher or equal to the previous value.
int prevValue = 0;
for(int i = 0; i < list.size(); i++){
if(!list.get(i) >= prevValue){
return false;
}
prevValue = list.get(i);
}
return true;

You'll want to read the next line as a String, split that line, and walk through the list of elements of that line, adding the positive elements to a LinkedList as you go. Make sure you store your scanner. You'll want to use nextLine, not nextInt.
import java.util.*;
public class List {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Please type positive integers one by one separated by a space.");
System.out.println("When you are done, please type a negative integer.");
String line = input.nextLine();
String[] integers = line.split(" ");
LinkedList ll = new LinkedList<Integer>();
for (int i = 0; i < integers.length; i++)
{
if (Integer.parseInt(integers[i]) > 0)
ll.add(integers[i]);
}
System.out.println(isSorted(ll));
input.close();
}
Then, just use a simple comparison function for LinkedList to check if it's sorted:
public static <T extends Comparable<T>> boolean isSorted(LinkedList<T> iterable)
{
Iterator<T> iter = iterable.iterator();
if (!iter.hasNext()) {
return true;
}
T t = iter.next();
while (iter.hasNext()) {
T t2 = iter.next();
if (t.compareTo(t2) > 0) {
return false;
}
t = t2;
}
return true;
}

storing multiple user inputs into an integer array

So here's what i'm trying to do. Have user input numbers, store them in an array, then print all those numbers out on one line. I must use a method/function as well. I think i have done quite well so far, it shows no errors in eclipse. What im stuck on is storing their input into an array. i want them to keep inputting numbers one at a time until they're satisifed and type 'quit'. i just havent read about how to store things in an array despite looking around, particularly storing more than one thing, progressively. here is my code so far, and thank you in advance for any help!
import java.util.Scanner;
public class intarray {
public static void main(String[] args) {
System.out.println("Enter a number then hit enter. You may then enter another number or end by typing quit.");
String x;
Scanner input = new Scanner(System.in);
while (true) {
x=input.next();
if (x.equalsIgnoreCase("quit")) {break;}
if (x.equals(null)) throw new Error("You entered nothing. Try again.");
int stringLength = x.trim().length();
if (stringLength == 0) throw new Error("Seems you only entered spaces. Try again.");
isNum(x);
int goingintoarray = Integer.parseInt(x);
int array[];
}
}
public static String isNum(String t) {
int user=Integer.parseInt(t);
String convertback = Integer.toString(user);
return convertback;
}
}

Since you don't know how many elements there will be an array is a bad idea since you will have to resize it quite often as new elements appear (copying arrays is expensive!) or instantiate a large enough array at the beginning (which is a waste and still doesn't protect you in 100% from having to resize it eventually).
Instead using Java's List (preferably LinkedList) sounds like a good idea since you can add elements dynamically without resizing the data structure.
List<Integer> numbers = new LinkedList<>();
while(true) {
// something
numbers.add(goingintoarray);
// something
}
Be careful of other implementations - for instance ArrayList uses an array (d'uh ;-) ) to store the elements so you would have the same problem but the resizing part would be taken care of for you by the implementation.
#Edit: by convention classes in Java are written using CamelCase starting with an uppercase letter.

ArrayList<Integer> inputs = new ArrayList<Integer>();
while (true) {
Scanner input = new Scanner(System.in);
x=input.next();
if (x.equalsIgnoreCase("quit")) {break;}
if (x.equals(null)) throw new Error("You entered nothing. Try again.");
int stringLength = x.trim().length();
if (stringLength == 0) throw new Error("Seems you only entered spaces. Try again.");
inputs.add(Integer.parseInt(x));
}
You don't want the isNum method since it gives same exception here if it gets wrong input for x.

import java.util.Scanner;
public class intarray {
public static int initSize = 5;
public static void main(String[] args) {
System.out.println("Enter a number then hit enter. You may then enter another number or end by typing quit.");
int array[] = new int[initSize];
int pos = 0;
int maxSize = initSize;
String x = null;
Scanner input = new Scanner(System.in);
while (true) {
x = input.next();
if (x.equalsIgnoreCase("quit")) {
break;
}
//input empty string, java io can filter. So , x impossible "null" or null.
//if (x.equals(null))
// throw new Error("You entered nothing. Try again.");
int stringLength = x.trim().length();
if (stringLength == 0)
throw new Error("Seems you only entered spaces. Try again.");
Integer numX = isNum(x);
// if the array is full, extend it
if(pos == maxSize){
int[] newArray = new int[2 * maxSize];
System.arraycopy(array, 0, newArray, 0, maxSize);
array = newArray;
maxSize = array.length;
}
if(null == numX)
System.out.println(x + " isn't a number."); //choose notify or throw error
else
array[pos++] = numX;
}
printArray(array, pos);
}
public static Integer isNum(String t) {
try {
return Integer.parseInt(t);
} catch (NumberFormatException e) {
return null;
}
}
public static void printArray(int[] array, int pos) {
if(null == array || array.length == 0 || pos <= 0)
return ;
for(int i = 0 ; i < pos; i++)
System.out.print(array[i] + " ");
}
}