I'm supposed to write a program using for loops that print out the even indexes of my array. For example, if I create an array that has 10 numbers, it will have indexes from 0-9 so in that case I would print out the numbers at index 2, 4, 6 and 8. This is what I wrote so far but it doesn't work. Please note that I am not trying to print out the even numbers of the array. All I want are the even indexes.
Example I enter the following array: 3,7,5,5,5,7,7,9,9,3
Program output:
5 // (the number at index 2)
5 // (the number at index 4)
7 // (the number at index 6)
9 // (the number at index 8)
My Code:
public class Arrayevenindex
{
public static void main(String[] args)
{
int number; // variable that will represent how many elements the user wants the array to have
Scanner key = new Scanner(System.in);
System.out.println(" How many elements would you like your array to have");
number = key.nextInt();
int [] array = new int [number];
// let the user enter the values of the array.
for (int index = 0; index < number; index ++)
{
System.out.print(" Value" + (index+1) + " :");
array[index] = key.nextInt();
}
// Print out the even indexes
System.out.println("/nI am now going to print out the even indexes");
for (int index = 0; index < array.length; index ++)
{
if (array[number+1]%2==0)
System.out.print(array[number]);
}
}
}
You can just change your for loop and get rid of the inner IF...
for( int index = 0; index < array.length; index += 2) {
System.out.println(array[index]);
}
Just absolutely same thing using java 8 Stream API
Integer[] ints = {0,1,2,3,4,5,6,7,8,9};
IntStream.range(0, ints.length).filter(i -> i % 2 == 0).forEach(i -> System.out.println(ints[i]));
I assume this would be sufficient
// For loop to search array
for (int i = 0; i < array.length; i++) {
// If to validate that the index is divisible by 2
if (i % 2 == 0) {
System.out.print(array[i]);
}
}
This is what I did and it works:also I am not printing out index[0] because technically its not even thats why I started the for loop at 2. Your post did help me a lot. I also thank everyone else as well that took the time to post an answer.
import java.util.Scanner;
public class Arrayevenindex
{
public static void main(String[] args)
{
int number; // variable that will represent how many elements the user wants the array to have
Scanner key = new Scanner(System.in);
System.out.println(" How many elements would you like your array to have");
number = key.nextInt();
int [] array = new int [number];
// let the user enter the values of the array.
for ( int index = 0; index < number; index ++)
{
System.out.print(" Value" + (index+1) + " :");
array[index] = key.nextInt();
}
// Print out the even indexes
System.out.println("/nI am now going to print out the even indexes");
for ( int index = 2; index < array.length; index +=2)
{
System.out.print(array[index] + " ");
}
}
}
Related
Im trying to print out an array but only print out the distinct numbers in that array.
For example: if the array has {5,5,3,6,3,5,2,1}
then it would print {5,3,6,2,1}
each time i do it either i only print the non repeating numbers, in this example {6,2,1} or i print them all. then i didnt it the way the assignment suggested
the assignment wants me to check the array before i place a value into it to see if its there first. If not then add it but if so dont.
now i just keep getting out of bounds error or it just prints everything.
any ideas on what i should do
import java.util.Scanner;
public class DistinctNums {
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
int value;
int count = 0;
int[] distinct = new int[6];
System.out.println("Enter Six Random Numbers: ");
for (int i = 0; i < 6; i++)
{
value = input.nextInt(); //places users input into a variable
for (int j = 0; i < distinct.length; j++) {
if (value != distinct[j]) //check to see if its in the array by making sure its not equal to anything in the array
{
distinct[count] = value; // if its not equal then place it in array
count++; // increase counter for the array
}
}
}
// Displays the number of distinct numbers and the
// distinct numbers separated by exactly one space
System.out.println("The number of distinct numbers is " + count);
System.out.print("The distinct numbers are");
for (int i = 0; i < distinct.length; i++)
{
System.out.println(distinct[i] + " ");
}
System.out.println("\n");
}
}
Always remember - if you want a single copy of elements then you need to use set.
Set is a collection of distinct objects.
In Java, you have something called HashSet. And if you want the order to be maintained then use LinkedHashSet.
int [] intputArray = {5,5,3,6,3,5,2,1};
LinkedHashSet<Integer> set = new LinkedHashSet<Integer>();
//add all the elements into set
for(int number:intputArray) {
set.add(number);
}
for(int element:set) {
System.out.print(element+" ");
}
You can make this using help array with lenght of 10 if the order is not important.
int [] intputArray = {5,5,3,6,3,5,2,1};
int [] helpArray = new int[10];
for(int i = 0; i < intputArray.length ; i++){
helpArray[intputArray[i]]++;
}
for(int i = 0; i < helpArray.length ; i++){
if(helpArray[i] > 0){
System.out.print(i + " ");
}
}
I'm trying to find the majority element using Boyer and Moore approach. The program should ask the user to input "n" number of lines on the first line, then there will be "n" numbers followed "n" lines as input. (Ex: user input 5 on the first line, then there will be 5 numbers followed)
Next, use Boyer and Moore approach to find the majority element of an input array. If the majority element doesn't exist in the input array, then execute -1. My program output shows 0 no matter what input I entered. Would you please check it and correct my program?
Output example:
4
12
12
1
12
12
/Or: 3 11 2 13 -1
public static void main (String[]args)
{
int a[] = new int [1000000];
Scanner sc = new Scanner (System.in);
// User input n number of lines
int numberOfLine = sc.nextInt();
//Loop to have n elements as input
for (int i = 1; i<= numberOfLine; i++)
{
a[i] = sc.nextInt();
}
// Call method to display majority element
getMajorityElement(a);
sc.close();
}
//Method to Find M.E using Boyer & Moore approach
public static void getMajorityElement(int [] array)
{
int majorityElement = array[0];
int count = 1;
for (int index = 1; index<array.length; index++)
{
if(majorityElement==array[index])
{
count++;
}
else if(count==0)
{
majorityElement = array[index];
count = 1;
}
else
{
count --;
}
}
// Check if candidate M.E occur more than n/2 times
count = 0;
for (int index = 0; index<array.length; index++)
{
if(array[index]==majorityElement)
{
count++;
}
}
if (count > array.length/2)
{
System.out.println(majorityElement);
}
else
{
System.out.println("-1");
}
}
The reason you get this behavior is that your array of 1,000,000 elements has a majority element of zero: only the initial three or four items are set, while the rest of the items are occupied by zeros - the default value of an int in Java.
Fix this problem by allocating at size once the length is entered. You also need to fix the code that reads input to make sure that the data ends up at indexes 0..numberOfLine-1:
Scanner sc = new Scanner (System.in);
// User input n number of lines
int numberOfLine = sc.nextInt();
int a[] = new int [numberOfLine];
//Loop to have n elements as input
for (int i = 0 ; i < numberOfLine ; i++) {
a[i] = sc.nextInt();
}
I'm trying to make a Java program that uses the input value from a user to calculate and list the products of two numbers up to the entered number. Like if a user enters 2, the program should calculate the products between the two numbers (1 *1, 1*2, 2*1, 2*2) stores the products in a two-dimensional array, and list the products. I'm not sure that I totally understand arrays and so I feel as though my code is problem not right in many instances, can someone please tell me what I should do to my current code to make it work properly. Thanks in advance! :)
import java.util.Scanner;
public class ProductTable {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String inputString;
char letter = 'y';
// Prompt the user to enter an integer
while(letter != 'q') {
System.out.print("Enter a positive integer: ");
int integer = input.nextInt();
// Create an two-dimensional array to store products
int[][] m = new int[integer][3];
for (int j = 1; j <= m.length; j++) {
m[j][0] = input.nextInt();
m[j][1] = input.nextInt();
m[j][2] = input.nextInt();
}
// Display the number title
System.out.print(" ");
for (int j = 1; j <= m.length; j++)
System.out.print(" " + j);
System.out.println("\n--- ");
// Display table body
for (int i = 1; i <= m.length + 1; i++) {
System.out.print(i);
for (int j = 1; j <= m.length + 1; i++) {
System.out.printf("%4d", i * j);
}
System.out.println();
}
// Prompt the user to either continue or quit
System.out.print("Enter q to quit or any other key to continue: ");
String character = input.nextLine();
inputString = input.nextLine();
letter = inputString.charAt(0);
}
}
}
You can achieve your multiplication table by iterating over 2 counter variables as you already did in your output
// Display table body
// loops running out of bounds (until m.length + 1 instead of m.length-1)
for (int i = 1; i <= m.length + 1; i++) {
System.out.print(i);
for (int j = 1; j <= m.length + 1; i++) { // missed to increment j here
System.out.printf("%4d", i * j);
}
System.out.println();
}
Arrays should be based on 0, that means an array with 3 fields has the indexes 0, 1, 2. So the last index is length-1. Your condition <=m.length+1 runs out of bounds. index<length will work since 2 is less than 3, but not 3 less than 3.
You have also a typo: In the inner loop you are doing an increment of i instead of j, so the loop will run infinite.
Try to create an outer and inner loop as you did, but with start index 0 and end condition index<inputValue. Calculate multiTable[index1][index2] = (1+index1)*(1+index2).
Then do a similar loop and print the array fields. You don't need to use an array, you could output directly as you did. But you wanted to practice handling arrays.
Since you want to learn and understand, I don't like just to write the code. imagine an array with 3 fields having the indexes 0, 1, 2:
index 0 | 1 | 2
value 1 | 2 | 3
Now you would check the length at first, that's 3. You start with 0 and count while the counter does not reach the length since the zero based index is 1 below our natural order (0,1,2 vs. 1,2,3). That is the case as long the index is below the length, meaning index<length.
Try this logic (pseudo code). To simplify the problem we include the 0 in our product table. So we get 0*0, 0*1, 0*2... Doing so, we do not have to ignore index 0 or to calculate between index 0 should represent value 1.
maxNumber = userInput()
outer loop idx1 from 0 to maxNumber
inner loop idx2 from 0 to maxNumber
array[idx1][idx2] = (idx1) * (idx2)
end loop
end loop
Do the same to dump your generated array to screen.
First get it running. Afterwards you could try to alter the logic, so that it shows you only numbers from 1 to max.
I want to make a list going to n numbers depending on user input. I then want to put a second number in each place and print the entire table. As for testing I have tried with a length 4 and numbers 1,2,3,4 but I get a error: ArrayIndexOutOfBounds. I wanted it to print 1,2,3,4.
Scanner keyboard = new Scanner (System.in);
System.out.println("Whats the length of the table?");
int lengde = keyboard.nextInt();
int[] minTabell = new int[lengde];
for (int i =1; i <= lengde+ 1; i++) {
System.out.println((i) + (" give a number"));
minTabell[i] = keyboard.nextInt();
}
System.out.println(minTabell);
keyboard.close();
Indexes in Java arrays are 0-based, while your for-loop starts from 1. So,
for (int i =1; i <= lengde+ 1; i++) {
System.out.println((i) + (" give a number"));
minTabell[i] = keyboard.nextInt();
}
should be
for (int i =0; i < lengde; i++) {
// ^ ^^^^^^^^
System.out.println((i+1) + (" give a number"));
// ^^^
minTabell[i] = keyboard.nextInt();
}
As for printing the content of the array, I suggest you use
for (int i : minTabell)
System.out.println(i);
In Java array indexing starts at 0. The first element is positioned at minTabel1[0]. Your for-loop runs from 1 to lengde + 1, which means that you will try to fill a position outside of the array.
The first element of an array has the index 0. The last valid index of your array is lengde-1.
Try this:
for (int i=0; i < lengde; i++) {
System.out.println((i) + (" give a number"));
minTabell[i] = keyboard.nextInt();
}
To print the array, i suggest the following:
System.out.println(Arrays.toString(minTabell));
I have made a program that outputs the number of repeats in a 2D array. The problem is that it outputs the same number twice.
For example: I input the numbers in the 2D array through Scanner: 10 10 9 28 29 9 1 28.
The output I get is:
Number 10 repeats 2 times.
Number 10 repeats 2 times.
Number 9 repeats 2 times.
Number 28 repeats 2 times.
Number 29 repeats 1 times.
Number 9 repeats 2 times.
Number 1 repeats 1 times.
Number 28 repeats 2 times.
I want it so it skips the number if it has already found the number of repeats for it. The output should be:
Number 10 repeats 2 times.
Number 9 repeats 2 times.
Number 28 repeats 2 times.
Number 29 repeats 1 times.
Number 1 repeats 1 times.
Here is my code:
import java.util.Scanner;
public class Repeat
{
static Scanner leopard = new Scanner(System.in);
public static void main(String [] args)
{
final int ROW = 10; //Row size
final int COL = 10; //Column size
int [][] num = new int[ROW][COL];
int size;
//Get input
size = getData(num);
//Find repeat
findRepeats(num, size);
}
public static int getData(int [][] num)
{
int input = 0, actualSize = 0; //Hold input and actualSize of array
System.out.print("Enter positive integers (-999 to stop): ");
//Ask for input
for(int i = 0; i < num.length && input != -999; i++)
{
for(int j = 0; j < num[i].length && input != -999; j++)
{
input = leopard.nextInt();
//Check if end
if(input != -999)
{
num[i][j] = input;
actualSize++;
}
}
}
System.out.println();
return actualSize;
}
public static void findRepeats(int [][] num, int size)
{
int findNum;
int total = 0, row = 0, col = 0;
for(int x = 0; x < size; x++)
{
//Set to number
findNum = num[row][col];
//Loop through whole array to find repeats
for(int i = 0; i < num.length; i++)
{
for(int j = 0; j < num[i].length; j++)
{
if(num[i][j] == findNum)
total++;
}
}
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
//Display total repeats
System.out.println("Number " + findNum + " appears " + total + " times.");
total = 0;
}
}
}
I know why it is doing it, but I cannot figure out how to check if the number has already been checked for it to skip that number and go to the next number. I cannot use any classes or code that is not used in the code.
Since you cannot use anything other than this, lets say, basic elements of Java consider this:
Make another temporary 2D array with two columns (or just two separate arrays, personally I prefer this one). On the start of the algorithm the new arrays are empty.
When you take a number (any number) from the source 2D structure, first check if it is present in the first temporary array. If it is, just increment the value (count) in the second temporary array for one (+1). If it is not present in the first tmp array, add it to it and increase the count (+1) in the second at the same index as the newly added number in the first (which should be the last item of the array, basically).
This way you are building pairs of numbers in two arrays. The first array holds all your distinct values found in the 2D array, and the second one the number of appearances of the respective number from the first.
At the and of the algorithm just iterate the both arrays in parallel and you should have your school task finished. I could (and anyone) code this out but we are not really doing you a favor since this is a very typical school assignment.
It's counting the number two times, first time it appears in the code and second time when it appears in the code.
To avoid that keep a system to check if you have already checked for that number. I see you use check int array but you haven't used it anywhere in the code.
Do this,
Put the number in the check list if you have already found the count of it.
int count = 0;
check[count] = findNum;
count++;
Note: You can prefill you array with negative numbers at first in order to avoid for having numbers that user already gave you in input.
Next time in your for loop skip checking that number which you have already found a count for
for(int x = 0; x < size; x++) {
findNum = num[row][col];
if(check.containsNumber(findNUm)) { //sorry there is no such thing as contains for array, write another function here which checks if a number exists in the array
//skip the your code till the end of the first for loop, or in other words then don't run the code inside the for loop at all.
}
}
Frankly speaking I think you have just started to learn coding. Good luck! with that but this code can be improved a lot better. A piece of advice never create a situation where you have to use 3 nested for loops.
I hope that you understood my solution and you know how to do it.
All answers gives you some insight about the problem. I try to stick to your code, and add a little trick of swap. With this code you don't need to check if the number is already outputted or not. I appreciate your comments, structured approach of coding, and ask a question as clear as possible.
public static void findRepeats(int [][] num, int size)
{
int findNum;
int total = 1, row = 0, col = 0;
int [] check = new int[size];
while(row < num.length && col < num[0].length)
{
//Set to number
findNum = num[row][col];
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
//Loop through whole array to find repeats
for(int i = row; i < num.length; i++)
{
for(int j = col; j < num[i].length; j++)
{
if(num[i][j] == findNum) {
total++;
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
if(row < num.length - 1 && col < num[0].length -1)
num[i][j] = num[row][col];
}
}
}
//Display total repeats
System.out.println("Number " + findNum + " appears " + total + " times.");
total = 1;
}
}
you can use a HashMap to store the result. It Goes like this:
// Create a hash map
HashMap arrayRepeat = new HashMap();
// Put elements to the map
arrayRepeat.put(Number, Repeated);