I need some help with this assignment I've been given. Not asking anyone to do my work but I'm really honestly stuck on how to even do this.
I'm supposed to write a program that prompts the user to enter 10 numbers and then have it write all the numbers in reverse order.
Example:
Enter 10 Numbers: 23 89 21 55 67 89 99 13 98 78
Reverse Order: 78 98 13 99 89 67 55 21 89 23
So far all I have is how to get the user inputs. If anyone can push me in the right direction, i'd be very grateful!
import java.util.*;
public class ReverseNumbers
{
public static void main(String[] args)
{
Scanner keyboard = new Scanner(System.in);
int[] values;
values = new int[10];
//Ask the user to enter 10 integers
System.out.println("Please enter 10 numbers:");
for (int i = 0; i< values.length; i++)
{
values[i] = keyboard.nextInt();
}
int[] reverseNums;
reverseNums = new int[10];
for (int i = (values.length -1); i>= 0; i--)
{
reverseNums[ reverseNums.length -1 -i ] = values[ i ];
System.out.println( reverseNums[ i ] );
}
}
}
If you dont want to store the reversed values
for (int i = (values.length -1); i>= 0; i--)
{
System.out.println( values[ i ] );
}
Your code looks good for reading inputs into values. Now you can loop over that array in reverse and print the values:
for (int i = 0; i < values.length; i++)
System.out.println(values[values.length - i - 1]);
Think about it, when i == 0 it will print values[9] since 10 - 0 - 1 = 9. At the end, when i == 9 it will print values[0] since 10 - 9 - 1 = 0.
As you can see, there is no need for the extra reverseNums array.
PS: If you want, you can combine int[] values; and values = new int[10]; into a single line: int[] values = new int[10];
Read in the whole line as a string using Scanner.nextLine(), and then split it into an array using String.split(" ");. After this, you can simply iterate from the end of the array going backwards and print the numbers.
If it's not an assignment then why not try using http://commons.apache.org/proper/commons-lang/ library
ArrayUtils.reverse(int[] array)
I think your code is creating the array correctly. You are just printing out the numbers in the original order because your second for-loop is iterating over them backwards. You can see the correct result by adding the statement below after the last loop:
System.out.println(java.util.Arrays.toString(reverseNums));
You can also perform the complete task with only one iteration:
Scanner keyboard = new Scanner(System.in);
int[] reverseNums= new int[10];
System.out.println("Please enter 10 numbers:");
for (int i = 0; i< values.length; i++) {
reverseNums[values.length -1 - i] = keyboard.nextInt();
}
System.out.println(java.util.Arrays.toString(reverseNums));
To reverse it:
for (int i = 0; i < values.length; i++)
reverseNums[values.length - i - 1] = values[i];
Related
This question already has answers here:
What causes a java.lang.ArrayIndexOutOfBoundsException and how do I prevent it?
(26 answers)
Closed 5 years ago.
I am still new to Java, this question like: An ISBN-10 (International Standard Book Number)
consists of 10 digits: d1d2d3d4d5d6d7d8d9d10. The last digit, d10, is a checksum,
which is calculated from the other nine digits using the following formula:
(d1 * 1 + d2 * 2 + d3 * 3 + d4 * 4 + d5 * 5 +
d6 * 6 + d7 * 7 + d8 * 8 + d9 * 9) % 11
If the checksum is 10, the last digit is denoted as X according to the ISBN-10
convention. Write a program that prompts the user to enter the first 9 digits and
displays the 10-digit ISBN (including leading zeros). Your program should read
the input as an integer.
See my code below:
It is working but the result is not correct!
public static Scanner input;
public static void main(String[] args)
{
input = new Scanner(System.in);
System.out.print("Enter first 9 digit numbers: ");
int[] arr = new int[9];
int checkSum = 0;
for (int i = 0 ; i < arr.length; i++)
{
arr[i] = input.nextInt();
checkSum = (arr[i] * i) % 11;
}
System.out.print("The ISBN-10 Number is " );
for(int j = 0 ; j < arr.length; j++)
{
System.out.print(arr[j]);
}
if(checkSum == 10)
{
System.out.println("x");
}
else
{
System.out.println(checkSum);
}
I just want to use loop, make my method works. , i know how to use method without loop.
well. for JAVA
an array index start from 0. and Max index is length － 1.
if the index is not within this range. an arrayoutofbounds exception will be thrown
The problem is not that you are doing int i = 1 instead of int i = 0. The problem is that you changed i < arr.length; to i <= arr.length;. Since arr.length is 9, your code is trying to refer to arr[9] but arr only has elements arr[0] through arr[8].
First, review how we use arrays...
int[] I = new int[3];
The above creates an array with 3 spots in it. After instantiating the array, each element is based off of a 0-index. You only have I[0], I[1], and I[2] available (note this is three numbers) and trying to access I[3] will throw the error that you encountered in your program above: ArrayIndexOutOfBoundsException.
That being said, your loop is trying to access arr[9]. The highest index you have in the array is arr[8], because despite the array having 9 elements in it, it's 0-indexed.
Assuming you MUST have i starting from 1 for a homework assignment or something, change your code to:
public static Scanner input;
public static void main(String[] args)
{
input = new Scanner(System.in);
System.out.print("Enter first 9 digit numbers: ");
int[] arr = new int[9];
int checkSum = 0;
for (int i = 1 ; i <= arr.length; i++)
{
arr[i-1] = input.nextInt();
checkSum = (arr[i-1] * i) % 11;
}
System.out.print("The ISBN-10 Number is " );
for(int j = 1 ; j <= arr.length; j++)
{
System.out.print(arr[j-1]);
}
if(checkSum == 10)
{
System.out.println("x");
}
else
{
System.out.println(checkSum);
}
int[] arr = new int[9];
This means the array has 9 slots. And the numbering of those slots starts from 0. Thus,
arr[0] is the first slot
arr[1] is the second slot
The last slot would be arr[8]. But in the following code you are iterating till i is equal to 9 which does not exist.
for (int i = 1 ; i <= arr.length; i++)
{
arr[i] = input.nextInt();
checkSum = (arr[i] * (i+1)) % 11;
}
This results in the ArrayIndexOutOfBoundsException. Change for (int i = 1 ; i <= arr.length; i++) to for (int i = 0 ; i < arr.length; i++)
My problem is that I must create an array of random integers from 0-100 and count the number of single digit integers that are produced.
So far I have this:
public class randomarray {
public static void main(String[] args) {
int i;
int smallintegers;
int[] RandomArray = new int[10];
for(i = 0; i < RandomArray.length; i++) {
RandomArray[i] = (int)(Math.random() * 100);
System.out.print(RandomArray[i] + " ");
}
}
}
This works and creates an array of random digits like so:
17 56 44 41 42 96 8 42 21 86
Now I need to find out a way to count the number of single digit integers, and in this case it would have been 1.
The first method I thought of would be to use filter() to copy the single-digit integers, create a new array from filter(), and then produce the length of the new array.
Thanks in advance
How about checking for single digit integers while you are building the array:
int numSingle = 0;
for (i=0; i < RandomArray.length; i++) {
int num = (int)(Math.random() * 100);
if (num < 10) ++numSingle;
RandomArray[i] = num;
System.out.print(num + " ");
}
System.out.println("There are " + numSingle + " integers in the array.");
If, for some reason, you can't check the array as it's being built (e.g. you are receiving it from some API you can't control), then you can go with the Wombat's suggestion to just iterate the array. If you use a Java 8 stream to do this I expect it would also iterate over the array internally.
Here is how you could use a stream to answer the same question:
int numSingle = Arrays.stream(RandomArray).filter(n -> n < 10).toArray().length;
Another simple yet not as elegant way to accomplish this would be to convert each integer to a string and use string.length. I know I am not allowed to use lambda expressions in my classes so this is a route that doesn't include it.
I need to take a text file like the one below and create a 2d array from the numbers in it. However, it needs to be very general so that it could apply to a text file with more or fewer entries than this one.
1 1 11
1 2 32
1 4 23
2 2 24
2 5 45
3 1 16
3 2 37
3 3 50
3 4 79
3 5 68
4 4 33
4 5 67
1 1 75
1 4 65
2 1 26
2 3 89
2 5 74
This is what I have so far, but it just gives me all zeroes when I print it.
import java.util.*;
public class MySales11 {
//variables
private ArrayList<String> list = new ArrayList<>();
private int numberOfEntries;
private int [][] allSales;
//constructor
public MySales11 (Scanner scan) {
//scan and find # of entries
while (scan.hasNext()){
String line = scan.nextLine();
list.add(line);
}
//define size of AllSales array
allSales = new int[list.size()][3];
//populate AllSales array with list ArrayList
for(int a = 0; a < allSales.length; a++){
String[] tokens = list.get(a).split(" ");
for(int b = 0; b < tokens.length; b++){
allSales[a][b] = Integer.parseInt(tokens[b]);
}
}
}
}
You read all the lines when you wanted to create a array of size numOfEntries.
while (scan.hasNext()) {
scan.nextLine();
numberOfEntries++;//this reads all the lines but never stores
}
allSales = new int[numberOfEntries][3];
while (scan.hasNext()) {//input is empty
//the execution never comes here.
}
Now the input is empty. So it'll never add values to the array.
You can use an arrayList, dynamic - no need to calculate number of lines.
ArrayList<String> list = new ArrayList();
while (scan.hasNext()) {
String s = scan.nextLine();
list.add(s);
}
int [][] myArray = new int[list.size()][3];
for(int i = 0; i < myArray.length; ++i)
{
String[] tokens = list.get(i).split("\\s+");//extra spaces
for(int j = 0; j < tokens.length; ++j)
{
myArray[i][j] = Integer.parseInt(tokens[j]);
}
}
I need help printing the array, I need to print 6 items per line and switch to a next line for the seventh and following numbers. Also who do i Enter numbers into an array without defining how many numbers will be entered?
import java.util.Scanner;
public class NumberArray
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.println("How many grades do you want to enter?");
int num = input.nextInt();
int array[] = new int[num];
System.out.println("Enter the " + num + " grades now.");
for (int grades = 0 ; grades < array.length; grades++ )
{
array[grades] = input.nextInt();
}
System.out.println("These are the grades you have entered.");
printArray(array);
}
public static void printArray(int arr[])
{
int n = arr.length;
for (int i = 0; i < n; i++) {
System.out.print(arr[i] + " \t");
}
}
}
I need help printing the array, I need to print 6 items per line and switch to a next line for the seventh and following numbers.
From this question, it seems to indicate that you want the output to look like this:
1 2 3 4 5 6
7 8 9 ... n
This can be achieved quite simply.
Option 1 - The classic If statement
for(int x = 0; x < array.length; x++) {
System.out.print(array[x]);
if(x == 5) {
// 5 because we're counting from 0!
System.out.println();
}
}
Option 2 - Using the Ternary operator to keep it on one line
NOTE: This is more or less the same. It's just nice to be complete in these sorts of answers.
for(int x = 0; x < array.length; x++) {
System.out.print(array[x] + x == 5? "\n":"");
}
Edit
If you meant that you want 6 items on each line, like so:
1 2 3 4 5 6
7 8 9 10 11 12
...
Then you can use the % (The modulus operator) to print out a new line on every output. This is actually quite easy to change, but you'll need to make sure that you're checking the value before you're outputting the content. This can be shown in this IDEOne.
Use modulus operator (%) to break to a new line line:
public static void printArray(int arr[])
{
int n = arr.length;
for (int i = 0; i < n; i++) {
if(i % 6 == 0) // if you don't want the initial newline, check for i > 0
System.out.println()
System.out.print(arr[i] + " \t");
}
}
you can also use printf() method to format the line; which is probably better:
System.out.printf("%5d", arr[i]);
The reason why this is better, is because you can easily format the output to a specific justification, column width, etc., which will make your output look better.
I'm writing a program, that take 10 Integers from a keyboard and list them in an array indexed from 0-9, and reports the position of the lowest number in the array. If the lowest number has any other position than 0, then the program is supposed to switch the position of the lowest number input with the number in the first position in the array:
import java.util.Scanner;
public class q35 {
public static void main(String args[]) {
Scanner tastatur = new Scanner(System.in);
int[] helTall = new int[10];
int input;
int lowest = Integer.MAX_VALUE;
for(int i=0;i<helTall.length;i++) {
System.out.println("Integers? ");
input = tastatur.nextInt();
if (input < lowest) {
lowest = input;
}
helTall[i] = input;
}
for (int i = 0; i < helTall.length; i++) {
helTall[0] = lowest;
System.out.println(helTall[i]);
}
System.out.println("Lowest number is " + lowest);
}
}
The only problem is that instead of changing position with the lowest number with the number at helTall[0], it just completely replaces the first number in the sequence helTall[0] with the lowest Integer, that way if my input is 4 5 63 23 6 -4 7 33 23 99, then the output becomes -4 5 63 23 6 -4 7 33 23 99 (as you can see the first input number is completely erased), but it should have been -4 5 63 23 6 4 7 33 23 99 any tips/advice/solutions? Thanks in advance.
You should keep track of the index of the lowest number (each time you write lowest = input; you should add lowestIndex=i;.
Then helTall[lowestIndex] will be the lowest number.
So you swap helTall[lowestIndex] with helTall[0] instead of just overwriting the value of helTall[0].
I thought it was enough to describe the solution in words, but I guess it wasn't...
int lowest = Integer.MAX_VALUE;
int lowestIndex = 0;
for(int i=0;i<helTall.length;i++){
System.out.println("Integers? ");
input = tastatur.nextInt();
if (input < lowest){
lowest = input;
lowestIndex = i;
}
helTall[i]=input;
}
// swap the numbers
if (lowestIndex > 0) {
int temp = lowest;
helTall[lowestIndex] = helTall[0];
helTall[0] = temp;
}
// display output
for (int i = 0; i < helTall.length; i ++) {
System.out.println(helTall[i]);
}
This part is wrong:
for (int i = 0; i < helTall.length; i ++) {
helTall[0]=lowest;
System.out.println(helTall[i]);
}
First, you do not need to repeatedly (10 times) put lowest into helTall[0]. Let's move that outside first so it's only done once:
helTall[0]=lowest;
for (int i = 0; i < helTall.length; i ++) {
System.out.println(helTall[i]);
}
Next, the line we put outside the loop overwrites the helTall[0] without regard for what is already in there. We need to temporarily save that number in there elsewhere, then overwrite the spot, so that we can use it to overwrite the spot where the lowest number was:
int numberAtLocationZero = helTall[0];
helTall[0]=lowest;
// Now, on this line we need to write helTall[lowestNumberIndex] = numberAtLocationZero;
for (int i = 0; i < helTall.length; i ++) {
System.out.println(helTall[i]);
}
In the above code I wrote a comment. It relies on you either knowing where in the array the lowest number was, or to find it again. If you add a new variable to your code that takes in values called lowestNumberIndex and update that each time you detect a new lowest number, then you're pretty much done if you uncomment my comment.
You need a separate variable to store what is in the bottom of the array before you overwrite it with your new lowest input number.
helTall[0]=lowest;
should be proceeded by
int placeholder = helTall[0]
and followed by
hellTall[i] = placeholder;
This way, you'll end up swapping the two array elements, the first swapped with the lowest. Is this what you're trying to accomplish? This leaves most of the array unsorted.