I need to write a java method sumAll() which takes any number of integers and returns their sum.
sumAll(1,2,3) returns 6
sumAll() returns 0
sumAll(20) returns 20
I don't know how to do this.
If your using Java8 you can use the IntStream:
int[] listOfNumbers = {5,4,13,7,7,8,9,10,5,92,11,3,4,2,1};
System.out.println(IntStream.of(listOfNumbers).sum());
Results: 181
Just 1 line of code which will sum the array.
You need:
public int sumAll(int...numbers){
int result = 0;
for(int i = 0 ; i < numbers.length; i++) {
result += numbers[i];
}
return result;
}
Then call the method and give it as many int values as you need:
int result = sumAll(1,4,6,3,5,393,4,5);//.....
System.out.println(result);
public int sumAll(int... nums) { //var-args to let the caller pass an arbitrary number of int
int sum = 0; //start with 0
for(int n : nums) { //this won't execute if no argument is passed
sum += n; // this will repeat for all the arguments
}
return sum; //return the sum
}
Use var args
public long sum(int... numbers){
if(numbers == null){ return 0L;}
long result = 0L;
for(int number: numbers){
result += number;
}
return result;
}
import java.util.Scanner;
public class SumAll {
public static void sumAll(int arr[]) {//initialize method return sum
int sum = 0;
for (int i = 0; i < arr.length; i++) {
sum += arr[i];
}
System.out.println("Sum is : " + sum);
}
public static void main(String[] args) {
int num;
Scanner input = new Scanner(System.in);//create scanner object
System.out.print("How many # you want to add : ");
num = input.nextInt();//return num from keyboard
int[] arr2 = new int[num];
for (int i = 0; i < arr2.length; i++) {
System.out.print("Enter Num" + (i + 1) + ": ");
arr2[i] = input.nextInt();
}
sumAll(arr2);
}
}
public static void main(String args[])
{
System.out.println(SumofAll(12,13,14,15));//Insert your number here.
{
public static int SumofAll(int...sum)//Call this method in main method.
int total=0;//Declare a variable which will hold the total value.
for(int x:sum)
{
total+=sum;
}
return total;//And return the total variable.
}
}
You could do, assuming you have an array with value and array length: arrayVal[i], arrayLength:
int sum = 0;
for (int i = 0; i < arrayLength; i++) {
sum += arrayVal[i];
}
System.out.println("the sum is" + sum);
I hope this helps.
Related
First of all let me say I am quite new to programming its been my second week since I started so if you see any bad practice or error in code please accept my apologies.
I want to print sum of first n odd numbers. But so far I can only do the sum of odd number up to the given number. kindly help.
public static void main(String[] args)
{
Scanner userInput = new Scanner(System.in);
System.out.print("Please enter the number : ");
int num1 = userInput.nextInt();
int sum = sumOfOdd(num1);
System.out.println("sum of first " +num1 + " odd numbers is " + sum);
userInput.close();
}
static int sumOfOdd(int num)
{
int sum = 0;
for (int i = 0; i <= num; i++)
{
if(i % 2 != 0)
{
sum += i;
}
}
return sum;
}
}
You don't have to use a loop at all
static int sumOfOdd(int num) {
return num*num;
}
For Any Arithmetic Progression, the sum of numbers is given by,
Sn=1/2×n[2a+(n-1)×d]
Where,
Sn= Sum of n numbers
n = n numbers
a = First term of an A.P
d= Common difference in an A.P
Using above formula we can derive this quick formula to calculate sum of first n odd numbers,
Sn(odd numbers)= n²
Try this it uses a for loop that increments by two to only account for odd numbers.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Enter the value of n: ");
int n = scanner.nextInt();
System.out.println("The sum of the first " + n + " odd numbers is: " + sumOfOddNumbers(n));
}
public static int sumOfOddNumbers(int n) {
int sum = 0;
for(int i = 1; i < n*2; i+=2) {
sum += i;
}
return sum;
}
}
Example usage:
Enter the value of n: 5
The sum of the first 5 odd numbers is: 25
Try this:
static int sumOfOdd(int num) {
int sum = 0;
for (int i = 0; i < num; i++){
sum += i*2+1;
}
return sum;
}
It sums up all odd numbers until the limit is reached.
With i*2+1 you get the next odd number. Then you add it to the sum.
Tested with System.out.println(sumOfOdd(4)); and got the expected result 16 (1+3+5+7)
Change the counter to the number of times you add an odd number to the sum value...
static int sumOfOdd(int num) {
int sum = 0;
int i = 0;
int count = 0;
do {
if(i % 2 != 0) {
sum += i;
count++;
}
i++;
} while (count < num);
return sum;
}
Or even cleaner:
static int sumOfOdd(int num) {
int sum=0;
for (int i=1;i<num*2;i+=2) {
sum=sum+i;
}
return sum;
}
You should count how many numbers have you added and in the condition to check if count of numbers you summed is less or equal than your n. Just add the counter in your for loop and set condition to: count <= num and it should work.
Every time you add a number to the sum increment the count by count++.
The code suppose to look like this:
static int sumOfOdd(int num)
{
int sum = 0;
int count = 0;
for (int i = 0, count= 0; count <= num; i++)
{
if(i % 2 != 0)
{
sum += i;
count++;
}
}
return sum;
}
I haven't checked it, but it should be correct
Since you don't know how many loop cycles are required you have to change the exit condition of the for loop.
Or you can use a while loop exploiting the same exit condition.
static int sumOfOdd(int num){
int sum = 0;
int counter = 0;
int currentNumber = 0;
while (counter<num){
if(currentNumber % 2 != 0){
sum += currentNumber;
counter++;
}
currentNumber++;
}
return sum;
}
Here is the complete code you'd be using:
public class YourClass {
public static void main(String[] args)
{
Scanner userInput = new Scanner(System.in);
System.out.print("Please enter the number : ");
int num1 = userInput.nextInt();
int sum = sumOfOdd(num1);
System.out.println("sum of first " +num1 + " odd numbers is " + sum);
userInput.close();
}
static int sumOfOdd(int num)
{
int counter = 0;
for (int i = 0;; i++)
{
int sum = 0;
if(i % 2 != 0)
{
counter++;
sum += i;
}
if(counter == num) return sum;
}
}
}
Another alternative.
static int sumOfOdd(int num) {
int sum = 0;
int last = 2*num-1;
for (int i = 1; i <= last; i+=2){
sum += i;
}
return sum;
}
Obviously return num*num; is the most efficient but if you're obliged to use a loop then this method avoids a * inside the loop.
This will be a tiny (tiny) bit more efficient than:
for (int i = 0; i < num; ++i){
sum += 2*i+1;
}
import java.util.*;
class karan{
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int i = n;
int sumOddNumber = n * i;
System.out.println(n*i);
}
}
I am trying to return the sum of all the values in the the array while also trying to return the largest value to the main method, however, the program states that I have an error at return total and at return number. The error states, "Type mismatch: cannot convert from int to int[]."
public static void main(String[] args) {
Scanner number = new Scanner(System.in);
int myArray[] = new int[10];
for(int i = 0; i <= myArray.length-1; i++ ) {
System.out.println("Enter Number: ");
int nums = number.nextInt();
myArray[i] = nums;
}
int [] sum = computeTotal(myArray);
System.out.println("The numbers total up to: "+sum);
int [] largest = getLargest(myArray);
System.out.println("The largest number is: "+largest);
}
public static int[] computeTotal(int myArray[]) {
int total = 0;
for (int z : myArray){
total += z;
}
return total;
}
public static int[] getLargest(int myArray[]) {
int number = myArray[0];
for(int i = 0; i < myArray.length; i++) {
if(myArray[i] > number) {
number = myArray[i];
}
}
return number;
}
The methods computeTotal and getLargestshould be changed the return types to int. Please refer this:
public static void main(String[] args) {
Scanner number = new Scanner(System.in);
int myArray[] = new int[10];
for(int i = 0; i <= myArray.length-1; i++ ) {
System.out.println("Enter Number: ");
int nums = number.nextInt();
myArray[i] = nums;
}
int sum = computeTotal(myArray);
System.out.println("The numbers total up to: "+sum);
int largest = getLargest(myArray);
System.out.println("The largest number is: "+largest);
}
public static int computeTotal(int myArray[]) {
int total = 0;
for (int z : myArray){
total += z;
}
return total;
}
public static int getLargest(int myArray[]) {
int number = myArray[0];
for(int i = 0; i < myArray.length; i++) {
if(myArray[i] > number) {
number = myArray[i];
}
}
return number;
}
Hope this help.
Probably in java8 there're easier way to get the max and sum.
int sum = Arrays.stream(new int[] {1,2, 3}).sum(); //6
int max = Arrays.stream(new int[] {1,3, 2}).max().getAsInt(); //3
This question already has answers here:
Need Java array help using scanner class to output an average and sort method
(4 answers)
Closed 6 years ago.
I have a method to output the highest value, lowest value, average value and I need a sort method. I have tried to put what is called a "bubble method" but it isn't working out. Anyone know other sort methods I can use?
import java.util.Scanner;
public class Arrayassignment {
public static void main(String[] args) {
Scanner sin = new Scanner(System.in);
System.out.println("Enter an intiger for array size.");
int number = sin.nextInt();
int array[] = new int[number];
System.out.println("Array size " + number + " initiated.\n");
System.out.println("Now enter the array intigers.");
for (int i = 0; i < number; i++) {
array[i] = sin.nextInt();
}
//System.out.println ( "\nLargest " + max (1, 3, 5) );
System.out.println("sorting" + sort(array));
System.out.println("The highest number in the array is " + max(array));
System.out.println("The smallest number in the array is " + min(array));
System.out.println("The average of the numbers in the array is " + avg(array));
}
public static int sort(int[] arg) {
for (int i = 1; i < arg.length - 1; i++) {
for (int j = i + 1; j < arg.length; j++) {
if (arg[i] > arg[j]) {
int arrange = arg[i];
arg[i] = arg[j];
arg[j] = arrange;
}
}
}
return arrange;
}
public static int max(int[] arg) {
if (arg.length == 0) {
System.out.println(" empty arguement list ");
return 0;
}
int largest = arg[0];
for (int i = 1; i < arg.length; i++) {
if (arg[i] > largest) {
largest = arg[i];
}
}
return largest;
}
public static int min(int[] arg) {
if (arg.length == 0) {
System.out.println(" empty arguement list ");
return 0;
}
int smallest = arg[0];
for (int i = 1; i < arg.length; i++) {
if (arg[i] < smallest) {
smallest = arg[i];
}
}
return smallest;
}
public static double avg(int... arr) {
int sum = 0;
for (int i = 0; i < arr.length; i++) {
sum += arr[i];
}
double average = (double) sum / arr.length;
return average;
}
}
There are many other sort methods you can use. The one you were trying to use is called "Bubble Sort" and is very expensive on large data sets unless they are somewhat ordered. I would recommend using selection sort or insertion sort for what you are trying to accomplish.
Here is a link to the many sorting algorithms you can implement: Sorting Algorithms
Here are some animations showing the process of these sorts (Highly recommend you look at these before implementing your algorithm):
Helpful animations
You can use any sorting method as your convenient and according to your requirement. After sorted the array you can easily pick up the minimum and maximum value from the sorted array, first element and the last element of the array.
For calculate the average you have to use separate method as you used or you can use static variable to calculate the total inside the sorting method.
Refer this code.
public class Arrayassignment {
public static void main(String[] args) {
Scanner sin = new Scanner(System.in);
System.out.println("Enter an intiger for array size.");
int number = sin.nextInt();
int array[] = new int[number];
System.out.println("Array size " + number + " initiated.\n");
System.out.println("Now enter the array intigers.");
for (int i = 0; i < number; i++) {
array[i] = sin.nextInt();
}
sin.close();
System.out.println("sorting");
printArray(array); //Before sort
sort(array);
printArray(array); //After sort
System.out.println("The highest number in the array is " + array[array.length - 1]);
System.out.println("The smallest number in the array is " + array[0]);
System.out.println("The average of the numbers in the array is " + avg(array));
}
public static void sort(int[] arg) {
int arrange;
for (int i = 0; i < arg.length - 1; i++)
for (int j = i + 1; j < arg.length; j++) {
if (arg[i] > arg[j]) {
arrange = arg[i];
arg[i] = arg[j];
arg[j] = arrange;
}
}
}
public static double avg(int... arr) {
int sum = 0;
for (int i = 0; i < arr.length; i++) {
sum += arr[i];
}
double average = (double) sum / arr.length;
return average;
}
public static void printArray(int[] arr) {
for (int value : arr) {
// print elements according to your convenient
System.out.println(value);
}
}
To print the array you have traversal through the array. See Above code method.
I was practicing with some exercises from UVA Online Judge, I tried to do the Odd sum which basically is given a range[a,b], calcule the sum of all odd numbers from a to b.
I wrote the code but for some reason I don't understand I'm getting 891896832 as result when the range is [1,2] and based on the algorithm it should be 1, isn't it?
import java.util.Scanner;
public class OddSum
{
static Scanner teclado = new Scanner(System.in);
public static void main(String[] args)
{
int T = teclado.nextInt();
int[] array = new int[T];
for(int i = 0; i < array.length; i++)
{
System.out.println("Case "+(i+1)+": "+sum());
}
}
public static int sum()
{
int a=teclado.nextInt();
int b = teclado.nextInt();
int array[] = new int[1000000];
for (int i = 0; i < array.length; i++)
{
if(a%2!=0)
{
array[i]=a;
if(array[i]==(b))
{
break;
}
}
a++;
}
int res=0;
for (int i = 0; i < array.length; i++)
{
if(array[i]==1 && array[2]==0)
{
return 1;
}
else
{
res = res + array[i];
}
}
return res;
}
}
Your stopping condition is only ever checked when your interval's high end is odd.
Move
if (array[i] == (b)) {
break;
}
out of the if(a % 2 != 0) clause.
In general, I don't think you need an array, just sum the odd values in your loop instead of adding them to the array.
Keep it as simple as possible by simply keeping track of the sum along the way, as opposed to storing anything in an array. Use a for-loop and add the index to the sum if the index is an odd number:
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.println("Enter minimum range value: ");
int min = keyboard.nextInt();
System.out.println("Enter maximum range value: ");
int max = keyboard.nextInt();
int sum = 0;
for(int i = min; i < max; i++) {
if(i % 2 != 0) {
sum += i;
}
}
System.out.println("The sum of the odd numbers from " + min + " to " + max + " are " + sum);
}
I don't have Java installed right now, however a simple C# equivalent is as follows: (assign any values in a and b)
int a = 0;
int b = 10;
int result = 0;
for (int counter = a; counter <= b; counter++)
{
if ((counter % 2) != 0) // is odd
{
result += counter;
}
}
System.out.println("Sum: " + result);
No major dramas, simple n clean.
this code doesn't function,
it said that lessthaAverage(int) in calculateArray cannot be applied to (), I'm a beginner so I still don't understand this coding yet, this is the question ask, Write an object oriented program that randomly generates an array of 1000 integers between 1 to 1000.
Calculate the occurrences of number more than 500 and find the average of the numbers.
Count the number which is less than the average and finally sort the numbers in descending order.
Display all your output. Please do HELP ME!!!,Thank You...
import java.util.*;
import java.io.*;
//import java.util.random;
public class CalculateArray
{
//declare attributes
private int arr[] = new int[1000];
int i;
//generates an array of 1000 integers between 1 to 1000
public void genArr()
{
Random ran = new Random();
for(i = 0; i < arr.length; i++)
{
arr[i] = ran.nextInt(1000) + 1;
}
}
//Calculate the occurences of number more than 500
public int occNumb()
{
int count;
for(i = 0; i < arr.length; i++)
{
if(arr[i] > 500)
{
count++;
}
}
return count;
}
//find the average of the numbers
public int average()
{
int sum, aver;
for(i = 0; i < arr.length; i++)
{
sum += arr[i];
}
aver = sum/1000;
return aver;
}
//Count the number which is less than the average
public int lessthanAverage(int aver)
{
int cnt;
cnt = 0;
for(i = 0; i < arr.length; i++)
{
if(arr[i] < aver)
{
cnt++;
}
}
return cnt;
}
//finally sort the numbers in descending order.
public void sort(int[] num)
{
System.out.println("Numbers in Descending Order:" );
for (int i=0; i <= num.length; i++)
for (int x=1; x <= num.length; x++)
if (num[x] > num[x+1])
{
int temp = num[x];
num[x] = num[x+1];
num[x+1] = temp;
}
}
//Display all your output
public void display()
{
int count, aver;
System.out.println(arr[i] + " ");
System.out.println("Found " + count + " values greater than 500");
System.out.println("The average of the numbers is " + aver);
System.out.println("Found " + count + " values that less than average number ");
}
public static void main(String[] args)
{
CalculateArray show = new CalculateArray();
show.genArr();
int c= show.occNumb();
show.average();
int d=show.lessthanAverage();
show.sort(arr);
show.display();
}
}
Your method lessthanAverage is expecting a int parameter. You should store the result of the average method call into a int variable and pass it to the call to lessthanAverage.
int avg = show.average();
int d=show.lessthanAverage(avg);
Your lessthaAverage() method expects an average to be passed in as a parameter, but you are not passing it in when you call it.
It seems that your method lessthanAverage needs an int as a parameter, but you are not passing it in main
public int lessthanAverage(int aver)
In main aver is not being passed:
int d=show.lessthanAverage(); // where is aver?
But if you wanted to know the average inside the method you could call your average method inside lessthanAverage:
if(arr[i] < this.average())
and not pass any parameter at all:
public int lessthanAverage()