I nedd to find the fatest and shortest way to calculate the sum of positive integer that the user input in.
else if(num1<0 && num2 >0 && num3>0 && num4>0 && num5>0 &&num6>0){
totalPositiveNumber =num2 + num3 + num4 + num5 + num6;
System.out.println("The sum of positive integer is: " + totalPositiveNumber);
}
You can read the numbers and store them in an array. Later iterate over the array and sum only, when number is greater than 0. Check below for sample code for reading input from System.in using Scanner class.
import java.util.Scanner;
public class Sum {
public static void main(String args[]) {
Scanner read = new Scanner(System.in);
int array[] = new int[6];
int sum = 0;
for (int i = 0; i<6; i++) {
array[i] = read.nextInt();
}
for (int i = 0; i<6; i++) {
if (array[i] > 0) {
sum = sum + array[i];
}
}
System.out.println(sum);
}
}
Related
I'm working on a code which has a JSpinner, which gives an output and moves it to int = n, I want to save every JSpinner output and add it into an array to check for duplicates and once found the JPanel should close itself or say you lost.
Scanner s = new Scanner(System.in);
int n = (Integer) spinner.getValue();
if (isPrime(n)) {
Input.setText(n + " is a prime number");
score++;
Highscore.setText("Score: " + score);
int[] array = new int[4];
for(int i=0; i<4;i++)
{
array[i]= (Integer) spinner.getValue();
}
for (int i=0; i < 4;i++)
{
System.out.println(array[i]);
}
Spinner output into array but it fills the whole array with one number example: [3,3,3,3].
private <T> boolean duplicate(T... array)
{
for (int i = 0; i < array.length; i++)
{
for (int j = i + 1; j < array.length; j++)
{
if (array[i] != null && array[i].equals(array[j])) {
return true;
dispose();
}
}
}
return false;
}
Duplicate check
I tried adding the user input from the JSpinner into an array and from there to check for duplicates.
Searched Online but could'nt find anything.
This if my first post so if you need anything more you can tell me.
public class test {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int a;
for(int i = 1; i > 0; ++i) {
a = scanner.nextInt();
if(a <= 0 & i != 1) {
System.out.println("The largest number is ");
break;
}else if(a <= 0 & i == 1){
System.out.println("No Number entered.");
break;
}else if(a <= 0){
System.out.println("No Number entered.");
break;
}else
System.out.println("Number "+i+": "+a);
}
}
}
QUESTION
I want to scanf some numbers, for example, 3, 5 and 6.
after I scanf 0, it should printf the biggest number, in this example, 6.
How do I write the code for that without using arrays (I am not allowed to use arrays)
I wrote the code that if I scanf 0 first, it tells me "No numbers entered".
and after I scanf several numbers, 0 is the number that breaks the loop and ends the program
try this code :
int n;
scanf("%d",&n);
int a,result;
//result=-infiniti
result=-100000000;
for(int i=1;i<=n;i++)
{
scanf("%d",&a);
if(a>result)
{
result=a;
}
}
printf("%d",result);
Here's a sample solution for Sliding Window Maximum problem in Java.
Given an array nums, there is a sliding window of size k which is
moving from the very left of the array to the very right. You can only
see the k numbers in the window. Each time the sliding window moves
right by one position.
I want to get the time and space complexity of this function. Here's what I think would be the answer:
Time: O((n-k)(k * logk)) == O(nklogk)
Space (auxiliary): O(n) for return int[] and O(k) for pq. Total of O(n).
Is this correct?
private static int[] maxSlidingWindow(int[] a, int k) {
if(a == null || a.length == 0) return new int[] {};
PriorityQueue<Integer> pq = new PriorityQueue<Integer>(k, new Comparator<Integer>() {
// max heap
public int compare(Integer o1, Integer o2) {
return o2 - o1;
}
});
int[] result = new int[a.length - k + 1];
int count = 0;
// time: n - k times
for (int i = 0; i < a.length - k + 1; i++) {
for (int j = i; j < i + k; j++) {
// time k*logk (the part I'm not sure about)
pq.offer(a[j]);
}
// logk
result[count] = pq.poll();
count = count + 1;
pq.clear();
}
return result;
}
You're right in most of the part except -
for (int j = i; j < i + k; j++) {
// time k*logk (the part I'm not sure about)
pq.offer(a[j]);
}
Here total number of executions is log1 + log2 + log3 + log4 + ... + logk. The summation of this series -
log1 + log2 + log3 + log4 + ... + logk = log(k!)
And second thought is, you can do it better than your linearithmic time solution using double-ended queue property which will be O(n). Here is my solution -
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || k <= 0) {
return new int[0];
}
int n = nums.length;
int[] result = new int[n - k + 1];
int indx = 0;
Deque<Integer> q = new ArrayDeque<>();
for (int i = 0; i < n; i++) {
// remove numbers out of range k
while (!q.isEmpty() && q.peek() < i - k + 1) {
q.poll();
}
// remove smaller numbers in k range as they are useless
while (!q.isEmpty() && nums[q.peekLast()] < nums[i]) {
q.pollLast();
}
q.offer(i);
if (i >= k - 1) {
result[indx++] = nums[q.peek()];
}
}
return result;
}
HTH.
I was helped earlier in creating this code that would create a histogram of a randomint. Everything looks good except I accidently had the output as a sum instead of a mean of all the numbers that were randomly chosen.I dont want to mess anything up so I was just going to ask, How can I convert this sum into a mean output instead?
import java.util.Random;
class Assignment4
{
public static void main(String[] args)
{
Random r = new Random();
int sum = 0;
int[] bins = new int[10];
for(int i = 0; i < 100; i++)
{
int randomint = 1 + r.nextInt(10);
sum += randomint;
bins[randomint-1]++;
//System.out.print(randomint + ", ");
}
System.out.println("Sum = " + sum);
System.out.println("Data shown below: ");
for (int i = 0; i < bins.length; i++)
{
int binvalue = bins[i];
System.out.print((i+1) + ": ");
for(int j = 0; j < binvalue; j++)
{
System.out.print('*');
}
System.out.println(" (" + binvalue + ")");
}
}
}
Never mind figured it out.... just turned System.out.println("Sum = " + sum); into System.out.println("Mean = " + sum/100);
i am trying to increment ipv6 address based on mask.
i am getting problem when there is F in place of increment.
could any one plz check this
public String IncrementIPV6ForPrefixLength (String IPv6String, int times) throws UnknownHostException
{
int result , carry = 0, i;
int bits;
int mask=0;
int index=IPv6String.indexOf("/");
mask=Integer.parseInt(IPv6String.substring(index+1, IPv6String.length()));
IPv6String=IPv6String.substring(0, index);
InetAddress iaddr=InetAddress.getByName(IPv6String);
byte[] IPv6Arr=iaddr.getAddress();
if(mask > 128 || mask < 0)
return null;
i = mask/8;
bits = mask%8;
if(bits>0)
{
result = ((int)(IPv6Arr[i]>>(8-bits))) + times;
IPv6Arr[i] =(byte) ((result << (8-bits)) | (IPv6Arr[i] & (0xff >> (bits))));
carry = (result << (8-bits))/256;
times /= 256;
}
i--;
for(;i>=0;i--)
{
result = ((int)IPv6Arr[i]) + ((times + carry)& 0xFF);
IPv6Arr[i] = (byte)(result % 256);
carry = result / 256;
if(carry == 0)
{
iaddr=InetAddress.getByAddress(IPv6Arr);
String s=iaddr.toString();
if(s.indexOf('/') != -1){
s = s.substring(1, s.length()).toUpperCase();
}
StringBuffer buff =new StringBuffer("");
String[] ss = s.split(":");
for(int k=0;k<ss.length;k++){
int Differ = 4 - ss[k].length();
for(int j = 0; j<Differ;j++){
buff.append("0");
}
buff.append(ss[k]);
if(k!=7)buff=buff.append(":");
}
return buff.toString()+"/"+mask;
}
times /= 256;
}
return null;
}
input like this:
FD34:4FB7:FFFF:A13F:1325:2252:1525:325F/48
FD34:41B7:FFFF::/48
FD34:4FBF:F400:A13E:1325:2252:1525:3256/35
output like this
if increment by 1
FD34:4FB8:0000:A13F:1325:2252:1525:325F/48
FD34:41B8:0000::/48
FD34:4FC0:0400:A13E:1325:2252:1525:3256/35
if increment by 2
FD34:4FB8:0001:A13F:1325:2252:1525:325F/48
FD34:41B8:0001::/48
FD34:4FC0:1400:A13E:1325:2252:1525:3256/35
can u plz find where i am doing wrong.
Disregarding the posted code, try to model the operation as a direct numerical operation on the 128-bit number that the IPv6 address really is. Convert to BigInteger and use BigInteger.add.