Will this solution really fix the problem?

The list method reverse reverses the elements in the list. Define a function named reverse that reverses the elements in its lis

Alisiya [41]

Answer:

public class Reverse {

public static void reverseList(int list [], int n)
{
int[] reversedList = new int[n];
int k = n;
for (int i = 0; i < n; i++) {
reversedList[k - 1] = list[i];
k = k - 1;
}
//printing the reversed list
System.out.println("The Reversed list \n");
for (int j = 0; j < n; j++) {
System.out.println(reversedList[j]);
}
}

Explanation:

Using Java, An array is implemented to hold a list of items

A method reverseList() is created to accept an array as parameter and using a for statement reverses the elements of the array and prints each element of the list

See below a complete code with a main method that calls this method

public class Reverse {

public static void main(String[] args) {

int [] arr = {10, 20, 30, 40, 50};

reverseList(arr, arr.length);

}

public static void reverseList(int list [], int n)

{

int[] reversedList = new int[n];

int k = n;

for (int i = 0; i < n; i++) {

reversedList[k - 1] = list[i];

k = k - 1;

}

//printing the reversed list

System.out.println("The Reversed list \n");

for (int j = 0; j < n; j++) {

System.out.println(reversedList[j]);

}

}

7 0

4 years ago

How to be professional in graphic design without CAD?

Mama L [17]

Answer:

There are two types of image processing software, raster-based, and vector-based. The CAD and the CAID are just the two vectors based software. However, there is much other software like Adobe Illustrator, Microsoft Publisher, Corel Draw, Photoshop on which you can embark upon, and become an equally good Graphics designer. And hence, you can certainly become a Graphics designer with the CAD experience, however, you should keep in mind that all the great Graphic designers use CAD and CAID. And hence, you should use it sometimes certainly in the future.

Explanation:

Please check the answer section.

4 0

3 years ago

Create a Java program with threads that looks through a vary large array (100,000,000 elements) to find the smallest number in t

olchik [2.2K]

Answer:

See explaination

Explanation:

import java.util.Random;

public class Sample{

static class MinMax implements Runnable{

int []arr;

int start,end,min,max;

MinMax(int[]arr, int start,int end){

this.start=start;

this.end=end;

min=Integer.MAX_VALUE;

max=Integer.MIN_VALUE;

this.arr=arr;

}

atOverride

public void run() {

for(int i=start;i<=end;i++){ //search min and max form strant to end index

min=Math.min(min,arr[i]);

max=Math.max(max, arr[i]);

}

public static void main(String[] args) throws Exception{

long beginTime = System.nanoTime();

Random gen = new Random();

int n=100000000;

int[] data = new int[n]; //generate and fill random numbers

for(int i = 0; i < data.length; i++) {

data[i] = gen.nextInt()%1000000;

}

long endTime = System.nanoTime();

System.out.println("Done filling the array. That took " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 1 thread"); //1 thread

MinMax m1=new MinMax(data,0,n-1); //class object

Thread t1=new Thread(m1); //new thread

beginTime=System.nanoTime(); //start timer

t1.start(); //start thread

t1.join(0); //wait until thread finishes

endTime=System.nanoTime(); //end timer

System.out.println("Min,Max: "+m1.min+","+m1.max); //print minimum and maximum

System.out.println("Time using 1 thread " + (endTime - beginTime)/1000000000f + " seconds."); //print time taken

//-----------------------------------------

System.out.println("Using 2 thread");

m1=new MinMax(data,0,n/2);

MinMax m2=new MinMax(data,n/2+1,n-1);

t1=new Thread(m1);

Thread t2=new Thread(m2);

beginTime=System.nanoTime();

t1.start();

t2.start();

t1.join(0);

t2.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(m1.min,m2.min)+","+Math.max(m1.max,m2.max));

System.out.println("Time using 2 thread " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 3 thread");

m1=new MinMax(data,0,n/3);

m2=new MinMax(data,n/3+1,2*n/3);

MinMax m3=new MinMax(data,2*n/3+1,n-1);

t1=new Thread(m1);

t2=new Thread(m2);

Thread t3=new Thread(m3);

beginTime=System.nanoTime();

t1.start();

t2.start();

t3.start();

t1.join(0);

t2.join(0);

t3.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(Math.min(m1.min,m2.min),m3.min)+","+Math.max(Math.max(m1.max,m2.max),m3.max));

System.out.println("Time using 3 thread " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 4 thread");

m1=new MinMax(data,0,n/4);

m2=new MinMax(data,n/4+1,2*n/4);

m3=new MinMax(data,2*n/4+1,3*n/4);

MinMax m4=new MinMax(data,3*n/4+1,n-1);

t1=new Thread(m1);

t2=new Thread(m2);

t3=new Thread(m3);

Thread t4=new Thread(m4);

beginTime=System.nanoTime();

t1.start();

t2.start();

t3.start();

t4.start();

t1.join(0);

t2.join(0);

t3.join(0);

t4.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(Math.min(m1.min,m2.min),Math.min(m3.min,m4.min))+","+Math.max(Math.max(m1.max,m2.max),Math.max(m3.max,m4.max)));

System.out.println("Time using 4 thread " + (endTime - beginTime)/1000000000f + " seconds.");

}

6 0

3 years ago

In a non-price rationing system, consumers receive goods and services first-come, first served. Give me an example of a time whe

Anna35 [415]

When someone may be giving away something for free.

7 0

3 years ago

You need a VPN to connect to a private, remote network in order to access some files. You click the network icon in your taskbar

sineoko [7]

Answer:

Buy a vpn

Explanation:

NordVPN is a good one I use.

7 0

3 years ago