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Crank
3 years ago
6

Write a method called printRange that accepts two integers as arguments and prints the sequence of numbers between the two argum

ents, separated by spaces. Print an increasing sequence if the first argument is smaller than the second; otherwise, print a decreasing sequence. If the two numbers are the same, that number should be printed by itself. Here are some sample calls to printRange:printRange(2, 7);printRange(19, 11);printRange(5, 5);The output produced from these calls should be the following sequences of numbers:2 3 4 5 6 719 18 17 16 15 14 13 12 115Test the method using the following main program:import java.util.*; // for Scannerpublic class Lab4Q2 {public static void main(String[] args) {Scanner console = new Scanner(System.in);System.out.print("Enter a positive integer: ");int num1 = console.nextInt();System.out.print("\nEnter a second positive integer: ");int num2 = console.nextInt();System.out.println();printRange(num1, num2);}
Computers and Technology
1 answer:
Papessa [141]3 years ago
4 0

Answer:

import java.util.*;

// for Scanner

public class Lab4Q2{

     public static void main(String[] args){

           Scanner console = new Scanner(System.in);

           System.out.print("Enter a positive integer: ");

           int num1 = console.nextInt();

           System.out.print("\nEnter a second positive integer: ");

           int num2 = console.nextInt();

          System.out.println();

          printRange(num1, num2);

}

     public static void printRange(int a, int b){

           if(a == b){

               System.out.print(a);

}           else if (a < b){

                for(int i = a; i <= b; i++){

                     System.out.print(i + " ");

}

}

           else if (a > b){

                for(int i = a; i >= b; i--){

                    System.out.print(i + " ");

}

}

}

}

Explanation:

In the printRange method that is called from the main method; we pass the two parameters of numbers entered as 'a' and 'b'. First, we check if 'a' and 'b' are the same, then we output a single instance of the input.

Next, we check if the first input is less than the second input then we output in ascending order.

Lastly, we check if the first input is greater than the second input then we output in descending order.

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MariettaO [177]

Answer:

ab2

Explanation:

3 0
3 years ago
What is the worst case time complexity of insertion sort where position of the data to be inserted is calculated using binary se
laiz [17]

Answer:

O(n²)

Explanation:

The worse case time complexity of insertion sort using binary search for positioning of data would be O(n²).

This is due to the fact that there are quite a number of series of swapping operations that are needed to handle each insertion.

4 0
3 years ago
Choose the 3 Points in good story telling
Sidana [21]

Answer:

1.Choose a clear central message 2. Embrace conflict 3.Have a clear structure  

Explanation:

8 0
2 years ago
Nothin to see here just browsin
velikii [3]

Answer:

What????

What does that mean??????????????????????????????????

4 0
3 years ago
In general, a unit test may need drivers and stubs, but not both.<br><br> TRUE OR FALSE
Ymorist [56]

Answer:

True

Explanation:

If a module or code is not ready then the unit test can use the Stubs to simulate a called upon function to test the process. On the other hand if the main unit is not ready the test can use Drivers to simulate the calling of said function to test the rest of the modules.

Therefore, a unit test will use either Drivers or Stubs at a given moment for testing but not both simultaneously.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

5 0
3 years ago
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